For me, I go Warriors in 6. Golden State knows how to use its crowd, which will be the key factor in game 6.
How 'bout you?

mavs in 7. of course. baron davis double-T's in game 6 and from that point on can't penetrate the zone.
Mavs win, 106-95 (game 6)

and then game 7 is gonna be a blowout

since when is telling the truth spam?
and ratings don't count in G&FF

[quote="footballrocks41237"]Mavs in 6.[/quote]
umm...
is this a joke?

[quote="footballrocks41237"]Mavs in 6.[/quote]

yeah...this kinda can't possibly happen

[quote="i_like_pie"][quote="madskillz1"]I rated that post as spam :rotfl:[/quote]
If you don't agree with someone, that doesn't mean their posts are spam. That just means they have a different opinion, and you should respect that opinion. I think the Warriors are going to win, but I'm not going to rate sapphyre's post as spam just because I don't agree with him/her.[/quote]

ok that was a joke. a joke.

um
i dont really care much for basketball
but since im from sanfrancisco
WARRIORS FTW!
in 6.2

[quote="ra5249"]halftime, warriors up by two (or three? i forget...  :roll: )[/quote]
Go warriors!!!! Up 23 at end of 3rd.

warriors win!!!
aka houston has now made it to the conference finals

[quote="sapphyre571"]Nowitzsucked
and the Warriors are the luckiest 3-point shooting team ever. ever.[/quote]
whatever. u wish.
in ur face, sapphyre!  :P
jkjk
but the warriors are good. not lucky.
and kleb, not yet. but yeah, its going to happen.

[quote="sapphyre571"]Nowitzsucked
and the Warriors are the luckiest 3-point shooting team ever. ever.[/quote]
Not really, the Mav's defense really sucked. Watch the replays of Jackson's shots. He was almost always wide open.

Now, how do you justify a 67-15 season?

[quote="Fanatic"]The only PHO vs. GS game I saw this year resulted in a GS shooting spree, which they barely sqeaked out but at one point was winning by like 20.

Eastern:

Cavs over Nets in 6
Bulls over pistons in 5
Bulls over Cavs in 4

GGNORETHX :P[/quote]

Woah...woah..woah...Bulls over pistons????  Even with all your Chicago pride, isn't that going a *bit* too far   :D

What are these teams?
I don't follow sports.  :lol:

um... NBA (basketball)
and PHOENIX WILL NOT BEAT SAN ANTONIO
Spurs in 6 or 7

Hopefully the rockets will beat the Jazz...and i dont think the rockets will lose to golden state...

[quote="Aznwithabrain"][quote="Art of Owna"][quote="baseballfreak2"]The Warriors have an offhand shot to make the NBA finals.

It will be:

Warriors over Houston in 6
Phoenix over SA in 6

Warriors over Phoenix in 6.[/quote]

No Pheonix will beat the warriors because they play similar styles so the warriors cant utalize their speed and full court play anymore..
the only reason they beat the mavs is because the mavs were way too slow(devin harris was the only fast player for the mavs), not as athletic, and they could not stop baron davis driving ability with no shot blocker in the middle..

suns-
they are faster then the warriors
they have a deeper bench so the fast paced game wont fatigue them as much
and they actually have a legit post player in stoudamire instead of having dampier.

Oh yea if the rockets play the warriors in the next round, theres no way the rockets will win unless they dont play yao...he jsut the slowest player on the floor, he cant block shots, and he cant even rebound despite his size because he has no strengh and aggresiveness, which means the warriors will probalby, depite undersized, outreboud the warriors.  Although he is a pretty good offensive shooter and an amazing free throw shooter he will get tired so easily with the way the warriors play..the jazz is the warriors play them (i think houston will beat the jazz), the jazz are very unathletic although the jazz do have a very good point guard and all 5 guys can shoot the ball well (their center is their best 3 pt shooter)[/quote]

Do I hear doubt about Houston?  :rotfl:
just kidding

Phoenix and Golden State would be close:

Steve Nash vs. Baron Davis
Baron Davis

Raja Bell vs. Jason Richardson
Jason Richardson

Boris Diaw vs. Matt Barnes
Matt Barnes

Shawn Marion vs. Stephen Jackson
Shawn Marion

Amare Stoudamire vs. Al Harrington
Amare Stoudamire

Phoenix Bench vs. Golden State Bench
Phoenix Bench

3-3

If Houston advances (please help god)

Houston vs. Golden State:

Rafer Alston vs. Baron Davis
Baron Davis

Tracy McGrady vs. Jason Richardson
Tracy McGrady

Shane Battier vs. Matt Barnes
Shane Battier

Chuck Hayes vs. Stephen Jackson
Stephen Jackson

Yao Ming vs Al Harrington
Yao Ming

Houston Bench vs. Golden State Bench
Houston Bench

4-2[/quote]
Houston bench pretty much sucks;  I don't know too much about the warriors bench though-is it worse?

[quote="Art of Owna"]Aznwithabrain, you cannot compare individual players one on one.  Its a team sport.  If you want to compare talent alone, the jazz wouldve easily lost already instead of a game 7.  Also, the warriors and houston play totally differnt games.  the warrios have no big men, but they dont need it as their game plan is to shoot as many times and get as many [b]fast[/b] breaks as they can.  And if they play houston, yao's lack of speed will hurt houston.  But if yao can get going  on offense, he will be unstopable because the warriors cneter, al harriongton, is 6'9", yao is 7'6"..as for benches, the warriors have one prominent bench player (mikeal pietris, matt barnes maybe..), while houston has like 3 or 4, but the warriors dont get tired so it wont matter.

ANd the east, the bulls are almost an exact replica of the pistons except younger (less experience) and with a little deeper bench

AND change the subject to something else besides warriors vs mavs, because thats alreayd over.[/quote]

Yeah...i kinda figured that baskeball was a team sport by now  :rotfl:
Thats why they don't play the games on paper :P

About the Houston Bench (Klebian):
Dikembe Mutumbo (the key veteran to the future Houston Golden State series)
Luther Head (an awesome three point shooter)
Juwan Howard (a veteran role player)

The warriors bench really depends on who they start.

In game 6 they started: Davis, Richardson, Barnes, Jackson, and Biedrins.
In game 5 they started: Davis, Ellis, Richardson, Jackson, and Biedrins.

So theyre bench is mainly Ellis/Barnes, Harrington, and Pietrus.

Ellis and Harrington were literally nonexistent in the Dallas series.

even though i live in san francisco
GO SUNS
steve nash is awesome


Solve in positive numbers the system

$ x_1\plus{}\frac{1}{x_2}\equal{}4, x_2\plus{}\frac{1}{x_3}\equal{}1, x_3\plus{}\frac{1}{x_4}\equal{}4, ..., x_{99}\plus{}\frac{1}{x_{100}}\equal{}4, x_{100}\plus{}\frac{1}{x_1}\equal{}1$

[quote="outback"]Solve in positive numbers the system

$ x_1 \plus{} \frac {1}{x_2} \equal{} 4, x_2 \plus{} \frac {1}{x_3} \equal{} 1, x_3 \plus{} \frac {1}{x_4} \equal{} 4, ..., x_{99} \plus{} \frac {1}{x_{100}} \equal{} 4, x_{100} \plus{} \frac {1}{x_1} \equal{} 1$[/quote]

[hide="solution"]
Let $ y_i \equal{} \frac {x_i}{2}$(if $ i$=odd)and $ y_i \equal{} 2x_i$(if $ i$ is even),the system is equivalent to

$ y_1 \plus{} \frac {1}{y_2} \equal{} 2, y_2 \plus{} \frac {1}{y_3} \equal{} 2, y_3 \plus{} \frac {1}{y_4} \equal{} 2, ..., y_{99} \plus{} \frac {1}{y_{100}} \equal{} 2, y_{100} \plus{} \frac {1}{y_1} \equal{} 2$.

Sum all each hands,we have

$ (y_1 \plus{} \frac {1}{y_1}) \plus{} (y_2 \plus{} \frac {1}{y_2}) \plus{} \cdots \plus{} (y_{99} \plus{} \frac {1}{y_{99}}) \plus{} (y_{100} \plus{} \frac {1}{y_{100}}) \equal{} 200$.

By AM-GM inequality, $ y_i \plus{} \frac {1}{y_i}\ge 2$ holds for all $ i$,thus we have

$ (y_1 \plus{} \frac {1}{y_1}) \plus{} (y_2 \plus{} \frac {1}{y_2}) \plus{} \cdots \plus{} (y_{99} \plus{} \frac {1}{y_{99}}) \plus{} (y_{100} \plus{} \frac {1}{y_{100}})\ge 200$

therefore,the equality of the AM-GM must holds for all $ i$,so we have $ y_1 \equal{} y_2 \equal{} \cdots \equal{} y_{100} \equal{} 1$ and we see the solution $ \boxed{x_1 \equal{} x_3 \equal{} \cdots \equal{} x_{99} \equal{} 2, x_2 \equal{} x_4 \equal{} \cdots \equal{} x_{100} \equal{} \frac {1}{2}}$.[/hide]

[quote="t0rajir0u"]Was $ x_{100} \plus{} \frac {1}{x_1} \equal{} 1$ a typo?[/quote]

Why do you think it is?


AMC AIME inequalities USA(J)MO USAMO \/closed

This year we are considering running both the AIME A and the AIME B Problem Series beginning in November.  Currently on the AIME B Problem Series is on the schedule because more of the material will be new to more students.

The Problem Series are at about the same difficulty level and both focus on a broad spectrum of AIME material.  The problems discussed are about 95% distinct, though the concepts covered are about 95% the same.

In order to get a feel for whether or not we should hold both classes, I would like students to let us know which they would take.  Some of this will have to do with whether or not students have already taken the A, enrolled in the B, or which fits their schedule:

AIME A Fridays from 7:30 to 9:00 PM ET
AIME B Mondays from 8:00 to 9:30 PM ET

I've already taken A.....
would taking B be a good idea? The time is fine for me, though.

I am choosing between Olympiad Inequalities and AIME B next semester.

Some students will be taking the B who have already taken the A.  Of course, students take these classes at different times in their development.

Whether you take one course or another depends on what you want to get out of the course.

If your goal is to make the USAMO, the AIME B might be best.

If your goal is to continue along with your algebraic skills at a proof-writing level and get a solid footing with inequalities, the Olympiad Inequalities is probably the better option.

My son has submitted a registration for the AIME B course, having taken the AIME A course last year.

In my personal view as a parent, one can never get too much practice in AIME problems, so I would urge math-likers who haven't tried the AIME course yet, but who have qualified for AIME before, to try the course (A or B).

Is the deadline already over? I saw the "courses in session" list and it didn't include AIME B, but it's not in the enrollment section either.

Yes, but you can call the office and register (619) 659-1612.


calculus
integration
function
email
Euler
calculus computations

Find Integral of x/(e^x-1) dx  from 0 to infy.

Find Integral of x^2/(e^x-1) dx from 0 to infy.

perhaps you mean :
 $\int^{\infty}_0 x / ( e^x - 1) \ dx$ and
 $\int^{\infty}_0 x^2 / ( e^x - 1) \ dx$

Most of this has been posted before on this site.

Let me start with two facts. One is the geometric series. For $x>0,$

$\frac1{1-e^{-x}}=\sum_{n=0}^{\infty}e^{-nx}.$

The other is a computation related to the Gamma function: for $m\in\mathbb{N}, a>0$

$\int_0^{\infty}x^me^{-ax}\,dx = \frac{m!}{a^{m+1}}.$

Now:

$\int_0^{\infty}\frac{x^m}{e^x-1}\,dx=\int_0^{\infty}x^me^{-x}\cdot\frac1{1-e^{-x}} \,dx$

$=\int_0^{\infty}x^me^{-x}\sum_{n=0}^{\infty}e^{-nx}\,dx =\sum_{n=0}^{\infty} \int_0^{\infty}x^me^{-(n+1)x}\,dx$

$=\sum_{n=0}^{\infty}\frac{m!}{(n+1)^{m+1}}=\sum_{k=1}^{\infty}\frac{m!} {k^{m+1}} = m!\zeta(m+1).$

So your first problem is $\zeta(2),$ which is rather famously equal to $\frac{\pi^2}6.$

Your second problem is $2\zeta(3),$ which doesn't have an elementary representation.

[quote="Kent Merryfield"]Your second problem is $2\zeta(3),$ which doesn't have an elementary representation.[/quote]

Has that been proven? I haven't been reading about this problem much lately, but the most recent thing about it I can remember is Apery's proof that it is irrational...

Not proven. I should have said "which has no [i]known[/i] elementary representation."

We have the following representation:

For $\text{Re}(z)>1,\,\Gamma(z)\zeta(z)=\int_0^{\infty}\frac{x^{z-1}}{e^x-1}\,dx$

The details are exactly what I said above, freed from the requirement that $m$ be an integer.

Thanks a lot for posting the soln.
Xevarion you said there is a proof that Zeta(3) is irrational.
Where can I find it?

PM me with your email address and I'll send it to you. That goes for anyone else who's interested, as well.

Just a note of caution: it is a difficult proof. When Euler can't do it, it has to be hard.


The side of the square is 3, and the circles' centers are evenly spaced.

[hide]r=1[/hide]

Call the square $ABCD$.
The center of the circle that is closest to $A$ be $A'$, etc.
Let $A'_{AB}$ be the intersection of $AB$ and the line perpindicular to $AB$ that intersects $A'$, etc.

Before I start, I think I need a clarification on this: by "the centers are equally spaced," do you mean that $A'_{AD}A'=A'B'=B'B'_{BC}$, etc. for all the points?

I'm sorry, but I don't understand the statement regarding even spacing of the centers.

What is the common radius?

Dear Administrator:

Can you please move this into the Geometry Unsolved Problems section?

It was prestty complicatedwhen I solved it, and I didn't finish.

To other people:

Think of the smaller square with the center of the circles are vertices. Start from there.

This is definitely not supposed to be in the advanced section.

[hide]Focus in on the middle. You can draw a 1 by 1 square with a flower-ish design that's made by making 4 quarter circles, one from each vertice. On any circle, the places where it overlaps other quarter circles if connected to the origin of the first quarter circle, it will make a 30 degree angle. We can divide the shaded figure into a square, and then 4 little slivers. Now it's a simple task of finding the side of the square and the area of the slivers.[hide]

[/hide][/hide]


AMC
USA(J)MO
USAMO
AMC 12
AMC 10
AIME
AIME I

We try to understand the USAMO selection rules (see URL link and the detailed rules below) and are concerned about the 9th and 10th graders who took AMC12 and they will be in disadvantage compared with those who took AMC10.

According to rule #4, “If there are more than 160 young students with a score above the floor value, then approximately 160 students will be selected from this group by using the USAMO index.”. This seems to suggest there is no difference between the AMC12 and AMC10 scores in the USAMO index calculation, and is it correct?

Let’s use last year as an example. For two Sophomores, one got 122.5 on AMC10 while the other got 120 on AMC12, and both got AIME score of 8,. According to the rule, the first student was selected while the second was not. But, in reality, the student who took AMC12 would most likely get higher score if taking AMC10. His chance was eliminated mainly because he took AMC12.

According to what we know, a lot of schools do not let students to choose which one to take. It seems unfair to equate AMC10 and AMC12 scores in the USAMO index in the situation where AIME is the same.

In our opinion, using purely the AIME scores for 10th graders and below will be fairer to all. I believe this was the selection criteria until 2006 (http://en.wikipedia.org/wiki/United_States_of_America_Mathematical_Olympiad), and we are not sure why the rule was changed. If USAMO index is to be used, at least some credit should be given to the 10th graders and below who took AMC12 when comparing with AMC10 score.

Can someone explain to us our thinking is wrong here?

Thanks in advance.....

USAMO Selection rules: http://www.unl.edu/amc/f-miscellaneous/faq.shtml

“Selection for the USAMO will be made according to the following rules:
1.	The goal is to select about 500 of the top scorers from this years’s AIME and AMC 12A, AMC 12B, AMC 10A and AMC 10B contests to participate in the USAMO.
2.	Selection will be based on the USAMO index which is defined as 10 times the student’s AIME score plus the student’s score on the AMC 12 or the AMC 10.
3.	The first selection will be the approximately 330 highest USAMO indices of students taking the AMC 12A or AMC 12B contest.
4.	The lowest AIME score among those 330 first selected will determine a floor value. The second selection of approximately 160 USAMO participants will be among students in the 10th grade and below who received an AIME score at least as high as the floor value. If there are more than 160 young students with a score above the floor value, then approximately 160 students will be selected from this group by using the USAMO index.
5.	The student with the highest USAMO index from each state, territory, or U.S. possession not already represented in the selection of the first and second groups will be invited to take the USAMO.
6.	To adjust for variations in contest difficulty, the number of students selected from A & B contests will be proportional to the number of students who took the A & B Contests. This also applies to the students who tale AIME I vs. AIME II.
7.	In advising young students (in grade 10 or below) who desire to be selected for the USAMO whether to take the AMC 12 contest or the AMC 10 contest, please be aware of the following facts:
a.	In 2008, among 503 students invited to take the USAMO, 232 were in 10th grade and below. Those students had scored 6 or greater on the AIME.
b.	Among those 232 students, 69 had their AIME qualifying high score based on the AMC 12 and 163 had their AIME qualifying high score based on the AMC 10.
c.	In 2008, among 8,702 students who took the AIME, 3,039 were in grades 10 and below. Of those, 845 qualified for the AIME from the AMC 12 and 2,194 qualified from the AMC 10. “

These are all true and are all valid concerns. However, you can't use just AIME as it's far too low-resolution. There was an extremely long thread a few years ago with many proposals, but none of them were better than the current one. It's annoying, sure, and it prevented me from qualifying for USAMO in freshman year, but we haven't found anything better.

Everything you said makes sense, and this has increasingly become an issue since more and more 9th and 10th graders do well on the AIME every year.  In 2006, there was a huge controversy about this because there was a clause in the USAMO selection rules saying that 9th and 10th graders who want to qualify for USAMO should be advised to take the AMC 12, even though the selection process itself favored the AMC 10.  Since then I believe the AMC has removed that clause from the selection rules and replaced it with facts about the previous year's USAMO qualifiers.

To be honest, the selection rules are there, so it's the students' job to keep those selection rules in mind when choosing which AMC they want to take.  The system is fair, since all the rules are laid out up front.

Also by the same logic if you give a little "booster" for the 12 and some schools don't let students take the 12 they'll be at a disadvantage since if students take the 10 the maximum possible index derived from the AMC would be lower

To worthawholebean:

"These are all true and are all valid concerns. However, you can't use just AIME as it's far too low-resolution." - was not USAMO selection purely based on AIME score prior to 2007? Why the change?

http://en.wikipedia.org/wiki/United_States_of_America_Mathematical_Olympiad

I agreed with 236factorial that "It would be fairer if everyone took the AMC 12 and then the AIME, so that the index would have a common basis".

I am not sure about your school, in my school, all 10th graders have to take AMC12 because teachers did not order AMC10 at all, so that is not even an option for us.

[quote="kfeng40"]To worthawholebean:

"These are all true and are all valid concerns. However, you can't use just AIME as it's far too low-resolution." - was not USAMO selection purely based on AIME score prior to 2007? Why the change?
[/quote]

I'm not sure if that's true. Can you quote the place where that is said? The index table listing the min. qualifying scores shows indices, so the AMC 12 was used before.

[quote="kfeng40"]To worthawholebean:

"These are all true and are all valid concerns. However, you can't use just AIME as it's far too low-resolution." - was not USAMO selection purely based on AIME score prior to 2007? Why the change?[/quote]

There was no actual change in rules - the rule is and has been that students selected for the USAMO must have at least the lowest AIME score of the people selected in the "first group."  However, when there are too many people below 11th grade with at least this score, index is used as a tiebreaker.

The difference is that in the past, almost every year, there were few enough people below 11th grade with less than the floor AIME score each year, that all of them qualified for USAMO.  Now it seems that students are getting better at a younger age, so we may never see a year like that again.

I think kfeng is referring to the fact that before 2007, the floor cutoff was "pure" - everyone 10th and below who got at least the floor was invited.  But in 2007 and 2008, there was an index attached to the floor that needed to be met in addition to reaching the floor.

Thus, prior to 2007, the AMC 10 was irrelevant for USAMO qualification.

As a 10th grader, I have been wondering myself about this topic.

The AMC stats for 2009 was just out (http://www.unl.edu/amc/e-exams/e6-amc12/e6-1-12archive/2009-12a/StateStats/IL.HTML) and it is very interesting to see and compare. For example, for state of Illinois, there are only 24 papers above 120 in AMC12 A/B while there are 57 papers above 120 in AMC10 A/B. When comparing AMC12 score with AMC10 score, it does sound very fair for those 10 graders who took the AMC 12 as the AMC10 scores seem to be inflated.

Part of it is the expansion. Expansion thus lower floor thus more kids who get the floor thus they need index to distinguish between them.

But as a junior, I don't really care anymore.

[quote="worthawholebean"]There was an extremely long thread a few years ago with many proposals, but none of them were better than the current one. [/quote]
There may have been more than one thread, but [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=142430]this[/url] is probably the one worthawholebean was referring to. If you go back and read it, you'll see that I was an active participant. To be honest, I haven't changed my mind and I haven't been convinced that my idea of a single index (with some tweaks to the numbers to keep from shutting out too many 11th and 12th graders) wouldn't be better than the system in place. I think the system in place is "broken,"  and that it performed particularly badly in 2008.

The worst aspect of the system in place is that it provides strong students an incentive to take the AMC 10 when they probably should be challenging themselves with the AMC 12.

Oddly, that's one reason why you'd have to tweak the numbers a little. At the time of that thread, I had a correspondence with one of the people in the AMC office who ran a model of my proposal on what was then the current year - and the first-cut qualifying index went up (confounding my initial guess that it would stay the same), which cut out some 11th and 12th graders who would have qualified.

My proposal was this: define index as

(AMC score) - (minimum AMC score to reach AIME) + 10*AIME + 100

(The "+100" is not a logical necessity.) In practice, this would usually deduct about 20 points from an AMC 10 based index as compared to an AMC 12 based index. Then base all selection on this index. Take a first cut of all eligible students, then go back and take a second cut to include younger students.

If you make the first cut about as big as the current AMC 12 index only first cut (about 330) people, then the required index would go up. Why? Because some overwhelmingly qualified students (USAMO under any system) who took only the AMC 10 would displace some 11th-12th graders. Probably the solution to that is to make the first cut a little bigger, maybe 350-360 and the second cut correspondingly smaller. But there would be a lot of tools to play with, a lot of knobs to twirl.

No matter how many knobs have to be twirled, wouldn't a fair contest be worth it?

What is the AMC's opinion on this?

[quote="PI-Dimension"][quote="236factorial"]Despite what the AMC says, "The selection process is designed to favor students who take the more mathematically comprehensive AMC 12A and AMC 12B contests", in past years the AMC 12 takers have a clear disadvantage. Since many bright younger students take the AMC 12 anyway, the top 330 in the AMC 12 is a much more competitive pool than the top 160 in the AMC 10 group.

As your example shows, the selection is quite unfair, favoring the AMC 10 students (I guess this isn't a bad idea, to give them more opportunity).

It would be fairer if everyone took the AMC 12 and then the AIME, so that the index would have a common basis. The AMC 10 was created (I guess) to make it more accessible to less able students, but in my opinion it is still much too hard. There is good reason to suspect that it is impossible to make a 75 minute test that can provide a good experience for normal people and differentiate among the best, so why not just have everyone take one test?[/quote]

they should make it 25 minutes :P <--time I finished in[/quote]
and then you got #1 wrong lolsz


[quote="probability1.01"]I have never ever taken calc, but wouldn't the answer to that be 0?[/quote]

No that is incorrect. You might in fact find the answer by just graphing and pluggin into your TI-89, by plugging in a large number. But remember all Calc Challenge problems posted by me are non-calculator unless otherwise stated.

I would suggest writing it as (something)/(1/x), (hoping that) it approaches 0/0, so L'hopitals rule could be used.

[quote="belenos"]Near zero (1+y)^a = 1 + y/a + 1/2a(a-1)y + O(y)

So (x^6+x^5)^1/6 = x (1+1/x)^1/6 = x + 1/6 + O(1/x)
and  (x^6-x^5)^1/6 = x (1-1/x)^1/6 = x - 1/6 + O(1/x)

So limit is 1/3. Correct ??[/quote]

Belenos, Great Solution! It's short and to the point, very nice... for those of you wondering, here is the (much longer) calculus solution.

Zableman is correct, but I'll do it out.

[tex]\displaystyle{\lim_{x\rightarrow\infty} \sqrt[6]{x^6+x^5} - \sqrt[6]{x^6-x^5}}[/tex]
[tex]\begin{eqnarray*}
\quad &=& \lim_{x\rightarrow\infty} x\cdot\sqrt[6]{1+\frac{1}{x}} - x\cdot\sqrt[6]{1-\frac{1}{x}} \\
&=& \lim_{x\rightarrow\infty} \frac{\sqrt[6]{1+\frac{1}{x}} - \sqrt[6]{1-\frac{1}{x}}}{\frac{1}{x}}\\
&=& \lim_{x\rightarrow\infty} \frac{\frac{1}{6x^2\cdot\sqrt[6]{1+\frac{1}{x}}^5} + \frac{1}{6x^2\cdot\sqrt[6]{1-\frac{1}{x}}^5}}{\frac{1}{x^2}}\\
\end{eqnarray*}[/tex]
[tex]\begin{eqnarray*}
\quad &=& \lim_{x\rightarrow\infty} {\frac{1}{6\cdot\sqrt[6]{1+\frac{1}{x}}^5} + \frac{1}{6\cdot\sqrt[6]{1-\frac{1}{x}}^5}}\\
&=& \frac{1}{3}
\end{eqnarray*}
[/tex]

[quote="churchilljrhigh"]Since I did OK on this year's Calc Challenge I decided to post some more problems.
[tex]\lim _{x\rightarrow \infty }\sqrt [6]{{x}^{6}+{x}^{5}}-\sqrt [6]{{x}^{
6}-{x}^{5}}[/tex][/quote]

Don't forget to put \displaystyle{........} around your tex, it will change it from:

[tex]\lim _{x\rightarrow \infty }\sqrt [6]{{x}^{6}+{x}^{5}}-\sqrt [6]{{x}^{
6}-{x}^{5}}[/tex]

To:

[tex]\displaystyle{\lim _{x\rightarrow \infty }\sqrt [6]{{x}^{6}+{x}^{5}}-\sqrt [6]{{x}^{
6}-{x}^{5}}}[/tex]

Dang it! I don't have enough time to check this forum often enough so someone always beats me to the answer! Good answer though spoon.


modular arithmetic
induction
number theory unsolved
number theory

I've been trying to solve the following problem from a Taiwan TST: let $\{a_n\}_{n\geq 0}$ be a sequence of positive integers, given by $a_0=2$ and for all positive integers $n$, \[ a_n = 2a_{n-1}^2 - 1 . \]
Prove that if $p$ is an odd prime such that $p \mid a_n$ then $p^2 \equiv 1 \pmod{ 2^{n+3}}$.

[hide="What I have manage to prove until now is that"]
we have $a_1=7$, $a_2=97=32\cdot 3 +1$, and through induction $2^{2n+1} \parallel a_n$ for all $n\geq 2$.
Thus we have that $p^2 \mid a_{n+1} -1 \equiv 0 \pmod {2^{2n+3}}$, but I haven't been able to continue from this on (maybe I'm on the wrong path :? ). [/hide]

I'm sure it's been posted before (from the same source, but I'm not sure about this) and widely discussed. I'll try to find it.

Edit:

Here's a link, but there's another one around, which I cannot find right now:

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=lecture&t=5937[/url]

So the problem is indeed hard, as I thought :) At least a good news :D


group theory
superior algebra
superior algebra unsolved

Let $ G$ be a non-commutative group of order $ 8$. Show that $ G$ is isomorphic to the group of symmetries of the square, or to the quaternion group.

$ G$ is not cyclic, nor is every square the identity, for in either case $ G$ would be abelian. So let $ a$ be an element of order 4. We have two cases: either we can find $ b\not\in <a>$ of order 2 or not.

[hide="if yes"]
Again, $ G\equal{}<a><b>$ where $ <a>,<b>$ have trivial intersection. So $ a,b$ do not commute. But $ <a>$ normal in $ G$ so $ bab^{\minus{}1}$ is a power of $ a$ of order 4 other than $ a$. This forces $ bab^{\minus{}1}\equal{}a^3$. We get the set of relations for the dihedral group (i.e. the method of last problem can be used).
[/hide]

[hide="if no"]
The only element of order 2 is $ a^2$. Let $ c\not\in <a>$: then $ c^2\equal{}a^2$ and (as $ ac\not\in <a>$) $ (ac)^2\equal{}a^2$. This gives the relations
\[ a^4\equal{}1,c^2\equal{}a^2,cac^{\minus{}1}\equal{}a^3\]
which uniquely determine the quaternion group.
[/hide]


calculus
integration
quadratics
logarithms
trigonometry
special factorizations
calculus computations

You need some linear/affine/whatever + trigonometric change of variable, and then it's a rational fraction you will have to deal with.

[quote="julien_santini"]You need some linear/affine/whatever ... change of variable[/quote]
After first completing the square in $ax^2+bx+c$ so that you can see what needs to be be done. Expect your solution to have a three-way branch based on the sign of the discriminant in the quadratic formula.

My computation for $b^2 - 4ac > 0$ give $\frac{1}{\sqrt{a}} \ln \left | \frac{2 \sqrt{a}}{\sqrt{b^2-4ac}}\left ( \frac{b+2ax}{2 \sqrt{a}} + \sqrt{ax^2 +bx +c} \right ) \right | +C$

Maple computation, the integral takes form $\frac{1}{\sqrt{a}} \ln \left | \frac{b+2ax}{2 \sqrt{a}} + \sqrt{ax^2 +bx +c} \right | +C$

actually they are same in term of antiderivative. since $\frac{1}{\sqrt{a}} \ln \left | \frac{2 \sqrt{a}}{\sqrt{b^2-4ac}} \right |$ is a constant

for $b^2 - 4ac =0$ we have $\frac{1}{\sqrt{2}} \ln \left | x+ \frac{b}{2} \right | +C$

the other case deal with complex number

You did the first part by hand Beginner!?  That is impressive, how did you do it by hand?  Woah!  Nice job again.

Using similar Idea we can find $\int { \frac{1}{ax^2+bx+c} dx}$
write $\frac{1}{\sqrt{ax^2+bx+c}} = \frac{1}{\sqrt{a}} \frac{1}{\sqrt{\left (x+ \frac{b}{2a} \right )^2 + \frac{4ac-b^2}{4a^2}}}$ for convenient define $D= b^2-4ac$
for $D>0$
use trig subtitution $x= \frac{\sqrt{D}}{2a} \sec \theta - \frac{b}{2a}$. Then
$\int { \frac{1}{\sqrt{ax^2+bx+c}} dx} = \frac{1}{\sqrt{a}} \int {\sec \theta \, d \theta}$


For class we are learning proofs. on the tests i do horrible because i cannot remember all of the theorums. can any one help.

Well, all I can say is use your book wisely.  It has lots of practice problems and examples.  Also, most books have the answers in the back, so you can go check that.

Also, if you can't remember some theorems that you feel you need, try to familiarize yourself with the proofs of the theorems, or specific theorems that lead to others.

To remember them better, you should read the theorems and their proofs, and then redo the proofs until you understand them completely. Also, you could ask your teacher for extra problems so you can practice the theorems. You need to really understand the theorems to be able to remember them, and the most efficient way to do this is to practice as much as possible.

yeah try to understand them and see some basic examples and theyll start to seem more trivial


geometry
circumcircle
trapezoid
trigonometry
geometry unsolved

1. In a circle $ AB$ and $ AC$ are two chords equal in length. How many chords are there, such that $ AB$ and $ AC$ divide the chords into three parts equal in length?

$ \triangle ABC$ is isosceles with AB = AC and circumcircle $ \mathcal O \equiv (O, R).$ Let a circumcircle chord XY cut AB, AC at U, V, such the $ XU \equal{} UV \equal{} VY.$ Powers of U, V to (O) are equal, $ \overline {UX} \cdot \overline {UY} \equal{} \overline {VX} \cdot \overline {VY}$ $ \Longrightarrow$ U, V are on a circle $ \mathcal P \equiv (O, \varrho)$ concentric with the circumcircle $ \mathcal O$ and radius $ \varrho \equal{} OU \equal{} OV.$ This circle intersects AB at U, Q and AC at at V, S, forming an isosceles trapezoid with bases parallel to BC and circumcircle $ \mathcal P.$ UV can be its base or its diagonal.

Assume $ UV \parallel BC$ is a base of an isosceles trapezoid UVSQ. By construction, exactly one chord $ XY \parallel BC$ always exists: Let B', C' be points on the line BC, such that B', B, C, C' follow in this order and B'B = BC = CC'. Then the lines AB', AC' cut (O) again at X, Y and $ XY \parallel BC$ cuts AB, AC at U, V, such that XU = UV = VY.

Assume UV is a diagonal of an isosceles trapezoid USVQ. Then $ UV \equal{} 2 \varrho \sin B.$ Power of U to $ \mathcal O$ is $ \varrho^2 \minus{} R^2 \equal{} \minus{} 2 \cdot UV^2 \equal{} \minus{} 8 \varrho^2 \sin^2 B$ $ \Longrightarrow$ $ \varrho \equal{} \frac {_R}{^{\sqrt {1 \plus{} 8 \sin^2 B}}}.$ However, the circle $ \mathcal P$ must intersect AB, AC (not just touch), therefore $ \varrho > OM \equal{} ON \equal{} R \cos B,$ where M, N are midpoints of AB, AC $ \Longrightarrow$ $ \frac {_1}{^{1 \plus{} 8 \sin^2 B}} > \cos^2 B,$ $ \cos^2 \frac {_A}{^2} \equal{} \sin^2 B > \frac {_7}{^8},$ $ \cos A \equal{} 2 \cos^2 \frac {_A}{^2} \minus{} 1 > \frac {_3}{^4}.$ Under this condition, 2 more chords XY, ZT of $ \mathcal O$ exist, the 2 diagonal lines of the isosceles trapezoid USVQ, cutting AB at U, Q and AC at V, S, such that $ XU \equal{} UV \equal{} VY$ and $ ZQ \equal{} QS \equal{} ST.$


Is there any way we could get some more moods for the blogs?  It seems they never have the right one :P  Or, in fact, a custom option would be nice...you wouldn't have a smiley next to it though.  Maybe you could make it such that you can type your mood in and choose one of the emoticons.

I know the admins are racked with work.  This isn't of very high importance.

We'll put it on the list.  If someone wanted to put together a list of moods and associated smileys, I think we could add them to the list.

(I agree, can't ever find just the right one.)

I just started out and put all the emoticons and started putting down moods related with them...  We probably don't need to have all the emoticons but here's what I have anyway:

Hmm, the smiley limit might make this hard...

Happy :)
Sad  :(
Confused :?
Smirky :D
 :lol:
 ;)
Embarrassed (1) :blush:
 :P
Shocked :o
Mad :mad:
Embarrassed (2) :oops:
Adamant :!:
Puzzled :?:
Innovative :idea:
Neutral :|
Laughing Uncontrollably :rotfl:
Baffled :huh:
 :roll:
Furtive :ninja:
Solemn :no:
In Love :love:
Disoriented :wacko:
Dumbfounded :what?:
Alienated :alien:
Cool :cool:
Triumphant :first:
Dry :dry:
Amused :laugh:
 :coolspeak:
Regretful :oops_sign:
 :whistling:
  :yinyang:
 :w00t:
 :pilot:
 :play_ball:
Judicious :police:
 :read:
 :showoff:
 :sleep2:
 :sleeping:
 :spam:
 :spidy:
 :starwars:
 :stink:
Crushed :stretcher:
 :cleaning:
 :surf:
Submissive :surrender:
 :thumbup:
 :trampoline:
 :spider:
 :w00tb:
Fed up :wallbash:
Really fed up :wallbash_red:
 :weightlift:
 :welcome:
 :welcomeani:
 :winner_first:
 :winner_second:
 :winner_third:
 :wow:  :huuh:
 :yankchain:
 :yup:  :10:
 :agent:
 :bomb:
 :bruce:
 :bye:
 :censored:
 :chief:
 :clap:
 :clap2:
 :coool:
 :cursing:
 :ddr:
Devilish :diablo:
 :evilgrin:
 :ewpu:
Strong :flex:
Stupid :fool:
Football :football:
Furious :furious:
Gregarious :gathering:
 :gleam:
 :harhar:
Distraught :help:
 :icecream:
 :juggle:
 :jump:
 :moose:
 :nhl:
 :noo:
 :omighty:
Very Wise :yoda:

The last post didn't have moods attached with each one...here is one with just the ones I have moods for already:

Happy :)
Sad  :(
Confused :?
Smirky :D
Embarrassed (1) :blush:
Shocked :o
Mad :mad:
Embarrassed (2) :oops:
Adamant :!:
Puzzled :?:
Innovative :idea:
Neutral :|
Laughing Uncontrollably :rotfl:
Baffled :huh:
Furtive :ninja:
Solemn :no:
In Love :love:
Disoriented :wacko:
Dumbfounded :what?:
Alienated :alien:
Cool :cool:
Triumphant :first:
Dry :dry:
Amused :laugh:
Regretful :oops_sign:
Judicious :police:
Crushed :stretcher:
Submissive :surrender:
Fed up :wallbash:
Really fed up :wallbash_red:
Devilish :diablo:
Strong :flex:
Stupid :fool:
Football :football:
Furious :furious:
Gregarious :gathering:
Distraught :help:
Very Wise :yoda:

Or perhaps if there was a blank space to type in your mood, and you could choose an appropriate smiley if you wanted.

Keep the ideas coming guys, because soon there will be a major upgrade on the whole site :)


Calculate this product: $\prod_{k=0}^{n}\cos(2^{k}\alpha)$

[quote="aviateurpilot"]$=\frac{sin(2^{n+1}\alpha)}{2^{n+1}sin(\alpha)}$[/quote]
Can u show me the solution please? thx a lot!


[quote="Allnames"]Find all positive integers $ m$ satisfying the property $ m^2 \minus{} 1$ is a divisor of $ 3^m \plus{} 5^m$[/quote]
No ideas!  :maybe: . It is really nice and not easy!. My result is $ m\equal{}3$.I am waiting

[quote="TTsphn"]See here ;)
http://www.mathlinks.ro/viewtopic.php?t=275139[/quote]
In the fact, This problem should be due to Gabriel Dospinescu. he proposed " Find all pairs of positive integers $ (m;n)$ such that : $ m^2\minus{}1| 3^,\plus{}(n!\minus{}1)^m$". After some arguments I got $ n\equal{}3$ and the equation above. In your link, maybe there was no right solution  :maybe: .

It's some mistakes but I can't press " Edit".
We can easy see that $ 3|m; m$is odd. Setting $ m\equal{}3k. ( k\in N*).$
Suppose$ k>1.$
If $ 5|m$ then $ m^2\minus{}1$ will be the form of $ 15r\minus{}1$. Because $ \left( {\frac{{ \minus{} 15}}{{15r \minus{} 1}}} \right) \equal{} \left( {\frac{{ \minus{} 1}}{{15r \minus{} 1}}} \right)^2 .\left( {\frac{{ \minus{} 1}}{3}} \right).\left( {\frac{{ \minus{} 1}}{5}} \right) \equal{}  \minus{} 1$, there exist
$ p\in P; p|m^2\minus{}1 st (\minus{}1/p)\equal{}\minus{}1$. That gives contradiction.
If $ m \equiv 2[5] \Rightarrow m \equal{} 15r \minus{} 3\equal{}>m\minus{}1\equal{}15r\minus{}4$. From $ \left( {\frac{{ \minus{} 15}}{{15r \minus{} 4}}} \right) \equal{} \left( {\frac{{ \minus{} 1}}{{15r \minus{} 4}}} \right)^2 .\left( {\frac{{ \minus{} 4}}{3}} \right).\left( {\frac{{ \minus{} 4}}{5}} \right) \equal{}  \minus{} 1$. That gives contradiction.
If$ m \equiv 3[5] \Rightarrow m \equal{} 15r \plus{} 3\equal{}>m\minus{}1\equal{}15r\plus{}2.$ But $ \left( {\frac{{ \minus{} 15}}{{15r \plus{} 2}}} \right) \equal{} \left( {\frac{{ \minus{} 1}}{{15r \plus{} 2}}} \right)^2 .\left( {\frac{2}{3}} \right).\left( {\frac{2}{5}} \right) \equal{}  \minus{} 1.$
Thus $ k\equal{}1.$


algebra
polynomial
number theory proposed
number theory

$ f(x)$ is an polynomial with integer coefficient and is not constant .
a) Prove that for any natural number n  exist infinite $ (x,y)\in \mathbb{N}$ such that :
\[ f(x\plus{}j)|f(y\plus{}j),
\forall j\equal{}1,..,n
\]
b) Suppose $ f(x\plus{}n)|f(y\plus{}n),\forall n\in \mathbb{N}$
Prove that $ x\equal{}y$

Hmm, a bit easy, isn't it? The first one is a joke: just take $ x\equal{}y$. If you still insist taking $ x\ne y$, take any $ y$ which is congruent to $ x$ modulo the product of all $ f(x\plus{}j)$ for $ j\equal{}1,2,...,n$. You get for every $ x$ (actually, for those such that the previous product is nonzero...) infinitely many $ y$. The second one: the quotient of the two numbers, considered as a sequence in $ n$ converges to $ 1$ and so it becomes constant equal to $ 1$ from a certain point. Thus $ f(t\plus{}x)\equal{}f(t\plus{}y)$ and now everything is obvious, since $ f$ would be periodic if $ x\ne y$. Of course, one can argue without analysis, but I like this argument.


This problem is very intelligent:

Solve the ES: $ \left\{\begin{array}{l} 2x^2 \plus{} y^2 \minus{} 8x \minus{} 4y \plus{} 12 \equal{} 0 \\
x \plus{} y \equal{} 4 \\
\end{array} \right.$

Oh, no! I think that isn't intelligent! :D :D

[quote="isabella2296"]We can write $ x$ as $ 4 \minus{} y$. Then we have:

$ 2(4 \minus{} y)^2 \plus{} y^2 \minus{} 8(4 \minus{} y) \plus{} 12 \equal{} 0$

$ 2(16 \minus{} 8y \plus{} y^2) \plus{} y^2 \minus{} 32 \plus{} 8y \plus{} 12 \equal{} 0$

$ 32 \minus{} 16y \plus{} 2y^2 \plus{} y^2 \minus{} 10 \plus{} 8y \equal{} 0$

$ 22 \minus{} 8y \plus{} 3y^2 \equal{} 0$

$ 3y^2 \minus{} 8y \plus{} 22$ doesn't have any real roots.[/quote]

That's wrong solution, do you find the error?

oh
i see
izzy forgot the $ \minus{} 4y$ part near the end

so, if we add that in to the equation, we get $ 3y^2 \minus{} 12y \plus{} 22$ right?
but that doesn't work in the quadratic equation either

edit: gasp! also, $ \minus{} 32 \plus{} 12 \equal{} 20$, not $ 10$
so, solving now, we have $ 3y^2 \minus{} 12y \plus{} 12$

therefore, we have our answer x=2, y=2
right?

[quote]oh
i see
izzy forgot the $ \minus{} 4y$ part near the end[/quote]

Yes, yes!

[quote]so, if we add that in to the equation, we get $ 3y^2 \minus{} 12y \plus{} 22$ right?
but that doesn't work in the quadratic equation either[/quote]

No no, that's wrong?  :D  :D  Do you find the error?

ya, i know
that's what i said in my edit
after i saw that $ \minus{}32\plus{}12\equal{}20$ :D

[quote="vallon22"]ya, i know
that's what i said in my edit
after i saw that $ \minus{} 32 \plus{} 12 \equal{} 20$ :D[/quote]
 :D  :D

But I have a great solution, do you find it?

hello, you can write your first equation in the form $ 2(x\minus{}2)^2\plus{}(y\minus{}2)^2\equal{}0$.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, you can write your first equation in the form $ 2(x \minus{} 2)^2 \plus{} (y \minus{} 2)^2 \equal{} 0$.
Sonnhard.[/quote]

Oh, Yes Yes, Very Good! That is my solution as I want you to find it! :)  :)  :wink:

Oh wow, I was thinking combined factoring. :P

We rewrite as $ 2(x \minus{} 2)^2 \plus{} (y \minus{} 2)^2\equal{}0$, or $ 2(x \minus{} 2)^2 \equal{} \minus{} (y \minus{} 2)^2$. Because perfect squares can only be non-negative, both sides can ONLY equal zero otherwise there is no way for one side to be the negative of the other. so we have $ x \minus{} 2 \equal{} 0$, and $ y \minus{} 2 \equal{} 0$. So, $ x \equal{} 2$, and $ y \equal{} 2$.

No need for the second equation.  :lol:

Check: 2+2=4. It works  :D  :D

[quote="Nguyen Ngoc Linh"]He teased us. hey Euler, what did u mean by 'intelligent' ?.
You think we are silly, huh?
(vietnamese people are sly)
Mathlinker, this equation can be solved by almost Vietnamese student.[/quote]

No no, I don't have that idea, you can't speak that! :(  :(


invariant
combinatorics unsolved
combinatorics

Given a $ 2007$x$ 2007$ square. Take out 1 cell of the square. Can you cover all the entire cells of the square by pieces with L shape??

you can colour  the columns ,alternatively with black and white ,hence we have $ 1004$ black columns and $ 1003$ white columns. Consider the number of black L shape (which has three black cells) be $ x$ and the number of white L shape be $ y$ then $ 3x\plus{}y\equal{}x\plus{}3y\equal{}\frac{(2007)^{2}\minus{}1}{2}\Longrightarrow x\equal{}y$ and it's contradiction.

also the following related problem holds which you can prove it using induction...
[hide="related problem"]
given a $ 2^{n}\times 2^{n}$ square,Take out 1 cell of the square. Can you cover all the entire cells of the square by pieces with L shape?[/hide]

An other problem(Japan....)
Let regular m*n cut off 4 corner .Find all (m,n) such that can cover this regular use L shape.


Find the greatest $N$ for which $12^{N}$ evenly divides $20!$.

Find the greatest $N$ for which $2^{n}$ evenly divides $100!$.

For the second question, it is basically asking how many factors of 2 there are in the product $100\cdot 99\cdot 98\cdot\ldots\cdot 1$.

The answer is:
 [hide]
$\left\lfloor\frac{100}{2}\right\rfloor+\left\lfloor\frac{100}{4}\right\rfloor+\left\lfloor\frac{100}{8}\right\rfloor+\left\lfloor\frac{100}{16}\right\rfloor+\left\lfloor\frac{100}{32}\right\rfloor+\left\lfloor\frac{100}{64}\right\rfloor=\boxed{97}$.[/hide]

[hide]You are finding how many 12s are in 20!. 12=2*2*3, and there are more 2s than 3s so just look for 3s.
20/3=6
20/9=2

So N=$8$[/hide]


geometry
geometric transformation
homothety
power of a point
radical axis
geometry solved

i have reduced a problem to the following one:

we have two circles $C_1$ and $C_2$ of centers $O_1$ and $O_2$ We draw a common tangnet of the circles its intersections are $B_1$ and $B_2$ the line $0_1$ and $0_2$ intersects the circle in $A_1$ and $A_2$ Lines intersect $A_2B_1$ and $A_1B_2$ INTERSECT IN c  PROVE THAT C IS ON THE RADICAL AXIS

Am I the only one who doesn't understand this? :? Could ou please rephrase it very carefully? :) For instance, what does the following mean:

"the line $0_1$ and $0_2$ intersects the circle in $A_1$ and $A_2$"? You have two circles, and the line $O_1O_2$ (if this is indeed the line you're referring to) cuts each one of these circles in two points, so I think you can see why the whole thing is very confusing.

sorry grobber: yes im refering to the line $O_1O_2$ moreover this piece of segment only intersects the circles in two points (one each) however the extension of this lines intersects the circles each in one more point  :P  is the problem understandable now?? :lol:

Okay, I got it now:

[color=blue][b]Problem.[/b] Given two circles $C_1$ and $C_2$ with centers $O_1$ and $O_2$. Let a common tangent of these circles $C_1$ and $C_2$ touch these circles at the points $B_1$ and $B_2$, and let the segment $O_1O_2$ intersect the circles $C_1$ and $C_2$ at the points $A_2$ and $A_1$, respectively. Let the lines $A_2B_1$ and $A_1B_2$ intersect at a point C. Prove that the point C lies on the radical axis of the two circles $C_1$ and $C_2$.[/color]

[i]Solution.[/i] Since the line $B_1B_2$ touches the circles $C_1$ and $C_2$ at the points $B_1$ and $B_2$, we have $O_1B_1\perp B_1B_2$ and $O_2B_2\perp B_1B_2$, so that $\measuredangle O_1B_1B_2=90^{\circ}$ and $\measuredangle O_2B_2B_1=90^{\circ}$. Hence, by the sum of angles in the quadilateral $O_1B_1B_2O_2$, we have

$\measuredangle O_1O_2B_2=360^{\circ}-\measuredangle O_2B_2B_1-\measuredangle O_1B_1B_2-\measuredangle O_2O_1B_1$
$=360^{\circ}-90^{\circ}-90^{\circ}-\measuredangle O_2O_1B_1=180^{\circ}-\measuredangle O_2O_1B_1$.

But since the points $A_2$ and $B_1$ both lie on the circle $C_1$, whose center is $O_1$, we have $O_1A_2=O_1B_1$, so that the triangle $A_2O_1B_1$ is isosceles, and thus its base angle equals

$\measuredangle O_1A_2B_1=\frac{180^{\circ}-\measuredangle A_2O_1B_1}{2}=\frac{180^{\circ}-\measuredangle O_2O_1B_1}{2}=\frac{\measuredangle O_1O_2B_2}{2}$.

In other words, $\measuredangle A_1A_2B_1=\frac{\measuredangle O_1O_2B_2}{2}$.

Similarly to how we showed that the triangle $A_2O_1B_1$ is isosceles, we see that the triangle $A_1O_2B_2$ is isosceles, and thus its base angle equals

$\measuredangle A_1B_2O_2=\frac{180^{\circ}-\measuredangle A_1O_2B_2}{2}=\frac{180^{\circ}-\measuredangle O_1O_2B_2}{2}$.

Thus,

$\measuredangle A_1B_2B_1=\measuredangle O_2B_2B_1-\measuredangle A_1B_2O_2$
$=90^{\circ}-\frac{180^{\circ}-\measuredangle O_1O_2B_2}{2}=\frac{\measuredangle O_1O_2B_2}{2}=\measuredangle A_1A_2B_1$.

Hence, the points $A_1$, $A_2$, $B_1$, $B_2$ lie on one circle. Thus, by the intersecting chords theorem, $CA_2\cdot CB_1=CA_1\cdot CB_2$. But $CA_2\cdot CB_1$ is the power of the point C with respect to the circle $C_1$, and $CA_1\cdot CB_2$ is the power of the point C with respect to the circle $C_2$. Thus, the point C has equal powers with respect to the circles $C_1$ and $C_2$, and hence it lies on the radical axis of these two circles. Qed..

  Darij

darij : your solution is great¡¡ apart from the typo in paragraph five . also did you change the wording of my problem??
because i think you proved the result from another C? wich leads me to think my original problem was wrong?

[quote="marko avila"]darij : your solution is great¡¡ apart from the typo in paragraph five .[/quote]

Which typo? I don't see anything wrong...

[quote="marko avila"]also did you change the wording of my problem??[/quote]

Your wording allowed two possible interpretations (depending on the order of the points $A_1$ and $A_2$ - which one lies on $C_1$ and which on $C_2$?), but one of them led to a wrong problem, so I only considered the other one. Did you mean something completely different?

  Darij

[quote="darij grinberg"]

Which typo? I don't see anything wrong...[/quote]

I think he means $\measuredangle A_1B_2B_1 = 180^{\circ} - \measuredangle A_1A_2B_1$

[quote="marko avila"]also did you change the wording of my problem??[/quote]

Well, can you post the original problem (before you reduce) so that we can understand it  :)

[quote="warut_suk"]I think he means $\measuredangle A_1B_2B_1 = 180^{\circ} - \measuredangle A_1A_2B_1$[/quote]

Ah, this actually depends on the arrangement of the points $O_1$, $O_2$, $A_1$, $A_2$ on their common line. I assumed that these points lie there in the order $O_1\to A_1\to A_2\to O_2$; your equation holds when the points lie in the order $O_1\to A_2\to A_1\to O_2$.

  Darij

You say: in other words $2$$A_2B_1=$O_1O_2B_2$ IS THIS TRUE??  sorry if i missed it.yes i think you proved for another C but i will check my problem and see if its correct . anyway thanks darij

I think you have interchanged the points $A_1$ and $A_2$. Please, draw a sketch where the points are named according to my definitions:

[quote="darij grinberg"][color=blue][b]Problem.[/b] Given two circles $C_1$ and $C_2$ with centers $O_1$ and $O_2$. Let a common tangent of these circles $C_1$ and $C_2$ touch these circles at the points $B_1$ and $B_2$, and let the segment $O_1O_2$ intersect the circles $C_1$ and $C_2$ at the points $A_2$ and $A_1$, respectively. Let the lines $A_2B_1$ and $A_1B_2$ intersect at a point C. Prove that the point C lies on the radical axis of the two circles $C_1$ and $C_2$.[/color][/quote]

  Darij

OK thanks.

So when writing a solution to this, do I have to write both arrangements?

With Darij's notation, with $A_2$ on $C_1$, $A_1$ on $C_2$:
We have $B_1A_2$ and $A_1B_2$ are antiparallel in the quadrilateral $B_1A_2A_1B_2$. Due to the homothety between the circles $O_1$ and $O_2$, It follows from:The line $O_2O_1$ cut $O_1,O_2$ at $A'_2,A'_1$, now $A'_2B_1//A_1B_2$ and $A'_1B_2//A_2B_1$, now $<A_1A'_1B_2=<A_1B_2B_1=<A'_2A_2B_1$ so $A_2A_1B_2B_1$ is cyclic.Hence $CA_2.CB_1=CA_1.CB_2$, i.e. $C$ is on the radical axis of $O_1$ and $O_2$.

darij: i get it now. ohh¡¡ see you didn´t consider the case when the circles dont intersect

I think that should be the same as the case considered, or isn't it?

no have you seen the figure?' no i dont think so


number theory
Diophantine equation
number theory unsolved

Is it possible for a perfect square (which is not a power of $10$) to contain only $0$s and $1$s in its decimal representation ?

I think someone asked this question on the forum before [Fedor Petrov?], and no one managed to give an answer. (But I may be wrong, I couldn't find anything in the search). I'd be very surprised if there isn't such perfect square. In my opinion, the answer to this question should be positive, or else the solution should be (at least, from scratch) extremely difficult (and nonelementary). What I think I can show is that, for every $n > 1$, the equation $y^{2}= 10^{k_{1}}+\cdots+10^{k_{n}}$ with $0 \leq k_{1}< \cdots < k_{n}$ and $k_{1}\in \{0,1\}$ has only finitely many solutions. However, my proof uses Schmidt's subspace theorem (the multidimensional generalization of Roth's theorem on diophantine approximations), and hence doesn't produce any bound on such solution $(y; k_{1},\ldots,k_{n})$. So even the following special case is open to me:

Does the equation $y^{2}= 1+10^{a}+10^{b}+10^{c}$ have any solution in nonnegative integers $y; a < b < c$?

As I said, I know that there are only finitely many potential solutions, but without any bound on such potential solution. So let's try this problem first; for $n \leq 3$ it is easy to see that there are no solutions.

I'm not sure if this is right so if there is an error in this proof someone should point it out  :D

$y^{2}=\sum_{i}10^{k_{i}}$ where $k_{1}<k_{2}<...<k_{i}$
assume there are more than 1 term otherwise we are done and there exists solutions of the forum $10^{2*a}$
we then want to factor out $10^{k_{1}}$ so
$y^{2}=10^{k_{1}}\sum_{i}10^{k_{i}-k_{1}}$
We should then divide this into 2 cases
Case 1:$k_{1}$ is odd
This implies that $10^{k_{1}}|y^{2}$ so then we can write $y^{2}=10^{k_{1}+1}s^{2}$ which means that $10s^{2}=1+\sum_{i!=1}10^{k_{i}-k_{1}}$
so that implies that 10 divides the RHS but 10 does not divide the right hand side so its impossible
Case 2:$k_{1}$ is even
This implies that $10^{k_{1}}|y^{2}$ so we can set $y=10^{k_{1}/2}*s$
so we just need to solve the equation $s^{2}=1+\sum_{i!=1}10^{k_{i}-k_{1}}$
now factor $(s+1)(s-1)=\sum_{i!=1}10^{k_{i}-k_{1}}$
now pull out the lowest term $(s+1)(s-1)=10^{k_{2}-k_{1}\sum_{i!=1,2}10^{k_{i}-k_{2}}}$
Now $Gcd(10^{k_{2}-k_{1}},\sum_{i!=1}10^{k_{i}-k_{2}})=1$
so then we can set $s-1=10^{k_{2}-k_{1}}$ and $s+1=\sum_{i!=1}10^{k_{i}-k_{2}}$
This implies that $2=-10^{k_{2}-k_{1}}+\sum_{i!=1,2}10^{k_{i}-k_{2}}$ which is obviously impossible. Now you can set $s+1=10^{k_{2}-k_{1}}$ and $s-1=\sum_{i!=1,2}10^{k_{i}-k_{2}}$ and you can go through the same steps and find that you can't satisfy the conditions.
Hence there are no solutions except the trivial ones $10^{2*a}$

[quote="bishnu13"]$(s+1)(s-1)=10^{k_{2}-k_{1}\sum_{i!=1,2}10^{k_{i}-k_{2}}}$
Now $Gcd(10^{k_{2}-k_{1}},\sum_{i!=1}10^{k_{i}-k_{2}})=1$
so then we can set $s-1=10^{k_{2}-k_{1}}$ and $s+1=\sum_{i!=1}10^{k_{i}-k_{2}}$
This implies that $2=-10^{k_{2}-k_{1}}+\sum_{i!=1,2}10^{k_{i}-k_{2}}$ which is obviously impossible. Now you can set $s+1=10^{k_{2}-k_{1}}$ and $s-1=\sum_{i!=1,2}10^{k_{i}-k_{2}}$ and you can go through the same steps and find that you can't satisfy the conditions.
Hence there are no solutions except the trivial ones $10^{2*a}$[/quote]
This is wrong :P

[quote="ali666"][quote="bishnu13"]so we just need  to solve the equation $s^{2}=1+\sum_{i!=1}10^{k_{i}-k_{1}}$
[/quote]note that:$S^{2}=0,1\mod{4}$and$1+\sum_{i!=1}10^{k_{i}-k_{1}}=3\mod{4}$[/quote]
??...

For bishnu13:
$gcd(a,b)=gcd(c,d)=1$ and ab=cd doesn't imply a=c b=d or a=d b=c
for example 2*3=6*1  :P

[quote="Albanian Eagle"][quote="ali666"][quote="bishnu13"]so we just need  to solve the equation $s^{2}=1+\sum_{i!=1}10^{k_{i}-k_{1}}$
[/quote]note that:$S^{2}=0,1\mod{4}$and$1+\sum_{i!=1}10^{k_{i}-k_{1}}=3\mod{4}$[/quote]
??...
[/quote]
sorry :blush:

Actually, I was wrong  :blush:  that the equation $y^{2}= 10^{a}+10^{b}+10^{c}$ is easy to solve (but, yes, my proof that it had at most finitely many solution was correct).

For such questions, see

Corvaja, Zannier: On the diophantine equation $f(a^{m},y) = b^{n}$, in Acta Arithm. [b] 94[/b] (2000), 25--40.

You will quickly find that this question is much too deep for an elementary discussion. Even the equation $y^{2}= 1+10^{a}+10^{b}$ is extremely difficult.

It is in general believed that, for a fixed $n$, an equation $x^{2}= u_{1}+\cdots+u_{n}$ has "essentially" (in a sense that can be made precise) finitely many solutions in $S$-units $u_{i}$. And indeed this can be shown in certain cases, like the present $x^{2}= 10^{k_{1}}+\cdots+10^{k_{n}}$. But in general, it is already an open problem to prove the finiteness of solutions to $x^{2}= 1+2^{a}+3^{b}$!

[quote="vess"]Actually, I was wrong  :blush:  that the equation $y^{2}= 10^{a}+10^{b}+10^{c}$ is easy to solve [/quote]

 :laugh: But of course it's obvious, modulo $9$, that this particular equation doesn't have a solution...

As far as I know $x^{2}=1+t^{a}+t^{b}$ is an open problem for most $t$
2 included.

Actually, the equation $x^{2}= 1+2^{a}+2^{b}$ has been completely solved by L. Szalay in:

L. Szalay: [i]The equations $2^{m}\pm 2^{n}\pm 2^{l}= z^{2}$, Indag Math.,[/i] [b]13[/b] (2002), pp. 131--142.

2002?

well sorry my memory was apparently referring to
http://lsze.cosam.calpoly.edu/wcntc/problems2002.pdf
 :oops:

but thank you cuz now I found
http://arxiv.org/PS_cache/math/pdf/0608/0608796v1.pdf
which is quite nice.. :)


Σε μια καλοκαιρινη συζητηση με καλη παρεα σε μερος δροσερο ( 950 μ υψομετρο) ρωτηθηκα απο καποιον της παρεας για αν μπορει καποιος  να αντιμετωπισει αξιοπρεπως  θεματα των Ολυμπιαδων χωρις ιδιαιτερες γνωσεις μαθηματικων.Η απαντηση μου ηταν ,χωρις να το σκεφθω ιδιαιτερα, δεν εχει τυχη .Η επομενη ερωτηση ηταν μπορεις να μας πεις μια ασκηση επιπεδου Ολυμπιαδας και οχι κατα προτιμηση Combinatorics (δεν το προφερε ακριβως ετσι γιατι δεν ηξερε την ορολογια αλλα καταλαβα οτι δεν ηθελε αυτου του τυπου τα προβληματα) που να μπορει να λυθει με απλες γνωσεις μαθηματικων;
Θυμηθηκα τοτε ενα παλιο και πολυ ομορφο θεμα Ολυμπιαδας του οποιου η λυση ειναι στοιχειωδης και κατανοητη απο πολλους που δεν εχουν ιδιαιτερη σχεση με το αντικειμενο η οποια και αρεσε  στους παρεβρισκομενους και για αυτο την θυμιζω και στο forum
[b]Να βρεθει το μεγιστο γινομενο των θετικων ακεραιων που εχουν αθροισμα $ 2008$[/b]

Το πρόβλημα δεν είναι δύσκολο. Όση δυσκολία αντιμετώπισα ήταν καθαρά πάνω στην προσπάθεια να το λύσω σε μια πιο γενική μορφή πχ να το περιγράψω μαθηματικά και να το λύσω με κάποια μέθοδο βελτιστοποίησης (πχ παραγώγους). Αρχικά πήγαινα να το λύσω για πραγματικούς, αλλά τελικά φαίνεται πως η λύση για ακεραίους είναι τελείως διαφορετική.

Με απλή ψηλάφηση για μικρά νούμερα φαίνεται πως ο τρόπος γραφής του αριθμού με σταθερό άθροισμα που μεγιστοποιεί το γινόμενο αποτελείται μόνο από 2αρια και 3αρια.

Αυτό αποδεικνύεται πολύ εύκολα:
Το 1 δεν παίζει καθόλου γιατί αυξάνει το άθροισμα χωρίς να αυξάνει το γινόμενο. Οι αριθμοί από το 4 και πάνω (έστω κ) μπορούν να σπάσουν σε 2 + (κ-2) έτσι ώστε το άθροισμα να παραμείνει σταθερό αλλά το γινόμενο να αυξηθεί. Ακραία περίπτωση είναι το 4 που σπάει σε 2+2 που αφήνει άθροισμα και γινόμενο ίδια. Η συνταγή σταματάει στο 4. Πιο κάτω αυτό δεν ισχύει πια. Άρα οι παράγοντες είναι μόνο 2 και 3.

Το θέμα είναι να δούμε πόσα 3αρια και πόσα 2άρια θα έχουμε. Ισχύει $ 2\plus{}2\plus{}2\equal{}3\plus{}3$ και $ 2*2*2<3*3$, άρα για κάθε 3 δυάρια προτιμούμε τα 2 τριάρια. Έτσι συμφέρει να γεμίσουμε με 3αρια και ότι περισσέψει για να βγει το άθροισμα να γίνει με 2άρια. Εξαίρεση είναι αν υπολείπεται 1 οπότε μας συμφέρει το 1 μαζί με ένα 3άρι να γίνουν δύο 2άρια (ίδιο άθροισμα αλλά μεγαλύτερο γινόμενο).

Τελικά το 2008 (που θα μπορούσε να είναι κάθε αριθμός αλλά φαντάζομαι ότι σχετίζεται με το τρέχον έτος) γράφεται σαν 3+3+3+3+3… +2+2.
Δηλαδή μέχρι το 2004 θα πάμε με 3άρια και μετά θα έχουμε δύο 2άρια στο τέλος. Έχουμε 668 τριάρια και 2 διάρια. Άρα το άθροισμα είναι 668 * 3 + 2 * 2= 2008.
Το μέγιστο γινόμενο είναι $ 3^{668}*2^2$

Ελπίζω να μην ξέφυγε κάτι.

Δεν ξέρω αν έχει νόημα να ρωτήσουμε πως λύνεται αυτό το πρόβλημα στους πραγματικούς. Αξίζει ή θα μπλέξουμε με κάτι πολύ δύσκολο;

Πολυ ωραια αυτη ακριβως ειναι η λυση
Το ερωτημα σου τι θα γινοταν στους πραγματικους εχει ενδιαφερον

Λοιπον απο $ AM\minus{}GM$ εχουμε $ \prod_{i\equal{}1}^n{xi}\leq(\frac{2008}{n})^n$ οπου $ n$ το πληθος των προσθετεων .Η παρασταση $ (\frac{2008}{n})^n$ ευκολα με παραγωγιση γινεται μεγιστη για $ n\equal{}739$ (o πλησιεστερος ακεραιος στον $ 2008/e$ που δινει μεγαλυτερη τιμη στο$ (\frac{2008}{n})^n$ ) δηλαδη οταν $ x_1\equal{}x_2\equal{}.......\equal{}x_{739}\equal{}\frac{2008}{739}$ περιπου $ e$ kai to μεγιστο ειναι το $ (\frac{2008}{739})^{739}$ και αν δεν εκανα λαθος υπολογισμο ειναι  μεγαλυτερο απ το προηγουμενο μεγιστο


calculus
integration
calculus computations

hello, your first integral can't expressed by the known elementary functions.
Sonnhard.

hello, you can convert the integrand of the second integral in the following form
$ {\frac {6\,{x}^{3}\plus{}18\,{x}^{2}\minus{}30\,x\plus{}42}{{x}^{4}\plus{}4\,{x}^{3}\minus{}4\,{x}^{2}
\plus{}4\,x\plus{}12}}\minus{}3\, \left( x\plus{}1 \right) ^{\minus{}1}$
Sonnhard.


please say if you're going to do the hartford chapter for mcounts
edit:or if you're in new haven
------------
grade?
proj. rank
school

me:
8
top 8
smith middle school

That's not the Hartford Chapter is it then?

i heard u 2 made great friends with dan tobias!
at nats  :lol:
-jorian

great, i'll love to c you guys at states

our school district makes it so we don't learn mathcounts math fully until 8th grade

last year i knew alot but stunk up in geometry so this year im trying to cram everything from these math sites until february

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

i really do think i'm 99% going to states and 20% going to nats
-jorian

(will u guys help me?)

try refreshing once u see this page
-jorian

as in the hartford chapter for mathcounts
-jorian

[quote="now a ranger"]hartford as in what?[/quote]

As in this is why this is a local forum as most people know what the capital of their own state is.


[img]http://img.photobucket.com/albums/v194/mukmasterxps/formula.gif[/img]

 :maybe:

Cool.

Can it be done without induction (derived)?


I need to learn how to construct a graph using the equilateral method.
i am not given any mesurements. and i dont have any nubers to plug into any equations.
aparently this is possible so please, please help me!

PLEASE PLEASE.
if you know anything please help.
its so fusterating.

You don't need to post the same thing 3 times.


Putnam
probability
calculus
function
integration
geometry
geometric transformation

Let $n$ be given, $n\ge 4,$ and suppose that $P_1,P_2,\dots,P_n$ are $n$ randomly, independently and uniformly, chosen points on a circle. Consider the convex $n$-gon whose vertices are the $P_i.$ What is the probability that at least one of the vertex angles of this polygon is acute.?

It is a piece of cake to see that the problem is equivalent to finding the probabilty that at least one of
the xi's (n>=i>=1) will exceed one in the following:
x1+x2+...+xn =4.

No, it isn't. That's not a trivial problem, and it's not equivalent either. You have to pay attention to what the probability distributions are, and the interesting ones are neither uniform nor independent.

I'm writing up my solution now, and will probably post it tomorrow.
It's a very nice problem; at least the calculus is quite a bit easier than A5.

The [hide="answer"] $\frac{n(n-2)}{2^{n-1}}$[/hide]

Are those exterior angles? How do you account for the fact that $x_1+x_3=2$ and $x_2+x_4=2$ when $n=4$? It might be the same problem, if you set up the right probability distributions. Without details, I just don't believe you.

okay ...
just to convince you...

take angles Pi O Pi+1 = ai

then Pi-1 Pi Pi+1 is equal to pi - 1/2( ai-1 +ai)
so the angle Pi-1 Pi Pi+1 is obtuse iff
ai-1 + ai < pi

Now let yi = ai + ai-1 for n>=i>=1

Then sum yi = 2 sum ai = 4 pi
and we want to find the probability that all yi's are less than pi.

define xi = yi/ pi the result follows.

You've just redefined into a probability distribution that's much harder to understand than the original. That's useless- see my earlier posts. We need to understand the correlation between adjacent angles, not sweep that under the rug.

Here's a real solution:

Notation: $P()$ is a probability, $\rho()$ is a probability density]

This uses one key fact repeatedly; if $m$ random variables are independently and uniformly distributed on the interval $[0,a]$, the density function for the smallest of the $m$ random variables at $x$ is $m\frac{(1-x)^{m-1}}{a^m}.$

Let $O$ be the center of the circle, and renumber $P_1,P_2,\dots,P_n$ to be in clockwise order. Let $\angle P_1OP_2=2\pi\alpha,\angle P_2OP_3=2\pi\beta,\angle P_3OP_4=2\pi\gamma$. $\angle P_1P_2P_3$ is acute when $\alpha+\beta>\frac12,$ and $P_2P_3P_4$ is acute when $\beta+\gamma>\frac12$. Note that if two vertex angles are both acute, they must be adjacent. Consequently, there are at most two acute angles, and these two are adjacent if they exist. The total probability that some vertex angle is acute is then $P(P_1P_2P_3\text{ is acute})+P(P_2P_3P_4\text{ is acute})+\cdots-P(P_1P_2P_3\text{ and }P_2P_3P_4\text{ are acute})+\cdots$
$=nP(P_1P_2P_3\text{ is acute})-nP(P_1P_2P_3\text{ and }P_2P_3P_4\text{ are acute}).$

Now, we calculate these probabilities.
$\rho(\alpha=x)=(n-1)(1-x)^{n-2}$ since we can treat $P_1$ as fixed and let the other $n-1$ points vary freely. Similarly, $\rho(\beta=y|\alpha=x)=(n-2)\frac{(1-x-y)^{n-3}}{(1-x)^{n-2}}$. Therefore,
$P(\alpha+\beta>\frac12)$
$=\int_{\frac12}^1(n-1)(1-x)^{n-2}\,dx+\int_0^{\frac12} \int_{\frac12-x}^{1-x}(n-1)(1-x)^{n-2}\cdot(n-2)\frac{(1-x-y)^{n-3}}{(1-x)^{n-2}}\,dy\,dx$
${=\left[-(1-x)^{n-1}\right]_{x=\frac12}}^{x=1}+ \int_0^{\frac12} \int_{\frac12-x}^{1-x}(n-1)(n-2)(1-x-y)^{n-3}\,dy\,dx$
$=\frac1{2^{n-1}}+\int_0^{\frac12}(n-1)\left[-(1-x-y)^{n-2}\right]_{y=\frac12-x}^{y=1-x}\,dx =\frac1{2^{n-1}}+\int_0^{\frac12}(n-1)\frac1{2^{n-2}}\,dx$
$P(\alpha+\beta>\frac12)=\frac1{2^{n-1}}+\frac{n-1}{2^{n-1}}=\frac{n}{2^{n-1}}$

Now we calculate the probability that two adjacent angles are both acute.
$\rho(\beta=y)=(n-1)(1-y)^{n-2},\rho(\alpha=x|\beta=y)=(n-2)\frac{(1-y-x)^{n-3}} {(1-y)^{n-2}},$ and $\rho(\gamma=z|\alpha=x,\beta=y)=(n-3)\frac{(1-y-x-z)^{n-4}} {(1-y-x)^{n-3}}$, so $\rho(\alpha=x,\beta=y,\gamma=z)=(n-1)(n-2)(n-3)(1-y-x-z)^{n-4}$ and
$P(\alpha+\beta>\frac12,\beta+\gamma>\frac12)=P(\beta>\frac12)+$
$\int_0^{\frac12}\int_{\frac12-y}^{\frac12}\int_{\frac12-y}^{1-y-x} (n-1)(n-2)(n-3)(1-y-x-z)^{n-4}\,dz\,dx\,dy$
$=\frac1{2^{n-1}}+\int_0^{\frac12}\int_{\frac12-y}^{\frac12} (n-1)(n-2)\left(\frac12-x\right)^{n-3}\,dx\,dy$
$=\frac1{2^{n-1}}+\int_0^{\frac12} (n-1)x^{n-2}\,dy$
$P(\alpha+\beta>\frac12,\beta+\gamma>\frac12)=\frac1{2^{n-1}}+\frac1{2^{n-1}} =\frac{2}{2^{n-1}}$.
In the above, note that we cannot have $\alpha>\frac12$ because then $\beta+\gamma$ would have to be less than $\frac12$.

The probability that there is some acute angle is now
$nP(\alpha+\beta>\frac12)-nP(\alpha+\beta>\frac12,\beta+\gamma>\frac12)= \frac{n^2}{2^{n-1}}-\frac{2n}{2^{n-1}}=\frac{n(n-2)}{2^{n-1}}$

I've just violated one of the great unwritten rules of problem solving- I wrote down multiple integrals, and I evaluated them in the same order I wrote them down!

Wow! I sort of gave up when I understood what exactly it was I needed to evaluate. I figured that a lot of integration was required, and I was right! Good thing I gave up on it. Now I'll proceed to spend the next two hours trying to understand this solution.
Or maybe I'll do my math homework instead. Or study for my language final.

It looks like a lot of integration, but they're all easy (unless you make a mistake); all the  work is in the setup.

If you know the general order statistics for the uniform distribution, you can do the first part directly as a single integral.

for this question, the important thing to note is that you cannot have 2 acute angles which are not adjacent to each other. (otherwise, they mark out >2 pi of the circumference)

then, the rest of the problem is much easier. There is no need to resort to integration.

So, how do you find the probabilities we need? Let's see your method.

I agree that the rest of the problem is easier after the key insight that at most two adjacent angles are acute, but the integration was the easiest way to find the probabilities of acute angles and acute angle pairs for me.
Those are easy integrals; they're only worth avoiding if your alternate method is nice.

The probability that two adjacent angles $P_1$ and $P_2$ are acute requires just one integral. Note that the "circular" distance between $P_1$ and $P_2$ is uniformly distributed on the interval $[0,\tfrac 12]$ (I assume that the length of the entire circle is $1$) and that, once $P_1$ and $P_2$ are fixed, the only situations in which both angles are acute are when all other points lie on the shorter arc between $P_1$ and $P_2$ or when they all lie on an arc symmetric to this arc with respect to the diameter parallel to $P_1P_2$. This gives the integral $2\cdot 2\int_0^{\frac 12}x^{n-2}\,dx=\frac 4{(n-1)2^{n-1}}$.
The evaluation of the probability that $P_1$ is acute requires no integration at all. Just denote by $C_+$ the half-circle with one endpoint at $P_1$ and note that the angle $P_1$ is acute if either all other $n-1$ points belong  to $C_+$ (which is an event of probability $\frac 1{2^{n-1}}$) or there is a point on the other half-circle $C_-$ closest to $P_1$ and all other $n-2$ points lie on the half-circle with one endpoint at that point not containing $P_1$. Since that closest point may be any of the remaining $(n-1)$ points, we get the probability $(n-1)\frac 12\frac 1{2^{n-2}}=\frac{n-1}{2^{n-1}}$ for this event. So, the total probability of the event that the angle $P_1$ is acute is $\frac n{2^{n-1}}$. Still, I suspect that this is not as different from jmerry's solution as it seems (perhaps, it is equivalent to evaluating the integrals in a different order :)). The final formula for the probability is almost the same as jmerry's: $\mathcal P=n\mathcal P\{P_1\text{ is acute}\}-\frac{n(n-1)}{2}\mathcal P\{P_1,P_2\text{ are acute}\}$, which, of course, results in the same $\mathcal P =\frac{n(n-2)}{2^{n-1}}$

heres a quick idea.

first, probability that 2 points are diameter of circle is 0, so we shall ignore such cases.

Next, consider the point P (-1,0) on the unit circle.
Consider (n-1) points S_2, S_3, .. S_n, that are on the top semicircle.
Reflect them abt the x-axis to get points R_2, R_3, ... R_n

Consider all polygons formed from point P, and one of the pair (S_i, R_i). There are 2^(n-1) such polygons.
We want to find the polygons, where P is the starting vertex of the actue angle taken anticlockwise.
All such polygons must have the form P R_i, S_n, S_(n-1) ... S_(either 2 or 3). There are (n-2) possibilities for i

Thus, probability that P is the starting acute vertex is (n-2)/2^(n-1)

Since P is any vertex, total probability is n(n-2)/2^(n-1)

hmz, having looked at this again, im not so sure that this is completely correct.
Din get a solution on putnam bacause i made a stupid carless mistake right at the start.

That last explanation doesn't quite work, because it doesn't handle the double-counting issue. We actually have $n$ of those polygons with an acute angle counterclockwise from $P$; this happens if either exactly 1 or exactly 0 of the $R_i$ are included.

Now, to handle double-counting: The angle at $P$ is also acute when we included none of the $R_i$, and is also acute whenever $R_i$ is closer to the antipode of $P$ than $S_2$ is to $P$ or when $i=2$ and $S_3$ is closer to the antipode of $P$ than $R_2$ is to $P$.

That's an expected value you need to evaluate, and it's much more complicated than the rest of the argument.

To fedja:
The first version I ran used the second order statistic to calculate the probability that a single angle is acute. I used an integral there, but it's easy to shortcut that; either exactly one or zero points are in that semicircle, and that probability is easy to calculate. The version using the second order statistic may be the only way to do this without splitting those two cases.
Your method does feel legitimately different than mine; note how I assumed order first to calculate everything.

A step-by step explanation of how I chose the limits for my integrals:
$P(\alpha+\beta>\frac12):$ Either $\alpha>\frac12$ or $\alpha<\frac12$. The first possibility gives the first integral; $\alpha$ is between $\frac12$ and $1$, and any value of $\beta$ works. The second gives the second integral; $0<\frac12<\alpha$, and we must have $\frac12-\alpha<\beta$ for the given condition. Since $\alpha+\beta<1$ always, we have the upper and lower limits.

$P(\alpha+\beta>\frac12,\beta+\gamma>\frac12):$ If $\beta>\frac12$, we are done already. This probability has already been calculated above. If $\beta<\frac12$, we have $\alpha>\frac12-\beta$. For the upper limit, note that $\beta+\gamma<1-\alpha$, so $\alpha<1-(\beta+\gamma)<\frac12$. On the inner integral, $\gamma>\frac12-\alpha$ to satisfy the condition, and $\gamma<1-\beta-\alpha$ to fit in the circle.

You do not need any integration to do this problem, there is a so-called "hidden independence" here, if you consider that with probability 1 no two points are diametrically opposed, and also consider if we take n random points and their antipodal points, how many of these configurations satisfy the condition. This provides a clean way of looking at the problem.

[quote="Nukular"] consider if we take n random points and their antipodal points, how many of these configurations satisfy the condition. This provides a clean way of looking at the problem.[/quote]
I tried this approach but couldn't  find a neat way to count the configurations with at least one acute angle. Can you show how this can be done?

Yeah, I see now it needs no integration. But I'm still happy because I wouldn't have had enough time to do it anyway. It's basically what Calvin posted, but by "starting acute vertex", he means the non-acute vertex just before the acute one, counting anti-clockwise.

Then you just consider the pairs of antipodal points and blah blah blah.

I mean the numerical value of course not the method.

No, the probability that at least one angle is obtuse is $\frac34$ for $n=3$, and $1$ for $n>3$. Your alleged answer goes to zero, which is ridiculous.

We have this one right.

what's wrong with you ...  :?: don't u have another thing to do rather than reply to me as soon as i put it there
okay I know I got it wrong, but believe me I was about to edit it ...

The equivalent problem that I talked about is not useless. Infact any one who lets himself talk about an
approach like that should be a stupid head because I'm about to write the perfectly valid solution to A6 with no integral
and based on looking at
sum xi =4

There is either no acute angle, a single acute angle, or two consecutive acute angles (since a corner is acute if the two gaps on either side are more then half the circle). Thus the probability of having at least one is $n$ times the probability that $P_1$ is acute and $P_2$ is not, when $P_n$ is the point to the right of $P_1$ along the circle.

Clearly the location of $P_1$ can be fixed. To chose the other points we can first choose $n-1$ diameters, and as a second step choose an end-point of each. Given the diameters, there are $2^{n-1}$ possibilities for the end-point selection. Of those $n$ result in an acute angle at $P_1$, and of those, 2 result also in an acute angle at $P_n$.

It follows that $P_1$ is acute and $P_n$ is not acute with probability $\frac{n-2}{2^{n-1}}$ and the answer is $\frac{n(n-2)}{2^{n-1}}$.

Finally, a really nice solution. This was already hinted at earlier.

The $n$ configurations that work are those in which the $n-1$ other vertices are consecutive among the $2n-1$ possible points. The two we subtract are the one in which $P_n$ is the only vertex in its half of the circle, and the one in which the arc between $P_1$ and $P_n$ is more than half of the circle.

The following contains not only the answer but also a clear explanation of how to do it.

Let $q=P[P_1\text{ acute}]$, and let $r=P[P_1\wedge P_2\text{ acute}]$. Note that given $n\ge 4$, we have at most two of the angles are acute.

Note that
\[q=P[\max(x_1,\ldots,x_{n-1})-\min(x_1,\ldots,x_{n-1})\le\tfrac{1}{2}]\]
where $x_1,\ldots,x_{n-1}$ are iid chosen uniformly in $[0,1]$. We compute this probability to be
\[q = \frac{\int_0^{1/2}da\int_a^{a+1/2}db\frac{(b-a)^{n-3}}{(n-3)!} + \int_{1/2}^1 da\int_a^1 db\frac{(b-a)^{n-3}}{(n-3)!}}{\frac{1}{(n-1)!}} = \frac{n}{2^{n-1}}.\]

Now, we see that $P_1$ and $P_2$ are acute if and only if all the other points lie in the minor arc $P_1P_2$, or in its reflection about the center. Thus, we see that
\[r = 4\int_0^{1/2}dp_2\, p_2^{n-2} = \frac{1}{n-1}\frac{1}{2^{n-3}}.\]
Thus, the desired probability is
\[nq-\binom{n}{2}r = \boxed{\frac{n(n-2)}{2^{n-1}}}.\]


Let $ m>1$ and $ n>1$ odd integers. On each square of an $ mxn$ board ($ m$ rows, $ n$ columns) we write distinct real numbers. A number is good if:

$ 1:$ The number is the greatest of its row and /or column.
or/and
$ 2:$ The number is the middle number (magnitude) of its column and/or row

 ¿What is the maximum number of good numbers?

You can have $ 2(m\plus{}n\minus{}2)$ good numbers, because there is an array that fulfills it. But I haven't proof it is the maximum.

I think that maximum is $ m\plus{}n\minus{}1$ yet I don't know how to prove it...
Any ideas?
Thank's :)

Daniel

Does the number have to satisfy "condition 1 or condition 2" OR " condition 1 and condtion 2" to be good??

The number has to be either
1)Bigger in the column,  middle in the row
2)Bigger in the row ,middle in the column


calculus
integration
logarithms
calculus computations

Let $ k\geq 0$ be a natural number. Calculate the integrals

$ I_{k}\equal{}\int\limits_{0}^{\infty}x^{k}\left(\ln(1\plus{}e^{x})\minus{}x\right)dx.$

$ J_{k}\equal{}\int\limits_{0}^{\infty}x^{k}\ln(1\minus{}e^{\minus{}x})dx.$

$ J_k\equal{}\minus{}\int_0^{\infty}x^k\sum_{j\equal{}1}^{\infty}\frac{e^{\minus{}jx}}{j}\,dx$

$ \equal{}\minus{}\sum_{j\equal{}1}^{\infty}\frac1j\int_0^{\infty}x^ke^{\minus{}jx}\,dx\equal{} \minus{}\sum_{j\equal{}1}^{\infty}\frac{k!}{j^{k\plus{}2}}\equal{}\minus{}k!\cdot\zeta(k\plus{}2)$

=========

As for $ I_k,$ I've pushed it around in several directions without deciding how to finish it off yet:

$ I_k\equal{}\int_0^{\infty}x^k\int_{e^x}^{e^x\plus{}1}\frac1t\,dt\,dx$

$ \equal{}\int_1^{\infty}\int_{\ln(t\minus{}1)}^{\ln t}\frac{x^k}{t}\,dx\,dt$

$ \equal{}\frac1{k\plus{}1}\int_1^{\infty}\frac1t \left[\ln\left(\frac{t}{t\minus{}1}\right)\right]^{k\plus{}1}\,dx$

$ \equal{}\frac1{k\plus{}1}\int_0^{\infty}\frac{u^{k\plus{}1}}{(e^u\minus{}1)^2}\,du$

I should probably introduce a series in there somewhere.

[quote="Kent Merryfield"]$ \frac1{k + 1}\int_0^{\infty}\frac {u^{k + 1}}{(e^u - 1)^2}\,du$[/quote]

Here is a way to attack this integral. Put $ I(k, \alpha) = \int_0^{\infty}\frac {u^{k}}{e^{\alpha u} - 1} \, du$ for $ \alpha > 0$. This integral is easy to calculate, and actually we have $ I(k, \alpha) = \frac{\Gamma(k+1) \zeta(k+1)}{\alpha^{k+1}}$.

Now differentiate $ I$ w.r.t. $ \alpha$. Then $ \frac{\partial}{\partial u}I(k, \alpha) = - \int_0^{\infty}\frac {u^{k+1} \, e^{\alpha u}}{(e^{\alpha u} - 1)^2} \, du = - \int_0^{\infty}\frac {u^{k+1}}{(e^{\alpha u} - 1)^2} \, du - I(k+1, \alpha)$.

From this, we have $ \int_0^{\infty}\frac {u^{k+1}}{(e^{\alpha u} - 1)^2} \, du = \frac{\Gamma(k+2) (\zeta(k+1) - \zeta(k+2))}{\alpha^{k+2}}$. Plugging $ \alpha = 1$ gives an explicit formula for the given integral.

But I suspect that what you gave is not a right formula for $ I_k$, for

\begin{eqnarray*}
I_k
& = & \int_{0}^{\infty} x^k \log (1 + e^{-x}) \, dx\\
& = & \left[ \frac{x^{k+1}}{k+1} \log (1 + e^{-x}) \right]_{0}^{\infty} + \frac{1}{k+1} \int_{0}^{\infty} \frac{x^{k+1} e^{-x}}{1 + e^{-x}} \, dx\\
& = & \frac{1}{k+1} \sum_{n=0}^{\infty} (-1)^n \int_{0}^{\infty} x^{k+1} e^{-(n+1)x} \, dx\\
& = & \frac{1}{k+1} \sum_{n=0}^{\infty} \frac{(-1)^n \Gamma(k+2)}{(n+1)^{k+2}}\\
& = & k! \, \eta(k+2)\\
& = & k! \, (1 - 2^{-k-1}) \zeta(k+2)
\end{eqnarray*}

where $ \eta$ is the Dirichlet eta function.


geometry
circumcircle
incenter
geometry solved

[quote="Stun"]Quadrangle $ABCD$ has circumcircle $(O)$. $(I)$ is the circle which touch $(O),AC,BD$ at K,E,F respectively. ($(I)$ cuts $CD$)
Prove that : EF pass through the incenter of $\triangle{ACD}$ and $\triangle{BCD}$[/quote]

This actually is the Sawayama-Thébault theorem in disguise. Look at the paper [url=http://forumgeom.fau.edu/FG2003volume3/FG200325index.html]Jean-Louis Ayme, [i]Sawayama and Thébault's Theorem[/i], Forum Geometricorum 3 (2003) p. 225-229[/url], especially at the Theorem 2 there. Applying this theorem to the triangle ACD, you see that the line EF passes through the incenter of triangle ACD, and applying it to the triangle BCD, you see that the line EF passes through the incenter of triangle BCD. And the problem is solved.

  Darij

[quote="Andrei"][b]Darij[/b], in the link you mentioned, there is a note about [b]Reim's Theorem[/b], and [i]They[/i] use it to prove that 4 points are concyclic. Could you please post this theorem here ?[/quote]

I'm sorry, but I really don't know what Reim's theorem is. I didn't ever read the paper, I just remembered that the theorem was proven there and gave the link. Anyway, as Ayme writes, you don't need any special theorem to show that the points A, K, F and J are concyclic. You can just do some angle chasing:

< FJA = < NMA            (since EF || MN)
= < NKA         (cyclic)
= < FKA,

and thus, the points A, K, F and J are concylic.

  Darij

[quote="Andrei"][b]Darij[/b], in the link you mentioned, there is a note about [b]Reim's Theorem[/b], and [i]They[/i] use it to prove that 4 points are concyclic. Could you please post this theorem here ?[/quote]
I believe it's the following result (in fact trivial):

Given four concyclic points $A,B,C,D$ and points $P\in AC$, $Q\in BD$ then $PQ\parallel CD$ if and only if $A,B,P,Q$ are concyclic.

[b]Sorry[/b], I observed that in elementary geometry many trivial results have a name ... as [b]Reim's theorem[/b].

Mr. Nicula wrote

..............
I observed that in elementary geometry many trivial results have a name ... as Reim's theorem.
.................

    How true!

  Thanks to V.Nicula for his acute observation.

  T.Y.
  M.T.


limit
algebra
function
domain
calculus
logarithms
calculus computations

kindly tell me what is: lim x--->0 [ln(a)]
and if u can explain the reason then i will be thankful to u. thanx.

umm this question isnt appropriate for this forum, but I'll answer it anyways

[hide]I dont think the limit exists. In order for $\displaystyle \lim_{x\rightarrow 0}\ln(a)$ to exist, we need to show that $\displaystyle \lim_{x\rightarrow 0^{-}}\ln(a)=\lim_{x\rightarrow 0^{+}}\ln(a)$. We can see immediately that $\displaystyle \lim_{x\rightarrow 0^{+}}\ln(a)=0$ But we know that the domain of any $\ln(a)$ is $[0,\infty)$. So we can see that $\lim_{x\rightarrow 0^{-}}\ln(a)$ isn't real so we can conclude that $\displaystyle \lim_{x\rightarrow 0}\ln(a)$ doesnt exist. Next time you post a limit question, I think it'd be better if you put it in the calculus section  ;) [/hide]

edit- lol this is as wrong as i can get haha, ignore this post :blush:

[quote="dangvh"]
[hide]We can see immediately that $\displaystyle \lim_{x\rightarrow 0^{+}}\ln(a)=0$[/hide][/quote]

That's not true :). The $\displaystyle \lim_{x\rightarrow 0^{+}}\ln(x)=-\infty$.

-BLt

the brokenlight is correct

ln(0)=y

$e^y=0$

$y$ has to be a really large negative

this is pretty straitforward so i am not sure if i should move it.

thanx boy.
 by the way thebrokenlight is also correct, but instead precisely correct.

Indeed, please take care to post in the correct forum.  Calculus and limits do not belong in getting started; I am moving this thread.

This is nitpicking but you have to write:

$\lim_{x\to 0} \ln x$

not $\ln a$. The relationship between $a$ and $x$ is not known.


integration
function
geometry
geometric transformation
limit
calculus
inequalities

Suppose $f \in L^{p}(\mathbb{R})$ and let $f_\epsilon(x) = \frac{1}{2\epsilon}\int_{x-\epsilon}^{x+\epsilon}f(t) \, dt$. Show that $f_{\epsilon}$ is absolutely continuous and that $f_{\epsilon}$ tends to $f$ in $L^{p}(\mathbb{R})$ as $\epsilon \to 0$.

For the second part the problem turned out to be more difficult than the examinator thought when he was asked. One could use Hardy-Littlewood's result to get a bound on $f_\epsilon$ that is in $L^{p}$ to justify convergence easily, but that theorem is far beyond my course. Incidentally, can someone tell me what the name of the result is? We just said "Hardy-Littlewood". Anyway, I'd like to see your ideas about the problem.

You do not really need the fact that the maximal function of an $L^{p}$ function is in $L^{p}$ here if that is what you mean by Hardy-Littlewood. Just note that $\|f_\varepsilon\|_{p}\le\|f\|_{p}$ and that convergence "obviously" takes place when $f$ is a compactly supported continuous function. It remains to use the fact that such functions are dense in $L^{p}$.

You don't need a maximal function here - and I assume that you're talking about a maximal function when you mention Hardy-Littlewood. The maximal function is needed for pointwise [i]a.e.[/i] convergence results, but here we're asking about $L^{p}$ convergence, and that's easier.

[b]Step 1:[/b] For $1\le p<\infty,\ L^{p}(\mathbb{R})$ has various "nice" dense subspaces. To name one that I'll use, let's take $C_{c},$ the set of continuous compactly supported functions.

[b]Step 2:[/b] Translation is continuous in $L^{p}(\mathbb{R}).$ That is: define $f_{t}(x)=f(x-t).$ Then $\lim_{t\to 0}\|f_{t}-f\|_{p}=0.$

Proof: Fix $\epsilon>0.$ Find $g\in C_{c}$ such that $\|f-g\|_{p}<\frac{\epsilon}4.$ Then $\|f_{t}-f\|_{p}\le\|g_{t}-g\|_{p}+\|(f_{t}-g_{t})-(f-g)\|_{p}.$

As $t\to0,\ (g_{t}-g)$ is supported within some fixed compact set and tends uniformly to zero - with the uniform convergence a consequence of the uniform continuity of $g.$ Hence $\|g_{t}-g\|_{p}\to0$ as $t\to 0.$ So pick $\delta$ such that if $|t|<\delta,$ then $\|g_{t}-g\|_{p}<\frac{\epsilon}2.$ Since $\|(f_{t}-g_{t})-(f-g)\|_{p}\le2\|f-g\|_{p}<\frac{\epsilon}2,$ that shows that for $|t|<\delta,\ \|f_{t}-f\|_{p}<\epsilon.$

[b]Step 3:[/b] I'll prove a more general theorem that includes your problem as a specific case. Let $\varphi$ be a fixed $L^{1}$ function such that $\int_{\mathbb{R}}\varphi=1.$ Define $\varphi_{\epsilon}(x)=\frac1{\epsilon}\varphi\left(\frac{x}{\epsilon}\right).$ Note that $\varphi_{\epsilon}$ also has integral 1 and the same $L^{1}$ norm as $\varphi.$ In your particular case, $\varphi$ is the normalized characteristic function of $[-1,1].$

[b]Claim:[/b] With this setup, $\lim_{\epsilon\to0^{+}}\|\varphi_{\epsilon}*f-f\|_{p}=0.$

Proof:

$\varphi_{\epsilon}*f(x)-f(x)=\int_{\mathbb{R}}\varphi_{\epsilon}(y)f(x-y)\,dy-f(x)$

$=\int_{\mathbb{R}}\varphi_{\epsilon}(y)(f(x-y)-f(x))\,dy$ (This uses that the integral is 1.)

$=\int_{\mathbb{R}}\frac1{\epsilon}\varphi\left(\frac{y}{\epsilon}\right)(f(x-y)-f(x))\,dy$ Let $u=\frac{y}{\epsilon}.$

$=\int_{\mathbb{R}}\varphi(u)(f(x-\epsilon u)-f(x))\,du$

Now use the translation notation used in Step 2, and use [i]Minkowski's integral inequality:[/i] For suitable $g,$

$\left[\int\left|\int g(x,y)\,dy\right|^{p}\,dx\right]^{\frac1p}\le \int\left[\int |g(x,y)|^{p}\,dx\right]^{p}\,dy$

$\left[\int_{\mathbb{R}}|\varphi_{\epsilon}*f(x)-f(x)|^{p}\,dx\right]^{\frac1p}\le \int_{\mathbb{R}}\left[\int_{\mathbb{R}}\left|\varphi(u)(f(x-\epsilon u)-f(x))\right|^{p}\,dx\right]^{\frac1p}\,du$

$\le\int_{\mathbb{R}}|\varphi(u)|\,\|f_{\epsilon u}-f\|_{p}\,du.$

We know that for each $u,$ as $\epsilon\to0^{+},$ then (by Step 2), $\|f_{\epsilon u}-f\|_{p}\to 0.$ It's also bounded by $2\|f\|_{p}.$ Hence, using $2|\varphi(u)|\,\|f\|_{p}$ as a dominating function, we may quote the Dominated Convergence Theorem to show that this last integral tends to zero.

---

This is a standard argument that those of my "tribe" of mathematicians were brought up with. It works just as well in $\mathbb{R}^{n}$ as in $\mathbb{R}.$ You could have used a different "nice" subspace in Step 1 - for instance, step functions - and still, with suitable modification of the details of Step 2, make the whole thing work.

Note also that for $p=\infty,$ Step 1 is false, the rest of the proof collapses, and indeed the result is false.

---

[Added after I saw fedja's repsone: he knows the ways of my tribe - I've just written a more expanded version of the same thing.]

Stupid question: Why is $\| f_\epsilon \|_{p}\leq \| f \|_{p}$?

I was thinking maybe this was due to that that "Minkowski integral inequality" but it doesn't seem to work out. Perhaps I am making a mistake.

It is precisely because of that inequality; more generally, $\|f*g\|_{p}\le \|f\|_{p}\cdot\|g\|_{1}$. In even more generality, $\|f*g\|_{r}\le \|f\|_{p}\cdot\|g\|_{q}$, where $1\le p,q,r\le\infty$ and $\frac1p+\frac1q=\frac1r+1$.

I would have posted the same argument, but I couldn't remember how the proof went myself.

[quote="jmerry"]... more generally, $\|f*g\|_{p}\le \|f\|_{p}\cdot\|g\|_{1}$.[/quote]
It should be pointed out that that much is true even for $p=\infty.$ (Which jmerry did state in the next line.) The convergence result fails for $p=\infty$ because $L^{\infty}$ fails to have "nice" dense subspaces (which is essentially the same as it not being separable.)

I am somehow botching the calculation, as I seem to be getting the wrong variables or something. Doesn't the inequality give the following:

$\|f_\epsilon\|_{p}\leq \int_\mathbb{R}\left( \int_\mathbb{R}\left|\chi_{[x-\epsilon,x+\epsilon]}(t)\frac{1}{2\epsilon}f(t)\right|^{p}dx \right)^{1/p}dt$

and that would be some constant times $\|f\|_{1}$ ... Am I misunderstanding the inequality? Sorry for this stupidity.

Aha ... Do I have to change variables so that I am integrating over $[-1,1]$ first to make it work? I REALLY don't understand why it works better just because you do that. It is past midnight and I am now very confused. Sorry for not getting more out of your post Kent, I am not really sure what  convolution even is. I guess I'll take a closer look some other time.

Let's make it this: $f\in L^{p},g\in L^{1}.$

$\|f*g\|_{p}=\left[\int \left|\int f(x-y)g(y)\,dy\right|^{p}\,dx\right]^{\frac1p}$

$\le \int\left[\int |f(x-y)g(y)|^{p}\,dx\right]^{\frac1p}\,dy$

$=\int |g(y)|\left[\int |f(x-y)|^{p}\,dx\right]^{\frac1p}\,dy=\int |g(y)|\left[\int |f(u)|^{p}\,du\right]^{\frac1p}\,dy=\|g\|_{1}\|f\|_{p}$

---

Or, here's another way to argue it, a step close to the raw proof of the Minkowski inequality. Let $\frac1p+\frac1q=1.$ (And yes, we'll have to go back to recheck the cases of $p=1$ and $p=\infty.$)

$\|f*g\|_{p}=\sup_{\|h\|_{q}=1}\int f*g(x)h(x)\,dx\le \sup_{\|h\|_{q}=1}\int \int |f(y)g(x-y)h(x)|\,dy\,dx$

$=\sup_{\|h\|_{q}=1}\int |f(y)|\left[\int |g(x-y)h(x)|\,dx\right]\,dy$

But by Hölder's inequality, $\int |g(x-y)h(x)|\,dx\le\|g\|_{p}\,\|h\|_{q}=\|g\|_{p}.$

So  $\|f*g\|_{p}\le \int |f(y)|\,\|g\|_{p}\,dy=\|f\|_{1}\|g\|_{p}.$

Ok. I was confused about the following: I thought that I could show that for my particular $f$, $\| f_\epsilon \|_{p}\leq \| f \|_{p}$ by applying Minkowski's inequality, by writing $f_\epsilon$ in the form in my original post. However, it sems it is necessary to make a change of variables to write $f_\epsilon$ in the convolution form for the argument to work. This I do find a little bit confusing.

Does applying the inequality directly to my $f_\epsilon$ really give that $\|f_\epsilon\|_{p}$ is less than some constant times $\|f\|_{1}$, or did I make a mistake there too? This seems weird too.

For the particular problem I posted, we knew from the course that $f_\epsilon$ tends to $f$ a.e. so once we have that $|\ f_\epsilon \|_{p}\leq \|f\|_{p}$ it is easier to motivate convergence with that in this particular case. Indeed, I do find the general argument quite weird and tricky!


One root of the equation $ 5x^2\plus{}4x\equal{}k$ (for some real $ k$) is 2. What is the other?

It really doesn't matter what $ k$ is. By Vieta's the sum of the roots is $ \minus{}\frac{4}{5}$. Since we know one root is 2, the other is $ \minus{}\frac{4}{5} \minus{} 2 \equal{} \minus{}\frac{14}{5}$.

Vieta's formulas are some formulas that, when you have an equation, you can determine stuff like sums and products of roots.

For example, consider the equation $ x^2 \minus{} x \plus{} 6 \equal{} 0$, and suppose you wanted to find the sum of the roots. You [i]could[/i] actually find the roots and add them, but this can get tedious, especially when you have yucky roots.

But let's look at the general form of a quadratic, $ ax^2 \plus{} bx \plus{} c$. We know that $ x$ is $ \frac{\minus{}b \plus{} \sqrt{b^2 \minus{} 4ac}}{2a}$ and $ \frac{\minus{}b \minus{} \sqrt{b^2 \minus{} 4ac}}{2a}$. If we add those expressions together, we get $ \minus{}\frac{b}{a}$.

From that, we can conclude that the sum of the roots is just the negative of the $ b$ value divided by $ a$. So in that example in the second paragraph, the sum of the roots is just $ \minus{}\frac{\minus{}1}{1} \equal{} 1$.

Similarly, we can multiply the general forms of the roots together to get $ \frac{c}{a}$, so that's how you find the product of the roots.

These can be generalized for polynomials of higher powers, but that's a little harder.


Ninety-six girls were surveyed at Euclid High School. There were 48 softball players and 45 track athletes. If $ \frac16$ of the girls surveyed did not play either sport, how many girls played both sports?

By PIE, total=softball+track-Both+Neither.
Plugging in, we have $ 96\equal{}48\plus{}45\minus{}\text{Both}\plus{}16 \implies \text{Both}\equal{}\boxed{13}$.

This "PIE" gauss speaks of is the Principle of Inclusion Exclusion.

http://www.artofproblemsolving.com/Wiki/index.php/Principle_of_Inclusion-Exclusion

ill answer this in a series of steps

[hide="this is my explanation"]1.we assign variables

2.the BOTH for each sport we wil put that for X

3.there are 48 in softball  and there are 45 track

4.now these softball and track some of them take both so by using PIE we'll need to subtract each of them by X

5.so we get the equations  $ 48-x $ and $ 45-x $ and we know that people who share is $ x $

6.and also the problem said that the $ \frac{1}6 $ of children DOES NOT participate in either one so we use the equation 96*$ \frac{1}6 $ and get 13

7.so all the answer we got equals 96 so we get the equation

8.$ 48-x+x+45-x+16=96 $
   $ -x+109=96 $
   $ x=13 $

9.[hide="answer"]13[/hide][/hide]

 :10: i have 10/10 confindence that im correct

[color=#0080FF][b][i][u]PLEASE RATE MY POST THANK YOU SO MUCH!!!!![/u][/i][/b][/color]