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Let R+ be the set of positive reals and let F be the set of all functions f : R+ → R+ such that f(3x) ≥ f( f(2x) ) + x for all x. Find the largest A such that f(x) ≥ A x for all f in F and all x in R+.

I solved this problem some time ago. Search for it.

Let S be the set S = {a >= 0 ; f(x) >= a*x for all x > 0 and for all f in F}.
S is not empty because 0 is in S. Manifestly g(x) = x/2 is in F , so for all a in S we have a :le: 1/2 .
So the superior bound of S exists. Let A be this superior bound. Manifestly A :le:  1/2.
Put x/3 in the initial inequality, so f(x) >=  f(f(2x)/3) + x/3  >=  A*f(2x/3) + x/3  >=  (2*x/3)*A2  + x/3  =  (2*A2/3 + 1/3)*x for all x > 0 and all f in F
So (2*A2 +1)/3 is in S , so (2*A2 +1)/3 :le:  A , so 1/2 :le:  A :le:  1
Finally the asked value is A=1/2 .
 :cool:   vietnamese team is good :cool:

I think it isn't the most difficult problem .

[b]Problem 3:[/b] Let Sn be the number of permutations (a1, a2, ... , an) of (1, 2, ... , n) such that 1 ≤ |ak - k | ≤ 2 for all k. Show that (7/4) Sn-1 < Sn < 2 Sn-1 for n > 6.

I can't solve it , it is really hard  !

[Moderator edit: Moved to http://www.mathlinks.ro/Forum/viewtopic.php?t=3710 .]


I was just wondering if someone could do a couple of questions for me. Just with step by step solutions, thanks a lot. I'm learning a lot from these forums, thanks peoples!

1) If x = 1 is a double root of the equation x^4 + a(x^3) + b(x^2) − 5x + 1 = 0, find a and b.

2) Given P(x) = 2(x^4) − 20(x^3) + 74(x^2) − 120x + 72 = 0 has two double roots. Find all the roots of the equation
P(x) = 0. Factor P(x) completely.

Thanks a million!

If $a$ is a double-root of our polynomial, then surely $W(1)=0$
$1+a+b-5+1=0 \Rightarrow a+b=3$

Now substitute $3-a$ for $b$

$W(x)=x^4+ax^3+(3-a)x^2-5x+1$
We divide $W(x)$ by $x-1$
We obtain:
$W(x)=(x-1)(x^3+(a+1)x^2+4x-1)$
$1$ must be a root of $R(x)=x^3+(a+1)x^2+4x-1$
$1+a+1+4-1=0 \Rightarrow \boxed{a=-5}$ and $\boxed{b=8}$

$W(x)=(x-1)(x^3-4x^2+4x-1)=(x-1)^2(x^2-3x+1)$

Second:
Theorem:
If polynomial has integer coefficients and root $x_0$, then $x_0|a_0$, where $a_0$ is "the last coefficient"
$W(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots a_2x^2+a_1x+a_0$

Let's go back to our problem. We must solve the equation:
$2x^4-20x^3+74x^2-120x+72=0$ Divide by $2$
$x^4-10x^3+37x^2-60x+36=0$
We look for integer roots among divisors of $36$ it turns out, that $2$ is OK!
We divide polynomial $R(x)=x^4-10x^3+37x^2-60x+36$ by $x-2$
$R(x)=(x-2)(x^3-8x^2+21x-18)$
Now we consider polynomial:
$Q(x)=x^3-8x^2+21x-18$ and we look for roots among divisors of $18$ We found that $3$ is a root
So we divide $Q(x)$ by $x-3$ and obtain $x^2-5x+6=(x-2)(x-3)$

Finally:
$\boxed{2x^4-20x^3+74x^2-120x+72=2(x-2)^2(x-3)^2}$


What is the concentration of ions in 1.2 M aluminum nitrate $Al(NO_3)_3$?

I can't decide if it's 0.40 M or 4.8 M...

Thanks

Each mol of $Al(NO_3)_3$ contributes $4$ mols of ions in solution, so multiply by $4$.

Al(NO3)3 ---> Al3+ +3NO3-
therefore there are 4 ions


How many perfect squares are there (mod $ 2^{n}$)?

I think this means that there are two squares $ (mod 2^{1})$: $ 0, 1$; there are three squares $ (mod 2^{3})$: all squares are either $ 0, 1$ or $ 4 (mod 8)$;  there are four $ (mod 16)$, .... But I'm not sure.

Thanks r3d30m3j. How did you arrive at this?


Let $ a,b,c$ be sides of a triangle. Prove that

$ \sqrt{3(\sqrt{ab}\plus{}\sqrt{bc}\plus{}\sqrt{ca})}\geq \sqrt{a\plus{}b\minus{}c}\plus{}\sqrt{b\plus{}c\minus{}a}\plus{}\sqrt{c\plus{}a\minus{}b}$

Let $ x,y,z>0$ such that $ a=x+y,b=y+z,c=z+x$. Then our inequality is equivalent to
\[ 3\sum_{cyc}\sqrt{(x+y)(y+z)}\ge 2\left(\sum_{cyc}\sqrt x\right)^2\]
From Cauchy-Schwarz inequality,
\begin{align*}3\sum_{cyc}\sqrt{(x+y)(y+z)}&\ge 3\sum_{cyc}(y+\sqrt{zx})\\
&\ge 2\sum_{cyc}y+4\sum_{cyc}\sqrt{zx}\\
&=2\left(\sum_{cyc}\sqrt x\right)^2\end{align*}

[quote="jbmorgan"]Let $ a,b,c$ be sides of a triangle. Prove that

$ \sqrt {3(\sqrt {ab} \plus{} \sqrt {bc} \plus{} \sqrt {ca})}\geq \sqrt {a \plus{} b \minus{} c} \plus{} \sqrt {b \plus{} c \minus{} a} \plus{} \sqrt {c \plus{} a \minus{} b}$[/quote]
Let $ a \equal{} x \plus{} y$, $ b \equal{} y \plus{} z$, $ c \equal{} z \plus{} x$, where $ x,y,z \in \mathbb{R}$. Since $ a,b,c$ are sides of a triangle $ x,y,z \ge 0$.

Then the problem becomes:
$ \sqrt {3\sum_{cyc}\sqrt {(x \plus{} y)(x \plus{} z)}} \ge \sum_{cyc}\sqrt {2x}$

Then by cauchy-schwartz: $ \sqrt {(x \plus{} y)(x \plus{} z)} \ge x \plus{} \sqrt {yz}$, so it suffices to prove:

$ 3\sum_{cyc} x \plus{} 3\sqrt {yz} \ge 2\sum_{cyc}x \plus{} 4\sum{cyc}\sqrt {yz}$ $ \iff$
$ x \plus{} y \plus{} z \ge \sqrt {xy} \plus{} \sqrt {yz} \plus{} \sqrt {zx}$ $ \iff$
$ \sum_{cyc} (\sqrt {x} \minus{} \sqrt {y})^2 \ge 0$. QED :)

Edit: Too late.. Beaten :P


calculus
integration
algebra
function
domain
Ring Theory
superior algebra

I have a question..

    If R is an integral domain then R is unique factorization domain if and only if  every nonzero prime ideal in R contains a nonzero principal ideal that is prime .

May you help me please?

A principal ideal generated by a prime element is a prime ideal. If $R$ is UFD and $P$ is a non-zero prime ideal, let $x\in P$, $x\neq 0$. Then $x=p_{1}^{a_{1}}\cdot p_{k}^{a_{k}}$ for some primes $p_{*}$ and integers $a_{*}$. Because $P$ is prime, it must then contain one of the $p_{*}$s and we are done.

Conversely, assume every prime ideal contains a prime element.
Consider $S$ the multiplicative system of elements gerated by the invertible elements of $R$ and by the primes (elements not ideals) in $R$.
If $S=R\setminus\{0\}$, then $R$ is an UFD. Otherwise, let $P$ be a non-zero prime ideal of $S^{-1}R$. This corresponds to a nonzero ideal of $R$ such that $P\cap S=\emptyset$. By the hypothesis, $P$ contains a prime element which must belong to $S$. We have a contradiction, so $S\cup 0=R$ and $R$ is UFD.


Find all functions $ f(n)$ defined on the non-negative integers with values in the set $ \{0, 1, 2, ... , 2000\}$ such that: (1) $ f(n) \equal{} n$ for $ 0\leq n\leq 2000$; and (2) $ f( f(m) \plus{} f(n) ) \equal{} f(m \plus{} n)$ for all $ m, n$.


For which integer a does $ x^{2}-x+a$ divide $ x^{13}+x+90$?

Let $ x^{13}+x+90=(x^{2}-x+a)p(x).$ Because the olderst coefficient of $ x^{2}-x+a$ equals 1 the polinom $ p(x)$ has integer coefficients. It follows that given any $ x$ we must have that integer $ x^{2}-x+a$ must devide integer  $ x^{13}+x+90$
Take
$ x=0 \Rightarrow a | 90$
$ x=1 \Rightarrow a | 92$
The only possibilities are possible: $ a=\pm 1, \ a=\pm 2$
$ x=-1 \Rightarrow a+2 | 90$ This kills $ \ a=\pm 2$
To kill $ a=1$ take $ x=2$.
To kill $ a=-1$ take $ x=-2$.

So who have mistaken?

$ x =-1$ should be $ a+2 \mid 88$, not $ 90$, so I believe $ a = 2$ works. Nice solution.

[quote="MellowMelon"]$ x =-1$ should be $ a+2 \mid 88$, not $ 90$ ... [/quote]
 :oops_sign: It was my mistake, thanks. The case $ a=2$ works:

$ x^{13}+x+90=(x^{2}-x+2)p(x)$ with
$ p(x)=x^{11}+x^{10}-x^{9}-3x^{8}-x^{7}+5x^{6}+7x^{5}-3x^{4}-17x^{3}-11x^{2}+23x+45$
The other approach to verify is to check that $ x^{13}+x+90=0$ for $ x=\frac{1+\sqrt7i}{2}$ but it doesn't look to be more simple.

[b]Added[/b]: I've changed my mind: to check $ x=\frac{1+\sqrt7i}{2}$ is rather simple if we conseaquently calculate
$ x^{2}, \ x^{4}, \ x^{8}, \ x^{12}=x^{4}x^{8}, \ x(x^{12}+1)$.

But perhaps the most easy way is the following:

Let $ x^{2}-x+2=0$. Then
$ x^{2}=x-2 \Rightarrow x^{4}=x^{2}-4x+4 = x-2-4x+4 =-3x+2 \Rightarrow x^{8}=...=-3x-14 \Rightarrow x^{12}=x^{4}x^{8}=(3x-2)(3x+14)=...=45x-46 \Rightarrow x^{13}+x+90=x(45x-45)+90=45(x^{2}-x+2)=0$


We have an infinite set $ \{x_{i} / i \in \mathbb{N}\}$ and we know the following relation between it's elements:

$ (n \plus{} 1) \times x_{n \plus{} 1} \equal{} n \times x_{n \plus{} 2} \plus{} x_{n \plus{} 2} \times x_{n \plus{} 1}^{2}$.

Find $ x_{n}$ in terms of $ n$.

1. If $ \exists x_i \equal{} 0$, then all $ x_n \equal{} 0$. Proof by induction.

Assume $ x_n \neq 0$.

2. $ x_n \ge n$. Proof by triviality.

Then $ nx_n \equal{} x_{n\plus{}1}(n \minus{} 1 \plus{} x_n^2)$.
$ x_n > n \minus{} 1 \plus{} x_n^2$.
$ x_n^2 < x_n$.
$ x_n < 1$.

Contradiction.

what is a contradiction? this is a well defined sequence of numbers, I think that your wrong assumption is that $ x_{n} > n$, there must be some mistake in your proof of that. Can you write it down?

Something that is obviously is that $ x_{1} < x_{2} < .... <x_{n} < ... < 1$. But I can't see your statement.


inequalities
floor function
number theory unsolved
number theory

Prove that for all positive integers $n$,

\[0 < \sum_{k=1}^n \frac{g(k)}{k} - \frac{2n}{3} < \frac{2}{3},\]

where $g(k)$ denotes the greatest odd divisor of $k$.

What we have to do is clear, isn't it?

We need to show $\frac 23\cdot n<S_n=\sum_{k=1}^n\frac 1{2^{ord_2k}}<\frac 23\cdot(n+1)$, where $ord_2k$ is the exponent of $2$ in the decomposition of $k$. Let $a_n(2^k)=\lfloor\frac n{2^k}\rfloor-\lfloor\frac n{2^{k+1}}\rfloor$.

Between $1$ and $n$ there are $a_n(1)=n-\lfloor\frac n2\rfloor$ odd numbers, $a_n(2)=\lfloor\frac n2\rfloor-\lfloor\frac n4\rfloor$ multiples of $2$ which are not multiples of $4$, and so on. We have $S_n=\sum_{k\ge 0} \frac 1{2^k}\cdot a_n(2^k)$. When we expand this sum we get $n-(\frac 12\cdot\lfloor\frac n2\rfloor+\frac 1{2^2}\cdot\lfloor\frac n{2^2}\rfloor)+\ldots\ (*)$. All we need to do now is use the inequality $x-1<\lfloor x\rfloor\le x$ on each of the $\lfloor x\rfloor$ in $(*)$. The result will follow.

Indeed, lower bound is clear, but grobber is slightly wrong concerning upper bound ;) , since his method gives $\frac{2n}{3}+1$ instead of $\frac{2n}{3}+\frac{2}{3}$.

Actually, we need more accurate estimate. Note that $[\frac{a}{b}]\geq\frac{a+1}{b}-1$ for all $a,b\in\mathbb{N}$, so
\[n-\sum_{k\geq 1}\frac{1}{2^k} \lfloor \frac{n}{2^k} \rfloor < n-\sum_{k\geq 1}\frac{1}{2^k}(\frac{n+1}{2^k}-1)\leq \frac{2n}{3}+1-\frac{1}{3}.\]

What is wrong with my \rfloor and \lfloor ?  :?

[code]\letf\lfloor[/code]

I think it's fairly obvious what's wrong now :) .

Oh, sorry about that. I don't know why it seemed to me like we were gettring $\frac 23\cdot n+\frac 13$, which was even better. Obviously, I was wrong :).


limit
logarithms
function
inequalities
AoPSwiki
complex analysis
number theory proposed

Does anyone have an elegant proof for

\[ \prod_{p\in \mathbb{P}, p<x} p\approx e^x.\]

$ \Leftrightarrow\lim_{n\to \infty}\frac {1}{n}\sum_{p\in P,p < n}\log p \equal{} \lim_{n \to \infty}\frac {\theta (n)}{n} \equal{} 1$ where $ \theta (n)$ is the first [url=http://en.wikipedia.org/wiki/Chebyshev_function]Chebyshev function[/url].

Of course, this is quite unsatisfying, since your question is ultimately [i]why[/i] $ \frac {\theta (n)}{n}\to 1$.  A good place to start might be [url=http://www.artofproblemsolving.com/Wiki/index.php/Prime_Number_Theorem]this AoPSWiki[/url] page on the Prime Number Theorem.  Note especially Lemma 5.  The function $ \psi$ is closely related to $ \theta$ ($ \psi$ is the [i]second[/i] Chebyshev function), and the Lemma ties the statement $ \frac {\psi (x)}{x} \to 1$ to PNT (a proof is only given in one direction, but in fact they are equivalent).

Since the statement is equivalent to PNT, I don't think there's an elegant and elementary (i.e. not involving complex analysis) proof to be had.

[quote="JoeBlow"]Since the statement is equivalent to PNT, I don't think there's an elegant and elementary (i.e. not involving complex analysis) proof to be had.[/quote]

Well, that's debatable.  There are [url=http://en.wikipedia.org/wiki/Prime_number_theorem#Elementary_proofs]elementary proofs[/url] available, but whether they're more or less elegant than the complex-analytic ones...

I agree the 'elegant' part is definitely subjective to begin with, so I gave my subjective opinion that the existing elementary proofs are not very elegant. :)


1. f’(x)=4x^3-12x^2. find out in which interval it increases and decreases. I make the function equals to 0, so I got x=0, x=3. Then I make a table for each interval: x < 0, 0 < x < 3, 3 < x.  After all, I know that f is decreasing in x<0, however my book writes this as (- infinity, 0], and 0<x<3 [0, 3], and increasing in 3<x [3, infinity)  I mean why my book uses "]" to include the numbers? It should not include since I didn't make it include when I did the interval. Does it matter if I didn't include the numbers?
2. This is a piecewise function: y=2x, x<1; y=2-x^2, x>=1. Find the local extremes values.  My book says: none. But I think when x=1 (it's the critical point which makes f' doesn't exist), there is a maximum 1.

Thanks for help.

#1: The definitions: $ f(x)$ is increasing on an interval $ I$ if $ f(x)\le f(y)$ whenever $ x,y\in I$ and $ x<y$. It's strictly increasing on $ I$ if $ f(x)<f(y)$ whenever $ x,y\in I$ and $ x<y$. Decreasing and strictly decreasing functions are also easily handled.

For any continuous function $ f$ defined on $ \mathbb{R}$ and any $ x_0\in \mathbb{R}$, the largest possible interval containing $ x_0$ on which $ f$ is increasing must be closed (This interval may be the single point $ \{x_0\}$). The same is true for strict increase, decrease, and strict decrease.

#2: If the function isn't defined on an open interval containing $ x_0$, $ x_0$ is not a local extremum.

[quote="jmerry"]#1: The definitions: $ f(x)$ is increasing on an interval $ I$ if $ f(x)\le f(y)$ whenever $ x,y\in I$ and $ x < y$. It's strictly increasing on $ I$ if $ f(x) < f(y)$ whenever $ x,y\in I$ and $ x < y$. Decreasing and strictly decreasing functions are also easily handled.

For any continuous function $ f$ defined on $ \mathbb{R}$ and any $ x_0\in \mathbb{R}$, the largest possible interval containing $ x_0$ on which $ f$ is increasing must be closed (This interval may be the single point $ \{x_0\}$). The same is true for strict increase, decrease, and strict decrease.

#2: If the function isn't defined on an open interval containing $ x_0$, $ x_0$ is not a local extremum.[/quote]

But for the second question, the second function of the piecewise function is defined when x=1.

I didn't notice that- actually, that means that there isn't a maximum of any kind. The supremum of the values near 1 is 2, but this isn't achieved.

Or, in other words, $ x \equal{} 1$ isn't a local maximum because there are points nearby at which the function takes larger values than it does at $ x \equal{} 1$.

Sry, I don't understand some parts. What do you mean by 'increasing' or 'strictly increasing'? What is X0? What is f(y)?


Prove that $ \tan\frac x6\cdot \sin x\equal{}\left(1\minus{}\cos \frac x3 \right)\left(4\cos^2\frac x3\minus{}1\right),\ \forall\ x\in\mathbb{R}$.

[hide="Hint"]
Take $ \frac{x}{3}\equal{}\theta$
$ \cos \theta\equal{}y$
$ tan \frac{\theta}{2}\equal{}\sqrt{\frac{1\plus{}y}{1\minus{}y}}$
Use $ \sin 3\theta\equal{}3\sin \theta\minus{}4\sin^{3}\theta\equal{}\sin \theta(y^{2}\minus{}1)$
And $ \sin \theta\equal{}\sqrt{1\minus{}y^{2}}\equal{}\sqrt{(1\minus{}y)(1\plus{}y)}$
[/hide]

Hope this will help you

Actually, tangent is not defined when $ x\equal{}6k\pi\plus{}3\pi$ for all $ k\in\mathbb{Z}$.


What is the probability that a randomly drawn positive factor of 60 is less than 7?

$\left(A\right) \frac{1}{10}$
$\left(B\right) \frac{1}{6}$
$\left(C\right) \frac{1}{4}$
$\left(D\right) \frac{1}{3}$
$\left(E\right) \frac{1}{2}$

[hide]
$7^{2}< 60 < 8^{2}\implies 7 < \sqrt{60}< 8$
Every factor less than 7 has a corresponding factor greater than 7, so the answer is $\boxed{\text{E}}$
[/hide]


geometry
symmetry
probability
analytic geometry

I have openned this topic because, i think, that's the most important thing that preoccupates theoritical physicists nowadays
i wish everybody put here what he finds, thinks or proposes  :lol:  to solve the ultime physics delimma

[quote="hello"]I think that is a more appropriate question to advanced physicists since they are doing the advanced work.[/quote]
ok , but don't forget that Einstein that you have his photo as avatar, had developped his first ideas about relativity when he was still a teen with his friends in a forum they called "olympia" .
I don't ask for the equation, i just want to discuss things generally : i want that everybody exposes what he knows so as to discuss it,  and trust me that's very intresting ;)

[quote="inteluser"]I have discovered a truly remarkable theory of everything which this post is too small to contain.[/quote]

That seems exciting, just give the general idea , can you?

[quote="inteluser"]No, because perhaps posterity will thank me for having shown it that the ancients did not know everything.[/quote]

I see ... :D

[quote="inteluser"]I have discovered a truly remarkable theory of everything which this post is too small to contain.[/quote]
haha  lol  :P

[quote="inteluser"]I have discovered a truly remarkable theory of everything which this post is too small to contain.[/quote]

Nice one Fermat. :D

[quote="11th dimension"]String theory/M-theory might be the unifying theory that physicsts are seeking for a long time.[/quote]

perhaps it is, but physicists wish not, because this theory requires lots of tough calculs.
there is an other thing, string theory involves many dimensions, and that seems a little 'artificial', and remembers the misadvanture of the "ether".
nevertheless, it has given no few good results, and advances notably the other theories candidates such as : non commutative geometry ( alain connes), fractal relativity ... but nothing is clear nor sure up to now.

[quote="hello"]Maybe there is no unifed theory, unlikely that's its true though lol. Well, I like this conversation if it involves the origin of the universe. Want to discuss that?[/quote]
of course, but please open a new topic if you want so

In my perspective, in order to formulate an unified theory of everything a lot of different areas of physics would have to be included. Firstly, we need to group the Universe into two main categories, Matters and Energy, and from each category there had be other sub-categories such the weak, strong, electromagnetic and gravitational interactions.

In this unified theory, we must be able to observe the symmetry and parity proposed by various theorists such as T.D. Lee. The theory should be able to govern the behaviours of things from Macroscopic objects (such as how will a car be affected when a certain amoount of force is applied) to microscopic objects (like how leptons, quarks, neutrino, positron etc. behaves).

A probable proposal is that the mechanics of Quantum Mechanics, Quantum Electrodynamis, Quantum Chromodynamics, Quantum Field Theory, String Theory, M-Theory etc. All the behaviours of the Universe proposed by these theories should be able to be explained (regardless of the fact of correctness).

If one could somehow combine the Theory of General Relativity and Quantum together, we might be on the verge of the breakthrough to a true Grand Unified Theory. The elimination of the infinite value that pops up when trying to combine gravitational force with Quantum will enable theoretical physicists to calculate the probability of particle behaviour more accurately without undefined and meaningless solutions.

Just thought I'd post a stupid little quote:

[i]"There is a theory which states that if ever anybody discovers exactly what the Universe is for and why it is here, it will instantly disappear and be replaced by something even more bizarre and inexplicable. There is another theory which states that this has already happened. "[/i]
-Douglas Adams, author of "The Hitchiker's Guide to the galaxy"

oof! Every single thing that man's ever written is my favorite quote

You guys can start out by explaining why (not how) things move, that has always troubled me. What is movement and why can't two particles occupy the same time/space coordinates? Mind you, i'd like an explanation of movement that doesn't make use of movement. :)

argh... well, for one, we're not the professionals, so it's unlikely we'd know on this board.  http://www.physicsforums.com has some very knowledgeable people, though
and two...
well, doesnt incompleteness mean you cant really prove anything without self-reference somewhere?
Can't we just say "things move because they do"

It has been theorized that such a GUT existed in the first $10^{-46}$ seconds of the universe. At this point in time, there was such a high internal energy of the universe that everything was split into its component parts. This brings a question up for me. Taking a limit as time approaches the beginning of the universe, wouldn't internal energy of the system be high enough as to break apart even the components of everything? Wouldn't this imply a theory that there are no indivisible units?

Here's something about M-Theory, including some of my own logical observations.  M-Theory, the unification of the 5 string theories, is what the theory of everything is all about.  A purely logical and mathematical theory based on physics, is currently known as the theory of everything which is hopefully going to unify the 4 forces, Weak Force, Strong Force, Electromagnetism, and Gravity from Enistein's theory of relativity.  This will allow us, humans, to explain the mysterious actions within a blackhole's singularity.  M-Theory combined with my own logical thoughts can also explain the formation of everything as one type of energy, no a days known as variable vibrating strings of energy within quarks.  Hitherto, we consider matter as anything other than energy, $x\neq E$ :).  If you consider matter as tiny things being held together with the forces of electromagnetism, you will be abe to reverse your thoughts and explain energy in the same way.  Let's say that matter as we can touch is composed of bits of vibrating strings that are D-branes (connecting to the membrane of the dimensions we live in).  This means that energy is composed of p-branes$\neq$D-branes$\rightarrow$ they can travel between dimensions.  If so, energy is something that travels between dimensions, and does not necessarily exist in our dimension...we just sense its presece.  Gravitons may be considered as particles travelling between dimensions, however, the effects of gravitons in the magnificent world of Einstein's spacetime always take place by the warpage of the fabric of spacetime caused by an energy density. We know that even one branch of the theory of big bang is known as nucleosysthesis, the procees in which phtons were created into matetr as we today.  Now, photons are quite hard to explain with M-Theory because if we disregard the fact that [b]everything[/b] is composed of waves, we will have to deal with the complexity of the phtons' particle-wave duality. :)

Masoud Zargar

Are you aware of a book that treats the basics of String theory mathematically???

No.  But I have a pdf file around 80 pages that has all the mathematics of M-Thery in it.  Would you like to have it just to take a look over it?  You may be able to understand it now or even in the future. :)

Masoud Zargar

Yes!!!!! Mail it to me at rushilgoel@yahoo.com   If you are aware of any other book etc which treats String theory properly, please tell
!!!!!1 :lol:  :lol:  :lol:  :D  :D  :D

I'll mail it to, and I think I have some othe rphysics books too. :)

If you have them online, please mail them to me too!!! If you have their names, please discuss them in the Physics Books topic I have in the Physics Forum


1) One store sold red pens at four for a dollar and yellow pens at three for a dollar.  A second store sold red pens at four for a dollar and yellow pens at six for a dollar. You bought [i]m[/i] red pens and [i]n[/i] yellow pens from each store, spending a total of ten dollars. How many pens did you buy?

A) 10
B) 20
C) 30
D) 40
E) Not enough information

[hide="I did..."]
Red Store A & B - 25 cents
Yellow Store A - 33 cents
Yellow Store B - 67 cents.

So, you could have 40 red pens. Or you could have like one yellow from each store to make a dollar, and then 36 red pens to make the other 9 dollars. So, I put E. Why then, is the answer D?[/hide]

2) A small tree 5 feet from a lamp post casts a shadow 4 feet long.  If the lamp post were 2 feet taller the shadow would only be 3 feet long. How tall is the tree?

A) 9/2 ft.
B) 16/3 ft.
C) 14/3 ft.
D) 18/5 ft.
E) 24/5 ft.

[hide="I did..."]
I drew the figure, and made it into similar triangles.
4 /x = 9/12

But, I don't remember where I got the twelve. I solved for 'x' and got B, but the coreect answer is 'E.' Anyone? [/hide]

3) Let f(x) be a funstion such that f(x) = 2 f(x - 1) + 4 and f(0) = 16.  What is the value of f(-2) + f(2)?

A) 0
B) 16
C) 32
D) 77
E) 152

[hide="I did..."]
I found that f(1) =36, and with that wounf that f(2) = 76. From there, I guessed D, that the answer was 77, and it was correct, but I have no idea how.  :blush:  [/hide]

4) How many real zeros does the function
f(x) = x :^4: + x :^3: - 2x 2 + 4x - 24 have?

A) 0
B) 1
C) 2
D) 3
E) 4

[hide="I did..."]
I tried to factor it, but couldn't get it to work. Any ideas?[/hide]

This was a contest that I took this morning. The ICTM (Indiana Council of Teachers of Mathematics) was what it was called. Anyone else take it? If so, what was your score? I got a 28/36. These were some of the problems that I missed, and don't know how. Thnx in advance for the help!

4. Try synthetic division.

3. f(0) = 2f(-1) + 4... Hopefully that'll give you an idea.

1) Yellow pens cost 25 cents each, and we have to buy $m$ of them from each store, and red pens cost 1/3 dollars or 1/6 dollars, and we have to buyt $n$ at each price, so they sell at an average price of 25 cents, and we only spend 10 dollars, so we have $\frac{1}{4}m+\frac{1}{4}n=10$ if we multiply both sides by 4 we get $m+n=40$, and thus we need to buy 40 pens.  As for your question, you would have 2 yellows from each store, and thus 4 total, then 18 reds from each store for a total of 40 pens.

2) In the first case, let $y$ represent the height of the lamp post, and let $x$ represent the height of the tree, then by the similar triangles we have $\frac{y}{9} = \frac{x}{4}$, from the second case we have $\frac{y+2}{8}=\frac{x}{3}$, solving this system gives $x=\frac{24}{5}$, and the height of the post happens to be $\frac{54}{5}$.

3) $f(2) = 2f(1)+4$
    $f(1) = 2f(0)+4=36$
    $f(2) = 2(36)+4=76$
    $f(0) = 2f(-1)+4 \Rightarrow f(-1) = 6$
    $f(-1) = 2f(-2)+4 \Rightarrow f(-2) = 1$
Thus $f(2) + f(-2) =77$.

4)$x^4+x^3-2x^2+4x-24=(x-2)(x+3)(x^2+4)$
The only real zeros are $x=2$ and $x=-3$.


Let $a,b,c\geq0$.If \[ac\geq b^{2}\] and \[ab\geq c^{2}\] then prove that $a=b=c$ [or find a counter-example]

[quote="paladin8"]What if you add the condition $bc \ge a^{2}$? (It's easy, but at least it's true).[/quote]

[hide]just add them up and use the commonly known $a^{2}+b^{2}+c^{2}\ge ab+bc+ca$[/hide]

[quote="D.P.L"][quote="paladin8"]What if you add the condition $bc \ge a^{2}$? (It's easy, but at least it's true).[/quote]

Can deduce that $bc \ge a^{2}$, just multiply the first and second condition and simplify. :wink:[/quote]

Multiplying the first and second condition gives $bc \le a^{2}$.  In any case, a counterexample was already given.

$ab+bc+ac\ge a^{2}+b^{2}+c^{2}$

 :ninja:  :rotfl:  :rotfl:

If you were given $bc \ge a^{2}$, you could just multiply the three to get they're all equalities...


geometry
geometric transformation
dilation
quadratics
logarithms

Here are some problems from the NCIML. I didn't want to post in the New York section, because there isn't anything for Long Island. But a lot of them are close in similarity to the AMC.

1. When 3 times a number is subtracted from 8, and the result is multiplied by the sum of twice the number and (16/3), the result is 12. Compute the square of the number.

2. John has a gallon of sulfuric acid solution in water. How many gallons of water must he add to the solution to dilute the solution to 70%?

3. Compute # of digits in 175 to the 2009.

4. The base of an isosceles triangle is 65 cm. An altitude to one of the legs is 60 cm. Compute the length of one of the legs of the triangle.

5. The sum of the squares of the digits of a two digit # is 18 more than the number formed by reversing the digits. The original number is 18 more than twice the products of its digits. Compute the original number.

6. It takes an electrician 6 hours to wire a house. It takes an apprentice 10 years to do the same job. The electrician started the job alone, then turned it over to three apprentices, working together to finish the job. If the total amount of time for the entire job was 5 hours, compute the amount of time that the electrician worked alone.
 :D  :first:  :rotfl:

[b]1. When 3 times a number is subtracted from 8, and the result is multiplied by the sum of twice the number and (16/3), the result is 12. Compute the square of the number.[/b]
[hide="My Solution"]Let the number be $ x$. According to given conditions,
$ (8 \minus{} 3x)(2x \plus{} \frac{16}{3}) \equal{} 12$
=> $ (8 \minus{} 3x)(6x \plus{} 16) \equal{} 36$
=> $ (8 \minus{} 3x)(8 \plus{} 3x) \equal{} 18$
=> $ 8^{2} \minus{} (3x)^{2} \equal{} 18$
=> $ 9x^{2} \equal{} 64 \minus{} 18 \equal{} 46$
=> $ x^{2} \equal{} \frac{46}{9}$[/hide]

[b]2. John has a gallon of sulfuric acid solution in water. How many gallons of water must he add to the solution to dilute the solution to 70%?
[hide="My Solution"]Let $ x$ gallons of water is to be added to inital 100 gallons of sulfuric acid.[/b]
So as given, $ \frac{100}{100 \plus{} x} \equal{} \frac{70}{100}$
=>$ 70x \equal{} 3000$
=>$ x \equal{} \frac{300}{7} gallons$[/hide]

Hi;

[hide]

3) 2009 log(175) = 4506.263439 implying there are 4507 digits in 175 ^ 2009.

5) the original number is 74[/hide]

Hi,
[hide=" Question 4"]
Fist find Point of intersection of alt. and leg to base
$ 65^2\minus{}60^2\equal{}c^2 c\equal{}25$
Then set up quadratic
Let side =x
$ (x\minus{}25)^2\plus{}60^2\equal{}x^2$
$ x\equal{} 84.5$
[/hide]

For Problem 6, do you mean $ 10$ years or hours?

I'm going to assume he meant $ 10$ hours.
[hide="6."] If it takes the electrician $ 6$ hours to do the job, then he completes $ \frac16$ of it in one hour. In the same way, the apprentice completes $ \frac{1}{10}$ of the job in an hour. Let $ x$ be the amount time the electrician spends, and $ y$, the amount of the three apprentices. Then, we have

$ \frac{x}{6}\plus{}\frac{3y}{10}\equal{}1$

$ x\plus{}y\equal{}5$

Solving, we find that $ x\equal{}\boxed{\frac{15}{4}}$.[/hide]

[hide="5"]
Let the desired number be $ ab$. Then, $ a^{2}\plus{}b^{2}\equal{}10b\plus{}a\plus{}18$ and $ 2ab\plus{}18\equal{}10a\plus{}b$
Adding the two equations, we have: $ (a\plus{}b)^{2}\equal{}11(a\plus{}b) \Rightarrow a\plus{}b\equal{}11$

Substituting, $ a^{2}\plus{}(11\minus{}a)^{2}\equal{}10(11\minus{}a)\plus{}a\plus{}18 \Rightarrow 2a^{2}\minus{}13a\minus{}7\equal{}0 \Rightarrow a\equal{}7, b\equal{}4 \Rightarrow$ The number is $ \boxed{74}$.[/hide]

problems 1,3,5,6 are correct

i have to check the others. 2 is definitely not correct. .....

[quote="kamallohia"]
[b]2. John has a gallon of sulfuric acid solution in water. How many gallons of water must he add to the solution to dilute the solution to 70%?
[hide="My Solution"]Let $ x$ gallons of water is to be added to inital 100 gallons of sulfuric acid.[/b]
So as given, $ \frac {100}{100 \plus{} x} \equal{} \frac {70}{100}$
=>$ 70x \equal{} 3000$
=>$ x \equal{} \frac {300}{7} gallons$[/hide][/quote]

You solved the question with $ 100$ gallons of sulfuric acid, instead of $ 1$ gallon, as stated in the question.

[hide="Solution"]
Add $ x$ gallons of water to the initial $ 1$ gallon of sulfuric acid, to get $ \frac {1}{1\plus{} x} \equal{} \frac {70}{100} \Rightarrow x\equal{}\boxed{\frac{3}{7}}$ gallon.[/hide]

I'm not sure how doable #3 is. Tossing the multiple of 7 around isn't going to be fun if it's supposed to be done with no calculator.
I wish the Suffolk County math league was more like this. It seems like it got nerfed this year, even though the pressure of sending our first team to division A in NYSML should be encouraging more selectiveness. From what I can tell, the #1s on your contests are easy, but still challenging to the average student. Meanwhile, question #1 in today's meet was to find $ \log_{2009} 2009$...  :rotfl:

2 and 4 are correct

i am not reconsidering 6. i have to check.

@rpond lol. if only the first question was as easy as that always. but we have a pretty ridiculous problem maker this year. pretty tough in general. nice to see we have some long islanders on this site :) [/quote]


Find $ a,b,c,x,y,z$ such that
$ ax \plus{} by \plus{} cz \equal{} 1$
$ ax^2 \plus{} by^2 \plus{} cz^2 \equal{} 2$
$ ax^3 \plus{} by^3 \plus{} cz^3 \equal{} 3$
$ ax^4 \plus{} by^4 \plus{} cz^4 \equal{} 5$
$ ax^5 \plus{} by^5 \plus{} cz^5 \equal{} 8$
$ ax^6 \plus{} by^6 \plus{} cz^6 \equal{} 14$

Is it possible that 14 should be 13? Because then you can take $ z\equal{}c\equal{}0, a\equal{}\minus{}b\equal{}\frac{1}{\sqrt{5}}$ and $ x\equal{}\frac{1\plus{}\frac{1}{\sqrt{5}}}{2}, y\equal{}1\minus{}x$

Thanks alot for replying, everyone.

I showed my friend this topic ,  and said

[quote="bambaman"]I think the original problem has no solution -
Define $ a_n \equal{} ax^n \plus{} by^n \plus{} cy^n$. $ a_n$ is a linear recurrence sequence, i.e., there exist constant $ s,t,u$ such that $ a_{n \plus{} 3} \equal{} sa_{n \plus{} 2} \plus{} ta_{n \plus{} 1} \plus{} ua_{n}$. We will try to solve:
$ 5 \equal{} 3s \plus{} 2t \plus{} u$
$ 8 \equal{} 5s \plus{} 3t \plus{} 2u$
$ 14 \equal{} 8s \plus{} 5t \plus{} 3u$
But it can be shown that there are no solutions (add the first 2 equations and subtract the third).[/quote]

is right.

And he said " This is a good example that does not have any solutions."

Thank you.

More easier problem from AIME 1990.

[url]http://www.mathlinks.ro/Forum/viewtopic.php?p=459558#p459558[/url]


calculus
integration
trigonometry
function
calculus computations

This is the product of boredom in math class...

Solve for $ x$ exactly

$ \left(\int_{\pi/2}^{x^{2}\minus{}6x\plus{}9}\sin t\;dt\right)^{3}\:\plus{}\;\cot^{2}x\cos^{2}x\;\plus{}\:\int_{\pi/2}^{x^{2}\minus{}6x\plus{}9}3\sin t\; dt\; \equal{}\left(\int_{\pi/2}^{x^{2}\minus{}6x\plus{}9}\sqrt{3}\sin t\;dt\right)^{2}\plus{}\:\plus{}\:\cot^{2}x\:\plus{}\:\sin^{2}x$

Gosh I hope I haven't made any mistakes

Edit Note: I fixed one mistake. I changed t->dt next to one of the sine functions.

Please!! someone try to solve my problem. I'll give a hint

 [hide]group the trigonometric functions together[/hide]

I believe the answer is "never", since $ \pi$ is transcendental. We have two conditions involving $ x$ and multiples of $ \pi$ or $ \frac{\pi}{2}$; if both were satisfied, it would make $ \pi$ the root of a rational quadratic.

I don't particularly like the question; it's too easy to see how it was built, and the complications don't feel natural.

woah I forgot about that property of $ \pi$. Despite its quality, thanks for looking at the problem.


prove:
if $\phi(p^a)=2(6n+1), n>0,$ then $p>6$ is a $(12k+11)$-prime and $a>0$ is even.

Well, lets find all integers $n$ with $\phi (n) = 2(6k+1)$.
Since $\phi (n)$ is not divisible by $4$, it can just have one odd prime factor, and when having an odd factor, then just one times the prime $2$ in its decomposition (the powers of $2$ don't work trivialy). So $n$ is of the type $p^a$ or $2p^a$ with an odd prime $p$.

Now $\phi(2p^a) = \phi (p^a) = p^{a-1}(p-1)=12k+2$ and it only depends (for given $a$) on the residue class of $p \mod 12$.
Now all primes except $2;3$ are $\equiv 1;5;7;11 \mod 12$, and $p^2 \equiv 1 \mod 12$ for all four given types. So just check all $8$ possibilities to get the result.


hey guys ,try this
a^5+b^5+c^5 >= 5abc(b2-ac)
with a,b,c being positive reals

Let [tex] f(a,b,c)=a^5+b^5+c^5-5abc(b^2-ac)[/tex]
Easily see that:[tex] f(a,b,c)\geq f( \sqrt{ac},b,\sqrt{ac}) [/tex].
So take $a=c$ ,and we need to prove that:
[tex] 2a^5+b^5-5a^2b(b^2-a^2)\geq 0[/tex]
Take a=1 then we have:
[tex] b^5-5b^3+5b+2 \geq0 [/tex]
And it is easily now. :)

Please, a solution for

$ b^5 + 5b +2 \geq 5b^3 $

You can assume $ b > 1 $


A positive integer M divided by 6 gives a remainder of 3. A positive integer N divided by 6 gives a remainder of 5. What is the remainder when the sum M+N is divided by 6?

[hide=" Answer"]
226
[/hide]

How would you do this problem?
Solutions too, please.

Thanks,
LHM

how is there a remainder of 226 when a number is divided by 6?

226 is obviously wrong since it can't even be a remainder for a mod 6. :noo:

Basically, M can be 9 and N can be 11. 11+9 = 20, so the answer is 2.

that's what I thought it should be, so i guess that is a bug, not me, and i'm sure that's what it said so maybe a moderator could fix that or someone if it ever comes up again....

Yeah, I'm pretty sure it's 2.
If you don't want to do the plug in numbers way
[hide]$ n\equiv3\pmod{6}$ and $ m\equiv5\pmod{6}$. Adding, we have $ m \plus{} n\equiv8\pmod{6}\equiv2\pmod{6}$, so 2.[/hide]

EDIT:New post before me.

or you could just add 5 and 3 and find the remaider when it is divided by 6

that's pretty much what people have been saying
lmao... 226?  :rotfl:

Reading this thread saved me in one of the games a couple days ago :P

i agree, 226 is absurd

try pluging in 5 s m and 3 as n

then m + n= 8

8/6 leaves a remainder of 2


probability
search
function
induction
algorithm
calculus
graph theory

What is usually taught in this course?  Thanks.

graph theory, set theory, methods of proof, combinatorics, probability, counting. as discrete math is also one of the main math courses needed for programming, sometimes algorithmic stuff is taught as well. a sample syllabus that came up from a google search (first match) is here: http://www.rowan.edu/mars/depts/math/Syllbabi/UND.COURSES/DiscMath.html

Fierytycoon

thanks.  Have you or anyone else here taken it?  Is it worthwile?

Of course it's worthwhile!  All math is worthwhile!  :D

Whether it's an appropriate level for you may be more relevant.  What courses have you taken?

i just finished precalc.  But the course description from my brochure sounds kind of easy... topics include: logic, boolean algebra, sets, functions, sequences, and summations, matrices, induction, algorithms, recursion, combinatorics, graphs, and trees.  Have you taken this?  I want to be challenged, and its expensive, so im just seeing if it would be a good class.

Well, looking at my Exeter course catalog (no, I don't go there, I just have a catalog...as do a lot of AMC participants, right?) it says its prerequisite is Precalculus.

Personally, I haven't taken it, but I've looked at syllabuses and that sounds about right.  Except...there's a lot of real-life-problem-type analysis (i.e. fair-division and network problems like map-coloring and scheduling).  I believe this is called optimizaiton.  Someone correct me.

I think it'd be pretty challenging - in particular, the discussion of combinatorics, graph (theory?), matrices, etc.  For instance, it would probably go more into depth in matrices (precalculus doesn't cover everything about matrices...there's Eigenstructures/eigenvalues, lots of theorems and properties of matrices, etc).

But it's really up to you.  There may be pros and cons to either decision.  If you take it, there's the financial expense, but you might get more challenge.  If you feel this would challenge you enough to offset the financial cost (oops, redundancy!) then go right ahead.  But keep looking out for more about discrete math before you decide.

Also, I have a feeling you'll need to take it at some point, so if you're ready for it, you might want to take it right now.  But on the other hand, it might be helpful to know calculus before submerging yourself in discrete math.  You might take a bit calculus before doing discrete math.

I think the class sounds really interesting. The quality of any class though will depend largely on how good the teacher is. But the topics covered sound really good, and there are lots of interesting mathematical problems that arise in discrete math. Not to mention it's a very very useful subject.

Calculus should not be needed though. All you should need is a healthy understanding of Algebra II.


Using Newton method,  solve  the equation   $f(x): =x^{3}-x+\frac{1}{2}\; ,\; x\in{\mathbb R},$ [u]using as "start-point [/u]"    $\; \; \; \; x_{0}=1 .$

[size=75] In case of an equation $f(x)=0\; ,\; f$ being derivable,  Newton-method says    $x_{n+1}=x_{n}-\frac{f(x_{n})}{f^{\prime}(x_{n})}\; ,\; n\in \{0,1,...\}.$
[/size]

[quote="bus"]It won't work, you'll get back to the starting point after three iterations.[/quote]
Indeed !Thanks for interest !/Alex


algebra
polynomial
superior algebra
superior algebra unsolved

Find the dimension of the splitting fields over $Q$ of the following polynomials:
1).$x^p-1$ ($p$-prime)
2).$x^n-1$.
3).$x^p-a$ ($a\in Q$ and it isn't $p$'th power in $Q$, $p$-prime)

[b]2)[/b]
The field must contain the primitive roots of unity and their minimal polynomial has degree $\phi (n)$. But adjunction of one of these roots already gives the desired splitting field, so the answer to this is $\phi (n)$ and $\phi(p) = p-1$ gives [b]1)[/b]

[b]3)[/b]
It again must contain all $p$-th roots of unity. But since $x^p-a$ is irreducible, the splitting field must have a degree that is a multiple of $p$, but by the previous said also of $p-1$, so of $p(p-1)$. But adjunction of any root of $x^p-a$ and a $p$-th root of unity already splits the polynomial, so we are done.


geometry
incenter
circumcircle
analytic geometry

A boy walks from the incenter to the centroid and then to the circumcenter of a triangle with sides 9, 12, and 15. In simplest radical form, he walks $\frac{w-k\sqrt{p}}{2}$ farther than a girl who walks directly from the incenter to the circumcenter of the same triangle. Find the value of $k+p+w$.

Is the answer:

[hide="Answer"] 15[/hide]

[quote="deej21"][hide="hint"]
coordinates
[/hide][/quote]

That's how I did it.  :D


geometry
circumcircle
trigonometry
geometry unsolved

Let $ \triangle ABC$ and let $ BF \perp AC$, $ CE \perp AB$. Let $ \angle BAC \equal{} \alpha$ and area of $ \triangle ABC \equal{} S$. Find area of $ \triangle AEF$ by $ \alpha$ and $ S$.

$ S_{\triangle AEF}\equal{}cos^2\alpha S$ :D

let $ M$ be the midpoint of side $ BC$,now we know that $ BEFC$ is cyclic,let $ w$ be it's circumcircle,so $ w$ is a circle with center $ M$ and radius $ \frac a2$.

now let $ P$ be the power of point $ A$ wrt circle $ w$ we have:

$ P = AE.AB = AF.AC = AM^2 - \frac {a^2}4$ (*)

now we know that:

$ AM^2 = \frac {b^2}2 + \frac {c^2}2 - \frac {a^2}4$

$ \Rightarrow P = \frac {b^2 + c^2 - a^2}2$

on the other hand according to $ \cos$ law in $ \triangle ABC$ we have:

$ a^2 = b^2 + c^2 - 2bc\cos\alpha$

$ \Rightarrow\frac {b^2 + c^2 - a^2}2 = bc\cos\alpha$

$ \Rightarrow\boxed{P = bc\cos\alpha}$

now according to (*) we have:

$ AE = \frac {P}{c} = b\cos\alpha$

and similary we conclude that $ AF = c\cos\alpha$

also we know that:

$ S = \frac 12.bc.\sin\alpha$

now we have:

\begin{eqnarray*}S_{AEF} = \frac 12.AE.AF.\sin\alpha = \frac 12.(b\cos\alpha).(c\cos\alpha).\sin\alpha = \underbrace{(\frac 12.bc.\sin\alpha)}_{ = S}.\cos^2 \alpha=S\cos^2 \alpha\end{eqnarray*}

$ \Rightarrow\boxed{S_{AEF} = S\cos^2 \alpha}$

This is my solution:
$ \bullet$ We have $ AE \equal{} AC.cos\alpha$ and $ AF \equal{} AB.cos\alpha$.

$ \bullet$ $ S_{\triangle AEF} \equal{} \frac {1}{2}AE.AF.cos\alpha \equal{} \frac {1}{2}AC.cos\alpha.AB.cos\alpha.cos\alpha \equal{}$ $ \cos^2\alpha.\frac {1}{2}AB.AC.cos\alpha \equal{}$ $ \boxed {\cos^2\alpha.S}$  .

Oke!

I think this problem is easy. But it's nice.

in your solution thnhnam 2902,there is a small mistake :lol: ,you have to change a 'cos' by 'sin',but in general i followed the same reasoning of you :wink: .

$ \bullet$ We have $ AE \equal{} AC.cos\alpha$ and $ AF \equal{} AB.cos\alpha$.

$ \bullet$ $ S_{\triangle AEF} \equal{} \frac {1}{2}AE.AF.sin\alpha \equal{} \frac {1}{2}AC.cos\alpha.AB.cos\alpha.sin\alpha \equal{}$ $ \cos^2\alpha.\frac {1}{2}AB.AC.sin\alpha \equal{}$ $ \boxed {\cos^2\alpha.S}$  .


Ναι υπαρχει τετοια ακολουθια...

Ποια ακολουθία νομίζεις ότι δουλεύει;

[quote="Durandal"][quote="cmad"]Μπορούμε να βρούμε ακολουθία ανοικτών συνόλων $ U_i \subseteq \mathbb{R}$ έτσι ώστε $ \bigcap_{i \equal{} 1}^\infty U_i \equal{} \mathbb{Q}$ ?[/quote]

Baire... αλλά χωρίς αυτό, δε νομίζω ότι το χωράω σε ένα χαρτομάντηλο ;)

Έχεις πιο στοιχειώδη λύση?

Durandal 1707[/quote]

Και 'γώ αυτό έχω. Η απόδειξη του Baire είναι αρκετά σύντομη όμως, ούτως η άλλως.


A car is driving a one mile-long drive

The car drives 1/2 of a mile going 30 mph

How fast must the car drive the other half of the mile in order to average 60 mph on the drive?

Note: The answer is not 90 mph :ninja:

He completes the entire mile in 1 minute, and half of the mile in 45 seconds.  THus, he drives 120 mph for the last half.

In order for him to complete one mile at an average of 60 mph, he must do it in 1 minute. At 30 mph, he can do $ \frac{30}{60}\equal{}\frac12$ miles per minute, or $ \frac{1}{120}$ miles per second. Thus to go half of a mile he takes 1 minute. Therefore, he has to go $ \frac{1}{2}$ of a mile in 0 seconds. This is clearly impossible.

Alternatively, one could find his average speed in terms of the [url=http://en.wikipedia.org/wiki/Harmonic_mean]harmonic mean.[/url]

Calling his speed on the second "half" $ x$ mph, we get $ \frac{2}{\frac1x\plus{}\frac{1}{30}}\equal{}\frac{60x}{x\plus{}30}$

We set this equal to 60 mph (the desired speed).

$ \frac{60x}{x\plus{}30}\equal{}60 \implies 60x\equal{}60x\plus{}60\cdot30 \implies 0\equal{}1800$ which clearly never occurs. Thus, there is no solution.

whoa Mewto55555 is scaring me because now hes acting like t0rajirou

Don't you DARE compare Mewt0rajirou55555 to Obama!  :cursing:

Happy New year's, by the way.  :D

As for the problem...it's impossible because you can't suddenly speed up- or go anywhere at all- in the time required.

Mewto55555 got it right.

Unless it were possible for the car to travel at the speed of light therefore according to the theory of relativity time would stop and since 1/2 of mile at 30 mph takes one minute and the other half going at the speed of light takes o minutes he would have traveled one mile in one minute thus going 60 mph.  :ninja:


inequalities
calculus
derivative
inequalities proposed

Regrettably, your answer is incorrect. :(

kunny

[quote="kunny"][quote="zhaobin"]oh,sorry,it should be $[\frac 1{\sqrt 3},1]$[/quote]

 :(  This problem seems to be difficult.[/quote]
Sorry,I am wrong.I will try again. :oops:

No problem. I am looking forward to your answer. :lol:

Oaps,i think it's again an not hard problem,i have only one solution but from my opinion it's the strongest way to completely solve it.
Let we consider : |x| is the maximum value of the set { |x|,|y|,|z|} so x must be non zero
From the given condition:
$x+y+z=0$ we divide both 2 sides of the equality by x we have:
$\frac{y}{x}+ \frac{z}{x}+1=0$
By putting $\frac{y}{x}=u$ and $\frac{z}{x}=v$so we get : $u+v=-1$ or $u=-(1+v)$(1)
Also: $P= \frac{1+u^2+v^2}{1+|v|+|u|}$ (2)
By subtituting (1) into (2) we get:
$P=2 \frac{1+v+v^2}{1+|v|+|v+1|}$ i think the rest are quite easier and sorry for my laziness :D

$\sqrt{\frac{1}{2}}$ : $x=a, y=-a, z=0$ and all of the symmetric forms.
$\sqrt{\frac{3}{8}}$ : $x=a, y=a, z=-2a$ and all of the symmetric forms.

Could you give us full solution,  Sung-yoon Kim?

kunny

[quote="jianuovidiu"][b]1)[/b] For minimize it just use square mean - arithmetic mean ineq. :

$\frac{\sqrt{x^2+y^2+z^2}}{|x|+|y|+|z|} = \sqrt{3}\frac{1}{|x|+|y|+|z|} \sqrt{\frac{|x|^2+|y|^2+|z|^2}{3}}\geq$
$\geq\sqrt{3}\frac{1}{|x|+|y|+|z|} \frac{|x|+|y|+|z|}{3}=\frac{\sqrt{3}}{3}$

[b]2)[/b]  Now let's maximaze it:
$\frac{\sqrt{x^2+y^2+z^2}}{|x|+|y|+|z|} = \frac{\sqrt{|x|^2+|y|^2+|z|^2}}{|x|+|y|+|z|} \leq$
$\leq\frac{\sqrt{(|x|+|y|+|z|)^2}}{|x|+|y|+|z|}=1$

q.e.d.

...so we don't need $x+y+z=0$[/quote]

If there is no condition of $x+y+z=0$, then your answer is correct. However, there is the condition $x+y+z=0$ and $|x|+|y|+|z| \neq 0$. The equality conditions of your answers are all $x=y=z$, which means $x=y=z=0$: contradiction to $|x|+|y|+|z| \neq 0$.

Without the condition $|x|+|y|+|z|=0,$ you are right, jianuovidiu. :)

Incidentally the very your problem has set as follows.Now there is a flood of Japanese university entrance exam.

From Feb 25, national Universities such as Tokyo University and Kyoto University are to start entrance exam.

[b]Tokyo University of Agriculture entrance exam 2006 Feb.[/b]

(1) For all real numbers $a,b,$ find the maximum value which satisfies $k(|a|+|b|)\leq \sqrt{a^2+b^2}.$
(2) For all real numbers $a,b,c,$ find the minimum value which satisfies $\sqrt{a^2+b^2+c^2}\leq k(|a|+|b|+|c|).$

[quote="Sung-yoon Kim"][quote="kunny"]Let $x,y,z$ be real numbers such that $x+y+z=0,\ |x|+|y|+|z|\neq0.$
Find the range of the following equation which can be valued.

\[ \frac{\sqrt{x^2+y^2+z^2}}{|x|+|y|+|z|}.  \][/quote]

Well, the answer is $[\sqrt{\frac{3}{8}},\sqrt{\frac{1}{2}}]$. As $z=-x-y$, the original value is equal to $\frac{\sqrt{2x^2+2xy+2y^2}}{|x|+|y|+|x+y|}$. Then take the four case: $(1)x \geq 0, y \geq 0, (2) x \geq 0, y<0, x+y \geq 0 (3) x \geq 0, y<0, x+y<0 (4) x<0, y<0$. (the case of $y<0, x \geq 0$ is equal to (2) and (3).) Then all of the signs are determined.

For each case, express that by $\frac{x}{y}$ or $\frac{y}{x}$, and solve the range of $\frac{x}{y}$ and $\frac{y}{x}$. Then it can be reached that the answer is $[\sqrt{\frac{3}{8}},\sqrt{\frac{1}{2}}]$.[/quote]

(1) $x \geq 0, y \geq 0$ : $\sqrt{\frac{x^2+xy+y^2}{2(x^2+2xy+y^2)}}$. Let $\frac{x}{y}$ be $k$, then the value is $\sqrt{\frac{k^2+k+1}{2(k^2+2k+1)}}$ when $k \geq 0$. I reached the answer via differentiation, but maybe there is another way to get that. $k=0, \inf : \sqrt{\frac{1}{2}}$, $k=1 : \sqrt{\frac{3}{8}}$.

(2) $x \geq 0, y<0, x+y \geq 0$ : $\frac{\sqrt{2x^2+2xy+2y^2}}{2x}$. Let $\frac{y}{x}$ be $k$, then the value is $\sqrt{\frac{k^2+k+1}{2}}$. As $x+y \geq 0$, $1+k \geq 0$. So $0 \geq k \geq -1$, and we can get the answer easily.

(3) Similar to (2).

(4) Same as (1).

Thank you for your solution, Sung-yoon Kim.Your answer is correct. :)

kunny

for kunny's problems on post #14:

1) setting $a=b$ yields $k\leq \frac{1}{\sqrt2}$. the proof for $k=\frac{1}{\sqrt2}$ follows from cauchy-schwarz

2) setting $b=c=0$ yields $1\leq k$. the proof for $k=1$ comes from squaring both sides.


Let   $ A  $ an exterior point of a circle  , the tangents  $ AB , AC $  of the circle and the lines $  ADE , AZH $ which meet the circle at points $  D , E , Z , H  $. Prove that the lines  $ BC , DH  $ and $ EZ  $ are concurrent ( or parallel ).

It's trivial when you use poles and polars...

The polar of $DH \cap EZ$ is a line passing through $A = DE \cap ZH$ and $DZ \cap EH$, so $DH \cap EZ$ lies on the polar of $A$ with respect to the circle, which is just the line $BC$.

$AB=AC,\ \triangle ABD\sim \triangle AEB,\ \triangle ACZ\sim \triangle AHC,\ \triangle ADZ\sim \triangle AHE\Longrightarrow$

$\frac {BD}{BE}=\frac {AB}{AE},\ \frac {CZ}{HC}=\frac {AZ}{AC},\ \frac {EH}{DZ}=\frac {AE}{AZ}\Longrightarrow \frac {BD}{BE}\cdot \frac {CZ}{HC}\cdot \frac {EH}{DZ}=1\Longrightarrow$

$(BDZCHE),\ BD\cdot ZC\cdot HE=DZ\cdot CH\cdot EB\Longleftrightarrow DH\cap EZ\in BC.$

For the last equivalence see http://www.mathlinks.ro/Forum/viewtopic.php?p=276617#p276617
But this equivalence didn't solve still, although it is easy !

dear stergiu,read it carefully .its a usefull theorem. ;) .

let ABCD be a quadrilateral.AB and CD intersect each other in point Y.and AD and BC intersect each other in pointX.

theorem:(polar of a point to an angle).suppose L is a mobile line wich goes through

Y,and intersects XD andXC in points P and Q respectivly.and Z is a point on PQ is wich we have:(YZPQ)=-1

whats the locus of point Z?

first we prove another theorem and we will use it to prove the theorem above.

theorem2:

suppose (YZPQ)=-1. from point X wich isnt on line PQ , we draw lines XY and XP and XZ and XQ.another line inter

sects XY, XP, XZ, XQ in points R, S, T, U respectivly.prove that :(RTSU)=-1

to prove this theorem we should prove another theorem.

theorem3:we have (YZPQ)=-1.and X is a point wich isnt on line PQ.if aline wich is paralell to XY and goes through

Z,intersects XP and XQ in points M and N respectivly,prove that: ZM = ZN.

now we dont need any theorem to prove this theorem :D .

lets prove it.

according to the two simillar triangles XYP and NPZ .

and according to the two simillar triangles QZM and QYX we have:

YX/NZ=YP/PZ    and    YX /MZ=YQ/ZQ

and because of (YZPQ)=-1, we have:

YP/PZ=YQ/ZQ.

then we have : ZM=ZN

and the prove is complete.

now we want to prove the second theorem.

draw a line wich is paralell to MN,inwich intersects XP and XQ in points N' and M' respectivly.

now if we prove that TN'=TM' the the theorem will be proved.

and its clear why they are equal.because the triangles XNM and XN'M' are simillar to each other and it means that:

TN'=TM'

according to the second theorem the first theorem will be proved very easy.

the locus of Z will be a line .each line that intersects lines XY,XP,XZ,XQ , will make four harmonic points.and the first

theorem will be the special form of the second one.

the theorem is proved .but its still another fact wich is very im portant.

let O be the inter section of the lines BD and AC.

i claim that XZ and AC and BD are concurrent.

soppose they arent concurrent ,then O wont be on line XZ.let XZ intersect AB and DC in points F and G respectivly.

and let OF intersect DC in point G'.

according to the second theorem  (YFAB)=-1

then we should have (YG'DC)=-1 but we have (YGDC)=-1 .

this fact will be true if and only if G' be G.

then BD and AC and XZ will be concurrent.

now the quadrilaterall in u problem is cyclic and it means that  not only XZ is the polar of Y to the angle PXQ ,but

 also XZ the polar of Y to the circle.

i think by what i said the answer of u problem will be clear.but if u have questions about each part ,just tell me ;) .

Ashegh , thank you very much   for giving so time to write this for me.After Arne's solution I studied again polars and harmonic ratios  in some old books  books. I saw many good problems to be solved with method. You have written very well  these theorems , which are really very usefull.You are a good solver too.Thanks again. Which area in Maths do you like more ?Geometry , numbers, combinatorics , or algebra?And which books do you study(in english I mean)?

Babis

Babis, if possible, it would nice if you would continue this particular conversation with ashegh using email or PM :)

ure wellcome stergui .
i love geometry very very very much.
i have read an old book about combinattorics.wich its name is PTC(principles and technics in combinatorics)


but we must have $ a_i\minus{}1\ge 0$
gives $ \sum_{i\equal{}1}^{2024}\sqrt{\sqrt{a_i}\minus{}\sqrt{a_i\minus{}1}}\ge 2024>45$

[quote="aviateurpilot"]
 $ \sum_{i \equal{} 1}^{2024}\sqrt {\sqrt {a_i} \minus{} \sqrt {a_i \minus{} 1}}\ge 2024$[/quote]
why?

I don't think that's right. I got that it was greater than or equal to $ \sqrt{2024}$ through a couple applications of Jensen.

If $ a_i \equal{} (2m_i \plus{} 1)^2$, then $ \sqrt {\sqrt {a_i} \minus{} \sqrt {a_i \minus{} 1}} \equal{} \sqrt {m_i \plus{} 1} \minus{} \sqrt {m_i}$.
Therefore for $ m_i \equal{} i$ we had $ \sum_{i \equal{} 0}^{2024}\sqrt {\sqrt {(2i \plus{} 1)^2} \minus{} \sqrt {(2i \plus{} 1)^2 \minus{} 1}} \equal{} \sum_{i \equal{} 0}^{2024}\sqrt{i\plus{}1}\minus{}\sqrt i \equal{} \sqrt {2025} \equal{} 45$.


algebra
polynomial
algorithm
Ring Theory
function
domain
superior algebra

Hi everyone,

I have the following doubt. Suppose K is a field of characteristic p>0 and f(X_1,...,X_n) a multivariate polynomial with coefficients in K such that deg(f) < p.

Is it true that f admits a square-free descomposition? I mean, with such hypothesis: can the usual algorithm for characteristic 0 be adapted to prove this without asking for the field K to be perfect?

Thanks a lot,
Peter

May I ask you what you mean by square-free de(s)composition? In a UFD (unique factorization domain) like $ K\left[X_1,X_2,...,X_n\right]$, every element admits a prime factor decomposition, and thus also a decomposition into a product of a squarefree element and a square. Probably you mean something different than this?

  darij

He want's to handle inseperable polynomials, I think. But I don't completely understand what he means, too.


How can 100 mL of sodium hydroxide solution with a pH of 13 be converted to a sodium hydroxide solution with a pH of 12?

a. by diluting the solution w/ distilled water to a total volume of 108 mL

b. by diluting the solution w/ distilled water to a total volume of 200 mL

c. by diluting the solution w/ distilled water to a total volume of 1.00 L

d. by adding 100 mL of .10 M HCl

e. by adding 100 mL of .10 M NaOH

thanks! :)

pH scale is logaritmic and 1 unit means tenfold change in concentration.


analytic geometry
real analysis
real analysis unsolved

at first let me write some definition.
suppose we have Gomeomorphic:
$ F : M \minus{} > M$
where $ M$ is Manifold.
Point $ x$ called  periodically if there exist $ k$ such as $ F^{k}(x) \equal{} x$ where $ F^{k}(x)$ is $ F(F(F...(F(x)...)$ $ ...)))...$-$ k$ times.  :lol: Ok. and the set of periodically points defined as $ PerF$
Let denote by $ \Omega$ set of the point $ x$ such as for any $ U$ which $ x \in U$ and for all $ N$ there exist $ n > N$ such as :
$ F^{n}(U)*U$ is not equal to $ 0$(or zero set ).
 $ *$ mean Intersection. ( I cannot write intersection by Latex.  :( ) .
let by $ \omega (x_{0})$ denote the set of points $ x$ such as there exist $ n_{k} \minus{} > \plus{} oo$ : $ F^{n_{k}}(x_{0}) \minus{} > x$
Problem:
suppose we have gomeomorphic

$ F : M \minus{} > M$
where $ M$ is Manifold.then the next fact is true :
$ PerF \subseteq R \subseteq \Omega$
where $ R$ is set of points $ x$ such as $ x \in \omega (x)$

Question:
Show example when :
$ PerF \subset R \subset \Omega$
(I mean each set is subset but is not equal ).

[quote="Extremal"]Show example when :
$ PerF \subset R \subset \Omega$
(I mean each set is subset but is not equal ).[/quote]

Consider the following system in polar coordinates
\[ \begin{gathered}
  \dot r \equal{} r\left( {1 \minus{} r^2 } \right) \hfill \\
  \dot \theta  \equal{} 1 \hfill \\
\end{gathered}
\]
With the initial condition, e.g., $ \theta \equal{} 0$, $ r\equal{}2$, we can consider the solution curve of the Cauchy problem above as mapping $ F$. Then $ \operatorname{Per}F$ is empty. Otherwise $ R$ is unit circle.

Can you show example when $ PerF$ is not empty? :maybe:


Let ABC be a right angled triangle(<B=90 ).Let BH is perpendicular to AC side at H point.Construct BCDE square on BC side. [BH] and [AD] line segments are intersection at  point P.
Show that

1/BP=1/BH+1/AC

Well, you can just slap down a coordinate system and actually calculate those three distances ... kinda ugly, though.

Hmm...  I don't know exactly how to go about proving it, but the relationship you need to prove is the same formula you use for the two-pole problem.  Could there be a relationship?


Once again Asymptote images are not showing up in AoPS Wiki. Is this a side effect of the server change?

Yeah, an executable path is messed up. I'm working on getting it fixed.


Randomly select a whole number such that $0 < x < 100$. What is the probability that x equals the sum of two perfect squares? Express your answer as a fraction.

[quote="Zenox"][hide]
There are 9 perfect squares between 0 and 100, so therefore 8+7+6+5... ways to add them up. That is (8*9)/2 so 36 different sums. There are 99 numbers between 0 and 100, so that is 36/99, which reduces to:

4/11
[/hide][/quote]

However, some of these numbers are double counted, like 65=7^4+4^2=8^2+1^2.

So that is a good explanation, but some minor flaws within it.

Um... there are 46 ways to add perfect squares from 1-81 and be less than 100, not counting double-counts...

The answer book is screwed up, so we have absolutely no clue what the right answer is.

Ah, then I see,

so the probability is 46/99.

just take the average of u two and you arrive at the correct answer

jw, but how would you know all of the overlaps?

-jorian

[quote="jhredsox"]how would you know all of the overlaps?[/quote]

The numbers that can be expressed as a sum of two squares in more than one way must be a product of numbers expressed as a sum of two (nonzero) squares in one way (except $2$).  In this case,

$25 = 5 \cdot 5 = (1^{2}+2^{2})(1^{2}+2^{2})$
$50 = 5 \cdot 10 = (1^{2}+2^{2})(1^{2}+3^{2})$
$65 = 5 \cdot 13 = (1^{2}+2^{2})(2^{2}+3^{2})$

The reason is the identity

$(a^{2}+b^{2})(c^{2}+d^{2}) = (ac \pm bd)^{2}+(ad \mp bc)^{2}$

[quote="davidlizeng"]to neal: there are 46, not counting overlaps......
count them again, you missed some. me and Ubemaya
both worked on this problem today, we listed every one out.
the answer key said 46/99, but the answer was actually 14/33
according to our coach.............

to jorian: i don't think theres any other way than to list them all out and see where numbers repeat........ :([/quote]

neil is my name but w/e

What am I missing then?

sorry...i always mess up between neil and neal
i don't know what's missing, which is the dumb thing about this problem..................

[quote="davidlizeng"]in that case, it is 14/33
thanks to neil, who figured out the error...[/quote]

But still, what am I missing?

After all, you actually wrote out a list?

Wow, I really suck at counting.

I mean, I thought I had 6 per row, but instead I had 7 per row in the first row.

Thanks people.

oh yeah, this one...

My mathcounts team worked on this one last Friday. We coudn't get to the right answer because we had no time and the answer in the book was wrong.

However, my version of this problem (1997 state team) said two or [b]More[/b] perfect squares.  Might the answer to that one be 46/99?  Then, how would you use torajir0u's method of correcting for overcounting?

[quote="myspacesucks"]Easier way: Just see that the numbers there are inclusive 0-9 as can be squares less than 100.  You add them up, inclusive, and you get 45.  Because you have 4 overlapping, and one on a perfect square, you subtract by 3 (counted perfect square already when did inclusive).  Therefore it is 42/99 or 14/33.

Hope this helps[/quote]

First of all: This is EXACTLY the method that has been posted in this thread already.
Second: How did you get 45, not 46?

[quote="t0rajir0u"][quote="kyyuanmathcount"]two or [b]More[/b] perfect squares.  [/quote]

It would have to be two or more [b]distinct[/b] perfect squares, or otherwise $n$ is the sum of the $n$ perfect squares $1^{2}, 1^{2}, ...$[/quote]

Yeah, I noticed that when I was solving that the problem should have said distinct.  How would you have solved it if it said distinct?

Casework on sum of two distinct squares, three, four...

Ouch, that's deadly.  So, would you have 2-3 people of your team start listing off numbers, then crossing off overlaps?  How would we develop a strategy for those problems?

In my opinion, the "two or more" variant is too hard for a team round.  Here's the exact wording of my copy of 1997 State Team Round Problem 4:

Randomly select a whole number $x$ such that $0 < x < 100$.  What is the probability that $x$ equals the sum of two perfect squares?  Express your answer as a fraction.

Ubemaya, the way I did it was completely different.  I said 1-9 inclusive, not counting or whatever.  You add 1-9 inclusive and you get 45, and you subtract overlaps (already counted 25 overlap) so you get 42.  Its out of 99 numbers, so its 42/99 or 14/33.


Is correct that the point F is inside the rectangle?


When I insert a picture into the PDF file I want it in, it takes up the whole page when the picture is only about two inches on the computer. Is there any way I can correct this?


superior algebra
superior algebra unsolved

Let R, T, S are noncommutative rings. We heve a homomorphism $ \varphi: R\to T, \psi: R \to S$and$ \lambda : T\to S.$ And we know that $ \lambda(\varphi (x))\equal{}\psi(x)$for all $ x \in  R$ And we know that $ \varphi(y)$ is invertible in T and $ \psi(y)$ is invertible in S. Why $ \lambda([\varphi (y)] ^{\minus{}1}) \equal{}[\psi(y)] ^{\minus{}1}$?

What is $ \lambda([\varphi(y)]^{ \minus{} 1}) \times \psi(y)$?


[quote="Zenith Xan"][quote="hoshattack"]I will probably be taught this in school later, but being inthused about math, and learning about polynomials and factoring them and stuff; [b]I was wondering how you factor numbers with square roots, like $\sqrt{28}$...[/b] You know, taking factors of numbers out of the radical to get it to an easier to understand solution and in simplist form[/quote]

One of the properties of radicals is that $\sqrt{xy}=\sqrt{x\cdot y}=\sqrt{x}\cdot\sqrt{y}$

In your example:

$\sqrt{28}=\sqrt{4 \cdot{7}}=\sqrt{4}\cdot \sqrt{7}=2\sqrt{7}$[/quote]

be careful: $\sqrt{x\cdot y}=\sqrt x\sqrt y$ only works when at least one of $x$ and $y$ are positive. Otherwise you could do the following:

$1=\sqrt{1}=\sqrt{(-1)(-1)} = \sqrt{-1}\sqrt{-1} = i^2 = -1$ which obviously isn't true.

[quote="Hamster1800"]

be careful: $\sqrt{x\cdot y}=\sqrt x\sqrt y$ only works when at least one of $x$ and $y$ are positive. Otherwise you could do the following:

$1=\sqrt{1}=\sqrt{(-1)(-1)} = \sqrt{-1}\sqrt{-1} = i^2 = -1$ which obviously isn't true.[/quote]

Also be careful when combining radicals like $\sqrt{-6}\sqrt{-8}\neq{\sqrt{48}}$ It equals $i\sqrt{6}*i\sqrt{8}=i^2\sqrt{48}=-4\sqrt{3}$ :D

Hes obviously still working in early math, no need to confuse with imaginary.

Thanks...

I know some basic stuff about imaginary... My math teacher sometimes talks about stuff like that, and I pick onto a fair amount quick, but I think he doesn't want to confuse some of the class,  I'm almost finished with Algebra 1.


Perhaps a carbonyl compound is expected to form.

how come none of the FIITJEE students over here are trying this?? :huh:  :|

Because i think the question is wrong and it shd be $ \ce{C5H5CN}$ rather than $ \ce{C2H5CN}$ as i think it was typed wrongly in the fiitjee paper  :)

1-cyano cyclopenta-2,4-diene???????? :huh:  :huh:

oh, thats why my answer dint match theirs, i had a feeling the question was wrong. gawd, fiitjeepapers surely suck!

swearing not allowed generally but this being a special case...permission granted :D

[quote="VARUNARASU"]pardesi, you are not the mod over here. i can swear as much as i want n you cant do a thing :D[/quote]
no actually u got  over excited there ecists something called 'moderator's lounge' u all may not be able to view and a post there can lead to this being delted by the adminstrator  :D  :P  :P

oh dint know that :oops_sign: . is that what happened to atomikphisic or whatever his name was?

1-cyanocyclopentadiene does not exist due to degenerate sygmatropic rearrangement: better would be simply cyanocyclopentadiene. Also, deprotonation is too easy so perhaps that compound exists primarily as $ C_5H_4CN^\plus{}$.


[quote="GameBot"]Evaluate $ \sqrt [3]{27^2} \minus{} \sqrt [4]{16^2}$.[/quote]

The cube root of 27 is 3, so the cube root of $ 27^2$ is $ 3^2 \equal{} 9$. Similarly, the fourth root of 16 is 2, so the fourth root of $ 16^2$ is $ 2^2 \equal{} 4$. Thus our answer is $ 9 \minus{}  4 \equal{} \boxed{5}$.

[hide="answer"]the cube root of 27 is 3 then 3^2=9

the 4 th root of 16 is 2  then 2^2=4

9-4=5

so the answer is 5 or $\boxed{5}$[/hide]

[b][i][u]please rate my post thanks![/u][/i][/b]

please be fair though thanks [=

 :yup:  :yup:  :yup:  :yup:


Let $ f(x) \equal{} 3x\minus{}8$ and $ g(f(x)) \equal{} 2x^2 \plus{} 5x \minus{} 3$.  Find $ g(\minus{}5)$.

To find $ g(-5)$, we first have to find a value of $ x$ such that $ f(x)=-5$.  It's easy to see that $ x=1$ works.  So now we solve $ g(f(1))=g(-5)=2+5-3=4$.

PS: In the future, don't bother using $ \text{\LaTeX}$ syntax in your title, it doesn't render.  Instead say what the problem concerns.

[quote]This is a bot. The only way of fixing that is telling levans.
[/quote]
Why would it be programmed to just randomly post FTW problems?

They are not randomly posted, I think it uses some sort of cgi or php and generates these topics when someone clicks the ask forum button when reviewing.

[quote="isabella2296"]Yes; when you click "Ask on Forum", a new tab or window opens and the bot posts the problem.[/quote]

A very convenient bot.  So you don't have to copy the question and write it as a new post.

[quote="myyellowducky82"]i dont get functions.[/quote]

Please stop saying that "you don't get" something. It's considered spam. What, exactly, don't you get about functions?

for f(x)=.........., you plug in the x value into the variables on the right side.

I gave more detail in my post, Kali, let me know if it's helpful!

[quote="isabella2296"][quote="AIME15"]Yet only about 600 peopel have used it..[/quote]

When I click the button, nothing happens. So when there's problems with diagrams that I don't understand, I have no way of asking.[/quote]

Except by typing out the whole problem.  Maybe because of your browser?

[quote="isabella2296"]Well, no, I can't type pictures. I think it might have something do with Linux? Because I know other people with Firefox and it works fine for them.[/quote]

Using Firefox on my dad's desktop right now!  :wink:

[quote="isabella2296"]Go Firefox!

Sorry for spamming... :([/quote]

Yay!  Go chat threads!

I think we need to let this thread die now.

[quote="dragon96"][quote="GameBot"]Let $ f(x) \equal{} 3x \minus{} 8$ and $ g(f(x)) \equal{} 2x^2 \plus{} 5x \minus{} 3$.  Find $ g( \minus{} 5)$.[/quote]

No one has even posted a solution for the topic yet, so I'll do it.

This problem plugs in -5 in the parenthesis, so f(x)=3x-8, or x=1. Then we plug in x back into the 2nd function and we get 2+5-3 or 4.[/quote]

Well actually, a solution was posted. :)

Thanks, I'll use your suggestion #2 in the future.

And I doubt it's my browser, since I have FF 2, so it's probably my operating system, Fedora. The Linux Flash Player sometimes causes rather strange things to happen...

Linux seems to work perfectly for me...
i have ubuntu

[quote="isabella2296"][quote="myyellowducky82"]i dont get functions.[/quote]

I wrote a long, hopefully-helpful explanation on functions(like the systems of equations one.)

Here's the link, you'll have to scroll down a little.

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=211396[/quote]
thank you.

plz stop spamming
two word posts are sort of pointless....
pm or something if u must

[quote="mathemagician1729"]@Isabella if for any reason you cant ask the topic on the forum via the bot and there is a diagram involved here are your options.
1. If you are skilled in Asymptote then you can try reproducing it (More friendly to the aops server, but sometimes difficult to do)

2. You can also take a screen capture of the whole problem and attach it, if you use vista then there is a built in snip tool to do this. If you use xp then press prt sc button and paste it to paint.

EDIT: I know for a fact that the ask forum button works for FF 2&3 not sure about IE[/quote]
It works on IE


I was typing in the address of the login page (artofproblemsolving.com/Forum/login.php) except I typed in an extra / at the end.  The URL was this:
http://www.artofproblemsolving.com/Forum/login.php/
The buttons and alignment are different as well as the redirect.  Once you log on, it goes to
http://www.artofproblemsolving.com/Forum/login.php/index.php
And Firefox says it will never load the page.
Just thought I'd let you guys know.
 :)

That's probably a directory containing the scripts needed for login.


linear algebra
matrix
geometry
geometric transformation
homothety
linear algebra unsolved

Let $ A$ be a complex diagonalizable matrix such that for some natural $ n$ and  for each square $ X$ we have $ (XA-AX)^{n}=0.$ Prove $ AX=XA.$

I rewrite the question as follows:
Let $ A\in\mathcal{M}_{n}(\mathbb{C})$ s.t. $ \forall{X},XA-AX$ is nilpotent; then $ A$ is an homothety.
Proof: case 1: $ A$ isn't diagonalizable. Then, to prove the contradiction, we may assume that $ A=J_{k}$ where $ J_{k}$ is the Jordan's block (stictly upper triangular) of dimension $ k\geq2$. $ J_{k}J_{k}^{T}-J_{k}^{T}J_{k}=diag(1,0,\cdots,0,-1)$ isn't nilpotent.
Case 2: $ A$ is diagonalizable and isn't an homothety. Then to prove the contradiction we may assume that  \[ A=\begin{pmatrix}0&0\\0&1\end{pmatrix}\]. We choose \[ X=\begin{pmatrix}0&1\\1&0\end{pmatrix}\].

loup blanc, I didn't get your point: are you trying to disprove the statement? In your example, AX-XA isn't nilpotent.

My proof (please correct if I missed something): wlog $ A$ is diagonal. Then $ AX = (XA)^{T}\implies B = AX-XA =-B^{T}$, i.e. $ B$ is antisymmetric, hence normal, therefore orthodiagonalisable by the spectral theorem: $ B = QDQ^{T}$. $ B^{n}= 0 \implies D^{n}= 0 \implies D = 0 \implies B = 0$, as wanted.

$ \textbf{EDIT}$: PROOF WRONG.

Hi hedgehuogis.
1) I prove that  if $ A$ isn't an homothety then $ \exists{X}$ s.t. $ XA-AX$ isn't nilpotent what is equivalent to my asked question.
2) Your proof is false because $ AX\not=(XA)^{T}$ except if $ X$ is a symmetric matrix.
3) $ \exists{X}$ s.t. $ XA-AX$ is nilpotent doesn't imply $ AX=XA$; take for example :\[ X=\begin{pmatrix}1&1\\0&2\end{pmatrix},A=\begin{pmatrix}2&0\\0&3\end{pmatrix}\].

I should've been more careful  :oops:

1) I see your point now. But why can you assume $ A = \begin{pmatrix}0&0 \\ 0&1\end{pmatrix}$?
2) I totally agree. It would be too good if multiplication by a diagonal matrix on different sides would result in transposes of each other. It was very stupid of me to invent such a thing  :blush:
3) If you refer to my "proof" here, $ D$ is (supposedly) diagonal representation of $ B$, which then would be OK. The problem is, of course, that the "proof" does not contain any reason why $ B$ should (probably it doesn't need to) be diagonalisable

Hi hedgehuogis.
 I refer to my first post, case 2:
$ \exists{P}$ an invertible matrix s.t. $ A=Pdiag(d_{1},\cdots,d_{n})P^{-1}$ where $ d_{1}\not=d_{2}$. $ B=\frac{1}{d_{2}-d_{1}}(A-d_{1}I)=Pdiag(0,1,\cdots)P^{-1}$. Let \[ X=Pdiag(\begin{pmatrix}0&1\\1&0\end{pmatrix},0,\cdots,0)P^{-1}\]. $ X$ works for $ B$ , thus $ X$ works for $ A$.


absolute value
real analysis
real analysis unsolved

Can someone help me with this problem? Suppose f: [a,b] into R. Prove or give counter-example
to the assertions:
1) If f^3 is Riemann integrable, then f is Riemann integrable.
2) If f^2 is Riemann integrable, then f is Riemann integrable . What about the converse?
If f is Riemann integrable, then absolute value of f is Riemann integrable. What about the converse?

Do you know this fundamental result:?:  :D :
Let $ D(F)$ be  the set of discontinuity points of $ F : [a,b]\rightarrow R$ . Then  $ F$ is Riemann integrable on $ [a,b]$  iff  $ F$ is bounded and the measure of $ D(F)$ is zer0 : $ \mu(D(F))\equal{}0$

1) $ (f^3 \ continuous\ for\ x_0)\Longleftrightarrow (f\  continuous\ for\ x_0)$ , so $ D(f^3)\equal{}D(f)$, so  $ ( f^3 Riemann\ integrable)  \Longleftrightarrow  (f \ Riemann\ integrable)$.
2)Contre-exemple : $ f(x)\equal{}1$ if $ x\in Q$(rational) $ , f(x)\equal{}\minus{}1$ if $ x\notin Q$ . Manifestly $ f^2\equal{}1$ is Rieman integrable ($ D(f^2)\equal{}\emptyset$)  but $ f$ id not Rieman integrable ($ D(f)\equal{}[a,b])$
$ (f \ continuous\ for\ x_0) \Longrightarrow (f^2 \ continuous\ for\ x_0)$, so $ D(f^2)\subset D(f)$, so $ ( f \ Riemann\ integrable)  \Longrightarrow  (f^2 \ Rieman\ integrable)$
And so on .. :cool:

[quote="Diogene"]Do you know this fundamental result:?:  :D :
Let $ D(F)$ be  the set of discontinuity points of $ F : [a,b]\rightarrow R$ . Then  $ F$ is Riemann integrable on $ [a,b]$  iff  $ F$ is bounded and the measure of $ D(F)$ is zer0 : $ \mu(D(F)) \equal{} 0$

1) $ (f^3 \ continuous\ for\ x_0)\Longleftrightarrow (f\ continuous\ for\ x_0)$ , so $ D(f^3) \equal{} D(f)$, so  $ ( f^3 Riemann\ integrable) \Longleftrightarrow (f \ Riemann\ integrable)$.
2)Contre-exemple : $ f(x) \equal{} 1$ if $ x\in Q$(rational) $ , f(x) \equal{} \minus{} 1$ if $ x\notin Q$ . Manifestly $ f^2 \equal{} 1$ is Rieman integrable ($ D(f^2) \equal{} \emptyset$)  but $ f$ id not Rieman integrable ($ D(f) \equal{} [a,b])$
$ (f \ continuous\ for\ x_0) \Longrightarrow (f^2 \ continuous\ for\ x_0)$, so $ D(f^2)\subset D(f)$, so $ ( f \ Riemann\ integrable) \Longrightarrow (f^2 \ Rieman\ integrable)$
And so on .. :cool:[/quote]

Thanks a lot for your help. So, if f is R-integrable, then f is continuous and the absolute value of f is also continuous, which implies absolute value of f is R-integrable.
I think I can use the same counter example you showed for the case if absolute value of f is R-integrable, then f is not necessarily R-integrable.


LaTeX
trigonometry
geometry unsolved
geometry

Prove that tan(3π/11)+4sin(2π/11)=$ \sqrt11$
(π means pi(radian))

hello, why don't you use $ \text{\LaTeX}$?
Prove that $ \tan\left(\frac{3\pi}{11}\right)+4\sin\left(\frac{2\pi}{11}\right)=\sqrt{11}$
($ \pi$ means $ \pi$ (radian))
Sonnhard.

Thanks,I don't know how to type pi,tan,..,so it isn't very beautiful at first.But I know it now.

[quote="Dr Sonnhard Graubner"]Prove that $ \tan\left(\frac {3\pi}{11}\right) \plus{} 4\sin\left(\frac {2\pi}{11}\right) \equal{} \sqrt {11}$
($ \pi$ means $ \pi$ (radian))
[/quote]
Does anyone have a solution?

$0< \tan \frac {3\pi}{11} \plus{} 4 \sin \frac {2 \pi}{11} \equal{} \sqrt {11} \Longleftrightarrow$

$0< \sin \frac {3\pi}{11} \plus{} 2 \sin \frac {5 \pi}{11} \minus{} 2 \sin \frac {\pi}{11} \equal{} \sqrt {11} \cos \frac {3 \pi}{11}\  \Longleftrightarrow$

(squaring it)

$ \sin^2 \frac {3\pi}{11} \plus{} 4 \sin ^2\frac {5 \pi}{11} \plus{} 4 \sin^2 \frac {\pi}{11} \plus{}$
$ \plus{} 4 \sin \frac {3 \pi}{11} \sin \frac {5 \pi}{11} \minus{} 4 \sin \frac {\pi}{11} \sin \frac {3 \pi}{11} \minus{} 8 \sin \frac {3 \pi}{11} \sin \frac {5 \pi}{11} \equal{} 11 \minus{} 11 \sin^2 \frac{3 \pi}{11} \Longleftrightarrow$

$ 6 \minus{} 6 \cos \frac {6 \pi}{11} \plus{} 2 \minus{} 2 \cos \frac {10 \pi}{11} \plus{} 2 \minus{} 2 \cos \frac {2 \pi}{11} \plus{}$
$ \plus{} 2 \cos \frac {2 \pi}{11} \minus{} 2 \cos \frac {8 \pi}{11} \minus{} 2 \cos \frac {2 \pi}{11} \plus{} 2 \cos \frac {4 \pi}{11} \minus{} 4 \cos \frac {4 \pi}{11} \plus{} 4 \cos \frac {6 \pi}{11} \equal{} 11 \Longleftrightarrow$

$ 2 \cos \frac {2 \pi}{11} \plus{} 2 \cos \frac {4 \pi}{11} \plus{} 2 \cos \frac {6 \pi}{11} \plus{} 2 \cos \frac {8 \pi}{11} \plus{} 2 \cos \frac {10 \pi}{11} \equal{} \minus{} 1 \Longleftrightarrow$

$ \sum_{k \equal{} 1}^5 2 \cos \frac {2 \pi k}{11} \equal{} \minus{} 1 \Longleftrightarrow \sum_{k \equal{} \minus{} 5}^5 \cos \frac {2 \pi k}{11} \equal{} 0 \Longleftrightarrow \sum_{k \equal{} \minus{} 5}^5 \exp \frac {2 \pi i k}{11} \equal{} 0$

which is zero sum of the 11 complex roots of $ z^{11} \minus{} 1 \equal{} 0.$


trigonometry
limit
calculus
derivative
geometry
inequalities unsolved
inequalities

Let $ x\in(0;\frac{\pi}{2})$. Prove that
a)$ \sin x>\frac{1\minus{}\cos x}{x}$
b)$ (\frac{\sin x}{x})^3>\cos x$

[hide="a)"]The stronger one is also true: $ \sin\frac {x}{2} > \frac {1 \minus{} \cos x}{x}$

[u]Proof:[/u]

 [geogebra]a823f227a4ef1323a2343b0809de7ec62a57e5ab[/geogebra]

$ \sin \frac {x}{2} \equal{} \frac {EB}{DB} \equal{} \frac {AB \minus{} AE}{DB} > \frac {AB \minus{} AE}{\text{arc }DB} \equal{} \frac {\frac {AB \minus{} AE}{AB}}{\frac {\text{arc }DB}{AB}} \equal{} \frac {1 \minus{} \frac {AE}{AB}}{x} \equal{} \frac {1 \minus{} \frac {AE}{AD}}{x} \equal{} \frac {1 \minus{} \cos x}{x}$[/hide]

[hide="b)"]There's (perhaps) no geometric interpretation of cubing. So lets go straight into calculus.

Let $ f(x) \equal{} \left(\frac {\sin x}{x}\right)^3 \minus{} \cos x$
Since $ \lim_{x\to 0}f(x) \equal{} 0$ it is sufficient to prove that $ f'(x) > 0$ when $ x > 0$
$ f'(x) \equal{} \minus{} \frac {(3\sin x)^3}{x^4} \plus{} \frac {3\cos x(\sin x)^2}{x^3} \plus{} \sin x$
$ f'(x) > 0\iff g(x) \equal{} x^4 \plus{} 3x\sin x\cos x \minus{} 3(\sin x)^2 > 0$
Since $ g(0) \equal{} 0$, it is sufficient to prove that $ g'(x) > 0$ when $ x > 0$
we need to prove $ g'(x) \equal{} 3x\cos 2x \plus{} 4x^3 \minus{} \frac {3}{2}\sin 2x > 0$
Again $ g'(0) \equal{} 0$, it is sufficient to prove that $ g''(x) > 0$ when $ x > 0$
$ g''(x) \equal{} 6x(x \minus{} \sin x\cos x)$
We know that $ x > \sin x$. Hence $ g''(x)>0$ for $ x$ in the given interval.
QED.
(Hope there's no mistake in calculation.)[/hide]

[quote="hsbhatt"]The derivative approach works for the first problem too[/quote]
I know. But I always prefer more elementary approaches than calculus, euclidean geometry than any other form of geometry, ... etc.
Anyway if anyone has a non-calc solution to the second, please post it here.


Hey guys,

I'm releasing the complete 2008 AwesomeMath Program Team Contest as a .pdf file.  It contains all the problems from each round, compiled so that each problem appears exactly once (as opposed to, say, on all three rounds), with all necessary corrections appended.  Due recognition is finally given to the original proposers of each problem.

Thanks for all your hard work.  As I'm sure I've said many times, it was my pleasure to serve you as counselor, TA, and czar of team contest, and I looking forward to taking up these roles again in future years.

Please note: I've been asked to hold problems 34, 51, 52, 54, 55, 56.

anybody want to work together to try to come up with a set of solutions for these problems?

Why is Algebra 8 on first round not in here?

b/c the stronger format is on the second week's test. NO DUH

Unimpossible, it sounds like a good idea. I'm up for it, especially since I'd like to see solutions to certain problems from that set.

Will the rest of of the problems be put back after certain events have finished?

I'll type up a few of the trickier problems that I solved if I have time.

What happened to A8, C8, and N8? (Rnd 1)

Oh wait... They appeared in later rounds too.
Sorry...

can someone email me a copy?
i tried to download it and my computer gave me a "this file cant be found"
aurgh stupid computer
my email is
blackbelt14253@gmail.com


geometry
projective geometry
geometric transformation
homothety
analytic geometry
vector
LaTeX

[Moderator edit: This is going to be the topic for all recommendations of literature about olympiad and advanced Euclidean geometry.]

hi im trying to learn more on some more geometric  theorems, to attack my problems. im specifically looking for more advanced things, not the usual (menelaos, ceva,angle bisector, desargues, power of point, etc); more like  projective geometry , homothecy , similarity and all these new things that i could use to solve olympiad problems. so could anyone tell me what i should know and where can i find it??? thanks  :D

well, there are transformation, inversion, projective, and stuff using complex coordinate plane and vectors (don't ask me, i don't know any of these stuff. I'm noob)

you can find them at: [url]http://www.unl.edu/amc/a-activities/a4-for-students/s-index.html[/url]

go to the part where it says download the notes for geometry, download the post script one, i think the latex one is corrupted (some part gives u weird things)

postscript download: [url]http://www.cs.wisc.edu/~ghost/index.htm[/url]

download the gsview and the ghostscript. and open the file.

there is the right arrow button somewhere on top to scroll the page, but you can use gsview to convert it to pdf.

I'm using this notes to geometry, although i hardly can solve any practice problem in the notes....

nathan altshiler court ,has a good book about inversion wich its name is inversive geometry.

some easy geometry theorems have disscused here, in this forum:

http://www.mathlinks.ro/Forum/viewtopic.php?t=77530

[quote="Megus"]Just check the topic a little bit above:

http://www.mathlinks.ro/Forum/viewtopic.php?t=70654

this book should help you[/quote]

When I try to load the pdf file, it says it's damaged and could not be repaired.

I've checked it and you're right. I don't know what happened  :maybe:

Can anyone recommend a book teaching how to use inversion in Olympiad standard problems through  many examples and exercises?

check http://www.imomath.com study materials

The transformations in geometry are nice methods. Especially good books are.

Geometric Transformation 1, 2, 3 and 4 ( note there are four volumes) by I.M. Yaglom.
Other good problem books are by Vicktor Prasolov. Problems in plane geometry.
Another classical is Nathan Altshiller Court, College Geometry.
Please give some more rescources to study. i am looking for some recent ones.

[quote="Bugi"]Could you tell us more about Court's book (College Geometry)? I've seen it mentioned multiple times here, is it good for some kind of preparation and why?[/quote]

A copy of this available from [url]http://wvw.downarchive.com/?do=search&subaction=search&story=college+geometry&x=0&y=0[/url]
Read it and decide why its so good (or thought to be).


There are three glasses of what looks to be water, but only one is safe to drink. There is a guard who knows intermediate logic and also knows which glass is safe. You may ask him one yes/no question, to which he will respond "yes", "no", or "paradoxial". If both "yes" and "no" provide a truthful answer, he will choose one at random. The guard always tells the truth, and you must drink from one of the glasses. What do you do?

Stupid variation: exact same question, but with 20 glasses and 7 are safe. This one is pretty lame, though...

Does this work?

[hide]Suppose the glasses are 1, 2, and 3. I would ask the guard, if glass 1 is not poisoned, is glass 2 poisoned?[/hide]

But Im only telling the guard to answer it [b]if[/b] number 1 wasn't poisoned. Thus, the guard wouldn't say anything if #1 was poisoned.

When this was originally posted, I thought about it for a while and got this.
[hide]Tell the guard, "If you say yes, I will drink cup 1.  If you say no, I will drink cup 2.  If you say paradoxical, I will drink cup 3.  Will I die?  Please don't say yes unless you need to."

If he says no, then drink cup 2.  If he says paradoxical, then drink cup 1.  If he says yes, then drink cup 3.

Do you see why this works?[/hide]
I think that's right.


[quote="vidyamanohar"]A circle has 2006 points chosen so that the arcs between any two adjacent points are equal.  Three of these points are chosen at random.  Let the probability that the triangle is right be R, and the probability that the triangle formed is isosceles be I.  Find |R-I|.[/quote]

I think 0 is reasonable answer.

Namdung

My solution:
[hide="Start with R"]First, let's find R.  Any right triangle inscribed in a circle has its hypotenuse as the diameter of the circle.  Among 2006 points spaced evenly about the circle, 1003 pairs will form a circle.  With two points that form a diameter are picked, there are 2004 remaining points to form the next vertex.  Thus:
$ R=\frac{1003\cdot2004}{\binom{2006}{3}}$
[hide="Denominator..."]This is the number of total triangles that we can choose... 3 vertices out of the 2006[/hide][/hide]
[hide="Now to I"]Consider the vertex joining the two equal sides.  There are 2006 places to put that.  The next two vertices can be chosen together in 1002 places.  Note that no two isosceles triangles have been counted twice because 3 doesn't divide 2006 evenly so there are no equilateral triangles.
$ I=\frac{1003\cdot2004}{\binom{2006}{3}}=R$[/hide]
Finally:
[hide="answer"]
$ \abs{R-I}=0$[/hide]


inequalities
integration
function
calculus
real analysis
real analysis unsolved

This came up while trying to do something else: Let $ a_i \geq 0$ and $ b_j \geq 0$ be $ \ell^1$-sequences (that is, they should be summable). Is it then true that

$ \int_0^1 \sqrt{\sum_{i,j} a_i b_j (1\minus{}t^i)^2j^2 t^{2j}} \, dt < \infty$ ?

I think the answer is no, but it is frustrating that I cannot show this.

I assume the range of summation is $ i,j\ge 0$. Now, the expression under the square root  splits into the product $ f(t)\equal{}\sum_i a_i(1\minus{}t^i)^2$ and $ g(t)\equal{}\sum_j j^2b_j t^{2j}$.The first function can be estimated by $ \sum_i a_i$ and nothing better than that because you can always shift the sequence $ a_i$ very far to the right. Thus the question reduces to whether you can estimate $ \int_0^1 \sqrt g$ in terms of $ \sum_j b_j$. There are two fighting things here: the concavity of the square root and fact that the terms $ j^2t^{2j}$ in the weighted sum are very different from each other. So, it is natural to look at the terms that are the same and neglect everything that is much less. Since $ j^2 t^{2j}\approx j^2$ when $ 1\minus{}t\approx \frac 1j$ on the multiplicative scale, the dyadic partition is due. So, let us estimate $ I_m\equal{}\int_{1\minus{}2^{1\minus{}m}}^{1\minus{}2^{\minus{}m}}$. In this integral only the terms with $ j\in[2^{m\minus{}1},2^m\minus{}1]$ seem to matter, so $ g(t)\ge c 2^{2m}B_m$ where $ B_m\equal{}\sum_{j\equal{}2^{m\minus{}1}}^{2^m\minus{}1}b_j$. Thus $ I_m\ge c\sqrt{B_m}$. Now, the only restriction on $ B_m$ is that their sum is bounded, which doesn't allow to estimate the sum of their square roots. Thus, the inequality is false.


Hello,

I am writing a document for which I would need the \boxslash symbol.  My system does not seem to support this, and it turned out I needed a certain "stmaryrd" package.  Now I was told all of I needed was a certain file "stmaryrd.sty", a file that I already have a copy of.

But... what to do now? I haven't got any experience with introducing new packages.  I've tried several things.  I read that putting my sty file just in the same directory where the other ones were would have been enough, but it isn't.

Thank you,
fredbel6

[quote="stevem"][quote="fredbel6"]I read that putting my sty file just in the same directory where the other ones were would have been enough, but it isn't.[/quote]That's true for many style files but stmaryrd needs tex fonts installing as well. If you are using MiKTeX then it's easy. For versions 2.5+ just put \usepackage{stmaryrd} in your document and it will install it automatically. Otherwise go to Browse Packages (or Package Manager in older versions) right-click on stmaryrd and choose install.[/quote]
Thank you for your reply.  I am actually using "Texnic center"   :maybe:  but there seems to be no "Browse Packages" or "Package Manager" option there? Or did you mean something else?

[quote="stevem"]TeXnicCenter is just the editor, the LaTeX system is provided by MiKTeX. You'll find Browse Packages/Package Manager in XP Start->All Programs->MiKTeX.[/quote]
Ok thx, so I found this program Package Manager, saw a huge list, stmaryrd was in it, when I rightclicked it I got "install".
He told me he could't complete the installation, because he couldn't find the file.  So uhm, what could be wrong now?

[quote="stevem"]You are connected to the internet? MiKTeX may need to download the files. If that doesn't work then try MiKTeX->Settings->Packages->Change... and choose Packages should be installed from the internet etc then try again.[/quote]
Hmm, I found this stmaryrd.cab file on the net..but whatever I do with it it still doesn't work... :|

You MUST get MIKTeX to install it since the installation is non-trivial (see below) which is complicated unless you know what you are doing and happy to use the command line. MiKTeX does it all for you and downloads it from the net from its own server for you. Just tell it to do so, using the method I gave you in the last post (use Settings and tell it where to download from), and it will do the installation.

-------------
These are the install instructions which you won't like:
[quote]To install the fonts, you should:

 * run METAFONT on the fonts, for example under Unix you say:

      mf "\mode=localfont; input stmary5;"
      mf "\mode=localfont; input stmary6;"
      mf "\mode=localfont; input stmary7;"
      mf "\mode=localfont; input stmary8;"
      mf "\mode=localfont; input stmary9;"
      mf "\mode=localfont; input stmary10;"

 * move stmary*.tfm into a TeX font metrics directory

 * move stmary*.*pk into a TeX bitmaps directory

If you want to use the St Mary's Road font in LaTeX2e, you should:

 * run LaTeX2e on stmaryrd.dtx

 * move stmaryrd.sty and Ustmry.fd into a TeX inputs directory

 * read the documentation in stmaryrd.dvi[/quote]


i don't give it thought now but it should be nice.
For a rational number x that equals $\frac{a}{b}$ in lowest term we define:$f(x)=ab$
a)show that $\sum_{x\in Q^+}\frac{1}{f(x)^2}=\frac{5}{2}$ where the sum extend over all positive rational.
b)more generally;exhibite an infinite sequence of distinct rational  exponenet s such that $\sum_{x\in Q^+}f(x)^{-s}$ is rational.

Indeed, a very nice problem. For part b), we can take s the sequence of even numbers and use bilzhao's argument from a) and the fact that $ f(2n)=\frac{2^{2n-1}\cdot B_n }{(2n)!}\cdot (pi)^{2n} $, where f is the Riemann zeta function and $ B_n $ is the n'th Bernoulli number.

Alternatively,

$t = \Sigma_{(a, b) = 1} \frac{1}{a^2b^2}$
\begin{eqnarray*}
\frac{\pi^4}{90} t = \\
\Sigma_d{1/d^4} t = (\Sigma_d(\Sigma_{(a, b) = 1} \frac{1}{a^2b^2d^4})) \\
= \Sigma_d (\Sigma_{(a, b) = d} \frac{1}{a^2b^2} \\
= \Sigma_{a, b} \frac{1}{a^2b^2} \\
= (\frac{\pi^2}{6})^2
\end{eqnarray*}

(pardon my badly formatted TEX)


Let f:R-->R such that |f(x)-f(y)|\leq |x-y|. If, for all x in
R {f (n) (x)} is an aritmetic progresion then exists a in R such that f(x)=x+a

sol:

Let x,y in R. because {f (n) (x)} is p.a., exists w such that f (n) (x)=x+nw. In the same way, there is k such that f (n) (x+y)=x+y+nk.
With the hipotesis we have
|f (n) (x+y)-f (n) (x)|\leq
|f (n-1) (x+y)-f (n-1) (x)|\leq...\leq
|x+y-x|\leq |y|, so
|y+n(k-w)|\leq |y|, for all n in N. We conclude that w=k.
We can write
f(x+y)=x+y+w=f(x)+y, for all x,y in R. take x=0. :D

Sorry, I dont understand ! For ex. f(x) =0 verifies the conditions  ...

Wrong! if f=0, then $f^n(x)_{n\in N}$ is x,0,0,..., which is not an aritmetic progresion!

Again, is not a good example: for negative numbers we have, for example x=-2
$ -2=f^{(0)}(-2), 2=f^{(1)}(2), 2,... $ which is not an aritmetic progresion!

Sorry I didnt see , you start  from n=0 and note  f<sup>(0)</sup>(x) = x ...


Evaluate the radical, if possible, for $n \in \mathbb{Z^+} \ , \ n \geq 1$:

$\sqrt{n+\sqrt{(n-1)+\sqrt{(n-2)+ \sqrt{ ... +\sqrt{1}}}}}$

(Example: For $n=4$, the radical would be $\sqrt{4+\sqrt{3+\sqrt{2+\sqrt{1}}}}= \ ?$.  Is there a more simplified way to write it?

[hide="Hint: Not sure if this will help, but..."]
$\sqrt{1+\sqrt{1 + \sqrt ...}}= \phi$ [/hide]

[hide="Note"]
I don't know how to do it, and I'm not sure if it is possible (though it seems like there's a way to do it somehow. Any advice or solutions are welcome! Thanks :) [/hide]

[hide="Added note"]
I just noticed that the sequence for $n=1,2,3...$ is recursive. Can we do something with this?[/hide]

[quote="JBL"]This isn't going to evaluate very nicely at all, I'm afraid.  The nice recursive definition makes it easy to calculate many values on a computer, and the end behavior is going to be just like that of $\sqrt n$, or even more like $\sqrt{n +\sqrt n}$, unsurprisingly.[/quote]

ok.. well, thanks :)

also, why won't it evaluate nicely? i'm just wondering :P


In the figure, Lines AB and BC are radii of the circle with center B. What is the value of X? (see attached)

I am lost. I don't know how to begin to solve this problem.

<ABC is a right triangle. Does that factor into the equation?

It would appear that the answer has to be a negative number.

[quote="cf249"]Because B is the center of the circle, BC is a radius.  The length of this radius is 5.  If you don't understand why, it's because of the [url=http://www.artofproblemsolving.com/Wiki/index.php/Distance_formula]distance formula[/url].  To find the x-coordinate of point A, all you need to do is subtract 5 from the x-coordinate of point B, which gives -4.

The fact that ABC is a right triangle doesn't have anything to do with the problem, just the fact that AB and BC are both radii.

Extension:  Point D lies on circle B.  Its x-coordinate is 5.  What are the two possible y-coordinates for Point D?[/quote]

It's easy when you explain it that way.

Thank you

Basicly theres no need to use the formua that is square root of (x1-x2)^2+(y1-y2)^2. IF bc=ac where bc= 5 units then ac will be 5 units(Radius). (1,2) and (x,2) are 5 units away. Since they are on the same y axis u jz need to find out the diffrence in the x axis. So the answer is (1-5,2) that is (-4,2)


Let {an} is an increasing seqence of positive numbers such that a1 \leq 2003 and |an+1-an| \leq 2003. Prove an | am for some n and m, n \neq m.

[i]unknown source[/i]

This problem was used in Vietnamese Team selection 2001 (and ofcourse, 2003 is replaced by 2001). But it was published before in not very well-known russian journal: "Impery of Mathematics" (http://old.rcd.ru/em/). Is good journal with intersting articles and problems. But unfortunately there are only two numeros. I think this problem is very challenging.

Namdung

The source of this problem is All-Russia olympiad '99.
The aim of publishing this problem is to obtain solution, not links.

Here you are

Put A = {a1, a2, ...}. For every n, let call the number among n, n+1, ..., n+2002 which diveded by some element of A "characteristic" of n. Hence char(n) <= 2003 for all n, we can choose n0 such that char(n0) is maximal. Let char(n0) = p and n0+k1 is divided by a_i1 from A, n0+k2 is divided by a_i2, ... , n0+kp is divided by a_ip. Consider number y = n0 + a_i1.a_i2...a_ip. It is clear that y + ki is divided by a_ij, j=1, 2, ..., p. But by condition, there are a from A and s from 0, 1, ..., 2002 such that  a = y + s. From char(y) <= p follows that  a = y + kj for some j. But then we have a is divided by a_ij and we are done.

Namdung

" From char(y) <= p follows that a = y + kj for some j"

You've lost me here. Could you explain it a bit? How does this follow?

We know aij | y+kj for all j=1,...,p ==> char(y) \geq p ==> we need s=kj for some j otherwise char(y) \geq p+1.


group theory
abstract algebra
superior algebra
superior algebra solved

It is known that all the subgroups of an abelian group are normal. If all the subgroups of a group G are normal does it follow that G is abelian?

I hope this question is not too old to answer  :)

Unfortunately the converse is not true with the Quaternion group of order 8 being a counter-example. Non-abelian groups, all of whose subgroups are normal, are called Hamiltonian and have been completely classified. See http://en.wikipedia.org/wiki/Hamiltonian_group

In fact, we can say that:-

Group is Abelian => Every subgroup is normal =>Evey maximal subgroup is normal (or no subgroup is self-normalising)

The first lcass is Abelian groups, the second includes Abelian and Hamiltonian groups and the third class is that of nilpotent groups.