inequalities
calculus
derivative
inequalities proposed

[quote="quykhtn-qa1"]Given $ a,b,c \ge 0$ and $ a \plus{} b \plus{} c \equal{} 3$.Prove that:
$ \sqrt {a \plus{} 3} \plus{} \sqrt {b \plus{} 3} \plus{} \sqrt {c \plus{} 3} \ge \frac {\sqrt {3}}{2}\sqrt {ab \plus{} bc \plus{} ca \plus{} 45}$
I hope you will like it,my friend :)[/quote]
By Jensen we have:
We consider the $ f(x)\equal{}\sqrt{x\plus{}3}$
$ f''(x) \geq 0$
So we have that:
$ f(a)\plus{}f(b)\plus{}f(c)\geq 3f(\frac{a\plus{}b\plus{}c}{3})\equal{}3f(1)$
So we have that the $ LHS\geq 3 \sqrt{1\plus{}3}\equal{}6$
So we have to prove that:
$ 12 \geq \sqrt{3}\sqrt{ab\plus{}bc\plus{}ca\plus{}45}$
Which is:
$ 3 \geq ab\plus{}bc\plus{}ca \rightarrow 3^2\geq3(ab\plus{}bc\plus{}ca) \rightarrow (a\plus{}b\plus{}c)^2\geq 3(ab\plus{}bc\plus{}ca)$ which is true :)
Please check it...

[quote="Dimitris X"]
So we have that the $ LHS\geq 3 \sqrt {1 \plus{} 3} \equal{} 6$
[/quote]

it's not true ....... try $ a\equal{}1 ; b\equal{} 0,9 ;c\equal{}1,1$ :D

[quote="Dimitris X"][quote="quykhtn-qa1"]Given $ a,b,c \ge 0$ and $ a \plus{} b \plus{} c \equal{} 3$.Prove that:
$ \sqrt {a \plus{} 3} \plus{} \sqrt {b \plus{} 3} \plus{} \sqrt {c \plus{} 3} \ge \frac {\sqrt {3}}{2}\sqrt {ab \plus{} bc \plus{} ca \plus{} 45}$
I hope you will like it,my friend :)[/quote]
By Jensen we have:
We consider the $ f(x) \equal{} \sqrt {x \plus{} 3}$
$ f''(x) \geq 0$
So we have that:
$ f(a) \plus{} f(b) \plus{} f(c)\geq 3f(\frac {a \plus{} b \plus{} c}{3}) \equal{} 3f(1)$
So we have that the $ LHS\geq 3 \sqrt {1 \plus{} 3} \equal{} 6$
So we have to prove that:
$ 12 \geq \sqrt {3}\sqrt {ab \plus{} bc \plus{} ca \plus{} 45}$
Which is:
$ 3 \geq ab \plus{} bc \plus{} ca \rightarrow 3^2\geq3(ab \plus{} bc \plus{} ca) \rightarrow (a \plus{} b \plus{} c)^2\geq 3(ab \plus{} bc \plus{} ca)$ which is true :)
Please check it...[/quote]
Are you sure?By Cauchy-Schwarz inequality we have:
$ LHS \le \sqrt{3(a\plus{}3\plus{}b\plus{}3\plus{}c\plus{}3)} \equal{}6$ :wink:
Now,try it :oops:

[quote="Dimitris X"][quote="quykhtn-qa1"]Given $ a,b,c \ge 0$ and $ a \plus{} b \plus{} c \equal{} 3$.Prove that:
$ \sqrt {a \plus{} 3} \plus{} \sqrt {b \plus{} 3} \plus{} \sqrt {c \plus{} 3} \ge \frac {\sqrt {3}}{2}\sqrt {ab \plus{} bc \plus{} ca \plus{} 45}$
I hope you will like it,my friend :)[/quote]
By Jensen we have:
We consider the $ f(x) \equal{} \sqrt {x \plus{} 3}$
$ f''(x) \geq 0$
So we have that:
$ f(a) \plus{} f(b) \plus{} f(c)\geq 3f(\frac {a \plus{} b \plus{} c}{3}) \equal{} 3f(1)$
So we have that the $ LHS\geq 3 \sqrt {1 \plus{} 3} \equal{} 6$
So we have to prove that:
$ 12 \geq \sqrt {3}\sqrt {ab \plus{} bc \plus{} ca \plus{} 45}$
Which is:
$ 3 \geq ab \plus{} bc \plus{} ca \rightarrow 3^2\geq3(ab \plus{} bc \plus{} ca) \rightarrow (a \plus{} b \plus{} c)^2\geq 3(ab \plus{} bc \plus{} ca)$ which is true :)
Please check it...[/quote]

Are you sure that the second derivative is positive?

[quote="Dimitris X"][quote="Γιώργος"]
Are you sure that the second derivative is positive?[/quote]
Yes i had made a mistake :oops:
Sorry for misleading...[/quote]
I hope you will like it,thanks :)

[quote="quykhtn-qa1"]Given $ a,b,c \ge 0$ and $ a \plus{} b \plus{} c \equal{} 3$.Prove that:
$ \sqrt {a \plus{} 3} \plus{} \sqrt {b \plus{} 3} \plus{} \sqrt {c \plus{} 3} \ge \frac {\sqrt {3}}{2}\sqrt {ab \plus{} bc \plus{} ca \plus{} 45}$
[/quote]
There is a proof by AM-GM, which bases on $ \sum_{cyc}\sqrt{(a \plus{} 3)(b \plus{} 3)}\geq3\sqrt [3]{(a \plus{} 3)(b \plus{} 3)(c \plus{} 3)},$ but it's very ugly.

[quote="arqady"][quote="quykhtn-qa1"]Given $ a,b,c \ge 0$ and $ a \plus{} b \plus{} c \equal{} 3$.Prove that:
$ \sqrt {a \plus{} 3} \plus{} \sqrt {b \plus{} 3} \plus{} \sqrt {c \plus{} 3} \ge \frac {\sqrt {3}}{2}\sqrt {ab \plus{} bc \plus{} ca \plus{} 45}$
[/quote]
There is a proof by AM-GM, which bases on $ \sum_{cyc}\sqrt {(a \plus{} 3)(b \plus{} 3)}\geq3\sqrt [3]{(a \plus{} 3)(b \plus{} 3)(c \plus{} 3)},$ but it's very ugly.[/quote]
Thanks,arqady.My solution very nice :)

The best result is:
Given $ a, b, c \geq\ 0$ satisfy $ a\plus{}b\plus{}c\equal{}3$. Prove that;
$ \sqrt{a\plus{}1} \plus{} \sqrt{b\plus{}1} \plus{} \sqrt{c\plus{}1} \geq\ \sqrt{\frac{3}{2}} \sqrt{ab\plus{}bc\plus{}ca\plus{}9}$
Both of them can be solved easily by Am-Gm  :)

[quote="nguoivn"]The best result is:
Given $ a, b, c \geq\ 0$ satisfy $ a \plus{} b \plus{} c \equal{} 3$. Prove that;
$ \sqrt {a \plus{} 1} \plus{} \sqrt {b \plus{} 1} \plus{} \sqrt {c \plus{} 1} \geq\ \sqrt {\frac {3}{2}} \sqrt {ab \plus{} bc \plus{} ca \plus{} 9}$
Both of them can be solved easily by Am-Gm  :)[/quote]
Yes, my friend :lol:

Indeed, it's the best result with my way (that's interesting when with your way, it's easier :lol: )
Here is my proof for quykhtn-qa1's ineq:
[quote="arqady"]
nguoivn, why you think that your inequality is "The best result" here? Thank you![/quote]
Because $ 4.1\minus{}2^2\equal{}0$  :oops:

[quote="nguoivn"]Indeed, this is also quykhtn-qa1's nice hint. And that's interesting, this hint is same to my hint in my old article about Russia MO 2000 :oops:
Congratulation again, quykhtn-qa1  :)
[hide](Hoa' ra nick em la` quykhtn-qa1, moi. khi anh toan` nhin` nham` thanh` quykhtn-ga1. Sr em nhe', hom nay anh moi' de? y' ki~. Ki~ nang sang' ta'c cua? em tot' that. Co' gang' phat huy tiep nhe'  ^^![/hide][/quote]
Thanks,my brother :lol:
@ nguoivn: [hide] a, anh hoc. truong nao vay. a.?-ngay xua va- bay gio? :)  [/hide]

[quote="quykhtn-qa1"][quote="nguoivn"]Indeed, it's the best result with my way (that's interesting when with your way, it's easier :lol: )
Here is my proof for quykhtn-qa1's ineq:
[quote="arqady"]
nguoivn, why you think that your inequality is "The best result" here? Thank you![/quote]
Because $ 4.1 \minus{} 2^2 \equal{} 0$  :oops:[/quote]
Nice proof,nguoivn.Here 's my proof,my brother :)
Put $ a \plus{} 1 \equal{} 6x,b \plus{} 1 \equal{} 6y,c \plus{} 1 \equal{} 6z$,we have $ x \plus{} y \plus{} z \equal{} 1$ and this inequality become:
$ \sum \ \sqrt {x} \ge 3 \sqrt {\sum \ xy}$
Put $ x \equal{} \frac {1}{m^2}$,etc...,this inequality become:
$ mn \plus{} np \plus{} pm \ge 3 \sqrt {m^2 \plus{} n^2 \plus{} p^2}$
We can easyli prove that:$ m,n,p$ be the sidelengths of a triangle.By IRAN 96,we have:
$ \frac {1}{m^2} \plus{} \frac {1}{n^2} \plus{} \frac {1}{p^2} \ge \frac {9}{2(mn \plus{} np \plus{} pm) \minus{} m^2 \minus{} n^2 \minus{} p^2}$
Then, $ 2(mn \plus{} np \plus{} pm) \ge (m^2 \plus{} n^2 \plus{} p^2) \plus{} 9 \ge 6 \sqrt {m^2 \plus{} n^2 \plus{} p^2}$
Then,we have Q.E.D :)[/quote]
Amazing work...Congratulations quykhtn-qa1 :!:

[quote="Dimitris X"][/quote][quote="quykhtn-qa1"][quote="nguoivn"]Indeed, it's the best result with my way (that's interesting when with your way, it's easier :lol: )
Here is my proof for quykhtn-qa1's ineq:
[quote="arqady"]
nguoivn, why you think that your inequality is "The best result" here? Thank you![/quote]
Because $ 4.1 - 2^2 = 0$  :oops:[/quote]
Nice proof,nguoivn.Here 's my proof,my brother :)
Put $ a + 1 = 6x,b + 1 = 6y,c + 1 = 6z$,we have $ x + y + z = 1$ and this inequality become:
$ \sum \ \sqrt {x} \ge 3 \sqrt {\sum \ xy}$
Put $ x = \frac {1}{m^2}$,etc...,this inequality become:
$ mn + np + pm \ge 3 \sqrt {m^2 + n^2 + p^2}$
We can easyli prove that:$ m,n,p$ be the sidelengths of a triangle.By IRAN 96,we have:
$ \frac {1}{m^2} + \frac {1}{n^2} + \frac {1}{p^2} \ge \frac {9}{2(mn + np + pm) - m^2 - n^2 - p^2}$
Then, $ 2(mn + np + pm) \ge (m^2 + n^2 + p^2) + 9 \ge 6 \sqrt {m^2 + n^2 + p^2}$
Then,we have Q.E.D :)[/quote][quote="Dimitris X"]
Amazing work...Congratulations quykhtn-qa1 :!:[/quote]
Thank you very much,dimitris X :lol:

I think, nguoivn, your proof is wrong because $ S_b$ can be negative. :wink:
I have proved this inequality 5 minutes before by another way. Indeed it is true! :)

Please read it again, my friend  :) . We only need to prove $ Sx \plus{} Sy \geq\ 0$ because  $ Sx \leq\ Sy \leq\ Sz$
And notice that $ 1 \leq\ x, y, z \leq\ 2$. So, $ Sy \equal{} 2xz \minus{} y^2 \geq\ x^2 \minus{} y^2 \geq\ 0$  :P

OK, nguoivn! Your proof is true and very nice. In my proof I am using the Mixing Variables Method.

[quote="nguoivn"][quote="arqady"]It seems the following inequality is true.
Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ a \plus{} b \plus{} c \equal{} 3.$ Prove that:
\[ \sqrt {a \plus{} 1} \plus{} \sqrt {b \plus{} 1} \plus{} \sqrt {c \plus{} 1}\geq\sqrt {ab \plus{} ac \plus{} bc \plus{} 15}
\]
[/quote]
Yes, it's trues, arqady. quykhtn-qa1's way can't effect but my way still solved it easily  :lol:
See my proof here:
@quykhtn-qa1:[hide] Cap' ba anh hoc. Ams em a. Bay gio` anh dang hoc bach khoa nam thu' 2. ^^![/hide][/quote]
Nice proof,my brother :lol:

[quote="arqady"][quote="nguoivn"]The best result is:
Given $ a, b, c \geq\ 0$ satisfy $ a \plus{} b \plus{} c \equal{} 3$. Prove that;
$ \sqrt {a \plus{} 1} \plus{} \sqrt {b \plus{} 1} \plus{} \sqrt {c \plus{} 1} \geq\ \sqrt {\frac {3}{2}} \sqrt {ab \plus{} bc \plus{} ca \plus{} 9}$
[/quote]
After squaring your inequality is equivalent to
$ 4\sum_{cyc}\sqrt {(a \plus{} 1)(b \plus{} 1)}\geq\sum_{cyc}(3ab \plus{} 5)$
 and after using AM-GM we need to prove that
$ 4\sqrt [3]{(a \plus{} 1)(b \plus{} 1)(c \plus{} 1)}\geq ab \plus{} ac \plus{} bc \plus{} 5.$
Let $ a \plus{} b \plus{} c \equal{} 3u,$ $ ab \plus{} ac \plus{} bc \equal{} 3v^2,$ $ abc \equal{} w^3$ and $ u^2 \equal{} tv^2.$
Hence, we need to prove that $ 4u\sqrt [3]{w^3 \plus{} 3uv^2 \plus{} 4u^3}\geq5u^2 \plus{} 3v^2,$ which is equivalent to
$ 131u^6 \minus{} 33u^4v^2 \minus{} 135u^2v^4 \minus{} 27v^6 \plus{} 64u^3w^3\geq0.$
If $ 131u^6 \minus{} 33u^4v^2 \minus{} 135u^2v^4 \minus{} 27v^6\geq0$ then, our inequality is true.
Hence, it remains to prove him for $ 131t^3 \minus{} 33t^2 \minus{} 135t \minus{} 27\leq0,$ which gives $ 1\leq t\leq1.227...$
$ (a \minus{} b)^2(a \minus{} c)^2(b \minus{} c)^2\geq0$ gives $ w^3\geq3uv^2 \minus{} 2u^3 \minus{} 2\sqrt {(u^2 \minus{} v^2)^3}.$
Thus, it remains to prove that $ 3t^3 \plus{} 159t^2 \minus{} 135t \minus{} 27\geq128t\sqrt {t(t \minus{} 1)^3},$
which is obviously true for $ 1\leq t\leq1.228$ and very not strong.
nguoivn, why you think that your inequality is "The best result" here? Thank you![/quote]
sorry arqady, but here, I think after using AM-GM we have to prove that,
$ 4\sqrt [3]{[(a \plus{} 1)(b \plus{} 1)(c \plus{} 1)]^2}\geq ab \plus{} ac \plus{} bc \plus{} 5$ not $ 4\sqrt [3]{(a \plus{} 1)(b \plus{} 1)(c \plus{} 1)}\geq ab \plus{} ac \plus{} bc \plus{} 5$

[quote="peine"]
sorry arqady, but here, I think after using AM-GM we have to prove that,
$ 4\sqrt [3]{[(a \plus{} 1)(b \plus{} 1)(c \plus{} 1)]^2}\geq ab \plus{} ac \plus{} bc \plus{} 5$ not $ 4\sqrt [3]{(a \plus{} 1)(b \plus{} 1)(c \plus{} 1)}\geq ab \plus{} ac \plus{} bc \plus{} 5$[/quote]
If so we obtain $ 4\sqrt [6]{[(a \plus{} 1)(b \plus{} 1)(c \plus{} 1)]^2}\geq ab \plus{} ac \plus{} bc \plus{} 5,$ which is the same.:wink:

I think this is nice.

If $ a,b,c$ are nonnegative real numbers such that $ a\plus{}b\plus{}c\equal{}3$, then

$ \sqrt{6a\plus{}3}\plus{}\sqrt{6b\plus{}3}\plus{}\sqrt{6c\plus{}3} \ge \sqrt{8(ab\plus{}bc\plus{}ca)\plus{}57}$,

with equality for $ a\equal{}b\equal{}c$, and again for $ a\equal{}0$ and $ b\equal{}c$.

[quote="nguoivn"][quote="Vasc"]I think this is nice.

If $ a,b,c$ are nonnegative real numbers such that $ a \plus{} b \plus{} c \equal{} 3$, then

$ \sqrt {6a \plus{} 3} \plus{} \sqrt {6b \plus{} 3} \plus{} \sqrt {6c \plus{} 3} \ge \sqrt {8(ab \plus{} bc \plus{} ca) \plus{} 57}$,

with equality for $ a \equal{} b \equal{} c$, and again for $ a \equal{} 0$ and $ b \equal{} c$.[/quote]
Nice ineq, Vasc  :) . I haven't tried but I think my way still works here  :maybe:[/quote]
nguoivn, by your way we need to prove that $ \sum_{cyc}(x \minus{} y)^2(2xy \minus{} z^2)\geq0$
when $ x^2 \plus{} y^2 \plus{} z^2 \equal{} 9$ and $ \{x,y,z\}\subset[1,\sqrt7],$ which is not so obvious. :wink:
By the way, MV kills it easy enough.  :P


function
integration
calculus
calculus computations

Find function with the property

$ f(x)\equal{}x\minus{}\int^{x}_{0} \int^{t}_{0} f(u)dudt$


linear algebra
matrix
algebra
polynomial
linear algebra unsolved

[quote="bchui"]I am looking for the prove of the identity

$\det (A)={1\over 2}({\rm Tr}(A)^{2}-{\rm Tr}(A^{2}))$

where can I find the prove?[/quote]

This is true only for $2\times 2$ matrices and can be proven by direct computation:

$\frac{1}{2}\left(\left( \text{Tr}\left( \begin{array}[c]{cc}a & b\\ c & d \end{array}\right) \right)^{2}-\text{Tr}\left( \left( \begin{array}[c]{cc}a & b\\ c & d \end{array}\right)^{2}\right)\right)$
$=\frac{1}{2}\left( \left( a+d\right)^{2}-\left( \left( bc+a^{2}\right)+\left( bc+d^{2}\right) \right) \right) =\frac{1}{2}\left( 2ad-2bc\right)$
$=ad-bc=\det\left( \begin{array}[c]{cc}a & b\\ c & d \end{array}\right)$.

For an $n\times n$ matrix $A$, the determinant $\det A$ is the product of the eigenvalues (with multiplicities), while for any $k\in\mathbb{N}$, the number $\text{Tr }A^{k}$ is the sum of the $k$-th powers of the eigenvalues. Thus, we can compute $\det A$ from $\text{Tr }A^{1}$, $\text{Tr }A^{2}$, ..., $\text{Tr }A^{n}$ by the formulas used to compute the product of $n$ numbers from the sums of their $k$-th powers for $k\in\left\{1;\;2;\;...;\;n\right\}$ (there are such formulas, since the product of $n$ numbers is a symmetric polynomial of them, and every symmetric polynomial is an algebraic polynomial of the power sums). However, these formulas are pretty obscure in general (see [url=http://en.wikipedia.org/wiki/Newton%27s_identities]Wikipedia on Newton's identities[/url]).

  Darij


function
real analysis
real analysis unsolved

Ascertain the functions $f,g\ : \ R\rightarrow R$ for which $\left \{ \begin{array} {cc} f^2(x)+g^2(x)=1+x^2\\ \left[f'(x)\right]^2+\left[g'(x)\right]^2=x^2\ \end{array}\right.$.

$f(x)=\sqrt{1+x^2}\cos(x-\arctan x +C)$
$g(x)=\pm \sqrt{1+x^2}\sin(x-\arctan x +C)$

can you post your demonstration please...?


In an arithmetic sequence, the 7th term is 30, and the 11th term
is 60. What is the 21st term of this sequence?

4 term difference is 30, so 2 term difference is 15 => 10 term difference is 75 => 21st term is 60+75=[b]135[/b]


function
induction
analytic geometry
algebra unsolved
algebra

Let $\textit {P}$ be the plane and $\emph{L}$ be the set of all lines in $\textit {P}$. Does there exist an [b]injective[/b] function $f: \emph{L}\rightarrow \textit {P}$ such that
      $f(a)\in a,\ \forall a\in \emph{L}$?

Well, an hour after I posted this problem I managed to solve it. So I should move it to "Algebra Proposed & Own Problems", I guess... :).

 [hide="The answer is:"]Yes, there are such functions.[/hide]

By a simple transfinite induction construction you can even make such a function bijective, but this is not exactly an "olympiad-style" approach :).

[quote]a simple transfinite induction[/quote]
What's that?

I have a "digestible" function, but only injective.

What's Google? :)

[url]http://planetmath.org/encyclopedia/TransfiniteInduction.html[/url]

...google...google...it seems rather familiar... (thanks for the link  :) ).

[hide="My function"]If the equation of line $a$ is $y=mx+n$, consider the following cases:
1). if $m\cdot n>0$, then $f(a)$ is the point of coordinates $x=\frac 1{\sqrt(1+m^2)},\ y=\frac m{\sqrt(1+m^2)}+n$ (one of the intersection points of line $a$ with the circle $x^2+(y-n)^2=1$);
2). if $m\cdot n<0$, then   $f(a)$ is the point of coordinates $x=-\frac 1{\sqrt(1+m^2)},\ y=-\frac m{\sqrt(1+m^2)}+n$ (one of the intersection points of line $a$ with the circle $x^2+(y-n)^2=1$);
3). if $m=0,\ n\neq 0$, then $f(a)$ is the point of coordinates $x=0, \ y=n$;
4). if $m=n=0, \ f(a)$ is the point of coordinates $x=2, \ y=0$.

If the equation of line $a$ is $x=p$, consider the following cases:
1). if $p\neq 2$ then $f(a)$ is the point of coordinates $x=p, \ y=0$;
2). if $p=2$ then $f(a)$ is the point of coordinates $x=2, \ y=2$
[/hide]


http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2008-ua/2008usamoqual.shtml#ca . Wow congratulations.  :|

Brian and I got 126/7, 127.5/7...I feel like smashing things.

WE WIN!!!!!!!!

Not much of a victory, considering how incredibly, ridiculously unfair a 6 floor is to underclassmen who take the AMC-12...I'm seriously conflicted about whether I should still have any respect whatsoever for the CAMC.

Yeah, well, the floor was nonexistent this year. Literally.
Whoot only person from my school...actually, first person ever to qualify from my school.  :P

Congratulations people :D

edit: nice, david

it's kind of odd how i only recognize about 5 ca qualifiers as aopsers.

Whether Paly or Gunn won doesn't matter.  What makes me happy is that you both beat out TJ.

Of course, when I was in USAMO, TJ was the one who could always qualify an ARML team worth of students to USAMO.  Now I think Exeter has taken over that distinction from them.  Coincidentally, Exeter has also taken over the ARML championship from TJ.

YES I MADE IT
138 IS GOOD. So i doubled my index? lol

Wait I always though Exeter was a beast at AMC. Also what's 5% club? That school is also good I think

[quote="serialk11r"]Nice David :)
I think Andrew Liu actually had a chance too, except all of our AMC scores were too low. Gunn's AMC scores sucked *** this year...Jim had 138, the highest, which isn't even that incredibly high. More importantly, the next highest score was 130.5 on the 10, Jeff Han...
BTW Gunn actually has 7 qualifiers, since David Shi and Aakash Prasad took AMC10/12B somewhere else, and therefore took AIME there too. So Paly doesn't win. We're tied.
Saratoga has 7 qualifiers, plus 2 more Redwood people associated with them.[/quote]

We should have had 9 (plus 2 from Redwood), but for some reason two of our people either misbubbled or got their AIMEs misplaced -_- .

Aragon FTW, 2 USAMO qualifiers/ 3 AIME qualifiers!
lol, we finally got freshmen to qualify for USAMO and have trippled the number of students who have qualified in the history of our school YAY!  :rotfl:

yay Quatto made it
hmmm
the only issue is that I happen to be screwed for USAMO....
ughhh...

by the way braces still suck

Nice, bowei made it.

But I was all like HOLY EFF, LEWIS CHEN IS USAMO QUALIFIED? Hehe, look at what grade he's in. I know this dude (not personally, just seen him). Wow, that's impressive.

Who's ready for the USAMO proofs? I'm not. I still have to work on them.

[quote="no nastEy"]hello moogra[/quote]

AND Who the heck is this?
I don't believe I know a no nastEy.

Yeah, I'm going to get killed since I haven't really practiced my proofs at all, and I haven't really looked at any old olympiad problems. I hope they can read my handwriting, since it's really, really bad...I'm also horrible at explaining my solutions, since my mind works weirdly...

Try doing the old USAMTS problems. They're quite easy but they're good practice.

no nastEy seems to be aakash prasad, judging from the email.
WTF I should've chickened out and took the 10B...

Really? If that's true...you should be able to figure out who I am...

[quote="Ubemaya"]
We should have had 9 (plus 2 from Redwood), but for some reason two of our people either misbubbled or got their AIMEs misplaced -_- .[/quote]

I wonder who those two missing people are

gah
Ill kill someone if our proctor/ AMC people misplaced my test

[quote="miller4math"][quote="Carbon57"]Yay...made it finally... :)

Lynbrook had 7 qualifiers (including me) plus 2 from Miller, which is associated with it.[/quote]
what?? who said miller was associated with lynbrook?  :huh: (just cuz we're two blocks away...)
anyways our school (miller) did ok. (two qualifiers)
this guy with a 126/7 didn't make it...[/quote]
Associated in the released listing.

Homestead got 4 qualifiers this year, including yours truly. Which leaves me kind of pissed. We had two seniors who should have made it, and I_like_pie somehow didn't make it because he epic failed AIME (7)

[quote="miller4math"][quote="Carbon57"]Yay...made it finally... :)

Lynbrook had 7 qualifiers (including me) plus 2 from Miller, which is associated with it.[/quote]
what?? who said miller was associated with lynbrook?  :huh: (just cuz we're two blocks away...)
anyways our school (miller) did ok. (two qualifiers)
this guy with a 126/7 didn't make it...[/quote]

Well, nearly everyone student from Miller goes into Lynbrook, and Miller students take the USAMO at Lynbrook. That's better than we did last year, with only 1 qualifier...

I thought Lynbrook won last year as well?
Hmmm maybe I'm wrong.

so, who here received their USAMO qualifying email already?

i havent  :(  yet

lol, i haven't, and i kinda doubt i ever will. oh well, i know my usamo number already, and we're getting letters anyway, right?

I have :P Last year I had a large prime, this year I have a small abundant number. Interesting. Not really.

I have the sixth smallest prime to a prime.

I also got it.... multiple of 25. I wonder how do they assign the student #s?

It's probably random...I got mine...multiple of 157...you'll probably see a 0 or a 1 next to it  :( ...

I got mine too... 5 times a perfect square. Expect to see 0 or 1 for me too. But maybe I'll Ubey-fail and get over 9000. You never know :P

hmmm
so like why aren't we saying our actual numbers?
mines between 0 and 500 exclusive

[quote="The QuattoMaster 6000"][quote="CircleSquared"]My number is an integer a, where a=x and 0<x[/quote]
DUDE MINE IS TOO
WE HAVE THE SAME NUMBER[/quote]

Oh noes something are borked.

Update: Finally got my email past 9 AM today.

i got mine about 5 minutes ago
i wish theyd tell us they send the emails late gah
well anyway im happy.

[quote="miller4math"]i got my email!!!  :lol:
k well my number is going to have a [size=200]0[/size] next to it.. so you'll all know who it is.[/quote]

If you got a 10 on the AIME I doubt you'll get 0. :P I'll probably get 7 max.

Yeah, I got my letter, too. Forgot to send in form S-III by the deadline. Oops. I sent it in anyway.

I kinda want to know whether my teacher's sent in my T-II form yet. Because that updated pdf says he hasn't, but he swears he has.


Choose any $ n$ positive integers $ a_1$ through $ a_n$, not necessarily distinct, so that $ a_1 \plus{} a_2 \plus{} a_3 \plus{} ... \plus{} a_n \equal{} 2007$

Maximize $ P$, where $ P \equal{} \prod_{i \equal{} 1}^{n} a_i$.

You want [b]positive[/b] integers.

[hide="Hints"] $ 3 \cdot 3 > 2 \cdot 2 \cdot 2$
$ 5 < 2 \cdot 3$ [/hide]

[hide]For a constant value $ n\equal{}a_1\plus{}a_2\plus{}a_3\plus{}\ldots\plus{}a_n$, the maximum of value of $ a_1a_2a_3\cdots a_n$ is $ 2^{n\minus{}\lfloor n/3\rfloor}\cdot 3^{\lfloor n/3\rfloor}$.

Even more effective would be $ e$, but the problem asks for integers, and $ 3$ works best (this isn't inferred from rounding but rather from a proof).[/hide]

A notational suggestion to both the original poster and the poster of the previous post: don't make the same letter (in this case $ n$) mean two different things.  It can be very confusing.

A variation: Suppose that $ n$ (the number of variables) is fixed rather than variable.  What's the optimal choice of the values of the $ a_i$?

[quote="i_like_pie"][hide]For a constant value $ n \equal{} a_1 \plus{} a_2 \plus{} a_3 \plus{} \ldots \plus{} a_n$, the maximum of value of $ a_1a_2a_3\cdots a_n$ is $ 2^{n \minus{} \lfloor n/3\rfloor}\cdot 3^{\lfloor n/3\rfloor}$.
[/hide][/quote]

Wait, are you sure? I thought it was $ 3^k$ for $ n \equal{} 3k$, $ 2 \cdot 3^k$ for $ n \equal{} 3k \plus{} 2$, and $ 2^2 3^{k\minus{}1}$ for $ n \equal{} 3k \plus{} 1 (k \geq 1)$.

[quote="MellowMelon"][quote="i_like_pie"][hide]For a constant value $ n \equal{} a_1 \plus{} a_2 \plus{} a_3 \plus{} \ldots \plus{} a_n$, the maximum of value of $ a_1a_2a_3\cdots a_n$ is $ 2^{n \minus{} \lfloor n/3\rfloor}\cdot 3^{\lfloor n/3\rfloor}$.
[/hide][/quote]

Wait, are you sure? I thought it was $ 3^k$ for $ n \equal{} 3k$, $ 2 \cdot 3^k$ for $ n \equal{} 3k \plus{} 2$, and $ 2^2 3^{k \minus{} 1}$ for $ n \equal{} 3k \plus{} 1 (k \geq 1)$.[/quote]
You're right. My attempt at combining that into one expression failed. :ninja: Having a formula is actually unnecessary; the important part is the idea behind it: the use of as many 3's as possible, and then 2's if $ n\not\equiv0\bmod3$.

@ JBL - Thanks for the suggestion; I didn't realize I used $ n$ twice. :oops:


Um I think the post count thingy isn't working. Everytime I post, the total amount of posts I did doesn't change. :huh:

[quote="not_trig"]What? Mine does...[/quote]

Mine doesn't sometimes.

Oh, thanks SplashD. I never knew about the rules in G&FF.

Don't post for the post count. If you do that I might just set it up to be 0 always! :)


integration
trigonometry
calculus
complex analysis
real analysis
real analysis unsolved

Evaluate $ \int_0^{\infty} \frac{\cos x}{\cos h(\pi x)}e^{\minus{}x^2}dx.$

Are you sure that there is a closed expression for this integral?

It's a definite integral, so it can have a simple answer without the indefinite integral having a closed form for a solution.

My question is, can this problem be done without using tools from complex analysis?


[quote="PedSoar"]How do you find $a,b,c$ knowing that:

$a+b+c+9\equiv 0\ mod\ 2$[/quote]

[hide]There are an infinite number of solutions, as long as $a+b+c\equiv1\pmod{2}$. [/hide]

[quote="vishalarul"]$a+b+c+9\equiv0\pmod 2$
$a+b+c\equiv{-9}\equiv 1\pmod 2$[/quote]

So let's look at possible values of $a+b+c$:  It could be: 1, 3, 5,...etc.  Therefore, we can say that $a+b+c$ can be equal to any odd integer and satisfy the original congruence.


Hello everybody,
I'm a beginner in geometry, and i'd like to increase my knowledge in it.
If you know some nice tutorials or courses on it, could you possibly give me the links? I will be so grateful!

Have you tried the [url=http://www.artofproblemsolving.com/Classes/AoPS_C_Enroll.php]online school[/url] section of AoPS?  There is a course on geometry and a geometry textbook available there.

Other than that, however, I cannot think of any good introductory geometry books...

I've got a nice pdf on geometry basics... but it's in German :D
I also don't have the copyright of this pdf, so I don't really feel good uploading it. But if you want to have it give me a message  :maybe:

b-flat>>The courses cost a lot !  :o $345

What i would like is some FREE tutorials...i mean some PDF documents, even if they are not really courses.


function
analytic geometry
calculus
derivative
algebra
domain
absolute value

If F(x)=Abs(4x-5) what is the x-coordinate of the critical number.  I don't understand why I can't get the answer to this.  Is there a certain way to do the derivative?  F'(x) just equals 4 and the domain is (-infinity,infinity). Does this problem have a critical number? I've tried 5,0,4 and the negative values.  What am I doing wrong?

The absolute value function can be most practically understood as a piecewise defined function. That is,

$ |u|\equal{}\begin{cases}u,&\text{if }u\ge0\\ \minus{}u,&\text{if }u<0\end{cases}$

So for $ F(x)\equal{}|4x\minus{}5|,$ the cases divide at the point at which $ 4x\minus{}5\equal{}0,$ which is $ x\equal{}\frac54.$

Following the model above, $ |4x\minus{}5|\equal{}\begin{cases}4x\minus{}5,&x\ge\frac54\\ \minus{}4x\plus{}5,&x<\frac54\end{cases}.$

We have $ F'(x)\equal{}4$ for $ x>\frac54$ and $ F'(x)\equal{}\minus{}4$ for $ x<\frac54.$ So there are no critical numbers either for $ x>\frac54$ or for $ x<\frac54.$

At this point, I'm not even going to address the question of whether $ F'\left(\frac54\right)$ exists - in this case, it doesn't, but we can proceed without knowing that. I'm going to label $ x\equal{}\frac54$ with a label that I would use for any such boundary point for a piecewise defined function: I'll call it "a point suspected of possibly being a critical point."

This works as a practical designation for the following reason: it's already true that a critical point is a point that might or might not be a maximum or minimum - we still have to investigate each critical point to determine what it is. If we throw a point onto the list of suspects, then it will be investigated as if it were a critical point, and we'll still find out the information.

In this case, $ F\left(\frac54\right)\equal{}0$ and $ F(x)>0$ for all other $ x.$ Hence we have a minimum at $ x\equal{}\frac54.$ After the fact, that must have been a critical point after all.

[quote]I've tried 5,0,4 and the negative values. What am I doing wrong?[/quote]
What were you doing right? That was just fishing blindly. Have a valid reason for what you're doing, please.

[quote]That was just fishing blindly[/quote]

Undoubtedly,  I really didn't have any idea how to approach the problem because usually I find the derivative and then set the derivative equal to zero.  Because I couldn't do that in this case, I was at a loss.  I'm not really sure that was necessary for the explanation.  My valid reason for doing that is that people complain if you don't say what you tried, and sometimes you haven't tried anything because you don't know how to begin, so this was the best I could do.

Anyways, Thanks for the explanation.

My one sentence summary: any point at which a quantity inside an absolute value is zero is a "suspect point" that might be a critical point.

For instance, if $ f(x) \equal{} |x^2 \minus{} 2| \minus{} 2|x|,$ then $ \minus{} \sqrt {2},0,$ and $ \sqrt {2}$ are all "suspect points" that should be investigated. There might be other critical points as well - a graph might suggest where to look for those.

[quote="Kent Merryfield"]My one sentence summary: any point at which a quantity inside [b]an absolute value[/b] is zero is a "suspect point" that might be a critical point.[/quote]
[color=green][Ack! Proofreading! (Or the lack thereof). Thanks. - K.M.][/color]


Applying Euler's formula, you get $3^{\phi{1000}}\equiv 1 \bmod 1000$, so $3^{400}\equiv 1 \bmod{1000}$.

But it turns out $3^{100}\equiv 1 \bmod{1000}$. How would a person go about finding that? Applying Euler's formula on the prime factors of $1000$?

Yes.  This is known as the [url=http://mathworld.wolfram.com/CarmichaelFunction.html]Carmichael Function[/url] $\lambda(n)$, and it gives $\lambda(1000) = 100$.  It is essentially an application of the Chinese Remainder Theorem along with the fact that $\mathbb{Z}_{2^{n}}, n > 2$ does not have a primitive root.

This is the smallest exponent such that $a^{ \lambda(n) }\equiv 1 \bmod n$ for [b]all[/b] $gcd(a, n) = 1$.


[quote="chess64"]George Washington was born 11 years before Thomas Jefferson. In 1770 Washington's age was 3 more than 7 times the age of Jefferson in 1748. What was the sum of their ages in 1750?

This problem is so confusing... and wouldn't someone have a huge advantage if they knew there birthdates?[/quote]
Whats preventing setting  up a system of equations???

[hide="hint"]I set up two equations with variable w, and j, where w and j are Washington's and Jefferson's birthyears respectively.[/hide]

[hide="Hint"]
I also set up two equations but where w and j are Washington and Jefferson's age in 1750, it's easier that way.
[hide="Click if you want to see the two equations"]
$w-11=j$
$w+20-3=7(j-2)$
From here you can solve and get $w=18$ and $j=7$ so the sum of their ages is $\boxed{25}$[/hide][/hide]

[quote="chess64"]George Washington was born 11 years before Thomas Jefferson. In 1770 Washington's age was 3 more than 7 times the age of Jefferson in 1748. What was the sum of their ages in 1750?

This problem is so confusing... and wouldn't someone have a huge advantage if they knew there birthdates?[/quote]
[hide]
x=washingtons age in 1750
y=jeffersons age in 1750

x+20=7(y-2)+3
x-11=y

x+20=7(x-11-2)+3
x+20=7x-91+3
x+20=7x-88
108=6x
x=18,
y=7
18+7=[hide]25[/hide]
[/hide]

isn't this from AoPS Basics?( the book )

I wish moderators can search  :roll:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=43071


trigonometry
real analysis
real analysis theorems

When I was in high school and took calc my sr. year (well I took calc and trig so this could be a trig prob) there were a specific set of problems in which I had to solve an equation and it was always necessary to check the result b/c sometimes, even though the math was done right, the answer would be wrong. Does anyone remember what this type of problem was called?

Ummm.... well one could say that extraneous solutions were obtained... is that what you're thinking of?


[quote="Arvind_sn"]Billy Bob looked at the clock and saw that it was exactly six o' clock. He sat down again because he only needed to leave when both hands of the clock are right angles to each other. Exactly how much time will he need to wait?[/quote]

[hide]He will need to wait $\frac{180}{11}$ minutes.[/hide]

[hide]They are 180degrees to start with. For every minute, the hour hand moves 1/2degree and the minute hand moves 6degrees. So we want x such that $180+\frac{x}{2}-6x=90$. Solving, $x=\frac{180}{11}$ minutes.[/hide]

[quote="Arvind_sn"]Billy Bob looked at the clock and saw that it was exactly six o' clock. He sat down again because he only needed to leave when both hands of the clock are right angles to each other. Exactly how much time will he need to wait?[/quote]

[hide]Let x be the number of minutes that have passed by.

Notice that after x minutes, the hour hand moves $\frac{x}{2}$ degrees, and the minute hand moves $6x$ degrees.  Also, notice that after 6:00 until the hands coincide with each other the angle between the hands decreases.

So we need to solve the equation $180+\frac{x}{2}-6x=90$.

So $x=\frac{180}{11}$

$\frac{180}{11}$[/hide]


inequalities
search
inequalities unsolved

suranyi's inequality. You can search its proof on the forum. Albania Eagle generalized this inequality.


analytic geometry
algebra unsolved
algebra

Given XOY coordinate system, point sequence {A_n} lie on the positive half of Y axis, point sequence {B_n} lie on curve y =  :sqrt: (2x), (x >=0), and |OA_n| = |OB_n| = 1/n. Let the intercept of line A_nB_n with X axis be a_n and the abscissa of point B_n be b_n, where n is positive integer.
(1) Prove that a_n > a_(n+1) > 4.
(2) prove that there exists positive integer n_0, such that for every n > n_0, sum_k=1^n (b_(k+1)/b_k) < n - 2004.

Does anybody know nice solution without direct calculations?
I don't want to bother mathlinkers with computations.


modular arithmetic
linear algebra
matrix
algebra
system of equations
algebra unsolved

Find the total number of solutions to the following system of equations:

$ \{\begin{array}{l} a^2 + bc\equiv a \pmod{37} \\
b(a + d)\equiv b \pmod{37} \\
c(a + d)\equiv c \pmod{37} \\
bc + d^2\equiv d \pmod{37} \\
ad - bc\equiv 1 \pmod{37} \end{array}$

If $ a + d\equiv 1$ then by the first and sixth:

$ ad - bc \equiv ad - (a - a^2) \equiv a(d + a - 1) \equiv 0$ contradiction!

So $ a + d$ is not congruent to 1, so $ b\equiv c\equiv 0$

$ a(a - 1) \equiv 0$
$ d(d - 1) \equiv 0$
$ ad \equiv 1$

So by the third, neither a nor b is congruent to 0, so they're both congruent to 1. Hence, $ a\equiv\b\equiv 1$

And we're done.

I think I got another solution:
Consider $ 2 \times 2$ matrix$ A\equal{}$\[ \left( \begin{array}{ccc}a & b  \\c & d  \\\end{array} \right)\]
From first four equation, it gives that $ A^2\equal{}A$,
From last equation, it dives that $ det(A)\equal{}1(\neq0)$. It means $ A^{\minus{}1}\equal{}$\[ \left( \begin{array}{ccc}d & \minus{}b  \\\minus{}c & a  \\\end{array} \right)\]
Then, $ A^2A^{\minus{}1}\equal{}AA^{\minus{}1} \Rightarrow A\equal{}I$
So, $ a\equal{}d\equal{}1$ and $ b\equal{}c\equal{}0$


I'm probably missing something basic, but I'm really missing a basic functionality present on the TI-89 that would be extremely useful in Mathematica: the with ($|$) operator.
If you don't have the 89:

[i]The "with" (|) symbol serves as a binary operator.  The operand to the left of | is an expression.  The operand to the right of | specifies one or more relations that are intended to affect the simplification of the expression.  Multiple relations after | must be joined by a logical "and".

The "with operator provides three basic types of functionality: substitutions, interval contraints, and exclusions.

Substitutions are in the form of an equality, such as $x=3$ of $y=\sin x$.  To be most effective, the left side should be a simple variable.
[code]                           x | x = 3
                           3
[/code]
Interval constraints take the form of one or more inequalities joined by logical "and" operators.  Interval contraints also permit simplification that otherwise might be invalid or not computable.
[code]                           abs(x) | x < 0
                           -x
[/code]
Exclusions use the "not equals" relational operator to exclude a specific value from consideration.
[code]                           x/x | x != 0
                           1
[/code]
[/i]
If anyone knows of ways to do these things, it would be greatly appreciated.

This is probably a little late to be helpful, but the two things you could try are the substitution character /.
(so [code]x^2 /. x->7[/code] outputs 49) and the Reduce command.


trigonometry
calculus
integration
calculus computations

Hey,
[quote="Mikecf10"]
dx/dt=rad(1+sin(t^2)) and x(0)=3 what is x(2)?
[/quote]
Did you mean,
$ {\frac {{{d}^{}}{x}}{{{d}^{}}{t}}} \equal{} {\sqrt{{1} \plus{} {\sin{{t}^{2}}}}}{\,}{\,}{\text{?}}$

Thanks,

-[u][b][url=http://www.artofproblemsolving.com/Forum/profile.php?mode=viewprofile&u=33715]APSStudent[/url][/b][/u]

By the Fundamental Theorem of Calculus, $ x(2) \equal{} x(0) \plus{} \int_0^2 x'(t) \,dt \equal{} 3 \plus{} \int_0^2 \sqrt {1 \plus{} \sin{t^2}} \,dt$
The value of the integral can be calculated (approximated actually) on a calculator. These kind of problems are common on the AP exam, so if you are preparing for it make sure you understand this.


geometry trigonometry inequalities AMC AIME MATHCOUNTS USA(J)MO \/closed

[quote="1234567890"]OK. I am planning to take Intermediate Algebra over the summer, and hopefully a mixture of intermediate advanced classes after that.
So, in the fall, will you offer Intermediate Counting and Intermediate Geometry?
In the winter, will you offer Olympiad Prob Solving and Olympiad Inequalities.
Also, I heard that the trig class is tougher than the Inequalities course. Is that true?

Out of these three intermediate classes, which will help most in competitions: Trig, Geo, or Counting?
Also, would you recommend Trig over Inequalities and Problem Solving? The reason I don't think Trig/ Complex Numbers are useful because their problems are few on the AMC/AIME.

If I choose not to take Trig, anyone know of any good self-studdy books for it?[/quote]

We have not yet scheduled school year classes, though we plan to offer any class we are not offering this summer during at least one of the semesters (plus some repeats from the summer).

We do not yet have an Intermediate Geometry class, though we could offer one sometime next year depending on how other AoPS projects go.

I don't know if the Complex Numbers/Trig class is tougher than Olympiad Inequalities.  It might seem that way for some students because most of the students taking the Inequalities class are those with the greatest amount of prior experience.  The two classes are not all that different in difficulty right now and we plan to eventually split them both into two pieces, add more material, and have both Intermediate and Olympiad level classes in those subjects.

The Complex Numbers/Trig class is helpful both with the AMC 12 and with the AIME and even with the USAMO.  There are only 25 problems on the AMC 12 and 15 on the AIME, so trig and complex numbers problems don't add up to a lot each, but they are often problems that make a difference in the scores of the kinds of students who qualify for the AIME.  Learning more about trig and complex numbers also helps both your algebraic and geometric problem solving skills and the skill of translating between those kinds of math, so the applications are broader than you might at first think.  We wouldn't offer classes we didn't feel were important for student development.  I don't know of any books that cover those subjects the way we do, though we plan to eventually write one.  It might be worth asking other students or Googling.

We just discussed our school year course offerings.  Currently we have plans to run all four problem series classes (before the contests except that MATHCOUNTS might run up to the state constest time period), all three Olympiad classes ending before the USAMO, all of our intermediate classes except perhaps Algebra (and including a Number Theory course we are working on), and at least 2 of our 3 Introductory classes (we may might run Introductory Counting & Probability, but we might might offer it in February).

We are not yet to the point of assigning dates and times to those classes.

We will of course also be offering all 5 levels of the Independent Study furing each semester.

Course descriptions can be found [url=http://www.artofproblemsolving.com/Classes/AoPS_C_About.php]here[/url] or by clicking on the Online School tab in the left sidebar.

After I'll take Algebra, I'm pretty sure I'll take 4 more courses next year:
Intermediate Trignometry (I'm just bad with this)
Intermediate Counting (I'm okay at it, but I heard there are IMO level problems in there)
Olympiad Inequalities (I audited the class for the first day. Some of it seemed easy, but other things seemed way too hard for me)
Olympiad Problem Solving (I think I should take this class last, as its the hardest out of all the classes)

Could everyone give me info if they have taken any of those courses on how hard they were?

I don't want to discourage student response (please respond if you have taken any of these classes), but you can probably find some commentary in AoPS forums using the search feature at the top of the forums area.


if (2^n)+1 is prime & n>=1:
[b]prove that:[/b]
n=2^k

reasoning by the absurdity and let us suppose that n is not a power of 2 .
then n admits at least an odd divider D higher then 2 and we have n=d.q
thus : a^n=a^dq+1=(a^d)^q+1=(a^q+1)(a^{q(d-1)}-a^{q(d2)}+...+1)
and consequently a^n+1 is not prime.
(this last equality is consequence  that d is impair)

[quote="Leao"][hide="Is this solution ok?"]
If $2^{n}+1$ is prime, then $n$ must be even (except for 1), because if $n$ is odd, then $2^{n}+1^{n}$ can be factored. Thus, we let $n=2m$. We get $4^{m}+1^{m}$ after subsitution. With similiar reasoning, we see that $m$ must also be even, and you just go on and on...so $n=2^{k}$.
[/hide][/quote]yes of course your answer is OK. :D

[quote="Leao"][hide="Is this solution ok?"]
If $2^{n}+1$ is prime, then $n$ must be even (except for 1), because if $n$ is odd, then $2^{n}+1^{n}$ can be factored. Thus, we let $n=2m$. We get $4^{m}+1^{m}$ after subsitution. With similiar reasoning, we see that $m$ must also be even, and you just go on and on...so $n=2^{k}$.
[/hide][/quote]

The reasoning is fine, but there is a shorter way using essentially the same logic:

Suppose $n$ has an odd factor, $m$.  Let $n = km$.  Then $\left( 2^{k}\right)^{m}+1^{m}$ can be factored.

:)

[quote="Gyan"][quote="yash_17"]if (2^n)+1 is prime
[b]prove that:[/b]
n=2^k[/quote]
Assuming that you mean $(n,k)$ to be integer, this is [b]NOT [/b]true.
[hide]For example,  when $n=0$ , $2=2^{n}+1$ is prime  yet you can not write 0 as $2^{k}$

You have to  make it  that clear  ... ( eg . $n \ge 1$ )[/hide][/quote]you are completly right im so sorry. :blush:


LaTeX
inequalities proposed
inequalities

Because i'm the newcomer so i don't know how to write in LATEX so can anyone help me?

a,b,c,d >0 prove that:
(a+b)^5.(a+d).(b+c)/(c+d)^4+(b+c)^5.(c+d).(b+a)/(a+d)^4+(c+d)^5.(d+a).(b+c)/(b+d)^4+(a+d)^5.(a+b).(d+c)/(b+c)^4

>= ((a+b+c+d)^3)/2

[quote="erdos"][quote="CTLHP"]Because i'm the newcomer so i don't know how to write in LATEX so can anyone help me?[/quote]
Ok Let's $\LaTeX$ it :
$a,b,c,d$ are positive real numbers. Prove that :
$(a+b)^5(a+d)\frac{(b+c)}{(c+d)^4} +(b+c)^5(c+d)\frac{(b+a)}{(a+d)^4}
+(c+d)^5(d+a)\frac{(b+c)}{(b+d)^4} +(a+d)^5(a+b)\frac{(d+c)}{(b+c)^4} \geq \frac{(a+b+c+d)^3}{2}$[/quote]

any typo? should it be
$(a+b)^5(a+d)\frac{(b+c)}{(c+d)^4} +(b+c)^5(c+d)\frac{(b+a)}{(a+d)^4}
+(c+d)^5(d+a)\frac{(b+c)}{(b+a)^4} +(a+d)^5(a+b)\frac{(d+c)}{(b+c)^4} \geq \frac{(a+b+c+d)^3}{2}$[/quote]


inequalities
trigonometry
inequalities unsolved

$ 0<\alpha_1<\alpha_2<...<\alpha_n<\pi\div2$  so prove that
$ (\tan\alpha_1)^2<(\tan\alpha_1\plus{}\tan\alpha_2\plus{}...\plus{}\tan\alpha_n)\div(\cot\alpha_1\plus{}\cot\alpha_2\plus{}...\plus{}\cot\alpha_n)<(\tan\alpha_n)^2$
not too hard and it's my first post so if there is a mistake please forgive me  :lol:

Note $ f(x) = \tan{x}$ is increasing on $ (0,\frac {\pi}{2})$, $ \tan{0} = 0,\tan{\frac {\pi}{2}} = 1$ and so $ 0 < \tan{\alpha_1} < \tan{\alpha_1} < ... < \tan{\alpha_n} < 1$ so $ n\tan{\alpha_1} < \sum_{i = 1}^{n}\tan{\alpha_i} < n\tan{\alpha_n}$;
$ \frac {1}{\tan{\alpha_1} > \frac {1}{\tan{\alpha_2}} > ... > \frac {1}{\tan{\alpha_n}$ so $ \frac {n}{\tan{\alpha_n}} < \sum_{i = 1}^{n}\frac {1}{tan{\alpha_i}} < \frac {n}{\tan{\alpha_1}}$.

Therefore:
$ \frac {\sum_{i = 1}^{n}\tan{\alpha_i}}{\sum_{i = 1}^{n}\cot{\alpha_i}}$
$ = \frac {\sum_{i = 1}^{n}\tan{\alpha_i}}{\sum_{i = 1}^{n}\frac {1}{tan{\alpha_i}}}$
$ > \frac {n\tan{\alpha_1}}{\frac {n}{\tan{\alpha_1}}} = (\tan{\alpha_1})^2$ which proves the first part.
$ \frac {\sum_{i = 1}^{n}\tan{\alpha_i}}{\sum_{i = 1}^{n}\cot{\alpha_i}}$
$ = \frac {\sum_{i = 1}^{n}\tan{\alpha_i}}{\sum_{i = 1}^{n}\frac {1}{tan{\alpha_i}}}$
$ < \frac {n\tan{\alpha_n}}{\frac {n}{\tan{\alpha_n}}} = (\tan{\alpha_n})^2$ which proves the second part.

QED


[quote="G-UNIT"]With a 1998 meter long fence, I want to make the biggest rectangular shape garden with sides that are measured in whole number of meters. What is the area of this garden?

answer  [hide] 249,500 [/hide][/quote]

[hide]You want it to be as close to a square as possible... 1998/2=999. 999=499+500. 499*500=249500.[/hide]

[quote]You want it to be as close to a square as possible... 1998/2=999. 999=499+500. 499*500=249500. [/quote]
I agree.

Isn't the area or a four sided figure of one given perimeter the biggest when it's closest to a square, and smallest when it is the thinnest rectangle?

Yeah, it is. That was stated in the 2nd post.

[quote="superman"][quote="G-UNIT"]With a 1998 meter long fence, I want to make the biggest rectangular shape garden with sides that are measured in whole number of meters. What is the area of this garden?

answer  [hide] 249,500 [/hide][/quote]
why did you ask the question, and give the answer[/quote]

It was obviously a challengle, or maybe he knew the answer but didnt know how to get there.

lol, i knew how to do the problem (really who wouldnt), i simply posted the answer so people could verify their solutions validity

Just like teachers say you're supposed to do with the textbooks.

but what if youre wrong?

which youre not

that would be weird

[quote="Treething"]Aren't you supposed to copy down the answers in the back of the book and do the even ones on your own :P ?[/quote]

Wait, was that sarcasm? So, you mean that's not what you're supposed to do? Oh, I knew that  :blush:

unless you have to show your work on something like finding arc measures (really easy) then its easier just to do it in your work and not have to flip pages


Let S be a set with (n^2+n-1) elements, and n is a positive integer. All n-element subsets of S are partitioned into two groups. Prove that there exists one of the two groups such that there are n n-element subsets of S in the group which are pairwisely disjoint.


Hi, I have a TI-84+, and for the project I'm doing, I think I might need more than 27 variables. Is there a way to get more?

If you go to Apps and then Finance, there are a lot of variables there.

You could also use lists or matrices.

if you could use a computer it'd be a lot cooler and easier.

I don't know if this counts as bringing back an old topic as this is halfway through the first page, but here is a solution.

For games/projects that I use that require many variables, I save them in a matrix.

Use them as follow:
Storing: 1->[A](1,1)
Getting: [A](1,1)

Then you can do [A](1,2) etc. Make sure you don't use up too much memory though since matrices take up a lot of space.
You can also loop matrices
[code]for(I,1,50)
b+[A](1,b)->b
end[/code]


What country will have the most medals this year?

[quote="dynamo729"]I voted USA for patriotic reasons and because they seem unbeatable in swimming (the only thing I pay attention to).[/quote]

I don't watch much Olympic stuff.  But yeah, I'm with dynamo on that, vote USA for patriotism!

75% chance U.S.
24% chance China
1% chance other

something like that
and yeah, be patriotic!

WHOOOOT GO CHINA!!!!!!!

what? i'm being patriotic!

GO USA!!!!!!

too bad TYson Gay got injured, or else they would win the 200 meter dash also :P

hey! go china!

(yay dreambold!)

besides, its in china anyway this year and they should do well! :P

US will definitely have most medals, most gold will be very close between Us and CHina


quadratics
number theory proposed
number theory

Solve in integers $y^{2}=x^{3}-p^{2}x$ where $p$ is a prime congruent to 3 modulo 8.

This isn't quite finished, but I think it's pretty quick from here. Can someone else finish it off?

The only solution is x=p and y=0. We have $y^{2}=x(x^{2}-p^{2})$. So for any prime $q$ that divides $x$ an odd number of times, it must divide $x^{2}-p^{2}$ an odd number of times. For all $q\neq p$, this is clearly impossible, so either $x$ is of the form either $z^{2}$ or $pz^{2}$. In the first case, we have $y^{2}=z^{2}(z^{4}-p^{2})$, so $z^{4}-p^{2}$ is a perfect square, which we know is impossible (there are no nontrivial solutions to $a^{2}+b^{2}=c^{4}$).

The other possibility is $x=pz^{2}$. Then $y^{2}=p^{3}z^{2}(z^{4}-1)$, so $p(z^{4}-1)$ is a perfect square. In particular, this means $z^{4}\equiv 1$ (mod p), or $z^{2}=pw+1$ for some w (not minus because -1 is not a quadratic residue mod p). Hence $p(z^{4}-1)=p^{2}w(pw+2)$, so $w(pw+2)$ is a perfect square. Clearly no prime may divide w an odd number of times except possibly 2, since it must then divide $pw+2$. Then we have two cases: $w=v^{2}$ or $w=2v^{2}$. The former is impossible since $v^{2}(pv^{2}+2)$ may not be a perfect square (as 2 is not a quadratic residue mod p). Hence $w=2v^{2}$, and we want $2v^{2}(2pv^{2}+2)$, that is $pv^{2}+1$ to be a perfect square. However at the same time, $z^{2}=pw+1=2pv^{2}+1$ is a perfect square. We now have two Pell Equations. Maybe someone else can wrap it up.

Anyway, working backward from any v that satisfies both of those Pell Equations, the value of x is $2p^{2}v^{2}+p$. I suspect the only possible value of v is 0, making x=p and y=0.

I actually misread this as $y^{2}= x^{3}-px$, which is also an interesting problem.

Yeah I also got that far(MoL atleast). I don't know how to wrap it up though,have been trying to do so for a while.

Hmm.  It turns out that this is a special case of a well-understood class of elliptic curve:

http://www.math.umd.edu/~eve/cong_num.html
http://en.wikipedia.org/wiki/Congruent_number

In particular, since $p \equiv 3 \bmod 8$ there are no solutions with $y \neq 0$.

[quote="jb05"]This isn't quite finished, but I think it's pretty quick from here. Can someone else finish it off?[/quote]

I tried to solve it in the same way as you and I stopped exactly at your point, so maybe it is not so easy from this point on, but I'm not sure.

[quote="t0rajir0u"]In particular, since $p \equiv 3 \bmod 8$ there are no solutions with $y \neq 0$.[/quote]

Do you have some reference about it?

However thanks for the 2 links, they were very interesting.


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $ H$ be a subgroup of a finite group $ G$ and $ m$ a strictly positive integer with the property $ \vert$ G $ \vert$=m $ \vert$H$ \vert$.Prove that $ g^{m!}\in H$, $ \forall g\in G$.

For normal $ H$, we get $ g^m \in H$ for all $ g \in G$ by Lagranges theorem.
So let $ H$ now be any subgroup:
Now the intersection $ I$ of all conjugates of $ H$ has index dividing $ m!$ and is normal. Thus $ g^{m!} \in I \subset H$.


[quote="Chocolate milk"]My answer is different.
[hide] It's 232  days[/hide]
Maybe I was wrong[/quote]
You probably were right...
As eheeheehee once said, "I think I am right, but I know I am probably wrong..."

[quote="moogra"][quote="nutz_for2.718281828"]Anne Shirley is studying for the Queens course. The course will be held in 123 days. She will have to do 243 lessons in all. She also can do 0-12 lessons in a day. What is the sum greatest number of days that she will not have to study a single lesson and the greatest number of days that she can only study one lesson?[/quote]
[hide]
she will take at least 21 days
so 102 days no lessons
she can study 1 lesson for less than 122 days(too lazy)
by the way, no way can it be 200 days. THERE AREN"T EVEN 200 days[/hide][/quote]
Read the problem ;)


I dont seem to have a solution manuel for this question.

  If $\frac{a!}{b!}$ is a multiple of 4, but not a multiple of 8, what is the maximum value of $a-b$?

Im too lazy to write up my solution, but Ive gotten [hide] 3 [/hide]

My solution:
[hide]I got 3 as the solution as well.

In order for $\frac{a!}{b!}$ to be a multiple of 4 and not 8, $a!$ must be greater than $b!$ and contain exactly 2 more factors of 2.

$a$ cannot be 4 greater than $b$ because it would have 3 factors of 2.
$a$ must be at least 2 greater than $b$, or it will have less than 2 factors of 2.

Therefore, the greatest value of $a$ must be be 3 greater than $b$ and $a-b=3$[/hide]

Yea the factorization of $\frac{a!}{b!}$ would look like this:

(odd number) $\cdot (2\cdot$odd number$) \cdot$ (odd number)

  The next number would be $(2 \cdot$ even number$)$  which means the factorization would have 3 2's making an 8.  So the maximum must be 3.


inequalities
function
inequalities proposed

Consider the function
$f(y)=\frac{1}{e^{y}+1}$  where $y\geq 0$   and  use Jensen  :wink:


Intergrate this x(3^(x^2)dx

Thankx alot

[hide="here"]
$\displaystyle \int{x(3^{x^2})dx}$

$u=x^2$
$du=2x$

$\displaystyle \frac{1}{2}\int{3^u du}$

$\displaystyle \frac{3^{x^2}}{2\ln{3}}+C$
[/hide]

I dont understand the last two steps. how can you get 3^(x^2)/ln(3) ? what kind of rule did you apply?

[hide="Well, its not so much a rule as just a clever substitution..."]
$\int e^v \, dv = e^v$
$v = x \ln 3$
$e^v = 3^x$
$\int e^{x \ln 3} \ln 3 \, dx = e^{x \ln 3} = 3^x$
$\int 3^x dx = \frac{ 3^x }{\ln 3}$
[/hide]

[quote="Kle"]You answer is different it doesn't have the 3^(x^2) on the numerator ..? I am still not getting it[/quote]

I wasn't trying to answer the original question, I was trying to show you the 'rule' he was using, namely:  $\int a^u \, du = \frac{a^u}{ \ln a}$, with $a = 3, u = x^2$.  :)

The solution is quite simple.

$\\\int x3^{x^2}\ dx\hspace{.3in}u=x^2\hspace{.3in}du=2x\ dx\implies dx=\frac{du}{2x}\\ =\int x3^{u}\ \frac{du}{2x}\\ =\frac{1}{2}\int 3^u\ du=\frac{3^{x^2}}{2\ln 3}$

Masoud Zargar


Given the equation $(x^2-3x-2)^2-3(x^2-3x-2)-2-x=0$, show that the roots of the equation $x^2-4x-2=0$ are the roots of the initial equation and find all real roots of the given equation.

expand the first equation, x^4-6x^3+2x^2+20x+8=(x^2-4x-2)(x^2-2x-4), the following is quite simple.

Or say that the remainder when you divide the given equation by $x^2-4x-2$ is:

$x^2-3x-2-x$ or $x^2-4x-2$.

Therefore the original equation in divisible by $x^2-4x-2$


Find the solution $ \in Z$ of this equation?
$ \sqrt {x \plus{} \sqrt {x \plus{} \sqrt {x \plus{} ... \plus{} \sqrt {x \plus{} \sqrt {x}}}}} \equal{} y$
There are 100 roots !

PS:srry 4 my english  :(

[quote="Zhero"]Could you please be more precise? The problem statement is ambiguous.[/quote]

Find the integer solution(x=?,y=?) of equation ?

My apologies, I mistook "roots" to be solutions, rather than the actual radicals.

I'll work on it.

im sorry I don't understand what the [b]"100 roots of this equation"[/b] mean !?!?

Do you mean that the solution set will be something like

(x, y) $ \equiv$ ( 0, 0 ) , ( 2, 2 ) , ( 6, 3 ), ( 12, 4 ) .... etc ?

No, he means that there are 100 radicals there. So, for exampe, $ \sqrt{x \plus{} \sqrt{x \plus{} \sqrt{x}}}$ would have "3 roots."

oh O.K.

so (0, 0) is 1 trivial solution.

[b]Let me work on it...[/b]

Take the problem in a smaller scale...

Prove that for any positive nonsquare integer $ x$, $ x\plus{}\sqrt{x}$ is never an integer (belonging to a higher surd). Once, you do that, it's just a tower of surds and Nora.91's observation should be trivial indeed.

If you show $ x \plus{} \sqrt{x}$ isn't an integer for integral $ x$, how does that say anything about $ x \plus{} \sqrt{x \plus{} \sqrt{x}}$?

If $ x \plus{} \sqrt{x}$ isn't an integer, neither is $ \sqrt{x \plus{} \sqrt{x}}$.  (So $ \sqrt{x}$ must be an integer.  Then...)

[hide="Solution"]Continuing this, we see that in order for the left-hand side to be an integer, so must $ x \plus{} \sqrt {x \plus{} \sqrt {x}}$, which is possible only if $ x$ and $ x \plus{} \sqrt {x}$ are both perfect squares. We let $ x \equal{} k^2$ and $ j^2 \equal{} x \plus{} \sqrt {x} \equal{} k^2 \plus{} k$. Therefore, $ 4k^2 \plus{} 4k \plus{} 1 \equal{} 4j^2 \plus{} 1$, that is, $ (2k \plus{} 1)^2 \minus{} (2j)^2 \equal{} 1$, that is, $ (2k \plus{} 2j \plus{} 1)(2k \minus{} 2j \plus{} 1) \equal{} 1$. The only solutions to this are $ (0, 0)$ and $ ( \minus{} 1, 0)$. In the latter case, we get that $ x \equal{} 1$, and $ j^2 \equal{} 2$, a contradiction, so the only solution is $ \boxed{(x, y) \equal{} (0, 0)}$. [/hide]


On a circle there are written 2005 non-negative integers with sum 7022. Prove that there exist two pairs formed with two consecutive numbers on the circle such that the sum of the elements in each pair is greater or equal with 8.

[i]After an idea of Marin Chirciu[/i]

If any of the intgers is greater than or equal to $8$ we are done with its two neighbors. Thus, we may assume that none of the numbers is greater than $7$. Thus, omit on of them. It leaves $2004$ integers which can be divided into $1002$ pairwise disjoint pairs of consecutive numbers (in terms of position). Thus sum of these $1002$ pairs is at least $7022-7 = 7015 > 7 \times 1002.$ Therefore at least one of the pairs, say $(a,b)$, gives a sum not less than $8$.
Now, omit $a$, and use the same reasoning as above to dtermine another pair, and we are done.

Pierre.

[hide="does this work"]

Consider all 2005 pairs of consecutive numbers on the circle.

Suppose only one of them sums to at least 8 (or none, but a similar argument works). Call the sum x and the pair (m,n).

Then the sum of all the pairs is at most $2004(7)+x=2(7022)=14044$. So $x \ge 16$. WLOG, assume $m \ge n$ so then $m \ge 8$. So the two pairs formed by $m$ both sum to at least than 8. Contradiction.[/hide]


function
algebra
domain
polynomial
trigonometry
geometry
geometric transformation

Suppose if possible f(x) is neither even or odd.
Then there exists $a$ in the domain of f such that |f(a)| != |f(-a)|.
Thus f(a) ^ 2 != f(-a) ^ 2, f(x)^2 is not even, contradiction.
Q.E.D.

I'm not sure if it's right since I really lack practice in problems that deal with functions. (like f(x) or such) :maybe:

Don't worry, anipoh, it's all right. Nothing particular is asked in this question. BTW, when I read 'symmetric functions' I was thinking of Newton's symmetric polynomials, which got me into this topic :D ; however, it wasn't  :( .

Spoon wrote:If f(x)^2 is even, prove that f(x) must be even or odd.

[hide]f(x) is even.  Two odd numbers mltiplied together are odd, two even numbers multiplied together are even, and an even number multiplied by an odd number is eveen.  Since f(x)^2 is even, and f(x) = f(x), f(x) has to be even.[/hide]

I'm not sure that works.  Just because f(x)^2 isn't even if f(x) isn't even or odd doesn't necessarily make f(x)^2 even if f(x) is even or odd.

The way to prove it would be to say that since f(x)^2 is even, f(x)^2=f(-x)^2.  Then, by taking the square root of both sides, f(x)=f(-x) or f(x)=-f(-x), i.e. it is even or odd.

Edit: My post was in response to the first answer, not H_S_Potter's.  I think that the original question meant "even" and "odd" as properties of functions and not being divisible by 2 or not, which is what H_S_Potter's answer is for.

sppon, I think you shouldn't post a problem in 2 sections. And Wumbate I think you should llok at the posts on the same question in the Pre-Olympiad section. BTW: $f(x)$ is even iff $f(x)=f(-x)$ for ${\bf all}$ $x$.

Misha

Thanks, I looked there and found the counterexamples.  My answer fails when I say that f(x)=:pm:f(-x).  While this is true, that doesn't mean that FOR ALL X, f(x)=f(-x), or FOR ALL X f(x)=-f(-x).  It is one or the other for every x, but is not necessarily consistant for all x.

A counterexample would be the function f(x)=1 for x=1 and f(x)=-1 for x?1, as zscool said in the other post.

Misha, I posted this in two sections because I needed a quick response. This was not meant to be a challenge. I know that it is true, I was asked to show that it was true, and I can see it graphically but didn't know how to prove it, and I wanted to see the prove.

Your "counterexamples" go against what the question asks for.

f(x)^2 must be defined and even for all values of x so specifiying 0<x<1 doesn't quite cut it. I was not challenging you. I know for a fact that this property is true. I just wanted to see a proof.

Thanks for all of your "anti-help" and thanks for your negative attitude.

Now I remember why I stopped visiting this forum after the worst decision in the world was made to merge the two "communities".

Sorry if you see it as negative attitude. I didn't intend this. But my counterexample works though: I defined the function:

$f(x)=x+1$ for $x\leq0$
$f(x)=1-x$ for $0<x\leq 1$
$f(x)=x-1$ for $x>1$.

It's clear that it's continuos. Now I prove $f(x)^2=f(-x)^2$ for all positive $x$:

if $0<x\leq 1$ then $f(x)=1-x$ and $f(-x)=-x+1$. So $f(x)^2=f(-x)^2$.
if $x>1$ then $f(x)=x-1$ and $f(-x)=-x+1$ and $f(x)^2=f(-x)^2$.

I don't know why you were so sure that your statement is true but it is not (I hope you see it now).
It's not intended personally against you it is only a mathematical argument.

Misha

sigh, why do people complicate things... I clearly thought that it was implied, but f(x) must also differentiable AND CANNOT BE PIECEWISE.

and, for good measure, just to be a bit more meticulous incase it causes a problem...

Assume f(x)^2 is normalized on $\mathbb{R}$

Spoon, your original claim was false.  Misha gave you a perfectly valid counterexample.  And yet you treat his counterexample as "anti-help", a "negative attitude", and a reason to rant about the forum.

Then you decide to revise the question by saying the function has to be differentiable and cannot be piecewise.  Misha already pointed out that adding a differentiability requirement doesn't change the falseness of the claim  -- we can come up with counterexamples that are differentiable everywhere.  And I have no idea what your definition of "piecewise" is.

You owe Misha an apology.

I want to see it, then... show me one...

f(x) can have only one definition on the interval $(-\infty,\infty)$

I want to see your counter example. Then I will appologize.

[quote="Spoon"]Misha, I posted this in two sections because I needed a quick response.[/quote]

Sorry but 2 sections = spam. 2 sections for your personal needs is quite selfish considering the fact that this is a community forum.

[quote="Spoon"]Your "counterexamples" go against what the question asks for.[/quote]

Thanks Misha for the counterexample.

[quote="Spoon"]Thanks for all of your "anti-help" and thanks for your negative attitude.[/quote]

He took a math argument too personal as an offending comment.

[quote="Spoon"]Now I remember why I stopped visiting this forum after the worst decision in the world was made to merge the two "communities".[/quote]

Well, since you think the merging of two communities was like the end of the world, you can save yourself, pack up, and well... stop visiting our forums [b]until[/b] your mind is cleared up and come back to visit us when you are feeling better, OK?

And once again, thanks Misha for the nice counterexample. One more thing, h_s_potter got the wrong idea about "even functions" and "even numbers". Hopefully, he can learn something here  :)  Stay cool kiddos :cool:

3X.lich

Here's another example which seems differentiable and continuous:
f(x)=g(x)*(.5cos(x)+.5cos(x))
where g(x)=1 for all x except intervals from -pi to -3pi, -5pi to -7pi, etc... on those intervals it =-1.
There is probably a way to write it without having the g(x) in there.

It looks like a normal cosine curve when x is positive, but when x is negative it is alternately the cosine curve and it's reflection over the x-axis.
It meets the diff. requirement since the derivative at the parts where it switches is 0, and is continuous, and someone else here may be able to put it into one function without the embedded g(x), I don't have time right now.

Define
\[f(x) = x (x-1) (x+1) (x - |x| + |x - 1|).\]
Then f is differentiable everywhere, its square is even, and yet f itself is neither even nor odd.

Thank you all: 3X.lich, Ravi B. and Wumbate for you support. I like it a lot to meet people who can discuss about math using only mathematical arguments. I think that it was a great idea to unite both communities since the main aim of the forum, in my opinion, is to talk to very different people from different countries to learn mathematical approaches which are not usual in your country.

Misha

Cool... Ravi, how did you come up with that example?

The expression $x - |x| + |x - 1|$ is just Misha's function written as a single expression.  To smooth the function when x is 0 or 1, I multiplied by x(x - 1)(x + 1).

The nature of this problem stems from a proof in my quatum physics class, and rather than go into any of the physics, which is most likely too advanced for most people here, I just made it simplified...

Fine, I will except that under circumstances as that obscure function, it won't work, but that function certainly has nothing to do with quantum physics.

I didn't realize that this would turn into something so technical.

I didn't want to specify that f(x) (which is really $\Psi(x,t)$) had to satisfy the schrodinger equation.

Thanks for all of your help.

I apologize to all of the smart minds out there far beyond me.

Well, your 'apology' is clearly given only grudgingly, so let me ask you this:

How did you want us to prove something which couldn't be proven?

*notargueingjustpostingaboutthemathematicsbehindit*

i guess a function satisfying the schrder equation has to be holomorphic everywhere(which means that for any real $x$, $f$ can be put into a taylor series around $x$ (which means roughly that $f$ is very smooth and may have no poles)). so let's assume $f$ is holomorphic.

there is a theorem, if two holomorphic functions $f$ on a subset A of $\mathbb{C}$, $g$ on a subset B of $\mathbb{C}$ agree on an uncountable subset of the intersection of A and B, they agree on the whole intersection(which roughly means, that any holomorphic function on $\mathbb{C}$ is given by the values on some uncountable set).

so, let's assume $f$ has a taylor expansion around 0 which converges on the set U(which is clearly uncountable). so, at the moment we're calculating in the ring of formal power series in $\mathbb{C}$(the set of power series $a_0+a_1x+a_2x^2+...$ with possibly infinitely many terms). it is trivial to show that it has no zero divisors(just look at the two terms of minimal degree-their product is nonzero and the only term of the minimal degree in the product of the power series, thus the product is nonzero). thus, since $0=f(x)^2-f(-x)^2=(f(x)+f(-x))(f(x)-f(-x))$, either $f(x)=-f(-x)$ or $f(x)=f(-x)$. this holds on the set U where the taylor series converges. now, consider in the first case the two holomorphic functions $f(x)$ and $-f(-x)$. they agree on U, thus everywhere by the above theorem, thus $f$ is odd. the other case is similar.

Peter


calculus
integration
inequalities
function
linear algebra

I know a kid who, without a doubt, cheated on the AMC, and by some fluke or attempt to remove suspicion, changed one of the copied answers to a correct one, so he has the highest score in our school. It's certainly not the first time he's cheated: for the past three years, whenever someone mentions cheating on a test, they mention his name. He practically cheated his way into an Ivy League school, and still gets away with it. He would've had no reason to cheat on this test, had the math teacher not offered extra credit for good scores as an incentive to make an effort. He certainly made an 'effort.'

What angers me most is that he continues to do this, and nobody stops him. One of his teachers jokes about calling his college and telling them he cheats on every biology test, but for now it's just a joke. I can't believe we live in a world where cheating seems to be perfectly acceptable.

That's all I have to say.

What kind of school do you go to, and what score did he get?

I'm not sure what your question intended to ask, but I go to a public high school. The true solver of the problems got around 130, so I imagine the cheater managed to get slightly more than that.

[color=green]This post was slightly modified by a moderator.[/color]

Just think about how he'll be completely owned when he gets to real life. Then laugh. I know it sucks, but it's not that big a deal.

1. If he's in an "Ivy League" school, then how is he taking the AMC officially? In fact, why is your college distrubuting AMC?

2. Report his actions to the head person, I'm pretty sure he/she will take it more seriously than the "teacher."

Sadly, the same thing happened to me last year. Although it wasn't as serious, it still got on everyone's nerves and when I think about the girl going to middle school, where the desks will be separate, I laugh and so do my friends.

Eh, the cheater is the one that will get hurt, not you. Don't worry about it.

Also, 7h3.D3m0n, maybe torrentlord was referring to the past. OR the cheater is now in college, but torrentlord is still a highschool senior.

if you hate this cheater that much, then i believe you can report it to the amc office, or his college?
there is no reason to hate this cheater, because he'll be screwed for life.
once he gets into his job, he'll be fired almost instantly

so instead of wasting your time on complaining, we should all concentrate on improving our own lives
 :)

Ah, to clarify, he was accepted early this year. I'm not concerned about me, I just don't like the fact that people find it acceptable.

He has at least a three year long history of cheating. Why is it hard to believe that he cheated off the smartest kid in the school? Everyone knows who he is, I wouldn't put it past him to sit near him with the intent of cheating. He did better than the smartest kid in the school completely by chance, changing one of his wrong answers, leaving it blank or possibly changing it to a correct one. None of your points are as far-fetched as your post implies. I'm positive that he doesn't have the mathematical knowledge to answer more than 15 of the questions, let alone more than 20.

So in short, I am not jealous, because I do not envy cheaters.

[quote="mhisno1"]It is impossible to cheat and get the highest score.

Are you seriously trying to convince us that:
1.  You saw this kid cheat
2.  He cheated off of the smartest kid in the school
3.  He did better than the smartest kid in the school?

Sounds a bit like jealousy...

How would YOU know?[/quote]

No reason to envy a cheater, as torrentlord said. Possibly the cheater did one of the few problems he DID know how to do, but the person he was cheating off of made a careless error. He saw this, as did not erase his answer. Yeah, just a random theory.

Don't worry, the guy is hurting himself, not you. You can't cheat your way through life forever.

[quote="Nerd_of_the_Ages"]Don't worry, the guy is hurting himself, not you. You can't cheat your way through life forever.[/quote]

Well said. Since he got into an Ivy League college, that means he would've near impossibly cheated through every test in high school. Perhaps he would've studied at home and not put forth the effort in class? You don't know everything if you haven't talked to him.

[quote="Twiz"]There will be a point in his life when he can't cheat, and then he'll be in trouble.[/quote]

Do you really think that's true?

I am not so sure what the big deal is. Does the AMC really matter that much? If somebody copied off me this year and somehow got a better score than me, good job him for getting a 145.5 or a 150 on the AMC12a. It is not the end of the world for anybody. Study more, cry less (and bear in mind that there is pretty much no way to copy answers on the USAMO).

And congratulations to him for getting into an Ivy League school. Contrary to what you imply, it isn't possible to get in solely through copying; he must have some degree of talent and perseverance. Just keep in mind that you become the tool if he manages to succeed in life with Ivy League credentials and education - regardless of your  integrity and his oh-so-terrible cheating ways.

Sure, the fact that he cheated may catch up to him. But does this cheating really affect you in any way at all? Nothing but your own vindictive self is telling you to be pretentious and "despise cheaters"; alright, go ahead and condemn the winners in a world where cheating to get ahead is largely acceptable. I hold myself to a standard of academic integrity - unlike many of my friends - but this does not mean that I appoint myself judge, jury, and executioner of whomever I view as having less integrity than me. I just bear it and work harder.

just my two cents.

What's the point of getting into a great college if your supposed knowledge turns out to be of no use? The more you cheat, the more vulnerable you are (and the harder you fall). What if, one day, he takes a very important and difficult test alone in a confined room? If he became world-famous, and flunked that test, it would create quite a rumor.
     I dislike cheaters the same way because people tend to rely on my answers in hard classes because of my reputation from being advanced in math (my school is pretty easy). I love helping others by explaining them the procedures, but copying answers is simply taking an unfair advantage, even if it doesn't hurt me.

[quote="zephyredx"]What's the point of getting into a great college if your supposed knowledge turns out to be of no use? The more you cheat, the more vulnerable you are (and the harder you fall). [quote]

Because any Ivy League college is a nice thing to have on your diploma + plus its an easy way to springboard into a high paying job regardless of how you did in college? I know thats why most people go to top colleges.

[quote]What if, one day, he takes a very important and difficult test alone in a confined room? If he became world-famous, and flunked that test, it would create quite a rumor. [/quote][/quote][/quote]

Yes, as if that would ever happen. Is that really the best counter-example you could think of?

As someone who increasingly sees this from the teacher's side: most cheaters are bad at it. You recognize cheaters by what they get wrong.

This is especially true on the long-answer tests I deal with. The right answer is not enough.

[color=green]Note from moderator: there is [i]never[/i] a good reason to block-quote the entire post preceding yours.[/color]

Are you a lecturer at a university? You make it sound like people happen to cheat often. How do they dare to? What are the consequences of being caught cheating? Here it is suspension for at least a semester.

I'm a TA, and I don't see it often. Most of my knowledge is second-hand. It's probably also true that I'm not very good at catching it- I don't really know what to look for during the act.

Most offenses are dealt with entirely on a class level- the relevant grade is zeroed out, and the teacher doesn't take it any farther. If he or she does, the punishments do include suspension- but that's very rare. Even at the level of a formal investigation, most cases only lead to warnings, and the number of formal cases is about 40 in a year, among tens of thousands of students.

I also do TA work. I mostly have sessions where we go through their assigned problems, but I have corrected exams for high school competitions. I've never even thought of the possibility of cheating. Oops.

Derailing a bit (this topic is sort of meaningless anyway), what do you think about cheating in take home exams in mathematics? If I recognize a problem as something I've seen proven in a book or on the internet, is it plagiarism to write that proof down? I mean, I can't undo that solution in my head, and I know it works. Should you cite the source and get full points for it? On a take-home in integration one of the questions was to prove the schur test. I didn't recognize it, but if I had, I don't think I would have hesitated to read about it on wikipedia or whatever.

I would credit the source and rewrite the proof. The important thing is to understand how to prove it; rewriting in your own words helps with that. Using the source without credit does feel like plagiarism.

business people cheat all the time, and the more they cheat the more money they get, unless they're really bad at cheating and get arrested. people who say "he's ultimately going to get what he deserves" well, it's true sometimes but not most of the time.

[color=green]Note from moderator: there is [i]never[/i] a good reason to block-quote the entire post preceding yours.[/color]

Errm? What?

Business people cheat all the time.

Can you explain what you mean?

[quote]What if, one day, he takes a very important and difficult test alone in a confined room? If he became world-famous, and flunked that test, it would create quite a rumor. [/quote]

Yes, as if that would ever happen. Is that really the best counter-example you could think of?[/quote]

I'm not saying that's the absolute possibility that could foil the cheater. But it does show how that creating a cheating history doesn't just fade away. If the cheater eventually chose to relearn what he cheated on during school, it would result in the same amount of work to achieve the same results. Sure, it would get him into a better college and better jobs before the effort, but he would then be learning to escape vulnerability, and not for the sake of understanding new things. And even if he doesn't get cornered, the mental guilt will continually hurt him, unless he has absolutely no sense of justice and only cares for gratification. Still, he won't reach happiness, because the million dollars he earned comes from mere luck, so whether or not the rest of the world recognizes it as a fruit of talent, the memory of cheating will always be locked inside him. Hence, it's like already taking the test in a confined room. And being unable to leave until you fail or confess.

yeah, i think you're being FAR too idealistic here.

i'm not arguing that a cheating history doesn't "fade away" in the sense that yes, you never really learned those things that you cheated through in HS/college. On the other hand, most people do NOT care about the fact taht they cheated on something in school, but they do care about how much cash they make and how successful they are at that. Moreover, most people don't get any mental guilt from cheating. They're just trying to get through life reasonably well and your idealistic notion of trying to learn things for the sake of learning is somethign that usually doesn't cross their mind.

Yeah, something very similar happened to me... I cannot even say how angry I was at the time.
Now that I have cooled down... I realize how I wasted a lot of my time thinking about the girl who cheated (quite a bit). Who did this affect? Me. It bothered me, not her. She got away with it as usual, and I seemed to be the only one that cared. People are not willing to waste their time and energy into tracking down some cheater... They've got other things to do. Just focus on yourself and do what [i]you[/i] have to do. Or... tell someone that might actually do something about this.
This person cannot go a long way with cheating. It will catch up sooner or later.
You should have the satisfaction of knowing that mostly [i]everyone[/i] [b]knows [/b] him to be a cheater.

I'm finding it hard to believe that:

-A second term senior in an Ivy League school early still cared about extra points in math
-The smartest kid in the school let the well known cheater copy off him; it's extremely easy to make it so someone can't see what you bubbled
-Out of the twenty five answers, he decided to change one
-That one turned out to be a wrong one
-He changed it to the right one
-You are the only one who caught him

I could understand one or two on their own, but all six events happening at the same time suggests more of a grudge by the OP which he already admitted exists.

lol
the sooner you stop caring about marks, the better for you
exam halls have long since turned into pleasant chat rooms for us
some teachers are a bit troublesome, true
but most people here are established mathematicians busy with their own work
and they can hardly be bothered by such petty matters as exam hall decorum

for my part, i can't copy to save my life
there are things I do not care to keep in my head (the poisson mass function for example)
as and when required, I use learned neighbours for reference

some people keep notes and books hidden in the toilet
if necessary, they consult them
i admire their ability, but that is something I can't do
it takes a lot of focus to learn something up in the toilet in ten minutes with exams going on
and determination too
i'm never that fussy about marks

but of course, if two of us can slip out simultaneously, some analysis of solutions always come handy
but the research scholars who invigilate often turn out uncooperative  :(
i wonder what the world is coming to

we are not the competitive types you know
we believe in brotherhood
competition breeds ill feelings like the one demonstrated by the thread starter
and the world could do with a few ill feelings less

i was also a puritan in school and college freshman year, you know
but expositions into much deeper problems of the society somehow changed me
and the final nail was driven home by the Linear Algebra paper of the endsem exams, 2nd semester  :P

yes, that day was born a new me!

Well, not being ultra-competetive would be great. At my university there are very few paid phd student openings, so we're so competitive we don't share solutions with one another unless we're close friends.

[color=green]Note from moderator: there is [i]never[/i] a good reason to block-quote the entire post preceding yours.[/color]

True, but competitiveness often brings out the best in people, a desire to do well is only ignited by other people of a similar mentality, which is why being in classes which are too easy for you is a bad idea.

The girl last year who cheated off everyone is now back. She seems to have a nice nature, but she still has that "cheater" aura around her. I wonder if she's changed, after half of fifth grade in Florida, but cheaters stay cheaters. Don't they? If they aren't able to control the knowledge they have and just cheat all of the time, I'll know because she won't be placed in GT Reading or Math. That's my assumption.

This guy will fail at cheating on the AIME. If he got caught once, that's epic fail.

Yeah, okay. You despise cheaters. So do I. Not much to debate, is there?


If all the denominators are positive, prove that

      $\frac{x-\gamma}{\beta - \alpha} + \frac{x-\alpha}{\beta - \gamma} + \frac{\beta - x}{2x - \alpha - \gamma} \geq \frac{3}{2}$

 [u]Babis[/u]

say that $\beta - x = a$, $x-\gamma = b$ and $x-\alpha = c$

the inequality go to exactly the nesbitt inequality  :)

[quote="e.lopes"]say that $\beta - x = a$. In this you don't know if it is positive or negative. In my solution in this point I make two cases. In one $\beta >x$ and
$\beta <x$ because Nesbitt is only for positive numbers.

BTW, the problem was proposed by Sotiris Louridas, I think.

[b]This is the standard method.[/b]

  Put $\beta - \alpha = k , \beta - \gamma = l , 2x - \alpha - \gamma = m$ .Then

a)	$k , l , m > 0$
b)	$\beta = \frac{ k + l + \alpha + \gamma}{2} , x = \frac{ m + \alpha + \gamma}{2} , \beta - x = \frac{k +l - m}{2}$

c)	$x- \gamma = \frac{ m + \alpha - \gamma}{2} , x- \alpha = \frac{m - \alpha + \gamma}{2}$
d)	$l-k = \alpha - \gamma$
e)	The inequality is equivalent to:

$( \frac{m}{k} + \frac{k}{m} ) +( \frac{m}{l} + \frac{l}{m} ) + \frac{ \alpha - \gamma }{k} + \frac{ \gamma - \alpha }{l} \geq 4$

 or

$( \frac{m}{k} + \frac{k}{m} ) +( \frac{m}{l} + \frac{l}{m} ) + \frac{( \alpha - \gamma)^2 }{k\cdot l} \geq 4$

 But the last is obvious. We have equality only if

$\alpha = \gamma$ and $k = l = m$ , that is if

     $\gamma = \alpha$ and $x = \frac{\alpha + \beta}{2}$ .

 [u]Babis[/u]

Could anyone find another solution ?Mr Stergiou, Nickolas ?The others ?I am waiting


[quote="myyellowducky82"]please don't post. PM an admin.[/quote]That's a really dumb advice. [b]DO NOT[/b] PM us with these requests, we will simply ignore them.

[quote="Dojo"][quote="myyellowducky82"]please don't post. PM an admin.
EDIT: can someone please delete this post??[/quote]
I think you can delete your own post: just click on the little x button by the edit and quote buttons.[/quote]

you have to be the last post to do so

[quote="PowerOfPi"][quote="tcjy"]Maybe you could make a CHANGE USERNAME topic where anyone can post why they want their new username and what it is.  :D[/quote]
The admins don't encourage this, so a CHANGE USERNAME topic would not be good.
[quote="r15s11z55y89w21"]Wait...I don't get what is the difference between PMing and posting. Why is it differentiated?[/quote]
If you post, you can delete it. If you PM, you can't. Also, PM's can be cluttering. If you were supposed to PM admins, then this forum wouldn't have a purpose.[/quote]

It is as easy to delete a PM, if not easier, than to delete a post. I don't understand that first statement. And why did my previous post get marked as spam?

People stop! First, there WERE topics with username changing, and they said it was forbidden, except in special cases (mewto, rofler). But should now everyone change his username?! Put yourself in admin situation, getting a lot of ''can you change my username'' every day...

And, of course, from ''change username'' it went to ''deleting posts''. Instead of putting a link to a topic, or answering in one post, there obviously must be more posts...

@Admins: Lock this, 'cause people will just post more and more problems...

And one general problem: QS&A forum is getting a lot of topics which are useless or not appropriate...


function
advanced fields
advanced fields unsolved

Knowing that the function f is defined on [0,3] and
       f(0)=1 , f(1)=2 , f′(1) = −1, f(3) = f′(3) = 0.

(a) Estimate f(2)  using Hermite intepolation polynom

(b) Estimate de biggest error posibil from point (a) if we know that :
   f ∈ $ C^5$[0, 3] and |$ f^{(5)}$(x)| ≤ M on [0, 3].
That answer show be dependent on M

Thanks a lot

So am i right, a confirmation will really help me a lot,
Thanks


function
induction
algebra
functional equation
algebra proposed

Find all functions $f: \mathbb{Z}\rightarrow\mathbb{Z}$ such that for all $x,y \in \mathbb{Z}$:
\[f(x-y+f(y))=f(x)+f(y).\]

Let $g(x) = f(x)-x,$ and you're (quickly) done ^^.

I will write the solutions with more details.
Using the above subtitution we obtain that
$g(x+g(y))=g(x)+y$  (1). For $x=0$ the last relation becomes $g(g(y))=g(0)+y$
From this we find that $g$ is surjective ,so also f is surjective so there exist
u such that $f(u)=0$ .Setting in the first relation where x,y the u we obtain
$f(f(u))=2f(u)$ so $f(0)=0$ so $g(g(y))=y$ (2) .Now put in (1) where y the g(y)
to have $g(x+y)=g(x)+g(y)$ (3) due to (2) and (3) holds for all $x,y\in Z$ due to the surjectivity of g.
So $g(x)=cx$ where $c=g(1)$ so $f(x)=x+xc'$ . But subtituting this in the first relation gives c'=-1 or c'=1 so the fuctions are $f(x)=2x$ and $f(x)=0$

[quote="silouan"]I
From this we find that $g$ is surjective ,so also f is surjective [/quote]
I don't think it is true

[quote="silouan"]$ g(g(y)) \equal{} g(0) \plus{} y$
From this we find that $ g$ is surjective ,so also f is surjective so there exist
u such that $ f(u) \equal{} 0$ .Setting in the first relation where x,y the u we obtain
$ f(f(u)) \equal{} 2f(u)$ so $ f(0) \equal{} 0$ so $ g(g(y)) \equal{} y$[/quote]
From $ g(g(y))\equal{}g(0)\plus{}y$ we find that $ g$ is injective,hence by taking $ x\equal{}0$ in (1):we obtain that $ g(0)\equal{}0\Rightarrow g(g(y))\equal{}y$.

For $ x\equal{}y$ we have  $ f(f(x))\equal{}2f(x)$(1). For $ x\equal{}y\plus{}1$ $ \equal{}>$  $ f(1\plus{}f(y))\equal{}f(y\plus{}1)\plus{}f(y)$ and by induction obtain $ f(k\plus{}f(x))\equal{}f(x)\plus{}f(x\plus{}k)$(2).Now for $ x\minus{}>f(x)$ and $ y\minus{}>f(y)$ we have  $ f(f(x)\minus{}f(y)\plus{}f(f(y)))\equal{}f(fx))\plus{}f(f(y))$  $ \equal{}>$   $ f(f(x)\plus{}f(y))\equal{}2f(x)\plus{}2f(y)$  (1)  $ \equal{}>$  $ f(x)\plus{}f(f(y)\plus{}x)\equal{}2f(x)\plus{}2f(y)$ (2) $ \equal{}>$ $ f(x)\plus{}f(y)\plus{}f(x\plus{}y)\equal{}2f(x)\plus{}2f(y)$(2) $ \equal{}>$ $ f(x\plus{}y)\equal{}f(x)\plus{}f(y)$  $ \equal{}>$ $ f(x)\equal{}kx$, k-integer. Now we have $ k(x\minus{}y\plus{}f(y))\equal{}k(x\plus{}y)$ and we have $ k\equal{}0$ or $ k\equal{}2$ $ \equal{}>$ $ f(x)\equal{}0$ and $ f(x)\equal{}2x$.

For x=y, we get f(f(x))=2f(x).
For $y \mapsto f(y)$, we have f(f(y)+x)=f(x)+2f(y) (*), and for $x=f(y), 2f(y), ...$ in (*), we get by induction that $f(nf(y))=2nf(y)$, $(\forall n\in \mathbb{N})(\forall y\in \mathbb{Z})$.
Hence, for $x=2y,3y,...$ in the functional equation, and using (*), we obtain by induction f(ny)=nf(y), $(\forall n\in \mathbb{N})(\forall y\in \mathbb{Z})$.
Taking y=1 and y=-1, we have $f(n)=nf(1) (\forall n\in \mathbb{N})$ and $f(n)=-nf(-1) (\forall n\in \mathbb{Z}- \mathbb{N})$
We deduce that f(n)=2n or f(n)=0 after reporting in the functional equation.

[quote="AlexanderMusatov"]By substituting $ x=y $ we have $ f(f(x))=2f(x) $.Define $ a_1=f(x) $, $ a_n=f(a_{n-1})=2 \cdot a_{n-1} $,then $ a_n=a_1 \cdot 2^{n-1},n>1 $.If the function is not injective,let $ a,b $ be numbers such that $ a\neq b $ and $ f(a)=f(b)=c $.By substituting these values into the original equation we have $ f(x-a+c)=f(x)+c=f(x-b+c) $,therefore $ f(x) $ is periodic.However,it is an integer function,and since it is periodic it must be bounded.However $ a_n=a_1 \cdot 2^{k-1} $ is not bounded unless $ f(x)=0 $.Therefore the function is injective and we have $ f(x-y+f(y))=f(y-x+f(x))$ which is equivalent to $ x-y+f(y)=y-x+f(x)\Rightarrow f(x)-2x=f(y)-2y $,thus $ f(x)=2x+c $,where $ c $ is a constant.Plugging into the original equation we find that $ f(x)=2x $.[/quote]

i know it's a stupid question but i still have so some questions about fct.equations...why can't we proove injectivity like this?

suppose $f(f(x))=f(f(y))$

$P(x,x)$ we have $f(f(x))=2f(x)$ thus from our assumption $f(f(x))=2f(x)=2f(y)=f(f(y))$

so from $f(a)=f(b)$ follows $a=b$ with $a=f(x)$ , $b=f(y)$ ?????????????????

[quote="Varal7"][quote="dudeldai"]

i know it's a stupid question but i still have so some questions about fct.equations...why can't we proove injectivity like this?

suppose $f(f(x))=f(f(y))$

$P(x,x)$ we have $f(f(x))=2f(x)$ thus from our assumption $f(f(x))=2f(x)=2f(y)=f(f(y))$

so from $f(a)=f(b)$ follows $a=b$ with $a=f(x)$ , $b=f(y)$ ?????????????????[/quote]

$f: \mathbb{Z}\rightarrow\mathbb{Z}$

What you need to prove, in order to prove injectivity :
$\forall (x,y) \in \mathbb{Z}^2, f(x)=f(y) \Rightarrow x=y$.

What you have proven :
$\forall (x,y) \in (f(\mathbb{Z}))^2, f(x)=f(y) \Rightarrow x=y$.[/quote]

yeah i told you it was a stupid question.. btw thanks :)

Let $p(x,y)\Rightarrow f(x-y+f(y))=f(x)+f(y)$.

$p(x,x)\Rightarrow f(f(x))=2f(x).....(i)$

$p(x+y,y)\Rightarrow f(x+f(y))=f(x+y)+f(y).....(ii)$

$p(x,f(y))\Rightarrow f(x+f(y))=f(x)+2f(y)......(iii)$(By using $(i)$)

So $f(x+y)+f(y)=f(x)+2f(y)\Rightarrow f(x+y)=f(x)+f(y)\; \; \forall x,y\in \mathbb {Z}$

Remember this function is from $\mathbb {Z}$ to $\mathbb {Z}$.

So $f(x)=\zeta x\; \; \forall x\in \mathbb {Z}$ with an integer constant $\zeta $.

By plunging in the main equation we find only $\zeta =0$ and $\zeta =2$ satisfies the equation.So all the functions are $f(x)=0$ and $f(x)=2x$.

From x=y we have f(f(x))=2f(x)
Now when y=f(z), we got f(x-f(z)+f(f(z)))=f(x)+f(f(z)) -> f(x+f(z))=f(x)+2f(z)
When x=x-z -> f(x-z+f(z))=f(x-z)+2f(z) -> f(x)+f(z)=f(x-z)+2f(z) -> f(x)=f(x-z)+f(z)
Therefore f is an additive function -> f(x)=cx where c=const.
From f(x-y+f(y))=f(x)+f(y) ----> c=0 or c=2 :)

Let $f(0)=c$
$x=y$ gives $f(f(0))=2f(0)$ so $f(c)=2c$
If $x=c,y=c$ we get $f(2c)=2f(c)=4c$
If $x=c,y=0$ we get $f(2c)=3c$
So $c=0$ and $f(0)=0$
Now let $g(x) = f(x)-x$ so $g(0)=0$ and $g(x+g(y))=g(x)+y$ and we easily get that $g(x)$ is injective so
$g(g(x))=x$ that gets $g(x)=x$
So $f(x)=2x$ for all $x$.

The only solutions to the equation are $f(x)=2x$ and $f(x)=0$.

Let $P(x,y)$ be the given assertion.
From $P(x,x)$, we get that $f(f(y))=2f(y)$.
From $P(-f(0),0)$ we get that $f(0)=f(-f(0))+f(0) \implies f(-f(0))=0$.
From $P(x,-f(0))$, we get that $f(x+f(0))=f(x)=f(x)+f(0)$.
Thus we must have that $f(0)=0$.

From $P(0,y)$ we get that $f(f(y)-y)=f(y)$.
Now we must have that $f(x+f(y)-y)=f(x)+f(f(y)-y)$.
From $P(x,f(t)-t)$, we get that $f(x+t)=f(x)+f(t)$.
Thus we get that $f$ is additive on $\mathbb{Z}$, i.e. $f(x)=cx$.

When we plug that in $f(f(y))=2f(y)$, we get that $c\in \{ 0,2\}$


group theory
abstract algebra
superior algebra
superior algebra solved

Prove that Q is not a reunion of a finite number of nontrivial subgroups.

Let G1, .., Gn be the n groups which 'cover' Q completely (we assume this CAN be done). We have n>=2 because all G's are different from Q. Then there exist rational numbers p1/q1,.., pn/qn, s.t. pi/qi only belongs to Gi (and not to any other G) for all i from 1 to n (pi are integers and qi are positive integers s.t. (pi,qi)=1). Then the element A=1/(q1*q2*..*qn) can't belong to any of the n groups, because, for example, if it belongs to G1, then pn/qn=pn*q1*q2*..*q(n-1)*A also belongs to G1, but this is false since pn/qn only belongs to Gn.

I'm not entirely sure I was right here. Instead of taking pi/qi s.t. pi/qi is ONLY in Gi (I don't know if that's always possible), we should take pi/qi s.t. pi/qi ISN'T in Gi. This should work now. Sorry for posting it wrong. It's the same from here: E=1/q1*..*qn can't belong to Gi because then pi/qi would belong to Gi (this holds for all i from 1 to n, so the element E isn't in Q at all, which is obviously a contradiction).


[quote="yanglunj"][quote="stephencheng"][quote="Adharsh Raghavan"]easy one...

$ P(x) \equal{} x^2 \plus{} 1$[/quote]

I ask for a system of polynomials (which contains all polynomials satisfying the condition), not a single one.[/quote]

Don't be mean when you are asking politely for solutions from random people over the internet.

I'm pretty sure $ x^2 \plus{} 1$ is the only polynomial solution for this up until degree 9.[/quote]

My question ask for a system of polynomials , the polynomials can be of degree higher than $ 9$ .

May be it will be more appropriate to say ''Determine the set of polynomials satisfying the condition : .....''
rather than to say ''Determine the system of polynomials ...''.

In that case, don't you need Lagrange interpolation?

Well if you expand out $ k(x\minus{}1)(x\minus{}2)...(x\minus{}10)\plus{}x^2\plus{}1$ where $ k$ is either constant or a polynomial in $ x$, then you get all polynomials with that property.

[quote="tenniskidperson3"]Well if you expand out $ k(x \minus{} 1)(x \minus{} 2)...(x \minus{} 10) \plus{} x^2 \plus{} 1$ where $ k$ is either constant or a polynomial in $ x$, then you get all polynomials with that property.[/quote]

Correct.


Determine the three missing letters inclusive of the undernoted series provided hereunder as follows:

A, B, C, E, D, O, F, G, Q, __ , __ , __

Is this the solution?

[hide] J, U, R [/hide]

Not exactly.

HINT:

[hide]Think about the structure of the individual letters.[/hide]

Well this was what I did:

[hide]What I did was a quick sketch of each letter equaling a number like A = 1 B=2 C=3 etc...and worked it up like a pyramid but it was real fast so I might of made some miscalculations.[/hide]

Forge wrote :
[quote]What I did was a quick sketch of each letter equaling a number like A = 1 B=2 C=3 etc.....[/quote]

The word "structure" inclusive of the last hint does not imply that values should be assigned to the letters and the subsequent geometric structure considered.

[u]Hint 2:[/u]

[hide]Think about which letters are made by straight lines, by curved lines and which letters are made by a combination of both.[/hide]

[hide="The pattern is..."]all straight lines, some straight some curvy, all curvy. Chocoboom is right, I am wrong.[/hide]


geometry
vector
parallelogram
cyclic quadrilateral
geometry proposed

Let $ ABCD$ be a cyclic quadrilateral $ M,N,P,Q$ are midpoins of $ AB,BC,CD,DA$

a) Prove that the lines is through $ M,N,P,Q$ which are perpendicualr to $ CD,DA,AB,BC$ concurrent at point $ H$. It called be orthocenter of $ ABCD$.

b) Let $ d$ and $ d'$ be two perpendicular lines through $ H$. $ d$ cuts $ AB,BC,CD,DA,AC,BD$ at $ X,Y,Z,T,U,V$ and $ d'$ cuts $ AB,BC,CD,DA,AC,BD$ at $ X',Y',Z',T',U',V'$. Let $ E,F,K,L,I,J$ be midpoints of $ XX',YY',ZZ',TT',UU',VV'$. Prove that $ EK,FL,IJ$ are concurrent.

Sorry I have a typo and I edited my problem!

for the part (a) we use vectors :
we choose a point $ K$ in the plane.Let the perpendiculars from $ N,Q$ meet at $ S$.
Let the perpendiculars from $ M,P$ meet at $ L$.
we 'll prove that $ \vec{KS}\equal{}\vec{KL}$ which leads to what the problem asked.
Let the perpendiculars from $ M,P$ to $ DC,BA$ meet them at $ F,J$ respectively.
Let $ G$ be the centroid of $ ABCD$.
then... obviously $ MOPL$ is a parallelogram and we have
$ \vec{KL}\equal{}\vec{KM}\plus{}\vec{ML},\vec{KL}\equal{}\vec{KP}\plus{}\vec{PL}$
so we get
$ \implies 2.\vec{KL}\equal{}\vec{KM}\plus{}\vec{ML}\plus{}\vec{KP}\plus{}\vec{PL}\equal{}\frac{1}{2}(\vec{KA}\plus{}\vec{KB})\plus{}\vec{OP}\plus{}\frac{1}{2}(\vec{KC}\plus{}\vec{KD})\plus{}\vec{OM}\equal{}\frac{1}{2}( \vec{KA}\plus{}\vec{KB}\plus{}\vec{KC}\plus{}\vec{KD})\plus{}\frac{1}{2}(\vec{OD}\plus{}\vec{OC}\plus{}\vec{OA}\plus{}\vec{OB})$
and finally
$ \vec{KL}\equal{}\vec{KG}\plus{}\vec{OG}$.
doing the same thing for point $ S$ we'll get:$ \vec{KS}\equal{}\vec{OG}\plus{}\vec{KG}$ so $ \vec{KS}\equal{}\vec{KL}$ as we wanted.