limit
trigonometry
Euler
calculus
calculus computations

Exist  or no   limit of  sequence:
$ \lim\limits_{n \to \infty} \sin n$

No, it oscillates infinitely between -1 and 1.

[quote="Hong Quy"]Exist  or no   limit of  sequence:
$ lim_{n \to \infty} sin n$[/quote]
yes,it is old and easy ,my friend,
We let $ x_n\equal{}(\minus{}\frac{\pi }{2}\plus{}2n \pi)$ then $ lim_{n\to \infty} sin x_n\equal{}\minus{}1$
and $ x_n \equal{}\frac{\pi }{2}\plus{}n\pi$ thus $ lim_{n\to \infty} sin x_n\equal{}1$ it is impossiable
so the ansewr is not exist

Allnames is showing that the expression approaches 2 values from different sides, so the limit can't exist.

The question is NOT: Does $ \lim_{n\to \infty} \sin(x_n)$ exist for a sequence $ (x_n)$ with $ \lim_{n\to \infty} x_n \equal{} \infty$,
but:
Does $ \lim_{n\to\infty}\sin(n)$ exist.

Well, when the limit exists and $ \lim_{n\to\infty}\sin(n) \equal{} \alpha$,
then
$ \alpha \equal{} \lim_{n\to\infty}\sin(n \minus{} 1) \equal{} \lim_{n\to\infty} (\sin(n)\cos(1) \minus{} \cos(n)\sin(1))$
From this we can conclude that $ \lim_{n\to\infty}\cos(n) \equal{} \frac {\alpha \cos(1) \minus{} \alpha}{\sin(1)}$
and
$ \alpha \equal{} \lim_{n\to\infty}\sin(n \plus{} 1) \equal{} \lim_{n\to\infty} (\sin(n)\cos(1) \plus{} \cos(n)\sin(1))$
From this we can conclude that $ \lim_{n\to\infty}\cos(n) \equal{} \minus{} \frac {\alpha \cos(1) \minus{} \alpha}{\sin(1)}$
Then $ \lim_{n\to\infty}\cos(n)$ has to be 0.
and so
$ \alpha \equal{} \lim_{n\to\infty}\sin(2n) \equal{} \lim_{n\to\infty}2\sin(n) \cos(n) \equal{} 0$
But then 1 = $ \lim_{n\to\infty}(\sin^2(n) \plus{} \cos^2(n)) \equal{} 0$
A conflict. Thus the limit does not exist.

If the limit exists, it is contained in [-1,1] and it exists for the analogous cos limit.

but $ e^{ix} \equal{} \cos x \plus{} i\sin x$

Contradiction.

;)

[quote="§outh§tar"]If the limit exists, it is contained in [-1,1] and it exists for the analogous cos limit.[/quote]
Why should the analoguous cos limit exist?

You mean $ \lim_{n\to\infty}\sin(\pi n) \equal{} 0$ exists, so the limit exists for the analoguous cos limit,

but $ e^{ix} \equal{} \cos x \plus{} i \sin x$

Contradiction???????

It is clear that if the limit in question exists, it is contained in [-1,1]. You know how to express cos in terms of sin and you know your basic limit properties so you can see why the analogous limit for cos exists. I'm not sure where you're getting the pi factor from because I never mentioned that. Given these remarks, is it possible to take the limit of both sides in the Euler equation?


linear algebra
matrix
trigonometry
function
calculus
derivative
geometric series

I know that for $ f(x) = e^{x}$, the Maclaurin series is given by:

\[ 1 + x + \frac {x^{2}}_{2!}} + \frac {x^{3}}_{3!}} + . . .  = \sum_{n = 0}^{\infty}{\frac {x^{n}}_{n!}}\]

The general term is readily obvious here and I could figure out easily. But for something like $ f(x) = sin x$ or $ f(x) = tan^{ - 1}x$, is there a systematic approach to finding the general term of the series? Something like $ \frac {( - 1)^{n - 1}x^{2n - 1}}_{(2n - 1)!}$ just isn't obvious to me. Sure, I have some of them memorized but I'd just like to know how they're derived and wikipedia and other sources aren't helping me in that regard  :(

The two other examples you mentioned are not particularly hard to figure out since there is a definitive pattern.  Usually though, the series that you will be asked will produce a noticeable pattern after the first few entries so you should be able to make it out.  Try this one:
$ f(x)\equal{}e^{\minus{}x^{2}}$

Yeah. Often times it is all in how you organize how you are writing it down. When I was starting these, I found it helpful to just make a column down the left of the page of:

$ f(0)\equal{}$
$ f'(0)\equal{}$
$ f''(0)\equal{}$

etc. This will help in things that are giving you $ (\minus{}1)^n$ type terms. Then plugging in down the right side the actual term $ \frac{f^n(0)x^n}{n!}$.

$ f(x)\equal{}\sin(x)$.

$ \begin{matrix}
f(0)\equal{}0    & &             0\\
f'(0)\equal{}1    & &           x\\
f''(0)\equal{}0     & &           0\\
f'''(0)\equal{}\minus{}1   & &           \frac{\minus{}x^3}{3!}\\
\end{matrix}$

You could do more, or figure out at this point that you are only picking up odd terms and they alternate in sign. Hence $ 2n\minus{}1$ for odds, and then experiment to get the exponent $ n\minus{}1$ on the $ (\minus{}1)$ for proper sign. Try the example by JRav.

Don't forget that you can use infinite series to prove demoivres theorem.

$ e^x\equiv \sum^{\infty}_{n\equal{}0} \frac{x^n}{n!}\equal{}1\plus{}x\plus{}\frac{x^2}{2!}\plus{}\frac{x^3}{3!}\plus{}\ldots \Leftrightarrow$
$ e^{ix}\equiv \sum^{\infty}_{n\equal{}0} i^n\frac{x^n}{n!}\equal{}1\plus{}ix\minus{}\frac{x^2}{2!}\minus{}i\frac{x^3}{3!}\plus{}\frac{x^4}{4!}\plus{}i\frac{x^5}{5!}\plus{}\ldots$

Which is clearly $ e^{ix}\equal{}\cos{x}\plus{}i\sin{x}$; that may help you remember the pattern for some functions. Also, parity notations can be very useful when constructing a general term; note that $ i^{2n}\equal{}(\minus{}1)^n$ in the above series. From there the equations for cosine and sine become much clearer, as sine is to odds as cosine is to evens. I personally prefer notation with $ i$ because it shows the relationship in demoivres. Written this way,

$ \sin{x}\equiv \frac{1}{i}\sum^{\infty}_{n\equal{}0} \frac{(ix)^{2n\plus{}1}}{(2n\plus{}1)!}$

PS: You can paste off the proof by adding an exponent to both sides to show that $ [f(\theta)]^n\equal{}f(n\theta)$

The standard series are generally derived by just differentiating the functions repeatedly and noting the patterns. In the case of trig/exponential functions, we can also reasonably take the series as the definition.

The other important tool is the geometric series $ \frac1{1\minus{}z}\equal{}1\plus{}z\plus{}z^2\plus{}\cdots$. That can be derived by repeated differentiation, but the algebraic trick is easier.

How does $ \arctan x$ work? If we differentiate once, we're at $ \frac1{1\plus{}x^2}$. That's a geometric series $ 1\minus{}x^2\plus{}x^4\minus{}x^6\plus{}\cdots$, and we integrate termwise for the arctangent series.

Don't be afraid to substitute simple expressions into known series; if we know the power series for $ f(x)$, the series for $ f(5y^3)$ is obtained by substituting $ 5y^3$ for $ x$ everywhere. In this case, the terms even correspond 1-1.
This substitution has limits; we can't use something like $ 1\plus{}x$ that moves the base, since we would need to add infinitely many terms for each coefficient. Expressions like $ x\plus{}x^2$ work, but require increasingly large sums to find coefficients.


Let $A$ be a $8\times 8$ array with entries from the set $\{-1,1\}$ such that any $2\times 2$ sub-square of the array has the absolute value of the sum of its element equal with 2. Prove that the array must have at least two identical lines.

a minor question on the wording: "lines" --> is a row the same as a column enough as a requirement, or do it have to be 2 rows or 2 columns

(not that it really matters here but it could be better to explain this)

Please, give me a solution for this problem..  :D

i find good results.. but nothing that conclude the problem  :(

Can you not begin constructing the square since you must have 3 1's and a -1 or 3 -1's and a 1 in order for the absolute value of the sum of the elements in the 2x2 square to be 2.

I will work on it, that just came to mind after reading the question.

DPopov: You complete your solution ?  :D

sorry, i haven't been able to write a decent solution yet. I will respond ASAP

Ummm... don't there have to be at least 2 pairs of identical rows and 2 pairs of identical columns?

Chavez, could you please not double or triple post?  What I mean is don't post one message and then another message without someone responding to your last message.  There are certain instances where multi posting is acceptable but not what you've done in this thread.  And could you also try to cut down on the extra posts like where you just posted a smiley?

Thanks

I'm afraid I don't quite understand the problem :(.  But I found some somewhat interesting results looking at this row by row or column by column...

I have a proof but it is a bit brute forced. I see that once you construct a square which can be done in only two ways, 3 -1's and a 1 or 3 1's and a -1, not counting the permutations of the entries, then you can begin constructing the larger square by just moving each row up two spaces and each column to the right two spaces (assuming you start in the lower left corner). However, this can be broken down into several cases since after shifting the bottom two entries up two rows you can then form the opposite type of square. For example: Consider the 2x2 square in the bottom left corner with top entries {1,1} and bottom entries {1,-1}. After shifting the {1,-1} to the third row from the bottom you can then make the fourth row either {1,1} as I stated in my original construction or {-1,-1}. These cases also follow for shifting the entries in the columns.


ARML AMC AIME email USA(J)MO USAMO algorithm

does anyone know how to join the ARML team for GA? such as how to qualify. i cant find any info, so if someone could give me a link or something, that would be very helpful.

Well, basically you have to participate in some math competitions throughout the year.  Mr. Sigur, coach of the GA ARML team, is usually there at every one of them and takes note of who's winning what.  AMC and AIME are the biggies.  At the end of the year, he invites potential team members to the GCTM state tournament, at the end of which the ARML members are announced.  If you want to get in touch with him now, his email is s.sigur@comcast.net .

So do you get invited to go to ARML? Or do the top GCTM scorers become ARML people? Or should I just ask Mr. Sigur?  :P

My teacher, who is one of the ARML coaches, says that the ARML people are the people with the top 15 USAMO indices, plus the top 15 finishers at Varsity State.  Obviously, there is quite a bit of overlap, so then the rest of the team is decided without a set algorithm (she says it's a bit of a brouhaha), which, I expect, would be what lyf3xb3xsw33t was referring to.

What I heard from Mr. Sigur was the top 15 people at the State Varsity tournament are the A team, and then they talk to people who have beaten people who got on the A team for the B team. Technically, they are not supposed to use the results of non-GA competitions to determine the ARML team, but you never know.

[quote="Boy Soprano II"]My teacher, who is one of the ARML coaches, says that the ARML people are the people with the top 15 USAMO indices, plus the top 15 finishers at Varsity State.  Obviously, there is quite a bit of overlap, so then the rest of the team is decided without a set algorithm (she says it's a bit of a brouhaha), which, I expect, would be what lyf3xb3sw33t was referring to.[/quote]

Is the top 15 USAMO indices AMC10 indices also, or just AMC12?

How many people get on a GA ARML team?

im guessing no one comes by and checks this post, but ill ask my question under this topic anyway, i know for states like california and colorado (as far as i understand it) that the ARML practices are open to anyone who is interested in joining or just improving. is there a prac like that in GA for ARML or some kind of prac for ppl who just want to improve themselves? oh yeah, thanks for the responses to my first question!

@Santhosh (lol whoever you are):

How do you plan to study for ARML, AMC, etc.?

[quote="Boy Soprano II"]Santhosh—as you have the highest AMC 10 score in the state, I wouldn't worry, especially if you're coming to varsity state.  Keep in mind that the most USAMO qualifiers GA has ever had (since 1999, anyhow) was eight.  An ARML team is 15, and GA sends two.  Besides, Vova was on GA ARML team last year.

I don't know about the practices being open to anyone.  From what my teacher said, they rotate the venue each week to make it more balanced travel-wise, so I kind of doubt it, but I don't know.[/quote]

I'm not coming to Varsity State as of now. My school's last tournament is tommorrow and that is the trimathelon: an excuse to hand out awards and eat pizza.

Then again: If Vova got on the ARML team, I'm sure I could.

To answer Karth's question, I'm not sure. I don't study for the AMC10. For the AMC12, I do a few practice tests. For the ARML, I do a few of last years problems.

@ Santosh:

Did you go to the Cobb County Invitational tournament and get 1st place? (If so congrats  :) ) I just wanted to know if I knew the right Santosh  :D

>_< vova went to my old school. not bad for a little freshman, i must say.
anyways sort of a correction to the previous posts, the A and B teams aren't decided until like a night or two before the actual ARML competition, so don't even worry about those now. they'll split you into groups for the practices, but theyre mostly random. to get on the team in the first place, you gotta impress the coaches (or be on the team before) top scores on AMC and varsity state help, but aren't any guarantee, because they also take into account things like age and experience and things like that. but most important is probably the varsity state competition, unless you've been on the team for a few years and have a reaaaaaally good excuse not to go.

Hopefully Wheeler qualifies for the Varsity State Tournament.
The top 10 from the Lassiter Invitationals are probably going according to Mr. Slater.
I forgot what Wheeler got.

W00T!
[quote]
Because of our outstanding* season, Wheeler's JV Math Team has been invited to participate in the State Tournament at Mercer University in Macon GA.
[/quote]

*By outstanding we mean getting totaly pwned by Miles... I mean Lassiter.

So I guess I am going after all.

BTW: Is this a Varsity level tournament or a JV?

[/quote]

oooh, yay, congrats!

Yeah, this one's varsity level.  :P

One of my friends that is going said that I was wrong about JV state being last month and that this was the JV state tournament.

He also got a 16 out of 250 on the Cobb County Fall Tournament.

:lol:

I dunno...I'm pretty sure the Varsity state tournament is this month.  April 29th, to be exact.

Yes, Linda, it is indeed that date. :) I'm looking forward to it quite eagerly, it being the last competition of the year and all. But I haven't gotten an official invitation...

I'm looking foreward to it too.
Finally, a state tournament I actually know about before it occurs.

What is the format of the tournament?

Wait, so the top 10 people from Lassiter go to Varsity State? And the top 10 people there are on ARML?  :huh:

The way invitations work is for each tournament a team attends, it's "score" for that tournament is calculated by taking the number of teams that attended the tournament and dividing by your rank.

Ex. Wheeler's Varsitiy team placed 6th out of 11 teams at the Spring tournament. So it's "score" is: $\frac{11}{6}\approx1.83$ Your highest "score" at any one tournament is what is used to determine who gets invited.

According to Mr. Slater, Invitations go out to the top 50 or so teams in the state. He says that the cuttoff is a little less than 4 normally.

Since there were about 40  teams at the Lassiter Invitational, the top 10 will probably be invited. Your school's team can get a "score" at any tournament.

I just reffered to the Lassiter Invitational since that was our Varsity teams best chance to get an invitation. (I had no clue that JV teams could be invited.)

This is all according to what Mr. Slater told me at the Trimathelon. He could be wrong though.

BTW: a search on Google for "GA ARML" gives this thread as the #1 result and Billy's Blog as the #2 result.

Got my official invitation now via email. And I got all of the good math people's email addresses :D :P. Now I have bunches of people's emails.

How did you get all the good math people's e-mail addresses?

Also, Does anyone know where I can find previous year's tests?

Mr. Sigur sent the invitation to all us individuals (I think) together, so all the people's emails were on the Recipients list.

And there are some old ARML problem books; you can see some of them on the link at chess64's blog.

Our score at Furman was a 7. If only Furman were in GA . Next year....

I mean for the Varsity state tournament.

I guess 138 emails are not enough for you.

I have a copy of the 2005 Varsity State test... it's really easy  :|  I scored 47/50 in 90 minutes  :D

EDIT: Woohoo! Our school got invited (for the first time, because our varsity was really bad last year  :) )

Where can I find a copy of the 2005 Varsity State test  (or other Varsity tests for that matter)?  If it's nowhere online, mind scanning me a copy?

Yeah I got invited to this thing too.  Werdly unexpected, since I'm an alien :ninja: .  Another kid from my school got invited too...so hopefully we'll have a fun time at Mercer!

I'm more in tune with Canadian Mathematics contests, so I'm not sure how the ARML works.  Can anyone tell me about it? :oops:

Oh, this is kinda off topic, but did any of you guys make ghp?

Hmm, well, Xantos, Billy and I are freshmen :) .

[url=http://www.arml.com/]ARML[/url] is pretty much a national version of a math tournament, held at three locations simultaneously.  Team selection varies highly from region to region (you can read about it in the ARML forum), and I think most of the several opinions on how the GA team is selected have been voiced in this thread.  I can't tell you much about the problems, as I haven't looked at them much, but they seem perhaps something like AIME?  From what I hear, it's a lot of fun.

(Wow, this is the first thread in the GA forum to attain more than one page.)


trigonometry
geometry
AMC
AIME
angle bisector
trig identities
Law of Sines

Hey everyone!

Here is the anger venting medium where we post the stupid mistakes that we made on the Mock AMC 11B and get angry about them!!!  ;)

#4: Thought that a 3 or 5 digit number could have 0 for its first digit.

#8: Used the radius as 5 instead of 25.

GAAAAAAAAAHH!!!!  :twisted:

12 thought it said m+n instead of mn :-x

18.  Thought that cos pi = 1.  So I got an answer that wasn't one of the choices.  At the last minute I figured out that I had that mistake and fixed it.

1.  Thought the average of 19 and 19 was 19.5

2. I thought the solution to 50x-1=0 is .2...

1st mistake.  Doing the contest at all...lol

missed4 times the reciprical in number 2.

didn't pay enough attention on 3-5.

couldn't read on 7&8

Shouldn't have even attempted 9 and 10.

Well at least I tried and I did better than I expected to.[/i]

Ah I'm so stupid  :blush:

7.  I forgot that 6*1 = 6 AAHHH SO STUPID

Somehow missed #1, I think I forgot the definition of mode

can someone explain #16 to me... because i did law of sines and then used angle bisector and absinC/2 triangle area formula and i got [tex]36\sqrt3+108[/tex]

I forgot THE DEFINITION OF MEDIAN ON #1, I figured out the average.

#12: somehow thought that the set included all number 3-2005 :blush:
got an answer that wasn't any of the choices, so i picked e.  turns out i got it right.   :cool:

#4: didn't realize that 0 could be a digit

#16: [tex]$sin(75)=\sqrt2(sqrt3+1)$[\tex]

#18: the lines from the fly to the wall of length 8 and 1 must be perpendicular to their respective wall faces?

Oh yeah...I typed the wrong letter for one of my answers. Would have "qualified" then too. Oh well.

1. For a while I was looking for the mean, and not the mode.

8.Thought it said [b]inside[/b] and not [b]on[/b] the circle.

[quote="takuanitromars36"][quote="yif man12"]Please don't type one word so long. It makes the line not scroll to a new line on my computer and makes it really annoying to view. Since it's one word, it doesn't scroll and just extends the whole window.[/quote]

I sincerely apologize. Sorry. I've noticed that my window also is very long.[/quote]

No problem, man. Just edit your post and cut it down to reasonable size.  ;)

8. Forgot (25,0) is also a coordinate.

22, 23, 24 ,25 <- mistake for not attempting them.

I haven't ever taken them before. My first time will be next year, or this fall depending on when the actual test is going to be.


triangle ABC (AB=AC) ,
point M inside the triangle satisfying angle MBC=angle MCA . N is the midponit of segment BC , prove that angle BMN + angle AMC $ \equal{} 180^{o}$

Let $ O$ be the circumcenter of triangle $ MBC$
     $ D$ be the intersection of line $ AM$ and the circumcircle of triangle $ MBC$
We have $ \frac{BM}{BD}\equal{}\frac{AB}{AD}$
$ \frac{MN}{NC}\equal{}\frac{MN}{OD}.\frac{OC}{NC}\equal{}\frac{AN}{AD}.\frac{AC}{AN}\equal{}\frac{AC}{AD}\equal{}\frac{AB}{AD}$
So $ \frac{BM}{BD}\equal{}\frac{MN}{NC}$
Then $ BMD$ and $ NMC$ are similar triangles
So $ \angle BMD\equal{}\angle CMN$
Thus $ \angle BMN\equal{}\angle CMD\equal{}180\minus{}\angle AMC$

i have another solution by using the lemma
triangle ABC , M and N on the segment BC , angle NCA = angle MAB  $ \leftrightarrow\frac{MB.NB}{MC.NC}\equal{} (\frac{AB}{AC})^{2}$

sure , here is my solution
let K be the intersection of AM and BC , S and T on the segment BC satisfying MS parallel to AB and MT parallel to AC ,
first , we see that BSM and BMC are similar triangles and so are CTM and CMB
so we obtain $ MB^{2}\equal{} BS.BC and MC^{2}\equal{}TC.BC$
so $ (\frac{MB}{MC})^{2}\equal{}\frac{BS}{TC}\equal{}\frac{KS}{KT}\equal{}\frac{KB}{KC}$
applying the above lemma in triangle MBC , we have angle BMS = angle KMC -->QED

See that $AM$ is the symmedian of $\Delta BMC$ hence, if $P=BC\cap AM$, then $\angle PMC=\angle BMN$ and we are done!

Best regards,
sunken rock

As shown in the figure.

Extend PM to F such that
MF=PM
E is a point on AC such that ED//AB
G is the intersection of BF and CP

It's obvious
AE=BD

△ADE∽△BAP
so
AE/BP=DE/AP
         =DE/BA×AB/AP
         =AD/CB×AB/PB×PB/AP
         =PD/GB×BD/PD×PB/BF
         =PD/GB×BD/PD×PB/AP
         =BD/GB×PB/BF
Thus
$PB^2=BG\times BF$
△PBG∽△PAF
∠BPG=∠APF

Namely
∠APM+∠BPC=180


factorial
floor function
number theory proposed
number theory

Prove that for every composite number $ n>4$, numbers $ kn$ divides $ (n\minus{}1)!$ for every integer $ k$ such that $ 1\le k\le \lfloor \sqrt{n\minus{}1} \rfloor$.

Well, we know that n can be expressed as $p*q$ where $p>=\sqrt{n}>=q>=0$. If $k$ is not equal to $q$ then the statement is obviously true. If $k=q$, then we must have replace $q$ wit some multiple of $q$. Since we have that $q<\sqrt{n}$ there are atleast $q-1$ other multiples that work.
Case 1: $q$ is not equal to 2. Then the result is obvious since we are guaranteed that we won't select $q$ for the next choice
Case 2: $q=2$. Then since n>4 we always have a multiple of 2 less than $\sqrt{n}$.


Let $ D$ be a point on the side $ AC$ of triangle $ ABC$.Let $ E$ and $ F$ be points on the segments $ BD$ and $ BC$,respectively,such that $ \widehat{BAE}\equal{}\widehat{CAF}$.Let $ P$ and $ Q$ be points on $ BC$ and$ BD$.respectively,such that $ {EP}\parallel{DC}$ and $ {FQ}\parallel{CD}$.Prove that $ \widehat{BAP}\equal{}\widehat{CAQ}$

[color=indigo][size=117][b]This nice problem is an interesting and instructive application of the [u]Steiner's theorem[/u][/b] ![/size][/color]

[quote="apollo"][color=darkred][b]In $ \triangle ABC$ consider a point $ D\in (AC)$. For a point $ E\in (BD)$ construct the points $ \{\begin{array}{ccc}F\in (BC) & , &\widehat{BAE}\equiv\widehat{CAF}\\ \\ P\in (BC) & , & EP\parallel AC\\ \\ Q\in (BD) & , & QF\parallel AC\end{array}$ $ \implies$ $ \widehat{BAP}\equiv\widehat{CAQ}$.[/b][/color][/quote]
[color=darkblue][b]Proof.[/b] Denote the intersections $ \{\begin{array}{c}S\in AF\cap BD\ .\\ \\ R\in AQ\cap BC\ .\end{array}$ Observe that $ \{\begin{array}{c}EP\parallel AC\implies\frac{PB}{PC}=\frac{EB}{ED}\ .\\ \\ QF\parallel AC\implies\frac{FB}{FC}=\frac{QB}{QD}\ .\end{array}$ Apply the [b]Menelaus' theorem[/b] in $ \triangle BCD$ :

$ \|\begin{array}{cccc}\overline{ASF}\blacktriangleright &\frac{AC}{AD}\cdot\frac{SD}{SB}\cdot\frac{FB}{FC}= 1 &\implies &\frac{AC}{AD}\cdot\frac{SD}{SB}\cdot\frac{QB}{QD}= 1\ \ (1)\ .\\ \\ \overline{AQR}\blacktriangleright & & &\frac{AC}{AD}\cdot\frac{QD}{QB}\cdot\frac{RB}{RC}= 1\ \ (2)\ .\end{array}$ Apply the [b]Steiner's theorem[/b] for the isogonal rays $ [AE$, $ [AS$ in $ \triangle ABD\ :$

$ \frac{EB}{ED}\cdot\frac{SB}{SD}=(\frac{AB}{AD})^{2}\implies$ $ \frac{PB}{PC}\cdot\frac{SB}{SD}=(\frac{AB}{AD})^{2}\ \ (3)\ .$ The product of the $ (1)$, $ (2)$, $ (3)$ $ \implies$ $ \frac{PB}{PC}\cdot\frac{RB}{RC}=(\frac{AB}{AC})^{2}$ $ \implies$ $ \widehat{BAP}\equiv\widehat{CAQ}\ .$

In the last implication I use reciprocally the [b]Steiner's theorem[/b].

[b]Remark.[/b] See an extension of this problem at http://www.mathlinks.ro/Forum/viewtopic.php?t=167147 [/color]

[hide="My solution"]draw $ XE\parallel AQ$ then draw $ XP\because\widehat{BAE}\equal{}\widehat{FAC}$. assume $ \widehat{BAE}\equal{}\alpha\equal{}\widehat{FAC}$
$ \because AC\parallel QF\therefore\widehat{CAF}\equal{}\widehat{QFA}\equal{}\alpha$
from $ XE\parallel AQ\therefore\widehat{XED}\equal{}\widehat{AQD}...................(1)$
,from $ EP\parallel QE\therefore\widehat{PED}\equal{}\widehat{FQD}...................(2)$
$ (1)\plus{}(2) ;\widehat{XED}\plus{}\widehat{PED}\equal{}\widehat{AQD}\plus{}\widehat{FQD}$
$ \therefore\widehat{XEP}\equal{}\widehat{AQF}................................................(3)$
assume $ \widehat{EAP}\equal{}Z,\widehat{FAQ}\equal{}Z'$
from $ XE\parallel AQ$,$ EP\parallel QE$ and from $ (1),(2)$
we have $ \triangle{XEP}\sim\triangle{AQF}\therefore\widehat{XPE}\equal{}\widehat{AFQ}\equal{}\alpha\therefore\widehat{XPE}\equal{}\alpha$ from $ \widehat{XAE}\equal{}\alpha\therefore\widehat{XPE}\equal{}\widehat{XAE}\equal{}\alpha\therefore XAPE$ is concyclic
from $ \widehat{EXP}\equal{}\widehat{PAE}\therefore Z\equal{}Z'\therefore\widehat{EAP}\equal{}\widehat{FAQ}$
from $ \widehat{BAE}\equal{}\alpha\equal{}\widehat{FAC}\therefore\widehat{EAP}\plus{}\widehat{BAE}\equal{}\widehat{FAQ}\plus{}\widehat{FAC}\therefore\widehat{BAP}\equal{}\widehat{CAQ}$[/hide]


[b]
[hide]
Here is a little hint:
Notice that $6860=6859+1=19^3+1^3$[/hide][/b]


Συμφωνα με οσα εχουμε διδαχτει, το ανωτερο οριο μιας ακολουθιας ειναι το μεγιστο απο τα ορια ολων των υπακολουθιων της.
Στο βιβλιο της Αναλυσης ομως, βρηκα χτες τον παρακατω συμβολισμο τον οποιο δεν μπορω να καταλαβω:
Αν $ l$ ειναι το ανωτερο οριο της ακολουιας $ a_n$, τοτε: $ l \equal{} \inf_{n \in N}{\sup_{k \geq n}{a_k}}$
Γνωριζω πολυ καλα την εννοια των supremum & infimum, αλλα αυτο που καταλαβαινω απο τον παραπανω συμβολισμο ειναι οτι το ανωτερο οριο της ακολουθιας ειναι το μεγιστο κατω φραγμα του συνολου που περιεχει τα ελαχιστα ανω φραγματα των συνολων $ \{a_n, a_{n\plus{}1}, ...\}, n \in N$, πως σχετιζεται αυτο ομως με τον ορισμο που εχουμε διδαχτει για το ανωτερο οριο (οτι δηλαδη ειναι το μεγαλυτερο απο τα ορια των υπακολουθιων)???

[quote="NickNafplio"][quote="Papadopoulos"][quote="NickNafplio"]Συμφωνα με οσα εχουμε διδαχτει, το ανωτερο οριο μιας ακολουθιας ειναι το μεγιστο απο τα ορια ολων των υπακολουθιων της.
Στο βιβλιο της Αναλυσης ομως, βρηκα χτες τον παρακατω συμβολισμο τον οποιο δεν μπορω να καταλαβω:
Αν $ l$ ειναι το ανωτερο οριο της ακολουιας $ a_n$, τοτε: $ l \equal{} \inf_{n \in N}{\sup_{k \geq n}{a_k}}$
Γνωριζω πολυ καλα την εννοια των supremum & infimum, αλλα αυτο που καταλαβαινω απο τον παραπανω συμβολισμο ειναι οτι το ανωτερο οριο της ακολουθιας ειναι το μεγιστο κατω φραγμα του συνολου που περιεχει τα ελαχιστα ανω φραγματα των συνολων $ \{a_n, a_{n \plus{} 1}, ...\}, n \in N$, πως σχετιζεται αυτο ομως με τον ορισμο που εχουμε διδαχτει για το ανωτερο οριο (οτι δηλαδη ειναι το μεγαλυτερο απο τα ορια των υπακολουθιων)???[/quote]

Πρόκειται για δύο ισοδύναμους ορισμούς. Ο δεύτερος λέει
Εξετάζουμε την ακολουθία $ A_n \equal{} \sup_{k \geq n}{a_k} \equal{} sup\{a_n, a_{n \plus{} 1},a_{n \plus{} 2}, ...\}$
 Μετά βρίσκουμε το L = inf{$ A_1, A_2, A_3, ...$} (το οποίο ισούται και με το
lim$ A_n$ καθώς η ($ A_n$) είναι φθίνουσα).
 Αποδεικνύεται ότι το L αυτό ισούται με τον αριθμό που προκύπτει από τον πρώτο ορισμό.
Υπάρχουν και άλλοι ισοδύναμοι ορισμοί. Π.χ. το ανώτερο όριο μιας ακολουθίας
ισούται με εκείνον τον αριθμό λ για τον οποίο για κάθε ε >0 υπάρχουν άπειροι
όροι της ακολουθίας που είναι μεγαλύτεροι από λ - ε και πεπερασμένοι που είναι μεγαλύτεροι από λ+ε.

Αυτά μπορείς να τα βρεις σε προχωρημένα βιβλία Ανάλυσης.  Π.χ. Bartle, The Elements of Real Analysis.

καλό διάβασμα,

Τάκης Π.[/quote]

Ευχαριστω κυριε Παπαδοπουλε
Αυτο που θελω να βρω ειναι μερικες αποδειξεις που να συνδεουν αυτους τους ορισμους, γιατι δεν θα ηθελα απλα να αποστηθισω τους ορισμους απο το βιβλιο της Αναλυσης 1 χωρις να γνωριζω πως συνδεονται αυτοι. Απο προχωρημενα βιβλια Αναλυσης, αγορασα προσφατα αυτο του Luis Brand της ΕΜΕ, εχω ακουσει τα καλυτερα γι αυτο το βιβλιο και με μια ματια που του εριξα μου εδωσε την εντυπωση οτι ειναι πραγματικα παρα πολυ καλο, αλλα λιγο που κοιταξα το κεφαλαιο με τα ορια των ακολουθιων δεν βρηκα κατι σχετικο με τα ανωτερα ορια.[/quote]

Έχω την εντύπωση ότι το βιβλίο του Brand είναι κάπως πρακτικό παρά θεωρητικό, ταιριάζει νομίζω σε σπουδαστές Πολυτεχνείου παρά σε μαθηματικούς. Πάντως καλύπτει σε σχετικά μικρό όγκο μεγάλη πληθώρα θεμάτων αλλά πολύ δύσκολα θα το χαρακτήριζα πλήρες

[quote="NickNafplio"]Αυτο που θελω να βρω ειναι μερικες αποδειξεις που να συνδεουν αυτους τους ορισμους, γιατι δεν θα ηθελα απλα να αποστηθισω τους ορισμους απο το βιβλιο της Αναλυσης 1 χωρις να γνωριζω πως συνδεονται αυτοι. Απο προχωρημενα βιβλια Αναλυσης, αγορασα προσφατα αυτο του Luis Brand της ΕΜΕ, εχω ακουσει τα καλυτερα γι αυτο το βιβλιο και με μια ματια που του εριξα μου εδωσε την εντυπωση οτι ειναι πραγματικα παρα πολυ καλο, αλλα λιγο που κοιταξα το κεφαλαιο με τα ορια των ακολουθιων δεν βρηκα κατι σχετικο με τα ανωτερα ορια.[/quote]

Προσωπικά, θα σου πρότεινα να (προσπαθήσεις τουλάχιστον να) αποδείξεις μόνος σου τα παραπάνω πριν στραφείς στη βιβλιογραφία. Δεν είναι ιδιαίτερα δύσκολο και θα σου δώσει σημαντικά insights στο τί σημαίνει $ \limsup,\liminf$.

Cheerio,

Durandal 1707

Edit: Ως παράδειγμα, δοκίμασε επίσης να εφαρμόσεις τους ορισμούς και να καταλήξεις στην ισοδυναμία τους για μία ακολουθία που απαριθμεί τους ρητούς του $ [0,1]$. Εκεί, τα limes extrema λάμπουν :lol:

[quote="NickNafplio"]
Αυτο που θελω να βρω ειναι μερικες αποδειξεις που να συνδεουν αυτους τους ορισμους, γιατι [color=red]δεν θα ηθελα απλα να αποστηθισω τους ορισμους[/color] απο το βιβλιο της Αναλυσης 1 [color=red]χωρις να γνωριζω πως συνδεονται[/color] αυτοι. [/quote]

 Συγχαρητήρια. Η άποψη αυτή, δηλαδή η επιθυμία της κατανόησης των ιδεών και της εύρεσης εσωτερικής συνάφειας μεταξύ τους, είναι ότι πιό υγιές στα Μαθηματικά.

 Ειλικρινά, στο κλείσιμο του 2008 μου έδωσες, μεγάλη χαρά. Δεν υπάρχει μεγαλύτερη ικανοποιήση σε έναν δάσκαλο από το να βλέπει την νέα γενιά να έχει φιλομάθεια, επιστημονική περιέργεια και διάθεση να σκέπτεται.

Τάκης Παπαδόπουλος

[quote="r_boris"]
Έχω την εντύπωση ότι το βιβλίο του Brand είναι κάπως πρακτικό παρά θεωρητικό, ταιριάζει νομίζω σε σπουδαστές Πολυτεχνείου παρά σε μαθηματικούς. Πάντως καλύπτει σε σχετικά μικρό όγκο μεγάλη πληθώρα θεμάτων αλλά πολύ δύσκολα θα το χαρακτήριζα πλήρες[/quote]

Συμφωνώ απόλυτα με τον κ. Μπόρη.
Κατά τα άλλα συμβουλεύω να πάρεις τα πράγματα με τη σειρά, χωρίς βιασύνη.

1) Ξεκίνημα με τον Brand ή άλλο παρεμφερές: Ο Brand είναι καλό εισαγωγικό βιβλίο στον Απειροστικό. Έχει μεγαλύτερη έμφαση στις εφαρμογές παρά στη θεωρία.

[b]Μετά[/b] τον Brand μετάβαση σε ένα πιο προχωρημένο και λίγο πιο ανεβασμένο βιβλίο:

2) Μου αρέσει ιδιαίτερα ο Spivak, [i]Απειροστικός Λογισμός[/i], Παν/κες Εκδόσεις Κρήτης. Η θεμελίωση που κάνει είναι αυστηρή, αλλά το ύφος γλαφυρό. Ένας καλός πρωτοετής φοιτητής του Μαθηματικού οφείλει, στο τέλος της χρονιάς, να ξέρει τον Spivak απέξω και ανακατωτά. Παράλληλα μπορεί να διαβάζει τον [i]Απειροστικό[/i] του Νεγρεπόντη. Πρόκειται για πιο δύσκολο βιβλίο, που έχει "τα πάντα" στον Απειροστικό μιας μεταβλητής.

Μετά είναι έτοιμος κανείς να κάνει απρόσκοπτα το άλμα σε πιο προχωρημένα βιβλία Ανάλυσης.

3) Βιβλία όπως του Rudin ή του Bartle ή οποιδήποτε άλλο βιβλίο [i]Ανάλυσης[/i], είναι καλό. Τα περί limsup, liminf υπάρχουν σε αυτό το επίπεδο. Τέτοια βιβλία συνήθως τα διαβάζουν μόνο οι Μαθηματικοί ή οι Θεωρητικοί Φυσικοί.

Τάκης Π.

[quote="Papadopoulos"]3) Βιβλία όπως του Rudin ή του Bartle ή οποιδήποτε άλλο βιβλίο [i]Ανάλυσης[/i], είναι καλό. Τα περί limsup, liminf υπάρχουν σε αυτό το επίπεδο.[/quote]

+1 για τον Rudin! Είναι καταπληκτικό βιβλίο: τόσο ο μικρός όσο και ο μεγάλος Rudin θεωρούνται (δικαίως) βίβλοι :)

Επίσης, από καθαρά μαθηματικό standpoint, το Mathematical Analysis του Browder είναι πολύ καλό και, σε δεύτερο επίπεδο, το κλασικό Introductory Real Analysis των Kolmogorov-Fomin (μαζί με τον μεγάλο Rudin).

Spivak -> Browder/Rudin -> Kolmogorov etc. είναι μια πολύ ωραία πορεία :)

[quote]Τέτοια βιβλία συνήθως τα διαβάζουν μόνο οι Μαθηματικοί ή οι [color=red]Θεωρητικοί Φυσικοί[/color].[/quote]

Ha :( Μακάρι να το έκαναν, αλλά η νόρμα δύστυχως σίγουρα δεν το κάνει... :(

Καλή χρονιά παιδιά :)

Durandal 1707

[quote="Durandal"]
[quote]Τέτοια βιβλία συνήθως τα διαβάζουν μόνο οι Μαθηματικοί ή οι [color=red]Θεωρητικοί Φυσικοί[/color].[/quote]

Ha :( Μακάρι να το έκαναν, αλλά η νόρμα δύστυχως σίγουρα δεν το κάνει... :(

[/quote]

Καλή Χρονιά.

Έχεις δίκιο.

Αυτοί που διαβάζουν τέτοια βιβλία είναι [b]δυστυχώς[/b] γνήσιο υποσύνολο των Μαθηματικών  :mad:  ή των Θεωρητικών Φυσικών :huh: . Για τους δεύτερους θα έλεγα ότι είναι γνήσιο-γνήσιο υποσύνολο. :roll:

Λοιπον, αν και δεν εχω εξασκηθει πολυ με εψιλοντικους ορισμους, νομιζω οτι καταφερα να βρω μια αποδειξη για την ισοδυναμια των ορισμων και αλλη μια για την ισοτητα των supinf & infsup οταν η ακολουθια συγκλινει. Γραφω εδω τις αποδειξεις και θελω να μου πει καποιος αν ειναι σωστες (Εχω αμφιβολιες επηδη μου φαινονται αρκετα απλες για να ειναι σωστες).

Εστω $ L$ το ανωτερο οριο ακολουθιας $ (a_n)$ και $ a = \lim_{n\rightarrow + \infty}{a_n}$
Ειναι $ L = \inf{A_n} = \lim_{n\rightarrow + \infty}{A_n}$ με $ A_n = \sup{\{a_k, k \geq n\}}:$ γνησιως φθινουσα ακολουθια.

Εστω $ a_{k_n}$ τυχαια υπακολουθια της $ a_n$, με ${ \lim_{n\rightarrow + \infty}{a_{k_n}}} = l$, για την ισοδυναμια των 2 ορισμων αρκει νδο $ l \leq L$ (για το κατωτερο οριο εργαζομαστε αναλογος). Ειναι $ a_{k_n} \leq \sup{\{a_{k_n}, a_{k_n + 1}, ...\}} = A_{k_n}$, ομως η ακολουθια $ k_n$ ειναι γνησιως αυξουσα ακολουθια φυσικων αριθμων, αρα αποκλινει, οποτε παιρνοντας ορια για $ n \rightarrow + \infty$ στην προηγουμενη ανισοτητα παιρνουμε: $ l \leq L$ και τελειωσαμε.

Τωρα θα δειξω οτι αν η ακολουθια $ (a_n)$ συγκλινει στο $ a$, τοτε $ a = L$ (για το κατωτερο οριο εργαζομαστε αναλογος). Ειναι $ A_n = sup{\{a_n, a_{n + 1}, ...\}}$, απο τον ορισμο του ελαχιστου ανω φραγματος μιας ακολουθιας (θεωρουμε την ακολουθια $ b_m = a_{n + m})$), εχουμε οτι για καθε $ \epsilon > 0$, υπαρχει $ k > n$, ωστε $ A_n - \epsilon \leq a_k \leq A_n$, επηδη ομως $ k > n$, το $ k$ αυξανεται απεριοριστα καθος το $ n$ αυξανεται απεριοριστα, οποτε παιρνοντας ορια εχουμε οτι για καθε $ \epsilon > 0$ ειναι: $ L - \epsilon \leq a \leq L$, απο οπου ευκολα προκειπτει $ a = L$. Αντιστροφα, αν το ανωτερο και το κατωτερο οριο της ακολουθιας $ (a_n)$ ειναι μεταξυ τους ισα και ισα με $ a$, τοτε απο τον ορισμο του ανωτερου/κατωτερου οριου ευκολα προκειπτει οτι καθε υπακολουθια της $ (a_n)$ συγκλινει στο $ a$, οποτε εφαρμοζοντας τον ορισμο του οριου για καθε υπακολουθια ευκολα εχουμε οτι και η ακολουθια $ (a_n)$ συγκλινει στο $ a$.

Θα ηθελα να με διορθωσετε αν εχω καπου καποιο λαθος, και αν μπορει καποιος να μου εξηγησει τι ακριβως θα μπορουσα να κανω με την ακολουθια που απαριθμει τους ρητους στο [0,1] που ανεφερε ο Durandal γιατι εχω μπερδευτει εκει...

Ευχαριστω.

[quote="NickNafplio"]Λοιπον, αν και δεν εχω εξασκηθει πολυ με εψιλοντικους ορισμους, νομιζω οτι καταφερα να βρω μια αποδειξη για την ισοδυναμια των ορισμων και αλλη μια για την ισοτητα των supinf & infsup οταν η ακολουθια συγκλινει. Γραφω εδω τις αποδειξεις και θελω να μου πει καποιος αν ειναι σωστες (Εχω αμφιβολιες επηδη μου φαινονται αρκετα απλες για να ειναι σωστες).

Εστω $ L$ το ανωτερο οριο ακολουθιας $ (a_n)$ και $ a = \lim_{n\rightarrow + \infty}{a_n}$
Ειναι $ L = \inf{A_n} = \lim_{n\rightarrow + \infty}{A_n}$ με $ A_n = \sup{\{a_k, k \geq n\}}:$ γνησιως φθινουσα ακολουθια.

Εστω $ a_{k_n}$ τυχαια υπακολουθια της $ a_n$, με ${ \lim_{n\rightarrow + \infty}{a_{k_n}}} = l$, για την ισοδυναμια των 2 ορισμων αρκει νδο $ l \leq L$ (για το κατωτερο οριο εργαζομαστε αναλογος). Ειναι $ a_{k_n} \leq \sup{\{a_{k_n}, a_{k_n + 1}, ...\}} = A_{k_n}$, ομως η ακολουθια $ k_n$ ειναι γνησιως αυξουσα ακολουθια φυσικων αριθμων, αρα αποκλινει, οποτε παιρνοντας ορια για $ n \rightarrow + \infty$ στην προηγουμενη ανισοτητα παιρνουμε: $ l \leq L$ και τελειωσαμε.

Τωρα θα δειξω οτι αν η ακολουθια $ a_n$ συγκλινει στο $ a$, τοτε $ a = L$ (για το κατωτερο οριο εργαζομαστε αναλογος). Ειναι $ A_n = sup{\{a_n, a_{n + 1}, ...\}}$, απο τον ορισμο του ελαχιστου ανω φραγματος μιας ακολουθιας (θεωρουμε την ακολουθια $ b_m = a_{n + m})$), εχουμε οτι για καθε $ \epsilon > 0$, υπαρχει $ k > n$, ωστε $ A_n - \epsilon \leq a_k \leq A_n$, επηδη ομως $ k > n$, το $ k$ αυξανεται απεριοριστα καθος το $ n$ αυξανεται απεριοριστα, οποτε παιρνοντας ορια εχουμε οτι για καθε $ \epsilon > 0$ ειναι: $ L - \epsilon \leq a \leq L$, απο οπου ευκολα προκειπτει $ a = L$. Αντιστροφα, αν το ανωτερο και το κατωτερο οριο ειναι μεταξυ τους ισα με $ a$, τοτε απο τον ορισμο του ανωτερου/κατωτερου οριου, ευκολα προκειπτει οτι καθε υπακολουθια της $ (a_n)$ συγκλινει στο $ a$, οποτε εφαρμοζοντας τον ορισμο του οριου για καθε υπακολουθια ευκολα εχουμε οτι και η ακολουθια $ (a_n)$ συγκλινει στο $ a$.

Θα ηθελα να με διορθωσετε αν εχω καπου καποιο λαθος, και αν μπορει καποιος να μου εξηγησει τι ακριβως θα μπορουσα να κανω με την ακολουθια που απαριθμει τους ρητους στο [0,1] που ανεφερε ο Durandal γιατι εχω μπερδευτει εκει...

Ευχαριστω.[/quote]

Η απόδειξη είναι απολύτως σωστή.

Θέλω να σε συγχαρώ γιατί από ότι αντιλαμβάνομαι είσαι φοιτητής Πολυτεχνικής Σχολής που μπήκε στα χωράφια του Μαθηματικού, σε θέματα που δεν διδάχθηκες και που είναι δύσκολα. Αυτό δείχνει ότι έχεις σπάνιο Μαθηματικό ταλέντο.

Συγχαρητήρια και πάλι.

Τάκης Παπαδόπουλος.

[quote="NickNafplio"][quote="Papadopoulos"]
Η απόδειξη έχει κάποιες ατέλεις, αλλά μπορεί να διορθωθεί (χρειάζεται όμως κάποια δουλειά). Το κύριο ελάττωμα είναι ότι υπέθεσες (βλέπε το σημειωμένο με κόκκινο) ότι η $ a_n$ συγκλίνει, πράγμα που χρησιμοποίησες ουσιαστικά (βλέπε σημειωμένο με πράσινο).

Όταν μιλάμε για limsup, liminf εξετάζουμε γενικές ακολουθίες (ή το πολύ πολύ φραγμένες). Για τις συγκλίνουσες τα πράγματα είναι "απλά" και οι τρεις αριθμοί
limsup$ a_n$ , liminf$ a_n$ και lim$ a_n$   είναι ίσοι.

Παρά τις ατέλειες της απόδειξης θέλω να σε συγχαρώ. Από ότι αντιλαμβάνομαι είσαι μαθητής Λυκείου που μπήκε σε χωράφια που δεν διδάχτηκες και που είναι δύσκολα. Αυτό δείχνει ότι έχεις σπάνιο Μαθηματικό ταλέντο.
Είμαι βέβαιος ότι αν θα ακούσεις συστηματικά περί σύγκλισης ακολουθιών σε ένα Πανεπιστημιακό επίπεδο, όλα αυτά θα σου φανούν παιχνιδάκια.

Συγχαρητήρια και πάλι.

Τάκης Παπαδόπουλος.[/quote]

Ευχαριστω για τα καλα λογια :) . Βασικα αυτο που γραφω δεν ειναι μια αποδειξη αλλα δυο διαφορετικες. Πρωτα αποδεικνειω την ισοδυναμια των 2 ορισμων, μετα (απο το σημειο που σημειωσατε με κοκκινο και κατω) γραφω μια αποδειξη σε μια βασικη ιδιοτητα των infsup & supinf η οποια αποδειξη μου ηρθε τυχαια ενω προσπαθουσα να βρω κατι αλλο. Στη 2η αποδειξη, δειχνω οτι αν η $ (a_n)$ συγκλινει σε καποιο οριο (ειτε πραγματικο ειτε απειρο) τοτε τα infsup, supinf ειναι ισα με το οριο, και μετα και το αντιστροφο. Νομιζω ομως, οτι αφου απο την υποθεση γνωριζουμε οτι υπαρχουν τα 3 ορια, μπορουμε αφοβα να παρουμε ορια στην ανισοτικη σχεση.

ΥΓ: Ειμαι πρωτοετης στο ΕΜΠ και οχι μαθητης Λυκειου :P[/quote]

Έχεις δίκιο. Θα διορθώσω όσα έγραψα.

Τ.Π.

Μετα απο αρκετες ωρες βρηκα κατι που χρειαζεται ωστε να ειναι η πρωτη απο τις 2 αποδειξεις που εβαλα παραπανω πιο ολοκληρωμενη. Επηδη ομως παλι εχω καποιες αμφιβολιες, βαζω εδω την αποδειξη ωστε να μου πειτε αν εχω πουθενα καποιο λαθος.

Επηδη η ακολουθια $ (a_n)$ μπορει να μη συγκλινει, για να πουμε οτι το infsup ειναι το μεγιστο απο τα ορια των υπακολουθιων, θα πρεπει να δειξουμε οτι υπαρχει υπακολουθια της $ a_n$ η οποια συγκλινει στο $ L$. Εστω οτι δεν ισχυει αυτο, τοτε αν $ L'$ ειναι το μεγιστο απο τα ορια των υπακολουθιων, ευκολα εχουμε $ a_n \leq L' < L \forall n \in N$, αρα για καθε $ n \in N$ εχουμε $ |a_n \minus{} L| \geq |L' \minus{} L|$. Αν η ακολουθια $ A_n \equal{} sup{\{a_n, a_{n \plus{} 1},...\}}$ σταθεροποιειτε για αρκετα μεγαλο $ n$, τοτε για αρκετα μεγαλο $ m$ ειναι $ L \equal{} A_m$, οποτε για καθε $ \epsilon > 0$ υπαρχει $ n_0 > m$ ωστε $ L \minus{} \epsilon < a_{n_0} < L \Rightarrow |a_{n_0} \minus{} L| < \epsilon$, και για $ \epsilon \equal{} |L' \minus{} L|$ εχουμε: $ |a_{n_0} \minus{} L| < |L' \minus{} L|$ το οποιο ειναι ατοπο, αρα σε αυτη τη περιπτωση υπαρχει υπακολουθια της $ (a_n)$ η οποια συγκλινει στο $ L$. Αν η ακολουθια $ A_n$ δεν σταθεροποιειτε για αρκετα μεγαλα $ n$ τοτε προφανος υπαρχουν απειρες θετικες διαφορες: $ A_n \minus{} A_{n \plus{} 1}$, εστω οτι αυτες ειναι οι $ A_{k_n} \minus{} A_{k_n \plus{} 1}, n \equal{} 1,2...$, εστω τωρα $ a_{b_k}$ ενας απο τους ορους της ακολουθιας $ (a_n)$ για τον οποιο ισχυει: $ A_{k \plus{} 1} < a_{b_k} < A_k, k \equal{} 1,2,...$ (ο $ a_{b_k}$ υπαρχει παντα, οπως προκειπτει αν παρουμε τον ορισμο του $ A_k$ με $ \epsilon \equal{} A_k \minus{} A_{k \plus{} 1} > 0$), η ακολουθια $ a_{b_k}$ τωρα ειναι υπακολουθια της $ a_n$ (μπορουμε να αριθμισουμε τις διαφορες $ A_k \minus{} A_{k\plus{}1}$ ετσι ωστε η ακολουθια $ b_k$ να ειναι γνησιως αυξουσα) και βρησκεται αναμεσα στις ακολουθιες $ (A_k), (C_k) \equal{} (A_{k \plus{} 1})$ οι οποιες συγκλινουν στο $ L$ αρα συγκλινει και αυτη στο $ L$, οποτε η αποδειξη ολοκληρωθηκε.

[quote="NickNafplio"]Μετα απο αρκετες ωρες βρηκα κατι που χρειαζεται ωστε να ειναι η πρωτη απο τις 2 αποδειξεις που εβαλα παραπανω πιο ολοκληρωμενη. Επηδη ομως παλι εχω καποιες αμφιβολιες, βαζω εδω την αποδειξη ωστε να μου πειτε αν εχω πουθενα καποιο λαθος.

Επηδη η ακολουθια $ (a_n)$ μπορει να μη συγκλινει, για να πουμε οτι το infsup ειναι το μεγιστο απο τα ορια των υπακολουθιων, θα πρεπει να δειξουμε οτι υπαρχει υπακολουθια της $ a_n$ η οποια συγκλινει στο $ L$. Εστω οτι δεν ισχυει αυτο, τοτε [color=red]αν[/color] $ L'$ [color=red]ειναι το μεγιστο απο τα ορια των υπακολουθιων, ευκολα εχουμε [/color]$ a_n \leq L'$< L $ \forall n \in N$, αρα για καθε $ n \in N$ εχουμε $ |a_n \minus{} L| \geq |L' \minus{} L|$. Αν η ακολουθια $ A_n \equal{} sup{\{a_n, a_{n \plus{} 1},...\}}$ σταθεροποιειτε για αρκετα μεγαλο $ n$, τοτε για αρκετα μεγαλο $ m$ ειναι $ L \equal{} A_m$, οποτε για καθε $ \epsilon > 0$ υπαρχει $ n_0 > m$ ωστε $ L \minus{} \epsilon < a_{n_0} < L \Rightarrow |a_{n_0} \minus{} L| < \epsilon$, και για $ \epsilon \equal{} |L' \minus{} L|$ εχουμε: $ |a_{n_0} \minus{} L| < |L' \minus{} L|$ το οποιο ειναι ατοπο, αρα σε αυτη τη περιπτωση υπαρχει υπακολουθια της $ (a_n)$ η οποια συγκλινει στο $ L$. Αν η ακολουθια $ A_n$ δεν σταθεροποιειτε για αρκετα μεγαλα $ n$ τοτε προφανος υπαρχουν απειρες θετικες διαφορες: $ A_n \minus{} A_{n \plus{} 1}$, εστω οτι αυτες ειναι οι $ A_{k_n} \minus{} A_{k_n \plus{} 1}, n \equal{} 1,2...$, εστω τωρα $ a_{b_k}$ ενας απο τους ορους της ακολουθιας $ (a_n)$ για τον οποιο ισχυει: $ A_{k \plus{} 1} < a_{b_k} < A_k, k \equal{} 1,2,...$ (ο $ a_{b_k}$ υπαρχει παντα, οπως προκειπτει αν παρουμε τον ορισμο του $ A_k$ με $ \epsilon \equal{} A_k \minus{} A_{k \plus{} 1} > 0$), η ακολουθια $ a_{b_k}$ τωρα ειναι υπακολουθια της $ a_n$ και βρησκεται αναμεσα στις ακολουθιες $ (A_k), (C_k) \equal{} (A_{k \plus{} 1})$ οι οποιες συγκλινουν στο $ L$ αρα συγκλινει και αυτη στο $ L$, οποτε η αποδειξη ολοκληρωθηκε.[/quote]

Βλέπω κάποιο πρόβλημα. Μπορεί να κάνω λάθος γιατί διάβασα μεν το κείμενό σου, αλλά χωρίς "χαρτί και μολύβι".

Στο κοκκινισμένο χωρίο δεν βλέπω γιατί ισχύει η ανισότητα. Π.χ. αν είχαμε την μη-συγκλίνουσα ακολουθία

  0,  1 + $ \frac{1}{2}$,   0 ,   1 + $ \frac{1}{3}$, 0 ,  1 + $ \frac{1}{4}$,  0 ,   1 + $ \frac{1}{5}$,  ...

τότε οι συγκλίνουσες υπακολουθίες συγκλίνουν είτε στο 0 είτε στο 1. Δηλαδή L' = 1, καί όμως δεν ισχύει $ a_n \leq L'$, όποιο και αν είναι το n.
Το βήμα αυτό φτάχνει. Απλά χρειάζεται να παρατηρήσεις ότι για ε > 0 (χωρίς βλάβη ε < L - L',  οι όροι της $ a_n$ είναι τελικά μικρότεροι του L' + ε < L.

Τ.Π.

Εχετε δικιο. Το οριο δεν εχει απολυτος καμια σχεση με το ανω/κατω φραγμα σε μη μονοτονες ακολουθιες, ηταν ενα πολυ χαζο λαθος απο κεκτημενη ταχυτητα... Απλα προσπαθουσα αρκετες ωρες χτες να δειξω οτι καθε ακολουθια εχει μια υπακολουθια που συγκλινει στο $ L$, και επηδη δεν ειχα πολυ χρονο γιατι επρεπε να παω καπου, εγραψα στα γρηγορα αυτη τη ψιλολανθασμενη αποδειξη παραλειποντας μια τοσο βασικη εννοια...

Edit: μολις τωρα ειδα τη διορθωση σας στο λαθος που ειχα κανει στην αρχη της αποδειξης. Νομιζω πως η αποδειξη ειναι ολοσωστη τωρα :D
Ευχαριστω για τη διορθωση.

[quote="NickNafplio"]Θα ηθελα να με διορθωσετε αν εχω καπου καποιο λαθος, και αν μπορει καποιος να μου εξηγησει τι ακριβως θα μπορουσα να κανω με την ακολουθια που απαριθμει τους ρητους στο [0,1] που ανεφερε ο Durandal γιατι εχω μπερδευτει εκει...[/quote]

Well... το πρόβλημα είναι να φανταστείς ποια είναι τα $ \limsup, \liminf$ της ακολουθίας αυτής ή στο να το αποδείξεις?

Hint: το ένα δεν είναι (ιδιαίτερα) πιο δύσκολο από το άλλο ;)

Cheerio,

Durandal 1707

Βασικα αν παρουμε τον αλλον ορισμο που λεει οτι το infsup ειναι ο αριθμος κ για τον οποιο εχουμε απειρους ορους της ακολουθιας > κ - ε, και πεπερασμενους ορους >κ + ε για καθε ε>0, τοτε προκειπτει αμεσα οτι το infsup και το supinf ειναι το 1 και το 0 αντιστοιχα, αφου αναμεσα σε 2 οποιουσδηποτε πραγματικους αριθμους υπαρχουν απειροι ρητοι αριθμοι. Απλα δεν μπορουσα να καταλαβω πως θεωροντας τη συγκεκριμενη ακολουθια βοηθαει στο να δειξουμε οτι καθε υπακολουθια που συγκλινει, συγκλινει σε αριθμο $ \leq$ infsup και $ \geq$ supinf...

[quote="NickNafplio"]Απλα δεν μπορουσα να καταλαβω πως θεωροντας τη συγκεκριμενη ακολουθια βοηθαει στο να δειξουμε οτι καθε υπακολουθια που συγκλινει, συγκλινει σε αριθμο $ \leq$ infsup και $ \geq$ supinf...[/quote]

To μόνο που είπα είναι ότι μία τέτοια ακολουθία είναι αρκετά διδακτικό παράδειγμα για τις ιδιότητες των $ \liminf, \limsup$, δε βοηθάει να δείξεις κάτι γενικότερο.

Sorry if I confused you,

Durandal 1707

[quote="Durandal"]
Spivak -> Browder/Rudin -> Kolmogorov etc. είναι μια πολύ ωραία πορεία :)

[quote]Τέτοια βιβλία συνήθως τα διαβάζουν μόνο οι Μαθηματικοί ή οι [color=red]Θεωρητικοί Φυσικοί[/color].[/quote]

Ha :( Μακάρι να το έκαναν, αλλά η νόρμα δύστυχως σίγουρα δεν το κάνει... :(

Durandal 1707[/quote]

Φίλοι καλή χρονιά με υγεία

Η φυσική (η θεωρητική εννοώ) είναι χόμπυ μου και επειδή θέλω να το κάνω στη σωστή βάση το συγκεκριμένο θέμα με ενδιαφέρει πολύ… όπως ήδη ξέρει ο Παναγιώτης.

Δεν έχω καταλάβει γιατί οι θεωρητικοί φυσικοί πρέπει να δώσουν μεγάλη έμφαση στη θεμελίωση των μαθηματικών και ενδεχομένως και στην αυστηρότητα. Πχ διαβάζω αυτή τη στιγμή μηχανική από 2 βιβλία: του Feynman και του Landau.
Και οι 2 τους υπήρξαν εξέχουσες μορφές στο χώρο της θεωρητικής φυσικής, ο καθένας με το δικό του στυλ και είναι γνωστή η μαθηματική τους ικανότητα. Παρόλα αυτά δεν έχω αποκομίσει την εντύπωση ότι δίνουν τόση έμφαση σε θέματα θεμελίωσης και θέματα αυστηρότητας, όσο το να μπορεί κανείς να χρησιμοποιεί αυτά που ξέρει. Πχ ενδιαφέρει τον Landau το να ξέρει κανείς να υπολογίζει οποιοδήποτε αόριστο ολοκλήρωμα μπορεί να εκφραστεί με στοιχειώδεις συναρτήσεις . Επίσης και οι 2 χρησιμοποιούν κατά κόρον τις στοιχειώδης ποσότητες αδιαφορώντας για το ότι δε θεμελιώνονται αυστηρά.  Δηλαδή δείχνουν να ασχολούνται με το πρακτικό μέρος των μαθηματικών, πως εφαρμόζονται στην περιγραφή και επίλυση προβλημάτων φυσικής και όχι στη θεμελίωσή τους.

Τελικά πόσο αναγκαία είναι η θεμελίωση και η αυστηρότητα των μαθηματικών για ένα φυσικό και πόσο μακρυά μπορεί να φτάσει χωρίς τις στοιχειώδεις ποσότητες;

Με ενδιαφέρει αυτό το θέμα γιατί ο χρόνος που αφιερώνω στο χόμπυ μου είναι δυστυχώς λίγος (λόγω δουλειάς) και πρέπει να τον διανείμω σωστά. Μια λάθος επιλογή ως προς το τι  θα διαβάσω θα μου στοιχήσει σε χρόνο (που δεν έχω).

Γιώργος

[quote="gpapargi"]Τελικά πόσο αναγκαία είναι η θεμελίωση και η αυστηρότητα των μαθηματικών για ένα φυσικό και πόσο μακρυά μπορεί να φτάσει χωρίς τις στοιχειώδεις ποσότητες;

Με ενδιαφέρει αυτό το θέμα γιατί ο χρόνος που αφιερώνω στο χόμπυ μου είναι δυστυχώς λίγος (λόγω δουλειάς) και πρέπει να τον διανείμω σωστά. Μια λάθος επιλογή ως προς το τι  θα διαβάσω θα μου στοιχήσει σε χρόνο (που δεν έχω).[/quote]

George, τις απόψεις μου τις ξέρεις: αν θέλεις να ασχοληθείς με τη φυσική για χόμπυ και δεν έχεις το χρόνο, άστο να πέσει :D Όσο μαθαίνεις ακόμα τις πρώτες αρχές της μηχανικής, της σχετικότητας or whatnot και όταν βλέπεις πρώτη φορά Euler-Lagrange, εννοείται συμφωνώ μαζί σου: άσε τον αυστηρό ορισμό της δέλτα του Dirac στο ντουλάπι και κάντην ό,τι θές, κάνε τους μετ/μούς Fourier σου σε ό,τι συνάρτηση θέλεις και μην ανησυχείς για σύγκλιση και γενικά... keep on going! I can't stress this enough! :)

Επίσης, και ερευνητικά αν θέλεις να ασχοληθείς, υπάρχουν πολλοί τομείς που δε χρειάζεται η θεμελίωση. Ακόμα και  γενική σχετικότητα μπορείς να κάνεις σε μεγάλο βαθμό χωρίς να χρειάζεται να ξέρεις ανάλυση (πέραν του πρακτικού λογισμού). Πολλοί άνθρωποι προσπαθούν να λύσουν εξισώσεις Einstein με τις μεν ή τις δε παραδοχές και η δουλειά τους είναι υπολογιστική, δεν τους χρειάζεται να ξέρουν πότε εναλλάσσεται το όριο και το ολοκλήρωμα. Και δουλεύουν σε θεωρητική κοσμολογία, inflation, deflation or whatever...

Ωστόσο, δεν είναι για μένα οι γνώσεις που κερδίζεις όσο η νοοτροπία, και αυτή η νοοτροπία έχει IMHO πολύ μεγάλη σημασία αν θέλεις να μιλήσεις για τα [i]θεμέλια[/i] της φυσικής: για QFT, string theory or whatnot, δηλαδή για τα πιο "λεπτά" σημεία. Δεν υποστηρίζω να χρησιμοποιούμε rigged Hilbert spaces για να μιλήσουμε για ελεύθερα σωμάτια όταν πρωτοκάνεις Κβαντομηχανική... αλλά το γεγονός ότι τέτοιες αθώες "παρεκτροπές" έρχονται και μας δαγκώνουν μετά στη Θεωρία Πεδίου... ε, υπάρχει κάποιο πρόβλημα ;)
[hide]Και ναι, θα προτιμούσα να υπάρχει κάποιο solid mathematical footing στα ολ/τα διαδρομών του Feynmann αντί να γράφουμε ένα μέτρο στην τύχη και να το ξεχνάμε μετά όταν κάνουμε renormalization... και να έχουμε όσα προβλήματα προκύπτουν από αυτήν (και που χτυπάνε ακόμα και απτές εφαρμογές στη στερεά κατάσταση, όπως τη δουλειά του Dyson).[/hide]
Hope I'm making some sense :) If not, don't worry about it, για μένα η πρώτη παράγραφος που γράφω έχει περισσότερη σημασία από την τελευταία ;)

Απομακρυνθήκαμε όμως από το original post. Cheerio,

Durandal 1707


How would one prove that $x^{2}\ge 0$ for all $x \in \mathbb{R}$?

Show that if $x^{2}$ is negative, $x$ is imaginary, I guess...

[quote="i_like_pie"][hide]
\[x^{2}\ge 0 \]

\[|x|\ge0 \]
[/hide][/quote]

Er... what?

Anyway, if you're familiar with the axiomatic definition of the real numbers then it's sufficient to show that $(-1) \cdot (-1) = 1$.

Well, you would first show that $x^{2}\geq 0$ for nonnegative $x$, then for negative $x$ you would factor it as $(-1)^{2}(-x)^{2}$, where $(-x)$ is positive and $(-1)^{2}=1$, so $(-1)^{2}(-x)^{2}$ is positive. And we can prove that $(-1)^{2}=1$ using some of the elementary properties of fields and real numbers:

$(-1)(-1) = (-1)(-1)+(-2)+2$
$= (-1)(-1)+(-1)(2)+2$
$= (-1)(-1+2)+2$
$= (-1)(1)+2$
$=-1+2$
$= 1$

T.A.O., in your proof you used the fact that $-2 = (-1)\cdot 2$, or in other words that $(-1)(x) =-x$.  This fact is not direct from the field axioms, and indeed if it were it would give you $(-1)(-1) =-(-1)$ directly, and $1 =-(-1)$ follows from uniqueness of additive inverses.

Just to clarify what's going on, for those who don't know: the goal is to prove this from the most basic possible properties of the real numbers, in this case from the fact that the real numbers form an ordered field with certain other properties.  In such a situation, $-x$ means "the additive inverse of $x$" and so $-1$ means "the additive inverse of 1."  The fact that $(-1)(x) =-x$ is therefore not a definition and needs to be proven, hence this post.


Find out the maximum value of $ p$ such that

$ \sqrt \frac {x_1}{p \minus{} x_1} \plus{} \sqrt \frac {x_2}{p \minus{} x_2} \plus{} ... \plus{} \sqrt \frac {x_n}{p \minus{} x_n} \ge \frac n{\sqrt {p \minus{} 1}}$ ,

for all nonnegative real numbers $ x_1,x_2,...,x_n <p$ with $ x_1 \plus{} x_2 \plus{} ... \plus{} x_n \equal{} n$.

If we don't fix $ n$ then $ p\equal{}2$. For fixed $ n$ we can take slightly bigger $ p$. By any way $ p \leq 2\plus{}\frac{1}{n}$.

[quote="Alexander Khrabrov"]If we don't fix $ n$ then $ p \equal{} 2$. For fixed $ n$ we can take slightly bigger $ p$. By any way $ p \leq 2 \plus{} \frac {1}{n}$.[/quote]
Indeed, $ 2\plus{}\frac 1{n}$ is about right. :)


logarithms
calculus
integration
floor function
induction
function
calculus computations

This is kind of false. Take $ n \equal{} 1000$, for example. Clearly $ \log 1000 \equal{} 3$, and $ \sum_{i \equal{} 1}^{1000} \frac 1i \approx 7.485 > 7$. So $ \sum_{i \equal{} 1}^{1000} \frac 1i \minus{} \log 1000 \approx 4.485 > 4$.

Therefore, your claim is false.

[b]Conclusion[/b]: The harmonic sum diverges, and at a faster rate than $ \log x$.

I think he was talking natural logs.

The standard way to do this involves integration, so I think it would be better in Calc&Comp

[quote="apratimdefermat"]lets concentrate on how to solve it , plz . I clearly state that i meant ln,i.e.,log base e

plz post a solution[/quote]

Solution follows *almost* trivially from $ \int_1^n \frac {1}{x} \, dx \equal{} \ln n$

Having said that, you don't state it clearly in the first post (that is you're version of $ \log x$)

[quote="SimonM"]Solution follows *almost* trivially from $ \int_1^n \frac {1}{x} \, dx \equal{} \ln n$[/quote]
Not really.  The claim  here is that $ \int_{1}^{\infty} \left( \frac{1}{\lfloor x \rfloor} \minus{} \frac{1}{x} \right) \, dx$ is less than $ 4$, and it's not immediately obvious that this series has a sum that small or that it even converges without at least an observation.

Well, I haven't plugged it through, but I was going to give an argument considering the 'steps' above and below to bound the harmonic sums

Hint: with $ \ln$, the sum is descending.

How about this?

Let f(n)=$ \sum_{k \equal{} 1}^{n}\frac {1}{k} \minus{} ln(n)$

f(1)=1

$ f(n \plus{} 1) \minus{} f(n) \equal{} \frac {1}{n \plus{} 1} \minus{} ln(1 \plus{} n^{ \minus{} 1})$
Using the taylor series for the log, it is clear that the right side is always negative

so by induction

$ f(n) \leq 1$ for all positive integer n

That's not enough; you need to show that $ |f(n)| < 4$, which means you need to show that the sum doesn't get too negative.

Of course, the simplest method I know gives that $ 0<f(n)\le 1$. I don't know where $ 4$ is coming from, or what could possibly give such a weak estimate.

[quote="jmerry"]Of course, the simplest method I know gives that $ 0 < f(n)\le 1$. I don't know where $ 4$ is coming from, or what could possibly give such a weak estimate.[/quote]

What approach did you use? The best I'm getting is about 2.5, although I haven't worked at it very hard.

$ f(n) \equal{} \left( \int_{1}^{n} \frac{1}{\lfloor x \rfloor} \minus{} \frac{1}{x} \right) \plus{} \frac{1}{n} < \left( \sum_{k\equal{}1}^{n\minus{}1} \frac{1}{k} \minus{} \frac{1}{k\plus{}1} \right) \plus{} \frac{1}{n} \equal{} 1 \minus{} \frac{1}{n} \plus{} \frac{1}{n} \equal{} 1$; on the other hand, $ f(n) > \frac{1}{n}$.  Here we are using the coarse estimate

$ 0 < \int_{k}^{k\plus{}1} \frac{1}{\lfloor x \rfloor} \minus{} \frac{1}{x} < \frac{1}{k} \minus{} \frac{1}{k\plus{}1}$

since $ \frac{1}{x} > \frac{1}{k\plus{}1}$ on this interval except at an endpoint and $ \lfloor x \rfloor < x$ on this interval except at an endpoint.

The simple method I speak of: left and right endpoint estimates for $ \int_1^n \frac1x\,dx$. That's a decreasing function, so we get $ 1 \plus{} \frac12 \plus{} \frac13 \plus{} \cdots \plus{} \frac1{n \minus{} 1}\ge \ln n$ from the left endpoints and $ \frac12 \plus{} \frac13 \plus{} \cdots \plus{} \frac1n\le \ln n$ from the right endpoints. That can be rearranged to $ \frac1n\le \sum_{k \equal{} 1}^n \frac1k \minus{} \ln n\le 1$.

This is essentially the same as t0rajir0u's post, arranged slightly differently.

Here's a better one I came up with

$ f(x)\equal{}\sum_{k\equal{}1}^{\infty}\frac{x}{k(k\plus{}x)}\minus{}ln(x)$

$ f(x)\equal{}h(x)\minus{}ln(x)$

$ h'(x)\equal{}\sum_{k\equal{}1}^{\infty}\frac{1}{(k\plus{}x)^2}$

$ \int_{x}^{\infty}\frac{du}{u^2}>\sum_{k\equal{}1}^{\infty}\frac{1}{(k\plus{}x)^2}>\int_{1\plus{}x}^{\infty}\frac{du}{u^2}$

$ \frac{1}{x}>h'(x)>\frac{1}{x\plus{}1}$

$ 0>f'(x)>\minus{}\frac{1}{x(x\plus{}1)}$

since $ f(1)\equal{}1$

$ 1\minus{}ln(2)<f(x)<1$

in fact, since f(x) is strictly decreasing and bounded from below, it must approach some number $ \gamma$, which can be approximated to any desired degree of precision through various methods


The spindle speed of a CD player is $ 200$ revolutions per minute. How many revolutions does the spindle make while playing a song that lasts exactly $ 3$ minutes?

Use a ratio.
If there are 200 revolutions in 1 minute, multiply by 3 to get the # of revolutions for 3 minutes.
200*3=[b]600.[/b]


probability
geometry
function
analytic geometry
ratio

Im having problems finding the answer for the following problem. It is a continuous space so it involves primitive functions.[code]Two people agreed to meet each other on a particular day, between 5 and 6 pm. They arrive (independently) at a uniform time between 5 and 6, and wait for 15 minutes. What is the probabilty that they meet each other?[/code]I might have posted this in the wrong forum section, my apologies for that.

i will give you a hint:

let the times of their arrivals be x and y, what inequalitiy do they have to satisfy in the problem? graph them and find the satisfactory area over the total area

If x would be the arrival time of the first person, then y-x<15
That would equal the function y=15+x
Now the arrival of person x would be y=x
Now I would compute the surface between these functions in the interval 0<=x<=60 0<=y<=60

Would that be correct?
In that case the surfaces would be (ive drawn the two graphs):
y=15+x : 15*60 + 15*45 + (45*45)/ 2 = 2857.5
y=x : 60*60/2 = 1800
So the surface between the two graphs: 2857.5 - 1800 = 1057.5
So the total probability would be: 1057.5/60^2 = 0.29375

[b]Can somebody confirm this![/b]. Thanks for the hint by the way

You forgot that $x-y<15$ is also valid (it doesn't matter who comes first.)

In what way would that change the answer? (im not even sure if what ive done is correct, so you have to hint me a little, or give the computation you used)

In the coordinate system draw the square with the vertices $(0,0),(0,60),(60,60),(60,0)$. Then draw two lines we talked about and see how those divide the square. Determine which part of the square corresponds to what we want to happen, and then calculate the ratio of that area to the area of the square.

In that case you are talking about the function y=x+15 and y = x-15, the area they divide = 60^2 - 45^2
So the chance would be (60^2 - 45^2) / 60^2 which is 0.4375. [b]Need confirmation[/b]


Assume that $a,b,c,d$ are positive real numbers such that \[a^{2}+b^{2}+c^{2}+d^{2}=1.\] Prove the inequality \[101(a^{3}+b^{3}+c^{3}+d^{3})+125abcd\geq \frac{101}{\sqrt{3}}.\] This is a special case of [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=108398.[/url]

Another one similar:

Assume that $a,b,c,d,e$ are positive real numbers such that \[a^{2}+b^{2}+c^{2}+d^{2}+e^{2}=1.\] Prove the inequality \[62(a^{3}+b^{3}+c^{3}+d^{3}+e^{3})+183abcde \geq 31.\]


modular arithmetic
number theory
number theory proposed

Find the smallest $x \in\mathbb{N}$ for which $\frac{7x^{25}-10}{83}$ is an integer.

Currently, I am teaching students as MO trainers. (I don't related to Japan MO (official)).
So, I felt that this problem was just a good difficulty as a training of EGMO-Pre in Japan.
My edited [url=https://artofproblemsolving.com/community/c6h148363p838732]#2[/url] is more than 10 years old, I used Chinese Remainder Theorem, Fermat's Little Theorem, Quadratic Residue, etc. to solve this...
Will we have so much trouble solving this problem? (or appropriate?)
I think there is a more elegant solution.

Does anyone know a good solution?


Does all quadratic equations can be solve by factorizeing?

well if you have

ax^2+bx+c=0 and solutions

x1= (-b+sqrt(D))/2a
x2= (-b-sqrt(D))/2a

then factorization would be

(x-x1)(x-x2)=0

so in a way yes ... but it isn't necessary that x1 and x2 are integers

(and it's much more easy to just use

x=(-b(+-)sqrt(D))/2a

Yes, all quadratics ($ ax^2\plus{}bx\plus{}c$) can be factored in the form:
\[ \left(x \minus{} \frac { \minus{} b \plus{} \sqrt {b^2 \minus{} 4ac}}{2a} \right)\left(x \minus{} \frac { \minus{} b \minus{} \sqrt {b^2 \minus{} 4ac}}{2a}\right)
\]


Hello,

I was solving a problem that had a proof by contradiction, which was

Prove that sqrt(2) + sqrt(3) + sqrt(5) is an irrational number.

This was done by assuming that it is rational and squaring a couple of times led to

sqrt(30) being rational, which is false.

I tried extending the problem by adding an another square root of next prime number

(which is 7) and I found out that the same method does not extend to this one. (or am i mistaken?)

So does anyone know an answer to the extended problem and if yes can you prove irrationality of the following:
[ sqrt(2) sqrt(3) sqrt(5) sqrt(7) sqrt(11) sqrt(prime) ...] (dot product) [a1 a2 a3 a4 a5 a6 ... ]

p.s. Sorry about the messy notation, this is my first post and i haven't learned latex yet.

Well first of all, the square root of a number is irrational if the number is not a perfect square (doesn't have to be a prime). So to prove that the square root of some non-perfect square is irrational, yes, contradiction is very handy. Example:

Proof that $ \sqrt {7}$ is irrational:

Lets proceed by contradiction, so lets assume that $ \sqrt {7}$ is rational, or it can be written as $ \frac {p}{q}$, for some coprime integers $ p$ and $ q$. Then,

$ \sqrt {7} \equal{} \frac {p}{q}$

$ 7 \equal{} \frac {p^2}{q^2}$

$ p^2 \equal{} 7q^2 \rightarrow 7 | p^2 \rightarrow p \equal{} 7k$

$ \sqrt {7} \equal{} \frac {7k}{q}$

$ 7 \equal{} \frac {49k^2}{q^2} \rightarrow 7k^2 \equal{} q^2 \rightarrow 7 | q^2$

So 7 divides both $ p$ and $ q$, but it was given that $ p$ and $ q$ are coprime. Contradiction. This brings us to the conclusion that $ \sqrt {7}$ cannot be written as the ratio of 2 integers,AKA it is irrational.

Your next step should be proving that the sum of irrationals is irrational.

[quote="modularmarc101"]
Your next step should be proving that the sum of irrationals is irrational.[/quote]

Not all sums of irrationals are irrational

Take for example: $ \sqrt {2} \plus{} (1 \minus{} \sqrt {2}) \equal{} 1$

But a rational number plus an irrational number is irrational.

Proof, let $ a$ be a rational number and $ b$ be an irrational number,

We will prove by contradiction, assume that $ a\plus{}b$ equals a rational number $ c$.
such that $ a\plus{}b\equal{}c\implies b\equal{}c\minus{}a$, but since $ b$ is irrational and $ c\minus{}a$ is rational we have a contradiction.

Well is it possible to exclude the radical conjugate as a case? Or are there other cases where it is still possible?

EDIT: Nvm your proof says it all  :)

[hide="Proof"]We want to prove that for all sets of positive rational numbers $ \{a_1,a_2,\ldots,a_n\}$ such that $ \sqrt {\frac {a_i}{a_j}}$ is irrational for all $ i\ne j$, there are no rational $ c_1,c_2,\ldots$ not all equal to zero such that $ c_1\sqrt {a_1} \plus{} c_2\sqrt {a_2} \plus{} \cdots \plus{} c_n\sqrt {a_n} \equal{} 0$. To make this simpler, multiply by the smallest $ q$ possible so that $ a_1,a_2,\ldots$ become integers (and just treat them as integers now).

Define the [url=http://en.wikipedia.org/wiki/Field_%28mathematics%29#Definition_and_illustration]field[/url] $ \mathbb Q(\sqrt {a_1},\sqrt {a_2},\ldots,\sqrt {a_n})$ to contain exactly elements of the form $ \sum_{b_1,b_2,\ldots,b_n \equal{} 0}^{b_1,b_2,\ldots,b_n \equal{} 1}C_{b_1,b_2,\ldots,b_n}\sqrt {a_1}^{b_1}\sqrt {a_2}^{b_2}\cdots\sqrt {a_n}^{b_n}$ for rational $ C_{b_1,b_2,\ldots,b_n}$ (there are $ 2^n$ summands; each $ b_i$ can be $ 0$ or $ 1$). If we can prove that $ \mathbb Q(\sqrt {a_1},\sqrt {a_2},\ldots,\sqrt {a_n})$ has dimension $ 2^n$ (that is, there are $ 2^n$ products $ \sqrt {a_1}^{b_1}\sqrt {a_2}^{b_2}\cdots\sqrt {a_n}^{b_n}$ such that no linear combination of some of them equals a linear combination of a different group of them), then the original problem is proven.

We proceed by induction. For the base cases, when $ n \equal{} 0$, the dimension is $ 1$ because the field simply becomes the rational numbers which has only the "unit" $ 1$. When $ n \equal{} 1$, it's not hard to show that the dimension is $ 2$.

So assume that this is true for $ n \minus{} 1$ and $ n$ (the dimensions are $ 2^{n \minus{} 1}$ and $ 2^n$ respectively). Then consider the fields $ F_1\equal{}\mathbb Q(\sqrt {a_1},\sqrt {a_2},\ldots,\sqrt {a_{n \minus{} 1}})$ and $ F_2\equal{}\mathbb Q(\sqrt {a_1},\sqrt {a_2},\ldots,\sqrt {a_{n \minus{} 1}},\sqrt {a_n})$. Now we need to prove that $ \sqrt {a_{n \plus{} 1}}$ is not in $ F_2$.

For the sake of contradiction assume that $ \sqrt {a_{n \plus{} 1}}$ is in $ F_2$, which means that $ \sqrt {a_{n \plus{} 1}} \equal{} r \plus{} s\sqrt {a_n}$ where $ r,s$ are both in $ F_1$ (and of course, both sides are in $ F_2$). Squaring both sides, $ a_{n \plus{} 1} \equal{} r^2 \plus{} s^2a_n \plus{} 2rs\sqrt {a_n}\implies2rs\sqrt {a_n} \equal{} a_{n \plus{} 1} \minus{} r^2 \minus{} s^2a_n$. Note that the RHS is in $ F_1$, so either the LHS is in $ F_1$ and nonzero, $ r\equal{}0$, or $ s\equal{}0$.

[b]Case:[/b] $ \sqrt{a_n}$ is in $ F_1$. But this means that $ F_2\equal{}\mathbb Q(\sqrt {a_1},\sqrt {a_2},\ldots,\sqrt {a_{n \minus{} 1}},\sqrt {a_n})$ has the same dimension as $ F_1$, contradiction.
[b]Case:[/b] $ r \equal{} 0$. This means that $ \sqrt {a_{n \plus{} 1}} \equal{} s\sqrt {a_n}\implies\sqrt {a_na_{n \plus{} 1}} \equal{} sa_n$, so $ \sqrt {a_na_{n \plus{} 1}}$ is in $ F_1$. But $ \mathbb Q(\sqrt {a_1},\sqrt {a_2},\ldots,\sqrt {a_{n \minus{} 1}},\sqrt {a_na_{n \plus{} 1}})$ has dimension $ 2^n$ by the inductive hypothesis, even though $ \sqrt {a_na_{n \plus{} 1}}$ being in $ F_1$ means that the dimension should still be $ 2^{n \minus{} 1}$.
[b]Case:[/b] $ s \equal{} 0$. This means that $ \sqrt {a_{n \plus{} 1}} \equal{} r$, so $ \sqrt {a_{n \plus{} 1}}$ is in $ F_1$, so the dimension of $ \mathbb Q(\sqrt {a_1},\sqrt {a_2},\ldots,\sqrt {a_{n \minus{} 1}},\sqrt {a_{n \plus{} 1}})$ is $ 2^{n \minus{} 1}$. But by the inductive hypothesis, the dimension is $ 2^n$.[/hide]

[b]Edit:[/b] This is incorrect; I'm pretty sure I didn't actually know what fields were back then. See the "correct" statement [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=589&t=290946]here[/url].

I gave another proof [url=http://qchu.wordpress.com/2009/07/02/square-roots-have-no-unexpected-linear-relationships/]here[/url], which includes a link to Iurie Boreico's survey article on the subject.


Find all surjective functions $f: \mathbb{N}_{0}\rightarrow \mathbb{N}_{0}$ so that for every $m,n \in \mathbb{N}_{0}$: \[f(m)|f(n) \iff m|n\] To avoid misinterpretations, $\mathbb{N}_{0}$ is the set $\left\{1,2,3,...\right\}$

Nice question! :D I think I can do it, but I'll take it away and check.

Btw, I think $\mathbb{N}_{0}$ generally denotes $\mathbb{N}\cup \{0\}$. Somebody please correct me if this is not convention. To play it safe, N without zero should probably be notated $\mathbb{N}_{>0}$ or something.


Besides studying previous tests and doing the AoPS Vol. 1, what else is helpful to raise your score by 20-30 points?

i would just try to look at previous problems and just practice the formulas and techniques used to solve them

I would do the problems you know how to do, then check, then do the ones you don't know how to do, and check again. That should give you just a couple of minutes before the test ends.

Do contests a bit harder than AMC 10, like AMC 12, or beginning AIME or something like that.

Find out what subjects that you are lacking in, and bring up those subjects. (Like if your C&P was bad, use Alcumus)

Or, if your speed is bad, you can try playing FTW or simply practicing to see how fast you can do it. More speed+accuracy=more checking time.

If you do Volume 2, it might also help, since while you won't need all the advanced techniques, they may help, possibly providing a faster solution.


The reals x and y satisfy log8x + log4(y2) = 5 and log8y + log4(x2) = 7. Find xy.

Help please

hint: [hide]

you may notice that both 8 and 4 are powers of 2, so it may be convienent to change the log bases from 8 to 2 and from 4 to 2, and it shouldn't be too hard from there (knowing that log(a)+lob(b) = log(a*b))

[/hide]

[hide]Add the two equations together, we have log_8(xy)+2log_4(xy)=12.  Note that 4^(3/2)=8 so log_8(xy)=(2/3)log_4(xy). (8/3)log_4(xy)=12. log_4(xy)=9/2. xy=4^(9/2)=2^9=512[/hide]

The solution to your problem is:

[hide]

Using some of the basic properties of logs, including change of base and powers inside the log, we change the first condition to:

log2(x)/log2(8)+2log2(y)/log2(4)=5

log2(x)/3+log2(y)=log2(32)

log2(y* :^3:  :sqrt: x)=log2(32)

y*( :^3:  :sqrt: x)=32=2:^5:

xy :^3:=2^15

Doing the same with the second condition, we get:

x:^3:y=2^21

Raise the first equation to the 3 power to get x:^3:y^9=2^45, divide the left side by x:^3:y and the right side by 2^21 to get y^6=2^24, y=2 :^3:

substitute in to get 2^9x=2^15, x=2^6,

xy=2^6*2 :^3:=2^9=512

[/hide]

I was going to post a solution, but it was almost identical to Beta's. Good job, Beta.

Thanks, I was having trouble seeing it for some reason.


Let $n,k\in\mathbb{N}^{+}$, $m\in\mathbb{N}$, $a_{0},\ldots,a_{n-1}>0$ and $a_{i}$ be numerated modulo $n$. Prove that \[\sum_{0\leq i\leq n-1}\frac{a_{i}^{m+1}+\ldots+a_{i+k-1}^{m+1}}{a_{i}^{m}+\ldots+a_{i+k-1}^{m}}\geq \sum_{0\leq i\leq n-1}a_{i}.\]

[quote="Soarer"]Chebyshev on L.H.S. for each term, then sum them up.[/quote]
Nice solution :) .
My solution was Cauchy-Schwarzing the left hand side to obtain $\mathrm{LHS}(m)\geq\mathrm{LHS}(m-1)$. The rest is clear.
[hide="Full solution"]
At first we will prove that for any $\alpha_{0},\ldots,\alpha_{n-1}>0$ numerated modulo $n$ following inequality holds: \[\frac{\alpha_{0}^{m+1}+\ldots+\alpha_{k-1}^{m+1}}{\alpha_{0}^{m}+\ldots+\alpha_{k-1}^{m}}\geq \frac{\alpha_{0}^{m}+\ldots+\alpha_{k-1}^{m}}{\alpha_{0}^{m-1}+\ldots+\alpha_{k-1}^{m-1}}.\] Let's put $x_{i}=\alpha_{i}^{\frac{n+1}{2}}$ and $y_{i}=\alpha_{i}^{\frac{n-1}{2}}$. Then we have \[\frac{\alpha_{0}^{m+1}+\ldots+\alpha_{k-1}^{m+1}}{\alpha_{0}^{m}+\ldots+\alpha_{k-1}^{m}}\geq \frac{\alpha_{0}^{m}+\ldots+\alpha_{k-1}^{m}}{\alpha_{0}^{m-1}+\ldots+\alpha_{k-1}^{m-1}}\iff\]  \[\iff\left(\alpha_{0}^{m+1}+\ldots+\alpha_{k-1}^{m+1}\right)\cdot\left(\alpha_{0}^{m-1}+\ldots+\alpha_{k-1}^{m-1}\right)\geq \left(\alpha_{0}^{m}+\ldots+\alpha_{k-1}^{m}\right)^{2}\iff\]  \[\iff\left(x_{0}^{2}+\ldots+x_{k-1}^{2}\right)\cdot\left(y_{0}^{2}+\ldots+y_{k-1}^{2}\right)\geq\left(x_{0}y_{0}+\ldots+x_{k-1}y_{k-1}\right)^{2}\] which is true by Cauchy-Schwarz.
Using proved inequality we get \[\sum_{0\leq i\leq n-1}\frac{a_{i}^{m+1}+\ldots+a_{i+k-1}^{m+1}}{a_{i}^{m}+\ldots+a_{i+k-1}^{m}}\geq \sum_{0\leq i\leq n-1}\frac{a_{i}^{m}+\ldots+a_{i+k-1}^{m}}{a_{i}^{m-1}+\ldots+a_{i+k-1}^{m-1}}\geq\]  \[\geq\sum_{0\leq i\leq n-1}\frac{a_{i}^{m-1}+\ldots+a_{i+k-1}^{m-1}}{a_{i}^{m-2}+\ldots+a_{i+k-1}^{m-2}}\geq\ldots\geq \sum_{0\leq i\leq n-1}\frac{a_{i}^{1}+\ldots+a_{i+k-1}^{1}}{a_{i}^{0}+\ldots+a_{i+k-1}^{0}}=\]  \[=\frac{1}{k}\cdot \sum_{0\leq i\leq n-1}a_{i}+\ldots+a_{i+k-1}= \frac{1}{k}\cdot k\cdot\sum_{0\leq i\leq n-1}a_{i}= \sum_{0\leq i\leq n-1}a_{i}\] which ends the proof. $\blacksquare$
[/hide]

This reminds me of a problem in the inequalities sectio of aops vol 2 (its from Mandelbrot #3):
[quote="Problem"]Let $S_{k}=\sum_{i=1}^{n}x_{i}^{k}\ \ x_{i}>0$. Prove \[S_{19}S_{93}\geq S_{20}S_{92}\] (The problem didn't include the S notation)[/quote]
And the proof is simple, it uses the same technique as yours of setting x's into cauchy with powers of (n+1)/2 and minus...
This idea is fun to play with in generalizations...


[quote="guile"]Using properties of E and log, show that the limit of $\frac{a^{k}-1}{k}= log(a), a > 0$ as $k \mapsto \infty$[/quote]
No, I know $\lim_{k\to+\infty}\frac{a^{k}-1}{k}=+\infty$ for all $a>1$.

That's obviously meant to be the limit as $k\to 0$. The argument depends on the choice of definitions.


Notice that everything in that world is an ACME product.

Acme also produces Klein bottles in the real world:

http://www.kleinbottle.com/


function
algorithm
algebra unsolved
algebra

Suppose that we want to generate a sequence of integers X0, Xi, Xz, . . . ,
n the range 0 < X, < m. Let f(z) be any function such that 0 < z < m implies
0 < f(z) < m. Consider a sequence formed by the rule X(n+1) = f(X(n)).  a) Show that the sequence is ultimately periodic, in the sense that there exist numbers
X and p for which the values X0, Xi, , X,, . . . , Xp+x-i are distinct, but
X %+x = X, when n 2 CL. Find the maximum and minimum possible values of p
and X.
b) (R. W. Floyd.) Show that there exists an n > 0 such that X, = Xzn; and the
smallest such value of 72 lies in the range p < n 5 p + X. Furthermore the value
of X, is unique in the sense that if X, = X2, and X, = Xzr, then X, = X,.
c) Use the idea of part (b) to design an algorithm that calculates p and X for any
given function f and any given X0, using only O(k + X) steps and only a bounded
number of memory locations.


calculus
integration
function
algebra
polynomial
quadratics
absolute value

evaluate $ \int_0^{\pi/2}\frac1{a\plus{}sin^2\theta}\,d\theta$ where absolute value of a is greater than 1.

the  above integral is $ 0.5$ times the integral from 0 to $ 2\pi$ of the same function.upon substution i get a messy 4th degree and i dont seem to know how to get on from there. any help appreciated.

Take a hint from where you put it and look at $ z\equal{}e^{i\theta}$. It should become the integral of a rational function around the circle, which can be done by residues.

ya thats what i did and i got a 4th degree polynomial in the deoniminator. it split up as $ (z^2-p)\times(z^2-q)$ but p and q were messy and i couldnt figure out which of the roots lie in the unit circle. the integral i was talkin abt was the following

$ \int_{C}\frac{z}{z^4-2(2a+1)z^2+1}\,dz$ where C is the unit circle.

Let $ z \equal{} e^{i\theta}$.  Make that substitution and note the integral is just 1/4 the integral of same over $ 2\pi$, then I get:

$ \int_0^{\pi/2}\frac {d\theta}{a \plus{} Sin^2[\theta]} \equal{} \minus{} i\oint \frac {z}{4az^2 \minus{} (z \minus{} 1)^2}dz$

Solving $ 4az^2 \minus{} (z^2 \minus{} 1)^2 \equal{} 0$ for z using quadratic formula twice I get:

$ z \equal{} \pm\sqrt {(2a \plus{} 1)\pm 2\sqrt {a^2 \plus{} a}}$

Only:

$ z_1 \equal{} \minus{} \sqrt {(2a \plus{} 1) \minus{} 2\sqrt {a^2 \plus{} a}}$

$ z_2 \equal{} \sqrt {(2a \plus{} 1) \minus{} 2\sqrt {a^2 \plus{} a}}$

are inside the unit circle for $ a > 1$

Then for a>1:

$ \int_0^{\pi/2}\frac {1}{a \plus{} Sin^2[\theta]}d\theta \equal{} \minus{} i(2\pi i)\sum_{n \equal{} 1}^{2} \mathop\text{Res}\limits_{z \equal{} z_n}\left\{\frac {z}{4az^2 \minus{} (z^2 \minus{} 1)^2}\right\} \equal{} \frac {\pi}{2\sqrt {a^2 \plus{} a}}$

I haven't checked this rigorously.


A positive integer with exactly four positive factors is called "quadly."  Compute the least $ n$ for which each of $ n$, $ n\plus{}1$, and $ n\plus{}2$ is quadly.

--

Although the answer is [hide="relatively small"]33[/hide], does anyone have a way to do it without brute force?  Thanks!

http://www.arml.com/2008_Contest/ARML_2008_Individual_Questions_Solutions.pdf

#2 (see the second paragraph).

I don't get how the fourth sentence was gotten:

[quote]Then n + 1 = p3+ 1 cannot be quadly.[/quote]


Find max K such that for all $ a,b,c>0$
$ \sum_{cyc}\frac{(b\minus{}c)^2(b\plus{}c)}{a}\geq K(\sum_{cyc}a^2\minus{}\sum_{cyc}ab)$

I've never encountered this notation before; what does $ \displaystyle\sum_{cyc}$ mean?

Let $ b \equal{} 1,c \equal{} 1,$we have $ k\leq2$;And if $ K \equal{} 2$,then
\[ \sum\frac {(b \minus{} c)^2(b \plus{} c)}{a}\geq{ K\sum(a^2 \minus{} ab)}
\]

\[ < \equal{} \equal{} > \sum{a^2(b \plus{} c)(a \minus{} b)(a \minus{} c)}\geq0,
\]
the inequality holds.Thus $ K_{max} \equal{} 2.$


a capella math singing.  if you have some background in algebra you'll be able to understand this, but it's funny either way:

http://www.collegehumor.com/?movie_id=149448

im not sure if this is the right forum to put this in

i dont think this is the right forum for this either :rotfl:  ;)  :noo:  barnacle.

this was already posted here once.  i don't get the movie.  Well, i get some of it

Ah hahahaha I love it!  Hmm this is something me and my math friends do a lot...too bad we never did film any of it.

Wow that is really talented!! The guy on the right was making drum noises!! That was really cool!


linear algebra
matrix
geometry
geometric transformation
dilation
algebra
polynomial

Hello, I have a matrix question.

Below is a matrix describing linear map G: R^2 to R^2.

G = the matrix with
First row = 1, x
2nd row = 3, 4

I have to figure out every value of x for which the map G would be equal to both a shear and stretch by a single factor n in every direction.

I'm aware that a shear-dilation matrix has the form:

First row:  E E
Second row  0 E

where the E's are repeated eigenvalues, but I'm unsure of how to put G into the above format.

Thanks in advance to anyone who helps.

You want the characteristic polynomial to be of the form $ (x\minus{}c)^2$. What should $ c$ be? You can tell by the trace of $ 5$; $ c$ must be $ \frac52$. The determinant is thus $ \frac{25}{4}$, and $ x$ is $ \frac13\left(4\minus{}\frac{25}{4}\right)\equal{}\minus{}\frac34$.

Thanks jmerry!  That helps a lot.

What would happen in a case when the eigenvalues are not repeated?

For example, if you are trying to find every value of x for which the map G would be equal to a strain, then you would have to use the fact that a strain matrix is of the form

First row:  E1, 0
Second row:  0, E2

The characteristic polynomial would be of the form X^2 - trace(G) + det(G).

You would have that the trace = 5 like before.  But how would you figure out the determinant?

What I have so far is
X^2 - 5X + (4-3x)

(where the little x is the variable from the matrix and big X is the variable in the characteristic polynomial).


Solve these equation with $ x\in R$:

[u][b]equation 1[/b][/u]

$ \sqrt {x^2 \plus{} 2x} \plus{} \sqrt {2x \minus{} 1} \equal{} \sqrt {3x^2 \plus{} 4x \plus{} 1}$

[u][b]equation 2[/b][/u]

$ \frac {4}{x} \plus{} \sqrt {x \minus{} \frac {1}{x}} \equal{} x \plus{} \sqrt {2x \minus{} \frac {5}{x}}$

[u][b]equation 3[/b][/u]

$ \sqrt {4x \minus{} y^2} \minus{} \sqrt {y \plus{} 2} \equal{} \sqrt {4x^2 \plus{} y}$

[u][b]equation 4[/b][/u]

$ \sqrt {(x \plus{} 2)(x^3 \plus{} 2x^2 \plus{} 4x \plus{} 8)} \equal{} 1 \plus{} \sqrt {x^4 \minus{} 16}$

[color=red][u][size=150][b]Note:[/b][/size][/u][/color] [b][color=darkblue]Please post full solution. Thank you very much.[/color][/b]

[quote="thanhnam2902"]Solve these equation with $ x\in R$:

[u][b]equation 1[/b][/u]

$ \sqrt {x^2 \plus{} 2x} \plus{} \sqrt {2x \minus{} 1} \equal{} \sqrt {3x^2 \plus{} 4x \plus{} 1}$ [/quote]

Let $ y\equal{}\sqrt {x^2 \plus{} 2x} , z\equal{}\sqrt {2x \minus{} 1}$. The equation

$ \Leftrightarrow (y\plus{}z)^2\equal{}3y^2\minus{}z^2$ and $ y,z \ge 0$
$ \Leftrightarrow y^2\minus{}yz\minus{}z^2\equal{}0$ and $ y,z \ge 0$
$ \Leftrightarrow y\equal{}\frac{1\plus{}\sqrt{5}}{2}z$ and $ y,z \ge 0$
$ \Leftrightarrow x^2\plus{}2x\equal{}\frac{3\plus{}\sqrt{5}}{2}(2x\minus{}1)$ and $ y,z \ge 0$
$ \Leftrightarrow x\equal{}\frac{1\plus{}\sqrt{5}}{2}$


Putnam
algebra
polynomial
vector
calculus
integration
complex analysis

Say that a polynomial with real coefficients in two variable, $ x,y,$ is [i]balanced[/i] if the average value of the polynomial on each circle centered at the origin is $ 0.$ The balanced polynomials of degree at most $ 2009$ form a vector space $ V$ over $ \mathbb{R}.$ Find the dimension of $ V.$

The dimension is $ 1\plus{}2\plus{}\cdots\plus{}2010\minus{}1005\equal{}2020050$.

Split the polynomial into its homogeneous parts; for example, $ x^2\plus{}xy\plus{}3x\minus{}y\plus{}2$ becomes $ (x^2\plus{}xy)\plus{}(3x\minus{}y)\plus{}2$. The average value of a degree-$ n$ homogeneous polynomial on a circle of radius $ r$ is a constant multiple of $ r^n$, by similarity with the circle of radius $ 1$. Adding them up, the average value of the whole polynomial is a polynomial of degree at most $ 2009$.
This average value is a linear map from $ V$ to the space $ P_{2009}[r]$ of polynomials of degree at most $ 2009$ in $ r$. We seek the nullity of this map, which is the dimension of $ V$ minus the dimension of its image.

What can we get in that image? Well, the circle of radius $ \minus{}r$ is also the circle of radius $ r$, so everything in the image is even. Conversely, since $ (x^2\plus{}y^2)^n$ has average value $ r^{2n}$, the image contains every even power of $ r$ up to $ r^{2008}$. Including $ r^0$, there are $ 1005$ of these, and the nullity is $ \text{dim }V \minus{} 1005$. Finally, there are $ d\plus{}1$ monomials $ x^d,x^{d\minus{}1}y,\cdots,y^d$ of each degree $ d$, so $ \text{dim }V\equal{}1\plus{}2\plus{}3\plus{}\cdots\plus{}2010\equal{}\frac{2010\cdot 2011}{2}\equal{}2021055$. Subtract $ 1005$ to get the final answer.

Here's what I had. (I botched the arithmetic in the last step, but then I'm not a competitor either.)

As jmerry said, partition $ V$ into independent subspaces $ V_k$ where each polynomial in $ V_k$ is homogeneous of degree $ k.$

Consider the monomial $ x^iy^j.$ If either $ i$ or $ j$ is odd, then this averages to zero around a circle. If $ k$ is odd, then all of the monomials in $ V_k$ have an odd exponent. Hence for $ k$ odd, every polynomial is balance.

Now suppose $ p(x,y) \equal{} \sum_{j \equal{} 0}^ka_jx^{k \minus{} j}y^j\in V_k.$ When is it balanced? We can calculate the integral of this by summing the integral of each monomial; always zero for odd $ j$, some positive constant for each even $ j.$ We can even explicitly compute that constant, not that it matters for this argument. $ p(x,y)$ is balanced iff
\[ \sum_{i \equal{} 0}^{k/2}a_{2i}\Gamma\left(\frac {k \minus{} 2i \plus{} 1}2\right)\Gamma\left(\frac {2i \plus{} 1}2\right) \equal{} 0\]
(As I said, it doesn't really matter what those coefficients are, as long as at least one of them is nonzero.)
That's a null space of a nontrivial linear transformation from $ V_k$ to $ \mathbb{R}.$ By rank-nullity, the dimension of that null space is one less than the dimension of $ V_k$, namely $ k.$

The dimension of $ V$ should be the sum of the dimensions of $ V_k,$ namely:

$ 0 \plus{} 2 \plus{} 2 \plus{} 4 \plus{} 4 \plus{} 6 \plus{} 6 \plus{} \cdots 2008 \plus{} 2008 \plus{} 2010.$

We can write this as

$ 4\sum_{j \equal{} 1}^{1004}j \plus{} 2010 \equal{} 2\cdot 1004\cdot 1005 \plus{} 2010 \equal{} 1004\cdot 2010 \equal{} 2\cdot1005^2 \equal{} 2020050.$

(jmerry's way of doing the arithmetic is easier).

I think my solution was essentially equivalent to the second one, but I was off by 1 due to a counting error on my part. Bah! Frustrating!

Just plug in (x,y) = (rcos(t),rsin(t)) and integrate over t, you get a polynomial in r that equals 0 everywhere so this tells you that all the coefficients must be 0.  So this imposes a linear relation on the coefficients of your original polynomial in x and y for each degree.  Now just see that if the degree is even at least one terms in the linear relation is nonzero, and if the degree is odd they are all 0.  So for degree n you get n+1 if n is odd and n if n is even.  Then just sum it up.

Nateharman's way is how I did it, except I simplified the arithmetic to reduce calculations. Seeing the integral as a polynomial in r was the key insight for me. For a while I was tripped up since, seeing the apparent mean value property, I thought that complex analysis might lead to a solution. More likely it would have led to severe undercounting.

How I saved a step in the arithmetic: 2011*2010/2 - 1005 is 2011*1005-1005, or 2010*1005. As I like to think of it, (2x+10)(x+5) evaluated at x = 1000.


ratio
MATHCOUNTS
irrational number
continued fraction

How many irrational numbers do you know (with special symbols, duh!)

I like the Golden Ratio... 1.6180314 or something like that

yea - The golden ratio is  known to be the "most" irrational  of the irrational numbers .. okay can some one explain  why?

(PS - It is not a "wise crack" and the statement does make some sense..)

Gyan! Following theorems may give answer to question:

Theorem 1.
Let $\alpha$ be irrational number. Then there are [i]infinitely many[/i] rational numbers $\frac{a}{b}$ s.t. $\left|\alpha-\frac{a}{b}\right|\leq\frac{1}{\sqrt{5}b^2}$.

Theorem 2.
If $\lambda<\frac{1}{\sqrt{5}}$ and $\alpha=\frac{1+\sqrt{5}}{2}$ (golden ratio) then there are [i]finitely many[/i] rational numbers $\frac{a}{b}$ s.t. $\left|\alpha-\frac{a}{b}\right|\leq\frac{\lambda}{b^2}$.

It means that golden ratio cannot be stiffly accurately approximated with rational numbers (I hope you can understand my english).

That reminds me of a question I once had.  Let $\phi$ be the golden ratio, $(\sqrt{5}+1)/2$.  Then the following theorem is related to Theorem 2, except that it doesn't allow finitely many exceptions.

Theorem 3: If $a$ and $b$ are integers (and $b$ is nonzero), then
\[\left| \frac{a}{b} - \phi \right| \geq \frac{1}{\phi^2 b^2}.\]

This theorem is not hard to prove, and is tight when a = 2 and b = 1.  My question is whether the golden ratio is "most irrational" in this sense too.  In other words, is it possible to replace the golden ratio on the left side with another number?

Yes, of course! $\varphi+1$, $\varphi+2$, and so on ;)

Yes, sorry, I meant to exclude the golden ratio plus an integer.

There are estimate:
\[\left|\sqrt{D}-\frac{a}{b}\right|\geq\frac{\sqrt{D+1}-\sqrt{D}}{b^2},\]
where $D$ is not a perfect square. For $D=2$ we have $\sqrt{2}-1>\frac{1}{\varphi^2}$.

When $D = 2$, doesn't the right side of your bound involve $\sqrt{3}-\sqrt{2}$, not $\sqrt{2} - 1$?

By the way, in addition to the golden ratio, I have to exclude the negation of the golden ratio too (plus an integer).

I have just found ultimate theorem:

Theorem (Markov)
Consider arbitrary irrational number $\alpha$ s.t. $\alpha$ cann't be represented as $\frac{A\varphi+B}{C\varphi+D}$, where $AD-BC=\pm 1$, $A,B,C,D$ are integers. Then there are infinitely many rational numbers $\frac{a}{b}$ s.t.
\[\left|\frac{a}{b}-\alpha\right|<\frac{1}{\sqrt{8}b^2}.\]

Thanks for the reference to Markov's theorem.  I guess that still leaves open the cases $(A \phi + B) / (C \phi + D)$ for my problem, but perhaps we can take care of those cases separately.

Very interesting discussion!

Another way to look is to see the continued fraction .... it never has a number greater than 1 ...( A  larger  number in the continued fraction component  means a better approximation with a rational number)  :)

$\phi$ would make a better president than George W.

Happy?

Of Course We Can't Forget $\pi$, which would be a good vice president.

hey bob123 u were in miller last year.  who were you again?

gameworld7 wrote:[hide]you math nerds[/hide] seriously!  this is the f&g section.  relate something other than math once in a while

Hmmm. Kinda flamish, no? Might not be to some people, but you can detect a little of it.

mhm yeah right i didn't realy want to insult anyone...

This is way more interesting than any topic you've ever created.

I think $e$ and its running mate $\phi$ will win the election, athough $\pi$ and $i$ will give them a run for their money.  The independents, $1$ and $0$, will be defeated because they are too simple.

However, the election will be thrown into disarray with the revelation that all six candidates are involved in the largest scandal in the history of time: $e^{i\cdot\pi}+1=0\cdot\phi$!

[quote="Athena"]it DOES exist but were just blocked from it.  everyone LOVES the MILLER forum. u should have seen it last yr......[/quote]

Miller is a stupid school. It used to be good, but it turned bad. I go to lynbrook now anyways.

WhaT ia the miller forum?

Anyway, vote for the democratic one?

HEY! DON'T CALL MILLER A BAD SCHOOL!!!!!! and athena could u get the site for miller students (u know last year) u could post it and people could check it out (hopefully)

huh? u mean [url]http://www.myschoolonline.com/ca/miller[/url] or the mathcounts forum?

[quote="gameworld7"]hey yif man u shoudlve seen my questions last year.  they were *ahem* great.  u can't check em out though cause the miller forum doesn't exist anymore :([/quote]

I'm going by what I've seen here, and they aren't too "*ahem* great"

[quote="gameworld7"]yeah that one. sadly its been blocked out :([/quote]

huh? it IS still there. u can check it.

no no no no no! i'm meant the artofproblemsolving site for miller students.  i think its http://www.artofproblemsolving.com/miller or something... i don't really know. i forgot

we cant access it smart one. but just get sumone in math enrich this year to post something from it.


linear algebra
matrix
combinatorics proposed
combinatorics

Denote $ M\equal{}\{\minus{}1,0,1\}$. Find the number of collections $ (a_1,a_2,\cdots,a_n)$ satisfying
(i) $ a_i\in M$ for every $ i \equal{} 1,2,\cdots,n$
(ii) $ a_i \minus{} a_{i \plus{} 1}\in M$ for every $ i \equal{} 1,2,\cdots,n \minus{} 1$

we take $ U_{n,h}\equal{}$the number of collections $ (a_1,...,a_n)$ satisfying i,ii such that $ a_n\equal{}h$
let $ \vec{B_n}\equal{}(U_{n,0} ,U_{n,1} ,U_{n,\minus{}1})$
easly we find $ B_{n\plus{}1}\equal{}A.B_{n}$ for some matric $ A\in M_3(\mathbb{R})$
so $ B_{n}\equal{}A^{n\minus{}1}B_1$ where $ B_1\equal{}(1,1,1)$
and we have $ U_{n,0} \plus{}U_{n,1} \plus{}U_{n,\minus{}1}$ is the number of collections $ (a_1,...,a_n)$ satisfying i,ii


it's quite a nice card game, does someone here play it?
if so, maybe you a nice softwares of it? I have Bridge Master ,Bridge Baron,  and Bridge Base, any suggestions? thanks

I play, but I don't have a computer-based version.

I play with random people that know. When I went to visit my cousins, they are all "avid" bridge players and I played with them for a long time, even when I got bored lol

I tried to get people in my school to play it but no go.

But when I play with people who know what they're doing I always get creamed.

Yes, I know bridge.  I've asked everyone that I know, and I have only one friend who knows how to play bridge.  It's quite depressing really... Bridge is such an advanced and great game, too bad hardly anyone knows how to play.  I end up playing with my family whenever we go visit them (I can't play with my parents 'cause there's only three of us).  I enjoy bridge very much.

I do NOT know of any computer software with bridge.  I've always wanted one, but when I asked people about bridge and stuff, they say, "Is this a board game or something?"  Lol, no a card game...  :| But if you know some good bridge software, tell me!!!  :lol:

i have a software. it also includes lots of other card games too. its called hoyle card games 2005. my whole family loves bridge. my grandfather taught us it. he plays in online bridge tournaments

Does canasta count????

I'm currently having a mega-match against my dad to get to 10000 points... I'm almost ahead 2000 points.

Canasta does not count. Period.
I play, but I'm really not that good. We played so much on the bus ride back from ARML last year, it wasn't even funny.

[quote="CheeseIsGood"]Never heard of it. Is it hard to learn?[/quote]

Well, I found it pretty hard to learn, but maybe that's just me. In any case, the American Contract Bridge League's website (http://www.acbl.org I believe) has free learning software you can download.

i used to know all the rules. never actually played though....so i quit...obviously


Prove the following equation.
\[ x\plus{}\sqrt[4]{x\plus{}15}\equal{}\sqrt{x\plus{}8}\]

too much work to simplyfy....
...but it is true when x=1

[hide="The usual approach"]For $ x \leq \minus{} 2$, $ \sqrt {x \plus{} 8}\geq 0$ and $ x \plus{} \sqrt [4]{x \plus{} 15} < 0$

For $ x > \minus{} 2$, it is easy to see that $ (x \plus{} \sqrt [4]{x \plus{} 15})' > (\sqrt {x \plus{} 8})'$
- We know $ (\sqrt {x \plus{} 8})' < \text{avg slope from ( \minus{} 8,0) to this point} \equal{} \frac {1}{\sqrt {x \plus{} 8}} < 1$
- We know that $ (x \plus{} \sqrt [4]{x \plus{} 15})' > (x)' \equal{} 1$

Thus, there can be at most one solution...[/hide]

There is a nice solution without a derivative.

[hide="We know that $ x\equal{}1$ is the solution so"]
$ x\plus{}\sqrt[4]{x\plus{}15}\equal{}\sqrt{x\plus{}8}$
$ \leftrightarrow (x\minus{}1) \plus{}(\sqrt[4]{x\plus{}15}\minus{}2)\equal{}\sqrt{x\plus{}8}\minus{}3$
$ \leftrightarrow (x\minus{}1) \plus{} \frac{x\minus{}1}{\sqrt[4]{(x\plus{}15)^3}\plus{}2\sqrt{x\plus{}15}\plus{}4\sqrt[4]{x\plus{}15}\plus{}8}\minus{}\frac{x\minus{}1}{\sqrt{x\plus{}8}\plus{}1}\equal{}0$
Clearly, from here we get $ x\equal{}1$
since $ 1\minus{}\frac{x\minus{}1}{\sqrt{x\plus{}8}\plus{}1}>0$.[/hide]

[quote="bibobeo"][hide="We know that $ x \equal{} 1$ is the solution so"]
$ x \plus{} \sqrt [4]{x \plus{} 15} \equal{} \sqrt {x \plus{} 8}$
$ \leftrightarrow (x \minus{} 1) \plus{} (\sqrt [4]{x \plus{} 15} \minus{} 2) \equal{} \sqrt {x \plus{} 8} \minus{} 3$
$ \leftrightarrow (x \minus{} 1) \plus{} \frac {x \minus{} 1}{\sqrt [4]{(x \plus{} 15)^3} \plus{} 2\sqrt {x \plus{} 15} \plus{} 4\sqrt [4]{x \plus{} 15} \plus{} 8} \minus{} \frac {x \minus{} 1}{\sqrt {x \plus{} 8} \plus{} 1} \equal{} 0$
Clearly, from here we get $ x \equal{} 1$
since $ 1 \minus{} \frac {x \minus{} 1}{\sqrt {x \plus{} 8} \plus{} 1} > 0$.[/hide][/quote]

$ \boxed{\frac {x \minus{} 1}{\sqrt {x \plus{} 8} \plus{} 1}}$

$ \plus{}1$ in denominator , how comes?

Sorry, it's $ \frac{x\minus{}1}{\sqrt{x\plus{}8}\plus{}3}$
And this doesn't affect at all  :D

My congratulations, bibobeo! :lol:
It's also my idea.


Prove that the equation $x^{2}-2y^{2}=0$ has no integer solution other than $(x,y)=(0,0)$.

That's pretty obvious.
$x=\pm y\sqrt{2}$
If $y$ is integer, $x$ is irrational because of $\sqrt{2}$.

[quote="Franz Joseph"]We can also solve this by an infinite descent.[/quote]
[hide] $x^{2}=2y^{2}$. Therefore, $x^{2}$ must be even and $x$ must be even. We can write $x$ as $2a$. $4a^{2}=2y^{2}\Longrightarrow 2a^{2}=y^{2}$. Therefore, $y$ must be even and can be written as $2b$. $2a^{2}=4b^{2}\Longrightarrow a^{2}=2y^{2}$. And so on. [/hide]


how do u prove that $\prod_{k=1}^{n-1}\sin\left(\frac{k\pi}{n}\right)=\frac{n}{2^{n-1}}$?
and
can you make a similar equation for $\prod_{k=1}^{n-1}\cos\left(\frac{k\pi}{n}\right)$?

This problem has been discused before; just search for it. [hide="(The trick is just)"]Consider $\frac{x^{n}-1}{x-1}$ as a function, set it equal to $(x-r_{1})(x-r_{2})$ etc. where we the roots are the nth roots of unity. Now plug in x=1 and simplify the resulting identity to get the desired property.[/hide]
For the second one, $\cos(x)=\sin\left(\frac{\pi}{2}-x\right)$  :D

sorry i dont understand your solution. and what should i search for?

Sine product would be good... also try the word identity
no time now to type solution, sry but I posted it really nicely someplace a couple months ago

Perhaps these threads might help:

http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=pi+frac+sin&t=112904

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=108650

Go me and my awesome search skillz.

EDIT: Whoops, that's a summation not a product.

The first one isn't the same problem btw
the last link is the one I was referring to


probability
calculus
calculus computations

[b]The campus coffee shop is trying to determine how many workers that they should have during their peak times. They have found that the average number of customers that arrive in the 10-15 minute break between classes is 9 and follow a Poisson distribution.

a/ What is the probabilty that during a break that exactly 9 customers show up?

b/ What is the probability that less than 7 customers show up?

I found:

a/ Lambda =9 so p(x=9)= 0.587

b/ P(x<7)=  1- ( p(x=1)+........ P(x= 6)= 0.594

Iam not sure, so could you tell me if my answers are right or wrong? :o

Thanks a lot!  :P [/b]

As far as I see it a) is correct.. although b) dhoulf be calculated as P(x<7)=p(0)+...+p(6) ..

[b]I think taht my answers are wrong but i do not know what my errors are....

But thanks for our answer!!  :blush:

[/b]

That value $0.587$ is not $P(X=9).$ It's actually $P(X\le9).$ The correct value for $P(X=9)=\frac{e^{-9}9^9}{9!}\approx 0.1318.$

The Poisson distribution lends itself extremely well to recursive calculation (and recursive calculation is in turn extremely well-adapted to spreadsheets).

Start with $P(X=0)=e^{-\lambda}.$ Then for $k\ge1,$ use this:

$P(X=k)=P(X=k-1)\cdot\frac{\lambda}{k}.$

Do that with this problem, and get the following table:

[code]k       P(X=k)		P(X≤k)
0       0.00012341      0.00012341
1       0.00111069      0.00123410
2       0.00499810      0.00623220
3       0.01499429      0.02122649
4       0.03373716      0.05496364
5       0.06072688      0.11569052
6       0.09109032      0.20678084
7       0.11711612      0.32389696
8       0.13175564      0.45565260
9       0.13175564      0.58740824
10      0.11858008      0.70598832[/code]

From this chart, $P(X<7)=P(X\le6)\approx 0.20678084.$

Running the table down a little further, we can say that we're over 99% sure we won't have more than 17 customers.

[b]Thanks a lot for your explanations!! I got it!!  :blush: [/b]