ratio MATHCOUNTS geometry rectangle algebra polynomial quadratics

So how many of you participate in CML and what are your running scores?

I quit cuz I had to do my project during lunch :P

This is our school's first year in CAML so I can't give an opinion on whether this year is too easy or not. But I would like to point out that the 40 perfect scores this year are only for the first 3 rounds. There were actually 6 perfect scores at the end the 3rd round last year. It's still a big difference though.

[quote="DataBox"]^Your project?

And who is of the opinion that the contest is far too easy this year?  Last time I checked there were almost 40 perfect scorers vs last year's three...[/quote]
I actually will be agreeing with Mr. Saxena. This year's contest seems really trivial for the ones I did at least.
FYI I got 6,5 (blew the last problem with an incorrect similar triangles ratio), 6, n/a

Wait is this California Math League or Continential Math League? B/c CA math league is CAML, not CML

Whoops blew that, sorry, yeah it's caml not cml....

CAML keeps owning me because I'm too busy eating.

^MOPpers should be able to own CAML even when they are asleep....

Heh, what was the answer to number 6 on the latest one?  I guessed 1.8 :blush:

There was a Canadian who got USAMO in 4th grade (and failed to make in this year at 6th)

Yes.

How come he didn't qualify this year?

Maybe he couldn't take amc for some reason?

I've never had my name posted on AoPS before, haha, sweet.

yeah Amador Valley!

[quote="moogra"][quote="Ubemaya"]stop complaining guys, I'm the most random person here

in fact, I bet only 5ish people on this forum know my actual name[/quote]

I think [i]everyone[/i] knows your name.[/quote]

Even I know your name, and I'm from SoCal.

Silly rabbit, walruses don't have names.

Sorry Ubemaya, I am going to spill your real name:
[hide]
[size=200]TEAPOTS[/size]
[/hide]

[quote="r15s11z55y89w21"][quote="moogra"][quote="Ubemaya"]stop complaining guys, I'm the most random person here

in fact, I bet only 5ish people on this forum know my actual name[/quote]

I think [i]everyone[/i] knows your name.[/quote]

I don't.  :([/quote]

Good, because anyone who says he does is a liar (unless I say you're allowed to know).

Also, that was your 666th post... maybe this is a sign... hmm... does it mean my name cannot be said anymore?

[quote="Ubemaya"][/quote][quote="r15s11z55y89w21"][quote="moogra"][quote="Ubemaya"]stop complaining guys, I'm the most random person here

in fact, I bet only 5ish people on this forum know my actual name[/quote]

I think [i]everyone[/i] knows your name.[/quote]

I don't.  :([/quote][quote="Ubemaya"]

Good, because anyone who says he does is a liar (unless I say you're allowed to know).

Also, that was your 666th post... maybe this is a sign... hmm... does it mean my name cannot be said anymore?[/quote]

That was actually not my 666th post.


[b][size=100][color=DarkBlue]Δια της κορυφής $ A$ τετραπλεύρου $ ABCD$ περιγεγραμμένου περί κύκλο $ (K),$ φέρνουμε τις κάθετες ευθείες επί των πλευρών του $ AB,\ AD,$ οι οποίες τέμνουν αντιστοίχως τις διχοτόμους των γωνιών $ \angle B,\ \angle D,$ στα σημεία $ E,\ F.$ Εάν $ P$ είναι το σημείο τομής των ευθειών δια των σημείων $ E,\ F,$ καθέτων επι των ευθειών $ BC,\ CD$ αντιστοίχως, αποδείξτε ότι $ AP\perp BD.$[/color][/size][/b]

Εύχομαι από καρδιάς, ειρήνη παντού, υγεία και χαρά στις οικογένειές σας, προσωπική επιτυχία στις επιδιώξεις και τους στόχους σας για τη νέα χρονιά.

Κώστας Βήττας.

[quote="April"]Thank you dear [b]Kostas Vittas[/b] for your nice problem.

Denote by $ M$ the intersection of $ BC$ and $ PE$, $ N$ the intersection of $ CD$ and $ PF$. We have to prove that $ BP^2 \minus{} BA^2 \equal{} DP^2 \minus{} DA^2$ $ \iff BP^2 \minus{} BM^2 \equal{} DP^2 \minus{} DN^2$, $ \iff MP^2 \equal{} NP^2$ $ \iff CM^2 \equal{} CN^2$ $ \iff CM \equal{} CN$. Indeed, we have $ CM \equal{} BM \minus{} BC \equal{} BA \minus{} BC \equal{} DA \minus{} DC \equal{} DN \minus{} DC \equal{} CN$. (I used a well-known equation in a circumscribed quadrilateral $ ABCD$ that $ AB \plus{} CD \equal{} AD \plus{} BC$.)

[b]Remark.[/b] 1. We also can prove a result which was mentioned by [b]Darij Grinberg[/b] in his paper [url=http://www.cip.ifi.lmu.de/~grinberg/CircumRev.pdf]Circumscribed quadrilaterals revisited[/url] that $ EF\perp AC$ by the same argument.

2. Happy New Year! I wish you a joyful celebration and an amazing year ahead...[/quote]

Dear Son,

 the Greek Community sends  to you [color=red][b]the best wishes for  a happy new Year ![/b][/color]

 Babis


inequalities
group theory
number theory
relatively prime
superior algebra
superior algebra unsolved

[quote][b]

Suppose we have a group $G$ of order $|G|.$ For $d \hspace{.1in}|\hspace{.1in}|G|,$ let $\psi (d)$ be the number of elements in $G$ of order $d.$ And $\phi (d)$ is the Euler's function defined to be the number of integers between 1 and $d$ relatively prime to d.

Then we have that
\[\psi (d) \leq \phi (d) \hspace{.5in}\text{for all}\hspace{.5in}d\hspace{.1in}|\hspace{.1in}|G| \]
[/b][/quote]

Can anyone explain why this is true? Thanks!

Thank you for your reply. The reason why I'm asking this question is that when I was reading a proof that "$U(\mathbb{Z}/p\mathbb{Z})$ is a cyclic proof" the author first assert that

[quote][b]
From group-theoretic view,
\[\psi (d) \leq \phi (d) \hspace{.5in}\text{for}\hspace{.5in}d|p-1 \]
[/b][/quote]

Then eventually he proves that $\psi (d)$ is actually equal to $\phi (d).$ So I thought that he might used the lemma in my first entry.

$\psi(d) \leq \phi(d) \iff \psi(d) \leq \phi(d) \iff G$ is cyclic.

But you get $\psi(d) \leq \phi(d)$ from the number of solutions of $x^{d}=1$ in your field, and using the equivalence above, or better: only one direction (the author should show it), the result follows.

[quote][b]
But you get $\psi(d) \leq \phi(d)$ from the number of solutions of $x^{d}=1$ in your field,
[/b][/quote]

I'm sorry but why? Can you explain this more clearly?

[quote][b]
It's wrong, take any non-cyclic group.
[/b][/quote]

Well, consider the non-cyclic group $\mathbb{Z}$ under addition.
For any integer $d$, the number of elements of order $d$ is $1$, becuase only the additive identity $0$ has order $d.$

On the other hand, the number of elements between $1$ and $d$ relatively prime to $d$ can be greater than $1$ for any $d$. Hence the non-cyclic group $\mathbb{Z}$ does not surve as a counterexample.

ZetaX, is there anything wrong with this argument?

we're talking of finite groups, of course :)
the property doesn't even make sense for non-finite groups..

Ok, but how do you prove that in $U(\mathbb{Z}/p\mathbb{Z})$
\[\psi (d) \leq \phi (d) \hspace{.5in}\text{for}\hspace{.5in}d|p-1 \]

suppose the contrary, and take $d$ such that the inequality is reversed and strict. now consider the equation $x^{d}= 1$. take a primitive root $y$, and the set $\{y^{k}\}$, with $k$ coprime with $d$, and less than it. they're all primitive $d$-th roots, but, because of that inequality, they're not [i]all[/i] primitive $d$-th roots, so there's $y'$ outside of that set, which is still such root. now the set $\{y^{h}, y'^{k}\}$, as $0<h,k<d+1$ is a subset of the set of roots, and has cardinality strictly bigger than $d$. but that's not possible, since we're in a field.

quite a long way to prove it, though nice :)

To ma_go,

Thank you. Now suppose you don't know that U(Z/pZ) is cyclic, so that there isn't a primitive root y, can you still prove this?

Can anyone else think about this too?

i just proved that if $\psi(d) \le \phi(d)$, in $(\mathbb{Z}/p\mathbb{Z})^{*}$, which is what you needed for the proof of ciclicity, right?
i didn't use ciclicity at all.


[quote="various"][quote="lightrhee"]This is problem #3 from second round of Korean Mathematics Olympiad.

Show that the equation, $2x^4+2x^2y^2+y^4=z^2$, does not have integer solution when $x \neq 0$.[/quote]
stupid question: x,y,z are in Z?[/quote]

[quote="lightrhee"]...does not have[b] integer solution [/b]when ...[/quote]

I Hope it helped. :D

[quote="lightrhee"]This is problem #3 from second round of Korean Mathematics Olympiad.

Show that the equation, $2x^4+2x^2y^2+y^4=z^2$, does not have integer solution when $x \neq 0$.[/quote]

We have $(x^2)^2+(x^2+y^2)^2=z^2$  :lol:

Note that the equation can be rewritten as $(x^2)^2 + (x^2+y^2)^2 = z^2.$

Assume, for contradiction, that there exist solutions to the given equation where $x \neq 0.$ Then, we can WLOG assume that $x, y, z > 0$ (as $y = 0$ clearly doesn't work).

Consider the solution $(x, y, z)$ to the given equation which minimizes $z$ (take an arbitrary one if there are more than one).

If $\gcd(x, y) \neq 1$, then divide $x, y$ by $\gcd(x,y)$ and $z$ by $\gcd(x,y)^2$ to get a smaller solution.

Henceforth, we must have that $x, y$ are relatively prime.

By the well-known parameterization of Pythagorean triples, since $(x^2, x^2+y^2, z)$ is a Pythagorean triple with all side lengths pairwise relatively prime, we have that there exist $a, b$ so that either $a^2 - b^2 = x^2, 2ab = x^2+y^2$ or $a^2 - b^2 = x^2+y^2, 2ab = x^2.$

Note that if $x$ is odd, we have $(x^2)^2 + (x^2+y^2)^2 \equiv 2, 5$ (mod $8$). This is absurd, as $z^2 \equiv 0, 1$ (mod $8$). We therefore have that $x$ is even and so $y$ is odd (because $\gcd(x, y) = 1$). Hence, $x^2 +y^2$ is odd and therefore we must have $a^2 - b^2 = x^2+y^2, 2ab = x^2.$

We have now:

$$a^2 - 2ab-b^2 = y^2,$$

$$2ab = x^2.$$

As $\gcd(x, y) = 1$, we also must have $\gcd(a, b) = 1.$ Furthermore, because $a^2 - b^2 \equiv 0, 1, 2$ (mod $4$), we must have that $a$ is odd. We can't have $b$ odd because $4|2ab$, hence $a$ is odd and $b$ is even. Therefore, $\gcd(2b, a) = 1$ and so since $(2b)(a)$ is a perfect square, we know that there exist $v, t \in \mathbb{N}$ so that $2b = (2t)^2, a = v^2.$ By $\gcd(a, b) = 1$, we have $\gcd(t, v) = 1.$ Then, we have that $x = 2tv$ and so we have:

$$y^2 = a^2 - b^2 - x^2 = v^4 - 4v^2t^2 - 4t^4.$$

This can be rewritten as:

$$y^2 + (v^2+2t^2)^2 = 2v^4,$$

which implies that

$$(v^2 + 2t^2 - y)^2 + (v^2 + 2t^2 + y)^2 = 4v^4.$$

Observe that $v$ and $y$ are both odd, and so $2|v^2 + 2t^2 - y, v^2 + 2t^2 + y, 2v^2.$ Hence, we have that $\left(\frac{v^2 + 2t^2-y}{2}, \frac{v^2 + 2t^2 +y}{2}, v^2\right)$ is a Pythagorean triple.

We claim it's a primitive Pythagorean triple. Since $4 \nmid 2y$, it's easy to see that $2$ doesn't divide all the terms. If some prime $p$ divided all the terms, then we'd have that $p|v$ since $p|v^2$, $p|y$ since $p$ divides the difference of the first two numbers, which implies $p|t$ since $p|v^2 + 2t^2 - y.$ This implies that $\gcd(t, v) \neq 1$, which is absurd.

So hence this is a primitive Pythagorean triple, and so there exist relatively prime $\alpha, \beta \in \mathbb{N}$ so that:

$$\{\alpha^2 - \beta^2, 2 \alpha \beta\} = \{\frac{v^2 + 2t^2 -y}{2}, \frac{v^2 + 2t^2 + y}{2}\},$$

and
$$\alpha^2 + \beta^2 = v^2.$$

The first equation implies that $\alpha^2 - \beta^2 + 2 \alpha \beta = v^2 + 2t^2,$ and so subtracting the second equation gives $2t^2 = 2 \beta(\alpha - \beta) \Rightarrow (\beta)(\alpha - \beta) = t^2.$ Now, $\gcd(\alpha, \beta) = 1$ implies that $\beta, \alpha - \beta$ are both squares, say $\beta = m^2, \alpha = m^2 + n^2.$

Finally, we have from $\alpha^2 + \beta^2 = v^2$ that $m^2 + (m^2 + n^2)^2 = v^2.$ This means that $(m, n, v)$ is a solution with smaller third number, since $v< z$ is obvious.

This is a contradiction. Hence, our assumption that solutions with $x \neq 0$ exist was incorrect.

$\square$

$(x^2)^2+(x^2+y^2)^2=z^2\quad,x>0\text{ and }y,z\geq 0$
WLOG $\gcd(x,y)=1\Rightarrow\gcd(x^2,x^2+y^2)=\gcd(x^2,y^2)=1$
Lemma: $a^2+b^2=c^2\Rightarrow 4\mid a\lor 4\mid b$
$\because\gcd(x,y)=1\Rightarrow 4\nmid x^2+y^2\quad\therefore 4\mid x^2$ by lemma and $2\nmid y,z$
$\gcd(z+x^2,z-x^2)=\gcd(z+x^2,2z)=\gcd(z+x^2,z)=\gcd(x^2,z)=1$ since $z+x^2$ is odd
$\Rightarrow (x^2+y^2)^2=(z+x^2)(z-x^2)\quad\therefore z\pm x^2$ are square number
Let $(a^2,b^2)=(z+x^2,z-x^2)\Rightarrow a^2-b^2=2x^2$
$\because a^2-b^2\equiv 0,\pm1\mod 4\quad\therefore 2\mid a,b\quad\to\leftarrow$


calculus derivative quadratics function inequalities algebra polynomial

Let $ f(x)\equal{}ax^2\plus{}bx\plus{}c$ for some $ a,b,c \in \mathbb{R}$ such that $ |f(x)| \leq 1$ for $ |x| \leq 1.$
Prove that $ |f'(x)| \leq 4$ for $ |x| \leq 1.$

Actually is that true ?

Consider $ f(x) \equal{} x^2 \minus{}1$ , it does not satisfy $ | f'(x) | \leq   4$ for   $ |x|\leq 1.$

[b]vijay[/b], I'm not sure how you came to that conclusion.

[hide="Hint"]An equality case is $ 2x^2\minus{}1$[/hide]

Because $ f'(x)\equal{}2ax\plus{}b$ is linear, it suffice to prove only two inequality
$ |2a\plus{}b|\equal{}|f'(1)|\le4, |2a\minus{}b|\equal{}|f'(\minus{}1)|\le4.$
We consider three inequality $ |c|\equal{}|f(0)|\le1, |a\plus{}b\plus{}c|\equal{}|f(1)|\le1, |a\minus{}b\plus{}c|\equal{}|f(\minus{}1)|\le1$
$ \Rightarrow|2a\pm b|\equal{}|\minus{}2c\plus{}\frac{3}{2}(a\pm b\plus{}c)\plus{}\frac{1}{2}(a\mp b\plus{}c)|$
$ \le|\minus{}2\parallel{}c|\plus{}|\frac{3}{2}\parallel{}a\pm b\plus{}c|\plus{}|\frac{1}{2}\parallel{}a\mp b\plus{}c|\le 2\plus{}\frac{3}{2}\plus{}\frac{1}{2}\equal{}4.$

By [url=http://www.mathlinks.ro/viewtopic.php?t=317490]Markov's theorem of polynomials[/url], $ |f'(x)| \leq 4$ for $ |x| \leq 1$.!


Bombykol is a pheromone released by the female silkworm moth to attract mates. It was discovered by Adolf Butenandt in 1959 - who also named the pheromone after the moth's latin name [i]Bombyx mori[/i] -, and it was the first pheromone to be characterized chemically. Bombykol has been prepared by the following sequence of reactions:

$ 1-pentyne+n-C_{4}H_{9}MgBr \longrightarrow A$

$ A+HCHO \longrightarrow B$ (after treatment with acid)

$ B+PBr_{3}\longrightarrow C$

$ C+PPh_{3}/base \longrightarrow D$

$ D+OHCCH_{2}(CH_{2})_{7}CO_{2}C_{2}H_{5}\longrightarrow E$

$ E+H_{2}/Lindlar\,\, catalyst \longrightarrow F$

$ F+LiAlH_{4}\longrightarrow Bombykol$

Derive the structure of Bombykol.

carcul can i know the source  please  :maybe:

It was adapted from problem 22 in the end of chapter 21 (Carbanions 1) of Morrison. However, the version of Morrison that I have (and the only available in my country) was translated from the original, so there are probably differences in problems/chapters numbering.

thanks actuall there was a question in our country's premier entrance test IIT-JEE  about this Bombykol but quite different from this

Ok, so why don't you post it on this thread?

so here is it
A biologically active compound ,Bombykol($ C_{16}H_{30}O$) is obtained  from a natural source .the structure of the compound is determined in the following ways.

A) On hydrogenation Bombykol gives a compound A,$ C_{16}H_{34}O$ which reacts with acetic anhydride to give an ester.

B)Bombykol also reacts with acetic anhydride to give another ester whcih on oxidative ozonolysis($ O_{3}/H_{2}O_{2}$) gives a mixture of butanoic acid,oxalic acid and 10-acetoxy Decanoic acid.Determine the number of double bonds in Bombykol.Write the structure of compound A and Bombykol.How many geometrical isomers are possible for Bombykol?

Nice problem. Let's leave it for others to try to solve.


function
quadratics
algebra
quadratic formula

Let $ f(x)\equal{}\sqrt{ax\plus{}a^2}\ (x>\minus{}a,\ a>0)$.
Solve the equation $ f(x)\equal{}f^{\minus{}1}(x)$.

[quote="kunny"]Let $ f(x) \equal{} \sqrt {ax \plus{} a^2}\ (x > \minus{} a,\ a > 0)$.
Solve the equation $ f(x) \equal{} f^{ \minus{} 1}(x)$.[/quote]

[hide]
$ \frac{x^2}{a}\minus{}a\equal{}\sqrt{ax\plus{}a^2}$
$ \frac{x^4}{a^2}\minus{}2\plus{}a^2\equal{}ax\plus{}a^2$
$ \frac{x^4}{a^2}\minus{}2\equal{}ax$
$ x^4\minus{}2a^2\minus{}a^3x\equal{}0$
$ x(x^3\minus{}a^3)\equal{}2a^2$

uh oh this looks bad
[/hide]

[quote="kunny"]Let $ f(x) \equal{} \sqrt {ax \plus{} a^2}\ (x > \minus{} a,\ a > 0)$.
Solve the equation $ f(x) \equal{} f^{ \minus{} 1}(x)$.[/quote]

Since $ f^{ \minus{} 1}(x)$ is $ f(x)$ reflected over the line $ y\equal{}x$, if ever there is an intersection it would have to occur on that line. So:
$ x\equal{}\sqrt {ax \plus{} a^2}$
$ x^2\equal{}ax\plus{}a^2$
$ x^2\minus{}ax\minus{}a^2\equal{}0$
Quadratic formula
$ x\equal{}\frac{a\pm\sqrt{a^2\plus{}4a^2}}{2}\equal{}\frac{a(1\pm\sqrt{5})}{2}$

What does that have to do with economics?

We should get one solution. :D

[quote="caffeineboy"]What does that have to do with economics?[/quote]

I don't know it.

[quote="kunny"]We should get one solution. :D[/quote]

Oops, reject the negative root of that quadratic since $ f(x)$ is always positive.

That's right. sketching the graph of $ y\equal{}f^{\minus{}1}(x)$ gives a direct solution, That's the very; Seeing is to believing.

I would add one thing: $ f^{\minus{}1}(x)\equal{}\frac{1}{a}x^2\minus{}a$, note that $ x>0$.

[quote="CircleSquared"][quote="kunny"]Let $ f(x) \equal{} \sqrt {ax \plus{} a^2}\ (x > \minus{} a,\ a > 0)$.
Solve the equation $ f(x) \equal{} f^{ \minus{} 1}(x)$.[/quote]

[hide]
$ \frac {x^2}{a} \minus{} a \equal{} \sqrt {ax \plus{} a^2}$
$ \frac {x^4}{a^2} \minus{} 2 \plus{} a^2 \equal{} ax \plus{} a^2$
$ \frac {x^4}{a^2} \minus{} 2 \equal{} ax$
$ x^4 \minus{} 2a^2 \minus{} a^3x \equal{} 0$
$ x(x^3 \minus{} a^3) \equal{} 2a^2$

uh oh this looks bad
[/hide][/quote]
i think you made a mistake when raising f(x)^-1 to the second power shouldn't it be x^4/a^2 - 2x^2 + a^2
this gets messy later on.... annoying

$ f^{\minus{}1}(x)\minus{}1|f^{\minus{}1}(f^{\minus{}1}(x))\minus{}x$...then it is easy to solve

[quote="caffeineboy"][quote="kunny"]Let $ f(x) \equal{} \sqrt {ax \plus{} a^2}\ (x > \minus{} a,\ a > 0)$.
Solve the equation $ f(x) \equal{} f^{ \minus{} 1}(x)$.[/quote]

Since $ f^{ \minus{} 1}(x)$ is $ f(x)$ reflected over the line $ y\equal{}x$, if ever there is an intersection it would have to occur on that line. So:no think about y=-x
$ x\equal{}\sqrt {ax \plus{} a^2}$
$ x^2\equal{}ax\plus{}a^2$
$ x^2\minus{}ax\minus{}a^2\equal{}0$
Quadratic formula
$ x\equal{}\frac{a\pm\sqrt{a^2\plus{}4a^2}}{2}\equal{}\frac{a(1\pm\sqrt{5})}{2}$

What does that have to do with economics?[/quote]


group theory
abstract algebra
superior algebra
superior algebra theorems

Let P be a normal Sylow p-subgroup of G and H<G. Prove H /int P is the unique Sylow p-subgroup of H.

Thanks.

Let P be a normal Sylow p-subgroup of G and H<G. Prove (H intersection P) is the unique Sylow p-subgroup of H.

Thanks.

It is known that for every subgroup $H$ of a finite group and every $p$-sylow subgroup $P$ of $G$ there's $g \in G$ s.t. $H \cap g^{-1}P g$ is a $p$-sylow subgroup of $H$. In particular, if $P$ is normal. $H \cap P$ is a $p$-sylow subgroup. If $U$ is another $p$-sylow subgroup of $H$, then it is a $p$-subgroup of $G$ and may therefore be extended to a $p$-sylow subgroup of $G$, which must be $P$. Hence, $U \subseteq H \cap P$. These groups have the same order, thus $U = H \cap P$.


number theory
greatest common divisor
superior algebra
superior algebra unsolved

How do you prove that if f=gcd(a,b) then f/d divides a/d and b/d?

$ f/d \mid a/d, b/d$ in $ \mathbb{Z}  \iff f \mid a,b$ in $ \mathbb{Z} \iff f \mid (a,b)$ in $ \mathbb{Z}$

Isn't the first part of the if and only if statement just repeating my question? I am trying to figure out how to prove this relationship.

The standard definition of divisibility is $ d \mid c$  in $ D \iff c/d \in D.$  Are you using some other definition?


calculus
integration
logarithms
calculus computations

[quote="kunny"]Evaluate $ \int_1^e \left(\frac {1}{\sqrt {x\ln x}} \plus{} \sqrt {\frac {\ln x}{x}}\right)\ dx$.[/quote]

Using parts on $ \int_1^e \sqrt {\frac {\ln x}{x}} \, dx$ we get

$ \int_1^e \sqrt {\frac {\ln x}{x}} \, dx \equal{} 2\sqrt {x\ln x} \bigg{|}_1^e \minus{} \int_1^e \frac {1}{\sqrt {x\ln x}} \, dx$

$ \int_1^e \left(\frac {1}{\sqrt {x\ln x}} \plus{} \sqrt {\frac {\ln x}{x}}\right)\ dx \equal{} 2\sqrt {e}$ .

I hope i havent made computation mistakes  :P


The published answer is 5.  But I would say you can't partition the following n:  2, 3, 5, 6, 8, 11 and 14.  So largest = 14.  I would be appreciative if anyone could illustrate how to partition 6, 8, 11 and 14

National Team 1999 #9
As shown below, a square can be partitioned into four smaller squares, or nine smaller squares or even into seven smaller squares provided the squares do not have to be congruent. (Note that overlapping squares are not counted twice).  What is the greatest integer n such that a square cannot be partitioned into n smaller squares?

1st picture:  square with 1 vertical and 1 horizontal line connecting midpoints
2nd picture:  square with 2 vertical and 2 horizontal lines connecting trisection points
3rd picture:  same as first plus 1 of the 4 smaller squares is itself divided into 4 (ie horizontal and vertical lines connecting its midpoints)[/img]

still confused...
sorry, I don't know how to draw a picture (this is my first post) but if you add a smaller square partition to the upper corner of your picture and then split that smaller square into 4: you get 5 squares (your large 1 plus 4 smaller ones and 2 rectangles)

copying your code:
asy]
draw((0,0)--(0,50)--(50,50)--(50,0)--(0,0));
draw((0,0)--(0,30)--(30,30)--(30,0)--(0,0));
draw((30,30)--(30,50)--(50,50)--(50,30)--(30,30));
draw((30,40)--(30,50));
draw((40,30)--(40,50));
[/asy][/img]

you are correct that 2, and 3 are impossible partitions. however they asked for the largest number which is 5, since 5 squares is impossible.

any even number greater than 2 can be made, by having one "big" square in the bottom right corner, and n+1 squares across the top, and n on the left side. so for the partition of 6,
you would have one "big square in the bottom right, and 3 across the top, and 2 on the left. similar for 14
also, notice that if a partition of n can be made, n+3 can be made by putting a cross in one of the squares. since 4 can be made, 7 can be made, and everything in the list of 4,7,10,13....
since 6 can be made, you get the list
6,9,12,15
since 8 can be made(it's even) you get the list 8,11,14...

the largest number not on any of these lists is 5. you can check for yourself that 5 is impossible

Thanks.  My difficulty was that I read "partition into squares" to mean that each operation of partitioning was into squares and only squares (as well as thinking that a partition was obtained through a series of partitions). Drawing the first big square did not satifsy my definition.


JBL just made 10000 posts! Who do you think will achieve a five-digit quantity next?

grobber's last post was on January 31, 2009. It is between kunny and AIME15. About a 900 post difference, but AIME15 has 17 (17.00 at the time of this post) posts a day. I say AIME15.

I don't know how this works. It says JBL has over 10000 posts in JBL's profile, but when you click "Find all posts by JBL", it says there are less than 10000 posts. Maybe some where deleted, but do deleted posts count?

Some posts might be from private forums.

[quote="algebra2000"]aime15...hes the only person on the list i know[/quote]
Reason fail.

AIME15, Because he's in Middle School, hence he is More talkative than kunny, who is either in College or Has graduated. :)

probably kunny, i saw that recently, he's been posting, that means that he might reach 10,000 soon, if he posts enough in the summer. :P after all, he already has nearly 9,000 posts.

[quote="r15s11z55y89w21"]JBL just made 10000 posts! Who do you think will achieve a five-digit quantity [b]next[/b]?[/quote]

Reading is a virtue.

Haha sorry for reviving but I thought this was kinda funny (2 years later and I'm at 98xx)

Lesson: Minority wins. kunny, with two votes, reached 10,000 faster than AIME15, with 21 votes. (No-vote is still a fail.)

Bumping twice for a year...you have 9999 posts at the time of my post.

AIME15, why arent you posting your 10000th post?

Well, he only has 1 more post. (This is just a joke, but tell me if it is insulting or not funny.) His post could be spam lol

true but i think hes doing it on purpose

Congrat to AIME15 on making a 10000th post.

[quote="AIME15"][quote="Dr Sonnhard Graubner"]hello, do you mean $ 8\odot \cos(20^{\circ})\cdot \cos(40^{\circ})\cdot \cos(80^{\circ})=1$
Why don't you use $ \text{\LaTeX}$?
Sonnhad.[/quote]

hello, is your name Sonnhard or Sonnhad ?
Graubner.[/quote]

is his 10000th post.

^.^ Thread necromancy...  OH well...

No, it's going to be:

hello, i have used wolframalpha to calculate that this is my 10000th post,
Sonhad

He posted his 10000th post in a private forum I think.

grobber is on that list hahahahaahahahhaahhaahahahah

[quote="pi_Plus_45x23"]Wasn't
[quote="Dr Sonnhard Graubner"]hello, see here
http://en.wikipedia.org/wiki/Trigonometry
Sonnhard.[/quote]
His 10000th post?
(I looked at his posts.)[/quote]

He posted in a forum that does not show in that section. I know his 10000th post was in the Alcumus Problems Discussion Forum.

Lol. Hello, this is my 10000th post, Sonnhard.

But to be serious:

Don't poke fun at Dr. Graubner. He is a very nice person, who always helps people out with math problems. He worked hard for 8 years at it to do it, OK, so DO NOT POKE FUN AT DR. GRAUBNER!!!

[quote="Dr Sonnhard Graubner"] hello, this is my 10000th post,
Sonnhard[/quote]

[quote="sinhaarunabh"]Dr Sonnhard Graubner is about to get his 10000th post. My prediction of his post.

hello, this is my 10000th post,
Sonnhard[/quote]

My 10000th post, Sonnhard? That is making fun of him. Dr. Graubner is a very nice person who's been helping everyone with math for 8 years!

Please stop the petition against him.


conics
ellipse
geometry unsolved
geometry

Let $ABC$ be a non-equilateral triangle. Let $A_{1}$, $B_{1}$, $C_{1}$ be three points on the segments $BC$, $CA$, $AB$ such that the lines $AA_{1}$, $BB_{1}$, $CC_{1}$ concur at a point $P$ interior to triangle $ABC$ and satisfy $AA_{1}=BB_{1}=CC_{1}<\min\left\{BC,\ CA,\ AB\right\}$.
Prove that there exist three points $A_{2}$, $B_{2}$, $C_{2}$ on the segments $BC$, $CA$, $AB$ such that the lines $AA_{2}$, $BB_{2}$, $CC_{2}$ concur at a point $Q$ interior to triangle $ABC$ which is distinct from $P$, and such that $AA_{2}=BB_{2}=CC_{2}$.

[i]Alternative formulation.[/i] Let $ABC$ be a non-equilateral triangle. Let $P$ be an interior point of triangle $ABC$ such that the three cevians of the point $P$ have equal length $L$ satisfying $L<\min\left\{BC,\ CA,\ AB\right\}$. Prove there is another interior point $Q$ of triangle $ABC$ distinct from $P$ such that the three cevians of the point $Q$ also have equal length.

In another thread, [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15470]I have provided a link to a construction of the points P and Q[/url].

As for the points P and Q, note that the cevians of the points Q also have the length L. This, together with the fact that the points P and Q exist for every triangle ABC and coincide only if the triangle ABC is equilateral, is proven on pages 73-76 of H. Dörrie, [i]Mathematische Miniaturen[/i], §70. I have never been patient enough to read the solution, but it looks like a messy application of Stewart's theorem. The points P and Q are called [b]Bickart points[/b] of triangle ABC, and they are the focii of the Steiner circumellipse of triangle ABC. [The Steiner circumellipse is the ellipse passing through the vertices A, B, C and having its center at the centroid of triangle ABC.]

  Darij

:D its due to c r pranesachar  its not hard,will post sol. if you dont get ,idea consult dr.trig :roll:


function
conics
hyperbola
Asymptote
algebra
rational function

We haven't gone over this in class yet but it's a part of the homework. Can someone explain to me how one is suppose to Sketch the rational function of:

f(x)=

x-4
----                      (thats x-4 OVER x+5)
x+5

I've read the book but I'm still completly confused.

[quote="José"]$f(x)=\frac{x-4}{x+5}$  Now... what do you want to know about it???[/quote]

how to go about sketching it on a graph.

$y=\frac{x-4}{x+5}=\frac{x+5}{x+5}-\frac{9}{x+5}=1-\frac{9}{x+5}$
$\Rightarrow y-1=\frac{-9}{x+5}\Rightarrow (y-1)(x+5)=-9$

This is like the hyperbola $xy=-9$, except it's been translated so that its center is $(-5,1)$.

[hide="Some basic info"]
Vertical asymtote: $x=-5$ ($\because$ if you plug in $x=-5$, the denominator becomes $0$, which means the fraction is undefined.)

Horizontal asymptote: $y=1$ ($\because$ the degrees of the numerator and the denominator are the same, so dividing the leading coefficients gives $y=1$.)

$x$-intercept: $(4,0)$ ($\because$ Solving $\text{(numerator)}=0$ gives $x=4$)

$y$-intercept: $(0,-\frac45)$ ($\because$ if you plug in $x=0$ into the expression, you get $y=-\frac45$)

After finding these, I would just plot some points and see how the graph would look like.
[/hide]

so essentially it looks similiar to this:

http://mathworld.wolfram.com/images/eps-gif/HyperbolaAsymptotes_900.gif


function
integration
inequalities
real analysis
real analysis unsolved

Let $ \zeta (x) \equal{} \sum_{k\equal{}1}^\infty \frac{1}{k^x}$ is Riman`s zeta function. Prove that, if $ p\geq 3$:
\[ \zeta(p) < 1 \plus{} \frac {1}{p \plus{} 1}\]

since $ \frac {1}{x^p}$ is a decreasing function, $ \zeta(p) < 1 \plus{} \int_{2}^{\infty} \frac {1}{x^p} dx \equal{} 1 \plus{} \frac {1}{(p \minus{} 1)2^{p \minus{} 1}}\leq 1 \plus{} \frac {1}{p \plus{} 1}$, where the latter inequality follows from the truth of $ (p \minus{} 1)2^{p \minus{} 1}\geq p \plus{} 1$ for $ p\geq 3$.


geometry
geometric transformation
reflection
trapezoid
homothety
geometry proposed

The incircle of a triangle $ABC$ touches its sides $BC$, $CA$, $AB$ at the points $X$, $Y$, $Z$, respectively. Let $X^{\prime}$, $Y^{\prime}$, $Z^{\prime}$ be the reflections of these points $X$, $Y$, $Z$ in the external angle bisectors of the angles $CAB$, $ABC$, $BCA$, respectively. Show that $Y^{\prime}Z^{\prime}\parallel BC$, $Z^{\prime}X^{\prime}\parallel CA$ and $X^{\prime}Y^{\prime}\parallel AB$.

[quote="Sailor"]Since $ZXX'Y$ is a trapezoid it follows that $YX'=ZX$.
Since $XZ'ZY$ is a trapezoid it yields that $ZX=Z'Y$.
Thus $YX'=YZ'$ $\Longleftrightarrow$ $X'Z'||AC$.
The other relations can be proved in a similar way.[/quote]

[quote="darij grinberg"]Let $X^{\prime}$, $Y^{\prime}$, $Z^{\prime}$ be the reflections of these points $X$, $Y$, $Z$ in the [b]external[/b] angle bisectors of the angles $CAB$, $ABC$, $BCA$, respectively.[/quote]

Your proof would work if these angle bisectors were internal. In fact, in this case, the problem would be equivalent to the homothety of triangles $A_1A_2A_3$ and $S_1S_2S_3$ in [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=1307]IMO 1982 problem 2[/url]. My problem, with external angle bisectors, is slightly harder; you can use the internal bisector case as a lemma (this is Peter Scholze's solution) or solve it directly.

  darij

Sorry  for the above.:(

Here is the proof for the initial problem.
Actually I will continue my writings(with the same notations) and denote your notations by $X_1$ etc.
Then points $Y',B,Y_1$ are collinear(since $BI$ is on the mid-line of $\triangle{YY_1Y'}$) and analogously $Z',C,Z_1$ are collinear.
Now, let $Z'Z_1\cap{Y'Y_1}=\{K\}$; using the fact that $Z'Y'||BC$ we get:
\[ \frac{Z'K}{KY'}=\frac{Z'C}{Y'B}=\frac{Z'Z_1}{Y'Y_1}  \]  and the conclusion follows.


linear algebra
matrix
linear algebra unsolved

M_n is nxn matrix with elements the numbers from 1 to n^2, written clockwise. for example M_4 is

1     2      3      4
12   13    14     5
11   16    15     6
10   9      8      7    .

find the determinants of
a)M_3 and M_4.
b)M_n for any n.

$\text{det}(M_{3})=-48$ and $\text{det}(M_{4})= 660.$

As for $M_{n},$ this is problem #11270 in the AMM, proposed by Sergey Sadov, and
$\text{det}(M_{n})= (-1)^{n(n-1)/2}4^{n-1}\frac{3n-1}2 \prod_{k=0}^{n-2}\left(\frac{1}{2}+k \right).$

(However, I couldn't find a proof for it till now.)


I have two questions here that I am unsure of how to answer. I realize that the first just needs to be shown, but I don't believe that an example is sufficient. If anyone can help get me started on these it would be appreciated.

#1 Let k be an integer  2. Show that if n+1 integers are chosen from the set {1,2,3,...,kn} then there are always two integers which differ by at most k-1.

#2 The line segments joining 10 points are arbitrarily coloured red or blue. Prove that there must exist 3 points such that the 3 line segments joining them are all red, or 4 points such that the 6 line segments joining them are all blue (that is, r(3,4)10).

no you cant just say an examnple
you must prove that for any 2 numbers...

you can ignore the first one I've found a way to prove it, now for the second one I have the proof for the r(3,3), is there a way to partially incorporate it into my proof for r(3,4)?


Show that $S$ is set of positive integers not divisible by $4$.
\[ S = \{x^2 + (x + 1)^2 + y^2 + z^2: x, y, z \in Z \} \]

Well, one part is obvious, but how to express number $n$ not divisible by $4$ in given form?
Thx

The set of positive integers $n$ not divisible by $4$ is given by $n \equiv 1, 2, 3, \bmod 4$.  Then use the theorem that any integer is representible as the sum of at most four squares (I think?).

You didn't help me... :(
Can anybody help?


limit
real analysis
real analysis unsolved

If the series $a_{n}$ is positive and $\lim a_{n}=0$ (as $n$ tends to zero) then show that there is sequence $e_{n}=+1 or-1$ s.t  the series
                                           $e_{n}a_{n}$
is converges.


Find a solution of the system $\large \displaystyle { \{\begin{array}{ll} x^y=y^x \\ x^5=y^3 \end{array}}$,  $\large x,y>0$ different of the solution $\large x=1, y=1$.

[quote="kunny"]You can take logarithm with base $10$.[/quote]

That's not necessary. :idea:

[hide]just substitute $y=x^{\frac53}$ (assuming $x^\frac13$ would be commonly defined for negatives ..)
but then u have take the $\log_x$ as well ...

solution: $(x;y)|(\left(\frac53\right)^\frac32;\left(\frac53\right)^\frac52)$
[/hide]

[quote="peeta"][hide](assuming $x^\frac13$ would be commonly defined for negatives ..)[/quote]

[hide] :?:

$x>0, y>0$ for the initial hypothesis[/hide]

Your answer is correct peeta, but i still don't think that logs are necessary.
[/hide]


In a triangle $ABC$, $O$,$I$,$H$ are the circuncenter, incenter and orthocenter, respectively. Find with proof all the values of $\measuredangle ABC$ for which $OI=IH$.


algebra
polynomial
number theory
prime numbers
superior algebra
superior algebra unsolved

Show that no finite field of odd characteristic is algebraically closed. (Hint: show by counting that some polynomial x^2-a, for some a in F, has no zero in F).

If the field is Z_p, where p is odd, then the result is obvious because if a is in Z_p, then a^2 = (-a)^2, so there are at most (p-1)/2 + 1 = (p+1)/2 squares. (p+1)/2 < p since 1 < p, so there must be some element in Z_p that is not a square. How do you extend that argument to finite fields of characteristic odd characteristc?

At what point do you need that $ p$ is prime...¿

By the way, a proof that I would prefer: $ x^q\minus{}x\plus{}1$ has no roots in $ \mathbb F_q$.

finite fields $ K$ are not algebraically closed because $ 1\plus{}\prod_{a \in K} (x\minus{}a)$ evidently has no roots in $ K$. I don't understand the effort being made for alternative proofs   :wink:

Well, sometimes, new proofs give new insights.

The probably most suitable on is: There are $ \infty$ many primes.

Euclid's proof (principially the same as the proof above):
Assume that $ p_1, ..., p_n$ are all primes. Then $ p_1\cdot ... \cdot p_n \plus{} 1$ has a new prime factor, contradiction.

Euler's proof:
The sum $ \sum_{p \text{ prime}} \frac 1p$ diverges. Thus it isn't a finite sum.

And the latter somewhat led to Dirichlet's proof for his theorem on primes in arithmetic progressions, whereas the first attempt works for some special cases only.

[quote="-oo-"]finite fields $ K$ are not algebraically closed because $ 1 \plus{} \prod_{a \in K} (x \minus{} a)$ evidently has no roots in $ K$. I don't understand the effort being made for alternative proofs   :wink:[/quote]

Very nice! There is probably a clever way to show it just for fields whose characteristic is an odd prime by using the hint, but you're more general proof is good enough for me.

@Fan: as ZetaX indicated, your argument in the first post was correct; just replace Z/p be a finite field with q elements and odd characteristic ...
@ZetaX: good point

How exactly is proving that there infinitely many primes the same as proving what I asked about?

[quote="ZetaX"]Just look at the proof  :wink:
The corresponding theorem for arbitrary fields $ K$ is: $ K[x]$ has $ \infty$ many prime elements.
For algebraically closed fields, prime elements correspond to field elements.[/quote]

I am so confused. What does it mean that K[x] has infinitely many prime elements. K[x] is a field of polynomials, not numbers. Do you mean that there are infinitely many elements in K[x] that have prime numbers as zeros?

EDIT: that should be a ring of polynomials

[quote="FibonacciFan"]I am so confused. What does it mean that K[x] has infinitely many prime elements. K[x] is a field of polynomials, not numbers.[/quote]

$ K\left[x\right]$ is a ring, not a field. Rings have prime elements. The prime elements of $ K\left[x\right]$ are the irreducible polynomials of $ x$ over $ K$ (and $ 0$; I am not sure whether one usually calls $ 0$ irreducible and/or prime).

  darij

$ 0$ is not irreducible. not every ring has a prime element. :P

I see. Each of the prime elements of the ring K[x] has to have a different set of zeros since every $ irr(\alpha,K)$ is well-defined.

this is wrong and does not have to do with the rest here.


function
limit
calculus
calculus computations

Can something be said about the finite difference between consecutive terms in an infinite sum, when setting the sum equal to a constant?

What do you mean?  What is changing?

If this is a question that you are asking as part of the solution to another question, I have yet to find an instance where it was not clearer just to post the original question.

I'm asking whether or not the change in term of an infinite sum has to equal a certain something relative the function and the constant the sum equals.

For instance, if

$ \sum^\infty_{k} a_n \equal{} c_i$

In order for the sum to equal $ c_i$, does

$ a_n \minus{} a_{n \minus{} 1} \equal{} f(a;c_i)$

Well, yes, of course.  $ c$ can't change unless some of the terms $ a_n$ change, and the terms can't all increase or decrease by the same constant.

The relationship is clear via generating functions:  if $ A(x) \equal{} \sum a_k x^k$ then $ A(1) \equal{} c$ and $ (x \minus{} 1) A(x) \equal{} \sum (a_{k\minus{}1} \minus{} a_k) x^k$, so the behavior of $ \lim_{x \to 1} (x \minus{} 1) A(x)$ changes as $ c$ changes.


topology
counting
distinguishability
real analysis
real analysis unsolved

Let $ X$ be a set with a discrete metric defined by

\[ d\left(p,q\right)\equal{}\begin{cases}0 & p\equal{}q \\ 1 & p\neq q\end{cases}.\]
Since this clearly makes $ X$ into a metric space, which subsets of $ X$ are dense in $ X$?

Only $ X$ itself,Because the closure of any subset $ A$ of $ X$ is $ A$.

Here's another question about dense subsets. You are given a set $ X$ and a family of its subsets $ \mathcal{D}\subset 2^X$. Somebody claims that there is a topology on $ X$ under which the elements of $ \mathcal{D}$, and only they, are dense in $ X$. How can you find if this is true or not?

The first property to check would be $ A\in\mathcal{D}\ \& \ A\subset B \implies B\in\mathcal{D}$. Anything else?

It seems like Krishnamurthy ( [b]http://links.jstor.org/sici?sici=0002-9890(196602)73%3A2%3C154%3AOTNOTO%3E2.0.CO%3B2-V
[/b]) proved that the number of different topologies on an $ n$-point set is only about $ 2^{n^2}$ ("seems" because I am too greedy to pay $ \$$ 12 for that old article in Monthly and too stupid to recover the content of the 3 remaining pages from the first page shown by JSTOR myself, so I cannot verify the proof :( ). On the other hand, considering all $ D$ consisting of all sets of cardinality $ >n/2$ and [b]some[/b] sets of cardinality $ n/2$, we see that the sets satisfying mlok's obvious restriction are far too many for large $ n$, so there should be plenty of other necessary conditions.

One way to approach this question is to use the [url=http://en.wikipedia.org/wiki/Kuratowski_closure_axioms]Kuratowski closure axioms[/url]. What properties of $ \mathrm{cl}^{\minus{}1}(X)$ do they imply?

Seems I finally figured out what Krishnamurtha might possibly mean. The topology on a finite set is completely determined by the smallest open sets containing each given point. Since we have $ n$ points and $ 2^{n\minus{}1}$ possible candidates for each point, we get $ \le 2^{n(n\minus{}1)}$ different topologies.

@mlok Are you driving at something particular or just making a wild suggestion? I'm asking because I do not see any non-trivial properties directly implied by closure axioms :(

Um, I thought that closure axioms could simplify investigation of specific examples of topologies on finite sets. For instance, what is the simplest/smallest example for which the trivial necessary condition is not sufficient?
Or maybe I'm just posting this from a conference wi-fi hotspot to spend my (and your) time between lectures.  :lol:

[quote="mlok"]For instance, what is the simplest/smallest example for which the trivial necessary condition is not sufficient? [/quote]
Well, it seems like you cannot have a topology on a 4 point set for which $ \{1,2\}$ and $ \{3,4\}$ are both dense but no other 2-element set is. I'm not sure whether you can do something with a 3-element set but this example seems small enough (not that I can figure out from it what is really going on :( )
[quote="mlok"]
Or maybe I'm just posting this ... to spend my (and your) time ...  :lol:[/quote]
Well, I should admit that you succeeded in keeping me occupied for the last half an hour :P

If $ A$ is a minimal dense set, then the induced topology on $ A$ is discrete, because the closure of $ A\setminus\{a\}$ in $ A$ cannot be $ A$ itself. This looks like a lot of information, but I don't know how to use it to say anything about other dense sets.

Let's assume that $ X$ is a finite set. Then any closure operator on $ X$ is determined by its values on singletons. Therefore, the closure operators are in one-to-one correspondence with reflexive and transitive relations on $ X$. (The relation being $ y\in \overline{\{x\}}$). Any such relation is a partial order on a set of equivalence classes [of topologically indistinguishable points]. This gives a recipe for constructing all systems of dense sets:

1) Let $ \sim$ be an equivalence relation on $ X$. Given two subsets $ A,B\subseteq X$, write $ A\subseteq_{\sim}B$ if for any $ a\in A$ there is $ b \in B$ such that $ a\sim b$. (This can be called $ \sim$-relaxed inclusion.)
2) Choose a nonempty set $ A_0\subseteq X$.
3) Let $ \mathcal D \equal{} \{B\subseteq X \colon A_0 \subseteq_{\sim}B\}$. The family $ \mathcal{D}$ is of the form $ \mathrm{Cl}^{ \minus{} 1}(X)$ for some closure operator, and this construction exhausts all possibilities for $ \mathrm{Cl}^{ \minus{} 1}(X)$.

Given $ \mathcal{D}$ and $ \sim$, it is easy to check if $ A_0$ exists, because it must a smallest element of $ \mathcal{D}$ with respect to $ \sim$-relaxed inclusion. But I don't want to try all equivalence relations $ \sim$. There ought to be a way to determine $ \sim$ from $ \mathcal D$. Let's write $ a\sim_{\mathcal{D}} b$ if either $ a \equal{} b$ or the following holds: for any set $ A\in\mathcal{D}$ that contains both $ a$ an $ b$ we have $ A\setminus\{a\}\in\mathcal{D}$ and $ A\setminus\{b\}\in\mathcal{D}$. This is in fact an equivalence relation as long as $ \mathcal{D}$ is upward closed (as it must be). Any other feasible candidate for $ \sim$ would have to be a relation contained in $ \sim_{\mathcal{D}}$.

Let's try this on [b]fedja[/b]'s example with $ \mathcal{D}$ generated by $ \{1,2\}$ and $ \{3,4\}$. In this case $ \sim_{\mathcal D}$ has only singletons as equivalence classes. And since $ \mathcal{D}$ does not have a smallest element with respect to inclusion, it's not a system of dense sets.

Question: is it enough to try $ \sim_{\mathcal D}$ in (1) to reach a conclusion about $ \mathcal{D}$?

This isn't relevant to the actual math, but in case anyone is interested, I might be able to provide a copy of that entire document.  :wink:


What is the sum of the roots of the equation $(x^2-3x)^2-(3x^2-9x)=4$?

[quote="Andreas"]$(x^2 - 3x)^2 - 3(x^2 - 3x) - 4 = 0$
$a^2 - 3a - 4 = 0$
So the sum is : $6$[/quote]
Let me clarify it a bit.

Since we have $a^2-3a-4=0\Longleftrightarrow (a+1)(a-4)=0$ where $a=x^2-3x$, we get two equations: $x^2-3x=-1$ and $x^2-3x=4$.

The sum of the roots of a quadratic equation $ax^2+bx+c=0$ is $-\frac{b}{2a}$.
So the sum is $\left(-\frac{-3}{1}\right)+\left(-\frac{-3}{1}\right)=6$


geometry
3D geometry
trigonometry
calculus
calculus computations

Sand falls from a hopper at a rate of $ 0.1$ cubic meters per hour and forms a conical pile beneath.

If the side of the cone makes an angle of $ \pi/6$ radians with the vertical, find the rate at which the height of the cone increases in terms of $ h$, the height of the pile in meters.

At what rate in terms of $ h$, the height of the pile in meters, does the radius of the base increase?

hello
$ \frac{r}{h}\equal{}\sqrt{3}$
$ Volume\equal{}\pi h^3\equal{}\frac{\pi r^3}{3\sqrt{3}}$
$ \frac{dh}{dt}\equal{}\frac{0.1}{3\pi h^2}$
$ \frac{dr}{dt}\equal{}\frac{0.1}{\pi \sqrt{3} h^2}$
thank u

Not quite...

[hide]$ V\equal{}\frac{1}{3}\pi ^2 h$

$ \pi/6$ angle of cone makes $ r\equal{}h\tan (\pi/6)\equal{}h/\sqrt{3}$

$ V\equal{}\frac{1}{3} \pi (\frac{h}{\sqrt{3}})^2h\equal{}\frac{1}{9} \pi h^3$

$ \frac{dV}{dh}\equal{}\frac{1}{3}\pi h^2$

$ \frac{dh}{dt}\equal{}\frac{dh}{dV}\frac{dV}{dt}\equal{}\frac{3}{\pi h^2}*.1$

$ \frac{dr}{dt}\equal{}\frac{dh}{dt}\frac{1}{\sqrt{3}}\equal{}\frac{1}{\sqrt{3}}\frac{3}{\pi h^2}*.1$[/hide]


geometry
perimeter
inequalities
congruent triangles
triangle inequality

How many non-congruent triangles with perimeter $ 7$ have integer side lengths?

$ \textbf{(A)}\ 1 \qquad
\textbf{(B)}\ 2 \qquad
\textbf{(C)}\ 3 \qquad
\textbf{(D)}\ 4 \qquad
\textbf{(E)}\ 5$

[hide]
Well there's at most 5, so it really isn't that hard to list them all out.  Keeping the triangle inequality in mind, you know that none of the sides can be greater than 3.

2,2,3
3,3,1

So (B) 2
[/hide]

Let the sides of the triangle be $x$, $y$, and $z$.

We know that all of $x$, $y$, $z$ must be less than $4$, because of this fact, the sum of two sides must be greater than $3$.

Now we do casework on the $x$ value.

If $x = 1$, then $y = 3$, so the only possible triangle is $(1, 3, 3)$.

If $x = 2$, then $y = 3, 2$, so the possible triangles are $(2, 3, 2), (2, 2, 3)$.

If $x = 3$, then $y = 1, 2, 3$, so the possible triangles are $(3, 1, 3), (3, 2, 2), (3, 3, 1)$.

Clearly the two triangles are $(1, 3, 3), (3, 2, 2)$. So $\boxed{2}$.

the possible outcomes: $(1, 3, 3)$ and $(3, 2, 2) $ so there are $\boxed{2}$ possible triangles

Let our first case be that the smallest side is $1$. We then can only have $(1,3,3)$ as a possibility. Let the second case be that the smallest side is $2$. We then can only have $(2,2,3)$ as a possibility. If the smallest side is $3$, it is impossible, and if the smallest side is $4$ or larger, the Triangle Inequality is not satisfied. This gives $2 \implies \boxed{\text{B}}$.

[quote=smartguy888][quote=TI84Math]When the two smaller sides are added up, they have to be bigger that the longest side.

So,
3,3,1
3,2,2
Which makes two[/quote]

How did you join 10 years ago and only make a post today? Anyways welcome back[/quote]

That post was made in 2008 :P

oh nvm

this is the problem with 13 year old bumps :P


group theory
abstract algebra
greatest common divisor
superior algebra
superior algebra unsolved

Let $ H$ and $ K$ be subgroups of a finite index in a group $ G$. Show, that if the greatest common divisor of $ [G: H]$ and $ [G: K]$ is $ 1$, then $ G\equal{}HK.$ (Help: Show $ [G: H\cap K]\equal{}[G: H][G: K]$).

 For a finite group $ G$ I know how to use a hint, since it follows $ [G: K]\equal{}[H: H\cap K]$ and then $ \phi: H/H\cap K \rightarrow G/K$ defined by $ \phi(h(H\cap K))\equal{}hK,$ which is injective, will be also surjective and therefore it will follow $ G\equal{}HK$. But I do not know how to prove the hint.

Study the map $ G/(H\cap K)\to G/H\times G/K, g(H\cap K)\mapsto (gH,gK)$.

Where have you used, that the index are coprime?

Yes i got it, i was not carefull enought at first,....may bad,...sorry.


limit
logarithms
function
algebra
binomial theorem
calculus
calculus computations

Can you solve them without using the l'Hôpital's rule and the Taylor series expansion ?

Check out how many other methods you can find .  :rotfl:

$ \lim_{n\to 0^{+}}n^{n}=1\;\;\;\;\;\;\lim_{x\to+\infty}\sqrt[n]{n}=1$

[b]Theorem:[/b] If $ (s_{n})$ is a sequence of non-zero terms and $ \lim \left| \frac{s_{n+1}}{s_{n}}\right| = L$ then $ \lim |s_{n}|^{1/n}= L$.

I am not proving that.

So consider $ s_{n}= n$ then $ \frac{s_{n+1}}{s_{n}}\to 1$ that means by above that $ \lim |s_{n}|^{1/n}= \lim n^{1/n}= 1$

[b]Alternate Proof:[/b] Let $ s_{n}= n^{1/n}-1$ we need to show $ \lim s_{n}= 0$. So we have $ n = (1+s_{n})^{n}$. For $ n\geq 2$ we can use the binomial theorem to get $ n=(1+s_{n})^{n}\geq 1+ns+\frac{1}{2}n(n-1)s_{n}^{2}> \frac{1}{2}n(n-1)s_{n}^{2}$ because $ s_{n}\geq 0$. Thus, $ s_{n}^{2}< \frac{2}{n-1}$ for $ n\geq 2$. So $ s_{n}< \sqrt{\frac{2}{n-1}}\to 0$ and so $ s_{n}\to 0$.

[quote="kg2100"]$ \lim_{n\to 0^{-}}n^{n}$[/quote]I am worried about this minus because, $ n^{n}$ is not continuous for negative $ n$, does that make any difference?

Sorry , my fault , it is corrected now .

In 2nd limit there is a $ x\to\infty$, shoud be ofc $ n\to\infty$  :)

1st of all, we can notice that if we proof the 2nd limit, we'll proof the 1st limit too, becouse we have:
$ n^{n}=\exp\ln n^{n}=\exp n\ln n$, and if we put $ n=\frac{1}{m}$, we'll have:
${ \exp \frac{1}{m}\ln\frac{1}{m}}=\exp-\ln\sqrt[m]{m}$, so:
$ \lim_{n\to 0^{+}}n^{n}=\lim_{m\to\infty}\exp{-\ln\sqrt[m]{m}}=\exp-\ln 1=\exp 0=1$

Thereafter we can proof the limit $ \lim_{n\to\infty}\sqrt[n]{n}$ for $ n\in\mathbb{N}$, becouse function $ f(x)=x^{\frac{1}{x}}$ is decreasing and continous for $ x>e$.

[b]1st proof:[/b]
$ 1=\sqrt[n]{1}\leq\sqrt[n]{n}=\sqrt[n]{\sqrt{n}\sqrt{n}\underbrace{1\cdot\ldots\cdot 1}_{n-2}}\leq\frac{\sqrt{n}+\sqrt{n}+(n-2)}{n}$, and for $ n\to\infty$ we'll receive:
$ 1\leq\lim_{n\to\infty}\sqrt[n]{n}\leq 1$ so $ \lim_{n\to\infty}\sqrt[n]{n}=1$
[b]2nd proof:[/b]
Let
$ \sigma_{n}=\sqrt[n]{n}-1$, then:
$ n=\left(\sigma_{n}+1\right)^{n}\geq 1+\sigma_{n}+\frac{n(n-1)}{2}\sigma_{n}^{2}\geq1+\frac{n(n-1)}{2}\sigma_{n}^{2}$, so:
$ n\geq 1+\frac{n(n-1)}{2}\sigma_{n}^{2}$, so:
$ 0\leq\sigma_{n}\leq\sqrt{\frac{2}{n}}$, and for $ n\to\infty$ we'll have:
$ \sigma_{n}\to 0$, so $ \sqrt[n]{n}\to 1$.


For all positive reals $ a$, $ b$, and $ c$, what is the value of positive constant $ k$ satisfies the following inequality?
$ \frac{a}{c\plus{}kb}\plus{}\frac{b}{a\plus{}kc}\plus{}\frac{c}{b\plus{}ka}\geq\frac{1}{2007}$ .

[quote="N.T.TUAN"][quote="Sung-yoon Kim"]it's really early, since this competition was held just today![/quote]
But Can we discuss it now?  :maybe: If not then I must lock it![/quote]

no problem. anyhow, the answer is $ k\equal{}6020$. First $ a\equal{}b\equal{}c\equal{}1$ makes $ \frac{3}{k\plus{}1}\geq\frac{1}{2007}$ so $ k\leq 6020$. For such $ k$, $ \sum\frac{a}{c\plus{}kb}\geq\frac{(a\plus{}b\plus{}c)^{2}}{(k\plus{}1)(ab\plus{}bc\plus{}ca)}\geq\frac{1}{2007}$.

Actually, the answer is $ k\leq 6020$, since the problem asks all possible values of k.

Apply the[b] Cauchy-Scwharz inequality[/b] to obtain:
\[(a/(c+kb)+b/(a+kc)+c/(b+ka))(a(c+kb)+b(a+kc)+c(b+ka))\geq(a+b+c)^2\]
or,\[(a/(c+kb)+b/(a+kc)+c/(b+ka))\geq (a+b+c)^2/(k+1)(ab+bc+ca)\]
Thus it suffices to show that,
\[2007(a+b+c)^2\geq (k+1)(ab+bc+ca).\]
But we know that
\[(a+b+c)^2\geq 3(ab+bc+ca).\]
Hence we get k= 6020 as the [b]limiting value[/b] for which equality holds for[b] a=b=c.[/b]


Θεωρούμε τις ευθείες $\epsilon_{1},\epsilon_{2}$ και τα σημεία $A_{1},A_{2},A_{3}$ που ανήκουν σε σταθερή ευθεία $\epsilon$. Αν οι κορυφές $B_{1},B_{2}$ του τριγώνου ${B_{1}B_{2}B_{3}}$ κινούνται πάνω στις $\epsilon_{1},\epsilon_{2}$ αντίστοιχα, και οι πλευρές $B_{2}B_{3},B_{3}B_{1},B_{1}B_{2}$ διέρχονται αντίστοιχα από τα σημεία $A_{1},A_{2},A_{3}$,να βρείτε το γεωμετρικό τόπο της κορυφής $B_{3}$.

Θεωρούμε τρεις διαφορετικές θέσεις για το $B_{1}$ άρα προκύπτουν άλλες τρεις για το $B_{2}$ . Έστω $B_{2}',\ B_{2}'' ,\ B_{1}',\ B_{1}''$ .

Τότε αφού οι ευθείες $B_{1}B_{3}$  , $B_{1}'B_{3}'$  και $B_{1}''B_{3}''$  περνούν από το σταθερό σημείο $A_{2}$  δημιοργούν δέσμη .
Άλλες δύο τέτοιες δέσμες σχηματίζονται και στα $A_{1}$  και $A_{3}$ .
Συνεπώς όπως είναι γνωστό οι δέσμες του $A_{1}$  και $A_{2}$ τέμνονται σε συνευθειακά σημεία .(πώς το λένε αυτό το θεώρημα ρε παιδιά ?έχει όνομα ? )
Άρα τα $B_{1},B_{1}',B_{1}''$  είναι συνευθειακά .
Άρα ο γεωμετρικός τόπος των σημείων $B_{1}$  είναι ευθεία . :wink:

Νίκο έχεις άλλη απόδειξη ? Αν ναι βάλτην ....

το ΘΕώρημα το  van rapanen :rotfl:  :rotfl:  :rotfl:  ωραία απόδειξη συγχαρητήρια Σιλουανέ

Δεν ξέρω αν έχει ιδιαίτερο όνομα αλλά είναι το παρακάτω:

''Αν δύο δέσμες έχουν ίσους διπλούς λόγους και μία ομόλογη ακτίνα τους κοινή, τότε τα σημεία τομής των υπόλοιπων ομόλογων ακτίνων τους, είναι συνευθειακά''.

Στην περίπτωσή μας, ο ζητούμενος γεωμετρικός τόπος είναι ευθεία που διέρχεται από το σημείο τομής των ευθειών ε1, ε2 (τα Β1, Β2, μπορεί να ταυτίζονται με αυτό το σημείο), ή είναι παράλληλη προς αυτές, αν είναι παράλληλες μεταξύ τους.

Κώστας Βήττας.

[quote="vittasko"]
''Αν δύο δέσμες έχουν ίσους διπλούς λόγους και μία ομόλογη ακτίνα τους κοινή, τότε τα σημεία τομής των υπόλοιπων ομόλογων ακτίνων τους, είναι συνευθειακά''.

[/quote]
Ακριβώς....... :)

Αυτή η άσκηση είναι γνωστή σαν [u][b]<<Το πόρισμα του Ευκλείδη>>[/b][/u].Η απόδειξη που έγραψες είναι και η καλύτερη...Υπάρχει άλλη μία απόδειξη με χρήση διανυσμάτων.Η ιδέα σε γενικές γραμμές είναι η εξής:
Θεωρώ στις ε1 και ε2 τα διανύσματα:$\vec{e_{1}}=\vec{OK},\vec{e_{2}}=\vec{OL}$.....Εφαρμόζω Θ. Μενελάου για το τρίγωνο $A_{2}A_{3}B_{1}$ και για τη διατέμνουσα $B_{2}A_{1}$  και από εκεί υπολογίζω το $OB_{3}$.

[i]Αν θέλετε, μπορώ να τη γράψω ολοκληρωμένη και τη δεύτερη απόδειξη, αλλα νομίζω οτι η πρώτη είναι αρκετά καλή.[/i]


$ f( x ) \equal{} 1$ satisfies the given conditions

Define $ f(\pm 2^{2^n}) \equal{} 1 \plus{} (\minus{}1)^n$ for $ n \in \bf Z$, and $ f(x) \equal{} 1$ if $ x \neq \pm 2^{2^n}$ for any $ n \in \bf Z$.  It's obvious how to extend this to get a huge family of solutions.

If you impose also that $ f$ be continuous at $ 1$, then $ f(x) \equal{} 1$ is the unique solution.

Note that continuity at a single arbitrary point is not enough: you need to be able to get from any value $ x$ to a point of continuity by a combination of squaring and square roots.  For example, the function I've described is continuous on $ (\minus{}1, 1)$.  But continuity at one point in $ (\minus{}1, 1)$ and one point outside this interval is good enough.


I do have some doubt for this proof and especialy for N°3, so please correct me if YES!

$f=0$ is solution then from now we sup $f<>0$

1) Take  $x=0$ and any  $y$     , $f(0)=(1-y)f(0)$ then $f(0)=0$

2) Put $f(1)=k$ & sup $k<>0$  Take $y=1$  then $f(kx)=kx^2$
Subtitute $kx$ by $x$ then $f(x)=\frac{x^2}{k}$
So $f(1)=\frac{1}{k}=k$ then $k=+-1$ But both $f(x)=x^2$ and $f(x)=-x^2$ are not indeed solution then $f(1)=0$

3) Take $y<>0 , 1$ and sup there exit $a<>0$ such  that $f(y)=a$ then $f(ax)=(1-y)f(xy)+ax^2y^2$
 Take $x$ such that $xy=1$  then  $f(ax)=a$ so $f(x)=cste$ but the only $cste$ which verify the original equation is $cste=0$

So the only function is $f=0$

 :?

[quote="AYMANE"]3) Take $y<>0 , 1$ and sup there exit $a<>0$ such  that $f(y)=a$ then $f(ax)=(1-y)f(xy)+ax^2y^2$
 Take $x$ such that $xy=1$  then  $f(ax)=a$ so $f(x)=cste$ but the only $cste$ which verify the original equation is $cste=0$

So the only function is $f=0$

 :?[/quote]
AYMANE, You are wrong .
$x=\frac 1y \Longrightarrow f(\frac{f(y)}y)=f(y)$
$x=1 \Longrightarrow f(f(y))=(1-y+y^2)f(y)$

Could you prove that for a real $r$ the possible values for $f(r)$ are $\{0,r^2,r-r^2\}$ ??
Warning : at this step it's possible to have for exemple $f(a)=a^2 , f(b)=b-b^2, f(c)=0$ ...

Could you find  the solutions  ?
 :cool:

yes. can you send it?
AYMANE where  are you from?

Like you see from Morroco and especialy from casablanca :whistling:

[quote="akhrif"]Find all functions $f: \mathbb R \rightarrow \mathbb R$ such that $\forall x,y\in\mathbb R: \[ f(xf(y))=(1-y)f(xy)+x^2y^2f(y)$[/quote]
Posted befor
[url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=f%28xf%28y%29%29%3D%281-y%29f%28xy%29%2Bx%5E2y%5E2f%28y%29&t=61219]www.mathlinks.ro/Forum/viewtopic.php?highlight=f%28xf%28y%29%29%3D%281-y%29f%28xy%29%2Bx%5E2y%5E2f%28y%29&t=61219[/url]
Nice URL,isn't it? :lol:


inequalities
2-variable inequality
AM-GM

For $a,b$ positive reals, prove that
\[a^4 + b^4 + 3  \geq a + b + 3\left(\frac{3ab+1}4\right)^{\frac43}\]


calculus
derivative
logarithms
calculus computations

Find all relative maxima or minima.
y = (logx)^2 , x > 0

Take the derivative and the derivative comes to be $ \frac{2\log{x}}{x}$ and eqyate it to zero we get it at x=1....

now calculate the second derivative and put x=1... it is greater than zero so the minima is x=1....

I don think there exists any maxima for this function...


In the song "100 bottles of beer on the wall, 100 bottles of beer. Take one down, pass it around, 99 bottles of beer on the wall. 99 bottles of beer on the wall...", what is the sum of all the numbers that are sung in the song when we reach 0 bottles of beer.

Basically, we are just trying to find the sum of numbers 1 ~ 100. We use the formula: $ \frac{n(n\plus{}1)}{2}$. Plugging in 100 yields $ \boxed{5050}$.

this is too easy for MC mods please move this to MOEMS or classroom Math

actually its not
its 100+100+99+99 in the first verse and 99+99+98+98 in the second verso... all the way to 2+2+1+1 in the next to last verso then 1+1 in the very last verse
so actually it is 20000:
2(100)+4(99)+4(98)+4(97)...+4(2)+4(1)
which can be found by multiplying izzy's answer by 4 and subtracting 2(100)

aha! some lateral thinking

actually, there is only 3 99, 98, 97 etc in each (100 bottles of beer on the wall, take one down, pass it around, [b]99[/b] bottles of beer on the wall. [b]99[/b] bottles of beer on the wall, [b]99[/b] bottles of beer....and it goes on)
only 3 of each after 100

to add to that, there is not 4 ones
there is not 3 ones either
there is 1 one in each verse (take one down) and 3 in the last verse

with that said, it should be quite easy to solve now

dang it seems this is a very tricky problem
but there are 2 1's in the next to last verse...
so the answer is 3 times 5050 (sum of all numbers 1-100) -100 (b/c 100 appears 2 times)+100 (b/c1 is said in each of 100 verses)=15150

This is a tricky problem. At first, I didn't notice the word "one" in each verse:

"<number> bottles of beer on the wall!

<number> bottles of beer!

(You) take [b]one[/b] down, and pass it around!
<number - 1> bottles of beer on the wall!"


What is the greatest whole number value of $ x$ for which $ 10!$ is divisible by $ 2^x$?

Why don't you google it? I didn't know what the specific name was, but I know what it says.

[hide="Other Solution"]
$10!=10\cdot9\cdot8\cdot7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1$
Prime factorizing, we get
$10!=2\cdot5\cdot3^2\cdot2^3\cdot7\cdot2\cdot3\cdot5\cdot2^2\cdot3\cdot2\cdot1$
Rearranging, and getting rid of the $1$ (Sorry $1$!), we get
$10!=2\cdot2^3\cdot2\cdot2^2\cdot2\cdot3^2\cdot3\cdot3\cdot5\cdot5\cdot7$
Then,
$10!=2^8\cdot3^4\cdot5^2\cdot7$
As we can see, the greatest power of $2$ that $10!$ is divisible by is $8$, since $10!$ has $2^8$ in its prime factorization. Therefore, the answer is $\boxed8$
[/hide]


function
quadratics
algebra
polynomial
algorithm
quadratic formula

Can anyone tell me if there's a formula or something that can be use to solve any cubic function, and whether there is one that can he used to solve any quartic funtion?

umm yeah there is but it's not worth remembering...it takes 100 times longer than factoring it...

um...ok, but can you tell me anyways...in the case where it can't be factored? Thank you.

http://mathworld.wolfram.com/CubicFormula.html
http://planetmath.org/encyclopedia/CubicFormula.html
http://planetmath.org/encyclopedia/QuarticFormula.html

Unlike the quadratic formula, the cubic and quartic formulas are useless in any practical situation, and not even computers use them (there are much faster numerical methods).  It is only the fact that they exist that is important (in a theoretical sense).  On any competition problem, you should instead be looking for tricky factorizations and the like.

Interestingly enough, no such formula exists for polynomials of degree $ 5$ or higher.  The reason why involves some strikingly deep mathematics, and is well worth investigating if you have some spare time.  :)

You could use a more practical algorithm to solve a cubic, it's not too difficult to remember.


[quote="jelly119"]If I got no replies for my post in some forum, should I post the same topic in another forum?? :|[/quote]You should post it in the right forum in the first place. If you do not get a response there, then you have to wait. Posting the same problems in different forums to get a quicker answer is definitely not allowed and it is considered spamming.

[quote="shobber"]But if people have completely forget the post, like it's already several weeks after, can we post again?[/quote]If it's been more than 1 month, then it would be resonable to bump it once, but that's it :)

I would say that you should wait to repost (or reply to your own post) until your original post has been shunted to the second page of the subforum you posted in.  That could take anywhere from a week to a month.

Is that appropriate?  Or should we wait longer?

I would say wait a week to bump your post.  Or if it goes to the 2nd page you should bump it.  I would stay away from reposting it.  If you decide it's too hard for the forum you posted it in tell a moderator of that forum to please bump it into a different forum.

Those are just suggestions for how long you should wait.  Ultimately you should just use common sense.  Don't post a problem and then after one day of waiting post again and say "No one can do it??"  You have to give it a little time.


Hey everyone, I was just recently linked to this mathworld page by a friend and I'd like to share it with everyone because I think it's pretty neat. http://mathworld.wolfram.com/TuppersSelf-ReferentialFormula.html
Does anyone have any idea about how they discovered this?
P.S. can someone validate whether or not it's a joke or real? I'm a bit skeptical about it.

that will be crazy...

did they plot it over xy-plane???which scale???

I think it's a 1/1 scale, but the portion of the graph that they're looking at is insane, with moderate values of x and then values of y which are too big to type on a line so the number took up several.

[quote="kokosai"]I don't think the inequality holds for all x and y...[/quote]

If it did, the plot would just be a giant blob of black...

You only shade the regions for which the pairs of $(x,y)$ in that region satisfy the inequality.

[quote="me@home"]why would someone bother to do that??[/quote]Not much of a fan of knowledge for the sake of knowledge, me@home? :) Anyway, in mathematics, it's often not the specific answers that matter, it's the techniques for solving them that are important; even if a problem and its answer don't immediately have enormously significant mathematical consequences, the methods for solving the problem can often be used in other situations.  In this case, for example, think about the theory behind the author's solution; his answer's not that significant alone, but perhaps his techniques will be helpful in solving other problems.  Also, even if a mathematics problem seems trivial or purely recreational now, in the future it may prove useful (that's part of the rationale behind theoritical studies, rather than purely practical ones): just consider that there was little application for the study of different bases until computer science came along; in fact, I think just a century ago the study of bases was considered recreational mathematics, without much chance of ever providing future practical aplications.

wow
this is some pretty amazing stuff.
Is there another one like it?

[quote="me@home"]you're quite right, my response wasn't well thought out. what do you think will be the next big aplication? What kind of math do you think will be used in teleportation?[/quote]No, we'll use it to amuse and distract the aliens when they come to destroy us.  :P

And to keeyounglee: I do not know of another one like it

Yes, but x=1 doesn't graph the x. Or the =. I think you're out of luck  :P

[quote="Chocoboom"]What is the volume of a pizza of thickness a and radius z? Pi z z a  :rotfl:  :rotfl:[/quote]

 :D  That's awesome, did you come up with that?

Pretty bad example: What's the circumference of a pie with diameter e? ;)

epi? I don't get it.

[hide][/sarcasm][/hide]

oh..............
[hide="dont click here if you dont get it"]pie...[/hide]


limit
integration
logarithms
symmetry
calculus
real analysis
real analysis unsolved

The following limit of a double series converges or diverges?

${\lim_{n\to \infty }\, \left(\sum_{i=1}^{n}\sum_{j=1}^{n}\left(\begin{cases}\frac{1}{(i+j) n}+\frac{1}{(i-j) n}& i\neq j \\ \frac{2}{n}& i=j \end{cases}\right)\right)}$

[hide="Solution I"]I Made this small approximation.
I tested numerically and this agree with the sum. but since i dont know if this converges. or slowly diverges, it's just a waste of bits...

${\lim_{n\to \infty }\, \left(\sum_{i=1}^{n}\sum_{j=1}^{n}\begin{cases}\frac{\frac{1}{i+j}+\frac{1}{i-j}}{n}& i\neq j \\ \frac{2}{n}& i=j \end{cases}\right)\approx 2+\Re\left(\lim_{n\to \infty }\, \lim_{s\to n}\, \int_{1}^{s}\int_{1}^{n}\left(\frac{1}{n (i+j)}+\frac{1}{n (i-j)}\right)didj\right)}$
${2+\Re\left(\lim_{n\to \infty }\, \lim_{s\to n}\, \int_{1}^{s}\frac{\log \left(j^{2}-n^{2}\right)-\log \left(j^{2}-1\right)}{n}\, dj\right)}$
${2+\Re\left(\lim_{n\to \infty }\, \frac{(n+1) (-i \pi+\log (4)-2 \log (n+1))+2 n \log (n)}{n}\right)}$
${2+\log (4)}$[/hide]

[hide="Solution II"]This is a Horrible Approach to the problem, is too long and complicate
Evaluating

${\sum_{j=1}^{n}\left(\frac{1}{n (i+j)}+\frac{1}{n (i-j)}\right)=\frac{i \pi \cot (i \pi )-i H_{n-i}+i H_{i+n}-1}{i n}}$

If $i$ is integer ${\cot (i \pi )=\infty }$

So, we can just take this factor out of the output. because this factor is irrelevant.
Remember $\frac{1}{i-j}$. is a error by division by zero. So, taking it out and simplifying ( this is the unique factor infinity in the equation)

${\sum_{j=1}^{n}\left(\frac{1}{n (i+j)}+\frac{1}{n (i-j)}\right)-\frac{\pi \cot (i \pi )}{n}=\frac{H_{i+n}}{n}-\frac{H_{n-i}}{n}-\frac{1}{i n}}$

Summing this in variable $i$

${\sum_{i=1}^{n}\frac{H_{i+n}}{n}=\sum_{i=1}^{n}\int_{0}^{1}\frac{1-x^{i+n}}{(1-x) n}\, dx=\int_{0}^{1}\left(\sum_{i=1}^{n}\frac{1-x^{i+n}}{(1-x) n}\right) \, dx=-\frac{n+(n+1) H_{n}-(2 n+1) H_{2 n}}{n}}$

${\sum_{i=1}^{n}\frac{H_{n-i}}{n}=\sum_{i=1}^{n}\int_{0}^{1}\frac{1-x^{n-i}}{(1-x) n}\, dx=\int_{0}^{1}\left(\sum_{i=1}^{n}\frac{1-x^{n-i}}{(1-x) n}\right) \, dx=\psi^{(0)}(n)+\frac{1}{n}+\gamma-1}$

${\sum_{i=1}^{n}\frac{1}{i n}=\frac{H_{n}}{n}}$

${\left(-\frac{n+(n+1) H_{n}-(2 n+1) H_{2 n}}{n}\right)-\left(\psi^{(0)}(n)+\frac{1}{n}+\gamma-1\right)-\left(\frac{H_{n}}{n}\right)=}$
${-\frac{(n+2) H_{n}-(2 n+1) H_{2 n}+n \left(\psi^{(0)}(n)+\gamma \right)+1}{n}}$

${\lim_{n\to \infty }\, \left(-\frac{\left(\psi^{(0)}(n)+\gamma \right) n+(n+2) H_{n}-(2 n+1) H_{2 n}+1}{n}\right)=\log (4)}$

And we add the term $2$ because

${\sum_{s=1}^{n}\frac{2}{n}=2}$[/hide]

But the solutions is complicated, long and hard.
And i still doesnt prooved that this sum converges, but is reasonable to think that, if this convergers is to $2+\ln(4)$  :D

It is immediately obvious that the $\frac1{(i-j)n}$ terms sum to zero by symmetry- which is good, since the sum of their absolute values tends to $\infty$.

The diagonal terms sum to exactly 2.
That leaves $\sum_{i\neq j}\frac1{(i+j)n}$. Group these by the value $k$ of $i+j$; there are $k-2$ terms when $k$ is even and $\le n+1$, $k-1$ terms when $k$ is odd and $\le n+1$, $2n-k$ terms when $k$ is even and $\ge n+1$, and $2n+1-k$ terms when $k$ is odd and $\ge n$.
Let $e(k)=\begin{cases}0& k\text{ is even}\\1& k\text{ is odd}\end{cases}$. We want to estimate the sum $\sum_{k=1}^{n}\frac{k-2+e(k)}{k}\cdot\frac1n+\sum_{k=n+1}^{2n}\frac{2n-k+e(k)}{k}\cdot\frac1n$ for large $n$.
As $n\to\infty$, $\sum_{k=1}^{2n}\frac{c}{kn}$ tends to zero for any constant $c$; we can ignore the constant and $e(k)$ terms in the numerator. Making this change, we want $\lim_{n\to\infty}\sum_{k=1}^{n}\frac1n+\frac1n\sum_{k=n+1}^{2n}\frac{2n-k}{k}= 1+\int_{1}^{2}\frac{2-x}{x}\,dx=1-1+2\ln 2=2\ln 2$.

Adding in the diagonal terms, the limiting sum is $2+2\ln 2$.

Oh Jeez... How stupid I am!  :rotfl:  I did not see that.

So the problem is simple. humpf

${\lim_{n\to \infty }\, \left(\int_{1}^{n}\int_{1}^{n}\frac{1}{(i+j) n}didj-\int_{1}^{n}\frac{1}{(i+i) n}\, di\right)=\log (4)}$

The last integral is needed, but it's the limit is zero. and we add 2.... aff


<временно удалено  :blush: >

? ты уверен, что везде ТурГор проходит в ?ти ?роки?

Уверен. В указани?х по уча?ти? города [b]на?то?тельно[/b] рекомендует?? проводить ТурГор не более, чем через неделю по?ле официальной даты.


function
topology
real analysis
Functional Analysis

[quote="JSRosen3"]A function f is defined as follows:

f(x) = 0, if x is irrational.
f(x) = 1/q, if x can be expressed in the form p/q where p and q are integers, p/q is in lowest terms, and q > 0.

Where on [0,1] is f continuous?[/quote]

Yeah, that's a neat function. This begs an interesting question (to which I do know the answer, but I don't know of a proof that doesn't require more advanced mathematics): The function JSRosen3 defined is continuous on some set S. Is there a function that is continuous on the complement of S? If so, find one. If not, prove it.

A not too rigourous try, but I think it's the basic idea :

. if x irrational then f is continuous at x
. if x rational then f is not continuous at x

. Let x rational (the easy part !), f(x) != 0 but we can find an irrational as close as we want to x, for which f = 0. So f can't be continuous at x.

. Let x irrational, let eps > 0,

The set {p/q, |x - p/q| < eps, q < 1/eps } is finite so we can find eps' <= eps such as |x - p/q| < eps' for all p/q such as q > 1/eps'. But then |f(x)-f(p/q)| = 1/q < eps' <= eps which proves that f is continuous at x.

Yeah, that's right. Good job. Can anyone find a simple solution to my problem? I'm not sure if one exists.

I know the answer to your question ComplexZeta (because it's related to the subject of my senior thesis), but I don't know if there's a "simple" solution to it either.

Gauss202, could you give us some idea about this function ? Is your thesis available online ? Thank you !

Hint: [hide]I never said that there was such a function [/hide]

Another Hint: [hide]You'll need to know something about the topology of the real line. In particular, the Baire Category Theorem: The real numbers cannot be written as a countable union of nowhere dense sets. Equivalently, if {Un} is a sequence of dense open subsets of the real line, then the intersection of those Un is non-empty. This theorem should give you a good start, but there's still work to be done to arrive at the solution.[/hide]

And no, my thesis isn't online. It wasn't about this question in particular, it was about something somewhat related - namely a way to study functions which are continuous but non-differentiable on dense subsets of the real line. I just came across this question in preparing to write it.

Ok, I'm very stupid [hide](I did not read carefuly what you wrote !)[/hide]

and

I'm a little disappointed [hide] (I would have been interested by your thesis)[/hide]

The Baire Category Theorem is one of the great theorems of Topology/Real Analysis. You can prove many many interesting things using it. I'm sure you will spend much time on it in any class on real analysis, and maybe some on topology as well.

If you'd like, I can break the problem up into smaller problems that are more managable, but that lead to the solution.

As a first step, assume that there is such a function, f(x). Let Un be the union of all intervals (a,b) such that |f(x) - f(y)| :le: 1/n for all x,y in (a,b). Show that the intersection of all these sets Un is exactly the set of points at which f(x) is continuous.

Hey CZ look what I found! http://www.academic.marist.edu/math/Seminar/Spring02/020405.htm

Apparently a discussion of thise can be found in [u]Beginning Functional Analysis[/u] by Karen Saxe.

"...In another section of this chapter, two arguments are given that no function can be continuous at each rational number and discontinuous at each irrational number. One is the usual Baire category argument; the other is a less well known and more elementary argument due to Volterra."


Just to let everyone know, there will be a mock AMC contest which will be held fairly soon. Check out [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=100362#100362]this[/url] thread in the AMC forum for details! Many of the later problems will be quite difficult for beginners, but I estimate the early problems to be easy AMC 10 level, so the contest will certainly be worth trying.

About half of the questions are from old contests. The rest are either original or from sercet sources  :lol:

The contest will take place on Friday, July 23, at 3:00 PM Eastern time. I hope you can make it! Further details on the contest have been posted in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=100659#100659]this[/url] thread.

Are you going to do a "mock AIME" for the people who do well on the mock amc?

I don't think I'll end up doing one, because it will be hard for me to create questions that are difficult enough to be considered second half of AIME level. There are also too few old AIMEs to use questions from there. If someone else wants to, though, it'd be appreciated!

[quote="white_horse_king88"]A mock AIME?

Dunno - Hard to find 3 hours of spare time just lying around, though I could do it, if the mock AIME exists.[/quote]

It could be split up into 3 one-hour sessions with 5 questions if anyone ever does a contest like this.  Hmm... mabye I can do a mock AIME.  Ahh, but I can't make a good problem for [tex]\$ \aleph_{\aleph\ddots}}[/tex]!  And why is the tex showing up weirdly?

[quote="nr1337"] Ahh, but I can't make a good problem for [tex]\$ \aleph_{\aleph\ddots}}[/tex]!  And why is the tex showing up weirdly?[/quote]
Um... [code]\$ \aleph_{\aleph\ddots}}[/code] What were you trying to do?

Maybe he wanted something like  \[ \aleph_{\aleph_\ddots} \]

I have no clue why he has the  \$ at the beginning.

[quote="joml88"]Maybe he wanted something like  \[ \aleph_{\aleph_\ddots} \]

I have no clue why he has the  at the beginning.[/quote]

That's what I wanted.  I had the dollar sign because I couldn't make a good problem for $\aleph_{\aleph_\ddots}$ dollars!

O, ok.  I get your humor now  ;)   I'm sure that you could make at least a few good problems.  15, now that would be pretty difficult (and time consuming).


Show that non-negative integers $ a\leq b$ satisfy $ a^2 \plus{} b^2 \equal{} n^2(ab \plus{} 1)$, where $ n$ is a positive integer, if they are consecutive terms in the sequence $ a_k$ defined by $ a_0 \equal{} 0,
a_1 \equal{} n, a_{k\plus{}1} \equal{} n^2a_k \minus{} a_{k\minus{}1}$.


Let $ a,b,c$ be positive real numbers. Prove or disprove
$ \sum_{cyc} \sqrt{ \frac{a^3\plus{}abc}{3}} \ge \sqrt{ \frac{ a^3\plus{}b^3\plus{}c^3}{3}} \plus{}\sqrt{abc}$.

[quote="FelixD"]Let $ a,b,c$ be positive real numbers. Prove or disprove
$ \sum_{cyc} \sqrt {\frac {a^3 \plus{} abc}{3}} \ge \sqrt {\frac { a^3 \plus{} b^3 \plus{} c^3}{3}} \plus{} \sqrt {abc}$.[/quote]

I'd really be interested in seeing a solution- could you please post one? :)
Thanks!! :roll:

really nice problem...
equality holds for $ (t,0,0), t \in \mathbb{R^\plus{}}$

[quote="pfanni"]really nice problem...
equality holds for $ (t,0,0), t \in \mathbb{R^ \plus{} }$[/quote]

Can you post solution ????

[quote="FelixD"]Let $ a,b,c$ be positive real numbers. Prove or disprove
$ \sum_{cyc} \sqrt {\frac {a^3 \plus{} abc}{3}} \ge \sqrt {\frac { a^3 \plus{} b^3 \plus{} c^3}{3}} \plus{} \sqrt {abc}$.[/quote]
Squaring both sides, we can rewrite the inequality as
$ \frac{2}{3}\sum \sqrt{(a^3\plus{}abc)(b^3\plus{}abc)} \ge \frac{2}{\sqrt{3}}\sqrt{abc(a^3\plus{}b^3\plus{}c^3)}.$
Now, using the Cauchy Schwarz Inequality, we have
$ LHS \ge \frac{2}{3} \sum \sqrt{abc}(\sqrt{a^3}\plus{}\sqrt{b^3}) \equal{}\frac{4}{3}\sqrt{abc} (\sqrt{a^3}\plus{}\sqrt{b^3}\plus{}\sqrt{c^3}) \ge \frac{4}{3}\sqrt{abc}\cdot \sqrt{a^3\plus{}b^3\plus{}c^3} \ge RHS.$
So, we have done. :)

Very nice solution.  :P

Can you prove the generalisation too?

$ \sum_{cyc} \sqrt [n] {\frac {a^3 \plus{} abc}{3}} \ge \sqrt [n] {\frac {a^3 \plus{} b^3 \plus{} c^3}{3}} \plus{} \sqrt [n]{abc}$,
where $ n$ is a positive integer.

Does nobody want to post a solution for the generalisation?   :huh:


Hmm... I'm not understanding the question. Try putting some of the expressions into Latex. For example, what does the last part mean?

[quote="Calculus"]If a number D divides A and B, will it also divide their difference, sum, multiplication, division and if we take A to a certain power and then add it?[/quote]

We're given that $A\equiv 0 \pmod D$ and $B\equiv 0 \pmod D$. Hence, $A\pm B\equiv 0 \pmod D$ and $AB \equiv 0 \pmod D$.

However, $A/B$ is not necessarily an integer, so $D$ doesn't always divide it.

If I understand you correctly, the last part is $(A^k + D) \pmod D = A^k \pmod D + D\pmod D = A \equiv 0 \pmod D$.

So yes for all of them except division. ;)

Thanks, that's exactly what I was asking for. Sorry for the lack of clarity.

Ah, that makes sense. Got it. I didn't realize that the second part was a separate question. ;)


How many isomorphic graphs are for graphs:

a) [img]http://img354.imageshack.us/img354/8886/graph1va0.png[/img]

b) [img]http://img354.imageshack.us/img354/9329/graph2ul8.png[/img]

For b), checking the degree, we see that the center must be preserved. Then all the other vertices must also be preserved. Hence, it has only 1 isomorphic graph.

For a), the midpoint must be preserved => the 3 points at the center must be preserved. The rest could be interchanged in any way. Therefore, 3! = 6 isomorphic graphs

P/S: I don't think that the right word to use is "isomorphic".

Thx :)
[quote="mathangel"]P/S: I don't think that the right word to use is "isomorphic".[/quote]
I didn't know which word should I use.
Could you correct my mistake?

"Automorphism" = "isomorphism from an object to itself."  Do you mean, "How many automorphisms are there of these graphs?"  If so, the first one has automorphism group $ {\bf Z}/2{\bf Z} \times S_3$ (we can switch the two vertices adjacent to the "middle vertex" and we can arbitrarily permute the three vertices of degree four independently) for a total of 12 automorphisms while the second one has automorphism group $ S_3 \times S_3$ (any independent rearrangements of the three vertices of degree one and the three "outer" vertices is good) for a total of 36 automorphisms.


Suppose $\omega$ is a primitive $p$ root of unity and $A,B$ are complex square matrices such that $AB=\omega BA$. Then $ A^p+B^p=(A+B)^p$.

It is straightforward. $(A+B)^p=\sum C_1C_2\dots C_p$, where every $C_i$ is either $A$ or $B$. Take any such monomial and replace every $AB$ to $\omega BA$, until we get $\omega^NB^kA^{p-l}$. A number $N$ is the sum of $i$-s, for which $C_i=B$ minus ($1+2+\dots+k$). Indeed, the above sum of indecies decreases by 1 after any replacement $AB\rightarrow \omega BA$. Fix $k$ from 1 to $p-1$. I claim that $S:=\sum \omega^N$ by all $C_p^k$ $p$-tuples of $k$ $B-$s and $p-k$ $A$-s equals 0. It suffices to prove the claim. Indeed, this sum $S$ is nothing but a coefficient at $t^k$ in a product $\prod_{j=1}^p (1+t\omega^j)$. Numbers $-t\omega^j$ are roots of $(-t)^p$. So, by Bezout, $\prod_{j=1}^p (1+t\omega^j)\equiv 1-(-t)^p$ and we are done.

My God,this theorem is one of the most beautiful theorems I have ever seen. :o  :o  :o

I liked its simplicity and the nice result it contains.


vector
LaTeX
linear algebra
matrix
linear algebra unsolved

Let T: V-->V be a linear transformation and V a finite dimensional vector space over R with an inner product <,>
Assume for any u,v  belong to V
<Tu,Tv> =0 Whenever <u,v> =0 and T(z) different from 0  for some z belong V

[b]SHOW THAT T IS INVERTIBLE [/b]

Maybe we can prove by contrapose ? I doubt that my proof is false
[hide="My solution"]
T is not invertible, that we have one or more $ 0$ eigenvalue. Then we can choose $ x$ different from zero so that $ <Tx,Tx>\equal{}0$ but $ <x,x> \geq 0$.
[/hide]
Am I right ?

$ \text{\LaTeX}$ tip for Qruni: use \langle and \rangle instead of < and > to get $ \langle Tx,Ty\rangle$, which looks much better than $ <Tx,Ty>.$

What does the word "whenever" mean, anyway? I read it as one-way implication: if $ \langle x,y\rangle =0$ then $ \langle Tx,Ty\rangle =0.$ If that's what it means, then the original statement is false.

Qruni read the word "whenever" as "iff," in which case the statement is true by his proof. But if that is what was meant, "iff" should have been stated more directly. On the other hand, if it was supposed to mean "iff", then the whole second phrase is redundant.

[quote="Kent Merryfield"]if $ \langle x,y\rangle = 0$ then $ \langle Tx,Ty\rangle = 0.$ If that's what it means, then the original statement is false. [/quote]
I don't think so. :maybe: (assuming the inner product $ \langle ,\rangle$ means positive definite)
Assume we have $ z\in V\setminus\{0\}$ with $ Tz\not=0$ and set $ \lambda: = {|Tz|\over |z|}>0$.
Then for all $ x\in V$ we have $ \langle x,y\rangle=0$ for all $ y\in (\mathbb{R}x)^{\perp}$.
This gives $ 0=\langle Tx,Ty\rangle=\langle T^TTx,y\rangle$ for all $ y\in (\mathbb{R}x)^{\perp}$.
So we get $ T^TTx\in (\mathbb{R}x)^{\perp}^{\perp}=\mathbb{R}x$ for all $ x\in V$.

Then $ T^TTz=\mu z$ for some $ \mu \in\mathbb{R}$ and we get
$ \lambda^2\langle z,z\rangle=\langle Tz,Tz\rangle=\langle T^TTz,z\rangle=\mu\langle z,z\rangle$ and $ \mu=\lambda^2$.

Then $ T^TTx=\lambda^2 x$ for all $ x\in\mathbb{R}z$.
But for all $ x\in V\setminus \mathbb{R}z$ we find $ \alpha,\beta\in \mathbb{R}$
with $ T^TTx=\alpha x$ and $ T^TT(x+z)=\beta (x+z)$.
Then $ 0=T^TT(x+z)-T^TTx-T^TTz=(\beta-\alpha)x+(\beta-\lambda^2)z$
and $ x,z$ linear independent over $ \mathbb{R}$ gives $ \alpha=\beta=\lambda^2$.

So $ T^TTx=\lambda^2 x$ for all $ x\in V$ and $ T^TT=\lambda^2 \text{id}_V$.
This gives $ T$ invertible and $ \lambda^{-1}T$ orthogonal.

Oops. The counterexample I thought I had - a simple projection - is not a counterexample at all.

you can use the singular value decomposition: there exist two orthogonal transformation $ U,V$ and a diagonal matrix $ D$ such that $ T\equal{}UDV$. Then one can notice that $ D$ has the same property as $ T$, so $ D \equal{} \alpha Id$. Finally one concludes that $ T\equal{}\alpha O$, where $ O$ is orthogonal.


MATHCOUNTS
geometry
number theory
\/closed

I think we should have the number of available problems show up on the math forums.

Tare, I think this is a cool idea.  It's kind of like an index.

Another interesting index might be an announcement post that summarizes all the MATHCOUNTS Challenge Problems to date.  The challenge problems can be seen just by scrolling down, but the moderators could classify the challenges by problem type: Geometry, Number Theory, Algebra, Counting, etc.  That way students could work on their weak points or even simply get a better understanding of the common thread between different types of problems.  If you like I will help advise in the indexing.

Does this sound like it would be worth the effort?

This idea could be applied to AMC Forum too. Still, there is not enough people interested in the MATHCOUNTS forum to make it worthwhile.
I'll be more than happy to make the index of AMC, or Intermediate Topics.

Cooler idea: The color of unsolved posts would be some other color like green
Is this possible?

Why not simply make this property for all of the problem boards? I like the idea for the coloring of the unaswered problems. Perhaps each new thread in a problem board should be colored by default, and the moderators would change the color back to normal once they see that the problem on the thread has been answered.

I'll take a look this fall at how hard this would be to program (this is an opensource program - I don't know how the whole thing works; I only modified parts of it).  However, this would also require a lot more work on the part of the moderators, so it's only something moderators would take on if they wanted to.  I do like the idea, though.  Makes open questions easier to find.

[quote]I'll take a look this fall at how hard this would be to program (this is an opensource program - I don't know how the whole thing works; I only modified parts of it).
[/quote]

If its opensource, where exactly can I download the code? I know that there is a link on the bottom of the page, but I couldn't find it on the website.

It is a cool idea, but since school is starting for many, and already started for some, the moderators might not have enough time to do this on their forums. I kno i can't, or at least it would take me a while...

That link down there is just someone who did some layout work that we used.  The source for the forum comes from [url=http://phpbb.com/]here[/url].  We've doctored the original code in a number of fairly simple ways.

Very cool idea...I'd be willing to help since I'm being partially homeschooled this year and would have a lot more time than the reg public school kids.

I'm [b]very[/b] dissapointed that this hasn't been updated while I was gone. I'm beginning to think this can't be updated manually...

I think I'll move this to Suggestions pretty soon...


inequalities
trigonometry
function
inequalities unsolved

let  $ A$, $ B$ and $ C$ are three angles of triangle

prove that:
1)  $ \sin(A/2)\sin(B/2)\sin(C/2)\leq 1/8$
2)  $ \cos(A/2)\cos(B/2)\cos(C/2)\leq (3\sqrt{3})/8$
3)  $ \cos(A)\plus{}\cos(B)\plus{}\cos(C)\leq 3/2$
4)  $ \tan^2(A/2)\plus{}\tan^2(B/2)\plus{}\tan^2(C/2) \geq1$

i'm gonna solve only one because the method is the same in all cases except that in 3) and 4) u don't even need to use AM-GM but only Jensen

denote $ x,y,z$ as half angels of triangle
$ sinx\geq 0$ for if $ x$ from $ [0,\frac {\pi}{2}]$. and function $ sin$ is concave on $ [0,\frac {\pi}{2}]$
$ \sqrt [3]{sin(x)sin(y)sin(z)}\leq \frac {sin(x) \plus{} sin(y) \plus{} sin(z)}{3} \leq sin(\frac {x \plus{} y \plus{} z}{3}) \equal{} sin(\frac {\pi}{6})\equal{}\frac {1}{2}$
first by AM-GM second by Jensen. the inequality is obviously true than


The ratio of two numbers is $ 3: 7$. When 5 is subtracted from each of those numbers, the new ratio is $ 1: 3$. What is the smaller of those two numbers before 5 is subtracted?

hello, let $ x$ and $ y$ the searched numbers, then we get from the problem
$ \frac{x}{y}\equal{}\frac{3}{7}$
and
$ \frac{x\minus{}5}{y\minus{}5}\equal{}\frac{1}{3}$
from the second equation follows
$ y\equal{}3x\minus{}10$
plugging this in the first equation we get
$ x\equal{}15$ and $ y\equal{}35$
so the smaller of those two numbers is $ 15$.
Sonnhard.

[quote="markoboy"]1st of all we must assign variables to this lets call these X and Y the ratio of X to Y is 3/7

x/y=3/7

they say subtract 5 from these # so it looks like

X-5/Y-5=1/3

(I WONT SOLVE THE WAY TO DO THE REST BECUASE I BELIEVE YOU KNOW HOW TO DO THIS ) : )

[hide="ANsWer"]15[/hide]

[size=200][color=#0080BF][b][i][u]PLEASE RATE MY POST THANK YOU SO MUCH IT IS GREATLY APPRECIATED THANK YOU!!!!!![/u][/i][/b][/color][/size]

THE FOLLOWING STATEMENT WILL BE INSCRIBED IN WHITE

[color=#FFFFFF]Pure mathematics is, in its way, the poetry of logical ideas.  ~Albert Einstein

[/color]

 :stretcher:  :stretcher:  :stretcher:  :stretcher:[/quote]

[img]http://i0.kym-cdn.com/photos/images/original/000/096/044/trollface.jpg?1296494117[/img]


Hi all I just found this forum and it's a god send.  I've been out of school way to long and need some help and figured it'd probably be pretty basic.  The question is as follows, how many cubic meters are in a tank that measures 5 m * 4 m * 3.5 m?  I'm not as concerned with the answer so much as how to get to the answer.  Thank you for any assistance.   :D

[hide]In a box with side lengths $l, w, h$ with those being the length, width, and height respectively, the volume is $V=lwh$.  Take a face of the box, and find the area.  How does this help you find the volume?  Well, you multiply it by the height to find how many times the area is in the box.  This ends up being the formula above.  Notice that if all measurements are in meters, you are multiplying meters by meters be meters, which will get you $m^{3}$.  In this problem, you don't need to worry about assigning variables to numbers, just multiply the sides together.  So

$V=5*4*3.5=20*3.5=70$[/hide]


The total variation distance between two measures on a countable set $ S$ is defined to be

\[ \|\mu\minus{}\nu\|\equal{}\sum_{z\in S} |\mu(z)\minus{}\nu(z)|
\]

Show that $ \|\mu\minus{}\nu\|\leq 2\delta$ iff there are random variables $ X,Y$ with distributions $ \mu,\nu$ such that $ P(X\not\equal{}Y)\leq \delta$.

The thing that I may be confused on:  What is $ P(X\not\equal{}Y)\leq \delta$ in terms of $ \mu$ and $ \nu$?

You can get one direction by using the identity
\[ \mu(z) \minus{} \nu(z) \equal{} \Pr(X \ne Y \land X \equal{} z) \minus{} \Pr(X \ne Y \land Y \equal{} z).
\]


floor function
logarithms
limit
algebra unsolved
algebra

[quote="pankajsinha"]Find the value of :

$ \lfloor\frac {x \plus{} 1}{2}\rfloor \plus{} \lfloor\frac {x \plus{} 2}{4}\rfloor \plus{} \lfloor\frac {x \plus{} 3}{6}\rfloor \plus{} \lfloor\frac {x \plus{} 4}{8}\rfloor$ +...........upto  $ \infty$[/quote]

[hide="Just some inermediate results :"]

Let $ f(x)\equal{}\sum_{k\equal{}1}^{\plus{}\infty}\left[\frac{x\plus{}k}{2k}\right]$

I consider $ x\geq 0$

Obviously, $ f(x)$ is constant over $ [n,n\plus{}1)$ $ \forall n\in\mathbb Z$ and so $ f(x)\equal{}f([x])$. So we just have to compute $ f(x)$ over $ \mathbb N$

$ f(n)\minus{}f(n\minus{}1)\equal{}$ number of elements of the sum reaching an integer value, so such that $ \frac{n\plus{}k}{2k}\equal{}p$ and so such that $ n\equal{}k(2p\minus{}1)$, so :

$ f(0)\equal{}0$
$ f(n)\minus{}f(n\minus{}1)\equal{}$ number of odd divisors of $ n$ $ \equal{}\frac{\sigma(n)}{v_2(n)\plus{}1}$ where $ \sigma(n)$ is the number of divisors of $ n$.

So $ f(n)\equal{}\sum_{k\equal{}1}^n\frac{\sigma(k)}{v_2(k)\plus{}1}$

It's not the answer to the question, since I just replaced a sum with another one, but we no longer have $ []$ expressions  :wink:

Hope it will give some ideas to somebody else.[/hide]

:blush:  :blush:  $ [x\plus{}\frac{1}{2}]\equal{}[2x]\minus{}[x]$

I jus saw a formula in one of my books its says $ \lfloor\frac{x\plus{}1}{2}\rfloor\plus{}\lfloor\frac{x\plus{}2}{4}\rfloor\plus{}\lfloor\frac{x\plus{}3}{6}\rfloor\plus{}\lfloor\frac{x\plus{}4}{8}\rfloor ...... \equal{} x$ for x=1,2,3,4,.............

So the answer is x , but can anyone prove the above formula ??

[quote="hophinhan"]:blush:  :blush:  $ [x \plus{} \frac {1}{2}] \equal{} [2x] \minus{} [x]$[/quote]

@Hophinhan :

So $ f(n) \equal{} \sum_{k \equal{} 1}^{ \plus{} \infty}\left[\frac n{2k \minus{} 1}\right ]$ but then ... ?

[quote="vijaymenon"]I jus saw a formula in one of my books its says $ \lfloor\frac {x \plus{} 1}{2}\rfloor \plus{} \lfloor\frac {x \plus{} 2}{4}\rfloor \plus{} \lfloor\frac {x \plus{} 3}{6}\rfloor \plus{} \lfloor\frac {x \plus{} 4}{8}\rfloor ...... \equal{} x$ for x=1,2,3,4,.............

So the answer is x , but can anyone prove the above formula ??[/quote]

@Vijaymenon :

This is wrong : Just take $ x \equal{} 3$ : $ \sum_{k \equal{} 1}^{ \plus{} \infty}\left[\frac {3 \plus{} k}{2k}\right]$ $ \equal{} \left[\frac 42\right] \plus{} \left[\frac 54\right] \plus{} \left[\frac 66\right]$ $ \equal{} 4\neq 3$

Change your book  :wink:

@both :
What is ok is $ n \equal{} \sum_{k \equal{} 1}^{ \plus{} \infty}\left[\frac {n \plus{} 2^{k \minus{} 1}}{2^k}\right]$ $ \equal{} \sum_{k \equal{} 1}^{ \plus{} \infty}\left[\frac 12\plus{}\frac {n }{2^k}\right]$, which is not the required problem ($ \sum_{k \equal{} 1}^{ \plus{} \infty}\left[\frac {n \plus{} k}{2k}\right]$ $ \equal{}\sum_{k \equal{} 1}^{ \plus{} \infty}\left[\frac 12\plus{}\frac {n}{2k}\right]$)

$ \left[\frac{x\plus{}n}{2n}\right]\equal{}\left[\frac{[x]\plus{}n}{2n}\right]\equal{}\frac{[x]}{2n}\plus{}\frac{1}{2}\minus{}\left\{\frac{[x]}{2n}\plus{}\frac{1}{2}\right\}$

$ \left\{\frac{[x]}{2n}\plus{}\frac{1}{2}\right\}\equal{}\left\{\frac{[x] \mod 2n}{2n}\plus{}\frac{1}{2}\right\}$

If $ [x] \mod 2n < n$ then $ \left\{\frac{[x] \mod 2n}{2n}\plus{}\frac{1}{2}\right\}\equal{}\frac{[x] \mod 2n}{2n}\plus{}\frac{1}{2}$

If $ [x] \mod 2n \geq n$ then $ \left\{\frac{[x] \mod 2n}{2n}\plus{}\frac{1}{2}\right\}\equal{}\frac{[x] \mod 2n}{2n}\minus{}\frac{1}{2}$

We alsow know that $ n\leq [x]$

Our sum will be:

$ S\equal{}\sum_{n\equal{}1}^{[x]}\left[\frac{x\plus{}n}{2n}\right]\equal{}\frac{[x]}{2}\left(1\plus{}H_{[x]}\right)\minus{}\sum_{n\equal{}1}^{[x]}\left\{\frac{[x]}{2n}\plus{}\frac{1}{2}\right\}$

Maybe this helps... Last sum we can calculate, i hope...

[quote="vijaymenon"]I jus saw a formula in one of my books its says $ \lfloor\frac {x \plus{} 1}{2}\rfloor \plus{} \lfloor\frac {x \plus{} 2}{4}\rfloor \plus{} \lfloor\frac {x \plus{} 3}{6}\rfloor \plus{} \lfloor\frac {x \plus{} 4}{8}\rfloor ...... \equal{} x$ for x=1,2,3,4,.............

So the answer is x , but can anyone prove the above formula ??[/quote]

WRONG.Try checking your formula for x=3,4,5....

The formula is $ \lfloor\frac {x \plus{} 1}{2}\rfloor \plus{} \lfloor\frac {x \plus{} 2}{4}\rfloor \plus{} \lfloor\frac {x \plus{} 4}{8}\rfloor \plus{} \lfloor\frac {x \plus{} 8}{16}\rfloor ...... \equal{} x$ for x=1,2,3,4,.............

ya thats correct actually it was a typo .

Some simplifications...

As we can see $ S(x) \equal{} S([x])$, so let $ [x] \equal{} a$ is an integer.

Then if $ n\geq\left[\frac {a}{2}\right] \plus{} 1$ then $ a \mod 2n \equal{} a$ and $ \left\{\frac {a}{2n} \plus{} \frac {1}{2}\right\} \equal{} \left\{\frac {a\mod 2n}{2n} \plus{} \frac {1}{2}\right\} \equal{} \frac {a}{2n} \minus{} \frac {1}{2}$

So, we get:

$ \sum_{n \equal{} \left[\frac {a}{2}\right] \plus{} 1}^{a}\left\{\frac {a}{2n} \plus{} \frac {1}{2}\right\} \equal{} \sum_{n \equal{} \left[\frac {a}{2}\right] \plus{} 1}^{a}\left(\frac {a}{2n} \minus{} \frac {1}{2}\right) \equal{} \frac {a}{2}\cdot \sum_{n \equal{} \left[\frac {a}{2}\right] \plus{} 1}^{a}\frac {1}{n} \minus{} \frac {1}{2}\cdot\left(a \minus{} \left[\frac {a}{2}\right]\right)$

Our final answer transform to:

$ S \equal{} a \minus{} \frac {1}{2}\left[\frac {a}{2}\right] \plus{} \frac {a}{2}H_{\left[\frac {a}{2}\right]} \minus{} \sum_{n \equal{} 1}^{\left[\frac {a}{2}\right]}\left\{\frac {a}{2n} \plus{} \frac {1}{2}\right\}$

This formula is true, checked by computer... so no error in calculations )))

As we know, for big $ a$: $ H_a\sim \ln a$ and ... $ \sum_{n \equal{} 1}^{\left[\frac {a}{2}\right]}\left\{\frac {a}{2n} \plus{} \frac {1}{2}\right\} < \left[\frac {a}{2}\right]$, so...

when $ a$ is very big $ S\sim a\ln a$  :)

or more exactly: $ \lim_{a\to\infty}{\frac {S(a)}{a\ln a}} \equal{} \frac {1}{2}$.

[quote="Yuriy Solovyov"]Some simplifications...

As we can see $ S(x) \equal{} S([x])$, so let $ [x] \equal{} a$ is an integer.

Then if $ n\geq\left[\frac {a}{2}\right] \plus{} 1$ then $ a \mod 2n \equal{} a$ and $ \left\{\frac {a}{2n} \plus{} \frac {1}{2}\right\} \equal{} \left\{\frac {a\mod 2n}{2n} \plus{} \frac {1}{2}\right\} \equal{} \frac {a}{2n} \minus{} \frac {1}{2}$

[/quote]

What does $ \{ \}$ denote and what does $ a \mod2n$ denote.

[quote="pankajsinha"][quote="Yuriy Solovyov"]Some simplifications...

As we can see $ S(x) \equal{} S([x])$, so let $ [x] \equal{} a$ is an integer.

Then if $ n\geq\left[\frac {a}{2}\right] \plus{} 1$ then $ a \mod 2n \equal{} a$ and $ \left\{\frac {a}{2n} \plus{} \frac {1}{2}\right\} \equal{} \left\{\frac {a\mod 2n}{2n} \plus{} \frac {1}{2}\right\} \equal{} \frac {a}{2n} \minus{} \frac {1}{2}$

[/quote]

What does $ \{\}$ denote and what does $ a \mod2n$ denote.[/quote]

$ \{\}$ - fractional part of then real number. $ \{x\} \equal{} x \minus{} [x]$ and $ 0\leq\{x\} < 1$ for all real $ x$.

$ a\mod 2n$ - remainder of division $ a$ to $ 2n$. For example $ 13 \mod 3 \equal{} 1$ and $ 3\mod 6 \equal{} 3$.

$ H_{m}\equal{}\sum_{k\equal{}1}^{m}\frac{1}{k}$ - harmonic series.