I think this conversion has taken me long steps....
Suggest short steps...

1.Sulphonate-we get SO3H group in the para position
2.fuse with KOH to get the OH group.
3.use Acidified KMnO4 to get the acid.
4.Sulphonate again to get SO3H group in the meta positionto COOH group.
5.fuse with KOH
6.reduce with NaBH4 to get the alcohol.
7.Use P/HI to get the product..

Is this right :( ?????????????????????????????

Valeriummaximum, you would have problems in your steps 3 and 5.

Chemrock, can you tell me how its possible to make a Dakin reaction on benzaldehyde?

I think this conversion has taken me long steps....
Suggest short steps...

1.use alk KMnO4
2.sulphonate
3.fuse with KOH
4.sulphonate
5.fuse with KOH
6.reduce using NaBH4
7.use P/HI to get the product..

Is this correct  :wink: ?????????????????????

[quote="chemrock"]1.Etard's reaction(benzaldenhyde)
2.dakin's reaction(phenol)
3.duff reaction(salicylaldenhyde)
4.dakin's reaction(catechol)
5.riemer tiemann reaction(4-formyl catechol)
6.clemmensen reduction(4-methyl catechol)[/quote]

I think the only thing to be done is change the first and second steps (benaldehyde to phenol)
and btw wont Clemmensen's redn change the OH to Cl (tho not in this case ) and wont Wolff Kishner or Mozingo rn be better?

[quote="Valeriummaximum"]3.fuse with KOH[/quote]

Fusion of sulphonic acids with base often leads to side products, specially in this case in which we have an acidic group.

[quote="Carcul"]How can we prepare 4-methylcatechol from toluene?[/quote]
sulphonate and attach SO3H to the para position and then fuse with NaOH to get OH there and the add CL2/hv and you will get the CH3 TO BECOME CCl3 group now it is a metadirecting group now again repeat the about steps to get another OH group in the req position and then hydrolyse to get the COOH from CCL3 and then after protecting the OH groups reduce COOH to CH3 is this right??? :D  :lol:

One more time, -OH is an acidic group, so fusing with NaHO is expected to give side products. Also, -OH is a nucleophile and -CCl3 have good leaving groups...

[quote="madness"]Can we protect the OH grp using DNP then?  :maybe:[/quote]
or with tosyl chloride???

[quote="madness"]Can we protect the OH grp using DNP then?  :maybe:[/quote]
i think you would wanna use DHP not DNP :wink:

What is DNP?

Amazingly, this synthesis can be done in only 4 steps.

[quote="pardesi"]well as varuna was pointing out it was probably DHP[/quote]
having fun changing my name to a girl's, arent you? :mad:

[quote="pardesi"]hey i am not having any kind of fun  :huh: ....it was all starte by madness ...i am not doing it to tease u. ur names are long...how do u expect me to type in such big names :rotfl:
anyway varuna is nowhere a girls name
see [url=http://en.wikipedia.org/wiki/Varuna]thsi[/url] it explains who varuna(after whom u were probably named) is . :wink:[/quote]

How dare you accuse me of such a dastardly act  :furious: . I shall prosecute you under the 13th act of injustice 24th line in the 5th para on page 241 according to the Japanese court of law  :P  . I dont derive any fun from such an act (except for people whom i know by person and not a person who is easily irritated)

And btw it was a typo error it was in fact only DHP  :D

btw, my name is varun.
p.s: madness, i am not easily irritated :no:

You can also use TMCS which is a better protecting group


calculus
integration
invariant
topology
function
search
linear algebra

I want to do a integration over a surface $S$ in $R^{n}$. suppose the surface can be parameterized by $[x_{1}(u_{1},...,u_{n-1}),..., x_{n}(u_{1},...u_{n-1})]$, where $[u_{1},....u_{n-1}] \in U \subset R^{n-1}$. Then is the following equation true (under certain conditions):

$\int_{S}f(x_{1},....x_{n}) dA=\int_{U}f(x_{1}(u_{1},...,u_{n-1}),..., x_{n}(u_{1},...u_{n-1})) \sqrt{\det{J \cdot J^{T}}}d u_{1}....d u_{n-1}$

where $J$ is the $(n-1) \times n$ jacobian matrix. $dA$ is the measure on the surface $S$.

If so, where can I find a reference for this conclusion? I searched some books on calculus but only find 2D 3D case.

How would you define the measure on the surface anyway? I think you would take this as a definition, and then show that it is invariant under (smooth enough) change of parametrization.

I'd write that Jacobian factor as $|\det J|$; it's really a better indication of what we're doing. Negative determinants indicate a reversal of orientation, and some other integrals we might define depend on orientation.

[quote="mlok"]If one uses the (n-1)-dimensional Hausdorff measure on the surface, the formula requires some proof. The proof can be found (for Lipschitz f) on p.99 of the book
L.C. Evans, R.F. Gariepy, [i]Measure theory and fine properties of functions[/i], Studies in Advanced Mathematics, CRC Press, 1992.
You probably [url=http://www.fetchbook.info/compare.do?search=0849371570]don't want to buy it[/url], though.[/quote]

Yes, it is (n-1)-dimensional Hausdorff measure on the surface. My library has that book so I will definitely check that out. Thanks!

[quote="mlok"]If one uses the (n-1)-dimensional Hausdorff measure on the surface, the formula requires some proof. The proof can be found (for Lipschitz parametrizations) on p.99 of the book
L.C. Evans, R.F. Gariepy, [i]Measure theory and fine properties of functions[/i], Studies in Advanced Mathematics, CRC Press, 1992.
You probably [url=http://www.fetchbook.info/compare.do?search=0849371570]don't want to buy it[/url], though.[/quote]

Most of the stuff in that book can be found in William Ziemer's 'Weakly Differentiable Functions' (a Springer, hence cheaper, book), although Evans is a little better and goes into more detail with the proofs.

e: I actually don't think that that particular theorem is in Ziemer.

[quote="remgenesis"]I want to do a integration over a surface $S$ in $R^{n}$. suppose the surface can be parameterized by $[x_{1}(u_{1},...,u_{n-1}),..., x_{n}(u_{1},...u_{n-1})]$, where $[u_{1},....u_{n-1}] \in U \subset R^{n-1}$. Then is the following equation true (under certain conditions):

$\int_{S}f(x_{1},....x_{n}) dA=\int_{U}f(x_{1}(u_{1},...,u_{n-1}),..., x_{n}(u_{1},...u_{n-1})) \sqrt{\det{J \cdot J^{T}}}d u_{1}....d u_{n-1}$

where $J$ is the $(n-1) \times n$ jacobian matrix. $dA$ is the measure on the surface $S$.

.[/quote]

If the surface $S:{\mathbb R}^{k}\mapsto{\mathbb R}^{n}$ is $k$ dimensional instead of $n$-dimensional, we have the same formula where the $J$ in $\sqrt{\det(JJ^{T})}$ is now a $k\times n$ matrix


limit
real analysis
real analysis unsolved

[quote="N.T.TUAN"][quote="April"]Let a sequence $\{u_{n}\}$ satisfying that: $\begin{cases}u_{1}=1\\ u_{n+1}=\frac{u_{n}^{2}}{2007}+u_{n}\end{cases}, n=1,\,2,\,3,\ldots$
Calculate: $\lim_{n\to\infty}\left(\frac{u_{1}}{u_{2}}+\frac{u_{2}}{u_{3}}+\cdots+\frac{u_{n}}{u_{n+1}}\right)$[/quote]
You are wrong!
If you edited then we can use telescopic sum!  :wink:[/quote]
what's wrong?

we can find equivalent of $u_{n}$ and establish $lim$.

[quote="§outh§tar"]The problem is wrong?  :rotfl:[/quote]
No, It is not beautiful :D .I think it must is $\frac{u_{1}}{u_{2}}+...+\frac{u_{n}}{u_{n+1}}$.


Let $\displaystyle \left( P_n \right)_{n \geq 1}$ be an infinite family of planes and $\displaystyle \left( X_n \right)_{n \geq 1}$ be a family of non-void, finite sets of points such that $\displaystyle X_n \subset P_n$ and the projection of the set $\displaystyle X_{n+1}$ on the plane $\displaystyle P_n$ is included in the set $X_n$, for all $n$.

Prove that there is a sequence of points $\displaystyle \left( p_n \right)_{n \geq 1}$ such that $\displaystyle p_n \in P_n$  and $p_n$ is the projection of $p_{n+1}$ on the plane $P_n$, for all $n$.

Does the conclusion of the problem remain true if the sets $X_n$ are infinite?

[i]Claudiu Raicu[/i]

Consider the following graph: the vertices are the elements of $\bigcup_{n\ge 1}X_n$, and a pair of points is connected iff it's of the form $\{p_n\in X_n,p_{n+1}\in X_{n+1}\}$ such that $p_{n+1}$ is projected orthogonally onto $p_n$.

Clearly, every vertex of this graph has finite degree. Each connected component of our graph contains an element of $X_1$, so there must be at least one such component which is infinite. Konig's infinity lemma states that given a connected graph in which every vertex has finite degree, and a distinguished vertex $p_1$ of this graph, there is an infinite simple path starting at $p_1$. Apply this to the infinite connected component found above, with $p_1$ being one of its vertices which belong to $X_1$, and we're done.

The answer to the last question is 'no'. For a counterexample, assume the planes to be parallel those given by the equations $P_n=\{(x,y,z)\in\mathbb R^3\ |\ z=n\}$, and taking $X_n=\{(k,0,n)\ |\ k\ge n\}$.


[hide]
I only know that the expression can be divisible by $2$ and $3$ meaning $2,3$ would be two of the prime . And i believe that $n$ can only take the value $n=2,3,6$ .But I dont have idea to prove yet  :P
[/hide]

[quote="1234567890"][quote="fourierseries"]Find all natural numbers n such that:

$n(n+1)(n+2)(n+3)$ has exactly 3 prime divisors.[/quote]

Two of the divisors must be 2 and 3; this follows from the fact there are at least three consecutive numbers.
From the conditions of the problem, we see that among the four consecutive numbers, there is only one other prime divisor in all four of them.
It is easy to see that n = 1,2,3,6 satisfy the conditions.
Now I am stuck: I have asomewhat unpolished argument using the fact that as n becomes larger, you'll have different divisors, but I can't prove it. :([/quote]

$n=1$ doesnt satisfy since the prime divisor are only $2$ and $3$ . :)

Hi,

The product of 4 consecutive integers is always divisible by 4! (= binomial(n+4,4)*4!) so we have two of our prime factors : 2 and 3

if p prime >= 5 divides one of the n,n+1,n+2 or n+3 it is prime with the other 3 which must be of the form 2^a*3^b with a or b eventually null and we must find p "alone".

We can't have 2 consecutive even numbers being powers of 2 (except 2 and 4 treated separately), so the 4 numbers are in some order :
p odd prime
2^a
3^b
2^c*3^d
with a,b,c,d > 0
So the only two possibilities are either 3^b,2^a,p,2^c*3^d or 2^c*3^d,p,2^a,3^b which is easy to solve.

Summary :

If find n = 2, 3 or 6 as only solutions.


When 700 students from Genius M.S. played SCOOZ, 90% of them scored “proficient or above.” What number of students at Genius M.S. did not score “proficient or above?”

i believe you do
[hide] .9*700=630
then you do
700 - 630= 70 students[/hide]

well 90% scored well then
[hide] 10% scored not that well .10*700= 70[/hide]

[hide]Since 90% scored proficient or above, that means 10% didn't. 10% of 700 students is $70$.[/hide]

What was the point of posting the same answer/solution 15 times?


Βρείτε το επίπεδο χ/α+ψ/β+z/c=1 που περνάει από το σημείο (χ1,ψ1,z1) στο πρώτο όγδοο του χώρου και αποκόπτει το μικρότερο όγκο από αυτό[/size]

Ο όγκος που ζητάς είναι $ V\equal{} \dfrac{1}{6} abc$ Αρκεί λοιπόν να ελαχιστοποιήσεις το $ abc$
Από την ανίσωση ΑΜ-ΓΜ για τους θετικούς αριθμούς $ x_1 /a,y_1 /b,z_1 /c$ παίρνεις την $ \dfrac{1}{3} \ge \sqrt[3]{\dfrac{x_1 y_1 z_1}{V}}$ οπότε αν ισχύει η ισότητα βρίσκεις το επιθυμητό $ V_{min}$

[quote="r_boris"]Ο όγκος που ζητάς είναι $ V \equal{} \dfrac{1}{6} abc$ Αρκεί λοιπόν να ελαχιστοποιήσεις το $ abc$
Από την ανίσωση ΑΜ-ΓΜ για τους θετικούς αριθμούς $ x_1 /a,y_1 /b,z_1 /c$ παίρνεις την $ \dfrac{1}{3} \ge \sqrt [3]{\dfrac{x_1 y_1 z_1}{V}}$ οπότε αν ισχύει η ισότητα βρίσκεις το επιθυμητό $ V_{min}$[/quote]ΔΙΟΡΘΩΣΗ μέσα στην ρίζα γράψε $ 6V$ αντί του $ V$

ευχαριστώ πολύ r_borris με βοήθησες πολύ

[quote="spinos"]r_boris σχετικά με το θέμα που μου είχες πει για τον ελάχιστο όγκο του,που πρέπει να πάρω την ανισότητα του αριθμητικού και γεωμετρικού μέσου το έκανα και μου βγήκε ό ελάχιστος όγκος V=9/2αbc
όμως πρέπει να προσδιορίσω και τα α, b,c συναρτήσει του σημείου (x1,ψ1,z1) που διέρχεται το επίπεδο
μήπως υπάρχει τρόπος να τα προσδιορίσω
εγώ το έκανα με δεσμευμένα ακρότατα, έβγαλα τον ίδιο όγκο
και τα α=3χ1, b=3ψ1 και c=3z1
αλλά ήθελα να δω αν υπάρχει πιο εύκολος τρόπος,αν δεν σου είναι κόπος κοίταξε το.
Σ΄ευχαριστώ![/quote]
Συγνώμη για την καθυστέρηση , είχα πρόβλημα με την σύνδεση μου.
'Οταν ισχύει το "=" στην ανισότητα πρέπει οι παράγοντες να είναι ίσοι άρα προκύπτει ίδιο αποτέλεσμα με αυτό που βρήκες

Aυτό όμως r_boris δεν πρέπει να το αποδείξω?
δηλαδή γιατί να ισχύει ότι το '=' σε μια ανίσωση που έχει γινόμενο , μόνο όταν οι παράγοντές του είναι όλοι ίσοι μεταξύ τους?
αυτό που μου έρχεται στο νου  τώρα είναι εκείνο το πρόβλημα από τα σχολικά βιβλία που μας έλεγε πιο είναι το ορθογωνιο με σταθερή περίμετρο που όμως έχει το ελάχιστο εμβαδό
που είναι βέβαια το τετράγωνο!
πως αποδεικνύεται r_boris ότι πρέπει να είναι ίδια τα a,b και c?
ελπίζω να νην σε έχω πρήξει πολύ!

Η απόδειξη που ζητάς είναι ουσιαστικά η απόδειξη της ανίσωσης ΑΜ-ΓΜ .Υπάρχουν αποδείξεις σε διάφορα βιβλία .Ρώτα καλύτερα τον Silouan. Η γνώμη μου είναι πως δεν χρειάζεται να το αποδείξεις θεωρείται γνωστό όσο η ίδια η ανισότητα

Λοιπόν κοίταξε ... (Καταρχήν αν θέλετε τις απορίες να τις εκράζετε σε ΡΜ)

Για να αποδείξεις ότι το ίσον ισχύει όταν και τα τρία είναι ίσα παρατήρησε ότι

$ x\plus{}y\plus{}z\minus{}3\sqrt[3]{xyz}\equal{}\frac{1}{2}(\sqrt[3]{x}\plus{}\sqrt[3]{y}\plus{}\sqrt[3]{z})((\sqrt[3]{x}\minus{}\sqrt[3]{y})^2\plus{}(\sqrt[3]{x}\minus{}\sqrt[3]{z})^2\plus{}(\sqrt[3]{z}\minus{}\sqrt[3]{y})^2)$

Το πρώτο μέλος είναι 0 όταν και τοδεύτερο είναι 0 οπότε ...

οκ παιδιά κατατοπίστηκα πλήρως σας ευχαριστώ πολύ και τους δυό σας!
ναι έχεις δίκιο silouan έχεις δίκιο υπαρχει αυτή η ταυτότητα και στο σχολικό βιβλίο της Α' Λυκείου, αλλά δεν είχα καταλάβει πως πρόκειται για την ανισότητα του α. και γ. μέσου,
όπως έχεις και δίκιο να λες πως καλύτερα να εκφράζουμε τις απορίες μας σε προσωπικό μήνυμα,
αλλά όμως τώρα δεν θα το έβλεπες εσύ και δεν θα μου έδινες την απάντηση!
Και πάλι σας ευχαριστώ πολύ!
έχω μάθει πολλά πράγματα τον τελευταίο καιρό κυκλοφορώντας μέσα στο forum, που είχα άγνοια για την ύπαρξη του!
είναι πραγματικά ένα ζωντανό σχολείο!


Let $ a,\ b,\ c > 0$  with $ abc \equal{} 1$
Prove folowing inequality :

$ \frac {1}{a(a \plus{} b)} \plus{} \frac {1}{b(b \plus{} c)} \plus{} \frac {1}{c(c \plus{} a)} \geq \ 3/2$

[quote="Smilley"]let a,b,c are real positive numbers with $ abc \equal{} 1$
Prove folowing inequality :

$ \frac {1}{a(a \plus{} b)} \plus{} \frac {1}{b(b \plus{} c)} \plus{} \frac {1}{c(c \plus{} a)} \geq \ 3/2$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=81588


logarithms
trigonometry
ratio
calculus
calculus computations

I have an exam tomorrow the bulk of which will be testing for divergecne/convergence of a function. So here are my problems...

a) i have difficulty knowing which test to use
b) why cant i just use test for divergence every single time by taking the limit as n approaches infinity?

Thanks for your help

(a) The most important thing is to recognize size quickly. Know that geometric terms grow or shrink faster than mere powers of $n$, but that logarithms grow or shrink slowly compared to powers of $n$. If you see something like $\sin n$, think of it as "1, only wigglier."

De-clutter. If you see, say, $\sqrt{n^3+2n+5},$ think $n^{3/2}.$ The workhorse of tests using this idea is the limit comparison test.

Nearly every test you know is a comparison test. All comparison tests are tests for absolute convergence; go ahead and take the absolute values first, testing first for absolute convergence.

This includes that the ratio test as root test are comparison tests. They are comparisons to geometric series. They are only of any use when there are things that grow or shrink geometrically fast or faster - $a^n, n!, n^n$ - things like that. If you've only got powers of $n$, do not waste your time using the ratio test or root test - it has no chance of telling you anything.

The only test you know that isn't a comparison test is the alternating series test. Thus, this is the only test that could prove convergence in the absence of absolute convergence (conditional convergence).

(b) Why not? Because you can't prove convergence that way. I can see it now, on your test: a third of a page of the elaborate finding of the limit of the sequence of terms (some complicated expression) - that limit exists. Perhaps that limit is even zero. You triumphantly announce on your paper that the series converges. And your professor gives you zero points credit for the problem.

Read the logic of what the "test for divergence" says: if the series converges, the terms go to zero. The converse is false, as shown by (among other examples), the harmonic series, $\sum\frac1n.$

$ \sum_{n \equal{} 1}^{ \plus{} \infty} \left(1 \plus{} \frac1n\right)^n$,

$ \sum_{n \equal{} 1}^{ \plus{} \infty} ( \minus{} 1)^n$,

among many others.


Find the greatest $ n$ so that $ a^2\plus{}b^2\plus{}c^2\plus{}n(ab\plus{}bc\plus{}ca)\geq 0$ for all $ a, b, c\in\mathbb{R}$.

[hide]For $ n\equal{}2$, this becomes $ a^2\plus{}b^2\plus{}c^2\plus{}2(ab\plus{}bc\plus{}ca) \equal{} (a\plus{}b\plus{}c)^2 \ge 0$ which is true for all $ a,b,c \in \mathbb{R}$.

If $ n \equal{} k\plus{}2$ where $ k>0$ then the inequality becomes: $ (a\plus{}b\plus{}c)^2\plus{}k(ab\plus{}bc\plus{}ca) \ge 0$.

However, for $ (a,b,c) \equal{} (\minus{}2t,t,t)$ with $ t > 0$ we have: $ (a\plus{}b\plus{}c)^2\plus{}k(ab\plus{}bc\plus{}ca) \equal{} (0)^2 \plus{} k(\minus{}3t^2) \equal{} \minus{}3kt^2 \ge 0$ which is not true.

Thus, the largest possible $ n$ is $ 2$.[/hide]


Im just curious what chapters in AoPs would be good to study for the AIME. I have about a month and a half to get from about a 3 average to my goal of a 6-7. I dont think taking a timed test is really an option, except for maybe once or twice.

[quote="rrusczyk"]Here's how I'd roughly prioritize the chapters for AIME preparation:

High: 1, 2, 3, 4, 6, 7, 9, 13, 15, 16, 17, 18, 19, 23

Medium: 5, 10, 11, 12, 14, 24

Low: Everything else.

I'd also review the geometry chapters of AoPS1.[/quote]

However, as both instructors and experienced test takers will attest, taking practice tests is VERY important. Just being able to sit down and do problems for those 3 straight hours is a big task, also you might need some time management and "not being sucked into a single problem for too long"

There's an AOPS book? if so...where can i find it?


Let  $ F(u,v)$be a function of two variables. Find $ f '(x)$ for $ f(x) \equal{} F(x, x)$.
I need to find the answer in terms of $ F_u$, how can I do this?

$ u\equal{}u(x)\equal{}x, v\equal{}v(x)\equal{}x$

$ f'(x)\equal{}[F(u(x),v(x))]_x\equal{}F_u(u(x),v(x))u_x(x)\plus{}F_v(u(x),v(x))v_x(x)$
$ \equal{}F_u(x,x)\plus{}F_v(x,x)$

im not sure how you can get  rid of $ F_v$ its pretty important..

especially if like $ F(u,v)\equal{}h(v)$ then $ F_u\equal{}0$ and its hard to express anything in terms of zero...


Given the polynomial \[ f(x)=x^n+a_{1}x^{n-1}+a_{2}x^{n-2}+\cdots+a_{n-1}x+a_n \] with integer coefficients $a_1,a_2,\ldots,a_n$, and given also that there exist four distinct integers $a$, $b$, $c$ and $d$ such that \[ f(a)=f(b)=f(c)=f(d)=5, \] show that there is no integer $k$ such that $f(k)=8$.

Let $g(x)=f(x)-5$ , then 23 have $g(a)=g(b)=g(c)=g(d)=0$ . Hence we can let

$g(x)=(x-a)(x-b)(x-c)(x-d)h(x)$ where $h(x)$ is a polynomial with integer coefficient .

If there exist one $k$ such that $f(k)=8$ then $g(k)=(k-a)(k-b)(k-c)(k-d)h(k)=f(k)-5=3$

From pigeon hole principle , at least three from $k-a,k-b,k-c,k-d$ must equal to $1$.WLOG let $k-a=k-b=k-c=1$ But this gives us $a=b=c$ which contradict.

The above solution misses the possibility of negative roots. The factors of 3 are 3, -3, 1, and -1. From the given conditions in the problem, they must be $k-a$, $k-b$, $k-c$, and $k-d$. However, multiplying all of these together give 9. Hence, $9*h(k) = 3$, from which impossibility is clearly seen when one considers that h(k) must be an integer.

One question. I was extremely close to solving this problem in the AoPS Intermediate Algebra textbook, but when I got down to proving that there were no integers that satisfied $(k-a)(k-b)(k-c)(k-a)h(k)=3$, I could not prove that $h(k)$ needed to have integer coefficients, so I gave up, looked at the solution, and realized that the solution just assumes what I was trying to prove. So apparently, it should be trivial that $h(k)$ has integer coefficients, but I can't see why.

When you divide an integer coefficient polynomial by another, you get another integer coefficient polynomial.

Does that only apply when one of the polynomials is a factor of the other? Is there a proof anywhere?

If one of the polynomials is not a factor of the other, then you can't divide them in the first place without remainder ;)

The proof is easy to see with Gauss Lemma (if an integer polynomial is reducible in Q, then it is reducible in Z).

[quote=bobthesmartypants]If one of the polynomials is not a factor of the other, then you can't divide them in the first place without remainder ;)[/quote]

Oh, I feel dumb.

$g(x)=f(x)–5$ and let
$g(x)=(x–a) (x–b) (x–c) h(x)$
And let $r\in\mathbb{Z}$ so that $f(r)=8$
$g(r)=8–5=3=(r–a) (r–b) (r–c) h(r)$
 Since lhs has three factors $–1,3,–3$ and this has three distinct factors and also $h(x)$ and
$a\neq b\neq c$

Can someone prove why the unknown factors of $g(x)$, where $g(x) = f(x) - 5$, do not affect the known factors? In other words, why is it assumed that the factors $(x-a), (x-b), (x-c), (x-d)$ are elements of $\{\pm 1, \pm 3\}$, and this claim is unaffected by the other factors of $g(x)$? Thanks.

Example. Complex conjugate roots won't always cancel into 1, so if you let one pair of unknown factors multiply to $-\frac{1}{3}$, then perhaps $(x-a)(x-b) = -9$? Why can't this happen?

Because $h(x)$ could equal $1$ for some $x$ or $-1$ and we don't know much about it. (or basically any integer)
So we're trying to reach a contradiction just by looking at the things we know, i.e. $(r-a)(r-b)(r-c)(r-d)$
(All we know is that $h(x)$ has integer coeff and this basically is the only thing we need from it)

Actually, to answer your "In other words, why is it assumed that the factors $(x-a), (x-b), (x-c), (x-d)$ are elements of $\{\pm 1, \pm 3\}$" question, we know that since $h(x)$ is going to give an integer, we're assuming that it's either $1$ or $-1$ to minimize $|h(x)|$ and to also maximize the number of choices for $(x-a)$, $(x-b)$...

[quote=AwesomeLife_Math]...[/quote]

Thanks for the response, but why can't $h(x)$ have complex/irrational roots, and what if those factors (which contain complex/irrational roots) multiply to some fraction, which allows $(x-a)(x-b)(x-c)(x-d)$ to all be distinct, and $h(x)$ is thus a fraction?
 (i.e. In short, how do you know that $h(k)$ yields an integer, while also considering complex/irrational roots?)

[quote=AwesomeLife_Math]So do you agree that $h(x)$ has integer coefficients?

No problem btw.[/quote]

Sure; however, even if $h(x)$ has integer coefficients, it can still have complex roots, and I would like to know why this case causes no issues with the proof. (Hopefully I am not missing something obvious.) And I don't quite understand why dividing two integer coefficient polynomials yields another integer coefficient polynomial, at an intuitive level.
I couldn't solve the problem because I couldn't resolve these two doubts.

[quote=AwesomeLife_Math]Yes but if it has integer coefficients, $h(k)$ will always yield an integer when $k$ is an integer.

Note that we do not care about the roots of $h(k)$, we're only looking at its value.

And also since $f(x)$ is monic, if we divide it by $(x-a)(x-b)(x-c)(x-d)$, it must give another polynomial $h(x) \in \mathbb {Z} [x]$[/quote]

Thanks so much for the clarification, I guess I was missing something obvious. I completely neglected to look at $h(x)$ as a polynomial, thinking too much about its factored form. :P

No problem! That happens quite a lot to me too  :blush: .

[quote=AwesomeLife_Math]Yes but if it has integer coefficients, $h(k)$ will always yield an integer when $k$ is an integer.

Note that we do not care about the roots of $h(k)$, we're only looking at its value.

And also since $f(x)$ is monic, if we divide it by $(x-a)(x-b)(x-c)(x-d)$, it must give another polynomial $h(x) \in \mathbb {Z} [x]$[/quote]

For the love of God, can someone explain to me, in a rigorous and elementary way, why h(x) has integer coefficients, please? I'm getting crazy.
Why cannot be for example $h(x) = x^3+\frac{x^2-x}{2} =x^3+\frac{1}{2}x^2 - \frac{1}{2}x$ ...?

[quote=MagicalToaster53][hide = For @above]$f(x)$ is an integer polynomial. Hence so is $g(x) = f(x) - 5$. Observe that as $a, b, c, d$ are roots of $g(x)$ we must have that $g(x) = (x - a)(x - b)(x - c)(x -d)h(x)$, for some arbitrary polynomial $h(x)$. Now why must $h(x)$ be an integer polynomial? Well first that we know $g$ is an integer polynomial, and the product $(x - a)(x - b)(x - c)(x - d)$ is an integer polynomial too when expanded. Therefore $h(x)$ must also be an integer polynomial. Now with the solution:[/hide]

Define $g(x) = f(x) - 5$. Then as $a, b, c, d$ are roots of $g(x)$, we have that \[g(x) = (x - a)(x - b)(x - c)(x - d)h(x),\] for some integer polynomial $h(x)$.
Now if for some $k$ we have that $g(k) = 8$, observe that the possible values of $x - a, x - b, x - c, x - d, h(x)$ must be $\pm 1, \pm 2, \pm 4$, with an even number of negative factors. However observe by the pigeonhole principle that there can only be at most three distinct factors, which yields a contradiction. $\blacksquare$[/quote]

Sorry, but I don't understand how this proves that $h(k) \in \mathbb{Z}$. You assumed what you wanted to prove, why, for example, in this case, it couldn't be $(x-a, x-b, x-c, x-d, h(k))=(1, 2, 4, 8, \frac{1}{8} )$?
*Note: this message has translated by Google from PT-BR to Eng.*

Use the rational root theorem. If $P(x) = Q(x)R(x) = (x-a_1)(x-a_2) \cdots (x-a_n) R(x)$, and we have that $R(x)$ has a rational root, say $p/q \neq a_i$, then by the rational root theorem we have to have $q \mid \pm 1$, so that every root of $R$ is an integer so that it is an integer polynomial. (Also note the difference between degrees and number of roots of $P$ and so on blah blah blah. Just for intuition hopefully, nothing rigorous).


If $x^2+2\cos \phi \cdot x - {\sin}^2 \phi=0$, what is the range of the variable $x$ ?


A fish tank contains eight guppies and six swordtails. Twenty-one fish are added to the take, smoe guppies and some swordtails. How many of the 21 additional fish must be guppies to keep the original ratio?

The current ratio is 4:7. If 21 fish are added, there is a new total of 14 + 21 = 35. To maintain the ratio, there needs to be 20 guppies. Thus we must add $ \boxed{12}$ guppies.

[hide="A much more detailed and clear to understand solution"]Before the twenty-one fish were added, there were the total number of fishes was
\[ 8 \plus{} 6 \equal{} 14
\]
Eight of them were guppies and six of them were swordtails. The ratio of guppies to the total number of fish was $ \frac {8}{14}$.

Now, we need to keep that same ratio with an additional number of fish. For this to happen, the number additional of guppies/the twenty-one additional fish must also be $ \frac {8}{14}$.

$ \frac {8}{14}$ simplifies to $ \frac {4}{7}$.

Now we have the equation
\[ \frac {4}{7} \equal{} \frac {x}{21} \implies \frac {4 \cdot 3}{7\cdot3} \equal{} \frac {x}{21}
\]
Thus,
\[ x \equal{} \boxed{12}
\]
[/hide] :P


If angle DEF is 20
angle GDF is 50
angle HFD is 60

what is angle DGF?

[quote="Wichking"]if DE=EF then DGF=30[/quote]
and why is that?

because if thats the case with DE=EF, then angle F is 80 and GDF is 50, so it should be 50 degrees...

However, I am sure he/she mistyped this problem and intended for DGH.

oooops  :blush:  yes you are corect ....and you are probably rigth about cascadee mistyping the problem too....

does anyone know any colection of intermediate geometry problems?

Yes DGH sorry

how did you do it?

I know it is 30

See [url]http://www.mathlinks.ro/viewtopic.php?t=183931[/url] (the same)
and [url]http://www.mathlinks.ro/viewtopic.php?t=84476[/url] (similar).


Our football team has 10 members, of which only 3 are strong enough to play offensive lineman, while all other positions can be played by anyone. In how many ways can we choose a starting lineup consisting of a quarterback, a running back, an offensive lineman, and a wide receiver?

There are 3 ways to choose the offensive lineman, and 9C3 = 84 to fill in the remaining positions. (how do you do combinations in LaTeX?) Thus, there are $ 84\times 3\equal{}252$ ways to choose a starting lineup.

In the future, please refrain from posting off-topic things in threads. You can private message the person - it is not necessary to post here. Thanks. (PS: I cleaned up the posts.)

Different answer:

Only 3 people can play offensive lineman, while anyone else can play the other positions. You do not need combinations, because these are [b]distinct[/b] positions. Thus: $ 3 \times 9 \times 8 \times 7 \equal{} \boxed{1512}$.

EDIT: Mewto, what did I just say?

Another way (sorry)

There's three spots so...

$ 10\cdot9\cdot8\equal{}720$

That doesn't work because You have to choose 3 people from the group of 9, not 10. So you do 9 x 8 x 7 instead, and multiply by 3 for the offensive person, like Izzy did.


trigonometry
inequalities
Euler
inequalities proposed

Prove that for any triangle $ ABC$ and $ k \le \frac{8}{7}$,we have:
$ \frac{3\sqrt{3}}{2cos{\frac{A}{2}}cos{\frac{B}{2}}cos{\frac{C}{2}}}+8ksin{\frac{A}{2}}sin{\frac{B}{2}}sin{\frac{C}{2} \ge 4+k}$

[quote="quykhtn-qa1"]$\{\begin{array}{ccc}
\triangle\, ABC \\\
k\, \le\, \frac 87\end{array}\|\ \implies\ \boxed{\ 4+k\, \le\, 8k\sin\frac A2\sin\frac B2\sin\frac C2+\frac {3\sqrt 3}{2\cos\frac A2\cos\frac B2\cos\frac C2}\ }$[/quote]
[color=darkred][u][b]Proof.[/b][/u] Using the well known identities: $\{\begin{array}{cccc}
\prod\sin\frac A2=\frac r{4R} \\ \\
\prod\cos\frac A2=\frac s{4R}\end{array}$ the given inequality rewrites as: $4+k\, \le\, 8k\cdot\frac r{4R}+\frac {3\sqrt 3}{2\cdot\frac s{4R}}$

$\iff\boxed{4+k(1-\frac r{2R})\, \le\, \frac {6R\sqrt 3}s}$ and since $1-\frac {2r}R\ge 0$ ([u][b]Euler[/b][/u]) it will be enough to prove the inequality for $k=\frac 87$.

Thus, we have to show that : $\frac {36}7-\frac {16r}{7R}\, \le\, \frac {6R\sqrt 3}s\iff\boxed{s\le\frac {42R^2\sqrt 3}{36R-16r}}$ , which is weaker than the following one :

$\boxed{s^2\, \le\, (2R+r)^2\cdot\frac {8R^2-5r^2}{8R^2-7r^2}}$ (see[/color] [url=http://www.artofproblemsolving.com/Forum/blog.php?u=61896&b=34067][color=red][b]here[/b][/color][/url][color=darkred] a proof based on [u][b]Blundon[/b]'s inequality[/u]). Indeed, taking $t=\frac Rr\ge 2$ ([u][b]Euler[/b][/u])

the inequality $(2R+r)^2\cdot\frac {8R^2-5r^2}{8R^2-7r^2}\, \le\, (\frac {42R^2\sqrt 3}{36R-16r})^2$ reduces to $(2t+1)^2\cdot\frac {8t^2-5}{8t^2-7}\, \le\, \frac {5292t^4}{(36t-16)^2}\iff$

$\iff 5292t^4(8t^2-7)-(2t+1)^2(36y-16)^2(8t^2-5)\, \ge\, 0$ which is equivalent, by factorization, with

$4(t-2)(216t^5-720t^4+355t^3+1686t^2-160)\ge 0$ , obviously true.[/color]


function
integration
linear algebra
linear algebra unsolved

Let $ V$ be the vectorial space of continuous functions $ R \rightarrow R$, it is, the space of continuous functions of real value in the real axis. Let $ T$ be the linear operator over $ V$ defined as $ (Tf)(x) \equal{} \int_{0}^{x} f(t) dt$. Show that $ T$ has not eigenvalues.

Well, let's see, if $ T$ had an eigenvalue $ f$, we would have $ \int_0^x f(t)dt\equal{}\lambda f(x)$ for all $ x$.

Differentiating, we have $ f(x) \equal{} \lambda f'(x)$.  What is the solution to this equation?  What happens when you plug it back in to the first equation?

For every $ \lambda \neq 0$ and constant $ c\neq 0$ function ${ f(x) =ce^{\frac{1}{\lambda}x}}$ is an eigenvector for $ \lambda$. Isn't it? :)

So if it is not true problem is solved :).


Να δειχθει οτι το πολυωνυμο $ P(x)\equal{}ax^3\plus{}bx^2\minus{}(4a\plus{}3b\plus{}2c)x\plus{}c$ ,οπου $ a,b,c$ ,$ 3a\plus{}2b\plus{}c$ ειναι μη μηδενικοι πραγματικοι ,δεν μπορει να εχει τρεις ακεραιες ριζες.
(προεκυψε απο συζητηση που εγινε την ωρα του μαθηματος πανω σε γνωστη ασκηση...)

Xrisimopoieis Viete's relations stin lysi pou exeis?

Βαλε την λυση σου και το συζηταμε

Πολύ ωραία κατασκευασμένη άσκηση  :)
Έχουμε ότι $ p(0) \equal{} c$ , $ p(1) \equal{} \minus{} 3a \minus{} 2b \minus{} c$ , $ p( \minus{} 1) \equal{} 3a \plus{} 4b \plus{} 3c$ , $ p(2) \equal{} \minus{} 2b \minus{} 3c$

Υποθέτουμε ότι το πολυώνυμο έχει τρεις ακέραιες ρίζες και επίσης δβτγ ότι $ c\geq 0$ . Τότε πρέπει $ \minus{} 3a \minus{} 2b \minus{} c > 0$ ( μηδέν δεν μπορεί να είναι και αν ήταν αρνητικό από bolzano θα είχαμε ρίζα στο (0,1) ) Όμοια πρέπει $ 3a \plus{} 4b \plus{} 3c\geq 0$ kai $ \minus{} 2b \minus{} 3c\geq 0$

Προσθέτοντας τις παραπάνω ανισότητες παίρνουμε $ 0 > 0$ που είναι άτοπο  :)

Sil τωρα βλεπω την ωραία λυση σου .(Πολυ μαμά στο διάλειμμα............)
Εγω την λυνω λιγο διαφορετικα χρησιμοποιωντας το αθροισμα $ 0$ και αποριπτοντας μετα το Β-W τις $ 0$ και $ 1$  απο ριζες.
Ειναι μια απ' αυτες του ΤΕΣΤ που λεγαμε........
Να περνας καλα στην πρωτευουσα.....


What would happen to flora (plants) if they lacks of $ Cl^{\minus{}}$?

Interesting question, to which I would now also like to know the answer. An equivalent question is "What is the function of chloride ions in plant biology"?

My textbook says that the role of $ Cl^{\minus{}}$ in plants is light-dissociation of $ H_{2}O$, and ions balance.
What I want to know is if a plant lacks of $ Cl^{\minus{}}$, what will happen to it?
I may take an example for my question. If plants lack of $ N$, then they cannot grow up, and their leaves become yellow. (Hope you understand what I mean  :D )

Sorry for my bad English.

[quote="Bibobeo"]My textbook says that the role of $ Cl^{\minus{}}$ in plants is light-dissociation of $ H_{2}O$, and ions balance.[/quote]

Very interesting. And what is the purpose of the light dissociation of water molecules?

Oh, I'm really sorry for this. I even don't understand what it means in my language  :blush: . Too many academic vocabulary.

Anyone of you study this in highschool or college, please help me.

Anyone help me??  :(
I really need this. Thank you very much in advance.

Here's the answer  :D
Light-dissociation of $ H_{2}O$ is a small but important stage in photosynthesis of plants. And $ Cl^{\minus{}}$ help to activate the catalyzator in that light-dissociation.
So, without $ Cl^{\minus{}}$, plants cannot photosynthesize, which means their leaves will be small and "yellow".  :)

Thanks for the explanation, Bibobeo. Just by curiosity, what is the catalyst in the light dissociation of the water molecules?

I guess  the word chlorosis (yellowing of the leaves) must come from the word chlorine...

#Carcul:
I was explained like this:
$ H_{2}O\longrightarrow H^{\plus{}}\plus{}OH^{\minus{}}$
$ OH^{\minus{}}\longrightarrow H_{2}O_{2}$
$ H_{2}O_{2}\longrightarrow H_{2}O\plus{}O_{2}$ $ (*)$

Catalyzator is used in $ (*)$.

#sludgethrower: maybe  :D


Another revival of the previous discussion, also sign-ups for tourney with ssbb, just post your wii friend code, and i will post mine. id like some people.

HMMMMMM.................. Fox, Falco, and Wolf absolutley pwn, hands down

Falco, Marth, are beast.  But I seem to do well with Bowser too, even though he's really slow.

HMMM... do you have wolf yet?? he rly pwns. you can get him by defeating classic on intense. im not a gamer though. i do study them a bit.

not yet.  Idk if I can beat classic on intense....

Or you could beat the secret levels in subspace.

Edit: move this into games, not FF.

oh shoot, i mean get him by beating boss battles on any level with fox, and use only two full healths. what do u mean secret levels?.............................

everyone knows that Pit, Ike, Kirby, and King Deedeede(i don't know if i spelt that right.)

My brother says Pikachu instead of King Deedeede.

how do I join this tournament?

pit/g&w are my primary main characters :)
Diddy/kirby/IC and rob are secondaries.

Mewtwo? You can get him? tourneys, list ur friend code for it,, and ill list mine, then we can fight it out :lol:

Eh hem...
Kirby is the beastiest.
Hands down.
He will absorb you like...
hard.

Kirby? He is okay, though I like Jigglypuff somewhat. The rolling attack really sends them flying. Literally.

the rock smash is pwnage i agree and guess what!!!!!
[color=red][size=200]i just beat tabuu w00t :wow: and charziard helped me do it so did ness pit falco king dedede and pikachu!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!1[/size][/color]

good job on beating tabuu.

rock smash is ownage.

try captain falcon, fox, falco, samus.

all fast.

use the gun attack with foz and falco.

kirby is good for dodging the wing atack

Good Job on Tabuu.

It at first took me 14 tries to beat him on EASY, but within a day of playing I was pwning him on Intense.

The characters I used for Tabuu were

1. Kirby
2. Meta Knight
3. Pit
4. Fox
5. Toon Link
6. Sonic

So far, my best record on Tabuu(in Intense) is defeating him on Meta Knight. Pit and Fox are uber pwnage on Tabuu, not so awesome on real brawls.

They are pretty good on regular brawl, especially Pit, but they aren't the BEST at regular Brawl, as with Tabuu.

[quote="Joe10112"]pit and fox are uber pwnage on regular. Really. Fox is fast and i use a-attacks to pwn people. Pit just uses his side b and kills in like 2 minutes. Pit is cheap though.[/quote]
lol may be agaisnt level ones and people like you

Meta Knight is good against Tabuu because he can attack while staying in the air for a while. Just be sure not to get hit by the more powerful attacks. His aerial attack combos can keep damaging Tabuu while staying in the air (where tabuu is) for as long as possible. His speed can help when Tabuu changes position. And when Tabuu becomes big, stand at the very edge of the platform. The lasers cannot harm you there, and just hold the A button to constantly damage Tabuu. If you can avoid getting caught in the tether and bracket attacks, which usually kill Meta Knight in one hit on intense mode, you can win the fight. As for your backup characters...well...try choosing those that are also good at aerial combos (mostly other lightweights and middleweights), along with some powerhouses (like Ike and Dedede).

Edit: I found out that Tabuu on intense mode was voted the 2nd hardest boss in videogame history. Explains why I and many others had lots of trouble at first.

lol where did you get that stat?

Also, Pit really isn't that good.  And I still remember the times when my best character was Link...lol but that was when I was like 5 and at someone else's house playing the original SSB.

Seriously, Tabuu on Intense isn't that hard. All you need to know is this:

His first move will either be his Dive Bomb, Shark Attack,  or Bracket attack(Bracket is, in my opinion, the most annoying since it is hard to dodge and will 1-hit KO you). You can dodge all of them by moving around a lot or sidestepping.

His second move will possibly be his Tether, his multiple Tabuu attack, or his purple smoke thingy attack( ok i don't know names). Dodge by jumping and dodge rolling carefully.

His third move is usually his laser or spike thingy. Dodge both of them by jumping; however, on Intense, they travel very fast, especially the spike thingy, so you have to jump ahead of time.

Every other 4th move will be his famed "Butterfly Wing" attack. This is the most infamous of Tabuu's attack, but honestly, for an experienced dodger, this might become your favorite attack. All you do is sidestep 3 times in a row. It's VERY EASY. Every 8th move, he will have his Laser Eyes attack. All you need to do is move right to the edge of the platform and pwn him.

Really????? 2nd hardest boss in videogame history???

what would be the 1st hardest in history, and how are these "facts"  chosen?
what makes it "hard"?

Butterfly attack is actually really easy to dodge. And 2nd hardest is because of people's votes. The 1st (in terms of votes) is Queen Larsa from *ugh I forgot the name of the game*

Meta Knight: Recovery
Kirby: Inhaling and Stone
Marth: Because he's awesome
Dedede: Power

correct. THose work

but charzard for king dedede. Charzard escapes the 1-hit ko by switching pokeball.

fox uber pwns against tabuu. just use his gun attack

yes. You do constant small damage. After a while, you own.


ωραίο πρόβλημα

Η συζήτηση εδώ λύνει το πρόβλημα
http://www.mathlinks.ro/Forum/viewtopic.php?t=46075

Αγαπητέ Σιλουανέ θα ήθελα να μου απαντήσεις στο ερώτημα της άσκησης με βάση την παραπομπή σου η οποία δεν δίνει ξεκάθαρη απάντηση στο πρόβλημα
Επίσης παρακλώ να απαντήσετε αναλυτικά σοτ ακόλουθο πρόβλημα (καθότι εδώ την απάντηση την ξερω :D )
Στο καρτεσιανό επίπεδο οχψ θεωρούμε 25 σημεία με συντεταγμένες (κ,λ) με 4>=κ,λ>=0 κ.λ  ακέραιοι δηλαδη κ=0,1,2,3,4 κι λ=0,1,2,3,4
να προσδιορίσετε το πλήθος των τετραγώνων που κατασκευάζονται με κορυφές 4 από τα 25 δεδομένα σημεία
enjoy

Είναι γνωστή ......
Δείτε εδώ http://www.hms.gr/gr/eme/index_pin6.html

Σιλουανε η σελιδα δινει κατι αλλο και οχι το γνωστο θεμα της ΕΜΕ.επειδη η λυση της ασκησης ετσι και αλλιως ειναι λιτη και απλη οποιος μπορει να την λυσει δινοντας περρισοτερες λεπτομεριες...

Nai δεν ξέρω τι έγινε με το λινκ .....Πάντως είναι Ευκλείδης Β Λυκείου 2002-2003 .
Η απάντηση είναι $1\cdot 4^{2}+2\cdot 3^{2}+3\cdot 2^{2}+4\cdot 1^{2}=50$ τετράγωνα .

Θέλω εδώ τώρα να σας θέσω ένα βασικό ερώτημα συνδυστικής κι να δούμε ποιός θα το λύσει
Γιατί δεν μπορούμε "απευθείας" τουλάχιστον με την πολλαπλασιαστική αρχή να υπολογίσουμε τους συνδυασμούς των ν ανα κ?

Και βεβαια μπορουμε ουσιαστικα οι διαταξεις ειναι εφαρμογη της πολλαπλ.αρχης.Οπου εμφανιζονται παραγοντικα εχει εφαρμοστει η αρχη. :)

φίλε μου μην ξεχνάς ότι είπα απευθείας δηλαδή μεμιας άρα σωστό μεν αυτό που λές χρησιμοποιώντας πολλαπλασιαστική αρχη ουσιαστικά για να υπολογίσεις ότι κ! διατάξεις αντοιστοιχούν σε 1 συνδυασμό αλλά άλλο είναι αυτό που λέω Γιατί δεν γίνεται απευθείας υπολογισμός των συνδυασμών μεσω πολλαπλασιαστικής αρχής? να βοηθήσω λέγονατς αν το δείτε συολοθεωρητικά
χαιρετώ

Λοιπόν μιας κι κανείς δεν ενδιαφέρθηκε (γιατί άραγε?) για την απάντηση της ερώτησης θα την δώσω εγώ Άρα στην ερώτηση γιατί δεν μπορούμε «απευθείας» να υπολογίσουμε τους συνδυασμούς των ν ανά κ με την πολλαπλασιαστική αρχή (σημείωση χρησιμοποιήστε στοιχεία συνολοθεωρίας) η απάντηση έχει ως εξής
Προκαταρκτικά
1 Καρτεσιανό γινόμενο
2 Πολλαπλασιαστική αρχή (Με βάση Καρτεσιανό γινόμενο)
1 Καρτεσιανό γινόμενο 2 συνόλων (εύκολα γενικεύεται) Α, Β είναι
Α χ Β={(α, β)/ α ανήκει στο Α κι β ανήκει στο Β} δηλαδή όλα τα διατεταγμένα ζεύγη (α, β) με την ζητούμενη ιδιότητα
2 Πολλαπλασιαστική Αρχή με χρήση Καρτεσιανού γινομένου
Ν(Α χ Β)=Ν(Α)*Ν(Β)  όπου Ν(Ψ) ο πληθικός αριθμός του Ψ
Άρα τέλος η απάντηση στο ερώτημα είναι ότι αφού η Πολλαπλασιαστική αρχή έχει Καρτεσιανό γινόμενο το οποίο εκφράζει διάταξη είναι αδύνατον να χρησιμοποιηθεί η πολλαπλασιαστική αρχή για απευθείας υπολογισμό των συνδυασμών
Κι όποιος θέλει (κι εγώ αν επιθυμείτε) να σας πω για του λόγου το αληθές την εύρεση του τύπου των συνδυασμών των ν ανά κ
Χαιρετώ

Αμβοσιε πολυ καλη η σκεψη σου. ομως τι εννοεις? η πολαπλασιαστικη αρχη προκυπτει απο το καρτεσιανο γινομενο?


AMC
AIME
search
quadratics
AIME II
arithmetic series
algebra

Find the 8th term of the sequences 1440, 1716, 1848,... whose terms are formed by multiplying the corresponding terms of two arithmetic sequences.

This is #8 on the 2003 AIME II.

This problem is extensively discussed in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1831027765]in this existing thread[/url]. Feel free to review the discussion in that thread, and if appropriate/necessary, reply.

Threads for past AIME problems (as well as many, many other contest problems) can be accessed through the Contests section of the forum, accessible through the horizontal navigation bar near the top of the site.

Also, the AoPS Wiki, accessible from the left-side navigation column, has problems and solutions for many past AIME problems. This problem's AoPS Wiki page [url=http://www.artofproblemsolving.com/Wiki/index.php/2003_AIME_II_Problems/Problem_8]can be found here[/url].

To give a working link to a topic: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=713212#713212

:P

Nice, here's another method thats a bit longer but allows you to evaluate larger terms
[hide]
We have two arithmetic series of the form $ t_n \equal{} a_0 \plus{} (n\minus{}1)d$
Multiplying these together we get a quadratic equation passing through pionts $ (1,1440), (2, 1716)$ and $ (3, 1848)$

Using legrange interpolation:
$ l_0 \equal{} \dfrac{x\minus{}x_1}{x_0 \minus{} x_1}.\dfrac{x\minus{}x_2}{x_0\minus{}x_1} \equal{} \dfrac{x\minus{}2}{\minus{}1}.\dfrac{x\minus{}3}{\minus{}2}$
$ l_1\equal{}\minus{}(x\minus{}1).(x\minus{}3)$
$ l_2\equal{}\dfrac{1}{2}.(x\minus{}1)(x\minus{}2)$

So $ f(8)\equal{}1440.l_0 \plus{} 1716.l_1 \plus{} 1848.l_2$ where $ x\equal{}8$

$ f(8) \equal{} (1440).(15) \plus{} (1716).(\minus{}35) \plus{} (21).(1848) \equal{} 348$
[/hide]


function
logarithms
modular arithmetic
algebra proposed
algebra

Determine all functions $ f: \mathbb{N}\to\mathbb{R}$ such that for all integers $n \geq 1$, we have
\[ \log_{n\plus{}1}f(n)\equal{}\log_{f(n\plus{}2)}(n\plus{}3).\]

[quote="Pirkuliyev Rovsen"]Determine the functions $ f\colon \mathbb{N}\to\mathbb{R}$   $ n \neq 1$ such that $ \log_{n \plus{} 1}f(n) \equal{} \log_{f(n \plus{} 2)}(n \plus{} 3)$[/quote]

I suppose that $ n\ne 1$ in the problem statement means that $ f$ : $ \mathbb N\backslash\{1\}\to\mathbb R$
In order $ \log_{n \plus{} 1}f(n)$ be defined, we may also say that $ f$ : $ \mathbb N\backslash\{1\}\to\mathbb R^ \plus{}$

The equation may be written : $ \frac {\ln(f(n))}{\ln(n \plus{} 1)} \equal{} \frac {\ln(n \plus{} 3)}{\ln(f(n \plus{} 2))}$

Let then $ g(n) \equal{} \frac {\ln(f(n))}{\ln(n \plus{} 1)}$. The equation becomes $ g(n \plus{} 2) \equal{} \frac 1{g(n)}$

Hence the solutions :

Let $ a,b\in\mathbb R^*$ :
$ n \equal{} 0\pmod 4$ $ \implies$ $ f(n) \equal{} (n \plus{} 1)^a$
$ n \equal{} 1\pmod 4$ $ \implies$ $ f(n) \equal{} (n \plus{} 1)^b$
$ n \equal{} 2\pmod 4$ $ \implies$ $ f(n) \equal{} (n \plus{} 1)^{\frac 1a}$
$ n \equal{} 3\pmod 4$ $ \implies$ $ f(n) \equal{} (n \plus{} 1)^{\frac 1b}$


We have the set $A=\{0,1,\ldots,9\}$ and a seqence $a_{1},a_{2},\ldots,a_{99}$.This seqence has the following properties:
if $a_{i}=1$ then $a_{i+1}\not= 2$ and
if $a_{i}=3$ then $a_{i+1}\not= 4$.
Prove, that there exist different numbers $k,l$ such that $a_{k}=a_{l}$and $a_{k+1}=a_{l+1}$.

Suppose that there is no $k,l$ that satisfy the condition..
If $x\not=1,3$ then a number of $x$ in the sequence must be $\leq 10$ because there are only $10$ numbers $(0,1,2,...,9)$ that can follow each $x$.
If $x=1,3$ then a number of $x$ in the sequence must be $\leq 9$.
So the length of the sequence is $\leq 10\cdot8+9\cdot2=98<99$ contradiction.


vector
function
real analysis
real analysis unsolved

45. Let $E$ be a compact metric space. Let $C(E) = max \{|f(p)-g(p)|: p\in E\}$. Show that $C(E)$ is a complete normed vector space if we add its elements in the usual way, multiply them by real numbers in the usual way, and take $\parallel f\parallel = max \{|f(p)|\}: p\in E\}$ for all $f\in C(E)$.

Any have an elegant proof of this?

Proof that that is a norm should be straightforward. The only part of the proof likely to need comment is the proof that $C(E)$ is complete. Here's a broad sketch of that proof.

Suppose you have a uniformly Cauchy sequence of continuous functions. Uniformly Cauchy implies pointwise Cauchy, implies pointwise convergent, so there exists a pointwise limit function. The original sequence, because it is uniformly Cauchy, converges uniformly to this limit function. Finally, the uniform limit of continuous functions is continuous, so the limit function is a member of $C(E).$


algebra
polynomial
Vieta
number theory proposed
number theory

Solve in positive integers:

$ a^{2}\plus{}b^{2}\plus{}1\equal{}3ab$

[quote="delegat"]Solve in positive integers:

$ a^{2} \plus{} b^{2} \plus{} 1 \equal{} 3ab$[/quote]
I guess the original problem was to prove that if positive integers $ a,b,c$ satisfies the equation
$ a^2 \plus{} b^2 \plus{} 1 \equal{} abc$
then $ c \equal{} 3$.See here: http://www.mathlinks.ro/viewtopic.php?t=150373
Your problem can be solved using vieta-jumping.

No! I am interested in positive integer solutions of: $ a^{2}\plus{}b^{2}\plus{}1\equal{}3ab$.

Maybe solution can be found by using Pell's equation or some similar methods.

As i told you before all solutions can be found by using vieta-jumping method.
Obviously $ a_0\equal{}b_0\equal{}1$ is the solution.
Consider the sequence $ \{a_n,b_n\},a_{n\plus{}1}\equal{}3b_{n}\minus{}a_{n},b_{n\plus{}1}\equal{}a_{n}$ for all $ n\geq 0$.
Obviously this sequence satisfies to the equation and to the statement too(from [color=red]vieta[/color] theorem).
We will prove that all the solutions of the equation are terms of our sequence.
Suppose for the contrary that there exist $ (x,y)\in\mathbb{N}$,such that $ a^2\plus{}b^2\plus{}1\minus{}3ab\equal{}0$ and the solution $ (x,y)$ is not the term of $ \{a_n,b_n\}$.
Obviously $ x\neq y$(otherwise they are both equal to $ 1$) and WLOG assume that $ x>y$,$ x>2$.(If $ x\equal{}2$ then $ y\equal{}1$,contradiction).
Consider another solution $ x_0\equal{}3y\minus{}x,y_0\equal{}x$.Obviously $ x_0\equal{}\frac{y^2\plus{}1}{x}<y$,hence the sum $ x\plus{}y$ became less and
since it can not be less than $ 2$ the process will stops,when $ x_n\equal{}y_n\equal{}1$.But then $ x,y$ are uniqely defined by $ x_n,y_n$.And obviously $ (x,y)\in\{a_n,b_n\}$.Contradiction.


log(m) = b- log(n), then m=? (In terms of n and b)

All logs are base 10

[hide]log(m)+log(n)=b

log(m*n)=b

10^b = m*n

m = 10^b / n [/hide]

Guideline: if I can get it right it really should not be in intermediate.  Oh and [hide] if I got this wrong I have really made myself look bad![/hide]


search
number theory proposed
number theory

Let $P$ be a finite set of primes and let $a$ be an integer. prove there are only finite numbers $n$ such that all prime divisors of $a^n-1$ belong to $P$

I have solved this nice iranian problem some time ago. Search for one of samn's posts.

We can first show that if $n_k$ is an infinite sequence (which we must show doesn't exist), then the members $n_k$ of teh seuqence have only finitely many prime factors. This means that it's enough to show that there are only finitely many $n$ for which the prime factors of $a^{p^n}-1$ belong to $P$ (call this condition $(*)$).

We have $(\frac{a^{p^{n+1}}-1}{a^{p^n}-1},a^{p^n}-1)=(a^{p^n}-1,p)$ so, if we assume there are infinitely many such $n$, then all $n$ satisfy $(*)$, so for $n\ge n_0$ for some $n_0$ we have $\frac{a^{p^{n+1}}-1}{a^{p^n}-1}=p$, and we easily get a contradiction from here (we get $a=1$).

[quote="grobber"]We can first show that if $n_k$ is an infinite sequence (which we must show doesn't exist), then the members $n_k$ of teh seuqence have only finitely many prime factors.[/quote]
Probably, you use $\left(a^x-1,a^y-1\right)=a^{(x,y)}-1$ for this? But how do you deal with the case when all $a^{n_k}-1$ have the same sets of prime factors as the $a-1$? I can't find a contradiction...

It's very uncomfortable for me to go back to solutions written ages ago and figure out what the whole thing was about, but I'll try :).

Assume we have an infinite sequence of primes $p_n$ s.t. the prime factors of $a^{p_n}-1$ are among those of $a-1$. Use the exact same type of argument as in my previous message: $(\frac{a^{p_n}-1}{a-1},a-1)=(a-1,p_n),\ \forall n$. I'm sure you can find a contradiction here.


inequalities
quadratics
algebra
quadratic formula
inequalities solved

$a^2+b^2+c^2+2abc+1 \ge 2(ab+bc+ca)$
$a^2+b^2+c^2+3 \ge 2(a+b+c)$
sum them up and done

[quote="minhkhoa"][quote="siuhochung"]$a^2+b^2+c^2+2abc+1 \ge 2(ab+bc+ca)$
[/quote]
Can I know why this is true[/quote]
This is posted before, but anyway,
$a^6+b^6+c^6+2a^3b^3c^3+1 \ge a^6+b^6+c^6+3a^2b^2c^2 \ge \sum_{sym} a^4b^2 \ge \sum_{sym} a^3b^3$

[quote="minhkhoa"]Let $ a,b,c$ be positive real numbers such that $ bc \plus{} ca \plus{} ab \equal{} 3.$ Prove that

$ a^2 \plus{} b^2 \plus{} c^2 \plus{} abc\ge a \plus{} b \plus{} c \plus{} 1.$[/quote]$ a^2 \plus{} b^2 \plus{} c^2 \plus{} abc \minus{} (a \plus{} b \plus{} c \plus{} 1) \minus{} (bc \plus{} ca \plus{} ab \minus{} 3)$

$ \equal{} abc \plus{} a^2 \plus{} b^2 \plus{} c^2 \minus{} bc \minus{} ca \minus{} ab \minus{} a \minus{} b \minus{} c \plus{} 2$

$ \equal{} (b \minus{} c)^2 \plus{} (a \minus{} 1)^2 \plus{} (a \plus{} 1)(b \minus{} 1)(c \minus{} 1)\geq 0,$

which is clearly true for $ (b \minus{} 1)(c \minus{} 1) \equal{} \max\{(b \minus{} 1)(c \minus{} 1),(c \minus{} 1)(a \minus{} 1),(a \minus{} 1)(b \minus{} 1)\}.$

See also : http://www.mathlinks.ro/viewtopic.php?t=82866

may be old
Let $ a, b, c>0$ such that $ abc\equal{}8$. Prove that:
$ \sum \frac{a^2}{\sqrt{(a^3\plus{}1)(b^3\plus{}1)}}\ge \frac{4}{3}$

[quote="fesha3t"]Let a, b, c > 0 such that abc = 8. Prove that:
$ \sum \frac {a^2}{\sqrt {(a^3 \plus{} 1)(b^3 \plus{} 1)}}\ge \frac {4}{3}$[/quote]
$ \sqrt{a^3\plus{}1}\equal{}\sqrt{(a\plus{}1)(a^2\minus{}a\plus{}1)}\le\frac{a^2\plus{}2}2$
...

$ \Rightarrow LHS\ge\sum\frac{4a^2}{(a^2\plus{}2)(b^2\plus{}2)}$

We notice that
$ \sum\frac{a^2}{(a^2\plus{}2)(b^2\plus{}2)}\minus{}\sum\frac{b^2}{(a^2\plus{}2)(b^2\plus{}2)}\\\equal{}\sum\frac{(a^2\plus{}2)\minus{}(b^2\plus{}2)}{(a^2\plus{}2)(b^2\plus{}2)}\equal{}\sum\left(\frac1{b^2\plus{}2}\minus{}\frac1{a^2\plus{}2}\right)\equal{}0\\\Leftrightarrow\sum\frac{a^2}{(a^2\plus{}2)(b^2\plus{}2)}\equal{}\sum\frac{b^2}{(a^2\plus{}2)(b^2\plus{}2)}$

So, $ LHS\ge\sum\frac{2(a^2\plus{}b^2)}{(a^2\plus{}2)(b^2\plus{}2)}$
We only have to prove $ \sum\frac{(a^2\plus{}b^2)}{(a^2\plus{}2)(b^2\plus{}2)}\ge\frac23$
After multiplying everything out, it remains:
$ a^2b^2\plus{}b^2c^2\plus{}c^2a^2\plus{}6(a^2\plus{}b^2\plus{}c^2)\ge72\plus{}4(ab\plus{}bc\plus{}ca)$
which is true because
$ a^2b^2\plus{}b^2c^2\plus{}c^2a^2\ge3\sqrt[3]{a^4b^4c^4}\equal{}48\\6(a^2\plus{}b^2\plus{}c^2)\ge6(ab\plus{}bc\plus{}ca)\\2(ab\plus{}bc\plus{}ca)\ge2.3\sqrt[3]{a^2b^2c^2}\equal{}24$
Adding up the 3 ineq above => QED

[quote="minhkhoa"][quote="siuhochung"]$ a^2 \plus{} b^2 \plus{} c^2 \plus{} 2abc \plus{} 1 \ge 2(ab \plus{} bc \plus{} ca)$
[/quote]
Can I know why this is true[/quote]
We can use [b]Schur's Inequality[/b] to prove this problem:

[u]My solution:[/u]
Using the AM- GM Inequality, we have
$ 2abc \plus{} 1 \equal{} abc \plus{} abc \plus{} 1 \ge 3\sqrt[3]{{a^2 b^2 c^2 }} \equal{} 3\frac{{abc}}{{\sqrt[3]{{abc}}}} \ge 9\frac{{abc}}{{a \plus{} b \plus{} c}}$
So, \[ a^2  \plus{} b^2  \plus{} c^2  \plus{} 2abc \plus{} 1 \ge a^2  \plus{} b^2  \plus{} c^2  \plus{} \frac{{9abc}}{{a \plus{} b \plus{} c}}\]
It suffices to prove that:
\[ a^2  \plus{} b^2  \plus{} c^2  \plus{} \frac{{9abc}}{{a \plus{} b \plus{} c}} \ge 2(ab \plus{} bc \plus{} ca)\]
But, which is true from Schur's Inequality

[quote="minhkhoa"]Let a,b,c be positive real numbers such that $ab+bc+ca=3$
Prove that
$a^2+b^2+c^2+abc\ge a+b+c+1$[/quote]
In three numbers $a, b, c$ always exist two numbers together with greater than 1 or smaller than 1. Without loss of generality, we assume the two of them are $a,b$. When it $(a-1)(b-1) \geq 0 \Rightarrow abc \geq ac+bc-c=3-ab-c$
So we only need to prove: $a^2+b^2+c^2+3-ab-c \geq a+b+c+1 \Leftrightarrow a^2-(b+1)a+b^2+c^2-b-2c+2 \geq 0$ (1).
The left side of (1) the three quadratic formula is:$ \Delta =-3b^2+6b-4c^2+8c-7=-3(b-1)^2-4(b-1)^2 \leq 0$.
Since then we have to prove


geometry
geometric transformation
arithmetic sequence

Find number of all infinite arithmetic progressions of integers in which between first 10 terms are numbers 1 and 2005.

Are you trying to say this:

"Find the number of infinite arithmetic progression that exist where the first ten terms are between 1 and 2005"

?

hmm... I'm new to this board, but I know a thing or two about math.

Anyway wouldn't you have to specify that all ten numbers are integers?  Other wise I could say 1,1.001,1.002,1.003... mayber I'm understanding the problem wrong.

[quote]...progressions of integers ...[/quote]

so the first ten terms have to be integers between 1 and 2005, inclusive or not??
i'm not sure i understand the problem

oops missed it :lol:
[hide]
1-10 can appear in 1995 variations (starting with one, two, three, etc.)
1,3,5,...19 can appear in 1986 ways (9 less than the last)
1,4,7,...28 can appear 1967 ways (again nine less)
and so on... and so on...
1,223,445,...1999 can appear 6 ways

add up all those variations together [(1+2+3...+221)*9]+(221*6)= 220779+1326=222105 (221 unique differences between numbers in algebraic model)

Final answer 222105.  I hope that wasn't too confusing :? [/hide]

[hide]
There are $2005 / 9 \approx 222$ different constants that can be added to 1 to produce the first 10 numbers between 1 and 2005, I divided by 9 because the first term is excluded.

These progressions can start anywhere if the constant is continually added and the sum stays less than 2005 for the first 10 terms.

There are 1996 ways that 1 can be the constant since the last term will be 2005.
There are 1987 ways that 2 can be the constant since the last term will be 2005.
There are 1978 ways that 3 can be the constant since the last term will be 2005.
...
There are 7 ways that 222 can be the constant since the last term will be 2005.

Now to sum all of these:
7 is the first term and 1996 is the 222nd term:
$1996 = 7 + (n-1)9$
$1996 = 7 + 9n - 9$
$1998 = 9n$
$222 = n$

$S_n = n(\frac{a_1 + a_n}{2})$
$S_{222} = 222(\frac{7 + 1996}{2})$
$S_{222} = 222,333$

So there are 222,333.

If its not inclusive I'm wrong; eh Im prolly wrong anyway.
[/hide]

oops, sorry, problem may be confusing  :D
I will write it properly.You should find number of infinite arithmetic progressions of integers $a_1,a_2,a_3,\ldots$ such that there exists two distinct integers $i,j$ for which $1\leq i,j \leq 10$ such that $a_i=1, a_j=2005$.
There was translation error on my side (maybe usage of "between" in Slovak is strange to you  :) )

[hide]
For 2004 there are 9 possibilities
For 1002 there are 8 possibilities
For 501 there are 7 possibilities
For 334 there are 6 possibilities

So that adds up to 30, I think.
[/hide]

snowstorm: you've forgotten some solutions, but you are on very good way to solve it completely  :)

Ok now maybe?
[hide]
I forgot 668 last time.

For 2004 there are 9 possibilities
For 1002 there are 8 possibilities
For 668 there are 7 possibilities
For 501 there are 6 possibilities
For 334 there are 4 possibilities

So that adds up to 34, I think.
[/hide]

You almost got it, just notice that integers can be also negative... so instead of 9,8,7,6,4 there should be 2 times more: 18,16,14,12,8. So the answer is 2*34=68.


Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ a \plus{} b \plus{} c \equal{} 1.$ Prove that:
\[ \sqrt {7a^2 \plus{} 1} \plus{} \sqrt {7b^2 \plus{} 1} \plus{} \sqrt {7c^2 \plus{} 1}\geq4
\]

Oops.

Sorry read it wrong. Here we go again.

[hide="Solution?"]

Let $ f(x) \equal{} \sqrt {7x^{2} \plus{} 1}$ and this clearly concaves up for $ x \in \mathbb{R}$

Using Jensen's Inequality:

$ \frac {1}{3} ( f(a) \plus{} f(b) \plus{} f(c)) \geq f(\frac {1}{3}) \equal{} \frac {4}{3}$

Hence $ \sqrt {7a^{2} \plus{} 1} \plus{} \sqrt {7b^{2} \plus{} 1} \plus{} \sqrt {7c^{2} \plus{} 1} \geq 4$.
[/hide]

[quote="BanishedTraitor"]Oops.

Sorry read it wrong. Here we go again.

[hide=Solution?]

Let $ f(x) \equal{} \sqrt {7x^{2} \plus{} 1}$ and this clearly concaves up for $ x \in \mathbb{R}$

[/quote]

You probably mean that $ f(x)$ is convex :wink: otherwise the inequality would be reversed...

I am pretty sure $ \text{convex}$ and $ \text{concave up}$ refers to the same thing ($ f''(x) >0$).

We have to prove that

$ \sum\sqrt {7a^2 \plus{} 1}\geq 4$ or $ \sum (\sqrt {7a^2 \plus{} 1} \minus{} \frac {4}{3} \minus{} \frac {7}{4}(a \minus{} \frac {1}{3}))\geq 0$

or $ \sum\frac {7(3a \minus{} 1)^2}{4(3 \plus{} 7a \plus{} 4\sqrt {1 \plus{} 7a^2})}\geq 0$  which clearly holds and we are done


[quote="anitsahu"]i have some scores to settle with mr pardesi,mr madness varunabasu .............. :mad:
 :D
mr madness is going to pay dearly for it
all of u will see it soon[/quote]
that's varunarasu :huh:

i forgot to name some more
shruthi bhaskaran
and yes soumyashant nayak
mr ghatu is gonna hear from me a lot next time i meet him :mad:
and something more for all who have challenged me i have never lost a challenge in my life :D

[quote="pardesi"][quote="anitsahu"]i forgot to name some more
shruthi bhaskaran
and yes soumyashant nayak
[/quote]
and that's Shruthi Baskaran :P[/quote]
hey what do u mean by thats "shruthi bhaskaran"

You spelled it wrong!! :D

And anit, well it seems like you have bets to settle with most of AOPS except me...

I'll be auctioneer then.

Who'll get the pizza and who'll get the viisa!! We shall find out on the March 3. Be Tuned to Homi Bhabha TV \

have a look at her name [url=http://www.mathlinks.ro/profile.php?mode=viewprofile&u=37402]here[/url]
that's Baskaran and [b] not[/b] BHaskaran :P  :rotfl:  :rotfl:

hey shreyas i know something
i gonna have the pizza
the visas r gonna go to mr soumyashant nayak
and varunabasu
and the baby

Cool down anit :jump:

we'll see :w00tb:  :w00tb:

pake thane porum :rotfl:  :rotfl:  :P ...rakesh un super translation le edhu translate pannidu :D

ella ella ....WE will OBSERVE :P  :P  :rotfl:  :rotfl:

Translation
We will see only see:D

Anit, no I'm not for a challenge. Whoever gets through congrats from me. :D
And I swear this thread sounds like Challenges and Thrill

i am sure i will win my bet :lol:  :lol:  :lol:
they are the most ob ones :P

cha i lost the bet  :(   :o  ~crying otw~  :(

Anit one hot spicy pizza ready. :rotfl:
What toppings do u prefer? :huh:

all come to me funeral today   :(   :o   :(

how can i when me mind is in dreams  :(   :o

anit, dont forget that i have to get a pizza. :)

[quote="anitsahu"]i won the pizza
so whos the nerd
and now the killings will start
one more thing
i am leaving aops till the end of jee :([/quote]

Come on man, you can come for 5 or ten mins once a day??? Don't tell me you are that big a mega-nerd.. Or perhaps you don't have any phenomenon called relaxation.

Ah he probably didnt understand what you said anyway

bye to all
best of luck for your boards and jee

i hope its useful to you..
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OMG!!!! Was that sandeep?  :D  How come i never noticed that  :rotfl:   :D

I really doubt if it was Sandman. But yeah. Maybe we should rename this thread as bets and challenges for the ides of march.. Might be consoling for me :(


LaTeX
vector
analytic geometry
limit
topology
real analysis
real analysis unsolved

I'm trying to find the can someone help me out? I need this to prove l^2 is separable metric space
hints appreciated :)

thanks

by $ I^2$ you mean $ [0,1]\times [0,1]$? then $ \{(a,b)|a,b\in [0,1], a,b \in \mathbb{Q}\}$ will do.

Sorry, I meant L^2.... i.e. all sequences of real numbers that have finite sum of all |a_i|^2 where a_i is the i-th element of the sequence...

$ \text{\LaTeX}$ tip: the code \ell gets you $ \ell,$ hence $ \ell^2,$

For topological vector spaces, while the explicit definition of separable is that the space has a countable dense sub[b]set[/b], the practical and usable consequence is that the space has a "nice" dense sub[b]space[/b].

Let's start with finding a nice dense subspace. For a sequence space like $ \ell^2,$ there's a natural candidate: the space of finitely supported sequences. That is, the set of sequences $ (x_n)$ such that $ \exists\,N$ ($ N$ depends on the sequence) such that $ x_n=0$ for $ n>N.$ (The set of finitely supported sequences is one realization of a vector space with a countably infinite Hamel basis.)

We can approximate any $ \ell^2$ sequence arbitrarily closely by such a finitely supported sequence simply by truncating far enough out in the series. Having done that, we can then approximate the finitely supported sequences arbitrarily closely by finitely supported sequences with rational coordinates - and that's a countable set.

Ok, so I start with any sequence in $ \ell^2$.
Then I approximate it to a finite sequence by truncating some elements (hm, why can I do that?).
Now that I have a finite sequence of real numbers I just approximate it with the sequence of similar length, just instead of real numbers I put rationals (and I can do that since Q is dense in R).
makes sense - as long as I can explain the truncating :)

thanks!

Suppose $ x\in\ell^2$ so that $ \sum_{i\equal{}1}^{\infty}|x_i|^2<\infty.$ Let
\[ y_i^{[N]}\equal{}\begin{cases}x_i&\text{if }i\le N\\0&\text{if }i>n\end{cases}\]
Then $ \|x\minus{}y^{[N]}\|_2^2\equal{}\sum_{i\equal{}N\plus{}1}^{\infty}|x_i|^2$ and simply as a consequence of having a convergent series, we have that
\[ \lim_{N\to\infty}\sum_{i\equal{}N\plus{}1}^{\infty}|x_i|^2\equal{}0.\]

Clearly, the exact same argument works in any $ \ell^p$ for $ 1\le p<\infty.$ It does not work in $ \ell^{\infty}$ and in fact $ \ell^{\infty}$ is not separable.


calculus
integration
function
calculus computations

What definite integral gives the length of the graph of y = e^(e^x) between x=0 and x=1?

The arc length of a function $ f$ on the interval $ [a,b]$ is defined with:

$ L\equal{}\int^b_a \sqrt{1\plus{}(f'(t))^2} \,dt$

so the answer is:

sqrt(1 + e^(2(x+e^x))


induction
AMC
USA(J)MO
USAMO
abstract algebra
vector
group theory

At a certain mathematical conference, every pair of mathematicians are either friends or strangers. At mealtime, every participant eats in one of two large dining rooms. Each mathematician insists upon eating in a room which contains an even number of his or her friends. Prove that the number of ways that the mathematicians may be split between the two rooms is a power of two (i.e., is of the form $ 2^k$ for some positive integer $ k$).

It is not a full solution, but I noted the following:
[hide]You can replace any vertex with at least 4 neighbors with a set of three vertices, one connected only to the other two, and the other two divying up the original vertices neighbors, without modifying the desired number of ways to partition the graph. Then kill all the degree 0 or 1 vertices (also possible), and you are left with only degree 2 and 3 vertices. That is all I could get.[/hide]

Well I don't know but it is very easy to prove for trees (it is a strong induction basically). Then I was hoping that there'd be a way to like bridge the gap between that and a graph that has cycles.

Same here; I proved it for all vertices of degree at most 2, and then remarked that mathematicians don't normally have more than 2 friends. :) Clearly the hardest problem on the USAMO this year.

We'll prove that if there exists a good configuration, then the number of good configurations is a power of two.  Here's a basic sketch of it.
[hide]
Define an "order" to be a set of instructions, one instruction given to each mathematician.  Each mathematician is told to either "move" or to "stay" (you can think of stay as 0, and move as 1).  Now take some good configuration.  Consider the set of all orders which, when performed on this configuration, give us another good configuration.  Note the identity order, (0, 0, ..., 0), is in this set.  I claim this set is an abelian group under composition.

Proof of this: First clearly each thing is its own inverse, there is the identity, and the operation is clearly associative and commutative (because it's equivalent to addition of n-dimensional vectors mod 2).  So it suffices to show this set of orders is closed under composition.

Consider any mathematician.  If, in one of these orders, he is told to stay, then the number of his friends who are told to move must be even.  Similarly, if he is told to move, then the number of his friends who are told to stay must be even.  Now just consider two orders A and B and you can show that in A*B the same property will hold using parity.

Okay, so now that you've proven it's a group, we'll prove it has order two.  Let I be the identity.  Call this group G.

Let $ \{I, T_1\}\equal{}H_1$, where $ T$ is some element of $ G$.  Now pick an element $ T_2$ of $ G$ which is not in $ H_1$.  Notice that because the elements of $ H_1$ are distinct, the elements of $ \{T_2X|X \in H_1\}$ are distinct (if two elements of that set were the same, multiply by each on the right by $ T_2^{\minus{}1}$ and you have a contradiction).  Now notice that for any $ A, B \in H_1$, if we were to have $ T_2A\equal{}B$, then $ T_2\equal{}BA^{\minus{}1} \in H_1$.  Therefore, $ H_1$ and $ \{T_2X|X \in H_1\}$ are disjoint and of the same size.  Moreover, the product of any element in the first group and any element in the second group is a member of the second group.  Therefore, these two groups together form a group of order $ 2^2$.  Call this $ H_2$.  Do you see where I'm going?  You progressively build larger and larger subgroups of $ G$ until you get to $ G$ itself, whose order must then be a power of two.  Therefore, the number of good configurations of the mathematicians was a power of two.

This, of course, was all assuming $ I$ existed and was in the group.
[/hide]

Here is a full solution:

[hide]
Make the obvious re-interpretation as a graph. Let $ f : G \to \{0,1\}$ be an indicator function with $ f(v) \equal{} 0$ if a vertex is in the first partition and $ f(v) \equal{} 1$ otherwise (this corresponds, in the actual problem, to putting a mathematician in the first or second room). Then look at $ f$ as a function into the field with two elements, $ F_2$. Let $ V$ be the vector space of all such functions. Define the linear operator $ A : V \to V$ as

$ (Af)(v) \equal{} \sum_{v \sim w} f(v) \minus{} f(w)$

where $ \sim$ denotes adjacency. (Note that we can also think of $ A$ as a matrix, which is essentially the adjacency matrix where the diagonal is changed to be $ 1$ whenever the degree is odd; in more technical terms, it is the Laplacian of $ G$ over $ F_2$). Then $ f$ is a valid partition iff $ (Af)(v) \equal{} d(v)$, where $ d$ is the degree of $ v$, for all $ v$ (this is taken over $ F_2$). So we want all solution to $ Af \equal{} d$. Note that if $ Af \equal{} d, Ag \equal{} d$, then $ A(f \minus{} g) \equal{} 0$, so $ f \minus{} g$ is in the nullspace. Thus in particular the number of solutions, if non-zero, is the size of the nullspace of $ A$, which is $ 2^{dim(Null(A))}$ by considering all linear combinations of any basis of $ Null(A)$ over $ F_2$. Also let $ i$ be such that $ i(v) \equal{} 1$ for all $ v$. Then clearly $ Ai \equal{} 0$, so $ dim(Null(A)) > 0$, establishing that the number of ways to do this is $ 2^k$, $ k > 0$. Thus we need only prove the existence of a solution.

Since we can add a new vertex connected to exactly one previously existing vertex without changing the problem, WLOG all vertices have odd degree. Then we want to show that $ i \in Col(A)$. But it is a well-known fact in linear algebra that $ Col(A) \equal{} Null(A^T)^{\perp} \equal{} Null(A)^{\perp}$ since $ A$ is symmetric. Thus we need to show that if $ f \in Null(A)$, $ f$ is perpendicular to $ i$ and we will be done. So let $ f \in Null(A)$. Take the submatrix of $ A$ consisting of the rows and columns $ v$ such that $ f(v) \equal{} 1$. Then, since $ f \in Null(A)$, the sum of each row in this submatrix must be $ 0$ in $ F_2$. Thus the total number of $ 1$s in this submatrix is even. But since it is symmetric, the total number of $ 1$s off of the diagonal is even, so the total number of $ 1$s on the diagonal is even. But since every vertex has odd degree, the entire diagonal of $ A$ consists of $ 1$s, so this says that the size of the diagonal of this submatrix is even. But this is also the number of $ v$ such that $ f(v) \equal{} 1$, so $ f(v) \equal{} 1$ for an even number of $ v$, thus is perpendicular to $ i$, and we have our result.
[/hide]

Hm...I wonder are we allowed to submit alternate solutions to the USAMO people? My partial work for #6 was the first half of this, which will get much more credit I think if they know it can be finished with the same ideas.

EDIT: Interestingly, the induction for the official solution is similar to a well-known process in algebraic graph theory called rank-two reduction of a binary adjacency matrix. It can be found in [i]Algebraic Graph Theory[/i] by Godsil and Royle, p. 183. This was somewhat annoying as I was aware of this but didn't really remember the process, and also wasn't sure it would work...but it seems to be more or less the same spirit of induction.

[quote="K81o7"]We'll prove that if there exists a good configuration, then the number of good configurations is a power of two.  Here's a basic sketch of it.
[hide]
Define an "order" to be a set of instructions, one instruction given to each mathematician.  Each mathematician is told to either "move" or to "stay" (you can think of stay as 0, and move as 1).  Now take some good configuration.  Consider the set of all orders which, when performed on this configuration, give us another good configuration.  Note the identity order, (0, 0, ..., 0), is in this set.  I claim this set is an abelian group under composition.

Proof of this: First clearly each thing is its own inverse, there is the identity, and the operation is clearly associative and commutative (because it's equivalent to addition of n-dimensional vectors mod 2).  So it suffices to show this set of orders is closed under composition.

Consider any mathematician.  If, in one of these orders, he is told to stay, then the number of his friends who are told to move must be even.  Similarly, if he is told to move, then the number of his friends who are told to stay must be even.  Now just consider two orders A and B and you can show that in A*B the same property will hold using parity.

Okay, so now that you've proven it's a group, we'll prove it has order two.  Let I be the identity.  Call this group G.

Let $ \{I, T_1\} \equal{} H_1$, where $ T$ is some element of $ G$.  Now pick an element $ T_2$ of $ G$ which is not in $ H_1$.  Notice that because the elements of $ H_1$ are distinct, the elements of $ \{T_2X|X \in H_1\}$ are distinct (if two elements of that set were the same, multiply by each on the right by $ T_2^{ \minus{} 1}$ and you have a contradiction).  Now notice that for any $ A, B \in H_1$, if we were to have $ T_2A \equal{} B$, then $ T_2 \equal{} BA^{ \minus{} 1} \in H_1$.  Therefore, $ H_1$ and $ \{T_2X|X \in H_1\}$ are disjoint and of the same size.  Moreover, the product of any element in the first group and any element in the second group is a member of the second group.  Therefore, these two groups together form a group of order $ 2^2$.  Call this $ H_2$.  Do you see where I'm going?  You progressively build larger and larger subgroups of $ G$ until you get to $ G$ itself, whose order must then be a power of two.  Therefore, the number of good configurations of the mathematicians was a power of two.

This, of course, was all assuming $ I$ existed and was in the group.
[/hide][/quote]

That is a pretty interesting application of group theory to a combinatorics problem. I was wondering where you learned about such applications of group theory. Do you know of any resources that teach about applications of group theory to elementary combinatorics problems?

@Jacob: As always, you are king of linear algebra.  This is why I think you really deserve to make the IMO: you're interested in the math itself more than most people.  A lot of people didn't really pursue the problem after the competition.

@calc rulz: I learned basically all of the group theory I know from the book Topics in Algebra (Herstein), I read about half of the group theory chapter.  As for applying it to combinatorics problems, I don't know...I kinda just played around with some stuff and noticed the group structure.  Once you have a set of things with a certain property (especially if those things are functions which create maps between objects of a certain kind, like this case), and a binary operation, if you have associativity, an identity, and the ability to invert, you might feel the group-ness and check if the thing is closed.  That was quite a run on.

[quote="K81o7"]@Jacob: As always, you are king of linear algebra.  This is why I think you really deserve to make the IMO: you're interested in the math itself more than most people.  A lot of people didn't really pursue the problem after the competition.[/quote]
Forgive me being negative, but it seems to me there are better things to do than go to the IMO if the math is [i]all[/i] you're interested in. In recent times the IMO seems to emphasize techniques and preparation over the math, like with everyone trivializing 2007/5 by "Vieta jumping" or whatever.

That said, I thought he was going to make the IMO because he's good at these kinds of problems anyway.

[quote="JSteinhardt"]Well, I haven't failed to make the IMO team *yet*. Let's wait until we see the black MOP cutoff before we jump to conclusions.[/quote]

How did you do actually?

I solved 1 and 4 pretty definitely. Number 5 I tried to prove for all reals instead of non-negative, so my proof was something like "in the special case that $ r_1,r_2,r_3 > 0$, we can do this...then in all other cases, (insert bogus proof)". The consensus seems to be that it's worth a 7 anyways. And number 6 I have what is said in this thread, and probably about 2 points of partial credit out of 2 and 3...so probably a 25ish.

[quote="JSteinhardt"]Well, I haven't failed to make the IMO team *yet*. Let's wait until we see the black MOP cutoff before we jump to conclusions.[/quote]

Who's jumping to conclusions here?

Anyways, I hear Lars solved this on the test itself, I wonder if he has the same solution as the official solution (which I haven't read yet, I want to find a combinatorial solution myself when I get time)...

And did he get all 6 then I wonder?

[quote="t0rajir0u"]Jacob, I'm honestly still wondering why you didn't make Intel Finalist.  You're clearly as qualified as the math kids I met there  :huh:

Anyway, there are some great ideas in this thread.  The only interesting one I had:  one way to interpret the partition we desire is as a partition into two graphs with Eulerian circuits.[/quote]

Well it's possible that my project wasn't interesting enough...I mean it was pretty much just a random problem I came up with that was interesting and tangentially related to some current research. I think the problem I'm working on now might have worked better, but alas I'm half a year too late (it has to do with a conjecture on the chromatic number of a certain infinite graph, and I got to use all sorts of high-powered stuff to prove that there exists no finite coloring...although it's only 5 pages instead of 20, so who it might have looked kind of silly next to all of the 20-page papers...I kind of wish I was still eligible for Intel so that I could see what would have happened).

Also, in response to Krishanu:

[quote]That said, I thought he was going to make the IMO because he's good at these kinds of problems anyway.[/quote]

The past tense there seems to imply that he doesn't still think I will make it. But I could be misinterpreting. Anyways it would be cool to make the IMO team still, though I'd be happy with just going to MOP again.

^^Bump^^

Same question as post 46 regarding solution in post 44.

What the "1" means on the LHS is the constant coefficient 1, or you can take it as an equation where you have successfully eliminated all variables. For instance, if I had $x_1 + x_2 = 0$ and $x_1 + x_2 + 1 = 0$, then if I added both of them I would have gotten the equation $1 = 0$ and then there would be no solutions. But if I only had $x_1 = 0$ and $x_1 + x_2 + 1 =0$, then no matter how you tried to add them, the LHS would always have a variable (either $x_1$ or $x_2$). The idea is then that a system of equations over mod $2$ will always have solutions, unless you are able to derive that $1 = 0$ by adding up a bunch of them.

But he assumed 1=0 and then proved "1=0, contradiction."that's what I'm confused about. Sorry I wasn't clear about my question.

[u]Lemma:[/u] Let $A\in\mathbb{F}_2^{n\times n}$ be symmetric, and let $d=\mathrm{diag}A\in\mathbb{F}_2^n$ be the diagonal of $A$. Then, $\mathrm{diag} A\in\mathrm{im} A$.

[u]Proof of Lemma:[/u] We show that $d\in(\mathrm{null}A)^\perp=\mathrm{im}A$. Suppose $Ax=0$. We want to show $d\cdot x=0$.

Let $I=\{i:x_i=1\}$, and let $B$ be the restriction of $A$ on to $I$. We have that $B1=0$ where $1$ us the all $1$s vector. Adding up all the rows of the equation $B1=0$, we see that the off diagonal terms cancel because $B$ is symmetric and we are working mod $2$, so we see that $d'\cdot 1=0$ where $d'$ is the restriction of $d$ on to $I$, so we have $d\cdot x=0$, as desired. Thus, the claim, and the problem are proven. $\blacksquare$

Let $L$ be the laplacian of the friend graph. It is not hard to see that the problem is equivalent to showing that the number of solutions to $Lv=d$ in $\mathbb{F}_2$ is a power of $2$, where $d$ is the vector of degrees. By the lemma, we have that there is at least one solution, call it $v_0$. Therefore, $L(v-v_0)=d$. Let the dimension of $\mathrm{ker}(L)$ be $k$. It is clear that the number of solutions is then $2^k$. Note that $(1,\ldots,1)\in\mathrm{ker}(L)$, so $k\ge 1$, so the problem is complete. $\blacksquare$

Similar to this 2003 Tuymaada problem.
https://artofproblemsolving.com/community/c6h1837571p12329351

Interpret the problem as a graph.

Consider the adjacency matrix of the graph, and then on $a_{i,i}$ change it from a $0$ to a $1$ if the degree of vertex $i$. Call this matrix $A$. Now, consider a vector $v$, where the ith entry is $1$ if the ith mathematician is in the first room, and $0$ otherwise. Clearly if $v$ was a working distribution, then $Av$ is equal to the main diagonal of $A$, call this main diagonal vector $u$. Therefore, $Av = u$, where $A,u$ are fixed. By grobber's lemma, there exists such a $v$. Furthermore, if we let $S$ be the kernel of $A$, then for any $t\in S$, we have $v + t$ is also a working solution; all solutions must be generated this way. By taking the basis of this kernel, we clearly get that the number of solutions is a power of $2$.

How do you motivate the inductive step for the first part

Suppose that the mathematicians are named $1,2,\ldots,n$. Name the two rooms "room 1" and "room 310", and let $x_i$ be the indicator that $i$ is in room 1. Work in $\mathbb{F}_2$.

Let $S_i$ be the set of $i$'s friends. Consider the system of $n$ equations of the form $\sum_{j\in S_i}(x_i+x_j)=|S_i|$; note that if they are simultaneously satisfied, we have a valid construction. Since the set of solutions is an affine subspace of $\mathbb{F}_2^n$, either we have no solutions or we are done (at least two solutions exist, if any, by swapping the rooms).

Suppose FTSOC that some linear combination of our equations sum to $0=1$. This means that there exists a subset $V$ of the friendship graph such that every vertex outside of $V$ is adjacent to an even number of vertices in $V$, and the number of edges between $V$ and other vertices is odd (every edge within $V$ is counted twice), which fails mod $2$.

I realize that nobody ever posted the combo-only solution (avoiding linear algebra altogether). I am copy-pasting what I have from the official USAMO 2008 solutions file.

[code]Let n be the number of participants at the conference. We proceed by induction
on n.

If n = 1, then we have one participant who can eat in either room; that
gives us total of 2 = 21 options.

Let n ≥ 2. The case in which some participant, P , has no friends is trivial.
In this case, P can eat in either of the two rooms, so the total number of ways
to split n participants is wice as many as the number of ways to split (n − 1)
participants besides the participant P . By induction, the latter number is a
power of two, $2^k$, hence the number of ways to split n participants is $2
\cdot 2^k = 2^{k+1}$, also a power of two. So we assume from here on that every
participant has at least one friend.

We consider two different cases separately: the case when some participant has
an odd number of friends, and the case when each participant has an even number
of friends.

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

Case 1: Some participant, Z, has an odd number of friends.

Remove Z from consideration and for each pair (X, Y) of Z’s friends, reverse
the relationship between X and Y (from friends to strangers or vice versa).

CLAIM.  The number of possible seatings is unchanged after removing Z and
reversing the relationship between X and Y in each pair (X, Y) of Z’s friends.

Proof of the claim. Suppose we have an arrangement prior to Z’s departure. By
assumption, Z has an even number of friends in the room with him.

If this number is 0, the room composition is clearly still valid after Z leaves
the room.

If this number is positive, let X be one of Z’s friends in the room with him.
By assumption, person X also has an even number of friends in the same room.
Remove Z from the room; then X will have an odd number of friends left in the
room, and there will be an odd number of Z’s friends in this room besides X.
Reversing the relationship between X and each of Z’s friends in this room will
therefore restore the parity to even.

The same reasoning applies to any of Z’s friends in the other dining room.
Indeed, there will be an odd number of them in that room, hence each of them
will reverse relationships with an even number of individuals in that room,
preserving the parity of the number of friends present.

Moreover, a legitimate seating without Z arises from exactly one arrangement
including Z, because in the case under consideration, only one room contains an
even number of Z’s friends.

Thus, we have to double the number of seatings for (n − 1) participants which
is, by the induction hypothesis, a power of 2. Consequently, for n participants
we will get again a power of 2 for the number of different arrangements.

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

Case 2: Each participant has an even number of friends.

In this case, each valid split of participants in two rooms gives us an even
number of friends in either room.  Let (A, B) be any pair of friends. Remove
this pair from consideration and for each pair (C, D), where C is a friend of A
and D is a friend of B, change the relationship between C and D to the
opposite; do the same if C is a friend of B and D is a friend of A. Note that
if C and D are friends of both A and B, their relationship will be reversed
twice, leaving it unchanged.

Consider now an arbitrary participant X different from A and B and choose one
of the two dining rooms.  (Note that in the case under consideration, the total
number of participants is at least 3, so such a triplet (A, B; X) can be
chosen.) Let A have m friends in this room and let B have n friends in this
room; both m and n are even. When the pair (A, B) is removed, X’s relationship
will be reversed with either n, or m, or m + n − 2k (for k the number of mutual
friends of A and B in the chosen room), or 0 people within the chosen room
(depending on whether he/she is a friend of only A, only B, both, or neither).
Since m and n are both even, the parity of the number of X’s friends in that
room will be therefore unchanged in any case.

Again, a legitimate seating without A and B will arise from exactly one
arrangement that includes the pair (A, B): just add each of A and B to the room
with an odd number of the other’s friends, and then reverse all of the
relationships between a friend of A and a friend of B.  In this way we create a
one-to-one correspondence between all possible seatings before and after the
(A, B) removal.

Since the number of arrangements for n participants is twice as many as that
for (n − 2) participants, and that number for (n − 2) participants is, by the
induction hypothesis, a power of 2, we get in turn a power of 2 for the number
of arrangements for n participants.  The problem is completely solved.

This problem was suggested by Sam Vandervelde.[/code]

Here is our proof for why it is always possible to split the mathematicians between the two rooms: [url]https://calimath.org/pdf/USAMO2008-6-corollary.pdf[/url]
And we uploaded the solution with motivation to: [url]https://youtu.be/rR3iyhQGZ6E[/url]

Next week, we will upload the finished proof for the full problem on our website and our Youtube channel.

Here is our solution: [url]https://calimath.org/pdf/USAMO2008-6.pdf[/url]
And we uploaded the solution with motivation to: [url]https://youtu.be/A5w90STbHec[/url]

The operation described in the second step of the solution @2above is equivalent  to$\,\oplus_A w \ominus A \ominus B \ominus w$.

[quote=v_Enhance]
[code]Since the number of arrangements for n participants is twice as many as that
for (n − 2) participants, and that number for (n − 2) participants is, by the
induction hypothesis, a power of 2, we get in turn a power of 2 for the number
of arrangements for n participants.  The problem is completely solved.
[/code][/quote]

I think the number of arrangements for $n$ participants should be equal to that for $(n - 2)$ participants.

Let the rooms be "hall of arts" and "stever lounge", and number the mathematicians $1$ through $n$. Work in $\mathbb{F}_2$; let $x_i=1$ iff mathematician $i$ is in hall of arts. Then the problem is equivalent to showing that the following system of $n$ equations has $2^k$ solutions for some $k \geq 1$. This system is described as follows: if some mathematician $i$ has an even number of friends, then $\sum_{i,j\text{ friends}} x_j=0$, and if $i$ has an odd number of friends, then $x_i+\sum_{i,j\text{ friends}}x_j=1$. For "linear algebra reasons", the number of solutions is automatically a power of $2$ or zero, and furthermore solutions come in pairs by swapping the rooms, so it suffices to show that a solution exists.

If no solution exists, some nonzero sum of some of the equations must yield $0=1$; in particular, we must sum an odd number of "odd friends" equations. Take the obvious graph theoretic interpretation and call a subset $S$ of the vertices [i]janitorial[/i] if, when we add all of the edges adjacent to all of the vertices $S$ [i]with multiplicity[/i], every vertex is adjacent to an even number of edges. It is clear that finding an aforementioned nonzero sum is equivalent to finding a janitorial subset with an odd number of odd-degree vertices; I will prove that no such set exists.

First off, any such set must have at most $n-2$ vertices: clearly the set with $n$ vertices (i.e. all of them) must contain an even number of odd-degree vertices by double-counting, and if a set with $n-1$ vertices exists, it therefore must exclude an odd degree vertex. But this odd degree vertex is adjacent to each of its edges once, hence adjacent to an odd number of edges: contradiction. If some set $S$ exists, we may then take two vertices $v \neq w$ not in it. Now, delete $v$, and take all of the edges connected to $v$ (of which there are an even number), and attach them to $w$. Then reduce edges "modulo 2": if some pair of vertices has more than $1$ edge between then, subtract an even number, so that there are either $0$ or $1$ edges between any pair of vertices. It is not hard to see that $S$ is still a janitorial set with an odd number of odd-degree vertices of the resulting graph after these changes. Since clearly no such set exists when $n=1$, by downwards induction we obtain the desired contradiction. $\blacksquare$

[b]Remark:[/b] I don't think the immediate parity contradiction from "odd number of edges + every vertex has even number of adjacencies" is valid? Among other doubts I have, on the graph which is a triangle and an extra vertex with degree $1$, by taking one vertex of degree $1$ and one of degree $2$ we find a set producing an odd number edges such that there are $2$ vertices with an even number of adjacencies and $2$ with an odd number of adjacencies. Unless the parity argument is deeper, this should be "indistinguishable" from all four vertices having an even number of adjacencies.

It's grobbin time.

Represent mathematicians as elements of the adjacency matrix $M$ in ${\mathbb F}_2^{n^2}$.
Define the matrix \[ N = \begin{bmatrix} m_0 & 0 & \dots & 0 \\ 0 & m_1 & \dots & 0 \\ \vdots & \vdots & \ddots & 0 \\ 0 & 0 & 0 & m_n \end{bmatrix} \] where $m_i$ is the $i$th element of $M \cdot 1$. In other words, $N = M \cdot 1 \cdot \operatorname{id}$.
Then we claim that $v \in {\mathbb F}_2^n$ is a valid assignment of mathematicians to rooms if and only if \[ (M + N) \cdot v = M \cdot 1. \]
Let $A$ represent the room of $1$s in $v$. Elementwise, this follows since $M \cdot v$ calculates the number of friends $\pmod{2}$ that are in $A$, and $N \cdot v$ is the total number of friends when not in $A$ and else $0$.
Since $M \cdot 1$ is the diagonal of $(M + N)$, it follows by grobber's lemma that at least one solution $s_0$ exists.
Then, if $S$ is the solution set, it then follows that $\{s - s_0 \mid s \in S\}$ forms a vector space and thus has $2^k$ elements as desired.

It's not grobbin time

Assume the are $n$ mathematicians, represent the $i$'th mathematician $M_i$ as a vector in $v_i \in \mathbb{F}_2^n$. The $j$'th entry in $v_i$ is $1$ iff $M_i$ and $M_j$ are friends, and the $i$'th entry in $v_i$ is $1$ iff $M_i$ has an odd number of friends. Waiving some detail, we can get all arrangements of mathematicians by subsets adding $v_1,v_2, \ldots, v_n$ to the vector $0$, and a set of vectors work if they add up to $0$. But, all the sets of vectors that add up to $0$ create an $\mathbb{F}_2$ vector space $V$ and thus have magnitude some power of $2$, assuming it is non-empty.

So, all that is left is to show $V$ is non-empty. To say, there is at least one way to arrange the mathematicians so that each mathematician has an even number of other mathematicians in their room. We induct on the number of mathematicians. For basis, if we have $1$ mathematician then it is trivial. For the inductive step, assume we have some graph $G$ of $k$ mathematicians and we want to add one new mathematician to the graph $m$. We can do this by splitting into two cases.

 Case 1: Some vertex of $G \cup \{m\}$ has odd degree. Remove that vertex, arrange the mathematicians into the two rooms that work which we can do by the hypothesis, and then add the removed mathematician back. There will always be a room for the removed mathematician since one room has an even number of friends of the removed mathematician while the other has an odd number of friends of the removed mathematician.

Case 2: No vertex of $G \cup \{m\}$ has odd degree. We can just put all mathematicians in the same room

So, we are done. $\blacksquare$


[quote="wangzishan"]Let  $ a\geq0$,$ b\geq0$,$ c\geq0$  and  a+b+c=1,proof that$ (1 \plus{} a)(1 \plus{} b)(1 \plus{} c)\geq(1 \minus{} a^2)^2 \plus{} (1 \minus{} b^2)^2 \plus{} (1 \minus{} c^2)^2\geq2$[/quote]

$ (1\plus{}a)(1\plus{}b)(1\plus{}c)\minus{}(1\minus{}a^2)^2\minus{}(1\minus{}b^2)^2\minus{}(1\minus{}c^2)^2$
$ \equal{}(a\plus{}b\plus{}c)(2a\plus{}b\plus{}c)(a\plus{}2b\plus{}c)(a\plus{}b\plus{}2c)\minus{}((a\plus{}b\plus{}c)^2\minus{}a^2)^2\minus{}((a\plus{}b\plus{}c)^2\minus{}b^2)^2\minus{}((a\plus{}b\plus{}c)^2\minus{}c^2)^2$
$ \equal{}a^3b\plus{}ab^3\plus{}b^3c\plus{}bc^3\plus{}c^3a\plus{}ca^3\minus{}2abc(a\plus{}b\plus{}c)$
$ \equal{}(ab\plus{}c^2)(a\minus{}b)^2\plus{}(bc\plus{}a^2)(b\minus{}c)^2\plus{}(ca\plus{}b^2)(c\minus{}a)^2$
$ \ge 0$

$ (1\minus{}a^2)^2\plus{}(1\minus{}b^2)^2\plus{}(1\minus{}c^2)^2\minus{}2$
$ \equal{}((a\plus{}b\plus{}c)^2\minus{}a^2)^2\plus{}((a\plus{}b\plus{}c)^2\minus{}b^2)^2\plus{}((a\plus{}b\plus{}c)^2\minus{}c^2)^2\minus{}2(a\plus{}b\plus{}c)^4$
$ \equal{}2(a^2b^2\plus{}b^2c^2\plus{}c^2a^2)\plus{}8abc(a\plus{}b\plus{}c)$
$ \ge 0$

done.