floor function
inequalities
number theory
greatest common divisor
least common multiple
triangle inequality
algebra unsolved

For $ a_i \in \mathbb{Z}^ \plus{}$, $ i \equal{} 1, \ldots, k$, and $ n \equal{} \sum^k_{i \equal{} 1} a_i$, let $ d \equal{} \gcd(a_1, \ldots, a_k)$ denote the greatest common divisor of $ a_1, \ldots, a_k$.
Prove that $ \frac {d} {n} \cdot \frac {n!}{\prod\limits^k_{i \equal{} 1} (a_i!)}$ is an integer.

[i]Dan Schwarz, Romania[/i]

[quote="greentreeroad"][hide="hint"]write d as a linear combination of all the a's, you will find that you write the value as the sum of multiples of multinomial coefficients, which is known to be integers.

This is the official solution, but I've not heard of anyone do this during contests. Most people solved it by counting prime factors.[/hide][/quote]
When I saw the problem I solved it the same way as the official solution. But that was only because i had seen the nice solution to  $ \frac{(n,m)}{n} \binom{n}{m}$ is an integer. :)

[quote="greentreeroad"][hide="hint"]write d as a linear combination of all the a's, you will find that you write the value as the sum of multiples of multinomial coefficients, which is known to be integers.

This is the official solution, but I've not heard of anyone do this during contests. Most people solved it by counting prime factors.[/hide][/quote]
That's just too cool....
For the sake of completeness, here's the prime factor counting solution:
We want to show that $ \frac {n}{d} | \binom{n}{a_1,a_2,\cdots,a_k}$. Let $ p$ be a prime divisor of the LHS.
It is sufficient to show that for each $ 1 \le i \le k$ we have: $ v_p(n) \minus{} v_p(a_i) \le v_p(n!) \minus{} \sum_{i \equal{} 1}^{k} v_p(a_i!)$
Recall that $ v_p(t!) \equal{} \sum_{i \equal{} 1}^{\infty} \left\lfloor \frac {t}{p^i} \right\rfloor$, and notice that $ v_p(t) \equal{} v_p(t!) \minus{} v_p((t \minus{} 1)!) \equal{} \sum_{i \equal{} 1}^{\infty} \left\lfloor \frac {t}{p^i} \right\rfloor \minus{} \left\lfloor \frac {t \minus{} 1}{p^i} \right\rfloor$
So it is enough to show that for any $ i \in \mathbb{N}$ and $ 1 \le j \le k$, we have:
$ \left\lfloor \frac {n}{p^i} \right\rfloor \minus{} \left\lfloor \frac {n \minus{} 1}{p^i} \right\rfloor \minus{} \left\lfloor \frac {a_j}{p^i} \right\rfloor \plus{} \left\lfloor \frac {a_j \minus{} 1}{p^i} \right\rfloor \le \left\lfloor \frac {n}{p^i} \right\rfloor \minus{} \sum_{i \equal{} 1}^{k} \left\lfloor \frac {a_i}{p^i} \right\rfloor$
Or:
$ \left\lfloor \frac {n \minus{} 1}{p^i} \right\rfloor \ge \sum_{i \neq j} \left\lfloor \frac {a_i}{p^i} \right\rfloor \plus{} \left\lfloor \frac {a_j \minus{} 1}{p^i} \right\rfloor$

But this is true - it is the opposite of the triangle inequality for the floor function.
Proof of this property of $ \left\lfloor x \right\rfloor$:
[hide]
We want to show that $ \sum_{i \equal{} 1}^{n} \left\lfloor x_i \right\rfloor \le \left\lfloor \sum_{i \equal{} 1}^{n} x_i \right\rfloor$ (*)
Notice that if we add any integer to $ x_i$, it doesn't affect the validity of  (*). So we can assume WLOG that $ 0 \le x_i < 1$ for any $ i$.
The LHS becomes 0 and the RHS is obviously non-negative, and we're done.
[/hide]

My solution uses  a well known problem in number theory  $ d\equal{}\gcd(a_1,..,a_n)$ ,we can find $ (x_1,..,x_n)\in \mathbb{Z}^n$ such that $ \sum_{i\equal{}1}^n a_ix_i\equal{}n$ .Thus we can write this number in the form :
\[ \frac{d.(n\minus{}1)!}{\prod_{i\equal{}1}^n (a_i)!} \equal{}\sum_{i\equal{}1}^n x_i \binom{n}{a_i\minus{}1,a_{i\plus{}1},\cdots,a_{i\minus{}1}}
\]
and obvious this number is an integer .

[quote="TTsphn"]My solution uses  a well known problem in number theory  $ d \equal{} \gcd(a_1,..,a_n)$ ,we can find $ (x_1,..,x_n)\in \mathbb{Z}^n$ such that $ \sum_{i \equal{} 1}^n a_ix_i \equal{} n$ .Thus we can write this number in the form :
\[ \frac {d.(n \minus{} 1)!}{\prod_{i \equal{} 1}^n (a_i)!} \equal{} \sum_{i \equal{} 1}^n x_i \binom{n}{a_i \minus{} 1,a_{i \plus{} 1},\cdots,a_{i \minus{} 1}}
\]
and obvious this number is an integer .[/quote]
You mean $ \sum_{i \equal{} 1}^n a_ix_i \equal{} d$. It was the same solution greentreeroad stated  :lol:

We don't even need Bezout's Lemma (the fact that the GCD of some numbers is a linear combination of them over $ \mathbb{Z}$)!
$ \frac{n}{a_i} | \binom{n}{a_1,\cdots,a_k}$ holds for each $ i$ from the identity $ \frac{a_i}{n}\binom{n}{a_1,\cdots,a_k}\equal{}\binom{n\minus{}1}{a_1,\cdots,a_{i\minus{}1},a_i\minus{}1,a_{i\plus{}1},\cdots,a_k}$.
Thus, $ [\frac{n}{a_1},\cdots,\frac{n}{a_k}] | \binom{n}{a_1,\cdots,a_k}$. But this lcm is actually $ \frac{n}{d}$!

([] means LCM)

Analyzing my solution again, It seems as if it is not so different than greentreeroad or TTsphn solutions:
I showed that the problem is equivalent to showing that for any $ i$:
$ \sum_{k\equal{}1}^{\infty} \left\lfloor \frac {n \minus{} 1}{p^k} \right\rfloor \ge \sum_{k\equal{}1}^{\infty} \sum_{i \neq j} \left\lfloor \frac {a_i}{p^k} \right\rfloor \plus{} \left\lfloor \frac {a_j \minus{} 1}{p^k} \right\rfloor$
Which is the same as $ \binom{n\minus{}1}{a_1,\cdots,a_{i\minus{}1},a_i\minus{}1,a_{i\plus{}1},\cdots,a_k}$ being an integer...

Let $ Z\equal{}\frac{d \cdot (n\minus{}1)!}{\prod^k_{i\equal{}1} (a_i!)}$. Also let $ b_i\equal{}\frac{a_i}d$ for all $ i$.

$ (a_1\plus{}\ldots\plus{}a_{k\minus{}1})!$ is divisible by $ a_1!a_2!\ldots a_{k\minus{}1}!$, and $ (a_1\plus{}\ldots\plus{}a_{k\minus{}1}\plus{}1)(a_1\plus{}\ldots\plus{}a_{k\minus{}1}\plus{}2)\ldots(n\minus{}1)$ is divisible by $ (a_k\minus{}1)!$. So $ Zb_k$ is an integer. Similarly, $ Zb_i$ is integer for any $ i$. But $ gcd(b_1,b_2,\ldots,b_k)\equal{}1$, so $ Z$ must be an integer.

I'm not sure with this solution, especially the last part (it's just intuitive). Please correct me if I'm wrong.

Use $v_p(a!)=(n-s_p(a))/(p-1)$ and then use $s(a_1)+...+s(a_k)=s(n)-(p-1)X$ where $X$ is number of carryovers

First lemma : $v_p(n!) = \frac{n - s_p(n)}{p - 1}$ where $v_p(x) =$ the exponent of $p$ in $x$ and $s_p(x) =$ the sum of digits of $x$ in base $p$.
Second lemma : $(p - 1)v_p(a) + s_p(a) = 1 + s_p(a - 1)$ .
Proof of lemma 2: $v_p(n!) = v_p(n) + v_p((n - 1)!)$ and using lemma 1 we have the conclusion .
Now it is obvious that : $v_p(d) = min(v_p(a_i))$ .Let's suppose that we have this minimum for $a_1$ .
Using lemma1 we conclude that we need to solve this:
$(p - 1) v_p(a_1) + s_p(a_1) + ... + s_p(a_n) \ge (p -1)v_p(n) + s_p(n)$
Using lemma2 we have to show that:
$1 + s_p(a_1 - 1)  + s_p(a_2) + ... + s_p(a_k) \ge 1 + s_p(n - 1)$ so we need to show that :
$s_p(a_1 - 1) + s_p(a_2) + ... + s_p(a_k) \ge s_p(a_1 - 1 + a_2 + ... + a_k)$
Using that $s_p(x) + s_p(y) \ge s_p(x + y)$ we have the conclusion.

[hide=Solution][quote=RMM 2009 P1]For $ a_i \in \mathbb{Z}^ \plus{}$, $ i \equal{} 1, \ldots, k$, and $ n \equal{} \sum^k_{i \equal{} 1} a_i$, let $ d \equal{} \gcd(a_1, \ldots, a_k)$ denote the greatest common divisor of $ a_1, \ldots, a_k$.
Prove that $$ \frac {d} {n} \cdot \frac {n!}{\prod\limits^k_{i \equal{} 1} (a_i!)}$$ is an integer.
[/quote]
Choose a fixed prime $p$
\begin{align*}
&\nu_p \left( \frac{\gcd (a_1,a_2, \ldots, a_k )}{a_1+a_2+\ldots +a_k} \cdot \frac{ \left( a_1+a_2+\ldots +a_k \right) !}{\prod_{i=1}^{k} (a_k)!} \right) \\
&=  \nu_p \left( (a_1+a_2+\ldots +a_k)! \right) + \nu_p \left( \gcd(a_1,a_2,\ldots ,a_k ) \right) - \nu_p \left( a_1+a_2+\ldots +a_k \right) - \sum_{\ell =1}^{k} \nu_{p} \left( a_{\ell} \right) ! \\
&= \sum_{m \geq 1} \left( \left\lfloor \frac{a_1 +a_2+ \ldots +a_k}{p^m} \right\rfloor - \sum_{\ell =1}^{k} \left\lfloor \frac{a_{\ell}}{p^m} \right\rfloor \right)+\nu_p \left( \gcd( a_1,a_2, \ldots , a_k ) \right) - \nu_p \left( a_1+a_2+\ldots +a_k \right) \\
& - - - ~ ~ \left( \bigstar\right) \end{align*}
---------------------
$\textbf{Lemma (1)} ~$  Let $x,y,d \in \mathbb{Z}^+$
\begin{align*} \left\lfloor \frac{x+y}{d} \right\rfloor \geq \left\lfloor \frac{x}{d} \right\rfloor + \left\lfloor \frac{y}{d} \right\rfloor \end{align*}
---------------------
If there exists an index $i$, such that $\nu_p (a_i) \le \nu_p (a_b)$ for all $b \neq i$ $\le k$ and there exists an index $j$, such that $\nu_p (a_i)$ $<$ $\nu_p (a_j)$
\begin{align*} \therefore \qquad &\nu_p \left( a_1 +a_2+ \ldots + a_k \right) = \nu_p (a_i)\\
&= \min \{ \nu_p (a_1), \nu_p (a_2), \ldots , \nu_p (a_k) \} \\
&=\nu_p \left( \gcd(a_1,a_2, \ldots , a_k) \right) \\ \end{align*}
\begin{align*} \left( \bigstar \right)  =  \sum_{m \geq 1} \left( \left\lfloor \frac{a_1 +a_2+ \ldots +a_k}{p^m} \right\rfloor - \sum_{\ell =1}^{k} \left\lfloor \frac{a_{\ell}}{p^m} \right\rfloor \right) \geq 0 \end{align*}
as desired! Suppose otherwise $\nu_p (a_i) =\nu_p (a_j)=\chi $ for all $i,j \le k$ and let
\begin{align*} \nu_p \left( \frac{a_1+a_2+ \ldots +a_k }{p^{\chi }} \right) =y \end{align*}
Thus, we need to show
\begin{align*} \left( \bigstar \right)  =  \sum_{m \geq 1} \left( \left\lfloor \frac{a_1 +a_2+ \ldots +a_k}{p^m} \right\rfloor - \sum_{\ell =1}^{k} \left\lfloor \frac{a_{\ell}}{p^m} \right\rfloor \right) \geq y \end{align*}
Observe that
\begin{align*}  \sum_{m = 1}^{\chi} \left( \left\lfloor \frac{a_1 +a_2+ \ldots +a_k}{p^m} \right\rfloor - \sum_{\ell =1}^{k} \left\lfloor \frac{a_{\ell}}{p^m} \right\rfloor \right) =0 \end{align*}
And for all $\chi < m \le \chi +y $
\begin{align*} \left\lfloor \frac{a_1+a_2+ \ldots +a_k }{p^m} \right\rfloor = \frac{a_1+a_2+ \ldots + a_k}{p^m} > \sum_{\ell =1}^{k} \left\lfloor \frac{a_{\ell}}{p^m} \right\rfloor \end{align*}
Thus, summing up from $m = \chi +1$ to $m= \chi +y$
\begin{align*} \sum_{m \geq \chi} \left( \left\lfloor \frac{a_1 +a_2+ \ldots +a_k}{p^m} \right\rfloor - \sum_{\ell =1}^{k} \left\lfloor \frac{a_{\ell}}{p^m} \right\rfloor \right) \geq y \qquad \blacksquare \end{align*}
------------------------
If there's anything missing in this proof, might be the case when $\nu_p (a_i) =0$, but the details are easy to fix in, and I'm really lazy (:P), so I'll let it be[/hide]

By Bezout's lemma, let $d=\sum_{i=1}^{k}a_ib_i$ where $b_i\in\mathbb{Z}.$ Notice that $$\frac{d}{n}\cdot\frac{n!}{\prod_{i=1}^{k}{(a_i!)}}=\frac{n!}{\prod_{i=1}^{k}{(a_i!)}}\sum_{i=1}^{k}{\frac{a_ib_i}{n}}=\sum_{i=1}^{k}{\left(b_i\cdot\frac{n!}{n}\cdot\frac{a_i}{\prod_{i=1}^{k}{(a_i)!}}\right)}=\sum_{i=1}^{k}{b_i\binom{n-1}{a_1,.,a_i-1,.,a_k}}$$ which is an integer. $\square$

Take $v_p$ for an arbitrary prime $p$. Using the identity $v_p(n!)=\frac{n-s_p(n)}{p-1},$ this reduces to showing that $$(p-1)v_p(d)+s_p(a_1)\cdots+s_p(a_n)\geq s_p(n)+(p-1)v_p(n).$$ Note that $$s_p(a_1)\cdots+s_p(a_n)-s_p(n)$$ is $p-1$ times the number of carry-overs when adding $a_1+a_2\cdots +a_n$. Let this be $C$. Then, we just want to show that $$C\geq v_p(n)-v_p(d).$$ Consider adding $a_1+a_2\cdots +a_n$. We know that all digits after $v_p(n)$ are zero in the final sum. However, all places on or to the left of $v_p(d)$ will have at least one addend with a non-zero digit in that place. Hence, each place value in between these two must have a carry-over, so we are done.


If $ A \equal{} \{\minus{}4,\minus{}3,\minus{}2,\minus{}1,0\}$ and $ B \equal{} \{\minus{}3,\minus{}1,1,3\}$, then what is the sum of the numbers in $ A \cap B$?

That symbol means intersection, so everything in both A and B, so {-3,-1} so the sum is -4.


geometry
3D geometry
sphere
pyramid
search

What are the conditions to inscribe a sphere into a pyramid with a convex quadraliteral base??

Did you mean this?

http://artofproblemsolving.com/Forum/viewtopic.php?search_id=503322988&t=181591&sid=87339218da1a6e6bba261648a9562bf9


You should try looking for the email or number of someone involved with ARML.  They probably have extras you just need to find the right person.

Check with Don Barry, dbarry@andover.edu.  i think he has the remaining copies to send out.  if that doesn't work, i can put you in touch with someone else..

Tim Sanders
ARML Past President

There are instructions for how to order on the Resources page of our site (under problem books in the middle section).


modular arithmetic
number theory
relatively prime
greatest common divisor
algebra unsolved
algebra

For some relatively PRIME positive integers $ a$ and $ b$, find all integers that can be greatest common divisor of $ a \plus{} b$ and $ \frac {a^{2005} \plus{} b^{2005}}{a \plus{} b}$.

EDIT: Lol, didn't put 'prime' in there :D

What exactly do you mean by "relatively positive integers"? (Are they more positive than some other integer, or something like that? :wink:)

I'm assuming that you meant "relatively prime and positive".

[quote="FantasyLover"]For some relatively positive integers $ a$ and $ b$, find all integers that can be greatest common divisor of $ a \plus{} b$ and $ \frac {a^{2005} \plus{} b^{2005}}{a \plus{} b}$.[/quote]

If you mean for relatively prime positive integers $ a$ and $ b$, then

$ \frac{a^{2005}\plus{}b^{2005}}{a\plus{}b} \equiv a^{2004}\minus{}a^{2003}b\plus{}a^{2002}b^2\minus{}....\minus{}ab^{2003}\plus{}b^{2004}$
$ \equiv a^{2004}\minus{}a^{2003}(\minus{}a)\plus{}a^{2002}(\minus{}a)^2\minus{}....\minus{}a(\minus{}a)^{2003}\plus{}(\minus{}a)^{2004}$
$ \equiv 2005a^{2004} \pmod{a\plus{}b}$

So $ \gcd(a\plus{}b,\frac {a^{2005} \plus{} b^{2005}}{a \plus{} b})\equal{}\gcd(a\plus{}b,2005a^{2004})\equal{}\gcd(a\plus{}b,2005)$

$ 2005\equal{}5 \times 401$, so the possible positive integers are $ 1,5,401,2005$

Awesome solution :)

One question: How did you get $ \gcd(a\plus{}b,2005a^{2004})\equal{}\gcd(a\plus{}b,2005)$?

Since $ a,b$ are relatively prime, $ a\plus{}b$ does not divide $ a^{2004}$.

Thus, $ \gcd(a\plus{}b,2005a^{2004})\equal{}\gcd(a\plus{}b,2005)$


[b]If x is positive integer greater or equal to 8,
f:x->x,
f(x+y)=f(xy)
for all x>=4
y>=4
if f(8)=9
then what is value of f(9)?[/quote]

i think this is the very simple question
put x=8 and y=1
then
$ f(8\plus{}1)\equal{} f(8)$
$ f(9)\equal{} f(8)\equal{}9$
hence $ f(9)\equal{} 9$ answer..

[b]thanx...
the answer is correct...

Can u Pl tell Me whats the function satisfying given conditions?

The answer is correct, but the solution is not, since it specifies that $ x,y \geq 4$.

Also, this is a double post, so this thread can be deleted  :wink:

SORRY FOR DOUBLE POST.
SOME BODY PL GIVE A PROPER SOLUTION.
the question is given in some olympiad paper in same way..........

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=171634]Go to your other thread[/url] for a solution.

put y=0
we get the function satisfy this is $ f(x)\equal{}f(0)$ that is constant function satisfy this......

you cant put x=0 ,since x>0
think again...........

Since this thread appears to be the dominant version, I'll post my solution here, too:

[hide="Solution"]
We may use:

f(9) = f(4+5) = f(4*5) = f(20)
f(20) = f(4+16) = f(4*16) = f(64)
f(64)=f(8*8) = f(8+8) = f(16)
f(16) = f(4*4) = f(4+4) = f(8)

Hence, [b]f(9) = f(8) = 9[/b][/hide]

[b]Follow up:[/b] Can we assert that $ f(m) \equal{} f(n) \; \forall m,n \in \mathbb{N}^{\geq 8}$? If not, can we assert that $ f$ is a valid function? That is, can we prove that $ f(n)$ can take on only one unique value for any $ n$?

[quote="aayyaa"]i m not getting you....[/quote]
The Zuton Force is saying:
Prove or disprove: $ f(n)$ is constant for any $ n$ in the domain.

[quote="The Zuton Force"]
[b]Follow up:[/b] Can we assert that $ f(m) \equal{} f(n) \; \forall m,n \in \mathbb{N}^{\geq 8}$? If not, can we assert that $ f$ is a valid function? That is, can we prove that $ f(n)$ can take on only one unique value for any $ n$?[/quote]
[hide="Follow Up Solution"]
Without loss of generality, say that $ m\ge n$.  We have that $ f(n) \equal{} f(n*1) \equal{} f(n \plus{} 1) \equal{} f([n \plus{} 1]1) \equal{} f(n \plus{} 2)...$, so it is obvious that $ f(n) \equal{} f(n \plus{} x)$, and since x can be anything, we can set $ x \equal{} m \minus{} n$, thus giving us that $ f(n) \equal{} f(m)$. [/hide]
EDIT: Oh... sorry...  :oops:

Please recall that the function condition $ f(x\plus{}y)\equal{}f(xy)$ holds [b]if and only if[/b] $ x \geq 4$ [u]and[/u] $ y \geq 4$  :)

[i](This was, in fact, why the solution to the original question was not simply f(9)=f(1+8)=f(8)  :wink: ) [/i]

That's exactly The Zuton Force's solution, almost line by line.


LaTeX
inequalities proposed
inequalities

With $ a,b,c>0$ such that $ \sum a^2\equal{}3$.Prove that $ 1 \ge \sum \frac{a(a^2\plus{}b^2\plus{}c^2)}{(a^2\plus{}b^2\plus{}1)(a^2\plus{}c^2\plus{}1)}$

[quote="stevenwang1992"]you can  use latex![/quote]
Yes，it's quite easy！


2. Prove that the diameter of the Feuerbach circle is equal to the radii of the circle outside the triangle.

Why do you think it would be okay to ask for help on the questions to an ongoing competition?


linear algebra
matrix
geometry
3D geometry

I'm just curious on what everyone's favorite movies were...

My [b]ultimate[/b] favorite: Bend It Like Beckham

Seabiscuit
Pleasantville
Eternal Sunshine of the Spotless Mind
Finding Neverland
Finding Nemo
Shrek
Incredibles
To Kill a Mockingbird
Fiddler on the Roof
Forrest Gump
Harry Potter 3
Pirates of the Caribbean
What about Bob?
A Walk to Remember
Phantom of the Opera
Mean Girls
Gandhi
Aviator
West Side Story
Beauty and the Beast
Sound of Music

I really love movies, especially comedies and drama-ish.  :D

favorite movie, hmmm...we might've had a topic on this already but...

favorites:

mummy/mummy returns
Star Wars 1-6
LOTR 1-3
Pirates of the Carribbean
Back to the Future 1-3
Gladiator
The Patriot
Shrek 1-2
Happy Gilmore (i love the part with Bob Barker :rotfl: )
Finally, Monty Python and the Holy Grail

here's some good ones:

LoTR 1-3
HP 1-3 (can't wait for 4 to come out!)
PotC
Mean Girls
How to Lose a Guy in Ten Days
Shrek 1&2
Sound of Music
Fiddler on the Roof
Walk to Remember
Pay it Forward
Incredibles

more l8r when i can think straight...

Poor of the people who have only one favorite movie. Now, some of the ones I have liked the most are

Abre los Ojos (Open your Eyes, probably you have seen the remake, called Vanilla Sky)
Shrek
Thirteen Floor
Matrix
Fight Club
Braveheart
Life 101
Se7ven
The Game
The Cube
Requiem for a dream
American History X
The Score
Ocean's eleven
A River Runs Through It
Big Fish
Young Frankenstein
Magnolia
Monsters Inc.
Groundhog day
Brazil
Lean on me
Mr. Holland's Opus
Amadeus
Face Off

International Films

La vita e bella
Amores Perros
Hable con ella
Nadie conoce a nadie
La ciudad sin limites
Diarios de Motocicleta
Nueve Reinas
Lola Rennt
Nuovo cinema Paradiso
La Noche de los Lapices
Todo sobre mi madre
La Ley del deseo
Tacones Lejanos
Y tu mama tambien

And there are a bunch more.

The Green Mile
Amadeus
Shawshank Redemption
The Leopard (Il Gattopadro)

(not really a big movie fan)

I love to see the movies and my favourite movies are:
$Shrek1-2$
$\text{Mr.Holland's Opus}$

I really love the $\text{Mr.Holland's Opus}$. Can someone post some links related to that film?
Thanks!

I like any and all Monty Python movies :D

LOTR ROTK!!!!!!!!!!! :D
and the other 2 are good
matrix
matrix reloaded
pirates of the carribean
hero
once upon a time in china 1,2,3
monty python and the holy grail
Shrek 1 and 2
Star Wars 4,5,6
The One (maybe....)

elvenchamp777

Not in order: American History X, A beautiful mind, Million dollar baby, Aviator...

I heard Million dollar baby was a bad movie.....  Aviator was ok i guess.

elvenchamp777

[hide=" Best Movie Ever"]Happy Gilmore[/hide]

My Favorites movies are [url=http://www.imdb.com/chart/top]Top 250[/url] :D! Well i'm planning to see them all, of course. By now I saw about 35 of them...  [b]Forrest Gump, Amelie, Amadeus[/b]

i hated napolean dynamite, there was no plot and none of the jokes were funny :(

I like inside man
and a history of violence

sorta recent movies

SAW 1 - 2

scary movie 3
scary movie 4 sucks

hoodwinked, scary movie 3, underworld and underworld evolution, finding nemo, star wars, LOTR, harry potter (especially 3)....bunch more

my opinion: most oscar nominated movies are boring.

shrek 1 and 2 are my all time favorites.
i like comedy a lot more than anything else.

monty pyton stuff is great
napoleon dynamite was a "bestestest movie of 2004" prize from the captain sugar foundation.
scary movie 4 was insane, but funny.
knig kong was okay.
hoodwinked was hilarious and funny and stuff. mostly because it used old characters in stupid new plots. ;)  (it's my 6th favorite movie ever)
 finding nemo and forrest gump were okay, but ive seen them infinity number of times now. :)

one of my very favorite movies of ever is "It's a beautiful life." ( i can't remember what it was in italian.)  it was made in italy, so everyone spoke italian. i had to read the subtitles. it was about the holocaust, but it was really funny and sad at the same time. most holocaust films show how sad it was.  this one was so different, it was good.
a really good part was when the little family of mom and dad and the 6-year old son were captured and put in a concentration camp.  the dad told the kid that they were going to camp and said he had to hide in the barrack bed room thing so "the mean guys that yell" didn't find him. if they didn't find him, he got some points.  he said to his son, "once you get 1000 points, you get a prize." "what is it?" "it's a tank."  at the end the father gets shot by the germans the day before they are liberated by the americans.  the germans run off. when the americans come, the son is playing outside, and the first thing he sees is a tank coming around a corner.  he says "we did win a tank!" it's really good.

[quote="Captain Sugar"]
one of my very favorite movies of ever is "It's a beautiful life." ( i can't remember what it was in italian.)  it was made in italy, so everyone spoke italian. i had to read the subtitles. it was about the holocaust, but it was really funny and sad at the same time. most holocaust films show how sad it was.  this one was so different, it was good.
a really good part was when the little family of mom and dad and the 6-year old son were captured and put in a concentration camp.  the dad told the kid that they were going to camp and said he had to hide in the barrack bed room thing so "the mean guys that yell" didn't find him. if they didn't find him, he got some points.  he said to his son, "once you get 1000 points, you get a prize." "what is it?" "it's a tank."  at the end the father gets shot by the germans the day before they are liberated by the americans.  the germans run off. when the americans come, the son is playing outside, and the first thing he sees is a tank coming around a corner.  he says "we did win a tank!" it's really good.[/quote]

I saw recently [b]La Vita e Bella[/b]. After my opinion the plot was very simple and natural acted. The ideea if the fim I think is: Life is Beautiful as long as you have a reason for living it just as you are" :)

I love Pink Panther...omigosh so funny! We loved it over here?!?!??1

Yes, I realize that this topic is old and out of date, but I absolutely cannot believe that nobody has mentioned "The Rock"!  GREATEST MOVIE EVER!  The music, the explosions, the plot, the car chases. . . .AWESOME!  yes, it wins, no arguing.


function
linear algebra
algebra unsolved
algebra

Find all functions $f: \mathbb{Z}\leftarrow \mathbb{Z}$ such that $f(f(x)+y+1)=x+1+f(y)$ for all $(x,y) \in \mathbb{Z^2}$

[quote="chess64"]What is meant by $\mathbb{Z}^2$? :blush:[/quote]

$\mathbb{Z}^2=\mathbb{Z}\times\mathbb{Z}=\{(a,b)|a,b\in\mathbb{Z}\}$ (Generally, $S^n$ is short for Descartes' product $\underbrace{S\times S\times\dots\times S}_{n}$)


Given $ p\star$$ q\equal{}2p\minus{}\frac{q}{2}$, evaluate $ 13\star40$.

We have that $ p\equal{}13$ and $ q\equal{}40$ so $ 13\star 40\equal{}2(13)\minus{}\frac{40}{2}\equal{}26\minus{}20\equal{}6$.


[quote=hashtagmath][hide=Solution]
$f(2) = \frac{(6)(2)}{2^2-1} = \frac{12}{3} = \boxed{4}$.
[/hide][/quote]

wait.. so now you're bumping every single old thread??

edit: Not only that, you didn't make a proper contribution, as you just posted the exact same strategy that the first person to solve it used..

[quote=hashtagmath][quote=mobro] you just posted the exact same strategy that the first person to solve it used..[/quote]

OK, [hide=Different Solution]
We can begin by graphing the function. Doing so gives us [img]https://lh3.googleusercontent.com/-P0ibwBoTVpI/X8liW9-9AFI/AAAAAAAAMY8/9Re_bnMjT6gYd_LxyXUNhPDY0PxvmaGfACK8BGAsYHg/s512/2020-12-03.png[/img]

If we look closely at $x=2$, we notice that the $y$ value corresponds to the value of $4$. Thus, $f(2) = 4$. $\blacksquare$
[/hide][/quote]

Well this needs a graphing calculator, so technically not valid sol...


Let a,b,c,d  are postive real numbers, and   abcd=1
prove  that
$ \frac {1}{a(b \plus{} 1)} \plus{} \frac {1}{b(c \plus{} 1)} \plus{} \frac {1}{c(d \plus{} 1)} \plus{} \frac {1}{d(a \plus{} 1)}\geq2$
Thanks...


Let $ f$: $ \mathbb{N^*} \rightarrow \mathbb{N^*}$
satisfying :

a. $ f(ab) \equal{} f(a)f(b)$ for all positive integers
b. $ f$ is increasing ($ a < b \implies f(a) < f(b)$)
c. $ f(3)\geq 7$

Find the minimal value that $ f(3)$ might have.

Morocan 2nd test 08

Begining of a solution:
[hide]$ f \equal{} n^2$ satisfies all the conditions with $ f(3) \equal{} 9$ so possible values of min values of $ f(3)$ are $ 7,8,9$[/hide]

Someone continue please...
 :)

[quote="tchebytchev"]if $ f(3) \equal{} 7$ then $ f(2)^3 \equal{} f(8) < f(9) \equal{} f(3)^2 \equal{} 49$ so $ f(2) \le 3$ and so, $ f(12) \equal{} f(3)f(2)f(2) \le 7.3.3 \equal{} 42 < 49 \equal{} f(9)$ which it is a contradiction .
if $ f(3) \equal{} 8$ then $ f(2)^3 \equal{} f(8) < f(9) \equal{} f(3)^2 \equal{} 64$ so $ f(2) \le 3$ and so ,$ f(12) \equal{} f(3)f(2)f(2) \le 7.3.3 \equal{} 42 < 49 \equal{} f(9)$ which it is also a contradiction .
so$ f(3) \ge 9$ , anf for $ f(n) \equal{} n^2$ we have $ f(3) \equal{} 9$.[/quote]

For $ f(3)\equal{}8$, your demo is wrong, since we have :
If $ f(3) \equal{} 8$ then $ f(2)^3 \equal{} f(8) < f(9) \equal{} f(3)^2 \equal{} 64$ so $ f(2) \le 3$ and so ,$ f(12) \equal{} f(3)f(2)f(2) \le 8.3.3 \equal{} 72 > 64 \equal{} f(9)$ which is not a contradiction.

But, with the same idea we can write :
If $ f(3) \equal{} 8$ then $ f(2)^3 \equal{} f(8) < f(9) \equal{} f(3)^2 \equal{} 64$ so $ f(2) \le 3$ and so :
$ f(81)\equal{}f(3^4)\equal{}8^4\equal{}4096$
$ f(96)\equal{}f(3\times 2^5)\leq 8\times 3^5 \equal{} 1944$
And this is a contradiction.

[quote="pco"][quote="tchebytchev"]if $ f(3) \equal{} 7$ then $ f(2)^3 \equal{} f(8) < f(9) \equal{} f(3)^2 \equal{} 49$ so $ f(2) \le 3$ and so, $ f(12) \equal{} f(3)f(2)f(2) \le 7.3.3 \equal{} 42 < 49 \equal{} f(9)$ which it is a contradiction .
if $ f(3) \equal{} 8$ then $ f(2)^3 \equal{} f(8) < f(9) \equal{} f(3)^2 \equal{} 64$ so $ f(2) \le 3$ and so ,$ f(12) \equal{} f(3)f(2)f(2) \le 7.3.3 \equal{} 42 < 49 \equal{} f(9)$ which it is also a contradiction .
so$ f(3) \ge 9$ , anf for $ f(n) \equal{} n^2$ we have $ f(3) \equal{} 9$.[/quote]

For $ f(3) \equal{} 8$, your demo is wrong, since we have :
If $ f(3) \equal{} 8$ then $ f(2)^3 \equal{} f(8) < f(9) \equal{} f(3)^2 \equal{} 64$ so $ f(2) \le 3$ and so ,$ f(12) \equal{} f(3)f(2)f(2) \le 8.3.3 \equal{} 72 > 64 \equal{} f(9)$ which is not a contradiction.

But, with the same idea we can write :
If $ f(3) \equal{} 8$ then $ f(2)^3 \equal{} f(8) < f(9) \equal{} f(3)^2 \equal{} 64$ so $ f(2) \le 3$ and so :
$ f(81) \equal{} f(3^4) \equal{} 8^4 \equal{} 4096$
$ f(96) \equal{} f(3\times 2^5)\leq 8\times 3^5 \equal{} 1944$
And this is a contradiction.[/quote]
sorry I hade a mistake , but I think that it is true now.

[quote="pbornsztein"]It is a classical result from Erdös that if $ f: \mathbb{N}^* \rightarrow \mathbb{R}$ is increasing and multiplicative then $ f$ is the zero function or there exists $ c$ such that $ f(n) \equal{} n^c$ for all $ n$.

Pierre.[/quote]

Could you give me a link where I can see the proof of this please

It reminds to prove that $ \boxed{''\forall n: n^c \in \mathbb{N} ''  \implies c \in \mathbb{N}}$, wich I can't prove it

If it's true then for $ c \equal{} 2$ $ f(3) \equal{} 9$ and done : the min value is $ 9$

You do not need this second result, because you only have to test $ f(3)\equal{}7$ or $ f(3)\equal{}8$.

Pierre.

[quote="pbornsztein"]You do not need this second result, because you only have to test $ f(3) \equal{} 7$ or $ f(3) \equal{} 8$.

Pierre.[/quote]

but: $ f$:  $ \mathbb{N^*} \rightarrow \mathbb{N^*}$  :wink:

so $ n^c \in \mathbb{N^*} obligatory$

what do you mean by the second result

$ f(n)\equal{}n^c$
or
 $ \boxed{''\forall n: n^c \in \mathbb{N} '' \implies c \in \mathbb{N}}$

?

That one :

[quote="FOURRIER"]
 $ \boxed{''\forall n: n^c \in \mathbb{N} '' \implies c \in \mathbb{N}}$
[/quote]

Pierre.


hello, for 1) the left side is given by
$3(2(a^2+b^2+c^2)-2(ab+bc+ca))=3((a-b)^2+(b-c)^2+(c-a)^2)$
Sonnhard.

hello, for 2) we have
$(y^2+zy+x^2)(x^2-zy)$
Sonnhard.

(3)

This can be treated as a quadratic in $x$: $2x^2+x(6y+5z)+4y^2+6yz+2z^2$

After elementary procedures, the result is $(2x+2y+z)(x+2y+2z)$

#8.

LaTeX if $abc = 1$ prove that

$(a + b + c)\left(\frac{1}{a} - \frac{1}{b} - \frac{1}{c}\right) = 1 + (a + b)(b + c)(c + a)$

[hide="Proof"]

Brute force:

$1 - \frac{a}{b} - \frac{a}{c} + \frac{b}{a} - 1 - \frac{b}{c} + \frac{c}{a} - \frac{c}{b} - 1 = 1 + (ab + ac + b^2 + bc)(c + a) \Rightarrow$
$-2 - \left(\frac{a}{b} + \frac{b}{c} - \frac{c}{a}\right) - \left(\frac{a}{c} - \frac{b}{a} + \frac{c}{b}\right) = abc + a^2b + ac^2 + a^2c + b^2c + ab^2 + bc^2 + abc \Rightarrow$
$-2 - \left(\frac{a^2c}{abc} + \frac{ab^2}{abc} - \frac{bc^2}{abc}\right) - \left(\frac{a^2b}{abc} - \frac{b^2c}{abc} + \frac{ac^2}{abc}\right) = 2 + a^2b + ab^2 + ac^2 + a^2c + b^2c + bc^2 \Rightarrow$
$\frac{-a^2c - ab^2 + bc^2}{abc} + \frac{-a^2b + b^2c - ac^2}{abc} = 4 + a^2b + ab^2 + ac^2 + a^2c + b^2c + bc^2 \Rightarrow$
$-a^2c - ab^2 + bc^2 - a^2b + b^2c - ac^2 = 4 + a^2b + ab^2 + a^2c + ac^2 + b^2c + bc^2 \Rightarrow$
$4 + 2a^2b + 2ab^2 + 2a^2c + 2ac^2 = 0 \Rightarrow$
$a^2b + ab^2 + a^2c + ac^2 = -2$[/hide]

[hide="Counterexample"]

Let $a = \frac{2}{3}$, $b = \frac{3}{4}$, $c = 2$. ;) It can't be negative if $a$, $b$, and $c$ are all positive.[/hide]

[hide="Are you sure you didn't mean…"]
 If $abc = 1$ prove that

$(a + b + c)\left(-\frac{1}{a} - \frac{1}{b} - \frac{1}{c}\right) = 1 + (a + b)(b + c)(c + a)$?
[/hide]

[quote="Kingofmath101"][hide="Are you sure you didn't mean…"]
 If $abc = 1$ prove that

$(a + b + c)\left(-\frac{1}{a} - \frac{1}{b} - \frac{1}{c}\right) = 1 + (a + b)(b + c)(c + a)$?
[/hide][/quote]
This one wouldn't make sense either if all of them are positive, you have $LHS<0$ while $RHS>0$...

Let's figure out No. 8, by calculating

${1+(a+b)(b+c)(c+a)\over a+b+c}$ if $abc=1$

Put $S:=a+b+c$. Then

$E={1+(S-a)(S-b)(S-c)\over S}$

$E={\cancel{1}+S^3-S^2(a+b+c)+S(ab+bc+ca)-\cancel{abc}\over S}$

$E={S^3-S^3+S(ab+bc+ca)\over S}$

$E=ab+bc+ca$

But $ab={1\over c}, bc={1\over a}, ca={1\over b}$, hence

$E={1\over a}+{1\over b}+{1\over c}$

Thus, the correct form is:

If $abc=1$, then prove $(a+b+c)\left({1\over a}+{1\over b}+{1\over c}\right)=1+(a+b)(b+c)(c+a)$


quadratics
algebra
polynomial
superior algebra
superior algebra solved

Let $p$ be a prime of the form $8m-1$.

Define $Q_p$ to be the set of quadratic residues mod $p$ and $N_p$ to be the set of quadratic non-residues. (NB: $x$ is a quadratic residue mod $p$ iff there exists a $y$ belonging to $F_p$ such that $x=y^2$.)

Define $P(x)=1+\sum_{i\in N_p}x^i$ to be a polynomial in $F_2[x]$. (NB: $i\in F_p$ is an abuse of notation. Regard $i$ as an integer less than $p$.)

Let $\alpha$ be a $p$th root of unity in the algebraic closure of $F_2$. Prove that for a correct choice of $\alpha$, $\alpha^j$ is a root of $P(x)$ for all $j\in Q_p$.

Here it goes, but I don't think I need $p=8m-1$. Just $p=4n-1$, so that $-1$ is not a quadratic residue.

I assume that $Q_p$ is the set of nonzero quadratic residues and leave $0$ aside as "special".

Let $\alpha$ be a $p$-th root of unity over $\mathbb F_2$. Let $A(\alpha)=\sum_{i\in Q_p}\alpha^i$. Let $B(\alpha)=\sum_{i\in N_p}\alpha^i$. We notice that $\alpha^i$ makes perfect sense for $i\in \mathbb F_p$ because $\alpha^p=1$.
Another obvious remark is that $P(\alpha)=1+B(\alpha)$.
Since $-1$ is not a quadratic residue, $-N_p=Q_p$. (this is where I use the choice $0\not\in Q_p$). It is easy to see that $A(\alpha)=B(\alpha^{-1})$ for all $p$-th roots of unity $\alpha$. It even works for $\alpha=1$, because $|N_p|=|Q_p|=4m-1$.

$A(\alpha)B(\alpha)=\sum_{i\in\mathbb F_p}n(i)\alpha^i$, where $n(i)$ is an integer which counts the numbers of different ways of writing $i$ as the sum of a an element of $Q_p$ and an element of $N_p$.

$0=u+v$ with $u\in Q_p$ and $v\in N_p$ if and only if $u=-v$. We conclude $n(0)=|N_p|=|Q_p|=\frac{p-1}2=4m-1$.
If $i$ and $j$ are nonzero elements of $\mathbb F_p$, then $i=u+v$ with $u\in Q_p$ and $v\in N_p$ if and only if $j=\frac jiu+\frac jiv$. Let $k=\frac ji$. If $k$ is a quadratic residue, then $ku\in Q_p$ and $kv\in N_p$, otherwise $ku\in N_p$ and $kv\in Q_p$. From these it is not hard to prove $n(i)=n(j)$.

We have $n(0)+n(1)+\ldots+n(p-1)=\left(\frac{p-1}2\right)^2$ because $u+v\in\mathbb F_p$ for all $u\in Q_p$ and $v\in N_p$.
Then $n(0)+(p-1)n(1)=\left(\frac{p-1}2\right)^2$. This is an equality of integers. $n(0)=p-1\Rightarrow 1+n(1)=\frac{p-1}2\Rightarrow n(1)=\frac{p+1}2=4m$.

With these in mind, $A(\alpha)B(\alpha)=(p-1)+4m\cdot (\sum_{i\in\mathbb F_p^*}\alpha^i)=0$ in $\mathbb F_2$.
So $A(\alpha)B(\alpha)=0$.
$A(\alpha)+B(\alpha)=(\sum_{i\in\mathbb F_p}\alpha^i)-1=-1=1$ in $\mathbb F_2$.

We conclude that $A(\alpha)=0$ and $B(\alpha)=1$ or $A(\alpha)=1$ and $A(\alpha)=0$.
In the first case, we have $P(\alpha)=1+B(\alpha)=1+1=0$. In the second case, $P(\alpha^{-1})=1+B(\alpha^{-1})=1+A(\alpha)=1+1=0$.

Therefore there exists $\alpha$ such that $P(\alpha)=0$. Since the multiplication by an element of $Q_p$ just permutes the elements of $N_p$, we easily find $P(\alpha^i)=0$ for all $i\in Q_p$.
$P(1)=0$ because we add $1+|N_p|=1+4m-1=4m$ of $1$. With this we have also treated the "special" case $i=0$.


$ n$ is a positive integer. Calculate the sum $ \sum_{k = 1}^{n}{\sum_{1\le i_1 < \ldots < i_k\le n}^{}{\frac {2^k}{(i_1 + 1)(i_2 + 1)\ldots (i_k + 1)}}}$

it looks like the expansion of $ \prod_{i \equal{} 1}^{n} (\frac {2}{i \plus{} 1} \plus{} 1) \minus{} 1$  :maybe:
EDIT: it can be simplified: $ \prod_{i \equal{} 1}^{n} (\frac {2}{i \plus{} 1} \plus{} 1) \minus{} 1 \equal{} \prod_{i \equal{} 1}^{n} (\frac {i\plus{}3}{i \plus{} 1}) \minus{} 1\equal{} \frac{(n\plus{}3)!1!}{(n\plus{}1)!3!}\equal{}\frac{(n\plus{}2)(n\plus{}3)}{6} \minus{}1 \equal{} \frac{n^{2}\plus{}5n}{6}$


algebra
polynomial
superior algebra
superior algebra unsolved

Hello

Let $ \alpha \equal{} \sqrt{2\plus{}\sqrt 2}$ and let f the minimal polynomial of alpha over Q.

It is $ f \equal{} x^4\minus{}4x^2\plus{}2$ whose roots are $ \pm \alpha, \pm \beta$ where $ \beta \equal{} \sqrt{2\minus{}\sqrt{2}}$

Is there a more economic way to write the splitting field E for f over Q? I would write $ \mathbb E \equal{} \mathbb Q[\alpha, \beta]$

Is that right that $ [\mathbb E : \mathbb Q] \equal{} [ \mathbb Q[\alpha,\beta]: \mathbb Q[\alpha]][\mathbb Q[\alpha]: \mathbb Q] \equal{} 2 \cdot 4 \equal{} 8$ ??

Sorry I answered myself :)

It is $ \beta \in \mathbb Q[\alpha]$ hence $ [\mathbb E: \mathbb Q] \equal{} 4$  :D

Well the splitting field is $ \mathbb{Q}(\alpha,\minus{}\alpha,\beta,\minus{}\beta)$. By definition, given a field $ K$ and a non-constant polynomial $ f(x) \equal{} a_nx^n\plus{}...\plus{}a_1x\plus{}a_0$ we say $ F$ is a splitting field over $ K$ iff $ f(x) \equal{} a_n(x \minus{}\alpha_1)...(x\minus{}\alpha _n)$ for $ \alpha _i \in F$ and $ F \equal{} K(\alpha_1, ... ,\alpha_n)$. Since $ f(x) \equal{} x^4 \minus{} 4x^2\plus{}2$ splits when you introduce $ \pm \alpha, \pm \beta$ it means $ \mathbb{Q}(\alpha, \minus{}\alpha,\beta,\minus{}\beta)$ is the splitting field. Of course, we can simplify this field as $ \mathbb{Q}(\alpha, \beta)$.

Sure, but $ \mathbb Q[\alpha, \beta]$ is a tricky notation because one can make a mistake when computing the degree of the splitting field over $ \mathbb Q$ since it's not sparkling clear that $ \sqrt{2\minus{}\sqrt 2} \in \mathbb Q \left [\sqrt{2\plus{}\sqrt 2} \right ]$ :)

@sylow: please read the questions more carefully and stop posting these trivial things also known by the questioner (this also concerns other threads).


calculus
integration
trigonometry
calculus computations

OK, last integral from me today.....  :D

solve

$\int_{0}^{\frac{\pi}{4}}\;\frac{dx}{\sin^{4}x+\sin^{2}x\;\cos^{2}x+\cos^{4}x}$

[hide]
$answer: \;\;\;\quad \boxed{\frac{\pi}{2\sqrt 3}}$
[/hide]


function
limit
calculus
derivative
real analysis
real analysis unsolved

Let [x] be the greatest integer $ \leq x$

(a) Prove that $ \lim_{x\to 0}\frac {1}{x} - \left[\frac {1}{x}\right]$ does not exist.

(b) Define $ f(x) = 1 - x\left[\frac {1}{x}\right]$ for $ x > 0$ and $ f(0) = 0$. Is $ f$ continuous at $ 0$? Prove your answer.

(c) Define $ g(x) = x - x^2\left[\frac {1}{x}\right]$ for $ x > 0$ and $ g(0) = 0$. Where is $ g$ differentiable? Prove your answer.

My attempt:
[hide]Let $ n$ be a natural number. There are three possibilities for $ x$:
* $ x = 0$
* If $ \frac {1}{x}\in\mathbb{N}$ (i.e. $ x = \frac {1}{n}$), then $ \left[\frac {1}{x}\right] = \frac {1}{x}$.
* If $ \frac {1}{n + 1} < x < \frac {1}{n}$, then $ n < \frac {1}{x} < n + 1$, hence $ \left[\frac {1}{x}\right] = n$.

(a)
* If $ x = \frac {1}{n}$, then $ \frac {1}{x} - \left[\frac {1}{x}\right] = 0$.
* If $ \frac {1}{n + 1} < x < \frac {1}{n}$, then$ \frac {1}{x} - \left[\frac {1}{x}\right] = \frac {1}{x} - n$.
Furthermore, $ n < \frac {1}{x} < n + 1$, so $ 0 < \frac {1}{x} - n < 1$.

Don't know how I can (formally, with $ \epsilon, \delta$) show that the limit doensn't exist.

(b) If $ x = \frac {1}{n}$, then $ f(x) = 0$. If $ \frac {1}{n + 1} < x < \frac {1}{n}$, then $ f(x) = 1 - nx$. Furthermore, $ 0 < 1 - nx < 1 - \frac {n}{n + 1}$.

From this I see how the graph looks like (a 'saw' of less and less height as x->0), and it looks like $ f\to 0$, which would make $ f$ continuous at 0.
Indeed, $ 1 - \frac {n}{n + 1}\to 0$ as $ n\to \infty$. Again, I'm not sure about the proof:

Let $ \epsilon > 0$, and let $ x$ satisfy $ 0 < x < \delta$. If $ x = \frac {1}{n}$ for some natural $ n$, then $ f(x) = 0$, so $ |f(x) - 0| < \epsilon$.
If $ \frac {1}{n + 1} < x < \frac {1}{n}$, then ${ |f(x) - 0| = |1 - nx}$. What $ \delta$ should I pick?

(c) For $ x\neq a = 0$, we have $ \frac {g(x) - g(a)}{x - a} = \frac {g(x) - g(0)}{x - 0} = \frac {x - x^2\left[\frac {1}{x}\right]}{x} = 1 - x\left[\frac {1}{x}\right] = f(x)$.
Hence $ \lim _{x\to 0}\frac {g(x) - g(0)}{x - 0} = \lim _{x\to 0} f(x) = 0$, hence $ g$ is differentiable at $ a = 0$.

For $ x\neq a = \frac {1}{n}$ for some natural $ n$: $ \frac {g(x) - g(a)}{x - a} = \frac {g(x)}{x - a} = \frac {x - x^2\left[\frac {1}{x}\right]}{x - \frac {1}{n}} = ...$ ?
[/hide]
Thanks for your help!

I thought of something for part (b): we know that $ \frac {1}{x} \minus{} 1\leq \left[\frac {1}{x}\right]\leq \frac {1}{x}$, therefore $ 0\leq 1 \minus{} x\left[\frac {1}{x}\right] \leq x$.
So, by the squeeze theorem, we have $ \lim_{x\to 0} 1 \minus{} x\left[\frac {1}{x}\right] \equal{} 0$, hence $ f$ is continuous at 0. Correct?

For part (c), I got the following:

* If $ x$ is between two successive numbers of the form $ \frac {1}{n}$, i.e.  $ \frac {1}{n \plus{} 1} < x < \frac {1}{n}$ ($ n\in\mathbb{N}$), then
$ n < \frac {1}{x} < n \plus{} 1\Rightarrow \left[\frac {1}{x}\right] \equal{} n\Rightarrow g(x) \equal{} x \minus{} x^2n \equal{} x(1 \minus{} nx)$.
Hence, for $ x,a\in\left]\frac {1}{n \plus{} 1},\frac {1}{n}\right[$ we have
$ \frac {g(x) \minus{} g(a)}{x \minus{} a} \equal{} \frac {x(1 \minus{} nx) \minus{} a(1 \minus{} na)}{x \minus{} a} \equal{} \frac {x \minus{} nx^2 \minus{} a \plus{} na^2}{x \minus{} a} \equal{} 1 \minus{} n\frac {x^2 \minus{} a^2}{x \minus{} a} \equal{} 1 \minus{} n(x \plus{} a)$
which means $ g$ is differentiable on all intervals $ \left]\frac {1}{n \plus{} 1},\frac {1}{n}\right[$ ($ n\in\mathbb{N}$) with derivative $ 1 \minus{} 2nx$.

* If $ x > 1$ then $ 0 < \frac {1}{x} < 1$, so $ \left[\frac {1}{x}\right] \equal{} 0$, therefore $ g(x) \equal{} x$.
--------------------------------------------------------------------------------------------------------------------------------------

So $ g$ is differentiable at $ x \equal{} 0$, at $ x$ between two successive numbers of the form $ \frac {1}{n}$, and at $ x > 1$.
Now I have to show that $ g$ is NOT differentiable at numbers of the form $ \frac {1}{n}$.

Any help is appreciated (part (a) is also still open)!  :)

Thanks (again!) for your answer. Could you help me to prove $ g$ is not differentiable at every $ x \equal{} \frac {1}{n}$ in part (c)? My idea was to prove it's not even continuous at those points.

For (a), we actually haven't covered sequences yet, only the definition of limit, continuity and differentiability and some basic theorems. Do you know a different way?
Something like: let $ L\in\mathbb{R}$ be arbritrary, $ \delta > 0$. I have to pick an $ \epsilon > 0$ and a $ 0 < x < \delta$ such that $ |\frac {1}{x} \minus{} \left[\frac {1}{x}\right]\minus{}L|\geq \epsilon$.

Right: you need a sequence of values (because for each $ \delta$, you need a value smaller than it) at which your given expression is "not too near" 0 (where "not too near" means "differs by at least some fixed constant").   (In particular, what I'm saying doesn't require you to know anything at all about sequences :))  So, think about what this function is actually doing!  (It might help to draw a picture, for example.)  As a further hint: any value $ \varepsilon < 1$ will work nicely.

Your idea for part (c) is also right.  Try to figure out what the left and right limits are (and remember that in the limit, if $ x \to \frac {1}{n}$ then only the cases where $ \frac {1}{n \plus{} 1} < x < \frac {1}{n \minus{} 1}$ are relevant), and you'll see that they are indeed unequal.

Ah, part (c) is pretty straightforward I see:

Let $ n\in\mathbb{N}$, $ n\geq 2$. We want to calculate the left and right limits of $ g(x)$ as $ x$ approaches $ \frac {1}{n}$.
$ \frac {1}{n} < x < \frac {1}{n \minus{} 1}\Rightarrow g(x) \equal{} x(1 \minus{} (n \minus{} 1)x)$, so $ \lim_{x\uparrow \frac {1}{n}}g(x) \equal{} \lim_{x\uparrow \frac {1}{n}}x(1 \minus{} (n \minus{} 1)x) \equal{} \frac {1}{n^2}$

$ \frac {1}{n \plus{} 1} < x < \frac {1}{n}\Rightarrow g(x) \equal{} x(1 \minus{} nx)$, so $ \lim_{x\downarrow \frac {1}{n}}g(x) \equal{} \lim_{x\downarrow \frac {1}{n}}x(1 \minus{} nx) \equal{} 0$

Hence $ \lim_{x\to \frac {1}{n}}g(x)$ does not exist: $ g$ is discontinuous at every $ x \equal{} \frac {1}{n}$, and since differentiability impies continuity, $ g$ is certainly not differentiable at those points.

I'll think about (a) some more. I know the function 'oscillates' rapidly between 0 and 1 (or actually jumps), so it's obvious the limit doesn't exist. I'll try to make this into a rigorous answer.

[quote="JBL"]Your b and c look okay.  For (a), to prove a limit doesn't exist, find two sequences approaching 0 such that the function value along one sequence differs from the function value along the other sequence.
(...)
Right: you need a sequence of values (because for each $ \delta$, you need a value smaller than it) at which your given expression is "not too near" 0 (where "not too near" means "differs by at least some fixed constant").   (In particular, what I'm saying doesn't require you to know anything at all about sequences :))  So, think about what this function is actually doing!  (It might help to draw a picture, for example.)  As a further hint: any value $ \varepsilon < 1$ will work nicely.[/quote]
After some thought, I appreciate what you said about sequences. Let's say $ h(x) \equal{} \frac {1}{x} \minus{} \left[\frac {1}{x}\right]$.
I was thinking about $ x_n \equal{} \frac {1}{n}$, then $ \left[\frac {1}{x_n}\right] \equal{} [n] \equal{} n$, so $ h(x_n) \equal{} n \minus{} n \equal{} 0$.
For $ y_n \equal{} \frac {2}{2n \plus{} 1}$ we have $ \left[\frac {1}{x_n}\right] \equal{} \left[n \plus{} \frac {1}{2}\right] \equal{} n$, so $ h(y_n) \equal{} n \plus{} \frac {1}{2} \minus{} n \equal{} \frac {1}{2}$

In terms of limits, this becomes:
Take $ \epsilon \equal{} \frac {1}{4}$, and let $ \delta > 0$. Let $ x$ satisfy $ 0 < x \equal{} \frac {2}{2n \plus{} 1} < \delta$ for $ n\in\mathbb{N}$ (possible for every $ \delta$, just take $ n$ large enough). Then $ |h(x) \minus{} 0| \equal{} |\frac {1}{2}| > \epsilon$. Therefore, $ \lim_{x\to 0}h(x)\neq 0$. Now we must show that $ \lim_{x\to 0}h(x)\neq L$, for arbitrary $ L\neq 0$.
Take $ \epsilon \equal{} \frac {|L|}{2}$, and let $ \delta > 0$. Let $ x$ satisfy $ 0 < x \equal{} \frac {1}{n} < \delta$ for $ n\in\mathbb{N}$. Then $ |h(x) \minus{} L| \equal{} |0 \minus{} L| \equal{} |L| > \epsilon$.
Conclusion: $ \lim_{x\to 0}h(x)$ doesn't exist.

Correct?  :)

Looks good.  Another way this is that because on some sequence the function approaches 0, if the limit exists it can only be 0, while on some other sequence the function approaches 1/2, so if the limit exists it can only be 1/2.  At some point in the near future, you'll probably capture this by the following theorem: $ \lim_{x \to a} f(x) \equal{} L$ if and only if $ \lim_{n \to \infty} f(x_n) \equal{} L$ for every sequence $ x_n \to a$.  (We've just essentially done one direction of it.)

I like it how you provide me with just enough information to figure it out myself, it's really helpful (just like my last question on limits)!
And you're right: we just finished Ch. 1, while Ch. 3 will be about sequences and their relation to functions (Ch. 2 is about metric spaces, open and closed sets, etc.).


quadratics
inequalities
geometry
3D geometry
trigonometry
algebra
quadratic formula

There are a few problems that I am stuck on:

1. Solve for z:

z^2 - (2/sqrt(3) + i4sqrt(2)/sqrt(7))z + 1

How do I get the solutions:

z = (1/sqrt(3) +/- 2/sqrt(21)) + i(2sqrt(2)/sqrt(7) +/- sqrt(2))

I used the quadratic formula, but I can't get rid of those square roots of square roots.

2. Given that:

(x + y)^3 = z
(y + z)^3 = x
(z + x)^3 = y

Is it true that (x + y)^2 + (x + y)(y + z) + (y + z)^2 > 0?

How do you know?

(I'm not sure if knowing the system of equations is necessary, but it is this inequality that I am confused about)

3. What is SUM(k = 1, n) cos(kx) equal to?

The solution tells you to multiply by 2sin(x/2) to get

2 SUM(k = 1, n) sin(x/2)cos(kx) = SUM(k = 1, n) (sin(k +1/2)x - sin(k - 1/2)x)

= sin(n + 1/2)x - sin(x/2)

Although the solution makes sense, why do you need to multiply by sin(x/2)? Why doesn't the solution work if you just multiplied by sin(x)?

Suppose you want to find the square root of the following expression:

$D=-19+2\sqrt{42}~i$
[hide]
One method is to guess the form of the square root, and work backwards.  I will guess the real and imaginary parts each have a square root term.  Then I square my guess, and see if I can make that match the original term, D...

Guess $\sqrt{D}=\sqrt{s}+\sqrt{u}~i$

$D=(\sqrt{s}+\sqrt{u}~i)^2=s-u+2\sqrt{su}~i$

If my guess is correct, we should be able to find $s$ and $u$, such that:

$s-u + 2\sqrt{su}~i = -19+2\sqrt{42}~i$

Splitting real and imaginary parts:

$s-u=-19$

$2\sqrt{su}~i=2\sqrt{42}~i$

We need to find $s$ and $u$ such that $su=42$ and $s-u=-19$

A solution is $s=2$, $u=21$?
[/hide]
Now we have $\sqrt{D}=\sqrt{2}+\sqrt{21}~i$

This should help you find the answer to #1.  [What is the source of these problems, by the way?]

for 3 u can use complex numbers instead.

[quote="Soroban"][quote]2. Given that: $(x + y)^3 = z,\;(y + z)^3 = x,\;(z + x)^3 = y$

Is it true that $(x + y)^2 + (x + y)(y + z) + (y + z)^2 > 0$ ? [/quote][hide]
We note that: $(x+y)^3 - (y+z)^3 = z - x$

We further note that the above is a Difference of Cubes:
$[(x+y)-(y+z)]\;[(x+y)^2 + (x+y)(y+x) + (y+x)^2] \;= \;z - x$

or: $[x - z]\;[(x+y)^2+(x+y)(y+z)+(y+z)^2] \;= \;z-x$

Hence: $(x+y)^2 + (x+y)(y+z) + (y+z) \;= \;\frac{z-x}{x-z} \;= \;-1$

Therefore, the answer is $no.$[/hide][/quote]

Well, actually the answer is supposed to be yes, which is what I don't understand. Also, you can't divide by x - z because x - z may equal 0.

Mathnovice, take $x=y=z=0$ for 2, clearly, that quantity (call it $Q$) is equal to zero.

Alternately, we have:

$((x+y)-(y+z))(Q)=(x+y)^3-(y+z)^3$

therefore,
$(x-z)Q=(z-x)\rightarrow Q=-1$ if $x-z\not=0$

However, if $x=z$, we get $x=y=z=0$ (I think, don't have time to be 100% sure), so the answer is NO.

Well, the reason multiplying by sin x is not as nice is because then you would end up with
$\sum_{k=1}^n \left(\frac{1}{2}\sin ((k+1)x) - \frac{1}{2}\sin ((k-1)x)\right)$.
Then not everything telescopes and cancels out when you add them all together. When you do it with x/2, on the other hand, the gap between terms is x (instead of 2x), so all but the first and last terms cancel out. You should get the same answer after multiplying by sin x after some simplification, but its a bit messier.

where did you get these problems and solutions.  i would really like to find a decent site for this stuff.  AoPS is good but i need to find one with many probs


geometry
incenter
rhombus
circumcircle
geometry unsolved

It is given a rhombus $ ABCD$ with sidelength $ a$. On the line $ AC$ are chosen the points $ M$ and $ N$ in such a way that $ C$ lies between $ A$ and $ N$ and $ MA\cdot NC \equal{} a^2$. We denote with $ P$ the intersection point of $ MD$, $ BC$ and $ Q$ is the intersection point of $ ND$, $ AB$. Prove that $ D$ is the incenter of the triangle $ PQB$.

[quote="yetti"]Did not you forget to say that $ A$ lies between $ C$ and $ M$ :?:[/quote]
Actually this is what I found in the document that contained this problem. I am not sure. May be you are right :maybe:

Assume $ A$ is between $ C, M,$ just like $ C$ is between $ A, N,$ as you said. It would not work otherwise (you could not use SAS for $ \triangle MAB, \triangle BCN$).

$ MA \cdot NC \equal{} a^2$ $ \Longrightarrow$ $ \frac {MA}{AB} \equal{} \frac {BC}{CN}$ $ \Longrightarrow$ $ \triangle MAB \sim \triangle BCN$ are similar by SAS $ \Longrightarrow$ $ \angle QBM \equal{} \angle ABM \equal{} \angle CNB \equal{} \angle DNC \equal{} \angle QNM$ $ \Longrightarrow$ $ MBNQ$ is cyclic. In exactly the same way, $ \angle PBN \equal{} \angle CBN \equal{} \angle AMB \equal{} \angle DMA \equal{} \angle PMN,$ so that $ NBMP$ is also cyclic $ \Longrightarrow$ $ BMQPN$ is cyclic, with a circumcircle $ (O).$ $ NM, MN$ bisect the angles $ \angle BNQ, \angle BMP,$ respectively $ \Longrightarrow$ $ PM, QN$ bisect the angles $ \angle BPQ, \angle BQP,$ respectively. By definition of $ P, Q,$ the angle bisectors $ PM, QN$ of $ \triangle BPQ$ intersect at $ D.$


algebra
polynomial
inequalities
algebra unsolved

Find all polynomials
$ f(x)\equal{}x^n\plus{}a_1x^{n\minus{}1}\plus{}...\plus{}a_{n\minus{}1}x\plus{}a_n$
so that:
-all coefficients are either -1 or 1
-all roots are real

[hide]
Firstly I got that $ \sum(x_i)^2\equal{}3$ and then I assumed that there exist solutions for $ n >3$ and got the contradiction. But what I've done to get this seems quite long (I put all roots as $ (1\plus{}\alpha)$ and obtained it is impossible). I would like some other continuation of this solution starting from $ \sum(x_i)^2\equal{}3$
[/hide]

This is what I did. Hopefully I haven't made any mistakes.

[hide] I changed the notations a bit so that I am more comfortable with it:

$ f(x) \equal{} x^{n} \plus{} a_{n\minus{}1} x^{n\minus{}1} \plus{} ... \plus{} a_1x \plus{} a_0$ is the polynomial being considered.

First assume we can find a polynomial with $ deg(f(x)) >3$ satisfying the conditions of the problem. Let us write the roots of $ f(x)$ as $ r_1, r_2, r_3, ..... ,r_n \in \mathbb{R}$. We seek to deduce two values: $ A\equal{} r_1^2 \plus{} .... \plus{}r_n^2$ and $ B \equal{} \frac{1}{r_1^2} \plus{} .... \plus{} \frac{1}{r_n^2}$.

This is not difficult and we can deduce that $ A \equal{} (\minus{}a_{n\minus{}1})^2 \minus{} 2a_{n\minus{}2} \equal{} 1 \minus{} 2a_{n\minus{}2}$ and since all our roots are real $ A \geq 0$ and hence $ A \equal{} 3$.

Also we have $ B \equal{} \frac{a_1^2}{a_0^2} \minus{} 2\frac{a_2}{a_0}$ and since $ B \geq 0$ we have $ B\equal{}3$ as well.

From Cauchy's Inequality we know that $ AB \geq n^2 \Rightarrow 9 \geq n^2$ which is a contradiction and thus we only need to check $ n\equal{}1,2,3$.

I think the solutions are: $ x\plus{}1, x\minus{}1, x^2\plus{}x\minus{}1, x^2\minus{}x\minus{}1, x^3\minus{}x^2\minus{}x\plus{}1, x^3\plus{}x^2\minus{}x\minus{}1$[/hide]


inequalities
calculus
derivative
function
logarithms
search
inequalities unsolved

I can't solve that inequality.

\[ \frac{a}{\sqrt{b^2\plus{}c^2}}\plus{}\frac{b}{\sqrt{c^2\plus{}a^2}}\plus{}\frac{c}{\sqrt{a^2\plus{}b^2}}\geq\frac{3}{\sqrt2}\]

Please help me solve it or give me some hints.

Oni.

[quote="Oni"]I can't solve that inequality.
\[ \frac {a}{\sqrt {b^2 \plus{} c^2}} \plus{} \frac {b}{\sqrt {c^2 \plus{} a^2}} \plus{} \frac {c}{\sqrt {a^2 \plus{} b^2}}\geq\frac {3}{\sqrt2}
\]
Please help me solve it or give me some hints.

Oni.[/
It's no have condition.
I think, if it's true then derivative will solve it good.
$ \sum \frac {1}{\sqrt{x\plus{}y}} \geq 2\plus{}\frac {2}{\sqrt{2}}$
with $ xy\plus{}yz\plus{}zx\equal{}1$, $ x\equal{}\frac{b}{a}$ similar give $ y,z$

This inequality can be easily proofed using Jensen, and a well known inequality in the last step: $ \frac{a}{\sqrt{b^{2}\plus{}c^{2}}}\plus{}\frac{b}{\sqrt{c^{2}\plus{}a^{2}}}\plus{}\frac{c}{\sqrt{a^{2}\plus{}b^{2}}}\equal{}\sqrt{\frac{a^{2}}{b^{2}\plus{}c^{2}}}\plus{}\sqrt{\frac{b^{2}}{c^{2}\plus{}a^{2}}}\plus{}\sqrt{\frac{c^{2}}{a^{2}\plus{}b^{2}}}\geq3\sqrt{\frac{1}{3}(\frac{a^{2}}{b^{2}\plus{}c^{2}}\plus{}\frac{b^{2}}{c^{2}\plus{}a^{2}}\plus{}\frac{c^{2}}{a^{2}\plus{}b^{2}})}\geq3\sqrt{\frac{1}{3}\cdot\frac{3}{2}}\equal{}\frac{3}{\sqrt{2}}$

that's wrong
$ \frac{\sum_{i\equal{}1}^{n} x_{i}^{\alpha}}{n} \geq (\frac{\sum{x_{i}}}{n})^{\alpha}$
holds iff $ \alpha \geq 1$

Ok, i see it, but how come, that Jensen isnt correct, i mean this is a konveks funkcion.
Please explain.

taht's because the function ${ f(\alpha})=x^{\alpha}$ has double derivative negative  for $ \alpha <1$
due to the $ \ln \alpha$ factor

No, i am lookin the funkcion f(x)=x^1/2, thats not the same

sorry that's a mistake...i was busy solving this in another way.../

so any ways ur proof is wrong the reason being $ f(x) \equal{} x^{\frac{1}{2}}$  has derivative $ f^{''}(x) \equal{} \frac { \minus{} 1}{4x^{\frac {3}{2}}} < 0$

OK i understand my solution is wrong, bu i would like to know for wich funcions can i use Jense, because for this one Jensen is obviosu wrong,   i i also dont know what a derivate is or how you can canculate it.  Would you be so nice and explain it to me.

jensen's is actually true for all concave or convex functions....
actuallu u  can read best about inequalities [url=http://ultrametric.googlepages.com/tin2007.pdf]here[/url]

Normalize to $ \sum x^2\equal{}3$, then we note that we have the following inequality for $ x\in\left[\frac{1}{\sqrt{3}},\sqrt{3}\right)$

\[ \frac{x}{\sqrt{3\minus{}x^2}}\ge \frac{\sqrt{2}}{8}\cdot (3x^2\plus{}1)\]

summing these up gives the desired result, but if we don't have the variables in that range, we have to resort to other measures. But the inequality is weak in that case...

As it seems the inequality isnt true , but i still dont get it , why i got a wrong inequality using jensen, because, it is treu for al concave and convex funkcions.
Jensen implies:

$ \sqrt {\frac {a^{2}}{b^{2} \plus{} c^{2}}} \plus{} \sqrt {\frac {b^{2}}{c^{2} \plus{} a^{2}}} \plus{} \sqrt {\frac {c^{2}}{a^{2} \plus{} b^{2}}}\geq3\sqrt {\frac {1}{3}(\frac {a^{2}}{b^{2} \plus{} c^{2}} \plus{} \frac {b^{2}}{c^{2} \plus{} a^{2}} \plus{} \frac {c^{2}}{a^{2} \plus{} b^{2}})}$

but tha is obviously not corect, how if that possible, because Jensen should work for al funkcions.  Can somebody explain.

i stated that to u see my first post in this topic which clearly say the exponent $ \alpha >1$ and if u want the derivation i gave it to u .....since u don't know calculus what's the point discussing

OK, i know my mistake now:D, the funkcion isn convex, but concave.

[quote="arqady"][quote="Oni"]I can't solve that inequality.
\[ \frac {a}{\sqrt {b^2 \plus{} c^2}} \plus{} \frac {b}{\sqrt {c^2 \plus{} a^2}} \plus{} \frac {c}{\sqrt {a^2 \plus{} b^2}}\geq\frac {3}{\sqrt2}
\]
Please help me solve it or give me some hints.

Oni.[/quote]
Try $ a \equal{} b \equal{} 1$ and $ c \equal{} 0.$ :wink:[/quote]

Yes, this one has been posted many times without the op knowing that it's false. The best that we can do is
\[ \frac{a}{\sqrt{b^2\plus{}c^2}}\plus{}\frac{b}{\sqrt{c^2\plus{}a^2}}\plus{}\frac{c}{\sqrt{a^2\plus{}b^2}}>2\]
(if we allow one of $ a,b,c$ to be 0, then there can be equality). It can be proven with
\[ \frac{x}{\sqrt{y^2\plus{}z^2}}\equal{}\frac{x^2}{\sqrt{x^2(y^2\plus{}z^2)}}\geq \frac{2x^2}{x^2\plus{}y^2\plus{}z^2}\]
and the whole inequality is strict because if $ a\geq b\geq c$, we can't have $ c^2\equal{}a^2\plus{}b^2$.

The inequality
\[ \frac{a}{\sqrt{4b^2\plus{}bc\plus{}4c^2}}\plus{}\frac{b}{\sqrt{4c^2\plus{}ca\plus{}4a^2}}\plus{}\frac{c}{\sqrt{4a^2\plus{}ab\plus{}4b^2}}\geq 1\]
is also true.

yes surely that's holder...see the G of the keyboard is in the vicicnity of H :wink:

[quote="Oni"]I can't solve that inequality.
\[ \frac {a}{\sqrt {b^2 \plus{} c^2}} \plus{} \frac {b}{\sqrt {c^2 \plus{} a^2}} \plus{} \frac {c}{\sqrt {a^2 \plus{} b^2}}\geq\frac {3}{\sqrt2}
\]
Please help me solve it or give me some hints.

Oni.[/quote]
The general inequality is
$ \left(\frac {a}{b \plus{} c} \right)^k \plus{} \left(\frac {b}{c \plus{} a} \right)^k \plus{} \left(\frac {c}{a \plus{} b} \right)^k \ge \min \left(2;\frac {3}{2^k} \right)$
Let $ k \equal{} \frac {1}{2}$ we get
$ \frac {a}{\sqrt {b^2 \plus{} c^2}} \plus{} \frac {b}{\sqrt {c^2 \plus{} a^2}} \plus{} \frac {c}{\sqrt {a^2 \plus{} b^2}} \geq 2$.

[quote="scorpius119"][quote="arqady"][quote="Oni"]I can't solve that inequality.
\[ \frac {a}{\sqrt {b^2 + c^2}} + \frac {b}{\sqrt {c^2 + a^2}} + \frac {c}{\sqrt {a^2 + b^2}}\geq\frac {3}{\sqrt2}
\]
Please help me solve it or give me some hints.

Oni.[/quote]
Try $ a = b = 1$ and $ c = 0.$ :wink:[/quote]

Yes, this one has been posted many times without the op knowing that it's false. The best that we can do is
\[ \frac {a}{\sqrt {b^2 + c^2}} + \frac {b}{\sqrt {c^2 + a^2}} + \frac {c}{\sqrt {a^2 + b^2}} > 2
\]
(if we allow one of $ a,b,c$ to be 0, then there can be equality). It can be proven with
\[ \frac {x}{\sqrt {y^2 + z^2}} = \frac {x^2}{\sqrt {x^2(y^2 + z^2)}}\geq \frac {2x^2}{x^2 + y^2 + z^2}
\]
and the whole inequality is strict because if $ a\geq b\geq c$, we can't have $ c^2 = a^2 + b^2$.

The inequality
\[ \frac {a}{\sqrt {4b^2 + bc + 4c^2}} + \frac {b}{\sqrt {4c^2 + ca + 4a^2}} + \frac {c}{\sqrt {4a^2 + ab + 4b^2}}\geq 1
\]
is also true.[/quote]

Yeah I know if one of $ a,b,c$ is $ 0$ then the inequality I gave is not true. But if the condition $ a,b,c>0$ is given then the inequality $ \frac{a}{\sqrt{b^2+c^2}}+\frac{b}{\sqrt{c^2+a^2}}+\frac{c}{\sqrt{a^2+b^2}} \geq \frac{3}{\sqrt2}$
certainly follows. Here is the proof:

$ \frac{a}{\sqrt{b^2+c^2}}+\frac{b}{\sqrt{c^2+a^2}}+\frac{c}{\sqrt{a^2+b^2}} \geq \frac{3}{\sqrt2}......(1)$

$ \Leftrightarrow\frac{1}{3}\frac{a}{\sqrt{b^2+c^2}}+\frac{1}{3}\frac{b}{\sqrt{c^2+a^2}}+\frac{1}{3}\frac{c}{\sqrt{a^2+b^2}} \geq \frac{1}{\sqrt2}$

the L.H.S is a symmetric for the variables $ a,b,c$ and $ f(ta,tb,tc)=f(a,b,c)$ for all $ t>0$

So we normalize to $ a^2+b^2+c^2=1$

So L.H.S. becomes

$ \frac{1}{3}\frac{a}{\sqrt{1-a^2}}+\frac{1}{3}\frac{b}{\sqrt{1-b^2}}+\frac{1}{3}\frac{c}{\sqrt{1-c^2}}$

$ =\frac{1}{3}\sqrt{\frac{a^2}{1-a^2}}+\frac{1}{3}\sqrt{\frac{b^2}{1-b^2}}+\frac{1}{3}\sqrt{\frac{c^2}{1-c^2}}$

Now $ f(x)=\sqrt{x}$ is convex for all $ x>0$

So Jensens inequality for convex function can be applied to L.H.S. which gives:

$ \frac{1}{3}\sqrt{\frac{a^2}{1-a^2}}+\frac{1}{3}\sqrt{\frac{b^2}{1-b^2}}+\frac{1}{3}\sqrt{\frac{c^2}{1-c^2}}$

$ \geq \sqrt{\frac{1}{3}\frac{a^2}{1-a^2}+\frac{1}{3}\frac{b^2}{1-b^2}+\frac{1}{3}\frac{c^2}{1-c^2}}$

$ =\frac{1}{\sqrt3}\sqrt{\frac{a^2}{1-a^2}+\frac{b^2}{1-b^2}+\frac{c^2}{1-c^2}}......(2)$

Now the above statement is symmetric for variables$ a,b,c.$ So w.l.o.g. we may assume that ,
\[ a\geq b\geq c\]

$ \Longrightarrow \frac{1}{1-a^2}\geq \frac{1}{1-b^2}\geq \frac{1}{1-c^2}$

So Tchebyshef's inequality implies: \[ {\frac{a^2}{1-a^2}+\frac{b^2}{1-b^2}+\frac{c^2}{1-c^2}}\]
\[ \geq \frac{1}{3}(a^2+b^2+c^2)(\frac{1}{1-a^2}+\frac{1}{1-b^2}+\frac{1}{1-c^2})\]
\[ =\frac{1}{3}.1\times(\frac{1^2}{1-a^2}+\frac{1^2}{1-b^2}+\frac{1^2}{1-c^2})\]
\[ \geq \frac{1}{3}\frac{(1+1+1)^2}{1-a^2+1-b^2+1-c^2}\]
\[ =\frac{1}{3}.\frac{9}{2}\]
\[ =\frac{3}{2}\]
So $ {\frac{a^2}{1-a^2}+\frac{b^2}{1-b^2}+\frac{c^2}{1-c^2}}\geq \frac{3}{2}$

Plugging the above value into  (2) gives:

$ \frac{1}{\sqrt3}\sqrt{\frac{a^2}{1-a^2}+\frac{b^2}{1-b^2}+\frac{c^2}{1-c^2}}\geq \frac{1}{\sqrt3}.\sqrt{\frac{3}{2}}=\frac{1}{\sqrt2}$

And that proves the inequality....

I tried to write everything elaborately and I think there isn't any mathematical error here. But if u find anything plz reply...

(And thanks a lot for giving 20 replies in discussing my problem :P !!!)

Oni.

You might wanna check the second derivative of $ f(x) \equal{} \sqrt x$ :wink:

And $ a,b,c > 0$ doesn't change anything because we can have $ a \to 0$ :P

Oni, try $ a \equal{} b \equal{} 1$ and $ c \equal{} \frac {1}{5}.$ :wink:

I was completely wrong....
Thanks a lot to arqady and behemont for finding my error so elegantly!!!
behemont, when I was testing the convexity of $ f$, for some unknown reason I took $ f(x)\equal{}\frac{1}{\sqrt{x}}$ which is convex. And thats why I thought $ f(x)\equal{}\sqrt{x}$ is convex.

And I need the proof of the general inequality given by ehku earlier. Can ehku or someone can post the proof here or somewhere:

\[ \left(\frac{a}{b\plus{}c}\right)^k\plus{}\left(\frac{b}{c\plus{}a}\right)^k\plus{}\left(\frac{c}{a\plus{}b}\right)^k\geq \min\left(2,\frac{3}{2^k}\right)\]

Oni.

[quote="Oni"]I was completely wrong....
Thanks a lot to arqady and behemont for finding my error so elegantly!!!
behemont, when I was testing the convexity of $ f$, for some unknown reason I took $ f(x) \equal{} \frac {1}{\sqrt {x}}$ which is convex. And thats why I thought $ f(x) \equal{} \sqrt {x}$ is convex.

And I need the proof of the general inequality given by ehku earlier. Can ehku or someone can post the proof here or somewhere:
\[ \left(\frac {a}{b \plus{} c}\right)^k \plus{} \left(\frac {b}{c \plus{} a}\right)^k \plus{} \left(\frac {c}{a \plus{} b}\right)^k\geq \min\left(2,\frac {3}{2^k}\right)
\]
Oni.[/quote]
This problem was at Vietnam olympiad,but i forgot the exactly year.

[quote="arqady"][quote="scorpius119"]
The inequality
\[ \frac {a}{\sqrt {4b^2 \plus{} bc \plus{} 4c^2}} \plus{} \frac {b}{\sqrt {4c^2 \plus{} ca \plus{} 4a^2}} \plus{} \frac {c}{\sqrt {4a^2 \plus{} ab \plus{} 4b^2}}\geq 1
\]
is also true.[/quote]
Nice inequality, scorpius119, but easy!
By Golder we obtain:
$ \left(\sum_{cyc}\frac {a}{\sqrt {4b^2 \plus{} bc \plus{} 4c^2}}\right)^2\sum_{cyc}a(4b^2 \plus{} bc \plus{} 4c^2)\geq(a \plus{} b \plus{} c)^3.$
Thus, it remains to prove that $ (a \plus{} b \plus{} c)^3\geq\sum_{cyc}a(4b^2 \plus{} bc \plus{} 4c^2),$ which Schur.[/quote]

See here [hide]http://www.mathlinks.ro/viewtopic.php?search_id=870964413&t=116917[/hide]

[quote="Oni"]I can't solve that inequality.
\[ \frac {a}{\sqrt {b^2 \plus{} c^2}} \plus{} \frac {b}{\sqrt {c^2 \plus{} a^2}} \plus{} \frac {c}{\sqrt {a^2 \plus{} b^2}}\geq\frac {3}{\sqrt2}
\]
Please help me solve it or give me some hints.

Oni.[/quote]

Generalization see http://www.artofproblemsolving.com/Forum/viewtopic.php?t=187319

[quote="arqady"][quote="scorpius119"]
The inequality
\[ \frac {a}{\sqrt {4b^2 \plus{} bc \plus{} 4c^2}} \plus{} \frac {b}{\sqrt {4c^2 \plus{} ca \plus{} 4a^2}} \plus{} \frac {c}{\sqrt {4a^2 \plus{} ab \plus{} 4b^2}}\geq 1
\]
is also true.[/quote]
Nice inequality, scorpius119, but easy!
By Golder we obtain:
$ \left(\sum_{cyc}\frac {a}{\sqrt {4b^2 \plus{} bc \plus{} 4c^2}}\right)^2\sum_{cyc}a(4b^2 \plus{} bc \plus{} 4c^2)\geq(a\plus{}b\plus{}c)^3.$
Thus, it remains to prove that $ (a\plus{}b\plus{}c)^3\geq\sum_{cyc}a(4b^2 \plus{} bc \plus{} 4c^2),$ which Schur.[/quote]
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=524794：
If $a,b,c$ are positive real numbers, then \[\frac{a}{\sqrt{b^2+\lambda bc+c^2}}+\frac{b}{\sqrt{c^2+\lambda ca+a^2}}+\frac{c}{\sqrt{a^2+\lambda ab+b^2}}\ge\frac{3}{\sqrt{\lambda +2}}.(\lambda \ge\frac{1}{4})\]


function
algebra
functional equation
algebra open

I know

1.continuous
2.diffe......~
3.f(x^2)>0
.....

is there any other??

if you know any other method, explain,please

could you give me example?

****
if f is bound ?

(my english is very short... sorry.)

I'm not sure I understand.  Do you mean if we have a function $ f$ such that some functional equation $ f(x) \equal{} g(x)$ holds for all $ x \in \mathbb{Q}$ and you wish to prove it holds for $ x \in \mathbb{R}$?  If this statement is true, it's as a result of continuity (or a result of more carefully studying the original functional equation).  If you don't have continuity, you can arbitrarily define $ f$ on $ \mathbb{R} \minus{} \mathbb{Q}$ (again, depending on the nature of the original functional equation).

first, thank you for your reply.

I mean,

we use cauchy
i.e. f(x+y)=f(x)+f(y)=>f(x)=cx , x is Q

next step

we must prove f(x)=cx, x is R

I want to know when f is restriction (similarly f is bound)
i.e. f(x)<ax^2+bx

can we extend f(x)=cx , x is real?

Common restrictions include (and I am taking from the Wikipedia article here) continuous at a point, monotonic on any interval, and bounded on any interval.  The image of the pathological solutions is dense in the plane, so any condition that imposes any kind of "resaonable" behavior on $ f(x)$ (all of the above make it impossible to have dense images) allows you to conclude.


induction
inequalities
limit
inequalities solved

Let x1, x2, ... xn be real numbers between 0 and 1/2 inclusive. Prove that (1/x1 - 1)(1/x2 - 1)...(1/xn - 1) >= (n/(x1+x2+..+xn) -1)n.
I've a solution, and you?

We can easily prove it for n=2(just boring calculations and we use the fact that xi is in [0,1/2]). Then we can use a special kind of induction:n->2n and n+1->n:

Let's assume it's true for n numbers so:
(1/x1-1)..(1/xn-1)>=(n/(x1+..+xn)-1)^n and (1/x(n+1)-1)..(1/x(2n)-1)>= (n/(x(n+1)+..+x(2n))-1)^n but (x1+..+xn)/n and (x(n+1)+..+x(2n))/n are also in [0,1/2], so by multiplying the 2 inequalities and by using the fact that the inequality is true for n numbers and for 2 numbers we get the inequality true for 2n numbers, so n->2n (1)

Now let's assume that it's true for n+1 numbers, and we put x(n+1)=(x1+..+xn)/n. After writing the inequality and dividing by n/(x1+..+xn)-1 we get the fact that the inequality is true for n numbers, x1, x2,..,xn, so n+1->n (2)

By combining (1) and (2) we get a complete induction(we can see that the inequality is true for any natural number n). What's your soln?

Thank grobber. This is my soln:
inequaly is equivalent: ln(1/x1 -1) + ln(1/x2 -1) + ... + ln(1/xn -1) >=(n/(x1+ x2 + .. +xn) -1)n
 Consider f(x)=ln(1/x -1) with x on (0, 1/2]. f''(x)=(1-2x)/(x(1-x))2 >=0, then f(x) is concave on (0,1/2].
 Applying Jensen's inequality, we get what we want!!

[quote="hxtung"]Let $ 0 < x_1, x_2, \cdots, x_n\leq\frac {1}{2}$. Prove that

$ \left(\frac {1}{x_1} - 1\right)\left(\frac {1}{x_2} - 1\right)\cdots\left(\frac {1}{x_n} - 1\right)\geq\left(\frac {n}{x_1 + x_2 + \cdots + x_n} - 1\right)^n.$[/quote]The Problem 4 of Tuymaada Yakut Olympiad 2000 is due to Ky Fan, and be recored in

Edwin Ford Beckenbach, Richard Bellman, Inequalities, Springer, Berlin, 1961, p. 5.

Substitute $ \frac {1}{x_k} - 1 = a_k,(1\leq k\leq n)$, the Ky Fan inequality is equivalent to

$ 1 + G(a)\geq H(1 + a)$ for $ a_1, a_2, \cdots, a_n\geq 1$,

where $ G(a) = \sqrt [n]{a_{1}a_{2}\cdots a_{n}}$,

$ H(1 + a) = \cfrac{n}{\cfrac{1}{1 + a_1} + \cfrac{1}{1 + a_2} + \cdots + \cfrac{1}{1 + a_n}}.$

Denote $ M_p(x) = \sqrt [p]{\frac {x_1^p + x_2^p + \cdots + x_n^p}{n}},(p\neq 0)$,

and $ M_{0}(x) = \lim_{p\to0}M_p(x) = G(x)$,

the Ky Fan inequality can be generalized as follows :

Let real $ a_1, a_2, \cdots, a_n\geq 1$, then

$ 1 + M_{p}(a)\geq M_{2p - 1}(1 + a)$

valid for $ p\leq 1$; the inequality sign is reversed for $ p\geq 1$.

[b]Proof[/b] Substitute $ a_k = \sqrt [p]{x_k},(1\leq k\leq n)$.

Let $ f(x) = \left(1 + \sqrt [p]{x}\right)^{2p - 1},(\sqrt [p]{x}\geq 1)$, then

${{ f''(x) = \left(1 - \frac {1}{p}\right)\left(2 - \frac {1}{p}\right)x^{\frac {1}{p} - 2}\left(\sqrt [p]{x} - 1\right)\left(1 + \sqrt [p]{x}\right)^{2p - 3}\begin{cases} \geq 0 & \text{if}\ p < 0\ \text{or}\ 0 < p\leq\dfrac{1}{2}\ \text{or}\ p\geq1}, \\
\leq 0 & \text{if}\ \dfrac{1}{2}\leq p\leq 1}.\end{cases}$

Using the Jensen's inequality, we proved the inequality for $ p\neq 0$ and $ p\neq \frac {1}{2}$.

Let $ p\to 0$, we obtain the Ky Fan inequality; let $ p\to \frac {1}{2}$, we have

$ 1 + M_{\frac {1}{2}}(a)\geq G(1 + a)$.

Substitute $ a_k = x_k^2,(1\leq k\leq n)$, it's equivalent to

$ 1 + A^2(x)\geq G(1 + x^2)$ for $ x_1, x_2, \cdots, x_n\geq 1$,

which can be proved by the Jensen's inequality alone :

Let $ g(x) = \ln(1 + x^2)$ for $ x\geq 1$, then

$ g''(x) = \frac {2\left(1 - x^2\right)}{\left(1 + x^2\right)^2}\leq0$.


geometry
geometric transformation
reflection
circumcircle
angle bisector

[color=blue]If S is the circumcentre of triangle ABC and D,E,F are the feet of the altitudes of ABC, then prove that SB is perpendicular to DF  [/color]

[hide="Solution"]Clearly $\Delta DBF \sim \Delta ABC$. So a reflection through the angle bisector of $\angle B$, send $\Delta DBF$ into $\Delta D'BF'$ where $D'F' \parallel AC$.
So $BE \bot D'F'$.
On the other hand let $P$ be the point of intersection between the circumcircle of $\Delta ABC$ and the angle bisector of $\angle B$, and $M$ the midpoint of $AC$.
Then $P, M, S$ are collinear and $PS \bot AC$. Hence $\angle EBP = \angle BPS = \angle PBS$.
Then $BS$ is the isogonal line of $BE$, and finally $BS \bot DF$.[/hide]


What do you think the cutoff for MOP will be?

are you talking about red, blue, or black...

red - 8
blue 18-20 (hopefully it's the lower end and the graders like my #2)
black - 27

hopefully not high

that's as far as I'm going to go predicting because I have nooooooo clue

forgive me but what happens if you get MOP?

the USAMO is a test which determines whether you qualify for MOP, the AMC summer program.

if you get above the cutoff, you get to go to nebraska

[quote="noneoftheabove"]hopefully red...likes me but it seems most people got 1 and 4 so it will be the partials that go through[/quote]
You mean most people [i]think[/i] they got 1 and 4.  Most people overestimate their USAMO scores, especially people taking it for the first time (i.e., most freshmen).  But seriously, people, try to chill a little until they grade your papers.  I know that's kind of hard, but since on average people overestimate their scores by some unpredictable amount, you're probably not going to get any accurate information.

You also go to Nebraska if you make the cutoff exactly.  Never forget your equality cases.

ahh okay.

so then whats the difference between red, blue, black?

how many people make it, ish?

haha thanks miles

if only you were there to check my USAMO work

is 7-9 a realistic cutoff for red? Seems somewhat low considering number 1.

but just because you solve number 1 correctly does not mean everyone received full credit cutting the field by a considerable amount...i say 10 or so is reasonable

My prediction is that it's going to be:

[hide]OVER 9000!!!![/hide]

I think at least 24 freshman, and probably more, solved #1. I think I know about 10 of them.

I think it comes down to who solved 2/4 the most rigorous.

I think the assumption is that all red moppers will solve problem 1. 9-10 points seems like a pretty reasonable expectation for the red cutoff, IMO.

I wonder how much a semi-handwavey solution would be worth for number 4

I'm not entirely sure what you mean, but I would guess 2 if you also found the equality case and gave the example.

Wait there was an equality case for number 4??

[quote]
can you clarify what you mean when you say that the cross is an equality case?
[/quote]

As I am aware, all solutions to this problem first establish an upper bound on the number of cells a primitive dinosaur can have.

After a bound is set, all that remains is to show the "equality" case by drawing a dinosaur that actually reaches the bound.

For those who said a 30 cutoff for black lemme say noooooooooo way I'll be surprised if 5 people get 30. Not that this was necessarily that hard of a USAMO but #3 requires a fair amount of skill and #5 was just a terrible problem. #6 was like a Geometry bash but it really favors people with training. #1, 2, 4 were pretty easy, I think?

I would guess 22-24 for the top 12 cutoff ...? It will be very interesting to see....


calculus
integration
function
real analysis
real analysis unsolved

Given an arbitrary but fixed $ C > 0$. Are there non-trivial functions $ f,g\in L^2([ \minus{} 1,0],\mathbb{R})$ s.t. $ \forall\lambda > C$:
\[ \int_{ \minus{} 1}^0e^{\lambda t}(f(t)(\lambda \plus{} 1) \plus{} g(t)) \mathrm{d}t \equal{} 0\]
laxton

$ \int\dots dt$, I guess?

Anyway, differentiating w.r.t. $ \lambda$ leads to $ f\equal{}0$, whence $ g\equal{}0$.

This leads to
\[ \int_{\minus{}1}^0e^{\lambda t}(f(t)(\lambda t\plus{}t\plus{}1)\plus{}tg(t))\mathrm{d}t\equal{}0\]

laxton

Oh yes, sorry, I'm bad at differentiating, got $ (e^{\lambda t})'_\lambda = \lambda e^{\lambda t)}$ :)

Well, just take $ f$ smooth and compactly supported inside the interval $ [\minus{}1,0]$ and put $ g\equal{}f'\minus{}f$.


Our forum now accepts attachments, so you can add diagrams to your problems/solutions without having to have webspace of your own.

Please do not abuse this by attaching improper material.  If abuse occurs, we will have to restrict the ability to attach files to posts (and deal with the offender).  Also, please note that attachments have been disabled in private messages.

Here's a sample of what the attachments look like in posts:

Thank you - I think you originally suggested the idea.

i was wondering what the user attachment control panel is all about. so eventually (if we constantly post attatchment) we would use up the 150 MB and for teh attachment could we attach anyting or does it have to be specific things like (jpegs or bmps)

hmm well I can understand half of what you just wrote, the second half, click the "allowed extensions and sizes" to see what you can attach. as for the first half, I have no idea what you are talking about. i think I'd get it if there was a "?" somewhere :P

er...where's the "user attachment control panel"?...the only thing i see is the lil "Add an Attachment" box under "Post a Reply"

To add an attachment to your post, scroll down in the posting window after you click 'New Topic' or 'Post Reply'.  You should see an 'Add an Attachment' box.  Click Browse there to locate your attachment on your computer, then use the File Comment to add a note and 'Add attachment' to add the attachment to your post.

how come we can only attach picture realted attachments why not pdfs, adn the 100 kb limits is very small (forces you to use jpg)

Could you possibly add pdf, tex, or dvi files as an option?  Thanks.

You can include TeX files very easily already. PDF and DVI files may be quite difficult since I think people need to have the viewers already installed otherwise something bad might happen (I'm not very sure about this). If you don't play around with the formatting too much, people who have TeX editors should be able to compile your code without much trouble.

[quote="churchilljrhigh"]the user control panel can be accesed if you click post a message and then click the profile button. You will see the user control panel and be able to delete attachments (i think because i don't have anyu attachments)[/quote]
That's exactly what I'm doing...

[quote="pendulum"]I've noticed that attachments are invisible unless I'm logged in.  Is that intentional?[/quote]Only registred members can download attachements. It is normal.

Ok thanks.  I need to remember to write "attachment below" or something like that for people who may not be logged in.  Otherwise I'll be referring to something that isn't there!

[quote="Valentin Vornicu"]What exactly are you doing? ...  :?[/quote]

I was posting a response so I could click my username.


Two tangent lines of a circle at A and B intersect at C. The midpoint of segment AB is M. Let P be a point on arc AB. Line CP intersect the circle at C and Q. Line PM intersect the circle at C and R. Prove that BR=AQ.

[hide]APBQ is harmonic quadrilateral[/hide]

$ OMPQ$ is cyclic $ \rightarrow \angle OMR \equal{} \angle OQP \equal{} \angle OPQ \equal{} \angle OMQ$


Does anyone have some cool information on this?  I'm doing a research project on it.

Clay Math Institute has chunk of info on it. Other than that, I do not know... (and they have some Million-dollar thing going :? )

Yes.  I've read that.  I just wondered if anyone here had more on it.

I don't know much about what the P VS NP problem is. However, I know that NP problems are those problems in computer science which cannot be solved in polynomial time and hence, take too long a time to be solved for large test cases in current computers. Apparently, quantum computers would solve NP problems in polynomial times (not sure though).

An example of an NP problem is the Travelling Salesman Problem. We are given a series of roads between cities, each with a cost, and we have to visit all the cities and return to the origin with a total minimum cost. There is no known polynomial solution for it and the only solutions that runs in time involves an optimization or heuristics.

Possibly, the P VS NP problem is to prove that NP problems have no polynomial solutions.

Try the wikipedia entry: [url]http://en.wikipedia.org/wiki/Complexity_classes_P_and_NP[/url]. It's a good intro. There's also a book by Keith Devlin called "The Millennium Problems: The seven greatest unsolved mathematical puzzles of our time" that's pretty good and obvious includes the P vs NP problem. Hope this helps.

No... if it was proved that it couldn't be solved, people wouldn't be trying to solve it...

I have proved that P is different NP by creating a problem that can be done by NP but is unable to be done on a polynomial time period. But i don't know what to do.

I'm not sure what you mean, "I don't know what to do."  Ideally, you would write it in a paper and submit it to a mathematics journal, as that would be a significant discovery.  Perhaps you can post the problem here and the proof that it is NP but cannot be solved in P?


[quote="AHSME 1954"]
The straight line $AB$ is divided at $C$ so that $AC=3CB$. Circles are drawn with $AC$ and $CB$ as diameters and a common tangent to these meets $AB$ extended at $D$. Show that $DB$ equals the radius of the smaller circle.
[/quote]

See diagram!

[quote="chess64"][quote="Jan"]Your equation involving Pythagoras is wrong! Look at your triangle, $p$ is not the hypothenusa, so the equation will turn in to:$p^{2}=36r^{2}-9r^{2}=27r^{2}$
Which is the same as your equation involving power of a point.[/quote]

Wait...but it's opposite the $90^{\circ}$ angle, isn't it....  :?  :?  :?[/quote]
Try to magnify your figure and see what happen.
The point of the triangle you applied pythagoras is not the tangency point between p and the circle at the left.

Sure, you can make $p$ be the hypotenuse of that triangle, but then it wouldn't be a tangent.

If you want $p$ to be a tangent, then the right angle comes between $p$ and $3r$ (tangent and radius form $90^\circ$).

your diagram is incorrect, that hypotenuse is not tangent to the circles, look at the zoomed in pic i got for you

edit: just a little lesson, inaccurate geometrically drawn figures can lead to wacky things, many "seemingly fake" contradictions

So it's not a triangle, it only looks like one?

Hmmm, so far I have 9 votes against me :P

There is a triangle, but only where $p$ forms a right angle with the big circle, which would then mean it's also tangent to the small circle.  In that case, $3r + 2r + r$ is the hypotenuse.


Here is my comments for some problems:
Could you correct it for me?

For everyone, the chessboard is very simple and natural. But it doesn’t seem to be that, there are a lot of interesting things about chessboard that we want to present you:

We find the above mentions solutions impressive because they are distinguishable by the shortness and clarity. This proves the attractiveness and simplicity of the chessboard. The following not less interesting questions will be raised:

What‘s your opinion about the above mentioned example? It‘s started simply but can be applied to many other following problems, as well as the Ramsey numbers which is well-known to us. To close this question, let go to a more diffcult solution of the Ramsey numbers:

Thanks!

[quote="Kyoshiro"]For everyone, the chessboard is very simple and natural. But it doesn’t seem to be that, there are a lot of interesting things about chessboard that we want to present you:[/quote]
The first sentence is grammatically fine but stylistically awkward.  I might write, "The design of the chessboard is very simple and natural."  The intended meaning of the second sentence is unclear: it appears to indicate that there are [i]not[/i] any interesting things about the chessboard that you want to present.  Perhaps this sentence should be two sentences, the first of which suggests that there seems to be nothing to say about it, and the second of which says, "but actually we will present some interesting things."  (A more specific word than "things" would be nice.)

[quote="Kyoshiro"]We find the above mentions solutions impressive because they are distinguishable by the shortness and clarity. This proves the attractiveness and simplicity of the chessboard. The following not less interesting questions will be raised:[/quote] "The solutions above are impressive because of their shortness and clarity.  This shows the attractiveness and simplicity of the chessboard.  The following equally interesting questions [something]."

The "will be raised" is clearly wrong, but without full context it's not clear what an appropriate fix would be.  Perhaps just, "We now provide some other, equally interesting questions."

[quote="Kyoshiro"]What's your opinion about the above mentioned example? It‘s started simply but can be applied to many other following problems, as well as the Ramsey numbers which is well-known to us. To close this question, let go to a more diffcult solution of the Ramsey numbers:[/quote] "What is your opinion about the above-mentioned example?*  It is stated simply** but can be applied to many other problems, as well as to the Ramsey numbers which are well-known to us. ???"

What does "to close this question" mean?  What question?  "Let go" should probably be "let us go," but actually "go" is almost certainly not the best verb.  Also, "solution of the Ramsey numbers"?  What does that mean?  Finally, "diff[b]i[/i]cult" is the correct spelling.

* "Preceding example" or "example above" might be better.
** Or "It has a simple statement."


vector
superior algebra
superior algebra unsolved

Hi all,

I have a couple of questions.

Let $ V$ be a vector space of dimension $ m$ over $ \mathbb{R}$ with basis $ \{e_i \}_{i \equal{} 1, \ldots, m}$.

1.  Show $ \bigwedge^m V$ has dimension $ \leq 1$.
I know that, in fact, $ \bigwedge^m V$ has dimension $ 1$.  The question really only asks to exhibit a spanning element.  What I did was take an arbitrary element $ v_1 \wedge \cdots \wedge v_m \in \bigwedge^m V$ and express it as $ v_1 \wedge \cdots \wedge v_m \equal{} \left( \sum_{i \equal{} 1}^m a_{1,i}e_i \right) \wedge \cdots \wedge \left( \sum_{i \equal{} 1}^m a_{m,i}e_i \right)$.  I think I can use multilinearity to write this as sum of something times $ e_1 \wedge \cdots \wedge e_m$ so that $ e_1 \wedge \cdots \wedge e_m$ spans $ \bigwedge^m V$.  So my question is, what exactly are the coefficients of $ e_1 \wedge \cdots \wedge e_m$, and moreover, is $ e_1 \wedge \cdots \wedge e_m$ even the element that spans?  On a somewhat unrelated note, and hopefully this could help me understand this problem more, what is the dimension of $ \bigwedge^p V$ if $ p < m$?  What about if $ p > m$?

2.  Define a linear transformation $ \left( \bigwedge^p V \right) \otimes \left( \bigwedge^{m \minus{} p} V \right) \rightarrow \bigwedge^m V$.
I'm guessing that this probably has something to do with the determinant, but I really have no idea where to start.  Any help would be appreciated.

Thanks in advance.

Is my question confusing?  If so, I can re-ask it.

1. You do not need to care about what the coefficients are. Note, however, that you do not know a priori that every element of $ \bigwedge\!{}^mV$ has the form $ v_1\land\dots\land v_m$. Apart from this, you show that every element is a multiple of $ e_1\land\dots\land e_m$, so the dimension is at most 1. In general, the dimension of $ \bigwedge\!{}^pV$ is $ m\choose p$.

2. Just explain why $ (v_1\land\dots\land v_p)\otimes(w_1\land\dots\land w_{m\minus{}p})\mapsto v_1\land\dots\land v_p\land w_1\land\dots\land w_{m\minus{}p}$ defines a linear map. owk


Seven equal 10-inch-long pieces are cut from an eight-foot-long piece of lumber. If the waste per cut is $ \frac{1}{8}$ inch, how many inches long is the piece that is left over? Express your answer as a mixed number.

8 feet is $ 8 \times 12 \equal{} 96$ inches. $ 7 \times 10 \equal{} 70$ inches has been cut from it, leaving $ 96 \minus{} 70 \equal{} 26$ inches left. Each time a piece is cut, $ \frac{1}{8}$ inch is cut with it. So the number of inches left is:

$ 26 \minus{} \left (7 \cdot \frac{1}{8} \right ) \equal{} \boxed{25 \frac{1}{8}}$


if        $ 0 < a,b,c\in\Re$

Prove that
\[ \frac {2(a^3 \plus{} b^3 \plus{} c^3)}{abc} \plus{} \frac {9(a \plus{} b \plus{} c)^2}{a^2 \plus{} b^2 \plus{} c^2}\geq33
\]

Use SOS, we have
 $ 2\frac{a^3\plus{}b^3\plus{}c^3\minus{}3abc}{abc}\plus{} 9 \frac{(a\plus{}b\plus{}c)^2\minus{}3(a^2\plus{}b^2\plus{}c^2)}{(a^2\plus{}b^2\plus{}c^2)} \ge 0$

[quote="loc1991"]Use SOS, we have
 $ 2\frac {a^3 \plus{} b^3 \plus{} c^3 \minus{} 3abc}{abc} \plus{} 9 \frac {(a \plus{} b \plus{} c)^2 \minus{} 3(a^2 \plus{} b^2 \plus{} c^2)}{(a^2 \plus{} b^2 \plus{} c^2)} \ge 0$[/quote]

$ (a \plus{} b \plus{} c)^2 \leq 3(a^2 \plus{} b^2 \plus{} c^2)$ ;)

Clearing the denominator, and use $ \sum_{cyc} c(a\minus{}b)^4 \ge 0$ and $ \sum_{cyc} (a\plus{}b)(a\minus{}b)^4 \ge 0$

[quote="arqady"][quote="earldbest"]what's SOS?   :([/quote]
It's here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=80127
but the
Bayasgalan's inequality can be proven without SOS since,
it is equivalent to $ \frac {a \plus{} b \plus{} c}{abc}\geq\frac {9}{a^2 \plus{} b^2 \plus{} c^2}.$[/quote]

thanks a lot. :lol:  :idea:

setting
$ p \equal{} a \plus{} b \plus{} c \equal{} 1$
$ q \equal{} ab \plus{} bc \plus{} ca$
$ r \equal{} abc$
the problem turn to
$ (1 \minus{} 3q)(1\minus{}2q\minus{}9r)\geq 0$ which is obvious...


\[a\leq 1\Rightarrow a^{3}\leq 1\]  \[b\leq 1\Rightarrow b^{3}\leq 1\]  \[c\leq 1\Rightarrow c^{3}\leq 1\]  \[\Rightarrow a^{3}+b^{3}+c^{3}\leq 3\] .
But \[a^{3}+b^{3}+c^{3}=3\Rightarrow \]  \[\Rightarrow a=b=c=1\] Sorry :(


analytic geometry
combinatorics unsolved
combinatorics

$ ''Battleship''$

On a board $ 8$x$ 8$  is located a ship- a square $ 2$x$ 2$, which takes 4 cells of a board. A player makes a shot, naming any cell of a board, and then he's informed, whether he has got in the ship or not. Then he makes the following shot and so on. For what least number of shots it is possible to define for certain an exact place of the ship. (Unessentially having sunk it)

It i don't misunderstand your idea then my answer is $ 16$. We shout $ 16$ cells at $ (i;j)$ with $ i,j\equal{}1,3,5,7$ then a square $ 2*2$ must have consist of a cells from $ 16$ that cells.

Just for now, I found a way to destroy a ship with only 19 guesses , I don't have the proof that this is minimal, but I'll post it anyway, so if someone find solution with less guesses, tell me and I'll stop bothering trying to prove it :)

If you color all cells with both coordinates even (such as, (2,2), (2,4), ... ,(8,8)) in black, and the rest ones in white, you get that a ship must contain one black sell. So you have 16 guesses to find that black cell (because there are 16 black cells), and then 3 more to construct whole ship (it's enough to find two diagonal cells, then you know where the sheep is, and if you miss 3 times, then you know it in the 4th cell).
Bye

edit: well, i have the proof but it isn't really formal, and nice and all that. After all, mu english is awful so I could't write it here. I am sure that you can prove it yourself, when you know that 19 is minimal, it;s now hard...

Yes, i also found that we should make 19 shots. But i can't prove it

It isn't 19, it's 17... You guess in that way that you leave black cells that are on the border for the end (and you pick the cell (8,8) last). Then it's enough only one question to make sure where the ship is. Now, the proof is I think simple...

Bye

In some situations it's not enough to make 17 shots... We don't know where the ship is. Theoretically we can destroy it from first shoot:) We should solve this problem, when the player is in the worst situation.

[quote="pleurestique"]let me get this straight... the goal isn't to destroy it. the goal is to find exactly which squares it occupies, right?[/quote]

Yes you're right. I solved this problem once more and found that we should make 17 shots  :lol:

[quote="SpongeBob"]Yes! And the answer is 16 :)[/quote]

I don't know how did you get $ 16$?

Can anyone post a normal solution?

[quote="SpongeBob"]Here it is. You color the board like in my first post, and then you make guesses, but in that way that you first ask for (2,2), (2,4), (2,6), (4,2), (4,4), (4,6) and
(6,2),(6,4),(6,6). If in one of these cells there is a part of ship,  you need three more questions to be sure where is entire ship (you are looking for diagonal cell of that one you got right). If you still don't know where the ship is, then you ask for (2,8), (4,8), (6,8) and then (8,2), (8,4), (8,6). This cells are on the border, so if you guess one cell of the ship, you need only one question to know where the ship is, so maximal 16 guesses. If you still didn't atack the ship, then there is only one place where it can be, and that is bottom right corner (with one cell (8,8)).
So there you go, it's enough 16 guesses. And now it's simple to prove that this is really minimal. Ask if something isn't right.

Bye[/quote]

Ok, imagine, that we shooting at position (8, 6) and we attack the ship(it's our 15th shot). We don't know where the ship is, that is why, we can make a shot on (7, 6), so we attacked the ship again and we found 2 cells, but we still don't know where the ship is!! And making 17 shot on position (7, 5) or (7, 7) we finally know the place of the ship  :wink:
(In my solution the ship coordinates are (8, 6); (8, 7); (7, 6); (7, 7) )

[quote="cf249"][quote](In my solution the ship coordinates are (8, 6); (8, 7); (7, 5); (7, 7) )[/quote]

Don't you mean (8,6); (8,7); (7,[b]6[/b]); (7,7)? (because if you don't, your four coordinates don't form a square.)[/quote]

Yes (7, 6). sorry... :blush:

[quote="SpongeBob"][quote="Meatman"]
Ok, imagine, that we shooting at position (8, 6) and we attack the ship(it's our 15th shot). We don't know where the ship is, that is why, we can make a shot on (7, 6), so we attacked the ship again and we found 2 cells, but we still don't know where the ship is!! And making 17 shot on position (7, 5) or (7, 7) we finally know the place of the ship  :wink:
(In my solution the ship coordinates are (8, 6); (8, 7); (7, 6); (7, 7) )[/quote]

So, you attack in your 15th move cell $ (8,6)$ and there is one cell of your sheep. DIAGONAL to that cell in the sheep is either (7,7) or (7,5). After $ (8,6)$ you attack $ (7,7)$ that is your 16th move. If there is a cell of the sheep, then you know that shep is on 86,87,76,77; if there isn't then again you KNOW where the ship is 86,76,85,75; you only attacked one cell, but nevertheless, you KNOW where the ship is. OK?[/quote]

OK, i've understood you.


[quote="jelyman"]To anyone who has ordered from the AMC site, how long does it usually take to arrive?[/quote]

I can't remember an exact figure, but I remember being pleasantly surprised by the speed. Off hand, based on three different orders that I don't remember distinctly, I'd say a week or two to another part of the contiguous forty-eight states, unless your region is known for slow mail delivery.


algebra
polynomial
number theory
prime numbers
number theory proposed

Let $%Error. "mathGothicE" is a bad command.
{P}$ be a polynomial with integer coefficients that If $\mathfrak P(n)$ is Prime then $n$ is prime .Prove that if $n$ is prime then $P (n)$ is prime.

EDIT: $P$ must take a valua of a prime at least once.

and if $P(x)=4Q(x)$ where $Q(x) \in \mathbb{Z}[X]$ ?

lomos_lupin, alekk just gave you a counterexample since trivially 4*P(n) is never prime (for P in Z[X]). So (4*P(n) prime implies n prime) is true.
Homever n prime does not imply 4*P(n) prime. So if P is a polynomial such that P(n) prime => n prime , then the converse is not necessarily true. Hence modify your statement in order to make it correct :).

$\{x\}$ : i think i understand what alekk was trying to say ,there is no need for your translation.

Assume that $Q$ is bound less from above ,take $q$, a large enough prime then there exist a real number like $\alpha$ such that $Q(\alpha)=\frac{q}{4}$ thus $P(\alpha)=q$ .then $\alpha$ must be a prime [b]SO[/b]
$Q(x)$ is not just an arbitariry polynomial.

[b]Explicit counterexample.[/b]
Let $Q ( x ) = 4$, for all real $x$.
$\left[ Q ( p ) \mbox{ prime} \Rightarrow p \mbox{ prime} \right]$. Hence $Q$ satisfies the hypothesis.
But $[ p \mbox{ prime} \nRightarrow Q ( p ) \mbox{ prime} ]$.
Tell me [b]exactly[/b] where there is something wrong.

$P$ must at least once take a value of a prime number .

Actually your are not satisfying the hypothesis of the problem ,you are just U-turning it .

What about the Polynomial  P(n)= (n-1)*(1+n^2) ???

P(n) has two factors n-1 and n^2+1.
The first factor n-1 is non-trivial, except for n=2 and n=0.
The second factor n^2+1 is non-trivial, except for n=0.
Therefore, P(n) cannot be prime, unless n=0 or n=2.
We have P(0)=-1 (non-prime) and P(2)=5 (prime).

Therefore: P(n) takes a prime value at least once (for n=2).
And: [P(n) is prime] implies [n=2] implies [n prime]
So P(n) satisfies all desired properties.

But: P(3)=20.  Hence: [n prime] does not imply [P(n) prime].

[quote="lomos_lupin"]$P$ must at least once take a value of a prime number .

Actually your are not satisfying the hypothesis of the problem.[/quote]
No lomos_lupin you are wrong. $P$ must not take any prime value for the implication $\left[ P ( n ) \mbox{ prime} \Rightarrow n \mbox{ prime} \right]$ to be [b]formally[/b] true. As previously we let $P(x)=4$ for all real x. Suppose to the contrary $P(n)$ is prime, and $n$ is composite. But then $P(n)=4$, so $P(n)$ is not prime. A contradiction.
You can also remark that $\left[ P ( n ) \mbox{ prime} \Rightarrow n \mbox{ prime} \right]$ is [b]equivalent[/b] to $\left[ n \mbox{ not prime} \Rightarrow P(n) \mbox{ not prime} \right]$ and this is [b]equivalent[/b] to  $\left[ n \mbox{ composite or equal to 1} \Rightarrow P(n) \mbox{ composite or equal to 1} \right]$ which is clearly true for $P(x)=4$ for all real $x$. This last reasoning is probably easier to accept. Homever will you finally accept that you are wrong ?

Moreover it is easy to construct a polynomial $P(x)$ taking a prime value but not veryfying the hypotheses as the previous post shows.

[quote="lomos_lupin"] you are just U-turning it .[/quote]

I don't know what U-turning is.

EDIT: OK, sorry you've edited the problem. Homever the last post still gives you a counterexample.

In the problem, I don't think that $n$ is required to be an integer.  The problem is a bit  confusing, because the letter $n$ is usually reserved for integers.

Thus test20's polynomial is not a counterexample:  It does not satisfy the hypothesis.  For example, $P(n) = 2$ when $n$ is approximately $1.5747431$, which is not a prime.

[quote="Ravi B"]Thus test20's polynomial is not a counterexample:  It does not satisfy the hypothesis.  For example, $P(n) = 2$ when $n$ is approximately $1.5747431$, which is not a prime.[/quote]

The hypothesis doesn't say that $P(n)$ must take [b]any[/b] prime value (in fact, that's impossible, except for $P(n)\equiv n$, and then the implication is trivial.)

When I said "hypothesis", I was referring to "If $P(n)$ is prime then $n$ is prime".  test20's polynomial does not satisfy that hypothesis.  Remember that $n$ is not required to be an integer.

If so, then the problem is trivial, since we can require that for arbitrarily large prime $p$ there is a prime $n$ such that $P(n)=p$, because all the polynomials are unbounded. But no polynomial can cover all the prime numbers except $P(n)\equiv n$, and than we have the implication trivially.

I agree the problem is probably easy.  But don't you need an argument to show that $P(n) \equiv n$ is the only polynomial that covers all the primes?

Well, I take that as a long-proven fact in number theory (though I've never seen the proof)

[quote="Ravi B"]Remember that $n$ is not required to be an integer.[/quote]
That would explain in a part the "quiproquo" which the statement generated :)


how do u do venn diagram problems with 3 circles ?

Can you give an example?

Generally, if you are trying to organize data into Venn diagrams, you would start with the middle value, input that, and then move on to the values where two circles intersect (subtract the middle value from all of those, because you already put that down), and then fill in the rest.

I'm not sure if this was what you wanted though.

u should read the sections that explains that in the intro to counting/probability book. there's some nice examples of those problems in the book, including "venn diagrams with 2 circles" problems.

 :D