inequalities
LaTeX
geometry
geometric transformation
reflection
analytic geometry
graphing lines

If $ y \equal{} f(x)$ such that it has an inverse $ f^{ \minus{} 1}(x)$ and also that  $ y \equal{} f(x)$ is defined for all non negative $ x$  in the closed interval $ 0$ to $ 1$ such that $ f(0) \equal{} 0$ and  $ f'(0) \equal{} 0 ,f(1) \equal{} 1$ and $ f''(x) > 0$ show that $ f(x).f^{ \minus{} 1}(x) < x^2$ for the given interval

It's false.  Take $ f(x) \equal{} x^2$.  Then it satisfies your conditions, but for $ x\leq {1\over 4}$ we have $ f^{ \minus{} 1}(x) \equal{} \sqrt {x} \geq 2x$.

Also, when you type these questions, keep in mind that somebody has to read them--more whitespace would be nice. :)

Also, since you seem to use the correct $ \text{\LaTeX}$ syntax anyway, be it known to you that all you need is to include your expressions in dollar signs to have them appear on the forum as nice $ \text{\TeX}$ formulae. ;) I did it for you this time :).

Hey, you should post a comment when you change the problem, as it makes my solution look incorrect. :)

Claim: $ f(x) < x$ for $ x\in(0,1)$.  This follows from noting that $ f(x)\minus{}x$ is convex, so it takes its maxima at the endpoints of $ [0,1]$.

Now, fix any $ x\in (0,1)$, and set $ y\equal{}f(x)$.  We note that the average value of $ f'$ in $ [0,x]$ is $ y$, and $ f'$ is increasing, so $ f'(x) > {y\over x}$.  Again using the fact that $ f'$ is increasing, we find that $ f(z) > y \plus{} {y\over x} (z\minus{}x)$ for all $ z>x$.

We set $ z\equal{}{{x^2}\over y}$. (from the claim, $ x>y$, so $ z>x$)  We see that $ f({{x^2}\over y}) > y \plus{} {y\over x} ({{x^2}\over y}\minus{}x)\equal{}x$, and applying $ f^{\minus{}1}$ to both sides (which we can do since $ f$ is increasing) we find that $ f^{\minus{}1}(x) < {{x^2}\over y}$.  We conclude that $ f(x)f^{\minus{}1}(x) < y ({{x^2}\over y}) \equal{} x^2$.

[quote="Q.E.D"]@ LydianRain: I edited.There was a typo.[/quote]

Just let me know if you're planning to change it again. :)

My approach:

  Claim: $ f(x) < x$ for $ x\in(0,1)$. This follows from noting that $ f(x)\minus{}x$ is convex, so it takes its maxima at the endpoints of $ [0,1]$.
 Now, fix any $ x\in (0,1)$, and set $ y\equal{}f(x)$ and make a rough sketch of the graph of the function(It is similar to the graphs of $ y\equal{}x^n$ where n is odd).Consider the pt $ (x,f(x))$ on this curve.The corresponding point on the curve $ y\equal{}f^{\minus{}1}(x)$ which is just the reflection of the curve $ y\equal{}f(x)$ about the line $ y\equal{}x$ is (say P) $ (x,f^{\minus{}1}(x))$.Consider the secant connecting the origin to the point P.Also consider the secant connecting the origin to the reflection of the point on the curve $ (x,f(x))$ (call it Q).The slope of $ OQ>OP$ from the graph.Thus
$ {{x} \over f(x)}$ > $ {{f^{\minus{}1}(x)} \over x}$.hence $ f(x).f^{\minus{}1}(x)<x^2$.

I find this method quite useful.It can seriously give down to earth proofs of many problems. However is this method entirely correct and justifiable?

It's not quite rigorous, since you haven't used any properties about $ f$.  But it can be made rigorous.  Actually, this is exactly the picture I used in constructing my proof, but your explanation is more clear.

I think it's a little easier to think about the points $ A\equal{}(x,f(x))$ and $ B\equal{}(f^{\minus{}1}(x),x)$. (obviously the idea is the same)  We know that $ A$ lies to the left of $ B$ on the curve.  All we need to prove is that the slope of the secant line from the origin is an increasing function, and we are done.  I proved this implicitly above, and you can also arrive at this from the properties of convex functions, but you can also see it directly as follows:

Claim: $ g(x)\equal{}{f(x)\over x}$ is increasing in [0,1].

Proof: Since $ g'(x)\equal{}{(x f'(x) \minus{} f(x)\over {x^2}}$, it suffices to show that $ h(x) \equal{} x f'(x) \minus{} f(x) > 0$.  But $ h(0)\equal{}0$ and $ h'(x)\equal{}xf''(x) > 0$ for $ x\in (0,1)$, so we are done.


This is for all who love inequalities.
$\sum_{cyc}ab=3,a,b,c>0$.Prove that:$\sum_{cyc}\frac{1}{a^{2}+bc+1}\geq\frac{3}{\sum_{cyc}a^{2}b^{2}}$

Nobody tried??Or is it too hard??I also chalenge someone to find a sharper bound :D


Well here, just post how you did, and your team did.

Be sure to bide by these rules.
[quote="rrusczyk"]Do not discuss the 2008 MATHCOUNTS Chapter Competition until we say it is OK to do so.

Do not discuss problems from the contest.

Do not discuss your score.

Do not mention whether or not you (or anyone else) found it easy or hard.

You may simply state "I came in <whatever> place, and my team came in <whatever> place."  Nothing more than that.[/quote]

I got 1st place individual and my team got 1st place. I live in Southwest Region Connecticut. In countdown I lost in first round because I was nervous. (being in front of so many people.)

Haven't done it yet--will post.  :P
Have a lot of others already taken it?

well yesterday (February 2nd) was the first CHapter Round day.

I got first written, and cd, but my team isn't good at all, so we came 3rd.

Our team completely screwed up on the Team Round and got 2nd.
I wasn't on the team but I still made it to CD and such.

are we allowed to state what we got in comparison to other teams?

[quote="alanchou"]blargh, I just realized that the rule that richard put up seems to allow it because it doesn't explicitly state that you can't say team round scores, but it does say

"You may simply state "I came in <whatever> place, and my team came in <whatever> place." Nothing more than that.", and a team round score is more than that, so I wonder which way it would go.[/quote]

Howabout I clarify this, two additional rules:

1. When in doubt, say nothing
2. No scores of any type; no difficulty of ANY type; and no mentioning the problems in any way.

Blargh.

Not even top 10 individual. 3rd CD, in my chapter everyone participates in a strangely formatted countdown. 1st place team.

I got five trophies from MathCounts for...

2nd place Countdown Round

2... 3rd Place Team trophies (Milstadt, from the county south of us got 2nd and the spoiled brats from Liberty took first...) The top two go to state :( ...

4th Individual and Top Individual

Liberty Middle School and Milstadt Consolidated (weird name, huh?) went to state...

Me and Tim Chen are going to state, too... as individuals... I think.

We would've uber pwned the Liberty snobs if we another 8th grader was switched with 7th grader, becuz' that 7th grader doesn't know math very well. I got SO CLOSE TO GETTING THE TOP COUNTDOWN... we didn't do the written...

I wonder if we go to nationals...

My competition was this morning and I:

-was on the second place team
-5th individual (a slight improvement from my 6th last year, not as good as everyone else here)
-tied for 5th in countdown, but i'm technically 6th because I lost to the eventual second place winner
-top scorer from my team

Yay me! But everyone else here seems to have gone to states. But then again, my chapter is really tough to win in. People from this chapter make it to nationals every year.

[quote="policecap"][quote="lotsofmath"]Somewhere between 11th and 16th individually, 4th place team, and 2nd place countdown.[/quote]

Only top 10 get to COUNTDOWN, RIGHT?[/quote]

depends on the chapter, I think

Kings Chapter in New York
I got 2nd written and 1st team. 7th grader beat me, an 8th grader. I beat 3rd place. 3rd beat 4th. Top 7 were all from my school.

[quote="AlmightyBob"]Kings Chapter in New York
I got 2nd written and 1st team. 7th grader beat me, an 8th grader. I beat 3rd place. 3rd beat 4th Top 7 were all from my school.[/quote]

hmm, does that qualify as giving scores?  just wondering

Well, it seems to be enough, since bob's post is edited. (although alanchous quote is not...)

How do you get the top 7 in the countdown all from one school? The closest I've ever seen is 3 out of the top four, which happened this year. (I was the other one :D)

My first year participating in MATHCOUNTS as a coach the top 8 were from the same school, so yes, it's possible.  Of course, the school went on to win state.  Granted this was an unusual situation.  There were three coaches and kids practiced daily in place of regular math class.

[quote="trivea"]Our school was like half of the top ten this year.........[/quote]
Same here. I got first, and one of my teammates was second.

our team had all eight in the top ten

our team got first
i got first written
got beat on cd by one of my teammates UGGHH!

only four of our top 8 people were in the top ten in my chapter. :(  i was 3rd indiv. it totally sucks to have last year's national mathcounts written round winner in your chapter.

[quote="CircleSquared"]Top ten had five of ours I think, 3 reg teamers and 2 individuals[/quote]

Which is pretty surprising cause we could hear your Individual people cracking Chuck Norris jokes during Team Round.

 :huh:

#1 and #9 before countdown, #2 and #10 after.


[hide]$\tan 1$ is rational $\Rightarrow \tan(1+1)$ is rational $\Rightarrow \tan 3$ is rational.

Continuing in the same manner, $\tan 1$ is rational $\Rightarrow \tan 30$ is rational, which is contradiction.[/hide]

how does tan(1) is rational imply tan(1+1) is rational?

$\tan{(1^\circ+1^\circ)}=\frac{\tan{1^\circ}+\tan{1^\circ}}{1-\tan{1^\circ}\cdot\tan{1^\circ}}$

clearly, the product, sum, difference, and quotient of two rational numbers is also a rational number, so $\tan{(1^\circ+1^\circ)}$ would be irrational if $\tan{1^\circ}$ were irrational.

(note: the product, sum, difference, and quotient of two irrational numbers does NOT have to also be irrational..)

$\tan 1^{\circ}\in Q\Longrightarrow \tan 2^{\circ}=\tan \left(1^{\circ}+1^{\circ}\right)=\frac{2\cdot \tan 1^{\circ}}{1-\tan^2 1^{\circ}}\in Q\ .$
[b]Here is a simultaneousness ![/b] You are with all, some attractively !

haha yeah, cept i had to edit it to put in the degrees symbols  :blush:

[u]Haha, yeah,[/u] what understand you ?

i was agreeing with you that we posted at the same time.


Determine the only three-digit palindrome which, when multiplied by seven, results in a four-digit palindrome.

Answer:
[hide]858[/hide]

As far as I got:
[hide]1001a+110b=707x+70y...and that's as far as I got. Please tell me how to go beyond this step.[/hide]

oh sry

you mean that (a,x)=(0,0), (1,7), (2,4), (3,1) (4,8) (5,5) (6,2) (7,9) (8,6) (9,3) part, right?

ok

143x+110y=101a+10b

so regardless of the values of y and b, 110y and 10b will be multiples of 10, which means their last digit will be 0.

and since 143x+100y is equal to 101a+10b, their last digit must be same, too.

so 143x's last digit is equal to 101a's last digit.
thats how i got those numbers.
like
(2,4) -> 101a is 202, last digit is 2. - 143a is 572, last digit is 2.
(7,9) -> 101a is 707, last digit is 7. - 143a is 1287, last digit is 7.
etc etc.

EDIT :
um idk if theres an other way.

You already asked this in the Classroom Math Forum, [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1230461#1230461]where[/url] Textangle wrote a nice explanation.

actually, d. philly posted here first and THEN posted in classroom math.


integration
limit
logarithms
function
real analysis
real analysis unsolved

do you know dominated convergence theorem ?

Well, I'm not sure. Can you post it, please? ;)

[quote="cezar lupu"]Let $I_{n}=\int_{0}^{1} (1-x^{n})^{n} dx,$ $n\in\mathbb{N}.$ Find $\lim_{n\to\infty} I_{n}.$ [/quote]

How about the leading order term in $1 - I_n$?

[quote="hpe"]
How about the leading order term in $1 - I_n$?[/quote]
$\frac{\ln n}n$. :) How about the next term? :P

[quote="cezar lupu"]Let $I_{n}=\int_{0}^{1} (1-x^{n})^{n} dx,$[/quote]

with the change variable $y=x^{n}$ we get

$nI(n)=B(1/n;n+1)=\frac{\Gamma(1/n)\Gamma(n+1)}{\Gamma(n+1+1/n)}$

use asymptotic expansion of $\Gamma$ function

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

For another view of $B(1/n ; n)$ with n a positive integer

http://mathworld.wolfram.com/BetaFunction.html
see $B(p,n)$


This post is just to inform you about the [url=http://wpc.puzzles.com]U.S. Puzzle Championship[/url], occurring on June 19, which is used to select the U.S. (and maybe also the Canadian) team to the World Puzzle Championship.  I won't bore you with the details --  you can find them on the site.

Whoa-- I have never heard of this. Is this for high school students? How do people even prepare for this? Do you just do a lot of puzzles? They could ask you anything though!

The contest actually has no age limits. As for preparation, I'm going to guess that it does involve doing lots of puzzles.

Is anyone else doing this? Last day of registration is tomorrow....

If you look through the annals of the US team, you'll see some MOPpers.  I believe Wei-Hwa (mentioned a few times on that top page) is one...

man, it could just be me, but that was tough.
the 2.5 hours flew like a pigeon going after some rice.

I don't think I broke 50 points.

The test is [url=http://wpc.puzzles.com/gt-IV/index.htm]here[/url] and [url=http://wpc.puzzles.com/gt-IV/solutions.htm]solutions[/url] are up.

I got 96 points.

Wow I did a lot worse than I had expected. I only got 62. Too many mistakes... That's what I get for taking it without practicing.

i competed!  it has been a long time since i've been in a contest of any sort, and it was fun.  the time passed all too quickly of course, and there was a high negative correlation between the questions i answered correctly and those that were significantly mathematical.  all in all, barring input errors it looks like i got about a 142.  i'd be very interested to hear how other people did..

Ugh, yeah, I did horribly.

Did anyone else get 67 for #8. I was sure it was 67 and not 65.

Also, how did people do #16? I got 41 instead of 47.

Hm I got 35 for #16. I'm not sure what went wrong there. I didn't do #8. My favorite was #12. I thought that one was really neat.

[quote="Fierytycoon"]Did anyone else get 67 for #8. I was sure it was 67 and not 65.

Also, how did people do #16? I got 41 instead of 47.[/quote]

First of all, I thought these were all very high-quality puzzles.  As they suggest on their Web Site, this will be the fastest 2.5 hours you've ever spent.

I hadn't even read the Quadrilaterals puzzle when time ran out.  But now that I've seen it, it is my favorite.  My second favorite was the Pentomino puzzle, which I solved within the time limit.  If only they gave us the whole day to solve them!  :)

#8:  I counted 65 circles.  My brain thought the front and back gaskets were easy to count, but it had some trouble with the middle gasket.  So, with a colored pencil, I traced the circles in the middle gasket, counting them as I traced.

#16:  Divide the attached drawing into four quadrants.  First, convince yourself that you cannot skip over a quadrant.  (Otherwise, the quadrilateral will lop off a corner of the inner region, as shown in the first invalid example from the test).  Therefore, you must choose exactly one point from each quadrant.  This gives an upper bound of 3^4=81 possible quadrilaterals.

Now, if you consider the counterclockwise path around the quadrilateral, a blue point can never immediately follow a red point.  All other sequences are fine.  So you exclude 4*3^2=36 choices.  This leaves 81-36=45.  However, we excluded RBRB and BRBR twice, so add back one of each.  My final answer is 47.

#8: Hmm, how are you defining the front, middle, and back gaskets? I don't have the puzzle with me, but weren't there only the holes in the frontmost pink layer and the white circles on the black spheres behind that layer?

#16: Hmm, I did it a bit differently. I also divided the figure into four quadrants, but instead of subtracting the proper amount from 3^4, I set up a tree to show all of the possible routes. For example, I would have a topmost node of red, which could only connect to a node of red or green, etc... I got the wrong answer though. :(

[quote="Fierytycoon"]#8: ... weren't there only the holes in the frontmost pink layer and the white circles on the black spheres behind that layer?[/quote]

Pink?  Spheres?  There must be some sort of printing problem and/or optical illusion at work, here.  What I saw were three gaskets sandwiched together.

Speaking of printing problems, Problem 13 ("Puzzle Master") printed a bit "fuzzy" on my LaserJet.  In figure B, the cat's left eye was half-awake.  After the contest window was over, I zoomed in, using Acrobat Reader.  The eye was completely closed.

[quote]Pink? Spheres? There must be some sort of printing problem and/or optical illusion at work, here. What I saw were three gaskets sandwiched together.[/quote]

Mm, it probably was grayish, since for some reason, my printer prints some stuff as pink at times. Its weird. Anyway, I was refering to the frontmost layer with holes in it.


for which k
does the inequality
 1/(x^3+k)+1/(y^3+k)+1/(z^3+k)<=3/(xyz+k)
hold for all positive reals x,y,z?

(edited. It is 3/(xyz+k))

Take x = y = z = a.
Then for sufficiently large a, we have $a^{3}+k>0$. Now
\[\sum\frac{1}{x^{3}+k}=\frac{3}{a^{3}+k}>\frac{1}{a^{3}+k}=\frac{1}{xyz+k}\]

[quote="Amunaho"]It is wrong for k=1 (x,y,z)=(1/2,1/2,4) and thus for all k (use $\sqrt[3]k$)  :([/quote]

Haha it is wrong for $k=-1$ when $(x,y,z) \rightarrow (1,1,1)$ and thus for all $k<0$ too. And it is wrong for $k=0$ because of GM-HM inequality. :(  :P


Can someone tell me why this is true or provide a proof?

$ \text{The nth prime number is approximately}$ $ n \ln n.$

In other words,
$ \frac {p_n}{n \ln n}$ converges to 1 as $ n \to \infty.$

Thanks.  It seems like an amazing and interesting formula, so I thought I would ask here.

This is the [url=http://en.wikipedia.org/wiki/Prime_number_theorem]Prime number theorem[/url]. While the approximation is quite good, a closed-form formula for the prime numbers is, however, out of reach.


calculus
integration
geometry
3D geometry
LaTeX
calculus computations

Hi everyone,

Does anyone see what I'm doing wrong?

Find the area under the curves y=xsquared y=4xsquared-xsquared (0,0) (2,4)

I integrated with respect to x from 0 to 4

so..4x-xsquared-xsquared dx=

4xsquared/2-x cube/3-xcubed/3

2xsquared-2xcubed/3

2(xsquared-xcubed/3) 0 to 4

-224/3, but I can't have a negative area

Thank you

You have the wrong limits. What interval does $ x$ lie in?

Also, ^2 and ^3 are much more readable ways of indicating squares and cubes- that's the code for superscripts in $ \text{\LaTeX}$, and it's easy to understand in plain text as well.

Hi,

Thank you for answering. I was given that y=x^2 and y=4x-x^2 and they cross at (2,4) and (0,0). Wouldn't I take the integ. 4x-x^2-x^2 dx from 0 to 4?

Thank you

You are to calculate $ \int_{0}^{2}(4x\minus{}x^{2}\minus{}x^{2})\ dx$.

Thank you very much

I got 8/3 as my final answer. Could you please tell me if that is correct?

Thank you


superior algebra
superior algebra unsolved

How can i show that there are infinitely many prime which are not congurent to 1 mod n.

Suppose there are finitely many such primes $ p_1, p_2, ... p_k$.  The prime factors of $ np_1 p_2 ... p_k \minus{} 1$ cannot all be congruent to $ 1 \bmod n$; contradiction.

Harder (but still elementary) is to show that there are infinitely many primes which [i]are[/i] congruent to $ 1 \bmod n$.

You are saying that:

 The prime factors  cannot all be congruent to... ; contradiction.

How do you get that and how do you get the contradiction, I cant see it, please give some more details. Since this is first time I am working with number theory.

$ x\equal{}np_1\cdots p_k \minus{}1\equiv \minus{}1 \mod n$. Thus if every prime factor of $ x$ was congruent to $ 1\mod n$, then $ x\equiv 1\mod n$ which we just saw was $ \minus{}1\mod n$. Contradiction. (This will give us another prime not congruent to 1 mod n that was not in the list of all of them).


$ R$ is the real numbers and $ S$ is $ R$ excluding the point $ \frac23$. Find all functions $ f : S\rightarrow R$ such that $ 2 f(x) \plus{} f\Big(\frac {2x}{3x \minus{} 2}\Big) \equal{} 996x$ for all $ x$.

[hide="Hint"]$ g(x)\equal{}\frac{2x}{3x\minus{}2}$ then $ g(g(x))\equal{}x$[/hide]


[hide]\begin{eqnarray*}\cos{12^\circ}+i\sin{12^\circ}+\cos{48^\circ}+i\sin{48^\circ}&=&\text{cis}(12^\circ)+\text{cis}(48^\circ)\\ \text{cis}(12^\circ)+\text{cis}(48^\circ)&=&\text{cis}(30^\circ)(\text{cis}(-18^\circ)+\text{cis}(18^\circ))\\ &=& \text{cis}(30^\circ)(2\cos(18^\circ)) \end{eqnarray*}
And we know that $\text{cis}(30^\circ)^{6}=\text{cis}(180^\circ)$ hence the imaginary part is 0.[/hide]

could you explain that second line?  I dont see how to make that conclusion without factoring out $\text{cis}{12^\circ}$ and using the factorization of $x^{3}+1$ and then applying the cos double angle formula several times...

Well, we know that $\text{cis}(A) \times \text{cis}(B) = \text{cis}(A+B)$. We can also go the other way, breaking a sum into a product. Thus $\text{cis}(12^\circ)$ becomes $\text{cis}(-18^\circ+30^\circ)=\text{cis}(30^\circ) \times \text{cis}(-18^\circ)$. Doing that with the 48 as well gives a common factor of cis30, which you can then factor out, giving the RHS.

[quote="jhcreinhardt"]Find $\Im(\cos{12^\circ}+i\sin{12^\circ}+\cos{48^\circ}+i\sin{48^\circ})^{6}$[/quote]

[hide]$\Im(\cos{12^\circ}+i\sin{12^\circ}+\cos{48^\circ}+i\sin{48^\circ})^{6}$

Noting that $cis A \times cis B=cis (A+B)$ $cis \frac{\pi}{6}(cis \frac{-\pi}{10}+\frac{\pi}{10}$

Using the fact that $cos-x=cos x$

Since $Im(sin \frac{\pi}{6})^{6}=0$ thus $\Im(\cos{12^\circ}+i\sin{12^\circ}+\cos{48^\circ}+i\sin{48^\circ})^{6}=0$ :D [/hide]

Let $\{m,n,p,k\}\subset N^{*}$ and $z=\cos \frac{\pi}{k}+i\cdot\sin\frac{\pi}{k}\ .$ Prove that :

$\left(z^{m}+z^{n}\right)^{p}\in R\Longleftrightarrow$ $\left(\exists\right) s\in N$ so that $2ks=p(m+n)\ .$

[b][u]Particular case[/u][/b] : $k=15\ ,\ m=1\ ,\ n=4\ ,\ p=6$ and $2\cdot 15\cdot\boxed 1=6\cdot (1+4)\ .$


With a calculator, I get an answer of $8$.

[quote="jhcreinhardt"]Simplify:

$\csc^{2}{\frac{\pi}{7}}+\csc^{2}{\frac{2\pi}{7}}+\csc^{2}{\frac{3\pi}{7}}$[/quote]
I was considering using the fact that $\sin\frac{\pi}{7}$ is a root of the equation \[\textrm{Im}\left(\cos\frac{\pi}{7}+i\sin\frac{\pi}{7}\right)^{7}=0.\] However, I don't see a nice way of finding these roots.  Maybe someone else has a nicer way to do this.

well..

$\csc^{2}{\frac{3\pi}{7}}=\csc^{2}{\frac{4\pi}{7}}$

dunno if this helps..

Take a look at [url=http://mathworld.wolfram.com/TrigonometryAnglesPi7.html]this[/url].

Look at [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=106044]here[/url]

not that $\csc^{2}(x) = 1+cot^{2}(x)$ so you add 3 to 5 and get 8.


geometry
incenter
circumcircle
complex numbers
geometry theorems

Is there any way to obtain a nice formula in the complex plane for the Gergonne or Nagel points, for example by choosing a non-usual origin or smt. like that?

What is the definition of ''usual'' origin? The most suitable way (if I remember correctly this part of school math :)  ) to choose incenter as origin or do you need some ''nicer'' formulas?

Usual origin is circumcenter, as then $ |a|\equal{}|b|\equal{}|c|\equal{}1$

OK, then I gave 'unusual' one :lol:  To be serious, this is natural choice in most cases, when the problem related to  incircle of the triangle.

leshik,
I asked for a method to express the Nagel or the Gergonne point by a neat formula. The incenter is generally an often choice, but calcuting the asked points give us incredible difficulties in that case. That's why I created this topic, in the hope to find an answer.

Read Andreescu's Complex Numbers from A to Z:

Nagel point:
$ \alpha,\beta,\gamma$ - lengths of the sides of the triangle
$ a,b,c$ - complex numbers of vertexes
$ s$ - halfperimeter

$ z_N\equal{}\sum_{cyclic} a\left(1\minus{}\frac{\alpha}{s}\right)$


[size=150]Let a,b.c>0 Let a,b,c>0. Prove that $ /sqrt[3] {/frac{a}{b+c}+/sqrt[3] {/frac{b}{c+a}+/sqrt[3] {/frac{c}{a+b}>2}}}$[/size]
 :|

You should use [b][color=darkblue]latex[/color][/b]. Just put the expression between two $ \$$.

For example, if you want to write $ \sqrt[3]{\frac{4}{5}}$ then just put the text: \sqrt[3]{\frac{4}{5}} between two $ \$$.

$ \frac{{\sqrt 3 }}{{ab \plus{} c}} \plus{} \frac{{\sqrt 3 }}{{ac \plus{} b}} \plus{} \frac{{\sqrt 3 }}{{bc \plus{} a}} > 2;$

$ \mbox{Cred ca este vorba despre inegalitatea(parca am mai vazut aceasta inegalitate undeva, pe acest forum): } \\
\sqrt [3]{\frac {a}{b \plus{} c}} \plus{} \sqrt [3]{\frac {b}{a \plus{} c}} \plus{} \sqrt [3]{\frac {c}{a \plus{} b}} > 2;\ (\forall)\ a,b,c > 0.$


inequalities
function
symmetry
induction
combinatorics unsolved
combinatorics

$ \binom{n}{k} \le \frac{n^n}{k^k(n\minus{}k)^{n\minus{}k}}$

have you got any ideas ?

Hint:
[hide]
How many functions from $ \{1, \dots n\}$ to $ \{1, \dots n\}$ are there which map exactly $ k$ of the numbers to the interval $ \{1, \dots k\}$?[/hide]

(edited for proper english).

I don't understand you ...
Could you please give a more detailed hint ?

In a bit more detail (you may also want to check my edit above, which fixes the grammar problem in the original):
[hide]
You may want to try and solve the following two sub-problems:

1. How many functions are there (without any restriction) from $ \{1, 2, \dots n\}$ to $ \{1, 2, \dots n\}$?

2. How many functions are there once we add the additional restriction that [b]exactly[/b] $ k$ numbers get mapped to a number between $ 1$ and $ k$?

Further hint for #2:
[hide]First choose which $ k$ elements get sent to something between $ 1$ and $ k$, then choose where each element goes after that.  [/hide]

  [/hide]

[hide]
because of symmetry, we can consider only $ k\le\frac{n}{2}$.
We will use induction by $ k$. If $ k\equal{}1$, then $ \binom{n}{1}\equal{}n \le \frac{n^n}{1^1(n\minus{}1)^{n\minus{}1}}\equal{}n\left(1\plus{}\frac{1}{n\minus{}1}\right)^{n\minus{}1}$.
Let $ \binom{n}{k} \le \frac{n^n}{k^k(n\minus{}k)^{n\minus{}k}}$, then for $ k\plus{}1$:
\[ \binom{n}{k\plus{}1}\equal{}\frac{n\minus{}k}{k\plus{}1} \binom{n}{k}\le\frac{n\minus{}k}{k\plus{}1} \frac{n^n}{k^k(n\minus{}k)^{n\minus{}k}}\equal{}\frac{n^n}{k^k(k\plus{}1)(n\minus{}k)^{n\minus{}k\minus{}1}}\]
So we must prove that
\[ \frac{n^n}{k^k(k\plus{}1)(n\minus{}k)^{n\minus{}k\minus{}1}}\le\frac{n^n}{(k\plus{}1)^{k\plus{}1}(n\minus{}k\minus{}1)^{n\minus{}k\minus{}1}}\Leftrightarrow \\
\Leftrightarrow \left(\frac{k\plus{}1}{k}\right)^k\le\left(\frac{n\minus{}k}{n\minus{}k\minus{}1}\right)^{n\minus{}k\minus{}1} \Leftrightarrow  \left(1\plus{}\frac{1}{k}\right)^k\le\left(1\plus{}\frac{1}{n\minus{}k\minus{}1}\right)^{n\minus{}k\minus{}1}\]
But $ f(x)\equal{}\left(1\plus{}\frac{1}{x}\right)^x$ - increasing function, so it is true if $ k\le n\minus{}k\minus{}1 \Leftrightarrow k \le \frac{n\minus{}1}{2}$.
For $ k\equal{}\frac{n}{2}$ we have: $ \binom{n}{k}\equal{}\binom{2k}{k}\le\sum_{l\equal{}0}^{2k}\binom{2k}{l}\equal{}2^{2k}\equal{}\frac{(2k)^{2k}}{k^k\cdot k^k}\equal{}\frac{n^n}{k^k(n\minus{}k)^{n\minus{}k}}$
[/hide]


[quote="joejia"]Find the largest integer less than or equal to $2^{-3}\cdot 3^3$.[/quote]

[hide]$2^{-3}*3^3$

$\frac{1}{8}*27$

$\frac{27}{8}$

$\approx{3}$[/hide]

[quote="joejia"]Find the largest integer less than or equal to $2^{-3}\cdot 3^3$.[/quote]
[hide]
$ 2^{-3} \cdot 3^3 = \frac{1}{8} \cdot 27= \frac{27}{8}= 3 \frac{3}{8} \approx \boxed{3} $

[/hide]

[hide]
$2^{-3}\cdot 3^3=\frac{1}{8}\cdot 27=\frac{27}{8}=3.375$
Therefore, the largest integer less than or equal to $2^{-3}\cdot 3^3$ is $3$.[/hide]


Η ομάδα για τη ΒΜΟ 2007 είναι :

ΣΑΒΒΑΣ  ΜΙΧΑΗΛ           -------            3ο ΓΕΝΙΚΟ ΛΥΚΕΙΟ ΜΥΤΙΛΗΝΗΣ
ΒΑΦΕΙΔΗΣ ΑΝΑΣΤΑΣΙΟΣ  ------             ΙΔΙΩΤΙΚΟ ΛΥΚΕΙΟ ΒΕΡΟΙΑΣ
ΡΩΜΑΝΙΔΗΣ ΝΙΚΟΛΑΟΣ   ------            3ο ΓΕΝΙΚΟ ΛΥΚΕΙΟ ΓΙΑΝΝΙΤΣΩΝ
ΜΑΥΡΟΕΙΔΗ ΑΛΚΗΣΤΙΣ    ----               ΕΚΠΑΙΔΕΥΤΗΡΙΑ ΚΩΣΤΕΑ ΓΕΙΤΟΝΑ
ΡΑΠΑΝΟΣ ΝΙΚΟΛΑΟΣ         -----           ΠΕΙΡΑΜΑΤΙΚΟ ΛΥΚΕΙΟ ΠΑΤΡΩΝ
ΠΑΠΑΔΗΜΗΤΡΙΟΥ ΔΗΜΗΤΡΙΟΣ   ------   ΚΟΛΛΕΓΙΟ ΑΘΗΝΩΝ

Συγχαρητήρια παιδιά και καλή επιτυχία στη ΒΜΟ .

Εγώ δεύτερος αναπληρωματικός (από 18ος  τώρα 8ος   :D )
Καλή επιτυχία στην ΒΜΟ στα παιδιά αν και ελπίζω να μας πάνε και εμάς Ρόδο  :P  :wink:

Συγχαρητήρια σε όλα τα παιδιά κι όχι μόνο σε αυτά που περασαν :wink:
Είμαι πολύ χαρούμενος που είμαι ένας από αυτούς και αισθάνομαι πολύ τυχερός που είμαι ο τελευταίος :lol:
Θα ήθελα όμως να ρωτήσω: Αυτοί που πηγαίνουν στην ΒΜΟ δεν πηγαίνουν και στο IMO ή είναι διαφορετικοί οι μεν απ' τους δε :?:

Μπράβο σε   όλα τα παιδιά κι καλή τους επιτυχία στην Ρόδο
να ναι όλα τους καλά

συγχαρητηρια σε ολη την ομαδα και τους αναπληροματικους φυσικα,  :)

Συγχαρητήρια σε όλους που συμμετείχαν και [color=red]όλη η οκτάδα [/color]με ... ένα μετάλιο στις ΒΜΟ ή  ΙΜΟ !

                [u] Μπάμπης[/u]

Συγχαρητήρια στους βασικούς και στους αναπληρωματικούς και σε όλους όσους έδωσαν.
Καλή επιτυχία στη Βαλκανιάδα πρωτευόντως και στην Ολυμπιάδα δευτερευόντως.
Εγώ ατύχησα, πίκρα. :(
 Ήλπιζα για μια θέση στη δωδεκάδα (στον ήλιο  :D ) αλλά μου στοίχησε το άσχημο αποτέλεσμα στον Αρχιμήδη (22ος) σε συνδυασμό με ένα μέτριο αποτέλεσμα στο Μεσογειακό που οφείλεται κυρίως στο ότι δεν έλυσα το 4ο θέμα., το οποίο σημειωτέον έλυσα αμέσως μετά με ηρεμία στο σπίτι.  :mad:
Δεν πίστευα ότι μπορούσε να είναι τόσο εύκολο.
Θα προσπαθήσω τα δύο επόμενα χρόνια για ένα καλύτερο αποτέλεσμα με πιο μεθοδική δουλειά. :wink:

Φιλικά,
Κώστας Παππέλης

ΥΓ1: Θα συμβάλλω τώρα περισσότερο στο βιβλίο.
ΥΓ2: Θερμά συγχαρητήρια στο επιλεγμένο τιμ και ιδιαίτερα στο πολύ καλό φίλο μου Δημήτρη Παπαδημητρίου που κατάφερε από την πρώτη Λυκείου να μπει βασικός. :)

Συγχαρητήρια σε όσους πέρασαν αλλά και σε όσους δεν πέρασαν

Για τους πρώτους  καλή επιτυχία   :first: σε ό,τι ακολουθεί . Βάλτε τα δυνατά σας και παλέψτε το όσο μπορείτε :starwars:

γιατί δε λέτε και για τους μικρούς?

Sigxaritiria se ola ta paidia prwton gia tin simmetoxi tous kai eidikwtera sta paidia pou diakrithikan kai tha mas ekproswpisoun stin BMO kai (argotera) stin IMO. Euxomai kalo diavasma se olous wste na ginei epaksia h ekproswpisi stin Rodo.. Enas logos parapanw pou fetos ginetai sti xwra mas i Balkaniada, na diavasete parapanw... Prwta o Theos tha vriskomai ki egw ekei opote tha ta poume kai apo konta me kapoious apo esas.

Tha ithela ean kapoios exei ta onomata twn mikrwn diakrithentwn na ta dwsei kathws episis kai ta themata twn mikrwn gia ta opoia den egine akomi logos kai polloi apo emas den ta exoume... Fisika sigxaritiria anikoun kai se ekeina ta paidia pou tha mas ekproswpisoun fetos stin Voulgaria stin JBMO.

Alexandros

[quote="dr.koukoulas"]γιατί δε λέτε και για τους μικρούς?[/quote]

 :) Μάλλον πρέπει να μας τα πουν εκείνοι! Είχα ανοίξει παλαιότερα ένα χώρο για να βάζουν ασκήσεις για junior αλλά κανένας ΄μικρός ' δε συγκινήθηκε !
 Είναι κρίμα , αλλά δυστυχώς τους junior  στην Ελλάδα τους  εξωντώνουν κανονικά με τις ξένες γλώσσες και χιλιάδες άλλες βλακείες. Δέκα ώρες τη βδομάδα ξένες γλώσσες ! Στραβώνονται στο διάβασμα.Που να μείνει χρόνος να κάνουν λίγα μαθηματικά ! Και αυτοί που ασχολούνται, είναι γιατί κυρίως βρίσκεται κάποιος από πίσω και τους τα δίνει ... έτοιμα. Είναι φριχτό αυτό το πράγμα! Όλη η ελληνική κοινωνία έχει πέσει θύμα - της τεράστιας μπίζινες που έχουν στήσει τα ξενόγλωσσα τμήματα. Μια μαθήτριά μου χτες έδωσε για το Michigan , σήμερα μου λέει η μάνα της- φιλόλογος - θα συνεχίσει για το άλλο proficiency του Cambrige!  Βγήκα από τα ρούχα μου. Πάνω που το παιδάκι άρχισε να ζητάει ασκήσεις, γιατί της αρέσει τώρα η γεωμετρία , αρχίσαμε άλλον αγώνα. Θα με αναγκάσει να τη σταματήσω.
  Για να επανέλθω: Δεν παρακολουθεί κανένας junior αυτό το χώρο; Βλέπω στο αγγλικό φόρουμ ότι με τους junior γίνεται πανικός. ΜΑραθώνιοι στην άλγεβρα , στη γεωμετρία στους αριθμούς, ασκήσεις πάνε και έρχονται και τόσα άλλα! Εδώ που είναι οι ηλικίες από 13 έως 15 ; Μα διαβάζουν ξένες γλώσσες ! Καλά! Ας μην συνεχίσω γιατί θα πω και τίποτα βαρύ. Αν μου αρέσουν οι ξένες γλώσσες το γνωρίζετε. Σε ηλικία 29 χρονών και σε ένα εξάμηνο πήρα το Zertificat Deutsch  μόνο και μόνο για μπορώ να διαβάσω μερικά γερμανικά μαθηματικά βιβλία. Δε λέω για ρουμάνικα , ιταλικά , ισπανικά κλπ .Αλλά αυτό είναι άλλο πράγμα. Όποιος θέλει να μάθει ξένες γλώσσες μπορεί να μαθαίνει σε όλη του τη ζωή !
 Εδώ όλη η ζωή ενός παιδιού του Γυμνασίου χαραμίζεται στις ξένες γλώσσες ! Λες και σε αυτή τη χώρα όλοι θα γί νουν γκαρσόνια ! Εκτός και έχουν αποφασίσει να πάει εκέι το πράγμα , οπότε καλά μας κάνουν.
  Λοιπόν , παρακαλώ τους junior  να δραστηριποιηθούν στο φόρουμ , με τη βεβαιότητα ότι οι senior θα τους βοηθήσουν , όπου χρειαστεί.
 Να ανοίξουν δικά τους topic και να αρχίζουν να επικοινωνούν. Να λένε τον καυμό τους σε τελευταία ανάλυση και ότι άλλο τους αρέσει. Σιγά σιγά θα έρθουν και οι ασκήσεις.!

 [u]Μπάμπης[/u]

Θα πώ κι εγώ  σύντομα τον πόνο μου  :(   εγώ  έγραψα χάλια στο μεσογειακό κι για αυτό έχασα την 8η θέση μου κι την θέση για την ομάδα όμως σαν αναπληρωματικός θα δώσω τα πάντα για μια θέση στην ΙΜΟ είμαι Με βεβαιότητα κι παρρησία το λέω ο πιο πικραμένος από όλους μας αλλά πρέπει να το ξεπεράσω κι να παλέψω
 :starwars:  Καλή επιτυχία σε ¨ΟΛΟΥΣ ΑΝΕΞΑΙΡΕΤΑ ΣΤΗΝ ΒΜΟ
κι Συγχαρητήρια ΣΕ ΟΛΟΥΣ ΑΝΕΞΑΙΡΕΤΑ ΓΙΑ ΤΗΝ ΠΡΟΚΡΙΣΗ ΣΤΗΝ ΕΞΑΔΑ
Ιδιαίτερα μπράβο στον συμμαθητή μου Παπαδημητηρίου καθότι είναι Α λυκείου

[quote="ambros"]Ιδιαίτερα μπράβο στον συμμαθητή μου Παπαδημητηρίου καθότι είναι Α λυκείου[/quote]
Αγαπητέ Αμβρόσιε ευχαριστώ πολύ για τα καλά σου λόγια :) κι εύχομαι να είμαστε μαζί στον IMO
Ευχαριστώ επίσης τον καλό μου φίλο Κώστα (Protonio)και του εύχομαι καλή επιτυχία του χρόνου:wink:

Με την ευκαιρία εύχομαι Καλό Πάσχα σε όλους  :lol:

Gia τους Μικρούς :
1. Παγωνάκης Δημήτρης
2. Λογοθέτης Φώτης
3.Ντριάνκος Ιωάννης
4.Μπραζιτίκος Κωνσταντίνος (ο αδερφός μου  :D )
5.Λιάμπης Κωνσταντίνος
6.Παντέλα Στέλλα

Να αποδωσω κι εγω τα ευσημα απτην μερια μου σε ολους οσους διακρηθηκαν για την ομαδα και σε ολους οσους δεν τα καταφεραν να παλεψουν για κατι καλυτερο και να σκεφτουν οτι αν ηταν λιγοτερο αγχωμενοι και πιο συγκεντρωμενοι θα ειχαν παει σιγουρα καλυτερα. :wink:


Can anyone help with this – I’ve been struggling with it for days.  It relates to a sum in the sigma notation and if someone knew how to convert it into an explicit sum, it would help me hugely.  I have a sequence of 4 terms (‘n’) viz (in ascending order): 5.835, 6.221, 7.081 and 8.733.  I am trying to calculate three values based upon:

             n
a = n-1 Σ  (s -1) x(s)
               j=2     (n -1)

             n
b = n-1 Σ (s -1) (s - 2)  x(s)
               j=3   (n -1) (n - 2)
	  n
c = n-1 Σ  (s -1) (s - 2) (s - 3) x(s)
               j=3   (n -1) (n - 2) (n - 3)

where x(s) denotes the sth element in the sample as arranged in ascending order.

Any help would be very gladly received.

Thanks.

This is a triple-post.  Could a moderator lock or delete?  (Ditto the other College Playground copy of this -- I've already responded in PreOlympiad.)


trigonometry
limit
calculus
calculus computations

Let $x_{n}$ be the solution of the equation $\tan x = x$ from the interval $\displaystyle \left( (2n-1)\frac{\pi}{2}, (2n+1)\frac{\pi}{2} \right)$, $n \in N^{*}$. Compute the limit $\displaystyle \lim_{n \to \infty} n \pi \cdot \left[ (2n+1)\frac{\pi}{2} - x_{n} \right]$.

Ok,Nice Limit.
My appoarch:
First,Prove:
\[ \lim_{x \to +\infty} x({\frac{\pi}2-\arctan x})=1 \]
(It is not hard,use the L'hospital Rule.)
Secong,estimate:
\[ n \pi+\arctan n\pi \le x_n \le n \pi+\arctan (2n+1)\frac{\pi}2 \]
then prove
\[ \[\displaystyle \lim_{n \to \infty} n \pi \cdot \left[ (2n+1)\frac{\pi}{2} - n \pi-\arctan n\pi \right]=\lim_{n \to \infty} n \pi \cdot \left[ (2n+1)\frac{\pi}{2} - n \pi-\arctan (2n+1)\frac{\pi}2 \right]=1 \]
 :)


Hi everyone. I think everyone learns history in his or her school. But does your history textbooks only tell those "brilliant" parts of history? Does your history textbooks contain things that are obviously false?

When replying to the topic, please note which country do you come from.

Thanks.

[quote="h_s_potter2002"]Have you read [i]Lies My Teacher Told Me: Everything Your History Textbook Got Wrong[/i]?  It specifically talks about American History textbooks.  It's really suprising, all the stuff that's mentioned in it.

I'm from the USA.[/quote]

Is that seriously a real book?

[quote="SplashD"]Is that seriously a real book?[/quote]

It's an excellent book.  There was just an article in the New York Times about it the other day -- I think the author is coming out with an updated edition.

It's almost 400 pages and there's 2 editions. Which one is better?

The first edition (paperback) is from 1995.  It has the obvious advantange of being paperback and the obvious disadvantage of being more than a decade old.  The new (tenth anniversary) edition has the obvious disadvantage of being hardcover.  It has a "new preface by the author," but I can't tell just looking at the cover if it has newly researched material or not.  Maybe go to an actual bookstore and browse them to see (or, if you live near a bookstore where the clerks actually know things about the books (which is usually not true of the big chains), you can go and ask there).

I'll just check them both out from the local library. :D

Oh yeah, I've heard of that book, but I thought it'd be a real drag.  Is it actually... interesting?

the history book we used for AP was actually rather bold in its admission of american mistakes. i was impressed at parts. also it said that the underground railroad was all lies and that taft was very sad. we dubbed him eeyore. our history class was the best.

Please tell me more about the Underground Railroad being all lies.  Can you give a few more examples of things that that book claims are incorrect?

I don't really think my books (we didn't always use the book though) were incorrect, rather incomplete, which is also very important in history
1) USSR and China split due to ideological differences.  That was it.
2) Laos and Cambodia became communist too.  Nothing about the Pol Pot regime. So certainly nothing about the Vietnam-Cambodian war, giving me a somewhat twisted view of communist countries as nations that don't go to war with each other.
3) We never discussed which countries where predominantly Shia or Sunni muslim, it is only when I got out of high school that I learnt that the bulk of all shia muslims live in Iran.
4) No mention at all of the Nagorno Karabach conflict (well what do you expect,  it was ONLY TEN YEARS ago when we learnt modern history)
5)Very poor coverage of the Balkan conflict (not even an explanation of the differences in religion)

I do think I have a geography book from my third year in high school, that claimed that Israel took Gaza and Cisjordan from Egypt and Jordan respectively, just like the Golan heights.  Now as I see it, Egypt and Jordan took those areas themselves in 1948, but these areas were assigned to the Arabs in the UN Partition Plan.

O well, I like this topic, please do continue :wink:

our history books have too much text and a little pictures (only important ones)
really too much about ancient Greeks and Romans, medieval Europe (I've spoke with one polish girl, and she was amazed how much we learn about other parts of Europe)

in next grade I'll be learning modern history. it's going to be interesting. who knows what is written about WW2 and on.

Deimos, what grade is it that you will be going to next year (from earlier talks I remember, I suspect the last?) In that case it is the same form of history teaching I have had: in the last and sixth year of high school : modern history.

It is very interesting to compare this, I find this very important.  I once saw a Lituanian woman who had just found out a couple of years ago that USA was a serious player in the second World War, she was taught it was a Germany versus Russia thing.  Who knows what kind of weird ideas I have.

Our textbooks seem to leave out parts, they have actually never told an outright lie that I have found.  They do direct attention away from our failures and blame them on everything and everyone else.  That was just my US history (yes, I live in the USA).  I take world history next.  It will be interesting to see if their are any biases in that textbook.

I think that every history text has a bias, it is just some history textbooks display one more obviously.

Every textbook, while perhaps not outright lying, shades what
happens with its particular bias. That is not to say that it is always wrong, even a little bit.

For instance, a German and a Russian may both give true eyewitness accounts about what happened during WWII. However, talking to one or the other will leave you with a different impression about what happened in the war.

And again, a person may choose a certain perspective to look at an event in history, and decide that it best explains the event. Hence, the different possible interpretations and perspectives of events around cause debate about what really happened.

I recently read a book named "The lies of my theachers" ("Las mentiras de mis maestros") That talks about diferents parts of the history of Mexico. It has some great ideas and is realy light reading, but it talks only about a few "important" lies on History.
Personaly, I think that teaching a false history should make someone be as shame as if he/she were commiting a serious crime, but also should be that some who has more than 16 years old and has good education lets someone else tell him incredible lies about the history of his/her country

yes, it's going to be my last year in high school.

[quote="ShoeFactory"]I believe all of our textbook is true and without exaggeration.

USA[/quote]

All of a sudden I get this strange urge to order a history textbook from the USA.  :|

Bubka, I don't want this topic to diverge into the eternal battle about the lunar landings, but there are worse lies in history (if it's a lie!)


logarithms
number theory
prime numbers
number theory unsolved

Does there exist a natural number $ n$ wherefore we have $ ^2 \log(n)$ distinct prime numbers and their product is less than $ n/2$ ?

[quote="Jacobson"]$ ^2 \log(n)$ distinct prime numbers [/quote]
What is it?

I don't undestandt.
May be:
Does there exist n, suth that $ \tau (n)\ge ln(ln(n))$ and $ rad(n)<n/2$?
Were $ \tau(n)$ number of prime divisors n, $ rad(n)\equal{}\prod_{p|n} p$.

$ ^2 \log(n)$ is the natural logarithm with base $ 2$. Or what did you write above?

[quote="Jacobson"]$ ^2 \log(n)$ is the natural logarithm with base $ 2$. [/quote]

The natural logarithm is the logarithm with base $ e$.  We usually write the base-$ 2$ logarithm as $ \log_2 n$.

If I undestandt your question, it is obviosly:
$ P \equal{} p_1...p_k\ge 2*3*5*7*11... \ge 2^k$. If $ k \equal{} log_2(n)$, then $ P > 2^{[log_2(n)]} > 2^{log_2(n) \minus{} 1} \equal{} \frac n2.$
It work without "distinct".


Let $A=\left(\begin{array}{cc} s & 1 \\ 0 & t \end{array}\right)$ such that $|s|<1,\ |t|<1,\ s\neq t$ and $E=\left(\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right).$

Define $\sum_{n=0}^\infty A^n=\left(\begin{array}{cc} \displaystyle \sum_{n=0}^\infty a_n & \displaystyle \sum_{n=0}^\infty b_n \\\ \\ \displaystyle \sum_{n=0}^\infty c_n & \displaystyle \sum_{n=0}^\infty d_n \end{array}\right)$, Express $\left(\sum_{n=0}^{\infty} A^n\right)^{-1}$ in terms of $A,E.$

Sorry, I've got to use the letter $I$ instead of $E$ - force of habit, international tradition, and all of that.

Of course,  this is a geometric series.

$\sum_{n=0}^NA^n=(I-A)^{-1}(I-A^{N+1}).$

We now need to argue that $\lim_{N\to\infty}A^N=0.$ This follows from the fact that the spectral radius of $A$ is the maximum of $|s|$ and $|t|,$ which is in turn less than one.

So $\left(\sum_{n=1}^{\infty}A^n\right)^{-1}=I-A.$ (Or, to please the writers of the problem, $E-A.$)

If we want to be less technical than speaking of the spectral radius, the we do have enough information here to compute that

$A^n=\begin{pmatrix}s^n&\frac{s^n-t^n}{s-t}\\0&t^n\end{pmatrix}.$

From that explicit formula, it is clear that $A^n\to0$ as $n\to\infty.$


Functional Analysis
real analysis
real analysis unsolved

[i]Let $ (H_k)_{k\in\mathbb{N}}$ be a sequence of Hilbert spaces. Make $ \bigoplus_{k\in\mathbb{N}} H_k: \equal{} \left\{(x_k)_k\in \prod_{k\in\mathbb{N}} H_k\ |\ \sum\limits_{k\in\mathbb{N}}\|x_k\| < \infty\right\}$ into a Hilbert space by defining  $ \left < (x_k)_k|(y_k)_k\right > \equal{} \sum\limits_{k\in\mathbb{N}}\left < x_k|y_k\right >$.

Show that the direct sum has the following universal property: for every sequence $ (T_k)_k$ of uniformly bounded operators $ T_k\in L(H_k,F)$ between Hilbert spaces $ H_k,F$ there exists a unique bounded operator $ T: \bigoplus_{k\in\mathbb{N}} H_k\to F$ with $ T|_{H_n} \equal{} T_n$ for all $ n\in\mathbb{N}$[/i].

I was a bit confused about notation, so I introduced the following. Definine two 'projections on the $ n$-th coordinate':
$ \pi_n: \bigoplus_{k\in\mathbb{N}} H_k\to \bigoplus_{k\in\mathbb{N}} H_k: \ \ \ (x_k)_k \equal{} (x_1,x_2,...)\mapsto (0,...,0,x_n,0,...)$
$ \alpha_n: \bigoplus_{k\in\mathbb{N}} H_k\to H_n: \ \ \ (x_k)_k \equal{} (x_1,x_2,...)\mapsto x_n.$

Notice that  $ (x_k)_k \equal{} \sum\limits_{n\in\mathbb{N}} \pi_n(x_k)_k$ for every $ (x_k)_k\in\bigoplus_{k\in\mathbb{N}} H_k$. By $ T|_{H_n}$ we mean $ T\circ\pi_n: \bigoplus_{k\in\mathbb{N}} H_k\to F$. Now, the expression $ T|_{H_n} \equal{} T_n$ has to be read as $ T\circ\pi_n \equal{} T_n\circ\alpha_n$. [color=blue]Is this a correct interpretation? [/color]

[u]Existence[/u]: Let $ (x_k)_k\in\bigoplus_{k\in\mathbb{N}} H_k.$ Define $ T((x_k)_k) \equal{} \sum\limits_{k\in\mathbb{N}}T_kx_k$. It is clear $ T$ is linear. There is $ C > 0$ such that for all $ k\in\mathbb{N}$ we have $ \|T_k\| < C$. We show $ T$ is bounded:
$ \|T(x_k)_k\| \equal{} \|\sum\limits_{k\in\mathbb{N}}T_kx_k\|\leq\sum\limits_{k\in\mathbb{N}}\|T_kx_k\|\leq\sum\limits_{k\in\mathbb{N}}\|T_k\|\|x_k\| < C\sum\limits_{k\in\mathbb{N}}\|x_k\|$
[color=blue]I am having trouble with this estimate. I want to have $ \|T(x_k)_k\|\leq M\|(x_k)_k\|$, where $ \|(x_k)_k\|^2 \equal{} \sum_k\|x_k\|^2$. The squares don't seem to get right. Am I missing something obvious?[/color]
Since $ (T\circ\pi_n)(x_k)_k \equal{} T(0,..,0,x_n,0...) \equal{} T_nx_n \equal{} (T_n\circ\alpha_n)(x_k)_k,$ indeed $ T\circ\pi_n \equal{} T_n\circ\alpha_n$.

[u]Uniqueness[/u]: Assume $ S,T$ satisfy $ T\circ\pi_n \equal{} T_n\circ\alpha_n$ and $ S\circ\pi_n \equal{} T_n\circ\alpha_n$. Then

$ T(x_k)_k \equal{} T\sum\limits_{n\in\mathbb{N}} \pi_n(x_k)_k \equal{} \sum\limits_{n\in\mathbb{N}} (T\circ \pi_n)(x_k)_k$

$ \equal{} \sum\limits_{n\in\mathbb{N}} (S\circ \pi_n)(x_k)_k \equal{} S\sum\limits_{n\in\mathbb{N}} \pi_n(x_k)_k \equal{} S(x_k)_k.$ Hence $ T \equal{} S$.

[color=blue]Thanks in advance![/color]

It's probably cleaner this way: define the canonical injection $ \iota_n: H_n\to \bigoplus_{k\in\mathbb{N}} H_k$ by $ x_n\mapsto (0,...,0,x_n,0,...)$.
Then $ (x_k)_k\equal{}\sum\limits_{n\in\mathbb{N}}\iota_n(x_n)$ and $ T|_{H_n}: \equal{}T\circ\iota_n$.

It's clear how to define $ T$:

$ T((x_k)_k)\equal{}T\sum\limits_{n\in\mathbb{N}}\iota_n(x_n): \equal{}\sum\limits_{n\in\mathbb{N}}T_nx_n$.

My only problem is the boundedness.

$ \|T(x_k)_k\|^2\equal{}\|\sum_nT_nx_n\|^2\leq\left(\sum_n\|T_n\|\|x_n\|\right)^2<C^2\left(\sum_n\|x_n\|\right)^2$

...

$ \leq C^2\sum_n\|x_n\|^2\equal{}C^2\|(x_k)_k\|^2$ ??


algebra
polynomial
limit
real analysis
real analysis unsolved

[b][u]Problem:[/u][/b]
Cosider the sequence $\{ x_{n}\}: x_{1}=a;x_{n+1}=\frac{2x_{n}^{3}}{3x_{n}^{2}-1}\forall n\in N^{*}$
Find all $a\in \mathbb{R}$ such that $\{x_{n}\}$ is definite and convergence.

It has a long solution,so I want to find a short solution.

Tracing the fate of sequences defined by Newton's method for finding roots is notoriously delicate; I have on occasion generated interesting fractal patterns out of the basins of attraction for the various roots of cubic or quartic polynomials in the complex plane.

The sequence in this problem is precisely that for Newton's method for the sequence for solving the equation $g(x)=0$ where $g(x)=x^{3}-x.$

The only possible roots are $-1,0$ and $1.$

If $a=x_{1}=\pm \frac1{\sqrt{3}}$ then $x_{2}$ is undefined.

If $a>\frac1{\sqrt{3}}$ then $x_{n}\to1.$

If $a<-\frac1{\sqrt{3}}$ then $x_{n}\to-1.$

But the interval $\left(-\frac1{\sqrt{3}},\frac1{\sqrt{3}}\right)$ is going to require some more work.

[quote="Kent Merryfield"]But the interval $\left(-\frac1{\sqrt{3}},\frac1{\sqrt{3}}\right)$ is going to require some more work.[/quote]
Yes,let $g(x)=\frac{2x^{3}}{3x^{2}-1}$.$\forall c\in (-\infty,0]$ there is only one real number $x\in [0;\frac{1}{\sqrt{3}}]$ such that:$g(x)=c$

Let $a_{0}=\frac{1}{\sqrt{3}}$ there is $a_{1}\in [0;\frac{1}{\sqrt{3}}]: g(a_{1})=a_{0}$,there is $a_{2}\in [0;\frac{1}{\sqrt{3}}]: g(a_{2})=a_{1},...$
$\lim_{n\rightarrow \infty}a_{n}=\frac{1}{\sqrt{5}}$
with $|a|\not\in (a_{n})_{n=0}^{\infty}$ and $|a|\neq \frac{1}{\sqrt{5}}$,we have  $\{x_{n}\}$ is definite and convergence
but it's too long and ... :(


Let $a,b,c$ be positive reals, with $a+b+c=1$.

Prove that $\sqrt{\frac{1}{a}-1}\sqrt{\frac{1}{b}-1}+\sqrt{\frac{1}{b}-1}\sqrt{\frac{1}{c}-1}+\sqrt{\frac{1}{c}-1}\sqrt{\frac{1}{a}-1}\geq 6$

Let $a=\frac{x}{x+y+z}$  $b=\frac{y}{x+y+z}$  $c=\frac{z}{x+y+z}$
So the inequality becomes:
$\sqrt{\frac{(y+z)(x+z)}{xy}}+\sqrt{\frac{(x+y)(x+z)}{zy}}+\sqrt{\frac{(y+z)(x+y)}{xz}}\geq\ 6$
We use $AM-GM$ on each of the expressions under a radical:
$xy+yz+zx+z^{2}\geq\ 4\sqrt[4]{x^{2}y^{2}z^{4}}=4z\sqrt{xy}$
Similarly:
$xy+yz+zx+x^{2}\geq\ 4\sqrt[4]{x^{4}y^{2}z^{2}}=4x\sqrt{yz}$
$xy+yz+zx+y^{2}\geq\ 4\sqrt[4]{x^{2}y^{4}z^{2}}=4y\sqrt{xz}$
So that:
$\sqrt{\frac{(y+z)(x+z)}{xy}}+\sqrt{\frac{(x+y)(x+z)}{zy}}+\sqrt{\frac{(y+z)(x+y)}{xz}}\geq\ \sqrt{\frac{4z}{\sqrt{xy}}}+\sqrt{\frac{4x}{\sqrt{yz}}}+\sqrt{\frac{4y}{\sqrt{zx}}}$
So we have to prove that:
$\sqrt{\frac{z}{\sqrt{xy}}}+\sqrt{\frac{x}{\sqrt{yz}}}+\sqrt{\frac{y}{\sqrt{zx}}}\geq\ 3$
But, from $AM-GM$ we have
$\sqrt{\frac{z}{\sqrt{xy}}}+\sqrt{\frac{x}{\sqrt{yz}}}+\sqrt{\frac{y}{\sqrt{zx}}}\geq\ 3\sqrt[6]{\frac{xyz}{xyz}}$
And with this, the problem is solved.

$\left\{\begin{array}{c}\sum\sqrt{\frac{(a+b)(a+c)}{bc}}\ge 3\sqrt [3]{\frac{(a+b)(b+c)(c+a)}{abc}}\\\\ (a+b)(b+c)(c+a)\ge 8abc\end{array}\right\|$ $\Longrightarrow$ $\sum\sqrt{\frac{(a+b)(a+c)}{bc}}\ge 6\ .$


calculus
integration
function
trigonometry
algebra
partial fractions
trig identities

I'm stuck on this problem in Calculus II:

Integral of sec(x)*cot(x)

I think I have to do it with trigonometric identities.  I'd really appreciate the help.

P.S.
This is my first post in these forums. It's cool that I can get help with calculus on the internet.  I'll definitely start helping other people here as well.  Well, thanks in advance.

wait I got it:

secx=1/cosx
cotx=cosx/sinx

so it amounts to 1/sinx, which is cscx.  and the integral of that is just:

-ln |cscx+cotx| + C

For integration/antiderivative problems, I usually tell classes that the order in which I look at methods is (1) algebra, (2) substitution, (3) integration by parts. And with some subdivisions within that - partial fractions counts as algebra; "natural" substitutions get tried before you consider "forced" or "backwards" substitutions (including trigonometric substitutions).

With trigonometric functions, algebra is always likely. We have six names for what isn't really six "different" functions, so there's already a lot of redundancy built in. (Which is exactly what Shaun came around to.) And there are so many other trig identities to try.

One more algebraic idea: any trigonometric function can be the square root of something. So let a "$\sqrt{\cdot}$" symbol flicker in front of your eyes some of the time.

Example: $\int\frac{dx}{1-\sin x}$

Thought process: hmm... a square root in the denominator. I wouldn't hurt to try rationalizing that denominator. I wonder what would happen if I multiplied by $\frac{1+\sin x}{1+\sin x}$ ?

(I'm not saying this always works - just that it's worth trying some of the time.)


Thx very much...
but I put a wrong picture...
The question that I Wanted to do was:

How to solve this sum:

[img]http://img370.imageshack.us/img370/3658/cot2zi2.jpg[/img]

Thx

hello, prove that $ \sum_{k\equal{}1}^n\cot^2\left(\frac{\pi k}{2n\plus{}1}\right)\equal{}\frac{1}{3}(\minus{}n\plus{}2n^2)$.
Sonnhard.


Prove that if a prime number $ p\in\mathbb{P}$,can be expressed in the form $ p\equal{}x^2\plus{}y^2$,where $ x\geq y$,then there is only unique representation of $ p$,as the sum of two squares.

Suppose to the contrary, $ p \equal{} a^2\plus{}b^2 \equal{} c^2\plus{}d^2$.
Then $ a^2\minus{}c^2 \equal{} d^2\minus{}b^2$, so $ (a\minus{}c)(a\plus{}c)\equal{}(d\minus{}b)(d\plus{}b)$. Using famous factoring lemma, we can find $ k, l, m, n$ so that $ a\minus{}c \equal{} kl, a\plus{}c \equal{} mn, d\minus{}b \equal{} km, d\plus{}b \equal{} ln$. Since pairs $ (a,c)$ and $ (b,d)$ are distinct, $ k, l, m, n$ are non-zero. Then $ a \equal{} \frac {kl\plus{}mn}{2}, b \equal{} \frac {ln \minus{} km}{2}$, so $ p \equal{} a^2\plus{}b^2 \equal{} \frac { (kl\plus{}mn)^2 \plus{} (ln\minus{}km)^2 }{4} \equal{} \frac { (l^2 \plus{} m^2)(k^2 \plus{} n^2)}{4}$. However, this is clearly contradiction, since for $ p$ to be prime, $ l^2\plus{}m^2 \equal{} 4$ or $ k^2 \plus{} n^2 \equal{}4$ must occur, which cannot happen since $ k, l, m, n$ are all non-zero.

[quote="N.T.TUAN"]But $ 5 \equal{} 2^2 \plus{} 1^2 \equal{} ( \minus{} 1)^2 \plus{} ( \minus{} 2)^2$.  :lol:[/quote]
Uhh,of course $ x,y\in\mathbb{N}$

If so, it is well-known.  :P
http://tuan.nguyentrung.googlepages.com/ml.JPG

P.S. I can not upload this file.

Let p can be expressed by a^2 + b^2 and c^2 + d^2.(a more than b and c more than d) And a and c be distinct numbers. Without lost of generality, we can take a more than c. Product of two equation: a^2.c^2 + b^2.d^2 + a^2.d^2 + b^2.c^2. This is p^2. p^2 = (ac+bd)^2 + (ad-bc)^2 as well, p^2 = (ad+bc)^2 + (ac-bd)^2. By the equation(ac+bd)(ad+bc) = (a^2 + b^2)cd + (c^2+d^2)ab = p(ab+cd) we can say that, p divides at the least of one (ac+bd) and (ad+bc) (p is prime.). If p divides ac+bd so ad-bc=0. And then ad=bc and a/c = b/d. We took a more than c so b is more than d, and (a^2 + b^2) more than (c^2 + d^2). Contradiction. If p divides ad+bc so ac-bd=0 and a/b=d/c. a more than b so d more than c. Contradiction. So there is only unique representation of p, as the sum of two squares.

Sorry. I don't know how to use latex.

[quote="fallinlovewithmaths"]

Sorry. I don't know how to use latex.[/quote]
See here http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php  :wink:

[quote="Erken"][quote="epitomy01"] Using famous factoring lemma, we can find $ k, l, m, n$ so that $ a \minus{} c \equal{} kl, a \plus{} c \equal{} mn, d \minus{} b \equal{} km, d \plus{} b \equal{} ln$. Since pairs $ (a,c)$ and $ (b,d)$ are distinct, $ k, l, m, n$ are non-zero.[/quote]
It will be great if you post a proof to this lemma.[/quote]

[b] Lemma: Suppose $ xy \equal{} zw$ for integers $ x,y,z,w$. Then there are integers $ k,l,m,n$ with $ x \equal{} kl, y \equal{} mn, z \equal{} km, w \equal{} ln$. [/b]

It suffices to find $ k, m$ so that $ z \equal{} km$ and $ k|x$ and $ m|y$. Then we choose $ l \equal{} \frac {x}{k}$, $ n \equal{} \frac {y}{m}$ and since $ xy \equal{} zw$, it follows $ w \equal{} ln$. Since $ z | xy$, choose $ k$ to be the largest divisor of $ z$ so that $ k|x$. Choose $ m$ so $ z \equal{} km$. We need to prove $ m | y$, or $ \frac {z}{k} | y$, or $ z | ky$. If $ z|x$, then $ k \equal{} z$ and $ z|ky$ is clearly true. Otherwise, $ (z, \frac {x}{k}) \equal{} 1$ (since otherwise, we could replace $ k$ by a larger number). Then, since $ z | ky (\frac {x}{k})$, it follows $ z|ky$, as required.

another way:

assume that $ p \equal{} a^2 \plus{} b^2 \equal{} x^2 \plus{} y^2$

now we have:

$ (ax \plus{} by)(ay \plus{} bx) \equal{} xy(\underbrace{a^2 \plus{} b^2}_{ \equal{} p}) \plus{} ab(\underbrace{x^2 \plus{} y^2}_{\equal{} p}) \equal{} p(ab \plus{} xy)$

now we have two cases:

$ \boxed{\textrm{Case 1: } p\mid ax \plus{} by}$

first of all note that:

$ p^2 \equal{} (a^2 \plus{} b^2)(x^2 \plus{} y^2) \equal{} (ax \plus{} by)^2 \plus{} (ay \minus{} bx)^2$ (*)

in the other hand we have:

$ p\mid ax \plus{} by\Rightarrow ax \plus{} by \equal{} kp$ where $ k\in\mathbb{N}$

so we can rewrite (*) as follow:

$ p^2 \equal{} p^2k^2 \plus{} \underbrace{(ay \minus{} bx)^2}_{\geq 0}\geq p^2k^2$

so we must have $ k \equal{} 1$ which yields $ p \equal{} ax \plus{} by$,and also we must have $ ay \equal{} bx$ which yields $ a \equal{} \frac {bx}{y}$.

so we have:

$ p \equal{} ax \plus{} by \equal{} \frac {bx^2}{y} \plus{} by \equal{} \frac {b(\overbrace{x^2 \plus{} y^2}^{ \equal{} p})}{y}$

$ \Rightarrow p \equal{} \frac {b}{y}p\Rightarrow b \equal{} y$

now from $ ay \equal{} bx$ we get that $ a \equal{} x$

as wanted...

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

$ \boxed{\textrm{Case 2: } p\mid ay \plus{} bx}$

this case can be DONE in a similar way as the previous case...

so in either case $ p \equal{} a^2 \plus{} b^2 \equal{} x^2 \plus{} y^2$ yields $ a \equal{} x,b \equal{} y$...

QED


Given the transition matrices below, find all communicating classes and classify all states:
The probabilities don't matter for this exercise, so I'm just going to put ones for non-zero entries, even though I know this is absurd mathematically:
(A)
1 1 0 1
1 0 0 1
1 0 0 0
1 0 1 0
I believe that the communicating classes are {1,2,4} and {3} and all these are ergodic.

(B)
1 1 1 1
0 0 1 1
1 0 1 0
0 0 1 0
I think the communicating classes are {1,3}, {2}, and {4}.
I'm not quite sure, but I believe states 1,3, and 4 are ergodic, and 3 is of period 2.

(C)
1 1 0 0
1 1 0 0
0 0 0 1
0 0 1 1
The communicating classes are {1,2} and {3,4} with {1,2} of period 2 and {3,4} ergodic.

Is this right? I think I have the communicating classes right, but I'm not sure about the state classifications. Thanks in advance for any help.

(A) All four classes communicate (you can go from 1 to 2, from 2 to 4, from 4 to 3, and from 3 to 1).

(B) Again all four communicate - from any one of them, you can reach any other (try seeing for yourself why this is true). Also, since from 1 you can go to 1, the chain has period 1.

(C) You got the communicating classes right, but why do you think {1, 2} has period 2?

I'm having trouble figuring out these things. Can you classify these states or tell me which ones I did wrong?

Okay, I figured out the first two; I just don't know how to classify the states for the last transition matrix. I know they're closed, but are both classes aperiodic?

[quote="BrainMan1"]... so I'm just going to put ones for non-zero entries, ...[/quote]
That associates each of these matrices to a digraph (directed graph). Since these are fairly small matrices ($ 4\times 4$) the digraphs are quite small (4 nodes). Have you tried drawing pictures of the digraphs? I would find that to be very helpful here.

Yes, I drew the picture for the last one:

arrows from 1 to 2, 2 to 1, 1 to 1, and 2 to 2
arrows from 3 to 4, 4 to 3, 3 to 3

I just don't know if they're ergodic or periodic.

if it's a bipartite graph, period is a multiple of 2.
in general, if you can divide the states into n parts, such that from part 1 you can only go to part 2, from part 2 you can only go to part 3, and so on to part n and back, period is a multiple of n.
there are some helpful shortcuts you'll discover along the way; if a state in a class leads to itself, the class must be aperiodic (do you see why?). This trick works for example for your third problem.


i have to do my math fair project on KNOT THEORY, and basically i cant find any practical uses for it.

can somebody help me out?

The magic of Google:  http://www.google.com/search?hl=en&q=knot+theory+applications&btnG=Google+Search

Umm, speedcuber, please d0n'T tAlK LiKe tHiS, kay??!!! Please dont make a titled topic like that; it's distracting. (I'm saying when you typed the topic title KNoT THEORY, stating about using one lowercase letter when everything's supposed to be uppercase. I know it doesnt allow topic titles in all caps, so i think u should just use initial caps or all lowercase letters)


When two lines are drawn in a plane, they can create four non-overlapping regions. If five straight lines are drawn in the plane, what is the largest number of non-overlapping regions they can create?

is there an easy formula for this?
-jorian

[quote="Logg"]they're all triangle numbers...[/quote]
That is for the number of intersections ;)

[quote="jhredsox"]ya, that sounded really akward.

because.......

[code]
lines    most sections
1        2
2        4
3        7
4        11?
5        16?
[/code]

i guessed on the last two numbers, because this might be a pattern:

where u add one more than the last time

(how do you write that?)

-jorian[/quote]
If you're right then it would just be $\frac{n(n+1)}{2}+1$

[quote="nebula42"][quote="jhredsox"]ya, that sounded really akward.

because.......

[code]
lines    most sections
1        2
2        4
3        7
4        11?
5        16?
[/code]

i guessed on the last two numbers, because this might be a pattern:

where u add one more than the last time

(how do you write that?)

-jorian[/quote]
Well, if that happens, it means that the formula is simply a quadratic. Try to see if you can find that quadratic.
[hide="Hint"]Let the quadratic be $ax^{2}+bx+c$, and see if you can find the values of $a,b,$ and $c$ by plugging in values for $x$.[/hide][/quote]
Hey, um, we posted at the same time, but um the numbers are all triangular numbers plus 1.

[quote="jhredsox"][quote="nebula42"][quote="jhredsox"]ya, that sounded really akward.

because.......

[code]
lines    most sections
1        2
2        4
3        7
4        11?
5        16?
[/code]

i guessed on the last two numbers, because this might be a pattern:

where u add one more than the last time

(how do you write that?)

-jorian[/quote]
Well, if that happens, it means that the formula is simply a quadratic. Try to see if you can find that quadratic.
[hide="Hint"]Let the quadratic be $ax^{2}+bx+c$, and see if you can find the values of $a,b,$ and $c$ by plugging in values for $x$.[/hide][/quote]

how could you find 3 variables by doing that, wouldn't that take forever?
-jorian

p.s. i think i'm top 5 for the wiki competition, could someone check?[/quote]
Actually, it is a lot quicker than you would think, especially since the numbers are neat.

@bpms

could you show an example of the work to this problem?

-jorian

Sure.
[hide]when x=0 you get $c=1$. when x=1 you get $a+b+c=2$. Which is $a+b=1$. When x=2 you get $4a+2b+c=4$. Which is $4a+2b=3$ Solving this for a and b, we get $a=\frac{1}{2}$ and $b=\frac{1}{2}$. So we get our formula $\frac{x^{2}}{2}+\frac{x}{2}+1$ which can be reordered as the one I said earlier.[/hide]

i kind of get it, but could you make a table or something, i didn't really understand it too much
-jorian :blush:

Basically what he was doing was substituting values of $x$, for which he already knew $f(x)$. We know this is a quadratic function $ax^{2}+bx+c$, so we will be left equations in the form $na+mb+c=q$ satisfying $f(x)$. Then, if we have 3 equations, we can just solve for a, b, c.

Here is something algebraic that I suggest you know. If you have n  variables than you need n unparallel equations to find each variable.

[quote="jhredsox"]what carl said is a little bit confusing, but i get it

so, basically u isolate the 3 variables to the 3 equations?

-jorian[/quote]
SOlving is faster in this case than isolation, but isolation works for more complicated ones.

[hide]Well, we know that 1 line splits the plane into 2 regions and 2 lines split it into four regions.  Note, that ever time a line is created, a region occurs.  Now, that if every time a line hits another line a region also occurs.

So that is why we can say that, for the nth line added, n regions will be added.

So, that is how by induction the formula of n(n+1)/2+1 comes...[/hide]

If I'm not mistaken, this is a part of Algebra 2.


Point $ Z$ is the midpoint of $ \overline{MN}$. Point $ P$ is the midpoint of $ \overline{ZN}$. Point $ Q$ is the midpoint of $ \overline{PN}$ and Point $ R$ is the midpoint of $ QN$. If $ \overline{PR}$ is $ 24$ inches long, how many feet long is $ \overline{MN}$. Express your answer as a mixed number.

[hide="solution"]
On a number line, that would look like this.
[asy]import graph;
size(250);
draw((-0.5,0)--(4.5,0),Arrows);
dot("M",(0,0),S,black);
dot("N",(4,0),S,black);
dot("Z",(2,0),S,black);
dot("P",(3,0),S,black);
dot("Q",(3.5,0),S,black);
dot("R",(3.75,0),S,black);[/asy]
We can let the length of $ QR$ be $ n.$ It is obvious that $ PR\equal{}3n,$ so $ 3n\equal{}24\text{ inches}\Rightarrow n\equal{}8\text{ inches.}$ Since we divided segment $ MN$ into $ 16$ equal parts, the length of $ MN$ is $ 16\times8\text{ inches}\equal{}128\text{ inches}\equal{}\boxed{10\frac{2}{3}\text{ feet}}.$[/hide]


Can anyone help me with this proof please?

Suppose that $ \Omega \in \mathcal{F}$ and that $ A,B \in \mathcal{F}$ implies $ A \minus{} B \equal{} A \cap B^C \in F$. Show that $ \mathcal{F}$ is a field.

This question doesn't make sense as stated. What hypotheses do we have about the algebraic structure of $ F$?

Hi,

I am very sorry..

The definition of a field holds for a completely arbitrary nonempty space $ \Omega$.
A class $ \mathcal{F}$ of subsets of $ \Omega$ is called a [i]field[/i] if it contains $ \Omega$ itself and is closed under the formation of complements and finite unions:
(i) $ \Omega \in F$
(ii) $ A \in \mathcal{F}$ imples $ A^C \in \mathcal{F}$;
(iii) $ A, B \in \mathcal{F}$ implies $ A \cup B \in \mathcal{F}$.

But we can let $ \Omega$ be a finite set of of points $ \omega$?

Is this enough to be able to show it?

this is not a field in the usual algebraic sense, it's field of sets or algebra of sets instead - you should have posted this in analysis forum I think.

Anyway, where do you get stuck for this question? What are your attempts? It looks pretty straightforward to me.

I spoke with a friend of mine.

We just need to show that $ A^C \in F$ and $ A \cup B \in F$.
Now, $ \Omega \in F$, so $ \Omega \minus{} A \equal{} A^C \in F$ according to your definition.
And, $ A^C \in F$, so $ A^C \minus{} B \equal{} \Omega \minus{} (A \cup B) \in F$.
So, $ (\Omega \minus{} (A \cup B))^C \equal{} A \cup B \in F$.


Just wondering if there is some sort of correlation.

Post your nationality and how well you did.

I'm Eastern European and I got 51st (knocked from 41st due to tiebreakers).

I'm half Eastern European (Moscow, Russia) and half Asian (Taiwan).

5th place.

Eastern European, Polish, knocked from 12th to 14th on tiebreakers

Yeah, and just look in the yearbook. I can tell you right now that there are about 100 (obvious) Asians and many Europeans. I don't know about the others.

I'm Asian, only one on the Nebraska team. (Don't look at the yearbook picture, that's a crappy photo)

The majority of the kids there were Asian.

Asian.

50 bucks that one of the next ten posters says "AZN PRIDE!".

And Antarctican.
Antarcticans had the highest average written round score of all nationalities.

No.
Antarctican Average= 30pts/1penguin=[b]30[/b]
30 doesn't equal 0.

Yeah. You can extrapolate information really easily with one test subject. :roll:

Ok, the sampling size was a little small.
So what?

Western European. Specifically, 11/16 German, and a few other western European countries make up the remaining 5/16.

AZN PRIDE!

Please explain the 11/16 German part.
I am 1/2 German.  I think Germany is Eastern European

Eastern Europe is generally considered Slavic, like Ukraine/western Russia. Germany is kind of in the middle, but I'm considering it western.

11/16 means that 11 of my 16 great-great-grandparents were (approximately) fully German.

Yeah, Germany was half-and-half before the fall of the Berlin Wall, but it's probably considered Western now.

yeah im 1/4 german, 1/4 polish, 1/8 english, 1/8 welsh, 1/8 scottish, 1/8 norwegian......
i just went ahead and put western european.. i was 7th before coutndown..then  i got semifinalist

Can't we just say European? Who cares about which specific countries or part of Europe we're from?

Oh, yes, and I am European mostly.

Billy

[quote="jb05"]I'm 1/4 Russian and the rest is really mixed, but I think most of the rest is in Western Europe, so I put Western European.

I thought I may bring up this story. Last year, when we were discussing our background, a student said that he was 1/3 German. Incredibly, the discussion went on until I pointed out the error a while later. As I noted, this is possible only if humans have been around for an infinite amount of time. Then the family tree would still be a little strange (g is german, ? is anything):

?
??
g???
gg??????
ggggg???????????
etc.[/quote]

It's possible, but it involves multiple genitals.

[quote="dwx314"]Erm, I wasn't a competitor in the national competition, but I'd like to see what it is.
Could you modify the poll so that there is a option you can select that's like "Not There", or something so that we can see it though?
Thanks[/quote]

just click view results at the bottom.

[quote="biffanddoc"][quote="rorrimimage"]i am part polish/ukrainian/german/argentinian/canadian/... (3-4 generations...) i think, but still about 80% American...i got 115th...

-rorri :?[/quote]You can't be exactly 80% american[/quote]

1/2+1/4+1/32+1/64+1/512+1/1024+1/8192+1/16384+1/131072+1/262144+1/2097152+1/4194304+1/33554432 gets you close enough... although that's like 25 generations... but that's besides the point... and he said about 80% american not exactly 80%...

I'm 11/16 Deutsch,  and the other parts are W. Euro.  I did poorly...155th

[quote="bigHC"]im indian and i got 10th place[/quote]Pardha?

i 'm asian.... one of my teammate said that 90% of the ppl at Nationals were asian.... not quite... but still... i'm from South Dakota... i got 203(ya i know...) :blush:

I'm asian, but I did so badly I'm a disgrace to all asians.
[hide]At least the first digit on my rank starts with a 1, but it's 3 digits...[/hide]

Xantos you were not the #1 penguin. ShlomaDaPenguin was.

[quote="Xantos C. Guin"]Penguin! Pride.
I did great.[/quote]
If Penguin!(factorial)=Penguin then
Penguin = 1
Xantos C. Guin :!=: 1  so
Xantos C. Guin is not a real penguin
..................................................................
JK..... :P  :lol:  :P  :lol:  :P  :lol:  :P  :lol:


For a positive integer $ n\geq 5$, let $ a_i,b_i$ ($ i \equal{} 1,2,...,n$) be integers satisfying the following two conditions:

(i) The pairs $ (a_i,b_i)$ are distinct for $ i \equal{} 1,2,...,n$
(ii) $ |a_1b_2 \minus{} a_2b_1| \equal{} |a_2b_3 \minus{} a_3b_2| \equal{} ... \equal{} |a_nb_1 \minus{} a_1b_n| \equal{} 1$.

Prove that there exists indices $ i,j$ such that $ 1 < |i \minus{} j| < n \minus{} 1$ and $ |a_ib_j \minus{} a_jb_i| \equal{} 1$.

[hide="Dont know if it works"]We shall prove it by contradiction method.
Assume to the contrary that $ |a_ib_j \minus{} a_jb_i|\neq 1$ for all indices $ i,j$ satisfying $ 1 < |i \minus{} j| < n \minus{} 1$
\[ b_3(a_1b_2 \minus{} a_2b_1) \plus{} b_1(a_2b_3 \minus{} a_3b_2) \equal{} a_1b_2b_3 \minus{} a_2b_1b_3 \plus{} a_2b_1b_3 \minus{} a_3b_1b_2
\]

\[ \equal{} a_1b_2b_3 \minus{} a_3b_1b_2 \equal{} b_2(a_1b_3 \minus{} a_3b_1)
\]

\[ |b_3(a_1b_2 \minus{} a_2b_1) \plus{} b_1(a_2b_3 \minus{} a_3b_2)| \equal{} |b_2(a_1b_3 \minus{} a_3b_1)|(1)
\]

[b]Lemma[/b]: $ (a_1b_3 \minus{} a_3b_1)\neq 0$
if $ (a_1b_3 \minus{} a_3b_1)\equal{}0$
from $ (1)$:   $ |b_3(a_1b_2 \minus{} a_2b_1) \plus{} b_1(a_2b_3 \minus{} a_3b_2)|\equal{}0$
and since $ |a_1b_2 \minus{} a_2b_1|\equal{}|a_2b_3 \minus{} a_3b_2|\equal{}1$:
$ |b_1|\equal{}|b_3|$
Similarly it can be proven that
$ |a_1|\equal{}|a_3|$
But we have $ (a_i,b_i)$ are distinct for all $ i$ so we have contradiction
------(End Lemma)------
\[ |b_3(a_1b_2 \minus{} a_2b_1) \plus{} b_1(a_2b_3 \minus{} a_3b_2)| \le |b_3(a_1b_2 \minus{} a_2b_1)| \plus{} |b_1(a_2b_3 \minus{} a_3b_2)|
\]

\[ \equal{} |b_1| \plus{} |b_3|(2)
\]
By lemma $ |a_1b_3 \minus{} a_3b_1|\neq 0$ and by assumption $ |a_1b_3 \minus{} a_3b_1|\neq 1$. Hence:
\[ |a_1b_3 \minus{} a_3b_1|\ge 2
\]
Thus:
\[ |b_2(a_1b_3 \minus{} a_3b_1)|\ge 2|b_2|(3)
\]

From $ (1)$, with the help of $ (2)$, $ (3)$, we conclude that:
\[ |b_1| \plus{} |b_3|\ge 2|b_2|
\]
Similarly we'll have:
\[ |b_2| \plus{} |b_4|\ge 2|b_3|
\]

\[ |b_3| \plus{} |b_5|\ge 2|b_4|
\]

\[ \cdots \cdots \cdots
\]

\[ |b_n| \plus{} |b_2|\ge 2|b_1|
\]
Suppose $ |b_k|$ is the maximum of $ |b_i|$s.
$ ($The arithmetic mean of $ |b_{k \minus{} 1}|$ and $ |b_{k \plus{} 1}|)\ge$ $ |b_k|$.
Hence both $ |b_{k \minus{} 1}|$ and $ |b_{k \plus{} 1|}$ must be equal to $ |b_k|$
Continuing in this manner, it is evident that all $ |b_i|$s must be equal.
Similarly it can be proven that all $ |a_i|$s are equal.
then $ |a_1b_2| \equal{} |a_2b_1|$ and both integers.
Hence $ |a_1b_2 \minus{} a_2b_1| \equal{} 0\text{ or }2|a_1b_2|$, but this can't be $ 1$
Hence contradiction, as desired.[/hide]


A machine takes two consecutive integers, each 20 or under, multiplies them and adds 17. It then determines whether the output is prime or composite. If it is prime the printout reads P, if composite the printout reads C. There are two different sets of consecutive integers such that the printout will read C, what is the greater number in each set?

yay, i didn't post this one!  But there will be more in a few minutes.


inequalities
induction
combinatorics proposed
combinatorics

Let $ 0 < a_{1} < a_{2} < \ldots < a_{n}$ and let  $ e_{i} \equal{} \pm 1$ Show that
$ \sum_{i \equal{} 1}^{n}e_{i}a_{i}$ assumes at least $ \binom{n \plus{} 1}{2}$ distinct values
when the $ e_{i}$ range over the $ 2^{n}$ different possible combinations.
------------------------------------------------------------------------------------------------

[hide="Questions on the Solution"]
This can be splved by induction. Assuming it holds for say up to $ a_{k}$ we prove it holds for
$ a_{k \plus{} 1}$. So we have to find $ \binom{k \plus{} 2}{2} \minus{} \binom{k \plus{} 1}{2} \equal{} k \plus{} 1$ more sums that
are distinct from the others.
In the book by Larson, this is done by letting $ S \equal{} \sum_{i \equal{} 1}^{k}a_{i}$
and then these becomes the $ k \plus{} 1$ additional terms:
$ S \plus{} a_{k \plus{} 1}, S \plus{} (a_{k \plus{} 1} \minus{} 2a_{1}), S \plus{} (a_{k \plus{} 1} \minus{} 2a_{2}), \ldots , S \plus{} (a_{k \plus{} 1} \minus{} 2a_{k})$
Ok these are distinct from each other, [color=red]but what I don't get is how one shows that
they are greater than the $ \binom{k \plus{} 1}{2}$ rest.[/color].
In the book this is shown by an inequality like this:
$ S \plus{} (a_{k \plus{} 1} \minus{} 2a_{k}) > S \plus{} (a_{k \plus{} 1} \minus{} 2a_{k \plus{} 1}) \equal{} S \minus{} a_{k \plus{} 1}$

but still dont se why  :oops: . I mean isn't one of the $ \binom{k \plus{} 1}{2}$ sums equal to $ S$ in
the case when all $ e_{i} \equal{} 1$? In that case how does the inequality help?

[/hide]

please help on this

Ok, the reason that you're having a problem is because you are most likely envisioning the inductive step wrong.  Notice that before the inductive hypothesis is applied, the book is fixing $ e_{k+1}=-1$.  Then when the inductive step is applied on the rest of the numbers $ a_{1},a_{2},\dots,a_{k}$, we have our $ {k+1\choose 2}$ distinct sums.  But notice that the biggest of these distinct numbers must be $ S-2a_{k+1}$.  Now once we change $ e_{k+1}$ to $ 1$, we can obtain the $ k+1$ new values it enumerates, thereby finishing the induction, since as your final inequality states, all those numbers are indeed bigger than the biggest when $ e_{k+1}=-1$.  I think your fault was assuming that $ e_{k+1}=0$.  Does it make more sense now?


[hide]Well, if you assume no student studies more than two languages, since 90+75+42 is equal to 207, you have 107 students studying exactly two languages. Obviously you must have overcounted by at least 7, resulting in 7 students that must study all three languages.[/hide]

[hide]Edit: I think LynelleYe's method works... *thinks to self, "Why didn't I think of that?"*[/hide]

I don't know why but for some reason this seems to work.
[hide]Just take each number's difference from 100, add them up and subtract those from 100. I think it comes out 7.[/hide]

Try and test it out with other problems.

It worked with another mathcounts problem on some national test, but I'm just not sure why it works.

If I could write my complex thoughts into words, then I could tell you why.  But I can't think of the words to explain why.

[quote="ln(dx/dy)"]Is there an easy way to do this?

In a school of 100 students, 90 study English, 75 study Spanish and 42 study French. Every student must study at least one of the three languages. What is the least possible number of students who could be studying all three languages?[/quote]

[hide]If you want the least students, each student must take at least twoclasses

there are 207 classes in total so 207-200=7.[/hide]

True, but you still would have to do almost as much work, wouldn't you?


Find the value of $ \left(\frac{1}{4}\right)^{\log _2 3}$


function
integration
real analysis
real analysis unsolved

Let $ f$ and $ g$ be continuous positive functions defined on the interval $ [0, +\infty)$, and let $ E \subset[0,+\infty)$ be a set of positive measure. Prove that the range of the function defined on $ E \times E$ by the relation \[ F(x,y)= %Error. "dispalymath" is a bad command.
\int_0^xf(t)dt+ %Error. "dispalymath" is a bad command.
\int_0^y g(t)dt\] has a nonvoid interior.

[i]L. Losonczi[/i]

Let $A = \{ t >0 \colon t = \int_0^x f(s)ds \text{ for some } x \in E\},$ and define $B$ analogously with $g$ in place of $f.$ It suffices to show that $|A|, |B| > 0,$ and then classical methods with convolutions imply that $A+B$ contains an interval -- and hence the range of $F$ has nonempty interior.

Let $\tilde{f}(x) = \int_0^x f(s)ds.$ Then $\tilde{f}$ is strictly monotone on $[0,+\infty).$ Moreover, since $f>0$ is continuous, we have, for any compact set $K \subset [0,+\infty),$ that $\tilde{f}$ is bi-Lipschitz over $K$. That is, $\tilde{f}$ is invertible, and, letting its inverse be $\tilde{f}^{-1},$ then $|\tilde{f}'|,|(\tilde{f}^{-1})'| \le C$ for some constant $C.$

It is not hard to show that this condition implies that, for any set $S \subset K,$ we have $|S|/C \le |\tilde{f}(S)| \le C |S|,$ where $|\cdot|$ denotes Lebesgue measure. The claim then follows by approximating $E$ by a such interval, and using the analysis conducted above for both $f$ and $g.$


function
limit
calculus
calculus computations

1)put $ x \equal{} 0$ we get $ f(0) \equal{} 2f(0)$ so $ f(0) \equal{} 0$
2)putting $ x \to \frac {x}{2}$ and dividing by $ e^{x} \minus{} 1$ we get the required result
3)$ \lim_{h \to 0} \frac {f(h) \minus{} f(0)}{h} \equal{} [\lim_{h \to 0} \frac {f(h)}{e^{h} \minus{} 1}][\lim_{h \to 0} \frac {e^{h} \minus{} 1}{h}]\equal{} \lim_{h\rightarrow 0}\frac {f(h)}{e^h \minus{} 1}$ since each limit exists and the second one being $ 1$
4) this one intutively follows from 2)  :D

4) is key point, please justify your answer, pardesi. :)

[quote="kunny"]Let $ f(x)$ be a differential function at any real numbers $ x$ such that $ f(2x) = (e^x + 1)f(x)$ for any real numbers $ x.$

(1) Show that $ f(0) = 0.$

(2) Show that ${ \frac {f(x)}{e^x - 1} = \frac {f\left(\frac {x}{2}\right)}{e^{\frac {x}{2}} - 1}}$ for $ x\neq 0.$

(3) Show that $ f'(0) = \lim_{h\rightarrow 0}\frac {f(h)}{e^h - 1}$ in using the definition of differentiation.

(4) Show that $ f(x) = (e^x - 1)f'(0)$ holds.[/quote]

That's very nice kunny. Thanks for sharing!  :)

for 4 we see that

${{ \frac {f(x)}{e^x - 1} = \frac {f\left(\frac {x}{2}\right)}{e^{\frac {x}{2}} - 1}}=\frac {f\left(\frac {x}{2^n}\right)}{e^{\frac {x}{2^n}} - 1}}$ for $ x\neq 0$ and $ \forall n \in \mathbb{N}$.

So taking $ n\rightarrow \infty$ we get (due to 3) $ f(x)=(e^x - 1)f'(0)$ $ \forall x\neq 0$

since the last is true for $ 0$ as well, we are done.

 :)

Good! :)

[url=http://www.mathlinks.ro/viewtopic.php?t=188878]My solution is here.[/url], though which is written in Japanese.

[quote="socrates"][quote="kunny"]Let $ f(x)$ be a differential function at any real numbers $ x$ such that $ f(2x) = (e^x + 1)f(x)$ for any real numbers $ x.$

(1) Show that $ f(0) = 0.$

(2) Show that ${ \frac {f(x)}{e^x - 1} = \frac {f\left(\frac {x}{2}\right)}{e^{\frac {x}{2}} - 1}}$ for $ x\neq 0.$

(3) Show that $ f'(0) = \lim_{h\rightarrow 0}\frac {f(h)}{e^h - 1}$ in using the definition of differentiation.

(4) Show that $ f(x) = (e^x - 1)f'(0)$ holds.[/quote]

That's very nice kunny. Thanks for sharing!  :)

for 4 we see that

${{ \frac {f(x)}{e^x - 1} = \frac {f\left(\frac {x}{2}\right)}{e^{\frac {x}{2}} - 1}} = \frac {f\left(\frac {x}{2^n}\right)}{e^{\frac {x}{2^n}} - 1}}$ for $ x\neq 0$ and $ \forall n \in \mathbb{N}$.

So taking $ n\rightarrow \infty$ we get (due to 3) $ f(x) = (e^x - 1)f'(0)$ $ \forall x\neq 0$

since the last is true for $ 0$ as well, we are done.

 :)[/quote]
kunny i don't think this proof or urs(i don't understand japanese but .. :D) holds good
see ${ \lim_{n \to \infty} \frac {f\left(\frac {x}{2^n}\right)}{e^{\frac {x}{2^n}} - 1}} = \lim_{h \to 0} \frac {f(h)}{e^{h} - 1}$ iff $ \frac {f(x)}{e^{x} - 1}$is continious at $ x = 0$ but we have yet not sown it's continuity at that point

[quote="pardesi"]
see ${ \lim_{n \to \infty} \frac {f\left(\frac {x}{2^n}\right)}{e^{\frac {x}{2^n}} - 1}} = \lim_{h \to 0} \frac {f(h)}{e^{h} - 1}$ iff $ \frac {f(x)}{e^{x} - 1}$is continious at $ x = 0$ but we have yet not sown it's continuity at that point[/quote]
This statement is not true. What do you mean? We have shown the more general limit in $ h$ exists, and the first limit is a special case of that.

sorry u r right :rotfl:  :rotfl:  ..a misconception :oops:


inequalities
function
linear algebra
matrix

$(a+b-c)^a(b+c-a)^b(c+a-b)^c \leq a^ab^bc^c$ <->
$aln(a+b-c)+bln(b+c-a)+cln(c+a-b) \leq aln(a)+bln(b)+cln(c)$ <->
$aln(\frac{a+b-c}{a})+bln(\frac{b+c-a}{b})+cln(\frac{c+a-b}{c}) \leq 0$ <->
Now we can make $a=px$, $b=py$ and $c=pz$ where $p=\frac{a+b+c}{2}$. Then:
$xln(\frac{2-2z}{x})+yln(\frac{2-2x}{y})+zln(\frac{2-2y}{z}) \leq 0$ <->
$(x+y+z)ln(2)+xln(1-z)+yln(1-x)+zln(1-y) \leq xln(x)+yln(y)+zln(z)$ <->
But we have $x+y+z=2$ and we have to prove that:
$xln(1-z)+yln(1-x)+zln(1-y) \leq xln(x)+yln(y)+zln(z)-2ln(2)$
By Jensen's Inequalitie we have:
$\frac{2}{3}ln(\frac{2}{3}) \leq \frac{xln(x)+yln(y)+zln(z)}{3}$. And so:
$2ln(\frac{2}{3})-2ln(2) \leq xln(x)+yln(y)+zln(z)-2ln(2)$. And now I´ll prove that:
$xln(1-z)+yln(1-x)+zln(1-y) \leq 2ln(\frac{1}{3})$
And in this point I don´t have a nice idea and considered a function on the two variables x and y by making $z=2-x-y$. And I proved that a critical point occur in $x=y=\frac{2}{3}$ and that the eigenvalues oh the hessian matrix of this function are both negative and so it´s proved.

I found -9 and -3 the eigenvalues of the hessian. Somebody confirm? I was hurry... Thank´s!

[quote="leepakhin"][quote="stupidkid"]Prove that if $a,b$ and $c$ are sides of a triangle then \[ (a+b-c)^a(b+c-a)^b(c+a-b)^c \leq a^ab^bc^c  \][/quote]
Let $x=b+c-a$ and so on. Then $x,y,z$ are positive real numbers.
The inequality is equivalent to
$x^{\frac{y+z}{2}}y^{\frac{z+x}{2}}z^{\frac{x+y}{2}}\leq (\frac{y+z}{2})^{\frac{y+z}{2}}(\frac{z+x}{2})^{\frac{z+x}{2}}(\frac{x+y}{2})^{\frac{x+y}{2}}$
This is obviously true by AM-GM Inequality.[/quote]
Can you please explain how is it obviously true??  :huh:

I don't see how it's obvious either :oops:
But what I do know is that leepakhin probably made a mistake when subsituting $x$, $y$ and $z$. What he should get to is:
$x^{\frac{z+x}{2}}\cdot y^{\frac{x+y}{2}}\cdot z^{\frac{y+z}{2}}\le \left( \frac{z+x}{2} \right) ^{\frac{x+z}{2}} \cdot \left( \frac{x+y}{2} \right) ^{\frac{x+y}{2}} \cdot \left( \frac{y+z}{2} \right) ^{\frac{y+z}{2}}$

It would be better if you substitute
$a+b-c=2x$ $b+c-a=2y$ $c+a-b=2z$
Then $a=x+z$ $b=y+x$ $c=z+y$.


abstract algebra
superior algebra
superior algebra unsolved

hello i need solve this problem .
whether submodule every module generator finite one submodule is generator finite ?

As I said in http://www.mathlinks.ro/Forum/viewtopic.php?t=72518 , I have problems to understand it.
I interpret it now as:"Is a submodule of a finetely generated module finitely generated¿"

Answer: No
Take any ring $R$ that is not noetherian and see $R$ as module of itself. More or less by definition of noetharian rings, there is an ideal of $R$ that is not finetely generated, thus (when seen as submodule) you get a counterexample.

It's also not hard to show that the property of "every submodule of a finetely generated module over $R$ is finetely generated" is equivalent to $R$ being noetherian.


algebra
polynomial
Vieta
LaTeX
absolute value
Diophantine Equations

Determine all integers $a$ for which the equation \[x^{2}+axy+y^{2}=1\] has infinitely many distinct integer solutions $x, \;y$.

http://www.kalva.demon.co.uk/irish/soln/sol952.html

[ edit: new link since the death of Kalva: http://web.archive.org/web/20070611004423/http://www.kalva.demon.co.uk/irish/soln/sol952.html ]

Peter's link doesn't work anymore and chess64's link doesn't seem to provide a correct solution so I'll provide one here.

Clearly a = 0 doesn't work, and WLOG let a be negative.  Note that (x, y) = (1, a) is a solution and we can Vieta jump to infinitely many solutions, namely if (x, y) is a valid solution, then so is (ay - x, y).  For example, the solution (1, a) yields the solution (a, a^2 - 1) which in turn yields (a^2 - 1, a^3 - 2a) and so on.  It is clear that there are then infinitely many solutions for each negative integer a except a = -1, which clearly has finitely many solutions, so every integer a whose absolute value is greater than 1 works.

I don't know Latex, so if anyone could rewrite my solution (and add the actual method of vieta-jumping) that would be highly appreciated.

my solution : if we got \[x^{2}+axy+y^{2}=1 \] for $ x , y \in Z  , (x,y \neq 0)$   we make $ k , t $ respectively  \[x(1-a) + 2ay - a^{3}y  ,  -y + a(ay + x)\]  and we take $ 2y - a^{2}y - ax = n \Rightarrow $ we have : $ (na + x)^{2} + a(na + x)(y-n) + (y-n)^{2} = 1 $ and this sentence is right (try it !). for example : $(1,-a) ,  a > 1 $ is right in the equation $(1+ a(-a) + a^{2} = 1)$ and also $(1- 3a^{2} + a^{4} , 2a - a^{3}) .$
for $a = 2$ either $(1,-2)$  and  $(5,-4)$ are right .


Is there a linked-function $ f: R \rightarrow R$ such that :
f(rational)=irrational
f(irrational)=rational

What does it mean for a function to be [b]linked[/b]? Do you mean continuous?

[quote="mathnice1"]Is there a continuous function $ f: R \rightarrow R$ such that :
f(rational)=irrational
f(irrational)=rational[/quote]

No, there is not.
Else, consider the function $ g(x) \equal{} f(x) \plus{} x$ This function is continuous and $ g(x)\in\mathbb R\backslash\mathbb Q$ $ \forall x$ and so $ g(x) \equal{} c$ (else, if $ \exists a < b\in\mathbb R\backslash\mathbb Q$ such that $ g(u) \equal{} a$ and $ g(v) \equal{} b$, $ \exists d\in(a,b)\cap\mathbb Q$ and, since $ g(x)$ is continuous, $ \exists w$ such that $ g(w) \equal{} d\in\mathbb Q$)

So $ f(x) \equal{} c \minus{} x$ and $ f(0)\notin \mathbb Q$ $ \implies$ $ c\notin\mathbb Q$ but then it is always possible to find $ x$ non rational such that $ c \minus{} x$ is non rational too (else we would have an injection from $ \mathbb R\backslash\mathbb Q$ in $ \mathbb Q$).

So no such function exists.

My solution was the same as yours except for the second paragraph. I didn't discuss $ f(x)\equal{}c\minus{}x$..., would my solution be considered incomplete still? I figure in an analysis class what was written there goes unsaid. :maybe:


Let $ M$ be the intersection point of medians of a triangle $ \triangle ABC.$ It is known that $ AC \equal{} 2BC$ and $ \angle ACM \equal{} \angle CBM.$ Find $ \angle ACB.$

I think that this easy and nice problem is a old well-known problem which was "fresh painted". I shall use the same notations (I added only the middlepoint $E$ of the side $[AC]$) with the proposed problem for offer a [b]old equivalent enunciation[/b]:[quote="The old problem"][color=darkred]" Let $CBE$ be a isosceles triangle $(CB=CE)$for which $\widehat {CBE}\equiv \widehat {ECM}$,
where $M\in (BE)$ and $BM=2\cdot ME$. Ascertain $m(\widehat {BCE})$."[/color][/quote][hide="Answer."] $m(\widehat {BCE})=120^{\circ}$.[/hide][hide="Proof."]Denote $x=m(\widehat {CBE})$. Then $\sin 3x=2\sin x$. Therefore, $2\sin x=\sin x\left(3-4\sin^2x\right)$ $\Longrightarrow$ $4\sin^2x=1$ $\Longrightarrow$ $2(1-\cos 2x)=1$ $\Longrightarrow$ $\cos 2x=\frac 12$ $\Longrightarrow$ $x=30^{\circ}$. I am sorry cann't make the draw of this old problem. With all we will recognize this problem ![/hide]

I'm sorry for the problem appear to be old and well-known. :(

What about a solution without trig. formulas?

[hide="Here is the requested syntetical proof."]With the greatest pleasure, [i]Roque[/i] ! Define the middlepoint $N$ of the segment $[BM]\ (BN=NM=ME)$ $\Longrightarrow$ $\triangle BNC\equiv \triangle CME\ (S.A.S.)$ $\Longrightarrow$ $NC=MC=MN$ a.s.o.[/hide][hide="Here is a generalization with the another notations."]$\triangle ABC,\ AB=AC,\ M,N\in (BC),\ BM=CN,\ \frac{MN}{BC}=k\in \left(0,\frac 12\right)$$\Longrightarrow \cos A=\frac{k}{k-1}\ .$
The our particular case: $k=\frac 13\Longrightarrow A=120^{\circ}\ .$
[b]Remark.[/b] $k\searrow 0\ \Longrightarrow M\equiv N\Longleftrightarrow AB=AC\ \wedge\ A=90^{\circ}\ .$[/hide]

Define $E$ midpoint of $AC$, $\triangle EBC$ is isosceles: $EC=BC=a$ and $BE=3ME=x$, $\triangle EMC \sim \triangle EBC$: $EC^2=ME.BE$, so: $a^2= \frac{x^2}{3}$,then: $x=a.\sqrt{3}=BE$. Now: $<ECB=<ACB=120^0$.


What is the last nonzero digit of 10000!?

The On-Line Encyclopedia of Integer Sequences is always useful:
[url]http://www.research.att.com/~njas/sequences/A008904[/url]
It gives a very interesting link about this problem:
[url]http://www.mathpages.com/home/kmath489.htm[/url]

[hide]To get rid of all the 0s at the end of the number, you eliminate all numbers mod10 that would create 0mod10, and these are numbers that have one's digits of 0, 2, and 5 (cause 2 and 5 multiply together to get 10).  So then you have 1x3x4x6x7x8x9 = 8mod10

However, you have 1000 occurrences of this string, so next you find the integer n for which 8mod10 x n = 8mod10.  Through simple calculations n=6, so when you divide 1000 by 6 you get a remainder of 4.  So you multiply 8mod10 x 4 = 2mod10.

So the answer is 2.[/hide]

Sorry for not using Latex.


function
number theory unsolved
number theory

Find all pairs of positive integers $(a,b)$ such that $5a^b-b=2004$.  :lol:

We suppose $a>1$: $b$ must be less then $9$, because $5\cdot 2^9-9>2004$ and the function $5\cdot 2^n-n$ is increasing (and so also the function $5\cdot a^n-n$, when $a>2$). Moreover $b$ must be $\equiv 1$ (mod 5), so it can be only $1$ or $6$. So there aren't solutions. Am I wrong?

Your solution is correct but you forgot in case b=1 to see that then
              5a-1=2004;5a=2005;a=401
          So the solution is (a,b)=(401,1)  :D

You're all right, I'm getting more absent-minded each day more..it should be the heat.. :D

[quote="JohnConstantine"]Moreover $b$ must be $\equiv 1$ (mod 5)[/quote]

would you explain why this is a must ... I cant understand :blush:


integration
trigonometry
probability
function
algebra
domain
real analysis

When we define "mathematical expectation" , why we require $ \int_{\minus{}\infty}^{\infty}\left|x\right|f(x)dx < \infty$

not require $ \int_{\minus{}\infty}^{\infty}xf(x)dx$ instead ?

 :roll:

I suppose this is one way of phrasing your question: suppose $ f(x) \equal{} \frac1{\pi(x^2 \plus{} 1)}$ (the Cauchy distribution). Why is $ E(X)$ undefined when it is "obviously" zero?

Here's a numerical experiment to try. Work in some mathematical environment (spreadsheet, programming language, MATLAB/Maple/Mathematica, whatever) in which you have available a supply of random numbers (OK, pseudorandom) $ U$ with a uniform $ (0,1)$ distribution. For each $ U,$ let $ X \equal{} \tan(\pi(U \minus{} 0.5)).$ Then $ X$ will have a Cauchy distribution. Run up several thousand, or tens of thousands, of instances of $ X,$ and average them. Close enough to zero for you? Repeat with another few thousand instances. Is the result of this exercise stable, or does it hop around?

One way to answer your question is this: that expression, $ \int_{ \minus{} \infty}^{\infty}xf(x)\,dx,$ isn't directly the definition of expected value, and it doesn't even apply to all random variables with expectations - just the ones with absolutely continuous distributions. The real definition is this: a probability sample space is simply a measure space of total measure $ 1.$ A random variable is really a [i]function[/i] - that is, $ X$ is a (measurable) function whose domain is the probability sample space. Then $ E(X)$ is the Lebesgue integral of the function $ X$ over the probability sample space.

Now the thing about Lebesgue integrals: while you can take the Lebesgue integral of a nonnegative function and allow it to meaningfully equal $ \plus{} \infty,$ we only allow ourselves to use the word [i]integrable[/i] for integrals which are [i]absolutely convergent[/i]. Which is to say, for $ X$ not necessarily positive, we need $ E(|X|) < \infty$ in order to make sense of $ E(X).$

I should add this: suppose $ \int_{\minus{}\infty}^{\infty}|x|f(x)\,dx\equal{}\infty.$ Then $ \int_{\minus{}\infty}^{\infty}xf(x)\,dx$ cannot converge even as an improper integral. By the usual rules for improper integrals, that is improper for two reasons ($ \minus{}\infty$ and $ \plus{}\infty$) and you have to split it up to isolate each of those into a separate integral, and then only accept that as convergent if each piece converges.

$ \int_{\minus{}\infty}^{\infty}\frac{|x|}{\pi(x^2\plus{}1)}\,dx\equal{}\infty,$ and $ \int_{\minus{}\infty}^{\infty}\frac{x}{\pi(x^2\plus{}1)}\,dx$ is a divergent improper integral.

If you try to tell me that that latter integral is zero by symmetry, then you're not quoting the improper integral at all, but rather the principal value (the limit of balanced partial integrals), and the principal value is not the same thing as the improper integral.


Let $ a,b,c$ be non-negative numbers such that $ a^2\plus{}b^2\plus{}c^2\equal{}1$.Prove that
\[ \frac{1}{3\plus{}a^2\minus{}2bc}\plus{}\frac{1}{3\plus{}b^2\minus{}2ca}\plus{}\frac{1}{3\plus{}c^2\minus{}2ab} \leq \frac{9}{8}\]


Given a set of coins in the plane, all with different diameters, show that one of them is tangent to at most 5 of the others.

[hide="hint"]Extremal principle. Consider something extreme. Like extremely small. Like maybe even smaller than anything else. OK, also think about the problem and coins.[/hide]

Solution outlined by Xevarion:

Assuming there is a finite number of coins, consider the coin with the smallest diameter. Suppose, by way of contradiction, it is tangent to at least six coins. Then at least one of these six tangent coins must have a diameter less than or equal to the set coin, which is a contradiction to either all coins having different diameters or the set coin being the one with the smallest diameter.