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I've been working on this for long, however, till now I still cannot verify it except for the confidence of it being true.

Suppose $ A_n$ is n * n sized real matrix, and the (i, j) element $ a_{ij} \equal{} \int_0^1f ^ {(i)}f ^ {(j)}dx$, where i, j = 0, 1, ... , n - 1, f is an arbitrary real function in $ C^{\infty}[0, 1]$, $ f^{(i)} \equal{} \frac{d^if}{dx^i}$.

My problem is [b]how to verify the serie {$ \lambda_n$} tends to 0, where $ \lambda_n$ is the smallest eigenvalue of $ A_n$[/b], and how fast is the serie tending to 0? An estimation is most welcome.

By the way, by some numerical tests, I am almost sure it is true (tending to 0), however, if otherwise is correct, please also give me an example. Thanks a lot.

A first step: use the search function on this forum to search for the words "Hilbert Matrix." Do any of the links you find that way help you?

I checked the searching results, yet it seems not working. And it seems the matrix $ A_n$ is not a [b]Hilbert Matrix[/b] but a [b]Hanker Matrix[/b].

Let me suggest something, hope it could help.

Suppose $ \lambda_1 > \lambda_2 > \cdots > \lambda_n$ is the n eigenvalues of $ A_n$. Since $ \lambda_1$ might be much larger than $ \lambda_n$,

$ \lambda_n$ could be "covered" by the trace of $ A_n$, i.e. tr(A).

So let's consider tr($ A^{\minus{}1}_n$) which equals to $ \frac{1}{\lambda_1} \plus{} \cdots \plus{} \frac{1}{\lambda_n}$. Thus we have

$ tr(A^{\minus{}1}_n) \equal{} \frac{1}{\lambda_1} \plus{} \cdots \plus{} \frac{1}{\lambda_n} < \frac{n}{\lambda_n}$.

So $ \lambda_n < \frac{n}{tr(A^{\minus{}1}_n)}$.

And then comes our problem, who can give an estimate of tr($ A^{\minus{}1}_n$)?

Why $ \lim_{n\rightarrow\infty} \frac{n}{tr(A^{\minus{}1}_n)} \equal{} 0$?

Hope someone can help me. Thanks a lot.

Sorry, I couldn't understand you clearly.

But it is obvious that a lot of smooth function satisfy my assertion, for example $ \sin{x}$, and the $ A_n$ constructed by $ \sin{x}$ has the minimal eigenvalue which will tend to be 0 with the increase of n.

So my problem is [b]for any function[/b] $ f \in C^\infty[0, 1]$, the $ A_n$ constructed by the $ f$ will behave as I've said in my first post.

And also could you give some details on how to reduce my question to the inequality you suggested?

Thank you for your reply. I really appreciate it.

[quote="fedja"]Basically, the question boils down to whether we can have a smooth function $ f$ such that $ \int_0^1 \Bigl|\sum_k c_kf^{(k)}(x)\Bigr|^2\,dx\ge \sum|c_k|^2$ for every set of coefficients $ c_k$. The answer is "Yes, of course" because we can easily add fast oscillating bumps to the derivatives inductively to ensure sufficient amount of linear independence. On the other hand, the question whether one can have an [b]analytic[/b] function like that is a little bit more interesting ;)[/quote]

Sorry for the brevity of my previous post. The idea is to build a function whose derivatives are sufficiently linearly independent. What you can do for that is to add fast oscillating terms to high order derivatives that disappear after integration, so they do not show up in the low order derivatives. The simplest fast oscillating function is $ \cos \lambda x$ with big $ \lambda$. So, it is natural to build a series like
\[ f(x)\equal{}\sum_{k\ge 1}\lambda_k^{\minus{}k\plus{}\frac 12}\cos \lambda_kx\]
with very fast increasing $ \lambda_k$ ($ \lambda_k\equal{}2^{k!}$ should be enough). When you differentiate this series $ n$ times, you'll get $ \lambda_{n}^{1/2}$ times $ \sin$ or $ \cos$ of $ \lambda_n x$ as the dominating term in the series for the $ n$-th derivative. So, $ f^{(n)}$ will be large and essentially orthogonal on $ [0,1]$ and you can easily derive from here that you cannot find their linear combination that is small in the $ L^2$ norm. This function is, of course, horrible: it is $ C^\infty$ but very far from being analytic. That's why I asked my question about analytic functions. On the other hand, I suspect that even for as simple function as $ \frac 1{1\plus{}x}$, the eigenvalues do not tend to $ 0$: the factorials in the numerator far outweigh the powers of numbers less than $ 2$ in the denominator. Try to verify if it is true yourself. The functions like $ \sin x$ and polynomials are exceptional in the sense that their derivatives are linearly dependent, so quite soon the least eigenvalue becomes exactly $ 0$.

As to your last question, it is an immediate consequence of the standard theorem that the minimal eigenvalue of a non-negative definite real symmetric matrix is just the minimum of the corresponding quadratic form over the unit ball. In the direction you need it here, it is completely trivial: just consider $ \langle Ax,x\rangle$ where $ x$ is the eigenvector corresponding to the minimal eigenvalue.

Oops, don't know what happened to me :oops: : my remark about $ \frac 1{1 \plus{} x}$ is an ultimate nonsense (by the way, who can explain why?). As a matter of fact, I now believe that all analytic functions are OK. At least, if we consider the $ 1$-periodic analytic functions, we do have a positive result. The idea is to show that $ f'$ is in the linear span of the higher order derivatives. Suppose not. Then there is an $ L^2$ $ 1$-periodic function $ \psi$ such that $ \int_0^1\psi( \minus{} t) f^{(n)}(t)\,dt \equal{} 0$ for all $ n > 1$ but is not $ 0$ for $ n \equal{} 1$. But what is written is just the $ n$-th derivative of the convolution $ f*\psi$ at $ 0$. Since that convolution is analytic and has all but the first 2 derivatives $ 0$ at $ 0$, it must be a non-constant polynomial of degree $ 1$ but a polynomial of degree $ 1$ cannot be $ 1$-periodic! The question about non-periodic functions is more complicated. Let me think of it for a while :).

At least, it is clear now that the problem has hardly anything to do with linear algebra. So I'm moving it :P

Edit 1: We can also show that the statement is true for the functions analytic in a sufficiently large domain containing $ [0,1]$. Assume for instance that $ f$ is analytic in the disk of radius 3 around the origin. There are two possibilities:
a) $ f',f'',\dots$ span $ L^2[0,1]$. Then $ f$ can be approximated by their linear combinations and we are done.
b) There is a function $ \psi\in L^2[0,1]$ orthogonal to all derivatives. Let $ F\equal{}f'$. Consider $ \int_0^1 F(x\plus{}t)\psi(t)\,dt$. On one hand, we can write $ F(x\plus{}t)\equal{}\sum_{k\ge 0}\frac {x^k}{k!}F^{(k)}(t)$, which immediately implies that the integral must be $ 0$. On the other hand, we can also write $ F(x\plus{}t)\equal{}\sum_{k\ge 0}\frac {t^k}{k!}F^{(k)}(x)$, resulting in the identity $ \sum_k c_k F^{(k)}(x)\equal{}0$ with $ c_k\equal{}\frac 1{k!}\int_0^1 t^k\psi(t)\,dt$. Since $ t^k$ span $ L^2[0,1]$, not all $ c_k$ are zero and we are done again.

The more I think about this problem, the more I like it :) The right answer should, probably, be that the statement is true in every quasianalytic class and for every non-quasianalytic class there is a counterexample. But this is just a wild speculation and, honestly speaking, I don't have any more time to think of this problem today :(

Yes! You're right. In fact now I believe my statement is correct among the analytic functions. But there is still a little problem, that is about the $ C ^ \infty$ functions. To tell the truth, your counterexample seems not so persuading, or it just has something I haven't understood yet. Anyway thanks a lot.

And by the way, could you also help answer my second question in my initial post? The question is can we get an estimate about the speed of the minimal eigenvalue tending to zero? For example, it will be very nice if we can draw the conclusion that the minimal eigenvalue $ \lambda_n$ of $ A_n$ holds $ \lambda_n < \frac{1}{n}$. Of course, this is obviously not possible, and I guess the correct estimate should involve $ f ^ {(n)}$, and so this is just an example.

And at last I'd like to say your proof of the analytic functions is very useful for my current work. Thank you very much!

[quote="fedja"]Oops, don't know what happened to me :oops: : my remark about $ \frac 1{1 \plus{} x}$ is an ultimate nonsense (by the way, who can explain why?). As a matter of fact, I now believe that all analytic functions are OK. At least, if we consider the $ 1$-periodic analytic functions, we do have a positive result. The idea is to show that $ f'$ is in the linear span of the higher order derivatives. Suppose not. Then there is an $ L^2$ $ 1$-periodic function $ \psi$ such that $ \int_0^1\psi( \minus{} t) f^{(n)}(t)\,dt \equal{} 0$ for all $ n > 1$ but is not $ 0$ for $ n \equal{} 1$. But what is written is just the $ n$-th derivative of the convolution $ f*\psi$ at $ 0$. Since that convolution is analytic and has all but the first 2 derivatives $ 0$ at $ 0$, it must be a non-constant polynomial of degree $ 1$ but a polynomial of degree $ 1$ cannot be $ 1$-periodic! The question about non-periodic functions is more complicated. Let me think of it for a while :).

At least, it is clear now that the problem has hardly anything to do with linear algebra. So I'm moving it :P

Edit 1: We can also show that the statement is true for the functions analytic in a sufficiently large domain containing $ [0,1]$. Assume for instance that $ f$ is analytic in the disk of radius 3 around the origin. There are two possibilities:
a) $ f',f'',\dots$ span $ L^2[0,1]$. Then $ f$ can be approximated by their linear combinations and we are done.
b) There is a function $ \psi\in L^2[0,1]$ orthogonal to all derivatives. Let $ F \equal{} f'$. Consider $ \int_0^1 F(x \plus{} t)\psi(t)\,dt$. On one hand, we can write $ F(x \plus{} t) \equal{} \sum_{k\ge 0}\frac {x^k}{k!}F^{(k)}(t)$, which immediately implies that the integral must be $ 0$. On the other hand, we can also write $ F(x \plus{} t) \equal{} \sum_{k\ge 0}\frac {t^k}{k!}F^{(k)}(x)$, resulting in the identity $ \sum_k c_k F^{(k)}(x) \equal{} 0$ with $ c_k \equal{} \frac 1{k!}\int_0^1 t^k\psi(t)\,dt$. Since $ t^k$ span $ L^2[0,1]$, not all $ c_k$ are zero and we are done again.

The more I think about this problem, the more I like it :) The right answer should, probably, be that the statement is true in every quasianalytic class and for every non-quasianalytic class there is a counterexample. But this is just a wild speculation and, honestly speaking, I don't have any more time to think of this problem today :([/quote]


For abritary $ \alpha, \beta, x, y \in \mathbb{R}$ prove that:

$ (2 \alpha ^2 \plus{} 2 \alpha \beta \plus{} \beta ^2)(2 x ^2 \plus{} 2 xy \plus{} y ^2) \ge (2 \alpha x \plus{} \alpha y \plus{} \beta x \plus{} \beta y)^2$

yes easy but sometimes it's usful, just see $ (2 \alpha ^2 \plus{} 2 \alpha \beta \plus{} \beta ^2)(2 x ^2 \plus{} 2 xy \plus{} y ^2) \equal{} ( \alpha ^2 \plus{} ( \alpha \plus{} \beta )^2)( x ^2 \plus{} (x \plus{} y)^2)$
$ \ge (\alpha x \plus{} (\alpha \plus{} \beta)(x \plus{} y))^2$.

hello Mathias, your inequality is equivalent to $ (x\beta\minus{}y\alpha)^2\geq0$ and this is allways true for all
real $ x,y, \alpha$ and $ \beta$.
Sonnhard.

Nice solutions :)

The way I created the problem:
I chose an inner product space $ \left \langle \left ( \begin{array}{c} a_1 \\ a_2 \end{array} \right ) , \left ( \begin{array}{c} b_1 \\ b_2 \end{array} \right )  \right \rangle \equal{} 2a_1b_1 \plus{} a_1b_2 \plus{} a_2b_1 \plus{} a_2b_2$. Then choosing $ u \equal{} (\alpha, \beta)$ and $ v \equal{} (x,y)$ it just becomes $ \langle u, u \rangle ^{1/2} \langle v, v \rangle ^{1/2} \ge \langle u, v \rangle$ which is Cauchy Schwartz :)

Mathias, this simple inequality is really useful in killing tougher inequalties. I have killed one form a book using this.
Thanks anyway.


limit
real analysis
real analysis unsolved

Let sequences ${x_{n}}$ and ${y_{n}}$ be converge. Prove that
\[\lim_{n\to\infty}\frac1n\sum_{i=1}^{n}x_{i}y_{n+1-i}= \lim_{n\to\infty}x_{n}y_{n}\]

Here is an idea:

Let $z_{n}=\frac{x_{n}y_{1}+\cdots x_{1}y_{n}}{n}=\frac{(x_{1}-x)y_{n}+\cdots+(x_{n}-x)y_{1}}{n}+x\frac{(y_{1}-y)+\cdots (y_{n}-y)}{n}$

$+xy$, where $x=\lim x_{n}$ and $y=\lim y_{n}$.
Thus it is sufficient to prove that

the two sequences converge to 0.

Now, $|\frac{(x_{1}-x)y_{n}+\cdots+(x_{n}-x)y_{1}}{n}|\leq M\frac{|x_{1}-x|+\cdots+|x_{n}-x|}{n}\rightarrow 0$, where $M=\sup y_{n}$
The second sequence also converges to 0.
The result follows.


I am not sure where the appropriate place to post this is, if it's in the wrong spot then feel free to move it.

I was at a local math competition and one of the problems was something like "Find all values of X such that 9^X+3(1-2X)-4=0"

I found that X=0 and that 3^2X+3(1-2X)=4

The other answer is X=1/2, but I would like to know if there is a way in which I could've found them systematically and been sure that there were no other possibilities.

Do you mean $ 9^x \plus{} 3^{1 \minus{} 2x} \minus{} 4 \equal{} 0$?

Write as $ 9^x \plus{} \frac{3}{9^x} \minus{} 4 \equal{} 0$ and substitute $ y \equal{} 9^x$.

That is what I meant. Thanks for the answer, it works quite well


Let $ A$ be a set of real numbers has property for every distinct numbers $ a,b,c \in A$ then $ a^2\plus{}bc$ is rational numbers.Prove that there exist a positive number $ M$ such that $ a*M^{1/2}$ is rational for any $ a \in A$

This is INMO 2008 problem 3 (Though not original in that contest also).
[url=http://www.isid.ac.in/~rbb/inmosol_08.pdf]Click here[/url] for the solution.


geometry
3D geometry
sphere
search
number theory unsolved
number theory

The Unforgettable 6th June 1975

[b][color=darkred][Please don't remove these post and link. They are very important for our humanity!][/color][/b]

It was my ninth attempt to make contact.  The sky was perfectly clear.  Once again I observed the silence in the clearing.

There we have it.  They were there.  Everything was fine.

There was an agreed upon signal between us.  I knew that I could concentrate on sending my thoughts by telepathy because they had let me know that they were receiving them.  Then there was another shooting star which crossed the sky horizontally, once again a third one in the same direction.

My attention was drawn to lines of fire which stood out clearly against the night sky.  It was about half past ten at night.  I could feel a certain presence.  As observed, even studied.  I didn't take offence to their actions, even the contrary.  It was then that a small ball of light, which burst forth from the vastness of the sky, filled my heart with joy.

What a firework!

Half an hour went by without me seeing other closer signs.  The last one having still been very high in space.  Suddenly, as if to contradict me, I saw very clearly above me the shape of a space ship.  It was circular, equipped at the back with three small wings.  This block of light had just disappeared as if by magic.  In its view, I my gesturing signals but I quickly understood that it was useless.  The craft was far too far away to understand what I wanted.  However, it was responding well to my telepathic messages.  Everything was happening so high up that I too wanted to be there.  Why couldn't they land here where there was no one to disturb us?  I came to the tragic conclusion that they had no intention of contacting me but that they had shown themselves to encourage my actions.  I was angered by their attitude and I decided to break purely and simply any relationship with them.  I directed my terms to those who were high up there.

"I thank you for all that you have done and for showing yourselves to me at this time.  However, you must understand that I cannot consciously conceive your presence by such imprecise and distant apparitions.  I know that you are very good to us and that without such kindness we would have ceased to exist or would have become slaves.  But what is the use of my presence in these places if you don't do anything concrete to explain it to me.  I see my presence here as being useless.  If there is anything to be done later, do not count on me.  Therefore adieu (to God), or rather aux dieux (to the gods), to you"

I raised both my arms to the sky in a sign of peace.  Then, I prepared myself to leave the place.  That's when an extraordinary thing happened.  I had only just turned my head away when I noticed a strange orangey-red star above the hill in front of me.  However, I still hadn't totally understood this mysterious phenomenon.  My ability to understand was inhibited by its focusing effect. With great speed the star came straight at me and became a large sphere with a diameter of 3 metres.

I kept my feet voluntarily rooted to the ground because under no circumstances should I have run away from what was to happen.  The ball stopped implacably.  The object pivoted around itself.  It was now a small dome.  I hardly had enough time to notice its detail because an idiotic but very natural reflex made me protect my face with my left arm for just a moment during the approach, out of belief that it would crash into the ground and get me.  However, I had enough time to notice that there was at no point a change in colour or noise, even as little as there was.  The object was still in front of me, suspended in the air.

Quickly my understanding signalled this apparition as being caused by an extra-terrestrial and therefore conscious of what was within, I no longer remembered anything.  Then I saw again the object which was reversing several metres at a very slow pace.  Suddenly, it raised itself to an altitude of 300 metres, changed colour several times, and then left on a right angle, going horizontally before disappearing in a flash a few hundred metres away in the direction of the town.  When the object was no longer visible, there were in the sky four white trails.  Everything happened in such a fairytale way that I closed my eyes, pinched myself and slapped myself to know whether this was all real.  Therefore, only if when having opened my eyes I would have seen the trails left by the engine would I have believed in my observation.  I lowered my head, controlled the rhythm of my breathing and clenched my fists.  Then I opened my eyelids, my head straight and tilted towards the sky, I saw the impossible.  The four parallel and horizontal lines were really there.  My mental state following this wonderful encounter is impossible to describe, it was like being drunk.  For the first time in my life, I had seen while in possession of all of my faculties, a time and space belonging to the intersection of two worlds.  Mine and one from elsewhere. I had seen indisputable proof of its reality.  Who knows what happened during this encounter, I don't remember.  Was there a being inside?  All that I could know was that the ball could reappeared whenever it thought it to be a good time because while  going down the path, I saw it cross over once again high in the sky.

[img]http://www.antollma.info/astral/Colombier.jpg[/img]

Greetings,
Antoll MA
ALL THE PICTURES ON: [url=http://www.antollma.info]www.antollma.info[/url]

:rotfl:

@all moderators, admins: please don't delete it, it's cool :D :D :D

We need some fun between math ;) [and the G&FF is not funny]

I agree.Cool and funny!BTW, Lots of numbers and math terms in the story  :lol:
I have also a very funny one:
Google up for 'Polly Nomial".It's a first thing found by a search engine. :)

Even though he(or she) made an account just for this story, it is really funny! :rotfl:

Can you send me some pictures of the inside of the spaceship if you can?


LetABC  with $ A \leq B \leq 60^0 \leq C \leq90^0$
FIND MIN:P=cosA+cosB+cosC

hello, the searched minimum ist $ P_{Min}\equal{}1$ for $ A\equal{}B\equal{}\frac{\pi}{3}$ and $ C\equal{}\frac{\pi}{2}$.
Sonnhard.


Let  from $ f : R\to R$ such that $ f(x)\equal{}x^{3}\minus{}3x^{2}\plus{}3x$ .Prove the existance of $ f^{\minus{}1}$  and find $ f^{\minus{}1}(x)$

Well nice solution [b]gokhankececi[/b], a long way for proving the existence of $ f^{\minus{}1}$ we must prove that for all $ x_{1},x_{2}\in\mathbb{R}$,

$ f\left(x_{1}\right)\equal{}f\left(x_{2}\right)\Longrightarrow x_{1}\equal{}x_{2}$

$ x^{3}_{1}\plus{}3x^{2}_{1}\plus{}3x_{1}\equal{}x^{3}_{2}\plus{}3x^{2}_{2}\plus{}3x_{2}\Longrightarrow$

$ \left(x_{1}\minus{}x_{2}\right)\left(x^{2}_{1}\plus{}x_{1}x_{2}\plus{}x^{2}_{2}\right)\plus{}3\left(x_{1}\minus{}x_{2}\right)\left(x_{1}\plus{}x_{2}\right)\plus{}3\left(x_{1}\minus{}x_{2}\right)\equal{}0$

$ \Longrightarrow\left(x_{1}\minus{}x_{2}\right)\left(x^{2}_{1}\plus{}x_{1}x_{2}\plus{}x^{2}_{2}\plus{}3x_{1}\plus{}3x_{2}\plus{}3\right)\equal{}0$

Now we have $ x_{1}\minus{}x_{2}\equal{}0$ which gives $ x_{1}\equal{}x_{2}$ or we must have $ x^{2}_{1}\plus{}x_{1}x_{2}\plus{}x^{2}_{2}\plus{}3x_{1}\plus{}3x_{2}\plus{}3\equal{}0$

we assume this second equation as a degree$ 2$ polynomial with $ x_{2}$ its variable and $ x_{1}$ a constant number.

Now $ \Delta\equal{}\minus{}3\left(x_{1}\plus{}1\right)\leq0$ here for $ \Delta<0$ we have no solution and we have $ \Delta\equal{}0$ only if $ x_{1}\equal{}\minus{}1$ then we have $ x^{2}_{1}\plus{}x_{1}x_{2}\plus{}x^{2}_{2}\plus{}3x_{1}\plus{}3x_{2}\plus{}3\equal{}x^{2}_{2}\plus{}2x_{2}\plus{}1\equal{}0$ then $ x_{2}\equal{}\minus{}1$

which is the same reasult as $ x_{1}\equal{}x_{2}$ and Done! :)

this was the general way for roving the existence of $ f^{\minus{}1}$, actually it is not a nice solution for this function! solution given by [b]gokhankececi[/b] is really nice here! :)


In rectangle $ABCD$, angle $C$ is trisected by $\overline{CF}$ and $\overline{CE}$, where $E$ is on $\overline{AB}$, $F$ is on $\overline{AD}$, $BE = 6,$ and $AF = 2$. Which of the following is closest to the area of the rectangle $ABCD$?
[asy]
size(140);
pair A, B, C, D, E, F, X, Y;
real length = 12.5;
real width = 10;
A = origin;
B = (length, 0);
C = (length, width);
D = (0, width);
X = rotate(330, C)*B;
E = extension(C, X, A, B);
Y = rotate(30, C)*D;
F = extension(C, Y, A, D);
draw(E--C--F);
label("$2$", A--F, dir(180));
label("$6$", E--B, dir(270));

draw(A--B--C--D--cycle);
dot(A);dot(B);dot(C);dot(D);dot(E);dot(F);
label("$A$", A, dir(225));
label("$B$", B, dir(315));
label("$C$", C, dir(45));
label("$D$", D, dir(135));
label("$E$", E, dir(270));
label("$F$", F, dir(180));
[/asy]
$\textbf{(A)} \ 110 \qquad \textbf{(B)} \ 120 \qquad \textbf{(C)} \ 130 \qquad \textbf{(D)} \ 140 \qquad \textbf{(E)} \ 150$

[hide]

Trisecting $\angle BCE$ gives two $30-60-90$ triangles and one quadrilateral.  Looking at $\triangle ABC$, we see that $BC$ is $6\sqrt3$ and thus, $FD$ is $6\sqrt3 -2$.

By $\triangle FDC$, $DC$ is $\sqrt 3(6\sqrt3 - 2)$.

The dimensions of rectangle $ABCD$ is $6\sqrt3$ by $\sqrt 3(6\sqrt3 - 2)$ and the area is closest to $E) 150$[/hide]

Here's the diagram that was included with this problem, not to scale:
[asy]
size(150);
pair A, B, C, D, E, F, X, Y;
real length = 12.5;
real width = 10;
A = origin;
B = (length, 0);
C = (length, width);
D = (0, width);
X = rotate(330, C)*B;
E = extension(C, X, A, B);
Y = rotate(30, C)*D;
F = extension(C, Y, A, D);
draw(E--C--F);
label("$2$", A--F, dir(180));
label("$6$", E--B, dir(270));

draw(A--B--C--D--cycle);
dot(A);dot(B);dot(C);dot(D);dot(E);dot(F);
label("$A$", A, dir(225));
label("$B$", B, dir(315));
label("$C$", C, dir(45));
label("$D$", D, dir(135));
label("$E$", E, dir(270));
label("$F$", F, dir(180));
[/asy]
This is for the purpose of keeping the Contest Collections up-to-date with the diagrams that came with the original contest.


geometry
angle bisector
algebra solved
algebra

Prove sin 3 = (1+ \sqrt 5)* \sqrt (2+ \sqrt 3) + (1- \sqrt 3)* \sqrt (5+ \sqrt 5).

This is really strange. I wonder how on earth did someone find this value.

Well, I think it's just an application of the fact that 3 = 18 - 15,
and of course the values of sin(15), cos(15), sin(18) and cos(18) are well-known.

Boy, I'm really idiot. You're right, Arne, it's not so strange after all.  :blush:

Hi Arne,

I did not know that you know all these values. Otherwise it would not have been so easy...  :D

We learnt about the values of sin(15) and cos(15) at school.

The values of sin(18) and cos(18) were derived in our IMO training last year,
as a simple application of complex numbers.
However, there is also a clever geometric approach to find these values.

So please enlighten us regarding the geometric approach.  :)

Let ABC be a triangle such that <A = <B = 72 and <C = 36.
Let D be the intersection of AC and the angle bisector of <B.
Then <CBD = 36 = <BCD so triangle CBD is isosceles.
Also, <ADB = 72 = <BAD so triangle BAD is isosceles.
Hence CD = BD = AB.
The angle bisector theorem gives: CD/AD = BC/AB or AD/AB = CD/BC.
Now we have BC = 2CD sin(54) and AD = 2AB sin(18).
Hence: 2 sin(18) = 1/(2 sin(54)). So 4 sin(18) cos(36) = 1.
Since cos(36) = 1 - 2 sin2(18) it's easy from here.


geometry
3D geometry
sphere
induction
geometry solved

There are given mn + 1 points such that among any m + 1 of them there are two within distance 1 from each other. Prove that there exists a sphere of radius 1 containing at least n + 1 points.

It is an easy one :
Suppose, for a contradiction, that no sphere with radius 1 does contain more than n points.
Let M_1,...,M_(mn+1) be the points, and for each i, let S_i be the sphere with center M_i and radius 1.
Thus, S_1 contains non more than n points. It follows that there is a point which does not belongs to S_1. Wlog, say M_2. Then M_1M_2 > 1.
Since S_1 \cup S_2 does not contains more than 2n of the points, it follows that, if m > 1, there exists a point, say M_3 which is not in S_1 \cup S_2.
Note that, it follows that M_1M_3 > 1 and M_2M_3 > 1.
And so on, we construct by induction a sequence M_1,...M_m,M_(m+1) such that M_iM_j > 1 for each i < j. A contradiction.

Pierre.


A sequence is defined as follows:

$ t_1\equal{}3^8$
$ t_2\equal{}1000$

$ t_n \equal{} \frac{1\plus{}t_{n\minus{}1}}{t_{n\minus{}2}}$

What's the sum of the numerator and the denominator of $ t_{2008}$?

[quote="Gabuntu94"]A sequence is defined as follows:

$ t_1 \equal{} 3^8$
$ t_2 \equal{} 1000$

$ t_n \equal{} \frac {1 \plus{} t_{n \minus{} 1}}{t_{n \minus{} 2}}$

What's the sum of the numerator and the denominator of $ t_{2008}$?[/quote]

$ t_3\equal{}\frac{1\plus{}t_2}{t_1}$

$ t_4\equal{}\frac{1\plus{}t_3}{t_2}$ $ \equal{}\frac{1\plus{}\frac{1\plus{}t_2}{t_1}}{t_2}$ $ \equal{}\frac{1\plus{}t_1\plus{}t_2}{t_1t_2}$

$ t_5\equal{}\frac{1\plus{}t_4}{t_3}$ $ \equal{}\frac{1\plus{}\frac{1\plus{}t_1\plus{}t_2}{t_1t_2}}{\frac{1\plus{}t_2}{t_1}}$ $ \equal{}\frac{1\plus{}t_1\plus{}t_2\plus{}t_1t_2}{t_2(1\plus{}t_2)}$ $ \equal{}\frac{1\plus{}t_1}{t_2}$

$ t_6\equal{}\frac{1\plus{}t_5}{t_4}$ $ \equal{}\frac{1\plus{}\frac{1\plus{}t_1}{t_2}}{\frac{1\plus{}t_1\plus{}t_2}{t_1t_2}}$ $ \equal{}t_1$

$ t_7\equal{}\frac{1\plus{}t_6}{t_5}$ $ \equal{}\frac{1\plus{}t_1}{\frac{1\plus{}t_1}{t_2}}$ $ \equal{}t_2$

And so $ t_{n\plus{}5}\equal{}t_n$ $ \forall n$ and so $ t_{2008}\equal{}t_3\equal{}\frac{1\plus{}t_2}{t_1}$ $ \equal{}\frac{1001}{6561}$

Hence the result : $ \boxed{7562}$

Wow!  :10:

Thank you very much pco! I probably got something wrong with the calculations :ewpu:


because $(x-a)(x-b)=0\Longrightarrow x=a\ \ \text{or}\ \ x=b$ :wink:

[quote="bos1234"]$\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+....}}}}}$ = 3
$x+\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+....}}}}$ = 9
$x+3 = 9$
$x=6$

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

$3= 9-x$
$\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+....}}}}}$ = $(9-x)^{2}$
$x+\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+....}}}}}$ = $(9-x)^{2}$
$x+3 = 81-18x+x^{2}$
$x^{2}-19x-78=0$
$(x-13)(x-6) = 0$
$x = 13 or 6$

Why doesn't 13 work if you sub into original expression ?? :S[/quote]

It's called an [b]extraneous[/b] root.

Try to solve this equation: $x-\sqrt{x}= 6$. You should get $9$ and $4$ as solutions, but $4$ doesn't work. This sometimes happens when you square both sides of an equation.

[quote="bos1234"]$\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+....}}}}}$ = 3
$x+\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+....}}}}$ = 9
$x+3 = 9$
$x=6$

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

$3= 9-x$
$\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+....}}}}}$ = $(9-x)^{2}$
$x+\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+....}}}}}$ = $(9-x)^{2}$
$x+3 = 81-18x+x^{2}$
$x^{2}-19x-78=0$
$(x-13)(x-6) = 0$
$x = 13 or 6$

Why doesn't 13 work if you sub into original expression ?? :S[/quote]
Because $\sqrt{13}>3$
So obviously it doesn't work.
When you raise both sides of an equation to a power, often times extra roots are created.
For example, $x=-2 \rightarrow x^{2}=4$, but x=2 is false.

formally, extraneous roots exist because $f(x) = x^{2}$ is a non-bijective mapping from $\mathbb{R}$ to $\mathbb{R}^{+}$.  also, $f(x)x^{2k}$ for $k$ a positive integer also has this property.  But if $f(x) = x^{2k+1}$ for $k$ a positive integer, then this is a bijective mapping from $\mathbb{R}$ to $\mathbb{R}$.


function
calculus
derivative
real analysis
real analysis unsolved

The continuous function and twice differentiable function $f: \mathbb{R}\rightarrow\mathbb{R}$ satisfies $2007^{2}\cdot f(x)+f''(x)=0$. Prove that there exist two such real numbers $k$ and $l$ such that $f(x)=l\cdot\sin(2007x)+k\cdot\cos(2007x)$.

Just a second order differential equation...
 :)

refer this

[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=136624[/url]

That's a terrible contest problem. It's trivial to anyone with appropriate advanced material (uniqueness theorems for differential equations) and impenetrable to anyone else.

Terrible question indeed. ODE's of this exact form can be classified neatly without any DE theory, though. Let $u$ and $v$ be solutions to $y''+\alpha y=0$. Then there exists a real number number $k$ such that $u'v-uv'=k$. (Compute the derivative of $u'v-uv'$). Now given two solutions $u, v$ such that the Wronskian (let's be honest) above is not zero, it is easily shown by elementary means (use the above) that any solution of the equation is a linear combination of $u$ and $v$.

Outrageous problem indeed!  :lol:
However, I was thinking about some particular functions, maybe : \[g(x)=\frac{f^{\prime}(x)\cos(2007x)}{2007}+f(x)\sin(2007x).\]
Then, taking some derivatives  :maybe: stuff like that...
Maybe it's not that bad after all; what do you think guys? :)


An integer n is such that the tens digit of a^2  is odd. What is the units digit of a^2?

[hide=" answer"]

6

[/hide]

[quote="AndrewTom"]An integer n is such that the tens digit of a^2  is odd. What is the units digit of a^2?[/quote]

Consider $ \ a \equal{} 10m \plus{} n,m\in\mathbb N_0,\ n\in\{\ 0,\ 1,\ 2,\ 3,\ 4,\ 5,\ 6,\ 7,\ 8,\ 9,\}$

Now $ \ a^2 \equal{} 100n^2 \plus{} 20mn \plus{} n^2$

Note tens of $ \ 20mn$ will be even$ \ \ \implies n^2$ must give odd carryover to make tens of $ \ a^2$ odd.

$ \implies n^2 \equal{} 16,36$

$ \implies$ Units digit of $ \ a^2 \equal{} 6$

The units digit can be 0,4, or 6

The tens digit can be 1,3,5,7 or 9. Notice that if the units digit is 6, the number is of the form 4k in all cases whereas the same isnt true if the last digit is 0 or 4 (eg. take the last two digits as 34. then the number is of the form 4k+2 which cannot be a perfect square).

Hence the unit's digit is 6


The website of it :

http://www.math.su.se/bw2005/problems.shtml

Solutions


inequalities
triangle inequality
inequalities unsolved

how do u prove this
using the triangle inequality,
that 1) |A| -|B| <= |A-B|
2) max (a,b) = (|a +b| + |a-b|) /2
3) min (a,b) = (|a +b| - |a-b|) /2

asap plz

Don't post the same problem in multiple forums.  We're not here to do your homework for you.

1) Is $\vert a\vert\leq \vert a-b\vert +\vert b\vert$ true?

2,3) Consider the cases $a>b$, $b>a$ separately.

This, good pussy invader, is not a site for homework help.

[quote="blahblahblah"]yes, I decided to look the other way with respect to the name.[/quote]

Why?  It's so....brilliant?  Classy?  Charming?  I can't really think of enough nice words to describe it.

Someone should really ip ban this guy.

Edit: The man changed his name.


function
trigonometry
topology
real analysis
real analysis solved

Show that $f:R->R$ is continous if and only if $f(I)$ is an interval for every  interval $I$ in $R$  and $f$<sup>-1</sup>$(x)$ is a closed set$($Every limit of a sequence belongs to the set$)$.

It seems to me like $f$ can take open intervals to closed intervals as well :?.

:D  Really Killer ! Very interesting these new mathematics !! :D

Actually, I think this might have something to do with: $f$ is coutinuous iff $f^{-1}(U)$ is open for any open set $U$ of $\mathbb R$. This is actually used as the definition of a continuous function from $X$ to $Y$ (topological spaces in general).

I did not find a mistake till now, so here it goes:
   First, I will show that f takes intervals into intervals. It suffices to show that f(a) and f(b) are in the closure of $ f((a,b))$ for all a and b. Suppose we could find a and b such that this does not happen and assume for exampple (all other possibilities are similarly discussed) that $ f((a,b))=(m,M)$ and $ f(b)<m-r$ for a certain $ r>0$. Take any number $ c>b$ and write $ f((a,c))=f((a,b))|{f(b)}| f((b,c))$, where | means union. Since this union is an open interval, it is clear that $ m-r^$ must be in $ f((b,c))$. Thus making c closer and closer arround b we obtain a sequence $ c_n->b$ such that $m-r$ is in $ f((b,c_n))$. Write $ m-r=f(x_n)$ and deduce that $x_n$ is in the preimage of $ m-r$ for all r. Since this is closed, b is also in the preimage and thus $ f(b)=m-r$ false. Thus we have proved that f takes intervals to intervals and so it has Darboux property.
   Assume now that f is not continuous in $x_0$ thus there exists a sequence $ c_n$ convergent to $x_0$ and a $eps>0$ such that $|f(c_n)-f(x_0)|>eps$ for all n. Working with a subsequenceallows us to assume that $ f(c_n)>f(x_0)+eps$ for all n and thus by Darboux property there exists $y_n$  between $x_0$ and $c_n$ such that $ f(x_0)+eps=f(y_n)$. It follows easily that the limit of $y_n$ which is $x_0$ is also in the preimage of $f(x_0)+eps$ and thus $eps=0$, false. Hope it's correct.

Here's why I think there's a mistake (I didn't read your solution, Harazi, sorry :)).

One of the implications looks like this:

$f$ is continuous $\Rightarrow$ $f(I)$ is an open interval for every open interval $I\subset R$ [b]and[/b] blabla... Define $f(x)=\sin x$. We have $f((0,2\pi))=[-1,1]$, so already, one of the sentences in the conjunction is false, meaning that the conjunction is false.

Am I missing something?

Indeed, I did not discuss that implication simply because it is false!


modular arithmetic
number theory
relatively prime

m, n are positive integer where (m,n)=1, and {$ r_1, r_2, \cdots r_\phi n$},{$ s_1, s_2, \cdots s_\phi m$} are reduced residue systems.

Show that for i, j where $ 1 \le {i} \le {\phi {n}}$ and $ 1\le {j} \le {\phi {m}}$, show that there is only one x in mod mn $ x \equiv r_i \pmod{m}$  $ x \equiv s_j \pmod{n}$ gcd(x,mn)=1.

Yes, we can use the Chinese Remainder Theorem. By the Chinese Remainder Theorem, the system

$ x \equiv r_i \mod{m}$

$ x \equiv s_i \mod{n}$

has exactly one solution $ \pmod{mn}$. Because $ x$ is relatively prime to $ m$ and $ n$, it is relatively prime to $ mn$.


hi guys

any of you taking grade 12 data management?

how do you study for it? everyone in my teacher's class did bad in final exam last year, bascially everyone's mark dropped by 5% after final....

so how do you guys study for it? our tests are normal but exam is dead heard=(

any suggestions please?

thanks a lot!

as far as i know, data is one of the easiest gr.12 maths. and to review for the exam, just read your textbook, and you should be fine.


function
complex numbers
complex analysis
complex analysis unsolved

Find all the entire functions $f(z)$ such that $(Re f)^3+(Im f)^3=1$. Of course it would be nice to do it non-using Little Picard Theorem

It follows just from a judicious application of the Cauchy-Riemann equations, does it not?  Let $f(z) = f(x, y) = u(x, y) + iv(x, y)$.  Then we have $u^3 + v^3 = 1$ so $u^2 \frac{\delta u}{\delta x} + v^2 \frac{\delta v}{\delta x} = 0 = u^2 \frac{\delta u}{\delta y} + v^2 \frac{\delta v}{\delta y}$.  But since $\frac{\delta u}{\delta x} = \frac{\delta v}{\delta y}$ and $\frac{\delta u}{\delta y} = - \frac{\delta v}{\delta x}$, this gives us $u^2 \frac{\delta v}{\delta y} + v^2 \frac{\delta v}{\delta x} = 0$ and $-u^2 \frac{\delta v}{\delta x} + v^2 \frac{\delta v}{\delta y} = 0$.  However, multiplying each of these equations by the proper partial of $v$ and adding gives $((\frac{\delta v}{\delta x})^2 + (\frac{\delta v}{\delta y})^2)v^2 = 0$ from which we immediately get that $f$ is constant.


[quote="Nbach"]A $ 2n\times 2n$ chessboard is filled by  $ 4n^{2}$ real umbers with sum equal to zero.We know that absulute value of each of these numbers is not  grater than $ 1$ .Prove that absulute value of all the numbers from one column or a row doesn't exceed $ n$[/quote]

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=41119

See also the solution of C4 at

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=15622

[EDIT: Thanks to Mathmanman for the following link:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=54037 ]

  Darij


Through $B: ( 0, -b )$ draw a chord of $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ ($a>b>0$). Find the MAXIMUM of the length of the chord.

Editted.

[hide]$(0,-b)$ is on the graph. There is no definate answer, since chords cannot have a 0 length.[/hide]

[hide]Erm... not sure if that's possible, because the answer approaches 0 (and 0 can't be the length of a chord, as it's a point)...[/hide]

[quote="Farenhajt"]Or perhaps it should be MAXIMUM length...[/quote]
I can't believe that I made this mistake......Yes, MAXIMUM.

[quote="Farenhajt"][quote="paladin8"][hide]
Well...

We want to minimize $x^2+(y-b)^2$ given that $\frac{x^2}{a^2}+\frac{y^2}{b^2} = 1$.

Solve for $x^2$ in the second equation and substitute into the first.

We now have $a^2\left(1-\frac{y^2}{b^2}\right)+(y-b)^2 = \left(1-\frac{a^2}{b^2}\right)y^2-2by+(a^2+b^2)$.

Complete the square to get $\left(1-\frac{a^2}{b^2}\right)\left(y-\frac{b}{1-\frac{a^2}{b^2}}\right)^2+\frac{a^4}{a^2-b^2}$.

But the square term is negative (since $a > b$), so the maximum chord length occurs at $\frac{a^2}{\sqrt{a^2-b^2}}$.[/hide][/quote]

That seems fine, but when you substitute the obtained value for $y$ in the ellipse equation, you get $x^2=\frac{a^4(a^2-2b^2)}{(a^2-b^2)^2}$, which obviously must be discussed for sign, which means that the $y$ value might be outside of the ellipse.[/quote]

Hmm, you're right. If the condition were changed to, say, $a > b\sqrt{2}$, we'd be ok :D.

Well, your expression is $d^2=\frac{a^4}{a^2-b^2}-\frac{a^2-b^2}{b^2}\left(y+\frac{b^3}{a^2-b^2}\right)^2$, and obviously we must minimize $\left|y+\frac{b^3}{a^2-b^2}\right|$ when $y$ runs over $[-b,b].$ It's easily seen that the minimal value need not be zero, and if it's not, $y$ must be $-b$, hence it's either your solution if $a\geq b\sqrt{2}$, or $2b$ if $b\leq a\leq b\sqrt{2}.$

(Another note: Your final expression isn't even defined for a circular ellipse, in which case we know the solution must be the diameter :))


geometry
inequalities
email
geometry solved

Yes, it is a circle with $r\rightarrow \infty$

Let $d$ be a fixed line and two fixed points $A\in d$, $B\not\in d$, $AB\perp d$. For any mobile point $M$ which belongs to the ray $[AB$, we consider a circle $w(x)=C(M,x)$, where $x=AM$ (the circle $w(x)$ is always tangent to the line $d$ in the point $A$ and its center $M$ belongs always to the same semiplane $(d,B)$). If $x\rightarrow \infty$, then the circle $w(x)\rightarrow$ the line $d$ (the circle with the infinite radius). If $x\searrow 0$, then the circle $w(x)\rightarrow$ the point $A$ (the circle with the null radius).

yes,but i think point is not a circle actualy and in fact they are different.

[quote="Virgil Nicula"]I don't say you see a line like a circle, I say you feel it ![/quote]
very nice feeling.

[quote="shobber"]Why this kind of question is posted here? I see that the Geometry Open Questions is now full of that.

In my opinion, it belongs to the Other Problem Solving Section.[/quote]

It belongs into Geometry Theorems & Formulas, in my opinion. Anyway, I gave a rather detailed answer on this question in http://www.mathlinks.ro/Forum/viewtopic.php?t=65375 .

  Darij

jensen - please correct your email immediately.  You are choosing to be notified when topics are replied to, but your email is bad, so we are getting swamped with bounced emails.  I have PM'ed you 2 weeks ago about this - please correct it soon or we'll have to limit your access to the forum.


Find all x such that $ 2^x$ starts with x.

($ 2^{10}$ is of course a solution
Are there any other solutions??)


I know that this has been asked before, but I can't find the answer. Page numbers are by solution, not round, right?

That's correct - for example, if you use LaTeX, the template file always numbers by solution, i.e. "page one of problem 2", etc.


A bill, after being proposed by a member of Congress, must be passed (meaning it gets majority vote) in both the H.o.R. (House of Representatives) and the Senate before being sent to the President, who can either sign or veto the bill (a.k.a. approve/disapprove the bill). Let's assume that for a certain bill, the Senate has already voted and passed it 51-49. There are 435 members in the House of Representatives and only one president, and each member/president has an equal chance of voting for/against the bill. Find the probability that the President signs (approves) the bill, and it is also passed in the House of Representatives.

Note: There are no overrides, because the Senate only passed 51-49
Note 2: Don't come up with little details that could affect the problem (such as Lobbyists, or whatever.)

If it is unclear, just tell me. I originally made this problem for a Mock MathCounts Test, but I decided that I'm not actually gonna use the test. :ninja:

[hide]Use symmetry[/hide]

To be passed, it needs over 1/2 of the vote right?

There's a $ \frac{1}{2}$ chance that it passes through the House of Representatives (because the probability that it passes and doesn't pass are equal, and it cannot tie).  Then there is a $ \frac{1}{2}$ chance that the president approves it, so the total probability is $ \frac{1}{2} \times \frac{1}{2}\equal{}\boxed{\frac{1}{4}}$.

Dang. You got it. The original purpose of the problem was to intimidate people into skipping the problem.


[quote="SUPERMAN2"]Prove that if a,b,c≥0,then $ (a^3\plus{}b^3\plus{}c^3\plus{}3abc)^2$≥$ 2(ab\plus{}bc\plus{}ca)(ab(a^2\plus{}b^2)\plus{}bc(b^2\plus{}c^2)\plus{}ca(c^2\plus{}a^2))$[/quote]
Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, $(a^3\plus{}b^3\plus{}c^3\plus{}3abc)^2\geq2(ab+ac+bc)\sum_{cyc}(a^3b+a^3c)\Leftrightarrow$
$\Leftrightarrow(27u^3-27uv^2+6w^3)^2\geq6v^2(27u^2v^2-18v^4-3uv^2)\Leftrightarrow f(w^3)\geq0$,
where $f$ is an increasing function.
Hence, by [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=55&t=278791]uvw[/url] it's enough to check two cases:
1. $b=1$, $c=0$, which gives $a^6-2a^4+2a^3-2a^2+1\geq0$, which is obvious.
2. $b=c=1$, which gives $a^2(a^2-1)^2\geq0$. Done!


function
limit
real analysis
real analysis unsolved

[b]Theorem:[/b] Let $ f_n$ be cauchy in measure the there is a sub-sequence, $ f_{n_k}$ and a function f such that $ f_{n_k} \rightarrow f$ almost everywhere.

[b]Proof:[/b] Because $ f_n$ is cauchy in measure we can define a subsequnce $ f_{n_k} \equal{} g_k$ such that $ m\{ x : |g_{k \plus{} 1} \minus{} g_k| \geq 2^{ \minus{} k} \} < 2^{ \minus{} k}$. (So, if we take $ g_3$ the set of points where $ g_3$ fails to be within 1/8 of the rest of the functions after it has a measure smaller than 1/8.) Next, define some sets $ E_k \equal{} \{ x : |g_{k \plus{} 1} \minus{} g_k| \geq 2^{ \minus{} k} \}$. Then let $ F_k \equal{} \bigcup_{j \geq k} E_j$. ($ F_k$ is a set of values of x where $ g_k$ fails to be close enough to the function after $ g_k$ in the sequence.) $ m(F_k) \leq \sum_{j \geq k}^{\infty} m(E_j) < 2^{ \minus{} k \plus{} 1}$. (The measure of each of these $ F_k$ sets is smaller than the sum of the geometric series. It depends on k and we can make it as small as we want by making k larger.)

Next pick i, j with $ i \geq j \geq k$. Now [b]if x is not in [/b]$ F_k$ we have:

$ |g_i(x) \minus{} g_j(x)| < 2^{ \minus{} j \plus{} 1}$

(It is important that we are looking at g(x) since we are working on proving point-wise almost everywhere convergence.)

Next observe $ m(F) \equal{} m(\bigcap_{k \equal{} 1}^{\infty} F_k) \equal{} 0$ and when x is not in F $ g_k(x)$ is a Cauchy sequence.  Define f = lim $ g_i(x)$ when x is in F and 0 otherwise.  Now f is the function that the sub sequences converses to a.e.

OK. When I started typing this I had a questions about the proof (I've been looking at it for hours... ) but I think I almost get it now. We are proving point-wise a.e. so at some point in this proof we have fixed x... OK. [b]Why is it important that we have fixed x? [/b]

*sigh* at any rate...

------------------
Here is the open question:

Given $ f_n \rightarrow f$ and $ f_n \rightarrow g$ in measure. Prove f=g a.e.

It is not that $ x$ is fixed that it is important but that $ x\notin F_k$. You want to prove that $ g_k$ converges a.e. To this end, it suffices to explicitly produce a set $ F$ of measure $ 0$ such that $ g_k(x)$ converge for every $ x\notin F$. The idea is to define $ F\equal{}\bigcap_{k\ge 1} F_k$. In order to show that this set works, one needs to take an arbitrary point outside this set and to check convergence there. The first step is to note that if $ x\notin F$, then $ x\notin F_k$ for some $ k$. Then we show that for such $ x$ the sequence $ g_j(x)$ is Cauchy, etc.

Note that we've proved much more than we asked for, namely, we proved that $ g_k\to g$ uniformly outside each $ F_k$.

[quote="thefuturebird"][
Here is the open question:

Given $ f_n \rightarrow f$ and $ f_n \rightarrow g$ in measure. Prove f=g a.e.[/quote]

Is it an open question really?  :huh:
see, e.g. [url=http://books.google.pl/books?id=bfgSwxff_NwC&pg=PA121&lpg=PA121&dq=convergence+in+measure+uniqueness&source=bl&ots=5zi6JCgOHr&sig=SbI75pYRi1xha8J5benT123rYD4&hl=pl&ei=5MA_S5K1DoXp-Qb-jvyuCg&sa=X&oi=book_result&ct=result&resnum=8&ved=0CD8Q6AEwBw#v=onepage&q=convergence%20in%20measure%20uniqueness&f=false]th.1 on p. 121[/url]

The uniqueness of the limit is always proved in the same way. The first two words are "Suppose not". The next step is to give the negation of equality $ f\ne g$ a quantitative form. In the case of equality a.e., the first substep here is to say that the set ${ E=\{x: f(x)\ne g(x)}$ has measure $ \mu>0$. Unfortunately, this is not fully quantitative yet because the condition $ f(x)\ne g(x)$ is purely qualitative. We would like to replace it with $ |f(x)-g(x)|>\delta$ with some $ \delta>0$ independent on $ x$. We cannot do it on the entire set $ E$ but if we [i]define[/i] $ E_\delta=\{x: |f(x)-g(x)|>\delta\}$, then we'll get a monotone sequence of sets (the smaller $ \delta$ is, the larger $ E_\delta$ is) whose union is $ E$. Thus, $ \lim_{\delta\to 0+}|E_\delta|=|E|=\mu$, so $ |E_\delta|>\frac\mu 2$ for some $ \delta$. Now we are almost done. Just write
\[ \frac \mu 2<|E_\delta|=|\{|f-g|>\delta\}|\le |\{|f-f_n|>\tfrac \delta 2\}|+|\{|g-f_n|>\tfrac\delta 2\}|\]
and let $ n\to\infty$ to get a contradiction.

Thanks so much!

"the smaller $ \delta$ is, the larger $ E_\delta$ is"

...but in fact it will turn out to be 0 we are supposing otherwise for the proof.


linear algebra
matrix
superior algebra
superior algebra solved

Let $I=(1,2,..n)$and $(A_i)_{1\le  i \le   {2^n-1}}$ the nonempty subsets of $I$ ordered randomly.Let $B\in M_{2^n-1}(R)$ the matrix defined by $b_{ij}=1$ when $A_i \cap  A_j \neq \emptyset $ and $b_{ij}=0$ otherwise.Show that $det(B)$ does not depend on the chosen order and find it.

Strange.I never thought it comes from there.


FTW
rate problems
Pythagorean Theorem
geometry

[quote="GameBot"]Kelly drove north for 12 miles and then east for 9 miles at an average rate of 42 miles per hour to arrive at the town of Prime. Brenda left from the same location, at the same time, and drove along a straight road to Prime at an average rate of 45 miles per hour. How many minutes earlier than Kelly did Brenda arrive?[/quote]

Kelly drove a total of 21 miles, so it took her 30 minutes to get to Prime.

If Brenda left from the same location but drove along a straight road, that means she drove the hypotenuse.

 [geogebra]48d1e123a3a3672aae3133e1e2de0dee6a4176b4[/geogebra]

(The legs have length 12 and 9.)

By the Pythagorean Theorem, Brenda drove $ \sqrt{12^2 \plus{} 9^2} \equal{} 15$ miles, so it took her $ 33\frac{1}{3}$ minutes.

Thus Kelly arrived $ \boxed{3\frac{1}{3}}$ minutes earlier.

FTW says that the answer is 10, and the problem asks how much earlier Brenda arrived, not Kelly.

Solution: [hide]It took Brenda $\frac{15}{45}$ of an hour, or $ 20 $ minutes. Since it took Kelly $ 30 $ minutes, $ 30 - 20 = 10 $.[/hide]


logarithms
real analysis
real analysis unsolved

Hey, here's a differential equation that I didn't know how to solve.  It 'looks' easy, I bet some of you have seen it before:
$\frac{d^2y}{dx^2}=A+\frac{B}{y}-Cy$
A,B,C are positive but I'd be interested in if they were negative, too.

It's not simple.

This belongs to the following family of equations: $y''=f(y).$ A physical interpretation: an object moves under the influence of a force which depends only on position. (In one dimension, that has to be a conservative force, at least for smooth $f.$) We can solve such an equation using a physical insight based on the conservation of energy.

Use the symbol $v$ for $y'=\frac{dy}{dx}.$ Find a primitive for $f.$ That is, find $F(y)$ such that $F'(y)=f(y).$

Then define $u(x)=\frac12v^2-F(y).$ Then

$\frac{du}{dx}=v\frac{dv}{dx}-f(y)\frac{dy}{dx} - v\left(\frac{d^2y}{dx^2}-f(y)\right)=0.$

Thus $u$ is constant and we can write $\frac12v^2=F(y)+K$ or $\frac{dy}{dx}=\pm \sqrt{2F(y)+K}$ for some constant $K.$

That is an autonomous first order equation, hence separable. But it doesn't have to be pretty.

In this particular case, we get:

$\frac{dy}{dx}=\pm\sqrt{2Ay+2B\ln y-Cy^2+K}.$

And it's not pretty.


How do you basically find how many real roots are present in a cubic or fourth degree equation if the coefficients are too large enough for succesive division method and descartes sign rule only gives how many min or max real or positive ex are possible how do you find exactly how many are possible??? :?:  :maybe:  :ninja: is there any method by calculus ie by taking the nth differential and by doing something or so????  :)

For cubic [url=http://en.wikipedia.org/wiki/Cardano%27s_method]this[/url] is the method
For a Quartic [url=http://en.wikipedia.org/wiki/Quartic_equation]this[/url] is the method
For Quitic and hence Galois proved that the roots cannot be expressed as combination of cofficients and simple Algaebraic symbols

Youll be mad to use those eqns
And btw how did ya upsurge Rushil?  :maybe:
You didnt thank me for recommendation  :mad:

who said i didn't thank u
Manasalle nandri ye therivechinde  :D
A PM to Valentin
 Rushil $ \Rightarrow$ pardesi :wink:

And btw delete the post of Chappli abt me insultin me to the core if u r my friend new mod atleast out of gratitude  :D


algebra
polynomial
absolute value
algebra unsolved

1/ Let $P(x)$ be a cubic polynomial with rational coefficients.$q_{1},q_{2},q_{3}$... is a sequence of rationals such that $q_{n}=P(q_{n+1})$ for all positive $n$. Show that for some $k$, we have $q_{n+k}=q_{n}$ for all positive $n$.

2/$P(x)=(a+x)Q(x)$ is a real polynomial of degree $n$. The largest absolute
value of the coefficients of $P(x)$ is $h$ and the largest absolute value of $Q(x)$ is $k$. Prove that $k \leq nh$


Find n in the following equation:

n^n^n^...^n = N
(k times)

http://mathworld.wolfram.com/PowerTower.html

i think is impossible to find n in function of N

you can try inversion of series

but

this is impossible, huauh

To be clear as to what it is,

n^n^n^...^n means ((n^n)^n)^... [to a certain point] from

the order of operations, which is (n^(n^2))^... [to a certain

point].  But "(k times)" is not clear to me.  It appears the

number of exponentations is k.  If that is true, I have

x^(x^(k-1)) = N, with me not seeing any explicit solution for x

existing in terms of N as Thales418 has already stated

(thought).


inequalities
Functional Analysis
real analysis
real analysis unsolved

1) First One question

the canonical basis $ (e_{n})_{n\in \mathbb{N}}$ is NOT weakly null in $ \ell_{1}$.

Because if we suppose it is, this would imply (Schur property) that $ (e_{n})_{n\in \mathbb{N}}$ converges in norm to $ 0$, which is obviously not true. Am I right?

2) Another one.
Let $ p_{1} >
p_{2}
> \cdot \cdot \cdot > 1$ be a sequence of real numbers.
So consider the Banach space $ Z\equal{} (\bigoplus_{n\equal{}1}^{\infty} \ell_{p_{n}})_{1},$ If for all $ n\in \mathbb{N}$, $ (\overline{e}_{i,n})_{i\in \mathbb{N}}$ denote the canonical basis
for $ \ell_{p_{n}}$ AND let $ (e_{i,n})_{i\in \mathbb{N}}$ denote its natural copy in $ Z$:

\[ e_{i,n} \equal{} (\underbrace{0,\cdots,0}_{n\minus{}1},\overline{e}_{i,n},0,
\cdots) \in W.\]

How can I prove that $ e_{i,n}\rightarrow 0$ weakly, when
$ i\rightarrow \infty$.??

Im having problems because the norm on $ Z$ is the 1-norm. Although I know $ p_{n}>1$ for all $ n$.
I hope i made myself clear. Thx

Ups I just found the answer in kreyZsig book since $ p_{n} > 1$ the canonical basis converges weakly to $ 0$.


calculus
integration
real analysis
real analysis solved

f:[0,infinity)->(0, infinity), increasing, derivable, with a continuous derivative. If a_n=  \int 0 to n dx/(f(x)+f'(x)) <=1 for any n in N prove that there exists a>0 s.t.  \int 0 to n dx/(f(x)) <= a for any n in N.

I reckon that this is one of Cristinel Mortici's problems! :D

How did you know that? Maybe you have his book? (great book, by the way).
Bye! :D


calculus
integration
trigonometry
real analysis
real analysis unsolved

[hide]$I={\frac{\pi}4}$[/hide]

edit: i made a stupid mistake,i forgot to bring the 2 to the denominator

[quote="FieryHydra"]We first note that $\int_{-a}^{a}f(x)\, dx = \int_{0}^{a}\left( f(x)+f(-x) \right)\, dx.$ (I will leave this as an elementary exercise to the interested reader.)[/quote]
Thanks for the nice trick. I couldn't get started on the problem :blush:


limit
logarithms
algebra proposed
algebra

Find the limitation :\[ \lim_{n \rightarrow \infty}2\times \sqrt[3]{4}\times\sqrt[9]{16}\times...\times\sqrt[3^n]{2^{2^n}}. \]

Happy new year  :lol:  :D

Take logarithm and we want to find:
$\sum_{n=0}^\infty ln(\sqrt[3^n]{2^{2^n}}) = \sum_{n=0}^\infty \frac{2^n}{3^n}ln(2) = 3 ln(2)$
So the desired value is $2^3=8$.

In fact there is no need to take logarithm. ;)
$\lim_{n \rightarrow \infty}2\times \sqrt[3]{4}\times\sqrt[9]{16}\times...\times\sqrt[3^n]{2^{2^n}}$
$=\lim_{n \rightarrow \infty}2^1\times 2^{\frac23}\times 2^{\frac49}\times\cdots\times 2^{(\frac{2}{3})^n}$
$=\lim_{n \rightarrow \infty}2^{1+\frac23+(\frac23)^2+\cdots +(\frac23)^n}$
$=2^{\lim_{n \rightarrow \infty}1+\frac23+(\frac23)^2+\cdots +(\frac23)^n}$
$=2^3$
$=8$ :D


I thought I knew what independent and disjoint (mutually exclusive) events were until I ran across this question on an old AP Stat exam:

Which of the following statements is true for 2 events, each with probability greater than 0?

(A) If the events are mutually exclusive, they must be independent.

(B) If the events are independent, they must be mutually exclusive.

(C) If the events are not mutually exclusive, they must be independent.

(D) If the events are not independent, they must be mutually exclusive.

(E) If the events are mutually exclusive, they cannot be independent.

The correct answer is [hide](E)[/hide], but I don't really get why.

Try to prove the contrapositive of [hide](E)[/hide].

[hide]Can events A and B both be true at the same time?[/hide]

Here's the full solution, if you don't want to figure it out yourself: [hide]

A: Mutually exclusive means P(A&B) = 0. Independent means P(A&B) = P(A)P(B) > 0. X

B: Independent means P(A&B) = P(A)P(B) > 0. Mutually exclusive means P(A&B) = 0. X

C: Make event A that Bush "wins" the next election, and event B that the country falls to ruin within the next four years. They are clearly not mutually exclusive, but they are clearly not independent.

D: Make event A that Bush "wins" the next election, and event B that the country falls to ruin within the next four years. They are clearly not independent, but they are clearly not mutually exclusive.

E: Mutually exclusive means P(A&B) = 0. Independent means P(A&B) = P(A)P(B) > 0. P(A&B) cannot be both 0 and >0, so this is true.

[/hide]

Or if you want another way of thinking about it:

[hide]

Events are independent if one happening doesn't affect whether the other one happens. Being mutually exclusive means one happening means the other one cannot happen. So they can't be independent.

[/hide]


This problem is pretty easy, but it scared me at the beginning :D:

Prove that the following sentences are equivalent:

(1): There exist n odd numbers x_1, x_2,.., x_n s.t. their sum is equal to their product.

(2): n has form 4k+1.

If n=4k+1 then 1+1+1+...+1+3+(2k+1)=3(2k+1).

Suppose x1+x2+...+xn = x1x2...xn - it is an odd number ==> n is an odd number. Consider this equality (mod 4) ==> 3s+(n-s) = 3s where s is the number of xi = 3 (mod 4). So 2s+n = 3s (mod 4). If n = 3 (mod 4) then 2s+3 = 3s (mod 4) - it is wrong equality whenever s is odd or even.


Prove that: $ \sum_{i\equal{}1}^{n^2} \lfloor \frac{i}{3} \rfloor\equal{} \frac{n^2(n^2\minus{}1)}{6}$
For all $ n \in N$.

Let
\[ f(m)\equal{}\sum_{i\equal{}1}^{m}[\frac{i}{3}]\equal{}\frac{m(m\plus{}1)}{2}\minus{}\frac 13 ([\frac{m\plus{}2}{3}]\plus{}2[\frac{m\plus{}1}{3}])\equal{}\frac{m(m\minus{}1)}{6}\plus{}\theta (m),\] \[ \theta (m)\equal{}\frac 13 (m\minus{}[\frac{m\plus{}2}{3}]\minus{}2[\frac{m\plus{}1}{3}]).\]
Obviosly $ \theta (m\plus{}3)\equal{}\theta (m),\theta (0)\equal{}0\equal{}\theta (1),\theta (2)\equal{}\minus{}1$
Because $ n^2\equal{}0,1\mod 3$ we get $ \theta (n^2)\equal{}0.$

Clearly we need only prove that \[ \sum_{i\equal{}1}^{n^2}\left\{\frac{i}{3}\right\}\equal{}\frac{n^2}{3}\]
This is equivalent to showing that the number of $ i\equiv 2\mod{3},i\in[n^2]$ is equal to the number of $ i\equiv 0\mod{3}$, which is clear.

First note that if the claim holds for $n=3k$ then it holds for $n=3k+1,3k+2$. To prove the claim for $n=3k$, double count the number of lattice points in the interior of the triangle defined by $y=0, y=x/3,x=n^2$. Since the triangle has lattice vertices, we can apply Pick's Theorem: $A=0.5*n^2*(n^2/3)=n^4/6=I+B/2-1$. The LHS of the given equation is $I+B-(n^2+1)$ and it is easy to show that $B=(n^2+1)+n^2/3+(n^2-3)/3=5n^2/3$. The rest is simple algebra.


A circle with center $O$ passes through point $A$ and $C$ and intersects the sides $AB$ and $BC$ of the triangle $ABC$ at points $K$ and $N$, respectively. The circunscribed circles of the triangles $ABC$ and $KBN$ intersect at two distinct points $B$ and $M$. Prove that $\measuredangle OMB=90^\circ$

Let the circumcentre of circle(ABC) be X and the circumcentre of circle(AKN) be Y. Therefore $BY \perp BC$, $OX \perp BC$, $BX \perp KN$, $OY \perp KN$, thus $BY \parallel OX$ and $BX \parallel OY$, thus $BYOX$ is a parallelogram. So XY passes through the midpoint of BO. Since it also passes through the midpoint of BM, so $XY \parallel OM$. Since $XY \perp BM$, so $OM \perp BM$. Thus $\angle{OMB}=90^{o}$.


calculus
quadratics
geometry
geometric transformation
FTW

Err. I can crush pretty much everyone in my math class save maybe 2 people (that are AoPSers).

And Albert Einstein was involved a genius in physics not math  :P

Yeah, there definitely needs to be a bit between owning our maths group and Einstein. As in, I can take everyone in my maths class, but compared to people on here I'm a complete noob.

i totally agree.
i was so happy with my amc scores, but comparing to ppl here, they suck

[quote="BanishedTraitor"]Pawning Albert Einstein doesn't mean much seeing that he's not really a mathematician. Still.......[/quote]

Hmm... Diraic wasn't [i]really[/i] a mathematician, but I expect he could own any one on here. So could most of the people who worked on QT in the early days.

i'd say there's a big gap between 4 and 5..

i mean.. my class sucks at math... (its geometry.. D:)

and i can't pwn einsteing... i'd say i'm about 4.3 :D

4 was true when I was in school, but no longer, with the IMO bronze medallist in class, and another quite brilliant guy who for some reason never sat for olympiads.

seeing these, and the fact that I couldn't qualify for IMO, I think it's 1.

Well if u read abt Einstein youd knw that he knew advanced calculus also and even though he chose to leave his calculations to his assistants, he was very good in math so he could have pwned everyone here easily except maybe a few teachers in math in this forum  :)
If i was to go ranting like bubka, i have a fellow imotc guy in class but apart frm him i hope n think that i can pwn everyone else in olympiad math  :D

[quote="1=2"]Homeschooled, but at me math circle, I am one of the top students.

Notice the key word "one". I'm not the top student.[/quote]

That's a good thing. Being the best in your class/school/area means that you lose sight of the bigger picture

I've always sat back, until recently, satisfied with beating my classmates but now that isn't really enough

I can out-score all of the kids in my class, with some competition.

[quote="SimonM"]Being the best in your class/school/area means that you lose sight of the bigger picture[/quote]
Not really. For people who care more than just about being first, there are plenty of opportunities to network outside of the locality. For example, there aren't many high schoolers around me that i can have a nice math talk with. But that's not too big a problem because there's AoPS. Networks like AoPS remind people that true competition doesn't mean Jack across the hallway or Jill down the street, but Johann across the state or Jienzhang from across the country. It really gives a better perspective.

hmm, I think I chose the first option :huh:

I pwn everyone including Albert Einstein.

My math class is stupid bar 2 kids. AoPS however is beast.

It does seem as if the new trend is stupidity. (no offense to anyone xD)

Stupidity is the Key to Happiness.  :rotfl: See my blog.

They don't teach you to do math in school, or even think at all. They train you to become robots. They also don't do a very good job.

This is all OK up until about right here in Alg. 1:
Example: For the following quadratic equations, complete the square.
Translation: Apply the steps you hopefully memorized in class. If you forget one step, you get the question wrong.
And here usually the breaking point for # of steps to memorize is reached, and A+s go to D-s and no one likes math anymore (at least this is my experience in school math classes).

And so the people who have bad memory end up on the bottom, the people who have good memory end up in the middle, and the people who know the reasoning behind the methodology end up on top.

i'm a 4, because my class is full of idiots...no offense to them, but if they did the ftw, they would seriously have a rating of 1000...

but i cant pwn einstein, so i think im a 4.1

I could PWN Albert Einstein.

Two-year-old albert einstein, that is.

[quote=smartestpanda123]I hope I revived this???[/quote]

-_- gg well played troll lel xd

[quote=crazychinesemaniac]How good do you think you are at math?[/quote]

I voted 5!!!

I'm the only one who isn't failing his math class so...

Take a look at my alcumus rating and I've been grinding for 1 year. 14.4. :ewpu:
can you assume how bad i am at math??

why do the idiots think albert einstein is a mathmetician. He was genius at physics nurdz


floor function
number theory unsolved
number theory

[quote="knol"]Find all the real solutions to the equation

$ 4x^2 \minus{} 40 \lfloor x \rfloor \plus{} 51 \equal{} 0$[/quote]

Let $ [x]\equal{}n$. We have $ x^2\equal{}10n\minus{}\frac{51}4$ with the constraint $ n^2\le 10n\minus{}\frac{51}4 <(n\plus{}1)^2$ ($ n\ge 2$ and so $ x>0$) and so :

$ \iff$ $ n^2\minus{}10n\plus{}\frac{51}4\le 0$ and $ n^2\minus{}8n\plus{}\frac {55}4>0$

$ \iff$ $ (n\minus{}5)^2\le \frac {49}4$ and $ (n\minus{}4)^2>\frac 94$

$ \iff$ $ |n\minus{}5|\le \frac 72$ and $ |n\minus{}4|>\frac 32$

$ \iff$ $ n\in\{2,6,7,8\}$

Hence the solutions $ x\in\{\sqrt{20\minus{}\frac{51}4},\sqrt{60\minus{}\frac{51}4},$ $ \sqrt{70\minus{}\frac{51}4},\sqrt{80\minus{}\frac{51}4}\}$

And so $ \boxed{x\in\{\frac{\sqrt{29}}2,\frac{\sqrt{189}}2,\frac{\sqrt{229}}2,\frac{\sqrt{269}}2\}}$


Let $ P(x)$ be a polynomial of degree $ n$ whose roots are $ i\minus{}1,i\minus{}2,...,i\minus{}n$, where $ i^2\equal{}\minus{}1$. Let $ R(x),S(x)$ be the polynomials with real coefficients such that $ P(x)\equal{}R(x)\plus{}iS(x)$.

Show that the polynomial $ R$ has $ n$ real roots

$ P(x)\equal{}c\prod_{j\equal{}1}^n(x\minus{}i\plus{}j)$ with $ c\ne0$
Obviously, we have
$ 2R(x)\equal{}c\prod_{j\equal{}1}^n(x\minus{}i\plus{}j)\plus{}\bar{c}\prod_{j\equal{}1}^n(x\plus{}i\plus{}j)$.

Suppose $ R(a\plus{}bi)\equal{}0$ for some $ b\ne0$, we have
$ c\prod_{j\equal{}1}^n(a\plus{}j\plus{}(b\minus{}1)i)\equal{}\minus{}\bar{c}\prod_{j\equal{}1}^n(a\plus{}j\plus{}(b\plus{}1)i)$.
Both sides have same modulus, but
$ (a\plus{}j)^2\plus{}(b\minus{}1)^2<(a\plus{}j)^2\plus{}(b\plus{}1)^2$ when $ b>0$;
$ (a\plus{}j)^2\plus{}(b\minus{}1)^2>(a\plus{}j)^2\plus{}(b\plus{}1)^2$ when $ b<0$.
We also notice the two modulus can not be zero at the same time, because that requires $ b\equal{}1$ and $ b\equal{}\minus{}1$ at the same time.
Hence, we get contradiction and $ b$ has to be $ 0$.


This looks like it involves circumcircles.

No I wrote exactly true.
There is no circumcircles .

yea.... according to my graphing program, the angle varies, a similar problem was posted in the intermediate geometry marathon that might be helpful to look at problem #58

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=70276&start=220[/url]


function
logarithms
analytic geometry
graphing lines
slope
calculus
calculus computations

Find the necessary and sufficient condition of $ p,\ q$ for which the line $ y\equal{}px\plus{}q$ and the function $ y\equal{}\ln x$ have no intersection points.

The line and the curve (which is concave down) have the same slope when $ x\equal{}\frac1p.$ At that point, $ y\equal{}1\plus{}q$ on the line and $ y\equal{}\ln\left(\frac1p\right)\equal{}\minus{}\ln p$ on the curve. If the line is still above the curve, meaning that $ 1\plus{}q>\minus{}\ln p,$ then they will never intersect.

The necessary and sufficient condition for no intersection is $ \ln p\plus{}q\plus{}1>0.$

Ah, yes. I was tacitly assuming $ p>0.$ If $ p\le0,$ there will be an intersection. So the condition should be stated as $ \ln p\plus{}q\plus{}1>0$ and $ p>0.$

O.K. Let $ f(x)\equal{}px\plus{}q\minus{}\ln x\ (x>0)$, can anyone solve the problem by examining the behaviour of $ f(x)$?


I know I may have posted this before, but just in case.

Given that there is a regular n-gon inscribed within a circle of radius 1, how do you find the product of all the sides and diagonals of the triangle? What about sum??

[hide="Giveaway"] Roots of unity. [/hide]


Coach Grunt is preparing the 5-person starting lineup for his basketball team, the Grunters.  There are 12 players on the team.  Two of them, Ace and Zeppo, are league All-Stars, so they'll definitely be in the starting lineup.  How many different starting lineups are possible?  (The order of the players in a basketball lineup doesn't matter.)

We already know Ace and Zeppo will be on the team, so we only need to choose 3 players out of 10. Thus, our answer is $ \binom{10}{3} \equal{} 120$.

[quote=V_V][quote=Possible]This would be a more realistic question if they said that all of the other people are the same skill level  :dry:[/quote]

no cuz in sports some team members are better than others[/quote]

Exactly but the starting lineup could be the worst 3 players and the 2 best players

[hide=Solution]Coach Grunt has to choose 3 players from the 10 players that are remaining after Ace and Zeppo have been placed in the lineup. The order in which the players are chosen doesn't matter, so the answer is$$ \binom{10}{3} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = \boxed{120}. $$[/hide]

Why, oh, why, are problems in the textbook not level 1??????????


this is gonna be dumb for you physics geniuses :blush:

A particle initially at rest is subject to 2 forces, one is constant, the other is a retarding force proportional to the particle velocity. Draw the graph of its motion.

m*dv/dt = a - bv

ln (a - bv) = -b/m * t + C

bv =  a - Ce^(-b/m * t)

if v = 0 at time t=0, then C = a

integrate again to get

x(t) = a/bt + am/b^2 * e^(-b/m * t) + C

From there, you can graph it on any graphing calc

aw come on,  calculus :(  ! how about trying simple forces stuff , i mean insstead of this heavy math , how about some staraight forward physics ?? :)

ok,if not the graph ,can someone just tell how would it's motion be?


[u][b]Problem (zaizai-hoang):[/b][/u]
Let a,b,c are positive number such that: $ab+bc+ca+abc=4$. Prove that:
$3(a^{2}+b^{2}+c^{2})+abc\ge 10$

Here's my proof.

[b]Lemma.[/b] for $a,b,c \in R^{+}$, $a^{2}+b^{2}+c^{2}+2abc+1 \geq 2(ab+bc+ca)$

Lemma-pf) Let $a=x^{3}, b=y^{3}, c=z^{3}$. then ineq is changed to

$x^{6}+y^{6}+z^{6}+2x^{3}y^{3}z^{3}+1 \geq 2\sum_{cyclic}x^{3}y^{3}$

using AM-GM, we get $x^{3}y^{3}z^{3}+x^{3}y^{3}z^{3}+1 \geq 3x^{2}y^{2}z^{2}$

so ETS $x^{6}+y^{6}+z^{6}+3x^{2}y^{2}z^{2}\geq 2\sum_{cyclic}x^{3}y^{3}$

Here, from Schur ineq case $n=1$ and AM-GM, we get $x^{6}+y^{6}+z^{6}+3x^{2}y^{2}z^{2}\geq \sum_{sym}x^{4}y^{2}\geq 2\sum_{cyclic}x^{3}y^{3}$

so the lemma is proved.

[b][color=blue]Let's start to prove the problem ! [/color][/b]

If $abc>1$ then $ab+bc+ca\geq 3(abc)^{\frac{2}{3}}> 3$

so the condition $ab+bc+ca+abc=4$ can't be satisfied. Therefore, $abc \leq 1$

$3(a^{2}+b^{2}+c^{2})+abc \geq 10$

$\Leftrightarrow$ $3(a^{2}+b^{2}+c^{2}+2abc+1 ) \geq 13+5abc$

Using lemma, ETS $6(ab+bc+ca) \geq 13+5abc$

$\Leftrightarrow$ $24-6abc \geq 13+5abc$

$\Leftrightarrow$ $1 \geq abc$

$Q.E.D.$

I was fool...  :blush:

[b]Second proof.[/b]

$abc \leq 1$ as I showed above.

cuz $a^{2}+b^{2}+c^{2}\geq ab+bc+ca$

ETS $3(ab+bc+ca)+abc \geq 10$.

$\Leftrightarrow$ $12-2abc \geq 10$

$\Leftrightarrow$ $1 \geq abc$

$Q.E.D.$

 :blush:  :blush:

A simpler way ;) (it helps sometimes to know nothing complicated)

Change a 4 on the RHS to $ab+bc+ac+abc$. We get
$2(a^{2}+b^{2}+c^{2}) \ge 6$

But we need $ab+bc+ac+abc = 4$. If $ab+bc+ac < 3$, then we have $3 > ab+ac+bc \ge 3(abc)^{2/3}$, whence $abc < 1$ and our condition is contradicted. $ab+bc+ac \ge 3$.

We conclude by $2(a^{2}+b^{2}+c^{2}) \ge 2(ab+ac+bc) \ge 6$.

edit: I see [b]Chang Woo-JIn[/b] has a better solution also, faster than mine! :P

[quote="zaizai-hoang"]Let a,b,c are positive number such that: $ab+bc+ca+abc=4$. Prove that:
$3(a^{2}+b^{2}+c^{2})+abc\ge 10$[/quote]

From $ab+bc+ca+abc=4$, there's positive $p,q,r$ such that $a=\frac{2p}{q+r}, b=\frac{2q}{r+p}, c=\frac{2r}{p+q}$. Applying this, the original one becomes \[\sum \frac{6p^{2}}{(q+r)^{2}}+\frac{4pqr}{(p+q)(q+r)(r+p)}\geq 5\] or \[6\sum_{cyc}p^{6}+12\sum_{sym}p^{5}q+\sum_{sym}p^{4}q^{2}+14\sum_{cyc}p^{4}qr \geq 10 \sum_{cyc}p^{3}q^{3}+14 \sum_{sym}p^{3}q^{2}r+24p^{2}q^{2}r^{2}\] which can be obtained by summation of the following: \[6\sum_{cyc}p^{6}\geq 6\sum_{cyc}p^{3}q^{3}\]  \[2\sum_{sym}p^{5}q \geq 4\sum_{cyc}p^{3}q^{3}\]  \[10\sum_{sym}p^{5}q \geq 10 \sum_{sym}p^{3}q^{2}r\]  \[\sum_{sym}p^{4}q^{2}\geq \sum_{sym}p^{3}q^{2}r\]  \[6\sum_{cyc}p^{4}qr \geq 3\sum_{sym}p^{3}q^{2}r\]  \[8\sum_{cyc}p^{4}qr \geq 24p^{2}q^{2}r^{2}.\]

[quote="Sung-yoon Kim"][quote="zaizai-hoang"]Let a,b,c are positive number such that: $ab+bc+ca+abc=4$. Prove that:
$3(a^{2}+b^{2}+c^{2})+abc\ge 10$[/quote]

From $ab+bc+ca+abc=4$, there's positive $p,q,r$ such that $a=\frac{2p}{q+r}, b=\frac{2q}{r+p}, c=\frac{2r}{p+q}$. Applying this, the original one becomes \[\sum \frac{6p^{2}}{(q+r)^{2}}+\frac{4pqr}{(p+q)(q+r)(r+p)}\geq 5\] [/quote]
Then just use: $8pqr\ge (p+q)(q+r)(r+p)$ and $\sum\frac{p^{2}}{(q+r)^{2}}\ge \frac{3}{4}$[/quote]

[quote]
Then just use: $8pqr\ge (p+q)(q+r)(r+p)$ and $\sum\frac{p^{2}}{(q+r)^{2}}\ge \frac{3}{4}$[/quote]

I think you have a mistake. $(p+q)(q+r)(r+p)\ge 8pqr$ by AM-GM

[quote="zaizai-hoang"][quote="Chang Woo-JIn"]Here's my proof.

[b]Lemma.[/b] for $a,b,c \in R^{+}$, $a^{2}+b^{2}+c^{2}+2abc+1 \geq 2(ab+bc+ca)$

Lemma-pf) Let $a=x^{3}, b=y^{3}, c=z^{3}$. then ineq is changed to

$x^{6}+y^{6}+z^{6}+2x^{3}y^{3}z^{3}+1 \geq 2\sum_{cyclic}x^{3}y^{3}$

using AM-GM, we get $x^{3}y^{3}z^{3}+x^{3}y^{3}z^{3}+1 \geq 3x^{2}y^{2}z^{2}$

so ETS $x^{6}+y^{6}+z^{6}+3x^{2}y^{2}z^{2}\geq 2\sum_{cyclic}x^{3}y^{3}$

Here, from Schur ineq case $n=1$ and AM-GM, we get $x^{6}+y^{6}+z^{6}+3x^{2}y^{2}z^{2}\geq \sum_{sym}x^{4}y^{2}\geq 2\sum_{cyclic}x^{3}y^{3}$

so the lemma is proved.

 :blush:  :blush:[/quote]
Nice solution. I have a solution to prove that your [b]Lemma[/b]
WLOG we assume
 $(a-1)(b-1)\ge 0 \Leftrightarrow ab+1-a-b\ge 0 \\ \Leftrightarrow abc+c-ac-bc\ge 0$ (because $c\ge 0$)
Now the inequality equavalent:
$(a^{2}-2ab+b^{2})+2(abc+c-ac-bc)+(c^{2}-2c+1)\ge 0 \\ \Leftrightarrow (a-b)^{2}+2c(a-1)(b-1)+(c-1)^{2}\ge$
It true.  :P I think this lemma as same as VMO 2006, group B.
This problem my solution use mixing variable  :D[/quote]

The lemma was posted by Darij:
http://www.mathlinks.ro/Forum/viewtopic.php?t=19666

If $ a,b,c$ are [color=red]real[/color] numbers such that $ abc \plus{} bc \plus{} ca \plus{} ab \equal{} 4,$ then

$ abc \plus{} 3\left(a^2 \plus{} b^2 \plus{} c^2\right) \geq 10.$

[b]Proof[/b] From $ abc \plus{} bc \plus{} ca \plus{} ab \equal{} 4,$ there's [color=blue]real[/color] numbers $ p,q,r$ such that

$ a \equal{} \frac {2p}{q \plus{} r}, b \equal{} \frac {2q}{r \plus{} p}, c \equal{} \frac {2r}{p \plus{} q}.$

Applying this, the original one becomes

$ \frac {4pqr}{(q \plus{} r)(r \plus{} p)(p \plus{} q)} \plus{} \sum{\frac {6p^2}{(q \plus{} r)^2}}\geq 5,$

which can be obtained by the following :

$ \frac {4pqr}{(q \plus{} r)(r \plus{} p)(p \plus{} q)} \plus{} \sum \frac {6p^{2}}{(q \plus{} r)^{2}} \minus{} 5$

$ \equal{} \frac {(p \plus{} q \plus{} r)^2\{5(q \minus{} r)^2(r \minus{} p)^2(p \minus{} q)^2 \plus{} 3[(q \plus{} r)(q \minus{} r)^2 \plus{} (r \plus{} p)(r \minus{} p)^2 \plus{} (p \plus{} q)(p \minus{} q)^2]^2\}}{(q \plus{} r)^2(r \plus{} p)^2(p \plus{} q)^2[(q \minus{} r)^2 \plus{} (r \minus{} p)^2 \plus{} (p \minus{} q)^2]}$

$ \plus{} \sum{\frac {(q \minus{} r)^2(q \plus{} r \minus{} 2p)^2}{(q \plus{} r)^2[(q \minus{} r)^2 \plus{} (r \minus{} p)^2 \plus{} (p \minus{} q)^2]}}\geq 0.$

By the way, if $ a,b,c$ are [color=green]positive[/color] numbers, then

$ (a\plus{}b\plus{}c)\left(abc \plus{} 3a^2 \plus{}3 b^2 \plus{} 3c^2\minus{}10\right)\geq 7(abc \plus{} bc \plus{} ca \plus{} ab \minus{}4).$

$ 2(a^2\plus{}b^2\plus{}c^2) \ge 10\minus{}abc\equal{}6\plus{}ab\plus{}bc\plus{}ca$
$ (a\minus{}b)^2\plus{}(b\minus{}c)^2\plus{}(c\minus{}a)^2\plus{}2(a^2\plus{}b^2\plus{}c^2) \ge 12$
it's enough to prove ineq:
$ a^2\plus{}b^2\plus{}c^2 \ge 6$
we have $ ab\plus{}bc\plus{}ca\plus{}abc\equal{}4$ so
$ \frac{ \sqrt{ab} }{2}\equal{}cos \alpha$
$ \frac{ \sqrt{bc} }{2}\equal{}cos \beta$
$ \frac{ \sqrt{ca} }{2}\equal{}cos \gamma$
with $ \alpha \plus{} \beta \plus{} \gamma \equal{} \pi$
now we have inequality:
$ \sum_{cyc}^{}  \frac{(cos \alpha)^2 (cos \beta)^2}{(cos \gamma^2)} \ge  \frac{3}{4}$
it is well known that $ cos^2 \alpha\plus{}cos^2 \beta\plus{}cos^2 \gamma \ge  \frac{3}{2}$
so
$ \sum_{cyc}^{}  \frac{(cos \alpha)^2 (cos \beta)^2}{(cos \gamma^2)} \ge cos^2 \alpha\plus{}cos^2 \beta\plus{}cos^2 \gamma \ge  \frac{3}{2}$
Done!


integration
logarithms
calculus
calculus computations

The expression $ \frac{4x}{(x\minus{}5)(x\minus{}3)^{2}}$ is denoted by $ f(x)$.

(i) Express $ f(x)$ in the form $ \frac{A}{x\minus{}5} \plus{} \frac{B}{x\minus{}3} \plus{} \frac{C}{(x\minus{}3)^{2}}$, where $ A$, $ B$ and $ C$ are constants.

(ii) Hence find the exact value of $ \int_{1}^{2} f(x)\,dx$.

[quote="AndrewTom"]
(i) Express $ f(x)$ in the form $ \frac {A}{x \minus{} 5} \plus{} \frac {B}{x \minus{} 3} \plus{} \frac {C}{(x \minus{} 3)^{2}}$, where $ A$, $ B$ and $ C$ are constants.
[/quote]

comparing the coefficients, you get $ A \equal{} 5, B \equal{} \minus{}5 , C \equal{} \minus{}6$

$ \int f(x).dx \equal{} \frac {5}{x \minus{} 5} \plus{} \frac {\minus{}5}{x \minus{} 3} \plus{} \frac {\minus{}6}{(x \minus{} 3)^{2}}$

$ \equal{} 5 ln|x \minus{} 5| \minus{} 5 ln|x \minus{} 3| \plus{}  \frac{6}{(x\minus{}3)} \plus{} C$

[b](ii)[/b]

$ \int_1^2 f(x)dx \equal{} 5 ln3 \minus{} 15ln2 \minus{} 9$

Thanks Nora.91.

When I put in the limits, I get $ 5\ln(\frac{3}{2}) \minus{} 3$.

Yeah.. you are correct, AndrewTom

my mistake.. sorry  :blush:


The product of two whole numbers is 100000. If neither of the numbers are multiples of 10, what is their sum?

[hide]100,000 =
10^5=
2^5 * 5^5

Since neither 2^5 nor 5^5 are multiples of 10, we add
32
and 3125

to get 3157, our answer.[/hide]


integration
calculus
derivative
trigonometry
real analysis
real analysis unsolved

Find the sum of the following series and prove it:

$\sum_{n=1}^{\infty}\frac{1}{\left(\begin{array}{cc}2n\\\ n\end{array}\right)}$

If i'm not doing wrong in my computation, it is $\frac{\sqrt{3}\pi}{9}+\frac{5}{3}$

correct me if i'm wrong.

[quote="Kalle"]You can compare it with the power series for arcsin and find that that the sum is $\frac13 + \frac{4}{3\sqrt{3}}\arcsin\left(\frac12 \right)$, but I am too lazy to write down the details (it takes me a long time to write down things in latex).[/quote]

You are right!But i'd like to see your proof... ;)

I've check my computation and found an error on the last line.
I 'll post it here.

Let $S=\sum_{n=1}^{\infty} \frac{1}{\binom{2n}{n}}$. lets compute
$\sum_{n=0}^{\infty} \frac{1}{\binom{2n}{n}}=S+1$

We have that \begin{eqnarray*}\sum_{n=0}^{\infty} \frac{1}{\binom{2n}{n}} &=& \sum_{n=0}^{\infty} \frac{(n)!(n)!}{(2n)!} \\ &=& \sum_{n=0}^{\infty} (2n+1) \frac{\Gamma(n+1) \Gamma(n+1) }{\Gamma(2n+2)} \\ &=& (2n+1)B(n+1,n+1) \end{eqnarray*}

and we need to compute $\sum_{n=0}^{\infty} \int_{0}^1 (2n+1) \, t^n (1-t)^n \, dt$

Notice that the integran $t^n (1-t)^n$ with $n\geq 0$ is converges uniformly when $t \in [0,1)$ by Weierstrass M-Test since we have
\[ |t^n (1-t)^n | \leq M^n  \] for $0\leq t <M<1$
for the case of endpoint 1 the integran is zero. so in converges uniformly on $[0,1]$

now for $n t^n(1-t)^n$. it just derivative of the later series times $(t)(1-t)$.
Since the derivative of the convergent uniform series does convergent uniform on every interval inside of the convergence interval of original series.
so we just need to verified in the case $t=0$ and $t=1$ but it is identically zero on both of these points.

So the series is convergen uniform for $t \in [0,1]$.

Now the change of integral and sumation is alllowed. And we have that

\begin{eqnarray*}\int_{0}^1 \sum_{n=0}^{\infty} (2n+1) \, t^n (1-t)^n \, dt &=& \int_0^1 \frac{2t(1-t)}{(t^2-t+1)^2} + \frac{1}{(t^2-t+1)} \, dt \\ &=& \int_0^1 \frac{t+1-t^2}{(1+t^2-t)^2} \, dt \\ &=& \int_0^1 \frac{-1}{1+t^2-t} + \frac{2}{(1+t^2-t)^2} \, dt \end{eqnarray*}

Subtitution $t=\frac{\sqrt{3}}{2} \tan u +\frac{1}{2}$ will yield

\begin{eqnarray*} \int_{\frac{- \pi}{6}}^{\frac{\pi}{6}} \frac{-2\sqrt{3}}{3} + \frac{8 \sqrt{3}}{9} 2 \cos^2 u \, du &=& \frac{-2 \sqrt{3} \pi}{9} + \int_{\frac{- \pi}{6}}^{\frac{\pi}{6}} \frac{8 \sqrt{3}}{9} \cos 2u + 1 \, du \\ &=& \frac{-2 \sqrt{3} \pi}{9} + \frac{8 \sqrt{3}}{9} \left( \frac{1 \sqrt{3}}{2} + \frac{\pi}{3} \right) \\ &=& \frac{2 \sqrt{3} \pi}{27} + \frac{4}{3} \end{eqnarray*}

Then $S = \frac{2 \sqrt{3} \pi}{27} + \frac{4}{3} - 1 = \frac{2 \sqrt{3} \pi}{27} + \frac{1}{3}$

I hope you understand why i make an errors somewhere  :P


If $ N$ is a submanifold of $ M$ and $ V$ is an open subset of $ M$,possibly connected,
then show that $ N\cap V$ is the union of a countable collection of connected open subsets of $ N$(with its submanifold topology)

$ M$ is utterly irrelevant. However weak the submanifold condition is, inclusion of $ N$ into $ M$ is continuous, and hence the preimage $ N\cap V$ of $ V$ is open in $ N$.

We now forget about that. $ N$ is a topological manifold, and $ U$ is an open subset of it. We want to show that $ U$ is a countable union of connected open sets. The relevant definitions:
A $ n$-dimensional topological manifold is a second countable Hausdorff space which is locally Euclidean of dimension $ n$. Second countable? There's a countable base for the topology. Locally Euclidean? Each point has a neighborhood homeomorphic to an open ball in $ \mathbb{R}^n$.
Combining those bits of the definition, there's a countable base for the topology consisting of sets homeomorphic to open balls in $ \mathbb{R}^n$. They're connected. Every open set is a union of some of those, and we're done.


Is given the triangle $\ ABC$ whith $\ AB=BC=CA$ and its cirqlue $\ C(O)$.Let $\ D$ a point which lies on
$\ C(O)$.Prove that the greatest  between $\ (DA),(DB),(DC)$ is equal whith the sum of  distances from   $\ O$ to the others.
  [i] Example[/i](because I don't know some words in English)
If $\ (DC)$ is the smallest, then prove that $\ DC=DA+DB$ etc...

[quote="stancioiu sorin"]Is given the triangle $\ ABC$ whith $\ AB=BC=CA$ and its cirqlue $\ C(O)$.Let $\ D$ a point which lies on
$\ C(O)$.Prove that the smallest  between $\ (DA),(DB),(DC)$ is equal whith the sum of  distances from   $\ O$ to the others.
  [i] Example[/i](because I don't know some words in English)
If $\ (DC)$ is the smallest, then prove that $\ DC=DA+DB$ etc...[/quote]

This Chadu Theorem, it's easy using Ptolemy's theorem.

I think you should say $DC$ is the greatest distance and then: $DC=DA+DB$

[quote="stancioiu sorin"]Yes, excuse me. :oops: I did a mistake.But can you prove the conclusion?[/quote]

Like I said just use Ptolemy's theorem and simplify the equal sides.

[quote="stancioiu sorin"]I realised that I must use Ptolomy's Theorem, but I can't solve it...Maybe I am tired and I don't have no idea where I can apply it.Can you post a full solution?I mean, where I must apply the famous theorem? :?[/quote]
By Ptolemy's Theore we have: $DC \cdot AB = DA \cdot BC + DB \cdot AC$. Since the triangle is equilateral, so $DC=DA+DB$.

I think my conclusion is wrong.So, please to prove that $\ OG=OE+OF$, where $\ OG\bot CD$,$\ OF\bot BD$ and
$\ OE\bot AD$.Sorry for my mistake. :oops:


Sophie Germain identity
number theory
composite number
factorization
IMO
IMO 1969

Prove that there are infinitely many positive integers $m$, such that $n^4+m$ is not prime for any positive integer $n$.

If $5|n$ then $m=5k$ for $k=1,2,...$. If $n$ is not divisible by $5$ then $n^4 \equiv 1 \mod5$ so we take $m=5k+4$ for $k=1,2,...$.

But the result must be composite for ANY n, not for a GIVEN one. Maybe something like this works:
$(m^2+m+mk+\frac{k^2} 2+k+\frac 1 2)(m^2-m-mk+\frac{k^2} 2+k+\frac 1 2)=$
$=m^4+2k^2+k^3+\frac 1 4 k^4+\frac 1 4+k+\frac 3 2 k^2$.
All the numbers are integers for k odd, so each $n=2k^2+k^3+\frac 1 4 k^4+\frac 1 4+k$ works.

[i]Kondr[/i], you have the correct idea, which is to express $n^4+m$ as a product of two expressions, but your expressions are a little complicated. Since any product of two expressions without any $n^3, n^2$ or $n$ terms works, $(n^2 + 2kn + 2k^2)(n^2 - 2kn + 2k^2) = n^4 + 4k^4$ would be much easier to write.  :P

"But the result must be composite for ANY n, not for a GIVEN one"

Do you mean that it must be one set of integers $m$ for every $n$? It seems strange to me that we can't consider two cases depending on $n$...

Sorry [i]TomciO[/i], I was just skimming through and didn't really look at your post. Your solution is correct and is more elegant!  :)

Tomci0's solution may be elegant, but I still think the choice of m can't depend on n.
(if m could depend on n, we can just put m odd for n odd and m even for n even).
Because of that only Dblues' nice solution and my ugly solution work.

Right... anyway, the problem statement could be better: "Prove that there exists infinite set of positive integers $m$ such that for every $n$...".

For those who don't know:
Sophie Germain's problem:
When is $ m^4\plus{}4n^4$ prime ($ m,n\in \mathbb N$)?
[hide="Answer"]Never[/hide]
[hide="Solution"]
Add and subtract $ 4m^2n^2$ from $ m^4\plus{}4n^4$. Using binomial formula and $ (a\minus{}b)(a\plus{}b)\equal{}a^2\minus{}b^2$ we get:
$ m^4\plus{}4n^4\equal{}(m^4\plus{}4m^2n^2\plus{}4n^4)\minus{}4m^2n^2\equal{}(m^2\plus{}{2n}^2)^2\minus{}(2mn)^2\equal{}(m^2\minus{}2mn\plus{}{2n}^2)(m^2\plus{}2mn\plus{}{2n}^2)$
If the expression is a prime, then: $ (m^2\minus{}2mn\plus{}{2n}^2)\equal{}1$.
But that's impossible.
The end!!![/hide]

Sorry for the bump, but is this solution correct:
If $n$ is odd, let $a$ be an odd integer greater than $1$. Then, $n^4+a$ is an even number greater than $2$ so its not prime.
If $n$ is even, let $a$ be an even natural number. Then, $n^4+a$ is an even number greater than $2$ so its not prime.

In this problem, $m$ is fixed, so the solution is wrong.

[quote=Raja Oktovin]if $ n\equal{}4k^4$ and applying sophie germain identity to $ m^4\plus{}n$, we have done  :D[/quote]

now we are finding $m$, not $n$.

[quote=AdithyaBhaskar]Well this is Sophie Germain unmasked. We take $m = 4a^4$ for some positive integer $a.$ Done.[/quote]
I just want to add that a>=2 as for a=1 $z_1$=5 is a prime.

Take $a=4k^4$, for some non-zero integer $k$. We have
$$n^4+4k^4=(n^2+2k^2)-4n^2k^2=(n^2+2nk+2k^2)(n^2-2nk-2k^2)$$
If $(n^2-2nk-2k^2)=1$, we have that $(n-k)^2+k^2=1$, so since $k$ is positive, we have that $n=k$. Plugging in, we have that $(n^2+2nk-2k^2)=5n^2$, and which is composite due to how we defined $k$. If $(n^2-2nk-2k^2)$ does not equal one, than $n^2+a^4$ is still prime so we win, as desired. $\square$

[quote=Garou]Let $m=p^{2}$ where $p$ is a prime and $p \equiv -1 \pmod 4$. If , $n^{4}+p^{2}=q$ where $q$ is a prime then  by Fermat's christmas theorem  $ q|n$ and $q|p$ which is clearly contradictory $\square$.[/quote]

Something is totaly wrong here:

Choose $p=11$. Then $p$ is a prime and $p \equiv -1 \pmod 4$.
Choose $n=2$.
Then $q=n^4+p^2=16+121=137$ is prime,
but your conclusion $137|2$ and $137|11$ does not hold and there is no contradiction.

People seem to be missing the fact that $n^4+4m^4$ is a prime for $n=m=1$.
So better take $a=4m^4 \forall m \in \{2, 3 \dots \}$


Αν για τους πραγματικούς x,y,z isx;yoyn oi ισότητες:
\[ {x^2} \minus{} y \equal{} {z^2}\],\[ {y^2} \minus{} z \equal{} {x^2}\],\[ {z^2} \minus{} x \equal{} {y^2}\]

Να ποδειχθεί οτι \[ {x^3} \plus{} {y^3} \plus{} {z^3} \equal{} 0\]

[quote="Spribo"]Δοκίμασε να προσθέσεις τις τρεις σχέσεις. Θα πάρεις ότι x+y+z=0. Πάρε έπειτα την ταυτότητα Euler-Cauchy (την έχει και το σχολικό Άλγεβρας της Α΄Λυκείου), και παίρνεις ότι x^3+y^3+z^3=0.

Σημείωση: Αυτό είναι το 4ο θέμα του φετινού Θαλή της Α' Λυκείου.[/quote]
to kserw aplws me enan filo prospathousame na to lisoume etsi.Alla kai pali de vgainei.

[quote="Spribo"]Μα, βεβαίως και όπως έδειξα βγαίνει. Εάν θες πιο απλό τρόπο, πάρε ακριβώς την απόδειξη του σχολικού της Άλγεβρας για x^3+y^3+z^3=0, όταν x+y+z=0.
Απ' όσο γνωρίζω, τρεις ισότητες μπορούμε να τις προσθέσουμε![/quote]
μα δεν ισχύει αυτό που λές
Έστω x=5,y=-3,z=-2, τότε x+y+z=0
Άρα \[ {x^3} \plus{} {y^3} \plus{} {z^3} \equal{} 125 \minus{} 27 \minus{} 8 \ne 0\]

[quote="Spribo"]Μα την αλήθεια, έκανα γκάφα τρομερή! Επανορθώνω όμως άμεσα:
Από το x+y+z=0, έχουμε ότι x^3+y^+z^3=3xyz. Τώρα, θα αποδείξουμε ότι ένας τουλάχιστον από τους x,y,z είναι ίσος με 0.
Πάρε δυο-δυο τις σχέσεις (χρησιμοποιώ λόγο κι όχι σύμβολα, γιατί δεν ξέρω latex) και θα βγάλεις ότι (z+y)(y-z-1)=0, και ομοίως τις άλλες. Εάν τις συνδυάσεις παίρνοντας περιπτώσεις, θα καταλήγεις πάντα στο ότι ένας από αυτούς θα είναι 0.
Τουλάχιστον αυτήν την απόδειξη παρουσίασα στο Θαλή! (Τώρα δεν παρουσιάζω βέβαια τις πράξεις).
Είδες των πράξεων η βιασύνη, η αδύνατη μνήμη και η ισχυρογνωμοσύνη...[/quote]
Φιλε μου δεν πειράζει. Αλλά αυτο που ζητάω είναι χωρίς περιπτώσεις να βγεί ότι \[ {x^3} \plus{} {y^3} \plus{} {z^3} \equal{} 0\]
συμφωνα με τις δοσμένεσ σχέσεις

[quote="Spribo"]Πάρε την πρώτη σχέση. Πολλαπλασίασε με y^2. Έργασαι ομοίως και στις άλλες. Πρόσθεσε κατά μέλη. Κι έχεις x^3+y^3+z^3=0! (Πρέπει να υποθέσεις ότι x,y,z μη μηδενικοί, αλλά, εάν κάποιος ήταν μηδενικός, το ζητούμενο θα ίσχυε).[/quote]
Μιλας για το θέμα που έπεσε φέτος?

[quote="Spribo"]Μα βέβαια! Αυτό έπεσε τέταρτο, με δύο όμως υποερωτήματα. Το πρώτο έλεγε ν.δ.ο. x+y+z=0 και το δεύτερο ν.δ.ο x^3+y^3+z^3=0, εάν βέβαια θυμάμαι καλά...[/quote]
Το πρώτο έλεγε νδο χ^3+^3+ζ^3=3χυζ
>Και το δευτερο νδο ενας απο χυζ είναι ισος με 0

Εν πάσει περιπτώσει...

[quote="Spribo"]Εν πάσει περιπτώσει...[/quote]
Όμως αυτό που πήρες με περιπτώσεις γίνονταν πιο ευκολα απλἄ.
Αφού απο δεικνυεις οτι \[ {x^3} \plus{} {y^3} \plus{} {z^3} \equal{} 0\] τότε απο την ισότητα που απέδειξες \[ {x^3} \plus{} {y^3} \plus{} {z^3} \equal{} 3xyz\]
έπεται οτι \[ 3xyz\equal{}0\] άρα ή χ ή υ ή ζ είναι ισος με 0

[quote="Spribo"]Το ξέρω, αλλά εν τη ώρα του διαγωνισμού δε σκεφτόμουν πολύ την κομψότητα...[/quote]
Και η εμε λύση με περιπτωσεις εδωσε

Καληνυχτίζω!

[quote="Spribo"]Καληνυχτίζω![/quote]
Επίσης

Η λύση του Spribo είναι η σωστή.

Θέλουμε να αποδείξουμε ότι

x^3+y^3+z^3=0
Αφού x+y+z=0 από Euler έχουμε x^3+y^3+z^3=3xyz.

Άρα αρκεί 3xyz=0 <=> xyz=0 <=> x=0 ή y=0 ή z=0. Αυτό αποδεικνύεται όπως είπε ο Spribo.

[quote="Spribo"]Ευχαριστώ, αλλά αρχικά είχα πράγματι κάνει λάθος![/quote]
Basika ti lisi egw tn proteina kat auton tn tropo. Ostoso tn arxiki apodeiksi tn ekane o spirbo kai tn euxaristo


Let ABC be a triangle with sides lengths are a,b,c. It's well known that  if $a^2+b^2=c^2$  then ABC is a right angled triangle. I wonder is there such a triangle exists with a,b,c as its sides satisfying the equation  $a^n+b^n=c^n$ and 0<n<1. What is the shape of triangle in that case? I mean obtuse,acute or others.

Suppose that $a^t+b^t=c^t$, with $t>0$. Then, of course, $c$ is the longest side.

If $t>2$, the triangle is acute.
If $t=2$, the triangle is right.
If $1<t<2$, the triangle is obtuse.
If $0<t\le 1$, there's no triangle, because $a+b\le c$


search
number theory unsolved
number theory

A long time ago, [b]ZetaX[/b] told us the following generalisation of $ PEN \minus{} 11$:
Let $ a_{1},a_{2},...,a_{n}$ be integers. Then $ \prod_{i < j} \left( a_{i} \minus{} a_{j} \right)$ is divisible by $ \prod_{k \equal{} 1}^{n \minus{} 1}k!$.
In other words, $ \prod_{i < j} \left( \frac {a_{j} \minus{} a_{i}}{j \minus{} i} \right)$ is an integer.
Nobody sent a nice proof of that fact so I think it deserves a topic apart.
(Moderators: so I reposted here, if it is wrong delete it.)

Have a look here :
http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1321442849&t=4858

Pierre.


quadratics
Gauss
number theory unsolved
number theory

For distinct primes $ p$ and $ q$.Show that
$ (\frac{p}{q})(\frac{q}{p})\equal{}(\minus{}1)^{\frac{p\minus{}1}{2}\frac{q\minus{}1}{2}}$

There are many proofs of the [url=http://en.wikipedia.org/wiki/Quadratic_reciprocity]quadratic reciprocity law[/url].  One of the more elementary (although somewhat long) uses Gauss' lemma and then a geometric approach using lattice points.


Write as a common fraction: \[ \frac{1}{1\plus{}\frac{1}{1\plus{}\frac{1}{1\plus{}1}}}\]


Αν a,b,c,d > 0  και  ab(c+d)+cd(a+b) = 2abcd  δείξτε ότι

$ (a\plus{}b)(b\plus{}c)(c\plus{}d)(a\plus{}c)(b\plus{}d)(a\plus{}d)\le a^3b^3c^3d^3$

 Υ.Σ  Μανιέ το ξεπέταξες το ασκησάκι vm6 :)

 Αυτή για εσένα ...

Ευχαριστώ πάρα πολύ για αυτή την όμορφη αφιέρωση.

$ ab(c \plus{} d) \plus{} cd(a \plus{} b) \equal{} 2abcd \Leftrightarrow \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c} \plus{} \frac {1}{d} \equal{} 2$
που είναι συμμετρική ως προς οποιαδήποτε αναδιάταξη των $ a,b, c, d$ και άρα και η αρχική ισοδύναμή της σχέση είναι συμμετρική ως προς $ a,b, c, d$

$ 2abcd \equal{} ab(c \plus{} d) \plus{} cd(a \plus{} b)$ (από AM - GM) $ \ge 2\sqrt {abcd(a \plus{} b)(c \plus{} d)} \Leftrightarrow$
$ (abcd)^2 \ge abcd(a \plus{} b)(c \plus{} d) \Leftrightarrow abcd \ge (a \plus{} b)(c \plus{} d)$

Λόγω της συμμετρίας έχουμε ομοίως:
$ abcd \ge (b \plus{} c)(a \plus{} d)$ και
$ abcd \ge (a \plus{} c)(b \plus{} d)$

πολλαπλασιάζοντας τις τρεις αυτές κατά μέλη έχουμε το ζητούμενο
$ (a \plus{} b)(b \plus{} c)(c \plus{} d)(a \plus{} c)(b \plus{} d)(a \plus{} d) \le a^3b^3c^3d^3$ :lol:

Σημείωση:Μπορούμε να εφαρμόσουμε την AM - GM και να κάνουμε τους πολλαπλασιασμούς
επειδή έχουμε θετικούς αριθμούς.

Μπράβο μανιέ ... :)

Μια άλλη ιδέα είναι να χρησιμοποιήσουμε τη βασική ανισότητα $ (x \plus{} y)^2\ge 4xy$.
σε συνδυασμο με τη σχέση $ \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c} \plus{} \frac {1}{d} \equal{} 2$,  οπότε και έχουμε:

$ \left(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c} \plus{} \frac {1}{d}\right)^2\ge4\left(\frac {1}{a} \plus{} \frac {1}{b}\right)\left(\frac {1}{c} \plus{} \frac {1}{d}\right)$  η οποία δίνει $ abcd\ge (a \plus{} b)(c \plus{} d)$ (1)

Εφαρμόζουμε το ίδιο άλλες δύο φορές ομαδοποιώντας τα ζευγάρια κατάλληλα και έχουμε :  $ abcd\ge (a \plus{} c)(b \plus{} d)$ (2) και  $ abcd\ge (a \plus{} d)(b \plus{} c)$ (3)

Πολ/ζοντας τις (1),(2),(3)  προκύπτει το ζητούμενο... :)


linear algebra
matrix
linear algebra unsolved

$ A,B \in M_{3}(Z)$ , $ det(A)\equal{}det(B)\equal{}0$.Prove that $ \frac{det(A^3\plus{}B^3)\plus{}det(A^3\minus{}B^3)}{2}$ are cub perfect .

I think : Let $ A^{3}=X ,B^{3}=Y$

$ X,Y \in M_{3} \mathbb{(R)}$
$ det(X)=0 ,det(Y)=0$
Then : rank(Y) $ \in$ {1,2}
If Rank(Y)=1 then : $ det(X-Y)+det(X+Y) = 2.det(X)=0$
If Rank(Y)=2 then :$ det(X-Y)+det(X+Y) = 2.x_{33}$, $ x_{33}$ is colum 3 ,row 3 of X
You can prove  :roll:

[quote="posabogdan"]If Rank(Y)=1 then : $ det(X \minus{} Y) \plus{} det(X \plus{} Y) \equal{} 2.det(X)$
If Rank(Y)=2 then :$ det(X \minus{} Y) \plus{} det(X \plus{} Y) \equal{} 2.x_{33}$, $ x_{33}$ for every matrix $ \in M_{3}(C)$?
Can you prove this pls ?[/quote]

No, because it's not true. :)

$ X\equal{}\left(\begin{array}{ccc} 0 & 0 & 0 \\
0 & 0 & 1 \\
0 & 1 & 0\end{array}\right), Y\equal{}\left(\begin{array}{ccc} 1 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0\end{array}\right)$

Then $ \det(X\plus{}Y)\plus{}\det(X\minus{}Y)\equal{}\minus{}2$.

Similarly if $ Y\equal{}\left(\begin{array}{ccc} 1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 0\end{array}\right)$ for the rank 2 case.

Actually, the whole problem is wrong.  This is the simplest counterexample I could find:

$ A\equal{}\left(\begin{array}{ccc} 1 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0\end{array}\right), B\equal{}\left(\begin{array}{ccc} 1 & 1 & 0 \\
1 & 1 & 0 \\
0 & 0 & 1\end{array}\right)$

Then $ \frac{\det(A^3 \plus{} B^3) \plus{} \det(A^3\minus{}B^3)}{2} \equal{} 4$.

[quote="LydianRain"][quote="posabogdan"]If Rank(Y)=1 then : $ det(X \minus{} Y) \plus{} det(X \plus{} Y) \equal{} 2.det(X)$
If Rank(Y)=2 then :$ det(X \minus{} Y) \plus{} det(X \plus{} Y) \equal{} 2.x_{33}$, $ x_{33}$ for every matrix $ \in M_{3}(C)$?
Can you prove this pls ?[/quote]

No, because it's not true. :)

$ X \equal{} \left(\begin{array}{ccc} 0 & 0 & 0 \\
0 & 0 & 1 \\
0 & 1 & 0\end{array}\right), Y \equal{} \left(\begin{array}{ccc} 1 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0\end{array}\right)$

Then $ \det(X \plus{} Y) \plus{} \det(X \minus{} Y) \equal{} \minus{} 2$.

Similarly if $ Y \equal{} \left(\begin{array}{ccc} 1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 0\end{array}\right)$ for the rank 2 case.[/quote]

Something about me and Mathematica... anyway, this was not actually a counterexample to what posabogdan posted originally, since I mixed up the order of the matrices.

What posabogdan posted about rank 1 matrices was correct, though his statement about rank 2 matrices was wrong.  Here is a correct counterexample:

$ X \equal{} \left(\begin{array}{ccc} 1 & 0 & 0 \\
0 & 0 & 1 \\
0 & 1 & 0\end{array}\right), Y \equal{} \left(\begin{array}{ccc} 1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 0\end{array}\right)$


function
algebra
domain
calculus
calculus computations

A fixed point of a function f is a number c in its domain such that f(c) = c. (the function doesn't move the input value c; it stays fixed.) Suppose that f is a continuous function with domain [0,1] and range contained in [0,1]. Prove that f must have at  least one fixed point.

Let $g(x)=f(x)-x$. Since $f(x)$ is continuous, then $g(x)$ is continuous. Note that $g(0)\ge 0\ge g(1)$.

Case 1: $g(0)=0$ or $g(1)=0$.
This implies one of the endpoints is a fixed point.

Case 2: $g(0),g(1)\neq 0$.
By the intermediate value theorem, there exists $c$ with $0<c<1$ such that $g(c)=0$. At that point, $f(c)=c$, so $c$ is a fixed point.


How to prove there are infinetely many primes of form $ 2^{2^n}\plus{}1$ (fermat's prime number)

[quote="N.T.TUAN"][quote="Mathpro"] primes of form $ 2^{2^n}$ [/quote]
Are you sure?  :P[/quote]

Oh sorry I edited :wink: