inequalities
trigonometry
inequalities solved

Hey, Who can solved this problems?
x,y,z be positive real numbers such that
x+ y+ x+ 2=xyz.
Prove that 5(x + y + z)+ 18>=8( (xy)^1/2 + (yz)^1/2 + (xz)1/2)

Yes
x+ y+ z +2 = xyz is true
Can you solv? Lagrangia.
He he

Try to solv this problem:
 x,y,z be positive real numbers such that : x+ y+ x+ 2=xyz.Prove that
 4(x + y + z)+ 24 >=8( (xy)^1/2 + (yz)^1/2 + (xz)1/2)

perhaps what grobber posted would help you! have you thought about that?  :?

I didn't think about it yet, but I suppose we can at least divide the whole thing by 4?

I used the same method as above and I got this(which is harder to prove than the previous one):
 \sum(cosA^2)+ \sum (cosA*cosB)<=3/2

Try to prove this one. I couldn't solve it in a nice way, but I think you can prove it by calculating A LOT  :D ...

It's trivial. Rewrite it as sum (cos(A)+cos(B))^2 <= 3 , but cos(A)+cos(B)=2*cos(A+B)/2 cos(A-B)/2 <= 2 * sin(C/2) , then use Jensen and you're done .

[quote="hxtung"] $ x,y,z$ be positive real numbers such that  $ x \plus{} y \plus{} z \plus{} 2 \equal{} xyz$. Prove that

$ x \plus{} y \plus{} z \plus{} 6 \geq 2\left(\sqrt {yz} \plus{} \sqrt {zx} \plus{} \sqrt {xy}\right)$.[/quote]By the substitution $ x \equal{} \frac {v \plus{} w}{u},y \equal{} \frac {w \plus{} u}{v},z \equal{} \frac {u \plus{} v}{w}$, the original inequality is equivalent to

$ \sum{\frac {v \plus{} w}{u}} \plus{} 6\geq 2\sum\sqrt {\frac {(w \plus{} u)(u \plus{} v)}{vw}}$

for positive real numbers $ u,v,w$.

[hide="Proof."]$ (2vw \plus{} wu \plus{} uv)^2 \minus{} 4vw(w \plus{} u)(u \plus{} v) \equal{} u^2(v \minus{} w)^2\geq 0\Longrightarrow$

$ 2\sum\sqrt {\frac {(w \plus{} u)(u \plus{} v)}{vw}}\leq\sum{\frac {2vw \plus{} wu \plus{} uv}{vw}} \equal{} 6 \plus{} \sum{\frac {v \plus{} w}{u}}$.[/hide][quote="iandrei"]Rewrite it as $ \sum(\cos{B} \plus{} \cos{C})^2 \leq 3$ for $ \triangle{ABC}$, but

$ \cos{B} \plus{} \cos{C} \equal{} 2\cos{\frac {B \plus{} C}{2}}\cos{\frac {B \minus{} C}{2}}\leq 2\sin{\frac {A}{2}}$,

then use Jensen[/quote]$ \sum{\sin^2{\frac {A}{2}}}\geq \frac {3}{4}$ and you're done nothing.

By the way, if $ x,y,z$ be nonnegative real numbers such that  $ 9(x \plus{} y \plus{} z) \plus{} 10 \geq 8xyz$, then

$ x \plus{} y \plus{} z \plus{} 6 \geq 2\left(\sqrt {yz} \plus{} \sqrt {zx} \plus{} \sqrt {xy}\right)$.

[b]Remark[/b]  $ x \plus{} y \plus{} z \plus{} 2 \equal{} xyz\Longrightarrow9(x \plus{} y \plus{} z) \plus{} 10 \geq 8xyz$.

[quote="grobber"]I was going through some magazines and I found the following theorem:
If x,y,z>0 and x^2+y^2+z^2+x*y*z=4 then there are A,B,C in [0, pi\2) so that x=2*cosA, y=2*cosB, z=2*cosC and A+B+C=pi (it's not hard to prove, you just try to solve the equation as if it were a second degree equation in x, having y=2*cosB and z=2*cosC because y^2 and z^2 <=4, you get the positive solution x=2*cos(pi-B-C)).

Here's how you use it:
x=2/a,y=2/b,z=2/c. After some calculations you get: a*b+b*c+c*a+a*b*c=4 so (a*b)^1/2 and the others satisfy the conditions in the theorem above, so you have: (ab)^1/2=2*cosA, (b*c)^1/2=2*cosB and (c*a)^1/2=2*cosC, A,B,C being the angles of a triangle. So x=cosA/cosB*cosC and the other analogous relations. The inequality you want to prove is, after some more calculations:
 sum(cosA*cosB)-cosA*cosB*cosC<=5/8(you just replace x,y,z with their
found values). But   \sum(cosA*cosA)+2*cosA*cosB*cosC=1 so what you're trying to prove is:
 (sum(cosA))^2<=9/4 iff  sum(cosA)<=3/2, but we know that sum(cosA)=1+r/R, and R>=2*r so the last inequality is correct.[/quote]

Hmm...I don't know why sum of cosA is 1+r/R. Can you post why?

[quote="mathVNpro"]Here is my solution:
[b]Problem[/b]:
Let $ x,y,z$ be postive real numbers such that $ xyz \equal{} x \plus{} y \plus{} z \plus{} 2$. Prove that:
$ 2(\sqrt {xy} \plus{} \sqrt {yz} \plus{} \sqrt {zx})\le x \plus{} y \plus{} z \plus{} 6$.
[i]Proof[/i]:
The inequality is enquivalent to:
$ (\sum \sqrt {x})^2 \minus{} (x \plus{} y \plus{} z)\le x \plus{} y \plus{} z \plus{} 6$ or $ \sum \sqrt {x}\le \sqrt {2(x \plus{} y \plus{} z \plus{} 3)}$
Denote $ x \equal{} \frac {b \plus{} c}{a}$, $ y \equal{} \frac {c \plus{} a}{b}$, $ z \equal{} \frac {a \plus{} b}{c}$
Therefore, we just need to prove:
$ \sum \sqrt {\frac {b \plus{} c}{a}}\le\sqrt {2(\sum \frac {b \plus{} c}{a} \plus{} 3)} \equal{} \sqrt {2(a \plus{} b \plus{} c)(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c})} \equal{} \sqrt {[\sum {(b \plus{} c)}](\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c})}$
But this is obviously true due to [b]Cauchy-Schwartz[/b], which ends our proof. :P[/quote]

Very nice;mathVNpro  :P ..It's old problem.

[quote="mathVNpro"]Here is my solution:
[b]Problem[/b]:
Let $ x,y,z$ be postive real numbers such that $ xyz \equal{} x \plus{} y \plus{} z \plus{} 2$. Prove that:
$ 2(\sqrt {xy} \plus{} \sqrt {yz} \plus{} \sqrt {zx})\le x \plus{} y \plus{} z \plus{} 6$.
[i]Proof[/i]:
The inequality is enquivalent to:
$ (\sum \sqrt {x})^2 \minus{} (x \plus{} y \plus{} z)\le x \plus{} y \plus{} z \plus{} 6$ or $ \sum \sqrt {x}\le \sqrt {2(x \plus{} y \plus{} z \plus{} 3)}$
Denote $ x \equal{} \frac {b \plus{} c}{a}$, $ y \equal{} \frac {c \plus{} a}{b}$, $ z \equal{} \frac {a \plus{} b}{c}$
Therefore, we just need to prove:
$ \sum \sqrt {\frac {b \plus{} c}{a}}\le\sqrt {2(\sum \frac {b \plus{} c}{a} \plus{} 3)} \equal{} \sqrt {2(a \plus{} b \plus{} c)(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c})} \equal{} \sqrt {[\sum {(b \plus{} c)}](\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c})}$
But this is obviously true due to [b]Cauchy-Schwartz[/b], which ends our proof. :P[/quote]
very nice solution,mathVNpro ;)

[quote="mathVNpro"]Here is my solution:
[b]Problem[/b]:
Let $ x,y,z$ be postive real numbers such that $ xyz \equal{} x \plus{} y \plus{} z \plus{} 2$. Prove that:
$ 2(\sqrt {xy} \plus{} \sqrt {yz} \plus{} \sqrt {zx})\le x \plus{} y \plus{} z \plus{} 6$.
[i]Proof[/i]:
The inequality is enquivalent to:
$ (\sum \sqrt {x})^2 \minus{} (x \plus{} y \plus{} z)\le x \plus{} y \plus{} z \plus{} 6$ or $ \sum \sqrt {x}\le \sqrt {2(x \plus{} y \plus{} z \plus{} 3)}$
Denote $ x \equal{} \frac {b \plus{} c}{a}$, $ y \equal{} \frac {c \plus{} a}{b}$, $ z \equal{} \frac {a \plus{} b}{c}$
Therefore, we just need to prove:
$ \sum \sqrt {\frac {b \plus{} c}{a}}\le\sqrt {2(\sum \frac {b \plus{} c}{a} \plus{} 3)} \equal{} \sqrt {2(a \plus{} b \plus{} c)(\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c})} \equal{} \sqrt {[\sum {(b \plus{} c)}](\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c})}$
But this is obviously true due to [b]Cauchy-Schwartz[/b], which ends our proof. :P[/quote]
very nice ! :lol:

[quote="Ji Chen"][quote="hxtung"] $ x,y,z$ be positive real numbers such that  $ x \plus{} y \plus{} z \plus{} 2 \equal{} xyz$. Prove that

$ x \plus{} y \plus{} z \plus{} 6 \geq 2\left(\sqrt {yz} \plus{} \sqrt {zx} \plus{} \sqrt {xy}\right)$.[/quote]By the substitution $ x \equal{} \frac {v \plus{} w}{u},y \equal{} \frac {w \plus{} u}{v},z \equal{} \frac {u \plus{} v}{w}$, the original inequality is equivalent to

$ \sum{\frac {v \plus{} w}{u}} \plus{} 6\geq 2\sum\sqrt {\frac {(w \plus{} u)(u \plus{} v)}{vw}}$

for positive real numbers $ u,v,w$.

[/quote]
by AM-GM
$ 2\sum\sqrt {\frac{(w \plus{} u)(u \plus{} v)}{vw}} \le \sum \left( \frac{w \plus{} u}{w}\plus{}\frac{u \plus{} v}{v} \right)\equal{}\sum \left( \frac{u}{w}\plus{}\frac{u}{v}\plus{}2 \right)\equal{}\sum{\frac {v \plus{} w}{u}} \plus{} 6$

[quote="hxtung"] $ x,y,z$ be positive real numbers such that  $ x \plus{} y \plus{} z \plus{} 2 \equal{} xyz.$ Prove that

$ x \plus{} y \plus{} z \plus{} 6 \geq 2\left(\sqrt {yz} \plus{} \sqrt {zx} \plus{} \sqrt {xy}\right).$[/quote][quote="Ji Chen"]By the substitution $ x \equal{} \frac {v \plus{} w}{u},y \equal{} \frac {w \plus{} u}{v},z \equal{} \frac {u \plus{} v}{w}$, the original inequality is equivalent to

$ \sum{\frac {v \plus{} w}{u}} \plus{} 6\geq 2\sum\sqrt {\frac {(w \plus{} u)(u \plus{} v)}{vw}}$

for positive real numbers $ u,v,w$.[/quote][quote="kuing"]by AM-GM

$ 2\sum\sqrt {\frac {(w \plus{} u)(u \plus{} v)}{vw}} \le \sum \left( \frac {w \plus{} u}{w} \plus{} \frac {u \plus{} v}{v} \right) \equal{} \sum{\frac {v \plus{} w}{u}} \plus{} 6.$[/quote]This problem is introduced as Example 2.2.13 on page 47 in new book :

Ji Chen, Chao-Cheng Ji, [url=http://www.99read.com/product/detail.aspx?proid=499903]Algebraic Inequalities (Mathematical Olympiad Proposition people Lecture)[/url], Shanghai Scientific & Technological Educational Publishing House, 2009, 225p. (ISBN 978-7-5428-4848-2/O·613)
[quote="Ji Chen"]By the way, if $ x,y,z$ be nonnegative real numbers such that  $ 9(x \plus{} y \plus{} z) \plus{} 10 \geq 8xyz$, then

$ x \plus{} y \plus{} z \plus{} 6 \geq 2\left(\sqrt {yz} \plus{} \sqrt {zx} \plus{} \sqrt {xy}\right).$[/quote]This problem is record as Exercise 2.23 on page 87, which is proved by Yi-Chao Ye on pp. 209~210.

See also : http://forum.cnool.net/topic_show.jsp?id=4164763&thesisid=494


floor function
number theory unsolved
number theory

proof that $ \forall p$ odd prime st : $ p\equiv 3 \;(\textbf{mod}\;4\ )$ we have:
$ \prod_{1\leq x < y \leq \frac{p\minus{}1}{2}} (x^2\plus{}y^2) \equiv (\minus{}1)^{\lfloor\frac{p\plus{}1}{8}\rfloor} \;(\textbf{mod}\;p\ )$


Chris' running schedule dictates that he never runs more than two days in a row, nor does he go more than two days in a row without running. In two weeks, what is the positive difference between the fewest number of days and the greatest number of days that he could have run?

[hide]Simple counting shows 10 for the max, 4 for the minimum.
answer is 6. I hope.[/hide]

Whoa, triple post! And I posted right in the middle of your triple post =P.
So I still win.
XD  :rotfl:

yeah, it's 6. I didn't read the MORE THAN two days part, so I got 0. ><

[hide]
The most they can run is (R=run X=dont run)
RRXRRXRRXRRXRR=10 days

The least they can run is
XXRXXRXXRXXRXX=4 days

Subtract to get 6[/hide]

Woah, how did that happen?
moderator, please delete two of the posts.

I took care of the triple post.
[hide]The most days he could run in 14 days is 10, and the least is 4. So the answer is 6[/hide]

[hide]You can get a max of 10 days running, and a minimum of 4 days.  10-4 is 6[/hide]

[hide]if you have chris run the day before the two weeks starts, you can have
NNYNNYNNYNNYNN
N being no running, Y being running.
change all the N's to Y's and all the y's to N's to get 10-4, which is 6.[/hide]


I'm a prime number less than 30 and greater than 20 such that if you divide me by 8, the remainder is 4. What number am i?

28, because the only number divisible by 8 20-30 is 24 so add 4=28 :idea:

28 isn't prime,

4+multiple of 8=answer

I think there's no answer

[hide]
prime numbers between 20 and 30 are:
23
29
none of them work so the answer is "there is no answer"[/hide]

Hehe, the prime tricked someone  :P
Although the original problem (remainder 5 instead of 4) was possible.

no answer... I was tricked, at first, too.


calculus integration trigonometry function

Find the period of f(x)=sin2x+cos3x. It says in one full period of length T of f(x), both sin2x and cos3x must go through an integral number of periods. Can someone explain that to me? Thank you come again.

Hi,

Both going through an integral number of periods means that, for instance, if the sine function's graph goes through one period, and the cosine function's graph goes through 1/5 of an period, then that would not be the period of the graph that incorporates both, the sine and the cosine function. However, if the graph of sin was 4 periods, and then the graph of cosine was 5 periods for that 4 periods of the sine graph, then that is the period of the whole graph (the graph of both).

So, basically, you can think of it as the 'LCM' of the period of the graph of the since function and cosine function. (So, it is basically where both graphs (of sine and cosine, individually) meet up after each of them have completed an integral number of periods themselves.)

Hope that was clear  :maybe:


The line joining the midpoints of the diagonals of a trapezoid has length 3. If the longer side base is 97, then the shorter base is ... (94, 92, 91, 90, or 89)

AHSME 1959, Problem 22

[hide]Draw the diagonals and the line connecting the midpoints. Let the height of the little triangle in the middle by $x$ and the height of the trapezoid be $h$. We then have

$\frac {\frac h2 +x}{x} = \frac {97}3$

$x = \frac {3h}{188}$

Then we have

$\frac {\frac h2 - \frac {3h}{188}}{\frac h2 + \frac {3h}{188}} = \frac k{97}$ where $k$ is the length of the shorter base.

$k=91$
[/hide]

Notice that if the shorter length is 2x, then the length of the median is 2x+3. (This can be seen by looking at the pair of triangles formed by one leg of the trapezoid, the shorter base of the trapezoid, and a diagonal of the trapezoid.) Now, draw a line inside the trapezoid parallel to one of its legs, forming a parallelogram and consider the left out triangle. The base of the triangle will be double the length of the "extra segment" sticking out from the median (of length 3). Thus, the length of the base of the trapezoid is 2x+6=97, so the shorter length is just 91.

There's an easy formula for this. (In fact, proof is easy, too)

$x$: Length of the line joining the midpoints of trapezoid
$a$: Length of the shorter base
$b$: Length of the longer base

$x=\frac{b-a}{2}$

Simple! ;)

Yeah, that's what I was planning to post when I could access Geometer's Sketchpad. So, here it is:
http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=215
The distance between the midpoints of the diagonals is labeled HI. Let the length of this be x. Consider triangles FAB and EBA. Since K and J are the midpoints of AF and BE respectively, KH and IJ are each half the length of AB. Thus, KH+IJ=AB. At the same time, KH+HL=AB and HI=LJ. Then consider triangle BGE. Since L and J are the midpoints of BG and BE respectively, GE=2LJ. Thus, GE=2HI and FE=AB+2HI.
Actually, this might be okay for a proof without words illustration.

Do you need the extra line BG?  You could say the midline was the average of the bases, or something using KH=AB/2 and KI=FE/2.  Just wondering if there was something important about the line.

I used it to prove that the distance between the midpoints of the diagonals is the "extra" length when subtracted from the shorter base. Your method sounds a lot more straightforward though.


I was wondering where is the center of the universe, or what is at the center of the universe. So I want to see you guys' ideas.  :D Where do you think is the center of the unvierse or what do you think is at the center of the universe?

Everything is at the same time the center of univers, do you beleive it?


I have the following problem:

Find the plane perpendicular to the line $0=z=\frac{\partial f}{\partial x}x+\frac{\partial f}{\partial y}y$.

I know the answer, but the way I know is to do it by inspection (dot product = 0) or cross product between two vectors.  How would one find the perp. plane to this line?

Thanks a lot,

Alex

The perpendicular to the line $(x,y,z)=t(a,b,c)+(x_0,y_0,z_0)$ through $(x_0,y_0,z_0)$ is the plane $ax+by+cz=ax_0+by_0+cz_0$. Convert to this form, and the answer is clear.


superior algebra
superior algebra unsolved

A complex C is called split if there are maps Sn:Cn→Cn+1 such that d=dsd. If in addition C is acyclic(exact as a sequence),we say that C is split exact.
My question is"C is a split exact chain complex, then the identity map on C is null homotopic".
Thank for you help!

$ t_{n}= s_{n}-d_{n+2}s_{n+1}s_{n}$ is the desired homotopy.


algebra
polynomial
induction
algebra unsolved

Let $q_1$, $q_2$, ..., $q_n$, ... be a sequence of positive numbers. Define a sequence $f_n$ of polynomials by formulae
\[
f_0\left(x\right) \&= 1 \\
f_1\left(x\right) \&= x \\
\dots \dots \& \dots \\
f_{n+1}\left(x\right) \&= \left(1+q_n\right)xf_n\left(x\right) - q_n f_{n-1}\left(x\right).
\]
Show that the (real) roots of every polynomial in this sequence are in $\left[-1,1\right]$.

[url=http://imageshack.us][img]http://img294.imageshack.us/img294/763/algebravm7.png[/img][/url]
Shot at 2007-07-29

Unless I have made a big error, this is a standard induction problem.

[b][u] Claim:[/u][/b] [i]  For all non-negative integers $ n$, $ |f_{n+1}(x)|\ge |f_{n}(x)|$ for all $ x\le-1$ and $ x\ge 1$. [/i]

To prove this claim, it is a very simple induction.  It is trivial to see that the polynomials $ f_{0}(x)=1$ and $ f_{1}(x)=x$ satisfy the induction hypothesis, and to be cautious, it is not difficult to prove that $ |f_{2}(x)|\ge |f_{1}(x)|$ for any choice of $ q_{1}>0$.  So suppose that it is true for $ n$.  Then we have that for all $ |x|\ge 1$

$ |f_{n+1}(x)|=|(1+q_{n})xf_{n}(x)-q_{n}f_{n-1}(x)|$
$ \ge |(1+q_{n})xf_{n}(x)|-|q_{n}f_{n-1}(x)|=(1+q_{n})|x|\cdot |f_{n}(x)|-q_{n}|f_{n-1}(x)|$
$ \ge (1+q_{n}) |f_{n}(x)|-q_{n}|f_{n-1}(x)|=|f_{n}(x)|+q_{n}(|f_{n}(x)|-|f_{n-1}(x)|)\ge |f_{n}(x)|$.

Thus the induction is complete.  Hence, because $ |f_{0}(x)|\ge 1$ for all $ |x|\ge 1$, we know that $ |f_{n}(x)|\ge 1$ for all $ |x|\ge 1$.  Therefore the only real roots of $ f_{n}$ for any $ n$ must lie in the interval $ (-1,1)$.

Okay, I am a little rusty with real roots.

How do you know that it falls within the (-1,1) interval. The induction part is fine, but I'm still not sure how you are showing that the real roots are less than or equal to 1.

Ok, well the last part is easy.  We don't care what happens to $ f_{n}$ in the interval $ [-1,1]$ since if it has roots there it doesn't matter.  But look, we know that for $ |x|\ge 1$ that either $ f_{n}(x)\le-1$ or $ f_{n}(x)\ge 1$, always.  But if $ f_{n}(x)$ had a root outside the interval $ [-1,1]$, then $ f_{n}(r)=0$ for some $ |r|>1$, but this obviously contradicts the fact that $ |f_{n}(r)|\ge 1$.


function
calculus
derivative
algebra
polynomial
floor function
real analysis

Find the number of the extreme points of the function:
\[
f:\mathbb{R}\to \mathbb{R} \ \ \ \ \ \   f(x)=(x-1)(x-2)^2(x-3)^3....(x-n)^n
\]
$ \textrm{for a given}\ \ \ n\geq 1$

This problem probably requires more knowledge than I have, but as a contribution I notice off the start that $f(x)$ has an infinite number of zeroes at all $\{ x \in \mathbb{N} \mid x \ge 1 \}$

And also that the derivative for any term of the series is $n(x-n)^{n-1}$ which leads us to how to find extrema, or $(x-n)^{n-1} = 0$ Would the answer not be $n$? I'm really still learning harder math stuff, so I'm probably not correct, I'm just giving it some thought.

Adidasty: note the phrase "for a given $n$." This function is a polynomial and does not have infinitely many zeros. In fact, $f$ itself has exactly $n$ zeros: the integers $1,2,\dots,n.$

In fact, $f'$ has $2n-2$ zeros. Proving this takes two steps.

Proof that $f'$ has at least $2n-2$ zeros:

$f$ has a root of multiplicity at least $2$ for each integer $k,\,2\le k\le n.$ The derivative has a zero at such a point. So that's $n-1$ zeros of the derivative.

Then, we have Rolle's Theorem: for a smooth function, between any two adjacent zeros of the function must lie at least one zero of the derivative. This shows that $f'$ has at least one root on each of the intervals $(k,k+1)$ for $1\le k\le n-1.$ So that's another $n-1$ zeros of the derivative.

Proof that $f'$ has at most $2n-2$ zeros:

$f$ is a polynomial of degree $\frac{n(n+1)}2,$ so $f'$ is a polynomial of degree $\frac{n(n+1)}2-1.$ If we consider the points $2,3,\dots,n$, they account for $\frac{(n-1)n}2$ zeros (counting multiplicities). That leaves at most $\frac{n(n+1)}2-1-\frac{(n-1)n}2 = n-1$ other zeros, and we have accounted for all of those.

Ahhhh that makes more sense. Thanks for the clarification, Kent.

Furthermore it must be shown that these points are actually extremes of the function ;)

[quote="ikap"]Furthermore it must be shown that these points are actually extremes of the function Wink[/quote]

Ah, yes. Of the critical points counted, the ones at $x=3,5,\dots$ are not extreme points. Thus there are $\lfloor\frac{n+1}2\rfloor$ critical points that are not extreme points and the actual count of extreme points is $2n-2-\lfloor\frac{n+1}2\rfloor.$

I think that the extreme point are:  $ n-1+\displaystyle[n/2]$
Infact
\[
f'(x)=\frac{n(n+1)}{2}(x-c_1)(x-c_2)\cdots (x-c_{n-1})(x-2)(x-3)^2\cdots(x-n)^ {n-1}
\]
where $ c_k\in (k,k+1) $ and k  is a root of multiplicity k of f.

Nené is right. To find the source of my error, quoting myself:

[quote]Ah, yes. Of the critical points counted, the ones at $x=3,5,\dots$ are not extreme points.[/quote]
That's correct, but next I said:

[quote]Thus there are $\lfloor\frac{n+1}2\rfloor$ critical points that are not extreme points[/quote]
There's the mistake. The count of the odd numbers $k$ such that $3\le k\le n$ is $\lfloor\frac{n-1}2\rfloor.$

So I should have said that the number of extreme points is $2n-2-\lfloor\frac{n-1}2\rfloor.$ This is the same as $n-1+\lfloor\frac{n}2\rfloor,$ agreeing with what Nené said.


($a,b,c > 0$) \[\frac{1+\sqrt{3}}{3\sqrt{3}}(a^{2}+b^{2}+c^{2})\left( \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \geq a+b+c+\sqrt{a^{2}+b^{2}+c^{2}}\] I have a solution and wanna find other solution :) Let's try.

same solution  :blush:
The inequality is equivalent to \[\left(\frac{1}{3}\left( a^{2}+b^{2}+c^{2}\right)\left( \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\right)+\left( \frac{\sqrt{3}}{9}\left( a^{2}+b^{2}+c^{2}\right)\left( \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\right)\]  \[\geq \left( a+b+c \right)+\left( \sqrt{a^{2}+b^{2}+c^{2}}\right)\] First, the left side can be proved by
$\left( a+b+c \right)\left( \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\left( a^{2}+b^{2}+c^{2}\right)$
$\geq 9\left( a^{2}+b^{2}+c^{2}\right) \geq 9\cdot\frac{\left( a+b+c \right)^{2}}{3}$

Which means \[\frac{1}{3}\left( a^{2}+b^{2}+c^{2}\right)\left( \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \geq a+b+c\] Second, the remaining part is equivalent to \[\left( a^{2}+b^{2}+c^{2}\right)\left( \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \geq 9\sqrt{\frac{a^{2}+b^{2}+c^{2}}{3}}\] Because \[\left( a+b+c \right)\left( \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \geq 9 \qquad (1)\] and \[\frac{a^{2}+b^{2}+c^{2}}{a+b+c}\geq \sqrt{\frac{a^{2}+b^{2}+c^{2}}{3}}\]  \[\Longleftrightarrow a^{2}+b^{2}+c^{2}\geq \frac{\left( a+b+c\right)^{2}}{3}\qquad (2)\] This inequality is true by Power--Means inequality.
Multiply (1) and (2), we get the desired ienquality. And last, Adding them then get the result.

[quote="Fitim"][quote="Mathsniper"][quote="Fitim"]Mathsniper: you're albanian? :wink:[/quote]
no  :P
I live in China, Macao. And my nation is China, Macao  :blush:[/quote]

hahahahaaa! OK.
How you knows Albanian TST 2002?[/quote]
I do the Hojoo Lee inequality(i2006.pdf). oh? it's Albanian TST?

yeah you shouldn't expect hard problems :)
this year we had in the BMO TST the following inequality:
$a+b+c=1$
prove that:
$\sum_{cyc}\sqrt{(\frac{1}{a}-1)(\frac{1}{b}-1)}\geq 6$
 :blush:

Albanian Eagle click here:
http://www.mathlinks.ro/Forum/viewtopic.php?t=104617

Fitim I didn't say that I didn't know the solution but that it was extremely easy!...

[quote=sqing][b]ALBANIA - 2015[/b]         Problem 5
Let $x,y>0,x+y=1$ .Prove that
$$\frac{(3x - 1)^2}{x}+\frac{(3y-1)^2}{y}\geq 1.$$
For what values of $x$ and $y$ the equality takes place?[/quote]

ie to show $\frac{9x^2+1}{x}+\frac{9y^2+1}{y}\geq13$ i e $9(x+y)+\frac{1}{x}+\frac{1}{y}\geq13$ ie$ \frac{1}{x}+\frac{1}{y}\geq4$ which is true by titu's lemma
another proof is to apply CS directly

dear sping from where did you get albanian contests

Let $a,b,c$ are positive real numbers.Prove that
$$(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\geq9\sqrt[5]{\frac{a^3+b^3+c^3}{3abc}}$$
[url=https://artofproblemsolving.com/community/c6h1261720p6552069]h[/url] [url=https://math.stackexchange.com/questions/3459709/if-x-y-z0-prove-xyz-left-frac1x-frac1y-frac1z-right?noredirect=1]h[/url]
[quote=sqing][b]ALBANIA - 2015[/b]         Problem 5
Let $x,y>0,x+y=1$ .Prove that
$$\frac{(3x - 1)^2}{x}+\frac{(3y-1)^2}{y}\geq 1.$$
For what values of $x$ and $y$ the equality takes place?[/quote]
Proof of Zhangyunhua:


Given $ a^4\plus{}b^4\plus{}c^4\equal{}3$.Prove that:
$ \frac{(2a^2\minus{}bc)^2}{a^5(b^5\plus{}c^5)}\plus{}\frac{(2b^2\minus{}ca)^2}{b^5(c^5\plus{}a^5)}\plus{}\frac{(2c^2\minus{}ab)^2}{c^5(a^5\plus{}b^5)} \ge\ \frac{3}{2}$
It very nice and I hope you will like it, my friend :)
I am solving two inequalities of you. It very interesting :)

Indeed, I don't think it' trues. I'm sure that your way to created is: using Cauchy Schwarts and the well=known:
$ \sum\ a^5b^5 \leq\ 3$, we'll have to prove:
$ (2\sum\ a^2 \minus{}\sum\ ab)^2 \geq\ 3(a^4\plus{}b^4\plus{}c^4)$
But this ineq is not trues. I think you have just checked it when $ a\equal{}b$; $ c\equal{}0$ and you think that it's trues  :wink:

[quote="nguoivn"]Indeed, I don't think it' trues. I'm sure that your way to created is: using Cauchy Schwarts and the well=known:
$ \sum\ a^5b^5 \leq\ 3$, we'll have to prove:
$ (2\sum\ a^2 \minus{} \sum\ ab)^2 \geq\ 3(a^4 \plus{} b^4 \plus{} c^4)$
But this ineq is not trues. I think you have just checked it when $ a \equal{} b$; $ c \equal{} 0$ and you think that it's trues  :wink:[/quote]
Sorry  :blush: , it must be:
$ \frac{(3a^2\minus{}2bc)^2}{a^5(b^5\plus{}c^5)}\plus{}\frac{(3b^2\minus{}2ca)^2}{b^5(a^5\plus{}c^5)}\plus{}\frac{(3c^2\minus{}2ab)^2}{c^5(a^5\plus{}b^5)} \ge\ \frac{3}{2}$
now,it is true :)

[quote="nguoivn"]Use the way above, it's becomes:
$ [3a^2 \plus{} 3b^2 \plus{} 4c^2](a \minus{} b)^2 \geq\ 0$ (obviously trues)[/quote]
very nice :)


I believe in many of you like math, and maybe computer science very much. It's really powerful in helping of training in problem solving. And what have I to do now in order to achieve same favourness in math? I have tried but it's look like that it's very hard for me.  :blush:

You don't like math, but have the desire to like math...? That's a new one.

What is math to you? If you answer this question, it will increase the (very very slight) chances that someone will be able to hypnotize you and rewire your brain into liking math.

Perhaps there is something about mathematics that appeals to you at a fundamental level, but you're looking at textbooks or whatever and trying to study and your enthusiam dies out a bit. If this is true:

0. Find out what that something is and tell us -- it will help us answer your question.
1. Try looking at the Art of Problem Solving books as an alternative
2. Try out some math competitions. This will introduce the problem-solving aspect, and inject a bit of friendly competition
3. Hang out here a bit. Look at some of the problems that people post, and some of the creative solutions people come up with. It's much more interesting than most classes. But don't spend too much time here until you've tried out problem solving a bit yourself, or you'll either get lost or bored.

I love life and all its parts. It turns out to be that life is made up of math (and chemistry and physics, and more). Math is involved in cooking, playing cards, jumping rope, or painting a landscape. If you drop a coin in one of those "vortex" funnels and look at it spiral on its way down, you are enjoying math. When you look at a starfish or a flower, you can envision idealized forms showing radial symmetry, and that is math. When you notice that a word or number is a palindrome, or when you figure out how many arrangements of four letters you can use in playing scrabble or doing a crossword puzzle, you are also contemplating math. Math is all around for you to enjoy for your own satisfaction. The dislike of math begins when you try to please others; when you let others tell you what math is,what to learn, and how.
Math is not a large collection of rules and formulas that must be memorized. Math is not painting by numbers. When you sit in a class they call math, you are just being indoctrinated in special tactics to solve specific problems, and it is hard to love that. Your class requires that you memorize formats and recipes for problem solving. Mentioning code words is encouraged. Some of the formality may be needed to enhance communications, but I partially agree with Lockhart that it is overemphasized. What's worse, "the one right way" to solve each problem is so zealously preached, that you often see people trying to apply some complicated procedure to a simple problem that could be solved faster invoking a lot less definitions and theorems.

there's nothing you do to get yourself to like math.

it's like putting emotion into music. you can't listen to someone's advice, follow it, and then suddenly have soul flowing out of your fingertips.

and it's the same thing with math. it comes. if you want to like math, it will come with time. in the meanwhile, do math problems, open your mind to the intricacies of beautiful math problems, and it'll hit you like jesus.

If you want to like math, then you probably do like math, it's just not the math that you wanted to like. In other words, you don't think you like math enough, but you actually like math (in an area different than you thought). See if you enjoy math that is more of the area that you enjoy.

It really depends on what you mean when you say math. Do you mean that hopelessly boring mess of formulas and processes and vocabulary and symbols that are forced upon you during the so-called math class? Or are you referring to that wonderful, awe inspiring, amazing [i]thing[/i] that is everywhere, all around you? That thing that has caused uncountable numbers of people to study it and learn about it and learn [i]from[/i] it?

If you mean the first one I mentioned, don't bother with it. That is not math.

Mathematics is the very thing that shapes the world. It's the way seashells have spirals, and the paths that planets take, and everything you can imagine.

When you were a very small child, you probably knew what math was. As a young child, have you ever been to the ocean? Did you examine shells and their beautiful spirals, and the symmetry of starfish, and the pretty way the light bounced off waves? Have you ever watched an eclipse, or seen a shooting star, or watched the moon and other planets, whether with your naked eye or through a telescope or binoculars? Most importantly, did you think it was simply beautiful?

If you did, you loved mathematics. And maybe you forgot that breath-taking feeling of its beauty, but it's still in you. It's inside every one of us human beings.

Some people never see and appreciate mathematics. They don't even know what it truly is. And when most of them reach a certain age, they never will.

It's like music(which is, in essence, mathematics, by the way). There are millions of people, like myself, who will never appreciate music for the sound and thrills it may bring. You can't force yourself to love it with the same fervor Beethoven, or Mozart did, any more than you can force yourself to love mathematics with the same fervor as Archimedes or Isaac Newton.

And so, the only advice I have for you, is to remember any wonderment you experienced as a child, and see if you still have the ability to see mathematics around you.

[quote="isabella2296"]It really depends on what you mean when you say math. Do you mean that hopelessly boring mess of formulas and processes and vocabulary and symbols that are forced upon you during the so-called math class? Or are you referring to that wonderful, awe inspiring, amazing [i]thing[/i] that is everywhere, all around you? That thing that has caused uncountable numbers of people to study it and learn about it and learn [i]from[/i] it?

[/quote]

where have i heard that one before....

if you've seen olympiad math, and you still dont like math, then i think u'll probably not like math any time soon, cause olympiad math is as good as it gets

so you have 2 options

1. find something else that you're good at and you like, if you can
2. suck it up and keep practicing

I suggest you read the book Math Charmers by Alfred S. Posamentier.

School "math" is very repetitive. In fact, I don't think that I can even call it math. It really should be called "work."


sorry if i've already posted this in the intermediate section, but we would also like it if BRAVE middle schoolers would like to take the contest :D , and i mean BRAVE as in the questions will be tough for middleschoolers but we want more middleschoolers signing up! :) , click on the link on my signature!


graph theory
combinatorics unsolved
combinatorics

In the unit squre For the given natural number $n \geq 2$ find the smallest number $k$ that from each set of $k$ unit squares of the $n$x$n$ chessboard one can achoose a subset such that the number of the unit squares contained in this subset an lying in a row or column of the chessboard is even

It's easy to show that $k>2n-1$. Consider the $2n-1$ squares situated on the first row and the first column. It's easy to see that no subset satisfies the condition we want.

Let's try to show that $k=2n$ suffices.

Consider a bipartite graph with $n$ vertices in each class. The vertices in one class represent rows of the board, while the vertices in the other class represent columns of the board. Two vertices are joined iff there is a point at the intersection of the corresponding row and column (so the $2n$ squares we select play the role of edges in our graph). Our problem can be restated thus: given a bipartite graph with $n$ vertices in each class and $2n$ edges, show that it has at least an Eulerian subgraph. Actually, this will follow if we show that it has at least a cycle. This, however, is true because we have $2n$ vertices and $>2n-1$ edges, so our graph can't be a forest (a graph having trees for its connected components).

From any graph which has a cycle we can extract an elementary cycle, and in this cycle, as a subgraph, all vertices have degree $2$, which is even.

it is from poland 1999 a nice and easy problem!

I can't understand the meaning of the problem.Can somebody help me? :(


superior algebra
superior algebra solved

Sorry if this isnt in the right place but im looking for the proof that says 2 + 2 = 5 I saw it once but cant seem to find it, any help?

One proof I've seen (through my all-seeing eye) is the following:

$2+2=5$
[i]For extremely large values of 2.[/i]


When you make a poll, submit, and then edit your post, your poll gets deleted.  It's really annoying, and I don't see why it should get deleted.  Is there a reason why, or is it a bug/glitch?

Yeah, I get this same exact problem. Great you brought it up. I'm wondering the same thing.....

Something else related: When you try to remake your poll after it is deleted, all the choices/options get doubled up. :(


geometry
circumcircle
incenter
geometry solved

In the equilateral $\bigtriangleup ABC$, $D$ is a point on side $BC$.  $O_1$ and $I_1$ are the circumcenter and incenter of $\bigtriangleup ABD$ respectively, and $O_2$ and $I_2$ are the circumcenter and incenter of $\bigtriangleup ADC$ respectively.  $O_1I_1$ intersects $O_2I_2$ at $P$.  Find the locus of point $P$ as $D$ moves along $BC$.

[quote="orl"]In $\bigtriangleup ABC$, $D$ is a point on side $BC$.  $O_1$ and $I_1$ are the circumcenter and incenter of $\bigtriangleup ABD$ respectively, and $O_2$ and $I_2$ are the circumcenter and incenter of $\bigtriangleup ADC$ respectively.  $O_1I_1$ intersects $O_2I_2$ at $P$.  Find the locus of point $P$ as $D$ moves along $BC$.[/quote]

My dynamic sketch shows that the locus is an algebraic curve of degree $\geq 3$. Thus, the problem cannot be an olympiad problem. Was it translated correctly?

  Darij

It should be equilateral $\triangle ABC$. Note that now the answer is still an algebraic curve of degree 2.
(the locus is the curve $y^2-\frac{x^2}{3}=1$, where BC=2 is the x-axis with the midpoint as the origin)

As the above post notes, the locus is the curve $y = - \sqrt{\frac{x^2 +3}{3}}$ where $x \in (-1, 1)$ and $A$, $B$ and $C$ lie at $(0, \sqrt{3})$, $(-1,0)$ and $(1,0)$, respectively.. Without loss of generality, assume $BD \geq CD$ and let $\angle ADB = \alpha$.

[b]Claim. [/b] Points $A, B, D, O_2, I_2$ are all concyclic.
[i]Proof.[/i] First, note that $\angle AO_2D = 2 \angle ACD = 120^{\circ}$. On the other hand $\angle A I_2D = 90^{\circ} + \frac{1}{2} \angle ACD = 120^{\circ}$. Hence $A,O_2, I_2, D$ are concyclic.

Next, we show that $O_2$ lies on the circumcircle of $ADC$ centered at $O_1$. Note that $\angle AO_1B = 2\alpha = \angle AO_2C$. It follows that there is a spiral similarity $\triangle AO_1B \sim AO_2C$. This implies that $\triangle AO_1O_2 \sim ABC$, so $O_2O_1 = AO_1 = DO_1 = BO_1$. Hence $A, D, C, O_2$ are concyclic.

Combining both these steps proves the claim. By symmetry, it follows that $A, C, D, O_1, I_1$ are also concyclic. Next we show that $D$ is the center of the circumcircle of $PO_1O_2$. First observe that $PO_1 \parallel AI_2$ since
\[ \angle AO_1P = \angle AO_1I_1 = 180^{\circ} - \angle ADI_1 = 180^{\circ} - \frac{1}{2} \alpha \]
and
\[ \angle I_2 A O_1 = 90 - \angle I_2 D A = \frac{1}{2} \alpha . \]
Thus
\[ \angle O_2 P O_1 = \angle O_2 P O_1 = 180^{\circ} - \angle AI_2P = \angle AI_2O_2 = \angle ADO_2 = 30^{\circ} . \]
Since $DO_1O_2$ is equilateral, that is precisely half $\angle O_2DO_1$, so the claim is proven.

Now $\angle PDO_1 = 180^{\circ} - 2\angle DOP_1$. We can very easily compute
\[ \angle DO_1P = \angle DO_1I_1 = \angle DAI_1 = \frac{1}{2}DAB, \]
so $\angle PDO_1 = 180^{\circ} - \angle DAB$. However, $\angle O_1DB = 90^{\circ} - \angle DAB$, so $\angle BDP = 90^{\circ}$.

This leaves us with two crucial facts:
1. $PD \times BC$
2. $PD = DO_1$.
These are enough for us to determine the precise coordinates of $P$. Let $D = (x,0)$. By the Law of Sines
\[2 DO_1 = \frac{AD}{\sin \angle ABD} = \frac{\sqrt{3 + x^2}}{\frac{\sqrt 3}{2}} \implies PD = \sqrt{\frac{x^2+3}{3}} \]
Clearly $P$ is on the opposite side of $BC$ from $A$, so the conclusion is immediate.


college
Harvard
MIT
Princeton
Stanford
University of Chicago
calculus

Hello,
THe next year I will finish my bachelor in mathematics. I wish to finish my master in the USA. I was thinking of Berkely. WHere else can I go? Where are the best educations and so on?

Well, you might want to polish up your English first. That will help you. And, it is Berkeley, not Berkely. There are tons of great math schools at US. To choose the best one depends on what do you want to specialize at. I don't know the best place for each field though. So, ask around. In general, Harvard, MIT, Princeton, Stanford, UChicago, ... are great schools for math.

[quote="IReallyHateMath"]blahblahblah: Why is that? Is it because of research?[/quote]

In USA a master is more like of an investment to find a job in industry. Within a PhD you are contributing to the teaching and research output of the university and thus get paid some modest salary (more in the case of fellowships). Most PhD programmes have an initial mandatory course phase that corresponds to a master degree.

[quote="blahblahblah"]i'm not an algebraist so you should try to speak to someone who is, but you should know that it is very hard to get funding to do a masters in the US.  A Ph.D is a different matter, though.[/quote]This may not necessarily be the case.  Big state universities (like, for example, the University of Washington, where I used to teach) have lots of master's students, and have lots of need for teaching assistants (to teach calculus to hordes of undergraduates).  So if you are willing to teach, they will pay you to get your Master's degree.

For non-US students, however, this is more difficult, because unless you can convince them that you have a strong command of the English language (strong enough, in particular, to teach), they won't want to put you in the classroom, which may mean that they won't take you on as a TA.

The top private programs (Harvard, MIT, Princeton, etc.) generally do not take students who are only doing a Master's; they only admit Ph.D. students.


Hi! Is anybody will come to Isef 2006 in maths? If you know about something please write.

Im going to ISEF!
in the math category also

I'm going! I'm in the Computer Science category though. Hopefully we can trade project numbers as May approaches, so I can see what computer characters look like. :p

I'm going! But for an actual science project, it is about artificial lungs.


Given sec(x)=-8
with 90o < x < 180o. Find  sin(2x)
step by step please!

Not really calculus and doesn't belong here.

$\sec x =-8 \Leftrightarrow \cos x =-\frac{1}{8}$.  Now,

$\sin 2x = 2 \sin x \cos x = 2 \sqrt{1-\cos^{2}x}\cdot \cos x = 2 \sqrt{ \frac{63}{64}}\left(-\frac{1}{8}\right) =-\frac{ \sqrt{63}}{32}$

We choose the negative answer because for $90^{\circ}< x < 180^{\circ}$ the value of $\sin 2x$ is negative.


Did you hear about it?   Of course, it only serves the purpose of shaking some people's minds but as it is, that might be quite something. By the way, the  Panel on Climate Change of the United Nations, has just dropped conclusions in Paris, I know global warming has been discussed before, they've shed new light we might want to consider

Ursula

i think that global warming is just a bad investigated problem. There are 2 completely different conclusions about the  human´s interference on it. It´s good to remember that thw planet self-change also, like in 18th century, the period where there was a mini-glaciation and the temperature and the climate in general changed a lot  but it has acchieved normality after some decades. The problem is that we don´t have in general non-intentioned  researches about it. Most of them are sponsered by intentioned institutions like governements and non-governement institutions, and believe, institutions like greenpeace or many others institutions with the same principles have dark secrets some times more dark then the governements.
So, i will be with the most serious conclusion taken by non-intentioned researches, and the most convincent until now that i have seen says that human interference theory is idiot, since the oceans are the number one in emission of CO2. I don´t remember exactly the numbers (yes, the paper had numbers!) but the ocean was responsible for more then 80% of the emission, the human was responsible for about 7 or 10 percent and other stuff what lacks. Since is well known about the ocean emission ( only the percent of its emission  wasn´t enlighted until now) and the proportions... i think that this is correct the research which i believe. You may consider also that this deals with a lot of money, what is a not well known  term of Kyoto´s treatise.
This stuff also that men´s is fu..ing up the world as it is showed, has too much drama also. It´s showed like a novel, not like reality.

[quote="Brambila"]i think that global warming is just a bad investigated problem.....
[/quote]

And I think you should go and read at least excerpts of the report of the [url=http://www.ipcc.ch/]PCC[/url] recently announced. It is a UN report, not a government report, neither an NGO report, so your comment doesn't apply unless you are saying the UN has some dark interests behind it. Also it is a very comprehensive report, particularly there are about nine hundred pages dedicated solely to the scientific bases of global warming and why the overwhelming majority of the scientists claim that is caused by humans. Also know that the IPCC reports have been accused before of being [i]too conservative[/i] in for under assesing the global warming problem.
Your other argument about the emission of CO2 by the ocean is simply not an argument against anything -  this also goes to the other argument waved a lot about water vapour being the main source of greenhouse effect on Earth- No one wants to banish greenhouse gases, without the greenhouse effect there would beno life on earth, at least as we know it. What it is claimed is that the human actions are altering the balance very quickly. Oceans and forests absorb greenhouse gases and at the present rate of emission. mainly due to the burning of fossils, they can not uptake it all. Therefore, there is every day more CO2 in the atmosphere than what there should be. For instance, the present concentration of CO2 in the atmosphere with certainty "has not been exceded during the past 420 000 years" and "likely not during the past 20 million years," likely means here with 60-90& chance. The problem is humans have achieved this basicly in three hundred years. [i][b]mainly[/b][/i] in the las one hundred years so the concentration keeps rising quickly.

[quote]
This stuff also that men´s is fu..ing up the world as it is showed, has too much drama also. It´s showed like a novel, not like reality.[/quote]

Where it is shown to you like a novel? read some serious treatise, people supporting this are not idiots and I would dare to say that the majority of the scientific community is convinced of the importance of addressing climate change. Unfortunately, scientists are at the mercy of politicians, which in turn are at the mercy of businessmen when it comes to confront this problems.

Ursula

[quote="tim1234133"][quote="Brambila"] ... [/quote]

I'm only 16 and yet the climate/weather where I have lived has changed noticeably during my lifetime, so there is no doubt to me that there is some sort of climate change going on. There is just about entirely conclusive evidence that rising CO2 causes a rise in temperature; and the levels of CO2 have soared in the last few decades. This rise in carbon dioxide has (as you seem to imply, coincidentally) coincided with the aftermath of the industrial revolution and the population explosion. Can there be much doubt that human interference has done [i]something[/i] in all of this?

All that said, I'm not too alarmist, and I think we'll find ways round the problem without too drastic consequences to society. Though it is important to make real efforts, rather than token ones- e.g. in Britain there is a commonly held view that the aviation industry are the big sinners, despite the fact that the contribute between 2-5% of all greenhouse gas emissions, compared with around 18% from deforestation, 24% from power generation, and 20% from road transport. So I see no great reason to give up flying just yet, perhaps building nuclear/wind power plants would be more helpful  :)

Lastly, oceans [b]absorb[/b] carbon dioxide, not emit it.[/quote]

Damn you must be good noticing that 0.01 degree change in your lifetime.

Well, I don't claim to record changes in average global temperature, but, since the first heatwave in 2003, we have been having summer temperatures that often get into the mid-high thirties (Celsius); when I was younger, that never happened.

The past month here has been over 3 degrees above average for January; in fact we have hardly any frost this winter at all ( [url=http://www.metoffice.gov.uk/climate/uk/]link[/url]) and if you just google things like warmest month on record or warmest year they are all happening over the last ten years.


A function $ f$ has domain $ [0,2]$ and range $ [0,1]$. (The notation $ [a,b]$ denotes $ \{x: a\le x\le b\}$.) What are the domain and range, respectively, of the function $ g$ defined by $ g(x)\equal{}1\minus{}f(x\plus{}1)$?

$ \textbf{(A)}\ [\minus{}1,1],[\minus{}1,0] \qquad
\textbf{(B)}\ [\minus{}1,1],[0,1] \qquad
\textbf{(C)}\ [0,2],[\minus{}1,0] \qquad
\textbf{(D)}\ [1,3],[\minus{}1,0] \qquad
\textbf{(E)}\ [1,3],[0,1]$

[hide]
[b](B)[/b]
This is pretty straightforward. The domain is the values that "fit" into [i]f[/i]:
because you are adding 1 in [i]g[/i], you must subtract 1 in in your final domain.
The range of the values can be found by considering the range of [i]f[/i]. If [i]f[/i](x) is 0, the [i]g[/i](x) is 1, and vice versa.

Thus your answer is [b](B)[/b] [-1,1], [0,1]
[/hide]

[hide]
$ g(x)$ is defined iff $ f(x\plus{}1)$ is defined. Thus the domain is all $ x|x\plus{}1 \in [0,2] \rightarrow x \in [\minus{}1,1]$.
Since $ f(x\plus{}1) \in [0,1]$, $ \minus{}f(x\plus{}1) \in [\minus{}1,0]$. Thus $ g(x)\equal{}1\minus{}f(x\plus{}1) \in [0,1]$ is the range of $ g(x)$.
Thus the answer is $ \boxed{[\minus{}1,1],[0,1] \Rightarrow B}$.
[/hide]


Find all positive integers k for which the number $3^{k}+5^{k}$ is a power an integer with the exponent greater than $1$

[nice TeXing ;); and the title "nice equation" is also not good]


[quote="quykhtn-qa1"]Let $ 0 < a,b,c,d < \frac {1}{2}$.Prove that:
$ \frac {a^4 \plus{} b^4 \plus{} c^4 \plus{} d^4}{abcd} \ge \frac {(1 \minus{} a)^4 \plus{} (1 \minus{} b)^4 \plus{} (1 \minus{} c)^4 \plus{} (1 \minus{} d)^4}{(1 \minus{} a)(1 \minus{} b)(1 \minus{} c)(1 \minus{} d)}$[/quote]
Dear friend,
I am doing a job of collecting so much olympiad inequalities (from MO,TST or Mathematical Programs). In the work, we need to find so much solutions as we can. And this problem is a really interesting one, in the collection I have two proofs for it, and when it is possible I will try to present them. I post here with hoping that you will post your solution, it will help my collection becomes better. Thank you very much.

See also http://www.mathlinks.ro/viewtopic.php?t=280934  :)

Dear Can,
Could you post your nice proofs :lol:

[quote="quykhtn-qa1"]Let $ 0 < a,b,c,d < \frac {1}{2}$.Prove that:
$ \frac {a^4 \plus{} b^4 \plus{} c^4 \plus{} d^4}{abcd} \ge \frac {(1 \minus{} a)^4 \plus{} (1 \minus{} b)^4 \plus{} (1 \minus{} c)^4 \plus{} (1 \minus{} d)^4}{(1 \minus{} a)(1 \minus{} b)(1 \minus{} c)(1 \minus{} d)}$[/quote]
I dedicate the following proof to hxy09. It must be an ugly proof but I think the idea is natural and I hope you can guess it.

First, we claim that for any $ x,y,z,t \in \left(0,\frac{1}{2}\right)$, the inequality holds
\[ \frac{(1\minus{}x)^4\plus{}(1\minus{}y)^4}{(1\minus{}x)(1\minus{}y)} \plus{}\frac{(x^2\minus{}y^2)^2[zt(1\minus{}x)(1\minus{}y)\minus{}xy(1\minus{}z)(1\minus{}t)]}{x^2y^2zt} \le \frac{(1\minus{}x)(1\minus{}y)(x^4\plus{}y^4)}{x^2y^2}.\]
Indeed, since $ z,t \in \left(0,\frac{1}{2}\right),$ we have
\[ \frac{zt(1\minus{}x)(1\minus{}y)\minus{}xy(1\minus{}z)(1\minus{}t)}{zt} \equal{}(1\minus{}x)(1\minus{}y)\minus{}xy\cdot \frac{(1\minus{}z)(1\minus{}t)}{zt} <(1\minus{}x)(1\minus{}y)\minus{}xy\equal{}1\minus{}x\minus{}y,\]
and thus, it suffices to prove that
\[ \frac{(1\minus{}x)(1\minus{}y)(x^4\plus{}y^4)}{x^2y^2}\minus{}\frac{(1\minus{}x)^4\plus{}(1\minus{}y)^4}{(1\minus{}x)(1\minus{}y)} \ge \frac{(x^2\minus{}y^2)^2(1\minus{}x\minus{}y)}{x^2y^2},\]
which is equivalent to (to have easy computations, we note that $ mn(u^2\plus{}v^2)\minus{}uv(m^2\plus{}n^2)\equal{}(um\minus{}vn)(un\minus{}vm)$)
\[ \frac{(x\minus{}y)^2(1\minus{}x\minus{}y)(x\plus{}y\minus{}x^2\minus{}y^2)(x\plus{}y\minus{}2xy)}{x^2y^2(1\minus{}x)(1\minus{}y)} \ge \frac{(x^2\minus{}y^2)^2(1\minus{}x\minus{}y)}{x^2y^2}.\]
This can be simplified to
\[ (x\plus{}y\minus{}x^2\minus{}y^2)(x\plus{}y\minus{}2xy) \ge (x\plus{}y)^2(1\minus{}x)(1\minus{}y).\]
By the Cauchy Schwarz Inequality, we have
\[ (x\plus{}y\minus{}x^2\minus{}y^2)(x\plus{}y\minus{}2xy)\equal{}[x(1\minus{}x)\plus{}y(1\minus{}y)][x(1\minus{}y)\plus{}y(1\minus{}x)] \ge (x\plus{}y)^2(1\minus{}x)(1\minus{}y).\]
So the last inequality is valid and our claim is proved.

Now, turning back to our problem. Without loss of generality, we may assume that $ d \ge c \ge b \ge a.$ Then, applying the above claim by replacing $ (x,y,z,t)$ by $ (a,b,c,d)$ and $ (c,d,a,b),$ we get
\[ \frac{(1\minus{}a)^4\plus{}(1\minus{}b)^4}{(1\minus{}a)(1\minus{}b)(1\minus{}c)(1\minus{}d)} \le \frac{(1\minus{}a)(1\minus{}b)(a^4\plus{}b^4)}{a^2b^2(1\minus{}c)(1\minus{}d)} \minus{}\frac{M(a^2\minus{}b^2)^2}{a^2b^2cd(1\minus{}c)(1\minus{}d)},\]
and
\[ \frac{(1\minus{}c)^4\plus{}(1\minus{}d)^4}{(1\minus{}a)(1\minus{}b)(1\minus{}c)(1\minus{}d)} \le \frac{(1\minus{}c)(1\minus{}d)(c^4\plus{}d^4)}{c^2d^2(1\minus{}a)(1\minus{}b)}\plus{}\frac{M(c^2\minus{}d^2)^2}{abc^2d^2(1\minus{}a)(1\minus{}b)},\]
where $ M \equal{}cd(1\minus{}a)(1\minus{}b)\minus{}ab(1\minus{}c)(1\minus{}d) \ge 0$. From these two inequalities, we can see that the original inequality is deduced from
\[ \frac{cd(1\minus{}a)(1\minus{}b)(a^4\plus{}b^4)}{ab(1\minus{}c)(1\minus{}d)}\plus{}\frac{ab(1\minus{}c)(1\minus{}d)(c^4\plus{}d^4)}{cd(1\minus{}a)(1\minus{}b)} \plus{}\frac{M(c^2\minus{}d^2)^2}{cd(1\minus{}a)(1\minus{}b)} \minus{}\frac{M(a^2\minus{}b^2)^2}{ab(1\minus{}c)(1\minus{}d)} \le a^4\plus{}b^4\plus{}c^4\plus{}d^4.\]
Since
\[ a^4\plus{}b^4\minus{}\frac{cd(1\minus{}a)(1\minus{}b)(a^4\plus{}b^4)}{ab(1\minus{}c)(1\minus{}d)}\equal{}\minus{}\frac{M(a^4\plus{}b^4)}{ab(1\minus{}c)(1\minus{}d)},\]
\[ c^4\plus{}d^4\minus{}\frac{ab(1\minus{}c)(1\minus{}d)(c^4\plus{}d^4)}{cd(1\minus{}a)(1\minus{}b)} \equal{}\frac{M(c^4\plus{}d^4)}{cd(1\minus{}a)(1\minus{}b)},\]
one can see that the above inequality is equivalent to
\[ \frac{M(c^4\plus{}d^4)}{cd(1\minus{}a)(1\minus{}b)} \minus{}\frac{M(c^2\minus{}d^2)^2}{cd(1\minus{}a)(1\minus{}b)} \ge \frac{M(a^4\plus{}b^4)}{ab(1\minus{}c)(1\minus{}d)} \minus{}\frac{M(a^2\minus{}b^2)^2}{ab(1\minus{}c)(1\minus{}d)},\]
i.e.
\[ \frac{2cdM}{(1\minus{}a)(1\minus{}b)} \ge \frac{2abM}{(1\minus{}c)(1\minus{}d)},\]
which is true because $ M \ge 0$ and $ x(1\minus{}x) \ge y(1\minus{}y)$ for all $ \frac{1}{2}>x \ge y>0.$ Our proof is completed.

Dear can_hang2007,please post second proofs.
Thank you very much

Thank you very much for your nice proof :lol:

[quote="quykhtn-qa1"]Let $ 0 < a,b,c,d < \frac {1}{2}$.Prove that:
$ \frac {a^4 \plus{} b^4 \plus{} c^4 \plus{} d^4}{abcd} \ge \frac {(1 \minus{} a)^4 \plus{} (1 \minus{} b)^4 \plus{} (1 \minus{} c)^4 \plus{} (1 \minus{} d)^4}{(1 \minus{} a)(1 \minus{} b)(1 \minus{} c)(1 \minus{} d)}$[/quote]
$ < \equal{} > LHS \minus{} 4 \geq RHS \minus{} 4$
$ < \equal{} > \frac {(a^2 \minus{} b^2)^2 \plus{} (c^2 \minus{} d^2)^2 \plus{} 2(ab \minus{} cd)^2}{abcd} \geq \frac {((1 \minus{} a)^2 \minus{} (1 \minus{} b)^2)^2 \plus{} ((1 \minus{} c)^2 \minus{} (1 \minus{} d)^2)^2 \plus{} 2((1 \minus{} a)(1 \minus{} b) \minus{} (1 \minus{} c)(1 \minus{} d))^2}{(1 \minus{} a)(1 \minus{} b)(1 \minus{} c)(1 \minus{} d)}$
And now, we only need prove
$ (a^2 \minus{} b^2)^2(1 \minus{} a)(1 \minus{} b)(1 \minus{} c)(1 \minus{} d) \geq ((1 \minus{} a)^2 \minus{} (1 \minus{} b)^2)^2abcd (1)$
$ (ab \minus{} cd)^2(1 \minus{} a)(1 \minus{} b)(1 \minus{} c)(1 \minus{} d) \geq ((1 \minus{} a)(1 \minus{} b) \minus{} (1 \minus{} c)(1 \minus{} d))^2abcd (2)$
We have
$ (1 \minus{} c)(1 \minus{} d) \geq cd$ (because $ 0 < c, d < 1/2$ )
$ \equal{} > LHS(1) \minus{} RHS(1) \geq (a \minus{} b)^2((a \plus{} b)^2(1 \minus{} a)(1 \minus{} b) \minus{} (2 \minus{} a \minus{} b)^2ab)cd \equal{} (a \minus{} b)^4(1 \minus{} a \minus{} b)cd \geq 0$
and assume $ a \geq b \geq c \geq d$
$ \equal{} > a(1 \minus{} a) \geq b(1 \minus{} b) \geq c(1 \minus{} c) \geq d(1 \minus{} d)$ (because $ 0 < a,b,c,d < 1/2$ )
$ ab \geq cd, (1 \minus{} c)(1 \minus{} d) \geq (1 \minus{} a)(1 \minus{} b)$
$ \equal{} > LHS(2) \minus{} RHS(2) \equal{} (ab(1 \minus{} a)(1 \minus{} b) \minus{} cd(1 \minus{} c)(1 \minus{} d))(ab(1 \minus{} c)(1 \minus{} d) \minus{} cd(1 \minus{} a)(1 \minus{} b)) \geq 0$
So, we are done! :)


Let a be an algebraic number (above Q) of degree n,
and p(x) some rational polynomial. Is it true
that p(a) must be an algebraic number of degree
at most n?

I have proved it too but I want to see your
proof.


Para quien esta leyendo este post: La discusion es acerca de temarios en el area indicada por el titulo. Se pretende que si usted pone algo aqui, que sea referente al temario, discutiendolo (criticas completamente aceptadas), haciendo preguntas sobre el o proponiendo otros posibles resultados dentro del tema. Por favor, solo por mantener el orden, si su post no cae en esta categoria, pues inicie un nuevo topic para ello. Esta peticion es unicamente para conservar el orden para quienes estamos interesados en seguirle la pista al temario!!

Solo cuestion de notacion, las secciones, subsecciones y subsubsecciones

estan indicadas asi:
- Seccion
* Subseccion
** Subsubseccion

ACLARACION: Por Superbombas me refiero a esa clase de resultados que parecen profundamente poderosos y que pueden ser utiles para asecinar un problema, pero que de alguna manera el usarlos equivale a usar un canion para matar una mosca. Es decir, casi que un ultimo recurso. Cosas sumamente utiles, pero no realmente vitales.

[b]NOTA: Aqui tambien estoy un poco falto de ideas de momento![/b]

[b]Superbombas[/b]
- Teorema de Hoehn
- Polinomios de Chebishev
- Desigualdad de Erdos Mordell

[quote="Pascual2005"]que es el [b]Teorema de Hoehn[/b]?[/quote]

Tengo mucho suenio y mucha pereza de escribir, asi que lo que te voy a decir es: Anda a [url=http://mathworld.wolfram.com/HoehnsTheorem.html]esta pagina[/url] de [url=http://mathworld.wolfram.com/]MathWorld[/url] donde dan una buena explicacion al respecto. Ahora si, como dice el presidente de Costa Rica despues de cada cadena televisiva, [i]Muchas Gracias, y... Buenas Noches![/i]

[quote="Leonardo"]Jeje, solo un detalle, como tu siempre eres tan exigente con la cuestion de la ortografia, no es de menos q tu tengas buena ortografia :P, en especial para escribir el nombre de grandes matematicos

[quote="djimenez"] Superbombas
- Teorema de Hoehn
- Polinomios de Chebishev
- Desigualdad de [b]Erdosh Morden[/b][/quote]

Se escribe Erdös Mordell.
La otra cosa que queria preguntar es, en que problema has usado polinomios de Chebyshev y para q?...buehj, solo curiosidad. Gracias[/quote]

Upss!!, perdon, bueno, los simbolos especiales me da pereza, sin embargo las otras faltas, estaban todavia por alli en el archivo donde lo tengo.

Recuerdo por lo menos tres ejercicios de IMOs y al menos uno de Putnam que yo he resuelto con Polinomios de Chebishev. Basicamente son esos que te dicen que proves alguna igualdad o desigualdad con valores (usualmente producto o suma de valores) de expresiones trigonometricas. Los ejercicios de las IMOs creo que son de los 60s. Tambien para resolver ecuaciones!

Otro teorema bueno y bonito que deberiamos tener aqui es el Teorema de Ramsey.


I need to proof that
$F_{n}=\sum_{k=1}^{n} {n-k\choose{k-1}}$ assuming that ${p\choose l}=0$ if $p<l$ where $F$ is Fibonacci sequence {1,1,2,3,5,8,...} and $F_{n}$ is the nth term of sequence.

Thanks for help.

Induction.
$\sum_{k=1}^n\binom{n-k}{k-1}+ \sum_{j=0}^n\binom{(n+1)-(j+1)}{(j+1)-1}= \sum_{k=0}^n\left(\binom{n-k}{k-1}+\binom{n-k}{k}\right)$
$=\sum_{k=0}^n\binom{n+1-k}{k}=\sum_{j=1}^{n+2}\binom{n+2-j}{j-1}$

Actually, the sum indices are irrelevant- we're using the convention that everything outside the standard range is zero, and some of those are included in the sum anyway.


[hide]Let the difference of her age then , and her age now (12) be x.
Then, $ 12 \plus{} x \equal{} 2(4 \plus{} x)$ or $ 4 \equal{} x$. Therefore, she will be 16.[/hide]

[quote="binomial_4eva"]emily is $ 4$ years old and her sister amy is 3 times older then she is. how old will amy be when she is twice as old as emily ?[/quote]
[hide="different way"]
Amy is $ 12$ years old.
Amy is $ 8$ years older than emily.
Let $ z$ = amy's age when she is twice as old as emily
$ z \equal{} 2(z \minus{} 8)$
$ z \equal{} 2z \minus{} 16$
$ z \equal{} 16$
Amy is $ 16$ years old when she is twice as old as emily.
[/hide]


function
integration
calculus
geometry
induction
real analysis
real analysis unsolved

Take a looped rope of length $1$. If you cut the rope independently and randomly in $n$ places, one will get $n$ smaller pieces of rope.
Determine the avergae size of the biggest piece as function of $n$.

this is a beautiful problem.
a formula for the average is:
\[ \frac{1}{\text{vol}(D_n)}\int_{D_n}f(y_1,\dots, y_{n-1})dy_1\dots dy_{n-1} \]
where $D_n=\{(y_1,\dots, y_{n-1})\in R^n: y_i>0, y_1+\dots+y_{n-1}<1\}$ and $f(y)=\max\{y_1,\dots, y_{n-1}, 1-(y_1+\dots+y_{n-1})\}$.

The question is wether one should compute this elementary integral or if there are smarter ways around it.

[quote="cimabue"]a formula for the average is:
\[ \frac{1}{\text{vol}(D_n)}\int_{D_n}f(y_1,\dots, y_{n-1})dy_1\dots dy_{n-1}  \]
where $D_n=\{(y_1,\dots, y_{n-1})\in R^n: y_i>0, y_1+\dots+y_{n-1}<1\}$ and $f(y)=\max\{y_1,\dots, y_{n-1}, 1-(y_1+\dots+y_{n-1})\}$.

The question is wether one should compute this elementary integral or if there are smarter ways around it.[/quote]

$D_n=\{(y_1,\dots, y_{n-1})\in R^{n-1}: y_i>0, y_1+\dots+y_{n-1}<1\}$ and $f(y)=\max\{y_1,\dots, y_{n-1}, 1-(y_1+\dots+y_{n-1})\}$ more specific...
But it is not easier to compute it like a surface integral considering in $R^n$ the hyper-plane $y_1+\dots+y_{n}=1$ as:
\[ \frac{1}{\text{area}(H_n)}\int_{\delta H_n}\max\{y_1,\dots, y_n\}dy_1\dots dy_n  \]
where $H_n=\{(y_1,\dots, y_n)\in R^n: y_i \geq 0, y_1+\dots+y_n \leq 1\}$

Do you think it is easy to compute such integrals?!? Then what is your answer? I am a little confused about...
A smarter way to compute might be using discrete calculus - splitting the rope in $kn$ allowed cutting points - which leads to even more complicated computations - then tend $k$ to $+\infty$ but in fact is the same... elementary approach of those integrals...

the two integrals are the same, and up to a factor they are the same as
\[ \int_{x_i\geq 0, \; x_1+\dots, x_n\leq 1}\max\{x_1,\dots, x_n\}dvol \]
I tried to compute this integral by induction: let $F_n(a,b)=\int_{x_1+\dots+x_n\leq b}\max\{a, x_1,\dots, x_n\}$ and one can easily find reccurence formulas between $F_n$ and $F_{n-1}$ but it's not clear to me how to sort it out. I think I tried to compute this integral before but i couldnt. perhaps there is a clever way around it. i dont know about the discrete method.


WELCOME TO AOPS IS KOTW!

And I am the new millennium for the Art of Problem Solving. Now for those of you who don't know me, I am The King of the World, your, your new hero, your party host and most importantly the most charismatic showman to ever enter your living rooms via a computer screen.
Now when you think of a new millennium, you think of an even so gigantic that it changes the course of history. You think of a dawning of a new era. In this case, the dawning of a new era in the AopS.
Thank you, thank you.

And a new era is what this once proud and profitable forum sorely needs. The King of the World has come to save the AopS.

You people have been lead to believe that mediocrity is excellence. Uh-uh. KOTW IS EXCELLENCE. And now for the first time in AopS history, you have a man who can entertain you, you have a man who is good enough for you, you have a man who can make you jump up off your chairs, raise your filthy fat little hands in the air and scream, "Go King Go! Go King Go!".

The new millennium has arrived to the AopS and now that the KOTW problem is here, this forum from the front office idiot to all the amateurs in the different forums will NEVER. EVVVVER. BE THE SAME. AGAIN!

-------

Ok that was just my bright idea of a grand debut. Most of it was copied from this: http://youtube.com/watch?v=pM0O7NyDGd4

Well, to be honest I am surely a misfit in this forum, being that I am the last person you would classify as a nerd, but I only came to this place to see what's so great about it. I mean if it has some mystical powers that can magically force my friends to get up and park their asses in front of their computers and actually go online (I've been trying to accomplish this for 8 years now) it must be great.

That apart, I have no business here. Maybe a funny comment here and there, that's it.

Note to mods: I request you not to move this thread or delete it, atleast for some time please :)

are u an indian???
where do ya live????
come on speak up!!!!!!!!!!!11

Ah this is Vikku isnt it? What with Jericho etc. it isnt hard to guess... !!

My psychic skills tell me that the above poster is an "emerging mind" and a year younger than me.

The poster above that is uh- canadian moose-piss. Subtract 48 from it too.

nice try!!!!
King of the World....
i 've become resistant to that....
try something else(just joking)

ok!!!!Welcome and what is ur name????

[quote="Cyber-Shreyas"]His name is probably V.(Culsec+BigHead) Vignesh. Ofcourse the Jericho and psychic skills confirm that. :D Anyways you are the last person I expected here. So how did ya get this link.[/quote]

Weren't you the one who told me that you had scrapped all ur friends in orkut this link shreyas?  :huh:   :D

[quote="The King of the World"]I am from India, yes.

And I totally forgot to tell this to one of my friends: Ranga don't cry!!!![/quote]

 :welcome:  :welcome:

Hi da vikku! Continue posting something and u urself will feel good about this forum(in particular).

thank u Rakesh for promoting the Forum :D
hi vikku(if that's what ur name is).. :D
and i am ur MOD :P

VIKKU Please read the project document for chemistry project viva very well ok :D  :D  :D

Hey Vikku!!! Long time no see.. Lol... Nice to see you here... now like Rohit said, Sandu should be here too. And then the fun begins...

@Rockesh: Chem Best Project hopeful I am thinking :)

All the best for your practs btw

Probably we won't get the best project award but surely we will get the 'SIMPLEST PROJECT' award because all our materials are Lab Reagents and ofcourse 'H2O' the simplest chemical.

I think they are going to award that to one of sundari miss class students. But our project is very simple and so no hard questions in viva.

Ah well Mr.Rock, you can never be sure about that... My project involved pretty much the same.. How much easier can u do a project than testing hardness in water? ;)

U would have atleast wanted 'EDTA' and ofcourse Buffers and stuff like that whereas we just needed 'MnSO4' and the rest were 'TABLE REAGENTS'. :wink:  :wink:

U should know by now that 'MnSO4' is readily available in school and very very cheap.

Yep! Geeeez thanks..... I never knew that you would become so hmmm..... i dunno!!!!


algebra polynomial function calculus integration derivative inequalities

Assume $ f$ is an entire function (analytic in $ \mathbb{C}$). Assume further that $ |f(z)|\leq |z|^{\frac52}$
if $ |z|\geq R$, where $ R$ is a fixed real. Show that $ f(z)$ is a polynomial of at most degree 2.

A hint: Use the Cauchy integral formula to calculate the third derivative of $ f$, integrating around very large circles.

Jmerry gave a great hint, here is more if you need it.

[hide="More"]
If we want to show $ f(z)$ is a polynomial on $ \mathbb{C}$ of degree at most two. We can show that $ f(z)$ is an entire function [b]and[/b] $ f'''(z)=0$ (note: entire functions are infinitely differentiable). Now to show that $ f'''(z)=0$ we basically mimic the proof of [url=http://en.wikipedia.org/wiki/Liouville%27s_theorem_%28complex_analysis%29]Liouville's Theorem[/url], that is what Jmerry ment by "integrating over large circles".
[/hide]

Amusingly, is does not matter which way the inequality goes.  :)

Assume $ f$ is an entire function (analytic in $ \mathbb{C}$). Assume further that $ |f(z)|\geq |z|^{\frac52}$ if $ |z|\geq R$, where $ R$ is a fixed real. Show that $ f$ is a polynomial.

Sorry, if its a little late. But here is a nice easy result that proves this immediately.

[b]Theorem:[/b] Let $ f(z)$ be an entire function such that $ |f(z)|\leq K$ on $ |z|\equal{}R$. If $ a_{k}$ are coefficients in the power series $ f(z) \equal{}\sum_{n\equal{}0}^{\infty}a_{n}z^{n}$ then $ |a_{k}|\leq\frac{K}{R^{k}}$.

[b]Proof:[/b] We know that $ a_{n}\equal{}\frac{f^{(n)}(0)}{n!}\equal{}\frac{1}{2\pi i}\oint_{|z|\equal{}R}\frac{f(z)}{z^{n\plus{}1}}dz$. And we also know that on the circle $ |z|\equal{}R$ we have $ |z|\leq K$ so $ \left|\frac{f(z)}{z^{n\plus{}1}}\right|\leq\frac{K}{R^{n\plus{}1}}$ on $ |z|\equal{}R$. So that means $ |a_{k}| \equal{}\left|\frac{1}{2\pi i}\oint_{|z|\equal{}R}\frac{f(z)}{z^{n\plus{}1}}dz\right|\leq\frac{1}{2\pi i}\cdot\frac{2\pi R}{1}\cdot\frac{K}{R^{n\plus{}1}}\equal{}\frac{K}{R^{n}}$.

[b]Corollary:[/b] Let $ f(z)$ be an entire function so that $ |f(z)|\leq A\plus{}B|z|^{r}$. If $ a_{k}$ are the coefficients in the power series expansion then $ a_{k}\equal{}0$ for $ k>r$.

[b]Proof:[/b] Working on the circle $ |z|\equal{}R$ we find $ |a_{k}|\leq\frac{A\plus{}B\cdot R^{r}}{R^{k}}$ which goes to $ 0$ as $ R\to\infty$.

So in your problem, $ |f(z)|\leq |z|^{5/2}$ which means $ f(z) \equal{} a_{0}\plus{}a_{1}z\plus{}a_{2}z^{2}$ since $ a_{k}\equal{} 0$ for $ k\equal{}3,4,...,\infty,\infty\plus{}1,...\aleph_{1},...,\aleph_{\infty}$.
Because $ 3 >\frac{5}{2}$.


I just started a homework assignment on latex and I get an error message that I don't know how to fix.

[code]

\documentclass[oneside,letterpaper,10pt]{book}
\usepackage{amsmath,amssymb}
\newdimen\jot \jot=10pt
\usepackage[pdftex]{graphicx}
\usepackage{asymptote}
\newcommand{\eqnnumber}{\stepcounter{equation}\tag{\arabic{equation}}}
\newcommand*{\bx}[2]{\rlap{\boxed{#1#2}} \phantom{\hskip\fboxrule\hskip\fboxsep #1}&\phantom{#2}}
\newenvironment{mcen}%
{\par\centering}{\par}
\pagestyle{headings}
\allowdisplaybreaks
\begin{document}

\title{MAT-LCX(E) Assignment 2} \author{Eric Grosse} \maketitle

Note: I will often use the formula $nPr = r! nCr$

Proof:
\begin{align*}
nPr = \frac{n!}{(n-r)!}
\frac{nPr}{r!} = \frac{n!}{r!(n-r)!} = nCr
\end{align*}
\\[5pt]
In addition, I usually express $nCr$ as ${n\choose r}$. With that in mind, here are my solutions to the assignment. \\[10pt]
1) In how many ways can 3 different cars be places in 5 garages, one car per garage?

Since there are 5 distinct cars and 3 distinct garages, the order in which we place the cars in garages matters. Therefore, the answer must be ``5 permute 3'' or $3! {5 \choose 3}$.

\end{document}

[/code]

Any ideas?

[quote="Prime factorization"]Any ideas?[/quote]Possibly, if only I knew what the error message is - you forgot to post it  :)

Sorry about that. It says

"package ams warning: foreign command \atopwithdelims;
(amsmath)                  \frac or \genfrac should be used instead
(amsmath)                  on input line 24"

Ah, that's a warning not an error, so not so important and won't stop the file being compiled. The problem is that you are using \choose which is an old TeX command and has been replaced by \binom as in \binom{n}{r} $ \binom{n}{r}$. The reference to \genfrac is purely because \binom is defined in amsmath in terms of \genfrac.

Ok but I'm accustomed to using {n \choose r} so I'll just stick with that until my next document. Thanks as always!


Solve for m such that $ m^{17}$ = 2 (mod 41989), Given that 41989 = 211 x 199. Has to use Chinese Remainder Thereom and Fermat Little Thereom.

AN De

hello, i have derived $ m\equal{}38775$.
Sonnhard.

Can show me or give me some guide how to do it?
An De


Find the largest positive integer $n$ such that the following equations have integer solutions in $x, y_{1}, y_{2}, ... , y_{n}$ :
$(x+1)^{2}+y_{1}^{2}= (x+2)^{2}+y_{2}^{2}= ... = (x+n)^{2}+y_{n}^{2}.$

$n\leq 3$ Suppose that there is solution for n=4. Let $(x+1)^{2}+y_{1}^{2}=....t$ Looking modulo 4 we see that t is odd. Now among numbers x+1,x+2, x+3 and x+4 there are two even exactly one of which is divisible by 4. From here using the fact that $z^{2}\equiv 1(mod 8)$ for odd z we get a contradiction.
For n=3 there must be some easy example but i can't see it now.


geometry
inradius
circumcircle
inequalities
trigonometry
geometric transformation
reflection

Let $ABC$ be a triangle with the inradius $r$ and $O$ the center of the circumcircle.Let $C_{1,a}$ , with radius $r_{b,c}$, the circle tangent to the circumcircle and to the sides $AB$ and $AC$, and $C_{2,a}$ , with radius $r_{a}$, the circumcircle of triangle $BOC$.Prove the following ineguality:
\[2(\frac{1}{r_{b,c}}+\frac{1}{r_{c,a}}+\frac{1}{r_{a,b}})-\frac{3}{r}\geq\frac{1}{r_{a}}+\frac{1}{r_{b}}+\frac{1}{r_{c}}\]

I think there are some problems in your problem here we need triangle is acute and the inequality must be
\[2(\frac{1}{r_{bc}}+\frac{1}{r_{ca}}+\frac{1}{r_{ab}})-\frac{1}{r}\geq\frac{1}{r_{a}}+\frac{1}{r_{b}}+\frac{1}{r_{c}}\]
Indeed
$r_{bc}=\frac{S}{(s-a)\cos^{2}\frac{A}{2}}=\frac{bcr}{(s-a)^{2}},r_{a}=\frac{a}{2\sin 2A}=\frac{R}{2\cos A}$ thus we need to prove $2\sum\frac{(s-a)^{2}}{bcr}-\frac{1}{r}\ge 2\sum\frac{\cos A}{R}\Leftrightarrow 2\sum\frac{(s-a)^{2}}{bc}\ge 1+\frac{r}{R}8\prod\sin\frac{A}{2}$
But we easily seen
$2\sum\frac{(s-a)}{bc}\ge\frac{3}{2}$ and $\frac{r}{R}8\prod\sin\frac{A}{2}\le\frac{1}{2}$
we are done.

Yes :D.Sorry i forgot to say that the triangle $ABC$ is acute .
But in my opinion, i think that $r_{b,c}=\frac{r}{cos^{2}\frac{A}{2}}$.
Denote M,N the tangent points of $C_{1,a}$ to the sides $AB$ , respectivelly $AC$.So , from Casey applied to $A,B,C,C_{1,a}$ (where $A,B,C$ are degenerated circles) we find :
\[AB\cdot CN+AC\cdot BM=BC\cdot AM\]
so
\[c(b-AM)+b(c-AM)=a\cdot AM\] where we obtain $AM=\frac{bc}{p}$
If we denote $r$ the inradius we have $\frac{r}{r_{1}}=\frac{p(p-a)}{bc}=cos^{2}\frac{A}{2}$
Back to our problem we have :
\[2\sum \frac{1}{r_{bc}}-\frac{3}{r}=\sum( \frac{2}{r_{bc}}-\frac{1}{r})=\sum \frac{2cos^{2}\frac{A}{2}-1}{r}=\sum\frac{cosA}{r}\geq \sum \frac{2cosA}{R}=\sum\frac{1}{r_{a}}\] where i used the ineguality $R\geq 2r$

maybe you are wrong it must be here $r_{a}$ is exradii $r_{bc}=\frac{r_{a}}{\cos^{2}\frac{A}{2}}$ you mean $C_{1a}$ , with radius $r_{bc}$ is the circle tangent to the circumcircle and to the sides $AB$ and $AC$?

I'll refrase:
$r_{bc}$ is the radius of the circle $C_{1a}$ wich is tangent to the circumcircle and the sides $AB$ and $AC$ where i have just proven that $r_{bc}=\frac{r}{cos^{2}\frac{A}{2}}$
$r_{a}$ is the radius of the circumcircle $C_{2a}$ of the triangle $BOC$, where you yourself just said that $r_{a}=\frac{R}{2cosA}$
Sorry if my notations seem odd , because you're right..because $r_{a}$ is the notation for exradii..so i shouldn't have used this notation

Sorry but I think you must check your result again I still see (on computer)
\[r_{bc}=\frac{r_{a}}{\cos^{2}\frac{A}{2}}\]
 with $r_{a}$ is $A$-exradii , It isn't $r_{bc}=\frac{r}{cos^{2}\frac{A}{2}}$ as you said (you mean $r$ is inradii ?) :?:

Thus I believe you are wrong, it must be
\[r_{bc}=\frac{r_{a}}{\cos^{2}\frac{A}{2}}\]
with $r_{a}$ is $A$-exradii. Can you anybody check for us? :|

Maybe i am wrong :maybe: , i have seen my proof that $r_{bc}=\frac{r}{cos^{2}\frac{A}{2}}$ on number 10/2006 in Gazeta Matematica (series B).Hopefully someone will check this for us  :blush:

with inversion one can prove that $r_{bc}=r_{a}\frac{bc}{p^{2}}$ where $r_{a}$ is the radious of the ex-circle respect to $A$ and $p$ is the half of the perimeter...

Maybe you mean $r_{bc}=\frac{r_{a}bc}{p(p-a)}$BaBaK Ghalebi :?:

no...
invers the shape with center $A$ and power $AB.AC$ and then reflect it respect to the bisector of $A$ then one can easily see that:
$i)AB \mapsto AC$
$ii)I \mapsto I_{a}$ (because $\triangle AIB \sim AI_{a}C \Rightarrow AI.AI_{a}=AB.AC$)
$ii)$the ex-circle of $A$ $\mapsto C_{1a}$
from the last fact we get that $\frac{r_{a}}{r_{bc}}=\frac{AP^{2}}{k}$ where the ex-circle is tangent to $AC$ at point $P$ and $r_{a}$ is the radius of the ex-circle,so $AP=p$
$\Rightarrow r_{bc}=r_{a}.\frac{AB.AC}{p^{2}}$

I don't understand your solution but my solution is so simple(in my figure)
$r_{bc}=KO_{a}=KA\tan\frac{A}{2}=\frac{I_{a}A}{\cos\frac{A}{2}}\tan\frac{A}{2}=\frac{I_{a}A\sin\frac{A}{2}}{\cos^{2}\frac{A}{2}}=\frac{r_{a}}{\cos^{2}\frac{A}{2}}=\frac{r_{a}bc}{p(p-a)}$
Note that ${\cos^{2}\frac{A}{2}=\frac{p(p-a)}{bc}},r_{a}=\frac{S}{p-a}$ that's all.

oh,here is the reason why you and Vrajitoarea Andrei had trobles...
you see there is two circles $C_{1a}$ one inside the circumcircle and one outisde...
in my soluion I ment the inside one,and you ment the outside one...
(the circle that you ment is the inverse of the incircle in my solution...)

ok I understand, but I think there is one more problem in the problem of Vrajitoarea Andrei he must show clearly what is external tangent or internal tangent, ok with internal tangent I thin $r_{bc}=\frac{r}{\cos^{2}\frac{A}{2}}$ is true.  :)

You are wright :).Srry i was wrong  :( .Indeed i forgot to say that the tangent is internal.Looking at the bright side , we  also have an ineguality with the external tangent also  :rotfl: , wich gemath had just proven:\[2\sum \frac{1}{r_{bc}}-\frac{1}{r}\geq \sum\frac{1}{r_{a}}\]

In conclusion : [quote][color=darkred]Let $ABC$ be an acute triangle with the inradius $r$ and the circumcircle $w=C(O,R)$.

Denote : the radius $\theta_{a}$ of the circumcircle of the triangle $BOC$ a.s.o. ;

the radius $\beta_{a}$ of the circle which is tangent to the lines $AB$, $AC$ and [b]internally[/b] to the circle $w$ a.s.o. ;

the radius $\gamma_{a}$ of the circle which is tangent to the lines $AB$, $AC$ and [b]externally[/b] to the circle $w$ a.s.o.

Prove the following inequalities : $\{\begin{array}{cc}1\blacktriangleright &{ 2\cdot\sum \frac{1}{\beta_{a}}\ge \frac{3}{r}+\sum \frac{1}{\theta_{a}}}\\\\ 2\blacktriangleright & 2\cdot\sum\frac{1}{\beta_{a}}\ge 3\cdot\sum\frac{1}{\theta_{a}}\\\\ 3\blacktriangleright & 2\cdot\sum\frac{1}{\gamma_{a}}\ge \sum\frac{1}{\theta_{a}}\\\\ 4\blacktriangleright & 3\cdot\sum\frac{1}{\gamma_{a}}\ge \sum \frac{1}{\beta_{a}}\\\\ 5\blacktriangleright & 2\cdot\sum\frac{1}{\gamma_{a}}\ge \frac{1}{r}+\frac{1}{3}\cdot\sum\frac{1}{\theta_{a}}\end{array}$, with equality iff $\triangle ABC$ is equilateral.[/color][/quote]
[color=darkblue]I"ll use the following well-known relations :

$\{\begin{array}{c}a+b+c=2p\\\ ab+bc+ca=p^{2}+r^{2}+4Rr\\\ abc=4RS=4Rpr\end{array}\|$ $\implies$ $\{\begin{array}{c}\sum a^{2}=(\sum a)^{2}-2\cdot\sum bc=2(p^{2}-r^{2}-4Rr)\\\\ \sum a^{3}=(\sum a)^{3}-3\sum a\cdot\sum bc+3abc=2p(p^{2}-3r^{2}-6Rr)\end{array}$.

[b]Two exercises.[/b] $\sum a(p-a)=p\sum a-\sum a^{2}=2p^{2}-2(p^{2}-r^{2}-4Rr)$ $\implies$ $\boxed{\sum a(p-a)=2r(4R+r)}$.

$\sum a(p-a)^{2}=p^{2}\sum a-2p\sum a^{2}+\sum a^{3}=$ $2p^{3}-4p(p^{2}-r^{2}-4Rr)+2p(p^{2}-3r^{2}-6Rr)$ $\implies$ $\boxed{\sum a(p-a)^{2}=2pr(2R-r)}$.

[b][u]Proof of the proposed problem.[/u][/b] Prove easily that $\{\begin{array}{c}R=2\cos A\cdot\theta_{a}\\\\ p\cdot\beta_{a}=(p-a)\cdot\gamma_{a}=\frac{bc}{p-a}\cdot r\end{array}$ a.s.o. Therefore,

$\sum \frac{1}{\theta_{a}}=\frac{2}{R}\cdot\sum \cos A$ and $\sum\cos A=1+\frac{r}{R}$ $\implies$ $\boxed{\sum\frac{1}{\theta_{a}}=\frac{2(R+r)}{R^{2}}}$.

$\sum \frac{1}{\beta_{a}}=$ $\frac{p}{r}\cdot\sum\frac{p-a}{bc}=$ $\frac{p}{r}\cdot\frac{1}{abc}\cdot\sum a(p-a)=$ $\frac{p}{r}\cdot\frac{1}{4Rpr}\cdot 2r(4R+r)$ $\implies$ $\boxed{\sum\frac{1}{\beta_{a}}=\frac{4R+r}{2Rr}}$.

$\sum\frac{1}{\gamma_{a}}=$ $\frac{1}{r}\cdot\sum\frac{(p-a)^{2}}{bc}=$ $\frac{1}{r}\cdot\frac{1}{abc}\cdot \sum a(p-a)^{2}=$ $\frac{1}{r}\cdot \frac{1}{4Rpr}\cdot 2pr(2R-r)$ $\implies$ $\boxed{\sum\frac{1}{\gamma_{a}}=\frac{2R-r}{2Rr}}$.

$1\blacktriangleright$ $2\cdot\sum\frac{1}{\beta_{a}}-\frac{3}{r}\ge\sum\frac{1}{\theta_{a}}$ $\Longleftrightarrow$ $\frac{4R+r}{Rr}\ge \frac{3}{r}+\frac{2(R+r)}{R^{2}}$ $\Longleftrightarrow$ $R^{2}-Rr-2r^{2}\ge 0$ $\Longleftrightarrow$ $(R-2r)(R+r)\ge 0$, what is truly.

$2\blacktriangleright$ $2\cdot\sum\frac{1}{\beta_{a}}\ge 3\cdot\sum\frac{1}{\theta_{a}}$ $\Longleftrightarrow$ $2\cdot\frac{4R+r}{2Rr}\ge 3\cdot\frac{2(R+r)}{R^{2}}$ $\Longleftrightarrow$ $4R^{2}-5Rr-6r^{2}\ge 0$ $\Longleftrightarrow$ $(R-2r)(4R+3r)\ge 0$, what is truly.

$3\blacktriangleright$ $2\cdot\sum\frac{1}{\gamma_{a}}\ge \sum\frac{1}{\theta_{a}}$ $\Longleftrightarrow$ $\frac{2R-r}{Rr}\ge \frac{2(R+r)}{R^{2}}$ $\Longleftrightarrow$ $2R^{2}-3Rr-2r^{2}\ge 0$ $\Longleftrightarrow$ $(R-2r)(2R+r)\ge 0$, what is truly.

$4\blacktriangleright$ $3\cdot\sum\frac{1}{\gamma_{a}}\ge \sum\frac{1}{\eta_{a}}$ $\Longleftrightarrow$ $3\cdot\frac{2R-r}{2Rr}\ge \frac{4R+r}{2Rr}$ $\Longleftrightarrow$ $3(2R-r)\ge 4R+r$ $\Longleftrightarrow$ $R\ge 2r$, what is truly.

$5\blacktriangleright$ $2\cdot\sum\frac{1}{\gamma_{a}}\ge \frac{1}{r}+\frac{1}{3}\cdot\sum\frac{1}{\theta_{a}}$ $\Longleftrightarrow$ $2\cdot\frac{2R-r}{2Rr}\ge \frac{1}{r}+\frac{1}{3}\cdot\frac{2(R+r)}{R^{2}}$ $\Longleftrightarrow$ $3R^{2}-5Rr-2r^{2}\ge 0$ $\Longleftrightarrow$ $(3R+r)(R-2r)\ge 0$, what is truly.

[b][u]Remark I.[/u][/b] $\{\begin{array}{c}6\cdot\sum\frac{1}{\gamma_{a}}\ge 2\cdot\sum\frac{1}{\beta_{a}}\ge 3\cdot\sum\frac{1}{\theta_{a}}\\\\ 6\cdot\sum\frac{1}{\gamma_{a}}-\sum\frac{1}{\theta_{a}}\ge 2\cdot\sum\frac{1}{\beta_{a}}-\sum\frac{1}{\theta_{a}}\ge \frac{3}{r}\end{array}$.

[b][u]Remark II.[/u][/b] An more interesting inequality : $\boxed{\ \beta_{a}+\beta_{b}+\beta_{c}\ge 4r\ }$.[/color]


Let O be the circumcenter of ABC.Let D be the intersection of AO with BC.
Knowing that OD=BD=1/3 * BC, find the measure of the angles of ABC!

This problem was given today at the district phase in Bucharest (7th grade).

Denote $M$ be the midpoint of $[BC]$
We have:
$\begin{cases}DM=\frac{1}{6}BC=\frac{1}{2}OD\\ \angle OMD=90^\circ\end{cases}\Longrightarrow\begin{cases}\angle ODM=60^\circ\\ \angle DOM=30^\circ\end{cases}\Longrightarrow\begin{cases}\angle DBO=\angle DOB= 30^\circ\\ \angle BOD=\frac{1}{2}\angle BOC=60^\circ\end{cases}$
$\Longrightarrow\begin{cases}\angle OAB=\angle OBA=15^\circ\\ \angle AOC=360^\circ-\angle BOC-\angle AOB=90^\circ\end{cases}\Longrightarrow \begin{cases}\angle B=45^\circ\\ \angle OAC=\angle OCA=45^\circ\end{cases}$
Hence: $\angle A=60^\circ;\,\angle B=45^\circ;\,\angle C=75^\circ$


At a party, there are only married men and their wives, and single women. The probability of picking a women that is single out of all the women is 2/5. What is the probability of picking a married man?

I got 3/5, but that was incorrect.  :(  Apparently, the answer is 3/8.  :huh:  How do you get this? Please help.  :maybe:

Say there were only five women. Two of them would be single, because the chance is 2/5. That would mean that three women are married, so there are three men. So there are three men and five women all together. Thus, the chance of picking a man is 3/8.


Doesn't it follow because (triangle inequality) has $a+b\geq \sqrt{a^{2}+b^{2}}$
then rearrangement?

We have

$\sqrt{a^{2}+b^{2}}<a+b,\sqrt{b^{2}+c^{2}}<b+c,\sqrt{c^{2}+a^{2}}<c+a$

which gives

\begin{eqnarray*}&&{a+b \over \sqrt{b^{2}+c^{2}}}+{b+c \over \sqrt{c^{2}+a^{2}}}+{c+a \over \sqrt{a^{2}+b^{2}}}\\ &>& \frac{a+b}{b+c}+\frac{b+c}{c+a}+\frac{c+a}{a+b}\\ &\stackrel{\text{AM-GM}}{>}& 3\end{eqnarray*}

EDIT: [b]me@home[/b] beat me to it.


What's your future career, or current career?

Me - hopefully a cardiosurgeon someday.

whatever i can do to be pretty close to math and have enough money to buy math books :lol:


Let $ ABC$ be an triangle. Prove or disprove that
$ \cos(A \minus{} B)\plus{} \cos(B \minus{} C) \plus{}  \cos(C \minus{} A)\le \ 3$

___________________
Ligouras

[quote="Ligouras"]Let $ ABC$ be an triangle. Prove or disprove that
$ \cos(A \minus{} B) \plus{} \cos(B \minus{} C) \plus{} \cos(C \minus{} A)\le \ 3$

___________________
Ligouras[/quote]
$ cos(x) \le 1$ ;)

Was this problem supposed to make any sense?

I believe it was a mistake, but the inequality cant even be improved. In an equilateral triangle the equality holds.


1)  How many perfect square factors does the number 46,656 have?

2)  There are 8 boys and 6 girls in a band.  A quintet is being formed.  If the quintet must consist of 3 boys and 2 girls, how many quintets are possible?

[hide="clicky"]1) $ 46,656=2^6\times3^6 $  In order to have a perfect square, both the 2 and the 3 must have an even exponent.  The possible values are 0, 2, 4, and 6.  That means that there are 4 choices for each number.  $ 4\times4=\boxed{16 \mbox{ factors}} $[/hide]

[hide="part 2"]The boys can be chosen in $ {8\choose3}=56 $ ways.  The girls can be chosen in $ {6\choose2}=15 $ ways.  $ 56\times15=\boxed{840\mbox{ arrangements}} $[/hide]

math92, you posted the first question twice :

http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=52747


limit
real analysis
real analysis theorems

True or False: If $ \lim_{x \to a} f(x) \equal{} \lim_{x \to a} g(x) \equal{} 0$ and $ f$ and $ g$ are positive in the neighborhood of $ a$, then $ \lim_{x \to a} f(x)^{g(x)} \equal{} 1$

False. Let $ f(x)\equal{}e^{\minus{}1/x^2}$ and $ g(x)\equal{}x^2.$ Then as $ x\to 0,$ $ f(x)\to 0,g(x)\to 0,$ but $ f(x)^{g(x)}\to e^{\minus{}1}.$

The statement that $ f(x)^{g(x)}$ is a $ 0^0$ indeterminate form is the same as saying that $ g(x)\ln(f(x))$ is a $ 0\cdot\infty$ indeterminate form - we can still play with the relative sizes and get anything we want to to happen.

What if we also impose the condition that f and g are analytic in a neighborhood of $ x\equal{}a$?


Find all pairs $(a; b)$ of positive integers for which the numbers $a^{3}+6ab+1$ and $b^{3}+6ab+1$ are cubes of positive integers.

[quote="gotztahbeazn"]errr
[hide]

does expanding $(a+1)^{3}$ and $(b+1)^{3}$ work?

then we get the system:

$3a^{2}+3a = 6ab$
$3b^{2}+3b = 6ab$

solving gives $(1,1)$  as only solution

[/hide][/quote]

Well, you'd have to show that the cubes must be $(a+1)^{3}, (b+1)^{3}$.


function
linear algebra
matrix
calculus
derivative
integration
algebra

Consider the second order linear equation (x^2)y'' + (-4x + 5x^3)y' + 3y = 0 with a fundamental set of solutions (y1, y2) such that

y1(1) = -5
y1'(1) = 4
y2(1) = 4
y2'(1) = 2

Find explicitely the Wronskian function.

We know the Wronkian function is w(x) = y1(x) y2'(x) - y2(x) y1'(x) , but how to use it?

There are two parts here: calculate the value of the Wronskian at the point you know, and then solve the (first-order, linear, homogeneous) differential equation for the Wronskian with that initial condition.

[hide="Explanation, in way too much generality"]
In general, if we have a linear differential equation $y^{(n)}=p_{n-1}(x)y^{(n-1)}+\cdots+p_1(x)y'+p_0(x)$ and $n$ solutions $y_1,y_2,\dots,y_n$ to this equation, the Wronskian is
$\det M$ where $M=\begin{bmatrix}y_1&y_2&\cdots&y_n\\y_1'&y_2'&\cdots&y_n'\\ \vdots&\vdots&\ddots&\vdots\\y_1^{(n-1)}&y_2^{(n-1)}&\cdots&y_n^{(n-1)}\end{bmatrix}$.
We have $M'=PM$, where $P=\begin{bmatrix}0&0&0&\cdots&0&p_0(x)\\1&0&0&\cdots&0&p_1(x)\\ 0&1&0&\cdots&0&p_2(x)\\ \vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\ 0&0&0&\cdots&0&p_{n-2}(x)\\0&0&0&\cdots&1&p_{n-1}(x)\end{bmatrix}$
Now, we want to take the derivative of the determinant. $\det(M+PM\cdot dx)=\det M\cdot\det(I+P\cdot dx)=\det M\cdot(1+\mathrm{tr} P\cdot dx+O(dx^2))$ and the derivative of $\det M$ is $\det M\cdot\mathrm{tr} P=\det M\cdot p_{n-1}(x)$

This gives $W'=p_{n-1}(x)W$. Also, everything generalizes naturally to linear systems.
[/hide]
In this case, we have $W'=\left(\frac4x-5x\right)W$.

Also remember that the Wronskian is only dependent on p(x) in y'' + p(x)y' + q(x)y = 0.

$W=c\exp{\left(-\int p(x)\,dx\right)}$


CEA MAI NOUA METODA DE FACUT BANI PENTRU ROMANI!
BANI RAPID SI SIGUR PE INTERNET. TESTAT! EU AM FACUT
25.260.000 IN 3 SAPTAMANI!!! E RANDUL TAU!!!!!!
FUNCTIONEAZA!!!! SINGURUL CARE FUNCTIONEAZA CU
ADEVARAT

Nu este nici o gluma,crede-ma! Depinde numai de tine daca vrei sa citesti pagina asta sau nu, eu nu
pot spune decat ca 10 min. de citit iti pot schimba viata (si portofelul!), depinde numai de tine "POTI
FACE BANI USOR SI REPEDE" Crede-ma: am incercat toate metodele posibile (si legale).Aceasta este cea mai
USOARA si mai RAPIDA cale. Poti sa faci o gramada de bani, repede si legal. Merge! E testat! Tot ce trebuie
sa faci deocamdata este sa citesti cu atentie si pana la final textul. Eram sceptic, totusi... In urma cu
ceva vreme navigam aiurea pe Internet (probabil la fel cum faci si tu acum) cand am descoperit un articol
asemanator cu acesta in care se spunea ca pot sa castig milioane in doar cateva saptamani cu o
investitie de numai 60.000 de lei. Imediat m-am gandit "chestia asta trebuie sa fie vreo escrocherie". Dar,
curios fiind, am continuat sa citesc. Articolul spunea ca trebuie sa trimiti cate 10.000 de lei vechi sau 1 leu nou la fiecare
din cele 6 adrese mentionate in articol . Apoi sa iti treci numele si adresa la sfarsitul listei in pozitia
a 6-a si sa publici articolul la minim 200 de forum-uri/adrese cu anunturi pe Internet sau prin mail
la cat mai multe adrese. Pur si simplu! Dupa ce am analizat chestiunea si dupa ce am discutat cu cativa
prieteni m-am hotarat sa incerc. La urma urmei ce aveam de pierdut in afara de 60.000 de lei si 6
timbre? Cum este si normal, la inceput eram sceptic si chiar ingrijorat in privinta aspectului legal. Asa ca
m-am interesat la PTTR si mi-au confirmat ca intr-adevar e legal! Deci am facut infima investitie
de 60.000 de lei. Surpriza!...Ghiciti ce s-a petrecut !!! .. Intr-un interval de 7 zile am inceput sa
primesc bani in cutia postala. Eram uluit! M-am gandit ca probabil se va opri curand, dar banii au continuat
sa vina! In prima saptamana am primit 370.000 de lei. La sfarsitul celei de-a doua aveam 1.090.000. In a
treia saptamana stransesem 25.470.000 de lei! Si banii continua sa vina cu repeziciune. In mod cert a meritat
sa cheltui cei 60.000 de lei si cele 6 timbre (Am cheltuit mult mai mult jucand la loto si la bingo!).
Sa va spun cum functioneaza si mai ales de ce functioneaza. In primul rand va sfatuiesc sa salvati
pe hard sau sa printati articolul ca sa il aveti la indemana. Cum?
PASUL 1: Iei 6 coli de hartie si scrii pe fiecare din ele : "M-AM ADAUGAT LA LISTA TA", astfel incat
cel ce primeste plicul sa te adauge la listele pe care urmeaza sa le posteze si asa adresa ta va aparea pe
alte sute si sute multiplicate de forumuri, apoi impaturesti fiecare coala (cat sa intre in plic). Acum
iei 6 bancnote de 10.000 de lei si pui CATE UNA in FIECARE din cele 6 coli (ca sa nu se vada prin plic si
sa nu se fure), apoi pui FIECARE coala in CATE UN PLIC si inchizi plicurile. Ar trebui sa ai acum 6 plicuri
sigilate, cu numele si adresa ta pe ele (la expeditor, pentru a ajunge in continuare pe forumuri!), fiecare
dintre ele continand o coala de hartie si o bancnota de 10.000 de lei. Acum trimiti plicurile la
urmatoarele adrese:

1) Barbu Adrian , Str. Uzinei bl. 3 Ap. 6 , cod 550253 , Sibiu

2) Olarica Alin, str. N. Filimon,nr.28,bl.18,sc.B,ap.63,sect.6,Bucuresti,cod 060302

3) Balogh Robert, str.Oituz, bl. 6A, sc.A, ap 7, cod 430202, Baia Mare, jud. Maramures

4) Urseiu Adrian,Str Caporal Vasilescu Constantin,nr 3,bl. 169,sc. 3,ap. 37,sect. 4,Bucuresti.

5) Becheanu Costin,Str Mioritei ,nr 31,bl. 31,ap 14,Bacau, judetul Bacau, cod 600241

6) Bitea Septimiu Mircea, O.P. nr.1,Timisoara, C.P. nr. 518, loc. TIMISOARA, jud. TIMIS

PASUL 2: Acum tai din lista pe numarul 1 si muti celelalte adrese cu o pozitie mai sus (6 devine 5, 5
devine 4,etc...). Astfel numarul 6 devine liber : completezi cu numele si adresa TA aceasta
pozitie.Trebuie foarte clar sa intelegi ca DACA TE VEI PUNE IN ALTA POZITIE DECIT CEA DE-A 6-a, vei primi cu
mult mai putini bani, pentru ca factorul de multiplicare se diminueaza. Pentru a intari increderea
in joc, trebuie SA NU TRISEZI si SA TRIMITI PLICURILE, deoarece altminteri, te excluzi singur din circuit
semanind neincrederea.
PASUL 3: Schimba ce doresti dar, pe cat se poate, incearca sa mentii articolul cat mai aproape de
original. Acum publica anuntul la cit mai multe forum-uri/newsgroup-uri(sunt o groaza). 200 este atat
cat iti trebuie dar cu cat publici in mai multe locuri cu atat mai multi bani vei castiga! Nu e nevoie sa
scrii din nou tot acest articol; poti sa selectezi textul cu cursorul mouse-ului si sa il copiezi in
memoria calculatorului(\'Copy\' din meniul \'Edit\'); deschizi un fisier gol cu
\'notepad\'(\'File\'...\'New\') si inserezi textul ca sa iti adaugi numele in pozitia a 6-a (\'Edit\'...\
Paste\'); salvezi fisierul cu extensia .txt. Acum cauti pe Internet newsgroup-uri (forum-uri on-line,
message board-uri, site-uri de chat, de discutii) romanesti bineinteles, sau la diverse adrese e-mail pe
care le cunosti si publici articolul. Va sugerez sa treceti ca subiect un titlu cat mai atractiv pentru ca
acesta il vor vedea cei ce viziteaza aceste site-uri GATA! Felicitari... Nu ar trebui sa iti ia mai mult de
30 de secunde pentru fiecare newsgroup! TINE MINTE : MAI MULTE NEWSGROUP-URI INSEAMNA MAI MULTI BANI! TOT
CE TREBUIE ESTE UN MINIM DE 200! Calcule...Vei incepe sa primesti bani in cutia postala in cateva zile! Daca
vrei poti sa iti inchiriezi o casuta postala din cauza numarului mare de scrisori pe care le vei primi. De
ce? Din cele MINIM 200 de publicari pe care le-am facut sa spunem ca primesc 5 raspunsuri (un exemplu
FOARTE-FOARTE mic). Deci am castigat 50.000 de lei cu numele meu pe pozitia a 6- a. Acum fiecare din cele 5
persoane care tocmai mi-au trimis cate o scrisoare, publica la randul lor MINIMUL de 200 de articole cu
numele meu pe pozitia a 5-a si primesc si ei cate 50.000 de lei.Eu primesc astfel 5*5*10.000 de lei =
250.000 de lei. Continuand exemplul cu acest numar mic (5) de persoane care raspund la articol, castig
1.250.000 de lei cu numele meu pe pozitia a 4-a, 6.250.000 cu numele meu pe pozitia a 3-a, 31.250.000
de lei pe pozitia a 2-a si ... DA! 156.250.000 DE LEI cu numele meu pe pozitia intai. UIMITOR! Daca fiecare
participant publica articolul de MINIM 200 de ori! Si totul cu o investitie initiala de 60.000 de lei si 6
timbre. Cand numele tau nu se mai afla in lista poti sa repeti procedura : iei ultima varianta a
articolului si trimiti cate 10.000 de lei la adresele care se afla in lista, punandu-ti numele pe pozitia a
6-a Acum publici articolul din nou de MINIM 200 de ori. Va dati seama ca in fiecare zi mii de oameni
intra pe Internet si citesc aceste articole ASA CUM CITESTI TU ACUM !! Asa ca iti poti permite 60.000 de
lei sa vezi daca merge cu adevarat? Cred ca da... Unii au spus \"dar daca lumea se satura si nu mai trimite
nimeni nici un ban?\" Ce sanse sunt sa se intample asta cand ZILNIC se alatura 20- 50.000 de useri noi la
Internet dintre care foarte multi citesc astfel de anunturi si sunt dispusi sa incerce ?

JUCATI CINSTIT SI CORECT SI ADRESA TE VA AJUNGE PE MII DE FORUMURI, ASTEFEL VEI PRIMI MII DE PLICURI. IN
FIECARE ZI INTRA SUTE SAU MII DE OAMENI FE FORUM! VETI AVEA NUMAI DE CASTIGAT!!! JOCUL SE BAZEAZA PE O
BINECUNOASCUTA LEGE DIN ECONOMIE SI ANUME LEGEA MULTIPLICARII PE CARE CRED CA POTI SA O INTELEGI. CRED
CA MARILE MAGAZINE SE IMBOGATESC PT CA TU AI CUMPARAT O STICLA DE VIN. NU... SE CUMPARA INDUSTRIAL .. DECI
DACA JOCI CORECT SI TRIMITI PLICURI ADRESA TA VA PAREA PE MULT MAI MULTE SITE-RI SI VREI PRIMI CU MUUULT MAI
MULTI BAI! BAFTA! !!!!!!!! p.s. FOARTE IMPORTANT: O SA INTALNESTI OAMENI CARE NU SUNT DE ACORD CU ASTA SI
CARE NU AU CURAJ SA INCERCE SA SE IMBOGATEASCA INTELIGENT. ACESTIA SE MULTUMESC DOAR SA INJURE (MAI
MULT NU POT). TREBUIE DOAR SA-I IGNORI ASA CUM FAC EU, JOACA SI O SA VEZI CUM PLICURILE INCEP SA SOSEASCA.
DACA AI INCREDERE SI PERSEVEREZI O SA-MI MULTUMESTI CAT PENTRU CEI CARE NU JOACA... VA RACITI GURA
DEGEABA, EU PRIMESC ZECI DE PLICURI CU BANI SI O SA CONTINUI SA JOC... BAFTA

Isn't this exact same thing in the Games & Fun Factory?


Choose six points $ D;E;F;G;H;K$ in that order,on a circle wiht radius $ R$ and center at $ O$ such that $ DE \equal{} FG \equal{} HK \equal{} R$.$ KD$ and $ EF$ meet at $ A$;$ EF$ and $ GH$ meet at $ B$ and $ GH$ and $ KD$ meet at $ C$.Prove that:
$ OA.BC \equal{} OB.CA \equal{} OC.AB$


Congratulations to the selected 6. Special congrats to Akashnil and Gaurav patil.
Great to see someone from TN{Ananth Shankar} this year for the IMO.
Congratulations for those who attended the camp too.
Wishing all the best to those selected. Try to make India proud of you 6 by winning golds.

Yeah he is having a fantastic year.Akashnil is just a genius no doubt.

congrats to all selected :lol:
and best of luck for the IMO
hope you get gold :first:

Akashnil..a real genius

so 2 aopsians in the team ..great
and 1 frm delhi too

congrats and best of luck..!!

CONGRATS EVERYONE :D
AKASHNIL SUCH THINGS ARE EXPECTED FROM PEOPLE LIKE YOU. :)  :)

Congrats to all.
Be sure to return with gold medals :first:

[quote="ashwath.rabindranath"]Hi,

This is just to inform all of you about the Indian team that will take part in IMO 2009. The names are:

1. Subhadip Chowdhury (West Bengal)
2. Akashnil Dutta (West Bengal)
3. Akshay Mittal (Delhi)
4. Gaurav Patil (Maharashtra)
5. Ananth Shankar (Tamil Nadu)
6. Sameer Wagh (Maharashtra)

The team leader is Dr. CR Pranesachar and the deputy leader is Dr. Zafar Ahmed. Dr. SA Katre will be going as observer.

Congratulations to everyone! Come back with golds and make us proud.

with warm regards and the heartiest congratulations to everyone,

Ashwath[/quote]

My best wisjes to everyone and pilz win at atleast 5 Gold medals.... :)

When does the IMO start ?

 :D  All The Best  :D

congrats everyone,(esp. akashnil)  :P

best of luck for IMO :)

hey congrats to everyone especially akashnil and gaurav patil....
sabki faaad ke aana...
all the best ahead

kick their asses!!!

im in 11th ..  :(  Last chance , plus what nuclear said is right .
IIT gets into the way of everything , i too am unable to top tests here , was even 34th in one of them  :P

[quote="franklampard8"]IIT gets into the way of everything[/quote]
u shud forget everything else and concentrate on iit.
u r in evening batch at mpa so u hv a gr8 advantage coz prakash sir is the best and the students in evening batch are better of the lot.
[hide]sunil sir is also gr8 genius but he seems to be more interested in chatting abt off-the-topic things with students. [i]vaifal gappa[/i]  :P
[hide]don't tell him i said this  :blush: [/hide] [/hide]
evident from the fact that 18 students from evening batch were ahead of me in iit but only two from morn batch  :rotfl:
if i were in evening batch the competition wud hv surely prompted me to work harder and my rank wud hv been better by atleast 1000.
u hv gr8 opportunity which u shudn't waste  :)

Great work guys. :thumbup:  Improvement in the results compared to last year. Special congrats to ananth whom i met in last year camp. :coolspeak:

U can do it Franky!!!!!!!!!!!! :thumbup: 11th gives u max chance

[quote="Scary math"]Wanna ask abt mpa
I mean,it's pretty good,right? :)[/quote]
No, its THE best.

I really didnt mean to ask ur comment :D
Could u provide details,I tried the net search,they showed NDA :rotfl:
Yah,any of u trying for NDA?I mean,army's a good option,right? :D

@scary
mpa=m prakash academy
[url]http://mprakashacademy.co.in/[/url]

Yah,it's cool
But any of u guys trying for NDA,or havin it as an backup plan? :D

indeed , MPA is THE best .... 2 from mpa got medals this year ... that proves the fact.

Cheers for the IMO team.

Special congrats to Ananth, Sameer and Gaurav!  :)

More specially to Ananth!  :lol:

[quote="Agr_94_Math"]z.z. I think is Sai Mali. I am sorry if I am wrong.[/quote]
Wasn't Sai Mali sulfur oxide??
It would be great if he comes here. :)