limit trigonometry induction function integration calculus logarithms

Prove $ \lim_{n\to\infty}\sum_{k\equal{}1}^n \frac{1}{2^k} \tan \frac{x}{2^k} \equal{} \frac{1}{x} \minus{} \cot x, \quad 0<x<\frac{\pi}{2}$

Let $ f(x) \equal{} \sum_{k \equal{} 1}^{\infty}\frac1{2^k}\tan\left(\frac {x}{2^k}\right).$ (The left hand side.)

Let $ g(x) \equal{} \frac1x \minus{} \cot x.$ (The right hand side.)

If we double $ x,$ we can shift the index of summation by one and find that

$ f(2x) \equal{} \frac12\tan x \plus{} \frac12f(x).$

Try that with the other side:

$ g(2x) \equal{} \frac1{2x} \minus{} \cot(2x).$ But $ \cot(2x) \equal{} \frac12\cot x \minus{} \frac12\tan x,$ so

$ g(2x) \equal{} \frac12\tan x \plus{} \frac12g(x).$

So $ f$ and $ g$ satisfy the same functional equation.

Now let $ h(x) \equal{} f(x) \minus{} g(x).$ By combining these functional equations, we find that $ h(2x) \equal{} \frac12h(x).$

(I should add that the identities I've been deriving are valid for $ x$ and $ 2x$ both strictly between $ 0$ and $ \frac {\pi}2.$)

Rearrange that last identity: $ h\left(\frac x2\right) \equal{} 2h(x),$ so by induction $ h\left(\frac x{2^k}\right) \equal{} 2^kh(x).$

Taking the limit as $ k\to\infty,$ we see that either $ h(x) \equal{} 0$ or $ \limsup_{x\to 0}|h(x)| \equal{} \infty.$

But it is evident that as $ x\to 0,$ we have both $ f(x)\to 0$ and $ g(x)\to 0.$ We can be even more specific with a little attention to power series: $ f(x) \equal{} \frac x3 \plus{} O(x^3)$ and $ g(x) \equal{} \frac x3 \plus{} O(x^3).$

Hence, we have $ \lim_{x\to0}h(x) \equal{} 0.$ From what we've derived above, we must have $ h(x) \equal{} 0$ for all $ 0 < x < \frac {\pi}2,$ which is what we wanted to prove.

Having examined $ \sum_{k \equal{} 1}^{\infty}\frac {1}{2^{k}}\tan\frac {x}{2^{k}}$ uniformly converge to $ f(x)$ for any $ x$, excluding the points that the function is not defined,
we know that $ f(x)$ is integrable and its antiderivative converges to $ \sum_{k \equal{} 1}^\infty\frac {1}{2^{k}}\int_0^x\tan\frac {t}{2^{k}}\,dt \plus{} C.$
where C is a constant, try to make use of a trigonometric identity to simplify the sum.
Finally you can differentiate both sides to get desired result.

In the past my classmate created an identity that indeed solves the problem immediately.
That is $ \tan x \equal{} 2\cot 2x \minus{} \cot x$

[quote="Aile"]In the past my classmate created an identity that indeed solves the problem immediately.
That is $ \tan x \equal{} 2\cot 2x \minus{} \cot x$[/quote]
I don't know that I'd say "immediately" - there are other steps to the process. Note that if you look carefully at my earlier post, you'll see this identity used in the discussion of the right hand side.

[quote="Kent Merryfield"][quote="Aile"]In the past my classmate created an identity that indeed solves the problem immediately.
That is $ \tan x = 2\cot 2x - \cot x$[/quote]
I don't know that I'd say "immediately" - there are other steps to the process. Note that if you look carefully at my earlier post, you'll see this identity used in the discussion of the right hand side.[/quote]
(previously I mistyped the identity, it should be $ \tan x = \cot x - 2\cot 2x$)
I am sorry I haven't typed the detail, our approaches may be different. From the identity, we get
\[ \frac {1}{2^k}\tan \frac {x}{2^k} = \frac {1}{2^k}\cot \frac {x}{2^k} - \frac {1}{2^{k - 1}}\cot \frac {x}{2^{k - 1}},\]
taking summation on both sides, we get a result from the telescoping property,
\[ \sum_{k = 1}^n\frac {1}{2^k}\tan \frac {x}{2^k} = \frac {1}{2^n}\cot \frac {x}{2^n} - \cot x,\]
and we are left to use the fact that $ \lim_{x\to 0}\frac {\sin x}{x} = 1$, this way really gives me a feeling  "immediate". :(

Here is how integration works.

[hide="Click for details"]Since for any $ x$, excluding the points at which $ f(x)$ is not defined, $ \lim_{n\to\infty}\tan\frac {x}{2^n} = 0$, hence there exists an $ N$ such that
\[ n > N \implies \left|\frac {1}{2^k}\tan\frac {x}{2^k}\right| < \frac {1}{2^{k + 1}}.\]
By comparison test, $ \sum 1/2^{k + 1}$ converges implies $ \sum \frac {1}{2^n}\tan \frac {x}{2^n}$  converges absolutely, and hence $ \sum \frac {1}{2^n}\tan \frac {x}{2^n}$ converges uniformly.
\begin{align*} \sum\limits_{k = 1}^\infty {\frac {1}{{2^k }}\int_0^x {\tan \frac {t}{{2^k }}\,d t} } & = - \sum\limits_{k = 1}^\infty {\ln \cos \frac {x}{{2^k }}} = - \ln \left( {\prod\limits_{k = 1}^\infty {\cos \frac {x}{{2^k }}} } \right) = - \ln \left( {\mathop {\lim }\limits_{n \to \infty } \frac {{\sin x}}{{2^n \sin \frac {x}{{2^n }}}}} \right) \\
& = - \ln \left[ {\left( {\frac {{\sin x}}{x}} \right)\left( {\mathop {\lim }\limits_{n \to \infty } \frac {{\sin (x/2^n )}}{{x/2^n }}} \right)}^{ - 1} \right] = \ln \frac {x}{{\sin x}}, \end{align*}
Now since $ \sum\limits_{k = 1}^\infty \dis \frac {1}{2^k }\int_0^x \tan \frac {t}{2^k }\,d t$ converges and, as proved, $ \sum \dis\frac {1}{2^n}\tan \frac {x}{2^n}$ converges uniformly, we differentiate both sides to get that
\[ \sum_{k = 1}^\infty {\frac {1}{{2^k }}\tan \frac {x}{{2^k }}} = \frac {{\sin x - x\cos x}}{{x\sin x}} = \frac {1}{x} - \cot x.\]
[/hide]


My friend's having trouble with her Yahoo e-mail accountant because she can't get into her email because of phrase saying "cookie is rejected." What should she do?

No idea... I would try deleting all the cookies and the cache.  Tools>Internet Options>Delete Cookies

I think prolly that means Yahoo [b]wants[/b] to send a cookie and can't.  So I'd recommend doing the opposite and permitting cookies.  Go to security or something like that, most likely under Tools --> Internet Options, unless for some reason you really don't trust Yahoo...

Actually deleting cookies/enabling cookies aren't the opposite, sometimes you can just get old stuff stored on your computer and deleting them lets Yahoo/etc send you a new one. But most likely they will already be enabled. I had this problem with hotmail once or twice when like over a few weeks it just said that message for no reason - I tried everything and still don't know why it happened. It fixed itself up after a while.


logarithms
function
algebra
system of equations
algebra unsolved

Solve the following system of equations in real numbers:
\[ \begin{cases} \sqrt{x^2-2x+6}\cdot \log_{3}(6-y) =x \\ \sqrt{y^2-2y+6}\cdot \log_{3}(6-z)=y \\ \sqrt{z^2-2z+6}\cdot\log_{3}(6-x)=z \end{cases}.  \]

Let $f(x)=\frac{x}{\sqrt{x^2-2x+6}},$ This function and $-log_3(6-x)$ are increase when x<6. x=y=z=3 -solution. Therefore system have not other solutions.

[color=purple] Hi linkgreencold ! A long time don't meet you  :D now how are you and Abel :)  ?
 @ About this problem if my knowledge isn't so bad it's similar with a VMO 2004 problem isn't it ? [/color]

[b]Rust[/b]'s solution is not complete.But the idea is right and will lead to solution,but we need to do some work with $ x,y,z$,for example to consider maximum of them and after that to use conditions of our function,mentioned by [b]Rust[/b].If some needs I can explain solution in details,though the problem is easy and with hints above-even obvious...


induction
inequalities unsolved
inequalities

[quote="IMO 1980"]Define the numbers $ a_0, a_1, \ldots, a_n$ in the following way:
\[ a_0 \equal{} \frac {1}{2}, \quad a_{k \plus{} 1} \equal{} a_k \plus{} \frac {a^2_k}{n} \quad (n > 1, k \equal{} 0,1, \ldots, n \minus{} 1).\]
Prove that
\[ 1 \minus{} \frac {1}{n} < a_n < 1.\]
[/quote]

IMO 1980 link [hide][url]http://www.mathlinks.ro/resources.php?c=1&cid=16&year=1980[/url][/hide]
Suggested solution [hide][url]http://www.mathlinks.ro/viewtopic.php?p=18075#18075[/url][/hide]

But what if we are to prove that  $ \frac {n \plus{} 1}{2n \minus{} k \plus{} 2} < a_k < \frac {n}{2n \minus{} k}$, for $ k \equal{} 1,2,3,...,n$ ?

For $ k\equal{}1$ it just says
\[ \frac{n\plus{}1}{2n\plus{}1}<\frac12<\frac{n}{2n\minus{}1}\]
which is trivial. for higher $ k$ it follows by induction  :roll:


see a proof here: http://www.mathlinks.ro/viewtopic.php?p=1479101#1479101

[asy]size(300);
import geometry;

pair B=(0, 0), C=(1, 0);
pair A=intersectionpoint(B--(B+dir(40)), C--(C+dir(140)));
pair D=intersectionpoint(A--C, B--(B+dir(20)));

draw(A--B--C--cycle);
draw(B--D);

draw("$100^\circ$",arc(D,A,B,0.1),S);
draw("$20^\circ$",arc(A,B,D,0.1),NE);
draw("$20^\circ$",arc(D,B,C,0.1),E);

label("$A$", A, N);
label("$B$", B, SW);
label("$C$", C, SE);
label("$D$", D, SSW);

pair E=intersectionpoint(B--(B+dir(20)), C--(C+dir(100)));

draw(D--E^^C--E^^A--E,red);
label("$E$", E, NE);

draw("$40^\circ$",arc(E,C,D,0.1),red);

fill(E--arc(A,E,D,0.1)--cycle,blue);
label("$?$",E,10*dir(185),blue);[/asy]

Show this graphic.
It includes hint!

Do not double post:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=273198[/url]


An arithmetic progression has 5 terms that have a sum of 200. The sum of the last three terms added to 6 times the sum of the first two terms equals zero. What is the first term? :?:

[hide="Solution"]
Let the common difference be $ a$.

Then, since the middle number is $ 200\div5\equal{}40$, the five numbers are $ 40\minus{}2a, 40\minus{}a, 40, 40\plus{}a, 40\plus{}2a$.

By the condition of the problem, we then have $ 120\plus{}3a\plus{}6(80\minus{}3a)\equal{}0$.

Solving the equation, $ a\equal{}40$.

Hence, the first term is $ 40\minus{}40\cdot2\equal{}\minus{}40$.

Answer: -40[/hide]


algebra
polynomial
superior algebra
superior algebra unsolved

suppose $ %Error. "display" is a bad command.
f \in \mathbb{C} [x,y]$is a polynomial that the degrees of every it`s term is equal.and the total degree of f is positive .we  suppose the total degree of f is d.total degree is the maximom degree of every terms of f.proove that we can write f to product d linear form :$ %Error. "display" is a bad command.
\ a_i x + b_i y$

$ \mathbb{C}$ is algebraically complete. Set $ y \equal{} 1$ and factor $ f$ as a polynomial in $ x$. Put $ y$ back in by homogenizing each factor.

Also, please work on your English. That first sentence, for example, should be something like "Suppose $ f\in \mathbb{C}[x,y]$ is a nonconstant homogeneous polynomial." or "Suppose each term of $ f\in \mathbb{C}[x,y]$ has the same total degree, and this $ f$ has positive degree $ d$."

but how I put y back in by homogenizing each factor?

Something like $ x\minus{}1$ isn't homogeneous. $ x\minus{}y$ is; we multiply by the appropriate power of $ y$ in each monomial to do this.

There is one case this doesn't handle: a power of $ y$ multiplying the whole polynomial. In that case, we just factor it out so that the leading term is purely a power of $ x$ when we substitute.

I have not seen this kind of proof.thank you for your answer.I coud not find an example that this method does not work.I think it is correct.


LaTeX
inequalities
inequalities unsolved

Let $a,\;b,\;c$ be positive real numbers such that $abc = 1$.
Prove that \[\frac{1+ab}{1+a} + \frac{1+bc}{1+b} + \frac{1+ca}{1+c} \geq 3.\]

Very easy???

Now the real problem:

Let $a,\;b,\;c,\;d$ be positive real numbers such that $abcd = 1$.
Prove that
\[\frac{1+ab}{1+a} + \frac{1+bc}{1+b} + \frac{1+cd}{1+c} + \frac{1 + da}{1 + d} \geq 4.\]

Don't even think about applying the same technique again... It's very different.
After all $3 \neq 4$ right? Hehehe  ;)

Oh yeah can someone write those in latex please... I can't write in latex.  :oops:

[quote="Detective"]Oh yeah can someone write those in latex please... I can't write in latex.  :oops:[/quote]

Done.

I suggest you try to learn it, it's not hard. :)

1.Let $a,b,c>0$ such that $abc=1$.Prove that:

         $\frac{1+ab}{1+a}+\frac{1+bc}{1+a}+\frac{1+ca}{1+c}\geq 3$. ;)

2.Let $a,b,c,d>0$ such that $abcd=1$.Prove that:

              $\frac{1+ab}{1+a}+\frac{1+bc}{1+b}+\frac{1+cd}{1+c}+\frac{1+da}{1+d}\geq 4$. ;)

    Just make the substitutions $a=\frac{x}{y},b=\frac{y}{z}$.....and so on and then apply AM-GM directly ;)

Why don't you ever write full solutions?

Sometimes I write full solutions, but i'm too lazy most of the time.Now, it's kind of late, so i have to get some sleep. ;)

... Oh well. It's a bad habit, really (I am not the only one who thinks so, ask Harazi, for example). Please write full solutions in the future.

Arne, from now on I will let you bother people with the very bad habit of writing full solutions. You know, I'm just sick of what happens in the latest months with this part of the forum, so I will let the New Age coming and showing their values. Bye!

I can insist that : both 2 ineqs can be proved by the same way : Just apply $ \frac{1}{x}+\frac{1}{y} \geq \frac{4}{x+y} $
  For the second ineq :
 $ \frac{1+ab}{1+a}+\frac{1+bc}{1+b}+\frac{1+cd}{1+c}+\frac{1+da}{1+d}
  = \frac{1+ab}{1+a}+\frac{1+bc}{1+b}+\frac{abcd+cd}{abcd(1+c)}+\frac{abcd+da}{abcd(1+d)}
  = (1+ab)(\frac{1}{1+a}+\frac{1}{ab+abc}) +(1+bc)(\frac{1}{1+b}+\frac{1}{bc+bcd})
  \geq 4.(\frac{(1+ab)}{1+a+ab+abc}+\frac{1+bc}{1+b+bc+bcd})
  = 4.(\frac{1+ab}{1+a+ab+abc}+\frac{a+abc}{1+a+ab+abc}) = 4 $  ;)

Using cezar lupu's method and apply AM-GM, we can get the conclusion of general case.

At least show your work, instead of saying stuff like "Just use this and that and it is obvious".

[quote="cezar lupu"]Just make the substitutions $a=\frac{x}{y},\;b=\frac{y}{z}$.....and so on and then apply AM-GM directly ;)[/quote]

This doesn't even work. (Not even in the case n = 4)

Moreover, it's wrong:

[quote="nickolas"]Do  you think  we can  generalize  like  that ?

[tex]a_1,a_2,...,a_n>0  [/tex]  such  that [tex]a_1a_2...a_n=1[/tex]  .Proove  that  [tex]\frac{1+a_1a_2}{1+a_1}+\frac{1+a_2a_3}{1+a_2}+...+\frac{1+a_na_1}{1+a_n} \geq n [/tex][/quote]

Use the substitution $a_1=\frac{x_2}{x_1}$, $a_2=\frac{x_3}{x_2}$, ..., $a_n=\frac{x_1}{x_n}$; in other words, $a_i=\frac{x_{i+1}}{x_i}$ for any i (with indices cyclic modulo n). (This substitution is possible since $a_1a_2...a_n=1$: in fact, just let $x_1=1$, $x_2=a_1$, $x_3=a_1a_2$, ..., $x_n=a_1a_2...a_{n-1}$). Then, for any i, we have \[\frac{1+a_ia_{i+1}}{1+a_i}=\frac{1+\frac{x_{i+1}}{x_i}\cdot\frac{x_{i+2}}{x_{i+1}}}{1+\frac{x_{i+1}}{x_i}}=\frac{x_i+x_{i+2}}{x_i+x_{i+1}}=1+\frac{x_{i+2}-x_{i+1}}{x_i+x_{i+1}}.\] Hence, the inequality in question, $\sum_{i=1}^n\frac{1+a_ia_{i+1}}{1+a_i}\geq n$, becomes $\sum_{i=1}^n\left(1+\frac{x_{i+2}-x_{i+1}}{x_i+x_{i+1}}\right)\geq n$, what simplifies to $\sum_{i=1}^n\frac{x_{i+2}-x_{i+1}}{x_i+x_{i+1}}\geq 0$. But this inequality [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15102]holds not for all n[/url].

Sorry for the messy explanation, but I'm really busy right now.

  Darij

Great job nttu!!!

I'm very sorry cezar lupu, but your solution is wrong!!!
if it's that easy, i wouldn't propose it here.


[quote="Mathx"]Show that there are two real solutions to:
$x + y^2 + z^4 = 0 $
$y + z^2 + x^4 = 0$
$z + x^2 + y^5 = 0$.[/quote]
$x=y=z=0$ is one of the two solutions.
Yeah! We are half done! :D

[quote="Mathx"]do rest of it. :?:  :!:[/quote]
another solution
x=-1, y=-1 and z=0
but  i have no prove that there is no other real solution


geometry
incenter
circumcircle
analytic geometry
Euler
geometry proposed

Define as usual, let I,O,H be the incenter, circumcenter and orthocenter of $\triangle ABC$, where a>b>c. Let D,E be the points on sides BC,CA respectively s.t. AE=BD and CD+CE=AB. Let $BE\cap AD=K$. Prove that KH//OI and KH=2OI.

I don't know why it was required to prove KH=2OI, maybe they want to give a hint to the contestants :?
Anyway, this problem is even easier than the geometry of this year CMO :(

From conditon directly we obtain that $K$ is Nagel point of triangle $ABC$.

Nagel point = $(p-a,p-b,p-c)$ in triliniar coordinates

$N_a = (z_a+z_b+z_c) -2(az_a+ bz_b+ cz_c)/(a+b+c) =3G- 2I$

From Euler line theorem:  $H-O= 3(G-O)$ or $H=3G-2O$

Hence $N_a- H = 2(O- I)$ or $ KH=2OI$ and $KH$ is parallel to $OI$ c.t.d


What scores do college count as equal to college course? I know AB is 4 or more but what about BC? Same?

I'm talking about general number.

I think it varies too much on a college-by-college basis.  Some colleges will accept a $4$ and others won't accept AP scores for credit at all.

There is no such thing as a general number. Each institution has its own rules.

For instance, the rules at my campus (Cal State Long Beach, which is on a semester system):

We accept a 3 or better on the AB, or a 3 or better on the AB subscore of the BC, as credit for our Calculus I course, and allow such students to begin with our Calculus II course. (Personally, I often teach a fall section of Calculus II with the section rigged to be nearly entirely freshmen with AB credit.)

We accept a 3 or better on the BC as credit for both our Calculus I and Calculus II courses, and allow such students to begin with Calculus III (our multivariable calculus course).

As for what the rules are at the college you plan to attend (or the several colleges you will choose among)? You'll just have to ask them.

The College Board website has a listing of which colleges accept what as AP credit:

http://apps.collegeboard.com/apcreditpolicy/index.jsp

It also seems to provide links (if available) to the policy on the college's website.


search
calculus
derivative
calculus computations

Find all local extrema of:
f(x)=(x^2)(e^-x)  and classify each as a local max or a local min

When you look for local extrema you search for the zeros of the derivative

$f(x)=(x^2)(e^{-x})$

$f'(x)=(x^2)'(e^{-x})+(x^2)(e^{-x})'$

$f'(x)=2x(e^{-x})+(x^2)(-e^{-x})$

$f'(x)=(2x-x^2)(e^{-x})$

$f'(x)=x(2-x)(e^{-x})$

$e^{-x}$ is always positive. So we have $f'(x)=0 \Leftrightarrow x(2-x)=0\Leftrightarrow x=0$ or $x=2$

So the points $x=0$ and $x=2$ are [u]possible[/u] local extrema

Now, the sign of $f'(x)$ is equal with the sign of $x(2-x)\Rightarrow$

$f'<0,$ if $x \in (-\infty, 0)$
$f'>0,$ if $x \in (0, 2)$
$f'<0,$ if $x \in (2, \infty)$

$f'$ changes its sign left and right of the points  $x=0$ and $x=2$, so these points are indeed positions of local extrema
(for example, $f'(x)<0$ if $x<0$ and $f'(x)>0$ if $x>0$)

accurately, $0$ is a point of local minimum and $2$ is a point of local maximum

(for example $f$ is decreasing from $-\infty$ to $0$ and is increasing from $0$ to $2$)

the local min is $f(0) = 0^2e^0=0$

and the local max is $f(2) = 2^2e^{-2}=4e^{-2}$


Let $a,b,c \in \mathbb{R}^{+}$ and
               $a+1\geq b+c$
               $b+1\geq c+a$
               $c+1\geq a+b$
 Show that $a^{2}+b^{2}+c^{2}\leq 2abc+1$


geometry
parallelogram
analytic geometry
graphing lines
slope
AMC
AIME

Consider the parallelogram with vertices $(10,45),$ $(10,114),$ $(28,153),$ and $(28,84).$  A line through the origin cuts this figure into two congruent polygons.  The slope of the line is $m/n,$ where $m$ and $n$ are relatively prime positive integers.  Find $m+n.$

[hide="Solution"]For a line to cut the parallelogram into two congruent polygons, it must go through the center of the parallelogram.  The diagonals of a parallelogram bisect each other at the center of the parallelogram.  Therefore since $(10,45)$ and $(28,153)$ are opposite vertices, the center is $(19,99).$  The slope of the line through the origin, $(0,0)$, and $(19,99)$ is $\frac{99}{19}.$  The answer is thus $99+19=118.$[/hide]

Alternatively consider the line $ y\equal{}mx$ then $ m$ must satisfy $ 10m\minus{}45\equal{}153\minus{}28m$.

Anyone else notice how this was on the 1999 AIME and the center point was $(19,99)$?

[b]Solution:[/b] Notice that, for two congruent polygons to be formed, the line must pass through the center (midpoint of two opposite vertices) of the parallelogram. This can be calculated to be $\left(\frac{10+28}{2}, \frac{45+153}{2}\right) = (19, 99) \Rightarrow \boxed{118}$.


Let $ S(n)$ is the sum of digits of $ n$ $ ( n \in N^*)$. If : $ S(n^3) \equal{} S(n)^3$ then find $ n$.

[quote="hophinhan"]Let $ S(n)$ is the sum of digits of $ n$ $ ( n \in N^*)$. If : $ f(n^3) \equal{} f(n)^3$ then find $ n$.[/quote]

I suppose we must read  $ S(n^3) \equal{} S(n)^3$

And I wonder what could be the general solution since there are a lot of such numbers. For example :

$ n\equal{}10^k$ $ \forall k\ge 0$ (e.g. $ 100$)

$ n\equal{}2\cdot 10^k$ $ \forall k\ge 0$ (e.g. $ 20$)

$ n\equal{}10^k\plus{}10^l$ $ \forall k\ne l\ge 0$ (e.g. $ 11$)

$ n\equal{}(10^k\plus{}10^l)(10^m\plus{}1)$ $ \forall m>3k>3l\ge 0$ (e.g. $ 11\cdot 1001\equal{}11011$)
...
and so on, infinitely many solutions, with any length we want, and as many non-zero digits we want.

So, hophinhan, I will be very very interested in your general solution.
Could you give it, please ?


[quote="Bugi"]Let $ a,b,c$ be positive real numbers such that $ abc \equal{} \frac {9}{4}$. Prove the inequality:
$ a^3 \plus{} b^3 \plus{} c^3 > a\sqrt {b \plus{} c} \plus{} b\sqrt {c \plus{} a} \plus{} c\sqrt {a \plus{} b}$

Jury's variant:
Prove the same, but with $ abc \equal{} 2$[/quote]
(I'll just prove Jury's variant. The other follows directly from it)
Obviously $ a^3 \plus{} b^3 \plus{} c^3 \ge 6$ by AM-GM. So:
$ a^3 \plus{} b^3 \plus{} c^3 \ge \sqrt {6(a^3 \plus{} b^3 \plus{} c^3)} \ge \sqrt {2(a^2 \plus{} b^2 \plus{} c^2)(a \plus{} b \plus{} c)}$ By rearrangement, AM-GM, muirhead, something...
And $ \sqrt {(a^2 \plus{} b^2 \plus{} c^2)((b \plus{} c) \plus{} (c \plus{} a) \plus{} (a \plus{} b))} \ge a\sqrt {b \plus{} c} \plus{} b\sqrt {a \plus{} c} \plus{} c\sqrt {a \plus{} b}$ and it is proved.

[quote="dr_Civot"][quote="Bugi"]Let $ a,b,c$ be positive real numbers such that $ abc\equal{}\frac{9}{4}$. Prove the inequality:
$ a^3 \plus{} b^3 \plus{} c^3 > a\sqrt {b \plus{} c} \plus{} b\sqrt {c \plus{} a} \plus{} c\sqrt {a \plus{} b}$

Jury's variant:
Prove the same, but with $ abc\equal{}2$[/quote]

Also we can prove same, but with $abc=1$.
use $ab(a+b)\leq a^3+b^3$.[/quote]

Actually we can't...
For $a=b=c=1$ we have to prove that $3>3\sqrt{2}$ which apparently is not

For abc = 2,we can prove it like this.First by AM-GM we have a^3+b^3 >= ab(a+b) (a^3+a^3+b^3 >= 3*a^2*b and b^3+b^3+a^3 >= 3*b^2*a).Then ab(a+b)+2*c^3 >= 2*sqrt(abc*2*c^2*(a+b)) =
4*c*(sqrt(a+b)) and then cyclic .

For $ abc=2 $ it's very easy. By $AM-GM$ we prove that $ \frac{a^3+b^3}{4}\ge \frac{ab(a+b)}{4}=\frac{2ab(a+b)}{4*abc}=\frac{(a+b)}{2c}.$ So again by $AM-GM$ $\frac{a^3+b^3}{4} +\frac{c^3}{2}\ge\frac{(a+b)}{2c}+\frac{c^3}{2}\ge c\sqrt{a+b}$ and by repeating this we get the wanted inequality. $\Box$

Another approach with AM-GM Schur's inequality
$a\sqrt{b+c}+b\sqrt{a+c}+c\sqrt{a+b}=a\sqrt{abc(\frac{b+c}{2})}+b\sqrt{abc(\frac{a+c}{2})}+c\sqrt{abc(\frac{a+b}{2})}\leq(AM-GM)\leq\frac{3abc+ \frac{ab^2+ac^2+bc^2+b^2c+a^2b+a^2c}{2}}{2}\leq(Schur's)\leq \frac{3abc+\frac{a^3+b^3+c^3+3abc}{2}}{2}\leq(AM-GM)\leq\frac{a^3+b^3+c^3+\frac{a^3+b^3+c^3+a^3+b^3+c^3}{2}}{2}=a^3+b^3+c^3$
as desired. Equality occurs when $a=b=c=\sqrt[3]{2}$.


search
function
number theory proposed
number theory

Proove that between n and 2n there is at least one prime number.

That's Bertrand's Postulate, it was already posted before, search function is helpfull here.

http://www.mathlinks.ro/Forum/viewtopic.php?t=40143
Sorry Zetax :lol:


Hi, I've landed on the [url=http://www.artofproblemsolving.com/USAMTS/AoPS_U_MyProfile.php]USAMTS My Profile page[/url] logged on with my own user ID (tokenadult), and I see some information entered in the fields there pertaining to my son A+MATH, who indeed does intend to participate in USAMTS this year. Will it snarl up the AoPS system for administering USAMTS to have his USAMTS profile linked to my AoPS username, or doesn't that matter. We have printed out all the paper forms, and can use the USAMTS user ID number there to fill out the forms and get them in the mail, if that's okay.

Thanks for taking over the administration of USAMTS. It looks like my son's first year of participation in USAMTS will be an exciting year.

So long as the user ID and username match up (meaning the userID is the ID number for the name username), all should be fine.

My son's USAMTS user number is associated with MY AoPS username, so I will have to send you an email about that.


α) Να βρεθει (να κατασκευασθει) το βαρυκεντρο ενος τραπεζιου και γενικοτερα ενος κυρτου τετραπλευρου
β) Ενα πιο γνωστο
Να εκφρασθει το εμβαδο τραπεζιου συναρτησει των πλευρων του

 ( Το τελευταιο το θυμηθηκα οταν ρωτηθηκαμε πριν απο καποιες μερες απο ομαδα μηχανικων για την ευρεση του. Εκει σκεθφηκα μηπως καποιος πρεπει να τους μιλησει για τον ...  VITΤASKO ...)

Μια περιληπτική λύση για το 2ο...

Εστω $ ABCD$ το τραπέζιο με $ AB//CD$(κι $ AB$ η μικρότερη βάση). Φέρουμε το σημείο $ O$ ως τομή των ευθειών $ AD,BC$.

Τότε από όμοια τρίγωνα έχουμε $ \frac{DC}{AB}\equal{}\frac{OD}{AO}\equal{}1\plus{}\frac{AD}{AO}$ που σημαίνει ότι μπορούμε να βρούμε το $ AO$ κι αντίστοιχα βρίσκουμε και το $ OB$.

Τέλος το εμβαδόν που ψάχνουμε ισούται με $ (ODC)\minus{}(OAB)$ των οποίων το εμβαδόν μπορούμε να βρούμε αφού γνωρίζουμε τις πλευρές τους :wink:

Καλη λυση .Βεβαια εχει πολυπλοκο τυπο λογω Ηρωνα
Θα βαλω τη λυση μου μετα απο λιγες μερες μηπως και δω και αλλη λυση .Εξ αλλου γνωση θελουμε..

μια λυση για το βαρυκεντρο τετραπλευρου ειναι η εξης
Το βαρυκεντρο του ΑΒΓΔ βρικεται συμφωνα με το θεωρημα των ροπων-βαρων στην ευθεια που ενωνει τα βαρυκεντρα των τριγωνων ΑΒΓ και ΓΔΑ και ομοιως και στην ευθεια που ενωνει τα βαρυκεντρα των τριγ. ΑΒΔ και ΒΓΔ.Αρα ειναι το σημειο τομης των 2 παραπανω ευθειων


Let $ a,b,c,x,y,z$ be positive real numbers such that
\[ (a\plus{}b\plus{}c)(x\plus{}y\plus{}z)\equal{}(a^2\plus{}b^2\plus{}c^2)(x^2\plus{}y^2\plus{}z^2)\equal{}4\]
Prove that
\[ abcxyz \le \frac{16}{729}\]
With equality holds for example when $ (a,b,c,x,y,z)\equal{}\left( \frac{1}{3},\frac{1}{3},\frac{4}{3},\frac{1}{3},\frac{1}{3},\frac{4}{3}\right)$
:)

[quote="can_hang2007"]Let $ a,b,c,x,y,z$ be positive real numbers such that
\[ (a \plus{} b \plus{} c)(x \plus{} y \plus{} z) \equal{} (a^2 \plus{} b^2 \plus{} c^2)(x^2 \plus{} y^2 \plus{} z^2) \equal{} 4
\]
Prove that
\[ abcxyz \le \frac {16}{729}
\]
With equality holds for example when $ (a,b,c,x,y,z) \equal{} \left( \frac {1}{3},\frac {1}{3},\frac {4}{3},\frac {1}{3},\frac {1}{3},\frac {4}{3}\right)$
:)[/quote] Very nice, Can_hang2007.  :lol: The following is also true.

Let $ a,b,c,d,x,y,z,t$ be positive real numbers such that

$ (a \plus{} b \plus{} c\plus{}d)(x \plus{} y \plus{} z\plus{}t) \equal{} (a^2 \plus{} b^2 \plus{} c^2\plus{}d^2)(x^2 \plus{} y^2 \plus{} z^2\plus{}t^2) \equal{} 9.$

Prove that

$ abcdxyzt \le \frac {9}{256},$

with equality for $ (a,b,c,d,x,y,z,t) \equal{} \left( \frac {1}{2},\frac {1}{2},\frac {1}{2},\frac 3{2},\frac {1}{2},\frac {1}{2},\frac {1}{2},\frac 3{2}\right)$.

CONJECTURE

Let $ a_1,a_2,...a_n$ and $ b_1,b_2,...,b_n$ be positive real numbers such that

$ (a_1\plus{}a_2\plus{}...\plus{}a_n)(b_1\plus{}b_2\plus{}...\plus{}b_n)\equal{}(a_1^2\plus{}a_2^2\plus{}...\plus{}a_n^2)(b_1^2\plus{}b_2^2\plus{}...\plus{}b_n^2)\equal{}(n\minus{}1)^2$.

If $ n\ge 7$, then the product

$ P_n\equal{}a_1a_2...a_nb_1b_2...b_n$

is maximal for either $ a_1\equal{}a_2\equal{}...\equal{}a_n$ or $ b_1\equal{}b_2\equal{}...\equal{}b_n$.

Vasc, can_hang2007, would one of you two post a solution to your problems?
Hope they are nice and natural.

[quote="nguyenvuthanhha"][i]My younger brother , Vo Thanh Van posted a very good solution for this  nice one here  :lol:  :[/i]

  http://diendantoanhoc.net/forum/index.php?showtopic=48578[/quote]
[quote="vo thanh van"]
$ \Rightarrow t\le \frac{9}{4} \Leftrightarrow ax\le \frac{9}{4}$
...........
$ \Rightarrow 9\equal{}(a\plus{}b\plus{}c\plus{}d)(x\plus{}y\plus{}z\plus{}t)\ge 36 \sqrt[6]{\frac{abcdxyzt.(ax)^2}{3^6}}$
$ \Rightarrow \frac{3^6}{2^{12}(ax)^2}\ge abcdxyzt$
$ \Rightarrow abcdxyzt\le \frac{9}{256}$
[/quote]
I think his proof is not correct. Please read carefully these lines.

[i]Oh , sorry for my mistake :(  :blush:

   Can you post your solution for that nice problem , Can ?[/i]


geometry
parallelogram
circumcircle
angle bisector

A parallelogram $ ABCD$ with an acute angle $ BAD$ is given. The bisector of $ \angle BAD$ intersects $ CD$ at point $ L$, and the line $ BC$ at point $ K$. Let $ O$ be the circumcenter of $ \triangle LCK$. Prove that quadrilateral $ DBCO$ is inside the circle.

I'm having trouble seeing how the angle bisector can intersect both lines. Am I drawing the thing wrong?

It means the extended lines of CD and BC, I think...

[color=darkblue]Call $ \angle{BAK} \equal{} \angle{KAD} \equal{} \angle{LKC} \equal{} \angle{ALD} \equal{} \phi$.

Triangles $ ABK$ and $ KCL$ are isosceles, so $ BK \equal{} AB \equal{} AD$. Also $ OC \perp KL$, so $ \left\| \begin{array}{cc}
\angle{KOC} \equal{} 2\phi \\ \angle{OKC} \equal{} \angle{OCK} \equal{} 90^\circ \minus{} \phi \\ \angle{BKO} \equal{} \angle{OCD} \equal{} 90^\circ \plus{} \phi
\end{array} \right\|$.

In $ \triangle{BKO}$ and $ \triangle{DCO}$ we have:

$ \left\| \begin{array}{cc} BK \equal{} CD \\
\angle{BKO} \equal{} \angle{OCD} \\
OK \equal{} OC \end{array} \right\| \Rightarrow \triangle{BKO}\equiv \triangle{DCO} \Rightarrow \angle{OBC} \equal{} \angle{ODC} \Rightarrow OBDC$ - cyclic.

[hide="Image"][img]http://ipicture.ru/uploads/080611/6aVQeUD3zm.png[/img][/hide][/color]

#H34N1, lies inside which circle?  In any circle, or in circle $ O$?  Because Ahiles showed that DBCO is cyclic, but the picture makes it clear that it isn't in the circumcircle.

By the way, (if the problem is correctly stated) the problem says "lines" and not "line segments."  It would be impossible to intersect both segments; it is impossible for the bisector not to intersect both lines.

Even though this is completely not the case for this problem...it could intersect both line segments if it intersects the intersection of those 2 lines segments  :P

ok, Carbon, I forgot about that technicality.  But then L, K and C are the same point so the triangle LCK is degenerate and has no circumcenter.  If you want to then say that O=C then quadrilateral DBCO is actually a triangle, and all triangles can be circumscribed.

facis... Why it is not cyclic???
[img]http://ipicture.ru/uploads/080612/kycCOrmvLg.png[/img]

It is cyclic, Ahiles.  You proved that pretty well.  I just saying that the circle that quadrilateral OBDC fits in is not circle O, as the problem seems to suggest.

the only thing I did notice: why did you write $ AB\equal{}AD$?  that is not necessarily true, and also irrelevant to your proof.

[quote="facis"]
the only thing I did notice: why did you write $ AB \equal{} AD$?  that is not necessarily true, and also irrelevant to your proof.[/quote]

He meant $ CD$..

Right... that makes more sense.  I was a bit confused when I was reading through this earlier when he claimed that $ BK\equal{}CD$ later on and it took me a few minutes to realize it was the transitive property on the isosceles triangle legs and the sides of the parallelogram.

But anyway, I am still wondering why the problem is worded as if $ DBCO$ is supposed to be inside circle $ O$.  Any ideas?  Perhaps just a poorly worded question?


what's the integer x?

1.3.6.5.1.2.4.1.6.0.1.6.x

copied off from a math book

it's an interesting book.It talks about the secret of right brain

the author ( head of the MIT math department ) says you have

to use right brain when you do math...

talk to me if you are interested about this book

I always wanted to share info about the book

happy new year

yes!!!

it is 0

how did u figure it out?


abstract algebra
linear algebra
matrix
superior algebra
superior algebra theorems

Prove that any projective finitely generated module over a local ring is free.

Let $M$ be a finitely generated $R$-module, $R$ local. Let $0 \rightarrow K \rightarrow A^{n}\rightarrow M \rightarrow 0$ a minimal presentation. Because $M$ is projective, the sequence is split and so it remains exact after $-\otimes_{R}R/\mathfrak{m}$. Now, the last map becomes an isomorphism because the presentation was chosen minimal, so $K/\mathfrak{m}K = 0$, so $K = 0$ by Nakayama.

I needed these further explanations:
1. Since the sequence is split, $K$ is also finitely generated so it is a quotient of a free $R$-module $L$ of finite rank.
2. Minimally presented may mean that the entries of the matrix of the linear application from $L$ to $R^{n}$ induced by the application from $K$ to $R^{n}$ are all in $m$. It can be proved that this implies the other more natural definition of a minimal presentation.
3. (2) Shows that after we tensor the exact sequence by $R/m$ the application $K\to R^{n}$ becomes $0$ because its image is in $mR^{n}$.
4. We need (1) to see that we can apply Nakayama.


[quote="Labiliau"]Let matrix $B \in K^{m \times n}$ and $C \in K^{r \times n}$ such that $rang B = r$ an $rang C = r$. Show that $rang BC = r$.[/quote]

B has n columns, C has r strings, right? u cannot multiply BC.

Thats normal, everybody can make error in notations.

[quote="Labiliau"]Let matrix $B \in K^{m \times r}$ and $C \in K^{r \times n}$ such that $rang B = r$ an $rang C = r$. Show that $rang BC = r$.[/quote]

It means matrices $B$ and $C$ have whole rank (we write rank) and m,n=r or m,n>r. For square matrices m,n=r answer follows directly from $det BC$=$det B$$det C$. For other matrices we can consider (in multiplication) only r linear independent rows of $B$ and r linear independent columns of $C$, i.e. we easily go to the previous case.

Besides, it's well known property of whole rank matrices. We say, these matrices have whole rank r, so Rank$BC$=r. Is whole enough :)


HMMT calculus geometry ARML AMC USA(J)MO USAMO

Quagga
Saiem  	Gilani  	 	 Calculus 	 Combinatorics
Arnav 	Tripathy 		Algebra 	Geometry
Yannik 	Pitcan 		Calculus 	Geometry
Wendy 	Hou 		Combinatorics 	Algebra
Marcel 	Nunez 		Algebra 	Combinatorics
Jason 	LaRue 		Calculus 	Algebra
Rui 	Yang 		General 1 	General 2
Gaku 	Liu 		Geometry 	Combinatorics

I have you on general, unless you'd like something else.

Walter  Cravero      Algebra 	     Combinatorics
David 	Jia 		Algebra 	Combinatorics
Johnathan Mei 		Geometry 	Algebra
Louie 	Wu 		Combinatorics 	Geometry
Jimmy 	Wong 		General 1 	General 2
Ambar 	Mehta 		Calculus 	Geometry
Carolyn Kim 		Algebra 	Combinatorics
Andrew 	Lin 		General 1 	General 2

These topics are just a guess really, I haven't talked to most about HMMT yet.

Cheran  Wu  	 	 Algebra 	 Geometry
Gene 	Katsevich 	Geometry 	Combinatorics
Frank 	Rodriguez 	Algebra 	Combinatorics
Juan 	Bustos 		Algebra 	Combinatorics
Roshan 	Goli 		Calculus 	Algebra
Pratik 	Shah 		Calculus 	Geometry
Vincent Dang 		Calculus 	Algebra
Yang 	Ge 		Calculus 	Algebra

GO GO GO!!! FLORIDA A!!! THAT IS, if Alabama doesn't go. I doubt they will.

vestavia represents alabama, since the school is in alabama  :lol:

same for ARML

but the arml team is technically "alabama"

hmmt is vestavia

that's true

for example, they might say

Vestavia Hills High School of Alabama placed 2nd

and Alabama is technically a term for the region they are from

What is this non-sense?

Btw. These teams aren't at all like that anymore

Quagga
Arnav
Yannik
Saiem
Jason LaRue
Wendy
Gaku Liu
Tony Liu (omg, same last name.)
Marcel Nunez

B
Jonathan Mei
Louie Wu
Carolyn Kim
Edward Zheng (NJ guy)
Youlian Simidjiyski (NJ guy)
Alex Anderson (IL guy)
Ambar Mehta
Andrew Lin

C
Vincent Dang
Preet Patel
Carson Banov
Chris O'Bryan
Siraj Iqbal
Gene Katsevich
Cheran Wu
Carlos Giron

[quote="eyefragment"]Wow, we're quite stacked... even our B team has several USAMO qualifiers (of course, I doubt if that'll stop Exeter... we'll see...).  This should be fun.  Look forward to seeing y'all there.

~Youlian Simidjiyski (One of the NJ guys)[/quote]

i am pretty sure that only edward zhang qualified [at least for 2006 USAMO]

but yea...the A team should so win...d*** though...i have to redeem myself on the geometry test...

Hah, I suppose I should have checked my sources before posting...  I thought that you qualified last year, but I guess I was wrong.  Given how awesome you were at Olympiad Geometry, I figured that you'd get 4 or 5 problems for free on the AIME, so I'm quite surprised that you didn't make it.  Ah well, there's always this year...

As a side note, I'm assuming the redemption is With Regards To the PUMAC?

yea...and for the aime i sucked so bad...i was like wtf? i got a 7...and w/o careless errors, i would have gotten a 13...which is like what the **** i got the first two problems wrong


Find all functions $f: \mathbb R\to \mathbb R$ such that
\[f(f(x))=5f(x)-4x, \quad \forall x \in \mathbb R.\]

1) You need use LaTex, please see here http://www.mathlinks.ro/LaTeX/AoPS_L_About.php
2) Bad title http://www.mathlinks.ro/Forum/viewtopic.php?t=144404

[quote="Modul"]Find all functions $f: R\to R$ such that
$f(f(x))=5f(x)-4x,\forall x \in R.$[/quote]

Besides the two immediate trivial solutions $f(x)=x$ and $f(x)=4x+b$, this problem has infinitely many solutions.

From $f(f(x))=5f(x)-4x$, it is easy to see that :

1) $f(x)$ is a bijective function
2) $f^{\circ n}(x)=\frac{4^{n}-1}{3}f(x)-\frac{4^{n}-4}{3}x$ (where $f^{\circ n}(x)=f\circ f\circ\ldots\circ f(x)$ $n$ times)
3) $f(f(x))-f(x)=4(f(x)-x)$ and so $f(x)>x\Leftrightarrow f(f(x))>f(x)$
4) $g^{\circ n}(x)=\frac{4}{3}(1-\frac{1}{4^{n}})g(x)-\frac{1}{3}(1-\frac{1}{4^{n-1}})x$ where $g(x)=f^{\circ(-1)}(x)$ is the reciprocical function of $f(x)$
5) $g(g(x))-g(x)=\frac{1}{4}(g(x)-x)$ and so $g(x)>x\Leftrightarrow g(g(x))>g(x)$

With these 5 statements, it is easy to build a lot of solutions :

a) Take any couple $(a,b>a)$ and any bijective function $h(x)$ $[a,b]\rightarrow [b,5b-4a]$
b) Define $f(x)=h(x)$ for any $x\in[a,b]$
c) Using points 2) and 3) above, it's easy to build $f(x)$ on $[b,+\infty)$
d) Using points 4) and 5) above, it's easy to build $f(x)$ on $(\frac{4a-b}{3},a)$
e) Take $f(\frac{4a-b}{3})=\frac{4a-b}{3}$
f) Take any couple $(c<\frac{4a-b}{3},d<c)$ and any bijective function $k(x)$ $[d,c]\rightarrow [5d-4c,d]$
g) Define $f(x)=k(x)$ for any $x\in[d,c]$
h) Using points 2) and 3) above, it's easy to build $f(x)$ on $(-\infty,d]$
i) Using points 4) and 5) above, it's easy to build $f(x)$ on $[c,\frac{4a-b}{3})$

Such construction gives infinitely many continuous increasing solutions.

If you add the constraint that $f(x)$ must be differentiable, this constraint for $x=\frac{4a-b}{3}$ would imply $f(x)=x$ or $f(x)=4x+c$


Find an integer $ a$ such that $ (x \minus{} a)(x \minus{} 10) \plus{} 1$ can be written as a product $ (x \plus{} b)(x \plus{} c)$  of two factors where  $ b,c$ $ \in$ $ \mathbb{Z}$

$ (x \minus{} a)(x \minus{} 10) \plus{} 1 \equal{} x^2 \minus{} (a \plus{} 10)x \plus{} 10a \plus{} 1$
$ (x \plus{} b)(x \plus{} c) \equal{} x^2 \plus{} (b \plus{} c) x \plus{} bc$
So,
$ \left\{\begin{array}{l} \minus{} (a \plus{} 10) \equal{} b \plus{} c \\
10a \plus{} 1 \equal{} bc \end{array}\right. \iff \left\{\begin{array}{l} b \equal{} \frac{1}{2} \left( \minus{} a \minus{} 10\pm\sqrt {(a \minus{} 8)(a \minus{} 12)}\right) \\
c \equal{} \frac{1}{2} \left(\minus{} a \minus{} 10\mp\sqrt {(a \minus{} 8)(a \minus{} 12)}\right) \end{array}\right.$

So, $ a \equal{}\boxed{8 \ \text{or} \ 12}$.

$ b\plus{}c\equal{}\minus{}10\minus{}a \\ \\ bc \equal{} 10a\plus{}1$

$ \Rightarrow \ bc \equal{} 10(\minus{}10\minus{}b\minus{}c)\plus{}1 \\ \\ bc \plus{}10b\plus{}10c\plus{}100\equal{}1$

$ (b\plus{}10)(c\plus{}10) \equal{} 1$

since $ b,c \in Z$,

$ (b\plus{}10)(c\plus{}10) \equal{} 1 \times 1 \ or \ (\minus{}1)\times (\minus{}1)$

this gives $ b \equal{} c \equal{} \minus{}11 \ or \ \minus{}9$

$ a\equal{}\minus{}10\minus{}b\minus{}c \equal{} \boxed{12} \ or \ \boxed{8}$


:twisted:
Who's seen the new Matrix movie? What do you think of it? Why do you think Neo can affect the machines in the end? I'm very curious as to what you all are speculating about it!

I have yet to see it  :cry:  :cry:  :cry:  :cry:  :cry:  :cry:  :cry:  :cry:  :cry:

But I have read many many spoilers  :twisted:  :twisted:

I have a question....why does he need to fight ppl if he's infinitely times better than them...my friend suggests he likes to play in god mode but...

I enjoyed the spoilers.... any one else have any opinions?

(This was primarily just to check if my avatar worked...I was bored...anyone recognize him?)

As far as the plot goes, I really have nothing to say...I thought the movie was pretty cool.  It did seem like they stuck in extra scenes to include various special effects that didn't affect the plot or anything at all, but I guess that's the point of the Matrix.  I enjoyed the scene with the Architect - I thought all the screens were neat =)

well, i definately thought the first one was better, but i guess thats just to be expected.  some people say that there was too much fighting, some people say there wasnt enough plot, some people say that stuff is weird.. yeah well there just talkin... thats a movie to enjoy as BRAIN CANDY... my favorite  :wink: fight scenes (where else can you see people fight in trench coats?! yeah, no where... thats what i thought!) that only the matrix can do...  i hafta admit it was lackin a bit on the plot stuff tho... it just means that the next one is gonna be extra super good.  yay!   :D   anyways, thats what i thought, if any of you guys were wondering :)

I most definitely agree that the first one was better. I had trouble understanding some of the plot when it came to the architect scene. Half of the words he said in that scene werent a part of my vocabulary, so he might as well be speaking gibberish than explaining why Neo was the 6th "one" to appear. How confusing.

The special effects were all good though. Those made it worth it to see it, I'd say.

I went into the matrix reloaded not expecting it to be very good at all after what I had heard, but it was way better than I expected. RE the end bit where he stops the machines, the main theory going around here at the moment is that there is a matrix inside a matrix, perhaps more than one..

i was pondering the more than one matrix thing also, it seems to fit, but i enjoyed matrix reloaded to the max, speaking of max, ddrmax and ddrmax2 are awesome games, also if anyone of you have the chance check out enter the matrix and check out the animatrix coming on dvd June 3, wooooooooooooooot. :P

ddrmax as in dance dance revolution? I suck as that but I still spend lots of money on it...

I also went into Matrix Reloaded with low expectations, so I was pleasantly surprised.  My only problem, well two problems, are that the car chase scene was ridiculous and the bad guys are way more interesting than the good guys ( although this isn't necessarily negative).

About the car chase - I'm sorry, but either these guys have extremely bad aim with their guns, or else everyone in that cute little Caddy should be dead several times over.  I did really like the motorcycle driving the wrong way...

About the bad guys - loved Agent Smith, but what is his motivation?  I'm curious to see in the third movie.  I hope it's something good.  Loved the "French" guy too.  Morpheus, Neo, and Trinity are all just cardboard cutouts - no personality.

I'm also wondering where the "machines" that seem to have broken from the system fit in (the Oracle, Agent Smith, the key guy, the "french" guy...)  Are they the previous "ones"?

Can't wait to see the last movie - I hope it has a satisfying and thought=-provoking end...

I don't think the Oracle did break out of the system, from what I gathered the oracle was there to lead the One (hey just realised Neo .. One .. same letters?) along the same path as they lead every other One so the loop goes on. However about Smith I really don't know, perhaps he is part of a big loop that goes around in the 'outer' matrix..like the oracle is in the inner one.. I think everything there will have a purpose at the end. Will be interesting to see how well they explain things.

For one thing...the oracle actress died b4 the third movie so they had to fix that...and about the lack of animation...The Wachowski Brothers used waaaaaay too much CGI animation and they never change the expressions of the characters in the fights....just pose them like action figures...

um mad i can't see it. parents sayz itz rated R so i cant cuz im not 17 yet. doesn't that just suck. anywaz, kinda new to this site, but i really wish they'll change the color, itz really contrasting w/ the dark background but ok just my opinion. w/e. im gonna look at whatz coool bout this............ later dayz

Wuzzup people? I just joined :lol:  :lol:  so how do u get a picture with a post? The Matrix Reloaded was good. However, the plot line was horrible and I couldnt distinguish Neos dreams and reality.  :cry: I cant wait for the 3rd  one to come out in sept.

yoy, to put a picture next to your post, click on profile (which is next to log out), then scroll all the way to the bottom. where it says "upload avatar from your machine" click on browse and upload a previously saved picture. the correct name for the picture is avatar. hope this helps. ajods

While we're on that subject, how do you make the avatars move?

[quote="The Southpaw"] I had trouble understanding some of the plot when it came to the architect scene. Half of the words he said in that scene werent a part of my vocabulary, so he might as well be speaking gibberish than explaining why Neo was the 6th "one" to appear. How confusing.
[/quote]
I agree. I hope in the DVD version they explain to us what the architect said. (I can just hope the people explaining are at the same level of English vocab. as us...)

About Matrix Reloaded in whole...
In the car chase sequence: :shock:
When the architect was talking; :?  :?:
And the rest... :roll:

well, no programming needed on your side. you just have to get an avatar thats already moving

Oh.

[hide]*whacks myself with a foam clue bat*[/hide]

I'm beginning to get the hang of it (using asterisks and such)...

Reloaded was interesting enough...rereading the many transcripts of the architect scene on the web may help one to understand.

And I STILL haven't seen Reloaded.

-Isaac

[quote="andy17null"]WHAT? How did Tare get 395 posts in a month? Oh, I am just clueless... I've missed too much...

-Isaac[/quote]
Technically, 400 posts in 10 days. I joined a month ago, but started posting 10 days ago.
At this rate, I'll be the top poster in about a month...

That is freaking disturbing.... I wondered why I've never seen you around.

"disturbing"? "weird" maybe, but "disturbing"!?
... :(


I believe you're assuming that the maximum area occurs when the side lengths are as large as possible, which is [b]not[/b] true.

I think he's using this argument:

Make the base be AB, and we have an [i]almost[/i] isosceles triangle. NOW he's making the sides as big as possible to get the greatest height.

I can't see where that logic fails.

I suppose that having the sides be as large as possible doesn't give you greatest height. [geogebra]280dbf3bc08d44008fe04adf9375a89b27c9dd9f[/geogebra]

In the diagram, you have two triangles with the same base. One has larger sides but has a smaller height.

Even worse is that by making two sides almost sum to the third, you're putting an almost $ 180$ degree angle in. Pancake triangles don't have much area.

Look at some of the area formulas to get a better idea of how sides and area relate.
$ [ABC] \equal{} \frac {ab}{2} \sin C \equal{} \frac {abc}{4R}$


[hide="solution"]

$x\not = 0$ hence divide the equation with $x^2$ to get

$x^2-4x+1+\frac{4}{x}+\frac{1}{x^2}=0$

Letting $y=x-\frac{1}{x}$ , we get

$y^2-4y+3=0$

$y=1,3$

So solving for $x-\frac{1}{x}=1$ or $3$ to get $x=\frac{1\pm\sqrt{5}}{2},\frac{3\pm\sqrt{13}}{2}$
[/hide]

[quote="Silverfalcon"][quote="Altheman"][hide="hint "]if r is a root, then -1/r is a root[/hide][/quote]

Hmm. How did you figure this out? I think I'm missing something obvious though.  :?[/quote]

If $x_i$ are the roots of an equation, then $\frac 1{x_i}$ are the roots of the equation which is obtained by reversing the order of the coefficients of the initial equation, and $-x_i$ are the roots of the equation which is obtained by changing the sign of non-leading odd-power coefficients of the initial equation. (Both can be proved using Vieta's formulas.)

Combine the two: reversing the order gives $x^4+4x^3+x^2-4x+1=0$, and now changing the sign of odd-power coefficients gives $x^4-4x^3+x^2+4x+1=0$, which is the original equation.


If 1 pint of paint is needed to paint a statue 6 ft. high, then the number of pints it will take to paint (to the same thickness) 540 statues similar to the original but only 1 ft. high, is
\[ \text{(A)}\ 90 \qquad \text{(B)}\ 72 \qquad \text{(C)}\ 45 \qquad \text{(D)}\ 30 \qquad \text{(E)}\ 15 \]

[hide]For every 6 foot satue, there is 1 pt. of paint.  So, for every 1 ft. statue, there's ${{\frac{1}{6}}^2}$ pt. of paint, since this is area.  So mulitply 540 $\frac{1}{36}$ and get 15, or e as te answer.[/hide]

[quote="eryaman"][hide]The ratio between linear measures, surface area, and volume of a shape is: $r:r^{2}:r^{3}$. Using this we get $\frac{1}{36}\times540=15$.[/hide][/quote]

[hide="Question"]
I simplified the question and pretended that it's a cube, and I got the same answer of (E) 15. But how do we demonstrate that $r:r^{2}:r^{3}$ for all statues, including irregular shaped ones?
[/hide]

the shape doesn't matter than ratio still holds true.

Unless it's a statue with the graph 1/x, from 1 to infinity, rotated around the x-axis :P .

That actually can (in theory) exist... it's called Gabriel's Horn...

in theory many things can exist, but only in theory :lol:

In theory, I could nuke George Bush with his own weapons of mass destruction. So never insult theory :P .

[quote="random_mathematician"][hide]that question is absurdly simple...

surely 1/6*540 = 15 is the answer[/hide][/quote]
dude, the question has been answered like four times :lol:

[quote="random_mathematician"][hide]that question is absurdly simple...

surely 1/6*540 = 15 is the answer[/hide][/quote]

Yeah, and 1/6*540 :!=: 15

[quote="NoSoupForYou"]I think he meant $\frac{1}{36}\cdot15$.[/quote]

1/36*15=5/12


It's very weak because if assume $ a \plus{} b \plus{} c \equal{} 3.$
$ LHS \equal{} \sum\ \frac {a}{\sqrt [9]{\sqrt []{a}(b \plus{} c)}}$
Using Am-Gm, $ \sqrt []{a}(b \plus{} c) \leq\ 2$
=> $ LHS \geq\ \frac {3}{\sqrt [9]{2}}$
And obviously, with $ a\plus{}b\plus{}c\equal{}3,$ $ RHS \leq\ \frac{3}{\sqrt[9]{2}}$
=> q.e.d
Now, please think about my problem, same to you but nicer and harder.
With it, this method can't solve it:
http://www.mathlinks.ro/Forum/viewtopic.php?t=248430
 :)


Given triangle $ ABC$. $ M,N$ lie on segments $ AB,AC$ respectively, and $ P$ lies on segment $ MN$. Denote the area of triangles $ ABC, BMP, CNP$ by $ A, A_1, A_2$ respectively.
Prove that $ \sqrt[3]A_1\plus{}\sqrt[3]A_2\le\sqrt[3]A$

[quote="Kyoshiro"]Given triangle $ ABC$. $ M,N$ lie on segments $ AB,AC$ respectively, and $ P$ lies on segment $ MN$. Denote the area of triangles $ ABC, BMP, CNP$ by $ A, A_1, A_2$ respectively.
Prove that $ \sqrt [3]A_1 \plus{} \sqrt [3]A_2\le\sqrt [3]A$[/quote]
It's a Holder's inequality! :lol:

[quote="Kyoshiro"]Given triangle $ ABC$. $ M,N$ lie on segments $ AB,AC$ respectively, and $ P$ lies on segment $ MN$. Denote the area of triangles $ ABC, BMP, CNP$ by $ A, A_1, A_2$ respectively.
Prove that $ \sqrt [3]A_1 \plus{} \sqrt [3]A_2\le\sqrt [3]A$[/quote]
It's in the CGMO 2003, Problem 1
http://www.mathlinks.ro/Forum/viewtopic.php?search_id=23406617&t=247757


Let $p(x)=2x^{3}-3x^{2}+2$ and let $S={p(n)|n \in N,n \leq 1999}$
$T={n^{2}+1|n \in N}$ and $U={n^{2}+2|n \in N}$.Prove that the set S intersection with T (I don´t know to write it at TEX) and the set S intersection with U have the same number of elments.

hallo,
[quote]Prove that the set S intersection with T (I don´t know to write it at TEX) and the set S intersection with U have the same number of elments.[/quote]
do you mean :
$|S \cap T| = |S \cap U|$ ?

It is possible that I have missunderstood the problem or made a silly mistake.

$S%Error. "capT" is a bad command.
=\left\{ p(n) | n\in N\wedge\exists(m)\in N(p(n)-1=m^{2}) \right\}$
$S%Error. "capU" is a bad command.
=\left\{ p(n) | n\in N\wedge\exists(m)\in N(p(n)-2=m^{2})\right\}$

But  $p(n)-1= (n-1)^{2}(2n+1)$,  so  $p(n)\in S%Error. "capT" is a bad command.
\;\leftrightarrow\;\exists (m)(\;m^{2}=(2n+1)\;)$ and  $p(n)-2= n^{2}(2n-3)$, so  $p(n)\in S%Error. "capU" is a bad command.
\;\leftrightarrow\;\exists (m)(\;m^{2}=2n-3\;)$. But since ecah of those conditions is satisfied for an infinite number of  $n$  and the number of pairs $x,y\in N$ such that  $p(x)=p(y)$$\wedge \;x>y$  is finite, is it not possible to conclude that both intersections are infinite?

Consider the element of $S \cap T$. We'll find $|S \cap T|$
In other words find the number of $1 \leq n \leq 1999$ satisfying
$A^{2}+1 = 2n^{3}-3n^{2}+2, A \geq 1$
$A^{2}= 2n^{3}-3n^{2}+1 = (n-1)^{2}(2n+1)$
Hence, $2n+1$ must be a square or $n = 1$. Let $K^{2}= 2n+1$ or $n = \frac{K^{2}-1}{2}, 3 \leq K^{2}\leq 3999, 3 \leq K \leq 63, K$ odd.
TOTAL : $\frac{63-3}{2}+1 = 31$

Consider the element of $S \cap U$. We'll find $|S \cap U|$
In other words find the number of $1 \leq n \leq 1999$ satisfying
$B^{2}+2 = 2n^{3}-3n^{2}+2$
$B^{2}= 2n^{3}-3n^{2}= n^{2}(2n-3)$
Hence, $2n-3$ must be a square. Let $L^{2}= 2n-3$ or $n = \frac{L^{2}+3}{2}, 1 \leq L^{2}\leq 3995, 1 \leq L \leq 63, L$ odd.
TOTAL : $\frac{63-1}{2}+1 = 32$

They're different...  :blush:
But as I see the official solution, they permit $n = 0, A=0$ to complete the proof...
Really weird..  :P

Uupps   :oops:  I didn't see the $n\leq 1999$. Sorry.


calculus
function
derivative
integration
calculus computations

Hey,

[b]1. The problem statement, all variables and given/known data.[/b]
Let,
$ {{f(x)}, {{{f}^{\prime}}{(x)}}, {{{f}^{\prime\prime}}{(x)}},...,} \equal{} {{f}, {{f}^{\prime}}, {{f}^{\prime\prime}},...,}$

Given,
$ {{f} \plus{} {{f}^{\prime\prime}}} \equal{} {0}$
$ {{{f}^{2}} \plus{} {{{f}^{\prime}}^{2}}} \equal{} {0}$
$ {f(0)} \equal{} {0}$
$ {{{f}^{\prime}}(0)} \equal{} {0}$

[u][b]Prove (with Mathematical Rigour),[/b][/u]
$ {f} \equal{} {0}$

[b]2. Relevant equations.[/b]
Knowledge of Calculus and solving Differential Equations.

[b]3. The attempt at a solution.[/b]
[hide="My Attempt at a Solution."]
I'm looking for a [u]mathematically rigorous proof[/u] for the above. My guess is that the answer can be proved by solving a 2nd order ODE? Lastly, I am well aware the answer can be "verbally explained" by reasoning what values the sum of two squares can take on, reasoning from there that $ f \equal{} 0$. However, that solution lacks mathematical rigour and I am asking for a mathematically rigorous proof here.
[/hide]
Thanks,

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

Well, you got so totally confused that couldn't even state the problem right. The correct problem statement is [i]"Prove that every twice differentiable on the real line function $ f$ satisfying $ f\plus{}f''\equal{}0$, $ f(0)\equal{}f'(0)\equal{}0$ is identically $ 0$ by considering $ f^2\plus{}f'^2$ and showing that it is identically $ 0$"[/i] (I do not guarantee the exact words or the grammatical structure of the sentence, of course, but I'm 100% sure about the meaning).

Hey,
[quote="fedja"]
Well, you got so totally confused that couldn't even state the problem right. The correct problem statement is [i]"Prove that every twice differentiable on the real line function $ f$ satisfying $ f \plus{} f'' \equal{} 0$, $ f(0) \equal{} f'(0) \equal{} 0$ is identically $ 0$ by considering $ f^2 \plus{} f'^2$ and showing that it is identically $ 0$"[/i] (I do not guarantee the exact words or the grammatical structure of the sentence, of course, but I'm 100% sure about the meaning).
[/quote]
Not quite. If you want the background of the problem it is the following,
[hide="Click to See the Original Problem Statement and Solution."]
[img]http://d.imagehost.org/0948/Lemma_2_4.jpg[/img]
[/hide]
As for the proof demonstrated, I did not think it was mathematically rigorous enough, particularly at the end to satisfactorily show that $ f \equal{} 0$.

Thanks,

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

[hide="Maybe this will clear it up?"]
I assume you know why $ (f')^2 \plus{} f^2 \equal{} 0$ (It is shown that its derivative is 0, and so it must be constant).

For the last part, you know that for all $ g(x)$,
$ (g(x))^2 \geq 0$
by the trivial inequality.

This only holds in the above case when we have $ f(x) \equal{} 0$. If $ f(x) \neq 0$, then $ (f')^2 \plus{} f^2 \neq 0$ since $ (f')^2 \geq 0, \ f^2 > 0$

And as far as I know, fedja stated the problem correctly?
[/hide]

APSStudent, it appears you have taken a page from lecture notes that I at some point posted on this site.  I can tell you that everything is correct and mathematically rigorous because that proof comes directly from Spivak.  Why do you not consider rigorous?  What does rigorous mean to you?

Hey,
[quote="JBL"]
For those keeping score, this is the fourth thread the OP has started on this question.

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=234283
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=237091
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=237877
[/quote]
I do not make it a point to explain my posts, but to clear up the assumption that all four threads are mindless repostings of the same question, I will.

The earliest post was a general question about how to prove the original stated Lemma, in that post JRav was kind enough to provide a possible solution via the characteristic equation (however the validity of using the characteristic equation still needed to be proved) and t0rajir0u helped me understand a mid-proof step in the lecture notes which I subsequently (and I may add logically) reformed as an integration step.

In the next post following my reforming of the integration step I was (at the time) perplexed with how to find the constant of integration (within that problem). Since, my original thread was dedicated to providing some general help with the proof, I decided to create a new thread to answer this specific question.

In the third post, now having understood nearly all the steps of lemma, I was annoyed by how the author of the post simply jumps from,
[hide="Click to See the Last Lines."]
[img]http://d.imagehost.org/0790/line.jpg[/img]
[/hide]
to,
$ f \equal{} 0$ concluding at the end that the previous lines "implies," -- where is the proof of that?
In that thread t0rajir0u referred to the following theorem to give some insight,
[hide="T0rajir0us Suggested Lemma/Proof."]
[quote="t0rajir0u"]
[b]Lemma:[/b]  A sum of non-negative real numbers $ a_1, a_2, ... a_k$ is zero if and only if $ a_1 \equal{} a_2 \equal{} ... \equal{} a_k \equal{} 0$.

[i]Proof.[/i]  Suppose that, without loss of generality, $ a_1 > 0$.  Since $ a_2, a_3, ... a_k \ge 0$, summing the inequalities gives $ a_1 \plus{} a_2 \plus{} ... \plus{} a_k > 0$; contradiction.
[/quote]
[/hide]
However, while that theorem is nice, it really didn't help that much since I would have had to show $ {f} \equal{} {{f}^{\prime}}$ (which is not true).

Upon realizing the above I immediately thought of JRav's characteristic equation idea and even hinted to it in my "My Attempt At A Solution" collpase tag above.

@[url=http://www.artofproblemsolving.com/Forum/profile.php?mode=viewprofile&u=1430][b]JBL[/b][/url], I should make it a point, that I don't go around digging up how many times you post certain problems simply to skew the legitimacy of your question or you as the poster. I suggest you not do so to mine, particularly when you choose not to contribute anything positive to mine. Thank you.
---------------------------------------------------------------------------------------------------------
Continuing on with my post,
[quote="JRav"]
APSStudent, it appears you have taken a page from lecture notes that I at some point posted on this site. I can tell you that everything is correct and mathematically rigorous because that proof comes directly from Spivak.
Why do you not consider rigorous?
[/quote]
I did, thanks again for posting those lecture notes. To be clear, I agree that the proof is correct. However, how much mathematical rigour is in the proof - is what I take issue with, because mathematical rigour in my experience is really something that differs quite a bit from perspective to perspective. In other words, some may look at the proof and agree it is mathematically rigorous, others may look at it and say well the author could have elaborated here and there in the interest of clarity.

In addition, I would also say that for me as a mathematics major memorizing proofs/theorems in mathematics is not my style of learning. I prefer to learn how to prove as opposed to memorizing proofs. :wink:
[quote="JRav"]
What does rigorous mean to you?
[/quote]
Exhaustive. I'd like for a proof to be extremely complete and that all the details of the proof are explicitly addressed.
[quote="stuck"]
I assume you know why $ (f')^2 \plus{} f^2 \equal{} 0$ (It is shown that its derivative is 0, and so it must be constant).
[/quote]
Ok, I got that.
[quote="stuck"]
For the last part, you know that for all $ g(x)$,
$ (g(x))^2 \geq 0$
by the trivial inequality.

This only holds in the above case when we have $ f(x) \equal{} 0$. If $ f(x) \neq 0$, then $ (f')^2 \plus{} f^2 \neq 0$ since $ (f')^2 \geq 0, \ f^2 > 0$
[/quote]
Ah, Proof by Contradiction. That looks good, since that is the easiest way to finish the proof.

Lastly,
Can $ {{f} \equal{} {0}}$ be also proved by considering the following equations,
$ {{f} \plus{} {{f}^{\prime\prime}}} \equal{} {0}$
$ {{{f}^{2}} \plus{} {{{f}^{\prime}}^{2}}} \equal{} {0}$
That is, by setting both equations equal to one another we have a second-order differential equation.
$ {{{f}{\left(f \minus{} 1\right)}} \plus{} {{{f}^{\prime}}^{2}} \plus{} {{f}^{\prime\prime}}} \equal{} {0}$

Can the above be solved for $ {f}$?

Thanks,

-[url=http://www.artofproblemsolving.com/Forum/profile.php?mode=viewprofile&u=33715][b]APSStudent[/b][/url]
P.S.: To further bring clarity to this post, I made this post 4-5 months after my three previous threads (which were in Oct./Nov.). I would also add that I make it a point to actually thoroughly [i][u]attempt[/u][/i] my problems before bothering to post to ensure I have as best of a firm grasp of the problem as I can (this is always indicated by my "My Attempt at a Solution" collapse tag) and to indicate to the readers that I have already put in effort towards a solution to the problem.

[quote="APSStudent"]
[quote="JRav"]
What does rigorous mean to you?
[/quote]
Exhaustive. I'd like for a proof to be extremely complete and that all the details of the proof are explicitly addressed.
[/quote]
Well, you should keep in mind that proofs are built upon other proofs and some basic stuff is assumed to be known when writing them in the books. The fact that if the sum of squares of several real numbers is $ 0$, then each of them is $ 0$ is a part of that basic stuff (so basic that it was completely unclear to me that it was this step that gave you trouble). I admit that you may have to see it proved once (like the construction of real numbers, Peano axioms for integers, and what not) but nobody will go into such details every time in an advanced (or even beginner's) calculus textbook.

But nevermind, what bothers me is not this but your inability to extract your difficulty from the problem and to present it in its pure form. Any minimally trained mathematician having the same question as you had would just post something like:

$ a,b\in\mathbb R$, $ a^2\plus{}b^2\equal{}0$. Prove that $ a\equal{}0$.

Instead, you not only carry on the no longer relevant information, but even try to incorporate it into the final step thus erasing all the gains of the previous parts of the argument. :roll:

Mathematics is about rigor, and understanding means going into details, all right, but both are also about seeing the general picture and being able to extract its composing parts and to treat them separately, which is the skill you currently seem to lack entirely.

This is not to discourage you or anything like that, just to attract your attention to your weakest side. As long as you don't do something about it, you'll face a lot of misunderstanding, just as it happened in these 4 threads you started :).

Hey,
[quote="fedja"]
[quote="APSStudent"]
Exhaustive. I'd like for a proof to be extremely complete and that all the details of the proof are explicitly addressed.
[/quote]
Well, you should keep in mind that proofs are built upon other proofs and some basic stuff is assumed to be known when writing them in the books. The fact that if the sum of squares of several real numbers is $ 0$, then each of them is $ 0$ is a part of that basic stuff (so basic that it was completely unclear to me that it was this step that gave you trouble).
[/quote]
I would point out that I addressed  in my "My Attempt at a Solution" collapse tag in this post and the previous one that it was already obvious to me that $ f \equal{} 0$. [u]However[/u], a proof explicitly showing $ f \equal{} 0$ within the context of proving the overall Lemma was not obvious. It wasn't until Stuck's post demonstrated a Proof by Contradiction approach that I saw this proof.

[quote="fedja"]
I admit that you may have to see it proved once (like the construction of real numbers, Peano axioms for integers, and what not) but nobody will go into such details every time in an advanced (or even beginner's) calculus textbook.
[/quote]
True, this is not the standard for most calculus textbooks nor most "standard" mathematical textbooks, however I find this annoying none the less. At heart this issue is a difference of opinion.

[quote="fedja"]
But nevermind, what bothers me is not this but your inability to extract your difficulty from the problem and to present it in its pure form. Any minimally trained mathematician having the same question as you had would just post something like:

$ a,b\in\mathbb R$, $ a^2 \plus{} b^2 \equal{} 0$. Prove that $ a \equal{} 0$.
[/quote]
Agreed, but this problem is a [u]differential equation[/u] problem and I reasoned that towards the end proving $ f \equal{} 0$ came down to proving it directly (as Stuck's post indicated) or solving a second-order differential equation as I indicated in the post above. In other words, I [u]intentionally[/u] did not ask for the following to be proved,
$ a,b\in\mathbb R$, $ a^2 \plus{} b^2 \equal{} 0$. Prove that $ a \equal{} 0$.

[quote="fedja"]
Instead, you not only carry on the no longer relevant information, but even try to incorporate it into the final step thus erasing all the gains of the previous parts of the argument. :roll:

Mathematics is about rigor, and understanding means going into details, all right, but both are also about seeing the general picture and being able to extract its composing parts and to treat them separately, which is the skill you currently seem to lack entirely.
[/quote]
I'll take this as a difference of opinion and leave it at that. Thanks.

[quote="fedja"]
This is not to discourage you or anything like that, just to attract your attention to your weakest side. As long as you don't do something about it, you'll face a lot of misunderstanding, just as it happened in these 4 threads you started :).
[/quote]
Quite the contrary, the feedback is appreciated as that helps me work on getting better. Thanks. :)

Thanks,

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


modular arithmetic
algebra
difference of squares
special factorizations

Find all $\large n\in N*$ satisfy $\large 3^{n}$-1 is divisible by $\large n^{3}$.

[hide]
We claim $\boxed{1,2}$ are the only solutions (clearly they work). Suppose true for some $n>2$. Then let $p$ be the least prime divisor of $n$. We have $3^{n}\equiv 1\pmod{p}$ and $3^{p-1}\equiv 1\pmod{p}$. Let $e$ be the order of 3 modulo $p$, then $e$ divides both $p-1$ and $n$ (which has no prime divisor less than or equal to $p-1$) so $e=1$. But $3\equiv 1\pmod{p}$ implies $p=2$.

Now write $n=2^{k}\cdot m$ where $k,m$ positive integers and $m$ odd. Then $2^{3k}\cdot m^{3}$ divides $3^{2^{k}\cdot m}-1$.

First, $2^{3k}$ divides $3^{2^{k}\cdot m}-1$ which factors by difference of squares into
\[(3^{m}-1)\prod_{j=0}^{k-1}(3^{2^{j}\cdot m}+1)\]
$3^{m}-1$ is congruent to 2 mod 4, $3^{m}+1$ is congruent to 4 mod 8, and $3^{2^{j}\cdot m}+1$ is congruent to 2 mod 4 for positive $j$. That means the above product is divisible by $2^{k+2}$ but not $2^{k+3}$. This means $k+2\geq 3k$, so $k$ must equal 1.

Now with $k=1$, $n=2m$ so $8m^{3}$ divides $3^{2m}-1$. Now $n>2$ means $m>1$, and we do the same thing as before: let $q$ be the smallest prime divisor of $m$, so that $3^{2m}\equiv 1\pmod{q}$ and $3^{q-1}\equiv 1\pmod{q}$, from which we deduce $3^{2}\equiv 1\pmod{q}$. But this means $q=2$ which contradicts $m$ odd! Therefore, no other solutions.
[/hide]


This should be in High School Basics.

Use the Binomial Theorem, $(a+b)^{n}=\sum_{k=0}^{n}(\binom{n}{k}a^{k}b^{n-k})$


advanced fields
advanced fields unsolved

[quote="chungyc"]A subset E of a convex set K is called an extreme subset of K if:
(i) E is convex and nonempty,
(ii) whenever a point x of E is expressed as (y+z)/2 where y,z belong to K, then both y and z belong to E.

Give an example to show that the image of an extreme subset under a linear map need not be an extreme subset of the image.
i.e. if E is an extreme subset of K and f is a linear map on K, then f(E) need not be an extreme subset of f(K).[/quote]
Consider $E = [0,1]\times[0,1] \subset \mathbb{R}^2$ and a suitable projection onto a 1-dimensional space.

I get the idea. I took the unit disk as a convex subset of $\mathbb{R}^{2}$. Then any point on the unit circle is an extreme point. I took the projection onto the horizontal axis as the linear map. Although the point (0,1) is an extreme point, its image 0 is not an extreme point of the image of the disk, which is the interval [-1,1].


calculus
derivative
trigonometry
calculus computations

Let $ f(x)\equal{}2x\minus{}3, g(x)\equal{}\frac12x^2\minus{}x, h(x)\equal{}x^2\plus{}2$, find $ \frac{d}{dx}f(g(h(x)))$.

[quote="EastyMoryan"][hide="Solution 1"]
Let:
$ z \equal{} 2y \minus{} 3$
$ y \equal{} \frac {u^2}{2} \minus{} u$
$ u \equal{} x^2 \plus{} 2$

$ \frac {d}{dx}f(g(h(x))) \equal{} \frac {dz}{dx}$

$ \frac {dz}{dy} \equal{} 2$
$ \frac {dy}{du} \equal{} u \minus{} 1$
$ \frac {du}{dx} \equal{} 2x$

$ \frac {dz}{dx} \equal{} \frac {dz}{dy}\cdot\frac {dy}{du}\cdot\frac {du}{dx}$

$ \frac {dz}{dx} \equal{} 2(u \minus{} 1)(2x)$
$ \frac {dz}{dx} \equal{} 4x(x^2 \plus{} 1)$
$ \frac {dz}{dx} \equal{} 4x^3 \plus{} 4x$

$ \frac {d}{dx}f(g(h(x))) \equal{} 4x^3 \plus{} 4x$[/hide]
[hide="Solution 2"]
$ f(g(h(x))) \equal{} 2\left [\frac {(x^2 \plus{} 2)^2}{2} \minus{} (x^2 \plus{} 2)\right ] \minus{} 3$
$ f(g(h(x))) \equal{} (x^4 \plus{} 4x^2 \plus{} 4) \minus{} (2x^2 \plus{} 4) \minus{} 3$
$ f(g(h(x))) \equal{} x^4 \plus{} 2x^2 \minus{} 3$
$ \frac {d}{dx}f(g(h(x))) \equal{} 4x^3 \plus{} 4x$[/hide][/quote]

I wonder which solution would you prefer?

Solution 2 is a better choice if it's easy to just write out $ f(g(h(x)))$.
Otherwise, solution 1 would be better.

For example, if the question were as such:

$ f(x)\equal{}2x^3\plus{}3e^x$
$ g(x)\equal{}4^x\plus{}5\ln(x^3\plus{}2x^2\plus{}x\plus{}3\cos x)$
$ h(x)\equal{}\frac{7x^3}{2\minus{}x^2}$

Then solution 1 would be far easier.


function
integration
probability and stats

Let $ X, Y$ be independent and identically-distributed real valued random variables with distribution function $ F$. Prove that
\[ \mathbb{E}|X\minus{}Y| \equal{} 2\int_{\minus{}\infty}^{\plus{}\infty} F(t)(1\minus{}F(t)) dt\]

\begin{align*}
E|X-Y|&=\int_R\int_R |x-y|dF(x)dF(y)=2\int_R dF(y)\int_y^\infty (x-y)dF(x) =-2\int_R dF(y)\int_y^\infty (x-y)d(1-F(x))\\
&=-2\int_R dF(y)\Big((x-y)(1-F(x))|^\infty_y - \int_y^\infty (1-F(x))dx\Big) = 2\int_R \int_y^\infty (1-F(x))dxdF(y)\\
&= \int_R (1-F(x))dx\int_{-\infty}^x dF(y) = \int_R F(x)(1-F(x))dx.
\end{align*}


Let $p > 5$ be a prime number and define \[ S = \{p - n^2 | n \in \mathbb{N}, n^2 < p\}.\]
Prove that there exist $a,b$ in $S$ such that $1 < a < b$ and $a|b$.

This is problem #2 of Balkan 1996. A solution is available here :
http://www.kalva.demon.co.uk/balkan/bsoln/bsol962.html

Pierre.


algebra
polynomial
inequalities
search
algebra solved

Let $ P(x)$  be a polynomial with a non-negative coefficients. Prove that if the inequality $ P\left(\frac {1}{x}\right)P(x)\geq 1$ holds for $ x \equal{} 1$, then this inequality holds for each positive $ x$.

This isn't too hard.

If $ f(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_0 $, then the given condition gives $ f(1)^2 = (\sum a_i )^2 \geq 1 $. So for any x, Cauchy gives
\[
f(x) f(1/x) = (\sum a_i x^i) (\sum \frac{a_i}{x^i}) \geq (\sum a_i)^2 \geq 1.
\]
And since everything's positive, the result follows.

Yes, I know, this was one of the proposed solutions. Another one, not using Cauchy, goes as follows: With $P\left( x\right) =\sum_{k=0}^{n}a_{k}x^{n}$, where all $a_{i}$ are positive, we have

$P\left( x\right) \cdot P\left( \frac{1}{x}\right) =\sum_{k=0}^{n}a_{k}x^{k}\cdot \sum_{k=0}^{n}\frac{a_{k}}{x^{k}}=\sum_{0\leq i\leq n,\;0\leq j\leq n}\left( a_{i}x^{i}\cdot \frac{a_{j}}{x^{j}}\right) $
$=\sum_{0\leq i\leq n,\;0\leq j\leq n}\left( a_{i}a_{j}x^{i-j}\right)  =\sum_{i=0}^{n}a_{i}^{2}+\sum_{0\leq i<j\leq n}\left( a_{i}a_{j}x^{i-j}+a_{j}a_{i}x^{j-i}\right)$
$=\sum_{i=0}^{n}a_{i}^{2}+\sum_{0\leq i<j\leq n}\left( a_{i}a_{j}\left( x^{i-j}+\frac{1}{x^{i-j}}\right) \right) $
$\geq \sum_{i=0}^{n}a_{i}^{2}+\sum_{0\leq i<j\leq n}\left( 2a_{i}a_{j}\right) \text{\ \ \ \ \ \ \ \ \ \ (since }x^{i-j}+\frac{1}{x^{i-j}}\geq 2\text{)} $
$=\left( \sum_{k=0}^{n}a_{k}\right) ^{2}=\left( P\left( 1\right) \right)^{2}\geq 1$

(but I guess it's just the same).

  Darij

Darij wrote:
[quote]
(but I guess it's just the same)
[/quote]
In fact you're right. What you exactly do is to show Cauchy in middle of the problem. I think it's just easier to assume it true and use it :D

[quote="Leonardo"]He is the Author of a book called "Problem-Solving Strategies", wich I think is very interesting, it helps a lot :D. Also I think he is (I'm not sure) one of the trainers of USA IMO team.[/quote]

Well, if you replace "USA" by "German", then it's completely right...

The book is indeed very good, but unfortunately there are some errors ranging from typos up to incorrect proofs... but this is "comme il faut" nowadays.

  Darij

Can i find this book on the internet anywhere?
Thx a lot...
If it's interesting... it may surely help
:):):)

_______________________
"Our days are never coming back... " (Highway Song -> System of a down)

I think you'll have to buy it. It's quite expensive btw.

Ah....
that's a bad news for a Moldavian... :):):)
It seems quite an interesting book at first sight...

           Palosh alias Downito

____________________________________
"Our days are never coming back... " (Highway Song -> System of a down)

[quote="darij grinberg"]Let P be a polynomial with positive coefficients. Prove: If the inequality $P\left( \frac{1}{x}\right) \geq \frac{1}{P\left( x\right) }$ is valid for x = 1, then it is valid for all x > 0, too.[/quote]

This is also Baltic Way 2004 Problem 2:

Let P(x) be a polynomial with positive coefficients. Prove that if $P\left(\frac{1}{x}\right)P\left(x\right) \geq 1$ holds for x = 1, then the same inequality holds for each positive x.

  Darij

[quote="Palosh"]Can i find this book on the internet anywhere?
[/quote]
Search for keyword '' Problem-Solving Strategies'' on Google , You will have that book!  :wink:


prove that there exist infinitly many natural numbers like $k$ which :

 \[ \pi(k) | k  \]

$\pi(k)$ is the number of primes $\leq k$

Consider the sequence of primes $p_i$.  For $p_i \le n < p_{i+1}$ we have $\pi(n) = i$.  We desire that $i | n$ for some $n$ in this range.  If we can show that $p_{i+1} - p_i + 1 \ge i$ for infinitely many $i$ then this follows.  This follows from a theorem, whose name I cannot recall, that the gap between primes is arbitrarily large.

:)

Though the gap of primes is arbitrarily large,we can't say $p_{i+1}-p_i+1\geq i$ holds for infinite $i$
So what can we do?

Lomos, try searching on the forum, it has already been posted by Omid Hatami and I remember having posted a solution.

Well, suppose $p_{i+1} - p_i + 1 \ge i$ a finite number of times.  Consider the maximal value $i_{max}$ at which this occurs.  For $M > i_{Max}$ we will have $p_{M+1} - p_M + 1 < M$, meaning the sequence of primes grows at most quadratically from this point (Edited)... and this must form a contradiction somehow.  It can't possibly be right.    :|

I doubt about the "grows grows at best (what does best mean here¿) quadratically"-part as long as best means at least (and otherwise there is no contradiction).

Aha, Find it at last ,The solution is Marvelous,Beautiful, Indeed.

[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=5132[/url]

How on the Earth did Someone Come up with this Solution ?

Marvelous.

At least grobber came up with the same ideas as me :)  I would be interested in seeing how to derive a contradiction from the statement I made.  It seems that it would imply the existence of far too many large primes.

Edit:  The statement I derived implies $\pi(n) \approx \sqrt{n}$, clearly false since $\pi(n) \approx Li(n)$?


limit
trigonometry
function
integration
real analysis
real analysis unsolved

Prove or disprove: There exists an increasing sequence $\{a_{n}\}$ of positive integers such that $\lim_{n\to\infty}\cos (a_{n}x) = 0$ for almost all $x$ in $[0,1]$.

Suppose this were true. Consider the sequence of functions $f_{n}(x)=\cos^{2}(a_{n}x).$ Our assumption is that $f_{n}(x)\to 0$ for almost every $x.$ The sequence is also dominated by the integrable function $1$ and hence the Dominated Convegence Theorem applies.

By the DCT, $\lim_{n\to\infty}\int_{0}^{1}\cos^{2}(a_{n}x)\,dx=0.$

However, we know better. By direct calculation,

$\lim_{n\to\infty}\int_{0}^{1}\cos^{2}(a_{n}x)\,dx=\frac12.$

So the hypothesis is false. There exists no such sequence.


logarithms
calculus
calculus computations

Solve:

$ \exp\left(\frac{z+i}{2}\right)=1-i$

Use the fact tht $ 1-i = \exp \left[ \frac{1}{2}\ln 2+\left(2n-\frac{1}{4}\right)\pi i \right]$.

Take the $ log$ of both sides to get:

$ \frac{z+i}{2}=\log(1-i)=\ln(\sqrt{2})+i arg(1-i)=\ln(\sqrt{2})+i\left(\frac{-\pi}{4}+2n\pi\right)$ Now we have

$ z+i=\ln 2+2i\left(\frac{-\pi}{4}+2n\pi\right)\Rightarrow z=\ln 2+i\left(\frac{-\pi}{2}+4n\pi-1\right)$

There doesn't seem to be a nice form for this? Factoring some stuff out maybe?


ABCD is a cyclic quadrirateral. S is the intersection of diagonals (AC and BD). Points P, Q are orthogonal projections of S onto AB and CD. The midpoints of segments AD and BC are M, N. Prove that the lines MN and PQ are perpendicular.

See at http://www.mathlinks.ro/Forum/viewtopic.php?t=213068

Kostas Vittas.


[size=150]Let $ x,y,z,u,v,w$ are positive numbers,prove that:

$ (x\plus{}y\plus{}z)(u\plus{}v\plus{}w)(\frac{1}{(ux\plus{}vy\plus{}wz}\plus{}\frac{1}{uy\plus{}vz\plus{}wx}\plus{}\frac{1}{uz\plus{}vx\plus{}wy}) \geq9.$
BQ[/size]

[quote="xzlbq"][size=150]Let $ x,y,z,u,v,w$ are positive numbers,prove that:

$ (x \plus{} y \plus{} z)(u \plus{} v \plus{} w)(\frac {1}{(ux \plus{} vy \plus{} wz} \plus{} \frac {1}{uy \plus{} vz \plus{} wx} \plus{} \frac {1}{uz \plus{} vx \plus{} wy}) \geq9.$
BQ[/size][/quote]

yes, that's so easy

$ \frac {1}{ux \plus{} vy \plus{} wz} \plus{} \frac {1}{uy \plus{} vz \plus{} wx} \plus{} \frac {1}{uz \plus{} vx \plus{} wy}$

$ \ge \frac{9}{ux \plus{} vy \plus{} wz\plus{}uy \plus{} vz \plus{} wx\plus{}uz \plus{} vx \plus{} wy}$

$ \equal{}\frac{9}{(x\plus{}y\plus{}z)(u\plus{}v\plus{}w)}$

 :wink:

[size=100]replace it:

Let x,y,z,u,v,w are positive numbers,prove that:

$ \frac {x}{y \plus{} w} \plus{} \frac {y}{z \plus{} u} \plus{} \frac {z}{x \plus{} v} \plus{} \frac {u}{v \plus{} z} \plus{} \frac {v}{w \plus{} x} \plus{} \frac {w}{u \plus{} y} \geq 3$

BQ[/size]

$ \frac{x}{y\plus{}w\plus{}u}\plus{}\frac{y}{z\plus{}u\plus{}v}\plus{}\frac{z}{x\plus{}v\plus{}w}\plus{}\frac{u}{v\plus{}z\plus{}x}\plus{}\frac{v}{w\plus{}x\plus{}y}\plus{}\frac{w}{u\plus{}y\plus{}z} \geq 2;$

[size=150]easy?

$ (\frac{1}{x}\plus{}\frac{1}{y}\plus{}\frac{1}{z})(\frac{1}{u}\plus{}\frac{1}{v}\plus{}\frac{1}{w}) \geq 9(\frac{1}{ux\plus{}vy\plus{}wz}\plus{}\frac{1}{uy\plus{}vz\plus{}wx}\plus{}\frac{1}{uz\plus{}vx\plus{}wy})$

BQ[/size]

[quote="xzlbq"][size=100]replace it:

Let x,y,z,u,v,w are positive numbers,prove that:

$ \frac {x}{y \plus{} w} \plus{} \frac {y}{z \plus{} u} \plus{} \frac {z}{x \plus{} v} \plus{} \frac {u}{v \plus{} z} \plus{} \frac {v}{w \plus{} x} \plus{} \frac {w}{u \plus{} y} \geq 3$

BQ[/size]

$ \frac {x}{y \plus{} w \plus{} u} \plus{} \frac {y}{z \plus{} u \plus{} v} \plus{} \frac {z}{x \plus{} v \plus{} w} \plus{} \frac {u}{v \plus{} z \plus{} x} \plus{} \frac {v}{w \plus{} x \plus{} y} \plus{} \frac {w}{u \plus{} y \plus{} z} \geq 2;$[/quote]

the frist ineq I try to use the similar way to prove:

$ \frac {x}{y \plus{} w} \plus{} \frac {y}{z \plus{} u} \plus{} \frac {z}{x \plus{} v} \plus{} \frac {u}{v \plus{} z} \plus{} \frac {v}{w \plus{} x} \plus{} \frac {w}{u \plus{} y}$

$ \equal{} \frac {x^2}{x(y \plus{} w)} \plus{} \frac {y^2}{y(z \plus{} u)} \plus{} \frac {z^2}{z(x \plus{} v)} \plus{} \frac {u^2}{u(v \plus{} z)} \plus{} \frac {v^2}{v(w \plus{} x)} \plus{} \frac {w^2}{w(u \plus{} y)}$

$ \ge \frac {(x \plus{} y \plus{} z \plus{} u \plus{} v \plus{} w)^2}{x(y \plus{} w) \plus{} y(z \plus{} u) \plus{} z(x \plus{} v) \plus{} u(v \plus{} z) \plus{} v(w \plus{} x) \plus{} w(u \plus{} y)},$

just need to prove that

$ (x \plus{} y \plus{} z \plus{} u \plus{} v \plus{} w)^2 \ge 3(x(y \plus{} w) \plus{} y(z \plus{} u) \plus{} z(x \plus{} v) \plus{} u(v \plus{} z) \plus{} v(w \plus{} x) \plus{} w(u \plus{} y)),$

this ineq holds, but I can prove it,  :( help....... :maybe:

and the second ineq, still use above way, that's OK:

$ \frac {x}{y \plus{} w \plus{} u} \plus{} \frac {y}{z \plus{} u \plus{} v} \plus{} \frac {z}{x \plus{} v \plus{} w} \plus{} \frac {u}{v \plus{} z \plus{} x} \plus{} \frac {v}{w \plus{} x \plus{} y} \plus{} \frac {w}{u \plus{} y \plus{} z}$

$ \equal{}\frac {x^2}{x(y \plus{} w \plus{} u)} \plus{} \frac {y^2}{y(z \plus{} u \plus{} v)} \plus{} \frac {z^2}{z(x \plus{} v \plus{} w)} \plus{} \frac {u^2}{u(v \plus{} z \plus{} x)} \plus{} \frac {v^2}{v(w \plus{} x \plus{} y)} \plus{} \frac {w^2}{w(u \plus{} y \plus{} z)}$

$ \ge \frac {(x \plus{} y \plus{} z \plus{} u \plus{} v \plus{} w)^2}{x(y\plus{}w\plus{}u)\plus{}y(z\plus{}u\plus{}v)\plus{}z(x\plus{}v\plus{}w)\plus{}u(v\plus{}z\plus{}x)\plus{}v(w\plus{}x\plus{}y)\plus{}w(u\plus{}y\plus{}z)},$

so we just need to prove that

$ (x \plus{} y \plus{} z \plus{} u \plus{} v \plus{} w)^2 \ge 2(x(y\plus{}w\plus{}u)\plus{}y(z\plus{}u\plus{}v)\plus{}z(x\plus{}v\plus{}w)\plus{}u(v\plus{}z\plus{}x)\plus{}v(w\plus{}x\plus{}y)\plus{}w(u\plus{}y\plus{}z))$

$ \iff (x\minus{}u)^2\plus{}(y\minus{}v)^2\plus{}(z\minus{}w)^2 \ge 0,$

clear true.

[quote="kuing"]
just need to prove that

$ (x \plus{} y \plus{} z \plus{} u \plus{} v \plus{} w)^2 \ge 3(x(y \plus{} w) \plus{} y(z \plus{} u) \plus{} z(x \plus{} v) \plus{} u(v \plus{} z) \plus{} v(w \plus{} x) \plus{} w(u \plus{} y)),$

this ineq holds, but I can prove it,  :( help....... :maybe:
[/quote]
It's equivalent to $ (x\plus{}u\minus{}w\minus{}z)^2\plus{}(x\plus{}u\minus{}y\minus{}v)^2\plus{}(y\plus{}v\minus{}w\minus{}z)^2\geq0.$

[quote="xzlbq"][size=150]easy?

$ (\frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z})(\frac {1}{u} \plus{} \frac {1}{v} \plus{} \frac {1}{w}) \geq 9(\frac {1}{ux \plus{} vy \plus{} wz} \plus{} \frac {1}{uy \plus{} vz \plus{} wx} \plus{} \frac {1}{uz \plus{} vx \plus{} wy})$

BQ[/size][/quote]

yes, that's easy

by carlson ineq, we have

$ (\frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z})(\frac {1}{u} \plus{} \frac {1}{v} \plus{} \frac {1}{w})(ux \plus{} vy \plus{} wz)\ge (1 \plus{} 1 \plus{} 1)^3,$

so we have

$ (\frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z})(\frac {1}{u} \plus{} \frac {1}{v} \plus{} \frac {1}{w})\ge \frac {27}{ux \plus{} vy \plus{} wz},$

similarly, we get

$ (\frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z})(\frac {1}{u} \plus{} \frac {1}{v} \plus{} \frac {1}{w})\ge \frac {27}{uy \plus{} vz \plus{} wx},$

$ (\frac {1}{x} \plus{} \frac {1}{y} \plus{} \frac {1}{z})(\frac {1}{u} \plus{} \frac {1}{v} \plus{} \frac {1}{w})\ge \frac {27}{uz \plus{} vx \plus{} wy},$

adding them together, we get the result.

7#'s ineq is similar.

[quote="arqady"][quote="kuing"]
just need to prove that

$ (x \plus{} y \plus{} z \plus{} u \plus{} v \plus{} w)^2 \ge 3(x(y \plus{} w) \plus{} y(z \plus{} u) \plus{} z(x \plus{} v) \plus{} u(v \plus{} z) \plus{} v(w \plus{} x) \plus{} w(u \plus{} y)),$

this ineq holds, but I can prove it,  :( help....... :maybe:
[/quote]
It's equivalent to $ (x \plus{} u \minus{} w \minus{} z)^2 \plus{} (x \plus{} u \minus{} y \minus{} v)^2 \plus{} (y \plus{} v \minus{} w \minus{} z)^2\geq0.$[/quote]

thank you very much!

 :lol:

[quote=spanferkel][quote="kuing"][quote="xzlbq"]If this,must be To the n-variables
BQ[/quote]

I will try to reseach it, but I think, for n-variables, my above way should be hard work,
 the last S.O.S is so hard to transformation...[/quote]
The method with C-S doesn't work for $ n\ge 5$. Let $ s \equal{} \sum x_i$ and $ t \equal{} \sum y_i$. For the ineq
\[ \sum_{cyc}\left(\frac {x_i}{x_{i \plus{} 1} \plus{} t \minus{} y_{i \plus{} 1}} \plus{} \frac {y_i}{y_{i \plus{} 1} \plus{} s \minus{} x_{i \plus{} 1}}\right)\ge2\]
which is the general version of the second one in post #5 ($ n \equal{} 3$) and of post #9 ($ n \equal{} 4$),
after C-S, we would have to prove, putting $ z_i: \equal{} x_i \minus{} y_i\in\mathbb R$,
\[ (\sum z_i)^2\ge 2\sum z_iz_{i \plus{} 1}\]
But this is false for $ n\ge5$, e.g. $ (z_i) \equal{} (0,\dots,0,1,1,0, \minus{} 2)$.  :([/quote]


What place did you get at the state round. Remember to abide by these rules:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=191755

.I got 2nd individual, lost 3-0 in countdown against Alan. My team placed 1st in state.

dude sweet
i am assuming you are jason li

who were 3rd and 4th

3rd was this guy I know from my Piano class or something, his name's like Young-yi or something.
Fourth was Allison Burkys, but she wanted to go to Mexico instead of Nats, so Alan's going to Nats instead.

hold on, what? speak spanish instead of do math? if i remember correctly the mexico trip last year didn't concur with the nationals. oh well.

wait so

1. kevin wen?
2. jason li
3. piano person who you are referring to
4. allison bukys
5. alan chou

man this is so random

what is up with the third place person i hope it wasn't some random

gah a team with allison, alan, jason would be pretty beastly

What?
Mr. Frost told me he was shaun ong...
 :|
Or maybe his real name just sounded like that.
Anyways, what place are you in?
Alan told me you're from MA.

I don't even know what state you are talking about..
Anyways, from Michigan, results were:
1. David Lu
2. Peter Su
3. Annie Xiao
4. Allen Chen

Still our team owned everyone, even with me on it...
I like lowered the average by $ x$ points., where $ x\ge1$
Johnathan got a high fever, so he couldn't come.

For real :

MASSACHUSETTS :

Teams:
1st Jonas Clarke
2nd RJ Grey
3rd Diamond

Individuals :

1st Xiaoyu He  (RJ Grey)
2nd Jason Li (Clarke)
3rd Yongyi Chen (John Glenn Bedford)
4th Allison Bukys(Clarke, but not going to Nats due to prior commitment)
5th Alan chou (Clarke, going to Nats in Allison's place)

Countdown, Alan went from 5th to 1st, so if countdown had been official we'd have the same Nats team anyway :lol:

Great job to everyone in Mass!!

Yeah, Peijin, you got x points where x is also less than 47.
Also, Carl, you posted a question just like this is and I gave the placements. Its only Lexington though.

[quote="CatalystOfNostalgia"]hmmmm individually it seems that you were more successful than we were last year.

so what were you, 2, 4, 5, 25? last year we were like 1, 12, 14, 15.

lexington apparently had 6 of the top 11?[/quote]
2 4 5 23
Doesn't really make a difference really, since I still failed miserably.
And for some reason, it seemed to me that #10-#5 were better at countdown than the top 4.
EDIT: OH, When Mr. Frost said Xiao-Yu, I thought he said Shau-Young, so I thought he was the shau ong guy from AoPS.

Wait, were you the guy in the yellow shirt with the first place individual trophy from chapter?

no, the guy in the yellow shirt was alan chiao!
he had the trophy cuz he got 1st place individual in chapter but there was a mistake and he didnt get the trophy at chapter.
Im the guy in the red sox shirt

Ugh........
Red sox shirt?
.....
My memory is so bad.
I only remember the diamond kids, the two chinese guys, first place and the guy with the trophy, and yeah, that about it.

Yeah, nice job to the top 4 in MA. My rank was...  :maybe: . And yeah I got 20th. Now I'm in the same club as perfect628. GL for nats.

BTW, the guy in 10th was Matthew Arbesfeld.

Please remember that the forum is not a chat room :lol:

Question and answers are cool, but a whole conversation????  Say bye bye to those posts :lol:

AHA YOU'RE KEVIN!
Now I know your username...

err... No option for teams? I'm gonna go make a better poll =D... but... too many options so. nvm =D

Ok, I got 1st Countdown, and 2nd Written. We got 5th place team by a tiny bit
:(
I wish countdown counted.

[quote="1=2"][quote="CatalystOfNostalgia"]5. alan chou[/quote]

It's so annoying when you get 5th, isn't it?

I'll quit the phail club if and only if I make nats this year.[/quote]

Less annoying when your teammate is 4th and they decide to go on a school trip to mexico that same weekend and give you their spot to Nationals!!

so you think alan's annoyed that he gets to go to nats instead of me then?  did i want to have to choose?  (i'm allison, for those who don't know)

oh, nvm, i get it. oops :oops:

Now you start posting:-)  ( I must have missed the other 11)

Welcome Allison.  And congrats on making Nats...... hey now Alan owes you a big favor.

I hope you continue posting and working on AoPS even if you everyone here will think you are nuts for giving up your spot at nats....well everyone but Alan of course :D

Oh, did we mention on this forum "no hablo espanol"

lol, nothing on the poll really applies to me...

well alan, you did make top 4 countdown and did move on... but thats probably not what it means...  :lol:

btw, mr frost: "on this forum i dont speak spanish."  congrats.  you don't speak spanish regardless of where you are, online or not.  you meant [i]no hablamos espanol[/i]

i would hablar en espanol, but you'd just delete it anyway... so pointless to even try

[quote="Yongyi781"]How funny it is for Peijin to call me "The Piano Guy"... now everybody will expect me to play piano at the Nationals :mad:

Anyway, I got third written, fourth Countdown (beaten by Alan). :blush:[/quote]

huh, I missed that part involving the piano... I can't play the piano for my life :(

Eh, I know him from Piano class...
So Carl called him the piano guy.
Anyways, Congratulations on joining AoPS!
:lol:

I made countdown but not nats. Countdown was unofficial, so I was 6th in my state. I was very close, though... :oops:


group theory
abstract algebra
superior algebra
superior algebra unsolved

If $ G$ is a finite group that doesn't have a subgroup of index two, prove that the subgroup of index $ 3$ of $ G$ is a normal subgroup. Thanks in advance.

The next step is also true: If G is a group without subgroups of index 2 and 3, then a subgroup with index 4 is normal.

But then the following thought is false: $ A_5$ is a group without subgroups of order 2,3,4 but has non normal subgroups of index 5 (in fact all such subgroups are not normal).

I guess the appropriate generalization (well known) would be this one:
Let $ p$ be the smallest prime dividing $ |G|$. If $ H$ is a subgroup of index $ p$, then it is normal.

The proof is the same as the one ZetaX has given.

But note that this is not a true generalisation: there are groups of even order not having a subgroup of index $ 2$.
And for the original problem, $ G$ needs not to be finite.

[quote="Gockel"]The next step is also true: If G is a group without subgroups of index 2 and 3, then a subgroup with index 4 is normal[/quote]
Note that this implies that no such groups exists, as the quotient group would have order $ 4$, thus a normal subgroup of order $ 2$. So the $ G$ would have a subgroup of index $ 2$ (by taking the preimage of the projection map).

[quote="ZetaX"]Let $ H$ have index $ 3$. Consider the set $ G/H$ of left cosets $ {aH}$. Now you get some homomorphism $ G \to S_3$ by $ G$ acting on $ G/H$ (by left multiplication). Now go on...[/quote]
Sorry, you mean : $ G\minus{}>S_3$, that means $ |G|\equal{}6$?? (If yes, can you please determine the elements of $ G$ ??)
or, you mean that $ G/H\minus{}>S_3$, that means $ |G/H|\equal{}6$ ?? (If yes, can you please determine the element of $ G/H$ please ? Thanks in advance.

The existence of a homomorphism says nothing (ok a bit, but not what you say) about the group order.


analytic geometry
geometry
rectangle
combinatorics unsolved
combinatorics

On a plane we have coordinate system. Let M be a set of points (x,y) where $x,y \in N$ and $x, y \leqslant 12$. Each point is blue, red or white. Prove, that there exists a rectangle with sides parallel to coordinate axes and and his vertex are same coloured.

Does $\mathbb N$ contain $0$? If it does, I can solve it :). If it doesn't, we'll have to wait for someone to refine the following:

On each column, there are at least $5$ points with the same color. Call such a color the characteristic color of the column. Since there are $13$ columns, we have $5$ columns witht the same characteristic color (say it's white).

From here it's easy: we have to show that there are two columns with two common positions among the five white ones we choose on each of the five columns. I didn't find a smart argument so I actually tried a few cases (about three), but it works out just fine (try it :)).

[quote="grobber"]I didn't find a smart argument so I actually tried a few cases (about three), but it works out just fine (try it :)).[/quote]

It may be not so smart but assume that there are no two columns with a same white pair :
Then, among the rows, the first columns uses at least 5 of them. The second column uses at least 4 more (at least 5 white points and not more than one in an already chosen row). The third column uses at least 3 non already chosen rows. The fourth uses at least 2 non already chosen row. Without considering the last column, we already need at least 5+4+3+2 = 14 pairwise distinct rows, which is a bit too much.

Pierre.