geometry
3D geometry
ARML
Stanford
college

okay, so i'm gonna probably get some good stickers for my cube on cubesmith
just checking, this is mostly towards albert, are the normal stickers there the good ones that don't easily come off, like the ones on the good cubes at ARML?
thanks

No, the normal ones are the ones that I have. They don't come off but they chip, but they're much better than most stickers. The one's that Andy and other people have are tiled stickers. Although there is a trick that some people use for regular stickers (and Alexei and Timothy both have cubes like this), which is cover all the stickers with a thin coat of clear nail polish.

I'm probably going to get new stickers sometime... I need a new cube also.

Eh.

No.

I keep all my Squares. Although Square I is lying broken into pieces somewhere in my room (I believe). Actually I might have thrown him away to keep space. Actually his axis is quite good.

I'm getting a DIY cube and a stackmat, and I'm finally going to learn the entire Fridrich method.

So yeah... I have my work cut out for me for the entire summer. :)

wtf

I want a DIY cube *and* a stackmat.

Unfortunately I've heard that stackmats cost too much, so screw that, I'll just use JNetCube and hope I can accurately hit the spacebar.

If I was 7 * 191 more than you, I would be 48132 seconds.

[quote="DiscreetFourierTransform"]My little brother got the little cup stacking game for his birthday, and I mooched off the stackmat from it.
He doesn't use it anymore, but I do  :lol:

It's really far superior to using the spacebar.
I've learned almost half the PLLs and almost 2/3 the OLLs and I'm now down to a 36 sec average.
I don't do F2L, I STILL do layer by layer.

Eric knows how godlike I am at cubing ;-)
Except Ubemaya is more 7 * 191 than I am.[/quote]
oh my
you're better than me now
almost

[quote="DiscreetFourierTransform"]Junggi, I could never beat you...
You're too 7 * 191 for me.[/quote]
um my best average is 33...
hmm
$36-33=...$
$\boxed{3}$
OHMYGOSHOMGOMASDGLAKSJLHKJAPWEO,ZB,XMBADSLFKJ
MY LAYER BY LAYER GODLINESS IS THREATENED

The really super-nice stickers that cubesmith has are the "tiles".  They come in both smooth and textured varieties, and which you get is a matter of personal preference.  I'd recommend them; they're a lot nicer than normal stickers and not much more expensive.

Woah how do you know so much about cubes? I thought you were anti-cube...?

Me?  No way!  I'm only anti-cubing-in-the-middle-of-ARML.  In fact, the only thing I was concerned about was that at the end of the Power round, people would get bored and start cubing when they could be working on proofreading/rewriting solutions.  When I was competing, we had major problems with people sitting around being bored (and distracting!) at the end of power, and got yelled at by the coaches a couple of times.  And they were right--everyone should be doing something at any given time during the power round.  So I thought keeping cubes out of the team room would help facilitate that.

You just didn't see me doing any cubing this year for some reason, though my cube was in my backpack the whole time.  I'll show off my mad skillz at ARML or SMT or Stanford ARML (stay tuned for a super-cool announcement...) next year.

On a good day, just under a minute with inspection and just over a minute without.

hey, you're just about my speed....we should race sometime :)

[quote="DiscreetFourierTransform"]My little brother got the little cup stacking game for his birthday, and I mooched off the stackmat from it.
He doesn't use it anymore, but I do  :lol:

It's really far superior to using the spacebar.
[/quote]
I'm not a cubing person. So I'm confused as to why you would use a cup stacking mat instead of a spacebar to solve a Rubik's cube.

The mat has a timer with a little pressure sensitive zone where any contact with both hands will stop the clock.
You start with your hands on the timer, and when you pull them off, the clock starts. You solve the cube as fast as you can, and slam your hands back down on the pads to stop the clock.

The alternative is using a keyboard with a space bar as follows:
Open timer program...
Set up timer program...
With your wrists, awkwardly hit the space bar really hard.
Grab your cube with your hands and solve it frenetically.
Smash your cube back down on your keyboard.
Smack the living daylights out of the space bar.
Most likely purchase a new keyboard.

As you can see, the stackmat timer is much better, and it is designed specifically for this sort of thing, while the space bar is not.

I just smash my whole hand down on the keyboard and hope I get lucky.

New average: 23.83.

Improved by 0.16 from last week!

Darn... just got 24.13.

My sister was being annoying partway through and drumming her stomach really loudly (she's as fat as me, almost) and I got two 27s and a 35... ugh. So 24.13 is pretty good.

----- JNetCube Session Statistics for Tue Jun 12 20:54:00 PDT 2007 -----

Cubes Solved: 12
Total Pops: 0
Average: 23.46

Fastest Time: 17.95
Slowest Time: 27.22
Standard Deviation: 02.73

Individual Times:
1)          25.05          B' F D2 L2 R2 B' F' L U2 L' R' F2 U2 F D F' B U2 D2 B L2 R2 D' B' L'
2)          27.22          F U2 B2 U' R F' R2 B R2 U2 L2 R U R2 F2 B' R' D' F2 D2 L2 U D2 L2 R2
3)          25.47          U' F' R D2 U L' U R2 B' R2 F' L D2 U R' L2 D F2 D F U2 F R F' B'
4)          24.39          U D F2 B R U' L F B L' D' L2 D' F2 R L2 D L2 D B' F' U' F2 L D'
5)          26.76          F2 U' B D2 R F2 B' R B' D' U' R2 B' D L D2 R' L' U' D' L R' U' D2 L'
6)          24.16          F2 R2 L D2 B F2 L2 D' L F' B2 U' F2 U D' B' R B' F2 R' U D' L B' U
7)          21.84          D R2 B2 F' D' F2 L D R2 U' D B F' L2 R' F2 U L D' U' L2 F2 D F2 D
8)          17.95          F U2 D R L2 D' F' L2 U' L R' U2 L2 F D B2 R' L D2 R' D' F' R L' B'
9)          23.27          F D F' R' D' F2 D' U' R2 L' F U D2 F D2 R' L2 U2 B' R B' D2 B L R'
10)          24.11          F' U' B' L D B' D U2 L' B' L2 U R L B' U' L R U L' B2 L R2 U2 B
11)          20.44          U2 F2 U2 F D L U2 L U2 L2 B U2 B' D' U' R' U' D2 R2 D B R F2 B L2
12)          20.89          U2 D2 R2 D' U R' B D F2 L' U R' B' U2 L2 B D2 B D' U R B' L2 D2 U'


A certain company produces two items, although they only advertise for Product A. Product B costs them less than a penny to make each, it isn't really used for anything; in fact, every customer is allowed as many as they want for free. However, their stores of Product B are kept under just as tight security as Product A, and aren't allowed to be passed freely around the office for fear of theft. What are they producing?

Yes-no questions, right?  Accordingly,

- Is Product A a physical item (as opposed to a service)?
- Is Product B a physical item (as opposed to a service)?

- Is Product A a physical item (as opposed to a service)? [b]yes[/b]
- Is Product B a physical item (as opposed to a service)? [b]yes[/b]

Is it possible to carry around Product A in one's hand?
Is it possible to carry around Product B in one's hand?

- Is Product B dangerous if used incorrectly?
- Is the reason that Product B is kept under security related to Product A?

Is it possible to carry around Product A in one's hand? [b]No[/b]
Is it possible to carry around Product B in one's hand? [b]Yes[/b]
- Is Product B dangerous if used incorrectly? [b]No[/b]
- Is the reason that Product B is kept under security related to Product A? [b]Yes[/b]

is B needed to use A?

I have a good idea what it might be.

-Is A living? [b]No[/b]
-Is B living? [b]No[/b]

is B needed to use A? [b]Not really, but it helps[/b]

[quote="archimedes1"]Does having A enhance the experience of having B?
Or vice versa?[/quote]

This was already pretty much asked.

I thought of something:

Is A liquid?
Is A solid?
Is A gas?

Ditto for B.

[quote]...less than a penny to make each, it isn't really used for anything; in fact, every customer is allowed as many as they want for free... [/quote]

Can't really see a key/lock set doing that.

Are either A or B highly flammable? [b]No[/b]
Does having A enhance the experience of having B? [b]Yes[/b]
Or vice versa? [b]Yes[/b]
[b]A and B are both solids.[/b]
is b a key and a lock [b]No[/b]

Is B a license for a person to do something that would otherwise be illegal?

Is B something printed on a piece of paper?

Do most people have A?
Do most people have B?

Is B used in either a school or something other organization that meets regularly?

After looking at the thread I think I've figured out the answer.

[hide="Answer"]A=Gun, B=Bullet[/hide]

[quote="randomdragoon"]Product B costs them less than a penny to make each, it isn't really used for anything; in fact, every customer is allowed as many as they want for free. [...] for fear of theft. [/quote]
And B is a form of identification, remember.

I thought maybe email or something (for B), but then what would A be?

[quote="randomdragoon"][quote="Hamster1800"]Is  B some sort of identification?[/quote]
[size=200][color=orange][b]YES![/b][/color][/size][/quote]

I'll make it big, bold, and colorful so that everyone can see it.  :)

[hide="?"]A bullet is a way to identify a gun. Tho' they don't give those for free ...[/hide]
[hide="Another thing that verifies"]A=cell phone B=sim card[/hide]

A could be a computer game, and B is the CD key....

Question:

Does the company ONLY make products A and B?

@ Temperal - Email [i]is[/i] a form of ID. :roll:

@ Valentin - Cell phones can be held in one's hand, but it has been said that A cannot be.

Does the manufacturing of A and B fall under any sort of copyright law?

"Clarification: A is a product, but a service is usually needed to make the product work (except for maybe DIY junkies)"

Could I or the average person make the product work without needing the service?
Would I be able to do it legally?

I guess copyright/serial numbers work too ...

well I don't think this is getting anywhere (and plus I'm getting kind of busy so I can't come on here that often so

[hide="answer"]A=home security system, B=the "protected by ____" sticker. Many people want to get the sticker to deter thieves without actually buying the home security system.[/hide]

Well... that's a very strange answer. Why would B be kept under heavy guard? Anyone could make their own....

i_Like_pie:I got the impression that he's not talking about anything that's unique to someone such as an email (or AoPS!) account, more like those corny student ID cards you get in middle and high school. And I was right! The home security system sign is not like email at all, more like my original ideas.

EDIT: And plus, email is a service!


MATHCOUNTS
geometry
3D geometry
calculus
integration
bump
factorization

for being an idiot and being confused on what i was actually asking on that "n" topic, ill post this one, though its geometry, but meh

What is the total number of cubes of any size with integral side lengths in a 5 x 5 x 5 cube

nice laziness Neal... just put up all the problems that Mrs. Nelson gave us... lol

yup yup

can you verify the question in the thread "Mathcounts Challenge: N"

i did it off the top o my head, and i dont have the paper

Please tell me you two aren't spamming...[b]please[/b]! :lol:

Kind of weird this problem is actually more of a combinatrics problem...

The number of cubes with integral side lengths in any n * n * n cube is (n(n+1)/2) :^2: ...the same as the  n :Sigma: i=1 n :^3:.  Since it's a cube with side lengths of five, the answer will be (5(5+1)/2) :^2: which is 225.

You make interesting statements, but you give me no reason to believe that they are true . . .
prove your formula.

Okay.  Now explain why the sum of the cubes (like, 3^3) equals the number of cubes (like, 2x2x2) in the big cube.

the answer is 225

i used n(n+1)

[hide]QQQQQlll[/hide]2

and then square the equation. The answer is 15sqr or 225

[color=violet]if you use finite differences the you can find the pattern which leads to the equation of the garcian sum squared.[/color]

[quote=HamstPan38825]chungus bump ... i wish i had a comment on this[/quote]

why are you bumping like crazy?

Ha ha he bumped a thread that has been dead for 17 years


If a² + b² + c² = ab + bc + ac

Prove that a = b = c.

Couldn't prove it.

Thank you.

$ a^2\plus{}b^2\plus{}c^2\equal{}ab\plus{}bc\plus{}ca \iff \frac{1}{2}\{(a\minus{}b)^2 \plus{} (b\minus{}c)^2 \plus{} (c\minus{}a)^2 \}\equal{}0$.
Thus, $ a\minus{}b\equal{}b\minus{}c\equal{}c\minus{}a\equal{}0 \iff \boxed{a\equal{}b\equal{}c}$.


hi guys im stuck at practcing repeated sqauring. i have this question:

"If we want to calculate the remainder of 14^27 using repeated squaring:
27 = 1+2+8+16

14^1 = 14(mod55)
14^2 =196=31(mod55)
14^4=(14^2)=31^2=961=26(mod55)
14^8=(14^4)^2=26^2=676(mod55)

etc...

the thing that i dont get here in this example is where did get the number 31 and mod 55??? the example doesnt explain how it got the these numbers...please help guys

thanks

$14^1\equiv 14 \mod(55)$
$14^2=(14^1)^2=196\equiv 31 \mod(55)$
This is because $\frac{196}{55}=3\frac{31}{55}$
When the problem asks, "We want to calculate the remainder of $14^{27}$", what modulo are they referring to?  They probably mean mod 55.
To complete the problem, we have
$14^4=(14^2)^2\equiv 31^2 =961\equiv 26 \mod(55)$
$14^8=(14^4)^2\equiv 26^2 =676\equiv 16 \mod(55)$
$14^{16}=(14^8)^2\equiv 16^2 =256\equiv 36 \mod(55)$
$14^{27}=14\cdot 14^2 \cdot 14^8 \cdot 14^{16} \equiv 14\cdot 31 \cdot 16\cdot 36 = 249984 \equiv 9 \mod(55)$
Let me know if this helps.


find all values of reel x for which the following series coverge;

$ \sum_{n = 1}^{infinity}({\frac {1.3...(2n - 1)}{2.4...(2n)})^x}$

$ \frac{1\cdot3\cdots(2n\minus{}1)}{2\cdot4\cdot(2n)} \equal{} \frac{(2n)!}{4^n(n!)^2} \sim \frac{1}{\sqrt{\pi n}}$ by Striling's formula. So by comparison test with p-series, the given series converges if and only if $ x > 2$.

Stirling formula referes that $ n!$ can be aproximated with $ n^ne^{\minus{}n}\sqrt{2n\pi}$ in fractions


logarithms
real analysis
real analysis unsolved

using the product and quotient laws of logarithms to prove the power law of logarithms

${log}_a\, p^c\,$= c ${log}_a\, p\,$
where a element (1,infinity),
p, q element (0, infinity),
c element (-infinity, infinity)

hello, here you will find the proof
http://pc40sf06.blogspot.com/2006/11/scribin-laws-of-logarithms.html
Sonnhard.

[b]Def:[/b] $ log_a{x} \equal{} y\Leftrightarrow x \equal{} a^y$;

[b]cons.1:[/b] $ log_a{a^y} \equal{} y$;

[b]cons.2:[/b] $ a^{log_a{x}} \equal{} x$.

[b]Proof for: [/b]  $ log_a{p^c} \equal{}c.log_a{p}.$
Ussing [b]cons.2[/b]  and [b]cons.1[/b] we:
$ log_a{p^c} \equal{} log_a\left(a^{log_a{p}}\right)^c \equal{} log_a{a^{c.log_a{p}}} \equal{} c.log_a{p}.$


function
group theory
abstract algebra
algebra unsolved
algebra

find all function f:Q---> Q such:
                   $f(x+y+f(x))=f(x)+f(y)+x \forall x \in Q$

x=0 give $f(y+a)=f(y)+a, \ a=f(0),$ therefore $f(na)=na+a \ \forall n\in Z$.
1.If $y=-x-f(x)$, then $a=f(y)-y$. If g(x)=x+f(x) is surjective, then solution is $f(x)=x+a$.
2. solution $f(x)=-x \ \forall x .$

Rust, can you prove that $x+f(x)$ is surjective?

OK, here is the general solution.
Set $g(x)=f(x)+x$ for $x\in\mathbb{Q}$. Then
(*)    $f(x+y+f(x))=f(x)+f(y)+x$
transforms into
(**)  $g(y+g(x))=g(y)+2g(x)$  for all $x,y\in\mathbb{Q}$
Now if we set $A= \sum_{x\in\mathbb{Q}}\mathbb{Z}g(x)$ the additive subgroup of $(\mathbb{Q},+)$ generated by the images of $g$.
Then (**) gives
(***) $g(y+a)=g(y)+2a$ for all $y\in \mathbb{Q},a\in A$.
If $R$ is system of coset representitives of $A$ in $\mathbb{Q}$, then (***) shows, that $g$ is uniquely determined by its restriction $R\rightarrow A$.
Conversely if $A$ is any additive subgroup of $\mathbb{Q}$ with system of coset representitives $R$, and if $g: R\rightarrow A$ is some map.
Then $g$ has a unique extension $\mathbb{Q}\rightarrow A$ satisfying (***) and (**).
And with $f(x)=g(x)-x$ we get a well defined solution $f: \mathbb{Q}\rightarrow \mathbb{Q}$ of (*) with
$f(r+a)=g(r)-r+a$ for all $r\in R$ and $a\in A$.

In case $A=\{0\}$ we get $g=0$,$R=\mathbb{Q}$ and $f(x)=-x$ for all $x\in \mathbb{Q}$.
In case $A=\mathbb{Q}$ we have $|R|=1$ and $f(x)=x+c$ for some $c\in \mathbb{Q}$.
But there is much more.
For example $A=\mathbb{Z}$,$R=[0,1[$,$g(R)\equiv c\in \mathbb{Z}$ gives $f(x)=2[x]-x+c$ for all $x\in \mathbb{Q}$.

In order to understand the general solution a little better, it's necessary to classify all additive subgroups of $\mathbb{Q}$
and find a "nice" system of coset representitives for each of them.
Therefore if $P$ is a set of primes, we define
$\mathbb{Z}_{P}: =\{\frac{a}{b}| a,b \in\mathbb{Z}, b \text{ has all prime factors in }P\}$
and
$R_{P}: =\{\frac{a}{b}| a,b \in\mathbb{Z},0\leq a<b, b \text{ has no prime factor in }P\}$
Then $\mathbb{Z}_{P}$ is an additive subgroup of $\mathbb{Q}$ with system of coset representitives $R_{P}$, and it's not hard to prove,
that every additive subgroup of $\mathbb{Q}$ has the Form $\alpha \mathbb{Z}_{P}=\{\alpha x: x\in\mathbb{Z}_{P}\}$ for some set of primes $P$ and some $\alpha \in\mathbb{Q}$,
with system of coset representitives $\alpha R_{P}$ for $\alpha \not =0$.
.
(Hint: $\frac{a}{b},\frac{c}{d}\in A\text{ with }gcd(a,b)=gcd(c,d)=1\Rightarrow \frac{gcd(a,c)}{lcm(b,d)}\in A$)

Now if $P$ is a set of primes, $\alpha \in\mathbb{Q}\setminus\{0\}$ and $g: R_{P}\rightarrow \mathbb{Z}_{P}$ some map,
we get a well defined solution $f: \mathbb{Q}\rightarrow \mathbb{Q}$ of (*) by
$f(\alpha(r+a))=\alpha(g(r)-r+a)$ for all $r\in R_{P},a\in\mathbb{Z}_{P}$
and we have seen, that those cover all solutions of (*), except $f(x)=-x$ (given by $\alpha=0,A=\{0\}$).

A is subgroup Q. I give solution for all cases.
A=Q $f(x)=x+c$,
$A=\{0\}$ $f(x)=-x$.
If $\{0\}\not =A\not =Q$ exist $b\not =0$, suth that $A=bZ$. Solution for these case is $f(x)=-x+b[x/b]$.

Sorry I want to know that '' what is $additive$ $subgroup$ ?

[quote="Rust"]If $\{0\}\not =A\not =Q$ exist $b\not =0$, suth that $A=bZ$. Solution for these case is $f(x)=-x+b[x/b]$.[/quote]

Sorry, but $\{\frac{m}{2^{n}}\text{ : }m,n\in\mathbb{Z},n\geq 0\}$ has not the form $b\mathbb{Z}$ for some $b\in\mathbb{Q}$. You forgot the localizations.

[quote="hunter_cvp_ic"][quote="olorin"]Now if we set $A= \sum_{x\in\mathbb{Q}}\mathbb{Z}g(x)$ the additive subgroup of $(\mathbb{Q},+)$ generated by the images of $g$. [/quote]
I can't understand what A?
Can't you show me olorin ? thanks
and what is your sol??[/quote]

An additive subgroup of $\mathbb{Q}$ means simply a subgroup of $(\mathbb{Q},+)$, while a multiplikative subgroup of $\mathbb{Q}$ is a subgroup of $(\mathbb{Q}^{*},\cdot)$.
And a subgroup of the group $(\mathbb{Q},+)$ is simply a non empty subset closed under summation and subtraction.
(i.e. $A\not = \oslash$ and $x,y\in A \Rightarrow x+y,x-y\in A$)
Now for every subset $M\subseteq\mathbb{Q}$ we get a subgroup $A= \sum_{x\in M}\mathbb{Z}x: =\{n_{1}x_{1}+\cdots+n_{r}x_{r}\text{ : }r\geq 0,n_{1},\cdots,n_{r}\in\mathbb{Z},x_{1},\cdots,x_{r}\in M\}$ of $(\mathbb{Q},+)$,
which is indeed the smallest subgroup containing $M$.

[quote="olorin"]
Sorry, but $\{\frac{m}{2^{n}}\text{ : }m,n\in\mathbb{Z},n\geq 0\}$ has not the form $b\mathbb{Z}$ for some $b\in\mathbb{Q}$. You forgot the localizations.
[/quote]
Yes, you are right. Solution is $f(b+a)=-b+a+a_{0}$, were $a\in A,b\in Q/A, a_{0}\in A$ It give $g(x)=f(x)+x\in A$ and $f(b+a)=f(b)+a$ for all $a\in A$. $a_{0}=\phi (b), \ \phi: Q/A\to A$.


Hello,

Given is:

[img]http://img56.imageshack.us/img56/8520/veekkz7.jpg[/img]

Now They ask me to find $f^{39}(\vec{v})$ for $\vec{v}=(x,y,z)$

Who can you find that? Greets.

[b]Answer :[/b] For any $n\in N\ ,\ f^{n}(\vec v)=\vec v\cdot A^{n}\ ,$ where $A=\left(\begin{array}{ccc}-4 & 3 & 6\\\ 0 &-1 & 0\\-3 & 3 & 5 \end{array}\right)\ .$

oké thanks but do you need to do this 39 times?

Nope if you have $A^{2}$ , you can find $A^{4}$ by squaring $A^{2}$, then you can find $A^{8}$in the same way

and then you could use $A^{39}=A^{32}A^{4}A^{2}A$

or...( and this is probably what your teacher wants you to do)
you can use diagonalization (I checked the eigenvalues, they are nice numbers and that is no coincidence :wink: )

[hide]Je hebt - denk ik -een foutje gemaakt.  Ofwel bedoelde je "Where can I find that?" ofwel "Who can find that?"  Er zijn hier veel mensen die nog Engels aan het leren zijn, en die soms vreemde zaken schrijven, en ik ben zelf ook nog aan het verbeteren, maar ik geloof in de kracht van correctie. :| [/hide]

[hide]oké bedankt voor de verbeteringen fijn dat je dat wil doen trouwens één van de redenen dat ik hier graag eens kom voor mijn engels te verbeteren.[/hide]

So If I calculate the eigenvalues you can write this as a diagonal matric and I need only multiply the diagonal?

Greets.

matric->matrix (just as in Dutch)

well, it comes down to that

write $A$as$P^{-1}D P$with $D$ diagonal and$P$ invertible

then$A^{n}= P^{-1}D^{n}P$ and calculating $D^{n}$ is easy, just replace every diagonal element with its nth power

You do have to calculate the eigenvectors too though.


inequalities
integration
calculus
calculus computations

Suppose $ a>0,b>0$,$ f(x)$ be a twice continuously differentiable in the interval [$ o,a$],$ f(0)\equal{}f(a)\equal{}b,f(x) \geq 0 ,f^{"}(x) \geq 0$ ,$ \forall x \in [o,a]$ Prove inequality :
          $ 2\int_{0}^{a}{f(x)[1\plus{}[f^{'}(x)]^2]^{\frac{1}{2}}dx} \leq b(a^2\plus{}b^2)^{\frac{1}{2}}$

[b]Wait a solution [/b]!!! Thanks  :oops:

Multiply the both sides by $ \pi$, you would see something.


calculus
integration
Support
LaTeX
logarithms
trigonometry
calculus computations

[quote="MrMagician"]$ \frac {7\pi^2}{6(3)^{1/2}}$[/quote]
Yes, that's the correct answer.

Here's my solution:

It's routine to show that $ \int_{0}^{\frac{\pi }{2}}{\frac{dt}{1\minus{}\sin (t)\cos (t)}}\equal{}\frac{4\pi }{3\sqrt{3}},\,\int_{0}^{\frac{\pi }{2}}{\frac{dt}{1\minus{}\sin ^{2}(t)\cos ^{2}(t)}}\equal{}\frac{\pi }{\sqrt{3}}.$

$ I\equal{}\int_{0}^{\pi }{\frac{t}{1\plus{}\sin (t)\cos (t)}\,dt}\equal{}\int_{0}^{\frac{\pi }{2}}{\frac{t}{1\plus{}\sin (t)\cos (t)}\,dt}\plus{}\int_{\frac{\pi }{2}}^{\pi }{\frac{t}{1\plus{}\sin (t)\cos (t)}\,dt},$ now on the second integral put $ t\mapsto t\plus{}\frac\pi2$ then $ I\equal{}2\int_{0}^{\frac{\pi }{2}}{\frac{t}{1\minus{}\sin ^{2}(t)\cos ^{2}(t)}\,dt}\plus{}\frac{2\pi ^{2}}{3\sqrt{3}}.$ Now to evaluate the last integral let's put $ t\mapsto\frac\pi2\minus{}t$ so $ I\equal{}\frac{\pi }{2}\int_{0}^{\frac{\pi }{2}}{\frac{dt}{1\minus{}\sin ^{2}(t)\cos ^{2}(t)}}\plus{}\frac{2\pi ^{2}}{3\sqrt{3}}\equal{}\frac{\pi ^{2}}{2\sqrt{3}}\plus{}\frac{2\pi ^{2}}{3\sqrt{3}}\equal{}\frac{7\pi ^{2}}{6\sqrt{3}}.$


Let $ \mathbb{B}_t$ be $ d$ dimension standard Brownian motion. Show that:
\[ P(\lim_{t\rightarrow \infty}\parallel{}\mathbb{B}_t\parallel{}=\infty),d\geq 3\]
\[ P(\exists t>0,s.t. \mathbb{B}_t=\mathbb{O})=0,d=2\]
And for arbitrary $ \epsilon>0$
\[ P(\forall t>0,\exists s>0,s.t. \parallel{}\mathbb{B}_s\parallel{}<\epsilon)=1,d=2\]
Thanks in advance. :)

[quote="sss"]for arbitrary $ \epsilon > 0$
\[ P(\forall t > 0,\exists s > 0,s.t. \parallel{}\mathbb{B}_s\parallel{} < \epsilon) = 1,d = 2\]
[/quote]
have you forgotten something like $ s  > t$ ?


MATHCOUNTS
AMC
AMC 10
AIME
USA(J)MO
USAMO
analytic geometry

[quote="monkeyman"]MathCounts nationals.  Well did anyone actually go to the contest?  Well I did and I was on the arizona team :).  Anyway it was really fun, and we got every thing we could poosibly want paid for.  Also everyone who went got a free TI-83 Plus silver edition calc compliments of texas instruments.  Oh ya and I met gregory (the person who got grand national champion), and he is, um, nice I guess.  Anyway did anyone else go, cause I might recognize you.  Oh ya and there were 228 students, so there was quite some competition.[/quote]
if you'd looked at other topics, you would've seen that many people from AoPS went to nationals. it was pretty nice, except the calculators weren't as spiffy as in recent years. just plain TI83 silvers. call me spoiled, but before, they gave out 89s or 83 silvers with mathcounts written on the back in red.

Nationals must have been fun! Missed it by one place, but that's all right. I'll do better in AMC! :)

I made soooooooo many stupid mistakes at state I didn't even make it into the countdown round.   :(

[quote="tandjsnell"]No I didn't go to nationals. But I probably live in 3rd toughest state(New Jersey). I got a 36 combined and I was ranked something like 34th  :? ! I better move before next year :D .[/quote]

How did you rank New Jersey as the 3rd toughest state, even though they were 10th in nationals? (btw, a 36 would've gotten you like 6th in my state  :D )

What I meant by 3rd toughest state is 3rd toughest state to qualify in.
Your btw statement backs that up.  I didn't know the national results either.

I didn't (got stuck at state because of too many stupid mistakes), but my friend did in 2004, and for CA too.

[quote="tandjsnell"]No I didn't go to nationals. But I probably live in 3rd toughest state(New Jersey). I got a 36 combined and I was ranked something like 34th  :? ! I better move before next year :D .[/quote]

The kid that won was fast!  I was in a team, we didn't do so good, 24 out of 28 i think

wow... people have been responding so much to the mathcounts section.  i didn't go to nationals , but i sorta tied to be on california team.  not exactly, i was 6th place.  but i had the same overall score as dark_shadow

lol i suck, i got a 40 which was 10th in CA

[quote="captcha000"]i went for georgia, and i definitely think the 89's we got the previous year were better than the 83 silvers we got this year but wutever, a little variety couldn't hurt. i didnt do so well at state, my score was pretty low even though i did get 3rd, and well i was hoping to do better at nats but yea[/quote]

I didn't like the variety since I had lost my TI89  :(

Gaah, I would post Nationals pictures, but the set of pictures my teammate sent me were all screwed up :( .

Only 4 6th graders made it to Nats last year
(me and 3 other people)

i'm in 8th... first year in MATHCOUNTS.. just wonderin what its like in comp.  and what chapter is  :P

Chapter is the regional competiton amoung schools in your area.  Seems like every chapter runs their competitions differently, but you'll take a 30 question/40 minute, no calculator "Sprint Round", there will be 4 two question/6 minute w/ calculator "Target Rounds".  After that there is a 10 question/20 minute "Team Round" where the 4 team members get to work together with calculators.

Depending on your chapter you may have a "Countdown Round which is a head to head speed competition with the top scorers from the Sprint and Target Rounds.

Good luck

[quote="frost13"]Depending on your chapter you may have a "Countdown Round which is a head to head speed competition with the top scorers from the Sprint and Target Rounds.[/quote]

All the chapters in a state do the Countdown Round officially, or some chapters do it for fun and some don't do it at all.

The winning team and the top two non-team people go to state. However, less than six people may go if the winning team has less than four people, and more than six may go if the state director says so. The policy for additional teams and mathletes must be the same for all the chapters.

[quote="eryaman"]I made it to state in 7th and 8th grade.  Unfortunately, I live in Texas so I only got 31st or 32nd in 7th grade with a score of 38 or 39, and like 55th in 8th grade with a 33.  I should have gotten 37 though, made two stupid accidental errors on target.[/quote]
Oh ya For Arizona.  I got first with 42 I believe (hey its good for me) 2nd 40 3rd 38 and 4th 37 so ya...

[quote="shake9991"][quote="RC-7th"]
The winning team and the top two non-team people go to state. However, less than six people may go if the winning team has less than four people, and more than six may go if the state director says so. The policy for additional teams and mathletes must be the same for all the chapters.[/quote]
actually: (being a math contest as it is) the state coordinater comes up with a proportion that depicts number of students and number of teams per chapter... so say chapter A has 3 teams ad 24 individuals, then only like the top team and top individual can go...  but say: chapter A has 200+ people and 25 teams then like 5 teams and 5 top individuals go
also:
a team consists of 4 people and only 4 people.... a school can not go to state if it has an official team with less than 4 people...[/quote]

In my state 3 teams and 4 additional kids from each chapter go to state regardless of the size of the chapter.  I'm trying to get the coordinator to change it to be proportional to the number of teams in a chapter, but as it turns out almost 1 in 3 teams advance to State.  It really is different in every state.

Also, I am pretty sure you can have a team with only 3 kids, but when calculating the team score, you still divide the sum by 4.  Last year if my 4th kid had not showed up and I had no alternates we would have been a team with a 4th score of 0.(and we still would have won states since the other 3 kids were 3rd,4th, and 6th with a 10 on the team)

For Connecticut, I'm pretty sure it's a certain number of teams and then those who aren't on school teams that are going to states and are in the top 32 individual for the chapter.

[quote="tcs09"][quote="Treething"][quote]No I didn't go to nationals. But I probably live in 3rd toughest state(New Jersey). I got a 36 combined and I was ranked something like 34th  ! I better move before next year :D .[/quote]

I'm from New Jersey too. I made it to Nationals and placed in the top 40.[/quote]

You're the sixth grader, right?[/quote]
Well now I'm a 7th grader :P .

I made our school second team each time tell now when i won but didn't compete


not just the pure math ones, but your reflexes :D

http://www.mathlinks.ro/fun/dodge.html

have fun! you can see my max. below :).

22.57

I do suspect that the score is affected by the RAM of the computer... :?

wow. I could only get up to 25, it's moving waaayyy to fast to counter-act at that point. Charlie how much DDRAM do you have?

i was trying this game at the cybercafe and had the slowest computer there for me and lasted 82.17 secs.sure,RAM is playing it's part

i'm on a pentium 3, 2.6ghz, 512 DDR ram, but still I lasted for 35 sec :D

:D  :D  :D Hey now I just came up with this, what is your fastest losing time? In other words, what's the smallest time you can get in this game? I can't get lower than 0.02 seconds.

Try a good score on a good computer. The speed is very important. If the computer is slow, than you can easily resist 100 sec. :)
To get high scores run a CPU killer (if your computer is too good :D )
Or another one: change the time while you play (777,000 sec :D  :D )

the blue blocks arent even moving on my computer.

:lol:  It's time for a new computer.  :lol:

[quote="Valentin Vornicu"]0.0172 fastest losing time. I guess my PIV 2,8 Ghz, 1Gb DDram comp helps out here ;) :D:D[/quote]

my losing time is 0.  No one can top that

[quote="3X.lich"]Hey guess what? Anonymous means lack of confidence. By the way, we all already topped you, 0 is unreachable considering the reaction time of human nerves are limited, so the time should be non-zero and the real loser is you. :D Oops I just flamed :P[/quote]
thats what it says on my computer.  BTW how do you add images?(like the one valintin added)

you must be a registered user first. then you can do (almost) everything you like :D.

[quote="3X.lich"]Hey guess what? Anonymous means lack of confidence. By the way, we all already topped you, 0 is unreachable considering the reaction time of human nerves are limited, so the time should be non-zero and the real loser is you. :D Oops I just flamed :P[/quote]

geez.  You're certaintly rude.  Dissing someone for reading what it said on the screen. :(

[quote="3X.lich"]Hey guess what? Anonymous means lack of confidence.[/quote]

im anonymous simply because i didnt send the form in yet.  -sigh-.  Non Sequitur.

[quote="Anonymous from above :D"]im anonymous simply because i didnt send the form in yet.  -sigh-.  Non Sequitur.[/quote]
well if you are waiting for a special invitation, here it is: join us. :P

-687624.905 !!!  :)  :D   :D (I've switched the system time during playing!)

Amazingly, this is what I got on my computer, when I saved the webpage
:)
I cheated , but still :))))))))))))

[quote="3X.lich"]Hey guess what? Anonymous means lack of confidence. By the way, we all already topped you, 0 is unreachable considering the reaction time of human nerves are limited, so the time should be non-zero and the real loser is you. :D Oops I just flamed :P[/quote]
ill just mention the sad part is, that u havent even considered  a slow computer or one, which, the timer doesnt work upon (sorry to revive up an old thread, but i was the guest who said 0)

[quote]sorry to revive up an old thread, but i was the guest who said 0[/quote] So you finally decided to join us, months later? :) Great!
Now we can really patch this up, and move on ;)


[quote="2006"]Let $ a,b,c,a$ be positve real numbers
  prove that $\sqrt[2006]{fra{a}{b+c}}$+$\sqrt[2006]{fra{b}{a+c}}$+$\sqrt[2006]{fra{c}{b+a}}$ $\>>2$[/quote]

i think you meant $\sqrt[2006]{\frac{a}{b+c}}$ + etc

[quote="Hawk Tiger"]How clever you are,pohoatza !But I think it has been posted,.[/quote]

 :rotfl:


Determine all triples of real numbers $(a,b,c)$ such that \begin{eqnarray*} xyz &=& 8 \\ x^2y + y^2z + z^2x &=& 73 \\ x(y-z)^2 + y(z-x)^2 + z(x-y)^2 &=& 98 . \end{eqnarray*}

Expanding the third equation we get $xy^2-2xy+z^2x+yz^2-2xyz+yx^2+zx^2-2xy+y^2z=98$ Now by substituing the first two equations we get: $xy^2+yz^2+zx^2=73$ and by the second equation we obtain that $xy^2+yz^2+zx^2=x^2y+y^2z+z^2x$.  So since this eq. is symmetric and we can't have $x>y>z$, two of them must be equal, say $x=y$. So we have from the first eq. $x^2z=8$.  Plugging this and $x=y$ in the second eq. we have that $x^3+64/(x^3)-65=0$, So put $x^3=a$, and we get $a^2-65a+64=0$. Solving this quadratic we get that $a=64$ or $a=1$, therefore $x=4$ or $x=1$.  So the solutions are $(1,1,8)$ and $(1/2,4,4)$, and of course all the permutations of these.

[hide]
Expanding the third equation we get
\[xy^2+yz^2+zx^2-6xyz+xz^2+yx^2+zy^2=98\]
substituting the first two equations we achieve $xy^2+yz^2+zx^2=xz^2+yx^2+zy^2 \rightarrow xy(y-x)+yz(z-y)+xz(x-z)=0$ Substituting the first equation again;
\[\frac{8(y-x)}{z}+\frac{8(z-y)}{x}=xz(z-x)\]
(remark: this equation is well defined since $z\neq 0$ and $x\neq0$.)
Multiplying the equation by $xz$ and rearranging the LHS;
\[8y(x-z)+8(z-x)(z+x)=x^2z^2(z-x)\]
Now we put all the terms to one side of the equation and factor $z-x$ resulting in $z=x$ or $x^2z^2+8y-8(z+x)=0$.
If $x^2z^2+8y-8(z+x)=0$ then $x^2y^2=8(x+z-y)$ and by the first equation, $xz=y(x+z-y)$ finally put all in one side and factor ; $(x-y)(z-y)=0$. We then know that at least two variables are the same.
WLOG $z=x$
The first equation becomes $x^2y=8$, the second $x^2y+xy^2+x^3=73$, and third equation $x(x-y)^2=49$. Substituting the first in the second and then plugging that into the third equation results in
\[8+\frac{64}{x^3}+x^3=73\]
which is a quadratic equation. Solving we see that the solutions are $(4, 1/2,4), (1,8,1)$ and their permutations.


Can someone give me a tutorial on how to use asymptote?

Have you checked in AoPSWiki? I remember one person had a tutorial on his blog...

Well, basically, to draw a line from Cartesian (x1,y1) to (x2,y2), use

draw((x1,y1)--(x2,y2));

and to draw a point, use

dot((x,y));.

You can store points in variables, like this...

pair a= (x,y);

and you can also store paths.

path b=(x1,y1)--(x2,y2);

and paths can also cycle...

path c=(0,0)--(0,1)--(1,1)--(1,0)--cycle;

and then you can fill...

fill((0,0)--(0,1)--(1,1)--(1,0)--cycle,red);

and you can also use if statements and for- and while- loops...

and a bunch of other stuff.

Aha my blog finally comes to use!!!
Go here:
[url=http://www.artofproblemsolving.com/Forum/weblog.php?ppage=5&w=1157&start=5]YAY FOR MY BLOG!!![/url]


The side lengths of tetrahedron $ABCD$ are:

$AB=3$, $AC=4$, $BC=5$, $AD=6$, $BD=7$, $CD=8$

Find the volume of the tetrahedron.


So who came this year?  What did you think?  I liked the problem sets this year (afterall, I did write several of the problems), and I think the rounds were really interesting.  I had fun grading papers this year, although it is a lot of work.

So what were your thoughts?

It probably belongs in "Other US contests and programs"; that's a general forum as well as a category.


Ann and Bob play a game where someone continues to roll a dice until Ann or Bob wins. If the dice rolls a 9, Ann wins, if it is $ n$, Bob wins.
For what value of $ n$ is the game fair?
With this value of $ n$, what is the probability that the game ends on the first roll?

Wouldn't any value of n be fair?   :maybe:

No.

Hint 1: what's the probability you roll a 9?

Hint 2: now find another no. for which the probability is the same ( in the {7,...,12} set)

Hint 1: 1/9

Hint 2:
7 - 1/9
8 - 1/9
9 - n/a
10 - 1/10 (for 10 - 12 the probability of rolling a 9 becomes the same as them)
11 - 1/11
12 - 1/12

I have a feeling I might be misunderstanding the problem now...  :?:

Explanation of the problem: A and B have two 1-6 dice. And they roll them. If they roll a 9 (first die+second die), A wins, and if they roll a n, B wins.

yes, the odds are 1/9, as the no of possible results is 36, and no of desired results is 4. (simplest way to derive it is to write out all combinations* (pairs (First die, Second die)), and then count). This method will quickly give you all other probabilities: please check them the ones you wrote don't seem correct.

Can you answer the second question?

____
*not really combinations, it's rep-variations  :)

Oh ok now I get it.

The ways to get 9 are
(3, 6)
(6, 3)
(4, 5)
(5, 4)

Now we just need another number that you get with only 4 ways.  Pretty quickly after going through you can see that 5 works.  So n = 5

Thanks a lot hsijlak!  :D


LaTeX
geometry
3D geometry
analytic geometry
graphing lines
AMC
coordinate geometry

I would like to start a WORD GAME Marathon.
This is how it works:
Someone starts out with a certain word.
The next person makes a new word by switching 2 letters of the word, adding a letter, or taking out a letter.
All words must be from the English Language.

When there exist no more continuations, the next user may pick a new word.

FIRST WORD:  $ CAT$

$ BIT$ (hope I understood the rules right :P ).

(Am I allowed to rearrange [i]and[/i] remove a letter?)

$ BALE$

Doing 2 things at once defeats the purpose of the game. I'll just go off of the previous word.

$ FLAME$

Geeks :winner_first: <-geek A :winner_second: <-geek B :winner_third: <-geek C

seek[hide][hide][hide]Enough with hiding things like this!!!!![/hide][/hide][/hide]

ze

la la la
you and me
can sing this melody
so come on
dance to the beat
wave your hands together
come feel the heat forever and forever
listen and learn now is time for prancing
now we are here with
caramelldansen

ben

Mai ya hi!
Mai ya hoo!
Mai ya ha!
Mai ya ha-ha!


How do you solve

5^x = x^2

for all values of x?

thanks!

i thought it looked easy but i don't know what to do lol

when you graph 5^x-x^2 and then find the zeros on a graphing calculator you get -.6113995 for x when y = 0, which is the only answer.... I also tried playing around with it by hand but I couldn't get an x term on one side and the answer on the other... probably someone else will solve it for you though

ya with graphing calculator it can be solved but i got no idea how to do it w/o graphing calculator

thx tho :)

5^2.5 is 55.9, which is not the solution... are you sure that works?  How did you go from xlog55=2log5x to 5=2log5x???  Please explain if I am incorrect...

no it doesn't, i doubt you can solve it with logs i tried it for quite a while...

thanks tho

This type of question (except it was $2^x=x^2$) came up at one of my Algebra 2 competitions and I had no idea how to solve it [actually, it asked how many real solutions does this equation have].

[quote="José"][quote="Agent ';..;'"]5^x = x^2[/quote]

[hide]$5^x=x^2$

$5^{\frac{1}{2}x}=x$, from where $x>0$

$5^{\frac{1}{2}x}\geq 1$.

If $5^{\frac{1}{2}x}=1$, then $x=0$, but when $x$ start being greater, the difference between both sides of $5^{\frac{1}{2}x}=x$ become greater (the left side is bigger and bigger). Hence, the're no solutions (but no sure, please if someone can correct me...)[/hide][/quote]

yea that approach work but you can do the same thing with $5^x=x^2$ like say "if x gets bigger then the left side is greater than the right side".  but if you examine the graph of $x^2$ and then examine $5^x$ there's a solution in the negative domain.
thanks for the help

[quote="kokosai"]I don't think these types of problems can usually be solved exactly; 2^x=x^2 is solvable exactly, of course, but I think that's an exception.  You can approximate the answer as closely as you want, but I don't think the means exist to find the exact answer.[/quote]

I think that maybe you're right...

[quote="Agent ';..;'"]

 if you examine the graph of $x^2$ and then examine $5^x$ there's a solution in the negative domain.[/quote]

Where exactly????

I would just graph the equations and see where they intersect, whether it is on a graphing calculator or, if possible, by hand...

[quote="Agent ';..;'"] if you examine the graph of $x^2$ and then examine $5^x$ there's a solution in the negative domain.
thanks for the help[/quote]

Thinking about it: how can graphs of $x^2$ and $5^x$ cross in the negative part, if, in $y=x^2$, $y$ is always non-negative?????

Oh sorry, I see: graph it and they cross... if I can find the solution more o less I'll post it!!!

[quote="José"]And another: $x=4$[/quote]
yah I put two. but there is apparently another solution, some negative irrational number.

yah ti-89 only gives you -.611399

$5^x=x^2$. There's only one solution, and I got it, but aproximatly: $x=-0,61$

I'm interested to know if there is an actual way to solve these types of problem

I don't think it can be solved through normal algebra ... but I typed it into Maple and got this solution (don't know what LambertW means):
$x=-\frac{2 \cdot \text{LambertW}\left(\frac{1}{2}\ln{5}\right)}{\ln{5}}\approx-0.6113994672$

Edit: Here's a link to the function:
[url=http://mathworld.wolfram.com/LambertW-Function.html]mathworld.wolfram.com/LambertW-Function.html[/url]

[quote="Agent ';..;'"]yea is there a way to solve these kinda problems without graphing calculators..[/quote]

know how?? please share it...


conics
parabola
function
integration
calculus
trigonometry
analytic geometry

I don't know if this is a famous problem or if there's just a formula you can apply.

You graph the function $f(x)=x^{2}$ and you strech out the portion of the parabola in between $x=0$ and $x=2$ straight. How long would it be?

Let $f(x)$ be a smooth function in the interval [a,b]. Then, the lenght of the graphic of $f$ between points [a,f(a)] and [b,f(b)] is given by

$L = \int^{b}_{a}\sqrt{1+[f'(x)]^{2}}\,dx$.

(you can prove this by a geometrical argument and then using Lagrange's Theorem).

So in your case $L = \int^{2}_{0}\sqrt{1+4x^{2}}\,dx$. To evaluate this integral you can, for example, change variables according to $x = \frac{1}{2}\tan t$ and then

$\int\sqrt{1+4x^{2}}\,dx = \frac{1}{2}\int\sec^{3}t\,dt$,

and $\int\sec^{3}t\,dt = \int\sec^{2}t \sec t\,dt = \tan t \sec t-\int\tan^{2}t \sec t\,dt$
$= \tan t \sec t-\int\sec^{3}t\,dt+\int\sec t\,dt$
$\Rightarrow  2\int\sec^{3}t\,dt = \tan t \sec t+\ln|\sec t+\tan t|+C$

and so

$\int\sqrt{1+4x^{2}}\,dx = \frac{1}{4}\tan t \sec t+\frac{1}{4}\ln|\sec t+\tan t|+C$
$= \frac{x\sqrt{1+4x^{2}}}{2}+\frac{1}{4}\ln|2x+\sqrt{1+4x^{2}}|+C$.

Now doing the calculations, the definite integral L above equals

$L = \sqrt{17}+\frac{1}{4}\ln(4+\sqrt{17})$.

If it won't bother you too much, can you post the proof?

The line segment connecting the points $(t,f(t))$ and $(t+h,f(t+h))$ has length $\sqrt{h^{2}+(f(t+h)-f(t))^{2}}$ which is equal to
\[h\sqrt{1+\left( \frac{f(t+h)-f(t)}{h}\right)^{2}}\]
So if we approximate the curve from $x=a$ to $x=b$ by $n$ line segments whose $x$-coordinates differ by $h$ where $h=\frac{b-a}{n}$, the the total length of the segments is equal to
\[h\sum_{k=0}^{n-1}\sqrt{1+\left( \frac{f(t_{k}+h)-f(t_{k})}{h}\right)^{2}}\]
where $t_{k}=a+kh$.

To find the length of the curve itself, send $n\rightarrow \infty$ so the length of every segment is sent to 0. We have the definition of the derivative $\lim_{h\rightarrow 0}\frac{f(t_{k}+h)-f(t_{k})}{h}=f^\prime (t_{k})$ along with the definition of the integral, which converts the sum into the definite integral
\[\int_{a}^{b}\sqrt{1+[f^\prime (t)]^{2}}\ dt \]
A similar argument can be used to show that the length of a parametric curve from $t=a$ to $t=b$ and $x$ and $y$ in terms of $t$ equals
\[\int_{a}^{b}\sqrt{[x^\prime (t)]^{2}+[y^\prime (t)]^{2}}\ dt \]


It looks like Holder on

$ \sum{x_i^2} * \sum{\sqrt{x_i}} * \sum{\sqrt{x_i}}$

I didn't use Holder.

I'm really not used to Holder, so I don't see how to start with that.

Could you post a complete solution please?

 :)

[quote="ElChapin"]For positive $ x_i$ such that   $ \sum_{i \equal{} 1}^{n}{\sqrt {x_i}} \equal{} 1$

Prove that the following holds:

          $ \sum_{i \equal{} 1}^{n}{x_i^2}\ge(\sum_{i \equal{} 1}^{n}{x_1})^3$

 :)[/quote]
I think:(Using Holder)
$ (\sum_{i \equal{} 1}^{n}{x_i^2})(\sum_{i \equal{} 1}^{n}{\sqrt{x_i}})(\sum_{i \equal{} 1}^{n}{\sqrt{x_i}}) \ge (\sum_{i \equal{} 1}^{n}{\sqrt[3]{x_i^2.\sqrt{x_i}\sqrt{x_i}}})^3\equal{}(\sum_{i \equal{} 1}^{n}{x_1})^3$


What do you want us to solve for?
You've given one equation in two variables.
We can't find a unique value, but we can solve for one in terms of another.
Here's a start.
[hide="Possible Hint"]
Get a common denominator by multiplying the first fractiong by $\frac{a^{x-3}}{a^{x-3}}$, the second by $\frac{a^{x-2}}{a^{x-2}}$, and the third by $\frac{a^{x-1}}{a^{x-1}}$

Now we have, after clearing the denoms,
$a^{x-3}+a^{x-2}= 2a^{2}a^{x-1}$
$a^{x-3}+a^{x-2}= 2a^{x+1}$
Multiply everything through by $a^{3}$
$a^{x}+a^{x+1}= 2a^{x+4}$
What substitution(s) can you make to solve this problem?
[/hide]

EDIT: Hint is invalid due to multiplication issues.

the same question bothered me too . i found it somewhere and it seemed strange to me ... i thought that maybe there's something i missed . i'm supposed to find x .

[quote="DiscreetFourierTransform"]Get a common denominator by multiplying the first fractiong by $\frac{a^{x-3}}{a^{x-3}}$, the second by $\frac{a^{x-2}}{a^{x-2}}$, and the third by $\frac{a^{x-1}}{a^{x-1}}$[/quote]

Unfortunately, you seem to have forgotten your rules of exponentiation.  This only complicates the situation.

[hide="What you probably want to do is..."] Just multiply by $a^{(x-1)(x-2)}$.  This gives, after some simplification,

$1+a^{x-1}= 2a^{2x-2}$

Make the substitution $u = a^{x-1}$.  Then

$2u^{2}-u-1 = 0 \Leftrightarrow$
$(2u+1)(u-1) = 0 \Leftrightarrow$
$u =-\frac{1}{2}, 1$

Since $a > 0$ the first case is impossible, leaving only the case $a^{x-1}= 0 \Leftrightarrow \boxed{x = 1}$

Regardless of the value of $a$. [/hide]

If the question allows for complex solutions, then wouldn't x = 1+log[size=75]a[/size](-1/2) have an infinite number of complex values?

Auugh! Blasted rules of expoentiation!
By doing what I had done, you end up adding an exponent to the denominators, not multiplying the two.

Yes, that was the solution I intended.
I completely messed that one up, though.
Tor, nice solution.

tor that's the solution for sure, but how come if you get  a common denominator ( like we said ) you get that contradiction ?

You mean DFT's original solution?  The denominator doesn't end up being common.  That's the problem.

seems pretty common to me ... what's wrong with it ?

By multiplying like I did, you'll end up with a random fraction like
$\frac{a^{x-3}}{a^{(x-1)(x-2)+(x-3)}}$, not with the exponents in a product like T0r suggested.

now it's crystal clear ... seems like we both made the same mistake . thanks guys for clearing that out


For every $a,b,c,d$ in $\left[ 1,2\right] $ prove that :
$\frac{a+b}{b+c}+\frac{c+d}{d+a}\leq 4\frac{a+c}{b+d}$

This is easy ,use convex I will prove when [tex]a,b,c,d \in \{1,2\}[/tex]

[url=http://www.math.ust.hk/excalibur/v9_n5.pdf]www.math.ust.hk/excalibur/v9_n5.pdf[/url]
See the solution to problem 211, page 3.


Hello!

can you find all the functions $f:{\Bbb R}\to{\Bbb R}$ such that for all real numbers $x,y$ we have: $f(xy)=f\left(f(x)+f(y)\right)$? :P

[quote="t0rajir0u"]$y = 0 \implies f(0) = f( f(x)+f(0))$.  Let $f(0) = k$; then $f( f(x)+k) = k$ independent of $x$.  So $f$ must be constant.[/quote]
It is not true. Equation had many solutions. For example $f(x)=g(x)>a, x<b$ any function, and $f(x)=k, x\ge b$. Then f(x) satisfyed your equation $f(f(x)+k)=k$ if $b=2k,a=k$.


logarithms
analytic geometry
graphing lines
slope
calculus
calculus computations

This problem was proposed for the entry exam at a faculty in our country and I strongly feel that it could be wrong.
Let $m\in R$. Determine $m$ such that the equation $|\ln (x)|=mx$ has exactly $3$ different real solutions.
[hide="Their answer"]$m\in (0,\frac{1}{e})$[/hide]

You're right; $-\frac1e<m<0$ also works.

From the graph, it is clear that the answer must be $(0, c)$ for some $c \in \bf R$.  So, we just need to find the line $y = cx$ which is tangent to $y = \log x$ at some point $(x_{0},y_{0})$ with $x_{0}> 1$.  Thus, we have $\frac{\log x_{0}}{x_{0}}= \frac{1}{x_{0}}$ so $x_{0}= e$ and the slope is indeed $c = \frac{1}{e}$.  What do you think is wrong with it?

jmerry -- how so?  The graph of $y = |\log x|$ lies entirely in the first quadrant.

What I did wrong was the graph. I had the strange impression that $\ln (x),x>1$ was convex :blush: . I got that out of my system  :ewpu:


Let $a,b,c$ be positive real numbers such that \[ a+b+c\geq abc. \] Prove that at least two of the inequalities \[ \frac{2}{a}+\frac{3}{b}+\frac{6}{c}\geq6,\;\;\;\;\;\frac{2}{b}+\frac{3}{c}+\frac{6}{a}\geq6,\;\;\;\;\;\frac{2}{c}+\frac{3}{a}+\frac{6}{b}\geq6 \] are true.

This one shoule be a problem of USATST 2001.

Does anyone have a solution to this problem?

[hide = "Elegant Solution"]

Let $x, y, z$ denote $\frac1a, \frac1b, \frac1c,$ respectively, which are clearly positive reals. Then, the problem wants us to show that that if $xy+yz+zx \ge 1$, then at least $2$ of $2x + 3y + 6z \ge 6, 2y+3z+6x \ge 6,$ and $2z + 3x + 6y \ge 6$ are true. We'll prove this by showing its contrapositive. Suppose that $2x+3y+6z < 6, 2y+3z+6x < 6$ are true, then we claim that $xy+yz+zx < 1.$ The problem would then following by proving two symmetric statements. To show this, it suffices by scaling to show that when $2x+3y+6z = 6, 2y+3z+6z \le 6,$ then we have $xy+yz+zx \le 1.$ Then, let $t = 2y+3z+6x \le 6$. We can now easily solve for $y, z$ in terms of $x, y$:

$$y = (2t-6) - 10x, z = (4-t) + \frac{14}{3}x.$$
Then, we have that:

$$xy+yz+zx = -52x^2 + (\frac{61}{3}t-70)x + (2t-6)(4-t),$$

and so since $t = 2y+3z + 6x \ge \frac12 (2x+3y+6z) \ge 3 \Rightarrow \frac{61}{3}t > 70$, we have that this is maximal when $x = \frac{\frac{61}{3}t - 70}{104},$ which we know is positive by $\frac{61}{3}t - 70 > 0.$  Plugging this in for $x$, we have that:

$$xy + yz + zx \le -52 (\frac{\frac{61}{3}t - 70}{104})^2 + (2t-6)(4t-6).$$
This implies that:

$$208(xy+yz+zx) \le -(\frac{61}{3}t - 70)^2 + 208(8t^2 - 36t+36).$$

This is an upwards parabola with a vertex at $\frac{20886}{11255} < 2$, and so hence is maximized when $t = 6$. When $t = 6$, we know that $x = \frac{\frac{61}{3}t - 70}{104} = \frac{122 - 70}{104} = \frac12$, and so plugging $x, t$ into our equations for $y, z$ give that $y = 1, z = \frac13.$ In this case, $xy + yz + zx = 1$, and so since this was our maximal case, we are done.

$\square$

[/hide]


group theory
abstract algebra
superior algebra
superior algebra unsolved

Suppose $G,H$ be two finite group such that $(\mid H \mid ,\mid G \mid)=1$ prove that:
$Aut( G \times H) \simeq Aut(G) \times Aut(H)$

Obviously $Aut(G)\times Aut(H)$ is isomorphic to a subgroup of $Aut(G\times H)$. Show that $G\times\{1\}$ and $\{1\}\times H$ are characteristic subgroups of $G\times H$ if $gcd(|G|,|H|)=1$ and conclude that the injection is in fact surjective.


[url=http://img132.imageshack.us/my.php?image=71591588wz3.jpg][img]http://img132.imageshack.us/img132/9683/71591588wz3.th.jpg[/img][/url]

prove these triangles are similar

[hide="Basic stuff"]
- Identify angles with $ 90^{\circ}$, $ 180^{\circ}$, $ 360^{\circ}$.
- Name the other angles.
- The sum of some of the angles might be another known angle (or even better: $ 90^{\circ}$, $ 180^{\circ}$, $ 360^{\circ}$). Write down these equations.
- Finally, you must know that the sum of the angles of a triangle is $ 180^{\circ}$.
Now try to prove that the two triangles in your problem have the same angles: $ 90^{\circ}$, $ \alpha$, $ 90^{\circ}\minus{}\alpha$.[/hide]


calculus
integration
trigonometry
calculus computations

Find the general i ndefinite integral:

[img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/5f800b9178b904f0b3b972b23677f999.gif[/img]

i first thought to try and make it look a bit more simplestic so i did this:
[img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/ff499a7c9eb9175c8c9176fee4621c04.gif[/img]

so then i get:

[img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/81bc0b1960330dae9117e2ecf0119e2e.gif[/img]

i have no clue where to go from here....any thoughts?

thanks!

You can't integrate part of a product like that.

The correct approach is to rewrite this as

$\int \frac{\sin x}{\cos^{2}x}\, dx$

By Pythagorean identity.  Then you might want to consider a substitution.

If you're still having trouble
[hide]Let $u= \cos x$[/hide]

$\int \frac{sinx}{1-sin^{2}x}\,dx = \int sinx cos^{-2}x\,dx$
$=-\frac{cos^{-1}x}{-1}+C= \frac{1}{cosx}+C$

Carcul


AMC USA(J)MO USAMO inequalities function search MIT

Well this is a very simple problem if viewed from the Titu's Lemma...

Using that we have
the LHS of the inequality to be greater than or equal to 9/2{(a^3+b^3+c^3)+3abc} and then it is just a consequence of the AM GM inequality to prove that the newly obtained expression is greater than or equal to the RHS>

We use $ a^3\plus{}b^3\geq a^2b\plus{}b^2a$ and we can assume abc=1 =>
$ \sum \frac{1}{a^3\plus{}b^3\plus{}abc}\leq \sum \frac{1}{a^2b\plus{}ab^2\plus{}abc}\equal{}\frac{1}{a\plus{}b\plus{}c}\sum \frac{1}{ab}\equal{} \frac{1}{abc}$ QED

A little alteration to the last part of my soln:
The obtained expression can be proved to be less than or equal to the RHS.. :)

To limes 123
How can you assume abc=1
?

The inequality is homogeneous of degree $ \minus{}3$, so each variable can be multiplied by the same constant without affecting the truth or falsity of the inequality.  In other words, we are trying to prove that $ f(a, b, c) \ge 0$ for some function $ f(a, b, c)$ with the property that $ f(ta, tb, tc) \equal{} t^{\minus{}3} f(a, b, c)$, so we can set $ t \equal{} \frac{1}{ \sqrt[3]{abc} }$ to simplify our calculations.  This is a common tactic in solving inequalities.

to Aravind

please post titu's lemma and it wud be gud if u can explain me the bunching thing

here's the reference link

kedlaya page 13 adobe reader

http://www.artofproblemsolving.com/Resources/Papers/KedlayaInequalities.pdf

to t0rajir0u

can u tell me how will u proceed then i'm unable to get it...

thank you in advance

@ sfsreporter

See I don't know what u mean by that bunching thingy.....
Anyway, if u r in US , u can get Mathematical Olympiad Treasures by Titu Andreescu....The lemma is in that book.

Anyway
if a, b are reals and x and y are positive reals ,
then we have
a^2/x + b^2/y >=     (a+b)^2/(x+y) ...
The proof is very simple and left to u as an exercise for algebraic computain and then inductivel prove for  any n no. of terms. :)

The inequality
$ \frac {a_1^2}{b_1} \plus{} \frac {a_2^2}{b_2} \plus{} ... \plus{} \frac {a_n^2}{b_n}\geq \frac {(\sum_{i \equal{} 1}^n a_i)^2}{\sum_{i \equal{} 1}^n b_i}$ can be proved by Cauchy Schwarz.

Why r u complicating the proof???
it can be proved inductively...
Anyway  it is another form of the Cauchy Schwarz??

Aravind,

bunching is nothing else but the process of applying Muirhead...so can u explain how shud we use it...

to b more precise... please explain the one given in te link below on page 12 example 11..

most importantly i would like to understand how do we write the expressions for the symmetric summation..that will clear all my doubts...the one given in the link is not that simple for a 11th std in CBSE can easily understand..i hope u understand my level now ...so please keep the thing such  that i can get what you say.

link:  http://www.artofproblemsolving.com/Resources/Papers/KedlayaInequalities.pdf[/url]

Sorry if this is a spam or something as it out of topic somewhat....

@sfsreporter...
I am a junior to u. I am in 9th CBSE Chennai.
I am not getting opened to the link given by u..But anyway Sigma symmetrics and all those stuff are quite easy to comprehend....Try ur best once more :)

Aravind,

 i'm not ashamed to learn from someone who is junior to me and knows more then i do ':yup:'  ...well now over the days the symmetric summation has become more clear to me..so i think you can ignore the topic..

the link if u want.. go to yahoo...search for muirhead's inequality... go to the AoSPS search result and scroll down to the external links..click on the one By Kiran S Kedlaya...the MIT professor

another thing... you in 9th and know so much.... it's good,you'll do well in life dude!!':coolspeak:'
continue the same way...pm me ur id (gmail or yahoo) we'll sit and chat some day..  ':gathering:'

$ (\sum_{i\equal{}1}^n b_i)(\sum_{i\equal{}1}^n \frac{a_i^2}{b_i})\geq (\sum_{i\equal{}1}^n a_i)^2$ Aravind do you think it is a complicated proof...?

limes123 i think this is what Aravind referred to as Titu's lemma...

here's the attatchment of the lemma


Exploding Dots is an alternative method for examining place value. We’ll look at some exotic methods of representing numbers and see a few applications. Some methods make use of antidots and some require black holes.
[url]
http://metroplexmathcircle.wordpress.com/2009/11/04/nov-7-dr-harold-reiter-exploding-dots/[/url]

There's a stray "[/url]" at the end of the link that's getting included in the web address.


Question, How would you do the problem "What is the sum of the Digits 1 through 1,000?" :oops:  :blush:

Consider how many times each digit appears in each place.  In other words,

- How many times does $0$ appear as the ones digit?  $1$?  $2$?  $3$? ...
- How many times does $0$ appear as the tens digit?  $1$?  $2$?  $3$? ...
- ...

Edit:  Alternate solution.  I'm not really sure how these two approaches compare in terms of how hard they are to understand.

[hide="Solution"]Ignore $1000$ for now and just sum from $000$ through $999$.  Given a number, say $123$, find the number such that each of the second number's digits is $9$ minus the first number's digits (in this case, $876$).  The two numbers have a combined digit total of $(1+8)+(2+7)+(3+6) = 3 \cdot 9 = 27$.  This works no matter what number you start with.

We can pair all $1000$ numbers from $000$ through $999$ this way.  There will be $500$ pairs, each with a digit sum of $27$, so our sum will be

$500 \cdot 27 = 13500$

Now we add in $1000$, whose digits sum to $1$, for our final answer

$13500+1 = \boxed{ 13501 }$. [/hide]

[hide]
From 0-999, the units digit repeat every 10 and go from 0-9, so there are 100 sets of 45 which is 4500.
The tens digit repeats every 100 so there are 10 sets, and each set have 10 (because in each set each one appears 10 times) * 45 (sum of 0-9), then times 10 to include all sets to get 4500.
The hundreds digit is 100 * 45 because each digit appears 100 times and 45 is the sum of 1-9.

So totally there are 4500 * 3 + 1 = 13501.[/hide]

[quote="jhredsox"]wait, where'd the plus 1 come from?
jorian[/quote]

its for the sum of 1000, the first part is just the sum of digits in 1-999


vector
limit
combinatorial geometry
topology

The convex hull of any closed subset of $ \mathbb{R}^{n}$ is closed. True or false?

Hope you all like it.  :wink:

This is false.

Counterexample: $ A : = \left(\mathbb{R} \times \{0\}\right) \cup \{(0,1)\}$. $ A$ is closed.

However, ${ \text{conv.hull}(A) = \mathbb{R} \times [0,1[ \,\,\,\cup \,\,\{(0,1)}\}$, is not closed.

how about closed bounded instead of just closed?

To [b][color=red]Tyl[/color][/b]:

Your answer is perfect. Good work, [i]man[/i]!

[quote="ma_go"]How about closed-bounded instead of just closed?[/quote]

In that case we ascertain that the convex hull of the set is compact, inclusive.  :D

Does anybody here feel like proving this claim of mine?

Of course, "in $ \mathbb{R}^n$" matters there as well. In an infinite-dimensional topological vector space, even compactness might not save you.

yes, actually i meant to extend the question to the topological vector spaces setting (say Banach or Hilbert spaces, if you prefer).

The convex hull of a compact subset of $ \mathbb{R}^n$ is indeed compact. Boundedness follows trivially, and for it being closed, you can prove that by applying Caratheodory's theorem and Bolzano-Weierstrass a bunch of times. I can fill in the details if you'd like, [color=red]coquitao[/color].

[quote="Tyl"] I can fill in the details if you'd like, [color=red]coquitao[/color].[/quote]

Would really like to a look at this proof of yours, [i]man[/i].

Caratheodory's theorem says that in $ \mathbb{R}^d$, any convex combination of points from a subset $ X \subset \mathbb{R}^d$ is a convex combination of at most $ d \plus{} 1$ points (of $ X$).

Let $ A$ be our compact subset of $ \mathbb{R}^n$. Then, as $ A$ is bounded, there exists some ball $ B$ centered around the origin such that $ A \subset B$. Now, as $ B$ is is convex, by definition of the convex hull, we have that $ \text{conv.hull}(A) \subset B$. So $ \text{conv.hull}(A)$ is bounded.

To show $ \text{conv.hull}(A)$ is closed, take some series $ (x_j)_{j\in\mathbb{N}} \subset \text{conv.hull}(A)$ such that $ x_j \to p \in X$. We have to show that $ p \in \text{conv.hull}(A)$.

For this, fix some $ m \in \mathbb{N}$. Then, by Caratheodory's theorem, there exists $ x^m_1, \dots, x^m_{n \plus{} 1} \in A$ and $ \lambda^m_1, \dots, \lambda^m_{n \plus{} 1} \in [0,1]$ such that:

$ x_m \equal{} \sum\limits_{i \equal{} 1}^{n \plus{} 1} \lambda^m_i \, x^m_i$   and  $ \sum\limits_{i \equal{} 1}^{n \plus{} 1} \lambda^m_i \equal{} 1$.

Because both $ A$ and $ [0,1]$ are compact, we can find by Bolzano-Weierstrass a convergent subsequence of the $ x^m_1$. That is, $ \lim\limits_{k\to\infty} x^{m(k)}_1 \equal{} p_1$ for some $ p_1 \in A$. We can do this for all of the $ x^m_i$ and $ \lambda^m_i$, which after repeated applications of the Bolzano-Weierstrass theorem gives some subsequence (which I will again denote by the indices $ m(k)$ instead of something like $ m(k(k_2(k_3(k_4(\dots)))))$), such that as $ k \to \infty$, we have that $ x^{m(k)}_i \to p_i \in A$ and $ \lambda^{m(k)}_i \to \lambda_i \in [0,1]$ for all $ i \in \{1, \dots, n \plus{} 1\}$.

We have for each $ m$ that $ \sum\limits_{i \equal{} 1}^{n \plus{} 1} \lambda^m_i \equal{} 1$, so that also $ \sum\limits_{i \equal{} 1}^{n \plus{} 1} \lambda^{m(k)}_i \equal{} 1$, which gives after letting $ k$ go to infinity that $ \sum\limits_{i \equal{} 1}^{n \plus{} 1} \lambda_i \equal{} 1$.

We can conclude from this that we must have that $ x_{m(k)} \equal{} \sum\limits_{i \equal{} 1}^{n \plus{} 1} \lambda^{m(k)}_i \, x^{m(k)}_i \to \sum\limits_{i \equal{} 1}^{n \plus{} 1} \lambda_i \, p_i$.

As $ x_m \to p$, we have $ x_{m(k)} \to p$, so we conclude that $ p \equal{} \sum\limits_{i \equal{} 1}^{n \plus{} 1} \lambda_i \, p_i$.

As for all the $ i$, we have $ p_i \in A$, indeed we have that $ p$ is a convex combination of elements of $ A$, i.e. we have $ p \in \text{conv.hull}(A)$, which in turn shows that $ \text{conv.hull}(A)$ is closed.

A bounded and closed subset of $ \mathbb{R}^n$ is compact.

(Note: did not fully make this proof up myself. I remembered having proven this before in some course I took, and checked that proof for the idea of applying Bolzano-Weierstrass like this. Also, it seems to be pretty standard.)


trigonometry
absolute value
calculus
calculus computations

The infinite series-- sum from n = 1 to infinity of [n(-1)^n]/[n^2 +2]. When I do the limit as n goes to infinity for the absolute value of the series, and then use L'Hopital's Rule, doesn't that equal the limit as n approaches infinity for 1/(2n), which equals zero, proving absolute convergence?

You are looking at the sum not the sequence.

If you are trying to say that the terms go to zero implies convergence, this is not correct. Famous example $\sum \frac{1}{n}$

Thanks! I now understand why it converges conditionally. But a question about absolute convergence-- with the sum from n = 2 to infinity of [(-1)^n * sin(3n)] / n^2, how can I prove absolute convergence? I know using the nth term divergence test proves that the alternating series does converge, but how can i prove that it does so absolutely?

That's [i]not[/i] an alternating series, because the $\sin$ changes sign. The only tests you have available for a series like this are absolute convergence tests.

Hint: $|\sin(3n)|\le1$. For the purpose of determining absolute convergence, it might as well be 1.

Wow, thanks, I never even thought of that!


I  follow   this    guide    to  include      my  figures        in    my  LaTeX      document:
[url]http://www.mathlinks.ro/Wiki/index.php/Asymptote:_Advanced_Configuration[/url]
Well,
1)I  want      to       include     2   or     3        pictures     in        my      pdf,      like        this:

figure1..........................figure2...................................figure3

In   other       words:    the        3          figures   stand           in           a             row.

2)I      have         only       one          figure         to      include      in     my               pdf.
the     figure   is       on       the      right       of    the    page      and       some      words     on       the        left,       like            this:

wordwordword
wordwordword..............................figure
wordwordword

How        to         get          this?

Many             thanks!

[b][color=red]Thank  you very much! especially  stevem and  fedja[/color][/b]

[url]https://sourceforge.net/forum/forum.php?thread_id=3329976&forum_id=409349here[/url]


$ p$ is odd,
then $ p| C_p^k , k\equal{}1,\cdots,p\minus{}1$ ?

thanks

you know that $ C_p^k \equal{} \frac {p(p \minus{} 1)\cdots (p \minus{} k \plus{} 1)}{k!}$ is an integer. every one of the numbers $ 1, ..., k$ is relatively prime to $ p$, so $ k!$ must divide $ (p \minus{} 1)\cdots(p \minus{} k \plus{} 1)$. so when reducing the fraction, the $ p$ on top remains and thus $ p|C_p^k$.


inequalities
triangle inequality
inequalities proposed

I think every method have some  interesting dots . And If we want use it , I think that is hard word . Must more training . With Cauchy Schwarz is  nice & popular , but want to use it a expert way ,we must have creative for example can_hang2007  :lol: .
In past & today , I also only know to use C-S a direct way so I still use a amatuer way and don't know make it to stronger because i'm a stupid student  :( . Who can help me about it ?
And who can share your experience  , pls ?
Thanks! :)

[quote="Sunkern_sunflora"]I have seen almost all the links provided by u, can_hang2007. U r working pretty good, as onlylove_math mentioned. :)
I remember that earlier I proved $ A.M.\geq G.M.$ with help of $ Jensen$, it too seems a pretty strong ineq for me. However, in spite of proving many elementary ineq from $ Jensen$, I failed to prove $ C.S.$ from it, and thereby I believe that $ C.S.$ is the strongest of all ineqs.
I even believe that we might deduce $ A.M.\geq G.M.$ from $ C.S.$. What do u say, can_hang?[/quote]
Dear sunkern,
In my opinion, Cauchy Schwarz is a nice and strong inequality and I like it very much (the most in all inequalities I like). But I dont think it is the strongest, since we can prove it using AM-GM inequality.

I know, but still I think someday we might do the reverse, I mean $ A.M.\geq G.M.$ from $ C.S.$.
Don't take it otherwise respected can_hang sir, I told that it is my personal belief, it may be false, but I love the ineq and I love to think it as the STRONGEST of all. Thank u.
Can u proof $ C.S.$ from $ Jensen$? Can u discuss about some generalisation of it?

I edit my post, sorry, but did u say can_hang sir that we can proof $ C.S.$ from $ A.M.\geq G.M.$? Can u post the proof here please?

If we want to prove Cauchy-Schwarz by AM-GM, we may use the induction.
When $ n\equal{}1$ the equality of C.-S. hold. When $ n\equal{}2$, the inequlity we have to prove is

$ a_1b_1\plus{}a_2b_2\leq\sqrt{a_1^2\plus{}a_2^2}\sqrt{b_1^2\plus{}b_2^2}$

$ (a_1b_1\plus{}a_2b_2)^2\leq(a_1^2\plus{}a_2^2)(b_1^2\plus{}b_2^2)$

$ \Longleftrightarrow(a_1b_1\plus{}a_2b_2)^2\leq(a_1^2\plus{}a_2^2)(b_1^2\plus{}b_2^2)$

$ \Longleftrightarrow2a_1a_2b_1b_2\leq a_1^2b_2^2\plus{}a_2^2b_1^2$

The last inequality follows from the AM-GM inequality. Thus the inequality holds for $ n\equal{}2$.

The remainder is left to others.

[hide="Hint"]When you prove the inequality holds for $ n\equal{}k\plus{}1$, you may use the inequalities for $ n\equal{}2$ and $ n\equal{}k$.[/hide]

[quote="ifai"]   $ \Longleftrightarrow(a_1b_1\plus{}a_2b_2)^2\leq(a_1^2\plus{}a_2^2)(b_1^2\plus{}b_2^2)$
$ \Longleftrightarrow2a_1a_2b_1b_2\leq a_1^2b_2^2\plus{}a_2^2b_1^2$[/quote]
Will u plz detail this step, I understand everything else pretty clear.

What about $ C.S.$ from $ Jensen$?
 :)

[quote="Sunkern_sunflora"][quote="ifai"]   $ \Longleftrightarrow(a_1b_1 \plus{} a_2b_2)^2\leq(a_1^2 \plus{} a_2^2)(b_1^2 \plus{} b_2^2)$
$ \Longleftrightarrow2a_1a_2b_1b_2\leq a_1^2b_2^2 \plus{} a_2^2b_1^2$[/quote]
Will u plz detail this step, I understand everything else pretty clear.

What about $ C.S.$ from $ Jensen$?
 :)[/quote]
Just do some algebraic computation:

$ (a_1b_1 \plus{} a_2b_2)^2\leq(a_1^2 \plus{} a_2^2)(b_1^2 \plus{} b_2^2),$

$ \{a_1^2b_1^2 \plus{} a_2^2b_2^2\} \plus{} 2a_1a_2b_1b_2\leq \{a_1^2b_1^2 \plus{} a_2^2b_2^2\} \plus{} a_1^2b_2^2 \plus{} a_2^2b_1^2,$

$ 2a_1a_2b_1b_2\leq a_1^2b_2^2 \plus{} a_2^2b_1^2.$

[quote="kunny"]$ (a_1^2 + a_2^2)(b_1^2 + b_2^2)\geq (a{}_1b{}_1 + a{}_2b{}_2)^2$.

By using with $ 2ab\leq a^2 + b^2$ with ${{ a = \frac {a_i}{\sqrt {a_1^2 + a_2^2}}}, b = \frac {b_i}{\sqrt {b_1^2 + b_2^2}}}$ for $ i = 1,\ 2$.[/quote]
In general, we can prove $ \left|\sum_{i=1}^n {a_i b_i}\right| \leq \sqrt{\sum_{i=1}^n{a_i^2}}\sqrt{\sum_{i=1}^n{b_i^2}}$ by the same way.

Assume $ \Sigma a_i^2, \Sigma b_i^2$ are nonzero. It follows from $ st\leq\frac{s^2}2+\frac{t^2}2$ for $ x,y\in\mathbb R$ with $ s=|a_i|$ and $ t=|b_i|$ that

$ \sum_{i=1}^n |a_i b_i|\leq\frac12\sum_{i=1}^n a_i^2+\frac12\sum_{i=1}^n b_i^2.$  (*)

Now we let $ a'_i={a_i}/{\sqrt{\Sigma a_j^2}}$ and $ b'_i={b_i}/{\sqrt{\Sigma b_j^2}}$, then

$ \sum_{i=1}^n a'_i^2=\sum_{i=1}^n b'_i^2=1.$

By (*) we have $ \Sigma |a'_i b'_i|\leq 1$, i.e.

$ \sum_{i=1}^n \left| \frac{a_i}{\sqrt{\Sigma a_j^2}} \frac{b_i}{\sqrt{\Sigma b_j^2}}\right|\leq 1.$

Thus $ \sum |a_i b_i| \leq \sqrt{\sum a_j^2}\sqrt{\sum b_j^2}$.

Finally, by using the triangle inequality $ |\Sigma a{}_ib{}_i|\leq \Sigma |a{}_ib{}_i|$, we are done.

[quote="can_hang2007"]I created this topic not to solve the problems, I just want us to discuss how useful is the Cauchy Schwarz Inequality?
In my opininion, this and AM-GM Inequality are the nicest in inequality, they provided us a very elementary and classical solution. The readers can easily check the solution and understand the idea. It does not need any nonelementary knowledge when proving inequality using this. And as you see, in many problems, I stated that they could be proved by Cauchy Schwarz...
I think it is a nice inequality and we need to do more research about it.
What do you think?[/quote]
I remember one more very nice application of C-S by Zhaobin:
http://www.mathlinks.ro/viewtopic.php?t=109029

[quote="can_hang2007"]I created this topic not to solve the problems, I just want us to discuss how useful is the Cauchy Schwarz Inequality?
In my opininion, this and AM-GM Inequality are the nicest in inequality, they provided us a very elementary and classical solution. The readers can easily check the solution and understand the idea. It does not need any nonelementary knowledge when proving inequality using this. And as you see, in many problems, I stated that they could be proved by Cauchy Schwarz...
I think it is a nice inequality and we need to do more research about it.
What do you think?
Here is some links:
http://www.mathlinks.ro/viewtopic.php?t=208794
http://www.mathlinks.ro/viewtopic.php?t=218503
http://www.mathlinks.ro/viewtopic.php?t=216132
http://www.mathlinks.ro/viewtopic.php?t=218850
http://www.mathlinks.ro/viewtopic.php?t=216986
http://www.mathlinks.ro/viewtopic.php?t=218380
http://www.mathlinks.ro/viewtopic.php?t=218465
.............
and my solution for IMO 2008: http://www.mathlinks.ro/viewtopic.php?t=215222&start=20[/quote]
I agree with you,my dear
But in my opinion,i feel C-S is the nicest ineq in Math,in the past,i really dílike it because i am very stupid,but now,since i have known you ,i found (and will find) many nice features of CS ( Although i cant use it like you :blush: )
Every body, i know Can ,the person who develop CS to a strong way,(i think he will mention it in his new book)
The proof of CS,ofcourse,we dont need AMGM( but if we use it,we will have a nicer one :wink: )
Can_hang2007,we are waiting your CS technique and your interseting book,thanks

Hello can_hang sir! Here's my proof of $ A.M. \geq G.M.$ from $ C.S.$. I am trying to prove it for two variables. I think we can generalize for $ n$ variables as well.
 :)

Let $ a,b$ be two positive real nos.
From $ C.S.$ we get;

$ a^2\plus{}b^2 \geq \frac{(a\plus{}b)^2}{2}$
$ \Rightarrow a^2\plus{}b^2 \geq \frac{a^2\plus{}b^2}{2}\plus{}ab$
$ \Rightarrow a^2\plus{}b^2 \geq \frac{(a\plus{}b)^2}{4}\plus{}ab$
$ \Rightarrow a^2\plus{}b^2 \geq \frac{a^2\plus{}b^2}{4}\plus{}\frac{1}{2}ab\plus{}ab$

Continuing likewise after $ n$ steps;

$ a^2\plus{}b^2 \geq \frac{a^2\plus{}b^2}{2^n}\plus{}ab(1\plus{}\frac{1}{2}\plus{}\frac{1}{4}\plus{}...\plus{}\frac{1}{2^{n\minus{}1}})$

Thus, when $ n\rightarrow \infty$ we get;

$ \frac{a^2\plus{}b^2}{2^n}\rightarrow{0}$ and the infinite G.P. tends to $ 2ab$. Basically it is,

$ a^2\plus{}b^2 \geq 2ab$ which is nothing but $ A.M. \geq G.M.$. :)

[quote="Sunkern_sunflora"]Hello can_hang sir! Here's my proof of $ A.M. \geq G.M.$ from $ C.S.$. I am trying to prove it for two variables. I think we can generalize for $ n$ variables as well.
 :)

Let $ a,b$ be two positive real nos.
From $ C.S.$ we get;

$ a^2 + b^2 \geq \frac {(a + b)^2}{2}$
$ \Rightarrow a^2 + b^2 \geq \frac {a^2 + b^2}{2} + ab$
$ \Rightarrow a^2 + b^2 \geq \frac {(a + b)^2}{4} + ab$
$ \Rightarrow a^2 + b^2 \geq \frac {a^2 + b^2}{4} + \frac {1}{2}ab + ab$

Continuing likewise after $ n$ steps;

$ a^2 + b^2 \geq \frac {a^2 + b^2}{2^n} + ab(1 + \frac {1}{2} + \frac {1}{4} + ... + \frac {1}{2^{n - 1}})$

Thus, when $ n\rightarrow \infty$ we get;

$ \frac {a^2 + b^2}{2^n}\rightarrow{0}$ and the infinite G.P. tends to $ 2ab$. Basically it is,

$ a^2 + b^2 \geq 2ab$ which is nothing but $ A.M. \geq G.M.$. :)[/quote]

 :rotfl: you could have spared everything between the first and the last line... just multiply the first line by 2  :lol:

[quote="Sunkern_sunflora"]Hello can_hang sir! Here's my proof of $ A.M. \geq G.M.$ from $ C.S.$. I am trying to prove it for two variables. I think we can generalize for $ n$ variables as well.
 :)

Let $ a,b$ be two positive real nos.
From $ C.S.$ we get;

$ a^2 \plus{} b^2 \geq \frac {(a \plus{} b)^2}{2}$
$ \Rightarrow a^2 \plus{} b^2 \geq \frac {a^2 \plus{} b^2}{2} \plus{} ab$
$ \Rightarrow a^2 \plus{} b^2 \geq \frac {(a \plus{} b)^2}{4} \plus{} ab$
$ \Rightarrow a^2 \plus{} b^2 \geq \frac {a^2 \plus{} b^2}{4} \plus{} \frac {1}{2}ab \plus{} ab$

Continuing likewise after $ n$ steps;

$ a^2 \plus{} b^2 \geq \frac {a^2 \plus{} b^2}{2^n} \plus{} ab(1 \plus{} \frac {1}{2} \plus{} \frac {1}{4} \plus{} ... \plus{} \frac {1}{2^{n \minus{} 1}})$

Thus, when $ n\rightarrow \infty$ we get;

$ \frac {a^2 \plus{} b^2}{2^n}\rightarrow{0}$ and the infinite G.P. tends to $ 2ab$. Basically it is,

$ a^2 \plus{} b^2 \geq 2ab$ which is nothing but $ A.M. \geq G.M.$. :)[/quote]
I think that youre thinking too far, sunkern.
Since at the first line of your proof, the AM-GM for 2 variable imediatly follows...

I guess there can never be a better name for me than $ IDIOT$! Sorry! :(  :(  :(


In a trapezoid a parallel is drawn through the point of intersection of diagonals to the
parallel sides to be terminated by the other two sides at X and Y.Then

a)XY is the A.M of the lengths of the parallel sides
b)XY is the G.M of the lengths of the parallel sides
c)XY is the H.P of the lengths of the parallel sides
d)none of these

note:please tell me when the line passes through the point is it the midpoint or not

$AB \parallel CD$
The diagonals intersect at $O$

$\triangle DXO \sim \triangle DAB \Rightarrow \frac{XD}{AD}= \frac{XO}{AB}$

$\triangle AXO \sim \triangle ADC \Rightarrow \frac{XA}{AD}= \frac{XO}{CD}$

$\frac{XO}{AB}+\frac{XO}{CD}= \frac{XD}{AD}+\frac{XA}{AD}= \frac{AD}{AD}= 1 \Rightarrow$

$XO (\frac{1}{AB}+\frac{1}{CD}) = 1 \Rightarrow$

$XO \frac{AB+CD}{AB \cdot CD}= 1 \Rightarrow$

$XO = \frac{AB \cdot CD}{AB+CD}$

similarly we find that $OY = \frac{AB \cdot CD}{AB+CD}$ (So $O$ is the midpoint of $XY$)

Finally $XY = 2 \frac{AB \cdot CD}{AB+CD}\Rightarrow \frac{1}{XY}= \frac{AB+CD}{2\cdot AB \cdot CD}\Rightarrow $

$\frac{1}{XY}= \frac{1}{2}\left(\frac{1}{AB}+\frac{1}{CD}\right)$

So, it is the harmonic mean of $AB,CD$


polynomial
function
number theory
complex numbers

I know that we can easily prove that $ (a^2\plus{}b^2)(c^2\plus{}d^2)\equal{}(ac\minus{}bd)^2\plus{}(ad\plus{}bc)^2$ by simply multiplying it all out, but my dad says there is a way to factor the LHS and rearrange to prove the equation to be true. Does anyone know how you would go about doing that?

um sorry if I sound dumb, but how is it different? Wouldn't I have to multiply out the LHS to factor it anyway?

Well, let me post the original problem that my Dad presented (its from a number theory book):

"Prove that $ (a^2\plus{}b^2)(c^2\plus{}d^2)\equal{}(ac\minus{}bd)^2\plus{}(ad\plus{}bc)^2$. HINT: If you know about complex numbers, one approach is to factor the left side and rearrange."

The main thing to note is that if $ z=a+bi$, then $ z\bar{z}=|z|^2=a^2+b^2$, where $ \bar{z}=a-bi$, the complex conjugate of $ z$. Then: \begin{eqnarray*}(a^2+b^2)(c^2+d^2)&=&(a+bi)(a-bi)(c+di)(c-di)\\
&=&(ac-bd+adi+bci)(ac-bd-adi-bci)\\
&=&(ac-bd)^2+(ad+bc)^2\end{eqnarray*}

Oh I see what you mean. Somebody recently posted a problem (I can't find it anymore) on the forum that went something like this:

If a real integer $ k$ can be represented as the sum of two squares, and $ q$ can be represented likewise, prove that $ kq$ can also be represented as such.

That would require the same setup. Anyway the non-complex way:

\begin{eqnarray*}(a^2 + b^2)(c^2 + d^2)&=&a^2c^2 + a^2d^2 + b^2c^2 + b^2d^2\\
&=&(ac)^2 - 2(ac)(bd) + (bd)^2 + (ad)^2 + 2(ad)(bc) + (bc)^2\\
&=&(ac - bd)^2 + (ad + bc)^2\end{eqnarray*}

The problem is posted here:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=199769[/url]

[b]In the same vein:[/b]  Prove that if $ p, q$ are two integers that can be represented in the form $ x^2 \plus{} xy \plus{} y^2$ for integers $ x, y$, then so can $ pq$.

[hide="Hint"] Think [url=http://mathworld.wolfram.com/EisensteinInteger.html]Eisenstein integers[/url].  The original problem is about [url=http://mathworld.wolfram.com/GaussianInteger.html]Gaussian integers[/url].  :) [/hide]

[hide="ehh I fail.."]
my non-Eisensteinian, non-Gaussian, limited ramblings:

\begin{eqnarray*}(a^2 + ab + b^2)(c^2 + cd + d^2) & = & (ac)^2 + (ad)^2 + (bc)^2 + (bd)^2 + a^2cd + b^2cd + c^2ab + d^2ab + abcd \\
& = & (ac + bd)^2 + (ad + bc)^2 + (ad)(ac) + (bc)(bd) + (ac)(bc) + (ad)(bd) + abcd - 4abcd \\
& = & (ac + bd)^2 + (ac + bd)(ad + bc) + (ad + bc)^2 - 3abcd \end{eqnarray*}

So are there no ways to do this so I don't end up with that $ 3abcd$ at the end unless I use Eisenstein integers?
[/hide]

EDIT: [hide="lmao"]
Using some of the values given on mathworld,
\begin{eqnarray*}(a^2 + ab + b^2)(c^2 + cd + d^2) & = & (a - b\omega)(a - b\omega^2)(c - d\omega)(c - d\omega^2) \\
& = & [(a - b\omega)(c - d\omega^2)][(a - b\omega^2)(c - d\omega)] \\
& = & [(ac + bd\omega^3) - (bc\omega + ad\omega^2)][(ac + bd\omega^3) - (bc\omega^2 + ad\omega)] \\
& = & (ac + bd)^2 - (ac + bd)(ad\omega + ad\omega^2 + bc\omega + bc\omega^2) + (bc\omega + ad\omega^2)(bc\omega^2 + ad\omega) \\
& = & (ac + bd)^2 - (ac + bd)(ad(\omega + \omega^2) + bc(\omega + \omega^2)) + (bc)^2\omega^3 + (ad)(bc)(\omega^2 + \omega^4) + (ad)^2\omega^3 \\
& = & (ac + bd)^2 - (ac + bd)( - ad - bc) + (bc)^2 + (ad)(bc)((\omega + \omega^2)^2 - 2\omega^3) + (ad)^2 \\
& = & (ac + bd)^2 + (ac + bd)(ad + bc) + (ad)^2 + 2(ad)(bc) + (bc)^2 - 3(ad)(bc) \\
& = & (ac + bd)^2 + (ac + bd)(ad + bc) + (ad + bc)^2 - 3abcd \\
\end{eqnarray*}

I'm definitely doing something wrong :rotfl:
[/hide]

The resulting linear combinations are not straightforward like they are in the Gaussian integer version; this is not an easy problem to do without the motivation from the Eisenstein integers.

Edit:  In response to your edit, you need to multiply $ a \plus{} b \omega$ by $ c \plus{} d \omega$ and then write the answer in the form $ e \plus{} f \omega$ using the identity $ \omega^2 \equal{} \minus{} \omega \minus{} 1$.  To help you along, here is a [hide="cleaned up proof for Gaussian integers"] Define $ \overline{a \plus{} bi} \equal{} a \minus{} bi$.  This function is multiplicative, and so is $ N(a \plus{} bi) \equal{} (a \plus{} bi)( \overline{a \plus{} bi} ) \equal{} a^2 \plus{} b^2$.  So since

$ zw \equal{} (a \plus{} bi)(c \plus{} di) \equal{} (ac \minus{} bd) \plus{} (bc \plus{} ad) i$

we know that

$ N(zw) \equal{} (a \plus{} bi)(c \plus{} di)(a \minus{} bi)(c \minus{} di) \equal{} (a^2 \plus{} b^2)(c^2 \plus{} d^2) \equal{} (ac \minus{} bd)^2 \plus{} (bc \plus{} ad)^2$. [/hide]


Find natural numbers x and y which satisfy 79x + 179y = 7319.

I don't know how to solve this.

hello, see here:
http://mathworld.wolfram.com/DiophantineEquation.html
Sonnhard.

[quote="AIME15"]Find natural numbers x and y which satisfy 79x + 179y = 7319.[/quote]

Take this $ \bmod{79}$ to get

$ 21y\equiv51\equiv525\pmod{79}\implies y\equiv25\pmod{79}$

$ \implies y\equal{}79n\plus{}25$ for some integer $ n$.

Now, just plug this into the original equation and solve for $ x$

to get $ 79x\plus{}14141n\plus{}4475\equal{}7319\implies 79x\equal{}2844\minus{}14141n\implies x\equal{}36\minus{}179n$.

Since $ x,y\geq1$, we have

$ 79n\plus{}25\geq1\implies n\geq\minus{}0.303\ldots\implies n\geq0$, and

$ 36\minus{}179n\geq1\implies n\leq 0.1955\ldots\implies n\leq 0$,

so the only solution occurs when $ n\equal{}0\implies(x,y)\equal{}(36,25)$.


Prove that $ \left (\frac{a\plus{}1}{b\plus{}1}\right)^{b\plus{}1}>\left(\frac{a}{b}\right )^b$, $ a,b>0$, $ a\neq b$.

[hide]$ \frac{ b \frac{a}{b} \plus{} 1 \cdot 1}{b\plus{}1} \geq \sqrt[b\plus{}1]{\left(\frac{a}{b}\right)^b \cdot 1}$ by weighted AM-GM. Equality iff $ \frac{a}{b} \equal{} 1$ which doesn't occur if $ a \neq b$ so the inequality is strict.[/hide]

[quote="Ognjen09"]Prove that $ \left (\frac {a \plus{} 1}{b \plus{} 1}\right)^{b \plus{} 1} > \left(\frac {a}{b}\right )^b$, $ a,b > 0$, $ a\neq b$.[/quote]

You may go to
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1310634#1310634[/url]

as reference.


polynomial
number theory unsolved
number theory

i could do with some help with this one if someone has a nice proof.
is it true or not?

let $ a$ and $ b$ are two positive integers that can be represented in a form: $ a\equal{}x^2\plus{}xy\plus{}y^2$ (where $ x,y$ are integers)
can their product always be represented in the desired way?

Tnx in advance

Yes.  The underlying principle here is the norm in the [url=http://mathworld.wolfram.com/EisensteinInteger.html]Eisenstein integers[/url].

ah yes! i was thinking about eisenstein integers just didn't realize it's so easy. forgot to realize norm is multiplicative. tnx for reminding
anyway we get the following result

$ (a^2\plus{}ab\plus{}b^2)(x^2\plus{}xy\plus{}y^2)\equal{}(ax\minus{}by)^2\plus{}(ax\minus{}by)(ay\plus{}yb\plus{}bx)\plus{}(ay\plus{}bx\plus{}by)^2$


To convert Kelvin temperatures to Celcius, subtract 273.15. The lowest recorded temperature on the surface of Earth was $ \minus{}88^\circ$ C, recorded on Antarctica. How many degrees was this on the Kelvin scale? Express your answer as a decimal to the nearest hundredth.

[quote="$LaTeX$"]You add 273.15 to -88 go get degrees Kelvin. Thus the answer to the nearest hundredth is 185.15.[/quote]

Hi $LaTeX$, lemme $\LaTeX$ify that for ya!

You add $273.15$ to $-88$ get degrees in Kelvin. Thus the answer, to the nearest hundredth, is $\boxed{185.15}$.


Isolate the boy $ B_{1}$ and girl $ G_{1}$ who must not sit together.

Arrange the $ 5$ boys in $ 5!$ ways.

Select $ 3$ of the $ 6$ spaces between the boys or beyond them in $ \binom{6}{3}$ ways. Then arrange the $ 3$ girls in thise $ 3$ places in $ 3!$ ways.

We now have $ 9$ spaces in which we can put the pair and then arrange them in $ 2!$ ways, giving a total of $ 5! \times \binom {6}{3} \times 3! \times 9 \times 2! \equal{} 259200$

With no restrictions, there are $ 10! \equal{} 3628800$ ways to seat the $ 10$ children.

Therefore the required number is $ 3628800 \minus{} 259200 \equal{} 3369600$

[quote="marcopolo"]6 boys and 4 girls are to be seated in a row. How many ways are there if a particular boy and a particular girl do not sit beside each other? Please explain with solution :)[/quote]

Lets use complementary counting. Lets do the problem with no restrictions first. So, basically, 10 people to be seated in a row:
$ 10x9x8x7...x2x1\equal{}10!$

Now, lets make it so that one particular girl and one particular boy must sit together:
The way to think about this is, we mold those two kids into one superkid. and then arrange nine people as $ 9!$. Now, we account for the interchanging the two kids and we get $ 9!x2!$ as the answer.

Now, we subtract: $ 10!\minus{}(9!\minus{}2!)$ and the answer is: 3265922


function
algebra
floor function
number theory open
number theory

for $ n\in\mathbb{N}$ let $ f(n)\equal{}$ the highest number of consecutive natural numbers such that none of them has sum of digits divisible of $ n$.

find explicit formula or any reccurent relation for $ f(n)$.

Assuming the numbers are written in base 10, we can see that if n is a digit, f(n) = n - 1. And that is achieved  If n > 9, then write n=9k+a, where a is between 1 and 8 (a is a residue modulo 9). We can see that the minimum number whose sum is n is (a+1)(10^k) - 1. In a set of (a+1)(10^k) - 1, the k last digits will have taken every value from 0-9, and the last digit (the leftmost) will have taken at least a value, that implies that in every (a+1)(10^k) - 1 consecutives there will be an element which sum of digits will divide n=9k+a. So, (a+1)(10^k) - 2 is the highest value for a collection of consecutives and then the function would be:

If n is a digit, f(n) = n - 1
If n>9, 0<a<9 f(9k+a) = (a+1)(10^k) - 2. There can be a function with only n involved, by substituting
a= n - 9(floor function of n/9) and k = (floor function of n/9).

thanks for reply, but I think something is wrong there :(

$ f(5)\equal{}4$ is not true because we have 8 numbers: 6,7,8,9,10,11,12,13,
obviously 5 does not divide the sum of digits of this numbers
so $ f(5)\ge 8$ ..

f(1)=0 because 1 divides every integer.
f(2)=2 because the sum of consecutive digits in the same ten's, will have consecutive sums of digits. So we can have at most 1 number in one ten. We can take as an example 29 and 30.
f(3)=2 because every 3 consecutive integer there is a multiple of 3. we can take 1,2
f(4)=6 we can have at most 4 in every ten, a max would be 6. It can be achieved with 997, 998, 999, 1000, 1001, 1002.
f(5)=8 we can have at most 5 consecutives in every ten, that leaves us with a max=8 with can be achieved with the example you gave.
f(6)=10 if we want to involve more than 5 digits by including a ten, max=10, it can be achieved with 16, 17, 18,..., 25.
f(7)=12. we can only have less than 7 consecutives per ten, so our max= 12, is achieved with 994 - 1005.
f(8)= 14. max would be 14, achieved with 9999993 - 10000006
f(9)=8 because 9 divides the sum of digits of a number if and only it's sum is divisible by 9. since in every 9 consecutives there is a multiple of 9, max=8, it can be achieved with 10, 11,...,18.


Find all function $ f: \mathbb R^\plus{}\rightarrow\mathbb R^\plus{}$ satisfying $ f(x\plus{}f(y))\equal{}f(x\plus{}y)\plus{}f(y)$ for every $ x,y\in \mathbb R^\plus{}$

$ x \rightarrow f(x)$ give us:
$ f(f(x) \plus{} f(y)) \equal{} f(f(x) \plus{} y) \plus{} f(y) \equal{} f(x) \plus{} f(y) \plus{} f(x \plus{} y)$, and then, in the original equation:
$ f(f(x \plus{} f(y))) \equal{} f(f(y \plus{} x) \plus{} f(y)) \equal{} f(y \plus{} x) \plus{} f(y) \plus{} f(y \plus{} x \plus{} y) \equal{} f(y \plus{} x) \plus{} f(y) \plus{} f(2y \plus{} x)$.
It is easy to check that $ f(f(y)) \equal{} 2f(y)$ by replacing $ x \equal{} 0$ in the original equation, wich means $ 2f(x \plus{} f(y)) \equal{} f(y \plus{} x) \plus{} f(y) \plus{} f(2y \plus{} x)$, but $ f(x \plus{} f(y)) \equal{} f(y) \plus{} f(y \plus{} x)$, and we get:
$ f(y) \plus{} f(y \plus{} x) \equal{} f(2y \plus{} x)$, wich is Cauchy's function (put $ y \equal{} a$ and $ y \plus{} x \equal{} b$, assuming without lost of generality that $ b \geq a$), and we are done! $ f(x) \equal{} cx$, and replacing it in the original equation, we get $ c \equal{} 2$.

Oh, Cauchy's function is to integer $ x$. I'm  stupid!
I am convinced that we are able to prove that $ f(x)\equal{}2x$, since it is true to integers (perhaps using inequalities).It's too late and I must sleep, I can't go further. I'm really sorry by my mistake.

Morever, you replaced x=0, but $ x\in\mathbb R^+$!!!  :(

I though $ \mathbb R^\plus{}$ is about $ \mathbb{R} \geq 0$, the non-negative real numbers (wich means including 0).
http://www.springerlink.com/content/w141805x12613537/
Explain me, please, if I'm wrong!

In my country, $ \mathbb R^\plus{}$ is the set of positive real numbers. Please solve it with this convention.

Making $ x : \equal{} f(y)$ (substituting $ x$ by $ f(y)$), we get

$ f(2f(y)) \equal{} f(f(y) \plus{} y) \plus{} f(y) \equal{} f(2y) \plus{} 2f(y)$

Now let us make $ y : \equal{} 2f(y)$.

$ f(x \plus{} 2f(y) \plus{} f(2y)) \equal{} f(x \plus{} 2f(y)) \plus{} f(2y) \plus{} 2f(y)$
$ \Longrightarrow f(x \plus{} 4y) \plus{} 2f(y) \plus{} f(2y) \equal{} f(x \plus{} 2f(y)) \plus{} 2f(y) \plus{} f(2y)$
$ \Longrightarrow f(x \plus{} 4y) \equal{} f(x \plus{} 2f(y))$
$ \Longrightarrow f(x \plus{} 4y) \equal{} f(x \plus{} 2y) \plus{} 2f(y)$

Then $ f$ is increasing. In particular, $ f$ is injective. The third line in the chain above gives us the answer.

Any mistake?


Prove that $\frac{1}{n}$ for $n\ge2$ is a terminating decimal if and only if $n$'s only distinct prime factors are $2$ and $5$.

If $\frac{1}{n}$ terminates at some digit $k$ after the decimal point, then the number $\frac{10^{k}}{n}$ is an integer, but this is only true if $n|10^{k}$, ie. $n=2^{a}5^{b}$ for some positive integers $a$, $b$.


4. An arc is drawn in a square with center D at one of the vertices of the square and the arc is
tangent to the opposite two sides of the square. The arc is also tangent to the hypotenuse of $ \triangle BFE$ as shown, where $ \overline{BF} \equal{} 1$. What is the radius of the circle?

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Let the radius of the circle be $ r$ and let the point of tangency be $ T$. Draw $ \overline{DE}$, $ \overline{DT}$, and $ \overline{DF}$. We know that the side length of the square is $ r$ and we also know that $ DT \equal{} r$ and that $ \overline{DT} \perp \overline{EF}$. We have $ BE \equal{} \sqrt{3}$ and $ EF \equal{} 2$ because of the 30-60-90 right triangle. It follows that $ AE \equal{} r \minus{} \sqrt{3}$ and that $ FC \equal{} r \minus{} 1$. Suppose we consider using the area to our advantage. The sum of the areas of the triangle is equivalent to the area of the square. Therefore,

$ \frac{2r \plus{} r(r \minus{} 1) \plus{} r(r \minus{} \sqrt{3}) \plus{} \sqrt{3}}{2} \equal{} r^2\\
2r^2 \plus{} (1 \minus{} \sqrt{3})r \plus{} \sqrt{3} \equal{} 2r^2\\
r \equal{} \frac{\sqrt{3}}{\sqrt{3} \minus{} 1}\\
r \equal{} \boxed{\frac{3 \plus{} \sqrt{3}}{2}}$
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Thanks I just wanted to make sure I had it right.

I did it in a more simplistic manner and just equated it after using POP and the 30-60-90 properties.

If G is the tangent point of the hypotenuse, then the problem becomes $ \overline{EG}\plus{}\overline{GF}  \equal{}2$

[quote="mathemagician1729"]Thanks I just wanted to make sure I had it right.

I did it in a more simplistic manner and just equated it after using POP and the 30-60-90 properties.

If G is the tangent point of the hypotenuse, then the problem becomes $ \overline{EG} \plus{} \overline{GF} \equal{} 2$[/quote]

Then, it is almost solved. Since $ (r \minus{} \sqrt {3}) \plus{} (r \minus{} 1) \equal{} 2$, $ r \equal{} \frac {3 \plus{} \sqrt{3}}{2}$.

Yes, that was my method, I just wanted to see if there was a more elegant way...

There's another method based on the excircle formula.

I used it for this same problem before.
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1380786#1380786

you can also see page 4 of [url=http://www.math.uncc.edu/~hbreiter/AHSME/1997so.pdf]this[/url].


$ \cos(B \minus{} C) \equal{} \frac {2\sin(B)\sin(C)}{\sin^2(A)} \Rightarrow ABC$ is a right triangle

$ \sin^2(B)\equal{}\cos (C) \cos(A\minus{}B)  \Rightarrow ABC$ is a right triangle


induction
number theory proposed
number theory

By induction we immediately get $\frac{n}{n+k}$.