articles
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geometry
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Is there? I know that all particles stop at |zero|, but nothing can travel faster than the speed of light.

Well at the speed of light, relativity says the object will have infinite mass => infinite energy => impossible. So the speed of light cannot be attained (that's as much as I know from my physics book at least).

EDIT: maybe this thread should go under the new Physics forum?

[quote="paladin8"]Well at the speed of light, relativity says the object will have infinite mass => infinite energy => impossible. So the speed of light cannot be attained (that's as much as I know from my physics book at least).
[/quote]

Well, I don't see a problem in this...  :?

But is there any estimate for the mass/energy in the whole universe¿

Now that the forum exists, this post should be moved to the Physics forum (I think!)

[quote="denizuzun"]not will be discovered i think
and the maximum velocity is light, but i don't think it is proved that we cannot go up the speed of light
(sorry if it is proved :D )[/quote]
maybe there is another world where the speed of light is the infimum of the speed.

[quote="shobber"]just wondering, why there is a |zero| of $-273$? Why can't we get to $-274$?[/quote]

At absolute zero, all the particles of a material have stopped moving.  Furthermore, this temperature never has really been attained, and is a theoretical value for the point at which all matter stops moving.

[quote="shobber"]just wondering, why there is a |zero| of $-273$? Why can't we get to $-274$?[/quote]
$E=\frac{3}{2}K_{B}T.$Where $t=T-273$.
Ok?

[quote="Danbert"]Yeah, the highest possible temperature is essentially infinite.  To get it do the following:
1) Convert all matter in the Universe to energy (remember, E=mc 2 )
2) Compress all of this energy into a single subatomic particle

Of course, this is impossible with today's technology, and wouldn't be a good idea even if we could do it :)[/quote]

yes of course, that's completely true
scientists think that this once really happened, at the begining of the univers, at the first second - and even before- of the big bang
so the maximun tempereture must actually be that one, unfortunatly scientists nowadays cannot mesure such temperature since the matter does no longer exist after a definite limit and becomes "plasma"  ..  very difficult in deed!

[quote="ZetaX"][quote="Al-Khawarizmi"]
yes of course, that's completely true
scientists think that this once really happened, at the begining of the univers, at the first second - and even before- of the big bang
so the maximun tempereture must actually be that one, unfortunatly scientists nowadays cannot mesure such temperature since the matter does no longer exist after a definite limit and becomes "plasma"  ..  very difficult in deed![/quote]

I would say the 'reality' is much more complicated:
even if there was the big bang, I think it was a lot more 'different' than only some energy.
And why shouldn't you be able to measure the heat of plasma, it is possible![/quote]

But we are not talking about the nature of the singularity that gave birth to big bang, whathever its nature was , it was surely very very hot, it attained at the begining the temperature we cannot théoritically overtake (10^32 KELVIN), but it's not necessery the maximal one we can get, since this later can be lower or in other words just limited by that one.
concening plasma, it's the stage where protons  and neutrons are separeted from electrons,  followed by another plasma where protons and neutrons are transformed to quarks ,but the problem is : how can we check there is not any stage after plasma, which should be hotter  of course?that's the question i wanted to shed light on.

Yeah. It makes more sense to talk about it in terms of energy.
But with regards to the topic's title... Take an arbitrarily large amount on energy in a given mass, and you should be able to reach an arbitrarily large temperature. It shouldn't be hard to reach large temperatures. Blast it with lasers till it makes a plasma. Whip it around and confine it in a magnetic field. Blast it some more. Watch the lil' photons shred the nuclei apart. Blast it some more. Laugh evily. Get shredded to bits by neutrons escaping your confinement field halfway through your thrid BWAHAHA.

[quote="ZetaX"]Where is the problem at the limits given by speed of light¿
I have as good as no knowledge about 'temperature', but isn't it defined by the energy in the 'caotical moving' particles¿
(Because this would enable a very high temparature, only limited by the energy that exists in the whole universe)[/quote]

Given a system having internal energy $E$ , and entropy $S$, temperature is defined as:

$T=\frac{\partial E}{\partial S}$

[quote="Devendra Patel"]Just wondering if a body was moving with half the velocity of light and was in a projectile path towards a black hole and was able to just skid off the pull of the black hole ,woldnt it be possible that the gravity of the balck hole  accelerate the body to maybe more the speed of light. :?:  :!: and maybe because of the speed the body would also escape the black hole.[/quote]
I also have such questions...
If a body has an constant acceleration, will it finally have speed faster than light?? It seems that this isn't true but why?

Whoa. There is some bad physics going on here. That's one of the problem with physics, is that sometimes even simple questions can have elusive answers. That's because so many of the equations you learn early on in physics are either approximations, or apply only under "well-behaved" situations.

But its always good to try and do a little research before you try and answer a question like this, as its probably been asked by someone before.

In particular there's something called the Planck Temperature. Its the highest theoretical temperature in the Universe. Its basically the temperature that you would get if the entire universe were squeezed into a space smaller than the Planck length ($1.616 \times 10^{-34} \, m$). To understand why requires some ideas of quantum mechanics and thermodynamics.

The Planck temperature turns out to depend only on physical constants of the universe (the speed of light, the planck constant, the universal gravitational constant, and the Boltzmann constant). $T_p = \sqrt{\frac{hc^5}{2 \pi G{k}^2}}$. The Planck Temperature is approximately $1.416 \times 10^{32} \, K$.

Using modern classical laws of physics this would have been the temperature of the Universe approximately $5.4 \times 10^{-44} \, sec$ after the Big Bang (with there being no physically meaningful measure of time smaller than that).

[quote="shobber"]
I also have such questions...
If a body has an constant acceleration, will it finally have speed faster than light?? It seems that this isn't true but why?[/quote]

Hmm .. if a body has constant acceleration, it WILL finally have speed faster than light. BUT this situation CAN NOT occur due to physic I am familiar with.

For example, acceleration is always produced by a force, an then it is equal to ratio of the force and mass of the body. Since the mass of the body increases to infinity, when its speed approaches the speed of light, the acceleration decreases to zero, with some finite force.
And therefore you will never be able to accelerate something beyond the speed of light, since you'd need infinite force.

[quote="fdump"]since we are talking about energy in universe, can anybody tell me how is graviation manifested? I mean, ok, so two objects in space atract eachother with a force given by a well known formula. But how can you prove it? what causes the attraction?[/quote]

Newton's Law of gravitational attraction is an observational law. It is not derived from other laws.

An alternate explaination of gravity can be found in General Relativity, from which Newton's Law comes as a consequence. The mathematical basis for General Relativity is too advanced to explain in detail. I'm sure you've heard loosely though, that matter (and even pressure and energy!) cause a curvature of space-time. You have to wait until you learn Differential Geometry before this can be made more precise.

For a good layman's explaination of gravity, look for a book called [u]The Fabric of the Cosmos[/u] by Brian Greene.

[quote="fdump"]since we are talking about energy in universe, can anybody tell me how is graviation manifested? I mean, ok, so two objects in space atract eachother with a force given by a well known formula. But how can you prove it? what causes the attraction?[/quote]

Gravitation is the result of mass, or energy if you like thanks to Albert Einstein ($E=mc^2$), distorting (warping) the space around it. Sort of like having four friends hold the corners of a bed sheet while dropping a bowling ball in the middle. Any smaller object you then put on the sheet will slide (roll) toward the bowling ball.

[quote="shobber"]just wondering, why there is a |zero| of $-273$? Why can't we get to $-274$?[/quote]

1) I don't think anyone answered this question. The Celcius scale was formed by taking 0 to be the freezing point of water, 100 to be the boiling point and then  evenly dividing 100 sections between the numbers. Basically it is a relative scale, which extends beyond those two points in a linear fashion. Now, someone did some basic experiments with the ideal gas law and found that most physical properties go to zero at -273.15 C. They then created a NEW scale, an absolute scale, which we refer to as the Kelvin scale. Thus 0 Kelvin was where "stuff" goes to zero. Anything less than 0 K or -273.15 C have no realistic significance. I tried very hard to put that at the most basic level possible.

2) Why can't we ever get to absolute zero? 1) Because the third law of thermodynamics proves that we cannot reach 0 K in a finite number of processes. To truly understand this, you should have a firm undergraduate understanding of thermodynamics, in my opinion. Otherwise, I'd say "just because we can't".

3) I started to freak out about all of this nonsense I was seeing until I came to Gauss's post (Thank the lord). There are a few things I would like to clear up. It IS POSSIBLE for something travel FASTER than the speed of light. Theoretically it is perfectly acceptable to have a world line outside of the real time cone in Minkowski space, you merely cannot transmit signals across the 45 degree line (cone). Again, to understand what I'm talking about, you have to have some experience with special relativity. This brings me to my 4th and final point.

4) I have noticed very quickly that this mess of a forum is a direct result of people discussing physics who are at drastically different levels and I think it is frustrating people on both sides of things. You wouldn't as a high school student studying for the AMC go over to the advanced analysis section of this forum and start asking some vastly theoretical questions, yet that is exactly what is happening on this Physics forum. I think we definitely need at least SOME subdivision of the subject so that at least people can have discussions on their own level. Otherwise it becomes a big mess like this thread. Either that, or we should announce what level we expect an answer to be at. For example, if you are in high school, tell us before you ask a question and someone goes into an explanation that involves differential manifolds, it would be a waste of time for both parties. Just my observation.

I agree with you. I have suggested to have some sub-forums here... ...
I got a question, why there is HUGE gap between high school physics and college ones, is that only because of the participation of Calculus and other super hard math stuff??

[quote="shobber"]why there is HUGE gap between high school physics and college ones, is that only because of the participation of Calculus and other super hard math stuff??[/quote]

Its largely because of the gap in mathematical sophistication. But also its the number of years of study. In high school you might have one, or maybe two, years of physics study. Basically an introduction. In college you may take a physics course every semester.

But yes, the main obstacle to the study of modern physics is its level of mathematics. If you think about it chronologically - in high school you only learn math known up until about the time of Newton (the early 1700s). So you likewise only learn physics up until the time of Newton (Newtonian Mechanics).

As you learn more modern mathematics (vector calculus, linear algebra, calculus of variations, differential geometry, group theory, topology) you can learn more modern physics.

We have four years of physics in High School but we also learn the basics...

there is no energy in an absolute 0 particle. go to google and type in BEC (Bose Einstein condensates) or absolute zero.

actually theres an infinite amount of energy in an absolute 0 system. Its one of those zero-point energy, quantum paradox thingies.


calculus
integration
trigonometry
calculus computations

hello, Maple can integrate this.
Sonnhard.


[quote="gendion"]Hi, i've got problem with this 3 problems.
We have to find unknows in real.
$ \begin{cases} x_1 x_2 x_3 \plus{} x_4 \equal{} 2 \\
x_1 x_2 x_4 \plus{} x_3 \equal{} 2 \\
x_2 x_3 x_4 \plus{} x_1 \equal{} 2 \\
x_1 x_3 x_4 \plus{} x_2 \equal{} 2 \end{cases}$

Prove that equation $ xy \plus{} yz \plus{} zx \equal{} 1$ have infty solution in integers

Find minimum:
$ \min_{m,n } \left|12^m \minus{} 5^n \right|$

Sorry for my English ^^ ;)[/quote]

I think for the first one 1,1,1,1 works. But I might be exceedingly stupid here.

Yes, 1,1,1,1 works, but it's just only one solution? nothing more?
We have:
$ \begin{cases}  \left(x_2 x_4 \plus{}1 \right) \left(x_3 \plus{} x_1 \right)\equal{}4   \\  \left(x_2 \plus{} x_4 \right) \left( x_1 x_3 \plus{}1\right)\equal{}4   \end{cases}$
so
$ 4|\left(x_2 x_4 \plus{}1 \right)$
$ 4|\left(x_3 \plus{} x_1 \right)$

$ 4|\left(x_2 \plus{} x_4 \right)$
$ 4|\left( x_1 x_3 \plus{}1\right)$

All beauty, but we don't have to have solutions in integers, but in rational. So, maybe here is more solutions?

[hide="1, messy"]Any variable $ 0$ clearly fails. Let $ x_1 x_2 x_3 x_4 \equal{} C$, so that each equation is $ \frac {C}{x_i} \plus{} x_i \equal{} 2$. Then $ x_i^2 \minus{} 2x_i \plus{} C \equal{} 0$. If $ C \equal{} 1$ all $ x_i$ must be $ 1$. If $ C > 1$ no solutions to this. Otherwise, some case work. (we will be implicitly using $ x_1 x_2 x_3 x_4 \equal{} C$ and that the product of the roots of that quadratic is $ C$ by Vieta's; these facts work very well together)

1. All four of the $ x_i$ are the same root of this equation. Then $ x_i^4 \equal{} C$, so the two roots must be $ \pm C^{3/4}, \pm C^{1/4}$. Note $ 0 < C < 1$ from $ x_i^4 \equal{} C$. The roots of the equation are
$ 1 \pm \sqrt {1 \minus{} C}$, both positive. We need $ (1 \minus{} \sqrt {1 \minus{} C})^3 \equal{} 1 \plus{} \sqrt {1 \minus{} C}$. Let $ a \equal{} \sqrt {1 \minus{} C} > 0$, then this reduces to $ a^2 \minus{} 3a \plus{} 4 \equal{} 0$, with no solutions.

2. Three of the $ x_i$ are the same root, and the last is the other. WLOG $ x_4$ is the lone one. Then $ x_3 x_4 \equal{} C$, so $ x_1 x_2$ must be $ 1$. Note $ x_1 \equal{} x_2$, so $ x_1 \equal{} x_2 \equal{} 1$ (only if $ C \equal{} 1$, which is taken care of), or $ x_1 \equal{} x_2 \equal{} \minus{} 1$. This happens when $ C \equal{} \minus{} 3$.

3. Two of the $ x_i$ are one root, and two are the other. Their product must be $ C^2$, so $ C^2 \equal{} C$ and $ C \equal{} 0$ or $ C \equal{} 1$. $ C \equal{} 1$ taken care of, and $ C \equal{} 0$ is not possible.

From before the case work: $ (1,1,1,1)$ works.
From case 2: $ ( \minus{} 1, \minus{} 1, \minus{} 1,3)$ and cyclic permutations work.[/hide]

[hide="2"]Let $ x \equal{} 1 \minus{} z$ and $ y \equal{} 1 \minus{} z \plus{} z^2$ for any $ z$.

Motivation: Equation can be reduced to $ (x \plus{} z)(y \plus{} z) \equal{} 1 \plus{} z^2$.[/hide]


trigonometry
calculus
integration
logarithms
algebra
linear equation
calculus computations

How do you solve this:
\[ \int (\ tan x)^2* sec x dx\]
I can't get to an integral I know.  Thanks

I know it becomes  \[ \int ((sec^3 x) \minus{} sec x )dx\] but how do you integrate (sec x)^3. Is the only answer to the original problem \[ \int \sec^3 x \, dx \equal{} \frac{1}{2}\sec x \tan x \plus{} \frac{1}{2}\ln|\sec x \plus{} \tan x| \plus{} C.\] then subtract \[ \int \sec{x} \, dx \equal{} \ln{\left| \sec{x} \plus{} \tan{x}\right|} \plus{} C\]

The "infinite loop" is actually the key to working the problem.

Do one integration step and one trig identity step. You should get an equation in which the unknown integral you want appears on both sides of the equation, but with different coefficients. You'll also have another term or two on one side. Now, consider this an equation to be solved for the unknown integral - a linear equation, so you don't have to do anything unusual. Just add or subtract to get all copies of the unknown integral on the same side of the equation, combine them, and then divided by the appropriate constant.

Hello

$ I = \int sec^3x dx$
=$ secx.sec^2x dx$
=$ secx.tanx - \int secxtan^2x dx$
=$ secxtanx - \int secx(sec^2x - 1)dx$
=$ secxtanx - I + \ln{\left| \sec{x} + \tan{x}\right|} + C$
$ 2I = {secxtanx + \ln{\left| \sec{x} + \tan{x}\right|} + C}$
$ I = \frac {1}{2}[secxtanx + \ln{\left| \sec{x} + \tan{x}\right|}] + c$

thank u

[quote="mikesdaniel"]
$ \int \tan^2x \sec x dx$

[/quote]

$ =\int (\tan x)(\sec x \tan x dx)
\\
\\
\\=\int \left\{\sqrt{\sec^2x-1}\right\}\{\sec x\}'
\\
\\
=\int \sqrt{t^2 - 1}dt, \; \text{where } t=\sec x
\\
\\
\;
\text{Now, use: }
\\
\\
\int \sqrt{x^2-a^2}dx = \frac{x}{2}\sqrt{x^2 - a^2} - \frac{a^2}{2}\ln \left(x+\sqrt{x^2-a^2\right)}$

A lot of times, when doing integration by parts, you'll do it twice and end up right back where you started, except with a negative.  And this is the key since you can now combine like terms and divide.  In particular, $ \int\sec^{3}x\,dx$ is a common integral.
[hide="Exercise"]Evaluate the indefinite integral $ \int\sqrt{x^{2}\plus{}a^{2}}\,dx$.
[Hint: Do the right trig substitution and you'll wind up with that secant integral, except in a different variable.  Then it's just re-inserting your original variables.][/hide]


what's up with the march 10 olympiad transcript isn't that in the future. maybe the instructors can predict the future

Actually I believe that one is just the transcript for the Olympiad Geometry Math Jam hosted by Mr. Rusczyk on December 3rd. I think someone accidentally double-uploaded the transcript for it again mixing it with the upcoming Olympiad Math Jam.

I believe there was also a March 10 in the past? Although that Math Jam was Dec. 3.

Impossible. AoPS did not exist in March 2003.

I didn't say it did. I said that I thought there was a March 10th in the past...

Alas, my future-predicting powers are most meager.

Fixed this.


calculus
integration
algebra
polynomial
calculus computations

Find the integral

${ \int^1_0\frac{x^{4n}(1-x)^{4n}}{1+x^2}dx}$ , $ n\in\mathbb{N}$

I would do this:

Let $ I_{4n,4n}\equal{}\int\limits_{0}^{1}\frac{x^{4n}(1\minus{}x)^{4n}}{1\plus{}x^{2}}dx$.

Then we have that

$ I_{4n,4n}\equal{}\int\limits_{0}^{1}x^{4n\minus{}2}(1\minus{}x)^{4n}dx\minus{}\int\limits_{0}^{1}\frac{x^{4n\minus{}2}(1\minus{}x)^{4n}}{1\plus{}x^{2}}dx$

We get the following recurrence formula:

$ I_{4n,4n}\equal{}\beta(4n\minus{}1,4n\plus{}1)\minus{}I_{4n\minus{}2,4n}$.

and now simply play with it.....

$ I_{4n,4n}\equal{}\sum^{2n}_{k\equal{}1}(\minus{}1)^{k\plus{}1}\frac{(4n\minus{}2k)!(4n)!}{(8n\minus{}2k\plus{}1)!}\plus{}I_{0,4n}$

How would you further simplified ?  :ninja:

Use the substitution $ x\equal{}1\minus{}t$, and you get that:

$ I_{0,4n}\equal{}\int\limits_{0}^{1}\frac{(1\minus{}x)^{4n}}{1\plus{}x^{2}}dx\equal{}\int\limits_{0}^{1}\frac{x^{4n}}{x^{2}\minus{}2x\plus{}2}dx$.

Now, simply divide the two polynomials, namely $ x^{4n}$ by $ x^{2}\minus{}2x\plus{}2$ and then integrate.

If you are looking for better approximations of $ \pi$ perhaps this integral should be studied:

$ \int\limits_{0}^{1}\frac{x^{4n}(1\minus{}x)^{4}}{1\plus{}x^{2}}dx\equal{}$

$ \equal{}4\left(1\minus{}\frac{1}{3}\plus{}\frac{1}{5}\minus{}\cdots\minus{}\frac{1}{4n\minus{}1}\right)\plus{}\frac{5}{4n\plus{}1}\minus{}\frac{4}{4n\plus{}2}\plus{}\frac{1}{4n\plus{}3}\minus{}\pi$.


inequalities
function
limit
real analysis
calculus
integration
real analysis unsolved

$ (X,M,\mu)$ is a measure space with $ \mu(X)<\infty$ and suppose a measurable function $ f$ is in $ L^p(\mu)$ for $ p\geq 1$.

$ 1.$ Verify that $ \|f\|_q\leq \|f\|_p^{(p/q)}\|f\|_p^{(1\minus{}p/q)}$ for all $ p\leq q\leq \infty$

$ 2.$ Show that $ f\in L^\infty$

$ 3.$ and $ \|f\|_\infty\equal{}\lim_{q\to\infty}\|f\|_q$

ABM, I've told you before that problems about Lebesgue measure and integration (or any other topic normally covered in graduate courses) do not belong in the the "computations and tutorials" section.

As for this problem, [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=208924]this topic[/url] may be relevant - although it's not quite the same problem.

Most of what you've said above is either misstated or false. Certainly $ f\in L^p$ does not imply that $ f\in L^{\infty}.$


I've only seen Star Wars, so.... Star Wars.

there are alot of very good movies out there.

How can you expect us to pick from a list of just nine?
Some of those movies aren't even that good :ninja:

Schindler's List but no Titanic?

No Silence of the Lambs?

FAIL.

MONTY PYTHON AND THE HOLY GRAIL

And any othe Python. But I voted for LoTR. Actually I think I like Star Wars more, but the new three polluted the old three. So many movies are awesome though! Banlieue 13 anyone?

Hey, has anyone seen JSA (Joint Security Area)?
It's a korean movie, and a really good one

thats my pick.

i believe "korean movie" and "good" are mutually exclusive

jackass 2 is rather good, as is office space
and euro trip is 23.99 karat gold.

[quote="Treething"]i believe "korean movie" and "good" are mutually exclusive[/quote]
not really, I think there's an intersection set of size one. Tae guk gi is a nice movie(my fav war movie actually)

i like harry potter myself. read them all    seen all the movies!         harry potter fans pm me!!!!!! :D                                            harry potter is number :first:

i htougt you hated this forum and you liked ur art skillz

Come on moogra, that was his first civilized post, and you insult him. Lets show some positive feedback. You go cubedude! Keep on truckin!

there are 4 correct answers to this poll:

1. Holy Grail
2. Fight Club
3. Almost Famous
4. Life of Brian

sure there's other good movies, but those are the best, so ummmmm.........  yeah.

"Neo I am your father."

ok where was IJ he is the best!!!! Someday maybe even your bones will be worth something to sort of quote him!

Schindler's list is not bad but overrated as well...
A bit over the top at times :|

I voted Star Wars, but my other favorite it Monty Python and the Holy Grail!!!

ALL HAIL MONTY PYTHON! :omighty:

And now for something completely different and the Holy Grail are the best, even though I voted for the Lord of the rings (the scenes of Minas Tirith are absolutely amazing!)

Judge: and what do you burn   [i]besides[/i] witches?
Crowd: MORE WITCHES!

Person in the crowd: She turned me into a newt.
Judge (sees that theres nothing wrong with the person) A newt?
Person:.........I got better.

The Prestige!! Yea

I'm just kidding, any Christian Bale movie is the best and they are all equally good

how come cubedude's post count is zero?
EDIT - he posted in the G&FF.  lol


Καθορίστε την ελάχιστη τιμή της παράστασης
$ S \equal{} \sqrt {(a \plus{} 1)^{2} \plus{} 2(b \minus{} 2)^{2} \plus{} (c \plus{} 3)^{2}} \plus{} \sqrt {(b \plus{} 1)^{2} \plus{} 2(c \minus{} 2)^{2} \plus{} (d \plus{} 3)^{2}} \plus{} \sqrt {(c \plus{} 1)^{2} \plus{} 2(d \minus{} 2)^{2} \plus{} (a \plus{} 3)^{2}} \plus{} \sqrt {(d \plus{} 1)^{2} \plus{} 2(a \minus{} 2)^{2} \plus{} (b \plus{} 3)^{2}}$ όπου $ a,b,c,d$ πραγματικοί αριθμοί...

[hide]θεωρω $ k\equal{}\sum a$
τοτε απο Miknowsky ισχυει,
$ S\geq \sqrt{ (a\plus{}b\plus{}c\plus{}d\plus{}4)^2\plus{} 2(a\plus{}b\plus{}c\plus{}d\minus{}8)^2\plus{}(a\plus{}b\plus{}c\plus{}d\plus{}12)^2} \equal{} \sqrt {4(k^2\plus{}72)} \geq 12\sqrt{2}$
ισοτητα για a=b=c=0 αρα $ S_{min}\equal{} 12\sqrt{2}$[/hide]


geometry
circumcircle
geometric transformation
reflection
inequalities
inequalities proposed

Let $ ABC$ be a triangle, the its circumcircle $ C(O,R)$ , the its incircle $ C(I,r)$ ,

the orthocenter $ H$ and the reflection $ A'$ of the point $ A$ w.r.t. the point $ O$ .

Prove that there is the inequality $ AH\cdot AI \plus{} A'H\cdot A'I\ge 2R^2\ .$

Nobody ? I don't think it is very hardly !


can ne 1 please help me with kvpy.......????
all my teachrs r expectin a lot from me.........
but i dont kno from where to prepare for it........???

please help.......!!!

[url]http://www.mathlinks.ro/viewtopic.php?t=212237[/url]

it would be better if you post in Indian Forum about kvpy.
I am also giving the same. :)


Two blocks - http://www.mathlinks.ro/weblog_entry.php?p=949758#949758

what's up with that?

[quote="jmerry"]I can't access your link. Can you provide the code here?[/quote]hope you can shed some light for me, thanks!!! :-]  much appreciated.

$ \\
\uparrow \equal{} \mbox{Plug into equation above} \\
 \\
a) \\
 \\
\Sigma{F}_{x_{A}} \equal{} 0 \\
\mu_{s}N \equal{} 0 \\
 \\
\Sigma{F}_{y_{A}} \equal{} 0 \\
N \minus{} w_{A} \equal{} 0 \\
 \\
\Sigma{F}_{x_{B}} \equal{} 0 \\
\mu_{k}N \plus{} \mu_{s}N \minus{} F \equal{} 0 \\
 \\
\Sigma{F}_{y_{B}} \equal{} 0 \\
N \minus{} (w_{A} \plus{} w_{B}) \equal{} 0 \\
 \\
\mu_{k}N \minus{} F \equal{} 0 \\
\uparrow N \minus{} (w_{A} \plus{} w_{B}) \equal{} 0 \\
 \\
F \equal{} \mu_{k}(w_{A} \plus{} w_{B}) \\
 \\
F \equal{} 4.80N \\
 \\
b) \\
 \\
\Sigma{F}_{x_{A}} \equal{} 0 \\
T \minus{} \mu_{s}N \equal{} 0 \\
 \\
\Sigma{F}_{y_{A}} \equal{} 0 \\
N \minus{} w_{A} \equal{} 0 \\
 \\
\Sigma{F}_{x_{B}} \equal{} 0 \\
\mu_{k}N \plus{} \mu_{s}N \minus{} F \equal{} 0 \\
 \\
\Sigma{F}_{y_{B}} \equal{} 0 \\
N \minus{} (w_{A} \plus{} w_{B}) \equal{} 0 \\
 \\
\mu_{k}N \plus{} \mu_{s}N \minus{} F \equal{} 0 \\
\uparrow_{\mu_{k}} N \minus{} (w_{A} \plus{} w_{B}) \equal{} 0 \\
 \\
\mu_{k}(w_{A} \plus{} w_{B}) \plus{} \mu_{s}N \minus{} F \equal{} 0 \\
\uparrow_{\mu_{s}} T \minus{} \mu_{s}N \equal{} 0 \\
\uparrow N \minus{} w_{A} \equal{} 0 \\
 \\
F \equal{} \mu_{k}(w_{A} \plus{} w_{B}) \plus{} \mu_{s}w_{A} \\
 \\
F \equal{} 1.80N$

[quote="stevem"]The image runs out of room so the last line F = 1.80N can''t decide whether to go at the end or the next page, except that there isn't a next page. So either remove some of the blank lines or split the code into separate tex areas.[/quote]gotcha thanks


Given that $ x < 5$, rewrite $ 5x \minus{} |x \minus{} 5|$ without using absolute value signs.

hello, if $ x<5$ then we have $ x\minus{}5<0$ and our term is $ 5x\minus{}(5\minus{}x)\equal{}5x\minus{}5\plus{}x\equal{}6x\minus{}5$.
Sonnhard.


integration
trigonometry
calculus
algebra
binomial theorem
real analysis
real analysis theorems

[quote="benorin"][quote="greengrass"]come on, nobody?[/quote]

I played with this for a while... nothing solid yet. It would be easier as a definite integral, say -pi.2 to pi/2. Did you try converting it to complex exponentials and expanding by the binomial theorem, multiplying the series as a Cauchy product and then integrating? I got it to a double sum that could be integrated but the result was not pretty...[/quote]

I tried and I get the same or a similar result  :(  not a recurrence formula. Tks.

[quote="benorin"]I found an exercise in a text book which says to evaluate $\int_{-\pi/2}^{\pi/2}(1+e^{2ix})^\alpha (1+e^{-2ix})^\beta\, dx$ by use of the binomial theorem and theorem giving $_{2}F_{1}$ as an integral to show that

$\int_{-\pi/2}^{\pi/2}\cos^{\alpha+\beta}x\cos\left[ (\alpha-\beta) x \right] dx = \frac{\pi\Gamma (\alpha+\beta+1)}{2^{\alpha+\beta+1}\Gamma (\alpha+1)\Gamma (\beta+1)}$[/quote]

Tks benorin. I left this. Actually I think it does not exist.

[quote="greengrass"]$\int \sin^{n}ax \cdot \sin^{m}bx \cdot dx$

Tks.[/quote]
I think there not exist some formula

By the way, this is my recurrence formulae list

http://es.wikipedia.org/wiki/F%C3%B3rmulas_de_reducci%C3%B3n


1. How do you know if a decimal number can be expressed in the binary system?

2. According to my computer science teacher, 1/10 can't be expressed in the binary system. What does he mean?

3. For my homework, I got the question that says that Java uses a 32 bit integer type. How do I determine the largest integer and the smallest integer in that system?

4. Referring to the above problem about Java, what number is represented internally by $ 11111111111111111111111111111111_{2}$ (that should be 32 1s if I typed it correctly)? What does it mean for it to be represented internally?

For some reason, I think that the 1 denotes the number is negative. Does it?

5. Another of my homework questions was: "In an 8-bit system the base 16 number 9A is input. What base 10 number is output?"
How do I tell if 9A is in the range of numbers in an 8 bit system? If it isn't, how do I find out the output?

6. One of my problems required me to find the value of a fraction with a base 9 number as the numerator and a base 4 number as the denominator and express that value as a base 5 number. Is there any faster way to do the problem than converting the numbers to base 10 and converting the result to base 5?

1) For fixed-width numbers (i.e. 32-bit ints, it just has to be in the range specified in #3). For fractions, the base has to be a power of two for it to terminate in binary.

2) 1/10 repeats in binary, so computers need to truncate it somewhere, unless they use special software such as symbolic manipulation libraries which would literally store it as 1/10, rather than its decimal form.

3) For an n-bit type, the max value is $ 2^{n\minus{}1}\minus{}1$ and the min value is $ \minus{}2^{n\minus{}1}$.

4) $ 1111\ 1111\ 1111\ 1111\ 1111\ 1111\ 1111\ 1111_{2}$ would represent $ \minus{}1$ in java because if you add $ 1$ to it, you get $ 0000\ 0000\ 0000\ 0000\ 0000\ 0000\ 0000\ 0000_{2}\equal{} 0$ (the leftmost bit went out of the range)

5) $ 9A_{16}\equal{} 1001\ 1010_{2}$, then do your thing to figure out which (negative) value it is.

6) fastest (although perhaps not for by hand) is to convert them to base five and do the division in base five. Alternatively convert to base 9 or 4, so that you only have to do 2 base conversions instead of 3 (but dividing in bases other than 10 can be tricky by hand)

Thanks.

1. About your answer for question 5, what is that thing to figure out which negative value it is?

I have written down for class notes that after finding the number in base 2, I switch the ones and zeros and then add 1. Then, I convert the resulting number to base 10. Is that correct?

2. Where can I find the algorithms for converting numbers from the output to what the computer sees internally to the output?

3. Where can I find similar practice problems to the ones I found in my homework?

4. In any bit system, how do we find out how a number is represented as 1s and 0s?

5 In an 8-bit system, how is -64 represented as 1s and 0s? The minimum value of an 8-bit system is $ \minus{}2^{7}\equal{}\minus{}128$ so -64 should be in the range... so how do we got about finding it?

6.In an n-bit type, the max value is $ 2^{n\minus{}1}\minus{}1$ and the min value is $ \minus{}2^{n\minus{}1}$. (See Below)

a. How will a number that falls out of the range be represented?

b. Can I just add or subtract $ 2^{n}$ from the number so it falls in the range?

c. If it still doesn't fall in the range, can I keep on adding or subtracting $ 2^{n}$ until it does? - Why or why not?


When does your school start the season, and when is your school competition?

Our season starts either 9/29 or 10/6, and our school competition is probably on 1/13 or 1/20.

I have no idea. Our Mathcounts is pretty bad. I think we started in November last year. ( last year we only had 3 7th graders and not a single 8th)

Our school doesn't really have "practices" or anything. You can just get warm-ups, work-outs, and prior competitions if you ask certain teachers.

My friends and I would organize practices on our own, but that was pretty much the extent of it. Yet, our school always manages to snag first in the state.

my school practice started today!!!
i don't know when the competition is though.

January...dunno when exactly tho. when do u start?

My coach is obsessed. This year he got about 45 people to try-out. We started 9/14 this year, and our school competition is always right before Thanksgiving. But the work pays off... we've won the PA states probably more than any other team :lol:

yah our team always won either 1st or 2nd in states and sometimes won 1st place in nationals...i hope we get first place again this year... :D

Sign up for my math club just started. I forgot to got to the office to sign up today. I'll do it on Monday. Math Club starts on Oct. 3 for me.

I can't wait! :D

Our math club starts next Tuesday.  Everyone in the 'accelerated' math classes [i]must[/i] try out....


The following inequality comes from Vietnam Forum. I don't know author's inequality  :) But It is really nice and easy:D
Let $ a,b,c$ be positive real numbers. Prove that:
\[ \sqrt[3]{\frac{a^{2}+bc}{b^{2}+c^{2}}}+\sqrt[3]{\frac{b^{2}+ca}{c^{2}+a^{2}}}+\sqrt[3]{\frac{c^{2}+ab}{a^{2}+b^{2}}}\ge \frac{9\sqrt[3]{abc}}{a+b+c}\]

Zaizai, see here [url]http://reflections.awesomemath.org/2006_5/2006_5_problems.pdf[/url], it's problem S27. The solution I sent along with the proposal is as follows:

By the AM-GM inequality,
\[ \frac{a(b^{2}+c^{2})+b(c^{2}+a^{2})+c(a^{2}+b^{2})}{a^{2}+bc}=\frac{a(b^{2}+c^{2})}{a^{2}+bc}+b+c\ge 3\sqrt[3]{\frac{abc(b^{2}+c^{2})}{a^{2}+bc}}\]
Hence
\[ \sum_{cycl}\sqrt[3]{\frac{a^{2}+bc}{abc(b^{2}+c^{2})}}\ge\frac{3(a^{2}+b^{2}+c^{2}+ab+bc+ca)}{a(b^{2}+c^{2})+b(c^{2}+a^{2})+c(a^{2}+b^{2})}\ge\frac{9}{a+b+c}\]
where we use the fact
\[ (a+b+c)(a^{2}+b^{2}+c^{2}+ab+bc+ca)-3\sum_{cycl}a(b^{2}+c^{2})=\sum_{cycl}a(a-b)(a-c)\ge 0 \]
which is none other than Schur inequality. Anyway, glad that you think it's nice ;-)

Very nice solution, Ductrung. I really like it.

really nice... very very nice :D A nice solution and a nice problem. Thank you for a nice problem. Hungkhtn, Can you post your?  :D


algorithm
combinatorics open
combinatorics

I would like to ask you whether there are any simple algorithms to recognize whether a given simple non-oriented graph (with less than 15-20 vertices) is planar.

Thanks for any hints.

[url=http://en.wikipedia.org/wiki/Planar_graph]Kuratowski's theorem[/url] can be turned into an algorithm:  one searches for copies of $ K_5$ or $ K_{3, 3}$ and if there aren't any the graph is planar.

I don't think that's as simple as you suggest -- in particular, Kuratowski tells you that you can't have any subgraphs that are subdivisions of $ K_5$ or $ K_{3,3}$, and this is a lot harder than just finding these as subgraphs.  There is, however, an entire Wikipedia article devoted to [url=http://en.wikipedia.org/wiki/Planarity_testing]planarity testing[/url].


solve for max and min of:

[img]http://latex.codecogs.com/gif.latex?y=\sqrt{sinx+tanx}+\sqrt{cosx+cotx}[/img]

Is this:

Find the maximum and minimum of

$ \sqrt{\sin x \plus{} \tan x} \plus{} \sqrt{\cos x \plus{} \cot x}$?

There is no max (sup = $ \infty$), since $ \lim_{0<x \to 0} \equal{} \infty$. It seems min = $ 2\sqrt{1\minus{}\sqrt{2}/2}$, obtained e.g. for $ x \equal{} 5\pi/4$, but I'm reluctant to do the computations!


Let $ a,b,c\ge 0$ Prone that
${ \sqrt {\frac {a^2 - ab + b^2}{c^2 + ab}} + \sqrt {\frac {b^2 - bc + c^2}{a^2 + bc}} + \sqrt {\frac {c^2 - ca + a^2}{b^2 + ca}}\ge \frac{3}{\sqrt 2}}$

The similar can be solved easily with SOS:
$ \sum\ \frac {(a \plus{} b)^2}{ab \plus{} c^2} \geq\ 6$
 :)

[quote="arqady"][quote="Vasc"][quote="Allnames"]Let $ a,b,c\ge 0$ Prone that
${ \sqrt {\frac {a^2 - ab + b^2}{c^2 + ab}} + \sqrt {\frac {b^2 - bc + c^2}{a^2 + bc}} + \sqrt {\frac {c^2 - ca + a^2}{b^2 + ca}}\ge \frac {3}{\sqrt 2}}$[/quote]
This is stronger, but is also easy:

$ \sqrt {\frac {a^2 + b^2}{c^2 + ab}} + \sqrt {\frac {b^2 + c^2}{a^2 + bc}} + \sqrt {\frac {c^2 + a^2}{b^2 + ca}}\ge 3$[/quote]
Yes! AM-GM works here.[/quote]
So you meant the last thing we have to finish off is to prove this following inequality
$ (a^2+b^2)(c^2+b^2)(a^2+c^2) \geq (a^2+bc)(b^2+ac)(c^2+ab)$
which is obvious after expanding. :)

[quote="great math"]
So you meant the last thing we have to finish off is to prove this following inequality
$ (a^2 \plus{} b^2)(c^2 \plus{} b^2)(a^2 \plus{} c^2) \geq (a^2 \plus{} bc)(b^2 \plus{} ac)(c^2 \plus{} ab)$
which is obvious after expanding. :)[/quote]
It's easy even without expanding: $ (a^2\plus{}b^2)(a^2\plus{}c^2)\geq(a^2\plus{}bc)^2.$ :wink:

[quote="Allnames"]How about  my problem?Sorry but I cant see what are they similar to :blush:[/quote]
\[ x^2-xy+y^2 \ge \frac{x^2+y^2}{2} \quad \for all x,y\]

$ \sum\sqrt {\frac {a^2 \minus{} ab \plus{} b^2}{c^2 \plus{} ab}}\geq\sqrt {\frac {a^2 \plus{} b^2}{2(c^2 \plus{} ab)}}\geq\frac {3}{\sqrt {2}}$


I think I found a nice solution:

Call $G$ the intersection point of $EF$ and $BC$.
Now, because $AD,BE,CF$ are concurrent in Gergonne's point, we may conclude that $G,D$ divide $B,C$ harmonically.

Therefore, $HG,HC,HD,HB$ is a harmonic bundle. Because $EF \bot HD$, they have to be the bisectors of $HB,HC$.
$\mathbb{QED}$!

[img]http://img407.imageshack.us/img407/6841/aprilmathlinksol7.jpg[/img]

It’s as same as mine :lol: . And I have another solution.
[quote="April"][color=darkred]The circle inscribed a triangle $ABC$ meets the sides $BC,\,CA,\,AB$ at $D,\,E,\,F$ respectively. Let $H$ be the projection of $D$ on $EF.$
Prove that: $\angle BHD=\angle CHD$[/color][/quote]
[color=indigo][b]Proof.[/b]
Let $I,\,K$ be the projections of $B,\,C$ on $EF$ respectively.
We have: $\frac{BF}{CE}=\frac{BD}{DC}=\frac{IH}{HK}\quad (1)$
On the otherhand, we have: $\angle BFI=\angle AFE=\angle AEF=\angle CEK\Longrightarrow \triangle BIF\sim\triangle CKE$ $\Longrightarrow\frac{BI}{CK}=\frac{IF}{KE}=\frac{BF}{CE}\quad (2)$
From $(1)$ and $(2)$ we have: $\frac{BF}{CE}=\frac{IH}{HK}=\frac{IF}{EK}=\frac{BI}{CK}\Longrightarrow \triangle BIH\sim\triangle CKH$ $\Longrightarrow\angle BHI=\angle CHK$$\Longrightarrow\angle BHD=\angle CHD$[/color]

[color=red]Nice Problem ! It was a 'Greek National Olympiad' problem ![/color]

See also

 http://www.mathlinks.ro/Forum/viewtopic.php?search_id=2067745109&t=59664

 where you can read beautifull solutions.There I have a solution  , exactly the same as above.

[u] Babis[/u]


I don`t understand what needs to be proven because $\ln(1+x_2)\leq \ln(1+x_1)$ and $\ln(1-x_1)\leq \ln(1-x_2)$.


Ross Mathematics Program
MIT
college
search
Columbia
Stanford
CMU

Any students have any plans for college yet? I may be only a freshman in high school, but I'm already interested in picking a good college. I live in upstate NY, and the best colleges hereabouts seem to be RPI and Skidmore. I like the campuses, anyway, don't know much about their education. My goal now is that if I can take 1 college course per semester, I can skip my freshman year in college. I've already earned 8 college credits... Feel free to post about how smart you are here...

-(very) Egotistically, Isaac

I have 3 college credits (GO ROSS!!!) ... and i would love to goto an ivy league...or MIT (my dream)...any one going to liberal arts colleges or anything?

No liberal arts for me... math or CSCI, rather... I'd also love to get into MIT. I don't necessarily want to get into an Ivy League.

-Isaac

i think he means AP classes which mean Advanced Placed classes. some high schools dont have them though. nuff said

Actually I did well enough on the SAT that RPI (a college near me) considered and accepted me for scholarship courses. I am also taking AP courses.

-Isaac

I live in a university town and the school district has a deal with the U that qualified students can take classes at the university for free, getting collage and high school credit.  I'm planning on finishing my last two years in high school... in collage.  Still not sure where I want to go, don't want to stay here though.

Hey there MIT-philes...

My husband and I both graduated from there. If you have any questions I'll be glad to answer 'em as best I can.  Feel free to post here, but since I'm only an every other week browser of the forum, you can also write me at:

emfermier@alum.mit.edu

Lots of luck!

Elise

I'm going to take a summer course at FLCC ([url=http://www.flcc.edu/]click here[/url]).

Out of curiosity, andy17null, how much did you need to get on the SAT for the college to offer classes to you? I might see if I could do that.

Erm... I don't remember at this moment.  :oops:. When I find my scoresheet I'll tell you. I took it in 8th grade with the Johns Hopkins Talent Search. They referred me to the college. Check out  [url=http://www.jhu.edu/gifted/ts/index.html]this[/url] site for specifics. Unfortunately, it's only open for students up to 8th grade. If you're in high school, I reccommend that you call the Admissions Office for your local colleges. Ask them if they have any programs that will let you take courses. Give them a copy of your transcript and a letter from your guidance counselor. That's what I did, anyway. Hope this helps, and I'll get you the scores ASAP.

-Isaac

I got college courses offered to me twice through the Johns Hopkins Talent Search too.  I don't know what the score to get one is, but I can tell you what I got, maybe that can help.  In seventh grade I got a 720 math (710 verbal) and in eight I got a 700 math (800 verbal).  (I think that the second time probably wasn't for my math score.)

Don't be kiding me, but 800 verbal is quite an accomplishment. Unlike 800 math. The SAT I math is a complete junk, as I have stated already somewhere else on this forum. It's basically a test of your not making any stupid mistakes. That was the reason why I got 760 on my first try.

you can still miss 1 or 2 questions and get a perfect score on the math. if youre lucky that is.

The SAT I math sections are quite easy (especially if you've trained for math in math competitions).  The verbal sections are a lot harder for me.  You can miss like 5 or 6 questions and still get an 800, but it's really hard.  Even with a more leanient scale, it still poses a big challenge for me.

-interesting_move

I've got an idea. If ComplexZeta joins the conversation, it wont be 2-way anymore.

Yet it'd be a paradox...
If ComplexZeta joins, then it won't be a two-way conversation anymore, thus he'd contradict himself
If he doesn't, he proves his point

Wasn't there something like this in "Minority Report"? :)

Anyways, maybe it's because it's kind of getting late...(s'pecially here in the East Coast...it's almost midnight and I [b]still[/b] have homework to finish...*sigh*)

lol...yea...i think we should get off and continue this spam tomorrow

yea, but i still can't go to sleep...I'm almost done, but [b]for some reason[/b] time just flies, and I don't have anything done. Any idea where my little friend "time" went? :)

it's a joke...i mean, i know it's a bad one. i'm not even bothering to capitalize the is any more and the glue stick i'm using smell's really good, like lavender.

help, pleese :)

well, more phisical since something that starts with the letter c has let me stay up this late :twisted:

and it's perfectly legal, don't worry, because it's not what you think it is.
Even more confusing fact: Coke has tons of this stuff

hmm...caffeine is good for u!!...especially in the writing process

ok, now it's running out...quickly depleting...WARNING: Blood-Coc...Caffiene level reaching <5% :WARNING :)

I'm going to bed...

this revive is older than some people

nice


What is the units digit of the sum of the squares of the first nine
positive integers?


calculus
integration
function
inequalities
real analysis
real analysis unsolved

Let $f: [0,\infty)\to [0,\infty)$ bea continous function such that $f(f(x))=x^{2}$.
    Prove that \[\int_{0}^{1}f^{2}(x)dx\geq\frac{3}{13}.\]

[quote="Cezar Lupu"]Let $f: [0,\infty)\to [0,\infty)$ bea continous function such that $f(f(x))=x^{2}$.

[/quote]

From this condition $f$ is bijective $R+$ to $R+$

Do we need use Cauchy-Schwarz inequality Cezar ?

First,I will prove f is strictly increasing on $[0,+\infty)$
By observe,It is easy to see $f$ is bijective $[0,+\infty)$ to $[0,+\infty)$(As Moubinool said).
If there exist $a>b \ge 0$,such that $f(a) \le f(b)$
because $f$ is bijective,so we have:$f(a) < f(b)$
Then using $f$ is continous and continous.We can prove:$f(x)<f(b),\forall x>a$.
Let $M=\max\{f(x)|x \in [0,a]\}$,Thus $f(x) \le M,\forall x \in [0,+\infty)$
Which is contradiction with $f$ is bijective.
Thus we proved that  f is strictly increasing on $[0,+\infty)$.
For proving the original problem,we should use another lemma:
$Lemma:$If $u(x),v(x)$ are monotonic continous function on $[a,b]$ then we have:$u(x)v(x)|_{a}^{b}=\int_{a}^{b}u(x)d v(x)+\int_{a}^{b}v(x)d u(x)$
The lemma is not hard to prove.

For the original problem:
By  f is strictly increasing on $[0,+\infty)$.
It is easy to see $f(0)=0,f(1)=1$,let $x=f(t)$ in the inequality.
We have:$\int_{0}^{1}f^{2}(x)dx=\int_{0}^{1}f^{2}(f(t))df(t)=\int_{0}^{1}t^{4}df(t)$
$=t^{4}f(t)|_{0}^{1}-\int_{0}^{1}f(t)dt^{4}=1-4\int_{0}^{1}t^{3}f(t)dt \ge 1-4\sqrt{\int_{0}^{1}t^{6}dt \int_{0}^{1}f^{2}(t)dt}=1-4\sqrt{\frac{1}{7}\int_{0}^{1}f^{2}(x)dx}$
With it,it is easy to prove:$\int_{0}^{1}f^{2}(x)dx \ge \frac 3{13}$
 :)


[quote="Anto"]Prove or disprove the following inequality.
$(x^2+y^2)(y^2+z^2)(z^2+x^2)(xy+yz+zx) \geq \frac{3}{8}(3+x^2)(3+y^2)(3+z^2)xyz $ ,
where x,y,z are positive reals such that $x^2 +y^2+z^2 = 3$[/quote]
Calculate the terms for $x=\sqrt{2},y=z=\frac{\sqrt{2}}{2}$ and you will see that the inequality is wrong.


algebra
polynomial
number theory proposed
number theory

problem 4
a,b,c are real number such that $ \ a^n \plus{} b^n \plus{} c^n$ is a integer number for all positive integer n
prove that there exists 3 integer number p,q,r such that a,b,c are 3 roots of the polynomial $ \ x^3 \plus{} px^2 \plus{} qx \plus{} r$

Hey, i have this solution, but I don't think it's quite good
Prove this lemma: If
\[ a^n \plus{} b^n \plus{} c^n \in Z \] with\[ n\in Z \plus{}\] then there must be 1 interger (a or b or c)
Let c is an interger
So \[ c^n, a^n \plus{} b^n \in Z\]
with \[ n\in Z \plus{} \]
Then we have \[ a \plus{} b, a^2 \plus{} b^2, a^4 \plus{} b^4 \]
are intergers -> \[ ab \] is an interger
Now, we will solve the problem
Applying Viete we have a,b,c are roots of that equation if and only if
\[ a \plus{} b \plus{} c \equal{} \minus{} p,ab \plus{} bc \plus{} ca \equal{} q,abc \equal{} \minus{} r \]
Moreover, we have
\[ q \equal{} ab \plus{} bc \plus{} ca \equal{} ab \plus{} c(a \plus{} b \plus{} c) \] and \[ \minus{} r \equal{} abc \]
are intergers
So there exists p,q,r -> Done  :P
P/s: I'm vietnamese. I'm dissapointed because i'm not old enough to take part in my school's team :mad:

The lemma is wrong.

Let take $ a, b, c$ are roots of equations $ x^3 \minus{} 3x \plus{} 1 \equal{} 0$ then no one of $ a, b, c$ is integer, but

     $ a^n \plus{} b^n \plus{} c^n$ is integer for all $ n$ from $ N$.

You must to prove lemma before using it.

Oh man, i'm sry. I was wrong
Here is my new solution (I think it won't be wrong anymore ....  :( )
The first, we have $ a \plus{} b \plus{} c \equal{} \minus{} p$ so then there exists p
Then we prove that  $ 2(ab \plus{} bc \plus{} ca) \equal{} (a \plus{} b \plus{} c)^2 \minus{} (a^2 \plus{} b^2 \plus{} c^2) \in Z$ and $ 2(a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2) \equal{} (a^4 \plus{} b^4 \plus{} c^4) \minus{} (a^2 \plus{} b^2 \plus{} c^2)^2\in Z$
So, we have $ 6abc \equal{} 2(a^3 \plus{} b^3 \plus{} c^3) \minus{} (a \plus{} b \plus{} c)[2(a^2 \plus{} b^2 \plus{} c^2) \minus{} 2(ab \plus{} bc \plus{} ca)] \in Z$
and $ 6a^2b^2c^2 \equal{} 2(a^6 \plus{} b^6 \plus{} c^6) \minus{} (a^2 \plus{} b^2 \plus{} c^2)[2(a^4 \plus{} b^4 \plus{} c^4) \minus{} 2(a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2)] \in Z$ ->$ abc \in Z$
Moreover, $ 2(ab \plus{} bc \plus{} ca)^2 \equal{} 2(a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2) \plus{} 4abc(a \plus{} b \plus{} c) \in Z$ so we have $ ab \plus{} bc \plus{} ca$ is an interger
-> There exists p,q,r
 :oops:

Nice and neat solution! Congratulations!

But you should explain in details about the facts:
    1) 6z and 6z^2 in Z => z in Z
    2) 2z and 2z^2 in Z => z in Z

Oh thank you
It's not too hard to explain  :)
Let $ z\equal{}\frac{k}{2}$ with k is odd then $ 2z^2\equal{}\frac{k^2}{2}\not \in Z$ -> wrong !!!!
We do the same way to prove that 6z and 6z^2 in Z => z is in Z
P/s: Are you a teacher of PTNK ???

[quote="mathboy2710"][quote="Mathias_DK"][quote="mathboy2710"]Prove this lemma: If
\[ a^n \plus{} b^n \plus{} c^n \in Z
\]
with
\[ n\in Z \plus{}
\]
then there must be 1 interger (a or b or c)
[/quote]
But that is clearly wrong... Let $ a,b,c$ be roots in $ x^3 \minus{} 901x^2 \plus{} 260000x \minus{} 24000000 \equal{} 0$...[/quote]
Hey, try n=5, $ a^5 \plus{} b^5 \plus{} c^5 \equal{} 13669406804497,5......\not\in Z$  :lol:[/quote]
Are you sure.. ;)


A model train is made by connecting cars with possible lengths of $ 2''$ and $ 5''$. What is the number of inches in the greatest impossible whole-number length of such a train?

Use the Chicken McNugget Theorem. This yields

$ 2 \times 5 \minus{} 2 \minus{} 5 \equal{} 3$

Another way w/o formula: Every even can be made, since we have 2's. 5 can also be made. To make any other odd greater than 5, you can just add 2's. 1 and 3 obviously cannot be made, so the answer is 3.


LaTeX
binomial coefficients
combinatorics unsolved
combinatorics

Consider the multiset {n*a, n*b, 1,2,3,...n+1} of size 3n + 1.  Determine the number of n combinations.

Do I use

n + k - 1 choose n  where k = 3n + 1

giving me 4n choose n

So for n = 2, I would have 8 choose 2, given me 28 different 2 combinations?

Does this logic make sense?

By "number of $ n$ combinations" I assume you mean the number of sub-multisets of size $ n$, right?

Your logic doesn't make sense because you haven't used any logic: you've taken a formula and misapplied it.    $ \binom{n \plus{} k \minus{} 1}{n}$ is the number of ways to choose $ n$ things from a collection of size $ k$ where (1) each of the $ k$ things is distinct and (2) each thing may be chosen as often as you like.  Neither (1) nor (2) applies here, leaving us no reason whatsoever to expect that formula to work.  In particular, it's clear that the multiset $ \{1, 2, 3, a, a, b, b\}$ doesn't have nearly $ 28$ distinct sub-multisets of size 2.

There is no convenient formula for the number of sub-multisets of a given multiset of a given size.  In this case, I would count by considering how many $ a$s and $ b$s you choose.  That will give you a sum of binomial coefficients to simplify.  At the end you'll get a nice, simple formula, so there should be a straightforward combinatorial argument to arrive at it (but I haven't thought of one yet :wink: ).

Thanks for the advice Joel, I'm stuck in figuring out the simple formula; so if anyone else has any other help that would greatly be appreciated.

Why don't you write down what you have done (the forum is equipped with [[LaTeX]] -- see for example http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690 -- so it's easy to write mathematical formulas -- the key things you'll need are \binom{blah}{halb} for $ \binom{blah}{halb}$ and \sum_{bottom}^{top} for $ \sum_{bottom}^{top}$, then just put them in between two dollar signs) and then we can see about helping you along?

Joel,

What I've done is in the 1st post, there is nothing beyond that I have done; I can't figure anything else.

And I've already given you a suggestion on how to proceed in my first post on this thread.

Ok so what i've tried so far is breaking it up into sets

basically

(n*a , n*b)  * (1,2,....,n+1)

I can make a n-combination of the  multiset (n*a, n*b) by $ \binom{n \plus{} 1}{n}$, and I do understand the (1,2,3,....,n+1) will be a summation, I just don't know of what.  Any advice?

Let me give you a more directed hint: let $ i$ be the number of $ a$s and $ b$s total that you use in your $ n$-submultiset.

(1) What is the range of possible values of $ i$?
(2) For a fixed $ i$, how many ways are there to choose the $ i$ total $ a$s and $ b$s for our subset?
(3) For a fixed $ i$, how many ways are there to choose the other (non-$ a$, non-$ b$) elements of the subset?

$ \sum_{i \equal{} 0}^{n}$ $ \binom{i \plus{} 1}{i}$  $ \binom{n \plus{} 1}{n \minus{} i}$

Thank you for the helpful hints.  I think this is right?

Indeed it is.  As I mentioned, it happens that we can evaluate this sum into a nice form:
[hide="calculation"]\begin{align*}\sum_{i = 0}^{n} \binom{i + 1}{i} \binom{n + 1}{n - i} & = \sum_{i = 0}^{n} (i + 1) \binom{n + 1}{n - i} \\
& = \sum_{i = 0}^{n} (n + 1)\binom{n}{i} \\
& = (n + 1)2^n
\end{align*} (using some standard binomial coefficient magic).[/hide]

Joel, thank you so much for your help

I spent 2 hours on it before I posted it here, I was making it much more difficult then it needed to be.


Suppose a and b are real numbers such that the roots of the equation
         ax^3 – x^2 +bx -1 = 0 are all positive real numbers. Prove that
(i) 0<3ab<=1 and (ii) b>= 3^(1/2)

PLZ provide an answer...

thanx.....

pleez tell me how to prepare for RMO?
i urgently need some advice abt that... :|


The basis of a  topology on a set $ \mathbf{S^{\sim }}$ obtained from one-point compactification of set $ \mathbf{S}$ with $ \infty$ is as given in the book i read the union of the basis of the topology on $ \mathbf{S}$ and the complements of closed compact subsets of $ \mathbf{S}$.
Though i do understand the motivation of defining the topolgy basis in such a way by imagining $ \mathbf{S} \equal{} \mathbf{R^{2}}$.But isnt simply the union of basis sets in $ \mathbf{S}$ and $ \infty$(that is if we remove the condition that the sets have to be complements of closed compact subsets of $ \mathbf{S}$) considered as a point set also a basis though of course it will be very 'boring' example to consider

Also, that construction is not a compact space. "Compactification" implies otherwise.

[quote="t0rajir0u"]It doesn't have the right properties; the closure of $ S$ needs to contain $ \infty$, but the topology you've given makes $ \infty$ an isolated point.[/quote]
My construction was just an example to form a basis without the property of the subset being closed and compact.I just wanted to know if the definition was just a motivated one from the real line/plane(to include the point $ \infty$) or whether the condition of closed and compactness was needed to prove that the new "basis" set was indeed a basis set which clearly is not true.Thanks!

[quote="t0rajir0u"]The definition is easy to motivate: if $ X$ is a compact Hausdorff space and $ p$ a point in $ X$ which is not isolated, then $ X \minus{} p$ is locally compact Hausdorff.  We can recover the topology of $ X$ from the subspace topology on $ X \minus{} p$ precisely by constructing the one-point compactification.[/quote]
yep! nicely put across.thanks!


well, i don't know what's happened, but i have had this one problem repeat to me about three times now and it has the same ratings as i put on the first time. i don't know if that's a glitch or not, but it's weird to me.

second, somehow right now i cannot do any problems on anyother subjects other than what i have already passed. i can watch the vids, but i cant do the problems. i dont know what's happening.

What happens when you let Alcumus choose the topic?

it just chooses a topic that i have already passed and most problems are the one's i already solved

I am having similar problems. It kepts repeating the same problem and it has totally forgotten my progress in a topics.

Can you post here when the problem repeat issue happens again?  We've added some code to help us diagnose the problem.

Hello, everyone!

We've tried our best to fix this problem. Please tell us if this continues to happen. :)

today i had the problem there are 6 letters ABCDEF and it must begin with C and always has a letter B in it popup over and over
by the way the ans is 36


Can you show me how to solve this problem?
If 1/3<= a,b,c<=3 then a/(a+b)+b/(b+c)+c/(c+a)>=7/5

Thanks.

see http://www.mathlinks.ro/Forum/viewtopic.php?t=134826  :wink:


The first basic step is to use the search function and look into the forum before posting. Just a couple of posts below, there was this topic: http://www.mathlinks.ro/Forum/viewtopic.php?t=28268


what's the name for an 11-sided polygon?  :wallbash:  :|  :wallbash_red:

hendecagon

Google and Wikipedia are your friends. :)

Wow, hey, could I add you on my friends list?  :lol:

thanks, llama and i_like_pie, appreciate your help!  :)

[quote="buzzer11"][quote="i_like_pie"][quote="Walk Around The River"]Its also commonly referred to as an "undecagon" as well as hendecagon[/quote]
That is true, but [url=http://en.wikipedia.org/wiki/Hendecagon]Wikipedia[/url] specifically states that the prefix 'un' (Latin) is incorrect, and 'hen' (Greek) should be used instead.[/quote]

That is probably why Walk Around The River said "commonly".[/quote]

um, i think that Walk Around The River said that both undecagon and hendecagon are used commonly.  :lol:


geometry
rectangle
parallelogram
perimeter
geometry proposed

Post next problem after you solve one - full solution please (problems preferrably not from the resource section already with well-written solutions since everyone has access to it plus there is no point in reposting them when there are already solutions there).

Problem 1: In [u]concave[/u] hexagon $ ABCDEF$, $ m\angle A \equal{} m\angle B \equal{} m\angle C \equal{} 90^\circ$, $ m\angle D \equal{} 100^\circ$, and $ m\angle F \equal{} 80^\circ$.  Also, $ CD \equal{} FA$, $ AB \equal{} 7$, $ BC \equal{} 10$, and $ EF \plus{} DE \equal{} 12$.  Compute the area of the hexagon.

I haven't solved the problem, but I think I did make progress on it.

[hide]

Construct line CF, and label $ m\angle BCF \equal{} x$.  Then $ m\angle FCD \equal{} 90 \minus{} x$ (complimentary angles), $ m\angle CFE \equal{} 90 \plus{} x$ (sum of angles in a quadrilateral add up to 360), and $ m\angle AFC \equal{} 180 \minus{} x$.

Write the area of Quadrilateral $ FCDE$ in two different ways.  I leave out .5's because they don't matter.
\[ [FCDE] \equal{} [CFE] \plus{} [CED] \equal{} (FC)(FE)sin(m\angle CFE) \plus{} (CD)(DE)sin(m\angle CDE) \equal{} (FC)(FE)sin(90 \plus{} x) \plus{} (CD)(DE)sin(100) \equal{} (FC)(FE)cos(x) \plus{} (CD)(DE)sin(80)\]

\[ [FCDE] \equal{} [FED] \plus{} [CDF] \equal{} (CD)(FC)sin(m\angle FCD) \plus{} (EF)(ED)sin(m\angle FED) \equal{} (CD)(FC)sin(90 \minus{} x) \plus{} (EF)(ED)sin(80) \equal{} (CD)(FC)cos(x) \plus{} (EF)(ED)sin(80)\]
Subtract these two area formulations by each other to get
\[ 0 \equal{} (FC)(FE)cos(x) \minus{} (CD)(FC)cos(x) \plus{} (CD)(ED)sin(80) \minus{} (EF)(ED)sin(80) \equal{} (FC)cos(x)(FE \minus{} CD) \plus{} (ED)sin(80)(CD \minus{} EF) \equal{} (FE \minus{} CD)((FC)cos(x) \minus{} (ED)sin(80)) \equal{} 0\]
So either $ FE \equal{} CD$ or $ (FC)cos(x) \equal{} (ED)sin(80)$

I would go with the former case being correct, because it seems simpler.

I will add in the two other observations I have made.  $ m\angle AFE \equal{} 360 \minus{} (180 \minus{} x) \minus{} (90 \plus{} x) \equal{} 90$.

$ (FC)cos(x) \equal{} 10 \minus{} AF$

[/hide]

edit: Oops.  Most of this is wrong, because I read angle F = 80 as angle E = 80... so I was essentially working a different problem  :maybe: .  Sorry about that  :blush:

SOLUTION TO PROBLEM 1:

extend AF to X such that $ \angle AXD\equal{}90^\circ$

area of hexagon=area of rectangle-area of FEDX

$ FE\equal{}a, ED\equal{}12\minus{}a$

$ FX\equal{}b, XD\equal{}b\minus{}3$

$ FD^2\equal{}EF^2\plus{}ED^2\equal{}FX^2\plus{}XD^2$

$ a^2\plus{}(12\minus{}a)^2\equal{}b^2\plus{}(b\minus{}3)^2$

$ 144\minus{}2a(12\minus{}a)\equal{}9\plus{}2b(b\minus{}3)$

$ 135\equal{}4(\frac{a(12\minus{}a)}{2}\plus{}\frac{b(b\minus{}3)}{2}$

$ area(FEDX)\equal{}33.75$

$ area(ABCDEF)\equal{}70\minus{}33.75\equal{}36.25$

PROBLEM 2
Prove that the area of triangle formed by the lengths of medians of a given triangle is three fourth the area of the given triangle.

I'm feeling extremely uncreative tonight, so here's a solution with analytical geometry

[hide="Solution"]
This might be a bit hard to follow. Let $ \vec{m_1}$ and $ \vec{m_2}$ be two medians, and $ \vec{a}$ and $ \vec{b}$ be the two sides that the "feet" of the medians lie on.

It's not hard to see that $ \vec{m_1} \equal{} \vec{a} \plus{} \frac {\vec{b}}{2}$ and $ \vec{m_2} \equal{} \minus{} \left(\frac {\vec{a}}{2} \plus{} \vec{b}\right)$.

Then $ |\vec{m_1}\times \vec{m_2}| \equal{} \frac {1}{4}\left|4(\vec{b}\times\vec{a}) \minus{} \vec{b}\times\vec{a}\right| \equal{} \frac {3}{4}|\vec{b}\times\vec{a}|$, and we're done.

(Note that $ \vec{u}\times\vec{v}$ is the cross product. I left some stuff out, but if you're having trouble following the last step, note that the magnitude of the cross product is the area of the parallelogram enclosed by the vectors).
[/hide]

Edit: Oh, someone else feel free to post another problem. Sorry, I don't really know of any good ones that haven't been posted here already. If I find something I'll put it up.

Is there a synthetic solution to the median problem?
[b]Problem 3[/b]:Two square pieces of paper are made subject to the constraint that the sum of their areas is 2 square units. These pieces of paper are to be put inside a rectangle such that the sides of the sqauares are parallel to the sides of the rectangle. What is the least possible value for the area of a rectangle if we want to be sure tht this rectangle will enclose these two pieces of paper? Prove that your answer is the minimum.

[b]Problem 3:[/b]Two square pieces of paper are made subject to the constraint that the sum of their areas is 2 square units. These pieces of paper are to be put inside a rectangle such that the sides of the sqauares are parallel to the sides of the rectangle. What is the least possible value for the area of a rectangle if we want to be sure that this rectangle will enclose these two pieces of paper? Prove that your answer is the minimum.
[b]Problem 4:[/b]Prove that of all triangles inscribed in a given triangle, the one with least perimeter connects the feet of the given triangle.


For $ \minus{} 90^\circ \leq \theta \leq 90^\circ$, find the maximum and minimum value of $ \sin \theta \plus{} 4\sqrt {2}\cos (\theta \plus{} 45^\circ)$.

$ \cos (\theta \plus{} 45^\circ)\equal{}\frac{1}{\sqrt{2}}\cos\theta\minus{}\frac{1}{\sqrt{2}}\sin\theta$

$ \sin\theta \plus{} 4\sqrt{2}\cos (\theta \plus{} 45^\circ)\equal{}4\cos\theta\minus{}3\sin\theta\equal{}5\left(\frac{4}{5}\cos\theta\minus{}\frac{3}{5}\sin\theta\right)\equal{}5\cos (\theta\plus{}\alpha)$, where $ \alpha\equal{}\arcsin\frac{3}{5}$. So maximum is $ 5$ when $ \theta\equal{}\minus{}\alpha$ and minimum is $ \minus{}3$ when $ \theta\equal{}\frac{\pi}{2}$.

[quote="Yuriy Solovyov"]$ \cos (\theta + 45^\circ) = \frac {1}{\sqrt {2}}\cos\theta - \frac {1}{\sqrt {2}}\sin\theta$

$ \sin\theta + 4\sqrt {2}\cos (\theta + 45^\circ) = 4\cos\theta - 3\sin\theta = 5\left(\frac {4}{5}\cos\theta - \frac {3}{5}\sin\theta\right) = 5\cos (\theta + \alpha)$, where $ \alpha = \arcsin\frac {3}{5}$. So maximum is $ 5$ when $ \theta = - \alpha$ and minimum is $ - 3$ when $ \theta = \frac {\pi}{2}$.[/quote]

I can not this : $ \cos(\theta + 45^\circ) = \frac {1}{%Error. "sqer" is a bad command.
{2}}\cos\theta - \frac {1}{\sqrt {2}}\sin\theta$
As I know from my worth Trigonometry:  $ \cos(\theta + 45) = \cos\theta\cos\ {45} - \sin\theta\sin\ {45}$

[i]Note: I don't know why LaTeX looks like this  :wink: [/i]

hello, we have $ \sin(\theta)\plus{}4\sqrt{2}\cos(\theta\plus{}45^{\circ})\equal{}4\cos(\theta)\minus{}3\sin(\theta)$.
Sonnhard.

[quote="enndb0x"]
I can not this : $ \cos(\theta + 45^\circ) = \frac {1}{%Error. "sqer" is a bad command.
{2}}\cos\theta - \frac {1}{\sqrt {2}}\sin\theta$
As I know from my worth Trigonometry:  $ \cos(\theta + 45) = \cos\theta\cos\ {45} - \sin\theta\sin\ {45}$
[/quote]
 :o

$ \sin 45^{\circ}=\cos 45^{\circ}=\frac{1}{\sqrt{2}}$ ;)


New to this Math Forums but I already need some help. My teacher has me doing interest formulas that are compounded and I need some help. Here are the formulas:

1123.60=[(1+ x/12)^(12*2)]*1000

I get this far: 1126.60= [(1+x/12)^24]*1000

Let's call that one problem number 1.

Now for problem number 2...

1123.60= [(1+4.75/24)^24*x]*1000

I get this far= 1123.60= [115/96)^24x]*1000

But I don't know how to solve for "x" on either one. Any help would be greatly appreciated.

Thanks,
Jajou

These don't look like Olympiad problems....

Are these homework problems?


Let $a,b,c>0$ such that $a+b+c=1$.prove that:

      $a^2+b^2+c^2 +1\geq 4(ab+bc+ca) $ ;)

Cezar, do you really see any interest of this type of problems? I mean, these say absolutely nothing and the only reason for their existence is to occupe space.  :)

Ok,you are right.Promiss next time if i going to post such inequalities,it will be in Classroom math or I won't post this "kind" of things.I think the second option is the best. :blush:  :oops:


calculus
integration
trigonometry
limit
logarithms
function
calculus computations

What is the value of $ \int_0^\infty\dfrac{cos(t)\minus{}e^{\minus{}t}}{t}dt$ ?
Give an elementary proof.

[quote="Fermat's Little Turtle"]What would be elementary? Could we use Parseval's?[/quote]
What?.Who can tell me more about Parseval's?. And give a solution. It is really hard. :(

\begin{eqnarray*}\int_{0}^{\infty} \frac {\cos t - e^{ - t}}{t} \, dt & = & \int_{0}^{\infty} (\cos t - e^{ - t}) \left( \int_{0}^{\infty} e^{ - tu} \, du \right) \, dt \\
& = & \int_{0}^{\infty} \int_{0}^{\infty} (\cos t - e^{ - t})e^{ - ut} \, dt du \\
& = & \int_{0}^{\infty} \left( \frac {u}{1 + u^2} - \frac {1}{1 + u} \right) \, du \\
& = & \lim_{R \to \infty} \int_{0}^{R} \left( \frac {u}{1 + u^2} - \frac {1}{1 + u} \right) \, du \\
& = & \lim_{R \to \infty} \log \left( \frac {\sqrt {1 + R^2}}{1 + R} \right) \\
& = & 0 \end{eqnarray*}

One can question the validity of interchanging the order of integration, but happily this is the case.

A simliar but slightly different approach is also possible by exploiting some properties of Laplace transform. Let $ F(s) = \int_{0}^{\infty} \frac {\cos t - e^{ - t}}{t} \, e^{-st} \, dt$. Then Leibniz's integral rule allows us to evaluate $ F'(s)$, from which an explicit formula for $ F(s)$ follows. Then just take $ s \to 0$. In fact, this argument can be used to prove the former one.

I'm thinking out loud here, could we turn this into a complex contour integral?  I'm pretty sure we can do $ \int_{0}^{\infty}\frac{\cos{t}}{t}\,dt$ as a contour.  I haven't really tried the second one.

Hi sos440,
About your first proof:
you cannot use Fubini because $ t\rightarrow\int_0^\infty{(cos(t)-e^{-t})e^{-tu}du}$ is not in  $ L^1$.
About your second proof:
If $ s\geq{0}$ then let $ F(s) =\int_{0}^{\infty}\frac{\cos t-e^{-t}}{t}\, e^{-st}\, dt$. You proved that $ F(s)$ tends to $ 0$ but you have not proved that $ F$ is continuous in $ 0$.

I can answer loup blanc's objection about lack of absolute convergence and fix sos440's first proof by integrating by parts first. Also, to get the correct behavior as $ t\to0,$ we need the antiderivative of $ \minus{}e^{\minus{}t}\,dt$ to be $ e^{\minus{}t}\minus{}1.$

$ \int_0^{\infty}\frac{\cos t\minus{}e^{\minus{}t}}{t}\,dt\equal{}\left.\frac{\sin t\plus{}e^{\minus{}t}\minus{}1}{t}\right|_0^{\infty}\plus{}\int_0^{\infty}\frac{\sin t\plus{}e^{\minus{}t}\minus{}1}{t^2}\,dt.$

Once we establish that $ \sin t\plus{}e^{\minus{}t}\minus{}1\equal{}O(t^2)$ as $ t\to0$ we have both that the boundary terms evaluate to zero and the second integral is absolutely convergent.

We then use that $ \frac1{t^2}\equal{}\int_0^{\infty}xe^{\minus{}tx}\,dx.$

$ \int_0^{\infty}\frac{\sin t\plus{}e^{\minus{}t}\minus{}1}{t^2}\,dt\equal{}\int_0^{\infty}\int_0^{\infty}(\sin t\plus{}e^{\minus{}t}\minus{}1)xe^{\minus{}tx}\,dx\,dt$

$ \equal{}\int_0^{\infty}\int_0^{\infty}(\sin t\plus{}e^{\minus{}t}\minus{}1)xe^{\minus{}tx}\,dt\,dt$

$ \equal{}\int_0^{\infty}x\left(\frac1{x^2\plus{}1}\plus{}\frac1{x\plus{}1}\minus{}\frac1x\right)\equal{}\int_0^{\infty}\frac{x}{x^2\plus{}1}\minus{}\frac1{x\plus{}1}\,dx$

$ \equal{}\frac12\ln\left(x^2\plus{}1\right)\minus{}\ln(x\plus{}1)\bigg{|}_0^{\infty}\equal{}\left.\ln\left(\frac{\sqrt{x^2\plus{}1}}{x\plus{}1}\right)\right|_0^{\infty}\equal{}0.$

The use of Fubini's Theorem is fully justified by the absolute convergence of the double integral.

sos440's second proof needs only the continuous analogue of Abel's theorem for series.

Hi loup blanc, some special properties of Laplace transform validates interchange of the order of integration under a weaker condition: If the improper integral $ \textstyle \int_{0}^{\infty} f$ exists, then for any integer $ n \geq 1$,

\begin{eqnarray*}\int_{0}^{\infty} f(x) \, dx
& = & \int_{0}^{\infty} x^n f(x) \left( \frac{1}{\Gamma(n)} \int_{0}^{\infty} t^{n-1}e^{-xt} \, dt \right) \, dx\\
& = & \int_{0}^{\infty} \frac{t^{n-1}}{\Gamma(n)} \left(  \int_{0}^{\infty} x^n f(x) e^{-tx} \, dx \right) \, dt
\end{eqnarray*}

where each integral appearing is improper. So we don't need Fubini's theorem. Click [url=http://cfile23.uf.tistory.com/image/1369010A49EE8BC4890ED7]here[/url] to see the proof. Well, but one can question if it is elementary...  :D

-JRav: $ \int_0^{\infty}\frac{\cos t}{t}\,dt\equal{}\infty$. We need to handle the singularity at zero before splitting. The simple idea of splitting as $ \minus{}\frac{(1\minus{}\cos t)}{t}$ and $ \frac{1\minus{}e^{\minus{}t}}{t}$ doesn't work either, since both of those don't go to zero fast enough at $ \infty$, and have infinite integral.

There really isn't a simple way around this to split it.

consider $ f(z) \equal{} \frac {e^{iz} \minus{} e^{ \minus{} z}}{z}$
let's integrate this function on the contour $ \gamma$
$ \int_{\gamma} f(z) dz \equal{} 0$
note that :
$ \int_{\gamma}f(z)dz \equal{} \int_{\varepsilon}^{R}f(z)dz \plus{} \int_{bigcircle}f(z)dz \plus{} \int_{iR}^{i\varepsilon}f(z)dz \plus{} \int_{smallcircle} f(z)dz$
note that
$ \int_{smallcircle} f(z)dz \approx \mathcal{O}(\varepsilon) \to 0$ as $ \varepsilon \to 0$.
$ \int_{bigcircle} f(z)dz \to 0$ as $ R\to \infty$
thus we have :
\[ 0 \equal{} \int_{\varepsilon}^{R}f(z)dz \plus{} \int_{iR}^{i\varepsilon}f(z)dz \equal{} 2\int_{\varepsilon}^{R}\frac {\cos t \minus{} e^{ \minus{} t}}{t} \to 2\int_{0}^{\infty}\frac {\cos t \minus{} e^{ \minus{} t}}{t}
\]
thus :
\[ \int_{0}^{\infty}\frac {\cos t \minus{} e^{ \minus{} t}}{t}\equal{}0
\]


limit
function
ratio
algebra unsolved
algebra

I don't know how hard (or easy) this problem is...

Let $f:Z\rightarrow Z$ be a bijection. Let
$A=\lim_{m,n\rightarrow \infty}\frac{{\displaystyle \sum_{k=-n}^m f(k)-k}}{m+n+1}$

What are all possible values of A? Does it always exist?

It is certainly possible that the limit does not exist.
Consider a rapidly growing sequence of positive and negative integers, let the permutation equal identity outside this set and scramble the elements of the set.

I don't understand your counter-example... could you explain why it works?
I'm pretty sure that A (assuming it exists) has to be in Z, but I don't see a simple proof...

Here is a possibly simpler construction using the same idea.
Let $g(1), g(2), ... $ be some rapidly increasing function from positive integers to positive integers.  We will specify in a moment what is sufficiently "rapid" for this problem.

Consider the permutation $f$ that exchanges $g(1)$ with $g(2)$, $g(3)$ with $g(4)$, and so on.
Then the ratio $A(f,m,n)$ appearing in the limit will be zero for $m = n = f(2k)$ but when $m=n=f(2k-1)$, the ratio is much larger:   $(f(2n) - f(2n-1))/(2n+1)$.
So $f(n) = n^2$ suffices.

I have some doubts about the integrality of $\lim A$, when it exists.


In proportion to its total body weight, an ant's brain is one of the largest of all the insect world. An ant's brain weighs 27 mg and its total body weight is 453 mg. Express the ratio of the brain weight to total body weight as a simplified common fraction.


How many distinct four-digit numbers are divisible by $ 3$ and have $ 23$ as their last two digits?

$ \textbf{(A)}\ 27 \qquad
\textbf{(B)}\ 30 \qquad
\textbf{(C)}\ 33 \qquad
\textbf{(D)}\ 81 \qquad
\textbf{(E)}\ 90$

[hide]The sum of the digits must be divisible by $ 3$, so if we label the first two digits $ a$ and $ b$, then $ a\plus{}b\plus{}2\plus{}3\equal{}a\plus{}b\plus{}5$ is divisible by $ 3$, so $ a\plus{}b\minus{}1$ is divisible by $ 3$.

Since $ 1 \le a \le 9$ and $ 0 \le b \le 9$, we have that $ 0 \le a\plus{}b\minus{}1 \le 17$. If we want $ a\plus{}b\minus{}1$ to be divisible by $ 3$, we can have
\[ a\plus{}b\equal{}1,4,7,10,13,16.\]
Thus we have
\[ 1\plus{}4\plus{}7\plus{}9\plus{}6\plus{}3\equal{}5\plus{}16\plus{}9\equal{}21\plus{}9\equal{}30 \rightarrow \boxed{\textbf{B}}.\][/hide]

Yeah, I was kinda surprised that this was a #25...

First is 1023, last is 9723, and it cycles every 300. So $ (9723\minus{}1023)/300\plus{}1\equal{}29\plus{}1\equal{}\boxed{30}$.

[quote="123456789"][hide]The sum of the digits must be divisible by $ 3$, so if we label the first two digits $ a$ and $ b$, then $ a \plus{} b \plus{} 2 \plus{} 3 \equal{} a \plus{} b \plus{} 5$ is divisible by $ 3$, so $ a \plus{} b \minus{} 1$ is divisible by $ 3$.

Since $ 1 \le a \le 9$ and $ 0 \le b \le 9$, we have that $ 0 \le a \plus{} b \minus{} 1 \le 17$. If we want $ a \plus{} b \minus{} 1$ to be divisible by $ 3$, we can have
\[ a \plus{} b \equal{} 1,4,7,10,13,16.
\]
Thus we have
\[ 1 \plus{} 4 \plus{} 7 \plus{} 9 \plus{} 6 \plus{} 3 \equal{} 5 \plus{} 16 \plus{} 9 \equal{} 21 \plus{} 9 \equal{} 30 \rightarrow \boxed{\textbf{B}}.
\]
[/hide][/quote]

 may distinct 4-digits numbers be no digit repeated in the number?
 :P

Hm, well the wording "distinct four-digit numbers" strongly suggests that the numbers as a whole are distinct, not the individual digits. Otherwise, they would have to accept 30 because of the poor wording.

[hide="Not quite brute-force"]If $ \overline{abcd}$ is the number, we have $ a\plus{}b\plus{}c\plus{}d\equiv 0\pmod{3}$ and $ 10c\plus{}d\equal{}23$.

Obviously $ c\equal{}2$ and $ d\equal{}3$. Now, $ a\plus{}b\equiv \minus{}5\equiv 1\pmod{3}$. Suppose $ a\equiv 1\pmod{3}$ and $ b\equiv 0\pmod 3$. $ a$ can be $ 1, 4, 7$, and $ b$ can be $ 0, 3, 6, 9$.

That's $ 3\cdot 4\equal{}12$ solutions so far.

Now if $ a\equiv 0\pmod 3$ and $ b\equiv 1\pmod 3$, $ a$ can be $ 3, 6, 9$ and $ b$ can be $ 1, 4, 7$, so we have $ 3\cdot 3\equal{}9$ more solutions. The total is $ 21$ so far.

Now suppose $ a\equiv 2\pmod 3$ and $ b\equiv 2\pmod 3$. Then $ a$ and $ b$ can be $ 2, 5, 8$, so we get $ 3\cdot 3\equal{}9$ more. Now we have $ 30$.

There are no other solutions, so the answer is $ \boxed{\textbf{(B)}\ 30}$.[/hide]

Hm, has anyone noticed that the sequence of numbers which fit the above property is perhaps... an [i]arithmetic[/i] sequence? :roll:

Yeah, given the condition $ a \plus{} b\equiv1\pmod3$, one can simply find all two digit integers $ \overline{ab} \equal{} 3k \plus{} 1$ for some integer $ k$, or equivalently,

$ 10\le3k \plus{} 1\le99\implies9\le3k\le98\implies3\le k\le32\implies 32 \minus{} 3 \plus{} 1 \equal{} \boxed{\textbf{(B)}\ 30}$.

[quote="123456789"]
$ 1 \plus{} 4 \plus{} 7 \plus{} 9 \plus{} 6 \plus{} 3 \equal{} 5 \plus{} 16 \plus{} 9 \equal{} 21 \plus{} 9 \equal{} 30 \rightarrow \boxed{\textbf{B}}$.
[/quote]
Why does adding these give you the number of distinct solutions?

When $ a\plus{}b\equal{}r$, there are $ r$ pairs of digits $ (a,b)$ with $ a \ne 0$ that satisfy the equation.

#25
Divisability rule for 3 is if the digits add up to a multiple of three than it is divisible by three so lets say our number is ab23 then 5+a+b has to be a multiple of three so the possible values for a+b are 1,4,7,10,13,16 and there are 30 different ways to make these numbers.
This is what I did, but I haven't looked at the solution

Or to expand on that, $ a$ can be any number from 1 to r, while $ b$ can be any from r-1 to 0.

You can find a pattern. If you have 1 in the thousands place, you can see (using the divisibility rule) for three, their are 3 possible numbers.  Doing the same for 2 and 3, you see that there are 3 for each. Then going to 4, you can see that there are 4 for each. Continuing, this creates a pattern of 4 numbers, 3 numbers, 3 numbers, which repeats 3 times. 4 + 3 + 3 = 10, and 10 x 3 = 30.


Did anyone notice that all the triples (generators) that start with odd number have the next two numbers differing by 1.  For example:

3-4-5
5-12-13
7-24-25
etc.

Does anyone have any thoughts on this, or any way to prove that there is a consecutive pair of integers [i](b,c)[/i] for every odd integer [i]a[/i].  (In the formula [b][i]a 2 +b 2 = c 2[/i][/b] of course)

This would be a good Olympiad question; maybe I should post it somewhere else.....

[quote]Did anyone notice that all the triples (generators) that start with odd number have the next two numbers differing by 1.[/quote]

This is not quite true, although I was surprised at how far you have to go to find an example of a primitive (no common factor > 1) Pythagorean triple where the smaller leg is odd and the other two sides do not differ by 1.

[hide="Click here for the smallest example"]
33,56,65 is the smallest example (odd leg is as small as possible)
[/hide]

But there is a triple :  33,544,545 that does fit the bill

[quote]Um, 33-544-545 is a right triangle. It does not say that the ONLY right triangle possible will have the larger leg and hypotenuse differ by 1[/quote]

The initial sentence of the post did sound like it was saying that, though ("*all* the triples that start with odd number have the next two numbers differing by 1").  I agree that by the end of the post, it is more clear that only a single pair (b,c) differing by 1 was claimed.

Yakko

[quote="Treething"]Just square the odd number, divide it by two, and add/subtract 1/2 to get the three sides.

For example...

11:^2: = 121
121/2 = 60.5
60.5 +/- 0.5 = 60, 61

11, 60, 61[/quote]

Exactly what I was looking for!

Sorry Smartmouth, I should have said that my first post, that's how I got the triple I posted.  For evens the rule is similar:

if one leg is even, square it, divide by 4, and then Add and Subtract one to get the other leg and hypotenuse.

Example:  8^2 =64 , 64/4 =16, + or - 1 = 15,17

so 8,15,17 is your triple.

Now I'm not saying these are all the triples, just a way to find some


limit
linear algebra
matrix
algebra
polynomial
function
complex numbers

Let $ a(1), a(2), \cdots$ be a sequence of non-negative real numbers such that $ \sum_{n\equal{}1}^{\infty} a(n)$ diverges. Suppose that the power series:
$ f(z) \equal{} \sum_{n\equal{}1}^{\infty} a(n)z^n$
converges for all complex numbers $ |z| < 1$ and that:
$ \lim_{r \to 1^{\minus{}}} f(re^{i\theta})$
exists and is finite for every $ 0 < \theta < 2 \pi$. Prove that the sum of $ a(n)$ for any $ n \equiv a (\bmod q)$ diverges.

In other words, prove that:
$ a(1) \plus{} a(1 \plus{} q) \plus{} \cdots$
$ a(2) \plus{} a(2 \plus{} q) \plus{} \cdots$
$ \vdots$
$ a(q\minus{}1) \plus{} a(q\minus{}1 \plus{} q) \plus{} \cdots$
diverge.

The "roots of unity filter" gets you

$ \frac {f(r) \plus{} f(r \zeta_q) \plus{} f(r \zeta_q^2) \plus{} ... \plus{} f(r \zeta^{q \minus{} 1}_q)}{q} \equal{} \sum_{q |n} a(n) r^n.$

On the other hand, as $ r \to 1^{ \minus{} 1}$ all these sums converge except the first one.  Now you need to shift the index of the sum to cover $ n \equiv a \bmod q$; this is done by weighting the sum with the appropriate row in the [url=http://en.wikipedia.org/wiki/Discrete_Fourier_transform]DFT[/url] matrix of order $ q$.  These entries are indeed related to Dirichlet characters.

uh... I have no idea what any of that means... can you explain it a little further? I'm not familiar with DFT stuff either.

Let me work through an example with $ q \equal{} 3$.  Here $ \zeta_q \equal{} \omega \equal{} e^{ \frac{2 \pi i}{3} }$ is a primitive third root of unity.  Given a power series $ f(z) \equal{} a(0) \plus{} a(1) z \plus{} a(2) z^2 \plus{} a(3) z^3 \plus{} ...$, we have the following identities:

$ f(\omega z) \equal{} a(0) \plus{} a(1) \omega z \plus{} a(2) \omega^2 z^2 \plus{} a(3) z^3 \plus{} ...$
$ f(\omega^2 z) \equal{} a(0) \plus{} a(1) \omega^2 z \plus{} a(2) \omega z^2 \plus{} a(3) z^3 \plus{} ...$.

Since $ \omega^3 \equal{} 1$ but $ \omega \neq 1$, it follows that $ 1 \plus{} \omega \plus{} \omega^2 \equal{} 0$, hence summing the above three identities gets you

$ f(z) \plus{} f(\omega z) \plus{} f(\omega^2 z) \equal{} 3a(0) \plus{} 3a(3) z^3 \plus{} 3a(6) z^6 \plus{} ...$

where all the terms with index not divisible by $ 3$ have dropped out.  This is referred to as a "roots of unity filter"; applied to polynomials, it is a useful way to extract coefficients from [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=220351]certain generating functions[/url].  See also the connection to DFTs hinted at [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=212490]here[/url].

But ignoring all that theoretical stuff, the next question to ask is how to modify the above sum to extract the terms with index congruent to $ 1$ and $ 2 \bmod 3$, respectively.  Well, we probably want to take advantage of $ 1 \plus{} \omega \plus{} \omega^2 \equal{} 0$ again, but this time we want the $ 1$ coefficients to appear in a different spot.  We can do this by computing the sums

$ f(z) \plus{} \omega^2 f(\omega z) \plus{} \omega f(\omega^2 z) \equal{} 3a(1) z \plus{} 3 a(4) z^4 \plus{} 3 a(7) z^7 \plus{} ...$

and

$ f(z) \plus{} \omega f(\omega z) \plus{} \omega^2 f(\omega z) \equal{} 3a(2) z^2 \plus{} 3 a(5) z^5 \plus{} 3 a(8) z^8 \plus{} ...$.

Even if you aren't confident that this idea generalizes, you should be able to work out coefficient-by-coefficient that the above three sums are valid for this particular case.  Now what happens if you take $ z \equal{} r$ and let $ r \to 1^{\minus{}}$?

ah okay i get it now thanks a lot t0raji0u!


My team talked about this kind of problem today.  Can anyone give a good systematic way to solve:

What is the least positive multiple of a number m, with n positive integer divisors?

I can always get the right answer, but I don't have a nice way of doing it.

Sample problem: What is the least positive multiple of 10 with exactly 10 positive integer divisors.

I took the first 5 integers:1,2,3,4,5.  Then multiplied 5 x 6 = 30.  Continued with 5 x 7 = 35, 5 x 8, etc.  until I got the number that was a multiple for (1,2,3,4,5,10, & the #).  It was 5 x 12 = 60.

That method doesn't always work though.  Any ideas?

60 isn't the answer, 80 is

if it has 10 factors then you know that when in prime factorized form a^b x c^d would have (b+1)(d+1) divsors.  so then 10 can be 2x5 or 10x1, it is obvious that 10x1 would have too big of a number, so discount that.  look at 2x5 now, b and d would then be 1 and 4.  You know that it is in the form of 2^b x 5^d, the smallest value appearing when b=4 and d=1, so then 2^4x5^1=80

Thanks, we were really stinking up this problem today.  That helps a lot.


I know most of these people are high schoolers, but we do have some middle schoolers here rite?

anyway how did u guys do on amc 8 ppl?

I think i got a perfect, but even though the answers are online, i don't have my test booklet...

I got a perfect score! (And so did Yubo Su and Nakkul Sreenivas...)  All three people, including me, were in my math class and period.  OMVG!! :o

Yup, I got a perfect score all rite.

But were there really only 120 perfects nationwide? Cuz that's what my parents told me and i'm afraid that I'm getting too much self-esteem XD

???
ZOMG there's a bunch of people I know...Sitan, Arvind, Yubo...who else (who is younger than I am) will show up?


Of the 6 students who make it to the IMO, how many problems of the USAMO did they usually solve? Also is there a site I can go to to find previous USAMO papers by the ones who made it to the IMO?

It depends on the year, but it generally takes 4 problems to be a winner, and some of the eventual team members will usually solve 5 or 6 problems.  I think to realistically be in contention for the IMO team, in practice you need to be able to solve USAMO #1/2/4/5's almost every time, and you need to be able to solve USAMO #3/6's at least about half the time.  On the actual test, you should shoot for at least 4 problems plus some partial credit, or 5 problems if you can, since if you solve exactly 4 problems, you're likely to lose a few points here and there and miss the cutoff.

About the USAMO papers, that's a good question.  I haven't heard of any such resource, but maybe it's out there.


logarithms
integration
calculus
calculus computations

Let T(t) be the temperature of an object with time t, and let A be the constant temperature to the environment. Newtons Law of Cooling says: T'(t) = k(T(t) - A), where k is a constant. Let T0 be the temperature of the object by time t. Now rewrite the differensial equation by substituting u(t) = T(t) - A, and use this to find T(t) expressed by T0, A and k.

thanks!

The question tells you how to proceed (although I assume there's a typo and it means to say that $ T_0 \equal{} T(0)$ is the temperature of the object at time 0?) -- so what have you done, and where are you having trouble?

[hide=" Click to reveal hidden content"]

$ \frac{dT}{dt} = k(T-A)$

$ u=T-A \implies \frac{du}{dt} = \frac{dT}{dt}$

$ \frac{du}{dt} = ku$

$ \ln u = kt + c$

$ u= Ce^{kt}$

$ t=0$ gives $ T_{0} - A = C$

Therefore ${ T=A + (T_{0}-A)e^kt}$.

[/hide]

Thanks for your reply!
Sorry but I don't quite see what's happening here:

$ \frac{du}{dt}\equal{} ku$

$ \ln u \equal{} kt\plus{}c$

$ u\equal{} Ce^{kt}$

Tho i know this last one is a formula for growth and decay.

Hi Protonics,

$ \frac{du}{dt} \equal{} ku$

$ \frac{du}{u} \equal{} k \ dt$

$ \int \frac{du}{u}\ du \equal{} \int k\ dt$

$ \ln u \equal{} kt \plus{} c$

$ e^{\ln u} \equal{} e^{kt \plus{} c}$

$ u \equal{} e^{c}e^{kt}$

$ u \equal{} Ce^{kt}$.

Hope this helps.


algebra
binomial theorem
Pascal\u0027s Triangle

How do you factorize those two:
${1), (a+b)^n}$
$(2), a^n-b^n$

[quote="ineedmathhelp"]For the first one, do you want us to expand it?[/quote]

Yes.......if there is a nice way to do so.

For the first one, it is just the binomial theorem:

\[(a+b)^n=\binom{n}{n}a^n+\binom{n}{n-1}a^{n-1}b+\cdots+\binom{n}{k}a^kb^{n-k}+\cdots+\binom{n}{0}b^n.\]

You can write this in sigma notation as:

\[\sum_{k=0}^n\binom{n}{k}a^kb^{n-k}.\]

The second question depends a whole lot on $n$ but what I can tell you is that $a-b$ will always be a factor.

Yeah, the Binomial Theorem will do the trick.

But if you're like me and you're very lazy (which you probably aren't) you could just fill in rows of Pascal's triangle until you reach the nth row.

Actually I think it takes more work to fill in Pascal's triangle.  Like say there was a question that asked what the coefficient of $x^3$ was in the expansion of $(x+1)^{98}.$


linear algebra
matrix
algebra
polynomial

let A be square matrix such that :exist$ n >2$ then $ A^n\equal{}0 .$
  prove that :$ A^2\equal{}0$

this is wrong, e.g., set $ A\equal{}\left(\begin{array}{ccc}0 & 1 & 0\\
0 & 0 & 1\\
0 & 0 & 0\end{array}\right)$, then $ A^3\equal{}0\not\equal{}A^2$.

This is the correct version. If $ A$ is an $ n\times n$ matrix and $ A^m\equal{}0$, then $ A^n\equal{}0$.

Let $ m_A,\chi_A$ be the minimal polynomial resp. the characteristic polynomial of $ A$, then $ m_A\,|\,\gcd(x^m,\chi_A)$, hence, $ m_A\equal{}x^t$ with $ t\leq\min\{m,\deg\chi_A\}\leq n$ and we have $ A^n\equal{}A^{n\minus{}t}A^t\equal{}0$


LaTeX
number theory
least common multiple

How do I  use \mathbb in the \LaTeX document? I know it works on this forum, but not on my PC. I have TeXnicCenter and MiKTeX installed on my PC.

By the way, what is the command for the least common multiple, there is no \lcm?

Still learning...

Thanks

On my computer I have put brackets after \mathbb.  For example, to get [tex]f: \mathbb{C} \to \mathbb{Z}[/tex], I would have to put f: \mathbb{C} \to \mathbb{Z}.  If that doesn't work, it may be in some package, probably amsmath.

1. You need to include this line before your \begin{document}:

\usepackage{amssymb}

2. \lcm doesn't exist (at least, in normal LaTeX). I just use

\mathrm{lcd}~

If you use amsmath (\usepackage {amsmath}), you can do

\DeclareMathOperator{\lcm}{lcm}

and then you can use \lcm.

Alternatively, you can use

\operatorname{lcm} to achieve the same effect.


number theory
prime numbers
number theory unsolved

Find all pairs $(p,q)$ of prime numbers such that $pq$ divides $2^p + 2^q$.

I think problem is open.
Just for $q = 1$
We've got $p | 2^p + 2$ and $n-1 | 2^n+1 \Longrightarrow 2^p+2 |2^{2^p+2} + 2$
f.e. $2 | 2^2 + 2 = 6$, $6 | 2^6 + 2 = 66$, $2^{66} | 2^{66} + 2$ and so on..
but we've got other sequence starting from $p = 945$
and dunno how many sequence there is

This problem was on Bulgarian math competition 1995, but not tottaly the same:

[color=blue] If $k>1$, show that $k$ doesn't divide $2^{k-1}+1$. Use this to find all prime numbers $p$ and $q$ such that $pq|2^p+2^q$.[/color]

If this doesn't help, let ne know, and I will post complete solution

bye

It is easy problem. Solution is (2,2),(2,3)(3,2).
Let $n_p,n_q$ are minimal numbers suth that $2^{n_p}=-1(mod p),2^{n_q}=-1(mod q)$.
We have $\frac{|p-q|}{n_p} \ and \ \frac{|p-q|}{n_q}$ are odd numbers.
$2^p+2^q=2(2^{p-1}+1) (mod q) \Longrightarrow \frac{p-1}{n_q} \ odd$. And $\frac{q-1}{n_p} \ odd$. It give contradition if p and q odd primes. Therefore one of them equal 2, then another equal 2 or 3.

@Big Bolzano: To be open for $q=1$ doesn't mean that it is open for primes $q$ ;)
Please read the statement again (and what is $n$ in your post¿).

My approach: $p=q$ implies $p=2=q$.
So WLOG $p>q$, then we want that $p,q|2^{p-q}+1$ or $q=2$ and the condition for $p$ only.
1. case $q=2$: then we want $p|2^{p-2}+1 \equiv 2^{-1}+1 \mod p$, impossible for $p>3$.
2. case $q>2$: $p,q|2^{p-q}+1$, thus $2^{q-1} \equiv -1 \mod p$, $2^{p-1} \equiv -1 \mod q$.
Now $ord_p(2) \nmid q-1$ and $ord_p(2) \mid 2(q-1)$, thus $v_2(ord_p(2)) = 1 + v_2(q-1)$.
Also $v_2(ord_p(2)) \leq v_2(p-1)$, giving $v_2(p-1) \geq 1 + v_2(q-1)$; since symmetrically also $v_2(q-1) \geq 1 + v_2(p-1)$, we get a contradiction.
So all solutions are $(2,2),(2,3),(3,2)$.

I know that it's similar to Rust's, but I started typing it some hours ago, was chased away from the PC before I could end it and don't want to delete it now ;)


[quote="Shinraehee"]Let $ a,b,c > 0$ . Prove that:
$ 9(a^2 \plus{} bc)(b^2 \plus{} ca)(c^2 \plus{} ab) \le 8 (a^3 \plus{} b^3 \plus{} c^3)^2$[/quote]

My stronger inequality: http://mathscope.org/forum/showthread.php?t=1463


Let $ a,b,c$ be nonnegative real numbers such that $ ab\plus{}bc\plus{}ca\equal{}3$. Prove that

$ (3a^2\plus{}1)(3b^2\plus{}1)(3c^2\plus{}1) \ge 64$.

Setting $ p \equal{} a \plus{} b \plus{} c,r \equal{} abc$ then the ineq becomes $ \left( {3r \minus{} p} \right)^2  \ge 0$
which is trivial  :lol:

[quote="Vasc"]Let $ a,b,c$ be nonnegative real numbers such that $ ab \plus{} bc \plus{} ca \equal{} 3$. Prove that

$ (3a^2 \plus{} 1)(3b^2 \plus{} 1)(3c^2 \plus{} 1) \ge 64$.[/quote]
The solution of Loc is right and also the best constant for the inequality
\[ (a^2 \plus{} k)(b^2 \plus{} k)(c^2 \plus{} k) \ge (k \plus{} 1)^3
\]
where $ a,b,c \in \mathbb R, ab \plus{} bc \plus{} ca \ge 3$

is $ k \equal{} \frac {1}{3}$ :)

PS. It's Phan Thanh Nam's problem.

[quote="Vasc"]Let $ a,b,c$ be nonnegative real numbers such that $ ab \plus{} bc \plus{} ca \equal{} 3$. Prove that

$ (3a^2 \plus{} 1)(3b^2 \plus{} 1)(3c^2 \plus{} 1) \ge 64$.[/quote]
See here: http://canhang2007.wordpress.com/2009/11/16/inequality-70-v-cirtoaje-p-t-nam/

[quote="Vasc"]Let $ a,b,c$ be nonnegative real numbers such that $ ab \plus{} bc \plus{} ca \equal{} 3$. Prove that

$ (3a^2 \plus{} 1)(3b^2 \plus{} 1)(3c^2 \plus{} 1) \ge 64$.[/quote]

$ (x^2a^2 \plus{} 1)(x^2b^2 \plus{} 1)(x^2c^2 \plus{} 1) \equal{}[x^3abc\minus{}x(a\plus{}b\plus{}c)]^2\plus{}[x^2(ab\plus{}bc\plus{}ca)\minus{}1]^2$

$ \ge [x^2(ab\plus{}bc\plus{}ca)\minus{}1]^2$


algebra
polynomial
induction
algebra unsolved

Prove that for all natural $ m$, $ n$ polynomial $ \sum_{i \equal{} 0}^{m}\binom{n\plus{}i}{n}\cdot x^i$ has at most one real root.

It is easy to see these polynomials have no positive roots.
Now we assume $ x<0$ and denote the polynomial as $ P(n,m)$.
$ (x-1)P(n,m)\\=\sum_{i=1}^{m+1}\binom{n+i-1}n x^i-\sum_{i=0}^m\binom{n+i}n x^i\\
=\binom{n+m}m x^{m+1}-\sum_{i=0}^m\binom{n+i-1}{n-1}x^i
\\=\binom{n+m}m x^{m+1}-P(n-1,m)$

or

$ P(n,m)-\frac1{1-x}P(n-1,m)=-\binom{n+m}nx^{m+1}\frac1{1-x}$.

Then we can prove by induction
$ P(n,m)=\frac1{(1-x)^n}P(0,m)-\sum_{i=1}^n\binom{m+i}ix^{m+1}\frac1{(1-x)^{n+1-i}}$

or

$ P(n,m)=\frac1{(1-x)^n}+\sum_{i=0}^n\binom{m+i}i(-x^{m+1})\frac1{(1-x)^{n+1-i}}$

When $ m$ is even, we see each term on RHS is positive. So $ P(n,m)$ has no real root for even $ m$.
Also $ P(n,m)>0,\forall 2|m$.

We notice that
$ P^'(n,m)\\=\sum_{i=1}^m\binom{n+i}n ix^{i-1}
\\=(n+1)\sum_{i=1}^m\binom{n+i}{n+1} x^{i-1}
\\=(n+1)\sum_{i=0}^{m-1}\binom{n+1+i}{n+1}ix^i
\\=(n+1)P(n+1,m-1)$.

If $ m$ is odd, we see $ P^'(n,m)>0$ implies $ P(n,m)$ is monotone, then $ P(n,m)$ has exactly one real root.

Q.E.D.


calculus
integration
calculus computations

What region R gives the maximum value?

$ \displaystyle\int^{ }_R \int (9\minus{}x^2\minus{}y^2)\,dA$

Use \iint for double integral: $ \iint_R$ looks much better. As to the question itself, just think first of the following: which subset of the set $ \{\minus{}2,3,5,\minus{}8,12,4,\minus{}3\}$ has the largest sum of its elements?


I got a different answer.
It seems to be that $x=3$

[quote="sharkman"]Twenty-four dogs are in a kennel.  Twelve of the dogs are black, six of the dogs have short tails, and fifteen of the dogs have long hair.  There is only one dog that is black with a short tail and long hair.  Two of the dogs are black with short tails and do not have long hair.  Two of the dogs have short tails and long hair but are not black.  If all of the dogs in the kennel have at least one of the mentioned characteristics, how many dogs are black with long hair but do not have short tails?[/quote]

For this problem, I think using Venn Diagram is simpler than using PIE as i_like_pie did because of it's different wording than usual problems.

[hide="Without drawing the actual Venn Diagram."]Let the set of black dogs be $B$, set of short-tailed dogs be $S$, and set of long-haired dogs be $L$. Then:
$n(B \cap S \cap L)=1$
$n(B)=12$
$n(S)=6$
$n(L)=15$
$n(B \cap S-L)=2$
$n(S \cap L-B)=2$
$n(B \cup S \cup L)=24$.

We need to find $n(B \cap L-S)$. This is equal to
\[n(B)+n(S)+n(L)-n(B \cup S \cup L)-n(B \cap S-L)-n(S \cap L-B)-2 \cdot n(B \cap S \cap L) \]

\[= 12+6+15-24-2-2-2=\boxed{3}. \]
So it was like visualizing the Venn Diagram...
[/hide]


If $ a,b,c$ be real numbers such that $ abc \equal{} 1,$  then

$ \frac {a \plus{} 1}{7a^2 \plus{} 6a \plus{} 7} \plus{} \frac {b \plus{} 1}{7b^2 \plus{} 6b \plus{} 7} \plus{} \frac {c \plus{} 1}{7c^2 \plus{} 6c \plus{} 7}\leq\frac {3}{10},$

with equality if and only if $ a\equal{}b\equal{}c\equal{}1.$

[quote="Ji Chen"]If $ a,b,c$ be real numbers such that $ abc = 1,$  then

$ \frac {a + 1}{7a^2 + 6a + 7} + \frac {b + 1}{7b^2 + 6b + 7} + \frac {c + 1}{7c^2 + 6c + 7}\leq\frac {3}{10},$

with equality if and only if $ a = b = c = 1.$[/quote]
Because
$ \frac{6(x+1)}{7x^+6x+7} =1-\frac{7x^2+1}{7x^2+6x+7} \le 1-\frac{7x^2+1}{7x^2+6|x|+7},$
one can see that it suffices to prove the inequality holds for $ a,b,c>0.$

Now, for $ a,b,c>0,$ we have
$ \sum \frac{a+1}{7a^2+6a+7} \le \sum \frac{a+1}{\frac{20}{3}(a^2+a+1)}=\frac{3}{20}\sum \frac{a+1}{a^2+a+1},$
and thus, it is enough to check that
$ \sum \frac{a+1}{a^2+a+1} \le 2,$
i.e.
$ \frac{a^2}{a^2+a+1}+\frac{b^2}{b^2+b+1}+\frac{c^2}{c^2+c+1} \e 1.$
Substituting $ a=\frac{x^2}{yz},b=\frac{y^2}{zx},c=\frac{z^2}{xy}$ ($ x,y,z>0$) in this inequality, it becomes
\[ \sum \frac{x^4}{x^4+x^2yz+y^2z^2} \ge 1,\]
which is true according to the Cauchy Schwarz Inequality
$ \sum \frac{x^4}{x^4+x^2yz+y^2z^2} \ge \frac{\left(\sum x^2\right)^2}{\sum (x^4+x^2yz+y^2z^2)},$
and the well-known
$ \sum x^2yz \le \sum y^2z^2.$

[quote="can_hang2007"][quote="Vasc"]What is Vasc Inequality, Ji Chen?[/quote]
I think he meant the following inequality of you, Vasc:

If $ x,y,z$ are positive numbers such that $ xyz \equal{} 1,$ then
$ \sqrt {\frac {2}{x^2 \plus{} 1}} \plus{} \sqrt {\frac {2}{y^2 \plus{} 1}} \plus{} \sqrt {\frac {2}{z^2 \plus{} 1}} \le 3.$

The reason is because we have
$ \sqrt {\frac {2}{x^2 \plus{} 1}} \le \frac {10(x \plus{} 1)}{7x^2 \plus{} 6x \plus{} 7}$
for positive real number $ x.$[/quote]
Indeed, this is my old

$ \sum\sqrt\frac{2a}{a\plus{}b} \le 3$.  :idea:


:?:

 This is the last question concerning factoring in my textbook.

It can be factored.
[hide="hint 1"]Factor d out of the equation.[/hide]

[hide="hint 2"]Substitute 2k for a and treat it like a quadratic.[/hide]

[quote="Arcenas"]Factor d^4k+1 - 5d^2k+1 + 4d . . . (my edit)

 This is the last question concerning factoring in my textbook.[/quote]

Arcenas,

you typed this incorrectly for horizontal style.  Because of the order of operations,
it needs grouping symbols around the exponents like this:

"Factor d^(4k+1) - 5d^(2k+1) + 4d."

Otherwise, it could be shown as "Factor $ d^{4k\plus{}1} \minus{} 5d^{2k\plus{}1} \plus{} 4d$."

Or use $ LaTeX$ aka put dollar signs around :)
$ \equal{} d^{4k \plus{} 1} \minus{} 5d^{2k \plus{} 1} \plus{} 4d$
$ \equal{} d(d^{4k} \minus{} 5d^{2k} \plus{} 4)$
$ \equal{} d(d^{2k} \minus{} 4)(d^{2k} \minus{} 1)$
$ \equal{} (d^k \minus{} 2)(d^k \minus{} 1)(d)(d^k \plus{} 1)(d^k \plus{} 2)$

[quote="Arrange your tan"][quote="addikaye03"]Factor $ d^{4k \plus{} 1} \minus{} 5d^{2k \plus{} 1} \plus{} 4d$.

d[d^4k-5d^2k+4]

d[d^4k-d^2k-4d^2k+4]

d[d^2k(d^2k-1)-4(d^2k-1)]

d(d^2k-4)(d^2k-1)

d(d^k-2)(d^k+2)(d^k-1)(d^k+1)[/quote]

You need grouping symbols around the exponents of more
than one variable or more than one constant (or a product
of them) for the order of operations.  Also, you should space
out your addition and subtraction signs for easier readability:

d^(4k+1) - 5d^(2k+1) + 4d

= d[d^(4k) - 5d^(2k) + 4]

= d[d^(4k) - d^(2k) - 4d^(2k) + 4]

= d{d^(2k)[d^(2k) - 1] - 4[d^(2k) - 1]}

= d[d^(2k) - 4][d^(2k) - 1]

= d(d^k - 2)(d^k + 2)(d^k - 1)(d^k + 1)[/quote]

yeah, sorry. Usually i bracket everything but was in a hurray


Is there a part of the site that shows all of the AoPS animated proofs? Is there a way to view all of them?

http://usamts.org/Gallery/G_Gallery.php

They have them on AoPS somewhere too, but I forget where.