MATHCOUNTS AMC AIME geometry algebra polynomial Vieta

Trend. As people get more practice, harder problems come.

[quote="Arachnotron"]Has the IMO become harder in recent years, or have the past few years just been anomalies for the most part? I look at the 90's and even early 2000's problems and they are much easier than the stuff from 2006 onwards.

Is this a trend or just coincidence?[/quote]

Should be.

A lot of the other contests sort of did the same.
Ex: MATHCOUNTS, AHSME/AMC 12, AIME

Also, the math community has been rapidly growing in size since websites like AoPS opened up.  So now that there's a bigger pool and more interest, the best students are now better than the best were before.

As an example, take say the 2003 State Sprint Round.

On that round, most people on this website would easily get 25+ sprint and 5+ target very easily, but now, attaining such a score is significantly harder.

Now, if anyone has a 1984 School Sprint Round, I'd like to see the difficulty(for free please  :lol: ) :rotfl:

[quote="Maybach"]On that round, most people on this website would easily get 25+ sprint and 5+ target very easily, but now, attaining such a score is significantly harder.[/quote]

I'm pretty sure that the 2003 MATHCOUNTS wasn't any more difficult (if so, very subtle changes) than 2009 :wink:

1984 School Sprint #3: Subtract 27.03 - 15.234
#4 : Simplify $ \sqrt{121}$.
.
.
.
#39: Find $ 76.076 \div 76$.
#40: What is the product of the digits of the sum $ 41874\plus{}3938\plus{}14676$?

[quote="AIME15"][quote="Maybach"]On that round, most people on this website would easily get 25+ sprint and 5+ target very easily, but now, attaining such a score is significantly harder.[/quote]

I'm pretty sure that the 2003 MATHCOUNTS wasn't any more difficult (if so, very subtle changes) than 2009 :wink:

1984 School Sprint #3: Subtract 27.03 - 15.234
#4 : Simplify $ \sqrt {121}$.
.
.
.
#39: Find $ 76.076 \div 76$.
#40: What is the product of the digits of the sum $ 41874 \plus{} 3938 \plus{} 14676$?[/quote]

I was saying 2003 MC was EASIER

and wow, wouldn't #4 have two solutions(-11 and 11).  I'm sure though the answer was just 11.

Also, I assume they weren't even trying to escalate difficulty as the test goes on.  That stuff is truly stuff taught in normal 6th grade math classrooms.

I meant to say "I'm pretty sure that the 2003 MATHCOUNTS wasn't really any easier (if so, very subtle changes) than 2009"

[quote="Maybach"][quote="AIME15"]#4 : Simplify $ \sqrt {121}$.[/quote]
and wow, wouldn't #4 have two solutions(-11 and 11).  I'm sure though the answer was just 11.[/quote]
$ \sqrt{121}$ indicates the principal square root of 121, which is just 11. (This is why $ \sqrt{x^2}\equal{}|x|$.)

[quote="Arachnotron"]So, do you guys think there will eventually be no more (good) IMO problems to ask? I mean, will there always be good questions, or will there be a point where we've asked all that can be asked?[/quote]

There will always be more problems (or variations of problems).  As mathematics continues to evolve, so do the problems that can be asked.

[quote="Arachnotron"]So, do you guys think there will eventually be no more (good) IMO problems to ask? I mean, will there always be good questions, or will there be a point where we've asked all that can be asked?[/quote]

There will always be good questions to ask. There have been good mathematics questions to ask for the last 2000+ years. I don't think they are going to run out anytime soon. But what may be a good question in one era may be different that what is considered a good question in another era.

Well, mathematics evolves, but aren't competition problems drawing from a pretty static (geometry, basic number theory, elementary combinatorics, algebra) subset of mathematics?

I would guess that new formulas and techniques would make old problems trivial. Some problems that were once thought too long or tedious can now be put in because of these new techniques.

Ihatepie is right, example:

IMO'88/6 => killing NT, now everyone can solve it because everyone knows Vieta Jumping

Really the states started getting harder to where we are now, about 1990s

By the way, we went from IMO to Mathcounts :P

Or, looking at it over even more long range, tastes in what are considered "good" problems change over time. Competitive/Recreational problem solving was not born with the IMO. It has been a sport since ancient times.

In ancient times, solving what are today considered routine, though sometimes computationally tedious, word problems was considered cool.

In the 1500's, solving hard algebraic word problems (basically word problems that reduced to solving cubic or quartic polynomials by hand) was considered interesting and cool. Today, such problems would be considered mundane and boring.

In the 1800's, problems in computing difficult integrals were found on the Cambridge Tripos exams but today, you won't find such problems on either the modern Tripos exam or the Putnam.

Basically, contest questions reflect, to some degree, what is taught in schools as well as tastes in what is considered interesting. Both of those things change with the times.

Even though the current pool of "theorems" to base problems upon for high school contests is finite (yet slowly changing), there is a large number of ways to combine skills or to disguise problems. So there won't be any shortage of interesting problems to ask. And since there are so many problems out there - even if a question is repeated a few times between contests somewhere in the world, it might be unlikely that students sitting for a given contest are familiar with that question already. Because no one can solve every question posed in every contest from every year around the world.

[quote="Arachnotron"]Has the IMO become harder in recent years, or have the past few years just been anomalies for the most part? I look at the 90's and even early 2000's problems and they are much easier than the stuff from 2006 onwards.

Is this a trend or just coincidence?[/quote]
While this is certainly not true on a year by year basis, there is no doubt that around every 10 years the contest gets harder.

Also. there are still many difficult shortlist problems in the 1999-2006 range, they just don't make it on to the test.

It's nice to know that there will always be a beautiful IMO Shortlist to look forward to each year :)

[quote="AlphaBetaTheta"]It must, everything must get harder. As the world population increases the competition to be the best math person increases. Every year people must work extra harder to win because of the competition.[/quote]
The problem difficulty is increasing at a much faster rate than world population is growing, methinks

As for beautiful IMO shortlist.... for some of the problems, I would not be so sure about "beautiful"  :P

Well, on the AHSME, AHSME grew easier because the more elementary problems were deleted/changed

Because now they are considered more mundane or routine

What would you guys say about the quality of the shortlist in general though?


function
calculus
complex analysis
topology
advanced fields
advanced fields unsolved

A differentiable function on the real line is continuous. The example of $f(x) = |x|$ shows that a continuous function may fail to be differentiable at a point. The experience of a first calculus course may lead a student to intuitively suspect that a continuous function must be differentiable "most of the time". This is false, as can be shown by the continuous but nowhere differentiable Weierstrass example:

$\displaystyle f(x) = \sum_{k=1}^\infty (\frac34)^k \cos(4^kx)$.

Somewhat later, you learn about analytic functions. A function is analytic at a point if there is a neighborhood of that point on which the function is given by a power series centered at that point. An analytic function must be $C^\infty$, but the converse is not true, as can be shown by such examples as $f(x) = e^{-1/x}$ for $x>0$ and zero otherwise. That example is $C^\infty$ everywhere but fails to be analytic at 0.

So here's the question: is there a $C^\infty$ function on the line that is not analytic at any point?

One example, very much like yours and also from Weierstrass I think, is :

f(x) =  :Sigma: sin(a^n*x)/n!

with a odd integer > 1.

I think you're probably right, belenos, but there's still some heavy lifting to be done. I'll start with a small question: what purpose (or convenience) in the proof is served by insisting that [i]a[/i] be an [i]odd[/i] integer?

Sorry for late answer !

I'll try to find again this example. I don't remember why 'a' has to be even  :? Maybe I just wrongly remember ....

[quote="Kent Merryfield"]
So here's the question: is there a $C^\infty$ function on the line that is not analytic at any point?[/quote]

From Rudin's book chapter 19 Holomorphic Fourier Transform page 385
exercice 13

$ f(z) = \sum_{k\geq 1}c_k.e^{i.\lambda_k.x}$ where
$ c_k $, $ \lambda_k $

are sequences of positive numbers, chosen so that

$ \sum c_k\lambda_k^{n} < \infty $

and $ c_n\lambda^n_n $

increases   very
rapidly and is much larger than the sum of all the other terms in the serie

$ \sum c_k \lambda^n_k $

$f$ is C-oo such that the power series $ \sum \frac{f^{(n)}(a)}{n!}(x-a)^n $
has radius of convergence $ 0 $ for every $ a\in R $

My own favorite version (and I've got it written up somewhere, with all the details) is
$\displaystyle \sum_{k=1}^\infty 2^{-2^{k/2}}e^{i2^kx}$.

Now who can tell me why this works? It's not that hard to see why such a function is $C^\infty$ (just differentiate the series term by term), but why can't it be analytic anywhere? What's so special about lacunary trigonometric series, anyway?

Well, in fact it can be prove that the "most" $C^\infty$ functions in and interval are not analitic in any point, in the sense that in certain topology the analitic in almost  one point functions form an empty interior set.
This fact you can found in James Dujungis's topology book.
If I'm not wrong he use the distance \[d(f,g)=\dis\sum\frac{\sup_{x\in [a,b]}(f^{n}(x)-g^{n}(x))}{2^n}\]
Prove that $C^\infty[a,b]$ is complete with this metric, so it can be use the baire' s theorem that says that in such spaces the union of empty interior closed sets is empty interior too. Well finally he prove that the one point analitic functions is this kind of set,...

A natural question is if this result is true in higher dimensions, in any open set?
I think it is, anyway if generaly it isn't in which open sets it is?
If anybody knows something about it, please let me know....


Given cos(x)=-1/9
with 90o < x < 180o. Find cos(1/2x)

could you help me step by step.. much appreciated

Try using $\text{\LaTeX}$.
Your question is ambiguous. I don't know if you want me to find $\cos(\frac{1}{2x})$ or $\cos(\frac{x}{2})$.

I can do the latter:
[hide="Click me!"]We know that $45<\frac{x}{2}<90$. So the cosine is positive.
Since $\cos(\frac{x}{2}) = \sqrt{\frac{1+\cos(x)}{2}}$,
$\cos(\frac{x}{2})$
$= \sqrt{\frac{1+\cos(x)}{2}}$
$= \sqrt{\frac{\frac{8}{9}}{2}}$
$= \sqrt{\frac{4}{9}}$
$= \frac{2}{3}$
[/hide]

[quote="vishalarul"]Try using $\text{\LaTeX}$.
Your question is ambiguous. I don't know if you want me to find $\cos(\frac{1}{2x})$ or $\cos(\frac{x}{2})$.

I can do the latter:
[hide="Click me!"]We know that $45<\frac{x}{2}<90$. So the cosine is positive.
Since $\cos(\frac{x}{2}) = \sqrt{\frac{1-\cos(x)}{2}}$,
$\cos(\frac{x}{2})$
$= \sqrt{\frac{1-\cos(x)}{2}}$
$= \sqrt{\frac{\frac{10}{9}}{2}}$
$= \sqrt{\frac{5}{9}}$
$= \frac{\sqrt{5}}{3}$
[/hide][/quote]

I believe that $\cos(\frac{x}{2})=\sqrt{\frac{1+\cos(x)}{2}}$ instead of $\sqrt{\frac{1-\cos(x)}{2}}$ ;)

Just got done with a long arduous Precalc unit over that.


advanced fields
advanced fields unsolved

find the next two numbers in this sequence

594, 487, 566, 493, 310, 447, ____, ____


What`s going on with it?Is there going to be a new edition anyway?

This is posted in the wrong forum. However, the answer is yes there will be a 6th edition. I will make the necessary announcement very soon on the board (either today or tomorrow) about it ;)


calculus
integration
function
calculus computations

Find

$ \sum_{k\equal{}1}^{\infty}\frac{1}{{n \choose k}}$.

If I am not making a mistake, this pretty well known problem does not require the use of Calculus, that is, this problem can be moved to the Olympiad section.

$ \frac {1}{{n \choose k}}\equal{}\int_0^1 (1\minus{}x)^kx^{n\minus{}1}\ dx$.

[quote="Mr.Doe"]Find

$ \sum_{k \equal{} 1}^{\infty}\frac {1}{{n \choose k}}$.[/quote]
doe, i don't think this is summable in a [b]closed [/b]form, at least i can't do it. :|

$ \sum_{k \equal{} 1}^{\infty}\frac {1}{{n \choose k}}\equal{}\sum_{k \equal{} 0}^{\infty}\frac {1}{{n \choose k}}\;\minus{}\;1$

$ \equal{}\left(\sum_{k \equal{} 0}^{\infty}\;\dfrac{\Gamma{(k\plus{}1)}\cdot\Gamma{(n\minus{}k\plus{}1)}}{\Gamma{(n\plus{}1)}}\right)\;\minus{}\;1$
$ \equal{}(n\plus{}1)\sum_{k \equal{} 0}^{\infty}\left(\int_0^1\;x^k\,(1\minus{}x)^{n\minus{}k}\;dx\right)\;\minus{}\;1$
$ \equal{}\minus{}(n\plus{}1)\left(\int_0^1\dfrac{1}{(1\plus{}x)\cdot(1\minus{}x)^{n\plus{}2}}\;dx\right)\;\minus{}\;1$

i get an answer in terms of a hypergeometric function,
$ \sum_{k \equal{} 1}^{\infty}\frac {1}{{n \choose k}}\;\equal{}\;\boxed{\left(\;_2\,\textbf{HG}\,_1\,(1\,,1\,;\minus{}n\,;\minus{}1)\;\right)\;\minus{}\;1}$


A MUCH simpler solution exists for this inequality.

[hide] Using Cauchy:

(a + b + c + d)(1 / a + 1 / b + 1 / c + 1 / d) :ge: 16

(1 / a + 1 / b + 1 / c + 1 / d) :ge: 1

Now, using Cauchy again,

4 (LHS) :ge: (a + b + c + d + 1 / a + 1 / b + 1 / c + 1 / d)2

Which means, LHS :ge: (16 + 1)2 / 4 = 289 / 4.[/hide]

[quote="Osiris"]A MUCH simpler solution [/quote]

Indeed !


I know many colleges require at least two or three SAT IIs. Since my strengths are in math, is it OK if I take both Math I and Math II (then maybe chemistry for my third one)? I don't think there's a rule against it but would it look bad?

Any input is welcome, thanks.

I've seen a big shift in the last year or so from people having the Level 1 test to them having the Level 2. (Incidentally, some research I was doing this year into the grades of freshmen taking either Precalculus or Calculus I suggest that the SAT II Level 2 does a better job of predicting grades than the SAT I.) I think (I'm not totally sure of this) that the University of California system made a decision to stop accepting the Level 1 test as meeting their requirements. That's not binding on universities elsewhere in the country, but it may be the start of a trend.

My advice. Take the Level 2. Don't take the Level 1. Then take something else, maybe Chemistry, as a second. You may or may not need a third - check on that.

[quote="Kent Merryfield"]My advice. Take the Level 2. Don't take the Level 1. Then take something else, maybe Chemistry, as a second. You may or may not need a third - check on that.[/quote]

It used to be that the writing SAT II was required at most "top-tier" programs (along with two others of your choice), but since that's now been envelopped into the SAT I, I imagine many programs have dropped that extra requirement.

There is a small group of colleges that still require three SAT II tests for admission, and a larger group that require two. Homeschooled applicants or any other applicants who want to demonstrate that their high school course work was rigorous are well advised to take more SAT II tests than are strictly required. One college, Stanford, only "recommends" SAT II testing, but admits rather few students without good SAT II scores. A College Confidential thread

http://talk.collegeconfidential.com/showthread.php?t=229607

has more details about recent requirements.

Check individual colleges for WHICH tests they prefer or require among SAT II Subject Tests. I agree with the advice that it is more convincing to take the SAT II Math Level 2 test (which is still an EASY test for most AoPSers) than the SAT II Math Level 1 test.


linear algebra
matrix
limit
linear algebra unsolved

Let $ \lambda_{1},...,\lambda_{n}$ be eigenvalues of $ n\times n$ matrix $ A$ and  $ \mid \lambda_{1}\mid \ge\mid \lambda_{k}\mid$  ($ k \in \{2,...,n\}$).Show that:
$ \lim_{m\to\infty}tr(A^m)^\frac{1}{m} \equal{}\lambda_{1}$

payman_pm,
Take $ A\equal{}\begin{pmatrix}
\minus{}1 &0\\
0 & 0
\end{pmatrix}$
$ (tr(A^{2m\plus{}1}))^{1/(2m\plus{}1)}\equal{}\minus{}1$ and not $ 1$,
[quote="payman_pm"]...Show that:
$ \lim_{m\to\infty}tr(A^m)^\frac {1}{m} \equal{} \lambda_{1}$[/quote] so there is at least a small typo.

[quote="JC_math"]payman_pm,
Take $ A \equal{} \begin{pmatrix} \minus{} 1 & 0 \\
0 & 0 \end{pmatrix}$
$ (tr(A^{2m \plus{} 1}))^{1/(2m \plus{} 1)} \equal{} \minus{} 1$ and not $ 1$,
[quote="payman_pm"]...Show that:
$ \lim_{m\to\infty}tr(A^m)^\frac {1}{m} \equal{} \lambda_{1}$[/quote] so there is at least a small typo.[/quote]
I think
$ \lim_{m\to\infty}\mid tr(A^m)^\frac {1}{m}\mid \equal{} \mid  \lambda_1 \mid$
works.
we have:
$ tr(A^m)\equal{}\lambda_1^m\plus{}...\plus{}\lambda_n^m$ so
$ \lim_{m\to\infty}\mid tr(A^m)^\frac {1}{m}\mid\equal{}\lim_{m\to\infty}\mid(\sum_{i\equal{}1}^{n}(\lambda_i)^m)^{(\frac{1}{m})} \mid\equal{}\mid\lambda_1\mid$


let H be a finite set of positive integers none of which has a prime factor grester than 3. show that the sum of the reciprocals oh the elements of H is smaller than 3

The sum of the reciprocals of the elements of $ H$ is bounded above by

$ \left( 1 \plus{} \frac{1}{2} \plus{} \frac{1}{2^2} \plus{} \frac{1}{2^3} \plus{} ... \right) \left( 1 \plus{} \frac{1}{3} \plus{} \frac{1}{3^2} \plus{} \frac{1}{3^3} \plus{} ... \right) \equal{} 2 \cdot \frac{3}{2} \equal{} 3$.


You are given the plan for a five-room house, as shown, and you are asked to draw a continuous line that passes through each of the 16 wall-segments once and only once.
Is there a path that solves the puzzle?  If there is no such path, prove that this is the case.

[quote="7h3.D3m0n.117"][hide]

No there is no linear line that exists.

Proof: Well a single, linear line cannot cross 3 walls in a row (like 15, 12, and 6) [b]and[/b] cross the other walls as well.

[/hide][/quote]

i dont think he means like straight linear line.  just like connected line segments.

Random guess:

Put the diagram on a sphere, and then draw a line that goes around the sphere infinitely and eventually intersects every segment.

[hide="Complete Solution"]Consult the famous bridges of Konigsburg.
I'll post a picture which translates this into terms of the famous problem.
So basically, wherever there is a room, my picture has an island, and wherever there is a wall to cross, my picture has a bridge. I've counted the exterior as one island, because we can move around outside the shape at will.
[img]8438[/img]
Basically, we see that if a network has more than two odd vertices, it does not have an Euler path.
From my picture, we can see that the two big islands and the one small middle island all have an odd number, and finally the outside "box" island has 9 edges connect to it, which makes more than two odd vertices, so by Euler's famous solution, this network has no Euler path and thus no line can be drawn as requested.
[/hide]
EDIT: Click my picture to see all of it. Apparently it zooms it out on the page, so several lines are missing. Just zoom in to see what I meant...

Cool problem by the way.


algebra
system of equations
number theory proposed
number theory

There are 3 equations:
1) $x^{4}-y^{4}=z^{2}$ (Pherma prove, that this equation had not integer solution $xyz\not =0$)
2)$x^{4}-y^{4}=2z^{2},$
and
3)$x^{4}-y^{4}=3z^{2}$.
For 2 and 3 works method Pherma.

[quote="Fitim"][quote="Albanian Eagle"]show that the equations
$x^{4}-y^{4}=z^{2}$
$x^{4}-y^{4}=2z^{2}$
$x^{4}-y^{4}=3z^{2}$
do not have solutions in $\mathbb{N}$[/quote]

$x^{4}-y^{4}-x^{4}+y^{4}=3z^{2}-2z^{2}$
$z^{2}=0$, so $z=0$.
Now $z$ doesn't is, in natural numbers![/quote]

It's not a system of equations but three different problems.

Let me add also another similar equation to be solved.
$x^{4}-y^{4}=2007 z^{2}$  :)
Maybe someone will complete the solutions of the equations in the first post (or I will do it after one week if no one is interested).
remember they were lemmas [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=hard&t=28590]here[/url] :wink:

[quote="nguyentrang"]Can you show how to solve the equations, Albanian Eagle?[/quote]

So, can you?   :lol:

I can but I don't have time right now,
why don't you try to. This article here will help for sure:
http://reflections.awesomemath.org/2007_6/factoring_lemma.pdf

[quote=jUaNi]What about the equation $ x^4\plus{}y^4\equal{}2z^2$?[/quote]

Trivial solutions over the integers.

If $x^4+y^4=2z^2$ then

$(x^4-z^2)^2=z^4-(xy)^4$  this is the first equations at first post.


The expression $ 21x^2 \plus{} ax \plus{} 21$ is to be factored into two linear prime binomial factors with integer coefficients. This can be one if $ a$ is:

$ \textbf{(A)}\ \text{any odd number} \qquad\textbf{(B)}\ \text{some odd number} \qquad\textbf{(C)}\ \text{any even number}$
$ \textbf{(D)}\ \text{some even number} \qquad\textbf{(E)}\ \text{zero}$

[hide="Click for solution"]
Let the factors be $ Ax\plus{}B$ and $ Cx\plus{}D$.  Then $ (Ax\plus{}B)(Cx\plus{}D)\equal{}ACx^2\plus{}(AD\plus{}BC)x\plus{}BD\equal{}21x^2\plus{}ax\plus{}21 \implies AC\equal{}21,BD\equal{}21$.  Since $ 21$ is odd, all its factors are odd.  Therefore, $ A$ and $ C$ are odd.  The same is true for $ B$ and $ D$.  Since the product of two odd numbers is odd, $ AD$ and $ BC$ are odd numbers, and $ a\equal{}AD\plus{}BC$, so $ a$ is $ \boxed{\textbf{(D)}}$.
[/hide]


integration trigonometry function algebra domain calculus limit

[quote="coffeym"]Find

$\int^{\beta}_{\alpha}{\frac{1}{\sqrt{(x-\alpha)(\beta-x)}}}\,dx$

Where $\alpha,\beta\in\Re$ and $\beta>\alpha$

If you can't get started, I suggest [hide="this substitution"]
$x=\alpha\cos^{2}{\theta}+\beta\sin^{2}{\theta}$[/hide][/quote]

It looks like a beta function, so make the linear substitution that transforms the domain of integration into $0\leq y\leq1,$ namely let $y=\frac{x-\alpha}{\beta-\alpha}\Rightarrow dy=\frac{dx}{\beta-\alpha}$ so that $x-\alpha = (\beta-\alpha )y$ and that $\beta-x = (\beta-\alpha )(1-y)$ and the integral becomes

$\int^{\beta}_{\alpha}{\frac{1}{\sqrt{(x-\alpha)(\beta-x)}}}\,dx = \int_{0}^{1}\frac{(\beta-\alpha )dy}{\sqrt{(\beta-\alpha )^{2}y(1-y)}}= \int_{0}^{1}y^{-\frac{1}{2}}(1-y)^{-\frac{1}{2}}\ dy$
$= \mbox{B}\left( \frac{1}{2},\frac{1}{2}\right) = \frac{\Gamma^{2}\left( \frac{1}{2}\right) }{\Gamma(1)}= \frac{(\sqrt{\pi})^{2}}{1}= \pi$

[color=darkblue]Denote $\frac{\beta+\alpha}{2}=u$, $\frac{\beta-\alpha}{2}=v>0$, i.e. $\alpha=u-v$, $\beta=u+v$ and $I\equiv \int_{\alpha}^{\beta}\frac{1}{\sqrt{(x-\alpha )(\beta-x)}}\ dx\ .$

Therefore, $I=\int_{u-v}^{u+v}\frac{1}{\sqrt{(x-u+v)(u+v-x)}}\ dx\ .$ With the substitution $x: =\phi (t)=t+u$ obtain

$I=\int_{-v}^{v}\frac{1}{\sqrt{(v+t)(v-t)}}\ dt=$ $\int_{-v}^{v}\frac{1}{\sqrt{v^{2}-t^{2}}}\ dt=$ $2\cdot\int_{0}^{v}\frac{1}{\sqrt{v^{2}-t^{2}}}\ dt=$ $2\left[\arcsin\frac{t}{v}\right]_{0}^{v}=\pi$ $\Longrightarrow$ $\boxed{\ I=\pi\ }\ .$

[b]A similar exercises.[/b] Let $a,b$ be two real numbers for which $a<b\ .$ Then :[/color]

$1.\blacktriangleright\ \int_{1}^{3}|(x-1)(x-2)(x-3)|\ dx=\frac{1}{2}\ .$

$2.\blacktriangleright\ \int_{a}^{b}\sqrt{(x-a)(b-x)}\ dx=\frac{(b-a)^{2}\cdot\pi}{8}\ .$

$3.\blacktriangleright\ \lim_{n\to\infty}n\cdot\sum_{k=0}^{n-1}\int_{\frac{k}{n}}^{\frac{k+1}{n}}\sqrt{\left(x-\frac{k}{n}\right)\left(\frac{k+1}{n}-x\right)}\ dx=\frac{\pi}{8}\ .$

$4.\blacktriangleright\ \lim_{n\to\infty}\left(\int_{a}^{b}[(x-a)(b-x)]^{n}\ dx\right)^\frac{1}{n}=\left(\frac{b-a}{2}\right)^{2}\ .$


calculus
derivative
calculus computations

How can you find the rate of change of a population of size $ P(t)\equal{}t^3\plus{}4t\plus{}1$ at time $ t\equal{}1$?

Wouldn't you just find the derivative and plug in $ t$?

So
$ P'(t) \equal{} 3t^2 \plus{} 4$
$ 3(1) \plus{} 4 \equal{} \boxed{7}$


$ Ag(NO)_3 \plus{} 3HCL \rightarrow AgCl_3 \plus{} 3HNO$
Then just factor label.

i think the equation would be something like:
AgNO3 + HCl = AgCl + HNO3
i don't think there is some formula like AgCl3 because silver have only one oxidation number which is positive 1.

my mistake you are correct.
chem is a little rusty  :oops:

The correct chemical equation is

$ \ce{AgNO3} \plus{} \ce{Cl\minus{}} \longrightarrow \ce{AgCl} \plus{} \ce{NO3\minus{}}$.

And although the most common oxidation number of silver is indeed +1 there are some species where its oxidation number is different.

just curious, but how would one usually detect the end point of this reaction?

With $ \ce{K2CrO4}$. Now as an exercise explain how this compound acts as an indicator in this precipitation titration.

[quote="mr. math"]What mass of chloride ion is present in a sample of water if 15.7 mL of 0.108 M AgNO3 (Silver Nitrate) is required to titrate the sample?

I'm having trouble getting the chemical formula. After I get this, I'm pretty sure I can solve the problem.[/quote]
Silver and chloride ions form the insoluble salt silver chloride, AgCl, using one mole of silver (or of silver nitrate) per mole of chloride ion. Setting aside the issue of how you'll know when all the chloride is titrated (used up), you will need one mole of silver nitrate per mole of chloride ion titrated.
How many moles of silver nitrate are in 15.7 mL of a 0.108 M solution (containing 0.108 moles per liter)?
That's the number of moles of chloride ion titrated. The mass of one mole of chloride ion is 35.45g.


function
calculus
integration
calculus computations

I was just wondering a few things...
If the function f is an antiderivative of g then d/dx*g(x)=f(x) is this right or wrong?
Also
If d/dx f(x)=g(x) then F(X)+4 is an antiderivative of g (X) I think this is wrong am I right?

Also I am having a hard time with these two problems I believe I have the correct answer but if I don't could you solve them for me?

#1 evaluate the integral of (-6u-5)(3u-8)du
I think the answer is (3u-8)(6u-5)x

#2 evaluate the integral (6+-1)du
I think the answer is (6+-1)x but it must be harder than that!

Again I would like to thank you all for your help!

the function $ f$ is an antiderivative of $ g$ means $ \frac{df}{dx}(x)\equal{}g(x)$....


geometry
incenter
circumcircle
analytic geometry
geometry proposed

ABC is a triangle. P and Q are the feet of the angle bisectors from A and B; D and E are the feet of the altitudes from A and B. I is the incentre, and O is the circumcenter of triangle ABC. Prove: D, E and I are collinear iff P, Q and O are collinear.

[color=red][Moderator edit: Later discussed at http://www.mathlinks.ro/Forum/viewtopic.php?t=50582 .][/color]

This was the problem 4 of the 2nd Round of the Bundeswettbewerb Mathematik (German National Mathematics Competition) 2002. It is very easy with trilinear coordinates (or with barycentric coordinates) and very hard without these. (In fact, there is a (looong) synthetic solution in German on [url=http://www.bundeswettbewerb-mathematik.de/aufgaben/aufgaben.htm#Loesungen]the BWM website[/url] (see "Lösungen2002.2"). I have never had the patience to read it through...)

  Darij

Hey, do anybody can explain me what are barycentric coordinates?. Is it the same as trilinear coordinates?, where can i find a place with some information about this?:?  . Thanks  :)

I can't remember where I found this file, so I'm attaching it to this message. It's a very good introductory paper by Paul Yiu.

[quote="Leonardo"]Hey, do anybody can explain me what are barycentric coordinates?. Is it the same as trilinear coordinates?[/quote]

Not exactly: If ( x : y : z ) are the trilinear coordinates of a point with respect to a triangle ABC, then its barycentric coordinates with respect to triangle ABC are ( ax : by : cz ).

Thanks, Grobber, for mentioning Paul Yiu's paper. It can be found at http://www.math.fau.edu/yiu/barycentricpaper.pdf

Also see his website http://www.math.fau.edu/yiu/Geometry.html - it contains some more texts about barycentric coordinates and triangle geometry.

  Darij


calculus
integration
calculus computations

Evaluate the indefinite integral

\[ \int \frac{\sqrt{1\minus{}x^{2}}\minus{}x}{x^{3}\minus{}x^{2}\minus{}x\plus{}1\minus{}\sqrt{1\minus{}x^{2}}\plus{}x\sqrt{1\minus{}x^{2}}}dx\]

$ \int\frac {\sqrt {1 \minus{} x^{2}} \minus{} x}{x^{3} \minus{} x^{2} \minus{} x \plus{} 1 \minus{} \sqrt {1 \minus{} x^{2}} \plus{} x\sqrt {1 \minus{} x^{2}}}dx$

Group the terms and the integral becomes :

$ \int\frac {\sqrt {1 \minus{} x^{2} } \minus{} x}{ ( \sqrt { 1 \minus{} x^{2} } \minus{} 1)(1 \minus{} x)(\sqrt {1 \minus{} x^{2} } ) }dx$

$ \equal{} \int\frac {dx}{ ( \sqrt { 1 \minus{} x^{2} } )(1 \minus{} x) } \plus{} \int\frac {dx}{( \sqrt { 1 \minus{} x^{2} } \minus{} 1)(\sqrt {1 \minus{} x^{2} } ) }$

$ \int\frac {dx}{( \sqrt { 1 \minus{} x^{2} })(1 \minus{} x) }$ ; put $ 1 \minus{} x \equal{} \frac {1}{t}$ and proceed .

$ \int\frac {dx}{ ( \sqrt { 1 \minus{} x^{2} } \minus{} 1)(\sqrt {1 \minus{} x^{2} } ) } :$ here put $ \sqrt {(1 \minus{} x^2) } \equal{} t^2$ . doing this reduces this to the same form as the previous integral .

I hope this is correct . if anything is wrong please point out i shall post my entire solution.

THankyou


modular arithmetic
induction
number theory
relatively prime
number theory unsolved

Prove that every two members of the sequence $ F_k\equal{}2^{2^k}\plus{}1,k\equal{}1,2,..,n;(n \in N)$ are relatively prime between them.

We have $ F_n\equal{}F_1\dots F_{n\minus{}1}\plus{}2$. That can be proved using induction on $ n$. Hence $ (F_k$, $ F_l) \in \{1$, $ 2\}$. But as $ F_n$ is odd, the conclusion follows.

[quote="FelixD"]We have $ F_n \equal{} F_1\dots F_{n \minus{} 1} \plus{} 2$. That can be proved using induction on $ n$. Hence $ (F_k$, $ F_l) \in \{1$, $ 2\}$. But as $ F_n$ is odd, the conclusion follows.[/quote]

$ F_n \equal{} F_1\dots F_{n \minus{} 1} \plus{} 2$,should be $ F_n \equal{} F_1\dots F_{n \minus{} 1} \plus{} 1$ because the last supposed member of the set has the form $ F_n\equal{}2^{2^k}\plus{}1$, right ?

[quote="Obel1x"]

$ F_n \equal{} F_1\dots F_{n \minus{} 1} \plus{} 2$,should be $ F_n \equal{} F_1\dots F_{n \minus{} 1} \plus{} 1$ because the last supposed member of the set has the form $ F_n \equal{} 2^{2^k} \plus{} 1$, right ?[/quote]

No. Note that $ F_n \minus{} 2 \equal{} 2^{2^n}\minus{}1 \equal{} \left(2^{2^{n\minus{}1}} \minus{} 1\right)\left(2^{2^{n\minus{}1}} \plus{} 1\right)$ and we can carry on with this process.

[quote="Obel1x"]
$ F_n \equal{} F_1\dots F_{n \minus{} 1} \plus{} 2$,should be $ F_n \equal{} F_1\dots F_{n \minus{} 1} \plus{} 1$ because the last supposed member of the set has the form $ F_n \equal{} 2^{2^k} \plus{} 1$, right ?[/quote]

No its $ F_n \equal{} F_1\dots F_{n \minus{} 1} \plus{} 2$ as stated, just try the first few cases $ 5 \equal{} 3\plus{}2$, $ 17 \equal{} 3.5\plus{}2$ etc.
Perhaps you were mixing up with another relation $ F_n \equal{} 2.(F_1 \minus{}1)\dots (F_{n \minus{} 1} \minus{}1) \plus{} 1$, but that does not help here.

[quote="PhilG"][quote="Obel1x"]
$ F_n \equal{} F_1\dots F_{n \minus{} 1} \plus{} 2$,should be $ F_n \equal{} F_1\dots F_{n \minus{} 1} \plus{} 1$ because the last supposed member of the set has the form $ F_n \equal{} 2^{2^k} \plus{} 1$, right ?[/quote]

No its $ F_n \equal{} F_1\dots F_{n \minus{} 1} \plus{} 2$ as stated, just try the first few cases $ 5 \equal{} 3 \plus{} 2$, $ 17 \equal{} 3.5 \plus{} 2$ etc.
Perhaps you were mixing up with another relation $ F_n \equal{} 2.(F_1 \minus{} 1)\dots (F_{n \minus{} 1} \minus{} 1) \plus{} 1$, but that does not help here.[/quote]
yes,u're right. I completely messed up with that :( .
Can you post full solution by induction?

@Obel1x: the proof using induction is very easy. I'm sure that you can do that by yourself  :wink:


Ring Theory
linear algebra
linear algebra unsolved

] Do there exist matrices $A$, $B$, $C$ which have no common eigenvectors and satisfy the condition $AB=BC=CA$?
(V. Brayman)

Hi Moubi,
what is exactly the condition to check? I think that it is A,B have no common eigenvector and the same for B,C and C,A.
If not we can take (line by line) these (2,2) matrices over C:
a:=[i,1,0,i],b=[0,1,1,2i],c=[-i,4,1,3i]. b and c have no common eigenvector because (bc-cb)^2=identity.

I had not finished the calculations in my preceding post!
We have also (ab-ba)^2=(ac-ca)^2=id and a,b,c have no common eigenvector (2 by 2).
 The following is an example in the real (2,2) matrices:
a:=[1,1,0,1],b=[2,1,-1,0],c=[1,0,-1,1].

Loup blanc you found three matrices A,B,C with no common
eigenvectors with AB=BC=CA. So the answer is yes.
Thank you for your solution.

Experience of solving math problem, experience Peter.

Hi Peter and Moubi,the following is a proof without using the intuition except for the choices: c invertible, a non diagonalizable.
Suppose that a,b,c are (2,2) matrices, ab=bc=ca,(ab-ba)^2=id, c is invertible, a non diagonalizable.
   Then a and b are similar; we may suppose a=x.id+P, b=x.id+Q where P=[0,1,0,0] (line by line) and Q^2=0. (ab-ba)^2=(PQ-QP)^2=0 gives Q=+/-[v,v^2,-1,-v]; we take the signum +.
bab=aba gives by equalizing the terms in position [2,1]:x^2=1.
Remark:if we would choose the signum -  the result would be x^2=-1.
For x=1: a=id+P;b=id+Q; c=(b^-1)ab=(id-Q)(id+P)(id+Q).


Euler
logarithms
calculus
integration
complex numbers

Use any method yo can think of to determine the square root of i.

[hide]

$i = \text{cis} 90$

hence $\sqrt{i} = (\text{cis} 90)^{1/2} = \text{cis} 45 = \pm(\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} i)$

[/hide]

oops good catch zeta haha

There is not [u]the[/u] square root of $i$. There are two roots.

I'm interested in this question as well, however I do not understand your answer. Could you provide a link to a resource explaining cis90 or do so yourself? We havn't yet covered that topic. I had the exact same question as king_23 and we were supposed to solve it using a + bi = sqrt(i), then squaring both sides and expanding. I'm lost after that however.

[hide="Euler"]
$i=e^{i(\pi/2+2k\pi)}$
$\sqrt{i} = e^{i(\pi/4+k\pi)}$
[/hide]

[hide="LegendsEnd"]
$\sqrt{i}=a+bi$
$i = a^2-b^2 + 2abi$
$a$ and $b$ are real numbers, so we can equate the real and imaginary parts, giving $a^2-b^2=0$ and $i=2abi$.

$a^2=b^2 \Rightarrow |a|=|b|$ and since $ab = \frac{1}{2}$, $a=b=\sqrt{\frac{1}{2}}$ or $a=b=-\sqrt{\frac{1}{2}}$[/hide]

It's possible to properly define one and only one square root of a complex number; the same holds for any $k$-th root with $k$ any complex number.
But:
You need to make (not everytime canonical) choices what root to take.
Especially, you could define $a^b : = e^{b \cdot \log (a)}$, and $\sqrt[b]a : = a^\frac{1}{b}$ for $b \neq 0$. And here the problem occurs: what is $\log$¿
Whereas $e^x$ is canonically defined on $\mathbb{C}$ on a unique way (this does not hold on $\mathbb{C}_p$ for example), the $\log$ has some bad properties: $e^{x}=e^{x+2\pi i}$, so $\log$ can only be defined $\mod 2\pi i$ (and conversly, every such definition works).
What is $\mathbb{C} \mod 2 \pi i$¿ Well, it is for example represented by those $x+iy$ with $0 \leq y < 2 \pi$ or also by those with $-\pi<y \leq\pi$; thus for every single point of such a "strip", we take any value that works.

Another definition could be as integral of $\frac{1}{x}$, but again, cause of $\int_{|x|=1} \frac{1}{x} dx= 2 \pi i$ (meaning: when starting point = ending point, the integral is not necessary $0$), the same problems occure.
Most often one requires $\log$ to be continuous (it is clearly when defined as integral), then one gets what is called a "branch" of the logarithm.

When taking for example tha "main branch", meaning to take $l=\log(z)$ such that $-\pi<\Im(l) \leq\pi$ (can be archieved with integrals by avoiding to integrate over the negative reals in the plane), then $\sqrt[i]i = \sqrt{e^\pi}$.

A bit difficult for Int. Topics in my opinion

I don't think so.
At first, it contains the reason why it is not very straightforward what root to take. At second, it is meant to give some preview.
(it is especially not meant as proof of something and doesn't want to be totally understood)

[quote="mna851"]Find the $\ i$ th root of $\ i$.[/quote]Well, $i=e^{\frac{i\pi}{2}}$ so taking the ith root gives $i^{\frac{1}{i}}=e^{\frac{i\pi}{2i}}=e^{\frac{\pi}{2}}$.

well, you know that the set complex numbers is closed with square roots, (i don't know the exact set term, but the square root of a complex number is a complex number, i.e. if $z\in\mathbb{C}$, then $\sqrt{z}\in\mathbb{C}$)

so then we can say that:

$\sqrt{i}=a+bi$

$i=a^2+2abi-b^2$, equating real and imaginary parts,

$a^2-b^2=0$ and $2ab=1$, which is easily solved to get

$(a,b)=\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right), \left(-\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\right)$

now obviously, the cis way is more practical, but if you don't know demoivre's theorem, this is a low tech way to approach it

also

zeta, what do you mean by canonical? the only definition i know is related to religion, do you mean if you choose one output for complex roots, then you are just making a choice of preference?

@canonical: yes, I meant that none of the roots is special, so there is no reason to choose that one and not another (same for the $\log$); it's different in the reals where it is practical to take the positive root (and the real value of the $\log$).


algebra
polynomial
linear algebra
matrix
linear algebra unsolved

Let $ V$ be a finite-dimensional inner product space, and let $ E$ be an idempotent linear operator on $ V$, i.e, $ E^{2}= E$. Prove that if $ EE^{*}= E^{*}E$, then $ E$ is self-adjoint, i.e $ E=E^{*}$.

(from Hoffman & Kunze, 8.4)

This is really easy, using the fact that since $ E$ is normal, it is diagonalizable in an orthogonal basis. But since it satisfies the polynomial equation $ E^{2}-E=0$ hence its minimal polynomial is a divisor of $ x^{2}-x$ and hence all its eigenvalues are roots of $ x^{2}-x$ which are $ 0$ and $ 1$. Hence all its eigenvalues are real, which shows that the operator is self-adjoint.

Yup. Exactly.

Here's a follow-up, also from Hoffman & Kunze's:

Prove that if $ A$ is a matrix which commutes with $ AA^{*}$ ($ AAA^{*}= AA^{*}A$) then $ A$ is normal.


Math League

Can someone help me on #6 of the IML today?  I have no clue where to start

me too
that was probably the hardest one so far on all the math leeagues

We've decided not to discuss the problems on the Math League contests until at leats a day after they are done. Thus, I'm locking this thread for today. I agree on the fact that it was the most interesting question yet.

Also, from now on, please refer to the questions as "Math League." That way we avoid the confusion of all the separate states (its the same contest everywhere.)


I saw that there used to be discussions like this when the Wii version came out....

Favorite tracks, best times, time trials, etc.

Gamecube: Mushroom Bridge
Ds: Wario stadium (only had for 2 days)

Post times when I can/or in response

Favorite tracks (DS): Mario circuit (star cup), bowser castle (leaf), and rainbow road

I assume you have beaten all the missions (they are simple), but what are your ratings?

Mine are *,a,b,b,b,c

Meh. Need to work on that.

(2 or 3 stars), (star), A, C, C, C

For 3, 4, and 5, I have one mission that's causing me to not have a (star) rating.

DS: Wario Stadium, Rainbow Road, Airship Fortress.

Wii: Rainbow Road, Moonview Highway, Coconut Mall, DK Summit.

For DS missions: star on level 1, A on levels 5 and 2, and B on levels 3, 6, and 4.

I know there is a Level 7, but one needs to get at least on star on each of the first 6 levels.

I worked on my missions and now they are **,*,*,A,B,C

@spaceguy: yeah, those are my favorites too.

I have one mission on Level 5 in DS missons that is really hard for me. It's the mission with you driving as Bowser in GBA Bowser Castle 2 and you have to get through all 10 gates.

For level 4, I hate King Boo... I got a B on that level.

Level 6: Those Rocky Wrenches on Airship Fortress, I hate that.

YES

Only two unstarred missions now.....6-6 and 6-7

I agree the Bowser castle is hard took me like 10 tries but I got it with 16:689 left and got one star

I also have the GBA version of Mario Kart.

If you try to drift on it, don't hold the R button. It will make turn more qickely

GOT MISSION 7

Won't ruin any suprises, but one event I finished with .083 left and got an A!.

Overall B.

I'm a huge Mario Kart fan, sadly however I am not that good so I will not embarrass myself by posting track times.

Some of my favorite tracks though would have to be Waluigi's Stadium, Rainbow Road, and Baby Park.

I only have the N64 version, although I guess I'm ok at it. Favorite track is hard to say. Wario Stadium and Rainbow Road aren't bad. Luigi Raceway actually is pretty good, too. Easier to say are the tracks I know I don't like: Toads Turnpike (especially in extra, where you have to drive against the traffic), Bowser's Castle, etc.

I haven't played for a long time, but I used to be pretty good. Missions: ***,***,**,**,**,*,* I think... maybe 7 is an A or 5 is a *... on wi-fi I'm something about 10-0 (undefeated, but I haven't played any snakers (read cheaters) yet)

It isn't really cheating....

Aren't there snaking missions? Like 14 drifts in Yoshi circuit?

I unlocked the staff ghost for baby park with a time of like 52.6 or so

You can do $ 14$ drifts on Yoshi Circuit.

I raise Level 2 to a *, so my ratings are:

*,*,b,b,a,b.

I have an "A" on 6-7

and a "B" or "A" on 6-6.

At 5 seconds, it always seems that I am halfway through gates 4 and 5.

Why the 13101 copies of spam, Mr. Spammer?

HOW DID YOU COUNT THAT???
um ...
because ...

duh!

I copied it onto a word document and used the word count

I finally have 3 stars on the 8-gate one!

Yay, drifted through all 8 gates with 2 seconds left remaining :)

I got on my third attempt with 1.584 left. ***! yay

ugh...

I still have an A on that mission :(

Sorry to *revive*...  but it hasn't been a month yet, so...

Does anybody know how to ace the Level 3 mission "Collect the 20 Coins"?  The one with Mario in the Twilight House?  The best rating I ever got was a *.

You just have to try to pick up EVERY coin  you pass very quickly! Practicing will make you better. I would suggest to try picking an order of rooms and get the coins in that order.

Cool, thanks.  I just got a *** rating on it.  Yay!   :D

Just picked up Mario Kart 8 Deluxe from gamestop!  It's better than the others!

My best times are 30 seconds on Grumble Volcano and a 130 on Wario's Gold Mine

My best time is 15 sec on grumble volcano.


floor function
induction
number theory
prime factorization

That's false. It's easy to find integers such that the denominator has a prime factor that the numerator does not have. For example, $a_{1}=1,a_{2}=2$.

edit. You probably mean the reciprocal. In that case, it's just a multinomial coefficient.

Heres what I did:
[hide]
Let S be a string of letters, with $a_{1}, a_{2}, a_{3}, \cdots, a_{k}$ each representing the number of repeated elements for a certain letter.

(For example, if S = aabbbbccc, then  $a_{1}$ = 2 (2 repetitions for a), $a_{2}$ = 4 (4 repetitions for b), and $a_{3}$ = 3 (3 repetitions for c)

Therefore,

$\frac{(a_{1}+a_{2}+a_{3}+\cdots+a_{k})!}{a_{1}!a_{2}!a_{3}!\cdots a_{k}!}$

is another way of finding the number if distinct permutations of the string of letter S. Since permutations are always integers (you cant find half of a permutation/arrangement of something), this shows that the fraction will always be an integer.
[/hide]

[hide="Non-combinatorial"]

For an integer prime $p$, the exponent in the prime factorization of this rational (extending UPF to rationals over integers, ie negative exponents are allowed) is

\[\sum_{i=1}^{\infty}\left(\lfloor{\frac{\sum_{j=1}^{k}a_{j}}{p^{i}}\rfloor}-\sum_{j=1}^{k}\lfloor{\frac{a_{j}}{p^{i}}\rfloor}\right)\]

Hmm...the floor of the sum is greater than or equal to the sum of the floors.  For two elements it's just $\lfloor{a+b\rfloor}\ge \lfloor{\lfloor{a\rfloor}+\lfloor{b\rfloor}\rfloor}=\lfloor{a\rfloor}+\lfloor{b\rfloor}$, and for any general number of elements, you use induction using the same expression. (ie substitute $b=a_{1}+a_{2}+...+a_{n}$, and let $a=a_{n+1}$.  Then the above expression gives us an inductive step from n to n+1.)[/hide]

More about that technique:

http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=141316


calculus integration trigonometry quadratics function limit inequalities

Using the fact that $cos^{2}(x) = \frac{1+cos(2x)}{2}$
We get $\int_{0}^{\pi}\frac{cos^{2}(x)}{e-cos^{2}(x)}dx = \int_{0}^{\pi}\frac{1+cos(2x)}{2e-1-cos(2x)}dx = \frac{1}{2}\int_{0}^{2\pi}\frac{1+cos(x)}{2e-1-cos(x)}dx$

Write $z = e^{ix}(0\leq x \leq 2\pi)$
Now use the relations $cosx = \frac{z^{2}+1}{2z}$ and $dx = \frac{dz}{iz}$
So we get $\frac{i}{2}\int_{C}\frac{z^{2}+2z+1}{z(z^{2}+(2-4e)z+1)}dz$

$f(z) = \frac{z^{2}+2z+1}{z(z^{2}+(2-4e)z+1)}$ has singularities at $z = 0$, the other two can be found using quadratic equation. Then consider only the one that lie in the interior of $|z| = 1$ and then apply residue Theorem

The answer
[hide]The primitive function is:

\[\frac{e\cdot\arctan\left[\frac{e\cdot\tan{x}}{\sqrt{e^{2}-e}}\right]}{ \sqrt{e^{2}-e}}-x \]

if we evaluate this at the limits we get $\boxed{-\pi}$

[color=red]BUT How can this be shown?[/color][/hide]

Yes you are right. Perhaps I evaluated it wrong, but is the antiderivative correct?
Mathematica gave me that answer.

Yeah, I used The Integrator, it gave me the same  :rotfl:

It is possible that we are integrating over something bad perhaps the function is not defined for some $x \in[0, \pi]$, making it useless to apply residue theorem.

Your antiderivative has a jump discontinuity as written. If you want it to make sense globally, you need to switch branches of $\arctan$ when you cross the singularity you introduced.

In other words: it's an antiderivative on $[0,\frac{\pi}2)$ and an antiderivative on $(\frac{\pi}2,\pi]$. It's not an antiderivative on the whole interval $[0,\pi]$.

you're absolutely right!

i should have stated [i]instead [/i](as my last line):

$\hspace* \quad \;Cauchy\;PV\;\int_{0}^{\pi}\; f(x)\;dx \;=\;\lim_{\xi \rightarrow 0,\eta \in \mathbb{R}}\; ( \;\int_{0}^{\frac{\pi}{2}-\eta\xi}\; f(x)\;dx \;+\;\int_{\frac{\pi}{2}+\eta\xi}^{\pi}\; f(x)\;dx\; ) \;$

that would then validate and complete the integration.  :maybe:

How about this question then?
[quote="Carcul"]Just a question: the function to integrate is always positive on the interval $[0,\pi]$, then shouldn't the integral also be positive?[/quote]

I think we don't have to worry about the singularity of its antiderivative, because the original integral can be modified to avoid such problem, like this:

$\int_{0}^{\pi}\frac{\cos^{2}{x}}{e-\cos^{2}{x}}\, dx = 2\int_{0}^{\frac{\pi}{2}}\frac{\cos^{2}{x}}{e-\cos^{2}{x}}\, dx$

(maybe this is a similar argument to misan's)

[quote="Carcul"]Just a question: the function to integrate is always positive on the interval $[0,\pi]$, then shouldn't the integral also be positive?[/quote]

not necessarily... it is fairly easy to show [size=150]I[/size] could be [b]negative[/b]...
use the [b]ML[/b] approximation from real analysis.

$\int_{0}^{\pi}\frac{\cos^{2}x}{e-\cos^{2}x}\;dx\;=\;-\;\int_{0}^{\pi}\;dx \;+\;e\;\int_{0}^{\pi}\frac{dx}{e-\cos^{2}x}\\ \\ \geq\;-\pi\;+\;e\;\cdot\;min_{(x\in\Gamma)}|(\frac{1}{e-\cos^{2}x})|\;\cdot\;\ell\;(\Gamma)\\ \\ \geq\;-\pi\;+\;\lim_{\xi \rightarrow 0}\;\xi\;\cdot \;e\;\cdot\;\frac{1}{max_{(x\in\Gamma)}|(e-\cos^{2}x)|}\qquad(since\;\;\;\lim_{\xi \rightarrow 0}\;\xi\;\leq\;\ell\;(\Gamma)\;\leq\;\pi)\\ \\ \geq\;-\pi\;+\;\frac{\lim_{\xi \rightarrow 0}\;\xi\cdot e}{e}\\ \\ \geq\;-\pi\qquad\Rightarrow\;I\;\leq 0\;\;\;i.e.\;\;\;-ve$

That is nonsense- the inequality points in the other direction.

[quote="jmerry"]Of course, that singularity is entirely artificial- it comes from your substitution, not the original function.[/quote]

that is incorrect. my substitution had [i]nothing [/i]to do with introducing a singularity. divide the original function by $\cos^{2}x$ and the discontinuity is clearly apparent at $\frac{\pi}{2}$... my substitution was merely aimed at solving the [b]original [/b]integral (latter part uses Cauchy PV to cancel the effects of that discontinuity from either ends).

There was no singularity until you divided by $\cos^{2}$, which is zero there. Would you say that $\frac{x^{2}}{1+x^{2}}$ has a singularity at zero?

There's also no need for a principal value; this isn't even an improper integral. We do need to split the interval if we use that form of the antiderivative, but that's different.

[b]jmerry[/b] You said that the argument above was nonsense
and I agree (have a hard time imagine the opposite) but how do we
explain the negative answer? Could someone please comment this particular aspect?


Let $x,y,z >0$.Prove that the following inequality holds:

    $(x^{y}y^{x})^{\frac{2}{x+y}}+(y^{z} z^{y})^{\frac{2}{y+z}}+(z^{x}x^{z})^{\frac{2}{z+x}}\leq xy+yz+zx$ ;)

[quote="silouan"]What ineq do you use?[/quote]

weight AM-GM if im not wrong for variable $(x,y)$ and their respective weight $(y,x)$

 :)


Excavation starts at Stonehenge
http://news.bbc.co.uk/1/hi/sci/tech/7322134.stm
Stonehenge dig starts today
http://www.wiltshiretimes.co.uk/news/latestheadlines/display.var.2158016.0.stonehenge_dig_starts_today.php
Archaeologists begin historic Stonehenge dig
http://www.telegraph.co.uk/news/main.jhtml?xml=/news/2008/03/31/nstonehenge131.xml
Excavation begins at Stonehenge
http://news.bbc.co.uk/2/hi/uk_news/england/wiltshire/7322238.stm
Archaeologists dig deep to find origins of Stonehenge
http://afp.google.com/article/ALeqM5hX2SySFzNGr0UEhtZPzPYb0KIrVQ
First Stonehenge dig in 44 years
http://www.independent.co.uk/news/uk/home-news/first-stonehenge-dig-in-44-years-802950.html
Unlocking Stonehenge's secrets
http://news.bbc.co.uk/2/hi/science/nature/7321882.stm

Not only are Moses and Aaron's corpses beneath Heelstone but the gold Mercy Seat, the gold Ark of the Testimony, the gold Table of Shewbread, the gold Candlestick, the gold Ephod, the gold Breastplate and the gold Altar of Incense also. According to DI's new 2004 ground-penetrating radargrams all seven (7) are packed inside of the brasen Altar of Burnt Offering 4 ft (1.2m) under Heelstone, with that southernmost gold Ark of the Covenant long axis being east-west. G-D old 1656 core samples indicate the gold Table of Shewbread as northernmost and gold Altar of Incense inside the brasen Altar of Burnt Offering in the center, 4 ft (1.2m) below the Heelstone's base, at Stonehenge.

http://en.wikipedia.org/wiki/Heelstone
http://en.wikipedia.org/wiki/Scroll_Trench
http://en.wikipedia.org/wiki/Heelstone_Ditch

A large 8'x8'x4' artifact below Heelstone at Stonehenge? Other than a core sample of it (1656), a magnetic survey of it (1984), an electromagnetic (EM) conductivity survey of it (1984), and a seismic refraction survey of it (1984); no other tangible proof of it being in Scroll Trench where it is circled by Heelstone Ditch exists, save and except for: an electrical resistivity survey of it (1994), a ground-penetrating radargram of it (2004), confirmation geophysical surveys of it (2004), and a confirmation core sample of it (2004). Should this 8'x8'x4' artifact below Heelstone be removed before being stolen? It is right next to a paved highway for God's sake!

http://www.panoramio.com/photo/7529895
http://www.flickr.com/photos/garrydenke/979647239
http://groups.msn.com/StonehengeGeologyandGeophysics
http://groups.yahoo.com/group/stonehengegeologyandgeophysics/

Garry Denke

[b]A303 / A344 / A360 Stonehenge Improvement Programme[/b]

There are Three (3) Tunnels to dig (The Pagan, The Druid, and The Wiccan) and an Airman's Corner Stonehenge Visitor Centre to build by the 2012 Olympics.

http://maps.google.com/maps?q=Stonehenge

[b]Stonehenge Three (3) Tunnels' Names:[/b]
A303 Stonehenge Tunnel Name - The Pagan Tunnel
A344 Stonehenge Tunnel Name - The Druid Tunnel
A360 Stonehenge Tunnel Name - The Wiccan Tunnel

[b]Stonehenge Three (3) Tunnels' Lengths:[/b]
The Pagan at Stonehenge - 5.5 kilometres (3.4 miles)
The Druid at Stonehenge - 3.0 kilometres (1.9 miles)
The Wiccan at Stonehenge - 1.5 kilometres (1.0 miles)

[b]Stonehenge Visitors' Centre Name:[/b]
Airman's Corner Stonehenge Visitor Centre

National Trust's going to Avenue Banks to withdraw Construction Funds from Heel Stone Vault for these Three (3) Tunnels and New Visitors' Centre next.

[url=http://www.thisiswiltshire.co.uk/news/headlines/display.var.2158518.0.new_dig_probes_wiltshires_stonehenge_mystery.php]New dig probes Wiltshire's Stonehenge mystery (from This Is Wiltshire)[/url]

[b]Heelstone dig Next:[/b]

[url=http://www.garrydenke.com/stonehenge.htm]Garry Denke - Stonehenge[/url]
[url=http://www.garrydenke.com/seismic.htm]Heel Stone Refraction Seismic[/url]
[url=http://www.denocoinc.com/stonehenge.htm]Denoco Inc. - Stonehenge[/url]
[url=http://www.denocoinc.com/seismic.htm]Heelstone Refraction Seismic[/url]

 :)


Four packages are delivered to four houses, one to each house. If these packages are
randomly delivered, what is the probability that exactly two of them are delivered to
the correct houses? Express your answer as a common fraction.

There are $ 4!\equal{}24$ ways to distribute the packages.

After choosing the two packages delivered correctly with $ \binom{4}{2}\equal{}6$ ways to do so, the other two are fixed, so the probability is $ \frac{6}{24}\equal{}\boxed{\frac{1}{4}}$

[hide=Solution]Since there are 4 houses and 4 packages, we can choose ${4 \choose 2} = 6$ pairs of houses to be the pair that will receive the correct package. In that case, the other two houses must have one another's package. The probability of this occuring for any arrangement is $\frac{1}{4} \cdot \frac{1}{3} \cdot \frac{1}{2}$, as the first fraction represents the probability of a given house getting the correct package, and the second fraction the subsequent probability that the other given house gets the correct package, and the final fraction the probability that the last two houses have each other's packages. So, the probability is $6 \cdot \frac{1}{2 \cdot 3 \cdot 4} = \boxed{\frac{1}{4}}$.[/hide]

And yes I revived this 12 year old thread.


Find all natural numbers $n$ such that it is possible to construct a sequence in which each number $1, 2, 3, \ldots, n$ appears twice, the second of the appearances of each integer $r$ being $r$ places beyond the first appearance.  For instance, for $n = 4$,
\[ 4, 2, 3, 2, 4, 3, 1, 1. \]
Also, for $n = 5$,
\[ 3, 5, 2, 3, 2, 4, 5, 1, 1, 4. \]

Let a_k be the place of the first appearance of k.  Example:  in 42324311,  a_4 = 1, a_2 = 2, a_3 = 3, a_1 = 7.

Then $\sum_{j=1}^{2n} = \sum_{j=1}^n 2a_j + j$  implying $\sum_{j=1}^n a_j = \frac{3n^2 + n}{4}$.

If n is $2$ or $3$ mod 4, then $\sum a_j$ is not an integer, contradiction.

Now we just need an example for n = 0, 1 mod 4.


For which positive $n$ is there a permutation $(x_{1},x_{2},...,x_{n})$ of $\{1,2,...,n\}$ such that all the differences $|x_{k}-k|$,$k=1,2...,n$,are distinct ?

It was the August 2002 Problem of the Month at [url=http://mathcentral.uregina.ca/MP/previous2001/aug02sol.html]Mathcentral[/url]. :)
There's a constructive solution for $n=4m$ and $n=4m+1$, and two different ways to prove that there's no solution for $n=4m+2$ and $n=4m+3$.
Notice that an AOPS moderator, Pierre Bornsztein, was behind this problem! :P


$ ABCD$ is a cyclic quadrangle. $ AB$,$ CD$ intersect each other at $ E$ and $ AD$,$ BC$ intersect each other at $ F$. $ M,N$ are midpoints of $ AC,BD$ and $ AC > BD$.

prove that:

$ \frac {MN}{EF}$ $ \equal{}$ $ \frac {1}{2}$  $ (\frac {AC}{BD} \minus{} \frac {BD}{AC})$


Source: ACoPS
Find all real values of $x$ that satisfy $(16x^2-9)^3+(9x^2-16)^3=(25x^2-25)^3$.

[quote="Farenhajt"]Note that $16x^2-9+9x^2-16=25x^2-25$, hence the equation is of the form $a^3+b^3=(a+b)^3$, which directly simplifies to $3ab(a+b)=0$, giving $(16x^2-9)(9x^2-16)(25x^2-25)=0$, and that in turn yields the set of solutions $x\in\{\pm{4\over 3},\pm{3\over 4},\pm 1\}$[/quote]

You returned!!! Why haven't you posted for some much time????

[quote="kimby_102"][quote="boxedexe"]Just use the fact that if $a+b+c=0$, then $a^3+b^3+c^3=3abc=0\implies abc=0$. Done. :)

Masoud Zargar[/quote]
Why would $a^3+b^3+c^3=3abc$ equal to zero?
I think it's just $a^3+b^3+c^3=3abc$.[/quote]

But also $a^3+b^3+c^3 = 0$    :)

You move the cube on the right to the left and get $a^3+b^3+c^3=0$.  ;)

[quote="José"][quote="Farenhajt"]Note that $16x^2-9+9x^2-16=25x^2-25$, hence the equation is of the form $a^3+b^3=(a+b)^3$, which directly simplifies to $3ab(a+b)=0$, giving $(16x^2-9)(9x^2-16)(25x^2-25)=0$, and that in turn yields the set of solutions $x\in\{\pm{4\over 3},\pm{3\over 4},\pm 1\}$[/quote]

You returned!!! Why haven't you posted for so much time????[/quote]

I guess he posts more in the olympiad forum :)  Solves all the sums on earth

[quote="shadysaysurspammed"]Solves all the sums on earth[/quote]

What does it mean?

[quote="shadysaysurspammed"]That means he solves all the questions you give him :lol:[/quote]

Jajaja got it


$\text{Find the value of the following if } \sin54^\circ  = \displaystyle\frac{\sqrt{5} + 1}{4}$

$(\tan18^\circ )(\sin36^\circ )(\cos54^\circ )(\tan72^\circ )(\sin90^\circ )(\tan108^\circ )(\cos126^\circ )(\sin144^\circ )(\tan162^\circ )(\cos180^\circ )$

The answer is [hide]${\displaystyle\frac{15-5\sqrt{5}}{32}}$[/hide]

[hide="basically answer"]The tans cancel.  The rest converts to $\cos^4 54^\circ$.  Use $1-\sin^2 = \cos^2$[/hide]


Δίνεται κύκλος , μια διάμετρος $ AB$  , μια χορδή $ CD$ κάθετη στη διάμετρο και πιο κοντά στο $ B$ , η εφαπτομένη του

κύκλου στο $ C$ που τέμνει την $ AB$ στο $ E$ , η προβολή $ K$ του $ C$ στην $ AD$ και το μέσο $ M$ του $ CK$.Αν η ευθεία

$ AM$ τέμνει τον κύκλο στο $ Z$ , να αποδείξετε ότι :

[b]a) [/b]η ευθεία  $ CZ$ διχοτομεί το τμήμα $ HE$ , όπου $ H$ είναι η τομή των  $ AB,CD$.

[b] b)[/b] Η ευθεία $ AE$  εφάπτεται του περιγεγραμμένου κύκλου του τριγώνου $ CZE$

 [u]Μπάμπης[/u]

[b] Σχόλιο[/b]

Ας βρούμε και καμιά απλή λύση χωρίς προωθημένα θεωρήματα και αρμονικότητες(αλλά και αυτές ! )

Πολύ μου αρέσουν τέτοιου είδους ασκήσεις :)

Εγώ θα δώσω λύση με αρμονικά, αν και θα ψάξω να βρω και με άλλο τρόπο (παρεπιμπτόντως η άσκηση μου θυμίζει πολύ μία που είχα προτείνει εγώ και δεν ξέρω μήπως είναι και ίδια... :huh: πάντως δεν έχω Χρόνο να το ψάξω...)

Προεκτείνουμε την $ EZ$ ώστε να τέμνει τον κύκλο στο $ Z'$. Εύκολα το $ ZCZ'D$ είναι αρμονικό τετράπλευρο. Αρα η δέσμη $ A(AD,AZ,AC,AZ')$ είναι αρμονική κι οπότε αν πάρουμε την τομή της με την $ KC$ επειδή $ M$ το μέσο θα είναι $ AZ'//KC\perp AD$.

Αρα $ \angle{Z'CD}\equal{}Z'AD\equal{}90\iff ZC//AE$.

Από εδώ το αποτέλεσμα είναι πολύ άμεσο. Δηλαδή κατ' αρχάς επειδή η δέσμη $ CC,CZ,CD,CZ'$ είναι αρμονική (οπου $ CC$ η εφαπτομένη) και επειδή $ CZ'//AE$ προκύπτει ότι $ CZ$ τέμνει την $ HE$ στο μέσο της (που φαντάζομαι είναι το ζητούμενο, γαιτί υπάρχει typo στην εκφώνηση...).

Επίσης από την παραλληλία και το θεώρημα υπό χορδής κι εφαπτομένης παίρνουμε:
$ \angle{ECZ}\equal{}CZ'Z\equal{}Z'EL$ και έπεται το δεύτερο ζητούμενο

Q.E.D. :)

[hide]$ \bullet$ Στα όμοια ορθογώνια τρίγωνα $ \bigtriangleup KAC$ και $ \bigtriangleup HCE$ $ ($ γιατί έχουν $ \angle KAC \equal{} \angle HCE$ $ ),$ αποδεικνύεται εύκολα ότι $ \angle ZCE \equal{} \angle MAC,$ $ \angle ZCH \equal{} MAK$ και συμπεραίνουμε έτσι ότι η ευθεία $ CZ,$ είναι διάμεσος του τριγώνου $ \bigtriangleup HCE,$ αφού και η ευθεία $ AM,$ είναι διάμεσος του τριγώνου  $ \bigtriangleup KAC.$

Άρα, $ NE \equal{} NH,$ όπου $ N\equiv AE\cap CZ.$

$ \bullet$ Για το δεύτερο ζητούμενο, αποδεικνύεται εύκολα ότι οι ευθείες $ CB,$ $ CA,$ είναι διχοτόμοι δια της κορυφής $ C,$ του τριγώνου $ \bigtriangleup HCE$ $ ($ γιατί $ \angle BCE \equal{} \angle BAC \equal{} \angle BAD \equal{} \angle BCD$ και $ BC\perp CA$ $ ),$ και επομένως έχουμε ότι τα σημεία $ A,\ B,$ είναι αρμονικά συζυγή των $ H,\ E.$

Επειδή τώρα το σημείο $ N$ είναι το μέσον του $ HE,$ σύμφωνα με το [b][size=100]θεώρημα Newton[/size][/b], έχουμε ότι $ (NE)^{2} \equal{} (NB)\cdot (NA)$ $ ,(1)$

Αλλά από το εγγράψιμο τετράπλευρο $ ABZC,$ έχουμε $ (NB)\cdot (NA) \equal{} (NZ)\cdot (NC)$ $ ,(2)$

Από $ (1),$ $ (2)$ $ \Longrightarrow$ $ (NE)^{2} \equal{} (NZ)\cdot (NC)$ $ ,(3)$

Από την $ (3)$ συμπεραίνουμε ότι η ευθεία $ NE\equiv AE,$ εφάπτεται στον περίκυκλο του τριγώνου $ \bigtriangleup CZE,$ στο σημείο $ E$ και η πρόταση έχει αποδειχθεί. [/hide]
Κώστας Βήττας.

Δανείζομαι το πρώτο ερώτημα από τον κ Βήττα(επειδή έκανα ακριβώς το ίδιο) και στο δεύτερο βάζω μια άλλη λύση . Πιο elementary δεν γίνεται  :)

$ \bullet$ Στα όμοια ορθογώνια τρίγωνα $ \bigtriangleup KAC$ και $ \bigtriangleup HCE$ $ ($ γιατί έχουν $ \angle KAC \equal{} \angle HCE$ $ ),$ αποδεικνύεται εύκολα ότι $ \angle ZCE \equal{} \angle MAC,$ $ \angle ZCH \equal{} MAK$ και συμπεραίνουμε έτσι ότι η ευθεία $ CZ,$ είναι διάμεσος του τριγώνου $ \bigtriangleup HCE,$ αφού και η ευθεία $ AM,$ είναι διάμεσος του τριγώνου  $ \bigtriangleup KAC.$

Άρα, $ NE \equal{} NH,$ όπου $ N\equiv AE\cap CZ.$

gia to b) Είναι $ HM//KD$ άρα $ \angle{CHM}\equal{}\angle{CZA}$ άρα το CZHM εγγράψιμο άρα $ \angle{CZH}\equal{}90$ άρα
από το θεωρ προβολών έχουμε $ NE^2\equal{}NH^2\equal{}NZ\cdot NC$ and we are done  :D

[color=red][b]Πολύ ωραία![/b][/color]

 Με εξαίρεση την πρώτη λύση , στα ίδια συμεράσματα είχα φτάσει και γω.Νομίζω ότι την άσκηση την έχουμε προσεγγίσει καλά και δε χρειάζεται περισσότερο ψάξιμο.
 Εϊναι έτοιμη για πληκτρολόγιση  !

[u] Μπάμπης[/u]


Please see the attachment. That is supposed to be a proof of the associativity of the wedge product. I included most of it for context but the part I don't understand is from where the red underline is onwards (scroll all the way down to see it). I don't understand how the author concluded that the permutations must be the same. I even made an example and it didn't work. Very frustrating.

I have figured it out (I am an idiot). Good luck to all who would like to try too!

I remember a problem  in that book (in the first part) when you´re asked to prove that the inverse of a Hilbert Matrix is made of integers. Anyone knows an elementary proof ???

See that button at the top of the page marked "Search"? Try it. See what pops up if you search for "Hilbert matrix."


Who on this year's (2006) Minnesota teams to ARML in Iowa City is here?

Go Maroon and Gold.

I am, for AMC(did rather miserably at the state tournament).

Congratulations. Maybe my son will meet you at a training session, as I just heard today that he will be a "student in training" invited to practices although not (at the moment) part of either Minnesota team.

For any MN ARML folks reading this, one of this year's team members is planning a preseason social gathering this Saturday.  We did a similar thing last year; the idea is to hang out, meet teammates, etc.  Unfortunately, he doesn't have contact information for most of the team members.  So if you're interested, PM me with your email and I'll forward you the invitation.


Has anyone gone to the Honors Summer Math Camp in Texas? I want to know what it's like and if it's worth going to.

I plan on applying to Texas HSMC this year. Unfortunately, however, I have never attended this summer camp before. However, many of my classmates/friends have and they all have very good things to say about it. It's basically 6 weeks of advanced math and fun. This summer camp is also known for giving second year campers the privilege to complete a Siemens research project and thus compete in the Siemens competition, which is one of America's most prestigious research competitions for high school students.

I attended 3 summers ago, feel free to ask or PM me any questions you might have.


Try to find an appropriate modulus in which the LHS cannot be equal to the RHS.  (This problem has been posted before.)

[hide="Moduli to try"] $ 5, 16$.  (Why?) [/hide]


Use a combinatorial proof to show:

$k \displaystyle\binom{n}{k} = n \displaystyle\binom{n-1}{k-1}$

A combinatorial proof involves coming up with a question or example that uses both sides to answer the question. I cannot come up with a convincing real life example to account for the extra $k$ on the left hand side and the $n$ on the RHS.

Could some one help?
Thanks.

Pick $k$ people for a commitee, one of whom is president.

Either take $\binom{n}{k}$ and pick one of the $k$ for president, so $\binom{n}{k}(k)$ ways

or

take one of the $n$ to be president. Select $\binom{n-1}{k-1}$ for the remainder of the commitee. Thus

$\binom{n}{k}(k) = \binom{n-1}{k-1}(n)$


if f(x)=x:^2:-4x+8, what expression describes f(2x)? Express you answer in terms of x. simplified also

this is open to anyone i guess

A note to students: This type of problem examines what is called a functional transformation.  I will be teaching functional transformations as part of my Algebraic Problem Solving course.

in response to darkquantum's question

[hide] substitute x/2 into f(2x), which gets you  x:^3:/8 - 3x:^2:/4 + 5 [/hide]

[quote="Syntax Error"]awesome possom everyone![/quote]

awesome possum???

Anyway, you guys are correct. Hope you guys all get the drift of how functional notation works.

Syntax Error wrote:Chinaboy wrote:darkquantum wrote:Yup, that's correct.

How about this one?

f(2x)=x^3-3x^2+5

Express f(x) in terms of x.

[hide](1/8)x^3 - (3/4)x^2 + 5[/hide]

you didnt post a solution

O, Chinaboy and s0mp are correct

[quote="MCrawford"]Here's another functional transformation problem:

Let  f(x) = 2x :^2:  + 4x + 3

Let g(x) = f(x) + 4
Let h(x) = f(x + 4)

How do the graphs y = f(x), y = g(x), and y = h(x) differ?[/quote]

y=g(x) is y=f(x) shifted 4 units up.

y=h(x) is also a parabola, but with a different vertex, or something like that....

See if you can figure out what these things mean rather than recalculating everything. you should be able to do these problems MCrawford posted without any computation.

[quote]
O, Chinaboy and s0mp are correct[/quote]

i know theyre correct. but solutions are always goodly[/quote]

P.S.
dont post more than over a triple quote. otherwise ill either 1)delete it, or 2)edit it

ill most likely do the latter, unless some other circumstances come into play

Mmm...I disagree with you, since I don't think it's a big deal as long as it doesn't get out of hand.
Oh well, it's your call.

we'll pm it if we have that case. but if its necessary, then itll flow wit me

hmmm figure out what f(ax) looks like compared to f(x). Then figure out what f(x+b) looks like. Then figure out (if I recall correctly, this is hard) what f(ax+b) looks like.

I'm sorry, Mystic. I've failed you...even though I supported your cause...*sigh*

sigh, ill let it flow, but dont get too out of hand


Good then, then those powers probably shouldn't give me problems in the future, or else i will get back to this post :) Thx for your help.


geometry
geometric transformation
rotation
graph theory
combinatorics unsolved
combinatorics

In a group of $n$ people, each one knows exactly three others. They are seated around a table. We say that the seating is $perfect$ if everyone knows the two sitting by their sides. Show that, if there is a perfect seating $S$ for the group, then there is always another perfect seating which cannot be obtained from $S$ by rotation or reflection.

This is one of the most beautiful problems I've ever seen. In fact, it states that if a $3$-regular graph has a hamiltonian cycle, it has another one too. As a hint, try proving this Lemma:

Let $X$ be the set of vertices of the graph $G$ whose degrees are even and let $v$ be an arbitrary vertex of the graph. Then the number of hamiltonian paths starting at $v$ and ending in a vertex of $X$ is even.


[img]http://www.artofproblemsolving.com/Community/MJImages/MJ11/Ineq.gif[/img]

This is a warmup problem in one of the Olympiad [url=http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=11]MathJam[/url]. There is a simple proof without using AM-GM.

[tex]\displaystyle{(2-{1\over m})(2-{3\over m})\cdots(2-{2m-1\over m})[/tex]
[tex]\displaystyle{={(2m-1)(2m-3)\cdots 1\over m^m}[/tex]
[tex]\displaystyle{={(2m)(2m-1)(2m-2)(2m-3)\cdots2\cdot1 \over (2m)(2m-2)\cdots2\cdot m^m}[/tex]
[tex]\displaystyle{={(2m)!\over m!\codt (2m)^m}[/tex]
[tex]\displaystyle{={(2m)(2m-1)\cdots (m+1)\over (2m)^m}[/tex]
[tex]\displaystyle{={2m\over 2m}\cdot{2m-1\over 2m}\cdots {m+1\over 2m} \leq 1\cdot1\cdots1 = 1}[/tex]

[img]http://www.artofproblemsolving.com/Community/MJImages/MJ11/Ineq.gif[/img]

[color=red][b]And using AM-GM:

(2 - 1/m)(2 - 3/m)(2 - 5/m)...(2 - (2m-1)/m)  :le: [((2 - 1/m)+(2 - 3/m)+(2 - 5/m)+...+(2 - (2m-1)/m))/m]^m = [(2*m - m :^2:/m)/m]^m = [(2m - m)/m]^m = [m/m]^m = 1^m = 1

(2 - 1/m)(2 - 3/m)(2 - 5/m)...(2 - (2m-1)/m)  :le: 1

QED

Btw, that's a nice way of doing it without AM-GM fuzzylogic.  Did you find that out by yourself?[/b][/color]

This is pretty much the same proof provided by [b]MysticTerminator[/b] in the MathJam [url=http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=11]transcript[/url]. By the way, the m-th root is missing on the GM side (LHS) in your proof.

In any case, this is an easy problem which has many solutions. For example, a ncie solution using pairing was discussed in the MathJam.

I always try to give a try myself first when I see the problem. Actually, I had another proof using induction which was considered messy and ugly in the MathJam. It's actually not that bad, here it is:

Let f(m) be the LHS, f(1)<=1 is trivial. Suppose it is true for k, i.e. f(k)<=1. Then
f(k+1)=f(k) * f(k+1)/f(k) <= f(k+1)/f(k). Simplify, we get

[tex]\displaystyle{{f(k+1)\over f(k)} = {(2k+1)k^k\over (k+1)^{k+1}}={(2k+1)\over(k+1)}/ \bigg(1+ {1\over k}\bigg)^k} \leq 1}[/tex], since [tex]\displaystyle{{2k+1\over k+1}\leq2}[/tex] and [tex]\displaystyle{\bigg(1+ {1\over k}\bigg)^k}\geq2}[/tex] by the Binomial Theorem.
Q.E.D.


Λοιπον, αυτη η ασκηση ειναι μια που εφτιαξα πριν λιγο με βαση μια παρομοια που ειδα σε ενα βιβλιο.

Αν $ x, y, z, w$ ειναι θετικοι πραγματικοι αριθμοι, να βρεθει η ελαχιστη τιμη της παραστασης:

$ A \equal{} \sqrt [2008]{\left(\frac {x}{z \plus{} w}\right)^{2008} \plus{} \left(\frac {z}{x \plus{} y}\right)^{2008}} \plus{} \sqrt [2008]{\left(\frac {w}{x \plus{} y}\right)^{2008} \plus{} \left(\frac {y}{z \plus{} w}\right)^{2008}}$

Απο την Minkowski εχουμε:

$ A\geq \sqrt [2008]{\left(\frac {x \plus{} y}{z \plus{} w}\right)^{2008} \plus{} \left(\frac {z \plus{} w}{x \plus{} y}\right)^{2008}}$

Θετουμε $ a\equal{}(x \plus{} y)^{2008}$, $ b\equal{}(z \plus{} w)^{2008}$ και ισχυει

$ \frac {a}{b} \plus{} \frac {b}{a}\geq 2$

Αρα τελικα $ A\geq\sqrt [2008]{2}$

qed

[quote="mailo"]Απο την Minkowski εχουμε:

$ A\geq \sqrt [2008]{\left(\frac {x \plus{} y}{z \plus{} w}\right)^{2008} \plus{} \left(\frac {z \plus{} w}{x \plus{} y}\right)^{2008}}$

Θετουμε $ a \equal{} (x \plus{} y)^{2008}$, $ b \equal{} (z \plus{} w)^{2008}$ και ισχυει

$ \frac {a}{b} \plus{} \frac {b}{a}\geq 2$

Αρα τελικα $ A\geq\sqrt [2008]{2}$

qed[/quote]

σωστος, αυτη ακριβως ειναι η λυση!  :)

[quote="Athinaios"]ρε παιδια απ'οσο γνωριζω ο minkowsky  ισχυει για θετικους αριθμους...το προβλημα nick μας τους δινει πραγματικους. :|[/quote]

ωχ, ναι εχεις δικιο :maybe:
Τωρα κανω edit


let $ \log2\equal{}a$ and $ \log3\equal{}b$ find  $ \log\frac{64}{15}$ in terms of $ a$ and $ b$.

[hide="Solution"]$ \log\frac {64}{15} \equal{} \log64 \minus{} \log15 \equal{} \log2^6 \minus{} \log\frac {30}{2} \equal{} 6a\minus{} (\log3 \plus{} \log10 \minus{} \log2) \equal{} 6a \minus{} (b \plus{} 1 \minus{} a)$

Answer: $ 7a \minus{} b \minus{} 1$[/hide]

EDIT: Oops, typo :oops:

Thanks for pointing out, modularmarc101.  :)

EDIT2: Wow, fail...

Thanks sunehra and tonypr for pointing out.

ohh, wow nice solution.

i knew how to do the problem, but i never thought of changing the 15 into a 30/2. :oops:

$ nice$

Fantasy, I think there's a little error where it says:

$ a^6 \minus{} (\log{3} \plus{} \log{1} \minus{} \log{2})$

It should be $ \log{10}$ instead of $ \log{1}$ but I see that for the next step you simplified it to $ 1$, so it was probably just a typo. Anyways just clearing that up.

Isn't it $ 6a$ and not $ a^6$? You are not performing the operation $ (\log{(2)})^6$; you are doing $ \log{(2^6)} \equal{} 6\log{(2)}$.

Well, there might be another little error there, $ \log 2^6 \equal{} 6a$ not $ a^6$, it's an easy mistake to make.

So the answer is $ 7a \minus{} b \minus{} 1$

Edit: darn sunehra beat me by a minute