USAMTS
geometry
geometric transformation
reflection
inequalities
trapezoid
power of a point

[color=cyan]Okay, problems of this type (geometric construction) are relatively unpopular nowadays, although the Etvos (did I write that correctly?) contests from the earlier parts of the 20th century were full of them.  But I thought I'd give some, just for fun:

1) Given a unit length, construct the length which is the square of any given length.

2) Given a unit length, construct te length which is the square root of any given length.

3) Circle inversion is a transformation that identifies every point in the plane with another point, with one exception, according to the following rule:  given a circle with center O and radius r, a point P is transformed to the point P' according to the rules OP * OP' = r^2 and OPP' or OP'P.  Give a construction for inversion of a point inside or outside a given circle.

That's all I have for now.  I'll think of more later.[/color]

Aren't these constructions well-known? I think Mascheroni constructions are really interesting personally. Here's a (hard) example: given a circle, construct the center of the circle with just a compass. That was the first Mascheroni construction I ever saw. It was in the advanced section of a prize book I got from the USAMTS a few years ago.

[color=cyan]"Well-known" is relative.  I would venture to guess that most of the other people on this website are not familiar with these -- I just learned the first 2 very recently.[/color]

I have no clue for any of these...I guess ComplexZeta's could be:[hide]Set a compass to a particular length less than the diameter of the circle. Choose any point on the circle and write a half circle with the point on the circle as its center. Repeat this until you get a smaller circle inside the original circle. Continue these steps until you get an extremely small circle or a dot. That is the center.[/hide]

Of course, this would take a long time and it won't be that accurate, but it would be better than guessing, I guess.

It has to be exact, and it must be done in a finite number of steps.

JBL wrote:Okay, problems of this type (geometric construction) are relatively unpopular nowadays, although the Etvos (did I write that correctly?) contests from the earlier parts of the 20th century were full of them.  But I thought I'd give some, just for fun:

1) Given a unit length, construct the length which is the square of any given length.

2) Given a unit length, construct te length which is the square root of any given length.

3) Circle inversion is a transformation that identifies every point in the plane with another point, with one exception, according to the following rule:  given a circle with center O and radius r, a point P is transformed to the point P' according to the rules OP * OP' = r^2 and OPP' or OP'P.  Give a construction for inversion of a point inside or outside a given circle.

That's all I have for now.  I'll think of more later.

These are fairly well-known. Hint for the first two:[hide]think power of a point is the way I read it[/hide]. Um, if you use the inversive plane, then there is no exception to the circle inversion rule. (Inversive plane for those who don't know if the normal plane, but with a point at infinity.)

I'm really having a difficult time...more hints would be greatly appreciated.
*sigh*

ok...[hide]think power of a point inside a circle, when two of the lengths are equal.....meaning when you multiply them....you get a square....[/hide]

ah, I see what you mean, but technically, if you reflect the semi-circle over it's diameter, it's the same as mine, no?

It could be, depending on what you plan to do with yours.

[quote="ComplexZeta"]Hm. I guess that's one way of doing it. I was thinking of using a semicircle and the fact that sqrt(a) is the geometric mean of 1 and a. See if you can finish the solution from there.[/quote]

Hmm, reminds me of trapezoids....

Ah yes, that reminds me. If you have [i]An Introduction to Inequalities[/i] from the Anneli Lax New Mathematics Library series, there's a great diagram of a trapezoid with the root-mean-square, arithmetic mean, geometric mean, and harmonic mean. That's probably the best part of the book though.

[quote]Ah yes, that reminds me. If you have An Introduction to Inequalities from the Anneli Lax New Mathematics Library series, there's a great diagram of a trapezoid with the root-mean-square, arithmetic mean, geometric mean, and harmonic mean.[/quote]

Waiit...what do those inequalities have to do with trapezoids?

Suppose you take a trapezoid with bases a and b (with a>b). Then if you draw a segment parallel to the bases such that the two parts have equal areas, then that line is the RMS of a and b. If the segment is equidistant from the bases, it's the AM. If the two pieces are similar, then it's the GM. If it's at the intersection of the diagonals, then it's the HM.

And is this for any trapezoid, or only isosoceles?

[color=cyan]Think about whether or not that would be relevant . . . if you slide the bases around, should any of those lengths change at all?

I just went and posted some easier construction problems on the Intermediate Problems forum.[/color]


Find al integer solutions of
$2^x=3^y+5$.

I wonder why some problems are posted at least twice from different people at quite the same time...Where did you get this one ?

Have a look here :
http://www.mathlinks.ro/Forum/viewtopic.php?t=27477

Pierre.

Oh.. I'm sorry
I have it from a list of problems and it says that it is from USA 2004.. but I didn't find it on the forum (keyword:"USA 2004") so.. I posted it :(
thx pbornsztein


logarithms
limit
calculus
calculus computations

Find
$ \sum_{p=1}^{\infty}\left( \sqrt[p]{p}-1 \right)^{4}$

Its interesting to note that, contrary to the series $ \sum_{n=1}^{+\infty}(\sqrt[n]{n}-1)$ (which diverges), the present series
$ \sum_{n=1}^{+\infty}(\sqrt[n]{n}-1)^{4}$ is convergent. To see this start by noting that $ \sum_{n=1}^{+\infty}\frac{(\ln n)^{4}}{n^{4}}$ converges, because, for some $ n$ big enough we have $ (\ln n)^{4}< n$, and so

$ \frac{(\ln n)^{4}}{n^{4}}< \frac{n}{n^{4}}= \frac{1}{n^{3}}$

and this last series converges (its a p-series with p > 1).

Now, $ (\sqrt[n]{n}-1)^{4}= (e^{\frac{\ln n}{n}}-1)^{4}$ and because $ \lim_{n\to\infty}\frac{\ln n}{n}= 0$ we have

$ \lim_{n\to\infty}\frac{(e^{\frac{\ln n}{n}}-1)^{4}}{\frac{(\ln n)^{4}}{n^{4}}}= \lim_{n\to\infty}\left(\frac{\frac{\ln n}{n}}{\frac{\ln n}{n}}\right)^{4}= 1$

and so our series also converges.


Here is something that looks intuitively trivial, but which I couldn't prove. Let $a \in (0,1)$ be arbitrary. Does there exist a positive integer $m \in \mathbb{N}$ such that $m^a \notin \mathbb{Q}$? (Of course the answer should be 'yes,' but ...  :D ) [b]Any[/b] ideas?  :?

What is Putnam problem you talking about?

And how is it possible that both $2^a$ and $3^a$ are rational?

Sorry if my question are very stupid. :blush:

The Putnam problem (A6 from 1971) is this: if $x$ is a real number such that $1^x,2^x,3^x,\ldots$ are all positive integers, show that $x \in \mathbb{N}_0$.

To me it too seems very strange that $2^a$ and $3^a$ can be both rational, but here is a quote from jhaussmann5 from [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=Putnam+1971&t=24042]this[/url] thread: [i]I'm trying to recall a result that says something like if $2^x, 3^x$ and $5^x$ are integers, then $x$ is a positive integer, but $2^x$ and $3^x$ are not enough. Does anyone else remember this result?[/i]

Vess, I don't think that it is possible to have $2^x, 3^x$ integers and $x$ not integer. At least, I managed to prove that the four exponentials conjecture implies that $x$ is also integer. And I really doubt that someone has found a counter-example for the 4 exponentials theorem. ;)

You're right, it is an immediate consequence of the 4-exponentials conjecture that at least one of $2^x$ and $3^x$ is transcendental provided that $x \notin \mathbb{Q}$ (and the case $x \in \mathbb{Q}$ is, of course, trivial) ... I wonder if this special case has been proven (maybe that's what jhaussmann5 meant -- this may well be an open problem) and, if so, whether there is a relatively simple proof?

Anyway, the implication $2^x, 3^x, 5^x \in \mathbb{Q} \,  \Rightarrow \, x \in \mathbb{N}_0$ follows from the 6-exponential theorem, a weaker version of the 4-exponential conjecture, proven by Siegel, Schneider, Lang, and Ramachandra. So, this is known for sure to be true.

Anyway, I don't even have a hint about the ideas in the proof of the 6-exponential theorem (which is probably an inhuman proof  :lol: ), so I would really be interested in seeing a proof of the weaker assertion that for every $a \in (0,1)$ there is an $m \in \mathbb{N}$ with $m^a \notin \mathbb{Q}$. I have to admit that I don't have any ideas at the moment of how to handle this ...


Write two specific survey questions that you would ask voters in the next senatorial election in your state. Choose the type of question and response (yes/no, scale 1 to 5, numerical responses, etc{ that would be most appropriate for the issues involed. What relationships would be expected when the responses are analyzed?

Can someone give me an example please because I dont understand what they mean by relationships. It just seems really vague.

This doesn't belong in HSB as it isn't math-related. Try posting in a more appropriate forum to get an answer.


geometry
trapezoid
AMC
AIME
angle bisector

Here's the problem:

$ 65.$ Side $ AB$ of square $ ABCD$ is also the hypotenuse of right triangle $ ABP$($ ABP$ lies outside $ ABCD$). Prove that the angle bisector of $ \angle APB$ bisects the area of $ ABCD$.

I'll hide the solution if anyone wants to try the problem. I have a feeling its pretty easy with cyclic quadrilaterals (also, its at the beginning of the chapter problems).

[hide="My solution"]

Call the points of intersection of the bisector of $ \angle APB$ with AB point $ Q$ and with CD point $ R$.

Take triangle $ APB$ and translate it along the angle bisector of $ \angle APB$, to triangle $ A'P'B'$ Next, flip the triangle over its altitude to base $ A'B'$ to triangle $ A''P''B''$, and it is observed that the original angle bisector remains the bisector of $ \angle A'''P'''B'''$ Also, $ A'''B''' \equal{} AB \equal{} CD$

It is easily proven that $ PQB \equiv P'''B'''R$ and $ PAQ \equiv P'''A'''R$ This implies that the bases of trapezoids are equal, and since their heights are equal, their areas are equal and $ ABCD$ is equally divided into two parts.

Uh, here $ \equiv$ means "congruent to." Sorry about that.

[/hide]

Since my solution doesn't make use of the fact that $ \angle APB$ is right, it holds for any $ \angle APB$. What I'm wondering is:

1. Is this a valid method (by method, I mean the translating and flipping part)?
2. Is this a useful method (particularly for AIME?)

Thanks for any help!

Definitely not correct, as the result is not true for general angle (see what happens when the angle $ \angle APB$ approaches 0.

Anyways, Take the center of the square to be $ K$. Show $ APBK$ is concyclic, so then the angle bisector passes through $ K$, and we are done since it is easy to see that any line that passes through the center of the square bisects the area of the square by symmetry.


geometry
perimeter
trigonometry
inequalities
triangle inequality
inequalities unsolved

Let $a,b,c$ be the sides of a triangle who's perimeter doesn't excede $2\pi$. Prove that $\sin a, \sin b, \sin c$ can also be the side lengths of a triangle.

It is quite easy problem. Why do you say "strange problem"?

Because usually one doesn't take the sines of the sides of a triangle... but times change, everything is possible on olympiads nowadays  :)

  dg

I really wonder why the staff put this problem in the final TST AS PROBLEM 6.

maybe they have a quick one row solution ? :)

isn't this problem obvious? and i dont understand why the perimeter has to be less than 2*pi? since sin a = a / 2R, etc, if a+b>c, etc, then a/2R+b/2R>c/2R, i.e. sin a + sin b > sin c? or is there something very wrong with that?

Well,suppose that:
[tex] sina <sinb<sinc [/tex]
If $ c> \pi $ then $ a+b+c>2c>2 \pi $
So $ c \leq \pi $.
If [tex] \frac{a+b}{2}< \frac{\pi}{2}\ [/tex] then [tex] sin\frac{a+b}{2}\ > sin\frac{c}{2} [/tex]
If [tex] \frac{a+b}{2}\geq \frac{\pi}{2} [/tex] then [tex] sin\frac{a+b}{2}>sin\frac{2\pi-c}{2}=sin\frac{c}{2}\ [/tex]
So
[tex] sina+sinb=2sin\frac{a+b}{2}\ cos \frac{a-b}{2}\ > 2sin\frac{c}{2}\ cos \frac {c}{2} \ =sinc [/tex] :)

Well, My solution is same as above :D  :D
P.S. use [tex]\sin[/tex] for trigonometric functions. (\sin)

My solution, because I don't believe in using trigonometry to solve trigonometric inequalities ;-):
wlog $a \le b \le c$.
First, $\sin a + \sin b > \sin b$ because $a$ is in (0, pi].
Also, because sin is concave on [0, pi], $\sin a + \sin b > \sin (a+b) + \sin 0 = \sin (a+b)$, by majorization.

We also know that $b \le c < a + b$, and $c < pi$, as Anh Cuong noted.

Case 1: a + b is in the interval [0, pi/2]
Then $\sin a + \sin b > \sin(a + b) > \sin c$ because sin is increasing on [0, pi/2] and pi/2 > a + b > c.
Case 2: b is in the interval [pi/2, pi].
Then $\sin a + \sin b > \sin b \ge \sin c$ because $b \le c$ and b, c are both in [pi/2, pi] where sin is decreasing.
Case 2: $b < \pi/2 < a + b$.  Then $b > \pi/ 2 - a$, so because sin is increasing on [0, pi/2], $\sin b > \sin (\pi /2 - a) = \cos a$.  So $\sin a + \sin b > \sin a + \cos a$.  But $\sin a + \cos a > 1$ by the triangle inequality on the triangle with sides $\sin a$, $\cos a$, $1$.  Henc $\sin a + \sin b > \sin a + \cos a > 1 \ge \sin c$.

(Edit):
Grrr.  I am stupid.  I didn't realize that a < b < c does not imply sin a < sin b < sin c.


Find all solutions in positive integers $ a, b, c$ to the equation:
\[ a! b!=a!+b!+c! \]

I think the problem has no solution.
Suppose $ a\leq b\Rightarrow a!b!=a!+a!(a+1)\ldots b+c!\Rightarrow a!|c!\Rightarrow a\leq c\Rightarrow$
$ \Rightarrow a!b!=a!+a!(a+1)\ldots b+a!(a+1)\ldots c\Rightarrow b!=1+(a+1)\ldots b+(a+1)\ldots c$.
If $ b\leq c$ then $ b!=1+(a+1)\ldots b+(a+1)\ldots b\cdot\ldots\cdot c\Rightarrow b|1\Rightarrow b=1$.
Then $ a=0$ or $ a=1\Rightarrow a!b!=1$ but $ a!+b!+c!=c!+2>1$
If $ b>c$ then $ c|1\Rightarrow c=1\Rightarrow a=0\Rightarrow b!=1+b!+1$, false.


Find all functions $f: \mathbb{Z}\to \mathbb{Z}$  such that for all $m\in \mathbb{Z}$: \[f(f(m))=m+1.\]

There are no such functions. Indeed, plugging $m=f(l)$ in our equation we get \[f(f(f(l)))=f(l+1)\] but on the other hand \[f(f(f(l)))=f(l)+1\]
and thus \[f(l+1)=f(l)+1\] which means that our function has form \[f(l)=l+k\] where $k\in N$.
Plugging this into the equation we obtain $2\cdot k=1$, which is contradiction.

So what you say is, $k=f(0)$, am I right?

My solution: You have f^2(m)=m+1; ( here f^2 denotes f(f()) ), so if you instead of m put f^2(m) you get
f(f^2(m))=f^2(m)+1
And then if you put for f^2(m) m+1 you get
f(m+1)=m+2 =>
f(x)=x+1 for every x e Z. End of proof.

Actually in 4 th line you have f(f^2(m))=f(m)+1 =>
f(x+1)=f(x)+1 and than if we put f(0)=z where z is fro
 Z we get f(x)=f(x-1) +1 we get z+f(m)=m+1=> 2z=1 => z=1/2 => contradiction.

[hide="Solution"]WLOG, let $f(0) = a$. It follows by the identity that $f(a)=1$, $f(1) = a+1$, and so on. In general, we have that $f(n) = a+n$, and $f(a+n) = n+1$ for nonnegative integers $n$. We have two cases. $a$ cannot equal $0$ as if $a=0$, we would have $f(f(0)) = f(0) = 0$ which contradicts the identity. If $a<0$, we take the value $n=-a$. This gives us $f(-a) = 0$, and $f(0) = 1-a$. However, we have that $f(0) = a$, and this gives us $1-a = a$ or $a=\dfrac{1}{2}$, which is not an integer. The other case is when $a>0$. In this case, we take $n = a$. This gives us $f(a) = 2a$, but we also have that $f(a)=f(f(0)) = 1$ by the identity. This gives us $2a = 1$ or $a=\dfrac{1}{2}$ which is also not an integer. Because all integer values which $f(0)$ could take on lead to contradictions, it follows that no such functions exist.[/hide]


geometry
3D geometry
pyramid
trigonometry

[quote="onlylove_math"]Let pyramid $ M.ABC$ and $ MA \equal{} MB \equal{} MC \equal{} 1, \angle BMC \equal{} x,\angle CMA \equal{} y,\angle AMB \equal{} z.$

Denote $ S$ is total area of top $ ( S \equal{} S_{MAB} \plus{} S_{MBC} \plus{} S_{MCA}))$. Prove that:

      $ 2\sqrt {3}S \le\ 9 \minus{} 2(cosx \plus{} cosy \plus{} cosz)$[/quote]

[hide]
$ S \equal{} \sum \frac {1}{2} \sin x$
(Repeated use of $ A \equal{} \frac {1}{2}ab \sin C$
Therefore it remains to show

$ \sqrt {3} ( \sum \sin x ) < \equal{} 9 \minus{} 2 ( \sum \cos x )$

Which can be rearrange to:

$ \sqrt {7} ( \sum \sin a ) < \equal{} 9$ (Use $ a\sin x \plus{} b \cos x \equal{} \sqrt {a^2 \plus{} b^2} \sin ( x \plus{} m)$

But $ 3 \sqrt {7} < \equal{} 9$ and we are done
[/hide]

[quote="SimonM"][quote="onlylove_math"]Let pyramid $ M.ABC$ and $ MA \equal{} MB \equal{} MC \equal{} 1, \angle BMC \equal{} x,\angle CMA \equal{} y,\angle AMB \equal{} z.$

Denote $ S$ is total area of top $ ( S \equal{} S_{MAB} \plus{} S_{MBC} \plus{} S_{MCA}))$. Prove that:

      $ 2\sqrt {3}S \le\ 9 \minus{} 2(cosx \plus{} cosy \plus{} cosz)$[/quote]

[hide]
$ S \equal{} \sum \frac {1}{2} \sin x$
(Repeated use of $ A \equal{} \frac {1}{2}ab \sin C$
Therefore it remains to show

$ \sqrt {3} ( \sum \sin x ) < \equal{} 9 \minus{} 2 ( \sum \cos x )$

Which can be rearrange to:

$ \sqrt {7} ( \sum \sin (x\plus{}m) ) < \equal{} 9$ (Use $ a\sin x \plus{} b \cos x \equal{} \sqrt {a^2 \plus{} b^2} \sin ( x \plus{} m)$

But $ 3 \sqrt {7} < \equal{} 9$ and we are done
[/hide][/quote]
Thank you , It 's nice :)


Prove that if a,b,c are greater than 3/10 and a+b+c=1, then
 2/3<= a*sqrt(a+bc)+b*sqrt(b+ca)+c*sqrt(c+ab)<3/4

the right hand side is really a weak inequality.
suppose 4/10>=a>=b>=c>3/10
hence( a-3/10)(a-4/10)<0 and therefore a^2<7/10 a -12/100
hence sum a \sqrt a+bc =1/2 sum 2a \sqrt a+bc <= 1/4sum 4a^2+a+bc
= sum a^2 +sum a/4 +1/4 sum bc < 34/100 +1/4+1/12  =0.673 < 3/4

Yes, you're right, but what about the other part of the inequality? Anyway, as I said, the problem is very simple.

Very nice inequality, Harazi. When do you post your solution?

Thank you very much.

P.S. If you don't want to post your solution, please a hint.

The left part is very easy and I will skip it. For the second part, let us observe that sqrt(a+bc)=sqrt((a+b)(a+c))>=a+sqrt(bc), so sum a*sqrt(a+bc)>= sum a^2+sum a*sqrt(bc)=1-2(ab+bc+ca)+sum a*sqrt(bc). So we must prove that 1/3-2(ab+bc+ca)+sum a*sqrt(bc)>=0, which can be written as sum (a-b)^2/3>=sum c*(sqrt(a)-sqrt(b))^2. We will prove that
(a-b)^2/3>=c*(sqrt(a)-sqrt(b))^2, which is the same as sqrt(a)+sqrt(b)>=sqrt(3c), or a+b+2sqrt(ab)>=3-3a-3b, and this one is true since a,b,c>=3/10.Summing up three inequalities of this type solves the problem.
   I must mention that I wrote "Very simple inequality" reffering to the left part, since the second part is in my opinion quite difficult.


function
probability
search
inequalities
calculus
induction
number theory

Prove that the phi-function satisfy the following property:

phi(1)/1 + phi(2)/2 + phi(3)/3 +... +phi(n)/n>= 7n/12

a guy in my school solved it... he said that it can be proved by noticing that phi(n)/n is the probability that a number k so that 1<=k<=n is relatively prime with n ....

[quote="alekk"]phi(n)/n is the probability that a number k so that 1<=k<=n is relatively prime with n ....[/quote]

that's the definition of phi(n)  :D

yeah that's it but if you notice that, you can prove that
sum(phi(k)/k,k=1..n)>=6*n/(pi^2) ...
think that the probability that 2 numbers a and b are relatively prime is 6/(Pi^2)

[quote="alekk"]sum(phi(k)/k,k=1..n)>=6*n/(pi^2) ...
think that the probability that 2 numbers a and b are relatively prime is 6/(Pi^2)[/quote]
also a well-known conclusion. :)

so how do you prove the pb ? i haven't seen a complete proof yet ..

isn't 6n / \pi 2 > 7n/12 ?!!! it seems so to me :?

Hi,

Sorry folks, but I don't get the trivial link between  \sum  phi(k)/k and the density of the primes...

Indeed here one is asked to proove an equality standing for all n. An equivalent in +oo (of a different quantity!) might be helpful but definitely isn't enough to proove the statement.

Cheers,

George

I concur with George...
but a quick computer search shows no number satisfying equality for n<=10000?

Well, inequalities involving  \emptyset and the n'th prime number and
\pi (n) and things like these are usually weak. Remember this: p_n <= 4^n, where p_n is the n'th prime. You can see how absurdly weak this is, yet, it's not trivial; it widely discussed on this forum, as I recall.

About this problem, I agree with George. I'd also like to see the detailed proof of that thing with 6n/Pi^2.

[quote="yifeic"]I was referring to the original problem, letting f[n]=\sum_{i=1}^n Phi[i]/i but inf({f[1],f[2],...}) seems to be > 7n/12 (at least no equality for n=1 through 10000), so this seems like weak inequality.[/quote]

phi(1)/1 + phi(2)/2 + phi(3)/3 +... +phi(n)/n>= 7n/12

...

>= means  \geq

I can give some proof for that thing about 6n/ \pi  2.

The total number of pairs (a,b) of numbers  \leq n is n 2.
For a given prime, p, the number of pairs (a,b) with a,b \leq n and p|a, p|b is n 2/p 2 .
For 2 distinct primes p,q the number of pairs (a,b) with a,b \leq n and pq|a, pq|b is (n/pq) 2.

Apply inclusion-exclusion principle to obtain the number of pairs of coprime numbers (a,b) from 1..n:
n 2(1- sum{p prime} 1/p 2 +sum{p<>q primes} 1/(pq) 2-...+(-1) ksum{p_1,...,p_k distinct primes  \leq n}1/(p_1p_2...p_k) 2+...).

Notice that the last sum is S=n 2(1-1/p1 2)*...*(1-1/pm 2) where p_1<...<p_m are all the primes  \leq n.
We know that 1-1/k=1/(1+1/k+1/k 2+1/k 3+...)
So S=n 2/(1+1/p_1+1/p_1 2+...)(1+1/p_2+...)*...={ :D the nice part :D } n 2/(1+1/2 2+1/3 2 +1/4 2 +...).
It is well known that  \sum 1/k 2  tends to  \pi  2/6, so the probability (remember that we divide by n 2 - the total number of cases) is 6/ \pi  2.

Now the question is how to link this to the problem.

So does this mean that inequality holds with 6n/ \pi  2 for all sufficiently large n's? Or not? I'm confused.. :D

I should after someone links that talk about probabilities to the problem.

THis problem it's easy except case $n=2^k$ but have a little complication

Solution: for $n<6$ the calculus is simple, and it's clear that for every prime $p>3$ then $1/2+(1-1/p)\ge 7/6$, so by induction on $n$, if $n=1$ holds, and if it's true for $n$ let's see thast holds for $n+1$. Now, we need to consider two cases:

Case 1. $n+1 \ne 2^k$ then if $n+1=p_1^{\alpha_1}\cdots p_k^{\alpha_k}$ with $p_1<p_2<p_k$ primes and $k\ge 1$ [obviously if $k=1$ then $p_1>2$]  then

\[\frac{\varphi(n+1)} {n+1} \ge (1-\frac 1 {p_1})^{\alpha_1}>\frac 7 {12} \ \Rightarrow \ \sum_{k=1}^{n+1} \frac{\varphi(k)}k\ge \frac {7n} {12} + \frac 7 {12}=\frac {7(n+1)}{12}. \]

Case 2. $n+1=2^k$, then as $n>5$ (prime) then $\frac{\varphi(n+1)} {n+1}=\frac 1 2$ but by before inequality we take the last prime before $n+1$, namely $q_n>3$, then $\frac{\varphi(q_n)} {q_n}-\frac 7{12}=A_n$ and then
\[
\sum_{k=1}^{n+1} \frac{\varphi(k)}k\ge  \frac {7n} {12}+A_n+\frac 1 2\ge  \frac {7(n+1)}{12}.
\]
[Here I have used the Erdos Theorem that exists a new prime between $2^n$ and $2^{n+1}$].

I hope you like the simple proof.

Best regards.
RX the Crazy Math

Well, is FALSE, OOOoooppss, sorry! I made, at least,  a mistake. But is not my speciality  :blush: .


Answer is 2.

I thinking of method. But I think in objective paper i will score.

yes that's right...actually it should eb asubejective question

1)

My ans was because (2,3) being adjacent primes. :D

Anyway I know the logic is because of the continuity of the function, the fact that it can take only rational values and because primes don't follow any pattern, will post a proper logic ASAP>

Sorry if i'm wrong but isn't it a constant function?? :huh:

It is.

Consider the space between any 2 prime no's. Since the function is continuous it curve however it wants but has to return back to 2 at the next prime. But the range is rational as well. So, there must be no value of x for which y can be irrational. So the function has to be constant.

yes that's just what the logic is ....if u put in $ \epsilon,\delta$ form u get the logic

[quote="madness"]Basically the lemma is:

Lemma: There are infinite irrational nos btwn 2 rational nos.  :D

I like the word lemma after seeing Peter Scholes use it  :D[/quote]
everyoine does...
i used it in RMO,INMO have been using it always

Can you give a proper proof the kind that would be accepted in RMO.

Which basic definitions are u talking about??

the definitions of limits and continuity

The question may be posed in a better fashion perhaps....The domain could well be just positive reals instead of the entire real line.
Sathej

it's actually a nice tricky question...just a fun


How do you skip a large block of space on the page? For example, I want to describe a picture then skip a couple of lines so that I can  draw in the picture (with a pencil) later.

Use \vspace{5in} or whatever amount you want. You can use cm or mm for the units if you wish.


algebra
polynomial
group theory
abstract algebra
Galois Theory
superior algebra
superior algebra unsolved

Let $ f(x)\equal{}x^3\minus{}2$ be defined over $ {\bf Q}$.  Find the Galois group of its splitting field extension.

A quick proof: Notice that the order of the Galois group for some degree $ n$ polynomial is always divisible by $ n$, because the splitting field must also be an extension of $ \mathbb{Q}(\alpha)$ for some root $ \alpha$.  The Galois group must also be a subgroup of $ S_n$, since it is a permutation group on the roots. (these two facts are very standard Galois theory)

Now, since the degree is at least 3 and the group is a subgroup of $ S_3$, which has 6 elements, there are only two possibilities for the group: $ C_3$ or $ S_3\equal{}D_3$.  But we can eliminate one by looking at $ \mathbb{Q}(\sqrt[3]{2})$.  This extension is degree 3 but only contains one root, since the other two are complex.  So the order must be greater than 3, and the result follows.

(you could also observe that since complex conjugation is a degree 2 automorphism that swaps two of the roots of $ f$, the Galois group must contain an element of order 2, hence have even order)


Euler
number theory
prime numbers
algebra proposed
algebra

Find [b]simple proofs[/b] (no routine) of  the identities (between real numbers) :
  \[ (*)\; \; \;\; \; \displaystyle (a^2+b^2)(A^2+B^2)= \displaystyle (aA+bB)^2 +(aB -bA)^2 \; ,  \]
\[ (**)\; \; \; \;\; \; \begin{array}{c} \displaystyle (a^2+b^2+c^2+d^2)(A^2+B^2+C^2+D^2) =\\ = \displaystyle (aA+bB+cC+dD)^2 + \displaystyle (aB-bA-cD+dC)^2 +\\ + \displaystyle (aC+bD-cA-dB)^2 + \displaystyle (aD-bC+cB-dA)^2 \; . \end{array}  \]
Applications ?

For the first let we can see that the left side is:

$N(a+bi)N(A+Bi)$ which equals to $N(aA-bB+(Ab+Ba)i)$ hence the result. Analogically for the second using norm of quaternions.

And as for applications: I've seen first one to be used in proving that if two numbers can be written as the sum of two squares then their product too. The second one is used in a proof of Four Squares Theorem to reduce it to the case when number is a prime.

The second one can also be used to show that prime numbers are sums of four squares.
Norms of number-fields and similar ones are all nice :)


inequalities
integration
real analysis
real analysis solved

Let $f\in C^\infty(R,C)$. For every $k$ in $[0..n]$ we take $M_k=\sup_{x\in R}| f^{(k)}(x)|$.We suppose that $M_0$ and $M_n$ are finite.Show that for every $k$: $M_k$ is finite and $M_k \le 2^{\frac{k(n-k)}{2}}M_0 ^{1-\frac{k}{n}} M_n^{\frac{k}{n}}$.

My lectures tell me that it is Landau-Hadamare inequality.
I will prove inequality for $n=2$, i.e. $M_1\leq \sqrt{2M_0M_2}$.

We have for all $h$:
\[f'(x)=\frac{1}{2h}\int_{-h}^{h}(f'(x)-f'(x+t))dt+\frac{f(x+h)-f(x-h)}{2h}.\]
Thus
\[|f'(x)|\leq \frac{1}{2h}M_2\int_{-h}^{h}|t|dt+\frac{M_0}{h}=\frac{h}{2}M_2+\frac{1}{h}M_0.\]
We take $h=\sqrt{\frac{2M_0}{M_2}}$, so we are done.

You should read $M_k=\sup_{x\in R}| f^{(k)}(x)|$

And I was hoping that I posted at last a nice problem without forgeting something.

[quote="Myth"]
\[f'(x)=\frac{1}{2h}\int_{-h}^{h}(f'(x)-f'(x+t))dt+\frac{f(x+h)-f(x-h)}{2h}.\]
Thus
\[|f'(x)|\leq \frac{1}{2h}M_2\int_{-h}^{h}|t|dt+\frac{M_0}{h}=\frac{h}{2}M_2+\frac{1}{h}M_0.\]
[/quote]

I'm sorry but I find this problem interesting and I don't understand your answer, could you explain me (one or two lines intermediates from equality to inequality) ?
thanks

I just used Lagrange formula $g(a)-g(b)=g'(c)(a-b)$ and simple estimates using module.


I think the IMO is happening these few days.  I was wondering why Tiankai didn't go.  I've heard some say that he prefered to go to RSI.  If any of you have any imputs on the 2003, I've love to know about it.

-interesting_move

supposedly he chose to go to RSI. you can read more about it on the mathcounts forum. actually twas eve who put it on the mathcounts forum.

Maybe im mistaken, but it doesnt seem as if there is a post by eve on the mathcounts forum.... DId u mean a different one? If so, please link this other forum.

hmm...ill just quote it.
basically the question arose about why Tiankai Liu wasnt part of the IMO team. then Eve answered:[quote]The people I know who've skipped MOP (thereby giving up a potential IMO team spot) and opted to attend RSI instead all feel they made the right decision. One went on to win the Siemens Westinghouse competition (which comes with a $100,000 scholarship) and another came in third in Intel ($50,000, I believe) with their RSI research projects. RSI is also an incredible program that I doubt anyone would regret experiencing, even if they didn't go on to do well in science competitions.

Having four IMO gold medals (as opposed to three) is certainly a nice record to hold... but really, Tiankai's already proven that he's one of the best in the world. I didn't understand it at first, but now I definitely can see why he'd go to RSI after having already been successful at two IMOs.[/quote]

hope that helped.

Sincerely
High Inquisitor Syntax Error

The first contest day was/will be today, anyone who has got hold of the problems, results etc?

They certainly haven't been graded yet. I imagine the questions and results will be available in a week or slightly less, since the 19th is the last day on the schedule at [url]http://www.imo2003.com[/url]. I want to tackle the problems too... (waiting not so patiently...)

wow, complexzeta, 7 more posts till you get 200. yah for complex.

Sincerely,
High Inquisitor Syntax Error (self-appointed)

yes, and as of right now (198 posts), you only need:
116 posts to be cool
2944 posts to be pretty cool
31218 posts to be really cool
313961 posts to be way up there

-AND-

!!!!!!!!!

3141395 posts to be the utmost pinnacle of cool.

Unless you like e better than Pi, in which case, you descend into the pits of wierdness as you post more. jk

Actually I have a lot more posts left before I reach Skewes' Number (10^10^10^34), which is the second-largest number ever used in a proof. It was an upper bound for the first time pi(x)>Li(x). pi(x) is the number of primes not exceeding x, and Li(x)=integral from 2 to x 1/ln(t) dt. The bound was later reduced to 1.39822*10^316 or so (from [u]Gamma[/u], by Julian Havil).

you misspelled weird mystic terminator, i dont care though, careless mistake. i just wanted to post somat.

Yes, Skewes' Number is quite large. At that time (1930s I think), it was the largest number ever used in a proof (infinity doesn't count). Graham's Number, which is too large to be expressible in standard (exponential) notation, is larger, but I think Graham's paper wasn't published until the 1980s or 1990s. Graham's Number is also an upper bound on something (I believe in Ramsey Theory). Interestingly, it is generally thought that the number Graham was bounding is actually six, so his bound was very slightly too large, but sometimes it is convenient to prove that something actually exists rather than find out its exact value. A professor once told me about the existance of a certain class of numbers (that is, they're noninfinite), and I found a bound that gave me about 10 million for a number that should have been around 35. That was amusing.

Well, what Mathworld says is that he proved that his number was an upper bound, and also proved that 6 was a lower bound.  However, that lower bound has since been raised to 11, and there is "experimental evidence suggesting that it is actually even larger."  Haha.
It also gives a nice explanation to what that is the upper bound of.
mathworld.wolfram.com

Wow... I really ought to read the forum more often.  My name is all over the place!

And yes, Tiankai is at RSI right now.  Any messages for him?  I'll be there in a week.

I'm going to try to bug some of the IMO guys and see if they wouldn't mind e-mailing me the problems.  I'll post them as soon as I get them.

egads. i wish i were popular. lol. why shouldnt the High Inquisitor be popular. everybody loves me. i have no trouble making friends. *sobs*. no really.

I stuck the IMO problems online here: [url]http://web.mit.edu/drucker/Public/IMO_2003.html[/url]

Enjoy!

Sweet, glorious problems!

Unfortunately, I haven't the faintest idea how to solve them.  That's why I'm here at artofproblemsolving.com.  To get better!

Thanks Eve!

(Cross posted to Other Contests)

There was a small (but deadly) typo in problem 3.  The sqrt(3/2) should have been a sqrt(3)/2.  It's since been corrected.  I hope I didn't cause any unnecessary confusion.


For $ a,b,c$ be positive real number, show that
\[ \frac1{a\sqrt{2(a^2\plus{}bc)}}\plus{}\frac1{b\sqrt{2(b^2\plus{}ca)}}\plus{}\frac1{c\sqrt{2(c^2\plus{}ab)}}\ge\frac9{2(ab\plus{}bc\plus{}ca)}\]

It is enough to check that
\[ \sum_{cyc}\frac{b\plus{}c}{(a\plus{}b)(a\plus{}c)} \ge \frac{3\sqrt{3}}{2\sqrt{ab\plus{}bc\plus{}ca}}\]

which is true by using various methods including supplementary inequality such as Iran 1996.

[quote="dduclam"]For $ a,b,c$ be positive real number, show that
\[ \frac1{a\sqrt {2(a^2 \plus{} bc)}} \plus{} \frac1{b\sqrt {2(b^2 \plus{} ca)}} \plus{} \frac1{c\sqrt {2(c^2 \plus{} ab)}}\ge\frac9{2(ab \plus{} bc \plus{} ca)}\]
[/quote]
See here: http://canhang2007.wordpress.com/2009/11/02/inequality-32-d-d-lam/

Nice, Can. My solution using the lemma
\[ \sum\sqrt{\frac{2(b\plus{}c)}a}\ge\frac{27(a\plus{}b)(b\plus{}c)(c\plus{}a)}{4(a\plus{}b\plus{}c)(ab\plus{}bc\plus{}ca)}\]

[quote="great math"]It is enough to check that
\[ \sum_{cyc}\frac {b \plus{} c}{(a \plus{} b)(a \plus{} c)} \ge \frac {3\sqrt {3}}{2\sqrt {ab \plus{} bc \plus{} ca}}\]
which is true by using various methods including supplementary inequality such as Iran 1996.[/quote]

I would like to see your full solution, great math :)  Can you post it?

Here is the full proof ( sometimes I wonder what distinguishes a detailed proof and a full one)

We have

\[ \frac{1}{a\sqrt{2(a^{2}+bc)}}=\frac{\sqrt{b+c}}{\sqrt{2a}.\sqrt{(ab+ac)(a^2+bc)}} \ge \frac{\sqrt{2(b+c)}}{\sqrt{a}.(a+b)(a+c)}\]

Hence, it suffices to show that

\[ \sum_{cyc}\frac{\sqrt{2}(b+c)}{\sqrt{a(b+c)}.(a+b)(a+c)} \ge \frac{9}{2(ab+bc+ca)}\]

Now, WLOG, we may assume that $ a \ge b \ge c$ and thus,

\[ \{ \begin{array} {cc}
\frac{b+c}{(a+c)(a+b)} \le \frac{a+c}{(b+c)(b+a)} \le \frac{b+a}{(c+a)(c+b)} \\ \\
\frac{\sqrt{2}}{\sqrt{a(b+c)}} \le \frac{\sqrt{2}}{\sqrt{b(a+c)}} \le \frac{\sqrt{2}}{\sqrt{c(b+a)}}
\end{array}\]

Again, using Chebyshev inequality for 2 monotonic sequences as shown above, we complete our proof.

[quote="manlio"][quote="dduclam"]Nice, Can. My solution using the lemma
\[ \sum\sqrt {\frac {2(b \plus{} c)}a}\ge\frac {27(a \plus{} b)(b \plus{} c)(c \plus{} a)}{4(a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca)}\]
[/quote]

Dear dduclam, can you please post the solution of this lemma and the full solution?

Thank you very much.[/quote]
I think Ddlam's proof is the same as this: http://canhang2007.wordpress.com/2009/11/02/inequality-32-d-d-lam/ (see second proof). Of course, there will be some difference points in the way of writing proof (Because he is himself, and I am myself  :lol:)

[quote="manlio"][quote="dduclam"]Nice, Can. My solution using the lemma
\[ \sum\sqrt {\frac {2(b \plus{} c)}a}\ge\frac {27(a \plus{} b)(b \plus{} c)(c \plus{} a)}{4(a \plus{} b \plus{} c)(ab \plus{} bc \plus{} ca)}\]
[/quote]

Dear dduclam, can you please post the solution of this lemma and the full solution?

Thank you very much.[/quote]

My solution.

By AM-GM Inequality we have
\[ \frac{1}{a\sqrt{2(a^{2}\plus{}bc)}}\equal{}\frac{\sqrt{b\plus{}c}}{\sqrt{2a}.\sqrt{(ab\plus{}ac)(a^{2}\plus{}bc)}}\ge\frac{\sqrt{2(b\plus{}c)}}{\sqrt{a}.(a\plus{}b)(a\plus{}c)}\]
It suffices to show that
\[ \sum (b\plus{}c)\sqrt{\frac{2(b\plus{}c)}a}\ge\frac{9(a\plus{}b)(b\plus{}c)(c\plus{}a)}{2(ab\plus{}bc\plus{}ca)}\]
By Chebyselv Inequality we have
\[ \sum (b\plus{}c)\sqrt{\frac{2(b\plus{}c)}a}\ge\frac2{3}(a\plus{}b\plus{}c).\sum\sqrt{\frac{2(b\plus{}c)}a}\]
Hence, it suffices to show that
\[ \sum\sqrt{\frac{2(b\plus{}c)}{a}}\ge\frac{27(a\plus{}b)(b\plus{}c)(c\plus{}a)}{4(a\plus{}b\plus{}c)(ab\plus{}bc\plus{}ca)}\]
By Cauchy-Schwarz Inequality, we get
\[ \left(\sum\sqrt{\frac{2(b\plus{}c)}{a}}\right)^2\left(\sum a(b\plus{}c)^2\right)\ge16(a\plus{}b\plus{}c)^3\]
And by AM-GM Inequality,
\[ 27(a\plus{}b)(b\plus{}c)(c\plus{}a)\le8(a\plus{}b\plus{}c)^3\]
Finally, we need to show that
\[ \frac{16(a\plus{}b\plus{}c)^3}{\sum a(b\plus{}c)^2}\ge\frac{27.8(a\plus{}b\plus{}c)^3(a\plus{}b)(b\plus{}c)(c\plus{}a)}{16(a\plus{}b\plus{}c)^2(ab\plus{}bc\plus{}ca)^2}\]
or
\[ 32(a\plus{}b\plus{}c)^2(ab\plus{}bc\plus{}ca)^2\ge27(a\plus{}b)(b\plus{}c)(c\plus{}a)\left((a\plus{}b)(b\plus{}c)(c\plus{}a)\plus{}4abc\right)\]
or
\[ 5x^2\plus{}32y^2\ge44xy\]
where we setting $ x\equal{}(a\plus{}b)(b\plus{}c)(c\plus{}a)$, $ y\equal{}abc$ and using the equality $ (a\plus{}b\plus{}c)(ab\plus{}bc\plus{}ca)\equal{}x\plus{}y$
The last inequality is true because it equivalent $ (x\minus{}8y)(5x\minus{}4y)\ge0$, obviously.

[quote="dduclam"]For $ a,b,c$ be positive real number, show that
\[ \frac1{a\sqrt {2(a^2 \plus{} bc)}} \plus{} \frac1{b\sqrt {2(b^2 \plus{} ca)}} \plus{} \frac1{c\sqrt {2(c^2 \plus{} ab)}}\ge\frac9{2(ab \plus{} bc \plus{} ca)}\]
[/quote]

I think the following inequalities are also true for all $ a,b,c$ be sidelengths of a triangle:
\[ \frac1{a\sqrt {2(a^2 \plus{} bc)}} \plus{} \frac1{b\sqrt {2(b^2 \plus{} ca)}} \plus{} \frac1{c\sqrt {2(c^2 \plus{} ab)}}\ge\frac2{(a\plus{}b)^2}\plus{}\frac2{(b\plus{}c)^2}\plus{}\frac2{(c\plus{}a)^2}\]
and
\[ \frac1{a\sqrt {a^2 \plus{} 3bc}} \plus{} \frac1{b\sqrt {b^2 \plus{} 3ca}} \plus{} \frac1{c\sqrt {c^2 \plus{} 3ab}}\ge\frac2{(a\plus{}b)^2}\plus{}\frac2{(b\plus{}c)^2}\plus{}\frac2{(c\plus{}a)^2}\]


probability
geometry
3D geometry
rotation
geometric transformation
counting
distinguishability

[hide="Solution"]
Put the cube down on the surface first.

Four faces are blue with probability $ \frac {16}{81}$.
Four faces are red with probability $ \frac {1}{81}$.

Bottom and top face whatever color, so it's 1.

So, the answer is $ \frac {16 \plus{} 1}{81}*1 \equal{} \frac {17}{81}$

Answer: $ \frac {17}{81}$[/hide]

EDIT: Oops, 'can be placed'...

[quote="mathwizarddude"]Each face of a cube is painted either red or blue, with a probability of $ 1/3$ for red and $ 2/3$ for blue. The color of each face is determined independently. What is the probability that the painted cube can be placed on a horizontal surface so that the four vertical faces are all the same color?[/quote]

well, it all depends. if you paint it and you can rotate it and whatnot then fanfan is wrong. if you just put it flat down the same direction you painted it then fanfan is right (thats the more likely possibility)

GASP fanfan got a probability question right? sign of the apocalypse!! oh noez!

sorry just had to include that

I believe that FantasyLover's interpretation is wrong, as the wording implies rotations are possible. Successful cases: RRRRRR, RRRRRB, RRRRBB, RRBBBB, RBBBBB, BBBBBB, which have probabilities $ 1\left(\frac13\right)^6$, $ 6\left(\frac13\right)^5\left(\frac23\right)$, $ 3\left(\frac13\right)^4\left(\frac23\right)^2$, $ 3\left(\frac13\right)^2\left(\frac23\right)^4$, $ 6\left(\frac13\right)\left(\frac23\right)^5$, and $ 1\left(\frac23\right)^6$, respectively (assuming that rotations are distinguishable to weight it correctly), so adding, we get $ \boxed{\frac {329}{729}}$.

This was probably based off of [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1346258471&t=242640]AMC12 2004B #20[/url].

Yep, but can you solve it without resorting to case work?


n the game of darts you get 11 points for hitting the inner-most circle, 7 for the ring surrounding it, 5 for the next, and 2 if you miss. What is the least number of throw needed to earn exactly 50 points?
Generalize the method (Explain how you get the answer)

You get 2 if you miss?
I like this game :)

Well...it's kind of like the AMC :P  :lol:

[hide]TripleM, you can lose two points when you miss and solve the problem and still get the same answer. I'm sure you knew that though.[/hide]

Yup, but just in case I put that in spoiler because it is a hint... :wink:


function
IMO Shortlist
functional equation
algebra

The function $ F$ is defined on the set of nonnegative integers and takes nonnegative integer values satisfying the following conditions: for every $ n \geq 0,$

(i) $ F(4n) \equal{} F(2n) \plus{} F(n),$
(ii) $ F(4n \plus{} 2) \equal{} F(4n) \plus{} 1,$
(iii) $ F(2n \plus{} 1) \equal{} F(2n) \plus{} 1.$

Prove that for each positive integer $ m,$ the number of integers $ n$ with $ 0 \leq n < 2^m$ and $ F(4n) \equal{} F(3n)$ is $ F(2^{m \plus{} 1}).$

I get F(0)=0 from n=0 in i), then F(1)=1 from n=1 in iii). Then putting n=1 in i) and ii), I get F(4)=F(2)+F(1)=F(2)+F(1), and also F(2)=F(4)+1, implying F(2)=F(4), implying 0=1???

Another edit: [hide="Progress"]I find that if $ n \equal{} (d_id_{i \minus{} 1}\cdots d_1d_0)_2$, then $ F(n) \equal{} \sum d_if_i$, where $ f_1 \equal{} f_2 \equal{} 1, f_{i \plus{} 1} \equal{} f_i \plus{} f_{i \minus{} 1}$. This can be proven by induction easily. Then it remains to prove that there's $ f_{m \plus{} 2}$ integers $ n$ satisfying $ F(3n) \equal{} F(4n)$ and $ 2^m\leq n \leq 2^{m \plus{} 1}$.
Edit again again:
Looking for $ n$ such that $ F(3n) \equal{} F(2n) \plus{} F(n)$, we first prove that $ F(3n)\leq F(2n) \plus{} F(n)$ with equality iff the binary representation of $ n$ has no two adjacent 1's. This is because having two adjacent ones will make a carry-over, and what was expected to be $ 2f_k$ becomes $ f_{k \plus{} 1} \equal{} f_k \plus{} f_{k \minus{} 1} < 2f_k$. So then it becomes a counting problem, which is very trivial by recursion.[/hide]

I am not following the logic used to state that F(2)=F(4)+1.

I will try to get a proof for this problem in the next few days.

[hide="Hint"]
if $ n\equal{}a_k2^k\plus{}a_{k\minus{}1}2^{k\minus{}1}\plus{}...\plus{}a_12\plus{}a_0$

then $ F(n)\equal{}a_kF_{k\plus{}1}\plus{}a_{k\minus{}1}F_k\plus{}....\plus{}a_0F_1$

with $ (F_n)$ fibonacci sequence.

[/hide]

Well, I am having a heck of a time actually proving the statement to be true. I know what is going on in the problem, but can't actually develop a proof. Here is a tiny bit of what i have uncovered.

$ F(2^{m\plus{}1}) \equal{} F_{m\plus{}1}$

Where $ F_{n}$ follows the fibonacci sequence.

Also, putting all values of n such that $ F(4n) \equal{} F(3n),$ into binary creates an obvious pattern as follows.

0
1
10
100
101
1000
1001
1010
10000
10001
10010
10100
10101
100000
100001
100010
100100
100101
101000
101001
101010

in essence, it is creating every combination such that no two 1's touch each other within the numbers.

This is very strongly tied into the fibonacci sequence, so if you are going to attempt a proof, you will definitely have to use that.

I'm not going to be extremely detailed in this, only main ideas:

By induction we can show the following result: If $n = \sum_{i = 0}^\infty{\epsilon_i \cdot 2^i}$ ($\epsilon_i = 0, 1$) then $F(n) = \sum_{i = 0}^\infty{\epsilon_i \cdot f_i}$, where $f_0 = 1, f_1 = 1, f_{i+2} = f_{i+1}+f_i$, so the Fibonacci's shifted by 1.

Nextly, the following inequality holds, and equality only when $n$ has no 2 consecutive 1's in binary: $F(3n) \le F(4n)$. You can show this directly by computing the binary expression of $n, 3n, 4n$ and using the above formula.

So the number of number $n < 2^m$ (let's call it $u_m$) with no consecutive 1's, satisfy the recurrence $u_{i+2} = u_{i+1}+u_i$, where $u_1 = 1, u_2 = 2$, so $u_m = f_{m+1} = F(2^{m+1})$.

(8 year bump)
Does this work?

Let $n$ be good if $F(4n)=F(3n)$.

We claim that $F(2^n)=F_{n+1}$, where $F_n$ is the nth Fibonacci number, and $F(n) = F(n-2^m) + F(2^m)$ for all $n$ otherwise.

The proof of the first part is obvious by induction. For the second part, assume this holds until $2^m$. Then,

$F(4k+4) - F(4k) = F(2k+2) - F(2k) + F(k+1) - F(k)$

$F(4k + 4 - 2^m) - F(4k - 2^m) = F(2k+2-2^{m-1}) - F(2k - 2^{m-1}) + F(k + 1 - 2^{m-2}) - F(k - 2^{m-2})$

and by our induction hypothesis, we get that $F(4k) = F(4k - 2^m) + c$ for some constant $c$. Substituting $k=2^m$, we find that $c=F(2^m)$. Combining that with the fact that $F(4k+1)=F(4k+2)=F(4k)+1$ and $f(4k+3) = F(4k)+2$, this completes the proof for the second part.

Now, once again we use induction to prove the main problem statement. By our induction hypothesis, we know that $0\leq n < 2^{m-1}$ has $F(2^m)$ good numbers. We wish to show that $2^{m-1}\leq n < 2^m$ has $F(2^{m-1})$ good numbers. Suppose $k$ is a good number from $0 \leq n < 2^{m-1}$, then

$F(k+2^{m-1}) = F(k) + F(2^{m-1})$

$F(\frac 34(k+2^{m-1}) = F(\frac 34 k) + F(2^{m-2}) + F(2^{m-3})$

and since $F(2^{m-1}) = F(2^{m-2}) + F(2^{m-3})$ we are done.

[b]This is merely a sketch (it takes a lot more tedious explanations of cases in the part about consecutive ones), but it works. You can work out the details if you want.[/b]

Suppose $n$ can be respresented in base two as $(a_ka_{k-1}a_{k-2}\dots {a_0})_2.$ Let $F$ be the fibonacci sequence: $P_0=P_1=1$ and $P_n=P_{n-1}+P_{n-2}$ for $n\ge 2.$ We claim that \[F(n)=\sum_{i=0}^{k}{a_0P(0)}\]
First, note that from the first relation $F(0)=0$ and so we have $F(1)=1,F(2)=1,F(3)=2.$ Note that these all satisfy our claimed function.

Now, we proceed with induction:

[b][color=#0B5394]Base Case:[/color][/b] Obviously, $n=0,1,2,3$ satisfies.

[b][color=#0B5394]Induction Hypothesis:[/color][/b] Suppose the claim is true for all $1,2,\dots,4n-1$.

[b][color=#0B5394]Induction Step:[/color][/b] Now, look at $4n,4n+1,4n+2,4n+3:$

Note that $F(4n)=F(2n)+F(n).$ Note that $4n=(a_ka_{k-1}a_{k-2}\dots {a_0}00)_2.$ Then \[F(4n)=\sum_{i=0}^{k}{P(i+1)a_i}+F(4n)+\sum_{i=0}^{k}{P(i)a_i}=\sum_{i=0}^{k}{P(i+2)a_i}\] as desired. Now, the conclusion follows for $F(4n+1),F(4n+2),F(4n+3)$ very simply.

Now, we claim that for all $n$ such that its base two representation has no consecutive ones, $F(3n)=F(2n)+F(n).$

Given the function of multiplying a base two number by three, we note that if there are no consecutive ones, the result copies the input, but the digit to the left of each one there is another one. Therefore, $F(3n)$ can be expressed as the sum of $P(i)a_i$ of the input and $P(i+1)a_i$ of the input, so $F(3n)=F(n)+F(2n)$ as desired.

Note that for each one in the input, the maximum amount of contribution it can give to the output is just $P(i)a_i+P(i+1)a_i$ and any consecutive ones will deviate from this maximum. Thus, $F(3n)=F(n)+F(2n)$ iff there are no consecutive ones in the representation of $n.$ It's easy to count the number of these that work, the result follows.

Let $F_n$ be the Fibonacci numbers where $F_0=F_1=1$. Denote $\&$ as bitwise AND and $|$ as bitwise OR.

[b]Claim.[/b] Let $n$ be $(\dots a_2 a_1 a_0)_2$ in binary. Then $F(n)=F_0 a_0+F_1 a_1+F_2 a_2+\dots$
[b]Proof.[/b] Just verify the three properties.

[b]Claim.[/b] For all $a,b$, $F(a)+F(b)\ge F(a+b)$. Equality occurs when $a\&b=0$, or $a\&b=2$ and $(a|b)\&4=0$.
[b]Proof.[/b] Just check this with the previous claim.

Now note that $F(3n)=F(2n+n)\le F(2n)+F(n)=F(4n)$, so $n\&(2n)=0$(this implies that there is no $11$ substring in the binary representation of $n$), or $a\&b=2$(this implies that $n\equiv 3\pmod 4$) and $(a|b)\&4=0$(this is therefore impossible).

Therefore, $F(3n)=F(4n)$ if and only if there is no $11$ substring in the binary representation of $n$. Thus if the number of $n$ with exactly $m$ digits in binary is $a_m$, then $a_0=1$(where $n=0$), $a_1=1$, $a_2=1$, and $a_m=a_{m-2}+a_{m-3}+\dots+a_1$ for all $m\ge 3$. Thus $a_m=a_{m-2}+a_{m-1}$. The number of $n<2^m$ that works is then
\[
a_0+a_1+\dots+a_m
=1+F_0+F_1+F_2+\dots+F_{m-1}
=F_{m+1}
=F(2^{m+1}).\;\blacksquare
\]


Is there any method for finding perfect numbers, numbers that are equivalent to the sum of their distinct factors except for the number itself (e.g., 6 = 1 + 2+ 3)?

Does anyone know of any other mathematical intricacies related to perfect numbers?

[hide]Here is a formula for finding perfect numbers, it was developed by Euclid i think.

Given that k is a natural number, and (2^k -1) is prime, a perfect number can be defined as:

(2^(k-1))(2^k -1)

Edit - some other things that I have noticed about perfect numbers (meaning i am not sure if they always hold) is that they all seem to end in a 6 or a 28, and these are preceded by an odd digit (except for 6 and 28 of course). they can all be expressed as the sum of natural numbers beginning with one. (ex. 496= 1 + 2 + 3 + 4 + ... + 28 + 29 + 30 + 31) And to show a little more i think every perfec tnumber after 6 is the partial sum of the series 1^3 + 3^3 + 5^3 + 7^3 +... (ex. 28 = 1^3 + 3^3 and 496 = 1^3 + 3^3 + 5^3 + 7^3)[/hide]

I believe Euclid worked on this and came up with:

If, for some $k > 1$, $2^k - 1$ is prime then $2^{k-1}(2^k - 1)$ is a perfect number.

If a number is a Mersenne prime, that is say the number equals $M$.  This means \[M = 2^a - 1\] If M is a prime number then the Mth triangular number is a perfect number.  Since the formula for triangular numbers is T = n(n+1)/2 then if you plug M in for n you get P = (2^a - 1)(2^a)/2 or  P = 2^(a-1)(2^a - 1).

While this result seems weird plug in the first mersenne prime, 3 so a = 2, and notice you get 6.  Since a = 3 produces M = 7 which is also prime plugging in a = 3 yields P = 28.  This will continue for every M that is prime.

[quote="YSJeong"][quote="paladin8"]Yeah, the $2^p-1$ primes are called Mersenne Primes btw.[/quote]
yea... i remember this came up in the recent contest
.[/quote]

I noticed that you are from WA. Are you referring the the NT test at MAO state? And do you have any relation with xxreddevilzxx?

wow. you guys really DO have some connections around here.... yes i am related.. although i didnt go to the state contest, i got the test from Young. BTW, we are sibling and we DO have the same initials. except the middle initial.


calculus
integration
quadratics
algebra
calculus computations

Let $k$ be real constants.How many real roots can the following quadratic equation have? \[x^{2}=-2x+k+\int_{0}^{1}|t+k|\ dt.\]

the discriminant is

$D=1+k+\int^{1}_{0}|t+k|dt$

if $k\geq 0$ , then
$D=1+k+\int^{1}_{0}t+kdt\\=\frac{3}{2}+2k>0$

if $-1<k<0$ , then
$D=1+k+\int^{1}_{-k}t+kdt-\int^{-k}_{0}t+kdt\\=\frac{1}{2}(2k^{2}+4k+3)>0$

if $k\leq-1$ , then
$D=1+k-\int^{1}_{0}t+kdt\\=\frac{1}{2}>0$

So all cases yields two real roots for the quadratic equation .

[quote="kunny"]Thank you for your reply.Please correct your answer, for $k\leq-1.$ and

rewrite for the third case, $k\leq-1.$[/quote]

ok ,done  :D
Is it ok now ?


Find the equation of a circle which ahs as diameter the segment joining the points (3,-5) and (7,-3).

[hide]Finding the midpoint of the points we get $(5,-4)$ which is the center. Finding the distance between the center and one of the points given is the radius which is $\sqrt{5}$. Putting this into the circle equation we have $\boxed{(x-5)^2+(y+4)^2=5}$[/hide]

nice here's another
find the equation of a circle which passes through the origin, radius = 10, and the abscissa of the center is -6.

[hide]Use midpoint formula to find center $=(5,-4)$ then by distance formula, $r=\sqrt{5}$.
The equation for the circle is then $(x-5)^2+(y+4)^2=5$.[/hide]

[hide]The ordinate is $\pm8$, from 3-4-5 r-triangle, so the equation for the circle is
$(x+6)^2+(y\pm 8)^2=100$[/hide]

abscissa= x-coordinate
ordinate= y-coordinate

[hide]
Since the problem says that the circle goes through (0,0), the equation works for $x$ and $y$ when they are 0. The equation of the circle is $(0-(-6))^2+(0-y_0)^2=100$
$36+(y_0)^2=100$
$y_0=\pm8$
So, the equation is:
$(x+6)^2+(y\pm8)^2=100$[/hide]

I'm pretty sure that's above MathCounts level.


Prove that for any a,b,c positive reals we have:

1/(a^2+ab+b^2)+1/(b^2+bc+c^2)+1/(c^2+ca+a^2) >= 9/(a+b+c)^2.

Cauchy didnt work for me immediately, but straight-up multiplying it out isn't too bad. It turns into

[tex]\displaystyle \frac{1}{2}\sum_{sym}a^6+3 \sum_{sym}a^5 b
\ge
\frac{1}{2}\sum_{sym}a^3 b^3+ 3\sum_{sym}a^4 b^2[/tex]

and those of you who are familiar with Miurhead can at this point say "we're done." If you're not familiar with that beautiful theorem, I'll go ahead and prove the above inequality with AM-GM:

[tex]\displaystyle \frac{a^6+b^6}{2}\ge a^3b^3[/tex] directly from AM-GM, and adding symmetrically yields

[tex]\displaystyle \sum_{sym}a^6\ge\sum_{sym}a^3 b^3[/tex].         (*)

Again from AM-GM, we have

[tex]\displaystyle \frac{a^5 b+a^5 b+a^5 b+a b^5}{4}\ge a^4 b^2[/tex],

and adding this symmetrically yields

[tex]\displaystyle \sum_{sym}a^5b \ge \sum_{sym}a^4b^2[/tex].         (**)

These 2 inequalities (* and **) should make the desired inequality apparent.

[quote="Vialli"]Prove that for any $ a,b,c$ positive reals we have:

$ \frac {1}{b^{2} \plus{} bc \plus{} c^{2}} \plus{} \frac {1}{c^{2} \plus{} ca \plus{} a^{2}} \plus{} \frac {1}{a^{2} \plus{} ab \plus{} b^{2}}\ge\frac {9}{(a \plus{} b \plus{} c)^2}.$[/quote]Using Cauchy inequality,

$ \left(\sum{\frac {1}{b^2 \plus{} bc \plus{} c^2}}\right)\left[\sum{(4a \plus{} b \plus{} c)^2(b^2 \plus{} bc \plus{} c^2)}\right]$

$ \geq\left[\sum{(4a \plus{} b \plus{} c)}\right]^2 \equal{} 36\left(\sum{a}\right)^2,$

we must prove that

$ 4\left(\sum{a}\right)^4\geq \sum{(4a \plus{} b \plus{} c)^2(b^2 \plus{} bc \plus{} c^2)}.$

This comes down to

$ 4\left(\sum{a}\right)^4 \minus{} \sum{(4a \plus{} b \plus{} c)^2(b^2 \plus{} bc \plus{} c^2)}$

$ \equal{} 2\sum{a^4} \plus{} 5\sum{a^3(b \plus{} c)} \minus{} 12\sum{b^2c^2}$

$ \equal{} \sum{(b^4 \plus{} c^4 \minus{} 2b^2c^2)} \plus{} 5\sum{(b^3c \plus{} bc^3 \minus{} 2b^2c^2)}$

$ \equal{} \sum{(b \minus{} c)^2(b^2 \plus{} 7bc \plus{} c^2)}\geq 0.$

A stronger inequality due to Vasc

$ \frac {23}{8b^{2} \plus{} 7bc \plus{} 8c^{2}} \plus{} \frac {23}{8c^{2} \plus{} 7ca \plus{} 8a^{2}} \plus{} \frac {23}{8a^{2} \plus{} 7ab \plus{} 8b^{2}}\ge\frac {27}{(a \plus{} b \plus{} c)^2}$

see here : http://www.mathlinks.ro/viewtopic.php?t=28706

[quote="Ji Chen"]A stronger inequality due to Vasc

$ \frac {23}{8b^{2} \plus{} 7bc \plus{} 8c^{2}} \plus{} \frac {23}{8c^{2} \plus{} 7ca \plus{} 8a^{2}} \plus{} \frac {23}{8a^{2} \plus{} 7ab \plus{} 8b^{2}}\ge\frac {27}{(a \plus{} b \plus{} c)^2}.$[/quote][quote="zabelman"]Cauchy did not work for me immediately, but straight up multiplying it out isn't too bad. It turns into[/quote]$ 1472\sum{a^6} \plus{} 4232\sum{a^5(b \plus{} c)} \minus{} 5360\sum{a^4(b^2 \plus{} c^2)} \minus{} 297\sum{a^4bc} \minus{} 688\sum{b^3c^3}$

$ \plus{} 1125\sum{a^3(b^2c \plus{} bc^2)} \minus{} 1443a^2b^2c^2$

$ \equal{} 1472\left[\sum{a^6} \minus{} \sum{a^4(b^2 \plus{} c^2)} \minus{} \sum{a^4bc} \plus{} \sum{a^3(b^2c \plus{} bc^2)}\right]$

$ \plus{} 4232\sum{a^5(b \plus{} c)} \minus{} 3888\sum{a^4(b^2 \plus{} c^2)} \plus{} 1175\sum{a^4bc} \minus{} 688\sum{b^3c^3}$

$ \minus{} 347\sum{a^3(b^2c \plus{} bc^2)} \minus{} 1443a^2b^2c^2$

$ \equal{} 1472(a \plus{} b \plus{} c)\sum{a^3(a \minus{} b)(a \minus{} c)}$

$ \plus{} 3888\left[\sum{a^5(b \plus{} c)} \minus{} \sum{a^4(b^2 \plus{} c^2)}\right] \plus{} 344\left[\sum{a^5(b \plus{} c)} \minus{} 2\sum{b^3c^3}\right]$

$ \plus{} 347abc\left[2\sum{a^3} \minus{} \sum{a^2(b \plus{} c)}\right] \plus{} 481abc\left(\sum{a^3} \minus{} 3abc\right)\geq0,$

which is true by [url=http://www.mathlinks.ro/viewtopic.php?t=1166]Schur inequality[/url] and [url=http://www.mathlinks.ro/viewtopic.php?t=15896]Muirhead  Theorem[/url].

See also here : http://www.mathlinks.ro/viewtopic.php?t=29347


Given a bijection $ f: {\mathbb Q}\rightarrow{\mathbb Q}$ verify if there are two bijections $ \alpha$ and $ \beta$ of $ {\mathbb Q}$ such that $ f\equal{}\alpha\cdot\beta$.

for example $ \alpha \equal{}f, \beta\equal{}id$.
Exist non trivial solution. If $ f\equal{}id. \alpha \equal{}\beta \equal{}\minus{}id.$ Else exist $ x_0\equal{}x$, suth that $ f(x_0)\equal{}x_1\not \equal{} x_1$.
Let $ x_{n\plus{}1}\equal{}f(x_n),f^{\minus{}1}(x_n)\equal{}x_{n\minus{}1},n\in Z$. If it had finite period $ x_N\equal{}x_0$, then work
 $ \alpha (x_i)\equal{}x_{i\plus{}1},\alpha(x)\equal{}f(x),x\not \equal{} x_i,\beta (x_i)\equal{}x_{N\minus{}1\plus{}i},\beta (x)\equal{}x\not \equal{}x_i$.
If $ \{x_i\}$ infinite, then work $ \alpha (x_i)\equal{}x_{\minus{}i}\equal{}\beta (x_i)$.

@ Rust: I think you misunderstood something. By $ f\equal{}\alpha\cdot\beta$ I meant $ f\left(x\right)\equal{}\alpha\left(x\right)\cdot\beta\left(x\right)$ for all $ x\in{\mathbb Q}$ [b]and not[/b] $ f\equal{}\alpha\circ\beta$.


function
floor function
number theory unsolved
number theory

Prove that for any positive integer $n$:

a) ${\lfloor \frac{n}{3}\rfloor+\lfloor \frac{n+2}{6}\rfloor}+\lfloor \frac{n+4}{6}\rfloor=\lfloor \frac{n}{2}\rfloor+\lfloor \frac{n+3}{6}\rfloor$

b)${\lfloor \frac{n+1}{2}\rfloor+\lfloor \frac{n+2}{4}\rfloor}+\lfloor \frac{n+4}{8}\rfloor+\lfloor \frac{n+8}{16}\rfloor+... =n$

c)$\lfloor \sqrt{n}+\sqrt{n+1}+\sqrt{n+2}\rfloor = \lfloor \sqrt{9n+8}\rfloor$

a) Check it by hand, after recognizing you can admit $n = 0, 1, \ldots, 5$ wlog (just set $n = 6q+r$, where $q = \lfloor n/6 \rfloor$ and $r \in \overline{0, 5}$, and make the necessary computations).


algebra
polynomial
inequalities
inequalities proposed
111

Well, this problem can be done not really hard by use $ABC$ consequence:
"Give $x,y,z \in R^{+}$ , then $P(x,y,z)$ where $P$ is a symmetric polynomial whose degree is equal or less than five take minimum and maximum when $x=y$ or $z=0$".
Even a more powerful consequence is given:
"Give $x_{1},x_{2},...,x_{n}\in R^{+}$ , then $P(x_{1},x_{2},...,x_{n})$ where $P$ is a symmetric polynomial whose degree is equal or less than five take minimum and maximum at $(a,a..,a,b,b,..b,0,0,...,0,)$"  :)

[quote="Vasc"]Let $a,b,c$ be positive numbers such that $a+b+c=3$. Prove that

$3(a^{4}+b^{4}+c^{4})+a^{2}+b^{2}+c^{2}+6 \geq 6(a^{3}+b^{3}+c^{3})$.[/quote]
It is a good exercise for S.O.S. :lol:

[quote="Soarer"]We note that $12\sum (1-c)(a-b)^{2}\ge 0$[/quote]
It is not true (see $a=b=1.5$ and $c=0$).

It is equivalent to $\sum (a-1)^{2}(3a^{2}-2)\geq 0$ .
I have solution for this but it is ungly .Is there a nice way to end it ?

[quote="Ji Chen"] The inequality is equivalent to
$(y+z-2x)^{2}(y+z-8x)^{2}+3(y-z)^{2}(27(y-z)^{2}+58(y+z)^{2}-100x(y+z)+4x^{2}) \ge 0,$
which is clearly true for $x=\min\{x,y,z\}$.
[/quote]
Very nice result !

[quote="Ji Chen"][quote="Vasc"]Let $x,y,z$ be positive numbers such that $x+y+z=3$. Prove that
$3(x^{4}+y^{4}+z^{4})+x^{2}+y^{2}+z^{2}+6 \geq 6(x^{3}+y^{3}+z^{3})$.[/quote]
The inequality is equvalent to
$16\sum x^{4}-20\sum x^{3}(y+z)+9\sum y^{2}z^{2}+15\sum x^{2}yz\geq0$.
[/quote]
A generalization of this inequality is valid:

For any real numbers $u,v,x,y,z$, we have

$u^{2}\left(\sum{x^{4}}-\sum{y^{2}z^{2}}\right)+v^{2}\left(\sum{y^{2}z^{2}}-\sum{x^{2}yz}\right)\geq uv\left[\sum{x^{3}(y+z)}-2\sum{x^{2}yz}\right]$,

with equality if $x=v-u,y=z=u$.

Specially, let $u=4, v=5$, we obtain the original inequality.
[hide="Proof."]
$4\left(\sum{x^{4}}-\sum{y^{2}z^{2}}\right)\left(\sum{y^{2}z^{2}}-\sum{x^{2}yz}\right)-\left[\sum{x^{3}(y+z)}-2\sum{x^{2}yz}\right]^{2}$

$=3(y-z)^{2}(z-x)^{2}(x-y)^{2}(x+y+z)^{2}\geq0$.[/hide]

I used to [b]U.C.T[/b] and Vornicu Schur's inequality to solve it  :D
$\bigstar \sum (a-1)^{2}(3a^{2}-2)\ge 0$
It is mean:
$\bigstar \sum(a-b)(a-c)(4a^{2}+b^{2}+c^{2}-4)\ge 0$
Now WLOG $a\ge b\ge c$ and by Vornicu Schur inequality we are easy to prove above inequality :P

[quote="zaizai-hoang"][quote="arqady"][quote="zaizai-hoang"]
It is mean:
$\bigstar \sum(a-b)(a-c)(4a^{2}+b^{2}+c^{2}-4)\ge 0$
Now WLOG $a\ge b\ge c$ and by Vornicu Schur inequality we are easy to prove above inequality :P[/quote]
But why $S_{a},$ $S_{b}$ and $S_{c}$ are non-negative? :wink:  :P[/quote]
Dear Ảgardy  :D
If you use following lemma, you will easy to prove that $\bigstar \sum(a-b)(a-c)(4a^{2}+b^{2}+c^{2}-4)\ge 0$ :P
$\bigstar$[color=blue]Lemma[/color]
Let $a\ge b\ge c$ and $S_{a},S_{b},S_{c}\ge 0$.
Put $N=S_{a}(a-b)(a-c)+S_{b}(b-c)(b-a)+S_{c}(c-a)(c-b)$. Then
$\bigstar N\ge 0$ if $S_{a}\ge S_{b}$ or $S_{c}\ge S_{b}$

I sure you will easy to prove above lemma :P

Now, I will use above lemma to prove the first inequality.
WLOG we assume $a\ge b\ge c$ then $S_{a}\ge S_{b}\ge S_{c}$. Therefore:
$\bigstar S_{c}=4c^{2}+a^{2}+b^{2}-4\ge 4c^{2}+\frac{(a+b)^{2}}{2}-4\ge \frac{(3c-1)^{2}}{2}\ge 0$
It is true :D Then we have the first inequality  :P I am very keen on above technique. It help me solve many problem. Example: Romania TST 2006 :D If you are new to the technique, you should try more. I hope you will like it, it very interesting. :D[/quote]
Wow , i'll try it ... but i saw it , in some way , similar to Vornicu-Shur , but stronger

[quote="HTA"][/quote][quote="zaizai-hoang"][quote="arqady"][quote="zaizai-hoang"]
It is mean:
$\bigstar \sum(a-b)(a-c)(4a^{2}+b^{2}+c^{2}-4)\ge 0$
Now WLOG $a\ge b\ge c$ and by Vornicu Schur inequality we are easy to prove above inequality :P[/quote]
But why $S_{a},$ $S_{b}$ and $S_{c}$ are non-negative? :wink:  :P[/quote]
Dear Ảgardy  :D
If you use following lemma, you will easy to prove that $\bigstar \sum(a-b)(a-c)(4a^{2}+b^{2}+c^{2}-4)\ge 0$ :P
$\bigstar$[color=blue]Lemma[/color]
Let $a\ge b\ge c$ and $S_{a},S_{b},S_{c}\ge 0$.
Put $N=S_{a}(a-b)(a-c)+S_{b}(b-c)(b-a)+S_{c}(c-a)(c-b)$. Then
$\bigstar N\ge 0$ if $S_{a}\ge S_{b}$ or $S_{c}\ge S_{b}$

I sure you will easy to prove above lemma :P

Now, I will use above lemma to prove the first inequality.
WLOG we assume $a\ge b\ge c$ then $S_{a}\ge S_{b}\ge S_{c}$. Therefore:
$\bigstar S_{c}=4c^{2}+a^{2}+b^{2}-4\ge 4c^{2}+\frac{(a+b)^{2}}{2}-4\ge \frac{(3c-1)^{2}}{2}\ge 0$
It is true :D Then we have the first inequality  :P I am very keen on above technique. It help me solve many problem. Example: Romania TST 2006 :D If you are new to the technique, you should try more. I hope you will like it, it very interesting. :D[/quote][quote="HTA"]
Wow , i'll try it ... but i saw it , in some way , similar to Vornicu-Shur , but stronger[/quote]
Oh, why stronger? It's a particular case of Vornicu-Schur's inequality!

Yes, It is also Vornicu Schur's inequality :P But in here I don't want talk about Vornicu Schur's inequality. I want talk about without expanding expression come to Vornicu Schur's inequality :P I and my brother (toanhocmuonmau) have a technique call U.C.T (Underfined Coefficient Technique), in this note we consider use U.C.T and Vornicu Schur to solve some problem with some new idea. Important thing of all technique is U.C.T open :D It is very simple but very useful. Example a hard inequality of Vasc :P

scorpius119 thank you very much for completing this one .Great work  :D
So now we have and a third very nice solution . Nice collection  :wink:

[quote="Vasc"]Let $ a,b,c$ be positive numbers such that $ a \plus{} b \plus{} c \equal{} 3$. Prove that

$ 3\left(a^{4} \plus{} b^{4} \plus{} c^{4}\right) \plus{} a^{2} \plus{} b^{2} \plus{} c^{2} \plus{} 6 \geq 6\left(a^{3} \plus{} b^{3} \plus{} c^{3}\right).$[/quote][quote="arqady"]It is a good exercise for S.O.S. :lol:[/quote]$ 3\left(a^4 \plus{} b^4 \plus{} c^4\right) \plus{} \left(a^2 \plus{} b^2 \plus{} c^2\right)\left(\frac {a \plus{} b \plus{} c}{3}\right)^2 \plus{} 6\left(\frac {a \plus{} b \plus{} c}{3}\right)^4$

$ \minus{} 6\left(a^3 \plus{} b^3 \plus{} c^3\right)\left(\frac {a \plus{} b \plus{} c}{3}\right)$

$ \equal{} \frac {1}{27}\sum{(b \minus{} c)^2(4b \plus{} 4c \minus{} 5a)^2}\geq0.$

This inequality is introduced as Example 3.1.4 on page 91 in new book :

Ji Chen, Chao-Cheng Ji, [url=http://www.bookuu.com/kgsm/ts/2009/09/18/1602993.shtml]Algebraic Inequalities (Mathematical Olympiad Proposition people Lecture)[/url], Shanghai Scientific & Technological Educational Publishing House, 2009, 225p. (ISBN 978-7-5428-4848-2/O·613)
[quote="Ji Chen"]For any real numbers $ u,v,x,y,z$, we have

$ u^{2}\left(\sum{x^{4}} \minus{} \sum{y^{2}z^{2}}\right) \plus{} v^{2}\left(\sum{y^{2}z^{2}} \minus{} \sum{x^{2}yz}\right)$

$ \geq uv\left[\sum{x^{3}(y \plus{} z)} \minus{} 2\sum{x^{2}yz}\right],$

with equality if $ x \equal{} v \minus{} u,y \equal{} z \equal{} u$.[/quote]This generalization is recorded on page 15.

[quote="Ji Chen"]For any real numbers $ u,v,x,y,z$, we have

$ u^{2}(\sum{x^{4}} - \sum{y^{2}z^{2}}) + v^{2}(\sum{y^{2}z^{2}} - \sum{x^{2}yz})$

$ \geq uv[\sum{x^{3}(y + z)} - 2\sum{x^{2}yz}],$

with equality if $ x = v - u,y = z = u$.[/quote]
To prove this inequality, it suffices to prove that
$ 4(\sum x^4 - \sum y^2z^2)( \sum y^2z^2 - \sum x^2yz) \ge [ \sum x^3(y + z) - 2\sum x^2yz]^2.$
Notice that
$ 2\sum x^4 - 2\sum y^2z^2 = \sum (y^2 - z^2)^2,$
$ 2\sum y^2z^2 - 2\sum x^2yz = \sum x^2(y - z)^2,$
and
$ \sum x^3(y + z) - 2\sum x^2yz = \sum x(y^3 + z^3 - y^2z - yz^2) = \sum x(y - z)^2(y + z),$
so the inequality can be written as
$ [\sum (y^2 - z^2)^2] [ \sum x^2(y - z)^2] \ge [ \sum x(y - z)^2(y + z)]^2.$
In this form, we can see that it is a direct consequence of the Cauchy Schwarz Inequality
$ LHS \ge [ \sum |x(y - z)(y^2 - z^2)|]^2 \ge RHS.$

[quote="Vasc"]This is like my old inequality
$ 4(\sum{x^{4}} - \sum{y^{2}z^{2}})(\sum{y^{2}z^{2}} - xyz\sum{x}) \ge 3(\sum{x^{3}y} - xyz\sum{x})^{2}$[/quote]
We have
$ 2\sum x^4 - 2\sum y^2z^2 = \sum (y^2 - z^2)^2,$
$ 6\sum y^2z^2 - 6\sum x^2yz = \sum (xy + yz - 2zx)^2,$
and
$ \sum x^3y - \sum x^2yz = \sum zx(z^2 - y^2) = \sum zx(z^2 - y^2) - \frac {1}{3}(xy + yz + zx)\sum (z^2 - y^2) = \frac {1}{3}\sum (y^2 - z^2)(xy + yz - 2zx).$
Therefore, the inequality can be written in the form
$ [ \sum (y^2 - z^2)^2][ \sum (xy + yz - 2zx)^2] \ge [ \sum (y^2 - z^2)(xy + yz - 2zx)]^2,$
and in this form, we can see immediately that it is a direct consequence of the Cauchy Schwarz Inequality.

Your solution are very nice, Can_hang.
I remember you the following nice identity:
$ AB \equal{} C^2 \minus{} CD \plus{} D^2$,
where
$ A \equal{} \sum x^4 \minus{} \sum x^2y^2, \ \ \ \ B \equal{} \sum x^2y^2 \minus{} xyz\sum x,$
$ C \equal{} \sum x^3y \minus{} xyz\sum x, \ \ \ \ D \equal{} \sum xy^3 \minus{} xyz\sum x.$


inequalities
function
integration
advanced fields
advanced fields unsolved

Let $ B_r\equal{}B(0,r)$ be the ball in $ \mathbf{R}^n$. Assume that $ u\in C^2(B_1)\cap C(\bar{B}_1)$ is harmonic in $ B_1$. Prove that there is a constant $ C\equal{}C_n$ such that for all $ 0<r<R<1$, it holds that
$ \frac{1}{r^n}\int_{B_r}|u|^2\le \frac{C}{R^n}\int_{B_R} |u|^2.$

Thanks in advance.

I suspect that you want to mimic the proof of the Mean Value Property, using u^2 in place of u, obtaining an inequality instead of an equality when you differentiate in r.  After that, I'm less confident about what to do, but I believe that one should use some sort of ODE comparison principle.

Let's assume $ u$ is real-valued. If not, do this separately for real and imaginary parts.

$ \Delta(|u|^2)\equal{}\nabla\cdot\nabla(u^2)\equal{}2\nabla\cdot(u\nabla u)\equal{}2\nabla u\cdot\nabla u\plus{}2\nabla\cdot\nabla u\equal{}2|\nabla u|^2\plus{}2\Delta u$

Since we're assuming that $ \Delta u\equal{}0,$ it follows that $ \Delta(u^2)>0.$ In other words, $ u^2$ is subharmonic.

Find a way of using that. (As yenlee suggested.)


integration
probability
function
algebra
polynomial
probability and stats

Let $ (X_n)$ be i.i.d. random variables with standard Cauchy distribution.
Find a succession $ (b_n)$, with $ b_n \to \plus{}\infty$, such that
$ \frac{Y_1\plus{}Y_2\plus{}\cdots\plus{}Y_n}{b_n} \to N(0,1)$ in distribution,
where $ Y_n\equal{}X_n$ if $ |X_n| \le\frac{1}{n}$, $ Y_n\equal{}0$ if $ |X_n|>\frac{1}{n}$.

$ F_n(\xi) \equal{} E(\exp(i \xi Y_n)) \equal{} 1 \minus{} \frac {2}{3 \pi} \frac{\xi^2}{n^33} \plus{} O(\frac {\xi^2 \plus{} \xi^4}{n^5})$.
(just compute the asymptotic development of $ n \mapsto \int_{|x| < \frac {1}{n}} \frac {\exp(i \xi x)}{1 \plus{} x^2} dx$). This is enough to conclude since it suffices that
$ \prod_1^n F_k(\frac {\xi}{b_n}) \to \minus{} \frac {\xi^2}{2} \equal{} E[\exp(i \xi N)]$
One sees that this is equivalent to:
$ \frac {2}{3 \pi b_n^2} \sum_1^n \frac {1}{k^3} \to \minus{} \frac {1}{2}$
which gives the answer: $ b_n^2 \equal{} C \equal{} \frac {4 \zeta(3)}{3 \pi}$. (modulo computation mistakes).

[Edit:] Finaly I think that what I wrote is false because the term $ O(\frac {\xi^2 \plus{} \xi^4}{n^5})$ cannot be ignored. (it would be the case if $ b_n \to \infty$). I have the feeling that the sequence $ Y_n$ converges too fast towards $ 0$. If $ Y_n$ had been defined by something like:
$ Y_n \equal{} X_n$ if $ |X_n|<\frac{1}{n^{\alpha}}$, $ Y_n \equal{} 0$ if $ |X_n|>\frac{1}{n^{\alpha}}$ with $ 0< \alpha\leq \frac13$
the argument would have worked,I think. This is not the case.

The variance of your partial sums does not tend to infinity, so $ b_n$ can't tend to infinity.

yes, this is what I was saying. To have an infinite variance one has to replace $ \frac{1}{n}$ by something like $ \frac{1}{n^{\alpha}}$. So is this true that $ \sum_1^n Y_k$ converges towards a normal distribution ?

I am afraid that the initial problem, stated as it is, is false. Here is a rough estimate of the density probability of
\[ \sum_1^n Y_k\]
based on $ 10^5$ simulations - it is very far from a normal distribution, even if it is not a proof -

Just a side comment: the characteristic function of a single Cauchy distribution is $ e^{\minus{}|\xi|}.$ Of course, any computations based on Taylor polynomials of this centered at zero are doomed, since this function isn't even differentiable at zero.

but here, in the above computation, $ \xi$ is fixed. One are trying to do a Taylor expansion of $ n \mapsto E[ \exp(i \xi Y_n )]$. (and these are troncated Cauchy variables, so even $ \xi \mapsto E[ \exp(i \xi Y_n )]$ is analytic)


Salam
in topic baraye bahs dar morede doreye emsal hastesh. :)

hala ki midoone darse panjome emsal chiye? :D

salam bar hameye azizan
be hameye doostani ke dar marhale dovom pazirofte shodan tabrik arz mikonam va omidvaram dore ye khoob va bedone havashi ro dar pish dashte bashim va ye time toop baraye imo az hamin bacheha be vietnam ferestade beshe ke baz name azize iran ro dar in reghabate elmi be gooshe hameye farzanegane jahan beresoone.
kasayi ham ke ghabool nashodan ziad khodeshuno narahat nakonan inshala baraye sale bad ya konkur  ;)

aghaye hatami etelaate raje be shorue dore va ... ra bayad az koja begirim?

dore rooze 9 tir shoroo mishe.
8 tir rooze sabte name

hamoontor ke too weblogam goftam; darse panjom jabre khatie :)

Ketabe "Mathematical Olympiads in China" (ke too ketabkhooneye bashgah hast) ye ketabe kheili aalie va dar fasle avvalesh nesbate nahamsaz ra be trore kamel dars dade, va masalehaye mofid ham kheili dare. Kameltarin marja dar morede nahamsaz ke man didam hamin ketabe. (Man az oon ketab nesbatr nahamsaz ra yad gereftam.)

[quote="mohamadhosein"][quote="Ashegh"][quote="mohamadhosein"]salam
dar morede nesbat haye nahamsaz che ketabi khoobe?[/quote]

boro natan ro bekhoon. ;)[/quote]
manzooret korte?too kurt nahamsaz nadare.[/quote]
ekhtiar darin, in che farmaieshie?

albathe rast migi, man hamsazo ba nahamsaz eshtebah geresftam.
valla man be gheir az matlabi keh omid goft, chizeh digeh e balad nistam,

[quote="maryam rashin"]baraye masale too nahamsaz,khorak faghat sanaze(sanaz fotoohi)
albatte eslah pazir ham ghabele paziroftane.[/quote]

felam keh manbae behtari az chape irani , be gheir az in nadarim. ;)


[quote="Quevvy"]In terms of $N$, a natural number, how many more positive integral factors does $24^{N}$ have than $16^{N}$? Express your answer in simplest form.[/quote]

[hide]
24 has 2^3 times 3. 16 has 2^4
$24^{N}$ has (3N+1)(N+1) factors, and $16^{N}$ has $4N+1$ factors
so the 24 minus the 16 is $3N^{2}+4N+1-4N-1$, so the answer is $3N^{2}$

[/hide]


inequalities
algebra
polynomial
inequalities proposed

Let $ x,y,z$ be positive real numbers, show that $ \frac {xy}{z} \plus{} \frac {yz}{x} \plus{} \frac {zx}{y} > 2\sqrt [3]{x^3 \plus{} y^3 \plus{} z^3}.$

Set $ a^2 = \frac{yz}{x},\, b^2 = \frac{zx}{y},\, c^2 = \frac{xy}{z}\implies a,b,c>0$ the inequality can be rewritten as
\begin{align*}
a^2+b^2+c^2 &> \sqrt[3]{a^3b^3+b^3c^3+c^3a^3} \\
\left(a^2+b^2+c^2\right)^3 &> 8\sum_{cyc} a^3b^3 \\
\sum_{cyc} a^6 + 3\sum_{sym} a^4b^2 + 6a^2b^2c^2 &> 8\sum_{cyc} a^3b^3
\end{align*}

Schur's inequality implies $ \sum_{cyc} a^6 + 3a^2b^2c^2 \ge \sum_{sym} a^4b^2$ and $ 4\sum_{sym} a^4b^2 \ge 8\sum_{cyc} a^3b^3 = 4\sum_{sym} a^3b^3$ is a consequence of Muirhead (or AM-GM). Using $ 3a^2b^2c^2>0$, we have proven the inequality.

[quote="Fang-jh"]Let $ x,y,z$ be positive real numbers, show that

$ \frac {yz}{x} + \frac {zx}{y} + \frac {xy}{z} > 2\sqrt [3]{x^3 + y^3 + z^3}.$[/quote][quote="bertram"]Set $ a^2 = \frac {yz}{x},\, b^2 = \frac {zx}{y},\, c^2 = \frac {xy}{z}\implies$ the inequality can be rewritten as

$ \left(a^2 + b^2 + c^2\right)^3 \& > 8\left(b^3c^3 + c^3a^3 + a^3b^3\right).$[/quote]$ \left(a^2 + b^2 + c^2\right)^3 - 8\left(b^3c^3 + c^3a^3 + a^3b^3\right)\equiv F(a,b,c) = F(a,a + s,a + t)$

$ = 3a^6 + 6(s + t)a^5 + 15(s^2 + t^2)a^4 + 4(s + t)(7s^2 - 10st + 7t^2)a^3$

$ + 3(7s^4 - 10s^2t^2 + 7t^4)a^2 + 6(s - t)^2(s + t)^3a + (s - t)^2(s^4 + 2s^3t + 6s^2t^2 + 2st^3 + t^4) > 0$,

which is clearly true for $ 0 < a = \min\{a,b,c\}$.

By the way, exist real polynomials $ p_1,p_2,p_3,p_4$ in $ a,b,c$ such that $ F(a,b,c) = p_1^2 + p_2^2 + p_3^2 + p_4^2$, see:

http://www.mathlinks.ro/Forum/viewtopic.php?t=202683

Let $a,b,c$ are positive real numbers. Prove that \[\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}\ge \sqrt[3]{8(a^3+b^3+c^3)+3abc}.\]
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&p=2895265

Set $ a=\sqrt{\frac{y^3z^3}{x^3}}, b=\sqrt{\frac{x^3z^3}{y^3}},c=\sqrt{\frac{x^3y^3}{z^3}} $.The inequality can be rewritten as
               $ \sqrt[3]{a^2}+\sqrt[3]{b^2}+\sqrt[3]{c^2}> 2\sqrt[3]{ab+bc+ca} $
Using the holder inequality $ LHS^3(a^2+b^2+c^2)\geq (a+b+c)^4 $.Therefore, it suffices to show that
$ \frac{(a+b+c)^4}{a^2+b^2+c^2}\geq 8(ab+bc+ca) $
Using the AM-GM inequality $ 8(ab+bc+ca)(a^2+b^2+c^2)=4(2ab+2bc+2ca)(a^2+b^2+c^2) \leq (a+b+c)^4 $
                                                                                Q.E.D


superior algebra
superior algebra unsolved

find the element u so that $ \mathbb{Q} (\sqrt[3]{7} ,\sqrt{2}) \equal{}\mathbb{Q}(u)$.


geometry
trigonometry
parallelogram
rhombus
rectangle
trapezoid
area of a triangle

I don't think you can calculate the area of a quadrilateral given only the lengths of the sides, if that's what you mean, unless you know it's inscribed in a circle (i.e. the opposite angles are supplementary). Then there's this formula that looks a lot like Heron's Formula for triangles for those types of "cyclic" quadrilaterals. But if you know one of the diagonals, you can use Heron's for the separate triangles in the quadrilateral and add them up. If you know the angles, you can use trig to find the length of the diagonal. But I don't know of a formula in terms of the four sides, if all you know is the four sides. Unless it's special in some other way like it's a parallelogram or something.

Hope that helps.

[quote="randallnash"]Can anyone give me an easy formula to calculate the area of a  quadrilateral?
Thanks[/quote]

You could break it up into triangles and use other little tricks like that, but I dont think that there is a "perfect" formula for quadrilaterals. You could use some trig.

There is [url=http://mathworld.wolfram.com/BrahmaguptasFormula.html]Brahmagupta's formula[/url]:

\[A=\sqrt{(s-a)(s-b)(s-c)(s-d)-abcd\cos^2\left[\frac 12(A+B)\right]}.\]

[quote="randallnash"]Can anyone give me an easy formula to calculate the area of a  quadrilateral?
Thanks[/quote]

Depends which quadrilateral it is, if it's a square or rectangle, base times height. If it's trapezoid, add the two bases and divide by $2$ and multiply by the height. If it's a rhombus, it's the multiplication of the two diagonals and then divide by $2$. There are many formulas, but just incase you find "strange" quadrilaterals then just break it into triangles and other shapes.

For quadrilateral $ABCD$, we say that the angle between $AC$ and $BD$ is $\theta$. Then
\[S=\frac{1}{2} AC\cdot BD \sin{\theta}\]

[quote="mathnerd314"]Here's why there's no formula for a quadrilateral given only the lengths of its sides:

Consider a trivial example of a rhombus with side length 1. Now suppose there was a formula to calculate the area just from the side lengths. We draw a diagram of the rhombus and it looks like a square. OK, so the area is 1. We are not convinced of this and draw a different diagram that looks more like a line segment. OOPS! the ara is closer to zero, and in fact approaches zero. That means 1=0. Contradiction. QED.[/quote]
nice attempt on a 'proof' but you don't need to use all that area approaching zero and use contradiction  :rotfl:

randallnash, there is no general formula that only uses lengths of sides but as joml88 pointed out, there is a general formula if you know some of the angles...


AMC
AMC 10
geometry
AMC 12
number theory

I've been taking old AMC 10s under timed conditions and have been getting roughly 120s, but I haven't been getting 120 or over consistently. I don't feel like I have a single weak area such as number theory or probability...If you were me, how would you study for the few weeks coming before the AMC 10 after you finished going through all the previous AMC 10s?

look at all the problems that you got wrong/ didn't know how to do and think about why you got them wrong and what you could do to get them right the next time that you take the test

contrary to a lot that you hear, the amc tests can be 'studied for,' of course you need ingenuity, but at least the first 20 problems fall into various categories that you can master

it is your job to find how to 'algorithmize' the amc for yourself

As Altheman said, go over the problems you don't know. Since there's like a month before the test, and you don't think you're "weak" in a subject, maybe you should try some harder problems in subjects. Also, since you've been roughly getting 120s, you should aim a lot at the last 5 questions or so, since [i]most[/i] of the time they are the hardest.

This is more for future preparation, but I'd suggest that you don't rush through as many tests as possible, since then you run out of stuff to study.  Personally, unless you're in your last year of AMC, I'd take it slowly.

After looking at the solutions to the questions you didn't know how to do, go back two weeks later and do the questions again. It might seem redundant, but really learning the solutions to unfamiliar problems will help you.

If you can take the AMC12 in addition, you might want to give it a shot. In my freshman year, I got a 107.5 on the 10 but a 108 on the 12. Then again, I know that some people score much higher on the 10. But I would take a practice 12 to see if you find qualifying from the 12 easier than qualifying for the 10.

Also, I used to be intimidated by the later questions. I later found that some of the last five questions are easier than expected. On last year's AMC10, I got all of the last five questions correct, but missed 4 other questions (and not all due to careless mistakes). Many times, I find that number 25 is easier than many other problems on the test.

These are just my opinions, but some of them may apply to you...

Maybe someone said this in an earlier post, but u should work through the art of problem solving intro series for your topic(s) of discomfort. And Work through old tests.

[quote="Narcissa"]lol yeah what would u suggest me doing...i've been getting a HUGE range in AMC12's...90s-130s, wtf??![/quote]

This is probably due to your inconsistencies in certain areas of math. Try to identify them and revise them.

Or it could be inconsistencies in the times that you take the tests.  For example, some people, like me, simply can't think at all if it's really late at night.  Thinking levels fluctuate a lot with different times of day.


Juan takes a number, adds 2 to it, multiplies the answer by 2, subtracts 2 from the result, and finally divides that difference by 2. If his answer is 7, what was the original number?

Let the number=x.
1st step=x+2
2nd step=2(x+2)=2x+4
3rd step=2x+2
4th step: (2x+2)/2=x+1=7
x=6.

$ 7 \plus{} 2 \equal{} 9$

$ 9 \times 2 \equal{} 18$

$ 18 \minus{} 2 \equal{} 16$

$ 16 \div 2 \equal{} \boxed{8}$

EDIT: Beaten again...

EDIT 2: Jongao, $ 7 \plus{} 1 \equal{} 8$.

Note Ernie, READ THE QUESTION!
The answer to his problem is 7, not his original number.


What is the number of square centimeters in the area of an isosceles triangle with base length 20 centimeters and side length 26 centimeters?

Draw the altitude to the base. Each of the triangles formed is right, with side lengths 10, 24, and 26. Thus the altitude is 24, and the area of the isoceles triangle is $ \frac{1}{2} \cdot 24 \cdot 20 \equal{} 240$.

One could also find the area of each right triangle formed, then double this area.

You could always use the pythagorean theorem to find the height of the triangle


No matter where I look, I can't find any information about the sum 1^1+2^2+3^3+....+n^n. Does anyone know the closed form of this sum?

There isn't one.

I don't have a proof for that, but it's the only answer which makes sense.


How many times does 24 divide 100!..
please help and explain ..thank you...

BTW, it is Number theory problem ;)

You are to determine the power of 2 and 3 in the expansion of 100!.

There is a nice formula for that. The maximal power of $p$ ($p$ is a prime), which divides $n!$, is \[\left[\frac{n}{p}\right]+\left[\frac{n}{p^2}\right]+\left[\frac{n}{p^3}\right]+...\]

For your case it gives that $100!$ is divisible by $2^{97}$ and by $3^{48}$. Since $24=2^3\cdot 3$ we conclude that 100! is divisible by $24^{32}$ and is not divisible by $24^{33}$.

I dont understand the part where you got 2 raised to the power of 97 and 3 raised to the power of 48..
and does the formula have something to do with those numbers below...You said P is the prime number ..so which one is P in the problem ..is it 23?
please explain to me...thank you

The reasoning behind it:  take for example the number 20!, and the prime 2.

We see that there is a two every two numbers:
2, 4, 6, 8, ...20 all have a factor of two.

But we haven't accounted for fours, because those have two twos.
4, 8, 12, 16, 20 all have a four.  We already have counted a two, so we need to count one more two.

But we haven't accounted for eights, because those have 3 twos.
8, 16 all have an 8.  We already counted a four, so we need to count one more two.

But we haven't accounted for sixteens, because those have 4 twos.
16 has a 16.  We already counted an 8, so we need to count one more two.

The number of two $\leq 20$ is $\left\lfloor\frac{20}{2}\right\rfloor$.
The number of fours is $\left\lfloor\frac{20}{4}\right\rfloor$.
The number of eights is $\left\lfloor\frac{20}{8}\right\rfloor$.
The number of sixteens is $\left\lfloor\frac{20}{16}\right\rfloor$.

So we want the sum of these numbers.  Compare this specific case to Myth's general formula.

ummm...where did he use 48?  I don't see it anywhere... :?

[quote="Kle"]I still dont get this ...  these two sequences below .. where do they stop ... and its hard to understand that where you got 97 and 48 .. are those the number when you add them together.... I'm kinda lost .. please explain more clearly..[/quote]You must think, do not [b]only[/b] read. If you only read Myth's answer and you don't write the solution on paper and do not think about it, you will never understand.
Your question is: how many times does 24 divide 100! ?
First organise. What is the question looking for? Highest exponent in 24, so you are looking for $\alpha$ such that $24^\alpha$ divides 100! but $24^{\alpha+1}$ does not divide 100!

This question involves primes. What primes are you working with? $24=2^3\cdot 3^1$ So every factor of 24 have primes 2 and 3 in a 3:1 ratio.

Next, what is Myth trying to say? Largest exponent of prime number $p$ that divide $n!$ is \[\left[\frac{n}{p}\right]+\left[\frac{n}{p^2}\right]+\left[\frac{n}{p^3}\right]+\ldots\]

Did you try more numbers? Don't be lazy.

$\left[\frac{100}{2}\right]=50$

$\left[\frac{100}{2^2}\right]=25$

$\left[\frac{100}{2^3}\right]=12$

$\left[\frac{100}{2^4}\right]=6$

$\left[\frac{100}{2^5}\right]=3$

$\left[\frac{100}{2^6}\right]=1$

$\left[\frac{100}{2^7}\right]=0$

You can stop here because if you try higher exponents $t$, $2^t>100$ and $\left[\frac{100}{2^t}\right]=0$

So Myth sequence give 50+25+12+6+3+1+0+0+0+0+0+0+0+...=97

Now for practice, do the same with prime 3 and DO IT LIKE I DID

[hide="Click here to check if you got it right"]Myth sequence for prime 3 give 33+11+3+1+0+0+0+...=48[/hide]

Now count which is the least one:

If you use all primes 2, prime 2 is 3:1 ratio to prime 3. But you have 97 primes of 2, you can use at most 96/3 or 32 primes of 3.

If you use all primes 3, prime 3 is 1:3 ratio to prime 2. But you have 48 primes of 3, and you need at least  48*3=144 primes of 2. This case is not possible

So highest exponent is 32. Answer is 32

[quote="Kle"]according to iversonfan...why do we need to find 48 when it is not used.... cuz I didn't see any use of it in the formular[/quote] How would you know which number is smaller in quantity of prime factors? You need to find 48 because you need to compare with primes of 2 and it happens 32<48 and you can discard 48 [b]after[/b] you make compare

[quote="Kle"]according to iversonfan...why do we need to find 48 when it is not used.... cuz I didn't see any use of it in the formular[/quote]

You need to see which one is smaller, obviously $3$ would be less quantity, because it's less quantity in $24$, but still, just to make sure... :)


:maybe: Find:
$ lim_{n \to \infty}[cos(2\pi n!x)]^n$
a) if $ x$ is rational.
b) if $ x$ is irrational.

http://www.mathlinks.ro/Forum/viewtopic.php?t=164508

There you can find my weak attempt so solve it  :oops: , and jmerry's  great explanation.


calculus
derivative
ratio
real analysis
real analysis unsolved

Let $ f(x) = 1 + \sum_{n=1}^{\infty}\frac{\alpha(\alpha-1) . . . (\alpha-n+1)}{n!}x^n$

Prove that
$ (1+x)f^'(x)=\alpha f(x)$

Assume uniform convergence of the series on some interval. There, we can exchange the differentation and summation. We should check whether this is true of course, but I leave that as given.

Define $ \alpha_n : \equal{} \alpha(\alpha\minus{}1)\dots(\alpha\minus{}n\plus{}1)$.

We get:

$ f'(x) \equal{} \sum\limits_{n\equal{}1}^{\infty} \frac{\alpha_n}{(n\minus{}1)!}x^{n\minus{}1}$

$ (1\plus{}x)f'(x) \equal{}  \sum\limits_{n\equal{}1}^{\infty} \frac{\alpha_n}{(n\minus{}1)!}x^{n\minus{}1} \plus{}  \sum\limits_{n\equal{}1}^{\infty} \frac{\alpha_n}{(n\minus{}1)!}x^n$

$ \alpha f(x) \equal{} \alpha \plus{} \sum\limits_{n\equal{}1}^{\infty} \frac{\alpha \alpha_n}{n!}x^n$

Split off the first term from the first summation in the expression of $ (1\plus{}x)f'(x)$ (which is equal to $ \alpha$), then change indices $ n \rightarrow n \minus{} 1$ to get:

$ (1\plus{}x)f'(x) \equal{}  \alpha \plus{} \sum\limits_{n\equal{}1}^{\infty} \frac{\alpha_{n\plus{}1}}{n!}x^n \plus{}  \sum\limits_{n\equal{}1}^{\infty} \frac{\alpha_n}{(n\minus{}1)!}x^n$

Now, unicity of the power series gives that we can suffice by showing that the coefficients of $ x^n$ match for both expressions, thus we should check whether:

$ \frac{\alpha\alpha_n}{n!} \equal{} \frac{\alpha_{n\plus{}1}}{n!} \plus{} \frac{\alpha_n}{(n\minus{}1)!}$

Divide by $ \alpha_n$ to get:

$ \frac{\alpha}{n!} \equal{} \frac{\alpha \minus{} n}{n!} \plus{} \frac{1}{(n\minus{}1)!}$ which is easy.

So we get that indeed $ (1\plus{}x)f'(x) \equal{} \alpha f(x)$.

[quote="Tyl"]Assume uniform convergence of the series on some interval.[/quote]
You've got that - and uniform convergence of the series for the derivative, as well - on any interval of the form $ [\minus{}r,r]$ with $ r<1,$ since you can show by the ratio test that the series has radius of convergence $ 1.$ Hence your final conclusion - the differential equation - is valid for $ x\in(\minus{}1,1).$


integration
calculus
complex numbers
real analysis
real analysis unsolved

This is an exercise I solved some time ago, but recently found a nicer proof of.  It proceeds in two parts.  The first part, which may be of independent interest, is as follows:

1)  Show that if $ f\in L^1$, $ f$ is continuous at $ 0$, and $ \widehat{f}\geq 0$, then $ \widehat{f}\in L^1$

Next, say that $ f$ is continuous and integrable with nonnegative fourier transform.

2)  Show that $ f$ is a positive definite function.  That is, show that for any real numbers $ x_1,\ldots,x_k$ and any complex numbers $ \alpha_1,\ldots,\alpha_k$, $ \sum^k_{i,j\equal{}1} \alpha_i\overline{\alpha_j}f(x_i\minus{}x_j)\geq 0$.

we have
\[ \widehat{f}(x)\equal{}\frac{1}{(2\pi)^n}\int e^{i(x,y)}\widehat{f}(y)dy\]
Hence
\[ \begin{aligned}
\sum_{j,k}\alpha_j\bar{\alpha}_k f(x_j\minus{}x_k)&\equal{}\frac{1}{(2\pi)^n}\int\sum_{j,k}\alpha_j\bar{\alpha}_k e^{i(x_j\minus{}x_k,y)}\widehat{f}(y)dy\\
&\equal{}\frac{1}{(2\pi)^n}\int\sum_{j,k}\alpha_j e^{i(x_j,y)}\bar{\alpha}_k e^{\minus{}i(x_k,y)} \widehat{f}(y)dy\\
&\equal{}\frac{1}{(2\pi)^n}\int|\sum_j\alpha_je^{i(x_j,y)}|^2\widehat{f}(y)dy\geq 0
\end{aligned}\]
it follows that f is positive definite.

Sure.  Of course, to use that representation for $ f$ in terms of $ \widehat{f}$, you need the inversion formula, which means you need $ \widehat{f}\in L^1$, which is what a solution to the first part would give you :)

to prove $ \widehat{f}\in L^1$, we consider the integral $ \int\widehat{f}(x)e^{\minus{}\epsilon|x|^2}dx$ which $ \epsilon > 0$.
we have
\[ \begin{aligned}
\int\widehat{f}(x)e^{\minus{}\epsilon|x|^2}dx&\equal{}\int\int e^{\minus{}i(x,y)}f(y)dy e^{\minus{}\epsilon|x|^2}dx\equal{}\int f(y)\int e^{\minus{}i(x,y)}e^{\minus{}\epsilon|x|^2}dxdy\\
&\equal{}\epsilon^{\frac{\minus{}n}{2}}\pi^{\frac{n}{2}}\int f(y)e^{\frac{\minus{}|y|^2}{4\epsilon}}dy
\end{aligned}\]
we will show that $ \epsilon^{\frac{\minus{}n}{2}}\pi^{\frac{n}{2}}\int f(y)e^{\frac{\minus{}|y|^2}{4\epsilon}}dy\leq C$ for all $ \epsilon$ is small. Indeed, f is continue at 0, then there is $ \delta > 0$ such that $ |f(y)| \leq |f(0)|\plus{}1$ for all $ |y|\leq\delta$. Therefore
\[ \begin{aligned}
\epsilon^{\frac{\minus{}n}{2}}\pi^{\frac{n}{2}}\int f(y)e^{\frac{\minus{}|y|^2}{4\epsilon}}dy&\equal{}\epsilon^{\frac{\minus{}n}{2}}\pi^{\frac{n}{2}}\int\limits_{|y|\leq\delta} |f(y)|e^{\frac{\minus{}|y|^2}{4\epsilon}}dy\plus{}\epsilon^{\frac{\minus{}n}{2}}\pi^{\frac{n}{2}}\int\limits_{|y|>\delta} |f(y)|e^{\frac{\minus{}|y|^2}{4\epsilon}}dy\\
&\leq (2\pi)^{n}(|f(0)|\plus{}1) \plus{} \epsilon^{\frac{\minus{}n}{2}}e^{\minus{}\frac{\delta^2}{4\epsilon}}\pi^{\frac{n}{2}}\int|f(y)|dy.
\end{aligned}\]
the last term tend to 0 when $ \epsilon\to 0$. Hence there exists $ C>0$ such that
\[ \int\widehat{f}(x)e^{\minus{}\epsilon|x|^2}dx\equal{}\epsilon^{\frac{\minus{}n}{2}}\pi^{\frac{n}{2}}\int f(y)e^{\frac{\minus{}|y|^2}{4\epsilon}}dy\leq C\]
for all $ \epsilon$ is small enough. Let $ \epsilon$ tend to 0, and apply the monotone convergence theorem (note that $ \widehat{f}\geq 0$), we have
\[ \int\widehat{f}(x)dx\leq C\]
that is $ \widehat{f}\in L^1$.\\
Note that:
\[ \int e^{\minus{}i(x,y)}e^{\minus{}|y|^2}dy\equal{} \pi^{n/2}e^{\minus{}|x|^2/4}\]


Prove that there exists a quotient map from (0,1) to [0,1], but not from [0,1] to (0,1).

Quotient maps are continuous? That seems to work for why there is none from [0,1] to (0,1) since the image of a compact set under a continuous map has to be compact.

In the other direction, there's an "obvious" quotient map from $ (\minus{}1,1)$ to $ [\minus{}1/2,1/2]$
that identifies all $ x \le \minus{}1/2$ together and all $ y \ge 1/2$ together. Compose with the standard
homeomorphisms on each end and you'll get a quotient map  $ (0,1) \to [0,1]$.


well observe

if $2^{x}-1$ $<$ $0$ then this question is absolutely meaningless
but since $x$ $>$ $0$ is given $u$ $=$ $2^{x}-1$ $> 0$

this gives $1$ $+$ $log(u+\frac{1}{u})$ $=$ $sin(\frac{\pi}{2}x)$ $+$ $log2$

but we know $u+\frac{1}{u}$ $\geq$ $2$
also $1$ $\geq$ $sin(\frac{\pi}{2}x)$
also $log$ is an increasing function
so equality clearly holds only if $x$ $=$ $1$


function
limit
calculus
calculus computations

$ a>o$
$ a_{1}\equal{}ln a$
$ a_{2}\equal{}ln(a\minus{}a_{1})$
$ a_{n\plus{}1}\equal{}a_{n}\plus{}ln(a\minus{}a_{n})$
Find $ lim a_{n}$

Please read http://www.mathlinks.ro/Forum/viewtopic.php?t=125085 . Moved!

[hide="Answer"]The limit is a-1. Take limit in both sides of $ a_{n\plus{}1}\equal{} a_{n}\plus{}\ln(a\minus{}a_{n})$. Let the limit be L: then $ L \equal{} L\plus{}\ln(a\minus{}L)\Rightarrow L \equal{} a\minus{}1$.[/hide]

Exists $ \lim a_{n}$. Are you sure?  :maybe:

Indeed,I'm sure that there exist.
Here is my hint:
First,please prove that if $ f(t)=t-\lnt$ with $ t>0$ then $ f(t)>1$ (by cosider $ f'(t)$) (*)
Since that,we have $ a_{2}<a_{1}<a$
$ a_{3}=a_{2}+\ln(a-a_{2}) <a_{2}+(a-a_{2})=a$
...
Hence this sequence $ (a_{n})$ is defineded and we have $ a_{n}< a$ for all $ n\in N$
Now we consider function $ g(x) = x+\ln(a-x)$ with $ 0<x<a-1$
$ g'(x) = 1-\frac{1}{a-x}=\frac{a-x-1}{a-x}>0$.Then $ g(x)$ is increasing in $ (0;a-1)$
We have $ a_{n+1}=g(a_{n})$ for all $ n\geq 2$
 Since (*) ,$ a-a_{n}> 1$ for all $ n\in N$
There for sequence $ (a_{n})$ is increasing when $ n\geq 2$ and $ a_{n}< a$.hence there exist limit.
(note that $ a_{2}<a-1$ so $ a_{3}=a_{2}+\ln(a-a_{2})>a_{2}$)

See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=162858]here[/url] for a general version of the argument.


vector
superior algebra
superior algebra unsolved

Is $ Aut(D_n)\cong D_n, \forall n$ ?

 :)

[quote="Gockel"]No. $ Aut(D_2) \equal{} Aut(V_4) \equal{} S_3$.[/quote]

how to prove that $ Aut(V_4)\cong S_3$ ?

I know it must be six elements in $ Aut(V_4)$, but I don't know how to find the bijection mapping

 [quote="-oo-"]in general, we have $ Aut(D_n) \equal{} \mathbb{Z}/n \rtimes (\mathbb{Z}/n)^*$[/quote]

Sorry , I don't understand the symbols u used , what do u mean by $ \mathbb {Z}/n$ and the $ *$.

 :)

To understand the automorphisms of $ (\mathbb{Z}/2\mathbb{Z})^2$, look at it as a vector space. The automorphisms are the $ 2\times 2$ invertible matrices over $ F_2$.
The other way to look at it: what can you do to the three non-identity elements? Permute them.

The * is "the invertible elements in this ring". See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=255364]here[/url].


real analysis
integration
function
calculus
percent
real analysis solved

Fix an enumeration of the rationals $r_1,r_2,\dots$. For $x\in\mathbb R$ define $f(x)$ as the cardinal number of $\{n : |x-r_n|<2^{-n}\}$.

1. Show that $f$ is Lebesgue Measurable.
2. Evaluate $\int_{\mathbb R}f(x)\;dx$.

Let $g_n(x)=1$ for $|x-r_n|<2^{-n}$ and zero otherwise. This is a step function; clearly measurable and integrable.

Then $f(x)=\sum_{n=1}^{\infty}g_n(x).$

This clearly makes $f$ measurable, and we can calculate its integral by the Monotone Convergence Theorem.

$\int f = \int\sum_n g_n = \sum_n\int g_n = \sum_{n=1}^{\infty}2\cdot 2^{-n}=2.$

Note that $f$ must be unbounded in any interval and is clearly not Riemann integrable.

A hundred percent right. Actually that is basically my same solution. :first: