Are there any vegetarians on here? If so, why are you a vegetarian. If not, why not? My little brother is, for no reason at all. Every few weeks, he makes up a new reason, which I prove to be ridiculous :rotfl: . They range from being "I won't eat cute animals" to "I won't eat ugly animals" to "Because". For people that have a good reason, such as "I like animals", I respect. The problem is, my brother doesn't like animals :rotfl: . So anyway, how do you feel about vegetarianism.

Animals have no souls.

People who don't eat meat will grow little you know what's all over their body and eventually turn into a giant one.

I'm vegetarian becuase I am against the harsh conditions animals are put into to give someone a big belly.

You can get your nutrients from dead plants or by causing a tree a twinge of pain rather than killing a living being. Always do what causes the least harm. Of course, if I was stranded on an island with no edible food, I would eat fish.

Also, in scientific studies, meat-eaters are shown to have a greater rate of memory loss. Not sure if this study was looking at correlation or causation, but yea...

I have seen a study saying that people that eat lots of vegetables have better memory. I think that it might not be that eating meat causes memory loss, as eating vegetables causes a gain of memory.

[quote="neelnal"]I'm vegetarian becuase I am against the harsh conditions animals are put into to give someone a big belly.
[/quote]
sorry, but IMO, that's just stupid.
just because YOU don't eat them, doesn't mean that the dead animals will suddenly revive and live happily ever after. Also, what does you not eating animals have to do with animals put in harsh conditions and getting killed to give ppl food? They're not going to kill less animals for food, just because some people don't eat them.
besides, they taste good.

You could be partially vegetarian also. I don't, for instance, eat at McDonalds. When I eat cow, I eat cow that has been humanly treated, on my uncle's farm. (Which disproves my brother's 3rd reason for being vegetarian. Him "I don't eat cows because McDonalds chops down rainforests for cows" Me "We eat cows from Uncle Tom's farm"

[quote="Ignite168"]Animals have no souls.[/quote]
It's okay to be of the opinion that animals don't have souls, but don't state it as fact. I feel that whether animals have souls is a meaningless question. What matters to me is that they can feel pain, so I don't think it's right to torture them.
"The question is not, 'Can they reason?' nor 'Can they talk?' but 'Can they suffer?'" -- Bentham

[quote="junggi"]They're not going to kill less animals for food, just because some people don't eat them.[/quote]
Actually, they are. It's the law of supply and demand.

[quote="Ignite168"]People who don't eat meat will grow little you know what's all over their body and eventually turn into a giant one.[/quote]
I've heard of many studies showing that vegetarianism is healthy. In fact, based on our biology, we're supposed to be herbivores, like most other primates.

[quote="jmadsen"][quote="junggi"]They're not going to kill less animals for food, just because some people don't eat them.[/quote]
Actually, they are. It's the law of supply and demand.
[/quote]
are you telling me that they will let go the animals they have, just because a few people refuse to eat animals?

No, but they will breed fewer animals in the future.

[quote="jmadsen"][quote="Ignite168"]Animals have no souls.[/quote]
It's okay to be of the opinion that animals don't have souls, but don't state it as fact. I feel that whether animals have souls is a meaningless question. What matters to me is that they can feel pain, so I don't think it's right to torture them.
"The question is not, 'Can they reason?' nor 'Can they talk?' but 'Can they suffer?'" -- Bentham
[/quote]

If they don't have souls, they can't interpret pain as anything- which would mean they don't suffer.  :maybe:
This point may be seen as just plain stupid but, How do you know that animals don't find what we think of as pain (in terms of nervous response and all that) enjoyable under some circumstances? Have you ever asked one? We can't make presumptions about what it feels like to be kept as a battery hen or slaughtered in an abbatoir, even in the (IMO, very unlikely) event that animals do have souls and have a way of feeling.

One final point that may be laughed off, but I think it relevant nonetheless: Your soul could have resided in any one of trillions of animals in the worlds history, millions of species. By chance, it seems, you are a homo sapiens, the most advanced of all of these millions of species. Does this not suggest that, perhaps, only the most advanced animals (like us and perhaps our most recent ancestors) have souls?

(I expect dozens of reasons why my logic is flawed, but I believe that anyway)

Im a vegetarian for almost 10 years since primary school(now a pre-u student).

I started off by following my parents and then influenced by abit of religious problem(im a buddhism) and move on to health problem (well , meats nowadays are not more safe to be eaten....) and now im used to it . :lol:

But the most important is , I wont be having  dreams like beef,pork or mutton chasing after me for my life when im asleep. :D

i'm a vegetarian.  its only because ppl at home dont eat non veg plus my mom doesnt know to cook animals  :)

[hide] I am quoting "EmoryABrown" here. [/hide]
[hide]COWS RULE: EAT UM UP
 [/hide]
MEAT IS AWESOME! it's not that i dont eat vegetables, but red meat is awesome, i eat beef pork lamb chicken, and a lot of seafood. I cant live without meat.[/hide]

I think as an obligation of our high place on the food chain, we have the right and the OBLIGATION to ourselves to kill all animals around us and eat them. Anyways, why eat plants? Nothing tastes better than a hearty steak along with some veal cutlets and spicy shrimp. The ultimate meal -- leaves and roots just don't compare.

I think that being a vegetarian is a very good thing and simply not something that people should make fun of. Vegetables are the sustenance of life. When I crunch down on juicy celery and succulent asparagus, the exhilirating taste flows down my mouth and travels to all parts of my body. I feel as if I were on fire; it's SOOOO good. I love vegetables and the stamina that they provide me.
When I eat bittersquash, I feel as if I were superman or a superhero, and I generally tend to run around my neighborhood naked like a madman afterwards. I like to take big bites out of the Yggdrasil when I have time, and flowers are always nice too. My parents say that it's strange when I bend down and eat grass growing out of the ground, but I think it's perfectly normal. My stomach is like that of a cow's. Uh so good.

Vegetarianism is very good.

im in scouts, so i've had a lot of raman :P

sign up for the food program in college

at least you KNOW you have a meal

-jorian

I'm not an all out vegetarian, but I have vowed to eat a lot less meat than I used to.  I realized that I don't need to be eating as much meat as my heart desires, it's probably not the healthiest diet for me, and the more animals that can be spared the better.

Being vegetarian in college depends on your college.  I can say from experience that it seems easier to be a vegetarian at MIT (and in the Boston area in general).  There are definitely a lot of people here who are vegetarian.  I'm guessing that at state schools, for example, you can still be a vegetarian, but doing so restricts your options and food quality immensely.

I went to a camp at Northwestern University, and I made a couple of vegetarian friends there that had no trouble eating

[quote="Ignite168"][quote="neelnal"][quote="asdf2345"]I think it kinda depends on how much of a vegetarian you are. Vegetables are good for you and all, but I don't know if a strict vegetarian would be able to get nutrients like protein and such.[/quote]

Beans, eggs, milk, bamboo etc. can give you all of the protein you can want.

I eat two eggs every day along with 4 cups of milk and a serving of beans. Plenty enough.[/quote]

Eggs have so much fat and cholesterol compared to the protein they give you. Meats such as tuna or chicken breast are much better for you.[/quote]

Depends on how the chicken breast is prepared.

And when people say that somethings taste so good and how vegetarians miss out on it is kinda stupid cuz people have different tastes. I know people who enjoy eating airline food okay!!!

[quote="most-wanted"]Does anyone believe that Vegetable are healthier than meats ?
[/quote]

Actually, does anyone believe they aren't? The highest risk to your health by eating only vegetables is the vitamin B12 deficiency I believe. Vegetarians take it with pills or something. In the other hand eating meat without control is associated to heart diseases, gout and some types of cancer.
Ultimately the best thing to do probably is to keep a balanced diet. This you can do vegetarian or not. The proteins arguments is not true. Animals synthesize rpoteins from aminoacids, and aminoacids we need are produced in our body except for the ones called essential. All of those can be found in vegetables. You're right in that vegetarians must worry about this though, like they must eat soy frequently and all that

As omnivores I understand humans eating meat for food. I see no one complaining about lions hunting zebras to eat so I don't see why humanity should stop eating cows. I also don't see how are we going to feed everyone in the world without breeding animals and so. What is wrong is hurting animals unnecesarily, when, at least in some cases like mammals, they obviously suffer. Also I think one should answer first the very related question, should we care about animals in a general sense?

Ursula

^ Largely agree, but....

"As omnivores I understand humans eating meat for food. I see no one complaining about lions hunting zebras to eat so I don't see why humanity should stop eating cows. I also don't see how are we going to feed everyone in the world without breeding animals and so."

I far better argument for vegetarianism is not that KILLING animals is wrong, but CAUSING SUFFERING for them is.  The treatment of animals in the meat industry is radically worse than the existence which they face in "nature".  While the death of an animal, whether it's throat is slit in captvity, or it's neck is bitten and broken in the wild is undoubtably painful, the life of the animal is generally the trully horrific part of mass-produced meat and creates a strong argument for veganism, or at least a  switch to more concientious farming, with less emphasis on mechanical efficiency.  As an example, the crates in which egg chickens are stored so small that these chickens can never, at any point in their life, even stand up, spread its wings, &c.

Additionally, feeding the world would be much easier if vegetarianism was adopted on a large scale.  The amount of energy and land required to grow plant food is dramatically less than for animal food (a number i've heard is 20 times more efficient).  This is why meat is generally more expensive to buy.

[quote="ursula"]As omnivores I understand humans eating meat for food. I see no one complaining about lions hunting zebras to eat so I don't see why humanity should stop eating cows. I also don't see how are we going to feed everyone in the world without breeding animals and so. What is wrong is hurting animals unnecesarily, when, at least in some cases like mammals, they obviously suffer. Also I think one should answer first the very related question, should we care about animals in a general sense?

Ursula[/quote]

[quote="kstan013"]oh my gosh. Do you think tigers feel bad when they eat their predators? We are at the top of the food chain for a reason folks. I would never ever ever give up meat. It tastes way too good[/quote]

The thing I don't like about these arguments are that most of the time, you aren't doing the actual killing.  When a tiger or a lion eat their prey, they have to chase it, and then kill it by themselves.  When humans eat animals, they get meat that's already been killed when farmers kill their animals.

no im not, because:
1. my whole family isn't
2. my mom makes me eat everything
3. meat like beef tastes good :)

[quote="PI"]The human race cannot support everyone on a diet that has no meat. Meat is just a more efficient method of transporting the energy we need. We don't have enough space to farm everything with the current population.[/quote]

I don't know where did you get this but I seriously doubt it. In developed countries there is an overproduction of food, particularly field products. Actually with all the european union common policy and stuff, farmers in some countries have had to sotp farming particular products that are being produced on other countries. And I would say that one way or another vegetables are already the fundamental ingredient of most people diet even those who like meat so much. Think of bread, spaghetti, rice, morning corn flakes... those are largely responsible for your health.
Besides to produce that meat you're talking about the chicken or cow had to be fed for months with some vegetables, of which he used a big part to sustain itself as a living being. In other words, having chicken implies having a whole farming field of of seeds so I don't see how you end up with more space.

[quote="h_s_potter"]The thing I don't like about these arguments are that most of the time, you aren't doing the actual killing. When a tiger or a lion eat their prey, they have to chase it, and then kill it by themselves. When humans eat animals, they get meat that's already been killed when farmers kill their animals.[/quote]

This argument doesn't stand. There's no way we can be all hunting our animals which would be in the other hand disastrous for nature. To maintain a human population this big we must breed our food, be it meat or carrots, not passively take it from the environment. If there were six and a half billion lions in the world they'd be eating themselves. One other thing is that I'm more productive to mankind when I research in physics than when I farm. It started happening a long time ago and it's called division of labour.

[quote="capybaralet"] far better argument for vegetarianism is not that KILLING animals is wrong, but CAUSING SUFFERING for them is...[/quote]

Of course I agree  with this. However, I'd say better, in a practical sense, to switch to concientious farming than everyone going vegan. What one should do in my opinion is start buying products which one knows, because the company certifies it, are being produced without caussing animals to suffer.
I know that if you asked why I want to do this, probably the answer would reduce to a matter of how we would liked the world to be. There's no actual neccessity of it in the same sense there's no actual necessity for us that there be whales. However I prefer a world with whales.

Ursula

[quote="ursula"][quote="h_s_potter"]The thing I don't like about these arguments are that most of the time, you aren't doing the actual killing. When a tiger or a lion eat their prey, they have to chase it, and then kill it by themselves. When humans eat animals, they get meat that's already been killed when farmers kill their animals.[/quote]

This argument doesn't stand. There's no way we can be all hunting our animals which would be in the other hand disastrous for nature. To maintain a human population this big we must breed our food, be it meat or carrots, not passively take it from the environment. If there were six and a half billion lions in the world they'd be eating themselves. One other thing is that I'm more productive to mankind when I research in physics than when I farm. It started happening a long time ago and it's called division of labour.[/quote]

I'm not saying that we should be hunting our food; however, it doesn't really make sense to say that it's okay to eat meat because there are animals out there that also eat other animals because our way of getting meat is a lot different from how a lion or a tiger gets their food.

[quote="PenguinIntegral"]The human race cannot support everyone on a diet that has no meat. Meat is just a more efficient method of transporting the energy we need. We don't have enough space to farm everything with the current population.[/quote]

But all those animals need food too, which takes up space.  (I dunno if I'm right about this), And also, only 10% of the energy that animals get from eating their food actually comes to us.  So shouldn't it take up less space if no one eats meat?

And just to let people know, I'm against forcing people to be vegetarian , or pressuring them to become one.

Let me go find the cite for that statement. In the meanwhile, could someone enlighten me on the statistics on mat in the diet? I was pretty sure that some meat was good for you, or you would be better off eating meat. Most Americans get too much though...

See here: http://www.second-opinions.co.uk/vegetarian.html

The gist is that most land can support grass for grazing but not crops for direct human consumption. Hence, to make efficient use of the land for the overpopulated earth, eat some meat!

[quote]
I'm not saying that we should be hunting our food; however, it doesn't really make sense to say that it's okay to eat meat because there are animals out there that also eat other animals because our way of getting meat is a lot different from how a lion or a tiger gets their food.[/quote]

Fortunately. If we were to get our food like lions and tiger do, soon there will be no lions or tigers or cows or deers. The percent of animals that reach adulthood in the wild without humans hunting them is already superlow. What we do when we put animals in farms is to guarantee an extremely high chance for their survival, (we even help them have their babies!) so the population growth matches our needs for food. That is the main advantage of farming. We also do not tend in general(i know the exceptions) to kill for instance young animals which lions and tigers prefer a lot and affects the survival of the species. In short, the only thing I see we can do to feed mankind with meat is farming, which by tthe way is exactly what we do with vegetables. What is to be argued is how we should do this farming.

Ursula

IF one assumes that it is right to decide what food to eat based on a desire to "help" animals, it isn't clear at all that an animal would prefer to not exist at all rather than live for a while and be killed to provide food.

How many of you would prefer not to exist rather than living until adulthood and then dying?  I suggest that most animals that die for food would not exist at all if there were not the food market to provide the money to pay for someone to provide the animals the space, food and protection to live.  (I don't eat road kill or non-domesticated derived meat.)

A good reason for preferring not to eat meat, that hasn't been given is that the burden on the planet is less.  It takes many times the calories meat provides in food to produce the meat.

Another good reason would be to avoid the various diseases that are associated with some kinds of meat.  However, as suggested in this thread, it appears that we were designed to eat plants and meat.  It is unusual for a person to only eat plants and select a diet that is not lacking in something needed.

Let's think a bit more critically before jumping to what would be viewed the PC position.

Those of you that said that you are following your parents, that's a fine, defensible reason.  That's also probably the real reason most of us meat eaters have the diet we do also.


A circular pizza with a radius of 6 inches is cut along radii into
three wedge-shaped slices. The measures of two of the central
angles are 80 degrees and 130 degrees. What is the number of
square inches in the area of the largest slice? Express your
answer in terms of $ \pi$.

The measure of the third central angle is $ 360 \minus{} 80 \minus{} 130 \equal{} 150^\circ$. Thus the area is $ \pi(6)^2\times{\frac{150^\circ}{360^\circ}} \equal{} \boxed{\frac {45\pi}{2}}$.

guass you got it wrong...

150/360 reduced is 5/12 and

36pi*(5/12) is 15pi.
thus our answer is

[size=134][b]15pi[/b][/size]

the answer is [b]not[/b] 45pi/2

 :lol:

Oh. I accidentally thought that $ \frac{150}{360}\equal{}\frac{5}{8}$ :blush: . You're right, it is $ 15\pi$.

Thanks.

But the problem is that the problem doesn't say how to input $\pi$. I got $15\pi$, but didn't know how to type it (I finally looked up how the LaTeX works.


Please vote for an ARML shirt if you are going.
The one I created is MRDraft, and ccy created the other one.
You should vote for my shirt (please?)  :P , but unfortunately I can't vote for my own design (MRDraft) because I can't make it to the competition this year.

Or vote for Charles. Or neither.
Whatever. They're both good!

I'm excited for the shirt either way.

What happened to combining the two, with ccy's design on the front, and your fractal thingy on the back?

Honestly, I see that as the best solution.
(maybe cuz i'm losing  :P )
But I really, really like his design too.
Can you extend the poll for a (both) option?

I can't, but send an email to Ted.  Mention that there was a fair bit of support on AoPS for it.

I'm not really sure if the combining would work. The colors don't really match in my opinion... bright for your design and dark for ccy's.

also, to add an option, ted would need to start over the poll, which he is pretty reluctant to do, i bet


The following is a problem that came from the 1995 Korean Math Olympiad, and I have been having some difficulty with it:

For any positive integer [i]m[/i], show that there exist integers [i]a, b[/i] satisfying

[i]|a|<=m
|b|<=m
0<a+b*sqrt(2)<=(1+sqrt(2))/(m+2).[/i]

This came from a list of practice problems for the pigeonhole principle, and I was able to solve the problems before and after it, so I must be missing something simple. Any ideas?

[color=cyan]I haven't tried to make a formal argument of any type, but I did play around a little.  I have a possible suggestion -- go read the thread "Powers of Certain Numbers" in the Challenge Others forum and also look at the possilbe choices of (a, b) in your question for small (from 1 to 10, say) values of m.[/color]

This seems like a nice classic Pigeonhole problem.

We usually attack these by creating a series of buckets of relevant width. What would the buckets be in this case?

Then we use pigeonhole to show that there are at least two numbers of the form a+b:rt2: that are in one bucket.  We then deduce that the difference of these two numbers gives us the desired a+b:rt2:.

So, see if you can fill in: (a) the buckets; (b) the definition of the a+b:rt2: that fall in these buckets such that we can both apply Pigeonhole as well as be guaranteed that the difference between the two that fall in the same bucket is also of the form a+b:rt2:.

(And post your sol'n here for others who can't get there with these hints - or post more hints)

The problem with Olypiad problems is that the answer is really obvious, but only after you know what it is. They are very inconvenient like that.

Alright, for anyone who wants to know a formal solution, I've put it below (you have to highlight).

[hide]

Lemma: There are m+3 ordered pairs (a,b) such that a+b*sqrt(2) is in the interval [0, sqrt(2)], a is nonnegative, and |a|,|b|<=m.

Proof: Clearly 0, sqrt(2), 1, and 1+sqrt(1) are on the interval.

For all a from 2 to m, we can find the smallest negative integer value of b such that a+bsqrt(2)>0. This must greater than -m, because otherwise we would have a+bsqrt(2)<=m(1-sqrt(2)<0, which is a contradiction. Now suppose a+bsqrt(2)>1+sqrt(2). Then a+(b-1)sqrt(2)>1>0, which is a contradiction of the minimality of b. Therefore for a from 2 to m, there is at least one value of b such that 0<=a+bsqrt(2)<=1+sqrt(2).

Thus we have at least 4+m-1=m+3 pairs (a,b) satisfying a>=0 and 0<=a+bsqrt(2)<=1+sqrt(2).    []

That little [] is a makeshift end-of-lemma symbol.

Now the problem follows almost immediately. We can divide the inteveral [0,1+sqrt(2)] into m+2 subintervals of length (1+sqrt(2))/(m+2), and by the pigeonhole principle there must be two values of a+bsqrt(2) with a nonegative in one of these intervals. Because the a terms are nonnegative and the b terms range from -m+1 to 1, the difference of these two value can be written in the form c+dsqrt(2), with |c|,|d| <= m and 0<c+dsqrt(2)<=(1+sqrt(2))/(m+2).[/hide]

QED.

If I am not mistaken, this inequality is really not that sharp. If we are willing to sacrifice the 1+sqrt(2) possibility, we could change 1 to an arbitrarily small value, and prove that there exist a,b such that 0<a+bsqrt(2)<=(sqrt(2)+epsilon)/(m+1), which, through a little reasoning about the concept of "arbitrarily small," leads to 0<a+bsqrt(2)<=sqrt(2)/(m+1), a sharper inequality for all positive values of m. Is this correct, or am I misinterpretting the nature of my own proof?


trigonometry
inequalities unsolved
inequalities

[quote="shobber"]Let $g(x)= \sum_{k=1}^{n} a_k \cos{kx}$, $a_1,a_2, \cdots, a_n, x \in R$. If $g(x) \geq -1$ holds for every $x \in R$, prove that $\sum_{k=1}^{n}a_k \leq n$.[/quote]
The hint see here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=65041

please post a perfect solution of this problem.thanks

[quote="shobber"]Let $g(x)= \sum_{k=1}^{n} a_k \cos{kx}$, $a_1,a_2, \cdots, a_n, x \in R$. If $g(x) \geq -1$ holds for every $x \in R$, prove that $\sum_{k=1}^{n}a_k \leq n$.[/quote]
Putting $x_j:=\frac{j\pi}{n+1}$, $j=1,...,n$ and adding it all up yields
\[-n\le\sum_{j=1}^ng(x_j)=\sum_{j=1}^n\sum_{k=1}^na_k \cos{\frac{jk\pi}{n+1}}=-\sum_{k=1}^na_k\] using $\sum_{j=0}^n \cos{\frac{jk\pi}{n+1}=0}$. (see $e^{2k\pi i}-1=0$) :)


Is there a way to write f(t) = Asin(t+k) + Bsin(t+p) in simplest form?

Assuming $ A,B,k,p$ are arbitrary constants,

we can get it to $ f(t)\equal{}A\sin(t\plus{}k)\plus{}B\sin(t\plus{}p)$

$ \equal{}A\sin t\cos k\plus{}A\sin k\cos t\plus{}B\sin t\cos p\plus{}B\sin p\cos t$

$ \equal{}\sin t(A\cos k\plus{}B\cos p)\plus{}\cos t(A\sin k\plus{}B\sin p)$.


algebra
polynomial
ceiling function
algebra proposed

Prove that polynomial $ P(x)\equal{}(x\minus{}a_1)^2(x\minus{}a_2)^2\dots (x\minus{}a_n)^2\plus{}1$,where $ a_1,a_2,\dots a_n\in\mathbb{Z}$,
is irreducible over $ \mathbb{Z}[x]$.

This has probably been posted several times before.

[hide]
If $ P(x)$ is reducible over $ \mathbb{Z}[x]$, then $ \exists x : P(x) \equal{} 0 \Longrightarrow [(x \minus{} a_1)(x \minus{} a_2)\dots (x \minus{} a_n)]^2 \equal{} \minus{} 1$, which is impossible. $ \boxed{Q.E.D.}$
[/hide]

Edit: nvm, I can't read.

[quote="Xantos C. Guin"]
If $ P(x)$ is reducible over $ \mathbb{Z}[x]$, then $ \exists x : P(x) \equal{} 0 \Longrightarrow [(x \minus{} a_1)(x \minus{} a_2)\dots (x \minus{} a_n)]^2 \equal{} \minus{} 1$, which is impossible. $ \boxed{Q.E.D.}$
[/quote]
Reducible does not mean that $ P(x)$ has a real root. :rotfl:
For example polynomial
$ P(x)\equal{}(x^2\plus{}x\plus{}1)(x^2\plus{}1)$,is reducible but it has no real root.

Suppose $ P(x) \equal{} Q(x) R(x)$ where $ \text{deg} Q \ge \text{deg} R$.  Then $ \deg{R} \le n$.  but

$ P(a_k) \equal{} 1 \equal{} Q(a_k) R(a_k) \implies R(a_k) \equal{} \pm 1, k \equal{} 1, 2, ... n$

WLOG $ R(a_k) \equal{} 1$ for at least $ \left\lceil \frac {n}{2} \right\rceil$ of $ k \in \{ 1, 2, ... n \}$.  Moreover, each root occurs with multiplicity $ 2$.  (Differentiate!  Even if the $ a_k$ are not distinct, it's not hard to show that the roots occur with the expected multiplicities.)  This implies that $ R(x) \equal{} 1 \forall x$.

I think there is another condition that $ a_1,\cdots,a_n$ are distinct.

Sorry; either $ R(x) \equal{} 1 \forall x$ or

$ R(x) \equal{} (\pm 1) \prod_{k \in S} (x \minus{} a_k)^2 \plus{} 1$

where $ S \subset \{ 1, 2, ... n \}$ and $ |S| \equal{} \frac {n}{2}$ when $ n$ is even.  But if $ P(a_k) \equal{} R(a_k) \equal{} 1$ then $ Q(a_k) \equal{} 1$, so we get $ Q(x) \equal{} R(x)$.  But clearly $ P(x)$ is not a square polynomial (consider $ P(0)$).  :)

In fact you can use $ R(x)$ has no real roots because $ P(x)$.Then you can conclude $ R(a_k) \equal{} 1$,for any k or $ R(a_k) \equal{} \minus{} 1$,for any k.

For example $ x^6\plus{}1\equal{}(x^2\plus{}1)(x^4\minus{}x^2\plus{}1)$ is not  irreducible over $ Z[x]$

[quote="Erken"][quote="t0rajir0u"]
Moreover, each root occurs with multiplicity $ 2$.[/quote]
Why?[/quote]

$ P(a_k) \minus{} 1 \equal{} 0$ with multiplicity $ 2$ and $ R(a_k) \equal{} 0$ implies $ P'(a_k) \equal{} 0$, hence

$ P'(a_k) \equal{} Q'(a_k) R(a_k) \plus{} Q(a_k) R'(a_k) \equal{} Q(a_k) R'(a_k) \equal{} 0$.

Note that $ Q(a_k) \equal{} \pm 1$, hence $ R'(a_k) \equal{} 0$.  :)


function
trigonometry
integration
calculus
derivative
calculus computations

How does one prove that $ \arcsin x \equal{} \sum\limits_{k \equal{} 0}^\infty  {\frac{{(2k)! \cdot x^{2k \plus{} 1} }}
{{4^k (k!)^2 \cdot(2k \plus{} 1)}}} ,\,\left| x \right| < 1$?

well, ummm... use   :maybe:

$ \sin^{\minus{}1}\;x\;\equal{}\;\int_0^x\;\dfrac{\textbf dy}{\sqrt{\; 1\minus{}y^2\;}}\;,\qquad \forall \; |x|\le 1$

Yeah, I thought about that too, but it's not the same as finding series representation for $ \arctan x,$ unless I'm missing something, a series for $ \frac1{\sqrt{1\minus{}y^2}}$ as you stated.

[quote="Black"]Yeah, I thought about that too, but it's not the same as finding series representation for $ \arctan x,$ unless I'm missing something, a series for $ \frac1{\sqrt {1 \minus{} y^2}}$ as you stated.[/quote]

i don't follow you...  :(

we have,
$ \dfrac{1}{\sqrt{1\minus{}4z}}\;\equal{}\;\sum_{k\equal{}0}^\infty\;\dfrac{(2k)!}{(k!)^2}\;z^k\qquad \forall |z| < \dfrac{1}{4}$

$ \Rightarrow \dfrac{1}{\sqrt{1\minus{}y^2}}\;\equal{}\;\sum_{k\equal{}0}^\infty\;\dfrac{(2k)!}{(k!)^2}\;\left(\dfrac{y^2}{4}\right)^k \qquad \forall |y| < 1$

so, $ \int_0^x\; \dfrac{1}{\sqrt{1\minus{}y^2}}\;\textbf dy\;\equal{}\;\int_0^x\; \sum_{k\equal{}0}^\infty\;\dfrac{(2k)!}{(k!)^2}\;\left(\dfrac{y^2}{4}\right)^k \;\textbf dy \qquad \forall |x| < 1$

using monotone, we get

$ \sin^{\minus{}1}\;x \;\equal{}\; \sum_{k\equal{}0}^\infty\;\dfrac{(2k)!}{(k!)^2}\;\dfrac{1}{4^k}\;\int_0^x\; y^{2k} \;\textbf dy$

$ \equal{}\; \sum_{k\equal{}0}^\infty\;\dfrac{(2k)!}{(k!)^2}\;\dfrac{1}{4^k}\;\dfrac{x^{2k\plus{}1}}{2k\plus{}1}$

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1056507#1056507 has something closely related that can be used to conclude this series.  It's also not hard to actually compute the derivatives of $ (1 \minus{} 4z)^{\minus{}\frac{1}{2}}$ and get the Taylor series that way.

[quote="Black"][quote="misan"]
we have,
$ \dfrac{1}{\sqrt {1 \minus{} 4z}}\; \equal{} \;\sum_{k \equal{} 0}^\infty\;\dfrac{(2k)!}{(k!)^2}\;z^k\qquad \forall |z| < \dfrac{1}{4}$[/quote]
This is what I meant in post #3, how do you prove this?[/quote]

you mean "how do you show this"..... not "how do you prove this" !  :maybe:

 i don't want to spend anymore time on this question, so i will give you a hint, which you can use to show the above relation:

$ \sum_{n\equal{}0}^{k} {k \choose n} ^2 \;\equal{}\; {{2k} \choose k}$

It's Newton's favorite tool, the binomial series. For $ |x|<1$, $ (1\plus{}x)^r\equal{}\sum_{n\equal{}0}^{\infty}\binom{r}{n}x^n$, where $ \binom{r}{n}\equal{}\frac{r(r\minus{}1)\cdots(r\minus{}n\plus{}1)}{1\cdot 2\cdots n}$

With $ r\equal{}\minus{}\frac12$ and $ x\equal{}t^2$, we get the series for $ \frac1{\sqrt{1\minus{}t^2}}$. Integrate term by term to get $ \arcsin t$.

As a historical note, Newton found this series [i]before[/i] the series for $ \sin$.


@problem 1:

Try to prove that $ a_n\equal{} 2n \minus{} \left\lfloor \frac{1\plus{} \sqrt{8n\minus{}7}} { 2}\right\rfloor$.

Proving is one thing; getting the expression is another. How did you figure that out?

The idea is that $ b_{T_n} \equal{} n$, where $ T_n \equal{} \frac {n(n \minus{} 1)}{2}$ is the $ n^{th}$ triangular number; hence the closed form of $ b_n$ should behave like the inverse of the function $ f(x) \equal{} \frac{x(x \minus{} 1)}{2}$.  It's only a matter of figuring out exactly how the floor function works from here.

[quote="ThetaPi"][b]Problem 1. [/b]The sequence $ 1,2,4,5,7,9,10,12,14,16,17,\cdots$ has one odd number followed by two evens, then three odds, four evens and so on. What number is the 2003th term? [hide="Outline"]

The solution given let the sequence $ \{a_n\}$ to be $ 1,2,4,5,7,9,10,12,14,\cdots$, and the sequence $ \{b_n\}$ to be $ 1,2,2,3,3,3,4,4,4,4,\cdots$. Using the fact that $ a_n \plus{} b_n \equal{} 2n$ for all $ n \in \mathbb{N}$, they computed $ b_{2003}$ and easily found $ a_{2003}$. However, this solution is difficult as it requires much analytical skills. Does anyone have an alternative solution?[/hide]

[b]Problem 2. [/b]What is the smallest possible integer $ n > 1$ such that $ 3^n$ ends with 003?[/quote]

[hide="Problem 1"][b]Problem 1:[/b] The $ 1st$ term$ \equal{}(2 \times 1\minus{}1)$. The $ (1\plus{}2)th$ term $ \equal{}(2 \times 1\minus{}1)\plus{}(2 \times 2\minus{}1)$. The $ (1\plus{}2\plus{}3)th$ term $ \equal{}(2 \times 1\minus{}1)\plus{}(2 \times 2\minus{}1)\plus{}(2 \times 3\minus{}1)$.   ....... So we observe that the $ (1\plus{}2\plus{}...\plus{}n)th$ term $ \equal{}(2 \times 1\minus{}1)\plus{}(2 \times 2\minus{}1)\plus{}....\plus{}(2n\minus{}1)$ which is not difficult to be proved by induction.

So the $ \frac{n(n\plus{}1)}{2}th$ term $ \equal{}2 \cdot \frac{n(n\plus{}1)}{2}\minus{}n\equal{}n^2$

The $ \frac{63(63\plus{}1)}{2}\equal{}2016th$ term $ \equal{}63^2$

So the $ 2003th$ term $ \equal{}63^2\minus{}2 \times 13\equal{}3943$[/hide]

[hide="Problem 2"][b]Problem 2:[/b]

[b]Case 1:[/b]$ n\equal{}2k$ is even. Then $ 3^n\equal{}(10\minus{}1)^k\equal{}(\minus{}1)^k(\mod 10)$, impossible.

[b]Case 2:[/b]$ n\equal{}2k\plus{}1$ is odd. Then $ 3^n\equal{}3(10\minus{}1)^k\equal{}3(\minus{}1)^k (\mod 10)$ gives $ k\equal{}2m$ is even.

So $ 3^n\equal{}3(10\minus{}1)^{2m}\equal{}\frac{6m(2m\minus{}1)}{2}(10)^2\minus{}6m(10)\plus{}3(\mod 1000)$

So $ 3^n\equal{}3(\mod 1000) \Leftrightarrow 1000|3m(2m\minus{}1)(10)^2\minus{}60m$
$ \Leftrightarrow 1000|120m(5m\minus{}3) \Leftrightarrow 25|3m(5m\minus{}3) \Leftrightarrow 25|m$

$ \Rightarrow$ the smallest possible $ m$ is $ 25$.

Therefore the smallest possible $ n$ $ \equal{}2k\plus{}1\equal{}4m\plus{}1\equal{}101$[/hide]


Let $ ABC$ be a triangle.Choose the pair of points $ A_1;A_2;B_1;B_2;C_1;C_2$ on the sides $ BC;CA;AB$,respectively such that $ A_1A_2B_1B_2C_1C_2$ is a convex hexagon with equal opposite sides and the triangle $ AA_1A_2: BB_1B_2;CC_1C_2$ have the same areas.Prove that the lines $ A_1B_2;B_1C_2;C_1A_2$ are concurrency

Lock the topic! I realied that it's a M&Y problem!


videos
MATHCOUNTS
email
summer program
MathPath

Awww.... we can't upload videos.

Any suggestions?

Please don't post the Andrew A video, urggh.

use this guy http://rapidshare.de/  and then u have to post the link here.  only 1 download/hr though unless you pay.

Use [url=https://upload.video.google.com/]this![/url]  Another fine product brought to you at no cost by the people at Google!

BUY THE STOCK NOW!   :D

Ahhhhhhhhhhhhhhh! EFuzzy, you have revealed hexadedron's true name. He must be known as R2D2.

Actually, you revealed it.  I had no idea Andrew was hexahedrom until you said so!

actually, my name got posted in the mathcounts section in the nats topics

By spelling your name wrong (hexadedron), I've managed to put this thread third when you google "hexadedron".

Mysmartmouth, how long does it take you to get a video verified?

I think chess64 volunteered to host the video.

I sent it to kadeel@gmail.com and I just sent it again

In the meantime, I've posted some of the better photos from mathpath in my photo album.

http://www.upal.ca/adeel/mathpath.MOV

Enjoy! :)

Lol, is that Andrew in MP photo 6? :rotfl:


A collection $ \mathcal{F}$ of subsets of a set $ X$ is said to have the finite intersection property if $ \cap_{j \equal{} 1}^n F_j\neq \phi$ for any finite subcollection $ \{F_1, F_2,...,F_n\}$ of $ \mathcal{F}$. Show that a topological space $ X$ is compact iff for every collection $ \{F_\alpha: \alpha \in A\}$ of closed subsets of $ X$ which has the finite intersection property, we have $ \cap_{\alpha\in A}F_\alpha \neq \phi$.

proof by contradiction, demorgans laws
http://planetmath.org/encyclopedia/4181.html


group theory
abstract algebra
induction
geometry
geometric transformation
linear algebra
matrix

[quote="dmitin"]Let  $ G$ be a finite non-commutative group of order $ t \equal{} 2^nm$, where $ n, m$ are positive and $ m$ is odd. Prove, that if the group contains an element of order $ 2^n$, then
(i) $ G$ is not simple;
(ii) $ G$ contains a normal subgroup of order $ m$.[/quote]

Rotman 1999, exercise 3.30,  gives this hint : Consider $ G$ as a permutation group via Cayley theorem and show that it contains an odd permutation.

I would say what follows. We prove (ii) by induction on $ n$. More precisely, we prove by induction on $ n$ that the elements of odd order of $ G$ form a subgroup of $ G$.

If $ n \equal{} 0$, it is trivial. Let $ n > 0$. Let $ a$ be an element of order $ 2^n$ of $ G$. Let us denote by $ L$ the injective homomorphism from $ G$ into the symmetric group $ S_{G}$ which sends $ x$ onto the left translation $ L_{x} : y \mapsto xy$. (The group structure we choose on  $ S_{G}$ is $ (f, g) \mapsto f \circ g$, with $ f \circ g (z) \equal{} f(g(z))$.)
By hypothesis, there is an element $ a$ of order $ 2^{n}$ in G. Then the canonical decomposition of the permutation $ L_{a}$ in cycles is a product of $ m$ disjoint cycles of length $ 2^{n}$. (Easy exercise. See Rotman 1999, exerc. 3.29, p. 54.) Since we assume $ n \geq 1$, such a cycle has even length and is thus an odd permutation. Since $ m$ is odd, the permutation $ L_{a}$ is thus odd. So $ L(G)$ contains at least an odd permutation. It implies that the even permutations in $ L(G)$ form a subgroup of index 2 of $ L(G)$ (Rotman, 1999, exerc. 3.22, p. 51.) Since $ G$ is isomorphic to $ L(G)$, $ G$ has thus a subgroup of index 2, let $ H$. The order of $ H$ is $ 2^{n \minus{} 1}m$. Since $ H$ is of index 2 in $ G$, $ a^{2}$ belongs to $ H$ (Rotman 1999, exerc. 2.16, p. 27; a subgroup of index 2 is always normal, so it is a consequence of a more general fact about normal subgroups, which is easily demonstrated by use of the factor group).
Since $ a$ is of order $ 2^{n}$, $ a^{2}$ is of order $ 2^{n \minus{} 1}$ (Rotman 1999, exerc. 2.11, p. 27). Thus, by induction hypothesis, the elements of odd order of $ H$ form a subgroup of $ H$. But each element of odd order of $ G$ is in $ H$. Indeed, in view of a fact already noted, we have $ x^{2} \in H$ for each $ x \in G$. If $ x$ is of odd order, it is easy to conclude that $ x$ lies in $ H$.
So the elements of odd order of $ G$ are exactly the elements of odd order of $ H$ and we saw that the elements of odd order of $ H$ form a subgroup of $ H$, and thus of $ G$. So we have demonstrated by induction that the odd elements of odd order of $ G$ form a subgroup $ K$ of $ G$.
Since every conjugate of an element of odd order is of odd order, $ K$ is a normal subgroup of $ G$.
Every Sylow subgroup of odd order of $ G$ is contained in $ K$, thus the order of $ K$ is divisible by $ m$ and clearly must be equal to $ m$. So $ K$ is a normal subgroup of order $ m$ of $ G$, which proves (ii). If $ n > 0$ and $ m$ > 1, (i) follows.

"that the odd elements of odd order of G"

Sorry. Read : "that the elements of odd order of G".

First we'll show $Sgn:S_n\to \{\pm 1\}$ is homomorphism. For that consider permutation matrix, i.e send $\sigma$ to $A_{\sigma}$ where $[A_{\sigma}]_{i,j}=1$ if, $\sigma(i)=j$ and zero otherwise. In other words $A_{\sigma}$ sends $e_i \to e_{\sigma_i}$. And also, $\sigma \to A_{\sigma}$ is again a homomorphism. Now one can verify if $\sigma$ is a cycle of length $m$ then $det(A_{\sigma})=(-1)^{m-1}$. Now also we know we can write any permutation from $S_n$ as product of some disjoint cycles, so now one can see why $Sgn$ is multiplicative map (more preciously $Sgn$ map is written as product two homomorphisms, given by det and permutaion matrix construction map). Now any group $G$ is isomorphic to a subgroup of $S_{|G|}$ via left regular action i.e $L:G \to S_{|G|} | g\to L_g: L_g(x)=gx$. Now suppose $|G|=n$ and let $g$ be and element. Now for any $\sigma\in S_n$ cycle type of $\sigma$ are sizes of orbits of $<\sigma>$ on $\{1,2,..,n\}$. Now $G$ acts freely on itself via this action, and so $<g>$ also acts freely on $G$ and so orbits of $<g>$ on $G$ are cosets of $<g>$. Now as there are $\frac{n}{O(g)}$ cosets, so $L_g$ is nothing but product of $\frac{n}{O(g)}$ cycles each of length $O(g)$. Now here take an element $x$ of order $2^n$. Now we know $L_x$ is product of odd number of even cycles and hence $Sgn(L_x)=-1$.  Now denote $f=Sgn \circ L$. Hence, $f:G\to \{\pm 1\}$ is again a homomorphism, and $ker(f)$ is non trivial and so $|im(f)|=2 \implies |G:ker(f)|=2$ and hence $ker(f)$ is a normal subgroup of $G$ of index $2$. This proves first part, for second part, if $n=1$`then clearly done, but when $n>1$ then if $O(a)=2^n\implies O(a^2)=2^{n-1}$ and so, $Sgn(L_{a^2})=1 \implies a^2\in ker(f)$. And so again considering $ker(f)$ and using same argument as before we boiled down to $n=1$ and that's what all we need to prove second part.

[quote=subham1729]for second part, if $n=1$`then clearly done, but when $n>1$ then if $O(a)=2^n\implies O(a^2)=2^{n-1}$ and so, $Sgn(L_{a^2})=1 \implies a^2\in ker(f)$. And so again considering $ker(f)$ and using same argument as before we boiled down to $n=1$ and that's what all we need to prove second part.[/quote]
No, that only proves the existence but not the normality.

Knowing (ii), it allows us to tell even more: $G\cong H\rtimes_\varphi (\mathbb{Z}/2^n)$, where $\varphi\colon\mathbb{Z}/2^n\to\operatorname{Aut}(H)$. Choosing $\varphi$ is the same as choosing an automorphism of $H$ of order dividing $2^n$.

I think it suffices to write down a correct proof for $n =1$ as the base case rather than $n =0$.
 Let $ |G| = 2m$, where m is odd and $m > 1$. Let G act on itself by left multiplication: $ \sigma_{g}: G \rightarrow G: \sigma_{g}(x) = gx$ for all $x \in G$. Then this action affords a homomorphism $ \varphi: G \rightarrow \mathrm{Sym}(G): \varphi(g) = \sigma_{g}$. The kernel which is a normal subgroup is either non-trivial or trivial. If the kernel is non-trivial then already we know that G is not simple. If the kernel is trivial, then $G$ can be embedded in $ \mathrm{Sym}(G) = S_{2m}$. Note that  by Cauchy Theorem, G has a non-identity element $ a$ of order 2. Then for each $x \in G$, we have $ \sigma_{a}(x) = ax$ and $ \sigma_{a}(ax) = x$. We see that $ \sigma_{a}$ is a product of m transpositions $(x, ax)$. Hence, $ \sigma_{a}$ is an odd permutation as m is odd. We construct a normal subgroup as follows: $ H = \{x \in G: \varphi(x) = \sigma_{x} \in A_{2m}\}$. Since $ a\in G -H$, we see that H has index 2 and so it is normal, G is not simple.

Use the above argument to show that there exists a composition series
$$G_0 \triangleleft G_1 \triangleleft \cdots \triangleleft G_n = G$$
where $|G_0| = m$, and $G_{i + 1}/G_i \cong \mathbb{Z}_2$. It suffices to show that $G_0 \triangleleft G$ (in fact, by Jordan-Hölder, if the problem were true, we must have $G_0 \triangleleft G$). We show via induction that $G_0 \triangleleft G_k \implies G_0 \triangleleft G_{k + 1}$ by showing that $G_0$ is the only subgroup of order $m$ in $G_k$. Since $G_k \triangleleft G_{k + 1}$, conjugation of $G_0$ by an element of $G_{k + 1}$ would bring $G_0$ to a subgroup of order $m$ in $G_k$. By showing that $G_0$ is the only subgroup of order $m$, we must have that $G_0$ is normal in $G_{k + 1}$.

To show that $G_0$ is the only subgroup of order $m$ in $G_k$, we mimic the proof of Sylow 2. Let $G$ be a group and $H$ a subgroup of $G$. If $G$ acts on $G/H$ (the left cosets) via left multiplication, we can show that each of the stabilizers are conjugates of $H$. So going back to the problem, suppose $H$ is a subgroup of $G_k$ of order $m$. Then $H$ acts on $G_k/G_0$ by the same action. The stabilizers are of the form $H \cap H_0$ where $H_0$ is a conjugate of $G_0$. However, by our induction hypothesis, $G_0$ is normal in $G_k$ and therefore, $H_0 = G_0$, so the stabilizer is $H \cap G_0$; i.e. it's constant. By the orbit-stabilizer theorem, the size of the orbits must all be the same as well and must divide $m$. However, as remarked in the proof of Sylow 2, if $p | m$, since $p$ does not divide $2^n$, we may choose an orbit of size $r$ such that $p$ does not divide $r$. Since the size of the orbits stay constant, for each prime $p$ that divides $m$, $p$ cannot divide the size of the orbit. Hence, $r = 1$, and so $H = G_0$ and so $G_0$ is the only subgroup of order $m$ in $G_k$. $\blacksquare$


The diagonal of a cube is represented by 2x. Find the total surface of the cube in terms of x.

The [url=http://www.google.com/search?q=define%3A+space+diagonal&ie=utf-8&oe=utf-8&aq=t&rls=org.mozilla:en-US:official&client=firefox-a]space diagonal[/url] of a cube is simply the side length multiplied by $ \sqrt{3}$, because of the Pythagorean theorem. Therefore, if the diagonal is $ 2x$, the side length is $ \frac{2x}{\sqrt{3}} \equal{} \frac{2x\sqrt{3}}{3}$ and so the surface area is $ \boxed{4x\sqrt{3}}$.

DUDE! Put all your problems in one post. You have 10 topics and are cluttering up forum space!

[quote="algebra2000"]izzy, i got a different answer [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=276026]here[/url]

same problem, two threads...[/quote]
I see what Izzy did wrong. When squaring $ \frac{2x\sqrt{3}}{3}$ then multiplying by $ 6$, Izzy made a mistake. It should be:
$ 6(\frac{2x\sqrt{3}}{3})^2\equal{}6*\frac{4*x^2*3}{9}\equal{}\frac{72x^2}{9}\equal{}8x^2$

Can a mod delete these questions?

We should make a k00lperson marathon. he posts the questions and we answer. It'd be even more active than the mathcounts marathon.

She.  :wink:

I don't see what's wrong with k00lperson asking questions (though I would suggest she put them all in one thread to reduce clutter); after all, isn't that what fora are meant for?

--

Oops, thanks for catching that, PowerOfPi.  :)

[b]She[/b] can ask questions, but she should seriously consider getting a thread just by herself. That many questions and threads seriously take up some space in the server. k00lperson: Please consider this.  :|


$a > 0$ and $b$, $c$ are integers such that $ac$ – $b^2$ is a square-free positive integer P. [hide="For example"] P could be $3*5$, but not $3^2*5$.[/hide] Let $f(n)$ be the number of pairs of integers $d, e$ such that $ad^2 + 2bde + ce^2= n$.  Show that$f(n)$ is finite and that $f(n) = f(P^{k}n)$ for every positive integer $k$.

[b]Original Statement:[/b]

Let $a,b,c$ be given integers $a > 0,$ $ac-b^2 = P = P_1 \cdots P_n$ where $P_1 \cdots P_n$ are (distinct) prime numbers. Let $M(n)$ denote the number of pairs of integers $(x,y)$ for which \[ ax^2 + 2bxy + cy^2 = n.  \] Prove that $M(n)$ is finite and $M(n) = M(P_k \cdot n)$ for every integer $k \geq 0.$ Note that the "$n$" in $P_N$ and the "$n$" in $M(n)$ do not have to be the same.

Let $\alpha=\frac{-b+\sqrt{-p}}a$ and $\bar\alpha=\frac{-b-\sqrt{-p}}a$. Now, $\alpha,\bar\alpha$ are the roots of the equation $at^2+2bt+c=0$, and $f(n)$ is the number of integer solutions $(x,y)$ to $a(x-\alpha y)(x-\bar\alpha y)=n\ (*)$.

Construct the lattice $\Delta_{\alpha}=\{u-\alpha v\ |\ u,v,\in\mathbb Z\}$ (we denote $u-\alpha v$ by $(u,v)$). This is a obviously a discrete set, so every circle contains finitely many points of $\Delta_{\alpha}$. In particular, the circle centered at the origin of radius $\sqrt{\frac na}$ contains finitely many points of our lattice, but these points $(x,y)$ are precisely those for which $(*)$ holds, so the finitude of $f(n)$ is proven.

The second part is equivalent to proving that for every positive integer $n$, we have $f(n)=f(pn)$. We define a map from the solutions of $(*)$ to those of $a(x-\alpha y)(x-\bar\alpha y)=pn\ (**)$, taking $(x,y)$ to $(bx+cy,-ax-by)$. It's easy to check that this function does indeed map solutions of $(*)$ onto solutions of $(**)$, and it's obviously injective. What we must prove now is that it's also surjective, i.e. given $u,v$ s.t. $au^2+2buv+cv^2=pn$, we can find [i]integers[/i] $x$ and $y$ with $bx+cy=u,-ax-by=v$. This reduces to proving that whenever $p|au^2+2buv+cv^2$, we have $p|au+bv$ and $p|bu+cv$. In turn, since $p$ is square-free, we must prove that if $q$ is a prime divisor of $p$ and $q|au^2+2buv+cv^2$, then $q$ divides both $au+bv$ and $bu+cv$.

First of all, if $q|a$, then $q|b,q\not|c$, so $q|a,b,v$, and we’re done. Something similar happens if $q|c$, so we assume $q\not|a,b,c$.

In $\mathbb F_q$, the equation $at^2+2bt+c$ has the double root $-ba^{-1}$, since the discriminant of this equation is $0$ (in the field with $q$ elements). This means that if $u,v$ satisfy $au^2+2buv+cv^2=0$ in $\mathbb F_q$, then $u-(-ba^{-1})v=0$, i.e. $au+bv=0$, and $bu+cv=0$ follows from $au+bv=0$ and $ac-b^2=0$.

[hide]
Write the equation as $(ad+be)^2+pe^2=an$. It is clear that the number of solutions to this is exactly the number of solutions $(x,y)$ to $x^2+py^2=an$. Now we wish to prove that the number of such solutions for $x^2+py^2=N$ is the same as for $x^2+py^2=pN$. This is obvious; just take $(x,y) <=> (py, x)$.
[/hide]


vector
linear algebra
matrix
linear algebra unsolved

Let V be a vector space of n dimension. If A be a "nXn" nilpotent matrix of power "n" then show that for any non-zero vector "b", {b,Ab,A^2b,....,A^(n-1)b} will be a basis of V

Since $ A^{n\minus{}1}\neq 0$, there is some $ x$ with $ A^{n\minus{}1}x\neq 0$. Then $ x,Ax,\cdots,A^{n\minus{}1}x$ will be a basis. It's easy to show they're linearly independent; if $ c_{k}A^{k}x\plus{}c_{k\plus{}1}A^{k\plus{}1}x\plus{}\cdots\plus{}c_{n\minus{}1}A^{n\minus{}1}x\equal{}0$, $ A^{n\minus{}k\minus{}1}(c_{k}A^{k}x\plus{}c_{k\plus{}1}A^{k\plus{}1}x\plus{}\cdots\plus{}c_{n\minus{}1}A^{n\minus{}1}x) \equal{}c_{k}A^{n\minus{}1}x\equal{}0$, and $ c_{k}\equal{}0$. Since there are $ n$ of them, they're a basis.

[quote="jmerry"]It's easy to show they're linearly independent; if $ c_{k}A^{k}x\plus{}c_{k\plus{}1}A^{k\plus{}1}x\plus{}\cdots\plus{}c_{n\minus{}1}A^{n\minus{}1}x \equal{} 0$, $ A^{n\minus{}k\minus{}1}(c_{k}A^{k}x\plus{}c_{k\plus{}1}A^{k\plus{}1}x\plus{}\cdots\plus{}c_{n\minus{}1}A^{n\minus{}1}x) \equal{} c_{k}A^{n\minus{}1}x \equal{} 0$, and $ c_{k}\equal{} 0$. Since there are $ n$ of them, they're a basis.[/quote]

can you please explain this sir? i couldnt understand......
thank you....

Look at the definition of linear dependence; take a linear combination that's zero, and look at the first nonzero coefficient. Oops, that coefficient's zero too (each other term in the sum after multiplication is zero).

To put into perspective: If S and T are linear operators on an n-dimensional linear space V and

1) $ T^{n\minus{}1}\ne 0$

2) $ S^{n\minus{}1}\ne 0$

3) $ S^{n}\equal{} T^{n}\equal{} 0$

then there is an invertible linear operator U on V such that $ S \equal{} U^{\minus{}1}TU$. You can figure out the corresponding statement for matrices from this. If square matrices A and B satisfy the analogues of 1 and 2, then $ A \equal{} L^{\minus{}1}BL$. What does L look like?

edit: thanks to owk for making sure I remembered #3

[quote="§outh§tar"]If S and T are linear operators on an n-dimensional linear space V and

1) $ T^{n\minus{}1}\ne 0$

2) $ S^{n\minus{}1}\ne 0$

then there is an invertible linear operator U on V such that $ S \equal{} U^{\minus{}1}TU$.[/quote]

Counter-example: $ S\equal{}\begin{pmatrix}1&1\\0&1\end{pmatrix},T\equal{}\begin{pmatrix}1&0\\0&1\end{pmatrix}$. owk

owk you are being needlessly picky  :roll:

The OP was talking about nilpotent matrices and so I (mistakenly) neglected to add that $ S^{n}\equal{} T^{n}\equal{} 0$. It should have been clear from context though  :P

My apologies. I should have read the entire thread. :blush: owk

[color=white][Blunder Alert]What about $ S \equal{}\begin{pmatrix}0 & 1 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 0\end{pmatrix}$ and $ T \equal{}\begin{pmatrix}0 & 1 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\end{pmatrix}$?  :maybe:[/Blunder Alert][/color]

In this example, $ S^{3}\equal{}T^{3}\equal{}0$. So what? owk


I have tried to solve this problem but am not sucessful. Please help me:
1) The casino gane of American Roulette consists of a circular wheel with 38 spots. Thirty six of the spots are numbered 1,2,3...36. Eighteen of these numbers are colored black and the other eighteen are colored red. The remaining two spots are colored green with no number assignment. If you play a color (either red or black), the casino will pay you even odds. That is, if you play black and wager 1 dollards, you will profit 1 dollards if black comes up on that particular spin of the wheel.
a) Let X= profit or loss for one spin of the wheel when playing black. Find the expected value of  X and explain what it means.
b) How much money is expected to be won or lost after 1000 games where the player wagers 10 dollards per game on black?
c) Would the answer to part b be different if the player wagers 10 dollards per game on red? Explain.

[hide="Answers"]
a. x/19 lost where x is amout wagered
b. you can expect to lose  $\$$526.32
c. it would be the same because P(black)=P(red)
[/hide]


There is a cube with length of side $n$.
The cube consists of $n^3$ boxes.
A rook in a cube attacks all the boxes placed in horizontal,vertical and height "line".
1)What is the minimal number of rooks that we need such that all the boxes will be under attack.
2) Where do we need to put the rooks to satisfy this condition?

I'm not sure that there is a general rule...but maybe there is one.

It is obvious that the minimum number when $n=2$ is 2 (at (1,1,1) and (2,2,2)).

I believe that the minimum number when $n=3$ is 5 (at (1,1,1), (2,2,1), (1,2,2), (2,1,2), and (3,3,3).)

I believe that the minimum number when $n=4$ is 8 (at (1,1,1), (2,2,1), (1,2,2), (2,1,2), (3,3,3), (4,4,3), (3,4,4), and (4,3,4)).

I found it helpful to use $n$ ${n}\times{n}$ grids.  I would place rooks and make a note of the squares that were not covered by the rooks on that particular grid. I would make sure that rooks were placed in those same squares on other grids.


Let a triangle $ABC$
prove that:
$\sqrt{p-a}+\sqrt{p-b}+\sqrt{p-c}+\sqrt{p} \geq \sqrt{a}+\sqrt{b}+\sqrt{c}$
with $p=\frac{a+b+c}{2}$

This seems to be a strict inequality.
Let $x=p-a$ and so on. We have to prove $\sqrt{x}+\sqrt{y}+\sqrt{z}+\sqrt{x+y+z} \ge \sqrt{x+y} + \sqrt{y+z} + \sqrt{z+x}$.
Squaring both sides, it remains to prove
\[ \sqrt{xy}+\sqrt{yz}+\sqrt{zx} + (\sqrt{x}+\sqrt{y}+\sqrt{z})\sqrt{x+y+z} \ge \sqrt{(x+y)(y+z)} + \sqrt{(y+z)(z+x)} + \sqrt{(z+x)(x+y)} \]
Now, $\sqrt{xy}+\sqrt{z}(\sqrt{x}+\sqrt{y}+\sqrt{z}) \ge \sqrt{(y+z)(x+z)}$ and so on completes the proof.


Have a name, but would prefer a different name?  Doesn't matter.  Your name tells your indentity...
[size=150]Anyways, list your favorite name(s) here![/size]  If you want, you can add why.  I wana see what name most people like. :D

P.S. Watch it come out to be like BillyBob The BillyBobBilly The fourth son of Bob the billy bob the hill billy.... Or maybe not hehe :D
 :blush:  :D

[quote="filletwho"][quote="lightspeed"]
P.S. Watch it come out to be like BillyBob The BillyBobBilly The fourth son of Bob the billy bob the hill billy.... Or maybe not hehe :D
 :blush:  :D[/quote]

Hehe... my friends and I made bob the subsitute word for a swear word. That's just wrong.[/quote]

Don't think of it that way.  I think it's a very clever way of not getting into trouble :D
It would sound wierd though.
Example:
You slam your finger in the door.  You shout, "BOB!, my finger!"
You lose something.  You say, " Oh BOB it!"
Lol get my point?

Nice, Nice.  James Brown and Ponce de Leon...  Not bad.  Any others?

[NOTHING ELSE IS BETTER:]

[hide]Dark V A N I S H E R  :D [/hide]

One of my school's history teachers named his son Justin Love. Heehee..

If I were to get another name... I'd definitely get one that's compatible in more than one language.
My name pronounced with Spanish accent sounds ABSOLUTELY horrible. (and I hear it all the time!! :( )

[quote="GoBraves"][quote="lightspeed"][quote="GoBraves"][hide="Perfect Name"]Ameem[/hide][/quote]
Lol, that actually sounds really nice.  Two questions:
1) Is that your name?
2) Is that a boy or a girl's name?[/quote]

Yes it is my name...and its a boys name[/quote]
I thought so!  That's good that you are proud of your own name.

[quote="peter"]Holly Crappo (this was a name my Biology teacher saw when she went to a high school graduation)

Whose parents would name their child that?

Imagine the teasing![/quote]

Are you serious?  That's wicked bad!  Her parents must've had a bad life or something.  There's no need to name your child that!  Are you sure its not a joke?

Yes, Brian is such a...beautiful name. I know. I own it.

Montoya...what movie was that? Wasn't the Spaniard from The Princess Bride named Inigo Montoya? That is one cool name! That and Fezzik...

I think there was an anti-pornography activist named Linda Lovelace or something like that (correct me if I'm wrong), which is pretty cool

And somebody told me about a race car driver named...[hide]Dick Trikle[/hide]

[quote="09husbri"]Montoya...what movie was that? Wasn't the Spaniard from The Princess Bride named Inigo Montoya? That is one cool name! [/quote]

Montoya is a [b]surname[/b], not the real name of the guy (well, at least I wouldn't say that the name of Albert Einstein is Einstein!)

In high school I knew a: Nathaniel Liberty (sounds very patriotic)

and at Cornell I met a kid named: James Maxwell (who actually dropped out of a physics E/M course)

But because I love higher algebra so much, I want to name my kids:

[b]Norm[/b]an and [b]Trace[/b]y. :)

My favorite name though is my sister's: Alizah (pronounced ah-leez-ahh about)

It is hebrew for merry/cheerful/sunshiny.

[quote="djimenez"]

Montoya is a [b]surname[/b], not the real name of the guy (well, at least I wouldn't say that the name of Albert Einstein is Einstein!)[/quote]

I know that. Inigo Montoya. That's what I said, wasn't it?

[quote="09husbri"]I know that. Inigo Montoya. That's what I said, wasn't it?[/quote]

By the way, are you sure that the name is Inigo, I have never heard it (and I am hispanic!) but I have heard Ingo!

our principal's name is Dickie Dingle (I imagine Dickie is a nickname for Richard though).

I feel sorry for the person whose name is:
[hide]Pamela Chu (if you don't get it, try switching the order)[/hide]
I love how the hispanics put a virtual "e" before every "s".  It makes every "s" name sound interesting. My favorite names are those that can be  abbreviated into two letters. Such as "Ed" or "Al" I don't like those that are one lettered, or initials, though.

[quote="h_s_potter2002"]Really cool name - Brfxxccxxmnpcccclllmmnprxvclmnckssqlbb11116 (pronounced "Albin")  :D[/quote]

 [hide="Woah That Name Definently Gets"]:first: First Place[/hide]

[quote="hmoonster"]I feel sorry for the person whose name is:
[hide]Pamela Chu (if you don't get it, try switching the order)[/hide]
[/quote]
Well, if we consider that Chu seems to be chinease, and that chineases use first their surname and after that their given name, it is even more interesting. Nevertheless, I am not sure if a non-hispanic person (or a person without any knowledge of spanish or portuguese) would get it!

[quote="hmoonster"]I love how the hispanics put a virtual "e" before every "s".  It makes every "s" name sound interesting.[/quote]
Well, that is if the hispanic person don't try to pronounce correctly. In spanish we don't have words starting with S+consonant, and most of the vowel+s+consonant, the vowel is an e (estatua, esperanza, esmeralda, esto, estilo, etcetera...)

[quote="stupidkid"]I like the name Gaylord  :D I wonder if anyone actually has that name.[/quote]
Well, it is even funnier when it is accompained by the surname Focker!

[quote="jpark"][quote="Sunny"]Kathlynn

I like it so much because its a mix of the two common names Kaitlynn and Kathryn, yet Kathlynn is very uncommon.[/quote]

Which also happens to be your middle name  :P[/quote]

 :D   Furthering my reasons to like it.

mackenzie 4 a girl, ethan for a boy(and i don't know why!)


Let $ a_1, a_2, a_3,\dots$ be an increasing arithmetic sequence of integers. If $ a_4a_5 \equal{} 13$, what is $ a_3a_6$?

The only way 13 factors is $ 13$ and $ 1$. It could be $ \minus{}13$ and $ \minus{}1$, but that doesn't affect anything. Thus the difference is $ 12$, and $ a_3\equal{}\minus{}11$, and $ a_6\equal{}25$. Thus $ a_3a_6\equal{}\boxed{\minus{}275}$.

Since the sequence is comprised of integers, $ a_{4}$ and $ a_{5}$ will have to be factors of $ 13$. There is only one solution that is increasing, $ (1,13)$. That is good since we know already what we're working with, as we derive the common difference to be $ 13 \minus{} 1 \equal{} 12$. Thus, $ a_{3} \equal{} 1 \minus{} 12 \equal{} \minus{} 11$ and $ a_{6} \equal{} 13 \plus{} 12 \equal{} 25$, and $ a_{3} \plus{} a_{6} \equal{} \minus{} 11(25)\equal{} \boxed{\minus{}275}$.

EDIT: Whoops. BTW, it's the number you put in your first post, -275.

We want $ a_3a_6$ not $ a_3\plus{}a_6$, so $ \boxed{\minus{}375}$ is our answer.


The perimeter of a right triangle is $ 15\plus{}20\sqrt{22}$. The Sum of the squares of all its sides is $ 396$. Find the area of the triangle....

hello, let $ a^2\plus{}b^2\plus{}c^2\equal{}396$ and $ a\plus{}b\plus{}c\equal{}15\plus{}20\sqrt{22}$, then we have $ 2c^2\equal{}396$ or $ c\equal{}3\sqrt{22}$, from here we get
$ a^2\plus{}b^2\equal{}198$
$ a\plus{}b\equal{}15\plus{}17\sqrt{22}$
But this system has no real solutions.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, let $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 396$ and $ a \plus{} b \plus{} c \equal{} 15 \plus{} 20\sqrt {22}$, then we have $ 2c^2 \equal{} 396$ or $ c \equal{} 3\sqrt {22}$, from here we get
$ a^2 \plus{} b^2 \equal{} 198$
$ a \plus{} b \equal{} 15 \plus{} 17\sqrt {22}$
But this system has no real solutions.
Sonnhard.[/quote]

Dr Sonnhard, You were right on those values though...

Hmm...but, I don't think it has no real solutions :)

[hide="My Solution."]

Since $ a \plus{} b \equal{} 15 \plus{} 17\sqrt {22}$

$ \Longrightarrow$                                                            $ a^2\plus{}b^2\plus{}2ab\equal{}225\plus{}510\sqrt{22}\plus{}6358$

$ \Longrightarrow$   $ 198 \plus{} 2ab\equal{}6583\plus{}510\sqrt{22}$

 $ \Longrightarrow$    $ ab\equal{}\frac{6385\plus{}510\sqrt{22}}{2}$

$ Area\triangle\equal{}\frac{ab}{2}\equal{}\frac{6385\plus{}510\sqrt{22}}{4}$

[/hide]

Well, if you look at it for a second, it's not possible. Assume that there are positive $ a,b$ that satisfy this. By Cauchy, $ (1\plus{}1)(a^2\plus{}b^2)\equal{}396\ge(a\plus{}b)^2\implies a\plus{}b\le6\sqrt{11}<6\sqrt{22}<15\plus{}17\sqrt{22}$.


In 3) 0, because first and second row linear dependent, the same reason in 1) about columns.
What concerns second you should use induction to prove that it is simething like this:
$ \frac{19f(5)\minus{}5f(19)}{19\minus{}5}$ or $ \minus{}\frac{19f(5)\minus{}5f(19)}{19\minus{}5}$ where $ f(x)$-is product of $ 2008\minus{}i\minus{}x$, $ i$ chandes from 0 to 118.

[b]Thanks you !!![/b]
[b]You had understand me !!![/b]


On each side of a unit square, an equilateral triangle of side length 1 is constructed. On each new side of each equilateral triangle, another equilateral triangle of side length 1 is constructed. The interiors of the square and the 12 triangles have no points in common. Let $ R$ be the region formed by the union of the square and all the triangles, and $ S$ be the smallest convex polygon that contains $ R$. What is the area of the region that is inside $ S$ but outside $ R$?

$ \textbf{(A)} \; \frac{1}{4} \qquad \textbf{(B)} \; \frac{\sqrt{2}}{4} \qquad \textbf{(C)} \; 1 \qquad \textbf{(D)} \; \sqrt{3} \qquad \textbf{(E)} \; 2 \sqrt{3}$

[quote="t0rajir0u"][hide="Solution"] I'm afraid I can't provide a diagram, but the equilateral triangles form trapezoids with side lengths $ 1, 1, 1, 2$ (half a unit hexagon) on each face of the unit square.  The four triangles "in between" these trapezoids are isosceles triangles with two side lengths $ 1$ and an angle of $ 30^{\circ}$ in between them, so the total area of these triangles (which is the area of $ S \minus{} R$) is

$ 4 \left( \frac {1}{2} \sin 30^{\circ} \right) \equal{} \sqrt {3}$

which makes the answer $ \boxed{D}$. [/hide][/quote]

[hide]Actually, $ \sin 30^{\circ} \equal{} \frac {1}{2}$, which the makes the answer $ 1 \equal{} C$.[/hide]

For the not-as-smart people like me, I calculated the area of S and subtracted R, which takes a substantially longer time.
[hide="my solution"]
First I took the smallest square that contains R, which has side length $ 1\plus{}\sqrt{3}$, and area $ 4\plus{}2\sqrt{3}$. Then I subtracted the four right isosceles triangles in the corners to find the area of the polygon. Each of their leg lengths is the side length of the large square, minus 2 and divided by 2, or $ \frac{\sqrt{3}\minus{}1}{2}$ Then, their total areas are $ 4\cdot\frac{(\frac{\sqrt{3}\minus{}1}{2})^2}{2}\equal{}\frac{4\minus{}2\sqrt{3}}{2}\equal{}2\minus{}\sqrt{3}$
Then we subtract that from the area of the large square to get the area of the polygon S: $ (4\plus{}2\sqrt{3})\minus{}(2\minus{}\sqrt{3})\equal{}2\plus{}3\sqrt{3}$
The area of the unit square is 1 and the area of all of the triangles is $ 12\cdot\frac{\sqrt{3}}{4}\equal{}3\sqrt{3}$, so when we subtract that from the area of S, we get
$ (2\plus{}3\sqrt{3})\minus{}(1\plus{}3\sqrt{3})$
which is 1.[/hide]

[quote="tenniskidperson3"]Here is a picture.[/quote]
Thanks for the picture... no wonder this problem initially looked so messy to me - I interpreted that the triangles are drawn in the INTERIOR the square instead of outside... I always seem to either misunderstand those contest problems or taking too long (like 2 min or more for lengthy word problems) to understand... or sometimes make some stupid mistake and then have to go back and take 5 min or even more to figure it out - but what make me angry is that I know all the steps to solve those problem but at the end not getting the right answer (plus stress on the time limit)... I guess this will get me into trouble if I'm really taking the test, so is there a way to avoid that (English was not my native language not until 2 years ago, so yeah...)?

For this problem, it says that the interior of the square and triangles have no points in common.
I think it is important to read the long word problems carefully and draw the correct diagram even if it takes a little longer. But everyone makes stupid mistakes(especially under pressure).

[quote="t0rajir0u"][hide="Solution"] I'm afraid I can't provide a diagram, but the equilateral triangles form trapezoids with side lengths $ 1, 1, 1, 2$ (half a unit hexagon) on each face of the unit square.  The four triangles "in between" these trapezoids are isosceles triangles with two side lengths $ 1$ and an angle of $ 30^{\circ}$ in between them, so the total area of these triangles (which is the area of $ S \minus{} R$) is

$ 4 \left( \frac {1}{2} \sin 30^{\circ} \right) \equal{} 1$

which makes the answer $ \boxed{C}$. [/hide][/quote]

Why is the area of the isoceles triangles 1/2 sin 30?

Area of a triangle is $bc\sin{30}$. All the sidelengths are 1, and the angle is 360-90-60*4 = 30.


integration
calculus
calculus computations


When a positive integer $ n$ is multiplied by 7, the result is 3 less than Maria's favorite number. When $ n\plus{}2$ is multiplied by 5, the result is 1 less than Maria's favorite number. What is Maria's favorite number?

Let Maria's favorite number be $ M$
We get the equations $ 7n \equal{} M\minus{}3$ and $ 5(n\plus{}2) \equal{} M\minus{}1$

Simplifying, $ 7n \plus{} 3 \equal{} M$ and $ 5n \plus{} 11 \equal{} M$
$ 7n \plus{} 3 \equal{} 5n \plus{} 11 \implies n \equal{} 4$
Plugging n back into the first equation, $ 7*4 \equal{} M\minus{}3$, and $ M \equal{} 31$
Checking with the second equation, $ 5(4\plus{}2) \equal{} M\minus{}1$, and $ M \equal{} 31$
So Maria's favorite number is $ \boxed{31}$


We have to conduct this survey for school.

It is located [url=http://www.mathprob.com/survey/]here[/url].

If you don't feel comfortable answering the questions, then don't take the survey.

EDIT: THE RESULTS WILL NOT BE POSTED.

You could've added some more options for each question and maybe more questions. :)

Well the other questions I had weren't fit for being put online :P

Please take the survey if you have not already. I must turn in the records by Tuesday!

I would object to body piercing AND getting tattoos.

For everyone who have not taken the survey yet, please take it! Results will be collected today at 11:59 PM EST. The survey will be taken down then.

Thank you


geometry
3D geometry
probability
search
function
number theory
prime factorization

#24 - Three distinct vertices of a cube are chosen at random. What is the probability that the plane determined by these three vertices contains points inside the cube?

My attempt:
[hide]Total number of planes = $ \binom{8}{3} \equal{} 56$ and for the plane to contain points inside the plain, all you need is for them NOT to be on the same face. So using complementery counting, we get the probability is
$ \frac{1\minus{}\binom{4}{3}(6)}{\binom{8}{3}}$ but it gives me a negative number, why??[/hide]

#25 - For $ k>0$, let $ I_k \equal{} 10...064$, where there are $ k$ zeroes between the $ 1$ and the $ 6$. Let $ N(k)$ be the number of factors of $ 2$ in the prime factorization of $ I_k$. What is the maximum value of $ N(k)$ ?

Er, it should be $ 1 \minus{} \frac {\binom{4}{3}(6)}{\binom{8}{3}}$, not $ \frac {1 \minus{} \binom{4}{3}(6)}{\binom{8}{3}}$.

Also, you should know that there is both a search function and a resource section.

[url=http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=43&year=2009]Here are the AMC 10A and 10B of this year[/url].


limit function trigonometry inequalities

[quote="madness"]Sorry i hadnt seen this till now and hence hadnt given the try earlier  :D

$ \lim_{x\to0} \frac {1}{x^{2}}$ - $ \cot^2x$

$ \leftrightarrow \lim_{x\to0} \frac {1}{x^{2}}$ - $ \frac {cos^2x}{sin^2x}$

$ \leftrightarrow \lim_{x\to0} \frac {sin^2x \minus{} x^2\cdot cos^2x}{x^{2}\cdot sin^2x}$

$ \leftrightarrow \lim_{x\to0} \frac {(x^2 \plus{} 1)\cdot sin^2x \minus{} x^2}{x^{2}\cdot sin^2x}$

Now i think we can apply series expansion and then L Hospital Rule  :D[/quote]

Do w/o using L'Hospital from now onwards. :D

Here is my solution
consider $ g(x) \equal{} \frac {1}{x^{2}} \minus{} \cot^{2} x$
observe that $ g$ is a even function ence it suffices to consider the RHL
so we have to get $ \lim_{x \to 0^{ \plus{} }} g(x) \equal{} L$
consider the function $ f(x) \equal{} \frac {1}{x^{2}}$
$ g(x) \equal{} \frac {(f(\tan x) \minus{} f(x))(\tan x \minus{} x)}{\tan x \minus{} x}$ by LMVT and that $ \tan x > x$ there esits a $ \alpha_{x} \in (x,\tan x)$
 such that $ f'(\alpha_{x}) \equal{} \frac {f(\tan x) \minus{} f(x)}{\tan x \minus{} x}$ we have $ g(x) \equal{} f'(\alpha_{x})(\tan x \minus{} x)$
also use the fact that $ \frac {1}{x^{3}}$ is decreasing for $ x \in \mathbb{R}^{ \plus{} }$
hence we get
$ \frac {2(\tan x \minus{} x)}{x^{3}} \geq f'(\alpha_{x})(\tan x \minus{} x) \geq \frac {2(\tan x \minus{} x) }{\tan^{3} x}$
also note that as $ x \to 0^{ \plus{} }$ $ \alpha_{x} \to 0^{ \plus{} }$
hence taking both limits from both the side of inequality and noting that $ \lim_{x \to 0^{ \plus{} }}\frac {\tan^{3} x}{x^{3}} \equal{} 1$
we get that $ L \equal{} \lim_{x\to 0^{ \plus{} }}\frac {2(\tan x \minus{} x)}{x^{3}}$
jsut use L'H now this gives $ L \equal{} \lim_{x\to 0^{ \plus{} }}\frac {2(\sec^{2} x \minus{} 1)}{3x^{2}} \equal{} \frac {2}{3}$

[url=http://www.mathlinks.ro/viewtopic.php?t=137241]Here[/url] is my other solution using my method to a wonderful problem

Is my method right is the question?  :D  :D

actually shreyas wanted a solution so i posted that...
as far as ur solution why don't u continue with LH and show the result ans is $ \frac{2}{3}$...ofcourse  LH is bound to give u the answer but it's time taking and takes a hell lot of time

obviously not  :D  I am a bit fast on mental math (dont count me as boasting  :P ) so the answer is very easy to think and do within a matter of 5 mins rather than assuming a fn. It may take many steps if written in detail but the idea is very very simple so..........  :D

not all are as 'gifted' as u in mental arithmetic... :P
i wrote my proof 'long'(which anyways is shorter than urs) so that people can know another method...for me it would have been a 3 step result... :D

[quote="Cyber-Shreyas"]Hey Pardesi, I saw this that day itself. I was more impressed by the use of your method in the other sum. That was very nice!! I again wish MO had calculus.

And you don't need to tell me to check my PM's. I regularly check, though I might take sometime to reply.

And I didn't want to interrupt when two 2 genii of mathematics were boasting.[/quote]

Yes i agree that his method is excellent, but dont u think it is less likely that for an idiot like myself, it is very tough to think of although thou shreyas (being brilliant) might find it easier.  :D

[quote="Cyber-Shreyas"]Hey Pardesi, I saw this that day itself. I was more impressed by the use of your method in the other sum. That was very nice!! I again wish MO had calculus.

And you don't need to tell me to check my PM's. I regularly check, though I might take sometime to reply.

And I didn't want to interrupt when two 2 genii of mathematics were boasting. And to genius rohit, it doesn't take your brilliance in mathematics to post the method you posted. Even lesser mortals  (Like myself) can do that.[/quote]
thanks da...actually the idea for this came from my other post...
sorry i told u so because u hadn't read the pm it was in outbox :(


You can find all of it [url=http://en.wikipedia.org/wiki/Set_theory#Basic_concepts]here[/url]. :) To remember the two that you asked, I just think of the U as [u]u[/u]nion and the the upside down one is uh...not the union, so the intersection. :P

U stands for Union. The union of sets $ \{a, b\}$ and $ \{c, d\}$ is $ \{a, b, c, d \}$. Say $ a \in S$ and $ P$. The union of S and P would only list $ a$ once.

The upside down U stands for intersection. The intersection of sets $ \{a, b, c \}$ and $ \{c, d \}$ is $ \{ c \}$, the element they have in common.

EDIT: Beaten

[quote="ernie"]Can someone explain what the U and the upside-down U, etc. are?  I never learned this...[/quote]

The symbol $ \cup$ means the [b]union[/b] of two sets, i.e. $ A\cup B \equal{}\{x\, | x\in A\,\textbf{or}\,x\in B\}$.

Likewise the symbol $ \cap$ means the [b]intersection[/b] of two sets, i.e. $ A\cap B \equal{}\{y\, |y\in A\,\textbf{and}\,y\in B\}$.

The expression $ z\in C$ is read as "$ z$ is a member of, $ z$ belongs to, $ z$ is in, or $ z$ is an element of, the set $ C$" or any equivalent phase.

Therefore, for example, the sentence $ \{y\, |y\in A\,\text{and}\,y\in B\}$ is read as, "The set of $ y$'s such that (any particular) $ y$ belongs to $ A$ and to $ B$."


inequalities
combinatorics unsolved
combinatorics

Suppose we have a sequence $x_1, \ldots, x_n$ such that two conditions hold:

1. For every 7 consecutive  terms $x_i, x_{i+1}, \ldots, x_{i+6}$ we have $x_i+x_{i+1}+\ldots+x_{i+6}>0$.

2. For every 12 consecutive terms $x_i, \ldots, x_{i+11}$ we have $x_i+\ldots+x_{i+11}<0$.

Find the largest possible value of $n$.

It is a very well known idea: Suppose the sequence has at least $18$ terms , then consider the sums:

$x_1+x_2+x_3+...x_{12}$
$x_2+x_3+x_4+...x_{13}$
$x_3+x_4+x_5+...x_{14}$
$.$
$.$
$.$
$x_7+x_8+x_9+...x_{18}$

Suming rows it is negative, but suming columns it is positive, contradiction, so we get it has at most $17$ terms.

But we also have to give an example of such a sequence with 17 terms.

I think the idea for this probem is something like this (to build an example):
Let $s_i = \sum_{j=1}^{n}x_i$. Then we have inequalities like :
$s_i<s_{i+6}$ and $s_i>s_{i+11}$. So, starting from $s_1$ and adding or substracting 6 or 11, we will get a contradiction. But this way, we can also build an easy example, just take some numbers that would respect the ordering for $s_i$ and find $x_i$ from them...
I hope it was clear!

[quote="Valiowk"]There was a very similar question on [url=http://www.kalva.demon.co.uk/apmo/asoln/asol925.html]APMO 92/5[/url] where the sum of every 7 consecutive terms was positive and the sum of every 11 consecutive terms was negative.  For that question, a construction could be obtained by constructing a sequence appropriately e.g. $aabaaabaabaaabaa$ and then using simultaneous equations, for instance.

I've tried to use a similar technique for this question but it doesn't seem to work, mainly because I can't find an initial sequence which has the same number of $a$'s and $b$'s for every consecutive sequence of 7 and 12 terms.[/quote]
$ababaabababaababa$

We may assume $a=30$ and $b=-41$.

:) Note to self: Do not attempt to construct sequences at 1 am. :blush:


[quote="computer_nuke"][quote="guile"]Find the solution sets of the following.
a.)x + |x – 2| = 1 + |x|
b.)| x – 3| + | x – 1| < 4
c.)|(x + 5)/(x – 3)| < 1[/quote]

A is x = 1, x = -1 and x = 3

I'm working on other ones right now  :)[/quote]

I already have the answer and the solution for 1 (But none for 2 and 3). I am just looking for other solutions. Can you show your solution for number 1 please?

[quote="guile"]Find the solution sets of the following.
1.  $x+|x-2| = 1+|x|$
2.  $|x-3|+|x-1| < 4$
3.  $\left|\frac{x+5}{x-3}\right| < 1$[/quote]

1. $x=0$ isn't a solution. Suppose $x>0$. Then $|x|=x$, so $|x-2|=1$. Therefore, $x=1, 3$. Suppose $x<0$. Then $|x|=-x$ and $|x-2| = 2-x$, so $x+2-x=1-x$, so $x=-1$. Thus $x\in\{-1,1,3\}$.

2. Because, for all real numbers $a,b$ we have that $|a|+|b| \geq |a+b|$, we have $4 > |x-3|+|x-1| \geq |x-3+x-1| = |-4| = 4$. Hence, no solutions.

3. Because, for all real numbers $a,b$ we have that $\left|\frac{a}{b}\right| = \frac{|a|}{|b|}$, we have that the inequality is equivalent with $|x+5| < |x-3|$. Square both sides to get $x^{2}+10x+25 < x^{2}-6x+9$, so $x<-1$.

If I am not wrong, shouldn't these problems be in the "Intermediate" section?

[quote="junggi"][quote="Kurt Gödel"]
2. Because, for all real numbers $a,b$ we have that $|a|+|b| \geq |a+b|$, we have $4 > |x-3|+|x-1| \geq |x-3+x-1| = |-4| = 4$. Hence, no solutions.[/quote]

$|x-3+x-1| = |-4|$

How did that happen? :huh:
Shouldn't it be:
$|2x-4|$?

Also counterexample to your answer:
$x=2$ works.[/quote]

Auwch. Yeah, it should be $|x-3-(x-1)| = |-2| = 2$, but that's irrelevant...
Let $x \geq 3$: $|x-3|+|x-1| = 2x-4 < 4$, so $x < 4$.
Let $1 \leq x < 3$: $|x-3|+|x-1| = 2 < 4$. Always true.
Let $1 > x$: $|x-3|+|x-1| = 4-2x < 4$, so $0<x$.

So, $0 < x < 4$.


let M,N are 2 point in ABC tringle that <MAB=<NAC and <MBA=<NBC
prove that :
AM.AN/AB.AC + BM.BN/AB.BC + CM.CN/AC.BC =1

Question: Do you know how many times this problem has been posted before???
Answer: Many times!

Best regards,
sunken rock

I have posted this three days ago you can see the solution.

http://www.mathlinks.ro/viewtopic.php?t=18479
Lock this topic!


induction
logarithms
function
inequalities unsolved
inequalities

Positive reals $x_{1},x_{2},\ldots,x_{n}$ satisfy the following condition:
$\sum_{i=1}^{n}{x_{i}}\leq \sum_{i=1}^{n}{x_{i}^{2}}$.Prove that
$\sum_{i=1}^{n}{x_{i}^{t}}\leq \sum_{i=1}^{n}{x_{i}^{t+1}}$ for each real $t >1$.

just use cauchy and proceed by induction...
use the notation:
S_k=x1^k+...+xn^k;
the condition says S_1<=S_2;
multiply it by S_3, use cauchy on the LHS, and you'll get S_2<=S_3;
proceed by induction and the result follows

But we can't apply induction because t is not an integer

Let $f(t)=\sum_{i=1}^{n}x_{i}^{t+1}-\sum_{i=1}^{n}x_{i}^{t}$
Then:
$f'(t)=\sum_{i=1}^{n}\ln{x_{i}}x_{i}^{t+1}-\sum_{i=1}^{n}\ln{x_{i}}x_{i}^{t}= \sum_{i=1}^{n}\ln{x_{i}}(x_{i}^{t+1}-x_{i}^{t})$
Now it's easy to see that each term of this sum is nonnegative, since if $x_{i}\geq 1$ then $\ln{x_{i}}\geq 0$ and $x_{i}^{t+1}\geq x_{i}^{t}$ and if $x_{i}< 1$ then $\ln{x_{i}}< 0$ and $x_{i}^{t+1}< x_{i}^{t}$. So, because for $t>1$ we have $f'(t) \geq 0$ we see that $f$ is non-decreasing on $(1,+\infty)$ and since $f(1) \geq 0$ conclusion follows.

Let $f: R\to R$, $f(t)=\sum_{k=1}^{n}x_{k}^{t}$
From CBS $\rightarrow$ $(\sum_{k=1}^{n}x_{k})^{2}\leq n\cdot(\sum_{k=1}^{n}x_{k}^{2})$ .But we also know that $\sum_{k=1}^{n}x_{k}\leq \sum_{k=1}^{n}x_{k}^{2}$ .From this two relations $\rightarrow$ $f(1)\leq n = f(0)$ .We also kwno that $f(1)\leq f(2)$ and because f is convex $\rightarrow$ $f(1)$ is the minimum value of the function $\rightarrow$ f is increasing on $[1,\infty)$ Therefore $f(t+1)\geq f(t)$


Someone once asked me, "What keeps you motivated to do math?

This is a question I have found difficult to render an answer to. I mean its mostly because I like the number 6 but otherwise there is little else to say.

On a more serious note, it interests me to know what keeps many of you students more interested in math. I mean, is there a piece of literature you read that sparked your interest in math? or... do you like the feel of solving a geometry problem at 2 in the A.M. ? For me, its definitely the latter, and the number 6.

Cheers,
The Idiot without a village.

For one thing, if it weren't for mathematics, I would be rather a useless person. I don't think I'm particularly outstanding at anything else, probably because I haven't worked that hard at anything else. But the reason that I work hard and always want to know more is that I find it exceptionally beautiful and surprising. I get bored or tired of everything else eventually, but I am never sick of math. I can't say that there is any particular work that made me so interested in math. Perhaps the Joy of Mathematics books by Theoni Pappas helped a lot since I was given the first one as a birthday present when I was about 8 years old since people around me saw that I liked to do math. There is an infinite amount of mathematics out there. It's our job to uncover as much of it as we can. Eventually we may find some theory of everything that answers all our questions about physics, for example, but the same can never be true in mathematics. I guess that is what keeps me going reading (currently) a book on the Euler-Mascheroni Constant at 1:00 in the morning, when everyone else in my house is asleep or solving an interesting problem that someone posts to this site. I don't know if that was what you wanted.

"Numbers have always fascinated people.
         At least, they have always fascinated some people."
           - Underwood

"Accurate reckoning. The entrance into the knowledge of all existing things and all obscure secrets" - Rhind Papyrus ca 1800 BC

Easy to answer this one. Because its FUN!!! Seriously. Thats the one and only reason. And basically this problem solving is so open, every single question is different and it can never get repetitive.

I agree with TripleM - furthermore, math has so many different fields that one can always find an agreeable niche. =)

Mathematics provides a meaning to my life.  What else do I need to say?

i agree with every post so far. except the number 6 doesnt keep me going.  its the thrill and exhileration of the competitions and the challenge. during the school year i would sometimes literally beg my parents to let me keep taking more tests and solving more problems into the wee hours of the morning. sure i would still be good at school and stuff without my math skittles, but i wouldnt be outstanding in anything. better in somethings. but not the best.  of course, i would work to be outstanding at something. most likely math. hehe. i think im confusing some peeps. there're lots of reasons. i just cant think of them right now.  ill post other stuff if i think of them. im really tired. i slept at 7 am. ack. ajods!

While we're on this topic.. has anyone worked out what to reply when someone asks "why are you so good at maths / when did you first get good at maths" or something along those lines. Seeing both my parents are maths teachers they usually think thats the reason, which I don't think is (one of them only became a maths teacher a year or two ago anyway!). Usually I say something along the lines of enjoying it and such.. how about you?

i dont want to sound offensive or provoking or summat like that Mr. MMM, but why in New Zealand do they call what we call here in the U.S. math, maths?  i know that in India where me padre and madre are from they call it maths also.  is it just an American thing? just like the metric system and such.  just wondering. like some of my indian math books say. maths on it. i find it cool. anywho

It has to do with how it is abbreviated. As far as I can tell, ``mathematics'' is always a singular noun. So in some places, such as the US, it is abbreviated as ``math'' since singular nouns don't usually have an `s' at the end. But in other countries, the idea is to preserve the nature of the original word, so the final `s' remains part of the abbreviation.

now that my question has been answered. i can go back to being a SUPER FREAK.........SUPER FREAK!

Yeah - I've travelled frequently (moved to the US from Singapore 2 years ago) and as far as I can tell, the US of A is the only place which says 'math' instead of 'maths'...

[color=white]Nobody else except for me and Idiot_Without_A_Village likes the number 6?  Or 692?  Personally that keeps my math interest going.  I also like fractals......[/color]

it does help if youre actually good. i dunno. just some calcium for thought. oh yeah. is it actually true according to the song that "girls dont like boys, girls like cars and monay!"

Well, 6 is the smallest perfect number, so it is certainly a special number.

Actually the fifth perfect number is 33550336, and the sixth is 8589869056, and yes, I did memorize them that far.

Oops, little slip of the key there.  But wow, I'm not [i]that[/i] fascinated by perfect numbers.  Still cool little things.

Oh, and the 7th is 137438691328
And the 8th, 2305843008139952128
Did you memorize those?

Wait, did I say LITTLE things?

I only memorized up to the sixth because once I saw the sixth in a book and showed it to a friend, and he found a neat mnemonic:

85 89 (differ by 4)
86 90 (add one to each above, thus also differ by 4)
56

(not that it's all that hard to memorize it anyway).

Cool.  8-) Mnemonics are good for certain school subjects requiring like <ahem> history.

Later,
mathFANATIC

P.S. Just wondering... what else did you memorize?  I find powers of two a very good thing because whenever I get mad, or sad, or anything not resembling a usual, blank expression, I start saying the powers under my breath.   :)

Anyone ever tried memorising digits of pi? I went up to 200 once, for some reason, forgot most of them now.

History is fun. Geography, on the other hand, would require some mnemonic devices. Hey Simon, Grace tells me you know pi up to 700 digits...really? Darn. I used to memorize powers of two...for no especial reason.

you too try to memorize twos?  ooops, too many similies make many smilies  :)   :)  :)

Oh well, that was my feeble attempt at hyoomur.  BTW, do any of you enjoy rambling about completely random topics?  Because you ought to know that I do... like that time I started rambling about this friend of mine whose pet hippopotamus was actually related to Nate, who happened to live in the same jungle as this other hippo whose grandfather's owner's best friend's nephew's second cousin, once removed was Napolean's uncle... :P

...

Now, mathfanatic, just step down this corridor with me...we've got some great accomodation just for you...you get your own room...with padded walls...we pass you meals through this barred window and we even supply you with a beautiful leather jacket. Sounds good?

Look closer at the Nothing below... you may see something.

[hide]Hey, I resent that... stop it, you're making me feel emotional so now I have to go 1 2 4 8 16 32 64 128 256 512 1024 2048 4096 8192 16384 32768 65536 131072 262144 524288 1048576 2097152 4194304...

Better.  Now what were you saying?  You like jalapeno bagels with chocolate syrup too?  Because if they don't have jalapeno bagels with chocolate syrup I'm not going to that "accommodation" whatever that is.  Besides, I'm too busy reading Douglas Adams and Lewis Carroll to have time to go.  Sorry.

BTW, does that mean that you run an asylum?  Because I personally think that would be a very cool job.  You get to socialize with all the wackos, get Michael Jackson's autograph, and everything!  Hey, that sounds like fun...

P.S. My apologies to any MJ fans...[/hide]

When I was in 5th grade I knew *precisely* 744 digits of :pi:. I now know about 100e digits (or 271). Actually I think it might be slightly more than that, but it's not really that many.

It occurred to me that we were writing in the wrong thread.  After all, jalapeno bagels with chocolate syrup. "nerd stalkers," and digits of pi have [i]nothing[/i] to do with the "Looking for Answers" thread, which is supposed to be for questions and answers to math problems.  In the future, we will post only math problems in here... agreed?  ComplexZeta?  Hann?

Jojobird, fractals are definitely one of the reasons why I keep on reading about math. Other than that, I just like math. I like numbers and I like pushing my brain to the limit. Only math can do something like that.


calculus
integration
calculus computations

How can i compute the integral :
$\int \frac{e^x(x^2-2x+1)}{(x^2+1)^2} dx$

[hide]You can use by parts (you have to use it a few times) and partial fraction decomposition.  The answer is $\frac{e^x}{x^2+1}$, which makes sense if you think about the quotient rule in calculus.[/hide]