[hide]The formula for the sum of angles is:

\[(n-2)\times 180^\circ\]

So you just set that equal

$(n-2)\times 180^\circ=436$

$n-2=\frac{436}{180}$

$n=\frac{436}{180}+2$

$n\neq \text{Integer}$

Therefore, you must clarify. Interior angles? :| [/hide]

This question was in my geometry book and this is all they have said.   :(

[quote="anirudh"]Well...what I demonstrated is what you have to do.

Therefore, the problem is probably wrong or something...unless i'm not getting something[/quote]

Ya thats what I thought when I saw the problem but maybe we both are missing something.  :D

Okay...but isn't that out of math counts level?

[quote="bpms"][quote="anirudh"]Well...what I demonstrated is what you have to do.

Therefore, the problem is probably wrong or something...unless i'm not getting something[/quote]
Haven't you heard of non-euclidean geometry?
The answer is 4.  :P[/quote]

Um does that mean that each angle is 109?? Anyway we arent studying non-euclidean geo.

non-euclidian sounds about right for this. are you sure about the problem?

Wait, so which geometry is this? Even if this isn't Math Counts level, I'd still like to know the proof of that answer. :lol:

[quote="nutz_for2.718281828"]Lemma: The sum of all angles in a polygon is a multiple of 10.

Proof: You can make n-2 triangles out of an n sided polygon. Therefore, the sum or the angles of any polygon is 180(n-2), which is a multiple of 10.[/quote]
More specifically, a multiple of 180, which will always end in 0.

[quote="purple plume"]that's an easy problem it was on the Florida Math league[/quote]
Then what is the solution/answer?

[quote="mathgeniuse^ln(x)"]I bet that your text was old, so it meant to say

4860, but that is just me.[/quote]
The funny thing is, we just did a problem with that sum of the interior angles   :rotfl: and then you could solve that...

[quote="tridecalogism_lover"]why don't you add 2 to  436/180?[/quote]
A polygon can't have an non-integral number of sides.

As for the original problem, what if the polygon is concave?

Is that possible to figure out then? :huh:

If the polygon is concave, there is more than one answer.

[quote="funcia"]Oh...I see.  :wallbash:
well...we know it has at least 4 sides and at least one angle is over 180 so
[hide]I get(since we know it most be bigger than what we get for convex polygons-)4 since (4-2)(180)=360 and that is for 4 sides and if we do 5 we get a big answer that is too big for this case[/hide][/quote]
You are correct about the four side minimum, but there is a much simpler reason: The sum of the interior angles of a triangle will always be 180, and triangles can't be concave, so $s\ge4$.

So far, we have established that the polygon must be concave and has a minimum of 4 sides. Does anybody have any ideas that might help us determine the exact number?

umm it thought that the exact answer was 4 i mean if it was 5 do the math with smaller ones...(n-2)(180) so we would get 540 so therefore if this is concave the polygon is 4 sided

[quote="i_like_pie"]No... First of all, I repeat: The formula does not apply for concave polygons. Second, think about this: You always say the measure of  an angle is $45^{\circ}$, not $315^{\circ}$. In a concave polygon, you don't measure the angle(s) that are "inverted" from the inside, but instead from the outside. A concave quadrilateral doesn't necessarily have interior angles sum to $360^{\circ}$. These special properties of concave polygons make it hard to determine answers to such questions.[/quote]
what? A concave squre has an angle sum of 360 because it does have 315 degree measures because you aren't measuring the smaller angle, because the smaller angle isn't part of the polygon.

[quote="i_like_pie"]No... First of all, I repeat: The formula does not apply for concave polygons. Second, think about this: You always say the measure of  an angle is $45^{\circ}$, not $315^{\circ}$. In a concave polygon, you don't measure the angle(s) that are "inverted" from the inside, but instead from the outside. A concave quadrilateral doesn't necessarily have interior angles sum to $360^{\circ}$. These special properties of concave polygons make it hard to determine answers to such questions.[/quote]
Is it possible to draw a concave quadrilateral where the sum of the interior angles is not 360?

I think it would have to be like in diffferent planes, but I don't know.

[quote="ProtestanT"][quote="i_like_pie"][quote="funcia"]Iff Euclidean:
This must be impossible if _ is 180 degrees-I mean if it is all interior and exterior 436 would be a multiple of 180, right?
Oh well, this is probably one of those trick questions.[/quote]
Concave polygons don't have a formula for the sum of their interior angles.[/quote]
[quote="i_like_pie"]No... First of all, I repeat: The formula does not apply for concave polygons. Second, think about this: You always say the measure of  an angle is $45^{\circ}$, not $315^{\circ}$. In a concave polygon, you don't measure the angle(s) that are "inverted" from the inside, but instead from the outside. A concave quadrilateral doesn't necessarily have interior angles sum to $360^{\circ}$. These special properties of concave polygons make it hard to determine answers to such questions.[/quote]
concave polygons have the same formulas as convex polygons for the sum of their [i]interior[/i] angles. all the angles must be measured from the [i]inside[/i], or the [i]interior[/i] of the polygon, so the 315 degree angle would count as a 315 degree, not a 45 degree[/quote]

Yes... it is counted as 45 degree angle in radians... i believe

the point was that it was for interior angles, and the interior angle is reflex. if it is counted the other way, it would be counted as exterior.

[quote="sonny"][/quote][quote="ProtestanT"][quote="i_like_pie"][quote="funcia"]Iff Euclidean:
This must be impossible if _ is 180 degrees-I mean if it is all interior and exterior 436 would be a multiple of 180, right?
Oh well, this is probably one of those trick questions.[/quote]
Concave polygons don't have a formula for the sum of their interior angles.[/quote]
[quote="i_like_pie"]No... First of all, I repeat: The formula does not apply for concave polygons. Second, think about this: You always say the measure of  an angle is $45^{\circ}$, not $315^{\circ}$. In a concave polygon, you don't measure the angle(s) that are "inverted" from the inside, but instead from the outside. A concave quadrilateral doesn't necessarily have interior angles sum to $360^{\circ}$. These special properties of concave polygons make it hard to determine answers to such questions.[/quote]
concave polygons have the same formulas as convex polygons for the sum of their [i]interior[/i] angles. all the angles must be measured from the [i]inside[/i], or the [i]interior[/i] of the polygon, so the 315 degree angle would count as a 315 degree, not a 45 degree[/quote][quote="sonny"]

Yes... it is counted as 45 degree angle in radians... i believe[/quote]
No, in radians a 315 degree angle is $\frac{7 \pi}{4}$
$180$ degrees $=\pi$ radians.

[quote="bpms"][/quote][quote="sonny"][/quote][quote="ProtestanT"][quote="i_like_pie"][quote="funcia"]Iff Euclidean:
This must be impossible if _ is 180 degrees-I mean if it is all interior and exterior 436 would be a multiple of 180, right?
Oh well, this is probably one of those trick questions.[/quote]
Concave polygons don't have a formula for the sum of their interior angles.[/quote]
[quote="i_like_pie"]No... First of all, I repeat: The formula does not apply for concave polygons. Second, think about this: You always say the measure of  an angle is $45^{\circ}$, not $315^{\circ}$. In a concave polygon, you don't measure the angle(s) that are "inverted" from the inside, but instead from the outside. A concave quadrilateral doesn't necessarily have interior angles sum to $360^{\circ}$. These special properties of concave polygons make it hard to determine answers to such questions.[/quote]
concave polygons have the same formulas as convex polygons for the sum of their [i]interior[/i] angles. all the angles must be measured from the [i]inside[/i], or the [i]interior[/i] of the polygon, so the 315 degree angle would count as a 315 degree, not a 45 degree[/quote][quote="sonny"]

Yes... it is counted as 45 degree angle in radians... i believe[/quote][quote="bpms"]
No, in radians a 315 degree angle is $\frac{7 \pi}{4}$
$180$ degrees $=\pi$ radians.[/quote]

Yes thats true but I meant the circle where you draw the 315 degree angle.... you are left with 45 degrees  :|

[quote="i_like_pie"]Apparently my understanding of concave polygons wasn't entirely correct. :maybe:

Either the sum of the interior angles of a polygon is always $180(n-2)$ and this problem has no solution, or the formula does not apply to concave polygons and therefore there can be more than one solution. Either this problem's solution is not unique or there is none.[/quote]

[quote="sonny"][quote="mathgeniuse^ln(x)"][quote="sonny"]How many sides does a polygon have if the sum of its angles is 436?[/quote]

There is definitely a typo, even if the sum was 4360 it still wouldn't work.[/quote]

Yes I noticed that too. It does look like a typo so I dont think anyone should waste their time trying to figure it out.  :lol:[/quote]

Here is a site that explains the sum of interior and exterior angles [url]http://mathforum.org/library/drmath/view/62228.html[/url]


ive tried to get my school to get an arml team but they ignored me and i was wondering if there was any arml team in VA that a guy from fairfax could join? I know there is the TJ team but im in 8th grade at the moment so cant go with them.

There was a team that competed in 2008 that went by the name "Fairfax Math Circle." Was anyone here on that and does anyone have contact information you could give to peregrinefalcon88?

ive heard something about a fairfax math circle also but when i googled it i got nothing can anyone tell me anything about that either?

The only thing I can tell you is that they're the only non-TJ ARML team ever to represent Virginia. TJ did call themselves Fairfax until 1992, and they may allow some outsiders in.

Trying to get an ARML team from one school is generally a bad idea. Thomas Jefferson can do it because they draw many of the best students from the whole state.

Anyone in Virginia who does not go to TJ is eligible for the Chesapeake team.  PM me if you are interested.


1) If f(x) = log_a (x),
show that f(x+h) - f(x)/h = log_a (1 + h/x)^1/h
where h does not = 0.  I see the difference quotient there, right?

Please use LaTeX. Someone told you before, and obviously you didn't listen.

\[\frac{f(x+h) - f(x)}{h} = \frac{\log_a((x+h)/x)}{h} = \frac{1}{h}\log_a(1 + h/x) = \log_a((1 + h/x)^{1/h}) \]

No, I did listen to the others about using Latex but my computer does not allow me to enter the pdf files where Latex is taught, you see. I always listen but my computer has a virus does not allow me to download pdf files and many other files.  I thank you for your help but what are the steps to find the answer on my own?

I've only applied the basic log laws... Most of your other questions can also be solved by simply applying the log laws and simplifying.

Kalle,

Can you show me how to solve the other questions?

[quote="symmetry"]No, I did listen to the others about using Latex but my computer does not allow me to enter the pdf files where Latex is taught, you see. I always listen but my computer has a virus does not allow me to download pdf files and many other files.[/quote]

You don't need to download anything to post in LaTeX on the forum.

Not even a pdf file.

Arne,

Okay.  I did not know that about Latex.


Convert quantitatively cylopentanone to 3-Alkyl cylopentanone

how abt using nbs and elimination? and then treatment wid R2CuLi :huh:

what do you suppose it will? :D i forgot to metnion but 2-benzhydryl-pyrrolidine shud be the solvent

can u link me to that reaction ...i haven't heard of it actuall y this is a beautiful problem corncerning "sterochemistry" real beauty

you can ask chemrock to shed some lite on his long search but you'll have to wait till he finishes his uru suthifying. see #6 http://www.mathlinks.ro/Forum/viewtopic.php?t=185285
i cant see any stereo chemistry in the questiong

then still ur yield won't eb that quantitaive  :D ...

dear par, wat yield are u expecting. 95+ :D

yeah that' why this method is used higly regioselective :)

is it a one step synthesis( i doubt that) or a long one?

no about three steps ...all normal standard reactions :)

1) LDA, THF

2) PhSeBr

3) H2O2, room temperature

4) Gilman reagent

However, a much better synthesis involves only two steps:

[hide]1. RMgBr (for example)

2. PCC[/hide]

my synethesis involved
1.RMgBr/H+
2.HBO(remember it's highly regio selective) and gives 4 equivalent products
3.Elimnation using Alc.KOH ,since HBO is syn and elimation trans so there will be high amount of 2- alhyl cylopentene

Source:Morrison & Boyd  .

1) Those are not three steps.

2) Others products formed, so not a good method.


Prove that for every natural $n$ there exists a natural $N$ such that $(\sqrt{2}-1)^{n}=\sqrt{N}-\sqrt{N-1}$.

Following from what Karth said

you can let $(\sqrt{2}-1)^{-1}=\sqrt{2}+1$, so then you square it and you get

$N=\frac{(3-2\sqrt{2})^{n}+(3+2\sqrt{2})^{n}}{4}+\frac{2}{4}$

Now N is an integer so that thing must be an integer.

let $a_{n}= (3-2\sqrt{2})^{n}+(3+2\sqrt{2})^{n}$. it's obvious $a_{1}= 6$ and $a_{2}= 34$, so we show that $a_{n}$ is 2 mod 4 and we're done. but a_1 = a_2 = 2 mod 4, and $a_{n}$ is the definition for the recurrence $a_{n+2}=6a_{n+1}-a_{n}$, and inductively $a_{n}$ is 2 mod 4 so $N$ is an integer for any $n$ and we're done.

If you've done work with simple continued fractions before, the solution is pretty obvious...but here's a simpler way.
[hide]
Clearly, $\sqrt{2}-1=\sqrt{2}-\sqrt{1}$.

Suppose $a\sqrt{2}-b$ is a power of $\sqrt{2}-1$, and $2a^{2}-b^{2}=1$.  Then \[(a\sqrt{2}-b)(\sqrt{2}-1)=(2a+b)-(a+b)\sqrt{2}\]  \[(2a+b)^{2}-2(a+b)^{2}=4a^{2}+4ab+b^{2}-2a^{2}-4ab-2b^{2}=2a^{2}-b^{2}=1\] Thus, the next power of $\sqrt{2}-1$, namely $(2a+b)-(a+b)\sqrt{2}$, has the desired property.

So I wasn't very thorough, but that's basically an inductive method...because $2a^{2}-b^{2}=1$, we have $2a^{2}=b^{2}+1$, so $a\sqrt{2}-b=\sqrt{2a^{2}}-\sqrt{b^{2}}$, which is the form we want...so $(2a+b)^{2}-2(a+b)^{2}=a$ implies that $\sqrt{(2a+b)^{2}}-\sqrt{2(a+b)^{2}}$ has the form we want as well.[/hide]


geometry
rectangle
3D geometry
tetrahedron
search
invariant
linear algebra

Two (convex) polygons are defined as being "good" if they can be cut up into smaller polygons which are pairwise congruent.
Prove: if two (convex) polygons have the same area then they are "good".

Well, if the number of pieces in the dissection is finite, the usual wording is not 'good' but 'equidecomposable'.
The result you are talking about is a famous theorem of Bolyai and Gerwein which states that :
If A and B are polygons of the same area then A and B are equidecomposables.
Since the dissection only uses polygons, the converse of the theorem is obvious.

Pierre.

oh ok
i didn't know it was a theorom :blush:
are there any elegant proofs for it? (the proof ive seen involves first obviously that if A and B are good, B and C are good then A and C are good - so it would suffice to prove it for triangle-rectangle - which isnt that elegant).

the same problem can be posed ... though for polyhedra of euqal volume... if i'm not mistaken it's the Hilberts third problem. And surprisingly the result is that two equal volume doesn't neccessarily decompose into congruent polyhedra. i think it was for tetrahedra of equal base and height.
the proof of that uses linear algebra i think

psykyk

I have recently read that Max Dehn proved in 1902 that the cube and the regular tetrahedron of equal volumes aren't equidecomposable. I wonder if the proof is really tough. I'd like to read it. I have a reference, but maybe there are better ones available.

Apparently, all I had to do in order to find more about Dehn's proof was to search "equidecomnposable polyhedra" with Google. I found something really cool and interesting:

[url]http://www.math.columbia.edu/~zare/dehninvstuff.html[/url]

This concept of Dehn invariant is beautiful!


If it is that easy, why did you post at "Unsolved Section"?  :)

Cezar would you like to post me some books related with Inequalities.
Because iam going to participate this year to the IMO i hope ...
Can you ? But not only Inequalities can be Geo and N-t ... :roll:

[quote="striker"]Let a,b,c be positive real numbers. Prove that a^4+b^4+c^4+(a+b+c)^4/27 >= 2(ab)^2+2(bc)^2+2(ca)^2
That is the easy problem. ;)[/quote]
Too easy.
We have: $(a+b+c)^4\ge 9(ab+bc+ca)^2$
$\Rightarrow$ We must prove:
$3\sum a^4+2\sum a^2bc\ge 5\sum b^2c^2$
But we have:
$\sum a^4+\sum a^2bc\ge\sum(a^3b+a^3c)$ (Schur) $\ge 2\sum b^2c^2$
$\Rightarrow 2\sum a^4+2\sum a^2bc\ge 4\sum b^2c^2$   (1)
$\sum a^4\ge\sum b^2c^2$     (2)
(1)+(2) $\Rightarrow$  :lol:


trigonometry
inequalities proposed
inequalities

Show that for all $ x$,  $ csc^2 x \plus{} sec^2 x \geq 4$

$ \frac{1}{sinx^2}\plus{}\frac{1}{cosx^2}\ge 2\frac{1}{\frac{sinx^2\plus{}cosx^2}{2}}\equal{}4$ because f(x)=1/x convex for $ x\ge 0$

hello, there is a small typo, it must be $ f(x)\equal{}\frac{1}{x}$ is convex for all $ x$
with $ 0<x<\plus{}\infty$.
Sonnhard.

For all $ x$, we can write,
$ \csc^2 x \plus{} \sec^2 x \geq 2{\sec x}{\csc x} \equal{} \frac {2}{{\sin x}{\cos x}}$

Now, $ {\sin x}{\cos x} \leq \frac {{\sin^2 x} \plus{} {\cos^2 x}}{2} \equal{} \frac{1}{2}$

Thus, $ {\csc^2 x} \plus{} {\sec^2 x} \geq \frac {2}{\frac {1}{2}} \equal{} 4$ ......Done!
 :)


Given a n-tuple, $ (a_1,a_2,\cdots,a_n)$ such as for every permutation of it $ (b_1,b_2,\cdots,b_n)$,  $ a_1 \cdot b_1 \plus{} a_2 \cdot b_2 \plus{}\cdots \plus{} a_n \cdot b_n$ is always the same number prove that all $ a_i$ are equal.

What is the maximum number of permutations ($ f(n)$), for the existence of an n-tuple of all different integers such as that scalar product is always the same number? (clearly $ 1< f(n) < n!$)


Alcumus
probability
complementary counting

On the Saturday of a weekend softball tournament, Team A plays Team B and Team C plays Team D. Then on Sunday, the two Saturday winners play for first and second places while the two Saturday losers play for third and fourth places. There are no ties. One possible ranking of the team from first place to fourth place at the end of the tournament is the sequence ACBD. What is the total number of possible four-team ranking sequences at the end of the tournament?

You should post this on the "Alcumus Problem Discussion" forum.

Actually, to go further, you should have the Bot do that.

Wouldn't that just be $ 4!$? Four letters how many arrangments?

Yes, i believe you are correct, SuperNerd123.

@myyellowducky82: How do you know this is an Alcumus problem?

@smellydabian please dont spam

[u]The Solution[/u]
I dont think it would be 4!...well maybe it would....:oops: Not a 100% confident

I thought the answer was 2^3= $ 8$ Possibilities

The problem says
[quote]
What is the total number of possible four-team ranking sequences at the [u]end[/u] of the tournament?
[/quote]
It doesn't matter who wins wat match just the ranking at the end.

Any team has equal probability to be 1st, 2nd, 3rd and 4th place.
So the answer should be 4!

no, the ranking couldn't be ABCD, CDAB, BACD, BADC, and some others, because since A and B are playing first, only one of them could win and play in the championship game. The same goes with C and D. So it is impossible for A and B to take first and second place, and C and D to take first and second place.

The total number of possible rankings is 16

either A or B wins in the first game (2 possible)
either C or D wins in the second game (2 possible)

The winners of each game play each other for 1st and 2nd place (2 possible)
The losers of each game play each other for 3rd and 4th place (2 possible)

so the total is $ 2\times 2\times 2\times 2 \equal{} 16$

This question was on a sprint round (i forgot which), and i remember the correct answer was 16.

Think of it this way:

There are 2 ways to arrange the winner of each of 4 matches.  Each arrangement leads to a unique placing.  The answer is thus $ 2^4\equal{}\boxed{16}$.

Is that overcounting though? Is that why someone said it was 8?

What do you mean by overcounting? I don't think you are counting something 2 times so...

Sorry, never mind. Yes, i agree with 16 as the answer.

Complementary counting method:

Ok so clearly A,B can't be in 1,2 or 3,4. So we find the ways that fail.

You can choose 2 ways for AB to be in12 or 34.

2 arrangements of AB within that.

2 arrangements of CD within the other two.

Thus, 2x2x2 fail so 24-8=16/

[hide="Alternative Solution"]
We know that both A and B or C and D can't be first and second. We can now use these restrictions to find the combinations.

We have 4 choices for the first team. For example, it can be C.
We have 2 choices for the second team because it can be any team but the team first place played with. For example, it can be B.
We have 2 choices for the third team. For example, it can be D.
We have 1 choice for the last team, because all other teams have a spot already. For example, it can be A.

Multiplying, $ 4\cdot 2 \cdot 2 \cdot 1 \equal{} \boxed{ 16 }$
[/hide]

Looking at the ranking sequence, we see that A and B cannot both win on Saturday, and so neither AB nor BA can be the first and second places. Similarly, CD and DC cannot be the third and fourth places. Thus, the first and second place may be (A or B) and (C or D) or vice versa. This makes 2 (for the order) $\times 2\times 2 = 8$ possibilities for the first and second places. Then the third and fourth places can be two possible arrangements of the losers. In total, this is $8\times 2 = \boxed{16}$ arrangements.

Also, There are 2 options for A and B win and 2 for C and D, leading to 4 possibilities for the finals. Then, there are 2 ways each one could end, giving us 8 possibilities for the 1st and 2nd place. The same goes for 3rd and 4th, giving us 8 + 8, or 16 arrangements

This is definitely Alcumus. Don't post it here.


[quote="Valentin Vornicu"][quote="paganinio"]A MATH christmas, cool.[/quote]I understand this is your first Math Christmas? :D[/quote]
It would big surprize if it is his very first Christmas.  :lol:  :lol:

[quote="paganinio"]A MATH christmas, cool.[/quote]

Dear paganinio, Everyone,

I strongly agree with paganinio, a [i]math[/i] Christmas [i]IS[/i] cool.
It says something when folks would take the time on Chirstmas day to visit this forum.   :)

Now only if there was a MathemaSanta to bring us all wonderful mathematical gifts for Christmas... :lol:

I wish Everyone all the best.

Yours truly,

M.E.B.
(AKA mathfreak)


Prove that if $ a,b,c,d\geq0$ then
\[ \sum\frac{1}{a^{2}\minus{}ab\plus{}b^{2}}\geq\frac{18}{ab\plus{}bc\plus{}cd\plus{}da\plus{}ac\plus{}bd}.\]

If $ \sum$ is cyclic, then the inequality is wrong for a=0 and b=c=d.


You're wrong,.It must be:
 $a,b,c \geq 0$.Prove that:$(ab+bc+ca)(a+b+c)^4 \geq 8(a^2+b^2+c^2)(a^2b^2+b^2c^2+c^2a^2)$ and this proble isn't hard,I will post my solution later (because,i'm very busy now) :)  :)

I am sorry for the mistypo. I have edited it.

PVThuan

You know that this problem is the problem in ussue of recent Math & Magazine. So you don't permit to post and discuss problem now. Please Admin locks this topic.


great article :D i'm really glad that there's no real connection between the name of the probe and the name of the movie

[quote="10000th User"][b]On Target: NASA Rocket Slams Into Comet[/b]
By ALICIA CHANG, AP Science Writer
3 hours ago

PASADENA, Calif. - It sounded like science fiction - NASA scientists used a space probe to chase down a speeding comet 83 million miles away and slammed it into the frozen ball of dirty ice and debris in a mission to learn how the solar system was formed.

The unmanned probe of the Deep Impact mission collided with Tempel 1, a pickle-shaped comet half the size of Manhattan, late Sunday as thousands of people across the country fixed their eyes to the southwestern sky for a glimpse.

The impact at 10:52 p.m. PDT was cause for celebration not only to scientists at NASA's Jet Propulsion Laboratory in Pasadena, but also for the more than 10,000 people camped out at Hawaii's Waikiki Beach to watch it on a giant movie screen.

"It's almost like one of those science fiction movies," said Steve Lin, a Honolulu physician.

The cosmic smash-up did not significantly alter the comet's orbit around the sun and NASA said the experiment does not pose any danger to Earth - unlike the scary comet headed for Earth in the 1998 movie, "Deep Impact."

Scientists at mission control erupted in applause and exchanged hugs as a voice on a speaker proclaimed, "Team, we got a confirmation."

It was a milestone for the U.S. space agency, because no other space mission has flown this close to a comet. In 2004, NASA's Stardust craft flew within 147 miles of Comet Wild 2 en route back to Earth carrying interstellar dust samples.

"A lot of people said we couldn't do this or wouldn't be able to pull it off," said Rick Grammier, the mission's project manager. "It happened like clockwork and I think that's something to be proud of on America's birthday."

Rough images by the mothership that released the probe on its suicide mission 24 hours earlier showed a bright white flash from the comet upon impact, which hurled a cloud of debris into space. When the dust settles, scientists hope to peek inside the comet's frozen core - a composite of ice and rock left over from the early solar system.

In Darmstadt, Germany, David Southwood of the European Space Agency congratulated NASA and controllers erupted into applause upon impact. "The Deep Impact mission brought the world together in an excellent opportunity to make a new step into the advancement of cometary science," he said.

The European agency was observing and photographing the comet collision with its Rosetta spacecraft, which will attempt to rendezvous with a comet in 2014.

"I had some doubts, quite frankly, but it was quite spectacular and a deserved success," said Manfred Warhaut, who heads ESA's Rosetta mission. "The whole thing was so flawlessly put in place and executed it deserves some respect."

The camera of the Deep Impact probe temporarily blacked out twice, probably from being sandblasted by comet debris, NASA scientists said. Still, the probe - on battery power and tumbling toward the comet, using thrusters to get a perfect aim - took pictures right up to the final moments, revealing crater-like features. The last image was taken three seconds before impact.

The energy produced from the impact was equivalent to exploding five tons of dynamite and it caused the comet to shine six times brighter than normal.

Scientists had compared the barrel-shaped probe's journey to standing in the middle of the road and being hit by a semi-truck roaring at 23,000 mph. They expect the crater left behind to be anywhere from the size of a large house to a football stadium and between two and 14 stories deep.

Soon after the crash on the comet's sunlit side, the mothership prepared to approach Tempel 1 to peer into the crater site and send more data back to Earth. The spacecraft was to fly within 310 miles of the comet before activating its dust shields to protect itself from a blizzard of debris.

Comets are frozen balls of dirty ice, rock and dust that orbit the sun. A giant cloud of gas and dust collapsed to create the sun and planets about 4.5 billion years ago and comets formed from the leftover building blocks of the solar system.

NASA's fleet of space telescopes, including the Hubble Space Telescope, Chandra X-ray Observatory and Spitzer Space Telescope, and dozens of ground observatories recorded the impact.

Deep Impact launched Jan. 12 from Cape Canaveral, Fla., on its 268 million-mile voyage. Scientists say the choice of the mission name was a coincidence and not inspired by the movie.[/quote]

Did you site this article?


$Pro1. a)$Find a  polynomial $P(x)\in Q[x]$such that : $min P(x)=\sqrt2$
       and Prove that $degP \geq 4$
      $b$ Find a  polynomial $P(x) \in Z[x]$ such that :$minP(x)=-\sqrt2$
$Pro2$ Find all polynomial$P(x)\in R[x]$ such that :
                $P(x)^2+1=P(x^2+1)$
$Pro3$ Let 2 polynomials $P(x),G(x)\in N[x]$ satisfying:there exist $a;b \in Z^{+} ;a<b$ and:$P(a)=G(a);P(b)=G(b)$
        Prove that:
        If all coeffcients of $P(x)$ are $< b$ then $P(x)=G(x)$.
 Please help me any problem if you can.Thank!

1. P(x) can be find satisfied P'(x)=(x^2-2)^(2k+1)*f(x), f(x) belongs Q[x] and f(sqrt(2))>0.
2. If P(x)^2+1=P(x^2+1), then for any x P(-x)^2=P(x)^2 or P(x)=P(-x) or P(x)=-P(-x).
2.1 Let P(x)=P(-x), then exist f(x): P(x)=f(x^2+1), degP(x)=2degf(x).
2.2. Let P(x)=-P(-x), then P(x)=x*f(x), f(x)=f(-x). Exist only one solution f(x)=1. P(x)=x.
Let P(0,x)=x,P(1,x)=x^2+1,P(n+1,x)=P(1,P(n,x))=P(n,P(1,x)). All solution is one of P(k,x).

[quote="Rust"]1. P(x) can be find satisfied P'(x)=(x^2-2)^(2k+1)*f(x), f(x) belongs Q[x] and f(sqrt(2))>0.
[\quote]
you're WRONG!Please read carefully...$minP(x)=\sqrt2$ not $minP(x)=P(\sqrt2$!
For problem 2 ,I need a proof  .I had known the answer.

Ok! 1a) Let $P(x_0 )=\sqrt 2 \ and \ P(x)>0 \ for \ any \ x\not=x_0.$
Since $P(x)\in Z[x] \ and \ P^{'} (x)\in Z[x] ,\ we \ have \ P^{'}(x_0)=0 \ and \ P(x_0)=\sqrt2,$ $\sqrt2\in Z[x_0].$
Therefore $degP^{'}(x)\ge2 \ (x_0 \ is \ not \ rational ) \ and \ degP(x) \ is \ even, \ then \ degP(x)\ge4.$
1b) Let $P^{'}(x)=c(x-a-b\sqrt2)(x-a+b\sqrt2)x, \ x_0=a+b\sqrt2$ , then $P(x)=cx^4/4-2acx^3/3+c(a^2-2b^2)x^2/2+d,$ and
$P(x_0)=A+B\sqrt2, \ where$
 $A=c(a^4+12a^2b^2+4b^4)+2ac(a^3+6ab^2)/3+c(a^2-2bb^2)(a^2+2b^2)+d$
and  B  rational  expression.  By chose rational c number  B ($B=c*B_1$) can be equal 1 (or -1), by use d  number A=0 ($A=A_0+d$).
2) is correct.
3) Let $P^{(1)}(x)=P(x+1)-P(x), \ P^{(n+1)}(x)=P^{(n)}(x+1)-P^{(n)}(x).$
$We \ have P^{(i)}(x) \in Z[x] \ and P^{(i)}(x)>0 \ if \ x\ge0.$ If $G(b)\geP(b), \ G(a)=P(a) \ We \ have \ G^{(b-a)}(a) \ge P^{(b-a)}(a) \ then$
$if \ G^{(b-a)}(a) \ge P^{(b-a)}(a)+1 \Longrightarrow \ one \ of \ coefficients \ of \ (G-P) \ greater \ than \ b.$

But if $\deg P =4$,I think P doesn't exist,am I wrong?

[quote="Rust"]Ok! 1a) Let $P(x_0 )=\sqrt 2 \ and \ P(x)>0 \ for \ any \ x\not=x_0.$
Since $P(x)\in Z[x] \ and \ P^{'} (x)\in Z[x] ,\ we \ have \ P^{'}(x_0)=0 \ and \ P(x_0)=\sqrt2,$ $\sqrt2\in Z[x_0].$
Therefore $degP^{'}(x)\ge2 \ (x_0 \ is \ not \ rational ) \ and \ degP(x) \ is \ even, \ then \ degP(x)\ge4.$
quote]
The problem haven't been solved...Be sad!!!
[b]pigfly[/b] is RIGHT!(We can prove it quite easily).
So that ,$degP \geq 6$
@Rust: Please check all your thinking before posting!

[quote="Rust"]$Yes, \ if \ deg(P(x))=4, \ then \ minP(x) \ is \ rational \ (P(x) \in Q[x]), \ but \ local \ maximum \ may \ be \ irrational.$
$You \ need \ to \ prove \ deg \ge4, \ what \ I \ do.$[/quote]
Oh yeah,Rust! I had solved the Pro 1 ...IO find a polynomial with $deg =8$ .A question we have :$min degP=?$( This equivalent :Can $depP=6$)
Your solution for Pro 3, I have some difficult to understand...Can you explain me:
[quote="Rust"]
 If $G(b)\ge P(b), \ G(a)=P(a) \ We \ have \ G^{(b-a)}(a) \ge P^{(b-a)}(a) \ then$
$if \ G^{(b-a)}(a) \ge P^{(b-a)}(a)+1 \Longrightarrow \ one \ of \ coefficients \ of \ (G-P) \ greater \ than \ b.$[/quote]

1. Prove, that deg(P(x)>=6.
$Let \ \deg P(x)=4, \ P(x_0)=min=m+\sqrt D, \ m \ and \ D \ rational, \ \sqrt D \ - \ irrational. \ We \ have \ P^{'}(x_0)=0, \ therefore \ x_0=a_0+a_1 \sqrt D, x_1=a_0-a_1 \sqrt D \ a_0, a_1 \ - are \ rational. \ Let \ R(x)=P(x)(mod P^{'}(x), \deg R(x) \le 2. \ We \ have \ P^{'}(x_0)=P^{'}(x_1)=0, \ therefore \ P(x_0)=R(x_0), \ P(x_1)=R(x_1)$
$Therefore, \ P(x_1)=m- \sqrt D<P(x_0) \ contradition.$

Find P(x) with deg(P(x))=6.
$Let P(x_0)=min=m+ \sqrt D, \ and \ P_1(x_0) \ is \ minimal \ degre \ polinom \ in \ Q[x], \ sucth \ that \ P_1(x_0)=0. \ If \ g \ is \ automorfism \ field \ Q(x_0), \ then \ P(g(x_0))=g(P(x_0))=m+g( \sqrt D). \ Therefore \ x_0 \ can \ not \ find \ vief \ a_0+a_1 \sqrt D. \ We \ fiend \ x_0 \ in \ field \$
$F: Q( \sqrt D)=2, \ for \ example \ P(x)=f(x^2).$ $Let \ P^{'}(x)=2xP_2(x^2-D), \ we \ have \ 5 \ extemums \ P( \pm \sqrt{y_1+D)}, \ P( \pm \sqrt{y_2 +D}) \ and \ P(0), \ if \ second \ root \ P_2 \ less \ than \ \sqrt D.$
$Therefore \ you \ can \ find \ P(x)=d(x^6-ax^4+bx^2+c),4a^2-6b=2*r^2.$

Rust!help me now!
[quote="anhminh"]Your solution for Pro 3, I have some difficult to understand...Can you explain me:
[quote="Rust"]
 If $G(b)\ge P(b), \ G(a)=P(a) \ We \ have \ G^{(b-a)}(a) \ge P^{(b-a)}(a) \ then$
$if \ G^{(b-a)}(a) \ge P^{(b-a)}(a)+1 \Longrightarrow \ one \ of \ coefficients \ of \ (G-P) \ greater \ than \ b.$[/quote][/quote]

$I %Error. "proved" is a bad command.
, \ that \ deg(P(x))\ge 8 \ for \ 1a.$


calculus
integration
analytic geometry
graphing lines
slope
trigonometry

Is there an equilateral triangle with all its vertices in the knots of the
integral square lattice? If the answer is YES, give an example, if the answer
is NO, prove it.

[hide="Solution"]Let $ ABC$ be an equilateral triangle with lattice points $ A,B,C$. Denote by $ m_1$ the slope of the line through $ A$ and $ C$ and by $ m_2$ the slope of the line through $ A$ and $ B$. If their angle of intersection is $ \alpha$ ist's obvious that $ \tan\alpha \equal{} \tan\left(\arctan m_2 \minus{} \arctan m_1\right) \equal{} \frac{m_2\minus{}m_1}{1\plus{}m_1m_2}\in\mathbb{Q}$ since by definition $ m_1$ and $ m_2$ are rational. But $ \tan\frac{\pi}{3} \equal{} \frac{1}{\sqrt{3}}\not\in\mathbb{Q}\implies$ There is no equilateral triangle with lattice points as vertices.[/hide]


Find the solution of the equation:
$ 2^{x}+2^{y}=2^{z}$
where $ x,y,z \in\mathbb{N}$ and $ x+y+z=2005$

WLOG let $ x\le y\le z$. We get after dividing by $ 2^{x}$, $ 1+2^{y-x}=2^{z-x}$, which is possible only if $ y-x=0$ and $ z-x=1$. Hence $ x+y+z=3x+1=2005\iff x=668$

So the only solution is $ (x,y,z)=(668,668,669)$.


geometry
number theory theorems
number theory

Can anyone state and prove me the Siegel Theorem on Integer Points?

[quote="puuhikki"]Look at Hindry, Silverman: Diophantine Geometry: An Introduction pages 353-360.[/quote]
Sorry I don't have that book ! Can you at least give me the statement?

Let $ K/\mathbb{Q}$ be a number field, $ R_S$ the ring of $ S$-integers in $ K$ for a finite set of places $ S$, and $ f\in K(C)$ a non-constant function. Assume $ C$ has genus $ g\geq 1$. Then the set $ \{P\in C(K)|f(P)\in R_S\}$ is finite.


[hide]
To get a sum of $0$, $i^x$ and $i^y$ should be $1$ and $-1$. The lowest powers to get this would be $i^2$ and $i^4$. So the answer is $(2*4)^3=8^3=512.$
$(D)$
[/hide]

[quote="la mafioso"]Edit: I have a bit more experience in it, so no need for fixing numerous mistakes anymore!

$i^x + i^y = 0$

If this is true...

$(xy)^3$

What is this simplest product possible? (If x does not equal y, and 0 is not a possible part of the variables $x$ or $y$)

Also, $x$ and $y$ are NATURAL NUMBERS!

A: $8$

B: $16$

C: $27$

D: $512$

E: $1024$

Show work as needed, and believe me, there is a simplest product in there.[/quote]

$1+i^{x-y}=0$
$i^{x-y}=i^2$
$x-y\equiv 2 \mod{4}$

since they are natural numbers, and we want lowest values, then y=1, x=3, so the answer is $\boxed{C=27}$

also, those are pretty bad answer choices, you can eliminate the first 2 with ease since we are dealing with natural numbers...

you don't even need to do much work for this one...

[hide]u know 4 things:
i to the first power is i.
i to the second power is -1.
i to the third power is -i.
i to the fourth power is 1.

thus, u can make two sets that could fit into the expression:
x=1, y=3

x=2, y=4

from glance, u can tell that xy for the first set is smaller than the xy for the latter set

so C is the answer[/hide]

Oh, I forgot about $i$ and $-i$... :blush:


Έστω $ X \equal{} \{x_1,\ldots,x_n\} \subseteq \ell_1^2$. Μπορούμε εύκολα να υπολογίσουμε την διάμετρο του Χ σε χρόνο $ O(n^2)$. Υπάρχει καλύτερος αλγόριθμος;

Τώρα που το ξαναβλέπω δεν είμαι σίγουρος αν είναι κατανοητή η ορολογία. Εννοώ $ X \subseteq \mathbb{R}^2$ με την 1-norm $ \| \cdot \|_1$.

[hide]
Μπορείτε να βρείτε ένα καλύτερο αλγόριθμο αν οι αποστάσεις δίνονται απο την  $ \| \cdot \|_{\infty}$ ;
[/hide]

Σίγουρα θες να αλλάξεις το norm σε $ \Vert \cdot \Vert_{\infty}$? Γιατί στο αρχικό ποστ δεν φαινόταν τέτοιο intention. In any case, με $ \Vert \cdot \Vert_{1}$ μπορεί να γίνει σε γραμμικό χρόνο, και το trick είναι ωραίο για όσους δεν το έχουν ξανασκεφτεί. Αν δεν υπάρξει ενδιαφέρον από τρίτους θα γράψω λύση.

Ο χρόνος είναι πράγματι γραμμικός. Ίσως όμως να σκεφτόμαστε διαφορετική λύση.

Θα δώσω τότε ένα [hide="hint"]Μπορεί να γίνει σε $ k$ διαστάσεις σε χρόνο $ \Theta(k 2^k n)$, δηλαδή γραμμικό στο $ n$. Σκεφτείτε ότι για $ x \in \mathbb{R}^k$ έχουμε $ \Vert x \Vert_1 \equal{} \max_{q \in \{\pm 1\}^k} \langle x, q \rangle$, όπου $ \langle \cdot, \cdot \rangle$ το κλασικό dot product.[/hide]


calculus
derivative
function
real analysis
real analysis unsolved

[color=red]One more challenge ![/color]
 After a similar problem(see : solved problems. in this section), I wonder if there exists a real function defined on the set of the real numbers with continous derivative and
 $(f^{-1}(x) )' = (f')^{-1}(x) , \forall x \in IR$

 [u]Babis[/u]

But for $f(x) = x$ the inverse of the derivative does not exist , because $f'(x) = 1$ and $f'(x)$ is not  $<< 1-1>>$.
 I have proved in the past that we can not find a function with  $f'(x) = f^{-1}(x)$ , and f defined on IR. Could this help ?I'm not sure.

  Babis

[quote="stergiu"]
 $(f^{-1}(x) )' = (f')^{-1}(x) , \forall x \in \mathbb R$
[/quote]
One trivial remark: if $(f')^{-1}$ is defined for all $x\in\mathbb R$, then there exists a point $x_{0}$ such that $f'(x_{0})=0$. But then $f^{-1}$ is not differentiable at $f(x_{0})$. So, probably, it is better to ask about such functions on $(0,+\infty)$ instead of $\mathbb R$.

Mmm !
I think that this note solves the initial problem and the answer is ''no''.Now , for other functions on $(0,+\infty )$ , it will be an open problem.
  Thanks for helping.

        [u] Babis [/u]


logarithms
inequalities
calculus
calculus computations

Prove that if $n>2$ then we have \[{ n^2 > \frac{n^2}{\sqrt[n]{2}} + \log_2 \log_2 \frac{n^2}{\sqrt[n]{2}}} .  \]

I don't whether this is the right forum for it, but the "Inequalities" forum seemed a little unappropriate for this type of problem.

yes, the problem asks for $n$ integer

however, it may be true for $n \geq 3$ real (considering $2^{1/n}$ instead of $\sqrt[n]2$).

This is a horrible problem, and nobody will ever solve it, right? I mean, i can't see any nice approach

 :blush:

[quote="perfect_radio"]I mean, i can't see any nice approach[/quote]
Then you should be looking for an ugly approach.

By the power series, $2^{-\frac1n}<1-\frac{\ln 2}n+\frac{\ln^2 2}{2n^2}$, so
${{n^2-n\ln 2+\frac{\ln^2 2}2+\log_2 \log_2 \frac{n^2}{\sqrt[n]{2}}} > \frac{n^2}{\sqrt[n]{2}} + \log_2 \log_2 \frac{n^2}{\sqrt[n]{2}}}$

If we can show that ${\log_2 \log_2 \frac{n^2}{\sqrt[n]{2}}}<n\ln 2-\frac{\ln^2 2}2$, we will be done.
First, ${\log_2 \log_2 \frac{n^2}{\sqrt[n]{2}}}<\log_2\log_2 n^2=\frac{\ln\ln n}{\ln 2}+1-\frac{\ln\ln 2}{\ln 2}$. Consolidating the constants, $1-\frac{\ln\ln 2}{\ln 2}\approx 1.529$ and $\frac{\ln^2 2}2\approx .240$, so we want to show that
$\frac{\ln\ln n}{\ln 2}<n\ln 2-1.529-.240=n\ln 2-1.769\approx n\ln 2-2.552\ln 2$.
That 2.552 is ugly, so we instead prove the stronger inequality
$\frac{\ln\ln n}{\ln 2}<(n-e)\ln 2$.

By concavity of the logarithm and its tangent at $e$, $\ln n\le\frac ne$. Therefore $\frac{\ln\ln n}{\ln 2}\le\frac{\ln n-\ln e}{\ln 2}=\frac{\ln n-1}{\ln 2}$. By the same inequality, $\ln n-1\le\frac{n-e}e$ and $\frac{\ln\ln n}{\ln 2}\le\frac{n-e}{e\ln 2}$. Since $\frac1{e\ln 2}<\ln 2$, $\frac{\ln\ln n}{\ln 2}<(n-e)\ln 2$ whenever $n>e$ and we are done.

The inequality is true for all real values $n>e$. It's not sharp anywhere.

And this finally solves this problem: http://www.mathlinks.ro/Forum/viewtopic.php?p=181667#p181667 , which had an official solution which spread over many pages.


Given that 6x + y= 15, the value of 3x can be written in terms of y as ay+b for some numbers a and b. What is the simplified value of a +b.

Simply use algebra.
$ 6x\equal{}\minus{}y\plus{}15$
$ 3x\equal{}\minus{}\frac{1}{2}y\plus{}\frac{15}{2}$

Now you have $ a\equal{}\minus{}\frac{1}{2}$ and $ b\equal{}\frac{15}{2}$. So $ a\plus{}b\equal{}\frac{15}{2}\minus{}\frac{1}{2}\equal{}\boxed{7}$.


Let a,b,c be real numbers such that a,b,c>1 and abc=2 \sqrt 2.
Prove that

   (a+1)(b+1)(c+1)>8(a-1)(b-1)(c-1)

When abc=p 3 and a,b,c>1 we have:
 (a+1)(b+1)(c+1) \geq (p+1)3  and
 (a-1)(b-1)(c-1) \leq (p-1)3

We prove that using...brute force :D
(a+1)(b+1)(c+1)=abc+ \sum a+\sum ab+1 \geq abc+
3(abc)1/3 +3(abc)2/3  \geq
p3 +3p2 +3p+1=(p+1)3

(a-1)(b-1)(c-1)=abc-\sum ab+\sum a-1=p3 -1+\sum a-\sum ab  (i)
But if a=1+x, b=1+y, c=1+z where x,y,z>0 we have that  \sum ab-\sum a=\sum xy+\sum x and also that p3=abc=(1+x)(1+y)(1+z)=1+xyz+\sum x+\sum xy=p3  (1)
=> p3\geq xyz+1+3(xyz)2/3+3(xyz) 1/3=(1+(xyz)1/3)3=>(p-1)3\geq xyz (2)
Returning in (1) we have that  \sum xy+\sum x=p3 -1-xyz
Using (2) now we obtain that \sum xy+\sum x\geq 3p2-3p.
But then  \sum ab-\sum a=\sum xy+\sum x\geq 3p2-3p
and (i) will imply (a-1)(b-1)(c-1)\leq p3-1- 3p2+3p=(p-1)3

In our case, p=\sqrt 2 and all we need to prove now is that  ( \sqrt 2+1) 3 \geq 8(\sqrt 2-1)3  and after a short computation we find that it is true.

Your proof is very nice and from it can be deduced that a better value
for that inequality is

 (a+1)(b+1)(c+1)>125(a-1)(b-1)(c-1)

Why using brute force for such a trivial inequality?? :)

 [i] Solution. [/i]

    We have \[ \frac{a+1}{a-1}=1+\frac{2}{a-1}\geq 2\sqrt{\frac{2}{a-1}}>\frac{2\sqrt{2}}{\sqrt{a}}. \]
     Writing the analogues ones, and by multiplying these inequalities one can easily get

            \[ \frac{(a+1)(b+1)(c+1)}{(a-1)(b-1)(c-1)}>\frac{8\cdot 2\sqrt{2}}{\sqrt{abc}}=8\sqrt{abc}>8 \]

  P.S. Mihnea, please from now on try not to talk like that to mister Gigel Militaru. If you don't know, he is a master
         in non-comutative algebra. Show some respect here!!! :idea:

This solution I've seen it in a book from GIL, Cezar.

I think a pupil in 7-th grade can find this solution easily. ;)

You are right.This problem is 2002 or 2003 Shortlist in Romania.It was easy!


For a,b,c positive  and a+b+c=1  prove that $ 2\plus{}9abc \ge 7(ab\plus{}bc\plus{}ac)$

Here's a "0% intelligence" solution:

Your inequality is equivalent to

$ 2 (a \plus{} b \plus{} c)^3 \plus{} 9 a b c \minus{} 7 (a \plus{} b \plus{} c) (a b \plus{} b c \plus{} c a)\geq 0$

$ 2 a^3 \minus{} a^2 b \minus{} a b^2 \plus{} 2 b^3 \minus{} a^2 c \minus{} b^2 c \minus{} a c^2 \minus{} b c^2 \plus{} 2 c^3\geq 0$

$ 2\sum a^3\geq\sum a^2 b$ which is true by AM-GM or Muirhead.

Similar, but a bit more sophisticated  :P
Observe that $ 1\equal{}(a\plus{}b\plus{}c)^{2}\geq 3\sum_{cyc}ab$
Now we have to prove $ 1\plus{}9abc\geq 4(ab\plus{}bc\plus{}ca)$, but
$ ((a\plus{}b\plus{}c)^{2}\plus{}\frac{9abc}{a\plus{}b\plus{}c}\geq 4(ab\plus{}bc\plus{}ca))\iff (\sum_{cyc}a(a\minus{}b)(a\minus{}c)\geq 0)$ is another version of Schur... :wink:

[quote="Elio (n)"] [color=darkred]For $ a\ ,\ b,\ c$ positive  and $ a \plus{} b \plus{} c \equal{} 1$  prove that $ 2 \plus{} 9abc \ge 7(ab \plus{} bc \plus{} ac)\ .$[/color] [/quote]
[quote="Virgil Nicula"][color=darkred][b][u]Generalization[/u].[/b] For $ a\ ,\ b,\ c$ positive, $ \alpha\ne 0$  and $ a \plus{} b \plus{} c \equal{} 1$  prove that

$ \boxed {\ 9abc \plus{} \alpha\cdot\max\ \left\{\frac 14\ ,\ \frac {\alpha \minus{} 1}{3\alpha}\right\}\ \ge\ \alpha\cdot (ab \plus{} bc \plus{} ca)\ }\ .$ [/color][/quote]

[color=darkblue][u]Particular cases[/u].

$ \begin{array}{cccc}
\blacktriangleright & \alpha\in (0\ ,\ 4) & \implies & 36abc \plus{} \alpha \ \ge\ 4\alpha (ab \plus{} bc \plus{} ca)\ .\\\\
\blacktriangleright & \alpha < 0\ \vee\ \alpha\ge 4 & \implies & 27abc \plus{} (\alpha \minus{} 1)\ \ge\ 3\alpha (ab \plus{} bc \plus{} ca)\ .\\\\
\blacktriangleright & \alpha \equal{} 4 & \implies & 4(ab \plus{} bc \plus{} ca)\ \le\ 1 \plus{} 9abc\ .\\\\
\blacktriangleright & \alpha \equal{} \frac 92 & \implies & ab \plus{} bc \plus{} ca \minus{} 2abc\ \le\ \frac {7}{27}\ \ \mathrm{(I.M.O., 1984)}\ .\\\\
\blacktriangleright & \alpha \equal{} 5 & \implies & 15(ab \plus{} bc \plus{} ca)\ \le\ 4 \plus{} 27abc\ .\\\\
\blacktriangleright & \alpha \equal{} \frac {45}{8} & \implies & 8abc \plus{} \frac {37}{27}\ \ge\ 5(ab \plus{} bc \plus{} ca)\ .\\\\
\blacktriangleright & \alpha \equal{} 7 &  \implies & \mathrm {proposed\ problem.}\\\\
\blacktriangleright & \alpha \equal{} 9 & \implies & (a \plus{} b \plus{} c)^3\ \le\ 9\left(a^3 \plus{} b^3 \plus{} c^3\right)\ .\end{array}$

[b]Sorry[/b], I"present my proof after two weeks.[/color]

[quote="Virgil Nicula"][quote="Elio (n)"] [color=darkred]For $ a\ ,\ b,\ c$ positive  and $ a \plus{} b \plus{} c \equal{} 1$  prove that $ 2 \plus{} 9abc \ge 7(ab \plus{} bc \plus{} ac)\ .$[/color] [/quote]
[quote="Virgil Nicula"][color=darkred][b][u]Generalization[/u].[/b] For $ a\ ,\ b,\ c$ positive, $ \alpha\ne 0$  and $ a \plus{} b \plus{} c \equal{} 1$  prove that

$ \boxed {\ 9abc \plus{} \alpha\cdot\max\ \left\{\frac 14\ ,\ \frac {\alpha \minus{} 1}{3\alpha}\right\}\ \ge\ \alpha\cdot (ab \plus{} bc \plus{} ca)\ }\ .$ [/color][/quote]

[color=darkblue][u]Particular cases[/u].

$ \begin{array}{cccc} \blacktriangleright & \alpha\in (0\ ,\ 4) & \implies & 36abc \plus{} \alpha \ \ge\ 4\alpha (ab \plus{} bc \plus{} ca)\ . \\
 \\
\blacktriangleright & \alpha < 0\ \vee\ \alpha\ge 4 & \implies & 27abc \plus{} (\alpha \minus{} 1)\ \ge\ 3\alpha (ab \plus{} bc \plus{} ca)\ . \\
 \\
\blacktriangleright & \alpha \equal{} 4 & \implies & 4(ab \plus{} bc \plus{} ca)\ \le\ 1 \plus{} 9abc\ . \\
 \\
\blacktriangleright & \alpha \equal{} \frac 92 & \implies & ab \plus{} bc \plus{} ca \minus{} 2abc\ \le\ \frac {7}{27}\ \ \mathrm{(I.M.O., 1984)}\ . \\
 \\
\blacktriangleright & \alpha \equal{} 5 & \implies & 15(ab \plus{} bc \plus{} ca)\ \le\ 4 \plus{} 27abc\ . \\
 \\
\blacktriangleright & \alpha \equal{} \frac {45}{8} & \implies & 8abc \plus{} \frac {37}{27}\ \ge\ 5(ab \plus{} bc \plus{} ca)\ . \\
 \\
\blacktriangleright & \alpha \equal{} 7 & \implies & \mathrm {proposed\ problem.} \\
 \\
\blacktriangleright & \alpha \equal{} 9 & \implies & (a \plus{} b \plus{} c)^3\ \le\ 9\left(a^3 \plus{} b^3 \plus{} c^3\right)\ .\end{array}$

[b]Sorry[/b], I"present my proof after two weeks.[/color][/quote]
My proof for the generalisation("arqady-style"):
Let $ 3u \equal{} a\plus{}b\plus{}c$, $ 3v^2 \equal{} ab\plus{}bc\plus{}ca$, $ w^3 \equal{} abc$, and $ u^2 \equal{} tv^2$, so $ t \ge 1$.
First the case $ \alpha \in (0;4)$. Then it's equivalent to:
$ 4w^3 \plus{} 3\alpha u^3 \ge 4\alpha uv^2$
From $ (a\minus{}b)^2(b\minus{}c)^2(c\minus{}a)^2 \ge 0$ we get $ w^3 \ge 3uv^2 \minus{} 2u^3 \minus{} 2\sqrt{(u^2\minus{}v^2)^3}$ (and equality can be obtained when $ t \le \frac{4}{3}$). Assume $ t < \frac{4}{3}$:
$ 4w^3 \plus{} 3\alpha u^3 \ge 12uv^2 \minus{} 8u^3 \minus{} 8\sqrt{(u^2\minus{}v^2)^3} \plus{} 3\alpha u^3 \ge 4\alpha uv^2 \iff$
$ 12 \sqrt{t} \minus{} 8\sqrt{t}t \minus{} 8\sqrt{(t\minus{}1)^3} \plus{} 3\alpha \sqrt{t}t \minus{} 4 \alpha \sqrt{t} \ge 0 \iff$
$ \sqrt{t}( (12\minus{}4\alpha) \plus{} (3\alpha\minus{}8)t ) \ge 8\sqrt{(t\minus{}1)^3}$
Then first prove that $ RHS \ge 0$:
$ 3\alpha\minus{}8 \ge 0 \Rightarrow (12\minus{}4\alpha) \plus{} (3\alpha\minus{}8)t \ge 4\minus{}\alpha \ge 0$
$ 3\alpha\minus{}8 < 0 \Rightarrow (12\minus{}4\alpha) \plus{} (3\alpha\minus{}8)t \ge (12\minus{}4\alpha) \plus{} \frac{4}{3}(3\alpha\minus{}8) \equal{} 12 \minus{} \frac{32}{3} > 0$
Then we just have to prove:
$ (\sqrt{t}( (12\minus{}4\alpha) \plus{} (3\alpha\minus{}8)t ))^2 \ge (8\sqrt{(t\minus{}1)^3})^2$
$ t( (12\minus{}4\alpha) \plus{} (3\alpha\minus{}8)t )^2 \ge 64(t\minus{}1)^3$
$ (9\alpha^2 \minus{} 48\alpha)t^3 \plus{} (\minus{}24\alpha^2 \plus{} 136\alpha)t^2 \plus{} (16\alpha^2 \minus{} 96\alpha \minus{} 48)t \plus{} 64 \ge 0 \iff$
$ (4\minus{}3t)((\minus{}3\alpha^2\plus{}16\alpha)t^2 \plus{} (4\alpha^2 \minus{} 24\alpha)t \plus{} 16) \ge 0 \iff$
$ g(\alpha) \equal{} \alpha^2(\minus{}3t^2\plus{}4t) \plus{} (\minus{}24t\plus{}16t^2)\alpha \plus{} 16 \ge 0$
Now $ g''(\alpha) \equal{} \minus{}3t^2 \plus{} 4t > 0$, so $ g(\alpha)$ is convex and so we just have to check the case where $ g'(\alpha) \equal{} 0$, to find the minimum. $ g'(\alpha) \equal{} 2t((\minus{}3t\plus{}4)\alpha \minus{} (12\minus{}8t)) \equal{} 0 \iff \alpha \equal{} \frac{12\minus{}8t}{4\minus{}3t}$
And $ g \left ( \frac{12\minus{}8t}{4\minus{}3t} \right ) \equal{} \frac{64(t\minus{}1)^3}{3t\minus{}4} \ge 0$. Equality iff $ t \equal{} 1$ and $ \alpha \equal{} \frac{12\minus{}8t}{4\minus{}3t} \equal{} 4$. But $ \alpha < 4$, so there cannot be equality when $ t < \frac{4}{3}$
For $ t \ge \frac{4}{3}$ we have:
$ 4w^3 \plus{} 3\alpha u^3 \ge 3\alpha u^3 \ge 4\alpha uv^2 \iff$
$ 3u^2 \ge 4v^2 \iff$
$ t \ge \frac{4}{3}$, which is our assumption. So equality when $ t \equal{} \frac{4}{3}$ and $ w\equal{}0$. (That is $ a\equal{}b\equal{}\frac{1}{2}, c\equal{}0$ or permutations)

That concludes the proof of the case $ \alpha \in (0; 4)$. Equality iff $ a\equal{}b\equal{}\frac{1}{2}, c\equal{}0$ and cyclic permutations.

The inequaltiy doesn't hold for $ \alpha < 0$. Try $ a\equal{}1,b\equal{}c\equal{}0$ then we have $ \alpha \minus{} 1 \ge 0$, which is clearly wrong. So the proof for the case $ \alpha \ge 4$: It is equivalent to:
$ (w^3 \plus{} 3u^3 \minus{} 4uv^2) \plus{} u(\alpha\minus{}4)(u^2\minus{}v^2) \ge 0$. But $ w^3 \plus{} 3u^3 \minus{} 4uv^2 \ge 0$ is Schur and $ u(\alpha\minus{}4)(u^2\minus{}v^2) \ge 0$ is obvious. So equality iff $ a\equal{}b\equal{}c\equal{}\frac{1}{3}$ or $ \alpha\equal{}4$ and $ a\equal{}b\equal{}\frac{1}{2},c\equal{}0$ and permutations.

Note: The proof for the case $ \alpha \in (0;4)$ includes a lot of calculations. But I think that we can quite easily prove that we only have to consider the cases where $ a\equal{}b$ and $ a\equal{}0$.

[quote="Elio (n)"]This is my solution :

Since $ a \plus{} b \plus{} c \equal{} 1$ than by squaring both sides we get $ a^{2} \plus{} b^{2} \plus{} c^{2} \equal{} 1 \minus{} 2ab \minus{} 2ac \minus{} 2bc$

So the inequality is equivalent to $ 2(a^{2} \plus{} b^{2} \plus{} c^{2}) \plus{} 9abc \ge 3(ab \plus{} bc \plus{} ac)$

Which is equivalent to $ (a \minus{} b)^{2} \plus{} (b \minus{} c)^{2} \plus{} (a \minus{} c)^{2} \plus{} 9abc \ge ab \plus{} bc \plus{} ac$

Then we  can transform the last inequality as $ (a \minus{} b)^{2} \plus{} (b \minus{} c)^{2} \plus{} (a \minus{} c)^{2} \minus{} a(b \minus{} c)^{2} \minus{} b(a \minus{} c)^{2} \minus{} c(a \minus{} b)^{2}\ge 0$ and finally we get $ (a \minus{} b)^{2}(a \plus{} b) \plus{} (b \minus{} c)^{2}(b \plus{} c) \plus{} (a \minus{} c)^{2}(a \plus{} c) \ge 0$ which is obsvually true . Equality holds for $ a \equal{} b \equal{} c \equal{} \frac {1}{3}$

I have one question : in what does Mourihead  consist ? I see that you guys  use it  very often.  Thank you[/quote]

May be I stupid, would you explain how to tranform

$ \boxed{(a \minus{} b)^{2} \plus{} (b \minus{} c)^{2} \plus{} (a \minus{} c)^{2} \minus{} a(b \minus{} c)^{2} \minus{} b(a \minus{} c)^{2} \minus{} c(a \minus{} b)^{2}\ge 0}$

form last ineq.


calculus
derivative
inequalities unsolved
inequalities

Find the minium of
$f(x)=|1+2sinx|+|1+2cosx|$

it also has maximum:

just by derrivative :D

it 's VietNam university entrance examination: ;)


integration
real analysis
real analysis solved

Let f:[0, 1]->R continuous s.t. for any x in [0, 1] we have  \int {x, 1} f(t)dt >= (1-x^2)/2. Prove that  \int {0, 1} (f(t))^2dt >=1/3.

This might be a little harder.

Because f is continuous we get that f admits primitives on [0,1] (hurray) :D
Let F: [0,1] --> R be a primitive of f. SO F is derivable (again hurray) with the continous derivate :  \int {x,1} f(t)dt>=(1-x^2)/2 for any x in [0,1] which is equivalent to F(1)-F(x)>=(1-x^2)/2 for any x in [0,1] and now we integrate it and get that:

 \int {0,1} (F(1)-F(x))dx >=  \int {0,1} (1-x^2)/2 dx =1/2 (x-x^3/3) (computed from 0 to 1) =1/3 [1].

On the onther hand we have that (xF(x))'=x*F'(x)+F(x)=xf(x)+F(x) for any x in [0,1] so we get that:

 \int {0,1} (xF(x))' dx =  \int {0,1} xf(x)dx +  \int {0,1} F(x)dx which is equivalent to saying that:
x*F(x) (computed from 0 to 1) =  \int {0,1} x*f(x)dx+ \int {0,1} F(x)dx which is equivalent to :

F(1)-  \int {0,1} F(x)dx=  \int {0,1} xf(x)dx which is :  \int {0,1} xf(x)dx=  \int {0,1} xf(x)dx=  \int {0,1} (F(1)-F(x)) dx [2].

From [1] and [2] we get that  \int {0,1}xf(x)dx>=1/3. Because (f(x)-x)^2>=0 , for any x in [0,1] which gives :
f^2(x)>=2xf(x)-x^2, for any x in [0,1] and integrating it gives us:
 \int {0,1} f^2(x)dx >= 2  \int {0,1} x *f(x)dx-x^3/3 (computed from 0 to 1)>=1/3 and it's all done.

cheers! :D :D

I've got it! You have the same book I have! :D Where did you find it? I've heard it's kind of hard to find. I mean "Probleme date la olimpiadele de matematica pentru licee"  by DM batinetu-Giurgiu (really cool teacher) and IV Maftei and others.

I don't have that book .. :( (gonna look for that anyway! )

I have the official solutions from the national stage on some paper!!

cheers!!


floor function
ceiling function
combinatorics open
combinatorics

First off, your sequence is [url=http://oeis.org/A005566]A005566[/url] with a somewhat simpler formula $P(n)=\binom{n}{\left\lfloor \frac{n}{2}\right\rfloor} \binom{n+1}{\left\lfloor \frac{n+1}{2}\right\rfloor}$.

Suppose that there are k vertical moves and n-k horizontal moves in a path.
Note that there are $f(k)$ different vertical subpaths and $f(n-k)$ different horizontal subpaths where $f(k)=\binom{k}{\left\lfloor \frac{k}{2} \right\rfloor}$ (see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=359854]that thread[/url]). Therefore, there are $\binom{n}{k} f(k) f(n-k)$ paths with k vertical moves and n-k horizontal moves, and the total number of path of length n is
\[ \sum_{k=0}^n \binom{n}{k} \binom{k}{\left\lfloor \frac{k}{2}\right\rfloor} \binom{n-k}{\left\lfloor \frac{n-k}{2}\right\rfloor} = \sum_{k=0}^n \frac{n!}{\left\lfloor \frac{k}{2}\right\rfloor! \left\lceil \frac{k}{2}\right\rceil! \left\lfloor \frac{n-k}{2}\right\rfloor! \left\lceil \frac{n-k}{2}\right\rceil!}.  \]
In particular, for even n, we have
\[ \sum_{k=0}^n \frac{n!}{\left\lfloor \frac{k}{2}\right\rfloor! \left\lceil \frac{k}{2}\right\rceil! \left\lfloor \frac{n-k}{2}\right\rfloor! \left\lceil \frac{n-k}{2}\right\rceil!} = \binom{n}{\frac{n}{2}} \sum_{k=0}^n \binom{\frac{n}{2}}{\left\lfloor \frac{k}{2}\right\rfloor} \binom{\frac{n}{2}}{\left\lceil \frac{k}{2}\right\rceil} = \binom{n}{\frac{n}{2}} \binom{n+1}{\frac{n}{2}}.  \]
For odd n, the formula is obtained similarly.

[quote="maxal"]First off, your sequence is [url=http://www.research.att.com/~njas/sequences/A005566]A005566[/url] with a somewhat simpler formula $P(n)=\binom{n}{\left\lfloor \frac{n}{2}\right\rfloor} \binom{n+1}{\left\lfloor \frac{n+1}{2}\right\rfloor}$.

Suppose that there are k vertical moves and n-k horizontal moves in a path.
Note that there are $f(k)$ different vertical subpaths and $f(n-k)$ where $f(k)=\binom{k}{\left\lfloor \frac{k}{2} \right\rfloor}$ (see [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=359854]that thread[/url]). Therefore, there are $\binom{n}{k} f(k) f(n-k)$ paths with k vertical moves and n-k horizontal moves, and the total number of path of length n is
\[ \sum_{k=0}^n \binom{n}{k} \binom{k}{\left\lfloor \frac{k}{2}\right\rfloor} \binom{n-k}{\left\lfloor \frac{n-k}{2}\right\rfloor} = \sum_{k=0}^n \frac{n!}{\left\lfloor \frac{k}{2}\right\rfloor! \left\lceil \frac{k}{2}\right\rceil! \left\lfloor \frac{n-k}{2}\right\rfloor! \left\lceil \frac{n-k}{2}\right\rceil!}.  \]
In particular, for even n, we have
\[ \sum_{k=0}^n \frac{n!}{\left\lfloor \frac{k}{2}\right\rfloor! \left\lceil \frac{k}{2}\right\rceil! \left\lfloor \frac{n-k}{2}\right\rfloor! \left\lceil \frac{n-k}{2}\right\rceil!} = \binom{n}{\frac{n}{2}} \sum_{k=0}^n \binom{\frac{n}{2}}{\left\lfloor \frac{k}{2}\right\rfloor} \binom{\frac{n}{2}}{\left\lceil \frac{k}{2}\right\rceil} = \binom{n}{\frac{n}{2}} \binom{n+1}{\frac{n}{2}}.  \]
For odd n, the formula is obtained similarly.[/quote]

Lol darnit I can't believe I spent so long thinking about this problem and didn't notice that the sum simplified. Thanks a lot for the info, it's interesting to see that somebody's already thought of this. I guess now the goal is to generalize it into arbitrary dimensionalities, heheheh.


Euler
search
integration
calculus
calculus computations

I'm in a numerical analysis class and I'm trying to approximate an initial value problem. I get the approximation using Forward Euler and other methods, but I'm not sure what the true answer is. I haven't done ODE's in a while.

Does anyone know how to solve the problem?
y'(t) = y - t^2 + 1, t in [0,2]
y(0) = 2

Thanks for any help in advance.

Let's use the  variation of constants method .  :D  . The solution of the homogenous equation $ y'\equal{}y$ is $ \ y \equal{} Ce^t$
Now you search the solution $ y(t) \equal{} K(t)e^t$ so $ K'(t)e^t \equal{} \minus{}t^2\plus{}1\Longrightarrow K(t)\equal{}K(0) \plus{}\int_0^t e^{\minus{}t}(\minus{}t^2\plus{}1) \equal{} ...$
Finally you determine $ K(0)$ from $ y(0) \equal{} 2$ and so on ..
 :cool:

hello, the solution of the homogenious part of your equation is $ y(t)\equal{}Ce^t$, for the particular part you can make the ansatz $ y(t)\equal{}At^2\plus{}Bt\plus{}C$ with the real numbers $ A,B,C$.
Sonnhard.

$ y'(t)\minus{}y(t)\equal{}1\minus{}t^2$

$ e^{\minus{}t}y'(t)\minus{}e^{\minus{}t}y(t)\equal{}(1\minus{}t^2)e^{\minus{}t}$

$ \Longleftrightarrow \frac{d}{dt}e^{\minus{}t}y(t)\equal{}(1\minus{}t^2)e^{\minus{}t}$

$ \therefore e^{\minus{}t}y(t)\equal{}(t\plus{}1)^2e^{\minus{}t}\plus{}C$.

$ t\equal{}0\Longrightarrow y(0)\equal{}C\equal{}2$, yielding $ y(t)\equal{}(t\plus{}1)^2\plus{}2e^t\ (0\leq t\leq 2)$.


algebra
polynomial
geometry
congruent triangles

:?:
How many triangles can be found in a triangular region of the triangular lattice? I mean, if the side length of the region is 1, then there is 1 triangle; if the side length of the region is 2, then there are 5 trianges; if the side length is 3, there are 13 triangles. Looking for a general formula...

How many triangles are there whose three vertices are each in a rectangular grid of dots? (I believe the case with 3x3 dots was a CD question at state...)

How many triangles are there from a triangular grid of dots?

How many non-congruent triangles for each of the above?

There is a set of n points in the plane. c of them are collinear. None of the others are. How many triangles can be drawn with three of these points as vertices?

for the first, I don't really think there is a formula. (I bet tor will prove me wrong) :lol:

Basically for the CD problem, it is 9C3, then just minus the cases where the points are collinear. so 73 for that.

So basically total amount of points  and make it Choose 3. (so if 8 points, 8C3,), and minus the times where  the points are collinear.

For the triangular shaped grid, use the same strategy as for the rectangular shaped grid.

For non-congruent, I would just use casework.

I don't understand your last question, could you clarify?

Nevermind about the last question, it's a simpler case of the other questions.

Is there a formula for the number of sets of collinear points?

I don't get what you guys are talking about. Could someone clarify what leoxnlin is asking?

Are you unclear about the questions in my first post or in my second post?

Rephrasing for tinytim:

1. Tile an equilateral triangle of side length n with equilateral triangles of side length 1. How many triangles are in the figure? (How many triangles are formed with the edges of the smallest triangles?)

Ex. n = 1, Answer = 1; n = 2, Answer = 5; etc.

2. Draw an MxN rectangular grid of lattice points. How many triangles can be drawn that have all three vertices as one of these points?

3. Instead of a rectangular grid of lattice points, use a triangular arrangement (the vertices in the figure from #1).

4. How many different non-congruent triangles can be formed in #'s 2 and 3?

In my second post, I am asking if there is a quick formula for the number of sets of three collinear points in #'s 2 and 3.

I would guess $ n^2\plus{}\sum_{i\equal{}1}^{n\minus{}1} t_n$ where $ t_n\equal{}\frac{n(n\plus{}1)}{2}$ simply by looking at the triangles for the first few examples.  It's easy to check this.  You end up adding triangular numbers after counting the equilateral triangles with side length 1.

Now, is there a quick formula for the sum of consecutive triangular numbers? :P

The sequence starts with $ 1, 4, 10, 20, \dots$.

Actually yes.  If you take differences between the differences of the differences of successive terms (=P), you can see that the terms correspond to a cubic polynomial.

So, we have:
$ y \equal{} ax^3 \plus{} bx^2 \plus{} cx \plus{} d$
$ 1 \equal{} a \plus{} b \plus{} c \plus{} d$
$ 4 \equal{} 8a \plus{} 4b \plus{} 2c \plus{} d$
$ 10 \equal{} 27a \plus{} 9b \plus{} 3c \plus{} d$
$ 20 \equal{} 64a \plus{} 16b \plus{} 4c \plus{} d$

Solve for $ a,b,c,d$. Have fun.

The answer turns out to be $ \frac {n(n \plus{} 1)(n \plus{} 2)}{6}$, which is really nice.

If you're curious, $ a \equal{} \frac {1}{6}, b \equal{} \frac {1}{2}, d \equal{} 0, c \equal{} \frac {1}{3}$

So I guess the formula is $ n^2 \plus{} \frac {n(n \minus{} 1)(n \plus{} 1)}{6}$

Edit: whoops fixed xD

Just note that $ t_n\plus{}t_{n\minus{}1}\equal{}n^2$
Continue.  If $ n$ is even, you end up going down to $ t_2\plus{}t_1\equal{}4$.

If $ n$ is odd, you end up going down to $ t_3\plus{}t_2\equal{}9$ and then adding 1 at the end.

Try and figure it out.

[quote="mathcrazed"]The answer turns out to be $ \frac {n(n \plus{} 1)(n \plus{} 2)}{6}$, which is really nice.[/quote]
Might that number look familiar?[hide="Yes"]$ \binom{n\plus{}2}{3}$. Now find motivation.[/hide]

Er, actually, $ d$ is 0 and $ c$ is $ \frac{1}{3}$.

So the cubic polynomial is
$ \frac{1}{6}x^3 \plus{} \frac{1}{2}x^2 \plus{} \frac{1}{3}x \equal{} \frac{1}{6}x(x^2 \plus{} 3x \plus{} 2) \equal{} \frac{1}{6}x(x \plus{} 1)(x \plus{} 2)$

Back up, mathcrazed... I think you've forgotten the upside-down triangles...

Well w/e. I don't care. =P.  Just count the upside down ones later, which can actually be found using a formula... Derive it yourself xD.

[quote="perfect628"][quote="mathcrazed"]The answer turns out to be $ \frac {n(n \plus{} 1)(n \plus{} 2)}{6}$, which is really nice.[/quote]
Might that number look familiar?[hide="Yes"]$ \binom{n \plus{} 2}{3}$. Now find motivation.[/hide][/quote]

Its called tetrahedral numbers XD


function
algebra
domain
trigonometry
complex numbers

[quote="nr1337"]Many function problems have statements like [tex]$f: \mathbb{C} \to \mathbb{C}[/tex].  What does this mean?[/quote]

It means f is a function which takes values in the complex numbers and spits out numbers in the complex numbers.  So, for example, we might take the function [tex]\displaystyle{f(x)=x^2}[/tex] as a function [tex]\displaystyle{f: \mathbb{C} \to \mathbb{C}, f: \mathbb{R} \to \mathbb{R}, f: \mathbb{R} \to \mathbb{R}^+, f: \mathbb{Q} \to \mathbb{Q}, f: \mathbb{Z}_{\leq0} \to \mathbb{Z}_{\geq0}}[/tex], etc., meaning respectively that f turns complex numbers into complex numbers, real numbers into real numbers, real numbers into positive real numbers, rational numbers into rational numbers, and negative integers into positive integers.

[quote="nr1337"]Many function problems have statements like [tex]$f: \mathbb{C} \to \mathbb{C}[/tex].  What does this mean?[/quote]

That means the function maps numbers from the field of complex numbers to the field of complex numbers. The "blackboard capital" letter [tex]\mathbb{C}[/tex] indicates the set of complex numbers, just as [tex]\mathbb{R}[/tex] indicates real numbers, [tex]\mathbb{Q}[/tex] indicates rational numbers (the Q comes from "quotient"), [tex]\mathbb{Z}[/tex] indicates integers (the Z comes from "Zahl," the German word for "number"), and [tex]\mathbb{N}[/tex] indicates natural numbers. The arrow notation you saw is mapping notation, which is often used for descriptions of functions in higher mathematics.

Hope this helps!

Another way to think about it is [tex] $f: Domain \to Range [/tex]. So the notation specifies the domain and range of the function.

Technically, a function isn't just the rule that maps one set to another (as is commonly taught in school). A function is a map [i]along with[/i] it's specified domain and range. For example [tex] x^2: \mathbb{R} \to \mathbb{R} [/tex] is actually a different [i]function[/i] from [tex] x^2: \mathbb{Z} \to \mathbb{Z} [/tex].

So two functions are the same [i]if and only if[/i] they have the same domain and the same range and they assign elements to each other in the same way. This may seem like a trivial distinction, but later on in math it turns out to be important.

No, it applies to any kind of function including trig.  It just has to do with domain and range.  Domain and range are characteristic of any function.


Χαίρετε! Σας επισυνάπτω μια όμορφη - κατά τη γνώμη μου- άσκηση!

Έχουμε $ a \equiv 1 \bmod 8,a \equiv 1 \bmod 3,a \equiv 1 \bmod 5$, άρα $ a \equiv 1 \bmod 120$. Άρα $ x \equiv (A\minus{}1) \bmod 120$ και έτσι βρίσκουμε το x (ακόμα και αν $ 1080 \leq x < 1200$ ή $ 1080 < x \leq 1200$).

[quote="Demetres"]Έχουμε $ a \equiv 1 \bmod 8,a \equiv 1 \bmod 3,a \equiv 1 \bmod 5$, άρα $ a \equiv 1 \bmod 120$. Άρα $ x \equiv (A \minus{} 1) \bmod 120$ και έτσι βρίσκουμε το x (ακόμα και αν $ 1080 \leq x < 1200$ ή $ 1080 < x \leq 1200$).[/quote]

Όμορφη απόδειξη. Εγώ είχα μια πιο εμπειρική, αλλά αυτή εδώ υπερτερεί!


For positive integers $a,b,c$ let $ \alpha, \beta, \gamma$ be pairwise distinct positive integers such that
    \[ \begin{cases}{c} \displaystyle a &= \alpha + \beta + \gamma, \\
    b &= \alpha \cdot \beta + \beta \cdot \gamma + \gamma \cdot \alpha, \\
    c^2 &= \alpha\beta\gamma. \end{cases} \]
    Also, let $ \lambda$ be a real number that satisfies the condition
    \[\lambda^4 -2a\lambda^2 + 8c\lambda + a^2 - 4b = 0.\]
    Prove that $\lambda$ is an integer if and only if $\alpha, \beta, \gamma$ are all perfect squares.

It is easy to check the equation has four roots as follows:
$ \sqrt{\alpha}\plus{}\sqrt{\beta}\minus{}\sqrt{\gamma},\sqrt{\alpha}\minus{}\sqrt{\beta}\plus{}\sqrt{\gamma},\minus{}\sqrt{\alpha}\plus{}\sqrt{\beta}\plus{}\sqrt{\gamma},\minus{}\sqrt{\alpha}\minus{}\sqrt{\beta}\minus{}\sqrt{\gamma}$.

Now sufficiency is trivial. To prove necessity, we assume, WLOG,
$ \sqrt{\alpha}\plus{}\sqrt{\beta}\minus{}\sqrt{\gamma}\equal{}\lambda$.
We can square $ \sqrt{\alpha}\plus{}\sqrt{\beta}\equal{}\sqrt{\gamma}\plus{}\lambda$ and see $ \sqrt{\alpha\beta}\minus{}\lambda\sqrt{\gamma}$ is rational.

Similarly, we see
$ \sqrt{\beta\gamma}\minus{}\lambda\sqrt{\alpha},\sqrt{\alpha\gamma}\minus{}\lambda\sqrt{\beta},\sqrt{\alpha\beta}\minus{}\lambda\sqrt{\gamma}$ are rationals.
Since $ c\equal{}\sqrt{\alpha\beta\gamma}$ is integer, we see
$ \sqrt{\alpha\beta\gamma}\minus{}\lambda\alpha,\sqrt{\alpha\beta\gamma}\minus{}\lambda\beta,\sqrt{\alpha\beta\gamma}\minus{}\lambda\gamma$ are integers.

Now it is trivial to prove $ \sqrt{\alpha},\sqrt{\beta},\sqrt{\gamma}$ are all rationals.
Since $ \alpha,\beta,\gamma$ are integers, we know they are also perfect squares.
Q.E.D.


I play like a mad man! Although I do need to get a new game sooner than later... I'm kind of bored of what I have.

Played for like... 2 hours today. LOL. Are you good? I'm ok.

Heh... I used to play every day but then i started reading more. :P I can just about every 9 ft song.

play irregularly when i go to a firiends house, say

[quote="shobber"]What is DDR? :?[/quote]

[url]http://www.ddrfreak.com/aboutddr.php[/url]

[quote="shobber"]What is DDR? :?[/quote]This must be the most often question asked in this particular forum  :lol:

If you like DDR then got to:
http://legendaryfrog.com/lite/movies.htm
And click on Legendary Tales,
And choose #4: Dance, Dance, Dance ;)

There were some ppl in crosscountry that ddred for practice on running.

i've played DDR like once and i didn't like it. i personally don't like the game.

[quote="captcha000"]and how long do they ddr for...?[/quote]

[i]almost[/i] as long as half the time I spend on math. In other words, a long time :D .

DDR is a game that I think you can play on most game consols. There's also a finger DDR version on the calculator.

[quote="Valentin Vornicu"]Can someone please explain to me how and where can you ddr?  :?[/quote]

DDR pads are available for the PlayStation series and computer (and perhaps some other consols). Most people play on those. Some malls, arcades, and etc. have slightly fancier pads to dance on. I like the fancier ones. ^-^

I got my pads cheap on some store online, and they are so great.

You can train on DDR. Trust me. Try heavy mode nonstop for 30 minutes. You will be sweating profusely.

i played for the first time last weekend at the cast party for our school's musical, and it was sooo much fun. i'd never played b4, so i only played on light and beginner, but it was great. it was really funny, there were these 3 brothers in the musical (a junior, a 6th grader and a 4th grader) and their 9th grade brother came to the party as well. 2 of them would be doing heavy on the pads that we had, and the other 2 would be doing it right behind them. it was sooo fun to watch the yoder brothers go at it. lol.

man I love DDR... had some great times with friends playin that game...

I remember spraining my arm playing that. Don't ask. Just... don't ask :P .

I used to be able to do most every song in DDR Max 2 on Heavy, but considering it's only at my cousin's house and I haven't played it in about a year...


Does there exist an infinite subset $T$ of the positive integers such that whenever $a, b$ are in $T$, then $a^2 - ab + b^2$ divides $a^2b^2$.

No,suppose that $gcd(a,b)=d,a=da_1,b=db_1$
Then $d^2(a_1^2-a_1b_1+b_1^2)|d^4a_1^2b_1^2$
Since $gcd(a_1,b_1)=1,gcd(a_1^2-a_1b_1+b_1^2,a_1^2)=gcd(a_1^2-a_1b_1+b_1^2,b_1^2)=1$
So $a_1^2-a_1b_1+b_1^2|d^2(*)$
But for any $a\in N$,there is finite couples of $(a_1,d)$
And for each couple,there is finite $b_1$ satisfied (*)
(For $b_1$ big enough,$a_1^2-a_1b_1+b_1^2>d^2$)
So for any $a\in N$,there is finite $b$ such that $a^2-ab+b^2|a^2b^2$.

That was the expected model solution given to this question. However someone on the UK squad had this ingenious solution.

Assume $a^2 - ab + b^2 | a^2b^2$, then $a^2 - ab + b^2 | a^2b^2 - b^2(a^2 - ab + b^2) \Rightarrow a^2 - ab + b^2 | ab^3 - b^4$. Now fix $b$ and make $a$ very, very large. The LHS is quadratic in $a$, the RHS only linear in $a$. Therefore eventually $a^2 - ab + b^2 > ab^3 - b^4 > 0$ and so there can be no infinite sequence.


[quote="M@ri@n"][quote="thanhnam2902"]$ cot^2x \plus{} 4tan^2x \plus{} cot^6x \equal{} 6$[/quote]

Let $ \cot ^2x \equal{} : t$. Then

$ \cot^2x \plus{} 4\tan^2x \plus{} \cot^6x \equal{} t \plus{} \frac {4}{t} \plus{} t^3 \equal{} 6$.

Hence

$ t^4 \plus{} t^2 \minus{} 6t \plus{} 4 \equal{} 0\quad\Leftrightarrow\quad (t \minus{} 1)^2(t^2 \plus{} 2t \minus{} 4) \equal{} 0$.[/quote]

It's nice and true. Thank you!

[color=darkblue]I have a solution of this problem by use AM-AG inequality.[/color]

[quote="thanhnam2902"]$ cot^2x + 4tan^2x + cot^6x = 6$[/quote]

$ LHS = (cot^2x + tan^2x) + (3 tan^2x + cot^6x)$

$ LHS \ge\ \sqrt {cot^2x.tan^2x} + \sqrt [4]{tan^6x.cot^6x} = 6$

Equality hold on $ cot^2x = tan^2x \leftrightarrow tan^4x = 1$

$ \leftrightarrow x = \frac {\Pi\}{4} + \frac {k\Pi\}{2}$

[quote="onlylove_math"][quote="thanhnam2902"]$ cot^2x + 4tan^2x + cot^6x = 6$[/quote]

$ LHS = (cot^2x + tan^2x) + (3 tan^2x + cot^6x)$

$ LHS \ge\ \sqrt {cot^2x.tan^2x} + \sqrt [4]{tan^6x.cot^6x} = 6$

Equality hold on $ cot^2x = tan^2x \leftrightarrow tan^4x = 1$

$ \leftrightarrow x = \frac {\pi\}{4} + \frac {k\pi\}{2}$[/quote]

I think so. Thanks onlylove_math.


If you attend Fairview Middle School in Tallahassee, Florida, please let me know (post here). I may know you!

wouldn't [url=http://www.artofproblemsolving.com/Forum/index.php?f=174]this[/url] PROBABLY be a better place to post it? ;)


[quote="GameBot"]Find an ordered pair $ (x,y)$ that satisfies both of the equations below: \begin{align*} 2x - 3y &= -5,\\ 5x - 2y &= 4. \end{align*}[/quote]

Multiply the first eq by 2, and the second by 3, then subtract.

$ 4x-15x=(-10)-(12)$
$ -11x=-22$
$ x=2$

Substituting back in gives $ y=3$

Thus, the answer is $ \boxed{(2,3)}$


calculus
trigonometry
number theory
real analysis
real analysis unsolved

almost one year ago, i got a problem from my calculus class. the problem was to examine the convergence of $\sum_{n=1}^{\infty}\frac{\tan n}{2^{n}}$.
of course, no one in that class was able to solve this. in fact, there was a typo and the original one was $\sum_{n=1}^{\infty}\frac{\tan^{-1}n}{2^{n}}$, a simple problem.
but next day, one of my friend said that he had achieved the [b]absolute[/b] convergence of that series. (huh!)
from then, i've been trying to prove it whenever i recall it.

but now i think it is the very time to call for a help!

[b]Q.[/b] examine the convergence of $\sum_{n=1}^{\infty}\frac{\tan n}{2^{n}}$

(does it need more advanced theory than calculus?)

I think this goes quite a bit beyond calculus. Even $\frac{\tan n}{2^{n}}\to 0$ is a nontrivial statement. Since $|\tan n|\approx \left(\min_{m\in\mathbb Z}|n-m\pi-\pi/2|\right)^{-1}$, the problem has more to do with number theory (rational approximations of $\pi$) than calculus.

This one does converge absolutely because $\pi$ is non-Liouvillian and, therefore, $|\tan n|\le n^{M}$ for some $M>0$. This is, indeed, a nontrivial fact but there exist several understandable proofs of it. You can find one of such proofs right here, on AoPS Wiki :).


inequalities
calculus
inequalities unsolved

Let $x,y,z$ be nennegative real numbers, find the greatest possible positive value of $k$ such that \[(x+y+z)^{5}\geq k(x^{2}+y^{2}+z^{2})(x^{2}y+y^{2}z+z^{2}x).\] The following is also nice \[(x+y+z)^{5}\geq 8(x^{2}+y^{2}+z^{2})(x^{2}y+y^{2}z+z^{2}x+xy^{2}+yz^{2}+zx^{2})?\]

I don't know if you have different solutions but they seem to be solved just by Muirhead or SOS which I don't like .

Nice solutions, Arqady.
For the first one, if we put $y=1$ and $z=0$, it follows by calculus that  $\frac{(x+y+z)^{5}}{ (x^{2}+y^{2}+z^{2})(x^{2}y+y^{2}z+z^{2}x)}$ attains its minimum $k\approx 11.353$ for $x\approx 3.26953$, the only  root of $x^{3}-4x^{2}+3x-2=0.$
It only remains to show (but that should not be too hard) that in an extremal case, one of $x,y,z$ must be $0$. :roll:


calculus
integration
geometry
absolute value
real analysis
real analysis unsolved

the integral of:

|(x+2)^1/2 -x| on the interval from 0 to 4.

I keep getting a negative area?!  Am I doing something wrong?  I keep getting -28.

Thanks in advance!

There is an absolute value sign...
From $\sqrt{x+2}-x=0$ we find that $x=2$.

So you need to write the integral as
$\int_0^2 (\sqrt{x+2} - x) dx + \int_2^4 (x - \sqrt{x+2}) dx$


1. which letter shows a bond where electrons are being transferred?
2. which letter shoes a bond where electrons are equally shared?
3. which letter shows a bond that would cause each atom to become partially charged
4. which atom can be considered an anion?
5. which bond above would cause the compound to have the highest melting point?

answers: Y,X,Z,4,Y

are they correct??
thanks!!

P.S. EN stands for electronegativity

I believe your answers are correct, though atom 4 is numbered as atom 1 (resulting in two atom 1s).

somehow my visual was inncorrect. i have updated it though.

thanks

could someone check this up with the new visual??

thanks

[quote="Eli"]I believe the answer to #4 is 3, because that is atom of the ionic bond that becomes positive.[/quote]

Isn't an anion negative though? So either atom 1 (if bonded with atom 3) or 4 (higher EN) can be anions.


[quote="@nn@"]Come on. It's just simple math. :spider:[/quote]

5900039110000000000000? Easy I thnk

next please put commas it gets a little hard for me when theres an infinite amount of zeros i need to deal with.


Hello my friends, i am new in this forum, and i have many problems we can solve together, here is the first one, is from de book Curso de Analise Vol II(Analysis Course VII)by Elon Lages Lima a portuguese mathematician, here we go

Let $ f: I\subseteq{\mathbb{R}}\rightarrow{\mathbb{R}^n}$ be a differentiable mapping. If $ a\in {I}$ is a limit point of the set $ f^{ \minus{} 1}(v)$, i.e. $ a\in{(f^{ \minus{} 1}(v))'}$ for some $ v\in{\mathbb{R}^n}$, prove that  $ f'(a) \equal{} 0$.

If $ v_n \rightarrow a$, $ v_n \in f^{\minus{}1}(v)$, then of course $ \frac{f(v_n) \minus{} f(a)}{v_n\minus{}a} \equal{} 0$, so the limit is 0.

[quote="enloalto"]Hello Soarer, nice to meet you, thank you very much for your help, I have a doubt, what happened with that $ v$? I do not understand what it is its purpose.[/quote]

$ f(v_n) \equal{} f(a) \equal{} v$ for all $ n$, so $ f(v_n) \minus{} f(a) \equal{} 0$.


Does AoPS encourage/discourage or acknowledge the use of the package from Wolfram called Mathematica? Like LaTeK (sic), it has math typesetting capabilities, not to mention lots of mathematical and educational functions.

BTW, I like this site!

Regards..

Is this an advertisement? For discussion, see
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=173626

$ \text{\LaTeX}$ cannot even compare to any other mathematical typesetting program. Mathematica is nice though for other features that it has (factorization of algebraic expressions and so on).

Sorry. I was not trying to advertise it. Actually, I wanted to use the search function, but I could not find it amongst all of this icons and buttons. Thanks for pointing this out.

Regards..


I find this problem in a book but there is no solution in the book.
Prove that if x,y,z>0,then
$ \frac{1}{x^2\plus{}xy\plus{}y^2}\plus{}\frac{1}{y^2\plus{}yz\plus{}z^2}\plus{}\frac{1}{z^2\plus{}zx\plus{}x^2}$≥$ \frac{9}{(x\plus{}y\plus{}z)^2}$

[quote="SUPERMAN2"]I find this problem in a book but there is no solution in the book.
Prove that if x,y,z>0,then
$ \frac {1}{x^2 + xy + y^2} + \frac {1}{y^2 + yz + z^2} + \frac {1}{z^2 + zx + x^2}$≥$ \frac {9}{(x + y + z)^2}$[/quote]

This inequaltiy is equivolent to:
${ \frac{1}{2}\sum_{sym}{x^6}+6\sum_{sym}{x^4y^2}+3\sum_{sym}{x^5y}+4\sum_{sym}{x^3y^3}+9\sum_{cyclic}{x^4yz}+18\sum_{sym}{x^3y^2z}+27x^2y^2z^2 \geq 9\sum_{sym}{x^4y^2}+\frac{9}{2}\sum_{sym}{x^3y^3}+9\sum_{cyclic}{x^4yz}}+18\sum_{sym}{x^3y^2z}+27x^2y^2z^2$

or
$ \frac{1}{2}\sum_{sym}{x^6}+3\sum_{sym}{x^5y} \geq 3\sum_{sym}{x^4y^2}+\frac{1}{2}\sum_{sym}{x^3y^3}$
which is right by Muirhead inequality.


Does anyone here watch anime, such as Cowboy Bebop, One-Peice, Full Metal Alchemist, or anyo ther anime series or movies? If you do, please state which ones, and which one's your favorite and why if possible.

All of those are over...

Yakitake Japan (recently ended)
Naruto
Bleach

and for long ago ended ones

Air
Mai Hime
Kodocha
Hikaru no Go
BECK
Full Metal Alchemist
Full Metal Panic
Psychic Academy
Inuyasha
The list is really long..

As for current favorite, I'd say it's Bleach, I hate Naruto fillers.

Anime is Art and therefore belongs there. :P

I have already made such a topic there : [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=81982[/url]

[quote="junggi"]personally, I don't really like watching animes, because the u.s. voice actors arent really that good....so I prefer mangas.
[/quote]

Well, there are a lot of fansubs out there  :wink:  that are a hundred times better.  Those people who dub Japanese animation simply stop at nothing, they change names like "Ten Shin Han" into "Tien" and "Kenshin" into "Kenshee" :?:  :|

Subtitles are FUN and GOOD!  :roll:   If I can watch Japanese stuff with English subtitles (while English is not my native) how come anglophonics can't :huh:   Why do they dub?!

[quote="fredbel6"][quote="Treething"]Anime is [b]awful[/b].

go watch pokemon or something (child friendly equivalent of FYAD)[/quote]
 :huh:

Do you care to elaborate?  Does FYAD stand for "f-word you and die"?

So basically you write to simply inform us that anime is awful, but nevertheless, you still encourage us to go watch Pokemon, which is anime.

Maybe I should deduce from your impressive collection of arguments that...Pokemon is the only anime you know? :!:[/quote]

your conclusions are as valid as my mathematical conclusions

[quote="Treething"]
your conclusions are as valid as my mathematical conclusions[/quote]
And that means? :?:

[quote="junggi"]personally, I don't really like watching animes, because the u.s. voice actors arent really that good....so I prefer mangas.
some nice mangas:
naruto
bleach
prince of tennis
death note(dont know if this has anime)
full metal alchemist
hikaru no go
rave master

thats about all I know :ninja:[/quote]

I've never seen Anime in English..

if u ask me animes from japan are the best:

while i was younger:
pokemon
digimon
and
PRINCESS MONONOKE (MONONOKE HIME)

now:
.Hack

and I'm trying to find somewhere Spirited Away but I haven't made it so far (I've heard it's really good)

[quote]and I'm trying to find somewhere Spirited Away but I haven't made it so far (I've heard it's really good)[/quote]

Well I watched it two years ago and...  The animation is simply unbelievable, it makes Dragonball Z look like South Park.  Without fancy 3D animation they make incredible scenes!
But the story isn't that good.  It's a fantastic world with lots of interesting characters, and the amazement never stops..but in the end you don't have the feeling it all added up.  Some people explain this by pointing out that in western movies, the emphasis is on plot, while in Japanese (and Asian in general) movies the emphasis is more on character development.  So I was kind of dissatisfied with the ending, but I would still recommend it.
[quote]no one here downloads anime?

bittorrent is very good[/quote]
You naughty boy :wink:   You're supposed to claim you don't.  Although I have to agree I don't own anything legally basically, except three videos of Neon Genesis Evangelion and the Akira movie.

[quote]if u ask me animes from japan are the best:[/quote]  Everybody likes manga, even politicians do  :P  :

[img]http://upload.wikimedia.org/wikipedia/en/2/2a/Leejapan.jpg[/img]

I never use BitTorrent to download anime.

Wow, I didn't know Pokemon was considered Anime.  What exactly defines Anime (Pokemon is still BEAST)?  A guy at my school did a project on it, but I don't remember all of it...

[quote="Iversonfan2007"]lemonwire!!![/quote]
I don't know that one.  Could you please describe it a little bit and tell us what you like. :)

or dl lw pro from lw itself... feel special

[quote="Iversonfan2007"]have u heard of limewire? Well, lemonwire is a branch of limewire that has turbospeed downloads for free :o  :lol:[/quote]I thought you meant an anime. :roll:

Does anybody know where can i download Naruto?

[quote="kh-"]Uh, I don't think giving explicit information on how to find (steal) content is allowed per the rules. It shouldn't be, at any rate.

(As for anime, I quite enjoyed FLCL. Gainax make some interesting animations when they feel like it. Love Hina was also amusing. I don't really see the appeal in Naruto).[/quote]
Downloading Free Fansubs is not stealing.

I'm sorry to have to contradict you, but it really is. I mean, if you're downloading the video alongside the subtitles, [i]then[/i] it is illegal, because the subtitles are free but the video is still copyright. A lot of fansubbers themselves admit this - quick Google search says as much, [url=http://en.wikipedia.org/wiki/Fansub#Recent_legal_action]as does Wikipedia[/url]. If you were to download just fan subtitles without the video (although it'd be pretty useless I suppose), it would be OK.

I don't actually care or think any less of people that do it, but you should at least be aware of what you are doing. I'm not a lawyer and I could be completely wrong, but, y'know, I feel better for saying it anyway.

HECK YES! YEAH FOR ANIME! I LIKE MOST BUT NOT THE POPULAR ONES LIKE naruto or bleach.

At the moment, I don't watch any.
But I used to watch

1. Pokemon
2. Yugioh

I have seen a couple epsidoes of Noein.  It was pretty cool but, not something I would watch repeatedly.

I've watched Bleach, Hikaru no Go, and Detective Conan.
I finished Hikaru no Go unfortunately (i wish it was a longer series)
I couldnt find any more fansubs for detective conan past ep 479
and i'm keeping up with the bleach fansubs that come out weekly.

Hey, watch anime on veoh now. All the anime uploaders are going there. Also I am uploading a series called Utawarerumono.

Explanation:
http://www.en.wikipedia.com/wiki/Utawarerumono

Naruto. I got into it about two months ago. I'm at 107 in regular and 7 in shippuden.

After i finnish all the fillers, i'll start watching bleach and death note.

About spirited away, that movie has a really bad ending.

I don't think very many people like it but Haikyuu [hide]I've never seen anyone else mention it[/hide]


Prove that there is no function f:R+--R+ such that:

(f(x)) 2  >= f(x+y)(f(x)+y)

Now that you know the source, you can look the official solution up on the Net, but here's mine (hope it's Ok :)):

Clearly, $f^2(x)>f(x+y)f(x)\Rightarrow f$ is strictly decreasing. Make $y=f(x)\Rightarrow f(x)\ge 2f(x+f(x))$. We repeat this process: $f(x)\ge 2f(x+f(x))\ge 4f(x+f(x)+f(x+f(x)))\ge\ldots$. If we consider the sequence $\displaystyle a_0=x,\ a_{n+1}=f(\sum_{i=0}^na_i)$, then $\displaystyle f(\sum_{i=0}^na_i)\to 0$ as $n\to \infty$. This can only happen if $a_0+a_1+\ldots=\infty$, but this can't happen because $a_{n+1}\le \frac {a_n}2$.


Prove $a^{2}+b^{2}+c^{2}\ge ab+bc+ac$ in as many ways as possible

EDIT: My $10^{2}$ post

[hide]$a^{2}+b^{2}\geq2ab$
$b^{2}+c^{2}\geq2bc$
$a^{2}+c^{2}\geq2ac$
adding and dividing by 2 we get the desired ineq.[/hide]

[quote="D.P.L"]Prove $a^{2}+b^{2}+c^{2}%Error. "geab" is a bad command.
+bc+ac$ in as many ways as possible[/quote]

Prove what?

I take it you mean $a^{2}+b^{2}+c^{2}\ge ab+bc+ac$. I see you edited it anyway.

[hide="Method 1"]
By AM-GM $\frac{a^{2}+b^{2}}{2}\ge ab$. Adding together two like inequalities gives the result.[/hide]
[hide="Method 2"]
It is clear that we can assume $a,b,c$ positive since making one or more negative does not increase the RHS or decrease the LHS.
By Muirhead on $(2,0,0) \succ (1,1,0)$ we get
$2(a^{2}+b^{2}+c^{2}) \geq 2(ab+bc+ca)$[/hide]
[hide="Method 3"]
Assume wlog $a \ge b \ge c$ (and again that all terms are positive).
We then get the desired inequality by rearrangement.[/hide]

[quote="tim1234133"]
Assume wlog $a \ge b \ge c$[/quote]
You don't need this assumption.

[quote]
You don't need this assumption.
[/quote]

Well, we don't need this assumption, but it makes it clearer to understand if we do, and since we can, then we might as well do so.

I remember that i did it a few weeks ago  :) .

[hide="what splash said"]

We multiply the both sides by 2.We will have
$2a^{2}+2b^{2}+2c^{2}\geq2ab+2bc+2ac\\2a^{2}+2b^{2}+2c^{2}-(2ab+2bc+2ac)\geq0\\(a^{2}-2ab+b^{2})+(a^{2}-2ac+c^{2})+(b^{2}-2bc+c^{2})\geq0\\(a-b)^{2}+(a-c)^{2}+(b-c)^{2}\geq0$
We know that any number like $x^{2}$ is  $\geq0$ so our previous relation is true.
I hope Splash won't mind because i wrote his complete solution :huh:  [/hide]

[quote="pkerichang"]One word solution: rearrangement[/quote]
This methos was already posted:
[quote="tim1234133"]
[hide="Method 3"]
Assume wlog $a \ge b \ge c$ (and again that all terms are positive).
We then get the desired inequality by rearrangement.[/hide][/quote]
Please read previous posts before writing your solution!

DIfferent[hide]Let $F(x)=(x-a)(x-b)+(x-a)(x-c)+(x-b)(x-c)$ the discriminant of $F(x)=0$ is $D=a^{2}+b^{2}+c^{2}-ab-ac-bc$ . WLOG $a \le b \le c$ then $F(b)=(b-a)(b-c) \le 0$ and we see that $F(x)=0$ always has real solutions so $D \ge 0$. Equality holds only when $a=b=c$.[/hide]


We say a positive integer $ n$ is [i]good[/i] if there exists a permutation $ a_1, a_2, \ldots, a_n$ of $ 1, 2, \ldots, n$ such that $ k \plus{} a_k$ is perfect square for all $ 1\le k\le n$. Determine all the good numbers in the set $ \{11, 13, 15, 17, 19\}$.


Εστω $ x,y,z$ πραγματικοί με $ x\plus{}y\plus{}z\equal{}1$. Να δείξετε ότι:

$ \frac {x}{x^2 \plus{} 1} \plus{} \frac {y}{y^2 \plus{} 1} \plus{} \frac {z}{z^2 \plus{} 1}\leq \frac {9}{10}$

[i]
Την απέδειξα για θετικούς αλλά για τους πραγματικούς δεν είμαι σίγουρος για τη λύση μου...[/i]

πραγματι πολυ ωραια.
[hide]λοιπον εστω η συναρτηση
$ f(x) \equal{} \frac {1}{1 \plus{} x^2}$
then,
$ f''(x) \equal{} \frac {2(x \minus{} 1)^3}{x^2 \plus{} 1} \leq 0 , x \leq1$
so,jensen(x+y+z=1)
$ xf(x) \plus{} yf(y) \plus{} zf(z) \leq f(x^2 \plus{} y^2 \plus{} z^2) \equal{} \frac {1}{(x^2 \plus{} y^2 \plus{} z^2)^2 \plus{} 1}\leq \frac {9}{10}$
$ \leftrightarrow \sum \frac {x}{x^2 \plus{} 1} \leq \frac {9}{10}$[/hide]
EDIT: η προσεγγιση αναφερεται μονο για μη αρνητικους x,y,z οπως αναφερεται παρακατω.

[quote="Athinaios"]πραγματι πολυ ωραια.
[hide]λοιπον εστω η συναρτηση
$ f(x) \equal{} \frac {1}{1 \plus{} x^2}$
then,
$ f''(x) \equal{} \frac {2(x \minus{} 1)^3}{x^2 \plus{} 1} \leq 0 , x \leq1$
so,jensen(x+y+z=1)
$ xf(x) \plus{} yf(y) \plus{} zf(z) \leq f(x^2 \plus{} y^2 \plus{} z^2) \equal{} \frac {1}{(x^2 \plus{} y^2 \plus{} z^2)^2 \plus{} 1}\leq \frac {9}{10}$
$ \leftrightarrow \sum \frac {x}{x^2 \plus{} 1} \leq \frac {9}{10}$[/hide]
ελπιζω να μην εχασα καπου λογω βιασυνης..[/quote]

den to eleksa....alla polu kali lusi!para polu mou arese.einai auto pou leme death by jensen.

to teleutaio einai stolen apo to fridays. ekei exoun to gluko death by chocolate.

ps. for the festivals o rapanos.....(ti mlkies leo....)

bravo kai pali!

Αθηναίε, η λύση σου δουλεύει μόνο για μη αρνητικούς $ x,y,z$.

Δεν ξέρω αν αυτή την λύση είχε υπόψη του ο bilot4, αλλά σίγουρα μπορείς να την κάνεις να δουλέψει και για πραγματικούς.

O Demetres έχει δίκιο!!! Ηλία αυτό είπα κι εγώ απ' την αρχή να προσέξετε! Αυτή η λύση ισχύει μόνο για θετικούς (αφού προυποθέτεις $ x\leq 1$)! Κι εγώ για θετικούς πάντως με Jensen την έβγαλα :P  απλώ πήρα αμέσως τη συνάρτηση $ f(x)\equal{}\frac{x}{x^2\plus{}1}$

Άκυρο, το original post deleted.

Εκεί που ήθελα να καταλήξω είναι ότι ένας έμμεσος τρόπος για να δει κανείς ότι δε μας ενοχλούν οι αρνητικές τιμές είναι με πολ/στές Lagrange. Η πρώτη παράγωγος της Lagrangian έχει μόνο $ x^2, y^2$ και $ z^2$, δηλαδή είναι συμμετρική σε οποιαδήποτε εναλλαγή $ x_i\to \minus{} x_i$. Άρα, αρκεί να μελετήσεις μόνο ένα από τα ογδοημόρια και το $ x,y,z\geq 0$ δουλεύει μια χαρά για αυτόν το σκοπό.

Cheerio,

Durandal 1707

PS: Nick, βάλε και μένα for the festivals :D

[quote="bilot4"]Κι εγώ λύση με Lagrange είχα στο μυαλό μου για τους πραγματικούς απλά επειδή ακόμα δεν την ξέρω καλά δεν ήμουν σίγουρος για τη λύση μου... Μπορεί κάποιος να βάλει αναλυτική λύση με τους πολλαπλασιαστές?[/quote]

Δεν έκατσα να εντοπίσω επακριβώς τις θέσεις των μεγίστων, αλλά το τρικ είναι το εξής: έστω ότι με τους πολ/στές βγάζεις μία λύση $ (a,b,c)$ για max. Λόγω της συμμετρίας των παραγώγων και κάθε άλλη τριπλέτα $ (\pm a, \pm b, \pm c)$ θα είναι [i]πιθανό[/i] max.

Όμως, η αρχική συνάρτηση έχει $ f(x,y,z)\leq f(|x|,|y|,|z|)$, οπότε... :lol:

Cheerio,

Durandal 1707


The edge length of each cube shown is 1 centimeter. Four cubes are added at each stage. What is the number of cubic centimeters in the volume of the figure formed in stage 20?

There is no pic, so this question needs to be corrected.


Prove

$ \sum_{k\equal{}1}^n \exp\left(\frac{2\pi i}{n}\, k^2\right)\equal{}\frac{\sqrt{n}}{2} \, (1\plus{}i) \, \left[1\plus{}\exp\left(\minus{}\frac{i\pi n}{2}\right)\right] \qquad \forall n\in\Bbb{N}$


calculus
integration
Euler
calculus computations

other than rewriting sums such as
$\sum_{n=1}^{\infty}\frac{1}{n^{2}+n}$
as
$\int_{0}^{1}\sum_{n=1}^{\infty}\frac{x^{n}}{n}dx$
how else could one rewrite a series as an integral (other than using $x^{n}$) for example, could one turn
$\sum_{n=1}^{\infty}\frac{ln(n)}{n^{2}}or \sum_{n=1}^{\infty}\frac{(-1)^{n}}{n^{2}+1}$ into a definite integral using a similar technique?

One formula gives an exact answer for any sum/series, and this is the [url=http://planetmath.org/encyclopedia/EulerMaclaurinSummationFormula.html]Euler-Maclaurin summation formula[/url].

can u run through an example, I'm not sure I fully understand the formula


integration
real analysis
advanced fields
advanced fields unsolved

Can we use laplace transform to solve an ODE with variable coefficients?

Like this one:

4x y" + 2 y' + y = exp (-x)

Yes, but.

For multiplication by powers of $x,$ we can use the following:

If $Y(s)=\mathcal{L}(y)(s)=\int_{0}^{\infty}y(x)e^{-sx}\,dx,$ then

$\frac{d}{ds}Y(s)=-\int_{0}^{\infty}xy(x)e^{-sx}\,dx=-\mathcal{L}(xy(x)).$

It would be possible to work out as a related formula what the Laplace transform of $xy''(x)$ would be, in terms of $Y(s)$ and $Y'(s).$ (I'm not going to do that right now, but all it would take is a little patience in working out the details.)

So if we apply the Laplace transform to your equation, we will obtain a differential equation for $Y(s).$ (Not merely an algebraic equation for $Y(s),$ which is what happens in constant coefficient cases.) Whether that's helpful would depend on what the new differential equation looks like.

For more general variable coeffiencents, we'll have to use some form of "the transform of the products is the convolution of the transforms." Again the result won't be a simple algebraic equation.


Let $d$ be a positive integer, and let $S$ be the set of all positive integers of the form $x^{2}+dy^{2}$,
where $x$ and $y$ are non-negative integers

(a) Prove that if $a \in S$ and $b \in S$, then $ab \in S$
(b) Prove that if $a \in S$ and $p \in S$, such that $p$ is a prime and $gcd(p,a)=p$ then, $\frac{a}{p}\in S$
(c) Assume that the equation $x^{2}+dy^{2}=p$ has a solution in non-negative integers $x$ and $y$,
 where $p$ is a given prime. Show that if $d \geq 2$, then the solution is unique, and if $d=1$,
then there are exactly two solutions.

(a) is very easy, but I think that it will be useful to prove (b). (just guess)

Sorry for my bad english  :oops:  :blush:

I don't know how to move this topic! help!

All where posted before...
b) http://www.mathlinks.ro/Forum/viewtopic.php?t=66248
c) http://www.mathlinks.ro/Forum/viewtopic.php?t=86455

[hide="Index"]prime represented case unique[/hide]

[quote="Euler-ls"]Let $d$ be a positive integer, and let $S$ be the set of all positive integers of the form $x^{2}+dy^{2}$,
where $x$ and $y$ are non-negative integers

(a) Prove that if $a \in S$ and $b \in S$, then $ab \in S$
(b) Prove that if $a \in S$ and $p \in S$, such that $p$ is a prime and $gcd(p,a)=p$ then, $\frac{a}{p}\in S$
(c) Assume that the equation $x^{2}+dy^{2}=p$ has a solution in non-negative integers $x$ and $y$,
 where $p$ is a given prime. Show that if $d \geq 2$, then the solution is unique, and if $d=1$,
then there are exactly two solutions.[/quote]

Hm. So you should keep in mind that I don't necessarily know what I'm doing, but in the ring $\mathbb{Z}[-\sqrt{d}]$, we define $N(a+bi\sqrt{d}) = a^{2}+db^{2}$, and note that $N(a+bi\sqrt{d})N(x+yi\sqrt{d}) = (a^{2}+db^{2})(x^{2}+dy^{2}) = (ax-dby)^{2}+d(ay+bx)^{2}$ $=$ $N((a+bi\sqrt{d})(x+yi\sqrt{d}))$, i.e. $N$ is multiplicative. Now, for part b, if $p = x^{2}+dy^{2}= (x+yi\sqrt{d})(x-yi\sqrt{d})$, we have that $N(x+yi\sqrt{d}) = N(x-yi\sqrt{d}) = p$, so both $x \pm yi\sqrt{d}$ must be primes (if they factored as $z \cdot w$, we would have $N(z) \cdot N(w) = p$, i.e. either $N(z)$ or $N(w)$ would be 1, which can only happen if the element is a unit in the ring). However, if $a = s^{2}+dt^{2}= (s+it\sqrt{d})(s-it\sqrt{d})$, each of the above primes must divide this product as $p|a$, i.e. $x+yi\sqrt{d}$ divides either $s \pm it \sqrt{d}$, i.e. if you do the division by rationalizing the denominator, etc, and write it in the form $\alpha+i \beta \sqrt{d}$, we have $\alpha, \beta \in \mathbb{Z}$, implying that $x-yi \sqrt{d}$ divides $s \mp it \sqrt{d}$, so $\frac{a}{p}= \alpha^{2}+d \beta^{2}$, as desired.


trigonometry
logarithms
calculus
calculus computations

Evaluate   $ S = \sum_{n=1}^{+\infty}\frac{1}{4^{n}}\sec^{2}\left(\frac{\alpha}{2^{n}}\right)$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=162080]Here[/url] you find a derivation of the formula $ \frac{\sin x}{x}=\prod_{k=1}^\infty \cos \frac{x}{2^{k}}$

Take the logarithm of it $ \ln(\sin x)-\ln x=\sum_{k=1}^\infty \ln \cos \frac{x}{2^{k}}$ and differentiate $ \cot x-\frac{1}{x}=-\sum_{k=1}^\infty \frac{\tan\frac{x}{2^{k}}}{2^{k}}$

Change sign and differentiate again $ \frac{1}{\sin^{2}x}-\frac{1}{x^{2}}=\sum_{k=1}^\infty \frac{\sec^{2}\frac{x}{2^{k}}}{2^{2k}}$

Very nice. We could also use the concept of telescoping series.


If $ f(x) \equal{} x \plus{} 1$ and $ g(x) \equal{} x^2 \plus{} 3$, what is the value of $ f(g(2))$?


linear algebra
matrix
geometry
geometric transformation
rotation
complex numbers

Does there exist a matrix A such that the entire family $ A \plus{} cI$ is invertible for all complex numbers $ c$? Find a real matrix with $ A \plus{} rI$ invertible for all real $ r$.

Can anyone explain this question to me? I don't really know what it's asking.

$ I$ is the identity matrix: ones on the main diagonal, zeros elsewhere. $ cI$ is its scalar multiply, it has $ c$ on the main diagonal and $ 0$ elsewhere.
You are asked if there exists $ A$ such that the sum $ A\plus{}cI$ is an invertible matrix for every complex number $ c$.

Hint: "eigenvalue".

A second hint:  a matrix invertible if and only if its determinant is not zero.

Thanks guys I think I've got it now. I just need to find a real matrix that has complex eigenvalues for the second one.

And examples of this sort exist only when the matrix size is an even number.


I want my website to convert latex that is coded in the XHTML to GIF files like it does here...

I know you can achieve this by creating a LaTeX file and then converting it to MathML, but browser support isn't good for that.

How can I implement what you guys do to my site?
Can it be found somewhere?

 :P  :P  Thanks :P  :P

I think it's best you generate your pictures with the TeXer and then copy them over and use them in your html code. To do what there is here on this site is not simple at all, and would require extended knowledge of php (and use of such system - which I see you don't on your site).