Interesting distribution - I think it points to a good test.

And, congrats, btw.

So how did the Canadians do? I also heard that Natee Pitiwan is like 20 years old and has already gone to IMO with bronzes from Thailand. [/quote]

i'm confused: the amc site says 17 people got between 0 and 1, and there are 18 stars next to 0 on the chart.

there are 265 numbers given, but 259 test takers by qualification and by adding up the numbers in the amc chart.  there are a total of 23 0-1 people on the star chart, that is 6 extra people

does that mean they throw in extra numbers randomly and assign them the score zero?

Time to work towards MOP 2006 I guess. Shouldn't have neglected this year...

i think in the individual scores by number, they throw in some random zeros to make the scores seem random or something so there would be 18 zeros listed individually, but only 12 in reality

[quote="probability1.01"]Yufei, did you [i]count[/i]?  :?[/quote]

Of course not. Who can possibly count all that and post it within 5 min of the release of the scores?

But if you do count it, there are 18 0's, for some reason.

Peng Shi and I (both Canadians) tied for third place with 34 pts.

So I really tied for 5th :P

Thanks Canada haha

I like that distribution.  Nice and sparse up top, so we can tell who's good.  No need letting everybody get almost 42's.

[quote="darktreb"]So I really tied for 5th[/quote]
Why would you say that? It's not like that we were unofficial.  :D

[quote="MithsApprentice"][quote="inteluser"]Another thread has it as Brian Lawrence (I think).[/quote]

Makes sense. He's the only one on last year's IMO list (alternate 2) that hasn't gone off to college.[/quote]

You're forgetting that Sherry Gong was an IMO silver medallist, which sure beats any contest standing ever achieved by any USAMO qualifier this year. How did she do anyway?

Also, how did Brian Lawrence become an IMO alternate last year? He wasn't on the USAMO winner/HM list.

he pwned the TST so zuming just chose him

[quote="billzhao"]
You're forgetting that Sherry Gong was an IMO silver medallist, which sure beats any contest standing ever achieved by any USAMO qualifier this year. How did she do anyway?

Also, how did Brian Lawrence become an IMO alternate last year? He wasn't on the USAMO winner/HM list.[/quote]

I was referring to the U.S. IMO team.

Anyhow, Brian got 5 questions (or something insane like that) right on the TST last year. That kinda moved him up a bit in the rankings.

Sherry gong is an IMO medallist?? when did that happen?

[quote]Why would you say that? It's not like that we were unofficial. [/quote]

You guys are definitely official, but you're not official team candidates :P.

As for Sherry Gong, she's been going to IMO as a member of the Puerto Rico team (yes they have a team, and yes you should move there now). I believe she got a 29.

Anyone know how Natee Pitiwan did?

Hmm...I think Sherry counts as an American this year since she's at Exeter now.

Btw...I wonder if these people that we talk about get freaked out by so many people knowing stuff about them.

Well I actually happen to know the people I talk about.


vector
function
topology
Functional Analysis
real analysis
real analysis solved

$T:X\to Y$ is a linear operator, where $X$ and $Y$ are normed linear spaces.
Is $T$ bounded if the nullspace of $T$ is closed?
Is the nullspace closed if $T$ is bounded?

The answer to the second question is 'yes', because $T$ being bounded implies that it is continuous, so the preimage of any closed set of $Y$ must be closed. Now, since $\{0_Y\}$ is closed, $T^{-1}(0)$ must be closed in $X$.

As for the first question, I believe the answer is negative here. Consider the real vector space of sequences which are eventually $0$, with the standard basis $e_1,e_2,\ldots$ (with the norm $\sup$, for example), and define $T$ to be $T(\sum\alpha_i e_i)=\sum k_i\alpha_ie_i$ where the $k_i$'s are some reals which define the operator $T$. If we choose $k_i\to\infty$, we have a counterexample (the null space consists of the null vector of $X$, so it's closed, but $T$ is not bounded).

An important feature to notice about grobber's counterexample: that space, the finitely supported sequences with whatever norm he decided to put on in (it didn't matter which one), is a normed vector space but not a Banach space. It's not complete.

The completion of the finitely supported sequences in the sup norm is the space $c_0$ of sequences that tend to zero. The operator grobber defined does not extend to a continous operator on that space.

So, tossing the question back to liyi: if $X$ is a Banach space, is $T$ bounded if its null space is closed?

Eh. I don't think so.

Let $X=C([0,1])$ and the normed is defined by $f\| = \max\limits_{x\in [0,1]} |f|$
Then $X$ is a Banach space.

Let $T$ be the differential operator, $Tf =\frac{df}{dx}$. Then $T(x^n) = nx^{n-1}$ and $\|T(x^n)\|=n$. $T$ is unbounded. But $\mathcal{N}(T)$ consists of constant functions, it is closed.

The problem with that example is that the differential operator is not defined on all of $X$- it's only defined on a dense subspace, where it is essentially similar to grobber's example.

I don't think the extra condition of $X$ being complete makes any difference. We can simply take any infinite-dimensional Banach space and define an operator pretty much as above from this space to itself, can't we? :?

grobber is right. Completeness makes no difference, and the essential features of grobber's first example control the situation.

Let $X$ be any infinite-dimensional normed vector space. Let $\{e_{\gamma}\}_{\gamma\in\Gamma}$ be a [b]Hamel[/b] basis for $X$. (Yeah, that's right - Hamel basis. #%!#&!) Let $f : \Gamma\mapsto(0,\infty)$ be an unbounded nonzero function. Define $T : X\mapsto X$ by $Te_{\gamma}=f(\gamma)e_{\gamma},$ extended to the whole of $X$ by linearity.

The operator is unbounded (not to mention such other attributes as being nonconstructible, unimaginable, and thoroughly useless), but the null space is the closed set $\{0\}.$

Note that in grobber's original presentation of the example, he was using a Hamel basis - he opted for an example of a vector space with a countable Hamel basis.

One condition that would work: the linear operator has finite-dimensional range.


[quote="jensen"]solve this equation:($x,y,x \in R$)
$x+y=5$
$y+2z=7$
$z+3x=4$ :D[/quote]I'm assuming you mean x,y, and z are real.

Subtracting the first from the second, we get $2z-x=2$.  Multiplying by 3 and adding to the 3rd, we get that $7z=10$ and $z=\frac{10}{7}$.  Therefore $x=\frac{6}{7}$ and $y=\frac{29}{7}$.

[quote="lotrgreengrapes7926"][quote="DanK"][quote="samath"]I'm assuming you mean x,y, and z are real.[/quote]

That's what $(x,y,z\in R)$ means...

Btw, I did the same thing as you...[/quote]

It says $(x,y,x\in R)$ in the question, not $(x,y,z\in R)$, and that was obviously a typo.[/quote]
yes you are right,soory it was a mistake. :D

[quote="lotrgreengrapes7926"][quote="DanK"][quote="samath"]I'm assuming you mean x,y, and z are real.[/quote]

That's what $(x,y,z\in R)$ means...

Btw, I did the same thing as you...[/quote]

It says $(x,y,x\in R)$ in the question, not $(x,y,z\in R)$, and that was obviously a typo.[/quote]
yes you are right,soory it was a mistake.but i think you got what i mean. :D

[quote="José"][quote="jensen"]solve this equation:($x,y,z \in R$)
$x+y=5$
$y+2z=7$
$z+3x=4$ :D[/quote]

$x+y=5 \Longleftrightarrow y=5-x$
$y+2z=7 \Longleftrightarrow y=7-2z$
$5-x=7-2z \Longleftrightarrow z=\frac{x+2}{2}$
$z+3x=4 \Longleftrightarrow z=4-3x$
$\frac{x+2}{2}=4-3x \Longleftrightarrow x=\frac{6}{7}$.
Then, the solution follows...[/quote]
nice job. :D


trigonometry
inequalities proposed
inequalities

My book P156，there is a Ugly ineq,in triangle:
$ \frac {1}{\sin\frac {A}{2}}\geq \frac {\sqrt {3}}{2}\frac {r}{s}(1 \plus{} \frac {1}{\sin\frac {A}{2}}) \plus{} \frac {1}{2}.$

By adjustment,get a nice:

$ \frac {w_a}{3r}\geq \frac {2w_a \plus{} r_a}{2r_a \plus{} w_a} \geq \frac {\sqrt {3}w_a}{s} .$---------------(*)
BQ

Note:
I fond out (*) anly useing time 5 minutes.

Yes,
I did not want to hide my own,
It is very nice!
BQ

The use of reorientation of thinking,
As well as the "scale",
Is not difficult to find them.
I hope to share with you to explore.

Although Chinese Text,
But with On-line translator,
We can be a very good exchange,
Is not it?
Look,I say you all understand it?

BQ

in order to Festival of space, put here a few days later delete.

next next next next next next next next next next next next

[size=200]in windows mspaint(Paintbrush)read is Clear.[/size]


[quote="Ignite168"]It is now past 3:00AM but we will just call it 3:00AM. Ignite168 realizes that it may be time to consider to go to bed. If Ignite168 falls asleep immediately at 3:00AM, has a 1/24 chance of waking up to go to the bathroom at 4AM, 2/24 one at 5AM, 3/24 one at 6AM.. etc. and will be cranky if he wakes up before 9:01AM, what is the probability Ignite168 will NOT be cranky when he wakes up?[/quote]

[hide]Add up the probabilities. There is a $\frac{21}{24}$ chance that Ignite will wake up before 9:01 AM therefore, there is a $\boxed{\frac{1}{8}}$ he will not be cranky.[/hide]

[quote="bpms"][quote="Ignite168"]It is now past 3:00AM but we will just call it 3:00AM. Ignite168 realizes that it may be time to consider to go to bed. If Ignite168 falls asleep immediately at 3:00AM, has a 1/24 chance of waking up to go to the bathroom at 4AM, 2/24 one at 5AM, 3/24 one at 6AM.. etc. and will be cranky if he wakes up before 9:01AM, what is the probability Ignite168 will NOT be cranky when he wakes up?[/quote]
[hide]$\frac{1}{24}+\frac{2}{24}+\frac{3}{24}+\frac{4}{24}+\frac{5}{24}+\frac{6}{24}=\frac{18}{24}=\frac34$
$1-\frac34=\frac14$[/hide][/quote]

You added the fractions wrong.

[quote="bpms"][quote="Ignite168"]It is now past 3:00AM but we will just call it 3:00AM. Ignite168 realizes that it may be time to consider to go to bed. If Ignite168 falls asleep immediately at 3:00AM, has a 1/24 chance of waking up to go to the bathroom at 4AM, 2/24 one at 5AM, 3/24 one at 6AM.. etc. and will be cranky if he wakes up before 9:01AM, what is the probability Ignite168 will NOT be cranky when he wakes up?[/quote]
[hide]$\frac{1}{24}+\frac{2}{24}+\frac{3}{24}+\frac{4}{24}+\frac{5}{24}+\frac{6}{24}=\frac{18}{24}=\frac34$
$1-\frac34=\frac14$[/hide][/quote]

I dont think yours is correct.
[hide]$\frac{1}{24}+\frac{2}{24}+\frac{3}{24}+\frac{4}{24}+\frac{5}{24}+\frac{6}{24}=\frac{21}{24}=\frac78$
$1-\frac78=\frac18$[/hide]

This is assuming that you could go to the bathroom every hour, correct?

[hide="if so"]Just add them up  :D .  $1+2+...+6=\frac{6\cdot7}{2}=21$, and each one is over 24, so the probability is $1-\frac78=\frac18$[/hide]

[quote="bpms"]:ewpu: 6*3.5 not 6*3.[/quote]

Err, what do you mean?  I did $\frac{6\cdot7}{2}=21$.

What I meant is you guys are right. For $1+2+3+4+5+6$ I did $6*3$ not $6*3.5$

Well, you realized your mistake.  Let us hope that this doesn't happen to you at National MathCounts, State MathCounts, or Chapter.  It happened to me. :mad:

[quote="Ignite168"]The problem was supposed to say once Ignite168 wakes up to use the bathroom, afterwards, he will remain awake, so there would be multiplication involved.[/quote]
But this way the problem is easier :D.
Here's the solution the way ignite intended for it to be
[hide]
$\frac{1}{24}=400 \\\frac{23}{24}\cdot\frac{2}{24}=\frac{23}{288}=500 \\\frac{27}{32}\cdot\frac{3}{24}=\frac{27}{256}$\\
this keeps getting uglier. If I had a calc then this would be doable but this is crazy. I give up unless this magically becomes a target problem in which case I will solve it later.[/hide]

[quote="Ignite168"][quote="mathgeniuse^ln(x)"][quote="Ignite168"]The problem was supposed to say once Ignite168 wakes up to use the bathroom, afterwards, he will remain awake, so there would be multiplication involved.[/quote]

But, it didn't say that and so multiplication isn't involved.  Eitherway, the multiplication is too messy.[/quote]

Not really, only 5 things have to be multiplied and 24 is a pretty nice number, goes into 2, 3, 4, and 6.[/quote]
Have you worked it out? If so, please post your solution. I must be doing something wrong because my numbers are ridiculously ugly.

[quote="bpms"][/quote][quote="Ignite168"][quote="mathgeniuse^ln(x)"][quote="Ignite168"]The problem was supposed to say once Ignite168 wakes up to use the bathroom, afterwards, he will remain awake, so there would be multiplication involved.[/quote]

But, it didn't say that and so multiplication isn't involved.  Eitherway, the multiplication is too messy.[/quote]

Not really, only 5 things have to be multiplied and 24 is a pretty nice number, goes into 2, 3, 4, and 6.[/quote][quote="bpms"]
Have you worked it out? If so, please post your solution. I must be doing something wrong because my numbers are ridiculously ugly.[/quote]
[hide=" :roll:"]
I thought it would be obvious the answer was $\frac{168245}{331776}$. Sorry.
[/hide]

[quote="Ignite168"][/quote][quote="bpms"][/quote][quote="Ignite168"][quote="mathgeniuse^ln(x)"][quote="Ignite168"]The problem was supposed to say once Ignite168 wakes up to use the bathroom, afterwards, he will remain awake, so there would be multiplication involved.[/quote]

But, it didn't say that and so multiplication isn't involved.  Eitherway, the multiplication is too messy.[/quote]

Not really, only 5 things have to be multiplied and 24 is a pretty nice number, goes into 2, 3, 4, and 6.[/quote][quote="bpms"]
Have you worked it out? If so, please post your solution. I must be doing something wrong because my numbers are ridiculously ugly.[/quote][quote="Ignite168"]
[hide=" :roll:"]
I thought it would be obvious the answer was $\frac{168245}{331776}$. Sorry.
[/hide][/quote]

How was that obvious? That answer would have never hit me when I looked at the problem.

Also, can you people not make little teeny little hide tags in thier signature that hides the quote button? It's so annoying trying to find it. At least with me, the hide tag is really big.

[quote="SplashD"][/quote][quote="Ignite168"][/quote][quote="bpms"][/quote][quote="Ignite168"][quote="mathgeniuse^ln(x)"][quote="Ignite168"]The problem was supposed to say once Ignite168 wakes up to use the bathroom, afterwards, he will remain awake, so there would be multiplication involved.[/quote]

But, it didn't say that and so multiplication isn't involved.  Eitherway, the multiplication is too messy.[/quote]

Not really, only 5 things have to be multiplied and 24 is a pretty nice number, goes into 2, 3, 4, and 6.[/quote][quote="bpms"]
Have you worked it out? If so, please post your solution. I must be doing something wrong because my numbers are ridiculously ugly.[/quote][quote="Ignite168"]
[hide=" :roll:"]
I thought it would be obvious the answer was $\frac{168245}{331776}$. Sorry.
[/hide][/quote][quote="SplashD"]

How was that obvious? That answer would have never hit me when I looked at the problem.

Also, can you people not make little teeny little hide tags in thier signature that hides the quote button? It's so annoying trying to find it. At least with me, the hide tag is really big.[/quote]
Yes, would you please not do that? It's starting to get really annoying, especially the signatures with the tiny single dots. In fact, it's better not to use hide tags in your signatures at all, because for example, there may be a new user who can answer a question you have, but he/she can't find the quote button and so can't post a reply to your question.

[quote="SplashD"][/quote][quote="Ignite168"][/quote][quote="bpms"][/quote][quote="Ignite168"][quote="mathgeniuse^ln(x)"][quote="Ignite168"]The problem was supposed to say once Ignite168 wakes up to use the bathroom, afterwards, he will remain awake, so there would be multiplication involved.[/quote]

But, it didn't say that and so multiplication isn't involved.  Eitherway, the multiplication is too messy.[/quote]

Not really, only 5 things have to be multiplied and 24 is a pretty nice number, goes into 2, 3, 4, and 6.[/quote][quote="bpms"]
Have you worked it out? If so, please post your solution. I must be doing something wrong because my numbers are ridiculously ugly.[/quote][quote="Ignite168"]
[hide=" :roll:"]
I thought it would be obvious the answer was $\frac{168245}{331776}$. Sorry.
[/hide][/quote][quote="SplashD"]

How was that obvious? That answer would have never hit me when I looked at the problem.

Also, can you people not make little teeny little hide tags in thier signature that hides the quote button? It's so annoying trying to find it. At least with me, the hide tag is really big.[/quote]He was being sarcastic.

[quote="bpms"][quote="Captain Sugar"]i was trying to quote him earlier! i looked all over his signature and couldn't find anything!  :blush:[/quote]
THere is a very tiny dot in his sig.[/quote]

i know that now. it was just... annoying.

This has been braught up at the following link:

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=566020#566020

There is another topic that was linked from it.

Wierd, I didn't need to put it in url tags to get the url to work  :o


From these five numbers select three different numbers for x, y, and z, in order to make this expression,
 x(y - z),  as large as possible

What is the largest possible value of x(y - z)?

A.60	B.85	C.90	D.105	E.120

You need to list the numbers.  :)

[hide="Proposed Solution"]$ z$ is the smallest integer possible, and $ y\minus{}z$ and $ x$ should be as close as possible (by AM-GM), the maximum given the sum is constant is when the two quantities are equal.
[/hide]


Let a,b,c be nonnegative real numbers
prove that
\[ \frac{{a\left( {b \plus{} c} \right)}}{{\mathop a\nolimits^2  \plus{} bc}} \plus{} \frac{{b\left( {c \plus{} a} \right)}}{{\mathop b\nolimits^2  \plus{} ac}} \plus{} \frac{{c\left( {a \plus{} b} \right)}}{{\mathop c\nolimits^2  \plus{} ab}} \ge 2
\]

[quote="csbonny"]Let a,b,c be nonnegative real numbers
prove that
\[ \frac {{a\left( {b \plus{} c} \right)}}{{\mathop a\nolimits^2 \plus{} bc}} \plus{} \frac {{b\left( {c \plus{} a} \right)}}{{\mathop b\nolimits^2 \plus{} ac}} \plus{} \frac {{c\left( {a \plus{} b} \right)}}{{\mathop c\nolimits^2 \plus{} ab}} \ge 2
\]
[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=76441


[hide="Solution"]
Simply plugging the values into the function then adding gives \begin{align*}f(-1)+f(0)+f(1)&=\left[(-1)^2-1+1\right]+\left[0^2+0+1\right]+\left[1^2+1+1\right]\\&=1+1+3\\&=5,\end{align*} which is the answer choice $ \boxed{\textbf{(E)}}.$[/hide]

[hide="Alternate"]For a slightly faster solution, note that $ \minus{}1\plus{}0\plus{}1\equal{}0$ so you can take out the $ x$. Since $ 1\plus{}1\plus{}1\equal{}3$, just take
\[ 3 \plus{} ( \minus{} 1)^{2} \plus{} 0^{2} \plus{} 1^{2} \equal{} 2 \plus{} 3 \equal{} 5\]
[/hide]


I hope it's okay that I post this here. If not, tell me and I will delete it.

[tex]
f(x) = \frac {\sqrt {x^{j_i}-x^{j_{i+1}}}} {\sin x}
[/tex]

[tex]
\lim_{x \to 0} \frac{3x^2 +7x^3}{x^2 +5x^4} = 3.
[/tex]

[tex]
\lim_{x \to +\infty} \frac{1}{x} = 0.
[/tex]

it's ok to try latex formulae in here :)


An isosceles trapezoid $ABCD$ is placed inside a semicircle such that they share the same base, $AD=4$, ($AD$ is a diameter) and the lengths $AB=DC=1$ are chords. Find the length $BC$.

[quote="frt"]That's not correct...

I don't understand this part:
[quote="Iversonfan2005"]then let us call point $P$ which is collinear to $AD$ and $PO=BX=\frac{x}{2}$ and thus the equations $(2-\frac{x}{2})^2+4-\frac{x^2}{4}=1$[/quote]
What's $X$, and how did you get the last equation? :?[/quote]

Sorry I don't think I gave a very good explanation: [hide]Here's what I did, I made $O$ the center of diameter $AD$ which is also the center of the entire circle. From there, I drew radius $BO=2$, then I made $M$ the midpoint of $BC$ and drew in $OM$ which is the height of the trapezoid. Since $BM=\frac{x}{2}$, and triangle $BMO$ is a right triangle, with $BO$ as the hypotenuse, we see that $MO=\sqrt{2^2-(\frac{x}{2})^2}$, now let us call point $P$ which is on segment $AO$ such that $PB$ is parallel to $MO$ and is also equal to $MO$, therefore $PB=\sqrt{2^2-(\frac{x}{2})^2}$, now $AP=2-\frac{x}{2}$, from here we obtain a right triangle with $AB$ as the hypotenuse and $AP$, $PB$ as legs, so we set up the equation: $2^2-(\frac{x}{2})^2+(2-\frac{x}{2})^2=1^2$ and we get $\boxed{x=\frac{7}{2}}$[/hide]
Sorry I made a mistake in my algebra, but I think that should be right now.If my answer or solution doesn't make sense, then someone can please tell me what I did wrong :)


I know that when defining a shape we list it's vertices in order, but are they supposed to be clockwise or counterwise, or does it not matter? Also, Ahould I started on a specific letter,like the one that comes first in the alphabet?

I was joking when I said this was a test post :?


superior algebra
superior algebra unsolved

Prove that the idèle class group of the rationals is isomorphic to $ \mathbb{R}^\times \times \prod_p \mathbb{Z}_p^\times$.

That's not completely correct I think:

The idele-group $ \coprod_{p\in P} \mathbb Q_p^\times$ consists of those $ (a_p)_p$ with all but finitely many $ a_p \in \mathbb Z_p^\times$ (for $ p\equal{}\infty$, this is ignored); I used $ \coprod$ because I couldn't find the correct TeX-code. Here $ P\equal{}\{2,3,5,...,\infty\}$ runs through the different valuations.
Note that a rational number is uniquely determined by it's non-archimedian valuations and it's sign.
Let $ (a_p)_p \in \coprod_{p\in P} \mathbb Q_p^\times$. Then we can find exactly one $ b \in \mathbb Q^\times$ with $ v_p(b)\equal{}v_p(a)$ for all non-archimedean valuations and same sign than $ a$. In the idele-class-group, we can replace $ a$ by $ ab^{\minus{}1}$, this is in $ \mathbb{R}^\plus{} \times \prod_p \mathbb{Z}_p^\times$. As $ b$ is unique, every such $ a$ corresponds to exactly one representant from $ \mathbb{R}^\plus{} \times \prod_p \mathbb{Z}_p^\times$, so the idele-class-group is isomorphic to this.


From a Point P outside a circle with center O, lines are drawn tangent to the circle at points A and B. A third line is drawn from P, intersecting the circle at M on minor arc AB such that m(MPA)=30 and m(MPB) = 60. Evaluate cos(measure of angle MOB)

Notice that PBOA is a square. Can you solve it from there?

Another hint:

[hide]Draw the diagonal of the square. You kno wat APO, BOP are, rite? Then you can find out wat OMP is. Can you figure out the answer now?[/hide]

[hide] 3/5? [/hide]

i had to use coordinates, and even though it didnt take to long, is there another way to do it?


function
calculus
derivative
real analysis
real analysis unsolved

I think you'd better include "$ f$ is continuous on $ \mathbb{R}$" in your hypotheses, or else we'll come up with some trivially silly counterexamples.

[quote="Kent Merryfield"]I think you'd better include "$ f$ is continuous on $ \mathbb{R}$" in ....[/quote]
I corrected, $ f$ is assumed differentiable.

Ah, making it differentiable everywhere cuts out my nontrivial example. I'll mention it anyway.

Let $ g(x)$ be the Cantor "devil's staircase" function, which takes $ (\frac13,\frac23)$ to $ \frac12$, $ (\frac19,\frac29)$ to $ \frac14$, $ (\frac79,\frac89)$ to $ \frac34$, and so on. It's continuous, (non-strictly) increasing, and constant on each subinterval of the complement of the Cantor set, for zero derivative there.
Let $ f(x)\equal{}x\minus{}g(x)$, extended to be periodic of period 1. It is differentiable almost everywhere, with its derivative equal to $ 1$ at those points. It is definitely not globally increasing, or even increasing on any interval with length greater than $ \frac13$.

[quote="Fleury"]Let $ f : \mathbb{R}\to \mathbb{R}$ differentiable, and $ \Omega\subsetneq \mathbb{R}$, with $ \overline{\Omega} \equal{} \mathbb{R}$. Assume that for all $ x\in \Omega$, $ f'(x) > 0$.
Is $ f$ increasing?[/quote]
No, $ f$ is not necessarily (strictly) increasing, but finding a counter example, with $ f$ differentiable over $ \mathbb{R}$, is not easy at all.

See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=2822]this thread[/url] and the reference therein ;).


rotation
combinatorics proposed
combinatorics

On an $n\times n$ board a token is placed in one of the corners. Two players each move the token to an adjacent square (sharing an edge). It’s forbidden for the token to visit a square it has visited before. The player who cannot move loses. Who has a winning strategy?

[I’ve modified it a bit to make it more challenging :)]

A really cute one :
The first player has a winning strategy if $n$ is even and the second player has a winning strategy if $n$ is odd.

- First assume that $n$ is even :
Then, we may cover the board with $\frac {n^2} 2 $ non-overlapping dominos. Now, the first player move the token from the initial unit square to the other unit square of the same domino. Then, the second player has to move the token (if possible) to the  unit square of a non already touched domino. Thus, player one can move it to the other unit square of this domino and so on.
At each of his turn, the second player has to move the token in a unit square which belongs to a domino whose two unit squares have not been visited. Thus, if the second player can play, the first play can too by moving the token ointo the other unit square of the same domino.
Therefore, the first player has a winning strategy.

- Now, assume that $n$ is odd :
Wlog, we may assume that the first player moves the token from the upper-left corner to the adjacent unit square which belongs to the same row (otherwise just rotate the board). Now, 'delete' the upper-left unit square. It is easy to verify that the remaining board can be covered with non-overlapping dominos. Then, the second player can use the strategy of the preceding case, and win the game.

Pierre.

Ok, this is really scary: I did exactly the same thing! :)

I think that this is the standard way to do such a problem :D

I know of an extension to this problem, but it's been proposed as a prize problem in an issue of one of the Singapore mathematics magazines (it's kind of funny because the prize problems are easier than most of our training problems, but we get book vouchers if we submit really good solutions for them :) ), so I won't post it up until 1st October when the submission dateline is over.

Just a word about it : Please create a new topic for this generalisation.

Pierre.


calculus integration logarithms limit factorial derivative algorithm

Hello!

I'm solving two integrals and I find myself stuck in the same place: Residues!
First one is:

Show that for $ n\in\mathbb{N}$
$ \int_0^{\infty} \frac{x}{(1\plus{}x)^{n\plus{}2}}dx\equal{}\frac{1}{n(n\plus{}1)}$

First I added a logarithm when setting real to complex integral:
$ \oint\frac{z \ln z}{(1\plus{}z)^{n\plus{}2}}dz$

Then I found poles:
$ (1\plus{}z)^{n\plus{}2}\equal{}0\Rightarrow z\equal{}\minus{}1$ the pole of n+2 order.

So residue is:
$ Res(f,z)\equal{}\frac{1}{(n\plus{}2\minus{}1)!}\lim_{z\to\minus{}1}\frac{d^{n\plus{}2\minus{}1}}{dz^{n\plus{}2\minus{}1}}\left[(z\plus{}1)^{n\plus{}2}\frac{z\ln z}{(z\plus{}1)^{n\plus{}2}}\right]$

And you see my problem: how to find: $ \lim_{z\to\minus{}1}\frac{d^{n\plus{}1}}{dz^{n\plus{}1}}\left[z\ln z\right]$

Even if I made mistake with logarithm, how to derive sth like that? Isn't that 0 or what :?

Second one:

Show that for: $ n\in\mathbb{N}, 0<p<1$ and $ a>0$

$ \int_0^{\infty}\frac{x^p}{(x\plus{}a)^n}dx\equal{}\frac{\Gamma(p\plus{}1)\Gamma(n\minus{}1\minus{}p)}{\Gamma(n)}a^{p\plus{}1\minus{}n}$

so integral in complex is (I'm assuming) this:
$ \oint\frac{z^p}{(z\plus{}a)^n}dz$ the poles are:
$ (z\plus{}a)^n\equal{}0\Rightarrow z\equal{}\minus{}a$ the pole of n-th order

Residue:
$ Res(f,\minus{}a)\equal{}\frac{1}{(n\minus{}1)!}\lim_{z\to\minus{}a}\frac{d^{n\minus{}1}}{dz^{n\minus{}1}}\left[(z\plus{}a)^n\frac{z^p}{(z\plus{}a)^n}\right]$

So that is:
$ Res(f,\minus{}a)\equal{}\frac{1}{(n\minus{}1)!}\lim_{z\to\minus{}a}\frac{d^{n\minus{}1}}{dz^{n\minus{}1}}[z^p]\equal{}
\frac{1}{(n\minus{}1)!}\lim_{z\to\minus{}a}[p(p\minus{}1)(p\minus{}2)\cdot\ldots\cdot(p\minus{}n\plus{}1\plus{}1)z^{p\minus{}n\plus{}1}]\equal{}$
$ \equal{}\frac{1}{(n\minus{}1)!}[p(p\minus{}1)(p\minus{}2)\cdot\ldots\cdot(p\minus{}n\plus{}2)(\minus{}a)^{p\minus{}n\plus{}1}]$

I have a feeling that I should get $ p!$ in there but I don't know how :(
So that are my problems. I would appreciate if someone could help.

Also I was wondering if there's a place, or a book, that has that factorial thing explained. For instance:
$ n(n\plus{}1)(n\plus{}2)\cdot\ldots\cdot(2n\minus{}2)$ I multiply that by $ \frac{(n\minus{}1)!}{(n\minus{}1)!}$ and that becomes: $ \frac{(2n\minus{}2)!}{(n\minus{}1)!}$.

One of my friend showed me that but I still don't quite understand why...

So if anyone can help I'll be very gratefull!

With regards to that derivative computation, it's actually really easy.  Note that $ f'(z)\equal{}1\cdot\ln(z)\plus{}\frac{z}{z}\equal{}\ln(z)\plus{}1$.  Then $ f''(z)\equal{}\frac{1}{z}$, and from here, the $ n$th derivative is straightforward:
$ \frac{d^{n}f}{dz^{n}}\equal{}\frac{(\minus{}1)^{n}(n\minus{}2)!}{z^{n\minus{}1}}$, for $ n\geq 2$.  Then $ \lim_{z\rightarrow \minus{}1}\frac{(\minus{}1)^{n\plus{}1}(n\minus{}1)!}{z^{n}}\equal{}(\minus{}1)^{n}(n\minus{}1)!$, assuming I didn't botch any arithmetic.

Please be advised that I did not check your work, I just computed what you told me to so it could still be wrong.

Thanks for the reply.

You took $ f(z)\equal{}z\ln(z)$, right?
How do you know the formula $ \frac{d^{n}f}{dz^{n}}\equal{}\frac{(\minus{}1)^{n}(n\minus{}2)!}{z^{n\minus{}1}}$? Is it intuitive or is there some algorithm to follow? I have problems with that :|

And that's n-th derivative, which is fine but I need $ \frac{d^{n\plus{}1}}{dz^{n\plus{}1}}$. Or can I just make it like n-th derivative $ \times$ one derivative?

Well first I took the ordinary derivative, and then, once seeing that it just reduced to $ \ln{z}$, I saw that it would be easy to work with.  Then I just took some more derivatives, noticed that the numerator was growing factorially, observed how the sign was changing, and what the exponent in the denominator was doing.  Then it was just a matter of matching signs.  So to answer your question, I didn't use any magic formulas, I just observed a pattern and tried to match it.

And I took the $ n\plus{}1$th derivative into account when I took the limit.

Wow, I was going to reply how I got a part of that factorial thing and then I realised the whole thing!! Yay :D
You got rid of that +1 derivative in $ \frac{d^{n\plus{}1}}{dz^{n\plus{}1}}$ and you got $ \ln(z)\plus{}1$ so you derived that the remaining n times, right? So by deriving $ \ln(z)\plus{}1$ (which is the same as $ \ln(z)$ because 1 is a constant) you got
$ \frac{(\minus{}1)^{n\minus{}1}(n\minus{}1)!}{z^n}$ for $ n\ge1$ and when putting $ z\equal{}\minus{}1$ you get $ \frac{(\minus{}1)^{n\minus{}1}(n\minus{}1)!}{(\minus{}1)^n}$.
But I don't understand how you got $ (\minus{}1)^n(n\minus{}1)!$? Shouldn't $ (\minus{}1)^n$ factors cancel out? :?

Yes, you're right, I canceled out the wrong power of -1, so it should just be $ \minus{}(n\minus{}1)!$.

I scrolled down to the bottom of your first post, and noticed that you questioned the expression
[quote="dingo_bastard"]$ n(n\plus{}1)(n\plus{}2)\cdot\ldots\cdot(2n\minus{}2)$ I multiply that by $ \frac{(n\minus{}1)!}{(n\minus{}1)!}$ and that becomes: $ \frac{(2n\minus{}2)!}{(n\minus{}1)!}.$[/quote]
This is correct.  When I look at the expression on the left, I immediately see the one on the right (so that extra step of multiplying is not even there).  I suppose you would like an explanation: I'm sure you have no trouble comprehending why the denominator is $ (n\minus{}1)!$ (because you multiplied by it!).  In the numerator, you end up with $ (n\minus{}1)!n(n\plus{}1)\cdots (2n\minus{}2)$.  If you rewrite this in its expanded form, you get $ 1\cdot 2\cdots (n\minus{}1)n(n\plus{}1)(n\plus{}2)\cdots(2n\minus{}2)$, which is the very definition of $ (2n\minus{}2)!$.

Oh, right I see it now! I have made myself a little drawing of how it works, and it's clear to me. I can be silly sometimes :D Thanks for the help! I really appreciate it!


Let $ a_1,...,a_n $ be positive numbers which add up to n. Prove that $\displaystyle\sum_{1\leq i\leq n}{\frac{1-a_i\cdot a_{i+2}}{a_i+a_i\cdot a_{i+1}}}\geq 0 $. Of course, you know that indices are taken mod n.

[quote="harazi"]Let $a_1,...,a_n$ be positive numbers which add up to n. Prove that $\displaystyle\sum_{1\leq i\leq n}{\frac{1-a_i\cdot a_{i+2}}{a_i+a_i\cdot a_{i+1}}}\geq 0$. Of course, you know that indices are taken mod n.[/quote]

 :D  :D


algebra
polynomial
search
algebra proposed

I found this problem in the book The IMO compendium:
Let $a$ be a number different from 0. For all integers $n$ define
$S_{n}=a^{n}+a^{-n}$. Prove that if for some integer $k$ both $S_{k}$ and $S_{k+1}$
are integers, then for each integer $n$ the number $S_{n}$ is an integer.

The problem doesnt look hard but unfortunately I didnt find any short solution.
I hope that you can find something more elegant.
And its important to note that problem doesnt seem to be absolutely correct:
for $a=\exp(i\pi/6)$ we have $S_{1}=\sqrt{3},S_{2}=1,S_{3}=0$.
But if you add assumption $a\neq\exp(\pm i\pi/6)$ and $a\neq\exp(\pm5i\pi/6)$,
I think it would be correct.

I think $S_{n}$ is integer if and only if $a=1$

suppose $n=1$

so $1/x+x$ is never integer.

 I did not get what u r saying

$a$ can be complex.  In other words, $S_{n}$ is a sequence satisfying the recurrence

$S_{0}= 2$
$S_{k+1}= S_{1}S_{k}-S_{k-1}$

Edit:  Working this recurrence backwards it's possible to express $S_{1}$ as a polynomial in $S_{1}$, and if $S_{1}$ can be shown to be integer the conclusion follows.  But this tack does not seem easy...

Have a look here
http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1077570309&t=6325

Pierre.


algebra
polynomial
algorithm
abstract algebra

This may be too easy for some.... I feel i am missing something...

Define $e_{p}(a)$ as the smallest exponent $e \geq 1$ so that $a^{e}\equiv 1 \mod p$.
Now, a number $g$ with maximum exponent $e_{p}(g) = p-1$ is called the primitive root $\mod p$.

Thus, $2, 3$ are primitive roots \mod $5$. (Since $2^{4}\equiv 1 \mod 5$ and $3^{4}\equiv 1 \mod 5$).

Prove:
If $g$ is a primitive root $\mod p$, then every non-zero number $\mod p$ can be represented as a power of $g$.

In other words, for any number $1 \leq a < p$, we can take one of the following:
$g, g^{2}, g^{3}, g^{4}, \cdots g^{p-3}, g^{p-2}, g^{p-1}$ as being congruent to $a$ $\mod p$.

How to prove this?

I feel that it should direclty follow from the definitions, but I am not able to put all things together...

[quote="K81o7"]What is your definition of primitive root?  Perhaps you have to prove it given "primitive root" means an element whose order is divisible by all of the other orders in the set.  There's a really nice proof of this...try to figure it out before looking.

[hide]Let's say $g$ has order $a$.  Given any element $u$ of order $y$, we have $y|a \implies a=yz$.  Then $g^{z}$ has order $y$.  Then consider the polynomial $x^{y}-1=0$.  $g^{z}, g^{2z}, ..., g^{yz}$ are all roots, and are distinct (because $g$ has order $yz$).  But by Lagrange's Theorem, we have at most $y$ roots to this polynomial mod $p$.  Then the above powers of $g$ form the solution set.  But $u$ is also a solution of the polynomial.  Thus, $u$ is in the solutions set, which means it is a power of $g$, as desired. [/hide][/quote]
[hide="quick question"]$x^{2}-1\equiv 0\mod 12$ has 4 roots $\mod 12$. Why does Lagrange hold for your case?[/hide]

[hide="My proof"]
If $|\langle g\rangle |=|\{g,g^{2},\ldots,g^{p-1}\}|=p-1$, we are done. Suppose on the contrary that the order of this set is less than $p-1$. Then, for some $0<i<j\le p-1$, we have $g^{i}\equiv g^{j}\mod p$. Then $g^{j-i}\equiv 1\mod p$. Since $ord(g)=p-1$, we have $j=i$. Thus each of the $p-1$ elements of $\langle g\rangle$ are unique and $\left(\mathbb{Z}/p\mathbb{Z}\right)^{*}=\langle g\rangle$.[/hide]

[quote="Bictor717"][hide="quick question"]$x^{2}-1\equiv 0\mod 12$ has 4 roots $\mod 12$. Why does Lagrange hold for your case?[/hide][/quote]

The problem fails for composite moduli, since then the nonzero elements $\bmod p$ include zero divisors, which cannot be the power of a primitive root.

The problem is implicitly stated with the assumption $p$ is a prime, although it holds in $U_{m}$ as long as that system has a primitive root.

Edit: I see that your proof uses the same assumption  :wink:

I think the problem here is defining "primitive root".  Usually, I see primitive root defined as what we are trying to prove in this problem.

And also, Lagrange is only for prime moduli...we can say this because the division algorithm holds only in $\mathbb{Z}_{p}[x]$.


modular arithmetic
function
number theory open
number theory

Find the squares of the form n^4+7^n+47:with n positive integer

if $ (n,2) \equal{} 1$  $ \implies$  $ x^{2} \equal{} n^{4} \plus{} 7^{n} \plus{} 47\equiv \minus{} 1\pmod{4}$   contradiction! $ \implies$  $ 2|n$

If n is odd there are no solutions mod 4
If n is even then $ 7^n \equal{} x^2$ is a square and for $ n \ge 8$ the function falls between $ x^2$ and $ (x\plus{}1)^2$
So only solution is $ n \equal{} 4$ which gives $ 52^2$


The sequence $ (x_n)$ satisifies $ x_{n\plus{}1}\equal{}ax_n\minus{}x_{n\minus{}1}$ for all $ n$. Prove that $ \frac{x_n^2\plus{}x_{n\plus{}1}^2}{1\plus{}x_nx_{n\plus{}1}}\equal{}a$ for all $ n$.

hello, your recurrence equation is fulfilled for
$ x_n\equal{}2^{\minus{}n}(a\minus{}\sqrt{a^2\minus{}4})^nC_1\plus{}2^{\minus{}n}(a\plus{}\sqrt{a^2\minus{}4})^nC_2$.
Sonnhard.

Downgrade:
[hide="just algebra"]
$ x_n^2\plus{}x_{n\plus{}1}^2\equal{}a(1\plus{}x_nx_{n\plus{}1}$
$ x_n^2\plus{}(ax_n\minus{}x_{n\minus{}1})^2\equal{}a(1\plus{}x_n(ax_n\minus{}x_{n\minus{}1}))$
$ x_n^2\plus{}a^2x_n^2\plus{}x_{n\minus{}1}^2\minus{}2ax_nx_{n\minus{}1}\equal{}a\plus{}a^2x_n^2\minus{}ax_nx_{n\minus{}1}$
$ x_n^2\plus{}x_{n\minus{}1}^2\equal{}a(1\plus{}x_nx_{n\minus{}1})$
[/hide]
And we see that it will satisfy for all terms.
How to write a rigorous proof based on above is beyond my capacity. Please complete.
Also, how will this sequence begin? Dont we need two numbers to start the sequence? Then these two numbers will have to follow $ x_n^2\plus{}x_{n\plus{}1}^2\equal{}a(1\plus{}x_nx_{n\plus{}1}$ and we have to get an $ x_{\minus{}1}$, therefore this sequence will run till negative infinity? Confused  :(

If we are given the first two values of the sequence, then you can rigorously prove it by induction. RobRoobiks, the steps you did would be the inductive step since you are showing that if the formula works of $ n \minus{} 1,n$, then it works for $ n, n \plus{} 1$ and therefore it would work for all $ n$ if it works for $ n \equal{} 1$ and $ n \equal{} 2$ (assuming the sequence starts at $ n \equal{} 1$).

TachyonPulse: Suppose that the first two integers are $ 1,2$ as you say, then $ a$ will [b]have [/b]to be $ \frac 53$ or else the relation I dont think will hold


I'm using

\begin{figure}
	\centering
		\includegraphics{C:/Documents and Settings/Desktop/ARML1999I8.gif}
	\label{fig:ARML1999I8}
\end{figure}

but it doesn't work.

It could be because of the space in the path to the file. Try either using quotes \includegraphics{"C:/Documents and Settings/Desktop/ARML1999I8.gif"} or better still move the gif to a path with no spaces.

It's always safer not to use spaces when doing anything with LaTeX.

If that's not the problem then what is the error message in the log?

I don't really understand the error log, but it says \includgraphics is an undefined control sequence.

[i]\includgraphics is an undefined control sequence[/i] tells you that LaTeX doesn't understand the command. \includegraphics is a command defined by the graphics package, so to use it you need something like \usepackage[dvips]{graphics,color} in the preamble.

I have the following packages:
\usepackage{amsthm, amssymb, amsmath}
What exactly do I have to type?

\usepackage[dvips]{graphics,color} didn't work.

Can you copy the complete code here for a minimal document that doesn't work? What I mean is a small document, with a few lines of text including includegraphics then perhaps we can see if there's some conflict somewhere.

[code]\documentclass[11pt]{article}
\usepackage{amsthm, amssymb, amsmath}
\usepackage[dvips]{graphics,color}
\begin{document}

\begin{figure}
	\centering
		\includegraphics{C:/Documents and Settings/Tony Feng/Desktop/ARML1999I8.gif}
	\label{fig:ARML1999I8}
\end{figure}

\end{document}[/code]

Two problems:
1. You have a spaces in the path so you must either move the image file or use quotes as
\includegraphics{"C:/Documents and Settings/Tony Feng/Desktop/ARML1999I8.gif"}
But as a general principle you should [i]never[/i] use spaces in paths for latex documents.

2. Once you've sorted that out so the undefined control sequence error goes away, you will get another one about undefined size of graphic. dvips won't process gif files; you must use eps files. There is plenty of conversion software around, but OpenOffice makes good eps images. Better still, you may be able to use jpg or pdf images provided you are compiling to a pdf rather than dvi. Have a read of AoPS's [url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_PictHow.php]Images in LaTeX[/url] which tells you how to go about it.

Is it possible to choose a directory without spaces?

Just move the image file from your desktop into somewhere else. But don't use anything in My Documents. Use a directory such as c:\images. Or, better still, you could start all your LaTeX documents in subdirectories of c:\texdocs (such as c:\texdocs\doc1, c:\texdocs\doc2) and then put image files in the same directory as the LaTeX document and then you won't need to bother with paths and can use \includegraphics{ARML1999I8.gif}.

I'm not convinced though that I understand your problem :)


inequalities
search
inequalities proposed

Let's think about infinite positive sequence $ a_1, a_2 , \cdots $.

For arbitrary positive integer N, $ \sum_{n=1}^{N} {b_n}^2 \leq 4 \sum_{n=1}^{N} {a_n}^2$

( $ b_n = $${a_1+ a_2 + \cdots + a_n} \over {n}$)

[quote="Chang Woo-JIn"]Let's think about infinite positive sequence $ a_1, a_2 , \cdots $.

For arbitrary positive integer N, $ \sum_{n=1}^{N} {b_n}^2 \geq 4 \sum_{n=1}^{N} {a_n}^2$

( $ b_n = $${a_1+ a_2 + \cdots + a_n} \over {n}$)[/quote]

i don' tunderstand it, what if all a_i are equal?

[quote="harazi"]We have discussed this inequality at least 5 times![/quote]

can you please post the links? :) Thanks.

Sorry, I do not remember the links. Try with search and Hardy inequality.


inequalities
linear algebra
matrix
vector
triangle inequality
inequalities proposed

No one interested? I post a similar inequality:
Let $a,b,c,d$ be positive reals satisfying $a+b+c+d=1$. Prove that \[\left|a^{4}+b^{4}+c^{4}+d^{4}-2a^{2}b^{2}-2b^{2}c^{2}-2c^{2}d^{2}-2d^{2}a^{2}+8abcd\right|<1.\]

The first inequality is unsolved for one week. Maybe [b]arqady[/b] can prove it?

[quote="vietnam2007"]for the second one I suggest that solution
[img]http://www.gnux.be/latex/data/a9a98a9496dd1d5c2162e66d6c450606.png[/img][/quote]
It's a really nice solution, but I'm afraid you proved only one part of the inequality, i.e. $a^{4}+b^{4}+c^{4}+d^{4}-2a^{2}b^{2}-2b^{2}c^{2}-2c^{2}d^{2}-2d^{2}a^{2}+8abcd<1$, because generally $x\leqslant y\implies |x|\leqslant |y|$ doesn't hold. I hope someone will prove the second part.

yes you are right.I m trying now to prove the second part of the problem.

independently of the other inequalities, I am very intersted by the last inequalities .
$\sum_{cyc}{a^{2}b^{2}}\leq \sum_{cyc}{a^{2}}\leq \sum_{cyc}{a}=1$

we have only the following inequalities

$\sum_{cyc}{a^{2}b^{2}}\leq \sum_{cyc}{a^{4}}$

 $\sum_{cyc}{a^{4}}=6(a^{4}+b^{4}+c^{4}+d^{4})$

we know that $a^{4}+b^{4}+c^{4}+d^{4}\leq a+b+c+d$
so we have
$\sum_{cyc}{a^{2}b^{2}}\leq 6$

I've mixed the two summation up.thank you for the signal.

yeah its not difficult, but i didn't get it at first

[quote="Albanian Eagle"]also the first inequality is not very difficult after we see that it is equivalent to:
$|\left((a-c)^{2}+(b-d)^{2}\right)(a+c-b-d)|<1$ :)[/quote]
It's a very nice solution and completing it with your idea is indeed very nice and simple :lol: . But unfortunately it is quite hard to find such factorisation without a computer.

Here is the $5$-variable version of this inequality:

For any $a,b,c,d,e>0$ satisfying $a+b+c+d+e=1$ prove that
\[0<\sum_{\mathrm{cyclic}}a^{5}-5(a^{3}cd+ab^{3}c)+5(a^{2}b^{2}d+a^{2}bc^{2})-abcde<1.\]

I'm sorry - it should be $0\leqslant\sum_{\mathrm{cyclic}}\cdots$.
Did anybody try to solve it? I can prove only the right inequality, but in a non-elementar way.

I proved also the left inequality. I wonder if someone can find an elementar solution, because my proof is rather ugly.

a nice proof for the left part:
$\sum_{cyc}\left[a^{5}-5(a^{3}cd+b^{3}ac)+5(a^{2}b^{2}d+a^{2}bc^{2})-abcde\right]=\sum_{cyc}a[(a^{2}-\frac{bc+de}{2})^{2}+3\left(\frac{bc-de}{2}\right)^{2}]\geq 0$ :)

BTW , I never use computer programs.

[quote="Albanian Eagle"]a nice proof for the left part:
$\sum_{cyc}\left[a^{5}-5(a^{3}cd+b^{3}ac)+5(a^{2}b^{2}d+a^{2}bc^{2})-abcde\right]= \sum_{cyc}a[(a^{2}-\frac{bc+de}{2})^{2}+3\left(\frac{bc-de}{2}\right)^{2}]\geq 0$ :)

BTW , I never use computer programs.[/quote]
Nice! :) But when you prove the second part, why not to think about the $6$-variable version of the inequality?
Let $a,b,c,d,e,f$ be positive numbers satisfying $a+b+c+d+e+f=1$. Prove that
\[\begin{array}{r}\left|\sum a^{6}+2a^{3}c^{3}+3\left(a^{2}c^{2}e^{2}-a^{4}d^{2}\right)+6\left(ab^{4}-a^{4}ce+ac^{3}d^{2}-a^{2}b^{3}d\right)-9a^{2}b^{2}c^{2}+\right.\\ \left. 12\left(a^{3}bcd+abcd^{3}\right)+18ab^{2}d^{2}e\right|<1, \end{array}\]
where $\sum$ means cyclic sum with the second, fourth and sixth constituents taken with sign $-$, i.e. $\sum F(a,b,c,d,e,f)= F(a,b,c,d,e,f)-F(b,c,d,e,f,a)+F(c,d,e,f,a,b)$ $-F(d,e,f,a,b,c)+F(e,f,a,b,c,d)-F(f,a,b,c,d,e)$.

For the second one, we can also solve it by using mixing variables. :)

It seemes that no one likes my beautiful $6$-variable inequality, so I will show the general case with a proof.
One can notice that $a^{4}-b^{4}+c^{4}-d^{4}+4ab^{2}c-4bc^{2}d+4cd^{2}a-4da^{2}b+2b^{2}d^{2}-2a^{2}c^{2}= \left|\begin{array}{cccc}a&b&c&d\\ d&a&b&c\\ c&d&a&b\\ b&c&d&a\end{array}\right|$ and similarly for more variables. Thus the general case of the inequality can be stated as follows:

[color=blue]Let $a_{0},\ldots,a_{n-1}$ be positive real numbers such that $\sum_{0\leqslant i\leqslant n-1}a_{i}=1$. The matrix $A$ satisfies $[A]_{i,j}= a_{j-i}$ (we numerate all indices modulo $n$). Prove that
[b]1.[/b] for any $n$ we have $\left|\det(A)\right|<1$,
[b]2.[/b] for odd $n$ we have $\det(A)\geqslant 0$. [/color]

Let us put $\vec{a}: = \begin{bmatrix}a_{0}\\\vdots\\a_{n-1}\end{bmatrix}$, $\omega: = \mathrm{e}^{\frac{2\pi i}{n}}$, $\left[\vec{v}_{k}\right]_{j}: = \omega^{kj}$. It is easy to see that $\frac{\left[A\cdot\vec{v}_{k}\right]_{j}}{\left[ \vec{v}_{k}\right]_{j}}= \frac{\sum_{0\leqslant s\leqslant n-1}[A]_{j,s}\left[\vec{v}\right]_{s}}{\left[\vec{v}_{k}\right]_{j}}= \frac{\sum_{0\leqslant s\leqslant n-1}a_{s-j}\cdot \omega^{ks}}{\omega^{kj}}$ $=\sum_{0\leqslant s\leqslant n-1}a_{s-j}\cdot \omega^{k(s-j)}= \sum_{0\leqslant s\leqslant n-1}a_{s}\cdot \omega^{ks}=: \lambda_{k}$, so the vectors $\vec{v}_{0},\ldots,\vec{v}_{n-1}$ are eigenvectors of matrix $A$ corresponding with eigenvalues $\lambda_{0},\ldots,\lambda_{n-1}$ respectively. Furthermore, these vectos are linearly independent: a matrix $V$ whose $k$-th column is $\vec{v}_{k}$ (i.e. $[V]_{r,s}: = \omega^{rs}$) is invertible, because it is a Vandermonde matrix whose columns are pairwise different. Therefore $\det(A)= \lambda_{0}\cdot\ldots\cdot \lambda_{n-1}$. By the triangle inequality: $\left|\lambda_{k}\right|= \left|\sum_{0\leqslant s\leqslant n-1}a_{s}\omega^{ks}\right|\leqslant \sum_{0\leqslant s\leqslant n-1}a_{s}\left|\omega^{ks}\right|= \sum_{0\leqslant s\leqslant n-1}a_{s}=1$ and besides the cases $k=0$ and $k=\frac{n}{2}$ for even $n$ the equality can not hold (since $a_{0},\ldots,a_{n-1}>0$, the equality case of triangle equality requires that the vectors $\omega^{0\cdot k},\ldots,\omega^{(n-1)\cdot k}$ are paralell). Thus $\left|\det(A)\right|= \left|\lambda_{0}\cdot\ldots\cdot\lambda_{n-1}\right|= \left|\lambda_{0}\right|\cdot\ldots\cdot\left|\lambda_{n-1}\right|< 1\cdot\ldots\cdot 1= 1$ and the first inequality is proved.
For the second inequality we notice that $\lambda_{n-j}= \sum_{0\leqslant s\leqslant n-1}a_{s}\omega^{-sj}= \sum_{0\leqslant s\leqslant n-1}a_{s}\overline{\omega}^{sj}= \overline{\left(\sum_{0\leqslant s\leqslant n-1}a_{s}\omega^{sj}\right)}= \overline{\lambda_{j}}$. Thus for odd $n$ we have $\det(A)= \lambda_{0}\cdot\ldots\cdot\lambda_{n-1}= \lambda_{0}\cdot\left(\lambda_{1}\cdot\lambda_{n-1}\right)\cdot\ldots \cdot \left(\lambda_{\frac{n-1}{2}}\lambda_{\frac{n+1}{2}}\right)$ $= \lambda_{0}\cdot\left|\lambda_{1}\right|^{2}\cdot \ldots\cdot \left|\lambda_{\frac{n-1}{2}}\right|^{2}\geqslant 0$ since $\lambda_{0}=1$ and the second inequality is proved.


Hello, i would appreciate if someone helps me out with this [url]http://www.mathlinks.ro/viewtopic.php?t=155096[/url]

thanks

Ben

Wait for one of the moderators of the intermediate forum to move the original problem here.


$ \dfrac{1}{576} \pi  \left( 2 \plus{} \cot \left( \dfrac{1}{12} \left( 9 \minus{} 2 i \sqrt{2} \right) \pi \right) \plus{} \cot \left( \dfrac{1}{12} \left( 9 \plus{} 2 i \sqrt{2}\right) \pi \right)\right)$
$ \approx 0.0061907817802936815\cdots$

The correct answer is $ S \equal{} \frac {29 \pi^3}{864} \approx 1.0407$, provided I have made no mistake.

Sorry.

$ \sum_{n\equal{}0}^\infty \frac{(\minus{}1)^n}{(6n\plus{}1)^3 \plus{} (6n\plus{}3)^3 \plus{} (6n\plus{}5)^3} \equal{} \dfrac{1}{576} \pi \left( 2 \plus{} \cot \left( \dfrac{1}{12} \left( 9 \minus{} 2 i \sqrt{2} \right) \pi \right) \plus{} \cot \left( \dfrac{1}{12} \left( 9 \plus{} 2 i \sqrt{2}\right) \pi \right)\right)$

$ \sum_{n\equal{}0}^\infty \left(\frac{(\minus{}1)^n}{(6n\plus{}1)^3} \plus{}\frac{(\minus{}1)^n}{ (6n\plus{}3)^3} \plus{} \frac{(\minus{}1)^n}{(6n\plus{}5)^3}\right) \equal{} \frac{29 \pi^3}{864}$


quadratics
function
algebra unsolved
algebra

If the quadratic $ ax^2\plus{}bx\plus{}c$ takes rational values for more than two rational values of x, then show that a, b, c are all rational numbers.

Use the Lagrange interpolating polynomial.


vector
projective geometry
geometry unsolved
geometry

Let $ ABCD$ is a quadrilateral. The line $ d$ through point $ A$ and parallel with $ BC$, $ d$ meet $ BD$ at $ M$. The line $ d'$ through point $ B$ and parellel with $ AD$, $ d'$ meet $ AC$ at $ N$. Prove $ MN$ and $ DC$ are parallel.

--------------------------------------------------------------------------
I want you slove it by vector!

Non-vector way...

Let $ AC$ and $ BD$ intersect at $ P$.  Then

$ \triangle{APD} \sim \triangle{NPB} \implies AP \cdot BP \equal{} NP \cdot PD$
$ \triangle{BPC} \sim \triangle{MPA} \implies AP \cdot BP \equal{} MP \cdot PC$
So $ \frac{MP}{NP}\equal{}\frac{PD}{PC}$
which yields the desired result.

Thank you! I see your solution, can you slove it by vector?

Dear Mathlinkers
applied "little Pappus" to the hexagone ADCBNMA build on AC and BC.
Sincerely
Jean-Louis

let $ AD$ and $ BC$ intersect at $ P$ and $ d$ meets $ DC$ at $ S$ then we have $ \frac{AM}{MS}\equal{}\frac{PB}{BC}$ and $ \frac{AN}{NC}\equal{}\frac{PB}{BC}$ and the easy problem solved :)


real analysis
advanced fields
advanced fields unsolved

i'm not sure how you are interpreting Borel sets or the Borel algebra of $ \mathbb{R}$, but if it isn't already part of your definition, you should show that the Borel algebra contains all open sets of $ \mathbb{R}$ and work from there

[quote="sxcxc"]Show that there exists two disjoint borel sets $ A$ and $ B$  of $ \mathbb{R}$ such that $ A\cap I$ and $ B\cap I$ both have positive Lebesgue measure for every nonempty interval $ I$.[/quote]

Hint: Let I_n be the open intervals with rational end points. In I_1, choose disjoint Cantor  sets A_1 and B_1 of positive measure. In I_2, do the same thing, making sure you miss the previous Cantor sets. Continue ...


geometry
real analysis
real analysis unsolved

Let $C$ be a closed curve with no self-intersections. Consider a segment $AB$ with $A,B$ belonging to the curve $C$, which has size $L$ fixed. Now, take a point $P$ belonging to the segment $AB$, and say that $AP=a, PB=b$. As we vary the segment along the curve C, the point $P$ describes a curve, say $C'$. Show that the area between $C$ and $C'$ is equal to $\pi ab$.

PS.: I'd really like to see a solution which parametrizes the desired region (because I have no idea how to parametrize this region =])! =]
PS2.: I hope i was clear in announcing the problem, because I had some problems in translating from portuguese =]
PS3.: Consider it is possible to satisfy the conditions of the problems, i.e, do not take in consideration the curves discribed by P which are discontinuous for example...=]

Hey people, at least try to help me here  :(

I think I can prove it for nice enough curves (like, say, smooth and convex, as it is stated [url=http://mathworld.wolfram.com/HolditchsTheorem.html]here[/url]). You use the formula $A=\frac12\int(xy'-x'y)\mbox dt$ for the area $A$ enclosed by a curve written as $c(t)=(x(t),y(t))$, and then it's just boring algebraic manipulation. I certainly won't write it here though: too long :).

I know...that was my solution too...

but my professor showed us one as parametrizing the region and I forgot that...i'd really like to see that solution again...


function
Support
real analysis
real analysis unsolved

Show that if $\phi$ (test functions) vanishes in a neighbourhood of the support of f, then $(f, \phi) = 0$ (inner product) in the sense of distributions.

[quote="fedja"]Hmm... Doesn' it follow immediately from the definition of support? If not, what definition of support are you using?[/quote]

I don't think so. The support here is just where the function is non-zero. At least I think it is, this problem is confusing.

[quote="swag"]The support here is just where the function is non-zero. At least I think it is, this problem is confusing.[/quote]
Well, but what does it mean for a generalized function to be non-zero? The usual definition of the (closed) support of a generalized function $f$ is the following: it is the complement of the union $V$ of all open sets $U$ such that $f$ is $0$ on $U$ and $f$ is $0$ on $U$ if $(f,\phi)=0$ for all test-functions $\phi$ whose support is a compact subset of $U$. From here it follows almost immediately (using partitions of unity) that $(f,\phi)=0$ for all $\phi$ whose support is a compact subset of $V$. And this immediately implies the property you asked about. Some different approaches to the notion of support are possible, but just to say that it is the set of points where $f(x)\ne 0$ doesn't make sense because there is no such thing as $f(x)$.

There's a mindset here: instead of asking what the distrubution $f$ "is", you can only ask what it does. The only way you can see $f$ is too look at it through a test function $\phi.$ Put another way, the distribution $f$ is its action on all possible test functions - nothing more, nothing less.

Given that point of view, how can you possibly define any concept (such as support) involving a distribution? That definition can only be made in terms of actions on test functions. Hence the definition fedja gave, or something very much like it, is all that you can do.


In a triangle $ABC$,Let $AB=c,BC=a,CA=b$,and $a=\max \{a,b,c\}$
Let M be the point on the plane coincide with the triangle $ABC$.Let $R_{1},R_{2},R_{3}$ be the distances from $M$ to $A,B,C$ respectively.Prove:
\[\frac{a^{2}R_{1}+b^{2}R_{2}+c^{2}R_{3}}{R_{1}+R_{2}+R_{3}}\ge bc \]

Thank you very much dear zhaobin for your nice result and interesting solution. But can you show me what problem of MR that you inspired it ?
Actually here is a strong result We well know the result \[ MAMB+MBMC+MCMA\ge\min\{ab,bc,ca\}\] forall aribitrary point $ M$ in plane.
But we can prove that
\[ MBMC+MCMA+MAMB\ge\frac{a^{2}MA+b^{2}MB+c^{2}MC}{MA+MB+MC}\ge\min\{ab,bc,ca\}\]
which is stronger. More interesting $ \min\{ab,bc,ca\}$ is minimum value of both
$ \frac{a^{2}MA+b^{2}MB+c^{2}MC}{MA+MB+MC}$ and $ MBMC+MCMA+MAMB$ equal iff $ M\equiv A,\ a\ge b\ge c$ :)

Not problem,but an acticle in MR.
See[url]http://reflections.awesomemath.org/2006_6/2006_6_minimum.pdf[/url] :)


Dimensions a,b,r are known, as is the linear distance from point A to
point B. I need a formula to determine the value of h.

Points A and B represent pressure sensors in fixed positions on the
base of a round tank. The chord through CD represents the water level
in the tank. Lines a and b are the heights of water registered by each
sensor.

So you're saying...

Given a tank (cylindrical) with radius $r$, find height $h$? You need to know the volume and the value of $r$ to determine $h$.

The volume of a cylinder is $\pi r^{2}h$.

If you know the radius and the volume, you can find the height.

$h=\frac{V}{\pi r^{2}}$


search
inequalities unsolved
inequalities

x,y,z>0 satisfy $ \sqrt{x}+\sqrt{y}+\sqrt{z}=1$, prove:
$ \sum_{cyc}{\frac{x^{2}+yz}{\sqrt{2x^{2}(y+z)}}}\geq 1$

[quote="hanschinaman"]x,y,z>0 satisfy $ \sqrt {x} \plus{} \sqrt {y} \plus{} \sqrt {z} \equal{} 1$, prove:
$ \sum_{cyc}{\frac {x^{2} \plus{} yz}{\sqrt {2x^{2}(y \plus{} z)}}}\geq 1$[/quote]
It is a problem in APMO 2007
See at http://www.mathlinks.ro/Forum/viewtopic.php?search_id=59923038&t=141394


AMC USA(J)MO USAMO inequalities logarithms function

For a, b, c > 0, prove that a^a*b^b*c^c >= (a*b*c)^((a+b+c)/3) :)

You apply the logarithm function and then everything is simple ... :D :D

one can even use chebyshev i could never until now  :P


inequalities
vector
trigonometry
algebra proposed
algebra

Prove that the system of inequations  :

$ x_1x_2 \plus{} y_1y_2 < 0$ , $ x_1x_3 \plus{} y_1y_3 < 0$ , $ x_1x_4 \plus{} y_1y_4 < 0$ , $ x_2x_3 \plus{} y_2y_3 < 0$ , $ x_2x_4 \plus{} y_2y_4 < 0$ and $ x_3x_4 \plus{} y_3y_4 < 0$ has no real solutions :) .

the RHS terms of the inequalities are a dot product of 2 vectors in $ R^{2}$: $ v_{i} \equal{} (x_{i},y_{i})$. the product is negative iff $ \cos\theta < 0$ where $ \theta$ is the angle between two vectors. Let $ \theta_{i}$ be the angle between $ v_{i}$ and $ v_{i \plus{} 1}$ (the index is being taken modulo 4) (it doesn't matter if we take $ \theta$ or $ 2\pi \minus{} \theta$, because the $ \cos$ is the same. I will take the angles clockwise, and assume WLOG that the "order" of the vectors, clockwise, is $ v_{1},...v_{4}$).
the following holds:
$ \frac {\pi}{2} < \theta_{i} < \frac {3}{2}\pi, i \equal{} 1,2,3,4 \equal{} > \sum_{i \equal{} 1}^{4}\theta_{i} > 2\pi$
$ \sum_{i \equal{} 1}^{4}\theta_{i} \equal{} 2\pi$
a contradiction, therefore no solution.


AMC
AIME
geometry
Stanford
college
ARML
AIME I

If everything goes well, I think I can qualify for AIME this year.

But, due to a scheduling conflict, I will not be able to take AIME I on 3/18.
My school does not offer AIME II though. So I am trying to find an AIME II site
in the Bay Area. Does anyone here know a site?

Thanks for any pointers.

If someone could find out the time, you might be able to take the AIME II at Stanford EPGY... they usually have some extra spots. But if you qualify, and you ask Ted Alper, the EPGY proctor, you might be able to take it there.

Thanks xscapezaer for the good info.

The AIME II will be on April 2nd, 2008.

I will contact Mr. Alper for that. He is a nice guy. (BTW, I was in last year's ARML so he knows me).


inequalities
linear algebra
matrix
floor function
linear algebra unsolved

For what values of $r,n$ does there exist $A\in\mathcal{M}_n(\mathbb{C})$ with $\text{rank}(A)=r$ such that $A^2=0$?

The pairs of positive integers $(r,n)$ satisfying $r\le\frac n2$.

The condition is necessary by Sylvester's inequality: $0=\text{rank}(A^2)\ge2\text{rank}(A)-n=2r-n$.

It's also sufficient: for $(r,n)$ with $r\le\frac n2$ consider the matrix having zero everywhere except for the positions $(1,n-r+1),(2,n-r+2),\ldots,(r,n)$.

Jordan canonical form. We need zeros all down the main diagonal, with some 1's on the superdiagonal. But for $A^2=0,$ we need all the Jordan blocks to be $2\times 2.$

So the answer is any $n\ge 2,\,0\le r\le\lfloor n/2\rfloor.$

Ah - grobber beat me too it. I'll post my answer anyway.


ARML AMC AIME Stanford college Harvard MIT

What are ways to prepare for the AMC 10? I am done with all the intro books, Vol 2 is in the USPS car as we speak. What would I do instead of the AMC 10 tests? I mean, there are only like 16 of em so i won't have any problems left by the time the contest comes around.

thanks

There are other contests out there-do some of those.  AoPS has links to many old contests.

like what for example? ARML (I'm thinking that'd be AIME level) ?

Then do ARML.  Several universities, such as Stanford, Rice, Harvard-MIT, and Hopkins all hold contests.  Do some of their questions.

Harvard-MIT is more of an end AMC12 - mid AIME difficulty;
just do art of problem solving 1 and you should be set.

No.  The intro books go way-more in depth, which is good if you're struggling with a particular topic.

any other ways? And, Once i took a look at the 2005 ARML indiv they seem at AIME level not at AMC 10 level. Should I still do them?

[quote="andersonw"]Just because you're preparing for the AMC 10 doesn't mean you're not allowed to do harder problems.
Doing past AMCs, AIMEs, ARMLs are all good if you want more math preparation, but I find that overloading yourself with problems is bad.
Also, to me, working on problems is more important than reading AoPS books, although books can help, I guess. (pretty much all of my preparation last year for AMC/AIME/USAMO was working on contest problems for fun, and not forcing myself to actually study)[/quote]

Working on random problems that were interesting was most of my preparation last year. Also, when I was preparing for AMC 10, I mostly did olympiad problems as preparation.

Hmm...I heard that systematic preparation is good for geometry. Unfortunately, I fail at geometry, and I am very bad at systematically preparing...

[quote="Poincare"]I'm not that bad at at geometry and can do almost all AMC 10/12 and AIME geometry problems that do not require enormous amounts of trig. Can anyone supply me with more ARML problems? Please do not recommend any books to buy as my piggy bank is empty :([/quote]

yes, but AMC 10/12 and AIME geometry problems seem entirely different from USAMO problems... I mean problems where you have construct random stuff (in geometry, not nt). Of course, its not really random, but it looks pretty random... like, not obvious things...

For ARML problems, if there is a Florida ARML team, then the coach there might have some ARML problems...

Is there a Florida ARML?

ARML problems are difficult to find because the ARML people don't let us post them on AoPS.

how would I find out if there's a Florida ARML?

@worthawholebean
I may or may not have had someone(not saying who though) give me the problems


abstract algebra
group theory
superior algebra

In the questions below: $G$ is a ﬁnite group; $H \leq G$ a subgroup of $G; |G : H |$ the index of $H$ in $G; |X |$ the number of elements of $X \subseteq G; Z (G)$ the center of $G; G'$ the commutator subgroup of $G; N_{G}(H )$ the normalizer of $H$ in $G; C_{G}(H )$ the centralizer of $H$ in $G$; and $S_{n}$ the $n$-th symmetric group.

Suppose $k \geq 2$ is an integer such that for all $x, y \in G$ and $i \in \{k-1, k, k+1\}$ the relation $(xy)^{i}= x^{i}y^{i}$ holds. Show that $G$ is Abelian.

$x^{k}y^{k}=(xy)^{k}=(xy)(xy)^{k-1}=xyx^{k-1}y^{k-1},$
from which $x^{k-1}y=yx^{k-1}$.  Since this holds for all $y$, $x^{k-1}$ lies in the center of $G$.  Similarly for $x^{k}$, so $x\in\mathcal{Z}(G)$ for all $x$, i.e., $G$ is abelian.

[quote="N.T.TUAN"]
Suppose $k \geq 2$ is an integer such that for all $x, y \in G$ and $i \in \{k-1, k, k+1\}$ the relation $(xy)^{i}= x^{i}y^{i}$ holds. Show that $G$ is Abelian.[/quote]
Instead of $x,y$ I will use $a,b$. Instead of $k$ I will use $n$. Because it is just unnatural for me to use that notation in group theory.

Let $a,b\in G$ then we know that, for $n-1,n,n+1$:
$(ab)^{n+1}=a^{n+1}b^{n+1}$ and $(ab)^{n}= a^{n}b^{n}$

Let us first work with the first equation and then with the second.

Write out the first equation,
$\underbrace{(ab)(ab)...(ab)(ab)}_{n+1}= a^{n+1}b^{n+1}$
So, so cancelling we get,
$\underbrace{(ba)(ba)...(ba)}_{n}= a^{n}b^{n}\to (ba)^{n}= a^{n}b^{n}$ by hypothesis, $(ba)^{n}= (ab)^{n}$

Similar reasoning with second equation, just instead of $n+1$ we have $n$, so,
$\underbrace{(ba)...(ba)}_{n-1}= a^{n-1}b^{n-1}\to (ba)^{n-1}= a^{n-1}b^{n-1}$ by hypothesis $(ba)^{n-1}= (ab)^{n-1}$.

That is it! It is almost finished.
We already shown that,
$(ab)^{n}= (ba)^{n}$ which we write as $(ab)(ab)^{n-1}= (ba)(ba)^{n-1}$
But we use the above derivation,
$(ab)(ab)^{n-1}= (ba)(ab)^{n-1}$
So,
$ab=ba$.
Q.E.D.


A certain number of students, let's say $n$ students, are standing outside a high school. In the school, there are $n$ closed lockers numbered from 1 to $n$. The first student walks into the school, opens every locker, and walks out. The second student walks in, closes every second locker ( i.e. locker #2, locker #4, etc.), and walks out. The third person walks in, opens (if the locker is closed) or closes (if open) every third locker. This process is continued all the way to the $n$th student. when the $n$th student walks out, which lockers are open, which are closed, and, most importantly, why?

The math requirement for this is quite low, but it can be difficult to find at first. If you want to simplify it a bit, let's say that $n=100$, but you can do it with any number, it doesn't matter what it is.

[quote="pontios"][hide]Sieve of Eratosthenes[/hide][/quote]

what is that supposed to mean? (no offense or anything...just curious)

Here's a nice website on the Sieve of Eratosthenes:[url=http://en.wikipedia.org/wiki/Sieve_of_Eratosthenes]en.wikipedia.org/wiki/Sieve_of_Eratosthenes[/url]

[quote="pontios"][hide]Sieve of Eratosthenes[/hide][/quote]

Anyway, it was a hasty conclusion.  :D

The correct answer was given by [b]jhollenbeck[/b]

[quote="jhollenbeck"]Only the perfect squares are open, because only perfect squares can have an odd number of factors.[/quote]

Please explain this.........

Sandeep

The lockers open and close for each divisor of its locker number. For it to be open at the end, it must have an odd number of divisors. Only perfect squares have an odd number of divisors, thus the lockers that are open at the end are the ones with perfect square locker numbers.


MATHCOUNTS
analytic geometry
function
calculus
AMC 8

Could someone post some optimization problems?  I'm not very good at those and need practice.

Optimization problems are great problems, but wouldn't they belong in a higher-level problem-solving forum than Getting Started?

Well, most optimization problems I've come across involve differential calculus....

Perhaps if you could give an example or two of the type of problem you are talking about, we would have a better sense of what you mean.  As vinny said, most optimization problems rely on the use of higher mathematics; however, there certainly are classes of such problems that don't.  If you could show a little bit of what you meant, that would be helpful.

[quote="nr1337"]Why would they be in higher level boards?  There are a few in AoPS Volume 1, so I would think they would be some in Getting Started.[/quote]

Much of AoPS volume 1 is really at the Intermediate level in terms of problem-solving contests and people's progression through curricular math.

If it's on the AMC 8, or if it's in the Mathcounts Coaches Handbook, it's at the Getting Started level. If the problem is "Prove . . . ," sad experience has taught that that problem is at least at Intermediate level, as a challenge problem.

What kind of optimization problems are we talking about here? When is the terminology "optimization" introduced in the school mathematics curriculum where you go to school?

Well I am not sure but there is two things you could be talking about.  One is Linear Programming.  The other is finding the coordinates of the relative extrema.  Here are examples of both:

Find the values for x and y that maximize the objective function P=3x+2y.  The restrictions are {2 :le: x :le: 6, 1 :le: y :le: 5, x+y :le: 8}.

This is a pretty famous physics problem.  Its called archimedes' challenge.  Note that you have to use a graphing calculator (or calc) to find the solution.  Anyway, the question is, if you are given a 6 in x 6 in piece of Aluminum foil, what should its dimensions be if you want it to hold the maximum amount of mass while floating in water?

[quote="JS1527"]Well I am not sure but there is two things you could be talking about.  One is Linear Programming.  The other is finding the coordinates of the relative extrema.  Here are examples of both:

Find the values for x and y that maximize the objective function P=3x+2y.  The restrictions are {2 :le: x :le: 6, 1 :le: y :le: 5, x+y :le: 8}.

[/quote]

Since x and y are independent, and P(x+k, y) increases faster than P(x,y+k), it follows immediately that we should maximize x, so x=6 and y=2 and P = 22.

But this is generally a more difficult problem (because of the intervals, not the constraint) than the more common calculus optimization problems.

Lagrangian multipliers are a very powerful optimization technique, but require (basic) knowledge of multivariable calculus.
.

I am not sure but I think that is incorrect.

[quote="vinny919"]There are some problems where you have a shaded region bounded by a few lines and you have to maximize the value of some function. In that case, evaluate the function at each intersection of two lines. (the reason for this involves some calculus related topics)[/quote]

Not really.  It's pretty intuitive.  Basically, z = [something using x and y].  f(x) = [something using x] + [a certain constant].  We want to make the constant big (or small sometimes) so that the graph of f(x) is just barely touching the bounded region.  If f(x) is linear, imagine a very large (or small sometimes) constant such that f(x) does not touch the bounded region.  Then, you have to keep decreasing the constant until f(x) first touches the bounded region, which is at either an entire side or at a corner.  It's similar for non-linear f(x).

Ahh I see. I always thought of it as calculus related because the maxima and minima have to occur where the value changes direction (i.e. dy/dx = 0). And you can also do maximizing and minimizing problems in more than 2 variables (actually you can do 3 if you want to do the 3-d thing) in a simplex tableau, but I don't know if thats in the scope of the Getting Started board.


algorithm
combinatorics solved
combinatorics

I supposed that you mean the vertices of such a triangle? Not the sides?
In the first case, Gallai's theorem states that more generally if V is a finite set of |R m then for each finite coloring of |R m there exists a monochromatic set W which is homothetic to V.

Pierre.

[quote="Cristi Sav"]If we color the points in the plane with a finit number of colors, then exist a triangle monocromatique with the angles of measures less the 75 degrees.[/quote]
Suppose no such triangle exists.
Because we have a finite number of colors choose 2 points A and B colored the same, say with color 1. Construct an equilateral triangle ABC. Now construct a circle wich passes through A and B such that the center O of the circle and the point C are on the same side of AB and for any point M on the arc AB mAMB=75. Denote by S the surface of the triangle and T the surface of the circle. Because 60<75 we have S \cap T<> Ǿ and S-T<>Ǿ. It's easy to check that for any point P \in S-T the measures of the angles of the triangle APB are <75. In fact 2 of them are <60. Because A and B are colored with color 1 by hypothesis we have that P is colored with one of the colors 2 3 ...n otherwise we have found a triangle with the desired property. So for every point P from S-T we have that P is not colored with color 1. So we have found a region of the plane colored with n-1 colors < n. Repeating this algoritm in S-T (finding two points colored the same, constructing the equilateral triangle ...) we can find a region of the plane colored with only one color. We can choose a triangle with the desired property from this region. contradiction.

This problem is created by dzeta and me. Just to put the source of the problem. And we didn't know Gallai's theorem.


[quote="joejia"]A tennis racquet originally sold for $120.00. For a special sale the price was reduced by 10%. After the sale it was then marked up 10%. What was the price of the tennis racquet after the sale?[/quote]

[hide]During the sale, the price of the tennis racquet went down by 12.00, so the price became 108.00

After the sale, the price was increased by 10.80, so the price became 118.80[/hide]

[quote="joejia"]A tennis racquet originally sold for $120.00. For a special sale the price was reduced by 10%. After the sale it was then marked up 10%. What was the price of the tennis racquet after the sale?[/quote]
[hide]
120*.9*1.1=108*1.1=118.8[/hide][/hide]

[hide]10% of 120 = 12; so \, the discounted price is 108.
10% of 108 = 10.80; so, the final price is: $\framebox{118.80}$[/hide]


logarithms calculus integration inequalities function algebra domain

Hello :),
I've accidentaly found out that according to [i]Mathematica[/i] the following is true
$ (*) \quad \mathcal{L} \{\ln t \} = - \tfrac{1}{s} ( \ln s + \gamma )$
The problem is that firstly integral $ \int_0^\infty \ln t \; \cdot \; e^{ - st} \; dt$ diverges;
secondly there is no such $ s_0$ that inequality $ | \ln t \; \cdot \; e^{ - s_0 t}| \leqslant M$ is true.
How's it possible?

What is more in one of my textbooks I found formula which is claimed to be true: $ \mathcal{L}^{ - 1} \{\tfrac{\ln s}{s} \} = - \ln t - \gamma$
From this formula it's not hard to conclude (*).

[quote="@petko"]The problem is that firstly integral $ \int_0^\infty \ln t \; \cdot \; e^{ \minus{} st} \; dt$ diverges;[/quote]
Actually, that integral converges. Sure, it's improper at $ 0,$ but the singularity is only logarithmic in size, and that's small enough for an integral to converge.

For instance, $ \int_0^1\ln t\,dt \equal{} \minus{} 1.$

The condition you quoted - that $ f(t)e^{ \minus{} s_0t}$ be bounded - is not a necessary condition for the existence of the Laplace transform, and as a sufficient condition it's stronger than you really need. You need for $ f$ to be locally integrable (as $ \ln t$ is locally integrable), and you need some control over the behavior as $ t\to\infty$ (not a problem for the logarithm.)

(None of what I said gives the specific computation - that's another question.)

Thank you!
You're right - the integral converges. I must have made some errors while defining it's convergence.

[quote="Kent Merryfield"]The condition you quoted - that $ f(t)e^{\minus{}s_0 t}$ be bounded - is not a necessary condition for the existence of the Laplace transform[/quote]
So what are a necessary conditions?

Two [i]sufficient[/i] conditions are piecewise continuity on $ [0, \infty)$ and exponential order. However, these are not [i]necessary[/i] conditions, as the functions $ f(t) \equal{} \frac{1}{\sqrt{t}}$, $ f(t) \equal{} 2te^{t^2} \cos(e^{t^2})$, and $ f(t) \equal{} \ln t$ demonstrate. For these, local integrability [convergence of $ \int_0^1f(t)\,dt$] assure existence of the Laplace transform (as Professor Merryfield already mentioned).

Great! Thanks :) .
One more question - could you specify what does local integrability mean? Is it always $ \int_0^1 f(t) dt$? Or the interval $ [0,1]$ is due to 'improperance' :) of logarithm near $ 0$  :?:

$ f$ is locally integrable if for every point in the domain, there is an interval $ I$ containing that point such that $ \int_I|f|<\infty.$ So, for instance, $ \frac{1}{\sqrt{|t\minus{}1|}}$ is locally integrable but $ \frac1{t\minus{}1}$ is not.

That's not sufficient for the Laplace transform, as you also need some condition concerning behavior at $ \infty.$

Perhaps the simplest way to say it is that there exists some $ s_0$ such that $ \int_{0}^{\infty}|f(t)|e^{\minus{}s_0t}\,dt<\infty.$

But there are cases of functions that do not satisfy this but still make sense as distributions (generalized functions) and for which you may go ahead and define the Laplace transform in the appropriate sense.


algebra
polynomial
function
number theory proposed
number theory

Give the polynomial $f(x)=ax^3+bx^2+cx+d$ ($a,b,c,d \in Z$ and different to 0) such that $xf(x)=yf(y)$ for infinite $x,y \in Z$ and x different to y.Show that the polynomial $f(x)$ has at least one root $\in Z$ :)

Denote $ g(x) = xf(x) $. Then $ g(x) $ is a polynomial of degree
$ 4 $, therefore $ g(x), g(-x) $ are increasing to $ +\infty
$,when $ x
> 0 $ is large enough. Also there exists some constant $ C > 0 $,
such that for $ x > 0 $ big enough we have $ g( x - C ) < g( - x)
< g(x+C) $.
 Therefore if we have that $
g(x_{n}) = g(y_{n}) $, when $ n=1,2,3, \ldots $, such that $
x_{n},y_{n} $ are integers $ | x_{n} | , | y_{n} | \rightarrow
+\infty $, then for big enough $ n $, we have that one of $ x_{n},
y_{n} $ is positive and another is negative. Assume that $ x_{n} <
0 < y_{n} $. Then for $ n $ big enough we have that $ -x_{n},y_{n}
> 0 $ are big enough, therefore we have that $ g( -x_{n} - C ) <
g( x_{n} ) = g(y_{n}) < g( -x_{n} + C) $, so  $ g( -x_{n} - C ) <
g(y_{n})  < g( -x_{n} + C) $, where $   -x_{n} - C, y_{n}, -x_{n}
+ C > 0 $ are big enough, therefore because of monotonicity of $
g(x) $ for large $ x > 0 $, we have that $  -x_{n} - C  < y_{n} <
-x_{n} + C $. Therefore for big $ n $, we have that $ | x_{n} +
y_{n} | < C $. Therefore $ x_{n} + y_{n} $ takes only finite
number of integer values , therefore some value $ c $ appears
infinite number of times. Take a subsequence $ x_{n}' , y_{n}' $,
of $ x_{n},y_{n} $, such that $ g(x_{n}') = g(y_{n}') $ and $
x_{n}' + y_{n}' = c $. Then for $ x = x_{n}' $ we have $ g(x) =
g(c-x) $. Therefore a polynomial $ g(x) - g(c-x) $ vanishes for
infinite number of values, therefore $ g(x) = g(c-x) $ for every $
x $, so $ xf(x) = (c-x)f(c-x) $ for every $ x $. Then if $ c = 0
$, then $ f $ is an odd function, so $ f(0) = 0 $. If $ c \neq 0
$, then we have that $ cf(c) = (c-c)f(-c) = 0 $, so $ f(c) = 0 $.

I hope it is correct.


I just took the new SAT the other day. It was long (4 hours - 10 sections). Does anybody have an idea what the new range of competitive score (like, to get into most selective colleges) will be and how it affects the admission process?

Thanks.

how was the new SAT? compared to the old one?

Since 1400's on the old SAT were decent, 2100+ would probably be good enough for most schools.

I want to know how the essay is graded....

I'm a senior, so it doesn't apply to me, but one of my friends told me that as long as you have a readable 5 paragraph essay, you will get high on that....

I'm lucky I don't got to take it, as you can tell, my grammar sucks.

i heard there's alg II stuff on it...

how is the vocab and the essay? the vocab was hard enough for me 2 years ago.

The essay was easy. You were asked about whether the opinion of the majority is good or bad. The vocab is fairly easy this time. The reading passages were a pain. They were long, random and horribly written for a standardized test in my opinion. But i think i did ok. Im only a sophomore anyway  :lol:

The instructions to essay graders are to ignore spelling and grammar mistakes, and to look for certain aspects of structure and content. The College Board has a document for counselors, [url=http://www.collegeboard.com/prod_downloads/sat/newsat_students_ask.pdf]What Students Will Ask about the New SAT[/url], with a lot of explanations about the new test.

Good luck in your applications.


The lengths of the diagonals of a rhombus are 6 inches and 8 inches. If a circle is inscribed in the rhombus, how many inches are in its circumference? Use 3.14 as an approximation for $ \pi$, and express your answer as a decimal to the nearest tenth.


Perhaps you've seen this one before... How do I find all sets of three distinct primes such that the sum of any two of them is a perfect square?

Hmm.. is there any?

A square is either $0$ or $1$ modulo $4$.
Takes any two odd primes whose sum is a square. Then they cannot be both $1$ modulo $4$ and they cannot be both $3$, thus one of them is $1$ and the other is $3$ modulo $4$.
It follows that there cannot be three distinct odd primes whose sums in pair is always a square.
Moreover, if the third prime is $2$ then with the one equals to $1$ modulo $4$ they add up to something equals to $3$ modulo $4$ leading to same impossibility as above.

Pierre.


If $a,b,c>0$ then we have:
$  \frac{9abc}{2 \sum a} \leq \sum \frac{ab^2}{a+b} \leq \frac { \sum a^2}{2}$.

(made by Moldovan C.)

here is what I have found:

$  \frac{9abc}{2 \sum a} \leq \sum \frac{ab^2}{a+b}$ is equivalent to
$ ( \sum (a+b) )( \sum \frac{ab^2}{a+b}) \geq 3 \prod \sqrt[3]{a+b}$ $ 3  \prod \sqrt[3]{a}$ $ \frac{1}{ \prod \sqrt[3]{a+b}} =9abc$ and we have equality iff $a=b=c$.

$ \sum \frac{ab^2}{a+b} \leq \ frac{ \sum a^2}{2}$
we have that $ \sum \frac{2ab}{a+b}b \leq \sum \frac{(a+b)b}{2}=\frac{ \sum ab + \sum a^2 }{2} \leq \sum a^2$.

cheers! :D :D

do you have other solutions? :D

[u][b]The author of this posting is : arosisi[/b][/u]
____________________________________________________________________

for the first one if we assume that abc = 1 and let x=a/b, y=b/c, z=c/a
your inequality will be more beautiful
x + y + z + 1/(x+1) + 1/(y+1) + 1/(z+1) >= 9/2
is this right?

Darij, what happened to you with "theauthor of this posting is..."?

He re-posts some old posts which were deleted...

why and when were any posts deleted?? :?


summer program
Mathcamp
Gauss
floor function
AoPS Books
group theory
graph theory

Many of us get out of school in early June and have about one month before the beginning of Mathcamp with many hours to spare.  I was wondering if anyone had any recommendations for “pre-camp reading.”

I might be one of the few on this site who [i]has not[/i] already purchased the AoPS books, but will probably do so soon.  Does anyone have other suggestions?

for math reading, just look at the resources section of this site.  It's very helpful!

As much of the math tought at mathcamp is far from the tranditional math of your highschool cirumlum, its hard to know whats gonna be there, and thus how to prepare. Reading a book a about proofs and knowing proof technique qould be a good thing though. Its alwasy tought the first week of mathcamp - but that tkaes up a weeks worth of classes for one slot, ist pretty easy to learn on you own - and thus if you want to kill time isnt a bad thing to do Other than that, i dont ahve much of any segguestions. :)

Oh I can do better than that--[Edit: -to the first reply-post.  Proofs can be a good thing, yes.}
Topics for reading should pertain to some of the subjects of interest you discover and expect to attend to at camp.  There's a wide variety of things there, but particularly for me was the fact that I had never had a mess in Group Theory.  I don't know if you intend to tackle a class in Geometric Group Theory (and become "an apparatus for turning coffee into theories"  :roll:) too soon or your first year at camp, but there are loads of things there, and having an introduction to any of them will certainly help.

The thing to remember is that if you are really ready to fit it in, just having a basic Idea of some of the Informaiton Covered will do a tremendous job of getting you at least into the right class and getting you excited once there.

Elsewise, I really enjoyed Graph Theory, Kenny's class, problem solving--and Gauss; I tried to (:laugh::heli:) follow along in Gauss.  You'll want to be sure to be familiar with some Number Theory when you leave at least.
By now, you're probably getting right confused as to what's there and the jive-mumbo.  Just remember to be OK with words like abellion, modulo, element ($\in$ Set values), iff--and to expect to see wierd lines and circles on things and all numbers written in the form of their relation to $Z\sub{+}$  There may also be an occasional $\neg \oint \widehat{z + 2a} \prod \lfloor q \rfloor$, etc.  ---Most of that was just to scare you.---Try to get a hold of a copy of some previous course(-s/ descriptions).  And in all, it's a blast.

btw, I can dig up all my class handouts and weekly schedules from last year if you need me--but that would probably take a PM and about a week and a half because it's graduation and I'm faking it due to my utter overanalytical lack of good competence.

  Good luck; I hope that wasn't too confusing.  The more familiar and therefore lost you are, the more questions [i]you will have[/i] to be answered in July (3rd?).

Eddy

If you went to Mathcamp before, then you just need to look through the yearbook.
If you never went to Mathcamp, then you need to read the instruction menu for a time machine in order to obtain a Mathcamp yearbook (Just kidding).
So yeah, just enjoy reading whichever kind

The things that you really need to know will probably be pretty simple, but if you don't know some terminology just ask the mentor/staff about it after the class, or actually even better, during class: Then you won't be lost for the class.

One thing to remember is that if you get lost in a class, you probably won't get found. So, if you don't get something, don't think that you'll just skip this part of the class. Ask questions!


If $ a\plus{}1\equal{}b\plus{}2\equal{}c\plus{}3\equal{}d\plus{}4\equal{}a\plus{}b\plus{}c\plus{}d\plus{}5$, then what does $ a\plus{}b\plus{}c\plus{}d$ equal?

[hide="Want to check your answer?"][hide="Click here!"][hide="Wrong!!!Click here!"]$ \minus{}\frac{10}{3}$[/hide][/hide][/hide]

[hide]Adding up the first 4 equations $ (a \plus{} 1 \equal{} a \plus{} b \plus{} c \plus{} d \plus{} 5,b \plus{} 2 \equal{} \cdots)$,
\[ a \plus{} b \plus{} c \plus{} d \plus{} 10 \equal{} 4(a \plus{} b \plus{} c \plus{} d) \plus{} 20
\]

\[ \minus{} 10 \equal{} 3(a \plus{} b \plus{} c \plus{} d)
\]

\[ a \plus{} b \plus{} c \plus{} d \equal{} \minus{}\frac {10}{3}
\]
[/hide]

[hide="slow way"]put everything in terms of a. $ a\minus{}1\equal{}b$,$ a\minus{}2\equal{}c$, $ a\minus{}3\equal{}d$, $ a\plus{}1\equal{}a\plus{}b\plus{}c\plus{}d\plus{}5$, $ b\plus{}c\plus{}d\equal{}3a\minus{}6\equal{}\minus{}4$ a=2/3 so $ a\plus{}b\plus{}c\plus{}d\equal{}1\plus{}2/3\minus{}5\equal{}\minus{}10/3$[/hide]

Lol why would anyone do it the slow way?

i thought i was the first one to post, so i thought someone would post a faster solution after, but you beat me to the post

hey
$ Fast>Slow$, so obviously it would be posted before yours...lol


trovare i valori di x per cui risulti

\[\sqrt{x^2-1}>\ x\]
vorrei sapere come vi torna a voi.. :)

la radice q arriva o dovrebbere arrivare fino a 1 :D

[quote="hammond"]$\sqrt{x^2-1}>x$
le condizioni per l'esistenza del radicale sono $x\le -1 \vee x \ge 1$;
per $x\le -1$ la disequazione è ovviamente verificata (primo membro positivo, secondo negativo), mentre se $x\ge 1$ possiamo elevare al quadrato e si vede che non è mai soddisfatta. Quindi la soluzione è $x\le -1$

[quote="EUCLA"]Source: esame di ammissione a pisa [/quote]
Intendi alla Normale?!? Mi sembra mooolto facile e 'scolastico' come esercizio... (a meno che non abbia fatto boiate io)[/quote]
allora è anche la mia soluzione...mi sembrano troppo facili anche a me ma gli ho presi da internet dal sito della Normale......maaaa :?

[quote="ma_go"]il livello dei problemi di ammissione è molto vario... in generale, per ogni anno, ci sono uno/due problemi mediamente facili, più un paio di abbastanza tosti, più un altro paio di problemi davvero tosti...
poi, il livello medio dei problemi è molto basso per i primi anni, si alza, ha delle punte negli anni 90, e ora è cominciato a scendere leggermente. l'anno scorso (prova del 2003) era già facilina, quest'anno (sylvester a parte) era assolutamente fattibile e piuttosto nota (se uno studia un po' di teoria olimpica, se la cava abbastanza facilmente).
comunque nell'ultima decina d'anni sono saltati fuori dei problemi davvero interessanti! prova a confrontare le prove vecchie, già con quelle dell'85/90... ti accorgerai della differenza... poi dal 90 in poi alcuni sono davvero difficili!
comunque, in 2/3 anni hai tutto il tempo di prepararti... se vuoi qualche consiglio, mi posso anche rendere disponibile, assieme a catraga. ci sono un paio di libricini assai utili...
a proposito: tu fai le olimpiadi di matematica? sarai a cesenatico?[/quote]
grazie x la disponibilità
non le faccio le olimpiadi..non ho mai fatto niente gare di matematica :( ...diciamo che prefersico farla cosi x conto mio senza dover gareggiare...anche se ammetto che mi piacerebbe...cmq io ho solo 15 anni :lol: quindi per ora preferisco studiare e poi magari tra qualche anno mi dedicherò alle gare....non lo sapevo che ci fosse anche in italia

ehm... a dire il vero, le olimpiadi si possono fare solo se si frequenta un istituto superiore (nel tuo caso, suppongo un liceo)... quindi non ti consiglio di aspettare!
e comunque vale la pena, ci si diverte un mondo e si conosce un sacco di gente interessante... e poi (vero, soprattutto a noi maschietti) la competizione stimola...
boh, poi vedi tu :)
ciao ciao!
m.
ps. ti sei registrata sull'altro forum?? che aspetti?!

ho provato l'altra volta nn mi funzionava la registrazione :(

$x^2-1\ge 0\Longrightarrow D=(-\infty ,-1]\cup [1,\infty).$

$0\le \sqrt{x^2-1}\le x\Longleftrightarrow x\in D,\ x\ge 0,\ x^2-1\le x^2\Longleftrightarrow x\ge 1.$

$\sqrt{x^2 -1} >x\Longleftrightarrow \overline {\sqrt{x^2 -1}\le x}\Longleftrightarrow x\in D,\ x<1\Longleftrightarrow x\in (-\infty ,-1].$

[i]With friendship[/i], [u]my dear brother[/u] ([i]ginta latina[/i]), [i]Virgil.[/i]

[b]P.S.[/b] Eucla, are you a Euclid's descendant ?

[quote="Ph-An"]

[b]P.S.[/b] Eucla, are you a Euclid's descendant ?[/quote]
i'd like to  :roll:

[quote="Virgil Nicula"]I would like be a descendant of the great Dante ![/quote]
really?why?

$y=\sqrt{x^2-1}$ is hyperbola and $y=x$ is assymptotes, thus the answer is obviously $x\leq -1.$ :D


According to the Bible, the Genesis Flood was a cataclysim caused by God to wipe out the sinful men that have inhabited the Earth.

Only Noah and his family and (relatives?) were safe on board an ark made of gopher wood and covered with tar. The ark was about the size of a small modern day aircraft carrier.

Science has many contradictions...yet so many truths lie about.

There is evidence of great floods that have occured on Earth thousands of years ago, but worldwide???

It is a fact that there is not enough water in the world for waters to rise and cover all of tops of every mountain in the world, yet, back then, the seabeds might have been higher and the mountains lower, adn then there would be enough water.

If the ark had landed on the mountains of Ararat, there should be evidence that life around the world started there.

There is none...or is there???

The ark could probably not have been able to hold every single type of animal in the world...millions of insects, thousands of animals...impossible. And the food to supply them? How are water animals to die in a flood?

Many civilizations have reported massive floods destroying much of their lands...could they all be referring to the same thing???

What do you think?

Science has many contradictions?

Where?

[quote="hwenterprise"]Many civilizations have reported massive floods destroying much of their lands...could they all be referring to the same thing???[/quote]
As our civilisation now does after the Tsunami... ;)

I think that the biblical flood has a real background, but it wasn't an earth covered with water, but just some (small¿) part.

what about this??????

EXPEDITION TO MOUNT ARARAT

They found the Coffer?

MOSCU
An expedition organized by the Russian association "Unknown Planet", that produces film on the Earth mysteries, assured to have discovered "the Coffer of Noah" in Monte Ararat. The announcement was given by the mission leader, Andrei Polyakov.

Already in the past, specialists affirmed that the rest of the "Coffer of Noah", an enormous petrified wood structure that would correspond to the measures indicated by the Bible, were photographed from the space.

Polyakov said that it obtained the permission of the Turkish authorities to raise Monte Aratat, closed for reasons of security in the Southeast of the Kurd region, and to have arrived at the place where would be the Coffer: "it is not in the top of the Ararat, but to 30 kilometers. It measures 150 meters in length, 25 of wide and 15 of stop ".

Then, watch this photos here: http://www.levantandoacristo.cl/arcadenoe.htm

There are also stories telling that China has experienced a huge flood about 5 thousand years ago.

I read somewhere that there was a massive flood involving the Black Sea long ago, leading perhaps to stories of the Flood.

[quote="José"][quote="Captain Sugar"]in the old testament, the writers only wrote about what they knew. hence, mesopotamia was their whole world. they wrote that their whole world was covered in water.  archaeologists say there was a giant flood that covered all of mesopotamia (and then some) before the book was written. so the giant flood really did exist, but it wasn't as giant as the bible makes it sound.[/quote]

Ohhh. I never thought about that... it's true... their "world" was only that part of the true world.... good thought Captain!!!!![/quote]

i read it out of a science magazine.  :lol:


calculus
integration
function
limit
induction
calculus computations

Let $ a$ be a positive constant number. For a positive integer $ n$, define a function $ I_n(t)$ by $ I_n(t)\equal{}\int_0^t x^ne^{\minus{}ax}dx$. Answer the following questions.

Note that you may use $ \lim_{t\rightarrow \infty} t^ne^{\minus{}at}\equal{}0$ without proof.

(1) Evaluate $ I_1(t)$.

(2) Find the relation of $ I_{n\plus{}1}(t),\ I_n(t)$.

(3) Prove that there exists $ \lim_{t\to\infty} I_n(t)$ for all natural number $ n$ by using mathematical induction.

(4) Find $ \lim_{t\to\infty} I_n(t)$.

Let's do the second one first. Using integration by parts, it's easy to get:

$ \int^t_0x^{n \plus{} 1}e^{ \minus{} a x}\,dx \equal{} \minus{} \frac {1}{a}t^{n \plus{} 1}e^{ \minus{} a t} \plus{} \frac {n \plus{} 1}{a}\int^t_0x^ne^{ \minus{} a x}\,dx$

$ I_{n \plus{} 1}(t) \equal{} \minus{} \frac {1}{a}t^{n \plus{} 1}e^{ \minus{} a t} \plus{} \frac {n \plus{} 1}{a}I_n(t)$

Now, the first one is easy. We get:

$ I_{1}(t) \equal{} \minus{} \frac {1}{a}t e^{ \minus{} a t} \plus{} \frac {1}{a}\int^t_0 e^{ \minus{} a x}\,dx \equal{} \frac {1}{a^2} \minus{} \frac {1}{a^2}e^{ \minus{} at} \minus{} \frac {t}{a}e^{ \minus{} a t}$

As for the third, denote $ L_n \equal{} \lim_{t \rightarrow \infty}I_n(t)$. Let $ t \rightarrow \infty$ and we have:

$ L_{n \plus{} 1} \equal{} \frac {n \plus{} 1}{a}L_n$

(this is basically induction because we did the $ n \equal{} 1$ part, and set the limit equality). As for the (4) it's easy to solve the recursion and get:

$ L_{n} \equal{} \frac {n!}{a^n}L_0 \equal{} \frac {n!}{a^{n \plus{} 1}}$


The Fibonacci sequence is the sequence 1, 1, 2, 3, 5, $ \ldots$ where each term is the sum of the previous two terms. What is the remainder when the $ 100^{\mathrm{th}}$ term of the sequence is divided by 4?

The first term is 1 (mod 4), second is 1 (mod 4), the third is 2 (mod 4), the fourth is 3 (mod 4), the fifth is 1 (mod 4), the sixth is 0 (mod 4), the seventh is 1 (mod 4), the eighth is 1 (mod 4). The pattern repeats every six numbers: 1, 1, 2, 3, 1, 0, 1, 1, 2, 3, 1, 0, 1, 1, 2, 3, 1, 0, etc.

We find 100 (mod 6) to be 4. That means it corresponds to the fourth number in the sequence, or 3, which is 3 (mod 4). So the answer is 3.

There's got to be an easier way.

Well, if you got a Pascal's triangle on hand, then you could do this problem.  It's tougher though.

1
1 1
1 2 1
1 3 3 1
1 4 6 4 1

If you make diagonals (i.e. 1-3-1), you get Fibonnaci numbers.


linear algebra
matrix
complex numbers
linear algebra unsolved

Let T be a normal operator on an inner product space, prove that T and its adoint have same range.

In the finite-dimensional case, if $ T$ is normal, then it is diagonalizable. WLOG, we may assume that $ T$ is diagonal, in which case $ T^*$ is just the complex conjugate of a diagonal matrix, and clearly has the same range as $ T.$

In general, can this be done without spectral theory? So far, I don't see it.

I have a question: only in complex inner product space a normal operator must have a diagonal matrix, right? How about real inner product space for this problem?

(Limiting my reply to the finite dimensional case.)

A real normal matrix is diagonalizable over the complex numbers; the obstacle to it being diagonalizable over the reals is that its eigenvalues might not be real.

Example: $ A\equal{}\begin{bmatrix}0&\minus{}1\\1&0\end{bmatrix}$ is normal, but cannot be diagonalized over $ \mathbb{R}.$

However, in this particular case, $ A$ and $ A^T$ have the same range.

In fact, we should be able to proceed as follows:

Show that if $ A\in M_n(\mathbb{R})$ and $ A^TA\equal{}AA^T,$ then we can find a real orthogonal matrix $ U$ such that $ U^TAU$ is in block-diagonal form, with the blocks on the diagonal being either $ 1\times 1$ or $ 2\times 2$ blocks, with the $ 2\times 2$ blocks being of the form $ \begin{bmatrix}a&\minus{}b\\b&a\end{bmatrix}$ for $ b\ne 0.$

If you can do that, you should be able to replicate my original argument.


I remember it was solved by Harazi in the forum.