geometry
3D geometry
inequalities
number theory
greatest common divisor
triangle inequality
algebra

[quote="immodestius"]Can you solve the equation $ x^2 \equal{} y^3 \minus{} 4$ in integers?[/quote]

(x,y)=(-2,2),(2,2),(11.5)............ :wink:
mr mohsmed

[hide]
First off, $ x$ cannot be 0 mod 4. Otherwise, we have $ x \equal{} 4n$ for some $ n$ for some $ n$, and $ 16n^2 \plus{} 4 \equal{} y^3$, implying that $ y \equal{} 2m$ for some integer $ m$. Substituting and simplifying, we have $ 4n^2 \plus{} 1 \equal{} 2m^3$, or $ 1 \equal{} 2m^3 \minus{} 4n^2$. But the RHS is even, so this is impossible.

Now, it is well-known that the ring of Gaussian integers is a unique factorization domain. Our equation is $ y^3 \equal{} x^2 \plus{} 4 \equal{} (x \plus{} 2i)(x \minus{} 2i)$.

Let $ d \equal{} \gcd(x \plus{} 2i, x \minus{} 2i)$. I claim that $ d \equal{} 1$, $ d \equal{} 2$, or $ d \equal{} 4$, up to multiplication by a unit. Indeed, we have that $ d | x \plus{} 2i \minus{} (x \minus{} 2i) \equal{} 4i$, so $ d | 4$. Furthermore, since $ x$ is an integer, we get that $ d$ must be real. Thus, $ d \equal{} 4$, $ d \equal{} 2$, or $ d \equal{} 1$. However, if $ d \equal{} 4$, then $ 4 | 2i$, a contradiction, so we must have that either $ d \equal{} 1$ or $ d \equal{} 2$.

If $ d \equal{} 1$, since $ x \plus{} 2i$ and $ x \minus{} 2i$ both multiply to perfect cubes in $ \mathbb{Z}[i]$, we get that $ x \plus{} 2i \equal{} w^3$ and $ x \minus{} 2i \equal{} z^3$, where $ w$ and $ z$ are nonzero Gaussian integers. We get that $ 4i \equal{} z^3 \minus{} w^3$, or $ r \equal{} |z \minus{} w\parallel{}z^2 \plus{} zw \plus{} w^2|$. Thus, we have that $ |z \minus{} w| \equal{} 1$ and $ |z^2 \plus{} zw \plus{} w^2| \equal{} 4$, $ |z \minus{} w| \equal{} 2$ and $ |z^2 \plus{} zw \plus{} w^2| \equal{} 2$, or $ |z \minus{} w| \equal{} 4$ and $ |z^2 \plus{} zw \plus{} w^2|$ = 1.

By the triangle inequality, $ |z^2 \plus{} zw \plus{} w^2| \equal{} |(z \minus{} w)^2 \plus{} 3zw| \leq |3zw| \plus{} |z \minus{} w|^2$. This tells us that $ 4 \leq 1 \plus{} |3zw| \equal{} 1 \plus{} 3|zw| \leq 4$ in the first equation, implying that $ w$ and $ z$ must both be units. But the difference of two units cannot be a unit, so $ |z \minus{} w| \equal{} 1$ is impossible. Therefore, we must have $ |z^2 \plus{} zw \plus{} w^2| \equal{} 1$ and $ |z \minus{} w| \equal{} 4$ or $ |z^2 \plus{} zw \plus{} w^2| \equal{} |z \minus{} w| \equal{} 2$.

If $ |z \minus{} w| \equal{} 4$, then $ z \equal{} w \plus{} 4i$, $ z \equal{} w \minus{} 4i$, $ z \equal{} w \plus{} 4$, or $ z \equal{} w \minus{} 4$ (a circle centered at $ z$ on the complex plane only passes through those 4 Gaussian integers.)

Recall now that $ z^3 \plus{} w^3 \equal{} 2x$, so $ z^3$ and $ w^3$ must, in fact, have a real sum. This, coupled with $ |z^2 \plus{} zw \plus{} w^2| \equal{} 1$, should show that this equation has no solutions.

In the case that $ |z \minus{} w| \equal{} |z^2 \plus{} zw \plus{} w^2| \equal{} 2$, we again have $ z \equal{} w \plus{} 2i$, $ z \equal{} w \minus{} 2i$, $ z \equal{} w \plus{} 2$, or $ z \equal{} w \minus{} 2$. We find the solutions $ (\pm 11, 5)$ from this system.

(I'm too lazy to actually do it out, sorry. :/)

Now, we have the case that $ d \equal{} 2$. Therefore, $ x$ must be even, i.e., $ x \equal{} 2u$ for some integer $ u$. Clearly, $ \gcd(x \plus{} 2i, x \minus{} 2i) \equal{} \gcd(2u \plus{} 2i, 2u \minus{} 2i) \equal{} 2 \gcd(u \plus{} i, u \minus{} i)$; since the gcd of $ x \plus{} 2i$ and $ x \minus{} 2i$ is 1, we get that $ \gcd(u \plus{} i, u \minus{} i)$ is 1.

We observe that $ u \plus{} i$ and $ u \minus{} i$ multiply to a number twice a perfect cube. In other words, $ (u \plus{} i)(u \minus{} i) \equal{} 2v^3$, for some complex $ v$. Therefore, $ u \plus{} i$ and $ u \minus{} i$ must both be two times, $ 1 \plus{} i$ times, $ 1 \minus{} i$ times, or be equal to a perfect cube. Neither $ u \plus{} i$ nor $ u \minus{} i$ could be twice a perfect square, as then $ 2 | u \plus{} i$, implying that $ 2 | i$, a contradiction. Furthermore, if any one of them were a plain perfect cube, the other would have to be twice a perfect cube, so we get that $ u \plus{} i$ and $ u \minus{} i$ must be $ 1 \plus{} i$ times and $ 1 \minus{} i$ times some perfect cube, not necesarily respectively. We will first consider the case in which $ (1 \plus{} i)w^3 \equal{} u \plus{} i$ and $ (1 \minus{} i)z^3 \equal{} u \minus{} i$.

Dividing our first equation by $ 1 \plus{} i$, we see that $ w^3 \equal{} \frac {x \plus{} 1}{2} \minus{} \frac {x \minus{} 1}{2}i$. Let $ w \equal{} a \plus{} bi$. Expanding and equating real/imaginary parts, we have that $ a^3 \minus{} 3ab^2 \equal{} \frac {x \plus{} 1}{2}$ and $ 3a^2 b \minus{} b^3 \equal{} \frac {x \minus{} 1}{2}$. Subtracting our two equations, we get that $ a^3 \minus{} 3ab^2 \minus{} 3a^2 b \plus{} b^3 \equal{} 1$, that is, $ (a \plus{} b)(a^2 \minus{} ab \plus{} b^2) \minus{} 3ab(a \plus{} b) \equal{} 1$, or $ (a \plus{} b)((a \plus{} b)^2 \minus{} 6ab) \equal{} 1$. Let $ p \equal{} a \plus{} b$ and $ q \equal{} ab$. Then $ p(p^2 \minus{} 6q) \equal{} 1$. Hence, $ p \equal{} p^2 \minus{} 6q \equal{} 1$ or $ p \equal{} p^2 \minus{} 6q \equal{} \minus{} 1$. In the second case, we have $ 1 \minus{} 6q \equal{} \minus{} 1$, so $ 6q \equal{} 2$, which clearly yields no solutions. The first equation gives $ 1 \minus{} 6q \equal{} 1$, so $ q \equal{} 0$, from which it follows that $ a \equal{} 1$ and $ b \equal{} 0$ or $ a \equal{} 0$ and $ b \equal{} 1$, that is, $ u \equal{} 1$ or $ u \equal{} \minus{} 1$.

Our only other possible candidates are obtained from the equation $ u \plus{} i \equal{} (1 \minus{} i)w^3$ and $ u \minus{} i \equal{} (1 \plus{} i)z^3$. However, letting $ u' \equal{} \minus{} u$, $ w' \equal{} \minus{} w$, and $ z' \equal{} \minus{} z$, it turns out that this "new" system of equations is exactly our old one, with solutions $ u' \equal{} 1, \minus{} 1$. However, this once again just leaves $ u \equal{} 1, \minus{} 1$ as our possible candidates, so our only solutions to this Diophantine are $ (x, y) \equal{} (2, 2), ( \minus{} 2, 2)$.
[/hide]
Dang apparently I'm missing a solution? I'll see what I did later. Excuse me for the iffy part, which may well be the thing that makes my proof fall apart...

EDIT: So I made a pretty stupid mistake in my proof before... I'm still a bit lazy about my other two cases when $ (x \plus{} 2i, x \minus{} 2i) \equal{} 1$, but hopefully it shouldn't be too hard to do. As in, we have finitely many cases to check. And they're quadratics.

[quote="Zhero"][hide]
First off, $ x$ cannot be 0 mod 4. Otherwise, we have $ x \equal{} 4n$ for some $ n$ for some $ n$, and $ 16n^2 \plus{} 4 \equal{} y^3$, implying that $ y \equal{} 2m$ for some integer $ m$. Substituting and simplifying, we have $ 4n^2 \plus{} 1 \equal{} 2m^3$, or $ 1 \equal{} 2m^3 \minus{} 4n^2$. But the RHS is even, so this is impossible.

Now, it is well-known that the ring of Gaussian integers is a unique factorization domain. Our equation is $ y^3 \equal{} x^2 \plus{} 4 \equal{} (x \plus{} 2i)(x \minus{} 2i)$.

Let $ d \equal{} \gcd(x \plus{} 2i, x \minus{} 2i)$. I claim that $ d \equal{} 1$, $ d \equal{} 2$, or $ d \equal{} 4$, up to multiplication by a unit. Indeed, we have that $ d | x \plus{} 2i \minus{} (x \minus{} 2i) \equal{} 4i$, so $ d | 4$. Furthermore, since $ x$ is an integer, we get that $ d$ must be real. Thus, $ d \equal{} 4$, $ d \equal{} 2$, or $ d \equal{} 1$. However, if $ d \equal{} 4$, then $ 4 | 2i$, a contradiction, so we must have that either $ d \equal{} 1$ or $ d \equal{} 2$.

If $ d \equal{} 1$, since $ x \plus{} 2i$ and $ x \minus{} 2i$ both multiply to perfect cubes in $ \mathbb{Z}[i]$, we get that $ x \plus{} 2i \equal{} w^3$ and $ x \minus{} 2i \equal{} z^3$, where $ w$ and $ z$ are nonzero Gaussian integers. We get that $ 4i \equal{} z^3 \minus{} w^3$, or $ r \equal{} |z \minus{} w\parallel{}z^2 \plus{} zw \plus{} w^2|$. Thus, we have that $ |z \minus{} w| \equal{} 1$ and $ |z^2 \plus{} zw \plus{} w^2| \equal{} 4$, $ |z \minus{} w| \equal{} 2$ and $ |z^2 \plus{} zw \plus{} w^2| \equal{} 2$, or $ |z \minus{} w| \equal{} 4$ and $ |z^2 \plus{} zw \plus{} w^2|$ = 1.

By the triangle inequality, $ |z^2 \plus{} zw \plus{} w^2| \equal{} |(z \minus{} w)^2 \plus{} 3zw| \leq |3zw| \plus{} |z \minus{} w|^2$. This quickly eliminates the second equation, and tells us that $ 4 \leq 1 \plus{} |3zw| \equal{} 1 \plus{} 3|zw| \leq 4$ in the second equation, implying that $ w$ and $ z$ must both be units. But the difference of two units cannot be a unit, so $ |z \minus{} w| \equal{} 1$ is impossible. Therefore, we must have $ |z^2 \plus{} zw \plus{} w^2| \equal{} 1$ and $ |z \minus{} w| \equal{} 4$.

If $ |z \minus{} w| \equal{} 4$, then $ z \equal{} w \plus{} 4i$, $ z \equal{} w \minus{} 4i$, $ z \equal{} w \plus{} 4$, or $ z \equal{} w \minus{} 4$ (a circle centered at $ z$ on the complex plane only passes through those 4 Gaussian integers.)

Recall now that $ z^3 \plus{} w^3 \equal{} 2x$, so $ z^3$ and $ w^3$ must, in fact, have a real sum. This, coupled with $ |z^2 \plus{} zw \plus{} w^2| \equal{} 1$, should show fairly easily that $ d \equal{} 1$ yields no solutions to our original equation (though I may be wrong; I'm currently just too lazy to do this part now. :/)

Now, we have the case that $ d \equal{} 2$. Therefore, $ x$ must be even, i.e., $ x \equal{} 2u$ for some integer $ u$. Clearly, $ \gcd(x \plus{} 2i, x \minus{} 2i) \equal{} \gcd(2u \plus{} 2i, 2u \minus{} 2i) \equal{} 2 \gcd(u \plus{} i, u \minus{} i)$; since the gcd of $ x \plus{} 2i$ and $ x \minus{} 2i$ is 1, we get that $ \gcd(u \plus{} i, u \minus{} i)$ is 1.

We observe that $ u \plus{} i$ and $ u \minus{} i$ multiply to a number twice a perfect cube. In other words, $ (u \plus{} i)(u \minus{} i) \equal{} 2v^3$, for some complex $ v$. Therefore, $ u \plus{} i$ and $ u \minus{} i$ must both be two times, $ 1 \plus{} i$ times, $ 1 \minus{} i$ times, or be equal to a perfect cube. Neither $ u \plus{} i$ nor $ u \minus{} i$ could be twice a perfect square, as then $ 2 | u \plus{} i$, implying that $ 2 | i$, a contradiction. Furthermore, if any one of them were a plain perfect cube, the other would have to be twice a perfect cube, so we get that $ u \plus{} i$ and $ u \minus{} i$ must be $ 1 \plus{} i$ times and $ 1 \minus{} i$ times some perfect cube, not necesarily respectively. We will first consider the case in which $ (1 \plus{} i)w^3 \equal{} u \plus{} i$ and $ (1 \minus{} i)z^3 \equal{} u \minus{} i$.

Dividing our first equation by $ 1 \plus{} i$, we see that $ w^3 \equal{} \frac {x \plus{} 1}{2} \minus{} \frac {x \minus{} 1}{2}i$. Let $ w \equal{} a \plus{} bi$. Expanding and equating real/imaginary parts, we have that $ a^3 \minus{} 3ab^2 \equal{} \frac {x \plus{} 1}{2}$ and $ 3a^2 b \minus{} b^3 \equal{} \frac {x \minus{} 1}{2}$. Subtracting our two equations, we get that $ a^3 \minus{} 3ab^2 \minus{} 3a^2 b \plus{} b^3 \equal{} 1$, that is, $ (a \plus{} b)(a^2 \minus{} ab \plus{} b^2) \minus{} 3ab(a \plus{} b) \equal{} 1$, or $ (a \plus{} b)((a \plus{} b)^2 \minus{} 6ab) \equal{} 1$. Let $ p \equal{} a \plus{} b$ and $ q \equal{} ab$. Then $ p(p^2 \minus{} 6q) \equal{} 1$. Hence, $ p \equal{} p^2 \minus{} 6q \equal{} 1$ or $ p \equal{} p^2 \minus{} 6q \equal{} \minus{} 1$. In the second case, we have $ 1 \minus{} 6q \equal{} \minus{} 1$, so $ 6q \equal{} 2$, which clearly yields no solutions. The first equation gives $ 1 \minus{} 6q \equal{} 1$, so $ q \equal{} 0$, from which it follows that $ a \equal{} 1$ and $ b \equal{} 0$ or $ a \equal{} 0$ and $ b \equal{} 1$, that is, $ u \equal{} 1$ or $ u \equal{} \minus{} 1$.

Our only other possible candidates are obtained from the equation $ u \plus{} i \equal{} (1 \minus{} i)w^3$ and $ u \minus{} i \equal{} (1 \plus{} i)z^3$. However, letting $ u' \equal{} \minus{} u$, $ w' \equal{} \minus{} w$, and $ z' \equal{} \minus{} z$, it turns out that this "new" system of equations is exactly our old one, with solutions $ u' \equal{} 1, \minus{} 1$. However, this once again just leaves $ u \equal{} 1, \minus{} 1$ as our possible candidates, so our only solutions to this Diophantine are $ (x, y) \equal{} (2, 2), ( \minus{} 2, 2)$.
[/hide]
Dang apparently I'm missing a solution? I'll see what I did later. Excuse me for the iffy part, which may well be the thing that makes my proof fall apart...[/quote]

what about ( 11 , 5 )
 :wink:

[quote="almonkez"]what about ( 11 , 5 )
 :wink:[/quote]
Yeah, that's the missing solution. (And (-11, 5) of course.) But where's the hole in my proof?

The thing is, $ 11 \plus{} 2i$ and $ 11 \minus{} 2i$ are relatively prime, and their product is a perfect cube in the Gaussian integers. Despite this, neither of them are perfect cubes in the Gaussian integers...?

Would anyone care to tell me why this is? I'm really confused...

$ 11 \plus{} 2i$ and $ 11 \minus{} 2i$ are cubes.

 $ 11 \plus{} 2i \equal{} (\minus{}2i \minus{} 1)^3$

I fail. Sorry.  :blush:

I'll fix my proof shortly.

m sorry..but it shall be more better if you could solve it in an elementary way..
i shall post a proof soon..in a hurry presently..~`

[hide]$ x^{2} \equal{} y^{3} \minus{} 4\Leftrightarrow x^{2} \minus{} 4 \equal{} y^{3} \minus{} 8\Leftrightarrow x^2 \minus{} 4 \equal{} (y \minus{} 2)(y^{2} \plus{} 2y \plus{} 4)$
$ \Leftrightarrow \frac {x^2 \minus{} 4}{y \minus{} 2} \equal{} (y \plus{} 1)^2 \plus{} 3; \Leftrightarrow (y \plus{} 1)^2 \equal{} \frac {x^2 \minus{} 3y \plus{} 2}{y \minus{} 2}$
Since $ (y \plus{} 1)^2$ is positive,the $ RHS$ must be positive,so:
1.$ x^2 \minus{} 3y \plus{} 2 > 0$ and $ y > 2 \Leftrightarrow \frac {x^2}{3} > y > 2$
2.$ x^2 \minus{} 3y \plus{} 2 < 0$ and $ y < 2 \Leftrightarrow \frac {x^2}{3} < y < 2$
Wolfram|alpha gives from case 2.
$ (x,y) \equal{} ( \minus{} 1,1);(1,1);(0,1)$ which aren't correct solutions for problem

Am I missing something ?[/hide]

the equation shall be solved by
1. considering divisibility by 3.

2. checking parity cases.

3. make very simple but minute observations in the following equations for further reduction (not even as complicated as gcd and modulus are needed :) )

4. simplify the equation (or rather descending it) lower and finally solve the equation as a quadratic.

5. further substitutions shall be needed.and an infinite descent shall be obtained giving no further solutions.

i apologise that i m not posting it..actually it is a part of my research papers, which contains infinite series for $ \pi$ and other infinite series and very elementary proofs of elliptic curve equations with cases $ k\in \{1,2,3,4,5\}$ and these are yet to be published.


Can someone explain to me what exactly this is and how to use it? Here's an example I know for sure that uses CRT:

Find the remainder when $2^{1990}$ is divided by $1990$.

Look here for an introduction to the theorem:

http://www.artofproblemsolving.com/Wiki/index.php/Chinese_Remainder_Theorem

Yes, that solves my problem then. Thanks! By the way, this was a question from the IRO (Indian Regional Olympiad).


inequalities
inequalities proposed
algebra

[quote=chien than]Let $ a;b;c$ be positive real numbers satisfying $ abc\equal{}1$
Prove that
a)$ \sum_{cyc} \frac{b\plus{}c}{a^2\plus{}1} \leq \sum_{cyc} \frac{1}{a^2}$
b)$ \sum_{cyc} \frac{b\plus{}c}{a^3\plus{}bc} \leq \sum_{cyc} \frac{1}{a^2}$[/quote]
Let $a,b$ be positive real numbers.[url=https://artofproblemsolving.com/community/c6h144254p817448]Prove that[/url]
$$\frac{a}{\sqrt{a^{2}+b^{2}}}+\frac{b}{\sqrt{9a^{2}+b^{2}}}+\frac{2ab}{\sqrt{a^{2}+b^{2}}\sqrt{9a^{2}+b^{2}}}\leq \frac{3}{2}$$
For $a,b,c$ are real numbers .[url=https://artofproblemsolving.com/community/c6h133929p757918]Prove that:[/url]
$$\frac{1}{(2a-b)^{2}}+\frac{1}{(2b-c)^{2}}+\frac{1}{(2c-a)^{2}}\geq \frac{11}{7(a^{2}+b^{2}+c^{2})}$$
Any one?


[quote="stvs_f"]For which k can the set {1990, 1991, 1992, ... , 1990 + k} be divided into two disjoint subsets with equal sums?[/quote]

http://www.mathlinks.ro/Forum/viewtopic.php?p=1225238#1225238


algebra polynomial function LaTeX trigonometry

[quote="chess64"]Please post the source of all problems that you post. There's a Source field under the subject field! If you don't have a source, we will be suspicious...this is the time of year when everyone is doing application quizzes for math camps and the like, so no cheating please. ;)[/quote]

When i try to post a new topic, there is no source field under the subject field.. It works on other certain forums, but for this one it doesnt work...  :?  :huh:

Hmm...it works for me. That could have something to do with my mod powers, but I'm not sure. Let me ask Valentin.

[quote="chess64"]This is the time of year when everyone is doing application quizzes for math camps and the like, so no cheating please. ;)[/quote]Oh, I'll definately be on the lookout for those.  ;)

Don't cheat or I will hunt you down.  :P (Just kidding)

Just to get the topic have fun :D  i say:"try to solve the second degree polynomial in function of x and study the discriminant"

[quote="fireforce"]Can I fifth the motion?  :D
But, seriously u have a source field when u reply so its a problem with the programing.  Is there anyway to do this so it shows up on the topic board?  :?:[/quote]Valentin might be working on that, but I'm not sure. As of now, only Admins and Mods seem to have source fields.

[quote="Jeff219"][quote="chess64"]OK then, just post it after the problem. Like this:

Find the sum $2+2$. (Source: Classic)

:D[/quote]
Two plus two make five :ninja:[/quote]

cuac

[quote="mathnerd314"][quote="buzzer11"]I just use the 'description' field under the title. It seems to work.[/quote]

That's because they added it onto the new server ;)

In the future, please refrain from reviving old topics. Thanks :)[/quote]

Um, it's a stickie. It'd be at the top regardless.

[quote="Jeff219"][quote="chess64"]OK then, just post it after the problem. Like this:

Find the sum $2+2$. (Source: Classic)

:D[/quote]
Two plus two make five :ninja:[/quote]

That is what is says in the book 1984. Where the government brainwashes the masses and yea...

Sorry had to comment. Plz don't ban me!

Source? oh okay. I usually just name it from a math league or from a past tournament.

Since I never remember which math league is which.

[quote="chess64"]OK then, just post it after the problem. Like this:

Find the sum $2+2$. (Source: Classic)

:D[/quote]

The sum is 11 :huh:

[hide]I was working in trinary. :rotfl: [/hide]

[quote="bpms"]Ok, it seems that in the past week or two this forum has been flooded with obvious homework questions. I have heard both that homework questions are ok, and that they aren't. I personally think we shouldn't be doing others homework, but no one cares what I think... Anyways, can a mod clarify on this?[/quote]

 :lol:  :lol: well sometimes during weekends i practice $\LaTeX$ by posting some of the homework answers (work included) in the test forum...if they are hard and some of my classmates need help i let them look at them...but if they just do that to copy i don't really care they'll pay for it during test time  :D

It is totally wrong to post [b]HOMEWORK[/b] but things like problems you can't solve or math team practice is okay...to me that is.

What should I do under the following conditions

A.  I remember a problem but forgot where it is from.
B.  I made the problem up.

Is it okay to say stuff like "my head" for problem B?

[quote="jli"][quote="bpms"]We all have our own opinions, but can I get the official word from a mod?

Thanks,
bpms[/quote]

no homework allowed.  i guess youre reffering to Interval's posts but he said they arent homework (hes over 40).  I think it said no homework was allowed in a sticky post, but if it is, only the replies should be hints, not full solutions[/quote]
My post came before interval's influx (it was made dec. 17) but no, there are some problems that people post that are obviously homework that they're too lazy to do themselves.

oh  :blush: , yeah.  i think homework shouldnt be allowed but im pretty sure it is :ninja:

If you post a homework question, post the source as "homework".  and again, if you reply to a homework problem, only give hints!

to jhollenbeck

A. just say i forgot and then maybe someone else may know
B. my head is fine.

[quote="nutz_for2.718281828"]SOURCE FIELD????? Whats that????????

Well, when I get to a topic, I can see the source if there is one...[/quote]

yeah...everyone can.  were talking about the source when you post something

[quote="jli"][quote="Altheman"]there should be some forum for the homework problems...although that forum would become a wasteland, you could just ignore the homework problems if you wanted to[/quote]

but then people would just post their whole homework assignment and not learn anything[/quote]

huh? why? people would post homework in some other forum, and if people wanted to answer them they can, but they would not infest these boards...it is not like somebody is going to answer all of the homework assignment...so the whole learning is not a problem, because i would be shocked to find somebody here to do a whole homework assignment...

This is more toward the homework problems, but if you want to learn how to solve a problem, please post the work you have done so far so that it will be much easier to see what you need help with.

[quote="ericlu149"]It would be ok to post old homework just to see what others do with it right? :?:[/quote]

That depends on the problems.  If they are of correct difficulty and you think people might enjoy them, then post them, by all means.  However, if they are simply routine homework problems, then there is no need to post them.

[quote="Poincare"]for the rules of this forum, it says we can only do *very* basic trig here then why are people posting stuff like double angle formula, half angle formula, roots of unity etc. is that all basic stuff?[/quote]

Double angle, half angle formulas are all just plug and chug equations. Not very difficult when you think about the more difficult topics. For the record, roots of unity type questions will be moved to Intermediate.


calculus
integration
limit
derivative
calculus computations

$ \frac{x^{3}(1\minus{}x)^{4}}{(1\minus{}2 x)^{5}}\equal{}\minus{}\frac{x^{2}}{32}\plus{}\frac{3 x}{64}\plus{}\frac{1}{128}\minus{}\frac{3}{128 (2 x\minus{}1)}\minus{}\frac{3}{128 (2 x\minus{}1)^{2}}\plus{}\frac{3}{128 (2 x\minus{}1)^{3}}\plus{}\frac{1}{128 (2 x\minus{}1)^{4}}\minus{}\frac{1}{128 (2 x\minus{}1)^{5}}$

we could use that if the integral convers, but, it diverges. so there is no result.

Proving that:

$ \lim_{\epsilon\to 0^{\plus{}}}\,\left(\int_{0}^{\frac{1}{2}\minus{}\epsilon }\frac{x^{3}(1\minus{}x)^{4}}{(1\minus{}2 x)^{5}}\, dx\plus{}\int_{\epsilon\plus{}\frac{1}{2}}^{1}\frac{x^{3}(1\minus{}x)^{4}}{(1\minus{}2 x)^{5}}\, dx\right)\equal{}\lim_{\epsilon\to 0^{\plus{}}}\,\frac{(1\minus{}2\epsilon )^{4}(4 (\epsilon\plus{}2)\epsilon\plus{}1)}{3072\epsilon^{3}}\equal{}\infty$

if the directional derivative were the opposite, the result would be opposite.

A wrong result, dont using convergence would show $ \frac{1}{768}(32\plus{}9 i\pi )$


Well, personally, I think the easiest ones are the log ones. Just fool around with the equations they give you (its all just algebra), and know the identities (look in AoPS 2, or any other textbook) like you know the alphabet.

Yeah...normally log questions are pretty straightforward.  They aren't like some problems that are really abtract (and impossible)  :oops: .

-interesting_move

No problems on the AIME are impossible, otherwise they wouldn't be there. Each problem there is a problem that is designed to take the top students in the country less than twelve minutes. I'm not sure how much difference all this psychological stuff makes while taking competitions, but for sure it isn't good to say that there's no way for you to get questions from 11 on or something like that. Just try them. Who knows what will happen? Maybe you will get them, and maybe you won't. But don't say that it isn't worth trying.


[hide]To start, $\gcd\left(a_{1},\dots ,a_{n}\right)$ divides the LHS.[/hide]

Show gcd(a,b,c)=gcd(gcd(a,b),c) and apply induction.

[hide="Three-line condensed proof"] Let $\mathcal{S}= \{ s \, | \, \exists k_{1}, k_{2}, ... k_{n}: s = k_{1}a_{1}+...+k_{n}a_{n}\}$ and let $d$ be its smallest positive element.

$d | a_{1}, a_{2}, ... a_{n}\implies d | gcd(a_{1}, a_{2}, ... a_{n})$ (consider the remainder)

$gcd(a_{1}, a_{2}, ... a_{n}) | d \implies d = gcd(a_{1}, a_{2}, ... a_{n})$[/hide]


I question about blogs. How will you figure out automatically whether your blog has been changed by others, e.g. within the shoutbox or as comment to one of my post entries? Automatic notification will be useful if I don't check the blog regularly or if it concerns an only old entry of mine.

It's on the wish list, and will be done ;)


function
Functional Analysis
real analysis
real analysis unsolved

Didn't find anything by searching. I'm semi-seriously going through a functional analysis book. I've gone through some exercises to prove that $ L^1$ is not the dual of anything, and some other stuff. But I can't figure out what the dual of $ L^\infty$ actually is? It's giving me a headache.

For instance, if $ \Omega$ is an open subset of $ \mathbb{R}^n$:
- $ L^{\infty}(\Omega,\mathbb{C})$ is a commutative C*-algebra.
- By the Gelfand transformation, $ L^{\infty}(\Omega,\mathbb{C})$ is isometrically isomorphic to the space $ C(K,\mathbb{C})$ of complex-valued continuous functions defined of the compact spectrum $ K$ of our algebra.
- So you can view the dual of $ L^{\infty}(\Omega,\mathbb{C})$ as the space of complex-valued Radon measures on some compact space.


inequalities
function
real analysis
real analysis theorems

Can any one prove the inequality

(a-b)/(lna-lnb)>(ab)^(1/2) which is same with LM > GM

Where a and b are distinct positive reals.

I proved AM>LM but couldn't do this one.

Since multiplication of $ a$ and $ b$ by a constant and exchanging them do not change anything, we may assume that $ ab\equal{}1$, $ a<b$. Putting $ a\equal{}e^{\minus{}x}$, we see that we need to prove that $ e^x\minus{}e^{\minus{}x}\ge 2x$ for $ x\ge 0$. The Taylor expansion of the exponential function yields the result in no time but, probably, there are plenty of more elementary ways to prove the last inequality as well.


(x - 1)(x - 5) = -4

Look at the factors (x - 1) and (x - 5).

The second factor is exactly 4 less than the first factor.

Look at products of two integers which equal -4:

-4*1, -2*2, -1*4

In the middle pair, -2 is 4 less than 2.

You could set (x - 1) = 2 and solve that.

Edit

x - 1 = 2
x = 3

3 is a repeated solution of the original equation,
so that is the only value that satisfies it.

Compare this to

$ (x \plus{} 2)(x \minus{} 8) \equal{} \minus{}24$.

It is satisfied by x = 2 or x = 4.

EDIT

$ (x\minus{}1)(x\minus{}5)\equal{}\minus{}4$
$ x^2\minus{}6x\plus{}5\equal{}\minus{}4$
$ x^2\minus{}6x\plus{}9\equal{}0$

There are 2 ways of solving this: quadratic equation, or square of a binom. I'll do the second way.

$ (x\minus{}3)^2\equal{}0$
$ x\equal{}3$

And there's no repeated solution, it's the same because you took x-1=a,x-5=a-4 and for any a it has one solution for x.

[quote="Bugi"]$ (x \minus{} 1)(x \minus{} 5) \equal{} \minus{} 4$
$ x^2 \minus{} 6x \plus{} 5 \equal{} \minus{} 4$
$ x^2 \minus{} 6x \plus{} 9 \equal{} 0$

There are 2 ?? ways of solving this: quadratic equation, or square of a binom.
I'll do the second ?? way.

$ (x \minus{} 3)^2 \equal{} 0$
$ x \equal{} 3$

And there's no repeated solution, it's the same because you took x-1=a,x-5=a-4 and for any a it has one solution for x.[/quote]

No, there aren't two ways of doing this.  You just happened
to demonstrate a second way.

$ x^2 \minus{} 6x \plus{} 9 \equal{} 0$ has a root of 3 with a multiplicity of 2.

[quote="Arrange your tan"][quote="Bugi"]$ (x \minus{} 1)(x \minus{} 5) \equal{} \minus{} 4$
$ x^2 \minus{} 6x \plus{} 5 \equal{} \minus{} 4$
$ x^2 \minus{} 6x \plus{} 9 \equal{} 0$

There are 2 ways of solving this: quadratic equation, or square of a binom.
I'll do the second way.

$ (x \minus{} 3)^2 \equal{} 0$
$ x \equal{} 3$

And there's no repeated solution, it's the same because you took x-1=a,x-5=a-4 and for any a it has one solution for x.[/quote]

No, there aren't two ways of doing this.  You just happened
to demonstate a second way.[/quote]

I never said there are 2 ways of solving any quadratics.

And there is:

$ ax^2 \plus{} bx \plus{} c \equal{} 0$
$ (ax^2 \plus{} bx \plus{} \frac {b^2}{4a}) \equal{} \frac {b^2 \minus{} 4ac}{4a}$
$ (\sqrt {a}x \plus{} \frac {b}{2\sqrt {a}}) \equal{} \pm\frac {\sqrt {b^2 \minus{} 4ac}}{2\sqrt {a}}$
$ x \equal{} \frac { \minus{} b\pm\sqrt {b^2 \minus{} 4ac}}{2a}$

So if you don't know quadratic formula, there is another way of doing (by the square of a binom). Although it is the same, and longer than using formula immediately, it's good for people who don't know that formula, and this is HSB...

So there is another way.

[quote="AIME15"]To the OP: Unfortunately, this method doesn't work on all quadratics - you aren't guaranteed the integer roots.[/quote]

You are claiming that for only those quadratics** with both
integer roots (I am meaning, let's not consider the
noninteger root ones at this time), that there is a subset
of those quadratics** that this method for which it will
not work.


Hey does anybody know if Emory holds a math contest...you know like HMMT, Duke, Princeton, etc? I might be going there next year (if i get rejected from Brown) and I thought it would be pretty cool if it started a really big math contest. Do you guys think you would be able to go to this contest if it was made (trying to see if a bunch of people from the deep south could possibly go)? THIS IS ALL HYPOTHETICAL OF COURSE JUST WONDERING IF THIS IS A GOOD IDEA HAH

Emory does not have a math contest, or it a very well-kept secret.
[url]http://www.gctm.org/math_competitions.htm[/url]

If it did start a contest, it would [i]probably[/i] be well-attended, since it's a pretty big school, as long as it didn't conflict with another major math tournament.  And if you do ever start a Georgia tournament at Emory, I beg of you to have it at some time other than mid-January or late February/early March (All-State Orchestra/Band/Chorus, in which several people are involved).

NOTE: ONCE AGAIN THIS IS ALL HYPOTHETICAL OK COOL

Yea, I really don't like these math contests being around October-November because of early decision stuff for colleges, and I don't like them being in December because of Christmas break, midterms, etc. Also, summer is not a good idea, especially right after ARML and well...um, it's summer?

I think the ideal time to have this contest would be in February. This if for obvioius reasons...through my extensive research in Greek, Ancient Chinese, Indian, Native American, Spanish, and Russian mythology, I have also found out that students tend to perform their best in February because they are not stressed from such things such as AP exam preparation and also because Feb. does not conflict with things like SAT and stuff.

So yea, February I would think would be the best.

And wtf, why does emory not have a math contest? That would be awesome if i helped make one though :D.

I was thinking Atlanta, GA was the perfect spot, because most of GA, NC, SC, other dirty south states and maybe FL can come...making great competition :wink:

lol...interesting you're from south carolina...but yeah i've never heard of emory having a math contest. however, as boy soprano said, all the math competitions seem to be packed into that late winter/early spring period, and there are like competitions every weekend. if you did start a competition at emory, you'd probably want to make sure there are no conflicts, a pretty difficult task during the february/march season.

january usually doesnt have as many contests, but you've also got to factor in the winter vacation. mid/late january should be an ok time though


Solve for $ x$: $ x \equal{} \frac{x}{6} \plus{} \frac{x}{12} \plus{} \frac{x}{7} \plus{} 5 \plus{} \frac{x}{2} \plus{} 4$.

Umm..i guess you would multiply both sides by the least common denominator, $ 84$, and get the equation

$ 84x \equal{} 14x \plus{} 7x \plus{} 12x \plus{} 5\times84 \plus{} 42x \plus{} 4\times84$

Simplify to get:
$ 84x \minus{} 14x \minus{} 7x \minus{} 12x \minus{} 42x \equal{} (5\plus{}4)\times84$
$ 9x \equal{} 9\times84$
$ x \equal{} 84$

Is there a way to do this quickly?  :|  :(  :what?:

We could multiply by 12 to get $12x=2x+x+\frac{12x}{7}+60+6x+48 \implies 12x=9x+\frac{12x}{7}+64 \implies 3x=frac{12x}{7}+108$. Multiply by 7 to get $21x=12x+7 \cdot 108 \implies 9x=7 \cdot 108 \implies x=7 \cdot 12 \implies x=84$. Multiplying by 12 and then 7 is probably quicker.


is it worth putting that you ranked 40th on general test and 20th on general test1 on college apps?

Aren't you a freshman? Don't worry about it.

No, it's not.

That doesn't mean that 40th on the general test is inadequate -- it just means that there are probably more "telling" things that you can put on your college apps. This is especially so because many admissions officers don't really understand the difficulty of HMMT, and they're not going to call up the math department to ask.

Remember, every neutral thing you put on your college apps [i]hurts[/i] you because it dilutes your application and makes it harder for the spectacular things to stand out!

That said, try to get a better rank. If your rank is posted on the internet (ie, top 10 in a round), then put it on your apps. Otherwise, focus on other stuff.

Ok thanks man. What about USAMTS silver + mandelbrot regional top10

I barely know anything about USAMTS -- I've never done it. But Mandelbrot regional top 10 is most likely worth putting.


logarithms
irrational number
number theory
number theory unsolved

Prove that there exist positive integer$N$ such that the decimal

representation of the number $2000^{N}$starts with the sequence of the
digits $200120012001$

Classical (using the density of Z+rZ in R where r is an irrational number, R are the reals and Z the integers). You can also prove the same result for p^N for any p not divisible by 10 ( the result holds for any "beginning string" ).

[quote="Megus"][b]Julien[/b], (or anybody else) could you please show full proof? I have no clue how to use your idea of the density (honestly saying I'm new to using density in number theory but as I've seen quite a lot problem solutions using it so far, I want to learn it eventually so I'll be very grateful for any links dealing with this topic)[/quote]
I have the same opinion. Please Julien post a solution with all the details

[quote="Megus"][b]Julien[/b], (or anybody else) could you please show full proof? I have no clue how to use your idea of the density (honestly saying I'm new to using density in number theory but as I've seen quite a lot problem solutions using it so far, I want to learn it eventually so I'll be very grateful for any links dealing with this topic)[/quote]

The densities in number theory have not much (or nothing indeed) to do with that problem. Here something different is meant: a set $S$ of real numbers is called 'dense' (in the real numbers), if for every $x\in \mathbb{R}$ and every $\epsilon > 0$ there is some $s \in S$ with $|x-s| < \epsilon$.
Then a well known theorem (prove is easy but boring) says that the numbers of the form $a+br$ with integers $a,b$ are dense iff $r$ is irrational. Even when assuming $b>0$ the same result holds.
Now consider the numbers of the mentioned form with $r=log_{10}(2000)$:
they are dense and so you can find for every $\epsilon > 0$ an $s$ from that numbers with $|log_{10}(200120012001)-s| < \epsilon$. Now if $\epsilon$ is small enough, the difference between $10^s$ and $10^{log_{10}(200120012001)} = 200120012001$ is smaller than $1$ (indeed you need $s>log_{10}(200120012001)$, but this can easily be fulfilled). So you have that $10^s = 10^{a+b \cdot log_{10}(2000)} = 10^a \cdot 2000^b$ is a little bit bigger than $200120012001$, and as $b$ can be assumed positive, you get a number $2000^b$ that starts with $200120012001$.

[quote="silouan"][quote="ZetaX"]Then a well known theorem (prove is easy but boring) says that the numbers of the form $a+br$ with integers $a,b$ are dense iff $r$ is irrational. [/quote]

Please someone prove it[/quote]
Here is a proof: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=171749#p171749

[quote="ZetaX"][quote="Megus"][b]Julien[/b], (or anybody else) could you please show full proof? I have no clue how to use your idea of the density (honestly saying I'm new to using density in number theory but as I've seen quite a lot problem solutions using it so far, I want to learn it eventually so I'll be very grateful for any links dealing with this topic)[/quote]

Then a well known theorem (prove is easy but boring) says that the numbers of the form $a+br$ with integers $a,b$ are dense iff $r$ is irrational. Even when assuming $b>0$ the same result holds.
Now consider the numbers of the mentioned form with $r=log_{10}(2000)$:
they are dense and so you can find for every $\epsilon > 0$ an $s$ from that numbers with $|log_{10}(200120012001)-s| < \epsilon$. Now if $\epsilon$ is small enough, the difference between $10^s$ and $10^{log_{10}(200120012001)} = 200120012001$ is smaller than $1$ (indeed you need $s>log_{10}(200120012001)$, but this can easily be fulfilled). So you have that $10^s = 10^{a+b \cdot log_{10}(2000)} = 10^a \cdot 2000^b$ is a little bit bigger than $200120012001$, and as $b$ can be assumed positive, you get a number $2000^b$ that starts with $200120012001$.[/quote]

How are you sure that at the same time $s > \log_{10}(200120012001)$,but still within $\epsilon$? The density theorem you're using only guarantees they can be made very close, but doesn't guarantee anything about which one will be bigger. But we need $s$ to be the bigger one, and we cannot conclude that for sure. Am I missing something?
Also, I was trying to find more proves that $2^n$ can start with any sequence of digits, but found only 1 more(from the Mediterranean olympiad), so if you guys have some more links, let me know.

P.S. About this problem, I was trying to prove we can always find positive integers $k, N$ such that if $A = log_{10}200120012001$, $log_{10}A < -k + Nlog_{10}2000 < log(A + 1)$. This would complete the proof. Does it follow somehow from the density theorem ZetaX mentioned? If we put $x = log_{10}A$, $s = -k + Nlog_{10}200$, we can ensure that $|s - x| < log(A+1) - log(A)$, but we also need to be sure that $s > logA$.


abstract algebra
algebra
function
domain
superior algebra
superior algebra unsolved

Let R be a noetherian commutative ring which is not domain and which has Krull dimension = 1. Let M be a cyclic R-module. Thus M \simeq R/I, where I is an annihilator of the generator of M. I need to prove that Krull dimension of R/I is equal to 0.
Thx.


The product of two of the four zeros of the quatratic equation $x^4-18x^3+kx^2+200x-1984=0$  is $-32$.  Find $k$

[hide]Let $a$ and $b$ be the two roots with product $-32$ and $c$ and $d$ be the other two roots. We are given
\begin{eqnarray*} a+b+c+d&=&18\\ ab+ac+ad+bc+bd+cd&=&k\\ abc+abd+acd+bcd&=&-200\\ abcd&=&-1984 \end{eqnarray*}
Substituting $ab=-32$ into the last equation, $cd=62$. Now let $x=a+b$ and $y=c+d$. The first three equations can be written
\begin{eqnarray*} x+y&=&18\\ xy+30&=&k\\ 62x-32y&=&-200 \end{eqnarray*}
Adding the first equation 32 times to the third, $94x=376\Rightarrow x=4$. Substituting into the first equation, $y=14$.

From $x=4,y=14$, we get $k=4\cdot 14+30=\boxed{86}$.
[/hide]


Go to Other US Contests and Programs forum under the Contest/Program forum.

[quote="drunner2007"]I received something in the mail from a mathcamp in portland oregon for doing well on the AMC. Does anybody know a lot about the program?[/quote]

This would be the board for that camp. You can post questions here.


Prove that the equation $ x^{2} \equal{} 3y^{3} \minus{} 1$ has no natural solutions.

Obviosly $ x^2\not \equal{}\minus{}1\mod 3$.


Prove that the arithmetic mean of any two consecutive primes (not including 2) is composite.

[hide="Rather obvious?"]The arithmetic mean of two primes must lie exactly between them on the number line, and you can't have a prime between two consecutive primes.

More formally, take $ p$ and $ q$ to be two consecutive primes such that $ p<q$, and suppose their arithmetic mean is prime. Then $ p\plus{}p<p\plus{}q$ by adding $ p$ to both sides, so $ 2p<p\plus{}q\implies p<\frac{p\plus{}q}{2}$. Similarly, $ p\plus{}q<q\plus{}q$, so $ \frac{p\plus{}q}{2}<q$.

Then $ p<\frac{p\plus{}q}{2}<q$, but there's no way a prime can be bounded between two consecutive primes, so we've reached a contradiction.
[/hide]

[hide]
Yep, that's a correct solution. I was wondering whether someone would try to find the solution using the definition of composite rather than the definition of consecutive, however, that seems extremely tedious.
[/hide]


This is yet another problem I came across in the Hwa Chong Challenge (Singapore).  (See my other 2 posts, Trigo and Trigo Part 2, for the other problems.)

Given that:
      sin a + cos b = 3/5,
      cos a + sin b = 4/5.
Find the value of cos a * sin b.

The answer is [hide]-11/100[/hide].  Not very nice, and again, this question is from the buzzer round (another part of it) and needs to be solved in under 2 minutes.  And again, i know of a long solution that would never be able to be solved in such a short period of time (see attached).

Anyone knows of a better way to solve this kind of questions?

[quote="Waynerius"] $ \|\begin{array}{c} \sin a + \cos b = \frac 35 \\
 \\
\sin b + \cos a = \frac 45\end{array}\ \ \implies\ \ \cos a \cdot \sin b = \ ?$ [/quote]

$ \|\begin{array}{cccc} 1\blacktriangleright & (\sin a + \cos b)^2 + (\sin b + \cos a)^2 = 1 & \implies & \sin (a + b) = - \frac 12 \\
 \\
2\blacktriangleright & (\sin b + \cos a)^2 - (\sin a + \cos b)^2 = \frac {7}{25} & \implies & \sin (b - a) = \frac {7}{25}\end{array}\|$ $ \implies$ $ \|\begin{array}{c}
\cos a\cdot\sin b = - \frac {11}{100}\\\\
\cos b\cdot\sin a=-\frac {39}{100}\end{array}$

[b]Remark.[/b]

$ (\sin a + \cos b)(\sin b + \cos a) = \frac {12}{25}\ \implies\ \cos (a - b) = \frac {24}{25}$ .

 $ \tan\frac {a-b}{2}=\frac {\sin (a-b)}{1+\cos (a-b)}\ \implies\ \boxed {\ \tan\frac {a-b}{2}=-\frac 17\ }$ .


function
advanced fields
advanced fields unsolved

find every function $ f(x,y): C^1( \mathbb{R}^2 \to \mathbb{R})$, that satisfies:
$ \| \nabla f \|_{2} \equal{} 1$ for all $ (x,y) \in \mathbb{R}^2$

I'll try the general case of $ \mathbb R^n$.

Such a function $ u$ must be unbounded from above and below (consider the gradient flow). Let $ Z \equal{} \{x\in \mathbb{R}^n\colon u(x) \equal{} 0\}$; this is a $ C^1$ hypersurface. Since $ u$ is $ 1$-Lipschitz, we have $ |u(x)|\le \mathrm{dist}(x,Z)$. On the other hand, gradient flow brings us from $ x$ to a point in $ Z$ in time $ |u(x)|$. Therefore, $ |u(x)|\equiv \mathrm{dist}(x,Z)$ and the gradient flow lines are simply normal lines to $ Z$. Since they do not intersect each other, $ Z$ is a hyperplane. Hence $ u$ is an affine function.

Any gaps?


I was helping a friend with his homework which is mainly conditional probaibilty, This seems like the basic problem. Can someone help me out with this one? I'm sure once you show me I'll remember.

lets say joe has a 50% chance to get accepted by harvard, 40% chance to get accepted by yale and 20% to get accepted by both, what is the chance he gets accepted by neither.

Thanks in advance, and maybe explain each step to help me refresh my memory. (Stupid probability)

He has 50% to go to harvard, 40% for yale, and 20% for both

Then 50+40-20 to get in either, so 70.  Then percentage that he gets accepted by neither is 100-70=30%.


limit
trigonometry
calculus
calculus computations

OK, i made this one up.... LOL :D

[u][b]without[/b][/u] using power series [i]or[/i] l'hospital , find the limit

$ \lim_{n\to 0} \; \left(\frac{\tan \; n}{n} \right) ^ { \frac {1}{ n^2 } }$

no, akech , it's not 1.  also, i made it slightly harder by imposing you cannot use power series or l'hospital.   :lol:

The limit is easy enough: it is $ e^{1/3}.$ I don't have any particular desire to play along with not using appropriate tools.

Its quite simple. Its of the form $ 1^{\infty}$
The limit is equivalent to $ e^{\lim_{n \to 0} \frac {\tan n \minus{} n}{n^3}}$
Now we just need to fins the limit L = \lim_{n \to 0} \frac {\tan n - n}{n^3} Replace n by 3n and get L in terms of L and also a constant!

[quote="akech"]For a small x we have:
$ \sin x\ \equal{} \tan x\ \equal{} x$.
Hence the limit in question is 1.[/quote]

$ \lim_{n\to 0} f(n) \equal{} 1 \not\implies \lim_{n \to 0} f(n)^{ \frac{1}{n} } \equal{} 1$ (and certainly not the same for $ n^2$).  Otherwise we would conclude $ \lim_{n \to \infty} \left( 1 \plus{} \frac{1}{n} \right)^n \equal{} 1$!


Given a function $f$ for which
\[f(x)=f(398-x)=f(2158-x)=f(3214-x) \]holds for all real $x,$ what is the largest number of different values that can appear in the list $f(0),f(1),f(2),\ldots,f(999)?$

[hide="Solution... :maybe: "]
hmm so we see that:
$f(999)=f(-601)=f(1159)=f(2225)$
$f(399)=f(-1)=f(1759)=f(2815)$

This tells us that from $f(399)$ to $f(999)$ all the values are different..

$f(0)=f(398)$
$f(1)=f(397)$
...
$f(198)=f(200)$

This incorporates $199$ numbers that have been duplicated ..
f(199)=f(199) therefore it is unique..
so the total number is:$1000-199=\boxed{801}$[/hide]

Sorry to revive an old thread, but Kalva gives a much smaller answer.

[hide="Solution"] $f(2518-x) = f(x) = f(3214-(2158-x)) = f(1056+x)$

$f(398-x) = f(x) = f(2158-(398-x)) = f(1760+x)$

Since $gcd(1056, 1760) = 352$ we can conclude that

$f(x) = f(352+x)$

So we need only to consider one period $f(0), f(1), ... f(351)$, which can have at most $352$ distinct values which determine the value of $f(x)$ at all other integers.

But we also know that $f(x) = f(46-x) = f(398-x)$, so the values $x = 24, 25, ... 46$ and $x = 200, 201, ... 351$ are repeated.  This gives a total of

$352-(46-24+1)-(351-200+1) = \boxed{ 177 }$

Distinct values. [/hide]

well thx for pointing that out...so i assume thats the correct answer!

[quote="t0rajir0u"]
[hide="Solution"]

Since $gcd(1056, 1760) = 352$ we can conclude that

$f(x) = f(352+x)$

 [/hide][/quote]

How do you know that?

t0....

can you explain how to use the Euclidean algorithm in this problem?

sorry for bringing up an old topic...

[b]Proposition:[/b]  If a periodic function has periods $ a$ and $ b$, where $ a, b$ are integers, then it has period $ \gcd(a, b)$.

Proof:  $ f(z) \equal{} f(z \plus{} a) \equal{} f(z \plus{} b) \implies f(z \plus{} ax \plus{} by)$ for any integers $ x, y$.  But by [url=http://en.wikipedia.org/wiki/B%C3%A9zout%27s_identity]Bezout's identity[/url], there exist integers $ x, y$ such that $ ax \plus{} by \equal{} \gcd(a, b)$.  $ x, y$ are computed by the Euclidean algorithm.

[quote="t0rajir0u"]
[hide="Solution"] But we also know that $ f(x) \equal{} f(46 \minus{} x) \equal{} f(398 \minus{} x)$, so the values $ x \equal{} 24, 25, ... 46$ and $ x \equal{} 200, 201, ... 351$ are repeated.  This gives a total of

$ 352 \minus{} (46 \minus{} 24 \plus{} 1) \minus{} (351 \minus{} 200 \plus{} 1) \equal{} \boxed{ 177 }$

Distinct values. [/hide][/quote]

How do we know that those are [u]all[/u] the repeats?  (And that among the 177 none are further repeated).

The "periods" given in the original problem were reduced $ \bmod 352$.  The function I describe then satisfies all three original requirements.

[quote="t0rajir0u"]Sorry to revive an old thread, but Kalva gives a much smaller answer.

[hide="Solution"] $ f(2518 \minus{} x) \equal{} f(x) \equal{} f(3214 \minus{} (2158 \minus{} x)) \equal{} f(1056 \plus{} x)$

$ f(398 \minus{} x) \equal{} f(x) \equal{} f(2158 \minus{} (398 \minus{} x)) \equal{} f(1760 \plus{} x)$

Since $ gcd(1056, 1760) \equal{} 352$ we can conclude that

$ f(x) \equal{} f(352 \plus{} x)$

So we need only to consider one period $ f(0), f(1), ... f(351)$, which can have at most $ 352$ distinct values which determine the value of $ f(x)$ at all other integers.

But we also know that $ f(x) \equal{} f(46 \minus{} x) \equal{} f(398 \minus{} x)$, so the values $ x \equal{} 24, 25, ... 46$ and $ x \equal{} 200, 201, ... 351$ are repeated.  This gives a total of

$ 352 \minus{} (46 \minus{} 24 \plus{} 1) \minus{} (351 \minus{} 200 \plus{} 1) \equal{} \boxed{ 177 }$

Distinct values. [/hide][/quote]
How do you know $ f(x)\equal{}f(46\minus{}x)$? Also how do you prove those are the only values repeated (what is the function you described)?

$ 46 \equal{} 398 \minus{} 352$.  Any function which has distinct values up to the constraints that I wrote down satisfies all of the original problem constraints, which are equivalent to the following two constraints (by the argument I've already given):  $ f(x) \equal{} f(46 \minus{} x)$ and $ f(x) \equal{} f(x \plus{} 352)$.

But how do you prove 46 is the minimum k for which $ f(x)\equal{}f(k\minus{}x)$? (I see how you proved 352 is the minimum period)
Also how do you prove those two conditions are sufficient?

$ 352 | 398 \minus{} 46$ and $ 2158 \minus{} 46$ and $ 3214 \minus{} 46$.  This is a consequence of the first part of my argument.

(p8) I did the same thing as the second poster - what's wrong with that argument?


calculus
integration
trigonometry
geometry
function
complex analysis
complex analysis unsolved

Hello!

I have this integral:

\[ \frac{1}{2i\pi}\int_{S}\frac{f(z)\sin(z\plus{}x)}{\sin z\minus{}\sin x}dz\]

Where S is the contour shown in the figure below:

[url=http://img254.imageshack.us/my.php?image=contornocm1pb4.jpg][img]http://img254.imageshack.us/img254/2208/contornocm1pb4.th.jpg[/img][/url]

and is composed by AB, S', CD and S''.

The author states that:

AB and CD are vertical lines passing through $ \minus{}\pi\plus{}\eta$ and $ \pi\plus{}\eta$.(in the image shown they pass through $ \minus{}\pi$ and $ \pi$)

$ f(z)$ is holomorph in the considered area and has period=$ 2\pi$.

$ \frac{f(z)\sin(z\plus{}x)}{\sin z\minus{}\sin x}$ has only one pole $ z\equal{}x$ in the interior of $ S$.

$ \eta<\frac{\pi}{2}$

And this last statement is what confuses me... why does $ \eta$ has to be $ <\frac{\pi}{2}$?

If it was bigger, it wouldn't change the period considered. I think it wouldn't change the number of poles of the funtion, or would it? :(

btw, the contour considered is the result of $ |\sin z|\equal{}c$, where $ c>1$.

Do you understand why this condition is needed?

Thank you!

Do you find this weird too?

Are you suspicious of any kind of mistake in the book?


prove that   with$ m\geq 1$     =>
           $sqrt{(2x+1)} + sqrt{(x+2)} \leq m \sqrt{(2x+3)} $ have roof

Your inequality is trivially false when $m=1$.
(What is "have roof" indeed?)


hey...how can i delete this account?
(if somebody can delete this account, it would be greatly appreciated :))


AMC
USA(J)MO
USAMO
function
combinatorics proposed
combinatorics

Find, with proof, the number of positive integers whose base-$n$ representation consists of distinct digits with the property that, except for the leftmost digit, every digit differs by $\pm 1$ from some digit further to the left. (Your answer should be an explicit function of $n$ in simplest form.)

See:

http://www.unl.edu/amc/a-activities/a7-problems/problemarchive.html

[color=red][Moderator edit: Also posted at http://www.mathlinks.ro/Forum/viewtopic.php?t=60243 .][/color]

[hide]Picture instead of a number base n, a numberline from 0 to n-1.

Place a point to represent the first number of n.  After that point, place additional points to represent the other numbers after the first.  There will be no gaps in the placement of the points (the points placed at any given time cover consecutive numbers).

Let the first digit of the number be x+1.  There are x numbers to the left of x+1 and n-x-1 to the right of it.  If the number is has all x numbers to the left of x+1 in it, the number of possibilities are (Let aCb be a choose b):
xCx+(x+1)Cx+(x+2)Cx...(n-1)Cx= by hockeystick identity(n)C(x+1)
counting the possibilities for only x-1 of the numbers to the left of x+1, x-2, x-3...0 numbers we have:
(n)C(x+1)+(n-1)C(x)...(n-x-1)C(0)=(n)C(n-x-1)+(n-1)C(n-x-1)...(n-x-1)C(n-x-1) (since aCb=aC(a-b))=(n+1)C(n-x)

Now we count all possibilities for x
(n+1)C(n-1)+(n+1)C(n-2)...+(n+1)C(2)=x^(n+1)-2n-2 (I need to go)[/hide]

[quote="Phelpedo"][hide]
Let $ f_n$ denote the number in question. We consider two cases:

Case 1: The first digit is $ n \minus{} 1$. Now there are only $ n$ possible numbers: $ n \minus{} 1$, $ (n \minus{} 1)(n \minus{} 2)$, ... , $ (n \minus{} 1)...(0)$.

Case 2: Let's suppose the first digit is $ m < n \minus{} 1$. Now there are two possibilities for the next digit: $ m \plus{} 1$ or $ m \minus{} 1$. In either case, combine these first two digits into a block. Now we see that this constitutes a bijection to the numbers of $ f_{n \minus{} 1}$. Thus this case contributes $ 2f_{n \minus{} 1}$ numbers.

Our recursion is $ f_n \equal{} n \plus{} 2f_{n \minus{} 1}$. This should be pretty standard to solve:

We create a sequence $ g_n \plus{} k \equal{} f_n$. We want this to satisfy $ g_n \equal{} \alpha g_{n \minus{} 1}$, so $ k \equal{} \minus{} n$. So we have the recursion $ g_n \equal{} 2g_{n \minus{} 1}$. That means $ g_n \equal{} 2^{n \minus{} 1}g_1$. Since $ g_1 \equal{} 1$, we have that $ g_n \equal{} 2^{n \minus{} 1}$ and $ f_n \equal{} 2^{n \minus{} 1} \plus{} n$.
[/hide][/quote]

sorry to bump the topic again, but isn't this not correct? suppose n = 2, then you have 1 and 10, but you have $ f_2 \equal{} 2^1 \plus{} 2 \equal{} 4$...

I think it has to do with what's in your "Now we see that this constitutes a bijection to the numbers of $ f_{n \minus{} 1}$. Thus this case contributes $ 2f_{n \minus{} 1}$ numbers."

Yeah, there is a problem with that bijection, although we can actually use that bijection to solve the problem.

Pretend that we can start a number with the digit 0.  Let $ f_{n}$ be the number of solutions in this case.  Then, if we have an base $ n$ number, it is either one digit long, or it is at least 2 digits long.

If it is a one digit number, there are $ n$ possibilities, since there are $ n$ digits including 0.

If the number has at least two digits, let the digits be $ a$ and $ a\plus{}1$.  Then, the number can be finished in the same number of ways as the base $ n\minus{}1$ number beginning with $ a$.  Since we can reach every $ a$ such that $ 0 \leq a < n\minus{}1$ in exactly two ways, we have $ 2f_{n\minus{}1}$ numbers with at least two digits.

Thus, we have $ f_{n} \equal{} 2f_{n\minus{}1} \plus{} n$ with $ f_{1} \equal{} 1$, so solving $ f_{n} \equal{} 2^{n\plus{}1} \minus{} 2 \minus{} n$, which is trivial to prove by induction.

Subtracting out the $ n$ numbers that start with the digit zero, we find that the number of solutions is $ 2^{n\plus{}1} \minus{} 2 \minus{} 2n$

Alternately, we could prove this without solving an ugly recursion.

Again allowing use of the leading digit 0, we note that in base $ n$ there is $ 1$ set of $ n$ consecutive digits, $ 2$ sets of $ n\minus{}1$ consecutive digits and so on.

If we have $ k$ consecutive digits, building our number from the units digit to the left, since we must choose either the greatest or least digit in our set, we see that there are 2 choices for the units digit, 2 choices for the next left-most digit and so on, until the last digit is forced.  Thus, there are $ 2^{k\minus{}1}$ possible numbers for any $ k$ consecutive digits.

Thus, we have $ 1*2^{n\minus{}1} \plus{} 2*2^{n\minus{}2} \plus{} 3*2^{n\minus{}3} \plus{} ... \plus{} n*2^{0}$ total numbers.

This collapses to be $ (2^{n} \minus{} 1) \plus{} (2^{n\minus{}1} \minus{} 1) \plus{} ... \plus{} (2^{1} \minus{} 1)$.

This is equal to $ 2*(2^{n} \minus{} 1) \minus{} n \equal{} 2^{n\plus{}1} \minus{} 2 \minus{} n$.

Subtracting the $ n$ numbers that start with the digit zero again, we find that the total number of solutions is $ 2^{n\plus{}1} \minus{} 2 \minus{} 2n$.

Bump!

Let the leading digit be $m.$ Note there are $m$ numbers less than it and $n-m-1$ numbers greater than it. Then we choose some amount of the $m$ numbers less than $m$ and the $n-m-1$ numbers greater than $m.$ Note that this is essentially the same as making the numbers to the left identical and the numbers to the right identical since their order is fixed, so we want to find
\[\sum_{a=0}^m\left(\sum_{b=0}^{n-m-1}\binom{a+b}{a}\right).\]
This is equal to
\[\sum_{a=0}^m\binom{a+n-m}{a+1}=\sum_{a=0}^m\binom{a+n-m}{n-m-1}\]
by the Hockey Stick Identity, and this further simplifies into
\[\binom{n+1}{n-m}-1=\binom{n+1}{m+1}-1.\]
Now note that
\[\sum_{m=1}^{n-1}\left(\binom{n+1}{m+1}-1\right)=\sum_{m=1}^{n-1}\binom{n+1}{m+1}-(n-1)=\left(2^{n+1}-\binom{n+1}{0}-\binom{n+1}{1}-\binom{n+1}{n+1}\right)-(n-1)=2^{n+1}-(2n+2),\]
which is our answer.


geometry
AMC
AIME
USA(J)MO
USAMO
MATHCOUNTS
AMC 10

Hi, I'm new to this forum. I'm from SC, and in Grade 8. I moved to the US from Canada, and I hadn't heard of MC until I came here. Anyways, as this was my first year, my coach recruited me to try out for school team. I took the school test (simple), moved on to team. I haven't so far taken the whole MC thing seriously, but at Chapter, I surprisingly came first in Individual and CD (sweep), and my team won the Team Round. Anyways, now my coach is pestering me about how I am a prospect for the state team and everything, and that you can get scholarships etc.

 So I now have decided to take this seriously. Are there any training strategies or anything that I could use? I've been averaging around 24ish on State sprints and 14 on targets. Is that good enough for the SC team?

Thanks for your compliance in editing it, vaele.

The best strategy is just to take old tests. Time yourself. Make every test seem like the real deal. Go beyond knowing how to do to problem and dive into knowing how to do it the best, and quickest way. This is for MC only... for higher level math it gets more complicated.

[quote="Fanatic"]QUICK: Edit out your score, and the team scores! Please read the stickies at the top.

YOU ARE NOT ALLOWED TO POST YOUR SCORE; OTHER STATES HAVE NOT TAKEN THIS TEST YET!

Edit: I will gladly reply to your question as soon as you do this. Luckily, I've already taken the chapter anyways, but 10-15 other people could've seen this and let's just say your post is a huge giveaway.[/quote]

what are you talking about ? he didn't say his score

he said his average practice score...  :oops:

lol Sorry; I didn't know about that rule. I don't think SC is all taht good; last year we mad 25th or something like that...with a math prodigy on the team. Too bad Ming (my friend) is in 9th grade now.

Hehe, sorry about the confusion guys, :blush:  I edited and forgot to sign.  Yeah, go ahead and discuss what was intended in this thread, but NO SCORES!!! thanks!

Be thankful that you don't live in California.

Well I was 5th in CA last year (mostly luck but still, 5th is pretty good) and those scores are what I usually get on some of the harder States, so I think you should be able to make Nats.

Okay. I also have questions on how to prepare for the state exam.

Also, we don't even have individual cd :(

Yeah-MA usually isn't that great, almost always in the middle of top 10. But this year, we have 4 people returning from last year, so we'll probably do better, and it will probably be tougher to make nats.

[quote="jhredsox"]omg, like 80 people have asked

do questions

study about the ones you don't get

jorian[/quote]

Grr Does anylne have any other ways? I ran out of problems...

find more. and what do u mean "other ways"? hypnosis?

lol...putting a geometry book under your pillow. :roll:

I wanted to know about which level (state, national, amc 10?, amc 12?, aime?, usamo?, imo shortlists?)

do nats. if u can do those, u can do states. of course, if u can do IMOs, do those. nvm, i'm pretty sure they wouldn't really prepare u for MC...

you spell surprise surprise not suprise.
practice old contests.
this seems like a trivial matter. moving to nd? come on.

[quote="jhredsox"]they either changed it or your eyes suck

jw

yah, don't do imos

jorian[/quote]
Or maybe you can't read...
Reread the title ;)

Please spell out your words entirely and don't use chatspeak. Also, the topic at hand is how to do well at state, and discussing IMOs doesn't fit. Don't frighten people who just made state for the first time with IMO problems. Even after I made Nat MC I was scared of USAMO problems, much less IMO problems.

[quote="solafidefarms"]May I please request that you spell out your words entirely and not use chatspeak? And mayhap ya'll could stick to the topic and not bother with discussing IMOs? You don't want to frighten people who just made state for the first time with IMO problems. Even after I made Nat MC I was astonishingly frightened of USAMO problems, much less IMO problems.[/quote]
I thought USAMO problems=IMO problems with a style difference

just a note to bpms because i can read: "Suprise Qualifying!"

back on subject: i think amc 10's also help quite a bit

jorian

[quote="Fanatic"]He/she edited it. When I originally posted, it had a very distinct individual chapter score.

Edited.

Lol am I the ONLY one that saw that?

EDIT: I don't know how good South Carolina is, so I can't help you out in that department.[/quote]

I saw that.

Okay. My dad just went to that contest resource page and printed out like 5 pages of Imo shortlists and is making me do those and I can't even get like more than 1 if I'm lucky :(

Where else can I get stuff since I have almost all states and nat'l questions?

[quote="Treething"]Wow, if you're studying for MathCounts, wouldn't it make sense to study MathCounts problems?[/quote]

You want to do other problems too.


Suppose $ a$ and $ b$ are constants greater than $ 1$.  How many solutions $ x$ are there to the equation $ (\log_{a}x)(\log_{b}x)\equal{}2007$?  Prove that your answer is correct.

$ (log_bx)^2\equal{}2007log_ba$

Now, if $ log_ba <0$ we have no solutions.

If $ log_ba\equal{}0$ then we have 1 solution ($ log_bx\equal{}0$ which gives $ x\equal{}b$)

If $ log_ba >0$ we have 2 solutions ($ log_bx\equal{}\pm\sqrt{2007log_ba}$ which gives $ x\equal{}b^{\pm\sqrt{2007log_ba}}$)


[b](i)[/b] Find the volume of the region described by $ \{ (x,y,z) \in \mathbb{R}^3 \; : \; |3x \plus{} 2y \plus{} z| \; \plus{} \; |x \plus{} 3y \plus{} 2z| \; \plus{} \; |2x \plus{} y \plus{} 3z| \leq 6 \}$

[b](ii)[/b] Is the following approach a valid way of solving the problem? (i.e., does it give the correct answer and is it a correct procedure)

[hide="Matrices"]
$ A \left( \begin{array}{c} x \\
y \\
z \end{array} \right) \equal{} \left( \begin{array}{ccc} 3 & 2 & 1 \\
1 & 3 & 2 \\
2 & 1 & 3 \end{array} \right) \left( \begin{array}{c} x \\
y \\
z \end{array} \right) \equal{} \left( \begin{array}{c} 3x \plus{} 2y \plus{} z \\
x \plus{} 3y \plus{} 2z \\
2x \plus{} y \plus{} 3z \end{array} \right) \equal{} \left( \begin{array}{c} r \\
s \\
t \end{array} \right)$

$ \iff \left( \begin{array}{c} x \\
y \\
z \end{array} \right) \equal{} A^{ \minus{} 1} \left( \begin{array}{c} r \\
s \\
t \end{array} \right)$

Then, the volume of $ |r| \plus{} |s| \plus{} |t| \leq 6$ is $ \text{abs det} \left( A^{ \minus{} 1} \right) \equal{} \frac {1}{\text{abs det} (A)} \equal{} \frac {1}{18}$ times the volume of $ |x| \plus{} |y| \plus{} |z| \leq 6$, and is thus:

$ \frac {1}{18} \cdot \left( \frac {1}{6} \cdot 6^3 \cdot 8 \right) \equal{} 16$
[/hide]

[quote="The Zuton Force"][b](i)[/b] Find the volume of the region described by $ \{ (x,y,z) \in \mathbb{R}^3 \; : \; |3x \plus{} 2y \plus{} z| \; \plus{} \; |x \plus{} 3y \plus{} 2z| \; \plus{} \; |2x \plus{} y \plus{} 3z| \leq 6 \}$[/quote]

Could you explain to me what this means!!! (sorry im kinda dumb!)

It's the region in the x-y-z space that satisfies that inequality. Dissection:
(x,y,z) - a point (or vector) with 3 components x, y, z
$ \in$ - in
$ \mathbb{R}^3$ - the set of triples of real numbers, aka. 3D space
: - "such that"

ahhh thanks so much!(like i said im kinda dumb--thats why im part of the "1m A PHA1LUR3" club!)


How many lines of symmetry does a regular octagon have?


Let us play THE GAME. If you want to know the rules, ask at your own risk. If you know how to play THE GAME, I dare you to post a ddr emoticon telling me you know how the play THE GAME. You are allowed to annoy others. Begin![/i]

holy **** a WOOTer actually posted this???

[quote="AIME15"]How do you play?[/quote] You just lost. I will PM you the rules.

Shoot, I just lost The Game.

I lost!!!!!

Anyway, how do you play this game?

wow
i just won

don't worry guys
practice makes perfect

Ah, but it IS possible! See here:

http://xkcd.com/391/

[quote="pythag011"]dude. It is not possible to win the game. If you win the game, you jsut lost it.

And I just lost.[/quote]

it was a joke, duh...

dude
you just made me lose the game

someone lock this please
so i don't keep losing the game...and my sanity

So...if you can't win, isn't this justspam?

people learn it from reading posts with little "i lost" comments on the bottom and PM them about them. Not from a topic that someone posted. Let this float to the bottom.

[url=http://en.wikipedia.org/wiki/The_Game_(mind_game)]Explained here.[/url]
Wiki forgets to mention that this is essentially an internet meme.
By the way, this topic has made me lose about 29 times.

[Moderator: locked to prevent more people from getting doomed]


trigonometry
LaTeX
function
calculus
algebra unsolved
algebra

[quote="lotharek"]Use derivates.[/quote]
Please write your solution with details. NOT SPAM

$f(x) = 100xcosx - 65$
$f'(x) = 100cosx - 100xsinx = 100cosx(1-xtgx)$
Because f is continuous and derivable on $[0, \frac{\pi}{2}]$, we have (Weierstrass) that it is bounded and it touches its upper margin(excuse my english) in a point $M\in[0, \frac{\pi}{2}]$.
If $M=0$ or $M=\frac{\pi}{2}$, then $f(M) = -65 < 0$
If $M\in(0, \frac{\pi}{2})$, then we have $f'(M)=0$
This means $f'(M)=0 \Rightarrow 100cosM(1-tgM)=0\Rightarrow 1-MtgM=0\Rightarrow MtgM=1$
So, if $X$ is the set of sollutions of the equation $xtgx=1 \Leftrightarrow cosx=xsinx$ in $(0,\frac{\pi}{2})$, because $xtgx$ is strictly monotoneus,so injective and $\lambda$ is the unique sollution, we have
$f(x)\leq f(\lambda)=100\lambda cos\lambda - 65=100\lambda^{2}sin\lambda-65\leq 100\lambda^{2}-65$
Because $xtgx$ is strictly increasing, we have $\frac{\pi}{4}tg\frac{\pi}{4}=\frac{\pi}{4}>0=\lambda tg\lambda$, so $\lambda<\frac{\pi}{4}$
So, $f(x)\leq100\frac{\pi^{2}}{16}-65<0$

your solution is correct I think but you should try to learn Latex better because someone read your solution with difficulty. try to improve it :)

I am wonder if there exist a different solution?

[quote="Frozen"]
So, if $X$ is the set of sollutions of the equation $xtgx=1$ in $(0,\frac{\pi}{2})$ and $\lambda$ is an arbitrary element from $X$ we have
$f(x)\leq max\left\{f(\lambda)\right\}=max\left\{100\lambda cos\lambda - 65\right\}=max\left\{100\frac{cos^{2}\lambda}{sin\lambda}-65\right\}=$
$max\left\{100-100sin\lambda-65\right\}=max\left\{35-100sin\lambda\right\}$ [/quote]

What's up with this: $max\left\{100\frac{cos^{2}\lambda}{sin\lambda}-65\right\}=max\left\{100-100sin\lambda-65\right\}$ ?

I wanted to say that the maximum is calculated after the elements of the X. But I realised the X has just an element, because $xtgx$ is a strictly monotoneus function, so it is injective. So there's no need for that max. I'll edit the second post.

[quote="silouan"]If $x\in [0,\frac{\pi}{2}]$ prove that
$100x\cos x<65$[/quote]
I have another calculus solution :blush: .
[hide]Simplifying, we have:
$x\cos x<\frac{65}{100}$

Now $\sqrt {x\cos x} \le \frac{x + \cos x}{2}$ by AM-GM

Yielding: $2 \sqrt{ x\cos x} \le x + \cos x$             (1)

We differentiate the RHS of (1) and are left with $1 - \sin x$.

In the interval $[0,\frac{\pi}{2}]$ this is always positive, so $x + \cos x$ is always increaing in the same interval. So the maximum value of $x + \cos x$ in this interval occurs at $\frac{\pi}{2}$. This maximum value is $\frac{\pi}{2}$, from pluging back into the RHS of (1).  So substituting back into (1), we have $2 \sqrt{ x\cos x} \le \frac{\pi}{2}$

$\sqrt{ x\cos x} \le \frac{\pi}{4}$

$ x\cos x \le (\frac{\pi}{4})^2 < \frac{65}{100}$

So $100x\cos x<\ 65$

[b]QED[/b][/hide]
Okay, I'm probably wrong somewhere, though. :oops:

But this step is quite suspicious :)
$100\frac{cos^{2}\lambda}{sin\lambda}-65=100-100sin\lambda-65$
because it implies $\lambda=\frac{\pi}{2}$

1234567890's solution i think is correct :)

Sorry, I repaired the error. Anyways, my solution sucks.. 1234567890's is better.

[quote="Frozen"]$f(x) = 100xcosx - 65$
$f'(x) = 100cosx - 100xsinx = 100cosx(1-xtgx)$
Because f is continuous and derivable on $[0, \frac{\pi}{2}]$, we have (Weierstrass) that it is bounded and it touches its upper margin(excuse my english) in a point $M\in[0, \frac{\pi}{2}]$.
If $M=0$ or $M=\frac{\pi}{2}$, then $f(M) = -65 < 0$
If $M\in(0, \frac{\pi}{2})$, then we have $f'(M)=0$
This means $f'(M)=0 \Rightarrow 100cosM(1-tgM)=0\Rightarrow 1-MtgM=0\Rightarrow MtgM=1$
So, if $X$ is the set of sollutions of the equation $xtgx=1 \Leftrightarrow cosx=xsinx$ in $(0,\frac{\pi}{2})$, because $xtgx$ is strictly monotoneus,so injective and $\lambda$ is the unique sollution, we have
$f(x)\leq f(\lambda)=100\lambda cos\lambda - 65=100\lambda^{2}sin\lambda-65\leq 100\lambda^{2}-65\leq100\frac{\pi^{2}}{4}-65<0$
So, $f(x)<0$[/quote]

Unfortunately $100\frac{\pi^{2}}{4}-65>0$ but maybe you can fix it finding more special $\lambda$ .

[quote="1234567890"]In the interval $[0,\frac{\pi}{2}]$ this is always positive, so $x + \cos x$ is always increaing in the same interval. So the maximum value of $x + \cos x$ in this interval occurs at $\frac{\pi}{2}$. [/quote]
I didnot understood it. Ok  $x{max}=\frac{\pi}{2}$ but $\cos x{max}$ attainted at 0.
So when $x$ increases $\cos x$ decreases. what do we have?

Final edit :P If I have a mistake now too, I'll give up :D
Well, $(x+cosx)'=1-sinx\geq 0$, so $x+cosx$ is increasing.

[quote="Frozen"]$f(x) = 100xcosx - 65$
...
$f(x)\leq f(\lambda)=100\lambda cos\lambda - 65=100\lambda^{2}sin\lambda-65\leq 100\lambda^{2}-65$
Because $xtgx$ is strictly increasing, we have $\frac{\pi}{4}tg\frac{\pi}{4}=\frac{\pi}{4}>0=\lambda tg\lambda$, so $\lambda<\frac{\pi}{4}$
So, $f(x)\leq100\frac{\pi^{2}}{16}-65<0$[/quote]

$\lambda tg\lambda=1$  so $\lambda>\frac{\pi}{4}$ :roll:
but if you gave the problem some time surely you'd find the proper $\lambda$ and your solution would be correct ;)


trigonometry
calculus
function
geometry
real analysis
real analysis solved

Find Closed Form:

$\sum_{n=1,3,5\ldots} \frac{1}{n}e^{-n\pi x/a}\sin{(n\pi y/a)}$

Please don't double post in multiple forums...I am going to delete your post in the Intermediate Forum.

with the 40 thousand forums these days, I never know where to post or what level my question falls under... is this problem too advanced for this forum?

Given the nature of the sum, this belongs somewhere in Calculus-Analysis; this is evaluated by interpreting it as a power series.

It's certainly impossible without at least some calculus material.

One observation:

If we let $u(x,y)=\sum_{\substack{n=1 \\ n \,\text{odd}}}^{\infty} \frac1ne^{-n\pi x/a} \sin(n\pi y/a)$ for $x>0$

then $\frac{\partial^2u}{\partial x^2}+\frac{\partial^2u}{\partial y^2}=0.$

In other words, this is a harmonic function on the right hand half plane with boundary values equal to a "square wave":

$u(0,y)= \frac{\pi}4$ for $0<y<a,$ $=-\frac{\pi}4$ for $-a<y<0,$ and continued to be periodic of period $2a.$

As for a closed form solution, I'll get it started.

Let $r=e^{-\pi x/a}$ and let $\theta = \frac{\pi y}a.$ Then we want to compute

$\sum_{\substack{n=1 \\ n \,\text{odd}}}^{\infty} \frac1nr^n\sin(n\theta) = \text{Im}\left(\sum_{\substack{n=1 \\ n \,\text{odd}}}^{\infty} \frac1nz^n\right).$

Let $g(z)=\sum_{\substack{n=1 \\ n \,\text{odd}}}^{\infty} \frac1nz^n.$ Differentiate that to get $g(z)=\frac1{1-z^2}.$

Integrate that and take the imaginary part. The sum we are looking for is

$\frac12\text{Arg}\left(\frac{1+z}{1-z}\right).$

Someone else can take over from here and finish this.

[quote="Myth"][quote="Spoon"]would an admin be so kind as to move this post?[/quote]
You should ask moderators for that ;)[/quote]

by "admin" I think I meant "anyone with the power to move a post"... I was not being so specific...

Anyway, Kent, good catch on Laplace's Equation... This solution actually resulted from solving laplace's equation to begin with :).

ok... so like could the moderator please put my post back where it was? Since it's been moved, nobody has even given it a look.

Thanks

[quote="Ravi B"][hide="Hint"]You can use the identity
\[\sin \theta = \frac{e^{i\theta} - e^{-i\theta}}{2i}.\][/hide][/quote]
Yes, this maybe the first proof traightforward....

I don't see how that leads to the solution...

Well, I've got an answer for you in #8, but I didn't finish it.

$u=\frac12(\arg(1+z)-\arg(1-z))$ where

$z=re^{i\theta},\, r=e^{-\pi x/a},\,\theta=\frac{\pi y}a.$

Of course, there's a long way to go.

The first step after that is the geometry-trigonometry problem of computing $\arg(1+re^{i\theta})$ and $\arg(1-re^{i\theta}).$

That should take some time; I would suggest starting by drawing pictures of triangles.

Then you can put in what $r$ and $\theta$ are.


algorithm
search
number theory
prime factorization

could someone please explain this code line by line for me, or at least show a different code for prime factorization with an explanation (Dr. Scheme please).

(define (primefactorization x)
  (let inner ((c 2) (p x))
    (if (> (* c c) p)
        (list p)
        (if (zero? (remainder p c))
          (cons c (inner c (quotient p c)))
          (inner (+ c 1) p)))))

I really don't understand some of the bits of code in the above program. What language is it in?? If you could give me the algo I'll analyze it.

I'm guessing this is what you want:
Given 72 - 2^3 x 3^2

An algorithm:
First search for a prime factor which divides. Once you have found it then start a sub loop and go on till you find the largest exponent of the prime factor which divides the number.

[hide="Sample Code"]
In C++
k=n;j=0;
for(i=2;i<=(n/2);i++)
{
 while(!(k%i))
 {
  k=k/i;
  j++
 }
 cout<<[i Power j] ;
 j=0;
 k=n;
}

  [/hide]

[quote="Cyber-Shreyas"]I really don't understand some of the bits of code in the above program. What language is it in?? If you could give me the algo I'll analyze it.
[/quote]

That's exactly what he's asking you to do :P And its in Scheme, as he said.


probability
geometry
Gauss
conics
ellipse
3D geometry
sphere

OK instead of SD III, welcome to my new competition, Survivor, going along with perfect628's reality TV theme.

THERE IS MATH INVOLVED, DON'T WORRY

I am limiting to 18 people, but I can accomadate more if they really want to do it. I doubt that many people will, in spite of SD II's surprising popularity (35-40 or so)

Soo here's how it will work:

First off there will be 2 tribes at the beginning. There might be reshufflings, and there will eventually be a merge. In reward challenges, the tribes will compete for some kind of advantage in the immunity challenge, and it might not just be points. In the immunity challenge, the teams will compete for immunity. The loser goes to tribal council, where someone will be voted out of the tribe. The cycle continues until the merge.

After the merge, the procedure will be similar. Reward challenge, Immunity, except the challenges will be individual. Tribal council will involve everyone, but you can't vote for one person. We continue until 2 remain, then the jury of 7 (7 most recent people voted out) votes to choose a THE SOLE SURVIVOR. No 1,000,000 dollars, sorry.

Here's the math:

Reward Challenge (not always will it happen, due to time constraints, i don't want this taking years): This will be a series of around 10 short answer questions, sprint level, approximately in increasing difficulty. Each will have a point value, 1-5 probably. Rewards vary, such as some number of extra points on the immunity, freezing people of the opposite tribe, etc. No calculators.

Immunity Challenge: Around 5 questions requiring solutions, target level mostly, but some sprint level questions. Calculators allowed. Winning tribe/person gets immunity

Exile Island:

At the beginning of the game I will have determined a number between 1 and 1000. At points during the game certain people will be sent to exile island. At exile island, you have 2 guesses as to which number it is. The first person to go there has no clues, the 2nd person gets one clue, the 3rd person gets that clue and a new clue, etc. If someone guesses the number correct, they win the hidden Immunity idol. If he/she gets the highest amount of votes at a tribal council, they play the idol AFTER the votes have been tallied and the person with the 2nd-most votes is voted out.

Giving the idol to other people IN YOUR TRIBE is legal, but you must tell me first, so I know where the idol is at all times.

More information will be given out later in the game. Registration is now closed.
[size=0]
HERE IS THE FIRST IMMUNITY CHALLENGE.
[hide]
CHALLENGE IS FOR IMMUNITY. The losing tribe will vote out one of its own “tonight.” The top THREE scorers in the losing tribe are immune from that vote. So basically, the losing tribe votes out one of its bottom four in this challenge. In addition, the person with the THIRD highest score on the winning tribe is sent to exile island (remember, it’s a good thing in this game.)

Each problem is worth 7 points. Solutions are required. Just an answer will receive at most one point. I am being strict on explaining, you must explain everything that you are doing. Once you submit a solution to a problem, you may NOT resubmit a solution to the same problem. Not submitting everything at the same time is discouraged, though allowed. Good luck!

1. Find the value of $1^{2}+2^{2}+3^{2}+.....+20^{2}$.
2. Find the number of ways to scramble the word OKALOOSA, so that it begins and ends with a consonant.
3. Define the Fibonacci sequence to be $f_{1}=1$, $f_{2}=1$, and $f_{n}=f_{n-1}+f_{n-2}$ for all integers $n$. A random integer is selected between 1 and 1600, inclusive. Find the probablility that $f_{n}$ ends in 0 or 5. Express your answer as a common fraction.
4. Find the area of the quadrilateral enclosed by the vertices (0,0), (1,1), (3,7), and (-4,2).

Bonus (3 points):

A man has 8 identical bars of metal, in that the volumes are the same. A man has a two-pan balance, which can be used to determine whether or not two objects are the same weight, or which is heavier. 7 of the bars are gold, 1 is fool’s gold. If he can only use the two-pan balance, describe the most efficient (least weighings) way for him to determine which is fool’s gold. Assume that fool's gold is lighter than real gold.[/hide][/size]

Great-OK a tip to people who aren't familiar with the format of the game...

Thursdays at 8:00 EST on CBS. You don't have to know the people and the tribes, just the format of the game/competition.

[quote="13375P34K43V312"]We continue until 2 remain, then the jury of 7 (7 most recent people voted out) votes to choose a THE SOLE SURVIVOR. [/quote]
imo, the winner should be the person who got the most correct answers after the merge, or decide by extra hard problems, or some otherway related to math or their performance in the game.

ps. this game sounds cool. I guess I'll have to check this forum once in a while to look at this thing :lol:

[quote="bpms"]Does the format and all really matter that much? The way I look at it, work your hardest on the problems, and you will do well. Is that too hard?[/quote]

The format does matter because in the survivor format, it's not about who's the best (it is if you can get all the immunities) but who doesn't get voted off the island. you can be the best and still get voted off, which isn't that fair in my opinion.

OK how about this:

The sole survivor is voted by the 7 jury members, BUT the number of individual immunities factors into their vote total. For example, if I had the jury's vote, 5-2, but my opponent had won 4 individual immunities while I won 0, he would win 6-5.

[quote="SplashD"]I'm also wondering, how do you get sent to Exile Island?
[/quote]
Sometimes, I'll have the losing tribe chose one member of the winning tribe to get sent to exile, or vice versa. I might say the person who comes in last in a reward challenge gets sent to exile, something like that.

Current survivors so far:

SplashD
bpms
goldendomer
anirudh
kstan013
ProtestanT
jhredsox
ragnarok23
meenamathgirl
slamchess
perfect628
easyas3.14159...

I need either 4 or 6 more.

are you going to randomly destroy someone?

All right, it doesn't look like we're going to get anymore, so we're going to start with 14. Confirmation PMs are going out, you MUST respond to them within 4 days. Tribal Assignments (subject to change) are below. Note: the ?? next to a person's name means that they have yet to confirm.

FERMAT TRIBE

ragnarok23
jhredsox
SplashD
perfect628
meenamathgirl
bryantacademy
easyas3.14159...

GAUSS TRIBE

bpms
goldendomer
slamchess
kstan013
anirudh
Fanatic
ProtestanT

I will still allow people to sign up, as long as we maintain an even number of people.

are we allowed to switch tribes with somebody from the other tribe?

This format is a little flawed-the winner is voted on, and so is who's eliminated. So, in the end, its not math-based. That's why I didn't use BB for mine. You should at least try to make it math related.

OK, that sounds like a good idea.

On the losing tribe, the two highest scores have immunity at tribal council.

Most likely 9 people will go to the merge, since 7 people will end up on the jury.

i know!!!! :lol:
i suck why not start one where the person with the lowest points wins?
how about i make it all in favor PM me.

I don't either but I thought it would be fun anyway. I am just eccentric no one pay attention to the funcia behind the curtain.

umm....i think this should be locked...

-jorian

You know you won't have to worry about that if you just let it drop to the bottom rather than deprive the thread of that opportunity by posting as it begins to fall to the bottom.

i just said that so people wouldn't discuss things (yes, i was trying to let it drop)

anyways, let it drop

and check out my blog :P

-jorian

So what you want is a comp. for people who aren't good at math? Sprry for posting.

No that's not what he said. He was expressing frustration that a comp like this, designed so that the best mathematician may not win (although probably will because most of us would agree that we would give our jury votes to the best performer), was just a popularity contest.

It didn't make sense to vote out jhredsox cause he was the best, obviously. But it also didn't make sense to vote him out cause he was the worst. Both weeks you guys voted him out there were people that performed worst.

It sucks that this is over :(. I wanted to see how me vs. perfect would've mulled over.

Oh, and perfect, it wasn't anything personal. Anirudh and I just went "ok we can't vote out splashD, anirudh knows meena, and the next highest is perfect so... let's vote perfect!". And if jhredsox had voted correctly, you'd bet out =/.

ya, we were forming an alliance, but i sent my choice in before i found that out :P

and i'm not really saying the most popular wins, but someone horrible at math, like me, could never win it. i sent a pm to perfect628 about what i mean.

---end of discussion---

-jorian

Wait. Now I'm confused. If so, then I agree w/ perfect and leetspeaketc. There isn't a point of having a math contest where the best at math doesn't win.

well...i mean it would be nice if i didn't just join competitions online and get nowhere...

-jorian

you get nowhere because you think you're bad at math compared to all these other people your age    :rotfl:

If anyone posts beyond this I'm seriously going to have to track you down and murder you.
Seriously, actually listen to me this time and ignore this thread.


You have a whole pizza in the refrigerator.  On your first trip to the refrigerator, you eat half of the pizza.  On each successive trip, you eat half of the remaining pizza.  After five trips to the refrigerator, what fractional part of the pizza have you eaten?

Yeah, that's what I get too... And btw- just tell that dude to finish the pizza already... it can't be that big because otherwise he couldn't eat half of it in the first place.  :)  :)  :)  :lol:  :P

he will [b]never[/b] finish the pizza...

[color=indigo]yeah, this is one of those never ending series (the limit is 1).

[hide]trip 1: 1/2
trip 2: 3/4
trip 3: 7/8
trip 4: 15/16
trip 5: 31/32[/color][/hide]

[hide]After an infinite amount of trips he will finish.
That will take a long time though.[/hide]


This is a famous series, Libniz or Gregory series, the answer is $\frac{\pi}{4}$.

[url=http://imageshack.us][img]http://img503.imageshack.us/img503/7348/seriescn1.jpg[/img][/url]
Shot at 2007-06-28

why didn't I think of that? thanks for the info.


Find the term free of y in the expansion of (3xy²-1/6x²y³)²°. Simplify the term.

[hide]Hmmm..... the term free of y, must have for the first term, say, exponent a, and for the second term, exponent b, such:

2a = 3b
a + b = 20

After simplification, it yields
a=12
b=8

Now, the other factors of the term, are 3, 1/6, x, 1/x2 and the factor (8 20) i think.

3^12 / 6 ^ 8 = 3^12 / (3 ^ 8) (2^8) = (3/4)^4

x^12 / x ^ 16 = 1/x^4

Then the final term is

(8 20) (3/4x)^4

Sorry, I don't know yet how to post with your notation.[/hide]
Angel "Java" Lopez


The director of the school cafeteria reported that $ 70\%$ of the girls
and $ 89\%$ of the boys in the school ate the school lunch on Friday.
This represents $ 80\%$ of the students in the school. What is the ratio
of girls to boys in the school? Express your answer as a common
fraction.

[hide="solution"]Let $ b\equal{}\text{total number of boys}$ and $ g\equal{}\text{total number of girls}.$ We have the equation \[ \frac{70}{100}g\plus{}\frac{89}{100}b\equal{}\frac{80}{100}(g\plus{}b).\] Clearing the denominator gives \[ 70g\plus{}89b\equal{}80b\plus{}80g\Rightarrow9b\equal{}10g.\] This gives the ratio \[ \frac{g}{b}\equal{}\boxed{\frac{9}{10}}.\][/hide]


group theory
abstract algebra
superior algebra
superior algebra unsolved

Suppose that $G$ is group and $p$ is the smallest prime number that there exist $H\leq G$ and $[G: H]=p$. Prove that $H$ is normal subgroup of $G$.

As it is stated, the problem is wrong. A simple counterexample would be $G=A_5$, with $H$ being the subgroup of $G$ consisting of those elements of $A_5$ which fix the symbol $5$. We have $[G: H]=5$, which is our prime $p$. $5$ is also the smallest prime which occurs as an index of a subgroup of $H$, because a simple group (such as $G$) cannot have subgroups of index $\le 4$ (an easy exercise). However, since $G$ is simple, $H$ cannot be normal.

I suspect that what you really meant was for $p$ to be the smallest prime divisor of the numbers occuring as finite indices of proper subgroups of $G$, in which case the problem is indeed correct:

Consider the action of $G$ by left multiplication on the $p$-element set of left cosets of $H$. This action induces a homomorphism $\varphi: G\to S_p$, so $|G/\ker\varphi|\ |\ p!$. If $|G/\ker\varphi|\ne p\ (*)$, we get a contradiction (notice that $\bigcap_{g\in G}gHg^{-1}=\ker\varphi\le H$, so $|G/\ker\varphi|>1$), which means that we must have equality in $(*)$. In turn, this means that $\ker\varphi=H\Rightarrow H\lhd G$, as desired.

That would imply that there are no simple groups not having prime order ;)
When you meant that: $p$ is the minimal integer $>1$ with that property and happens to be prime: it's also wrong.
Indeed, the last one works for $p=2,3,4$ (where $p=4$ immediatly gives a contradiction), but there are counterexamples for $p=5$, e.g. $A_4 \subset A_5$.

Is that a reply to Omid's message or to mine? :)

I started writing when your post wasn't there.


Given in the plane two non-parallel segments AB and CD, find the locus of the points P of the plane so that the area of the triangle ABP equals the area of the triangle CDP.

Extend segments $ AB$ and $ CD$. Let the resulting lines be represented by equations $ ax\plus{}by\plus{}m\equal{}0$ and $ cx\plus{}dy\plus{}n\equal{}0$, respectively. Let $ P_a$ represent the perpendicular from point $ P$ to line segment $ AB$ and let $ P_c$ represent the perpendicular from point $ P$ to line segment $ CD$. $ Let P\equal{}(x_o,y_o)$ From scalar projections, we know that

$ P_a\equal{}\frac{ax_o\plus{}by_o\plus{}n}{\sqrt{a^2\plus{}b^2}} \\
P_c\equal{}\frac{cx_o\plus{}dy_o\plus{}n}{\sqrt{c^2\plus{}d^2}}$

We have that
$ [ABP]\equal{}[CDP] \\
\frac{1}{2}(AB)(P_a)\equal{}\frac{1}{2}(CD)(P_c) \\
(AB)(\frac{ax_o\plus{}by_o\plus{}n}{\sqrt{a^2\plus{}b^2}})\equal{}(CD)(\frac{cx_o\plus{}dy_o\plus{}n}{\sqrt{c^2\plus{}d^2}}) \\ k\equal{}\frac{AB}{CD}\frac{\sqrt{c^2\plus{}d^2}}{\sqrt{a^2\plus{}b^2}}\equal{}\frac{cx_o\plus{}dy_o\plus{}n}{ax_o\plus{}by_o\plus{}n} $ for some constant $ k$.

$ (ak\minus{}c)x_o\plus{}(bk\minus{}d)y_o\plus{}(mk\minus{}n)\equal{}0$

Locus is some line with some convoluted variables.

Provide a geometric construction for that locus.

Your answer isn't correct anyway...you can't just drop the absolute value bars like that.

There's no need to use such heavy algebra... consider, for a particular value $ A$ of the area, the locus of points that produce that area for one of the segments, and then for the other segment.
[hide="Continued"]If $ L$ is the length of the segment and $ \ell$ is the line containing the segment, and $ A$ is the target area, then the locus is the set of points on either of the two lines parallel to $ \ell$ at a perpendicular distance of $ \frac {2A}{L}$[/hide]

If we do this with the other segment as well, we will get a similar construction
[hide="Continued"]Thus, for any particular target value of $ A$, we will have two pairs of parallel lines, and the four intersections mark the four possible points at which the two triangle areas are the same.[/hide]

Now, what happens if we vary $ A$?

Oh whoops. I wasn't paying attention and forgot about the absolute value signs.

So, you ultimately get two lines. So does one just plot the four points that correspond to an arbitrary $ A$, perhaps plot the single point where $ A\equal{}0$ and just connect?

[hide]Call the line containing segment $ AB,CD$ as $ l_1,l_2$ and denote the length of $ AB, CD$ as $ a,b$, respectively. The locus is then the two lines each of which divides one of the supplementary angles formed by $ l_1,l_2$ into $ \alpha, \beta$ such that $ \sin \alpha: \sin \beta \equal{} b: a$, where $ \alpha$ shares a side with $ l_1$ and $ \beta$ shares a side with $ l_2$. Briefly speaking, it's all the points $ P$ such that $ d(P,l_1): d(P,l_2) \equal{} b: a$.

Construction:
Denote the intersection of $ l_1,l_2$ as $ O$. Draw circles with radius $ a,b$ and center $ O$. Let one intersection on $ l_1$ be $ P$ and the two intersections on $ l_2$ be $ Q_1, Q_2$. Draw lines $ PQ_1$, $ PQ_2$. Finally, draw lines $ L_1 // PQ_1$ and $ L_2 // PQ_2$ past $ O$. $ L_1\cup L_2$ are the desired locus.[/hide]


factorial
floor function
induction
number theory proposed
number theory

Prove that $ \frac{(2n)!}{(n!)^2}$ is even if $ n$ is a positive integer.

Can someone check if my proof is correct. Thanks

[hide="My proof"]
Note $ n$ is a positive integer, so we're not dealing with non-integer factorials involving the Gamma function.

Rewrite $ \frac{(2n)!}{(n!)^2}$ as $ 2*\frac{(2n-1)!}{n!(n-1)!}$.

We need to show $ \frac{(2n-1)!}{n!(n-1)!}$ is an integer, and the proof will be finished (because any integers times two is an even integer).

This will be done by showing that every prime divides the numerator to at least as high a power as it divides the denominator (implying $ n!(n-1)! | (2n-1)!$ and hence $ \frac{(2n-1)!}{n!(n-1)!}$ is an integer). The power of a prime that divides the denominator is the sum of the power of the prime that divides $ n$ and the power of the prime that divides $ n-1$. Using de Polignac's formula (Let $ p$ denote a prime. Then the largest exponent $ e$ such that $ p^e | n!$ is $ e = \sum_{i=1}^{\infty}{\left\lfloor\frac{n}{p^i}\right\rfloor}$), we get the statement we want to prove:

$ \sum_{i=1}^{\infty}{\left\lfloor\frac{2n-1}{p^i}\right\rfloor\ge\sum_{i=1}^{\infty}{\left\lfloor\frac{n}{p^i}\right\rfloor + \sum_{i=1}^{\infty}{\left\lfloor\frac{n-1}{p^i}\right\rfloor}}}$

Recall the theorem that $ \lfloor{x}\rfloor+\lfloor{y}\rfloor \le \lfloor{x+y}\rfloor$.

Using this theorem, we get $ \left\lfloor\frac{n}{p^i}\right\rfloor + \left\lfloor\frac{n-1}{p^i}\right\rfloor \le
\left\lfloor\frac{n+(n-1)}{p^i}\right\rfloor = \left\lfloor\frac{2n-1}{p^i}\right\rfloor$.

Summing this over $ i$ we have our desired result.

$ \sum_{i=1}^{\infty}{\left\lfloor\frac{2n-1}{p^i}\right\rfloor\ge\sum_{i=1}^{\infty}{\left\lfloor\frac{n}{p^i}\right\rfloor + \sum_{i=1}^{\infty}{\left\lfloor\frac{n-1}{p^i}\right\rfloor}}}$[/hide]

[quote="ark_xm"]Prove that $ \frac {(2n)!}{(n!)^2}$ is even if $ n$ is a positive integer.

Can someone check if my proof is correct. Thanks[/quote]

There's an easier proof:

$ \frac {(2n)!}{(n!)^2}$ can be rewritten as $ 2\frac {(2n - 1)!}{n!(n - 1)!}$ which is the same as ${ 2{2n - 1 \choose n}}$. Since we know that all combinations are integers, $ {2n - 1 \choose n}$ is an integer and therefore $ 2{2n - 1 \choose n}$ is even.

But yes, your proof is correct

[quote="kunny"]$ \frac{(2n)!}{(n!)^2}\equal{}_{2n}C_n$[/quote]

How do you show from here that $ \frac{(2n)!}{(n!)^2}$ is even? :maybe:

The more general problem .
Show that exponent of $ p$ in $ \frac{(np)!}{(n!)^p}$ is $ S_p(n)$ where $ S_p(n)$ is the sum digits of $ n$ on base $ p$ .
It is clear that $ S_2(n)\geq 1$ therefore this number must be an event integer .

[quote="nayel"][quote="kunny"]$ \frac {(2n)!}{(n!)^2} \equal{} _{2n}C_n$[/quote]

How do you show from here that $ \frac {(2n)!}{(n!)^2}$ is even? :maybe:[/quote]

Induction in using $ _n C_r \equal{} _{n \minus{} 1}C_r \plus{} _{n \minus{} 1}C_{r \minus{} 1}$.

$ \sum_{k \equal{} 1}^{2n} \dbinom{2n}{k} \equal{} 2^{2n}$, but $ \dbinom{2n}{n \minus{} k} \equal{} \dbinom{2n}{n \plus{} k}$, so we get:
$ 2(\sum_{k \equal{} 1}^{n \minus{} 1} \dbinom{2n}{k}) \plus{} \dbinom{2n}{n} \equal{} 2^{2n}$, than we easily see that $ \dbinom{2n}{n}$ is even.


Calculate:
$ Lim_{x\minus{}>\infty}\frac{1}{lnx}(sin mx\minus{}sin nx);m,n$ belongs to N

[quote="quocbao153"]Calculate:
$ Lim_{x \minus{} > \infty}\frac {1}{lnx}(sin mx \minus{} sin nx);m,n$ belongs to N[/quote]

$ \frac{\minus{}1}{lnx} \le \frac {1}{lnx}(sin mx \minus{} sin nx) \le \frac{1}{lnx}$ for all $ x>e$

So by Sandwich Rule , $ \lim_{x \to \infty}\frac {1}{lnx}(sin mx \minus{} sin nx)\equal{}0$


conics
ellipse
analytic geometry
graphing lines
slope

Given an ellipse centred at the origin with essentricity $e=\frac{1}{2}$. One focus of the ellipse is located at $(-m,0)$ with $m$ being a positive contant number.

(1) Find the equation of the ellipse.
(2) Let $Q$ be a point on the ellipse. Line $l$ passes through $F$ and $Q$ and meets the y-axis at $M$. If $\overrightarrow{MQ}= 2 \overrightarrow{FQ}$, then find the slope of $l$.

First Part
[hide]There's infinitely many ellipses centered at the origin with eccentricity 1/2.
The eccentricity is c/a where c is the distance from the center to the foci and the a is the distance of the semi-major axis.  In this case c=m.  a=2m.  THere's also another equation that states, $b^2=a^2(1-e^2)$ or $b^2=4m^2(3/4)$ or $b^2=3m^2$.  Therefore, the equation for this family of concentric ellipses is $\frac{x^2}{4m^2}+\frac{y^2}{3m^2}=1$.[/hide]
Second Part
Is F on the ellipse, outside it, or inside it?


inequalities
logarithms
function
inequalities unsolved

let $ x$  , $ y$ , $ z$ $ \geq 0$ such that $ xyz\equal{}1$ , prove that :

$ x^2\plus{}y^2\plus{}z^2\plus{}x\plus{}y\plus{}z \geq 2(xy\plus{}yz\plus{}zx)$

[quote="y-a-s-s-i-n-e"]let $ x$  , $ y$ , $ z$ $ \geq 0$ such that $ xyz = 1$ , prove that :

$ x^2 + y^2 + z^2 + x + y + z \geq 2(xy + yz + zx)$[/quote]
setting $ F(x,y,z) = x^2 + y^2 + z^2 + x + y + z - 2(xy + yz + zx)$

$ F(x,y,z) - F(x,\sqrt {yz},\sqrt {yz}) = (\sqrt {y} - \sqrt {z})^2[(\sqrt {y} + \sqrt {z})^2 + 1 - 2x]$

assume $ x = min(x,y,z) %Error. "arrowright" is a bad command.
x\le 1/3$  ->$ 1 - 2x\ge 0$

$ %Error. "arrowright" is a bad command.
F(x,y,z)\ge F(x,\sqrt {yz},\sqrt {yz})$

setting $ \sqrt {yz} = t$ due to $ xyz = 1$ ->$ x = 1/t^2$

then, we need prove that $ F(1/t^2,t,t)\ge 0$ easy :wink:

The inequality is easy for me. :D

From $ xyz \equal{} 1$, we have $ x\neq 0$ and $ y\neq 0$ and $ z\neq 0$, thus it is suffice to think for $ x > 0,\ y > 0,\ z > 0$.

Let $ f(x) \equal{} x^2 \plus{} x \minus{} \frac {2}{x} \minus{} 5\ln x\ (x > 0)$, $ f'(x) \equal{} \frac {x \plus{} 2}{x^2}(x \minus{} 1)(2x \minus{} 1)$, yielding $ f(x)\geq f(1)\ \therefore f(x)\geq 0$.

$ f(x) \plus{} f(y) \plus{} f(z)\geq 0$ completes the proof.

Kunny, for 0<x<.348 with your new function, f(x)<0 (plug in values to see, I used my graphing calculator).

[quote="dgreenb801"]Bunhiacovski wrote "assume $ x = min(x,y,z) %Error. "arrowright" is a bad command.
x\le 1/3$  ->$ 1 - 2x\ge 0$"
This is not true. Take x=2/3, y=2/3, z=9/4, all greater than 1/2.

Kunny, I am confused, f(1) < 0, right?[/quote]
sorry. I am wrong   :) rectification as follows

$ (\sqrt{y}+\sqrt{z})^2+1-2x=y+z+2\sqrt{yz}+1-2x$

assume $ x = min(x,y,z)$ --> $ 2\sqrt{yz}\ge 2x$

ergo $ (\sqrt{y}+\sqrt{z})^2+1-2x \ge 0$
next.. done :wink:

[quote="dgreenb801"]Bunhiacovski wrote "assume $ x = min(x,y,z) %Error. "arrowright" is a bad command.
x\le 1/3$  ->$ 1 - 2x\ge 0$"
This is not true. Take x=2/3, y=2/3, z=9/4, all greater than 1/2.

Kunny, why do you have a -5 at the end of f(x)? Even still, f(x)<0 for all 0<x<1. (Plug in values to see).[/quote]

Sorry I made a typo. I have just edited.


Uhh, the Introduction to Number Theory Archive can still be posted in. What's up with that?

EDIT: Also, is the same thing happening to the other archived forums?

Its the same for Mathcounts Problem Series.  I think they make it that way so you can still post solutions and discuss with other students if they look at it at least once a week.

But I thought they would lock all topics? Oh well, fine with me, another topic to discuss on!  :D