I need help with the following inequalities:

Given triangle $ ABC$. $ A,B,C$ denote the angles, $ a,b,c$ denote the lengths of the sides of the triangle. Prove:

1) $ sin(2A)\plus{}sin(2B)\plus{}sin(2C) \le sin(A)\plus{}sin(B)\plus{}sin(C)$

2) $ \left(1\plus{}cos^2(A/2)\right) \left(1\plus{}cos^2(B/2)\right) \left(1\plus{}cos^2(C/2)\right)< \left(1\plus{}\frac{\sqrt{3}}{4}\right)^{3\sqrt{3}}$

3) $ \left(1\plus{}\frac{1}{(sin(A/2))^{1/n}}\right)\left(1\plus{}\frac{1}{(sin(B/2))^{1/n}}\right)\left(1\plus{}\frac{1}{(sin(C/2))^{1/n}}\right) \ge \left(1\plus{}2^{1/n}\right)^3$ for $ n\in N$

4) $ \left(4\plus{}\frac{\sqrt{3}}{cos(A/2)}\plus{}\frac{1}{cos(B)}\right) \left(4\plus{}\frac{\sqrt{3}}{cos(B/2)}\plus{}\frac{1}{cos(C)}\right) \left(4\plus{}\frac{\sqrt{3}}{cos(C/2)}\plus{}\frac{1}{cos(A)}\right) \ge 512$    given A, B, C are acute angles

5) $ 2\left(\frac{1}{cos(A)}\plus{}\frac{1}{cos(B)}\plus{}\frac{1}{cos(C)}\right) \minus{} \sqrt{3}(tan(A)\plus{}tan(B)\plus{}tan(C)) \ge 3$ given A, B, C are acute angles

6) $ \frac{1}{2^{tan(A)}} \plus{} \frac{1}{2^{tan(B)}} \plus{} \frac{1}{2^{tan(C)}} \ge 3\left(\frac{tan(A)}{2^{tan(A)}} \plus{} \frac{tan(B)}{2^{tan(B)}} \plus{} \frac{tan(C)}{2^{tan(C)}}\right)$ given $ tan(A)\plus{}tan(B)\plus{}tan(C) \equal{} 1$

[quote="kcstevonne"]I need help with the following inequalities:

Given triangle $ ABC$. $ A,B,C$ denote the angles, $ a,b,c$ denote the lengths of the sides of the triangle. Prove:

1) $ sin(2A) \plus{} sin(2B) \plus{} sin(2C) \le sin(A) \plus{} sin(B) \plus{} sin(C)$

[/quote]
We have:
$ \frac{3}{2} \geq cos(A)\plus{}cos(B)\plus{}cos(C)\equal{}1\plus{}4sin(\frac{A}{2})sin(\frac{B}{2})sin(\frac{C}{2})$
or
$ 1 \geq 8sin(\frac{A}{2})sin(\frac{B}{2})sin(\frac{C}{2})$
or
$ 4cos(\frac{A}{2})cos(\frac{B}{2})cos(\frac{C}{2}) \geq 4sin(A)sin(B)sin(C)$
So:
$ sin(A) \plus{} sin(B) \plus{} sin(C)\equal{}4cos(\frac{A}{2})cos(\frac{B}{2})cos(\frac{C}{2}) \geq 4sin(A)sin(B)sin(C)\equal{}sin(2A) \plus{} sin(2B) \plus{} sin(2C)$

Proof of (1).
Suppose triangle $ ABC$ is inscribed in a unit circle. If $ S$ is the area of the triangle, then $ \sin{2A} \plus{} \sin{2B} \plus{} \sin{2C} \equal{} 2S$.

Since $ R \equal{} 1$ and $ R \geq 2r$, so $ r \leq \frac12$. Hence:
$ 2S \equal{} r(a \plus{} b \plus{} c) \leq \frac {R}{2}(a \plus{} b \plus{} c) \equal{} \frac12 (2R\sin{A} \plus{} 2R\sin{B} \plus{} 2R\sin{C}) \equal{} \sin{A} \plus{} \sin{B} \plus{} \sin{C}$

So, $ \sin{2A} \plus{} \sin{2B} \plus{} \sin{2C} \leq \sin{A} \plus{} \sin{B} \plus{} \sin{C}$.

No6.[hide]In any triangle  we have the identity tannA +tanB +tanC =tanA.tanB.tanC.      This and the given condition, tanA +tanB +tanC=1 show that the triangle must be acute-angled.
We can therefore assume without loss of generality that tanA $ \leq$ tanB $ \leq$ tanC.
Therefore $ \frac{1}{2^{tanA}}$ $ \geq$ $ \frac{1}{2^{tanB}}$ $ \geq$ $ \frac{1}{2^{tanC}}$ .and by the Chebychev inequality we can write

$ \frac{1}{3}$.$ \sum{tanA}$.$ \frac{1}{3}$.$ \sum{\frac{1}}{2^{tanA}}$ $ \geq$ $ \frac{1}{3}$.$ \sum{\frac{tanA}{2^{tanA}}}$[/hide]


Let $D$ be a family of $s$-element subsets of $\{1.\ldots,n\}$ such that every $k$ members of $D$ have non-empty intersection. Denote by $D(n,s,k)$ the maximum cardinality of such a family.
a)	Find $D(n,s,4)$.
b)	Find $D(n,s,3)$.

it's is a nice problem.have anybody solved it?


inequalities
function
MIT
college
inequalities theorems

Hi.
This topic is for sharing your ultimate technique in solving difficult inequalities. For example [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=3048]www.artofproblemsolving.com/Forum/viewtopic.php?t=3048[/url] the 6th and 7th post and [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=180]www.artofproblemsolving.com/Forum/viewtopic.php?t=180[/url] the 5th post.
If everybody posts his favourite technique it will be great. :D  :D  :D

The technique's nice.  :lol: Arne's demonstration is great, but I still think what Myth has done is more impressive. http://www.mathlinks.ro/Forum/viewtopic.php?p=120395#p120395

[quote="Arne"]http://www.mathlinks.ro/Forum/viewtopic.php?highlight=elegant&t=13897

I like the thing I did there.

And of course, it is very nice to see that Thomas (Mildorf) was so impressed by the technique :)[/quote]

Indeed, it (weighted Jensen that does not use each entire fraction as its function argument) continues to be one of my all-time favorite ideas! Myth's is very nice as well. (In my mental Rolodex, the two are filed the same base idea.)

Incidentally, another thread to this effect (although a bit more introductory than most threads here) has surfaced in the Pre-Olympiad forum at [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=40857[/url].

I will introduce my latest summer project here too: A nice summary of Olympiad Inequalities (including many nice ideas.) It can be found online at [url]http://web.mit.edu/~tmildorf/www/Inequalities.pdf[/url].

I agree with you. myth's soln is really great.  ;)

i used that technique to solve an inequality recently posted on ML (which was trivial if one thought of holder).

[quote="Mildorf"][quote="siuhochung"]Nice work! I think some generalizations like vornicu-schur or generalized popoviciu could be mentioned though.

By the way, you entered MIT? congratulations![/quote]

Done! The $n$ variable Popoviciu, while an interesting result, looks quite unwieldy. In fact, I can't remember the last time I saw an Olympiad Inequality proven via any flavor of Popoviciu. (I heard rumors that such a problem showed up a long time ago in MOP, but that was before my first year in 2002.)

Thanks! In fact I leave home for MIT in a few days.[/quote]
Think about harazi's bloody inequality appeared in ML contest in which he proved it via one generalization of popoviciu.. (less generalized than http://www.mathlinks.ro/Forum/viewtopic.php?p=144036#p144036). Anyway, we seldom use it.... but I think it's a very beautiful inequality.


calculus
inequalities unsolved
inequalities

[quote="tranthanhnam"]find minimal of expression P :
$(x^2+1)\sqrt{x^2+1} -x.\sqrt{x^4+2x^2+5} +(x-1)^2$[/quote]

 Can you show me your minimal, Thanh Nam? I use caculus and come up solving a gain monster equation. I think it may have minimal but that value isn't nice at all.  :)

oh, i thinks  minnimal of expression is very nice
           i think that min P=0 at  x =1

you can prove stronger inquality
$(x^2+1)\sqrt{x^2+1} >x.\sqrt{x^4+2x^2+5}$ when "x" is not 1


let $ (a_1,a_2,...,a_n)$ $ \in$$ Z^n$ ,proce that $ P\equal{}(X\minus{}a_1)(X\minus{}a_2)...(X\minus{}a_n)\minus{}1$  is irreducible in $ Z$

suppose that $ P \equal{} FG$ where $ n \minus{} 1 \ge deg(G) \ge 1$  and  $ n \minus{} 1 \ge deg(F) \ge 1$ so we have $ F(a_i)G(a_i) \equal{} \minus{} 1$ so $ F(a_i) \plus{} G(a_i) \equal{} 0$ for every $ i$ .but $ deg(F \plus{} G) \le n \minus{} 1$ so $ F \plus{} G \equal{} 0$ so $ P \equal{} \minus{} F^2$ but $ C(P) \equal{} 1$and with this $ C(P) \equal{} \minus{} 1$ we get contradiction.

[quote="drapt@"]let $ (a_1,a_2,...,a_n)$ $ \in$$ Z^n$ ,proce that $ P \equal{} (X \minus{} a_1)(X \minus{} a_2)...(X \minus{} a_n) \minus{} 1$  is irreducible in $ Z$[/quote]
$ a_i,a_j$ must be different .
For example you chose $ a_1\equal{}..\equal{}a_n$ then easy to check that it is reducible.


In an oil leakage at sea, a circular slick of oil has a radius of 1km at 0700 hours on the 20th March and a radius of 2km at 0700 hours on the 21st March.

The area of the oil slick, A, varies with the time, t hours, since the oil first started leaking.

a) Calculate the time when the oil first started leaking.

b) Calculate when the oil slick will reach an area 5km away from the centre of the oil slick.

Welcome :welcome:

Does the area vary with the time or the square of the time (i.e. the area varying with the radius)? I don't know how it is in real life but I'd like this clarified please. Thanks.


quadratics
LaTeX
AMC
USA(J)MO
USAMO
algebra
quadratic formula

[hide]a) By the quadratic formula, the roots of the first are $\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}$  Substituting in the coefficients of the second one, we get roots of $\frac{-b\pm \sqrt{b^{2}-4ac}}{2}$, which are the roots of the first one divided by $\frac{1}{a}$.

b) We can simply use the quadratic formula again to find that the roots are $\frac{7\pm \sqrt{49+480}}{24}=-\frac{2}{3}, \frac{5}{4}$[/hide]

[size=75][color=red]Preview your posts before posting.
b-flat[/color][/size]

[quote="Ravi B"][b](a)[/b] Let $a$, $b$, and $c$ be numbers such that $a \ne 0$.  Show that the roots of the quadratic $a x^{2}+bx+c$ are $\frac{1}{a}$ times the roots of the quadratic $x^{2}+bx+ac$.

[b](b)[/b] Using part (a), solve the equation \[12 x^{2}-7x-10 = 0 \, .\] (This method could be useful if you find it easier to factor quadratics when they have a leading coefficient of 1.)[/quote]
[hide]Let $r$ be a root of the second quadratic. Then
$a\times \left(\frac{r}{a}\right)^{2}+b\times \frac{r}{a}+c = \frac{r^{2}}{a}+\frac{br}{a}+c = \frac{r^{2}+br+ac}{a}= 0$.

The roots of the equation are $\frac{1}{12}$ of the roots of
$x^{2}-7x-120 = 0$, which can be factored as $(x-15)(x+8)=0$. Its roots are $x = 15,-8$, so the roots of the orignal equation are $\frac{15}{12}=\boxed{\frac{5}{4}},-\frac{8}{12}=\boxed{-\frac{2}{3}}$.

Of course, you could also just factor it normally to get
$(4x-5)(3x+2) = 0$, and then work from there.[/hide]

@samath: You missed a '}' in your $\LaTeX$ somewhere.

[hide="a part"]Let the roots of $ax^{2}+bx+c$ be $r_{1}$. We have $r_{1}+r_{2}=\frac{-b}{a}$, so $ar_{1}+ar_{2}=-b$, and $r_{1}r_{2}=\frac{c}{a}$, so $(ar_{1})(ar_{2})=ac$. Thus the quadratic with roots $ar_{1}$ and $ar_{2}$ is $x^{2}+bx+ac$. Thus $ax^{2}+bx+c$ has roots $\frac{1}{a}$ times that of $x^{2}+bx+ac$.[/hide]
[hide="b part"]From part a, $x^{2}-7x-120=(x+8)(x-15)=0$ has roots 12 times that of $12x^{2}-7x-10=0$, so the roots are $\frac{15}{12},\frac{-8}{12}=\frac{5}{4},\frac{-2}{3}$.[/hide]

[quote="Ravi B"]nebula42, your solution is elegant.

samath, your solution to part (a) is fine.  In part (b), there is an error in your LaTeX.[/quote]
Didn't nebula write like a 12 on the USAMO (or something like that...)?


search
real analysis
real analysis unsolved

Show that \[ z \plus{} \frac{z^2}{2} \plus{} \frac{z^3}{3} \plus{} \cdots\] converges for any $ z$ with $ |z| \equal{}1$, except for $ z \equal{} 1$.

In fact, show that, for any $ \delta > 0$, the series converges uniformly on $ D_\delta \equal{} \{z \in \mathbb{C}: |z| \leq 1,\,|z \minus{} 1| \geq \delta\}$.

That's in there, too. Read my post, the one the follows jmerry's. (I only claimed $ |z|\equal{}1,|z\minus{}1|\ge\delta,$ but the argument works for $ |z|\le 1,|z\minus{}1|\ge \delta.$)


Let $ f : R \to R$ satisfying $ f(xy)\plus{}f(x\minus{}y) \geq f(x\plus{}y)$ all $ x;y \in R$
Prove that $ f(x) \geq 0$ with all $ x \in R$

[hide="Solution"]Let $ x \equal{} 0$. Then we have $ f( \minus{} y)\geq f(y)$. If we let $ y \equal{} \minus{} y$, then we have $ f(y)\geq f( \minus{} y)$. Therefore, $ f(y) \equal{} f( \minus{} y)$.
Now plug in $ \minus{} y$ for $ y$ into the original equation, giving $ f( \minus{} xy) \plus{} f(x \plus{} y) \geq f(x \minus{} y)$. Adding $ f(xy)$ to both sides gives $ f(xy) \plus{} f( \minus{} xy) \plus{} f(x \plus{} y) \geq f(x \minus{} y) \plus{} f(xy) \geq f(x \plus{} y)$, or $ f(xy) \plus{} f( \minus{} xy) \geq 0$. The result follows.[/hide]
Edit: Never mind, solution is wrong.

[quote="Brut3Forc3"][hide="Solution"]Let $ x \equal{} 0$. Then we have $ f( \minus{} y)\geq f(y)$. If we let $ y \equal{} \minus{} y$, then we have $ f(y)\geq f( \minus{} y)$. Therefore, $ f(y) \equal{} f( \minus{} y)$.
Now plug in $ \minus{} y$ for $ y$ into the original equation, giving $ f( \minus{} xy) \plus{} f(x \plus{} y) \geq f(x \minus{} y)$. Adding $ f(xy)$ to both sides gives $ f(xy) \plus{} f( \minus{} xy) \plus{} f(x \plus{} y) \geq f(x \minus{} y) \plus{} f(xy) \geq f(x \plus{} y)$, or $ f(xy) \plus{} f( \minus{} xy) \geq 0$. The result follows.[/hide][/quote]

Sorry I'm no good at these function equations, can you explain your fist step;
If $ x\equal{}0$ then we have
$ f(0) \plus{} f(\minus{}y) \ge f(y)$,
But if we let $ x\equal{}0, y\equal{}0$ then; $ 2f(0) \ge f(0)$ which implies $ f(0)\ge 0$
So how do we get
$ f(\minus{}y) \ge f(y)$ :?:

Okay, how about this then
[hide]
Let $ x \equal{} 1$
$ f(y) \plus{} f(1 \minus{} y) \ge f(1 \plus{} y) \qquad(1)$
Now let $ y \equal{} \minus{} y$
$ f( \minus{} y) \plus{} f(1 \plus{} y) \ge f(1 \minus{} y) \qquad(2)$
Adding $ (1)$ and $ (2)$ yields;
$ f( \minus{} y) \plus{} f(y) \ge 0\qquad(3)$

Subing $ x \equal{} 0$ into the origional equation and adding $ (3)$ yeilds
$ f(0)\ge f( \minus{} 2y)$

Now in the origoinal equation let $ x \equal{} \minus{} y$
$ f( \minus{} y^2) \plus{} f( \minus{} 2y) \ge f(0)$

$ \Rightarrow f( \minus{} y^2) \plus{} f( \minus{} 2y) \ge f(0) \ge f( \minus{} 2y) \Longrightarrow f( \minus{} y^2)\ge 0$

This covers all the negative numbers so; let $ x \equal{} y,y \equal{} x$ and using the orgional equation
$ f(xy) \plus{} f(x \minus{} y) \ge f(x \plus{} y)$
$ f(yx) \plus{} f(y \minus{} x) \ge f(x \plus{} y)$

$ \Rightarrow f(x \minus{} y) \ge f(y \minus{} x)$, but similarly
$ f(y \minus{} x) \ge f(x \minus{} y) \Rightarrow f(x \minus{} y) \equal{} f(y \minus{} x)$ or $ f(x) \equal{} f( \minus{} x)$
And since we have shown that $ f(x) \ge 0$ for $ x\le 0$ then it follows that
$ f(x) \ge 0$ for all $ x \in \mathbb{R} \qquad Q.E.D$  :lol:
[/hide]

[quote="ocha"][hide]$ f(xy) \plus{} f(x \minus{} y) \ge f(x \plus{} y)$
$ f(yx) \plus{} f(y \minus{} x) \ge f(x \plus{} y)$

$ \Rightarrow f(x \minus{} y) \ge f(y \minus{} x)$, but similarly
$ f(y \minus{} x) \ge f(x \minus{} y) \Rightarrow f(x \minus{} y) \equal{} f(y \minus{} x)$[/hide][/quote]

Explain? I hope you didn't subtract those inequalities.

I think my (revised) solution works now!
[hide="Solution"]We have $ f(x)\plus{}f(\minus{}x) \geq 0$ (proof omitted, see my or ocha's post). Now assume that there exists a value $ a$ such that $ f(a) < 0$. Now, solving the system $ xy\equal{}x\minus{}y\equal{}a$, and letting $ \sqrt{a^2\plus{}4a}\equal{}b$ we get $ x\equal{}\frac{a \pm b}{2}, y\equal{}\frac{\minus{}a\pm b}{2}$ as solutions. Plugging both sets in gives us $ f(a)\plus{}f(a) \geq f(b), f(a)\plus{}f(a) \geq f(\minus{}b)$. Since $ 0 < f(a)$, we have $ 0 < f(b),f(\minus{}b)$ This is a contradiction, since $ f(x)\plus{}f(\minus{}x) \geq 0$. Therefore, there is no value $ a$ such that $ f(a) < 0$, and we are done.[/hide]


Prove that:
\[ \frac{(x\minus{}b)(x\minus{}c)}{(a\minus{}b)(a\minus{}c)}\plus{}\frac{(x\minus{}c)(x\minus{}a)}{(b\minus{}c)(b\minus{}a)}\plus{}\frac{(x\minus{}a)(x\minus{}b)}{(c\minus{}a)(c\minus{}b)}\equal{}1\]
where $ a \neq b \neq c$

[hide="hint"]
Let $ F(x)$ and $ G(x)$ be two polynomials of degree $ n$.
Then if there exists $ n \plus{} 1$ [b]dinstict[/b] $ t_i$, where $ i \in \{1,2,... ,n \plus{} 1\}$ such that $ F(t_i) \equal{} G(t_i)$ then $ F(x) \equal{} G(x).$[/hide]

[quote="hasan4444"]Hello Dimitris X,

Sorry to say that but I didn't understand that much, does it have a name so I can search for it. Or can you give me an easy example.

Thanking you very much[/quote]

For you hasan:

[b]Theorem[/b]
If $ F(x)$ and $ G(x)$ are polynomials of degree not exceeding $ n$ and such that for $ n \plus{} 1$ dinstict numbers $ c_0,c_1,...c_n$ we have $ F(c_i) \equal{} G(c_i)$ ($ i \equal{} 0,1,...,n$),then $ F(x) \equal{} G(x).$

If you searching for a proof:

Let us assume that $ F(x)$ and $ G(x)$ are dinstict
Consider a polynomial $ h(x) \equal{} F(x) \minus{} G(x)$
Then $ h(x)$ is a nonzero polynomial of degree not exceeding $ n$ and has $ n \plus{} 1$ zeros zeros $ c_0,c_1,...,c_n$-which is contradiction.....

Now what can you say for $ F(x)\equal{}\sum\frac{(x\minus{}a)(x\minus{}b)}{(a\minus{}b)(a\minus{}c)}$ and $ G(x)\equal{}1$?????

$ F(a)\equal{}1,F(b)\equal{}1,F(c)\equal{}1$ and $ F$ is of degree 2....
I hope  i dint made a mistake....(as i used to lately..... :rotfl: )

put that $ P(x)\equal{}\frac{(x\minus{}b)(x\minus{}c)}{(a\minus{}b)(a\minus{}c)}\plus{}\frac{(x\minus{}c)(x\minus{}a)}{(b\minus{}c)(b\minus{}a)}\plus{}\frac{(x\minus{}a)(x\minus{}b)}{(c\minus{}a)(c\minus{}b)}$
$ P$ is a second degre plynomial function, then there are at most two distinct number $ x_1$ and $ x_2$ such that $ f(x_1)\equal{}f(x_2)$ but we have $ a,b,c$ are all distinct and $ P(a)\equal{}P(b)\equal{}P(c)\equal{}1$ then $ P$ is constant, from where we deduct that $ P(x)\equal{}1$
for all real number $ x$
N.B: a second degre plynomial function can never be periodic.


Irina stood at the center of a circular field that had a radius of 138 feet. She walked due north halfway to the circle. She then turned and walked due east half way to the circle. She turned again and walked due south half way to the circle. Finally she turned and walked due west half way to the circle. When she stopped, how many feet was Irina from the center of the circle? Express your answer to the nearest foot.

This has been posted several times before...
Search "Pedro Circle" and you will find results.


calculus
integration
calculus computations

dz/du=sin 2u/cos z

if u= pi/4, z= pi/6

then sin z=?

other question...:

dy/dx=4x/3(y^2)

if y=2, x=1 then y=?

last one:

if dy/dx=[img]http://upload.wikimedia.org/math/1/6/0/160a5b4ac79375bd4c5e13c0f3a95f73.png[/img]y

^square root of y

when y=4 and x=2 then y=?

lost...so please if you can help, please do so. :ninja:

thanks

is there some shortcut to do these? or no... i have no idea how to approach them... if someone could solve one with a short explanation that could really help and i dont need all to be solved then. Please if you could answer these....that would help a lot

thanks...much appreciated

The appropriate technique for these differential equations is separation of variables.

$\frac{dy}{dx}=\frac{4x}{3y^{2}}$, cross-multiply to get $3y^{2}dy=4xdx$, integrate both sides, $\int 3y^{2}dy=\int 4xdx\Rightarrow y^{3}+A=2x^{2}+B$ but you can combine these constants so that $y^{3}=2x^{2}+C$. According to your initial conditions $(2)^{3}=2(1)^{2}+C\Rightarrow 6=C$, so $y^{3}=2x^{2}+6\Rightarrow y=\sqrt[3]{2x^{2}+6}$. Others should be similar.


function
algebra unsolved
algebra
functional equation

Let $ a , b$ be fixed real numbers and $ a < 0$. Let $ f: \mathbb{R} \ \to \ \mathbb{R}$ be a function such that
\[f(f(x)) \ \equal{} \ ax \plus{} b ,\quad  \forall   x \in \mathbb{R}.\]
Prove that there are infinitely many value $x_0$ such that $f$ is discontinuous at $x_0$.

[quote="nguyenvuthanhha"][i]Let $ a , b$ be fixed real numbers , $ a < 0$

 $ f$ be a function , $ f: \mathbb{R} \ \to \ \mathbb{R}$  such that :

   $ f(f(x)) \ \equal{} \ ax \plus{} b \ \forall \ x \ \in \ \mathbb{R}$

Prove that there are infinite many value $ x_0$ such that :

  $ f$ is discontinuous at $ x_0$[/i][/quote]

First, get rid of $ b$ :
Let $ u\equal{}\frac{b}{a\minus{}1}$ and $ f(x)\equal{}g(x\plus{}u)\minus{}u$. The equation becomes $ g(g(x\plus{}u))\minus{}u\equal{}ax\plus{}b\equal{}a(x\plus{}u)\plus{}b\minus{}au$ and so $ g(g(x))\equal{}ax$

$ g(g(x))\equal{}ax$ $ \implies$ $ g(x)$ is a bijection
$ g(g(x))\equal{}ax$ $ \implies$ $ g(ax)\equal{}ag(x)$ and so $ g(0)\equal{}0$. So, since $ g(x)$ is injective, $ g(x)\neq 0$ $ \forall x\neq 0$

Let then $ u>0$ and $ v\equal{}g(u)$ and $ w\equal{}g(v)\equal{}au$ and $ t\equal{}g(w)\equal{}av$
If $ v>0$, we have $ u>0$, $ v>0$, $ g(u)>0$ and $ g(v)<0$ and so a discontinuity point in $ (u,v)$
If $ v<0$, we have $ v<0$, $ w<0$, $ g(v)<0$ and $ g(w)>0$ and so a discontinuity point in $ (v,w)$
And so we have at least one discontinuity point $ x_0\neq 0$

Then 2 cases :

1) $ a\neq \minus{}1$
Then $ g(a^nx)\equal{}a^ng(x)$ and so, if $ x_0$ is a discontinuity point of $ g(x)$, $ a^nx_0$ is too. and so we have infinitely many discontinuity points.

2) $ a\equal{}\minus{}1$
Suppose we have a finite number of discontinuity points.
Consider the set of intervals $ (u,v)$ (including $ (u,\plus{}\infty)$ and $ (\minus{}\infty,v)$) where $ g(x)$ is continuous (with $ u,v$ discontinuity points or $ \pm\infty)$)

It's easy to see that we have an even number of such positive intervals, so an odd number of positive discontinuity points.
But, if $ u$ is a discontinuity point, $ v\equal{}g(u)$ is too and so are $ \minus{}u$ and $ \minus{}v$, so the set of positive discontinuity points need to be even.
Hence the contradiction.
And so we have infinitely many discontinuity points.


Use the cases environment:
$ y\equal{}\begin{cases}
x^2 & x>0 \\
x\plus{}2 & x \leq 0
\end{cases}$

\[ y\equal{}
\left\{\begin{array}{ll}
x^2 & x>0 \\
x\plus{}2 & x \leq 0
\end{array}\right.
\]

[code]\[
y=
\left\{\begin{array}{ll}
x^2 & x>0 \\
x+2 & x \leq 0
\end{array}\right.
\][/code]

Yes, it automatically centers if you use

[code]
\[
latex code here
\]
[/code]

Also,

\[ y\equal{}
\left\{\begin{array}{ll}
x^2 & x>0 \\
x\plus{}2 & x \leq 0
\end{array}\right.\]

If you use

[code]
$$
latex code here
$$
[/code]

it also centers it.

[quote="AIME15"]If you use

[code]
$$
latex code here
$$
[/code]

it also centers it.[/quote]
Please don't use double dollar signs as they are obsolete and can cause problems. See [url=http://www.tex.ac.uk/cgi-bin/texfaq2html?label=dolldoll]Why use \ [ ...\ ] in place of $ \$\$$ ...$ \$\$$?[/url].


$ \text {Given that } 2^{10} \plus{}2^{13}\plus{}2^m \equal{} N^2  \text{ , where } n ,m \in \mathbb {N}$

$ \text {Find the possible value of m and N.}$

[quote="vinskman"]$ \text {Given that } 2^{10} \plus{} 2^{13} \plus{} 2^m \equal{} N^2 \text{ , where } n ,m \in \mathbb {N}$

$ \text {Find the possible value of m and N.}$[/quote]

Let $ S\equal{}2^{10}\plus{}2^{13}\equal{}2^{10}(1\plus{}2^{3})\equal{}2^{10}\cdot 9\equal{}(2^{5}\cdot 3)^2\equal{}a^2$ (where $ a\equal{}2^5\cdot 3$).
Now $ a^2\plus{}2^m\equal{}N^2\Rightarrow2^m\equal{}N^2\minus{}a^2\equal{}(N\minus{}a)(N\plus{}a)$, i.e: $ N\minus{}a\equal{}2^w$; $ N\plus{}a\equal{}2^z$ where $ w\plus{}z\equal{}m$. We deduce that $ 2^z\minus{}2^w\equal{}2a\equal{}2\cdot 2^5\cdot 3\equal{}2^6\cdot 3\equal{}2^6(2^2\minus{}1)$.
But $ 2^z\minus{}2^w\equal{}2^w(2^{z\minus{}w}\minus{}1)\equal{}2^6(2^2\minus{}1)\Rightarrow w\equal{}6; z\minus{}w\equal{}z\minus{}6\equal{}2\Rightarrow z\equal{}8$.
This gives $ m\equal{}z\plus{}w\equal{}6\plus{}8\equal{}14$, and $ N\equal{}160$.
Answer: $ (m,N)\equal{}(14,160)$.

$ \text {Well Done...   BinGo..}$ :roll:


Αν το $ ABCDE$ ειναι κυρτο πενταγωνο και $ E_1,E_2,E_3, E_4, E_5$ ειναι τα εμβαδα των τριγωνων $ ABC, BCD,CDE,DEA,EAB$  αντιστοιχα ν.δ.ο. $ t > \sqrt {2(\sum_{i \equal{} 1}^{5}{E_i})}$ οπου $ t$ η ημιπεριμετρος του $ ABCDE$

Πολύ ωραία η άσκηση και η ιδέα για την κατασκευή  :)

Έστω $ a_1,...,a_5$ οι πλευρές του πενταγώνου . Τότε

$ E_i \equal{} \frac {1}{2}a_ia_{i \plus{} 1}\sin \angle (a_i,a_{i \plus{} 1})\leq\frac {a_ia_{i \plus{} 1}}{2}$

όπου οι δείκτες είναι modulo 5 προφανώς .

Άρα μένει να αποδείξουμε ότι  $ (\sum_{i \equal{} 1}^5 a_i)^2 > 4(a_1a_2 \plus{} a_2a_3 \plus{} a_3a_4 \plus{} a_4a_5 \plus{} a_5a_1)$

Έστω δίχως βλάβη της γενικότητας ότι ο $ a_4$ είναι ο ελάχιστος  .

Αν $ f(a_1,...,a_5) \equal{} (\sum_{i \equal{} 1}^5 a_i)^2 \minus{} 4(a_1a_2 \plus{} a_2a_3 \plus{} a_3a_4 \plus{} a_4a_5 \plus{} a_5a_1)$

Τότε $ f(a_1,...,a_5) \minus{} f(a_1,a_2,a_3,a_4 \plus{} a_5) \equal{} 4(a_3a_4 \plus{} (a_3 \minus{} a_4)a_5 \plus{} a_1a_4) > 0$

άρα αρκεί να δείξουμε ότι $ f(a_1,a_2,a_3,a_4 \plus{} a_5)\geq 0 \Leftrightarrow f(x,y,z,w)\geq 0$  ή

 $ (x \plus{} y \plus{} z \plus{} w)^2\geq 4(xy \plus{} yz \plus{} zw \plus{} wx) \Leftrightarrow (x \plus{} y \plus{} z \plus{} w)^2\geq 4(x \plus{} z)(y \plus{} w)\Leftrightarrow (x \plus{} z \minus{} y \minus{} w)^2\geq 0$ που ισχύει  :)

Εκπληκτικος οπως παντα.....


Let's say we have the equation $K^{2}+t+sK = 0$ with roots $r_{1},r_{2}$ such that $r_{1}-2r_{2}= 2$ and $2r_{1}-3r_{2}= 5$. Find the value of $s$.

[hide]So lets call the first root x and the second root y. So we have two equations which are $x-2y=2$ and $2x-3y=5$. Multiplying the first equaiton by $-2$ we have $-2x+4y=-4$ and $2x-3y=5$ Addind so we can solve for the second root which is 1. The other root will be 4. Then since those are the roots we have (K-4)(K-1)=0. Multiplying it out we have $K^{2}-5K+4$. So the s=-5.[/hide]


Let x_i(i=1,..,n) be real positive numbers. Prove that

 sum(i=1,..,n)(x_i)^(n-1) \geq (x_1*x_2*..*x_n)(1/x_1+..+1/x_n)

For each j, from AM/GM, we have :
Sum{i=/=j} (x_i) n-1   \geq prod{i=/=j} x_i = P/x_j,
where P = x_1*...*x_n.
Then sum all these inequalities when j= 1,...,n to get the desired inequality.

Pierre.

[quote="pbornsztein"]4 minutes...aargh I did not beat Harazi's record...(I'm Joking!!!)

Pierre.[/quote]
Harazi has a T3 connection, in direct LAN with Mathlinks.ro, so maybe it's just the internet speed :D:D:D


A fly starts at one vertex of a cube with edge length 1 inch and randomly walks along edges of the cube. What is the number of inches in the longest distance that it could walk without retracing any edge?

[hide]
$\boxed{9}$ inches
Just draw a cube and try it. :D [/hide]

[hide]9. The reason is that the ant can walk around the top four, go down, and go around the bottom four. [/hide]

Use a system of networks to represent it to realize it can only touch nine arcs. Nice quadruple post daermon.


The incircle of $ \triangle ABC$ touches the its sides in $ D\in (BC)$ , $ E\in (CA)$ , $ F\in (AB)$ . Denote

$ \begin{array}{c}
P\in BE\cap DF\\\
R\in CF\cap DE\end{array}$ and the midpoints $ Q$ , $ S$ of $ [PE]$ , $ [RF]$ respectively. Prove that $ \widehat {EFQ}\equiv\widehat {FES}$ .

I like very much this problem. I have only a metrical with a tehnical trigonometry proof of it .

Can somebody find a synthetical proof ?! I"ll post later my ugly proof. Thank you !

Nice problem, dear Virgil Nicula. You can find a synthetic proof here: http://www.mathlinks.ro/viewtopic.php?t=193904
See also http://www.mathlinks.ro/viewtopic.php?t=202824 and http://www.mathlinks.ro/viewtopic.php?p=131967#p131967

[quote][color=darkred]The incircle of $ \triangle ABC$ touches the its sides in $ D\in (BC)$ , $ E\in (CA)$ , $ F\in (AB)$ . Denote

$ \begin{array}{c} P\in BE\cap DF \\
\ R\in CF\cap DE\end{array}$ and the midpoints $ Q$ , $ S$ of $ [PE]$ , $ [RF]$ respectively. Prove that $ \widehat {EFQ}\equiv\widehat {FES}$ .[/color][/quote]

[color=darkblue][b][u]Proof (metrical).[/u][/b] Denote $ m(\widehat {EFQ}) = \phi$ . Thus, $ m(\widehat {PFQ}) = 90^{\circ} - (\frac C2 + \phi)$ and observe that

$ \|\begin{array}{c} EF = 2(p - a)\sin\frac A2 \\
 \\
DF = 2(p - b)\sin\frac B2\end{array}$ . Therefore, $ QP = QE$ $ \Longleftrightarrow$ $ \frac {FP}{EF} = \frac {\sin\widehat {EFQ}}{\sin\widehat {PFQ}}$ $ \Longleftrightarrow$ $ \boxed {\ FP = \frac {2(p - a)\sin\frac A2\sin\phi}{\cos(\frac C2 + \phi)}}\ \ (1)$ .

$ \|\begin{array}{c} \frac {PF}{PD} = \frac {BF}{BD}\cdot\frac {\sin\widehat {PBF}}{\sin\widehat {PBD}} \\
 \\
\frac {BA}{BC}\cdot\frac {\sin\widehat {EBF}}{\sin\widehat {EBC}} = \frac {EA}{EC}\end{array}\|$ $ \implies$ $ \frac {PF}{PD} = \frac {a(p - a)}{c(p - c)}$ $ \implies$ $ \frac {PF}{a(p - a)} = \frac {PD}{c(p - c)} = \frac {FD}{a(p - a) + c(p - c)}$ . Hence

$ \boxed {\ FP = \frac {2a(p - a)(p - b)\sin\frac B2}{a(p - a) + c(p - c)}\ }\ \ (2)$ . From the relations $ (1)$ , $ (2)$ obtain $ \frac {a(p - b)\sin\frac B2}{a(p - a) + c(p - c)} = \frac {\sin\frac A2\sin\phi}{\cos(\frac C2 + \phi)}$ $ \implies$

$ \frac {a(p - b)\sin\frac B2}{a(p - a) + c(p - c)} = \frac {\sin\frac A2}{\cos\frac C2\cot\phi - \sin\frac C2}$ $ \implies$ $ \boxed {\ \cot\phi = \frac {a(p - b)\sin\frac B2\sin\frac C2 + [a(p - a) + c(p - c)]\sin\frac A2}{a(p - b)\sin\frac B2\sin\frac C2}\ }\ \ (3)$.

Since $ \|\begin{array}{c} \sin\frac B2\sin\frac C2 = \sqrt {\frac {(p - a)(p - c)}{ac}\cdot\frac {(p - a)(p - b)}{ab}} = \frac {p - a}{a}\cdot\sin\frac A2 \\
 \\
\sin\frac B2\cos\frac C2 = \sqrt {\frac {(p - a)(p - c)}{ac}\cdot\frac {p(p - c)}{ab}} = \frac {p - c}{a}\cdot\cos\frac A2\end{array}$ , the relation $ (3)$ becomes

$ \cot\phi = \frac {a(p - b)\cdot\frac {p - a}{a}\sin\frac A2 + [a(p - a) + c(p - c)]\cdot\sin\frac A2}{a(p - b)\cdot\frac {p - c}{a}\cos\frac A2}$ $ \implies$

$ \boxed {\ \cot\phi = \frac {(p - a)(p - b) + a(p - a) + c(p - c)}{(p - b)(p - c)}\cdot\tan\frac A2\ }\ \ (4)$ . Prove easily that

$ (p - a)(p - b) + a(p - a) + c(p - c) = 2a(p - a) + (p - b)(p - c)$ . Therefore, the relation $ (4)$ becomes:

$ \cot\phi = \frac {2a(p - a) + (p - b)(p - c)}{(p - b)(p - c)}\cdot\tan\frac A2$ $ \implies$ $ \boxed {\boxed {\ \cot\phi = \tan\frac A2 + \frac {2a(p - a)}{pr}\ }}$ .

The expresion of $ \cot\phi$ is symmetrically in a pair $ \{b,c\}$ , what means $ \cot\phi = \cot\psi$ , i.e. $ \phi = \psi$ .[/color]


umm.... wow.... this is really random.... :huh:

[quote="mathfreak77"][quote="Nebraska boy"]According to the current votes , Brazil remains on the top of the pool  then Venezuela in the second spot
My favorites are: Colombia , Dominican Republic , USA , Kosovo , China and Vietnam.

Will a latina girl win this year?

Will Venezuela repeat the crown ?

Could an European win the crown  ?

Will USA fall again on stage ?  :rotfl:

I wanna know your predictions[/quote]
huh???[/quote]

yeah, i agree with mathfreak77... what do you mean??
and why is this posted in this forum.. although this is "fun factory", this may be just a tad too unrelated.... :|

That was soooooooooooooooooooooooooooooooooooooooo random! I don't even know who ms. universe is! :rotfl:

[quote="hjoon0125"]GOOGLE IZ UR FRIEND[/quote]

+its FUN factory.


calculus
derivative
geometric series
calculus computations

Find the nth derivative of $ f(x)\equal{}\frac{x^n}{1\minus{}x}$.

What is a good substitution?

Do the long division, or think in power series. (expand it as a geometric series and differentiate term by term). There is a difference between the first $ n\minus{}1$ derivatives and later derivatives.

$ f(x) \equal{} \frac{1}{1 \minus{} x} \minus{} \frac{1 \minus{} x^n}{1 \minus{} x} \equal{} \frac{1}{1 \minus{} x} \minus{} \left( 1 \plus{} x \plus{} x^2 \plus{} ... \plus{} x^{n\minus{}1} \right)$

Hence $ f^{(n)} (x) \equal{} \frac{n!}{(1 \minus{} x)^{n\plus{}1}}$.


Let a,b,c be positive reals with sum 1.Prove that:
$ \ \sqrt {ab\plus{}c}\plus{}\sqrt{bc\plus{}a}\plus{}\sqrt{ca\plus{}b} \ge 1\plus{}\sqrt{ab}\plus{}\sqrt{bc}\plus{}\sqrt{ca}$

We have: $ \sqrt[]{(a\plus{}b)(a\plus{}c)} \geq\ a\plus{}\sqrt[]{bc}$
So, the problem was solved  :)


geometry
trigonometry
trig identities
Law of Sines

[img]http://img238.imageshack.us/img238/695/adszek4.jpg[/img]
Thanks and best regards!

You can apply the Law of Sines to each of the smaller triangles. Let's have angle BAD = A and CAD = B, such that A+B=X.

[hide="Left Triangle"]
$ \frac {3}{\sin x} \equal{} \frac {3\sqrt {26}}{\sin 90}$

$ 3 \equal{} 3 \sqrt {26} \sin x$

$ \frac {1}{\sqrt {26}} \equal{} \sin A$

$ \sin ^{ \minus{} 1} \frac {1}{\sqrt {26}} \equal{} A$

$ 11.3 \approx A$
[/hide]

[hide="Right Triangle"]
$ \frac {5\sqrt {13}}{\sin 90} \equal{} \frac {10}{\sin B}$

$ \sin B \equal{} \frac {10}{5\sqrt {13}}$

$ B \equal{} \sin ^{ \minus{} 1} \frac {10}{5\sqrt {13}}$

$ B \approx 33.7$
[/hide]

That's assuming I didn't make any mistakes in my calculations.

Another way

15*13=AB*BC*Sin A {Compare Area of triangle ABC}

=> Sin A = 15*13/3*sqrt{26}*5*sqrt{13}
=>Sin A = 1/sqrt{2}
=>A=45

[quote="SimonM"]Not jumping to any conclusions, but the "urgent" suggested homework to me, hence the hint[/quote]

Ah I didn't notice the time-stamp, I thought it was a thread that was left unanswered a while ago.

We can apply the tangent addition formula: \[ \tan\angle BAC\equal{}\tan\angle BAD\plus{}\angle CAD\equal{}\frac{\tan\angle BAD\plus{}\tan\angle CAD}{1\minus{}\tan\angle BAD\cdot\tan\angle CAD}\equal{}\frac{\frac{3}{15}\plus{}\frac{10}{15}}{1\minus{}\frac{3}{15}\cdot\frac{10}{15}}\equal{}1\] Therefore $ \angle BAC\equal{}\frac{\pi}{4}$.

Thank you.
I am from Turkey and i have finished high school.I have very big exam 70 days later.If i will pass the exam i will go to univercity.So all days i am solving problem and sometimes i didnt solve and i added here...they are not my homework.
Thanks and best regards!

Yeah, good to know - I think people were just asking because we get a lot of posts (I personally don't post a lot - but I do look quite a bit in the forums) asking for help with homework.

I hope you get into your university.

Best of luck.

Ok! you can post it to this forum. Every body will help you.

$ S_{\triangle ABC}=\frac{1}{2}AD.BC=\frac{1}{2}AB.AC.sinA$ $ \Longleftrightarrow 15.13=AB.BC.Sin A$

$ \Longleftrightarrow SinA= \frac{15.13}{AB.BC}$

But $ AB \sqrt{15^2+3^2}=3\sqrt{26}, AC= \sqrt{15^2+10^2}=5\sqrt{13}$

$ \Longrightarrow SinA= \frac{15.13}{15.13.\sqrt{2}}=\frac{\sqrt2}{2}$

$ \Longrightarrow A= 45^0$.


On a line $ p$ which does not meet a circle $ K$ with center $ O$, point $ P$ is taken such that $ OP$ is perpendicular to $ P$. Let $ X\neq P$ be an arbitrary point on $ p$. The tangents from $ X$ to $ K$ touch it at $ A$ and $ B$. Denote by $ C$ and $ D$ the orthogonal projections of $ P$ on $ AX$ and $ BX$ respectively.

(a) Prove that the intersection point $ Y$ of $ AB$ and $ OP$ is independent of the location of $ X$.
(b) Lines $ CD$ and $ OP$ meet at $ Z$. Prove that $ Z$ is the midpoint of $ PY$ .

For (a), $ Y$ is the polar point of the line $ p$. And hence, always fixed.


number theory
prime numbers
number theory unsolved

[quote="sam-n"]assume that p1,p2,...,pr are distinct prime numbers,qnd n1,n2,...,nr are  natural numbers . prove that thye number of (x,y) {x,y are integers}which x^3+y^3=p1^n1*p2^n2*...Pr^nr,is at most 2^(r-1)[/quote]

Sorry, the problem is WRONG because for $r=1$ then $2^3+1=9$, so has 2 solutions, more than 1.

RX


$ \triangle ABC$ $ \implies$ $ \{\begin{array}{cc}1\blacktriangleright &\min\{(\frac{b+c}{a})^{2}\ ,\ (\frac{c+a}{b})^{2}\ ,\ (\frac{a+b}{c})^{2}\}\le\frac{2R}{r}\\ \\ 2\blacktriangleright &\min\{\frac{b+c}{a}\ ,\ \frac{c+a}{b}\ ,\ \frac{a+b}{c}\}\le\frac{R}{r}\end{array}$ (standard notations).

[b]Remark.[/b] See http://www.mathlinks.ro/Forum/viewtopic.php?t=163531 .

[quote="13375P34K43V312"][quote="b-flat"][hide="second one??"]By the Euler Theorem (I think that is the one, anyway), $ R\geq2r\Rightarrow\frac{R}{r}\geq2$.

We can assume without loss of generality that $ a\geq b\geq c$, so thus, $ 2a\geq b\plus{}c\Rightarrow 2\geq\frac{b\plus{}c}{a}$.  We only have to show that it is true for one of

Combining the two, we are done.[/hide]

Or am I missing something?[/quote]

Your sign is switched, meaning you weakened it too much.[/quote]

I don't see anything wrong with that.  That seems fine....  But I think the point was to prove the stronger inequality $ \frac{R}{r}\ge 2$...though that anyways follows easily from the nine-point circle...

Moreover, note that the first inequality also follows easily from your proof.

Was it supposed to be the maximum of those terms, Virgil?


analytic geometry
graphing lines
slope
geometry
ratio

In the XY plane consider the L-shaped region bounded by the horizontal and vertical segments with vertices (0,0), (3,0), (3,3), (3,1), (5,1), AND (5,0). What is the slope of the line through the origin that divides the area of this region in half?

Please tell me the first few steps to do this problem, not the solution. Thanks. :)

[quote="ckck"]Is it 1/5?[/quote]
No, I believe the answer is
[hide]7/9[/hide]

If it's an L-shape, then the second point must be $(0,3)$ and not $(3,0)$.

[hide="And the answer is"]$y={7\over 9}x$[/hide]

[quote="ckck"]

Please tell me the first few steps to do this problem, not the solution. Thanks.[/quote]

So far that's what peoiple haven't said. after you make a sketch of the graph (and I do think you may have typed up one of the points wrong) find the area of the whole thing. That way you can find the area of 1/2. then you know the equation of the line has to be y=mx, (since it goes through the origin). realize that m is a ratio: the ratio of $\frac{rise}{run}$
so then you know the area that you want to have this line cut off: 1/2 the total area. you need that to equal 1/2*base*height because it will probably cut off a triangle. also, 'base' is the same length as 'run' above and 'height' is the same as 'rise'. (you know what the rise is) so with the numbers you have, solve first for 'run' or 'base', then plug taht in with the 'rise' or 'height' to get your slope.

of course, this is all assuming that the part it cus off is a triangle and not a triangle with a littler triangle cut out of it... if that were the case this would be much toughter... we can't really tell though, so I'd check after I got the line to make sure it didn't produce this case.

[hide][quote="ckck"]Can you explain.[/quote]
We have a line $y=ax$ that divides this figure in half. First, divide this figure into the 3x3 square and the 2x1 rectange. Let the area of the top part of the figure be $m$. Let the bottom part of the figure that is inside the 3x3 square be $n$. We know $m+n=9$ and $m=n+2$ We find that $n=\frac{7}{2}$. This represents the area of the triangle with base 3, and height unknown. If the line $y=ax$ intersects the height of the triangle at $x=3$, it intersects it at the point $3, 3a$ and the height is $3a$. So, we have the area, $\frac{7}{2}=\frac{1}{2}(3)(3a)$. Solve to get $a=\frac{7}{9}$[/hide]


[quote="VANCHANH123"]$4^{3x-2}+3^{x}=7^{x}$[/quote]

It's rather easy to show that $f(x)=3^{x}+4^{3x-2}-7^{x}$ just have two zeros on $%Error. "mathlabb" is a bad command.
{R}$ :
$x=1$
$x=a$ with $a\in(0,1)$ (in fact $a$ near of $0.0946$

--
Patrick

Bad title http://www.mathlinks.ro/Forum/viewtopic.php?t=144403 Locked!


[quote="zaizai-hoang"]Let $a,b,c$ are positive real number such that $a+b+c=3$. Prove that:
$\frac{1}{a^{3}+3}+\frac{1}{b^{3}+3}+\frac{1}{c^{3}+3}\ge \frac{3}{4}$[/quote]
Try $a=b=1.4$ and $c=0.2$ :wink:


what motivates you to do math

right now i don't feel like doing any problems

Usually finding something interesting as well as challenging. Particularly, problems that seem to have simple solutions, but in reality, are far more complex than expected.

Ditto that; I often look at a post and say "oh pfft"..."OH".

Also, I've always liked things I can improve in, and nothing allows that more than mathematics. When I see a problem I can't solve, I write it down. Then I learn the necessary material and slowly, but surely, I tick problems off my list.

If you're burned out though, just try going out for a bit. Relax. That should help.


linear algebra
matrix
LaTeX
vector
linear algebra unsolved

Let P be a nxn Markov matrix which can be decomposed into k matrices to be arranged on the main diagonal $P= \begin{bmatrix}P_1&0&\cdots&0\\0&P_2&\cdots&0\\ \vdots&\vdots&\ddots&\vdots\\0&0&\cdots&P_k\end{bmatrix}$. Define the P' matrix as follow: $P' = \alpha P+(1-\alpha)J/n$ where J is the matrix of all 1s. Matrix P' has an eigenvector $\pi =(\pi_{1},...,\pi_{n})$  corresponds to the eigenvalue $\lambda=1$ and $\sum^n_{i=1} \pi_i=1$. Let $P_i$ be $n_i\mbox{x}n_i$ matrices where $\sum^k_{i=1} n_i=n$ and define $P_{i}'= \alpha P_i+(1-\alpha)J/n$ with eigenvector $\pi^i$ repectively, $i=\overline{1,k}$. $\forall \delta = (\delta_1,...,\delta_k)$  where $\sum^k_1 \delta_i=1$, prove that $\pi=(\delta_{1}\pi_{1},...,\delta_{k} \pi_{k})$ is an eigenvector of matrix P. First please check the case where $\delta_{i}=n_{i}/n$.

$P\pi=\begin{bmatrix}P_1&0&\cdots&0\\0&P_2&\cdots&0\\ \vdots&\vdots&\ddots&\vdots\\0&0&\cdots&P_k\end{bmatrix}\cdot\begin{bmatrix} \delta_1pi_1\\\delta_2\pi_2\\\vdots\\\delta_k\pi_k\end{bmatrix}=\begin{bmatrix} \delta_1P_1pi_1\\\delta_2P_2\pi_2\\\vdots\\\delta_kP_k\pi_k\end{bmatrix}= \begin{bmatrix} \delta_1pi_1\\\delta_2\pi_2\\\vdots\\\delta_k\pi_k\end{bmatrix}$

That's really all there is to it. You have the matrices expressed in compatible block structure- use it to multiply. The matrix is even block diagonal, so each "element" is only one term.

Everthing else is pointless distraction- the special cases aren't any easier than the general version.

[quote="jmerry"]$P\pi=\begin{bmatrix}P_1&0&\cdots&0\\0&P_2&\cdots&0\\ \vdots&\vdots&\ddots&\vdots\\0&0&\cdots&P_k\end{bmatrix}\cdot\begin{bmatrix} \delta_1pi_1\\\delta_2\pi_2\\\vdots\\\delta_k\pi_k\end{bmatrix}=\begin{bmatrix} \delta_1P_1pi_1\\\delta_2P_2\pi_2\\\vdots\\\delta_kP_k\pi_k\end{bmatrix}= \begin{bmatrix} \delta_1pi_1\\\delta_2\pi_2\\\vdots\\\delta_k\pi_k\end{bmatrix}$

That's really all there is to it. You have the matrices expressed in compatible block structure- use it to multiply. The matrix is even block diagonal, so each "element" is only one term.[/quote]
Sorry, this is my first time using LaTeX so I did not express truly what I meant. Sorry, now I will provide the exact problem. Let P be a nxn Markov matrix which can be decomposed into k matrices to be arranged on the main diagonal $P= \begin{bmatrix}P_1&0&\cdots&0\\0&P_2&\cdots&0\\ \vdots&\vdots&\ddots&\vdots\\0&0&\cdots&P_k\end{bmatrix}$. Define the $P'$ matrix as follow: $P' = \alpha P+(1-\alpha)J/n$ where J is the matrix of all 1s. Matrix $P'$ has an eigenvector $ \pi =(\pi_{1},...,\pi_{n}) $ corresponds to the eigenvalue $\lambda=1$ and $\sum^n_{i=1} \pi_i=1$. Let $P_i$ be $n_i\mbox{x}n_i$ matrices where $\sum^k_{i=1} n_i=n$ and define $P_{i}'=  \alpha P_i+(1-\alpha)J/n$ with eigenvector $\pi^i$ repectively, $i=\overline{1,k}$. $\forall \delta = (\delta_1,...,\delta_k) $ where $\sum^k_1 \delta_i=1$, prove that $\pi=(\delta_{1}\pi_{1},...,\delta_{k} \pi_{k})$ is an eigenvector of matrix P. First please check the case where $\delta_{i}=n_{i}/n$.

[quote="Vagabon"]
Sorry, this is my first time using LaTeX so I did not express truly what I meant. Sorry, now I will provide the exact problem. Let P be a nxn Markov matrix which can be decomposed into k matrices to be arranged on the main diagonal $P= \begin{bmatrix}P_1&0&\cdots&0\\0&P_2&\cdots&0\\ \vdots&\vdots&\ddots&\vdots\\0&0&\cdots&P_k\end{bmatrix}$. Define the $P'$ matrix as follow: $P' = \alpha P+(1-\alpha)J/n$ where J is the matrix of all 1s. Matrix $P'$ has an eigenvector $ \pi =(\pi_{1},...,\pi_{n}) $ corresponds to the eigenvalue $\lambda=1$ and $\sum^n_{i=1} \pi_i=1$. Let $P_i$ be $n_i\mbox{x}n_i$ matrices where $\sum^k_{i=1} n_i=n$ and define $P_{i}'=  \alpha P_i+(1-\alpha)J/n$ with eigenvector $\pi^i$ repectively, $i=\overline{1,k}$. $\forall \delta = (\delta_1,...,\delta_k) $ where $\sum^k_1 \delta_i=1$, prove that $\pi=(\delta_{1}\pi_{1},...,\delta_{k} \pi_{k})$ is an eigenvector of matrix P. First please check the case where $\delta_{i}=n_{i}/n$.[/quote]
I am waiting for hints or complete answer for my problem from you all. Please.

Hic, couldn't someone help me? Pls just give me a hint, i couldn't have any ideas on it. Sorry if my English is so terrible.

I'm pretty sure you're wrong, but I'm still not sure what you're asking.

I'm working here with the matrices acting on column vectors, and having column sums of 1. Let $j$ be the vector with all elements equal to 1.
The normalized eigenvector associated to 1 for $P'$ is $\begin{bmatrix}u_1\\u_2\\\vdots\\u_k\end{bmatrix}$, where $u_i=\alpha P_iu_i+(1-\alpha)\frac jn$, giving $u_i=(I-\alpha P_i)^{-1}\cdot\frac jn$

Since $u_i$ is not the eigenvector for $P_i$ or $\alpha P_i+(1-\alpha)\frac J{n_i}$ and $\alpha P_i+(1-\alpha)\frac Jn$ is not a stochastic matrix, the problem is wrong.


What is the sum of the reciprocals of the roots of the equation
\[\frac{2003}{2004}x+1+\frac{1}{x}=0\]?

[quote="#H34N1"]What is the sum of the reciprocals of the roots of the equation
\[\frac{2003}{2004}x+1+\frac{1}{x}=0 \]
?[/quote]

[hide="solution"]Let the roots be $a$ and $b$.  The sum of the recipricals is $\frac{a+b}{ab}$.

$a+b=-\frac{2004}{2003}$
and
$ab=\frac{2004}{2003}$

by vetas theorem.

Dividing the 2 gives us $\boxed{-1}$[/hide]

[quote="#H34N1"]What is the vetas theorem? I don't get it[/quote]This has been discussed/mentioned/said/sliced/cooked/fried plenty of times. Try searching AoPS or googling [b]Vieta formulas[/b] or [b]Vieta equations[/b].

[quote="#H34N1"]Thanks. I googled Vetas theorem before I posted this and couldn't get anything.[/quote]That's because the spelling is wrong. No results for [b]VETAS[/b], try [b]VIETA[/b]

Wasn't this on an AMC 10?

[hide="solution"]We don't have to multiply by x.

Let the roots be a and b.

$\frac{1}{a}+\frac{1}{b}=\frac{a+b}{ab}$.

$a+b=-\frac{2004}{2003}$.

$ab=\frac{2004}{2003}$

$\frac{a+b}{ab}=-1$.[/hide]

It's Vieta's theorem.

Let $P(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots+a_{3}x^{3}+a_{2}x^{2}+a_{1}x+a_{0}$, with $a_{n}\neq 0$.

Let the roots be $r_{1}$ through $r_{n}$.

Vieta's formla states that

$a_{k}=(-1)^{n-k}a_{n}(r_{1}r_{2}\cdots r_{k}+r_{1}r_{2}\cdots r_{k-1}r_{k+1}+\cdots+r_{n-k+1}r_{n-k+2}\cdots r_{n})$.

Therefore,

$r_{1}r_{2}\cdots r_{k}+r_{1}r_{2}\cdots r_{k-1}r_{k+i}+\cdots+r_{n-k+1}r_{n-k+2}\cdots r_{n}=(-1)^{k}\frac{a_{n-k}}{a_{n}}$

Aww drat. 3 minutes late!

[hide]

$\frac{2003}{2004}x+1+\frac{1}{x}=0$
$\frac{2003}{2004}x^{2}+x+1=0$

It's a neat fact that if you reverse the coefficients of a polynomial, you'll get a polynomial whose roots are the recipricals of the original

$x^{2}+x+\frac{2003}{2004}$

And by Vieta's, the sum of the roots of this polynomial is $\boxed{-1}$

[/hide]


I have developed an interesting algorithm regarding integer factorization which may be of interest.

p and q are odd primes. pq = n.

Let $a_{0}= a \, small \, integer.$ (I usually start with $a_{0}= 4$)

$a_{1}= a_{0}^{a_{0}}(mod \, n)$

$a_{2}= a_{1}^{a_{1}}(mod \, n)$

$a_{3}= a_{2}^{a_{2}}(mod \, n)$
.
.
.
$a_{m+1}= a_{m}^{a_{m}}(mod \, n)$

Usually for some small i you will find,

$a_{i}\equiv 1 (mod \, p)$ or

$a_{i}\equiv 1 (mod \, q)$

So it is a matter of testing

$gcd(n, a_{i}-1)$ for each i.

Any comments on this?

Could you explain how this algorithm works, I don't quite get it :). You compute numbers $a_{i}$ and when you find one such that $gcd(n,a_{i}-1)=n$, you do what?

EDIT:

Oh, wait, no, you're looking for such $a_{i}$ that $1<gcd(n,a_{i}-1)<n$, right? I understand now :). Anyway, do you know how often this works and how fast you get the answer?

What is happening is that $a_{0}$ is being exponentiated very quickly. $a_{i}$ is simply $a_{0}^{M}$ for some very large $M$.

We are looking for $a_{0}^{k}\, \equiv \, 1 (mod \, p) \, or \, 1 (mod \, q).$

The period or cycle length of $k$ will be $ord_{p}(a_{0}) (or \, ord_{q}(a_{0})).$ Because $M$ itself exponentiating so quickly we can, for some fixed $m,$ say that $M = m^{s}$. We are now looking for

$m^{s}\, \equiv \, k \, (mod \, ord_{p}(a_{0}))$

Let $ord_{p}(a_{0}) = \beta$

Because $m$ is considered to be fixed the cycle length of $s$ will be $ord_\beta \, (m)$ So the system of congruences is reduced by two orders and this helps to find $s$ 'quickly'.

Since $\phi(n)$ can have very large prime divisors these orders can be high so the algorithm can be very hit and miss for such integers but, with different starting valuse for $a_{0}$ we can be very lucky.

Also, we have two chances of finding $s$ because we can also let Let $ord_{q}(a_{0}) = \beta$ and the cycle length of $s$ modulo $ord_\beta \, (m)$ will also apply.

12 digit integers can be factored in this way in say 14, 20 or 50 iterations. I don't have any exact figures at present but I will post some later.

The algorithm is very easy to write provided you have a good version of the Fast Exponentiation Algorithm. I would be interested in any results for 20+ digit integers as I am using Java (long) at present and am limited in the size of integers I can test.

Yes I'm still in the early stages with this and am still interested in other people's results. Very pressed for time at the moment...

[quote="Temperal"]You say "usually". What are the exceptions to this algorithm?[/quote]

The exceptions are when $\phi(n)$ has large odd divisors but even then such numbers can factor easily. The easiest to factor are ones which have many divisors for $\phi(n)$. The more it has the smaller they are likely to be.


geometry
geometric transformation
reflection
circumcircle
complex numbers
geometry unsolved

A convex hexagon $ABCDEF$ with all its sides equal satisfies: $\angle{A}+\angle{C}+\angle{E}=\angle{B}+\angle{D}+\angle{F}=2\pi.$Prove that lines $AD,BE,CF$ are concurrent.

[hide="PS"]A proof with complex numbers would be appreciated[/hide]

It is not difficult to show that B, D, and F are reflections of the circumcenter of ACE about AC, CE, and EA, respectively. It can then be seen that AD, BE, and CF concur at the circumcenter of the medial triangle of ACE.


I could not determine where to post this, so I would ask the mods/experienced members where to post this inequality..... I invented this when I read in 7th standard. :|
Here goes, for $ a, b, c\in \mathbb{R}$ such that $ a \plus{} b \plus{} c\ge1$, prove that:
$ 4(a \plus{} b \plus{} c) \minus{} \frac {a^2}{b \plus{} c} \minus{} \frac {b^2}{c \plus{} a} \minus{} \frac {c^2}{a \plus{} b}\ge\frac {9}{2}$
 :D
Hope you will enjoy it! :roll:
[edited]

[quote="Potla"]Solution? Anyone? :roll:[/quote]

First, don't be so impatient.  It's unreasonable to expect anyone to respond to your post so quickly.

[quote="Potla"]for $ a, b, c\in \mathbb{R}$ such that $ a \plus{} b \plus{} c\ge1$, prove that:
$ 4(a \plus{} b \plus{} c) \minus{} \frac {a^2}{b \plus{} c} \minus{} \frac {b^2}{c \plus{} a} \minus{} \frac {c^2}{a \plus{} b}\ge\frac {9}{2}$
[/quote]

Second, your inequality is quite flawed.  It doesn't even hold for $ a \equal{} b \equal{} c \equal{} 1/3$.

Extremely sorry to all mathlinkers.
As I was reviewing my posts I suddenly came across this inequality. In fact, it is quite flawed and the correct version would be:
For all $ a,b,c\in\mathbb R^\plus{}$ such that $ a\plus{}b\plus{}c\geq 1$;
Show that we have
\[ 4(a\plus{}b\plus{}c)\plus{}\frac{a^2}{b\plus{}c}\plus{}\frac{b^2}{c\plus{}a}\plus{}\frac{c^2}{a\plus{}b}\geq \frac 92\]

It is extremely easy as a problem in the Inequalities Proposed and own problems, and it can be moved to the pre-olympiad section.
I had the following idea in my mind when I posted this inequality such a long time ago; and here was my solution for it:
[hide="My previous solution"]
\[ \sum_{cyc}\left[ 2a\plus{}b\plus{}c\plus{}\frac{a^2}{b\plus{}c}\right] \equal{} \sum_{cyc}\frac{a^2\plus{}b^2\plus{}c^2\plus{}2ab\plus{}2bc\plus{}2ca}{b\plus{}c}\]
\[ \equal{}(a\plus{}b\plus{}c)^2\cdot \sum_{cyc}\frac 1{a\plus{}b}\geq (a\plus{}b\plus{}c)^2\cdot \frac 9 {2(a\plus{}b\plus{}c)}\]
\[ \equal{}\frac 92 (a\plus{}b\plus{}c)\geq \frac 92\]
[/hide]
In fact this is quite weak. We can use direct Cauchy-Schwartz inequality too in order to get
\[ \sum_{cyc}\frac{a^2}{b\plus{}c}\geq \frac{(a\plus{}b\plus{}c)^2}{2(a\plus{}b\plus{}c)}\]
\[ \therefore 4(a\plus{}b\plus{}c)\plus{}\sum_{cyc}\frac{a^2}{b\plus{}c}\geq \frac 92 (a\plus{}b\plus{}c)\geq \frac 92\]
I am really sorry that I had completely forgotten about this thread. Also I apologise to all the members for the inconvenience. I hope people don't mind it too much since I was a newbie to inequalities.

Sorry for the inconvenience.
Potla

I reckon that your inequality can be proved by the following estimation

\[ \frac{a^2}{b\plus{}c} \plus{} \frac{b\plus{}c}{4} \ge a\]


Given 3 circles on a plane. Consider a hexagon $ ABCDEF$ whose vertices are Intersection point of common internal tangents of circles. Prove that $ AD, BE, CF$ are concurrent

it's a hard problem,as far as I know there is no "known" solution yet...
it was given at Iran TST this year I think,and as far as I know no one solved it,anyway there is already two topics discussing it,one in this forum (which I can't find the link),and the other one in persian [url]http://www.mathlinks.ro/viewtopic.php?t=141482[/url] in the last topic,some one suggested the following link to study in order to solve this one (but I don't recommend that):
[url]http://mathworld.wolfram.com/FourConicsTheorem.html[/url]

[quote="BaBaK Ghalebi"]it's a hard problem,as far as I know there is no "known" solution yet...
it was given at Iran TST this year I think,and as far as I know no one solved it,anyway there is already two topics discussing it,one in this forum (which I can't find the link),...[/quote]
It is a challenge problem !!!  http://www.mathlinks.ro/Forum/viewtopic.php?t=140096

Kostas Vittas.


Could someone please post the answers to 10A?

I think these are the answers, but they may be wrong.  This is just what my dad said.

1.D
2.C
3.B
4.B
5.A
6.B
7.B
8.C
9.B
10.A
11.B
12.B
13.E
14.E
15.E
16.D
17.D
18.A
19.D
20.A
21.B
22.D
23.C
24.B
25.D

And again, these may be wrong.

[url=http://www.unl.edu/amc/e-exams/e5-amc10/e5-1-10archive/2005-10a/05AMC10AB-answers.html]Here's[/url] a link for those of you who can't find the AMC answers - they're in the archive, but not on the front page as of 3:45 on Thursday.

Additionally, I've typed them here:
1D
2C
3B
4B
5A
6B
7B
8C
9B
10A
11B
12B
13E
14E
15E
16D
17D
18A (not B - they aren't weighted the same! not D - conditional probability!)
19D
20A
21B
22D
23C
24B
25D


integration
function
vector
limit
geometry
rectangle
Functional Analysis

I was trying to proof the Picard Theorem using the Banach theorem of fixed point....
But it is not simple  :)
So, what do you think about this? Or maybe you've already did that?
Please write......
I really need this....

there is few simple elementary mistakes but I think you can understand.
proof is given by method of Landau

Actually I need a proof about Picard's 	existence theorem .
The one about existence and uniqueness of  the solutions of differential equations.
But only using the Banach Theorem of fixed point.

We have $ f(x,y)$ continuous in a neighborhood of $ (x_0,y_0)$ and satisfies a Lipschitz condition in $ y$ on that neighborhood.

Define $ T(\phi)(x)\equal{}y_0\plus{}\int_{x_0}^xf(t,\phi(t))\,dt.$

For $ \delta,\eta>0$ to be chosen later, let $ X$ be the set of continuous functions from $ [x_0\minus{}\delta,x_0\plus{}\delta]$ to $ [y_0\minus{}\eta,y_0\plus{}\eta].$ For $ \phi,\psi\in X,$ define $ d(\phi,\psi)\equal{}\sup_{|x\minus{}x_0|\le\delta}|\phi(x)\minus{}\psi(x)|\equal{}\|\phi\minus{}\psi\|.$ We claim that under this metric, $ X$ is a complete metric space. (It's not itself a Banach space, but it is a closed subset of a Banach space.)

Here's what we need: there exist $ \delta,\eta>0$ such that $ T$ maps $ X$ into $ X$ and $ \|T(\phi)\minus{}T(\psi)\|\le k \|T(\phi)\minus{}T(\psi)\|$ for some $ k<1.$ Once we have that, we can invoke the Banach fixed point theorem (also known as the contraction mapping principle).

I won't do all of it - there are some grubby details - but here's a start:

$ |T(\phi)(x)\minus{}T(\psi)(x)|\equal{}\left|\int_{x_0}^xf(t,\phi(t))\minus{}f(t,\psi(t)\,dt\right|$

$ \le\left|\int_{x_0}^x|f(t,\phi(t)\minus{}f(t,\psi(t)|\,dt\right|$

$ \le\left|\int_{x_0}^xM|\phi(t)\minus{}\psi(t)|\,dt\right|$ by the Lipschitz condition on $ f$

$ \le M|x\minus{}x_0|\|\phi\minus{}\psi\|$

We're not done - but you should be able to see where to go from here.

Oh - and for the full practical utility of the theorem, we should re-do this for (finite-dimensional) vector-valued $ y,\phi,\psi.$

thank u...
I'm going to finish the proof...

hi,
understanding your idea I succeeded in proving the theorem, but I can't fully explain why the metric space X is complete metric space.
this is my task, so  I want everithing to be good explained.
Could you help me?

Let $ C[a,b]$ be the set of all continuous functions on $ [a,b]$ with the supremum distance - that is, $ d(f,g)\equal{}\|f\minus{}g\|\equal{}\sup_{x\in[a,b]}|f(x)\minus{}g(x)|.$ We need to show that $ C[a,b]$ is a complete metric space. Once we've shown that, we then need to show that $ X$ as given above is a closed subset of a complete metric space and thus complete itself.

As for proving that $ C[a,b]$ is a complete metric space: I'll leave to you showing that the proposed metric is, in fact, a metric (including showing the triangle inequality). The proof of completeness should be covered in considerable detail in many different analysis textbooks. Here's an outline:

We want to show that if $ (f_n)$ is a Cauchy sequence, then there exists $ f$ in the space such that $ f_n\to f.$ So suppose $ (f_n)$ is a Cauchy sequence in this metric.

1. Show that for each fixed $ x\in[a,b],\ (f_n(x))$ is a Cauchy sequence of real numbers. Hence $ \lim_{n\to\infty}f_n(x)$ exists for each $ x.$ Define $ f(x)$ to equal this limit.

2. Show that $ f_n\to f$ uniformly on $ [a,b].$

3. The uniform limit of continuous functions is continuous. This classic $ \epsilon/3$ argument should be in any analysis textbook. Hence $ f\in C[a,b].$

I tried to use some theorems ( already proved) and this is my idea:
The rectangle D is closed and bounded, and that's why it is compact.
Now, the set X ( of continuous functions ) is closed subset of D.
Every closed subset of compact space is complite.
This is my explanation.
I would like to hear your opinion?
Thank you...

Now we're getting seriously off track.

The set $ X$ is [b]not[/b] compact.

The statement that closed bounded sets are compact is a theorem about $ \mathbb{R}^n$ that is completely false in infinite dimensional normed spaces.

If by $ D$ you mean the set $ [x_0\minus{}\delta,x_0\plus{}\delta]\times [y_0\minus{}\eta,y_0\plus{}\eta],$ then $ X$ is [b]not[/b] a subset of $ D.$

The word "compact" should not appear in the final step of proof of this theorem. We don't need it. The Banach fixed point theorem requires a [b]complete[/b] space; it doesn't care about compactness.

You are absolutely right. I confused everything....
How could I???
I'm ashamed ...  :blush:


function
integration
calculus
real analysis
real analysis unsolved

Prove that every Riemann-integrable function can be $\epsilon$-approximated by a continuous function, i.e. there exists a continuous function $f_{\epsilon}$ such that $\int_{a}^{b}{|f-f_{\epsilon}|}< \epsilon$.

What I would do is first show approximation by step functions (that's directly part of the definition of the integral, through upper and lower sums). Once I have step functions, I'd "bevel the corners" to get continuous approximations.


calculus
integration
quadratics
number theory
Diophantine equation
Ring Theory
number theory theorems

Hi,

Can someone help me to understand some technics which involve "gaussian intenger", some things like Q[V2] etc ... in order to solve some number theory problems like diophantine equations etc ...

Thanks !

The best way is probably to post your questions on the forum.

In fact, I have no knoweldges in algebraic number theory, so I stir up your help in order to have some documents to learn these technics which invole algebra in arithmetic.

Thanks.

For exemple, how can I solve $ 8y^2 \minus{} k^2 \equal{} 7$ with these technics ? (y and k intengers)

Your problem is a Pell type equation. (This is very well known)

Yes, but Pell-equation is $ x^2 \minus{} dy^2 \equal{} 1$, and here I've "7"

Just solve $ x^2 \minus{} 8y^2 \equal{} 1$ as a Pell equation, then find one solution to $ x^2\minus{}8y^2 \equal{} 7$. And then you obtain all solutions multiplying the solutions of the first equation. There's a pretty nice paper about these in the imo compendium site.

This method is in general not correct (you e.g. miss the conjugated solutions, but even more if $ b$ is not prime). It works here, but it doesn't work in general. Try it on $ x^2 \minus{} 15y^2 \equal{} 49$.
In fact, this can get rather complicated.

The principle is (only a sketch, and this requires some knowledge about number rings):
If you want to solve $ x^2 \minus{} ay^2 \equal{} b$, then you are looking for elements $ x \plus{} y \sqrt a$ of $ \mathbb Q [\sqrt {a}]$ having norm $ b$. Additionally, you want $ x,y$ to be integers (so far we got all rational $ x,y$ which is kind of different), thus we now consider it a bit different (and in a more general manner):
Given some number ring $ R$ and some integer $ b$, we want to find all $ z \in R$ such that $ N(z) \equal{} b$.

Note that this is not exactly the same (as $ \mathbb Z [\sqrt a]$ and it's integral closure need not to be the same). Nonetheless, after having solved the mentioned general problem, this is not much to do. So we tackle the general one now:

Let's rephrase it again: we are looking for all principal ideals $ \mathfrak z \equal{} (z)$ such that $ N(\mathfrak z) \equal{} b$ (here the norm is the ideal norm, i.e. the number of elements of $ R / \mathfrak z$). But instead of only doing this for the principal ideals, we can even solve it for all ideals $ \mathfrak z$. Reason: all such possible $ \mathfrak z$ are divisors of the ideal $ (b)$, and there explicit ways to describe all ideals dividing some given ideal (e.g. by reducing it to $ (b)$ being a prime ideal in $ \mathbb Z$). For the methods, see J. Neukirch, Algebraic Number Theory, chapter 1.

Now in case of $ b$ being prime and $ R$ being a quadratic extension, one gets that there are only two (conjugated) ideals having norm $ b$ (or there are none). If they are principal, the equation has a solution (otherwise not), and that solution is unique modulo units and conjugation.

PS: I hope I haven't scared you away  :D


inequalities
calculus
calculus computations

Hi guys. I'm new to the field of analysis, so I am quite confused with problems that may seem simple to you veterans out there. Can anyone help me with this problem, or at least give me a hint?

Prove that if $a \in R, a > 1$, then the set $\{a, a^{2}, a^{3}, \dots\}$ is not bounded from above.

The book mentions a hint:

First find a positive integer $n$ such that $a > 1+\frac{1}{n}$ and prove that $a^{n}> (1+\frac{1}{n})^{n}\ge 2$.

My confusion is that I don't know what to do with this hint. I am just wondering how to start this proof.

I've tried a contradiction proof. Assume the set $\{a, a^{2}, a^{3}, \dots\}$ is bounded from above. This means that there is a least upper bound $y$ such that $a^{n}\le y, n \in N$.

Then I have something else:

If $a > 1$, then $a-1 > 0$. Then obviously $|a-1| > 0$. Call $|a-1| = \epsilon$. This implies $\epsilon > 0$. There is a property of the least upper bound that says for any positive real number $\epsilon$, there exists an integer $n$ such that $\frac{1}{n}< \epsilon$. Thus, $|a-1| > \frac{1}{n}$. This implies $a-1 > \frac{1}{n}$, so $a > 1+\frac{1}{n}$. This solves the first part of the hint.

But then I don't know how to tie up the first part of the hint and the 2nd part of the hint :rotfl: .

Help is appreciated. Thanks so much.

Use Bernoulli's inequality - if $a>0$, then $(1+a)^{n}\geq 1+na$.  It should be easy to prove that $(1+na)_{n}$ is unbounded.

you can also go on with his method:
the fact of the existence of a supremum is that it's actually the least possible constant.. so for any $\epsilon$ there exist $n$ such that $y-\epsilon < a^{n}\le y$. but if you choose $\epsilon$ such that $y-\epsilon > \frac{y}{a}$, you get your contradiction...

Thanks for your two replies guys, but blahblahblah, I don't really know Bernoulli's inequality, so I would like a more common approach to this problem, using the hint the text gave, so on the professor's tests I can get credit. ma_go, can you explain your idea a bit more? Thanx

If $n$ is such that $a\ge 1+\frac{1}{n}$, then \[a^{n}\ge \left(1+\frac{1}{n}\right)^{n}=1+n\frac{1}{n}+\dots \ge 2,\] where the second step is the binomial formula (you only need two terms of it). Therefore, $2^{mn}\ge 2^{m}$ for any positive integer $m$. But you still have to convince yourself (and more importantly, your professor) that the sequence $2^{m}$ is unbounded...

Your idea of arguing by contradiction is very reasonable. Let $y$ be as in your post. By definition of the least upper bound for any $\epsilon>0$ there exists $n$ such that $a^{n}>y-\epsilon$. Then $a^{2n}>(y-\epsilon)^{2}$, but at the same time $a^{2n}\le y$. Hence $(y-\epsilon)^{2}\le y$ for any $\epsilon>0$. Let $\epsilon\to 0$, etc.

Well, assume the sequence $\{1, 2, 2^{2}, \ldots , 2^{m-1}, 2^{m}, ...\}$ is bounded from above, and call this least upper bound $y$. Then $2^{m}\le y$ for some positive integer $m$. Now consider $2(2^{m})$. We know $2 > 1$, so $2(2^{m}) > 2^{m}$. However, this implies $2^{m+1}\le y$ or $2^{m}\le y/2$. But the assumption was that $y$ was the least upper bound. Hence, a contradiction and therefore the original sequence is unbounded.

Am I right?

Also, now we have come this far, how do I tie this back to the original proof that the sequence $\{a, a^{2}, a^{3}, \ldots, ...\}$ is unbounded? Could I merely say that since $a^{n}\ge 2$, and by a term by term comparison to $\{1, 2, 2^{2}, \ldots , 2^{m-1}, 2^{m}, ...\}$, $\{a, a^{2}, a^{3}, \ldots, ...\}$  is unbounded too?

Yes, you could... but I don't see what is gained by switching from $a$ to $2$. The argument you gave works with $a$ directly: suppose $y$ is the least upper bound; then $y/a$ is also an upper bound, but $y/a<y$, a contradiction. No reason to involve $2$ and the binomials.


Heather places 5 read balls and 5 green balls in any combination into two identical boxes. A box is randomly selected, then a ball is selected from that box. Heather wins if a green ball is drawn. One paticular distribution of the balls between will maximize Heather's chance of winning. What is the probability that Heather will win if she uses the distribution that maximizes her chances? Express your answer as a common fraction.

Sweet, I remember that last time I did this problem I couldn't figure out how to do it, but I think I just got it. :)

[hide="Answer"]Since the balls can be in any combination, we want one box with all greens, since that will automatically give a $\frac{1}{2}$ total chance of winning. However, we also want as many green balls as we can get in the other box, to maximize the chances there. So, we place one green ball into one box, and the rest into the other. The chance of winning from the mixed box is $\frac{2}{9}$, so the total chance of winning is $\boxed{\frac{13}{18}}$.[/hide]


algebra
polynomial
function
real analysis
real analysis unsolved

The Muntz-Szasz theorem says that the closed linear span of $\{1,x^{\lambda_{1}},x^{\lambda_{2}},\ldots\}$ for $0<\lambda_{1}<\lambda_{2},\ldots$ is all of $C([0,1])$ precisely when
\[\sum_{i\geq 1}\lambda_{i}^{-1}= \infty \]
Find an elementary argument that suffices to show this result for the case where $\lambda_{i}=2i$, that is, show that the closed linear span of
\[\{1,x^{2},x^{4},x^{6},\ldots\}\]
is all of $C([0,1])$

Let $f\in C[0,1].$ Now, define $g(x)=f(\sqrt{x}).$ We also have $g\in C[0,1].$

Invoke the standard Weierstrass Approximation Theorem. Given $\epsilon>0,$ there exists a polynomial $P$ such that $|g(x)-P(x)|<\epsilon\,\forall\,x\in[0,1].$
Then $|g(x^{2})-P(x^{2})|=|f(x)-P(x^{2})|<\epsilon$ uniformly, which is precisely what is being called for.

It also follows immediately by Stone-Weierstrass.

I was looking for something more along the lines of what Dr. Merryfield did, of course the relevant functions on $[0,1]$ form an algebra that separates points and so on (is the Stone-Weierstrass theorem any less elementary than the Weierstrass Approximation theorem? I tend to say 'yes', because of the constructive proof of the Weierstrass theorem)

Does elementary refer to the level of the tools involved or the difficulty of the proof? The proof I've seen of Stone-Weierstrass does not use any advanced tools, but is quite involved. Anyone could find a proof for Weierstrass classical theorem given a little bit of time, but Stone-Weierstrass theorem seems much more difficult to come up with.

can someone exhibit a function that is not of the closure of $Span\{1, x, x^{4}, x^{9}, \ldots, x^{k^{2}}, \ldots\}$ ?


ratio
geometry
3D geometry
pyramid
algebra
polynomial
trigonometry

Hi to all!
now that the second round is completed, we can openly talk about the solutions  :)  :)
Here's what I've got:
Problem 1: 9
Problem 2: 3
Problem 3: 120
Problem 4: 7840
Problem 5: $ 72\varphi^2$ (where $ \varphi$ is the golden ratio) (sry guys...misread my own solution! Former number was the volume of the pyramid not of the bipyramid!)

It would be nice to have a little discussion here about the solutions!
regards
aaton

Yes, I got the same solutions. Hopefully we're correct!

But for number 5, I got double your answer. Did you calculate the volume of only one of the pyramids? Or did I do something wrong?

How did everybody do number 1? I couldn't find an elegant way, so I brute-forced it by writing a program.

[quote="123456789"]Yes, I got the same solutions. Hopefully we're correct!

But for number 5, I got double your answer. Did you calculate the volume of only one of the pyramids? Or did I do something wrong?[/quote]

Yes, I also got the double of aaton's answer.

The rest of my answers do match.

I believe #5 is $ 72\phi^2$. Also yeah... #1 was bashable with casework...

For #1, I first showed that there exists a sequence of length 8 with no monochromatic arithmetic sequence (RRBBRRBB), and then I did two cases: either both the first and the ninth numbers are colored the same color, or they are colored with different colors. Either way, a monochromatic arithmetic sequence is forced to be there.

How did you do number 2? I bashed it with system of equations, but I feel like there must have been a much easier way. Was there? Like using Newton's sums?

[quote="not_trig"]I believe #5 is $ 72\phi^2$.[/quote]

I got $ 108\plus{}36\sqrt{5}$ which equals $ 72\phi^2$, but did you get your answer directly as $ 72\phi^2$ or did you convert it from $ 108\plus{}36\sqrt{5}$?  If so, could you give me a brief overview of your solution?

[quote="123456789"][quote="lingomaniac88"]I got $ 72\varphi^2$ for #5.  All other answers, I agree with.

And I did #2 with Newton's Sums.[/quote]

How exactly did you apply Newton's Sums? I got halfway through and got stuck, so I decided to use algebraic manipulation.[/quote]
Let x, y, and z be the roots to the cubic equation $ p^{3}\plus{}ap^{2}\plus{}bp\plus{}c$. Define $ n_{k}$ to be $ x^{k}\plus{}y^{k}\plus{}z^{k}$. We have, by newton's sums, that $ n_{0}\equal{}3$, $ n_{1}\equal{}a\equal{}0$, $ n_{2}\equal{}\minus{}2b$ (using a=0), and that for $ k \geq 0$, $ n_{k\plus{}3}\equal{}\minus{}an_{k\plus{}2}\minus{}bn_{k\plus{}1}\minus{}cn_{k}\equal{}\minus{}bn_{k\plus{}1}\minus{}c_{k}$. We have $ n_{3}\equal{}3\equal{}\minus{}3c \Rightarrow c\equal{}\minus{}1$ allowing us to refine our recurrence to $ n_{k\plus{}3}\equal{}\minus{}bn_{k\plus{}1}\plus{}n_{k}$. We're also given that $ n_{5}\equal{}0\equal{}\minus{}bn_{3}\plus{}n_{2}\equal{}\minus{}3b\minus{}2b \Rightarrow b\equal{}0$. So, our polynomial is $ x^{3}\equal{}1$, telling us that $ x\equal{}1$, $ y\equal{}\text{cis} 60$, and $ z\equal{} \text{cis} 120$ (we can easily check that these values work). So, we can evaluate $ n_{2007}$ as $ x^{2007}\plus{}y^{2007}\plus{}z^{2007}\equal{}(x^{3})^{669}\plus{}(y^{3})^{669}\plus{}(z^{3})^{669}\equal{}\boxed{3}$.

[quote="diophantient"][quote="123456789"][quote="lingomaniac88"]I got $ 72\varphi^2$ for #5.  All other answers, I agree with.

And I did #2 with Newton's Sums.[/quote]

How exactly did you apply Newton's Sums? I got halfway through and got stuck, so I decided to use algebraic manipulation.[/quote]
Let x, y, and z be the roots to the cubic equation $ p^{3} \plus{} ap^{2} \plus{} bp \plus{} c$. Define $ n_{k}$ to be $ x^{k} \plus{} y^{k} \plus{} z^{k}$. We have, by newton's sums, that $ n_{0} \equal{} 3$, $ n_{1} \equal{} a \equal{} 0$, $ n_{2} \equal{} \minus{} 2b$ (using a=0), and that for $ k \geq 0$, $ n_{k \plus{} 3} \equal{} \minus{} an_{k \plus{} 2} \minus{} bn_{k \plus{} 1} \minus{} cn_{k} \equal{} \minus{} bn_{k \plus{} 1} \minus{} c_{k}$. We have $ n_{3} \equal{} 3 \equal{} \minus{} 3c \Rightarrow c \equal{} \minus{} 1$ allowing us to refine our recurrence to $ n_{k \plus{} 3} \equal{} \minus{} bn_{k \plus{} 1} \plus{} n_{k}$. We're also given that $ n_{5} \equal{} 0 \equal{} \minus{} bn_{3} \plus{} n_{2} \equal{} \minus{} 3b \minus{} 2b \Rightarrow b \equal{} 0$. So, our polynomial is $ x^{3} \equal{} 1$, telling us that $ x \equal{} 1$, $ y \equal{} \text{cis} 60$, and $ z \equal{} \text{cis} 120$ (we can easily check that these values work). So, we can evaluate $ n_{2007}$ as $ x^{2007} \plus{} y^{2007} \plus{} z^{2007} \equal{} (x^{3})^{669} \plus{} (y^{3})^{669} \plus{} (z^{3})^{669} \equal{} \boxed{3}$.[/quote]

Thanks.

How many points do we lose for a computation error? I accidentally did $ x^2 \equal{} a^2(1\minus{}2\cos \theta)$, instead of $ x^2 \equal{} 2a^2(1\minus{}\cos \theta)$, for the Law of Cosines. I feel dumb. (Moral: Don't leave writing the proofs to the last night)

Dude I thought my solution to #2 was pretty cool, but it had some computations.

$ x^{3} \plus{} y^{3} \plus{} z^{3} \minus{} 3xyz \equal{} (x \plus{} y \plus{} z)(x^{2} \plus{} y^{2} \plus{} z^{2} \minus{} xy \minus{} yz \minus{} xz)$
But $ x \plus{} y \plus{} z \equal{} 0$, so LHS=0 and $ xyz \equal{} 1$. $ z\neq0$, so

$ x \plus{} y \equal{} \minus{} z$
$ xy \equal{} \dfrac{1}{z}$.

Thus $ x,y$ are the two roots of $ r^{2} \plus{} rz \plus{} \dfrac{1}{z}$.

Now, compute $ x$ and $ y$ in terms of $ z$, then plug into $ x^{5} \plus{} y^{5} \plus{} z^{5}$. A bunch of stuff cancels, and we are left with $ 5z^{2} \minus{} \dfrac{5}{z} \equal{} 0$, and $ z\neq0$, so $ z^{3} \equal{} 1$. Similarly, we obtain $ x^{3} \equal{} 1,y^{3} \equal{} 1$. We can easily find a solution $ (x,y,z)$, permutations of (1,cis120,cis240), so our answer is 3.

diophantient, the roots of x^3 = 1 aren't 1, cis60, cis120.  :wink:

I casework bashed #1, and did a huge bash for #4 (4 pages). Seeing that the answer is so nice, $ 5{8\choose2}{8\choose3}$, is there a solution using some clever bijection? #5 was VERY easily researchable, for example, once you figure out what the solid looks like, [url]http://www.kjmaclean.com/Geometry/Icosahedron.html[/url] basically solves the problem for you.

I thought #3 was the easiest problem.

Wasn't number one's answer online everywhere about van der Waerden's Theorem?

van der Waerden's theorem is what I used for number 1

"One form of van der Waerden's theorem states that for all positive integers $ k$ and $ r$, there exists a constant $ n(k,r)$ such that $ n_{0}\geq n(k,r)$ and $ \{1, 2, \ldots, n_{0}\}\subset C_{1}\cup C_{2}\cup\ldots \cup C_{r}$, then some set $ C_{i}$ contains an arithmetic progression of length $ k$. The least possible value of $ n(k,r)$ is known as a van der Waerden number."

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

Also here: http://en.wikipedia.org/wiki/Van_der_Waerden%27s_theorem

It can be easily researched that the van der Waerden number is 9, which then can be shown that it works

On #5, I got $ 36\sqrt{6\sqrt{5}\plus{}14}$ which is the same as $ 72\phi^{2}$. That was from calculation of the volume of 4 tetrahedrons by the Cayley-Menger Determinant. (giant 5-by-5 matrix determinant calculation)

Same answers for all except #4, which was probably my error, because I suck at counting. For #4, I got 4352.

[quote="CatalystOfNostalgia"][quote]@carl: is lemma 2 in your #3 necessary? you proved that the greatest in the fourth row is at least 120, so if the greatest is not in the fourth row, its gonna be larger than 120 anyway (oh god 4 page brute force for 4 lol)[/quote]

Well, I guess it's not that necessary, but it's better to have a 5 or 6 extra lines with some substance than to be missing a step, and I was afraid if I didn't include it (I put it in on the last day), it would be too much to assume.

Is your feminine name necessary?[/quote]

(1)There always exist a number greater than or equal to 120 in the the array
(2)There does exist an array with 120 as the greatest number
(1), (2) is logically sufficient to imply that 120 is the smallest possible value as the greatest entry in the triangular array.

@Xantos C. Guin: your generalized solution for no.4 is amazing! I was doing lots of dumb stuff for "rigor" and spent 3 pages, not to even count the pictures...(I sent them separately)

For number 4, I got $ T(m,n)\equal{}2\left[\dbinom{m}{2}\dbinom{n}{2}\plus{}\dbinom{m}{2}\dbinom{n}{3}\plus{}\dbinom{m}{3}\dbinom{n}{2}\right]$
where $ m$ and $ n$ are the number of points on each arc.

[quote="ehehheehee"]1. Isn't it sufficient to show a single counterexample for the case $ n\leq 8$?[/quote]
I would think so.  It is obvious that if it's true for $ k$, it's true for $ k\plus{}1$.  Take its counterexample and you can see that if it's not true for 8, it's not true for any number from 1 to 7.

My solution for 2 was incredibly inelegant. It used a massive three-part induction to get that $ x^{3n}\plus{}y^{3n}\plus{}z^{3n}\equal{}3$.

[quote="calc rulz"]Hmm, I seem to have gotten everything except #4. What is wrong with my solution?[/quote]

I believe what happened is that you counted all Case B triangles that have their one vertex on the $ a$ arc, but you forgot to count the ones on the $ b$ arc.

[quote]Hmm, I seem to have gotten everything except #4. What is wrong with my solution?[/quote]

I think that you forgot to include vertiexis on the bottom arc.  If you double the last two summations you simplfied you would have what I got

Here are my solutions. I think my solutions are generally good. I was pretty stupid on that triangular array problem...I was just like omg casework, and I saved it for the last night. As you can see, that should loose points.

I haven't read other solutions yet, but in general, you don't use theorems that trivialize problems (in the case of problem one).

EDIT: 1/(ln x) I am pretty sure your proof for problem one is going to get crushed...

Here are my solutions. #5 is rather long and probably full of errors.

By the way, did anybody else not receive their checkmark by Wednesday (e-mail)?

[quote="ehehheehee"]By the way, did anybody else not receive their checkmark by Wednesday (e-mail)?[/quote]

Yeeah...

when does the third round actually begin???

[quote="ehehheehee"]
By the way, did anybody else not receive their checkmark by Wednesday (e-mail)?[/quote]

When did you submit your solutions?

I sent mine the Saturday before, Nov. 17, and no checkmark yet.

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=175670[/url]

I have not recieved my checkmark, but according to the above, that's not a problem.

Waah, messed up on 5. :( It's going to be really hard for me to get gold now. But I think I understand why I messed up, which is more important than the score (at least to me).


calculus
integration
inequalities
trigonometry
derivative
calculus computations

Prove that

$ \int_{0}^{\sqrt{2\pi}} sint^2dt > 0$

First, make the substitution $ t \equal{} \sqrt {x}$, turning the integral into

$ \int_0^{2\pi}\frac1{2\sqrt {x}}\,\sin x\,dx.$

Now integrate by parts, and use our free choice of constant to say that the antiderivative of $ \sin x\,dx$ is $ 1 \minus{} \cos x:$

$ \equal{} \left.\frac {1 \minus{} \cos x}{2\sqrt {x}}\right|_0^{2\pi} \plus{} \int_0^{2\pi} \frac1{4x^{3/2}}\,(1 \minus{} \cos x)\,dx$

The boundary terms are zero at both ends. (We have to interpret the evaluation at $ x \equal{} 0$ as the appropriate limit.) And the remaining integral is clearly positive.

In fact, this proof shows that $ \int_0^a\sin t^2\,dt > 0$ for all $ a > 0.$

[color=green][Edited to fix a mistake. The derivative of $ \color{green}\frac1{2\sqrt{x}}$ is $ \color{green}\minus{}\frac1{4x^{3/2}},$ not $ \color{green}\minus{}\sqrt{x}.$ It makes sense now; it didn't make sense before.][/color]

Split the integral into $ \int_0^\sqrt{\pi}$ and $ \int_{\sqrt{\pi}}^{\sqrt{2\pi}}$. On the second, substitute $ u^2+t^2=2\pi$ giving
\[ u=\sqrt{2\pi -t^2},\ t=\sqrt{2\pi -u^2},\ dt=-du\cdot \frac{u}{\sqrt{2\pi -u^2}}\]
\[ \int_{\sqrt{\pi}}^{\sqrt{2\pi }}\sin t^2\ dt=-\int_{\sqrt{\pi}}^0 \frac{u\sin (2\pi -u^2)}{\sqrt{2\pi -u^2}}\ du\]
\[ =-\int_0^{\sqrt{\pi}} \frac{u\sin u^2}{\sqrt{2\pi -u^2}}\ du\]
So the original integral is equal to
\[ \int_0^{\sqrt{\pi}} \sin t^2 \left( 1-\frac{t}{\sqrt{2\pi -t^2}}\right)\ dt\]
which is now clearly positive.


If a bag contains 6 red, 4 green, and 3 blue marbles, express as a common fraction the probability that one marble chosen randomly will not be green.

There are $ 6 \plus{} 3 \equal{} 9$ marbles that are NOT green.  The total number of marbles is $ 6 \plus{} 4 \plus{} 3 \equal{} 13$.  Therefore, the probability that one marble chosen randomly will not be green is
\[ \boxed{\frac {9}{13}}
\]

[hide="alternate"]
You could also use complementary probability. We have \[ P(\text{marble is not green})\equal{}1\minus{}P(\text{marble is green}).\] The probability the marble is green is $ \frac{4}{13},$ so the probability the marble is [b]not[/b] green is \[ 1\minus{}\frac{4}{13}\equal{}\boxed{\frac{9}{13}}.\][/hide]


Let   $\Re$      denote the set of all real numbers.Find all functions $f: \Re\to \Re$ such that
$f(x^2+f(y))=y+f(x)^2$ for all $x,y \epsilon \Re$/

First note that $f$ is injective: if $f(a)=f(b)$, then for any $x$, $f(x^2+f(a))=f(x^2+f(b))$, so $f(x)^2+a=f(x)^2+b, a=b$. $f$ is also surjective, because if given some $a$ we want $b$ such that $f(b)=a$, we can take $y=a-f(x)^2$ and $b$ what we get in the parentheses in the LHS.

Plugging $x=0$ we get $f(f(y))=y+f(0)^2$. Since $f$ is bijective, there's $r$ such that $f(r)=0$. Now take the original equation, apply $f$ to both sides, and we get: $f(f(x)^2+y)=f(f(x^2+f(y)))=x^2+f(y)+f(0)^2$. Put $x=r$, then $f(y)=f(y)+f(0)^2+r^2$. Since $f(0)^2, r^2$ are squares, they are non-negative, so to have equality we need $f(0)=r=0$. Then $f(0)=0$.

From this, $f(f(y))=y$. Putting $y=0$ yields $f(x^2)=f(x)^2$. If $x=1$, we get $f(1)=f(1)^2$, so $f(1)=0, 1$, but since $f(0)=0$, we have $f(1)=1$.

Also $f(x^2+f(y))=f(x)^2+y=f(x^2)+f(f(y))$. Put $f(y)=z$ and we get $f(x^2+z)=f(x^2)+f(z)$. In other words, $f(a+b)=f(a)+f(b)$ whenever at least one of $a, b$ is non-negative. It is easy to show that from this (and $f(0)=0, f(1)=1$) it follows $f(x)=x$ for every non-negative rational $x$. Since $f(x^2)=f(x)^2 \geq 0$, $f$ is non-negative for non-negative $x$ so, being additive, it is also non-decreasing, which leads to $f(x)=x$ for every non-negative $x$ by bounding $x$ with rationals. Finally, putting $y=-f(x)^2$ yields $-a=f(-a)$ for non-negative $a$, that is, $f(x)=x$ for all $x$.

Thank you Severius! ;) you have nice solution.

we can solve
find all$f: R\rightarrow R$
$f(x^n+f(y))=y+f(x)^n$


Let $ A \in M_n{(C)}$ such that : $ A^{3}\equal{}0$ prove that : $ 2Rank(A^{2}) \leq Rank(A)$

Looking at the Jordan blocks it should be easy to conclude.

[quote="QuyBac"]Let $ A \in M_n{(C)}$ such that : $ A^{3} \equal{} 0$ prove that : $ 2Rank(A^{2}) \leq Rank(A)$[/quote]
You can read : http://www.mathlinks.ro/viewtopic.php?t=33860
  :bruce:  :bruce:  :bruce:


geometry
incenter
circumcircle
inradius
geometry unsolved

In $\triangle ABC$, let $I$ be the incenter, $O$ be the circumcenter, $H$ be the orthocenter, $R$ be the circumradius, $E$ be the midpoint of $OH$, $r$ be the inradius, and $s$ be the semiperimeter.

[b][color=blue](a)[/color][/b] Find the distance $IH$ in form of $R,r,s$
[b][color=blue](b)[/color][/b] Find the distance $IE$ in form of $R,r$


Find a positive integrer number $n$ such that, if yor put a number $2$ on the left and a number $1$ on the right, the new number is equal to $33n$.

Let $n$ have $k$ digits.
Then we want to have $2 \cdot 10^{k+1}+10n+1 = 33n$.
Or in other words: $2 \cdot 10^{k+1}+1 = 23n$.
So lets find $k$ with $2 \cdot 10^{k+1} \equiv -1 \mod 23$ or then $10^{k+1} \equiv 11 \mod 23$.
$k=2$ works, giving $n=87$.


64 unit cubes are placed together to create a large cube. How many cubes with integer lengths are in the 4*4*4 cube?

How do you solve this? Please explain simply because I have hardly any background in geometry.


function
complex numbers
imaginary numbers

Multiplication rule (and division) only works when everything is real.
Since you're taking the roots of negative numbers, the answer is -4.

$\sqrt{a}\cdot\sqrt{b}=\sqrt{ab}$ only applies to positive a and b. So -4 is the right answer.

[quote]What is the right answer?
x= (:sqrt:-8)(:sqrt:-2)
x= (:sqrt:8)(:sqrt:2)(:sqrt:-1)(:sqrt:-1)
x= (:sqrt:16), by multiplying
x= 4

[i]or [/i]

x= (:sqrt:-8)(:sqrt:-2)
x= (:sqrt:-1)(:sqrt:-1)(:sqrt:8)(:sqrt:2)
x= (i)(i)(:sqrt:16)
x= -1(4)
x= -4

Which is the right answer?[/quote]

Both are the answer. Square root is a two-valued mapping. Thus, the square root of 25 is both +5 and -5. Since we prefer that functions have one value, we define the square root function, $\sqrt{n}$, to be the principal square root. For the square root of a positive number, the principal square root is the positive square root. For the square root of a negative number, there is no good way to distinguish between $i$ and $-i$, where $i$ is the imaginary number, so we can't distinguish the two square roots to pick a principal square root. For most cases, we just say $\sqrt{-n}=(\sqrt{n})i$, where n is positive, but that is fudging the case for convenience. The square root function of a negative number is not rigorously defined, so it can lead to apparent contradictions like the above. If you want to treat square roots of negatives rigorously,  you have to treat it as a two-valued function, $\pm\sqrt{-n}$. In that case, $(\pm\sqrt{-8})(\pm\sqrt{-2})=\pm 4$.

Erin Schram

Looking at the orginal problem posted in the title:

$\sqrt{\sqrt{-8}\sqrt{-2}}=\sqrt{\pm4}=\pm 2 \text{ or } \pm 2i \text{ ?}$


Finf all funtions $ f: \mathbb{R}\to\mathbb{R}$ such that
\[f(x\plus{}y)\plus{}f(x)f(y)\equal{}f(xy)\plus{}f(x)\plus{}f(y),\quad \forall x,y\in\mathbb{R}.\]

By observation, f(x) = 0 is one such function.
Now,plug in x=0,y=0…this yields f(0)= 0 or 2.
Then plug in y=0 to give f(x)=2 (if f(0) = 2).
Hence another possible function is f(x)=2 (constant function)…..

Another possibility is f(x+y) = f(x) + f(y) and f(xy)=f(x)*f(y)
This gives (on plugging x=y=0 and later y=0,x=1,2,3…..) f(x) =x

This is all I’ve been able to do…though I feel (intuitively) that there are no more functions, I haven’t been able to prove/disprove that….:oops:
Shyam

[quote="phuong"]Finf all funtions $ f: \mathbb{R}\to\mathbb{R}$ such that for arbitrary real numbers $ x$ and $ y$: $ f(x \plus{} y) \plus{} f(x)f(y) \equal{} f(xy) \plus{} f(x) \plus{} f(y),\forall x,y\in\mathbb{R}$[/quote]

$ 1)$ If $ f(x)\equal{}c$ constant, we have $ c\plus{}c^2\equal{}3c$ and we find immediatly two solutions $ f(x)\equal{}0$ and $ f(x)\equal{}2$

$ 2)$ If $ f(x)$ is not constant :
Let $ P(x,y)$ the property $ f(x \plus{} y) \plus{} f(x)f(y) \equal{} f(xy) \plus{} f(x) \plus{} f(y)$
$ P(x,0)$ gives $ f(x) \plus{} f(x)f(0) \equal{} f(0) \plus{} f(x) \plus{} f(0)$ and so $ f(0)\equal{}0$ (since $ f(x)$ is not the constant $ 2$).
$ P(x,1)$ gives $ f(x\plus{}1)\equal{}(2\minus{}f(1))f(x)\plus{}f(1)$ and so
$ P(x\plus{}1,1)$ gives $ f(x\plus{}2)\equal{}(2\minus{}f(1))f(x\plus{}1)\plus{}f(1)$ $ \equal{}(2\minus{}f(1))^2f(x)\plus{}f(1)(3\minus{}f(1))$
Since $ f(0)\equal{}0$, we can also write this last equation $ f(x\plus{}2)\equal{}(2\minus{}f(1))^2f(x)\plus{}f(2)$, with $ f(2)\equal{}f(1)(3\minus{}f(1))$
But $ P(x,2)$ gives $ f(x\plus{}2)\equal{}f(2x)\plus{}f(x)(1\minus{}f(2))\plus{}f(2)$

And so $ (2\minus{}f(1))^2f(x)\plus{}f(2)\equal{}f(2x)\plus{}f(x)(1\minus{}f(2))\plus{}f(2)$
implies $ (2\minus{}f(1))^2f(x)\equal{}f(2x)\plus{}f(x)(1\minus{}3f(1)\plus{}f(1)^2)$
implies $ f(2x)\equal{}(3\minus{}f(1))f(x)$

So we have $ f(2x)\equal{}af(x)$ and $ f(4x)\equal{}a^2f(x)$ with $ a\equal{}3\minus{}f(1)$.

Then $ P(2x,2y)$ gives $ af(x \plus{} y) \plus{} a^2f(x)f(y) \equal{} a^2f(xy) \plus{} af(x) \plus{} af(y)$.
We also have $ P(x,y)$ : $ f(x \plus{} y) \plus{} f(x)f(y) \equal{} f(xy) \plus{} f(x) \plus{} f(y)$ which implies $ a^2f(x \plus{} y) \plus{} a^2f(x)f(y) \equal{} a^2f(xy) \plus{} a^2f(x) \plus{} a^2f(y)$

Subtracting these two equations, we have :
$ a(a\minus{}1)f(x\plus{}y)\equal{}a(a\minus{}1)(f(x)\plus{}f(y))$

But, we have :
$ a\neq 0$, else $ f(2x)\equal{}af(x)$ would imply $ f(x)\equal{}0$, constant.
$ a\neq 1$, else $ f(1)\equal{}3\minus{}a\equal{}2$ and, since $ f(x\plus{}1)\equal{}(2\minus{}f(1))f(x)\plus{}f(1)$, $ f(x\plus{}1)\equal{}2$ and $ f(x)\equal{}2$, constant.

So we have $ f(x\plus{}y)\equal{}f(x)\plus{}f(y)$ and, as a consequence $ f(xy)\equal{}f(x)f(y)$

And this well known system has a unique solution $ f(x)\equal{}x$

So we just have three solutions, as Shyamkumar thought :
$ f(x)\equal{}0$
$ f(x)\equal{}2$
$ f(x)\equal{}x$

[i]Pco's solution is correct and I have just found an other one :lol: [/i]

[quote="greentreeroad"][quote="pco"]

So we have $ f(x \plus{} y) \equal{} f(x) \plus{} f(y)$ and, as a consequence $ f(xy) \equal{} f(x)f(y)$

And this well known system has a unique solution $ f(x) \equal{} x$
[/quote]

how do you show this, thanks :roll:[/quote]

$ f(x\plus{}y\equal{}f(x)\plus{}f(y)$ is Cauchy's equation and gives $ f(x)\equal{}f(1)x$ $ \forall x\in\mathbb Q$
$ f(xy)\equal{}f(x)f(y)$ gives $ f(x^2)\equal{}f(x)^2$ and so $ f(x)\geq 0$ $ \forall x\geq 0$

So $ f(x\plus{}y)\geq f(x)$ $ \forall x,\forall y\geq 0$ and so $ f(x)$ is monotonous (non decreasing).

And  $ f(x)\equal{}f(1)x$ $ \forall x\in\mathbb Q$ PLUS  $ f(x)$  monotonous implies $ f(x)\equal{}f(1)x$ $ \forall x\in\mathbb R$

Now $ f(xy)\equal{}f(x)f(y)$ $ \implies$ $ f(1)^2\equal{}f(1)$ and so two solutions :
$ f(x)\equal{}x$
$ f(x)\equal{}0$

And, since in paragraph 2 we supposed that $ f(x)$ was not a constant function : $ f(x)\equal{}x$


group theory
abstract algebra
number theory
relatively prime
least common multiple
superior algebra
superior algebra unsolved

Let $ G$ be an abelian group and let $ H$ and $ K$ be finite cyclic cubgroups with $ |H|\equal{}r$ and $ |K|\equal{}s$.
Show that $ G$ contains a cyclic subgroup of order the least common multiple of $ r$ and $ s$.

I proven that if $ r$ and $ s$ are relatively prime, then $ G$ contains a cyclic subgroup of order $ rs$.
But I am stuck on this problem. Please help me.

[quote="Stiger"]Let $ G$ be an abelian group and let $ H$ and $ K$ be finite cyclic cubgroups with $ |H| \equal{} r$ and $ |K| \equal{} s$.
Show that $ G$ contains a cyclic subgroup of order the least common multiple of $ r$ and $ s$.

I proven that if $ r$ and $ s$ are relatively prime, then $ G$ contains a cyclic subgroup of order $ rs$.
But I am stuck on this problem. Please help me.[/quote]

$ (hk)^n \equal{} h^nk^n$ because the group is abelian.
$ (hk)^n \equal{} 0 \Longrightarrow lcm(r,s) \mid n$
$ (hk)^{lcm(r,s)} \equal{} 0$ and so $ o(hk)\equal{}lcm(r,s)$

If $ x$ has order $ r$ and $ y$ has order $ s$, with $ (r,s)\equal{}d$, can you find an element $ y'$ of order $ s/d$?  Then you can use your result.

I disagree with vict85's solution.  In fact, we can easily set $ y\equal{}x^{\minus{}1}$ to see that $ xy$ needn't have the order we want.

[quote="LydianRain"]If $ x$ has order $ r$ and $ y$ has order $ s$, with $ (r,s) \equal{} d$, can you find an element $ y'$ of order $ s/d$?  Then you can use your result.

I disagree with vict85's solution.  In fact, we can easily set $ y \equal{} x^{ \minus{} 1}$ to see that $ xy$ needn't have the order we want.[/quote]

If $ s\equal{}20$ and $ r\equal{}50$, then $ d\equal{}(20, 50)\equal{}10$ ,$ s/d\equal{}2$ and hence $ (s/d,r)\equal{}(2,50)\equal{}2\ne 1$.
So I can't use my result.
How can I use my result??

You're right--you need something slightly more tricky.  Try dividing up the primes according to which of $ s$ and $ r$ is divisible by a higher power.

In other words, for $ s\equal{}20$ and $ r\equal{}50$, you want to find elements of order $ 4$ and $ 25$, then use your result.

[quote="LydianRain"]If $ x$ has order $ r$ and $ y$ has order $ s$, with $ (r,s) \equal{} d$, can you find an element $ y'$ of order $ s/d$?  Then you can use your result.

I disagree with vict85's solution.  In fact, we can easily set $ y \equal{} x^{ \minus{} 1}$ to see that $ xy$ needn't have the order we want.[/quote]

I simply write an hint, not the solution (and I have mixed the multiplicative notation with the additive one  :blush:, in this days I'm very distracted).

I have only considered the case $ H\cap K \equal{} \{1\}$.

$ (hk)^{lcm(r,s)}$ is always $ 1$ and so $ o(hk)\mid lcm(r,s)$.

If $ (hk)^{o(hk)} \equal{} h^{o(hk)}k^{o(hk)} \equal{} 1$ then $ h^{o(hk)} \equal{} k^{o(hk)} \equal{} 1$ or $ k^{o(hk)} \equal{} (h^{o(hk)})^{\minus{}1}$.

If $ h^{o(hk)} \equal{} k^{o(hk)} \equal{} 1$ then $ lcm(r,s)\mid {o(hk)}$ and so $ o(hk)\equal{}lcm(r,s)$ and we have find the cyclic subgroup.

The cases $ H\equal{}K$, $ H\subset K$ or $ H\supset K$ are banal.

The last case is more difficult but you have already write how to prove it.


I want to do a bachelor's program in math. Which is better between ISI and CMI?

This is perhaps one of the most difficult question i hav ever encountered!!

both r just equals :D
what is better just depends on ur demands :)

According to me, CMI is better than ISI, infact the best for pure maths and CSC courses in India.
So go for CMI. I am in Chennai and very well know about CMI and have talked to may seniors as well.

i agree with aravind with my suggession(although i hardly know how exactly it goes there)

I think the Algebra faculty of CMI is "relatively" better than ISI's and most olympiad guys tend to get attracted to Algebra because of its resemblance to what they have been loving doing in high school i.e. Olympiad Mathematics (and also because it is a beautiful subject). So if you take feedback from most Olympiad guys, they will say CMI is better...
Most ISI Bangalore Algebra profs (with some exceptions) are not presently active researchers whereas CMI Algebra profs are...

Also CMI has the plus-point of awarding a Comp. Sci. degree...
So Computer Science faculty of CMI is definitely way better than ISI Bangalore's which depends on IISc for comp science faculty...

But recently a new algebra prof. Upendra Kulkarni (IMO - 1989 bronze) has joined ISI Bangalore and is a very motivated teacher. We hope to see more good Algebra teachers in ISI in the near future.

In other subjects (like Analysis, Analysis-related subjects, Probability Theory, Statistics,etc) I think ISI is better than CMI. The level of mathematical rigority in what is taught in these other subjects is better than CMI (from what I have heard from people in CMI as well as ISI).

Overall, I guess, ISI will give you a stronger mathematical background, but of course, that can be attributed to my personal prejudice.

[quote="Akashnil"][hide="My opinion"]Very simple, no ifs and buts: [b]ISI is better[/b]. And yes, don't ask me why etc. to complicate things. :P [/hide][/quote]
Anyway do you know anything about CMI or are you supporting ISI only becausew you are a bengali?(I Guess) :)

[quote="murgi"]Anyway do you know anything about CMI or are you supporting ISI only becausew you are a bengali?(I Guess) :)[/quote]
[quote="I"]don't ask me why etc. to complicate things. :P [/quote]
 :P

[hide][hide]OK. Seriously. You are kinda right.
It's not right though that I don't know anything at all about CMI. But certainly not adequate enough to give someone advice. And I'm a bit biased about ISI not only cuz I'm a Bengali but also for the ppl (student or not) I know there (ISI Kolkata).
I didn't want anyone to take that post seriously, ending with a  :P[/hide][/hide]

I didn't want anyone to take that post seriously, ending with a  :P[/hide][/hide][/quote]
Anyway I also didnot take it seeriously anyway. :[hide] :P  :harhar:  [hide][/hide][/hide]


Given $ a, b, c \geq\ 0.$ Prove that:
$ \sqrt[9]{\frac{a^3}{b\plus{}c}} \plus{} \sqrt[9]{\frac{b^3}{c\plus{}a}} \plus{} \sqrt[9]{\frac{c^3}{a\plus{}b}} \geq\ \frac{1}{\sqrt[9]{2}}(\sqrt[9]{ab} \plus{} \sqrt[9]{bc} \plus{} \sqrt[9]{ca})$
 :)

Nearly, a friend asked me the proof for it. And my answer is: you can see the proof for it in our new book (of can_hang and me) which named "inequalities and the nice solutions" (this book'll appears about September). Thanks you very much, my friend.
 :)