I thought this was impossible, but here it is:

[url]http://video.google.com/videoplay?docid=-5120299724584837019&q=excellent+goal&hl=en[/url]

- what? a goal from just outside the box? oh i have seen many of those! what? the box on the [i]opposite[/i] side? what? WHAT??

[quote="ragnarok23"]That goalie is an idiot.[/quote]
This goalie is more of an idiot.

[url]http://video.google.com/videoplay?docid=2188358998697411341&q=worst+soccer&hl=en[/url]

Well, I can't say that the goalie was an idiot, but I can say that they were planning on that guy from coming out of nowhere and scoring a goal.

[quote="ragnarok23"]That goalie is an idiot.[/quote]

How is that goalie an idiot?

He was watching the ball and was trying to stop it but the ball bounced over his head and he immediately ran back as far as he could, that was a good attempt, I'd like to see anybody else do better.

[quote="kheg_k"][b]that[/b] was a good punt.[/quote]

I don't think that was a punt...all he did was just kick it when it was on the ground.

[quote="Koda"][quote="ragnarok23"]That goalie is an idiot.[/quote]

Why? He scored the goal! :rotfl:[/quote]
Uh... Exactly!? Own goal = idiot.

[quote="junggi"][quote="i_like_pie"][quote="Koda"]Who scored an own goal? I was talking of the first goal. And the second goalie was not supposed to have eyes behind the back of his head after all![/quote]
Ragnarok said the the second goalie was an idiot. You two are talking about different people. :roll:[/quote]
huh[/quote]
Ragnarok said the the second goalie was an idiot, but Koda thought Ragnarok was talking about the first one.

[quote="i_like_pie"][/quote][quote="junggi"][quote="i_like_pie"][quote="Koda"]Who scored an own goal? I was talking of the first goal. And the second goalie was not supposed to have eyes behind the back of his head after all![/quote]
Ragnarok said the the second goalie was an idiot. You two are talking about different people. :roll:[/quote]
huh[/quote][quote="i_like_pie"]
Ragnarok said the the second goalie was an idiot, but Koda thought Ragnarok was talking about the first one.[/quote]
ragnarok is talking about the first one isn't he?
he said it before you posted the second clip

[quote="junggi"][/quote][quote="i_like_pie"][/quote][quote="junggi"][quote="i_like_pie"][quote="Koda"]Who scored an own goal? I was talking of the first goal. And the second goalie was not supposed to have eyes behind the back of his head after all![/quote]
Ragnarok said the the second goalie was an idiot. You two are talking about different people. :roll:[/quote]
huh[/quote][quote="i_like_pie"]
Ragnarok said the the second goalie was an idiot, but Koda thought Ragnarok was talking about the first one.[/quote][quote="junggi"]
ragnarok is talking about the first one isn't he?
he said it before you posted the second clip[/quote]
Yes, they were both talking about the first clip, but Koda misunderstood Ragnarok.

[quote="Koda"]Ok here is how I see it:

ragnarok23 was talking about the first goalie.
Koda was talking about the second goalie.
i_like_pie was talking about the second goal.

So we have two goals and three goalies, exactly two of whom are idiots.[/quote]
only one idiot here.
ragnarok was talking of the 1st goalie
koda was talking about the 2nd goalie(here's the idiot)
i like pie was takling about 1st goal.(and well, the goaile technically scored own goal, he sort of pushed the ball in, well while trying to get it out)

(ok this spamming is getting stupid lets just stop)


Asymptote LaTeX articles integration trigonometry

I've been trying to include Asymptote codes in my Latex documents, you know, using the environment

\begin{asy}
...
\end{asy}

inside the Latex document. I followed the steps of first passing the tex code through Latex, then run Asymptote on the new asy file containing the code, and then passing the tex file through Latex again. But the picture does not appear at all! I just get an error message from Latex:

! LaTex error: File 'example' not found.

Can someone please help me fix this error? :(

Please could you post a minimal example of where this happens - that means a complete tex document with a short version of the code but which still causes a problem. Then we can find out what example file it is that LaTeX can't find.

Actually this is just an example I found in a Asymptote's FAQ at Art of Problem Solving forum (so it's supposed to work!):

\documentclass{article}
\usepackage[pdftex]{graphicx}
\usepackage{asymptote}
\begin{document}
Hello.
I like to make pics with Asymptote like this one:
\begin{figure}[h]
 \begin{asy}
  include graph;
  size(1inch);
  filldraw(circle((0,0),1),yellow,black);
  fill(circle((-.3,.4),.1),black);
  fill(circle((.3,.4),.1),black);
  draw(arc((0,0),.5,-140,-40));
\end{asy}
\end{figure}
\par It makes me happy,
since I can still type my normal LaTeX stuff around it:
\(\displaystyle\int_0^{\pi}{\sin{x}}\,dx=2\)
\end{document}

I really don't know why I get the error message saying that latex cannot find the file "example_1", because I searched the directory where the file "example.tex" is located, and the files "example.asy" and "example_1.ps" are right there! Perhaps LaTex does not recognize the ps file...

Thanks in advance for your help! :)

Are you compiling it with PDFtex or with LAtex?

When I was compiling with Latex, I realized that removing the line \usepackage[pdftex]{graphicx} made the Asymptote figure appear! :) So that was it!  :blush: Thanks for your illuminating comment Valentin! :P


Ring Theory
search
superior algebra
superior algebra solved

hi,

i am trying to prove that if p=1 mod 3 , it is of form x*x+3*y*y

I already knew that if p=1 mod 3, then -3 is square mod p

so there is an integer q such that p divides q*q+3

I used an argument  in Z[1/2*(1+/-sqrt(-3))]  to reach the fact that 4*p=a*a+3*b*b
how to proceed from this, because if a and b are odd, i still have a problem?

thx

Normaly, you get such representations as norms in such fields (after some calculations). At your case, does multiplying with a nontrivial root of unity on that number not work¿

But why not proving it more elementary:

Let be $z^2\equiv -3 \mod p$.
Now take integers $x,y$ with $x \equiv zy \mod p$ and $|x|,|y| < \sqrt p, y \neq 0$ by Thue's lemma.
But then $x^2+3y^2 \equiv 0 \mod p$ and $0< x^2+3y^2<4p$.
If $x^2+3y^2=p$ we are done.
If $x^2+3y^2=2p$ we easily get a contradiction.
If $x^2+3y^2=3p$, $3$ divides $x$, so $x=3x'$ and by this $y^2+3x'^2=p$ and done again.

Could You ZETAX  give me a link or explain what is $ Thue's lemma$? thanks

I'm not sure what ZetaX is referring to (googling Thue's Lemma appears to give the answer to a different question), but his solution can be arrived at from a modification of the lattice method given [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=219017]here[/url].

As for your question, note that if $ a, b$ are both odd then $ a^2 \plus{} 3b^2 \equal{} 4 \left( \left( \frac{a\minus{}3b}{2} \right)^2 \plus{} 3 \left( \frac{b\plus{}a}{2} \right)^2 \right)$.  This is just dividing by $ 1 \plus{} \sqrt{\minus{}3}$.

The PlanetMath entry for Thue's Lemma is stupid. The correct one is in http://www.cims.nyu.edu/~tschinke/teaching/Fall05/class4.pdf or directly here on MathLinks at http://www.mathlinks.ro/Forum/viewtopic.php?p=293201#293201 . Tip: always search on this forum first  :wink:


calculus
geometry
geometric transformation
reflection
conics
ellipse
vector

Can someone help me with the following proof? Thanks :)

Prove the following reflection property of prolate spheroids. Given an ellipsoid embedded in space as {(x, y, z) in R^3 | x^2/(a^2) + y^2/(a^2) + z^2/(c^2) = 1} with c > a. Show there exist two points F1 and F2 on the interior of the ellipsoid such that a particle ejected from F1 in any direction will bounce off the surface of the ellipsoid in such a manner that it will hit F2.

Does the same property hold for arbitrary ellipsoids {(x, y, z) in R^3 | x^2/(a^2) + y^2/(b^2) + z^2/(c^2) = 1}?

The prolate spheroid is obtained by revolving an ellipse around its long axis. This suggests that the points $ F_1$ and $ F_2$ are the foci of the ellipse, namely $ (0,0,\pm\sqrt {c^2 \minus{} a^2})$. The "bouncing" property can be expressed as follows: for any point $ P$ on the surface the difference of unit vectors $ \frac {\overrightarrow{F_1P}}{|\overrightarrow{F_1P}|}$ and $ \frac {\overrightarrow{PF_2}}{|\overrightarrow{PF_2}|}$ is parallel to the normal vector at $ P$. Indeed, if these unit vectors represent the velocity of particle before and after hitting $ P$, then they should differ only in the normal component.

A generic ellipsoid does not have the focusing property. Different cross-sections along its longest axis are ellipses with varying eccentricity and therefore varying position of foci.


advanced fields
advanced fields unsolved

Are the forms $ 3x^2 \plus{} 2xy \plus{} 23y^2$ and $ 7x^2 \plus{} 6xy \plus{} 11y^2$ equivalent?

the first form can't represent 7 so they are not eq (?)

I think that he is asking that the two values CAN be equal.

So...I am posting the solution. Let be the two values are A and B.

If $ A\minus{}B\equal{}4x^{2}\plus{}4xy\minus{}12y^{2}\equal{}0$, $ x^2\plus{}xy\minus{}3y^2\equal{}0$. Since $ D\equal{}13$, the two values can't be equal.


whats the most embarrasing moment of your life

[quote="hunter34"]do i have to tell you?

[hide]
When I was little, I ran into a college building in my pajamas
[/hide][/quote]

Wow, must have been embarrassing.

[hide]
When I was little, I mistaken some stranger for my dad. I called him dad and then he turned around and I ran away.[/hide]

Euggh.  I'm not going to tell you.  It's too embarrassing.[hide=" :blush: "]MathCounts.  I got to the countdown round and then froze in front of the entire audience.  Under pressure, I couldn't figure out what 40% of 50% was.  Still gives me the shivers thinking about it.  I even raised my hand at one point and said 400 but my voicy was all high and squeaky and the answer was 600 anyway.[/hide]

Errr... please, don't read...

[hide][hide][hide]Oh. That's embarassing. hahhaha[/hide][hide][/hide][/hide][/hide][hide][/hide]

[quote="archimedes1"][quote="BanishedTraitor"]
[hide]
When I was little, I mistaken some stranger for my dad. I called him dad and then he turned around and I ran away.[/hide][/quote]
haha that reminds me of one time when I was about 4...
[hide][hide="why are we hiding things if people seee them anyway?"]I didn't call my dad "Daddy" like most little American boys did.  When my dad once ran to go buy flight tickets(we were late) or something my mom suddenly remembered something important she wanted to tell him.  She told me to run after my dad and call him back, but call him "Daddy!" (to fit in) instead of what I usually call my dad.  I don't know why she told me to say "daddy"  but I asked no questions.  So I ran after my dad through the airport yelling "Daddy!  Daddy! Wait!!!"  and of course, my dad didn't turn around because he thought it was some other kid, he knew I never called him "Daddy."  I kept running after my dad and he kept running and people turned and stared. "Why is this idiot running away from his kid?"   :P  :oops:  [/hide] [/hide][/quote]

What did you call him?

It all comes from not listening to your mother.

[hide="mine"]My mom was carrying around, and when we came into the kitchen, I said randomly "That black burner is cold." My mom said "No, it's not. It was just on."

Then I screamed:

"BLACK MEANS COLD!!!"

and touched the hot burner.

You should have seen the look on my face.[/hide]


Find all functions $g: \mathbb R \to \mathbb R$ such that for all real numbers $x$ and $y$,
\[g(x+y)+g(x)g(y)=g(xy)+g(x)+g(y).\]

[hide="Solution"]
Plug in $ x\equal{}y\equal{}0$. We see that:
\[ g(0) \plus{} [g(0)]^2 \equal{} 3g(0)\]

Solving, we see that we have two cases: $ g(0) \equal{} 0$ or $ g(0) \equal{} 2$.

Case 1: $ g(0) \equal{} 2$

Plug in $ y \equal{} y$ and $ x \equal{} 0$. We see that:
\[ g(y) \plus{} 2g(y) \equal{} 2 \plus{} 2 \plus{} g(y)\]

Solving for that, we see that the only possible solution is that $ g(y) \equal{} 2$.

Case 2: $ g(0) \equal{} 0$

Plug in $ y\equal{}y$ and $ x\equal{}0$. We see that:
\[ g(y) \plus{} 0 \equal{} 0 \plus{} 0 \plus{} g(y)\]

Obviously, we see that this restriction that $ g(y) \equal{} g(y)$ holds for every function. So we see that as long as $ g(0) \equal{} 0$, any function $ g(x)$ is a solution to this functional equation.

To summarize, all of the solutions are: $ g(y) \equal{} 2$ and any function $ g(y)$ such that $ g(0) \equal{} 0$.
[/hide]

[quote="bokagadha"]
Plug in y=y and x=0. We see that:
$ g(y) + 0 = 0 + 0 + g(y)$[/quote]

you are right when you say that this holds for every function.
however note that this is only the case when you have plugged x = 0 and y = y

if you plug x = 1 and y = y we get

$ \& g\left( {y + 1} \right) + g\left( y \right)g\left( 1 \right) = g\left( y \right) + g\left( y \right) + g\left( 1 \right) \\
\& \Rightarrow g\left( {y + 1} \right) + g\left( y \right)g\left( 1 \right) = 2\cdot\left( {g\left( y \right)} \right) + g\left( 1 \right)$

can you say that this is true for all functions $ g\left( {y} \right)$?.
remember, a function is required which satisfies the give condition for [u][b]all real numbers $ x$ and $ y$[/b]
[/u]
hence your logic is incomplete and solution incorrect.

e.g. consider the function
\[ g\left( x \right) = \left\{ {
   6,\forall x \ne 0,  \\
   0,x = 0  \\
 } \right\}\]
$ \& g\left( 2 \right) + g\left( 1 \right)g\left( 1 \right) = 6 + 36 = 42  \\
  \& g\left( {1\cdot1} \right) + g\left( 1 \right) + g\left( 1 \right) = 6 + 6 + 6 = 18  \\
  \& LHS \ne RHS $

[quote="KMK00009"]Find all functions g:R->R such that for any real numbers x and y:
g(x+y)+g(x)g(y)=g(xy)+g(x)+g(y)[/quote]
Let $ P(x,y)$ be the assertion $ f(x\plus{}y)\plus{}f(x)f(y)\equal{}f(xy)\plus{}f(x)\plus{}f(y)$
Let $ f(1)\equal{}a$

$ P(1,1)$ $ \implies$ $ f(2)\equal{}3a\minus{}a^2$
$ P(2,2)$ $ \implies$ $ f(2)^2\equal{}2f(2)$ and so $ (3a\minus{}a^2)^2\equal{}2(3a\minus{}a^2)$ $ \iff$ $ a(a\minus{}1)(a\minus{}2)(a\minus{}3)\equal{}0$ and so :

1) either $ a\equal{}0$
===========
Then $ P(x,1)$ $ \implies$ $ f(x\plus{}1)\equal{}2f(x)$
So $ P(x,y\plus{}1)$ $ \implies$ $ f(x\plus{}y\plus{}1)\plus{}f(x)f(y\plus{}1)\equal{}f(xy\plus{}x)\plus{}f(x)\plus{}f(y\plus{}1)$ and so $ 2f(x\plus{}y)\plus{}2f(x)f(y)\equal{}f(xy\plus{}x)\plus{}f(x)\plus{}2f(y)$
And since $ P(x,y)$ $ \implies$ $ 2f(x\plus{}y)\plus{}2f(x)f(y)\equal{}2f(xy)\plus{}2f(x)\plus{}2f(y)$, we get (subtracting these two equalities) :
$ 2f(xy)\plus{}f(x)\equal{}f(xy\plus{}x)$

Set $ x\ne 0$ and $ y\equal{}\frac 1x$ in this equality and we get $ f(x\plus{}1)\equal{}f(x)$ and, since we algeady got $ f(x\plus{}1)\equal{}2f(x)$ :
$ f(x)\equal{}0$ $ \forall x\ne 0$
Then $ P(1,\minus{}1)$ $ \implies$ $ f(0)\equal{}0$ and so $ f(x)\equal{}0$ $ \forall x$
Which, indeed, is a solution of our equation.

2) either $ a\equal{}1$
==========
Then  $ P(x,1)$ $ \implies$ $ f(x\plus{}1)\equal{}f(x)\plus{}1$ and so $ f(n)\equal{}n$ $ \forall n\in\mathbb Z$
Then $ P(x,y\plus{}1)$ $ \implies$  $ f(x\plus{}y\plus{}1)\plus{}f(x)f(y\plus{}1)\equal{}f(xy\plus{}x)\plus{}f(x)\plus{}f(y\plus{}1)$ and so $ f(x\plus{}y)\plus{}f(x)f(y)\equal{}f(xy\plus{}x)\plus{}f(y)$
Subtracting $ P(x,y)$ from this equality, we get $ f(xy\plus{}x)\equal{}f(xy)\plus{}f(x)$

Setting $ x\equal{}u\ne 0$ and $ y\equal{}\frac vu$ in this equality, we get $ f(u\plus{}v)\equal{}f(u)\plus{}f(v)$ $ \forall u\ne 0,v$ and, since $ f(0)\equal{}0$ :
$ f(x\plus{}y)\equal{}f(x)\plus{}f(y)$ $ \forall x,y$. Setting this in $ P(x,y)$, we get also :
$ f(xy)\equal{}f(x)f(y)$
and this is a classical functional equation whose unique solution is $ f(x)\equal{}x$ $ \forall x$
Which, indeed, is a solution of our equation.

3) either $ a\equal{}2$
==========
Then  $ P(x,1)$ $ \implies$ $ f(x\plus{}1)\equal{}2$ and so $ f(x)\equal{}2$ $ \forall x$
Which, indeed, is a solution of our equation.

3) either $ a\equal{}3$
==========
Then $ P(1,1)$ $ \implies$ $ f(2)\equal{}0$
Then $ P(x,1)$ $ \implies$ $ f(x\plus{}1)\equal{}3\minus{}f(x)$ and so $ f(x\plus{}2)\equal{}3\minus{}f(x\plus{}1)\equal{}f(x)$
But $ P(x,2)$ $ \implies$ $ f(x\plus{}2)\equal{}f(2x)\plus{}f(x)$
And so $ f(2x)\equal{}0$, which is impossible since, for example, $ f(2\frac 12)\equal{}3$

4) Synthesis of solutions
=========================
We got three solutions :
$ f(x)\equal{}0$ $ \forall x$
$ f(x)\equal{}2$ $ \forall x$
$ f(x)\equal{}x$ $ \forall x$

[quote="pco"]
Setting $ x\equal{}u\ne 0$ and $ y\equal{}\frac vu$ in this equality, we get $ f(u\plus{}v)\equal{}f(u)\plus{}f(v) \forall u\ne 0,v$ and, since $ f(0)\equal{}0$ :
[/quote]

some questions:
why put the extra condition of $ u\ne v$
why would $ f(0)\equal{}0$ have any implications. because, since u,v are completely random real numbers, hence the result $ f(x\plus{}y)\equal{}f(x)\plus{}f(y)$ $ \forall x,y$. should be true anyway. is it just to account for the case of $ f\left( {0 \plus{} 0} \right) \equal{} f\left( 0 \right) \plus{} f\left( 0 \right)$

if someone could kindly explain, id be most grateful

[quote="Maharjun"][quote="pco"]
Setting $ x \equal{} u\ne 0$ and $ y \equal{} \frac vu$ in this equality, we get $ f(u \plus{} v) \equal{} f(u) \plus{} f(v) \forall u\ne 0,v$ and, since $ f(0) \equal{} 0$ :
[/quote]

some questions:
why put the extra condition of $ u\ne v$
why would $ f(0) \equal{} 0$ have any implications. because, since u,v are completely random real numbers, hence the result $ f(x \plus{} y) \equal{} f(x) \plus{} f(y)$ $ \forall x,y$. should be true anyway. is it just to account for the case of $ f\left( {0 \plus{} 0} \right) \equal{} f\left( 0 \right) \plus{} f\left( 0 \right)$

if someone could kindly explain, id be most grateful[/quote]
1) There is no extra condition $ u\ne v$. I just wanted to write : $ \forall u\ne 0,\forall v$

2) We need $ u\ne 0$ in order to allow $ y\equal{}\frac vu$ And so we established $ f(u\plus{}v)\equal{}f(u)\plus{}f(v)$ only for $ u\ne 0$. Then, $ f(0)\equal{}0$ implies $ f(0\plus{}v)\equal{}f(0)\plus{}f(v)$ and so the equatlity $ f(u\plus{}v)\equal{}f(u)\plus{}f(v)$, which was first established only when $ u\ne 0$ is also true for $ u\equal{}0$ and so is true $ \forall u,v$

[quote=pco][quote="KMK00009"]Find all functions g:R->R such that for any real numbers x and y:
g(x+y)+g(x)g(y)=g(xy)+g(x)+g(y)[/quote]
Let $ P(x,y)$ be the assertion $ f(x\plus{}y)\plus{}f(x)f(y)\equal{}f(xy)\plus{}f(x)\plus{}f(y)$
Let $ f(1)\equal{}a$

$ P(1,1)$ $ \implies$ $ f(2)\equal{}3a\minus{}a^2$
$ P(2,2)$ $ \implies$ $ f(2)^2\equal{}2f(2)$
[/quote]
How do you get rid of $f(4)$?

[quote=pco][quote=Delray][quote=pco]Let $ P(x,y)$ be the assertion $ f(x\plus{}y)\plus{}f(x)f(y)\equal{}f(xy)\plus{}f(x)\plus{}f(y)$
...$ P(2,2)$ $ \implies$ $ f(2)^2\equal{}2f(2)$
[/quote]
How do you get rid of $f(4)$?[/quote]
$f(2+2)$ in LHS is cancelled by $f(2\times 2)$ in RHS[/quote]

Thanks so much! Not sure why I missed that :)

[quote=pco]
2) either $ a\equal{}1$

$ f(x\plus{}y)\equal{}f(x)\plus{}f(y)$ $ \forall x,y$. Setting this in $ P(x,y)$, we get also :
$ f(xy)\equal{}f(x)f(y)$
and this is a classical functional equation whose unique solution is $ f(x)\equal{}x$ $ \forall x$
Which, indeed, is a solution of our equation.

[/quote]

Dear PCO. I haven't found source on the internet on what are the solutions for $f(x)f(y)=f(xy)$, if neither is $x,y \geq 0$ necessarily, nor is $f$ is continous. Are you sure this is true? Or did you combine it with $f(x)+f(y)=f(xy)$?

[quote=misinnyo]Or did you combine it with $f(x)+f(y)=f(xy)$?[/quote]
Indeed it's the combination of multiplicativity ($f(xy)=f(x)f(y)$) and additivity ($f(x+y)=f(x)+f(y)$) that makes the solution unique.
Indeed, it is well-known and rather easy to prove that any additive function $f:\mathbb{R} \to \mathbb{R}$ is either of the form $f(x)=cx$ for some $c$ (and here only $c=f(1)=a=1$ satisfies the assumptions) or the graph of $f$ in the plane must be dense (using the fact that you can always "rationally interpolate" between any two points on the graph).
This is why additive implies linear as soon as any of the conditions like boundedness on an interval, continuity at one point etc. are satisfied.
And here we can use multiplicativity: Indeed, $f(xy)=f(x)f(y)$ implies $f(x^2)=f(x)^2 \ge 0$ and hence $f(x) \ge 0$ whenever $x \ge 0$.
So $f$ is bounded from below on the positive numbers and hence must indeed be linear.

This is a wonderful FE problem!
Here's my somewhat-bashy solution:

Plugging in $(x,y)=(0,0)$ into the equation yields that $f(0)=0,2$. If $f(0)=2$, then plugging in $(x,y)=(x,0)$ yields $f(x)+f(0)f(x)=2f(0)+f(x) \implies f(x)=2$. We can easily show that this holds.

What happens when $f(0)=0$ is somewhat elusive, so here we go!  We plug in $(x,y)=(2,2)$ to find that $f(4)+(f(2))^2-f(4)=2f(2) \implies f(2)=0,2$.

[b][size=150]Case 1: $f(0)=0, f(2)=0$[/size][/b]
-----
Let $(x,y)=(1,1)$. Then, $f(2)+((f(1))^2=3f(1) \implies f(1)=0,3$.
-----
[b]Subcase 1: $f(0)=0, f(1)=0, f(2)=0$[/b]
Then, $f(x+1)+f(x)f(1)=2f(1)+f(x) \implies f(x+1)=f(x) \implies f(x)=0$.
-----
[b]Subcase 2: $f(0)=0, f(1)=3, f(2)=0$[/b]
Then, $f(x+1)+f(x)f(1)=2f(x)+f(1) \implies f(x)=3-f(x+1)$. This is equivalent to saying $f(x+2)=f(x)$ or $f(x+4)=f(x)$. Plugging in $(x,y)=(2,\frac{1}{2})$ into the original equation yields $f(2+1/2)+f(1/2)f(2)=f(-2)+f(2)+f(-1) \implies f(5/2)=f(1/2)+3$, which contradicts our claim. Hence, it is impossible to have a function that satisfies the given equation and has $f(0)=0, f(1)=3, f(2)=0$.
-----
[b][size=150]Case 2: $f(0)=0, f(2)=2$[/size][/b]
Letting $(x,y)=(1,1)$, we have that $f(2)+(f(1))^2=3f(1) \implies f(1)=1,2$.

[b]Subcase 1: $f(0)=0, f(1)=1, f(2)=2$[/b]
-----
Letting $(x,y)=(1,x)$, we have that $f(1+x)+f(1)f(x)=f(x)+f(x)+f(1) \implies f(1+x)=f(x)+1$. It's easy to prove that, by induction, $f(x)=x$ for all integer $x$. Let $(x,y)=(x,y+1)$. Then, $f(x+y+1)+f(x)f(y+1)=f(x(y+1))+f(x)+f(y+1)$. Subtracting $f(x+y)+f(x)f(y)=f(xy)+f(x)+f(y)$ from this inequality yields that $f(x(y+1))=f(xy)+f(x)$. Plugging in $(x,y)=(x,\frac{y}{x})$ yields $f(x+y)=f(x)+f(y)$ for all non-zero $x$. We can easily check that this holds when $x=0$ as well. Hence, remembering that if $f(x+y)=f(x)+f(y)$, then $f(x)=x$, $f(x)=x,0$ satisfies the above FE.
-----
[b]Subcase 2: $f(0)=0, f(1)=2, f(2)=2$[/b]
Letting $(x,y)=(x,1)$, we have that $f(x+1)=2$. Hence, $f(x)=2$ is a solution to the FE.

Putting this all together, we see that the only solutions are $f(x)=0,2,x$.

$f (x+y)=f (x)+f (y)\nRightarrow f (x)=x $
Note, $g (x)+g (y)=g (x+y)\implies g (x+y)=g (x) g (y) $
Two equations $\implies g (x)=x $, and $g (x)=0$
Solve $P (x,y+1) $ to get Cauchy function.

[quote=AopsUser101]
[b]Subcase 1: $f(0)=0, f(1)=0, f(2)=0$[/b]
Then, $f(x+1)+f(x)f(1)=2f(1)+f(x) \implies f(x+1)=f(x) \implies f(x)=0$.
[/quote]
Isn't $f(x+1)+f(x)f(1) = 2f(x)+f(1)?$ This implies that (for the assumptions of the subcase) $f(x+1)=2f(x),$ which I'm not sure how to disprove.

The only solutions are $f\equiv 0, f\equiv 2, f(x) = x$, which clearly work.

Let $P(x,y)$ denote the given assertion.

$P(0,0): f(0)\in \{0,2\}$. If $f(0) = 2$, then $P(x,0): 3f(x) = f(x) + 4$, so $f(x) = 2$ for each $x$. Henceforth assume $f(0) = 0$.

$P(2,2): f(2)\in \{0,2\}$.

$P(1,1): f(1)^2 - 3f(1) + f(2) = 0$. This gives four cases:
$f(1) = f(2) = 0$
$f(1) = 3$ and $f(2) = 0$
$f(1) = f(2) = 2$
$f(1) = 1$ and $f(2) = 2$

$P(x, y+1): f(x+y+1) + f(x)f(y+1) = f(x) + f(y+1) + f(xy + x)$.

$P(x,1): f(x+1) = 2f(x) + f(1) - f(x)f(1)$.

Thus, if $f(1) = f(2) = 2$, then $f\equiv 2$.

Case 1: $ f(1) = f(2) = 0$
$P(x,1)$ implies that $f(x+1) = 2f(x)$.

$P(x,y+1): 2f(x+y) + 2f(x)f(y) = f(x) + 2f(y) + f(xy + x)$.

Comparing this with the original FE, we get $2f(x) + 2f(y) + 2f(xy)  = f(x) + 2f(y) + f(xy +x)$, so \[f(x) + 2f(xy) = f(xy + x)\]

Set $y = 1/x$ here for $x\ne 0$. We get $f(x) + 2f(1) = f(x+1) = 2f(x)$, so $f(x) = 2f(1) = 0$ for all $x\ne 0$, and ofc $f(0)= 0$.

Case 2: $f(1) = 3$ and $f(2) = 0$.
$P(x,1)$ implies that $f(x+1) = 3-f(x)$.

$P(x, y+1): 3- f(x+y)  + f(x)(3-f(y)) = f(x) + 3-f(y) + f(xy + x)$, so \[ f(x+y)  + f(x)f(y) =2 f(x) +f(y) - f(xy + x)\] Comparing this with the original FE, we get $2f(x) + f(y)  - f(xy + x) = f(x) + f(y) + f(xy)$, which implies \[f(x)  - f(xy + x) = f(xy)\] Setting $x=2, y=1/2$ here gives $f(2) - f(3) = f(1)\implies 0 - 3 = 3$, which isn't true.

Case 3: $f(1) = 1$ and $f(2) = 2$.
$P(x,1)$ implies that $f(x+1) = f(x) + 1$.

$P(x,y+1): f(x+y)  + f(x)f(y)  =  f(y)  + f(xy + x)$. Comparing with the original FE gives $f(x) + f(xy) = f(xy + x)$.

For $x\ne 0$, $P(x, y/x): f(x) + f(y) = f(x+y)$. This also holds for $x=0$, so $f$ is additive. Using this in the original FE, we have $f(x)f(y) = f(xy)$, so $f$ is multiplicative. Since $f(1) = 1$, it is well known that $f(x) = x$ for all $x$.

This has some nice parts, and some terrible bashy parts. The terrible part comes first. $f(x)=0,2,x$ are our three solutions.

Set all $0$ to get $f(0) = 0$ or $f(0) = 2$. In the latter case, setting $y=0$ makes $f$ identically $2$, so throw that case out. $f(0) = 0$ now yay.

Plug in $(-2,1)$, $(-1,-1)$, and $(1,-1)$ to get a system of three equations in $f(1),f(-1),f(2)$, and solve. You get either $f(-1)=f(1)=f(2) = 0$, all of them map to themselves, or... (here's the hellish case) $f(1) = f(-1) = 3,f(2) = 0$.

We have to kill the nasty case first. We can get a recursion $f(n+1)+f(n)f(1) = f(n)+f(1)+f(n)$, and pump it (forwards and backwards) to get $f$ alternates between $3$ and $0$ on integers. We can now plug in $x=-\frac{1}{2},y=1$ to get $f(\frac{1}{2}) = 1.5$, and pumping our recursion here (forwards and backwards) gets us $f(k+\frac{1}{2})=\frac{3}{2}$ for all integers $k$. Now plug in $(-0.5,2)$ to get $f(-1.5)=f(-0.5)+3$, which is a contradiction. This kills the nasty case.

Now for the zero case: Pump the recursion to get $f(n) = 0$ for all integers $n$. Now, plugging in $x=1$ gives $f(y+1) = f(y)$ for all real $y$ now, so the function is $1$-periodic. Now here's the kicker; compare $P(x,y)$ to $P(x+n,y+n)$. EVERYTHING is identical but one tiny thing; we get $f(xy) = f(xy+n(x+y))$ for all integers $n$ and reals $x,y$. Now, consider $xy=1$; we can turn $x+y$ into anything in $[2,\infty)$ and $(-\infty,-2]$, so we have $0=f(1)=f(r)$ for any $r \geq 3$ or $r \leq -1$. Now look at $xy=16$; we can turn $x+y$ into anything at most $-8$, so we get $f(16) = f(r)$ when $r$ is at most $8$. This implies $f$ is constant everywhere, and then zero everywhere, as desired.

Now, for the $x$ case: Our recursion now gets $f(x+1) = f(x)+1$ for all $x$. Now, plug in $x=n$ instead; our $f(x+1) = f(x)+1$ already told us $f(x+n) = f(x)+n$, so we get the much-needed $f(nx) = nf(x)$ for any real $x$ and integer $n$. In particular, $f$ is odd and $f(2x) = 2f(x)$. We will now exploit INEQUALITIES to solve the problem.

Plug in $(x,x)$ to now get (after cancelling $f(2x)$ etc. :)) that $f(x^2) = f(x)^2$, or that $f$ is nonnegative on nonnegatives. Odd function implies nonpositive on nonpositives too.

Claim: $f(x) \geq x$ for all $x > 0$

Proof: Assume FTSOC $f(x_0) = x_0 -\epsilon$ for some positive $\epsilon$. Then, $f(Nx_0-\lfloor Nx_0 \rfloor) = Nf(x_0)-\lfloor Nx_0 \rfloor = Nx_0-\lfloor Nx_0 \rfloor-N\epsilon$. The LHS has to be nonnegative, but $Nx_0-\lfloor Nx_0 \rfloor) < 1$ so picking $N$ large with $N\epsilon > 1$ yields a negative RHS, contradiction! This implies the claim.

Claim: $f(x) \leq x$ for all $x > 0$

Proof: Assume FTSOC some $f(x_0) = x_0 + \epsilon$ for some positive $\epsilon$. Then, $f(-x_0+N) = N+f(-x_0) = -x_0+N-\epsilon$. Picking $N$ large such that $-x_0+N$ then implies some positive $x$ exists such that $f(x) < x$, contradicting our previous claim! This implies our desired claim.

This means $f(x) = x$ everywhere. PHEW.


In each row of $(n-2)\times n$ array numbers $1, 2,..., n$ are written (all of them, so there is no number which is written twice) and the numbers in each column are all diferent.
 Prove that this array can be expanded into an $n \times n$ array such that each row and each column consists of the numbers $1, 2,...,n$

Please, post only solution which isn't via graphs.....

It's pretty clear what one must do:

First place the two missing numbers $a_{n-11},a_{n1}$ in the first column as you wish. After that, go to the column $k$ which has to contain $a_{n1}$ and place $a_{n-1}k=a_{n1},a_{nk}$. Then, go to the column $k'$ which must contain $a_{nk}$, place the numbers $a_{n-1k'}=a_{nk},a_{nk'}$, and so on, until we complete a cycle. After that we simply choose another column, place the two missing numbers there arbitrarily, and repeat the process.

Sorry if it's unclear, but it's a lot easier to see than to explain :).

I don't think so...
Look at rows, in that way you place all the numbers in which columnt they have to be but what about rows...

Read it more carefully, man :). That procedure ensures that you never put two identical numbers in the same row. That means that when it finishes, the two rows will each have the numbers $1,2,\ldots,n$.


There is a m*n grid. there are many trees in it. given an integer L. you have to surround an area which has perimeter <= L and contains maximum possible trees. What is the number of trees in that area?

Note that, the trees are at grid points and the perimeter should be alog the grid line.

If you make 2d array of booleans where 1 represents a tree, you have to find maximum sum subrectangle with the w*h = L. Then just apply DP.

well i think you missed one thing,

you said, w*h=L.

But actually w*h <= L possible.

we need O(N^3) algorithm (as far as i can remember the bound of W and H) or O(N^2 lgN)

[quote]
well i think you missed one thing,

you said, w*h=L.

But actually w*h <= L possible. [/quote]

Oops, even that's wrong  :blush:

You said perimeter, so 2*(w+h) <= L.

[quote]we need O(N^3) algorithm (as far as i can remember the bound of W and H) or O(N^2 lgN)[/quote]

Finding maximum sum subrectangle takes O(n^3). Adding restriction on perimeter wouldn't change it very much.

you mean the normal O(N^3) algorihtm that we use for the maximum sub rectangle? how?

say you will restrict two rows a and b. and now you will loop through columns. when column length saturates say: i to j. then we will move i -> i+1. but j's value can be i+1<= j <current_j

so how it will get AC?

Wait I'm confused  :? I didn't get your point at all.

Are you saying that maximum sum sub rectangle is impossible in O(n^3) or it's impossible to apply it to that problem?

O(N^3) for maximum subractangle problem is surely possible.

But it can't be applied in this problem.

oh sorry! it is possible: say we select 2 rows. H distance. Now we select W so that: 2(W+H) <= L. and move it.

O(N^3)

I think that the problem will get more interesting if the selected area would not necessarily be a rectangle, while its edges still having to remain paralel to the axes. We had a similar one at the ioi selection, but only one guy managed to find a reasonable solution for it. What do you say?  :D

well... i am confused, u said edges parallel to the axis, but not rectangle, how???

did u mean concave polygon?

Theres plenty of shapes with sides parallel to the axes that aren't rectangles ;) They just have more than 4 sides.


[quote="rituraj007"]cud u plz suggest an event from some mathematicians life/related to some discovery which can b depicted as short play???????? :D[/quote]

copy the film of ramanujan :D  Where the goddess of namakkal comes and ....... :D

Many would remember that i suppose..........

And also, i thought that i heard someone saying that some mathematician had been depicted in a film :)

[quote="madness"]
And also, i thought that i heard someone saying that some mathematician had been depicted in a film :)[/quote]
that would be [url=http://en.wikipedia.org/wiki/A_Beautiful_Mind_%28film%29]A beautiful Mind[/url]  :)

hey man!!!i want a 3-minute skit!!!!!!!!

[quote="madness"]then make it like one second ramanujan is born and the next second he goes to school :rotfl: :rotfl: :D[/quote]

it can't be second ramanujan. it has to be third.
the second one was already born on september 5th, 1990  :rotfl:  :rotfl:   :P

[quote="hash_include"][quote="madness"]then make it like one second ramanujan is born and the next second he goes to school :rotfl: :rotfl: :D[/quote]

it can't be second ramanujan. it has to be third.
the second one was already born on september 5th, 1990  [/quote]

and who may that be  :huh:   :P  Well i said one second wrt time :rotfl:

now that you ask, let me make it clear to everyone  :D
rohit (aka madness aka sandeepramesh) was born on september 5, 1990  :P  :D

:wallbash:  :wallbash_red: Wah!!!!!!!!!!!!!!
I missed being RAMANUJAN by 4 days!!!!!!!!!!!!!!!!!!!!!!! :noo:


The rational numbers between 0 and 1 are listed in a sequence \[0.a_{11}a_{12}a_{13}\dots,\ 0.a_{21}a_{22}a_{23}\dots,\ 0.a_{31}a_{32}a_{33}\dots \] ie. $a_{nd}$ is digit number $d$ in the representation of the $n$th number in the list.

(Don't worry too much about the list, the rationals are countable etc, but if you don't know about countability, then just assume you *can* list them)

Now, given such a list, prove that \[0.a_{11}a_{22}a_{33}\dots\] will always be irrational  :)

well actually this is part of the proof of why the irrational numbers are not countable and the point of it is that $0.a_{11}a_{22}a_{33}...$ is not part of the numbers before. but i forgot why ...

Actually, the proof of uncountability uses the complement of $a_{ii}$.  This problem is harder.

o yeah thats what it was, thats the reason too why i didnt get why it wasnt in the set of numbers ...

Simple. Take $1-0.a_{11}a_{22}a_{33}\dots = 0.b_{11}b_{22}b_{33}\dots $.
Since $1 = .99999\dots$, $b_{nn} = 9-a_{nn}$.
So, $\forall n: b_{nn} \neq a_{nn}$.
Now,  $0.b_{11}b_{22}b_{33}\dots$ can't be on the list, because it disagrees in the nth place with the nth member of the list.
So, $0.b_{11}b_{22}b_{33}\dots$ irrational.
And since $\mathbb{Q}$ is closed under subtraction and whatnot,  $1-0.a_{11}a_{22}a_{33}\dots$ irrational.

Yep, that is the solution. It is a more generalized version of Cantor's method.

I will propose a second challenge: Can you construct [i]every[/i] irrational number like this? :)

i would say if u could, the set of irrationals would be countable as well, since u could make a list of numbers like this:

$a_{11}a_{22}a_{33}...$
$a_{12}a_{21}a_{33}...$
$a_{11}a_{23}a_{32}...$
$\vdots$
$a_{21}a_{12}a_{33}...$
$\vdots$

(dont know if that argumentation is correct  :blush: )

I'm not exactly sure what you mean by "can be constructed like this", but I think you mean whether for all irrationals (in (0,1)) there exists an ordering of $\mathbb{Q}$ n [0,1] such that the diagonal is that irrational. If so, the answer is no. Consider an irrational like
$A = \sum _{n=0} ^{\infty} 10^{n!}$ (I know this has a name but I forget what it is. I think it's transcendental.) Where can $\frac{7}{9}$ go on the list so that the nth digit of $\frac{7}{9}$ = the nth digit of A?

I wonder if all irrational numbers such that they contain an infinite number of each digit can be constructed like this. There's uncountably many orderings of $\mathbb{Q}$, so I don't think a countability argument can help, but I could be wrong.

[quote="tomkern"]
I wonder if all irrational numbers such that they contain an infinite number of each digit can be constructed like this. [/quote]

There are such numbers which cannot be constructed as described above. Choose some rational $r$ which does not have eventually constant decimal representation. Take $r=0.(01)$, for instance (the $(01)$ is its period). Now you can easily construct an irrational number whose decimal representation contains every digit infinitely many times, and whose $n$'th digit differes from the $n$'th digit of $r$ for all $n$.


inequalities
calculus
integration
induction
derivative
logarithms
calculus computations

Show that
\[ \sum^N_{k=1}\frac{2k}{1+k^2} \ge ln(1 + N + \frac12 N^2)  \]

Since $\frac{2x}{1 + x^2}$ is decreasing for $x \ge 1$, integral estimation gives that $\sum^N_{k=1}\frac{2k}{1+k^2} \ge \frac{2n}{1+n^2} + \int^n_1\frac{2x}{1 + x^2}dx = \frac{2n}{1+n^2} + \ln(1+n^2) - \ln2$

However, showing that this is larger than what's given in the question seems difficult. Induction gets messy really quickly, and comparing derivatives is also hard. Is there something obvious I am missing here?

edit: I am inconsistent with n and N, but I mean the same thing with them.

Note that $\frac{2n}{1+n^2} > \ln\left(\frac{n^2+2n+2}{n^2+1}\right)$.  This is not really a proof, but as $n$ gets large, the left hand side is essentially $\frac{2}{n}$ and the right is essentially $\ln\left(\frac{n^2+2n}{n^2} \right) = \ln\left(1+\frac{2}{n}\right)$.  Because $n$ is large, the $1$ is negligable and the left hand side is larger.  (I'm going to try to come up with a good proof for this).

After that, we have
$\sum^N_{k=1}\frac{2k}{1+k^2} \ge \frac{2n}{1+n^2} + \int^n_1\frac{2x}{1 + x^2}dx = \frac{2n}{1+n^2} + \ln(1+n^2) - \ln2 > \ln\left(\frac{n^2+2n+2}{n^2+1}\right)$$+ \ln(1+n^2) -\ln 2 = \ln\left(1+n+\frac{1}{2}n^2\right)$

I learned in class yesterday that there is another estimation which goes
\[ \sum^n_{k=1}\frac{2k}{1+k^2} \ge \int^{n+1}_1\frac{2x}{1 + x^2}dx \]

The problem is just a tiiiny bit easier using this  :D .

So, the proof doesn't work? Don't we want it to be smaller?

Yeah, that actually doesn't work.  You can probably take more terms of a taylor expansion to fix it, however.

[quote="blahblahblah"][quote="Jimmy"]Note that $\frac{2n}{1+n^2} > \ln\left(\frac{n^2+2n+2}{n^2+1}\right)$ (I'm going to try to come up with a good proof for this).[/quote]

Apply the mean value theorem

We know that

$\log (n^2+2n+2)-\log (n^2+1)=\frac{2n+1}{c}$

for some $c\in(n^2+1,n^2+2n+2)$.  Since $1/c$ is decreasing, it follows that it is at most $\frac{2n+1}{n^2+1}$.  But it is trivial to see that this is larger than $\frac{2n}{n^2+1}$.[/quote]

I'm confused, $\frac{2n}{1+n^2}$ is greater than $\ln\left(\frac{n^2+2n+2}{n^2+1}\right)$ for $n>1$. (at least that's what it looks like graphically).

[quote="blahblahblah"]As I said, I made a mistake.[/quote]

Oh, ok.  I thought you were saying what I wrote didn't work, not what you wrote.


[size=150][b]Equilateral triangles ABX and ACU are described on sides AB, AC of a triangle (ABC) externally to triangle (ABC). Prove that CX=BY.[/b][/size]

Dear Mathlinkers,
you can see also
http://jl.ayme.pagesperso-orange.fr/ vol. 16  Deux triangles semblables..., p. 10.

Sincerely
Jean-Louis


Let $ N$ be the Nagel point of $ \triangle ABC.$ Its incircle $(I)$ touches $BC$ at $X$ and line $ AN$ intersects $(I)$ at $Y$ (closer to BC). $H$ is the foot of the A-altitude $ AH.$ Draw through $ H$ a tangent to $(I)$ which touches it at $P.$ Define lines $s_a \equiv XY$ and  $ r_a \equiv XP.$ Similarly, define the pairs of lines $ s_b,r_b$ and $ s_c,r_c.$ Prove that $ s_a,s_b,s_c,r_a,r_b,r_c$  bound a hexagon circumscribed in a conic.

The said lines are obviously polars WRT the incircle (I) of the altitude feet and midpoints of the triangle sides, all on the 9-point circle (N). They are tangent to a conic with one focus I and pedal circle (N'), the inversion image of (N) in (I).


topology
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real analysis unsolved

let $ f: M\rightarrow\ N$  be a differentiable function and $ dimM \equal{} dimN$  prove that $ f(M)$ is an open subset of $ N$.

Obviously false without additional assumptions. Just take $ M\equal{}N\equal{}\mathbb R$, $ f(x)\equal{}\sin x$. (or, even better, $ f\equal{}\text{const}$)

it's correct if $ rg_x(f) \equal{} dim(N)$ for every $ x \in M$.


Let $n,k$ be positive integers such that $n^k>(k+1)!$ and consider the set
\[ M=\{(x_1,x_2,\ldots,x_n)\dvd x_i\in\{1,2,\ldots,n\},\ i=\overline{1,k}\}. \]
Prove that if $A\subset M$ has $(k+1)!+1$ elements, then there are two elements $\{\alpha,\beta\}\subset A$, $\alpha=(\alpha_1,\alpha_2,\ldots,\alpha_n)$, $\beta=(\beta_1,\beta_2,\ldots,\beta_n)$ such that
\[ (k+1)! \left| (\beta_1-\alpha_1)(\beta_2-\alpha_2)\cdots (\beta_k-\alpha_k) \right. .\]

Is (k+1)! supposed to divide that product, or...

So basically this is a restatement of the Pigeon-Hole principle ... :) (this was actually the official solution to this problem ...  :rotfl: ).


Fixed. Usage of \par in the code is no longer necessary.


articles probability calculus integration Princeton college Yale

John[b] Von[/b] Neumann[b] is my one of my favorite[u] Mathematical[/u] Heroes[u].Accordingly,his ideas ranged from ergodic[b] theorem[/b] to ways of Controlling[b] Weather[/b],from famous theory[b] of[/b] games[b] to the new[/b] theory[b] of[/b] algebra[b],and finally,to Quantum[/b] Physics[b].[/b] Can anyone of the A.O.P.S. Members[b] recommend useful internet[u] references[/u] that collects John Von Neumann collected[u] works[/u] related to his mathematical[u] contributions[/u]?[u]

                                             Respectfully,
                                             A.Eddington[/u][/b]

[b]John [Janos] von Neumann [1903-1957] [/b]

[b] [color=green]References for John von Neumann [/color][/b]

   1. J Dieudonne, Biography in Dictionary of Scientific Biography (New York 1970-1990).
   2. Biography in Encyclopaedia Britannica. [Available on the Web]
   3. Obituary in The Times [available on the Web]

Books:

   4. W Aspray, John von Neumann and the origins of modern computing (Cambridge, M., 1990).
   5. S J Heims, John von Neumann and Norbert Wiener: From mathematics to the technologies of life and death (Cambridge, MA, 1980).
   6. T Legendi and T Szentivanyi (eds.), Leben und Werk von John von Neumann (Mannheim, 1983).
   7. N Macrae, John von Neumann (New York, 1992).
   8. W Poundstone, Prisoner's dilemma (Oxford, 1993).
   9. N A Vonneuman, John von Neumann: as seen by his brother (Meadowbrook, PA, 1987).

Articles:

  10. A Adám, John von Neumann, Life and work of John von Neumann (Mannheim, 1983), 11-43.
  11. H Araki, Some of the legacy of John von Neumann in physics: theory of measurement, quantum logic, and von Neumann algebras in physics, The legacy of John von Neumann (Providence, R.I., 1990), 119-136.
  12. W Aspray, The mathematical reception of the modern computer: John von Neumann and the Institute for Advanced Study computer, Studies in the history of mathematics (Washington, DC, 1987), 166-194.
  13. W Aspray, John von Neumann's contributions to computing and computer science, Ann. Hist. Comput. 11 (3) (1989), 189-195.
  14. W Aspray, The transformation of numerical analysis by the computer: an example from the work of John von Neumann, The history of modern mathematics II (Boston, MA, 1989).
  15. W Aspray, The origins of John von Neumann's theory of automata, The legacy of John von Neumann (Providence, R.I., 1990), 289-309.
  16. H Baumgärtel, John von Neumann, aus Leben und Werk, Mitt. Math. Ges. DDR 3-4 (1982), 87-98.
  17. G Birkhoff, Von Neumann and lattice theory, Bull. Amer. Math. Soc. 64 (1958), 50-56.
  18. P R Halmos, The legend of John von Neumann, Amer. Math. Monthly 80 (1973), 382-394.
  19. P R Halmos, Von Neumann on measure and ergodic theory, Bull. Amer. Math. Soc. 64 (1958), 86-94.
  20. I Halperin, The extraordinary inspiration of John von Neumann, The legacy of John von Neumann (Providence, R.I., 1990), 15-17.
  21. T A Heppenheimer, How Von Neumann showed the way, American heritage of invention and technology 6 (2) (1990), 8-16.
  22. R V Kadison, Theory of operators, Part I. Operator algebras, Bull. Amer. Math. Soc. 64 (1958), 61-85.
  23. H W Kuhn and A W Tucker, John von Neumann's work in the theory of games and mathematical economics, Bull. Amer. Math. Soc. 64 (1958), 100-122.
  24. P D Lax, Remembering John von Neumann, John von Neumann: a personal view, The legacy of John von Neumann (Providence, R.I., 1990), 5-7.
  25. F J Murray, Theory of operators, Part I. Single operators, Bull. Amer. Math. Soc. 64 (1958), 57-60.
  26. L Rédei, John von Neumann's work in algebra and number theory (Hungarian), Mat. Lapok 10 (1959), 226-230.
  27. U Rellstab, New insights into the collaboration between John von Neumann and Oskar Morgenstern on the Theory of games and economic behavior, Toward a history of game theory (Durham, NC, 1992), 77-93.
  28. G Révész, John von Neumann und der Rechner, Life and work of John von Neumann (Mannheim, 1983), 99-113.
  29. C E Shannon, Von Neumann's contributions to automata theory, Bull. Amer. Math. Soc. 64 (1958), 123-129.
  30. F Smithies, John von Neumann, J. London Math. Soc. 34 (1959), 373-384.
  31. N Stern, John von Neumann's influence on electronic digital computing, 1944-1946, Ann. Hist. Comput. 2 (4) (1980), 349-362.
  32. J Szelezsán, John von Neumann, 'der Rechentechniker' (sein Werk auf dem Gebiet der numerischen Methoden), Life and work of John von Neumann (Mannheim, 1983), 115-151.
  33. J Todd, John von Neumann and the national accounting machine, SIAM Rev. 16 (1974), 526-530.
  34. C B Tompkins, John von Neumann, 1903-1957, Math. Tables Aids Comput. 11 (1957), 127-128.
  35. S Ulam, John von Neumann, 1903-1957, Bull. Amer. Math. Soc. 64 (1958), 1-49.
  36. L van Hove, Von Neumann's contributions to quantum theory, Bull. Amer. Math. Soc. 64 (1958), 95-99.
  37. N Vonneuman, The philosophical legacy of John von Neumann, in the light of its inception and evolution in his formative years, The legacy of John von Neumann (Providence, R.I., 1990), 19-24.
  38. M v N Whitman, John von Neumann: a personal view, The legacy of John von Neumann (Providence, R.I., 1990), 1-4.
  39. O P Yee, On the life and work of John von Neumann, Menemui Mat. 4 (3) (1982), 114-127.

[source=Internet]

=====================
I found also:

Subject:      [HM] von Neumann's prodigious mind (was: Oskar Morgenstern)
Author:       Antreas P. Hatzipolakis <xpolakis@otenet.gr>
Date:         Sun, 5 Dec 1999 14:33:23 +0300

John McKay wrote (in part):

> He [Morgenstern] also told of von Neumann's prodigious mind. That
> JvN was asked how thick a coin need be so that the probability of
> its landing on its edge is 1/3. This, he answered (using reasonable
> assumptions) in seconds.

Another problem posed to John von Neumann:

Two trains, 200 kilometres apart, are moving toward each other. Each one
is moving at a constant speed of 50 kilometres per hour. A fly starting
on the front of one of them flies back and forth between the trains at
a rate of 75 kilometres per hour. It does this until the trains collide
(with unfortunate results for the fly). What is the total distance the
fly has flown?

He immediately replied, "150 miles."

"Interesting," said the poser, "but nearly everyone tries to sum the
infinite series."

"What do you mean?" asked Von Neumann. "That's how I did it!"

        http://home.istar.ca/~brsears/oldprob.htm

Antreas

=======================================

Others...:
[b][color=green]Significant Publications[/color][/b]

Charney, J. G., R. Fjörtoft and John von Neumann. 1950. "Numerical Integration of the Barotropic Vorticity Equation", Tellus, Vol. 2, pp. 237-254.

Taub, A. H. (ed). 1961-63. John von Neumann: Collected Works, 1903-1957, 6 Vols., Pergamon Press, Oxford (UK).

von Neumann, John and Oskar Morgenstern. 1944. Theory of Games and Economic Behavior, Princeton Univ. Press, Princeton NJ.

von Neumann, John. 30 June 1945. First Draft of a Report on the EDVAC, Contract No. W-670-ORD-492, Moore School of Electrical Engineering, Univ. of Penn., Philadelphia. Reprinted (in part) in Randell, Brian. 1982. Origins of Digital Computers: Selected Papers, Springer-Verlag, Berlin Heidelberg, pp. 383-392.

von Neumann, John. 1946. "The Principles of Large-Scale Computing Machines", reprinted in Ann. Hist. Comp., Vol. 3, No. 3, pp. 263-273.

von Neumann, John. 1958. The Computer and the Brain, Yale Univ. Press, New Haven.

von Neumann, John and Arthur W. Burks. 1966. Theory of Self-Reproducing Automata, Univ. of Illinois Press, Urbana IL.


induction
function
floor function
strong induction

Mathematical induction for a set of statements indexed by the natural numbers is a simple consequence of the Well-Ordering Principle.  The Well-Ordering Principle states that for $A\subseteq\mathbb{N}$ with $A\neq\varnothing$, there exists $x\in A$ such that for each $y\in A$, $x\leq y$.  Basically what it means is that each subset of the natural numbers contains a least element.  Now consider a set of statements $\left\{S_{n}\,|\,n\in\mathbb{N}\right\}$.  Assume that $S_{1}$ is true, and $S_{n}\rightarrow S_{n+1}$.  If there existed an element $S_{m}$ which was false, then $S_{m}\subseteq A_{F}$ where $A_{F}$ contains a least element by the Well-Ordering Principle.  However, $\min\left\{m\right\}\neq 1$ since $S_{1}$ is true, so it must be $S_{k-1}$ for some $k$.  But since $S_{k-1}\rightarrow S_{k}$ and each $S_{k}$ is true (since a true implies a true), we have a contradiction.  Thus the conditions of $S_{1}$ being true and $S_{n}\rightarrow S_{n+1}$ are sufficient to say that $\left\{S_{n}\,|\,n\in\mathbb{N}\right\}$ contains only true statements.

I hope that wasn't too confusing, but that's the way I was taught to understand why the Principle of Mathematical Induction exists and works.  I'm sure someone may have another explanation which is a bit more clear.

[quote="nateharman1234"]suppose that if the statement is true for k then you prove that it works for k +1. this would mean that before k, k -1 must have worked in order to get to k, and k-2 before that, ect. so it must work for one term at the begining in order for every term after it to work[/quote]

But doesn't any value of $k$ being valid mean that $k$ can be any number from $(- \infty, \infty)$?

Or are you saying that even if $k$ is an infinitely small number, it still must have worked for $k-1$.

I think what I don't understand is that why inputting a variable is insufficient to proving it simply works for everything.  If $k$ [i]works[/i], doesn't that, with no doubt whatsoever, mean that [i]any[/i] $k$ will work?

[quote="Bictor717"][quote="brady00n"]Mathematical induction says that a base case has to be proven [i]before[/i] the inductive step.

Why wouldn't a proof such as the following be invalid?

1. First, try to prove that an assertion is true for $k$.

2. If the statement above is true, then also proceed by trying to prove that the assertion is true for $k+1$.

3. If the statement above is true, then any integer $k$, and $k+1$ is valid.  Therefore, $k$ can be replaced with any integer value.  If any integer value can be chosen for $k$, then every integer is valid.

Why do we have to prove something for a definite non-variable base case in the beginning?  Doesn't the fact that any integer work come naturally if $k$ and $k+1$ is valid?

Thanks.  This has been bugging me for a while...[/quote]
That is not induction. If you manage to prove a statement is true for a general k, as in your step 1, then you showed it's true for any k. Why do you need to go through the trouble of proving k+1 if the statement is true for any k?
Induction says you assume the truth of the statement for k and show that the assumption leads to the truth of the statement for k+1. Then you have to give some value of k for which the statement is true.
Sometimes, induction is presented with a domino analogy. When you show that the assumption for the k case leads to the k+1 case, you show that if a domino is knocked over, it will knock down the one immediately after it. At this point, nothing has actually been physically knocked down. Giving a base case is the knocking-down that begins the rest of the knocking-downs.
[quote="nateharman1234"]suppose that if the statement is true for k then you prove that it works for k +1. this would mean that before k, k -1 must have worked in order to get to k, and k-2 before that, ect. so it must work for one term at the begining in order for every term after it to work[/quote]
That is not what induction says. The classic induction "proof" can demonstrate: Suppose all groups of k horses are the same color. Consider a group of k+1 horses. Remove 1 of them; the remaining k horses are the same color. Put the removed horse back in and remove a different horse. Again, the remaining k horses are the same color. Then all groups of k+1 horses must be the same color.
However, note that this reasoning fails for k=1. You cannot "induct down".[/quote]

Hmm.  I think I get most of it now.  Please correct me where I make errors in my statements or in my logic.  First, a base case (say 43) is true for a given assertion.  Therefore, since one number worked, assume $k$ works.  If $k+1$ also works, then the fact that both $k$ and $k+1$ work is proven.  Therefore, any number and any number above it works, so that means any number works.

I still have a couple questions.
1) You can use any base case.  If the question asks to prove for all integers, then that is saying that all integers already work, and that the only task you must complete is to formally prove that it is indeed true.  Therefore, any base case can be used because any integer indeed works...?
2) So the logical chain is that:

1) A question asks for something to be proven for all integers.
2) If the statement wasn't true, then you wouldn't be asked to prove it.
3) Therefore, you can assume that any integer will work (however, you still must formally prove it).
4) So pick any integer.  It works, to no surprise!
5) Any $k$ should also work, by the premises above.
6) However, we're not absolutely sure if $k+1$ will work unless we prove it.  However, it does work.
7) Therefore, if a certain base case works, and a case higher than the base case also works, then this can be extended infinitely...

I hope I'm at least on the right track...

One more question: Is $k$ implying the base case, or any integer?

[quote="brady00n"]
1) A question asks for something to be proven for all integers.
2) If the statement wasn't true, then you wouldn't be asked to prove it.
3) Therefore, you can assume that any integer will work (however, you still must formally prove it).
4) So pick any integer.  It works, to no surprise!
5) Any $k$ should also work, by the premises above.
6) However, we're not absolutely sure if $k+1$ will work unless we prove it.  However, it does work.
7) Therefore, if a certain base case works, and a case higher than the base case also works, then this can be extended infinitely...

I hope I'm at least on the right track...

One more question: Is $k$ implying the base case, or any integer?[/quote]

We'll if you want to do it that way, then

just say its true because they wouldnt ask you to prove it if it wasnt true.

Obviously, thats a horrible way of going about the problem because you havent proven anything. If its true for k = k+1, that just means that k and k+1 will have the same output.

But we need something to be true to conclude that something else is true, and thats the base case.

I believe k is any integer.

[quote="brady00n"]
1. First, try to prove that an assertion is true for $k$.[/quote]

If you mean $k$ generally then you've already completed your proof!   ;)

And if you mean a specific value of $k$, then this is equivalent to induction, except that you haven't proven for values $< k$.

Edit:  The logic behind mathematical induction is that if you can show that an assertion is true for $1$, and if you can show that

$\text{True for k} \implies \text{True for k+1}$

Then it's true for $2$, meaning it's true for $3$, meaning it's true for $4$, ...

[quote="brady00n"][quote="scorpius119"]basically induction works like this:

- You first prove a statement is true for some integer $b$.
- You prove that if it is true for any $n$ between $b$ and $k$ inclusive (usually you just use $k$), then it is true for $k+1$

And then the statement is proven true for all integers greater than or equal to $b$. $b\Rightarrow b+1\Rightarrow b+2\Rightarrow \cdots$ (to $\infty$)

Also, anything that is proven by induction can also be proven by infinite descent:

[hide="proof here"]-Suppose it is false for some integer $k$ greater than or equal to $b$, then let $k_0$ be the smallest of these integers.

-Show that $k_0$ can't be $b$ because $b$ is true.

-Then show that the statement would have to be false for $k_0-1$ to be false for $k_0$.

-Since $b\leq k_0-1<k_0$, $k_0-1$ is an integer greater than or equal to $b$ for which the statement is false, but smaller than the least of these integers, contradiction.[/hide]

On the other hand, using induction to prove infinite descent is more complicated because there are lots of decreasing functions ($\lfloor \sqrt{n} \rfloor +2$ for $n\geq 3$, for example).[/quote]

So in a proof, you would prove for a base case, then simply plug in $k$, and then $k+1$?

Not quite sure about the $n$ integers between $b$ and $k$.[/quote]

What do you mean? k works as b and k +1 is b+1.

Then k is b+1 and k+1 is b+2 and it works this way until infinity.

Ok, I think scorpius119 is talking about generic induction method (a.k.a. strong induction).  Because the normal induction is proving if $k$ is true, then $k+1$ is true.  However, some induction proofs need more than that.  So here's strong induction (or generic induction method): prove that if it's true for $k, k+1, k+2, \dots, k+n$, then $k+n+1$ is true.  (Of course, usually $n = 0$).

For the base case, theoritically you can pick any integer as base case (like 43), however, that's more tiring.  Because then you only know that it's true for $n \ge 43$, but not the rest.  Then you have to prove that if $k$ is true it also implies that $k-1$ is true to finish off the proof.

[quote="brady00n"]
So in a proof, you would prove for a base case, then simply plug in $k$, and then $k+1$?

Not quite sure about the $n$ integers between $b$ and $k$.[/quote]

It's more complicated than just plugging in $k, k+1$.  If I wasn't being clear before:

You need to try to show that $\text{Assuming k works implies that k+1 also works}$.  And then you need to show that $\text{some base case b works}$.  Then, since you know $b$ works, $b+1, b+2, b+3, ....$ must also work.

Ok, final questions...

From what the posters above have said, I'll try to come to the conclusion.

1. We must prove an assertion for positive integers.  Assume that an integer $k$ satisfies the assertion, and then prove that $k+1$ also satisfies it.  This implies that any two consecutive integers satisfy the assertion, if and only if the original assumption is correct (that $k$ indeed works).

2. In order to verify that the original assumption is correct, we try a base case, most commonly $1$.  If $1$ works, a definite case to prove the assumption for $k$ has been proven.  This then proves that $k+1=1+1=2$ works as well.  Since $k$ can be any positive integer, suppose that $k$ now equals $2$.  Now $3$ can be assumed to work, etc.

[quote]Now, we show that if it is true for $k$, it must also be true for $k+1$.  If the assertion is true for $k$, we have

$1+2+3+...+k=\displaystyle\frac{k(k+1)}{2}$[/quote]

If the original problem is to prove that [quote]$1+2+3+...+n=\displaystyle\frac{n(n+1)}{2}$[/quote]

wouldn't these be equivalent expressions, with replaced variables?

Problems from AoPS Volume 1.

My view on induction is this. First you must show that if you can reach the $k\text{th}$ rung of a ladder, then you can reach the $(k + 1)\text{th}$. But then you must show you can actually start climbing the ladder in the first place (base case).

brady: No, they're not equivalent. To solve the problem you firstly assume $1 + \ldots + k = \frac{k(k + 1)}{2}$ and then try and prove $1 + \ldots + (k + 1) = \frac{(k+1)(k + 2)}{2}$, not simply show it by plugging it in. Then we know we can move up a rung. But to show it is actually true for all $n$ you need to get on the ladder by plugging a value in, not $k$. You don't plug $k$ in, you plug an actual number like $0$ or $1$. $k$ is just a hypothetical number you've assumed it works for. It may not exist. You must show its existence with the base case.

Example:

The induction step of $1 + 2 + \ldots + n = \frac{(n-1)(n+2)}{2}$ works, but there's no base case. There's no value you can plug in for $n$ to make the expression true.


I don't know if the following was already posted, but I find it very nice:
   Find all values of $x$ such that the sequence $ sin(2^nx)$ converges.

I haven't checked it properly, but I think that by looking at the binary expansion of $\frac x\pi$, we can show that it must be rational, with denominator a power of $2$. Sorry if this is not it :).


calculus
integration
function
limit
trigonometry
real analysis
algebra

Let \[(a,b) \subset R \] be an open interval with
\[(-\infty \leqslant a < b \leqslant \infty ) \] and let \[f: (a,b) \to R \] be a function such that the improper integral \[\int_{a}^{b}{f(x)dx}\] converges absolutely. Prove that
\[\mathop{\lim }_{\lambda \to \infty }\int_{a}^{b}{f(x)\sin (\lambda x)dx}= \mathop{\lim }_{\lambda \to \infty }\int_{a}^{b}{f(x)\cos (\lambda x)dx}= 0 \]

I think that this is the Riemann -Lebesque lemma and the proof of it is classical. Any integrable function can be approximated by step functions.

I like approximation by $C^{1}$ functions here- use integration by parts to prove it for functions in that class (and yes, I say $C^{1}$ because it's $f$ you want to differentiate).

I do believe this is the Riemann-Lebesgue lemma also, thanks. However, that is for the case where [a,b] is closed and I want to understand it in the (a,b) open inerval, how is it going to differ?

The only real difference is in the argument that our approximating functions are dense. On a closed interval, Riemann integrable functions are bounded- we allow unbounded functions here as long as their improper integrals converge.

Approximation by step functions within $\epsilon$ in the 1-norm:
Choose a subinterval $I$ so that $f$ is (properly) Riemann integrable on $I$ and the improper integral of $|f|$ outside $I$ is less than $\frac{\epsilon}2.$ Find a partition $P$ of $I$ such that the lower sums and upper sums with respect to $P$ differ by less than $\frac{\epsilon}2$. For each subinterval $J$ of $P$, let $g(x)$ be equal to the infimum of $f$ on $J$ for all $x\in J$. For $x\not\in I$, let $g(x)=0$. This $g$ is the desired approximation.

Approximation by $C^{k}$ (and continuous on the closed interval) functions within $\epsilon$ in the 1-norm:
Choose $I$ and $P$ as before, except that we replace each $\frac{\epsilon}2$ by $\frac{\epsilon}3$. On a subinterval $J=[a,b]$ of $P$, let $g(x)=f(b)h(\frac{x-a}{b-a})+f(a)h(\frac{b-x}{b-a})$, where $h(t)=\frac{\int_{0}^{t}x^{k}(1-x)^{k}\,dx}{\int_{0}^{1}x^{k}(1-x)^{k}\,dx}$; $g$ is then piecewise polynomial, and its first $k$ derivatives are zero at each such endpoint. It also lies between $f(a)$ and $f(b)$ on $[a,b]$, so the 1-norm of the error on $I$ is less than $\frac{\epsilon}3$.
Outside $I$, use appropriately scaled versions of $h$ spliced with zeros so that the absolute integral is at most $\frac{\epsilon}3$. The total error outside $I$ is then less than $\frac{2\epsilon}3$ by the triangle inequality, and $g$ is the desired approximation.

In both cases, we could also quote more general approximation theorems from the Lebesgue theory. My $h$ from the second version approaches the idea of a "bump function"; we could easily modify it to be $C^\infty$ if we wanted to.


geometry
incenter
analytic geometry
inradius
trigonometry
trig identities
tangent half angle

Hi Everyone,
  I'm having a dang time trying to solve this simple problem...think I've just lost it...I can work out the radius of the incircle, but can't work out the Coordinate center point I

The best example I could find on the net was:
http://www.terra.es/personal/rgonzal1/incentre.htm

But I have no idea what happens to m, when he derives k, and gets the answer :-/

I know the answer, but want to know how to get it:

I (Incenter of a triangle) = P|QR| + Q|RP| R|PQ| /  |QR| + |RP| + |PQ|

Where P, Q and R are the 3 corners of the triangle.

Be great for any help on this....or direction.

thanks,

Ben

Well since it's the intersection of the angle bisectors you can use the tangent half-angle formula :D

Or you can use $K=rs$ where $K$ is the area of the triangle, $r$ is its inradius, and $s$ is half its perimeter.

Tangent half-angle formula: $\tan \left(\frac{x}{2}\right) =\pm\sqrt{\frac{1-\cos x}{1+\cos x}} = \frac{1-\cos x}{\sin x} =\frac{\sin x}{1+\cos x}$


:rotfl:

[size=125]For $ x\in R$ such that $ 0\le x_1\plus{}x_2\plus{}x_3\le 3$ solve the equation:

$ e^{x_1}\plus{}e^{x_2}\plus{}e^{x_3}\equal{}e^3\plus{}2$[/size]

[quote="fredi"]:rotfl:

[size=125]For $ x\in R$ such that $ 0\le x_1 \plus{} x_2 \plus{} x_3\le 3$ solve the equation:

$ e^{x_1} \plus{} e^{x_2} \plus{} e^{x_3} \equal{} e^3 \plus{} 2$[/size][/quote]

If $ x_1,x_2,x_3$ can be negative, then there are infinitely many solutions.

0,0,3 or the 2 other combos are the only possible solutions.

I am sorry for the error!

I should be so:

$ x_1,x_2,x_3\in R^\plus{}$

If $ a\leq x_1 \plus{} x_2 \plus{} x_3\leq b$ for some $ a,b\in \mathbb{R^{ \plus{} }}$ maximum value of the $ e^{x_1} \plus{} e^{x_2} \plus{} e^{x_3}$ will be when one of the $ x_i$ is $ b$ and others is $ a$ because function $ f(x_1,x_2,x_3) \equal{} e^{x_1} \plus{} e^{x_2} \plus{} e^{x_3}$ is strictly increesing. ;) Maximum is $ e^{b}\plus{}2e^{a}$ and there are three solutions $ (3,0,0)$, $ (0,3,0)$, $ (0,0,3)$.

More detailed:
$ e^b \plus{} e^0 > e^{b \minus{} x} \plus{} e^x$ for $ 0 < x < b$ as $ e^{b \minus{} x}(e^x \minus{} 1) > (e^x \minus{} 1)$,
So $ e^3 \plus{} 2 \equal{} e^3 \plus{} e^0 \plus{} e^0 \ge e^{x_1} \plus{} e^{x_2} \plus{} e^{x_3}$ and equality holds iff one of the $ x_1,x_2,x_3 \equal{} 3$ and the others $ \equal{} 0$


1. http://www.mathlinks.ro/Forum/viewtopic.php?p=910659#910659
Consider the string of length 6 composed of three characters $ a$, $ b$, $ c$.
For each string, if two $ a$s are next to each other, or two $ b$s are next to each other,
then replace $ aa$ by $ b$, and replace $ bb$ by $ a$.
Also, if $ a$ and $ b$ are next to each other, or two $ c$s are next to each other,
remove all two of them(i.e. delete $ ab$, $ ba$, $ cc$)
Determine the number of strings that can be reduced to $ c$, the string of length 1,
by the reducing processes mentioned above.

2. http://www.mathlinks.ro/Forum/viewtopic.php?p=910658#910658
$ A_{1}B_{1}B_{2}A_{2}$ is a convex quadrilateral, and $ A_{1}B_{1}\neq A_{2}B_{2}$.
Show that there exists a point $ M$ such that;
$ \frac{A_{1}B_{1}}{A_{2}B_{2}}\equal{}\frac{MA_{1}}{MA_{2}}\equal{}\frac{MB_{1}}{MB_{2}}$

3. http://www.mathlinks.ro/Forum/viewtopic.php?p=910656#910656
Let $ S$ be the set of all positive integers whose all digits are $ 1$ or $ 2$.
Denote $ T_{n}$ as the set of all integers which is divisible by $ n$,
then find all positive integers $ n$ such that $ S\cap T_{n}$ is an infinite set.

4. http://www.mathlinks.ro/Forum/viewtopic.php?p=910653#910653
Two real sequence $ \{x_{n}\}$ and $ \{y_{n}\}$ satisfies following recurrence formula;

$ x_{0}\equal{} 1$, $ y_{0}\equal{} 2007$
$ x_{n\plus{}1}\equal{} x_{n}\minus{}(x_{n}y_{n}\plus{}x_{n\plus{}1}y_{n\plus{}1}\minus{}2)(y_{n}\plus{}y_{n\plus{}1})$,
$ y_{n\plus{}1}\equal{} y_{n}\minus{}(x_{n}y_{n}\plus{}x_{n\plus{}1}y_{n\plus{}1}\minus{}2)(x_{n}\plus{}x_{n\plus{}1})$

Then show that for all nonnegative integer $ n$, $ {x_{n}}^{2}\leq 2007$.

[quote="modal_soul"]could anyone please post solution for p4?[/quote]
Please link to problem 4 , I posted links help you!  :wink:


function
algebra
polynomial
complex analysis
complex analysis unsolved

Prove that if entire function $f$ attains all the complex values and $f'$ doesn't vanish anywhere then $f$ is linear

Perhaps, some conditions are missing because, as stated, the problem is wrong: the antiderivative of $e^{z^2}$, say, attains every complex value (if it weren't so, it would be representable as $e^{P(z)}+C$ where $P(z)$ is a polynomial of degree at most $2$, which is, clearly, impossible).

Thanks,Fedja,for you reply. It seems to me that i don't know some facts from comlex analysis: could you explain me why the antiderivative of $e^{z^2}$ if it doesn't attain every complex number should exactly be of the form $e^{P(z)}+C$ ?
BTW, i checked the problem conditions: i've written all of them, of course it maybe wrong...

The antiderivative $F(z)$ of $e^{z^2}$ grows not faster than $e^{|z|^2}$ itself. Therefore, if it never attains the value $0$, it must be representable as $e^{P(z)}$ where $P(z)$ is an entire function such that $\text{Re}\,P(z)\leqslant |z|^{2}$ for large $z$. It is a classical result (and a good exercise) that this condition implies that $|P(z)|\leqslant C|z|^2$, which, of course, implies that $P$ is a polynomial of degree $2$ or less. If the value that is omitted is $C\neq 0$, just condider $F(z)-C$.


vector
superior algebra
superior algebra unsolved

Let $ X$ be in the center of the general linear group over a field $ F$. Show that $ X$ has the standard basis as an eigenbasis.

In fact every non-zero vector is an eigenvector and the corresponding eigenvalues are all the same:

We show $ Z(GL(V)) \equal{} Z(V): \equal{} \{k\cdot id_V | k\in F^\times\}$. The one inclusion is obvious, so let us proof the other one. If is obvious too, if $ dim(V)\leq 1$ so consider just the other cases:

Let $ \zeta\in\ Z(GL(V))$, $ dim(V)\geq 2$ and let $ \tau$ be a linear map of the form $ \tau(v) \equal{} v \plus{} \phi(v)u$ with $ \phi\in V^\ast$ and $ u\in ker(\phi)$. All those $ \tau$ are linear and invertible with the inverse map $ \tau^{ \minus{} 1}(v) \equal{} v \minus{} \phi(v)u$ as one can easily see.

We have $ \zeta\tau \equal{} \tau\zeta$ for $ \zeta\in Z(GL(V))$
$ \Rightarrow\forall v: \zeta(v) \plus{} \phi(v)\zeta(u) \equal{} \zeta(v) \plus{} \phi(\zeta(v))u$
$ \Rightarrow\forall v: \phi(v)\zeta(u) \equal{} \phi(\zeta(v))u$

If we choose $ v\notin ker(\phi)$ we deduce that u is an eigenvector of $ \zeta$ (if it is not the zero-vector). Because there is a $ \phi\in V^\ast$ with $ u\in ker(\phi)$ for every non-zero u, we conclude, that every non-zero u is an eigenvector of $ \zeta$.

So for all non-zero u there is a $ k_u\in F^\times$ with $ \zeta(u) \equal{} k_u u$. Obviously $ k_u \equal{} k_{u'}$ if u and u' are lineary dependent, but non-zero vectors. If u and u' are linear independent then we have:
$ k_{u \plus{} u'} u \plus{} k_{u \plus{} u'} u' \equal{} k_{u \plus{} u'} (u \plus{} u') \equal{} \zeta(u \plus{} u') \equal{} \zeta(u) \plus{} \zeta(u') \equal{} k_u u \plus{} k_{u'} u'$
$ \Rightarrow k_u \equal{} k_{u \plus{} u'} \equal{} k_{u'}$ by linear independence

So $ k_u$ is in fact independent of u, which means that $ \zeta(u) \equal{} k u$ for all u, so $ \zeta \equal{} k\cdot id_V$. qed


calculus
integration
trigonometry
logarithms
calculus computations

Let $ I \equal{} \int_0^{\frac {\pi}{2}}\frac {\cos 2x}{3\cos 2x \plus{} 4\sin 2x \plus{} 5}dx, J \equal{} \int_0^{\frac {\pi}{2}}\frac {\sin 2x}{3\cos 2x \plus{} 4\sin 2x \plus{} 5}dx$,

$ K \equal{} \int_0^{\frac {\pi}{2}} \frac {1}{3\cos 2x \plus{} 4\sin 2x \plus{} 5}dx$, we are to obtain $ 3I \plus{} 4J \plus{} 5K \equal{} 1, 8I \minus{} 6J \equal{} \minus{} 2\ln2$.

The answer is $ I \plus{} K \equal{} ...$

Another approach:

$ \int_0^{\frac{\pi}{2}} \frac{\cos ^ 2 x}{(\sin x\plus{}2\cos x)^2}\ dx$

$ \equal{}\int_0^{\frac{\pi}{2}} \frac{1}{(\tan x \plus{}2)^2}dx$

we can set $ \tan x \equal{} t$ directly. We are to use $ \cos x \equal{} \frac {1 \minus{} t^2}{1 \plus{} t^2},\ \sin x \equal{} \frac {2t}{1 \plus{} t^2}$.


geometry
geometric transformation
rotation
search
AMC
AIME
linear algebra

Please don't yell at me, I tried using search but I did not find what I wanted. I also tried looking through the wiki. Does anybody know of any threads that teach about generating systems of equations that rotate points around other points on the xy plane. For example, on an AIME problem the solution involved creating a system of equations or a matrix that generated an (x',y') by rotating (x,y) around a certain point, and it said something like "using rotation formulas", X' = ... and Y' = ... but didn't explain further.

Does anybody have suggestions?

The formula for rotating a point $ (x,y)$ around the origin by $ \theta$ can be derived like this:

Let $ z\equal{}x\plus{}yi$.

Multiply by $ e^{i\theta}$ to rotate:

$ z'\equal{}(x\plus{}yi)e^{i\theta}\equal{}(x\plus{}yi)(\cos\theta\plus{}i\sin\theta)\equal{}x\cos\theta\minus{}y\sin\theta\plus{}i(x\sin\theta\plus{}y\cos\theta)$.

Then the rotated point is $ \left(\text{Re}\,(z'),\text{Im}\,(z')\right)\equal{}\boxed{(x\cos\theta\minus{}y\sin\theta,x\sin\theta\plus{}y\cos\theta)}$.

If you're familiar with complex numbers (and/or polar coordinates) this is easy; otherwise, rotation matrices give the exact same result.

You both gave me exactly what I was looking for. Thank you very much!


How many integers between 1 and 1000, inclusive, are divisible by neither 2, 3, or 5?

EDIT: Oops! I wasn't thinking very clearly when I typed part b)   :roll:

[hide="2,3,5"] Between 1 and 1000, inclusive, there are 500 integers divisible by 2, 333 divisible by 3 and 200 divisible by 5. BUT! We overcounted! There are 166 divisible by 6 (2*3), 66 divisible by 15 (3*5), 100 divisible by 10 (5*2) and 33 divisible by 30 (2*3*5).
Then $(500+333+200)-(166+66+100+33)=1033-365=668$
The answer is $1000-668$, so the final answer is 332![/hide]
[hide="2,4,8"] This one's easy! :P
Since 8 and 4 are divisible by 2, all you need is how many are divisible by 2 - which is 500. Then $1000-500$ gives the final answer of 500![/hide]

[quote="max1546"]I guess it's like Venn Diagrams. [/quote]

It is, in fact, exactly like a Venn Diagram, which is a great and easy way to visualize these kinds of problems  :)

[quote="max1546"]Correct me if I'm wrong but I think quevvy you got a) wrong.
Yes there are 500 multiples of 2; 333 multiples of 3; 200 multiples of 5 as:
1000/2=500
1000/3=333...1
1000/5=200
Therefore the integers not divisible by neither 2,3,5 is:
$1000-((500+333+200)-(166+100+66)+33)=1000-734=266$
I think you got the 33 part mixed up.  I guess it's like Venn Diagrams.  You add all 3 sets A,B,C.  Then AB,AC,BC are repeated once and ABC is repeated twice.  By taking out the (AB+AC),(AB+BC),(AC+BC)which exclude the overlap, ABC is taken out 3times.  Thus there is ABC less than the proper answer.  Therefore ADD the ABC to get the total set.  I think I explained it quite weakly but I think we all know already.  And by the way so a) is 266, b) is correct I think.
Thanks[/quote]

Yes, 266 is the correct answer. PIE (Principle of Inclusion-Exclusion) is really formally stating the information you obtain from a Venn Diagram. For three sets:

$|A\cup B\cup C|=|A|+|B|+|C|-(|A\cap B|+|B\cap C|+|A\cap C|)+|A\cap B\cap C|.$

I see how I messed up now. My picture shows it all.
[hide="Picture"]$267+67+134+133+67+33+33=734$ and then $1000-734=266$[/hide]
Sorry, I couldn't get the picture in any other way except through a word document.


when n is divided by 3 the remainder is 2, what is the remainder when 5n is divided by 3?

sorry this must be the question.

What's the definition of "remainder"?
I think the remainder upon divison by 2 should be 0 or 1
CMIIW

$ n\equiv 3\equiv 1\bmod 2\iff n\equal{}2k\plus{}1$ for some $ k\in\mathbb{Z}$. Then $ 5n\equal{}10k\plus{}5\equiv k\plus{}2\bmod 3$.
Therefore, any remainder is attainable, so there is no single answer.

I think you meant to say that when $ n$ is divided by 3, the remainder is 2. In that case.

$ n\equiv 2\implies 5n\equiv 10\equiv \boxed{1}\bmod 3$.


algebra
function
domain
calculus
integration
vector
advanced fields

1. Let $U=\mathbb{R}^3 - \{(x,y,z): x^2+y^2=1, z=0\}$. Give an example of a closed, non-exact 1-form $\omega \in \Omega(U)$.

2. Let $U=\mathbb{R}^2 - \{(-1,0),(1,0)\}$. Give an example of a symmetric $(\omega(x,y)=\omega(-x,y)$, closed,
non-exact 1-form $\omega \in \Omega(U)$.

Idea for #1: If you had a uniform electric current circling that ring (the set omitted from the domain), what would be the magnetic field induced by it?

A similar idea for #2 would be to take a cross-sectional look at the magnetic field generated by the current in two infinitely long straight wires. The only question: should the current in the two wires be moving in the same or in opposite directions?

To answer my own question for #2: opposite directions. Try

$\omega = \frac{-y\,dx+(x-1)\,dy}{(x-1)^2+y^2}+ \frac{y\,dx-(x+1)\,dy}{(x+1)^2+y^2}.$

There's an assumption here that when you send $x$ to $-x,$ you also mean to send $dx$ to $-dx.$

yes, the form you gave is the the one I was looking for; assuming this form is exact, we calculate the value of
the integral of $\omega$ on the sum of two circles with radiuses - suppose - $1$ and centers in those two points mentioned in the text (with standard orientation) ...the value is $4\pi$ ($2\pi+2\pi$) so we have a contradiction from Stoke's theorem (it cannot be nonzero...) (I know it's easy but I'm a begginer in forms so some things aren't so obvious for me)

ad. 1
I forgot a lot from physics :)...would you mind explaining this example in a different way?
(I know this part of analysis has a lot to do with physics or rather - physical interpretations, my friends have a slight advantage :) )

[quote]the integral of \omega on the sum of two circles with radiuses - suppose - 1 and centers in those two points mentioned in the text (with standard orientation) ...the value is $4\pi (2\pi+2\pi)$[/quote]
If you gave both circles the same orientation, the integral over the sum of the two circles would cancel and be zero. You can take just one of the circles to get a nonzero integral, or you can orient the circles in opposite directions (or take a figure-8 path, which is essentially the same) to get what you said.

A physicist would probably rather say "a vector field with zero curl that does not have a potential" or "that is not conservative" rather than "a closed 1-form that is not exact." Different language and different notation for the same problem.


function
abstract algebra
IMO Shortlist
group theory
research

As a straightforward generalization of IMO Shortlist 2001 problem C4, we can show the following fact:

Let $ u$ and $ v$ be two positive integers. A set of three integers $ \left\{ x,y,z\right\}$ with $ x < y < z$ is called [i]historic[/i] if $ \left\{ z \minus{} y,y \minus{} x\right\} \equal{} \left\{ u,v\right\}$. Then, the set of all nonnegative integers can be written as a union of pairwise disjoint historic sets.

My questions are:

$ \cdot$ Is there a [i]periodic[/i] way to write the set of all nonnegative integers as a union of pairwise disjoint historic sets?  Equivalently, is there a function $f:\left\lbrace  0,1,2,...\right\rbrace  \rightarrow\mathbb{N}$ such that $\left\lbrace x\in\left\lbrace 0,1,2,...\right\rbrace  \mid f\left(  x\right)  =d\right\rbrace  $ is a historic set for every $d\in\mathbb{N}$, and such that $f\left(x+u\right)=f\left(x\right)+v$ for some fixed positive integers $u$ and $v$ and all sufficiently large $x$ ?
Actually, the answer is [i]yes[/i] as can be deduced from the above problem using an additional argument.

$ \cdot$ Is there a [i]purely periodic[/i] way to do this? Equivalently, is there a function $f:\left\lbrace 0,1,2,...\right\rbrace  \rightarrow\mathbb{N}$ such that $\left\lbrace  x\in\left\lbrace  0,1,2,...\right\rbrace  \mid f\left(  x\right)  =d\right\rbrace  $ is a historic set for every $d\in\mathbb{N}$, and such that $f\left(x+u\right)=f\left(x\right)+v$ for some fixed positive integers $u$ and $v$ and all $x$?

$ \cdot$ Is there a positive integer $ n$ such that the set $ \left\{ 0,1,2,...,n\right\}$ is a union of pairwise disjoint historic sets?

$ \cdot$ What happens if we start with three positive integers $ u$, $ v$, $ w$, and call a set of four integers $ \left\{ x,y,z,t\right\}$ with $ x < y < z < t$ historic if $ \left\{ t \minus{} z,z \minus{} y,y \minus{} x\right\} \equal{} \left\{ u,v,w\right\}$ ?

Being unable to check even the simplest values without making stupid mistakes at adding little integers, I wouldn't be surprised to see obvious counterexamples to some of these questions.

[b]Discussion:[/b]

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=150869 (PEN problem O 26)
http://mathoverflow.net/questions/70501

This problem was where I got for the idea of my post "[Group Theory/ Comb] Group Tilings", I misread the ISL question at first and was very puzzled when I proved that the question need not be true. The case where $ G \equal{} \mathbb{Z}, |S| \equal{} 3$ is somewhat related to this (that is what I misread this ISL problem to be).
Perhaps you could merge the two topics and lump together the content of the posts or something, since they are related?

[quote="Vinoth"]Perhaps you could merge the two topics and lump together the content of the posts or something, since they are related?[/quote]

No, since they are not really related... As you said, you misread the problem and invented a different one ;)

  darij

Hmmm .. How does this sound for a common generalization:
Given two (or $ n$ if you want more pathological cases) sets $ S_1, S_2$ ($ ... S_{n}$), when is it possible to cover the group $ G$ by sets, each of which is some translate of some set $ S_{i}$?

Well that's a (arguably artificial) generalization of both questions. But of course I don't really have any thing about "periodic" or "purely periodic". And you could argue in some very vague sense, my question is more "general", but in the case of $ \mathbb{Z}$ or finitely generated abelian groups, it's not at all more general.

Question edited: now with 20% less mistakes per mistake.

Also posted it at http://mathoverflow.net/questions/70501 .


Let $ a, b$ and $ c$ be positive real numbers. Prove that

$ \frac ab \plus{} \frac bc \plus{} \frac ca  \le  \frac {a^2}{b^2} \plus{} \frac {b^2}{c^2} \plus{} \frac {c^2}{a^2}$

[hide="P.S."]
This forum has been idle since april 11. I thought that it would become more active once IIT-JEE
got over, but the opposite happened. So, I am posting this just to revive this forum.[/hide]

This can be solved by the re-arrangement inequality.

a/c*c/b + b/a*a/c + c/b*b/a <= $ \frac{a^{2}}{b^{2}}\plus{}\frac{b^{2}}{c^{2}}\plus{}\frac{c^{2}}{a^{2}}$

Equality when $ \frac{a}{c}\equal{}\frac{c}{b}\equal{}\frac{b}{a}$

We need to prove that

$ \frac {a^{2}}{b^{2}} \plus{} \frac {b^{2}}{c^{2}} \plus{} \frac {c^{2}}{a^{2}} \minus{} \frac {a}{b} \minus{} \frac {b}{c} \minus{} \frac {c}{a}\ge 0$

Multiply and divide by 2

so, to prove that :

$ \frac {2a^{2}}{b^{2}} \plus{} \frac {2b^{2}}{c^{2}} \plus{} \frac {2c^{2}}{a^{2}} \minus{} \frac {2a}{b} \minus{} \frac {2b}{c} \minus{} \frac {2c}{a}\ge 0$

or TPT(to prove that)
$ {(\frac {a}{b} \minus{} \frac {b}{c})}^2 \plus{} {(\frac {b}{c} \minus{} \frac {c}{a})}^2 \plus{} {(\frac {c}{a} \minus{} \frac {a}{b})}^2 \ge 0$

which is obvious....

[quote="befuddlers"]$ a/c*c/b \plus{} b/a*a/c \plus{} c/b*b/a <\equal{} \frac{a^{2}}{b^{2}}\plus{}\frac{b^{2}}{c^{2}}\plus{}\frac{c^{2}}{a^{2}}$[/quote]

i think rearrangement gives $ \frac{a^{2}}{c^{2}}\plus{}\frac{b^{2}}{a^{2}}\plus{}\frac{c^{2}}{b^{2}}$
not  $ \frac{a^{2}}{b^{2}}\plus{}\frac{b^{2}}{c^{2}}\plus{}\frac{c^{2}}{a^{2}}$

@adarsh:
I think u have made a mistake in simplification. check it.

I think i simply copied ur formula to make minor changes in it...now i've corrected it

Aaaaaaarrrrrrrrrggggggghhhhhhhhhh.............i got the question wrong!!   :wallbash_red:

Someone please post a solution. (this problem is simple)

\[ \begin{gathered}
  \frac{a}
{b} \equal{} x,\frac{b}
{c} \equal{} y,\frac{c}
{a} \equal{} z \Leftrightarrow xyz \equal{} 1 \hfill \\
  \sqrt {3(x^2  \plus{} y^2  \plus{} z^2 )}  \geqslant (x \plus{} y \plus{} z)...(i) \hfill \\
  Again, \hfill \\
  \frac{{(x^2  \plus{} y^2  \plus{} z^2 )}}
{3} \geqslant \sqrt[3]{{x^2 y^2 z^2 }} \hfill \\
  \sqrt {\frac{{(x^2  \plus{} y^2  \plus{} z^2 )}}
{3}}  \geqslant \left( {x^2 y^2 z^2 } \right)^{\frac{1}
{6}}  \equal{} 1.......(ii) \hfill \\
  (i)*(ii), \hfill \\
  (x^2  \plus{} y^2  \plus{} z^2 ) \geqslant (x \plus{} y \plus{} z) \hfill \\
   \Leftrightarrow \frac{{a^2 }}
{{b^2 }} \plus{} \frac{{b^2 }}
{{c^2 }} \plus{} \frac{{c^2 }}
{{a^2 }} \geqslant \frac{a}
{b} \plus{} \frac{b}
{c} \plus{} \frac{c}
{a} \hfill \\
\end{gathered}
\]

[quote="keyree10"]Let $ a, b$ and $ c$ be positive real numbers. Prove that

$ \frac ab \plus{} \frac bc \plus{} \frac ca \le \frac {a^2}{b^2} \plus{} \frac {b^2}{c^2} \plus{} \frac {c^2}{a^2}$
[/quote]

I got this completely random method.

By A.M. - G.M. we have that $ \frac ab \plus{} \frac bc \plus{} \frac ca \ge 3$

So add $ \frac ab \plus{} \frac bc \plus{} \frac ca$ to the left and add 3 to the right.

Then we observe that it is just completion of squares......................

we know that

$ { a^2/b^2 \plus{}1 >\equal{} 2a/b}$  (am,gm)

now similarly add the other three inequalities as well

then we have

$ {a^2/b^2 \plus{} b^2/c^2 \plus{} c^2/a^2 >\equal{} (2a/b \plus{}2b/c \plus{}2c/a) \minus{} 3}$

now we r left to prove that

$ {2a/b \plus{} 2b/c \plus{} 2c/a \minus{}3 >\equal{} a/b \plus{}b/c \plus{} c/a}$

or

$ {a/b \plus{} b/c \plus{} c/a >\equal{}3}$

which is trivial from am gm :P  :P  :P


My school won't have finished all of campbell by the time the USABO Open is administrated.

What chapters of Campbell should I read in order to take the open test?

I've read chapters 1-25 and chapters 50-55 at this point.

You'll probably be fine if you know it well. I qualified for the semfinal round with a good amount of room to spare after having covered less than material than that. The semifinal round is a bit more nasty, though.

I thought the Open Exam was more difficult than the semifinal exam. I think that's because it's meant to cut down the competitive field a lot. The semifinal exam seemed to have easier questions (except for a few brutal fill-in questions) but the curve is ridiculous to make the Top 20.

Did you read the whole book by the time of the open exam adidsty?

I actually agree. The open exam did have, for the most part, tougher questions, but you didn't have to get as many of them correct to pass on to the next round.

I had read the whole thing by the Open Exam last year. This year is kind of whack...I really haven't studied much because I don't want to read stuff I've already read like a zillion times. And they'll always frustrate you with some questions that address literally ONE sentence in that book. I really like biology...but I feel that doing pointless things like memorizing the amino acid chart is kind of counterproductive. If I get past the Open Exam I think I'll actually put more effort into it.

Actually, though, because biology is conceptually pretty easy, they kind of have to put a few pure memorization questions on there...I don't know.


Let $\alpha$ the greatest root of $x^{3}-3x^{2}+1=0$, prove that $[\alpha^{2004}]$ is divisible for $17$.


Has anyone worked through olympiads from around the world by Titu Andreescu?
I have the one from 1999-2000. The problem is I don't know how to work thru it, since there are so many national and regional contests...
Should I work from beginning to the end?
Thanks,

[quote="JBL"][quote="chess64"]If you're just starting with olympiad math, you should do all the #1's, and then when you finish you can do the #2's.[/quote]

No, I don't think so.  You should look through to find the countries with relatively easier olympiads (I personally recommend Canadian and Irish, although there are undoubtedly many others -- avoid FUSSR, China, probably Iran) and do the first few questions of those.  Also, regional olympiads tend to be easier than national olympiads.[/quote]
Yes, that's what Mr.rrusczyk recommended, but I'm having a difficult time....

I have total 4 books in this kind (95-96,96-97,97-98,99-2000,2000-2001) all of books in pdf format .If you need plz contact with me :)

[quote="Bugi"][quote="Altunshukurlu"]You can download National Math olympiads from 1995 to 2001 at http://www.4shared.com in search write math olympiads and then there will be folders near the results click on them.[/quote]

Please do not post illegal links (do that only by PM-s).[/quote]www.4shared.com is not an illegal link. It is a valid online storage site. :)

However, there is copyrighted material on the site...

There's also plenty of copyrighted material on say... YouTube. We even have a special button to embed links to there!

Hmm, I checked the book, and I think it's a nice collection of problems, but (1) the solutions are WAAAAY too close to the questions, and (2) if you are having trouble, maybe using a book such as Engel with examples and targeted problems may be more helpful?

It seems like these books may also be used for AIME preparation if you do Canada, Ireland and easy ones like that.

I agree.  I recognized some of the Canada problems from AIME class...


For any three positive reals a, b, c, we have

$\frac{a^2+3bc}{b^2+c^2}+\frac{b^2+3ca}{c^2+a^2}+\frac{c^2+3ab}{a^2+b^2} \ge \frac{7}{2}$

[quote="harazi"]Can't anyone find the most general form of this type of inequalities? I've seen till now at least 10 types of such inequalities and all the proofs were the same. Probably there is a very general inequality... :?  :?  :?[/quote]

I Agree with you,Harazi!
We have alot of this kind,such as:

$\frac{a^2+kbc}{(b^2+c^2}+\frac{b^2+kac}{(a^2+c^2}+\frac{c^2+kab}{(a^2+b^2} \ge 3\frac{1+k}{2}$

Or :

$\frac{a^2+kbc}{(b+c)^2}+\frac{b^2+kac}{(a+c)^2}+\frac{c^2+kab}{(a+b)^2} \ge 3\frac{1+k}{4}$

It looks quite nice,but so old with me....
 :)

Related to this Xtar's inequality, I think Harazi and Hungkhtn are not right.
This is a new special inequality.
Moreover, I think the folowing ineq are valid

$\frac{a^2+16bc}{b^2+c^2}+\frac{b^2+16ca}{c^2+a^2}+\frac{c^2+16ab}{a^2+b^2} \ge 10$ . ;)

For $a,b,c$ non-negative numbers, the inequalities hold:

$\frac{a^2+4bc}{b^2-bc+c^2}+\frac{b^2+4ca}{c^2-ca+a^2}+\frac{c^2+4ab}{a^2-ab+b^2} \ge 6$ ;

$\frac{a^2+16bc}{b^2+c^2}+\frac{b^2+16ca}{c^2+a^2}+\frac{c^2+16ab}{a^2+b^2} \ge 10$ ;

$\frac{a^2+36bc}{b^2+bc+c^2}+\frac{b^2+36ca}{c^2+ca+a^2}+\frac{c^2+36ab}{a^2+ab+b^2} \ge 14$ ;

$\frac{a^2+64bc}{(b+c)^2}+\frac{b^2+64ca}{(c+a)^2}+\frac{c^2+64ab}{(a+b)^2} \ge 18$ .

The general statement is the following: For $a,b,c$ non-negative numbers, $p>-2$ and $\frac{2p+1}{2}\le q\le 4(p+2)^2$, the inequality holds:

$\frac{a^2+qbc}{b^2+pbc+c^2}+\frac{b^2+qca}{c^2+pca+a^2}+\frac{c^2+qab}{a^2+pab+b^2} \ge 2+\frac{q}{2+p}$ ;

Remark: Now I have a solution for none of them. :P

[quote="hungkhtn"]Sorry Because of My Hurry.

It's really a new kind.
Last night I've check the inequal:

$\frac{a^2+16bc}{b^2+c^2}+\frac{b^2+16ac}{a^2+c^2}+\frac{c^2+16ab}{a^2+b^2} \ge 10$

It's not so difficult.Use expanding to have:
$M(a-b)^2+f(c) \ge 0$

I think this way can be used to prove general case.[/quote]

Today during math lesson I proved xtar's original inequality, expressing things like what hungkhtn mentioned, but it looks as if my solution does not work for vasc's inequality. Can you post your proof, hungkhtn?

[quote="Vasc"]For $a,b,c$ non-negative numbers, the inequalities hold:

$\frac{a^2+4bc}{b^2-bc+c^2}+\frac{b^2+4ca}{c^2-ca+a^2}+\frac{c^2+4ab}{a^2-ab+b^2} \ge 6$ ;

$\frac{a^2+16bc}{b^2+c^2}+\frac{b^2+16ca}{c^2+a^2}+\frac{c^2+16ab}{a^2+b^2} \ge 10$ ;

$\frac{a^2+36bc}{b^2+bc+c^2}+\frac{b^2+36ca}{c^2+ca+a^2}+\frac{c^2+36ab}{a^2+ab+b^2} \ge 14$ ;

$\frac{a^2+64bc}{(b+c)^2}+\frac{b^2+64ca}{(c+a)^2}+\frac{c^2+64ab}{(a+b)^2} \ge 18$ .

The general statement is the following: For $a,b,c$ non-negative numbers, $p>-2$ and $\frac{2p+1}{2}\le q\le 4(p+2)^2$, the inequality holds:

$\frac{a^2+qbc}{b^2+pbc+c^2}+\frac{b^2+qca}{c^2+pca+a^2}+\frac{c^2+qab}{a^2+pab+b^2} \ge 2+\frac{q}{2+p}$ ;

[/quote]
Nobody posted a solution so far !
.

[quote="spdf"]My solution for Vasc's inequality:
Assume that [tex]a\leq\ b\leq\ c[/tex];
since [tex](b+c)^2(b^2+c^2)\geq\ 8b^2c^2[/tex].It remains to prove:[tex]15(b^2+c^2)(b+c)bc+15a^2(b+c)(b^2+c^2)\geq\ 8(b-c)^2a(a^2+b^2+c^2}[/tex],which is obvious[/quote]

Dear spdf
Please, rewrite your solution in more details.

Thank you very much.

Does anyone now has solutions for problems posted by Vasc ?And you Vasc ?


The set $S$ consists of five integers. If pairs of distinct elements of $S$ are added, the following ten sums are obtained: 1967,1972,1973,1974,1975,1980,1983,1984,1989,1991. What are the elements of $S$?

[hide]
Let the five distinct element be $a>b>c>d>e$ . Then sum up all we get

$a+b+c+d+e=4947$ . Since $a+b$ is biggest , it is $a+b=1991$ . Similarly the smallest is $d+e=1967$ . From this two we can find $c=989$ . Also the second biggest is $a+c=1989$ so we can get $a=1000$ , $b=991$ . Also the different between $a+d$ and $c+d$ is similar with the different between $a$ and $c$ which is $11$ . The only remaining pair of numbers with different $11$ is $(1984,1973)$ and $(1983,1972)$ . But one of them must be $(a+d,c+d)$ and another is $(a+e,c+e)$ . So we get $(a+d,c+d)=(1984,1973)$ which is $d=984$ and $e=983$

Hence the five elements is $\{983,984,989,991,1000\}$

[/hide]


If $10^{2004} -2004$ is expressed as an integer , what is the sum of it's digits ??

Well, $10^{2004}$ is a one folllowed by 2004 zeros.  Thus, after subtracting 2004, we obtain 2000 9s followed by 7996.

Now the sum of the digits is $2000*9 +7+9+9+6=18000+31=18031$.

this is not Pre_Olympiad...
it's more like Intermidiate ;)


Let $ (C)$ be a circle and let $ R$ it's ray
and $ A,B,C,D \in (C)$ then
Find all quadrilaterals ABCD such that: $ AB.BC.CD.DA \ge 4R^4$

Thanks


Hi,

Find all pairs of natural numbers n and m, such that  $3^{n}-5^{m}=4$

:bye:

Hint: We prove that $n$ is even, then write $n=2k$ and we have $(3^{k}-2)(3^{k}+2)=5^{m}$,... :wink:

Moreover simply prove that $n=2(2\cdot k+1),k=0,1,2,3...$ and $n=2,m=1$ - single solution.


logarithms
inequalities unsolved
inequalities

Let $ a,b,c$ be the positive reals satisfying $ abc\equal{}1$.Prove that:
$ \frac{1}{a^2\minus{}a\plus{}1}\plus{}\frac{1}{b^2\minus{}b\plus{}1}\plus{}\frac{1}{c^2\minus{}c\plus{}1}$≤$ 3$

[quote="SUPERMAN2"]Let $ a,b,c$ be the positive reals satisfying $ abc \equal{} 1$.Prove that:
$ \frac {1}{a^2 \minus{} a \plus{} 1} \plus{} \frac {1}{b^2 \minus{} b \plus{} 1} \plus{} \frac {1}{c^2 \minus{} c \plus{} 1}$≤$ 3$[/quote]
It's equivalent to $ \sum_{cyc}\left(1\minus{}\frac{1}{a^2\minus{}a\plus{}1}\minus{}\ln a\right)\geq0,$ which is obvious.

[quote="arqady"][quote="SUPERMAN2"]Let $ a,b,c$ be the positive reals satisfying $ abc \equal{} 1$.Prove that:
$ \frac {1}{a^2 \minus{} a \plus{} 1} \plus{} \frac {1}{b^2 \minus{} b \plus{} 1} \plus{} \frac {1}{c^2 \minus{} c \plus{} 1}$≤$ 3$[/quote]
It's equivalent to $ \sum_{cyc}\left(1 \minus{} \frac {1}{a^2 \minus{} a \plus{} 1} \minus{} \ln a\right)\geq0,$ which is obvious.[/quote]
Thanks.But could you explain more clearly,please?

I have found mistake in this way.
See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=87574

[quote="nguoivn"][quote="SUPERMAN2"]Let $ a,b,c$ be the positive reals satisfying $ abc \equal{} 1$.Prove that:
$ \frac {1}{a^2 \minus{} a \plus{} 1} \plus{} \frac {1}{b^2 \minus{} b \plus{} 1} \plus{} \frac {1}{c^2 \minus{} c \plus{} 1}$≤$ 3$[/quote]
We can solve it easily by Am-Gm and CS  :maybe:[/quote]
Can you post your full solution dear friend???


email
combinatorics unsolved
combinatorics

There are 3 groups of people, each of which with $ n$. Each person knows at least $ n \plus{} 1$ people, none of which is in the same group. Prove that there exist 3 people all of each know each other.

See
http://www.mathlinks.ro/viewtopic.php?t=49347 for a slightly stronger version and
http://www.mathlinks.ro/viewtopic.php?t=40982 for the problem.

However, I am not sure whether any of these has a correct solution. If you want one, please PM me and I'll send it to you (valid email address required).

PS. The "subject description" field is for giving the source of the problem, and not for writing "standard" ;)

  darij


geometry
perimeter
trigonometry
ratio
similar triangles
geometric series

I'm not sure if these were ever seen before, most likely they were, I made the pictures and had come up with the questions but I do not take credit for it for it may and most likely is out there already. So enjoy!

Alright, onto the question. Here are the two questions:

1) In the picture below, a right triangle ▲ACD is shown. Angle ADB = θ, AB = 4, BC = 2, and CD = 3. Using trigonometric identities, show how you would find the exact value of tanθ.
(picture below)

[url=http://img397.imageshack.us/my.php?image=000000aaaaaaaarighttriangleide.jpg][img]http://img397.imageshack.us/img397/7992/000000aaaaaaaarighttriangleide.th.jpg[/img][/url]

2) An equilateral triangle with each side 4cm long is shown. The midpoints of each the sides are joined to form another equilateral triangle. The midpoints of this triangle are joined to form yet another triangle. If this process were to continue indefinitely, what is the sum of the perimeters of the triangles formed?
(picture below)

[url=http://img474.imageshack.us/my.php?image=00000aaaaaequlateraltriangleco.jpg][img]http://img474.imageshack.us/img474/4516/00000aaaaaequlateraltriangleco.th.jpg[/img][/url]

[hide="1"]The leg lengths of both right triangles are given; using these $\tan ADC=2$ and $\tan BDC=\frac 23$.

Using subtraction of tangents, $\tan \theta=\tan \left(\arctan 2-\arctan \frac 23 \right)=\frac{2-\frac 23}{1+2*\frac 23}=\frac{6-2}{3+2*2}=\boxed{\frac 47}$.[/hide]

[hide="2"]pretty obvious through similar triangles (SAS ~) that the new sides are half the length of the initial side, and the perimeter is also half, so the initial perimeter is 12, then 6, 3,3/2, etc

the sum is a geometric series with initial term 12 and ratio 1/2, so

$\sum P = \frac{12}{1-\frac{1}{2}}=24$[/hide]


When  you  write the  contents, how i  put  dots  and  at end  of the line  the page number?
Like: Problems............23
 solutions.........34
but  the page numbers at  end of  line?
Thank you.