I'm looking for a proof by induction that a set with n elements has 2^n subsets.

There's an easy way without induction but if you want it...
[hide="induction"]  Base case = 1 element, and it works. (empty set is the other subset)

Now assume it works for $ n$ elements.   In a set with $ n\plus{}1$ elements, we can separate one element from the set so that the remaining $ n$ has $ 2^{n}$ subsets.  Now, in each one of these subsets of the $ n$ set, the $ n\plus{}1$th element can be in or out. (2 possibilities)  So, the total subsets is $ 2^{n}\cdot 2 \equal{} 2^{n\plus{}1}$ completing the induction.[/hide]

I know of the other ways, and I don't particularly like this induction proof because it is basically just using the "each element has 2 choices" proof.

Is there another induction proof?

1.one way:

choose 0 element from the set that has N elements. there nC0 subset
choose 1 element from the set that has N elements. there nC1 subset
choose 2 elements from the set that has N elements. there nC2 subsets
...
choose n element from the set that has N elements. there nCn subsets

nC0+nC1+nC2+......+nCn=2^n

2.funny way: :rotfl:

 there are two choices for one element, either in the subset or not in the subset.

 so for n elements there are 2*2*2.... (there are n times for 2*2)=2^n

[quote="aspiringchemlete"]I don't particularly like this induction proof because it is basically just using the "each element has 2 choices" proof. [/quote]

The "2 choices" concept is essentially the content of the problem, so any alternate proof would either use this concept or use a stronger result...


function
search
algebra unsolved
algebra

Let f :N -->N be increasing function satisfying :
    f(f(n))=3n for all n is natural
   determine the value of f(2004). :D

Unless I am mistaken, the problem has already been discussed somewhat here:

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=962726304&t=108086[/url]


[hide="1"]16, but isn't any power of 1 equal to 1? :maybe: [/hide]
[hide="2"]We have $ 3a \plus{} 11b \equal{} 31$, and using guess and check, we find that $ a\equal{}\boxed{3}$.[/hide]

[quote="MCNational!"]Well jonathan

MOEMS is pretty(i mean really!!!) ez(i could do it in my SLEEP)

watch them put the second problem on one of their test

and yes i did get a perfect on MOEMS

thank you thank you thank you all :winner_first:[/quote]

lol, :D

Well, I suggest the mods close this small thread :D

[quote="007math"]yeah, don't really post problems unless you actually need help on them.  I don't think anyone here only does MOEMS, people from the MC forum just come and spam and the MOEMS forum. :D[/quote]

exactly :D

[quote="jonathanchou711"]Please don't post such easy questions here. I assume you know the answers to these questions tcjy. If you dont, I'm sorry. But I really assume people should know these in middle school. And please try them out yourself before posting the question.   :lol:[/quote] I agree I am in elementary and I could do this while I am sleeping.

[quote="cgyao15"]no need to revive this one-month old post :wink:[/quote]

[color=orange][size=200]Why am I here[/color][/size]


geometry
Alcumus
geometric transformation
reflection
rotation
congruent triangles

How many non-congruent triangles can be drawn, such that their vertices are lattice points around the edge of an 5 by 5 grid of lattice points shown?
[asy]size(50);
dot((0,0));dot((1,0));dot((2,0));dot((3,0));dot((4,0));
dot((0,1));dot((4,1));
dot((0,2));dot((4,2));
dot((0,3));dot((4,3));
dot((0,4));dot((1,4));dot((2,4));dot((3,4));dot((4,4));[/asy]

There are $ \binom{5}{3} \equal{} \boxed{10}$ ways to do so.

Umm I can't read. It should be $ \binom{16}{3}$. Sorry. I just saw the number "5" and "triangles".

EDIT: Oh. I really can't read.

you are over counting many triangles cause many are congruent. And if you pick 3 in a row, you don't get I triangle. I posted this problem....I get the solution but I think there has to be another way...

did you just list the possibilities?  that's the only way I can think of right now

[quote="myyellowducky82"]I think so...[/quote]

Then how do you do it?

This problem is ridiculous...  I got close, but I still missed some.  One of the hardest casework problems I have attempted.  Catching ll of the possibilities without over-counting would have taken me all evening.

2 years and still no solution? sadly i can't think of a good one either

This is hard.

I can't seem to account for all the casework.

You also have triangles that have edges passing through more than one point on the grid. This throws me out of whack.

Sorry for the double post.

I tried to use casework for the dimensions/types of triangles, but I always feel like I'm leaving something out. Is there any sure way to do this problem?

I'm not positive about my answer, but I'll try some of the casework.

[hide]Case 1: One vertex of the triangle is at the corner of the grid, and the other two are on the sides adjacent to it (creating a right triangle)

There are $4+3+2+1=10$ ways to do this.

Case 2:

[asy]
size(50); dot((0,0));dot((1,0));dot((2,0));dot((3,0));dot((4,0)); dot((0,1));dot((4,1)); dot((0,2));dot((4,2)); dot((0,3));dot((4,3)); dot((0,4));dot((1,4),red);dot((2,4));dot((3,4));dot((4,4));[/asy]

We can connect this red point with any other two points that are not from the same side. We can connect it with any two of these points.

[asy]
size(50); dot((0,0), red);dot((1,0), red);dot((2,0), red);dot((3,0), red);dot((4,0), red); dot((0,1), red);dot((4,1), red); dot((0,2), red);dot((4,2), red); dot((0,3), red);dot((4,3), red); dot((0,4));dot((1,4));dot((2,4));dot((3,4));dot((4,4));[/asy]

Thus, there are $ \binom{11}{2}=55 $ ways to do so.

Case 3:
[asy]
size(50); dot((0,0));dot((1,0));dot((2,0));dot((3,0));dot((4,0)); dot((0,1));dot((4,1)); dot((0,2));dot((4,2)); dot((0,3));dot((4,3)); dot((0,4));dot((1,4));dot((2,4), red);dot((3,4));dot((4,4));[/asy]

Now, we can connect this point with any point that is not on the same side, which gives us $ \binom{11}{2}=55 $ for this case.

In total, there are $10+55+55=\boxed{120}$ non-congruent triangles that can be made.[/hide]

Edit: Eh, never mind. I think I over-counted.

I think casework by area would work; maximum of 16 cases.
but 16 cases is still a lot :O

there are 62 triangles ... i didnt get it for a long time

So you used casework like the others did?

Alcumus has some crazy long solution, but the answer is [hide="dun dun dun"]$\boxed{59}$[/hide]

I have an alternative method of solution which unfortunately gives a different answer, 71.  I would be grateful if anyone can point out either the error in my method or in the calculations using it.

There are 16 C 3 (=560) ways of choosing three distinct points from the edges of the square.  Those three points are either collinear or the vertices of a scalene or isosceles triangle.  Let the number of ways of choosing three collinear points be L, and the number of ways of choosing three points that are the vertices of a scalene or isosceles triangle be S and I respectively. Then 560=L+S+I.

Now L=4*(5 C 3)= 40. Hence 520=S+I.

We want the number of non-congruent scalene and isosceles triangles, which we denote by S* and I* respectively.  For each scalene triangle we can form 7 other congruent triangles by reflection and rotation.  For each isosceles triangle we can form 3 other congruent triangles by rotation (reflection either leaves an isosceles triangle unchanged or generates a triangle obtainable by rotation). Hence S*=S/8 and I*=I/4.

All isosceles triangles are formed by choosing an apex at the corner or centre of a side.  To determine I* (the number of non-congruent ones) let the apex be either the top left corner or the centre of the top row.  If the apex is the top left corner, each of the other lattice points in the left hand column and three more in the bottom row can be a base with the other base fixed (by reflection in the leading diagonal).  If the apex is the centre of the top row, there are four lattice points in the left hand column and one more lattice point on the bottom row that can form a base with the other base fixed (by reflection in the vertical line through the center of the square).  Hence I*=7+5=12 and I=4(I*)=48.

Since 520=S+I and I=48 we have S=520-48=472.  Hence S*=S/8=59.

Our answer, S*+I*, is therefore 59+12=71.

What have I missed?

Regards,

Zax


calculus
integration
real analysis
limit
real analysis unsolved

I have a tiny problem, I think it's mostly conceptual in nature.

I have to evaluate the following integral:

$ \int_0^{2\pi}\sum_{_k\equal{}\minus{}\infty}^\infty\delta\left(\phi\minus{}\left(k\plus{}\frac{1}{2}\right)\pi\right)d\phi \ ;\quad k\in \mathbb{Z}$

Where $ \delta$ is the Dirac delta 'function'. Now the integral will give only two values (those within the interval of integration) but how to find them?

You don't need to find them.  This just counts how many of the points are in the interval.  Which is easy.  Except perhaps when an endpoint is involved.

More generally: Let $ f(x)\equal{}\sum_{k\equal{}\minus{}\infty}^{\infty}\delta
\left(x\minus{}\left(k\plus{}\frac12\right)\pi\right).$

Then, for any $ n\in \mathbb{Z},$ compute $ \widehat{f}(n)\equal{}\int_0^{2\pi}f(x)e^{\minus{}inx}\,dx.$

The question you asked was $ \widehat{f}(0)$ (well, technically $ 2\pi\widehat{f}(0).$)

If you can find the constant term, you should be able to find all of the Fourier coefficients. We're trying to find the Fourier coefficients for $ f,$ considering $ f$ to be periodic of period $ 2\pi.$ The fact that $ f$ is actually periodic of period $ \pi$ will show up as a certain property of those Fourier coefficients.

One other comment: the Riemann-Lebesgue Lemma does not apply to distributions that are not integrable functions. We don't have $ \lim_{|n|\to\infty}\widehat{f}(n)\equal{}0,$ and that doesn't contradict anything, since $ f\notin L^1(\mathbb{T}).$


ok here's the reaction:

$H_{2}C_{2}O_{4} + NaOH \rightarrow$   ????

it's oxalic acid plus sodium hydroxide...i know one product is water..not sure of others though and not sure how to go about doing this..

thanks!

I'm horrible at chemical reactions, but I guess Na2C2O4 would be the other product lol.

ohh yea lol thanks i can't believe i didn't see that...must've been too much reading cuz i've been reading 5 steps to a 5 for AP chemistry...

what grade r u in?


function geometry rectangle algebra domain limit integration

Let $ \Omega$ be a smooth bounded open connected set in $ \mathbf R^{n}$, $ \Gamma$ be an open smooth portion of $ \partial\Omega$. Prove that if $ u$ satisfies
\[ \begin{cases}-\Delta u = 0 & \text{ in }\Omega, \\ u = \frac{\partial u}{\partial\nu}= 0 & \text{ on }\Gamma, \end{cases}\]
then $ u=0$.

bump.  lets just assume that $ \Omega$ is a rectangle in $ \mathbb{R}^2$

i would love to see a proof of just this easier case =o

There is no real difference: extend $ u$ by $ 0$ across $ \Gamma$. You'll get a harmonic function in a connected domain that is $ 0$ on an open set. But harmonic functions are real-analytic.

One could also use the uniqueness of the solution of the PDE.

@Gockel:  All you've done is restate the problem.  The whole point is to prove uniqueness.

There is still a bit of work to be done with fedja's solution since the extended function is not $ C^2$ across $ \Gamma$.

That depends on how much one assumes "well-known" about harmonic functions. I assumed that it is known that if $ u$ is continuous and $ \lim_{r\to 0\plus{}}\frac 1{\pi r^2}\int_{D(x,r)}[u\minus{}u(x)]\equal{}0$ for all $ x\in \Omega$, then $ u$ is harmonic in $ \Omega$. This is, of course, not immediate from the definition but it is not so hard either and most textbooks I know cover this property. :)

fedja, just wanted to make sure that the detail was not ignored (for the sake of the children  :) ).  I would say that the "missing" part of your solution was not a proof of the Mean Value Property (which I agree is standard), but rather a sentence like, "One can use the Mean Value Property to show that the the extended function is harmonic."

Sorry for this general question, but what does it mean exactly to extend across an arc? Here I just assume it means to extend it to an open neighborhood of (a subarc of) $ \Gamma$? Is there a precise usage?

Your interpretation is fine except when you have a domain like this:
[asy]size(70);
MP("\Omega",(-0.5,0));
D(CR((0,0),1));
D((0,0)--(1,0));
MC("\Gamma", D( (0.3,0)--(0.6,0),orange+linewidth(1)));[/asy]
you actually would like to consider the function on one side of $ \Gamma$ and extend it to an open set containing (a subarc of) $ \Gamma$ entirely ignoring the values on the other side, etc.. In general you are talking about something similar to "analytic continuation" in complex analysis.


inequalities
geometry
geometric transformation
LaTeX
inequalities theorems

Hello all, the known topic of nhocnhoc in maths.vn forums contains more than 200 problems now, and I have been listing those problems, translating them , and creating a pdf file for more than a month. The [url=http://www.maths.vn/forums/showthread.php?t=9244]Topic Of Nhocnhoc[/url] or nguoivn is actually a vast collection of problems on inequalities in Vietnamese.

However, I have been translating them to english manually, compiled them, and have created a .pdf file that contains the log. I think it is suitable for posting in mathlinks.ro, because I have never contributed anything to mathlinks / AoPS before, and I feel happy to post something useful. :)

Also I did not know where to post this thread, so if this gets misplaced, mods please move it to the correct forum. :oops:

There might be several translation mistakes, and if anyone spots a mistake, please send me a pm notifying about the mistake. I will again post a newer version of this compilation set with correct translations and adjusted alignments. I will correct the alignments whenever I get time.

Any advice regarding improvement of this file is welcome through pm or via email.
If you have anything to comment, please comment here.
* Almost all of these have been discussed in Mathlinks before.

Thanks a lot to nguoivn, can_hang2007, dduclam and other memebrs who have posted the problems in the forum. :)

Eagerly waiting for your reply.

Sincerely,
Potla(S. Mukherjee)

[quote="broniran"]Thank you! It is really useful  :)[/quote]
Thanks, I will make the alignment of the problems using itemization as soon as possible (if I get time). The inequalities are too much unsorted, and I would also like to sort them and include some solutions too.

Potla :)

Hi all,

Sorry for being late in compiling the file with a better view.
I learnt some better Latex over time and tried to make a better PDF file, so it took a lot of time. In this file I have deleted all the unnecessary repetitions, aligned the formula in a better manner, and also I will add solutions as needed.

Solutions to any of these problems are welcome, because when compiling solutions I would need a variety of them. Right now I have solutions to about

This is a great work, Sayan  :)

Thank you :)


number theory
prime numbers
number theory solved

A positive integer is called fragile if there are positive integers a,b,x,y for which a + b = n and x/a + y/b = 1.
Determine the set of fragile numbers !

The fragile numbers are the non prime numbers ..
1) Let n be a non prime number : n = P*Q , with P > 1 , Q > 1
Take a = P , b = P*(Q-1) , x = P-1 , y = Q-1, so n is manifestly fragile.
2) Let n be a fragile number.
Note C = gcd(a,b) ===> a = C*A , b = C*B  with A,B co-prime , so :
x/a + y/b = 1 ===> x*B + y*A = C*A*B ===> x = U*A , y = V*B , U+V = C ===> C > 1
a + b = n ===> C*(A+B) = n , so n is manifestly non prime (q.e.d)


Jack picks 5 pounds of apples in 8 minutes.  Jill picks 6 pounds
of apples in 10 minutes.  Working together at these rates, how
many pounds of apples will they pick in 40 minutes?

First, we simply look at them individually.  Jack picks $ 5$ pounds every $ 8$ minutes, so he would pick $ \frac {40}{8}\cdot5 \equal{} 25$ pounds in $ 40$ minutes.  Similarly, Jill can pick $ 6$ pounds every $ 10$ minutes, so she picks $ \frac {40}{10}\cdot6 \equal{} 24$ pounds every $ 40$ minutes.  So, in total, they pick $ 25 \plus{} 24 \equal{} \boxed{49}$ pounds of apples in $ 40$ minutes.


Fie numerele reale $\ a,b,c$.Gasiti cea mai buna constanta $\pi$ astfel incat sa fie adevarata inegalitatea $\ a^{4}+b^{4}+c^{4}+ab^{3}+bc^{3}+ca^{3}-2(a^{3}b+b^{3}c+c^{3}a)\geq\pi$, $\forall a,b,c\in\mathbb{R}$.

1. De ce ai pus la sursa "own(?)"?? Sa nu zici ca tu ai creat-o :D

2. In [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=6026&start=40]thread-ul acesta[/url] s-au dat doua solutii la aceasta problema. A treia rezolvare ar folosi o ecuatie patratica in $a$, exact cum a fost rezolvata prima problema de acolo.


function
calculus
integration
derivative
calculus computations

Given that $ f(x)=\int_{0}^{x}f(t)dt$

find $f(x)$ in tems of elementary functions.

Hmm, I'm only getting $f(x) = 0$.

If we take the derivative of both sides, we get

$f'(x) = f(x) \Rightarrow f(x) = Ce^{x}$.

But then if we plug back into the equation we're given,

$Ce^{x}= \int_{0}^{x}Ce^{t}dt = Ce^{x}-C$,

which is only true if $C = 0$ so $f(x) = 0$.

You're right. The zero function is the only answer.


AMC USA(J)MO USAMO email probability conics AIME

If any of you guys receives a phone call from AMC or Mr.Feng, because you are selected to MOP, just sign up here, and share your happiness and experience.  :)
I hope the phone call may start from today, although I have very little chance to get in. :P

wow, you just scared the daylights outta me, i thought you actually *had* received the phone call and I should run and get mine and anxiously wait for the annoying little ring to start... *phew*

but yes, please, if somebody gets their phone call, *please!* post here so we all know to start getting anxious/excited!

I'm a freshman with a score of probably 8-9, so I will probably be right near the cutoff.  Hopefully above.  :D

The grading this year is Friday-Sunday, so they should finish grading today.  Who knows whether the calls will go out today, tomorrow, or some other day...

[quote="billzhao"]Hmm, Mr. Dunbar seems to be the one making the calls this time ...[/quote]

Wait calls are really going out or are you just kidding?

No (he's Yufei Zhao!).

Edit--in case it's not clear, I think billzhao might have [hide="personal experience"]UMMM AKA HE'S A USAMO WINNER GET IT??[/hide]

I would imagine the only way he could have found that out is if he recieved one.  :P

Last year, the calls for Winners and HMs were on Sunday, and the calls for MOPpers were on Monday, so don't worry if the calls don't come today.

Will the person who calls leave a message?  It may be difficult to catch somone from my house at home....

(That's if I made it, though!)

[quote="calc rulz"][quote="billzhao"]Hmm, Mr. Dunbar seems to be the one making the calls this time ...[/quote]

You found this out how?[/quote]

wiretapping

[quote="beta"][quote="calc rulz"][quote="billzhao"]Hmm, Mr. Dunbar seems to be the one making the calls this time ...[/quote]

You found this out how?[/quote]

Yup they have already called the USAMO winners![/quote]

You made it? Or somebody else made it (if so, how do you know that)?
Did they call the HMs? So fast!!!

-edit- The individual scoring isn't even up yet, but you can try to figure out what the address will be ... just don't refresh it every five seconds.

(It would slightly amuse me if Mr. Feng called...)

[hide=":D"]murderply[/hide]

[quote="randomdragoon"]No, they didn't call yet. The individual scoring isn't even up yet, but you can try to figure out what the address will be ... just don't refresh it every five seconds.[/quote]

I believe Winners have gotten calls (probability1.01, MysticTerminator). Individual scoring comes much later (next week I think) than the calls.

Congrats Boy Soprano!!  Hopefully the cutoff is $\le 8$, not $9$.

[quote="n^4+4"]Yeah it seems they sent everyone (Blue group included) an email, so if you haven't gotten a call yet check your email  :)[/quote]

Jeremy, did you make MOP?  If you did, congratulations!

That's weird . . . I got a 10 as a freshman, and I didn't get any email.

[quote="n^4+4"]Yeah!  I made MOP with room to spare  :)  :) .  It sure beats that 6 on the AIME last year.[/quote]
Good job!

I'm sure the heavy emphasis on NT and combinatorics didn't hurt you at all    ;)

(delete) - wow there's something wrong with my browser... sorry about this guys

Yayy!!

Yes, I did in fact get an e-mail. And I believe I might have gotten a phone call, but nobody was home... there were a few 866 numbers, so I bet one of them was AMC =D

No one called me. (or I didn't get it) :(  Oh well...there's this email though! :D

I got my phone call about 15 minutes ago.

no phone call but I got the email.  woohoo.

[quote="scorpius119"]I was originally called when I was away and my parents didn't tell me about it until half an hour after I returned home. At least I got the e-mail [i]and[/i] a second call, in case they "forgot".[/quote]

Lol, the first thing that happened when I got home was my parents told me I got the phone call. Too bad I wasn't home when they sent it, I wanted to talk to Zuming Feng  ;) .

Yay... just received my "official" phone call now too.... Talking to Zuming Feng was awefully.... exciting, except for the fact that I had to use both sides of my white-boy brain to understand his accent :D :D


What is the greatest four-digit number, "abcd",where a,b,c, and d are distinct integers such that a-c=b+d ?

srry, I don't think my question was clear. I failed to mention whether  a,b,c,d could be or couldn't be the same. But I meant that a,b,c,d had to be different. If they could be the same, then your answer would be correct.

[quote="wiseidiot"]What is the [b]greatest[/b] four-digit number, "abcd",where a,b,c, and d are distinct integers such that a-c=b+d ?[/quote]
$ 9810>9801$


calculus
search
college
real analysis
real analysis theorems

Guide me , please

I'm interesting in the book on this series : Calculus , Difference Equation , and want to know list of all book. After, I had try to search at Amazon,but It got shown list only some book.

Thanks you very much,

Phonghan Porana.

Student at Kmitl in Thailand.

"Berkery" = "Berkeley" and "Difference equation" = "Differential equation"?  (That might be the source of your problem ;) )

Thanks for suggestion , I need problem book that contain various type of problem and I think that in this series not only particular calculus, right.   I not sure that Calculus book in this volume in number, I think
 the volume may contain more than three books on calculus and I just need to see completed list .

Berkery = Berkeley = Berkley ?

thanks[/hide][/u]


riazi fek konam balaye 5 bashe yani taghriban motmaenam va computeram fek konam hododan 80 az 200

na baba como az bad tashih kardan ta chizhaye dige dar nazar gereftam,min nomram beine 110 ta 120 shod vali ghabool nashodam.rasti baraye eteraz mitoonim tozihe rahe hallemoono befrestim ta agar nafahmidandesh,motevajjeh beshan.

Be nazare manam kafe riazi 5+epsilon boode vali kafe com fekr konam beyne 100-110 boode :)

kaf 5 mehaye epsilon bud maa chand nafar dashtym ba 5 gabul shodan valy y
ek nafar ba 33.5 gabul nashod.


conics
ellipse
calculus
calculus computations

Let $ p,\ q$ be real numbers. Consider the ellipse $ E: (x\minus{}p)^{2}\plus{}4(y\minus{}q)^{2}\equal{}4$ on $ xy$ plane.
(1) Find the foci of the ellipse $ E$.

(2) Find the values of $ p,\ q\ (p>0)$ for which the ellipse $ E$ touches to the both lines $ y\equal{}x$ and $ y\equal{}\minus{}x$.

[quote="kunny"][color=darkred][b]Let $ p,\ q$ be real numbers. Consider the ellipse $ E: (x\minus{}p)^{2}\plus{}4(y\minus{}q)^{2}\equal{} 4$ on $ xy$ plane.
$ 1\blacktriangleright$ Find the foci of the ellipse $ E$.
$ 2\blacktriangleright$ Find the values of $ p,\ q\ (p > 0)$ for which the ellipse $ E$ touches to the both lines $ y \equal{} x$ and $ y \equal{}\minus{}x$.[/b][/color][/quote]
[color=darkblue][b]Proof.[/b]

$ 2\blacktriangleright$ The relation $ \boxed{\ q \equal{} 0\ }$ is evidently. Thus, the equations $ \left\{\begin{array}{c}(x\minus{}p)^{2}\plus{}4y^{2}\equal{} 4\\ \ y^{2}\equal{} x^{2}\end{array}\right\|$ $ \implies$

the equation $ (x\minus{}p)^{2}\plus{}4 x^{2}\equal{} 4$ has $ \Delta \equal{} 0$, i.e. $ \boxed{\ p \equal{}\sqrt 5\ }$ because $ p > 0$.

[b]Remark.[/b] The tangent points are $ T_{1,2}\left(\frac{\sqrt 5}{5},\pm\frac{\sqrt 5}{5}\right)$.[/color]


topology
function
advanced fields
advanced fields unsolved

Having a little trouble seeing how to go about this one:

If $ X$ is compact topological space and $ Z$ is a Hausdorff ($ T_2$) space, with continuous, bijective function
$ f: X \rightarrow Z$ then $ f$ is a homeomorphism.

let $ F$ be any closed set in $ X$,  $ \Rightarrow F$ is compact in $ X$.Thus, $ f(F)$ is compact in $ Z$.

$ Z$ is a Hausdorff space , then $ f(F)$ is closed in $ Z$.

Hence the result  :wink:


Divisibility
algebra
Sequence
IMO Shortlist

Given any integer $ n \geq 2,$ assume that the integers $ a_1, a_2, \ldots, a_n$ are not divisible by $ n$ and, moreover, that $ n$ does not divide $ \sum^n_{i\equal{}1} a_i.$ Prove that there exist at least $ n$ different sequences $ (e_1, e_2, \ldots, e_n)$ consisting of zeros or ones such $ \sum^n_{i\equal{}1} e_i \cdot a_i$ is divisible by $ n.$

[i]Lemma:[/i] Consider a binary matrix $k\times n$  with $k\le n-2$ such that on each row we have at least one zero and at least two ones. Then the columns of the matrix can be permuted such that there is no row having all of its ones in a contiguous block. [hide="Proof."]For $k=1$ and any $n$ is is obvious. Induct on $k+n$. A $1\times 1$ square is called special if it either is the only zero on its row, or one of the only two ones on its row. On each line there are at most $2$ special squares, so there is a column with at most $1$ special square. Suppose wlog that it is the first column. If there exists a special square on this column, suppose wlog that it is the top one. Apply the induction hypothesis on the matrix obtained by removing the first row and the first column and permute the rows and columns of the original matrix accordingly. If there is a zero in the top-left square, move the first column right after the column of the first one that appears in the first row. If there is an one in the top-left square, either keep the first column the same or move it after the column of the first zero that appears in the first row (depending on whether the second square in the first row contains a zero or not). This new matrix satisfies the lemma.  [/hide]

The problem is equivalent to finding $n-1$ sums of $a_i$s that are divisible by $n$ (as the trivial case when $e_i=0$ for all $i$ also satisfies the requirments). Note that given $x_1,...\ ,x_n$ integers, one can find $1\le i<j\le n$ such that $n$ divides $x_{i}+x_{i+1}+...+x_j$; by looking at the sums $x_1,x_1+x_2,...,x_1+...+x_n$ we either find one divisible by $n$, or we find two congruent sums which we then substract $(*)$. By applying $(*)$ for $x_i=a_i$s we find a sum $a_{i_1}+...+a_{i_j}$ divisible by $n$ (with $1<j<n$). Suppose that we have found $k\le n-2$ sums of $a_i$s divisible by $n$. By our lemma, we can find a permutation $\sigma \in S_n$ such that $\{a_{\sigma (i)},...,a_{\sigma (j)}\}$ does not have exactly the terms of one of the already $k$ found sums for any $1\le i<j\le n$. Apply $(*)$ to $x_i=a_{\sigma (i)}$ to find a new sum of $a_i$s divisible by $n$. By continuing this procedure, we obtain $n-1$ distinct sums of $a_i$s divisble by $n$.


[quote="anirudh"]Seven points are equally distributed on a circle. How many non-congruent quadrilaterals can be drawn with vertices chosen from among the seven points?

[img]http://i96.photobucket.com/albums/l167/adasarathy/CIRCLEPROBLEM.jpg[/img]

Sorry if its blurry. I drew it on Geometer's Sketpad on my calculator and tried unsuccessfully to transfer it.[/quote]

not sure but found 10

No. That's not right.  :(

re-read hte question.

[quote="anirudh"]Seven points are equally distributed on a circle. How many [b][i]non-congruent[/i][/b] quadrilaterals can be drawn with vertices chosen from among the seven points? [/quote]

So we can't use the center of the circle?

[hide]If so, I only found 3.[/hide]

I'm going to stick my neck out and say that there are no non-congruent quadrilaterals. Provided that they are all equally spaced.

[hide]
By pigeon hole there are two adjacent points must be selected.

Given that they are chosen, systematically count how many different quadrilaterals can be formed.

I got 6
[/hide]

I believe it's $\frac{{7\choose4}}{7}$, which is $5$.  :huh:  Can you show us how exactly you got 4, in some non brute force way, or explain to me how I overcounted?

Direction: Find how long the shortest side must be, and then find the limits for the longest side.

[quote="i_like_pie"][quote="JavaMan"]By naming the points starting from the top (in the diagram) clockwise from 1 to 7, with points 4 and 5 always selected (WLOG), show me the quadrilaterals they found.  :)[/quote]
[img]http://www.initdotd.com/~nikita/Pictures/Quadrilaterals.jpg[/img][/quote]

whoa! how did you make that?

Thanks, I see how I overcounted by two by not considering symmetry in two cases..

[quote="anirudh"]whoa! how did you make that?[/quote]
For the fifth (sixth?) time, Microsoft Visio (2003).


Why is \[ \left(\frac{n \plus{} 1}{2}\right)^n \geq n!\] equivalent to showing that \[ (n \plus{} 1)^n \geq 2n^n?\] Can you show me why those two are equivalent?

Thanks!

I am quite inexperienced so this is probably not the best way:

We need to prove $ 2n^n > \equal{} 2^{n}(n!)$. Rearranging gives $ (n / 2)^{n \minus{} 1} > \equal{} (n \minus{} 1)!$, and this is what we must prove. We know that $ (n / 2)^2 > \equal{} 1 * (n \minus{} 1)$, $ (n / 2)^2 > \equal{} 2 * (n \minus{} 2)$, etc. For n is odd, multiplying together all of those previous inequalities produces the desired result; for even n's, multiplying all the inequalities together gives $ (n / 2)^ {n \minus{} 2} > \equal{} (n \minus{} 1)! / (n / 2)$ and it is easy to go on from here.

[quote="bengal"]

We need to prove $ 2n^n > \equal{} 2^{n}(n!)$. [/quote]

Proving $ 2n^n\geq 2^nn!$ lets you prove the first inequality from the second. but $ 2n^n \geq 2^nn!$ is, as you pointed out, equivalent to the first. so we've gone in a circle. however, you did provide a proof of the first inequality without even using the second, though this isn't what was asked for. I don't yet see how the first can be used to prove the second (it's usually not very easy to turn a weaker inequality into a stronger one), but it's easy to see that you *can* directly use the second to prove the first by induction. so that at least demonstrates one side of the equivalence.

Thanks, Bengal. I found a pretty simple way to do it too. Just replace the original equation with $ (n \minus{} 1)$ and use the hypothesis to show that this must be equivalent. It was an exercise in Induction, guess I should have mentioned that.

Well, I might as well practice writing the proof.

Assume that $ n \equal{} k \minus{} 1$ for some $ k$, then \[ \left(\frac{k}{2}\right)^{k \minus{} 1} \geq (k \minus{} 1)!,\] and so \[ k^{k \minus{} 1} \geq 2^{k \minus{} 1} (k \minus{} 1)!.\]

Multiply both sides by $ 2k$ and you get \[ 2k^{k} \geq 2^{k}(k!)\] and this is still true due to the hypothesis. However, since our destination is $ (k \plus{} 1)^{k}$ and we have $ k^k$, we need to prove that $ (k \plus{} 1)^k \geq 2 k^k\geq 2^{k}(k!)$.

In other words, prove that $ \left(\frac{k \plus{} 1}{k}\right)^k \geq 2$, which can be done in many ways.

Thank you for your help!

so your original statement should have said something like, show that to prove the first inequality it is *sufficient* to prove the second. they're not actually equivalent per se.

Oh, you're right huh.  :oops: Thanks for your input though!


function
real analysis
real analysis theorems

We can have both. "or" is inclusive unless specified otherwise.

If we have both, we just pick one for the next step of the proof.


Hello,

prove that there are not positive intergers $ a,b,c,d$ such that

$ a^a \plus{} b^b \plus{} c^c\equal{}d^d$

On positive integers, $ n^n$ is strictly increasing, so wlog $ a \le b \le c < d$. Then $ 3c^c \ge d^d \ge d^{c\plus{}1} > d \cdot c^c$, so $ d \equal{} 1,2$, which do not work.

I did it as azjps but made a mistake in finishing it.


algebra
polynomial
quadratics
parameterization
analytic geometry

Find or describe all the polynomials of smallest possible degree that satisfy $P(1003) = 2007, P(1002) = 892, \text{and }P(1000) = 0$

$P(x)$ would be quadratic and we could use Lagrange's interpolation.

A generic polynomial of degree $D$ has $D+1$ parameters (one for the term of each degree and the constant).

Given $n$ points, no two of which share the same x-coordinate, it is possible to fit the points with degree $n-1$ or less.

In this case, we have $3$ valid points, so there is a unique quadratic that will fit this; in this case:
$f(x)=223(x-1000)^{2}$

[hide]If $f(x) = ax^{2}+bx+c$ then we have

$a(1003)^{2}+b(1003)+c = 2007\\ a(1002)^{2}+b(1002)+c = 892\\ a(1000)^{2}+b(1000)+c = 0$

So we have 3 equations, with 3 variables, which can tell us the quadratic that fits these points
[/hide]


The polynomial can be written as: $q(x-a_1)(x-a_2)(x-a_3)(x-a_4)(x-a_5) + 2$
with $q$ a polynomial $(p_1x^n+p_2x^{n-1}+...+p_n)$ and $p_i$ an integer, because the polynomial has integer coëfficients.

So for finding b in the expression $f(b)=9$, we let $q(x-a_1)(x-a_2)(x-a_3)(x-a_4)(x-a_5) + 2$ equal $9$
\[ \Rightarrow q(b-a_1)(b-a_2)(b-a_3)(b-a_4)(b-a_5) = 7 \]
(with $q = (p_1b^n+p_2b^{n-1}+...+p_n)$

Because the polynomial has integer coëfficients, $a_1, a_2, a_3, a_4, a_5$ have to be integers too, and because b has to be an integer, the LHS is a product of integers.

7 is a prime number, and can therefor not be factored, so there exist no such integer $b$.

[quote]for five integers a_1, \ldots a_4 one has [/quote]
It must be mentioned that $a_i$ are all distinct as well as the typo above must be removed.

Is this right?

For any polynomial $p$$(x)$; $x-y$ divides $p(x)-p(y)$.
So, then let there exist a value 'b' that satisfies that p(b)=9.
Then b-a1,b-a2,b-a3,b-a4,b-a5 divide 7.
But then only a maximum of 4 distinct values of k can be obtained such that b-k divides 7( 1,-1,7,-7).
Hence, atleast two values of a1,a2,a3,a4,a5 are same. But it is given that they are distinct. Hence , a contradiction.

[quote="Rushil"]Let $f(x)$ be a polynomial in $x$ with integer coefficients and suppose that for five distinct integers $a_1, \ldots, a_5$ one has $f(a_1) = f(a_2) = \ldots = f(a_5) = 2$. Show that there does not exist an integer $b$ such that $f(b) = 9$.[/quote]
My solution goes like this:
Take that $g(x) = f(x)-2$.
Then, $a_1, a_2,a_3,a_4,a_5$ are the roots of the equation $g(x)=0$.
Now, let me prove a simple lemma (many might have seen this).
If $p(x)$ = $b_nx^n+........+b_1x+b$ with $b \ne 0$  is a polynomial with integer coefficients and $\alpha$ is its root then for any integer $m$  $a-m | p(m)$.
Say for integer polynomial $d(x)$,   $ p(x) = (x-m)d(x) + p(m)$  $\Rightarrow p(\alpha) = (\alpha-m)d(\alpha)+p(m)$
$\Rightarrow (a-m)d(\alpha)=-p(m)$ $\implies a-m | p(m)$
--------------------------------------------------------------
Now, say for $\beta$  $(integer)$ $p(\beta) = 7 $.
According to our proven lemma $a_1,a_2,a_3,a_4,a_5$ divide 7 but 7 can have only 4 divisors as it is a prime. Since $a_1,.......,a_5$ are distinct  we got the contradiction!

Does nobody notice that it is enough to have $f(a_i)=2$ for only four distinct integers? (which explains the initial typo). This is because $7$ and $-7$ cannot simultaneously occur in the factorization of $f(b)-2 = 7$.

Given,
$f(a_i)=2$for $i=1,2,3,4,5$
Let $p(x)=f(x)-2$
So, $p(x) $ has 5 roots $a_1,a_2,a_3,a_4,a_5$
$p(x)=(x-a_1)(x-a_2)(x-a_3)(x-a_4)(x-a_5)q(x)$,where $ q(x)$ is an  polynomial with integer coefficients.
So,
$f(x)=(x-a_1)(x-a_2)(x-a_3)(x-a_4)(x-a_5)q(x)+2$
If $f(b)=9$
So,
$f(x)=(x-a_1)(x-a_2)(x-a_3)(x-a_4)(x-a_5)q(x)+2=9$
$(x-a_1)(x-a_2)(x-a_3)(x-a_4)(x-a_5)q(x)=7$
$7 $ is a prime number so, it has only $2$ factors  but we see $(x-a_1)(x-a_2)(x-a_3)(x-a_4)(x-a_5)q(x) $ as six distinct integers (as $q(x)$ is an integer polynomial) which is not possible.

Simple, write $P(x)$ as $Q(x)\prod_{k=1}^5(x-a_k)+2$. Then $9=P(b)=Q(b)\prod_{k=1}^5(b-a_k)+2$, so $Q(b)\prod_{k=1}^5(b-a_k)=7$. Then exactly one of these is equal to seven, since seven is prime, but two of $a_n$ must be equal, contradiction.

My First Ever subjective type question solved!!!!
We Have $f(x)$ is a polynomial with integer coefficient and we have given $$f(a_1)=f(a_2)=f(a_3)=f(a_4)=f(a_5)=2$$
Assume for contradiction!!
Let's assume that there exists an integer $b$ such that $\boxed{f(b)=9}$ Now Since, $$\boxed{f(a_1)=f(a_2)=....=f(a_5)=2}$$ We can write this as $f(a_1)-2=0....f(a_5)-2=0$ therefore, we can write this as $\boxed{f(x)-2=0}$ whose roots are $a_1,a_2,a_3,a_4,a_5$ Now By Factor Theorem, we get, $$\boxed{f(x)=(x-a_1)(x-a_2)(x-a_3)(x-a_4)(x-a_5)+2}$$ Putting $b$ both sides we get, $$f(b)=(b-a_1)(b-a_2)....(b-a_5)+2$$
We assume that $\boxed{f(b)=9}$ then $\boxed{(b-a_1)(b-a_2)....(b-a_5)}$ has to be $7$
Since $7$ is a prime number and $a_1,a_2,.....a_5$ are distinct integers therefore this product can't equals $7$ which gives us the contradiction!!


Any AIME I'ers or Instructors or just people in general have good advice to give to AIME II'ers?

:-D

Countdown is now at 7 days...

Double check.
Double check again.

Getting 5 right is a lot better than getting 10 half right.

--Dan

Bring a supply of rocks.  If anything makes an annoying noise, throw a  rock at it to silence it before it drives you insane.  This applies to clocks, librarians, stupid people who can't read "math competition: do not enter" signs, and hot girls.

politely ask your proctor not to make you change rooms during the aime, as happened at my school.

hmm...well i'm homeschooled and I'm taking it at a church. So i basically have the entire church to myself and so I doubt i'll need to change rooms.

the rocks...u realli cant silence the air conditioner with just a few rocks... -.- ...and there shouldn't be any annoying ppl trying to break into the room or anything... :-D


Set S is the set of all positive four-digit integers that have four distinct digits. How many of the integers in the set have a value less than 1235?

[hide="Solution"]The first digit must be 1. We split this up into cases of the second digit; 2 and 0.

[hide="Case 1: 2"]
[u]12--[/u]
If the third digit is 3, the last must be 0 or 4
The third digit cannot be 1 or 2.
If it is 0, the last is one of 3,4,5,6,7,8,9

$ \boxed{9}$[/hide]
[hide="Case 2: 0"]
[u]10--[/u]
The third digit can be one of 2,3,4,5,6,7,8,9, and the last has 10-2-1=7 other choices.

$ 7\cdot8\equal{}\boxed{56}$[/hide]
[hide="Conclusion"]Thus, there are $ 9\plus{}56\equal{}\boxed{65}$ such integers.[/hide][/hide]


ARML Stanford college email algebra polynomial AMC

All ARML discussion not related to Albert/Meggy-meg/etc. goes here.

Good...
Anyone have directions to building 420? (I assume this is relevant...)

You can get there via the basement of the math building, if you know where building 380 is.  380 and 420 have their basements connected, so the new practice room is down the hall from the old one.  If it's your first year on the team, building 380 is the "math corner" on Stanford maps; it's at the corner of the quad, by Palm Drive.

You can get campus maps here: [url]http://campus-map.stanford.edu/[/url]

Tip: DON'T go in the front entrance, the one opposite the parking lot.

Instead, take a right turn to the basement entrance.

Wait what's this? I don't want to be lost the next time I go...

Hey how did you all do on the tests we did today?  I got an 8 individually but made up for it by messing up the team round so that the 6 team beat us (right?).  But hey, nobody volunteered to check my work.  :P
serial, I think Circle is just talking about the exit everyone left through last time (the one by the bike rack where you have to climb some stairs once you're outside.

8 individual, claimed credit for #8 on team even though John did most of the work...6 team beat 7/8 team? Really?

I got 6 on individual, and we got seven team problems right (I think).

How many team probs did you guys get?

[quote="ccy"]Ha i love my 5 or so individual.  Turns out that 5- beat you all in team with an 8. :P[/quote]
Yeah, this is what I was talking about.  Someone told be that we got owned by the lower score division and I assumed it was the 6 group.

What did you think of the problems overall?

Haha was in the 7-8 group and wasn't allowed to help you guys out since I had already completed that team round before. :(
It sucked watching people put up incorrect answers and not being able to correct their work. :mad:

What type of score(s) would you need to get to make the A or B team?

wait
am i missing something
what is this so called
7-8 team?
kids with 7/8, 8/8 on indivs?
there were 15 of these guys?
what?

I think there were more like 20 people who got 7s or 8s on the individual round.

It was some random mathleague test...still epic-failed it though.

[quote="DataBox"]o dorney makes up for the lost potential  :lol: jk
if we could just not epic fail at indivs, i think we'd have a decent shot at winning[/quote]

rofl i think it means something if i could guess that youre amrit just from eric calling you a dirty whore

haha, well for some odd reason they call me dirty, i'm not the one that had snuggling time, ran around bottomless, or went to a certain type of website that is not intended for our use  :lol: *cough* daermon jk

[quote="farmertim"]yeah seriously, the people who didn't even bother to go this year won't make us do any worse by not going next year.  we had NO people graduating from our A or B teams, so we should only be better next year..[/quote]

That's what we said last year too... but then a bunch of people didn't return.

nice job, all, i'm impressed. hopefully i can make it next year.

[quote="Sly Si"][quote="Ubemaya"]I'm planning to join next year for sure, unless

1) overwhelmed by finals
2) another date with meggy-meg[/quote]

Next year, I'm pretty sure ARML falls [i]during[/i] Stanford finals.  I'm going anyway.  And if a girl wants me to go out with her that weekend, she can either come to ARML with me or schedule it for another time.[/quote]
SILAS
WE MISSED YOU

err
high school finals shouldnt take more than 2 hours of studying the day before the final
ie you can more than afford to miss the weekend
??

Unless you have a final on Friday.

I had a final on Friday the 23rd, so if ARML was a week earlier, I wouldn't have been able to go.

And... guys
I went digging in previous ARML results, and found something pretty interesting.
The last time a team got a perfect power was in
[hide]2000, and it was[hide][hide]SFBA Team A[/hide][/hide][/hide]:rotfl:
And the last time a team got perfect team and power was NYC A in 1994.

Then again, difficulty matters.

And you tell me to not waste my time on lists...hypocrite...jk  :D
That's pretty interesting, I would expect '94 to be loaded with perfect powers considering the amc scores and imo scores that year.

The usual reason for a lot of high scores is an easy contest. That's definitely the case for the 1994 AHSME. There is no reason to assume other contests from the same year would be easy.

yeah so I was sitting around and then I had to look up some paper by this economist guy and then BAM he was a 94 perfect scorer so I was like huh that's kind of weird do I know any other 94 perfect scorers and then BAM my econ professor at the time was also a 94 perfect scorer and then like BAM they are everywhere

[quote="Ubemaya"][quote="Sly Si"][quote="Ubemaya"]I'm planning to join next year for sure, unless

1) overwhelmed by finals
2) another date with meggy-meg[/quote]

Next year, I'm pretty sure ARML falls [i]during[/i] Stanford finals.  I'm going anyway.  And if a girl wants me to go out with her that weekend, she can either come to ARML with me or schedule it for another time.[/quote]

Well...
1) high school > Stanford
2) Meggy-Meg > all[/quote]

Sorry Ubey, I'm gonna have to disagree with both of those...  :ninja:


Let's say that an ordered triple of positive integers $(a,b,c)$ is [i]$n$-powerful[/i] if $a\le b\le c,\gcd (a,b,c)=1$ and $a^n+b^n+c^n$ is divisible by $a+b+c$. For example, $(1,2,2)$ is $5$-powerful.
$a)$ Determine all ordered triples (if any) which are $n$-powerful for all $n\ge 1$.
$b)$ Determine all ordered triples (if any) which are $2004$-powerful and $2005$-powerful, but not $2007$-powerful.

[b]Claim[/b]: if some triple, $ (a,b,c)$, is $ n$ and $ n \plus{} 1$-powerful, then it is $ n \plus{} 3$ powerful either.

[b]Proof:[/b]

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

$ \square$

Therefore, the answer to $ b$ is no.

If a triple $ (a,b,c)$ is $ 1,2,3$-powerful, then it is $ n$-powerful for any $ n$ due to the [b]Claim.[/b]

Clearly, any triple is $ 1$-powerful. As we know,

$ a^3 \plus{} b^3 \plus{} c^3 \equal{} (a \plus{} b \plus{} c)(a^2 \plus{} b^2 \plus{} c^2 \minus{} ab \minus{} bc \minus{} ca) \plus{} 3abc$, and $ a^2 \plus{} b^2 \plus{} c^2 \equal{} (a \plus{} b \plus{} c)^2 \minus{} 2(ab \plus{} bc \plus{} ca)$,therefore, $ 3\cdot abc$ and $ 2(ab \plus{} bc \plus{} ca)$ has to be divisible by $ a \plus{} b \plus{} c$. Besides, $ ab \plus{} bc \plus{} ca \equal{} a\cdot (a \plus{} b \plus{} c) \minus{} a^2 \plus{} bc$. So we also have that $ 2(a^2 \minus{} bc)$ is divisibly by $ a \plus{} b \plus{} c$.

Multiply the last expression by $ 3a$, we get that $ 6(a^3 \minus{} abc)$ is a multiple of $ a \plus{} b \plus{} c$. Recall that $ 3abc$ is divisible by $ a \plus{} b \plus{} c$, thus

$ a \plus{} b \plus{} c|6a^3$.

Analogously, $ a \plus{} b \plus{} c| 6b^3$ and $ |6c^3$. If $ (a \plus{} b \plus{} c,6) \equal{} 1$, then we get a contradiction to $ (a,b,c) \equal{} 1$.

Case1: Suppose that $ a \plus{} b \plus{} c \equal{} 3\cdot k$, where $ k$ is odd. It means that $ k|a^3,b^3,c^3$, so $ k \equal{} 1$ and $ a \plus{} b \plus{} c \equal{} 3$, it gives us one triple, $ a \equal{} b \equal{} c \equal{} 1$.

Case2: Suppose that $ a \plus{} b \plus{} c \equal{} 2\cdot k$, where $ (k,3) \equal{} 1$. Similarly, $ k \equal{} 1$ and $ a \plus{} b \plus{} c \equal{} 2$, which is quite impossible.

Case3: Suppose that $ a \plus{} b \plus{} c \equal{} 6$. Consequently, $ a \equal{} 1, b \equal{} 2, c \equal{} 3$, or $ a \equal{} b \equal{} c \equal{} 2$. Both cases are impossible. The case $ a\equal{}b\equal{}1$ and $ c\equal{}4$ - is possible as well.

The answer to a) is  $ a \equal{} b \equal{} c \equal{} 1$ and $ a\equal{}b\equal{}1, c\equal{}4$

While [b]Erken's[/b] solution is correct, I find the identity somewhat tricky, so here's an easier way for part $(a)$:

Consider all maximal prime powers $p^{k}$ dividing $a+b+c$.

Case 1: One of $a, b, c$ shares a prime factor with $p$.

In this case, we take a sufficiently large $m$ such that $e = m(p-1)p^{k-1} > k$. Observe that \[ 0 \equiv a^e + b^e + c^e \equiv a^e + b^e \equiv 1 + 1
\equiv 2 \pmod{p^k} \]

Case 2: $a$, $b$, and $c$ are all relatively prime to $p$.

In this case, just take $e = m(p-1)p^{k-1}$ and we get \[ 0 \equiv 3 \pmod{p^k} \]. Hence, $a+b+c \mid 6$ and it's a case check from here.


2. We use complementary counting...
1 2 3 4 5 6
There are $ 6!\equal{}720$ total ways to arrange them. In order for Daniel to be next to Celine, they need to be seated in the 1-2, 2-3, ..., or 5-6 slot. There are 5 of these slots, and Daniel and Celine can be arranged in two ways after picking which slot they're in. After that, there are 4 remaining seats, and $ 4!\equal{}24$ ways to arrange the other 4. So there are $ 5\cdot2\cdot24\equal{}240$ ways to make them next to each other, and $ 720\minus{}240\equal{}\boxed{480}$ ways for them not to sit by each other.

For number two, does "near" mean next to?

For #1 it's http://www.talpink.com/public/08/09/01/fraction834e0.jpg

or $ 1\div3\div5\div7\div9\div11\div13$ ?

I believe that he means $ 1\div3\div5\div7\div9\div11\div13$, if it confuses you.

um, the numbers are 1,3,5,7,11,13,17

[hide="hint"]
notice that each of these numbers are prime.
[/hide]

Only 1 value can be obtained - $ \frac {1}{2\cdot3\cdot5\cdot7\cdot11\cdot13\cdot17}\equal{}\frac {1}{510510}$

How about if you put parentheses around the $ 2/3/5/\dots/17$? The it becomes $ 1/(2/3/5/7/11/13/17) \equal{} 17/(2/3/5/7/11/13) \equal{} 17 \cdot 13/(2/3/5/7/11) \equal{} \dots \equal{} 17 \cdot 13 \cdot \dots \cdot 3 / 2$.

I didn't know what do parentheses mean, I thought it's callet "brackets"  :lol:

Confusing problem  :wacko:

I got the same thing as math154 and Bachukas for numbers 2 and 1, respectively.

[hide="#1"]
I don't know if this is right.

In that situation, the parenthesis basically changes the $ \div$ sign into a $ \times$ sign. Note that the numbers except for $ 1$ are all prime, and thus have no common factors other than $ 1$, which means the number of different ways we can arrange $ \div$ and $ \times$ signs is the answer. There are $ 6$ signs, and we can choose any subset of them, which leads me to believe the answer is $ 2^6\equal{}\boxed{64}$.
[/hide]

[hide="#2"]
The number of ways to arrange them randomly is $ 6! \equal{} 720$. Now, take out all the possibilities where Daniel sits by Celine. Assume that Daniel and Celine are put into a group. There are $ 6 \minus{} 1 \equal{} 5$ sets of chairs they can sit in, and this is multiplied by $ 2! \equal{} 2$ ways we can organize Daniel and Celine into a "group". There are also $ 4! \equal{} 24$ ways to seat the other students. Thus, the answer is $ 720 \minus{} 24 \cdot 2 \cdot 5 \equal{} \boxed{480}$ ways to arrange them.
[/hide]


Show that the product of the divisors of $ n$ is $ n^{\frac {1}{2}\tau(n)}$, where $ \tau(n)$ is the number of divisors of n.

this is easy. pair each divisor $ d$ of $ n$ with the divisor $ \frac{n}{d}$; the product in each pair is $ n$...


Stanford
college
geometry
MIT
Harvard
calculus
geometric transformation

My GPA is, to put it euphemistically, low. But I read somewhere on this forum that if your GPA is low, it's wise to explain what you were doing with your time. Is this correct? If so, could someone tell me anything more regarding this?
Another question. Just how much of community services do universities require? (I am applying to Caltech, Stanford, Berkeley,
UC San Diego, Irvine and CSULB). Luckily I have done some community service. I hope it helps.

How low is "low" for you? If it's below 2.0, then there's probably a good chance you  won't get into the colleges you want without a VERY good reason. If it's 3.x or so, I wouldn't worry. Sure, you could explain, but the GPA really isn't that bad to justify an explanation.

Also, I'm pretty sure none of those require any community service. I got into Stanford and Caltech with 0 hours of service.

At my own institution, CSULB, admission is automatic, "by the numbers," and quite inflexible. We don't care about community service, we don't ask for letters of recommendation, and we have no routine mechanism for even hearing your explanation.

For California applicants (not from the immediate Long Beach area) we have a formula: SAT I math + SAT I verbal + 800*(HS GPA). If that is above this year's cutoff level (probably somewhere near 3500) and if you have met the course pattern requirements (4 years English, one year visual or performing arts, etc., etc.,) then you'll be admitted. Note that the formula places a fairly heavy weight on GPA. We do use "weighted" GPA's with extra points for honors or AP courses.

For out of state U.S. applicants, we apply a very similar set of standards. For international applicants, we don't necessarily ask for the same course pattern, and we're used to not necessarily having SAT scores, so grades become even more important - I hear we'd look for something like a 3.10 GPA.

I'm glad that community service is not important for those colleges.
I haven't done much of it.
My GPA is around 3.0. It's not what I would call low but it certainly is what the universities would call low.

With a 3.0 GPA, it's very likely the colleges at the top of your list will reject you on the first pass, and never look closely at the rest of your application. Those schools have formulas too- they don't use them for everything, but any applicant who isn't at least close to their standards by formula has no chance.

Thanks for the reply. Now I know which colleges I should focus on for application purposes (i.e. san diego, irvine, csulb).
I was fully aware that Stanford and Caltech might reject me after looking at my GPA. But I was only hoping that my SAT score of 2250 (Math800, Reading760) could help me.

It certainly helps, but GPA is weighted heavily. When I went through, I think the index the UCs used was SAT I (two parts)+SAT II (three tests)+1000*GPA. They used weighted GPA, so there were a lot of students getting more than 4000 on that part. As far as I know, the only change they've made to this system is that the SAT II writing test has become part of the SAT I, and thus the two/three split goes the other way.

Thanks again.
I don't know how my GPA will be calculated, though. I have done my schooling in India. There never were any ranks. Just marks.
But what is a weighted GPA?

A relevant question. How are grades from different educational systems converted to GPA's ? For example I got 81% in the examination conducted by my state and I have heard that this is low by US standards. However the top score in my school was 87% and I was in top ten in a school of 150.

[quote="Kent Merryfield"]"Weighted GPA" is a system in which certain classes - AP classes and officially designated "honors" classes - have an A count for 5 grade points instead of 4 and B count for 4 grade points instead of 3. It's supposed to encourage students to take more challenging schedules. Whether it accomplishes its purpose ... who knows? That's another debate.[/quote]

At my school, we have the weighted GPA system with honors classes, as well as "Accelerated/Enriched (AE)" classes that count for 4.5. However, honors classes tend to give significantly more homework, and thus most teachers often recommend to to take more than 3 honors classes (Is anyone else's school like this?). Unfortunately, the weighted GPA has the effect of causing a huge number of students to take all honors classes, which is a negative because it produces pressure for students to take too heavy a course-load.

Last year (9th grade) I took 3 honors courses (5 courses total) and got a 4.63 GPA, and there year I took 5 honors courses (6 courses total) and got a 4.46 GPA. The average GPA's of students from my school who get into MIT and Harvard are around 4.55 and 4.61 respectively.

The problem that is arising in my school district with regard to "honors" and "AP" classes is that EVERYONE is trying to take them because it helps your weighted GPA (thus class rank) and looks good on your transcript. From the point of view of the administration, yay, power statistics for the school - we have one of the highest Newsweek AP classes per graduating senior ratios. From my point of view, this is only dumbing down the classes so that people can handle them without failing. The school got rid of regular freshman english this year; so now all freshman take "honors" english. But, really, what is an "honors" course that everyone takes? A regular course. I highly doubt that the student body as a whole is getting progressively more intelligent/capable. And for AP classes, many students go into the test expecting a 1 or a 2; honestly, in my opinion, there's no reason they should be there. Unfortunately, it seems that the actual AP scores do not influence as many statistics (maybe they are more difficult to obtain) as the sheer number of tests, so people do not care so much.

Basically, the emphasis on these courses is hurting the level at which they are tought; I hope to see this problem remedied when I go to college in the fall.

I am quite aware that the "weighted GPA" system is easily vulnerable to being gamed or corrupted, and paladin8 has described some of the mechanisms for that to happen.

One of the reasons that AP exam scores themselves are difficult to use in the admissions process is that by far the greatest number of such exams are taken by students in their senior year, and the results are available far, far too late to be used in the admissions process. (An aside: I'm going to be helping some incoming freshmen enroll tomorrow, and I got to talk to some of them tonight. "Did you take the AP Calculus AB exam? OK, now, best honest guess - do you [i]think[/i] you passed it?" And I'll have to place people based on that guess.)

The historical results and pass rates of a school can be obtained. The College Board has that data, and the school itself receives it, and I think that a college that insisted on knowing would be able to obtain it. But could a college declare that a school's AP course would not earn the bonus weighting unless the AP exam pass rate for that course and school met a certain threshold? There are two major problems with that idea. The first is that it penalizes an individual student for the actions of a group - and it's not even his own group, but a preceeding group. That's the smaller problem; the bigger problem is that it would be a racial/political minefield. There were lawsuits filed in recent years over the fact that schools in poor neigborhoods offered fewer AP classes than those in rich neighborhoods. There's been a great hue and cry to expand access to AP courses as widely as possible. If the bonus point were shut off for schools with low pass rates, which schools do you think would suffer? And what do you think would be said?

As for honors classes - I think some large state universities try to play some role in approving the "honors" designation, but I'm not sure how, or what schools have to demonstrate and to whom. I am a little surprised that a school could get away with what paladin8 described above for the freshman English course - but maybe I shouldn't be too surprised.

There would be a way to move the AP quality burden to the individual, but I doubt it would happen. A student could receive a letter saying something like this: "You have been conditionally admitted. This is conditional on you maintaining such-and-such a GPA throughout your senior year. Furthermore, in your case, that admission was partly based on your ambitious schedule which includes 5 AP courses in your senior year. You must pass at least 3 of those AP exams, or we will rescind your admission." In late July, no less. Look, I already know that colleges do rescind admissions for poor senior year grades - hey, even my college does that. But to try to do it for failure to pass AP exams - can you imagine the explosion, the huge pile of negative PR?

I understand that the faculty at one UC campus tried a few years ago to rein in the "weighted grades" machine. I think they wanted to keep the bonus but reduce its size from a full grade point to half a grade point. But I don't think that has happened.

The usefulness of GPA is eroded by the fact that it is context-dependent. Different high schools have different [i]de facto[/i] grading standards; different courses have different levels of difficulty; even individual teachers have, well, their own individual standards. A straight-A student who somehow avoided every difficult class in a school wouldn't be a very good prospect for a highly selective college. The idea of "weighted GPA" tries to correct for that, but that opens the door to all of the other distortions that we are talking about. I'm a baseball fan, a "sabermetric" or numbers-literate baseball fan - and I know that context matters. A pitcher earning an ERA of 3.00 for the Houston Astros in 1968 has a very different meaning than a pitcher earning an ERA of 3.00 for the Colorado Rockies in 1996. Baseball is sufficiently well-studied that that you can intelligently correct for the difference in context. ([url=baseball-reference.com]My favorite such source.[/url]) If GPA could be corrected for context half as well as baseball stats, it would probably be a far better predictor of college success than any standardized tests. But I don't have any practical ideas for taking a significant fraction of the noise out.

Your 3.0 GPA from India is essentially meaningless.  For all I know, that could be the highest GPA in your school.  Your class rank and class size, combined with some information about the quality of your school, might carry a little more information.  It seems that a lot of international students on this forum ask questions about admissions, but the sad fact is that very few people know much about how these work.  It's possible that *no one* does in the sense that the different schools might be evaluating international applicants in vastly different ways.

And all this time, I was thinking that I could at least get into some place like Irvine. I had no idea that the GPA was so much more important than the SAT reasoning test.
Another thing. I read in a lot of college websites that one should mention in his application all the advanced courses he has taken. Now, should these be only from the high school or can they be from some institute (as long as the institute is good enough to be giving doctorates)? I ask this because no school I ever went to had stuff like advanced placement programmes. Neither do any universities in my city provide opportunities for high school students to attend college lectures.

[quote="Abel"]I had no idea that the GPA was so much more important than the SAT reasoning test.

Another thing. I read in a lot of college websites that one should mention in his application all the advanced courses he has taken. Now, should these be only from the high school or can they be from some institute (as long as the institute is good enough to be giving doctorates)?[/quote]

The most important criterion is "strength of schedule," whether you have taken challenging courses before. If the courses you took were VERY hard, you could have a lower GPA than some other applicant with easy courses, and still get in.

Usually a college will expect to see a graded transcript for all formal courses you have taken, wherever they were taken. It is certainly to your advantage to have taken courses at a college or university as part of your "high school" program.

I defer to Kent Merryfield on details of UC and Cal State admission, as those systems have local rules with which I may not be familiar. But in general strong courses help, and it is routine for the college to which you apply to expect to see transcripts of all your previous course work.

I have another few questions while were at it.
yes, i am going into 9th, but i need to know what not to screw up in my high school years.

1. how much does class rank affect college admissions?  because the school im going to is like super smart (IB), and if i went to any other school, my class rank would no doubt be higher.

1 continued...
how much do they care about hard classes? for instance, is it better for me to get a B in an Ib or AP course, or an A in some regular course?

and AP vs IB, anyone?

this upcoming school year (freshman) im taking 4 APs, and 3 Pre-IBs, but as i get older, theyll make me take more ib classes.

2. my laziness in middle school, and not doing homework, resulting in me getting bad grades (Bs) in high school credit classes, including math courses.  would this hurt my GPA, or do they not care about middle school classes.

3.  do colleges look at extracurriculars at all?  for instance, i play baseball.  is it worth it for me to play varsity baseball, if i meant i would get, say, two B's where i could get As if i didnt.

4. same question as 3, but with academic extracurriculars, not math, like brain bowl, history fair, etc.

thanks.

For your #2, there's no way any college will be looking at middle-school grades. At least I'm about 99.9% sure colleges don't look at grades from before high school. It just wouldn't make sense.

My own questions:
My school offers only one AP English (AP Lit) and one AP history/social studies course (AP Gov), offers no honors English or history, and limits everyone to taking 3 AP's per year. Also, my school only ranks the valedictorian and salutatorian, so whoever would be third in the class would never know - I'll probably fall in somewhere between 3rd and 5th in the class. I currently have a 4.07 GPA, 3.94 unweighted, 4.0 on some scales. Is class rank really that important in admission/merit aid considerations? And will lack of AP humanities courses really cripple me in admissions - even at tech schools?

@Walk Around The River:

1. Colleges get some sort of school profile, so they at least try to take the school's student body into consideration when regarding class rank. I would definitely NOT change schools for the reason of improving class rank. As for hard classes, it shows that you are ready to handle college coursework or, even better, that you are willing to. The answer to your question is to get an A in the IB or AP course. And for the record, I like AP a lot more than IB, but I'm obviously biased since I never actually did IB, only heard about it. Your 4 AP and 3 pre-IB classes in freshman year is probably overkill already; that's great if you can handle it, but don't screw yourself over.

2. Just don't carry the middle school grades over to your high school transcript. Don't be lazy (or at least be lazy sparingly).

3. Yes, extracurriculars help, but you'll need to stand out for it to really impact your admission. Sports are slightly different, but again you'll need to stand out. Just having a little bit of everything isn't much good - leadership and excellent performance is where it's at.

4. Same as 3.

@sapphyre571:

Class rank helps colleges determine where you stand among your classmates as well as grade inflation/deflation. If your school only "ranks" valedictorian and salutatorion, it doesn't really count as ranking so I wouldn't worry about it. It would certainly help to be able to say that you are one of them, though. I don't know anything about merit aid, but I don't believe class rank is extremely significant in admissions. Lack of AP humanities when you had the opportunity will be looked down upon by admissions officers; it reflects your inability and/or unwillingness to handle difficult coursework. That said, it shouldn't be a huge influence at tech schools, depending on how your other strengths are played out.

EDIT: In case I get refuted, I didn't mean in any way to belittle the importance of class rank in admissions. It is important. But relative to taking challenging courses and maintaining a strong GPA and standardized test scores, it has less weight (AFAIK).

I plan on taking both AP Lit and AP Gov. I would probably take AP US history and AP European history and AP English language if I had the opportunity. I was just wondering if the lack of opportunity leading to lack of AP humanities would be a problem.

Not taking AP Humanities when you have the opportunity to won't hurt you that much. It didn't hurt me for Stanford, at least, which I would think is a huge humanities school. The only humanities I took is honors US history in 9th grade and AP English in 12th grade (what a waste of effort).

Whether or not to take AP Humanities courses depends a lot on the situation.

Example: At my old school, the AP US History teacher was a nightmare. All the AP pilers ended up spending about 2/3 of their homework time on APUSH, with frequent late-nighters. Now, if I had stayed there (which I'm glad I didn't), I would've been the only "AP piler" not taking APUSH.

But the amount of time I would've saved would probably have been enough to fully carry out two competitive science fair projects. Or to do volunteer work or intern or shadow a doctor or whatever else.

Ultimately, you need to show that you made good use of your time. People will forgive you if you don't care that much about history or such things, especially if you show that you used the saved time effectively.


An integer N consisting of five all-different nonzero digits has the curious property that it is equal to the sum of all the different three-digit integers formed by the three-digit permutations of its five digits. Find N.

[hide]Let N = abcde. There are 5C3 (3!) permutations. The average of all such numbers is 111(a + b + c + d + e) / 5. Thus, the total sum is: (5C3)(3!)(111)(a + b + c + d + e) / 5.

N = (9)(4)(37)(a + b + c + d + e).

9 | N implies 9 | (a + b + c + d + e). So let (a + b + c + d + e) = 9m.

N = 81(4)(37)(m) = 11988m. From here, we can try all 8 cases for m, and m = 3 works. Therefore, N = 35964.[/hide]


I'm pretty sure that Ohio State is going to win.  And that Florida pretty much has no chance.

well ohio state's biggest wins were over bowl losers... figures...


algorithm
real analysis
real analysis unsolved

By finding solutions as power series in $ x$ solve
\[ 4xy'' \plus{} 2(1\minus{}x)y' \minus{} y \equal{} 0.\]

Okay, so what is giving you trouble?

Assume a solution of the form $ y(x)\equal{}\sum_{n\equal{}0}^{\infty}a_{n}x^{n}$, so that $ y'(x)\equal{}\sum_{n\equal{}1}^{\infty}na_{n}x^{n\minus{}1}$, and so on.  Then play with the indices so that you can combine the sums, and go from there.

I think I'm okay with that algorithm - however, how do we know from the start that solutions in power series exist? Or is that out of the scope of differential equations?

Also, is it not $ y(x) \equal{} \sum_{x\equal{}0}^{\infty} a_nx^{n\plus{}r}$? That's what I get from [url=http://en.wikipedia.org/wiki/Frobenius_series]wikipedia[/url], but why is there an extra $ r$ in that case?

[quote="aidan"]I think I'm okay with that algorithm - however, how do we know from the start that solutions in power series exist? Or is that out of the scope of differential equations?[/quote]

There are some conditions to check that $ 0$ is a "regular singular point" of the differential equation.  Surely your textbook has this!

[quote]Also, is it not $ y(x) \equal{} \sum_{x \equal{} 0}^{\infty} a_nx^{n \plus{} r}$? That's what I get from [url=http://en.wikipedia.org/wiki/Frobenius_series]wikipedia[/url], but why is there an extra $ r$ in that case?[/quote]

You are right, this is the form in general.  There is an equation to solve to find $ r$ before you do the other stuff.

I don't use a textbook, but I have lecture notes that touch on regular singular points. However I don't understand them fully (or haven't tried enough problems), so I'm trying to brush that up!

hello, we get
$ y(x)\equal{}C_1\left(\sum_{n\equal{}0}^{\infty}(\frac{1}{2})^n\frac{(x\plus{}1)^n}{n!}\right)$.
Sonnhard.


[quote="AYMANE"]$\displaystyle a_{n+1}=\frac{11.3^n-2n-7}{4}$
 :?[/quote]
How did you get this answer??


quick percentage question

if I received 17/20 on a test, and that test was worth 5% of my final grade, how would i calculate to find my score out of 5%. thanks

[hide]If it's $ 5\%$ of your grade, that means a $ \frac {20}{20}$ would've gotten you $ 5\%$ of your grade. Since you scored a $ \frac {17}{20}$, that's $ \frac {17}{20} \cdot 5\% \equal{} \boxed{4.25\%}$.[/hide]


In triangle ABC, the internal angle bisectors of angles <B and <C cut the median AD (where D is on BC) in E and F respectively. Assuming that BE = CF, prove that triangle ABC is isosceles.

Let <BDA= x and AD= M

Applying sine rule in triangles BED & FDC we have

BE/sinx = ED/sin(B/2)------(1)     &    CF/sin(\pi-x) = FD/sin(C/2)----(2)

LHS of the two equations are equal hence we have

ED/sin(B/2) = FD/sin(C/2)---------(3)

We have ED= [a/(a+2c)]*M  & FD= [a/(a+2b)]*M as BE and CF are the angular bisectors of angles B and C in triangles ABD & ACD hence they divide AD in the ratio of the other two sides.....

Applying sine rule and rearranging the terms in eq (3) we have.......

sinA[sin(B/2) - sin (C/2)] = 4 sin(B/2)sin(C/2)[cos(B/2)- cos (C/2)]

If B and C are unequal then LHS and RHS bear the opposite signs.....

Hence B=C

I may be wrong but I think this is [b]a lot[/b] simpler: :D

*** will indicate the bisector theorem, or whatever it is called (about proportions of bisector lengths)

Construct M point on EB \cap FC.

applying *** on ADC and on ABD gives us:
MD/BD = AM/AB   and    MD/DC = MA/AC
as MD and MA are shared we get: AB/DB = AC/DC

but as AD is the median line, BD=DC
thus AB=AC and thus triangle ABC is isosceles.

Peter can you be a bit more clear on ***(The bisector proportion theorem)
I didn't get your point straight.

I think his solution is wrong - he assumes that M lies on AD, and you can't assume that.

yes, I know, I discussed this with Arne before. I counted M on AD because it is like that when 2 bisectors have the same length. I didn't think of the fact that that rule is only a consequence of the fact that its isosceles.

So in fact I used the formula we had to prove :D little mistake

but still I'm convinced there are shorter solutions than the first one...


I heard there's a math competition called MEANT, but I missed it.  I'd like to take part in it next year, but I have no idea what questions are on the test.  Can someone either post or PM me (or give a link) a few tests if you have them? I don't know, but if participants do not receive the test, then can someone try and remember a type of problem that was on there and post it?  Thanks!!

Last year's MEANT contest was great with grade specific exams from 6th through 12th grade.  They talked about doing it again this year but I never saw an official announcement.  Here is the reference on their web site:

[url]http://www.meant.org/matholympiad.html[/url]

I'll see if I can find out any more.  Here is information on last year's contest:

[url]http://metroplexmathcircle.wordpress.com/2008/10/28/meant-math-contest-november-15th/[/url]


Anybody interested in this? I made one last year and that was fun ( you can search for it...). I'm thinking of making another one for this year.

I didn't try out for ARML....but I'm interested too.

I'd definitely do it. Sounds good. I'm interested to see what you come up with for #8.

I'm interested.

I might write one of these myself . . . if I can come up with enough problems.

Yeah I would like one... I'm beginning to think I need a lot more practice in individual rounds...

8 is going to be geometry, isn't it... :help:

I'll do it although I can't go to ARML this year due to FPS Internationals.

Is there ever not interest in your mock tests?  :wink:

there's always interest in mock tests right?

especially in experienced ppl's who have made them before (altheman)

I'm interested, but make #8 slightly easier..

Hey everyone!

I have written my own Mock ARML Individual.  Rather than opening up a new thread, I'll ask here: does anyone want to testsolve the problems to make sure they're okay before I release the test?

I'll be a beta tester for e^(pi*i)=-1's test.[/i]

I will test them, though I can't guarantee I'll be able to do #8 or potentially others on the test.

im interested and will test them even though ill use brute force on all.

Multiple testers are probably unnecessary.  I have sent the problems to CatalystOfNostalgia.

Yeah, hopefully I'll have answers and comments to you by sometime late this afternoon. If not, probably tonight, because I have an orchestra concert.

[quote="Altheman"]hmm okay so i'm almost done (i keep coming up with ideas that turn into problems that are way too hard...)

worthawholebean, i hope that you can look over the test...i still have to change two problems because they are too hard i think (or the pairs don't work out), i'll pm the test to you tonight i hope[/quote]Sure, I'll look it over.

I'm interested in taking a Mock ARML Individual round.


inequality system
pigeonhole principle
n-variable inequality
linear algebra
IMO
IMO 1987

Let $x_1,x_2,\ldots,x_n$ be real numbers satisfying $x_1^2+x_2^2+\ldots+x_n^2=1$. Prove that for every integer $k\ge2$ there are integers $a_1,a_2,\ldots,a_n$, not all zero, such that $|a_i|\le k-1$ for all $i$, and $|a_1x_1+a_2x_2+\ldots+a_nx_n|\le{(k-1)\sqrt n\over k^n-1}$.

Without loss of generality we may assume the $ x_i$ are nonnegative, by an appropriate (consistent) change in the signs of the $ a_i$.

Now, take the set of $ k^n \minus{} 1$ nonzero vectors $ \textbf{a}$ formed by each possible combination of $ 0\leq a_i \leq k \minus{} 1$.
Since the $ x_i$ are nonnegative, $ \textbf{a.x} \leq (k \minus{} 1)\sum x_i \leq (k \minus{} 1)\sqrt {n\sum x_i^2} \equal{} (k \minus{} 1)\sqrt {n}$

We therefore take each vector in our set and can put its dot product with [b]x[/b] into one of the $ k^n \minus{} 1$ intervals $ \left[\frac {\lambda(k \minus{} 1)\sqrt {n}}{k^n \minus{} 1}, \frac {(\lambda \plus{} 1)(k \minus{} 1)\sqrt {n}}{k^n \minus{} 1} \right) (\lambda \equal{} 0,1,....,k^n \minus{} 2)$.

If the $ \lambda \equal{} 0$ interval contains one of our dot products, we are done. Otherwise, by the pigeonhole principle, there exists an interval containing no less than two [b]a[/b] vectors. Taking their difference gives a new vector lying in the desired $ |\textbf{a.x}| \leq \frac {(k \minus{} 1)\sqrt {n}}{k^n \minus{} 1}$ region, so such a vector always exists.

Apparently, this result was strengthened, by a paper of Moser (1970): The second moment method in combinatorial analysis. The following holds: Such $a_i's$ exist, so that
$$
|\sum_i a_i x_i|\leqslant \sqrt{\frac{k^2-1}{k^{2n}-1}}.
$$
Note that as $n$ grows, Moser's bound is significantly stronger than this IMO problem. Now, a paper by Karmakar-Karp-Lueker-Odlyzko from 1986, titled ``Probabilistic Analysis of Optimum Partitioning" references the aforementioned result by Moser, and notes that a significantly weaker bound can be achieved by a very simple Pigeonhole argument, and lays out details, which precisely happens to be this IMO problem!

I want to touch upon certain research problems revolving around this problem.

This problem is broadly known as optimal partitioning (or number partitioning). Formally, let [b]$X_1,\dots,X_n$[/b] be real numbers. Find a vector $\sigma=(\sigma_1,\dots,\sigma_n)\in\{-1,1\}^n$ of signs such that $|\sum_{1\leq i\leq n}\sigma_i X_i|$ is as minimal as possible. In the case $X_1,\dots,X_n\in\mathbb{R}^d$, the problem is now known as vector balancing, where the idea is to find a set of indices as above such that $\|\sum_i \sigma_i X_i\|_\infty$ is as small as possible, where for a $v=(v_1,\dots,v_d)\in\mathbb{R}^d$, $\|v\|_\infty = \max_i |v_i|$. Since the search space involves $2^n$ elements, such problems are NP-hard, in the sense of computational complexity.

Several interesting ideas   around this problem are as  follows. Let $X_1,\dots,X_n$ above be i.i.d. standard normal random variables. Then the optimal value of this partitioning is known to be  $\Theta(\sqrt{n}2^{-n})$ with high probability, whereas the optimal polynomial time algorithm achieves a performance of only $\Theta(2^{-(\log n)^2})$!  Closing such gaps, or showing evidence of computational hardness  are currently open problems.

[b] [color=#f00][i][u]Just we can use PHP and C-S Inequality
[/u][/i][/color][/b]


for all non-negative $a,b,c$ with constrain $a+b+c=3$ , find the minimal of

$a^3+b^3+c^3+8abc$

 :)

Find also the maximal of the expression

[quote="M4RI0"]Isn't it 0?[/quote]

yes,you are right , i forget to put the constrain $a+b+c=3$ .Now it is ok  :blush:  :P  :D

hi,

here is my solution :

$a^3+b^3+c^3 + 8abc \geq 11abc (1)$

The only case of equality is when $a=b=c (2)$

$(1),(2) \Rightarrow Min(a^3+b^3+c^3+8abc) = 11$

 :) .

Sorry , the equality case does not occur when $a=b=c=1$   :(

you can try something like $a=1.1$ , $b=1.2$ , $c=0.7$ and you will see that it is $<11$ . So lets guess again  :D

[quote="shyong"]for all non-negative $a,b,c$ with constrain $a+b+c=3$ , find the minimal of

$a^3+b^3+c^3+8abc$

 :)[/quote]

  I solve it when $min (a,b,c) \leq \frac{9}{11}$ and found its min is $6+ \frac{3}{4}$ occurs when $a=b=\frac{3}{2}$,$c=0$

[quote="paladin8"][quote="shyong"][quote="paladin8"][hide]

Well we can show $a^3+b^3+c^3+8abc \ge \frac{(a+b+c)^3}{4}$, which simplifies down to

$3(a^3+b^3+c^3) + 26abc \ge 3(a^2b+a^2c+b^2c+b^2a+c^2a+c^2b)$.

But by Schur we have

$3(a^3+b^3+c^3) + 9abc \ge 3(a^2b+a^2c+b^2c+b^2a+c^2a+c^2b)$

and obviously

$17abc \ge 0$

so it's true. Equality holds in both cases when $a = b, c = 0$ and permutations of it. So the minimum is $\frac{27}{4}$.

EDIT: This result seems to show that the minimum is the same when $8$ is replaced by $\frac{15}{4}$. [/hide][/quote]

yes , paladin8 , I think that the result for the minimal of $a^3+b^3+c^3+kabc$ is $\frac{27}{4}$ for all $k\geq \frac{15}{4}$ .And we can show this using Schur like what paladin8 did . Now what do you all think about the maximal of the expression ?

Find the maximal of $a^3+b^3+c^3+8abc$ for all non-negative $a,b,c$ satisfying $a+b+c=3$  :)[/quote]

I have the minimum result generalized [url=http://wangsblog.com/jeffrey/?p=169]here[/url]. The proof is the same idea.

For the maximum, we can just say $a^3+b^3+c^3+8abc \le (a+b+c)^3$, right?[/quote]

yes , I also think the same way  :)

As for the minimal i think we can even say that

$a^3+b^3+c^3+kabc\geq \text{min}\{k+3,\frac{27}{4}\}$ for non-negative $a,b,c$ and real $k$ , with $a+b+c=3$

Do you agree ? :)

Yup. We can homogenize, expand, and use Schur again to show that if $k \le \frac{15}{4}$ and $a+b+c = 3$, then

$a^3+b^3+c^3+kabc \ge k+3$.

[quote="shyong"]

yes , I also think the same way  :)

As for the minimal i think we can even say that

$a^{3}+b^{3}+c^{3}+kabc\geq \text{min}\{k+3,\frac{27}{4}\}$ for non-negative $a,b,c$ and real $k$ , with $a+b+c=3$

Do you agree ? :)[/quote]

This can be generalized for $n$ numbers and any degree. Look [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=108398]here[/url].

[hide="to find the max"]
i think you can use

$\frac{a+b+c}{3}\geq\sqrt[3]{abc}$

and $a^{3}+b^{3}+c^{3}\leq(a+b+c)^{3}$

(i am not 100% sure about the last inequality though)
[/hide]

[quote="rem"][quote="maokid7"]
and $a^{3}+b^{3}+c^{3}\leq(a+b+c)^{3}$

(i am not 100% sure about the last inequality though)
[/quote]
 :huuh:
If $a,b,c\ge 0$ just expand RS and you will get a sum of non-negative numbers left on RS and $0$ on left side. This is true[/quote]

More generally, for $p>q>0$ we have \[\sqrt[p]{x_{1}^{p}+\cdots+x_{n}^{p}}\leq\sqrt[q]{x_{1}^{q}+\cdots+x_{n}^{q}}.\]

Of course that is true, consider the general AM-GM then remove the constants :P
I think there is only equality when all the x's are 0


[hide="Solution"]First, we change all the logarithms to base 2 such that $\log_2 \sqrt[3]{a}+\log_2b=5$ and $\log_2 \sqrt[3]{b}+\log_2a=7.$ Hence, $\log_2 \sqrt[3]{a}\cdot b=5$ and $\log_2 \sqrt[3]{b}\cdot a=7.$ Exponentiating, we can take advantage of the symmetry of the equations by Getting multiplying them together. This yields $(ab)^\frac{4}{3}=2^{12}$, and $ab=2^9=\boxed{512}$. [/hide]

[hide="Answer"]By definition, we have $\log_4b^2=\log_2b$ and, similarly, $\log_4a^2=\log_2a$. Summing the equations, we have $\log_8ab+\log_2ab=12\Rightarrow \log_2(ab)^\frac{1}{3}+\log_2(ab)=12\Rightarrow \log_2(ab)^\frac{4}{3}=12\Rightarrow$
$\frac{4}{3}\log_2ab=12\Rightarrow \log_2ab=9\Rightarrow ab=\boxed{512}$.[/hide]

[hide]
Taking the power out of the bases, we get $\log_2 \sqrt[3] a + \log_2 b = 5$ and $\log_2 a + \log_2 \sqrt[3] b = 7$.  From here we see $ab^3 = 2^{15}$ and $a^3b = 2^{21}$, so $a^4b^4 = 2^{36} \rightarrow ab = 2^9 = 512$
[/hide]

[hide="another way"]

changing all the base to 8 we have

$\log_8a+3\log_8b=5$
$3\log_8a+\log_8b=7$

Solving above simultaneously we get $a=64,b=8$ . So $ab=512$  :)
[/hide]

i did it the way with changing to base 2. but very nice solutions shyong. i like it better than the log one and it takes less time as well.

Here's a more direct method:

Adding the two equations we get:

$ \log_{8}ab\plus{}2*\log_{4}ab \equal{} 12$

==> $ \log_{8}ab\plus{}3*\log_{8}ab \equal{} 12$

==> $ \log_{8}ab \equal{} 3$

==> $ ab \equal{} 8^{3} \equal{} 512$  :)

Sorry revive topic, but came across method another :)
[hide]Change to equation.
a=8^x, b^2=4^y, x+y=5.
a=2^3x, b^2=2^2y
b=2^y, ab=2^(3x+y)
Next log into equation.
b=8^m, a^2=4^n, m+n=7
b=2^3m, a^2=2^2n
a=2^n
Compare:
3x=n, y=3m. Plug into second.
3x+y/3=7. 9x+y=21.
Elimination with first equation.
8x=16, x=2.
Plug back into wanted expression, which is 3x+y.
4+5=9, and 2^9=[b]512[/b]
[/hide]

Adding, we see that $\log_8(ab)+\log_4(ab)^2=12$
Thus we can assign an arbitrary value to the first and second quantities of the LHS
$8^x=ab$
$4^y=(ab)^2$
$4^\frac{y}{2}= ab$
Thus $2^y=ab$
Setting these equal
$2^{3x}=2^y$
$3x=y$
Plugging back in into the original equation
$x+3x=12$
$x=3$
Therefore $8^3=\fbox{512}$ and we have our answer.

Note that:
$$\log_8 a + \log_4 b^2 = \log_8 a + \log_8 b^3 = \log_8 (ab^3) = 5$$
$$\log_8 b + \log_4 a^2 = \log_8 b + \log_8 a^3 = \log_8 (a^3b) = 7$$

Hence, we have the equations:
$$ab^3 = 8^5$$
$$a^3b = 8^7$$

Dividing the second equation by the first equation yields that:
$$\frac{a^2}{b^2} = 8^2$$
$$\frac{a}{b} = 8$$
$$a = 8b$$

Replacing this value of $a$ into $ab^3 = 8^5$ yields:
$$8b^4 = 8^5$$
$$b = 8$$
$$a = 8^2$$
$$ab = 8^3 = \boxed{512}$$

[quote=4everwise]Determine the value of $ab$ if $\log_8 a + \log_4 b^2 = 5$ and $\log_8 b + \log_4 a^2 = 7$.[/quote]
so you can convert them to base 2
$\frac{\log_2 a}{3} + \frac{\log_2 b^2}{2}=5$
multiply by 6 both side
so then $2\log_2 a + 3\log_2 b^2=30$
other equation is $\frac{\log_2 b}{3} + \frac{\log_2 a^2}{2}=7$
multiply by 6 again so you get $2\log_2 b + 3\log_2 a^2 = 21$
you have two equation $2\log_2 a + 6\log_2 b=30$ and then $2\log_2 b + 6\log_2 a = 21$ so then you can just solve for log_2 b and log_2 a and then multiply a and b

[hide=Solution]We write the first equation solely in base $2$. Note that we have $\log_8{a}=\log_2{\sqrt[3]{a}}$ and that $\log_2{b}$. Therefore, $\sqrt[3]{a} \cdot b = 32$ and similarly the second equation simplifies to $\sqrt[3]{b} \cdot a = 128$. Multiplying these equations gives $(ab)^\frac{4}{3} = 32 \cdot 128 = 2^{12} \implies \boxed{512}$.[/hide]

[quote=NealShrestha][quote=RP3.1415][quote=NealShrestha]Wasn’t this an AIME question?[/quote]

Yes this was AIME 1984 P5

Quite the weird coincidence. My family went on a trip today and i printed a problem set of a bunch of log problems my math circle director gave all of us to do. This was one of the problems on it that I solved in the car today lol[/quote]

Could you send me the packet or just post it here? Thanks[/quote]

Its only ten problems and they are all from contests. You can just search up logarithm problems from NYSML, ARML, AIME etc..
This was one of the resources that is only for my local math circle's members for ARML prep/practice so I can't post our resources.

We let $a=2^x$ and $b=2^y$.

We then get the equations $\frac{x}{3}+y=5$ and $\frac{y}{3}+x=7$.

We add to get $\frac43(x+y)=12$.

We solve to get $x+y=9$ so $ab=2^{x+y}=2^9=\boxed{512}$.

:)

[hide=Solution]Converting both equations into base $2$ gives us that $\log_2 ({b\sqrt[3]{a}})=5$ and $\log_2 ({a\sqrt[3]{b}})=7$. Adding these two equations and converting the logarithm into exponential form gives us the equation $\sqrt[3]{(ab)^4}=2^{12}$, so $ab=2^9=\boxed{512}.$[/hide]

Let $\log_2 a=x, \log_2 b=y$. Then we have $\frac{1}{3}x+y=5$ and $\frac{1}{3}y+x=7$

Then multiply by $3$ gives $x+3y=15$ and $\frac{1}{3}y+x=7$. Subtract gives $\frac{8}{3}y=8$ gives $y=3$. tThen $x=6$.

Now we can find $2^{x+y}=2^{9}=\boxed{512}$.

[hide=Sol]All of the bases of the logarithms different, but, we want them to be the same, we can apply some logarithm properties to them. However, notice that all of the bases of the logarithms are powers of $2$, namely $4$ and $8$, which are $2^2$ and $2^3$, respectively. We can convert them to the same base, $2$ by taking the square and cube roots of the arguments, as such

$$\log_2 \sqrt[3]{a}+\log_2b=5$$
$$\log_2 \sqrt[3]{b}+\log_2a=7$$

This is a lot simpler. Now, we can combine each equation into one logarithm

$$\log_2 \sqrt[3]{a}b=5$$
$$\log_2 \sqrt[3]{b}a=7$$

Furthermore, we can write the cube roots as fractional exponents

$$\log_2 a^{1/3}b = 5$$
$$\log_2 b^{1/3}a = 7$$

Adding the equation together gives us

$$\log_2 (ab)^{4/3} = 12$$

Bringing the exponent out of the logarithm

$$4/3\log_2 ab = 12$$
$$\log_2 ab = 9$$

Therefore, $ab = 2^9 = \boxed{512}$.[/hide]

[quote=NealShrestha]Wasn’t this an AIME question?[/quote]
I knew it felt like i'd already seen this

[hide=solution]
$\log_2 \sqrt[3]{a}+\log_2b=5 \implies \log_2 \sqrt[3]{b}+\log_2a=7.$ So $\log_2 \sqrt[3]{a}\cdot b=5$ and $\log_2 \sqrt[3]{b}\cdot a=7.$ Now we get that $(ab)^{\frac{4}{3}}=2^{12} \implies ab=\boxed{512}.$

[/hide]


Prove that: in ABC triangle
$ a\sqrt{\frac{(p\minus{}b)(p\minus{}c)}{bc}}\plus{}b\sqrt{\frac{(p\minus{}c)(p\minus{}a)}{ca}}\plus{}c\sqrt{\frac{(p\minus{}a)(p\minus{}b)}{ab}}\geq{p}$

Put $ a = y + z,b = z + x,c = x + y$, the inequality can be rewrite as

$ \sum{(y + z)\sqrt {\frac {yz}{(z + x)(x + y)}}}\geq \sum{x},$

with equality if $ x\to 0$. Using Holder inequality,

$ (\sum{(y + z)\sqrt {\frac {yz}{(z + x)(x + y)}}})^2(\sum{y^2z^2(y + z)(z + x)(z + y)})$

$ \geq[\sum{yz(y + z)}]^3$,

we only need to prove :

${{ [\sum{yz(y + z)}]^3\geq\prod{(y + z)}(\sum{x}}})^2(\sum{y^2z^2})$.
[hide="Proof."]${{ [\sum{yz(y + z)}]^3 - \prod{(y + z)}(\sum{x}}})^2(\sum{y^2z^2})$

$ = xyz[2\sum{x^5(y + z) - 4\sum{y^3z^3} + \sum{x^3(y^2z + yz^2)} - 6x^2y^2z^2}]\geq0$,

which is obvious.[/hide]


Find all functions $f: R\rightarrow R$ such that
$f(f(x-y))=f(x)-f(y)-f(x)f(y)-xy$

trivial

let x=y;

f(f(0)) = -f(x)^2 -x^2

so f(x)^2 = -f(f(0)) - x^2
note that since f(x)^2 > 0 for all x then no solution exist.

[quote="ali"]

so f(x)^2 = -f(f(0)) - x^2
note that since f(x)^2 > 0 for all x then no solution exist.[/quote]

But $f(f(0))$ might be negative

[quote="lordWings"][quote="Jumbler"]But $f(f(0))$ might be negative[/quote]
The problem is that you must have $-f(f(0))>x^{2}$ for all $x\in\mathbb{R}$, so $f(f(0))=-\infty$, which isn't possible if $f: \mathbb{R}\to\mathbb{R}$.
Maybe the problem meant $f: \mathbb{R}^{*}\to\mathbb{R}$.[/quote]

umm... right.

What about this one?

$f: R\rightarrow R$

$f(f(x+y))=f(x+y)+f(x)f(y)-xy$

[quote="Jumbler"][quote="Rust"]$f(x)=x$.[/quote]

Proof?[/quote]

...


Hey guys....I had my math exam today...I havent done that well... :( I can't say how many marks I'll lose until I get the answers to a few problems... :maybe:
So I would really appreciate it if you guys could help me out...Just post the answers you get for the problems I'm posting...thanks a lot...
1.rationalize the denominator-$\frac{2\sqrt{3}+3\sqrt{2}}{2\sqrt{3}-3\sqrt{2}}$
2.The total surface area of a hollow metal cylinder open at both ends of external radius 10cm and height 14cm is$578\pi$cm. Taking 'r' as internal radius, write down an expression for 'r' and find the value of 'r' as well as the thickness of metal in the cylinder.
3.Factorize $(1-x^{2})(1-y^{2})+4xy$
4.Water flows through a cylindrical pipe of internal dameter 7cm at the rate of 36 km/hr. Calculate the time in minutes it would take to fill a cylindrical tank of radius 35 cm and height 1 meter.
5.Simplify:$\frac{8^{3a}* 2^{5}* 2^{2a}}{4 * 2^{11a}*2^{-2a}}$
Hopefully I won't lose too many marks...please tell me asap...
Thanks a ton,
Padmavathi

@Glenn Man-Thanks...but I made a small mistake...sorry :blush:
In the last one, it's $2^{-2a}$ in the denominator...sorry :oops:
Any progress on the others?

[hide="#2"]If $r$ is the internal radius, then the surface area will be

$14 \cdot 2 (10+r) \pi+2 \cdot (10^{2}-r^{2}) \pi=578 \pi$

So let's find the value of $r$ here:

$280+28r+200-2r^{2}=578$

$2r^{2}-28r+98=0$

$r^{2}-14r+49=0$

$(r-7)^{2}=0$

So $r=\boxed{7}$. Then the thickness will be $10-7=\boxed{3}$ (depends on what you mean by "thickness").[/hide]

[hide="#4"]Diameter of the base(circle) was multiplied by 10, but the height decreased by 1m/36km, which is 1/36000.

$1 \cdot 100 \cdot \frac{1}{36000}= \frac{1}{360}$ of an hour, which is equal to... $\frac{1}{6}$ of a minute(, which is equal to... $10$ seconds).[/hide]

[hide="#5"]$\frac{8^{3a}\cdot 2^{5}\cdot 2^{2a}}{4 \cdot 2^{11a}\cdot 2^{-2a}}$
$=2^{3 \cdot 3a+5+2a-2-11a+2a}$ $=\boxed{2^{3+2a}}$.[/hide]

Thank you all so much... :D
I'm losing....
[hide]5 marks... :mad:  :(  [/hide]
But still...thanks!!


something im confused about, i heard from someone that a function and its taylor approximation (with a finite number of terms) only intersect once...is this true?

like, if u had f(x) = sin x, and ur approximation was like the first 5 (or any finite number) terms of the taylor expansion for sin x, would the 2 graphs intersect only once??

thanks

It's not true.  Easy to come up with counterexamples.

However, they only [i]have[/i] to intersect once.  For which you could find examples, I'm sure.  Probably any odd-degree approximation for exponential functions, would be my guess.

hmmm...try e^x with an odd-degree approximation...i think it still intersects only once...

anyways, i asked my calc teacher this question today, she was unsure, but said that i was probably right (that they only intersect once)...so i dunno...

Sorry, you misunderstood.  What Ravi B said was true -- it could intersect the function in more than one point.  However, I said that it only had to intersect in one point, and thus my example.

A simple example of functions whose Taylor series intersect them in more than one point -- look at Cos(x) with a taylor approximation of degree 0 centered at any point.  That gives you infinitely many intersections.

Right.  And if for some reason you want only higher-degree Taylor approximations, you could do something like f(x) = x5 sin x.  The fifth-degree Taylor approximation of f at 0 is just the all-0 function.  But f has infinitely many zeros, so it has an infinite number of intersections with the Taylor approximation.

Edited: Replaced cos x with sin x.


integration
Support
function
complex analysis
real analysis
real analysis unsolved

$A>1$ real
$g$:$[1,A]\rightarrow C$ continous
suppose there exist $M>0$ such that for any $n\in N$, $|\int_{1}^{A}x^{n}g(x)dx|\leq M$

Prove that $g=0$

Isn't this just a restatement of the problem in [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=45127[/url]?

I think one can solve it as follows:
extend $g(x)$ by zero when $x \in \mathbb{R} , x \notin [ 1,A ]$. Then of course $g$ is in $L^{2}( \mathbb{R} )$ and has
compact support, then if we consider the Fourier transform $f(t) = \int_{ - \infty }^{ \infty } g(x) \exp( itx ) dx$, then $f(t)$ is an entire function in $t \in \mathbb{C}$, and $f|_{ \mathbb{R} } \in L^{2}( \mathbb{R} )$. Also the fact that $| \int_{1}^{A} x^{n}g(x) dx | \leq M$ means that $| f^{(n)}(0) | \leq M$, therefore for every $t \in \mathbb{C}$ we have $| f(t) | = | \sum_{n=0}^{ \infty } \dfrac{f^{(n)}(0)}{n!} t^{n} | \leqslant \sum_{n=0}^{ \infty } \dfrac{| f^{(n) }(0) |}{n!} |t|^{n} \leqslant \sum_{n=0}^{ \infty } \dfrac{M}{n!} |t|^{n} = M \exp(|t|)$. Then by Paley-Wiener theorem, the support of $g$ must be in $[ -1,1]$, so $g = 0$.

I hope it is correct.


[quote="maokid7"]i am having trouble with tig problems that are AIME leve...are there any books or other resoruces that you guys would recommend? (other than AoPS volume 2 which i have already used)[/quote]
I suggest you
[b]103 Trigonometry Problems[/b] by Titu Andreescu and Zuming Feng... but[b] 50[/b]$\$$!(one dollar for $2,06$ problems) :mad:  :mad:
 Look for "trigonometry problems" on Google !

You probably just need to start working through hard trig problems.  Go through old aime's and do the trig problems.  Then try and find some other sources for the problems.  I agree [b]103 Trigonometry Problems[/b] is a great book.


calculus
integration
inequalities
real analysis
real analysis unsolved

Let $ f: [0,1] \rightarrow \mathbb{R}$ a continuous and increasing function. Prove that $ \forall n \in \mathbb{N^*}$ we have the inequality: $ \int_{0}^{1}(t^n\plus{}t^{n\minus{}1})f(t)dt \geq \frac{2n\plus{}1}{n(n\plus{}1)}\int_{0}^{1}f(t)dt$.

It suffices to prove that

$ \int\limits_{0}^{1}t^{n}f(t)dt\geq \frac{1}{n\plus{}1}\int\limits_{0}^{1}f(t)dt\equal{}\int\limits_{0}^{1}t^{n}dt\int\limits_{0}^{1}f(t)dt$

which holds, based on Cebysev's Integral Inequality.


hmm what am i doing wrong with solution 2?

[color=blue]Problem: Find the product of the $n$th roots of $1$. [/color]

[u]Solution 1 [/u]
$x^{n}= 1 \ \Rightarrow \ x^{n}-1 = 0$
So the product of the roots is clearly $(-1)^{n}(-1)=\boxed{(-1)^{n+1}}$.

[u]Solution 2[/u]
$x = 1^{1/n}$
From this, we can calculate the product of the $n$ roots of unity by
\[\prod_{k=0}^{n-1}e^{\frac{2{\pi}ik}{n}}=e^{\frac{2{\pi}i(1+2+\cdots+(n-1))}{n}}= \boxed{e^{(n-1)\pi i}}\]

Since $e^{\pi i}=-1$, your second solution is also correct.


integration
logarithms
trigonometry
calculus
calculus computations

hello dudettes.  :D

can you show

$ \int_0^{\dfrac{\pi}{4}}\;x\;\ln\,\sin\,x\;\textbf dx\;\equal{}\;\boxed{\minus{}\dfrac{3}{16}\zeta(2)\ln\,2\minus{}\dfrac{\pi}{8}\beta(2)\plus{}\dfrac{35}{128}\zeta(3)}$

I'm sure I can't.  I would like to know how though.

See :
http://pagesperso-orange.fr/gery.huvent/articlespdf/COMPLEXE.PDF
page 3 (sorry in french)
I proved that
$ \int _{0}^{x}\!t\ln  \left( 2\,\sin \left( 1/2\,t \right)  \right) {dt}\equal{}
\minus{}{\it polylog} \left( 3,{e^{ix}} \right) \plus{}ix{\it polylog} \left( 2,{e^{ix}} \right) \plus{}\zeta  \left( 3 \right) \minus{}i\left(1/6\,{x}^{3}\minus{}1/4\,\pi \,{x}^{2} \right)$

where $ {\it polylog} \left( n,x \right) \equal{}\sum _{k\equal{}1}^{\infty }{\frac {{x}^{k}}{{k}^{n}}}$

with $ x\equal{}\pi/2$
it's easy

[quote="G.Huvent"]See :
http://pagesperso-orange.fr/gery.huvent/articlespdf/COMPLEXE.PDF
[/quote]
nice.
We have already answer for sum like this $ \sum \frac{(2n)!}{(n!)^{2}}x^{n}\equal{}A(x)$
if you just integrate it like this :
$ \int_{0}^{x}\frac{\int_{0}^{x}\frac{\left(\int_{0}^{x}\left(\frac{\left( \int_{0}^{x} \frac{A(x)}{x}dx \right)}{x} \right)\right)}{x}}{x}$
you will get something like this
$ \sum \frac{x^{n}}{n^{4}C_{2n}^{n}}$

Have a look at
http://pagesperso-orange.fr/gery.huvent/articlespdf/Autour_primitive.pdf
page 8
Another way to prove the integral
(G is Catalan constant).