I understand what you want, but at this point you cannot do that, unless you want to completely hide the contents of the Middle School Tournaments section.

[quote="ThinkFlow"]Is there a way to disregard new posts in certain sections? Specifically, I participate in the Mathcounts forum but don't participate in the Middle School Tournaments, and I don't want new posts in the Middle School Tournaments section to make the general Mathcounts forum say that there are new posts. Any help is greatly appreciated.

ThinkFlow

P.S. How do you post screenshots? Can you only post images from a webpage?[/quote]
Sceenshot (I have windows; may be different on macs)
Hit Print Screen and hit Ctrl+C
Open Paint and copy your picture there
edit it however you'd like
SAVE!
Post it by looking at the "add an attachment" section, hit browse, select the image, and there you go!

On mac its command-shift-3, saves to desktop

adding on to my other post...
i don't think you always need to hit Ctrl+C but I do it anyways...just in case.

No, PrintScreen automatically saves the image to the clipboard for you. You should [b]NOT[/b] hit Ctrl+C; if you were happening to have selected some text, the image will be ERASED and replaced with the text that you just copied.


probability
calculus
integration
logarithms
expected value
geometric series

Continuous Probability (might require calculus, I'm not entirely sure on the solution.)

A number x is chosen from the set of all real numbers between 0 and 1, inclusive.
A real number y is also chosen from the same set. If $y \ge x$ then we are done. If $y < x$, then another y value is chosen. Find the expected number of y values we have to choose if we are to get $y \ge x$.

Ignore the fact that exactly 0 or 1 is picked or same x and y are picked (Probabilty is 0). There are only two outcomes that can happen x>y or y<x. But since both are random variables both are equaly likely so prob =.5. expected means 2(.5) =1

Are you sure that's the expected value?
So you can be fairly sure that you can get y > x in only one pick?
I suppose that makes sense, because if half the time X is less that 0.5, and the EV for that is 1, and half the time x > 0.5, so EV is more than than one, but with lower P.

Can anyone corroborate this solution?

Only y changes; x remains fixed instead of them both being chosen randomly each time. if y<x there is a greater chance of x being in [0.5, 1].

I think you may have to factor that in as well.

There's no way the expected value of this can be 1. This means there is a 100% chance of picking a suitable $y$ the first try. This is obviously not true (it's 50%).

[hide="My attempt with calculus near the end"]For a given $x$, we have a chance of $x$ of needing another number, and a $1-x$ chance of not needing another. So we have $E = \sum_{i=0}^{\infty}{x^{i}}$, since we need at least one pick (hence the $i=0$ term), and each time we pick a bad number ($x$ chance) we add another to the expected value. By geometric series formula, $E = \frac{1}{1-x}$.

If you integrate this from 0 to 1, with an improper integral, you get $\infty$ as the answer. This makes me think that there is something conceptually wrong with integrating here, because $\infty$ is a counterintuitive answer. But that's not the first time math has left me scratching my head (Banach-Tarski, anyone?).[/hide]

What if you index shifted this, so that you're taking it from the set 1 to 2?
Would the integral then be from 1 to 2?
Would that let us get $\ln 2-\ln 1$?

I don't think shifting the range of $x$ and $y$ make a difference...But you could try to make it more interesting by making the $x$ only go from 0 to $\frac{1}{2}$ or something.

Well if you did that then

$S=\frac{1}{1-(x-1)}=\frac{1}{2-x}$

integrating it from 1 to 2 is still infinite

If you want to "fix it" (write a problem for which the answer isn't infinite), you pick some $r < 1$ and you say, "Let $x$ be randomly chosen in $[0, r]$ and do the above process.  What is the expected number of steps necessary to choose a $y$ larger than $x$?"

Here's a non-calculus way of showing that the answer is infinite:

If $x$ lies between $1-\frac{1}{n}$ and $1-\frac{1}{n+1}$, then (as per the argument several others have given) we expect to take at least $n$ draws before we get a big-enough $y$.  The likelyhood that $x$ falls into this interval is $\frac{1}{n}-\frac{1}{n+1}= \frac{1}{n(n+1)}$.  So the expected contribution from this interval is $\frac{1}{n+1}$.  If you know enough about the harmonic series, you're done.

Here's a different fix: suppose that we still choose $x$ randomly but not uniformly, so that the probability that $x$ falls in the interval $[r, 1]$ is given by $(1-r)^{2}$.  (In the original problem, this probability is just $1-r$.)   What is the resultant expected value?  (This really will need calculus, I'm afraid.)

Opps! I totaly read this problem wrong. Doing 32 simulations on my graphing calculator confirms what ehehheehee said.


Given $ a, b, c \geq\ 0.$ Prove that: $ (a^4b^4\plus{}b^4c^4\plus{}c^4a^4)(a\plus{}b\plus{}c) \geq\ abc(ab^2\plus{}bc^2\plus{}ca^2)^2$
 :)

[quote="nguoivn"]Given $ a, b, c \geq\ 0.$ Prove that: $ (a^4b^4 \plus{} b^4c^4 \plus{} c^4a^4)(a \plus{} b \plus{} c) \geq\ abc(ab^2 \plus{} bc^2 \plus{} ca^2)^2$
 :)[/quote]
By Cauchy Schwarz Inequality, we have
\[ (ab^2\plus{}bc^2\plus{}ca^2)^2 \le (a^2b^3\plus{}b^2c^3\plus{}c^2a^3)(a\plus{}b\plus{}c)\]
It suffices to prove that
\[ a^4b^4\plus{}b^4c^4\plus{}c^4a^4 \ge abc(a^2b^3\plus{}b^2c^3\plus{}c^2a^3)\]
Put $ a\equal{}\frac{1}{x},b\equal{}\frac{1}{y},c\equal{}\frac{1}{z}$, then this inequality becomes
\[ \frac{x^4\plus{}y^4\plus{}z^4}{x^4y^4z^4} \ge \frac{1}{xyz} ( \frac{1}{x^2y^3}\plus{}\frac{1}{y^2z^3}\plus{}\frac{1}{z^2x^3})\]
\[ \Leftrightarrow x^4\plus{}y^4\plus{}z^4 \ge x^3y\plus{}y^3z\plus{}z^3x\]
which is obviously true by AM-GM.
Done!
:)

[quote="nguoivn"]The stronger is also trues and easy  :)
$ (a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2)^2(a \plus{} b \plus{} c) \geq\ 3abc(ab^2 \plus{} bc^2 \plus{} ca^2)^2$[/quote]
Yes, it deduce from the VasC's Inequality
\[ (x^2\plus{}y^2\plus{}z^2)^2 \ge 3(x^3y\plus{}y^3z\plus{}z^3x)\]
;)

Yes, it's the way I created it. It's a small joke  :)


Suppose $ \sum_{n \equal{} 1}^{\plus{}\infty} a_n$ converges and $ \{b_n\}$ is monotonic and bounded.

Prove that $ \sum_{n \equal{} 1}^{\plus{}\infty} a_nb_n$ converges.

It does looks like one of those many Abel-like propositions,
but I couldn't figure it out.

Hint: try using summation by parts. Sum the $ a_n$ and difference the $ b_n.$ The partial sums of the $ a_n$ are bounded and the differences of the $ b_n$ form an absolutely convergent series.

I don't think you need to use summation by parts.

\[ \left|\sum^n_{k \equal{} 1}b_ka_k \minus{} \sum^m_{k \equal{} 1}b_ka_k\right| \equal{} \left|\sum^n_{k \equal{} m}b_ka_k\right| \le \sum^n_{k \equal{} m}|b_ka_k| \le \ldots\]

Keeping in mind that $ (b_k)$ is bounded, the rest should fall into place. My only nagging worry is that I never used the fact that $ (b_k)$ is monotonic, so something tells me the solution I came up with wasn't the intended one.

$ \sum_{k \equal{} m}^{n} | a_k b_k |$ need not be bounded, so you cannot apply Cauchy criterion in that way. See the example $ a_n \equal{} \frac {( \minus{} 1)^n}{n}$ and $ b_n \equal{} 1$.


inequalities
limit
inequalities unsolved

[quote="Werwulff"][quote="crax"]$ a,b $  and $c$ are reals.
prove that

$ \frac{a^2}{a+bc}+\frac{b^2}{b+ac}+\frac{c^2}{c+ab} <= 2 $
thanx[/quote]
This Inequality is definitly wrong!
\[\lim_{c\rightarrow\infty} LHS=\infty \not\leq 2\][/quote]
i thought that it was wrong . i just didn't  know why

Of course this inequality is wrong.Crax,have tried to find its minimum. Just apply Cauchy and see what happens.

Isn't it obvious that minimum is minus infinity for reals $a,b,c$. ANd if you restrict to positive numbers, then it is surely 0, so it's useless to try Cauchy. ;)


We are given a n-gon in the plane. Choose a permutation t of {1;2;...;n-1}.
Now in step i, we will construct isosceles triangles with apex angels
$\frac {2 t ( i ) \pi }{n} $ on the sides of the (i-1)th polygon ( the 0-th polygon is our initial n-gon), and then the i-th polygon is the n-gon ( which may degenerate )formed by the free apices of the triangles (Note that the apex angles are ordered angles) . Prove that for all permutations t, we will always result in a single point, and the (n-2)-th polygon is a regular n-gon.

So an isosceles triangle is constructed so that it runs through the interior of the polygon?

The apex of an isosceles triangle is the common endpoint of the two equal sides.

I have smth like a solution but I'm not sure.
I get we reach one point at step (n-1). This proves that step (n-2) yields a regular polygon.
We use complex numbers.
To z_i the roots of unity assign the number w_i=z_i/(z_i+z_i') (z' its conjugate)
Then one can see a_1 will change to w_t(1)'a_1+w_t(1)'a_2 and so on.
Now one can easily prove at n-1 step we get
w_1..w_n*(a1*(z_1'/z_1+..)+a2*(z_1'/z1*z_2'/z2)+....) which is obviously simmetric in a_1..a_n so result


Anyone know how to get things like $ \boxed{\text{answer}}$, $ \underline{\text{answer}}$, et cetra, in Microsoft word using the equation editor [i]not[/i] importing the $ \text{\LaTeX}$ image?

Well Word 07 has more features but I'm not sure if it can do those things.  You might have to draw in the textbox yourself.

For the underline, you can just hit Ctrl + U.

For the box, go to the Autoshapes and draw a box around your answer.

IME the boxes don't usually come out that great on MS Word.

Also, it is so much more fun to export pdfs using actual LaTeX.

well..
my word can do a box around selected text..
but it looks really ugly if you have 2 or more lines...

A quick google search reveals a list of coverters [url=http://www.tug.org/utilities/texconv/textopc.html]here.[/url] I know that meenamathgirl used [url=http://elevatorlady.ca/]Aurora[/url] for a science paper, but  it's a free trial (if you want to use it any longer, it will set you back 35-45 dollars).


Let $ a,b,c\geq0$ s.t. $ ab\plus{}bc\plus{}ca\equal{}3$.Prove that:
$ \dfrac{a}{\sqrt{5a\plus{}4b}}\plus{}\dfrac{b}{\sqrt{5b\plus{}4c}}\plus{}\dfrac{c}{\sqrt{5c\plus{}4a}}\geq1$

I see no solution after 5 months...
Is this problem so hard?

Ok, but I think Ji Chen's solution is very hard.

If anyone has another solution, please post it here.


hello
please help me understand this question
thanx

What is the probability that a five-card poker hand has the following?

a) A straight (a set of five consecutive values with the initial card value between Ace and 9)

10-J-Q-K-A can also be a straight, BTW.  Aces can be either high or low.

10* 4^5= 10240 NOT 10200?

can you confirm the results please
thanx

[quote="lingomaniac88"][hide]Suppose the bottom card is an ace.  Then, allowing for rearrangement of cards, there are $4^{5}$ ways to have a straight.  However, we must subtract 4 ways (where the suits are the same), for that will result in a straight flush.  So that means 1020 ways.  Multiply by 10 for each of the other possible low card for 10200 ways.
Now divide that by $\binom{52}{5}$ to get $\frac{10200}{2598960}$ or $\boxed{\frac{5}{1274}}$.[/hide][/quote]

huh? isn't a straight flush still count as a straight? isn't it just a special form of a straight? Just like a square, it can say as a rectangle. I don't know, I may be wrong.

And.. for T-J-Q-K-A, the initial card is a T? There are 10 cases of straights

So.. $\frac{10*4^{5}}{\binom{52}{5}}=\boxed{\frac{128}{32487}}$

I am not sure.. please criticize if you spot some error  :D


inequalities
logarithms
Irrational numbers

Suppose that $p, q \in \mathbb{N}$ satisfy the inequality \[\exp(1)\cdot( \sqrt{p+q}-\sqrt{q})^{2}<1.\] Show that $\ln \left(1+\frac{p}{q}\right)$ is irrational.

There's also a double reference here, is this correct?


In a group of cows and chickens, the number of legs is $ 14$ more than twice the number of heads.  How many cows are there in the group?

[hide="Solution"]
Setting up an equation (c=cow and h=chicken):

$ 2(c\plus{}h)\plus{}14\equal{}4c\plus{}2h$

$ 2c\plus{}2h\plus{}14\equal{}4c\plus{}2h$

$ 14\equal{}2c$

$ c\equal{}7$

$ Answer\equal{}7$
[/hide]


Find all values of the positive integer $ m$ such that there exists polynomials $ P(x),Q(x),R(x,y)$ with real coefficient satisfying the condition: For every real numbers $ a,b$ which satisfying $ a^m-b^2=0$, we always have that $ P(R(a,b))=a$ and $ Q(R(a,b))=b$.

suppose the $ m\equal{}2k$.put $ R(x,x^k)\equal{}L(x)$
we have $ P(R(x,x^k))\equal{}x$ so as we know that $ P,R$ are not constant so we have $ deg(P)\equal{}deg(L)\equal{}1$
if we put $ L(x)\equal{}xu\plus{}v$.we have$ p(x)\equal{}\frac{x\minus{}v}{u}$ and also $ Q(x)\equal{}P(x)^k$ so we have $ b\equal{}a^k$ but b can also be$ \minus{}a^k$

for $ m\equal{}2k\plus{}1$.put $ R(x^2,x^m)\equal{}L(x)$
we have  $ P(R(x^2,x^m))\equal{}x^2$.
 if $ deg(L)\equal{}2$.as we have $ Q(L(x))\equal{}x^m$ and $ m$ is odd so contradicts .
so $ deg(L)\equal{}1$ but in  $ R(x^2,x^m)$ the minimum power of x is $ min(2,m)$ so $ m\equal{}1$
and for $ m\equal{}1$
$ R(a,b)\equal{}b$ and $ P(a)\equal{}a^2,Q(a)\equal{}a$


How did you guys do? I came in first in my county and second in the eastern New York section with a 54. Time to start studying Atkins.  :D

I got around a 57, but I think I may not have qualified, because there can be only two people from every school, and there may have been two that scored high so, yeah.


function
algebra
functional equation
algebra unsolved

Find all functions $f: Q\to Q$ such that:

$f(f(x)+y)=f(x)+f(y) \forall x;y\in Q$

Are you sure you haven't missed a condition?
There are at least two classes of functions which satisfies the functional equation:
 1). $f(x)=x+a, \forall x\in Q$;
 2). $f(x)=[x+b], \forall x\in Q$
(where $a, b\in Q$)
If $f$ had been injective or surjective we would have had only the first class.
Indeed, let $a=f(o)$
For $y=0$ we get
(*) $f(f(x))=f(x)+a, \forall x\in Q$
For $x=0$ we get $f(a+y)=a+f(y)$, so $f(a+x)=a+f(x), \forall x\in Q$
Then $f(f(x))=f(x+a)$.
If f is injective it follows that $f(x)=x+a$.
If f is surjective, for each $t$ in $Q$ there is an $x$ in $Q$ such that $t=f(x)$ and (*) yields to $f(t)=t+a$..


superior algebra
superior algebra unsolved

Let $ K$ an infinite field and $ K\leq K(\alpha)$ an algebraic extension. Prove that every extension $ E$ such that $ K\leq E\leq K(a)$, is simple, which means that $ E\equal{}K(b)$, for some $ b\in E$.

Alexandros

finite simple extensions are those which are algebraic and have a finite number of intermediate fields.


Express as a common fraction in simplest form:
\[ \frac{2 \plus{} 4 \plus{} 6 \plus{} \ldots \plus{} 50}{3 \plus{} 6 \plus{} 9 \plus{} \ldots \plus{} 75}\]

[quote="GameBot"]Express as a common fraction in simplest form:
\[ \frac {2 \plus{} 4 \plus{} 6 \plus{} \ldots \plus{} 50}{3 \plus{} 6 \plus{} 9 \plus{} \ldots \plus{} 75}
\]
[/quote]

Note that the numerator and denominator are both arithmetic sequences. In the first, there are $ 50 \minus{} 2 \equal{} 48 \implies 48 \div 2 \equal{} 24 \implies 24 \plus{} 1 \equal{} 25$ terms, with a common difference of 2, and in the second, there are $ 75 \minus{} 3 \equal{} 72 \implies 72 \div 3 \equal{} 24 \implies 24 \plus{} 1 \equal{} 25$ terms, with a common difference of 3.

The first sequence has a sum of $ (50 \plus{} 2) \times 25 \equal{} 52 \times 25 \equal{} 1300$, while the second sequence has a sum of $ (75 \plus{} 3) \times 25 \equal{} 78 \times 25 \equal{} 1950$. Thus our answer is $ \frac {1300}{1950} \implies \boxed{\frac {26}{39}}$.

EDIT: Ah yes, because generally $ 13 \times 3 \equal{} 39$, not $ 14 \times 3$. Thanks.

EDIT 2: Yeah. That too.

[quote="GameBot"]Express as a common fraction in simplest form:
\[ \frac {2 \plus{} 4 \plus{} 6 \plus{} \ldots \plus{} 50}{3 \plus{} 6 \plus{} 9 \plus{} \ldots \plus{} 75}
\]
[/quote]

I don't understand why we can't just do $ \frac{2(1\plus{}2\plus{}\hdots\plus{}25)}{3(1\plus{}2\plus{}\hdots\plus{}25)}\equal{}\frac{2}{3}$...  :huh:


Let a,b,c are three nonegative numbers satisfy: $ a^2\plus{}b^2\plus{}c^2\equal{}1$
Prove that: $ \frac{a}{1\plus{}bc}\plus{}\frac{b}{1\plus{}ca}\plus{}\frac{c}{1\plus{}ab} \geq 1$

[quote="dulldrake"]Let a,b,c are three nonegative numbers satisfy: $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 1$
Prove that: $ \frac {a}{1 \plus{} bc} \plus{} \frac {b}{1 \plus{} ca} \plus{} \frac {c}{1 \plus{} ab} \geq 1$[/quote]
$ \sum_{cyc}\frac {a}{1 \plus{} bc}\geq\sum_{cyc}\frac {2a}{2 \plus{} b^2 \plus{} c^2} \equal{} 1 \plus{} \sum_{cyc}\left(\frac {2a}{3 \minus{} a^2} \minus{} a^2\right) \equal{}$
$ \equal{} 1 \plus{} \sum_{cyc}\frac {a(a \plus{} 2)(a \minus{} 1)^2}{3 \minus{} a^2}\geq1.$

[quote="Toan_VN_LC"][quote="arqady"][quote="dulldrake"]Let a,b,c are three nonegative numbers satisfy: $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 1$
Prove that: $ \frac {a}{1 \plus{} bc} \plus{} \frac {b}{1 \plus{} ca} \plus{} \frac {c}{1 \plus{} ab} \geq 1$[/quote]
$ \sum_{cyc}\frac {1}{1 \plus{} bc}\geq\sum_{cyc}\frac {2a}{2 \plus{} b^2 \plus{} c^2} \equal{} 1 \plus{} \sum_{cyc}\left(\frac {2a}{3 \minus{} a^2} \minus{} a^2\right) \equal{}$
$ \equal{} 1 \plus{} \sum_{cyc}\frac {a(a \plus{} 2)(a \minus{} 1)^2}{3 \minus{} a^2}\geq1.$[/quote]
$ \sum \frac {a}{1 \plus{} bc} \ne \sum\frac {1}{1 \plus{} bc}$[/quote]
$ \sum_{cyc}\frac {a}{1\plus{}bc} \ge \sum_{cyc}\frac {2a}{2\plus{}b^2\plus{}c^2}$.....
The solution of arqady is true :wink:

[quote="Toan_VN_LC"][quote="arqady"][quote="dulldrake"]Let a,b,c are three nonegative numbers satisfy: $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 1$
Prove that: $ \frac {a}{1 \plus{} bc} \plus{} \frac {b}{1 \plus{} ca} \plus{} \frac {c}{1 \plus{} ab} \geq 1$[/quote]
$ \sum_{cyc}\frac {1}{1 \plus{} bc}\geq\sum_{cyc}\frac {2a}{2 \plus{} b^2 \plus{} c^2} \equal{} 1 \plus{} \sum_{cyc}\left(\frac {2a}{3 \minus{} a^2} \minus{} a^2\right) \equal{}$
$ \equal{} 1 \plus{} \sum_{cyc}\frac {a(a \plus{} 2)(a \minus{} 1)^2}{3 \minus{} a^2}\geq1.$[/quote]
$ \sum \frac {a}{1 \plus{} bc} \ne \sum\frac {1}{1 \plus{} bc}$[/quote]
It should be $ \sum_{cyc}\frac {a}{1 \plus{} bc}\geq\sum_{cyc}\frac {2a}{2 \plus{} b^2 \plus{} c^2}.$ I have fixed it. Thank you!

multiply numerator and denominator by a,b,c  to get
$ \frac{a^2}{a\plus{}abc} \plus{}\frac{b^2}{b\plus{}abc}\plus{}\frac{c^2}{c\plus{}abc}$ $ \ge$ 1
then use the following lemma

for reals a,b,c ,x,y,z and for x,y,z $ \ge$ 0
$ \frac{a^2}{x} \plus{}\frac{b^2}{y}\plus{}\frac{c^2}{z}$ $ \ge$ $ \frac {(a\plus{}b\plus{}c ^2)}{x\plus{}y\plus{}z}$
to simplify and get
$ a^2\plus{}b^2\plus{}c^2$ $ \ge$ $ ab\plus{}bc\plus{}ca$  and we r done

[quote="quykhtn-qa1"][quote="sophie germain"]multiply numerator and denominator by a,b,c  to get
$ \frac {a^2}{a \plus{} abc} \plus{} \frac {b^2}{b \plus{} abc} \plus{} \frac {c^2}{c \plus{} abc}$ $ \ge$ 1
then use the following lemma

for reals a,b,c ,x,y,z and for x,y,z $ \ge$ 0
$ \frac {a^2}{x} \plus{} \frac {b^2}{y} \plus{} \frac {c^2}{z}$ $ \ge$ $ \frac {(a \plus{} b \plus{} c ^2)}{x \plus{} y \plus{} z}$
to simplify and get
$ a^2 \plus{} b^2 \plus{} c^2$ $ \ge$ $ ab \plus{} bc \plus{} ca$  and we r done[/quote]
Are you sure? :wink: Here is my solution:
We have: $ a \plus{} abc \le a \plus{} \frac {a(b^2 \plus{} c^2)}{2} \equal{} a \plus{} \frac {a(1 \minus{} a^2)}{2} \equal{} 1 \minus{} \frac {(a \minus{} 1)^2(a \plus{} 2)}{2} \le 1$
Then $ LHS \equal{} \sum \ \frac {a^2}{a \plus{} abc} \ge \sum \ a^2 \equal{} 1$.
Then,we have Q.E.D[/quote]
So when the equality occour?
$ a\equal{}b\equal{}c\equal{}1 ?$

Equality occurs when $ \left(a, \, b, \, c \right) \in \left\{ \left(1, \, 0, \, 0 \right), \, \left(0, \, 1, \, 0 \right), \, \left(0, \, 0, \, 1 \right)\right\}$.

why? whats wrong wid it? :maybe:
if u r talking abt the lemma? yeah im sure of it

[quote="dulldrake"][quote="quykhtn-qa1"][quote="sophie germain"]multiply numerator and denominator by a,b,c  to get
$ \frac {a^2}{a \plus{} abc} \plus{} \frac {b^2}{b \plus{} abc} \plus{} \frac {c^2}{c \plus{} abc}$ $ \ge$ 1
then use the following lemma

for reals a,b,c ,x,y,z and for x,y,z $ \ge$ 0
$ \frac {a^2}{x} \plus{} \frac {b^2}{y} \plus{} \frac {c^2}{z}$ $ \ge$ $ \frac {(a \plus{} b \plus{} c ^2)}{x \plus{} y \plus{} z}$
to simplify and get
$ a^2 \plus{} b^2 \plus{} c^2$ $ \ge$ $ ab \plus{} bc \plus{} ca$  and we r done[/quote]
Are you sure? :wink: Here is my solution:
We have: $ a \plus{} abc \le a \plus{} \frac {a(b^2 \plus{} c^2)}{2} \equal{} a \plus{} \frac {a(1 \minus{} a^2)}{2} \equal{} 1 \minus{} \frac {(a \minus{} 1)^2(a \plus{} 2)}{2} \le 1$
Then $ LHS \equal{} \sum \ \frac {a^2}{a \plus{} abc} \ge \sum \ a^2 \equal{} 1$.
Then,we have Q.E.D[/quote]
So when the equality occour?
$ a \equal{} b \equal{} c \equal{} 1 ?$[/quote]
If $ a\equal{}0$, or $ b\equal{}0$,or$ c\equal{}0$,we have Q.E.D
If $ a,b,c > 0$ then $ LHS >1$ :)


[quote="G-UNIT"]I find it uttely aggravating these inconsiderate and ill-mannered group of individuals plaugue our poor and sincere forums with satirical wit that is the cause for extensive pain and suffering.  May their souls find eternal restlessness in the quintic equations that await us on the other side.[/quote]

I'd add G-Unit to your list. :lol:

samus - if their rudeness becomes a personal concern, notify the appropriate moderators for post removal. Remember to provide links when you do so. Also, if certain individuals have a continuous and notable trend of ridicule, I'm sure Valentin would like to hear about it.

And G - UNIT, be nice. -.-

I don't think anyone in AMC forums have actually posted anything offensive though

AHHH I AM NOT INCLUDED ON THE LIST I AM SO SAD

[quote="samus"]it seems like AoPS is filled with way too much sarcastic people, especially in the AMC forum. (yifman, probability, mysticterminator,bubala, among others). Personally, I think its annoying, and tiring. Sick of never seeing straight replies or anything. How about y'all?[/quote]

I didn't find any sarcasm on the AMC forum (although some of them do plenty of wierd things in the games and fun factory).

EDIT: nvm, I just found plenty of unnecessary speech

[quote="yif man12"]All right, first! :winner_first:

I'd like to thank the academy, my 7th grade mathcounts coach for introducing me to AoPS, all the sarcastic people in my life, and everybody who has supported me in this journey, thank you so much![/quote]

You probably need to specifically thank your friend who improved from 75 on AMC10 in December to 150 AMC12/14AIME/42USAMO :lol: .

Alright still yifman you need to teach me how to be more sarcastic so I can be on the list in future as being formally recognized to be the "champion" in AoPS premiere sarcasm competition.

[quote="beta"]I don't think anyone in AMC forums have actually posted anything offensive though[/quote]

samus did not bring this up because people were saying things that were blatantly offensive to certain people. Instead, he is saying that all the sarcasm is getting annoying.  Especially the sarcasm where only the people who know the poster personally will understand.  I can usually figure out when someone's being facetious.  But I imagine it would annoy other people.  I don't think all the sarcasm is necessarily that bad just as long as at some point someone explains the joke.

Yeah, just try to be respectful of others and everything will be fine  :)

Darn you yifman.. I guess I must settle for a mere  :winner_second:

Seriously though, I usually try to make sure my posts have some explicit content if sarcasm is there (I can only think of one recent instance where I slipped here). I don't think I should have to mark off what is sarcasm and what isn't. And also, of all the people you listed, I don't think any of them really post sarcastically that often, especially the bubala. But anyways very sorry if I've made your AoPS experience miserable.

The only topic that I find bubala post sarcastic comment is

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=65275[/url], where almost everyone else on the list except yifman (and not on the list =>me) has posted some thing sarcastic.

Yifman's friend won him the title though.

If you want to find some sarcastic comments by bubala look in the AHSIMC forum :D or probably in his blog.

Don't cut out the sarcasm completely (we need some unintelligeble mathy humor every so often :P ) but be mindful when it appears that other people seem to be left out.

[quote="beta"][quote="joml88"]If you want to find some sarcastic comments by bubala look in the AHSIMC forum :D or probably in his blog.
[/quote]

hahaha AHSIMC forum... how many recent posts there are not sarcastic nor offensive?

Sarcasm about AHSIMC is perfectly okay.[/quote]

Yes, sarcasm THERE is definitely allowed. :D In fact AHSIMC =  :spam:

But oh wait...I'm supposed to be nice.  :spider: *cough*

[quote="MithsApprentice"]

Yes, sarcasm THERE is definitely allowed. :D In fact AHSIMC =  :spam:

But oh wait...I'm supposed to be nice.  :spider: *cough*[/quote]

Ah spamming right now is unnecessary since that Metrock guy is gone for awhile(hopefully he will never be back), but if he ever comes back, then we'll go spam /make sarcastic comments.

  ;) spamming/sarcasm/ is not only allowed in AHSIMC but encouraged lol.

Why are we still bitter about the iTest? It was free and it was how long ago? Who cares anymore? And do we really not want him to go away and not come back?

If he takes our advice seriously, there wouldn't be so many protests. :)

Yes but um...be nice for now guys.

[quote="solafidefarms"]And do we really not want him to go away and not come back?[/quote]

yes that leaves us more time to spam fagjr

I want to say that some threads and posts are a whole bunch funnier when you know what they're talking about. As an example, the mock aime iii thread isn't that funny if you don't know about AHSIMC. Then it's hilarious!

While we're on the topic, check out the conflicting quotes at [url]http://en.wikiquote.org/wiki/Sarcasm[/url].

IMO, sarcasm is only appropriate if the audience understands it. It's up to the poster to determine that. And a note to anyone that feels confused: just ask a question. I found out about AHSIMC when I asked, and then it was funny to read about winning an ipod in the Mock USAMO. ;)


Hello!
  I was wondering if anybody happened to know why the term trace is used for the intersection of a space and a subspace?  If not, would you have any ideas on where I could look to find this info?
Thank you,
NICK

A lot of such information can be found [url=http://members.aol.com/jeff570/mathword.html]here[/url], but not what you are looking for. To me, this usage of "trace" seems self-explanatory: when a surface intersects a plane, it leaves a trace in it, i.e. the curve of intersection.


What was everyone wearing? Here are the colors from Las Vegas:

Colorado: purple
DSW: red (with collars)
Macau: pink
Philippines: black
NorCal: lime green
Oregon: white with black sleeves
Roseburg: black
San Diego: black
SFBA: purple
SoCal: blue
Utah: light blue
Washington: purple

Some of the ones I remember from Penn State:

AAST: Black
Chesapeake: Pink
Fairfax Math Circle: Red
Howard County: Pink
Lehigh Valley: Black
Maine: White
NYC: Purple
Ontario West: Light blue
Phillips Exeter: Red
Upstate New York: Pink

Iowa site : [the ones I remember - possibly wrong]:

Chicago: Yellow
Indiana: Yellow
Wisconsin: Red
Texas: Dusty Pink -ish?
Iowa : Dark Green
Michigan: Blue
Michigan ICAE : White

Connecticut: Purple/Orange Tie-Dye
TJ: Dark Blue

Georgia site:

Florida - Red
North Carolina - Pink
South Carolina - Also pink
Tennessee - Also red
Kentucky - Orange (?)
Georgia - White

From Georgia:

North Carolina: Pink
South Carolina: Pink (way too much pink!)
Kentucky: gold
Florida: red (I think)
Georgia: white

I'm attaching the pdf for the Georgia shirt. I rather like it this year. The basic design was created by Steve Sigur. Steve passed away last summer after many years of service to the Georgia team.

Tom

Iowa:

Minnesota: light blue
Missouri: green

PSU:
West Virginia: White

Although we weren't the only ones wearing white if I remember correctly.

To pianoforte:
Minnesota would be [i]cornflower[/i].

We did that since we didn't want wayzata colors (blue and gold).

btw, wayzata=26+ of our team (inside joke?)

[quote="AIME15"]SFBA was [b]supposed[/b] to be "dark blue"...but they quite failed that.[/quote]

:( Dark blue would have been nice. I would have liked something cooler than Gauss' Line on the back too (but I was never one for knowing random collinearities).

My deepest
apologies to MN

I have soiled your shirt with my
inaccuracies.

You MEANIE!

Are you from Missouri?  (the other one you said?)


[quote="SUPERMAN2"]Prove that for any acute triangle $ ABC$,we have:
$ cosA + cosB + cosC \geq 1 + 4cosAcosBcosC$[/quote]

It is true for all triangles :

$ cos A +cos B +cos C = 1 +\frac{r}{R}$

$ 4cosAcosBcosC= \frac{p^2-(2R+r)^2}{R^2}$

$ \Leftrightarrow p^2 \le 4R^2+5Rr +r^2$ which is true $  ( p^2 \le 4R^2+4Rr+3r^2$ and $ 2r \le R )$

I think we also can substitute $ cosA\equal{}\frac{b^2\plus{}c^2\minus{}a^2}{2bc},cos B\equal{}\frac{c^2\plus{}a^2\minus{}b^2}{2ca},cos C\equal{}\frac{a^2\plus{}b^2\minus{}c^2}{2ab}$
Then expand the ineq.

[quote="SUPERMAN2"]Prove that for any acute triangle $ ABC$,we have:
$ cosA \plus{} cosB \plus{} cosC \geq 1 \plus{} 4cosAcosBcosC$[/quote]
I think this ineq is old and very easy! :P
We have $ cosA\plus{}cosB\plus{}cosC\equal{}1\plus{}4sin{A/2}sin{B/2}sin{C/2}$
And $ cosAcosBcosC \leq sin{A/2}sin{B/2}sin{C/2}$

[quote="arqady"][quote="SUPERMAN2"]Prove that for any acute triangle $ ABC$,we have:
$ cosA \plus{} cosB \plus{} cosC \geq 1 \plus{} 4cosAcosBcosC$[/quote]
It's true for all triangle.  :wink:[/quote]
yes,my sulotion is true for all triangle! :P


The dates for the MATHCOUNTS school, chapter, state, and national competitions can be found [url=http://www.mathcounts.org/webarticles/anmviewer.asp?a=185&z=53]here[/url].

Until each competition is completely passed, students should not share any information about the competitions beyond their place.  No score information should be discussed and problems should not be posted.

[quote="biffanddoc"]According to the pdf file, the last chapters were 2/25.
Does that mean we can discuss them?[/quote]

No!  Not until MATHCOUNTS says you can.  The national office emailed us and said that it's still possible chapters could be making up contests late due to weather conditions.  This is why the go ahead to post must come from MATHCOUNTS.

[quote="biffanddoc"]I just noticed that the tests are posted online at mathcounts.org
Can we discuss it now?[/quote]

If they've posted, there's not reason you can't discuss it.  Have fun.


Here's the diagram and problem. Good Luck.


This shape is built from six unit cubes. How many square units are in the surface area of the shape, where the solid lines indicate the front face of the cubes?
[asy]size(150);
defaultpen(linewidth(0.9));
void drawCube(pair A){
 pair B = (A.x + 1,A.y), C = (A.x + 1, A.y + 1), D = (A.x,A.y + 1);
 draw(A--B--C--D--cycle);
 pair A2 = (A.x + 0.6,A.y + 0.3), B2 = (B.x + 0.6, B.y + 0.3), C2 = (C.x + 0.6, C.y + 0.3), D2 = (D.x + 0.6, D.y + 0.3);
 draw(A2--B2--C2--D2--cycle,dashed);
 draw(A--A2,dashed); draw(B--B2,dashed); draw(C--C2,dashed); draw(D--D2,dashed);
}
drawCube((0,0));
drawCube((1,0));
drawCube((2,0));
drawCube((0,-1));
drawCube((2,-1));
drawCube((1,1));[/asy]

If they didn't touch this would be $ 6 \times 6\equal{}36$.  Each time they touch we subtract $ 2$.  They touch $ 5$ times, so the answer is $ 36\minus{}2(5)\equal{}\boxed{26}$.

Or you can just count the number of faces. :)

[quote= when u see the thing it is 6x6=36 and when u touch it u subtract 2 and u move it 5 times so the equation is 36 - 2(5)


inequalities
geometry
complex numbers
graph theory
\/closed

I've noticed that WOOT is the main program to practice for olympiads. However, I'm wondering whether I should consider taking olympiad courses (ie inequalities, problem solving, geometry) first so that I have some of the helpful background for olympiads. So, are the olympiad courses highly recommended when preparing for olympiads? Or is just simply WOOT a better idea?

I've also been thinking that taking olympiad courses would teach some aspects of classical mathematics that would be useful later on in life as well as helping in olympiads, whereas WOOT would just simply give proof-writing/olympiad practice.

[quote="ffdbzathf"]I believe that the olympiad classes are being discontinued due to WOOT.[/quote]I'm not sure where you got this information -- we are still planning to have Olympiad classes in the fall.  We're not offering them in the summer because, in the past, summer Olympiad enrollment has been very low.

The Olympiad classes are probably at a higher difficulty level than WOOT (especially the Olympiad subject classes), so if anything I would recommend taking WOOT first, then taking an Olympiad-level class.  The main advantage of WOOT is you get to see lots more subjects (though not at the depth that a 12-week subject class can give) and you get several practice tests.

[quote="DPatrick"][quote="ffdbzathf"]I believe that the olympiad classes are being discontinued due to WOOT.[/quote]I'm not sure where you got this information -- we are still planning to have Olympiad classes in the fall.  We're not offering them in the summer because, in the past, summer Olympiad enrollment has been very low.
[/quote]

I think ffdbzathf may have saw that in this post (in one of the threads in this subforum):

[quote="rrusczyk"][quote="1234567890"]Is AoPS going to shut down the Olympiad classes and replace them with WOOT? I think you were mentioning that as a possibility before.[/quote]

This is highly likely.  At this point, we expect to end the Olympiad classes and expand/alter WOOT.  The olympiad classes will in part be folded into WOOT.[/quote]

That'd be great if the Olympiad classes are still going to be held. I was really hoping to take one of them when I saw the mention of the Olympiad classes likely being shut down.

[quote="Phelpedo"]Is there any chance that classes, especially WOOT, might be held on weekends? I want to take some courses, but none of the times work for me.[/quote]

You can review transcripts.

EDIT: Read Joe's comments.

I want to take WOOT and or Olympiad Problem Solving.

Well my entire goal is to improve in proof writing.

Is Olympiad Problem Solving more about like, different ways to write proofs and less about how to solve different subject proofs or what?

I think WOOT does like alot of subjects and how to as well.. right?

I'm just wondering if the like 550 dollar difference is worth it lol.

WOOT is not designed to teach about proofs.  Through test feedback you can gain experience with proofs though.  And if you ask on the WOOT message board I'm sure you will get detailed help with proofs.  Proving your results is emphasized in WOOT but I wouldn't say that is the only pupose of the class.  Neither would I say that is the major difference (or even a difference at all!) between WOOT and the other Olympiad classes since everything I said above could equally describe them as well (except replace "tests" with "midterm" and "final").

Phelpedo, WOOT classes are always offered twice a week.  Both times are weekdays though.  Are you certain you can't make at least one of these days?  If you are [i]that[/i] busy that you cannot spare a few hours during the week I doubt you would be able to devote enough time to the class to get the full benefit (though you should still gain a lot).

Saiem, [url=http://www.artofproblemsolving.com/Classes/AoPS_C_ClassesS.php#ps]here's[/url] the description of Olympiad Problem Solving.  As you can see each class emphasizes a different proof method and then there are a few classes which are just entitled "Challenging Problems'.  These have a bunch of problems which can be proved using one or more of the proof methods discussed in the class.  So they serve to solidify your knowledge and give you some experience into how to determine what method you might want to try.  The class doesn't focus on any particular areas such as Algebra, NT, Geometry, etc. but you will get a little bit of each throughout the course.  If you want to get good at geometry, Olympiad Geometry is unequaled around AoPS.

WOOT has a lecture on a different topic every couple weeks.  The topics vary and usually are on specific subjects such as collinearity/concurrency, inequalities, NT, complex numbers, graph theory, and so forth.  The lecture schedule is very likely to change depending on who lectures next year and the preferences of the students enrolled in the course.

BTW, when will the syllabus for next year's WOOT be released?

[quote="calc rulz"]Okay thankyou, it seems like I'm going to take olympiad classes rather than WOOT. Although I may want to practice proof-writing at some point, my main focus right now is problem solving. This year I'll take olympiad courses so as to master those subjects and learn skills I need for olympiads. Then, once I've learned about inequalities, olympiad problem solving, and geometry, and I start to get more serious about olympiads, I'll probably try WOOT. This way I do get the in-depth knowledge of these courses.

Also, WOOT is more expensive and takes more time  ;)  (busy sophmore year)[/quote]

[edit: clarified my post a little]

No, WOOT isn't all about proof writing.  There weren't any lectures dedicated to it.  The tests are the only things that will make you better at writing solutions to problems but they also provide you with the ability to have practice with taking the test under real (or close to real) conditions.  WOOT is definitely not about just proof writing though by all means you should become better at writing proofs since the graders of the tests will be critiquing you and your peers along with the instructors will be giving you feedback on your solutions to board problems (as long as you post solutions).  I've never been to MOP but I'd have to say that WOOT can probably offer a somewhat similar experience (though there are obviously some things that MOP offers that WOOT cannot and vice versa).

[quote="chess64"]Yeah Silverfalcon's comment was kind of misleading...[/quote]

Eh, that's what I thought WOOT was (proof only) heh. BTW, if you knew that I was misleading, why didn't you say that I was misleading (instead of waiting for Joe to do all the talking)?  ;)

For the olympiad classes, are there any recommended/required books or materials?  If so, what are they?

[quote="joml88"]

[edit: clarified my post a little]

No, WOOT isn't all about proof writing.  There weren't any lectures dedicated to it.  The tests are the only things that will make you better at writing solutions to problems but they also provide you with the ability to have practice with taking the test under real (or close to real) conditions.  WOOT is definitely not about just proof writing though by all means you should become better at writing proofs since the graders of the tests will be critiquing you and your peers along with the instructors will be giving you feedback on your solutions to board problems (as long as you post solutions).  I've never been to MOP but I'd have to say that WOOT can probably offer a somewhat similar experience (though there are obviously some things that MOP offers that WOOT cannot and vice versa).[/quote]

Hmm though I'd still like to get learn some more in-depth material first and do something like WOOT in another year or something. Then I can use that background while doing general olympiad stuff, in particular get a lot of practice in junior year when I'll probably have my best shot at MOP. It also works better regarding time and money (ie busy sophmore year).

In response to DPatrick's comment, aren't olympiad courses more of a 1-time thing while WOOT is a general training program that one would do again and again? So what do you mean by "go take WOOT" before doing olympiad courses? (maybe take WOOT for a year or something?) Anyway, I seem to be ready for olympiad courses so that shouldn't be a problem.

[quote="rrusczyk"][quote="chess64"]huh why would you take WOOT twice?[/quote]

The classes, tests, etc., are nearly all different every year (this is part of the reason WOOT is more expensive than our other classes - we have to recreate the course nearly from scratch each year).[/quote]

oh, i didn't know that

[quote="chess64"]I haven't taken WOOT, so I think somebody else could give a better answer... (BTW, you haven't taken WOOT either, have you? :?)[/quote]

Nope and I probably won't. Money is a big reason for me and other thing is that I need to study hardcore for AP exams.


A number of men enter a disreputable establishment and each one leaves a coat and an umbrella at the door. When a message is received saying that the establishment is about to be raided by the police, the men leave hurriedly, and no man gets both the right coat and the umbrella. If there are $ n$ men show that the number  of ways in which this can happen is $ n!\left( n!\minus{} \frac{(n\minus{}1)!}{1!}\plus{}\frac{(n\minus{}2)!}{2!}\minus{}...\plus{}(\minus{}1)^n\frac{1}{n!}\right)$.

This is direct from inclusion-exclusion: the number of ways that a given $ k$-set of the men get their things is precisely ${ (n - k)!^2}$, and there are $ \binom{n}{k}$ $ k$-sets of men, so our answer is
\[ \sum_{k = 0}^n (-1)^{n - k} \binom{n}{k} (n - k)! ^2 = n! \sum_{k = 0}^n (-1)^{n - k} \frac{(n - k)!}{k!},\]
as desired.

I will try  to give a more explicit answer to this problem..

I understand this problem as gangsters entering a building, and the police raid the place and they take the wrong umbrellas and hats. Let:
$ A_{0}$ - be the set with the collections of all sets in which ALL the gangsters fail to pick up both their items. So $ |A_{0}|$ is what we are looking for.
$ A_{1}$ - the collection of sets in which gangster 1 picks up both his items
$ A_{2}$ - the collection of sets in which gangster 2 picks up both his items
...
$ A_{n}$ - the collection of sets in which gangster n picks up both his items.

Because $ A_{0}$ and ($ {A_{1}\cup A_{2}\cup...A_{n}}$) are disjoint, we can write:
$ |A_{0}| = |{A_{1}\cup A_{2}\cup...A_{n}\cup A_{0}|}$ \ $ |{A_{1}\cup A_{2}\cup...A_{n}}$| (1)

[b]1)[/b]The cardinal of the first expression from the right can be deduced directly. It is the total posibilities of picking the hats & umbrellas:
gangster 1->n*n
gangster 2->(n-1)*(n-1)
gangster 3->(n-2)*(n-2)
...
gangster n->1*1    ways, which is
$ |{A_{1}\cup A_{2}\cup...A_{n}\cup A_{0}| = [n!]^2}$

[b]2)[/b]The cardinal of the second expression from the right will be deduced from the inclusion-exclusion principle:
$ |{A_{1}\cup A_{2}\cup...A_{n}| = \sum |A_{i}| - \sum |A_{i}\cap A_{j}| + \sum |A_{i}\cap A_{j}\cap A_{k}| - ... + ( - 1)^{n + 1}|A_{1}\cap A_{2} \cap... \cap A_{n}|}$
Each of the cardinals on the right can be deduced like in 1) , such that:
$ |A_{1}| = |A_{2}| = ... = |A_{n}| = [(n - 1)!]^2$
$ |A_{1}\cap A_{2}| = |A_{1}\cap A_{3}| = ... = [(n - 2)!]^2$
...
$ |A_{1}\cap A_{2} \cap... \cap A_{n}| = [1!]^2$,
and the sums have
$ \binom{n}{1}, \binom{n}{2},...\binom{n}{n - 1}$ terms. So we have
$ |{A_{1}\cup A_{2}\cup...A_{n}| = \binom{n}{1}[(n - 1)!]^2 - \binom{n}{2}[(n - 2)!]^2 + ... + ( - 1)^{n + 1}\binom{n}{n - 1}[1!]^2}$

Now we can rewrite (1):
$ |A_{0}| = [n!]^2 - (\binom{n}{1}[(n - 1)!]^2 - \binom{n}{2}[(n - 2)!]^2 + ... - ( - 1)^{n + 1}\binom{n}{n - 1}[1!]^2)$
$ |A_{0}| = [n!]^2 - \frac {n!(n - 1)!(n - 1)!}{(n - 1)!1!} + \frac {n!(n - 2)!(n - 2)!}{(n - 2)!2!} - ... + ( - 1)^{n}\frac {n!1!1!}{1!(n - 1)!}$
$ |A_{0}| = n!\left( n! - \frac {(n - 1)!}{1!} + \frac {(n - 2)!}{2!} - ... + ( - 1)^{n}\frac {1}{n!}\right)$
Done.


Could somebody solve this over $\mathbb{R^+}$? I have a solution but it is pretty complicated, and I hope there is an easier one.

$x_1+x_2^2+x_3^3=3$
$x_2+x_3^2+x_4^3=3$
$x_3+x_4^2+x_1^3=3$
$x_4+x_1^2+x_2^3=3$

hello, i have only found $ x_1\equal{}x_2\equal{}x_3\equal{}x_4\equal{}1$.
Sonnhard.

Can you explain why aren't there other solutions?


vector
calculus
integration
geometry
3D geometry
algebra
polynomial

Two problems from the final exam I just gave:

2. Let $C$ be the closed triangular curve from $(0,0,6)$ to $(0,2,0)$ to $(1,1,2)$ back to $(0,0,6).$ Let $\vec{F}(x,y,z)=(y+z)\mathbf{i}+(2x+z)\mathbf{j}+(2x+2z)\mathbf{k}.$

Compute the line integral $\oint_{C}\vec{F}\cdot \vec{dr}.$

4. Let $\vec{G}(x,y,z)$ be defined as the vector field whose direction at $(x,y,z)$ is always directly away from the origin and whose magnitude at $(x,y,z)$ is always the cube of the distance from the origin to $(x,y,z).$

(a) Write an explicit formula for $\vec{G}(x,y,z).$

(b) Let $E=\{(x,y,z)\mid x^{2}+y^{2}+z^{2}\le a^{2}\}.$ Let $dV$ be $dx\,dy\,dz.$ Compute:
\[\iiint_{E}\nabla\cdot\vec{G}\,dV \]

[quote="Kent Merryfield"]4. Let $\vec{G}(x,y,z)$ be defined as the vector field whose direction at $(x,y,z)$ is always directly away from the origin and whose magnitude at $(x,y,z)$ is always the cube of the distance from the origin to $(x,y,z).$

(a) Write an explicit formula for $\vec{G}(x,y,z).$

(b) Let $E=\{(x,y,z)\mid x^{2}+y^{2}+z^{2}\le a^{2}\}.$ Let $dV$ be $dx\,dy\,dz.$ Compute:
\[\iiint_{E}\nabla\cdot\vec{G}\,dV \]
[/quote]

For this problem, it seems that the [i]n[/i]-dimensional version is just as easy (except perhaps the integral):
Let $\vec{r}$ denote the usual position vector $\left< x_{1}, x_{2}, \ldots, x_{n}\right>$. Let $E^{n}=\{( x_{1}, x_{2}, \ldots, x_{n})\mid x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}\le a^{2}\}.$ Let $dV^{n}$ be $dx_{1}\,dx_{2}\,\ldots\, dx_{n}$. Then

(a) $\vec{G}(\vec{x})=\frac{\vec{r}}{\left|\vec{r}\right|}\left|\vec{r}\right|^{3}=\left|\vec{r}\right|^{2}\vec{r}$

(b) $\nabla\cdot\vec{G}=\nabla\cdot\left(\left|\vec{r}\right|^{2}\vec{r}\right)= \left|\vec{r}\right|^{2}\nabla\cdot\vec{r}+\vec{r}\cdot\nabla\left(\left|\vec{r}\right|^{2}\right)=n\left|\vec{r}\right|^{2}+\vec{r}\cdot\left(2\vec{r}\right) =(n+2)\left|\vec{r}\right|^{2}$

Let $E_{+}^{n}=\{( x_{1}, x_{2}, \ldots, x_{n})\mid x_{1}, x_{2}, \ldots, x_{n}\ge 0, x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}\le a^{2}\}.$ Then

\[\iint\cdots\int_{E^{n}}\nabla\cdot\vec{G}\,dV^{n}= 2^{n}\iint\cdots\int_{E_{+}^{n}}\nabla\cdot\vec{G}\,dV^{n}\]
\[=2^{n}\iint\cdots\int_{E_{+}^{n}}(n+2)\left|\vec{r}\right|^{2}\,dV^{n}\]
\[=2^{n}(n+2)\iint\cdots\int_{E_{+}^{n}}\left(x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}\right)\,dV^{n}\]

I'll use this theorem:

[url=http://www.imagehoop.com/view_image/340c92053/DirichletIntegralTheorem.JPG][img]http://www.imagehoop.com/view_image/340c92053/DirichletIntegralTheorem.JPG[/img][/url]

with $F(u)=u,t=a,\alpha_{i}=1,\beta_{i}=2,\gamma_{i}=1$ to get

\[2^{n}(n+2)\iint\cdots\int_{E_{+}^{n}}\left(x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}\right)\,dV^{n}= 2^{n}(n+2)\frac{\prod_{k=1}^{n}\left[\frac{1}{2}\Gamma\left(\frac{1}{2}\right)\right]}{\Gamma\left(\sum_{i=1}^{n}\frac{1}{2}\right)}\int_{u=0}^{a}u^{\sum_{i=1}^{n}\frac{1}{2}-1}u\,du \]
\[= 2^{n}(n+2)\frac{\frac{1}{2^{n}}\Gamma^{n}\left(\frac{1}{2}\right)}{\Gamma\left(\frac{n}{2}\right)}\int_{u=0}^{a}u^{\frac{n}{2}}\,du \]
\[= \frac{(n+2)\pi^\frac{n}{2}}{\sqrt{\pi}\frac{(n-2)!!}{2^{\frac{n-1}{2}}}\right)}\cdot\frac{2}{n+2}a^{\frac{n}{2}+1}\]
\[= \frac{a^\frac{n+3}{2}\pi^\frac{n-1}{2}2^\frac{n+1}{2}}{(n-2)!!}= \boxed{\frac{a}{\pi}\cdot\frac{\left( 2a\pi\right)^\frac{n+1}{2}}{(n-2)!!}}\]

so that for $n=3$ we have

\[\iiint_{E}\nabla\cdot\vec{G}\,dV = \frac{a}{\pi}\cdot\frac{\left( 2a\pi\right)^\frac{3+1}{2}}{(3-2)!!}=\boxed{ 4\pi a^{3}}\]

Suffice it to say that I only [i]expected[/i] to see the three dimensional version on my exam.

There were two ways to do it. The first is to compute the divergence and integrate that over the volume - which you've done. (Note that the vector field is polynomial, hence smooth everywhere.) The second is to quote the divergence theorem and calculate the integral of $\vec{G}$ over the sphere. I was hoping to see the second method, but either method would have worked.

Full credit for the problem was quite rare in my class.

[quote="Kent Merryfield"]Two problems from the final exam I just gave:

2. Let $C$ be the closed triangular curve from $(0,0,6)$ to $(0,2,0)$ to $(1,1,2)$ back to $(0,0,6).$ Let $\vec{F}(x,y,z)=(y+z)\mathbf{i}+(2x+z)\mathbf{j}+(2x+2z)\mathbf{k}.$

Compute the line integral $\oint_{C}\vec{F}\cdot \vec{dr}.$[/quote]

Label the verticies: $A(0,0,6),\,B(0,2,0),$ and $C(1,1,2)$ so that $\vec{AB}=<0,2,-6>$ and $\vec{BC}=<1,-1,2>$ and hence a normal to the surface $S$ (the triangle) is given by $\vec{AB}\times \vec{BC}= <0,2,-6>\times <1,-1,2> = <-2,-6,-2>$, and since this normal gives positive orientation for $C$ we choose $\vec{n}=<-2,-6,-2>$. Also,
\[\vec{\nabla}\times \vec{F}= \left|\begin{array}{ccc}\mathbf{i}&\mathbf{j}&\mathbf{k}\\ \frac{\partial}{\partial x}& \frac{\partial}{\partial y}& \frac{\partial}{\partial x}\\ y+z&2x+z&2x+2z \end{array}\right| =<-1,-1,1> \]
Now, by the Curl Theorem (a.k.a. Stokes' Theorem) with $C=\partial S$ we have
\[\oint_{C}\vec{F}\cdot \vec{dr}= \iint_{S}\mbox{Curl}\,\vec{F}\cdot d\vec{S}=\iint_{S}\left(\vec{\nabla}\times \vec{F}\right)\cdot \frac{\vec{n}}{|\vec{n}|}\, dS \]

\[=\frac{1}{|\vec{n}|}\iint_{S}<-1,-1,1>\cdot <-2,-6,-2>\, dS \]

\[= \frac{1}{|\vec{n}|}(2+6-2)A(S) = \frac{6}{|\vec{n}|}\cdot \frac{1}{2}|\vec{n}| =\boxed{3}\]

About a quarter of my class got full or near-full credit. What I didn't mention was that question 1 was to find an equation for the plane through exactly those three points - that was meant to function as a hint. At least one of the the correct answers was from someone who did not use Stokes but simply paramaterized the three line segments and computed the line integral directly. That was possible.

There are some small choices that could be made along the way that would make things look slightly different than what benorin said. You could divide his (non-unit) normal vector by $2.$ You could explicitly talk about the projection on the triangle onto the $xy$ plane.

$\LaTeX$ choice: Using < and > gives $<-1,-1,1>.$ Using \langle and \rangle gives $\langle-1,-1,1\rangle.$ I prefer the latter.


Euler algebra polynomial calculus trigonometry search

Prove that

$\frac 1{1^2}+\frac 1{2^2}+\frac 1{3^2}+...=\frac{\pi^2}{6}$
(my stupidity has been edited out)

I was hoping there was a way to do this without calculus.

thats not true

1/1 +1/2 +1/3 + 1/4 + ... increases without bound
approximalty logrithmically

Perhaps you meant
$\frac 1{1^2}+\frac 1{2^2}+\frac 1{3^2}+...=\frac{\pi^2}{6}$?

[quote="mathnerd314"]Prove that

$\frac 1{1^2}+\frac 2{2^2}+\frac 3{3^2}+...=\frac{\pi^2}{6}$

I was hoping there was a way to do this without calculus.[/quote]Well, the way you have written is equivalent to the harmonic series and it diverges!  :omighty:

For the series $\frac 1{1^2}+\frac 1{2^2}+\frac 1{3^2}+...=\frac{\pi^2}{6}$, even mighty Euler used calculus...

[EDIT]Ack! [b]0714446459923[/b] beat me to it  :yankchain:

There is definitely no elementary proof.  It is not an elementary result.

Oops--duh. That's what I meant. :roll:

Nice proof :thumbup:

[quote="shyong"]amirhtlusa gives a nice prove using power series :)

[quote="amirhtlusa"]let $P(x)=1-\frac{x^2}{3!}+\frac{x^4}{5!}-\frac{x^6}{7!}.....$
clearly $P(0)=1$. to find the roots of $P(x)$, note that for $x\not=0$:
$P(x)=x[\frac{1-\frac{x^2}{3!}+\frac{x^4}{5!}-\frac{x^6}{7!}...}{x}]=\frac{x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}...}{x}=\frac{\sin{x}}{x}$
so $P(x)=0 \Rightarrow \sin{x}=0 \Rightarrow x= \pm k\pi$ for $k=1,2,3,...$ note that $x=0$ is not a solution. because $P(0)=1$
now if you factor $p(x)$ you get:
$P(x)= (1-\frac{x}{\pi})(1-\frac{x}{-\pi})(1-\frac{x}{2\pi})(1-\frac{x}{-2\pi})= [1-\frac{x^2}{\pi^2}][1-\frac{x^2}{4\pi^2}][1-\frac{x^2}{9\pi^2}]+....$
now look what happens when you expand the above equation:
you get
$1-[ (\frac{1}{\pi^2})+(\frac{1}{4\pi^2})+(\frac{1}{9\pi^2})+....]x^2+....$now set this equal to the original function for $P(x)$ you  get
$P(x)=1-\frac{x^2}{3!}+\frac{x^4}{5!}-\frac{x^6}{7!}.....= 1-[ (\frac{1}{\pi^2})+(\frac{1}{4\pi^2})+(\frac{1}{9\pi^2})+....]x^2+....$ now take a look at coefficient for the $x^2$ then set them equal to each other, you get:
$\frac{-1}{3!}=-(\frac{1}{\pi^2}+\frac{1}{4\pi^2}+\frac{1}{9\pi^2}+....)=\frac{-1}{\pi^2}(1+\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+....)$
therefore:
$1+\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...=\frac{\pi^2}{6}$
$\mathbb{QED}$
so we are done :D[/quote][/quote]
That proof is nice but not really rigorous as written.  You can't just wave your hands and say that an infinite polynomial can be factored in the same way that a finite one can be - I believe to rigorously do so requires concepts from complex analysis.  It's the most "accessible" proof (and the one printed in aoPS V2), but unfortunately even it reduces one problem (summing the series) to another (proving the factorability of an infinite polynomial).

Not only that, look at that step where recalled the Taylor expansion of sin x at x=0. Taylor expansion is a topic within analysis and it requires calculus.

***EDIT*** Nevermind, I was thinking of Euler's formula instead of DeMoivre's formula. But, does taking the limit count as part of calculus? :P

[quote="0714446459923"]Wikipedia has a noncalculus solution

http://en.wikipedia.org/wiki/Basel_problem

Is it ok to post links like these?[/quote]

The expansion of the Taylor series of $\sin x$ is a calculus step  :P

[quote="10000th User"]That's the Euler part. There is also another proof below it that claims to have no calculus but I have doubts to two things:

1) It has a limit used in the proof

2) It has Squeeze Theorem, if you read carefully towards the bottom. Squeeze theorem doesn't look like a calculus theorem but I first heard of it earlier this year in my calculus class and never heard of it before when I was taking Trigonometry during summer.[/quote]

It uses de Moivre's, for heaven's sakes.  de Moivre's is a result that can't really be put on any firm foundation without calculus.

I recall someone using this result to prove that pi is irrational.

You need calculus for sure, but not for the prove but already to define the infinite sum!

But some simple search would give a lot of threads already concerning this!
E.g., most links to further ones are contained in http://www.mathlinks.ro/Forum/viewtopic.php?t=81439 .

I was just hoping I didn't need calculus because:

[quote="DPatrick"]
Please do not post calculus problems in this forum.
[/quote]

Should I try again somewhere else?

[quote="mathnerd314"]Should I try again somewhere else?[/quote]
No, because it was already posted ;)

[quote="t0rajir0u"][quote="10000th User"]That's the Euler part. There is also another proof below it that claims to have no calculus but I have doubts to two things:

1) It has a limit used in the proof

2) It has Squeeze Theorem, if you read carefully towards the bottom. Squeeze theorem doesn't look like a calculus theorem but I first heard of it earlier this year in my calculus class and never heard of it before when I was taking Trigonometry during summer.[/quote]

It uses de Moivre's, [b]for heaven's sakes[/b].  de Moivre's is a result that can't really be put on any firm foundation without calculus.[/quote]??? What do you mean by that? I hope you are not implying anything derogative there...  :o

Does the squeeze therom really require calculus?
I thought it would be self evident for any continous, elementary function.


Find minimum of $A=\frac{a^{n}}{(b+c)^{n}}+\frac{b^{n}}{(c+a)^{n}}+\frac{c^{n}}{(a+b)^{n}}$ where $n\geq 2$ is positive integer and $a,b,c$ are positive reals

Jensen's+Nesbitt's = solution. \[\frac{a^{n}+b^{n}+c^{n}}{3}\ge \left(\frac{a+b+c}{3}\right)^{n},\ n \ge 1.\]

Why did you think so? This problem don't sovle by your way? It's very difficult? I know there is a rusult that is special case of this problem:
$\frac{a^{3}}{(b+3)^{3}}+\frac{b^{3}}{(c+a)^{3}}+\frac{c^{3}}{(a+b)^{3}}\geq\frac{3}{8}$

Since you posted the unreal version of Vietnam TST 2005, \[\sum_{cyc}{a^{3}\over (a+b)^{3}}\ge{3 \over 8}.\]


function
calculus
derivative
real analysis
real analysis unsolved

The function f is differentiable on R and the limits $lim_{x\to \pm \infty} f'(x)$ both exist finitely. Show that f is uniformly continuous on R.

Is this as easy I think it is? You can cut off a large interval compact $[-a,a]$ where we know f is uniformly continuous, since it is continuous. Then for two points outside $[-a,a]$ we can use the mean value theorem $f(x) - f(y) = f'(\xi)|x-y|$. We just require $|x-y|$ to be small. Now we have a problem if we have one point outisde the interval and one inside it, but if we just consider the larger interval $[-a-d,a+d]$ then f will still be uniformly continuous inside the interval and if we choose two points outside of it we still require $|x-y|$ to be small. Finally if we choose a point outside it and one inside it $f(x) - f(y)$ will still be small enough because it is covered by the first case without the extended interval.

Of course, I didn't do any details here, but is this how easy an argument could be made? I still feel a little unsure.

$f$ satisfies a Lipschitz condition (bounded derivative) for large enough $x$ and small enough $x$ and hence is uniformly continuous there; the rest is the usual bit about continuity being uniform continuity on compact sets.  The idea is that $f$ is basically linear at infinity; such functions are uniformly continuous.


floor function
logarithms
real analysis
real analysis unsolved

Prove or disprove:

If $ b_{n}$ is a convergent sequence in $ R$ and $ c_{n} \equal{} n(b_{n} \minus{} b_{n \minus{} 1})$ then $ c_{n}$ is a bounded sequence.

False.

Let $ b_n\equal{}\sum_{k\equal{}1}^n\frac{(\minus{}1)^k}{\sqrt{k}}.$

Now: what happens if we add as an additional hypothesis that $ b_n$ is monotone?

If $ b_n$ is a partial sum of a convergent series, then the statement reads: every convergent series (say, with nonnegative terms) is $ O(n^{\minus{}1})$. Which is clearly false: all we know about the terms of a convergent series is that they are $ o(1)$.

Here's an explicit example:

Let $ b_n\equal{}\frac1{\lfloor \log_2n\rfloor}.$

$ b_n\to0,$ so it converges, but

$ c_{2^k}\equal{}2^k\left(b_{2^k}\minus{}b_{2^k\minus{}1}\right)\equal{}
2^k\left(\frac1k\minus{}\frac1{k\minus{}1}\right)$

is certainly not bounded.


Prove that in every convex polyhedron,if concurrently it doesn't have quadrilateral plane,pentagonal plane,then have at least four planes are triangle!

For a convex polyhedron

denote by $ v$ the number of its vertices, respectively $ v_k$ the number of vertices where $ k$ edges meet, for $ k \geq 3$, hence $ v \equal{} v_3 \plus{} v_4 \plus{} v_5 \plus{} v_6 \plus{} \cdots$;

denote by $ f$ the number of its facets, respectively $ f_k$ the number of facets with $ k$ sides, for $ k \geq 3$, hence $ f \equal{} f_3 \plus{} f_4 \plus{} f_5 \plus{} f_6 \plus{} \cdots \equal{} f_3 \plus{} f_6 \plus{} \cdots$ (we are given $ f_4 \equal{} f_5 \equal{} 0$);

denote by $ e$ the number of its edges, so by Euler's formula $ v \plus{} f \equal{} e \plus{} 2$.

Now a little bit of double counting yields $ 2e \equal{} 3v_3 \plus{} 4v_4 \plus{} 5v_5 \plus{} 6v_6 \plus{} \cdots$, hence $ 2e \minus{} 3v \geq 0$. On the other hand, another bit of double counting yields $ 2e \equal{} 3f_3 \plus{} 4f_4 \plus{} 5f_5 \plus{} 6f_6 \plus{} \cdots \equal{} 3f_3 \plus{} 6f_6 \plus{} \cdots$, hence $ 2e \minus{} 6f \geq \minus{}3f_3$.

Finally, $ \minus{}12 \equal{} 6(e \minus{} v \minus{} f) \equal{} (2e \minus{} 6f) \plus{} 2(2e \minus{} 3v) \geq \minus{}3f_3$, therefore $ 3f_3 \geq 12$, i.e. $ f_3 \geq 4$, qed.

Hi [b]mavropnevma[/b],i'm very glad and thanks so much for your solution!


The way the flags work in Opera (I have version 7.53 with Java) makes no sense.. There are new posts since I've last been here, but they're not marked with orange flags, and some messages which I've viewed a hundred times or so refuse to show a blank flag (it stays orange). Moreover, no flags go blank if I view the topic by pressing previous/next topic from another topic.

Please don't tell me something like "use Internet Explorer", because that's not really a solution :D:D (and besides, there were problems with that too, and I don't want to go through all existing browsers :)).

I think that thing you did (changing the cookie management :)) works: flags are Ok now.

I use IE, and as of this morning, every topic cannot be marked read. Even if I read a topic, or press "mark all read" I still get the orange flag.

I have a lot of topics still showing that bug...

It's all in the cookie and session management of phpBB2. Session management is known to have certain problems, and the view-topic-unread mod does not address those problems.

We will try to correct this problems in the future, but please continue to report abnormalities in the working of this feature.

I think it may be a good start finding patterns in which topic do and don't work, as well as updating phpBB if possible with the many hacks installed.

[quote="Peter VDD"][quote="Valentin Vornicu"]I think that we are better off waiting for the next major release of phpBB. :)[/quote]

Now that's 2 things you shouldn't do to get an errorless board:
- installing new software as soon as it's released
- installing hacks on newly released software
:D

Nah, I guess we're best off just sitting here and telling people they dreamt about the errors :cool:[/quote]The last upgrades of phpbb2 do not addess the session handling issue, but rather some security issues. They have no connection with the problem here.

[quote="Valentin Vornicu"][quote="Peter VDD"]Now that's 2 things you shouldn't do to get an errorless board:
- installing new software as soon as it's released
- installing hacks on newly released software
:D

Nah, I guess we're best off just sitting here and telling people they dreamt about the errors :cool:[/quote]The last upgrades of phpbb2 do not addess the session handling issue, but rather some security issues. They have no connection with the problem here.[/quote]
I don't get the relevance? :?

I am saying that upgrading to phpBB 2.0.10 will not solve the problem in this thread. In fact, the version of phpBB displayed at the bottom of each page is a variable that I input in some field, so I might make it phpBB 2.2, but it wouldn't change a thing ;) :)

[quote="Peter VDD"]I know that :) But I was replying to this: [quote="Valentin Vornicu"]I think that we are better off waiting for the next major release of phpBB. :)[/quote][/quote]The next major release of phpBB2, which is 2.2, will also address these issues, that we are talking about here.

Yes, and that's what I replied to:
[quote="Peter VDD"]Now that's 2 things you shouldn't do to get an errorless board:
- installing new software as soon as it's released
- installing old hacks on newly released software
:D[/quote]

The first one is usually not too worse for phpBB (only a couple fatal errors)

But the second one is a bit worse: Ok, phpBB 2.2 has drastic changes on sessions, but also on a lot of other stuff. And that may cause problems with certain hacks on this board... Honestly, I don't think this version of the unread flags mod will work with phpBB 2.2 :cool:

And until the all the mods are converted to phpBB2.2, that can take a while... possibly even till 2006.

The 2.2 release already contains 90% of the mods installed on this board, including the unread topics, categories subforums, global annoucements ;)

Most likely the only mods that will have to be adapted are the template mods, the latexrender (although I doubt there will be much to adapt to), and a couple of other minor look mods.

Oh? In that case much has changed since the last time I checked phpBB2.2, will have to update my experience :) Last time I checked they were polling about keeping unread flags but there was like 70% against :?


linear algebra
matrix
linear algebra unsolved

Let $ A,B \in M_{3}(C)$ s.t $ AB\equal{}BA$ and $ det(A\plus{} B)\equal{}det(A)\plus{}det(B)$.
Provet that $ det(A^3\plus{}B^3)\equal{}(det(A\plus{}B))^3$

Counterexample: $ A\equal{}I$, $ B\equal{}diag(1,\minus{}3,\minus{}5)$.

But if we assume $ \det(A\minus{}B)\equal{}\det(A)\minus{}\det(B)$ also, then it is true.


Hey all, I had no idea where to put this...

I have a table used to show the rate at which water flows through tubing given the tubing diameter and water column hight. Is there anyway to extract the formula used to calculate this from the infor mation given in the said table?


which particular one dya mean?
 :: Divergence Theorem, Gauss's Digamma Theorem, Gauss's Double Point Theorem, Gauss's Hypergeometric Theorem, Gauss's Theorema Egregium  :lol:   :lol:
 try to look for what you want
 at [url=http://mathworld.wolfram.com/GausssTheorem.html]mathworld.wolfram.com/GausssTheorem.html[/url]
the triple 's' is correct ,because it's supposed to be Gauss's theorem .

I think you mean the following theorem

[i]Gauss' theorem[/i] The midpoints of a complete quadrilateral's three diagonals are collinear. [/i]

@treego: what do you mean by saying the [i]midpoints of a complete quadrilateral [/i]?

let $ABCD$ be a quadrilateral ,

we call it complete, if we draw its diagoals , and also draw

lines $AB,CD$and,$AD,BC$. and let them intersect each other.

[quote="Tiks"][quote="Ashegh"][quote="Tiks"]Guys this is wrong place to discusse this kind of quetions :!:[/quote]

dear, these disscusions have relations to each other.

then why shouldnt us disscus them? ;)[/quote]
You should disscus them in the section
[color=blue]Geometry theorems and Formuls[/color]! :P[/quote]

is (the meaning of completequadrilaterall),a theorem?

disscussions are available in every where. :wink:  :D

Maybe you forget what was the purpose of this discution :roll: [quote="abdurashidjon"]Is there anyone who knows Gauss theorem?[/quote]
I think he was asking for Gauss' theorem,isn't he?

[quote="Tiks"]Guys this is wrong place to discusse this kind of quetions :!:[/quote]
u said, THIS KIND OF QUESTIONS
ofcourse my explanation about perfect quadrilateral,is one of them.

may be the other questions, should be disscuss in theorems and formulas

section. but mine is only an explanation,

and explanations dont need any places to say.

and i can post them every where,

for the sake of god plz stop :!:

Now,the most important thing is to help abdurashidjan.not any thing more.

 :wink:


JBL wrote:Rationalize the denominators:

1)  1/(:rt2: + :rt3: + :rt5:)

2)  1/(:rt2: + 3:rt2:)

(Sorry if you've seen these before.)

A harder question that just occurred to me -- those are both algebraic numbers, I believe.  Can you find a polynomial that has integer coefficients with either of those numbers as a root?  (I can definitely do it for the first one, but the second one will take more work for sure.)

[hide]

1) Multiply by (:rt:2 - :rt:3 - :rt:5)/(:rt:2 - :rt:3 - :rt:5) and then multiply by (-6+2:rt:15)/(-6+2:rt:15) (there was probably an easier way) and simplify to get (3:rt2: + 2:rt3: - :sqrt:30)/12, assuming I didn't blunder.[/hide]

Confuted did not blunder, and Tetrahedr0n could do us all a favor by multiplying that out so I know whether or not he's right.

Anyone have any thoughts on the harder question?  (Instead of giving a specific polynomial, you might just give a method for finding it.)

This exact problem appeared on our 2003 Collaborative Problem-Solving Contest.  If your school participated in the event, your coach received complete solutions to all problems, including this one.  Essentially, you can always generate a polynomial by multiplying terms of the form (x - rn) where rn is the nth root of the desired polynomial.  For a number like  :rt2: + :rt3: + :rt5:  you just have to include all of the "conjugates" as the other roots.  In this case there are eight roots total (all possible combinations of +/- for each of the irrational components).

If your school does not currently participate in our Collaborative Problem-Solving Contests, there is still time to register for this year.  The contest will occur from April 14th to April 21st, and requires your team to collaborate for one week on fifteen difficult-yet-accessible problems.  More information is available on our website.


Write $ 1,000,000$ as a sum of a prime number and a perfect square.

$ 1,000,000\equal{}10^6\equal{}k^2\plus{}p$
$ (10^3\minus{}k)(10^3\plus{}k)\equal{}p$

Because $ p$ is prime, $ 10^3\minus{}k\equal{}1$ which implies that $ k\equal{}10^3\minus{}1$ and $ p\equal{}1999$.

You can go check to see that $ 1999$ is prime.

$ 1,000,000\equal{}(999)^2\plus{}1999$.


advanced fields
advanced fields unsolved

conjecture: every uncountable set has a bounded uncontable subset.

of course, otherwise every ball of radius $n$ would be countable and thus the union of these balls would be countable.


suppose that {a_k} k=1 to infinity is a sequence with a_k >=0 and a_k >= a_k+1 for all k>=1. prove that if sum_{k=1} to infinity a_k converges then ka_k goes to 0 as k goes to infinity.

[quote="jmerry"]Condensation test: If $a_n$ is a sequence which decreases to 0, $\sum a_n$ converges if and only if $\sum 2^na_{2^n}$ converges.[/quote]

Can you elaborate, jmerry? Can someone help.... Hw is due this evening. Thanks in advance...

Suppose the sum $ \sum_{k=1}^{\infty} a_{k} $ converges. Then
clearly $ T_{n} = \sum_{k=n}^{2n} a_{k} \leqslant \sum_{k=n}^{\infty} a_{k} \rightarrow 0 $,
when $ n \rightarrow \infty $. But clearly $ T_{n} \geqslant (n+1)a_{2n} , na_{2n-1} $,
so $ 2T_{n} \geqslant 2na_{2n} , (2n-1)a_{2n-1} $ , so subsequence of elements of odd indexes
and subsequence of elements of even indexes of a sequence $ b_{n} = na_{n} $ converge to $ 0 $,
so the sequence $ b_{n} = na_{n} $ converges to $ 0 $.


inequalities
group theory
abstract algebra
vector
superior algebra
superior algebra unsolved

Let $ \mathcal G$ be the set of all finite groups with at least two elements.

a) Prove that if $ G\in \mathcal G$, then the number of morphisms $ f: G\to G$ is at most $ \sqrt [p]{n^n}$, where $ p$ is the largest prime divisor of $ n$, and $ n$ is the number of elements in $ G$.

b) Find all the groups in $ \mathcal G$ for which the inequality at point a) is an equality.

[quote="Valentin Vornicu"]... where $ p$ is the largest divisor of $ n$[/quote]
I assume: prime divisor¿

Finally, something nice and decent. Take $ a$ an element of order $ p$ and let $ d\equal{}n/p$, $ x_2,...,x_d$ a system of representatives for the left cosets except the coset $ H$ itself, where $ H\equal{}<a>$ is the subgroup generated by $ a$. Any morphism is determined by the image of $ a$ and of $ x_2,...,x_d$, thus there are at most $ n^d$ such morphisms. This settles $ i)$. Now, for $ ii)$ assume that we have equality, so for any choice of values of $ f(a),f(x_i)$ one gets a morphism. Let's assume $ |G|>2$ and $ d\ne 1$. First, I claim that $ ax_2$ must have the form $ x_2a^k$. If not, then $ ax_2\equal{}x_ja^k$ for some $ j>2$ and some $ k$. But then for any $ g,h,l$ one may find a morphism such that $ f(a)\equal{}g, f(x_2)\equal{}h, f(x_j)\equal{}l$ and so $ gh\equal{}lg^k$ for all $ g,h,l$, certainly not possible. So, $ ax_2\equal{}x_2a^k$ and repeating this argument we have $ gh\equal{}hg^k$ for all $ g,h$. Thus $ g\equal{}g^k$ for all $ G$ and $ G$ is abelian. Also, recall that for any choice $ g$ one can find a morphism $ f$ such that $ f(a)\equal{}g$ and so $ g$ has order $ 1$ or $ p$. Thus $ G$ is abelian of exponent $ p$ and so a $ F_p$ vector space, thus it is isomorphic to some $ F_p^k$. No, it is trivial to conclude.

Probably Marian Andronache. I can bet. And probably problem 3 from 11th grade is Ion Savu. I've seen similar stuff before from him.

Valentin, problem 3 from 11-th grade is proposed by Bogdan Enescu.

[quote="harazi"]Does it really matter who proposed them? I mean, when the problem is original and interesting, it matters, but for such problems absolutely anybody can be considered an author.  :D[/quote]

???

Anyway, when I saw this problem, I thought, "Oh, how original and interesting," though it's nothing I'm going to tell my future grandchildren about or anything.


In right triangle $ABC$ with $\angle B=90^\circ$, $\angle CAB=60^\circ$. Extend $BC$ and let $D$ be the point on ray $BC$ such that $BC: CD=2: 1$, find the value of $\tan x$ where $x=\angle DAC$.

Well there is absolutely no trig in this problem other than the definition of tangent in a right triangle...

[hide]
Let $CD=a$, then we have $BC=2a$. Since $ABC$ is 30-60-90, we have $AB=\frac{2}{\sqrt{3}}a$ and $AC=\frac{4}{\sqrt{3}}a$. Hence $\tan{x}=BC/AB=\frac{3\sqrt{3}}{2}$.[/hide]

[quote="13375P34K43V312"]Well there is absolutely no trig in this problem other than the definition of tangent in a right triangle...

Let $CD=a$, then we have $BC=2a$. Since $ABC$ is 30-60-90, we have $AB=\frac{2}{\sqrt{3}}a$ and $AC=\frac{4}{\sqrt{3}}a$. Hence $\tan{x}=BC/AB=\frac{3\sqrt{3}}{2}$.[/quote]

I get $\sqrt{3}/11$

[hide]
umm.. drawing an image would help, but from the information given, one can conclude that $\tan{x+60) = \frac{3\sqrt{3}}{2}}$.

well since $\tan{(a+b)}= \frac{\tan{a}+\tan{b}}{1-\tan{a}\tan{b}}$,

$\frac{\tan{x}+\tan{60}}{1-\tan{x}\tan{60}}= \frac{3\sqrt{3}}{2}$

cross-multiply to get

$2\tan{x}+2\tan{60}= 3\sqrt{3}-3\sqrt{3}\tan{x}\tan{60}$

$2\tan{x}+2\sqrt{3}= 3\sqrt{3}-3\sqrt{3}\tan{x}\cdot\sqrt{3}$

$11\tan{x}= \sqrt{3}$

$\tan{x}= \frac{\sqrt{3}}{11}$
[/hide]

[hide] Let CD=x so then BC=2x and AB=$\frac{2x\sqrt{3}}{3}$.

Therefore ${tan(60+x)=\frac{BD}{AB}=\frac{3\sqrt{3}}{2}}$.

Using Tangent Addition Theorem we get:

$\frac{tan(60)+tan(x)}{1-tan(60)tan(x)}$

Simplifying:

$\frac{\sqrt{3}+tanx}{1-\sqrt{3}tanx}=\frac{3\sqrt{3}}{2}$

Solving for $tanx$: $\boxed{tanx=\frac{\sqrt{3}}{11}}$[/hide]

Edit: I messed up with the curly brackets.


Z is a complex number
Prove that I Z-2-i I <1 only if I Z +3 +2i I >4

i think this works....

set $Z$=$a+bi$, to make the question $(a-2)^2+(b-1)^2<1$ (i) $\rightarrow (a+3)^2+(b+2)^2>16$ (ii).

ok so from (i) we know $1<a<3$ and $0<b<2$.

expand (ii), write it as $f(b)=b^2+4b+a^2+6a-3$.  note that it's a quadratic in $b$.  this function will be greater than zero if its discriminant $16-4(a^2+6a-3)\leq0$.  Check it and see that it is true for $1<a<3$.    then do the same thing writing as quadratic in $a$.

edits:

eh screw it, none of that helped cause if $1<a<3$ its pretty obvious that (ii) holds true given the range of $b$.  huh now i think its trivial... we know (i) can be true only if both conditions are held ($1<a<3$ and $0<b<2 $).  If $a$=1+epsilon (epsilon>0) then (ii) is obviously true.  But for (i) to be true $a$ must be in the range of $1<a<3$, so 1+epsilon is the smallest possible value for $a$.  For any $a$ in the given range that is greater than $1+epsilon$, (ii) is trivially true.  so i think thats actually complete.  will come back to it tomorrow.

yeah but why  is the question to show it's >4 when it's trivially true (we can even say >sqrt(20) so i dont know why it said 4).  the 2nd inequality is always true so the conclusion would follow immediately, without even analyzing the first inequality..... its just so weird

This question is best seen from a geometric standpoint.Working algebraically is not the most efficient way to proceed in problems like this.
In the complex plane:
IZ-2-i I=1 is the equation of a circle centre (2,1) radius 1 . Call it C(1)

IZ+3+2iI =4 is the equation of a circle (-3,-2) radius 4.Call it C(2)

The question asks us to show that all points lying inside C(1) are exterior to to C(2).This can only happen if the circles dont intersect and we dont have the situation of  C(1) lying entirely within C(2)..
The distance between the centres of the circles  is 6 while the sum of their radii is 5.On the strength of this observation we can conclude that C(1) must lie entirely outside C(2).


function
integration
search
trigonometry
calculus
derivative
real analysis

let $f$ be a continuous function such that $\int_0^{\infty}|f(t)|dt < +\infty$.
show that the following differential equation $y''+f(t)y=0$
has at least one unbouded solution.
(if you know the official solution, let others search :-) )

[quote="alekk"]
if you know the official solution, let others search [/quote]

ok

I think it is quite simple. Suppose all solutions are bounded and take $f_1, f_2$ two solutions linearly independent. Then the wronskian is a constant function, non-zero. On the other hand, it is clear that $ f_i'$ converge at infinity, since $ f_i$ are bounded and $ f$ is absoutely integrable. Thus $ f_i\"$ are integrable on the set of positive reals and thus $f_i'$ convege at infinity. But sine $f_i$ are also bounded, the derivative must in fact converge towars 0. But then the wronskian is a constant non-zero function that tends to 0 in infinity, false. Am I missing something?


quadratics
number theory unsolved
number theory

I wonder how can u find out that $5$ is a quadratic residue mod $239$ and the smallest $n$ so that $14^{n}\equiv 1 (mod 239)$ is $238$ (because maybe $n=119$ also works...)...
Do you have any formula to do those tasks ? Thx a lot  :)

More general:
If $p$ is a prime of the form $4k+3$ and $p|F_{n},$ then $n$ is even. This is easily consequence of the fact that $F_{2n+1}=F_{n}^{2}+F_{n+1}^{2}.$


Does there exist uncountable family $F$ of subset of countable set with finite $x\cap y$ for all distinct $x,y\in F?$

Without loss of generality, we can assume that our countable set is $E=\mathbb{N}$. Let $\Sigma=(A_{i})$ be our uncountable family of subsets of $E$. We assume the result does not hold.
Let $a_{0}$ be the smallest nonnegative integer which is contained in uncountably many $A_{i}$ (the existence of $a_{0}$ is obvious). Denote by $\Sigma_{0}$ those $A_{i}$ containing $a_{0}$.
By induction, define $a_{n+1}$ to be the smallest nonnegative integer greater than $a_{n}$ such that $a_{n+1}\in A_{i}$ for uncountably many $A_{i}\in \Sigma_{n}$. Denote by $\Sigma_{n+1}$ this subset of $\Sigma_{n}$.
Now write $\Sigma^{*}=\cap{\Sigma_{n}}$. Since by construction: $\forall n, \Sigma_{n}-\Sigma_{n+1}$ is countable, it follows that $\Sigma^{*}$ is uncountable, and in particular nonempty. For any two elements $X,Y$ of $\Sigma^{*}$, $X \cap Y=\cup \{a_{n}\}\neq \emptyset$, which is absurd.

[quote="julien_santini"]Since by construction: $\forall n, \Sigma_{n}-\Sigma_{n+1}$ is countable, ... [/quote]
Hey, why?

Because by construction, $\forall k \in ]a_{n},a_{n+1}[$, $a_{k}$ is contained in at most a countable number of elements of $\Sigma_{n}$.

Sorry, when writing "why?" I havn't mean you will axplain the point.
That was only to show the first incorrect point of your proof.

Indeed, what the matter forbids us  to assume  $a_{0}=0, \ a_{1}=1$ (just for simplesity) and both $\Sigma_{1}$ and $\Sigma_{0}-\Sigma_{1}$ to be  uncountable?

It would take me some time to remember my own solution of the problem and those of my friends all rather simple and short but I could never forget the other one because it is perfectly evident and perfectly nice.

The answer to the question is yes.

Here is one construction.  Let the countable set be the set of finite binary strings.  Given an infinite binary string $x$, let $S_{x}$ be the set of finite prefixes of $x$.  If $x$ and $y$ are two different infinite binary strings, then $S_{x}\cap S_{y}$ is finite.  Let $F$ be the set of $S_{x}$ for all infinite binary strings $x$.  Because there are an uncountable number of infinite binary strings, $F$ is uncountable.

Yes, this is almost the same I mensioned above:

For a given real $x$  fix any rational consequence which converges to $x$ and let $F_{x}$ be the all their members.  It is obvious now that $\{F_{x}\}$ is desired family.


is it true that bigger head are smarter? but seriousely which one looks better?

they both look good if you are confident

I don't think either has anything to do with how smart you are.

neither looks good
(i now counterexamples for both  :roll: )


Let $ A : X \to U$ be a bounded linear map between Banach spaces. Let $ x_n$ be a weakly convergent sequence in $ X$. Then it seems to me like $ Ax_n$ is weakly convergent also, because if $ l \in U'$

$ l(Ax_n) \equal{} (A'l)(x_n) \to (A'l)(x) \equal{} l(Ax)$

where $ A'$ is the transpose of $ A$.

This short proof or observation with the transpose seems so strange to me. What is the gist of it? I.e. what is it that really ensures that weak convergence is preserved?

I think your way of proving the result is the easiest one can give. Maybe you can write down explicitly what the transpose operator does to reveal the simplicity: $ l(A x_n) \equal{} (l\circ A)(x_n) \to (l\circ A)(x) \equal{} l(Ax)$. So what you need is the observation that the sequence $ (l(A x_n))_{n\in {\mathbb N}}$ obtained by applying any $ l\in U^{\prime}$ to $ (A x_n)$ gives a sequence that is obtained from $ (x_n)$ by applying $ l\circ A$, which is a bounded functional $ X\to {\mathbb K}$, hence an element of $ X^{\prime}$, to which you can apply the assumption that $ x_n$ converges weakly to $ x$. That's all, I think.
Hanno

Ohps yeah, I am being dumb. I get confused by notation. Thanks


$A,B \in M(n,C)$ such that
$AB=A$
$BA=B$

Prove that $(A-B)^2=0$

One more example that in Romanian magazines linear algebra= multiplications+additions. Here we have $A^2=(AB)A=A(BA)=AB$ and $B^2=(BA)B=B(AB)=BA$. Conclusion.


quadratics
real analysis
real analysis unsolved

[color=darkblue]Let $ (u_n)$ is a number sequence and let $ \begin {cases} u_1 \equal{} \alpha \\
u_{n \plus{} 1} \equal{} (u_n)^2 \plus{} \beta \end {cases}$
Find formula of $ u_n$ by $ \alpha$, $ \beta$, $ n$.[/color]

$ u_1 \equal{} \alpha$
$ u_2 \equal{} \alpha^2 \plus{} \beta$
$ u_3 \equal{} \alpha^4 \plus{}2\alpha^2 \beta \plus{} \beta^2 \plus{} \beta$
$ u_4 \equal{} \alpha^8 \plus{}4\alpha^6 \beta \plus{} 2\alpha^4(3\beta^2 \plus{}\beta) \plus{}4 \alpha^2 (\beta^3 \plus{}\beta^2) \plus{} [\beta^4 \plus{} 2 \beta^3 \plus{} \beta^2 \plus{}\beta]$

$ u_n \equal{} \sum_{k\equal{}0}^{2^{n\minus{}1}} \alpha^{2k} p_{k,n} (\beta)$

$ u_{n\plus{}1} \equal{} [\sum_{k\equal{}0}^{2^{n\minus{}1}} \alpha^{2k} p_{k,n} (\beta)][\sum_{j\equal{}0}^{2^{n\minus{}1}} \alpha^{2j} p_{j,n} (\beta)] \plus{} \beta$

$ \equal{}\beta\plus{}\sum_{k\equal{}0}^{2^{n\minus{}1}} \sum_{j\equal{}0}^{2^{n\minus{}1}} \alpha^{2(k\plus{}j)} p_{k,n}(\beta)  p_{j,n} (\beta)$

$ p_{0,n\plus{}1}(\beta) \equal{} p_{0,n}^2(\beta) \plus{} \beta$
and for $ 0< l <2^n, p_{l,n\plus{}1}(\beta) \equal{} \sum_{k\plus{}j\equal{}l} p_{k,n}(\beta) p_{j,n}(\beta)\equal{} 2\sum_{k\equal{}0}^l p_{k,n} (\beta) p_{l\minus{}k,n} (\beta)$

having problems solving for these p_k,n explicitly
i am not sure this is really what you wanted though..  =(

[quote="t0rajir0u"]A generic [url=http://mathworld.wolfram.com/QuadraticMap.html]quadratic map[/url] doesn't have a closed-form solution.[/quote]

Thanks t0rajir0u and kenn4000 very much.

If $ \beta \equal{} \minus{}2$  then the closed-form solution exists.

[color=darkblue]Yes, If $ \beta \equal{}\minus{}2$ then the closed-form solution exists.[/color]


This has been posted before.. http://www.mathlinks.ro/Forum/viewtopic.php?p=457344#p457344
And the sign is reversed


(1) Find all function $ f: [0;1] \rightarrow [0;1]$ which satisfies:
1/ $ f(x_1) \neq f(x_2) \forall x_1 \neq x_2$
2/ $ 2x \minus{} f(x) \in [0;1] \forall x \in [0;1]$
3/ $ f(2x \minus{} f(x)) \equal{} x$

(2) Does there exist a function $ f: \mathbb R \rightarrow \mathbb R$ satisfying these below conditions:
1/ There exists $ M \ge 0$ such that $ \minus{}M \le f(x) \le M$
2/ $ f(1) \equal{} 1$
3/ If $ x \neq 0$ then $ f(x \plus{} \frac{1}{x^2}) \equal{} f(x) \plus{} f^2(\frac{1}{x})$.

(3) Find all function $ f,g: \mathbb R \rightarrow \mathbb R$ satisfying
\[ f(x \plus{} g(y)) \equal{} xf(y) \minus{} yf(x) \plus{} g(x) , \quad \forall x,y \in \mathbb R.\]

(4) Find all functions $ f: \mathbb R \rightarrow \mathbb R$ satisfying
1/ $ f(x_1) \neq f(x_2)  \forall x_1 \neq x_2$, and
2/ For all real numbers $x$ and $y$ with $x \neq y$,
\[ f\left(\frac{x \plus{} y}{x \minus{} y}\right) \equal{} \frac{f(x) \plus{} f(y)}{f(x) \minus{} f(y)}.\]

[quote="ll931110"]Problem 1:
Find all function $ f: [0;1] \rightarrow [0;1]$ which satisfies:
1/ $ f(x_1) \neq f(x_2) \forall x_1 \neq x_2$
2/ $ 2x \minus{} f(x) \in [0;1] \forall x \in [0;1]$
3/ $ f(2x \minus{} f(x)) \equal{} x$
[/quote]

Conditions 2. + 3. are enough to conclude $ f(x) \equal{} x$ :

Let $ f(a) \equal{} b$ and the following sequence :
$ u_0 \equal{} a$
$ u_1 \equal{} 2a \minus{} b$
$ u_{n \plus{} 2} \equal{} 2u_{n \plus{} 1} \minus{} u_n$

We obviously have $ f(u_{n \plus{} 1}) \equal{} u_n$ $ \forall n$ and $ u_n\in[0,1]$ $ \forall n$

But clearly $ u_n \equal{} n(a \minus{} b) \plus{} a$ and so $ a \equal{} b$, else $ u_n\in[0,1]$ $ \forall n$ would be wrong.

Thanks pco. I thought it when I was solving this problem, but at that time I wasn't sure my solution is correct.

What about other problems? Who can solve them?

[quote="ll931110"]Problem 2:
Does there exist a function $ f: R \rightarrow R$ satisfying these below conditions:
1/ There exists $ M \ge 0$ such that $ \minus{} M \le f(x) \le M$
2/ $ f(1) \equal{} 1$
3/ If $ x \neq 0$ then $ f(x \plus{} \frac {1}{x^2}) \equal{} f(x) \plus{} f^2(\frac {1}{x})$[/quote]

$ f(1) \equal{} 1$ $ \implies$ $ f(2) \equal{} f(1 \plus{} \frac {1}{1^2}) \equal{} f(1) \plus{} f^2(\frac {1}{1}) \equal{} 2$

Let then $ u$ such that $ f(u)\geq 2$

If $ f(\frac {1}{u})\geq 1 \minus{} f(u)$, let $ v \equal{} \frac {1}{u} \plus{} u^2$.
$ f(v) \equal{} f(\frac {1}{u}) \plus{} f^2(u)$ $ \geq 1 \minus{} f(u) \plus{} f^2(u)\geq f(u) \plus{} 1$

If $ f(\frac {1}{u}) < 1 \minus{} f(u)$, let $ v \equal{} u \plus{} \frac {1}{u^2}$.
$ f(\frac {1}{u}) < 1 \minus{} f(u) < 0$ $ \implies$ $ f^2(\frac {1}{u}) > f^2(u) \minus{} 2f(u) \plus{} 1$
$ f(v) \equal{} f(u) \plus{} f^2(\frac {1}{u})$ $ > f^2(u) \minus{} f(u) \plus{} 1\geq f(u) \plus{} 1$

So, if $ \exists u$ such that $ f(u)\geq 2$, then $ \exists v$ such that $ f(v)\geq f(u) \plus{} 1$

So, since $ f(2) \equal{} 2$, $ f(x)$ may be as great as we want, which is impossible, since $ f(x)\leq M$ $ \forall x$

So, no such function exists.

[quote="ll931110"]Problem 4:
Find all functionj $ f: R \rightarrow R$ satisfying
1/ $ f(x_1) \neq f(x_2) \forall x_1 \neq x_2$
2/ $ f(\frac {x \plus{} y}{x \minus{} y}) \equal{} \frac {f(x) \plus{} f(y)}{f(x) \minus{} f(y)} \forall x \neq y$[/quote]
Let $ P(x,y)$ be the assertion $ f(\frac{x\plus{}y}{x\minus{}y})\equal{}\frac{f(x)\plus{}f(y)}{f(x)\minus{}f(y)}$

$ P(x,0)$ $ \implies$ $ f(1)\equal{}\frac{f(x)\plus{}f(0)}{f(x)\minus{}f(0)}$ and so $ f(x)(f(1)\minus{}1)\equal{}f(0)(f(1)\plus{}1)$ and so, since $ (f(x)$ cant be a constant, since injective) :
$ f(0)\equal{}0$
$ f(1)\equal{}1$
$ P(0,1)$ $ \implies$ $ f(\minus{}1)\equal{}\minus{}1$

Compare now $ P(x,1)$ and $ P(xy,y)$ for $ x\neq 1$ and $ y\neq 0$. The two LHS are equal, so are the two RHS : $ \frac{f(x)\plus{}1}{f(x)\minus{}1}\equal{}\frac{f(xy)\plus{}f(y)}{f(xy)\minus{}f(y)}$ And so :

$ f(x)f(xy)\minus{}f(x)f(y)\plus{}f(xy)\minus{}f(y)$ $ \equal{}f(xy)f(x)\plus{}f(y)f(x)\minus{}f(xy)\minus{}f(y)$ and so $ \boxed{f(xy)\equal{}f(x)f(y)}$  and this is still true for $ x\equal{}1$ or $ y\equal{}0$

Let then $ f(2)\equal{}a$. $ P(x,1)$ may then be written (using $ f(xy)\equal{}f(x)f(y)$) : $ f(x\plus{}1)\equal{}f(x\minus{}1)\frac{f(x)\plus{}1}{f(x)\minus{}1}$ and so allow us to compute :
$ f(2)\equal{}a$
$ f(3)\equal{}\frac{a\plus{}1}{a\minus{}1}$
$ f(4)\equal{}a^2$
$ f(5)\equal{}\frac{a^2\plus{}1}{(a\minus{}1)^2}$
$ f(6)\equal{}a(a^2\minus{}a\plus{}1)$
Writing then $ f(6)\equal{}f(2)f(3)$ and using the fact that $ f(x)$ is injective (so $ a\neq 0$), we get $ a\equal{}2$.

Let now $ A\equal{}\frac{x^2\plus{}y^2\plus{}2xy}{x^2\plus{}y^2\minus{}2xy}$ for $ x\neq 0$, $ y\neq 0$ and $ x\neq y$

Using $ P(x^2\plus{}y^2,2xy)$ we obtain : $ f(A)\equal{}\frac{f(x^2\plus{}y^2)\plus{}2f(x)f(y)}{f(x^2\plus{}y^2)\minus{}2f(x)f(y)}$ (we needed $ f(2)\equal{}2$ in order to write this).

But $ A\equal{}(\frac{x\plus{}y}{x\minus{}y})^2$ and so $ f(A)\equal{}f^2(\frac{x\plus{}y}{x\minus{}y})$ $ \equal{}\frac{f^2(x)\plus{}2f(x)f(y)\plus{}f^2(y)}{f^2(x)\minus{}2f(x)f(y)\plus{}f^2(y)}$

Equating these two expressions give us $ f(x^2\plus{}y^2)\equal{}f(x^2)\plus{}f(y^2)$, which is obviously also true if $ x\equal{}0$, or $ y\equal{}0$, or $ x\equal{}y$

So $ f(x\plus{}y)\equal{}f(x)\plus{}f(y)$ $ \forall x,y\geq 0$ and, writing $ f(x)\equal{}f(x\plus{}y)\minus{}f(y)$, we have $ \boxed{f(x\plus{}y)\equal{}f(x)\plus{}f(y)\forall x,y}$

So we have a classical Cauchy equation with the complementary condition $ f(xy)\equal{}f(x)f(y)$ which gives, for any $ a>0$ $ f(x\plus{}a)\equal{}f(x)\plus{}f((\sqrt a)^2)\equal{}f(x)\plus{}f^2(\sqrt a)>f(x)$. So $ f(x)$ is strictly increasing and so $ f(x)\equal{}xf(1)$

Hence the solution $ f(x)\equal{}x$

@pco: I think I have another way for Problem 4 (at first, thanks for your help)

From $ f(xy) \equal{} f(x).f(y)$, replacing $ y$ by $ \frac{1}{x}$ into 2/ gives
$ f(\frac{x^2 \plus{} 1}{x^2 \minus{} 1}) \equal{} \frac{(f(x))^2 \plus{} 1}{(f(x))^2 \minus{} 1}$ (1)

However, we have $ f(\frac{x^2 \plus{} 1}{x^2 \minus{} 1}) \equal{} \frac{(f(x^2) \plus{} 1}{(f(x^2) \minus{} 1}$ (2)
Combining (1) and (2) gives $ f(x^2) \equal{} (f(x))^2 \forall x$, which means $ f(x) > 0 \forall x > 0$, which yields f is strictly increasing.

So we have $ f(x) \equal{} x^a$, and return 2/ gives $ a \equal{} 1$. So $ f(x) \equal{} x$

[quote="ll931110"] From $ f(xy) \equal{} f(x).f(y)$, replacing $ y$ by $ \frac {1}{x}$ into 2/ gives
$ f(\frac {x^2 \plus{} 1}{x^2 \minus{} 1}) \equal{} \frac {(f(x))^2 \plus{} 1}{(f(x))^2 \minus{} 1}$ (1)

However, we have $ f(\frac {x^2 \plus{} 1}{x^2 \minus{} 1}) \equal{} \frac {(f(x^2) \plus{} 1}{(f(x^2) \minus{} 1}$ (2)
Combining (1) and (2) gives $ f(x^2) \equal{} (f(x))^2 \forall x$, [/quote]

You dont need this step (using $ y\equal{}\frac{1}{x}$) : $ f(xy)\equal{}f(x)f(y)$ immediately implies $ f(x^2)\equal{}f^2(x)$

[quote="ll931110"] $ f(x^2) \equal{} (f(x))^2 \forall x$, which means $ f(x) > 0 \forall x > 0$, which yields f is strictly increasing.

So we have $ f(x) \equal{} x^a$, and return 2/ gives $ a \equal{} 1$. So $ f(x) \equal{} x$ [/quote]

And you're right. Well done :)

Be careful about the distinction $ x>0$ and $ x<0$

Sorry, I'm so silly :D.

In fact, we only need to consider whether $ x > 0$, because we have $ f(\minus{}x) \equal{} \minus{}f(x) \forall x$.

Can you help me to solve Problem 3, pco? Thanks a lot


search
inequalities theorems
inequalities

Is there any substitution for $ x$, $ y$ and $ z$ if if we have condition $ xy\plus{}yz\plus{}zx\plus{}xyz\equal{}4$ :?:

$ x\equal{}\frac{2a}{b\plus{}c},y\equal{}\frac{2b}{a\plus{}c},z\equal{}\frac{2c}{a\plus{}b}$.
 :wink:

If the range is positive integer,the solution will be only $ (1,1,1)$ :(

[quote="Erken"]2 zeroεinteger:
I think that he meant to use such substitution in inequalities. :wink:[/quote]

Sorry. :ninja:

[quote="FOURRIER"][quote="stephen38"]$ x \equal{} \frac {2a}{b \plus{} c},y \equal{} \frac {2b}{a \plus{} c}$[/quote]

If $ x \equal{} \frac {2a}{b \plus{} c}$ Then $ y$ does not necessary satisfies $ y \equal{} \frac {2b}{a \plus{} c}$

Im I right?[/quote]
you're wrong:
first of all WLOG we can assume that $ x \equal{} \frac {2a}{b \plus{} c},y \equal{} \frac {2b}{c \plus{} a}$ then calculate $ z$ we will have:

$ \frac {4ab}{(b \plus{} c)(c \plus{} a)} \plus{} z\left(\frac {2a}{b \plus{} c} \plus{} \frac {2b}{a \plus{} c} \plus{} \frac {4ab}{(b \plus{} c)(c \plus{} a)}\right) \equal{} 4$

$ \Rightarrow z\left(\frac {2(a \plus{} b)(a \plus{} b \plus{} c)}{(b \plus{} c)(c \plus{} a)}\right) \equal{} \frac {4(b \plus{} c)(c \plus{} a) \minus{} 4ab}{(b \plus{} c)(c \plus{} a)}$

$ \Rightarrow z \equal{} \frac {2(c^2 \plus{} ac \plus{} bc)}{(a \plus{} b)(a \plus{} b \plus{} c)} \equal{} \frac {2c}{a \plus{} b}$

so if $ xy \plus{} yz \plus{} zx \plus{} xyz \equal{} 4$ then we can substitute $ x,y,z$ by:

$ \frac {2a}{b \plus{} c},\frac {2b}{c \plus{} a},\frac {2c}{a \plus{} b}$

[quote="BaBaK Ghalebi"][quote="FOURRIER"][quote="stephen38"]$ x \equal{} \frac {2a}{b \plus{} c},y \equal{} \frac {2b}{a \plus{} c}$[/quote]

If $ x \equal{} \frac {2a}{b \plus{} c}$ Then $ y$ does not necessary satisfies $ y \equal{} \frac {2b}{a \plus{} c}$

Im I right?[/quote]
you're wrong:
first of all WLOG we can assume that $ x \equal{} \frac {2a}{b \plus{} c},y \equal{} \frac {2b}{c \plus{} a}$
[/quote]

 :coolspeak:

I was asking if we can assume  $ x \equal{} \frac {2a}{b \plus{} c},y \equal{} \frac {2b}{c \plus{} a}$ the rest I know is easy, you search $ z$  :wink:
Thank you anyway  :yup:

[quote="FOURRIER"]
I was asking if we can assume  $ x \equal{} \frac {2a}{b \plus{} c},y \equal{} \frac {2b}{c \plus{} a}$
[/quote]
We can assume it because,if we  fix  the  number $ a$ and let $ p,q\in R$,such that $ p\plus{}q\equal{}r$=const and $ x \equal{} \frac {2a}{p \plus{} q}$.
Now let $ p\equal{}\frac{y(\frac{2a}{x}\minus{}p\plus{}a)}{2}$.so we can find $ p$ and also $ q$.Now let $ p\equal{}b,q\equal{}c$.Thus we are done.

[quote="Erken"][quote="FOURRIER"]
I was asking if we can assume  $ x \equal{} \frac {2a}{b \plus{} c},y \equal{} \frac {2b}{c \plus{} a}$
[/quote]
We can assume it because,if we  fix  the  number $ a$ and let $ p,q\in R$,such that $ p \plus{} q \equal{} r$=const and $ x \equal{} \frac {2a}{p \plus{} q}$.
Now let $ p \equal{} \frac {y(\frac {2a}{x} \minus{} p \plus{} a)}{2}$.so we can find $ p$ and also $ q$.Now let $ p \equal{} b,q \equal{} c$.Thus we are done.[/quote]

I see it now  :wink:

triangle substitution!!!!!
notice that $ x \equal{} 2a  ,     y \equal{} 2b  ,    z \equal{} 2c$
then we have $ ab \plus{} bc \plus{} ca \plus{} 2abc \equal{} 1$
let $ p \equal{} \sqrt {ab}  ,q \equal{} \sqrt {bc}   ,     r \equal{} \sqrt {ca}$
then we have $ p^{2} \plus{} q^{2} \plus{} r^{2} \plus{} 2pqr \equal{} 1$
so we can make the substitution:
$ A,B,C \in [0,\frac{\pi}{2}]$
we have:
$ p \equal{} cosA      ,q \equal{} cosB    ,    r \equal{} cosC$
yes!

[quote="canada"]Is there any substitution for $ x$, $ y$ and $ z$ if we have condition $ xyz \plus{} yz \plus{} zx \plus{} xy \equal{} 4$ :?:[/quote]
[quote="stephen38"]$ x \equal{} \frac {2a}{b \plus{} c},y \equal{} \frac {2b}{c \plus{} a},z \equal{} \frac {2c}{a \plus{} b}$.:wink:[/quote]
$ a: b: c \equal{} \frac {x}{x \plus{} 2}: \frac {y}{y \plus{} 2}: \frac {z}{z \plus{} 2}$. :yup:

[quote="canada"]Is there any substitution for $ x$, $ y$ and $ z$ if if we have condition $ xy \plus{} yz \plus{} zx \plus{} xyz \equal{} 4$ :?:[/quote]
what about this one
\[ a\plus{}b\plus{}c\plus{}abc\equal{}4\]
???

[quote="Wolfbrother"]Can I somewhere find all usefull substitutions?[/quote]

Hojo lee book

you can find more at:
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=55&t=404216
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=55&t=404216&start=20
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=55&t=404216&start=40
and that link:
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=55&t=551821

[quote="Wolfbrother"]Can I somewhere find all usefull substitutions?[/quote]
You also can find in Problem from the book of Titu.