probability
LaTeX
geometry
Pythagorean Theorem

captain sugar decides to go to a donut store. the donuts there are either stale, discounted, or stale and discounted. if he buys a dozen random donuts, what is the chance that less than half are stale?

is the chance of being stale or discounted $\frac{1}{2}?$

stale, 1/3, discounted, 1/3, stale and discounted, 1/3.

This answer seems too small... but...

[hide] Is the answer .0664476395 etc.? My calculator can't even fractionatize it. I got that answer using two different methods... one involving total number of ways and number of ways where 7 or more are discounted, and the other using sheer probability.
[/hide]

When I said "sheer probability" i meant the way that lotrgreengrapes did.

yeah i used lotrgreengrapes' method but I was too lazy to write out all the latex

[quote="Captain Sugar"]my modem is broken, so i can't read lotrgreengrapes's answer. i'm assuming it's right.

the next question:
horrified, Captain Sugar finds that the donut factory is manufacturing Vegemite, the renowned Blandest Substance in the Universe!  while sneaking around in the "employees only" part of the donut store, the bland smell knocks him out, but not before he calls the police.

there are four squad cars and a helicopter available. Captain Sugar and the donut shop are located at (0,0). all squad cars travel at 80 mph, and the helicopter travels at 250 mph.  the first squad car is located at (9, 17), the second at (4,24), the third at (16, 12), and the fourth at (6, 19). the helicopter is located at (40, 58).  the helicopter flies straight to the scene. the squad cars must follow the roads, which are the grid lines. which police unit gets there first?

edit: it's also kind of helpful to know that each city block is a mile long.[/quote]

[hide]OK, so the first car travels 26 miles, the 2nd 28, 3rd 28, 4th 25.

So 25 miles, at 80 mph is 5/16 of an hour

The helicopter travels about 70 miles, at 250 miles per hour, so it's 7/25 or .28 miles per hour. since 5/16 is greater than .3, the helicopter gets there first.[/hide]

btw, instead of assuming it's right, you [i]could[/i] just tell us what your answer was.

[quote="akalra1"][quote="Klebian"][quote="akalra1"][hide="unnecessary info"]since the the helicopter is located at (40, 58), we use the pythogean theom to get the distance it travels

$a^{2}+b^{2}= c^{2}$ $\text{find}$ $c$
$40^{2}+58^{2}= c^{2}$
$1600+3364 = 4964$
$\sqrt{4964}= 70.45565981523414$
$c\approx70$

the helicoptor has to travel about $70$ miles

[/hide]
[hide="answer"]
$\text{squad car}$ $1$ $\text{has to travel}$ $9+17 = 26$ $\text{miles}$
$\text{squad car}$ $2$ $\text{has to travel}$ $4+24 = 28$ $\text{miles}$
$\text{squad car}$ $3$ $\text{has to travel}$ $16+12 =28$ $\text{miles}$
$\text{squad car}$ $4$ $\text{has to travel}$ $6+19 = 25$ $\text{miles}$

$\text{squad car}$ $4$ $\text{will get there first. WoO hOo!}$[/hide][/quote]

But the helicopter travels more than 3 times faster than the cars... (Also, just saying, MCrawford has said not to use the \text command in LaTeX too much, because it makes the page load slower)[/quote]

srry about that

i was bored so i decided to add the helicopter part in :)[/quote]

OK... but... you [i]need [/i]to use the helicopter part

 and what I meant by the \text command was this:
[code]$\text{squad car}$ $4$ $\text{will get there first. WoO hOo!}$[/code]
and similiar

yeah, the helicopter gets there first.
next question:

even though the police helicopter got there really fast, the donut store still had time to hide the Vegemite. They cleaned up the bland smell, too. but now they know that Captain Sugar knows they are making Vegemite, so they send SUPERNINJA ninja robots after him.  they send out y robots to randomly patrol the city. Captain Sugar destroys one while walking to a hardware store to get some stuff for his car. before he destroys one, there are 2x SUPERNINJAs. after he destroys one, there are 1/2(x) + 5 SUPERNINJAs. find how many SUPERNINJAs were released by the evil donut store.

here's a diagram:

                                 G

              NYC                              S

(assume G is a right angle.)
nyc-s is the hypotenuse. 490000 - 202500 is about 536 miles.

you were both finding the hypotenuse of the triangle. you were supposed to find a leg of the triangle.

Oh sorry I was under the impression that we were supposed to find the total distance traveled (that is the sum of the legs). Then it is just [hide] $50$ $\sqrt{115}$[/hide], or about $536$ like you said. Ready for the next problem.

sorry,  that question was badly worded. :blush:

Captain Sugar makes it to Sugarton. he has a choice of two rental car companies, five restaurants, and three hotels. he picks one of each.

1) how many possible combinations of rental cars, restaurants, and hotels are there?

2) Captain Sugar hates british food. one of the restaurants is a british food restaurant. He also decides that he does not need a rental car. now how many combinations are there?

[hide="1"]2 rental cars$\cdot$5 restaurants$\cdot$3 hotels = 30 combinations. [/hide]
[hide="2"]4 restaurants$\cdot$3 hotels = 12 combinations[/hide]

Captain Sugar calls a meeting with the Sugarton Military to warn them about Dark Lord Aspartame. there are 15 people in the military.

Captain Sugar picks three of the people from the military. their names are bob, joe, and ted.  one of them is the helicopter pilot, one of them is the secret agent, and one of them is the general.

clue things:
bob is afraid of heights.
ted went on vacation in utah last summer.
joe's favorite helicopter is the VM-160.
the one with the correct answer to this equation is the general:
(2x+1)^2=4
bob:1/2      joe:1         ted:1/4

what is ted's occupation?

[quote="Captain Sugar"]the one with the correct answer to this equation is the general:
(2x+1)^2=4
bob:1/2      joe:1         ted:1/4[/quote]

Of course, there are 2 solutions to this equation. :P

[hide]Substitute in the values in the equation and you'll see that Bob is the general. Joe is the pilot, so Ted must be the secret agent[hide="."]P.S.Where do saccarine, splenda, equal, etc. come in?[/hide] [/hide]

There can't be a fraction of a person!

$104\div 10=10.4$, so $10%\times 104=10.4
$.  There are [i]10.4[/i] normal people?

[quote="mgao really"]There can't be a fraction of a person!

$104\div 10=10.4$, so $10\%\times 104=10.4$.  There are [i]10.4[/i] normal people?[/quote]

you're going about it wrong. the normal people buy magazines 10% of the time. if there were 100 normal people, they would buy 10 magazines.

try to make it into an equation. :)

[quote="Captain Sugar"]yes, he's the secret agent.  actually, joe could be the secret agent, but if you use common sense, a secret agent probably wouldn't have a favorite helicopter.

by the way, qwang, Dark Lord Aspartame's motorcycle is named Saccharin.

the next question:
Captain Sugar gets a call from The Telemarketer. he tries in vain to sell Captain Sugar some subscriptions to magazines by asking him to "answer a few questions."

suppose The Telemarketer calls everybody in Sugarton. there are 350 people living in Sugarton, including tourists.
tourists subscribe to magazines 30% of the time.
gullible people subscribe to magazines 60% of the time.
normal people subscribe to magazines 10% of the time.
The Telemarketer sells 104 magazine subscriptions.
1) how many tourists are there?
2) how many gullible people are there?
3) how many normal people are there?[/quote]

if you make a set of equations, there are 10 tourists, 140 gullible people, and 200 normal people. i used a set of equations to do it.
(i actually forgot what the answer was to my own problem.  :oops: )


how can we use Excell to compare datasheets to find errors according to a criterium.

Huh, I'm not sure what you mean... but generally working with data from different sheets is not easy. Perhaps if you specified your question a little more.


abstract algebra
algorithm
combinatorics unsolved
combinatorics

Hi Guys

I have a big problem which is bugging me. Basically the problem is about agreedy algorithm. I have attached the problem which would be similer to what I would get in the exam. I am really confused about the question I would like to get help from anyone please or example of how to solve the problem.

the problem is about

Each of ten students A-J have chosen 3 modules out of eight available as follows:  :(  :o

Discuss how the Greedy Algorithm can be applied to this problem and hence insert subjects in to slots in such a way as to minimise the number of students who have a timetable clash.

Ive attached the problem. I would be very greatful if me anyone can help me with this question it would be very helpful to

Hi Guys

I am still stuck on this question.

Can anyone help me understand how this would be dun or give an example of how this problem would be done.  :(  :(


Obiously I like da colts alot. They are 4-0!!!  :D What are your predictions? I think the colts are going to go 13-3. Who do you think is going to be in the Super Bowl? I think its Eagles v. Colts.  :huh:

[hide="clikifyoulikedacolts"]Where is the Super Bowl? Anyone Know? :? [/hide]

I think it will be Eagles vs. Bengals
isn't this topic sort of like [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=49874]this one[/url]

I'm not sure about Eagles...maybe Falcons.

Definitely not Patriots though. :P

i think the pats have a chance at the playoffs, but not the sb.

tampa bay all the way.
and right now, the dolphins are leading the AFC east. anyone want to go out on a limb - jets lost 2 QB's, bills offence and run D is struggling, patriots have a lot of injuries.
and the Super Bowl will be in Detriot this year

they had this interview with this guy about his predictions of the winning streaks on NPR today.

fugot it, tho :P

I don't know any football teams at all. well, barely any. 49'ers and raiders is about it. oh! and miami dolphins (i think...)

Colts better win this, I got money on them. Colts are my favorite team btw.

I would like to say Colts, but we need to realize that the Colts are not very playoff experienced, if you know what I mean, however, I guess it would be them UNLESS the Patriots get back on track or Rothlisburger's injury isn't very serious.

Ofcourse, in the NFC, the only two teams that really matter are Eagles and Falcons, and ofcourse the Eagles are gonna make it.

[quote="jhollenbeck"]Colts are going all the way!   I am rooting for the Bengals in the other league, but Colts win super bowl tis the main thing[/quote]

Which other league? Don't make things up.

[quote="furious"]i feel the colts will make it this year. their defense is much improved and they're winning their games without throwing for 4 touchdowns. the eagles would be even better if they used their run offence.[/quote]

Or if McNabb hadn't gotten injured :(

[quote="Iversonfan2005"][quote="h_s_potter2002"]I don't know that much about football, but I just keep thinking that the Bengals will make it.  But I'm probably wrong though  :([/quote]

Your wrong, they're not a playoff experienced team, remember, experience has a lot to do with winning...[/quote]

Well now, if by "make it", he means they'll make the playoffs, it's a big possibility. Their division's very strong but the team is showing a lot of achievement after having a lot of promise the last couple years.

But I agree with you that you need at least a little playoff experience to go deep into the playoffs. There have been examples of young teams though.

go Texans! oh wait. Texans haven't won a game.  :(  the game on sunday vs. the colts will be murder.... that is... colts will be the murderers. hahaha.  man. at least we have the astros. woooo.

There's no way the Colts go to the SB...Denver Broncos are going to finally beat the Colts after losing for the past two years in the wild card game. So...

Denver vs. Carolina with Denver 21-3 win.

Back to the top!!!

My new prediction:
Colts are going to be upset by the Bengals in the divisional playoff round and then the Bengals are going to beat the Broncos in the AFC championships, and in the NFC, the Eagles are going to find a way to go to the SB.  Final score:
Broncos 24- Eagles 27

[quote="Klebian"]exactly my thoughts.
I've set up a bet that the Texans suck so bad, they're gonna lose all their games.
I'll win.
But the astros are really doing well, They might win the world series.
Rockets are going to do well also. This year is looking awesome.

And to reply to the topic, I think Bengals will get to playoffs, and lose in like the second round.
Not sure about anything else, I haven't realy been paying much attention.[/quote]

Haha, they won one. I predict that the astros will not win the world series :D . Rockets are struggling because T-Mac is injured and their guards can't shoot, but they will pull it together they have a hard early schedule.


induction
calculus
integration
real analysis
real analysis solved

Here?s a nice one (the author is Cristinel Mortici, a really cool Romanian maths teacher, at least in my opinion):

f:[0, 1]->R, integrable (is this a word?) s.t. (f(x)+f(y))/2<=f((xy)^1/2). If f is continous at 1/e prove that integral from 0 to 1 of f(x)dx <= f(1/e). It?s not hard, I just thought it was cute enough to post it. Bye!

use induction to prove that:

(f(x1)+..+f(xn))/n<=f((x1*...*xn)^(1/n)).

then we have that an=1/n * sum (k=1,n) f(k/n)<=f((1/n*2/n*..*n/n)^(1/n)=f((n!)^(1/n)/n)

and now you use the fact that lim (n!)^(1/n)/n=1/e (when n -->oo)

and that's the problem

cheers! :D :D


Problem :
A person throws a fair dice and if it is odd number , he gets to throw a coin . If he gets an even number from the dice , he gets to throw 2 coins . What is the probablity that he will get at least one HEAD？

[hide="my doubt"]
Ok so i calculate it using tree diagram and get P($\geq$ head)=$(1/2)(1/2)+(1/2)(3/4)=5/8$ .

But there is another method : we find all the out come , which have total of $18$ outcomes .(6 from the odd ,12 from the even) . Then from this 18 cases , only 12 of them has head . SO the probability is

$\frac{2}3$ . So who is right ?
[/hide]

[quote="gauss202"]In your second method, each of the 18 outcomes that you counted are not equally likely. Imagine if you had an unfair coin that was tails 1/3 of the time and heads 2/3 of the time. You couldn't argue that the probability of getting heads is 1/2 because it accounts for one of the two possibilities.

Your first answer is correct.[/quote]

but in here , the coin and dice are both fair  :huh:

The 12 outcomes that result from an even number have to be weighed equally with the 6 outcomes that result from the odd numbers.  So each individual outcome of the 12 is weighed only half that of the individual outcomes of the 6.


the new playstation is about to come out and the new xbox too :clap:
yaaaaaaaaa
anyone plan to buy one(or both)?

I'm not sure about the price...but expect to pay more than you have in the past....My guess is that it will be more than $200 at least....

For the Xbox...the costs are more since it has that HD within it...plus demand will be way higher than supply at the release point

For the PS3....they can put it at whatever they want simply because they are the top console....

Gamestop is already taking reservations on the X Box 360. There is no official release date for either...but XBox 360 should be out Holiday Season '05....meaning sometime in November - December....

I'm not going to be buying either since I'll be in college....but if I was to buy one...I'd like a ps3....

The Nintendo Revolution doesn't seem to have much of a chance, imo....

I didnt even hear about these untill now...im probably not gonna get them though, I already have PS2 and XBox

the ps3 will have to be a great improvement to keep up with xbox 360. i think the xbox 360 will be better. yet, i will not get it anytime soon. i'll wait for the price to drop. i'm still happy with my xbox now. i'm playing forza motorsport and i'm also the only person who hasn't finished halo 2. :w00t:

It's a shame they're not releasing Halo3 with the Xbox 360.

Though it is a pretty good marketting move to release Halo3 right when PS3 gets released, to take away from the specialness of it a little and have consumers be more interested in Halo3.

Isn't Nintendo also making a new system? something like Nintendo Revolution?
but anyways I think PC are better due to better graphics cards.

they can't release halo 3 bc halo 2 just came out.
i wanna a PS3 but i don't have any money :(
nintendo have nooooo chance :rotfl: (little kids are their only market)

i think they're releasing halo 3 when PS3 comes out so they can release xbox 360 earlier.

I've heard (on [url=http://www.slashdot.org]slashdot[/url]) that the PS3 will have better stats in all areas than the Xbox BUT harddrive does not come standard.


[quote="joml88"]Would it be possible to get the AoPSWiki to compile dollar signs as LaTeX instead of having to type <math> </math> all the time?  This would save a ton of time porting things over from the forum into the wiki. (I remember Valentin said he might look at it only after IMO...so that's why I ask now :D )[/quote]
I agree, It took me around 7 min to post one AMC 12 question.

Here's a temporary solution. You guys probably use it already, but.

Step 1: Type <math></math>.

Step 2: Select the text and copy it.

Step 3: Paste when needed.

 :D

There's a button you can push -- up on the little toolbar at the top of the editing screen.  It's labelled "$\sqrt n$ or some-such, and it inserts the text "<math>Insert text here</math>" at the cursor.  "Insert text here" is even conveniently highlighted.  Also, if you've already typed the equation up, you can highlight it and then the <math>s appear on either side.

Yes, I am quite aware of the math tags, how to copy/paste them to make it go quicker, and the button on the toolbar.  But there are a few problems with those:

1. <math> </math> takes longer than just $ $.  Then, if you want it centered you have to add on <center> </center> instead of just doubling the dollar signs.  Multiply this by a thousand and you can easily see how this wastes a considerable amount of time and energy.

2. There is a lot of good stuff on the forum that people will want to copy and paste over.  Of course, everything on the forums has dollar signs.  So then you have to go and change every dollar sign to a <math> or </math>.  I transferred a large post over from the forum to the wiki today and it took me at least 10 minutes to change all the dollar signs to math tags.

Oh yeah, and I should also mention the "failed to parse" gets a bit annoying as well.  The whole TeX system needs to be changed big time...well, only if you want the wiki to really catch on and grow.

I know the admins are always busy working on a hundred different things at a time.  Thank you for your time and letting us have the wiki.

May I add a temporary solution?

1) Open up Microsoft Word, Notepad, or Wordpad.
2) Copy-paste whatever you want (in forum Latex)
3) Go to "Edit"->"Replace" (or CTRL+H)
4) 'Find: $'
5) 'Replace with: <math>'
Go through and add the "/" where the close tag $ should be.

Enjoy! :D

Even better solution: Give rrusczyk a free hour after a good workout on a Friday afternoon.

I may have broken a bunch of other stuff, but I believe the wiki now knows dollar signs.  Sorry I didn't do this a lot earlier, but I thought it was gonna be way tougher than this.  (And it might still - let me know what I have broken.)

Yes, thank you!!  This will make posting problems and everything easier, especially if you don't have a lot of $\LaTeX$ separated by other things.

[quote="rrusczyk"]An addendum: This basically works by converting your dollar signs into <math> </math>, so after you input something with dollar signs, it will be converted into <math> </math>.  Currently, the wiki converts single and double dollar signs thus (but doesn't do the standard centered thing with the double dollar signs).[/quote]
Its much simpler and faster, thanks. :D

Oh, one other request.  Would it be possible to have it throw a \displaystyle in when it converts the dollar signs to math tags?

Edit: I also just realized that \binom{n}{k} doesn't work on the wiki  :(  (you have to use {n\choose k} instead).  I'm guessing stuff like that wouldn't be very easy at all to fix.  Unless you could have it convert \binom{n}{k} to {n\choose k} (?)..I dunno?

Actually, I just looked on the MediaWiki page and they say that adding  \,\! will force it to render as PNG's all the time.

[quote]
It is probably like the TeXnicCenter (you can't use some commands you use on AoPS, like \binom, \therefore, etc); you need the correct packages to use those commands.[/quote]

The most ideal thing would be to use the same setup the forum has for the wiki.  Mr. Rusczyk mentioned a while back he might try this at some point but he has understandably been busy.  In addition, no one's really sure how hard this is or how long it would take to implement.


geometry
circumcircle
inequalities
geometry unsolved

$ABC$ is a triangle, with sides $a$, $b$, $c$ , circumradius $R$, and exradii $r_a$, $r_b$, $r_c$If $2R\leq r_a$, show that $a > b$, $a > c$, $2R > r_b$, and $2R > r_c$.

The given inequality can be rewritten as $\frac{abc} {2\triangle} \leq \frac{\triangle} {s-a}$
which gives us $abc\leq 2s(s-b)(s-c)$ by Heron's Formula.
If we substitute $s-a$ as $x$, $s-b$ as $y$,and $s-c$ as $z$ then the given inequality may be rewritten as $(x+y)(y+z)(z+x)\leq  2(x+y+z)yz$
and we get $xy(x+y) + zx(z+x) \leq yz(y+z)$_____1
  Now if $a\leq b$ then $xy(x+y)$ will be nonpositive. which is not true .
So $a>b$ and similary $a>c$.
  Let us suppose that $2R\leq r_b$.As above we get $xy(x+y) +yz(y+z) \leq zx(z+x)$
Adding this to 1 we get $2xy(x+y\leq 0$ which again is not true.Hence $2R>r_b$


geometry
geometric transformation
rotation

What is the necessary and sufficient condition for chirality?...
Does four different groups attached to a carbon atom ensure chirality?
Moreover does chirality and optical rotation signify the same thing?
what is the requisite for each of them?..
I would be grateful if anyone answers it...
thanks a lot..!!

Four different groups on a carbon is neither necessary nor sufficient for chirality. It is not necessary since we can achieve chirality in different ways (for example, BINOL, or an allene, or four different groups on a sulfur), and it is not sufficient since there exist meso compounds in which there exist such stereogenic carbons but are still achiral (such as, I dunno, (R),(S)-1,2-dimethylcyclohexane (I don't know if that nomenclature is correct but you should get the point)). The definition for chirality is simply that the compound cannot be superimposed upon its mirror image.

I think chirality and optical rotation are related concepts. Achiral molecules do not rotate polarized light; neither do racemic mixtures of chiral molecules. As far as I know, any enantiomerically pure chiral substance does rotate light.

I'll throw in my two cents:

Chirality is an intrinsic property of a single [i]molecule[/i].  It doesn't matter in what sort of mixture a molecule is present.  (Recall a chiral molecule is one that cannot be imposed on its mirror image.)  On the other hand, optical rotation is a property of a [i]solution[/i].  As different mixtures of one enantiomer or another is present, the optical rotation of that solution changes.

So no, chirality and optical activity are not the same.

Thanks a lot for your replies...!!
If i am right the a non superimposible mirror image is a NECESSARY and SUFFICIENT condition for chirality??....
And if a compound is chiral then it DEFINITELY rotates the plane of polarised light doesn't it?...
Moreover can someone please shed some light on how compounds rotate the plane of light??
thanks in advance.....


x*y=z^2
x*x=y
x+y+z=7*x
z/x=y/z
z*y*x=64*z
(z*x)+y=6*z
x*2=z
x*z=32
x, z, and y are all even numbers
y>z>x
x*x=z*2

What is (x*y*z)-(x+y+z) written as a number with no variables?
*Extra* What is x, y, and z?

[hide]484.
u gave way too much information.
z*y*x = 64*z
y*x = 64
x*y = z^2
64 = z^2
[b]z = 8[/b]

x*y = z^2
x*x = y
x^3 = 64
x = 4

4^2 = y
y = 16 [/hide]

vicseb: Please use the forums responsibly, if it is [b]a tough problem[/b], it shouldn't be in here, [u]even[/u] if it's way way way way too easy for you. If you want to challenge audiences of AoPS/ML, at least post it in Getting Started or Intermediate Topics, not here... This is a MIDDLE SCHOOL/CLASSROOM MATH forum. If you are not sure what should be posted here, don't post here, post in Getting Started. If you are in Middle School, no exceptions, post it in higher forums. Either way, this type of problem belongs to higher forums, the problem is way too intimidating to starters. If you want to appeal my reply, please contact a moderator or PM me

That was quite an easy problem, although long. X=4, y=16, and z=8. The answer is 484. When you said (x)x= y that helped. alot. When you told us that zxy=64z i found xy=64. From there i took the xx=y information and applied it. i plugged in x(xsqaured0=64 so x cubed eqauled 64. 4 cubed eqauls 64. Then you told us that x qaured equals y so 16 is y. THen xz=32 so 32/4=8.  ;)

It was incredibly easy. As in, do-it-in-your-head-easy. Just look at a few key pieces of the information given.
x*y=z^2
x*2=z
x*z=32
x, z, and y are all even numbers

Therefore x and z are powers of 2, and since x*2=z then x*x*2=32 -> x=4, x*2=z -> z=8 and x*y=z^2 -> 4y=64 -> y=16.

Just sort through the data and look at the easy give-it-away clues.


Always exits $ n_0 > x$ then
${ \lim_{n\rightarrow \infty}\frac {x^n}{n!} = \lim_{n\rightarrow \infty}\frac {x^{n_0 - 1}}{(n_0 - 1)!}\times\frac {x}{n_0}\times\frac {x}{n_0 + 1}}\times...\times\frac {x}{n}$
$ \leq\lim_{n\rightarrow\infty}\frac {x^{n_0 - 1}}{(n_0 - 1)!}\times(\frac {x}{n_0})^{n - n_0 + 1} = 0$
(Because $ \frac {x^{n_0 - 1}}{(n_0 - 1)!}$ is a constant and $ \lim_{n\rightarrow\infty}(\frac {x}{n_0})^{n - n_0 + 1} = 0$)

[quote="hell_ever"]Question:

For x>1 $ \lim_{n\rightarrow \infty} \frac {x^n}{n!}$[/quote]

This isn't rigorous or anything but a logical answer would be 0. Because whatever be the value of x, n! would always be greater than x^n.

And what is amc??


inequalities
algebra
polynomial
LaTeX
Support
algebra unsolved

Let f(x) is a polynomical with degree n defined by:
    f(x) = (2^(n-1))*(x^n) + (a_(n-1))*(x^(n-1))+...+(a_(1))*(x^1)+a_0.
 Prove that f(x)=T_n(x).
    with T_n(x) is the Chebyshev s polynomical with degree n.
( For admin: I don't know how to use latex, please support it to this problem)

I think the book's title is "Special researches on Polynomials".

And I want to add that Le Hoanh Pho was my teacher. Now, he teaches at Le Quy Don special school in Danang city.

The problem was solved in that book .But it is too long to write down here,and I don't have any short solution too.You may look for it by yourself,darkmaster.

Where is it ? Can you tell me the page where it is ?I found the solution for it with n=2,3 in this book. But I didn't find the one I type there.
     Sorry for the name of the book, because my English is not good

[quote="darkmaster"]Let f(x) is a polynomical with degree n defined by:
    f(x) = (2^(n-1))*(x^n) + (a_(n-1))*(x^(n-1))+...+(a_(1))*(x^1)+a_0.
 Prove that f(x)=T_n(x).
    with T_n(x) is the Chebyshev �s polynomical with degree n.
( For admin: I don't know how to use latex, please support it to this problem)[/quote]You defined arbitrary polynomial with leading coefficient $2^{n-1}$ and ask us to prove it is $T_{n}$???


$x,y,z > 0$. Prove that :

$(x+y)(y+z)(z+x)+4xyz \le \sqrt{(x+y)(y+z)(z+x)}(\sqrt{xy(x+y)}+\sqrt{yz(y+z)}+\sqrt{xz(x+z)}$

$(x+y)(y+z)(z+x)+4xyz\le$
$\leq\sqrt{(x+y)(y+z)(z+x)}\left(\sqrt{xy(x+y)}+\sqrt{yz(y+z)}+\sqrt{xz(x+z)}\right)\Leftrightarrow$
$\Leftrightarrow(x+y)^2(x+z)^2(y+z)^2+8(x+y)(x+z)(y+z)xyz+16x^2y^2z^2\leq$
$\leq(x+y)(x+z)(y+z)\left((x+y)(x+z)(y+z)-2xyz+2\cdot\sum_{cyc}x\sqrt{yz(x+y)(x+z)}\right)\Leftrightarrow$
$\Leftrightarrow(x+y)(x+z)(y+z)\sum_{cyc}\sqrt{x(x+y)(x+y)}\geq$
$\geq5(x+y)(x+z)(y+z)\sqrt{xyz}+8(xyz)^{\frac{3}{2}}\Leftrightarrow$
$\Leftrightarrow(x+y)(x+z)(y+z)\left(\sum_{cyc}\sqrt{x(x+y)(x+z)}-5\sqrt{xyz}\right)\geq8\sqrt{x^3y^3z^3}.$
But   $\sum_{cyc}\sqrt{x(x+y)(x+y)}-5\sqrt{xyz}\geq\sqrt{xyz}\Leftrightarrow\sum_{cyc}\sqrt\frac{(x+y)(x+z)}{yz}\geq6.$
This is true since
$\sum_{cyc}\sqrt\frac{(x+y)(x+z)}{yz}\geq\sum_{cyc}\sqrt\frac{4\sqrt{x^2yz}}{yz}=2\cdot\sum_{cyc}\sqrt[4]{\frac{x^2}{yz}}\geq2\cdot3=6.$
Hence remain to prove that $(x+y)(x+z)(y+z)\sqrt{xyz}\geq8\sqrt{x^3y^3z^3},$ but this is obviously true. :)

Nice proof, Arqady! Mine is longer, I post it just because it somewhat strengthens the inequality.

We have to prove the inequality

$\sqrt{\left(y+z\right)\left(z+x\right)\left(x+y\right)}\cdot\sum\sqrt{yz\left(y+z\right)}\geq\left(y+z\right)\left(z+x\right)\left(x+y\right)+4xyz$

for any three nonnegative reals x, y, z, where the $\sum$ sign stands for cyclic sums.

Note that

$\sqrt{\left(y+z\right)\left(z+x\right)\left(x+y\right)}\cdot\sum\sqrt{yz\left(y+z\right)}$
$=\sum\sqrt{\left(y+z\right)\left(z+x\right)\left(x+y\right)}\cdot\sqrt{yz\left(y+z\right)}=\sum\left(y+z\right)\sqrt{y\left(z+x\right)\cdot z\left(x+y\right)}$
$\geq\sum\left(y+z\right)\frac{2\cdot y\left(z+x\right)\cdot z\left(x+y\right)}{y\left(z+x\right)+z\left(x+y\right)}$         (by the GM-HM inequality, applied to the numbers y (z + x) and z (x + y))
$=\left(y+z\right)\left(z+x\right)\left(x+y\right)\cdot\sum\frac{2yz}{y\left(z+x\right)+z\left(x+y\right)}$.

Now,

$\sum\frac{2yz}{y\left(z+x\right)+z\left(x+y\right)}=\sum\frac{2yz}{2yz+zx+xy}=\sum\frac{\left(2yz\right)^2}{2yz\left(2yz+zx+xy\right)}$
$\geq\frac{\left(2yz+2zx+2xy\right)^2}{2yz\left(2yz+zx+xy\right)+2zx\left(2zx+xy+yz\right)+2xy\left(2xy+yz+zx\right)}$    (by [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15125]the Cauchy-Schwarz inequality in Engel form[/url])
$=\frac{\left(yz+zx+xy\right)^2}{y^2z^2+z^2x^2+x^2y^2+xyz\left(x+y+z\right)}$,

so we get

$\sqrt{\left(y+z\right)\left(z+x\right)\left(x+y\right)}\cdot\sum\sqrt{yz\left(y+z\right)}\geq \left(y+z\right)\left(z+x\right)\left(x+y\right)\cdot\sum\frac{2yz}{y\left(z+x\right)+z\left(x+y\right)}$
$\geq\left(y+z\right)\left(z+x\right)\left(x+y\right)\cdot\frac{\left(yz+zx+xy\right)^2}{y^2z^2+z^2x^2+x^2y^2+xyz\left(x+y+z\right)}$.

In order to prove our inequality

$\sqrt{\left(y+z\right)\left(z+x\right)\left(x+y\right)}\cdot\sum\sqrt{yz\left(y+z\right)}\geq\left(y+z\right)\left(z+x\right)\left(x+y\right)+4xyz$

now, it remains to show that

$\left(y+z\right)\left(z+x\right)\left(x+y\right)\cdot\frac{\left(yz+zx+xy\right)^2}{y^2z^2+z^2x^2+x^2y^2+xyz\left(x+y+z\right)}$
$\geq\left(y+z\right)\left(z+x\right)\left(x+y\right)+4xyz$.

Upon subtraction of (y + z) (z + x) (x + y), this becomes

$\left(y+z\right)\left(z+x\right)\left(x+y\right)\cdot\left(\frac{\left(yz+zx+xy\right)^2}{y^2z^2+z^2x^2+x^2y^2+xyz\left(x+y+z\right)}-1\right)\geq 4xyz$
$\Longleftrightarrow\ \ \ \ \ \left(y+z\right)\left(z+x\right)\left(x+y\right)\cdot\frac{xyz\left(x+y+z\right)}{y^2z^2+z^2x^2+x^2y^2+xyz\left(x+y+z\right)}\geq 4xyz$
$\Longleftrightarrow\ \ \ \ \ \left(y+z\right)\left(z+x\right)\left(x+y\right)\left(x+y+z\right)\geq 4\left(y^2z^2+z^2x^2+x^2y^2+xyz\left(x+y+z\right)\right)$
$\Longleftrightarrow\ \ \ \ \ \left(y+z\right)\left(z+x\right)\left(x+y\right)\left(x+y+z\right)\geq 4\left(yz+zx+xy\right)^2-4xyz\left(x+y+z\right)$
$\Longleftrightarrow\ \ \ \ \ \left(y+z\right)\left(z+x\right)\left(x+y\right)\left(x+y+z\right)+4xyz\left(x+y+z\right)\geq 4\left(yz+zx+xy\right)^2$
$\Longleftrightarrow\ \ \ \ \ \left(x\left(y+z\right)^2+y\left(z+x\right)^2+z\left(x+y\right)^2\right)\left(x+y+z\right)\geq 4\left(yz+zx+xy\right)^2$;

but this is clear from Cauchy-Schwarz:

$\left(x\left(y+z\right)^2+y\left(z+x\right)^2+z\left(x+y\right)^2\right)\left(x+y+z\right)$
$\geq\left(\sqrt{x\left(y+z\right)^2\cdot x}+\sqrt{y\left(z+x\right)^2\cdot y}+\sqrt{z\left(x+y\right)^2\cdot z}\right)^2=\left(x\left(y+z\right)+y\left(z+x\right)+z\left(x+y\right)\right)^2$
$=4\left(yz+zx+xy\right)^2$,

and the inequality is proven.

  Darij


Let the quadrilateral has side length $ a,\ b,\ c,\ d$ with $ p\equal{}\frac{a\plus{}b\plus{}c\plus{}d}{2}$,

we have $ [ABCD]\equal{}\sqrt{(p\minus{}a)(p\minus{}b)(p\minus{}c)(p\minus{}d)}\equal{}(\sqrt[4]{(p\minus{}a)(p\minus{}b)(p\minus{}c)(p\minus{}d)})^2$

$ \leq \left\{\frac{(p\minus{}a)\plus{}(p\minus{}b)\plus{}(p\minus{}c)\plus{}(p\minus{}d)}{4}\right\}^2\equal{}\frac{p^2}{4}$,

the equality holds when $ a\equal{}b\equal{}c\equal{}d$.

I think that formula holds only if the quadrilateral is cyclic.

But we can do like this:

[hide="Solution"]

P - the area

$ P\le \frac{ab\plus{}cd}{2}, P\le \frac{ad\plus{}bc}{2}$, so $ 2P\le (a\plus{}c)(b\plus{}d)$. That is derived by splitting the quadrilateral in 2 by diagonal, using that $ \sin\alpha\le1$ for all angles.

$ 2P\le (a\plus{}c)(b\plus{}d)\le \frac{(a\plus{}b\plus{}c\plus{}d)^2}{4}\equal{}\frac{p^2}{4}$, with equality for a rhombus, and the sines of angles are 1, so the equality holds just for square

[/hide]

[quote="Bugi"]I think that formula holds only if the quadrilateral is cyclic. [/quote]

You are right.

Related problem

1969 Czechoslovakia Mathematical Olympiad

A convex quadrilateral $ ABCD$ with sides $ AB \equal{} a,\ BC \equal{} b,\ CD \equal{} c,\ DA \equal{} d$ and angles $ \alpha \equal{} \angle{DAB},\ \beta \equal{} \angle{ABC},\ \gamma \equal{} \angle{BCD},\ \delta \equal{} \angle{CDA}$
is given. Let $ \displaystyle s \equal{} \frac {a \plus{} b \plus{} c \plus{} d}{2}$ and $ P$ be the area of the quadrilateral. Prove that
\[ P^2 \equal{} (s \minus{} a)(s \minus{} b)(s \minus{} c)(s \minus{} d) \minus{} abcd\cos ^ 2 \displaystyle \frac {\alpha \plus{} \gamma}{2}.\]
Now the name of the nation changed,\ didn't that?

[color=blue]It occured to me that Ellipse has something to do with the solution.[/color]

[quote="Leon"]Show that among all quadrilaterals of a given perimeter the square has the largest area.[/quote]
Lemma 1:
Given a triangle $ \triangle ABC$, with $ BC$ and $ AB\plus{}AC$ fixed, then the area of $ \triangle ABC$ is largest when $ AB\equal{}AC$.
Proof:
The locus of all points $ P$ such that $ PB\plus{}PC \equal{} AB\plus{}AC$ is an ellipse. Since the area of $ \triangle PBC$ is the distance from the line $ BC$ to $ P$ times $ BC$, we see that $ \triangle PBC$ has the maximal area when $ P$ are furthest away from $ BC$. Because of symmetry this must be when $ P$ lies on the bisector of $ BC$, and hence $ AB\equal{}AC$.

Let $ \square ABCD$ be the quadrilateral with the largest area. Using lemma 1 on $ \triangle ABC$ we can wlog assume that $ AB\equal{}BC$. And using on $ \triangle ADC$: wlog $ AD\equal{}DC$. Hence $ \triangle BCD$ is concurrent to $ \triangle DAB$. So it is enough to find the maximal area of $ \triangle BCD$ when $ BC\plus{}CD$ is fixed. Using Lemma 1, we see that we can wlog assume that $ BC\equal{}CD$, so $ \square ABCD$ must be a rhombus. Let $ E$ be the projection of $ C$ on $ BD$. Then $ \triangle BEC$ is a rightangled triangle, with $ BC$ fixed. Now $ BE^2\plus{}EC^2 \equal{} BC^2$ is also fixed, and the area of $ \triangle BEC$ is $ \frac{1}{2}BE \times EC$. Since $ BE^2 \plus{} EC^2$ is fixed we see that $ BE \times EC$ is largest whenever $ BE\equal{}EC$.(Wellknown) Doing this for $ \triangle DEC$ also we see that $ \angle BCD \equal{} 90^\circ$, and since we could use this symmetric argument for all the angles in the quadrilateral, $ \square ABCD$ must be a square.

[quote=Maharjun]I have made use of the result that if the sum of two, or more positive quantities is constant. then their product is maximum when they are all equal. For two quantities, it can be proven using A.M G.M as follows-

[hide="Click here if interested in proof of above inequality for two quantities"]
let $ a,b$ be the required variables. $ a + b = c$ where $ c$ is a constant
let $ d$ be the equal value of $ a$ and $ b$ such that $ a + b = c$

$ \implies d = {c \over 2} = {a + b \over 2}$
we must prove that $ d^2 \geq ab \text{ } \forall \text{ } a + b = c \text{ and } a,b \ge 0$

$ \& \implies {\left( {\frac {{a + b}} {2}} \right)^2} \geq ab \\
\& \implies \frac {{a + b}} {2} \geq \sqrt {ab}$

which is nothing but $ A.M \geq G.M$. hence we are done
[/hide]

Step 1:(prove that the quadrilateral is a rhombus,using hero's formula for area of triangle)
[hide="solution"]
let us assume that the figure with the largest area is not a rhombus

let the sides be $ a,b,c,d$ in cyclic order.
This means that atleat two of $ a,b,c,d$ are not equal. wlog, let $ a \ne b$

let $ e$ be the diagonal having $ a,b$ and $ c,d$ on opposite sides.

let $ {\left( {a,b,e} \right)_\triangle }$ represent triangle containing sides $ a,b,e$.
let semiperimeter $ S = {a + b + e \over 2}$
Now,

$ a + b + c + d = {\text{constant}}$
since we only consider $ {\left( {a,b,e} \right)_\triangle }$ we leave the value of $ c + d$ untouched

$ \implies a + b = {\text{constant}}$
also, we decide not to change the value of $ e$.

$ \implies {a + b + e \over 2} = S = {\text{constant}} \implies (S - a) + (S - b) = \text{constant}$ -------------------------------------------------------$ \left( i \right)$

Now, we have
\[ \text{Ar} \left[ {{{\left( {a,b,e} \right)}_\triangle }} \right] = \sqrt {S\left( {S - a} \right)\left( {S - b} \right)\left( {S - e} \right)}\]
Now we know that $ S$ and $ \left( {S - e} \right)$ are constant.

and we know that $ \left( {S - a} \right)\left( {S - b} \right)$ will be maximum when $ S - a = S - b$ [using $ \left( i \right)$ and initially mentioned inequality]

but we also know that $ a \ne b \implies s - a \ne s - b$
thus, if we replace $ a$ and $ b$ each by $ {a + b\over 2}$. the sum will remain constant as required but the product and hence area will increase.

thus it is possible to get a quadrilateral [$ \because \text{ar} \left[ {{{\left( {c,d,e} \right)}_\vartriangle }} \right] = constant$] of larger area than the one we took without upsetting any of the conditions.

but this is a contradiction since the quadrilateral we took was assumed to be the one with the largest area.

this contradiction arose because we assumed that such a quadrilateral was not a rhombus.

[b]Hence the quadrilateral is a rhombus[/b]
[/hide]

step 2 -(prove that the rhombus is a square, trivial)
[hide="solution"]
let it be a rhombus with side $ a$ and acute angle $ \theta$

$ \implies \text{altitude} = h = a\cdot sin\left(\theta\right) \\
\implies \text{area} = a^2 sin\left(\theta\right)$

thus area maximum if $ sin\left(\theta\right) = 1 \implies \theta = {\pi \over 2}$

[u][b]Thus the rhombus is a square[/b][/u]
[/hide][/quote]

I think that this solution is wrong (because according to me, step 1 of your solution is not correct). Basically, you are there trying to show that if we replace $(a,b)$ by $({a + b\over 2},{a + b\over 2})$ then the area of the quadrilateral will increase. But that is not always true. Since if you do this replacement, then the value of $e$ also gets effected. So you cannot guarantee that the area of the quadrilateral will increase.

imagine a person using jensen" s inequality to prove it area of quadrilateral is a convex function :D


Here is a problem from one of the local math competitions in Georgia:

What is the sum of zero and all of the solutions of:

[code]\[\log_2 x \log_4 x \log_6 x = \log_2x \log_4 x + \log_2 x \log_6 x + \log_4 x \log_6 x\][/code]
The Latex looks okay (thanks Quintillian), but there is still a big "code" box around that.  Any way to get rid of it?

Thanks in advance for both answers (math and "code" box).

     -- Pawel

You don't have to put the code tags around it.  I just did that because I was showing you how to do it.  You just have to put the dollar signs.

Btw,  Quintillian is the person who said that quote in my signature.  My name is Joe  :D .

[hide]
you can divide the entire thing by $\log_2x$, and since $\log_4x=\frac{\log_2x}{2}$, it becomes:

$\log_4x \log_6x=\log_4x+\frac{3\log_6x}{2}$
after changing everything into $\log_{10}$ using base change formula (or any base other than 10), you can divide the entire thing by $\log x$ and then multiplying the entire thing by $\log 4 \log 6$ to get:

$\log x=\log 6+\frac{3}{2} \log 4 \rightarrow \log x=\log{(6 \cdot 4^{\frac{3}{2}})} \rightarrow x=48$  [/hide]

[hide]x=1 also works, so the answer is 49.[/hide]

[quote="verneer"]What is the sum of zero and all of the solutions of:
$\log_2 x \log_4 x \log_6 x = \log_2x \log_4 x + \log_2 x \log_6 x + \log_4 x \log_6 x$[/quote]

[hide]Divide through by $\log_2x\log_4x\log_6x$ to get\[1=\frac1{\log_6x}+\frac1{\log_4x}+\frac1{\log_6x}\]
RHS is \[\log_x6+\log_x4+\log_x2=\log_x2\cdot 4\cdot 6=\log_x48\]
Therefore $x=\fbox{48}$.
Test $\log_2x\log_4x\log_6x=0$ to get $x=\fbox{1}$ as well.
Answer is \[\fbox{49}\]

Practicing my boxing skills :).

Now for my multiplication.[/hide]

That's very clear ... thanks!

     -- Pawel


induction
combinatorics unsolved
combinatorics

Standard apology:  Sorry if this has already been covered -- I couldn't find it...

For a given number, I need to find out a way to come up with all the ways that numbers can add up to it:

For 3, it can be 1+1+1, 2+1, 1+2, 3

For 4, it can be 1+1+1+1, 2+2, 2+1+1, 1+2+1, 1+1+2, 3+1, 1+3, 4

...And so on

It gets more complicated, in that I need to get to a formula:
A=1, B=2, etc...

So, for C, it would be A+A+A, B+A, A+B, C

And, to make this a completely unrealistic goal, my final operation would be to convert it:

For C, the final result would be A*A*A+B*A+A*B+C = A^4+2AB+C

I'm not even sure how to write a code to do this...

I'm trying to figure out a way to get solutions for up to the number 3,000.

Thanks for everyone's input!

The answer is $2^{n}$. This can be seen as follows: Let $f(n)$ be the number of ways of wrinting the number $n$ as a sum of smaller numbers. Then $f(1) = 1$. Now, assume that we have shown that $f(n) = 2^{n}$ for all $n \leq k$ for some $k$. The first number $m$ in a sum adding up to $k$ can be any number between 1 and $k$. If the first number is $k$, then the sum has to be just $k$. In case $1 \leq m < k$, we have $f(k-m)$ sums adding up to $k$ that begin with $m$. So, \[f(k) = 1+\sum_{m=1}^{k-1}f(k-m) = 1+\sum_{i=1}^{k-1}2^{i}= 2^{k}.\] By induction we then have that $f(n) = 2^{n}$ for all $n$.

Thank you, but I'm not quite sure what you solved...  The reason I need a formula that keeps A,B,C (etc) intact, is that these numbers arfe not acually 1,2,3 (etc)...  They're part of an unrelated series, and those numbers need to be plugged into the ultimate formula, whose relationship to each other is based on as if they were 1,2,3, (etc).

I'm not looking the number of possible permutations, I'm looking to create an actual list of those permutations, from which I can create the formula.

Sorry for the confusion...

:blush: Oh, sorry, I didn't read your post carefully enough. But I'm not quite sure what your problem is...do you have to actually create the list, or is it enough to calculate the number, like $A^{3}+2AB+C$ (or perhaps something else, in case I've misunderstood it again  :P )

If I can calculate the number, that's ideal.  I have a feeling that I'm going to need to create the list.

Unfortunate, because the list is 3,000 numbers long...  I know I'm going to have to write a code for it, I just am not sure how to go about generating each formula for each number.

I believe that I'm getting close to creating a spreadsheet that does this, but it's only for 10 starting digits, and carried out to 60 places.

I still don't understand what you want except of the list (which can be created by recursion).

Oh well. First of all, the answer is $2^{n-1}$, not $2^{n}$.

Second, I don't know if this helps you, but there is a rather natural bijection between the ways that numbers can add up to $n$ and the subsets of $\{1,2,3,\ldots,n-1\}$. Consider a row of $n$ ones. There are $n-1$ "spaces" between consecutive ones. Number these spaces $1,2,3,\ldots,n-1$. Put a plus sign or nothing in each space. Note that this generate a subset of $\{1,2,3,\ldots\}$ (the spaces with plus signs). Then sum the numbers and voila, you have a way that numbers can add up to $n$. For example, for $n = 3$ we have \[\emptyset \leftrightarrow 1,\ \ 1,\ \ 1 \leftrightarrow 1,1,1\]  \[\{1\}\leftrightarrow 1+1,\ \ 1 \leftrightarrow 2,1\]  \[\{2\}\leftrightarrow 1,\ \ 1+1 \leftrightarrow 1,2\]  \[\{1,2\}\leftrightarrow 1+1+1 \leftrightarrow 3\] I hope this'll help you.


function
induction
real analysis
real analysis solved

Is there a continous function $f$ on $\mathbb{R}$ such that $f(0), f(f(0)),... f^{(n)}(0)...$ are dense in $\mathbb{R}$ :(

Define the following function:
$ f: \mathbb{R} \rightarrow \mathbb{R} $.
First of all, for $ k \in \mathbb{Z} $, $ k \geqslant 0 $ define
$ f(2k) = 0 $, $ f(4k+1) = 4k+8 $, $ f(4k+3) = -8 $. Then define $ f(x) $ for all
$ x \geqslant 0 $, such that $ f(x) $ will be linear at each interval $ [ k,k+1 ] $
for $ k \in \mathbb{Z} $, $ k \geqslant 0 $. Then define $ f(-x) = - f(x) $ for
$ x > 0 $. First of all, $ f $ is continuous on $ \mathbb{R} $. Also, for every
$ k \in \mathbb{Z} $, $ f $ is linear on the interval $ [ k,k+1 ] $ with
inclination  at least $ 8 $ in absolute value. Also it is easy to see that for every
$ k \in \mathbb{Z} $, $ f( (2k-1,2k+1) ) \supseteq ( -8,8 ) $ and for every
$ k \in \mathbb{Z} $, $ k \geqslant 1 $ we have
$ f( (-4k,4k) ) \supseteq (-4(k+1) , 4(k+1) ) $. Now I claim that for every open
interval $ J \subset \mathbb{R} $ we have $ \bigcup_{n=1}^{\infty} f^{(n)}(J) = \mathbb{R} $.
Indeed, suppose we have an interval $ J = (a,b) $, where $ a < b $. Suppose first
that it does not contain any of integers. Then $ f(J) $ will be an interval,
such that it's length is at least $ 8 $ times the length of $ J $. Therefore if
we look on iterations of $ J , f(J) , f(f(J)) , \ldots $, then at some place
we get the interval $ J' $ , which contains an integer number. Therefore $ J' $ contains
some interval of the form $ (m',m'+t) $ or $ (m'-t,m') $ for some $ m' \in \mathbb{Z} $ and
$ t > 0 $. Then if $ t < 1 $, then we have that $ f((m',m'+t)) $ ( or, respectively,
$ f((m'-t,m')) $ ) will be of the form $ (m'',m''+kt) $ or $ (m''-kt,m'') $ for some
$ m'' \in \mathbb{Z} $ and $ k \geqslant 8 $. Therefore , after some number of iterations,
we will get that for some $ n $ that $ J'' = f^{(n)}(J) $ contains some interval of the form
$ (m,m+1) $, where $ m \in \mathbb{Z} $. Therefore  $ f(J'') $ contains an interval
of length $ 8 $ with integer ends. Therefore there exists some $ k \in \mathbb{Z} $, such that
$ f(J'') \supset (2k-1,2k+1) $. Then $ f(f(J'')) \supset (-8,8) $. Then because of
$ f( (-4k,4k) ) \supseteq (-4(k+1) , 4(k+1) ) $ for every $ k \geqslant 1 $, by induction
we see that union of iterations of $ J $ by $ f $ is equal to $ \mathbb{R} $.
Therefore for every open interval $ J \subset \mathbb{R} $, we have that
$ \bigcup_{n=1}^{\infty} f^{(n)}(J) = \mathbb{R} $. Let's now prove that there exists
some $ x \in \mathbb{R} $, such that the set consisting of $ f(x),f(f(x)) , \ldots $
is dense in $ \mathbb{R} $. For this take some sequence $ r_{1},r_{2}, \ldots $ of
rational numbers, such that each rational number appears in this sequence infinite
number of times. Then we will inductively build some sequence
$ n_{1},n_{2}, \ldots \in \mathbb{N} $, and sequence of open intervals
$ J_{1} \supset J_{2} \supset \ldots $. Take $ J_{1} = (-1,1) $ and $ n_{1} = 1 $.
Then suppose that we have found $ n_{1},n_{2}, \ldots , n_{k} $ and
$ J_{1},J_{2}, \ldots, J_{k} $. Consider $ J_{k} $ and we have that
$ \bigcup_{n=1}^{\infty} f^{(n)}(J_{k}) = \mathbb{R} $. Therefore there exist some
$ n_{k+1} \in \mathbb{N} $ and $ x_{k} \in J_{k} $, such that $ f^{(n_{k+1})}(x_{k}) = r_{k+1}$.
Then, by continuity, there exists an open interval $ x_{k} \in J_{k+1} \subset J_{k} $,
such that $ | f^{(n_{k+1})}(x) - r_{k+1} | < 1/(k+1) $ for every $ x \in J_{k+1} $.
Of course we can take such a $ J_{k+1} $, such that $ J_{k} $ will contain a closure of
$ J_{k} $.
Therefore we have that for every $ k>1 $, for every $ x \in J_{k} $, we have that
$ | f^{(n_{k})}(x) - r_{k} | < 1/k $. Now, if we take some $ y \in \bigcap_{k=1}^{\infty}
J_{k} $, then $ | f^{(n_{k})}(y) - r_{k} | < 1/k $ for every $ k >1 $, and
because of the sequence $ r_{1},r_{2}, \ldots $ contains each rational number
infinite number of times, we get that the sequence $ f(x),f(f(x)) , \ldots $
is dense in $ \mathbb{R} $.

I hope it is correct.

It is correct. Now I know why I can't solve this  :(


AMC AIME geometry trigonometry algebra polynomial conics

Although I haven't looked at V1 much in a while, I think that it covers enough euclidian geometry to get any of those types of problems in the first 10.  But for all the other topics you will need V2.  Also, for the geometry problems, you will need to look at some AIME level geometry problems.  All/most of the concepts are in V1 but they don't really have the types of problems that would build up your experience level to get any in the first 10.

I wouldn't say so, but I guess if you can master all of the topics, you should do pretty good.

[quote="h_s_potter2002"]Is the first AoPS book enough to get at least a 8 or 9 on the AIME?  Because I've heard some people say that it's only enough to do good in MathCounts/AMC, but then There's a lot of knew stuff I'm learning, and I was able to do most of the problems on the AMC 10 last year, before I even looked at the books.[/quote]

Depends on the year and how deeply you understand the books.  Last year's test, for example, you could have gotten 10 or so with a deep mastery of Volume 1.  But you'd have a harder time with tests which have more emphasis on complex numbers, trig, polynomials, conics, and/or logarithms.

Could somone tell me a cheap way to buy AOPS books?cause on amazon they're tooooo expensive and on half.com they are out of stock...please help :?

Any cheaper way than that?cause the vol1 will come for around 35$ by the shippping thing.Cess64 once said he got the 2ndvol for 10$ at half.com.but the tocks over i gues.any other way?

Hey,thats the first time i wrote in $latex$ on this site,and that too by mistake :rotfl:  :rotfl: .I just meant to write 35dollars and 10dolars :rotfl:  :rotfl:

chess64 got very lucky. I can't find another one that cheap.


inequalities
calculus
function
inequalities proposed

find the maximum value of $x(1-x^{3})$ where $0\leqq\leq1$

I suppose you mean $x \leq 1$
 I use calculus b/c it is a natural way to solve your problem:
  Let $f(x)= x(1-x^{3})$  $0 \leq x \leq 1$
        $f'(x)= 1-4x^{3}$
    $f'(x)= 0$ when $x= \frac{\sqrt[3]{2}}{2}$
Plug the value to your function f , you will find the result . :)

HI
This solution use only :Am-Gm inequality.


induction
combinatorics proposed
combinatorics

Let $P$ be a convex n-gon with a triangulation, i.e. a dissection into triangles with non-intersecting diagonals.
Prove that it is possible to label all vertices of the n-gon with the (decimal) digits of $2007$ so that the sum of these labels in each quadrilateral of two adjacent (along an edge) triangles is $9$.

Isn't it a straightforward induction on $n$, by isolating one triangle 'on the border' of the $n$-gon, and by noting that in each quadrilateral the four vertices have received $0,0,2,7$ in some order?

Pierre.

[quote="pbornsztein"]
Isn't it a straightforward induction on $n$, by isolating one triangle 'on the border' of the $n$-gon, and by noting that in each quadrilateral the four vertices have received $0,0,2,7$ in some order?
[/quote]

Yes, this clearly works.

Does "Austrian National Olympiad 2007, Final Round" mean that this problem is from the Austrian TST?

Yes, this problem is indeed straightforward...

[quote]Does "Austrian National Olympiad 2007, Final Round" mean that this problem is from the Austrian TST?[/quote] Yes, in Austria, the final round of the national olympiad is at the same time also the IMO-TST.


calculus
derivative
function
calculus computations

I'm reading Calculus Made Easy and I'm on page 69 and 70 where the explain the quotient rule. Now, they do it in a complicated (tedious) way and get a complicated answer. Here's my method:

Let $ w\equal{}\frac{1}{v}$
$ y\equal{}uw\equal{}\frac{u}{v}$
$ y\plus{}dy\equal{}(u\plus{}du)(w\plus{}dw)$
$ y\plus{}dy\equal{}uw\plus{}(u)(dw)\plus{}(w)(du)\plus{}(du)(dw)$
Last term is infinitesimal of infinitesimal, so
$ y\plus{}dy\equal{}uw\plus{}(u)(dw)\plus{}(w)(du)$
Subtract $ y\equal{}uw$,
$ dy\equal{}(u)(dw)\plus{}(w)(du)$
$ \frac{dy}{dx}\equal{}u*\frac{dw}{dx}\plus{}w*\frac{du}{dx}$
Substituting,
$ \frac{dy}{dx}\equal{}u*\frac{d(\frac{1}{v})}{dx}\plus{}\frac{1}{v}*\frac{du}{dx}$

The question is, is this correct or am i missing something?

It's essentially correct, although you do need to calculate the derivative of the reciprocal function somewhere. Once you do that, the way to formalize it is to invoke the (already proved) product and chain rules.


geometry
trapezoid
circumcircle
incenter
national olympiad

Hi everyone here are the problems:(i hope that they aren't posted though i searched)
1-Let $ \triangle ABC$ be a triangle and $ (O)$ its circumcircle. $ D$ is the midpoint of arc $ BC$ which doesn't contain $ A$. We draw a circle $ W$ that is tangent internally to $ (O)$ at $ D$ and tangent to $ BC$.We draw the tangent $ AT$ from $ A$ to circle $ W$.$ P$ is taken on $ AB$ such that $ AP\equal{}AT$.$ P$ and $ T$ are at the same side wrt $ A$.PROVE $ \angle APD\equal{}90^\circ$.
2-There is given a trapezoid $ ABCD$.We have the following properties:$ AD\parallel{}BC,DA\equal{}DB\equal{}DC,\angle BCD \equal{}72^\circ$. A point $ P$ is taken on $ BD$ such that $ AD\equal{}AK,K \neq D$.Let $ M$ be the midpoint of $ CD$.$ AM$ intersects $ BD$ at $ N$.PROVE $ BK\equal{}ND$.
3-There is given a trapezoid $ ABCD$  in the plane with $ BC\parallel{}AD$.We know that the angle bisectors of the angles of the trapezoid are concurrent at $ O$.Let $ T$ be the intersection of the diagonals $ AC,BD$.Let $ Q$ be on $ CD$ such that $ \angle OQD\equal{}90^\circ$.Prove that if the circumcircle of the triangle $ OTQ$ intersects $ CD$ again at $ P$ then $ TP\parallel{}AD$.
4-Point $ P$ is taken on the segment $ BC$ of the scalene triangle $ ABC$ such that $ AP \neq AB,AP \neq AC$.$ l_1,l_2$ are the incenters of triangles $ ABP,ACP$ respectively. circles $ W_1,W_2$ are drawn centered at $ l_1,l_2$ and with radius equal to $ l_1P,l_2P$,respectively. $ W_1,W_2$ intersects at $ P$ and $ Q$. $ W_1$ intersects $ AB$ and $ BC$ at $ Y_1( \mbox{the intersection closer to B})$ and $ X_1$,respectively. $ W_2$ intersects $ AC$ and $ BC$ at $ Y_2(\mbox{the intersection closer to C})$ and $ X_2$,respectively.PROVE THE CONCURRENCY OF $ PQ,X_1Y_1,X_2Y_2$.
5-Two circles $ S_1$ and $ S_2$ with equal radius and intersecting at two points are given in the plane.A line $ l$ intersects $ S_1$ at $ B,D$ and $ S_2$ at $ A,C$(the order of the points on the line are as follows:$ A,B,C,D$).Two circles $ W_1$ and $ W_2$ are drawn such that both of them are tangent externally at $ S_1$ and internally at $ S_2$ and also tangent to $ l$ at both sides.Suppose $ W_1$ and $ W_2$ are tangent.Then PROVE $ AB\equal{}CD$.
TIME : $ 3$ HOURS.
u will see that the problems are rather easy...

[quote]2-There is given a trapezoid $ ABCD$.We have the following properties:$ AD\parallel{}BC,DA \equal{} DB \equal{} DC,\angle BCD \equal{} 72^\circ$. A point $ P$ is taken on $ BD$ such that $ AD \equal{} AK,K \neq D$.Let $ M$ be the midpoint of $ CD$.$ AM$ intersects $ BD$ at $ N$.PROVE $ BK \equal{} ND$[/quote]
hi all
i think there is some thing wrong in the question :maybe:
where is point p here excactely !!!
or i missed something
notice : i searched the forum to find any discution about that or correction but didnt find any thing
 :spider:

mr mohamed

[quote="shoki"]2-There is given a trapezoid $ ABCD$.We have the following properties:$ AD\parallel{}BC,DA \equal{} DB \equal{} DC,\angle BCD \equal{} 72^\circ$. A point $ P$ is taken on $ BD$ such that $ AD \equal{} AK,K \neq D$.Let $ M$ be the midpoint of $ CD$.$ AM$ intersects $ BD$ at $ N$.PROVE $ BK \equal{} ND$.
[/quote]
sorry;it should be :"...A point $ K$ is taken on $ BD$ such that..."

here are their discussion threads:
1-[url=http://www.mathlinks.ro/viewtopic.php?t=300805]Problem 1[/url]
2-[url=http://www.mathlinks.ro/viewtopic.php?t=300806]Problem 2[/url]
3-[url=http://www.mathlinks.ro/viewtopic.php?t=300807]Problem 3[/url]
4-[url=http://www.mathlinks.ro/viewtopic.php?t=300808]Problem 4[/url]
5-[url=http://www.mathlinks.ro/viewtopic.php?t=300809]Problem 5[/url]

It is quite perplexing that the geo-problems in 2009 were so easy, while those in 2008 were extremely difficult. :(


ratio
geometry
circumcircle
incenter
function
geometry proposed

[color=darkred]For a given $ \phi\in\left[\frac {\pi}{3},\pi\right)$ consider all triangles $ ABC$ for which $ \max\{A,B,C\}\le\phi$ . Find $ \boxed {\ f(\phi ) \equal{} \max_{ABC}\frac {IH}{IO}\ }$ ,

where $ O$ , $ I$ , $ H$ are the circumcenter, the incenter ans the orthocenter respectively of $ \triangle ABC$ .

[b][u]Remark.[/u][/b] Is well-known that $ f\left(\pi \minus{} 0\right)\le 2$ , $ f\left(\frac {\pi}{2}\right)\le\sqrt 2$ , $ f\left(\frac {\pi}{3}\right) \equal{} 1$ .[/color]

I think ${ \max_{ABC}\frac {IH}{IO}\ }$ is not a function of $ \phi$  :?: don't I?

you may like this fact: $ IH^2 \equal{} 4IO^2 \minus{} (a^2 \plus{} b^2 \plus{} c^2 \minus{} ab \minus{} bc \minus{} ca)$  :D

O.K. I like very much your idea, [b]Campos[/b]  ! Thank you. I had know this relation.


can u use factorials? theres something like this in the middle school-basics forum called Klebian's 24. there u can also use factorials.
and use two 5's to make [color=green]5[/color][color=red]5[/color]

if we can use factorials: $(5-1)!+5-5$ and $(5-5+5-1)!$

No..you can only use + - * / (anytime you want.you can use 2 pluses etc..)..no powers,no factorials etc..anything else?!

[quote="georgech"]Try this one,
You should use the numbers 5 5 5 1 one time each and the symbols + - * / one time each also and arrange them in such a way that you will have result 24..(i.e 5+(5-1)*5)...(you can use parentheses ).
Try any other way than ((5-(1/5))*5).[/quote]

This reminds me of  the 24 Game we used to do in elementary school.

yes..it''s easy to find 25..i know..i can't think of anything else ..:)

[quote="georgech"]Try any other way than ((5-(1/5))*5).[/quote]Are you sure there's any other way than $\left( 5-\frac15\right)\cdot 5=24$? :?

is there a trick to this, like use negatives of the numbers?

[quote="joml88"]$5\cdot5-1_5$[/quote]

that is same as mine aneways

two evil ones:

1 3 4 6 to make 24
3 3 8 8 to make 24

same rules as above. I think there's only one solution to each of these, and most people never think of it.

another one someone showed me (to get back at me for showing them the first two):

7 7 7 7 1 to make 100.

Hi,

For the first one : $\frac{6}{1-\frac{3}{4}}$ Is there any other solution? :?

This one's been driving me crazy for the last 2 days!

1 6 8 9   (24)

And also,
9 6 6 3   (24) I got it, but takes some work...

[hide]Okay, so I'm just kidding around.  :D
On a more serious note, has anyone heard of a slight variant of this game? I've played this game where you have to use the number 4, four times to get each of the integers from 1 to 20. Very similar game, maybe we should find some combination of numbers that'd be hard to do it with.
[/hide]

OK. who can find the minimum number of 4s, that can be used to get 2004? Provide your solution, too.

Hey, that wasn't so bad.

[hide="Click for my solution"]$(4*4*4*4)(4+4) - 4*4*4 + 4*4 + 4 = 2004$ (12 4s, no exponentiation)[/hide]
[i]1 6 8 9 (24)
9 6 6 3 (24)[/i]

Look more closely. The sum of each of these groups is 24. :blush:

How about some harder stuff? Use as few 17s as you can to get the following numbers (using only +, -, *, /, and () - no powers, concatenation, decimals, factorials...):
- 1000 (I can do it in 12)
- 1337 (I can do it in 12)
- 2006 (I can do it in 10)
Good luck![/i]

What is the minimum number of $0$'s needed to make 2004?  :P

[hide]$0!=1$[/hide]


Show that for every natural number $n$ the product \[\left( 4-\frac{2}{1}\right) \left( 4-\frac{2}{2}\right) \left( 4-\frac{2}{3}\right) \cdots \left( 4-\frac{2}{n}\right)\] is an integer.

[quote="Peter"]Show that for every natural number $ n$ the product
\[ \left( 4 \minus{} \frac {2}{1}\right) \left( 4 \minus{} \frac {2}{2}\right) \left( 4 \minus{} \frac {2}{3}\right) \cdots \left( 4 \minus{} \frac {2}{n}\right)
\]
is an integer.[/quote]

We have that
\[ \prod_{i \equal{} 1}^{n}{(4 \minus{} \frac {2}{i})} \equal{} \frac {\prod_{i \equal{} 1}^{n}{2(2i \minus{} 1)}}{n!} \equal{} \frac {2^{n}(2n \minus{} 1)!!}{n!} \equal{} \frac {(2n)!}{n!n!} \equal{} C_{2n}^{n}
\]
Where $ C_{2n}^{n}$ is a binomial coefficient and, as we know, it is a positive integer.

You can also say that

$\frac{\left( 4-\frac{2}{1}\right)\left( 4-\frac{2}{2}\right)\left( 4-\frac{2}{3}\right)\cdots\left( 4-\frac{2}{n}\right)}{n+1}$ is an integer because it is the $n^{th}$ catalan number.

Some more proofs of it may be seen in http://www.artofproblemsolving.com/Forum/viewtopic.php?f=56&t=349870

[hide="A55"] Note that $p_n = \frac{2 \cdot 6 \cdot ... \cdot (4n-2)}{n!} = \frac{2^n(1 \cdot 3 \cdot ... \cdot (2n-1))}{n!} =\frac{(2n)!}{n!n!} = \dbinom{2n}{n}$, an integer. [/hide]


function
algebra
partial fractions
geometric series

[url=http://en.wikipedia.org/wiki/Partial_fractions#Distinct_first-degree_factors_in_the_denominator]Description of the relevant approach[/url]

Possible hint: does this sum telescope when broken into partial fractions?
Can you look at it as a geometric series, if it does?

Fishy, it's considered VERY bad form to do arithmetic "with infinity." What you're innately trying to do is make an argument about the limiting behavior of a function without rigorously using limits, a behavior which can lead to some false conclusions.

It's good that you've made the telescoping step there, but the notation you used is a bit off-putting.

[quote="sunehra"][quote="fishythefish"]$ \displaystyle\sum_{k \equal{} 1}^{\infty}\frac {1}{k(k \plus{} 1)(k \plus{} 2)}$

$ \frac {1}{k(k \plus{} 1)(k \plus{} 2)} \equal{} \frac {1}{2}\left[\frac {1}{k(k \plus{} 1)} \minus{} \frac {1}{(k \plus{} 1)(k \plus{} 2)}\right]$

So, $ \displaystyle\sum_{k \equal{} 1}^{\infty}\frac {1}{k(k \plus{} 1)(k \plus{} 2)} \equal{} \frac {1}{2}\displaystyle\sum_{k \equal{} 1}^{\infty}(a_k \minus{} a_{k \plus{} 1}) \equal{} \frac {1}{2}(a_1 \minus{} a_{\infty}) \equal{} \frac {1}{2}\left(\frac {1}{2} \minus{} \frac {1}{\infty}\right) \equal{} \frac {1}{4}$[/quote]

Can I have some clarification on some of those steps? I got lost at step 2.[/quote]

$ \frac{1}{k(k\plus{}1)(k\plus{}2)}$

$ \equal{} \frac{1}{2}\times\frac{(k\plus{}2)\minus{}k}{k(k\plus{}1)(k\plus{}2)}$

$ \equal{} \frac{1}{2}\left[\frac{k\plus{}2}{k(k\plus{}1)(k\plus{}2)} \minus{} \frac{k}{k(k\plus{}1)(k\plus{}2)}\right]$

$ \equal{} \frac{1}{2}\left[\frac{1}{k(k\plus{}1)} \minus{} \frac{1}{(k\plus{}1)(k\plus{}2)}\right]$

Sorry, guys. My cousin was looking over my shoulder, and he wanted to solve this one, so I let him borrow my account. I don't get most of this either. (But, then again, he's a Teaching Assistant at MSU...)

Anyways, I was still trying to figure out how to telescope this series. I noticed some of the problems too, such as the division by infinity.


inequalities
algorithm
probability
linear algebra
matrix
expected value

I just thought about this, and I think that it's a good idea.

Let's try starting a game where someone mentions a semi-specific math topic (so that they can learn about it), and the next person has to explain the topic as clearly as possible. Then the person that originally asked about the topic says if they understand or not. If they don't, someone else can try to explain it.

If you successfully describe a topic, then you can ask about a topic yourself.

I know it sounds corny, but this could be interesting.

I'll start by asking about....

Chebyschev's Inequality

<- Didn't understand it at all.

Then again, I normally don't understand t0rajir0u posts.

First a word on Chebyshev's inequality.
It states that in a univariate data set, at most one-kth of the data points can differ from the mean by more than k standard deviations.
Also, in a probability distribution, the probability that the random variable takes a value that differs from the expected value by more than k standard deviations is not more than one over k.
 :rotfl: Sorry, but this is the Chebyshev's inequality that we know. :roll:

Now (after a quick look at Graham-Knuth-Patashnik), Stirling numbers of the second kind, denoted [[n, k]], are the number of distinct decompositions of a set of size n into k disjoint cycles. eg, [[4, 2]] is five, because {a, b, c, d} can be broken into 2 cycles in eleven ways:
{[a, b, c], [d]}, {[a, c, b], [d]}, {[a, b, d], [c]}, {[a, d, b], [c]}, {[a, c, d], [b]}, {[a, d, c], [b]}, {[b, c, d], [a]}, {[b, d, c], [a]}, {[a, b], [c, d]}, {[a, c], [b. d]}, {[a, d], [b, c]}.

Now who will tell me about the Generalized Singular Value Decomposition of a matrix over the complex field?

[quote="t0rajir0u"]Chebyshev's Inequality is just the Greedy algorithm in disguise.  Consider:  Given that you can assign coefficients $y_{1}, y_{2}, ... y_{n}$ to a weighted sum of the values $x_{1}, x_{2}, ... x_{n}$, how would you assign them to maximize the sum?

Obviously the easy way to approach this problem is to look at the maximal element in $x_{i}$ and assign the corresponding maximal coefficient $y_{i}$ to it, and so forth.  This is the essence of the statement of Chebyshev.  It also suggests a quick proof by induction.

I want to ask about...

Stirling Numbers of the Second Kind.   :P[/quote]

so you are given the coefficients, and then assign them, or can you use chebyshev when there is merely a condition imposed upson the coefficients (such as $\sum y_{i}= c$, where $c$ is a given constant)?
can you show the actual inequality?

by the way, from the way i originally intended that the game proceed, i would have to say, ok i understand, before you can post. also, any number of people can try and explain a concept. when the person that originally asked the question understands one of the explanation, he should say "ok, i understand this explanation."

so basically, everytime somone posts something that they want to learn about, everyone else sees who can give the clearest and most informative explanation.

[quote="t0rajir0u"][quote="drunner2007"]can you show the actual inequality?[/quote]

Oops, sorry, I was actually explaining Rearrangement on accident.  :oops:  But Chebyshev's does follows from Rearrangement.

The two-sequence case of Chebyshev states that if two sequences of reals $a_{i}, b_{i}$ satisfy

$a_{1}\le a_{2}\le ... \le a_{n}$
$b_{1}\le b_{2}\le ... \le b_{n}$

Then

$\frac{1}{n}\sum_{i=1}^{n}a_{i}b_{i}\ge \left( \frac{1}{n}\sum_{i=1}^{n}a_{i}\right) \left( \frac{1}{n}\sum_{i=1}^{n}b_{i}\right)$

Why this follows from Rearrangement:  The LHS is the maximum possible value of the "weighted sum"

$a_{1}b_{\delta(1)}+a_{2}b_{\delta(2)}+...+a_{n}b_{\delta(n)}$

Where $\delta$ is some permutation.  Expanding the RHS gives $n$ different permutations of the above sum, all of which are less than the LHS.[/quote]

that is a pretty good explanation. i understand it now  :)

you can ask about a topic now. remember not too ask about something too obscure or else the game will be stalled. just think about some part of math where you have a gap in your knowledge but you know others on here can fill you in.


How do you write in bold numbers to like make the fractions look good(See below)?   Are there techniques?

 :sqrt: 2 :!=:  :pi: [/b][/quote]

[quote="jmerry"]That's $\LaTeX$, not bold. Enclose your code in dollar signs to make it work

\sqrt{2}\neq\pi
$\sqrt{2}\neq\pi$

There's a lot of material on the site on how to use $\LaTeX$, and most of it is pretty natural- you can often find the code for something by trial and error (use preview for this).[/quote]

or if you want to see how somebody else texed their stuff, move your cursor on top of the LaTex. It is either linked to a popup that shows the code, or it appears.

Take a good look at my signature. It quickly explains everything ;)

[quote="jmerry"]That's $\LaTeX$, not bold. Enclose your code in dollar signs to make it work

\sqrt{2}\neq\pi
$\sqrt{2}\neq\pi$

There's a lot of material on the site on how to use $\LaTeX$, and most of it is pretty natural- you can often find the code for something by trial and error (use preview for this).[/quote]
how do you make the latex look funky?  You did it twice.


trigonometry
calculus
calculus computations

Assume that $ \phi(x)\equal{}\left\{\begin{array}{ll}
x^3\sin\frac{1}{x} & x\neq0\\
0 & x\equal{}0
\end{array}
\right.$

$ f(x)$ is derivable when $ x\equal{}0$.

Find $ F'(x)$, here $ F(x)\equal{}f[\phi(x)]$

Which part is giving you trouble?

Here's some help: $ \phi'(x)\equal{}3x^{2}\sin\frac{1}{x}\minus{}x\cos{\frac{1}{x}}$.

[quote="JRav"]Which part is giving you trouble?

Here's some help: $ \phi'(x) \equal{} 3x^{2}\sin\frac {1}{x} \minus{} x\cos{\frac {1}{x}}$.[/quote]

Do you mean that we can simply apply the chain's rule and get the following answers?

$ (f[\phi(x)])'\equal{}f'[\phi(x)]\phi'(x)$

I just wonder why we don't use the condition $ f(x)$ is derivable when $ x\equal{}0$? Can't we find a proper $ f$ which is derivable iff $ x\equal{}0$? I wonder if something wrong with this problem :blush:  :blush:  :blush:

You still need $ F'(0)$, which you can't get from that formula.


I think time management is a big deal on the AIME, so how long do you suggest me to spend on first 5, second 5 and the last 5 problems during the contest?

How well do you expect to do?  If you're aiming for say, only 1-3 correct answers, then take as much time as you need to solve a problem if you think you can.  If you're aiming for 10+, then don't waste too much time on the first few problems.  Simple math tells you that you should spend on average 12 minutes per problem, but of course, the problems at the end will take much longer than the problems at the beginning.  So use 12 minutes as a guide.

At the same time, try not to rush through the first few problems. I missed number two because of this (and many other people missed number one) on the 2009 AIME I. Also, on the later problems (where it may be harder to correct mistakes when it's too late), try to mentally track your calculations and problem solving process to find potential errors. Also, it's not necessary to do the problems in any particular order--if #14, for example, looks really routine while #8 might require some minutes of thought, try to finish it first (you will probably be subconsciously thinking about #8).

I generally take about 20 minutes for the first 5, 40 for the next 5, and 60 for the last 5. I then use an hour to check all my work.

No work check, but 30 minutes for the first seven, 30 minutes for the next three, and 2 hours for the last five.

For me, an hour for the first 7-9, and then 2 hours picking off any of the rest that seem solvable. More importantly, spend enough time to make sure your answer is probably correct, don't trick yourself into moving on if you have a slight feeling that it might be wrong. Don't skip around more than you need to, you won't get anything done, and then you'll start getting anxious and get even less done. Don't spend too much time on a problem not progressing.

I don't leave time to check work, but I probably should.

Since an 8 would give me a pretty good chance, im shooting to solve 9 or 10, just in case.
So, first 7-8 in hour and 15 minutes. Try as many out of 8-11 for 45 minutes. See if any of the last problems are possible for 40 minutes. Check my work for 20.

i usually do first 5 in about 30 min. and then skip around 6-12 in order of increasing difficulty for 2 hrs. (i solved 11 and 12 before 10). after solving the first 12, look for the easiest of 13-15, attempt it for 15 min. If I'm getting knowhere, I just check the arithmetic on previous problems.

ugh AIME I this year took me 40 minutes for #1-8

then I got two of those wrong and only solved three more (getting one of those wrong)? lol

I'm aiming for a 6+, any specific suggestions?

Make sure that you solve at least the first 5 correctly and then hope that you get at least 1 of the next 10.  The most important thing is to budget your time.  If you look at a problem and you don't think you'll be able to solve it quickly, move on and come back to it later.

I say solve seven in and hour and a half, check them for 45 minutes, spend then next 45 minutes trying to do other problems.


number theory
relatively prime
greatest common divisor
IMO Shortlist

Let $ a > b > 1$ be relatively prime positive integers. Define the weight of an integer $ c$, denoted by $ w(c)$ to be the minimal possible value of $ |x| \plus{} |y|$ taken over all pairs of integers $ x$ and $ y$ such that \[ax \plus{} by \equal{} c.\] An integer $ c$ is called a [i]local champion [/i]if $ w(c) \geq w(c \pm a)$ and $ w(c) \geq w(c \pm b)$.

Find all local champions and determine their number.

[i]Proposed by Zoran Sunic, USA[/i]

in the original version, this problem did not ask for all local champions.

there were two questions

(a) show that there are only finitely many local champions

(b) show that there exists at least one local champion

both questions allow for many different (and even elegant) solutions.

in my opinion, the beauty of the problem was completely distroyed by asking for all local champions, since many of the interesting ideas and approaches that work for (a) and (b) do not work anymore and one is left with a somewhat tedious work od grinding out all solutions.

Not hard but long
[hide]These are just the steps and...
Lemma1: XY<0
First we notice that if C=ax+by then C=a(x-bk)+b(y+ak) for arbitrary k.
Then just consider a local champion C and W(c+a).
So then wlog c=ax-by which x,y have same sign(wlog positive).
Now two of local champion properties are obvious.
The other ones show that k>0 and |x|<b and |y|<a then if for a local champion c, we consider the minimum k which weight of a(x+1-bk)-b(y-ak) is smaller than |x|+|y| , writing all the 9 cases of y<ak , ... Shows that the minimum k should be 1.
And then it's easy to see that the only local champions are the ones which x+y = $\lfloor \frac {a+b}{2} \rfloor$ and they all work.
I'm sorry  that I skipped the details, I try to post the complete solution later.   [/hide]


1.) Find the amount that will be in an account, given that        P = 1000, t = 5, r = 12% compounded ANNUALLY

2.) Find the inverse of the function f (x) = x + 9

This isn't your homework is it? :huh:

This isn't intermediate.

[hide="2"]General way to find inverses: Replace $x$ with $f^{-1}(x)$ and $f(x)$ with $x$ in the original equation, and then solve for $f^{-1}(x)$.[/hide]

Thanks for the hints.  Does anyone know the answers? :mad:

The point of AoPS isn't to do your homework.

That being said, I'll do it anyways.

1) $1000(1.12)^{5}$, or $\boxed{1762.34}$
2) f(x) = x - 9

Obviously this is homework(#1 is especially suspicious)......
For #1, there's a formula that you can use to get the answer. Look in your textbook or google it.
For #2, Just switch x and y. In this case, we start with y=x+9, so we switch x=y+9. Solving for y, we get $y=x-9=f^{-1}(x)$. That should be in your textbook too.
p.s. wtf is up with you? an angry smiley bc he gave you a hint?
p.p.s. randomdragoon is right. this definitely does not belong in intermediate. more like in classroom math forum under middle school.
edit treething beat me.

Treething you rock the nation.  For number 2 did you mean
f-1(x) = x - 9

Also can I throw out a couple more and see if you can help?

I meant $f^{-1}(x)$, and okay, I'll do a couple more.


$P,Q,R$ are represented by complex number $p,q,r$ in a complex plane respectively . Prove that if

$Arg\left(\frac{r-p}{r-q}\right)=\pm\frac{\pi}{2}$

then

$R$ lies on the circle with $PQ$ as diameter .

The given condition means for fixed $P$ and $Q, \vec{PR}$ and $\vec{QR}$ are perpendicular to each other. This implies $R$ is a point on the circle with $PQ$ as diameter. $\pm{90^{\circ}}$ determines the "orientation" of the point $R$.

[quote="Farenhajt"]Since $\arg(\frac{r-p}{r-q})=\pm{\pi\over 2}$, there exists a real number $m$ such that $\frac{r-p}{r-q}=mi$ (for $m>0$ the argument will be positive, and for $m<0$ the argument will be negative).

Then

\begin{eqnarray*}|1+mi|=|-1+mi| &\implies& \left|\frac{r-p+r-q}{r-q}\right| = \left|\frac{r-p-r+q}{r-q}\right|\\ &\implies& |2r-(p+q)|=|p-q|\\ &\implies& \left|r-\frac{p+q}{2}\right|=\left|\frac{p-q}{2}\right|\end{eqnarray*}

Now observe that $\left|\frac{p-q}{2}\right|$ is the radius of the circle with diameter $PQ$ and that $|r-\frac{p+q}{2}|$ is the distance between $R$ and the midpoint of $PQ$ - hence the conclusion.[/quote]

nice job :) . Im curious how you figure out to use
$|1+mi|=|-1+mi|$ ?

working backwards ? :P


Three points are chosen randomly and independently on a circle. What is the probability that all three pairwise distances between the points are less than the radius of the circle?

Answer: 1/12

My question:

4 points are chosen randomly and independently on a circle. What is the probability that all 4 pairwise distances between the points are less than the radius of the circle?

The first question is easy to solve using geometric probability, but I don't know how to solve my problem. Somebody help please!

[hide="Method"]You do it with geometric probability in higher dimensions, pretty much. Imagine an 3-dimensional (or $ (n \minus{} 1)$-dimensional generally) hypercube being moved between two points, and finding the volume of the region it travels through. This is actually not too difficult because you can find the initial volume of the hypercube, then multiply the distance it travels by the amount of volume accumulated each time, equal to the surface area of half of its faces.[/hide]

[hide="Answer for n points"]$ n/6^{n \minus{} 1}$[/hide]

I don't understand the "hypercube" part, could you elaborate a little?

Not sure exactly what you're confused about. The three dimensions comes about by fixing one of the points you pick and defining the position of the others to be the angle they make with the center and that fixed point. The position of the other $ n \minus{} 1$ points then vary in $ n \minus{} 1$-dimensional space. Let those $ n \minus{} 1$ angles be $ \theta_1, \theta_2, \ldots \theta_{n \minus{} 1}$. The problem's condition is satisfied if $ \theta_i \in [ \minus{} \pi/3,\pi/3]$ (so that they are close enough to the fixed point) and $ | \theta_i \minus{} \theta_j | \leq pi/3$ (so that the others are close enough to each other) for all $ i,j$. The second condition can be described geometrically by the union of all cubes defined by $ [x,x \plus{} \pi/3]^{n \minus{} 1}$ (that's a cartesian product of closed intervals).

Attached a picture of the approximate shape for 3 dimensions, or $ n \equal{} 4$. Note that the yellow and red cubes share a vertex.

So the "good" region is the red/yellow/orange region?


calculus
invariant
algebra
system of equations
partial fractions

Compute the sum of the infinite sries: \[ \frac 23 + \frac 14 + \frac{2}{15} + \cdots + \frac{2}{n(n+2)} + \cdots  \]

Nice job. Just curious, what's the general way to decompose a fraction like that?

Heh, I missed one of those problems at the San Diego Math League, but after seeing the solution, learned how to apply calculus concepts to decompose a general fraction.

Here it is: Partial Fraction Decomposition

Suppose you want to decompose the fraction $\frac{x^2+2x-1}{x(2x-1)(x+2)}$. I made it inconvenient to show that it works for large fractions. :P

First, write $\frac{x^2+2x-1}{x(2x-1)(x+2)}=\frac{A}{x}+\frac{B}{2x-1}+\frac{C}{x+2}$.

To determine A, B, and C, find the common denominator: $\frac{A(2x-1)(x+2)+Bx(x+2)+Cx(2x-1)}{x(2x-1)(x+2)}$.

Now notice that in the numerator $\frac{x^2+2x-1}{x(2x-1)(x+2)}=\frac{A(2x-1)(x+2)+Bx(x+2)+Cx(2x-1)}{x(2x-1)(x+2)}=\frac{(2A+B+2C)x^2+(3A+2B-C)x-2A}{x(2x-1)(x+2)}$

Now, we set up a system of equations so that $2A+B+2C=1$, $3A+2B-C=2$, and $-2A=-1$.

Solving the system of equations, $A=\frac 1 2$, $B=\frac 1 5$, and $C=-\frac 1 {10}$.

Thus, we have decomposed the original fraction into $\frac{A}{x}+\frac{B}{2x-1}+\frac{C}{x+2}= \frac{1}{2x}+\frac{1}{5(2x-1}+\frac{1}{10(x+2)}$

What grade do I sound like I'm in? Just out of curiosity. ;)

Partial fractions:
$\frac{an+b}{(gn-c)(hn-d)}=\frac{e}{gn-c}+\frac{f}{hn-d}$, $e,f$ unknown.

Multiplying through by $(gn-c)(hn-d)$, we have:
$an+b=e(hn-d)+f(gn-c)$

Equating the coefficients of $n$ and the constant terms, we have:
$a=eh+fg$
$b=ed-fc$

Solve the system for $(e,f)$ and you are set.

Practice? :)

1. Decompose $\frac{1}{n(n-1)}$

2. Decompose $\frac{3x-1}{x(2x-1)}$

Edit: 4everwise beat me to it. :)

[quote="1234567890"]Partial fractions:
$\frac{an+b}{(gn-c)(hn-d)}=\frac{e}{gn-c}+\frac{f}{hn-d}$, $a,b,c,d,e,f,g,h$ constants.

Multiplying through by $(gn-c)(hn-d)$, we have:
$an+b=e(hn-d)+f(gn-c)$

Equating the coefficients of $n$ and the constant terms, we have:
$a=eh+fg$
$b=ed-fc$

Solve the system for $(e,f)$ and you are set.

Practice? :)

1. Decompose $\frac{1}{n(n-1)}$

2. Decompose ${3x-1}{x(2x-1)}$[/quote]Too slow 12345678980. :P

Don't forget to include all powers. \[ \frac{P(x)}{(x-a)^n(x-b)^m}=\frac{A_1(x)}{x-a}+\ldots+\frac{A_n(x)}{(x-a)^n}+\frac{B_1(x)}{x-b}+\ldots+\frac{B_m(x)}{(x-b)^m}. \]

[quote="Hokkage"]Cool, thanks guys. I got it at SDML by guessing since it was multiple choice. I might know you 4everwise. Are you in 10th grade?[/quote]I was wondering why an SDML problem was posted.

I'm in 9th grade. ;) My name is Edward, Vincent is Chesspro. :)

What's you name? I might know you too. :)

I'm George. 10th grade. Won the invariant subtract-a-factor game at the invariants lesson by Rusczyk.
Hey Vincent: E4


vector
function
topology
calculus
analytic geometry
calculus computations

Let $ F$ be a vector field in space. Let $ g \equal{} g(x, y, z)$ be a scalar function which does not take the value $ 0$ such that $ gF$ is a gradient of a function. Prove that $ F$ and $ curl F$ are perpendicular.

I like manifold language, so I'll start by putting this in those terms.The curl and gradient are both examples of the $ d$ operator; the gradient takes scalar functions ($ 0$-forms) to $ 1$-forms, while the curl takes $ 1$-forms to $ 2$-forms.
We have a scalar function $ H$ such that $ g\cdot F\equal{}dH$. We want to show that $ F$ and $ dF$ are perpendicular. Oops- those aren't even in the same space. What do we mean by "perpendicular"?
The definition of the curl you see in multivariable calculus depends on a map (specific to $ \mathbb{R}^3$) from $ 2$-forms to $ 1$-forms; we take $ dx\wedge dy$ to $ dz$, $ dy\wedge dz$ to $ dx$, and $ dz\wedge dx$ to $ dy$. Then, we interpret the dot product between one of these and a one-form; $ dx\cdot (dy\wedge dz)\equal{}1$, $ dx\cdot (dx\wedge dy)\equal{}0$, and so on. This can be interpreted in terms of forms; $ dx\wedge dy\wedge dz$ is a $ 3$-form, and every $ 3$-form is a multiple of those. Since $ dx\wedge dx\wedge dy\equal{}0$, and similarly anywhere else a coordinate repeats, this wedge acts the same as the dot product. We want to show that $ F\wedge dF\equal{}0$.

Since the wedge product is alternating, $ F\wedge F\equal{}0$. Multiply by $ g$, and $ gF\wedge F\equal{}dH\wedge F\equal{}0$. Now, differentiate it: $ 0\equal{}d(dH\wedge F)\equal{}d^2H\wedge F\minus{}dH\wedge dF\equal{}0\wedge F\minus{}gF\wedge dF\equal{}\minus{}g(F\wedge dF)$ and $ F\wedge dF$ is identically zero as desired. We used two key properties here: $ d^2$ is always zero, and $ d$ has a product rule. That - sign comes from the fact that $ dH$ is a $ 1$-form, and would be $ (\minus{}1)^k$ if it were a $ k$-form.

We can translate this back into the $ \mathbb{R}^3$ language: the wedge product of two $ 1$-forms is a cross product, the wedge product of a $ 1$-form and a $ 2$-form is a dot product or the negative of one, and $ d$ is gradient/curl/divergence as appropriate:
$ F\times F\equal{}0$, so $ gF\times F\equal{}0$. Differentiating, $ 0\equal{}\text{div }(\text{grad } H\times F)\equal{} \text{curl } (\text{grad } H)\cdot F\minus{}\text{grad } H\cdot \text{curl } F\equal{}\minus{}\text{grad } H\cdot\text{curl } F$
In symbols, that's $ 0\equal{}\nabla\cdot(\nabla H\times F)\equal{}(\nabla\times(\nabla H))\cdot F\minus{}\nabla H\cdot(\nabla\times F)\equal{}\minus{}\nabla H\cdot(\nabla\times F)$.

It's nice to see problems like this get killed by differential forms.  :lol:

It's all just a matter of perspective. The key identity is either (a) an arcane combination of every symbol you've ever seen in differential vector calculus, or (b) just another case of the product rule for derivatives.


If i= sqrt-1, then what is 8-4i/4+2i. I would appreciate it if you show me the steps.

multiply the fraction by $ \frac{4\minus{}2i}{4\minus{}2i}$ to rationalize the denominator.

$ \frac{(8\minus{}4i)(4\minus{}2i)}{(4\plus{}2i)(4\minus{}2i)}\equal{}\frac{24\minus{}32i}{20}\equal{}\frac{6\minus{}8i}{5}$

To ease calculations, first reduce it to
$ \frac {8 \minus{} 4i}{4 \plus{} 2i} \equal{} 2\left(\frac {2 \minus{} i}{2 \plus{} i}\right)$, and then multiply by $ \frac{2\minus{}i}{2\minus{}i}$.

Is the $ 6$ supposed to be there? Should it be $ 60$?

Divide odd and even terms:
Odd terms: $ 5,5,6,10,19,35,60\ldots$
If $ A_x$ denotes the $ x$th odd term, then $ A_{x \plus{} 2} \equal{} A_{x \plus{} 1} \plus{} x^2$ for $ x\geq0$, and the formula for the $ x$th odd term is $ A_x \equal{} \frac13x^3 \minus{} \frac32x^2 \plus{} \frac {13}6x \plus{} 4$.

Even terms: $ 6,9,15,26,44,71,109\ldots$
If $ a_x$ denotes the $ x$th even term, then $ a_x \equal{} \frac13x^3 \minus{} \frac12x^2 \plus{} \frac {13}6x \plus{} 4$.

Edit:
You can combine this to get that the $ x$th term of the sequence $ S$ is
$ S_x \equal{} \frac13x^2 \minus{} \frac {( \minus{} 1)^{\left\lfloor\frac n2\right\rfloor} \minus{} 2}2x^2 \plus{} \frac {13}6x \plus{} 4$

whoa howd u do that

theres a new topic about this problem...

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=237793[/url]

[url=http://mathworld.wolfram.com/FiniteDifference.html]Finite Difference.[/url]

[url=http://en.wikipedia.org/wiki/Finite_difference]Here as well.[/url]


integration
logarithms
calculus
absolute value
real analysis
real analysis unsolved

I don't think that liyi's answer is correct, and my calculator would seem to indicate as much (although it could be $\pi^2/3$...)

Also, why is cezar writing $d\lambda$?

well, the absolute value is correct. but I missed the negative sign. It has been fixed now.
The numerical integration given by Mathematica is -1.64493...
and $\pi^2/6$ is 1.64493...

I think you missed the square - I get

$\int^1_0 \frac{\log t}{1-t}dt=\frac{\pi^2}{6}$...

Oh...
Thanks for pointing out that.
I've removed the wrong solution

Your solution to the other problem WAS nice, though :)


Where do we get the results of the past nationals... I only found 2003 and 2004...

They might not publish them anymore, in which case you'd have to find someone with them.

Some [url=http://www.mathcounts.org/archives/newsletters.html]old MATHCOUNTS newsletters[/url], going back about 5 years, are available.

Highlights of the previous National competition are usually in the Fall newsletters.


Let be $ a,b>0$ . Solve in $ \mathbb{R}$ this equation :

$ a^{\frac{x}{b}}\plus{}b^{\frac{x}{a}}\equal{}(a^a\plus{}b^b)^{\frac{x}{ab}}$

[quote="alex2008"]Let be $ a,b > 0$ . Solve in $ \mathbb{R}$ this equation :

$ a^{\frac {x}{b}} \plus{} b^{\frac {x}{a}} \equal{} (a^a \plus{} b^b)^{\frac {x}{ab}}$[/quote]

This equation may be written $ (\frac{a^a}{a^a\plus{}b^b})^{\frac{x}{ab}}$ $ \plus{}(\frac{b^b}{a^a\plus{}b^b})^{\frac{x}{ab}}\equal{}1$

Consider then $ f(x)\equal{}t^x\plus{}(1\minus{}t)^x$ with $ t\in(0,1)$. We know that  $ t^x$ and $ (1\minus{}t)^x$ are decreasing functions, and so is $ f(x)$. So the equation $ f(x)\equal{}1$ has at most one root. Since $ f(1)\equal{}1$, this is the unique root.

As a consequence, $ x\equal{}ab$ is the unique root of  $ (\frac{a^a}{a^a\plus{}b^b})^{\frac{x}{ab}}$ $ \plus{}(\frac{b^b}{a^a\plus{}b^b})^{\frac{x}{ab}}\equal{}1$, and so of the requested equation.


calculus derivative trigonometry function

Let $ f(x) \equal{} (x^2 \minus{} x).sin(x \minus{} 1)$

Find $ f^{(n)}(x)$ ?

This shouldn't be in Intermediate Topics, but...
[hide="Hint"]Each successive derivative can be written in the form of $ a\sin (x\minus{}1) \plus{}b\cos (x\minus{}1)$, where $ a$ and $ b$ are polynomials.[/hide]

[hide="Solution"]
From the product rule, we can form a recursion:
$ f^{(n)}(x)\equal{}a_n$
$ a_0\equal{}f(x) \equal{} (x^2 \minus{} x)\cdot\sin(x \minus{} 1)\plus{}0\cdot\cos(x\minus{}1)$
If $ a_n\equal{}a\sin (x\minus{}1) \plus{}b\cos (x\minus{}1)$, then $ a_{n\plus{}1}\equal{}(a'\minus{}b)\sin(x\minus{}1)\plus{}(a\plus{}b')\cos(x\minus{}1)$.

I'm guessing, however, that you're looking for a at least somewhat-closed form (i.e. a piecewise function), not a pseudo-recursion. Let's just find some of the derivatives.

$ a_1\equal{}(2x\minus{}1)\sin(x\minus{}1)\plus{}(x^2\minus{}x)\cos(x\minus{}1)$
$ a_2\equal{}(\minus{}x^2\plus{}x\plus{}2)\sin(x\minus{}1)\plus{}(4x\minus{}2)\cos(x\minus{}1)$
$ a_3\equal{}(\minus{}6x\plus{}3)\sin(x\minus{}1)\plus{}(\minus{}x^2\plus{}x\plus{}6)\cos(x\minus{}1)$.
$ a_4\equal{}(x^2\minus{}x\minus{}12)\sin(x\minus{}1)\plus{}(\minus{}8x\plus{}4)\cos(x\minus{}1)$

A clear pattern emerges:
For odd $ n>1$, $ a_n\equal{}(\minus{}1)^{n\plus{}1}(2nx\minus{}n)\sin(x\minus{}1)\plus{}(\minus{}1)^{n\plus{}1}(x^2\minus{}x\minus{}2\cdot3^{n\minus{}2})\cos(x\minus{}1)$.
For even $ n$, $ a_n\equal{}(\minus{}1)^{n\plus{}1}(x^2\minus{}x\minus{}2\cdot3^{n\minus{}1})\sin(x\minus{}1)\plus{}(\minus{}1)^n(2nx\minus{}n)\cos(x\minus{}1)$.
(I'm kinda shaky on these; I'll have to double-check them later)
We can confirm using the recursion.
[/hide]


$ 2^x \minus{} 2^y \equal{} 4$ and
$ 8^x \minus{} 8^y \equal{} 160$

solve for real values of $ x$ and $ y$.

[quote="xxxyyyy"]$ 2^x - 2^y = 4$ and
$ 8^x - 8^y = 160$

solve for real values of $ x$ and $ y$.[/quote]

[color=darkblue]Let $ u = 2^x$, $ v = 2^y$, $ u>0$ and $ v>0$. We have systems: $ \begin {cases} u - v = 4 \\
u^3 - v^3 = 160 \end {cases}$

$ \Longleftrightarrow \begin {cases} u - v = 4 \\
(u - v)(u^2 + v^2 + uv) = 160 \end {cases}$

$ \Longleftrightarrow \begin {cases} u - v = 4 \\
(u - v)[(u - v)^2 + 3uv] = 160 \end {cases}$

$ \Longleftrightarrow \begin {cases} u - v = 4 \\
(u - v)[(u - v)^2 + 3uv] = 160 \end {cases}$

$ \Longleftrightarrow \begin {cases} u - v = 4 \\
uv = 8 \end {cases}$

$ \Longleftrightarrow \begin {cases} v = u - 4 \\
u^2 - 4u - 8 = 0 \end {cases}$

$ \Longleftrightarrow \begin {cases} v = u - 4 \\
u = 2 + 2\sqrt {3} \end {cases}$ OR $ \Longleftrightarrow \begin {cases} v = u - 4 \\
u = 2 - 2\sqrt {3} < 0 \end {cases}$

$ \Longleftrightarrow \begin {cases}u = 2 + 2\sqrt {3} \\
v = - 2 + 2\sqrt {3} \end {cases}$ $ \Longleftrightarrow \begin {cases}u = 2 + 2\sqrt {3} \\
v = - 2 + 2\sqrt {3} \end {cases}$

Hence easy we get $ (x; y)$ ![/color]

[quote="xxxyyyy"]$ 2^x - 2^y = 4$ and
$ 8^x - 8^y = 160$

solve for real values of $ x$ and $ y$.[/quote]

If substitute $ 2^x,2^y$ for $ X=2^x,Y=2^y$,the equation is changed as follow:

$ X-Y=4$
$ X^3-Y^3=160$

$ X-Y=4 \leftrightarrow Y=X-4$

Let \begin{eqnarray*}&X^3-Y^3=160&
 \leftrightarrow &X^3-(X-4)^3=160&
 &12X^2-48X-96=0&
\leftrightarrow &X^2-4X-8=0& \end{eqnarray*}

i.e. $ X=2(1\pm\sqrt3),Y=2(-1\mp\sqrt3)$

So,$ X=2^x,Y=2^y$,
$ 2^x=2(1\pm\sqrt3),2^y=2(-1\mp\sqrt3)$

Then,$ x=log_22(1\pm\sqrt3),y=log_22(-1\mp\sqrt3)$

[color=darkblue]Hi [u]Rosnovisky[/u], you know that $ X>0$ and $ Y>0$.

So $ 2^x\equal{}2(1\minus{}\sqrt3),2^y\equal{}2(\minus{}1\minus{}\sqrt3)$ is not true.[/color]


let x in [1.5,5].

Prove that:
( \sqrt (2x-3)+ \sqrt (15-3x)+2 \sqrt (x+1)) 2 <71.25.

(In the original problem the RHS is 76.)

By Cauchy

A^2 = (sqrt(2).sqrt(x-3/2) + sqrt(3/2).sqrt(10-2x) + 2.sqrt(x+1))^2 <= (2+3/2+4)(x-3/2+10-2x+x+1) = 71.25.

It easy to see that the equality can not be hold, so we have a strict inequalty.

Not that maximum of A^2 is ~ 71.2334 so our result is quite strong.

Namdung

My calculation using [i]Mathematica[/i] Shows that the maximum value of A 2 is 71.2496995

I bet 71.25 is the best we can do using elementary methods.

Yes, Im agree with you.
I also use Maple to find that max A ~ 71.24969945

Namdung


Hi everyone,
Can any one help me with this?

y"^2= A^2(1+y'^2)
and A is Constant

 :?:

Thanks a lot

[quote="Kent Merryfield"]Note that $y$ does not appear directly. This may be formally a second order ODE, but in practical terms it is first order.

Let $v=y'.$ You have $(v')^{2}=A^{2}(1+v^{2})$ or

either $v'=A\sqrt{1+v^{2}}$ or $v'=-A\sqrt{1+v^{2}}.$

(Which of these applies depends on the initial conditions.)

Now you have an autonomous, hence separable, first order equation for $v.$ Can you see how to proceed?[/quote]

y'=v  =>  y''=vv'

Isn't it right?

[quote="Kent Merryfield"][quote="AngelaS"]y'=v  =>  y''=vv'

Isn't it right?[/quote]
No, $y''=v'.$ Where are you getting that?[/quote]

if y'=v(y)
then  y''=dv(y)/dx=[dv(y)/dy] [dy/dx] = v'v

Then you suppose y'=v it means y(x)'=v(x). And not v(y).


The prime numbers p and q, and natural number n satisfy the following equality:
  $ \frac{1}{p}\plus{}\frac{1}{q}\plus{}\frac{1}{pq}\equal{}\frac{1}{n}$
find these numbers.
no guesses please and dont post them if u get lucky. do give full solutios only and hide them.
too many instructions? :spider:  :evilgrin:

[hide]$ \frac {1}{p} \plus{} \frac {1}{q} \plus{} \frac {1}{pq} \equal{} \frac {1}{n}$
$ \Rightarrow\frac {p \plus{} q \plus{} 1}{pq} \equal{} \frac {1}{n}$
$ \Rightarrow (p \plus{} q \plus{} 1)|pq$
$ p \plus{} q \plus{} 1 > p,p \plus{} q \plus{} 1 > q,p \plus{} q \plus{} 1 > 1, so\ p \plus{} q \plus{} 1 \equal{} pq$
$ \Rightarrow (n,p,q) \equal{} (1,2,3);(1,3,2)$[/hide]

here's how i did it
$ \frac{1}{p}\plus{}\frac{1}{q}\plus{}\frac{1}{pq}\equal{}\frac{1}{n}
\implies\frac{p\plus{}q\plus{}1}{pq}\equal{}\frac{1}{n}
\implies(p\plus{}q\plus{}1)(n)\equal{}pq$
using parity, it can be checked that if both are odd, parity of the equation is not satisfied.so, p or q is even.
lets take p as 2,
$ \implies(3\plus{}q)(n)\equal{}2q$
here,in right hand side of equation there are 3 factors(q,2,1)
q is odd so, 2 is factor of (3+q) and since it greater than 1 so, only q can be the second factor of the same.
hence n=1
so, answers are (n,p,q)=(1,2,3),(1,3,2)

[quote="analytic"]here's how i did it
using parity, it can be checked that if both are odd, parity of the equation is not satisfied.so, p or q is even.
[/quote]
I don't think so. :maybe:
If both are odd, $ (p\plus{}q\plus{}1),(pq)$ are odd.
So $ n$ is odd and the parity of the equation is satisfied.
$ p$ or $ q$ don't have to be even.


If the parabola $ y \equal{} \minus{}x^2 \plus{} bx \minus{} 8$ has its vertex on the $ x$-axis, then $ b$ must be:

$ \textbf{(A)}\ \text{a positive integer}\qquad \\
\textbf{(B)}\ \text{a positive or a negative rational number}\qquad \\
\textbf{(C)}\ \text{a positive rational number}\qquad \\
\textbf{(D)}\ \text{a positive or a negative irrational number}\qquad \\
\textbf{(E)}\ \text{a negative irrational number}$

[hide="Solution"]The vertex occurs at $ x\equal{}\frac{b}{2}$.

Plugging in, $ y\equal{}\frac{b^2}{4}\minus{}8\equal{}0\implies b^2\equal{}32\implies b\equal{}\pm 4\sqrt{2}$.

Therefore, $ \boxed{\textbf{(D)}}$.[/hide]

[hide="Also"]In $ Ax^2 \plus{} Bx \plus{} C$, $ \dfrac{C}{A}$ is the product of the roots.

Since the vertex is on the x-axis, its a double root, and the product is $ \dfrac{\minus{}8}{\minus{}1} \equal{} 8$

So, $ x^2 \equal{} 8$ and $ x \equal{} \pm 2 \sqrt{2}$

Then, $ \dfrac{B}{A}$ is the sum of the roots.

Therefore, $ B \equal{} \pm 4 \sqrt{2}$[/hide]


calculus
integration
quadratics
algebra
polynomial
analytic geometry
function

I'm having a problem with this parametric problem, I know the basic formula to put it in but I can't figure out how to set up the integral with the parametrics

Find the length of parametrized curve given by
x(t) = 0 t^(3) + 12 t^(2) + 24 t
y(t) = -4 t^(3) - 12 t^(2) + 0 t

where t goes from zero to one.
Hint: The speed is a quadratic polynomial with integer coefficients.

Please help! Thanks!

Your objection makes no sense- polar coordinates are irrelevant.
What is the speed as a function of $ t$?


Let $ d_{1}, d_{2},\ldots, d_{n}$ be real number greater or equal $ 1$. Prove that:
\[ \prod_{i\equal{}1}^{n}d_{i}^{d_{i}}\geqslant\left (\sqrt{\frac{\sum_{i\equal{}1}^{n}d_{i}^{2}}{n}}\right )^{\sum_{i\equal{}1}^{n}d_{i}}\]

Very nice inequality, I will add a term on the left hand:

$ \left(\frac{\sum_{i\equal{}1}^{n}{d_{i}}^{2}}{\sum_{i\equal{}1}^{n}d_{i}}\right)^{\sum_{i\equal{}1}^{n}d_{i}}\geq\prod_{i\equal{}1}^{n}{d_{i}}^{d_{i}}\geq\left(\sqrt{\frac{\sum_{i\equal{}1}^{n}{d_{i}}^{2}}{n}}\right)^{\sum_{i\equal{}1}^{n}d_{i}}$