PM me some random words in the right category to me and I will post the ridiculous stories!!!
It will be up to you to vote for whose is the best.

Adjective:
Person on AoPS:
Adjective:
Animal:
Adjective:
Number:
Part of body (plural i.e. feet):
Another person on AoPS:
Plural Noun:
Noun:
Noun:
Adjective:
Verb (Pass Tense):
Noun:
Noun:
Adjective:

These are due:

[u][b][color=red][size=200]WEDNESDAY NOV 26!!!![/color][/size][/b][/u]

[b]Sign Ups:[/b]
ProtestanT
kookamunga
james4l
dragon96
aleph0

Please vote [b]in bold[/b]:

1. ProtestanT

Adjective: vicious
Person on AoPS: i_like_pie
Adjective: tiny
Animal: snake
Adjective: delicious
Number: forty-one
Part of body (plural i.e. feet): shoulders
Another person on AoPS: Brut3Forc3
Plural Noun: mice
Noun: house
Noun: meritocracy
Adjective: painful
Verb (Pass Tense): whined
Noun: nation
Noun: cube
Adjective: frozen

Dear Principal,

I am sorry to have to tell you that my vicious son i_like_pie will be unable to attend your tiny school this week as he has caught a case of the snake pox. The delicious doctor says that it will be 41 weeks before he is health and back on his shoulders again.

Dear Math Teacher,

I was driving Brut3 Forc3 to school when the mice failed and my car crashed into a house. By the time the tow meritocracy finally arrived and the painful mechanic whined the nation and recharged the cube, we missed your frozen class.

2. kookamunga

Adjective: fried
Person on AoPS: kookamunga
Adjective: singing
Animal: tarantula
Adjective: helpful
Number: -16
Part of body (plural i.e. feet): fingers
Another person on AoPS: Dojo
Plural Noun: eggs
Noun: submarine
Noun: piano
Adjective: sly
Verb (Pass Tense): floated
Noun: fish
Noun: noise
Adjective: expensive

Dear Principal,

I am sorry to have to tell you that my fried son kookamunga will be unable to attend your singing school this week as he has caught a case of the tarantula pox. The helpful doctor says that it will be -16 weeks before he is health and back on his fingers again.

Dear Math Teacher,

I was driving Dojo to school when the eggs failed and my car crashed into a submarine. By the time the tow piano finally arrived and the sly mechanic whined the fish and recharged the noise, we missed your expensive class.

3. james4l

Adjective: Destructible
Person on AoPS: Dojo
Adjective: Perishable
Animal: Homo Sapiens
Adjective: Comestible
Number:  $ i$
Part of body (plural i.e. feet): Cerebral cortices
Another person on AoPS: dragon96
Plural Noun: Nuclear Bombs
Noun: Planet
Noun: Plasmodium
Adjective: Eukaryotic
Verb (Pass Tense): Epic-Failed
Noun: Praseodymium
Noun: Anathema
Adjective: Empty

Dear Principal,

I am sorry to have to tell you that my Destructible son Dojo will be unable to attend your perishable school this week as he has caught a case of the Homo Sapiens pox. The Comestible doctor says that it will be $ i$ weeks before he is health and back on his Cerebral cortices again.

Dear Math Teacher,

I was driving dragon96 to school when the Nuclear Bombs failed and my car crashed into a planet. By the time the tow Plasmodium finally arrived and the Eukaryotic mechanic Epic-Failed the Praseodymium and recharged the Anathema, we missed your empty class.

4. dragon96

Adjective: ferocious
Person on AoPS: AIME15
Adjective: vicious
Animal: dragon
Adjective: sharp
Number: 1.4142135623730950488016887242097
Part of body (plural i.e. feet): talons (or if it has to be human part, toenail)
Another person on AoPS: isabella2296
Plural Noun: wings
Noun: wyvern
Noun: serpent
Adjective: rapid
Verb (Pass Tense): slashed
Noun: firestorm
Noun: heat wave
Adjective: hot

Dear Principal,

I am sorry to have to tell you that my ferocious son AIME15 will be unable to attend your vicious school this week as he has caught a case of the dragon pox. The sharp doctor says that it will be 1.4142135623730950488016887242097 weeks before he is health and back on his talons again.

Dear Math Teacher,

I was driving isabella2296 to school when the wings failed and my car crashed into a wyvern. By the time the tow serpent finally arrived and the rapid mechanic slashed the firestorm and recharged the heat wave, we missed your hot class.

5. aleph0

Adjective: spammy
Person on AoPS: eggylv999
Adjective: spammy
Animal: penguin
Adjective: spammy
Number: 999
Part of body (plural i.e. feet): eyes
Another person on AoPS: gauss1181
Plural Noun: eyes
Noun: spam
Noun: spam
Adjective: spammy
Verb (Pass Tense): spammed
Noun: spam
Noun: spam
Adjective: spammy

Dear Principal,

I am sorry to have to tell you that my spammy son eggylv999 will be unable to attend your spammy school this week as he has caught a case of the penguin pox. The spammy doctor says that it will be 999 weeks before he is health and back on his eyes again.

Dear Math Teacher,

I was driving gauss1181 to school when the eyes failed and my car crashed into a spam. By the time the tow spam finally arrived and the spammy mechanic spammed the spam and recharged the spam, we missed your spammy class.

6. 33286

Adjective: actual
Person on AoPS: 1=2
Adjective: emblematic
Animal: humuhumunukunukuapuaa
Adjective: xenial
Number: five hundred twenty-two undecillion, three hundred eighty-one decillion, fifty-seven nonillion, nine hundred seven octillion, two hundred thirty-six septillion, three sextillion, eight hundred eighty-six quintillion, one hundred nineteen quadrillion, seven hundred fifty trillion, two hundred sixty-four billion, three hundred forty-seven million, nine hundred six thousand, five hundred fifteen and one half (522,381,057,907,236,003,886,119,750,264,347,906,515½)
Part of body (plural i.e. feet): cerebellums
Another person on AoPS: Dojo
Plural Noun: revelations
Noun: qat
Noun: Europe
Adjective: implicit
Verb (Past Tense): imploded
Noun: archiplasm
Noun: heliolatry
Adjective: factitive

Dear Principal,

I am sorry to have to tell you that my actual son 1=2 will be unable to attend your emblematic school this week as he has caught a case of the humuhumunukunukuapuaa pox. The xenial doctor says that it will be five hundred twenty-two undecillion, three hundred eighty-one decillion, fifty-seven nonillion, nine hundred seven octillion, two hundred thirty-six septillion, three sextillion, eight hundred eighty-six quintillion, one hundred nineteen quadrillion, seven hundred fifty trillion, two hundred sixty-four billion, three hundred forty-seven million, nine hundred six thousand, five hundred fifteen and one half weeks before he is health and back on his eyes again.

Dear Math Teacher,

I was driving Dojo to school when the revelations failed and my car crashed into a qat. By the time the tow Europe finally arrived and the implicit mechanic imploded the archiplasm and recharged the heliolatry, we missed your factitive class.

Har-har, james. Real funny.

[b]Vote: 33286[/b]

You forgot to replace a couple of the words. For instance, you left "whined" in the second entry from the first (it should have said "floated"), and mine said "the spammy doctor" when it should have read "the xenial doctor".

I put "i" for the number. It was in LaTeX...

kookamunga's should say "floated the fish", and mine should say "back on his cerebellums".


How many different three-digit positive integers can be formed if each digit is a prime number and digits cannot be repeated within a number?

Easy. The posible digits are $ 2,3,5$, and $ 7$. So there are $ 4 P 3$(permutations) possible numbers which equals $ 24$ possible numbers.


calculus
integration
calculus computations

Let $m,\ n$ be non negative integers.
Calculate
\[\sum_{k=0}^{n}(-1)^{k}\frac{n+m+1}{k+m+1}\ nC_{k}. \]

where $_{i}C_{j}$ is a binomial coefficient which means $\frac{i\cdot (i-1)\cdots(i-j+1)}{j\cdot (j-1)\cdots 2\cdot 1}$.

I think we can write it as $(n+m+1) \int_{0}^{1}x^{m}\cdot (x-1)^{n}dx$, but I'm not sure how this is calculated.

That's almost correct. With a little more attention paid to the powers of $-1,$ it should be written as

$(n+m+1) \int_{0}^{1}x^{m}\cdot (1-x)^{n}dx$

After that? There have been quite a few topics devoted to the Beta integral or Beta function. For instance, [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=145241]this topic[/url] from today.

The answer is $\frac{n!m!}{(n+m)!},$ which is the reciprocal of a binomial coefficient.

[quote="amitie"]could someone explain how paladin8 arrived at that integral? thx :)[/quote]

This is what I think he did. The portion $ \dfrac{1}{k \plus{} m \plus{} 1}$ in the series can be taken to be $ \int_{0}^{1}x^{k \plus{} m}dx$. And so our original series is

$ (n \plus{} m \plus{} 1)\sum_{k \equal{} 0}^{n}\binom{n}{k}( \minus{} 1)^{k}\int_{0}^{1}x^{k \plus{} m}dx$

We can interchange the integral and sum without problems since the series is finite. Doing that leads to

$ (n \plus{} m \plus{} 1)\int_{0}^{1}\sum_{k \equal{} 0}^{n}\binom{n}{k}( \minus{} 1)^{k}x^{k \plus{} m}dx \equal{} (n \plus{} m \plus{} 1)\int_{0}^{1}x^{m}\sum_{k \equal{} 0}^{n}\binom{n}{k}( \minus{} 1)^{k}x^{k}dx$

$ \equal{} (n \plus{} m \plus{} 1)\int_{0}^{1}x^{m}\sum_{k \equal{} 0}^{n}\binom{n}{k}( \minus{} x)^{k}dx$

The series within the integral is just the expansion of $ (1 \minus{} x)^{n}$ so we have

$ (n \plus{} m \plus{} 1)\int_{0}^{1}x^{m}(1 \minus{} x)^{n}dx$

Evaluating that looks to be a matter of integrating by parts repeatedly ($ n$ times). It also looks like alot is just zero in between except for the constants pulled out of each integral (if that makes sense?). In other words it's the product

$ \dfrac{n}{m \plus{} 1}\cdot\dfrac{n \minus{} 1}{m \plus{} 2}\cdot\cdot\cdot\dfrac{1}{m \plus{} n} \equal{} \dfrac{n!m!}{(m \plus{} n)!}$

The last integral evaluates to $ \dfrac{1}{m \plus{} n \plus{} 1}$ which cancels out that original constant we factored out.


[quote="EUCLA"]per chi volessere fare un po d'esercizio

Prove that

$ 1 \minus{} \frac {x}{1!} \plus{} \frac {x(x \minus{} 1)}{2!} \minus{} \frac {x(x \minus{} 1)(x \minus{} 2)}{3!} \plus{} .... \plus{} ( \minus{} 1)^{n}\frac {x(x \minus{} 1)(x \minus{} 2)......(x \minus{} n \plus{} 1)}{n!} \equal{} ( \minus{} 1)^{n}\frac {(x \minus{} 1)(x \minus{} 2)....(x \minus{} n)}{n!}$

for all non-negative integers $ n$[/quote]

Basta fare il denominatore comune...


$\sqrt{\sqrt{1}}$

-1 is also a square root of 1,
square root -1 gives you get i  :?

You can't really "disprove" anything, you just have to know that a number has an imaginary square root also.

We can't apply our fractional power rules to imaginary numbers as well.

There are quite a few ways you can make $1$. It's a weird number like that, but just because you can use two different numbers to make it doesn't mean that those two numbers are equal.

We can find the fourth roots of unity (all the numbers where $n^4=1$) using DeMoivre's theorem, and $i$ is one of them.

Strange...

Most of the times, you could argue with the (true) fact that $i = \sqrt{-1}$ can't be written, but only $i^2 = -1$.

[quote="shinwoo"]this is something my friend brought up, but I know it's wrong but I don't know how to disprove it

$i=\sqrt{-1}$

then we know that

$i^4=1$

so if we 4th root both sides wouldn't we get

$i=1$

?
how do you disprove it?[/quote]
The fourth root of $1$ can be $1$ or $i$.

Or $-1$ or $-i$. Keep in mind that $n^4$ always has four roots, whether real or imaginary, and the same with polynomials.

If you raise a number to the fourth power, and then take the fourth root of the result, you don't necessarily get the same number you started with.  For example, $\sqrt[4]{(-2)^{4}}=2$.  Just as this does not prove that $-2=2$, showing that $\sqrt[4]{i^{4}}=1$ does not prove that $i=1$.

If you use DeMoivre's theorem, you get that $\sqrt[n]{x}$ has n answers that, if x is in trigonometric form, then $x=r\cis (\cos\theta+i\sin\theta)$ and $\sqrt[n]{x}=\sqrt[n]{r}\cis (\cos\frac{\theta+2\pi k}{n}+i\sin\frac{\theta+2\pi k}{n})$ with $k=1, 2, 3 \cdots$ which gives n roots, which, plotted on the circle with radius r will be evenly spaced, by a measurement of $\frac{2\pi}{n}$. Hope this was helpful. Oh, and $\sqrt[4]{1} = 1, -1, i, -i$.


[quote="kenn4000"]$ \sum_{k \equal{} 0}^{N} (k \plus{} 1)( \minus{} 3)^{ \minus{} k}\equal{} \frac {9}{16}[1 \plus{} \frac {4N \plus{} 7}{( \minus{} 3)^{N \plus{} 2}}]$[/quote]

That's correct.

Here is my solution:

$ \sum_{k\equal{}0}^N (k\plus{}1)(\minus{}3)^{\minus{}k}\equal{}\frac{1}{16}\sum_{k\equal{}0}^N \left(\frac{4k\plus{}7}{(\minus{}3)^k}\minus{}\frac{4(k\minus{}1)\plus{}7}{(\minus{}3)^{k\minus{}1}}\right)\equal{}\frac{4N\plus{}7}{16(\minus{}3)^N}\plus{}\frac{9}{16}$

yes, the sum will reduce into a telescoping series


geometry
function
advanced fields
advanced fields unsolved

I dont know if this is even in the correct forum.. so feel free mods to move if not with my appolagies.

My problem:

I need to come up with a good way to measure the potential of profitability with many variables. For instance, I look on a map at a possible location to put a new restuarant and look at a set of many many variables to decide the potential of the location if the restuarant is opened there.

Variables would include things like the type of road traffic, traffic counts, competitors in the area, if there is a high school with open campus in the area, the demographics of the area and so on. Basically give each variable a weighted value and in the end you have an idea of the potential of the location.

What is my best bet in this.

I think you want to construct a detailed model of the situation. I mean, make up "reasonable" functions (based on evidence or your own reasoning or whatever) to describe how each of the variables affects your success.

For example, higher traffic ought to mean more customers (assuming that your restaurant is the kind people randomly stop at, and that there's decent parking). If the conditions I just gave in parentheses are true, you might even say that the total demand in the area will be a linear function of the traffic.

Other inputs, like the high school, would also affect the total demand.

Then you'd want to decide how that demand will be divided between you and the competitors -- you might use the demographic information to decide mathematically how to do that. For example, if it's a very rich area, the demand for food from a fancy restaurant would be high, whereas if it's a very poor area, demand for fast food will probably be much higher. Also the relative accessibility of competitors will help (do you have a large and easily accessible parking lot while they do not? Are you two blocks from the high school while they are four blocks? etc)

I can't give you a more specific answer than this. You need to figure out how to use all of the information you have in order to construct a model of the situation. I hope my comments help a bit. :)


1)Find all $ (a,b)\in$N such that $ 2^a\plus{}3^b$ is a perfect square.

[quote="TTsphn"]1)Find all $ (a,b)\in$N such that $ 2^a \plus{} 3^b$ is a perfect square.[/quote]

We have $ 2^a\plus{}3^b\equal{}x^2$.
In this equation, RHS is $ 0$ or $ 1$ modulus $ 3$ and LHS is $ 1$ (if $ a$ is even) or $ 2$ (if $ a$ is odd). So $ a\equal{}2c$ is even.

We have the $ 2^{2c}\plus{}3^b\equal{}x^2$ and so $ 3^b\equal{}(x\minus{}2^c)(x\plus{}2^c)$ and so $ x\minus{}2^c\equal{}3^u$ and $ x\plus{}2^c\equal{}3^v$, which implies $ 2^{c\plus{}1}\equal{}3^v\minus{}3^u$
In this last equation, and since $ 3$ does not divide LHS (power of $ 2$), we need $ u\equal{}0$ and we are looking for $ 2^{c\plus{}1}\equal{}3^v\minus{}1$.
Since $ c>0$, $ 4$ divides LHS and so $ v\equal{}2w$ is even and we have $ 2^{c\plus{}1}\equal{}3^{2w}\minus{}1\equal{}(3^w\minus{}1)(3^w\plus{}1)$ and so $ 3^w\minus{}1\equal{}2^p$ and $ 3^w\plus{}1\equal{}2^q$, which implies $ 2^q\minus{}2^p\equal{}2$ and so $ p\equal{}1$ and $ q\equal{}2$, so $ w\equal{}1$, $ c\equal{}2$, $ a\equal{}4$, $ v\equal{}2$, $ c\equal{}2$, $ x\equal{}5$ and the unique solution :

$ 2^4\plus{}3^2\equal{}5^2$


inequalities
number theory unsolved
number theory

Find all positive integers $x$ such that the product of all digits of $x$ is given by $x^2 - 10 \cdot x - 22.$

$x$ is a positive integer so the product of his digits must be not negative
So We have the inequality:
$x^2 - 10 \cdot x - 22\geq 0$
After the calculation and the fact that $x$ is integer we get:
$x \leq 1$ or $x \geq 12$
when $x=1$ it's not satisfying the condition so
$x \geq 12$
we know from here that for all $x \geq 12$ $x^2 - 10 \cdot x - 22$ is increasing
Suppose that $x$ is at least 3 digits long so:
$x^2 - 10 \cdot x - 22\geq {9^3} = 729$
but the minimum of $x^2 - 10 \cdot x - 22$ when $x$ is at least 3 digits long is :
${100^2} -10 \cdot 100 -22 = 8978$
So we conclude from here that $x \leq 99$
if $x \leq 99$ the maximum product that can be is ${9^2}=81$
So we get a new inequality:
$x^2 - 10 \cdot x - 22\leq 81$
After the calculation we get:
$7 \geq x \leq 16$
From here  the maximum product that can be is $9$
So again we bulid an inequality:
$x^2 - 10 \cdot x - 22\leq 9$
And we get:
$3 \geq x \leq 12$
And before we proved that $x \geq 12$
So $x$ can be only equals to $12$
we check if it's satisfying the condition:
$12^2 - 10 \cdot 12 - 22 = 144-120-22 = 2$
Conclusion:
The only integer  such that the product of all digits of $x$ is given by $x^2 - 10 \cdot x - 22$ is $12$

Is the term mentioned in the question $x^2-10x-22$
 or $(x^2-10)(x-22)$ ?

Seeing that the solution given by pavel25 seems correct, I'd wager that it's the former. In addition, following PEMDAS arrives at the same conclusion.

But did you really need to revive a 10 year old topic to ask that?


[quote="h_s_potter2002"]I was wondering, is there a forum like this one about computer programming and contests [/quote]

I'm pretty sure there isn't, although there are people on [i]this[/i] forum who are interested in computing competitions. If you manage to get invited to USACO camp, you'll then be part of a USACO email list.

BTW - the Central European Olympiad in Informatics is coming up in a couple of weeks. If you don't happen to be in Poland, there is a division for people who want to participate over the internet (it's all in English). http://www.oi.edu.pl/php/ceoi2004.php?module=show&file=links

[quote="h_s_potter2002"]For the Central Europe Olympiad, how do you register, or do you even need to?[/quote]

Sorry, I don't know any more than what's on the site. I do know that the Balkan Olympiad, which is this week, had a registration deadline for the internet contest which has just passed. If you have to pre-register for the internet CEOI, I'm sure that the website tells you that somewhere. If I hear anything more, I'll post it.

[quote="h_s_potter2002"]I was wondering, is there a forum like this one about computer programming and contests for it (such as the ACL, or the USACO)?  If there is, can anyone please tell me them.

Thanks.[/quote]
You mean like [url=http://www.cprogramming.com]cprogramming.com[/url]?

not a forum, but another place for internet contests is at http://www.topcoder.com

I found one, but It's not really that great.  All of the messages are like one line long.

Well, anyways the website is [url]ioiforum.org[/url].


I've been doing pretty good until I came across this problem:

$(x^5 + 3x + 2) / (x^3 + 1 + 2x)$

Answer is:
[hide]$x^2 - 2 + [(-x^2 + 7x + 4) / (x^3 + 2x + 1)]$[/hide]

I was going to go in depth about this, however, I don't see the latex symbol that closely looks like the square root symbol for division....

Use long division method. I believe they teach it in grade 6 or 7 where I live

It's very obvious to me that I need long division. I just didn't know how to solve the problem, that's all. I don't understand the long division processes used in this.

Long division in plain numbers is somewhat easy, however, this polynomial division isn't. I was wondering if someone could give me detailed steps on how to take on this problem.

Its pretty simple. For example,
X^2+4x+4 / X+2 would be like:
        _________
X+2  )X^2+4x+4

X+2 would go into (or the closest into) X^2+4x would be x,so you put x
       X
       ___________
X+2 )X^2+4x+4
     -  X^2+2x
        ______
                2x+4
Then X+2 would go into 2x+4 2 times,thus completing the division.
     X    +     2
     ___________
X+2 )X^2+4x+4
     -  X^2+2x
        ______
                2x+4
               -2X+4

SImple as that.


Prove that if $ a\geq 4 ,b\geq 5 ,c\geq 6$ and $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 90$
           then  $ a \plus{} b \plus{} c \geq 16$        :P

For a fixed a, b+c is minimized when c is as high as possible and b is as low as possible. So we can assume b=5. Then, given $ a^2 \plus{} c^2 \equal{} 65$, we have to show
$ a \plus{} c \ge 11$
Again, a+c is minimized when a is as small as possible, and c is as large as possible. So we find equality holds when a=4, b=5, c=7.
In summary, a+b+c is minimized when c takes all the excess.

It 's here
http://www.mathlinks.ro/viewtopic.php?t=223523


[quote="juicybooty911"]if i'm finished w/ the AIME, am i allowed to leave early?[/quote]

Yeah... but if you actually finish all the problems on the AIME before 3 hours, I'd admire you greatly :P.

trust me, i will be early, because i will be guessing them all.

Don't leave early.  3 hours is a lot of time, and sometimes problems that appear intractable at first get a lot more tractable when you've been looking at them for a long time.  So use those 3 hours; it might just pay off.

Hate to be picky...but this is getting closed. No AIME discussion means no AIME discussion. Unless I hear otherwise, exceptions don't exist. :-)


number theory unsolved
number theory
india

Let $ A$ be a set of real numbers such that $ A$ has at least four elements. Suppose $ A$ has the property that $ a^2 \plus{} bc$ is a rational number for all distinct numbers $ a,b,c$ in $ A$. Prove that there exists a positive integer $ M$ such that $ a\sqrt{M}$ is a rational number for every $ a$ in $ A$.

[quote="Rijul saini"]Let $ A$ be a set of real numbers such that $ A$ has at least four elements. Suppose $ A$ has the property that $ a^2 \plus{} bc$ is a rational number for all distinct numbers $ a,b,c$ in $ A$. Prove that there exists a positive integer $ M$ such that $ a\sqrt {M}$ is a rational number for every $ a$ in $ A$.[/quote]

Let $ A\equal{}\{a_1,a_2,... \}$ and WLOG $ a_1 \neq 0$.
We have $ x\equal{}a_m^2\plus{}a_ia_j$ and $ y\equal{}a_m^2\plus{}a_ia_k$ for all distinct $ m,i,j,k$

$ x\minus{}y \in Q \Rightarrow a_i(a_j\minus{}a_k) \in Q \forall$ distinct $ i,j,k$

$ \Rightarrow \frac{a_i(a_j\minus{}a_k)}{a_1(a_j\minus{}a_k)} \equal{} \frac{a_i}{a_1} \in Q \forall$ distinct $ i,j,k$ where $ j,k \neq 1$

Now let $ b\equal{}k_1a_1 , c\equal{}k_2a_1$ , where $ k_1,k_2 \in Q$ ,
we have $ a_1^2\plus{}bc \in Q \Rightarrow a_1^2 \in Q$

Now we let $ a_1^2\equal{}\frac{q}{p}$ for some $ p,q \in N$ and $ M\equal{}pq$.
We have $ a_1\equal{}q$ and because $ \frac{a_i}{a_1} \in Q \forall i$  we get $ a_1,a_2,... \in Q$

Indians should be more creative in copying !

We will show that each member of the set A is of the form $\sqrt{\alpha}$ for some $\alpha \in Q$. This would complete the proof.
First, choose any 4 members of the set, say $a,b,c,d$, such that $c \ne 0$ and $b \ne -c$. Clearly, we can find such $b,c$, else all would be 0 (if $c=0$), or all would be equal to $-c$, from where we get that $c$ must be the square of a rational, and hence we get $M=1$.
Hence, $b^{2}+ca \in Q$ and $c^{2}+ab \in Q$ gives
$$(b^{2}-c^{2})+a(c-b) \in Q : (1)$$
Now, $d^{2}+ab \in Q$ and $d^{2}+ac \in Q$ gives
$$a(c-b) \in Q : (2)$$
Thus, (1) and (2) give
$$b^{2}-c^{2}=p \in Q : (3)$$
Now we know that $c^{2}+db \in Q$, $a^{2}+bc \in Q$ and $a^{2}+bd \in Q$. Hence,
$$(c^{2}+db)+(a^{2}+bc)-(a^{2}+bd)=c(b+c) \in Q : (4)$$
Hence (3) and (4) give us
$$(c^{2}+bc)-(c^{2}-b^{2})=b(b+c) \in Q: (5)$$
Thus, by (4) and (5);
$$\frac{b}{c} \in Q: (6)$$
We now use (3) and (6) to conclude as
$$\frac{p}{c^{2}}=\frac{b^{2}}{c^{2}}-1 \in Q$$
Thus we get $c^{2} \in Q$.
Thus $ab, bc, cd, da \in Q$ which gives us that all the elements of the set $A$ are squares of rationals, as desired.


Some information about the mock contest that will be starting tomorrow...

1. QUESTION FORMAT: Multiple Choice

2. NUMBER OF QUESTIONS: 25, just like the real thing.

3. PROBLEM SOURCES: Undisclosed at this time.  I've written many of these problems myself, and a few are from other places.  If you have a problem that would be appropriate, PM/email it to me.  Thanks!

4. SCORING: As with the real AMC, you get 6 points for a correct answer, 2.5 points for no answer, and 0 points for an incorrect answer.

5. SCOREBOARD: There will be a scoreboard to announce the scores of participants.  'nuff said.

6. SUBMITTING ANSWERS/SOLUTIONS: Please PM your answers/solutions to me, mathfanatic, or e-mail them to soccerdude9869 at netscape dot net.  (Use @s and .s where necessary.)  I will try to get back to you as quickly as possible.

7. TIME LIMIT: There is no time limit; that would be hard to regulate.  Take as much time as you like; check your answers if you wish.  However, keep in mind that on the real test, you will have a time limit, so you might want to do all these problems under a time limit.  The standard time limit is 90 minutes.

8. NEXT ROUND: Yes, I plan to have more than one mock AMC contest, however I don't know exactly when.  Especially because this will depend on the amount of work I have, etc.  Stay tuned!

Good luck and enjoy!

Why don't you get people to pm you the answers so that you can grade it like a normal AMC?

Good point.  Awlright, amended.

You may PM them, or send them to 'soccerdude9869 at netscape dot net'.  (The latter is actually preferred, since my inbox is unlikely to fill up.)

Oh yes...as for problem sources, I am actually planning on writing the questions myself.  I already composed five such questions.

i thought 75 minutes was the norm for an amc

Oh....hehe.  Shows you how much I know.

Anyway, did I say tomorrow?  I meant Saturday.  :oops:

When is the last day we can e-mail you answers?

There is no deadline, really.  However, I'm hoping to open the competition up on the board in a week or two so that everyone can post their solutions.  That way, we can hear different methods, etc.

However, I will accept and grade stuff at any time.  (Well, maybe not years from now...but you know what I mean...)


(1) $a^2+b^2+c^2+abc\geq 4$
Homogenize it, $9(a+b+c)(a^2+b^2+c^2)+27abc\geq 4(a+b+c)^3$
Expand everything, this is equivalent to $5(a^3+b^3+c^3)+3abc\geq 3\sum_{sym}a^2b$
By Schur, $\sum_{sym}a^2b\leq a^3+b^3+c^3+3abc$
So we only need to prove $5(a^3+b^3+c^3)+3abc\geq 3(a^3+b^3+c^3+3abc)$
$\Leftrightarrow 2(a^3+b^3+c^3)\geq 6abc$
This is true by AM-GM, Muirhead, or whatever you like. :D

(2) $a^3+b^3+c^3+3abc\geq 6$
Homogenize it, $9(a^3+b^3+c^3)+27abc\geq 2(a+b+c)^3$
Expand everything, this is equivalent to $7(a^3+b^3+c^3)+15abc\geq 6\sum_{sym}a^2b$
By Schur, $\sum_{sym}a^2b\leq a^3+b^3+c^3+3abc$
So we only need to prove $7(a^3+b^3+c^3)+15abc\geq 6(a^3+b^3+c^3+3abc)$
$\Leftrightarrow a^3+b^3+c^3\geq 3abc$
This is true by AM-GM, Muirhead, or whatever you like. :D

(3) $a^4+b^4+c^4+7abc\geq 10$
Homogenize it, $81(a^4+b^4+c^4)+189abc(a+b+c)\geq 10(a+b+c)^4$
Expand everything, this is equivalent to $71(a^4+b^4+c^4)+69(a^2bc+b^2ca+c^2ab)\geq 40\sum_{sym}a^3b+60(a^2b^2+b^2c^2+c^2a^2)$
By Schur, $\sum_{sym}a^3b\leq a^4+b^4+c^4+a^2bc+b^2ca+c^2ab$
So we only need to prove $71(a^4+b^4+c^4)+69[\sum_{sym}a^3b-(a^4+b^4+c^4)]\geq 40(\sum_{sym}a^3b)+60(a^2b^2+b^2c^2+c^2a^2)$
$\Leftrightarrow 2(a^4+b^4+c^4)+29\sum_{sym}a^3b\geq 60(a^2b^2+b^2c^2+c^2a^2)$
This is true by AM-GM, Muirhead, or whatever you like. :D

(4) I think this can be proved in a similar way... :)  :lol:  :D  :P

One time in diendantoanhoc.net I had discussed the general problem:

Find the best constant $k$ for the inequality:

$a_1^2+a_2^2+...+a_n^2+ka_1a_2...a_n \ge n+k$

If $a_1+a_2+...+a_n=n$.

The general for this:

$a_1^m+...+a_n^m+ka_1a_2...a_n \ge n+k$

I think two of them have same way to solution. Use intruction we have $n-1$ number equal. Maybe VASC also use n-1 EV to solve it.

For:
$ a,b,c>0,a+b+c=3\Rightarrow a^2+b^2+c^2+1.5abc\ge 4.5 $ we have:
$ a,b,c>0,a+b+c=3\Rightarrow a^2+b^2+c^2+1.5abc\ge 4.5 \Leftrightarrow $
$ a,b,c>0,a+b+c=3\Rightarrow 2(a^2+b^2+c^2)+3abc\ge 9 \Leftrightarrow $
$ \sum{(b+c-a)(b-c)^2}\ge 0 \Leftrightarrow $
$ S=S_{a}(b-c)^2+S_{b}(c-a)^2+S_{c}(a-b)^2\ge 0 $ where:
$ S_{a}=b+c-a,S_{b}=c+a-b,S_{c}=a+b-c $
Suppose that $ a\ge b\ge c $, result $ S_{b}\ge 0,S_{c}\ge 0,S_{b}+S_{c}\ge 0 $
According to the method SOS inequality is true!
__________
Marin Sandu

[quote="arqady"]If $a,b,c$ are positive numbers such that $a+b+c=3$, then
(1) $a^2+b^2+c^2+1.5abc \ge 4.5$;
(2) $a^3+b^3+c^3+3.75abc \ge 6.75$;
(3) $a^4+b^4+c^4+7.125abc \ge 10.125$;
(4) $a^5+b^5+c^5+12.1875abc \ge 15.1875$. :)[/quote]
(1) After homogenization it is just Schur \[\sum_{\text{sym}} (a^3 + abc - 2a^2b) \geq 0\]
(2) After homogenization it is just Schur \[\sum_{\text{sym}} (a^3 + abc - 2a^2b) \geq 0\]
(3) After homogenization it is \[\sum_{\text{sym}}(7a^4 + 7a^2bc) \geq \sum_{\text{sym}}(8a^3b + 6a^2b^2)\] which is true because \[\sum_{\text{sym}}(7a^4 + 7a^2bc) \geq \sum_{\text{sym}}14a^3b \geq \sum_{\text{sym}}(8a^3b + 6a^2b^2)\] The first inequality is by Schur and the second inequality is by Muirhead.
(4) After homogenization it is \[\sum_{\text{sym}} (9a^5 + a^3bc + 8a^2b^2c) \geq \sum_{\text{sym}} (6a^4b + 12a^3b^2)\] WLOG let $a \leq b \leq c$, $b = a + u$, $c = a + u + v$, $u,v \geq 0$, then it is just \[a^3 \left(25 u^2+25 u v+25 v^2\right)+a^2 \left(20 u^3+30 u^2 v+120 u
   v^2+55 v^3\right)+a \left(4 u^4+8 u^3 v+137 u^2 v^2+133 u v^3+39
   v^4\right)+48 u^3 v^2+72 u^2 v^3+42 u v^4+9 v^5 \geq 0\] which is obvious. Is there a way to finish (4) by Schur/Muirhead without BW?

[quote="nikoma"] Is there a way to finish (4) by Schur/Muirhead without BW?[/quote]
Yes of course! We can use here SOS, $uvw$ and more...


quadratics
function
algebra
domain
calculus
calculus computations

Hello.

Let $ p$ be a positive number

$ S \equal{} \sqrt {p \plus{} \sqrt {p \plus{} \sqrt {p \plus{} ...}....}}....$

Whats the limit of that series? A hint was given that fixed point iteration technique for finding roots of none linear equations can solve that question.

Anyone have any idea?

Thanks.

Here's how I'd do it.

1. Prove the limit exists: monotonicity and boundness.

2. if the limit exists, then denote it with $ L$. Then, as $ n\to \infty$ we have $ \sqrt{p\plus{}L}\equal{}L$, so just square it, and solve the quadratic equation.

To fill in a couple of more details of what hsiljak said and to talk about that numerical analysis course:

The function we are iterating is $ f(x) \equal{} \sqrt {x \plus{} p}.$ It's possible fixed points are solutions to $ f(x) \equal{} x$ or $ \sqrt {x \plus{} p} \equal{} x.$ The quadratic equation that hsiljak referred to is $ x^2 \minus{} x \minus{} p \equal{} 0,$ which has roots $ \frac { 1\pm\sqrt {1 \plus{} 4p}}2.$

But the the values of $ f(x)$ are always positive, so the sequence cannot possibly converge to the negative root. In fact, if we look closely, we see that the negative root falls outside the domain of the function. So there is only one root, $ L \equal{} \frac { 1 \plus{} \sqrt {1 \plus{} 4p}}2.$

Does the sequence converge? A necessary (but not sufficient) condition for the convergence of a fixed point iteration to a limit $ L$ is that $ |f'(L)| < 1.$ In this case, $ f'(x) \equal{} \frac1{2\sqrt {x \plus{} p}}.$ Putting $ L$ into that and simplifying, we get that $ f'(L) \equal{} \frac1{1 \plus{} \sqrt {1 \plus{} 4p}}.$ We clearly have that $ 0 < f'(L) < 1,$ so the necessary condition is met.

To actually prove that the limit exists, it would help to do this: show that for $ 0 < x < L$ we have $ x < f(x) < L$ so that if we start a point to the left of $ L$ the sequence increases and is bounded above, while for $ x > L$ we have $ x > f(x) > L$ so that the sequence decreases and is bounded below.


Hey guys,

I haven't been very active with AMC stuffs lately so I'm bit out of date. But I do recall reading one of old threads that said something like AMC prohibiting the use of calculator in AMC-12 and AMC-10? Can anyone verify this?

Great appreciation.

Happy "Early" New Year.  :)

Starting last year, no calculators are allowed.

Thanks.

So, AMC-10, AMC-12, and AIME ---- no calculator? Correct?

What about AMC-8 or is that same also? Thanks.

someone back me up, but i think the amc 8, too.

Somehow I had the impression that you were never allowed a calculator on AMC 8.  I'm not sure, though.

we were allowed calculators for AMC8 until this year..
starting next year, no calculators allowed..

at least that's what i heard.. :D

Yeah, I think I read that you could use the calc on this year's AMC 8, but not anymore starting next year.

With respect to the Feb 2008 AMC 10 and AMC 12, see the Teachers' Manual
page 4, Section I, paragraph 4, in the capital letters.
http://www.unl.edu/amc/d-publication/d1-pubarchive/2007-8pub/1012TM/2008AMC1012A-TM.pdf
See also Frequently Asked Question AMC 10/12 #7
http://www.unl.edu/amc/f-miscellaneous/faq.shtml

With respect to the just finished Nov 2007 AMC 8, see the Teachers' Manual, page 4, Section I, paragraph 7.
http://www.unl.edu/amc/d-publication/d1-pubarchive/2007-8pub/AMC8TM/2007-amc8TM.pdf
See also Frequently Asked Question AMC 8#3
http://www.unl.edu/amc/f-miscellaneous/faq.shtml

Steve Dunbar
MAA Director, American Mathematics Competitions

Thanks.

I'll now lock the topic since my question has been fully answered.

Again, thanks to all who gave responses.


geometry
3D geometry
pyramid
circumcircle
sphere
vector
geometry unsolved

Let $R$ be the circumradius of a right pyramid with a square base. Let $r$ be the radius of the sphere touching the four lateral faces and the circumsphere. Suppose that $2R = (1+2^{0.5})r$. Determine the angle between adjacent faces of the pyramid.

Here is a solution that involves some rigorous calculation and I shall give the crux of the solution may be someone can continue from there...I guess there must be a better solution...using Vectors...anyway working on those line....I would definitely be interested in a solution with Vector Algebra...

Let the radius of the circumsphere be x, height of the pyramid be h...
then we have
x= (R 2 +h 2 )/2h.....(1)

Let y be the perpendicular distance of the center of the circle touching the lateral sides to the square that is parallel to the base passing through the points of contact....

Then we have h 2= 25R 2 (r 2 -  y  2 )/8 y2  ......(2)

(R 2 +h 2 )/h = r + (5r 2 -3 y  2)/2y

Giving
R 2 (25r 2 -17y2 )= 50R 2 (r 2 -y2 )(5r-3y) 2 (r+y) 2 .....(*)

from (*) the value of y can be obtained...:D..
hence the value of h can be determined...
And the side of the square base a= \sqrt 2R hence the angle between the faces can be determined....some how...:D


Whenever I try to run the Asymptote tool in Latex, I get the following error message: (in Command Prompt)

Cannot load module (name).

Does anyone know how to fix this?

Can you give us an example and show us the output?

i.e. using the example on the AoPs Wiki (saving and titleing it "test")

draw((0,0)--(100,100));
draw((0,100)--(100,0));
dot((50,50));
label("$ P$",(50,50),S);

Then when I try to run Asymptote on it in TeXniCenter, it says "Could not load module test" on the DOS window.

The same thing happens when I try to run Asymptote on a LaTeX document with Asymptote code in it.
Edit: It works now for pure Asymptote code, but not for the LaTeX stuff.

Are you sure you have saved test as test.asy and not test.asy.tex? See Problem 2 of [url=http://www.artofproblemsolving.com/Wiki/index.php/Asymptote:_Help]Asymptote Help[/url].

I fixed the .tex to .asy, so it works for pure Asymptote code, but when I try to run it on a LaTeX document with Asymptote in it, it gives the same error message.

In a LaTeX document, saved as .tex, you compile the tex document first (Ctrl+F7), then run Asymptote, then recompile the tex document.

I've done that, but it still gives the same error message.

Check that the first compilation has created the asy file with the right content and also look at the log file for any errors.

There aren't any errors and the .asy file created is fine, but the Asymptote still won't run

Check the commands you have put in Tools, Customise, Tools are correct. If you aren't sure then post a screenshot here.

The commands are the same as the ones in the AoPS Wiki.
I tried restarting my computer, but now it Asymptote won't even work on pure Asymptote code

[quote="piggyinc"]but now it Asymptote won't even work on pure Asymptote code[/quote]What does 'doesn't work' mean? You need to be more precise.

It won't hurt to reinstall Asymptote over the top of the old version.


geometry
geometric transformation
rotation

Eleven girls are standing around a circle. A ball is thrown clockwise around the circle.
The first girl, Ami, starts with the ball, skips the next three girls and throws to the
fifth girl, who then skips the next three girls and throws the ball to the ninth girl. If
the throwing pattern continues, including Ami's initial throw, how many total throws are
necessary for the ball to return to Ami?

12 throws.
you can list it starting with 0 and finding one thats a multiple of 11:
0,4,8,12,16,20,24,28,32,36,40,44
44 would be the 12th throw

Hello the answer  is 11 but I dont know why.
Can someone tell me how?
I know the answer from FTW.

THis is because the first throw is Ami to the fifth girl, not to Ami herself.  xoangiexo counted 0 as a throw, but it is actually not a throw.

We can observe the pattern that if the ball starts with person n, after one complete rotation (3 throws clockwise), it ends up with person n + 1. Thus, we are basically going going around the polygon, and since it is an 11-gon, the answer must be 11.


The answer to a physics problem gives the following solution, and I'm unclear
about how to subtract the equations for $ a$ - could someone give the step/s involved
please? Thanks!

$ F(cos\theta \minus{} zsin\theta) \minus{} T \minus{} zyg \equal{} ya$

$ T \minus{} zxg \equal{} xa$

Subtracting the two equations for a and solving for the tension, we obtain

$ T \equal{} \dfrac {x(cos\theta \minus{} zsin\theta)} {x \plus{} y}F$

(I've replaced the physics symbols with x, y and z to simplify the appearance btw.)

p.s. have I posted this in the right forum?

just dividing
and you get
$ xFcos\theta \minus{}xzFsin\theta\minus{}xT\minus{}xyzg\equal{}yT \minus{}yzxg<\equal{}>xF(cos\theta\minus{}zsin\theta)\equal{}(x\plus{}y)T$
$ <\equal{}> T\equal{}\frac{x(cos\theta\minus{}zsin\theta)}{x\plus{}y}F$

Thanks very much for that Athinaios!  :)


[quote="diane"]let (a_n)n>=1 be a sequence (a_n#Cte) such that a_{n+1}/3=1/n(a_1+a_2...a_n) for all  n E N*.let b_n=a_n/n
prove that b_n is arithmetic[/quote]

Is it
"let $(a_{n})_{n \in \mathbb{N}}$ be a sequence such that $\frac{a_{n+1}}{3}=\frac{1}{n(a_{1}+a_{2}+\ldots+a_{n})}$ for all  $n \in \mathbb{N}$. Let $b_{n}=\frac{a_{n}}{n}$
prove that $b_{n}$ is arithmetic"
or
 "let $(a_{n})_{n \in \mathbb{N}}$ be a sequence such that $\frac{a_{n+1}}{3}=\frac{1}{n}(a_{1}+a_{2}+\ldots+a_{n})$ for all  $n \in \mathbb{N}$.Let $b_{n}=\frac{a_{n}}{n}$
prove that $b_{n}$ is arithmetic"?

Sorry, but I have no idea what a_n#Cte means  :maybe:

[quote="Pi-J"]Sorry, but I have no idea what a_n#Cte means  :maybe:[/quote]

I should think it's $a_{n}\neq\mathrm{Constante}$ (French abbreviate $\mathrm{Constante}$ to $\mathrm{C}^\mathrm{te}$)


function
algebra
functional equation
algebra proposed

Find all function f: R -> R for which

       f(x+1) - (f(x))^2 = x for all x from R

I dont have sln of this prb.

Namdung

Are you sure there is no other hypothesis, like continuity or whatever, because there are many solutions of that functional equation :

Let R be the equivalence relation defined on |R by xRy iff x = y + n for som integer n.
Then, you only have to solve the functional equation on each equivalence class separately.
Let x be in [0,1[.  Let f(x) = a = a_0 be any real  \geq  -1. You may find a solution on the class of x by f(x +/- n) = a_(+/-n),
where a_(n+1) = (a_n) 2  + n and a_(-(n+1)) =  \sqrt (a_n + n + 1) for each nonegative integer n.

There are other ways to construct f on a class (using - \sqrt  instead of + \sqrt  for example, with probably some constraints...).

Thus, you will have an uncoutable number of solutions....

Pierre.


If you have only a compass and a straight-edge and you are given a circle, what is the minimum number of straight lines that you will need to draw using the straight edge in order to locate the center of the circle?

I think 4. Draw 2 chords, and their line bisectors, and in the spot where bisectors cut is the center.

4 is the correct answer. Is it always true that the intersection of the line bisectors of the chords is the center?

I think it is(of course, take $ 2$ non-parallel chords).

You can do any ruler and compass construction with a compass alone by inversion, so the answer should be 0. :huh:

[quote="Kyoshiro"]I think we only need 2 straight line.
Drow 1 chord and drow the midperpendicular of the chord ,take the mid-point of the midperpendicular, we get the center of the circle.[/quote]

But the mid-point of the midperpendicular can't be constructed without ruler.

Midpoints can be constructed; here's how if your compass can transfer lengths (i.e. it isn't collapsible, which is not a necessary condition). Suppose you have a segment of length $ 1$. You can construct a hexagonal lattice of points with a bunch of circles, with the result that any integer is constructible. Now construct an isosceles trapezoid with diagonals $ 2,2$, and three adjacent sides $ 1,2,1$ in that order. By Ptolemy's, the last side is $ 3/2$, which you can subtract $ 1$ from to get your midpoint.


probability
complementary counting
expected value

1. A game is played in which the player flips a fair coin. If a head shows on the first
flip the player gets nothing and the game ends. If a tail shows on the first flip,
the payout is 2 dollars and the player continues to flip the coin until a head shows or she has flipped the coin ten times at which time the game ends. The total payout doubles for each tail that shows in succession. What is the probability that a player earns at least 10 dollars? Express your
answer as a common fraction.

I got this using casework and some common sense. Is there a fast way to do this (less than a minute)?

Thanks in Advance

[hide="answer"] 1/16 [/hide]

[quote="mathemagician1729"]1. A game is played in which the player flips a fair coin. If a head shows on the first
flip the player gets nothing and the game ends. If a tail shows on the first flip,
the payout is $ 2$ and the player continues to flip the coin until a head shows or she has flipped the coin ten times at which time the game ends. The total payout doubles for each tail that shows in succession. What is the probability that a player earns at least $ 10$? Express your
answer as a common fraction.[/quote]

Fixed, and...

The player must get at least $ 4$ tails. It is easier to do complementary counting (i.e. count the number of ways to get $ 3$ or fewer tails and then a head), which has probability $ \frac12 \plus{} \frac14 \plus{} \frac18 \plus{} \frac1{16} \equal{} \frac {15}{16}$, so our final answer is $ 1 \minus{} \frac {15}{16} \equal{} \boxed{\frac1{16}}$. (Note that the "...the coin ten times..." part could have any number in the place of the $ 10$ and still give this answer, as long as it is greater than or equal to $ 4$).

[quote] It is easier to do complementary counting [/quote]

um.. Just $ \frac{1}{2}\times{\frac{1}{2}}\times{\frac{1}{2}}\times{\frac{1}{2}}\equal{}\frac{1}{16}$  :huh:

[quote="stevenmeow"][quote="math154"][quote="mathemagician1729"]1. A game is played in which the player flips a fair coin. If a head shows on the first
flip the player gets nothing and the game ends. If a tail shows on the first flip,
the payout is $ \$2$ and the player continues to flip the coin until a head shows or she has flipped the coin ten times at which time the game ends. The total payout doubles for each tail that shows in succession. What is the probability that a player earns at least $ \$10$? Express your
answer as a common fraction.[/quote]

Fixed, and...

The player must get at least $ 4$ tails. It is easier to do complementary counting (i.e. count the number of ways to get $ 3$ or fewer tails and then a head), which has probability $ \frac12 \plus{} \frac14 \plus{} \frac18 \plus{} \frac1{16} \equal{} \frac {15}{16}$, so our final answer is $ 1 \minus{} \frac {15}{16} \equal{} \boxed{\frac1{16}}$. (Note that the "...the coin ten times..." part could have any number in the place of the $ 10$ and still give this answer, as long as it is greater than or equal to $ 4$).[/quote]
really fixed, and i think the first tail gives you 2 dollars, the second 4 dollars, the third 8 dollars, the fourth 26 dollars, and so on
(meaning the answer is 1/8)[/quote]
Did you mean that the fourth one gives 16 dollars? And yes, it's 1/8. I wonder what the expected payoff of this game is.

No, it's not 1/8. You have to get 4 flips to be above 16. Just because you are at 8 after 3 flips doesn't mean you are guaranteed to get 10+--you could get a heads.

[quote="aleph0"][/quote][quote="stevenmeow"][quote="math154"][quote="mathemagician1729"]1. A game is played in which the player flips a fair coin. If a head shows on the first
flip the player gets nothing and the game ends. If a tail shows on the first flip,
the payout is $ \$2$ and the player continues to flip the coin until a head shows or she has flipped the coin ten times at which time the game ends. The total payout doubles for each tail that shows in succession. What is the probability that a player earns at least $ \$10$? Express your
answer as a common fraction.[/quote]

Fixed, and...

The player must get at least $ 4$ tails. It is easier to do complementary counting (i.e. count the number of ways to get $ 3$ or fewer tails and then a head), which has probability $ \frac12 + \frac14 + \frac18 + \frac1{16} = \frac {15}{16}$, so our final answer is $ 1 - \frac {15}{16} = \boxed{\frac1{16}}$. (Note that the "...the coin ten times..." part could have any number in the place of the $ 10$ and still give this answer, as long as it is greater than or equal to $ 4$).[/quote]
really fixed, and i think the first tail gives you 2 dollars, the second 4 dollars, the third 8 dollars, the fourth 26 dollars, and so on
(meaning the answer is 1/8)[/quote][quote="aleph0"]
Did you mean that the fourth one gives 16 dollars? And yes, it's 1/8. I wonder what the expected payoff of this game is.[/quote]

u guys are having a quoting fest

[quote="5849206328x"]the problem statement is that the [b]total[/b] payout doubles with each tail, so if you get three tails, you only have $ \$8$  :maybe:[/quote]

That's correct. Also, the answers say $ \frac1{16}$, so... :wink:

The expected turnout is (definition of expected value)

$ E\equal{}\frac12\cdot0\plus{}\frac14\cdot2\plus{}\frac18\cdot4\plus{}\cdots\plus{}\frac1{1024}\cdot512$

$ \equal{}0\plus{}\frac12\plus{}\frac12\plus{}\cdots\plus{}\frac12$ ($ 10$ numbers in this sum total, including the $ 0$)

$ \equal{}\frac92\implies\boxed{\$4.50}$.


real analysis
calculus
integration
function
trigonometry
real analysis theorems

Hi...please i have some questions

1- Is any function that can be integrable in the sense of Riemann be Lebesgue integrable.( Examples please)

2- If the function is negative or altering sign in the interval of integration, is this function may be Lebesgue.

3- can you give me any unbounded function which Lebesgue integrable.

4- how can i determine whether the function Lebesgue integrable or not (bounded, continuous, monotone, .....)

In order to be Lebesgue integrable, a function must satisfy two properties:

(1) It must be measurable.

(2) It must be [b]absolutely integrable[/b].

And if you're looking for examples, (1) is not what you should dwell on. Every [i]reasonable[/i] way of defining a function - as a formula, piecewise, as a limit or a sup or a max - these all result in measurable functions. You have to be completely unreasonable to define a nonmeasurable function - it won't happen by accident.

So that leaves (2): absolute integrability, which is the same thing - the exact same thing, in some sense - as the absolute convergence of a series. A couple of examples:

$\frac{1}{\sqrt{x}}$ is Lebesgue integrable on $[0,1],$ despite being unbounded (and despite not even being defined at zero - one point is a set of measure zero, so we don't care.)

$\int_{0}^{\infty}\frac{\sin x}{x}\,dx$ converges as an improper integral, but does not converge absolutely. (In series terms, we'd call it conditional convergence.) So that does not exist as a Lebesgue integral. (It's not really a Riemann integral, either - it's an improper integral.)

Some additional points and examples.

Every Riemann integrable function is Lebesgue integrable, and to the same value. Please note that by "Riemann integrable" I am referring only to [i]proper[/i] Riemann integrals of bounded functions over bounded intervals. Improper integrals may be considered limits of Riemann integrals, but they are not themselves Riemann integrals.

Here's a well-known example: Let $f$ be the function on $[0,1]$ such that $f(x)=1$ for $x$ rational and $f(x)=0$ for $x$ irrational. (For short, the "characteristic function of the rationals.")

Riemann attitude towards this function: That's a very bad function, because it is discontinuous everywhere. If we try to integrate it, all the upper sums are $1$ and all the lower sums are $0,$ so it is not integrable.

Lebesgue attitude towards this function: That function is equal to zero almost everywhere; as such it might as well be the zero function. Its integral is the same as the integral of the zero function, namely zero.

In (4), you has how you can tell whether a function is Lebesgue (that is, absolutely) integrable. Let me ask this: how can you tell whether a series is absolutely convergent? The answer to that is by comparison (or by some more automated test that relies on comparison). And that's the answer here: by comparison to integrals that you know, or sums you know. By various inequalities.


Let a sequence: $ a_1\equal{}1,a_2\equal{}a_3\equal{}7,a_{n\plus{}1}\equal{}\frac{a_n^2\plus{}a_{n\minus{}1}^2\minus{}1}{a_{n\minus{}2}}$. Prove that: $ \sqrt{3a_n^2\minus{}3}\in N\ \forall n\geq 1$.

[quote="tdl"]Let a sequence: $ a_1 \equal{} 1,a_2 \equal{} a_3 \equal{} 7,a_{n \plus{} 1} \equal{} \frac {a_n^2 \plus{} a_{n \minus{} 1}^2 \minus{} 1}{a_{n \minus{} 2}}$. Prove that: $ \sqrt {3a_n^2 \minus{} 3}\in N\ \forall n\geq 1$.[/quote]

It's easy to show with induction that $ a_n \equal{} \cosh(b_n)$ where : $ b_1 \equal{} 0,b_2 \equal{} \text{arcosh}(7)$ and $ b_{n \plus{} 2} \equal{} b_{n \plus{} 1} \plus{} b_n$

Let then $ c_n \equal{} \frac {\sinh(b_n)}{\sqrt 3}$, and so $ c_1 \equal{} 0$ and $ c_2 \equal{} 4$

We have the relations :
$ a_n \equal{} a_{n \minus{} 1}a_{n \minus{} 2} \plus{} 3c_{n \minus{} 1}c_{n \minus{} 2}$ (just develop $ \cosh(b_{n \minus{} 1} \plus{} b_{n \minus{} 2})$)
$ c_n \equal{} a_{n \minus{} 1}c_{n \minus{} 2} \plus{} a_{n \minus{} 2}c_{n \minus{} 1}$ (just develop $ \sinh(b_{n \minus{} 1} \plus{} b_{n \minus{} 2})$)

And so $ a_n$ and $ b_n$ are natural numbers.

And so $ \sqrt {3a_n^2 \minus{} 3} \equal{} \sqrt 3\sinh(b_n) \equal{} 3c_n$ is also a natural number (assuming, for $ n \equal{} 1$ that $ 0\in\mathbb N$ :) )


Geometer's Sketchpad

Try googling it up.

It's also not free. You can try Euklides, which has a free trial edition.


When a small number is above and to the right of another number, it's the exponent. What does it mean when it's a small number on the bottom and to the right of another number?

Kind of like how elements are written, like H2O, with the 2 lower than the H and O. That 2 is for 2 hydrogen atoms.
What about when it's numbers instead of elements?

Thanks.

$ a_b$ means $ a$ in base $ b$. For example, $ 1001_2$ would be the same as $ 9$ (the usual base $ 10$ is omitted, like the $ 2$ in a square root).

Or, a small number to the bottom-right (called a subscript) could also indicate different variables that share a letter. For instance, if I want to set two variables representing two different radii, I might choose to write $ r_1$ and $ r_2$. (Note that if we decide, for some reason, to square our measure for the first radii, we write $ r^{2}_{1}$.) An example in context: the formula for the area of an [url=http://en.wikipedia.org/wiki/Annulus_(mathematics)]annulus[/url]is
\[ \pi(r^{2}_{1}\minus{}r^{2}_{2}).
\]


I was fooling around in math class and came up with this probability question, but unfortunately I don't know how to solve it.  Could someone tell me if there is good way to without a [b]lot[/b] of casework?

Bob is taking a 10 question test.  The chance that he gets problem one right is 90%, for problem two is 80%, for problem three is 70%, all the way down to problem 10, which he has a 0% chance of answering correctly.  What is the probability that Bob passes the test, in which a grade 70% or higher is considered passing?

[size=150]
UPDATED SOLUTION.
get ready for algebra bash...really a lot of work required[/size]
let p be the probability that he passes. He passes means he is WRONG on 0, 1, 2 or 3 questions.
Probability that he is WRONG on 0 questions: p0=[u]0[/u]
Probability that he is WRONG on 1 question: p1=[u]1[/u]
Probability that he is WRONG on 2 questions:
if he is WRONG on question one&ten: p(1)=0.1*0.8*0.7*...*0.1*1=__
if he is WRONG on question two&ten: p(2)=0.9*0.2*0.7*...*0.1*1=__
...
if he is WRONG on question nine&ten: p(10)=0.9*0.8*...*0.2*0.9*1=__
Probability that he is WRONG on 3 questions:
if he is WRONG on question one&two&ten: p(1&2)=0.1*0.2*0.7*0.6*...*0.1*1=__
if he is WRONG on question one&three&ten: p(1&3)=0.1*0.8*0.3*0.6*...*0.1*1=__
...
if he is WRONG on question eight&nine&ten: p(8&9)=0.9*0.8*0.7*...*0.3*0.8*0.9*1=__

Now do the finishing touch, add up all those "__" results, and you'll get p.

Hmmm...can anyone think of a shorter method to this?


Hello,

Prove that : $ \forall n \in \mathbb{N}^{\large\ast}$ ,  $ [ (2\plus{}\sqrt 3 )^n]$ is a odd integer

([x] is the integer part of x )

We have that :

$ (2\plus{}\sqrt{3})^n\plus{}(2\minus{}\sqrt{3})^n\equal{}2a_n\ ,\ a_n\in \mathbb{N}$

$ 0<(2\minus{}\sqrt{3})^n<1\Rightarrow [(2\plus{}\sqrt{3})^n]\equal{}[2a_n\minus{}(2\minus{}\sqrt{3})^n]\equal{}2a_n\minus{}1$

[quote="alex2008"]We have that :

$ (2 \plus{} \sqrt {3})^n \plus{} (2 \minus{} \sqrt {3})^n \equal{} 2a_n\ ,\ a_n\in \mathbb{N}$

$ 0 < (2 \minus{} \sqrt {3})^n < 1\Rightarrow [(2 \plus{} \sqrt {3})^n] \equal{} [2a_n \minus{} (2 \minus{} \sqrt {3})^n] \equal{} 2a_n \minus{} 1$[/quote]

nice alex2008  :first:


[quote="Elemennop"]If $\sqrt{\sin x}+\sqrt{\cos x}$ is rational, which of these statements follows immediately from one of Euler's proofs ($0<x<\frac{\pi}{2}$).

A) $x$ is rational
B) $x$ is irrational
C) $\sin x$ is rational
D) $\sin x$ is irrational
E) NOTA

I tried searching on google and mathworld for this proof, but couldn't find anything. Anyone willing to put a name to this proof?[/quote]

[hide]
sqrts are rational  when theyare either perfectsquares or 0. and both terms must be rational because two irrationalt erms can't add up to  a rationalterm unless  those two irrational terms are negatives of each other. so $\fbox{C}$ should be true. [/hide]

See, that's the logic that I used too, but x is between 0 and 90 degrees, so you can't have both terms rational.

The answer is supposedly D.

edit> Hopefully Mathew Crawford knows, since he got a 95/100 on this test when he took it.

[quote="Pakman2012"]Yeah I really don't get how they see that a square root of a irrational plus the square root of a likely irrational will yield a rational.  Anyone care to shed some light, this has been puzzling me too.[/quote]

Try $\sqrt{7 + 4\sqrt{3}}$ + $\sqrt{7 - 4\sqrt{3}}$

The sum of two irrational numbers can be rational (even without being negatives of each other).

In case anyone wonders you can show that the above numbers satisfy a quadratic equation by squaring it, and then deciding which of the two roots it must be in order to decide the value of the expression.

Has anyone figured out how to solve this problem? I have no idea.

Let $\sqrt{\sin(x)} = \frac{s}{t}$, noting that the given bounds on $x$ imply that $0 < s < t$ (i.e., no equality on either end).

Since $\sin^2(x) + \cos^2(x) = 1$, we know that $\sqrt{\cos(x)} = \sqrt[4]{1 - \frac{s^4}{t^4}} = \frac{\sqrt[4]{t^4 - s^4}}{t}$.

Since $\sqrt{\sin(x)}$ is rational and the sum is rational, we also need $\sqrt{\cos(x)}$ to be rational. But $s$ and $t$ are integers, and by Fermat's Last Theorem there are no perfect fourth powers of the form $t^4 - s^4$ with $s \ne 0, t$. Contradiction. So $\sqrt{\sin(x)}$ is irrational.


function
calculus
algebra
polynomial
articles
binomial theorem

$ (1\plus{}x\plus{}x^2\plus{}x^3\plus{}....)^6\equal{}\frac{1}{(1\minus{}x)^6}\equal{} \binom{5}{5} \plus{} \binom{6}{5}x \plus{} \binom{7}{5}x^2\plus{}\binom{8}{5}x^3...$

How do you get from the second step to the third step?

First step: Convert $ \binom 55, \binom 65, \binom 75, \dots$ to $ \binom 50, \binom 61, \binom 72, \dots$. This brings out more the idea of Taylor expansions, since we now have a denominator of $ n!$ in each term $ x^n$. (This serves as the motivation, if you didn't know to use Taylor already.)

Let $ f(n)\equal{}\frac 1{(1\minus{}x)^6}$. When computing $ \frac {f^{(n)}(0)}{n!}$, we are only concerned about the effects of the chain rule on differentiating $ f(n)$ (the denominator will remain 1 when evaluated at 0.) More specifically, the numerator will remain constant because the chain rule will only kick out a $ \minus{}1$, and no $ x$ terms. The denomiator, on the other hand, will have its exponent continuously increase by 1. Therefore, $ f^{(n)}(x)$ will be in the form $ \frac k{(1\minus{}x)^{n\plus{}6}}$.

Let $ u\equal{}1\minus{}x$. Then $ \frac{du}{dx}\equal{}\minus{}1$. Then $ \frac d{dx} \left(\frac 1{u^6}\right) \equal{} \minus{}\frac 6{u^7} \cdot \frac {du}{dx} \equal{} \frac 6{(1\minus{}x)^7}$. So $ f'(0) \equal{} 6$. This will repeat for $ f''(x),f'''(x)$, etc.

$ f(0)\equal{}1$, $ f'(0)\equal{}6$, $ f''(0)\equal{}7 \cdot 6$, $ f'''(0)\equal{}8 \cdot 7 \cdot 6$...

We have a recurrence relation (not exactly, don't know the exact term):
$ f^{(n)}(0) \equal{} (n\plus{}5)f^{(n\minus{}1)}(0), n\geq 1$, with $ f(0)\equal{}1$.

Now, we don't have to deal with calculus anymore. This has just become an ordinary recurrence relation, and its solution is
$ f^{(n)}(0) \equal{} \frac {(n\plus{}5)!}{5!}$, thus $ \frac{f^{(n)}(0)}{n!} \equal{} \frac {(n\plus{}5)!}{5!\,n!} \equal{} \binom {n\plus{}5}5$.

And now... I'll let you finish :)

[b]@ OP:[/b] I don't think that's necessarily the best logic flow to follow unless you are keen on using Taylor series. It is much more elementary to go from the first step directly to the third. You are expanding a product of six polynomials. Let's look at what happens:

You want the sum of all possible ways of picking one of $ x^j$ from each of the six terms. Suppose you pick $ x^{a_1}$ from the first term, $ x^{a_2}$ from the second, etc. Then, this particular selection adds the term $ x^{a_1} \cdot x^{a_2} \cdot \ldots x^{a_6} \equal{} x^{a_1 \plus{} a_2 \plus{} \ldots \plus{} a_6}$ to the final product. Here, $ a_i \in \mathbb{Z} ^{\geq 0}$

To find the coefficient on $ x^n$ in the final product, you want the number of ways you can express $ x^n$ as $ x^{a_1 \plus{} a_2 \plus{} \ldots \plus{} a_6}$. In other words, you want the number of solutions to:

$ a_1 \plus{} a_2 \plus{} a_3 \plus{} a_4 \plus{} a_5 \plus{} a_6 \equal{} n$, where $ a_i \in \mathbb{Z}^{\geq 0}$

From here, try to use the standard balls-and-urns argument to find the number of solutions to this equation in terms of $ n$.

Is there an equivalent to the binomial theorem for larger polynomials e.g.

$ (a_0 \plus{} a_1x \plus{} a_2 x^2...a_nx^n)^r$

There is the [url=http://en.wikipedia.org/wiki/Multinomial_theorem]Multinomial Theorem[/url], which follows from the same logic.

Another way of looking at those terms is as [i]negative[/i] binomial coefficients.  From that viewpoint, this is just a generalization of the binomial theorem, which is indeed proven using Taylor series.  See the discussion [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=184510]here[/url].

What I'd be interested in is a totally intuitive explanation of what a negative binomial coefficient means.

I like the idea of looking at it as negative binomial coefficients.  If $ k \geq 0$, then
$ {n \choose k} \equal{} \frac {n(n \minus{} 1) \dots (n \minus{} k \plus{} 1)}{1 \cdot 2 \cdot 3 \dots k} \equal{} ( \minus{} 1)^k { \minus{} n \plus{} k \minus{} 1 \choose k}$

Then we have, for example, $ \binom{ \minus{} 6}{3} \equal{} \binom{8}{3} \equal{} \binom{8}{5}$.

By the way, the original post was from joml88's article [url=http://www.artofproblemsolving.com/Resources/Papers/LaurendiPartitions.pdf]Partitions of Integers[/url].


Ysing that (1-x)(1-y)(1-z)=<0 we get (*)xy +yz+zx>=x+y+z
Now transforming the inequality and denoting A=xy +yz+zx gives
(A-4)(x+y+z)+3>=0,if a<4 the inequality is obvious if not then using(*) it becomes
A(A-4)+3>=0 or (A-3)(A-1)>=0 which is obvious from AM-GM ineq.

[quote="hardsoul"]Ysing that (1-x)(1-y)(1-z)=<0 [/quote]

why ? when we take x = 8; y = 8 and z = 1/64, then xyz = 1, but
(1-x)(1-y)(1-z) > 0

Oh,u're right.
Then here's another solution:
denote A=x+y+z,B=xy+yz+zx,C=(x+y) 2 /xy+(x+z) 2 /xz+(z+y) 2 /zy.
If B>=A then the inequality can be trnsformed into
the form AB+3>=4A or so it's sufficient to show that
A 2 -4A+3>=0 which factorises as (A-1)(A-3)>=0 what is obvious.
If A>=B then the inequality can be transformed into C>=4A.
To prove this apply Cauchy to ((x+y)/ :sqrt: xy,(x+z)/ :sqrt: xz,(z+y)/ :sqrt: zy) and ( :sqrt: xy, :sqrt: xz, :sqrt: zy) to get BC>=4A 2  or AC>=4A 2 .That we had to prove.

A kind of mixing variables also works here:

Set

  $f\left( x,y,z\right) =\left( y+z\right) \left( z+x\right) \left( x+y\right) -4\left( x+y+z-1\right) $

Then we have to prove that $f\left( x,y,z\right) \geq 0$ for all positive numbers x, y, z satisfying xyz = 1. We will show $f\left( x^{2},y^{2},z^{2}\right) \geq 0$ instead (in fact, once this will be shown, you can put $\sqrt{x}$, $\sqrt{y}$, $\sqrt{z}$ for x, y, z and get the desired inequality).

Since xyz = 1, among the numbers x, y, z there is at least one greater or equal 1. WLOG, we will assume that this number is x, i. e. we have $x\geq 1$. Now

  $f\left( x^{2},y^{2},z^{2}\right) -f\left( x^{2},yz,yz\right) =\left( y-z\right) ^{2}\left( x^{2}\left( y+z\right) ^{2}-4+y^{2}z^{2}+x^{4}\right) $.

But since $\left( y+z\right) ^{2}\geq 4yz$, we have $x^{2}\left( y+z\right) ^{2}\geq x^{2}\cdot 4yz=x\cdot 4xyz=x\cdot 4\geq 4$ (here we have used $x\geq 1$). Therefore, $x^{2}\left( y+z\right) ^{2}-4\geq 0$ and we obtain $f\left( x^{2},y^{2},z^{2}\right) -f\left( x^{2},yz,yz\right) \geq 0$, so that $f\left( x^{2},y^{2},z^{2}\right) \geq f\left( x^{2},yz,yz\right) $.

Now, since xyz = 1, we have yz = 1 / x, so

  $\displaystyle f\left( x^{2},yz,yz\right) =f\left( x^{2},\frac{1}{x},\frac{1}{x}\right) =2\left( x^{4}-x^{2}+1+2x\right) \frac{\left( x-1\right) ^{2}}{x^{3}}$.

Since $x^{4}-x^{2}+1=\left( x^{2}-1\right) ^{2}+x^{2}\geq 0$, we have $f\left( x^{2},yz,yz\right) \geq 0$, and therefore $f\left( x^{2},y^{2},z^{2}\right) \geq f\left( x^{2},yz,yz\right) \geq 0$, and the inequality is proven.

  Darij

$\displaystyle (x+y)(y+z)(z+x) = \sum_{sym}x^2y + 2xyz$

$= (x+y+z)(xy+yz+zx) - xyz = (x+y+z)(xy+yz+zx) - 1$

The inequality $(x+y)(y+z)(z+x) \ge 4(x+y+z -1)$ is equivalent to:

$\displaystyle (xy+yz+zx) + \frac{3}{x+y+z} \geq 4$ --------------- (1)

Let $a=1/x=yz, b=1/y=zx, c=1/z=xy$, then $abc=1$ and (1) is equivalent to:

$\displaystyle (a+b+c) + \frac{3}{ab+bc+ca} \geq 4$ --------------- (2)

the LHS of (2)  $\displaystyle =3\cdot \frac{a+b+c}{3} + \frac{3}{ab+bc+ca} \geq 4 \sqrt[4]{\left(\frac{a+b+c}{3}\right)^3\cdot \frac{3}{ab+bc+ca} }$ by AM-GM.

It suffices to prove
$\displaystyle \left(\frac{a+b+c}{3}\right)^3\cdot \frac{3}{ab+bc+ca} \geq 1$ which is pretty obvious since $(a+b+c)^2\geq 3(ab+bc+ca)$ and $a+b+c\geq 3$.

Q.E.D.

[quote=georg]Prove that if x,y,z are real positive numbers with the property that xyz=1 :
(x+y)(y+z)(z+x)>=4(x+y+z-1).[/quote]
https://artofproblemsolving.com/community/c6h142479p806239


Evaluate $ \frac{7!\plus{}8!}{6!\plus{}7!}$. Express your answer as a simplified fraction.

or you can bash (when I do it it's faster)


Find the smallest natural number $n$ such that the first three digits and the last four digits of $n^3$ are 111 and 1111 repectively.

For $n^3 mod 10$ to equal $1$, $n mod 10$ must equal $1$.  For $n^3 mod 100$ to equal $11$ in addition to this, $n mod 100$ must equal $71$.  For $n^3 mod 1000$ to equal $111$ in addition to this, $n mod 1000$ must equal $471$.  For $n^3 mod 10000$ to equal $1111$ in addition to this, $n mod 10000$ must equal $8471$.

I don't really know where to go from here... I started plugging 8471, 18471, 28471... into my calculator but got tired of it after 208471...

Anyone else have ideas?

[hide] $1038471^3$ works according to my calculator. I don't know if it's the smallest one though. You only have to search numbers when the cube is around $1.11*10^n$. So, my answer is $\boxed{1038471}$ :D [/hide]


integration
calculus
algebra solved
algebra

Have you a very simple solution for this limit:

lim(n--->00)(1^1*2^2*3^3.....*n^n)^(n^(-3))=1

I think to have found one!!

Let P(n) be the product.
Take the logarithm. It gives (the sums are over k = 1 to n) :
Ln(P(n)) = (1/n 3 )*sum k*Ln(k) = (1/n 3 )*sum (k*Ln(k/n) + kLn(n) ) = (1/n 2 )*sum[ (k/n)*Ln(k/n)] + (n+1)*Ln(n)/(2n 2 ).
It is clear that lim (n+1)*Ln(n)/(2n 2 ) = 0.
In another hand, recognizing a Riemann's sum, lim (1/n)*sum[ (k/n)*Ln(k/n)]  =  \int {0,1} x*Ln(x)*dx = -1/4.
Thus lim (1/n 2 )*sum[ (k/n)*Ln(k/n)]  = 0.
Then lim Ln(P(n)) = 0, from which it follows that lim P(n) = 1.

Pierre.

My proof is based on the MVT(Mean Value Theorem).

Using the MVT with f(x)=((x^2)*log(x))/2 -(x^2)/4 we have

 f(k)-f(k-1) =clogc

where k \geq 2  and c E [k-1,k] Hence

(k-1)log(k-1)<f(k)-f(k-1)<klog(k)   or

sum(k=2..n)(k-1)log(k-1)<(n^2log(n))/2 -(n^2)/4+1/4<sum(k=2..n)klogk

and then defining  E=e^((-n^2+1)/4) we have

log(n^(n^2/2))*E<log2^2*3^3*..n^n<log(n^(n^2/2+n))E

But n^(1/2n),n^(1/n^2) and e^(-1/4n+1/4n^3) approach 1 as n-->00

so lim(n-->00) (1^1*2^2*3^3*...*n^n)^(1/n^3) =1

P.S. I think this proof if correct is more simple than yours because
       doesn't use the concept of Riemann's integral

I also have a simple solution:
   First, an observation:
  It is clearly enough to find lim 1/n^3*sum k*lnk and if limit is l, then the requested limit is e^l.
  Use Cesaro-Stolz to obtain that lim 1/n^3*sum k*lnk=lim (n+1)*ln(n+1)/(3n^2+3n+1) and the last one is obviously 0.

Harazi proof is more simple, Cesaro-Stolz is very powerfull theorem
for computing limit of sequence.
Well done Harazi.

I agree with you Moubinool, Harazi proof is the most simple and elegant of all. Congratulation Harazi.


geometry
trigonometry
MATHCOUNTS
ratio
area of a triangle

[img]http://i44.tinypic.com/t5j9t5.jpg[/img]

I tried inscribing the octagon inside another square, and that should work, however, the algebra becomes quite messy.  Is there another way?

[hide="would this work?"] Let the center of the octagon be O. Let OA=OB=OC=r. Use the fact that the area of a triangle is $ \frac {ab\sin{C}}{2}$, and get expressions for the area.[/hide]

You can find the altitude from B to AC quite easily, and you can find AC, so you can find the area of ABC. This instantly solves this.

[quote="JBoyd"]You mentioned putting another octagon inside the square.  What would happen if you put another square outside the octagon?  Would the ratio of the two squares be the same as the ratio of the two octagons?  If so, the squares would be much easier to work with.  I don't know if this will work but it is something to consider.[/quote]

That's what I meant; basically putting a square around the octagon.  The algebra became messy, though, and I double-checked and got the same (wrong) answer.

[quote="policecap"][hide="follow_pythag011_s_way"]let AC=2a
height from B to AC=h=(sqrt(2)-1)*a
==>ratio=sqrt(2)[/hide][/quote]

I believe the answer is actually $ \boxed{\frac{\sqrt2}2}$.

Can someone post the picture?  My computer blocks pictures from this website and therefore I don't see the problem.

[hide="Alternate Solution"]
Consider the diagram with the sides extended as shown. The side lengths can be found from $ 45^\circ\minus{}45^\circ\minus{}90^\circ$ triangle properties and the Pythagorean Theorem. It is now clear that the ratio of areas is $ \frac{s^2(2\plus{}\sqrt2)}{s^2(2\plus{}\sqrt2)\plus{}4\left(\frac12\cdot s\cdot\frac{s\sqrt2}2\right)}\equal{}\frac{2\plus{}\sqrt2}{2(1\plus{}\sqrt2)}\cdot\frac{\sqrt2\minus{}1}{\sqrt2\minus{}1}\equal{}\boxed{\frac{\sqrt2}2}$.
[/hide]

The octagon has the same area as the rectangle.  Computing the ratio of the square's area to the rectangle's area is straightforward.


Find all function $ f: R \rightarrow R$ satisfy $ f(xf(x)\plus{}f(y))\equal{}f(x)^{2}\plus{}y \ \forall x,y \in R$

Let $ x\equal{}0$ then
$ f(f(y))\equal{}y\plus{}f(0)^2$
Consider
$ a\equal{}f(0)^2$
so $ f(f(y))\equal{}y\plus{}a,\Longleftrightarrow f(f(y\minus{}a))\equal{}y$
Let $ y\equal{}0$ then
$ f(xf(x)\plus{}f(0))\equal{}f(x)^2$
Let $ x\to f(x\minus{}a)$
$ f(f(x\minus{}a)f(f(x\minus{}a))\plus{}f(0))\equal{}f(f(x\minus{}a))^2$
$ f(xf(x\minus{}a)\plus{}f(0))\equal{}x^2$
Let $ x\equal{}0$ then $ f(f(0))\equal{}0$ so $ a\equal{}0$
So $ f(x)^2\equal{}x^2$
Imply that $ f(x)\equal{}x,f(x)\equal{}\minus{}x$
It is quite easy to prove that only $ f(x)\equal{}x,f(x)\equal{}\minus{}x$ satisfy condition.

Yes!
$ a\equal{}0$ then $ f(f(x))\equal{}x \ \forall x$ then $ f(x)^{2}\equal{}f(xf(x))\equal{}f(f(x)f(f(x)))\equal{}f(f(x))^{2}\equal{}x^{2}$
Assume $ \exists u \neq v \neq 0$ so that $ f(u)\equal{}\minus{}u ,f(v)\equal{}v$
Let $ (x,y) \to (u,v)$ so $ u^{2}\plus{}v\equal{}f(\minus{}u^{2}\plus{}v)$, then $ u^{2}\plus{}v\equal{}\minus{}u^{2}\plus{}v$ or $ \minus{}(u^{2}\plus{}v)\equal{}\minus{}u^{2}\plus{}v$
It means $ v\equal{}0$ or $ u\equal{}0$.Wrong!So this problem is finished.
\\
Thank you very much,TTsphn!

Can you explain why that:$ f(f(x)f(f(x)))\equal{}f(f(x))^2$ nedo!

we know that $ f(xf(x))\equal{}f(x)^2$ now put $ x\equal{}f(x)$ we get that $ f(f(x)f(f(x)))\equal{}f(f(x))^2\equal{}x^2$


Sandy originally beat Terry by 82 votes in a school election.  In a re-vote, all of Terry's supporters voted for her again, but some of Sandy's supporters switched to Terry.  If Terry beat Sandy by four votes in the re-vote, how many voters switched?

[quote="GameBot"]Sandy originally beat Terry by 82 votes in a school election.  In a re-vote, all of Terry's supporters voted for her again, but some of Sandy's supporters switched to Terry.  If [b]Terry beat Sandy[/b] by four votes in the re-vote, how many voters switched?[/quote]

Not "Sandy beat Terry."

Basically we only care about the net value, so we can just assume Terry got $ 0$ votes in the first vote. Then if $ x$ is the number of people who switched, then we have $ x\equal{}4\plus{}(82\minus{}x)\implies2x\equal{}86\implies x\equal{}\boxed{43}$.


Support
geometry
floor function
perimeter

Using compass and straightedge, how do you divide a circle into 9 equal parts?  Maybe this should be in Getting Started.

[hide]Draw a circle, draw another circle with the same center and radius 1/3 of the original circle. Then draw a radius of the larger circle, then make construct 45 degree angle .... So we have 8 radius, now erase everything that's inside of the smaller circle and we are done.[/hide]

i dont see how that makes 9 [b]equal [/b] parts. maybe a figure to support?

Do you want equal area pieces or congruent pieces?

Call the center of the circle O.  Set your compass to radius r, and draw a circle of radius r centered at O.  Pick a point on circle O called P1.  From P1, draw an arc of radius r that intersects circle O.  Label this intersection P2.  Repeat to create P3, P4, P5, and P6.  The six points Pn are spaced evenly around the circle.  Trisect the angles P10P3, P3OP5, and P5OP1 to make 9 equal divisions of the circle.  Connect these divisions to O to divide the circle.

(Note: I don't remember how to trisect an angle with a compass and straightedge, but I think it can be done.)

Actually it has been proven that the trisection of an arbitrary angle is [b]impossible[/b] :D  .

I agree with Feiqi's solution.  By drawing the circle with 1/3 the radius of the bigger he made a smaller circle with 1/9 the area of the big circle.  Then by drawing radii 45 degrees apart you divide the original circle into 8 equal area portions.  Here's a pic:

[img]http://img.photobucket.com/albums/v213/joml88/circle-equal9.jpg[/img]

Is there a way to draw a circle 1/3 the size of the original without a ruler?  I know you have a straightedge, but it doesn't allow you to measure anything.

An excellent question!  The general proposition: given an arbitrary length, what multiples of that length are constructible?  The slightly less general version that's relevant here: given an arbitrary segment, is it possible to construct a segment of any rational multiple of its length?  (Positive integer multiples, of course, are easy -- but can you get fractional multiples?)

Given a segment $\overline{AB}$ of length $a$. To do: construct a segment of length $a\cdot\frac{m}{n}$.
[hide]First construct a segment $\overline{AC}$ of length $a\cdot n$. Connect $B$ to $C$. Mark off the point $D$ on $\overline{AC}$, such that $AD=a\cdot m$. Construct the line through $D$ parallel to $\overline{BC}$ that intersects $\overline{AB}$ at $E$. $\frac{AE}{AB}=\frac{m}{n}$, by similar triangles.
For $m>n$, tack on the $\displaystyle\lfloor{\frac{m}{n}\rfloor}$ after constructing the fractional part.
Square root lengths can be constructed with right triangles.

I think those are all.[/hide]

Right on for the rational part.  For the more general question, you're close -- it turns out the answer is slightly broader than that.  [hide]Some obvious numbers you missed are finite sums or differences of square roots.  Also, it turns out it's possible to construct the square root of an arbitrary length (given a segment of length 1, of course) -- if Joe Stern is at the NYSML training, we should ask him how to do it (I've seen him do it, but I don't remember how :blush: ), that is if no one here happens to know.[/hide]

You can do it using the concept of powers of a circle.
[hide]
on the attachment a * 1 = x^2[/hide][/hide]

Let $A_0A_n = 2r$ be the circle horizontal diameter. Using the known construction, divide the segment $A_0A_n$ into n equal parts $A_0A_1 = A_1A_2 = ... = A_{n-1}A_n = \frac{A_0A_n}{n} = \frac{2r}{n}$.
$(1)$ Construct the lower semicircle on the diameter $A_0A_1$ and the upper semicircle on the diameter $A_1A_n$.
$(2)$ Construct the lower semicircle on the diameter $A_0A_2$ and the upper semicircle on the diameter $A_2A_n \ $.
$(3)$ Construct the lower semicircle on the diameter $A_0A_3$ and the upper semicircle on the diameter $A_3A_n$.
...
$(n-1)$ Construct the lower semicircle on the diameter $A_0A_{n-1}$ and the upper semicircle on the diameter $A_{n-1}A_n$.

This construction divides the circle into n non-congruent regions. The perimeters of the 1st, 2nd, k-th and n-th regions are:

$P_1 = \pi r + \frac 1 n \cdot \pi r + \frac{n-1}{n} \frac r n \pi r = 2 \pi r$

$P_2 = \frac 1 n \cdot \pi r + \frac{n-1}{n} \cdot \pi r + \frac{2}{n} \cdot \pi r + \frac{n-2}{n} \cdot \pi r  = 2 \pi r$
...

$P_k = \frac{k-1} n \cdot \pi r + \frac{n-(k+1)}{n} \cdot \pi r + \frac k n \cdot \pi r + \frac{n-k}{n} \pi r = 2 \pi r$
...

$P_n = \frac{n-1} n \cdot \pi r + \frac 1 n \cdot \pi r + \pi r = 2 \pi r$

Thus all n regions have the same perimeter $2 \pi r$. The areas  of the 1st, 2nd, k-th and n-th regions are:

$A_1 = \left(\frac 1 n\right)^2 \cdot \frac{\pi r^2}{2} + \frac{\pi r^2}{2} - \left(\frac{n-1}{n}\right)^2 \cdot \frac{\pi r^2}{2} =$

$= \frac{1 + n^2 - (n-1)^2}{n^2} \cdot \frac{\pi r^2}{2} = \frac{2n}{n^2} \cdot \frac{\pi r^2}{2} = \frac{\pi r^2}{n}$

$A_2 = \left(\frac 2 n\right)^2 \cdot \frac{\pi r^2}{2} - \left(\frac 1 n\right)^2 \cdot \frac{\pi r^2}{2} + \left(\frac{n-1}{n}\right)^2 \cdot \frac{\pi r^2}{2} - \left(\frac{n-2}{n}\right)^2 \cdot \frac{\pi r^2}{2} =$

$= \frac{4 - 1 + (n-1)^2 - (n-2)^2}{n^2} \cdot \frac{\pi r^2}{2} = \frac{4n - 2n}{n^2} \cdot \frac{\pi r^2}{2} = \frac{\pi r^2}{n}$
...

$A_k = \left(\frac k n\right)^2 \cdot \frac{\pi r^2}{2} - \left(\frac{k-1}{n}\right)^2 \cdot \frac{\pi r^2}{2} + \left(\frac{n-(k-1)}{n}\right)^2 \cdot \frac{\pi r^2}{2} - \left(\frac{n-k}{n}\right)^2 \cdot \frac{\pi r^2}{2} =$

$= \frac{k^2 - (k-1)^2 + [n-(k-1)]^2 - (n-k)^2}{n^2} \cdot \frac{\pi r^2}{2} = \frac{2kn - 2(k-1)n}{n^2} \cdot \frac{\pi r^2}{2} =$

$= \frac{\pi r^2}{n}$
...

$A_n = \frac{\pi r^2}{2} - \left(\frac{n-1}{n}\right)^2 \cdot \frac{\pi r^2}{2} + \left(\frac 1 n\right)^2 \cdot \frac{\pi r^2}{2} =$

$= \frac{n^2 - (n-1)^2 + 1}{n^2} \cdot \frac{\pi r^2}{2} = \frac{2n}{n^2} \cdot \frac{\pi r^2}{2} = \frac{\pi r^2}{n}$

Thus all n regions have the same area $\frac{\pi r^2}{n}$. However, they are not congruent.


[quote="bpms"][quote="sonny"]Each day, two out of three teams in a class are randomly selected to participate in a MATHCOUNTS trial competition. What is the probability that Team A is selected on at least two of the next three days? Express your answer as a common fraction.[/quote]
[hide]
All 3 days $\frac{8}{27}$
2 days $\frac{12}{27}$
$\frac{20}{27}$[/hide][/quote]

Correct!! : :)  :)

[quote="dolphin"][quote="bpms"][hide]
All 3 days $\frac{8}{27}$
2 days $\frac{12}{27}$
$\frac{20}{27}$[/hide][/quote]
Why is it $\frac{12}{27}$ for 2 days?[/quote]

The probablity that they are [u]only[/u] picked for three days is 2/3*2/3*1/3=4/27. There are three ways they can be picked for two days thus 4/27*3=12/27


search
Support
MATHCOUNTS
geometry
geometric transformation
linear algebra
matrix

official romanization matlab? main kuchh samjha nehi!  :maybe:

[quote="akalra1"]i cant read hindi but i can understand it[/quote]

Would that be reading it in text as Gyan shpws or reading it in english form as bubka types or can u only understand it speaking?

I go as far up sadly as bubka types in english form.

* :o What? What's this fireforce knows hindi? Shock and apall  :o *

[quote="Gyan"][quote="fredbel6"][quote="bubka"]namste! main soch rah tha ki bhai - ek to bhasha hoga jo main janta hoo! cholo ho gaya![/quote]
Hindi has an official romanization like Japanese? :maybe:[/quote]

[size=150]नही॑[/size]
But that does not stop me here, and romanize and write the above as "nahin"
(which means "no"  :) )
Any way  who gets the answer to this riddle?

[size=150]तो हुआ ऐसा कि नारद जी एक बार अपनी ७ इन्च लम्बी सोने की छड़ी लेकर धरती पर भ्रमण को आए। अब उन्हें सारे चिट्ठों की जानकारी वाले सजाल बनाने में सप्ताह भर का समय लगना था। तो वे एक होटल में एक कक्ष लेने का विचार बना कर वहाँ पहुँच गए। अब इनके पास कोई धनराशि तो थी नहीं , पर सात दिन के किराए के रूप में होटल के मैनेजर ने नारद जी से ७ (सात) इन्च लम्बी सोने की छड़ी लेना स्वीकार कर लिया। पर मैनेजर ने हर रोज का किराया उसी दिन लेना चाहा। तो इस बार की पहेली यह है कि नारद जी को कम से कम छड़ी के कितने टुकड़े करने होंगे, ताकि वो,
होटल का किराया चुका सकें? और
उन टुकड़ों की लम्बाई क्या होगी?[/size][/quote]

I can read that, however I suck at typing in romanized hindi, and there were some words I couldn't understand as well...
How do you type in hindi, gyan?

न ि क ऐ हु त स ा ो आ न द र

Even I can type Hindi.. I just have no idea what I'm typing means

It is easier, if you have some nice  word processing package  (Say word in Hindi), or memory resident program (Google search  would give a list of many  free programs) to convert your keystrokes which insert appropriate characters.

But if you just want to type something in a hurry, (assuming you have windows XP, and enabled hindi language support etc.. which BTW comes standard) just use   say open MS word:

Got to menu  "insert"
choose "symbol"
Use font "Arial Unicode MS"   --- Now you have all Indian languages  (Subset "Devnagri" - for Hindi script)
Just double click and insert  the  characters you want.

(On some windows  computers   Arial  Unicode font - which used to be loaded by default - may not be there, if so you may have to load that font from MS website (or any other place)

Another good place, if you don't have time/money  to get all the packages .. use the online version, say one at:
http://www.aczoom.com/itrans/online/

[quote="bubka"]han tum hindi me kaise type karte ho gyan bhaiia?[/quote]

isnt gyan a girl?

No I am father of two sons Both are in college.

Hal chal to bahut kharap hain bahenji. Ya matrix-watrix to mujhe ate hi nahin hain, aur ye linear algebra class itne tez chal rahe hain ...  :(

lol.

mera alg2/trig teacher bahut kharab hai.  har samhe hot dog aur beer aur sitaron aur pata nahi aur kis kis cheezon ke baray may bat karta hai.

duniya ke sabhi math teacher paagal hote hain!
mere math teacher kehti hai ki PI ka value hai...kuch 3 se...use yaad nahi!! aur phir woh kehti hai..."jaaon tuitions le lo! meri kya zaroorat!!" aur phir...RO PADTI HAI!!!

Hey star99-mujhe tumhara signature BAHUT BAHUT achcha laga! way to go!!  :D  :lol:  :rotfl:  :)

star 99 ka signature ek-dam ghatia hain  :mad:

thanks, padmavathi

bubka, tum ko sumujh mein nahi aye ga.  Hum Tum dehki hai?  woh scene joh tha, jisme Saif Ali Khan gaata hai, ulta chalta hai, aur gir parta hai.  bilkul typical guy.

saray math teacher paagal nahi hain...meri algebra aur geometry teachers bahut bahut acchi thi.  meri geometry teacher indian bhi thi!

HIJRA!

HIJRA!

I'm going to call one of my Indian friends that tommorow and ask him what it means.

HIJRA!
HIJRA!
HIJRA!
HIJRA!

Thanks for teaching me it AoPS

HIJRA is supposed to be a derogatory name for a homosexual. But in our country's context, it usaually refers to groups of homosexuals with female appearance and male voice who go around people's houses collecting money from a house if it has had a baby recently, or someone got married etc. It is seen as a very bad thing in our country. But outside it may not have a meaning.

@star99: main HumTum nahin dekha, lekin mere doston ne dekhen hain. Unhe pata hoga. Mujhe Hindi Bollywood cinema ke bare me bahut kam gyan hain. Lekin uske saath tumhare sig ka kya connection?  :maybe:

not directly, bas yeh ki tum ko achaa nahi laga, to is vajaisay.  hindi films toh dekhni chaiye.  koi koi bahut acchi hain.

Woh baat nahin nai. Ye gaana suna hai kya?

Sochti hain zaada,
Kam samajhti hain,
Dil kuchh kehta hain,
Kuchh aur bhi karti hain,
Ladkin kiyun, na jaane kyun,
Ladkon si nahin hoti,
Ladkin kiyun, na jaane kyun,
Ladkon si nahin hoti.

 :rotfl:  :rotfl:

sunay ko nahi bola tha.  dekhne ko bola tha. ha, sahi gana hai.  film acchi hai

gaane ka 2nd verse dekha/suna??(jo rani gaa rahi hai):lol:
movie bhi dekhiye
bole toh kal Baapu ka birthday tha... hum sabhi ko Lage Raho Munnabhai dekhna chahiye tha aur GANDHIGIRI aazmana chahiye. Badiya hai!!
Maine apne class ko kaha ki meri dushman ko chemical locha hai...uske ghar mein 55 GET WELL SOON cards aur phool mile! Kitna mazaa aya! Usne khud sorry bola!
Yeh movie Oscar mein bhi ja raha hai!! Rang De basanti ke saath jo bhi bahut hi accha hai. They both rock!! :lol:  :D
KANK dekhkar main to so gai...kitna kharaab hai! sorry agar kisi ko pasand hai... :D lekin SRK ne khud kaha hai ki usey KANK acha nahi laga!
Aap sab ko dekhoongi (cya)!!
Baad mein!(Later!!) !!  :rotfl:


Let $ x,y,z$ be nonnegative and $ x\plus{}y\plus{}z\equal{}1$. Prove that

$ xy\plus{}yz\plus{}zx\geq 8(x^2\plus{}y^2\plus{}z^2)(x^2y^2\plus{}y^2z^2\plus{}z^2x^2)$

[quote="outback"]Let $ x,y,z$ be nonnegative and $ x \plus{} y \plus{} z \equal{} 1$. Prove that

$ xy \plus{} yz \plus{} zx\geq 8(x^2 \plus{} y^2 \plus{} z^2)(x^2y^2 \plus{} y^2z^2 \plus{} z^2x^2)$[/quote]
A problem of hungkhtn.The hint is using mixing variables

[quote="outback"]Let $ x,y,z$ be nonnegative and $ x \plus{} y \plus{} z \equal{} 1$. Prove that

$ xy \plus{} yz \plus{} zx\geq 8(x^2 \plus{} y^2 \plus{} z^2)(x^2y^2 \plus{} y^2z^2 \plus{} z^2x^2)$[/quote]

Let $ a\plus{}b\plus{}c\equal{}p\equal{}1;ab\plus{}bc\plus{}ca\equal{}q;abc\equal{}r$ . We have:

$ Ineq \leftrightarrow q \ge\ 8(p^2\minus{}2q)(q^2\minus{}2pr)$

$ \leftrightarrow 16q^3 \minus{}8q^2 \plus{}q \plus{} 16r(1\minus{}2q) \ge\ 0$

And we have $ r \ge\ 0$. Need prove:

$ 16q^3 \minus{}8q^2 \plus{}q  \ge\ 0$

$ \leftrightarrow (4q\minus{}1)^2 \ge\ 0$

Equation hold on $ a\equal{}b\equal{}\frac{1}{2},c\equal{}0$

We have done!  :)


Who can show me the formula calculate area of Morley triangle of $ABC$ follow triangle $ABC$?

Yes they can, see here [url=http://mathworld.wolfram.com/FirstMorleyTriangle.html]Morley triangle[/url]


function
advanced fields
advanced fields unsolved

Is there any function (if any) f: Z -> Z such that
f(f(n))=-n , for every n belongs to Z(integers) ??

I think that there is not any function like the one described above but how can we prove it. Any ideas??
Thanks in Advance

Squaring a four-cycle gives you two transpositions.  This suggests the following:

Set $ f(0) \equal{} 0$, and for $ n > 0$ set $ f(2n \minus{} 1) \equal{} 2n$, $ f(2n) \equal{} 1 \minus{} 2n$, $ f(1 \minus{} 2n) \equal{} \minus{} 2n$ and $ f( \minus{} 2n) \equal{} 2n \minus{} 1$.

A similar construction works for any involution on a set $ S$ that has an infinite number of non-fixed points.  (The preceding sentence is not the strongest possible.)

for any set of int paires $ U \equal{} \left\{(a_{2i},a_{2i \plus{} 1})|a_i\in N\right\}$  $ \left\{\cdots a_i\cdots \right\} \equal{} N$, let $ f(a_{2i}) \equal{} a_{2i \plus{} 1}$,$ f(a_{2i \plus{} 1}) \equal{} \minus{} a_{2i}$, $ f( \minus{} a_{2i}) \equal{} \minus{} a_{2i \plus{} 1}$, $ f( \minus{} a_{2i \plus{} 1}) \equal{} a_{2i}$ and $ f(0) \equal{} 0$.


Simplify $ \sqrt{128} \plus{} \sqrt{72}$


Determine $ f: (0,\infty)\rightarrow (0,\infty)$ so that $ f(1\plus{}xf(y))\equal{}yf(x\plus{}y)$ where x,y>0.

[quote="DocEmBr"]Determine $ f: (0,\infty)\rightarrow (0,\infty)$ so that $ f(1 \plus{} xf(y)) \equal{} yf(x \plus{} y)$ where x,y>0.[/quote]

$ f(1 \plus{} xf(y)) \equal{} yf(x \plus{} y) (*)$

If $ y > 1$ and $ f(y) > 1$, put $ x \equal{} \frac {y \minus{} 1}{f(y) \minus{} 1}$ in $ (*)$, we get contradiction.

So $ f(x) \leq 1$ for all $ x > 1$. By $ (*)$ $ yf(x \plus{} y) \leq 1$.

Put $ x \equal{} \varepsilon$ and $ y \equal{} x \minus{} \varepsilon$, we get $ xf(x) \leq 1 (**)$.

Put $ x \equal{} \frac {1}{f(y)}$ in $ (*)$, we get $ \frac {f(2)}{y} \equal{} f(\frac {1}{f(y)} \plus{} y) (***)$.

By (**), $ \frac {1}{f(y)} \plus{} y \leq \frac {y}{f(2)}$, so there is a positive constant $ c$ such that $ xf(x) \geq c$ for all $ x$. $ (****)$

By $ (***)$, $ f$ is surjective. (*****) So for all real $ a$, there is $ b$ such that $ f(a)^2 \equal{} f(b)$.

Put $ y \equal{} a$ in $ (*)$, we get $ f(f(a)x \plus{} 1) \equal{} af(x \plus{} a)$.

So if $ f(a)x > 1$,

$ bf(x \plus{} b) \equal{} f(f(b)x \plus{} 1) \equal{} f(f(a)^2x \plus{} 1) \equal{} af(f(a)x \plus{} a)$

$ \equal{} af((f(a)x \plus{} a \minus{} 1) \plus{} 1) \equal{} a^2f(\frac {f(a)x \plus{} a \minus{} 1}{f(a)} \plus{} a)$. (******)

By (******) and (**) and (****), $ b \equal{} a^2$ and $ x \plus{} b \equal{} \frac {f(a)x \plus{} a \minus{} 1}{f(a)} \plus{} a$. So we get $ f(a) \equal{} \frac {1}{a}$.

It is easy to check that $ f(x) \equal{} \frac {1}{x}$ satisfies the condition.

Where is this problem from?

[quote="ringos"]By (******) and (**) and (****), $ b \equal{} a^2$ and $ x \plus{} b \equal{} \frac {f(a)x \plus{} a \minus{} 1}{f(a)} \plus{} a$. So we get $ f(a) \equal{} \frac {1}{a}$.[/quote]
I'm sorry, but that's complete nonsense.
However I managed to complete the proof,
based on your finding $ c\le xf(x) \le 1$ for all $ x > 0$.
(Very nice  :) , though you missed to show the obvious $ c \equal{} {1\over {1\over f(2)} \minus{} 1} > 0$)

$ (*)$ $ f(1 \plus{} xf(y)) \equal{} yf(x \plus{} y)$
Take $ c: \equal{} \inf_{x\in\mathbb{R}_ \plus{} }xf(x) > 0$.

Then:
[b]1)[/b] for all $ \epsilon > 0$ we can find $ t > 1$ with $ c\le tf(t) < c \plus{} \epsilon$:
By definition of $ c$ we find $ z > 0$ with $ zf(z) < c \plus{} \epsilon$.
Then for $ x \equal{} y \equal{} {z\over 2}$ in $ (*)$ we get
$ f(1 \plus{} {z\over 2}f({z\over 2})) \equal{} {z\over 2}f(z) < {c \plus{} \epsilon\over 2}$
and then for $ t \equal{} 1 \plus{} {z\over 2}f({z\over 2}) > 1$ we get
$ tf(t) < {1 \plus{} {z\over 2}f({z\over 2})\over 2}(c \plus{} \epsilon)\le {1 \plus{} 1\over 2}(c \plus{} \epsilon) \equal{} c \plus{} \epsilon$.

[b]2)[/b] $ f(y) \equal{} {1\over y}$ for all $ 0 < y < 1$:
Take $ \epsilon > 0$. Then by 1) we find $ t > 1$ with $ tf(t) < c \plus{} \epsilon$.
Then for $ x \equal{} t \minus{} y > 0$ in $ (*)$ we get
$ {c\over 1 \plus{} (t \minus{} y)f(y)}\le f(1 \plus{} (t \minus{} y)f(y)) \equal{} yf(t) < y{c \plus{} \epsilon\over t}$.
Then
$ 1 \plus{} (t \minus{} y)f(y) > {t\over y}\cdot {c\over c \plus{} \epsilon} \equal{} 1 \minus{} {\epsilon\over c \plus{} \epsilon} \plus{} {t \minus{} y\over y}\cdot {c\over c \plus{} \epsilon}$
and therefore
$ f(y) > \minus{} {1\over t \minus{} y}\cdot {\epsilon\over c \plus{} \epsilon} \plus{} {1\over y}\cdot {c\over c \plus{} \epsilon} > \minus{} {1\over 1 \minus{} y}\cdot {\epsilon\over c \plus{} \epsilon} \plus{} {1\over y}\cdot {c\over c \plus{} \epsilon}$
for all $ 0 < y < 1$ and $ \epsilon > 0$.
So $ \epsilon \to 0$ gives $ f(y) \ge {1\over y}$ for all $ 0 < y < 1$, which proves 2).

And finally we get
[b]3)[/b] $ f(x) \equal{} {1\over x}$ for all $ x\in\mathbb{R}_ \plus{}$:
$ x \plus{} y < 1$ in $ (*)$ gives
$ f(1 \plus{} {x\over y}) \equal{} {y\over x \plus{} y} \equal{} {1\over 1 \plus{} {x\over y}}$ using 2).
But for all $ z > 1$ we can take $ 0 < y < {1\over z}$ arbitrary and $ x \equal{} y(z \minus{} 1) > 0$.
Then $ x \plus{} y \equal{} yz < 1$ and $ z \equal{} 1 \plus{} {x\over y}$, which gives $ f(z) \equal{} {1\over z}$ for all $ z > 1$.
This leaves to show $ f(1) \equal{} 1$.
But $ x \equal{} y \equal{} 1$ in $ (*)$ now gives $ {1\over 1 \plus{} f(1)} \equal{} f(1 \plus{} f(1)) \equal{} f(2) \equal{} {1\over 2}$, as $ 1 \plus{} f(1) > 1$.
This completes the proof.

[quote="olorin"][quote="ringos"]By (******) and (**) and (****), $ b \equal{} a^2$ and $ x \plus{} b \equal{} \frac {f(a)x \plus{} a \minus{} 1}{f(a)} \plus{} a$. So we get $ f(a) \equal{} \frac {1}{a}$.[/quote]
I'm sorry, but that's complete nonsense.
[/quote]

I think it's true.

(******) says that there are $ c,d,e$ such that $ f(x\plus{}c) \equal{} df(x)$ for all $ x \geq e$.

So $ f(x\plus{}nc) \equal{} d^nf(x)$, but by (**) and (****), $ d$ must be 1.

D***, you got me. :blush:
I'm really really sorry. I just couldn't see that final argument.
Maybe you could work out your proof a little more for everyone.
(don't forget the case $ a\equal{}1$)
Again, I'm sorry I doubted you.


Se dau 2 focare (F1 si F2) si un segment de lungime L.Sa se traseze (fara a folosi masuratori) locul geometric al punctelor P din plan a.i. PF1 - PF2 = L sau PF2 - PF1 = L.
Any ideas??? :blush:

Could anyone translate this problem into English?


algorithm
number theory
Euclidean algorithm

Find x, y which are integer numbers such that 1155x is congruent to 442 mod16, cwith 1 < x < 441 e 442y is congruent to1155 mod35 with 1 < y < 1155

[hide]$1155x \equiv 442 \bmod 16 \Longleftrightarrow 3x \equiv 10 \bmod 16$. We want to find solutions of $3x-16y = 10$. Since $10$ is a multiple of $(3, 16)$ we can say that there are infinite solutions satisfy $x = x^{*}+\frac{16r}{(3, 16)}$ where $x^{*}$ is the particular solution of $3x-16y = 10$. By the extended Euclidean Algorithm we can find $x^{*}=-50$. Then it's easy to find $1 <-50+16r < 441$.[/hide]


abstract algebra
topology
advanced fields
advanced fields unsolved

Hi.

Let $G$ be a topological group with neutral element $e$. Prove that the fundamental group $\pi_1(G,e)$ is abelian.

Hanno

besides the composition $ *$ of paths, $ \pi_1(G,e)$ carries another group structure $ .$ given by pointwise multiplication in $ G$. to see that this is well-defined, we just have to multiply homotopies. now for $ f,g \in \pi_1(G,e)$, we have $ f*g \equal{} (f*e).(e*g) \equal{} f.g \equal{} (e*g).(f*e)\equal{}g*f$, so $ \pi_1(G,e)$ is abelian and the two group structures actually coincide.


[quote="arqady"][quote="chien than"]Let $a;b;c;d>0$.Prove that:
$(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d})(\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}+\frac{1}{d+1}) \geq \frac{16}{1+abcd}$[/quote]
It's wrong! Try $a=b=c=d=0.9.$[/quote]
However, the following is true:
\[\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+d}+\frac{1}{d+a}\right) \geq \frac{16}{1+abcd}.\]
Maybe this is what you meant. :)
[Note: this is the more general form of the well-known three variable inequality $\sum \frac 1a\sum \frac 1{a+b}\geq \frac 9{1+abc}.$]
My proof to the four variables problem uses only AM-GM.(with a little calculation) :)
I will post it here if someone wants me to. Thanks for this nice 4-variables inequality, chien than. :)

[quote="Potla"]
However, the following is true:
\[\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+d}+\frac{1}{d+a}\right) \geq \frac{16}{1+abcd}.\]
[/quote]
Certainly! After $\frac{16}{1+abcd}\leq\frac{8}{\sqrt{abcd}}$ we can use known Can's inequality. :wink:

[quote="can_hang2007"][quote="chien than"]Let $a;b;c;d>0$.Prove that:
$(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d})(\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}+\frac{1}{d+1}) \geq \frac{16}{1+abcd}$[/quote]
This is wrong. But the following inequality is true and nice
$\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)\left(\frac{1}{1+a^2}+\frac{1}{1+b^2}+\frac{1}{1+c^2}+\frac{1}{1+d^2}\right) \geq \frac{16}{1+abcd}.$[/quote]
And the following inequality is also true and nice
$\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+d}+\frac{1}{d+a}\right)\ge \frac{16}{abcd+1}.$

[quote="arqady"][quote="CSS-MU"]
And the following inequality is also true and nice
$\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+d}+\frac{1}{d+a}\right)\ge \frac{16}{abcd+1}.$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1929102#p1929102[/quote]
Sorry. Thanks.


The same problem is posted in a link on this forum titled "mod".

Hello everyone !!

What does UMO - means, thanks anyway.

Davron


[quote="Lovasz"]Let $a,b,c>0$ and $a^2+b^2+c^2 \ge 4\sqrt{abc}$, prove that:
\[ a+b+c \ge 2\sqrt{abc}  \][/quote]
Lovasz, this is not true. Try $a=b=c=4.$ ;)

[quote="Meghrabi"]This is easy:

a²+b²+c²+a+b+c >= 6 sqrt(abc) = 6 (a^3b^3c^3)^(1/6)[/quote]

And that's wrong because $A + B \geq C + D$ and $A \geq C$ doesn't imply that $B \geq D$.


calculus
derivative
function
calculus computations

I think the function you're looking for is $ f(n)\equal{}\left(\frac{(2n)!}{2^nn!}\right)^2\equal{}\left(\frac{(2n\minus{}1)!}{2^{n\minus{}1}(n\minus{}1)!}\right)^2$. In words, it's the square of the product of the first $ n$ odd integers.

Very nice Diarmuid . . . guess I don't have to work on it anymore.  You're sharper than me. :)

Or even simplier $ f(n) \equal{} (2n\minus{}1)!!^2$.

on the question: how to find it ? answer is next:
there is no difference between your and this finction $ g(z): \equal{} \frac {1}{\sqrt {(1 \minus{} z)}}$ (for example say $ z \equal{} x^{2}$ and we have your function)
thus we need to find taylor series expansion for $ g(z)$ and this last is simple example for taylor sereis
$ g(z) \equal{} \sum_{k} \binom{ \minus{} \frac {1}{2}}{k}z^{k}$

. . . wonderful . . . I didn't see that . . . sorry for making a baby Calculus problem unnecessarily superfluous. :oops:

Thanks for helping me guys.


$O$ is inside triangle ABC. And we have $\vec{OA}+2\vec{OB}+3\vec{OC}=0$.
Find $\frac{\triangle{ABC}}{\triangle{AOC}}$.

[quote="probability1.01"]If you just consider y components, let the ratio be r, and WLOG altitude of ABC to AC is 1.  Then you have $r + 3r - (2-2r) = 0 \implies r = \frac{1}{3}$[/quote]
Can you explain it??

[quote="probability1.01"]Let AC be horizontal.  If the ratio is r, and you use the other stuff I said, you have that the y components of OA and OC being each -r (I think I may have reversed them all in my solution though...).  Similar thing for OB.[/quote]
Still don't get your meaning...

The answer is $\frac{\triangle ABC}{\triangle AOC}=3$

Oh yeah, I accidentally calculated the inverse of that if it helps shobber...

[quote="shobber"]Can you please show your work?? I am completely lost... :([/quote]
What probability1.01 means is just to consider the imaginary parts of A,B,C on the complex plane with O being the origin and probably the real axis parallel to AC.

$\overrightarrow{OA}+2\overrightarrow{OB}+3\overrightarrow{OC}=\overrightarrow 0\Longleftrightarrow 3\cdot \frac{\overrightarrow{OA}+2\overrightarrow{OB}}{2+1}+3\overrightarrow{OC}=\overrightarrow 0$

Let $\overrightarrow{OD}=\frac{\overrightarrow{OA}+2\overrightarrow{OB}}{2+1}$, which shows point $D$ is the point such that $AD:DB=2:1$,

so  we have $3\overrightarrow{OD}+3\overrightarrow{OC}=\overrightarrow 0$,

that is to say, point $O$ is the mid point of $DC$.


Find all continuous function $ f: R \to R$ satisfy:
1)$ f(1)\equal{}1$
2)$ f(f(x))\equal{}(f(x))^2$
3)$ f$ is either decreasing or increasing.

Let $ A\equal{}Im(f)$. Obviously $ f(x)\equal{}1$ is one solution when $ A\equal{}\{1\}$.

Suppose $ a\in A$ with $ a<0$. Then $ [a,1]\subset A$. We can find $ b\in(0,1)$ such that $ \minus{}b,b\in A$. Then $ f(\minus{}b)\equal{}b^2\equal{}f(b)>0$ and $ f(0)\equal{}0$. A contradiction with monotonicity.

Equation 2) implies $ a^2\in A$ if $ a\in A$. Hence, if there exists $ a\in A$ with $ a<1$, we have $ (0,1]\subset A$. Also $ f(x)\equal{}x^2$ whenever $ x\in(0,1)$. By continuity, $ f(0)\equal{}0$, i.e., $ [0,1]\subset A$. Similarly if $ a\in A$ with $ a.1$, $ [1,\infty)\subset A$. Now we only have three cases:

i) $ A\equal{}[0,1]$.
$ f(x)\equal{}0$ when $ x\le0$; $ f(x)\equal{}x^2$ when $ x\in[0,1]$; $ f(x)\equal{}1$ when $ x\ge1$.

ii) $ A\equal{}[1,\infty)$.
$ f(x)\equal{}1$ when $ x\le1$; $ f(x)\equal{}x^2$ when $ x\ge1$.

iii) $ A\equal{}[0,\infty)$.
$ f(x)\equal{}0$ when $ x\le0$; $ f(x)\equal{}x^2$ when $ x\ge0$

Of course, we have
iv) $ f(x)\equal{}1$.