USAMTS Harvard college MIT email AMC AIME

Unfortunately, I have some bad news regarding Round 2 of the USAMTS.

All of the Round 2 papers were lost by our shippers while en route between grading sites.  While there is still some possibility that they may turn up, realistically I am not optimistic that this will happen.

What we are now planning to do is grade both Rounds 2 and 3 simultaneously in late January.  This is how it will work:

If you submitted your Round 2 solutions electronically, there is nothing that you need to do.  We will reprint your Round 2 solution and have it graded along with your Round 3 solutions.

If you submitted your Round 2 solutions by mail, unfortunately we no longer have a copy of them.  Hopefully, you made a copy for yourself (as it says to do in the submission instructions on the website).  If so, please send us a copy (and again, keep a copy for yourself) together with your Round 3 solutions.  Like the entire USAMTS, this resubmission will be on the honor system: you are pledging the submit to us the same solutions that you submitted in November.

If you submitted your Round 2 solutions by mail and did not keep a copy, then sadly they are lost.  What we will do is use your scores from Rounds 1, 3 and 4 to give you a pro-rated score for Round 2.

We still have the list of all students who submitted Round 2 solutions by mail, and we will only be accepting mail resubmissions for Round 2 from those students.

I realize that this is an inconvenience for everyone involved.  I hope that this does not discourage you from continuing with Rounds 3 and 4 of the USAMTS.  Keep in mind that the primary purpose of the USAMTS is to have fun and to challenge yourself by attempting hard, interesting problems, and that the competitive aspect of the USAMTS is secondary to this.  The scoring will sort itself out in the end; the important goals of the USAMTS are to learn some new math and to develop your problem solving and technical writing abilities.

I'm sorry for the loss... but how does the grading work? It seems very interesting. Do you print out everybody's paper, then ship to a grading site to let a group of graders (while sipping some coffee and eating donuts... maybe) grade in a few days??

The NSA funds much of this, and they help with the grading process. There are also other grading sites for different problems. In the end, the solutions and grades are sent back to Cal. where AoPS is situated, and the grades are released. This involves shipping the solutions to the testing sites, and one of these shipments lost the papers.

The grading is done in Cambridge, MA at Harvard or MIT and at NSA headquarters in suburban Washington.  The papers are shipped from our office in California to these locations.

Please remember that the USAMTS is largely a volunteer effort and that we have no dedicated full-time staff.  Our Harvard/MIT graders (mostly undergraduate students, including many past USAMO/IMO and USAMTS participants) do get paid (a relatively small amount) but give up big chunks of their weekend time to grade.  The NSA generously gives its mathematicians time off from their regular duties to volunteer to grade.  Also the problems that we use are often "donated" by various people who receive nothing in return.  AoPS contracts out its employees (like me) on a limited part-time basis to run the administrative aspects of the USAMTS.  The end-of-the-year prizes are either donated by our generous sponsors (like Wolfram Research) or are purchased with our limited funds.

say we got all 20's but did not resumbit round 2, would you "pro-rate" or round 2 score, give us an 80, and then we would be silver winners?

Are you 100% sure that the papers were lost?

[quote="mathemonster"]say we got all 20's but did not resumbit round 2, would you "pro-rate" or round 2 score, give us an 80, and then we would be silver winners?[/quote]Most likely that's what will happen; we haven't exactly figured it out yet.

[quote]Are you 100% sure that the papers were lost?[/quote]No -- see the first full paragraph of the announcement.

I'd estimate the difficulty of the round by the web submitters as opposed to web submitters from other rounds, multiply the difficulty factors by the score and get the average, but I don't know how effective that would be in real life

I recommend that we should recreate the round 2 again with another problems.

The solution for round 2 are out, which could prevent those who submitted the solution by mail not to "rewrite" their solution? And even for those who submitted the solution by mail but did not keep a copy, that would not be fair for counting their points from 3 rounds. Some may criticize round 2 are harder so counting points from 3 rounds are beneficial for them. Some may argue round 2 are easier so counting points from 3 rounds are disadvantages.

For the timeline, we can decrease the time for each round left and we can assign the round 2 and round 4 for fewer problems (4 problems for each should be good).

Another advise is that we don't recreate round 2 but we give round 4 more problems (7-8 problems?) and extent the deadline slightly.

Well *that* was not what I was expecting when I saw the thread title. I'd like to take a moment to thank all the USAMTS graders and administrators, for their commitment to us students to bring us this opportunity, even when unfortunate events like these happen. I fully appreciate all the hard work that makes USAMTS possible.

First off, I too would like to thank everyone involved with USAMTS for all the hard work they put in :)

Now, I mailed my solutions, and I don't have [i]exact[/i], word for word copies of my solutions. I did, however, email myself a rough draft of all my solutions, but I edited some sentences before I wrote up my real solutions. Would it be possible for me to use my rough draft and try to make it as exact as my original solutions as I can?

Also, involving the pro-rating, I strongly object to taking just a regular average of one's scores for rounds 1, 3, and 4 and then finding a total score based off that average. I can't say anything about this round's difficulty, but I'm expecting round 4 to be quite more challenging than round 2 was, and I highly doubt that I would be able to solve as many problems correctly that I did in round 2.

 I realize that there is no way of being able to give an exact score to someone whose solutions were lost. Thank you for just giving us the means to resubmit. This obviously will cause the graders to work harder, which we all appreciate.

hopefully i submitted electronically  :)

Dang, I made copies of Round 1 at my libarary (because I dont have a copy machine at home).

However, since I turned Round 2 in at the last day, I did not make a copy of my solutions (and it took me like 14 pages and like 3 hours to do just problem number 5).

Hopefully, they werent totally lost and they can find mine...

I second all of the thanks-givings above my post.

Anyways,

[quote="AbrAcAdAbrA"]The solution for round 2 are out, which could prevent those who submitted the solution by mail not to "rewrite" their solution?[/quote]

Like Mr. Patrick said, the competitive nature of the USAMTS is very, very secondary.  Normally, those who would do choose to do that wouldn't really be in a position to do well on the rest of the rounds, so I doubt that any respectable USAMTS participant would decide to cheat.  Even if they did, they wouldn't win anything of consequence to them except a T-shirt.

It's hard to imagine that somewhere out there, our papers are sitting, with no one to read them. Poor guys...  :(

[quote="lalalalife"]will the AIME qualification based on USAMTS change this year then?[/quote]

My thoughts exactly. I was looking forward to having the USAMTS to fallback on in case something goes terribly wrong with the two AMC 12s. But now, who knows how the MAA will take this. (I don't like to speculate.)

For some reason, nothing has really worked out for me this year, and not only in math...  :wink:

At any rate, thanks for all of your efforts, AoPS! Things will work themselves out somehow.

To comment on a few of the very good points that you have all raised:

1. We don't have the resources to do a 5th round.  Indeed, it is already going to stretch our resources a bit to have to reprocess the 2nd round again, but we feel that it is important enough to do.

2. We will be in discussions with the AMC to find a way to still permit AIME qualification via the USAMTS.  I'll be seeing (in person) the AMC Director in a couple of weeks and we'll work something out.  In the short time, please don't inquire about the details because I simply don't know how it will work yet.

3. The entire contest itself is on the honor system.  I am choosing to have the Round 2 resubmissions conducted using that philosophy.

4. Regarding the "pro-rating" of the Round 2 scores for papers that don't get resubmitted: we have not yet figured out how that will work, but we will of course strive to come up with an algorithm that is as fair as possible.  It will likely not be something as simple as
Round 2 score = (Average of scores from Rounds 1, 3, and 4)
since that does not take into account the varying difficulty of the rounds nor the fact that some students don't submit all the rounds, plus we may need to come up with something to satisfy AIME qualification.

[quote="DPatrick"]To comment on a few of the very good points that you have all raised:

1. We don't have the resources to do a 5th round.  Indeed, it is already going to stretch our resources a bit to have to reprocess the 2nd round again, but we feel that it is important enough to do.

2. We will be in discussions with the AMC to find a way to still permit AIME qualification via the USAMO.  I'll be seeing (in person) the AMC Director in a couple of weeks and we'll work something out.  In the short time, please don't inquire about the details because I simply don't know how it will work yet.

3. The entire contest itself is on the honor system.  I am choosing to have the Round 2 resubmissions conducted using that philosophy.

4. Regarding the "pro-rating" of the Round 2 scores for papers that don't get resubmitted: we have not yet figured out how that will work, but we will of course strive to come up with an algorithm that is as fair as possible.  It will likely not be something as simple as
Round 2 score = (Average of scores from Rounds 1, 3, and 4)
since that does not take into account the varying difficulty of the rounds nor the fact that some students don't submit all the rounds, plus we may need to come up with something to satisfy AIME qualification.[/quote]

Just a small typo: In 2. It should be USAMTS not USAMO (I hope...)

Anyway, my main point, is there a deadline to resubmit?

[quote="BOGTRO"]Just a small typo: In 2. It should be USAMTS not USAMO (I hope...)[/quote]Oops.  Fixed.

[quote]Anyway, my main point, is there a deadline to resubmit?[/quote]The Round 3 deadline (Jan 12, 2009).

If that is so, and we submit Round 2 rewritten solutions with Round 3 solutions, how long would it take for both rounds to be graded?

[quote="DPatrick"]4. Regarding the "pro-rating" of the Round 2 scores for papers that don't get resubmitted: we have not yet figured out how that will work, but we will of course strive to come up with an algorithm that is as fair as possible.  It will likely not be something as simple as
Round 2 score = (Average of scores from Rounds 1, 3, and 4)
since that does not take into account the varying difficulty of the rounds nor the fact that some students don't submit all the rounds, plus we may need to come up with something to satisfy AIME qualification.[/quote]
The mathematically sophisticated approach would be to use a trainable [url=http://en.wikipedia.org/wiki/Statistical_classification]statistical classification algorithm[/url] that takes as its input the scores from Rounds 1, 3, and 4 and returns as its output an estimate of the score for Round 2. We could use the scores of the resubmitted Round 2 solutions as a training set to calibrate the algorithm.

Unfortunately, the Art of Problem Solving would need software to perform the classification. Thus, they would be limited by the software already available to them or the time and money needed to acquire the software. The $ k$-Nearest Neighbor algorithm or a spline approximation might possibly be calculated by hand, but either would take a lot of time.

Erin Schram

[quote="buzzer11"]Just for clarification: is the USAMTS going to use a "trainable statistical classification algorithm" to estimate round 2 scores for EVERYONE, calibrating the algorithm using the "real" scores for round 2?

OR

Is that algorithm only going to be used for those who sent their solutions in by mail, and are those who sent their solutions in electronically just going to receive their "real" score?

Because if they do it the second way, 1) This doesn't account for the possibility that round 2 was harder/easier than other rounds, and 2) This would be worse [i]if[/i] someone broke the honor code.[/quote]
I made a suggestion that the administrators of the USAMTS could use a statistical classification algorithm. Part of the fun of being a mathematician is being able to pull an algorithm out of my hat and say, "Try this. It could solve your problem." However, I am not an administrator of the USAMTS, so they have no obligation to follow my suggestion. Nor is there any reason for them to make their decision immediately.

A classification algorithm or an extrapolation algorithm would only estimate a likely score for Round 2. It would make no sense to throw away a real score for Round 2 derived from an electronically stored solution or a resubmitted solution and replace it with an estimate. So real scores would not use the algorithm.

A classification algorithm [i]would[/i] take into account the possibility that Round 2 was easier or harder than the other rounds. That is the point in using one--it adjusts to the actual data. It does have the weakness that if a person would violate the honor code and turn in an improved solutions for Round 2, he or she would introduce erroneous data in the training set. So please stick to the honor code. If the administrators worry about that, they could use only the scores of the electronically stored solutions as the training set and leave the scores of the resubmitted solutions out of the training set.

Erin Schram

Okay, I see. So nobody really knows what's going to happen for sure as of now?

Oops, is that a bad idea to forget about the round 2? No resubmitting, no rescoring, no stretching resources, and move forward to Round 3? We come to USAMTS for fun, so would it be a big deal for someone who needs that score to go to AIME?

I apologize if you think that is a bad idea.

or can they possibly just use the maximum score as 75 and re-rank golds, silvers, bronzes, and honorable mentions from there? (I mean forget about round 2 so that people don't have to resubmit answers and they don't have to waste time re-grading?)

Another thing, some people tend to reword/rephrase their solutions if they didn't make a copy of their round 2 solutions and they resubmit? (e.g. that's one of the problems for resubmitting)...Otherwise USAMTS can simply disregard the round 2 score or just take averages...

Um, just ignoring round 2 would not be fair for the people who remembered to keep a backup copy to turn in or who submitted their solutions online like I did.  To throw away the work of these people is a complete waste of their time spent thinking about and solving these questions.  Admittedly, this doesn't matter as much because the whole competition is a learning experience, but it does matter because of the amount of time some people have spent working on round 2 and by extension the amount of commitment they have put into the USAMTS.

In other words, ignoring round 2 phails, and finding an average score for those whose solutions were lost while grading the solutions submitted online wins. :)

I don't think it's a "complete waste of time" to have worked on the problems and not have the validation of a score.  However, we do want to recognize the work that people did on Round 2, and provide feedback, so we are planning on going ahead with Round 2 grading.

We have not decided (or even really thought about very much) about an algorithm for assigning Round 2 scores to people who don't resubmit their papers.  I'm not going to comment about it further until mid-January.

I'm guess I'm barely okay with the predicting of a score for round two, given the circumstances, as long as it takes into account whether it was easier/harder than other rounds, statistically of course. (In comparison to other USAMTS rounds in general, I actually found it on the harder, though not extremely hard (round 3 last year), side).

And AbrAcAdAbrA, I love your spirit. I agree with you. After all, for those that participated, the mathlinks contest (Spring/summer of this year) still has not been graded, but it was definitely worth participating.

[quote="AbrAcAdAbrA"]Dear Math Geek,

You think ignoring round 2 is throwing away the work of solving problems? Do you remember that for what reason we come here? We come here for fun of solving problems! So whether the problems are graded or not graded, we still had fun because we spent hours on the desk, read lots of resource, wrote hundreds of pages! It is wonderful if our ability is recognized but please do not forget the main mission of USAMTS.

If you are smart enough, 3 rounds or 4 rounds are the same, you will get to AIME, USAMO, IMO with no matter which problem occurs![/quote]

Um, I'm pretty sure that the definition of ignoring round 2 would be throwing away people's work.  Admittedly, the goal of USAMTS is to learn from solving math problems, but like Mr. Patrick said, it would be in the best interests of everyone if people's work on round 2 was recognized.  I did the USAMTS not so I could make the AIME, cuz that honestly isn't much of a challenge to most AoPSers, but so I could, as you say, "spend hours on the desk, read lots of resource, write hundreds of pages!"  So by extension, if we followed the idea of some of the posters above my previous post and refused to give feedback to the people who worked so earnestly on a round of problems to receive the suggestions, comments, and wisdom of the graders, no one would learn from "writing hundreds of pages," and like I said before, that would phail.

I'm just putting forth my own opinion, so please don't interpret this as a pathetic attempt at flaming you.  @Mr. Patrick, I'm sorry for the exaggeration when I said "complete waste of time."  I was trying to say that I didn't agree with the idea of not recognizing the work that USAMTS participants did on Round 2.

Well, I personally would want some feedback. I want to know where I can improve on my solutions. I mean, I could predict them, but who knows? There are places that could've been more or less clear, and I want to know about that.

[quote="hurdler"]Sorry, I misread the announcement the first time, and I resubmitted my round 2 solutions as soon as I could (several weeks ago). I just sent my round 3 solutions in today. Is this ok?[/quote]That's fine.

[quote="mathemonster"]i mailed my copied r2 with my r3 today. Is that fine?[/quote]

Yes, it is.

do you guys think round 2 was easier than round 1/round 3?

i didnt copy my solutions for round 2 because i did not want to go to the library during lunch (of the day it was due, dont have a copier at home)...so i did not turn in my copied round 2 solutions with the rd 3 solutions...

however, after looking at the solutions for round 3, i think i only got around 17ish...i got 19 on round 1...meaning if they average i only wouldve gotten 18 on round2...

when i first turned the round 2 solutions in (before they were lost), i though i got a 22ish (4/5 problems i had almost the exact solution as the solutions on the website)...but it would really suck because if they took the averages, i would basically be losing about 4 points...

so i hope they have some other formula which takes into account the difficulty/how people scored on these rounds?

so do you guys think round 3 was the hardest so far?

No, Round 1 was the hardest, but [size=59]that was only because I failed it and got a lowly 13.[/size]

They'll most likely use a
[quote]statistical classification algorithm[/quote] to figure it out.

When will Rounds 2 and 3's scores and feedback be done and available?

[quote="BOGTRO"]When are the Round 2 grades going to come out? With the Round 4?[/quote]
They came out before the Round 3 grades.

^I think he means for the people whos answers were lost and did not turn another one in?

Atleast I still have a TBD.

TBD = To Be Determined, which is still the case.  We will not be considering this until the first week of April at the earliest.


[quote="Kenshin24"][quote="BlackMax"]Let $ \gamma_1_$ and $ \gamma_2_$ be two fixed circles on the plane. $ {K}$ is a various point on $ \gamma_2_$. Circle  $ \gamma_3_$ is tangent to $ \gamma_2_$ at the point $ {K}$ and to  $ \gamma_1_$ in the point $ {A}$. Circle  $ \gamma_4_$ is tangent to  $ \gamma_2_$ in the point $ {K}$ and to  $ \gamma_1_$ at the point $ {B}$. Prove that every line $ {AB}$ contains one fixed point, which is not depend on $ {K}$.[/quote]

Can you clarify the tangency, both internal ?, both external ?,mixed tangency? or it does not matter?

thanks[/quote]

It's indifferent.


find all functions from the integers into theirselves, such that $f(x+y+f(y))=f(x)+2y$ for all integers, $x$,$y$.

may be i have the solution with your problem
first one ,it doesn't difficult to prove that $f(0)$=0
hence$f(x+f(x))=2*x$ and $f(2*x+f(x))=f(x)+2*x$  for all x in integer
give $y=f(x)+x$,$x=0$ thenwe have $f(f(x)+x+f(f(x)+x))=2*(f(x)+x)$
hence $f(f(x)+3*x)=2*f(x)+2*x$ for all x in integer
otherwise,$f(f(x)+3*x)=f(2*x)+2*x$ then $f(2*x)=2*f(x)$ for all x in integer
give $y=x$ and $x=-x$ then $f(f(x))=f(-x)+2*x$  for all x in integer
give $y=f(x)+x$,$x=-f(x)-x)$ then we have $f(2*x)=f(-(x+f(x))+2*(x+f(x))$
so $f(-(f(x)+x)=-2*x$ for all x in integer
give $y=-x+f(x)$ and $x=f(x)+x$ then $f(-x)=f(f(x)+x)-2*(x+f(x))$
so f(x)=-f(-x) for all x in integer
hence $f(f(x))=-f(x)+2*x$ for all x integer
give $x=2*f(x)$ then $-f(2*f(x))=-2*f(f(x))=f(f(2*x))-4*f(x)$ hence $f(f(x))=f(x)$
for all x in integer so $f(x)=x$ for all all x in integer
 :D  :D  :D  :D

[quote="nhat"]
give $x=2*f(x)$ then $-f(2*f(x))=-2*f(f(x))=f(f(2*x))-4*f(x)$ hence $f(f(x))=f(x)$
for all x in integer so $f(x)=x$ for all all x in integer
 :D  :D  :D  :D[/quote]
I think you were wrong here,nhat!
And we can prove that f is additive from the result:$f(2x)=2(x)$.Indeed,from it we will find funtion f define from set of even numbers to itshelf instead of integers.So we can take $y=-\frac{f(x)}{2}$ to obtain that:$x-\frac{f(x)}{2}+f(-\frac{f(x)}{2})=0$(1)
Take $x=0$ in the initial relation:$f(y+f(y))=2y$
So:$f(x+y+f(y))=f(x)+f(y+f(y))$
Till now take $y=-\frac{f(y)}{2}: f(x-y)=f(x)+f(-y)$ as well!
But it can't reach the desired result!!

May I  was so tired at that time not to relize it's the easiest part of this problem
From f is additive so $f(x)=xf(1)$.Using it in the initial relation we have:
$f(1)^2+f(1)-2=0$ so $f(1)=1$ or $f(1)=-2$.
  The final answer is :$f(x)=x$ or $f(x)=-2x$.
\


If one of the people in charge of this website reads this, I would just like to suggest putting a huge number of old math questions of every single contest available in a certain downloadable section of the site.

If you are looking for questions, you can't go any further than http://www.kalva.demon.co.uk. Has basically every question from every competition, with some solutions. It would be far too big to put into a file I think.. maybe just a link to that site !

Of course if you really want problems/tuts and have a bit of time/inginuity you could always go to the god of the internet at [url]http://www.google.com[/url]

I already knew of the site you gave Triple M but thanks anyways, help is appreciated.  I have already gotten a fair bit of problems from various places but just wondering if the people running the site here could post some other problems.  Specifically, I am very interested in AIME problems from like 1989-1999 with solutions.

THANK YOU TRIPLEM!!! That site is really helpful....i managed to solve quite a few, didnt think i could do a lot of those problems  :o

Thank John Scholes. That man is brilliant.

[quote="Lev918"] Specifically, I am very interested in AIME problems from like 1989-1999 with solutions.[/quote]
Lev918, you may be interested in this site: [url]http://www.gotmath.com/aime.pdf[/url]
Which has AIME questions from 1983 to 1989 and 1997.  There are answers, but sadly no solutions.  Consider buying copies of the old tests if you need the solutions that badly.  Or you could post questions on this site.

Some people (Eve, etc.) may recognize this as a site that [i]was[/i] (but no longer is) mentioned on the late mathcounts.org forum.

(I say [i]late[/i] because hardly anyone posts there anymore. :()

Actually <by trial and error> I figured out that it's gotmath.com and not gotmath.org. But what a wonderful site. Thanks!

Oh! Another slip of the key.  That can only mean one thing... I'm [b]drunk.[/b]
On math, of course.

My apologies for that slip-up.  Here's another site.  Lots of stuff, albeit unorganized (and the title at the top of the page makes not sense): [url]http://imo.math.ntnu.edu.tw/English/past-exams/past-exams.html[/url]
Happy solving!

[quote="mathfanatic"]
Some people (Eve, etc.) may recognize this as a site that [i]was[/i] (but no longer is) mentioned on the late mathcounts.org forum.
[/quote]

Hey look at that!  My name!  Do we know each other?  And yes, I remember when that was posted to the mathcounts boards.  I knew they'd take it down quickly.

I've got some pdf's with problems and such at my site: http://hem.bredband.net/lukhor/index.html. I try to update the site daily.

Thanks, mindspa.  Unfortunately, the page seems to be down.  What a sad day for problem hounds like me!  Alas.  Ah, such is the life.

Oooh, Eve, yes - that's your name (I think??).  But I doubt we know each other.  I didn't do ARML or anything (not many math nerds in Nevada!)  We just happened to post on the same threads on mathcounts.org.

Now somewhere on this site, there's another reference to a mathcounts.org thread...if only I could remember.

Well it seems to be down because the it links to http://hem.bredband.net/lukhor/index.html. instead of http://hem.bredband.net/lukhor/index.html , which differs by a dot at the end. :)


Find all numbers $ k$ so that $ a^3 \plus{} b^3 \plus{} c^3 \equal{} kabc$ for some integer $ k$, positive integers $ a,b,c$.
(I tried with standard infinite descent arguments but I'm not sure how ... but I'm probably missing something simple).

There is no solution for k<3 by AM GM inequality.
For k=3 => (a+b+c)((a-b)^2+(b-c)^2+(c-a)^2)=0 => a=b=c=any postive integer is the solution

Another cases:
$ (a,b,c)_k\equal{}(2,1,1)_5$
$ (a,b,c)_k\equal{}(3,2,1)_6$
$ (a,b,c)_k\equal{}(9,2,1)_{41}$
$ (a,b,c)_k\equal{}(14,3,1)_{66}$

Notice that if $ (a,b,c)_k$ is a solution of the equation then $ (ra,rb,rc)_k)$ also for real $ r$

I looked up the possible numbers in the OEIS:
[url]http://www.research.att.com/~njas/sequences/A072716[/url]
It doesn't look like there is a simple solution.


Let A,B\inM_3(R) so that A B-B A=A. Show that A^3=O_3..it should be simple.. :?

It is easy to show that $tr(A)=tr(A^2)=0$. If one succedes in proving $tr(A^3)=0$ then the problem is solved.

From Hamilton Cayley we have : $A^3-Tr(A)\cdot A^2 + Tr(A^*)\cdot A - det(A) \cdot I_n =0_3$ . we have to proof that $Tr(A)=0$ ,  $Tr(A^*)=0$ and that $detA=0$

$Tr(A)=0$ comes out right away if we apply the function trace in the hypothesis ...
$Tr(A^2)=0$ results right away if we multiply our identity with A, then apply trace ...
$Tr(A^*)=0$ comes out from $Tr(A)=0$ and from $Tr(A^2)=0$ because it is well known that $Tr(A^*)=\frac{{Tr(A)}^2-Tr(A^2)} {2}$

Let's presume that $detA\neq 0$ , so then $A^{-1}$ exists . if we multiply to the left by $A^{-1}$ the identity $AB-BA=A$, we shall get $B-A^{-1}BA=I_n$. if we apply trace, we get that $0=n$, impossible . so $detA=0$

so we prooved everything we needed, hence $A^3=0_3$

[quote="pureblack2003"]From Hamilton Cayley we have : $A^3-Tr(A)\cdot A^2 + Tr(A^*)\cdot A - det(A) \cdot I_n =0_3$ [/quote]

How did you apply Hamilton-Cayley and get this?

it's the theorem for a $M_3$ matrix ... it does have a general form for $M_n$ matrix, but this is a particular case .

I` ve though about my solution and I generalised it for a matrix in $M_n(R)$.
If $A=AB-BA$ then $A^k=ABA^{k-1}-BA^k=A\cdot BA^{k-1}-BA^{k-1}\cdot A=XY-YX$ thus $tr(A^k)=0,\forall k \in N$ from it results that $A$ is nilpotent.

[quote="spix"]$tr(A^k)=0,\forall k \in N$ from it results that $A$ is nilpotent.[/quote]
the matrix A is nilpotent if there exists  k s.t. $A^k=O$ isn't it? but how did u get that result?

write $tr(A^k)=\alpha_1^k+\alpha_2^k+....+\alpha_n^k$ and solve the system obtained( you will get that all the eigenvalues are zero, thus the matrix is nilpotent)


parameterization
limit
logarithms
real analysis
real analysis unsolved

Evaluate :
\[ \lim_{n \to \infty} \left[ \sqrt [3]{n^3 \plus{} n^2 \plus{} 1} \minus{} \alpha \frac {\ln \left( e^{n^2} \plus{} 2n \plus{} 3 \right) }{\sqrt {n^2 \plus{} n \plus{} 1}} \right],
\]
where $ \alpha$ is a real number.

[hide="solution"]Expanding about $ n \equal{} \infty$ the first term is $ n \plus{} \frac {1}{3} \plus{} \mathcal{O}\left(\frac {1}{n}\right)$ while the second term is $ n \minus{} \frac {1}{2} \plus{} \mathcal{O}\left(\frac {1}{n}\right)$.  Therefore, if $ \alpha < 1$ the limit is $ \infty$, if $ \alpha > 1$ the limit is $ \minus{} \infty$, and if $ \alpha \equal{} 1$ the limit is $ \frac {5}{6}$.[/hide]


How do you memorize multiplication table?

Since I am a Chinese, I used Chinese to memorize it. For instance, the "formula" 7 x 6 = 42 we say "qi liu si shi er" in Mandarin Chinese. ("qu"=7, "liu"=6, "si shi er"=42) But how do the students who speak English to memorize the multiplication table? I think "seven times six equals forty-two" is too long (for me, at least) to remember.

back in 2nd or 3rd grade, the textbooks had tables up to $ 12\times 12$.  I'm not good at making them, so someone else can make the table.

after a while, once you see $ 8\times 9$, you automatically know what it is and don't need the table.

I personally don't know how I myself managed to get the facts into my head, but w/e

I am Chinese too, and I learned them in English. If you keep doing math problems without your calculator, you'll get used to it.

Damn ,im also a chinese from malaysia. If im not mistaken , when I was small i memorize qi liu si shi er in english by saying seven six fourty two if im not wrong.

I just did a LOT of problems like "7*6=?". After enough counting on your fingers, you start to memorize them.

Some people will say it this way:

seven sixes are forty-two

three threes are nine

eight fours are thirty-two

etc.

Meh, I am chinese too, so basically, the teachers just drove the multiplication tables up to 9 into our brains by making us memorize them like, for 2 hours a day, just keep on repeating and repeating.
It works, but it's not effective.

We have a system at our school where it's "Math Stars" and we get a sheet where we have to do 100 questions in 3 minutes. When you're younger, they increase the limit. Anyway, they give us a paper to help us learn the facts, with flashcards and practice work sheets. It motivates a lot of people.

yah, my school did that too, but we called it minute math and it was like if you could do 40 multiplication, division, addition or subtraction (mixed) then you would get like a cookie.  I was pretty good at it! I think its better not to memorize but get comfortable with the facts e.g.  when you see 7*5 just kinda subconsciousely know that its 35...it might take a while though! if your like me (3 years!!!!)

[quote="zolojetto"]I think its better not to memorize but get comfortable with the facts e.g.  when you see 7*5 just kinda subconsciousely know that its 35...it might take a while though![/quote]

Yes, it's better not to memorize, it's better to actually just know it like your own name, "sub-consciously".

Oh yeah, thats how our school places us into the math classes for the first year.
Just give us 200 simple problems, like $ 5\cdot 8$ and 5 minutes.
I got like a 192 or something.

yes ive done things like that before!!!theyre kinda fun (well they were) but now i have a feeling that im a bit too old for them!!!

[quote="ajain"]Well...there is a math program called KUMON where you do A LOT of repetitions, it increases your speed and and your ability to memorize the tables faster.[/quote]

Kumon's annoying when its hard though...I'm doing logarithms in it now.

All these work, but after a while, the practice makes it so that 3 times 7 doesn't take time to think; you just visualize 21. Same with all the problems. And it's not just with the multiplication tables up to 12 or whatever, like other people have said. It applies elsewhere. For example, in trigonometry, calculating sin, cos, tan, csc, sec, and cot for 45, 30, and 60 might originally require you to draw the triangle to calculate it. But after a while, most of the facts will be memorized.


A large cube is dipped in red paint and made into 125 equivalent, smaller cubes.  One small cube is randomly selected.  What is the probability that the cube selected will have at least 25% surface area painted red?  Express your answer as a common fraction.

I keep getting different answers. :blush:   Help!

[hide]A cube has 6 faces. Let us assume that each of the smaller cubes have volume of 1 unit, which means that each of there sides is 1 unit. This means that the large cube has side length of $\sqrt[3]{125}$, which is 5.

For a cube to have more than 25% of its surface painted, it obvioulsy cannot be a cube that was not on the surface of the old cube, for they got no paint. It also cannot have only 1 side painted, because 1/6<25%. So it has to have at least 2 sides painted.

Each edge of the large cube has 3 cubes with exactly 2 faces painted, and each corner of the large cube has 3 faces painted. So there are 12 edges at 3 cubes apiece and 8 corners, for 36+8=44 cubes that have more than 25% painted sides, so the probability is 44/125.[/hide]

Billy

Way to explain your answer, that's what I got too.

[hide]44/125[/hide]

[hide]Because the volume is 125, then the edge length of the cube equals 5. Thus, the surface area equals 150. The total surface area of all the cubes after you cut the big cube up is 750. 150/750=1/5.

Answer:1/5 [/hide]

Whoops, sorry, I accidentally calculated the probability that the rolled face would be red if you rolled the cube.

I accidentally misread the problem.

This is my final answer (after carefully rereading the problem):

[hide] If the cube is at least 25% red, then it must have two red faces (at least). The number of cubes which have zero red faces is 3x3x3=27. The number of cubes with one face painted is 3x3=9, times another six (for each face) equals 54. Subtracting these two values from 125 gives 44. Thus, a randomly chosen cube has a 44/125 probability of being at least 25% red.

Answer: 44/125[/hide]


[quote="shortlist"]Give $ m\geqq$, $ n\geq 2$ , $ x,y,m,n \in Z +$, and $ m\geq n$ such that $ (mxy - n)| (x - y)^2$
Prove $ 2009|(x - y)$ :)[/quote]

hint: IMO 2007, pro 5.


calculus integration geometry conics parabola

For problems like these:

"A car is accelerating along a straight path. Its velocity is defined by $v=2t^2(\frac{m}{sec^3})$ where v is velocity in m/sec and t is time in seconds. What is the displacement of the car at t=10 seconds?"

Must we use calculus?

Hehe. I learned how to do that a few days ago.

Just find the complete integral, not worrying about positive or negative areas.  :)

Hmm... you don't [i]have[/i] to use calculus; you can use the formula for the area of a parabola $\left(\frac{2}{3}bh\right)$.


[quote="Math Geek"]Hi, I'm new to USAMTS, and I have a question about the Permission Form that lets the USAMTS people submit my work and info to "educational institutions."  If my score is lousy, will they submite the score anyway?  That would be awful, so I need to ask before I fax the info.  Thanks! :D[/quote]
I think the institutions only request high scorers' names.

Thanks, now I can feel safe in submitting the permission form! :lol:

I don't think you have to worry about a college docking you for a low USAMTS score.  A good USAMTS score may help, but a bad score generally won't be noticed, as you're unlikely to self-report it!


integration
calculus
inequalities
real analysis
real analysis unsolved

Let $l>0$ and $f,g: [-l.l]\to \mathbb{R}$ such that f is continuous and g is derivable and increasing on [0,1]. Moreover, $g(l)\geq 0$, $g(-l+x)=g(l-x)$ and $\int_{-l+x}^{l-x}f(t)dt\leq 0$ for any $x$ in $[0,l]$. Prove that $\int_{-l}^{l}f(x)g(x)dx\leq 0$.

Well...after I tried to solved it I looked at the end points. 2 solutions were given but both wrong in my opinion (in fact I had done the same thing but i stopped before that mistake - they said at some step in the proof that $g'(x)\geq 0$on $[-l,0]$ :maybe: Maybe I was confused by the $l-x,-l+x$ notations...However, can anyone help me ant say if the problem is correct? (maybe the statement is wrong and g must be decreasing on [0.l])  :maybe:

I guess that the solutions were something like this:

Let $F$ be an antiderivative of $f$. Then, we have that
\begin{eqnarray*}\int_{-l}^{l}f(x) g(x) \, dx &=& \int_{-l}^{l}F^\prime (x) g(x) \, dx \\ \ &=& \left[ F (x) g(x) \right]_{-l}^{l}-\int_{-l}^{l}F(x) g^\prime (x) \, dx \\ \ &=& g(l) (F(l)-F(-l))+\int_{0}^{l}g^\prime (x) (F(-x)-F(x)) \, dx .\end{eqnarray*}
The first term is $\leq 0$.
If $g$ increases on $[0,1]$, then the integral $\geq 0$ (since $F(-x) \geq F(x), \, \forall x \in \left[ 0, l \right]$). If we take $f$ such that $F(l) = F(-l)$, but somewhere along way the inequality $F(-x) \geq F(x)$ is strict, then we can find a counterexample.
If $g$ decreases, all is clear.


geometry
geometric transformation
reflection

A beam of light shines from point $S$, reflects at point $P$, and reaches point $T$ so that $PT\perp RS$. What is the measure of angle $x$?

[img]http://img258.imageshack.us/img258/5425/reflection7lh.png[/img]

[hide] Extend TP to hit RS and M.  This implies Angle SMP is 90 degrees.  Therefore, we have that Angle SPT=Angle SMP+Angle S=90+26=116.  Therefore, x+x+Angle SPT=2x+116=180.  Hence, 2x=64, so x=32.  [/hide]

[hide="Heres another way"]
Draw a line through point P, parallel to RS. Since alternate interior angles are congruent, the angle is 26. Since PT is perpendicular to RS, it is also perpendicular to the line through P (the one parallel to RS) so then that angle is 90 add them up to get 116. 2x+116=180, 2x=64, x=32.[/hide]
Edit: Picture added

My way is also slightly different:[hide]Extend line TP to hit RS at point M (I'm borrowing a little from The QuattoMaster 6000 here :P ) Also, lets say that the nameless line passing through P intersects RS at N, for the sake of it having a name. Now we have triangle PMS, where angle PMS is 90 degrees and angle PSM is 26 degrees. So angle MPS is 64 degrees. Angle MPN is equal to x because of vertical angles. So 2x=64 degrees, or x=32 degrees. [/hide]


Four positive integers $ A$, $ B$, $ C$ and $ D$ have a sum of 36. If $ A\plus{}2 \equal{} B\minus{}2 \equal{} C \times 2 \equal{} D \div 2$, what is the value of the product $ A \times B \times C \times D$?

[quote="FantasyLover"]let's express B, C, D with A.

A=A
B=A+4
C=$ \frac{A\plus{}2}{2}$
D=2A+4

so $ A\plus{}A\plus{}\frac{A\plus{}2}{2}\plus{}2A\plus{}4\equal{}36$.

and $ \frac{9}{2}A\plus{}9\equal{}36$

so $ \frac{9}{2}A\equal{}27.$

A=6.
B=10
C=4
D=16
6*10*4*16=3840

answer : 3840[/quote]

You forgot toput in one of the 4's in the A+A+(A+2)/2+2A+4+4! :)

[quote=FantasyLover]let's express B, C, D with A.

A=A
B=A+4
C=$ \frac{A\plus{}2}{2}$
D=2A+4

so $ A\plus{}A\plus{}\frac{A\plus{}2}{2}\plus{}2A\plus{}4\equal{}36$.

and $ \frac{9}{2}A\plus{}9\equal{}36$

so $ \frac{9}{2}A\equal{}27.$

A=6.
B=10
C=4
D=16
6*10*4*16=3840

answer : 3840[/quote]
you should hide your text but the solution is good! :)


Just out of curiosity...who all solved those problems? Yes, I'm too impatient to wait until November 7.

[quote="rrusczyk"]Here's the link to the Monthly Challenge Problems ARML is posting on the NCTM website.

[url]http://www.nctm.org/high/challenge/challenge.htm[/url][/quote]

I think the link is broken; I get a Page Not Found error.

http://www.nctm.org/high/challenge/index.htm

^--That'll take you to the current problems

Are they still posting new monthly challenge problems?


At McDonald's you can order Chicken McNuggets in boxes of 6, 9, and 20. What is the largest number of nuggets that it is not possible to obtain by purchasing some combination of boxes?

Geez I know it for two numbers, but three? And 6 is a multiple of 9? I can't remember the formula for three....

i just did this in like 5 seconds, so it's probably wrong
[hide="is it"]59?[/hide]

with 6 and 9 only, infinitly large hte numbers can get, though they can get all multiples of 3 above 6.
Thus we are now dealing with 20a+3b largest cannot be attained. Plugging this into chicken mcnugget, we get 20*3-20-3=37. Add 3 since this may have to have b being 1.
so the answer is either 37 or 40.

edit- whitehorse, ur right. forgot about the restriction of b having to be greater than 1. Thus it is 37 or some multiple of 20 +3. since 43 is not attainable, we can eliminate 37. THus now, the number must be in the form 20n+3. The highest unattaniable is now 43, since anything above 43, such as 20k+3=20a+3b where b>1 can be changed to 63=20(a-k)+3b, so setting a=k and b=21 solves everything above 43

That's not correct Scrambled. How do I get 43?

The answer is indeed 43 but I'm confused on the reasoning part. Anyone care to enlighten me?

[hide="Hope this helps....."]It is easy to see that 43 McNuggets can't be obtained:
-[b]Case 1[/b]: If you have no boxes of 20 McNuggets, you have to make 43 McNuggets with boxes of 6 or 9. Both are multiples of 3 but 43 is not divisible by 3. So you can't make 43 McNuggets with only boxes of 6 and 9 McNuggets.
-[b]Case 2[/b]: If you have only 1 box of 20 McNuggets, you have to make 23 McNuggets with only boxes of 6 or 9. Both are divisible by 3 but 23 is not divisible by 3. So you can't make 43 McNuggets with only 1 box of 20 McNuggets.
-[b]Case 3[/b]: If you have only 2 boxes of 20 McNuggets, you have to make 3 McNuggets with only boxes of 6 or 9. You can't, since 6>3 and 9>3. So you can't make 43 McNuggets with only 2 box of 20 McNuggets.

But the following number of McNuggets can be obtained:
[b]44[/b]: 6 + 6 + 6 + 6 + 20
[b]45[/b]: 6 + 6 + 6 + 6 + 6 + 6 + 9
[b]46[/b]: 6 + 20 + 20
[b]47[/b]: 6 + 6 + 6 + 9 + 20
[b]48[/b]: 6 + 6 + 6 + 6 + 6 + 6 + 6 + 6
[b]49[/b]: 9 + 20 + 20

So any number n, n > 49, of McNuggets can be obtained by adding a multiple of a box of 6 McNuggets to the 44,45,46,47,48,49 McNuggets you already have. So we have proven that you can get n > 43 McNuggets no matter what.

So the answer is: $\fbox{43}$ McNuggets.[/hide]

This is a special case of the Frobenius problem.

hey derek , is this problem from $mathproblems.info$??


[quote="hasan"]If nonnegative  real numbers a,b,c satisfy $1/(a^{2}+1)+1/(b^{2}+1)+1/(c^{2}+1)=2$ . Prove that $ab+bc+ca\leq 3/2$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=103762


I am a third year student studying computer science in India.I will be giving GRE by the end of this semester. I would like to know (if possible),what Gre scores guarantee an admission in the top tier schools..(say MIT,Harvard,Carneggie Mellon ,Cornell ) etc. I want to pursue computer science or mathematics. Please also tell me about the other good colleges for these course.Because the ones i mentioned i dont think i stand a chance of getting into.
How much does your GPA  matter . Does it matter more than the GRE score.
i would be grateful if someone could answer these questions
best regards

[quote="galois"]I would like to know (if possible),what Gre scores guarantee an admission in the top tier schools...
How much does your GPA  matter . Does it matter more than the GRE score.[/quote]

To paraphrase a professor of mine who was on the Harvard math department admissions committee last year:

"GRE scores are really only used if you want to have an excuse not to admit someone."


algebra
polynomial
AMC
USA(J)MO
USAMO
number theory open
number theory

For what irreducible polynomials $P(x) \in \mathbb{Z} [x]$ is it true that $P(n)$ takes on infinite prime values for $n \in \mathbb{N}$?

All linear ones by dirichlet, all others by open conjectures (positive leading coefficient when you want primes to be positive).
See http://www.mathlinks.ro/Forum/viewtopic.php?t=72867 too.

Any nonconstant polynomial involves infinitely many prime [i]factors[/i], or more accurately:

If $f\in Z[x]$ is not constant and $S$ is any finite set of primes, then there exist a prime $p$ and an integer $n$ such that
-- $p\notin S$
-- $p\vert f(n)$.

Proof by William Elliot:
If constant term of $f$ is zero, then $p | f(p)$ for any $p$.
So assume the constant term of $f$ is $c\ne 0$ and let $p = \prod_{x\in S}x$.
Notice $f(cx) = cg(x)$ for some $g\in Z[x]$ with $g(0) = 1$, i.e. constant term of $g$ is 1.
Find $n\in N$ for which $|c| < |f(cp^n)|$.
Thus $1 < |g(p^n)|$, and any prime which divides $|g(p^n)|$ cannot be in $S$.

i think t0rajir0u's question isn't simple .

 for example: we don't know if there are infinitly many primes in the sequence : $n^2+1$

Larry Hammick:  Yes, I am aware of that lemma, it was used in the solution of USAMO 2006 #3.  The question I posed remains, as ZetaX says, open.

Well, the conjectures mentioned by ZetaX must be something else, because e.g. $f(x)=x^2+x+2$ is always divisible by 2. One must pose some condition implying that there is no integer $N>1$ s.t. $\gcd(N,f(n))>1$ for all $n$.


If $ a(1\minus{}b^2)\equal{}b(1\minus{}c^2)\equal{}c(1\minus{}d^2)\equal{}d(1\minus{}a^2) > 0$,find all $ a,b,c,d \in R^\plus{}$

[quote="apollo"]If $ a(1 \minus{} b^2) \equal{} b(1 \minus{} c^2) \equal{} c(1 \minus{} d^2) \equal{} d(1 \minus{} a^2) > 0$,find all $ a,b,c,d \in R^ \plus{}$[/quote]

We have $ a,b,c,d\in(0,1)$
Let $ u\equal{}$ the common value.
Let $ f(x)\equal{}\frac{u}{1\minus{}x^2}$

We have $ f(f(f(f(a))))\equal{}a$ with $ a,f(a),f(f(a)),f(f(f(a)))\in (0,1)$
It's clear that this last condition implies that
either $ f(f(f(f(a))))>f(f(f(a)))>f(f(a))>f(a)>a$,
either $ f(f(f(f(a))))<f(f(f(a)))<f(f(a))<f(a)<a$,
either $ f(f(f(f(a))))\equal{}f(f(f(a)))\equal{}f(f(a))\equal{}f(a)\equal{}a$

And so the only solutions for $ (a,b,c,d)$ are $ (x_0,x_0,x_0,x_0)$ for any $ x_0\in(0,1)$


ok here's the 3 equations:

H2O(l) + NaOH(s) -->

HCl(aq) + NaOH(s) -->

HCl(aq) + NaOH(s) -->

[hide]Aren't the 2nd and 3rd equations the same?  ;)
Anyway, $HCl+NaOH \to NaCl+H_2O$
[/hide]

yes sorry the 2nd is the same

but the third is...

HCl (aq) + NaOH (l) -->

would the end products be in the same states as the 2nd products?

and what about reaction 1??

H2O (l) + NaOH (s) -->


calculus
derivative
function
limit
calculus computations

Find the derivative of the following function
$ f(x) = \{\begin{array}{ccc} x+\frac{x}{1+e^{\frac{1}{x}}} & , & x \ne 0 \\
 \\
0 & , & x = 0 \end{array}$

Well if $ x\equal{}0$, the derivative is obviously zero.  Otherwise, use the chain rule:
$ f'(x)\equal{}1\plus{}\frac{1\plus{}e^{\frac{1}{x}}\plus{}\frac{e^{\frac{1}{x}}}{x}}{(1\plus{}e^{\frac{1}{x}})^{2}}$

Um, actually $ f^\prime (0)\equal{}\lim_{h\rightarrow 0} 1\plus{}\frac{1}{1\plus{}e^{1/h}}$ if it exists. And it doesn't: consider the right-hand and left-hand limits. So $ f$ is not differentiable at 0.

[quote="JRav"]Well if $ x \equal{} 0$, the derivative is obviously zero.  Otherwise, use the chain rule:
$ f'(x) \equal{} 1 \plus{} \frac {1 \plus{} e^{\frac {1}{x}} \plus{} \frac {e^{\frac {1}{x}}}{x}}{(1 \plus{} e^{\frac {1}{x}})^{2}}$[/quote]

Oh, I was tried to calculate right-hand and left-hand limits but I can't find it. Please explain!

[quote="JRav"]Well if $ x = 0$, the derivative is obviously zero.  Otherwise, use the chain rule:
$ f'(x) = 1 + \frac {1 + e^{\frac {1}{x}} + \frac {e^{\frac {1}{x}}}{x}}{(1 + e^{\frac {1}{x}})^{2}}$[/quote]

Differentiation can't be done piecewise on endpoints of intervals  :|

scorpius means that ${ \lim_{h \to 0^+} \frac{1}{1 + e^{1/h}} = \lim_{x \to \infty} \frac{1}{1 + e^x}} = 0$ while $ \lim_{h \to 0^-} \frac{1}{1 + e^{1/h}} = \lim_{x \to \infty} \frac{1}{1 + e^{-x}} = 1$.


quadratics
inequalities
algebra
quadratic formula

expanding:
x^2-bx+x^2-xc-xa+ac=0.
rearranging:
2x^2-xb-xc-xa+ac=0.
quadratic formula form:
2x^2-(a+b+c)*x+ac=0.
a quadratic has real roots if a quadratic in the form of ax^2+bx+c has  b^2-4ac>0.
Thus,
we must prove that
(a+b+c)^2-8ac>0.
using am-gm; this is true, so we are done.

[quote="RAGHAV GROVER"]The problem is still open !![/quote]

Didn't Lukezli prove it?

[quote="lukezli"]expanding:
$ x^2 - bx + x^2 - xc - xa + ac = 0$.
rearranging:
$ 2x^2 - xb - xc - xa + ac = 0$.
quadratic formula form:
$ 2x^2 - (a + b + c)*x + ac = 0$.
a quadratic has real roots if a quadratic in the form of $ ax^2 + bx + c$ has  $ b^2 - 4ac > 0$.
Thus,
we must prove that
$ (a + b + c)^2 - 8ac > 0$.
using am-gm; this is true, so we are done.[/quote]

I guess what you want is to show that $ (a + b + c)^2 > 8ac$

So here it goes:

\begin{eqnarray*}
(a + b + c)^2 & > & (a + a + a)^2 \\
(a + b + c)^2 - 8ac & > & (3a)^2 - 8a^2 \\
(a + b + c)^2 - 8ac & > & 9a^2 - 8a^2 \\
(a + b + c)^2 - 8ac & > & a^2 \\
a^2 & > & 0 \\
(a + b + c)^2 - 8ac & > & a^2 > 0\\
(a + b + c)^2 - 8ac & > & 0
\end{eqnarray*}

Note: I can change $ c$ and $ b$ for $ a$ because it's smaller then $ c$ and $ b$ so the inequality stays the same.

[quote="RAGHAV GROVER"]If 0<a<b<c,then prove that roots of the equation
x(x-b) + (x-a)(x-c)=0 are real...  :P[/quote]

Let $ f(x) \equal{} x(x \minus{} b) \plus{} (x \minus{} a)(x \minus{} c)$.

Then for $ x_1 > c$, $ f(x_1) > 0$

for $ a < x_2 < b$, $ f(x_2) < 0$, so there must be at least one real root.

Since $ f(x)$ is quadratic with real coefficient, another root is real also.

This is the kind of solution that i wanted..
good job stephencheng!!!


calculus
linear algebra
calculus computations

Can anybody recomend a good introductory book on mathematical analysis.
I am electrical engineer and i have studied calculus (from limits to differential equations) and linear algebra as well.
Thanks.

helo, i think this will help you
http://wwwmaths.anu.edu.au/~john/Assets/Lecture%20Notes/B21H_97.pdf
Sonnhard.

i think   for real analysis   the one by apostol is really good. it starts

Thanks a lot Dr Sonnhard Graubner. You are so helpful!


calculus
function
integration
irrational number
real analysis
real analysis unsolved

i am a learner of calculus, there is an exercise about continuous problem in a text book i can not solve.

Define the function h:[1, 2]->R as follows: h(x) = 0 if the point x in [1, 2] is irrational; h(x) = 1/n if the point x in [1, 2] is rational and x = m/n, where m and n are natural numbers having no common positive integer factor other than 1.
(a)prove that h:[1, 2]->R fails to be continuous at each rational number in [1, 2].
(b)prove that if e > 0, then the set {x in [1, 2] | h(x) > e} has only a finite number of points.
(c)Use (b) to prove that h:[1, 2]->R is continuous at each irrational number in [1, 2].

This is a very famous example and has surely been posted here before.

I usually don't see it presented with step (b) spelled out. That's a hint, and a very strong one.

One way to use part (b): if $z$ is irrational, the distance from $z$ to the nearest point is the (finite!) set $\{x|f(x)>\epsilon\}$ is a positive finite number.

--

Have you studied the Riemann integral? Because when you do, this function reappears in another exercise: prove that it is Riemann integrable.

I can not find a post about this problem in early posts, may be I havn't catch the key word.

I Have not studied the Riemann integral before and I have only studied based calculus. I am now learning a book named Advanced Calculus, the anthor is Patrick M. Fitzpatrick. Could you recommend some books for me to learn mathematical analysis.

Any way, thank you for your help!

It is sometimes called the "Stars over Babylon" function. Try searching for that.

In russian textbooks it is called the [i]Riemann function[/i].


calculus
integration
limit
derivative
inequalities
calculus computations

Let $f$ be continuous and $B_{\epsilon}$ be the ball of radius $\epsilon$ centered at the point $\left(x_{0},y_{0},z_{0}\right)$.  Let $\textrm{vol}\left(B_{\epsilon}\right)$ be the volume of $B_{\epsilon}$.  Prove that

\[ \lim_{\epsilon\rightarrow 0}\frac{1}{\textrm{vol}\left(B_{\epsilon}\right)}\iiint_{B_{\epsilon}}f\left(x,y,z\right)\,dV=f\left(x_{0},y_{0},z_{0}\right). \]

[hide]
Another way of writing that is
\[ \lim_{\epsilon\rightarrow 0}\frac{\iiint_{B_{\epsilon}}f\left(x,y,z\right)\,dV}{\iiint_{B_{\epsilon}}\,dV}, \]
which is really just the average value of $f\left(x,y,z\right)$.  As $\epsilon\rightarrow 0$, $\left(x,y,z\right)\rightarrow\left(x_{0},y_{0},z_{0}\right)$.  Basically, the ball shrinks to a single point.

Can someone help me finish this?  Does this prove the statement?

Thanks :)
[/hide]

This is a higher-dimensional version of the "derivative of the integral" form of the Fundamental Theorem of Calculus. Your observations (about average values) are quite pertinent.

You've got to use that hypothesis of continuity. Given $\eta>0,$ there exists $\delta>0$ such that for $\epsilon<\delta,$ if $(x,y,z)\in B_{\epsilon}(x_0,y_0,z_0)$, then $f(x_0,y_0,z_0)-\eta<f(x,y,z)<f(x_0,y_0,z_0)+\eta.$

Find a way to make use of that.

I may be misinterpreting this, but as the radius of the ball goes to zero, it seems we would have
\[ f\left(x_{0},y_{0},z_{0}\right)<f\left(x,y,z\right)<f\left(x_{0},y_{0},z_{0}\right)\implies f\left(x,y,z\right)=f\left(x_{0},y_{0},z_{0}\right). \]

The only problem I see there, is that $\eta$ was already defined to be greater than zero.  However, if I ignore that, the rest seems to work out well since $f\left(x_{0},y_{0},z_{0}\right)$ is a constant.  It can be pulled out of the integral and the two remaining volume (of the integration region) integrals will cancel, leaving just the $f\left(x_{0},y_{0},z_{0}\right)$.  Is this what you were getting at?

Thanks a lot.

Next, you want to use the inequality chain
\[ \left|\int_\Omega g(x)dx\right|\leq \int_\Omega |g(x)|dx \leq \sup_{x\in \Omega}|g(x)|\int_{\Omega} dx  \]
written appropriately as a triple integral.

[hide]
If you apply it to $g(x,y,z)=f(x,y,z)-f(x_0,y_0,z_0)$, given $\delta>0$, we can pick $\epsilon$ small enough such that $|g(x,y,z)|<\delta$ on $B_{\epsilon}$ (this is just continuity).  Hence the above inequality gives that
\begin{eqnarray*} \left|\frac 1{\text{vol}(B_\epsilon)}\iiint_{B_\epsilon}g(x,y,z)dV\right|&\leq& \frac 1{\text{vol}(B_\epsilon)}\sup_{B_\epsilon}|g(x,y,z)|\iiint_{B_\epsilon}dV \\ &=& \sup_{B_\epsilon}|g(x,y,z)| \\ &\leq& \delta \end{eqnarray*}
But this means that
\[ \lim_{\epsilon\to 0}\frac 1{\text{vol}(B_\epsilon)}\iiint_{B_\epsilon}g(x,y,z)dV =0  \]
which is a restatement of what you want.
[/hide]


ABC is a right triangle(C=90) and CA not equal to CB.CH is an altitude and CL is an angle bisector.Show that for Y different from C on the line CH we have angle YAC different from YBC.

Suppose <YAC=<YBC.Let YA meet BC at D,and YB meet AC at E.Then by [b]Ceva[/b],(EC/EA)(HA/HB)(DB/DC)=1 (Never heard about this,what's the general form of this equality?). But CDA and CEB are similar, so CD/CE = CA/CB. Hence (CB/CA)(HA/HB)(DB/EA) = 1.
But HA/HB = (HA/HC)(HC/HB) = (CA/CB)2 (triangles ACB, CHB, AHC all similar), so CA/CB = EA/DB, so ED is parallel to AB, so ∠CDE = ∠B. But AEDB is cyclic, so [b]∠CDE = ∠A[/b](Why?.They are not angles intercepting a common arc). So CA = CB. Contradiction.

[quote="Iris Aliaj"][b]Ceva[/b],(EC/EA)(HA/HB)(DB/DC)=1 (Never heard about this,what's the general form of this equality?).[/quote]

The Ceva theorem, in its most elementary form, states:

If A', B', C' are three points on the sides BC, CA, AB of a triangle ABC, such that the lines AA', BB' and CC' meet at one point, then

  BA' / A'C * CB' / B'A * AC' / C'B = 1.

(This theorem also has a converse. However, in order to formulate this converse strictly, we need directed segments.)

For a proof, see

  http://www.cut-the-knot.org/Generalization/ceva.shtml

[quote="Iris Aliaj"]But AEDB is cyclic, so [b]∠CDE = ∠A[/b](Why?.They are not angles intercepting a common arc).[/quote]

In fact, they have let out some steps: Since AEDB is cyclic, < EAB = 180 - < EDB; but 180 - < EDB = < CDE, so that < EAB = < CDE or simply A = < CDE.

  Darij


Was wondering if anybody could help clear up this paragraph for me, I don't quite get it:

[quote]...Note that $a_{k}-(A-a_{j}) \ge a_{j}+(A-a_{j})$ so we ca apply our lemma with $(A-a_{j})$ as our $\epsilon$. By our lemma, the average of the numbers in the new set is the same, but the product is now higher. [b]If we continue this process, we make one of the members of the set equal to A with each application of the process. Hence, in some finite number of steps we will make all the numbers equal to A[/b]... [/quote]

My confusion is in the wording of the bolded part; up until this point in the proof the only discussion was of two specific terms in the set (the one closest but smaller than the average A, and the largest element in the set). But then there is talk of "repeating the process", with which elements? How does this process make one of the members of the set equal to A with each application of the "process"?

I realize I left out most of the proof so somebody is going to have to grab their book to help me out, but it would have been a long post if I hadn't done otherwise, much help appreciated!

EDIT: "one", not "none".

Well. It helps to understand where the author's getting at. I think the proof is a little sketchy, but it makes conceptual sense.

The author is implying that you make $a_{k}$ and $a_{j}$ equal to $A$ (remember, the purpose is to maximize the product by using the lemma). Then, you will have another max element and another close to the mean element. And then, you "repeat the process."

I don't understand your latter question, but do you get the conclusion of the proof? It helps to work backwards a lot, especially in math.

Aaaah I see now, I read over the line that explains the implication you speak of:

[quote]Replace $a_{j}$ with $A$ and $a_{k}$ with $a_{k}-(A-a_{j})$.[/quote]

Now I'm starting to see the reasoning. Thank you!


number theory
least common multiple
greatest common divisor
prime factorization

Find the greatest common factor of 9,009 and 14,014.

i wrote out all the multiples of 14014 and 9009 and saw which multiple was the same.
[b]252,252[/b]

[hide][color=indigo]9009 =9*1001
14014 = 14*1001
since 14 and 9 have no prime factors in common, the greatest common factor is 1001.
[/color][/hide]

oops! :blush: i get mixed up with lcm and gcf alot!

[hide]
1001[/hide]

I did prime factorization, then tried to find the number that was on both factorizations (not prime) that was highest up.  It was 1001.


function
search
algebra
functional equation
algebra proposed

Find all functions $ f : \mathbb{R}\rightarrow \mathbb{R}$ such that $ f(x^{2}+y+f(y)) = 2y+f^{2}(x)$ for all real numbers $ x$ and $ y$.

[quote="pohoatza"]Find every function $ f : \mathbb{R}\rightarrow \mathbb{R}$ such that $ f(x^{2}+y+f(y)) = 2y+f^{2}(x)$ for all real numbers $ x$ and $ y$.[/quote]

Here is a rather complex solution (I think some simplier one must exist) :

We have the property $ P(x,y)$ : $ f(x^{2}+y+f(y)) = 2y+f^{2}(x)$

1) $ f(x)$ is surjective :
$ P(x,\frac{a-f^{2}(x)}{2})$ : $ f(\ldots)=a$. Q.E.D.

2) $ f(-x)=f(x)$ or $ f(-x)=-f(x)$
$ P(x,y)$ : $ f(x^{2}+y+f(y)) = 2y+f^{2}(x)$
$ P(-x,y)$ : $ f(x^{2}+y+f(y)) = 2y+f^{2}(-x)$
So $ f^{2}(-x)=f^{2}(x)$.
Q.E.D.

3) $ f(x)=0$ $ \Leftrightarrow$ $ x=0$
Since $ f(x)$ is surjective (see 1). above), it exist $ a$ such that $ f(a)=0$
Then $ P(0,a)$ gives $ 0 = 2a+f^{2}(0)$ and so $ a\leq 0$
But $ f(a)=0$ implies $ f(-a)=0$ (from 2). above) and so, with the same method, $ -a \leq 0$
And so $ a=0$.
Q.E.D.

4) $ f(x^{2})=f^{2}(x)$ and so $ f(x)>0$ $ \forall x>0$
Immediate with $ P(x,0)$.

5) $ f(x+f(x))=2x$ $ \forall x$
Immediate with $ P(0,x)$

6) $ f(x)\leq-x$ $ \forall x\leq 0$
Since $ f(x)\geq 0$ $ \forall x\geq 0$ (see 4.), and since $ f(x+f(x))=2x$ (see 5.), $ f(x+f(x))\leq 0$ $ \forall x\leq 0$ and so $ x+f(x)\leq 0$ $ \forall x\leq 0$
Q.E.D.

7) $ f(-x)=-f(x)$ $ \forall x$
We know (see 2.) that $ f(-x)=f(x)$ or $ f(-x)=-f(x)$ $ \forall x$
Assume then there exists $ a>0$ such that $ f(a)=f(-a)$
$ P(\sqrt{a},-a)$ : $ f(f(a)) =-2a+f(a)$
And since $ a>0$, $ f(a)>0$ and so $ f(f(a))>0$. So $ -2a+f(a)>0$ and $ f(a)>2a$
But $ f(a)=f(-a)$ and $ f(-a)\leq a$ (see 6.). So $ 0<2a<f(a)\leq a$ which is impossible and no such $ a$ exist.
So $ f(-x)=-f(x)$ $ \forall x$
Q.E.D.

8) $ f(x)+x$ is a surjective function.
Let $ x\geq 0$. $ P(\sqrt{\frac{x}{2}},-\frac{x}{2})$ : $ -f(f(\frac{x}{2})) =-x+f(\frac{x}{2})$ and so $ f(f(\frac{x}{2}))+f(\frac{x}{2})= x$
So $ f(x)+x$ may have any nonnegative value, and since $ f(x)+x$ is an odd function, $ f(x)+x$ may have any real value.
Q.E.D

9) $ f(x+y)=f(x)+f(y)$ $ \forall x,y$
From 4. and 5. above, the functional equation may be written : $ f(x^{2}+y+f(y)) = f(x^{2})+f(y+f(y))$
And so, since $ x+f(x)$ is surjective : $ f(x+y)=f(x)+f(y)$ $ \forall x\geq 0$, $ \forall y$
And since $ f(x)$ is an odd function : $ f(x+y)=f(x)+f(y)$ $ \forall x,y$
Q.E.D.

10) $ f(x)=x$ $ \forall x$
From 9., we know that $ f(x)$ is a solution of Cauchy's equation.
From 4., we know that $ f(x)>0$ $ \forall x>0$ and so $ f(x)$ is continuous (all non continuous solutions of Cauchy's equation have neither upper, nor lower bound on any non empty interval).
So $ f(x)=ax$
And, putting back this solution in the original equation, we get $ a=1$

And so $ f(x)=x$
Q.E.D.

I also posted it some time ago without knowing that it had already been posted three times before
http://www.mathlinks.ro/viewtopic.php?search_id=1056481267&t=75846
 :)

Well the good part is that we would had not seen pco's beautiful solution if not reposting it  :)

:maybe:
Isn't it like Darij's solution at http://www.mathlinks.ro/Forum/viewtopic.php?t=68972 ??


calculus
integration
algebra
function
domain
polynomial
linear algebra

Let $ R$ be a commutative ring with $ 1$, and $ A\in R^{n\times n}$ be a nilpotent matrix. ("Nilpotent" means that there exists an $ m\in\mathbb{N}$ such that $ A^m\equal{}0$; we need not have $ m\equal{}n$ here.)
Prove that $ \text{Tr }A$ is a nilpotent element of $ R$.

  darij

[hide]Reducing mod $ P$ for a prime ideal $ P$, we obviously have $ Tr(A)\in P$. Since this is true for every prime, we get the result.[/hide]

[hide]And we prove the statement for $ R$ an integral domain by passing to the field of fractions and using an eigenvalue argument?  Do I have that right?[/hide]

[EDIT: minor flaw fixed, thanks Myan]

Both of you are right, this is the simplest argument if one does not go for constructivity.

Here is the constructive solution found by Peter Scholze:

We use a result proved in http://www.mathlinks.ro/Forum/viewtopic.php?t=89417 :

[color=blue][b]Theorem 1.[/b] Let $ f$ be a polynomial in $ R\left[X\right]$, where $ R$ is a commutative ring with $ 1$. Then, ($ f$ is invertible in $ R\left[X\right]$) if and only if (the coefficient of $ f$ before $ X^0$ is invertible in $ R$, and all other coefficients of $ f$ are nilpotent).[/color]

Another well-known fact:

[color=blue][b]Theorem 2.[/b] Let $ S$ be a (not necessarily commutative) ring with $ 1$. Let $ n\in S$ be nilpotent, and $ x\in S$ be arbitrary such that $ xn \equal{} nx$. Then, $ 1 \minus{} xn$ is invertible in $ S$.[/color]

Back to our problem: Consider the matrix $ A$ in the ring $ \left(R\left[X\right]\right)^{n\times n}$ instead of the ring $ R^{n\times n}$ (here we use the canonical embedding $ R^{n\times n}\to \left(R\left[X\right]\right)^{n\times n}$). Apply Theorem 2 to $ S \equal{} \left(R\left[X\right]\right)^{n\times n}$ (the $ 1$ of this ring is the identity matrix $ I_n$), $ n \equal{} A$ and $ x \equal{} XI_n$ (note that $ xn \equal{} nx$ follows from $ XI_n\cdot A \equal{} A\cdot XI_n$, what is obvious), and obtain that $ I_n \minus{} XI_n\cdot A$ is invertible. In other words, $ I_n \minus{} XA$ is invertible.

Thus, there exists $ B\in\left(R\left[X\right]\right)^{n\times n}$ such that $ \left(I_n \minus{} XA\right)B \equal{} I_n$. Hence, $ \det\left(I_n \minus{} XA\right)\det B \equal{} \det\left(\left(I_n \minus{} XA\right)B\right) \equal{} \det I_n \equal{} 1$, so that $ \det\left(I_n \minus{} XA\right)$ is invertible in $ R\left[X\right]$.

So $ \det\left(I_n \minus{} XA\right)$ is a polynomial in $ R\left[X\right]$ which happens to be invertible in $ R\left[X\right]$. Thus, by Theorem 1, the coefficient of this polynomial before $ X^0$ is invertible, while all its other coefficients are nilpotent. In particular, the coefficient of $ \det\left(I_n \minus{} XA\right)$ before $ X^1$ is nilpotent. But the coefficient of $ \det\left(I_n \minus{} XA\right)$ before $ X^1$ is $ \minus{} \text{Tr }A$ (in fact, working in the ring extension $ R\left[X,X^{ \minus{} 1}\right]$ of $ R\left[X\right]$, we have $ \det\left(I_n \minus{} XA\right) \equal{} \det\left(X\left(X^{ \minus{} 1}I_n \minus{} A\right)\right) \equal{} X^n\det\left(X^{ \minus{} 1}I_n \minus{} A\right)$, so that

(the coefficient of the polynomial $ \det\left(I_n \minus{} XA\right)$ before $ X^1$)
= (the coefficient of the Laurent polynomial $ \det\left(X^{ \minus{} 1}I_n \minus{} A\right)$ before $ X^{1 \minus{} n}$)
= (the coefficient of the Laurent polynomial $ \det\left(X^{ \minus{} 1}I_n \minus{} A\right)$ before $ X^{ \minus{} \left(n \minus{} 1\right)}$)
= (the coefficient of the polynomial $ \det\left(XI_n \minus{} A\right)$ before $ X^{n \minus{} 1}$)
= (the coefficient of the characteristic polynomial of the matrix $ A$ before $ X^{n \minus{} 1}$)
= $ \minus{} \text{Tr }A$

as known from linear algebra). Hence, $ \minus{} \text{Tr }A$ is nilpotent, so that $ \text{Tr }A$ is nilpotent. Problem solved.

Note that Theorem 1 is absolutely obvious if $ R$ is a field (or integral domain), so we now have an easy solution of the problem over fields without using eigenvectors or Cayley-Hamilton.

  darij


function
Euler
number theory
totient function

[quote="Los"]For the Euler totient function $ \phi(n)$ which gives the number of coprimes less than or equal to $ n$, how can $ \phi(n)$ be found for large $ n$, without actually finding the numbers coprime to $ n$?[/quote]

[url]http://en.wikipedia.org/wiki/Totient[/url]

The best thing I could recommend would be the Euler Product, see the page linked above.

However, if this is a coding question, and you are going to be working out a large number of them, use the fact that is a multiplicative function

$ \phi(n) \equal{} n(1\minus{}\frac{1}{p_1})(1\minus{}\frac{1}{p_2})...(1\minus{}\frac{1}{p_m})$

where $ p_a$ are the prime factors of $ n$.


An organic nitro compound $ \ce {RNO_2}$ is electrolytically reduced in an aqueous acetate buffer solution having total acetate ($ \ce {HOAc} + \ce {OAc^-}$) 0.500 and pH=5.0. 300 ml of the buffered solution containing 0.01M RNO2 was reduced completely. The dissociation constant of acetic acid is $ 1.75 \cdot 10^{-5}$ at 298 K. The reaction is

$ \ce {RNO_2} + 4\ce {H^+} + 4e^- \rightarrow \ce {RNHOH} + \ce {H_2O}$.

Calculate the pH of the solution on completion of the reduction.


I am having some trouble getting TeX into PDF. I can do the LaTeX ==> DVI fine, but the LaTeX ==> PDF randomly freezes every time I try it. I checked the contents of my folder afterwards but no .pdf was present. The only problem I can think of is the fact that I never actually set the DVI output profile, but it was already set for me (and this seems unlikely to screw up my PDF stuff). Anyone have any ideas why this isn't working?

Actually, I just got it to work a couple times, but it doesn't work again now. Maybe it's one of those random problems from Windows XP...

Didn't bookaholic have a similar problem here?: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=60033

Yep, looks like I have the same problem. But I can deal with it now--looks like rebooting offers a high chance of (albeit perhaps temporary) success.

I use a different LaTeX system from TeXnic Center, but I recall having similar problems on Windows XP.  For some reason, there used to be copies of the Acrobat Reader process (AcroRd32.exe) running even after I had exited the Adobe Reader application.  When I manually killed those zombie processes via the Windows Task Manager, the problem seemed to disappear.


function
integration
real analysis
real analysis unsolved

Let $f: [0,1]\to\mathbb{R}$ be a continuous function such that $f(1)=0$.
   Prove that there is $c\in (0,1)$ such that \[f(c)=\int_{0}^{c}f(x)dx.\]

Assume the contrary. Without loss of generality, by continuity of $f$, we can suppose that $F'(x)-F(x)>0$ (*) for all $x \in (0,1)$, where $F(x)=\int_{0}^{x}{f(u)du}$. But solving gives $\forall x \in (0,1), F(x)>0$, which implies that $\forall x \in (0,1), f(x)>0$.

Now consider $g(x)=f(x)-\int_{0}^{x}{f(u)du}$. We have $g(1)=-F(1)<0$, and $g(x)>0$ on $(0,1)$ by (*). Thus there must exist $x_{0}\in (0,1)$ such that $g(x_{0})=0$.


calculus
integration
function
trigonometry
Euler
geometry
geometric transformation

Prove that for every real number $r>1$,

$\int_{0}^{\infty}\frac{dx}{1+x^{r}}= \frac{\pi}{r}\csc \left( \frac{\pi}{r}\right)$

Of course I proved it, but my proof is not so simple nor beautiful... so I wanna know if there is another idea to prove this.

In terms of a function of a complex variable, wouldn't the integrand have non-isolated (dense) singularities when r is an irrational number?  Seems to me anyway in which case residue calculus would not apply.  Kindly explain if not so.

My approach is to define $I(r) = \int_{0}^{\infty}\frac{dx}{1+x^{r}}$, calculate its value when $r$ is rational, and show that $I(r)$ is a continuous function.

[quote="jmerry"]See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=21316]here[/url] for multiple approaches. The problem there was phrased as a sum, but it can easily be converted into the integral.[/quote]

Good grief . . . :oops:  what was I thinking.  I reviewed it and now understand.  Sorry SOS.

This Integral is also a special case of the Beta-Function. Since:

$B(x,y) = \frac{\Gamma(x) \Gamma(y)}{\Gamma(x+y)}= \int_{0}^{\infty}\frac{s^{x-1}}{(1+s)^{x+y}}\,ds$

$\Re(x), \Re(y) > 0$

and: $\Gamma(1-x) \Gamma(x) = \frac{\pi}{\sin(\pi x)}$ (Euler Reflection Formula)

So we have:
$\frac{1}{r}\frac{\pi}{\sin(\frac{\pi}{r})}= \frac{1}{r}\Gamma\left(1-\frac{1}{r}\right) \Gamma\left(\frac{1}{r}\right) = \frac{1}{r}B\left(\frac{1}{r}, 1-\frac{1}{r}\right) = \frac{1}{r}\int_{0}^{\infty}\frac{x^{\frac{1}{r}-1}}{1+x}\,dx$
$= \int_{0}^{\infty}\frac{1}{1+x^{r}}\,dx$

which is clear by substitution $x \to x^{r}$.


Given the equation \[4x^3 + 4x^2y - 15xy^2 - 18y^3 - 12x^2 + 6xy + 36y^2 + 5x - 10y = 0,\] fina all positive integer solutions.

Does anybody want to finsh the solution?  :)


function
algebra
domain
calculus
calculus computations

Study the continuity of the function
$ f(x,y)=\{
\begin{array}{ll}
xe^{-\frac{x^2}{y}}&; y\neq 0\\
0&; y= 0
\end{array}$
defined in a closed upper half plane.

I forgot to write the other question. The situation with the upper half plane is obvious, since near the critical points, where $ y\equal{}0$, $ \frac{x^2}{y}\geq 0 \Rightarrow \frac{x}{e^{\frac{x^2}{y}}}\to 0$, where $ (x,y)\to (x_0,0)$.

However, what about if the domain is $ \mathbb{R}^{2}$?

If the domain is $ \mathbb{R}^{2}$ , the function is not continuous in $ (x_0,0)$  :cool:


Find all polynomials $ P\in\mathbb{R}[x]$ such that for all $ r\in\mathbb{Q}$,there exist
$ d\in\mathbb{Q}$ such that $ P(d)\equal{}r$

Suppose P has degree n and $ P(x)\equal{}\sum_{i\equal{}1}^na_nx^n$. Obviously all $ a_n$ have to be rational. We notice that $ P(ax)\plus{}b$ holds when $ P(x)$ holds, where a and b are rationals, a is non-zero. We can assume, wlog, all $ a_n$ are integers and $ a_0\equal{}0$. Let q be a prime with $ gcd(q,a_n)\equal{}1$. To solve $ P(x)\equal{}q$, we have to have $ x\equal{}\frac yz$, where y=1 or y=q and z|a_n. Note here z can be negative. We can always find a $ z$ such that $ z|a_n$ and $ P(\frac qz)\equal{}q$ holds for infinitely many $ q$. In other words, $ P(x)\minus{}zx\equal{}0$ has infinitely many roots. Hence, $ P(x)\equal{}zx$ with $ z$ rational. Inverting the $ P(ax)\plus{}b$ operation, we know P(x)=ax+b, where a,b are rational and a is nonzero.

[quote="xxp2000"]Obviously all $ a_n$ have to be rational. [/quote]
I don't think this is obvious at all.  In particuar, we aren't given the other direction ($ d$ rational implies $ P(d)$ rational).

P(x) has degree of n. You can find n+1 distinct pairs of $ (d_i, r_i)$ such that $ P(d_i)\equal{}r_i$. then you can construct P(x) by the following formula
$ P(x)\equal{}\sum_{i\equal{}1}^nr_i\frac{(x\minus{}d_1)...(x\minus{}d_{i\minus{}1})(x\minus{}d_{i\plus{}1})...(x\minus{}d_{n\plus{}1})}{(d_i\minus{}d_1)...(d_i\minus{}d_{i\minus{}1})(d_i\minus{}d_{i\plus{}1})...(d_i\minus{}d_{n\plus{}1})}$.
Please let me know if it is still obvious.


It's okay, I know how to do it now. You can delete this if you like.

I'm interested. Can you post the solution(s)?

... but is my solution correct?...  :roll:


Let a,b,c,d be real numbers $\in [1,2]$.

Show that:

     $( \frac{a+b} {b+c}) + (\frac{c+d }{d+a})\leq 4( \frac{a+c}{b+d})$

Hi Mathlinkers ....... ;)

Any idea to solve this ??

carlos Bravo .
Lima - PERU


how are you suppose to do those and what are they?  :cry:

1) identify and discuss any trends.
2)identify and describe 3 features.

[quote="xxreddevilzxx"]i dont think this is appropriate for Pre-Olympiad. Unless, if it was an Olympiad problem, which i doubt. (I dont see that many graphs in Olympiad). So: relocation, please.[/quote]I would say this is inappropriate for the entire forums..

Why isn't there a statistics section? I guess statistics rarely show up in problem solving, but I'm sure there are many people who would appreciate somewhere to talk about interesting problems.

Perhaps you are to look at the residual plots and comment on the goodness of fit. Are there better models for these data? Look for outliers. Address the problem in context. What does it mean that the best fit line has a slope of -.0778?


algebra
polynomial
superior algebra
superior algebra unsolved

Let $K$ be a field of characteristic different from $2$ and let $a_{1},...,a_{n}$ in $K$ such that the equation $a_{1}x_{1}^{2}+...+a_{n}x_{n}^{2}=0$ has only the trivial solution in $K^{n}$. Now, let $L$ be a field extension of odd degree. Prove that the same equation has only trivial solutions in $L^{n}$. I will not give the hint which we had in the exam, to leave people come up with nice approaches.

hi,
its enough to consider simple extensions of $K$.
as if else we first set $L_{0}$ to be the subfield of $L$ containing $K$ and maximal with respect to having no solutions for the equation under consideration and ........ .

now Let $L=K(\beta)$,$\lambda_{1},....\lambda_{n}$ be non trivial solutions in
$L^{n}$(its our assumption).Let $\lambda_{i}= g_{i}(\beta)$,$F(X)=\Sigma(a_{i}(g_{i}(X)^{2})$
Now we have $F(X)$ has no roots in $K$,moreover  $F(X)=F(-X)$,
also the constant term is non zero .So every irreducible factor of $F$ is a polynomial of even degree,infact with all its nonzero terms have even exponents in $X$.So now our assumption implies one of these irreducible factors have a root in an odd degree extension of $K$ a contradiction........ :roll:

Hi!

I can't understand the above post, but here is a real solution:

Assume the contrary and consider all pairs of extensions $ K'\subset L$ of $ K$ of odd degree such that there is a solution over $ L$, but not over $ K'$. Take one such pair for which the degree $ m$ of the extension $ L/K'$ is minimal.

Take $ \beta\in L\backslash K'$. By the minimality condition, we easily see that $ L \equal{} K'(\beta)$. Let the minimal polynomial of $ \beta$ over $ K'$ be $ P$. Now take a solution $ \lambda_{1},...,\lambda_{n}$ over $ L$. Write $ \lambda_{i} \equal{} g_{i}(\beta)$ for some polynomials $ g_{1},...,g_{n}$ (over $K'$) of degree at most $ m \minus{} 1$. Then $ P(x)$ divides $ S(x) \equal{} \sum_{i \equal{} 1}^{n}a_{i}g_{i}(x)^{2}$ because $ S(\beta) \equal{} 0$ and $ P$ is irreducible. Now, $ S(x)$ has even degree at most $ 2(m \minus{} 1)$ since there is no solution in $ K'$ (look at the leading coefficient). Therefore the quotient $ \frac {S(x)}{P(x)}$ is a polynomial of odd degree $ \leq 2(m \minus{} 1) \minus{} m \equal{} m \minus{} 2 < m$. Taking a zero $ \gamma$ of an irreducible factor of odd degree of this polynomial, we also have $ S(\gamma) \equal{} 0$ and thus $ g_{1}(\gamma),...,g_{n}(\gamma)$ is a solution in a field $ L' \equal{} K'(\gamma)$ of degree $ < m$, contradiction.

I wonder where we use that the field does not have characteristic 2...


For  $a,b,c,x,y,z $ positive real numbers such that $a+b+c=x+y+z=1$ prove

$ \frac{ax}{a+x} +\frac{by}{b+y} +\frac{cz}{c+z} \leq \frac{1}{2} $

$\frac{ax}{a+x} \leq \frac{a+x}{4}$ by AM-HM (or simply factoring). Add its three analogues up and we get the desired ineq.

And now you can try this one very easy again.

For $a,b,c,x,y,z$ positive reals numbers prove that

$ \frac{ax}{a+x} +\frac{by}{b+y} +\frac{cz}{c+z} \leq \frac{(a+b+c)(x+y+z)}{a+x+b+y+c+z} $

This inequality is a direct consequence of the first one you posted. And following the other problem solution, you will end proving this one too. right?

A hint: prove that $\frac{ab}{a+b}+\frac{cd}{c+d}\leq\frac{(a+c)(b+d)}{a+b+c+d} $ using brute force. It is easy.


trigonometry
function
limit
real analysis
real analysis unsolved

Given $ \sum sin ( \frac {x}{n^2} )$.
for $ |x| \le a$ , $ a > 0$ ,we have uniform convergence since $ |sin ( \frac {x}{n^2} ) | \le | \frac {x}{n^2} | \le | \frac {a}{n^2} |$ and by Weistrass M-test.
What can we say when $ |x| > a$?
Can anyone help?

It doesn't converge uniformly on the whole line (which seems to be what you're asking). Each term, $ \sin\left(\frac x{n^2}\right),$ is equal to $ 1$ for some values of $ x,$ and cannot be made uniformly small.

The particular kind of convergence we have here - pointwise everywhere and uniform on compact subsets of the line - is natural and common. It's the way that power series for entire functions converge, of course. And we can draw useful conclusions  - for instance, the function defined by your series is continuous on the whole line.

By contrast, uniform convergence on the whole line is just kind of odd; it's fairly rare, it's not very natural, and there aren't many conclusions that it gets you that you didn't already have from uniform on compact subsets.

For $ \vert x \vert >a,$ If I understand, consider $ \vert x \vert \le a'$ and $ a'>a.$ The series converges uniformly
also on  $ [\minus{}a',a']$ by the same proof. Actually the convergence is uniform on any interval unless the values of $ x$ are restricted far some apriori reason.

Kent Merryfield,I am confused in some parts.We have $ sin (\frac {x}{n^2} )\equal{}1$ for some $ x$.But if by this argument,it means that for the particular $ x$ value,we will not have the convergence of the series at $ x$.
But $ \sum sin (\frac {x}{n^2})$ converges for every $ x \in \mathbb{R}$

Will it be any contradiction here?
Or if I am wrong in my argument,can you please point out to me?

I think that your understanding of uniform convergence is shaky.

A sequence of functions does not converge uniformly at a point; it converges uniformly on some set. It makes absolutely no sense to say that the sequence "converges uniformly at $ x$" - that would be a meaningless statement. And the order of quantifiers is everything.

Here's $ g_n(x)\to g(x)$ pointwise for every $ x\in S:$

$ \forall x\in S$ and $ \forall \epsilon > 0,$ $ \exists N$ depending on $ x$ and $ \epsilon$ such that if $ n > N,$ then $ |g_n(x) \minus{} g(x)| < \epsilon.$

Here's $ g_n(x)\to g(x)$ uniformly on $ S:$

$ \forall \epsilon > 0,$ $ \exists N$ depending only on $ \epsilon$ such that if $ n > N$ and $ x\in S,$ then $ |g_n(x) \minus{} g(x)| < \epsilon.$

There's actually another layer of quantifier here: $ \forall a < 0,$ the series converges uniformly on $ [ \minus{} a,a].$ Since uniform convergence on $ S$ implies pointwise convergence for every $ x\in S,$ this means that the series converges pointwse on all of $ \mathbb{R}.$ But it does not mean that it converges uniformly on $ \mathbb{R}.$

For the series $ \sum f_n(x)$ to converge uniformly on $ S,$ is it necessary (but not sufficient) that $ \lim_{n\to\infty}\sup_{x\in S}|f_n(x)| \equal{} 0.$ The comment that $ \sin(x/n^2) \equal{} 1$ for some $ x$ was meant to show that $ \sup_{x\in\mathbb{R}}\sin(x/n^2) \equal{} 1,$ which is enough to negate uniform convergence on $ \mathbb{R}.$


algebra
polynomial
number theory unsolved
number theory

Let $ f(x)\equal{}a_0\plus{}a_1x\plus{}\cdots\plus{}a_nx^n$ be a polynomial with integer coefficients, where $ a_n>0$ and $ n\geq 1$. Prove that $ f(x)$ is composite for infinitely many integers $ x$.


vector
real analysis
real analysis unsolved

Suppose $ A$ is an algebra. Then we can consider $ e$ a multiplicative identity element (as in an algebra) and also $ 0$ an additive identity (as in a vector space) right?
I am doing some calculation and came to the following line:
$ xz \equal{} 0 \; \forall z \in A$ where $ xz$ is multiplication which makes sense as $ A$ is an algebra. I need to somehow conclude that $ x \equal{} 0$ from this but I cannot see how. Maybe I made some mistake in my calculation?

I appreciate your help.

Note the $ \forall$ quantifier there.

If your algebra does have an identity, we can simply use $ e$ for $ z$ to conclude that $ x\equal{}xe\equal{}0.$

If your algebra does not have an identity, this is less trivial. In the Banach algebra setting, you might try asking whether you have sequences that converge in some (probably weak) sense to the identity that isn't there, and whether such sequences would do you any good.

thank you!
luckily the algebra I'm working with does have an identity $ e$. (It was part of the assumption in the question)
Thanks!


Gauss
integration
trigonometry
LaTeX
calculus
analytic geometry
symmetry

You can't use Gauss' Law but you can still set up an integral.  Consider an infinitesimal piece of rod of length $ dx$.  Pick an origin, define the distance from the origin to the point, and integrate along $ dx$ from one side to the other.  I got something looking like this:
$ \int_{\minus{}\frac{r}{\tan{\alpha}}}^{\frac{r}{\tan{\beta}}}\frac{\lambda L}{r^{2}\plus{}L^{2}}\ dL$
where $ \lambda$ is linear charge density and $ r$ is the distance from the rod to the point.  I would check it though since I'm not completely sure whether or not I messed something up.  The method is correct, though.

Ok, but how did you manage to work out this answer? Could you please make yourself clear?
Of course we should pick an infinitesimal piece, let it be dl.Hence dq=[i]t[/i]*dl (sorry for LaTex)
I guess we should pick an infinitesimal piece of angle da, too. We have dl=(r*da/cosa).

I don't know what to do further! :maybe:
How should the two angles be connected to each other?(except of the integration point of course)

I made some mistakes but here is how you should go about it.

1) Pick an origin.
2) Take your infinitesimal charge and treat it as a point charge obviously so
$ \vec{E}\equal{}\frac{dQ}{r^{2}}\equal{}\frac{\lambda x}{r^{2}}$
where $ r$ is the distance between your "point charge" and the point you picked.
3) Set up your integral and evaluate it.

The tricky part of this problem is making sure the coordinate system is set up properly and then finding the limits of integration.  However, once you've figured that out, the integral can be found in most tables or evaluated with a trig substitution.

[quote="10001"]The conductor [b]AB[/b] is charged uniformally with a charge density equal to [i]t[/i]
Calculate the intensity of the electric field at a point P(its distance from [b]AB[/b] is [i]a[/i]) as shown below.

Thanks in advance :)[/quote]
\[ V \equal{} \int_{Q(A)}^{Q(B)} k \frac {\mathrm{d}q}{r}
\]

\[ E_x \equal{} \minus{} \frac {\partial V}{\partial x} \equal{} \minus{} k\int_{Q(A)}^{Q(B)}\frac {\partial(\frac {1}{r})}{\partial x} \,\mathrm{d}q \equal{} k\int_{Q(A)}^{Q(B)}\frac {1}{r^2}\frac {\partial r}{\partial x}\,\mathrm{d}q
\]

\[ r^2 \equal{} z^2 \plus{} x^2 \\
\Rightarrow 2r\frac {\partial r}{\partial x} \equal{} 2x \\
\Rightarrow \frac {\partial r}{\partial x} \equal{} \frac {x}{r}
\]

\[ E_x \equal{} k\int_{Q(A)}^{Q(B)}\frac {x}{r^3}\,\mathrm{d}q \equal{} kx\int_A^B\frac {t}{(z^2 \plus{} x^2)^{3\over{2}}}\,\mathrm{d}z
\]
Let $ \displaystyle{x \equal{} a}$  then $ \displaystyle{E_x \equal{} ka\int_A^B\frac {t}{(z^2 \plus{} a^2)^{3\over{2}}}\,\mathrm{d}z}$ .
\[ E_x \equal{} \left. kta \frac {z}{a^2\sqrt {z^2 \plus{} a^2}}\right|_{z \equal{} A}^B \equal{} kt\left(\frac {B}{a\sqrt {B^2 \plus{} a^2}} \minus{} \frac {A}{a\sqrt {A^2 \plus{} a^2}}\right)
\]

Edit: There is also a component $ E_z$ that is calculated in a similar way. However, due to symmetry if the point $ P$ is at the mid-point of the wire then the $ E_z$ component vanishes.

Set up your integral:
$ \frac{\lambda R}{4\pi\epsilon_{0}}\int_{\minus{}\frac{R}{\tan{\beta}}}^{\frac{R}{\tan{\alpha}}}\frac{dx}{((d\minus{}x)^{2}\plus{}R^{2})^{3/2}}$
where $ d\minus{}x$ is the distance from your origin to the a point on the rod and $ R$ is the distance from the rod to the point.  Evaluating the integral gives
$ \frac{\lambda}{4\pi\epsilon_{0}R}\left(\frac{x\minus{}d}{\sqrt{(d\minus{}x)^{2}\plus{}R^{2}}}\right)\bigg|_{\minus{}\frac{R}{\tan{\beta}}}^{\frac{R}{\tan{\alpha}}}\equal{}\frac{\lambda}{4\pi\epsilon_{0}R}\left(\frac{R\minus{}d\tan{\alpha}}{\tan{\alpha}\sqrt{\left(d\minus{}\frac{R}{\tan{\alpha}}\right)^{2}\plus{}R^{2}}}\plus{}\frac{R\plus{}d\tan{\beta}}{\tan{\beta}\sqrt{\left(d\plus{}\frac{R}{\tan{\beta}}\right)^{2}\plus{}R^{2}}}\right)$
If you want, you can then make a substitution for $ \tan{\alpha}$, $ \tan{\beta}$.

The problem has a cylindrical symmetry and can be extended to R3.

As a matter of fact, it is solved in cylindrical coordinates in a Brazilian Book PROBLEMAS DE FÍSICA - ELETRICIDADE - (ELETROSTÁTICA), by Eduardo Wilner.

In your notation it comes out like that:

\[ E_x\; \equal{}\; {t \over {4 \pi \epsilon_o a}} (cos \alpha \minus{} cos \beta)\]

\[ E_y\; \equal{}\;{t \over {4 \pi \epsilon_o a}} (sin \alpha \minus{} sin \beta)\]


vector
superior algebra
superior algebra unsolved

I'm stuck on these problems...

1. Suppose that $ L$ has transcendence degree $ n$n over $ K$ and that $ L$ is algebraic over $ K(\alpha_{1},\dots,\alpha_{n})$. Show that $ \alpha_{1},\dots,\alpha_{n}$ is a transcendence basis for $ L$ over $ K$.

2. Show that the algebraic closure of $ \mathbb{Q}$ in $ \mathbb{C}$ is countable.

3. Let $ K$ be an algebraically closed field and let $ D$ be a division ring that is also  a $ K$-algebra. Suppose that $ D$ is finite dimensional as a vector space over $ K$, under the vector space structure induced by its $ K$-algebra structure. Show that the inclusion $ K \subset D$ given by the $ K$-algebra structure map is a bijection, so that $ K\equal{}D$.

Thanks in advance

[quote="futurus"]I'm stuck on these problems...[/quote]I hope that the following hints can be helpful for you:[quote]1. Suppose that $ L$ has transcendence degree $ n$n over $ K$ and that $ L$ is algebraic over $ K(\alpha_{1},\dots,\alpha_{n})$. Show that $ \alpha_{1},\dots,\alpha_{n}$ is a transcendence basis for $ L$ over $ K$.[/quote]What if the $ \alpha_i$'s are algebraically dependent over $ K$?
[quote]2. Show that the algebraic closure of $ \mathbb{Q}$ in $ \mathbb{C}$ is countable.[/quote]How can you inductively construct the algebraic closure of $ \mathbb Q$? Note that every finite extension of a countable field is countable again.
[quote]3. Let $ K$ be an algebraically closed field and let $ D$ be a division ring that is also  a $ K$-algebra. Suppose that $ D$ is finite dimensional as a vector space over $ K$, under the vector space structure induced by its $ K$-algebra structure. Show that the inclusion $ K \subset D$ given by the $ K$-algebra structure map is a bijection, so that $ K \equal{} D$.[/quote]Consider the sub-algebra $ K[a]$ of $ D$ generated by some element $ a\in D$.


Support
geometry
geometric transformation

I was studying for my philosophy class and this question popped into my mind... It seems that there are quite a few philosophers whose works are now pretty highly regarded; some of them have lived in antic times, some are our contemporaries. This is a question addressed to those who have at least a basic knowledge of the development of philosophical thought over these 2500 years: can you single out a philosopher whose thought appeals to you the most and explain why you have chosen that particular person?

"I think therefore I am." -Descartes

dunno why but he's all i can think of at the moment

If someone provides a list to choose from, I could choose one.

I haven't provided a list or set up a poll because it wouldn't have covered all the possible choices. Even supposing that I could've constructed a list, it would be so long nobody would have the time to go through it and think about every name on it. This is not a contest with a multiple-choice question; just say whether you have a philosopher whose work you approve of, and, if you do, who he/she is and why do you support his/her view.

I have made very little study of philosophy, but I have developed an appreciation for utilitarianism (particular fondness for J.S. Mill, and I know little about Peter Singer but he also seems quite promising), even if I fail to live my life in a "good" way from the utilitarian standpoint.

Dude Plato is pretty awesome. I don't really know enough about other philosophers to compare, but I like book 4 of The Republic. In many ways Plato's ideas are simple yet insightful, and I haven't really seen much of that in ideas from other philosophers.

Sartre and Nietzsche are on top of my list... Especially Sartre's final phase (the marxist phase). Djole, thank you for reminding me that I have to study for my philosophy test on Monday... :D

Nietzsche of course !

At liest for "You must have a lot of chaos inside yourself to give birth to a star wich is dancing" ! (I'm afraid I have not correctly translated my signature...)

That translation looks pretty good to me ... I might say "which dances" instead of "which is dancing."

Hegel has to be given some significant credit for directing attention to the concept of the dialectic, which has a kind of universality that transcends a particular philosopher's particular philosophy.

Nietzsche!

Sartre, too.

Of the ancient philosophers, I happen to like Zeno and Democritus.

Actually, I found it really hard to answer this question, although I'm the one who posted it. In our philosophy class we have so far systematically covered everything up to (and including) Kant, and I've read some works by Nietzsche and the existentialists (who do seem promising), but so far the only philosopher I agree with completely is Protagoras.

Let me elaborate on this: what few sayings are atributed to Protagoras are things I really couldn't object to: that man is a measure for everything, that gods are created by humans in order to instill the morale within humanity, that absolute truth doesn't exist (it's not said with these exact words, but these are the points). Later I read Plato's attempts to refute him (eg. "Why isn't a baboon a measure of everything"), but they all lead to pretty obvious mistakes in logic. So far, I haven't found a point on which I would disagree with him.

So, I'm wondering, have you even heard of him and do you have a view on his dialectics?

[quote="Djole"] so far the only philosopher I agree with completely is Protagoras.
So, I'm wondering, have you even heard of him and do you have a view on his dialectics?[/quote]

[url]http://en.wikipedia.org/wiki/Protagoras[/url] I'm just discovered him but I think that Nietzsche is closer of human's behavior (passion, etc.)then anyone ! I believe I am wrong but I haven't studied the Philosophy like pupil, only like a adult ! :roll:

Yes, I've checked Wikipedia, but there isn't much about him. We've learned more about him in school  :maybe: As for Nietzsche, he has certainly covered a broader set of topics, but I'm still not well acquainted with his works to be any judge.

I've found more about Protagoras here:
[url]http://www.iep.utm.edu/p/protagor.htm[/url] :roll:

Who is my favorite philosopher? Definitely my wife.

Nietzsche is the winner !
I love his comments on religions.

does Richard Dawkins count as a philosopher?
he's my favorite :)

Does Peter Nilson count? He would definitely be my favourite.

To bring in a new point: I disagree with the idea of human being the ultimate measure of everything. I am here by chance, and I strongly believe that we all are, and there might be some structure that "grasps" and "understands" the universe and things beyond it far better than we ever might be able to, and this structure would be a better measure of everything.  :o  [Edit: ok, this may still be conform with the idea that human is the only and hence best measure for everything, which is available [i]for us[/i].]

To me, the deepest philosophical question is due to Pascal. I don't know the correct formulation in English, but his question goes along the lines: why there is Anything rather than Nothing? So cool, so many have tried their best to find an answer and either given up or failed miserably  :lol:

[quote="uldis"]To bring in a new point: I disagree with the idea of human being the ultimate measure of everything. I am here by chance, and I strongly believe that we all are, and there might be some structure that "grasps" and "understands" the universe and things beyond it far better than we ever might be able to, and this structure would be a better measure of everything. Shocked [Edit: ok, this may still be conform with the idea that human is the only and hence best measure for everything, which is available for us.][/quote]

Well, I believe that Protagoras shouldn't be taken too literally. I think that his points are valid for subjective impressions (i.e. what we think as individuals). From that perspective, it would be absurd to claim that any absolute proposition exists; for example, how could you assert that [i]every[/i] human finds a painting beautiful? You can't because you have no idea what other people think. In this sense, there doesn't exist an absolute truth, and each human is his own measure.

On the other hand, I do believe in the objective reality, without getting into the question of the entity which understands it the best. But I think that this objective world shouldn't be the topic of philosophy itself (natural sciences handle this stuff much more persuasively with the scientific method).

[quote="uldis"]why there is Anything rather than Nothing?[/quote]

Kant had a nice point about such questions :) Thinking about them might be a good mental exercise, but they are still 'antinomies of the common sense'...

[quote="Aegor"]I have two different answers.  I like Kant because his philosophy seems to fuse empiricism and rationalism more than any other major philosopher, and I also see the advantages of both.  It's just very interesting.

I am religious, so I enjoy religious philosophers like Aquinas and Augustine.  However, I somehow really like Nietzsche.  His philosophy is so logical when you think about it...  How could an atheist not embrace it?[/quote]

I agree with what Aegor said. Nietzsche FTW. (I'm atheist :D )

And Foucault is an interesting read, too.

atheism is the logical path
but i'm christian regardless, so i have this tendency to embrace illogical ideas and various fallacies

so i'd say marx is cool

Hmm I think Wittgenstein is really cool =)

And although I have little knowledge of philosophy, I particularly like Plato/Aristotle and Descartes.

[quote="daermon"][quote="Djole"][quote="daermon"]
does Richard Dawkins count as a philosopher?
he's my favorite[/quote]

Hm, interesting... Which one of Dawkins' ideas do you like?[/quote]
um... the fact that religion is extremely improbable and harmful to society :)[/quote]

You need to watch more [url=http://www.southparkzone.com/episode.php?vid=1012]South Park[/url].  :wink:

In and of itself, faith is not a problem. It's the conflicting politics of religion that cause trouble. More recently, political idealogies such as capitalism, Marxism, etc. have been the roots of conflict. And perhaps all this conflict isn't necessarily harmful to society in the long run; rather it might be thought of as the evolution of civilization.

Perhaps it could be thought of that way, but the process of natural selection acting on species has very unpleasant consequences for the actual individual members of the species even if it's "good for the species" in the long run.  (Yes, yes, a (classical) liberal bias in favor of measuring things by their effects on individuals, I know.)

Mackie or Flew anyone? Plato was the original and his ideas haven't lost relevance. I admit Kant's ideas are hard to avoid but the cake goes to Nietzsche.
I never liked the whole religious philosophy debate so no to Descartes and Pascal, among others. Though it is fun to see arguments of all forms presented by religious philosophers broken into their fundamental absurdities! :rotfl:

Nietzsche and Pascal(though I love Pascal Blez more as a mathematican :D )!!!

Ayn Rand is fun to read because she is so crazy.

I know next to nothing about philosophy, but I enjoy reading Kierkegaard, Camus, and Solzhenitsyn.