hmm..9 votes and nobody has posted anything

nobody wants to say? :(

im atheist.

looks like there are more theists than athiests so far.

i am atheist too, as i have mentioned earlier.

I was agnostic before, but events in the middle east have led me to believe that religion is sometimes extremely silly.

So yes.

in actuality good atheists and good muslims both experience a minor level of perdition, whereas good christians bask in heaven

your choice, i'm doing my best to annoy people by taking the stance of a fairly fanatical christian

Yes, I am, and I do think some religions are quite silly.

that's wrong.
religions are defined by what is written in the scriptures, and i see no need to update that with time.

However, if what is written in the scriptures have no practical purposes anymore, I feel that they should be updated. Things that were pertinent 2000 years ago do not necessarily make sense today.

In a sense this has already happened -- how many people do you see actually observing the Sabbath? (Like, the way they're supposed to?)

make another topic

"do you beleve yourself are God or Luci_fer?"

kamisama

by symmetry, im god in the same proportion im luci_fer

I'm a Christian, but I don't agree with a lot of the organized religions out there- I find that a lot of self-proclaimed religious people, especially the really well-known ones, tend to twist their religion to fit their own pre-established beliefs.

[quote="now a ranger"]there should be a neither button[/quote]
I agree.  I'm somewhat religious but sometimes i question the existence of god.

I'm not exactly of any religion... but I do believe that there is a god or something of the sort, and I am somewhat christian.
Therefore, I'm not exactly atheist...
Looks like the atheist/nonatheist vote has balanced with my vote. joy joy.

wow...all tied up again

i was thinking that there would be more athiests since this was a math forum but whatever

go to endchristmas.com

(the website is not as bad as it sounds.  watch the second clip.)

[quote="randomdragoon"][quote="ProtestanT"]I'm Christian...

about the creationism against evolution, you may say creationism is outdated, but if you look at it, evolution is also outdated. it is a theory that is over one hundred years old and still has no valid proofs.
Creationism, however, has some logical concepts and valid statistics.[/quote]
On the contrary! Evolution is the one with logical concepts, while it is easy to poke holes into creationism.[/quote]
On the contrary! Creationism also has logical concepts. It is easier to poke holes into Evolution.

I suggest that unless you provide some Specific Examples or Evidence, I can deny everything you said the same way you denied everything I said.

God failed me after a history test so therefore i'm atheist

hm...is there anythign wrong with failing a history test?
don't you think God intentionally did that so it will benefit you in the future?

[quote="ProtestanT"]God created God. A logical explanation for this is that he created himself in the future and sent himself back in time, because he controls time and he is time. Therefore he is always there. "In the begining God created ...", so that implied that God was there from the begining of time.[/quote]

But, he wouldnt be there in the future because he wasnt there in the first place...

I believe mathematicians have a word for arguments about the creation of God that work like that. "Hand-wavy". It sort of produces a paradox...

There is another example of macroevolution: There is a species of mammal that lives in western North America (I need to look up the specifics...) that has sort of a U-shaped range. Variations within this species can be observed like a gradient across the U shape. However, at the top of the U shape, part of the two end ranges overlap, but the two varients do not interbreed. If the intermediate species were to die out, the two varients of what is now the same species (because they can still interbreed indirectly via the middle species) would become two different species.

[quote="ProtestanT"][quote="randomdragoon"]Darwin's Finches.

The moths in Europe during the Industrial Revolution.
There was a species of moth (I forget exactly what) that came in a white variety and a black variety, the white variety being more common as they more easily blended into the white (birch?) trees. However, during the Industrial Revolution, soot from industries turned the trees black, and it was observed that the frequency of black moths became greater than the white moths.

Bacteria will gain resistances to drugs after some time.

Fossil Records.[/quote]
that was evolution as in micro-evolution. it could not be applied to macro-evolution to compare with creationism. Creationism in no way opposes micro-evolution.
Evolution cannot cross the boundaries between species. If so, we wouldn't have different species. we would have one huge blend of one specie.
When something does evolve from one species to another through natural selection, natural selection will wipe out the middle form before the second species is reached, because the middle form is the least fit for the environment.
Because of how complex life is, evolving to it will be disasterous. Just a tiny aspect missing can be fatal, like for example if an enzyme was not right and digested the mucus of the stomach, or something went wrong with the reproduction organs, both cases could be fatal. Everything had to have been created at the same time by an intelligent being known as God, because evolving to the current stage could be fatal if even one part was missing.

Fossil records support Creationism more than Evolution. Fossil records show that most phyla were made near the same period of time, and fossil records show that micro-evolution cannot apply to macro-evolution. There were no fossil records that showed the connection between species, and some gaps in the evolution were never filled with fossil records.

People might say that what I said above actually supports ID instead of Creationism, but if you think about it from my point of view, there is no other Intelligent Being than God himself. ID doesn't in any way contradict the first chapter of Genesis for Creationism. So concepts and evidence that support ID can be used to support Creationism.[/quote]

The animals changed over millions of years, and layers, and layers of micro-evolution eventually created a huge change or as you say it, macro-evolution (monkeys became humans, and that took tens of millions of years) the monkeys changed to their environment, and slowly became humans.

One of the main arguments for inteligent design is that things are too complex for it to occur naturally, so they must have been created by god.    Living things started out as 1-celled, and slowly grew and became more complex, (of course some species stayed as they were or became extinct) until we have the wide variety of animals we had today.

Also, show me a site (a scientific one) that shows how fossil evidence supports inteligent design.

"If one thing was screwed up, it would have been fatal" (or something to that effect)

Things probably did screw up (mutations), they were fatal, and that's why they're not around anymore. The things that screwed up but were beneficial  are what we see today.

lol, i dont really no wat u mean by the cambrian explosion:D

But anyways, animals don't start changing to become a different one, (monkeys didn't try to change into humans.)
Macro-evolution is hundreds and hundreds of micro-evolution and it occurs because animals change according to their surroundings. What i'm tryin to say is that animals don't change to become a different one but it is rather the hundreds or millions of different small changes that make different species.

[quote="ProtestanT"]all the middle species were screwed up, and were fatal, and couldn't have got any further.

@ihatepie: have u looked at records for teh cambrian explosion? there were also "missing" fossil records in some evolution trees that were never found, however they form a perfect "lawn-shaped" diagram if they were extended down to new fossil records found around the cambrian explosion.
Also, microevolution cannot be applied on a larger scale. as i said, most middle species would miss vital parts and that would be fatal, or middle species are very weak in their environment, and would get wiped out by natural selection.[/quote]
What vital parts are you talking about here?

Also, there is the concept of punctuated equilibrium. That there is very little change until something happens to disturb the equilibrium, and then stuff happens. Just because we don't see anything happening today doesn't mean it didn't happen earlier.

[quote="randomdragoon"][quote="daermon"][quote="ProtestanT"]all the middle species were screwed up, and were fatal, and couldn't have got any further.

@ihatepie: have u looked at records for teh cambrian explosion? there were also "missing" fossil records in some evolution trees that were never found, however they form a perfect "lawn-shaped" diagram if they were extended down to new fossil records found around the cambrian explosion.
Also, microevolution cannot be applied on a larger scale. as i said, most middle species would miss vital parts and that would be fatal, or middle species are very weak in their environment, and would get wiped out by natural selection.[/quote]
as randomdragoon has said, not "all" of the middle species were eliminated, only the ones that got screwed up
even if most middle species missed vital parts and were eliminated by natural selection, over time there would be some that weren't
also, maybe not all of a species would be wiped out, only some[/quote]
Nice way to put it.

Also, think of it this way: Even if only 1 in a million mutations turns out to be non-fatal, if you have several million mutations over thousands of years (plus or minus), some of them won't be fatal.[/quote]
because of the very tiny chances of having a non-fatal middle speices, and the very high chance of being wiped out by natural selection because of its uselessness, millions of years won't be enough. not even billions would be enough. there just is not enough time for MacroEvolution to work

The chance is high enough. Especially with r-selected organisms, but even with K-selected organisms.

P.S.: How did this thread turn into this debate in the first place?

[quote="randomdragoon"]I think religions that do not update their worldviews with the times are silly. Example - The evolution vs. creationism debate.[/quote]
it's when you listed two theories that are both out of date.


Find the residue when $x^{1000}$ is divided by $x^3+x^2+x+1.$ and the coefficient of  $x^{100}$ for the quotient.

[hide="Solution"]Denote $P(x)=x^{1000}.$ We have $x^3+x^2+x+1=(x+1)(x+i)(x-i)$, hence

$P(x)=(x+1)(x+i)(x-i)Q(x)+R(x)$ for some polynomials $Q(x), R(x)$, where $\deg R(x)=2$,
hence we can write $R(x)=ax^2+bx+c$

Since $P(1)=1, P(i)=P(-i)=(-1)^{500}=1$, we have

$1=a\cdot 1^2+b\cdot 1+c\implies a+b+c=1$
$1=a\cdot i^2+b\cdot i+c\implies -a+c+ib=1$
$1=a\cdot (-i)^2+b\cdot(-i)+c\implies -a+c-ib=1$

Obviously $b=0$ and from $a+c=1\land -a+c=1$ we get $a=0, c=1$

Hence $R(x)\equiv 1$

Now

\begin{eqnarray*}Q(x) &=& \frac{P(x)-R(x)}{x^3+x^2+x+1}\\ &=& \frac{(x^{1000}-1)(x-1)}{x^4-1} \\ &=& (x^{996}+x^{992}+\dots+x^4+1)(x-1)\end{eqnarray*}

Obviously, the coefficient of $x^{100}$ is $-1$.[/hide]

$x^4-1=(x-1)(x^3+x^2+x+1)\Longleftrightarrow x^4=(x-1)(x^3+x^2+x+1)+1.$ :)

Your answers are correct.  :)


http://erdos.fciencias.unam.mx is down. why? will it go up again?


[hide] We use the principle of inclusion-exclusion. The desired number is $ \left\lfloor{\frac{2001}{3}}\right\rfloor\plus{}\left\lfloor{\frac{2001}{4}}\right\rfloor\minus{}\left\lfloor{\frac{2001}{12}}\right\rfloor\minus{}\left\lfloor{\frac{2001}{15}}\right\rfloor\minus{}\left\lfloor{\frac{2001}{20}}\right\rfloor\plus{}\left\lfloor{\frac{2001}{60}}\right\rfloor\equal{}801$. Our answer is $ \boxed{\text{(B) 801}}$.[/hide]

[quote="worthawholebean"]How many positive integers not exceeding $ 2001$ are multiples of $ 3$ or $ 4$ but not $ 5$?

$ \textbf{(A)}\ 768 \qquad
\textbf{(B)}\ 801 \qquad
\textbf{(C)}\ 934 \qquad
\textbf{(D)}\ 1067 \qquad
\textbf{(E)}\ 1167$[/quote]
[hide="Solution"]
Our answer is $\lfloor \dfrac{2001}{ 3} \rfloor + \lfloor \dfrac{2001}{4 } \rfloor - \lfloor \dfrac{2001}{ 12} \rfloor - \lfloor \dfrac{2001}{ 15} \rfloor - \lfloor \dfrac{2001}{ 20} \rfloor + \lfloor \dfrac{2001}{ 60} \rfloor = 667 + 500 - 166 - 133 - 100 + 33 = \textbf{(B)}\ 801  $
[/hide]

i have question, i get why we subtract the terms that were subtracted, however why do we add back 2001/60 ? 60 is multiple of 3,4, AND 5... ? so why would we include it?

When you have three pieces of PIE, it's:

$N(A \cup B \cup C) =$

$N(A) + N(B) + N(C) - N(A \cap B) - N (B \cap C) - N (C \cap A) + N(A \cap B \cap C)$

Where $N(X)$ is the number of elements in set $X$.

Why?  Well, if you look at an item in the small section $A \cap B \cap C$, you should ask how many times it's already counted previously, because it should only be counted once when you're counting $A \cup B \cup C$.  It's in $A, B, C, A\cap B, B \cap C, C \cap A, A \cap B \cap C$.  In other words, it's in all six sets.  Therefore, it gets added three times at the beginning in the single sets, and subtracted three times in the double sets.  Thus, it gets counted $3 - 3 = 0$ times so far if we don't include the last term.  So, we need to add it back in in a triple set.

Can you extend to $N(A \cup B \cup C \cup D)$?

Sarcastic note to self:  Search and destroy whoever thought cap/cup was a cute idea in $\LaTeX$.

but... wouldnt the intersection of the three sets (in the problem) include multiples 5 , which is what we dont want?

D'oh, I wasn't really reading the original question.  You're right!

There are $666$ multiples of $3$ and $500$ multiples of $4$, for $1166$ numbers.  Then subtract off the $166$ multiples of $12$ to leave $1000$ numbers.  (Huh.  That's half of $2000$.  Kinda cool, eh?  Can you think of a quick way to prove that?)

Now, out of those $800$ numbers, there's $\frac{1}{5}$ of them that are multiples of $5$.  That should leave $640$.  Am I nuts?  Where's my error?

One simple way to do this would be to simply write out the numbers from $1$ to $60$, and circle only numbers that are multiples of $3$ or $4$.  Then cross out numbers that are a multiple of $5$.  There are $33$ of these full cycles, and you have $\frac{1}{3}$ of a cycle (20 numbers) remaining that you can repeat from the first $20$ numbers.

Um sorry... I still don't get it. why do we add the intersection of all three sets, if it cant be a multiple of 5? sorry.

Because we subtracted it out too many times in our count; I'm sorry, it's the best I can do for an explanation. To see a much more elegant, better explanation, check out Chapter 3 in Intermediate Counting and Probability (AoPS of course), or just look at the following links:

[url]http://www.artofproblemsolving.com/Wiki/index.php/PIE[/url]

Also, the first couple lines of [url=http://en.wikipedia.org/wiki/Principle_of_inclusion-exclusion]PIE[/url] might help.

@talkinaway,

You said:

Now, out of those 800 numbers, there's  1/5 of them that are multiples of 5. That should leave 640. Am I nuts? Where's my error?

There aren't 800 numbers, there are 1000, as you said in the sentence above. Subtracting 1/5 of 1000, you get 800. You forgot to count 2001, which also is a multiple of 3. (you said there were 666 multiples of 3, there should be 667).

so 800 + 1 = 801 and we get the right answer.

Use principle of inclusion and exclusion


A croissant shop has plain croissants, cherry croissants, chocolate croissants, almond croissants, apple croissants, and broccoli croissants. How many ways are there to choose

1) two dozen croissants with at least five chocolate croissants and at most three almond croissants?

[hide]Since 5 must be chocolate, we can choose 19 more.  Break into cases:
CASE I: 0 almond croissants: $5^{19}$ ways (19,073,486,328,125)
CASE II: 1 almond croissants: $1\left(5^{18}\right)$ ways (3,814,697,265,625)
CASE III: 2 almond croissants: $1^{2}\left(5^{17}\right)$ ways (762,939,453,125)
CASE IV: 3 almond croissants: $1^{3}\left(5^{16}\right)$ ways (152,587,890,625)
So there are $\boxed{\text{23,803,710,937,500}}$ ways.[/hide]


I did them on photoshop heres the img link [url=http://i6.photobucket.com/albums/y208/outoflimitz/math1.jpg]CLICK HERE TO SEE[/url]

Don't you mean you scanned it?

I'll do them later if I remember this topic. You are aware you'll never learn that much if we constantly do them?

As mentioned probably a thousand time before, please don't post multiple problems in a post.  For reasons, [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=23215]Look Here[/url]

As a result, I am locking this thread.


Lol, it's so true.  I thought I was the only person who was suspicious about Dora, guess not :) .  By your signature and your post I'm hoping your Conservative ;)  Coulter's books rock.

[quote="Ignite168"]1) She speak perfect Spanish at like what, age 5? What kind of American speaks perfect Spanish at age 5?
2) That backpack of her's has EVERYTHING in it! And we're talking everything! life support, water/food, clothing for any weather, ropes, grappling hooks, shoes, I mean come on now, it's so obvious.
3) She's carrying a freaking ZOO with her! i mean she has a monkey, an band of insects, a bajillion other animals! What kind of legal immigrant has that many pets?
4) She's always on an "adventure" to transport a "package" to some destination and is always being stalked by a person trying to take her package. Swiper is so obviously some sort of border patrol person trying to collect evidence of Dora's entire narcotics trafficking buisness
Where is she now? I need my hedges trimmed.[/quote]
LOL  !!!

make sure there are not any 5 year olds on this site, yhey might cry because of your cruelness!

I am going to cry. Dora most def. lives in Mexico. And she taught me how to count to tan in spanish and how to say blue and she is so nice. I mean she always helps her grandmother and she has a nice cousin, who rescues animals and she is always nice to little kids oh and she rescued the town from the grumpy man who doesn't like music and she sings that coce or whatever song and she has friends not pets that are animals and she teaches kids that people can tell eachother "swiper no swiping"

[quote="Ignite168"]1) She speak perfect Spanish at like what, age 5? What kind of American speaks perfect Spanish at age 5?
2) That backpack of her's has EVERYTHING in it! And we're talking everything! life support, water/food, clothing for any weather, ropes, grappling hooks, shoes, I mean come on now, it's so obvious.
3) She's carrying a freaking ZOO with her! i mean she has a monkey, an band of insects, a bajillion other animals! What kind of legal immigrant has that many pets?
4) She's always on an "adventure" to transport a "package" to some destination and is always being stalked by a person trying to take her package. Swiper is so obviously some sort of border patrol person trying to collect evidence of Dora's entire narcotics trafficking buisness
Where is she now? I need my hedges trimmed.[/quote]

That's racist...

:rotfl:

She is left-handed!

And in one episode she had a Ferris Wheel in her backpack.

If i remember correctly, she has horrible accents

An immigrant is one thing, an illegal immigrant is another.
And a lefty is another yet.

Also, in one, she had a cell phone. WHAT 5-YEAR OLD HAS A CELL PHONE!?!?!? There's DEFINITELY something going on over at nick jr...

[quote="mustafa"]Swiper No Swiping![/quote]
I don't think Swiper is a very good criminal. He leaves when they say "Swiper no Swiping", he walks [i]really[/i] slowly, and in the open. And whenever he manages to swipe something, he throws it away...
Swiper should be using a really big weapon, like a machine gun, or a giant club, beat Dora up, and sell everything in her backpack. Considering that the backpack has [i]everything[/i] in it, it should make a nice profit.

Swiper isn't a criminal, he's a narcotics buster.

And what's with boots? Why must he wear the red boots? And why does he talk to someone like dora? And why is he not in captivity? He's a TALKING MONKEY.

:rotfl: You guys are reading too much into "Dora the Explorer".

Just like when my gifted class did a systems-theoretic analysis of a "I Love Lucy" episode... :roll:

[quote="Ignite168"]you cant compare oranges and apples.[/quote]yes you can

[quote="Ignite168"][/quote][quote="Scrambled"][/quote][quote="Ignite168"][quote="junggi"][quote="Ignite168"]
stop spamming with your bs logic.
you don't hear latin americans call themselves americans so therefore they aren't americans de facto.[/quote]
wow this is getting idiotic[/quote]
i'll accept that as a concession of defeat[/quote][quote="Scrambled"]
I haven't seen you calling yourself a human on a regular basis.
Therefore, you aren't a human.[/quote][quote="Ignite168"]

you cant compare oranges and apples.[/quote]
Well you can't compare that comparison to the totally bs one you gave, and thus, you are a troll, and should be executed for disturbing the peace.  You are a lousy, good-for-nothing troll, and furthermore, you are an example of what AoPS should not be.  Therefore, I condemn you to ten-thousand years of hard labor in the coal mines of Antartica, where you'll have fun with penguins.

[quote="Ignite168"]
stop spamming with your bs logic.
you don't hear latin americans call themselves americans so therefore they aren't americans de facto.[/quote]

Actually I do... but I don't know how this would help.

[quote="ccy"][/quote][quote="Ignite168"][/quote][quote="Scrambled"][/quote][quote="Ignite168"][quote="junggi"][quote="Ignite168"]
stop spamming with your bs logic.
you don't hear latin americans call themselves americans so therefore they aren't americans de facto.[/quote]
wow this is getting idiotic[/quote]
i'll accept that as a concession of defeat[/quote][quote="Scrambled"]
I haven't seen you calling yourself a human on a regular basis.
Therefore, you aren't a human.[/quote][quote="Ignite168"]

you cant compare oranges and apples.[/quote][quote="ccy"]
Well you can't compare that comparison to the totally bs one you gave, and thus, you are a troll, and should be executed for disturbing the peace.  You are a lousy, good-for-nothing troll, and furthermore, you are an example of what AoPS should not be.  Therefore, I condemn you to ten-thousand years of hard labor in the coal mines of Antartica, where you'll have fun with penguins.[/quote]

and if thats true, youre feeding it.

[quote="Ignite168"][/quote][quote="ccy"][/quote][quote="Ignite168"][/quote][quote="Scrambled"][/quote][quote="Ignite168"][quote="junggi"][quote="Ignite168"]
stop spamming with your bs logic.
you don't hear latin americans call themselves americans so therefore they aren't americans de facto.[/quote]
wow this is getting idiotic[/quote]
i'll accept that as a concession of defeat[/quote][quote="Scrambled"]
I haven't seen you calling yourself a human on a regular basis.
Therefore, you aren't a human.[/quote][quote="Ignite168"]

you cant compare oranges and apples.[/quote][quote="ccy"]
Well you can't compare that comparison to the totally bs one you gave, and thus, you are a troll, and should be executed for disturbing the peace.  You are a lousy, good-for-nothing troll, and furthermore, you are an example of what AoPS should not be.  Therefore, I condemn you to ten-thousand years of hard labor in the coal mines of Antartica, where you'll have fun with penguins.[/quote][quote="Ignite168"]

and if thats true, youre feeding it.[/quote]
well you're ugly

Edit: haha Happy new year's, everyone! :P... and let's stop spamming.

[quote="Ignite168"][quote="mathnerd314"]Stop taking yourselves so seriously :rotfl:

On the line of personal attacks, here's one from FoxTrot:

[quote]Oh yeah? Well, your momma couldn't integrate a nested trig function, even if you let her use a computer.[/quote]

:roll: Seriously, though, happy new year to all :)[/quote]

You are not on topic.[/quote]
Yes he is.

[quote="ccy"][/quote][quote="Ignite168"][quote="mathnerd314"]Stop taking yourselves so seriously :rotfl:

On the line of personal attacks, here's one from FoxTrot:

[quote]Oh yeah? Well, your momma couldn't integrate a nested trig function, even if you let her use a computer.[/quote]

:roll: Seriously, though, happy new year to all :)[/quote]

You are not on topic.[/quote][quote="ccy"]
Yes he is.[/quote]

No he isn't. What part of his post was on topic?

[quote="mathnerd314"][quote]You are not on topic.
Yes he is.
No he isn't. What part of his post was on topic?[/quote]

http://en.wikipedia.org/wiki/The_Argument_Skit

I think someone should lock this topic.[/quote]

go away

I request that the topic be moved to [url=http://www.stormfront.org/forum/forumdisplay.php?f=39]Stormfront's British section[/url], mainly because British people are funny.

And don't revive old topics. But your a new member! Welcome to AoPS  :D  :D  :D


Найдите все непрерывные функции , которые удовлетворяют неравенствам f(2x) – f(x) ≥ 3x² + x и f(3x) – f(x) ≤ 8x² + 2x при каждом .

Please translate me this problem in english. Thank you

[quote="Beat"]Найдите все непрерывные функции , которые...[/quote]
Find all continious functions f(x), for which the inequalities f(2x) – f(x) ≥ 3x² + x and f(3x) – f(x) ≤ 8x² + 2x are true for all x.

$ f(x)\equal{}g(x)\minus{}2x^2\minus{}x$
So $ g(2x)\geq g(x)\geq g(3x)$
Imply that $ g(x)\equiv g(0)$
So $ f(x)\equal{}2x^2\plus{}x\plus{}c$


:mad: :mad:  :mad:  :mad:  [color=red][size=200]here are my four "SMILIES" LOL  (I am limited to four, I'm sure you all know that though ... I mean four is a number RIGHT!) However, they dont look like they are smiling to me! Thanks for the complete lack of intrest in helping me out. Not one person even bothered to post. I hope one of you needs help one day and everyone passes you by![/size][/color]

KAJSAJ, I do not know what problem you are talking about (your only post on the forum is currently this one and it is not very revealing :roll:). Anyway, there may be several reasons why you might not get any response ranging from the problem being too difficult to posting in a wrong forum to its being completely uninteresting (the last one is also a valid reason here: don't expect people to solve for you something that doesn't sparkle their interest; this forum is not for helping with routine homeworks). The only thing that I can suggest now is to post your problem again in a separate thread and to patiently wait until somebody answers. In most cases, you get a response sooner or later but it is not very unusual to have to wait for a couple of weeks or more: this forum is not a math. hotline either. As to myself, I can only promise that if you post your question again, I'll look at it during the next couple of days and will try to help if the question is within my abilities. :)

Only post AND condescending towards the other users! Although I am very interested in seeing this problem.


geometry
3D geometry
combinatorics unsolved
combinatorics

[quote="cyshine"]Prove that there exists a positive integer $ n_0$ with the following property: for each integer $ n \geq n_0$ it is possible to partition a cube into $ n$ smaller cubes.[/quote]
We can partion any cube into $ 8$ smaller cubes or $ 27$ smaller cubes easily.($ 2^3\equal{}8,3^3\equal{}27$)

So if we can split a cube into $ m$ smaller cubes we can also split it into $ m\plus{}7$ and $ m\plus{}26$ smaller cubes. Since we can "split" it into $ 1$ cube we can split it into $ 1\plus{}7x\plus{}26y$ for any $ x,y \in \mathbb{N}_0$. $ 1\plus{}7x\plus{}26y \equal{} n \iff x \equal{} \frac{n\minus{}1\minus{}26y}{7}$. Let $ 0 \le y \le 6$ be such that $ 26y \equiv n\minus{}1 \bmod 7 \iff y \equiv 3n\minus{}3 \bmod 7$. Let $ n_0 \equal{} 1 \plus{} 26 \cdot 6$, then $ x \ge 0$ and we can split it into $ n$ smaller cubes whenever $ n \ge n_0$. QED

check [url]http://www.research.att.com/~njas/sequences/A014544[/url]

for minimum $ n_0\equal{}48$[/hide]

Mathias' argument gives us that $ n_0\equal{}1\plus{}(7\minus{}1)(26\minus{}1)\plus{}1\equal{}152$ works; $ 150$ is the largest positive integer that can't be written as a sum of sevens and 26s. Obviously, we can do better.

First, that $ 26$ could be $ 19$; rather than replacing a $ 1\times 1\times 1$ cube with $ 27$ $ \frac13\times\frac13\times\frac13$ cubes, we replace it with one $ \frac23\times\frac23\times\frac23$ and $ 19$ $ \frac13\times\frac13\times\frac13$ cubes. Since adding $ 26$ is possible by doing the above and then splitting the large cube into eight pieces, we don't need to include it separately. From sevens and nineteens, we get a new estimate of $ n_0\ge 6\cdot 18\plus{}2\equal{}110$.
For the next improvement, consider breaking a $ 1\times 1\times 1$ into a $ \frac34\times\frac34\times\frac34$ and $ 37$ $ \frac14\times\frac14\times\frac14$ cubes; this allows us to add $ 37$. This is very useful, since $ 19\equiv \minus{}2\mod 7$ and $ 37\equiv 2\mod 7$. The multiples $ 0,19,38,57,76,37,74$ form a complete system of residues mod $ 7$, and $ 69$ is the last number we can't get as a sum of copies of $ 7,19,37$. The new estimate is $ n_0\ge 71$.

Finally, there are ways to dissect a cube into $ 49$, $ 51$, or $ 54$ smaller cubes, which effectively replace my dissections into $ 77,$ $ 58,$ and $ 75$ cubes respectively. You can look them up at mszew's link, or try to find them yourself if you're feeling clever. That gets $ n_0$ down to $ 48$.

We can partion any cube into 8 and 27 cubes. So, we can partion any cube into 8+7k, 8k+26, 27+7K and 27+7k.
8+26k and 27+26k form a complete set of residue classes modulo 7. So, we form all numbers greater than 152 (n0=152).. If we can partion any cube into x cubes, we can partion into x+7k


[quote="SuperMath"]Let $\large x;y;z > 0$. Prove that:
$\large \sum x(y+z)^2 \geq\ 2 \sum z \sqrt{xy(x+z)(y+z)}$[/quote]
since
$y(x+z)+x(y+z) \geq\ 2 \sqrt{xy(x+z)(y+z)}$
therefore
$xz^2+yz^2+2xyz \geq\ 2 z\sqrt{xy(x+z)(y+z)}$


I'm frustrated with my homework and need help figuring out how to come up with an equation and solving for the following word problem.  Can someone please help me?

The word problem reads:
When Ruth emptied her bank, she found that she had 84 coins, all nickels and dimes.  The value of the coins was 7 dollars fifteen cents.  How many dimes did she have? Make a table to organize the information according to the formula to be used.  All numbers may be considered to be exact values.

You can email me directly at bballking32015@aol.com
That would be a great help.  Thanks!

Let x = nickels
Let y = dimes

$ x \plus{} y \equal{} 84$
$ 5x \plus{} 10y \equal{} 715$

That's how you set it up now you attempt to solve it.

When Ruth emptied her bank, she found that she had 84 coins, all nickels and dimes. The value of the coins was 7 dollars fifteen cents. How many dimes did she have? Make a table to organize the information according to the formula to be used. All numbers may be considered to be exact values.

$ Total: 7.15$
$ Nickel \equal{} .5$
$ Dime \equal{} .10$
$ Total Coins: 84$

$ .5(x) \plus{} .10(84 \minus{} x) \equal{} 7.15$

You should be able to do the rest on your own.
I would solve it as the person above did
but I think this is the way you learn it the very first time..

Wait...shouldn't nickels be .05? Another helpful thing you may want to do is multiply by 100 on both sides. It will cancel out the decimals.


Hi all,
This message to announce you the new version of K3DSurf, a program to visualize and manipulate Multidimensional surfaces by using Mathematical equations. It supports Parametric and Isosurfaces equations. Available for Windows, Linux, Mac...
All your comments and suggestions are welcome.
Regards,
Taha

Official website : http://k3dsurf.sourceforge.net/

Download K3DSurf for Windows :
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For More Infos :
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[img]http://k3dsurf.sourceforge.net/images/k3dsurf_060/25049-1.png[/img]

6D = 6 Dimension .
In our real world we work with 3 dimension, however, mathematicly speaking, we can work with dimension up to 3D. Theses shapes can't exist in our world but we can  use some 3D drawing technics with the power of today computers to show us what they might look if they existed in our world.

Hi all,
K3DSurf 0.6.2 is out and is available for Windows, MacOS and Linux from :
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or :http://k3dsurf.sourceforge.net/
Download K3DSurf for Windows :
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Also, there is a new forum where you can see some use of mathematics in art, ask questions, find new examples... :
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[img]http://k3dsurf.sourceforge.net/images/k3dsurf_062/062a.png[/img]
Cheers,
Taha


I don't don't know how to access the alcumus assignments. I can find the videos, but there is no work to do.I have done the diagnostic test (60 problems), but I don't know what to do from there.

I'm assuming you are in the Intro to Counting and Probability class. Click on My Classes, and if needed, your class name. At the bottom is a list of Alcumus topics. Just use Alcumus normally after that. Problems will be fed to you somewhat randomly, but will likely be somewhat targeted to those topics you have not yet passed.


Let $ a,b,n$ be natural numbers such that $ (b,n) \equal{} 1$.
Let $ A \equal{} \{a,a \plus{} b,a \plus{} 2b,...,a \plus{} (n \minus{} 1)b\}$ and $ B \equal{} \{1,2,...,n\}$
Let $ p$ be any prime not dividing $ b$.

Prove or disprove the following:
We can pair the elements in the two sets into pairs $ (a_1,b_1),(a_2,b_2),...,(a_n,b_n)$ such that the exponents of $ p$ dividing $ a_i$ and $ b_i$ are the same for all $ i \equal{} 1,2,...,n \minus{} 1$, and the exponent of $ p$ dividing $ a_n$ is not less than that of $ b_n$.

$ \exists r\in\{0,1,...,p \minus{} 1\}: \ p|a \plus{} rb$
we take $ B_h \equal{} \{x\in B: \ p^h|x\},A_h \equal{} \{x\in A: \ p^h|x\}$
if $ n \equal{} \overline{a_ma_{m \minus{} 1}....a_1a_0}^{(p)}$
so for $ h\in[|1,m|]$: $ |B_h| \equal{} \sum_{k \equal{} h}^{m}a_{k}p^{k \minus{} h} \plus{} [\frac { \minus{} r}{p^{h}}]$ and $ |A_h| \equal{} \sum_{k \equal{} h}^{m}a_{k}p^{k \minus{} h}$
if $ r\neq 0$: $ \forall h\ge 1: \ |B_h| \equal{} |A_h| \minus{} 1$ then $ \forall h\ge 1: \ |B_h \minus{} B_{h \minus{} 1}| \equal{} |A_h \minus{} A_{h \minus{} 1}|$
if $ r\equal{}0 0$: $ \forall h\ge 1: \ |B_h| \equal{} |A_h|$ then $ \forall h\ge 1: \ |B_h \minus{} B_{h \minus{} 1}| \equal{} |A_h \minus{} A_{h \minus{} 1}|$

so for $ h\in\{1,2,....,m \minus{} 1\}: \ |\{x\in B: \ v_p(x) \equal{} h\}| \equal{} |\{x\in A: \ v_p(x) \equal{} h\}|$
then for $ h\in\{1,2,..,m \minus{} 1\}$ we can pair the elements in the two sets into pairs $ (a_{1h},b_{2h}),....,(a_{gh},b_{gh})$ such that $ v_p(a_{ih}) \equal{} v_p(b_{ih}) \equal{} h$
and for $ h \equal{} m$ it's obvious that $ B_m \minus{} B_{m \minus{} 1} \equal{} \{y,x_1,x_2,...,x_k\}$ where $ v_p(x_i) \equal{} m$ and $ v_p(y)\ge m$
and $ A_m \minus{} A_{m \minus{} 1} \equal{} \{y_1,y_2,....,y_{k \plus{} 1}\}$ where $ v_p(y_i) \equal{} m$
we take the pairs $ (x_i,y_i)_{i \equal{} 1..k}$ and $ (y_{k \plus{} 1},y)$ and we have $ v_p(x_i) \equal{} v_p(y_i)$ and $ v_p(y)\ge v_p(y_{k \plus{} 1}) \equal{} m$

and we can write $ n!|\prod_{j \equal{} 0}^{n \minus{} 1}(a \plus{} jb)$

[quote="aviateurpilot"]
so for $ h\in[|1,m|]$: $ |B_h| \equal{} \sum_{k \equal{} h}^{m}a_{k}p^{k \minus{} h} \plus{} [\frac { \minus{} r}{p^{h}}]$ and $ |A_h| \equal{} \sum_{k \equal{} h}^{m}a_{k}p^{k \minus{} h}$
[/quote]

I don't think your solution is quite correct, such as here...  :wink:

Can anybody make it right?


[quote="Peter"]Find all triples $(a, b, c)$ of positive integers  to the equation \[a! b! = a!+b!+c!.\][/quote]
Assume WLOG that $a \geq b$
$a!b!-a!-b!=c!$
$(a!-1)(b!-1)=c!+1$
(3,3,4)
Not sure how to prove that it's the only solution.

It isn't the only solution either. ;) There's (2,2,1) and I think a few others.

http://www.mathlinks.ro/Forum/viewtopic.php?t=169922
I have post solution with other method.

[quote="Peter"]Find all triples $(a, b, c)$ of positive integers  to the equation \[a! b! = a!+b!+c!.\]

a!b!=a!+b!+c!
CASE 1
IF, a<b<c
Then a!q=b! and
a!k=b!i=a!qi=c!                                ………(1)
substituting in the equation,we get
(a!)^2 q=a!(1+q+k)
(a!)q =1+q+k
q divides 1+q+k
q divides k+1                                   ………..(2)
But by (1),
a!k=a!qi
k=qi
q divides k contradicting (2)
Similarly, for all other permutations of a, b and c we get the same contradiction.
CASE 2
a =b=c
Substituting in the equation we get,
(a!)2 =3(a!)
(a!) =3 ,and we know there is no such ‘a’.
CASE 3
b=c
Substituting, in the equation we get
a!b! = a! +b! + b!
a!b! = a! +2b!
b!(a!-2) =a!
a!-2 divides a!
which we may see as impossible.
Similar for a=c.
 Now a=b,
Substituting this in the equation we get,
(a!)^2 =2(a!) +c!
a!(a!-2) =c!
c>a
a! -2 =(a+1)(a+2)(a+3)…….(a+n)
As 2 divides a,
(a+1)……(a+n) must contain only 2 factor.
We know 4 consecutive numbers will have 4 as a factor .
If a+1=a!-2
a=a!-3
a divides 3
a =(1,3)
a =3	                        ……(3)
If a!-2 =(a+1)(a+2)
a!-2 =a^2+3a+2
a! -4=a(a+3)
a divides 4
a=  (1,2,4)
here a has no solution.
Similarly a! -2 =(a+1)(a+2)(a+3) will also have no solution.
Therefore solution set is (a,b,c) = (3,3,4) by (3)


$N$ is odd and $N\geq15$.
There are $N$ cards such that on each card is written his index.
Jack chooses any card from the $N$ cards.
There are $3$ magicians:
The first and the second magicians get $\frac{N-1}{2}$ cards(Any of them).
Any of them is looking on his cards and gives $2$ cards to the third magician that he decide.
The third magician is looking on his $4$ cards now, and decides what card was chosen by Jack.
Find a strategy for the magicians to do this.
I don't on which subject should I post it, I think that the solution has a some connection to combinatorics.

You have already posted this in 3 different sections.
Please, chose one and delete the others  :!:


How many 3-element subsets of the set $\{3,3^2,3^3,\ldots,3^{1000}\}$ consist of three numbers which form a geometric sequence?

Not that bad....
[hide]Bijection between statement and" How many 3-element subsets of the set $\{1,2,3,\ldots,1000\}$ consist of three numbers which form an arithmetic sequence?. Well, let difference $=d$, total subsets $=t$
$d=1 \rightarrow$ sets: $\{1,2,3\},\{2,3,4\}......\{998,999,1000\} \rightarrow t=998$
$d=2 \rightarrow$ sets: $\{1,5,7\},\{2,6,8\}......\{996,998,1000\} \rightarrow t=996$
$d=3 \rightarrow$ sets: $\{1,4,7\},\{2,5,8\}......\{994,997,1000\} \rightarrow t=994$
.
.
. (Applying similarly)
$d=499 \rightarrow$ sets: $\{1,500,999\},\{2,501,1000\}......\{998,999,1000\} \rightarrow t=2$

What do you know? They are basically asking for the sum of the first $499$ even counting numbers. That is $\fbox{249,500}$ different subsets.[/hide]
Edited: Dumb mistake in last line :oops:

[hide="hint1"]A geometric sequence would mean that the exponents increase arithmetically, so it is the same question as finding the number of 3-element subsets of ${1,2,3,...,1000}$ that form an arithmetic sequence.[/hide]
[hide="hint2"]This means that there is a constant difference between the elements of the new set.  So all you have to do is set up an appropriate equation.[/hide]
[hide="hint3"]$N(d) = 1000-2d$[/hide]
[hide="hint4"]$\sum_{i=1}^{499}{1000-2d} = 499*1000 - 499*500 = 499*500 = 250000-500 = 249500$[/hide]
Read it starting with the first hint and go from there, they give more and more away. hint4 is the answer.


What are the last-two-digits of $ 13^{2005}\times9999^4\times3^{24}$ and of $ 13^{2009}\times9999^{6}\times3^{12}$?
[The last-two-digit of 1234 is 34.]

[hide="Solution to #1"]
$ \text{First one}\equiv 13^{2005}\times 3^{24}\bmod 100$. (I threw out the $ 9999$ bit because it's $ 1\bmod 100$).

$ \lambda(100)\equal{}20$, so $ 13^{2005}\equiv 13^5\equiv 93\bmod 100$.

$ 3^{24}\equiv 3^4\equiv 81\bmod 100$.

Thus, the answer is $ \boxed{33}\bmod 100$.
[/hide]

[hide="Solution to #2"]
$ \text{Second one}\equiv 13^{2009}\times 3^{12}\equiv 13^9\times 3^{12}\bmod 100$.

I can't think of a better way than bashing this out (raising them to powers and continually modding by 100).

$ 13^9\equiv 13\cdot 69^4\equiv 13\cdot 61^2\equiv 73\bmod 100$.

$ 3^{12}\equiv 9^6\equiv 81^3\equiv \minus{}19^3\equiv \minus{}59\equiv 41\bmod 100$.

Answer: $ \boxed{93}\bmod 100$.
[/hide]


Euler trigonometry integration function calculus limit geometry

next to euler's proof of $ \sum_{k=1}^\infty\;\frac{1}{k^{2}}\;=\;\boxed{\frac{\pi^{2}}{6}}$,
i think the following one is just as clever, ingenius, and original. :)
(euler is the master, of-course...  :omighty:  )

here is the proof (some of you may have seen this already) :

we have, $ \sin^{-1}x\;=\;\sum_{k=0}^\infty\;\frac{1\cdot 3\cdot 5\cdot 7\cdots (2k-1)}{2\cdot 4\cdot 6\cdot 8\cdots (2k)}\;\;\frac{x^{2k+1}}{(2k+1)},\;\;\;\quad \forall |x|\leq 1$

put $ y=\sin^{-1}x$ to get

$ y\;=\;\sum_{k=0}^\infty\;\frac{1\cdot 3\cdot 5\cdot 7\cdots (2k-1)}{2\cdot 4\cdot 6\cdot 8\cdots (2k)}\;\;\frac{\sin^{2k+1}y}{(2k+1)},\;\;\;\quad \forall |y|\leq \frac{\pi}{2}$

so, integrating w.r.t. $ y$ between the limits $ 0$ and $ \frac{\pi}{2}$, we get

$ \frac{1}{2}\;y^{2}\bigg|_{0}^\frac{\pi}{2}\;=\;\sum_{k=0}^\infty\;\frac{1\cdot 3\cdot 5\cdot 7\cdots (2k-1)}{2\cdot 4\cdot 6\cdot 8\cdots (2k)}\;\;\frac{1}{(2k+1)}\;\int_{0}^{\frac{\pi}{2}}\;\sin^{2k+1}y\;\textbf dy$

now using reduction formula, one can easily show

$ \int_{0}^{\frac{\pi}{2}}\;\sin^{2k+1}y\;\textbf dy\;=\;\frac{2\cdot 4\cdot 6\cdot 8\cdots (2k)}{1\cdot 3\cdot 5\cdot 7\cdots (2k+1)}$

and so, we have

$ \frac{\pi^{2}}{8}\;=\;\sum_{k=0}^\infty\;\frac{1\cdot 3\cdot 5\cdot 7\cdots (2k-1)}{2\cdot 4\cdot 6\cdot 8\cdots (2k)}\;\;\frac{1}{(2k+1)}\;\cdot\;\frac{2\cdot 4\cdot 6\cdot 8\cdots (2k)}{1\cdot 3\cdot 5\cdot 7\cdots (2k+1)}$

cancelling numerous terms from the numerator and denominator,
we are finally left with
$ \frac{\pi^{2}}{8}\;=\;\sum_{k=0}^\infty\;\frac{1}{(2k+1)^{2}}$

$ \Rightarrow\;\sum_{k=0}^\infty\;\frac{1}{k^{2}}\;=\;\boxed{\frac{\pi^{2}}{6}}$

what an absolute beauty!  :coolspeak:  :yup:  :clap2:

[ proof is from b. r. choe's note in 1987's AMM magazine, taken from r. chapman's compilation ]

wonder if there are any such wonderful but succinct proofs of $ \zeta{(2)}$!!

Another proof of this remarkable result of Euler can be obtained by using Fourier analysis, namely:

$ x^{2}=\frac{\pi^{2}}{3}+4\sum_{n=1}^{\infty}(-1)^{n}\frac{\cos nx}{n^{2}}$, $ -\pi\leq x\leq \pi$.

or

$ \frac{3x^{2}-6\pi x+2\pi^{2}}{12}=\sum_{n=1}^{\infty}\frac{\cos nx}{n^{2}}$, $ 0\leq x\leq \pi$.

However a very nice proof, based entirely on a trigonometry technique, is given in I.M. Yaglom's book, I think is called,

Nonelementary problems treated elementary.

Complex Analysis is overkill but here it goes...

1)Let $ \gamma_{n}$ be a square contour with vertices at $ (n+1/2)(\pm 1 \pm i)$.

2)Define $ f(z) = \frac{\pi \cot \pi z}{z^{2}}$.

3)$ f(z)$ has simple poles at $ \pm 1, \pm 2, ...$ and a triple pole at $ 0$.

4)Inside the contor $ \gamma_{n}$ the poles are at $ \{ 0, \pm 1,...,\pm n\}$.

5)Now $ \mbox{res}(f,\pm k) = \frac{1}{k^{2}}$ for $ 1\leq k \leq n$ and $ \mbox{res}(f,0)=-\frac{\pi^{2}}{3}$.

6)Residue theorem: $ \oint_{\gamma_{n}}f(z) dz = 2\pi i \left( \sum_{-n \leq k \leq n}^{k\not = 0}\frac{1}{k^{2}}-\frac{\pi^{2}}{3}\right)$

7)Now take the limit $ n\to \infty$ then $ \oint_{\gamma_{n}}f(z) dz \to 0$ (this takes a little work to show, but not too bad).

8)We end up with $ \sum_{k\in \mathbb{Z}}^{k\not = 0}\frac{1}{k^{2}}= \frac{\pi^{2}}{3}$

9)Ta daa: $ \sum_{k=1}^{\infty}\frac{1}{k^{2}}= \frac{\pi^{2}}{6}$

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=80515]Here is another proof[/url]

[quote="didilica"]However a very nice proof, based entirely on a trigonometry technique, is given in I.M. Yaglom's book, I think is called,

Nonelementary problems treated elementary.[/quote]

hmmm..... does that like use the [i]squeezing [/i]theorem after bounding the partial sums
$ \sum_{k=1}^{n}\;\frac{1}{k^{2}}$ between 2 exprs derived from identities via the $ \cot$ and the $ \csc$ functions? :|

Yes, if I remember corectly, that is the elementary proof of Euler's result.

Indeed, and that is the one of the most, if not the most beutiful and best proof I've seen on this "basel problem"
It can be found on wikipedia,[url=http://en.wikipedia.org/wiki/Basel_problem#A_rigorous_proof]here is the link[/url]
Enjoy ! :)

(perhaps we are not discussing the same proof? Anyway wikipedia states another source.)

Euler's own proof (although not rigorous) is also given there and it is too very beautiful,
one of the best.

yes, that was the proof i was referring to. :|

however, the next best proof i have seen, after those of euler and choe,
is the one where we apply parseval's identity, i.e.

$ \sum_{n=-\infty}^\infty\;a_{n}\cdot \overline{b_{n}}\;=\;\frac{1}{2\pi}\;\int_{-\pi}^{\pi}\;f(x)\cdot \overline{g(x)}\;\;\;\textbf dx\;\;\;\qquad\cdots\;\;\;\cdots\;\;\; (1)$

( where $ f(x)\;=\;\sum_{n=-\infty}^\infty\;a_{n}\cdot e^{\imath nx},\;\;\;\text{and}\;\;\;g(x)\;=\;\sum_{n=-\infty}^\infty\;b_{n}\cdot e^{\imath nx}$,
and the horizontal bar $ \overline{\textbf X}$ indicates complex conjugate)

$ f(x)$ and $ g(x)$ are 2 riemann-integrable "complex-valued" functions,
defined as continuous and piece-wise smooth within the interval $ [-\pi,\pi]$

now let $ f(x)\;=\;g(x)$, we immediately get, from $ (1)$
$ \sum_{n=-\infty}^\infty\;|\;a_{n}\;|^{2}\;=\;\frac{1}{2\pi}\;\int_{-\pi}^{\pi}\;|\;f(x)\; |^{2}\;\;\;\textbf dx$

put $ f(x)\;=\;x$, and compute the fourier co-efficients $ a_{n}$, and one has

$ a_{n}\;=\;\frac{1}{2\pi}\;\int_{-\pi}^{\pi}\;x\cdot e^{-\imath nx}\;\;\;\textbf dx\;=\;\left\{\begin{array}{cl}-\frac{(-1)^{n}}{n}\;\imath\; ,\;\;\;\quad \forall n<0\\ 0\; ,\;\;\;\quad for\;\;\;n=0\\ \frac{(-1)^{n}}{n}\;\imath\; ,\;\;\;\quad \forall n>0 \end{array}\right.$

and so, $ \forall x\in[-\pi,\pi]$, we have
$ \sum_{n=-\infty}^{-1}\;\bigg|\;-\frac{(-1)^{n}}{n}\;\imath\;\bigg|^{2}\;+\;0\;+\;\sum_{n=1}^\infty\;\bigg|\;\frac{(-1)^{n}}{n}\;\imath\;\bigg|^{2}\;=\;\frac{1}{2\pi}\;\int_{-\pi}^{\pi}\;|x|^{2}\;\;\;\textbf dx$

i.e. $ 2\;\sum_{n=1}^\infty\;\frac{1}{n^{2}}\;=\;\frac{\pi^{2}}{3}$

$ \Rightarrow\sum_{n=1}^\infty\;\frac{1}{n^{2}}\;=\;\boxed{\frac{\pi^{2}}{6}}$

parseval's theorem is truly amazing, it allows us to equate a proper
and definite [b][u]integral[/u][/b] to an [i]infinite [/i][b][u]sum[/u][/b] and vice versa... :)

Here is a paper which contains 14 different proofs of Euler's result.
http://www.secamlocal.ex.ac.uk/people/staff/rjchapma/etc/zeta2.pdf

wow... nice. :)

seemed a little surprising to me he doesn't show yet another
classic and elementary proof (from hall and knight's text
on "higher algebra")

one can show,
$ 1+\frac{1}{3^{2}}+\frac{1}{5^{2}}+\frac{1}{7^{2}}+\cdots+\;=\;2\left( 1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\;-\cdots+\;-\; \right)^{2}$

$ =\;2\left( \frac{\pi}{4}\right)^{2}\;=\;\frac{\pi^{2}}{8}$

therefore,
$ 1+\frac{1}{2^{2}}+\frac{1}{3^{2}}+\frac{1}{4^{2}}+\cdots+\;=\;\frac{4}{3}\left( \frac{\pi^{2}}{8}\right)\;=\;\boxed{\frac{\pi^{2}}{6}}$

[ PS: i am also aware of 3 other elementary proofs that are [u]not[/u]
mentioned in his compilation... ]

[quote="misan"]...

[ PS: i am also aware of 3 other elementary proofs that are [u]not[/u]
mentioned in his compilation... ][/quote]

Please share them with us [b]misan[/b].

OK, here is one of them, uses only reduction formula(s): :|

one can easily show
$ \int_{0}^{\frac{\pi}{2}}\;\cos^{2k}x\;\textbf dx\;=\;\frac{1\cdot 3\cdot 5\cdots (2k-1)}{2\cdot 4\cdot 6\cdots (2k)}\;\frac{\pi}{2}$

now applying integration by parts twice, show:
$ \textbf I_{2k}\;=\;\int_{0}^{\frac{\pi}{2}}\;x^{2}\cdot \cos^{2k}x\;\textbf dx\;=\;\left(\frac{2k-1}{2k}\right)\;\textbf I_{2k-2}\;-\;\frac{1}{2}\cdot\left(\frac{1}{k^{2}}\right)\;\int_{0}^{\frac{\pi}{2}}\;\cos^{2k}x\;\textbf dx$

eventually this yields,
$ \textbf I_{2k}\;=\;\frac{1\cdot 3\cdot 5\cdots (2k-1)}{2\cdot 4\cdot 6\cdots (2k)}\;\textbf I_{0}\;-\;\frac{1\cdot 3\cdot 5\cdots (2k-1)}{2\cdot 4\cdot 6\cdots (2k)}\cdot \left(\frac{\pi}{4}\right)\;\sum_{j=1}^{k}\;\frac{1}{j^{2}}$

now $ \textbf I_{0}\;=\;\int_{0}^{\frac{\pi}{2}}\;x^{2}\;\textbf dx\;=\;\frac{\pi^{3}}{24}$

hence, we have $ \frac{\pi^{3}}{24}\;-\;\left(\frac{\pi}{4}\right)\;\sum_{j=1}^{k}\;\frac{1}{j^{2}}\;=\;\frac{2\cdot 4\cdot 6\cdots (2k)}{1\cdot 3\cdot 5\cdots (2k-1)}\;\textbf I_{2k}$

now show RHS goes to $ 0$ in the $ \lim_{k\to\infty}$.

since $ \forall x\in[\;0,\frac{\pi}{2}\;]$, we have $ \sin x\geq \frac{2}{\pi}x\;\;\;\;\;\Rightarrow \;\;\;\cos^{2}x\leq 1-\frac{4}{\pi^{2}}x^{2}$

and so, $ \textbf I_{2k}\;\leq\;\int_{0}^{\frac{\pi}{2}}\;x^{2}\cdot\left(1-\frac{4}{\pi^{2}}x^{2}\right)^{k}\;\textbf dx\;$

hence, one can easily see that $ \lim_{k\to\infty}\;\left[\frac{2\cdot 4\cdot 6\cdots (2k)}{1\cdot 3\cdot 5\cdots (2k-1)}\;\textbf I_{2k}\right]\;=\;0$

therefore,
$ \lim_{k\to\infty}\;\left[\frac{\pi^{3}}{24}\;-\;\left(\frac{\pi}{4}\right)\;\sum_{j=1}^{k}\;\frac{1}{j^{2}}\right]\;=\;0$

i.e. $ \sum_{j=1}^\infty\;\frac{1}{j^{2}}\;=\;\boxed{\frac{\pi^{2}}{6}}$

here is another one, sorry that i had a hard time recalling it
(i actually learnt it a long time ago in high school) :|

we know,
$ \frac{\pi}{4}\;=\;1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\frac{1}{11}+\cdots$

rewrite it as an infinite product of geometric series of the reciprocal of primes
([i]excluding [/i]$ 2$) (sign adopted is negative or positive according to being
$ -1$ or $ +1$ when doing a mod $ 4$ division (of the prime), see boxes below):
$ \frac{\pi}{4}\;=\;\left(1\boxed{-}\frac{1}{3}+\frac{1}{9}-\cdots\right)\;\left(1\boxed{+}\frac{1}{5}+\frac{1}{25}-\cdots\right)\;\left(1\boxed{-}\frac{1}{7}+\frac{1}{49}-\cdots\right)\;\left(1\boxed{-}\frac{1}{11}+\frac{1}{121}-\cdots\right)\;\cdots$

$ =\;\left(\frac{1}{1+\frac{1}{3}}\right)\;\left(\frac{1}{1-\frac{1}{5}}\right)\;\left(\frac{1}{1+\frac{1}{7}}\right)\;\left(\frac{1}{1+\frac{1}{11}}\right)\;\cdots\;\;\;\quad\;\;\;\quad\;\;\;\quad\cdots\;\;\;(1)$

now take the infinite product where the signs are [i]complementary[/i] to those in $ (1)$, i.e.
$ \left(\frac{1}{1-\frac{1}{3}}\right)\;\left(\frac{1}{1+\frac{1}{5}}\right)\;\left(\frac{1}{1-\frac{1}{7}}\right)\;\left(\frac{1}{1-\frac{1}{11}}\right)\;\cdots$

expand this and you get
$ 1+\frac{1}{3}-\frac{1}{5}+\frac{1}{7}+\frac{1}{9}+\frac{1}{11}-\cdots$
whose closed sum you can easily evaluate to $ \frac{\pi}{2}$

and so,
$ \frac{\pi}{2}\;=\;\left(\frac{1}{1-\frac{1}{3}}\right)\;\left(\frac{1}{1+\frac{1}{5}}\right)\;\left(\frac{1}{1-\frac{1}{7}}\right)\;\left(\frac{1}{1-\frac{1}{11}}\right)\;\cdots\;\;\;\quad\;\;\;\quad\;\;\;\quad\cdots\;\;\;(2)$

multiply $ (1)$ and $ (2)$ to get
$ \frac{\pi^{2}}{8}\;=\;\left(\frac{1}{1-\frac{1}{3^{2}}}\right)\;\left(\frac{1}{1-\frac{1}{5^{2}}}\right)\;\left(\frac{1}{1-\frac{1}{7^{2}}}\right)\;\left(\frac{1}{1-\frac{1}{11^{2}}}\right)\;\cdots$

but $ \text{RHS}$ essentially expands to
$ 1+\frac{1}{3^{2}}+\frac{1}{5^{2}}+\frac{1}{7^{2}}+\frac{1}{9^{2}}+\frac{1}{11^{2}}+\cdots\;\;\;\quad\Rightarrow\;\;\;\quad\frac{3}{4}\;\zeta{(2)}\;=\;\frac{\pi^{2}}{8}$

and so $ \zeta{(2)}\;=\;\frac{4}{3}\left( \frac{\pi^{2}}{8}\right)\;=\;\boxed{\frac{\pi^{2}}{6}}$

there is another rather nice proof that i actually learnt from prof. knuth,
it uses elementary geometry. but instead of posting the answer right away,
let me make it fun, i will provide a hint and you can then try and work out
the puzzle, it is pretty neat once you see how to arrive at the answer! :D

looking at the series $ \zeta{(2)}\;=\;1+\frac{1}{2^{2}}+\frac{1}{3^{2}}+\frac{1}{4^{2}}+\frac{1}{5^{2}}+\frac{1}{6^{2}}+\cdots$,
you can think of the sum as essentially the sum of the [i]area[/i](s) of
[b]squares [/b] $ ( \;\;\blacksquare\;\; )$ whose sides are $ 1,\;\frac{1}{2},\;\frac{1}{3},\;\frac{1}{4},\;\frac{1}{5},\;\frac{1}{6},\cdots\;\;\;\text{etc.}$

and so, the puzzle lies in arranging all of these squares in such a fashion
that they nicely fill up a [b]rectangle[/b] whose sides are $ \frac{\pi}{2}$ and $ \frac{\pi}{3}$;
the area of the rectangle then gives us the answer we need.

[quote="perfect_radio"]Isn't this just proof 11 from Chapman's compilation?[/quote]

oh !  :(  you are right, [i]sorry[/i] i didn't notice...  :huh:

this one is for sylow_theory since he (or she) seems to like complex variables
a lot ! :D

using a [b]square [/b]contour with $ \boxed{\pi\cot \pi z\cdot \textbf f(z)}$ to generally determine
$ \sum_{n=-\infty}^\infty\;\textbf f(n)$, to me, requires a fair decent amount of work, especially
w.r.t. step $ (7)$ in your list. as such, i find it more convenient to use
a [b]circular [/b]contour instead with $ \boxed{\frac{1}{e^{z}-1}\cdot \textbf f(z)}$, because you can
arrive at the required sum [i]faster[/i]. i looove circles, but i hate squares,
and i hate triangles even more... LOL !  :rotfl:

here are the steps, much like yours:

let $ \textbf C_{N}$ be the circle $ \left\{z: \;|z|=(2N+1)\pi\right\}$ with [i]counterclockwise [/i]
orientation, where $ N\in\mathbb N$.

let $ \textbf h(z)\;=\;\frac{1}{e^{z}-1}\cdot \textbf f(z)$, with $ \textbf f(z)\;=\;\frac{1}{z^{2}}$.

we can see
$ \textbf h(z)$ has $ \text{simple}$ poles at $ 2\pi k\imath$ [ where $ \left\{k\in-\mathbb N,\;\mathbb N\right\},\;\;\;\text{i.e.}\;\;\;2\pi k\imath$
lies inside $ \textbf C_{N}\;\;\;\Rightarrow\;\;\;-N\leq k\leq N$ ] and a pole of $ \text{3rd}$ order at $ z=0$.
[ all of the poles lie on the $ \text{Im}$ axis as opposed to the $ \text{Re}$ axis
as is the case when using $ \pi\cot \pi z\cdot \textbf f(z)$ ]

now, by cauchy's residue theorem, we have
$ \oint_{C_{N}}\;\textbf h(z)\;\textbf dz\;=\;2\pi\imath\;\left\{\left[\sum_{k=-N}^{N}\;\text{Res}_{z=2\pi k\imath}\;\textbf h(z)\right]\;+\;\text{Res}_{z=0}\;\textbf h(z)\right\}\;\;\;\quad \cdots (1)$

let us find $ \text{Res}_{z=0}$ first.
using long division, the laurent series for $ \frac{1}{e^{z}-1}$ can be obtained as
$ \frac{1}{z}-\frac{1}{2}+\frac{z}{12}-\frac{z^{3}}{720}+\cdots$ ( where $ 0<\;|z|\;<2\pi$ )

and so, $ \textbf h(z)\;=\;\frac{1}{z^{2}\cdot (e^{z}-1)}\;=\;\frac{1}{z^{3}}-\frac{1}{2\;z^{2}}+\frac{1}{12\;z}-\frac{z}{720}+\cdots$

therefore, $ \text{Res}_{z=0}$, given by the co-efficient of $ \frac{1}{z}$, is $ \frac{1}{12}$

now we find $ \text{Res}_{z=2\pi k\imath}$
since the poles there are all [b]simple[/b], we know,
$ \text{Res}_{z=z_{0}}\;\frac{\textbf f(z)}{\textbf g(z)}\;=\;\frac{\textbf f(z)\bigg|_{z_{0}}}{\frac{d}{dz}\;\textbf g(z)\bigg|_{z_{0}}}$,
where $ \textbf f(z)$ and $ \textbf g(z)$ are defined as both [i]analytic [/i]at $ z=z_{0}$, and
$ \textbf f(z)\bigg|_{z_{0}}\neq 0,\;\;\;\textbf g(z)\bigg|_{z=0}=0,\;\;\;\text{and}\;\;\;\frac{d}{dz}\;\textbf g(z)\bigg|_{z_{0}}\neq 0$

and so, $ \text{Res}_{z=2\pi k\imath}\;=\;\frac{\frac{1}{z^{2}}\bigg|_{z=2\pi k\imath}}{\frac{d}{dz}\;(e^{z}-1)\bigg|_{z=2\pi k\imath}}\;=\;-\frac{1}{4\pi^{2}\;k^{2}}$

plugging this in equation $ (1)$ above, we get
$ \oint_{C_{N}}\;\textbf h(z)\;\textbf dz\;=\;2\pi\imath\;\left\{\sum_{k=-N}^{N}\;\left[-\frac{1}{4\pi^{2}\;k^{2}}\right]\;+\;\frac{1}{12}\right\}$
$ =\;2\pi\imath\;\left\{ \frac{1}{12}\;-\;\frac{1}{2\pi^{2}}\sum_{k=1}^{N}\;\frac{1}{k^{2}}\right\}\;\;\;\quad \cdots (2)$

now using the estimation lemma,
$ \bigg|\;\oint_{C_{N}}\;\textbf h(z)\;\textbf dz\;\bigg|\;\leq\;\text{M}\cdot \text{L}\;=\;\left\{\;\;\textbf{max}_{\;(|z|=R)\;}\;\bigg| \frac{1}{z^{2}\cdot (e^{z}-1)}\bigg|\;\;\right\}\cdot 2\pi R$
( where $ R\;=\;(2N+1)\pi$ )

therefore, we can see, $ \lim_{N\to\infty}\;\oint_{C_{N}}\;\textbf h(z)\;\textbf dz\;=\;0$

and so, from the equation in $ (2)$, $ \lim_{N\to\infty}\;\;2\pi\imath\;\left\{ \frac{1}{12}\;-\;\frac{1}{2\pi^{2}}\sum_{k=1}^{N}\;\frac{1}{k^{2}}\right\}\;=\;0$

$ \Rightarrow \;\;\;\sum_{k=1}^\infty\;\frac{1}{k^{2}}\;=\;\boxed{\frac{\pi^{2}}{6}}$

( and so, as you can see, proving step $ 7$ is rather quick here...  :| )

$ [\;\textbf{Q.E.D.}\;]$

[quote="kunny"]misan, have you seen my post above? :wink:[/quote]

yes, pretty cool.  :cool:

and also nicely done by perfect_radio !  :coolspeak:

( it's funny... the time between Q presented and A posted is [i]almost [/i]6 months... LOL !  :rotfl: )

[quote="misan"]

( it's funny... the time between Q presented and A posted is [i]almost [/i]6 months... LOL !  :rotfl: )[/quote]

What do you mean?  :ninja:  6 months!? :maybe:

here is another proof i just came up with... LOL !  :D it is pretty trivial...

we look at the series:
$ \textbf C\;=\;\cos x-\frac{1}{3}\cos 3x+\frac{1}{5}\cos 5x-\frac{1}{7}\cos 7x+\cdots+-\cdots$

similarly, let $ \textbf S\;=\;\sin x-\frac{1}{3}\sin 3x+\frac{1}{5}\sin 5x-\frac{1}{7}\sin 7x+\cdots+-\cdots$

and so, $ \textbf C+\imath\cdot\textbf S\;=\;e^{\imath x}-\frac{1}{3}e^{3\imath x}+\frac{1}{5}e^{5\imath x}-\frac{1}{7}e^{7\imath x}+\cdots+-\cdots$
$ =\;\tan^{-1}(e^{\imath x})$
( since $ \tan^{-1}y\;=\;y-\frac{1}{3}y^{3}+\frac{1}{5}y^{5}-\frac{1}{7}y^{7}+\cdots+-\cdots$ )

likewise, $ \textbf C-\imath\cdot\textbf S\;=\;e^{-\imath x}-\frac{1}{3}e^{-3\imath x}+\frac{1}{5}e^{-5\imath x}-\frac{1}{7}e^{-7\imath x}+\cdots+-\cdots$
$ =\;\tan^{-1}(e^{-\imath x})$

adding the two gives $ 2\textbf C\;=\;\tan^{-1}\left(\frac{e^{\imath x}+e^{-\imath x}}{1-e^{\imath x}\cdot e^{-\imath x}}\right)\;=\;\tan^{-1}\infty\;=\;\frac{\pi}{2}$
[ since $ \tan^{-1}A+\tan^{-1}B\;=\;\tan^{-1}\left(\frac{A+B}{1-AB}\right)$ ]

and so, $ \textbf C\;=\;\frac{\pi}{4}$
$ \Rightarrow\;\;\;\cos x-\frac{1}{3}\cos 3x+\frac{1}{5}\cos 5x-\frac{1}{7}\cos 7x+\cdots+-\cdots\;=\;\frac{\pi}{4}$

now integrate both sides w.r.t. $ x$ between $ 0$ and $ \frac{\pi}{2}$ to get
$ \sin x-\frac{1}{3^{2}}\sin 3x+\frac{1}{5^{2}}\sin 5x-\frac{1}{7^{2}}\sin 7x+\cdots+-\cdots\bigg|_{0}^{\frac{\pi}{2}}\;=\;\frac{\pi}{4}\cdot\frac{\pi}{2}\;=\;\frac{\pi^{2}}{8}$

$ \Rightarrow\;\;\;1+\frac{1}{3^{2}}+\frac{1}{5^{2}}+\frac{1}{7^{2}}+\cdots+\cdots\;=\;\frac{\pi^{2}}{8}$
$ \text{i.e.}\;\;\;\quad 1+\frac{1}{2^{2}}+\frac{1}{3^{2}}+\frac{1}{4^{2}}+\cdots+\cdots\;=\;\boxed{\frac{\pi^{2}}{6}}$


analytic geometry
combinatorics proposed
combinatorics

There are $ m \plus{} 1$ horizontal lines and $ m$ vertical lines on the plane so that $ m(m \plus{} 1)$ intersections are made.
A mark is placed at one of the $ m$ points of the lowest horizontal line.
2 players play the game of the following rules on this lines and points.

1. Each player moves a mark from a point to a point along the lines in turns.
2. The segment is erased after a mark moved along it.
3. When a player cannot make a move, then he loses.

Prove that the lead always wins the game.

PS I haven't found a student who solved it. There can be no one.

Working on it. Gotten some win cases and nothing really substantial.

Definitely would not have solved this in 3 hours.

[hide]
Let the points be (1,1) to (m,m+1). Cases:

1. m is even. For all k and i, pair up (2k, i) with (2k-1,i). Then all the points will be in exactly one pair. Whenever the second player moves the mark, the 1st player will move it to the corresponding mark. (Count the start place as the 2nd player's first move.) Then, the 1st player will win.

2. m is odd. Proceed similarly.[/hide]

pythag: Suppose the counter starts at $ (2,1)$. You move to the paired point $ (1,1)$. I move to $ (1,2)$. You move to the paired point $ (2,2)$. I move back to $ (2,1)$ (that segment has not been erased). You can't move to $ (1,1)$; what do you do?

Also, if player 2 moves horizontally and player 1 also moves horizontally, player 2 can win if he uses perfect play.

EDIT: Problem solved. Solution will be posted before two days are up unless I find a flaw.

Sorry, I got sidetracked off of other cool problems. :) Guess my 2 day restriction didn't help much, although it did force me to finish this today instead of reading http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=219017 .

I am without a doubt the worst proof writer on AoPS, so I'm sure there are many flaws. I would highly appreciate it if you would point them out for me.

Note: I filled in an empty grid instead of erasing a filled grid. The problem is identical, except it's easier to picture and work with.

Proof:

Points not on an edge or corner are "interior" points, and edge/corner points are "exterior" points.
Define the "diamond" as the structure that arises with diagonal sides (with breaks if the second player creates them) that touches both the left and right edges.
A "completed" diamond is one in which neither player can continue moving inside of it.  As an inherent part of its structure, the top must be the endpoint and the bottom must be its start.

Assume that player 1 is playing with perfect play. If player 1 moves in a direction, and player 2 continues in that direction, and player 2 had a winning strategy, then player 1 could have moved to player 2's position and become player 2, utilizing that winning strategy. Thus, any player must change courses to win against a player who is using perfect play.

A player will always be able to turn in the interior points if he moves only one unit at a time.
A player can only lose on an exterior point, as each interior point contains an even number of paths leading to and from it.

Player 1's first move will be to move up one unit, making him the vertical player. He will only move one unit at a time, so there will be no problems with player 2 forcing him to become the horizontal player. Player 2 must then move horizontally.

Each turn, if player 1 can force a win, he will do that. Otherwise, he will move one down, staying above the diagonal of the first move (illustrated in diagram 1) if possible, or will move one up if possible. If the other player strays out of the confines of that diagram (illustrated in diagram 2) we will continue on with the new point as our diagonal basis. This strategy will continue until it is no longer possible to continue within the illustrated diamond.

(Lemma 1): Player 1 can avoid moving out of the diamond-shaped structure

The only way player 1 will be forced out is if there is a vertex that already has two paths leading it, and player 2 takes another one of the paths, and the only path left leads out of the diamond. However, the only way this is possible is if the mark has already been outside. As we start on the diagonal, this is impossible.

(Lemma 2): The diamond has a maximum height of the top of the board.

Player 2 can only change the diamond by lengthening the diamond's width at the top or the bottom.

Case 1: width is increased at bottom
The diamond will actually have a smaller vertical height.

Case 2: width is increased at top
Player 2 can only increase the width by moving two or more spaces, creating a lengthened segment. Then, when player 1 moves down one space (it is possible because player 1 is outside the diamond) player 2 will be stuck beneath the segment, and will be unable to force player 1 to go back above it because player 1 will be able to force a win. As there is a limited amount of moves below the lengthened segment and player 1 has an answer for each of them, player 2 will eventually run out of moves and player 1 will win.

(Lemma 3): The diamond, when completed, will have all its sides filled.

Instead of analyzing strategies or the like, we can simply note that each point must have an even number of paths going to and from it. Assume we have a completed diamond that does not have all its sides filled. Thus, at least one of the points on its sides does not have two paths going to and from it, with the exception of the start and the finish. This means an entire side must be missing, which leads to a lowered endpoint inside the diamond, which means the player whose turn it is can continue to move within the diamond. Thus, the diamond is incomplete, and we have a contradiction.

(Lemma 4): when the diamond is completed, player 1 can force a win.

The endpoint is a segment sticking up, so it must be player 2's turn. Player 2 must move right or left. As the diamond encompasses the entire horizontal grid, player 2 cannot end up over an empty space. There are 3 cases for player 2:

Case 1: player 2 ends up over a horizontal segment.
Player 1 has the option of moving down to that segment and forcing player 2 to move vertically. Thus, player 1 has the win.

Case 2: player 2 ends up over an edge of the diamond.
Player 1 can move all the way down to the edge, and player 2 can only move horizontally away from the diamond. The y coordinate of the mark decreases each time this case occurs. As the board size is finite, this must eventually come to a stop.

Case 3: player 2 ends up on the edge of the board.
Player 1 can move all the way down, and player 2 has nowhere to go. Thus, player 1 wins.

QED


Well I'm no counting master, but here's my method:

[hide]
Find the number of integers for which that is NOT true and subtract from ten million. Like so:

9+
(8)(10)+, because the ones can be anything and the tens can be any but 0 or the ones
(8)(9)(10)+ because building from bottom up, the ones can be anything, tens can be any but 1s, 100s can be anything but 0 or tens
(8)(9)(9)(10)+
(8)(9)(9)(9)(10)+
$80(9^4)+80(9^5)$
=5314409

10000000-(5314409)= $4685591$ is my answer. I'm not certain.
[/hide]

Count the ones that don't. Then subtract it from 9,999,999.

[quote="ch1n353ch3s54a1l"][quote="surge"][hide="my answer"]
$9999999-(10\cdot 9^6 + 10 \cdot 9^5... + 10 \cdot 9^0 + 1)=\boxed{4021288}$

First, you can pick any digit for the first one (10) and 9 digits thereafter (hence $10 \cdot 9^6$).

But one has undercounted because 00xxxxx, 000xxxx, etc. is counted as integers that have two numbers next to each other. Therefore, $10 \cdot 9^5 + ...$ needs to be added.

Err... this may be wrong.
[/hide][/quote]

I don't think you can pick any digit for the first one...because then you would be repeating other cases with less digits with zeroes before the number.[/quote]

How so? You can pick a zero for the first one, then it disallows the choice of a zero for the second one (which is incorrect but later added).

[hide]
ch1n353ch3s54a1l is right. I've also tested it via brute force, and the answer is indeed $4619160$
[/hide]

[quote="ch1n353ch3s54a1l"]But..0089325=89325
So then the options for the 7 digit number also include the 5 digit number...oh well.[/quote]

Oh, I see what you mean, now. thanks.

[quote="ch1n353ch3s54a1l"][quote="computer_nuke"]How many positive integers less than ten million have the property that at least two adjacent digits are the same? Such as 6612345 will work.

I can't figure this one out, could someone please guide me and provide a briliant solution.

Thanks a lot.

David Jia/Gainesville FL[/quote]

[hide]I just did $9^7+9^6+...9^1=5380839$. Just count out the possibilities for each digit in the seven different cases...?
then 9,999,999-theanswer=4619160[/hide]

Edit: Oops. Changed...probably still wrong[/quote]

Ok, I see where you are going. so 9^7 is where you have the first digit as anything but zero and then the next digit to the right has 9 choices to choose from and then the next one also has 9 choices and so on. But why do you have 9^6+...9^1 as well? I do not understand this part, if you could clarify. Thanks a lot.

Hey, NVM, I see what you mean now, the 9 to the 6, 9 to the 5, etc are for 6 digit numbers, 5 digit numbers etc. OK, thanks alot everyone for the help. I really appreciate it.


The perimeter of a regular hexagon is 48 inches. What is the number of square inches in the positive difference between the areas of the circumscribed and the inscribed circles of the hexagon? Express your answer in terms of $ \pi$.

The side of hexagon: 8 inches.
The area of the annulus = pick a random tangent line of the inner circle, the tangency point to the larger circle = t, and the area = pi*t^2
Since such a line segment = 4 inches, answer = 16*pi

Connect the radius of the hexagon with two adjacent vertices of the hexagon to to form a equilateral triangle with side lengths of 8. Cut the triangle into two $30-60-90$ degree triangles. Finding the missing length of the altitude, we get $4\sqrt{3}$ inches. Subtracting the the area of the inscribed circle from the circumscribed circle, we get:

$(8)^{2}\pi-(4\sqrt{3})^{2}\pi=\boxed{16\pi}$.

Apparently it says the answer is $\frac{16}{\pi}$


inequalities
algebra
polynomial
inequalities proposed

\[ a,\ b,\ c>0\Longrightarrow a\plus{}b\plus{}c\plus{}\frac{3}{\frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}}\ge4\sqrt[3]{abc}\]

[list]$ a \plus{} b \plus{} c \plus{} \frac3{\frac1a \plus{} \frac1b \plus{} \frac1c}$[/list]
$ \equal{} \ a \plus{} b \plus{} c \plus{} \frac {3abc}{ab \plus{} bc \plus{} ca}$

$ \geqslant\ a \plus{} b \plus{} c \plus{} \frac {9abc}{(a \plus{} b \plus{} c)^2}$

$ \equal{} \ \frac a3 \plus{} \frac b3 \plus{} \frac c3 \plus{} 6\cdot\frac {a \plus{} b \plus{} c}9 \plus{} 3\cdot\frac {3abc}{(a \plus{} b \plus{} c)^2}$

$ \geqslant\ 12\sqrt [12]{\frac {(abc)^4}{9^6}}$

$ \equal{} \ 4\sqrt [3]{abc}$

[quote="Mîndrilă Claudiu"]
\[ a,\ b,\ c > 0\Longrightarrow a \plus{} b \plus{} c \plus{} \frac {3}{\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}}\ge4\sqrt [3]{abc}\]
[/quote]
we will prove inequality with abc=1 :
$ ineq \Leftrightarrow a\plus{}b\plus{}c\plus{}\frac{3}{ab\plus{}bc\plus{}ca}\geq 4$
it's true because :
$ ab\plus{}bc\plus{}ca\leq \frac{1}{3}(a\plus{}b\plus{}c)^{2}$
and
$ 3.\frac{a\plus{}b\plus{}c}{3}\plus{}\frac{9}{(a\plus{}b\plus{}c)^{2}}\geq 4\sqrt[4]{\frac{a\plus{}b\plus{}c}{3}}\geq 4$
Done ! :lol:

The following stronger inequality is true too.
Let $ a,$ $ b$ and $ c$ are positive numbers. Prove that:
\[ a \plus{} b \plus{} c \plus{} \frac {6}{\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}}\ge 5\sqrt [3]{abc}\]
It seems that $ a \plus{} b \plus{} c \plus{} \frac {8}{\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}}\ge \frac {17}{3}\sqrt [3]{abc}$  is also true ( with the same conditions ).

[quote="arqady"]The following stronger inequality is true too.
Let $ a,$ $ b$ and $ c$ are positive numbers. Prove that:
\[ a \plus{} b \plus{} c \plus{} \frac {6}{\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}}\ge 5\sqrt [3]{abc}\]
[/quote]

 Denote p=a+b+c, q=ab+bc+ca , r=abc.

 Show that r=1 and inequality becomes $ p \plus{} \frac {6}{q}\ge 5$

 But , from Schur inequality $ p^3 \plus{} 9\ge 4pq$ so is suffices to prove that $ p \plus{} \frac {24p}{p^3 \plus{} 9}\ge 5$ or

  $ (p \minus{} 3)(p^3 \minus{} 2p^2 \minus{} 6p \plus{} 15)\ge 0$

  wich is true because $ p\ge 3$

It is easy to show that the best fit must belong to the following

$ a\plus{}b\plus{}c \plus{} \frac{9}{\frac{1}{a} \plus{} \frac{1}{b} \plus{} \frac{1}{c} } \ge 6\sqrt[3]{abc}$

Nice proof, Marius! :lol:
[quote="great math"]It is easy to show that the best fit must belong to the following

$ a \plus{} b \plus{} c \plus{} \frac {9}{\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c} } \ge 6\sqrt [3]{abc}$[/quote]
which is wrong. Try $ a\equal{}0.53$ and $ b\equal{}c\equal{}1.35.$ :wink:

[quote="great math"]It is easy to show that the best fit must belong to the following

$ a \plus{} b \plus{} c \plus{} \frac {9}{\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c} } \ge 6\sqrt [3]{abc}$[/quote]

No! If

$ a \plus{} b \plus{} c \plus{} \frac {k}{\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c} } \ge {\frac {9 \plus{} k}{3}}\sqrt [3]{abc},$

then  $ k_{max} \equal{} 8.10864271...,$ ti's  a  root  of the  polynomial $ \minus{} 18225 \minus{} 3402k \plus{} 567k^2 \plus{} 16k^3.$

[quote="Bugi"]How did you get k max?[/quote]
Let $ a\equal{}\frac{1}{b^2}$ and $ c\equal{}b.$ Hence, $ k\leq\frac{(2b\plus{}1)(b^3\plus{}2)}{b^2(b\plus{}2)}.$ :wink:

thanks for pointing out my mistake. By the way, how can you figure out the max of k since I find no relation between your expression with the result of fjwxcsl?

[quote="great math"] I find no relation between your expression with the result of fjwxcsl[/quote]
Because it can be $ k\leq\frac{3(2b\plus{}1)(b^3\plus{}2)}{b^2(b\plus{}2)}$ and $ b_{min}\equal{}1.3614...$
Now are you see?

thanks arquady for your reasonable explanation!

With $ k\equal{}6$, we can solve it by Am-GM. See it here:
http://www.maths.vn/forums/showthread.php?t=9244&page=17
But with $ k\equal{}8$, my way is not effect  :oops:

[quote="Mîndrilă Claudiu"]
\[ a,\ b,\ c > 0\Longrightarrow a \plus{} b \plus{} c \plus{} \frac {3}{\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}}\ge4\sqrt [3]{abc}\]
[/quote]my solution is quite similar to [b]ValentineA[/b]'s one:
$ a \plus{} b \plus{} c \plus{} \frac {3}{\frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}} \equal{} a \plus{} b \plus{} c \plus{} \frac {3abc}{ab \plus{} bc \plus{} ca}\ge a \plus{} b \plus{} c \plus{} \frac {9abc}{(a \plus{} b \plus{} c)^2} \equal{} \frac {a \plus{} b \plus{} c}{3} \plus{} \frac {a \plus{} b \plus{} c}{3} \plus{} \frac {a \plus{} b \plus{} c}{3} \plus{} \frac {9abc}{(a \plus{} b \plus{} c)^2}\ge 4\sqrt [4]{\frac {(a \plus{} b \plus{} c)abc}{3}}\ge4\sqrt [3]{abc}$

[quote="nguoivn"]
But with $ k \equal{} 8$, my way is not effect  :oops:[/quote]
For $ k\equal{}8$ it's true by  MV method. :wink:

Can you post your full proof? Thank you very much, arqady  :)

Let $ a,$ $ b$ and $ c$ are positive numbers. Prove that:
\[ a + b + c + \frac {8}{\frac {1}{a} + \frac {1}{b} + \frac {1}{c}}\ge \frac {17}{3}\sqrt [3]{abc}\]
Let $ abc=1,$ $ a=\min\{a,b,c\}$ and $ f(a,b,c)=a + b + c + \frac {8}{ab+ac+bc}- \frac {17}{3}.$
Hence, $ a\in(0,1]$ and $ f(a,b,c)-f(a,\sqrt{bc},\sqrt{bc})=$
$ =b+c-2\sqrt{bc}-8(\frac{1}{2a\sqrt{bc}+bc}-\frac{1}{ab+ac+bc})=$
$ =(\sqrt b-\sqrt c)^2(1-\frac{8a}{(2a\sqrt{bc}+bc)(ab+ac+bc)})\geq$
$ \geq(\sqrt b-\sqrt c)^2(1-\frac{8a}{(2a\sqrt{bc}+bc)^2})=(\sqrt b-\sqrt c)^2(1-\frac{8a}{(2\sqrt a+\frac{1}{a})^2})=$
$ =\frac{(\sqrt b-\sqrt c)^2(1+4a\sqrt a-4a^3)}{(2a\sqrt a+1)^2})\geq0.$
Id est, it remains to prove that $ f(\frac{1}{b^2},b,b)\geq0.$
But $ f(\frac{1}{b^2},b,b)\geq0\Leftrightarrow(b-1)^2(6b^4-5b^3-16b^2+12b+6)\geq0,$ which is true.


geometry
angle bisector
geometry proposed

in parallegoram ABCD points M and N are on sides AB and BC such that $AM=CN$ .
Prove that $AN\cap BM=P$ is on angle bisector of $\angle ADC$

See

http://www.mathlinks.ro/Forum/viewtopic.php?t=5873
http://www.mathlinks.ro/Forum/viewtopic.php?t=19182

  Darij

[hide="Here is the my solution (using the area) for the proposed problem."]

[b]Notations:[/b] the area $[XYZ\ldots T]$ of the polygon $XYZ\ldots T$ and

the distance $\delta_{d}(X)$ of the point $X$ to the line $d$. Thus,

$AB=CD=a$, $AD=BC=b$, $AM=CN=x$ ($BM=a-x$, $BN=b-x$),

$h_1=\delta _{CD}(A)$, $h_2=\delta_{BC}(A)$, $u=\delta_{AD}(P)$, $v=\delta_{CD}(P)$

($\delta_{AB}(P)=h_1-v$, $\delta_{BC}(P)=h_2-u$).

[b]Solution.[/b]

$2[ABCP]=2([BCM]+[AMP])=(a-x)\cdot h_1+x\cdot (h_1-v)=ah_1-xv.$

$2[ABCP]=2([ABN]+[CNP])=(b-x)\cdot h_2+x\cdot (h_2-u)=bh_2-xu.$

$[ABCD]=ah_1=bh_2\Longrightarrow xu=xv\Longrightarrow u=v\Longrightarrow$

the point $P$ belongs to the bisector of the angle $\widehat {ADC}.$
[/hide]


[quote="sharkman"]If a circle of radius 2 is made to roll along the x-axis, what is the equation for the path of the center of the circle?[/quote]
This is one of those fun interactive problems, create a revolving Paea field on a math program, and apply 2 Pa to the circle. Then label the center as a reference point and check out it's pathway, it's really cool.

Maybe he meant: Given a fixed point on the circumference of the circle, what is the path of that point as the circle rotates.

Hint: parametric equation [use radians]

No, it's not a parametric equation problem.

I will check my math book but I think the answer is y = 2.

I'll go confirm.

Alright then. Your problem was easier than I thought. You should still consider my problem; it is a classic result.

The parametric equation for the cycloid is:

$x = a(t-\sin(t))\\ y = a(1-\cos(t))$

Where $a$ is a constant. It describes the path of a particle that lies on the outer edge of a rolling circle [that, presumably, would be moving in one direction].

By the way, why is there a space in front of my x? Can't seem to get rid of it. :D

The " // " puts you in paragraph mode. Just do

[code]

$[equation 1]$
$[equation 2]$

[/code]


geometry
conics
ellipse
symmetry
trigonometry
real analysis
real analysis unsolved

maximum area of an isosceles triangle inscribed in the ellipse

x^2/4 + y^2 = 1

i am kind of lost on the geometry for this problem, can someone give me a hint?

By the symmetry of the figure, you can set the vertices of the triangle as $A(0,1),\ B(-2\cos \theta,\ \sin \theta),\ C(2\cos \theta,\sin \theta).$


function
complex analysis
geometry
3D geometry
sphere
complex analysis unsolved

Suppose $ f$ is a bounded entire function on $ \mathbb{C}$.  Since all holomorphic functions on the compact Riemann surface $ C_{\infty}$ (the Riemann sphere) are constant, can we find an extension of $ f$ on $ C_{\infty}$ to prove Liouville's Theorem?  If so, would it make sense to use the stereographic projection from the Riemann sphere to the complex plane?

To find that extension, you'll have to prove that the singularity at $ \infty$ is actually removable. Stereographic projection won't get you that argument, but looking at $ f\left(\frac1z\right)$ will.


group theory
abstract algebra
superior algebra
superior algebra solved

Find a simple proof for the following theorem:

Let $x\in A=\{1,2,\cdots\,n\}$ and $S_{x}=\{\sigma\in S_{n}: \sigma(x)=x\}.$ Prove that for all $x,y\in A,$ $S_{x}$ and $S_{y}$ conjugate in $S_{n}.$ Conclude that $\bigcap_{g \in S_{n}}{g^{-1}S_{x}g}= 1.$

Take $x,y \in A$. By definition of a permutation, $\exists \psi \in S_{n}, \psi(y)=x$. But then $\sigma(x)=x \Leftrightarrow \psi^{-1}o \sigma o \psi (y)=y$. This means that $S_{y}=\psi^{-1}S_{x}\psi$.

As for the second question, you should clarify what you mean by $G$, and what this sums actually means ...

The second formula seems to be corrected ;).

Every element $\sigma$ of the intersection lies in every $S_{y}$ (since $S_{y}$ is conjugate to $S_{x}$), so that $\sigma(y) = y$. This holds for every $y$, thus $\sigma = id$.

Another solution:

The following thorem concludes the problem immediately:
If $G$ acts on $X$ transitively then stabilizers $G_{x}$ and $G_{y}$ are conjugate in $G$ for all $x,y\in G.$

Thanks.


topology
logarithms
function
real analysis
real analysis unsolved

How to show that the $\ln 2/\ln 3$ dimensional Hausdorff measure of the Cantor $1/3$-function is finite?

Let $s=\log 2/\log 3$. The Cantor set is covered by the $n$th generation pre-Cantor set, which is the union of $2^{n}$ intervals of length $3^{-n}$. Since $2^{n}(3^{-n})^{s}=1$, we have $\mathcal{H}^{s}(C)\le 1$.

Actually, $\mathcal{H}^{s}(C)=1$, but this is harder to prove.


Here's another question I couldn't get:

Two farmers agree that pigs are worth $300 an that goats are worth $210.  When one farmer owes the other money, he pays the debt in pigs or goats, with "change" received in the form of goats or pigs as necessary.  (For example, a $390 debt could be paid with two pigs, with one goat received in change.)  What is the amount of the smallest positive debt that can be resolved in this way?  (source: AMC 10A 2006)

Thanks

[hide] We are looking for the smallest amount of positive debt that can be paid off. We can thus throw out the LCM as the easy way to the solution. Clearly, the debt number must end in $ 0$, which implies divisibility by $ 10$. Now, we realize that since both are multiples of $ 3$ they can never be separated by any integer but a multiple of $ 3$. The smallest positive integer multiple of $ 3$ is $ 3$. Combining what we found earlier, we find that the smallest positive debt that can be paid off is $ \boxed{\$30}$. (This can be done by paying $ 3$ goats and receiving $ 2$ pigs in change) [/hide]

You posted in two threads, but anyways, here.

[hide]It would be 30 dollars, because a farmer could pay 3 goats, which are$ 210\cdot3 \equal{} 630$, while the change can be payed with two pigs, which are $ 300\cdot2 \equal{} 600.$

Therefore, $ 630 \minus{} 600 \equal{} 30$.[/hide]


$ 16)$ $ x$ is directly proportional to the sum of the squares of $ y$ and $ z$ and inversely proportional to $ y$ and the square of $ z$. If $ x\equal{}8$ when $ y\equal{}1/2$ and $ z\equal{}\sqrt{3}/2$, find $ y$ when $ x\equal{}1$ and $ z\equal{}6$.

Now, what I remembered from general chemistry (I didn't understand this proportion from AoPS 1 at all when I tried to learn it.) is that in a direct proportion, a fraction x/y is equal to a constant. In an inverse proportion, the product of x and y is a constant. However, I'm not sure what I'm doing wrong, or what method I use that doesn't work.

Anyways, for the first part, x is directly proportional to the sum of the squares of y and z:

$ \frac{8}{y^2\plus{}z^2}\equal{}8$

$ \frac{x}{y^2\plus{}6^2}\equal{}8$ and yeah... the answer for y didn't look right.

For the 2nd part, I wasn't sure what the inversely proportional to y and the square of z meant. Does that mean x is inversely proportional to the sum of y and the square of z? Hmm...

Well, if someone could help that would be awesome!

I think it just meant that both $ xy \equal{} k_1$, and $ xz^2 \equal{} k_2$, ie proportional to both.  :)  That should give a nice answer.

So if x is directly proportional to the sum of y and z, which gives a constant, and also inversely proportional to z, then it could be written all together to equal one constant?

Also, what's the difference between varying jointly and varying directly?

1) No; one is a binomial, while the others are monomials. Plus, why would the constants be equal?

2) Do you mean inversely in directly? If not, then what does that have to do with the problem? Else, [[proportion]].

By the way, I added your problem as a problem there.

It seems that the given condition is too much, which makes a independent system for there're as many equations as variables. So the answers contradict each other...


1)
Let n be a positive integer and m be the number of factors of n (not including 1 or n). For what value of n is n/m the smallest?

2)
Set A = {1,2,3,4,5,2,4,6,8,10,3,6,9,12,15,4,8,12,16,20,5,10,15,20,25}.
Set B has 625 terms consisting of the products of each term in set A with each term in set A (or set A squared maybe.)
What is the average of all terms in set B?

EDIT: A variation on 2. If set B was only 300 terms, with the product of every distinct pair of terms from A, what would be the average then?

3)
Let x and y be integers, x is negative and y is positive. Then f(x,y) is the sum of all odd integers between x and y (inclusive if x or y are also odd.)
What is f(f(-2^5,5^2),f(-4^3,3^4))?

3.
[hide]
so u have you

$f(f(-32,25),f(-64,81))$
u know for
$f(-32,25)$ it is $-31 + -29 ... + 25$
u notice that the -25 and 25 cross out, same with the -23 and 23 so u only have to deal with -31, -29, and -27
same for the $f(-64,81)$
so now u have $f(-87, 657)$
that means all the odds from -87 to 87 cross out
so you have $89+91+93...657$
use sum of arithmetic sequence
\[ 285(89+657)(\displaystyle\frac {1}{2}) = 106305\]
[/hide]
i dunno if thats right but yea

no 1.
[hide][hide]even possible? there are infinitely many abundant numbers and all their sums go higher and higher above the number[/hide][/hide]

It's not asking for sums. It's asking for m/n, and there clearly is a smallest, though there's isn't a largest.

1
[hide]I get 4, or any multiple of 2 not counting 2 because then there would be zero factors not counting 2 and 1. I don't think a number can ever have more factors than the product of its factors... so the minimum value of $\frac nm$ must be 4. I'm not that sure though.[/hide]

I think it's 6 or 12, with m/n is 3.

NVM on what I said previously - the person above me changed his response.


I m thinking of starting an AoPS-Mathlinks AIM(AOL Instant Messenger) ring.  Basically what I am thinking of is just a list of all the members on this site who have AIM sn's.  The purpose would be that if someone needed quick help on a problem or they felt shy about posting it here etc.  they could just IM someone for help.  There can also be many other purposes (i.e. meeting new ppl, relieving boredom etc.) for this.  So whoever wants to be a part of it can please pm me or post here.

Here is my list so far.  I just have the ppl who have participated in mathchats(if someone for some reason doesn't want to be on this list and they are on it tell me and I will take you off).

[u][b][size=150]AoPS-Mathlinks AIM Directory[/size][/b][/u]

[u]AoPS-Mathlinks Screen Name[/u]....[u]AIM Screen Name[/u]

joml88.......................................BaseballJoe2002

white_horse_king88.................whitehorseking88

theone853................................reaperofthegrave

SilverFalcon...............................jwec05

BHorseMath...............................B Horse Math

beta...........................................Feiqi J1989

yifman12....................................caocoprez

confuted....................................confuted7

h_s_potter2002.........................hspotter2002

SingtheSorrow...........................KillinginSleep

lonelytraveler8...........................lonelytraveler8

Rep123Max................................MaxHarris08

MathFiend..................................NegativeEight04 or MathFiend21

tetrahedr0n...............................tetrahedr0n

js1527.......................................js1527k

Mathfanatic................................VandermondeNT

Magnara.....................................funkyman1887

JSRosen3....................................JSRosen3

Nukular.......................................FantasyChrono

I don't think AIM Screennames are case sensitive.

Mine is without the _'s.  It wouldn't let me do that when I signed up.

You should be aware that in Europe / Asia AIM is not as popular as it might be in US. In fact I have no non-american contact on AIM. Rather MSN / Yahoo MSG are more used IM clients.

Okay, if someone's interested, I'm [b]isotomion[/b] on [b]Yahoo Messenger[/b]. But I have removed this stuff from my autostart directory, hence if someone would like to chat with me - please send me an email or a PM via MathLinks.

  Darij

If you're going to have msn's and yahoo's, I'm h_s_potter2002 for yahoo and h_s_potter2002@yahoo.com for msn.

yeah why am I not on that list I was definitely in a lot of the mathchats.

My AIM is the same as my AoPS username..."JSRosen3".

What is your sn mystic?  For everyone else that I have left out or for those who want to let ppl know there sn please post here.

New screenname for me: VandermondeNT.  (I considered 'Schur Thing' for a moment.)

[quote="tetrahedr0n"]lol, i used to have a VandermondeCI sn. lol[/quote]

Interesting.

I'd like to make it clear that the SN I use in mathchats is not one you should try and contact me at. (Joml88, i don't want mine posted)


I don't know if this problem was posted before but:

$\text{Evaluate}$

$1000...0001*999...999-199...999$

Where in the first number, there are 2005 0's
The second number, there are 2006 9's
The third number, there are 2006 9's.


There are three six-letter words where $i^{th}$ letter is later in the alphabet than the $(i-1)^{th}$ letter.  Find them all.

[hide="Answers - Try first"]
ALMOST
BEGINS
CHINTZ
[/hide]

If you can still do it before the editing deadline,

EDIT your post with parentheses so it reads as "(i - 1)th" in this
way, or in your original Latex way.  Anyway, they're needed to make sense.


$ a,b,c>0$ and $ \frac{1}{a\plus{}4}\plus{}\frac{1}{b\plus{}4}\plus{}\frac{1}{c\plus{}4}>\equal{}1$ prove that $ 2(ab\plus{}ac\plus{}bc)<\equal{}a\plus{}b\plus{}c$

Fek konam soorate namosavio eshtebah neveshti ... chon ini ke neveshti asan emkan nadare :
$ a\plus{}4>4$ pas $ \frac{1}{a\plus{}4} < \frac{1}{4}$ . pas jame oon seta hadeaksar mishe $ \frac{3}{4}$ ke az $ 1$ kamtare .

Agha kamyar in bazam moshkel dare ... mesal naghzesh $ a\equal{}b\equal{}c\equal{}\minus{}1$ hast .

are dige az n be bad behtare ar vaght post mizaram khodam kamel check konam va be digaran kamtar etminan konam!! soal kamel ghalate vali bejash ye soale ghashang va ziba ro mizaram. :blush: