Putnam
integration
function
calculus
calculus computations

Evaluate

$\displaystyle\int_0^\infty\left(x-\frac{x^3}{2}+\frac{x^5}{2\cdot 4}-\frac{x^7}{2\cdot 4\cdot 6}+\dots\right)\left(1+\frac{x^2}{2^2}+\frac{x^4}{2^2\cdot 4^2}+\frac{x^6}{2^2\cdot 4^2\cdot 6^2}\right)dx$.

[hide="so far i have"]
well i got that $\left(x-\frac{x^3}{2}+\frac{x^5}{2\cdot 4}-\frac{x^7}{2\cdot 4\cdot 6}+\dots\right)=e^{\frac{-x^2}{2}}$.
but am not able to find a "closed form" for
$\left(1+\frac{x^2}{2^2}+\frac{x^4}{2^2\cdot 4^2}+\frac{x^6}{2^2\cdot 4^2\cdot 6^2}\right)$.
[/hide]
could you please help me (hints only please)

(Of course it's been posted before.)

Note: that's $xe^{-\frac{x^2}2}.$

[hide="The other hint."]You do not need a closed form for the other function.[/hide]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=91286]One very recent topic on this problem[/url]

yes i meant $xe^{\frac{-x^2}{2}}$.
but how do u evaluate it w/o knowing what the funtion is?  :(  :?

This is the easiest Putnam problem I've ever tried. I have one question about the official solution:
[hide]
When changing order of integration/summation at an exam I would have to justify it either by showing uniform convergence of the series or using that the partial sums of the series form a growing function series $f_s$ and use the monotone convergence theorem.

No mention of either is made in the official solution. Is this because it is very obvious at the level Putnam is at or because there are stronger theorems that I haven't yet learned?
[/hide]

[quote="jmerry"]You have the integral of a sum. Switching the order is always worth a try.[/quote]
exactly what do u mean?

He means that if you have $\int\sum$ it's worth trying to make that $\sum\int.$ Or if you have $\sum\int$ it's worth trying to make it $\int\sum.$ And while we're at it, if you have a double sum or a double integral, it's worth trying to switch the order of summation or the order of integration.

Kalle wants a justification. Uniform convergence won't do, becuase we're integrating over an infinite domain. In this particular case, we are interchanging an integral with a sum of nonnegative terms. That's always justified by the Monotone Convergence Theorem. On the other hand a Putnam problem's justification shouldn't rest on material usually taught in a graduate course. It should be possible  to justify it specifically given the material in this problem - you do have the advantages of rapidly converging series for which crude but useful estimates can be devised.

I have never seen that.
could you please show me how it would work with regard to this problem.  :)  :blush:

Let's start with one step: can you calculate

$\int_0^{\infty}x^{2k}\cdot xe^{-x^2/2}\,dx$ ?

no.

i did by-parts and got

$\displaystyle\frac{1}{2}\left(-2u^ke^{\frac{-u}{2}}-\int_0^\infty -2ku^{k-1}e^{\frac{-u}{2}}du\right)$

where $u=x^2$

but i dont think this is what you wanted.  :oops:

Don't forget to evaluate the integrated terms of the integration by parts at zero and infinity - this is a definite integral, and we're going to get a number.

In fact, the integrated terms will go away, since they go to zero at both ends. Now look at what you have: $k$ on one side, $k-1$ on the other. You've got a live, useful recursion. Run with it.

(You also have the Gamma function - we're not exactly discovering anything unknown at this point.)

[quote="Kent Merryfield"]
In fact, the integrated terms will go away, since they go to zero at both ends. Now look at what you have: $k$ on one side, $k-1$ on the other. You've got a live, useful recursion. Run with it.

(You also have the Gamma function - we're not exactly discovering anything unknown at this point.)[/quote]

sorry but I dont quite follow what you are saying at this point.  :(

Let $I_k=\int_0^{\infty}x^{2k}\cdot xe^{-x^2/2}\,dx.$

If you clean up and finish that integration by parts, you should have a formula which expresses $I_k$ in terms of $I_{k-1}.$ That's what I mean by a recursion. Since you should also be able to compute $I_0$ directly, that should give you a way of figuring out $I_k$ in general.

so I got

$I_k=I_0+kI_{k-1}$ where $I_0=\frac{1}{\sqrt{e}}$
is that right?

However, I am not able to find a formula for $I_k$ interms of $k$
I have tried generating functions but that did not help.

No, that's not right. What happens when you evaluate the boundary terms (assuming that $k\ge1$)?

Also, $I_0$ is not $\frac1{\sqrt{e}}$.

I have checked my work and am unable to find where i made a mistake and even used a calculator and got that $I_0=\frac{1}{\sqrt{e}}$

i assume that $I_0=\int_1^\infty xe^{\frac{-x^2}{2}}dx$.

Make that $\int_0^\infty$.

The lower limit should be zero for all of these integrals- that makes everything much easier.

oops i thought that the lower bound was 1 instead of 0. ill redo my work.   :lol:
thanks

maokid7, are you still there?

[hide="Formula for I sub k"]$I_k=2^kk!$[/hide]
Are we ready to move back to the original problem?

yeha thats what i got after going back and fixing my limits.  :lol:

and yes I am ready to see how this relates to the original problem.


Okay an old classic  problem
Just using number  $1,2,3,4$ what is the largest number you can make.
(No use of facny "$!$"  or other functions - You can use  $+ - \times  / ( ) . $    powers, decimal points ,  etc)
 :[img]http://www.metacard.com/calc.gif[/img]
IOW    $Power$ in addition to above buttons .. but    digits $1,2,3,4$   each only once... :)

(I know the above calculator would most likely overflow etc... it is just to give an idea..)

Actaully  one can make number much larger than suggested above .. :)

[quote="ComplexZeta"][hide]$2^{3^{41}}$?[/hide][/quote]
I think that's it.

[hide]Very good but you  one can get much *much*  larger numbers ....[/hide]

I don't think you can use negatives or decimal points.

[quote="Mildorf"]How about [hide]$(.3)^{-\left((.2)^{-\left((.1)^{(-4)}\right)}\right)}$[/hide][/quote]

[b]Nice  !!! [/b]! It is interesting to see how some of these problems remain in circulation.. I remember the problem being asked by my father many decades ago.. and about 10 years ago I asked it to my kids

The solutions were simillar/same  in each case ( some had bottom digit as  3 - in  that case the result would be a natural number, although  "somewhat"  smaller than what you posted)

[quote]I don't think you can use negatives or decimal points[/quote]
Well , The problem above actually explicily mentions that you can use  decimal points and "-" .. (Not to mention that traditionally they are allowed in such problems -  See the button below 2  and between "0" and "=" sign  in the
.[url=http://www.metacard.com/calc.gif]Image posted above.[/url]  :) ...

how about the smallest number using 1,2,3, and 4? eh? :D

haha yep!  :D what about the number closest to 0?

[quote="joml88"]1+4-2-3[/quote]

Hehe, for sure you cannot get any closer  :D  :D  :D  :D .


You and your friend are playing a game trying to see who can guess the closest to a whole number from 1 to 1000 that a computer is randomly generating. Your friend guesses 3 first. What whole number should you guess to maximize your chance of winning?

...4? You beat his 3  99.7% of the time unless I am misunderstanding something.


How many odd perfect squares are between 5 and 211?

[hide = Sol]
Just Make A List:
$$3^2, 5^2, 7^2, 9^2, 11^2, 13^2$$
As you can see, there are $\boxed{6}$ numbers.

[quote=ME1231917]But didn't they use 3 in their list? I got 5 because 3<5, so then you have 5, 7, 9, 11, and 13 squared, which is 5 squares.[/quote]

You forgot about the $3$.


Hi I was solving this problem that says:

For some n >= 1 prove that there's exist n consecutive composite numbers.

I was reading a solution for such a problem. But I barely understand the point of it.

see the solution says....

Be k a number such 2<= k <=n+1.   Then k is a proper factor of (n+1)! + k  as k appears at least as one factor of (n+1)! = (n+1)n(n-1)(n-2)....1. Hence each of these numbers (n+1)! + k is composite, and we've shown n consecutive numbers. It's proved.

I don't see the point that these numbers are consecutive.... :?: [/b]


[quote="pianoforte"]A 1 by 7 rectangle ABCD is marked along BD. The perpendicular from A to BD intersects CD at E and the perpendicular from C to BD intersects AB at F. Find EF.[/quote]

Can you clarify your wording, it's kinda confusing, especially the first line.

Ah, my bad.
ABCD is a rectangle such that AD=BC=1 unit, AB=DC=7 units. We draw the line BD. Then we draw the altitude from A to BD and extend it to CD at E, and draw the altitude from C to BD and extend it to AB at F. Find the length of EF.

[hide]I think the answer is somewhere approximately $6.79$. However, I have two solutions and both of them give me an answer around 6.79 but the answers are not exactly the same . [/hide]


calculus
integration
logarithms
calculus computations

Evaluate
$\int_{0}^{1}\frac{\ln\left(\sqrt{1+x^{2}}\right)}{1+x^{2}}dx$

[color=green][Edited to match maokid7's comment below - mod.][/color]

That's unlikely. I would point out that $\frac{\pi}2$ is a very strange upper limit for this integral - in fact, I'll ask maokid if that's what he even intended. Perhaps it was supposed to be the integral as $x$ goes from $0$ to $\infty?$

yes it was not supposed to be $0$ to $\frac{\pi}{2}$.
what i meant was for $0$ to $1$.
sorry.  :oops:  :blush:


Construct a family $ F$ of subsets of some set $ X$ such that for every integer $ n>0$, one can find $ n$ pairwise disjoint subsets in $ F$ but $ F$ contains no infinite subfamily of pairwise disjoint subsets.

Let $ X$ be the set of integers.  For every prime $ p$ and every residue $ r$ (where $ 0 \le r < p$), let $ M_{r, p}$ be the set of integers that are $ r \mod p$.  Let $ F$ be the set of $ M_{r, p}$.  The residue classes modulo a fixed prime are disjoint, but the residue classes mod different primes intersect by the Chinese Remainder Theorem.

Another example: Let $ X$ be the set of sequences whose $ i$th term is an integer between $ 1$ and $ i$, for every $ i$.  Let $ S_{i, j}$ be the set of sequences whose $ i$th term is $ j$.  Let $ F$ be the set of $ S_{i, j}$.

Great. The example I found was more complicated. Now the next question: is it possible to construct such an example using intervals on the real line?
Namely, can we have a family of open intervals out of which we can select an arbitrarily large but not infinite pairwise disjoint subfamily?


Which digit is in $63^{rd}$ place to the right of the decimal point for \[\frac{3}{7}\]  ?????

i have a tourny tomorrow i neeed help!!!!!!!!!!! :(  :(

well $\frac{3}{7}$

has 42871 repeated every 6 times..

$\frac{63}{6}$ has 10 repeats and half of 42871 repeated
so thats 8?

not sure

lets see.. if u divide 3 by 7, u get a quotient of 0.428571 repeating. this is the same 6 digits over and over again. if ure looking for the 63d place, divide 6 into 63. u get 10 remainder 3. so u count the third digit, which is 8. so this is the answer... i think :D

where did you get the 63... and \[\frac{3}{7}\] is

$.428571$ repeated not $.42871$

ahhhhhhhhhhhhhh

by the way, where r u from reaper? u sound like an alabamian. if this is true, im guessing i will see u at pizitz saturday.  :?:

dude how did you know???? i am from alabama

and you will see me tomorrow....

what school are you from

lol im smarter than i seem hard to believe. i belong to randolph, a national school of excellence

[quote="snipefreak"][quote="myspacesucks"]The decimal repeats 428571, so therefore 1 will be your 60th digit.
Now take the 3rd number afterward and you will get 8, and thats your answer.  Hope this helps.[/quote]
Man...I never knew the fraction repeated 428571. *Commits to memory*.[/quote]
The sevenths are basically a pattern.
$\frac{1}{7}=0.\overline{142857}$
$\frac{2}{7}=0.\overline{285714}$
$\frac{3}{7}=0.\overline{428571}$
$\frac{4}{7}=0.\overline{571428}$
$\frac{5}{7}=0.\overline{714285}$
$\frac{6}{7}=0.\overline{857142}$
See the pattern? Now memorize it. :P


On blog options, when I try to remove/add contributors, and I click submit, it gives me the following:

SQL requests not achieved

DEBUG MODE

SQL Error: 1064 You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ')' at line 3

SQL Request: DELETE FROM phpbb_weblog_contributors WHERE weblog_id = 470 AND user_id IN ( )

Line : 261
File : class_blog_options.php

I'm using a Windows XP, and IE.

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=112086

I think the admins are working on it...


Does anyone know of some books or online articles that discuss good methods for introducing gifted young students to math/problem solving?  When I say young I mean about 2nd or 3rd grade and by gifted I mean doing math 3 or 4 grade levels above their own level.

I ask because I help such a student for about 30 minutes to an hour each week.  I would like to start introducing him to more problem solving type stuff but I'm not sure how to do it.  I want to keep him interested in math.

I would suggest using bits and pieces of AOPS Volume 1.  I would also reccomend getting them into geometry at a young age.  That's one of the big flaws of American math education, and could really help him/her in the long run.

For a problem solving textbook, you can try [i]Creative Problem Solving in School Mathematics[/i] by George Lenchner, which can be ordered on the [url=http://www.moems.org]MOEMS[/url] web site.  By level, it is like AoPS volume 0, if there were such a thing.

For a problem book, MOEMS also offers a collection of its old contests for 5th and 6th graders.  The book has 80 contests with 5 problems each.

The [url=http://www.continentalmathleague.hostrack.com]Continental Math League[/url] also offers collections of its old contests.  Some of its books are for grades 2-3, and others are for grades 4-6.

Finally, your student might enjoy the novel [url=http://www.amazon.com/exec/obidos/tg/detail/-/0805062998/qid=1112701332/sr=8-1/ref=pd_csp_1/104-2151585-9405540?v=glance&s=books&n=507846]The Number Devil[/url] by Hans Magnus Enzensberger.  It explains interesting math in the guise of fiction.  You can use each chapter as a take-off point to discuss the math further.  (There is also companion software with the same name as the book.  I haven't tried it.)

I used these resources when my daughter was around that age, and she enjoyed them.  Good luck.

If they are very young (I hope, that means around 10 or so) one good book with different levels of difficulty could be Mathematical Circles: Russian Experience, of  Dmitri Fomin. You can get it on Amazon by around 30 bucks.


algebra
polynomial
number theory
relatively prime
algebra proposed

Let $f,g \in \mathbb{C}[x]$ be fixed polynomials with complex coefficients. Show that there exists an (integer) constant $M \in \mathbb{N}$, depending only on $f$ and $g$, such that for any constants $a,b \C$, and any $m,n \in \mathbb{N}_{0}$, the polynomial $af(x)^{m}+bg(x)^{n}$ is either identically $0$, or NOT divisible by $x^{M}$. Can you describe the best such constant $M$ (in terms of $f$ and $g$)?

Did you forget to add something like the polynomials to be relatively prime? I mean, take $f=g=X$, there is no uniform bound like you say.

Of course... we must have $f(0)g(0) \neq 0$; sorry about that...

This is very easy. However, I have the feeling that you're going to kill me again by posting the variant for an arbitrary number of polynomials. Suppose that $X^{M}$ divides $af^{m}+bg^{n}$. Then $X^{M-1}$ divides $amf^{m-1}f'+bng^{n-1}g'$, thus also $amf'f^{m}+bng'g^{n-1}f$, thus also $bng'g^{n-1}f-mf'bg^{n}=bg^{n-1}(ng'f-mgf')$ and now it's done, because $X^{M-1}$ must divide $ng'f-mgf'$ which has degree bounded by $\deg(f)+deg(g)-1$ and the only possibility is when $ng'f=mgf'$, which happens when $f^{m}$ and $g^{n}$ are linearly dependent and in this case the fact is obvious. I think using this approach and properties of the Wronskien, we can even generalize it as you have already probably done it.


number theory
greatest common divisor
relatively prime

Prove that if $ \gcd(a,b)\equal{}1$ then $ \gcd(ab,a\plus{}b)\equal{}1$.

Is this right?
a and b are relatively prime
the factors of ab are a and b
but a+b not $ \equiv$  0 $ \bmod a$ and a+b not $ \equiv$  0 $ \bmod b$
thus ab and a+b dont share any common factors
$ \implies$ gcd(ab,a+b) = 1  :)

Let $ g \equal{} (ab,a \plus{} b)$. Then $ g\mid a(a \plus{} b) \minus{} ab \equal{} a^2$ and $ g\mid b(a \plus{} b) \minus{} ab \equal{} b^2$. Thus $ g\mid (a^2,b^2) \equal{} 1\implies g \equal{} 1$.

[quote="madness"]Is this right?
a and b are relatively prime
the factors of ab are a and b[/quote]

This is wrong. For example $ 8$ and $ 9$ are relatively prime, but that does not mean that $ 72\equal{}8\times 9$ has the only divisors $ 8$ and $ 9$.

[b]Follow-up:[/b]  Prove or disprove:  $ \gcd(a, b, c) \equal{} 1 \Leftrightarrow \gcd(a \plus{} b \plus{} c, ab \plus{} bc \plus{} ca, abc) \equal{} 1$.  :)

The $ (ab\plus{}bc\plus{}ca,abc)\equal{}1$. Suppose for the sake of contradiction that $ >1$. Then there is a prime in the $ GCD$. $ p|abc$, so WLOG $ p|a$. But then we need that $ p|ab\plus{}bc\plus{}ca$...but that means that $ p|bc$, which is a contradiction since $ (a,b,c)\equal{}1$.

I don't know about the other direction.


If $ a$, $ b$, $ c$, $ d$, $ e$, $ f$ are nonnegative real numbers satisfying $ a \plus{} b \plus{} c \plus{} d \plus{} e \plus{} f \equal{} 1$ and $ ace \plus{} bdf \geq \frac {1}{108}$, then prove that
\[ abc \plus{} bcd \plus{} cde \plus{} de f \plus{} efa \plus{} fab \leq \frac {1}{36}
\]

Put A=ace+bdf  and B= abc+bcd+cde+def+efa+fab.

By AM-GM inequality

A+B=(a+d)(b+e)(c+f) \leq (((a+d)+(b+e)+(c+f))/3)^3=1/27

so B \leq 1/27 -A  \leq 1/27 -1/108=1/36

S=abc+bcd+cde+def+efa<=1\36
but (a+d)(b+e)(c+f)=S+ace+bdf
thats why by am-gm because ace+bdf>=1\108 and a+b+c+d+e+f=1


The numbers from $ 1$ to $ n^2$ are randomly arranged in the
cells of a $ n\times n$ square $ (n \ge 2)$. For any line or column
the ration of the greatest number to the smallest number is
calculated. Let us call the \textbf{characteristic} of the
arrangement the smallest of these $ 2n$ fractions. What is the
highest possible value of the characteristic ?

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=131848#131848

The problem is essentially this, and putting IMO Shortlist problems in Classroom Math forum might not be a good idea :wink:


induction
algebra
functional equation
algebra solved

try this functional equation ,it's really coool (like most other functional equations
f:R---->R
f(x 2 -y 2 ) = (x-y)[f(x)+f(y)]
best regards

Here's my soln:
f(x^2-y^2)=(x-y)(f(x)+f(y)), f(y^2-x^2)=(y-x)(f(x)+f(y))=-f(x^2-y^2) and since x and y are reals we get that for every x f(-x)=-f(x). In f(x^2-y^2)=(x-y)(f(x)+f(y)) we replace y by -y so f(x^2-y^2)=(x+y)(f(x)-f(y))=(x-y)(f(x)+f(y)). Let f(1)=a. and let y=1 in our last equality. So (x+1)(f(x)-a) = (x-1)(f(x)+a), so f(x)-ax=ax-f(x) for any x, so f(x)=ax. I don't think we can do what bugzpodder did, since we don't even know if f is continous. For examplke, bugzpodder assumed that the sequence f(x^(1/(2^n))) tends to 1 as n-> infinity and that's why he got f(x)=x. We don't even know if this sequence has a limit.

nice job,grobber.
my solution is based on proving f(nt)=nf(t) by induction for all n,integers and then proving it for the rationals.that finisshes the job.but it's a bit long

Where is this problem from? It looks familiar to me. Maybe from Vietnam? I'm not sure.


If it is the truth, then you must be lying, but if you're lying, then you're telling the truth but if that's the truth then you must be lying, but if you're lying, then you're telling the truth but if that's the truth then you must be lying, but if you're lying, then you're telling the truth but if that's the truth then you must be lying, but if you're lying, then you're telling the truth but if that's the truth then you must be lying, but if you're lying, then you're telling the truth but if that's the truth then you must be lying, .....

If it is a lie, then you don't lie that much. :D

That sentence is false. You don't always lie, just most of the time :P

[quote="i_like_pie"][quote="alex__ib"]IS this sentence a lie or not?  :D[/quote]
If you always lie, then it's the truth (but that can't be).
If you always tell the truth, it's a lie (but that can't be).

Both cases are contradictions, so it's undefined. :rotfl:[/quote]
Actually, i_like_pie, you're wrong.  You forgot a third case: you SOMETIMES lie.

But the answer is FALSE.

It's a lie
coz if it's true then u DON'T always lie
and if it's a lie then u HAVE to tell the truth...
In all cases...this IS a lie.

but if u always lie the title would be a llie so u would be telling the truth

In logic, there are three kind of statement: true, false, and meaningless.

Obviously, "I always lie!" is meaningless statement.

a lie lol, wow that made me think... lolz :rotfl:

[hide] This sentence is neither a lie nor a truth. It is a question.[/hide]

It is the [url=http://en.wikipedia.org/wiki/Epimenides_paradox]Epimenides paradox[/url]
In Philosophy, it's a paradox
In Mathematics, it's just a [u]false[/u] statement .Because we can prove it  :wink:

[hide="One way is:"]
Just take all the cases for the person who makes the sentence:
Is he a liar?
Is he a truth-teller?
Does he sometimes say lies and sometimes say the truth?

[b][color=#006699]i_like_pie[/color][/b] noticed that, at the first and second case we have a contradiction.

But at the third case we have not a contradiction.[/hide]
[hide="Another way is:"]
Think about the sentence
Is it true?
Is it false?

If it is true then it is false (contradiction)

If it is false:

[quote="randomdragoon"]"I do NOT always lie" (the opposite of the statement) does not imply "I always tell the truth". It implies "I do not always lie" (could be sometimes or never) The statement is false, case closed.[/quote]

NOT(I always lie) [size=142]≠[/size] I always tell the truth
So there is no contradiction!

If we had a contradiction at this case too (assumption that the sentence is false), then we would say that the sentence is meaningless
[/hide]
If the sentence was "[i]I lie right now[/i]", then it would be a meaningless sentence.

[color=brown]P.S. To my understanding, "meaningless sentence" (said by [b]lightrhee[/b]) means that we have a paradox [/color]

Wait, he's asing is "IS this sentence a lie or not?"

So in other words, is this sentence a lie, meaningless, or the truth? Every sentence has to be one of them, so yes, the sentence is either a lie or not a lie.

[u][b]Sentence:[/b][/u] I always lie
[u][b]Question:[/b][/u] Is [color=red]this[/color] sentence a lie or not?

The question reffers to the sentence above, not to itself!

Yes, but the sentance is still either a lie or not a lie...


ABCD -- convex quadrilateral. AC \cap BD =M. Let P and Q --
circumcenters of ABM and CDM respectively. Prove that
AB+CD\leq 4PQ.

The solution  is  at  page  172  at the book  LMO  1987 - 1991  by  Dmitry  Fomin  and  Alexey  kirichenko , mathpro  press 1994.
In  case  you  dont  have the  book  let me  know  so  i  will  tell  you  the  solution  ok?

Draw  srtaight  line  L , which  is  the  bisector  of angles  AMB , CMD. Denote  the  other  intersection  points of  L  with  the  given  circles  by  K  and  E , and drop  perpendiculars  PR  and  QN  to  line  L .
We  have  anlgeKBM + angle  KAM = 180  because  AKBM  is cyclic , and  therefore  one  of these  angles , say  KBM , is  not  acute . It  follows  that  KM>KB . Line  MK  is the  bisector  of  angle  AMB , therefore AK = BK , and we  obtain  that  2KM>2KB = KB + AK>AB . Similar reasons  show  that  2EM>CD . All  that  remains  is  to  notice  that  PQ>=RN . Hence , 4PQ >= 4RN = 2KE = 2KM + 2 EM >AB + CD.


inequalities
function
algebra proposed
algebra

Let $ n$ be a positive integer greater than $ 1$. Find all functions $ f: \mathbb{R}\to\mathbb{R}$ such that for all real numbers $ x_1,x_2,\dots,x_n$,
$ x_1f(x_1)\plus{}x_2f(x_2)\plus{}\dots\plus{}x_n f(x_n)\geq\left(\frac{x_1\plus{}x_2\plus{}\dots\plus{}x_n}{n}\right)\left(\frac{f(x_1)\plus{}f(x_2)\plus{}\dots\plus{}f(x_n)}{n}\right)\geq\frac{x_1f(x_1)\plus{}x_2f(x_2)\plus{}\dots\plus{}x_n f(x_n)}{n}$.

Apply the ineq to $ (x, \minus{} x,0,...,0)$ and we get
\[ x(f(x) \minus{} f( \minus{} x))\ge0\ge\frac1nx(f(x) \minus{} f( \minus{} x)),\]
so $ f(x) \equal{} f( \minus{} x)$.

Apply the ineq to $ (x,0,...,0)$ and we get $ xf(x)\ge\frac 1nxf(x)$ and $ \minus{} xf(x)\ge \minus{} \frac 1nxf(x)$ :D as $ f$ is even.
Now $ n\ge 2$, thus $ xf(x) \equal{} 0$ for all $ x\in\mathbb R$.

Knowing that, apply the ineq again to $ (x,0,...,0)$ and we get $ 0\ge\frac xn\frac {(n \minus{} 1)f(0)}n\ge 0$. So $ f(0) \equal{} 0$, too.

So $ f\equiv 0$ is the only solution. :(


$ r \minus{} s \minus{} (s\minus{}r) \implies r \minus{} s \minus{} s \plus{} r \implies 2r \minus{} 2s \implies \boxed{\text{C}}$

[quote="sri340"]By how much does  r - s  exceed  s - r ?

a (r-s)/(s-r)     b 2(r-s)       c -2        d 2(s-r)      e -1[/quote]

Simply subtract $ s \minus{} r$ from $ r \minus{} s$ to get $ 2r \minus{} 2s \equal{} 2(r \minus{} s)$.

EDIT: dang. it.


This is not a big help, but there could be a chance of solve your problem if $ y' \plus{} y^2 \plus{} x^2 \equal{} 0$ could be solved.

[quote="Mr.Doe"]$ I''(\alpha) \plus{} \alpha^2 I(\alpha) \equal{} 0,~I(0) \equal{} I(\pi) \equal{} 0$.[/quote]

The equation have solution in terms Bessel's functions.

[quote="Mr.Doe"]$ I''(\alpha) \plus{} \alpha^2 I(\alpha) \equal{} 0,~I(0) \equal{} I(\pi) \equal{} 0$.[/quote]

It seems that $ 0 \leq \alpha \leq \pi$. Using finite element method we can solve the problem numerically.

hello, Maple gives
$ y(\alpha)\equal{}C_1\sqrt{\alpha}BesselJ\left(\frac{1}{4},\frac{1}{2}\alpha^2\right)\plus{}C_2\sqrt{\alpha}BesselY\left(\frac{1}{4},\frac{1}{2}\alpha^2\right)$
as a solution.
Sonnhard.


Poate ca nu e locul potrivit :P  dar vroiam sa le urez tuturor de pe forumul roamnesc sarbatori fericite si un an nou cu cat mai multe gold medals :D

Sarbatori fericite!
Ne mai vedem pe la concursuri :D

hai ca va bag si eu o urare : merry x-mas , folks !  :oops:

[size=100][color=darkblue][b]Va doresc tuturor rezultate cat mai frumoase in concursurile viitorului an.[/b][/color][/size]

[quote][size=134][color=darkred][b]Sarbatori fericite. La multi ani, 2007 ![/b][/color][/size][/quote]


Harvard college geometry Duke Gauss MIT Princeton

I've noticed that the math section of the June 7, 2003 SAT was particularly harder than previous ones....Anyways, I took just for practice, but I think I did pretty poorly on it overall.

Math is not my primary concern, but verbal is.  I always seem to lurk around 550-600.  Building a large vocabulary is essential to the verbal section, but I'm not sure if keeping a vocab. journal is the best thing to do.  It takes much longer if I have to write the word and its definition than just memorizing (temporarily, at least) it.  However, perhaps the word will stick in my head if I do write it out.  It's kind of a gray area.  How do you memorize vocabulary words for the SATs?

-interesting_move

ya know the most people answering that poll dont know much about the schools. like they may put harvard cause its supposedly the best school. but it may not suit some peoples taste. in some areas harvard is good and all. but in others you could do better. also the poll should be called what school would you go to if you had an infinite amount of money. those schools dont come cheap, another reason why some may not want to go to them. some people may not want a $300,000 loan (just an estimate) on their backs after undergrad. and if they go to med or law school be ready to dish out some more money. anyways. whats the question. to just post your thoughts about the verbal and math parts of the sat. no offense or anything. im just severely bored. and realizing that summa break is shorter than i thought. and that im wasting it. anywho. nuff said

For the SATs I didn't really study much for the verbal.  I basically looked through a vocab book that my mom had gotten me to see if I already knew the words.  I guess the best preparation that I did was to read a lot.  That way, I didn't actually have to study the words; I knew them already.
I don't think that just memorizing the word it too effective because there tend to be a lot of words to remember.  However, I've always been too lazy to write them down or anything like that.

On the sentence completion and analogy section, i think the best way
to do well is to know all of the words shown in the choices (Duh). If you know them all it isn't that hard to pick the right one. To know them all, i
suggest memorizing all of the words on the back of a
sat book, which is about 3000. It'll get really boring after 1000 but once you
know them all, the verbal section will be easy as spinning a book.
Some suggest reading a lot, but that never worked for me. I hate reading
and I think encountering sat words in books are less probable and more
time consuming than just memorizing it from the back. I just memorized
and memorized and memorized and got a 760. The trick about memorizing all those words is to make a menumonic device for eveysingle one of them except the obvious ones like cat. It will help you cling on to the words even after quite of time has passed. After I went through all of the words the first time I went back and realized I didn't know 90percent of them. So I went through it all again making some stories or whatever for each of the words. ex) renitent- i thought of rain and tent and how the tent endures the rain. yes i am crazy.
You just look at the word that you don't know and the first thing that comes to your mind when reading the word, link it with the definition. Maybe this is all nonsense and doesn't help you in the long run..but i guess it worked for me.
Enough said. If you read this far I thank you for your time

I memorized only 150 of those and got a 670. The roots of the words are what I really looked at.

I don't know about it being unlikely to encounter SAT words in books.  I guess it depends on what you're reading.  I guess an obvious way to remember words are to use them.  When I was trying to remember a word I annoyed my family by making up sentences with the words in them and telling them the sentences.  Hearing myself say the words helped me remember them.  Also, for the SAT you don't really have to know the words completely.  You just need to kind of know the gist of the word to figure out the correct one.  (At least that's what I did on the SATs.  When taking the test I didn't know all the words, but I just picked the one that sounded the best.  I think it worked for me because I scored pretty well on the verbal section.)

Reading extensively is much better than cramming if you want to do well for the SAT verbal - at least it works better for me. I took the SAT in 8th grade and got 760 on the verbal...besides, when you read, you might come across a similar word and therefore be able to connect it with whatever you're staring down at. And elimination always helps.

They're taking out analogies next year...*sobs* I love analogies...

i always found sentence completions easiest. and reading comp just a tidbit ahead of analogies. i guess thats good cause with the elimination of analogies i wont miss it much. theyre making the test like the psat for the uninitiated. so the scoring will be 2400. so in a few years if a person (still in high school) tells you that they got a 1600. thats not that good

What do you guys think of the change to add a writing section to the SATs?

I think it's terrible. If writing is added, then the English section will be worth twice as much as the math section. Is English worth twice as much as math? I certainly don't think so.

i think its fine how it is, but since so many people are now getting 1600's they had to change it somehow. why they had to add more english puzzles me.  arent they also including algebra II to the test? if thats true then like the duke tip thingies one would take in the 7th grade like i took would be hard for most. it wouldnt matter that much for the 8th graders though. cause the ones who get around an 800 usually already have done or are doing algebra II if not higher

Yeah - SATs will now test Algebra 2 as well...and I heard that the writing section is graded separately from the rest of the verbal, so there's no need to worry, Simon...Basically, the SATs will turn into mini-GCEs...

They should scrap the math section of the SAT and use AMC scores instead. That way the 800 group would be broken down into about 80-150...

You guys are forgetting who the SATs are really designed for. They are not designed so much for the high end students, but more as a basic skills test. So using AMC type questions for the math section would defeat the purpose.

I think the SAT would be better if it included MORE subject categories though like the ACT does. There are more basic skill sets than just verbal and math. Writing is an important component of all subjects, math included. But it should be graded on a totally different scale than the rest of the exam, like how the MCAT is done, with some letter grade system instead of a quantitative number.

I don't think the SAT is meant to tell much about the really high end students though. It's not really designed for them. Anything over a 1400 or so and the test doesn't tell much. There are other tests/ activities/ achievements for that.

Heh. Examination war. Every so often in S'pore some 6th grader throws him/herself off a building b/c of PSLE (Primary School Leaving Examination) stress...In S'pore, what secondary school you go to depends [i]entirely[/i] on what score you got on the PSLE. It's terrible.

[quote="ComplexZeta"]That is horrible -- a society in which eleven-year-olds commit suicide because of stress? Yikes.[/quote]

It's important to note that the youth suicide rate in Singapore, according to World Health Organization statistics that are gathered by a standard methodology all over the world, are actually quite low by international standards. The youth suicide rate in the United States is much higher, and has more than tripled in the decades following the second world war. The way I put it is that the social atmosphere of United States schools (people worrying about who has a girlfriend, or who is on the football team) is more lethal than the academic atmosphere in east Asian schools. Youth suicides are newsworthy in east Asia precisely because they are comparatively rare. Even one young person killing himself is too many, but youth suicides rarely make the news in the United States, unless they occur in "clusters" in one school district, because they are regrettably commonplace, one of the highest causes of death in that low-mortality age group.

References to professional literature can be found in a Web page, several years old, on the subject of school atmosphere and suicide

http://learninfreedom.org/suicide.html

Years ago in online discussion on the Usenet misc.education newsgroup, a professor of computer science from Canada who sometimes teaches at MathCamp explained to me the surprising fact that the youth suicide rate in Canada is slightly higher than in the United States. (I would have guessed before LOOKING IT UP that the United States rate would be higher, guessing that firearms are more readily available to youth in the United States than in Canada.) He said that any country with a more polar geographical location on the earth is likely to have a higher suicide rate than a country near the equator--and that's when I looked up the WHO figures for Singapore. The possible explanation for this is that variation in daylight influences mood disorders, and mood disorders are the number-one risk factor for suicide. I am told, by people who have been there, that the region of Europe that includes Hungary (the country in the world with the highest suicide rate of all, for many decades) is usually cloudy and sometimes depressing even for people of normal mood stability. I haven't checked cloudiness figures for different parts of the world, however.

Bottom line: seek out a friend if you feel depressed, and don't give up hope.

[quote="ComplexZeta"]That is horrible -- a society in which eleven-year-olds commit suicide because of stress? Yikes.[/quote]
You should see Japan:
Student A: "Hey, did you know so and so suicided?"
Student B: "No, why?"
Student C: "Obviously he was being threatened for money, he thinks he's oh so special..."
Student A: "Yeah, happens to everyone. Hey, wanna get some ice cream?"
Student C: "Sure!"

...OK, maybe not [b]that[/b] often, but it still is one of the major reasons for adolescent death. In fact it overwhelms the number of people killed by guns each year. The most horrible thing is that most people don't care about depressed people anymore since it's so common, so people just ignore them. Sad, truly sad.

...shouldn't have mentioned the examination war...

Wow, from SAT scores------------------->>suicide.
How did it happen?

-interesting_move

Again, shouldn't have mentioned the examination war...

is this "examination war" for getting into high school or college (or both?) I know in Romania that you have to get a very good score on a "capacity exam" at the end of 8th grade. there is a final average made - exam is 25% and all other grades from 5th to 8th grades are 75%. this determines whether you go to the best high school or a terrible one. also, after 12th, students take the bacaulareate as well as college entrance exams (each college generally has their own exam for students to take, but this is ok since there arent so many colleges.) is this system similar to the one in japan?

Yup, in Japan the bacaulareate is just called "Center (or Central) Examination" and each colledge also has a test of their own. Usually the examination war was only for college, but a decade ago or so people started the examination war for high school too, and now it's just ridiculous (it starts as early as [b]preschool[/b], can you imagine that!? :) ) How do you figure out if the child is smart or not if they can't even do 1+1? It's just this ridiculous jealousy/vanity/[b]elitism[/b] thing going on at Japan, not to mention a few other countries as well. [b]This[/b], my u171m473 1337 m45734 friend, is the downside of elitism.

We start the elitism in 3rd grade...preschool?!? *blink*

Yup, we study in preschool to get into a good elementary/middle/high school (kindargarten is included in preschool; usually the number 1 reson we study for elementary school is to get into an elementary/middle/high combined school, and college/university too if possible)

Umm well im in 8th grade so i havent taken SAT for real yet (i took it for practice once)
but my sister improved her verbal score by in a couple weeks, she did it by listening to the SAT Hit Parade (has a bunch of vocab words on it) and she also had a bunch of flashcards that came with it i think that had a bunch of vocab on it

SAT hit parade?...is that a book or what?

is there a website we can go to?

Did anyone here take the SAT IIs? I took them in October; did rather well.

i think 'hit parade' is princeton review's name for their vocabulary list.  I dont think there's a website; pick up their book from the library or something and im sure it'll be in there.

kay...thanks...i knew the name was familiar...


What is the probability that flipping a fair coin three times produces fewer than two heads? Express your answer as a common fraction.

If the result is fewer than two heads, then there must be either 1 head or no heads at all.  The possibilities are:  HTT, THT, TTH, and TTT.  This is $ 4$ combinations.  Since there are $ 2^3 \equal{} 8$ combinations total, the probability is $ \frac{4}{8} \equal{} \boxed{\frac{1}{2}}$,

it id both equally likely to land on head or tails so the answer is 1/2.

$[hide="i love money"]THE ANSWER IS 1/2[/hide][/quote]$

[size=200][color=#0080BF][b][i][u]PLEASE RATE MY POST THANK YOU SO MUCH IT IS GREATLY APPRECIATED !![/u][/i][/b][/color][/size]

[size=50][color=#FFFFFF]THIS IS IN WHITE CAN YOU SEE THIS IF SO PLEASE RATE MY POST I HAVE 99.1516651% THAT IM RIGHT THANK YOU SO MUCH!!!!!![/color][/size]

 :stretcher:  :stretcher:  :stretcher:  :stretcher:


geometry
3D geometry
inequalities
number theory unsolved
number theory

Find all pairs of nonnegative integers $(x,y)$ such that $x^3 + 8x^2 - 6x + 8 = y^3$.

[quote="TomciO"]$x^3 < x^3 + 8x^2 - 6x + 8 < x^3 + 9x^2 + 27x + 27 = (x+3)^3$

So, it means that: $x^3 + 8x^2 - 6x + 8 = (x+2)^3 = x^3 + 6x^2 + 12x + 8$
so
$x(x-9)=0$

Therefore, solutions are: $(x, y) \in {(0, 2), (9, 11)}$[/quote]

I think you mean $(x+1)^3$ for the left inequality; the rest of the solution works the same way.


Let $ a,b,c>0$ distinct and with $ a\plus{}b\plus{}c\equal{}k>0$. Prove that

$ \sum\frac{a}{(b\minus{}c)^2}\geq\frac{4}{k}$

[quote="Ligouras"]Let $ a,b,c > 0$ distinct and with $ a \plus{} b \plus{} c \equal{} k > 0$. Prove that

$ \sum\frac {a}{(b \minus{} c)^2}\geq\frac {4}{k}$[/quote]
Hallo mr Ligoura.

$ \sum\frac{a}{(b\minus{}c)^2} \ge \frac{(a\plus{}b\plus{}c)^2}{ \sum_{sym} a^2b\minus{}6abc}$.
So we have to prove that:
$ (a\plus{}b\plus{}c)^3\plus{}24abc \ge 4\sum_{sym}a^2b$ which is true after expanding and Schur...

[quote="Ligouras"]Let $ a,b,c > 0$ distinct and with $ a \plus{} b \plus{} c \equal{} k > 0$. Prove that

$ \sum\frac {a}{(b \minus{} c)^2}\geq\frac {4}{k}$[/quote]
$ \left ( \sum_{cyc} a(b\minus{}c)^2 \right ) LHS \ge (a\plus{}b\plus{}c)^2$
So it suffices to prove that:
$ (a\plus{}b\plus{}c)^3 \ge 4\left ( \sum_{cyc} a(b\minus{}c)^2 \right )$
$ \sum_{cyc} a(b\minus{}c)^2 \equal{} (a\plus{}b\plus{}c)(ab\plus{}bc\plus{}ca)\minus{}9abc$.
So we have to prove:
$ (a\plus{}b\plus{}c)^3 \plus{} 36abc \ge 4(a\plus{}b\plus{}c)(ab\plus{}bc\plus{}ca)$
But $ (a\plus{}b\plus{}c)^3 \plus{} 9abc \ge 4(a\plus{}b\plus{}c)(ab\plus{}bc\plus{}ca)$ by Schur, so we are done.
Equality when $ a\equal{}0,b\equal{}c$ or permutations.

[quote="Mathias_DK"][quote="Ligouras"]Let $ a,b,c > 0$ distinct and with $ a \plus{} b \plus{} c \equal{} k > 0$. Prove that

$ \sum\frac {a}{(b \minus{} c)^2}\geq\frac {4}{k}$[/quote]
$ \left ( \sum_{cyc} a(b \minus{} c)^2 \right ) LHS \ge (a \plus{} b \plus{} c)^2$
So it suffices to prove that:
$ (a \plus{} b \plus{} c)^3 \ge 4\left ( \sum_{cyc} a(b \minus{} c)^2 \right )$
$ \sum_{cyc} a(b \minus{} c)^2 \equal{} (a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca) \minus{} 9abc$.
So we have to prove:
$ (a \plus{} b \plus{} c)^3 \plus{} 36abc \ge 4(a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca)$
But $ (a \plus{} b \plus{} c)^3 \plus{} 9abc \ge 4(a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca)$ by Schur, so we are done.
Equality when $ a \equal{} 0,b \equal{} c$ or permutations.[/quote]

Simple and nice
Thanks

[quote="Dimitris X"][quote="Ligouras"]Let $ a,b,c > 0$ distinct and with $ a \plus{} b \plus{} c \equal{} k > 0$. Prove that

$ \sum\frac {a}{(b \minus{} c)^2}\geq\frac {4}{k}$[/quote]
Hallo mr Ligoura.

$ \sum\frac {a}{(b \minus{} c)^2} \ge \frac {(a \plus{} b \plus{} c)^2}{\sum_{sym} a^2b \minus{} 6abc}$.
So we have to prove that:
$ (a \plus{} b \plus{} c)^3 \plus{} 24abc \ge 4\sum_{sym}a^2b$ which is true after expanding and Schur...[/quote]

Thank you Dimitri. Se efxaristo poli


AMC
AIME
probability
number theory
relatively prime

In a five-team tournament, each team plays one game with every other team. Each team has a 50% chance of winning any game it plays. (There are no ties.) Let m/n be the probability that the tournament will produce neither an undefeated team nor a winless team, where m and n are relatively prime positive integers. Find m + n.

[hide]There can't be 2 teams both win all the games or lose all games. The probability that one team to win all games is 1/16. there are 5 teams, so the total for one team to win is 5/16, same thing for any one team loses, total should be 5/8 so there probability of that not happening is 3/8. But there exist the case that one team loses all games, and another team wins all games. the probability of that happening is 5*4*1/16*1/8=20/128=5/32. Add that to 3/8 we have 17/32. [/hide]

Beta:

[hide]Just wanted to say that the last few lines were a bit unclear. The reason you added the final probability was because of inclusion-exclusion, which is 1 - (3/8 - 5/32).

[/hide]

[quote]But there exist the case that one team loses all games, and another team wins all games. the probability of that happening is 5*4*1/16*1/8=20/128=5/32.[/quote]

Why is that the probability of one team winning all the games and one team losing all the games?

There are 5 possibilities for the winner. Then there are 4 for the loser. The probability the winner will win 4 games is (1/2)^4 = 1/16. The probability the loser will also lose its 3 other games is (1/2)^3 = 1/8.


calculus
analytic geometry
geometry
graphing lines
slope
Pythagorean Theorem

$\triangle ABC$ is a right triangle with $\angle B=90^\circ$, $AB=12$ and $BC=25$. If $P$ is a point on $BC$, find the minimum possible value of $5AP+3PC$.

(Note: There exists a non-calculus solution.)

[hide]Well, calling $AP=x$, $5AP+3PC = 5x-3(\sqrt{x^2-144})+75$. I'm not sure where to go from there, but graphing it from $12$ to $\sqrt{769}$, it has a minimum. I'm stuck...  :P [/hide]

(edit: darn forward slash....)

very interesting problem!
[hide="hint"]locus[/hide]

[hide="Another hint"]
Coordinate geometry might help (although you don't need to use it).

[hide="More hint?"]
A line whose slope is $\frac34$[/hide][/hide]

oh wow...i am so stupid...

[hide="solution"]it is clear that minimizing $5AP+3CP$ is the same as minimizing $5AP-3BP$, this happens when $\frac{AP}{BP}=\frac{5}{3}$ (see if you can prove this... generally, it holds for $M*AP-BP$ is a minimum for $\frac{AP}{BP}=M$) then using pythagorean theorem, $AP=15$, $BP=9$, then $CP=16$, so $15*5+3*16=123$[/hide]


Consider the ring of real matrices such that any element which is not on the first row or on the main diagonal is 0. What are the bilateral ideals of this ring?

If $e_1,\ldots,e_n$ are the matrices having $1$ in position $(i,i)$ and $0$ everywhere else, and $f_1,f_2,\ldots,f_n$ are those having $1$ in position $(1,i)$ and $0$ everywhere else ($f_1=e_1$), then the ideals are constructed like this:

Pick some $e_{k_1},\ldots,e_{k_t}$ and some $f_{l_1},\ldots,f_{l_u}$, and the ideals are the subspaces generated by $\{e_{k_1},\ldots,e_{k_t}\}\cup\{f_{l_1},\ldots,f_{l_u}\}\cup\{f_{k_1},\ldots,f_{k_t}\}$ for all such collections of $e$'s and $f$'s. It's just a matter of checking stability by multiplication to the left and right with all the $e$'s and $f$'s.

Maybe there's a nicer characterization (or maybe I'm wrong)? :?


algebra
polynomial
limit
function
modular arithmetic
calculus
integration

[quote="amir2"]No one??? :([/quote]
[quote="spx2"]please write the problem so that we understand it[/quote]
[quote="dieuhuynh"]where don't you understand? :ninja:[/quote]
[quote="dieuhuynh"]now,I don't understand,too :([/quote]
[quote="mumath"]The problem is very clearly, i think why someone don't understand it. :P[/quote]
[quote="dieuhuynh"]yes ,have you solution? ;)[/quote]
[quote="dieuhuynh"]some one can write a fuction as this[/quote]

I have just one thing to say on his thread:
[quote="Beat"]EXPLAIN! ALL OF YOU![/quote]
(or even better: just don't spam anymore; especially: tell what you don't understand)

@dieuhuynh: it is fu[b]n[/b]ction, not fuction ;)

PS: I will delete this post and all spam here in one week.

What if you let $f(a) \equiv b \pmod{m}$ with the condition $0 \le b \le m-1$. Then f is bounded above by m thus the limit would be 0. Or am I missing something?

[quote="amir2"]let $f: \mathbb{Z}\longrightarrow \mathbb{Z}$ and $f$ isn't a polynomial such that $\forall a,b,m\in \mathbb{Z}\ \ \ ,\ a\equiv b(\mod m)$  we have $f(a)\equiv f(b) (\mod m)$ prove that for every natural $k$ we have:
$\lim_{n\rightarrow \infty}\frac{|f(n)|}{n^k}=\infty$[/quote]

Does this mean that $f$ is a polynomial that dosnt satisfy this condition, or that $f$ isnt a polynomial at all?

$f$ is not a polynomial but fulfills the conditions afterwards.

Here come some hints to solve the problem.

i) Let $(k_n)=1,2,6,12,60,60,420,840,\dots$ be the sequence of the least common multiples (denoted by $[,]$) $k_1=[1]$, $k_2=[1,2]$, $k_3=[1,2,3]$, $k_4=[1,2,3,4]$,... then the sequence of polynomials $1, k_1\binom x1, k_2\binom x2, k_3\binom x3, k_4\binom x4, k_5\binom x5, k_6\binom x6, \dots$, or explicitly $1, x, x(x-1), x(x-1)(x-2), \frac{x(x-1)(x-2)(x-3)}2, \frac{x(x-1)(x-2)(x-3)(x-4)}2, \frac{x(x-1)(x-2)(x-3)(x-4)(x-5)}{12}, \dots$ satisfy that $\forall a, b\in\mathbb Z\quad(a-b)\vert(p(a)-p(b))$

ii) Once proven the result of (i), it is really easy to prove that this polynomials are an integral basis of the space of polynomials $p: \mathbb Z\longrightarrow\mathbb Z$ satisfying that $\forall a, b\in\mathbb Z\quad(a-b)\vert(p(a)-p(b))$, and even more, that the sequence of polynomials above acts like a kind of "integral basis" for the functions $f: \mathbb Z\longrightarrow\mathbb Z$ satisfying $\forall a, b\in\mathbb Z\quad(a-b)\vert(f(a)-f(b))$, so that the following is true:
  There exists a unique sequence of integers $(a_n, n\geq0)$ such that $\forall x\in\mathbb Z, n\leq 0$
\[ f(x)=a_0+a_1k_1\binom x1+a_2k_2\binom x2+a_3k_3\binom x3+\cdots=\sum_na_ik_i\binom xi  \]
and $\lim_n a_n=0$ if and only if $f: \mathbb Z\longrightarrow\mathbb Z$ is a polynomial function.
So, in our case, as $f$ is not a polynomial function, then there are an infinite number of the $a_n$ distinct from zero.

iii) One additional result is required to complete the proof. It is the following divisibility relation between the product and the least common multiple of a sequence of consecutive integer numbers:
\[ x(x-1)\cdots(x-k)\vert k![x,x-1,\cdots,x-k]  \]
So we have $x\vert[x]$, $x(x-1)\vert[x,x-1]$, $x(x-1)(x-2)\vert 2[x,x-1,x-2]$, $x(x-1)(x-2)(x-3)\vert 6[x,x-1,x-2,x-3]$ and so on.

iv) Now we apply the result of (iii) to the infinite series of (ii) and obtain that for integer $x\geq0$:
$\sum_{n\geq1}a_ik_i\binom xi$ is an integral multiple of the number $x$
$\sum_{n\geq2}a_ik_i\binom xi$ is an integral multiple of the number $x(x-1)$
$\sum_{n\geq3}a_ik_i\binom xi$ is an integral multiple of the number $x(x-1)(x-2)/2$
$\sum_{n\geq4}a_ik_i\binom xi$ is an integral multiple of the number $x(x-1)(x-2)(x-3)/6$
$\sum_{n\geq5}a_ik_i\binom xi$ is an integral multiple of the number $x(x-1)(x-2)(x-3)(x-4)/24$
and so on.

v) COMPLETE THE PROOF!!
Now you have everything is required to prove the theorem.


geometry
circumcircle
trapezoid
parallelogram
projective geometry
incenter
trigonometry

Dear Mathlinkers,
let ABC be a triangle, H the orthocenter of ABC, (0) the circumcircle of ABC, P a point (for example inside ABC),
X, Y, Z the second points of intersection of AP, BP, CP with (0),
X', Y', Z' the symmetric of X, Y, Z, wrt BC, CA, AB,
U the point of intersection of XX' and BC,
(1) the circumcircle of PBC, Pa the second points of intersection of AP with (1),
A" the point of intersection of the perpendicular to BC through Pa
and A"' the symmetric of A" wrt U.

It is knowed that X', Y', Z' and H are cocyclic. Let (2) be this circle and X" the second point of intersection of AH and (2).

Prove: A"'X"//AA"

Sincerely
Jean-Louis

Dear Mathlinkers,
this circle is knowed as the P-Hagge's circle.
Any ideas about the parallelisme?
Sincerely
Jean-Louis

$ AP$ cuts $ BC$ at $ S.$ $ BC$ is radical axis of the circles $ (O), (1)$ $ \Longrightarrow$ $ \overline{SA} \cdot \overline{SX} = \overline{SP} \cdot \overline{SP_a}$ $ \Longrightarrow$ $ \frac {\overline{SP}}{\overline{SA}} = \frac {\overline{SX}} {\overline{SP_a}} = \frac {\overline{SU}}{\overline{SA''}}$ $ \Longrightarrow$ $ \triangle PUX \sim \triangle AA''P_a$ are centrally similar with center $ S$ and $ PU \parallel AA''.$ Done with circle $ (1),$ it can be discarded.

Let $ A_a, B_a, C_a$ be anticomplementary triangle of $ \triangle ABC.$ Let $ P^*$ be isogonal conjugate of $ P$ and $ Q$ anticomplement of $ P^*.$ $ X'$ is on circumcircle of $ \triangle A_aCB \cong \triangle ABC$ and $ \angle CAX = \angle BA_aX' = \angle C_aA_aX'$ $ \Longrightarrow$ $ A_a, X', Q$ are collinear. Let $ AH, AP^*, XX'$ cut $ (O)$ again at $ D, X_0, X_1.$ $ ADXX_1, HDXX'$ are both isosceles trapezoids $ \Longrightarrow$ $ AHX'X_1$ is parallelogram and $ AX_1 \parallel HX'.$ $ AX_0XX_1$ is cyclic and $ XX_0 \parallel BC, XX_0 \perp XX_1$ $ \Longrightarrow$ $ AX_0 \perp AX_1, AX_0 \perp HX'.$ But $ AX_0 \parallel A_aX'Q,$ hence $ QX' \perp HX'.$ Similarly, $ QY' \perp HY', QZ' \perp HZ'$ $ \Longrightarrow$ $ X', Y', Z'$ are on a circle with diameter $ HQ.$ This is circle $ (2).$

Let the altitudes $ AH, BH, CH$ cut the circle $ (2)$ again at $ A', B', C',$ respectively. (This is change of notation from $ X'', Y'', Z''$ to $ A', B', C'.$) Obviously, $ \angle B'A'C' = \angle B'HC' = \angle CAB$ and cyclic exchange, so that $ \triangle A'B'C' \sim \triangle ABC$ are also oppositely similar. It is also possible to show that $ \triangle X'Y'Z' \sim \triangle XYZ$ are oppositely similar as well, but neither similarity is asked for.
[hide]________________________________________

Altitudes $ AH, BH, CH$ cut $ (O)$ again at $ D, E, F$ and $ AO, BO, CO$ cut $ (O)$ again at $ A_0, B_0, C_0.$ Parallel to $ BC$ through $ Y$ cuts $ (O)$ again at $ A_1$ and parallel to $ ZX$ through $ A_1$ cuts $ (O)$ again at $ A_2.$ Likewise, parallel to $ CA$ through $ Z$ cuts $ (O)$ again at $ B_1$ and parallel to $ XY$ through $ B_1$ cuts $ (O)$ again at $ B_2.$ $ C_1, C_2$ can be defined cyclically.

$ \angle B_2A_1A_2 = \angle YA_1A_2 - \angle YA_1B_2 = \angle YA_1A_2 - \angle B_1YX =$
$ = \angle (YA_1, ZX) - \angle B_1ZX = \angle (YA_1, ZB_1) = \angle BCA$ $ \Longrightarrow$
$ A_2B_2 = AB = A_0B_0$ $ \Longrightarrow$ $ A_0A_2 = B_0B_2 = C_0C_2$ (cyclically).

$ \angle (X'H, YZ) = \angle (BC, YZ) - \angle (X'H, BC) = \angle A_1YZ - \angle XDA_0 =$
$ = \angle XA_1A_2 - \angle XA_1A_0 = \angle A_0A_1A_2 = \frac {A_0OA_2}{2}$

Similarly, $ \angle (Y'H, ZX) = \frac {B_0OB_2}{2},\ \angle (Z'H, XY) = \frac {C_0OC_2}{2}$ $ \Longrightarrow$ $ \angle (X'H, YZ) = \angle (Y'H, ZX) = \angle (Z'H, XY).$ But from circle $ (2),$ $ \angle X'Z'Y' = \angle X'HY' = \angle YZX$ and cyclic exchange $ \Longrightarrow$ $ \triangle X'Y'Z' \sim \triangle XYZ$ are oppositely similar.
________________________________________ [/hide]
Since $ A'$ is on circle $ (2)$ with diameter $ HQ,$ $ HA' \perp QA'.$ But $ (HA' \equiv HA) \perp (BC \parallel B_aC_a)$ $ \Longrightarrow$ $ QA' \parallel B_aC_a.$ Let $ Q_a \in B_aC_a$ be pedal of $ Q$ and let $ V \in BC$ be pedal of $ P^*.$ Then $ \overline{A'A} = \overline{QQ_a} = 2 \cdot \overline{VP^*}.$ If $ XP^*$ cuts $ BC$ at $ W,$ then $ PW \parallel AP^*$ (see the lemma below)$ \Longrightarrow$

$ \frac {\overline{PX}}{\overline{PA}} = \frac {\overline{WU}}{\overline{WV}} = \frac {\overline{UX}}{\overline{VP^*}}\ \Longrightarrow\ \frac {\overline{X'X}}{\overline{PX}} = \frac {2 \cdot \overline{UX} \cdot \overline{VP^*}}{\overline{PA} \cdot \overline {UX}} = \frac {\overline{A'A}}{\overline {PA}}$

It follows that $ \triangle PXX' \sim \triangle PAA'$ are centrally similar with center $ P$ $ \Longrightarrow$ $ P \in A'X'.$ Similarly, $ P \in B'Y', P \in C'Z'$ $ \Longrightarrow$ $ A'X', B'Y', C'Z'$ are concurrent at $ P.$ Since $ U$ is midpoint of $ X'X,$ $ UP$ also cuts $ A'A$ at its midpoint $ \Longrightarrow$ $ UP$ is midparallel of $ A''A \parallel A'''A'.$ $ \square$

[color=blue][b]Lemma:[/b][/color] Let $ P \in AX, Q \in AY$ be arbitrary points of sides of a $ \triangle AXY$ with circumcircle $ (O).$ Parallel $ p \parallel AY$ cuts $ XQ$ at $ U$ and parallel $ q \parallel AX$ through $ Q$ cuts $ YP$ at $ V.$ Then $ UV \parallel XY.$ Moreover, if $ UV$ cuts $ (O)$ at $ B, C,$ then $ P, Q$ are isogonal conjugates WRT $ \triangle ABC.$

For example, Let $ x$ be parallel  to $ p \parallel AY$ through $ X$ and let $ y$ be parallel to $ q \parallel AX$ through $ Y.$ Let $ X', Y'$ be infinite points of $ AX, AY.$ By Pappus theorem for lines $ PXX', QYY',$ the intersections $ R \equiv QX \cap PY,$ $ S \equiv QX' \cap PY'$ and $ T \equiv XY' \cap YX'$ are collinear. This means that the quadrilateral $ RUSV \sim RXTY$ with parallel sides and one diagonal $ RS \equiv RT$ are centrally similar and their remaining diagonals $ UV \parallel XY$ are also parallel. I don't see how to finish the 2nd part of the lemma.
___________________________________________
When $ P$ coincides with incenter $ I$ of $ \triangle ABC,$ identical with orthocenter of the circumcevian $ \triangle XYZ,$ circle $ (2) \equiv \odot(X'Y'Z'H)$ becomes Fuhrmann circle of the $ \triangle XYZ.$. The proof follows by adapting the Furhmann circle proof to the general case.

[color=blue][b]Supporting lemma:[/b][/color] Let $ P, Q$ be isogonal conjugaetes WRT $ \triangle ABC.$ Let $ AP, AQ$ cut the triangle circumcircle $ (O)$ again at $ X, Y.$ Let $ XQ, YP$ cut $ BC$ at $ U, V.$ Then $ PU \parallel AY, QV \parallel AX.$

[i][color=blue]Proof:[/color][/i] Let $ AP, AQ$ cut $ BC$ at $ E, F.$ Using $ XY \parallel BC,$

$ \angle CBY \equal{} \angle XYB \equal{} \angle PAB \equal{} \angle CAQ \equal{} \alpha,$
$ \angle ABP \equal{} \angle QBC \equal{} \beta,$

$ \angle QAB \equal{} 180^\circ \minus{} \angle BXY \equal{} \angle A \minus{} \alpha,$
$ \angle PBX \equal{} 180^\circ \minus{} \angle BQA \equal{} \angle A \minus{} \alpha \plus{} \angle B \minus{} \beta,$

and sine theorem for $ \triangle BXA, \triangle BYQ$ and then for $ \triangle BXY, \triangle BQA$:

$ \frac {\overline{PA}}{\overline{PX}} \cdot \frac {\overline{FY}}{\overline{FQ}} \equal{} \left(\frac {BA}{BX} \cdot \frac {\sin \widehat{ABP}}{\sin \widehat{PBX}}\right) \cdot \left(\frac {BY}{BQ} \cdot \frac {\sin \widehat{CBY}} {\sin \widehat{QBC}}\right)$ $ \equal{}  \left(\frac {BY}{BX} \cdot \frac {\sin \widehat{XYB}} {\sin \widehat{BXY}}\right) \cdot \left(\frac {BA}{BQ} \cdot \frac {\sin \widehat{QAB}}{\sin \widehat{BQA}}\right) \equal{} 1$

If parallel to $ AY$ through $ P$ cuts $ XY$ at $ U',$ then $ \frac {\overline{U'Y}}{\overline{U'X}} \equal{} \frac {\overline{PA}}{\overline{PX}} \equal{} \frac {\overline{FQ}}{\overline{FY}}$ $ \Longrightarrow$ $ U \in PU'$ or $ PU \parallel AY$ and similarly, $ QV \parallel AX.$ $ \square.$
____________________________________________________________
This is simple and works. Needless to say, a simple working proof without trigonometry would be appreciated.

Dear Vladimir,
thank you very much for your help. I have found all the principal results you have mentionned, but I haven't consider the following:
A'A = 2.VP*.

With this result all is possible.
You write :
"The proof follows by adapting the Furhmann circle proof to the general case."
Do you have any reference about your remark?
Sincerely
Jean-Louis

Dear Jean-Louis,

This proof follows very closely the proof at [url]http://www.mathlinks.ro/viewtopic.php?t=101687[/url] that a triangle and its Fuhrmann triangle have the same orthocenter. First part, A'A = 2r is taken from Honsberger's Episodes and the general case is only slightly more complicated. The similarity coefficient OH/R of the Fuhrmann triangle and the reference triangle is not applicable in the general case. The second part is also very similar, although simpler for the Fuhrmann triangle, because incenter is its own isogonal conjugate, so that the supporting lemma is not necessary. BTW, if you have a proof of the supporting lemma without trigonometry, would you share it ?

Sincerely, Vladimir.

Dear Vladimir,
after looking for
Treegoner, Isogonal conjugate, Mathlinks du 03/08/2005 ; http://www.mathlinks.ro/Forum/viewtopic.php?t=46718 ;

I have posted the same result
Ayme J.-L., Isogonal points and parallelism, Mathlinks du 04/12/2007 ; http://www.mathlinks.ro/Forum/viewtopic.php?t=177608.

The answer of Leonhard Euler is very interesting. (without trigonometry).
Sincerely
Jean-Louis

Dear Vladimir and Mathlinkers,
the result concerning the collinearity of A', P and X' comes from Karl Hagge.
Any reference?
Sincerely
Jean-Louis

Dear Mathlinkers,
you can see for example
http://perso.orange.fr/jl.ayme/   vol. 5  Le P-cercle de Hagge p. 11, 22.
Sincerely
Jean-Louis


function
real analysis
real analysis solved

We have a set A and an equivalence relation R on A(it is rflexive,symmetric and transitive).As it is known, the eqivalence classes form a partition of A.
Now we want to define a function on the set of classes A/R.How do we proceed?
I dont know if you realize what i mean.I've seen functions defined in the following way: f(Cx)=g(x), where Cx is the lass of x.Why do we need such a definition?Cannot we just put f(c)=g(s(c)), for every class c, where s is a function defined on the set of classes, and for each class c s(c) is a n element from c?
And another strange thing is with the ambigous proof of independence of the representers of a class.What does this want to mean?
Please help me understand this.

The two definitions are equivalent; I would say the former emphasizes the fact that we use a particular function which acts, for a given class, independently of the particular element (representative) we choose for the class.

Once you define a g : A -> S and f : A/R -> S as f(C) = g(x) where x is some representative of C, one must ensure this definition is well-defined (i.e., if x and y are both in C, then g(x) = g(y)).  For surely not *any* g will lead to a well-defined function on A/R...

- Jeff


calculus
integration
calculus computations

(1) 2009 Kansai University entrance exam

Calculate $ \int \frac{e^{\minus{}2x}}{1\plus{}e^{\minus{}x}}\ dx$.

(2) 2009 Rikkyo University entrance exam/Science

Evaluate $ \int_0^ 1 \frac{2x^3}{1\plus{}x^2}\ dx$.

hello
$ \int \frac{e^{\minus{}2x}}{1\plus{}e^{\minus{}x}}\ dx$
use either partial fraction or just substitute $ 1\plus{}e^{\minus{}x}\equal{}t$
we will get $ log(1\plus{}e^{\minus{}x})\minus{}e^{\minus{}x}\plus{}C$
$ \int_0^ 1 \frac{2x^3}{1\plus{}x^2}\ dx$
=$ \int_0^1(2x\minus{}\frac{2x}{1\plus{}x^2})dx$
=$ 1\minus{}log2$
thank u


Let $ a,b,c,d$ be integers.Show that the product $ (a\minus{}b)(a\minus{}c)(a\minus{}d)(b\minus{}c)(b\minus{}d)(c\minus{}d)$ is divisible by 12.

[hide="solution"]From Pigeonhole Principle, there are at least two numbers in $ \{ a, b, c, d \}$ that are congruent modulo 3. Therefore their difference is 0 modulo 3, and the given product is a multiple of 3.
From the Pigeonhole Principle, there are at least two pairs of numbers in $ \{ a, b, c, d \}$ that are congruent modulo 2. Therefore their differences are 0 modulo 3, and the given product is a multiple of $ 2^2\equal{}4$.
Therefore the given product is a multiple of $ \gcd (3,4)\equal{}12$.[/hide]

PEN problem,isn`t it.
Solution is Pigeon Principle, as [b]1=2[/b] says.


You changed a minus sign into a plus sign when you changed the base of the logarithm.

\[ \begin{gathered}
  \left( {2.\log 11} \right).\log \left( {x^2  \minus{} 4x \plus{} 11} \right) \geqslant \left( {3.\log 5} \right).\log \left( {x^2  \minus{} 4x \minus{} 11} \right) \hfill \\
  \because \log 121 < \log 125 \hfill \\
  \therefore \log \left( {x^2  \minus{} 4x \plus{} 11} \right) \geqslant \log \left( {x^2  \minus{} 4x \minus{} 11} \right) \hfill \\
\end{gathered}
\]

@anorithdialga
$ \log\left({x^{2}\minus{}4x\plus{}11}\right) > \log\left({x^{2}\minus{}4x\minus{}11}\right)$ does not need to be proven like that.

$ x^2\minus{}4x\plus{}11>x^2\minus{}4x\minus{}11$

@chrischris:From the inequality given in the problem , I arrived at the said inequality
$ \log\left({x^{2}\minus{}4x\plus{}11}\right) >\log\left({x^{2}\minus{}4x\minus{}11}\right)$, which is true for all x, so that should be the solution

@anorithdialga
The conclusion was right. But that wasn't the answer to the question.
Notice in
$ \log\left({x^{2}\minus{}4x\plus{}11}\right)\geqslant\log\left({x^{2}\minus{}4x\minus{}11}\right)$

LHS is bigger than RHS and that goes without saying.
What makes this problem a problem is that the LHS has been multiplied with a factor smaller than that in the RHS.

You can check the case when $ x^2\minus{}4x\equal{}1000$.

[quote="anorithdialga"]@chrischris:From the inequality given in the problem , I arrived at [inequality] which is true for all x, so that should be the solution[/quote]

Assume $ 2 \plus{} 2 \equal{} 5$.  Subtract $ \frac {9}{2}$ from both sides and square to arrive at the equation $ \frac {1}{4} \equal{} \frac {1}{4}$, which is clearly true.  Thus $ 2 \plus{} 2 \equal{} 5$ must be true.   :wink:  :roll:

:lol: then wat is the solution of this strange problem :huh:

[hide="Solution Part 1"]$ \log_{5}(x^{2} \minus{} 4x \plus{} 11)^{2} \ge \log_{11}(x^{2} \minus{} 4x \minus{} 11)^{3}$

Rearrange as:
$ \frac {\log (x^{2} \minus{} 4x \plus{} 11)}{\log (x^{2} \minus{} 4x \minus{} 11)} \ge \frac {\log 125}{\log 121}$

Multiply the RHS by 5.5:
$ \frac {\log (x^{2} \minus{} 4x \plus{} 11)}{\log (x^{2} \minus{} 4x \minus{} 11)} \ge \frac {\log 687.5}{\log 665.5}$

Set y=x^2-4x-11:
$ \frac {\log (y \plus{} 22)} {\log y} \ge \frac {\log 665.5 \plus{} 22}{665.5}$[/hide]

Now, we're at a fork in the road, of sorts. I'll take one path.
[hide="Solution Part 2"]Now, we've arrived at the question: what is the nature of the function $ \frac {\log (y \plus{} 22)} {\log y}$?

For $ 0 < y < 1$, the function is negative.
For $ y < 0$, the function does not exist.
For $ y > 1$, we have decreasing behavior, since the logarithm is sub-exponential.

Thus, the LHS is greater than the RHS iff. $ 1 < y \leq 665.5$[/hide]

Someone solve for the final answer for this method, please.

Multiply the RHS by 5.5:
$ \frac {\log (x^{2} \minus{} 4x \plus{} 11)}{\log (x^{2} \minus{} 4x \minus{} 11)} \ge \frac {\log 687.5}{\log 665.5}$

[/quote]

Can you explain this please

$ \frac{\ln(125)}{\ln(121)} > \frac{\ln(687.5)}{\ln(665.5)}$

Um, yeah, ignore that. Oops.

My basic idea there was to find an equivalent form of $ \frac {\log 125}{\log 121}$ such that the number inside the log of the numerator was 22 more than that inside the log of the denominator.

Obviously, I can't multiply the top and bottom by 5.5, since that would raise the numbers inside the log to the power 5.5, rather than multiplying them by 5.5! That was a nonsensical error on my part.

However, we can do something else -- we can look for a number $ k$ such that $ 125^k \minus{} 121^k \equal{} 22$

This way, if we multiply the numerator and denominator by $ k$, we get:

$ \frac {\log 125}{\log 121} \equal{} \frac {k\log 125}{k\log 121} \equal{} \frac {\log 125^k}{\log 121^k} \equal{} \frac {\log (121^k \plus{} 22)}{\log 121^k}$

It would be [i]necessary[/i] to solve the equation I gave for $ k$ to solve this problem, or some equivalent equation, since it results in a closed endpoint.

I don't have Mathematica on my new laptop (yet), but I find that $ k \approx 1.29977386$, so we then have:

$ \frac {\log (y \plus{} 22)}{\log y}\ge\frac {\log 121^k \plus{} 22}{121^k} \approx \frac {\log 509.5068 \plus{} 22}{509.5068}$

Then, the same argument method from solution part 2 applies.


vector
Functional Analysis
advanced fields
advanced fields unsolved

$H$ a separable Hilbert space, $x_n$ an orthogonal basis, $y_n$ an independant sequence of vectors such that $\sum_{n \geq 0} |x_n-y_n|^2 < \infty$. Prove that the $y$ is actually a basis.

The way I knew it: $(x_{n})_{n}$ is an orthonormal basis (not merely orthogonal), and $(y_{n})_{n}$ is also an orthonormal set (not just “independent”). I’ll solve it in this setting.

Consider the operator defined by $Tx_{n}=x_{n}-y_{n}$. The hypothesis $\sum\|Tx_{n}\|^{2}=\sum\|x_{n}-y_{n}\|^{2}<\infty$ says precisely that $T$ is [url=http://en.wikipedia.org/wiki/Hilbert-Schmidt_operator]Hilbert-Schmidt[/url], and hence compact.

Now consider the operator $1-T$, whose range is the space spanned by $y_{n}$. According to the [url=http://en.wikipedia.org/wiki/Fredholm_alternative]Fredholm alternative[/url], whenever $T$ is a compact operator on a Banach space, $1-T$ is invertible iff it's one-to-one. In our case, $1-T$ is an isometry, because it sends the orthonormal basis $(x_{n})_{n}$ onto the orthonormal set $(y_{n})_{n}$. In particular, it’s one-to-one. This means that it must be onto as well, i.e. $(y_{n})_{n}$ spans the space we’re working in: Q.E.D.

P.S.

This is problem number 7 in Halmos’s “A Hilbert space problem book” (wonderful book, by the way). It always looked a bit intimidating until today, because the first words the author wrote after stating the problem were “This is a hard one” :). I guess it wasn’t that hard.


limit
geometry
3D geometry
algebra
binomial theorem
calculus
calculus computations

Compute :

$\displaystyle \lim_{x \to \infty} x ( \sqrt[3]{x^{3}+x^{2}+1} + \sqrt[3]{x^{3}-x^{2}+1} - 2x).$

I have a solution.. but it is too long.... I hope you have a better one! :D

First step: factor $x^3$ out from under each cube root. This leaves us with

$x\left(x\sqrt[3]{1+\frac1x+\frac1{x^3}}+ x\sqrt[3]{1-\frac1x+\frac1{x^3}}-2x\right)$

$=x^2\left(\sqrt[3]{1+\frac1x+\frac1{x^3}}+ \sqrt[3]{1-\frac1x+\frac1{x^3}}-2\right)$

The next step is the binomial theorem:

$(1+y)^{1/3}=1+\frac13\,y+\frac{\left(\frac13\right)\left(\frac{-2}3\right)}2\,y^2 +\frac{\left(\frac13\right)\left(\frac{-2}3\right)\left(\frac{-5}3\right)}6\,y^3 +\cdots$

$=1+\frac13\,y-\frac19\,y^2+O(y^3).$

So our limit becomes the limit of

$x^2\left(\left(1+\frac1{3x}-\frac1{9x^2}+O(x^{-3})\right) + \left(1-\frac1{3x}-\frac1{9x^2}+O(x^{-3})\right)-2\right)$

$=x^2\left(2-\frac2{9x^2}+O(x^{-3})-2\right)=\frac29+O(x^{-1})\to\frac29$ as $x\to\infty.$


(17th International math olympiad, 1975, problem 4)
When 4444^4444 is written in decimal notation, the sum of its digits is A. Let B be the sum of the digits of A. Find the sum of the digits of B.

Hints: First show that the sum of the digits of B is fairly small (in fact: less than 16). Then use the fact that, for any natural number N,

N :equiv: (sum of the digits of N) (mod9)

A really really well-known problem, and it seems that you have a solution so I move it into the 'proposed and own problems' section...

Pierre.


logarithms
function
calculus
integration
trigonometry
calculus computations

Evaliate the integral:
\[ \int_1^e \frac{dx}{x\sqrt{4-\ln^2x}}.  \]

Actually, your substitution is not very good, I think we can replace $\ln x=t$,... but I had a trouble with it.

No my substituion can work.

$\int \frac{dx}{x\sqrt{4-(\ln x)^2}}=\int \frac{1}{\sqrt{4-(\ln x)^2}}\ d(\ln x)=\sin ^ {-1} \left(\frac{\ln x}{2}\right)+Const.$

Sorry for that my explanation was unkind.

Here is a detail solution.

Let $t=\ln x,$ we have $dt=\frac{1}{x}dx.$ Thus

$\int_1^e \frac{dx}{x\sqrt{4-(\ln x)^2}}\ dx=\int_0^1 \frac{1}{\sqrt{4-t^2}}\ dt=\left[\sin ^ {-1} \frac{t}{2}\right]_0^1$

$=\sin ^ {-1} \frac{1}{2}-\sin ^ {-1} 0=\frac{\pi}{6}.$

Here is a solution for Japanese high school students.

Let $t=2\sin \theta ,$ we have $dt=2\cos \theta d\theta.$ Thus

$\int_0^1 \frac{1}{\sqrt{4-t^2}}\ dt=\int_0^{\frac{\pi}{6}} \frac{1}{\sqrt{4\cos ^ 2 \theta}}\cdot 2\cos \theta d\theta$

$=\int_0^{\frac{\pi}{6}} d\theta=\frac{\pi}{6}.$

kunny

I think both solutions are for school student.
They're primary.

Regrettably, in japan high school students don't usually study arc sinx etc.


geometry
rectangle
trapezoid
analytic geometry
graphing lines
slope

Show that any line that passes through the center of a square divides it into half?

And im pretty sure this can be generalized. Like I think its true for triangles (with the median), hexagons, rectangles, etc.

Can anyone show this?

the area of the triangles when divided is the
length of the rec x width / 2
which is half of the area of the rectangle's area (length x width)

You can show the two shapes are congruent using SSS or however many sides needed,
then using algebra to show each disected shape is half of original shape

You are right. But the main point is to prove the shapes are congruent, so throw in an angle or two in the SSS and it should be all good  :oops:

[quote="Thunder365"]I mean any line, not just the specific line which passes through the two endpoints (the diagonals). It doesnt necessarily form two triangles, it can form two trapezoids or two rectangles. So SSS would not be applicable, since it only holds for triangles, and in quadrilaterals even if two figures have the same side lengths they arent necessarily congruent (eg. squares and rhombuses)[/quote]

What RobRoobik's original solution was is almost correct. I'm not writing out the full solution but here's a diagram that should explain it. Now, center point isn't clearly defined, but most people would define it (for a square) as where the diagonals meet.

Look at that:
[geogebra]d2ef69f419d4f6fb2d42354212258202177b4d7f[/geogebra]

Just make triangles by connecting the points where the line intersects the square. Then prove the triangles congruent.

Or you can prove it using slopes and shoelace.

Thanks guys.

Can this be generalized to triangles/pentagons/hexagons, etc.

take ABCD as the square with centre O take EF as the line where E is on AD  & F on BC and join AC [geogebra]c1a189ab9935cff0f9d75e9e6c9766839364fdf6[/geogebra]
O is the midpoint of AC.so, Ar(AEO)=arECO
ar(AFO)=ar(CFO)
and AEO and FOC are congruent ASA
all have equal area
and EO=OF
and ABF=DEC
so, ABFE=ABF +AFO +AOE=DEC+COE+COF=DEFC

this can be generalized to only regular polygons


Find all similarities between angle bisectors and medians.

[quote="anirudh"]They are both in the middle  :rotfl:[/quote]
how do you mean?

The median is in the middle of a sequence, and the bisector is in the middle of the angle (because it splits the angle in two).

Both are cevians, or they intersect a vertex and the side opposite that vertex.

There are 3 of each in a triangle.

If you draw all 3, they are concurrent, or they intersect at one point.

Median and angle bisectors:
All pologons have both??


a friend gave me this problem a while ago, it is a nice problem

we have 2 decks of 36 cards each.(each numbered 1 through 36)
we match every card in the first deck with its pair in the second deck (ie we match 1-1,2-2,etc)
then we shuffle one deck randomly and put the shuffled deck on top of the original (unshuffled) deck. find the sum of distances of matched cards from each other.
(the distance between two cards is defined as the number of cards in between)

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=46944

yea there was a similar problem like this posted

so basing it off my work before answer is

[hide]

$35(36) = 1260$

[/hide]


A group of n people are standing in a circle, numbered consecutively
clockwise from 1 to n. Starting with person no. 2, we remove every other person, proceeding
clockwise. For example, if n = 6, the people are removed in the order 2, 4, 6, 3, 1, and the
last person remaining is no. 5. Let j(n) denote the last person remaining. Find some simple
way to compute j(n) for any positive integer n > 1.

$ n=2^{m}+l$ where $ 2^{m}$ is largest power of $ 2$ smaller then $ n$, and $ l=n-2^{m}$ ( notice that $ l < 2^{m}$ ). Then the answer is number $ 2l+1$.

[color=darkblue][u]Proof:[/u] [/color]

For $ n=2^{k}$ it is obvious that the answer is $ 1$.
[hide="why?"] Well, when we for the first time remove people we will remove the last one because the $ n$ is even, now we will jump over the first person and again remove the last one because the "new" $ n$ is also even, and so one until we came to the trivially case for $ n=2$. [/hide]

Now for the $ n=2^{m}+l$ we will remove first $ l$ persons, the last removed person will be that with number $ 2l$ (notice that we will not come to the last person, because if we will that means $ 2l\geq 2^{m}+l$ which can not holds because $ l < 2^{m}$), and now we have $ 2^{m}$ persons, and the first one is person with the number $ 2l+1$.

Q.E.D.


Prove, for all n, that:

f(2n) is divisible by f(n)

where f(n) means f subscript n with fibonacci number n starting with f(1)=f(2)=1

Hi,

[hide]

A more general problem was posted here.

[/hide]

Sambit


number theory
prime numbers
number theory unsolved

we have $n$ integers $a_1 ;a_2 ;......a_n \not=0$    ($n\ge 2$)

show that :$\exists p$ prime  such that : $\forall i ; p+a_i$has  $n$  differente prime factors  :D

Let's take $n^2$ distinct prime numbers $q_1, q_2, \ldots, q_{n^2}$ such that $\forall i, j\ (a_i, q_j)=1$. By the Chinese Remainder Theorem $\exists k \in\mathbb{N}\colon k\equiv -a_i \mod (\prod_{j=n(i-1)+1}^{n\cdot i}q_j)\ \ \ \forall i=\overline{1,n}$. Moreover all such $k$ form an arithmetic progresion with difference $h=\prod_{j=1}^{n^2}q_j$ and clearly $(k, h)=1$. Hence by Dirichlet's theorem there exists a prime number $p$ in this arithmetic progresion. Note that $p$ has desired properties.

I hope it's correct  :)

Well I hope you weren't asking for such $p$ that all the numbers $p+a_i$ had exactly $n$ prime factors. It would get more complicated then...


Let p be a prime .For integers r,s such that rs(r-s)(r+s) is not divisible by p,let f(r,s) denote the number of integers n in {1,2,...p-1} such that {rn/p} and {sn/p} are either both less than 1/2 or both greater than 1/2.Prove that there exists N>0 such that  for p>=N and all r,s,
[img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/71f162abc73feb290e7954ca9db423de.gif[/img]

You can assume $r=1$ and $s<p/2$ by trivial arguments, now just try to calculate!
Well the numbers $r^{-1}a$ for $a=1,2,3,4...,p-1$ form a complete set of residues mod p, so we can work with $rr^{-1}$ and $sr^{-1}$ instead of $s,r$.

Some thoughts.

First of all, the condition [i]"rs(r-s)(r+s) is not divisible by p"[/i] is equivalent to the following: there is a non-zero element $t=r^{-1}s$ modulo $p$ of the order $q \geq 3$.

Consider a multiplicative group $\mathbb{Z}^{\star}_p = \{ 1,\dots, p-1 \}$ and let $A=\{ 1,\dots, \frac{p-1}{2} \}$ and $B=\{ \frac{p+1}{2},\dots,p-1\}$.
[i]"{rn/p} and {sn/p} are either both less than 1/2 or both greater than 1/2"[/i] means that $m=rn$ and $mt=sn$ (as elements of $\mathbb{Z}^{\star}_p$)  both belong either to $A$ or to $B$.

Consider multiplication by $t$ as an action on $\mathbb{Z}^{\star}_p$ splitting $\mathbb{Z}^{\star}_p$ into the orbits:
\[ \mathbb{Z}^{\star}_p = \{ a_1, a_1 t, \dots, a_1 t^{q-1} \} \cup \dots \cup  \{ a_k, a_k t, \dots, a_k t^{q-1} \},\]
where $k=(p-1)/q$ is the number of orbits and $a_1, \dots, a_k$ are representatives of the orbits. Note that each orbit contains $q\geq 3$ elements.

Elements $m$ and $mt$ are adjacent elements in some orbit. Hence, $f(r,s)$ is the number of adjacent elements belonging to the same set $A$ or $B$ in the orbits defined by $t$.


I'd like to produce something like

[url=http://www.imagehosting.com/][img]http://img155.imageshack.us/img155/5119/fgbjhvrs6.png[/img][/url]

with/without hyperlinks.

I've done some stuff but it's not what I want.

How would you produce the above result? Also, how can I change the word "Contents" to anyone I'd like to use?

[quote="Bipshula"]I've done some stuff but it's not what I want.[/quote]That's far too vague. We can't know what are the differences between what you get and what you want. \tableocontents [i]does[/i] produce the one you've illustrated so you will have to be much clearer.

To change Contents to (for example) Subjects use
\renewcommand{\contentsname}{Subjects}

You will find the [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=tocloft]tocloft package[/url] useful for controlling the format of the Table of Contents.

I have:

[code]\begin{document}
\tableofcontents
\newpage
\setcounter{secnumdepth}{0}
\subsection[title]{\textbf{\normalsize{title}}}
Text
\end{document}[/code]

this works fine, almost produces what I want, but, on my first page I'd like something like:

1. title........................................... 2

and in the second page it should just appear:

title

Text

-----------------------

Also, when clicking on "title" (first page) it directs to second page and it appears:

[url=http://www.imagehosting.com/][img]http://img147.imageshack.us/img147/1300/vfgnvfghqs3.png[/img][/url]

and "title" does not appear entirely, it should appear entirely. :)

You need to use hyperef's \addcontentsline to get the link to go to above the text.
[code]\begin{document}
\tableofcontents
\newpage
\section*{}
\addcontentsline{toc}{subsection}{\addtocounter{subsection}{1}\thesubsection.\quad\textbf{\normalsize{title}}}
\subsection*{\textbf{\normalsize{title}}}
Text
\end{document}[/code]

I just realized what I really want, I was forgetting something important.

I want to produce:

[url=http://www.imagehosting.com/][img]http://img136.imageshack.us/img136/1476/fgbjhvrs6ob5.png[/img][/url]

but "Introduction" should look normal, not as a hyperlink. As for 1.1 & 1.2 the hyperlink is fine.

So, when clickin' for example on "Overview," it directs to the second page and appears:

Overview

(here goes the text)

------

Sorry to ask many questions but I was confused about this stuff.

As far as I am aware hyperref makes all entries in the table of contents into links, and this is normal behaviour so probably can't be changed. See the [url=http://www.tug.org/applications/hyperref/ftp/doc/manual.html]hyperref manual[/url].

Okay, okay, but tell me how to yield that with hyperlinks anyway!  :lol:

Actually, I am wrong :) Hyperref won't convert text in \addtocontents to a hyperlink, so this will give what you are looking for:
[code]\begin{document}
\tableofcontents
\newpage
\section*{}
\addtocontents{toc}{Introduction}
\addcontentsline{toc}{subsection}{\addtocounter{subsection}{1}\thesubsection.\quad\textbf{\normalsize{title}}}
\subsection*{\textbf{\normalsize{title}}}
Text
\end{document}[/code]


In a convex quadrilateral $ ABCD$ the points $ P$ and $ Q$ are chosen on the sides $ BC$ and $ CD$ respectively so that $ \angle{BAP}\equal{}\angle{DAQ}$. Prove that the line, passing through the orthocenters of triangles $ ABP$ and $ ADQ$, is perpendicular to $ AC$ if and only if the triangles $ ABP$ and $ ADQ$ have the same areas.

Under inversion wrt $ A $, suppose, any point $ L $ maps to $ L' $.
Suppose, $ H_1,H_2 $ are the orthocenters of $ \Delta ADQ $ and $ \Delta ABP $.
$ AH_1' $ passes through the circumcenter of $ \Delta AD'Q' $(call it $ O_1 $) and circumcenter of $ \Delta AB'P' $ be $ O_2 $.
$ AC\perp H_1H_2 $ implies that the radical axis of $ \Delta AD'Q' $ and $ \Delta AB'P' $ passes through the circumcenter of $ \Delta AH_1'H_2' $. So $ H_1'H_2'O_1O_2 $ is cyclic.
So $ AH_1'.AO_1=AH_2'.AO_2\implies AD.AQ=AB.AP\implies [ADQ]=[ABP] $. So done.

$AC\perp H_1H_2$ implies that the radical axis of $\Delta AD'Q'$ and $\Delta AB'P'$ passes through the circumcenter of $\Delta AH_1'H_2'$.
I can't understand this..

Actually, it is enough to show that $AH_{1}^2 + CH_{2}^2 = AH_{2}^2 + CH_{1}^2$, to prove that the shown lines are perpendicular to each other. After that we just need to prove that this equality is equivalent to $AD*AQ = AB*AP$.
It can be done by some simple calculations.


You're really going to have to be more specific than that.

Well, I do not expect anyone to find the book for me  :lol:, I think that, if someone has read it, they should recognize the chapters.

[quote="Poincare"][quote="Peteris"]Hey,

I was introduced to an interesting book, but it was a copy and the owner did not know the name of the book.  All I know is that the first chapter was Combinatorics and the second was Combinatorial arithmetic. As the second topic is not so usual, maybe you remember this book?

Thanks[/quote]

It may be Mathematics of Choice by Ivan Niven....[/quote]

No its not. First chapter of Mathematics of Choice is "Introductory Questions" and the second chapter was "Permutations and Combinations."

[url=http://www.amazon.com/gp/reader/0883856158/ref=sib_dp_pop_toc?ie=UTF8&p=S008#reader-link]table of contents from amazon[/url]

Heh, I found it. It's Counting and Configurations. Still, thanks  :roll:.


trigonometry
geometry
AMC
AIME
calculus
integration
perpendicular bisector

P lies inside the triangle ABC, and angle PAB = angle PBC = angle PCA. If AB = 13, BC = 14, CA = 15, find tan PAB.

The point $P$ such that $\angle PAB = \angle PBC = \angle PCA = \omega$ is called the 1st Brocard point of the triangle $\triangle ABC$. This point is the intersection of 3 circles $(U), (V), (W)$ centered on the perpendicular bisectors of the triangle sides $BC, CA, AB$ with the central angles $\angle BUC = 2 \angle C$, $\angle CVA = 2 \angle A$, $\angle AWB = 2 \angle B$. The center $U$ of the circle $(U)$ is the intersection of the perpendicular bisector of the side $BC$ with a normal to the side $CA$ at the vertex $C$. Hence, the circle $(U)$ passes through the vertex $B$ and it is tangent to the line $CA$ at the vertex $C$. For any point $P$ on the minor arc $BC$ of the circle $(U)$, we have

$\angle PBC = 180^o - (180^o - \angle B + \angle PCB) = \angle B -  \angle PCB = \angle PCA$

Similarly, the circle $(V)$ passes through the vertex $C$, it is tangent to the line $AB$ at the vertex $A$ and for any point $P$ on the minor arc $CA$ of the circle $(V)$, we have $\angle PCA = \angle PAB$. Likewise, the circle $(W)$ passes through the vertex $A$, it is tangent to the line $BC$ at the vertex $B$ and for any point $P$ on the minor arc $AB$ of the circle $(W)$, we have $\angle PAB = \angle PBC$. Let the circles $(U), (V)$ intersect at a point $Z$. Then

$\angle ZBC = \angle ZCA = \angle ZAB$

and consequently, the point $Z$ also lies on the circle $(W)$.

Let $c'$ be a line through the vertex $C$ and parallel to the triangle side line $c= AB$. Let this line meet the circle $(U)$ at a point $Q$ different form the vertex $C$ and let the line $AQ$ meet the circle $(U)$ at a point $Z$. Then the angles $\angle ZBC = \angle ZCA$ are equal, because this is true for any point on the minor arc $BC$ of the circle $(U)$. Since the circle $(U)$ is tangent to the line $CA$ at the point $C$, the angles $\angle ZCA = \angle ZQC$ are equal, because the angle between the chord $CZ$ and tangent $CA$ is equal to the angle spanning the major arc $CZ$ of the circle $(U)$. Since the lines $c' \equiv CQ \parallel AB$ are parallel, the angles $\angle ZAB \equiv \angle QAB = \angle AQC \equiv \angle ZQC$ are equal and as a result, the angles $\angle ZCA = \angle ZQC = \angle ZAB$ are also equal. This means that the point $Z$ constructed in this way is the 1st Brocard point of the triangle $\triangle ABC$.

Drop normals from the points $C, Q$ to the line $AB$ and let $F, G$ be the feet of these normals. $CF$ is the altitude of the triangle $\triangle ABC$ from the vertex $C$ and $QG = CF, CQ = FG$, because the $CFGQ$ is a rectangle. The angles $\angle C \equiv \angle BCA = \angle BQC$ are equal, because $CA$ is a tangent of the circle $(U)$ at the point $C$ and the angle $\angle BQC$ spans the arc $BC$ of this circle. The angles $\angle BQC = \angle QBG$ are equal, because the lines $CQ \parallel AB \equiv BG$ are parallel. Hence $\angle QBG = \angle C$. But then the Brocard angle $\omega$ is given by

$\cot \omega = \cot{\widehat{QAG}} = \frac{AG}{QG} = \frac{AF + BF + BG}{QG} =$

$= \frac{AF}{CF} + \frac{BF}{CF} + \frac{BG}{QG} = \cot{\widehat A} + \cot{\widehat B} + \cot{\widehat C}$

Let $a = BC = 14, b = CA = 15, c = AB = 13$ be the triangle sides, $s = 21$ the semiperimeter and $R$ the circumradius.

$\tan \omega = \frac{\tan{\widehat A} \tan{\widehat B} \tan{\widehat C}}{\tan{\widehat B} \tan{\widehat C} + \tan{\widehat C} \tan{\widehat A} + \tan{\widehat A}\tan{\widehat B}} =$

$= \frac{\sin{\widehat A} \sin{\widehat B} \sin{\widehat C}}{\sin{\widehat B} \sin{\widehat C} \cos{\widehat A} + \sin{\widehat C} \sin{\widehat A} \cos{\widehat B} + \sin{\widehat A}\sin{\widehat B} \cos{\widehat C}} =$

$= \frac{abc}{8R^3} \cdot \frac{4R^2}{bc\ cos{\widehat A} + ca\ cos{\widehat B} + ab\ cos{\widehat C}} =$

$= \frac{abc}{R} \cdot \frac{1}{b^2 + c^2 - a^2 + c^2 + a^2 - b^2 + a^2 + b^2 - c^2} = \frac{4 |\triangle ABC|}{a^2 + b^2 + c^2}$

where we used the cosine theorem $2bc\ \cos{\widehat A} = b^2 + c^2 - a^2$, etc. The area of the triangle $\triangle ABC$ is given by Heron's formula:

${|\triangle ABC|} = \sqrt{s(s - a) (s - b) (s - c)} = \sqrt{21 \cdot 7 \cdot 6 \cdot 8} = 84$

$a^2 + b^2 + c^2 = 196 + 225 + 169 = 590$

$\tan \omega = \frac{4 |\triangle ABC|}{a^2 + b^2 + c^2} = \frac{4 \cdot 84}{590} = \frac{336}{590},\ \ \omega \doteq 29.66^o$

Seems pretty involved for an AIME problem.  Do you knwo what number it was on the AIME?

I think it's AIME 1999. By the way, there should be easier solution than this!

Here's my hopefully easier solution:
Let $AP=x$, $BP=y$, and $CP=z$.  Also let $AB=c$, $BC=a$, $CA=b$ and $\angle PAB=\angle PBC=\angle PCA=\theta$.  Now apply the law of cosines too the three interior triangles to get the equations
$y^2=x^2+c^2-2xc\cos{\theta}
z^2=y^2+a^2-2ya\cos{\theta}
x^2=z^2+b^2-2zb\cos{\theta}$
Adding the three together and simplifying we have
$\cos{\theta}=\frac{a^2+b^2+c^2}{2ay+2bz+2cx}$
Let the area of $\triangle ABC=K$ then
$K=\frac{1}{2}ay\sin{\theta}+\frac{1}{2}bz\sin{\theta}+\frac{1}{2}cx\sin{\theta}$.
Dividing by $\sin{\theta}$ and then multiplying by 4 we have
$\frac{4K}{\sin{\theta}}=2ay+2bz+2cx$
Substituting this into the denominator of the $\cos{\theta}$ equation we get
$\cos{\theta}=\frac{(a^2+b^2+c^2)\sin{\theta}}{4K}$
which implies that $\tan{\theta}=\frac{4K}{a^2+b^2+c^2}$
Inserting specifics, by heron's formula we know $K=84$.  Also $a^2+b^2+c^2=590$.  So
$\tan{\theta}=\frac{336}{590}=\frac{166}{295}$
which I believe would make the AIME answer 461

Thanks a lot!
(Still looks difficult!  :D )

I just saw you simplified the last fraction wrong. Guess you're in a hurry.
By the way, this problem really asks for tan PAB, not the sum of its numerator and denominator!

You're right it should be $\frac{168}{295}$.  Also although the problem asks for tan of angle PAB, the AIME only takes integral answers from 000 to 999 so for this they probably did ask for the sum of the numerator and denominator which would be 463 not 461.


Seems like the right section to post my problem. I know theres a solution but I cannot solve it.
How do you construct a circle tangent to a given circle, a given line and a given point.

Any help in the right direction is very much appreciated (any books that have the solution will also be a big help).

Thanks.

Take a look here :

http://www.mathlinks.ro/Forum/viewtopic.php?highlight=circle&t=46945


function
algorithm
calculus
integration
probability
parameterization
probability and stats

(I can't find a good section for this question.  Sorry if I posted in the wrong forum!)

I have a function $ f: A\to B$ where $ A\equal{}\{a\in\mathbb{Z^\plus{}}|a\le 2^{600000}\}$ and $ B\equal{}\{b\in\mathbb{Z^\plus{}}|b\le 2^{100}\}$.  (One might think of this as a sort of hash function.)  I wish to find the [b]mean (rounded) outcome in $ B$[/b] among all possible inputs in $ A$, in other words,

\[ \frac{\sum_{i\equal{}1}^{\parallel{}A\parallel{}} f(i)}{\parallel{}A\parallel{}}.\]

There is no time (computationally) to input $ 2^{600,000}$ values into $ f$ and figure out the exact mean.  Therefore, I want to use sampling and statistical methods to [i]estimate[/i] the mean.

How do we go about doing this?  How do we measure the variance?  What properties of $ f$ need to be known?  ($ f$ is actually a complicated algorithm that, if published here, I think would only complicate matters.  Intuitively, I know that the distribution of the range should be somewhat of a bell curve, but I have no proof.)

Thanks in advance!

If it's a hash function, there's no reason to expect it to be smooth, and calculus-inspired methods related to approximate integration are a bad idea. That makes this a statistical problem; we don't care about the structure of the domain. We do have a statistics forum.

One method: choose $ N$ input values randomly*, and treat the results as a sample from a probability space. The mean is a decent estimate for the overall mean, and the variance can be estimated as shown [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=216088]here[/url]. You can get a decent picture of the distribution as well just by recording the values, but that's not as clear-cut a question.

This assumes no structure at all for $ f$.

*OK, you can't do "random". You have a pseudo-random number generator, and you hope it's not somehow correlated with the hash function.

I don't know how I missed the probability forum!  If a moderator could move this thread, it'd be much appreciated.  Thanks. [color=green][Done. -mathmanman][/color]

Thanks jmerry, I see what you're saying.  I guess my next question, then, is this: Is there a way to determine how many samples are "enough" for a "good" estimate?  In other words, can we fix some definition of "good" (as a parameter, even), and determine how many random samples out of the $ 2^{600000}$ would be necessary to have a "good" estimate of the mean?

One idea I had was to continue sampling until some condition was met.  For example: keep sampling random input values until the mean fluctuates less than $ 0.05\sigma$, ten times in a row (implying some sort of convergence, I think).  And maybe we can add another condition like... the variance ($ \sigma^2$) must fluctuate less than 5%, ten times in a row (implying convergence in variance as well).  I guess one would have to prove that the mean and variance estimators do indeed converge.  Doesn't that have something to do with "consistency" in statistics?  I feel like I have the necessary knowledge, but I'm unable to bring together the relevant facts needed to solve this problem.  Please help me out! =)

The number of samples you need doesn't really depend on that $ 2^{600000}$, or on the results you get. The average error is proportional to $ \frac1{\sqrt{N}}$; the standard deviation of the sample mean is $ \frac{\sigma}{\sqrt{n}}$.

I'm not a statistician but I think that you can evaluate a confidence intervalle using some t-student test.

This problem makes me think about the problem of using monte carlo method to evaluate an integral. I think that you can find a huge amount of reference on the net.


inequalities
geometry
incenter
circumcircle
geometry proposed

Let $ABC$ is a triangle inscribed in the circle$(O)$.
Let the medians of the triangle intersecting sides at $A_1,B_1,C_1$,and $(O)$ at $A_2,B_2,C_2$ respectively.
Prove that $AA_2+BB_2+CC_2\geq \frac{4}{3}(AA_1+BB_1+CC_1)$

First, show that $AA_2 = \frac{b^2+c^2}{2m_a}$. The inequality now is easy to prove because we can replace $(a, b, c)$ by $(\frac{2}{\sqrt{3}}m_a, \frac{2}{\sqrt{3}}m_b, \frac{2}{\sqrt{3}}m_c)$ and $(m_a, m_b, m_c)$ by $(\frac{\sqrt{3}}{2}a, \frac{\sqrt{3}}{2}b, \frac{\sqrt{3}}{2}c)$.

[quote="barasawala"][quote="treegoner"]we can replace $(a, b, c)$ by $(\frac{2}{\sqrt{3}}m_{a}, \frac{2}{\sqrt{3}}m_{b}, \frac{2}{\sqrt{3}}m_{c})$ and $(m_{a}, m_{b}, m_{c})$ by $(\frac{\sqrt{3}}{2}a, \frac{\sqrt{3}}{2}b, \frac{\sqrt{3}}{2}c)$.[/quote]

Why can we replace like that? I don't get it.[/quote]

Notice that the medians of the triangle with sides $(\frac{2}{\sqrt{3}}m_{a}, \frac{2}{\sqrt{3}}m_{b}, \frac{2}{\sqrt{3}}m_{c})$ are $(\frac{\sqrt{3}}{2}a, \frac{\sqrt{3}}{2}b, \frac{\sqrt{3}}{2}c)$!

I hope you get it. :roll:

[quote="gemath"]An open question find the locus $M$ inside triangle such that your inequality,
In this inequality is centroid $G$, I see incenter $I$ is true, but circumcenter $O$ isn't. :)[/quote]

Just to make it clear,is this the question that you are thinking about?

[color=blue]$ABC$ is a triangle and $M$ is a point in it.$MA,MB$ and $MC$ intersect the sides at $A_{1},B_{1}$ and $C_{1}$ and the circumcircle at $A_{2},B_{2}$ and $C_{2}$.Find the locus of the points $M$ that the following inequality holds.

\[AA_{2}+BB_{2}+CC_{2}\geq \frac{4}{3}(AA_{1}+BB_{1}+CC_{1})\][/color]

[quote="Tiks"]
Just to make it clear,is this the question that you are thinking about?
[color=blue]$ABC$ is a triangle and $M$ is a point in it.$MA,MB$ and $MC$ intersect the sides at $A_{1},B_{1}$ and $C_{1}$ and the circumcircle at $A_{2},B_{2}$ and $C_{2}$.Find the locus of the points $M$ that the following inequality holds.
\[AA_{2}+BB_{2}+CC_{2}\geq \frac{4}{3}(AA_{1}+BB_{1}+CC_{1}) \]
[/color][/quote]

ok that's right, it is my meaning, what do you think about this question ? :)

[quote="gemath"]

ok that's right, it is my meaning, what do you think about this question ? :)[/quote]

I think it is a very hard problem!!! :D

More seriously here is a sugesstion concerning this question.

[color=blue]Conjecture: The locus of the points $M$ is a convex figure ,including its interior and boundaries.[/color]

That is if for two points $X$ and $Y$ the inequality holds then it holds for all the points of the segment $XY$. :wink:

Although I rather sure that this conjecture is true(due to the Geometry Scetchpad :) ),I can't figure out the how to constructe the convex figure from the conjecture.I think it is neithr a usual figure(i.e. n-gon,circle) nor a conic. :ninja:


function
limit
integration
real analysis
real analysis unsolved

Let $(X,\mu)$ be a measure space, $(f_{n}) \subset L^{1}(X,\mu)$ a sequence of functions $\ge0$. Suppose that $f_{n}\to f$ a.e. (resp., in measure) for some $f \in L^{1}(X,\mu)$, and that
\[\lim_{n\to\infty}\int_{X}f_{n}= \int_{X}f. \]
Given a measurable set $E \subset X$, prove that
\[\lim_{n\to\infty}\int_{E}f_{n}= \int_{E}f. \]
Also, show that the condition $f_{n}\ge 0$, $n \in \mathbf N$ cannot be omitted.

There is no way for this to hold if $f$ has infinite integral. So I assume $f\in L^{1}$ and that the convergence is pointwise a.e.

Let $g_{n}= f+f_{n}-|f-f_{n}| \geq 0$. Applying Fatou's lemma to $g_{n}$ you find that
$2\int f \, d\mu \leq \varliminf \int (f+f_{n}) \, d\mu-\varlimsup \int |f-f_{n}| \, d\mu = 2\int f \, d\mu-\varlimsup \int |f-f_{n}| d\mu$

That is, $f_{n}$ converges to $f$ in $L^{1}$ (and $\chi_{E}f_{n}$ to $\chi_{E}f$)

You've made a typo which doesn't bring any trouble with the proof: there should be
\[2 \int f d\mu \le \limsup \int (f+f_{n}) d\mu-\limsup \int |f-f_{n}| d\mu,\]
because
\[\liminf \int g_{n}d\mu \le \limsup \int (f+f_{n}) d\mu+\liminf-\int |f-f_{n}| d\mu.\]

Very good proof! I have more complicated one, and I didn't prove that stronger result, namely that $f_{n}\to f$ strongly in $L^{1}$.

In fact, the case $f_{n}\to f$ in measure may be reduced to the case $f_{n}\to f$ a.e. arguing by contradiction.

Hmm? I don't think I've made a typo. Fatou's lemma shows that
$\int \varliminf g_{n}\, d\mu \leq \varliminf \int g_{n}\, d\mu = \varliminf \left( \int (f+f_{n}) \, d\mu-\int |f-f_{n}| \, d\mu\right)$
$= \lim \int (f+f_{n}) \, d\mu-\varlimsup \int |f-f_{n}| \, d\mu$

In the second equality sign I am keeping in mind that the sequence $\int (f+f_{n}) \, d\mu$ converges (to a real number) so I don't need to use an inequality. Maybe your way of presenting it is a little clearer though.