Which of the following is equivalent to $ \displaystyle \sqrt {\frac {x}{1 \minus{} \frac {x \minus{} 1}{x}}}$ when $ x < 0$?

$ \textbf{(A) } \minus{} x \qquad \textbf{(B) } x \qquad \textbf{(C) } 1 \qquad \textbf{(D) } \sqrt {\frac x2} \qquad \textbf{(E) } x\sqrt { \minus{} 1}$

[hide="Answer"]Algebraically simplifying, we have $\sqrt{x^2}=x$. However, this is for $x<0$, and the square root function gives the positive root, so our answer is $\boxed{A}$.[/hide]

i got A, but some of my freinds messed up with the sign and got B. which is really suprising considering it's only number 7.

I don't mean to revive, but I have a question:

For $ \sqrt{x^{2}}=x $, doesn't squaring a negative make it positive? Then the square root of a positive is still positive, so I got $B$.... what's wrong with that?

Yes, but if $\sqrt{(-x)^2}=x$, the answer is not $x$, it's $-x$.

But x is already a negative number, so -(-x) = positive x

But $x$ is negative. Think about it, $\sqrt{ (-4)^2}=4=-(-4)$. It doesn't make sense for $\sqrt{x^2}=x$ when $x$ is negative because square roots are always greater than or equal to $0$.

So if it was to be negative, it would actually be $E$? I'm confused right now....

Can you rephrase on what you are confused with?
The definition of square root is
$\sqrt{x^2}=| x |$
If $x\ge 0$, then $\sqrt{x^2}=x$
If $x<0$, then $\sqrt{x^2}=-x$

$\sqrt{-x^2}=x\sqrt{-1}$

Right so the answer isn't $A$, it's $E$?

If you square (E) you get $16*-1=-16$, not 16.

Just think about it: a positive is the negative of a negative number.
1=-(-1), doesn't it?

Let $x=-1$
So $\sqrt{x^2}=\sqrt{(-1)^2}=\sqrt{1}$
We always take the positive root (it's a definition that when we see the square root of something it is positive), so $\sqrt{1}$ is equal to 1.
1 does equal $-x$, as we said before.

$ \sqrt{\frac{x}{1-\frac{x-1}{x}}}=\sqrt{x^2} $ NOT $\sqrt{-x^2}$

Look at the diference
$ \sqrt{\frac{x}{1-\frac{x-1}{x}}}=\sqrt{x^2}=| x |$ This would be $-x$ if $x<0$
$ \sqrt{\frac{x}{\frac{x-1}{x}-1}} =\sqrt{-x^2}=|x|\sqrt{-1}$ This would be $-x\sqrt{-1}$ if $x<0$

No, the answer is A. The expression simplifies to $\sqrt{x^2}$. Since $x=-k$ is negative, we have $\sqrt{(-k)^2}=k=-x$.

Technically, couldn't it be both A and B?

Remember, it's the principle square root, so it's positive.  And since $x$ is negative, we want $-x$ which is positive.


If $x,y,z > \frac{1}{2}$ and $xyz \geq 8$ prove that:

$xy+yz+zx + \frac{1}{\sqrt{2}}(yz\sqrt{x}+zx\sqrt{y}+xy\sqrt{z}) \geq 3xyz$

 :)  :)  :)

[quote="spider_boy"][quote="gemath"]If $x,y,z > \frac{1}{2}$ and $xyz \geq \frac{1}{8}$[/quote]

Isn't there any redundancy? :?[/quote]

OK OK!! I have fixed it Thank you!!  :)

Why noboby help me??  :? or can you show me is it wrong ??


Let $a,b,c \in (0,1)$ and $a+b+c=2$. Prove that  $4(a^{2}+b^{2}+c^{2}) \geq 8-9abc$.
When do we have equality?

homogenize and it's equivalent to schur's
 :D

What's about this problem ?

Let 0< a, b ,c < 1 . a + b + c = 2 .
  Prove that :
  a  2 + b 2  + c 2    \leq  2 - 2abc

Homogenize, then put  x = m+n, y = n+p, z = p+m, m, n, p>0 (why we can?) and expand both sides, we come to an trivial inequality.

Namdung


invariant
linear algebra
matrix
linear algebra unsolved

Consider $K$ an arbitrary field and let $\sigma_1, \sigma_2$ two permutations in $S_n$ and $P(\sigma_1), P(\sigma_2)$ the permutation matrices associated to these permutations. Prove that $\sigma_1, \sigma_2$ are conjugated in $S_n$ if and only if the two matrices are similar. It is not that easy problem in any field.

The non-trivial implication is showing that $\sigma_1,\sigma_2$ are conjugates of one another if $P(\sigma_i)$ are similar.

We can work in the algebraic closure of $K$ to conclude that the matrices have the same Jordan form. However, I think it's not that hard to see that for each $i\in\overline{1,n}$ the number of Jordan blocks of dimension $i$ in the Jordan form of $P(\pi)$ is precisely the number of cycles of length $i$ in $\pi$, so the conclusion follows, since two permutations are conjugate iff they have the same cycle structure.

Grobber, could you please be a little bit more explicit? I'm not familiar with Jordan form and I cannot prove your statement about the number of cycles. The solution I have is more difficult, but elementary. Your approach looks very nice, but I don't understand very well. Sorry for being dumb. :(

I've erased my previous response.

The whole space can be decomposed in subspaces invariant to the transformation represented by the matrix in a unique way s.t. the subspaces involved are indecomposable, i.e. cannot be further decomposed in invariant subspaces. The dimensions of these spaces are precisely the dimensions of the Jordan blocks of a matrix in Jordan form.

It's pretty clear now that for each cycle of length $i$ in the permutation, there is an indecomposable invariant subspace of dimension $i$, and that these spaces can be joined as a direct sum to form the large space, so the cycles correspond to the Jordan blocks.


vector
linear algebra
matrix
superior algebra
superior algebra unsolved

Prove that the vectors {a_1,a_2,....,a_k} satisfying the $\xi_1 a_1+\xi_2 a_2+...+\xi_n a_n=0 \text{ where } \xi_i\ne 0 \forall i$ make the linear (n-1)th-space  :blush:
sorry i dont know really how this problem should be writen in english...

Did you mean by this that $\textrm{rank}(a_1,a_2,\dots,a_n)=n-1$?

i have found a solution involving the $rank(\xi_1,...,\xi_n)=1$so  the fundamental system of solution have the rank=n-1 as a corrallary from the theorem that if the matrix of the coefficients $a_{ij}, i,j=1 \text{ to } n$ has the rank=k, then the rank of the vectors in the fundamental system of solutions( the system of  solutions,such that any other solution is the linear combination of this solution) equals to n-r .
 but i want to know is there another solution that doesn't  ******* to this theorem


real analysis
topology
advanced fields
advanced fields unsolved

Is this result true?

Let $ (\Omega,\mathcal{B})$ be a Borel-measure space and let $ \mu$ be a $ \sigma$-finite measure defined on it. If $ B\in\mathcal{B}$ such that $ \mu(B)> 0$ then there exists a non-empty open set $ V$ that is contained in $ B$.

It is false. Just take $ \mathbb R\setminus\mathbb Q$ and the Lebesgue measure on the line. It is even more apparently false if you take the $ \delta$-measure for $ \mu$ and the one point set it is supported by for $ B$.

A result which is true: $ B\minus{}B$ contains an open set.

@ZetaX: could you explain that a bit? I didn't get what is $ B\minus{}B$

and what if $ \Omega$ is locally compact and Hausdorff?

$ \Omega \equal{} \mathbb R \backslash \mathbb Q$ is still a counterexample. You always get counterexamples if there is a countable basis for the topology.

With $ B \minus{} B$ I meant the Minkowski-difference, meaning: $ B \minus{} B \equal{} \{b_1 \minus{} b_2 | b_i \in B\}$.


Solve the functional equation $ f(x\plus{}t)\minus{}f(x\minus{}t)\equal{}4xt$

how do you [i]solve[/i] something like this?

Consider in particular the case $ x \equal{} t$, then $ f(2x) \equal{} 4x^2 \plus{} f(0)$, which is satisfied only if $ f(x) \equal{} x^2 \plus{} c$ for some constant $ c$.

and what if t is something else? and hwo do u knwo its asking for f(x)?

The 'solution' to a functional equation is, by definition, an expression for the function in terms of its parameters.  So in this case, we are looking for $ f(x)$ in terms of $ x$.  Since the functional equation satisfies the equality for [i]any[/i] $ x$ and $ t$ it must satisfy it, in particular, for the case when $ x\equal{}t$.  Thus, any $ f$ that satisfies your functional equation must necessarily satisfy $ f(x)\equal{}x^2\plus{}c$ for some constant $ c$.  We say, then, that $ f(x)\equal{}x^2\plus{}c$ is [i]the[/i] family of solutions to the functional equation.

[quote="JoeBlow"]The 'solution' to a functional equation is, by definition, an expression for the function in terms of its parameters.[/quote]

oh...that helps a lot, wish i knew that earlier

(no sarcasm)

thanks

 :lol:

JoeBlow, can you show why that is the only solution?

He just did, in the post above.  :wink: (see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=191785]here[/url] also).


jmadsen, what is the formation of the sequence?????

Primes everywhere, but I don't know how he got them.

i dont understand... maybe it is like if you reverse it is still prime????

[quote] [b]Jmadsen wrote:[/b]
113,149,157,167
and
Try reversing the digits
[/quote]

That's right. The series contain primes whose reversal of digits is a different prime in ascending order.

[quote]
[b]JBL wrote:[/b]
You missed 97[/quote]

Oops! Sorry for that.


EVERYONE

conway may seem pretty boring at first
this is because it's the first lecture and you aren't really used to getting lectured so much
but he really knows what he's talking about
it's also pretty confusing at times, he goes on tangents often, but they're always related and relevant

MAKE SURE YOU LISTEN IN HIS LECTURES

hey wait a minute are you allowed to record lectures?

well, rainbows is pretty physics-y

like I'm pretty sure on one of my physics tests two years ago, one of the questions asked us to explain rainbows

and physics is kinda math?

like it's based on snell's law, I suppose

and you could derive snell's law based on fermat's principle

using calculus of variations

AND HA THAT'S MATH

[quote="Ubemaya"]ahem

always relevant?

why was he talking about like the earth or whatever when he was supposed to be explaining rainbows?

AND SINCE WHEN WAS RAINBOWS THAT IMPORTANT TO MATH?![/quote]

well yeah lol i believe i fell asleep during that lecture
BUT THAT WAS THE ONLY TIME DURING CAMP I SWEAR
I SAT THROUGH ALL OF BERNDT'S LECUTRES ("cue (q) cue cubed sub infinity, cue cubed cue cubed sub zero, cue squared cue cubed sub n cubed...")

but yeah he was pretty funny in that lecture
"i'm one of only two people who know how rainbows work"

and yeah watch out for the "i have one fault, i'm too modest" thing, he'll say that at least once

his other lectures were pretty good though (bernoulli numbers, n-D geometry, fibonacci/lucas/similar sequences, infinite numbers), it would have been pretty cool if he had done cellular automata but w/e

pinecone bombs? those are pretty awesome. my friends made one.

no, just relating them to Fibonacci Numbers


ΠΡΟΤΑΣΗ  3. Δίδεται τρίγωνο ΑΒΓ και έστω Δ τυχόν σημείο επί της πλευράς ΒΓ. Προσδιορίστε σημείο Μ επί της ευθείας ΑΔ, έτσι ώστε τα τρίγωνα ΑΜΒ, ΑΜΓ, να έχουν ίσους περιγεγραμμένους κύκλους.

Ουσιαστικά είναι ισοδύναμη με την πρόταση (1)  :wink:

Στις σημειώσεις μου οι προτάσεις (1), (3) είναι ενοποιημένες σε μία πρόταση, λόγω αυτής ακριβώς της παρατήρησης. Τις παρουσίασα εδώ ξεχωρισμένες, γιατί έτυχε και εγώ να ασχοληθώ με αυτές ξεχωριστά. Η (3) μπορεί να σταθεί μόνη της σαν πρόταση και η (1) είναι ουσιαστικά πόρισμα της (3).

Λοιπόν αφού μου ζητήθηκε να βάλλω τη λύση μου here it is  :wink:

Έστω Γ' το συμμετρικό του Γ ως προς την ΑΔ . Τότε έχουμε <ΑΓΜ = <ΑΓ'Μ = <ΑΒΜ
Και τότε τα Γ', Β βρίσκονται στο ίδιο μέρος επίπεδου που χωρίζει η ΑΔ άρα το ζητούμενο σημείο είναι το σημείο τομής του περιγγεγραμμένου κύκλου του ΑΒΓ' τριγώνου με την ΑΔ .  :)

Έτσι έχουν τα πράγματα, τα συμμετρικά των Β, Γ, ως προς την ΑΔ, κείται επί των ίσων κύκλων ( μία μικρή μόνο τυπογραφική διόρθωση: έχουμε <ΑΓΜ = <ΑΓ'Μ = <ΑΒΜ ).

     Αν και γενικά οι κατασκευές ( πιθανόν ) δεν έχουν την αίγλη των αποδεικτικών προτάσεων, όταν ζητείται ο προσδιορισμός της θέσης ενός σημείου, ώστε να ικανοποιείται κάποια συνθήκη, στην ουσία έχουμε πάλι μία αποδεικτική πρόταση.

     Σε ευχαριστώ για το ενδιαφέρον σου.

     Κώστας Βήττας.

Εγώ ευχαριστώ για τις ωραίες ασκήσεις .
Να σημειώσω επίσης ότι αφού κατασκευάστηκε αυτή άμμεσα κατασκευάστηκε και η πρόταση (1) όπως είχαμε προαναφέρει  :wink:


[quote="kingkong"]Let $A, B \in Mat_{n}(K)$. Prove that:
1) $|rankA-rankB| \leq\ rank(A+B) \leq\ rank(A)+rank(B)$
2) $rankA+rankB-n \leq\ rank(AB) \leq\ min (rank(A), rank(B))$[/quote]

No one can prove!!??!!

You should consider using the Search function.
Related to the second part: http://www.mathlinks.ro/Forum/viewtopic.php?search_id=252804477&t=43691


Right, I think some serious action needs to be taken over something as stupid as this.. perhaps a banning for a couple of weeks? And then a permanent one if it continues? Thats usually what happens in other forums..

yes...he's just trying to annoy me...he kept doing the same in Fun & Games, but I deleted them all. He started here because I can't get to them...

apparently... me too...

well...

do some contests, like the Getting Started Contest.

(yes, that was an ad, lol)

Ah you must have only seen this after it was edited.. mystic just came through posting about 3 separate messages with their subjects and contents just the word "spam".

Oh by the way.. as for the edited topic.. I don't think this should be in fun and games, either "support" or "advanced problems" would be very suitable for your spam problem..

If you spam in Advanced Problems or Other Problems Solving Topics, your spams will quickly be removed. I shall be sure to enforce that. I think TripleM will enforce it in Advanced Problems as well if he gets to it before I do.

I pledge my support.  I will try to remove any spam from AMC if it appears there as soon as I see it.

Maybe we should have a treaty of moderators or something.


Brahmagupta's formula gives the area of cyclic quadrilaterals. Is there a general formula which gives the area of cyclic polygons?

:?  this is the same as me... I've asked my self this and asked my self about what about the product of diagonals of a incribed 2n-gon (you know if n=4 ptolomey, if n=6 furhmann)

thx for the response... I would like to know the furhmann formula....


If $ a,b,c,d,k \ge 0$ prove that
\[ \sum_{cyc} \frac {a^k}{(a \plus{} b \plus{} c)^k} \le \max \{ 2, \frac {4}{3^k}\}
\]
I have solution only for $ k \ge 1$ and $ k \le \frac {1}{2}$. :(

I think there is an inequality

For $ a,b,c \geq 0 ,  s>r>0$
$ (\sum_{cyc} a^s)^ {\frac {1}{s}} < (\sum_{cyc} a^r)^ {\frac {1}{r}}$  (Is it Jensen?)

For $ k \geq 1$ we have  $ \frac {\sum_{cyc} a^k}{(a+b+c)^k} \leq 1 \leq RS$

For $ k<1$ use power-mean

${ \frac {\sum_{cyc} a^k}{(a+b+c)^k} \leq 3^{(1-k)}}$
Obviosly $ 3^{(1-k)} \leq \frac {4}{3^k} \leq RS$

i will try to come up with another solution if i can


$\begin{gathered} f: N \to N,p,k \in N. \hfill \\ f^{(1)} (n) = f(n),f^{(m + 1)} (n) = f^{(m)} (f(n)). \hfill \\ \end{gathered}$
If $f^{(p)} (n) = n + k$.Prove $p|k$.

It'a cute problem :) Let us define also $f^{(0)}(n)=n$. Now we have $f^{(p)}(n)=n+k$ Thus: \[ f(n+k)=f^{(p+1)}(n)=f^{(p)}(f(n))=f(n)+k \] so we've obtained $f(n+k)=f(n)+k$. Our first equation means that $f^{(p)}\equiv f^{(0)}mod (k)$ It can be shown ( I can show it later if you want) that in that case all numbers $f^{(0)}(n),f^{(1)}(n),...,f^{(p-1)}(n),$ have different residues modulo $k$. One can also show that if for some $o,l,m,n$ $f^{(l)}(n) \not\equiv f^{(o)}(m)$ then for any $x,y$ $f^{(x)}(n) \not\equiv f^{(y)}(m)$. Let's take numbers $f^{(0)}(1),f^{(1)}(1),...,f^{(p-1)}(1)$ modulo $k$ these are $p$ diferent numbers. Now let's take the smallest residue  modulo $k$ which wasn't  chosen earlier. Let it be residue $r$. Then numbers $f^{(0)}(r),f^{(1)}(r),...,f^{(p-1)}(r)$ are different and so we have now chosen  $2p$  distinct residues modulo $k$....and so on ...I'm sure  you can see this :) we are always able to chose $p$ different residues, different to all residues we've chosen earlier. And as we have in total $k$ residues then $p|k$. Iknow this proof was not clear so I if you want me to write sth. more precisely just ask :)

[quote="jastrzab"]It can be shown ( I can show it later if you want) that in that case all numbers $f^{(0)}(n),f^{(1)}(n),...,f^{(p-1)}(n),$ have different residues modulo $k$. One can also show that if for some $o,l,m,n$ $f^{(l)}(n) \not\equiv f^{(o)}(m)$ then for any $x,y$ $f^{(x)}(n) \not\equiv f^{(y)}(m)$. [/quote]
Would you please show it more particularly?
It is not so easy to catch for me. ;)  ;)

Sorry....I've totally forgotten about this post  :)  Well first of all if for some $x,y$ we have $f^{(x)}(n )\equiv f^{(y)}(m) \; mod \; k$ then for every $w$ we have  $f^{(x+w)}(n )\equiv f^{(y+w)}(m) \; mod \; k \; (*)$ Suppose that in the set $f^{(0)}(n), f^{(1)}(n),..., f^{(p-1)}(n)$ there are two numbers with the same residue $mod \; k$ From this fact, there exists such indice $r<p-1$ for which $f^{(0)}(n) \equiv f^{(r)}(n)$ But this means that $r|p$ Now consider  $f^{(r)}(n)$ if it's $f^{(r)}(n)=n$ we would also have  $f^{(p)}(n)=n$ which is not good for us... And if $f^{(r)}(n)=n+ak$ where $k\geq1$ then we would have $f^{(p)}(n)=n+ak(\frac{p}{r})>n+k$ so we have a contradiction which shows that in set
$f^{(0)}(n), f^{(1)}(n),..., f^{(p-1)}(n)$ there are exactly $p$ different residues $mod \; k$. Now the second part of this what was unclear...well actually I'think I haven't written precisely what I wanted to write...Maybe now I'll do it more clear :)  I want to show that if for some $l,o,m,n$ we have $f^{(l)}(n)\equiv f^{(o)}(m)$ then thet ser of residues $f^{(0)}(n), f^{(1)}(n),..., f^{(p-1)}(n)$ is the same as set $f^{(0)}(m), f^{(1)}(m),..., f^{(p-1)}(m)$of course terms may be in different order, but sets are the same. This result comes straightforward from $(*)$ Now we are able to choose this disjoints sets of residues as we made in final part of this unclear proof

thank you very much!!
two more questions:
(1)why $f^{(0)}(n )\equiv f^{(r)}(n)$
(2)why $f^{(p)}(n)=n+ak(\frac{p}{r})>n+k$
forgive my foolishness :P

Well when we assume that there exist two different indices $i,j$ and $i<j$ such that $f^{(i)}(n) \equiv f^{(j)}(n) \; mod \; k$ then from the fact we've mentioned earlier $f^{(i+(p-i))}(n) \equiv f^{(j+(p-i))}(n) \; mod \; k$ but as $f^{(p)}(n) \equiv f^{(0)}(n) \; mod \; k$  we obtain $f^{(0)}(n) \equiv f^{(p+(j-i))}(n) \equiv f^{(j-i)}(n) \; mod \; k$  so we have our first statement.... :)  now when we got that $r|p$ and $r<p$ we have also $p\geq 2r$ ,so $n+(\frac{p}{r})k \geq n+2k$


superior algebra
superior algebra solved

G is a group of order n and p is an odd prime. We know that there exist distinct elements of G, xi with i from 1 to p, s.t. xixi+1 is the same no matter what i we choose (the indices are considered modulo p). Show that 2p | n.

Looks the same like this one:
http://www.mathlinks.ro/Forum/viewtopic.php?t=35507


You're given 9 empty boxes and the digits from 1 to 9. Each digit may only be used once and only one digit can be used in one box. Find some permutation of the digits such that \[\frac{\square}{\square}\times \square+\frac{\square \times \square \times \square}{\square}+\square \times \square =100.\] Have fun. :D

[hide="Answer"]$\frac{6*4}{8}+\frac{2*5*7}{1}+3*9$[/hide]

Is that the only way to do it?

How would you arrive to that answer other than doing guess and check or maybe brute force?

 :ninja:  :ninja:

Nah, I've found 1 more.

[hide="Approach"]I considered the parity in two cases. You have

even+even+even=even or
even+odd+odd=even

 :)
[/hide]


algebra
polynomial
absolute value
algebra unsolved

Let $ a,b_1,b_2,\cdots,b_n,c_1,c_2,\cdots,c_n$ be real numbers such that

$ x^{2n}\plus{}ax^{2n\minus{}1}\plus{}ax^{2n\minus{}2}\plus{}\cdots\plus{}ax\plus{}1\equal{}(x^2\plus{}b_1x\plus{}c_1)(x^2\plus{}b_2x\plus{}c_2)\cdots(x^2\plus{}b_nx\plus{}c_n)$

for all real numbers $ x$. Prove that $ c_1\equal{}c_2\equal{}\cdots\equal{}c_n\equal{}1$.

A partial proof:
Let $ p(x)$ be the studied polynomial. $ f(x)\equal{}p(x)(1\minus{}x)\equal{}\minus{}x^{2n\plus{}1}\plus{}x^{2n}(1\minus{}a)\plus{}x(a\minus{}1)\plus{}1$.
$ f''(x)\equal{}x^{2n\minus{}2}(\minus{}\alpha{x}\plus{}\beta)$ with $ \alpha>0$. Thus $ f'$ is increasing then decreasing. Therefore $ f$ has at most 3 (with multiplicity) real roots and $ p$ has at most 2 (with multiplicity) real roots in the form $ u,1/u$. The $ p$- decomposition in irreducible over $ \mathbb{R}$ is in one of the following forms: $ (x^2\plus{}b_1x\plus{}c_1)\cdots(x^2\plus{}b_{n\minus{}1}x\plus{}c_{n\minus{}1})(x\minus{}u)(x\minus{}1/u)$ or $ (x^2\plus{}b_1x\plus{}c_1)\cdots(x^2\plus{}b_{n}x\plus{}c_{n})$. It remains to prove that if $ z$ is a non real root of $ p$ then $ |z|\equal{}1$.

Let $ z\equal{}e^{2i\alpha}.$ Then $ \frac{z^{2n}\plus{}1}{z^{2n\minus{}1}\plus{}...\plus{}z}\equal{}\frac{\sin(2n\plus{}1)\alpha}{\sin(2n\minus{}1)\alpha}\minus{}1.$ (Prove)
Now consider $ f(\alpha)\equal{}\sin(2n\plus{}1)\alpha \plus{} (a\minus{}1)\sin(2n\minus{}1)\alpha\equal{}0.$ Prove that it has $ n\minus{}1$ roots in $ (0,\pi /2).$ Thus the given polynomial has at least $ n\minus{}1$ roots on the unit circle with arguments in $ (0, \pi /2).$ Since $ p(x)\in\mathbb{R}[x]$ it has at least $ 2n\minus{}2$ roots on the unit circle which yields to the statement from the problem.

[quote="loup blanc"]and $ p$ has at most 2 (with multiplicity) real roots in the form $ u,1/u$[/quote]
Only if $ p(x)$ has a real root $ u\not \equal{}{1\over u}$.
Otherwise you need to show, that $ 1$ and $ \minus{}1$ can't be both roots pf $ p(x)$.
Also if $ z$ is a non-real root of $ p(x)$ of absolute value $ \not \equal{}1$,
then $ z,\bar{z},{1\over z}$ are $ 3$ different such roots, which contradicts bilarev's findings.

Incredible work, loup blanc and bilarev!

Hi olorin,
you are right; we must prove that $ \pm1$ can't be both roots of $ p$; it's easy because else $ a\equal{}2$ and $ n\equal{}0$.


calculus
integration
function
LaTeX
real analysis
real analysis unsolved

Here are two problems for those who are interrested:

 $\bb First Problem $

   Let $f:[0,1]\to\mathbb{R}$ a continuous function such that $\int_{0}^{1} f(x)dx=0$.Prove that there is $c\in (0,1)$ such that $\int_{0}^{c} f(x)dx=f(c)$.

  $\bb Second Problem$

    Let $f:[a,b]\to\mathbb{R}$ a continuous function such that $\int_{a}^{b} f(x)dx=0$.Prove that  there is $c\in (a,b)$ such that $\int_{a}^{c} f(x)dx=c\cdot f(c)$

Be patient. You only posted them 3 hours ago. :)

For the first problem consider the function $g(x)=e^{-x} \cdot \int_{0}^{x} f(t)dt$
Observe that $g(0)=g(1)=0$ and apply the Rolle's theorem.

Is this ok?? :)  :)

The second is solved by the same way: taking the auxiliary function $h(x)=\frac{\int_{a}^{x} f(t)dt}{x}$

By the way, could you give me a latex form for the power that has base $e$ and exponent $-x$

At first glance,Socrates,I think it is correct. ;)

Ok.
They are quite clever problems from analysis!!! :D  ;)  :)


If $ x\plus{}y\ge 0, y\plus{}z\ge 0, z\plus{}x\ge 0$, then only one of $ x,y,z$ may be negative. From $ xyz>0$ we get, that $ x>0,y>0,z>0$, therefore $ \frac 1x\plus{}\frac 1y \plus{}\frac 1z >\frac{1}{x\plus{}y\plus{}z}$.  :lol:

[quote="thanhnam2902"]Let $ x,y,z\in R, x, y, z \neq 0$ and $ \frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z} \equal{} \frac {1}{x \plus{} y \plus{} z}$
[/quote]
This is equivalent to $ (x\plus{}y)(y\plus{}z)(z\plus{}x)\equal{}0$ . Done!  :P

Yes. Thank N.T.Tuan very muck. It's easy.


i read the following problem's solution and did not
understand at some point allong the way:

statement:
let $p>0$ be prime and $p\equiv 3 (mod 4)$
proove that $p$ is prime also in $\mathbb{Z}[i]$

proof(that i partially understand):
if p would not be prime in $\mathbb{Z}[i]$ there would
exist $\alpha,\beta \in\mathbb{Z}[i]$ so that
$p^{2}=N(p)=N(\alpha)N(\beta)$ (N beeing the norm $N(a+bi)=a^{2}+b^{2}$)
and through hypothesis( ?? ),we conclude that
$N(\alpha)=p$ but that would not be possible since the norm $N$ is
a sum of 2 squares wich is never congruent to 3 modulo 4.

where ive posed the question marks is where i do not understand,
what from hypothesis is used to state that $N(\alpha)=p$,in my
oppinion $N(\alpha)$ could also be $p^{2}$.
where am i wrong ?

If $N(\alpha)=p^{2}$ then $N(\beta)$ is $1$ so $\beta$ is a divisor of $1$ and the factorization $p=\alpha\beta$ is trivial...


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $G$ be a finite group, and $p$ be a prime dividing $\left|G\right|$. Let $P$ be a $p$-Sylow subgroup of $G$, and let $H$ be a subgroup of $G$ such that $N_{G}\left(P\right)\subseteq H\subseteq G$. Show that $H=N_{G}\left(H\right)$.

Here, for any subset $T$ of $G$, we denote by $N_{G}\left(T\right)$ the normalizer of $T$ in $G$, defined by $N_{G}\left(T\right)=\left\{g\in G\mid gTg^{-1}=T\right\}$.

Even a combinatorics "genius" like me has solved it, so don't expect it to be particularly challenging, but it is one of our nicer homework problems.

  Darij

I remember mathmanman telling me this problem and having came with a nice solution. Here it is: take $g\in N_{G}(H)$, then $gPg^{-1}$ is a $p$-Sylow of $H$ (it is already one of $G$!), as well as $P$. By Sylow theorems, they are conjugated IN $H$. Then $h^{-1}g\in N_{G}(P)$ and thus in $H$, thus $g$ is also in $H$. Done. Even if a "genius of combinatorics" like me solved it quickly, I really found it interesting and nice (otherwise, I would have already forgot about it...).


search
combinatorics unsolved
combinatorics

g is  a simple graph with n vertices none of the vertices have n-1 degrees。and suppose any two of the vertices have and only have an adjacent vertex.prove
that every vertex has the same degree

If I understand the question correctly then the conclusion can in fact be much stronger: there is no such graph. My interpretation of the condition: $ G$ is a simple graph such that for every two distinct vertices $ u,v$ there is precisely one vertex which is adjacent to both $ u$ and $ v$ (this condition holds regardless of whether or not $ u$ and $ v$ are themselves adjacent. Since $ G$ is simple, $ z$ must be distinct from $ u$ and $ v$).

The only graphs which satisfy this condition are "windmills": graphs which are made up of triangles all fused at a single vertex. In other words, those are graphs where one vertex $ z$ is connected to all other vertices in the graph, which in turn are hooked up in pairs.

In particular, these graphs do have a vertex of degree $ n\minus{}1$ where $ n$ is the number of vertices. If you exclude those, as this question seems to do, there's nothing left.

I don't know any really "elementary" proof of the fact I quoted above. It's a nice example of a problem that can be effectively handled by algebraic methods. I'm willing to bet it was discussed here before, but if needed I can present a complete proof.

When you say "handled by algebraic methods," do you mean by an eigenvalue computation?  I can sort of see how the proof would go, but I'm curious about the details.


Prove that for any set of distinct positive integers $n_1, n_2, \dots, n_k$ there exists a set of integer coefficients $a_1, a_2, \dots, a_k$ such that
$\sum_{i=1}^k a_i {n_i p\choose p}$ is divisible by $p^{2k-1}$ for all large enough primes $p$.

Examples:
$2 {p\choose p} - {2p\choose p}$ is divisible by $p^3$ for all primes $p\geq 3$.
$12{p\choose p}  - 9{2p\choose p} + 2{3p\choose p}$ is divisible by $p^5$ for all primes $p\geq 7$.
$60{p\choose p}  - 54{2p\choose p} + 20{3p\choose p} - 3{4p\choose p}$ is divisible by $p^7$ for all primes $p\geq 11$.
$840{p\choose p}  - 840{2p\choose p} + 400{3p\choose p} - 105{4p\choose p} + 12{5p\choose p}$ is divisible by $p^9$ for all primes $p\geq 3$.

[url=https://www.artofproblemsolving.com/community/user/16261]Rust[/url] and I have proved this type of divisibilities in general in the preprint [url=http://arxiv.org/abs/1602.02632]On p-adic approximation of sums of binomial coefficients[/url].


trigonometry
integration
function
calculus
real analysis
real analysis unsolved

Prove that the series $\sum_{n\geq1}\frac{e^{i \sqrt n}}{n}$ converges

Aplying the condensation test we get that $\sum_{n\ge 1}\frac{e^{i\sqrt n}}{n}$convrges iff $\sum_{k\ge 1}2^k \frac{e^{i(\sqrt 2)^k}}{2^k}=\sum_{k\ge 1}e^{i(\sqrt 2)^k}$onverges. The last sum can be splited into $\sum \sin \sqrt n$ and $\sum \cos \sqrt n$ which diverges since $(\sin{\sqrt{n}})_n$ is dense in $[0,1]$ and , hence, do not tend to $0$. Am I wrong? :?

Are you sure that condensation test can be applied  :?: I think all summands need to be positive

For the condensation test, the terms have to be positive and decreasing. It's a test for absolute convergence, and the convergence here is not absolute.

Note that $\int_1^\infty \frac{e^{i\sqrt{x}}}{x}\,dx$ converges (Integrate by parts to get an equal absolutely convergent integral). We can show that $\left|\frac{e^{i\sqrt{n}}}{n}-\int_n^{n+1}\frac{e^{i\sqrt{x}}}{x}\,dx\right| =O(n^{-\frac32})$, so the series differs from a known convergent series by an absolutely convergent series, and the original series converges.

There are some details to fill in- I might get back to this later.

The [hide="full version"]

Lemma 1: $\int_1^\infty \frac{e^{i\sqrt{x}}}{x}\,dx$ converges.
[hide="Proof"]$\int_1^M \frac{e^{i\sqrt{x}}}{x}\,dx=\int_1^M \frac{e^{i\sqrt{x}}}{2\sqrt{x}}\cdot\frac2{\sqrt{x}}\,dx$
Integrating by parts (integrate the first term, differentiate the second),
$\int_1^M \frac{e^{i\sqrt{x}}}{x}\,dx= \left[-ie^{i\sqrt{x}}\cdot \frac2{\sqrt{x}}\right]_{x=1}^{x=M} -\int_1^M-ie^{i\sqrt{x}}\cdot-x^{-\frac32}\,dx$
Both terms converge as $M\to\infty$; the first converges because the function we are evaluating goes to zero, and the second converges by comparison to the absolutely convergent integral $\int_1^M x^{-\frac32}\,dx$.[/hide]

Corollary: The series $\sum_{n=1}^\infty \int_n^{n+1} \frac{e^{i\sqrt{x}}}{x}\,dx$ converges.

Lemma 2 (Error estimate for a Riemann sum): If $f$ is differentiable on an interval $I=[a,b]$ and $|f'(x)|\le M$ on $I$, $\left|\int_a^b f(x)\,dx-(b-a)f(a)\right|\le \frac M2(b-a)^2$. For this problem, allow $f$ to be complex-valued.
[hide="Proof"]Let $g(x)=f(x)-f(a)$, so $\int_a^b g(x)\,dx=\int_a^b f(x)\,dx-(b-a)f(a)$. If $\int_a^b g(x)\,dx=e^{i\alpha}K$ for some positive real number $K$, let $h(x)=e^{-i\alpha}g(x)$, so $\int_a^b h(x)\,dx=K$ is a positive real number.
Define $H(x)=\int_a^x \text{Re}(h(t))\,dt$. Taylor's theorem applied to $H$ says that $K=H(b)=H(a)+(b-a)(\text{Re}(h(a)))+\frac{(b-a)^2}{2}\cdot \text{Re}(h'(x))$
$=\frac{(b-a)^2}{2}\cdot \text{Re}(h'(x))$ for some $x\in I$. Since $\text{Re}(h'(x))\le |h(x)|\le M$, we have $K\le \frac M2(b-a)^2$ as desired.[/hide]

By this estimate,
$\left|\int_n^{n+1}\frac{e^{i\sqrt{x}}}{x}\,dx-\frac{e^{i\sqrt{n}}}{n} \right|\le \max_{n\le x\le n+1}\left|\frac{\frac{d}{dx}\frac{e^{i\sqrt{x}}}{x}}{2}\right| \le \max_{n\le x\le n+1}\left|\frac{ie^{i\sqrt{x}}}{4x^{\frac32}} -\frac{e^{i\sqrt{x}}}{2x^2}\right|$
$\left|\int_n^{n+1}\frac{e^{i\sqrt{x}}}{x}\,dx-\frac{e^{i\sqrt{n}}}{n} \right|\le \frac1{4n^{\frac32}}+\frac1{2n^2}$
Define $e_n=\int_n^{n+1}\frac{e^{i\sqrt{x}}}{x}\,dx-\frac{e^{i\sqrt{n}}}{n}$. By comparison to the absolutely convergent series $\sum_n \frac1{4n^{\frac32}}+\frac1{2n^2}$, $\sum_n e_n$ converges.

Lemma 3: If $\sum_n a_n$ converges to $A$ and $\sum_n b_n$ converges to $B$, $\sum_n a_n+b_n$ converges to $A+B$.
[hide=" Easy Proof"]
Let $A_n=\sum_{i=1}^n a_i$ and $B_n=\sum_{i=1}^n b_i$. By hypothesis, $A_n\to A$ and $B_n\to B$. It is clear that $\sum_{i=1}^n a_i+b_i=A_n+B_n$. By the standard arithmetic theorem on limits, this converges to $A+B$ and we are done.[/hide]

Let $I_n=\int_n^{n+1} \frac{e^{i\sqrt{x}}}{x}\,dx$.
Putting the pieces together,
$\sum_{n=1}^\infty \frac{e^{i\sqrt{n}}}{n}=\sum_{n=1}^\infty I_n-e_n$. Since $\sum_n I_n$ converges by the corollary to Lemma 1 and $\sum_n -e_n$ converges by the remarks following Lemma 2, their sum $\sum_n I_n-e_n$ converges by Lemma 3 and we are done.[/hide]


group theory
abstract algebra
ratio
superior algebra
superior algebra unsolved

The question:

If G is the additive group Q/Z, what are the elements of the subgroup G(2)? Of G(P) for any positive prime P?

Where G(n)={a e G| |a| = n^(k) for some k is greater than or equal to 0}...That is the set of all a in G, s.t. the order of a is some power of n. (But since it is the additive group, I suppose it would just a be a multiple of n)

How do I even begin with this? Aren't the elements of Q/Z sets? The collections of right cosets? and don't they have infinite order?....

Thanks..^^

$ \mathbb{Q}/\mathbb{Z}$ can be thought of as the set of rational numbers in $ [ 0, 1 )$ where the group operation is addition $ \bmod 1$.  It is true that the quotient is constructed abstractly as a collection of cosets, but the above is also perfectly valid.

Every element of this group does in fact have finite order (and in fact this group is the torsion subgroup of the [url=http://en.wikipedia.org/wiki/Circle_group]circle group[/url]).  For example, $ \frac {1}{2}$ has order $ 2$.  (This example should help you figure out the question pretty well now.)

The operation is addition mod 1? Wouldn't that mean that 1/2 is the only order 2 element in G(2)? 1/2 + 1/2 = 1 ... since 1 is the only number to divide 1...

Did you mean ration number in (0,1]? or really [0,1) ...anyway...I think I understand...actually, any element n/2 when n is an integer will have order 2... But what is so special about order p, where p is a prime?

[quote="Frank1987"]Did you mean ration number in (0,1]? or really [0,1)[/quote]
It doesn't matter; we just change the name of the identity, which is $ 1$ in the first range and $ 0$ in the second.

[quote="Frank1987"]any element n/2 when n is an integer will have order 2... [/quote]
Yes, but these elements are all identified with each other under the quotient.  There are only two distinct such elements; the identity and $ \frac {1}{2}$, and the identity has order $ 1$.  Now:  what are the elements of order $ 3$?

[hide="Answer"] $ \frac{1}{3}$ and $ \frac{2}{3}$. [/hide]
[quote="Frank1987"] But what is so special about order p, where p is a prime?[/quote]
Nothing here; elements of every order exist.  But usually it is easier to understand the elements of prime order, and in some sense one can build up elements of other order from them.


limit
calculus
integration
calculus computations

compute:

$ \lim_{n \to \plus{} \infty} \sum_{k \equal{} 1}^{n} coh(\frac {1}{\sqrt {n \plus{} k}}) \minus{} n$

with $ coh(x) \equal{} \frac {e^x \plus{} e^{ \minus{} x}}{2}$

This doesn't look like an infinite series.

$ \sum_{k \equal{} 1}^n Cosh(\frac {1}{\sqrt {k \plus{} n}}) \minus{} n \equal{} \sum_{k \equal{} 1}^n \sum_{j \equal{} 0}^{\infty}\frac {1}{(2j!)(k \plus{} n)^j} \minus{} n$

$ \equal{} \sum_{k \equal{} 1}^n \frac {1}{2(k \plus{} n)} \plus{} \frac {1}{24(k \plus{} n)^2} \plus{} ...$
$ \equal{} \frac {1}{2n}\sum_{k \equal{} 1}^n \frac {1}{\frac {k}{n} \plus{} 1} \plus{} \frac {1}{24n}\sum_{k \equal{} 1}^n \frac {1}{n(\frac {k}{n} \plus{} 1)^2} \plus{} ...$

As $ n \rightarrow \infty$, we see we have alternate representations of riemann integrals

All of the terms except the first will be dominated by the factors of n in the denominator, making them all goto zero. So all we are concerned with is the first term which is

$ \lim_{n\to \plus{}\infty} \frac {1}{2n}\sum_{k \equal{} 1}^n \frac {1}{\frac {k}{n} \plus{} 1} \equal{} \frac {1}{2}\int_0^1\frac {1}{1 \plus{} x}dx \equal{} ln{\sqrt {2}}$
$ \therefore \lim_{n \to \plus{} \infty} \sum_{k \equal{} 1}^{n} coh(\frac {1}{\sqrt {n \plus{} k}}) \minus{} n \equal{} ln{\sqrt {2}}$


inequalities
number theory unsolved
number theory

Let there $ S(N)$ be the sum of digits , for every $ N \in \mathbb{N}$

Example : $ S(\overline{a_1a_2 \dots a_n}) \equal{} a_1 \plus{} a_2 \plus{} \dots \plus{} a_n$ .

Prove that : $ \boxed{\frac {S(8N)}{S(N)} \geq \frac {1}{8}}$ .

obviosly, $ S(10^k N)\equal{}S(N)$. So i get
$ S(1000N)\equal{}S(800N\plus{}80N\plus{}80N\plus{}8N\plus{}8N\plus{}8N\plus{}8N\plus{}8N)$
also, $ S(a\plus{}b) \leq S(a)\plus{}S(b)$.
so i get $ S(N)\leq 8S(8N)$ and hence the inequality holds.

Naphthalin


Let A,B,C be three angles of triangle ABC with A,B,C \in [0, \pi /2]
Prove that :  \sum 1/cos(A/2) - \sum tg(A/2)  \geq  \sqrt 3

just the kinda problems i need to solve these days (with no time for maths :D  :D )
put X=90 - A/2 and similarly for the others.we see that X,Y,Z are also the angles of a triangle (with each less than 90)
inequality equivalent to proving
 \sum tan (X/2) \geq   \sqrt 3
isnt this known :D  :D


Given trapezoid $ABCD$, $E$ and $F$ are points on $BC$ such that $DE\parallel AB$ and $AF\parallel DC$. Let $G$ be the intersection of $DE$ and $AF$. The area of parallelogram $ABED$ is $36$. If $AD=5$, $AG=3$ and $DG=4$, find the area of trapezoid $ABCD$.

[quote="catcurio"][quote="riddler"]Is $BC\parallel{AD}$? :?[/quote]
That was the only way I was able to draw the diagram.[/quote]Yeah, I forgot to mention that :blush:

[quote="catcurio"][hide="I think..."]Using the given information, we find that $\triangle AGD$ is a $3-4-5$ right triangle. Because of congruent angles, $\triangle EFG$ is similar to $\triangle AGD$.

The height of $ABED$ is $7.2$. By drawing line segment $AE$, we can divide $ABED$ in half. So the area of $\triangle AED$ is $18$, and we can find that $EG$ is $8$. Thus, the length of $EF$ is $10$.

The area of the trapezoid is $\frac{1}{2}h(b_1+b_2)$. So it is $\frac{1}{2}\cdot7.2(5+10+5+5)=\boxed{90}$.[/hide][/quote]That's correct! ;)


Suppose that $ a$ is a positive integer. Is it always true that there is a prime $ p$ such that $ p^2\mid a^{p\minus{}1}\minus{}1$?

Heuristic arguments suggest there this statement is likely to be true.
But it may be very hard to prove it rigorously.

See also
http://oeis.org/A039951
http://oeis.org/A096082


i am glad with the added options!
CHINA!

INDIA!!!!
[hide="Indian Song"]Welcome to India where the cows eat hey,
and we drive auto-richshaws everyday,
Goat Meats, Yummy Sweets, Wild Monkeys roaming
The Roosters Dont Crow till 5 in the morning[/hide]

Hmm, I'm the first Russian! (AKA Soviet)

[quote="DarkKnight"]US! Illinois.[/quote]

Woot! Go Illinois!

I posted this on the other thread, but....ARIZONA!!!!!! Parents born in Taiwan, though.

[quote="xxreddevilzxx"]from Seoul, Korea.[/quote]
Whee~ Seoul! Me, too..

China!  But I live in Illinois right now.

[quote="hello"]Hi, Sponge. I am also Russian. Where you from? Myth, who posts on AoPS is russian and imo gold medalist.[/quote]
I was born in Kiev in the Soviet days. Now, I live in Massachusetts. Meh, I'm Russian by ancestry and first language (first as in first learned, not better volcabulary), Ukranian by former/partial citizenship, and Soviet/Ukranian by birth. Now, I'm living in America. I'm reminded of a Chinese immigrant from Australia friend of mine.

turkey-istanbul
this is better than my post and poll.

i'm from japan, moved to the us when i was 8

Wow, not many people in the "rest of north of america"  I'm sure there's gotta be some others!

[quote="MathFiend"]US of A...more specifically, good old INDIANA!!![/quote]
Do you still live in Indiana and if you do what school do you go to?
I was wondering if I might know you from one of these competitions.


(X with the power of 1 )(X with the power of 2)(X with the power of 3).......................(X with the power of 2n) what does this question wants ? I dun understand what is the power of 2n. And X with the power of 1 multiply X with the power of 2 isnt it X with the power of 1+2 that is X with the power of 3?

could you rephrase that and use the "^" sign?
i have no idea what u r talking about and it would be helpful if u replaced "with the power of" with "^" By the way, "^" basically translates into "to the power of"

3^4 means
3 to the power of 4
3*3*3*3
81

The first one: $ (x^1)(x^2)(x^3)(x^4)...(x^{2n})$.  This is equal to $ x^{1\plus{}2\plus{}3\plus{}...\plus{}2n}\equal{}x^{n(2n\plus{}1)}$.
The second one: $ x^1\times x^2$, which does equal $ x^3$.

In the meantime, I'd suggest going to the LaTeX guide by [url=http://www.artofproblemsolving.com/Wiki/index.php/LaTeX]clicking here[/url]

[quote="shaun ong"]What is latex? I dun understand the information stated  there[/quote]

LaTeX is a programming language that is used by mathematicians to express math symbols in an easier way.

You can say,

(x+y^5-2x-y)/x+2z) and we'll have no idea what you're saying(or if we do, then we'll have to spend 5 painful minutes trying to unravel the mess).

In LaTeX, it can be expressed as

$ \frac{(x\plus{}y^{5\minus{}2x}\minus{}y)}{x\plus{}2z}$.

Clarity and easy to understand.

there is a LATEX program where you can learn it


inequalities
algebra
polynomial
inequalities proposed

i)$a,b,c >0$.Prove that:
               $\frac{ab(a+b)}{a^3+b^3+2c^3}+\frac{bc(b+c)}{b^3+c^3+2a^3}+\frac{ac(a+c)}{a^3+c^3+2b^3} \leq \frac{3}{2}$
ii)Find the best contanst $k$ such that the inequality hold:
               $\sum \frac{ab(a+b)}{a^3+b^3+kc^3} \leq \frac{6}{k+2}$ $(a,b,c >0)$

How to solve this one by SOS?I have some trouble.

:rotfl:  ii) The maximum of $k\,$ is the smaller positive root of the following polynomial:

$23\,k^6+374\,k^5-3733\,k^4+7456\,k^3+1156\,k^2-8192\,k+3888,$

that is,  $2.77948796...$


牛顿定理：ABCD是圆内接四边形，直线AB与直线CD交于E，直线BC与直线AD交于F，M,N,Q分别为AC,BD,EF的中点，求证：M,N,O共线.

很奇怪. 我的一本书上说: 这三点的连线叫做"高斯线", 可为什么题目叫牛顿定理?

BTW: 关于圆的条件可以去掉. 题目的结论对于任意凸四边形都是成立的.

似乎大家現在都很忙. 這裡又再冷清起來.

消点吧. 把线段比换成面积比, 再用一次定比分点公式.

事实上可以用Menelaus定理解决。

这题可以结合面积法解决。
欲证O\M\N共线，只需证S(FMN)=S(EMN)
取AB中点P,S(FMN)=S(FPM)+S(FPN)+S(PMN)=S(BPM)+S(APN)+S(PMN)=S(ANMB)
=S(ADMB)/2=S(ABCD)/4
同理S(EMN)=S(ABCD)/4
从而S(FMN)=S(EMN),命题得证。


algebra
polynomial
logarithms
function
rational function
number theory unsolved
number theory

prove that there aren't polynominal $P ,Q$ such that:$\pi(n)=\frac {P(x)}{Q(x)}$

Are you referring to polynomials $P, Q$ with integer coefficients? Or the coefficients can take on any real values?

It doesn't really matter. The Prime Number Theorem says that $\pi(n)$ increases like $\frac n{\ln n}$, so it couldn't be represented as a rational function in $n$, since such functions behave like polynomials as $n\to\infty$.


probability
floor function
probability and stats

If m balls are distributed to a boys and m girls. Find the probability that even no. of balls are distributed to boys :(

[hide="hint"]The first ball is given to one of $ m \plus{} a$ people, what's the probability that it will be given to one of the $ m$ girls?  The second ball is given to one of the remaining $ m \plus{} a \minus{} 1$ people, what's the probability that it will be given to one of the remaining $ m \minus{} 1$ girls?  Keep going until there are no balls left.  These $ m$ events are pairwise independent; how does that help?[/hide]

# ways to distribute balls:

$ _{a\plus{}m}C_m \equal{} \frac{(a\plus{}m)!}{a!m!}$

# ways to give boys 2k balls:

$ _a C_{2k} \cdot  _m C_{m\minus{}2k} \equal{} \frac {a!m!}{2k! (a\minus{}2k)! 2k! (m\minus{}2k)!}$

prob of giving boys 2k balls:

$ P(X\equal{}2k) \equal{} \left(\frac{a!m!}{k!}\right)^2 \frac{1}{(a\minus{}2k)!(m\minus{}2k)!(a\plus{}m)!}$

$ P(X even) \equal{} \sum_{k\equal{}0}^{\lfloor \frac{a}{2} \rfloor} P(X\equal{}2k)$

i dont see this cleaning up nicely though...


inequalities
limit
inequalities unsolved

Let $ a,b,c$ are positive real numbers and $ n \geq1$ prove that

$ \frac{a^{2}}{b}\plus{}\frac{b^{2}}{c}\plus{}\frac{c^{2}}{a}\geq3\frac{a^{n}\plus{}b^{n}\plus{}c^{n}}{a^{n\minus{}1}\plus{}b^{n\minus{}1}\plus{}c^{n\minus{}1}}$

[quote="mestav"]Let $ a,b,c$ are positive real numbers and $ n \geq1$ prove that

$ \frac {a^{2}}{b} \plus{} \frac {b^{2}}{c} \plus{} \frac {c^{2}}{a}\geq3\frac {a^{n} \plus{} b^{n} \plus{} c^{n}}{a^{n \minus{} 1} \plus{} b^{n \minus{} 1} \plus{} c^{n \minus{} 1}}$[/quote]

(1), we can prove that

\[ \frac {{a^{n \plus{} 1} \plus{} b^{n \plus{} 1} \plus{} c^{n \plus{} 1} }}{{a^n \plus{} b^n \plus{} c^n }} \ge \frac {{a^n \plus{} b^n \plus{} c^n }}{{a^{n \minus{} 1} \plus{} b^{n \minus{} 1} \plus{} c^{n \minus{} 1} }}\]
holds for any $ a,b,c > 0,n\ge1$.

(2), we can prove that

\[ \mathop {\lim }\limits_{n \to \plus{} \infty } \frac {{a^n \plus{} b^n \plus{} c^n }}{{a^{n \minus{} 1} \plus{} b^{n \minus{} 1} \plus{} c^{n \minus{} 1} }} \equal{} \max \left\{ {a,b,c} \right\}\]
holds for any $ a,b,c > 0$

so, it's equivalent to prove that
\[ \frac {a^{2}}{b} \plus{} \frac {b^{2}}{c} \plus{} \frac {c^{2}}{a}\geq 3 \max \left\{ {a,b,c} \right\}\]
but this is not hold! :mad:

so, the original inequality must have a counter example!

I will try to find it...

OK, found, counter example:

$ a \equal{} 9, b \equal{} 15, c \equal{} 10, n \equal{} 4$

$ \frac {a^{2}}{b} \plus{} \frac {b^{2}}{c} \plus{} \frac {c^{2}}{a} \minus{} 3 \cdot \frac {a^4 \plus{} b^4 \plus{} c^4}{a^3 \plus{} b^3 \plus{} c^3} \equal{} \minus{} \frac {55019}{114840} < 0$

by the way, when $ n \equal{} 1,2,3$, inequality holds. I have proved that $ n \equal{} 1,2$, who can prove that $ n \equal{} 3$?

$ \frac {a^{2}}{b} \plus{} \frac {b^{2}}{c} \plus{} \frac {c^{2}}{a} \minus{} (a \plus{} b \plus{} c) \equal{} \sum{\frac {(a \minus{} b)^2}{b}} \ge 0$

$ \frac {a^{2}}{b} \plus{} \frac {b^{2}}{c} \plus{} \frac {c^{2}}{a} \minus{} 3\cdot \frac {a^2 \plus{} b^2 \plus{} c^2}{a \plus{} b \plus{} c} \equal{} \sum{\left(\frac {1}{b} \minus{} \frac {1}{a \plus{} b \plus{} c}\right)(a \minus{} b)^2} \ge 0$

and ... $ n_{\max}$ is ... ?   :huh:


This year I didn't take the AMC because basically nobody at my school could have qualified for AIME except possibly me with my special skill at guessing correctly without a clue.  But, I want to know how I would have done so does anyone know of a link to a site with the questions?  Thanks in advance!

No, but you could always buy the questions for $1 soon - [url]http://www.unl.edu/amc/d-publication/pubform.html[/url]


<= in the code is parsed to
<
=
effectively disabling the code in the posts containing it. (>=, probably too, != no)
Can we fix it?

I don't understand what's going on. Could you be more specific? What code? Asy code?

He's saying if we had something like

[asy]real i; path p = (0,0)--(1,1); for(i=1; i<=10; ++i){ draw((p=shift((1,0))*p)); }[/asy]

It won't work because <= gets broken up on new lines. It seems to work for me though.

[asy]real i; path p = (0,0)--(1,1); for(i=0; i<10; ++i){ draw((p=shift((1,0))*p)); }[/asy]

Oh, well that's just the display part (for the human eye :) ). It shouldn't break after < though, it's a good catch. I'll try to fix it.

Try to edit or quote Hamster1800 post without touching the code and resubmit. You'll get an error. The first time you post it works OK though and I accidentally discovered it when I tried to add a smile to a perfectly functioning post and got an error message. It took me half an hour to figure out what was going on  :mad: :lol:

Ok, I've fixed that error, and also fixed the presentation for the for. Happy now? :P

[quote="Valentin Vornicu"]Ok, I've fixed that error, and also fixed the presentation for the for.[/quote]
:coolspeak:
[quote] Happy now? :P[/quote]
Way up, but I still have to multiply by 50  :heli:


Feuerbach's theorem:
Prove that in any triangle, the inscribed circle and the 3 exinscribed circles are tangent to Euler's circle.

i think i have a proof for the incircle touching euler's circle part,two of them to be precise.they are long ,i'll post em later.btw,i believe u do have the proofs ,mate :? ,and u just wanna know sum different solutions,dont ya??
cheers

Well, the problem is that I remember having proved this, but I have no idea how.. :? So I have to say no, I don't currently have a proof. I could look in some books of course, but I hate reading proofs in books. :D

Okay .. let's see now...

we need two lemmas first:

 [b] Lemma 1 [/b]: Let A' be the middle of [BC] of triangle ABC. Using an inversion(grobber's favourite method :D ) T(A',k) (k is arbitrary) Euler circle of triangle ABC ( from which we take out point A') becomes a antiparalel with BC from A.

[b] Proof of lemma 1 [/b]: It is well known that A' is on Euler circle. So using the inversion T(A',k) Euler circle( from which like I said we take out A') transforms itsself into a line d which is perpendicular on A'w, where w is the center of Euler circle. We know that w is the middle of HO (H- ortocenter and O - the center of the circumscribed circle). Let H' be the simetrical of H from A' . Then H' is diametrally opposite of A.

In the triangle HOH' the segment A'w is the middle line, so A'w || AH'. Because d is perpendicular on A'w is perpendicular on AH', which means that d is paralel with the tangent in point H' of the circumscriebed circle of triangle ABC. Let E be the point in with the tangent in H' intersects AB.
Then m(AEH')=90-m(EAH')=m(BH'A)=n(ACB).

So we have that BC and EH' are antiparalel. Because d is paralel to EH' we get that d is antiparalel with BC.

[b] Lemma 2 [/b]: Let C(I,r) be the incircle of triangle ABC and let C(I',r') be the exincribed circle corespunding to A. Let A' be the middle of BC and let A1 be the point in which the bisector of angle BAC intersects BC. If A" is the projection of A on BC and F the projection of I on F and F' the projection of I' on BC then:
i) BF=CF'
ii) A'F^2=A'A"*A'A1

[b] Proof of lemma 2 [/b]: For more clarity on your drawings lets consider that the triangle ABC has angle m(ABC)>90.

[b] i) [/b]: If we denote F1=the projection of I on ABC, x=AF1, y=BF, z=CF, a=BC, b=CA, c=AB then:
c=x+y
a=y+z
b=z+x
we get that x+y+z=p which gives us that y=BF=p-b where p is the semiperimeter of ABC.
using the following notations: u=AF", v=BF', alpha=CF' we get that
a=v+alpha
b=u-alpha
c=u-v
which gives us that alpha=CF'=p-b so BF=CF'.

[b] ii) [/b] from the bisector theorem we get that BA1/CA1=c/b from which we get that: BA1/(BA1+A1C)=c/(b+c) so BA1=ac/(b+c)
so we get A'A1=a/2-ac/(b+c) or A'A1=a(b-c)/(2(b+c)) (*)
from the right angled triangles AA"B and AA"C we get that :
A"A^2=AB^2-A"B^2 and A"A^2=AC^2-A"C^2
from the last two equalities we get that:
b^2-c^2=(A"C+A"B)(A"C-A"B) or b^2-c^2=a(a+2A"B) so we get that
A"B=(b^2-c^2-a^2)/(2a); because A'A"=A'B+A"b we get that A'A"=(b^2-c^2)/(2a) (**)
Because BF=p-b we get that A'F=A'B-BF=a/2-(p-b) so A'F=(b-c)/2 (***)
so now if we look at (*), (**) and (***) we get that:
A'A1*A'A"=a*(b-c)/(2(b+c))* (b^2-c^2)/(2a)=(b-c)^2/4=A'F^2

Now we are done with the lemmas, back to the problem :D.

[b] Proof of the problem [/b]We keep the denotes we made in lemma 2 and now we transfor that drawing using an inversion of pole A' and we use the power k=p ( the power of the point A' from the inscribed circle) so k=A'F^2.

Using this inversion the circle C(I,r) is transformed into itself. Because A'F=A'F' we also get that the circle C(I',r') is transformed into itself.

The equality A'A1*A'A"=A'F^2 (lemma 2) shows that the point A" ( which is on Euler's circle) is transformed into A1. Using lemma 1 we get that EUler circle (from which we take out A') is transformed in the straight line d which is antiparalel with BC which goes through A1. Because d pases through A1 and because it is antiparalel with BC we get that this is the 2nd exterior common tangent  of circles C(I,r) and  C(I',r'). d the inverse of Euler circle (from which we take out A') because it is tangent to C(I,r) and C(I',r')  we get that also Euler circle will be tangent to C(I,r) and C(I',r') ( they are the invariants in the considered inversion)

Let E (E' respectively) be the point in which d is tangent to circle C(I,r) (C(I',r') respectively). Then we have that Euler's circle is tangent to circle C(I,r) in the transformation of point E, and to the circle C(I',r')  in the transformation of point E' with the considered inversion. We do the same to prove that Euler's circle is tangent to the exinscriebed circles coresponding to B and C.

Proof done! Hope you will enjoy it!  :D

Holy Mother of Christ! That is enormous! :D But it works anyway. I'll try to remember what I came up with.

hmm.. grobber that was the exact thing that I said too :D

guess that this is indeed(hopefully) the first and last solution I will give with inversions :D

I really hate geometrical transformations! they are ingenious sometimes and also long sometimes :D

there is geometry book which tells a lot of Euclidean geometrical knowledge including Feuerbach's theorem.

"Modern Geometry"
by Roger A. Johnson
(C) Houghton Mifflin 1929

well refer brindon's geometry series , though I don't have it

View:
[i]Nathan Altshiller-Court[/i]: [b]COLLEGE GEOMETRY [/b](second Edition); p.105; Theorem 215.

Dear Mathlinkers,
reading some old message, I found this one...
In 2003, I found an enterely synthetic proof that I put on my website in 2007.
http://perso.orange.fr/jl.ayme (vol. 1 le théorème de Feuerbach) in French
or : revistaoim directed by professor Francisco Bellot Rosado (in spanish). Revista Escolar de la Olimpiada
Iberoamericana de Matematica (Espagne) 26 (2006): http://www.oei.es/oim/revistaoim/index.html
Sincerely
Jean-Louis


I was wondering how many people here are going to participate in this year's competition... so here I am, the first one :). Anybody else?

Oh well, never mind :(. If there's anybody interested, here are the problems:

http://www.osu.cz/fpr/kma/dokumenty/vjaimc/year2007.html

Category 1 wasn't too hard, five people managed to score full points. I'm not quite competent to judge the difficulty of category 2 problems though.

:maybe: I think there will be none from Indonesia. (Sorrry if I do a junk, can the moderators make a forum for the junkers, about any news besides the topic that has been in the forum). But, Votech jarnik 2007 has ended. I hope I can join the competition in 2008, if I had a money  :rotfl:  :(

[quote="jytang"]What was the turnout and # of teams for this competition?[/quote]

This can be seen from the page with (non-zero) results:
http://jarnik.osu.cz/index.php?page=year2007


How do you graph the inequality x+y+|z| <= 3?

First, I would like to know, how do you draw a 3-d graph on 2-d paper?

If you have a way to do that, the best way would probably be to graph it for positive y, then take that graph, plus the graph resulting from it being reflected across the x-y plane, and you're done.

Do you have a graphing program on your computer?
If so you can plug it in, get the intercepts, and transfer it to paper.
I'd like to know how to gragh 3D on 2D paper too. :D


algebra
polynomial
calculus
integration
search
number theory proposed
number theory

Prove or disprove that: if $P(x)\in Z[x]$ and for each $n\in N*$  there exist an integer $k\in N*$ such that $n|P(k)$ then $P(x)$ has at least one integeral root.  :)

it's just my idea!

I posted it some time ago, with additional question: find the minimal degree of a counterexample. Try to search for this topic.

@Harazi: you are too slow  :P  :D

@Rust: no, try $\mod 16$.


1.  Given the polynomial identity

$ (x^6 \plus{} 1) \equal{} (x^2 \plus{} 1)(x^2 \plus{} ax \plus{} 1)(x^2 \plus{} bx \plus{} 1)$

What is the value of $ ab$?

EDIT: I see what I did  :oops:

[quote="Stokes93"]1.  Given the polynomial identity

$ (x^6 \plus{} 1) \equal{} (x^2 \plus{} 1)(x^2 \plus{} ax \plus{} 1)(x^2 \plus{} bx \plus{} 1)$

What is the value of $ ab$?

2.  What is the number of distinct real solutions to the equation

$ x^4 \plus{} 6x^2 \plus{} 9 \equal{} 36x^2 \minus{} 72x \plus{} 36$

I know you can solve it by factoring... but I don't see what is wrong with my solution (below)

[code]We have $x^4 - 30x^2 - 25 = 0$ [/code]

 let $ p \equal{} x^2$ $ \Rightarrow$ $ p^2 \minus{} 30p \minus{} 25 \equal{} 0$

$ \Rightarrow$ $ p \equal{} 15 \pm 5 \sqrt {10}$ $ \equal{}$ $ x^2$

$ \Rightarrow$ $ x \equal{}$ $ \pm \sqrt { 15 \pm 5 \sqrt {10}}$

Now we can have $ \plus{}$ or $ \minus{}$ on the outside, but for a real solution the $ \pm$ on the inside must be $ \plus{}$ so we have two real distinct solutions.[/quote]

How does this come?

Oh!   :blush: I totally left out the $ \minus{}72x$ when I did it...

Thus substitution will not work well.... so never mind  :oops:

[quote="Stokes93"]

2.  What is the number of distinct real solutions to the equation

$ x^4 \plus{} 6x^2 \plus{} 9 \equal{} 36x^2 \minus{} 72x \plus{} 36$

I know you can solve it by factoring... but I don't see what is wrong with my solution (below)

We have $ x^4 \minus{} 30x^2 \minus{} 25 \equal{} 0$

 let $ p \equal{} x^2$ $ \Rightarrow$ $ p^2 \minus{} 30p \minus{} 25 \equal{} 0$

$ \Rightarrow$ $ p \equal{} 15 \pm 5 \sqrt {10}$ $ \equal{}$ $ x^2$

$ \Rightarrow$ $ x \equal{}$ $ \pm \sqrt { 15 \pm 5 \sqrt {10}}$

Now we can have $ \plus{}$ or $ \minus{}$ on the outside, but for a real solution the $ \pm$ on the inside must be $ \plus{}$ so we have two real distinct solutions.[/quote]

You forgot about the 72x.

[hide="#1"]
Multiplying out we have

$ x^6\plus{}1\equal{}x^6\plus{}(a\plus{}b)x^5\plus{}(ab\plus{}3)x^4\plus{}(2a\plus{}2b)x^3\plus{}(ab\plus{}3)x^2\plus{}(a\plus{}b)x\plus{}1$

Matching we have

$ a\plus{}b\equal{}0$ and $ ab\plus{}3\equal{}0$

We see $ a\equal{}\sqrt3$ and $ b\equal{}\minus{}\sqrt3$ (or vice versa)[/hide]

[quote="Stokes93"]1.  Given the polynomial identity

$ (x^6 \plus{} 1) \equal{} (x^2 \plus{} 1)(x^2 \plus{} ax \plus{} 1)(x^2 \plus{} bx \plus{} 1)$

What is the value of $ ab$?

2.  What is the number of distinct real solutions to the equation

$ x^4 \plus{} 6x^2 \plus{} 9 \equal{} 36x^2 \minus{} 72x \plus{} 36$

I know you can solve it by factoring... but I don't see what is wrong with my solution (below)

We have $ x^4 \minus{} 30x^2 \minus{} 25 \equal{} 0$

 let $ p \equal{} x^2$ $ \Rightarrow$ $ p^2 \minus{} 30p \minus{} 25 \equal{} 0$

$ \Rightarrow$ $ p \equal{} 15 \pm 5 \sqrt {10}$ $ \equal{}$ $ x^2$

$ \Rightarrow$ $ x \equal{}$ $ \pm \sqrt { 15 \pm 5 \sqrt {10}}$

Now we can have $ \plus{}$ or $ \minus{}$ on the outside, but for a real solution the $ \pm$ on the inside must be $ \plus{}$ so we have two real distinct solutions.[/quote]

[hide="#1"]
$ (x^2 \plus{} 1)(x^2 \plus{} ax \plus{} 1)(x^2 \plus{} bx \plus{} 1)$
$ \equal{}x^2(x^2 \plus{} ax \plus{} 1)(x^2 \plus{} bx \plus{} 1)\plus{}(x^2 \plus{} ax \plus{} 1)(x^2 \plus{} bx \plus{} 1)$
$ \equal{}x^2(x^4\plus{}(a\plus{}b)x^3\plus{}(ab\plus{}2)x^2\plus{}(a\plus{}b)x\plus{}1)\plus{}(x^4\plus{}(a\plus{}b)x^3\plus{}(ab\plus{}2)x^2\plus{}(a\plus{}b)x\plus{}1)$
$ \equal{}x^6\plus{}(a\plus{}b)x^5\plus{}(ab\plus{}3)x^4$+2(a+b)x^3+(ab+3)x^2+(a+b)x+1$ Then we have$=x^6+(a+b)x^5+(ab+3)x^4$ \plus{}2(a\plus{}b)x^3\plus{}(ab\plus{}3)x^2\plus{}(a\plus{}b)x\plus{}1\equal{}x^6\plus{}1$
               $ ab\plus{}3\equal{}0$
                $ ab\equal{}\minus{}3$

[/hide]


Well can anyone please provide me the link to the proof , or the proof itself, for the Van Aubels;s Theorem (For Triangles) . I have been trying to prove it for past several Days but have failed to devise any proof

 :maybe:

hello, see here
http://planetmath.org/encyclopedia/ProofOfVanAubelsTheorem.html
Sonnhard.

Thank you very much !! Finally I got the proof !!! What a relief
\ :roll:  :oops:


$a,b\ge1 , c\ge0$and  $a,b$are real numbers and $n$ is natural number then prove it:
$(ab+c)^{n}-c\le((b+c)^{n}-c)a^{n}$


geometry
inradius
inequalities
inequalities proposed

let $ a$ , $ b$ , $ c$ be the lenghts of the sides of a triangle $ r$ inradius and $ R$ outradius , prove that :

$ \frac{(a\plus{}b)(b\plus{}c)(c\plus{}a)}{4abc} \le \frac{R}{r}$

this is from Iran 98...

first of all let $ p = \frac {a + b + c}2$ now note that:

\begin{eqnarray*}(a + b)(b + c)(c + a) = a^2(b + c) + b^2(c + a) + c^2(a + b) + 2abc = 2p(a^2 + b^2 + c^2) - (a^3 + b^3 + c^3) + 2abc\end{eqnarray*}

now note that:

$ a^2 + b^2 + c^2 = 2(p^2 - r^2 - 4Rr)$

$ a^3 + b^3 + c^3 = 2p(p^2 - 3r^2 - 6Rr)$

so we have to show that:

$ \frac {r}{4R}\leq\frac {4Rpr}{4p(p^2 - r^2 - 4Rr) - 2p(p^2 - 3r^2 - 6Rr) + 8Rpr}$

$ \iff 8R^2 - 2Rr - r^2\geq p^2$

but we have:

$ p^2\leq 4R^2 + 4Rr + 3r^2$

so it's sufficient to show that:

$ 8R^2 - 2Rr - r^2\geq 4R^2 + 4Rr + 3r^2$

$ \iff 4R^2 - 6Rr - 4r^2\geq 0$

$ \iff 2(R - 2r)(2R + r)\geq 0$

$ \iff R\geq 2r$

which is true...

[quote="y-a-s-s-i-n-e"]let $ a$ , $ b$ , $ c$ be the lenghts of the sides of a triangle $ r$ inradius and $ R$ outradius , prove that :

$ \frac {(a \plus{} b)(b \plus{} c)(c \plus{} a)}{4abc} \le \frac {R}{r}$[/quote]
$ \frac{a\plus{}b}{c}\equal{}\frac{sin(A)\plus{}sin(B)}{sin(C)}\equal{}\frac{2sin{((A\plus{}B)/2)cos((A\minus{}B)/2)}}{2sin(C/2)cos(C/2)}\equal{}$ $ \frac{cos((A\minus{}B)/2)}{sin(C/2)} \le \frac{1}{sin(C/2)}$
so : $ \frac {(a \plus{} b)(b \plus{} c)(c \plus{} a)}{4abc} \le \frac{1}{4sin(A/2)sin(B/2)sin(C/2)}\equal{} \frac{R}{r}$; :)

I need some help regarding my solution approach.:help:
We know that A=area of triangle=abc/4R
thus, R=abc/4A
again let S=(a+b+c)/2
thus, r=A/S
thus, R/r=abcS/4A^2=abc(a+b+c)/8[(a+b+c)(a+b-c)(b+c-a)(c+a-b)/16]
or, R/r=2abc/(a+b-c)(b+c-a)(c+a-b)
thus, the inequality becomes,
          (a+b)(b+c)(c+a)(a+b-c)(b+c-a)(c+a-b) <= 8a^2b^2c^2
Now, [(a+b)(b+c)(c+a)(a+b-c)(b+c-a)(c+a-b)]^1/6 <= [a+b+b+c+c+a+a+b-c+b+c-a+c+a-b]/6=3[a+b+c]/6=S
thus enough to show, S^3 <= 8^1/2*abc...............Hereafter I failed to continue.........:help:


Find out the first nonzero rightmost digit when $ 125!$ is represented in base $ 5$.


(a) Show that for every $n\in\mathbb{N}$ there is exactly one $x\in\mathbb{R}^+$ so that $x^n+x^{n+1}=1$. Call this $x_n$.
(b) Find $\lim\limits_{n\rightarrow+\infty}x_n$.

Equation can be rewritten $1+ x = 1/x^n$

LHS is strictly increasing from 1 to $+\infty$, RHS is strictly decreasing from $\infty$ to $0^+$ this proves existence and uniqueness of solution

$x^n+x^{n+1}$ is strictly increasong and $1^n+1^{n+1} > 1$ so $1 > x_n > 0$ and $(1+x)^{1/n} \rightarrow 1$ (typo corrected).

Equation can be rewritten $x(1+ x)^{1/n} = 1$ proving that $x_n \to 1$

[quote="Altheman"]i disagree since if you have n being odd, then the polynomial has an even lead coefficient, and since the roots pair up, there cannot be just 1 real root, there must be a multiple of 2 of real roots[/quote]yes, I forgot a detail: x should be positive. The second solution will be negative in your case.

Tµtµ is right anyway. :)

[quote="tµtµ"] and $(1+x)^{1/n} \rightarrow 0$.[/quote]
but $(1+x)^{1/n} \rightarrow 1$ no?!

yes. he meant 1 I guess, since the rest of his solution uses 1. :)

(a)
$x^n+x^{n+1}$ is strictly increasing and continuous on $\mathbb R^{+}$.And $x^n+x^{n+1}$ goes from $0$ to $+\infty$ on $\mathbb R^{+}$.Then by [b]intermediate value theorem,[/b]
$\exists ! x_n>0$ s.t. $x_n^n+x_n^{n+1}=1\blacksquare$ :coool:

(b)
Obviously $0<x_n<1$.If $x_{n+1}<x_n,$then
$1=x_n^n+x_n^{n+1}>x_{n+1}^{n+1}+x_{n+1}^{n+2}$ which is absurd.Thus $x_{n+1}\ge x_n$.Then $\{x_n\}$ is increasing and upper-bounded.Therefore $\exists \alpha$ s.t. $\lim_{n\to +\infty} x_n=\alpha$.$0<\alpha \le 1$.If $0<\alpha <1,$then $\lim_{n\to +\infty} \ln x_n^n=\lim_{n\to +\infty} n\ln x_n=-\infty$.Hence
$\lim_{n\to +\infty} x_n^n=0$.And $0<x_n^{n+1}<x_n^n,$by [b]squeeze theorem,[/b]
$\lim_{n\to +\infty} x_n^{n+1}=0$.Thus
$\lim_{n\to +\infty} x_n^n+x_n^{n+1}=0$ which is absurd.Therefore $\alpha =1\blacksquare$ :coool:


Find all positive integers $ m$ and $ n$ such that $ m^{2}\plus{} n^{5} \equal{} 252$.

$ n^5\le 252$

so $ n\le 3$

From the above we can easily say that "n" and "m" must be equal to 3


quadratics
calculus
derivative
function
geometry solved
geometry

[quote="chien than"]Let a;b;c;d are real numbers satisfying a;b;c;d<1.Prove that:
$x(1-t)+t(1-z)+z(1-y)+y(1-x) \geq 1$[/quote]

What are $x,y,z,t$ ?

Why can't you order a,b,c,d and use rearrangement then minimize the quadratics?

Or, if that fails, use derivatives? The function itself seems very tame

[quote="bilarev"][quote="chien than"]Let a;b;c;d are real numbers satisfying a;b;c;d<1.Prove that:
$x(1-t)+t(1-z)+z(1-y)+y(1-x) \geq 1$[/quote]
What about $x=y=z=t=0$?[/quote]
Moreover, taking $x=y=z=t$ shows that even if we assume $x,y,z,t>0$, the $\mathrm{LHS}$ can have any value from the interval $\langle 0;2\rangle$ (with the additional assumption [b]me@home[/b]'s method is an easy way to prove that $2$ is the maximum).


geometry
calculus
integration
trigonometry
calculus computations

How do I find the area for the intersection of 2 polar curves without converting to Cartesian?
For example how do I get the integral for the area between $ r\equal{} \sin \theta$ and $ r\equal{}\cos\theta?$

The area of a "slice" of a polar curve $ r(\theta)$ is given by $ \frac{1}{2} \int_{\theta_1}^{\theta_2} r^2 \, d \theta$.  You need to find intersections $ \theta_1, \theta_2$ and then calculate the difference in the areas.


AMC
USA(J)MO
USAMO
probability
videos
AMC 10

I probably could get 22 or higher but in a testing situation I'll probably end up 15-21.

He's a middle school math beast.

He also recently won the Scripps Spelling Bee, which earned him a lot of attention. A TV interview with some news woman earned him even more publicity (you can see it on youtube).

[quote="1=2"]There was this guy who made it to the IMO when he was 10. He lived in Australia at the time, I think.[/quote]
Yeah, I think you thought about Terence Tao :wink: Real genius  :)

[quote="The Zuton Force"]He's a middle school math beast.

[b]He also recently won the Scripps Spelling Bee, which earned him a lot of attention. A TV interview with some news woman earned him even more publicity (you can see it on youtube).[/b][/quote]

you're kidding right?

i remember all that publicity was very negative if you read all the comments...
people dont appreciate his talents.

i watched an interview on youtube of Evan O. He seems like he has some issues is he autistic? some times academic achievement is not the most important thing. did he make IMO?

IMO, Evan O'Dorney won't make IMO. O.K. he made MOP last year. That was great. Making MOP in eighth grade is great. Yet, I think that others (Krishnau, Alex Anderson, etc) are more pro. Evan O'Dorney may have a chance next year.

[quote="blackbelt14253"]Who's the youngest person ever to make the US IMO team? What was his/her name?

And do you think that Evan O'Dorney dude will make it onto the team this year?[/quote]

I'm not sure, but I'd think it would be either Reid Barton or Tiankai Liu? I'm guessing they were about 14 at the time. Actually, come to think of it, Ricky Liu and Po-Ru Loh are also possible.

Well it's impressive and cool that Evan is so successful, but I do agree that it is not good publicity. I feel like those interviews reinforce the negative stereotypes that people have about people who like math. Those interviews were the most awkward thing that I've ever seen. It portrays math students as lacking any social skills whatsoever.

USAMO top twelve last year who have not graduated:

- Sergei Bernstein (11)
- Eric Larson (11)
- Haitao Mao (12)
- Delong Meng (11)
- Krishanu Shankar (12)
- Jacob Steinhardt (12)
- Alex Zhai (12)

Honorable Mentions last year who have not graduated:

- David Benjamin (11)
- Yingyu (Dan) Gao (12)
- Yan Li (12)
- Gaku Liu (12)
- Jeffrey Manning (12)
- Palmer Mebane (12)
- [b]Evan O'Dorney (9)[/b]
- Alexander Remorov (12)
- Max Rosett (12)
- Qiaochu Yuan (that's me! :) ) (12)

which puts Evan O'Dorney in some pretty respectable company.  In any case, social skills (or lack thereof) notwithstanding, he's got three more years to make it onto the team and a pretty good head start.  :)

[quote="PI-Dimension"][quote="n^4+4"]Haha is Jeremy Kahn approximately equal to Jeremy Hahn?[/quote]

Lol sure.

I was like wow that looks familiar.[/quote]

This is on topic?

Besides, if you watched the videos of Evan being interviewed, you would get it.

[quote="mathcrazed"]Evan O'Dorney has Asperger's... [/quote]

What is Asperger's?

A human with Asperger syndrome displays intense interest in restricted interests or activities. This may lead to that human being extremely knowledgeable or good at that activity/interest. However, because this is still a brain disorder, it is labeled as autism.
I don't know whether Evan O'Dorney has Aspergers or not, but someone with autism [i]can[/i] make MOP. Actually, with some types of autism, you're more likely to.

Asperger's syndrome is a syndrome on the autism [i]spectrum.[/i]  This is not quite the same thing as being autistic.  In any case, the autism spectrum is quite wide and there are many gradations of autistic behavior.

Evan shows signs of Asperger's... Although he does attempt to engage in social activities, he is unable to understand some types of irony, sarcasm, etc.  He does have an intense interest in math, as uldivad9 implied, which is also a symptom of Asperger's.

I think we should all stop speculating, it's Evan's business, and by no means ours.

Also, see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=159137]this[/url] and [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=162184]this[/url]

Yeah these threads seem to get really popular about him. Guys it shouldn't be our business to speculate his mental condition, and he probably gets enough beef if a lot of people perceive a problem anyway. At least not on an online forum  :wink:. It isn't on topic anyway...

Anyway I think this year has a lot of really good seniors and that USA will do really well at IMO, in my opinion. We might even win!  :D

h341
for the last time many people with autism are very good at math

[quote="#H34N1"][quote="CatalystOfNostalgia"][quote="#H34N1"]As a rule of thumb, you can't have autism and make MOP in eighth grade[/quote]

I would take offense to that if I was close to someone with autism. I don't know what your intention was with that comment, but one could interpret it as you completely berating autistic people. As far as I know, some autistic people have extraordinary abilities, in things such as memorizing pi and music. Who knows, maybe there is an autistic person out there, if Evan O'Dorney doesn't have Asperger's as some people are oddly suggesting, that excels at olympiad math. While I still may be grossly wrong, unless you really know your stuff about autism, you probably shouldn't make a generalization like that.[/quote]

I'm not berating autistic people. I'm just saying that it doesn't make sense to claim that someone is autistic when they are pwnage at math skillz :|[/quote]

People weren't claiming that he had Asperger's, or he was autistic, just because he was smart, they were claiming that because of his personality as seen on television. And I don't exactly see how what you are/were "just saying" can ever be interpreted from your comment. So long as we're clear...

Alright guys let's get back on topic? Evan O'Dorney may or may not have a mental condition, but I think that is a really private thing.

Any subsequent posts off topic will be deleted. Feel free, of course, to discuss what you like over PM.

Agreed. I will be watching this topic as well.


it is well known that there is no known closed-form of the formula for partitions.
Has it been proven that there is no formula? If so, could someone prove it here if it isn't too complicated?

[quote="DiscreetFourierTransform"]That's not entirely true.
There is the Rademacher Series, which is asymptotically correct.
Here's the wikipedia link, but beware, the Rademacher series is pretty complex and involves Dedekind sums.
[url=http://en.wikipedia.org/wiki/Partition_(number_theory)#Rademacher.27s__series]Here[/url][/quote]
I meant a closed form.

Well, I suppose that's as close to a closed form as we have.
However, you're right that there is no exact closed form.
That link has all the information in it.

EDIT: The link doesn't like the URL tags.
Just copy+paste it into the browser.


function
calculus
derivative
real analysis
real analysis solved

prove that for every n\in N* the following ecuation
\sum {k=1,n}1/ \sqrt (nx+k) = \sqrt(n)
has a unique real root x=x_n and find lim{n->oo}x_n.

Let f(x) be the function f(x) =  \sum(k=1,..,n){1/\sqrt (nx+k)} ,  defined for x > -1/n .

It's easy to see that lim(x-->+\infty){f(x)} =  0 <  \sqrt(n) and lim(x-->-1/n){f(x)} = +\infty > \sqrt(n) , so there is at least a real root x(n) , -1/n < x(n) < 0

Manifestly the derivative f'(x) is ngative, so f(x) is decreasing , so the root x(n) is  unique.

-1/n < x(n) < 0  ===> lim(n-->\infty){x(n)} = 0
........

not very correct
x_n is in (-1/n,oo)
so its limit is not 0.
check again your proof
u can treat the more genrealy problem

[i]f(x)=a \sqrt n has a unique root x=x_n and find it's limit, where a>0[/i][/i]

Oh god, Diogene is really too  bad, he mistook 0 for  \infty    !  So I , his " alter-ego", decided to take his place . :D  :D
And now, come  back to the problem ..

S_n(x) = (1/n) \sum(k=1,..,n){1/\sqrt(x+k/n)} = A , A > 0 ; We know that there is an unique solution x(n), x(n) > -1/n .

Briefly, we can relate S_n(x) to the Reamann sums for I(x) = \int(0 < t <1){1/sqrt(1/(x+t)).dt} = 2*(\sqrt (x+1) + \sqrt(x)) = I(x) , for x>0.

Finally the result is :
0 < A =< 2 ===> I(x)=A  has an unique solution  ===> lim(n-->\infty){x(n)} = (4-A2 )2/(16*A2)
2  I(x)=A  has not  solution ===> lim(n-->+\infty ){x(n)} = 0

In general, if the  (1/n)* \sum (k=1,..,n){f(x,k/n)} = g(x) has a solution x(n) ,

then the study of the behavior of the x(n) sequence is relating to the equation   \int(0< t <1){f(x,t).dt} = g(x)
...


The farmer’s pivot has an effective irrigation radius of ¼ mile (640 acres in a square mile) and is positioned alongside a straight section of highway. He wants to apply a special fertilizer at the rate of 0.7 gallons per acre. How much does he need to get for the irrigation system?

A) 45 gallons B) 88 gallons C) 32 gallons D) 50 gallons

His lot is a semi-circle with radius $ \frac14$. The area is $ \frac\pi{32}$ square miles or $ 20\pi$ acres.\[ 20\pi\cdot0.7\approx43.98\]This is close to A.


floor function
induction
combinatorics unsolved
combinatorics

A word is a sequence of n letters of the alphabet {a, b, c, d}. A word is said to be complicated if it contains two consecutive groups of identic letters. The words caab, baba and cababdc, for example, are complicated words, while bacba and dcbdc are not. A word that is not complicated is a simple word. Prove that the numbers of simple words with n letters is greater than $2^n$, if n is a positive integer.

It is also Romanian TST 2003. And as I remember, this problem were discussed before on forum.

Let us denote by $S(n)$ the set of simple words with $n$ letters and by $s_n$ the number of elements in $S(n)$. If we put a letter at the end of each of the simple words from $S(n)$ we obtain a set $T(n+1)$ of $t_{n+1}$ words of length $n+1$, their number being $t_{n+1}=4s_n$. Obviously $S(n+1) \subset T(n+1)$, $S(n+1)\neq T(n+1)$.

Let $T_1(n+1)$ be the set of those words from $T(n+1)$ which have the last two letters the same, $T_k(n+1)$ the set of $T(n+1)$ which end in two consecutive identical groups of $k$ letters, for each $k\in\{1,2,\ldots,m\}$, where $m=\displaystyle \left\lfloor \frac { n+1 } 2 \right\rfloor$. Obviously
\[ f(n+1)\geq t_{n+1}-|T_1(n+1)|-|T_2(n+1)|-\cdots - |T_m(n+1)|\]
We have $t_{n+1}=4f(n)$, $|T_1(n+1)|=f(n)$, and furthermore $|T_k(n+1)|\leq f(n+1-k)$, because of the fact that the mapping of $S(n+1-k)$ into $T_k(n+1)$ given by adding to a simple word of $n+1-k$ letters its own last $k$ letters is obviously surjective.

We have $f(1)=4$, $f(2)=12>4$. By induction we want to prove $f(k+1)>2f(k)$. We have
\[ f(n+1)\geq 4f(n)-f(n)-\frac 12 f(n) - \frac 14 f(n)- \cdots > 2 f(n) \] from which the conclusion follows.

But still the obvious next questions are :
Q1: Is 2^n  a tight bound ?
Q2: Is the bound dependent on the alphabet size ?

 A precision : when I say obvious I mean the esayness to formulate such questions
 not that the answers are necessarily easy.

 REMARK :
 It seems that too often that thes olympiad/IMO problems concentrated on one fact.
 I believe it is nice to follow up , to give perspective to the result by showing how it
 hinges on the hypothesis, otherwise the solution as little meaning or constitutes merrely
 a useful technique. See and read Polya about problem solving in general and making
VARIATION on problems.

Valentin Vornicu wrote:

\box{
We have $t_{n+1}= 4f(n)$, $|T_{1}(n+1)| = f(n)$, and furthermore $|T_{k}(n+1)| \leq f(n+1-k)$, because of the fact that the mapping of $S(n+1-k)$ into $T_{k}(n+1)$ given by adding to a simple word of $n+1-k$ letters its own last $k$ letters is obviously surjective.}

The mapping of $S(n+1-k)$ into $T_{k}(n+1)$  defined latter is wrong defined. The mapping of $T_{k}(n+1)$ into $S(n+1-k)$ given by removing to a word of $T_{k}(n+1)$ its own last $k$ letters is obviously injective (and this mapping is well defined).

another solution I came across by [b]math154[/b]  from [url=https://artofproblemsolving.com/community/c285h400665p2237686]here[/url]

[hide="Solution"]Let $f(n)$ denote the number of simple words with $n$ letters. We'll show by strong induction that $f(n+1)>2f(n)$ for all $n\ge0$. Since $f(0)=1$ and $f(1)=4$, the base case is clear.

Now consider a simple word $W$ with $n\ge1$ letters, and prepend a letter $x\in\{a,b,c,d\}$ to it. There exist strings $S,T$ (possibly empty) such that
\[x+W=x+S+x+S+T \Longleftrightarrow W=S+x+S+T\]iff $x+W$ is complicated (note that $x+S+T\in W$ and is thus simple). Hence there is a surjection from simple words with $1\le k=|x+S+T|\le n$ letters to complicated words of the form $x+W$ (i.e. prepending the first $n+1-k$ letters of $x+S+T$ to get $x+W$), so by complementary counting,
\[f(n+1) \ge 4f(n) - \sum_{k=1}^{n}f(k) > 4f(n) - \sum_{k=1}^{n}\frac{f(n)}{2^{n-k}} > 2f(n),\]as desired.[/hide]

I hope it is not exactly the same as above


Let $x,y,z\geq 2$. Find the least $k$ such that
$(x^3+y)(y^3+z)(z^3+x)\leq kxyz$

For every $k$ we can add $y=z=2$ and $x$ can be a large number; so the left side will be greater than the right side.
I think you mean this one : $x,y,z\geq2$
$(x^3+y)(y^3+z)(z^3+x)\geq125xyz$

[quote="Ramtin Madani"]
I think you mean this one : $x,y,z\geq2$
$(x^3+y)(y^3+z)(z^3+x)\geq125xyz$[/quote]

Right, can you prove it?


Hello

Plz help me to find some free books about inequalities.

Thanks

hello, see this here
http://www.eleves.ens.fr/home/kortchem/olympiades/Cours/Inegalites/tin2006.pdf
Sonnhard.

Thank you a lot for the book
it's great

are there any other books ? not about inequalities only

^^

10x


solve in positive integers the following equation
$x^2y^2=z^2(z^2-x^2-y^2)$

hmm.. discriminant of this equation in z^{2} is x^{4}+y^{4}+6x^{2}y^{2} which is never a perfect square .i proved this yesterday.and btw wot is the source of this problem or u made it up?
nice one ,neway

This is a Bulgarian problem I think, i'm not sure :?

Bulgaria 1998 :D
On the site of the Bulgarian MO, another solution is given


Find all functions $ f: [0,1] \to  [0,1]$ and \[ f(2x - f(x)) = x\] for all $x \in [0,1]$.

[quote="MJ GEO"]$ f$ form $ [0,1]$ to $ [0,1]$ and $ f(2x \minus{} f(x)) \equal{} x$[/quote]

This immediately implies $ f((n\plus{}1)x\minus{}nf(x))\equal{}nx\minus{}(n\minus{}1)f(x)$ and so $ 1\ge nx\minus{}(n\minus{}1)f(x)\ge 0$ and so $ \frac n{n\minus{}1}x\ge f(x)\ge \frac{nx\minus{}1}{n\minus{}1}$ and so $ \boxed{f(x)\equal{}x}$ which, indeed, is a solution.

i cant understand your solution.coud you write more about it? its mine
$ g(x)\equal{}2x\minus{}f(x)$ so $ f(g(x))\equal{}x$. $ g(g(x))\equal{}2g(x)\minus{}f(g(x))\equal{}(2g(x)\minus{}2x)\plus{}x$ and similiary $ g^n(x)\equal{}(ng(x)\minus{}nx)\plus{}x$ so $ g(x)\equal{}x$ so $ f(x)\equal{}x$


floor function
logarithms
number theory unsolved
number theory

Please help me with this one:
$A=3^{2004}+25$
1. How many digits A have(in base $10$)? (easy, I found that $957$ digits)
2.  $S=$sum of all this digits. How much is S?
3. How many do zeroes digits appear in A?

I doubt whether this can be done without using a computer.

Yes, Arnel. But using caculor is not a good way and it is meanless.
   I hope some could post a good solution for it! I know that it seems difficult but let try it ! :D

A "good" (elegant) solution simply cannot exist...

I know the answer to 1, however, I'm still thinking about 2 and 3.  Here
s 1's solution that can somehow be done without a calculator.  I'm still thinking about doing it without a calculator...hopefully, it is possible.
[hide]
$\\\lfloor\log 3^2004+1\rfloor\\=\lfloor2004\log 3+1\rfloor\\=\lfloor 956.15+1\rfloor\\=957$
[/hide]

Masoud Zargar


Do you know of any free (preferably free as in freedom) non-demo non-trial non-shareware music notation programs for Windows that can save/export as MIDI?

I'm currently using Finale Notepad 2006, but it doesn't save/export as a MIDI file.

NoteWorthy Composer, I'm not sure you can save as a MIDI file, but I know it works.  You could try.

The website says that it's a 30-day evaluation trial, and according to download.com and other sites, it's shareware. :(

-_- I've been on 30-day trial for a few years then...

Is Finale Notepad 2006 good for just writing though?

[quote="Eli"]Yes.  Mr. Muer recommends it.

Does the demo have nag screens, etc.?[/quote]

nag screens?

Do we have to use notepad for the assignment (and it's optional, correct?)?

[quote="Eli"]What I mean is if there are major annoyances, etc.  Are there any important features you can't use?

Mr. Muer wants us to use Notepad (probably because he doesn't know of anything else), but if it's a music file, he'll probably be fine with it.[/quote]

No, I believe it does not have "nag screens."  But don't take my word for it.

what is a MIDI?

For future purposes, if you don't need a WYSIWYG music editor then use musixtex package in MiKTeX.  Ask on the TeX forum.  I don't know how to use it myself :P

It's a type of music file (see http://en.wikipedia.org/wiki/Musical_Instrument_Digital_Interface), like mp3, wav, etc.

I know.  I like to have a WYSIWYG editor for music.  I don't have to worry about the syntax with them.

Currently I use GenieSoft Score Writer v2.6. It costs money, but it's only about 30 dollars. It does all the MIDI import/export stuff with ease and multiple tracks. It has a WYSIWYG interface with easy-to-use mouse and keyboard controls. With the given price, I'd say it's one of the more economical software packages out there.

However, changing the scores to PDF give some problems (yes, I use PDF pro v7).


$ S$ is a set of functions such that $ f: N \rightarrow R$ and
(1)$ f(1) \equal{} 2$
(2) $ f(n \plus{} 1) \ge f (n) \ge \frac {n}{n \plus{} 1} f(2n)$ .
 Determine the smallest $ M$ such that  $ f(n) < M$  (for any $ f \in S$ and for any $ n \in N$)

Obviosly $ f(n)\le f(2n)\le (1\equal{}\frac 1n )f(n)$, therefore $ f(\infty )\le \prod_{k\equal{}0}^{\infty }(1\plus{}\frac{1}{2^k})\equal{}g(\frac 12$, were $ g(x)\equal{}\prod_{k\equal{}0}^{\infty }(1\plus{}x^k)$ (|x|<1).

$ g(\frac {1}{2})$=? It's not finally answer.

[quote="Svejk"]I also managed to prove that $ f(\infty) \leq g(1/2)$ , but why $ g(1/2)$ would be the minimum value ?[/quote]
We can define (maximal possible mean) $ f(n)\equal{}\prod_{i\equal{}0}^{k}(1\plus{}\frac{1}{2^i})$, were $ k\equal{}[log_2(n\minus{}1)],n>1$.
$ g(x)$ is special function considered by Euler.


Six people arrive to purchase $10 tickets for a movie. Three of them have only $10 dollar bills and 3 have only $20 dollar bills. The person selling the tickets initially has no change. What is the probability that the people line up in such a way that the person selling the tickets always has enough change?

[hide]THe probatilty is 1/4.
Obviously, the last person to enter the movie theater must have been a person with a 20 dollar bill.  THerefore, all three people with 10 dollar bills and 2 people with 20 dollar bills must have entered before he did.  3/6*2/5*1/4*3/3*2/2=1/20.  Of course, the people with 10 dollar bills didn't have to go first, but it makes no difference becuase the numerators and denominators are just switched around accordingly.  Now, we just need to figure out how many ways this can be done.  Obviously, the first person has a 10 dollar bill and the last has a 20 dollar bill.  $4 \choose 2$ is 6 so there's 6 ways to choose the remaining 4 people, except that 10,20,20 is not allowed.  THerefore, there's 6-1=5 middle possibilities and 5*1/20=1/4.[/hide]


Whats me to solve for this system

(  2x-4y=32
{3x+5y=4
(

this system can be turned into 6x-12y=96 and 6x+10y=8, leaving us with y = -4 and x = 8.

Multiply $2x-4y=32$ by $3$ to get $6x-12y=96$. Multiply $3x+5y=4$ by $2$ to get $6x+10y=8$. Now, subtract the first equaiton from the second to get $22y=-88$. Divide both sides by $4$ to get $y=-4$. Now, plug in y back into the first equation to get $2x-4(-4)=32$ or $2x=16$. Divide both sides by $2$ to get $x=8$.

Problems like these are more middle school level and not high school level and can be posted here: http://www.artofproblemsolving.com/Forum/index.php?f=299 instead.

[hide]I would just do substitution myself, because i'm wierd.  Add 4y to both sides to get $\displaystyle{2x=4y+32}$  Then divide by 2...
$\displaystyle{x=2y+16}$  Substitute that for x and solve for y... then sub y in for x in this equation and you get x=8 and y=-4...[/hide]

[quote]because i'm wierd[/quote]

ur also weird cuz u spelled weird wrong. :rotfl:
just kidding.
no offense.


hello, this equation has no real solution.
Sonnhard.

i think there is some mistakes because this equation has more than one real root

[quote="hoangvn.fix"]$ \sqrt [3]{x^3 \minus{} 12x^2 \plus{} 27x \minus{} 8} \equal{} 2x \plus{} \sqrt [3]{x^3 \minus{} 3x}$[/quote]

$ x_1 \equal{} 2\cos \frac {4\pi}9$

$ x_2 \equal{} 2\cos \frac {10\pi}9$

$ x_3 \equal{} 2\cos \frac {16\pi}9$

[hide="Why?"]Let $ u\equal{}\sqrt[3]{x^3\minus{}3x}$.

The equation is $ \sqrt[3]{u^3\minus{}12x^2\plus{}30x\minus{}8}\equal{}2x\plus{}u$

Raising to power $ 3$ gives : $ u^3\minus{}12x^2\plus{}30x\minus{}8\equal{}8x^3\plus{}12x^2u\plus{}6xu^2\plus{}u^3$

$ \iff$ $ u^3\minus{}12x^2\plus{}30x\minus{}8\equal{}8u^3\plus{}24x\plus{}12x^2u\plus{}6xu^2\plus{}u^3$

$ \iff$ $ 4(u^3\plus{}1)\plus{}6x^2(u\plus{}1)\plus{}3x(u^2\minus{}1)\equal{}0$

$ \iff$ $ (u\plus{}1)(4u^2\minus{}4u\plus{}4\plus{}6x^2\plus{}3ux\minus{}3x)\equal{}0$

$ \iff$ $ (u\plus{}1)\frac{87(8u\plus{}3x\minus{}4)^2 \plus{}(87x\minus{}12)^2 \plus{}4032}{87\cdot 16}\equal{}0$

$ \iff$ $ u\equal{}\minus{}1$ (since the other factor is always $ >0$

$ \iff$ $ x^3\minus{}3x\plus{}1\equal{}0$.

It's easy to see that this equation has 3 real roots all in $ (\minus{}2,\plus{}2)$ and so let $ x\equal{}2\cos v$ and so $ 8\cos^3v\minus{}6\cos v\equal{}\minus{}1$

$ \iff$ $ \cos 3v\equal{}\minus{}\frac 12 \equal{}\cos \frac{4\pi}3$

Hence the result[/hide]


Can anyone tell me how much money cost one year at MIT and is there some scholarships?

[url]http://web.mit.edu/sfs/[/url]
Hope this helps. :)

[quote="Gen8"]MIT sure is an 'expensive' destination for an international student.[/quote]
Well,there're scholarschips,right.

I am  sure that If you get  outstanding academic achivements,your government will support you with a large fund,quite enough for you to follow some higher degree in foreign countries.

[quote="barcelona"]I hope to have outstanding results,but I'm still at the beginning.What exams should I take to enter an American university?I know SAT 1 is obligatory, but how many  SAT subjects are reqiured?My major is maths but what should the other subjects be? Are they optional or obligatory?
What scores do I have to get to have to have chances for some grant or scholarships. I would appreciate any information.[/quote]

most schools require 2-3 subject tests and some have a requirement that you take Math II as one of them. What score should you get? As high as you can if you're applying to MIT. But an 800  on everything wont guarantee a scholarship. Scholarships are almost never based on standardized tests.


The point $ (a,b)$ is located in the third quadrant in the Cartesian plane.  The coordinates are changed according to the functions given: $ f(a)\equal{}a^2 \plus{} 3$ and $ g(b)\equal{}b\minus{}4$.  In which quadrant (1, 2, 3, or 4) is $ (f(a),g(b))$ located?

Third quadrant means (neg,neg). Say that it is (-1,-1). Using the functions, we would get (4,-5), which is in the fourth quadrant. Our answer is 4.

As Kiseki said, the third quadrant consists of the points that are (neg,neg).

The square of all negative numbers are positive numbers, and adding a positive number to a positive number always yields a positive number.

And negative number subtracted by a positive numbers is a negative number.

So, we have the answer (pos,neg). This is represented by the $ \boxed{4}$ the quadrant.


trigonometry
calculus
integration
calculus computations

u''(x)+c*u(x)=0 -differential equation.

I was studying differential equations and when i reached the equations of grade 2 i encountered this equation.But it was saying''it is verified that'' and it just gave me the form of u(x).
Could somebody please demonstrate the form of u(x)?

A solution of characteristic equation $\lambda^{2}+c=0$ is $\lambda=\pm i\sqrt{c}$.
So, a general solution to find is $y=C_{1}\cos (\sqrt{c}x)+C_{2}\sin (\sqrt{c}x)$.
($C_{1}, C_{2}$ : Constant of Integration)

[quote="radac_mail"]This is what i am trying to understand:why is y=C1cos(sqrt(c)*x)+....  ?...and i don't want to begin with the y and reach that y''+cy=0 .
There must be a way of reaching the general solution![/quote]
If it defines it as $z_{1}=\cos \sqrt{c}x+i \sin \sqrt{c}x, \ z_{2}=\cos \sqrt{c}x-i \sin \sqrt{c}x$,
$z_{1}'=-\sqrt{c}\sin \sqrt{c}x+i \sqrt{c}\cos \sqrt{c}x=i \sqrt{c}(\cos \sqrt{c}x+i \sin \sqrt{c}x)=\lambda_{1}z_{1}$
$z_{1}''=\lambda_{1}z_{1}'=\lambda_{1}^{2}z_{1}$
So, $z_{1}''+c\cdot z_{1}=(\lambda_{1}^{2}+c)z_{1}=0$.
Likewise, $z_{2}''+c\cdot z_{2}=(\lambda_{2}^{2}+c)z_{2}=0$
$z_{1}, z_{2}$is a complex number value solution of $U''(x)+c\cdot U(x)=0$.
These linear combination
$U_{1}=\frac{z_{1}+z_{2}}{2}=\cos \sqrt{c}x, \ U_{2}=\frac{z_{1}-z_{2}}{2i}=\sin \sqrt{c}x$
is solutions of $U''(x)+c\cdot U(x)=0$ again.
Therefore, $U(x)=C_{1}\cos (\sqrt{c}x)+C_{2}\sin (\sqrt{c}x)$.
($C_{1}, C_{2}$ : Constant of Integration)


Nimeni nu se baga la inegalitatea asta? :(

ma io nu pricep o chestie...
$a+x^{2}>a$.
numitorul membrului drept este enorm (mai mare decat $9^{12}$).
deci
facand cele mai directe aproximari, obtin
$LHS>abc>(a-x)(b-y)(c-z)>9^{-12}(a-x)(b-y)(c-z)>RHS$.
ori am baut eu gaz, ori ceva nu`i bine p`acolo...

porneste de la faptul ca $\frac{a+x^{2}}{a-x}=\frac{x^{2}+x}{a-x}+1$...

io intrebam altceva
ce e gresit in rezolvarea mea de mai sus ?
daca rezolvarea mea e corecta, atunci inegalitatea e al dracului de slaba, si conditia cu xyz e in plus

Nu te-nfuria mai...cred ca e slaba.. :roll:

nu ma infurii
dar am mai patit`o cu inegalitati sa dau o solutie "corecta" si sa ma prind dupaia ca de fapt nu e bine.
d`asta cand vad semnu $\geq$ ma gandesc ca trebuie sa fie un caz de egalitate, si daca mie imi da cu $>$ atunci ceva nu e bine..

Stii, unii propunatori mai pun si semnul asta $\geq$ ca sa mai deruteze pe elevi..chiar daca normal ar trebui sa puna $\ >$.Asta am vrut eu aici..pacalici. :)

Nu.O problema[sau parte din anuntul ei] nu  trebuie sa deruteze elevul..Si semnul $\geq$ nu are niciun sens dak egalitatea nu are loc..DEsi logic vorbind inseamna"mai mare[b] sau[/b] egal..parerea mea

Mai, eu asa am mai vazut.Inegalitati menite sa mai deruteze concurentul..intr-un fel, asta e lipsa de fair-play.. :)


modular arithmetic
number theory proposed
number theory

if   $\frac{p^a-1}{p-1}=2^n$  that $p$ is a prime number and $n,a$ are positive integers.
find number of divisor of $n.a$

If a=2 then 2^n-1 is prime so the number of divisors of n is 2 since n must be prime. For a=2,3 its fairly easy to see there are no solutions. I don't know about the number of divisors in general.

1) Obviously $p>2$ else LHS would be odd
2) We note that $a$ must be a power of $2$ in fact if it's not we would have let's say $d|a$ with $d$ an odd number $>1$. Then we will have $\frac{p^d-1}{p-1} | \frac{ p^a -1 } {p-1} = 2^n$ and so $p^{d-1} + p^{d-2} + ... + 1|2^n$. Thus $p^{d-1} + p^{d-2} + ... + 1$ must be a power of $2$ but it's a sum of an odd number of odd number and thus is odd and it's obviously $>1$.

3) Thus $a=2^k$. Rewriting we get:

$\frac{p^{2^k} -1}{p-1} = (p^{2^{k-1}}+1)(p^{2^{k-2}} +1 )... (p+1) = 2^n$

Thus all the factors in LHS must be powers of $2$.
But $p^2+1 \equiv 2 \pmod{4}$ and so it can't be a power of $2$.
Thus $k=1$

4) For $k=1$ and so $a=2$ we get $p= 2^n-1$ and so $n$ must be prime. So $an$ has always $4$ divisors