Prove that if $a,b,c,d$ are positive real numbers which add up to 4 then $ 3(ab+bc+cd+da+ac+bd)$ is at most $ 16+2abcd$.

I would just homogenise it and destroy it. But there -must- be a better method.

Really, Harazi, is there are an universal method how do you obtain all these inequalities? I remember numerous similar your problems. And you know I like general methods :)

It is exactly suranyi's inequality for n=4 case
$3(a^2+b^2+c^2+d^2)+4abcd \ge (a+b+c+d)^2 = 16$
is it your method, harazi?

[quote="harazi"]It does not seem to be Suranyi's inequality... Are you sure?
 To Myth: it is just a consequence of the general inequality posted by Peter and beautifully solved by iura. You know I like particular cases.[/quote]

oops..  :blush: i remember the inequality wrongly, sorry :(

Haha... in the end I whacked to ensure I can do it (and to raise my confidence). To my dismay, it required Schur's in 3 and 4 variables. But at least now I know that it exists - thanks harazi!

No, no, no! I made you a service, you make me one. Please, full solution to this problem.  ;)  :D

But. It is purely whack - reveal hidden content at your peril!

[hide]Add up the following in equalities which are true by general Schur's (when I say cyclic sum, I mean all distinct terms with those powers):
$a^4 + b^4 + c^4 + d^4 + 8abcd \geq \sum_{cyc} a^2bc$
$\sum_{cyc} d(a^3 + b^3 + c^3 + 3abc) \geq \sum_{cyc} d(a^2b + ab^2 + bc^2 + cb^2 + ac^2 + a^2c)$
Add $2\sum_{cyc} a^2b^2$, $\sum_{cyc} ab^2c$, $(a^2 + b^2 + c^2 + d^2)(ab + ac + bc + ad + cd + bd)$, $12abcd$ to both sides.

$LHS$
$= \sum_{cyc} a^4 + \sum_{cyc} a^3b + 2\sum_{cyc} a^2b^2 + \sum_{cyc} ab^2c + (a^2 + b^2 + c^2 + d^2)(ab + ac + bc + ad + cd + bd) + 32abcd$
$= (a^2 + b^2 + c^2 + d^2)(a^2 + b^2 + c^2 + d^2 + 2(ab + ac + bc + ad + cd + bd)) + 32abcd$
$= (a^2 + b^2 + c^2 + d^2)(a+b+c+d)^2 + 32abcd$
$= 16(a^2 + b^2 + c^2 + d^2 + 2abcd)$

$RHS$
$= 4\sum_{cyc} a^2bc +  2\sum_{cyc} a^2b^2 + 12abcd + (a^2 + b^2 + c^2 + d^2)(ab + ac + bc + ad + cd + bd)$
$=2(ab + ac + bc + ad + cd + bd)^2 + (a^2 + b^2 + c^2 + d^2)(ab + ac + bc + ad + cd + bd)$
$= (ab + ac + bc + ad + cd + bd)(a^2 + b^2 + c^2 + d^2 + 2(ab + ac + bc + ad + cd + bd))$
$= 16(ab + ac + bc + ad + cd + bd)$

Thus $a^2 + b^2 + c^2 + d^2 + 2abcd \geq ab + ac + bc + ad + cd + bd$
$\rightarrow (a+b+c+d)^2 + 2abcd \geq 3(ab + ac + bc + ad + cd + bd)$
$\rightarrow 16+ 2abcd \geq 3(ab + ac + bc + ad + cd + bd)$[/hide]

Which I dare say is not what you wanted!

By the way, I have just found that this inequality is completely equivalent to the following one, which I solved in the open question section, due to Vasc ( topic is Schur for 4 numbers, by Anh Cuong): if $a+b+c+d=4$ then $ 4abcd+3(a^2+b^2+c^2+d^2)$ greater than 16.

http://www.mathlinks.ro/Forum/viewtopic.php?t=5303 in any case (= Smoothing argument!

wow, it looks like my general inequality is REALLY sharp :) - i wasn't aware of this :D

You know how it is, you prove it, I use it. :D


Can you explain the last sentence more throughly, I'm not sure I understand the question.

I'm pretty sure the test said 25 times the thickness of the first disc. [/hide]

It said 25 times the thickness of the original disc. Sorry :blush:

[quote="#H34N1"]A set of discs is being placed in a stack. The thickness of each successive disc added to the stack is 1.5 times the thickness of the disc immediately beneath it. What is the minimum number of discs that must be placed in the stack in order for the thickness of the stack to be 25 times the thickness of the stack.[/quote]
[hide]
let the thickness of the first disk be 1
using a calculator (which is one of the fastest ways), $1.5^{8}> 25$, so it is known for sure it is less than or equal to 8 times.
it should take 6 times because the 6th time is 11 and the fifth time is ~8 and the fourth is ~5 plus the rest add up to over 25.
so the answer is $1.5$
[/hide]

[quote="myspacesucks"]Technically, this problem is with powers of 1.5, but I don't think anyone knows 1.5^6 off the top of their head.  So you have to work it out, and it takes about 1 minute.  I started with the first disk equaling 1, because the answer would then be 25 or above, which makes it easy.  After doing this 4 times, and finishing the 5th, you get 13.875+7.59375, or 21.46875.  Since you need above
25, and the last addition had 7.59375, its only going to get higher, so its the next number, or the 6th in the equation.[/quote]
But people might know 3^6 and 2^6...

I got 1.5^8! That equals 25.62890625 and probably more......
I might've read the problem wrong... but I don't think so.

I read the problem wrong :blush:  oops!
It is 1.5^6, and you can use a calculator(I prefer TI) and the answer is 31.171875 and maybe more...

___________________________________________________
Scoring 37 at the Geogetown HS Invitational! And i'm just a 6th grader................... but, only got an 8th place, 'cause of stupid evil St. Mark's 7th graders that entered the novice..... 7 of them!


Let $a,b,c$ positive reals such that $abc=1$. Then it is very easy to prove that

$\frac{bc-a}{1+a}+ \frac{ca-b}{1+b}+\frac{ab-c}{1+c} \geq 0$

But how to prove it for $abc>1$ ?

No problem!
Suppose a=max(a,b,c)>b,c; abc>1. Then $LHS'_a=\frac{b}{1+c}+\frac{c}{1+b}-\frac{bc+1}{(1+a)^2}>0$ !
So it is sufficient to consider inequality for case a=b>=c; abc>1. It can be done by the same technique by reducing to case a=b=c; abc>1 :)
I hope you can complete all small steps which I knowingly omited ;)


limit
complex numbers
real analysis
real analysis unsolved

Prove that the linear space $ S$ of finite sequences $ u: N\to  C$ (where $ n$ is the set of integers and $ C$ the set of complex numbers) such that $ u_{j}\equal{}0$ for $ j>n$ for some $ n$ is dense in $ l^{1}$.

$ l^{1}$ is the set of sequences such that the infinite sum of the absolute values of each term converges.

[hide]Firstly,
$ \ell^{1}$ is equived with norm $ \parallel (x_{1},x_{2},\ldots)\parallel_{\ell_{1}}: \equal{}\sum_{j \equal{} 1}^{\plus{}\infty}|x_{j}| <\plus{}\infty$.[/hide]

[hide]Second,
If $ \sum_{j \equal{} 1}^{\plus{}\infty}|x_{j}| <\plus{}\infty$ then $ \lim_{n\to\infty}\Big(|x_{n}|\plus{}|x_{n\plus{}1}|\plus{}\cdots\Big) \equal{} 0$[/hide]

[hide]The third, think about $ x^{n}\equal{} (x_{1},\ldots,x_{n},0,\ldots)$ and $ x \equal{} (x_{1},x_{2},\ldots)$. Is there anything between them?[/hide]

[hide]No more guys :D[/hide]

The name I would use for the set $ S$ is the "finitely supported" or "compactly supported" sequences. $ S$ is dense in $ \ell^{p}$ for $ 1\le p<\infty.$ The arguments for that aren't much different than what pluricomplex has already posted. But $ S$ is not dense in $ \ell^{\infty}.$ The closure of $ S$ in the infinity norm is the space $ c_{0}$ of sequences that tend to zero.

The space $ S$ has a countable Hamel basis. In fact, we can use the natural basis, $ e_{1}, e_{2},\dots,$ where $ e_{1}\equal{}(1,0,0,\dots,),\ e_{2}\equal{}(0,1,0,\dots),$ and so on.

Here's a claim we can make that uses the fact that $ S$ is dense in $ \ell^{p},$ along with that natural Hamel basis for $ S.$ Suppose that $ f$ is in the dual space of $ \ell^{p},$ for $ 1\le p<\infty.$ That is, $ f$ is a bounded linear mapping from $ \ell^{p}$ to $ \mathbb{C}.$ Then there exists a sequence $ y$ such that for all $ x\in\ell^{p},\ f(x)\equal{}\sum_{k\equal{}1}^{\infty}y_{k}x_{x}.$

Once you've established that, you can proceed to showing that $ y\in\ell^{q}$ where $ \frac{1}p\plus{}\frac{1}q\equal{}1.$

You can use the same argument to establish the dual of $ c_{0}.$ But the dual of $ \ell^{\infty}$ cannot be reached by such methods.


In quadrilateral ABCD, AB>CD. E and F are on the extensions of BC and AD such that $ \frac{AF}{FD}\equal{}\frac{BE}{EC}\equal{}\lambda$. Line EF meets AB and CD at P and Q respectively.

    Show that $ \frac{AB}{CD}\equal{}\lambda \Longleftrightarrow \angle{APQ}\equal{}\angle{PQD}$

:blush:
In some cases,\[ \angle PQD \ge 90^0 ,\angle APQ \le 90^0
\]

In what cases? In my sketch, there is nothing wrong with the problem.


AMC
AIME
USA(J)MO
USAMO
email
AMC 8
AMC 10

[quote="BOGTRO"]Will the forum be locked because of the AMC 8 like it ws for the other AMCs?[/quote]

No, this is for AMC 10/12/AIME/USAMO discussion

This forum has always been for discussion of all AMC contest levels:  AMC 8, AMC 10, AMC 12, AIME, USAMO, and next spring USAJMO.

As long as AMC 8 participants do not discuss the 2009 AMC 8 problems or solutions on this forum until November 25, I see no need to lock the forum.  Please the Teachers Manual, Section IV, Item 15, on page 4, and also the fine print at the bottom of the front page of the contest booklet itself:
"The publication, reproduction, or communication of the problems or solutions of the AMC 8 during the period when students are eligible to participate seriously jeopardizes the integrity of the results.  Dissemination via copier, telephone, email, web or media of any type during this period is a violation of the competition rules."

Steve Dunbar,
Director, American Mathematics Competitions

Are you allowed to say what you got?

Like if you just say "I got ___" would that be bad?

How would you even know what you got? I don't think you are even supposed to know what you got until some predetermined date.

[quote="AIME15"]Can't discuss anything about the test: this includes difficulty, relative difficulty, your expected score, the problems, or anything else related to the 2009 AMC 8.[/quote]
(from [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1685886#1685886]here[/url])

[quote="AMCDirector"][quote]Can we privately discuss with a friend who we know has already taken the test?[/quote]

No.

The rule is:
"The publication, reproduction, or communication of the problems or solutions of the AMC 8 during the period when students are eligible to participate seriously jeopardizes the integrity of the results. Dissemination via copier, telephone, email, web or media of any type during this period is a violation of the competition rules."

Steve Dunbar
Director, American Mathematics Competitions[/quote]

My teacher said that we were just not allowed to discuss on Nov. 17th, because that was when you were supposed to have taken the exam.

Some schools take it later because of conflicts.


Susie's parents paid $ x$ dollars for their house and Brian's parents paid $ \$30,000$ less for their house than Susie's parents paid. How many dollars did Brian's parents pay for their house? Express your answer as an algebraic expression in terms of $ x$.

um..if x dollars was paid for a house
and one was 30000 less

isnt it
$ \boxed{x\minus{}30000}$?


vector
algebra
polynomial
linear algebra

Hey.
I have teh following problem:
 	Apply the Gramm-Schmidt process to the standard (mononomial) basis vectors in polynomial space P3, with the inner product defined on the interval [-1,1]. Choose basis vectors so that p(1) = 1 and draw a sketch plot showing all four of them on the same axes over the interval [-1,1].

The question that i have is what does : Choose basis vectors so that p(1) = 1 mean?
I figure that the basis that you should should equal 1 at the point 1, but just to be sure i decided to post here, since that notation ive never seen.
So i have
1,x,x^2,x^3 as my basis(original) right?

Thx in advance.

That's the natural place to start, and your calculations will be helped by the fact that the odd powers are orthogonal to the even powers. However, the request that $ p(1)=1$ is meant to apply to the final, orthogonal polynomials and not to the starting set. One thing to note: if you do that, the set you finish with will be orthogonal but not orthonormal. It's not worth the hassle to make it orthonormal - who needs those square roots?

For a problem this small, the Gram-Schmidt process is reasonable. For a larger (higher degree) set of polynomials, there are much more efficient ways of finding this sequence of orthogonal polynomials, which is known as the Legendre polynomials.

Yeah, thx alot.
I made my basis orthonormal since i thought p(1) was talking about which polynomial to use for the beginning of the Gram Schmidt process.


algebra
polynomial
complex numbers
complex analysis
complex analysis unsolved

$a_{0} \geq \dots \geq a_{n} >0$.
Show that $a_{0}+a_{1}z+\dots+a_{n}z^{n}$ has no root in the open unit disc.

Hint 1:
[hide]
Let $b_0 = a_0 - a_1$, $b_1 = a_1 - a_2$, ..., $b_{n-1} = a_{n-1} -a_n$, $b_n =a_n$. Then all $b_i$ are non-negative, $b_n$ is positive. The polynomial equals
\[ f(z) = b_n (1 + z + \dots + z^n) + b_{n-1} (1 + z + \dots + z^{n-1}) + \dots + b_1 (1 + z) + b_0. \]
[/hide]
Hint 2:
[hide]
Multiply $f(z)$ by $1 - z$. Compute the real part of each summand.
[/hide]


I am auctioning off 10 hours of 1-1 online tutoring via EBay.

[url=http://cgi.ebay.com/ws/eBayISAPI.dll?ViewItem&item=5550171129]Click here[/url] to see the auction.

All proceeds from this auction will be donated to the Art of Problem Solving Foundation to support the USAMTS.

If you have any questions about the auction, please post them here.

No.  It probably closed sometime in 2005.


A quadrilateral ABCD is inscribed in a circle so that [AB] is a diameter.
Suppose that AB and CD intersects at I, AD and BC at J, AC and BD at K and  consider a point N on [AB].
[color=red]Prove that IK is perpendicular to JN if and only if N is the midpoint of [AB].
 [/color]

if you know polars, there is a one line soln using http://www.mathlinks.ro/Forum/viewtopic.php?t=18781 (last post) -- i never thought this lemma could be useful

if you don't know polars, i'm sure there is a trivial proof


As you have found the mathematical model can be expressed in the form H = ab^n.
The relationship between fatling distance(s) and time(t) can be expressed as s = 0.5g^t

g the acceleration due to gravity on Earth is considered to be constant at 9.8ms^-2

Using the above information, calculate the total distance that your ball will travel and how long this will take.

First of all, you shouldn't post the same problem in three different forums. Secondly, this is nowhere near pre-olympiad difficulty. Will a moderator please lock these and the other two identical topics in getting started and intermediate?


calculus
derivative
inequalities
trigonometry
inequalities proposed

Let x,y,z are positive reals.
Prove that
$x^6+y^6+z^6+15(x^3y^3+y^3z^3+z^3x^3)\geq 16xyz(x^3 +y^3 +z^3)$

WLOG $xyz=1$.
Take, $x=\sqrt[3]{a}$, etc., so $abc=1$.
So we get:
$a^2+b^2+c^2+15(ab+bc+ca) \geq 16(a+b +c)$
Can anyone continue from here? Or give another solution maybe

You can show using mixing variables the last inequality. Or at least, you can show it reduces to the case when $a=b$ and then it's just huge computation.

[quote="perfect_radio"]What do you mean by mixing variables?
Do you mean the following:

Take $f(a,b,c)=a^2+b^2+c^2+15(ab+bc+ca)-16(a+b +c)$.
Then prove that $f(a,b,c) \geq f(\sqrt{ab},\sqrt{ab},c)$

?
If yes, i don't know whether it is true, nor how to prove it[/quote]
it is true if c is the largest among a,b,c.
You can try to prove it.

I used derivatives to solve this problem.
I could not find triky method. :(

then why did you say it is not hard but tricky?

We can also solve this ineq by n-1 Equal Variable Principle.
We write it in the form
$15(a+b+c)^2-32(a+b+c) \ge 13(a^2+b^2+c^2)$,
where $abc=1$. For $abc=1$ and fixed $a+b+c$, the sum $a^2 +b^2+c^2$ is maximal for
$a=b \le c$.


conics
hyperbola
analytic geometry
graphing lines
slope
calculus
calculus computations

i need your help!

What is the equation for the tangent to the hyperbola x^2 - y^2 = 12 at the point (4, 2)?

The first part of the story is to find the slope. To this end, the easiest way is to use implicit differentiation. We get $2x-2yy'=0$, so $y'=\frac xy=\frac 42=2$ at our point. The second step is to find the equation of a line with slope $2$ passing through our given point $(4,2)$. It is $y-2=2(x-4)$ or, after opening the parentheses, $y=2x-6$.

Generally speaking, the equation for the tangent to the graph of $Ax^{2}+By^{2}=C\ (ABC\neq 0).$ at the point $(x_{0},\ y_{0})$ is $Ax_{0}x+By_{0}y=C.$
How about for the graph of $ax^{2}+2hxy+by^{2}+cx+dy+e=0?$

If my computations are accurate, then the tangent to $a x^{2}+2hxy+b y^{2}+cx+dy+e = 0$ at $\left( x_{0}, y_{0}\right)$ is
\[a x_{0}x+h \left( x_{0}y+x y_{0}\right)+b y_{0}y+c \cdot \frac{x+x_{0}}2+d \cdot \frac{y+y_{0}}2+e = 0 .\]
Setting $h=c=d=0$, we get the same answer as kunny.


algorithm
number theory
relatively prime
Euclidean algorithm

Prove that for each positive integer n, there are pairwise relatively prime integers $ k_{0}, k_{1},...,k_{n}$, all strictly greater than 1, such that $ k_{0}k_{1}...k_{n} \minus{} 1$ is the product of two consecutive integers.

you're going crazy here aren't you.

Find a sequence =D.

Umm yea Im pretty sure I got a 0 on this question on the real test, when I took it in 2008.

However, I think, now, I can atleast salvage 1 point on this:

Ok, so (k0)(k1)(k2)...(kn)= x(x+1)+1=x^2+x+1.  Since3*7-1=4*5, we will use this as our induction/assumption step.

Since we know that x^4+x^2+1=(x^2+x+1)(x^2-x+1), we just have to prove that x^2-x+1 is relatively prime to x^2+x+1---I think this will mean that x^2-x+1 will be relatively prime to each of the numbers that are multiplied together to x^2+x+1---this is just a intuitive thought, and I have no proof for this. Also, x^2+x+1 works because of our assumption.

Ok, now I need some help.  First of all, is the above assumption true^ (I tried some smaller cases and they seemed to work out).  If its not, then this proof is totally bogus (will probably be only 1/7 just for proving the basic steps of induction)...if it is, can someone give me some hints how to finish this.  Thanks.

well....if you want to prove $ (x^2\minus{}x\plus{}1,x^2\plus{}x\plus{}1)\equal{}1$, I guess I can do it for you if it makes you feel any better....
the difference of the two is $ 2x$ so we just have to make sure that both $ x^2\minus{}x\plus{}1$ and $ x^2\plus{}x\plus{}1$ are not divisible by $ 2$ or $ x$; in theory this isn't enough because we would have to check divisibility for all of the factors of $ x$, but it's ok because both $ x^2\minus{}x\plus{}1$ and $ x^2\plus{}x\plus{}1$ are congruent $ 1\mod x$, meaning all factors of $ x$ will also be congruent $ 1\mod k$, where $ k$ is a factor of $ x$....$ \mod 2$ is an easy case to check, I'll leave that to you  :lol:

Or: from the euclidean algorithm, $ gcd(x^2 \minus{} x \plus{} 1, x^2 \plus{} x \plus{} 1) \equal{} gcd(x^2 \plus{} x \plus{} 1, 2x)$. Clearly $ x$ does not divide $ x^2 \plus{} x \plus{} 1$...unless x=1, but then the two are consecutive numbers.

Then as $ x^2 \plus{} x \plus{} 1 \equal{} x(x \plus{} 1) \plus{} 1$, which is always odd, we conclude the two are relatively prime.


Take any point $P_1$ on the side $BC$ of a triangle $ABC$ and draw the following chain of lines: $P_1P_2$ parallel to $AC$; $P_2P_3$ parallel to $BC$; $P_3P_4$ parallel to $AB$ ; $P_4P_5$ parallel to $CA$; and $P_5P_6$ parallel to $BC$, Here, $P_2,P_5$ lie on $AB$; $P_3,P_6$ lie on $CA$ and $P_4$ on $BC$> Show that $P_6P_1$ is parallel to $AB$.

Basically, Construct C' so that C'ACB is a parallelogram, B' so that BC'B"A' is a paralleogram, ..

Dear Mathlinkers,
in th epast I have prove this result according the barycentric calculus...

Sincerely
Jean-Louis

Remark that $P_5P_6 C P_4$ and  $P_2P_1 C P_3$ are paralleogram besides$P_1P_2 B $ and $CP_3P_4$ are homothetic then $ P_2B=P_3P_4,P_5P_6=P_4C$ and $P_5P_6 \parallel  CP_4 (_1) \implies P_1P_2 B $ and $CP_3P_4$ are congruent  thus $ BP_1 =P_4C$  with $(_1)$  yield $P_5P_6P_1B$ is parallelogram

my solution doesn't require any clever search for parallelograms...
[b]follow carefully[/b]
$\frac{AP_6}{P_6C}$=$\frac{AP_5}{P_5B}$=$\frac{CP_4}{P_4B}$=$\frac{CP_3}{P_3A}$=$\frac{BP_2}{AP_2}$=$\frac{BP_1}{CP_1}$([b]all by using basic proportionality theorem[/b])
Now by converse of $B.P.T$ the result follows.

[quote=kk108]Actually its just got to do with angle -chasing .[/quote]

Can you post a solution using angle-chasing?

[hide=only one parallelogram and you are done!] mark angles of triangle as $A,B$ and $C$ notice we have $P_{2}P_{1}\parallel AC$ then $\angle{P_{2}P_{1}B}=C$ and we also have $P_{1}P_{6}\parallel AP_{1}$ and $P_{2}P_{1}\parallel AP_{6}$ then $P_{1}P_{2}AP_{6}$ a parallelogram\\
which gives $\angle{P_{2}P_{1}P_{6}}=A$, so we get $\angle{P_{6}P_{1}C}=B$\\
which makes $P_{1}P_{6}\parallel AB$ $\blacksquare$[/hide]


trigonometry
linear algebra
matrix
real analysis
real analysis unsolved

Find sum $ \sum_{n \equal{} 1}^{\infty}{\arctan\left( \frac {2}{n^2}\right)}.$

[hide]$ \frac {3}{4}\pi$[/hide]

[quote="Carcul"]This sum is easy. Subsitute that 2 for a 1 and you get something much more interesting.[/quote]
what is ur mean?can u elaborate it?thanks a lot

[quote="Yuriy Solovyov"]Find sum $ \sum_{n = 1}^{\infty}{\arctan\left( \frac {2}{n^2}\right)}.$
[/quote]
I think, it isn't very hard problem.

We have $ \sum_{n = 1}^{\infty}{\arctan{\frac {2}{n^2}} = \arctan 2 + \arctan 0.5 + \sum_{n = 3}^{\infty}{\arctan{\frac {2}{n^2}}} = \frac {\pi}{2} + \sum_{n = 3}^{\infty}{\arctan{\frac {2}{n^2}}}}$.

Let $ S_n = \sum_{k = 3}^{n}{\arctan\left( \frac {2}{k^2}\right)}$.
I will prove that ${ S_n = \arctan\left( \frac {n^2 - n - 2}{n^2 + 3n}\right)}$ by induction.
The base $ n = 3$ is clear.
I will use the formula: $ \arctan x + \arctan y = \arctan {\frac {x + y}{1 - xy}}$, if $ 0 < x,y < 1$.

So, $ S_{n + 1} = S_n + \arctan {\frac {2}{(n + 1)^2}} = \arctan {\frac {\frac {n^2 - n - 2}{n^2 + 3n} + \frac {2}{(n + 1)^2}} {1 - \frac {2n^2 - 2n - 4}{(n + 1)^2(n^2 + 3n)}}}$ = $ \arctan{\frac {n^4 + n^3 - n^2 + n - 2}{n^4 + 5n^3 + 5n^2 + 5n + 4}} = \arctan{\frac {(n^2 + 1)(n^2 + n - 2)}{(n^2 + 1)(n^2 + 5n + 4)}} = \arctan{\frac {n^2 + n - 2}{n^2 + 5n + 4}} = \arctan{\frac {(n + 1)^2 - (n + 1) - 2}{(n + 1)^2 + 3(n + 1)}}$.

So, it's true that ${ S_n = \arctan\left( \frac {n^2 - n - 2}{n^2 + 3n}\right)}$.

Thus, $ lim_{n\to\infty} {S_n} = \arctan 1 = \frac {\pi}{4}$ and the answer is $ \frac {\pi}{2} + \frac {\pi}{4} = \frac {3\pi}{4}$.

[quote="Huanggang"]what is ur mean?can u elaborate it?thanks a lot[/quote]

Compute $ S \equal{} \sum_{n\equal{}1}^{\infty}\arctan \left(\frac{1}{n^2}\right)$.

Use the formulas
$ \tan(x_1\plus{}...\plus{}x_m)\equal{}\frac{1}{i}\, \frac{\prod\limits_{k\equal{}1}^m \Big(1\plus{}i\tan x_k\Big)\minus{}\prod\limits_{k\equal{}1}^m (1\minus{}i\tan x_k)}{\prod\limits_{k\equal{}1}^m (1\plus{}i\tan x_k)\plus{}\prod\limits_{k\equal{}1}^m (1\minus{}i\tan x_k)}$
and

$ \prod_{k\equal{}1}^\infty \left(1\minus{}\frac{z^n}{k^n}\right)^{\minus{}1}\equal{}\prod_{k\equal{}0}^{n\minus{}1} \Gamma\left(1\minus{}e^{\frac{2\pi i k}{n}} \, z\right)$ .

$ \tan\left(\sum_{k\equal{}1}^m \arctan \frac{z^n}{k^n}\right)\equal{}\frac{1}{i}\, \frac{\prod\limits_{k\equal{}1}^m \Big(1\plus{}i\frac{z^n}{k^n}\Big)\minus{}\prod\limits_{k\equal{}1}^m \Big(1\minus{}i\frac{z^n}{k^n}\Big)}{\prod\limits_{k\equal{}1}^m \Big(1\plus{}i\frac{z^n}{k^n}\Big)\plus{}\prod\limits_{k\equal{}1}^m \Big(1\minus{}i\frac{z^n}{k^n}\Big)}$

Divide nominator and denominator by $ \prod\limits_{k\equal{}1}^m \Big(1\minus{}i\frac{z^n}{k^n}\Big)$ and $ \prod\limits_{k\equal{}1}^m \Big(1\plus{}i\frac{z^n}{k^n}\Big)$

and let $ m$ tend to $ \infty$. Then $ \tan\left(\sum_{k\equal{}1}^\infty \arctan \frac{z^n}{k^n}\right)\equal{}\frac{1}{i}\frac{A\minus{}B}{A\plus{}B}$ ,

whereat $ \begin{matrix} A \\ B \end{matrix}\equal{}\prod_{k\equal{}1}^\infty \Big(1\mp i\frac{z^n}{k^n}\Big)^{\minus{}1}\equal{}\prod_{k\equal{}0}^{n\minus{}1} \Gamma\left(1\minus{}e^{\frac{(4k\pm 1)i\pi}{2n}} z\right)$ .

In particular for $ z\equal{}1$ and $ n\equal{}2$

$ A\equal{}\Gamma\left(1\minus{}e^{i\frac{\pi}{4}}\right)\,\Gamma\left(1\minus{}e^{5i\frac{\pi}{4}}\right)\equal{}\frac{e^{i\frac{\pi}{4}}}{\sin\left(\pi e^{i\frac{\pi}{4}}\right)}$
and
$ B\equal{}\Gamma\left(1\minus{}e^{\minus{}i\frac{\pi}{4}}\right)\,\Gamma\left(1\minus{}e^{3i\frac{\pi}{4}}\right)\equal{}\frac{e^{\minus{}i\frac{\pi}{4}}}{\sin\left(\pi e^{\minus{}i\frac{\pi}{4}}\right)}$ .

So $ \tan\left(\sum_{k\equal{}1}^\infty \arctan \frac{1}{k^2}\right)\equal{}\frac{1}{i}\, \frac{A\minus{}B}{A\plus{}B}\equal{}\frac{1\minus{}t}{1\plus{}t}$ with $ t\equal{}\frac{\tanh \frac{\pi}{\sqrt{2}}}{\tan \frac{\pi}{\sqrt{2}}}$ .

For real numbers a and b, one has \[ \arctan a \minus{} \arctan b \equal{} \arctan\left(\frac{a\minus{}b}{1\plus{}ab}\right).\]Therefore, $ \arctan 2/n^2 \equal{} \arctan(n\plus{}1)\minus{}\arctan(n\minus{}1)$, and thus

\[ \begin{aligned}
\sum_{n\equal{}1}^{N}\arctan\left(\frac{2}{n^2}\right) &
\equal{} \sum_{n\equal{}1}^{N}\left(\arctan(n\plus{}1) \minus{} \arctan(n\minus{}1)\right) \\ &
\equal{} \sum_{n\equal{}2}^{N\plus{}1}\arctan(n) \minus{} \sum_{n\equal{}0}^{N\minus{}1}\arctan(n) \\ &
\equal{} \arctan(N\plus{}1)\plus{}\arctan(N)\minus{}\arctan(1)\minus{}\arctan(0)
\end{aligned}\]

Taking the limit $ N\to\infty$, this gives $ 2\arctan(\infty)\minus{}\frac{\pi}{4}\minus{}0 \equal{} \pi\minus{}\frac{\pi}{4} \equal{} \frac{3\pi}{4}$.

Could you derive that $ \arctan$ formula please? :)

What happens when you take the tangent of both sides, and use the appropriate addition or subtraction formula for the tangent?

[quote="Andyhuangofnj"][quote="CaptainCanada"]I think this can be solved by turning it into an appropriate contour integral. If you want, I can work out the details and post it.[/quote]
it is very hard,but i have a good idea:)[/quote]
this answer is good... ...


calculus
derivative
algebra
system of equations
algebra unsolved

Solve the system of equations $ \ln(1 \plus{} x) \minus{} \ln(1 \plus{} y) \equal{} x \minus{} y$; $ x^2 \plus{} xy \plus{} y^2 \equal{} 3$.

i think you use the equation 2 to observe x and y ( 2 case : x<0 and x>0);
and it's easy if  you see carefully, the equation 1 can write : ln(1+x) -x = ln(1+y) -y  ( derivative) . i think that's enough if you dont confuse.

It is easy to chek, that $ f(x)\equal{}x\minus{}ln(1\plus{}x)$ increase for x>0 and decrease for x<0.
Therefore if x and y positive, then x=y. It give $ x\equal{}y\equal{}1$.
These equation had 3 solutions.
2. $ x_0<0, 1<y_0<2$
3. $ (x,y)\equal{}(y_0,x_0.$


Pretty simple. See the sketch below for a construction without words.


algebra
polynomial
inequalities
quadratics
algorithm
complex analysis
complex analysis unsolved

Is there a simple condition for checking whether a polynomial such as $f(x)=a_{n}x^{n}+\ldots a_{1}x+a_{0}=0$ has complex roots or not? Any necessary and sufficient for $f(x)=0$ to have No Complex roots?

[quote="t0rajir0u"]WLOG $a_{n}= 1$.

Let $S_{i}= \frac{ (-1)^{i}a_{n-i}}{{n \choose i}}$ be the "$i^{th}$ symmetric mean" of the roots.

If all of the roots are real, then $S_{i}$ satisfies [b]Newton's Inequality:[/b]

$S_{i-1}S_{i+1}\le S_{i}^{2}$

The inequality (for all $i$) is a [i]necessary[/i] condition for the polynomial to have [i]no[/i] complex roots.  I do not know if it is [i]sufficient[/i], and was actually going to post this problem at some point.

(Edit:  By the way, for a quadratic this becomes the discriminant.  Isn't that cool?)[/quote]

Thanks a lot and that seems helpful. Where can I find the proof of the above?  :roll:

Hmm, I don't know.  I just always took it as a very complicated consequence of Rearrangement.  Apparently there's a proof using Rolle's Theorem.

Alternatively, you can use the [url=http://mathworld.wolfram.com/SturmFunction.html]Sturm algorithm[/url] to determine the exact number of real roots (I assume that your polynomial has real coefficients).


let S(n) be the sum of all different natural divisors of odd natural number n>1 (including n and 1).
Prove that (S(n))^3 < n^4..


A rectangle $ABCD$ lies on a semicircle $E$ of radius $r$ such that $AB$ is on the diameter $E$ and the points $C$ and $D$ lies on the circumference. Determine the maximum perimeter of $ABCD$.

let the sides of the rectangle be 2a, and b...we have the constraint that $r=\sqrt{a^{2}+b^{2}}$ and we want to maximize $A=2ab$

blah blah am-gm

$\frac{a^{2}+b^{2}}{2}=\frac{r^{2}}{2}\ge\sqrt{a^{2}b^{2}}=ab=\frac{A}{2}$

so the max is $A=r^{2}$ where equality occurs when $a=b$

but thats maximizing the area...does that does not necessarily maiximize the perimeter...

oh sorry i did not read it carefully...well we still have

a^2+b^2=r^2, and we want the max of 2(2a+b)

by cauchy, we have that

$(a^{2}+b^{2})(2^{2}+1^{2})\ge (2a+b)^{2}$

so the answer is $2\sqrt{5r}$, equality when $\frac{a}{b}=2$

[quote="SM4RT"]isnt it

$2r\sqrt{5}$ or am i miscalculating...nice solution though.. easy to follow..[/quote]

oh yea...woops...lol


[quote="#H34N1"]Two different positive numbers $a$ and $b$ each differ from their reciprocals by $1$. What is $a+b$?[/quote]

[hide]$x-\frac{1}{x}=1$

There are 2 solutions to this equation so those will be $a$ and $b$

$x^{2}-x-1=0$

The sum of the roots is $1$ by vetas thing.[/hide]

[quote="#H34N1"]What is the vetas thing? I don't know[/quote]

Vieta's formula. (Look it up.)

It says more generally that for a polynomial
\[P(x)=a_{0}x^{n}+a_{1}x^{n-1}+\ldots+a_{n}\]
we have that
\[{\Pi_{m}=(-1)^{m}( \frac{ a_{m}}{a_{0}}}) \]
where $\Pi_{m}$ is the $\text{m}^\text{th}$ [url=http://mathworld.wolfram.com/SymmetricPolynomial.html]symmetric polynomial[/url]. This follows from direct factorization into the form $P(x)=a_{0}(x-r_{1})(x-r_{2})\cdots(x-r_{n})$, and looking at the coefficients of the expansion.

[quote="Altheman"][quote="cincodemayo5590"]It says more generally that for a polynomial
\[P(x)=a_{0}x^{n}+a_{1}x^{n-1}+\ldots+a_{n}\]
we have that
\[{S(m)=(-1)^{m}( \frac{ a_{m}}{a_{0}}}) \]
where $S(m)= \sum_\text{Sym.}\prod^{m}r_{i}$. This follows from direct factorization into the form $P(x)=a_{0}(x-r_{1})(x-r_{2})\cdots(x-r_{n})$, and looking at the coefficients of the expansion.[/quote]

careful with symmetric sum, that is not correct as is...

[if we have a,b,c] sum_{sym} abc=6abc[/quote]

Oh, oops ok, I'm too lazy to type the correct one out so i'll just put the link to mathworld.  :D

[quote="ajacoby"]The problem specifies that the two numbers are both positive.  The solutions above give you one positive and one negative.  Add the positive solution to its reciprocal.

[hide]$x = \frac{1+\sqrt{5}}{2}$ or $\frac{2}{1+\sqrt{5}}$
$x+\frac{1}{x}= \sqrt{5}$[/hide][/quote]

Oh, right. The equation should be $\left| x-\frac{1}{x}\right| =1$.


group theory
abstract algebra
superior algebra
superior algebra solved

Let $H$ be a subgroup of a group $G$ s.t. $g^{-1}hg\in{H}\ \forall{g\in{G}}\ \forall{h\in{H}}$. Show that every left coset $gH$
is the same as the right coset $Hg$.

But this is obvious, isn't it ? $\forall h, hg \in gH \Rightarrow Hg \subseteq gH$, and same thing the other way round.

Cezar, aren't you a little bit to old to wonder about normal subgroups?


trigonometry
trig identities
Law of Cosines
geometry
Law of Sines
Pythagorean Theorem

$AB$ is the diameter of a circle centered at $O$. $C$ is a point on the circle such that angle $BOC$ is $60^\circ$. If the diameter of the circle is $5$ inches, the length of chord $AC$, expressed in inches, is:

$\text{(A)} \ 3 \qquad \text{(B)} \ \frac{5\sqrt{2}}{2} \qquad \text{(C)} \frac{5\sqrt3}{2} \ \qquad \text{(D)} \ 3\sqrt3 \qquad \text{(E)} \ \text{none of these}$

AHSMEs are cool!  :D

[hide]Notice that triangle $ABC$ is a right triangle, because one of the sides is the diameter of the circle and it is inscribed. Also, triangle $BOC$ is isoscoles, so $\angle CBO=60$.

Thus, $ABC$ is a 30,60,90 right triangle and $AC=5\cdot \frac{\sqrt3}{2}=\frac{5\sqrt3}{2}$[/hide]

I didn't understand your reasoning.  :huh:  If angle BOC is 60 degrees and angle BCO is also 60 degrees, then Triangle BCO is not isoceles it must be equilateral.  ;)

Anyway, I did it in this fasion::

Since, Angle BOC is equal to 60 degrees than Angle AOC is equal to 120. Since, the radius OC is half the diameter it is 2.5 and radius OA is 2.5. Since we have to adjacent sides and their angle, the law of Cosines can be used to determine the third side.

which is 4.330127 . . .
the same as Answer (C)

 :D

Equilateral triangles are just a special type of isosceles triangles.  Isosceles triangles just need at least two equal sides, not only two...

Anyway, I did it yet a third way.  I found that $\angle AOC=120$ first.  Then, since two sides of triangle $AOC$ are radii, it is isosceles, with angles $120$, $30$, and $30$, and sides with lengths $2.5$, $2.5$, and $AC$.  Then I used the Law of Sines, giving $\frac{sin30}{2.5}=\frac{sin120}{AC}$  Working this out, you get $AC=2.5\sqrt{3}=\frac{5\sqrt{3}}{2}$

[hide]Diameter is 5. Notice that $\triangle BOC$ is equilateral since, thus $CB$ is equal in length to the radius, $\frac{5}{2}$. Also $\triangle ACB$ is a right triangle, with $\angle C$ being right. By Pythagorean Theorem,
$AC=\sqrt{\frac{100}{4}-\frac{25}{4}}=\frac{\sqrt{75}}{2}=\frac{5\sqrt{3}}{2}$

$\boxed{C}$

EDIT:  :fool: 30-60-90 triangle[/hide]

[hide]Take triangle COA.  It is isosceles and it has a $120$ degree obtuse angle, thus it also has two $30$ degree angles.  Split it in half to get two 30-60-90 triangles.  Then, $2.5$ is the 2 side (hypoteneuse) and $1.25\sqrt{3}$ is the opposite side of 60 degrees, so the length of the chord AC is twice that, or $\boxed{C}$[/hide]

[hide]
Drop an altitude on $\triangle AOC$ so that you get a right triangle. The base of the right triangle is $b=\frac{5}{2}+\frac{5}{4}=\frac{15}{4}$ from the $30^\circ-60^\circ-90^\circ$ triangle and the radius of the circle. The height of the right trangle is $\frac{5\sqrt{3}}{4}$.
Using the pythagorean theorem:
$d^2=b^2+h^2=\left(\frac{15}{4}\right)^2+\left(\frac{5\sqrt{3}}{4}\right)^2=\frac{225}{16}+\frac{75}{16}=\frac{300}{16}$
$d=\sqrt{\frac{300}{16}}=\boxed{\frac{5\sqrt{3}}{2}}$
[/hide]


My Ti-89 TItanium won't divide into decimals, it stays in fraction form..

Can someone please help me to reverse this so that when i compute results i get a decimal result and not a fraction result.
I checked the Ti Guide but so far it doesnt help.

Press [<>] [enter], instead of just [enter] when you want to calculate stuff. ([<>] stands for the diamond key thingy)

Also, you can put a decimal point after a number to force the calculator to convert it all to decimal.  e.g. for 5/9, type `5./9' or `5/9.' (`5.0/9' and `5/9.0' work as well).

Well...it also depends on the what mode you have it in.  You can set it to either auto, exact, or approximate.  Exact will give you a fraction every time unless you click [<>] [enter].  Approximate always gives you a decimal.  Auto is a mix of the two and does what countersixte said.


I need free online travel brochures for the following cities:
Boston
New York City
Miami
Chicago
Phoenix
Denver
Seattle
Los Angeles
Honolulu
Las Vegas

Can somebody please tell me a website where I can find these?


function
geometry
geometric transformation
integration
real analysis
real analysis unsolved

Consider the space of $C^{\infty}$ real functions defined on $R$ and that are bounded, with the norm of uniform convergence. Is there a function $f$ such that the vectorial space of its natural translations $f(x+n)$ is dense in this space? Is there a function such that $f, ff, fff,...$ (composition here) is dense in this space?

there are many things I can't understand :
1/ what is the vectorial subspace of its natural translation
2/ the space of bounded $C^{\infty}$ function with the uniform norm is not separable ?

For your first question, alekk, I think harazi is simply referring to the space spanned by the function $f_n$ defined by $f_n(x)=f(x+n)$. As for the second question, I don't get that either :).

A quick way to see that the space $E$ of bounded continuous functions (it's the same thing, because the smooth functions are dense in $E$) with the $\sup$ norm defined on the real line is not separable would be to consider the map $T: E\to\ell^\infty,\ T(f)=\left(\int_0^1f,\int_1^2f,\ldots\right)$. This is a continuous map of $E$ onto a dense subspace of $\ell^\infty$, and $\ell^\infty$ is clearly not separable (just consider all binary sequences; there are uncountably many of them, and the distance between each two is $1$).

I'm blaming any mistakes I may have made on the fact that it's really late here :D. So sorry if I'm posting nonsense here.

[quote="grobber"]
A quick way to see that the space $E$ of bounded continuous functions (it's the same thing, because the smooth functions are dense in $E$) with the $\sup$ norm defined on the real line is not separable would be to consider the map $T: E\to\ell^\infty,\ T(f)=\left(\int_0^1f,\int_1^2f,\ldots\right)$. This is a continuous map of $E$ onto a dense subspace of $\ell^\infty$, and $\ell^\infty$ is clearly not separable [/quote]

Grobber, could you please explain why the image $T(E)$ is a dense subspace of $\ell^\infty$ and why must it be countable ?

Thanks


modular arithmetic
Euler
number theory unsolved
number theory

Let $n$ be a positive integer, and let $A_{n}$ be the the set of all positive integers $a\le n$ such that $n|a^{n}+1$.
a) Find all $n$ such that $A_{n}\neq \emptyset$
b) Find all $n$ such that $|{A_{n}}|$ is even and non-zero.
c) Is there $n$ such that $|{A_{n}}| = 130$?

OK, let's solve the first question ...

[b]For all odd $n$ we have $A_n$ non-empty[/b]
If $n = 1$ then clearly $A_1 = \left\{ 1 \} \right.$.  If $n > 1$ is an odd positive integer, we let $a = n - 1$. Then
\begin{eqnarray*} a^n + 1 & \equiv & \left( n - 1 )^n + 1 \right.\\ & \equiv & ( - 1 )^n + 1\\ & \equiv & \left. 0 ( \mbox{ mod } n \right) \end{eqnarray*}
Therefore for all odd positive integer $n$ we have $( n - 1 ) \in A_n$.

[b]For an even $n$ we have $A_n$ non-empty if and only if $n = 10$[/b]
Suppose that $n = 2 m$ for some positive integer $m$ and that
\begin{eqnarray*} 2 m & | & a^{2 m} + 1 \end{eqnarray*}
for some positive integer $1 \leq a \leq 2 m$.  For all positive integer $k$ we have $k^2 + 1 \not \equiv 0 ( \mbox{ mod } 4 )$.
Therefore $a^{2 m} + 1 \not \equiv 0 ( \mbox{ mod } 4 )$ and it follows that $m$ is odd (otherwise $4|2 m|a^{2 m} + 1$)
For a positive integer $k$ any odd prime divisor of $k^2 + 1$ is $\equiv 1 ( \mbox{ mod } 4 )$. (Legendre)
Therefore if $p$ is a prime such that $p|m$ then $p > 2$ and $p|m| \left( a^m )^2 + 1 \right.$ hence $p \equiv 1 ( \mbox{ mod} 4 )$.
It follows that
\begin{eqnarray*} 2^{2 \omega ( m )} = 4^{\omega ( m )} & | & \prod_{p^{\alpha} \| m} ( p - 1 ) \cdot p^{\alpha - 1} = \varphi ( m ) = \varphi ( 2 m )\\ 2^{2 \omega ( m )} & | & \varphi ( 2 m ) \end{eqnarray*}
Now we have
\begin{eqnarray*} a^{2 m} & \equiv & \left. - 1 ( \mbox{ mod } 2 m \right)\\ a^{4 m} & \equiv & 1 ( \mbox{ mod } 2 m ) \end{eqnarray*}
This implies that $4 m| \varphi ( 2 m )$ or $\varphi ( 2 m ) |4m$ (since $( a, n ) = 1$ hence $a \in \mathbb{Z}_n^{\times}$).
The case $4 m| \varphi ( 2 m )$ is clearly impossible since $4 m > 2 m > \varphi ( 2 m )$ for all positive integer $m.$
Therefore $2^{2 \omega ( m )} | \varphi ( 2 m ) | 4 m$, this implies that $2^{2 \omega ( m ) - 2} |m \Longrightarrow 2 \omega ( m ) - 2 = 0$ since $m$ is odd.
Therefore $\omega ( m ) = 1$ and $n = 2 p^{\alpha}$ for some prime $p$ and a positive integer $\alpha$.
Again this means that $\left. \varphi ( 2 p^{\alpha} \right) |4p^{\alpha}$ or that $\left. 4 p^{\alpha} | \varphi ( 2 p^{\alpha} \right)$. The later being impossible we must have
\begin{eqnarray*} \left. \varphi ( 2 p^{\alpha} \right) & | & 4 p^{\alpha}\\ ( p - 1 ) \cdot p^{\alpha - 1} & | & 4 p^{\alpha}\\ p - 1 & | & 4 p \end{eqnarray*}
Therefore $p = 5$. Hence $n = 10$. We check that $3 \in A_{10} .$

[b]We have $A_n \neq \varnothing$ if and only if $n$ is odd or $n = 10$[/b]

I wonder, if there is a shorter solution... :(

I don't know where, but I think something went wrong in your proof {x}. Notice that

$5^{26}\equiv 5^2\equiv -1\pmod{26}$

by Euler's Formula.  I haven't tried very hard yet, but there probably is a shorter solution.

Damn, the wrong part is
[quote] ... This implies that 4 m| $\varphi ( 2 m )$ or $\varphi ( 2 m )$ |4m ... [/quote]

Let $n=2m$, $m$ such that if $p|m$ then $p=4k+1$. Then $\exists q$, $m|q^2+1$. Then $q^{2m}\equiv(-1)^m\equiv-1\pmod m$. Also $(m-k)^{2m}\equiv-1\pmod m$. Either $q$ or $m-k$ is odd. Consider $q$ odd. Then $2m|q^{2m}+1$.

Obviously, $|A_n|$ for such $n$ is even (if $q\in A_n$ then $n-q\in A_n$).

Can someone please solve the last question?

Fun problem.  I don't think my solution is very tricky, maybe there is a shorter way.

For part a, we make the observation that we can always choose $a = n-1$ for when $n$ is odd.  The only exception is 1, but that clearly also works.  Now we consider $n$ being even.  We have two cases.

Assume $4|n$:

This means that $n = 4k$ so that $4k|a^{4k}+1$  Since we know that $gcd(a,n) = 1$, it is clear that $a^{4k}\equiv 1 \pmod 4$  Thus, we have a contradiction.

So we know that even $n$ are of the form $2k$.  Where $k$ is an odd integer.  Now notice that $a^{n}\equiv-1 \pmod n$ with $n$ being even.  So we have that $(a^{\frac{n}{2}})^{2}\equiv-1 \pmod n$  Thus we know that $-1$ is a square in mod $n$.  We conclude that the only prime divisors $p$ of $k$ must have $p \equiv 1 \pmod 4$.  I claim that that is indeed all we need.  Notice that $2k \equiv 2 \pmod 4$, since $k$ is odd.  Thus if we have that $a^{2}\equiv-1 \pmod n$, then $a^{2k}\equiv-1 \pmod n$.  So we would have at least those two solutions.  But of course we could find such an $a$.  It is guaranteed by exactly the condition that the only prime divisors of $k$ are congruent to 1 mod 4.  So we are done.  The answer is all odd $n$ or even $n$ of the form $2k$ where the only prime divisors of $k$ are 1 mod 4.

Now onto part b.  I think the arguement follows fairly naturally after you decide that they asked only whether or not the number of solutions is even, not to actually count the number.  The answer seems to be for all the $n$ of the second type listed above except 2.  The ones that are even and have at least one $a$ that work.  It is easy to prove it for those particular $n$.  Since $n$ is even we have $a^{n}\equiv (-a)^{n}\pmod n$  So each solution must have another corresponding solution.  The only way this could fail is if $a \equiv-a \pmod n$.  But since we have that $gcd(a,n) = 1$ we know this is only possible when $n=2$ which is indeed the only exception.  Now we consider the odd $n$.  We have to count some solutions by using CRT.  We first prime factor $n$ and then consider one of the prime powers at a time.  Find the number of solutions in each of those mods and multiply.  Clearly, we are done if we find that there are an odd number of solutions in each mod on its own.  Now we consider the equation we are solving $a^{n}\equiv-1 \pmod{p^{t}}$  Where we have that $p^{t}|n$ and that $n$ is odd.  We have also that $a^{2n}\equiv 1 \pmod{p^{t}}$.  Thus, we know that $ord(a)|2n$ but it doesn't divide $n$.  In fact that condition is sufficient since in mod $p^{t}$ where $p$ is odd, $\pm 1$ are the only square roots of 1 (that fact follows directly from factoring or from hensel's lemma).  In other words we are looking for all the elements of mod $p^{t}$ with certain orders and we are counting their parity.  We take advantage of the fact that the multiplicative group of mod $p^{t}$ has a generator to conclude that the number of elements with order $x$ is $\phi(x)$.  But we know that $\phi(x)$ is always even except when $x = 1,2$.  We since the order of $a$ divides $2n$ and not $n$, it is clear it is of the form $2k$ where $k$ is odd.  Clearly the only odd number of solutions we are going to get is when $k=1$.  But that actually works in this case.  So we are done!

To do part c, we need only be a little bit more careful about the arguement we just used for part b.  Based on part b we need only consider even $n$ that work.  We start counting solutions again.  So we have that the number of solutions to $a^{n}\equiv-1 \pmod n$ must be 130.  The way to find these solutions is to first consider each prime power divisor separately again and then multiply the number together using CRT.  Remember that in this case $n$ has exactly one factor of 2.  In mod $p^{t}$  the number of solutions is exactly the same order trick again. However, in this case, things with order 2 do not work anymore.  In other words, there is an even number of solutions in each prime power mod.  Thus we can actually only have 1 prime power, since 4 does not divide 130.  This simplifies the problem a lot and leaves us with solving the problem of finding the number of solutions to $a^{2p^{t}}\equiv-1 \pmod{p^{t}}$.  Doing a little bit of simplifying with euler we get that $a^{2p^{t-1}}\equiv-1 \pmod{p^{t}}$.  Or in other words we have $ord(a)|4p^{t-1}$ but not $2p^{t-1}$.  Which is also a sufficient condition again.  Therefore we conclude $ord(a) = 4, 4p, 4p^{2}, \cdots, 4p^{t-1}$.  Keep in mind that these are legitimate orders since $p \equiv 1 \pmod 4$.  How many solutions do we have total?  Based on what we concluded earlier it simply: \[\phi(4)+\phi(4p)+\cdots \phi(4p^{t-1}) = 2(1+(p-1)+(p^{2}-p)+\cdots (p^{t-1}-p^{t-2})) = 2p^{t-1}\neq 130\] Yay!

i think there are two solutions of $a^{2}=-1(mod p^{k})$ for $p=1(mod4)$
this can be made by induction. (or i think i have made)

I solved it last summer, by using primitive root.

This problem solved at Sierpinski "250 problems of elementary number theory" №22

Beautiful problem that doesn't get the love.

We will ignore $n=1,2$, as these are stupid useless numbers that make life challenging. They only work for part (a).

We claim the answer to part (a) is the following:
Let $\mathcal P_{\geq 3}$ be the set of odd primes, and $\mathcal P_{1\pmod 4}$ be the set of primes 1 modulo 4. Then, $n$ has all prime factors in $\mathcal P_{\geq 3}$ or $\{2\}\cup\mathcal P_{1\pmod 4}$.

Note that if $4\mid n$, then clearly
\[\{a^n+1\}\subseteq\{a^2+1\}\subseteq\{1,2\pmod 4\}\]
so thus $4\nmid n$. Now, if $n$ is odd, taking $a=n-1$ gives
\[a^n+1\equiv (-1)^n+1\equiv 0\pmod n\]
Now, if $n$ is even, suppose $p\mid n$. Then, $a^2+1\equiv 0\pmod p$ has a solution, so $p\equiv 1\pmod 4$. For the converse, note that $p=x^2+y^2$ for any $p\equiv 1\pmod 4$, and thus we can note $|x+yi|=\sqrt p$. Thus,
\[p^k=|(x+yi)^k|^2\]
and thus we can write all powers of $p$ as the sum of two squares. In particular, we can find a solution to $p\equiv -1\pmod{p^k}$, so the solution to (a) is indeed as mentioned.

For part b, we note that if $n$ is even (and greater than 2), then $a$ and $n-a$ both work (clearly $a=n,\frac n2$ fails). If $n$ is odd, then clearly note that there are actually $\gcd(n,p-1)$ solutions to $x^n+1\equiv 0\pmod p$ - let $g$ be a primitive root modulo $p$, and note this implies that the number of solutions is the number of $1\leq k\leq p-1$ such that
\[nk\equiv 0\pmod{p-1}\]
which has $\gcd(n,p-1)$ solutions. As $n$ is odd, this is odd. Now, by Hensel's Lemma, for a prime power $p^k\mid n$, the number of solutions to
\[x^n\equiv -1\pmod{p^k}\]
is $p^{k-1}\gcd(n,p-1)$, so applying the Chinese Remainder Theorem, we get the desired result.

For part c, the answer is "no". If $p_1\neq p_2\in\mathcal P_{\geq 3}$ divided $n$, by our above discussion, there are $4\mid p_1^{k_1-1}\gcd(n,p_1-1)p_2^{k_2-1}\gcd(n,p_2-1)$ solutions, a contradiction. Thus, $n=2p^k$, impossible.

Solution from [i]Twitch Solves ISL[/i]:

[b]Part (a)[/b]: The answer is odd $n$ and even $n$ with $\nu_2(n) \le 1$ and only $1 \bmod 4$ prime factors. [list] 	[*]For odd $n$, let $a = n-1$ for a construction. 	[*]For even $n$, the right-hand side is a sum of two squares, 	one of which is odd. 	So $\nu_2(n) \le 1$ and the $1 \bmod 4$ condition are both needed. 	Conversely, 	we can choose $n$ to be a square root of $-1$ 	modulo each prime power dividing $n$; this works. [/list]
[b]Part (b)[/b]: The answer is all even numbers in (a) other than $n=2$. As for evenness: [list] 	[*]When $n$ is even, pairing $a$ and $-a$ implies $|A_n|$ even, except $n=2$ 	(the only time $n \mid (n/2)^n+1$). 	[*]When $n$ is odd, the number of solutions to $a^n \equiv -1 \pmod{p^e}$ 	for odd prime power $p^e$ is odd: $a \equiv -1$ is a solution, 	$a \equiv 1$ is not, and any other solutions come in pairs $\{a,1/a\}$. [/list]
[b]Part (c)[/b]: No. If such $n$ existed, by (b) it is even. [list] [*]If $n$ has more than one distinct odd prime factor, then $|A_n|$ is divisible by four, by Chinese remainder theorem. [*]For $n = 2p^e$, \[ a^{2p^e} \equiv -1 \pmod{p^e} \] We have a primitive root, of order $(p-1) \cdot p^{e-1}$. So there are $2p^{e-1}$ solutions. [/list] Neither case coincides with $130$, end proof.


induction
group theory
abstract algebra
invariant

Suppose there is a "still" magnetic field (in your reference frame). So it won't generate induction electronic field. Now you move, and the magnetic field "becomes" a varied field (suppose it's not a homogenous magnetic field so that dB/dt !=0), so according to Maxwell's theory, you would observe induction electronic field, a contradiction.

So, how to explain this?

I fail to see the contradiction here...

Are you aware of the transformation properties of the electromagnetic fields?

For example, suppose there is a magnetic field, on x-ary, it is B(x,y,z)=ax. and now the observer begin to move along x-ary with speed v, so B(x,0,0)=a(x-vt), dB/dt=-av.
I'm not aware of the transformation property, can you explain it in detail, or link some reference? thanks in advance.

The transformation properties are best seen via the covariant formulation of electrodynamics.
We join the electric and magnetic fields into an antisymmetric tensor called the field strength tensor $ F^{\mu\nu}$.
This can be explicitely defined via the four-potential:

\[ F^{\mu\nu}\equal{}\partial^{\mu}A^{\nu}\minus{}\partial^{\nu}A^{\mu}\]

Now this transforms according to the Lorentz group.
So lets say $ \Lambda^{\mu}_{\nu}$ is a representation of the real Lorentz group (we leave out parity transformations and time inversion, so we are just concerned with SO(3) and lorentz boosts. This is basically a subgroup of the extended Lorentz group)

Then the field strength tensor transforms according to:

\[ F^{\mu'\nu'}\equal{}\Lambda^{\mu'}_{\mu}\Lambda^{\nu'}_{\nu} F^{\mu\nu}\]

So as we see the after the transformation the fields get "mixed". This means that there is no contradiction since after applying a lorentz boost we get the transformed fields in the reference frame moving with some velocity $ v$, and from the above transformation property we see that the fields "mix" this means that the magnetic field observed in the moving frame will be a mixture of an electric and magnetic field.
Basically thats the reason that there is no contradiction.
So just check the transformation properties of the fields and you can come to this conclusion.

cheers

Hey, that's actually a very interesting question you are asking, and relates a fundamental principle of relativity (only relative motion really matters).  So whether the field moves (like you move a magnet) or the observer moves, both should observe some form of E/B field, however, if you place a wire, both should record a form of current (whose magnitude may/may not match).

It has to do with the transformation of fields; the E, and B field aren't the same for 2 observers whose reference frames are moving relative to each other.  However, the results of such B/E field configurations (charge moving in a wire) should be the same...  Therefore, the fields transform, from one frame to another.
\[ \vec {E' \bot} \equal{} \gamma \vec {E} \bot \plus{} \gamma \vec {\beta} \times \vec {B \bot}\]
where
\[ \vec {\beta} \equal{} \frac {\vec {v}}{c}\]

\[ \vec {E' \parallel} \equal{} \vec {E \parallel}\]
Similarly,
\[ \vec {B' \bot} \equal{} \gamma \vec {B} \minus{} \gamma \vec {\beta} \times \vec{E \bot}\]
and
\[ \vec {B' \parallel} \equal{} \vec {B \parallel}\]

(BTW, yeahnigga probably has it right, his formulas are more elegant, but also employs more math.  If you want to get started on deriving how fields transform, I would suggest you read ch. 5-6 of Purcell Electricity and magnetism).

Thank you.
There is another related questions (not sure if they are related), suppose two electron a and b move in two directions at permanent speed. So they should take lorentz force caused by each other's magnetic field. now the observer move with a (i.e, a becomes still in the new reference), and observe that a and b do not take lorentz force.

In another situation, suppose there is a large piece of homogeneous magnetic field, and a electron q moving at permanent speed through it, so it shoud take lorentz force. However if the observer move with q, then he  should observe that q do not take any force.

My classmates say, whether an electron take lorentz force depends on whether it move RLETIVE TO THE MAGNETIC FIELD not mere to the observer. But under this view, even if a magnetic field (its strength) is given, it would still be uncertain whether or how much force will an electron take. In another word, we could detect whether an invariant magnetic field is moving by obseverving whether a still electron in it take force or not. This seems somehow unconfortable to me and seem to violate the definition of strength of magnetic field. (What's the definition of the strength of magnetic field?)


trigonometry geometry circumcircle inequalities

Let be given a triangle $ABC$. Prove that: $\sum\cos A\le 1+\frac{1}{6}\left(\sum\cos^{2}\frac{A-B}{2}\right)$

Let O ,I , H denote the circumcentre,incetre and orthocentre respectively.

Let AI cut the circumcircle at Q, and OQ cut the circumcircle at P ,so that QP=2R is a diameter.

We will use these well known results of triangle geometry:
1.$\angle OAH$ =$B-C$
2.AI bisects $\angle OAH$ so $\angle OQA$=$\angle OAQ$ =$\frac{B-C}{2}$
3.AI =$4R{sin\frac{B}{2}}{sin\frac{C}{2}}$ and IQ =$2Rsin\frac{A}{2}$

Now $cos\frac{B-C}{2}$ =$\frac{AQ}{2R}$
                                       =$\frac{AI+IQ}{2R}$
                                      $\geq$ $\frac{2(\sqrt(AI.IQ)}_{2R}$ by AM-GM inequality
                                     =$\frac{\sqrt(2Rr)}_{R}$ using (3)
                                     =$\sqrt\frac{2r}{R}$

It follows that $\sum{cos^{2}(\frac{B-C}{2})}$ $\geq$ $\frac{6r}{R}$ and recalling that 1+$\frac{r}{R}$ =$\sum(cosA)$ the required result follows

Thank you for your proof.
[hide="And here is my solution"]
\[\left(\cos\frac{B-C}{2}-2\sin\frac{A}{2}\right)^{2}\ge 0 \]
\[\Longrightarrow\cos^{2}\frac{B-C}{2}+4\sin^{2}\frac{A}{2}\ge 4\cos\frac{B-C}{2}\cos\frac{B+C}{2}\]
\[\Longrightarrow\cos^{2}\frac{B-C}{2}+2(1-\cos A)\ge2(\cos B+\cos C) \]
\[\Longrightarrow\cos^{2}\frac{B-C}{2}+2\ge 2\sum\cos A \]
\[\Longrightarrow\sum\cos\frac{B-C}{2}+6\ge 6\sum\cos A \]
\[\Longrightarrow\sum\cos A\le 1+\frac{1}{6}\left(\sum\cos^{2}\frac{B-C}{2}\right) \]
[/hide]


$ P$ and $ Q$ are points of tangency of  incircle and escribed circle relative to the side $ AC$ of triangle $ ABC$ and $ PJ//QL//AB$. If $ \overline{AB}\equal{}12$, $ \overline{AC}\equal{}8$ and $ \overline{BC}\equal{}10$ find perimeter of the trapezoid $ PQLJ$.

[quote="LoboSolitario"]$ P$ and $ Q$ are points of tangency of  incircle and escribed circle relative to the side $ AC$ of triangle $ ABC$ and $ PJ//QL//AB$. If $ \overline{AB} \equal{} 12$, $ \overline{AC} \equal{} 8$ and $ \overline{BC} \equal{} 10$ find perimeter of the trapezoid $ PQLJ$.[/quote]

Just use $ 2|CQ| \equal{} a\plus{}b\minus{}c$ and $ 2|CP| \equal{} b\plus{}c\minus{}a$ over and over...


Let  $ ABC$ be an acute triangle. For a mobile point  $ M\in (BC)$ denote $ \|\begin{array}{ccc} N\in (CA) & , & MN\perp AC \\
 \\
P\in (AB) & , & MP\perp AB\end{array}$

Ascertain the position of the point $ M$ for which the area of the quadrilateral  $ ANMP$ is maximum.

cyclic*

[hide]By symmetry its simply where NP is parallel to BC.[/hide]

If we want to maximize $ [APMN]$, we can note that $ [ABC]$ is fixed, so we can equivalently minimize $ [ABC]\minus{}[APMN]\equal{}[BPM]\plus{}[CNM]$. Denote $ BM\equal{}x$ and $ CM\equal{}y$. Note that $ x\plus{}y\equal{}a$ and $ x,y\ge 0$. Then the sum of those areas is \[ K\equal{}\frac{1}{4}\cdot x^2\sin 2B\plus{}\frac{1}{4}\cdot y^2\sin 2C\]

Lemma 1: \[ mx^2\plus{}ny^2\ge \frac{mn}{m\plus{}n}(x\plus{}y)^2\] where $ m,n,x,y$ are non-negative real numbers and $ a\plus{}b>0$.
Proof: \[ (n\plus{}m)(mx^2\plus{}ny^2)\ge (\sqrt{nm}x\plus{}\sqrt{mn}y)^2\equal{}mn\cdot (x\plus{}y)^2\] by the Cauchy inequality. Divide by $ a\plus{}b$ to get the result. Equality occurs when $ \frac{mx^2}{n}\equal{}\frac{ny^2}{m}\implies mx\equal{}ny$.

Apply Lemma 1 with $ m\equal{}\frac{1}{4}\cdot\sin 2B$ and $ n\equal{}\frac{1}{4}\cdot \sin 2C$.

Equality occurs when $ x \sin 2B\equal{}y\sin 2C$.

This probably has an easy geometric interpretation, but I cannot think of it now.


LaTeX function parameterization calculus derivative algebra polynomial

[quote="mathfriend.250293"]Solve for x:
 3^x(4^x+6^x+9^x)=25^x+2.16^x[/quote]
Can you please write in Latex????Just use the dollar sign to open and close it :wink:

In latex
Solve for x:
$ 3^x(4^x\plus{}6^x\plus{}9^x)\equal{}25^x\plus{}2.16^x$

If $ 2.16$ is supposed to be $ 2\times 16$, then the obvious answer is $ x\equal{}0$.
Otherwise, there exists some root around $ \minus{}.35$ (graphing it) but I can't think of a way to get it by hand.

Logs seem excessively complicated here...where was this problem from?

[quote="little,,Dragon!"][quote="tonypr"]In latex
Solve for x:
$ 3^x(4^x \plus{} 6^x \plus{} 9^x) \equal{} 25^x \plus{} 2.16^x$[/quote]
$ 3^{x} \equal{} a$
$ 2^{x} \equal{} b$
$ 5^{x} \equal{} c$
$ a(b^{2} \plus{} ab \plus{} a^{2}) \equal{} c^{2} \plus{} 2b^{4}$[/quote]

thanks!!!
can you resolve $ a(b^{2} \plus{} ab \plus{} a^{2}) \equal{} c^{2} \plus{} 2b^{4}$[/quote] ????

[quote="little,,Dragon!"][quote="tonypr"]In latex
Solve for x:
$ 3^x(4^x \plus{} 6^x \plus{} 9^x) \equal{} 25^x \plus{} 2.16^x$[/quote]
$ 3^{x} \equal{} a$
$ 2^{x} \equal{} b$
$ 5^{x} \equal{} c$
$ a(b^{2} \plus{} ab \plus{} a^{2}) \equal{} c^{2} \plus{} 2b^{4}$[/quote]Is there a reason for doing this kind of substitution? There are tons of substitutions that can be used, so why did you choose this substitution?

How about this? Rewrite the equation into $ (27^x\minus{}25^x)\plus{}(18^x\minus{}16^x)\equal{}16^x\minus{}12^x$. Show that eventually LHS>RHS. That may require precalc+ concepts.

[quote="b555"]f the reasons why  dragon might have  used this substitution  is  because we have   2^2 , 3^2  and  2 x3 .  therefore  that would  be what  anyone be first looking at[/quote]I would like to see a proposed solution with that substitution, please. I do not see how this is useful.

hello, is this the right equation $ 3^x(4^x\plus{}6^x\plus{}9^x)\equal{}25^x\plus{}2\cdot16^x$?
Sonnhard.

Begin by rearranging it as [b]10000th User[/b] cleverly suggested:
$ (27^x \minus{} 25^x) \plus{} (18^x \minus{} 16^x) \equal{} 16^x \minus{} 12^x$

We notice that we can write all the terms in a similar form as follows:
$ (27^x \minus{} 25^x) \plus{} (18^x \minus{} 16^x) \equal{} (16^x \minus{} 14^x) \plus{} (14^x \minus{} 12^x)$

I'll make this "similar form" more explicit by introducing $ f (v,x) \equal{} (v \plus{} 2)^x \minus{} v^x$. Then:
$ f(25,x) \plus{} f(16,x) \equal{} f(14,x) \plus{} f(12,x)$

$ \blacktriangleright$ For $ x > 1$ and $ x < 0$, note that $ f(v,x)$ is an increasing function of $ v$. Thus:

$ f(25,x) > f(14,x) \; \; \textbf{and} \; \; f(16,x) > f(12,x)$
$ \implies f(25,x) \plus{} f(16,x) > f(14,x) \plus{} f(12,x)$

$ \blacktriangleright$ For $ 0 < x < 1$, note that $ f(v,x)$ is now a decreasing function of $ v$:

$ f(25,x) < f(14,x) \; \; \textbf{and} \; \; f(16,x) < f(12,x)$
$ \implies f(25,x) \plus{} f(16,x) < f(14,x) \plus{} f(12,x)$

$ \blacktriangleright$ For $ x \equal{} 0,1$, $ (v \plus{} 2)^x \minus{} v^x$ is constant, and thus:

$ f(25,x) \equal{} f(14,x) \; \; \textbf{and} \; \; f(16,x) \equal{} f(12,x)$
$ \implies f(25,x) \plus{} f(16,x) \equal{} f(14,x) \plus{} f(12,x)$

Our unique solutions are therefore $ \boxed{x \equal{} 0,1}$

* - Where did all this increasing-decreasing stuff come from? I looked at the convexity of the function $ y \equal{} v^x$, a function of $ v$, for different values of the parameter $ x$. If you know calculus, you can just take a second derivative to formalize this, but should also come intuitively if you have a good grasp of what graphs of simple polynomials look like.

@T2F:   sir  do  u think  the solution u have given is in the scope of a  regular high school student? :?:

Don't get caught up in the notation -- the concept is actually very simple. The introduction of the function allows us to think of x as a parameter rather than as a variable. This way, we're just dealing with polynomials instead of exponentials. From here, you basically look at the graph of a parabola and say, hey, the "slope" is increasing. Then you notice this is true for any exponent over 1, and not just squares. You then perform a similar observation for 0<x<1 and x<0.

So yes, I do think the problem is within the scope of the high school curriculum. Do I expect the average high school student to try to treat x as a parameter? No. But math contests are not about what every high school student will be able to do.

[quote="ernie"][quote="BOGTRO"]It looks simple enough to me, and I'm 6th grade.[/quote]

Uhh... dude, trust me.  Don't show off.[/quote]Cut off the drama, ya? Leave that for the PM's. Stick to the topic please.

[quote="ernie"][quote="BOGTRO"]It looks simple enough to me, and I'm 6th grade.[/quote]

Uhh... dude, trust me.  Don't show off.[/quote]

I'm missing how I'm showing off. I was answering the question about high schoolers understanding the solution.


What is the number of centimeters in the length of EF if AB||CD||EF (parallel) and AB=150 and DC=100?

Also, BD and AC are straight (stop making fun of my paint skills)

There's a nice formula for this. I got it from AoPS 1 (triangles: similar triangles-exercise-?)
[hide]
Given that AB||CD||EF,
$\frac{1}{AB}+\frac{1}{CD}=\frac{1}{EF}$
[hide]
$\frac{1}{150}+\frac{1}{100}=\frac{1}{EF}$

$=\frac{250}{15000}=\frac{1}{60}=\frac{1}{EF}$

$EF=60$
[/hide]
[/hide]

[quote="pianoforte"]There's a nice formula for this.[/quote]
Why is the formula true? Although formulas simplify a lot of work, it doesn't show how to solve a problem.

[hide="Click me to reveal the secret behind that formula"]First, think of right triangle $ABC$. Since $AB\|EF$, use similar triangles and you should get $\frac{CF}{EF}=\frac{CB}{AB}$.

Now think of right triangle $BCD$. Since $EF\|DC$, $\frac{BF}{EF}=\frac{BC}{DC}$.

Add them together to get

$\frac{CF}{EF}+\frac{BF}{EF}=\frac{CB}{AB}+\frac{BC}{DC}$

But, $CF+BF=BC$ and so

$\frac{CF}{EF}+\frac{BF}{EF}=\frac{BC}{EF}=\frac{CB}{AB}+\frac{BC}{DC}$

Divide both sides of equation by $BC$ (because $BC\neq 0$) and you should get that 'formula'

$\frac1{EF}=\frac1{AB}+\frac1{DC}$

Demystify them! :)[/hide]


integration
linear algebra
matrix
calculus
real analysis
real analysis solved

Is it possible that the equation $y'=y^2+f(x)$ has two distinct periodic solutions $y_1$ and $y_2$ with the same period $T$ s.t. \[\int_0^T(y_1(x)+y_2(x))dx\neq 0\,?\]

The answer is "No", but I wouldn't call my solution simple. If nobody finds anything better in a week, I'll post it. Now I'll tell only that it starts with some little piece of "general theory" of the equation $y'(t)=y(t)^2+f(t)$ and ends with the following observation:
If a $2\times 2$ matrix $U(t)=\left[ \begin{matrix}a(t) & b(t) \\ c(t) & d(t) \end{matrix}\right]$ satisfies the equation $U'=FU-UF$ where $F(t)=\left[ \begin{matrix} 0 & f(t) \\ -1 & 0 \end{matrix}\right]$, $U(T)=U(0)$ and $c(t)$ doesn't vanish on $[0,T]$, then $\int_0^T \frac{a(t)-d(t)}{c(t)}\,dt=0$.
Is your solution similar, Myth?

No, my solution (khm, not mine, my team-mate was solving it) is fully elementary.

If $ y_{1} , y_{2} $ satisfy $ y_{i}' = y_{i}^{2} + f(x) $, then subtract the two equations
and we will get $ (y_{1} - y_{2})' = (y_{1} - y_{2})(y_{1} + y_{2}) $. Now denote
$ z(x) = y_{1}(x) - y_{2}(x) $. because $ y_{1} $ and $ y_{2} $ are distinct, then
$ y_{1}(x) \neq y_{2}(x) $ for every $ x $, because of uniqueness of solution of differential
equation. We can suppose that $ z(x) > 0 $ for every $ x $.
Then we have that $ y_{1}(x)-y_{2}(x) = \frac{z'(x)}{z(x)} $ therefore
$ \int_{0}^{T} (y_{1}(x) + y_{2}(x)) dx = \int_{0}^{T} \frac{z'(x)}{z(x)} =
\ln(z(T)) - \ln(z(0)) = 0 $, because $ y_{1}, y_{2} $ are periodic with period $ T $.

Why?  :(
Because I puted the "minus" instead of "plus" ?  :?

What conditions do you need for uniquness of solution?

Cute! That means I do not need to post my monster anymore. :lol:  Still, let me post an observation about the equation $y'=y^2+f(t)$ that may be useful in some other problem. Consider the auxiliary linear system $A'(t)=F(t) A(t)$, $A(t_0)=I$, where $A(t)=\left[\begin{matrix}\alpha(t) & \beta(t) \\ \gamma(t) & \delta(t) \end{matrix}\right]$ is a $2\times 2$ matrix and $F(t)$ is the matrix from my previous post. Then the solution of the initial value problem $y(t_0)=y_0$ for the original equation is given by $y(t)=\frac{\alpha(t)y_0+\beta(t)}{\gamma(t)y_0+\delta(t)}$. Probably that's very well known, but I was quite amazed when found it  :) .

If you have that for some $ x_{0} $ that $ y_{1}(x_{0}) = y_{2}(x_{0}) = a $
then $ y_{1}, y_{2} $ satisfy the same differential equation of
the first order and have the same initial conditions at $ x_{0} $,
therefore they must be equal, am I right ?  :blush:   :blush:   :blush:

I think it failed due to unknown nature of $f(x)$. The proof refers to the integral operators. Can we properly define it in this case?

And if I say that we can define $ w(x) = y_{1}(x) + y_{2}(x) $,
then $ z(x) $ satisfies the equation $ z'(x) = z(x) w(x) $
which is pretty good for the uniqueness ( because $ w(x) $ is
differentiable) .
Is it ok now?   :?  :(  :?  :(


Start with the theorem "If two angles of a triangle are equal, the triangle is isosceles," and the following four statements:

1. If two angles of a triangle are not equal, the triangle is not isosceles.
2. The base angles of an isosceles triangle are equal.
3. If a triangle is not isosceles, then two of its angles are not equal.
4. A necessary condition that two angles of a triangle be equal is that the triangle be isosceles.

Which combination of statements contains only those which are logically equivalent to the given theorem?

$ \textbf{(A)}\ 1,\,2,\,3,\,4 \qquad
\textbf{(B)}\ 1,\,2,\,3\qquad
\textbf{(C)}\ 2,\,3,\,4\qquad
\textbf{(D)}\ 1,\,2\qquad
\textbf{(E)}\ 3,\,4$

[hide="My Solution"]
(1) This doesn't necessarily follow.
(2) We are not given information on where the equal angles are located in the theorem.
(3) This is a negation of the theorem, which holds.
(4) This is basically rewording the theorem.

Therefore, the answer is $ \fbox{(E)}$.
[/hide]


A dead body is found at 2 PM at a temperature of 29 degrees. The temperature in the room is 22 degrees. It is known that in those room conditions it takes 2 hours for the temperature of the body to drop to 35 degrees. When was the murder committed?

Yes, I read your problem which is a little bit different from mine.
This is a initial value problem, but the initial condition was given in a twisted way, I don't know how to get the constants.

I assume these temperatures are given in degrees Celsius and that the body temperature at the time of the murder was $37^{\circ}.$

I find it simplest to just subtract the room temperature from everything. Let $u=T-22$ where $T$ is the temperature in $^{\circ}C.$ Measure time in hours. Then $u(0)=15,\ u(2)=13,$ and $u(t)=7$ where $t$ is the time the body was found. It follows that $15\left(\frac{13}{15}\right)^{t/2}=7.$ We can solve that for $t:$

$t=\frac{2(\ln7-\ln15)}{\ln13-\ln15}\approx10.65$ hours.

This gives an approximate time for the murder of 3:20 a.m.


calculus
derivative
function
calculus computations

If $ y$ is a function of $ x$ and $ x$ is a function of $ u$ is $ \frac{d}{du}(\frac{dy}{dx})\equal{}\frac{d^2y}{dx^2}.\frac{dx}{du}$?

The chain rule gives:

$ \frac{d}{du}\left(\frac{dy}{dx}\right)\equal{}\frac{dx}{du}\cdot\frac{d}{dx}\left(\frac{dy}{dx}\right)\equal{}\frac{d^2y}{dx^2}\cdot\frac{dx}{du}$.


abstract algebra
superior algebra
superior algebra theorems

Let $R$ --- ring. Let $M$ --- module over $R$, generated by the set $\{x_i\}_{i\in I}$ ($I$ --- index set). So, we can consider epimorphism $f$ from the free module $F$ (the direct sum of $|I|$ copies of the $R$ as left module over own) to module $M,$ consigning $i$-th element of the canonical basis for  $F$ to the $x_i.$

How can I accelerate a description of the emersion of epimorphisms of such kind? Can I call it "the natural epimorphism", implying the isomorphism $M\cong F/K$, where $K=\ker(f)?

I do not understand how this is a theorem or a formula. Could someone please clarify for me? Thank you.


Two numbers are such that the sum of their cubes is 5 and the sum of their squares is 3. Find the sum of the numbers.

No matter I tried it from whatever method, I always got x+y=5/(3-x.y). Anybody, who can tell how to get the product of two numbers?

Note that we have $ (3 \minus{} x^2)^3 \equal{} (5 \minus{} x^3)^2$
$ 2x^6 \minus{} 9x^4 \minus{} 10x^3 \plus{} 27x^2 \minus{} 2 \equal{} 0$
$ (2x^2 \minus{} 4x \plus{} 1)(x^4 \plus{} 2x^3 \minus{} x^2 \minus{} 8x \minus{} 2) \equal{} 0$
So x+y can equal 2, to get the other answer you'd have to do it Bovy's way, which is better. I used a computer to find the factorization.

Ok,now $ x^3 \plus{} y^3 \equal{} 5$ and $ x^2 \plus{} y^2 \equal{} 3$.

[hide="solution"]
Let $ a \equal{} x \plus{} y$,$ b \equal{} xy$,we see the system is equivalent to

$ a^2 \minus{} 2b \equal{} 3$
$ a^3 \minus{} 3ab \equal{} 5$

(If $ x,y$ are reals,you must add the condition that $ a^2 \minus{} 4b\ge0$ i.e $ a^2\le 6$.)

By eliminating $ b$ of the system we have

$ a^3 \minus{} 9a \plus{} 10 \equal{} 0\Longleftrightarrow (a \minus{} 2)(a^2 \plus{} 2a \minus{} 5) \equal{} 0$

so you get the $ x \plus{} y \equal{} a \equal{} \boxed{2, \minus{} 1\pm \sqrt {6}}$.(but if $ x,y$ are reals,$ a \equal{} \minus{} 1 \minus{} \sqrt {6}$ is ruled out by the condition $ a^2\le 6$).[/hide]


inequalities search trigonometry algebra polynomial calculus derivative

There is a problem with a wrong solution in a book of mine and I can't solve it in another way(or I can't disprove it either). Here is the problem:

Prove there exists a real constant $c$ that for arbitrary positive integer $n$ and arbitrary reals $a_{1}, a_{2}, \ldots , a_{n}$ this inequality holds

$\max_{0\leq x\leq 2}\prod_{j=1}^{n}{|x-a_{j}|}\leq c^{n}\max_{0\leq x\leq 1}\prod_{j=1}^{n}{|x-a_{j}|}$

[quote="Rust"]${\max_{0\leq x\leq 1}|2x-a_{j}|}\leq 3\max_{0\leq x\leq 1}|x-a_{j}|.$[/quote]
What do you mean, Rust?
"Problem with a Wrong Solution" doesn't mean you should give a wrong solution but a correct one.

Since when the maximum of a product is the product of the maxima?  :maybe:  Anyway, this has been posted by Iura, but I don't remember the name of the topic. Please search for it and it you don't find it, I'll post my solution, with the best constant $c$! I've just realized that it would take me too long to write the solution for the best constant, so I will let you look on the following http://www.mathlinks.ro/Forum/viewtopic.php?highlight=alex&t=15379

Wow, I'm also interested in the best constant $c$.

Here is the solution on my book which, I think, is not correct.

Let $f(x)=\prod_{i=1}^{n}{(x-a_{j})}$, then

$f(x)=\sum_{i=0}^{n}{f(\frac{i}{n})\prod_{j\neq i, j=0\ldots n}\frac{x-j/n}{i/n-j/n}}=\sum_{i=0}^{n}{f(\frac{i}{n})\prod_{j\neq i, j=0\ldots n}\frac{nx-j}{i-j}}$

(Lagrange Interpolation Formula)

Let $M=\max_{0\leq x\leq 1}{|f(x)|}$, then

$\max_{0\leq x\leq 2}{|f(x)|}=\max_{0\leq x\leq 2}{\sum_{i=0}^{n}{|f(\frac{i}{n}|)\prod_{j\neq i, j=0\ldots n}|\frac{nx-j}{i-j}}|}\leq M\sum_{i=0}^{n}\prod_{j\neq i, j=0\ldots n}|\frac{2n-j}{i-j}|$

but

$\sum_{i=0}^{n}\prod_{j\neq i, j=0\ldots n}|\frac{2n-j}{i-j}|=\frac{(2n)!}{(n-1)!i!(n-i)!(2n-i)!}=\binom{2n}{i}\binom{2n-1-i}{n-i}\leq \binom{2n}{i}\binom{2n-1}{n-i}$

so

$\max_{0\leq x\leq 2}{|f(x)|}\leq M\sum_{i=0}^{n}\binom{2n}{i}\binom{2n-1}{n-i}<M\sum_{i=0}^{n}\binom{2n}{i}\sum_{i=0}^{n}\binom{2n-1}{n-i}<M2^{2n}2^{2n-1}<M16^{n}$

ie, $c=16$

Do you find the mistake above?

Exactly $c_{n}=(\frac{T_{n}(3)}{T_{n}(1)})^{1/n}$, were $T_{n}(x)=cos(narccos x)$ is Chebechev's polinom. For example $c_{1}=3,c_{2}=\sqrt{17}$.

That's correct, Rust. However, Libra-Gold, I don't see any fundamental mistake in the above proof (except that at a certain moment there should be a $\leq$ instead of an $=$, but anyway...). It's exactly the way I did it in the link I indicated. What do you find wrong in this solution?  :maybe:

It is correct. Small mistake (2n-i) (not (2n-i)!).
[quote="libra_gold"]$\sum_{i=0}^{n}\prod_{j\neq i, j=0\ldots n}|\frac{2n-j}{i-j}|=\frac{(2n)!}{(n-1)!i!(n-i)!(2n-i)!}$
[/quote]

Yes, it is just a small mistake :P

Now I know how correct it. Thanks.

But I wonder how to get the best constant $c$, Rust.

I think we can use the way in my book to get the best one, right?

I will let you do the computations required for the following:

 Lemma .  If we put $t_{k}=\cos\frac{k\pi }{n}$, $0\le k\le n$, then
\[f(x)=\prod_{k=0}^{n}(x-t_{k})=\frac{\sqrt{x^{2}-1}}{2^{n}}[(x+\sqrt{x^{2}-1})^{n}-(x-\sqrt{x^{2}-1})^{n}]. \]
Also,
i)  $\prod_{j\ne k}(t_{k}-t_{j})=\frac{(-1)^{k}n}{2^{n-1}}$ if $1\le k\le n-1$;

ii) $\prod_{j=1}^{n}(t_{0}-t_{j})=\frac{n}{2^{n-2}}$;

iii)
$\prod_{j=0}^{n-1}(t_{n}-t_{j})=\frac{(-1)^{n}n}{2^{n-2}}$.

    Let us fix a polynomial $P$ whose maximum on $[0,1]$ (in absolute value)  is 1 and fix $x\in [1,2]$ andconsider the numbers
$x_{k}=\frac{1+t_{k}}{2}$.
 Using Lagrange interpolation formula, we deduce that
\[|P(x)|\le\sum_{k=0}^{n}\left|\prod_{j\ne k}\frac{x-x_{j}}{x_{k}-x_{j}}\right| =\sum_{k=0}^{n}\prod_{j\ne k}\frac{x-x_{j}}{|x_{k}-x_{j}|}\]

\[\le \sum_{k=0}^{n}\prod_{j\ne k}\frac{2-x_{j}}{|x_{k}-x_{j}|}=\sum_{k=0}^{n}\prod_{j\ne k}\frac{3-t_{j}}{|t_{k}-t_{j}|}. \]
Using lemma , we can write
\[\sum_{k=0}^{n}\prod_{j\ne k}\frac{3-t_{j}}{|t_{k}-t_{j}|}=\frac{2^{n-1}}{n}\sum_{k=1}^{n-1}\prod_{j\ne k}(3-t_{j})+\]

\[+\frac{2^{n-2}}{n}\left[\prod_{j=0}^{n-1}(3-t_{j})+\prod_{j=1}^{n}(3-t_{j})\right]. \]
Based on the expression of the derivative from the proof of the lema, we obtain:
\[\sum_{k=1}^{n-1}\prod_{j\ne k}(3-t_{j})+\prod_{j=0}^{n-1}(3-t_{j})+\prod_{j=1}^{n}(3-t_{j})= \]

\[=\frac{n}{2^{n}}[(3+2\sqrt 2)^{n}+(3-2\sqrt 2)^{n}]+\frac{3}{2^{n+1}\sqrt 2}[(3+2\sqrt 2)^{n}-(3-2\sqrt 2)^{n}]. \]
All we have to do now is to compute
\[\prod_{j=0}^{n-1}(3-t_{j})+\prod_{j=1}^{n}(3-t_{j})=6\prod_{j=1}^{n-1}(3-t_{j}). \]
But, according to lemma 1, we deduce immediately that
\[\prod_{j=1}^{n-1}(3-t_{j})=\frac{1}{2^{n+1}\sqrt 2}[(3+2\sqrt 2)^{n}-(3-2\sqrt 2)^{n}]. \]
Putting all these observations together and making a small
computation, that we let to the reader, we easily deduce that
$|f(x)|\le c_{n}$. This proves that $\|f\|_{[0,2]}\le c_{n}\|f\|_{[0,1]}$ and solves the problem.

I used bad notations. The first $f(x)$ is defined as the polynomial whose roots are $\cos(k\pi/n)$. This one is not the $n$-th Chebyshev polynomial. The other $f(x)$ is any polynomial whose maximum on $[0,1]$ was considered to be 1. And by doing the computations, we do get your constant, Rust. [/code]


topology
geometry
3D geometry
sphere
advanced fields
advanced fields unsolved

How to prove that there is no embeddings $f: \mathbb{S}^{1}\to \mathbb{R}$ or $g: \mathbb{S}^{2}\to \mathbb{R}^{2}$?

For the first one you need the well-known connectedness argument (remove one point from $\mathbb{S}$ and see what happens), and for the second one some homotopy theory.

Much more is true:

[url]http://mathworld.wolfram.com/Borsuk-UlamTheorem.html[/url]

And if someone wants to see its proof, have a look at [url=http://planetmath.org/?op=getobj&from=objects&id=3621]here.[/url]

Another approach I can see is to prove that any embedding from an $n$-manifold to another $n$-manifold must be an open map. This would imply our result, as the sphere is compact and $R^{n}$ is connected and noncompact.
You definitely don't have the tools for the general case, but you should be able to do $n=1$ by a connectedness argument.
$n=2$ is harder- I don't know the best way to do it myself.

[quote="jmerry"]This would imply our result, as the sphere is compact and $R^{n}$ is connected and noncompact.[/quote]
But if $g$ is such an embedding, we might have that $g(\mathbb{S}^{2})\subset \mathbb{R}^{2}$, where the subset is proper. I don't see any reason why $g(\mathbb{S}^{2})$ can't be homeomorphism. Can you give me more details?

Well, the case $n=2$ follows from the Borsuk-Ulam theorem so I know now why  such an embedding won't exists.

If we know that the embedding is an open map, the image of the sphere is open. It's also compact, and $\mathbb{R}^{n}$ has no nonempty open compact (closed) subsets by connectedness.


function
abstract algebra
real analysis
real analysis unsolved

In my Fourier analysis class we explicitly constructed a continuous function whose fourier series diverges at a point.

I'm wondering whether we can prove the existence of such a function using a baire category argument- (like the way we prove the existence of cont everywhere ,nowhere differentiable function). Does anyone know a proof of this?

Yes, of course. All you need for that is to construct for every $ A>0$ a continuous function $ f$ bounded by $ 1$ and such that some partial sum of its Fourier series exceeds $ A$ at some fixed point. Then you can use the category argument in the usual way. The function is, of course, a continuous approximation of the sign of the Dirichlet kernel of sufficiently large order.


Twenty 11 inch by 17 inch sheets were printed on both sides so that when the sheets were stacked and folded down the center, an 8.5 inch by 11 inch booklet with pages numbered 1-80 would be formed.  When stacked and folded, however, the sheets containing the pages 12 and 53 were accidentally ommited.  What is the sum of the other numbers of the pages that were missing?

Can someone please explain answer to this problem?

wait is the answer 259???

the pages i added were 22, 28, 54, 11, 69, and 70.

I think that the answer is 195.
I added the numbers 9,10,11,54,55, and 56.
I don't believe the dimensions actually matter, they'll just throw you off.
But having Twenty pages does matter.  Because there are 80 numbers, all four sides (front, two in the middle, and back) of each page must be used.  So 9,10,11, and 12 are contained in one, and 53,54,55, and 56 are the other one.
Hope this helps. :D


[quote="Tiks"]$P(x)=(x^2-1)((x^2+x+1)^2+(a-2))$ ;)[/quote]

i don´t know how to finish... can you help me?

[quote="JohnConstantine"][quote="IgorCastro"]It´s from a test that i saw today...
Let $P(x)= x^6 + 2x^5 + ax^4 - ax^2 - 2x - 1$
find all $a$ real such that $P(x)$ have only real roots.[/quote]

$P(x)$ is a symmetric polinomial of sisth degree, so it has got two costant roots, or rather $1$ and $-1$. By the remainder theorem and the Ruffini theorem, we obtain $(x^2+x+1)^2+(a-2)=0$. $x^2+x+1$ is always positive (look at the discriminant) and $x^2+2\cdot \frac{1}{2}x+\frac{1}{4}=(x+\frac{1}{2})^2$$=x^2+x+1-\frac{3}{4}$, so $x^2+x+1\geq \frac{3}{4}$. So $(x^2+x+1)^2\geq\frac{3^2}{4^2}=\frac{9}{16}$, so $a-2 \leq \frac{-9}{16}\to a\leq \frac{23}{9}$[/quote]
Sorry,but I think this is same solution :(  :rotfl:

OOop, sorry, I've read yours later but the end seemed me different so I didn't delete it...I'm getting old :(   :D


If $3^{x}=5$, what is the value of $3^{2x+3}$

[hide][url=http://imageshack.us][img]http://img478.imageshack.us/img478/6476/yuuwdo4.jpg[/img][/url][/hide]

[size=75][color=darkred]Added hide tags.
-nebula42[/color][/size]

[quote="kstan013"]Yes. That's what I got too. I wish more people would hide answers *sigh*[/quote]

Yeah. That should be made a rule.  :lol:


geometry
trapezoid
circumcircle
cyclic quadrilateral
geometry proposed

Given a circumscribed trapezium $ ABCD$ with circumcircle $ \omega$ and 2 parallel sides $ AD,BC$ ($ BC<AD$). Tangent line of circle $ \omega$ at the point $ C$ meets with the line $ AD$ at point $ P$. $ PE$ is another tangent line of circle $ \omega$ and $ E\in\omega$. The line $ BP$ meets circle $ \omega$ at point $ K$. The line passing through the point $ C$ paralel to $ AB$ intersects with $ AE$ and $ AK$ at points $ N$ and $ M$ respectively. Prove that $ M$ is midpoint of segment $ CN$.

Nice :)
Solution:
Since $ BCKE$ is harmonic quadrilateral,so if we consider the pencil $ A(BCKE)$ and its intersection with $ \overline{CMN}$,we will obtain
that $ (CN,M\infty) \equal{} \minus{} 1$,hence $ CM \equal{} MN$ and we are done.
BTW:As you can see from my solution,the property of $ ABCD$ is trapezium is not necessary,it is still correct if $ ABCD$ is just a cyclic quadrilateral :wink:

We use complex numbers.

Let $(ABC)$ be the unit circle. Since $ABCD$ is a trapezoid, $ad=bc\implies d=\frac{bc}a$ and $p=\frac{(a+d)c^2-2acd}{c^2-ad}=\frac{a^2c^2+bc^3-2abc^2}{ac(c-b)}=\frac{c(a^2-2ab+bc)}{a(c-b)}=\frac{2ce}{c+e}\\\implies (a^2-2ab+bc)c=e(2ac-2ab-a^2+2ab-bc)\implies e=\frac{c(a^2-2ab+bc)}{2ac-a^2-bc}$.
Additionally, $\frac{b-p}{b-k}=\frac{\bar b-\bar p}{\bar b-\bar k}=\frac{k(1-b\bar p)}{k-b}\implies  k=\frac{p-b}{1-b\bar p}=\frac{\frac{c(a^2-2ab+bc)}{a(c-b)}-b}{1-b\frac{bc-2ac+a^2}{ac(b-c)}}=\frac{c(c(a^2-2ab+bc)+ab(b-c))}{ac(c-b)+b(bc-2ac+a^2)}=\frac{c(ab^2+bc^2+ca^2-3abc)}{a^2b+b^2c+c^2a-3abc}$. For $N$, we have $\frac{a-n}{a-e}=\frac{\bar a-\bar n}{\bar a-\bar e}=\frac{e(1-a\bar n)}{e-a}\implies n+ae\bar n=a+e$ and $\frac{c-n}{a-b}=\frac{\bar c-\bar n}{\bar a-\bar b}=\frac{ab(\bar c-\bar n)}{b-a}\implies n+ab\bar n=ab\bar c+c$. By Cramer's rule, $n=\frac{ab-e(ab\bar c+c-b)}{b-e}$. Similarly, $m=\frac{ab-k(ab\bar c+c-b)}{b-k}$.
$n=\frac{ab-\frac{c(a^2-2ab+bc)}{2ac-a^2-bc}(ab\bar c+c-b)}{b-\frac{c(a^2-2ab+bc)}{2ac-a^2-bc}}=\frac{ab(2ac-a^2-bc)-c(a^2-2ab+bc)(ab\bar c+c-b)}{b(2ac-a^2-bc)-c(a^2-2ab+bc)}\\=\frac{3a^2bc-a^3b-a^3b+2a^2b^2-a^2c^2+2abc^2-bc^3-4ab^2c+b^2c^2}{4abc-a^2b-b^2c-a^2c-bc^2}\\
m=\frac{ab-\frac{c(ab^2+bc^2+ca^2-3abc)}{a^2b+b^2c+c^2a-3abc}(ab\bar c+c-b)}{b-\frac{c(ab^2+bc^2+ca^2-3abc)}{a^2b+b^2c+c^2a-3abc}}\\=\frac{a^3b^2-a^3bc-a^2b^3+2a^2bc^2-a^2c^3+2ab^3c-5ab^2c^2+3abc^3+b^2c^3-bc^4}{b^3c+b^2a^2-4b^2ac+4bac^2-bc^3-a^2c^2}\\=\frac{(c-b)(3abc^2-bc^3-a^2c^2-2ab^2c+a^2bc+a^2b^2-a^3b)}{(c-b)(4abc-a^2b-b^2c-a^2c-bc^2)}$
Indeed,
$c+n=\frac{3a^2bc-a^3b-a^3b+2a^2b^2-a^2c^2+2abc^2-bc^3-4ab^2c+b^2c^2+c(4abc-a^2b-b^2c-a^2c-bc^2)}{4abc-a^2b-b^2c-a^2c-bc^2}\\=\frac{-2a^3b+2a^2b^2+2a^2bc-2a^2c^2-4ab^2c+6abc^2-2bc^3}{4abc-a^2b-b^2c-a^2c-bc^2}=2m,$
as desired.


Hi everybody! Would you mind telling me the name of the hotel where the moroccan delegation is? thanks.

If you are reffering to where the leaders of the Morroco teams are staying, I am not sure if the list have been finalized. For now we are all staying in a nice place at the beach in the middle of nowhere :D


linear algebra
matrix
rotation
analytic geometry
trigonometry
geometry
geometric transformation

NOTE: this was an explanation from me to someone else i was with on the phone, that's why it looks so weird

Hey, for matrix discussion

Ok, this is the idea we're trying to get across. Rotated!

Polar coordinates:

$x=rcosX, y=rcosX$. Therefore trig :)

Here is why: we dictate points in this way, it's just as cool as coordinates

$(x,y)= rcosX, rsinX$
$(x`,y`)= rcos(X+A), rsin(X+A)$

where X is the original angle and A is the angle of rotation (negative in this case)

$x`=rcos(X+A)=r(cosXcosA-sinXsinA)$
$y`=rsin(X+A)=r(sinXcosA-sinAcosX)$

$x`=xcosA-ysinA$
$y`=xsinA+ycosA$

For x-prime, y-prime, you could use x', y' (or x^\prime , y^\prime), which will generate a standard prime as in $x',y'$, rather than x`,y` (and those are backwards anyway :P ).

$x^\prime$ hmm, thanks! makes stuff better


solve:
$x^2-y^3=5^k$
$k \in \mathbb{N} , y \in \mathbb {Z^+} , x \in \mathbb{Z}$

[quote="amin_68"]I think It has no answer for $k=5$. :)[/quote]
why??
as far as I know MR.JAMALI didnt solve it
have you any soloution for $k=5$?? :?

but M.r Jamali guess that it has no answer for $k=5$. :)


The number m is a three-digit positive integer and is the product of the three distinct prime factors x, y, 10x+y, where x ad y are each less than 10. What is the largest possible value of m?

[quote="h78reg"]do you mean x, ym, and 10+y?[/quote]

Sorry, I missed the comma and hit the m. It's x, y, and 10x+y.

[hide]i'm not sure, but i think the answer is 795 using x=5 and y=3. the number must be equal to 10x^2y+xy^2. our choices for x and y are 7,5,3, and 2. x cannot be 7 because our product will always be a 4 digit number. so, x=5 and y=3 should be the largest possibility........[/hide]

[hide] We want $y$ to be the biggest prime smaller than 10, and $x$ the next biggest prime smaller than 10.

  Thus we have $x=5$, $y=7$

$5 \cdot 7 \cdot 17=595$  Thus the largest possible value of $m$ is $\boxed{595}$ [/hide]

[quote="h78reg"]i think 236factorial wanted 10x+y, not 10+y.[/quote]

  ok, then your answer should be right

[quote="236factorial"]The number m is a three-digit positive integer and is the product of the three distinct prime factors x, y, 10+y, where x ad y are each less than 10. What is the largest possible value of m?[/quote]

This was nats target #8!?!?!? :?

[hide]
$17 \cdot 7 \cdot x < 1000$
$x < 9$ so that doesn't restrict us at all
$17 \cdot 7 \cdot 5 = 595$
That was easy...
[/hide]

[quote]Sorry, I missed the comma and hit the m. It's x, y, and 10x+y.[/quote]

so, that doesn't work.

[hide]

The four prime numbers less than 10 are 1, 3, 5 and 7.

The prime factors are x, y and 10x + y

The number of possible values for 10x + y are [tex] 2(_4C_2) = 12[/tex]

The numbers are 13, 15, 17, 35, 37, 57, 31, 51, 71, 53, 73 and 75

if 35 were multiplied by anything greater than 29 it would be over 1000.
so 57 and 75 are eliminated.

if 15 were multiplied by anything greater than 67 it would be over 1000.
so 35 and 53 are possibilities.
since (15)(53) > (15)(35), 35 would be eliminated.

if 21 were multiplied by anything greater than 48 it would be over 1000.
so 37 is a possibility and 73 is eliminated.

3, 5 or 7 would have to be multiplied by a number greater than 100 in order to
obtain a product that has the greatest three-digit number, but since 10x + y can't be greater than 100, 13, 15, 17, 31, 51 and 71 are all eliminated.

so the greatest number is either (15)(53) or (21)(37)

(15)(53) = 795 and (21)(37) = 777 so 795 is the greatest number.
[/hide]

very nice, but the primes less than 10 are 2,3,5,7 not 1,3,5,7. but i think that was just a typo. :)

sorry I typed the problem so wrong, and I don't think chess64's 595 answer is right (maybe tricked by the typo).

I lost the packet this morning  :( , I think it's 795. I always remember an 8 though.
when I find it, I'll tell you what I got.

you are correct h78reg
I checked the numbers 23, 25, 27, 32, 52 and 72
(14)(72) = 1008 so that eliminates 72.
All the other products fall short of 795 so that eliminates the other five numbers.
So 795 is still the greatest three-digit number.


Can/May someone move these threads?
[url]http://www.mathlinks.ro/viewtopic.php?t=193093[/url] to the High School Forum's Intermediate Topics.
[url]http://www.mathlinks.ro/viewtopic.php?t=192256[/url] to G&FF or Round Table.

 :) Thanks.

So- you're saying the astronomy forum needs a moderator?

I guess. Not necessarily. Just thought someone could move them. :)

Well, usually it is moderator's responsibility to move the topics to appropriate places and do similar routine things; admins normally take care of more significant issues. :) Seems like we need to find somebody to moderate a few Arts and Sciences subfora. As a possible solution, I suggest that we do the same as was done with College Playground where each moderator in the group can moderate each of the subfora (though, of course, not all of them bother even to visit all the subfora). I don't think it will make any harm to create the Arts and Sciences moderators group so that such requests could be PM'd to any of them.

I moved the second one to GFF. The first one belongs IMO in the Astronomy forums.


real analysis
trigonometry
function
integration
real analysis unsolved

Hello, I'm studying Chapter 8 of baby Rudin. Can anyone please show me how to do the following??

[hide="Problem 1"]
Compute the Fourier transform of $ e^{\minus{}2 \pi |x|}.$
[/hide]

[hide="Problem 2 (Rudin 8.13)"]
Put $ f(x) \equal{} x$ if $ 0 \leq x < 2 \pi$, and apply Parseval's theorem to conclude that
\[ \sum_{n\equal{}1}^{\infty} \frac{1}{n^2}  \equal{} \frac{\pi^2}{6}.
\]
[/hide]

[hide="Problem 3 (Rudin 8.14)"]

If $ f(x) \equal{} (\pi \minus{} |x|^2)$ on $ [\minus{}\pi, \pi],$ prove that
\[ f(x) \equal{} \frac{\pi^2}{3} \plus{} \sum_{n\equal{}1}^{\infty} \frac{4}{n^2} \cos nx
\]
and deduce that
\[ \sum_{n\equal{}1}^{\infty} \frac{1}{n^2} \equal{} \frac{\pi^2}{6}, \quad
             \sum_{n\equal{}1}^{\infty} \frac{1}{n^4} \equal{} \frac{\pi^4}{90}.
\]
[/hide]

[hide="For Problem 1:"]

The fourier transform, $ \hat{f}(x)$, of a function f is given by $ \hat{f}(x) = \int_{ - \infty}^{\infty} {f(t) e^{ - ixt} \ dt}$.

\begin{eqnarray*} \hat{f}(x) & = & \frac {1}{\sqrt {2\pi}} \int_{ - \infty}^{\infty} {e^{ - 2\pi |t|} \cdot e^{ - ixt} \ dt} \\
& = & \frac {1}{\sqrt {2\pi}} \int_{ - \infty}^0 {e^{t(2\pi - ix)} \ dt} + \frac {1}{\sqrt {2\pi}} \int_0^{\infty} {e^{ - t(2\pi + ix)} \ dt \\
& = & \ldots }\end{eqnarray*}

You should be able to handle it from here.[/hide]

[hide="For Problem 2:"]

The fourier coefficients, $ \hat{f}_n$ (n an integer), of a function f are given by $ \hat{f}_n = \frac {1}{2\pi} \int_0^{2\pi} f(x) \cdot e^{ - inx} \ dx$. Apply this to the function f(x) = x to get the fourier coefficients, $ \hat{f}_n$, of f.

Parseval's Identity states that $ \sum_{n = - \infty}^{\infty} |\hat{f}_n|^2 = \frac {1}{2\pi} \int_0^{2\pi} |f(x)|^2 \ dx$.

Compute the coefficients of the fourier series and put them into Parseval's Identity, and then see if you can get to the desired conclusion.[/hide]

for problem 3 you can get the result on $ \sum_n n^{\minus{}2}$ by evaluating both sides of the identity at an appropriate $ x$, but you will need parseval to get the result on $ \sum_n n^{\minus{}4}$


calculus
integration
calculus computations

here is one more integral problem that has been giving me trouble:

Compute:  [size=150]int(int(4*e^(x^2),x,2y,2),y,0,1)  [/size]

please help

thanks
-Chris

Change the order of integration,
$\int_0^1 \int_{2y}^2 e^{x^2} dxdy = \int_0^2 \int_0^\frac{x}{2} e^{x^2} dydx = \int_0^2 \frac{x}{2} e^{x^2} dx$

I think you can finish it.

[quote="adidasty"]Changing the order of integration took me a while to get used to. I used to get these double integrals and think, "Darn! I can't integrate this!" But then the idea comes to you like a wave out of darkness, "I'll switch the order of integration!" Then you see the light.[/quote]
Switching the order of integration is such a powerful tool that it can be worth introducing an extra variable into a problem just so you can use it. See what I did [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=45582]here[/url] in coliver14's other problem.

A line I've used in advanced classes: If you show an analyst of my training an iterated (double) integral (or a double sum, or an integral of a sum), it's like waving a red cape in front of a bull. The question isn't whether I will switch the order of integration of summation; it's merely the tactical question of what is the best moment for that switch.


The set of numbers between 0 and 180 inclusive is in one-to one correspondence with the set of rays, in the union of a half-plane and its edge, with end point A in the edge of the half-plane. \[ \vec{AJ}\] corresponds to 30 and \[ \vec{AM}\] corresponds to 70. \[ \vec{AK}\] which bisects angle JAM, corresponds to___?

[quote="DaReaper"]The set of numbers between 0 and 180 inclusive is in one-to one correspondence with the set of rays, in the union of a half-plane and its edge, with end point A in the edge of the half-plane.
\[ \vec{AJ}
\]
corresponds to 30 and
\[ \vec{AM}
\]
corresponds to 70.
\[ \vec{AK}
\]
which bisects angle JAM, corresponds to___?[/quote]

[hide="protractor postulate ftw"]
this is basically asking what $ \frac{30\plus{}70}{2}$ is
its just kinda hard to figure out[/hide]

easy
the fact that it was aligning the numbers from 1 to 180 with a straight angle (180 degrees)


algebra
polynomial
function
calculus
integration
real analysis
real analysis unsolved

Suppose I have $n+1$ points in the 2D space. And I made a function $f_{n}$ of degree $n$ that passes through all those points.

There is a specific point (or more) that the distance of this point to all others points are minimum.

Or mathematically
[quote]
${f_{n}(x)=\sum_{i=0}^{n}\beta_{i}x^{i}}$

the set of points
${1\leq s\leq n+1}$
${\left( \begin{array}{ll}\alpha_{s}& f\left(\alpha_{s}\right) \end{array}\right)}$

I want to find an $u$ that minimizate:
${\sum_{k=1}^{n+1}\left|\int_{\alpha_{k}}^{u}\sqrt{1+\left(\frac{\partial f_{n}(t)}{\partial t}\right)^{2}}\, dt\right|=K}$
[/quote]

I think I write correctly, hope that there is no error.

As written, you seem to mean the "distance along the curve" rather than the Euclidean distance on the plane. If so, the setting of the problem becomes completely irrelevant and the only thing that matters is that if you have an arbitrary curve $\gamma$ with points $a_{1},\dots,a_{n}$ on it in this order, then, for every point $a$ on the curve, $d(a,a_{k})+d(a,a_{n-k})\ge d(a_{k},a_{n-k})$ with equalities attained when there are equally many points $a_{k}$ on both sides of $a$. It means that one should have $a=a_{(n+1)/2}$ when $n$ is odd and one can choose $a$ to be anything between $a_{n/2}$ and $a_{(n+2)/2}$ when $n$ is even.

if it is irrelevant or not, i dont know,

i think is interesting.

If i was looking for euclidian distance, so, there was no need to have a function $f$

suppose i didnt said nothing about the mean, so, there is a easy way to find the minimum value of $K$?!

Yes, it is as easy as to add up a few integrals: $K=\frac{1}{2}\sum_{k}\Bigl|\int_{\alpha_{k}}^{\alpha_{n-k}}R(t)\,dt\Bigr|$ assuming that $\alpha_{1}<\dots<\alpha_{n}$ where $R(t)$ is your square root. By saying that the setting was irrelevant I meant that there was no need to assume that the function was a polynomial, etc. The whole problem is just a variant of an old question about where on a road to build a school to make the total distance travelled by the kids in several towns on the same road minimal. In this case the "road" is just the graph of $f$ and the "towns" are located at $(\alpha_{k},f(\alpha_{k}))$.

i didin't know this old problem.

but how can i find the minimum!?

and thanks for your explanation


You can write each k+1 digit number as n*10^k, where 1<=n<10, so taking the [log] of that will yield k. We then add one to that.

n is the number of digits of x iff 10^(n-1) <= x < 10^n.
Take log (which is strictly increasing) : n-1 <= log_10(x) < n.

A n-digit number N is a(10^(n-1))+b(10^(n-2))+...k. Log(N) equals n-1 since n-1 is the smallest power of 10<or = N. Then we add 1 to that.


Solve the equation
$(x+ 3\sqrt{x} + 2)(x+ 9\sqrt{x} + 18) = 168x.$

Solution[hide]Let sqroot(x) be y.y>= 0.Expanding LS and bringing 168x on other side, get:
y ^ 4+12 * y ^ 3 - 121 * y ^ 2 + 72 * y + 36 = 0.
Factoring,(y ^ 2-7 * y+6)(y ^ 2+19 * y+6)=0
Get y=1, y=6 from the first bracket.For the 2nd bracket both values of y are negative -- impossible.
x=1,x=36.[/hide]


I wasn't sure who to ask about this or where to post this, but it seems this is as good a place as any; I apologize if I'm mistaken.  In one of the examples under Basics on the LaTeX section of the site, the solution to USAMTS Problem 4/4/15, line number 74 contains an error - it states that abc = -(1/3)sin(3x), a statement contradicted later in the problem, where it is assumed (correctly) that abc = -(1/[b]4[/b])sin(3x).  It would seem a typo was made, and it would greatly clarify the example if this were corrected.

Thanks for pointing that out - I hope to clean that up by the end of the week.


What if my solution to a USAMTS problem is more than one page? Should I just continue on another page? What do I put under "problem" to make sure that it doesn't get mixed up and mistaken for an incomplete solution?

Thanks.

The solution template will print the correct problem number on every page of that problem (assuming you're using LaTeX; if you're handwriting then I'm not sure of the procedure, but I assume it's the same - just write the correct problem number at the top of every page of that problem).  Your solution can spill over onto as many pages as you wish, and there will be no problem.

If you're handwriting your solutions on our blank solution sheets, and a problem goes over more than one page, then the bottom of that problem's pages should say "Page 1 of 2" and "Page 2 of 2" (or something similar if it's more than 2 pages).

Also, our graders can almost always tell if a problem page is missing or out of order.  As long as your name, your USAMTS ID #, and the round and problem number is on the top of every page, we'll figure it out.

So... if we go over one side of a page, does it have to be on a new sheet of paper, or can we keep writing on the back?

We don't have a fixed rule about this, but it's a lot easier to grade if everything is written on the front side only of each page.

Well, I already did that with one. Do I need to change it, or can I just leave that one and do everything in the future that way?

just finish it as you are going now (unless you prefer to rewrite for neatness or something) then next time just write on one side.


vector
limit
real analysis
real analysis unsolved

After this one
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=22143

Here a version in $E$ normed vector space

Let $ (E, ||...||) $ normed vector space, $f:[0,1] \rightarrow E$
continous at $0$, and suppose $\lim_{x\rightarrow 0^{+}}\frac{f(bx)-f(x)}{x}=L$ exist, where $b\geq 2$ integer.
Prove that $f$ is derivable at $0$

We can suppose $f(0)=L=0$

consider

$g(x)=f(x)-f(0)-kx$

$k$ depend $L,b$


[i]Find all function $ f : \mathbb{R} \ \to \ \mathbb{R}$ such that :

   $ f(x\plus{}f(y)) \ \equal{} \ 3f(x) \plus{} f(y) \minus{} 2x \ \forall \ x;y \ \in \ \mathbb{R}$[/i]

Let $ g(x)\equal{}f(x)\minus{}x$. Then substituting this in, we have $ g(x\plus{}y\plus{}g(y))\equal{}3g(x)$. Letting $ g(0)\equal{}a$, we plug in $ y\equal{}0,a$ to obtain $ g(x\plus{}a)\equal{}3g(x)\equal{}g(x\plus{}a\plus{}g(a))$. Plugging in $ x\equal{}0$ into the first part of the equation, we have $ g(a)\equal{}3g(0)\equal{}3a$, so $ g(x\plus{}a)\equal{}g(x\plus{}4a)$, or $ g(x)\equal{}g(x\plus{}3a)$.
We have $ g(a)\equal{}3a$, so $ g(a\plus{}a)\equal{}3g(a)\equal{}9a$, and $ g(2a\plus{}a)\equal{}3g(2a)\equal{}27a$. Thus, we have $ g(0)\equal{}a, g(3a)\equal{}27a$, and since these two are equal, we conclude $ a\equal{}0$. Now plugging in $ y\equal{}0$ to the original equation gives us $ g(x)\equal{}3g(x) \iff g(x)\equal{}0$, so $ f(x)\equal{}x$ is the only solution.