probability geometry ARML blogs AMC AIME MATHCOUNTS

Hey everybody, I'm planning on sending in a list of reforms to my school, and one of my reforms is student-run clubs. Well, if all goes well, and this is allowed, I'm definitely planning on making a math club. I'm interested to find out exactly how they are run and what makes them exciting. So here are a few questions (they are directed at personal responses, as it's possible that activities and policies from different math clubs vary.):

(All questions refer to the math club, of course. When I originally typed all of the questions out, though, having the words 'your math club' in every question seemed somewhat repetitive.)

What's a typical meeting like?

What makes your math club meaningful/worthwhile?

Does your math club have any special events that make it extremely exciting?

Does your math club offer outreach to non-members (i.e, if they need help on a problem or can't understand a topic)?

How many times do you meet per week?

Do you have meetings outside of school?

Are students in all grades allowed to join?

Is everybody available to join (regardless of talent/ability)?

I hope I covered everything that I wanted to know! Thanks to everybody that answers, and feel free to relate some personal experiences about your math club!

(This is a cross post from College Confidential—I didn't get too much of a response there :()

I would try to find a teacher to sponsor the club because usually you need one to enter your team in competitions.  But if you can't find one, being student run should still work.

[quote="requiem"]
What's a typical meeting like?[/quote]

Well, we meet after school for about 2 hours.  We spend a little time just socializing and then we start doing math.  Usually we break into two group, one for JV and one for Varsity.  Our teacher also gets us food :)

[quote]What makes your math club meaningful/worthwhile?[/quote]

Well, this year we are pretty successful compared to other years.  So that would be one reason why it is worthwhile.  Also, I think it's a great way to obtain a deeper interest in math.

[quote]Does your math club have any special events that make it extremely exciting?[/quote]

Hmm...we go out to restaurants a few times.  Other than that...not really.

[quote]Does your math club offer outreach to non-members (i.e, if they need help on a problem or can't understand a topic)?[/quote]

We never really have people ask us stuff like that...

[quote]How many times do you meet per week?[/quote]

Once.

[quote]Do you have meetings outside of school?[/quote]

Not usually...but I've gotten together with one or two members before to practice a little...not very often though.

[quote]Are students in all grades allowed to join?[/quote]

Yes.

[quote]Is everybody available to join (regardless of talent/ability)?[/quote]

Yes...though most ppl who will want to join a math club will be 'good' at math (at least compared to the standard curriculum).

[quote="requiem"]What's a typical meeting like?[/quote]
Approximately 40 minutes of discussion of miscellaneous topics.
[quote]What makes your math club meaningful/worthwhile?[/quote]
Not much, just the small amount of math that exists in our school outside of the regular curriculum.
[quote]Does your math club have any special events that make it extremely exciting?[/quote]
We do a few competitions (AMC and some others), but not much else.
[quote]Does your math club offer outreach to non-members (i.e, if they need help on a problem or can't understand a topic)?[/quote]
Sure, but no one would ever ask us anyway, and there is no clear list of math club members.
[quote]How many times do you meet per week?[/quote]
Once.
[quote]Do you have meetings outside of school?[/quote]
Nope.  Except for some competitions and whatnot.
[quote]Are students in all grades allowed to join?[/quote]
Yeah, although we have separate "clubs" more or less based on grade level.
[quote]Is everybody available to join (regardless of talent/ability)?[/quote]
Yeah, although ditto to what joml88 said.

I would say interest is one of the hardest things to get and keep.  As you can probably tell, I'm not exactly satisfied with my school's math team, especially since only a few people take it at all seriously.  Good luck to you though; it is people like you who establish environments that are helpful to the development of future talented math students.

Umm, I'll just # the questions instead of quoting them all:

1. About 1.5 hours of math - usually a practice test for an upcoming competition followed by a brief (if existant at all) discussion of problems.

2. Nothing, really, except perhaps the largest concentration of intelligence in the school at any time ;)

3. Competitions all the time and I guess we play games a lot, especially now that all the competitions are over.

4. Yeah, I guess. I don't really understand the question.

5. Once; Wednesday afternoons because we have short days every Wednesday.

6. Not really.

7. Yes, all grades (we even have an 8th grader).

8. Yes, anyone can join regardless of ability (or we'd have a very small club ;)).

What's a typical meeting like?
-1hr meeting of just testing... since i am pretty advanced, i do my own thing. or skip ;)
What makes your math club meaningful/worthwhile?
-being the school's most achieving/smartest club... although there are only few smart people in the team.
Does your math club have any special events that make it extremely exciting?
-hmm...competitions? we host a few.
Does your math club offer outreach to non-members (i.e, if they need help on a problem or can't understand a topic)?
-yes... because almost all the smart, core of the math team do ind. study, they help out to people in the regular classes who need help.
How many times do you meet per week?
-Tuesday after school, Saturday 10-12
Do you have meetings outside of school?
-nope. I intend to hold some meetings in B&N with my friends to teach them.
Are students in all grades allowed to join?
-yes. however, we've only had 5 8th graders join in the history of the math team.

good luck starting a team. you'll need it.
i'm the president of the shs math team. joml88 is on the team. he pretty much summed up what our team is like. and like probability1.01 said, it's hard to get and maintain a team that [b]wants[/b] to compete and is also good enough to win. our lower (jv) level had 3 consistant members and 2 of them were just average. they are good at calculator programming tho. the varsity team was awesome. but the problem comes up when there are only 8 or so "regulars" out of a school of 3500 students. many competions require a larger team, so we had to get random people to participate of the math team. and with no experience, they were more or less a space filler and did not help the team out. not trying to be harsh, but we are trying to create one of the best teams in the region. and unfortunately, we can't clone joml88 multiple times. :D

I forgot to answer a question:

Is everybody available to join (regardless of talent/ability)?

Well, it depends. Our math team is divided by two groups... not really noticeable or like named or anything, but once you are on math team, you know it. There are about 10-15 or so of us who are really good. We are the core "math team". Everyone else is "math club member". what separates math team from the math club members is the different competitions. Ones where we can enter a lot of members, such as Mu Alpha Theta nationals or State convention, we intend to include everyone. (70 people...whew) however, more copmetitive, hard, and competitions that require less students, such as AMC/ARML/Math Is Cool!(Washington State only), we put our best students, and nada mas.(i dunno why i am typing spansih....)

Typicaly for our team, unless if the coach knows you from Middle School (my case), when u join you are a math clubber. In my case I was almost immediately compared to the best of the school... i should be now. The thing about our school is that nobody strives to be at the Math Team member status... by the time i am Junior/Senior, our school will suck...mmm

The thing I fear about starting a math club, like probability said, is being able to keep interest. As of now, I'm pretty sure there is not a local area competition league (town of 24000, three schools). Since some of you have said that the most interesting thing about math club is the competitions, I think it might be hard to keep members. Basically the club would only be  lectures/problem sets to study for the AMC, and I think it might become somewhat monotonous. Another unfortunate thing, the only really motivated mathematics teacher in my entire school is reitiring this year.

Although it would be an enjoyable social club, I'm unsure if it would really work.

[Note: My school is pretty small; about 500 people in grades 7-12. Furthermore, the school is dominated by sports, and there is only a small minority of kids that don't play them.  This might also pose a problem with mainting a math club  ;) ]

Don't give up before giving it a shot.

Also, at most schools (i.e. public, not so much private) sports dominate so don't really let that be much of a discouraging factor.

Don't necessarily limit yourself to the suggestions here either; although I think our team at Summerville is great, you are correct that it is difficult to attract members (though those that are most worthy usually make the effort).  If you want to try to have a greater level of social activity, do so.  Your resources will probably limit you somewhat, but plan activities outside school, or even math, if that's what you think will make people interested.

one thing i might add to that is:

Do a lot of fun activities associated with math team...

For example, our school goes to Mu Alpha Theta every year. We plan ahead, go there 3-4 days before, and have fun, barely any practice. Our coach uses that propaganda to bring in people who were otherwise afraid to become a math clubber. (Of course, we train ALL summer before going) Other competitions we go to, if it is an overnight thing, we plan to go the day before and just have a blast.

Get people involved in doing fun activiities is a great way to launch ur math team.

If you wanna catch some interest at the beginning, what I do is come up with some catchy titles.  To start off the year, I always do "Crayons, Dominos, and cutting things up"
Where I do the class coloring problems, like the chessboard with 2 corners missing, intro to the 4 color theorem and into harder coloring problems.  I use that because people are like..."That's math?  What?!  It's actually...interesting."  I try to get it away from the school math, as far as possible.

We also get people to bring in snacks, alternate.  And for the more serious contest math people we have a separate meeting that week, so 2 meetings a week.  One for enterainment and one for serious stuff.  But its a lot of work...good luck!

What's a typical meeting like?
We meet everyday. The math club in my school counts as a class and shows up on the transcript. we also get school service credit for participating in contests

What makes your math club meaningful/worthwhile? /Does your math club have any special events that make it extremely exciting?

for us, the teachers make sure that the things they cover are not part of the normal curriculum. it's more challenging and often fun. also, we ocassionally have parties, especially the ice cream party after arml (with Penn State ice cream:)), sometimes we also get guest speakers and get the whole day out of class, that is fun :lol:. also, some people get to go to outside competitions, but that is not everyone/not too often

Does your math club offer outreach to non-members (i.e, if they need help on a problem or can't understand a topic)?
we hang out in the math team room after school and people sometimes stop by for help, just by word of mouth, it's not advertised though!

How many times do you meet per week?
40 minutes (1 school period) every day + after school meets

Do you have meetings outside of school?
no

Are students in all grades allowed to join?
yup

Is everybody available to join (regardless of talent/ability)?
there is an initial screening for teams, but those who didn't get in are still allowed to be in the class, just not compete as starters

What's a typical meeting like?
do old AMC problems, discuss problems, etc.

What makes your math club meaningful/worthwhile?
hmmm...we do a lot of math problems

Does your math club have any special events that make it extremely exciting? go to competitions, practice for AMC (not that exciting)

Does your math club offer outreach to non-members (i.e, if they need help on a problem or can't understand a topic)? well, our school has VERY few math people.

How many times do you meet per week? 1 :P

Do you have meetings outside of school? No :P  :P

Are students in all grades allowed to join? yes, but only freshman come

Is everybody available to join (regardless of talent/ability)? yes

overall, math club is always a fun experience. our math club teacher was a MOPer so he knows his math stuff. we have a lot of fun just discussing math problems.

[quote="Sunny"][quote="xxreddevilzxx"][quote="Sunny"]
We also get people to bring in snacks, alternate.  And for the more serious contest math people we have a separate meeting that week[/quote]

we sorta have that distinction as well, except all the good people tend to skip meetings..[/quote]

Give'em an incentive to come.  At our school we need service hours to graduate, and I've talked it over with the principal and if they help out I can give them volunteer hours.  ITs a faster easier way for them to get their service hours done.[/quote]

We give out math extra credits for people who come to math club practices... and Math Honor Cord for Graduation, but I dunno what the requirement is for that.

Hmm, we give them... that's a good question. What do we give them? ;) I think most people come for three reasons: 1. they are an officer, 2. it's pretty fun, or 3. they have nothing better to do Wednesday afternoon.

food/pizza parties are always good. also, math books are always a poplular choice at my school

requiem--wow I'm trying to do the same thing in my school.  See my blog for more details.

Yeah, for some reason people seem to love food, especially pizza.

Thanks guys for all your good ideas.  I'll be sure to implement them.

[quote="Mariag"]math books are always a poplular choice at my school[/quote]

Not for my school... actually, math in general isn't a popular thing.

Our school already has a math club. You need to take their test and do well on it. (not too hard for most people on this forum)

I find math really exciting, I hope it gets across in your schools.

This is interesting. I'm trying to do something similar, but essentially different, at my school. There already is a "Math Club", but this past year we really didn't do much interactive. The biggest problem is our lack of involved people. We had about six semi-regulars, including myself, and even that decreased over the year. We did have the good fortune of having three fairly good math competitions in our general vicinity: one two-part buzzer team/individual olympiad competition, one two-round simple proof contest, and one true olympiad, in that order during the school year. After seeing our awesome success at the first contest (2nd place and four in top 10), I was very disappointed to be the only one in the top ~50 on each of the other contests. We did have four make the AIME, but again, I was the only one with an AMC 12 score above 101. Plus, two of the qualifiers I just happened to run into the morning of the 12. After making MOP, I feel that I could easily lead our math club next year. Last year, it was lead by the parent of one of the three juniors. We definitely need more advertisement. I already have ambitious plans for next year, such as doubling our hour-a-week and including a larger variety of problems.

Sorry for the long story, but does anyone have any comments/suggestions on my situation?

a good thing to do is to start from summer, you can make a website for the team, and post stuff on there.... :)

well i live in korea, and there is no such thing as math team in high school. we just study alone, or go to an AoPS-type academy where we learn all the stuff, pretty much equivalent of AoPS online classes, except it is real, not virtual.

i specifically went on this website today to look for new ideas for the math club that is at my high school. I was elected to be the president for next year's math club.

As i read the messages i find that many of your math clubs are just people that compete. At my high school the math clb are split into about 5 sections. We have the math team section who do the competions such as mu alpha theta and the AMC. We also have the tutoring committee where they stay after school to help those who need help in math. we also have the publicity committee and they get the filers and newsletters about math club done, while the fundrasing committee do fndrasiers to make money.

At my high school my math club is very very big mainly becuase of the fact that many math teachers give extra credit for being "active" in math club. So you can probabaly say that a majority of the students in the club becuase of the extra credit or just to buff up thier resume for college.

We meet about once a moth during lunch time which is about 30 minutes. and many people in math club just arent the excited people.

So i was wondering if there are any suggestions to make a math club more interesting and more appealing for people. I really don't like the fact that people are in math club to buff up thier resume or just to get extra credit and i really want to change thier veiws about math and allow them to see that math is actually really exciting and not the boring subject they think it is.

I was thinking about asking guest speakers to come in.. but that the best i can think of. i just dont know how to make math  or math club more appealing to those who think that math is boring. It will be very easy to please those who alreayd like math, but my goal as president next year is more than that. Or at least i hope to be able to do more than just allowing math club just be a regualr club.

so if you have suggestions please tell me.

A good way to sustain a math club is to make it a math team by looking in your area for regional contests that they might have. Perhaps Mu Alpha Theta might be the way to go, I don't particularly like the form of their tests, but it gets me to stay after school to learn other stuff, and it brings other people who want to take advantage of the oppurtunity.

Our math team practices about 2 times a week, sometimes more depending if there's a competition soon. I think we have a pretty good math team compared to those around us, but we don't do too well in state competitions heh. But it's all fun.

Here's an idea for making Math Club more exciting and interesting for those not so keen in math: Organize/create a few in-school competitions and invite everyone in the club. Add an excitement factor (like some sort of countdown round) and people will come. You can also use these competitions to choose teams for Mandelbrot, for example. Just an idea, though...

the competition within the school sounds like a good idea..but yet im afraid that there still going to be people who do not want to take any kind of test or something in that path....

i was thinking about getting speakers in..like professors..mathematicians... but maybe some people will also get bored with that....

please keep the ideas coming..i seriously do not know what else to do..for next year...

Not everyone has to compete. Just watching an oral competition like MATHCOUNTS, with friends and classmates in it of course, can get someone excited about math, if that's your goal. They might want to take part in the next contest, in-school or out of school, and that's what you want, isn't it?

SLeon, it sounds like you are trying to please everyone, aka will end up pleasing no one in the end. It's best to run with one idea, keep a few others nearby so if the one idea fails, you can try those out. But if you try to do a little bit of everything, you're just going to be overwhelmed and end up worse off than if you had only chosen one idea and it failed. As for competitions, I really do believe it is a great starting point for a math club. It helps everyone focus on something instead of having sparatic discussions on stuff that half of the class won't understand due to grade level differences or because someone didn't talk loud enough, etc. and the meeting wll end with a lot of people feeling like they didn't accomplish anything.

But I mean, if you're willing to build a curriculum for you math team to focus on, then that could be an alternative.

Also, it might not be a breeze getting people to actually come to the meetings, so if you start it up and there's low attendance, you might want to talk to some of the math teachers at your school (or if there is a haead of the math department.. talk to him/her) they can usually advertise to their students as a whole and personally to those who they know are interested in math.

I think it's a great idea to start a math team and I'm sure that with a bit of preparation and good decision making skills you'll be able to get yours off the ground :)

Oh and your idea about speakers is a good idea, but it's good to get a visual of how many students you will have who come consistently, and the level at which they are at. Because if you bring a college professor to come in and talk about vectors or something, and the majority of your population is 9th graders, they're gonna be like "O.o, WHAT?!"

in my school's math club is overated and filled with lots of people who literally can't do math.  And that's because all the popular people are in there.  But we still sent our team to state and a guy to nats for mathcounts.  We do it one hour on Tuesdays.  Starting with a warmup and a workout where we can do it in groups after 10 minutes.

[quote="jimli"]in my school's math club is overated and filled with lots of people who literally can't do math.  And that's because all the popular people are in there.  [/quote]
WOW. since when did all the popular people join the math club? that's certainly not the case with my school


integration
induction
calculus
calculus computations

Please, need help with some explaination, thx!!!

$f: [a,b]\rightarrow R$ is a continuous function. Define a sequence $F_{m}: [a,b]\rightarrow R$ for $m\geq0$ by setting $F_{0}(x)= f(x)$ and $F_{m}(x)= \frac{1}{(m-1)!}\int_{a}^{x} (x-t)^{m-1}f(t) dt$ for $m\geq1$

1. Show that $F_{m}^{(i)}(a)=0$ for all $m\geq$$0$

2. Show that $F_{m}^{(m)}(x)= f(x)$ for all $x\in[a,b]$

I am thinking if I can use the induction here, could I?
Need some explaination!!


Hello!
I was wondering if anybody here does any elementary/middle/high school math tutoring while in high school. I was thinking that I could work as a tutor during winter break, tutoring elementary, middle or basic high school math courses. I'm currently a sophomore at high school, but I take an advanced math course. I'm not wanting to be payed a lot, but I was looking for a payed job. Please talk about your experience/tips. Also, do you know who should I ask to find such a job?
Thank you.
Also, I didn't post it in the Career Section, because I don't think this is a "career" in high school, but just a way of earning extra cash.

Just remember one thing: elementary and high school students don't hire tutors themselves - their parents hire tutors.

People who get a lot of tutoring jobs tend to make their connections by word of mouth, as parent talks to parent. Where do parents meet and talk to each other? Community centers, churches, other organizations? Is there a particular ethnic group you'd like to connect with?

Sometimes parents call schools or certain teachers at schools asking if they know of or could recommend a tutor. (They also call colleges, but that's not your connection at the moment.) You might try asking some teacher you know if he gets calls like this.

And find out, if you can, what the "going rate" is in your area and go ahead and ask for that. That's part of convincing parents that you're serious about this. And think about how you would present your qualifications, remembering that it's parents that you have to convince.

[quote="7h3.D3m0n.117"]Tip: I find it easier to tutor students who are interested in math, not the snobby ones where they have to have a tutor because their parents said so.[/quote]

Nowadays, kids like these are rare. Sure, its nicer to tutor them, but I'm guessing you'll be stuck with about 1 of those not-so-bright in the end. Honestly, most parents don't view math as a very deep subject that can also be very fun. If they think their children are doing pretty decently in school curriculum, they most likely won't get a tutor.  :|

But don't listen to me, I don't have a lot of experience in things like this. This is just what I think by watching adults.  :ninja:

A good idea would be to set up a mini-AoPS tutoring session at a local high school to those who are interested. That way, you are pretty much guarenteed a nice bunch of students who are willing to work hard. Also, this session can be available to all students regardless of their current grade in math (usually math tutors are hired because one is failing). And if money is the issue, you should be able to get a better pay if you taught a small class rather than an individual.

I would look at SAT tutoring.  If you can manage a perfect SAT math score, it looks really good when you offer SAT math tutoring to parents.  The words "perfect score" turn on parents like magic, and the one thing every parent wants for their kid (academically) is to get into a good college, and a good SAT score is a big part of that.  Trust me, there is a market for SAT tutoring no matter what high school you go to.


Eight women of different heights are at a party. Each woman decides to only participate in a handshake with women shorter than herself. How many handshakes take place?

Answer is 0.  If person a shakes person b, then a is taller and b and b is taller than a. contradiction!

I remember "way back" when I was in middle school a math teacher gave me some sample problems to prepare for MATHCOUNTS and this was one of them. It made me sick: it's a trick problem, not really much else to it.

Funny, I was playing with Alcumus a few minutes ago and I saw it again: I gave it all low scores for ratings, on principle.  :P

However, if this were a 'normal' problem, it would be:

$ \dfrac{8(8\minus{}1)}{2}\equal{}\dfrac{8(7)}{2}\equal{}4(7)\equal{}28$

Each of the eight women shake with 7 other people but since AB is the same as BA, you divide by 2.

Another solution:

The shortest person cannot shake hands with anyone.

No, that is not a solution. That is true, but you did not say anything about the other 7 people.

Oh yeah, I forgot to say the other part. :)

Well using ernies method, the shortest cant shake. We might as well eliminate her. But now someone else is the shortest! By similar logic, no one can shake. Its like the following problem:

My teacher told us today (friday) that we will have a test next week, but we wont know what day it is until we arrive at school and she tells us. What day is the test?

[quote=Potato2017]what a troll[/quote]

[quote=NotMyRealNameII]"It made me sick". Yes, this. I can't believe I spent time working this out.[/quote]

ikr
I got it wrong.

[hide=Sol]0 because nobody is the same height.[/hide]

one word for this problem: BRUH
one emoji for this problem:  :rotfl:

When I got this problem, I entered 28 thinking “that’s too easy. It can’t be 0. I will report an error once I get it correct.” But then 28 was wrong so it put 0 and it was correct.  :rotfl:

[quote=brickmaster8]is it zero????[/quote]

Yes scroll though the thread before asking

If this were on ftw

big rating drop ouefs

I thought it was 28 bc 7+6+5+4+3+2+1=28 but it was 0

Also look here, it says Each woman decides to only participate in a handshake with women shorter than herself. it say herself, so it doesn’t really matter

[quote=spacetime_traveler]They all have to be the same exact height to shake hands according to their own rules, but the problem clearly says that is not the case. Lol?[/quote]

Even if the problem did not state they were of different height, it would not work. If you were the same height as someone else, nobody is shorter than one another, and since you need to handshake with someone shorter anyways, it wouldn't work.

It's definitely 0. If the shortest one doesn't shake, then the next one doesn't, until all of them don't.

Double posting (posting two times in a row without a one day gap) is discouraged, so use the edit button instead.

[hide]The answer is $0$ because if a woman only shakes hands with a woman shorter than herself, the shorter woman would be shaking hands with a taller woman. This is impossible. So it is $\boxed{0}$[/hide]

[quote=JacobJB][quote=Mewto55555]
My teacher told us today (friday) that we will have a test next week, but we wont know what day it is until we arrive at school and she tells us. What day is the test?[/quote]

No idea... Monday?[/quote]

It's a paradox :P

[quote=Onafets]thank you for bumping a 12 year old post[/quote]
And thank you for telling us some random thing that we already know and also isn't true as it was bumped in 2016, and then 2020 again

[quote=Onafets]thank you for bumping a 12 year old post[/quote]

Your welcome. :)

[quote=Onafets]thank you for bumping a 12 year old post[/quote]

It really doesn't matter in Alcumus

If the taller person tries to initiate the handshake, then the shorter one refuses, because of their height??

Yeah, pretty much.

I think I got this problem forever ago and it is amazing. My only comment: bruh. Just bruh.


algebra
polynomial
quadratics
calculus
derivative
real analysis
real analysis unsolved

I'm a little stumped on this one. Here's the question:

dy/dx = Ln( x + 1 )  + sin( pi / 4y^2 ) with y=1 when x=0

Estimate y at x=0 to 4 decimal places using a quadratic (2nd degree) Taylor polynomial about x=0.

Now from what I can understand, you need to integrate the above equation to find 'y' because you are
asked for a point of x other than at x=0 which was given to you. So I broke it into two parts

  Ln( x + 1 )  and   sin( pi / 4y^2 )

to integrate separately. However I was stumped when it came to integrating the second part with respect
to 'x' (because it contained y). Now maybe I'm going about this the wrong way so any help provided would be greatly appreciated.
Thanks for your time,

seifer_md

You know I think I got it. I was just making it more complicated than it actually was (this happens a lot...)

It seems I only needed to find the second derivative (d^2y/dx^2) and I could just sub in for 'y' when I take the value for x=0 of the derivatives. Then I use that to form my Taylor polynomial to the second degree. Then substituting x=0.1 I can then find y to four decimal places.

Unless I'm wrong, this topic can be deleted.


is there any biology contest besides the olympiad?

i was interested in the same thing, but for college students

To your question idupda,

I have heard of such a biology contest as the International Brain Bee, which I myself hope to partake in when I grow older. The elligibility I always heard is for 9th-12th grade students.

First one must participate, in a school bee (optional). Then if you succeed in the school bee or skipped it you can go to the local bee (mandatory). Next, is your national bee(mandatory). Finally, is the international brain bee(mandatory)

It's a great competition, I would assume if you enjoy the field of neuroscience, and of course, are good at memorizing by rote.

-arb95

There's always AoPSWiki and you can search for science contests there. There's some science contests, where you could probably do something related to biology: http://www.tryscience.org/parents/se_7.html
Otherwise, I don't know  :?:

You can always view the AP Exam as a competition between your classmates as to see who got a better score  :P


calculus
derivative
function
advanced fields
advanced fields unsolved

Let $f: \mathbb{R}\rightarrow\mathbb{R}$, $f$ differentiable.

is the function $g$, defined by $g(x)=\max\{0,f(x)\}^{2}$ differentiable at poins $x$ such that $f(x)=0$ ? (you may assume $f'(x)\ne0$ at points $x$ such that $f(x)=0$).

I can't figure that out, apreciate any help.

I think it is. This is because $\frac{d}{dx}f(x)^{2}= 2f(x)f'(x)$, which is 0 when $f(x) = 0$. And either $g(x)$ is 0 in a neighborhood of the zero of $f$ or it is equal to $f(x)^{2}$ in a neighborhood of the zero on one side and 0 on the other. But clearly $g'(x)$ may not be differentiable at those points even if $f \in C^{2}$.

$g$ is differentiable at those points.

For simplicity, suppose $f(0)=0$. By differentiability, there are positive numbers $\epsilon$ and $M$ such that $|f(x)|<M|x|$ for all $|x|<\epsilon$. Then $|g(x)|<M^{2}|x|^{2}$ for all $|x|<\epsilon$, and $\left|\frac{g(x)-g(0)}{x-0}\right|<M|x|$ for all small enough $x$. By the definition, this makes $g$ differentiable at zero, with derivative zero.

Yet another solution: the function $h(x)=\max\{0,x\}^{2}$ is differentiable, being an antiderivative of the continuous function $2\max\{0,x\}$. By the Chain Rule $h\circ f$ is differentiable.

thank you all, you were really helpfull.


floor function
combinatorics proposed
combinatorics

consider n>=3 collinear points and distance bet. these points.its given that each such distance appears atmost twice.show there atleast [n/2] distances that appear once.

You can find a nice solution to this problem in the book [i]A path to Combinatorics for undergraduates[/i].

Can you post the solution please Jutaro?

Let $x$ be the number of distances that appear once, and $y$ the number of distances that appear twice. We wish to find a lower bound of the number $x+y$, the total number of distinct distances. For the sake of this, we count the segments from left to right, according to their left endpoints. Let's label the points $P_{1},\ P_{2},\ldots,\ P_{n}$ from left to right. Then $P_{1}$ is the left endpoint of $n-1$ segments with different lengths. Now we focus on $P_{2}$: This point is the left endpoint of $n-2$ segments, but some may have lenghts already counted by $P_{1}$. But this can be the case for no more than one length. Indeed, if we had $P_{1}P_{i}=P_{2}P_{j}$ and $P_{1}P_{k}=P_{2}P_{l}$ then $P_{1}P_{2}=P_{i}P_{j}=P_{k}P_{l}$, contradicting the assumption that each distance can appear at most twice. Hence no more than one length is repeated from $P_{1}$'s count, and we have at least $n-3$ new lengths. Moving to $P_{3}$ we can apply the same argument: There are $n-3$ segments with $P_{3}$ as left endpoint, but one distance may already have been counted by $P_{1}$ and another by $P_{2}$. So we have to change our lower bound to $n-5$. We may continue in this fashion to arrive to the lower bound \[x+y \geq (n-1)+(n-3)+(n-5)+\cdots=\left\lfloor \frac{n^{2}}{4}\right\rfloor\] We know that the total number of segments is \[x+2y=\binom{n}{2}=\frac{n(n-1)}{2}\] But from the former equation we get \[2x+2y \geq \left\lfloor \frac{n^{2}}{2}\right\rfloor\] Therefore, \[x \geq \left\lfloor \frac{n^{2}}{2}\right\rfloor-\left(\frac{n^{2}}{2}-\frac{n}{2}\right)=\left\lfloor \frac{n}{2}\right\rfloor\]


[color=green]1. Use Gauss Theorem to prove that an angle of 20 degree is not contructible.

2. Use Gauss Theorem to decide whether or not an angle of 6 degree is constructible.[/color]

What's Gauss Theorem?
For 2.We can construct 30 degree(By equailateral triangle)
And 36 degree(By regular pentagon).So we can get 6 degree.

Maybe you meant Galois instead of Gauss?  :?

This Gauss theorem states that a regular n-gon is constructible with straightedge and compass iff $n = 2^k p_{m_1} p_{m_2} ...$, where $p_{m_1},\ p_{m_2},\ ...$ are distinct Fermat primes $p_m = 2^{2^m} + 1,\ m = 0, 1, 2, ...$. Not all $p_m$ are primes, only five are known: 3, 5, 17, 257, 65537, Euler factored $p_5$ by 641. Gauss proved only sufficiency. Angle $20^\circ$ requires $n = \frac{360^\circ}{20^\circ} = 18 = 2 \cdot 3^2$, regular 18-gon, hence $20^\circ$ is not constructible. Angle $6^\circ$ requires $n = \frac{360^\circ}{6^\circ} = 60 = 2^2 \cdot 3 \cdot 5$, regular 60-gon, hence $6^\circ$ is constructible.


algebra
polynomial
advanced fields
advanced fields unsolved

a(n) = C(-3)^n + D(5)^n

What is the original second order linear recurrence relation with constant coefficients that was derived based on this explicit formula?

Thanks.

Just to check, the answer I got is:

a(k) = 2 ak-1 + 15 ak-2

Since -3^0, -3^1, -3^2 produces sequence of 1, -3, 9, -27, ... , (-3)^n
and since 5^0, 5^1, 5^2, producces sequence of 1, 5, 25, ... , 5^n

We then know the roots are:
(t-5)(t+3) = 0
t^2 + 2t - 15 = 0
t^2 = 2t + 15

so therefore:
a(k) = 2 ak-1 + 15 ak-2

Its characteristic polynomial is $P(z)=(z+3)(z-5)=z^2-2z-15$, so the recurrence relation is $a_{n+1}=2a_n+15a_{n-1}$.


Prove the following inequality for every natural number n.

[1/(n+1)]*{1+1/3+1/5...1/(2n-1)} > [1/n]*{1/2+1/4+1/6...1/(2n)}

[Regional mathematics olympiad (India) 1998]

[quote]Prove that for every natural number $ n > 1$
\[ \frac {1}{n \plus{} 1} \left( 1 \plus{} \frac {1}{3} \plus{} \frac {1}{5} \plus{} \ldots \plus{} \frac {1}{2n \minus{} 1} \right) > \frac {1}{n} \left( \frac {1}{2} \plus{} \frac {1}{4} \plus{} \ldots \plus{} \frac {1}{2n} \right) .\]
[/quote]
We can use either induction or prove it in the following way:(my induction proof is very long)
[hide="without induction"]
Let $ H_n \equal{} 1 \plus{} \frac {1}{2} \plus{} \cdots \frac {1}{n}$, then the given ineq $ \iff$
$ \frac {1}{n \plus{} 1}(H_{2n} \minus{} \frac {1}{2}H_n) > \frac {1}{2n}H_n$ After cross multiplication we are only required to prove:
$ \frac {2n}{n \plus{} 1}H_{2n} > \frac {2n \plus{} 1}{n \plus{} 1}H_n$
$ \iff \frac {1}{n\plus{}1}\plus{}\frac1{n\plus{}2}\plus{}\cdots \frac1{2n} > {{H_n}\over {2n}}$ which is obviously true.[/hide]


hi. hw do i start a blog in AOPS? I went to the blogs window but there is no new button.

thanks.

There should be a "Create a new blog" option from your profile.


I am wondering, what will happen if i put $S$ into $H_2O$??

Not much. IIRC there will be some very slow reaction yielding H2S and SO2. It is fast enough for a pile of sulfur (and I am talking about tons) to stink, but too slow for any practical purposes.

Borek

So there will be very little $H_2S$ and $H_2SO_3$ in it??

Yes. Pure sulfur is only kinetically stable.

But these reactions are VERY slow. One of the methods of sulfur mining uses hot steam to melt the sulfur deep under the surface - and even when treated with hot steam sulfur reacts slow enough to be pumped and transported. But it stinks like hell, I have been to such mine in Poland.

Borek


Point O is the beginning point of n unit vectors such that in any half plane
bounded by a straight line through O there are contained not less than k vectors (we assume
that the boundary line belongs to the half-plane). Prove that the length of the sum of these
vectors does not exceed n − 2k. :huh:

[quote="pqrs"]Point O is the beginning point of n unit vectors such that in any half plane
bounded by a straight line through O [b]there are contained not less than k vectors[/b] (we assume
that the boundary line belongs to the half-plane). Prove that the length of the sum of these
vectors does not exceed n − 2k. :huh:[/quote]
What do you mean with [b]there are contained not less than k vectors[/b]?


Are people generally good, bad or some mix of the 2? Personally, I think that human nature is mostly good and altruistic because we have sympathy for each other and are hurt by others' pain. I don't believe other animals feel the same way about each other, except for elephants and other animals that understand death. Other people's input is appreciated.

I recently read [i]Dr. Jekyll and Mr. Hyde[/i] and I found that humans, no matter how good on the outside, can always have a tiny spec of bad in them.

If you were to look at humans in a poor environment, like places with extreme poverty, you would be surprised.

Haha. Humans are born [b]selfish[/b], period.
Really, whatever we do, it is only for our own benefit that we do it.
Yes, there are some genuinely good people out there... but most have a selfish/bad streak.  :)

This poll is interesting, but can't be a measure for anything as it has variables of what a person defines to be good, bad, and human nature.

If I define human nature to be the built in instincts of humans:
And if I define good to be helping my chances of survival and bad to be dampening them then human nature is most certainly mostly good.

And if I define good to be happy and bad to be unhappy then I'd say human nature is mostly good as we are programmed to be social.

And if I define good to be helping the human race as a whole and bad as dampening it, then I have to define what helping the human race is. Does culture help it? Does increasing overall survival help it? Is overall survival longevity of the human race or average longevity of lives? If it's longevity of the human race, then destroying half the human population and the industrial sector would be GOOD as we would be prolonging the drain on this planet.

If I define human nature to be what humans today would generally do:
And if I define good to be helping my chances of survival and bad to be dampering them then human are most certainly mostly good.

etc. etc.

[quote="davidyko"]I might be pessimistic, but I view human nature as fundamentally self-centered. Whether that's good or evil, you decide.[/quote]

That's not pessimism, that's cynicism.
Get your terms right.

Yeah, I agree, though.
Cynicism works best when determining human nature.
Communist theory works on having all of the self-centered people work together to reach a goal for all.

AKA, those with similar interests please move to the right.
Ok, good, now work together to obtain your similar goal.
Next?

America at the moment is like this:

Ok, hmm, everyone with similar interests, please move to the left.

*People aren't too sure what they want. And they sway back in forth, acting as double spies, screwing up the communist effort in their idiocy.* w2go

And then you have rational egoists whom are a totally different story.

I've seen the American culture as deconstructive more than constructive.

I was talking to two professors: a philosophy prof. and sociology prof.
We both discussed how America has become so self-centered on obtaining success.
And at the same time, that idea can bring people into a state of solipism, detached reality, and social isolation.
With these things in mind, the idea of an individualisitic society has corrupted itself.
If you think about it enough, it's true.

Here is a box.
Here is a book.
Read the book while in this box.
Here is a test.
Take this test, while in the box.
Here is a job.
A job in a box.
Work while in the box.

- If you do all of those things correctly, you'll be successful.

Here is a piece of paper. Read it.
Here is a job that requires you to read this label and do what it says.
The label tells you to read the labels of these other objects and sort them.
You know how to sort them because the boxes they are sorted into have labels, too.
After you are done with work, you get paid something with words on it.
You bring that piece of paper with words and letter to a place with a currency sign.

Don't get it yet?
here: http://en.wikipedia.org/wiki/Allegory_of_the_cave

Way to go, you wasted your life looking at letters of the English alphabet.

Perhaps it's some left over psychic energy from the past thinking communism is bad.
Then again, marriage fluxes so much, I don't even think Americans can really work together.

I feel humans are generally good by nature but they are misguided by the miniscule wrong sort n end up getting bad  :(   :o . Well its just like one rotten apple in a basket that spoils them all  :rotfl:

Read East of Eden by Steinbeck. Enough said.

[quote]
This poll is interesting, but can't be a measure for anything as it has variables of what a person defines to be good, bad, and human nature. [/quote]

I realize this, but I am mostly going for other people's opinions to why they think one way or another. I'm not really looking for an answer.

I hang around my nice community and there are great people. However, compared to the amount of gangsters, emos, goths, and people who just feel depressed a lot (me), human nature can be a bit violent, offensive, and very unpredictable. From what I see, human nature is just cruel to me and is expressed in [fake] nice ways. It's a shame to have false emotions, isn't it?


we all know $\sum_{n=1}^{\infty}\frac{1}{n^{2}}=\frac{\pi^{2}}{6}$

But what is $\sum_{n=1}^{\infty}\frac{1}{n^{2}+1}$  ???

I got the answer from a fourier series, and wondering what other ways ye guys might have for it. Given the result in kinda complicated, im not sure as to elementry methods.

[hide="Answer"]$\frac{\pi \cosh{\pi}-\sinh{\pi}}{2 \sinh{\pi}}$

Where $\sinh{x}=\frac{e^{x}-e^{-x}}{2}$

And $\cosh{x}=\frac{e^{x}+e^{-x}}{2}$[/hide]

Can you explain that?  It seems to be the same thing he posted...

i think boxedexe pointed out that seamusoboyle forget the square in the first equation

Neither of those results is likely to have a nice, elementary explanation.  This post would be much more at home in a calculus forum.  Also, as long as you're putting an extra 1 there, you should try to see what happens when you put an $x^{2}$ in place of the 1.

And mathematica reveals all!  :lol:

Ps thanks boxedexe


Different people have told me different thing...
Which one of these is true?

$ sin^2x\equal{}sin(sinx)$

or

$ sin^2x\equal{}(sinx)^2$

The standard is that $ \sin^2 x \equal{} (\sin x)^2$.

However, for an arbitrary function $ f(x)$, the standard is that $ f^2(x) \equal{} f(f(x))$.

Standards are like that sometimes.   :oops:

I'm not sure, but I believe they put the "squared" above the sin because if someone wrote it after the whole thing, and forgot the parentheses, it would look like the sine of x squared. It is pretty annoying at first.


If x and y are positive integers such that x^2-2xy-3y^2 = 21 then the largest possible value for x is in the range:

a) x < 10 b) 10 [u]<[/u] x < 20 c) 20 [u]< [/u]x < 30 d) 30 [u]< [/u]x < 40 e) 40 [u]< [/u]x

[hide="partial solution"]$ x^2-2xy-3x^2=(x+y)(x-3y)$

Since $ x,y$ are positive integers, $ (x+y)$ is a positive integer too. Since $ (x+y)(x-3y)=21$, $ (x-3y)$ is also a positive integer.

The only ways to write $ 21$ as a product of 2 positive integers are $ 1 \cdot 21$ and $ 3 \cdot 7$. So we have four possibilities for $ (x,y)$.

\begin{align}
(x+y,x-3y)&=(1,21)\\
(x+y,x-3y)&=(3,7)\\
(x+y,x-3y)&=(7,3)\\
(x+y,x-3y)&=(21,1)
\end{align}

Just solve these four systems to find the largest value of $ x$.[/hide]

[quote="TheShadowWithin"]Sorry, so I got the highest possible value of x being 22, (21+1) which puts the answer as c, but the answer key is b. I'm a little confused.[/quote]

$ x$ and $ y$ are intigers so $ x \neq 22$ otherwise we would have $ (x\plus{}y)(x\minus{}3y)> 21$
If you read [b]Kitsune's[/b] answer, we dont need to evaluate all four equation because $ x$ wil be the greatest in $ (4) \rightarrow x\equal{}16$


trigonometry
trig identities
Law of Cosines

Point $G$ is where the medians of the triangle $ABC$ intersect and point $D$ is the midpoint of side $BC$. The triangle $BDG$ is equilateral with side length 1. Determine the lengths, $AB$, $BC$, and $CA$, of the sides of triangle $ABC$.

[asy]
size(200);
defaultpen(fontsize(10));
real r=100.8933946;
pair A=sqrt(7)*dir(r), B=origin, C=(2,0), D=midpoint(B--C), E=midpoint(A--C), F=midpoint(A--B), G=centroid(A,B,C);
draw(A--B--C--A--D^^B--E^^C--F);
pair point=G;
label("$A$", A, dir(point--A));
label("$B$", B, dir(point--B));
label("$C$", C, dir(point--C));
label("$D$", D, dir(point--D));
label("$E$", E, dir(point--E));
label("$F$", F, dir(point--F));
label("$G$", G, dir(20));
label("1", B--G, dir(150));
label("1", D--G, dir(30));
label("1", B--D, dir(270));[/asy]

[hide]$DC = 1 \Rightarrow \boxed {BC = 2}$

$BG = 2GE \Rightarrow GE = \frac{1}{2}$

$AG = 2GD \Rightarrow AG = 2$

$\angle AGE = 60^\circ$. Applying law of cosines in $\bigtriangleup AGE$, we get $AE = \frac{\sqrt{13}}{2} \Rightarrow \boxed {CA = \sqrt{13}}$

$\angle AGB = 120^\circ$. Applying law of cosines in $\bigtriangleup AGB$, we get $\boxed {AB = \sqrt{7}}$[/hide]

[hide]$G$ is the centroid of triangle $ABC$, so $GD=\frac{AD}{3}$ $\implies AD=3$.

In triangle $ABD$, $\angle {ADB}=60degrees$, $BD=1, AD=3$. By using cos law:

$AB^2 = 3^2 + 1^2 - 2*3*1*cos60$

$AB = \sqrt{7}$

Similarly with triangle $ADC$ with $\angle {ADC}=120degrees, DC=1, AD=3$ will give us $AC = \sqrt{13}$

$BC=2BD=2$[/hide]


Prove : $ cos3x \equal{} cosx( 1 \minus{} 4sin^2x)$

Starting from the left...

[hide]$ cos3x\equal{}cos2xcosx\minus{}sin2xsinx$
$ (cos^2x\minus{}sin^2x)cosx\minus{}2sin^2xcosx$
$ cosx(cos^2x\minus{}3sin^2x)$
$ cosx(cos^2x\plus{}sin^2x\minus{}sin^2x\minus{}3sin^2x)$
$ cosx(1\minus{}4sin^2x)$[/hide]

Use \cos instead of cos to clean up the LaTeX.

By De Moirve,

$ \cos 3x$
$ \equal{} \cos^3 x \minus{} 3\cos x \sin^2 x$
$ \equal{} \cos x((\cos^2 x \minus{} \sin^2 x) \minus{} 2 \sin^2 x)$
$ \equal{} \cos x(1 \minus{} 4\sin^2 x)$

Derived from Euler's identity, it says that

$ cis n\theta \equal{} (cis\theta)^n$

where

$ cis\theta \equal{} \cos \theta \plus{} i \sin \theta$.

From there you can just match up the real and imaginary parts.

Question: How do you type $ cis$ in Latex? \cis doesn't work

EDIT: SORRY didn't see that you wanted me to use addition identity  :oops:

\text{cis}.  de Moivre's and the addition formula are equivalent, and de Moivre's can be proven without Euler's identity.

Sorry i dont understand where the $ i$ is coming from and how it is applicable to the proof

We have the identity

$ (\cos \alpha \plus{} i \sin \alpha)(\cos \beta \plus{} i \sin \beta) \equal{} (\cos \alpha \cos \beta \minus{} \sin \alpha \sin \beta) \plus{} i (\sin \alpha \cos \beta \plus{} \sin \beta \cos \alpha \equal{} \cos (\alpha \plus{} beta) \plus{} i \sin (\alpha \plus{} \beta)$.

This is just an application of the addition formulas for sine and cosine and tells us is that the function $ \text{cis } z \equal{} \cos z \plus{} i \sin z$ behaves like the exponential function, i.e. $ \text{cis}(\alpha \plus{} \beta) \equal{} \text{cis } \alpha \plus{} \text{cis } \beta$.  Repeated application of the above fact tells us that

$ \left( \text{cis } z \right)^n \equal{} \text{cis } nz$

which allows us an easy algebraic technique to find the $ n$-angle formulas.  For example,

$ (\cos x \plus{} i \sin x)^3 \equal{} \cos 3x \plus{} i \sin 3x \equal{} (\cos^3 x \minus{} 3 \cos x \sin^2 x) \plus{} i(3 \cos^2 x \sin x \minus{} \sin^3 x)$

which gives us the two identities

$ \cos 3x \equal{} \cos^3 x \minus{} 3 \cos x \sin^2 x$
$ \sin 3x \equal{} 3 \cos^2 x \sin x \minus{} \sin^3 x$.


function
linear algebra
matrix
algebra unsolved
algebra

Let $f(x)=\frac{ax+b}{cx+d}$, $f(0)\neq0$, $f(f(0))\neq0$, $f(f(...(f(0))...))=0$ (n times). Prove that $f(f(...(f(x))...))=x$ (n times) for all x where the expression is well determined.

Let \[A=\begin{pmatrix}a&b\\ c&d\end{pmatrix}.\]
Then $f(x)=A(\frac{x}{1}), \ f^{(n)}(x)=A^{n}(\frac{x}{1})$.
Let $A=C^{-1}JC$ were $J$ is Jordanov's form. If $J=\begin{pmatrix}t&1\\ 0&t\end{pmatrix}$ then we have contradition with $f^{(n)}(0)=0$. Therefore $J=\begin{pmatrix}t_{1}&0\\ 0&t_{2}\end{pmatrix},\ t_{1}=exp(\frac{2\pi i k}{n})t_{2}$. It give $J^{n}=A^{n}=E$ and $f^{(n)}(x)=x \ \forall x$.

[quote="Rust"]Let
\[A=\begin{pmatrix}a&b\\ c&d\end{pmatrix}. \]
Then $f(x)=A(\frac{x}{1}), \ f^{(n)}(x)=A^{n}(\frac{x}{1})$.
Let $A=C^{-1}JC$ were $J$ is Jordanov's form. If $J=\begin{pmatrix}t&1\\ 0&t\end{pmatrix}$ then we have contradition with $f^{(n)}(0)=0$. Therefore $J=\begin{pmatrix}t_{1}&0\\ 0&t_{2}\end{pmatrix},\ t_{1}=exp(\frac{2\pi i k}{n})t_{2}$. It give $J^{n}=A^{n}=E$ and $f^{(n)}(x)=x \ \forall x$.[/quote]
Jesus! thx a lot... i will try to understand!

If $f(x)=\frac{ax+b}{cx+d}, A=\dbinom{a \ b}{c \ d}, g(x)=\frac{mx+n}{px+q}, B=\dbinom{m \ n}{p \ q},C=AB$, then $f(g(x))=\frac{(am+bp)x+(an+bq)}{(cm+dp)x+(cn+dq)}=C(\frac{x}{1}).$


inequalities
integration
combinatorics unsolved
combinatorics

let $P$ be a point in a plane. There is finitely many line segment in the plane, and sum of length of that segments are $1989$. Prove that there exist a line which has no connecting point with all segments and the distance from $P$ is less than $704$

it is related to pigeon hole principle..

I don't quite understand the question....
I think we can locate $P$ very far from any lines....

Can all the lines be parallel ???  :maybe:

[quote]I think we can locate $P$ very far from any lines.... [/quote]

We want to show that, for all points P, we can draw a line that

a) passes within a distance of 704 from P
b) does not intersect any of the line segments already given in the question.

This line is one that we draw: not one of the line segments.

How do you attack such a problem... it seems much too vague. And how is this combinatorics? (Unless there is a really ingenious solution...)

It seems like you could strengthen the sum of the lengths even more, because you would consider a circle of radius $704$ around $P$, and you want a chord of that circle which intersects no line segments...but you could really make the chord as short as you wanted to...hmm this is a good one

Suppose there is no such line. Draw a circle, $C$, of radius $704$ with center at $P$. Let $AB$, and $CD$ be two diameters of $C$ perpendicular to each other. Suppose that there are $k$ segments, $l_{1},\ldots,l_{k}$, in total and $a_{i}, b_{i}$ be the lengths of projections of $l_{i}$ onto $AB$ and $CD$ respectively.  We claim that,  (call this inequality [b](1)[/b] - for some reason I'm unable to number stuff) \[a_{1}+a_{2}+\ldots+a_{k}\geq 2\cdot 704 .\] For if not, not there is a point $Q$ on $AB$ not lying on any of the projections, so a line perpendicular to $AB$ from $Q$ will not intersect any line segment - contradicting our assumption. Similarly we can show that (call this inequality [b](2)[/b]) \[b_{1}+\ldots+b_{k}\geq 2 \cdot 704 .\] We will need the following elementary inequality.
[b]Claim[/b]: If $0 \leq a_{i}, b_{i}$ for $1 \leq i \leq k$, then (call this inequality [b](3)[/b]) \[\sum_{i=1}^{k}\sqrt{a_{i}^{2}+b_{i}^{2}}\geq \left[ (\sum_{i=1}^{k}a_{i})^{2}+(\sum_{i=1}^{k}b_{i})^{2}\right]^{1/2}.\] Now, by hypothesis \[1989 > \sum_{i=1}^{k}\sqrt{a_{i}^{2}+b_{i}^{2}}\] (the length of $l_{i}$ is $\sqrt{a_{i}^{2}+b_{i}^{2}}$). From the above inequality, [b](1), (2), (3) [/b]we get that - $1989 \geq [ (2\cdot 704)^{2}+(2\cdot 704)^{2}]^{1/2}$, which implies that $1989 > 1991$, a contradiction. So our assumption must be false.
We can prove [b](3)[/b] easily by considering a right angled triangle with edge lengths $a_{1}+a_{2}+\ldots+a_{n}$ and $b_{1}+b_{2}+\ldots+b_{n}$. I'll leave out the details.

Is it not possible to improve the boud to  $\frac{19989}{\pi}$ by considering not only those two directions, but the whole range of angles   $0\leq \alpha\leq \pi$ and then integrating?
Maybe something like
$\int_{0}^{\pi}(\sum(l_{i}|sin(\alpha+\alpha_{i})|))d\alpha\geq \int_{0}^{\pi}(2R)d\alpha$

$\sum(\int_{0}^{\pi}l_{i}|sin(\alpha+\alpha_{i})|d\alpha)=\sum(l_{i}\int_{0}^{\pi}|sin(\alpha+\alpha_{i})|d\alpha)=$
$\sum(l_{i}\int_{0}^{\pi}sin\alpha.d\alpha)=\sum(2l_{i})=2\times1989\geq \int_{0}^{\pi}(2R)d\alpha=2 \pi R$
$R\leq \frac{1989}{\pi}$


This sounds weird, I know, but do you wear glasses?  When I told a guy in my physics class that I didn't wear contacts like he thought, he thought I was joking b/c I like math and science.  I'm also the only person in my grade on the math team who doesn't wear glasses, so i thought i'd do this poll.

I only wear glasses if I'm going to somewhere other than the pool, such as school, or the library, etc.  I feel like I don't need them (they help me see better though), so I only wear them because people expect me to wear glasses.

Um... I have contacts, but most of the time I'm too lazy to put them on. Only when I do sports do I wear my contacts. Otherwise, glasses are fine with me. I just hate it when I play tennis and basketball and my glasses get fogged up.

same here I have contacts but I don't always wear them except when I'm doing a sport or something.

I wear contacts.  Usually I leave them in too long!  I am just too lazy to take them out.  For example, weekdays I have cross country in the morning.  I put my contacts in because I really wouldn't want to run with glasses.  On weekends I hardly ever put my contacts on unless I am going out somewhere.  Besides my glasses are old and since the time I got them my eyes have gotten worse  :(

Well, I have awful vision, but it just developed in the last few years of my life.  I am not used to wearing glasses so I usually don't wear them...they feel awkward and I am just not used to it.  I do have trouble seeing the board in class, but if I squint it's fine.  I am saving up to get contacts though before I go to college.  They are not a big deal (a lot of work), right?  Do they feel awkward?

The first couple weeks of wearing contacts are awkward.  Once you get used to them they become like brushing your teeth.  You put them in in the morning, and don't think about them until you want to take them out.  I have soft contacts.  I don't think that there is much of a difference b/w hard or soft in the ways of how long it takes to get used to them.

Awkward, how so?  Like they feel awkward in your eyes, or it's awkward to put them in every morning?

They are awkward putting in.  They might feel a little awkward wearing but I don't remember anything like that when I first started wearing them.

[quote="elston"]I wear glasses. Contacts are kind of annoying because you have to keep cleaning them and stuff. When my glasses get dirty, I wipe them on my shirt.  :P The only bad thing about them is that they aren't good for running and sports.[/quote]

I dispute that.  You have to constantly wipe your glasses with your shirt, as you said.  You can leave contacts in all day without touching them.  Glasses can also get scratches.  The only time you have to clean contacts is when you take them out and before you put them in.

This is the last post I am making in this thread... it is kinda silly to argue over contacts and glasses

Does anyone know if glasses make your eyes worse?

What about contacts? I haven't gotten contacts yet because I am afraid that putting something in my eye might hurt my vision.

I got the best eyes from all the people I know :D so no I really don't know at you guys are talking about.

i wear contacts. i used to wear glasses, but switched due to sports. it kinda hurts to get hit in the face with a soccer ball when u have glasses on. lol.

how about coloured contacts??? does anyone of u guyz wear those? mine are royal blue. they're prettiful. lol

[quote="MithsApprentice"]I doubt there's any sort of link between ability to do any type of academic activity and eyesight...

but perhaps the lifestyles of many kids who do participate in math and physics competitions would result in them wearing glasses[/quote]

may be a more reasonable explanation indeed

however, I've been sitting 8+ hours per day in front of my computer since I was a little lad, and still my eyes are undamaged :cool:

so it's not lifestyle alone either.

I think another reasonable explanation is: they have to read most books, so they care most about their sight. If you hardly ever read a book, and you don't need to drive a car, or whatever, then you got no much reason to wear glasses either... which doesn't mean your eyesight is better.

I think near-sightedness is mostly hereditary. Most of the people I know who where glasses have parents who where glasses too. Lifestyle isn't such an important factor, unless of course you are constantly exposed to damaging radiation. :D

[quote="starrynightskies"]wow... this is really pointless[/quote]

Of course it is... this is the spam forum...umm, i mean fun and games :lol:

lol... yea most posts in here are spam.. actually pretty much all... including this one

oh this whole topic is spam??? In that case I could just delete the whole thing....

[quote="RC-7th"]You could, just as you could with virtually everything else. How often does someone actually post a math problem in here?[/quote]

but, the point of this forum is not to be mathy. that's what the other forums are. note that the name is 'games and fun factory.' if this forum was meant to be mathy, then it woulr be. it, however, is not.

did i already mention that im near-sighted because of my reading habits as a child, and not because of my parents genetics? i read compulsively as a small child, and my eyesight suffered because of that.

If your eyes are good and you wear glasses that are worse than your eyes, then probably your eyes will get worse.

I wear glasses all the time. My eyesight is just plain horrible. A bunch of papers on the other side of the room (short way) looks like a piece of poster board without my glasses.

There is NO link between eyesight and your enjoyment for math/sciences.
I dont wear glasses (Thx for inventing LCD's ) :D


geometry
geometric transformation
homothety
geometry proposed

Let DM be the diameter of the incircle of a triangle ABC where D is the point at which the incircle touches the side AC. The extension of BM meets AC at K. Prove that AK = CD.

Well-known indeed...

The problem is contained in the assertions << BX = CX' >> and << S lies on the line AX' >> in my posting in the following thread: http://www.mathlinks.ro/Forum/viewtopic.php?t=5811

  Darij

I found anhoter way to prove this fact, which make use of a Brianchon's theorem.

If A' and B' are the points where the tangent to the incircle at M cuts BC and BA respectively, it results that MD, AA' and CC' are concurrents(*), let's say at H.

Then by similarity, CK/A'M = KB/MB = AK/MC' and from this CK/AK = A'M/MC' = AD/CD. So AK/CK+1 = CD/AD+1 ==> CK = AD.

(*) Probably, in this case, is possible to prove  the concurrence, using the fact that AC//A'C', whitout involving Brianchon. But when time is lacking is rare found the simplest solution.

[quote="sprmnt21"]
(*) Probably, in this case, is possible to prove  the concurrence, using the fact that AC//A'C', whitout involving Brianchon. But when time is lacking is rare found the simplest solution.[/quote]

Ok! now I have it.
Is possible to derive only from Thales the fact.

just note that by homothethy, K is the tangency point of the excircle , from which it follows that AK=CD=1/2(AC+BC-AB)


geometry
circumcircle
cyclic quadrilateral
exterior angle
perpendicular bisector
geometry proposed

We are given a convex cyclic quadrilateral ABCD. Let
P, Q, R, S be the intersections of the exterior angle bisectors of angles
ABD and ADB, DAB and DBA, ACD and ADC, DAC and DCA,
respectively. Show that the four points P, Q,R, S are concyclic.

P is the intersection of exterior angle bisectors of triangle ADB then AP is the bisector of $ \angle DAB$.
So $ \angle QAP\equal{}90^o$. Similarly, $ \angle QDP\equal{}90^o.$
$ \Rightarrow A, D$ lies on (PQ)
Denote I the circumcenter of (PQ).
We have $ \angle AIQ\equal{}2\angle APQ\equal{}2\angle ADQ\equal{}\angle ADB$ thus I lies on $ (ABCD)$. Moreover I lies on perpendicular bisector of AD.
Similarly, A, D lies on (RS) with circumcenter I.
We are done.


T men with one wife each all go to a dance. How many ways can the men pick dancing partners such that no man is paired with his wife?

[hide="This is wellknown"]The nearest integer to $ \frac{T!}{e}$[/hide]

It is simpler for me to write solution than to find the link.

This is the number of permutations $ \sigma$ s.t. $ \sigma (x)=x$ for no $ x$.
Consider the set $ M_{i}$ of permutations $ \sigma$ for which $ \sigma (i) = i$ and their union $ \cup M_{i}$.
By inclusion and exlusion  we have:

$ |\cup M_{i}| = \sum |M_{i}|-\sum |M_{i}\cap M_{j}|+... =$

$ \sum_{k=1}^{T}(-1)^{k-1}\binom{T}{k}(T-k)! = T! \sum_{k=1}^{T}\frac{ (-1)^{k-1}}{k!}=T! (1-e^{-1}+R)$,

where $ |R|<\frac{1}{(T+1)!}$

The sicking number is $ T!-|\cup M_{i}| = \frac{T! }{e}-R\cdot T!$ and because it is [b]integer[/b] we are done.


function
number theory
greatest common divisor
calculus
integration
Putnam
logarithms

Evaluate the sum of the infinite series

$S = \sum_{m, n = 1 \, | \, gcd(m, n) = 1}^{\infty}\frac{1}{mn(m+n)}$

No takers?

[hide="Musings"] I don't exactly have a solution for this, but breaking it up into an inner and outer summation:

$S = \sum_{m=1}^{\infty}\left( \sum_{n = 1, gcd(m, n) = 1}^{\infty}\frac{1}{mn(m+n)}\right)$

And evaluating the inner summation first,

$\sum_{n = 1, gcd(m, n) = 1}^{\infty}\frac{1}{mn(m+n)}= \frac{1}{m^{2}}\sum_{n=1, gcd(m, n) = 1}^{\infty}\left( \frac{1}{n}-\frac{1}{m+n}\right)$

If $gcd(m, n) = 1$ then $gcd(m, m+n) = 1$ so this summation telescopes so that it can be written

$\frac{1}{m^{2}}\sum_{n=1, gcd(m, n) = 1}^{m-1}\frac{1}{n}$

It seems like that sum would have a nice form, but I can't think of it. [/hide]

Maybe, but maybe not.  When I tried this problem, I did this:

Let $S(d)=\sum_{m,n=1;(m,n)=d}^{\infty}\frac{1}{mn(m+n)}$.  Then $S(d)=S(1)\cdot\frac{1}{d^{3}}$ and we also have

$\sum_{m,n=1}^{\infty}\frac{1}{mn(m+n)}=\sum_{d=1}^{\infty}S(d)=S(1)\cdot\sum_{i=1}^{\infty}\frac{1}{i^{3}}$.

But this means $\sum_{m=1}^{\infty}\frac{1}{m^{2}}H_{m}=S(1)\cdot\sum_{i=1}^{\infty}\frac{1}{i^{3}}$ where $H_{k}=\sum_{i=1}^{k}\frac{1}{i}$.  I don't know if $S(1)$ has an explicit form.  Has this problem been solved?

For evaluating the sum (without gcd)
\[\sum_{m=1}^\infty \sum_{n=1}^\infty \frac{1}{mn(m+n)}\, , \]
maybe we can use the integral identity
\[\frac{1}{m+n}= \int_{0}^{1}x^{m+n-1}\, dx . \]

We evaluated a similar sum this way in
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=121384[/url].

Yes, both of your approaches are part of the solution given in the text.  (The solution is definitely a bit beyond Pre-Olympiad level, actually... )

i think this is a putnam problem. :wink:

edit: or maybe ive seen it in Larson's book.

[quote="zgorkster"]Huh.  I think this means that the LHS above is $2\cdot\sum_{i=1}^{\infty}\frac{1}{i^{3}}$, and so that would mean $S(1)=2$.  Right?[/quote]

Yes :)

[quote="amirhtlusa"]i think this is a putnam problem. :wink:[/quote]

Yes :P   Perhaps I should've checked the solution to see how difficult it was before posting...

Does anybody see an easy way to show that

$\int_{0}^{1}\frac{(ln(1-x))^{2}}{x}dx=2\zeta (3)$?

[hide="The book gives the following..."] Let $x = 1-e^{-u}$.  This substitution produces

\begin{eqnarray*}\int_{0}^{1}\frac{\ln^{2}(1-x)}{x}\, dx &=& \int_{0}^{\infty}\frac{u^{2}e^{-u}}{1-e^{-u}}\, du \\ &=& \int_{0}^{\infty}u^{2}\sum_{n=1}^{\infty}e^{-nu}\, du \\ &=& \sum_{n=1}^{\infty}\int_{0}^{\infty}u^{2}e^{-nu}\, du \\ &=& \sum_{n=1}^{\infty}\frac{2}{n^{3}}\\ \end{eqnarray*}

It's quite ingenious.  I suppose you would arrive at that substitution by guessing at the substitution $1-x = e^{u}$ and then trying the inverse. [/hide]

Wow.  That is amazing.  That also gives the equality that

$\sum_{m=1}^{\infty}\frac{1}{m^{2}}\cdot H_{m}=2\zeta (3)$.

I think that the problem can be generalized in the following way.

Let $k\geq 2$ be a natural number.

Prove that

$\sum_{m,n=1, gcd(m,n)=1}\frac{1}{n^{k}m^{k-1}(n+m)}=\frac{\xi^{2}(k)}{2\xi(2k)}$.

In particular for $k=2$ we get that

$\sum_{m,n=1,qcd(m,n)=1}\frac{1}{n^{2}m(n+m)}=\frac{5}{4}$.

Am I right?


geometry
inequalities
calculus
derivative
combinatorics proposed
combinatorics

Edited problem:

Does there exists a configuration of an infinite number of lines in the plane with the following three properties?:
i) no three concurr
ii) no two are parallel
iii) the areas of all minimal polygons are equal (minimal polygons= polygons without segments inside)

Probably you forgot to say that any two lines must intersect.

Yes, with no two parallel and with not all identified in a point.

Are you sure you mean [b]every[/b] polygon formed? I mean, if this what you're saying, then not even four such lines exist :? First of all, no three can be concurrent (because a certain fourth line would then form three triangles with the three concurrent lines, and at least one of the triangles will be strictly included in another triangle). On the other hand, if we have four lines, no three concurrent, they form a convex quadrilateral which is strictly includes in some triangle.

Maybe you mean all minimal polygons, i.e. those which do not contain other polygons.

[quote="grobber"]Maybe you mean all minimal polygons, i.e. those which do not contain other polygons.[/quote]
Surely it was his idea.

yes of course sorry it is all minimal polygons

I have discussed this with manuel, and we have added a condition: no 3 lines concur. why? imagine a line with marks every unit away, and a point P outside the line, then the lines joining P with the marks, all triangles have equal areas.

Call a minimal polygon a cell. Clearly are cells are convex. Now there is a cell which is in fact a triangle, just choose the triangle with minimal area (it exists since all areas are integer multiples of some value)Let this triangle be $ABC$ and wlog assume it's equilateral or else apply an affine transformation to make it equilateral. Let $C_1=YABZ\cdots, C_2=TBCU\cdots, C_3=VCAX\cdots$ be the cells adjacent to the sides of the triangle. Since $YABZ$ belongs to $C_1$ its area is at most 1 thus $|CY| |CZ|\leq 2$ hence by CBS $|AY| |BZ| \leq (\sqrt 2-1)^2$. Combining this with analogous relations we deduce that $|AY| |AX| |BZ| |BT| |CU| |CV| \leq (\sqrt 2-1)^6$ thus one of $|AX| |AY|, |BZ| |BT|, |CU| |CV|$ is at most $\sqrt 2-1$, say $|AY| |AX|$. Now let $BY$ intersect $CX$ in $S$ with $A$ inside $\delta SBC$- this is easy to prove. Let the lines passing through $Y, X$ other then the sides of the triangle intersect in $R$ and $R$ is inside $SYAX$. Then the area of $RYAX$ is at most the area of $SYAX$ but we will prove this area is less than the area of $\delta ABC$ providing a contradiction since $RYAX$ is either itself a cell or contains a cell inside it. Now this is algebra: Assign masses $x,y,z$ to $B,C,S$ and let $AY=m, AX=n$. Then easy mass computations show that $\frac{[SYAX]}{[ABC]}=\frac{\frac{x+y+z}z-1-\frac x{y+z}-\frac y{x+z}}{1}$. To prove this is at most 1 we must prove $\frac{x+y+z}z\leq 2+\frac x{y+z}+\frac y{x+z}$. Nut now note that $\frac x{y+z}=m, \frac{y}{x+z}=n$ therefore $\frac z{x+y+z}=1-\frac m{m+1}-\frac n{n+1}$  thus our inequality transfoms to  $(\frac 1{m+1}+\frac 1{n+1})(2+\frac 1m+\frac 1n)\geq 1$. However we may note that if $mn \leq (\sqrt 2-1)^2$ then $\frac 1{m+1}+\frac 1{n+1}-1\geq \sqrt 2-1$ (trivial derivatives) and $2+\frac 1m+\frac 1n\geq 2+\frac 2{\sqrt 2-1}$ and thus we can easily conclude the desired inequality.


whatś the rest of division of 3^2^n by 2^n+1


ye soal functin equ az ireland 1995 $ f: R\longrightarrow R$ $ xf(x)\minus{}yf(y)\equal{}(x\minus{}y)f(x\plus{}y)$

Avval az hame age tarif konim $ g(x)\equal{}f(x)\minus{}f(0)$ dar in soorat $ f(x)\equal{}g(x)\plus{}f(0)$ pas age too rabete gharar bedim :
$ xg(x)\minus{}yg(y)\plus{}(x\minus{}y)f(0)\equal{}(x\minus{}y)g(x\plus{}y)\plus{}(x\minus{}y)f(0)$ pas :
$ xg(x)\minus{}yg(y)\equal{}(x\minus{}y)g(x\plus{}y)$
yani tabeye $ g$ ham khasiate masale ro dare va alave bar oon : $ g(0)\equal{}0$ pas kafie hameye tavabeye $ g$ ba sharte $ g(0)\equal{}0$ va $ xg(x)\minus{}yg(y)\equal{}(x\minus{}y)g(x\plus{}y)$ ro peyda konim :
tavvajoh konin age too rabeteye bala gharar bedim $ y\equal{}\minus{}x$ be dast miad : $ g(x)\plus{}g(\minus{}x)\equal{}0$ yani tabeye $ g$ farde .
hala age $ y\longrightarrow \minus{}y$ pas :
$ xg(x)\plus{}yg(\minus{}y)\equal{}(x\plus{}y)g(x\minus{}y)$ va az oonjayi ke $ g(\minus{}y)\equal{}\minus{}g(y)$ pas :
$ (x\minus{}y)g(x\plus{}y)\equal{}(x\plus{}y)g(x\minus{}y)$ yani : $ \frac{g(x\plus{}y)}{x\plus{}y}\equal{}\frac{g(x\minus{}y)}{x\minus{}y}$ . midoonim dastgahe moadelate $ x\plus{}y\equal{}a$ va $ x\minus{}y\equal{}b$ baraye har $ a$ va $ b$ ye javab mide pas baraye har $ a$ va $ b$ haghighi ke mokhalefe $ 0$ hastand darim :
$ \frac{g(a)}{a}\equal{}\frac{g(b)}{b}\equal{}c$
pas baraye har $ x$ ke $ x\neq 0$ darim $ g(x)\equal{}cx$ ama in rabete baraye $ x\equal{}0$ ham bargharare pas baraye har $ x$ darim : $ g(x)\equal{}cx$
dar natije : $ f(x)\equal{}g(x)\plus{}f(0)\equal{}cx\plus{}d$

rahe ghashangi bud. rahe manam taghriban shabihe in bud amma akharash.ye soal ham gozashtam male singapore 2008 khodam hanuz halesh nakardam.yeki hal kone


geometry
geometric transformation
rotation
homothety
function
ratio
geometry solved

In the plane, there are two circles $\Gamma_1, \Gamma_2$ intersecting each other at two points $A$ and $B$. Tangents of $\Gamma_1$ at $A$ and $B$ meet each other at $K$. Let us consider an arbitrary point $M$ (which is different of $A$ and $B$) on $\Gamma_1$. The line $MA$ meets $\Gamma_2$ again at $P$. The line $MK$ meets $\Gamma_1$ again at $C$. The line $CA$ meets $\Gamma_2 $ again at $Q$. Show that the midpoint of $PQ$ lies on the line $MC$ and the line $PQ$ passes through a fixed point when $M$ moves on $\Gamma_1$.

[color=red][Moderator edit: This problem was also discussed on http://www.mathlinks.ro/Forum/viewtopic.php?t=21414 .][/color]

The pencil of lines through $K$ determines an involution on the pencil of lines through $A$. I'm referring to the mapping $AM\leftrightarrow AC$. This involution is hyperbolic (it has two fixed elements, the lines $AB,\ AK$).

Take $D=AK\cap \Gamma_2$ and let $L$ be the intersection point of the tangents to $\Gamma_2$ in $B$ and $D$. The pencil of lines through $L$ determines an involution on the pencil of lines through $A$, having fixed elements $AD=AK$ and $AB$, so it's the same involution as the one in the first paragraph, so it maps $AP=AM$ to $AQ=AC$, so $PQ$ passes through $L$, which is one of the conclusions we were trying to reach.

I didn't show that the midpoint of $PQ$ is on $MC$. I'll keep trying :).

I hope I didn't make any stupid mistakes up there :).

hi

your idea seems great... though i can't understand what do you mean by pencils and involutions, could yo explain a little or tell me where can i find the theoretical foundations to understand it?

also i found solution using rotations and homotheties, denote rotomothety as a composition of a rotation and a homothety, then M and C are carried through a rotomothety centered at B to P and Q respectively, from there it follows that PQ passes through the antirotomothetical point of K.

Also for the second part we can use Menelaus in triangle APQ and line KCM, so that it rests to prove something about A C Q and M which also should follow from the rotomothety

"A pencil of lines through $A$" is just the fancy name for the set of lines passing through $A$ :).

An involution on this set is a function defined on the pencil with values in the same set, which maps the line $d$ to the line $d'$ and vice-versa (meaning that $f\circ f=1_P$, where $P$ is the pencil) s.t. $(d_1,d_2;d_3,d_4)=(d'_1,d'_2;d'_3,d'_4)$ for any four lines $d_i\in P$.

However, your solution seems to be much shorter :).

Also, I think there's a standard name for your "rotomothety" :). It's usually called "spiral similarity" or "similitude".

it's a very interesting transformation!
but what's (d1,d2;d3,d4) ? double ratio or something like that?

psykyk

It's the cross-ratio. Assume the four lines $d_i$ cut another line in the points $A,B,C,D$. Then $(A,B;C,D)=(\frac{\overline{CA}}{\overline{CB}}):(\frac{\overline{DA}}{\overline{DB}})$, where $\overline{XY}$ is the oriented segment $XY$.

When we change the line which cuts $d_i$ this value doesn't change, so we can speak about the cross-ratio of the four lines (this happens for four [b]concurrent[/b] lines).

Excellent Problem!
[quote=Vietnam TST 2004 P3]In the plane, there are two circles $\Gamma_1, \Gamma_2$ intersecting each other at two points $A$ and $B$. Tangents of $\Gamma_1$ at $A$ and $B$ meet each other at $K$. Let us consider an arbitrary point $M$ (which is different of $A$ and $B$) on $\Gamma_1$. The line $MA$ meets $\Gamma_2$ again at $P$. The line $MK$ meets $\Gamma_1$ again at $C$. The line $CA$ meets $\Gamma_2 $ again at $Q$. Show that the midpoint of $PQ$ lies on the line $MC$ and the line $PQ$ passes through a fixed point when $M$ moves on $\Gamma_1$.[/quote]
[b]Solution:[/b] Let $AK \cap \Gamma_2=E$. Let the tangents to $\Gamma_2$ at $B,E$ intersect at $D$
$$-1=(M, ~C ; ~ A, ~B) \overset{A}{=} (P, ~Q ; ~E, ~B)$$
Hence, $PQ$ passes through the fixed point $D$. Let $\odot (BDE) \cap PQ=F$. Let $BF \cap \Gamma_2=L$
$$\implies \angle BLE=\angle BAE=\angle BED=\angle BFD \implies FD||EL$$
Hence,
$$-1=(B, ~E ; ~Q, ~P) \overset{L}{=} (F, ~ \infty_{PQ} ; ~Q, ~P) \implies F \text{ is the midpoint of } PQ$$
$\angle DEB=\angle BAK \implies \Delta BAK \stackrel{+}{\sim} \Delta BED$ $\implies$ $K \in \odot (BDE)$ $\implies$ $\angle BKD=\angle BED=\angle BAK$ $\implies$ $FD$ tangent to $\odot (BAK)$,
$$\angle PFL=\angle BFD=\angle BKD=\angle BAK=\angle BMP \implies MBFP \text{ is cyclic}$$
Hence, $\angle BAC=\angle BPF=\angle BMF \implies$ $\overline{MCK}$ bisects $PQ$

[hide=Solution]Projective Cookie :)
Let $l$ be the tangent to $M$ at $\Gamma_1$.Let $T=KA\cap l$.
 $\textbf{Claim} :$[color=#00f]$AC\cap MB$ lies on $\Gamma_2$[/color]
$\textbf{Proof}$ : Using $l \parallel PQ$ we get $\angle  ABM + \angle QBA = \angle AMT + \angle QBA = \angle APQ + \angle ABQ = 180^{\circ}$.Hence the claim.$\square$
 Reim's theorm $\implies MM \parallel PQ \implies l \parallel PQ$.Therefore since $ABMC$ is harmonic $$-1=(A,B,M,C)\overset{M}{=}(P,Q,\infty_{PQ},PQ\cap MK)$$.Hence $PQ\cap MK$ will be the midpoint of $PQ$ and we are  done.

Unfortunately I haven't tried its part (b) so I'll post its solution later.
Oops Just realsed that the second part was much easier than the first one :P [/hide]
EDIT-This is completely wrong just a round about :wallbash:

Here is my solution for this problem
[b]Solution[/b]
Let $O_1$, $O_2$ be center of $\Omega_1$, $\Omega_2$, respectively; $N$, $U$ be midpoint of $PQ$, $CM$, respectively
We have: $(QP; QB) \equiv (AP; AB) \equiv (AM; AB)$ (mod $\pi$) and $(PB; PQ) \equiv (AB; AQ) \equiv (AB; AC) \equiv (MB; MC)$ (mod $\pi$)
Then: $\triangle{BCM}$ $\stackrel{+}{\sim}$ $\triangle{BQP}$
But: $U$, $N$ are midpoint of $CM$, $PQ$, respectively so: $\triangle{BCU}$ $\stackrel{+}{\sim}$ $\triangle{BQN}$
Hence: $(NB; NQ) \equiv (UB; UC) \equiv (MB; MA) \equiv (CB; CA) \equiv (CB; CQ)$ (mod $\pi$) or $B$, $C$, $N$, $Q$ lie on a circle
Then: $(CM; CN) \equiv (CM; CB) + (CB; CN) \equiv (QP; QB) + (CB; CN) \equiv 0$ (mod $\pi$) or $CM$ passes through midpoint $N$ of $PQ$
Let $S$ be a point which satisfies $KS$ $\perp$ $O_1O_2$ and $BS$ tangents $(O_2)$ at $B$
We have: $(SK; SB) \equiv (BA; BS) \equiv (PA; PB) \equiv (O_1O_2; O_2B)$ (mod $\pi$) and $(KB; KO_2) \equiv (KB; KO_1) \equiv \dfrac{\pi}{2} - (O_1K; O_1B) \equiv \dfrac{\pi}{2} - (MA; MB) \equiv \dfrac{\pi}{2} - (CA; CB) \equiv \dfrac{\pi}{2} - (CQ; CB)$ $\equiv \dfrac{\pi}{2} - (NQ; NB) \equiv (NB; NO_2)$ (mod $\pi$)
So: $B$, $K$, $S$, $O_2$, $N$ lie on a circle
But: $BS$ $\perp$ $BO_2$, $BK$ $\perp$ $O_1O_2$ then: $O_2N$ $\perp$ $SN$
Combine with: $O_2N$ $\perp$ $NQ$, we have: $P$, $N$, $Q$, $S$ are collinear
Hence: $PQ$ passes through fixed point $S$


now we know $ \minus{}\Delta G^{0} \equal{} RT\ln K$
now suppose the reactants are in their standard states and the temp. being 298 K and the pressure being 1 atm..
.Let  equilibrium be acheived then clearly $ \Delta G^{0}$ is 0 but not necessarily $ K$ is 1 so why do we get this contradiction???

$ \text{Standard State} \neq \text{Equilibrium}$

Consider $ \Delta G = \Delta G^{\circ} + RT \ln{Q}$ where $ Q$ is the reaction quotient. Standard state occurs at $ Q=1$ giving $ \Delta G = \Delta G^{\circ}$. When $ Q=K$ at equilibrium we have $ \Delta G = 0$.

that's ok...but what id the reactants at standard state give products and the reaction is in equilibrium???

[quote="pardesi"]now we know $ \minus{} \Delta G^{0} \equal{} RT\ln K$
now suppose the reactants are in their standard states and the temp. being 298 K and the pressure being 1 atm..
.Let  equilibrium be acheived then clearly $ \Delta G^{0}$ is 0 but not necessarily $ K$ is 1 so why do we get this contradiction???[/quote]

Well, atleast one non-organic question. :D Hmmm....From the formula it does appear that any equilibrium at 298K and 1atm will have to have Keq=1. Is this true??

[quote="pardesi"]now we know $ \minus{} \Delta G^{0} \equal{} RT\ln K$
now suppose the reactants are in their standard states and the temp. being 298 K and the pressure being 1 atm..
.Let  equilibrium be acheived then clearly $ \Delta G^{0}$ is 0 but not necessarily $ K$ is 1 so why do we get this contradiction???[/quote]

I think it may be bcos at standard state the eqbm cannot be achieved as it is the most stable state @ that temp.

no...u have equilibrium reactions at 298K

is it necessary that the compound is most stable in standard state :maybe:
i have never heard of this ...i though since 298K is the most common lab condition so this was logically chosen as standard

[quote="madness"]no wat i meant was maybe since the compd is most stable in its standard state (max thermodynamic stability) it might be the major compound @ the std condns and hence the eqbm may not ata all practically exist (or can it?)  :)[/quote]

Paethel. You are illogically trying to explain the product.

im only seconding Banished Traitor's opinion saying that both the std. state and the eqbm may not coexist together  :)

[quote="madness"]wat i meant was for ex: Graphite is the most stable compound (and not buckminsterfullerene) so that is the major thermodynamic prdt.[/quote]
but if my std. temp would have been higher for the case of sulphur it would have been monoclinic sulphur not rhombic sulphur

Pardesi, tell us again: why is $ \Delta G^o \equal{} 0$?

if u have caught the fallacy this may sound useless......
Consider a reaction carried out at 298K and 1atm pressure.Now suppose the reaction acheives equilibrium at some point then we have $ \Delta G^{0}\equal{}0$ since at equilibrium change in Gibb's Energy is 0

[quote="Carcul"]No. At equilibrium $ \Delta G \equal{} 0$, not $ \Delta G^o$.[/quote]

What he means is that what if the reaction attains equilibrium at 298 K and 1 atm.

[quote="pardesi"]yes that's right shreyas....(until carcul got the fallacy)[/quote]

Do you mean to say that that is the fallacy  :o   :huh:

yes of course delta G0 is the value at the std state only and it is a fixed value. Its not = 0 ofc.
Stupid me  :D  :D  :D

why is $ \Delta G^{0}$ fixed and not other $ \Delta G$ .....after all it is the gibb's energy at a specific temperature and it's behaviour should be similar to all other gibb's energies

bcos $ \Delta G^0$ is defined so. It is the value of $ \Delta G$ at a standard temp.  :D

[quote="Cyber-Shreyas"]What he means is that what if the reaction attains equilibrium at 298 K and 1 atm.[/quote]

What's so special about 298 K and 1 atm? Regarding the standard Gibbs energy variation for a reaction, the standard states are defined for 1 bar and [i]for each temperature[/i] (and not for 298 K as you all keep saying). Only by coincidence we will have $ \Delta G^o \equal{} 0$.

[quote="pardesi"]so what all $ \Delta G$ 's are measured at some temp. or the other....it doesn't take much pain to define some other temp.say 92489327589 K  :D  as the standard temp.[/quote]

But i say that $ \Delta G^0$ is a constant value as Carcul said  :D

ok actually we are taught it's for constant temp...
so if for a particular temp let's carry out the reaction with elemnts in standard state now when the attain equilibrium the $ \Delta G^{0}$ is 0 so we have $ K$ for the reacion 1.but $ K$ ofcourse depends only on temperature and this logic holds for any reaction... :D

[quote="Carcul"]
What's so special about 298 K and 1 atm? Regarding the standard Gibbs energy variation for a reaction, the standard states are defined for 1 bar and [i]for each temperature[/i] (and not for 298 K as you all keep saying). Only by coincidence we will have $ \Delta G^o \equal{} 0$.[/quote]
read this rohit it's defined for each temperature and not for specific one....
and my argument even hold if it is just defined for 298 K.
just react the reactants at 298 K...at some point equilibrium will be attained at that point $ \Delta G^{0}\equal{}0$ hence $ K\equal{}1$ hence for all reactions $ K\equal{}1$ at 298 K :P

[quote="madness"]No dear boy the value is a [u]constant independent of the temp  [/u]:D
Its $ \Delta G$ which is temp dependent and becomes equal to $ \Delta G^0$ at std. temp  :D[/quote]

He he dont try to confuse me  :D  Read my above post  :)

[quote="Carcul"]What a big confusion you are making here. Pardesi, why do you keep saying that $ \Delta G^o \equal{} 0$ when the reagents are in their standard states? And what about the products?[/quote]
no i am telling $ \Delta G^{0}$ is 0 when the reaction is at equilibrium with the reactants origianally being in their standard states

But can you explain why that must happen (according to you, of course)?

that's because when reactions are at equilibrium we say the change in Gibb's energy si 0
similarly for a reaction with reactants in standard states which attains equilibrium $ \Delta G^{0}$ is 0

[b]But what makes you think that the products are also in their standard state?[/b]

is it necessary that to get $ \Delta G^{0}$ the products should aso be in their standard states :D
well that pretty much is the fallacy...
i bet without u this would have taken another 2-3 pages of fight between me and madness

Yeah maybe  :D  [b]Between you and [u]MADNESS[/u]  [/b]:D  :D

no the actual reason...is that standard gibb's energy is pretty much ahypothetical cocept...it is the freee energy change  when the reactants and products(as carcul said) are in standard states nad the reaction is complete and irreversible

[quote="madness"]no wat i meant was maybe since the compd is most stable in its standard state (max thermodynamic stability) it might be the major compound @ the std condns and hence the eqbm may not ata all practically exist (or can it?)  :)[/quote]

Maybe that's what i after all meant above before you detracked me  :D  :D

[quote="madness"]
Maybe that's what i after all meant above before you detracked me [/quote]
 :huh:
pavom koyendai(baby) par ....vayu la veral vechallum kadikuka theriyathu :P

Dei vandam veena vambhu iyyukkadhe da  :D . MOD Edit: Called for. Plz delete those derogatory comments(In Tamil) abt me   :mad:
Osama ve onna close panna solliduvein   :P

Whatever Madness said about stable products is absolute rubbish.

I think this entire thing of Standard state came into place only so that the calculation part becomes easier. Enthalphy of formation of elements is 0. DO YOU BELIVE IN THAT FACT THEN???

there is nothing there not to beleive...
since we assume we have been provided with the requisite elemnets os their enthalpy of formation is 0

Sorry to revive an old thread. But can anybody please clearly define std state.

See this  :)
http://en.wikipedia.org/wiki/Standard_state


[color=darkblue]Let $ABC$ be a triangle for which $A=60^{\circ}$ and let $S$ be a point which belongs to the side $(BC)$.

Prove that $\frac{SB}{SC}=\frac{a+b}{a+c}\Longleftrightarrow m(\widehat{SAB})=\frac{B}{2}\Longleftrightarrow m(\widehat{CAS})=\frac{C}{2}\ .$[/color]

it's enough to show that for a point $S$ on $(BC)$ such that $m(\widehat{SAB})=\frac{\hat{B}}{2},m(\widehat{SAC})=\frac{\hat{C}}{2}$
we have:
$\frac{|SB|}{|SC|}=\frac{a+b}{a+c}$
to see this notice that :$\sin\frac{\hat{B}}{2}=\sqrt{\frac{(a+b-c)(c+b-a)}{4ac}}$
so since from sin rule we know:
$\frac{|SB|}{\sin \frac{\hat{B}}{2}}=\frac{c}{\sin ASB}$ and
$\frac{|SC|}{\sin \frac{\hat{C}}{2}}=\frac{b}{\sin ASC}$ we need to show that :
$\frac{c\sin \frac{\hat{B}}{2}}{b\sin \frac{\hat{C}}{2}}=\frac{a+b}{a+c}$
$\leftrightarrow (\frac{b(a+b)}{c(a+c)})^{2}=\frac{b(a+b-c)}{c(a+c-b)}$
iff$b(a+c-b)(a+b)^{2}=c(a+b-c)(a+c)^{2}$
iff $(b-c)(a^{2}-b^{2}-c^{2}+bc)(a+b+c)=0$ which is true if $A=60^{\circle}$

Sorry, you made a mistake because $SB+SC\ne BC$ for $A=60^{\circ}$, i.e. $a^{2}=b^{2}+c^{2}-bc$. Verify, please !

sorry for the mistake ! :blush:

I corrected above.


modular arithmetic
graph theory
combinatorics unsolved
combinatorics

The columns of an $ n\times n$ board are labeled $ 1$ to $ n$. The numbers $ 1,2,...,n$ are arranged in the board so that the numbers in each row and column are pairwise different. We call a cell "good" if the number in it is greater than the label of its column. For which $ n$ is there an arrangement in which each row contains equally many good cells?

is it too hard that nobody can solve it? :lol:

uhh... isn't this one pretty simple? The ith column has i bad cells, so there are n(n+1)/2 bad cells which is too many.

yes.in every row we must have $ \frac {n \minus{} 1}{2}$ good cells . then we know that n is odd  we consider $ n \equal{} 2k \plus{} 1$

by the under diagram. suppose that the diagram is the tble $ n*n$ and by the good arrangement we have  $ k$  good cells in every row.

2k+1 ,  2k  , 2k-1 ,   . . .  ,1

  2k    ,   2k-1,  2k-2,......  ,2k+1

  2k-1, 2k-2 , ....       ,    2k

  .
  .
  .

  .

  1 ,2k+1,2k, 2k-1,.... ,     2
   then we are done. :wink:

Hello. I got all odd integers $n$. Is this right?

[b]Solution[/b]
We claim that $n$ is odd. First we will show that $n$ cannot be even. In the $i$th column, there are $i-1$ good cells. Therefore there are in total $n(n-1)/2$ good cells. If $n$ is even, then it follows that since $n| n(n-1)/2 \Rightarrow 2|n-1$ which is a contradiction.

Now label the rows from top to bottom as $1,2, \dots, n$. Examine the following arrangement. In the $i$th column, each natural number $k \in (1,2, \dots, n)$ is in row $k-2i \pmod{n}$. For example, such arrangement is shown for $n=5$.

[b]1 3 5 2 4
2 4 1 3 5
3 5 2 4 1
4 1 3 5 2
5 2 4 1 3[/b]
In these arrangements, there are $(n-1)/2$ good cells in each row and each of $1,2, \dots, n$ appears only once in each row and each column.

Therefore there exists such an arrangement for each $n$ if and only if $n$ is odd.

First,we must have $n*(n-1)/2$ good numbers at the whole table,so we get that $n-1/2$ must be an integer,so $n$ is odd.Now,I will explain the motivation how to come up wit an example(in fact it is not hard to come up with but still I think it is better with motivation).Consider two numbers,$i,j$.The idea is,when they switch places(I mean,when $i$ is at $j$ column and vica-versa and they are in the same row) then we produce exactly one good number.Now,the problem is equivalent to the following:We need to color the edges of a complete graph with $2k+1$ edges in $2k+1$ colors,such that for each color exists a vertex who doesn't have an edge of that color and every other vertex has exactly one edge of that color.Now,this goes easy,label the edges $1,2,...2k+1$,if tho edge $i,j$ let $m$ such that $i+j=2*m(mod2k+1)$ color the edge $i,j$ with color $m$,so we are finished.

nice and easy
[hide = solution]
We show that the answer is all odd $n$.

First, let us show the necessity. If there are $k$ good cells in each row, then there are $nk$ good cells in all. Since the $i$th column contains exactly $n-i-1$ good cells summing over all $1 \leq i \leq n$ yields that $nk = {n \choose 2}$. As $k \in \mathbb{Z}$ this forces $n$ odd.

Now for the construction we color the cell $(i, j)$ with the unique number $m \in \{1, 2, \dots, n\}$ such that $m-1 \equiv j-i \pmod n$. Basically diagonals have the same color and descending a cell downward increases the number by one. To see that this work, fix any row with number $1 \leq x \leq n$ and remark that the number of good cells is \[\left\lfloor \frac{x}{2} \right\rfloor + \left\lfloor \frac{n-x}{2} \right\rfloor = \left\lfloor \frac{n}{2} \right\rfloor\] which is independent of $x$ (remember that $n$ is odd), so we're done.
[/hide]


function
limit
algebra
polynomial
search
complex numbers
real analysis

Let $ f: \mathbb{R}\rightarrow\mathbb{R}$ differentiable such that $ f(x \plus{} 1) \minus{} f(x) \equal{} f^\prime(x)$                          $ ( \forall)x\in\mathbb{R}$(***)

  a)   If $ (\exists)$   $ \lim_{n\to\infty}f^\prime(x) \equal{} L\in\mathbb{R}$ , then f is polynomial function with deg(f)$ \leq1$.
  b)  Prove that exists non-polynomial function differentiable with property (***).

English notes: it's "differentiable" and "polynomial". Does the "ne" prefix mean "not"? The English version would be "non-".
Also, "grad" doesn't mean anything in a polynomial context; you probably want the degree, abbreviated "deg".

I'm pretty sure this has appeared before, but it's not easy to search for.

An example of a solution for (b): the real or imaginary part of $ e^{\alpha x}$, where $ \alpha$ is a root of $ e^z\minus{}z\minus{}1\equal{}0$. The smallest nontrivial root is approximately $ 2.0888\plus{}7.4615i$.


I have a question about a textbook problem:

determine if the vectors $ (1,1,3), (2,3,6), (1,4,3)$ form a basis for $ R^3$.

I solved Ax = b to find these vectors are linearly dependent (ie, there's one free variable), so they can't be a basis.

However, the book also says, "when a square matrix is invertible, its columns are independent, and they are a basis for $ R^n$". However, I don't know if the logic follows that we can check of n vectors are a basis for $ R^n$ by checking if the n by n matrix is invertible. In this case, it seems to work b/c the 3 by 3 matrix formed by the 3 vectors given above is not invertible.

Can any1 clarify?

[b]Theorem:[/b]  If $ n$ vectors are linearly independent in $ \mathbb{R}^n$, then they form a basis in $ \mathbb{R}^n$.


calculus
real analysis
real analysis unsolved

$f$ is continuous on interval $\left[0,1\right]$ and $f(0)=f(1)$. Prove that for $n>2$ exists $c\in \left[0,1-\frac{1}{n}\right]$ such that : $f(c)=f(c+\frac{1}{n})$.

Note: i'm sinajackson, this is just my another user.... :D

Let $g(x)=f(x)-f(x+\frac{1}{n})$. we have:
$g(0)=f(0)-f(\frac{1}{n})$
$g(\frac{1}{n})=f(\frac{1}{n})-f(\frac{2}{n})$
$g(\frac{2}{n})=f(\frac{2}{n})-f(\frac{3}{n})$
                    $\vdots$
$g(\frac{n-1}{n})=f(\frac{n-1}{n})-f(\frac{n}{n})$ $\longrightarrow$ $g(1-\frac{1}{n})=f(1-\frac{1}{n})-f(1)$
and so $g(0)+g(\frac{1}{n})+g(\frac{2}{n})+\cdots+g(1-\frac{1}{n})=0$, as a consequence there exist $i,j$ $\in$ $\left\{1,2,...,n-1\right\}$ such that $i>j$ and $g(\frac{i}{n})g(\frac{j}{n})\leq 0$. This show's that there exists a $c\in\left[\frac{i}{n},\frac{j}{n}\right]$ such that $g(c)=0$ or $f(c)=f(c+\frac{1}{n})$. And we're done.....


Suppose that $p,q,r,s,t,u,v,w$ are nonnegative real numbers, not all equal to zero. Prove the inequality
\[\frac{88\sqrt{pqrstuvw}}{113(p^{3}+q^{3}+r^{3}+s^{3}+t^{3}+u^{3}+v^{3}+w^{3})}\geq \sum_{cyclic}\frac{p}{7}-\sqrt[3]{\sum_{cyclic}\frac{p^{3}}{7}}. \]

What happen if $p=0$ and $q=r=s=t=u=v=1$?


ratio
limit
function
search
real analysis
real analysis unsolved

We have the arithmetical progression: a_n, with a_1>0 and ratio r>0, show that :

 \lim oo [(a_1*a_2*...*a_(n+1))^(1/(n+1))-(a1*a2*..*a_n)^(1/n)]=r/e.

cheers!!!

Do you know how to solve it for r=1? It's called Lalescu's sequence:
((n+1)!)^1/(n+1) - (n!)^1/n. This has limit 1/e. If you know how to prove this one you can easily extend the soln to the general case. Give me some time to remember how it's proved.

Thanks .. but you don't have to bother posting it cause now I found it too and I know how to prove it!! cheers!

First of all, can you tell me more about Lalescu, is he a teacher ?
Where ?

Thank's
----------------------------------------------------------------------------

I observe that in Lalescu's sequence there is n!

but n! come from gamma function n! = gamma(n+1)

where gamma(x) = int_[0;+oo( t^(x-1)exp(-t)dt  for x>0

Lalescu's sequence with gamma become

(x) =(gamma(x+2)^(1/(x+1)) - (gamma(x+1))^(1/x)

So my question

1)Compute lim (x) when x-->+oo

2) Give an asymptotic expansion of (x) when x-->+oo
--------------------------------------------------------------------

Do a Yahoo search or a Google search on Traian Lalescu and look into the first two search results there.
He passed away in 1929.

you may find a paper on Lalescu sequences
at [url]http://matematika.etf.bg.ac.yu/publikacije/pub/PUB09.htm[/url]

[quote]
5. Gh. Toader: Lalescu sequences
(P09_05.dvi, P09_05.pdf.zip)
[/quote]
the links of the files are
[url]http://matematika.etf.bg.ac.yu/publikacije/pub/P09(98)/P09_05.DVI[/url]
and
[url]http://matematika.etf.bg.ac.yu/publikacije/pub/P09(98)/P09_05.ZIP[/url]

[quote="Moubinool"]
1)Compute lim ?(x) when x-->+oo
[/quote]

\[ \displaystyle\lim_{x\to\infty}(\Gamma(x+2)^{x+1}-\Gamma(x+1)^x) = \frac{1}{e}  \]

Here's the reciporical of that:

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=394351#p394351]http://www.mathlinks.ro/Forum/viewtopic.php?p=394351#p394351[/url]


[quote="epitomy01"]If $ U_{1}$, $ U_{2}$ and $ W$ are subspaces of vector space $ V$, such that, the direct sum of $ U_{1}$ and $ W$ is $ V$, and the direct sum of $ U_{2}$ and $ W$ is $ V$, then $ U_{1}$ and $ U_{2}$ are identical.[/quote]

Wrong (counter-example: $ V \equal{} K^2$, $ W \equal{} \text{span}\left(\left(0,1\right)\right)$, $ U_1 \equal{} \text{span}\left(\left(1,1\right)\right)$, $ U_2 \equal{} \text{span}\left(\left(1,0\right)\right)$). Maybe you want to claim that they are isomorphic (but this is trivial, since both $ U_1$ and $ U_2$ are isomorphic to $ V\slash W$).

  darij

Oh ok. I checked my proof and I made a silly error indeed  :blush:.


LaTeX
calculus
derivative
function
calculus computations

Help please:

differentiate e^(-x^2)

I haven't learnt latex yet.

Let $ u \equal{} \minus{} x^2$. By the chain rule:

$ \frac {d(e^u)}{dx} \equal{} \frac {d(e^u)}{du} \cdot \frac {du}{dx} \equal{} e^u \cdot u'$

$ \equal{} \boxed{ \minus{} 2x e^{ \minus{} x^2}}$

thank you; it was part of a larger (and harder), problem but I did this at the start of my A levels two years ago and I couldn't remember! You know how it is, use it or lose it. :)

Usage of the chain rule leads to a property of derivatives of functions such as $ e^{f(x)}$:
$ \frac{d}{dx}e^{f(x)}\equal{}e^{f(x)}f^\prime(x)$

Which gives you the derivative somewhat more quickly than using the chain rule.
$ \frac{d}{dx}e^{x^{\minus{}2}}\equal{}e^{x^{\minus{}2}}(\minus{}2x)$

Functions such as $ e^{f(x)}$ have many interesting properties in calculus  :)


group theory
abstract algebra
superior algebra
superior algebra solved

$\mathfrak{S}_n$ is isomorphic to a subgroup of $\mathfrak{A}_{n+2}$, but could we refine by the following statement?

Is there $N\in \mathbb{N}$ such that $\forall n >N$, $\mathfrak{S}_{2n+1}$  is isomorphic to a subgroup of $\mathfrak{A}_{2n+2}$?

There are no such $n$, and the result has been discussed before in this forum. Try searching.