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Functional Analysis
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real analysis unsolved

let $ A$ be a self adjoint operator on a Hilbert space $ H$. Prove that $ \exp\{i(A\minus{} \lambda )t\}\phi$ has a norm limit as $ t$ goes to infinity iff $ \phi$ is an eigenvector of $ A$ with eigenvalue $ \lambda$ .
(Hint: Start proving that, if $ \lambda$ is not an eigenvalue of $ A$, the operator integral $ \int_0^T\exp\{i(A\minus{} \lambda )t\}/T dt$ tends weakly to $ 0$ as $ T$ goes to infinity. For this, use the spectral representation of $ A$. Next, show that the existence of the norm limit would yield a contradiction.) Remark: Though the statement is true also for an unbounded self-adjoint operator, in this exercise consider $ A$ a bounded selfadjoint operator.)

 :o  :o  :o

Hmm... The hint is informative enough. Where exactly is your difficulty? :? (By the way, I $ \text{\LaTeX}$-ed your post. Check that it is what you meant it to be :)).

I do see how you can prove that the integral converges  weakly to zero. And moreover if you really think is easy why you didn't bother to write the answer? ???
If I posted means I do not know it... so if you really wanted to help, answer!, if not do not bother to send me your worthless comments.

thanks for latex-ing my problem. I have problems using this site since I am new...
I will figure out soon how things work .

Thanks .
mikiduta

I understand that you cannot solve the problem but I do not understand where exactly you are stuck and my aim is to give you a reasonably good hint, not a full solution. (that is what you may expect here when posting something that looks like a typical college course assignment). My post was rather an invitation for you to show what you've done already and where your troubles start than an attempt to explain anything outright. Sorry if it was not clear. Let's start from the beginning. Do you know what "spectral representation" means?

first it is not an assignment, since I am just in my second year of college and I didn't even start do to spectral theory. I think I have an idea about spectral representation, but I guess it is not enough since I do not know to solve the problem using a hint...
The point was to understand the problem ...I do not expect the solution... I just do not know.
Anyways I will try hader to solve it. Do not worry. If you can explain to me , ok..if not that's it.

Thanks

Now I at least see what the trouble is. Without knowing the spectral theorem well, it all becomes rather hard.

The spectral theorem that you need says that for every continuous $ f$, we have $ f(A)\equal{}\int_\mathbb R f(t)\,dE(t)$ where $ E$ is an operator-valued Borel measure on the line such that $ E(S)$ is an orthogonal projection to some subspace of $ H$ that commutes with $ A$ for every Borel $ S\subset\mathbb R$. If $ x,y\in H$, then $ S\mapsto \langle E(S)x,y\rangle$ is a usual finite (signed) measure on $ \mathbb R$, which I'll call $ \mu_{xy}$.

Now, take $ f_t(x)\equal{}e^{i(x\minus{}\lambda)t}$ and set $ B_T\equal{}\int_0^T f_t(A) \frac{dt}{T}$. Then $ \langle B_T x,y\rangle\equal{}\int g_T\,d\mu_{x,y}$ where $ g_T(x)\equal{}\int_0^T e^{i(x\minus{}\lambda)t}\frac{dt}{T}$. The function $ g_T$ is not hard to compute exactly, but all we need to know is that it never exceeds $ 1$, tends to $ 0$ for $ x\ne\lambda$, and is $ 1$ for $ x\equal{}\lambda$. Hence, by the dominated convergence theorem, $ \langle B_T\phi,\psi\rangle\to \mu_{x,y}\{\lambda\}\equal{}\langle E\{\lambda\}x,y\rangle$.

Now $ E\{\lambda\}H$ is exactly the space of the eigenvectors of $ A$ corresponding to the eigenvalue $ \lambda$ (do you see why?) So, if $ A$ has no such eigenvectors, then $ E\{\lambda\}H\equal{}\{0\}$ and we obtain that $ \langle B_T x,y\rangle\to 0$ for all $ x,y\in H$ as $ T\to\infty$, which is exactly what the weak convergence means in this case.

This is the first part. See if you understand it and can continue from here :).

Thanks a lot! I will try to see if I can understand it.The part that E{(\lambda)}H is  exactly the space of the eigenvectors of A corresponding to the eigenvalue \lambda is is OK. Now I am trying to see how this contradicts the norm convergence of my guy:)

I will try it and I have problems I will ask you . I cannot thank you enough for your help!You are a nice person. sorry I was rude before... I am just trying to understand things...and I guess there are to much for a 19 year old boy:( But I will not give up!

Thank you very very much!
I will come back!

Hi,

I do not know to continue the problem. I think is like that. I assume that B_{T} goes strongly to zero, since it converges weakly to zero, it means that in case it converges strongly meant is must to converges to zero. But that means  its strong limit goes to zero, and this implies the norm limit goes to zero (the limit the I know from the statement that exists)... but where is the contradiction?

It is ok?

Please let me know how it works if you have time!
Thank you very much!

Sincerely,
mikiduta

It is a bit fancier than that. You decompose $ \phi\equal{}E\{\lambda\}\phi\plus{}E(\mathbb R\setminus\{\lambda\})\phi\equal{}y\plus{}z$. Now $ f_t(A)y\equal{}y$ for all $ t$. In exactly the same way as before, we can see that $ g_T(A)z$ weakly tends to $ 0$. Hence, arguing as you did, we conclude that $ f_t(A)z$ goes to $ 0$ in norm. But $ f_t(A)$ is a unitary operator! (this is what you, probably, didn't notice :)). Thus $ \|z\|\equal{}0$ and we are done.

Hi,

 You are right saying that code $ f_{t}(A)$ is unitary and this indeed solves the problem!!You are very smart !  I liked your idea with decomposing $ \phi$ like that.
There are many things I need to learn...Thank you !
Apropos, for the reverse problem, if I know that $ \phi$an eigenvector of $ A$then $ A\lambda\equal{}\phi \lambda$. I write $ exp{i(\lambda \minus{}A)t}$ like $ e^i(\lambda \minus{}A)t \equal{} \sum_{k\equal{}0}^\infty{1 \over k!}{i(\lambda \minus{}A)t}^k$.and I see that all the terms in the RHS vanish when I apply to$ \phi$, but i do not know if it works. I get just one single thing on the RHS.I guess this solves the problem...Are you agree?

Thank you very much for all your help!
Sincerely,
Mikiduta

Hi,

Can you explain more why that integral that converges weakly to zero implies that exp{it(A- \lambda)} converges in norm when t goes to infinity. I think you did some mistakes...or I thought I got it ....
Anyways if you fell you have time, if no, I will try my luck again:))

Thanks,
mikiduta

done i solve it:))
I was confused but you were write.

thanks


The problem of a paritcular country is 25 million, where P represents population in millions and t represents time in years. What will be the population 20 years from now

formula       P = 25e^0.022t

Im not sure how to do this one but i think...

LN P/25 = 0.022(20)    then what?

------

simplify       log[size=67]2[/size] 4^x

-------

if log[size=67]4[/size] x = a, determine log[size=67]16[/size] x in terms of a

I'm not sure about the first one, but here are the other two:

$log_2{4^x}=2x$

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

If $log_4{x}=a$, then $log_{16}{x}=\frac{a}{2}$

how do you do the first question with calculator??
I don't really get what you do to get that answer

for the second question, could you please show work?

For the first problem, if you are allowed to use calculator, then as cincodemayo5590 wrote, all you need to do is calculate $25e^{0.022\times 20}$.

For the second one, use $\log_b p^q=q\log_b p$ ;)

frt, do you know how to simplify further or get an approximation for $e^{0.44+ln(25)}$?

Not really... We can rewrite it as $25e^{.44}$, which is what you've already had.

(Maybe we can approximate $e^{.44}$ using $e^{.5}=\sqrt{e}=\sqrt{2.718...}$ (which is between $1.6$ and $1.7$) etc.) :roll:

[quote="frt"]For the first problem, if you are allowed to use calculator, then as cincodemayo5590 wrote, all you need to do is calculate $25e^{0.022\times 20}$.

quote]

this is what im confused about!!

do you ln 25 then multiply 0.44 or what?!!

P = 25e^0.44

i know you divide both sides then Ln and get   Ln P/25 = 0.44 how do i use the calculator from that point on?

this is whats confusing me

Aren't you just trying to find $P$?

What I'm saying is, you don't even have to worry about $\ln$... Just type "25e^(0.022*20)" in your calculator, and that's the population 20 years from now, which is your final answer...

[quote="cincodemayo5590"]For the first problem--

With calculator:
    $P = 25e^{0.022(20)} = \boxed{38.82}$ million people.

Without calc:

$P = 25e^{.022(20)}$ so $ln(P/25) = 0.44$,

$ln(P/25) = ln(P) - ln(25)$ so $ln(P) = 0.44 + ln(25)$.

So $P = \boxed{e^{0.44 + ln(25)}}$ million people.

I don't know how to simplify that expression further, though i'm sure that there is some way w/o a calculator. But i don' t think that you are supposed to do this without a calculator.[/quote]

it is just like our math class...plugging in numbers  :lol:

[quote="Altheman"]

it is just like our math class...plugging in numbers  :lol:[/quote]

haha yea, i kno  :lol:

For the last problem:
log(base 16)x =
(log(base 4)x) / (log(base4)16)
= a/2.


HMMT MIT college Harvard geometry search

There will be plenty of math going on during HMMT but many of you will be arriving the day before the contest and leaving the day after.  In other words, there will be time for you to do more than just math while you're here!

The competition should take place from 8 a.m. - 5 p.m.  So you'll have all the time before/after the competition to explore.  It'll also still be winter.

First I want to that it is possible to arrange for you to stay in an MIT or Harvard undergrad's dorm instead of getting a hotel.  The benefit of this is: (a) you get an idea of college life at either MIT or Harvard and (b) it's free!  E-mail hmmt-request@mit.edu if you are interested.

Basically, the primary purpose of this thread is if you are interested in doing anything additional then you should post here and I'll see if we can organize a group or something.  For example, if you wanted to eat at a particular type of place or have a game of frisbee.  I've never played frisbee in February in MA, but I'm willing to try it if there's enough interest!

joe can I crash in your dorm

or will you be BUSY with your male prefrosh who decided to stick around due to your incredible hospitality

Hm, if SC goes.  Last time I checked it was a "probably not unless we get funding". Which really means "probably not unless Shobhit and Sean's dads donate more money".  So it depends on whether Sho gets into Clemson or something and whether Sean...uh...?

Anyway, hosting that prefrosh was a really great decision!  Now I don't have a roommate.  The last one must not really have been keen on those types of things ;) ;)  Hopefully I'll have a single then still in which case the whole SC team could crash in as long as they don't mind sleeping on the floor...or sleeping in the same bed  :o

Btw, I was informed that frisbee might not be a great idea, darn.

[quote="Xevarion"][quote="joml88"]Btw, I was informed that frisbee might not be a great idea, darn.[/quote]
My uncle is a frisbee fanatic and he lives in Lexington and he says that frisbee in the winter is great -- you can lay out much more often because you don't have to worry about holding yourself when you hit the ground![/quote]

Yeah, that's what I was thinking.  I'm sure Samson will play...well...if he can stop playing warcraft long enough :(

OK, I'll try to throw out some ideas for dining.  Giving a comprehensive list of food places in Boston would probably overload the server.

On MIT's campus, there's the student center which has a few food places:

[b]At the student center (and open on the weekend):[/b]
$\bullet$ Anna's Taqueria - really good burritos; very fast; good prices (about $\$$ 3.50 for a burrito); an MIT favorite (I'd say)
$\bullet$ Cambridge Grill - "BYOB - that's 'Build Your Own Burger' at The Cambridge Grill. Enjoy hot-off-the-grill specailities, stone-oven pizzas, salads and more. Located on the first floor of the Stratton Student Center (W20)."
$\bullet$ La Verde's - mini grocery store that makes subs, soups, and other things; my favorite (as well as many others') is the "All Day Everyday" breakfast sandwich for less than $\$$ 3
$\bullet$ Dunkin Donuts - donuts, muffins, croissants, pastries, coffee, and breakfast sandwiches.  If you don't like this Dunkin Donuts there's at least one on every street in Boston (or at least it seems so).

[b]Around campus:[/b]
$\bullet$ [url=http://web.mit.edu/dining/locations/foodtrucks.html]Food trucks[/url] - cheap, quick, and usually pretty good.
$\bullet$ Baker House Dining - dining hall food; very overpriced unless you get an MIT student with the preferred dining program to buy for you (and pay him/her back)...for example, it's about $\$$7-8 for stir-fry in my dorm but I get it for about 3.50-4.00 since I have preferred dining; btw, Baker dining is the only one open on weekends which is why it's the only one I have listed

[b][url=http://www.yelp.com/list_details?userid=vec4p1BHgEQVpxoE-70G8Q&list_id=-Iz9Pjc_1Jr8k_0QTZbLqQ]Kendall Square[/url] (very close to campus)[/b]

$\bullet$ Legal Sea Foods - pricy, but it's Legal; very famous around here; its clam chowder has been served at every presidential inauguration in the past 25 years if that helps
$\bullet$ Kendall Food Court - there's a food court right next to the Marriot; Au bon Pain, Sbarro's, a chinese place, a japanese place, a salad bar that serves other random stuff.  If you don't like the Au bon Pain here then there's one [i]right across the street[/i].
$\bullet$ there are a few other things around Kendall Square that I can't think of off the top of my head.

[b]Down Mass Ave (towards Harvard)[/b]

There are restaurants every which way on Massachusetts Avenue.  If you walk far enough you'll get to Central Square.  You shouldn't have to go farther than this to find a place you like, but if you do you'll end up in Harvard Square which is about 1.4 miles from MIT.

[b][url=http://www.cambridgesidegalleria.com/]Cambridgeside Galleria Mall[/url] (and surrounding areas)[/b]

This is down past Kendall Square.  There's a [url=http://www.cambridgesidegalleria.com/go/Poolb.cfm?MallID=238&FPURLID=2129954062]free bus[/url] that runs from Kendall Square to the Cambridgeside Galleria until 7 p.m. every day.  So you'd probably be able to get there using the bus but would havce to find a way back on your own (walk...personally I don't think it's that bad of a walk).   There's tons of places to eat at around and in the mall.

[b][url=http://www.prudentialcenter.com/]The Prudential Center[/url][/b]

This is on the Boston side of the Charles River.  There are tons of options in the center and around it.

[b][url=http://www.northendboston.com/]The North End[/url][/b]

This is on the Boston side.  It's basically a little Italy.  The best Italian food in Boston is found here.  It's about 2 miles from MIT though.  So you'd probably want to take the T depending on how much you mind walking.

[b][url=http://boston.about.com/cs/government/a/chinatown4698.htm]Chinatown[/url][/b]

Never been here but there's supposedly good, cheap Chinese food.  You'll probably want to take the T.

[b][url=http://www.harvardsquare.com/directory.php?parent=RESTAURANTS&cat=hotels]Harvard Square[/url][/b]

Harvard Square is a very nice place to visit.  There are definitely some great places to eat at.  It's about 1.5 miles from MIT though.  You can take the T or a bus or just walk.

I should also briefly explain transportation in Boston and Cambridge.  The T (aka metro) is the easiest way to get around.  It's quick and efficient.  It costs $\$$ 1.25 for every time you get on.  So if you take it to the North End and then back that will cost you $\$$ 2.50.  Depending on where you want to go you'll still have to walk a little bit from the station.  It's most helpful when you cut down, say, a 2.5 mile walk to a .4 mile walk...however, the closest station to MIT is the MIT/Kendall station.  This isn't too far from people who live on the East Side of Campus but for those of us on the westside it's almost a mile walk!  Then again, there are shuttles that circle campus, so you can plan things out for minimal walking.

I just talked to Samson and he'd be absolutely delighted to play frisbee!  Nothing (especially not the cold!) will stop him.

So who else is in?

[quote="joml88"]Sho gets into Clemson or something
[/quote]

uh...u mean Brown University?...

So tell me, how is the MIT cafeteria!!!

 :lol:

We don't really have a cafeteria...well, not like the usual ones at other colleges I've been to for math competitions.  There's a bunch of stuff in the student center but it's more a food court rather than a cafeteria.  Some of the dorms have their own dining halls but those are only open for dinner Sunday-Thursday so those won't be open (with the exception of Baker Dining which is open every night).

[quote="PTynan89"]I'll play, if we end up sending a team.

My school is cheap, and idk if we have enough good people.

I [i]think[/i] we can pull together a halfway decent team, but funding/transportation is still a big issue.[/quote]

same here. my school is good enough to give a bus, but people in my school are not that motivated to get up early and go on a 4 hour bus ride. also, me and another kid, the only 2 ppl that want to go are also the only two qualifiers for usamo. we don't even have a mopper.

[quote="13375P34K43V312"]I live near this area, I got to MIT all the time, everything is great. The MIT Student Center food is ok, not great. But Legal's Sea Food is awesome. BUT GO TO THE PRUDENTIAL CENTER it's by far the best in the area.[/quote]

Which places in there do you like?  I've been to Q'doba which is awesome.  And I know there's a Legal's in there as well.

But, yeah, the Prudential Center is on the Boston side of the river.  There's all kinds of places to stop between MIT and the "Pru" (its nickname).  This area is called the Back Bay.  However, the Pru is about 1.5 miles from MIT.

Also go to Chinatown, but when you get to the random dark alleys and the parking lots run by white people, turn around.

A small update: subway and bus fares have risen in Boston; it's now 2 dollars to ride the subway and 1.50 to take a bus (unless you get a plastic Charlie Card, in which case it's 1.70 and 1.25, respectively, but you might have to order it in advance of your arrival, which hardly anyone is going to do).

[quote="Elemennop"]How about SMASH!!

SUPER SMASH BROTHERS

yea[/quote]

yes we should do smash...or brawl...i ono Wii s are hard to get...but yea...we should play smash

Er... so like it's only three nights but people are bringing their GameCubes? Haha...

Anyway, Joe I'll do ultimate if you aren't concerned about having a rookie participate. I pretty much suck at it... but that's cool. It's an awesome game.

I say bughouse. I've been dying to play some. Seriously. I got a set and a clock. Anyone?

BTW, what time are the rest of you getting there? I'm taking a train from Philly, so for me it's like six hours, if I leave in the morning I should be there after lunch which is plenty to time to explore.

Thanks Joe for all the information. I'll be sure to share it with my team.

[quote="white_horse_king88"]Er... so like it's only three nights but people are bringing their GameCubes? Haha...[/quote]

uh
DUDE

are you kidding

I bring my gamecube to tournaments where we dont even stay in a hotel...

Ha, I'm probably a bigger rookie than you at ultimate :D  Though, I kinda doubt I'll try starting a game anymore.  Unless Samson shows more enthusiasm about playing  :roll: but that would require him to purhcase something warmer than a long-sleeve t-shirt most likely.

There will definitely be enough people interested in bughouse around.  It should be easy to get that going.  Samson would be a good one to talk with about that.

Are you a grader Joe? I'm just wondering.

I'm coming to Boston in the morning on Friday, so I might drop by later in the day... I'm not sure how many people are coming that early. What else is there to do on campus besides eat and eat? Haha...

That store is awesome, and there are like 3 in Boston.

[quote="paladin8"][quote="white_horse_king88"]Are you a grader Joe? I'm just wondering.[/quote]

Perhaps a trader Joe :wink:.[/quote]

Perhaps a traitor Joe.  :wink:  :wink:  :wink:

Ofcourse... I'm just kidding. The pun was irresistable.

I'm not sure what I will be yet.  I'll help our wherever we need help.

There's a Trader Joe's not too far from MIT.  I walked there about a month ago and realized that there was a computer store across the parking lot from it.  In case you were wondering, I got blueberry juice from Trader's and it was quite good :D  There are a few others across the river in Boston as well that I know of.

Yeah if you don't want to eat on campus there's plenty of other places to go like trader joe's. Ja, jk ;)  So, um, let's see.  There's definitely plenty of things for pre-frosh to do off of campus.  While [i]on[/i] campus you probably won't have access to everything that students have access to.  But...the libraries are open and you can check them out.  I can lead you to the math section ;)  By the way, the search feature for the library's online database is horrible.

Anyway, you can look at the [url=http://events.mit.edu/index.html?date=2007/2/23]MIT Event Calendar[/url] to see if there's anything that interests you.  Not all things on that list are necessarily available to you so you'll have to make sure.  I know some are though such as the breakdancing.

If you like World of Warcraft, then talk to towersfreak2006.  He's always playing that anyway.  He also likes bughouse and just chess in general.  There should be tons of people at the contest that like chess though.  In my dorm there are TV's, a pool table, a ping pong table, and a (small) weight room.  There's an outdoor track, tennis courts, field hockey turf, and fields for baseball/softball/football/soccer/etc that you can use.  Unfortunately you have to have a membership to the Z center to use the basketball courts which are located indoors.  You can take a self-guided tour of campus or get someone to show you around.  You can tool :D  There's a game room in the student center, but I advise finding a better way to spend your time than wasting it on arcade games.  There are tons of things to do...you just have to know who to ask.  That's one of the perks of staying on campus.  You can ask the person who's hosting you what there is to do.  Their answer will depend on their interests and their dorm, mostly.

Sorry I couldn't give you a better answer!  I'll see if I can produce a better answer later...

[quote="white_horse_king88"]I say bughouse. I've been dying to play some. Seriously. I got a set and a clock. Anyone?[/quote]

I'll bring a set and a clock too ;)

Thanks Joe! And yeah - I'll definitely talk to Samson - or towersfreak rather.

I wonder how cold it's going to be. If it's too cold I can imagine myself staying indoors the entire time. Maybe. Ping-pong sounds sweet. I'm bringing a poker set.

Oh, yeah, poker's definitely an option also (as well as other card games).  I also have a really nice set which I've only used for once which was for this year's duck konundrum problem on the MIT Mystery Hunt...

So since HMMT is this weekend I'll start trying to make some concrete plans or give more ideas for things to do while you're here in Boston/Cambridge.

$\bullet$ You can watch a movie shown by [url=http://lsc.mit.edu/index.shtml]MIT LSC[/url]. This weekend they'll be showing [i]Les Enfants du Paradis[/i], [i]Flags of Our Fathers[/i], and [i]stranger than fiction[/i].  Tickets are $\$$ 3.  See the website for the exact times.  LSC shows movies in 26-100 which is a large lecture hall at MIT.  The audio and visual are quite good.

$\bullet$ MIT Events Calendar: [url=http://events.mit.edu/index.html?date=2007/2/23]Friday Feb 23[/url], [url=http://events.mit.edu/index.html?date=2007/2/24]Saturday Feb 24[/url]...probably not too much of interest listed there but you never know

$\bullet$ Campus tours are Monday-Friday at 10:45 am and 2:45 pm.  You can also just explore the campus on your own.  There are maps in lobby 7 for this.

$\bullet$ Boston!  There's plenty to do and see in Boston.  Or Cambridge for that matter.

$\bullet$ There are plenty of colleges in the area if you want to visit one.  Besides MIT, there's Harvard, Boston University, Boston College and Northeastern which are pretty close and pretty big schools.  You can see a more comprehensive list [url=http://www.bostonvisit.com/college/collegeprofiles/index.shtml]here[/url].

As far as getting to MIT with airport backups, I don't really know much about the status of that.  If you're concerned I suggest contacting the airline you booked with.  Today is really warm here in Boston as compared to the last week.  I see there is the chance of some snow in the next few days but I don't see any signs of another "snow->sleet->hail->ouch, i shouldn't be riding my bike in this->rain->my feet are going numb walking back in 2 inches of almost frozen water->water freezes over night leaving the whole city a giant skating rink" kind of storm.

Just a question, is it too late to get into a dorm room for 2 nights?

[quote="jelyman"]K8, how many do you need to put in a dorm?[/quote]

Just me.

My team is going individually, then we're meeting up there in a hotel.  I'm catching a ride there and back with a group from my school (who happens to be going there...turns out we'll get there thursday night, so I guess that makes it 3 nights), but my parents are suddenly worrying that I'll be staying alone there.  So I naturally considered the dorm option, because it seems a lot of HMMT people will be in dorms, and there'll be some sort of meet up.

I'm still asking the members of the team and other people though to confirm that this would be the best choice.

If absolutely necessary, I could house you at Harvard.  I'd even have a mattress for you ;)

To give you two more options, there should be room in my room.  There's already one person who I believe is staying over in my room but there's certainly room for at least one more if needed.  The other option is to stay at ZBT which is my frat which is an incredibly awesome, fun place to be.

Awww, we were supposed to split our team between Theta Chi and ZBT...anyway, what's the difference between one frat and another?

I also have a room to offer. PM me if someone wants my cell phone number.

I do play some Warcraft III and various custom maps, such as DotA, in my spare time, but never have I played World of Warcraft. Super Smash Brothers or Mario Kart works for me too. ^_^

I would love to play bughouse though. I also have a spare racket if anyone wants to badminton with me. Ultimate frisbee would be nice too, but the weather probably will not allow. Many dormitories and living groups have ping-pong tables.

Different fraternities are different living groups. They live separately, and you should find that each fraternity has its own personality. I believe both Theta Chi and Zeta Beta Tau are in Boston, across the river from MIT. Nonetheless, transportation should not be an issue at all. There's a shuttle for the evening, and I think your host should provide transportation Saturday.

I haven't been to that many fraternities, but I'd imagine that most of them would be a lot of fun to be at.  I know a few people who pledged Theta Chi this year and they're cool.  I'd think that all of the frats would be pretty cool.  I mentioned ZBT because I'm inherently biased.  But also, I wouldn't have joined it myself if I didn't think I'd like to hang out with all of the people there.

If the ice melted and there wasn't a sopping amount of water on the ground, I'd still be up for ultimate :D


quadratics
number theory unsolved
number theory

Prove that exist such $a$ for all $k$ that $4k^2-5|a^2-a-1$. I know that this problem can be done using Jacobi symbol, but I don't know anything about Jacobi symbol except definition. Do you know any links where I can read about it.

Well, nothing besides definition of quadratic/non quadratic residue is needed:

We want to have that for all $p | 4k^2-5$ we have $p|a^2-a-1$ or equivalently $\frac{1+\sqrt{5}}{2}$ and $\frac{1+\sqrt{5}}{2}$ $\in \mathbb{Z_p}$ or equivalently $\sqrt{5}$ is quadratic residue modulo $p$ which is true from definition of $p$:  $5 \equiv (2k)^2 \mod p$


The points $ A,B,C$ are on a circle $ O$. The tangent line at $ A$ and the secant $ BC$ intersect at $ P$, $ B$ lying between $ C$ and $ P$. If $ \overline{BC} \equal{} 20$ and $ \overline{PA} \equal{} 10\sqrt {3}$, then $ \overline{PB}$ equals:

$ \textbf{(A)}\ 5 \qquad\textbf{(B)}\ 10 \qquad\textbf{(C)}\ 10\sqrt {3} \qquad\textbf{(D)}\ 20 \qquad\textbf{(E)}\ 30$

[hide]Let $ PB \equal{} x$ for the sake of cleaner algebra. Using power of a point:

$ x(x\plus{}20) \equal{} (10\sqrt{3})^2$

$ x^2 \plus{} 20x \minus{} 300 \equal{} 0$
$ x \equal{} 10, \minus{}30$

We can throw away $ \minus{}30$, thus the answer is $ 10 \rightarrow \boxed{B}$[/hide]


limit
integration
trigonometry
inequalities
real analysis
real analysis unsolved

[color=red]Here is a problem:[/color]
Assume$ f(x)\in C([0 \plus{}\infty))$, and $ f(x)$ is monotony tends to $ 0$ as $ x\rightarrow \plus{}\infty$
show that:$ \lim_{n\rightarrow \plus{}\infty}\int_{0}^{\plus{}\infty}f(x)\sin nxdx\equal{}0$

I want to use Bonnet mean value theorem to get $ \int_{0}^{b}f(x)\sin nxdx\equal{}f(0)\int_{0}^{c}\sin nxdx\plus{}f(b)\int_{c}^{b}\sin nxdx$
then:$ \left|\int_{0}^{\plus{}\infty}f(x)\sin nxdx\right|\equal{}\left|\lim_{b\rightarrow \plus{}\infty}\int_{0}^{b}f(x)\sin nxdx\right|$
$ \leq \lim_{b\rightarrow \plus{}\infty}\left|f(0)\right|\left|\int_{0}^{c}\sin nxdx\right|\plus{}\lim_{b\rightarrow \plus{}\infty}\left|f(b)\right|\left|\int_{c}^{b}\sin nxdx\right|$
$ \equal{}\lim_{b\rightarrow \plus{}\infty}\left|f(0)\right|\left|\int_{0}^{c}\sin nxdx\right|$

to prove the problem only need to prove:

$ \lim_{n\rightarrow \plus{}\infty}\lim_{b\rightarrow \plus{}\infty}\left|f(0)\right|\left|\int_{0}^{c}\sin nxdx\right|\equal{}0$     $ (*)$

obviously,we have: $ \lim_{n\rightarrow \plus{}\infty}\left|\int_{0}^{M}\sin nxdx\right|\equal{}0$, for $ \forall M>0$

but how to prove: $ (*)$?  when $ b\rightarrow \plus{}\infty$, what about $ c$? thanks very much

Note that we do not know whether the limit $ \lim_{b \to +\infty}|f(0)| \left| \int_{0}^{c}\sin nx \, dx\right|$ exists or not, so we have to avoid using it.

But just a simple observation can be used to make amend your argument and complete the proof: For any $ n = 1, 2, 3, \cdots$ and $ a < b$, we have $ \left| \int_{a}^{b} \sin nx \, dx \right| \leq \frac {2}{n}.$

The proof is very simple. All we have to know is that $ | \cos x | \leq 1$ for all $ x$ and $ \int_{a}^{b} \sin nx \, dx = \left[ - \frac {\cos n x}{n} \right]_{a}^{b} = \frac {\cos na - \cos nb}{n}.$

Then we have

\begin{eqnarray*}\left| \int_{0}^{b} f(x) \sin nx \, dx \right| & \leq & |f(0)| \left| \int_{0}^{c} \sin nx \, dx \right| + |f(b)| \left| \int_{c}^{b} \sin nx \, dx \right| \\
& \leq & \frac {2}{n} (|f(0)| + |f(b)|). \end{eqnarray*}

Taking $ b\to\infty$ to both sides, we have

$ \left| \int_{0}^{\infty} f(x) \sin nx \, dx \right| \leq \frac {2|f(0)|}{n}$

and applying squeezing theorem to this inequality completes the proof.


calculus integration email Putnam factorial

I am probably going to lassiter. who else is going? does anyone have a copy of the past tests? I only have the 2006 Varsity tests. JV and Varsity are ok...thanks! For anyone who doesn't know, it's on December 2nd.

I'll probably be grading ciphering and/or proctoring written tests, as I go to Lassiter.  I obviously have access to old tests, but right now I'm unfortunately too busy working on new ones to type up old ones at the moment.

I look forward to having you guys there, though I'll probably be too busy to talk to anybody.

Or you might run into me like last time.

BTW: How many questions did you give Mrs. Poss?

Sorry you had to miss the Cobb County Winter Tournament. Your rank will go down from #2 to like #6.

Lassiter Varsity still pwned all other schools without you.

I got 3 hours of sleep and scored really low.

btw, i know that lassiter is on the 2nd of december, but what time to what time is it? does anyone have the full schedule?
*cough cough* Boy Soprano II

hmm yeah i just got a form today. I think that it starts at 9:15 or 9:45 or something like that. But, I am sure that it ends at 3:20.  :)

I think we're probably going. Depends if we can put together a team or not

[quote="Boy Soprano II"]Schedule?  All I know at the moment is that [i]I[/i] am supposed to be there rougly from 8 until 4, but you guys won't have to arrive that early or stay that late.  So I'm not sure of the schedule, but your teacher should probably be getting tournament details from Mrs. Poss or Mr. Slater some time before it happens.  I have a recollection that last year, it started at something like 9:00 or 9:30, but we delayed, because some schools were coming late.  (Don't do that if you can help it; it is bad for everyone!)  Also, awards were supposed to conclude before 2:30, but we got really bogged down scoring all the tests, and we were still working in the score room at 3:00.  I think, however, that awards were finished before then; it doesn't take as much time to find the top so many teams and whatnot as it does to put together score reports for each school...

I'm not sure how many questions I submitted, but I think I wrote both of the JV power questions.  (Phillip took the varsity power first.)[/quote]

Please don't tell me that the power questions were some sort of puzzle like last year where there was a circular magic circle :-(

i'll be swimming...do you guys have a schedule so i can plan ahead?  :maybe: i mean it is hard for me to get some of the places  :blush: so any north atlanta comps?(or atlanta  :maybe: )

[quote="Karth"]is there generally any calc on the lassiter test? for example, would limits and integrals be on there? Also, i know these things: there is a 90 minute individual written test and a power round. what else?[/quote]
Ciphering.

We are discouraged from submitting calculus problems for the tournament.

How exactly do we submit problems, anyway?

You don't.

Students at Lassiter give them to Mrs. Poss.

I think they already have a test, and they don't need any more problems. Atleast I hope so because the tournament is in 2 days.

[quote="drunner2007"]Miles, is it possible to sign up on the day of the test as an individual?
This is just in case one of our teachers can't take a team.[/quote]

Erm, if I recall last year, we had people registering the week of the contest.  But to tell the truth, I don't really know the answers to questions about scheduling, registration, or logistics in general, so I suggest you send an email to Mrs. Poss/Mr. Slater.  Now.  Because the contest is less than eighteen hours away. :wink:

And yes, we do have a test by now. :wink:   And we would certainly not accept questions from potential competitors.  (By the way, I had to revise the JV power question... :( )

My school came in and sat down just as they called him up.  I give him a...                :10:

[quote="drunner2007"]Who else thought Billy's acceptance of his award to be really funny (in a good way)?

Out of all the tournaments I've been to, that's the most interesting/funny acceptance I've seen.

Well, maybe I should take that back. I was at a local tournament one time, when they called this guy's name out for an award. After saying his name about 3 times, no one came up, so they were going to move on (thinking he had left). Then all of a sudden this dude just gets up, walks up there, and accepts the award. It turns out he wasn't the guy that was supposed to win the award, and he wasn't even from the same school! Can't believe he had the guts to do that lol.[/quote]

Haha  :rotfl:

Agreed. That was probably the funniest thing I've seen (Billy's enthusiasm unleashed) (in a good way). He was literally jumping. And Sam's imitation of him was even funnier.

yeah...i got tenth jv. billy's...er...happiness was kind of weird. oh yeah and if anyone was wondering who i am i was the one who tripped...jog ur memory, huh?

[quote="Xantos C. Guin"][quote="drunner2007"]Who else thought Billy's acceptance of his award to be really funny (in a good way)?[/quote]

My thoughts at that moment: "How many sugar packets had Billy consumed during lunch? I guess Mr. Fulton wasn't there to stop Billy from consuming sugar."

Billy's Answer to my thought: "None, but I did have some Sweet Tea."[/quote]
wait a minute sugar? why would anyone eat it out of those packets

[quote="solafidefarms"]It tastes good, esp. when the semi-Yankee restuarant doesn't know how to sweeten their tea properly.

Meh, well, I agree that in hindsight I was fairly amusing. The last 2 times I was at Lassiter though before Sat. I had hurt myself, so I was careful not to hurt myself a third time :). Can someone explain what Sam did exactly? I didn't manage to see it...[/quote]

Er.. you don't wanna know

[quote="cleansmyheart"]haha we told sam we'd stop calling him gay for a day if he made fun of your jumping award acceptance thing, billy.[/quote]

lol thats pretty messed up

[quote="Karth"][quote="cleansmyheart"]haha we told sam we'd stop calling him gay for a day if he made fun of your jumping award acceptance thing, billy.[/quote]

LOL

and he actually did it?

that's sad  :lol:[/quote]

yes that was pretty mean to billy and sam. why are we making so much fun of billy? personally i would be that happy if i won like that...in fact i was so happy i tripped and fell on the stairs...so dont make fun of them like that

well i in no way am trying to make fun of billy. its just one of those things where you laugh with a person because they do something silly.

and i have nothing against sam either. not trying to make fun of him.

Oh you know that was just being me. So what if ya'll laugh at me? Feel free to make fun of me :D....

@Amy: Get Sam to tell you about me at state science fair last year, when I was wearing a coat and tie...


Find two sets $E,F$ in $\mathbb{R}^n$ (better in $\mathbb{R}^1$) such that
(1)$E\cap F=\emptyset$
(2) $m^*(E\cup F) < m^*(E)+m^*(F)$, where $m^*$ denotes Lebesgue outer measure.

Check out the following paper for an example of non-measurable set:


[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=80556]Problem 1[/url]

Take 10 points inside the circle with diameter 5. Prove that for any these 10 points there exist two points whose distance is less than 2.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=146]Problem 2[/url]

Let $ a,b,c$ be positive numbers. Prove that $ \frac{(b+c-a)^{2}}{(b+c)^{2}+a^{2}}+\frac{(c+a-b)^{2}}{(c+a)^{2}+a^{2}}+\frac{(a+b-c)^{2}}{(a+b)^{2}+c^{2}}\geq \frac{3}{5}$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=80563]Problem 3[/url]

Call the Graph the set which composed of several vertices $ P_{1},\ \cdots P_{2}$ and several edges $ ($segments$ )$ connecting two points among these vertices. Now let $ G$ be a graph with 9 vertices and satisfies the following condition.

Condition: Even if we select any five points from the vertices in $ G,$ there exist more than two edges whose endpoints are included in the set of 5 points.

What is the minimum possible numbers of edges satisfying the condition?

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=80564]Problem 4[/url]

Let $ A,B,C,D$ be points in space which are not on a same plane and no any 3 points are not colinear.
Suppose that the sum of the segments $ AX+BX+CX+DX$ is minimized at $ X=X_{0}$ which is different from $ A,B,C,D.$ Prove that $ \angle{AX_{0}B}=\angle{CX_{0}D}$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=145436]Problem 5[/url]

The letters $ A$ or $ B$ are assigned on the points divided equally into $ 2^{n}\ (n=1,\ 2,\cdots)$ parts of a circumference.If you choose $ n$ letters from any succesively arranging points directed clockwise, prove that there exists the way of assignning for which the line of letters are mutually distinct.

To Mods: Please move this topic to National Olympiad of Olympiad Section, I think so.

To mods: please also replace the problem 2 link by the link http://www.mathlinks.ro/Forum/viewtopic.php?t=146 .

  darij

[i]Done. Orlando.[/i]


probability
floor function
number theory
modular arithmetic
combinatorics unsolved
combinatorics

Two numbers X and Y chosen randomly with replacement from the set $ {1,2,...,n} (n > \equal{} 4)$.
 whats the probability p that $ (X^2 \minus{} Y^2) mod 3 \equal{} 0$ ?

Thanks.  :maybe:

this is just the probability that both are divisible by 3 or both aren't divisible by 3. the number of multiples of 3 in $ \{1, ..., n\}$ is $ r \equal{} \lfloor \frac{n}{3} \rfloor$. the probability in question is $ \frac{r^2\plus{}(n\minus{}r)^2}{n^2}$.

1: Why are we dividing by $ n^2$ ?

2: Why do we add the case where the numbers arent divisble by 3 ?

Probability means number of good ways divided by number of total ways.  The number of total ways is $ n^2$.  The good ways fall into two groups: (1) if $ x$ and $ y$ are both divisible by 3 and (2) if neither $ x$ nor $ y$ is divisible by 3.  (You have to do a little bit of modular arithmetic to see why exactly these are the only cases.)


Suppose you want to prove a statement, $P(n)$ is true for all positive integers.

Induction means proving two things:
1) Showing $P(1)$ [base case]
2) Showing $P(k)$ implies $P(k+1)$ [inductive hypothesis]

After we proved 1) and 2), we notice that $P(2)$ is true, since $P(1)$ implies $P(2)$, according to 2). Then $P(3)$ is true, since $P(2)$ implies $P(3)$, again according to 2). We can do this over and over again to show that $P(n)$ is true for all positive integral values of $n$.
[quote="rnwang2"]
Are you asking how to prove

$1+(k-1)\cdot 2^{k}+k\cdot 2^{k}= 1+k\cdot 2^{k}$?

If so, it is very simple:

$1+(k-1+1)\cdot 2^{k}=1+k\cdot 2^{k}$.

$1+k\cdot 2^{k}= 1+k\cdot 2^{k}$, which is true.[/quote]

You proved $1+(k-1)\cdot 2^{k}+2^{k}=1+k\cdot 2^{k}$, not $1+(k-1)\cdot 2^{k}+k\cdot 2^{k}= 1+k\cdot 2^{k}$. I think the original statement is false, though... :maybe:

[hide="Using induction on the problem"]$P(n)$ is the statement $ \sum_{i=1}^{n}i \cdot 2^{i-1}= 1+(n-1) \cdot 2^{n}$

In your case, we do the same two steps:
1) $P(1)$ is obviously true.

2) Suppose $ \sum_{i=1}^{k}i \cdot 2^{i-1}= 1+(k-1) \cdot 2^{k}$. You want to show that: $ \sum_{i=1}^{k+1}i \cdot 2^{i-1}= 1+k \cdot 2^{k+1}$.

Add $(k+1) \cdot 2^{k}$ to both sides of what we assumed:
$ \sum_{i=1}^{k}i \cdot 2^{i-1}+(k+1) \cdot 2^{k}= 1+(k-1) \cdot 2^{k}+(k+1) \cdot 2^{k}$

$ \sum_{i=1}^{k+1}i \cdot 2^{i-1}= 1+(k-1) \cdot 2^{k}+(k+1) \cdot 2^{k}$

$ \sum_{i=1}^{k+1}i \cdot 2^{i-1}= 1+((k-1)+(k+1)) \cdot 2^{k}$

$ \sum_{i=1}^{k+1}i \cdot 2^{i-1}= 1+2k \cdot 2^{k}$

$ \sum_{i=1}^{k+1}i \cdot 2^{i-1}= 1+k \cdot 2^{k+1}$

The inductive proof is complete.  :) [/hide]


limit
real analysis
real analysis solved

${a_{n+1}-1=(a_n-1)a_n\ (n\geqq 2)\Longleftrightarrow \frac{1}{a_{n+1}-1}=\frac{1}{a_{n-1}}-\frac{1}{a_n}}\ (n\geqq 2)$

$\Longleftrightarrow \frac{1}{a_n} =\frac{1}{a_n-1}-\frac{1}{a_{n+1}-1}\ (n\geqq 2)$

Therefore $\sum_{k=1}^n \frac{1}{a_k}=\frac{1}{a_1}+\sum_{k=2}^n \left(\frac{1}{a_{k}-1}-\frac{1}{a_{k+1}-1}\right)$

$=1-\frac{1}{a_{n+1}-1}<1$

maybe the answer is 1. I can't find the reason now.

[quote="kunny"]

$=1-\frac{1}{a_{n+1}-1}<1$

maybe the answer is 1. I can't find the reason now.[/quote]

Is it not pretty clear that $a_n\to\infty$, so $\frac 1{a_{n+1}-1}\to 0$? :? I do, however, think you missed a $1$, and the limit is $2$. Think about it: it surely is larger than $1$, since it's larger than $\frac 1{a_1}=1$.

Hmmm.I think Grobber is correct.My answer came out to be 2.
I am not very sure of my calculation... :(

[quote="grobber"]Is it not pretty clear that $a_n\to\infty$, so $\frac 1{a_{n+1}-1}\to 0$? :? [/quote]

WHAT???? It is clear as water. I mean, I really think it is quite easy to see it. See that by the definition, all the $a_n$ are positive integers, if you want, prove it by induction. After that, note that $a_n\leq (a_1\cdots a_{n-1})a_n < (a_1\cdots a_{n-1})a_n+1=a_{n+1}$, therefore the sequence is strictly increasing... in the integers...

[quote="grobber"]I do, however, think you missed a $1$, and the limit is $2$. Think about it: it surely is larger than $1$, since it's larger than $\frac 1{a_1}=1$.[/quote]

And you are 100% right in that. The mistake is quite easy. Note that if you have a telescopic sum, what you do is the following $\sum_{k=j}^n(u_{k}-u_{k+1})=u_j-u_{n+1}$, therefore

[quote="kunny"]$\sum_{k=1}^n \frac{1}{a_k}=\frac{1}{a_1}+\sum_{k=2}^n \left(\frac{1}{a_{k}-1}-\frac{1}{a_{k+1}-1}\right)$

$=1-\frac{1}{a_{n+1}-1}<1$
[/quote]

The correct sum is

$\sum_{k=1}^n \frac{1}{a_k}=\frac{1}{a_1}+\sum_{k=2}^n \left(\frac{1}{a_{k}-1}-\frac{1}{a_{k+1}-1}\right)$

$=\frac{1}{a_1}+\frac{1}{a_2-1}-\frac{1}{a_{n+1}-1}\rightarrow2$

because $a_{n+1}-1\rightarrow\infty$.

Best,


trigonometry
calculus
ratio
integration
logarithms
geometric series
calculus computations

We are given this series
\[ \sum_{n \geq 0} a^n \tan \frac {\alpha}{2^n}, \ a > 0, \ \alpha \in \left( 0, \pi \right) .
\]
In case of convergence, we are asked to estimate the sum with an error $ \epsilon$.

How to do this cleanly? The ideas involved are pretty straightforward, but somehow I feel I can't come up with a completely formal proof.

Also, the same thing for
\[ \sum_{n \geq 0} \frac {n! a^n}{n^n}, \ a > 0 .
\]
The thing in this case is that the approximation varies greatly with $ a$ (especially near the "special" point). Is there anything to do?

PS: I hope I won't bother you anymore with such questions. Not really my kind.

Tangent is so easy to estimate from below (its Maclaurin series has nonnegative coefficients, which is an interesting calculus exercise in itself), but from above... It's concave up on $ (0,\pi/2)$: indeed, $ (\tan x)'' \equal{} 2\sec^2 x\tan x > 0$. So, for $ n\ge n_0$ you have the estimate $ \tan\frac {\alpha}{2^{n}}\le 2^{n_0 \minus{} n}\tan\frac {\alpha}{2^{n_0}}$. Thus, the remainder is estimated from above using the geometric series formula.

The remainder of the second series can be estimated in the same way. Find $ n_0$ such that $ \frac{a_{n\plus{}1}}{a_n}\le r<1$ for $ n\ge n_0$. Then $ \sum_{n\equal{}n_0}^{\infty} a_n\le\frac{a_{n_0}}{1\minus{}r}$.

For the second one...
[hide] the ratio of the $ n\plus{}1$st term to the $ n$th is
\[ \frac{(n\plus{}1)!n^n}{n!(n\plus{}1)^{n\plus{}1}}\equal{}\left( \frac{n}{n\plus{}1}\right)^n \equal{}\frac{1}{\left( 1\plus{}\frac{1}{n}\right)^n}\]

However, $ \left( 1\plus{}\frac{1}{n}\right)^n$ can be shown to be increasing, so if $ a_k\equal{}\frac{k!}{k^k}$, then for $ N\geq n$,
\[ a_N\leq \frac{a_n}{\left( \left( 1\plus{}\frac{1}{n}\right)^n\right)^{N\minus{}n}}\]
[/hide]

in first $ a < 2$ and in seccond $ a < e$
IN FIRTS problem : what's about to find exact answer of this sum ? :)
I find that it's equal to :
I'm not sure but I promise that later I will explain my answer:
answer is :
$ (\frac {\sin \alpha}{\alpha})^{\frac {1}{1 \plus{} 2a}}(\frac {\frac{\alpha}{\tan \alpha} \minus{} 1}{(1 \plus{} 2a)\alpha})$

I have made some mistake in my calculatuion sorry ..
Here is what I do:
Let's consider serie :
$ \sum_{n \equal{} 0}a^{n}\tan(\frac { x}{2^{n}})$ when $ x \in (0,\pi)$ or let;'s choose some good interval where we can itegrate our serie  :) .
let's integrate this :
$ \int \sum_{n \equal{} 0}a^{n}\tan (\frac { x}{2^{n}}) dx \equal{} \sum_{n \equal{} 0}a^{n}\int\tan( \frac { x}{2^{n}} )dx \equal{} \minus{} \sum_{n \equal{} 0}(2a)^{n}\ln \cos (\frac {x}{2^{n}})$ let's consider case when $ a \equal{} \frac {1}{2}$
then we have :
$ \int \sum ..your serie.. \equal{} \minus{} \sum_{n \equal{} 0}\ln \cos (\frac {x}{2^{n}}) \equal{} \minus{} \ln \prod\cos (\frac {x}{2^{n}}) \equal{} \minus{} \ln \frac {\sin x}{x}$
I note that I use formula $ \prod \cos (x/2^{n}) \equal{} (\sin x) /x$
and in finally we have :
$ \sum .. your serie ... \equal{} \minus{} (\ln \frac {\sin x}{x})'|_{x \equal{} \alpha}$ and we exact value of your serie when $ a \equal{} \frac {1}{2}$
case when $ a < \frac {1}{2}$
then by same way prove that :
$ \int \sum ..your serie.. \equal{} \minus{} \sum_{n \equal{} 0}(2a)^{n}\ln \cos (\frac {x}{2^{n}}) \equal{} \minus{} \ln \prod (\cos (\frac {x}{2^{n}}))^{(2a)^{n}} \equal{} \minus{} \ln (\frac {\sin x}{x})^{\frac {1}{1 \plus{} 2a}}$ (but I'm not sure in my last step )
and in finally we get :

$ \sum ..your serie .. \equal{} \minus{} (\ln (\frac {\sin x}{x})^{\frac {1}{1 \plus{} 2a}})'|_{x \equal{} \alpha}$

[quote="milin"]You're transforming $ \prod_{n \equal{} 0}^{\infty}cos\frac {x}{2^n}$ into $ \frac {sin(x)}{x}$ but it's actually $ \frac {sin(2x)}{2x}$.

$ \prod_{n \equal{} 1}^{\infty}cos\frac {x}{2^n} \equal{} \frac {sin(x)}{x}$.[/quote]
Sure. thanks.


A bike is traveling at $ 6\pi$ meters per second.  Consider a point on the outside rim of the wheel of the bike.  Determine the velocity of the point relative to the axle of the wheel at the 3:00, 6:00, 9:00, and 12:00 positions on the wheel.  Also, determine the velocity of the point relative to the ground at the 3:00, 6:00, 9:00, and 12:00 positions on the wheel.

[b]Relative to the axle of the wheel: [/b]
Magnitude of each of the four is $ 6\pi$ meters per second,in clockwise tangential direction.

[b]Relative to the ground[/b]
Assume the radius of wheel is $ R$. For each of the four points, $ v\equal{}r\cdot\omega$, where $ r$ is the distance from that point to the point that touches the ground, and $ \omega$ can be obtained from $ \omega\equal{}\frac{v_{bike}}{R}\equal{}\frac{6\pi}{R}$
We have $ r_{3: 00}\equal{}\sqrt{2}R, r_{6: 00}\equal{}0, r_{9: 00}\equal{}\sqrt{2}R, r_{12: 00}\equal{}2R$, so
$ v_{3: 00}\equal{}6\sqrt{2}\pi,v_{6: 00}\equal{}0,v_{9: 00}\equal{}6\sqrt{2}\pi,v_{12: 00}\equal{}12\pi$, direction is perpendicular to each $ r$.


No, not really. The Poincare conjecture is a conjecture about the classification of objects called 3-manifolds, which are 3-d equivalents of 2-d surfaces.

See [url]http://en.wikipedia.org/wiki/Poincare_conjecture[/url] for more details.

Poincare was so prolific that his name has become attached to results in many different areas of mathematics. Euler and Cauchy have the same "problem".


[quote="samus"]How difficult are the USAMTS problems as per the AMC/AIME/USAMO program?[/quote]

Two threads in this forum have discussed that topic:
A Februrary 2004 discussion and poll at
AoPS - MathLinks Forum Index -> US Contests & Programs -> Other US Contests & Programs -> which one is harder?
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=10586[/url]

and a June 2004 discussion at
AoPS - MathLinks Forum Index -> US Contests & Programs -> USA Mathematical Talent Search -> How does the USAMTS compare?
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=13105[/url]

The consensus is that the USAMO is more difficult.

My suggestion, as someone involved in preparing the USAMTS, is that you can look at the USAMTS questions before deciding whether to participate. Judge for yourself whether the USAMTS problems are interesting enough to you to spend a weekend solving them before you spend a weekend solving them.

Erin Schram


geometry
geometric transformation
MIT
college

I'm taking chemistry in the class year 05-06 (first chemistry  :D ) and I'm hoping to study some of it during the summer. But I have NO clue on which book to look or choose.

Anyone have a good idea? I'll appreciate it.

If you're looking at a pre-AP level class, I can't say that I'm sure; for AP chemistry, I usually recommend [u]General Chemistry[/u] by Linus Pauling.  My copy's loaned out right now, so I can't check to see if it would be appropriate for someone with no chemistry background (although it certainly might be).  My school's AP class uses [u]Chemistry[/u] by Zumdahl.  It's appropriate for someone just beginning chemistry, as well as at AP level, but it's probably more expensive than Pauling, as it's a real hardcover textbook.

Hope that helps.

[quote="inteluser"]...[u]General Chemistry[/u] by Linus Pauling...[/quote]
i have this one!!! its good just a bit outdated on some topics, but lots of properties about elements not found elsewhere!

[quote="amirhtlusa"]General Chemsitry by George Mortimer. Vol 1&2[/quote]
is this rare book? i tried amazon.com and bn.com but they both showed no George Mortimer, instead it showed Robert Mortimer about physical chemistry...

akh...
sorry for the wrong name, its called, General chemsitry by Charles Mortimer Vol1&2
this is my Favorite chem book for Olympiad Prep. :)

[quote="Silverfalcon"]FIRST YEAR CHEMISTRY! ;)[/quote]
Don't worry Silverfalcon, it's not as hard as physics, I just finished first year chemistry in my Burma school and when i moved here with my cousin, im attending a general chem class in his university!!!

The two books I named should still be approachable for someone with no chem experience.  But, if you finish them... first-year chem will not present any kind of challenge.

I've already taken 1st year chem, and al taking AP next year. What book would you recommend to prepare for Olympiad?

AP class uses Zumdahls book--which i don't really like...
but as i said before, charles Mortimers book is a very good book...
also consider reading the three volumes of Merrill Chem... they are very simple, but it is worthed to read them, it's by:
Roberc C. smoot/Richard G.smith/Jack Price

I can't find the Mortimer book on amazon...

it's actually called Chemistry by Charles E.Mortimer.
it's a rather old book, i have a translation of it that was published in 1995.

BTW, if you cant find Mortimers book,
GET THIS ONE, it's an EXCELLENT guide to solve chem problems:

[u]How to Solve General Chemistry Problems[/u], by [b]R.S.Boikess& C.H.Sorum[/b]

[quote="Silverfalcon"]FIRST YEAR CHEMISTRY! ;)[/quote]
Read your text book carefully. That helps.

we used holt modern chemistry i think.... i thought it was pretty good and ap used zumdahl. i believe even if you start reading zumdahl with no chem bg except what random tidbits you might have learned in school, you should do really well.

What's wrong with Zumdahl's? Can someone tell me, because I plan to use that to study for the olympiad  this year

nothings wrong with it...
i just dont know if thats a good olympiad book...(i don't think it is...)

my first year chemistry in Burma (Myanmar) used Modern Chemistry by Metcat? i think it was Metcat (funny name  :P ). but my honors chemistry class used older version of Metcat and the other lower section used a newer version. when i asked teacher why we use older version, she replied: "you guys have a greater imagination and therefore challenge speaks out over newer version". i was  :mad: . our book was old textbook with no colors and crappy orange highlights but newer version had nice pictures and colors. i later found out that school didn't have enough money to purchase enough books and we were unlucky ones :(
[quote="pkothari13"]yeah but silverfalcon never mentioned any intentions of an olympiad book.. and if he wants to even consider studying for olympiad, he should first learn the basics, and i think zumdahl does a pretty good job at teaching you that much.[/quote]
hehe Silverfalcon is going to have a great summer :D
if the high school that im going to attend after this summer uses Zumdahl in AP, that would save me some bucks to get another supplemental one :)
so in the meantime, hey Silverfalcon get Linus Pauling's book, it's a classic. skip those parts that are hard but this text has a lengthy introduction and i found it easy reading. well i guess it depends on who is reading it thou, sorry  :blush:  :whistling:

I hear the Brown book is really good. I got it and Im spending one week of summer just studying it.  :)

The best chemistry book by far is Chemistry by Raymond Chang.

I'm actually a fan of Chemical Principles, by Atkins and Jones--if you can do everything in there, you will dominate the olympiad.  Also, it's the book used for gen chem at MIT, for what it's worth.

This is probably inappropriate for personal study (unless you are a graduate student or just an advanced student of Organic Chemistry) but it is an interesting book just for general reference of synthesizing huge molecules:

Classics in Total Synthesis- KC Nicolaou,EJ Sorensen

It describes synthesis of things like Vitamin $ B_{12}$, Progesterones and other large molecules.

[hide="Hide"]

Hopefully I will be a Chemistry Graduate one day.[/hide]

How good is Adkins for an intro, sorry for the necro.


abstract algebra
Ring Theory
superior algebra
superior algebra solved

Is $\left(\frac{\mathbb{Q}[x]}{\left(x^3\right)},\ +,\ \cdot\right)$ an Artinian ring?

A ring is said to be Artinian if any decreasing chain of ideals $I_0 \supseteq I_1 \supseteq I_2 \supseteq \cdots$ for this ring becomes "constant", that is, if $I_n = I_{n + 1} = I_{n + 2} = \cdots$ for some positive integer $n$.

Your ring is a local ring and the third power of its maximal ideal is 0. This is a proof :)

[edit] There is a theorem that goes like this:
Let $A$ be a local Noetherian ring. Then $A$ is artinian if and only if a power of its maximal ideal is 0.

Another theorem: Let $A$ be a ring. $A$ is artinian iff $A$ is noetherian and every prime ideal is maximal.

Here is a more elementary proof - at last.
One can easily show (although I needed quite some time to see this) that the invertible elements of the ring are precisely those having a nonzero constant term. Hence, the only maximal ideal of the ring is $(x)$, and $R$ has only 4 ideals: $0$, $(x)$, $\left(x^2\right)$ and the ring itself. Hence it is Artinian and Noetherian.


Divisor Functions
pen
imo shortlist 2001

Find all positive integers $n$ such that ${d(n)}^{3} =4n$.

Peter, your link does not work  :roll:  However i try with the problem..

Let $ n \equal{} \prod_i{p_i^{\alpha_i}}$ be the canonical factorization of $ n \in \mathbb{N}$, and $ p_n$ is the $ n$-th prime and $ \alpha_i \equal{} \upsilon_{p_n}(n) \ge 0$.
The equation becomes $ \prod_i{(\alpha_i \plus{} 1)^3} \equal{} 4n \implies \alpha_1 \ge 1$. Now there exist some integers $ \beta_1,\beta_2,...$ s.t. $ \alpha_1 \equal{} 3\beta_1 \minus{} 2, \beta_1 > 0, \alpha_i \equal{} 3\beta_i \forall i > 1$. So we obtain $ (3\beta_1 \minus{} 1)\prod_{i > 1}{(3\beta_i \plus{} 1)} \equal{} \prod_i{p_i^{\beta_i}} \implies \beta_2 \equal{} 0$. Since $ p^x \equal{} p(p^{x \minus{} 1}) > pe^{x \minus{} 1} \ge p(x \minus{} 1) \plus{} p \equal{} px > 3x \plus{} 1, \forall p \ge 5, x \ge 1$ we need to have $ 2^{\beta_1} \le 3\beta \minus{} 1 \implies 1 \le \beta_1 \le 3$.
If $ \beta_1 \equal{} 2 \pm 1 \implies \prod_{i > 2}{(3\beta_i \plus{} 1)} \le \prod_{i > 2}{p_i^{\beta_i}}$, with equality iff $ \beta_i \equal{} 0, \forall i > 1$.
If $ \beta_1 \equal{} 2 \implies \beta_3 > 0 \implies (\frac {3\beta_3 \plus{} 1}{4 \cdot 5^{\beta_3 \minus{} 1}})\prod_{i \ge 4}{(3\beta_i \plus{} 1)} \le \prod_{i \ge 4}{p_i^{\beta_i}}$, with equality iff $ \beta_3 \equal{} 1, \beta_i \equal{} 0, \forall i > 3$.
So we can conclude that $ n \in \{2,2^7,2^4 \cdot 5^3\}$.

Cheers

After some playing around I think we can show that n can have no more than 2 distinct prime divisors, and this reduces it to studying all numbers of the form $n=2^xp^y$ where $ p $ is a prime not equal to $2$.


calculus
integration
real analysis
function
derivative
real analysis theorems

I was preparing my course in Analysis based on W.R.Wade's book "An introduction to analysis" and everything was fine until I came to the change of variable theorem (Theorem 5.34 in section 5.3), whose formulation I didn't like. It seems that the statement should be that if $\varphi:[a,b]\to \mathbb R$ is continuously differentiable and $f$ is Riemann integrable on $\varphi([a,b])$, then $f(\varphi(x))\varphi'(x)$ is Riemann integrable on $[a,b]$ and $\int_{\varphi(a)}^{\varphi(b)}f(t)\,dt=\int_a^b f(\varphi(x))\varphi'(x)\,dx$. Surprisingly, I spent about one hour and couldn't find a decent proof of this statement (decent=not using Lebesgue measure and integral). Any ideas?

If we have a monotonic increasing function $\alpha$ with $\alpha'$ Riemann integrable on $[a,b]$ (this certainly meets the continuously differentiable requirement), and $f$ is bounded on $[a,b]$, then $f$ is integrable w.r.t $\alpha$ iff $f\alpha'$ is Riemann integrable, and we have

$\int^b_a fd\alpha=\int^b_a f(x)\alpha'(x)dx$

The proof is a pretty straightforward application of the mean value theorem.

Next, if we have a strictly increasing continuous function $\varphi$ which maps $[A,B]$ onto $[a,b]$, and a monotonically increasing function $\alpha$ on $[a,b]$ and a function $f$ which is integrable w.r.t $\alpha$ on $[a,b]$, then define $\beta$ and $g$ on $[a,b]$ as $\beta(y)=\alpha(\varphi(y))$ and $g(y)=f(\varphi(y))$.  Then $g$ is integrable w.r.t $\beta$, and

$\int^B_A gd\beta=\int^b_a fd\alpha$

The proof is automatic - just choose appropriate partitions of $[a,b]$ and $[A,B]$ respectively.

Now, set $\alpha(x)=x$, so that $\beta(x)=\varphi(x)$.  If $\varphi$ is continuously differentiable, or even just has a Riemann integrable derivative on $[a,b]$, then the first theorem shows that

$\int^b_a f(x)dx=\int^{\varphi(b)}_{\varphi(a)} f(\varphi(t))\varphi'(t)dt$

Yes, but the whole problem is what to do if $\varphi$ is not monotone...


a nice, portable puzzle game that I'm currently playing. Wouldn't it be fun if every math problem were presented this way? :oops:

I've played the first one, and it was really good.  I am going to borrow the next one when my friend finishes it.


A 75 kg person escapes from a burning building by jumping from a window 25 m above a catching net.  Assuming that air resistance exerts a 95 N force on the person during the fall, determine the person's velocity just before hitting the ground.

First I found the force of the man falling (F=mg) by multiplying 75 kg by 9.81 m/s^2 and got 735.75 N.  Then I subtracted 95 N from 735.75 N and got the overall force to be 640.75 N.  Next, I found acceleration by using a=F/m: 640.75N/75kg = 8.543333 m/s^2.

Then I found time by using the formula t = sqrt(2y/g), and got 2.257618205 seconds.  Then I used the formula V(final) = at, and found that the final velocity equals 19.28758486 m/s.  Is this the correct answer?

The answer in the back of my book says 20 m/s.  Did I mess up somehow?

I have no idea about the correct answer, but I can see where the apparent discrepancy comes from.

Somewhere along the line, rounding probably played a part. Your answer and the book's answer are close enough that somewhere someone rounded. Also, the number of significant digits in your answer is supposed to be equal to the number of significant digits in the data you have, so that's what the book probably did.

Actually, there is one thing you did wrong - in formula $t=\sqrt{2y/g}$, number $g$ should be the actual acceleration and not the gravitational acceleration so it should look like this: $t=\sqrt{2y/a}$. You'll still get 20.67 though...

You don't even need to use time:

The net force on the person is $640 N$ downward, so his downward acceleration is $\frac{128}{15}$, then use the equation ${v_f^2=v_i^2+2ad \Rightarrow v_f = \sqrt{1280/3} = \frac{16}{3}\sqrt{15}} \approx 20.6559 \ m/s$ (downward).


search
number theory proposed
number theory

Determine all triples (m, n, p) of positive rational numbers such that the
numbers
m + 1/np , n + 1/mp , p + 1/nm
are integers. (Romania)

Try a search before posting. See here: [url]http://www.mathlinks.ro/viewtopic.php?p=498992#498992[/url]


Μιας και πλησιάζει ο Θαλής βάζω μια ανισοτητούλα που θα 'ταν ένα ωραίο τέταρτο θέμα (ίσως τρίτο :P ) για τον διαγωνισμό.

Αν $ a,b,c,d$ θετικοί αριθμοί με $ \frac{1}{1\plus{}a^4}\plus{}\frac{1}{1\plus{}b^4}\plus{}\frac{1}{1\plus{}c^4}\plus{}\frac{1}{1\plus{}d^4}\equal{}1$ να δείξετε ότι $ abcd\geq 3$

Συμφωνα με την ανισοτητα αντρεεσκου εχουμε οτι $ \frac{1}{1\plus{}a^4}\plus{}\frac{1}{1\plus{}b^4}\plus{} \frac{1}{1\plus{}c^4}\plus{}\frac{1}{1\plus{}d^4}\ge\frac
{16}{4\plus{}a^4\plus{}b^4\plus{}c^4\plus{}d^4}$. Αρα $ 1\ge\frac{16}{4\plus{}a^4\plus{}b^4\plus{}c^4\plus{}d^4}$.Αρκει λοιπον $ a^4\plus{}b^4\plus{}c^4\plus{}d^4\ge12$.Ξανα απο την ανισωτητα αντρεεσκου εχουμε οτι$ 4abcd\ge12$ αρα $ abcd\ge3$.

Νομίζω κάτι δεν πάει καλά στη λύση... Πράγματι $ a^4 \plus{} b^4 \plus{} c^4 \plus{} d^4\geq 12$ αλλά από αυτό αυτό πώς προκύπτει το ζητούμενο (βασικά δεν προκύπτει :wink: ) ?

Συγγνωμη αλλα δεν ειμαι σωστος.Στη δευτερη φορα εννοουσα AM-GM  Αλλα και παλι ειναι λαθος.

just a hint!
[hide]a^2=tanx etc   :wink:  δοκιμαστε την με αυτον τον μετασχηματισμο![/hide]

Εστω$ \frac{1}{1\plus{}a^4}\equal{}x$,$ \frac{1}{1\plus{}b^4}\equal{}y$,$ \frac{1}{1\plus{}c^4}\equal{}w$,$ \frac{1}{1\plus{}d^4}\equal{}z$.Iσχυει $ x\plus{}y\plus{}w\plus{}z\equal{}1$ και $ a\equal{}\sqrt[4]{\frac{1\minus{}x}{x}}$.Ομοιως για τα b,c,d.Αρκει $ (1\minus{}x)(1\minus{}y)(1\minus{}w)(1\minus{}z)\ge81xywz$.Δηλαδη αρκει $ (y\plus{}w\plus{}z)(x\plus{}w\plus{}z)(x\plus{}y\plus{}z)(x\plus{}y\plus{}w)\ge81xywz$.Η σχεση φανερα ισχυει με χρηση AM-GM για καθε παρενθεση.Ελπιζω αυτη τη φορα να ειμαι σωστος.

Σάκη αυτή ακριβώς ήταν και η λύση μου.

Η ιδέα του Ηλία είναι επίσης πολύ σωστή και τέτοιου είδους αντικαταστάσεις είναι πολύ συχνά πολύ χρήσιμες :wink:


Buenos dias.

esto es para informarles ... que durante esta semana..
se esta realizando el proceso de admision de la Universidad Nacional de Ingenieria en PERU..
[url=http://WWW.UNI.EDU.PE]WEB[/url]

como ya muchos sabran es un evento (3 examenes) de una semana ..
Lunes [20 de feb]:  Aptitud Academica y Cultura Genral
Miercoles [22 de feb]: Matematica .
Viernes [24 de feb]: Fisica y Quimica.

La Sociedad de Ciencias y Humanidades, a travez de la Academia Cesar Vallejo, presenta en un version web los examens y sus soluciones ..
[url=http://www.aduni.com.pe]aduni web[/url]

Saludos,

Carlos Bravo ;)

Sería interesante que el miercoles subamos al foro algunas de la preguntas del examen de matemática, de hecho que va a haber varias preguntas interesantes.

$Tipe$

Pero de vez en cuando hay preguntas interesantes de Matemática...y por lo de la publicidad no creo que sea asi... Carlos no trabaja para Aduni ni algo por el estilo (eso creo  :? ) si no... que se manifieste Carlos...

$Tipe$

[quote="DarthTenebrus"]Por lo que a mi experiencia respecta el examen de la UNI es un examen serio pero no deja de ser un simple examen de opcion multiple .

pero lo que si me parece gracioso , Que hace el link de la ADUNI en el stick ? es acaso esto publicidad ?[/quote]

[color=blue]Respecto a este comentario...
les quiero decir que en ningun momento mi intencion a sido de publicitar a una u otra de estas instituciones de preparacion preuniversitaria, unicamente, la considero como una fuente mas en la web .... que cuenta con alguna informacion importante para el proposito de la Comunidad ...

Espero que esto deje claro el tema !!

Carlos Bravo ;)
Lima - Peru[/color]

[quote="DarthTenebrus"]Bueno ya viste la pagina de ADUNI y descargue tanto el examen como el solucionario . Por lo poco que he visto fijaos bien en las preguntas 13 y 20 :

El estilo de solucion carece de matematica seria :

En el problema 13 : la Induccion esta super incompleta
y en el problema 20 : no se dieron cuenta que el problema esta mal propuesto . ( la optimizacion era maximizar y no minimizar ademas que a y b deberian pertenecer a lso enteros )

Bueno y no es al primera vez eh .[/quote]

Darth Tenebrus tienes toda la razón:  la solución de problema 13 está incompleta y es el punto débil de la mayoría de las instituciones de preparación preuniversitaria (al menos en Lima): la rigurosidad matemática, para la inducción consideran que solo hay que darse cuenta de la propiedad y ya está y no hacen lo más importante que es el paso inductivo, así que es difícil que en un solucionario resuelvan ese tipo de problemas de forma rigurosa... es algo que ya no me sorprende.
En cuanto al problema 20... cuando lo comenzé a resolver estaba super confundido en cuanto al enunciado y a lo que piden y no había considerado que $a$ y $b$ eran enteros... después cuando no llegaba a la respuesta asumí (entre otras cosas)  que eran enteros y salió... que mal por la UNI  :mad: a... y por ahí en otro problema, no me acuerdo que número, ... pero que es de estadística asumen directamente que dos variables $p$ y $q$  son enteras positivas (podrián ser enteros negativos o cero) ...   :mad:

$Tipe$

De acuerdo contigo Jorge Tipe en efecto todos los años se cometen errores como estos en el examen de admision UNI y la verdad que tambien ya estoy acostumbrado .

Gracias a Dios que estos pequeños errores no se han visto reflejadas en el desempeño de los estudiantes peruanos en las olimpiadas Matematicas . quizas se deba al trabajo de los preparadores olimpicos .


linear algebra
matrix
linear algebra unsolved

let $ A,B$ be two diagonalizable matrices of the same size,  prove that if $ AB\equal{}BA$  then there exists an matrix $ P$ st : $ PAP^{\minus{}1}$ and $ PBP^{\minus{}1}$ are diagonal matrices.

thanks so much

Let $ A \equal{} PJP^{ \minus{} 1}$, with $ J$ a Jordan normal form for $ A$. Then

$ AB \equal{} BA < \equal{} > PJP^{ \minus{} 1} B \equal{} B PJP^{ \minus{} 1} < \equal{} > JY \equal{} YJ$ (1),

with $ Y \equal{} P^{ \minus{} 1}BP$. But if $ \lambda_1, \lambda_2, ..., \lambda_s$ are the distint eigenvalues of $ A$, with multiplicities $ \alpha_1, \alpha_2, ..., \alpha_s$, then A being diagonalizable implies that

$ J \equal{} diag(\lambda_{1_1}, \lambda_{1_2}, ..., \lambda_{1_{\alpha_1}}, \lambda_{2_1}, ..., \lambda_{2_{\alpha_2}}, ..., \lambda_{s_1}, ..., \lambda_{s_{\alpha_s}})$,

and so from (1) we see that $ Y \equal{} diag(Y_1, Y_2, ..., Y_s)$, with $ Y_j \in \mathbb{C}^{\alpha_j \times \alpha_j}$, $ j \equal{} 1, 2, ..., s$. Suppose now that each $ Y_j$ is reduced to Jordan normal form by $ Y_j \equal{} Q_j J_j Q_j^{ \minus{} 1}$, $ j \equal{} 1, 2, ..., s$. Therefore, defining $ Q \equal{} diag(Q_1, Q_2, ..., Q_s)$ we have

$ Y \equal{} Q diag(J_1, J_2, ..., J_s) Q^{ \minus{} 1} \equal{} Q\hat{J} Q^{ \minus{} 1}$,

and so

$ B \equal{} PYP^{ \minus{} 1} \equal{} (PQ) \hat{J} (PQ)^{ \minus{} 1} \equal{} T \hat{J} T^{ \minus{} 1}$,

where we see that $ \hat{J}$ is a Jordan normal form for $ B$. But $ B$ is also diagonalizable, which means that $ \hat{J}$ is a diagonal matrix and the columns of $ T$ are $ n$ linearly independent eigenvectors for $ B$. Now, we also have

$ TJT^{ \minus{} 1} \equal{} P(QJQ^{ \minus{} 1})P^{ \minus{} 1} \equal{} P(Q diag(\lambda_1 I_{\alpha_1}, ..., \lambda_s I_{\alpha_s}) Q^{ \minus{} 1})P^{ \minus{} 1}$

$ \equal{} P diag(\lambda_1 Q_1Q_1^{ \minus{} 1}, ..., \lambda_s Q_sQ_s^{ \minus{} 1}) P^{ \minus{} 1} \equal{} P diag(\lambda_1 I_{\alpha_1}, ..., \lambda_s I_{\alpha_s}) P^{ \minus{} 1}$

$ \equal{} PJP^{ \minus{} 1} \equal{} A$,

which means that $ T$ also diagonalizes $ A$. $ \Diamond$

You can also show the converse: if there is a matrix $ T$ that diagonalizes both $ A$ and $ B$, then $ AB \equal{} BA$.


Starting at the M in the center, you may
move left, right, up or down to an adjoining
letter. How many distinct paths can be
followed to spell the word MATH?

\begin{tabular}{ccccccc}
&&&H&&&\\
&&T&H&T&&\\
&H&T&A&T&H&\\
H&T&A&M&A&T&H\\
&H&T&A&T&H&\\
&&H&T&H&&\\
&&&H&&&\\
\end{tabular}

Starting at the M in the center, you may
move left, right, up or down to an adjoining
letter. How many distinct paths can be
followed to spell the word MATH?

$ \begin{tabular}{ccccccc}
&&&H&&&\\
&&T&H&T&&\\
&H&T&A&T&H&\\
H&T&A&M&A&T&H\\
&H&T&A&T&H&\\
&&H&T&H&&\\
&&&H&&&\\
\end{tabular}$

there we go

is there a special formula for that one? or just counting?

[quote="alanchou"]count the ways going from M to a certain A, then multiply by 4 by symmetry[/quote]
Actually, the figure is not symmetrical. If you look at the 2nd row and the 6th row, you will see. The answer is $\boxed{19}$ ways (I just counted them all).

$ \begin{tabular}{ccccccc}
&&&H&&&\\
&&T&H&T&&\\
&H&T&A&T&H&\\
H&T&A&M&A&T&H\\
&H&T&A&T&H&\\
&&H&T&H&&\\
&&&H&&&\\
\end{tabular}$

We have to start with the middle M being our point in the center.
$M$ is our first number and we have $1$ way to chose the M.
For $A$ we have $\binom{4}{1}=4$ ways to chose the A.
For $T$ we have $\binom{2}{1}=2$ ways to chose the T for the top A, and $\binom{3}{1}*3\implies \frac{3!}{2!1!}*3\implies \frac{6}{2}*3\implies 9$ ways to chose the T.
For H starting from the top $T$ we have $\binom{2}{1}=2$ ways to chose the H.  For H on the other three choices we have $\binom{3}{1}*3=9$ ways to chose the H.
$4+2+9+2+2+9\implies 6+9+2+2+9\implies 15+2+2+9\implies 19+9\implies \boxed{28}$

You don't really need that choose notation but still works


trigonometry
trig identities
Law of Cosines
geometry proposed
geometry

Suppose that two line segments with the lengths $d$ and $r$ , and an angle with the size of $\theta$ are given, where $d\in [0,r]$ and $\theta \in [0,\frac{\pi}{2}]$. Using only a compass and a ruler construct an angle with the size of $\alpha$ so that:

$\cos \alpha=\frac{d}{r}\sin^{2}\theta+\cos \theta \sqrt{1-\left( \frac{d}{r}\sin \theta\right)^{2}}$

Here's an outline of a solution:

[hide]First, using a compass, draw a circle with the center $O$ and radius $r$. Next, using a ruler, draw any diameter of the mentioned circle, like $AB$. Then, choose a point like $P$ on $OA$, so that $|OP|=d$.  Now draw a line that passes from $P$ and crosses the circle in two points like $X,Y$ such that $\angle APX=\angle BPY=\theta$. I claim that $\angle AOX=\alpha$.

To prove this, find the length of $AX$ in two ways: First write the law of cosines in $\triangle APX$  and then find the length of $PX$ by writing the cosine law in $\triangle POX$. Next, by writing the cosine law (a third time!!) in $\triangle AOX$, find another amount for $|AX|$. By solving these three equations the value of $\cos \alpha$ can be found.[/hide]


Two people peeled 400 potatoes .One compleated three pieces a minute ,the other two.The second worked 25 minutes longer than the first.
How long did each work? :rotfl:

yeah, i got the same answer as gr8sk8r (i did the problem with him)
[hide]27.5 minutes and
52.5 minutes[/hide]

hmm i got something different

[hide]
Say first guy worked for t1 minutes.  Then second guy worked for t1 + 25 minutes.  Together they completed 400 potatoes:

400 = guy1 + guy 2 = 3 pieces/min * t1 + 2 pieces/min *(t1+25)

Solve that equation to get t1 = 70 min, t2 = 95 min.
[/hide]

hope im not doing anything wrong[/hide]

You have to set second guy = first guy + 25, like Barnacle did.

oh, i see now...[hide="clik dis"]i get it :)  :) [/hide]


The number 210 can be written as the sum of consecutive positive integers in several ways. When written as the sum of the greatest possible number of consecutive positive integers, what is the largest of these integers?

Note that \[ 1\plus{}2\plus{}3\plus{}\cdots \plus{}\boxed{20}\equal{}\frac{20\cdot 21}{2}\equal{}210\]
Clearly no other way has more positive integers


He's a savage beast. What more of a story do you expect--one of sex, drugs, and rock and roll?

He has an account on AOPS, you know. You could message him and ask him =P

[quote="ic2149"]:thumbup:

Anybody know the story about Chen Sun--Winner of this year' CMO ?[/quote]

NOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOBBB!!!!!

Chen is a freaking MOOSE.

-Danny Shi

[quote="ic2149"]:thumbup:

Anybody know the story about Chen Sun--Winner of this year' CMO ?[/quote]
noooob!  :|

Chen is a moose.
Jon is a doll.
Cynthia is a monster.

[quote="shalevbd"]You guys are so mean!
he was just asking an innocent question, and the next thing he knows the whole IMO team is out to get him.

tsk tsk you guys. =P

and LOL where did you get those nicknames =S[/quote]

I don't think you would want to know. I for one am not curious.

And to shamelessly hijack this thread, I congratulate the above-mentioned members of this year's team on their performance, which can be viewed [url=http://www.imo-official.org/]here[/url].

[quote="sunchips"][quote="ic2149"]:thumbup:

Anybody know the story about Chen Sun--Winner of this year' CMO ?[/quote]

HAHAHAHHAHAA What a NOOB!!

I am a moose.
Jon is a doll.
Cynthia is a monster who ripped off our arms.
Danny is a pervert.
Alex is an amplifier.
Jarno is a normal male.[/quote]

Erm, Jarno is not a normal male. He is a skeleton.

-Danny Shi


geometry rectangle function quadratics MATHCOUNTS articles algebra

well wait, wat are the numbers on the dice [for example, are they consecutive integers startting with one, or otherwise?], and how many faces does the dice have? And also, how many times are you throwing the dice?

I assume the dice are normal dice :P

As for the probability.. I'm not sure. There is a simple formula for the number of ways of getting x numbers to add to n if the x numbers can range from 0-n each, but I'm not sure what you get when you restrict the values each one could take..

(200 posts, btw.. I was hoping for a great post that I could make to do it, but I'm getting impatient :P)

[color=cyan]Just looking at it, I'd say if anything it's pretty ugly.  You can recursively define it, of course, which will work for actual computation.  And you can get plenty of relationships using the function.  Actually, you should be able to model this with path-walking, much as the binary number nCr can be modeled as the number of ways to get from one corner to another of an
(n-r) x r rectangle, moving only one unit up or to the right at any given time.  You might try modeling smaller dice first -- a 1-sided die is fairly simple.  A 2-sided die looks like it might give you something related to Fibonacci numbers, maybe.  And so on.

Also, it isn't that hard at all to generate the functions for the first couple of rolls.  But by the third roll, you've really got a quadratic equation with absolute values thrown in in odd ways that I haven't fully worked out yet.

P.S.  With respect to your signature there, Yes, our counting system is based on an axiom.  But so is all of life -- you're basing your life on the principle that your observations are correct and meaningful.  Along with hundreds of other unfounded principles.  So complaining about a counting system that works just fine is rather silly, I think.[/color]

[quote="JBL"]With respect to your signature there, Yes, our counting system is based on an axiom.  But so is all of life -- you're basing your life on the principle that your observations are correct and meaningful.  Along with hundreds of other unfounded principles.  So complaining about a counting system that works just fine is rather silly, I think.[/quote]

I have to heartily agree with JBL and mathfanatic (below)

[quote="mathfanatic"]Think of your basic beliefs as your base. If you don't have any beliefs, you can't build anything off those (non)beliefs, just like you can't build a tower without a base. And if you can't build anything off your beliefs, then you'll never figure anything new out. How could Einstein or Euler or anyone else have figured out all the things they figured out if they had decided that 1 + 1 doesn't equal two? I'd rather figure out interesting stuff based on something I have to assume is true rather than sit on my butt and wonder all day what if we used the numeral "3" in place of the "2" and vice versa...[/quote]

(from [url=http://www.artofproblemsolving.com/wforum/viewtopic.php?p=2851&highlight=#2851]here[/url])

Yes, rep123max.  I think you [i]are[/i] the only one who absolutely [i]hates[/i] axioms.  So what if they aren't proven?  We haven't proven that cockroaches aren't the rulers of the universe, but do you think we could, even if we wanted to?  We can get a lot more done if we don't consider the cockroaches our master...you know, we don't have to worry about constantly looking at the ground to avoid stepping on one...instead, we can work on math problems and the like.  So the fact that 1 = 1 can't be proven, but how much could you get done if 1 equalled 2 on another day, and then 17.21, and other random numbers?  For this reason, I prefer to be possibly wrong but find out cool stuff about how the universe works.

We aren't so concerned that what we say is true. We are more concerned that what we say is consistent. Is it? Well, we don't know. We think it is.

[quote="Rep123max"]it seems all my posts are "controversial". the AMC  one got people mad, my signature gets people mad, im really a nice person. also, for those who do mathcounts, im a 7th grader in texas (was now will be an 8 ball) and am the fastest (according to a state number sense test-tied with some kid from houston) mental math kid in state (thats nothing compared to everybody else here), but we dont do an official countdown round. im working really hard on sprint to get me in the countdown next year but i dont think it is fair that countdown doesnt counts. shouldnt every round count? anyways, it will still have almost the same 4 kids go to nationals (except 1 person might not) so why wont they make every state either have an official one or not?[/quote]

Then again, you've started about 20 threads (most of them with things like "is there a formula for this or that..."), so I wouldn't say that all of your posts are controversial.

And please don't brag to us about how fast you are or complain about scoring.  We understand that you're a good math student, and you won't ever have to worry about proving that to us.

Anyways, mathcounts is somewhat inconsistent with respect to offical Countdown Rounds.  Like Simon said, we are concerned with things being consistent.  As long as an axiom stays the same, we're fine.  And if it changes - well, then, we've discovered something new.

Nothing wrong with being curious about formulas and stuff.  I was just stating a fact.

Anyway, I didn't see the parentheses, but I'm know for a fact that everyone around here is just as good as everybody else.  Everyone has potential for math.  Some students are more senior, and that's what makes them seem so good at math.  Don't get discouraged...believe me, you're doing great.  I was just suggesting to you that next time, if you want to keep in friendly relations with ohters, don't ever say anything about yourself.  It just makes things simpler and even if you don't think you're bragging, others may percieve it as so.

However, the statement "im trying to learn unlike you who cares about the names of people. am i the crazy one or am i missing somethingg" bothers me.  I never said I wasn't trying to learn.  In fact, I am learning a lot, both my asking questions and by answering them.  I just happen to enjoy hearing about people like me who have done a lot better than me in math.  It both motivates me to do better, and sometimes, if I see one of these people on AIM or something, I can congratulate them.  Yes, it seems like a waste of time, but it's an...interest of mine.  Besides, if you're unwilliing to focus on anyone other than yourself, you will never enjoy being good at math, or anything, for that matter.

And if you're calling me crazy, I've already heard it.  Most of the time it's from my friends.  So if you want to call me crazy, you will have to be my friend.  Will you be my friend?

It's not so good to ask for formulae for everything. For one thing, if you only know a bunch of formulae, you won't be able to solve problems unless you have seen them before, which isn't so good. Also, this website especially frowns upon memorizing formulae. (Read the article on that for example.) The idea is that you should be able to reason your way through a problem, although of course it is nice to know a few formulae that come up a lot.

Just to add "so thats about all."?
 :lol:

Edit: Well, thanks for replying rep123max.  Yup, let bygones be bygones.

Interesting about that formula...it's not half as messy as I thought it might be.  Which is a good thing.

It's a lot of fun to figure out a formula by yourself, as you did with the x in squares problem.  Plus, you learn by making the discoveries yourself.

[color=cyan]Geez, why did a philosophical discussion about axioms and such devolve into an argument containing personal attacks?  Watch how you talk to each other -- there's no need for that kind of thing at all.  My apologies for bringing it up.

As for the original question, I can give you one small fact that I worked out just by writing down the first few cases.  I haven't proven it yet, but I'd put money on its actually being true.  I think it's probably provable by induction.  Anyhow:
for n rolls, n greater than or equal to 2, the number of ways of getting k total is (k-1)*(k-2)*. . .*(k-n+1)/(n-1)! = (k-1)C(n-1), provided k is between 1 and n+5, inclusive.
Also, of course, you get equal values at the top end of the sums.  But what happens in the middle looks messy, and from the first 4 cases I didn't see a pattern.  I recommend writing a computer program to generate a lot of data, and even better (but more difficult), tell you what the polynomials are that give the values you want.  This is doable by solving a system of simultaneous equations which result from assuming the answer is a polynomial and plugging in the values you know (for example, that there are 6 ways to get a 5 in three rolls) to solve for the coefficients.  If you want more information on this, just ask.[/color]

The generating function for this is (x+x^2+x^3+x^4+x^5+x^6)^n for n rolls. Then you can use the multinomial theorem to find the x^k coefficient. I don't know why I didn't think of this before. But that will be really messy too.

[quote="Rep123max"]im not calling you crazy at all[/quote]

I don't think mathfanatic minds being called crazy.  He kind of likes it...

Anyway, it's nice that we managed to end that argument.  Thanks for the edit, mathfanatic - PMing helped that time.  As for what I was thinking when I was talking about cockroaches, who knows?  Heehee... cockroaches are the rulers of the universe?  I musta been high on altoids.

[quote]I don't think mathfanatic minds being called crazy. He kind of likes it... [/quote]

Too true, too true.  :lol:  And it's nice to know that I'm not the only one who can get high on altoids.

Let's all spread some joy!  On the count of 3, sing the "I Love You" song from Barney.  1...2...<cough cough choke strangle> Then again, maybe not.  Still, glad we're all happy now.   :D

I'm not so sure about what numbers to plug into the formula, so I'll solve this the way I usually do:

10 = 1+3+6 (6 ways to get this)
10 = 1+4+5 (6)
10 = 2+2+6 (3)
10 = 2+3+5 (6)
10 = 2+4+4 (3)
10 = 3+3+4 (3)

Which is 27 ways.

27/6^3 = 27/216 = 1/8

I think that's right...have to check.  How would you do this using one of the formulas?

While the generating function will be messy for building the general case, it does take care of specific cases pretty handily.  You can use it to get to mathfanatic's 27 (which I agree with) in a couple lines and without the fear of missing cases. (In fact, some folks would say that having a generating function for the answer is essentially equivalent to having the answer.)

I like this approach enough to this problem that I suspect those of you who take intermediate counting will be seeing it again....  We'll be spending a couple days on generating function problems...

The way that generating functions work is that instead of looking for the number of ways to roll a sum of some number directly, you look at the coefficient for a certain term. For example, in this case, if you want to fine the number of ways to roll a sum of 14 from 4 dice, you can look at the coefficient of the x^14 term in the expansion of (x+x^2+x^3+x^4+x^5+x^6)^4. I recommend Generatingfunctionology by Herbert Wilf. To get that book, you can either spend $56 on amazon.com, or you can do a Google search for Herbert Wilf and download it. It's very kind of him to put it there. After reading a few chapters, it should make sense why this method works and why it's so powerful.

You have to add a lot of terms together. The binomial coefficient (n choose k) = n!/(n-k)!k!, and the multinomial coefficient (n choose a_1, a_2, a_3, ..., a_k) with a_1+a_2+a_3+...+a_k=n =n!/a_1!a_2!a_3!...a_k!. The aforementioned multinomial coefficient is the number of ways to select a_1 of object 1, a_2 of object 2, ..., a_k of object k out of a pool of n, much like (n choose k) is the number of ways to divide n objects into two groups: group 1 of k and group 2 of n-k.

A bit of warning before plowing forward.  Generating functions are a pretty advanced tool.  If you've never seen them before, the following is going to be pretty confusing.  They're not really necessary for anything shy of AIME, and even AIME/USAMO problems which can be tackled with generating functions can usually be done other ways.  We do give them a pretty thorough treatment in our intermediate counting class, primarily because they're awfully cool.

You're gonna really like generating functions once you learn about them.  If you just want one coefficient, you can get there a little faster than rampant multiplication:

(x+x^2+...+x^6)^3 = x^3(1+x+x^2+...+x^5)^3 = x^3[(1-x^6)/(1-x)]^3

That last step is just summing a finite geometric series, or factoring.  Now we take the numerator and denominator separately:

1/(1-x)^3 = C(2,2) + C(3,2) x + C(4,2) x^2 + C(5,2) x^3 + ...

(Cube the infinite geom. series relationship 1/(1-x) = 1 + x + x^2 + x^3 ...  and you'll see the above.  Once you're more familiar with generating functions and aspects of advanced counting, this is automatic - you'll know this as clearly as you know factorials.)

Numerator:

(1-x^6)^3 = 1-3x^6 + ...  terms with x higher than x^10.  We don't care about them, since we want the coeff. of x^10.

So now our problem is finding the coefficient of x^10 in

x^3[(1-x^6)/(1-x)]^3 = x^3 (1-3x^6+...)(C(2,2) + C(3,2)x + C(4,2) x^2 +...).

We see that there are only two terms to worry about: x^3(1)(C(9,2)x^7)and x^3(-3x^6)(C(3,2) x).  Putting these together gives us our 36x^10-9x^10 = 27x^10.

This looks like a lot more work than it is.  An experienced counter scribbles this:

x^3[(1-x^6)/(1-x)]^3 = x^3 (1-3x^6+...)(C(2,2) + C(3,2)x + C(4,2) x^2 +...).

upon reading the problem and is essentially done.

(There are also clever counting ways to jump to the answer C(9,2) - 3(C(3,2); see if you can find those!  We cover these in intermediate counting, as well...)

If this is all Sanskrit to you, don't worry about it for now.  Find a beginning point for generating functions (we give an introduction in AoPS2, and will delve more deeply in our Interm. Counting class) and come back after you understand the basics a little better.  Starting with this problem to learn generating functions is a little like starting algebra with the quadratic formula rather than 'what is a variable'.

The multinomial theorem will cause you no end of grief if you want a specific case. You have to count the number of (1, 3, 6); (1, 4, 5); (2, 2, 6); (2, 3, 5); (2, 4, 4); (3, 3, 4) you can get. It's only useful if you want a (really ugly most likely) general formula.


1. What is the units digit of 1997^1997?

2. What is the hundreds digit of 2003^2004?

3. What is the tens digit of 1111^1234?

4. What is the units digit of 1234^5678?

I'll show you how to do the first one, and then you try the next 3.

Since we only want the units digit, we can disregard everything but the last digit. So this becomes finding the units digit of $ 7^{1997}$. It kind of like how if you wanted to know the units digit of $ 123453234543 \times 2$, you can see right away that it's simply $ 3 \times 2 \equal{} 6$.

Well, it's obviously way too computational to do $ 7^{1997}$, but we can maybe find a pattern.

$ 7^1 \equal{} 7$

$ 7^2 \equal{} 9$ (units digit)

$ 7^3 \equal{} 3$

$ 7^4 \equal{} 1$

$ 7^5 \equal{} 7$

$ 7^6 \equal{} 9$

Aha! A pattern. It goes in cycles of 4, 7-9-3-1. Since when 1997 gives a remainder of 1 when divided by 4, it corresponds to the first one, so our answer is $ \boxed{7}$.

Challenge: Prove the units digits always repeat in cycles of 4.

To solve the problems above, simply find a pattern. With experience, you can solve these easily. First, try the problems above by yourself by finding the first few exponents. Then, you will find a pattern. Apply it and move on.

With experience, you will get better.  :D


algebra
polynomial
calculus
integration
Rational Root Theorem
number theory unsolved
number theory

Find all integer polynomials $P(x)$, the highest coefficent is 1 such that: there exist infinitely irrational numbers $a$ such that $p(a)$ is a positive integer.

All polynomials of degree $\ge 2$, I think (I assumed $P$ o be non-constant).

Assume we're working on a neighbourhood of infinity on which $P$ is a bijection to another neighbourhood of infinity, and which contains only numbers large enough for our purpose (that is, suppose we're working only with naturals $n$ for which we assume that $P^{-1}(n)$ has exactly one rational element).

For all $n$, if the root of $P_n(x)=P(x)-n$ is to be rational, then it is integral, since its denominator divides the leading coefficient of $P_n(x)$, which is $1$. This means that we have $P^{-1}(n+1)-P^{-1}(n)\ge 1$. However, if the degree of $P$ is $\ge 2$, then $P^{-1}(n+1)-P^{-1}(n)\to 0$ as $n\to\infty$, and we have a contradiction.

Conversely, it's easy to see that all polynomials of degree $1$ do not satisfy the required condition, so those of degree $\ge 2$ are precisely the ones we're looking for.

This problem is very similar to an old problem which i solved.
Let $Q(x)=P(x)-a_0-p$
with p is a prime number.
$Q(1)=?<0$
$Q(\propto )=\propto>0$
so we have Q have a real root(crossing the x-axis). in interval $(1,\propto)$
So exist $x_p$ such that $P(x_p)=p+a_0$
We have two case:
1)If exist infinite $x_p$ be long to $\mathbb{I}$ so we are done.
2)if have finite mp such that $x_p$ is irrational so exits $p_0$ such that $x_p$ is rational when $p>p_0$ use rational root theorem we have
 $x_p$ is an integer and it divide $p$ so $x_p=p$ because $x_p>1$
so $P(p)=P(0)+p$ for all $p>p_0$ or $P(x)=P(0)+x$ or p have degree 1 contracdition :)


trigonometry
geometry
trig identities
Law of Sines
algebra unsolved
algebra

Find all ordered triples $ (x, y, z)$ of real numbers such that

\[ 5 \left(x \plus{}  \frac{1}{x} \right) \equal{} 12 \left(y \plus{}  \frac{1}{y} \right) \equal{} 13 \left(z \plus{}  \frac{1}{z} \right),\]

and \[ xy \plus{} yz \plus{} zy \equal{} 1.\]

Sketch: Make the trig substitution $x=\tan\left(\frac{_A}{^2}\right), y=\tan\left(\frac{_B}{^2}\right), z=\tan\left(\frac{_C}{^2}\right)$ with $A+B+C=180$ (since $xy+yz+zx=1$). Then plugging this into $x+\frac1x$ and simplifying gives $x+\frac1x=\frac{1}{\sin \left( \frac{A}{2} \right) \cos \left( \frac{A}{2} \right) }=\frac{2}{\sin A}$, and similarly for $y+\frac1y$ and $z+\frac1z$. This means that $\frac{\sin A}{\sin B}=\frac{5}{12}$ and $\frac{\sin A}{\sin C}=\frac{5}{13}$.
As $A,B,C$ are the angles of a triangle, it follows by the law of sines that they are the angles of a triangle similar to a 5-12-13 triple.
Therefore, $\tan A=\frac{5}{12},\tan B=\frac{12}{5}, \sin C=1$. So $z=\tan\left(\frac{C}{2}\right)=\tan 45=1$, and we can get the other two values by using double angle formulas for tangent, and we end up getting $x=\frac{1}{5},y=\frac{2}{3},z=1$.

Can you explain (prove) why that substitution works?

just remark $tan(\frac A 2)=\frac 1 { tan(\frac{B+C}2)}$ and the formula.
Also, the negative values for $x,y,z$ work too obviously.

Shouldn't the equation be \(xy + yz + zx = 1\)?

5 12 13 whispers a geometric interpretation

There is a completely algebraic solution:
Just homogenize the first equation using the second:
\[x+\frac{1}{x}=\frac{x^2+1}{x}=\frac{x^2+xy+xz+yz}{x}=\frac{(x+y)(x+z)}{x}.\]
So we get
\[\frac{xy+xz}{5}=\frac{yx+yz}{12}=\frac{zx+zy}{13}\]
from where we can easily solve for $xy=\frac{2}{15}, yz=\frac{2}{3}, zx=\frac{1}{5}$ and then easil find that $(x,y,z)=\left(\frac{1}{5}, \frac{2}{3}, 1\right)$ or $(x,y,z)=\left(-\frac{1}{5}, -\frac{2}{3}, -1\right)$.
(Incidentally, the solution in #2 misses this second solution, which shows how careful one has to be with geometric substitutions.)


why does the class always freeze on me?
is it b/c my computer or the actual classroom? :?:

probably your computer..that happened to me a few times when i was taking the mathcounts problem series and my internet was messed up..so go figure

If you are on wireless, then try moving to a non-wireless connection.  If not, then try using the Opera browser, which you can download for free here:
http://www.opera.com

Ideas:

1. Check your internet connection.

2. Press the "Get class transcript" button, and then if there are messages that havent shown yet but are in the transcript, try pressing the clear chatroom button.


floor function
limit
inequalities
real analysis
real analysis unsolved

i think this problem is not completed,, but..can anyone help me?

it's known that $ a_n$ is a series so that
$ a_n\equal{}a_{n\minus{}1}\plus{}\frac{1}{a_{n\minus{}1}}$...$ \forall n\geq 2$
if $ a_1\equal{}1$,, find the value of $ \lfloor {a_{100}}\rfloor$

Computer always help you ;)

$ a_{100} \approx 14.213709429126533$

And more generally: $ a(10^{4} n)\approx 10^{2}a(n)$

This gives $ a_{n} \equal{} \textbf{O}(\sqrt{n})$ and $ \lim_{n\to\infty}a_n \equal{} \infty$

This problem was in the book about olympiads published in USSR, but I can't remember the name of he book. The solution is that:
 We have:
\[ a_n^2\equal{}a_{n\minus{}1}^2\plus{}2\plus{}\frac{1}{a_{n\minus{}1}^2} > a_{n\minus{}1}^2\plus{}2 \Rightarrow a_n^2>2n\minus{}1(if\, n>1)\]
Also
\[ a_n^2\equal{}a_{n\minus{}1}^2\plus{}2\plus{}\frac{1}{a_{n\minus{}1}^2}\equal{}a_{n\minus{}2}^2\plus{}4\plus{}\frac{1}{a_{n\minus{}1}^2}\plus{}\frac{1}{a_{n\minus{}2}^2}\Rightarrow
a_n^2\equal{}2n\minus{}1\plus{}\sum_{k\equal{}1}^{n\minus{}1}\frac{1}{a_{k}^2}\]
But using the previous inequality we have:
\[ a_n^2\equal{}2n\minus{}1\plus{}\sum_{k\equal{}1}^{n\minus{}1}\frac{1}{a_{k}^2}<2n\minus{}1\plus{}\sum_{k\equal{}1}^{n\minus{}1}\frac{1}{2k\minus{}1}<2n\minus{}1\plus{}\sum_{l\equal{}1}^{2n}\frac{1}{l}<2n\minus{}1\plus{}\ln(2n)\plus{}1\equal{}2n\plus{}\ln(2n)\].
So for n=100
\[ 199<a_{100}<200\plus{}\ln(200)\equal{}205.29831737 \Rightarrow \lfloor a_{100}\rfloor \equal{}14\]

that's very nice... and we get
\[ \lim_{n\to\infty}{\frac{a_n}{\sqrt{n}}}\equal{}\sqrt{2}\]