logarithms trigonometry algebra function domain

Which of the following is equivalent to $ \frac{\log{(1\minus{}\cos{(x)})} \minus{} \log{2}}{2}$?

(1) log(tan(2x))

(2) $ \log{\left(\sin{\left(\frac{x}{2}\right)}\right)}$

(3) $ \log{\left(\cos{\left(\frac{x}{2}\right)}\right)}$

(4) log(cos(2x))

Thanks is advance.

[hide]
Definitions
$ \sin \frac {\alpha}{2} \equal{} \sqrt {\frac {1 \minus{} \cos \alpha}{2}}$.
$ \log a \minus{} \log b \equal{} \log \frac {a}{b}$.
$ \frac { \log c}{2} \equal{} \log c^{\frac {1}{2}} \equal{} \log \sqrt {c}$.
[/hide]

Something should probably be said about the domains of these expressions.

[hide="Solution"]Because $ 1 \minus{} \cos{x} \equal{} 2\sin^2{\frac {x}{2}}$, the given equation becomes:
\[ \frac {\log{\left(1 \minus{} \cos{x} \right)} \minus{} \log{2}}{2} \\
\equal{} \frac {1}{2} \cdot \left(\log{2\sin^2 {\frac {x}{2}}} \minus{} \log{2}\right) \\
\equal{} \frac {1}{2} \log{\frac {2\sin^2{\frac {x}{2}}}{2}} \\
\equal{} \log{\sqrt {\sin^2{\frac {x}{2}}}} \\
\equal{} \boxed{\log{\left(\sin{\frac {x}{2}}\right)}}
\]
[/hide]


quadratics
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Gauss
trigonometry
number theory
number theory unsolved

Prove or disprove:

$-3$ is a quadratic non-residue of primes of the form $6a-1$ and a quadratic residue of primes of the form $6a+1$

When I search gauss lemma in wikipedia I got this identity

$(\frac{2}{p})=(-1)^{\frac{p^{2}-1}{8}}$

how we can get number $\frac{p^{2}-1}{8}$ ?

And why $(\frac{a}{p})=\prod_{n=1}^{\frac{p-1}{2}}\frac{\sin 2\pi \frac{an}{p}}{\sin 2\pi \frac{n}{p}}$ ?

Can anyone give an explanation?

thx..

Remember the Legendre symbol is a multiplicative function. Thus applying the quadratic reciprocity theorem (they say, Gauss's favourite one from number theory):

$\left(\frac{-3}{6a \pm 1}\right) = \left(\frac{-1}{6a \pm 1}\right) \left(\frac{3}{6a \pm 1}\right) = \left(\frac{6a\pm 1}{3}\right) = \left(\frac{\pm 1}{3}\right) = \pm 1$.

[quote="s.tringali"]
$\left(\frac{-1}{6a \pm 1}\right) \left(\frac{3}{6a \pm 1}\right) = \left(\frac{6a\pm 1}{3}\right) = \left(\frac{\pm 1}{3}\right) = \pm 1$.[/quote]

Sorry,but I don't understand why $\left(\frac{-1}{6a \pm 1}\right) \left(\frac{3}{6a \pm 1}\right) = \left(\frac{6a\pm 1}{3}\right)$ and $\left(\frac{6a\pm 1}{3}\right)= \left(\frac{\pm 1}{3}\right)$

Can you give more explanation?
Thx

[quote="Yosh..."]When I search gauss lemma in wikipedia I got this identity

$(\frac{2}{p})=(-1)^{\frac{p^{2}-1}{8}}$

[/quote]
This is based on Gauss' Lemma.

[b]Lemma:[/b] Let $a$ be a integer with $\gcd(a,p)=1$ and $p$ an odd prime. Let $n$ denote the number of elements in the set:
$\left\{ a,2a,3a,...,\left(\frac{p-1}{2}\right)a\right\}$
Whose remainder modulo $p$ exceeds $p/2$ then:
$(-1)^{n}= (a/p)$.

Now we can say why $(2/p)=(-1)^{(p^{2}-1)/8}$

We consider the set:
$\left\{ 2,2\cdot 2,2\cdot 3,....,2\cdot \left( \frac{p-1}{2}\right) \right\}$
And we argue that (details omitted) that if $p\equiv \pm 1 (\bmod 8)$ then $(2/p)=1$ and if $p\equiv \pm 3 (\bmod 8)$ then $(2/p)=-1$.

Now if $p=8k\pm 1$ then $\frac{p^{2}-1}{8}= 8k^{2}\pm 2k$ which is even.
And if $p=8k\pm 3$ then $\frac{p^{2}-1}{8}= 8k^{2}\pm 6k+1$ which is odd.

Hence, $(2/p) = (-1)^{(p^{2}-1)/8}$

[quote="Yosh..."]
Sorry,but I don't understand why $\left(\frac{-1}{6a \pm 1}\right) \left(\frac{3}{6a \pm 1}\right) = \left(\frac{6a\pm 1}{3}\right)$ and $\left(\frac{6a\pm 1}{3}\right)= \left(\frac{\pm 1}{3}\right)$
[/quote]

He used Quadradic Reciprocity.

[b]Quadradic Reciprocity[/b]: $(p/q) = (q/p)$ for odd primes $p\not =q$ if either $p$ or $q$ is congruent 1 mod 4. Otherwise $(p/q) =-(q/p)$.

---
Next,
If $a\equiv b (\bmod p)$ and $\gcd(a,p)=\gcd(b,p)=1$
Then,
$(a/p) = (b/p)$
Since,
$6k\pm 1 \equiv 1 (\bmod 3)$
You can ignore the $6k$ thing by the theorem.

[quote="Yosh..."]
Sorry,but I don't understand why $\left(\frac{-1}{6a \pm 1}\right) \left(\frac{3}{6a \pm 1}\right) = \left(\frac{6a\pm 1}{3}\right)$ and $\left(\frac{6a\pm 1}{3}\right)= \left(\frac{\pm 1}{3}\right)$

Can you give more explanation?[/quote]
As to the first question, mind that i) $\left(\frac{-1}{p}\right) = (-1)^{\frac{1}{2}(p-1)}$; and ii) $\left(\frac{p}{q}\right) = (-1)^{\frac{1}{2}(p-1) \cdot \frac{1}{2}(q-1)}\left(\frac{q}{p}\right)$, for any couple of (rational) primes $p, q \ge 3$. In particular, ii) is the quadratic reciprocity theorem.

As to the second, just consider that $\left(\frac{a}{p}\right) = \left(\frac{b}{p}\right)$, if $p \ge 3$ is a (rational) prime and $a,b \in \mathbb{Z}$ are such that $a \equiv b \bmod p$.

And yes, all of these are well-known facts/properties about the Legendre symbol. You can simply prove them by yourself or try to search for any references even in the forum or outside.


If $ A$ is an open subset of $ \mathbb R^n$ and $ f : A \to \mathbb R^m$ is Lipschitz, then $ f$ is differentiable almost everywhere in $ A$.

What do you expect us to do with it? Post our favorite proofs? But they are all quite long... :?

[quote="fedja"]Post our favorite proofs?[/quote]
It's just a theorem that is worth mentioning.

I kind of knew that you'll feel obliged to answer, so I should have said something in the first post :blush:


group theory
abstract algebra
invariant
pigeonhole principle
induction
superior algebra
superior algebra theorems

Most proofs of the Sylow Theorems are rather ugly which is why some people despise them. I offer the way I learned them, perhaps among the smoothest of all proofs.

The only pre-requiste to this are an understanding of [url=http://en.wikipedia.org/wiki/G-set]G-sets[/url]. I will state an important theorem about G-sets that we need to know just for reference.

[b]Theorem:[/b] Let $ X$ be a $ G$-set with $ x\in X$. Then $ |Gx| = (G: G_{x})$.

The above theorem shows that if $ |G|$ is finite (important for us since we will work with finite groups) then $ |Gx|$ divides $ |G|$.

----------------------------------------------------------------------------
The following theorem is the foundation of all our work.

[b]Definition:[/b] Let $ X$ be a $ G$-set. We denote $ X_{G}$ to be set $ \{x\in X| gx = x \ \forall g\in G\}$. So this is really similar to the set $ X_{g}= \{ x\in X| gx = x\}$. Except here we are talking about [i]all[/i] elements $ g\in G$ instead of some specific element $ g$. (Another way to view this, is the intersection of all $ X_{g}$).

[b]Theorem*:[/b] Let $ G$ be a group of order $ p^{n}$ and $ X$ a finite $ G$-set then $ |X| \equiv |X_{G}| (\bmod p)$

[b]Proof:[/b] For $ x\in X$ notice that, the orbit, $ |Gx| = \{gx|g\in G\}$ by definition. Say there are $ m$ orbits in $ X$ under $ G$, choose $ x_{i}\in X$ to be any element in every single one of these orbits, and consider the elements $ x_{1},x_{2},...,x_{m}$. Notice that every $ x\in X$ must be contained in exactly one of these orbits $ Gx_{i}$. That means we have the sum $ |X| = \sum_{k=1}^{m}|Gx_{k}|$. Now there may be [i]single[/i] element orbits in $ X$ under $ G$. Let us say there are $ 0\leq q\leq n$ such single elements orbits. Reordering $ x_{i}$ if necessary we get $ |X| = |X_{G}|+\sum_{k=q+1}^{m}|Gx_{k}|$ if $ 0\leq q < n$ otherwise (if $ q=n$) $ |X| = |X_{G}|$. If the latter case ($ q=n$) then $ |X|-|X_{G}|=0$ and the proof is automatically complete. If the former case ($ q<n$) then $ |X|-|X_{G}| = \sum_{k=q+1}^{m}|Gx_{k}|$. By the important divisiblity statement above each $ |Gx_{i}|$ divides $ |G|=p^{n}$. But $ |Gx_{i}|>1$ (that is why we cleaverly seperated single elements orbits and multiple elements orbits above). That means each $ |Gx_{i}|$ is divisible by $ p$ because $ p$ is prime. As a consequence this shows $ p|(|X|-|X_{G}|) \implies |X| \equiv |X_{G}| (\bmod p)$. Q.E.D.

The proof of the first Sylow theorem depends on induction. That is a special case of Sylow's theorem called "Cauchy's Theorem".

[b]Theorem:[/b]Let $ p$ be a prime number. And let $ G$ be a finite group divisible by $ p$. Then there exists an element of order $ p$.

[b]Proof:[/b]The proof is truly quite element with the powerful counting theorem above. Define $ X = \{(g_{1},g_{2},...,g_{p})|g_{i}\in G \mbox{ and }g_{1}...g_{p}= 1\}$. The claim is that $ p$ divides $ |X|$. This is easily since because we can choose any $ g_{i}$ for the first $ p-1$ places and the last place is [i]uniquely[/i] determined as $ (g_{1}...g_{p})^{-1}= g_{p}^{-1}... g_{1}^{-1}$. Hence $ |X|= |G|^{p-1}$ since $ p$ divides $ |G|$ it is easily seen that $ |X|$ is divisible by $ p$ as well. Let $ \mu = (1,2,...,p)$ be a cycle in the symettric group $ S_{p}$. Define the action of $ \mu$ on $ X$ to act as $ \mu (g_{1},...,g_{p}) = (g_{2},g_{3},...,g_{p},g_{1})$ and this new $ p$-tuples is in $ X$ because $ (g_{2}g_{3}...g_{p})g_{1}=1$ since $ g_{1}(g_{2}...g_{p})=1$. Now define the group $ < \mu >$ with $ |<\mu >|=p$ we define the action of an element of $ <\mu >$ on $ X$ as iteration of the above action. It is easily seen that $ X$ is now a $ <\mu>$-set under the action described above. We also know that $ |X| = |X_{<\mu>}| (\bmod p)$. Since $ p$ divides $ |X|$ is means $ p$ divides $ |X_{<\mu>}|$. This is almost complete the proof because since $ p>1$ it means there are at least two elements in $ X_{<\mu>}$. For example, $ (1,1,...,1)$ is certainly an element in $ X_{<\mu>}$ because $ 1\cdot 1...1=1$ and it is left fixed under the action of $ <\mu>$. But there exists another elements $ (a,a,...,a)$ (it has to be a "homogeneous" $ p$-tuple because otherwise it will not be fixed under the action of the group) with $ a\not = 1$ and the condition $ a^{p}=a$. Since $ p$ is prime is means that has to be the order. Q.E.D.

[b]Corollay:[/b] If $ p$ divides order of a finite group $ G$ then $ G$ has a subgroup of order $ p$.

[b]Proof:[/b] Simply choose $ <a>$ as in the above proof.

[b]Definition:[/b] A group $ G$ is called a $ p$-group iff each element has prime power order.
A subgroup of $ G$ is called $ p$-subgroup iff it itself is a $ p$-group.

Here is an alternate way to view that definition above.

[b]Theorem:[/b] A finite group $ G$ is a $ p$-group if and only if $ |G|$ is a power of $ p$.

[b]Proof:[/b] This proof is almost trivial with Cauchy's theorem. Assume that $ G$ is a $ p$-group but it is [b]not[/b] a power of a prime $ p$. Then there exists a prime $ q\not =p$ such that $ q$ divides $ |G|$, by Cauchy's theorem there is an element of order $ q$ which is impossible because it is a $ p$-group. The other way around is similar. Q.E.D.

A very important concept for us is the "normalizer".

[b]Definition:[/b] Let $ G$ be a group. And let $ S = \{H|H\leq G\}$, i.e. the set of all subgroups of $ G$. We can turn $ S$ into a $ G$-set by letting $ G$ act on $ S$ by "conjugation" (also known as the "inner automorphism"). For $ H\in S$ and $ g\in G$ we define this action to be $ gHg^{-1}$. Note this maps subgroups into subgroups so it has the closure property. And the conditions for being a $ G$-set are satisfied. Hence $ S$ is a $ G$-set. Now given a specific subgroup $ H$ we define $ N_{G}[H]$ to be the isotropy subgroup, meaning, $ N_{G}[H] = \{g\in G | gHg^{-1}= H\}$. That is called "the normalizer of $ H$ in $ G$".

Notice that $ N_{G}[H]$ is the maximal group that contains $ H$ as a normal subgroup because it contains all the elements that leave $ H$ invariant under conjugation.

[b]Theorem:[/b] If $ ghg^{-1}\in H$ for all $ h\in H$ then $ g\in N_{G}[H]$.

[b]Proof:[/b] The map $ \phi_{g}: H\to H$ defined as $ \phi(h) = ghg^{-1}$ is injective. By the Pigeonhole Principle (since $ H$ is finite), this map is bijective. So $ \phi_{g}[H] = H$ and which means $ g\in N_{G}[H]$ by definition.

Here is a famous lemma.

[b]Lemma:[/b] Let $ H$ be a $ p$-subgroup of a finite group $ G$. Then $ (N_{G}[H]: H)\equiv (G: H) (\bmod p)$.

[b]Proof:[/b] Let $ S$ be the set of all left-cosets of $ H$ in $ G$ and note that $ (G: H) = |S|$. Define the action of $ H$ on $ S$ as $ h(xS) = (hx)S$, this makes $ S$ an $ H$-set. The idea now is to use starred theorem above. In order to do that we need to determine $ S_{H}$, i.e. the left cosets that are fixed under the action of all the elements of $ H$. Notice that $ h(xH) = xH$ iff $ x^{-1}hx\in H$ for all $ h \in H$. We have shown that the statement $ xH = h(xH)$ for all $ h\in H$ is equivalent to saying $ x^{-1}h (x^{-1})^{-1}\in H$ for all $ h\in H$, thus (and conversely) $ x\in N_{G}[H]$ by the theorem just proved. Thus the number of such left cosets $ S_{H}$ is the number of left cosets of $ N_{G}[H]$. And hence $ |S_{H}| = (N_{G}[H]: H)$ hence by starred theorem $ (N_{G}[H]: H) \equiv (G: H)(\bmod p)$. Q.E.D.

Now we are ready to proof the Sylow Theorems.

[b]First Theorem:[/b] Let $ G$ be a finite group and $ |G|=p^{n}m$ where $ n\geq 1$ and $ p\not | m$. Then for for each $ p^{k}$ ($ 1\leq k\leq n$) there exists $ H\leq G$ with $ |H|=p^{k}$. And (second part of thereom) for each $ |H|=p^{k}$ ($ 1\leq k<n$) there exists $ \bar H\leq G$ with $ |\bar H| = p^{k+1}$ such that $ H\triangleleft \bar H$.

[b]Proof:[/b] We do this by induction. The induction step is Cauchy's theorem stated in the begining. We will show that $ p^{k}$ implies existence of $ p^{k+1}$ for all $ 1\leq k <n$. Let $ H$ be the subgroup of $ G$ of order $ p^{k}$. By Lemma we see that $ (G: H)\equiv (N_{G}[H]: H) (\bmod p)$. Now $ (G: H)\equiv 0 (\bmod p)$ because $ k<n$, so $ p$ divides $ (N_{G}[H]: H)$. Since $ H\triangleleft N_{G}[H]$ (basically by definition) we can form the factor group $ N_{G}[H]/H$ and its order is the index of $ H$ under $ N_{G}[H]$. Thus, $ p$ divides $ |N_{G}[H]/H|$ and by Cauchy's theorem again there exists a subgroup (of this factor group) $ K$ with order $ p$. Form the canonical homomorphism (see below) $ \gamma: N_{G}[H]\mapsto N_{G}[H]/H$, since $ K\leq N_{G}[H]/H$ we find that $ \gamma^{-1}[K]\leq N[H]$. But the order of $ \gamma^{-1}[K]$ is $ p^{k+1}$ because $ \gamma^{-1}[K] = \{x\in N_{G}[H] | \gamma(x) \in K\}$ so $ \gamma^{-1}[K]$ contains $ p$ sets of $ p^{k}$ elements, which in total gives $ p^{k+1}$ elements in $ \gamma^{-1}[K]$. This completes the first part. The second part is immediate, note that by construction $ H < \gamma^{-1}[K]\leq N_{G}[H]$. Since $ H\triangleleft N_{G}[H]$ then for certainly $ H\triangleleft \gamma^{-1}[K]$. Q.E.D.

It is easily seen by the First Theorem, that the largest possible $ p$-subgroup (meaning not contained in another $ p$-subgroup) is $ p^{n}$ where $ n$ is the largest dividing exponent of the order. These subgroups have a special name.

[b]Definition:[/b] Given $ G$ a Sylow $ p$-subgroup is the maximal $ p$-subgroup of $ G$.

The next supprising theorem shows how Sylow subgroups are related to eachother.

[b]Second Theorem:[/b] Let $ P_{1},P_{2}$ be Sylow $ p$-subgroups of finite group $ G$. Then $ P_{1}$ and $ P_{2}$ are conjugates, i.e. $ P_{1}= gP_{2}g^{-1}$ for some $ g\in G$.

[b]Proof:[/b] The trick here is to create $ G$-sets and use the starred theorem mentioned above. Let $ S$ be the set of all cosets of $ P_{1}$, and let $ P_{2}$ act on $ S$ as $ p_{2}(p_{1}P_{1})=(p_{2}p_{1})P_{1}$ for $ p_{2}\in P_{2}$ and $ p_{1}\in P_{1}$. This makes $ S$ into a $ P_{2}$-set. Hence by starred theorem (and Sylow's First Theorem) $ |S|\equiv |S_{P_{2}}| (\bmod p)$. But $ |S|=(G: P_{1})$ and is not divisible by $ p$. So $ |S_{P_{2}}|\not = 0$. Let $ xP_{1}\in S_{P_{2}}$ then $ yx P_{1}= xP_{1}$ for all $ y\in P_{2}$. Which tells us that $ x^{-1}yx\in P_{1}$ for all $ y\in P_{2}$ implies $ y\in P_{1}$ and hence $ x^{-1}P_{2}x\leq P_{1}$ but since $ |P_{2}|=|P_{1}|$ (and they are finite) must mean that $ x^{-1}P_{2}x = P_{1}$. Q.E.D.

The next theorem is extremely useful in showing that finite groups of certain orders are not simple.

[b]Third Theorem:[/b] Let $ G$ be a finite group and $ p$ divides $ |G|$ then the number of Sylow $ p$-subgroups is congruent to $ 1$ modulo $ p$.

[b]Proof:[/b] We use the typical $ G$-set approach to this problem. Let $ S$ be the set of all Sylow $ p$-subgroups. And let $ P$ be a Sylow $ p$-subgroup. We define the action of $ P$ on $ S$ as conjugation, this makes $ S$ a $ P$-set. So by the starred theorem $ |S|\equiv |S_{P}| (\bmod p)$. Now it remains to show that $ |S_{P}|=1$ and the theorem is complete. Let $ P_{1}\in S$ then $ xP_{1}x^{-1}= P_{1}$ for all $ x\in P$. Thus, $ P \leq N_{G}[P_{1}]$, and certainly $ P_{1}\leq N_{G}[P_{1}]$. Since $ P$ and $ P_{1}$ are Sylow $ p$-subgroups in $ G$ they are Sylow $ p$-subgroups in $ N_{G}[P_{1}]$, but they are conjugate in $ N_{G}[P_{1}]$ by Sylow's second theorem. But since $ P_{1}\triangleleft N_{G}[P_{1}]$ it is [i]invariant under conguation[/i], and so $ P = P_{1}$. Thus, $ |S|\equiv 1(\bmod p)$. Q.E.D.

Here is an important corollary that is part of this theorem.

[b]Corollary:[/b] Let $ G$ be a finite group divisible by $ p$ then the number of Sylow $ p$-subgroups divides $ G$.

[b]Proof:[/b] Here we use the theorem mentioned all the way on top about counting orbits. Let $ G$ act on $ S$ by conjugation. Since all Sylow $ p$-subgroups are conjugates there is only one orbit in $ S$ under $ G$. If $ P\in S$ and $ GP$ represents the orbit in $ S$ under $ P$ we have $ |S| = |GP| = (G: G_{P})$ by theorems about $ G$-sets. But $ (G: G_{P})$ is a divisor of $ |G|$ so $ |S|$ divides $ |G|$. Q.E.D.

[b]Canonical Homomorphism:[/b] In case you are not familar with this terminology the map $ \gamma: G\mapsto G/H$ defined in the natural way as $ \gamma (g) = gH$ is called the [i]canonical homomorphism[/i].

[quote="cuopbientcd"]Thanks you.Do you know books so i can learn more?[/quote]
I used Fraleigh. This is also similar to Hungerford.


inequalities
function
calculus
derivative
real analysis
real analysis unsolved

A very beautiful problem: find all real functions $g$ for which there are continuous functions $f$ such that $ \frac{f(x)+f(y)}{2}\geq f(\frac{x+y}{2})+|g(x)-g(y)|$.

Wow! Mark given:0000!!! Cezar, we post solutions here!!!

Ok,here it is: let $g$ be a function with the property given.Well,clearly $f$ is continuous and $f$ is semiconvex,we deduce that $f$ is convex.Then we have the existence of $f^'_+(x)$ and $f^'_-(x)$. Let $h$ and $x$ real number.
We take the inequality given for $x=x+h$ and $y=x-h$.We obtain $f(x+h)+f(x-h)\geq 2f(x)+2|g(x+h)-g(x-h)|$
 or $f(x+h)-f(x)\geq f(x)-f(x-h)+2|g(x+h)-g(x-h)|$ or $f(x+h)-f(x)/h\geq f(x)-f(x-h)/h+2|g(x+h)-g(x-h)/h|$.
By passing to limit when $h$ tends to zero we deduce $f^'_+(x)-f^'_-(x)\geq 4|g^'(x)|$.  Since $f$ is convex we have
f diferentiabile except  a set at most countable.Well,we have $0\geq 4|g^'(x)|$ and since $g^'$ has Darboux property the conclusion follows.

I'm sorry once again for bad Latex.I hope you do understand what I wrote.

Please edit your post. Cezar, do you really know to prove that a convex function is differentiable allmost everywhere? ;) And also, I did not know that a convex function is differentiable except a countable set. I knew it for sets of zero measure.

I just want to know if it is sure that a convex function is differentiable on the complementary of a set which is at most countable. I have some doubts about that... And I would also want to see the proofs of two of your statements: the one which refers to the fact that a convex function is differentiable of such a large set and the second one:  why can we assume that g is derivable? I have some guesses about this assumption you made. ;)

Excuse me,but I assumed from the beginning that $g$ is derivable.

I know that the first statement you've asked me is true.I found it in a article in R.M.T. long time ago,I don't
 remember what year.There is the proof.

[quote="harazi"]I just want to know if it is sure that a convex function is differentiable on the complementary of a set which is at most countable.[/quote]

I think this is true, but I'm reluctant to attempt a proof, since there are already 10 messages here but almost nothing regarding the solution to the problem which was actually posted (and Cezar, assuming that $g$ is differentiable even though nobody said it was does not count :)). Anyway.. :)

$f$ has left and right derivatives in every point, and both the right derivative and the left derivative are increasing functions, so the union of the sets of discontinuities of $f'_+$ and $f'_-$ is at most countable (denote it by $A$). Moreover, if $x<y$, then $f'_+(x)\le f'_-(y)$.

Let $A_n$ the set of points $x$ for which $f'_+(x)-f'_-(x)\ge\frac 1n$. It's easy to see, from the observations above, that $A_n\setminus A$ has no limit points in itself, so it's at most countable, meaning that $A_n$ is at most countable. The set of points where $f$ has no derivative is contained in $\bigcup_n A_n$, so we're done.

Think it's ok, harazi?

At first glance, I think it is ok. Thanks, grobber.

As for the problem itself, all we need is a slight modification of what Cezar did (it turns out he did almost all the work :), and we don't even need the differentiability of convex functions on a countably complemented set):

I'll replace $y$ with $x$ and $x$ with $x+2h,\ h>0$. We get $|g(x+2h)-g(x)|\le \frac{f(x+2h)-f(x)}2-(f(x+h)-f(x))$. We divide everything by $h$ to get $2\left|\frac{g(x+2h)-g(x)}{2h}\right|\le\frac{f(x+2h)-f(x)}{2h}-\frac{f(x+h)-f(x)}h\ (*)$.

As $h\to 0$ from above, the RHS in $(*)$ goes to $0$ because both the terms tend to $f'_+(x)$, so we find that $g$ has a right derivative in all points, and it's equal to $0$ everywhere. If we work with $h<0$ instead, we get the same thing, except that the order of the two terms of the RHS in $(*)$ is reversed. This makes no difference to the result: $g$ also has a left derivative equal to $0$ everywhere.

We have thus shown that $g$ is differentiable with null derivative, so it's constant.

thank you very much,grobber.I'm sure it is much clear now.


The quadrangle $ABCD$ is cyclic . $AC\cap BD\equiv E$.
Let $I_1,I_2,I_3,I_4$ be the incenter of 4 triangle $EAB,EBC,ECD,EDA$.
Prove that $I_1I_2I_3I_4$ cyclic <--> $AB+CD=AD+BC$

Since the condition AB + CD = AD + BC is clearly equivalent to the assertion that the quadrilateral ABCD has an incircle, your problem reduces to showing that the quadrilateral $I_1I_2I_3I_4$ is cyclic if and only if the quadrilateral ABCD has an incircle. This was proved on http://www.mathlinks.ro/Forum/viewtopic.php?t=21758 (see the PDF file attached to my post #2).

  Darij


algebra
polynomial
induction
group theory
abstract algebra
Galois Theory
superior algebra

Let [b]K[/b] be a some field, [b]f[/b] in [b]K[X][/b] -  some polynomial with degree [i]n[/i] and with coeffitients in [b]K[/b], and let [b]L[/b] be the splitting field of [b]K[X][/b]. Prove that [b][L:K][/b] [u]<[/u] [i]n[/i]!

Hmm, can we prove that when we adjoin a single root, the dimension is at most $ n$?

And then can we use induction? :)

Actually, we have $ [L: K]|n!$. Why? Because $ Gal(K/L)\subset S_n$. The group of automorphisms, when viewed as permuting the roots of $ f$, is a subgroup of $ S_n$ which has order $ n!$.

Something tells me that our friend has not gotten to Galois groups yet.

Though I like your way of looking at it, it's also possible to see that $ [L: K]|n!$ directly from the construction of the splitting field.

You can't use Galois groups as it's not necessary a Galois extension.
And see http://www.mathlinks.ro/Forum/viewtopic.php?t=31288 .

Is it not a Galois extension because we are not assuming $ char\equal{}0$?

Artin makes the assumption $ char\equal{}0$ and concludes that "$ K$ is the splitting field of a polynomial $ f(x)\in F[x]$ is equivalent to "$ K/F$ is Galois.

The poster said nothing about characteristics or separability. So there is no reason to assume this as the result is true nonetheless.
Artin's description is rather bad, he should at least talk about perfect fields. And in fact, a lot of authors define "Galois extensions" to be normal and seperable (and normal ones are easily characterised as comming from splititng fields of a set of polynomials).

Thanks. I guess I'll have to learn some Galois Theory from another source as well. Of course we cannot assume that $ char\equal{}0$.


If $ a,b,c$ are nonnegative numbers such that $ a^{2}\plus{}b^{2}\plus{}c^{2}\equal{} 3$, then

$ \frac{1}{1\plus{}a^{5}}\plus{}\frac{1}{1\plus{}b^{5}}\plus{}\frac{1}{1\plus{}c^{5}}\ge\frac{3}{2}$.

In my regard, there are solutions without using RCF-Theorem.

[quote="hitek"]Given $ a,b,c,d$ such that $ a^{2}\plus{}b^{2}\plus{}c^{2}\plus{}d^{2}\equal{} 4$, prove that
\[ \frac{1}{1\plus{}a^{5}}\plus{}\frac{1}{1\plus{}b^{5}}\plus{}\frac{1}{1\plus{}c^{5}}\plus{}\frac{1}{1\plus{}d^{5}}\geq2.\]
[/quote]

It is wrong.  :P Try a=0 and b=c=d.

if 2 refered by 4 i can solve it with elementary methode. :)


algebra
polynomial
calculus
integration
pigeonhole principle
superior algebra
superior algebra unsolved

hi all,
I've just seen a wonderfull solution to the following problem:
find all rational r so that cos(r*Pi) is a rational. The solution uses field extension but is really amazing. Could you find it ??

bye.

Here's the first thing that came to my mind. I know there are lots of other more beautiful solutions, but this will do for now :).

Let $z=\cos(r\pi)+i\sin(r\pi),\ t=\cos(r\pi)$. Assume we have eliminated the cases $z=\pm 1$. We have $P(z)=z^2-2tz+1=0$. There is some $n$ s.t. $z^n=1$, so $P(x)|\prod_{d|n}\phi_d(x)$, where $\phi_d$ is the $d$'th cyclotomic polynomial. However, the cyclotomic polynomials are irreducible, so $P$ must actually be a cyclotomic polynomial, so it has integral coefficients, and in this case we have $2t\in\{0,\pm 1\,\pm 2\}$ (this actually includes the cases we have eliminated, $z\in\{\pm 1\}$).

We have found $t=\cos(r\pi)\in\{0,\pm\frac 12,\pm 1\}$.

You can prove it without field extensions too. Hint in spoiler: [hide]Start by showing that if  is rational, then so is . Then use the pigeonhole principle.[/hide]

Here's a faster way to do what I did above:

$z,\bar z$ both satisfy the equation $x^n-1=0$, so they're algebraic integers. This means that $2\cos(r\pi)=z+\bar z$ is also an algebraic integer, but since it's rational, it must be integral, so $2\cos(r\pi)\in\{0,\pm1,\pm 2\}$, and that's that.

hi grobber,

do you have a natural proof to show that the cyclotomic polynomials are irreducible over Q. I don't have a natural solution ..

thanx

I don't think there are any solutions which don't use higher algebra. However, we don't need that for this problem, as you can see from my previous post.

You can use my hint to find a solution that only uses number theory and no abstract algebra.

Alek

There is another proof in R.M.S Revue de Mathmatiques Spciales
Mai 2004 page 89


Hi, folks!

Instead of doing a billion replies to all of the threads, I figured I would kill several birds with one e-mail (*squawk!* *thud*)

1.  The top 8 teams in Division A and the top 6 teams in Division B will be receiving a TI-Nspire calculator as well as a copy of Mathematica, along with something else (either a book or software prize, or Tangoes).   You should have received the certificate for Mathematica from your coach, and the calculators and other prizes will be mailed.  Incidentally, if you already have Mathematica from a previous year, there's nothing saying you can't pass on the certificate to another one of your friends or fellow teammates.  Site winners should have received their books at the contest.

2.  I thought the super relay was going to be a single submission thing, too, but Tom said folks could submit as many answers as they wanted right before the event, and I figured it wasn't worth it to correct him.  Next year, we'll make sure the final answer is not easily guessable.

3.  Apologies to the folks who had to do extra tiebreakers (unnecessarily).  The tiebreaker procedure should have been one correct answer, then done.  For the 7s, the first person at each site to get the tiebreaker correct should have been the site winner (since, conveniently, we had exactly 10 8s from the US).

I think that covers most of the woes submitted so far.
Paul

[quote="Kent Merryfield"]I suppose WWTBAM was a one-time deal? But we'll have something planned for Friday evening after the team and power rounds next year, right?[/quote]

I'm not sure if WWTBAM is coming back next year.  They might go to another site, for all I know.  I'm going to follow up with them about it.  However, yes, we'll definitely have something next year, whether it is a speaker, a song contest, or something else.  From all accounts, Cynthia was well-received as a speaker for HCSSiM, so we may continue to have that in the future.  Having both the WWTBAM practices and the speaker in the afternoon ended up being a bit difficult for the WWTBAM students who wanted to get dinner prior to the team round, but I really wanted to have both things happen this year.  Next year will (hopefully) be a bit less time constrained.


geometry
geometric transformation
reflection
geometry proposed

Let $ \triangle ABC$ , $ AB\perp AC$ . For a point $ D\in (AC)$ denote the reflection $ E$ of the point $ A$ w.r.t. $ AD$

and the point $ F\in CE$ for which $ DF\perp BC$ . Prove that the lines $ AF$ , $ DE$ , $ BC$ are concurrently.

See and [url=http://www.mateforum.ro/viewtopic.php?t=1639][color=red][b]here[/b][/color][/url] .

See also here: [url]http://www.mathlinks.ro/viewtopic.php?p=860114#p860114[/url].


score:38 rank:1
score:33 rank:2
score:31 rank:3
score:30 rank:4
score:29 rank:8
score:28 rank:9
score:27 rank:11
score:26 rank:13
score:25 rank:18
score:24 rank:20
score:23 rank:22
score:22 rank:26
score:21 rank:31
score:20 rank:36
score:19 rank:41
score:18 rank:44
score:17 rank:51
score:16 rank:55
score:15 rank:59
score:14 rank:67
score:13 rank:81
score:12 rank:102
score:11 rank:116
score:10 rank:125
score:9 rank:140
score:8 rank:166
score:7 rank:197
score:6 rank:221
score:5 rank:237
score:4 rank:248
score:3 rank:299
score:2 rank:347
score:1 rank:441
score:0 rank:461

[quote="funtwo"]the optimistic side of me says i beat 60% of people
the pessimistic side says 40% of people beat me
the prevailing side is too tired to care right now[/quote]

Correction: of the people [b]who actually qualified for USAMO[/b].

[quote="daine"]Getting a 0 cannot be 6th percentile. 6th percentile means that you did BETTER than 6% of all the people.[/quote]
I agree. It probably means 6% of USAMO qualifiers score 0

I think Putnam-style means that it puts you smack in the middle of the percentile range of that score, i.e. for someone score of 0, 0% are below that person, but 88% are above that person, meaning that person can be said to be anywhere from the 0th to 12th percentile, so he or she is given a percentile of 6, even though 12% got a 0.

$ n$th percentile means you beat $ n\%$ of all the people.

If it bothers you that much, you can find the actual percentile by looking at the entry in the next row. For example, 1 should be in the 6th percentile.
Edit: Ooh, never mind.

The Putnam style is to average the ranks, and I converted that average into the percentile. The way to look at it is that the zeros were the lowest 12% of the scores; the average of that is 6%.

i have not seen evan o'dorney name, he had a silver medal the other year

He doesn't appear in the winner/HM list, so all we know is he fell a bit below 23. No one knows how he did, until he posts it somewhere (unlikely). Asking repeatedly isn't going to get the answer faster.

I did some calculations, which MAY have errors.  But I think they're right:

Average Score of a USAMO Participant: 6.95 (effectively, 7)
Converted to Percents: 16.56%

You have a small arithmetic error. The mean is precisely 7 - which is also what Sly Si said on post #126 in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=274335&start=120]this thread.[/url]


1. If x^2 +3<4 and 2x^2+3>4, what is one possible value of x?
I just solved the two inequalities. first one, I got -1<x<1. Second one, I got -radical 1/2 <x< radical 1/2.
Then combined the two, I got: -radical 1/2<x<radical 1/2, but it's wrong. The answer was radical 1/2<x<1. BUt it doesn't make sense.

Thanks for help.

[quote="MIA6"]Second one, I got -radical 1/2 <x< radical 1/2.[/quote]

This isn't correct.

The correct form is

$ x>\sqrt{\frac{1}{2}}$ or $ x<\minus{}\sqrt{\frac{1}{2}}$

Use number-lines if you are not sure.

[quote="iniquitus"][quote="MIA6"]Second one, I got -radical 1/2 <x< radical 1/2.[/quote]

This isn't correct.

The correct form is

$ x > \sqrt {\frac {1}{2}}$ or $ x < \minus{} \sqrt {\frac {1}{2}}$

Use number-lines if you are not sure.[/quote]

But the thing is if  1>x>-1, x>0.7, or x<-0.7, when you combine them, it's still x>0.7 and x<-0.7.  I am still confused, I mean you have to combine two inequalities. Can you explain it to me?

Indeed, good job!  :lol: . Don't trust the book's answer. Use common sense.. try plugging in -0.8.

The correct answer is $ \minus{}1<x<\minus{}\sqrt{\frac{1}{2}}$ or $ 1>x>\sqrt{\frac{1}{2}}$


The next problem is like this:

There is a cylinder on the table.The length of this cylinder is L=44cm.There is a wall in the middle of this cylinder.Left part of this cylinder contains $O_{2}$ and the right part contains $N_{2}$ gas.All processes are isothermal in this problem.We take out the wall between these gases.What will be the displacement of this cylinder and to which side?Neglect all frictions.Molar masses of gases are as followings:$M{}_{O}{}_{2}=32g/mol$ $M{}_{N}{}_{2}=28g/mol$

Um, don't you need one additional fact, ie. the ratio of the pressures of oxigen and nitrogen, or the ratio of the respective mole numbers?

if we consider  the wall is free than the ratio of pressure is 1 and the cylinder will move L/30=1.46 cm to the right

Somehow, the term 'wall' suggests that its not movable :) But ok, supposing this than its trivial, you just equate the positions of the center of mass...

Djole,yes you are right.The ratio of pressures of gases is 2.But I dont remember exactly $N_{2}$ over $O_{2}$ or $O_{2}$ over $N_{2}$.If you can please answer for both situations.I'm waiting.

:roll: ...well 3 cases never killed somebody.i'll just add corresponding pressures to my already made calculations.
is Po=2Pn then the cylinder mooves 4.3 cm to the right
is Pn=2Po then the cylinder mooves 3 cm to the left

The center of mass doesn't move, so
\[d = \frac{m_{N}L/4-m_{O}L/4}{m_{N}+m_{O}}. \]
From the gas laws we have
\[\frac{m_{O}}{m_{N}}\frac{M_{N}}{M_{O}}= \frac{1}{2}, \]
so
\[d = \frac{L}{4}\frac{1-\frac{m_{O}}{m_{N}}}{1+\frac{m_{O}}{m_{N}}}= \frac{L}{4}\frac{1-\frac{M_{O}}{2M_{N}}}{1+\frac{M_{O}}{2M_{N}}}. \]
Now you can simplify and calculate...

Btw, what level of competition is this? Is it the final, where you choose the olympic team, or not?

Djole and radac_mail,both of your answers are correct.
We had some problems with our jury.They said that the answer is 6cm.But I and one of my friends had the same answers.Now we will prove to our teacher.It was National Olympiad of my country.Team selection exams will be in may.But I got Silver Medal in the last IPHO that's why I was selected to team without team selection exam.
I wanted to know what are your occupations.(Djole , Radac_mail)

I won a silver medal in Singapore as well, but I still have to go through the entire competition cycle :( I'm now in my senior (fourth) grade in high school.

Any other problems?


[quote="Leon"]Suppose $ A_1,\dots, A_6$ are six sets each with four elements and $ B_1,\dots,B_n$ are $ n$ sets each with two elements, Let $ S \equal{} A_1 \cup A_2 \cup \cdots \cup A_6 \equal{} B_1 \cup \cdots \cup B_n$. Given that each elements of $ S$ belogs to exactly four of the $ A$'s and to exactly three of the $ B$'s, find $ n$.[/quote]

$ |S|\equal{}\frac{6|A|}{4}\equal{}6\equal{}\frac{n|B|}{3}\equal{}\frac{2n}{3}$ and so $ n\equal{}9$

[quote="gouthamphilomath"][b][u]Generalistation:[/u][/b]

Let $(A_1)_1, (A_1)_2, (A_1)_3, \cdots, (A_1)_{m_1}$  be $m_1$ sets each with $a_1$ elements.
Let $(A_2)_1, (A_2)_2, (A_2)_3, \cdots, (A_2)_{m_2}$ be $m_2$ sets each with $a_2$ elements.
and define analogously for $(A_i)_1, \cdots, (A_i)_{m_i}$ to be $m_i$ sets each with $a_i$ elements.
And let these sets be defined till $i=n$
Let $S=\bigcup_{q=1}^{m_k} (A_k)_q$ for all $k\in\{1, 2, \cdots, n\}$
Suppose each element of $S$ occurs in exactly $k_i$ sets of $(A_i)_1, \cdots, (A_i)_{m_i}$ for all $1\le i\le n$

Then, prove that there exists an arbitrary integer constant $\lambda$ such that $\frac{m_ia_i}{k_i}=\lambda\forall i\in\{1, 2, \cdots, n\}$

[u][b]Proof:[/b][/u]

Suppose we just list out all the elements of $(A_i)_k$ not caring about repetition where $1\le k\le m_i$ for some fixed $i$, then we have $m_ia_i$ elements in all. Out of these, $k_i$ elements are repeating.

So, the cardinality of $S$ is $\frac{m_ia_i}{k_i}\forall 1\le i\le n$

Now, if we choose $\lambda=|S|,$ we have $\frac{m_ia_i}{k_i}=\lambda\forall i\in\{1, 2, \cdots, n\}$ as required.

[u][b]Application:[/b][/u]
In this problem, we have $n=2, m_1=6, m_2=x$(replacing $n$ by $x$), $a_1=4, a_2=2, k_1=4$ and $k_2=3$

So, $\lambda=\frac{6\times 4}{4}=\frac{x\times 2}{3}\Longrightarrow x=9$ as required.[/quote]

from where did u get this proof, gouthamphilomath .Is it valid for all values  $A_i$ in any case , only for $A_i\cup$ or
intersections of $A_i\cap$ :help:
$\frac{m_ia_i}{k_i}=\lambda\forall i\in\{1, 2, \cdots, n\}$
what is meant by $a_i$ , is it magnitude of the given $A_i$ set

[quote="siddharthanand"]

from where did u get this proof, gouthamphilomath?[/quote]

I got it on my own.

[quote="siddharthanand"]

Is it valid for all values  $A_i$ in any case , only for $A_i\cup$ or
intersections of $A_i\cap$t[/quote]

It works only for unions, I didn't still work out for intersections.

[quote="siddharthanand"]
what is meant by $a_i$ , is it magnitude of the given $A_i$ set[/quote]

$a_i$ is the cardinality of $(A_i)_q$ for all $q=1,2, \cdots m_i$


If the latitude of Chicago is about 42 degrees and the radius of the earth is about 4000 miles, how far is Chicago from the equator? (use pi= 3.1416)

[hide]The radius is 4000, so the circumference (I know it's a sphere) would be $8000\pi$.  You need only $\frac{42}{360}=\frac{7}{60}$ of the circumference, which is $2932.151$ miles.[/hide]

It's probably wrong.

[hide]Let the radius be 4000 mi and assume that the earth is a perfect circle
If Chicago is 42 degrees above the equater, then out of 360 degrees Chicago is .11 2/3 of the way around the circumference of the earth.  Since C = 2pir, the distance would be:
d = 2(3.1416)(4000mi)(.11 2/3)
d = 2932.16 mi
I hope that is correct...[/hide]


limit
linear algebra
matrix
real analysis
superior algebra
superior algebra unsolved

Let $ v$ be a discrete valuation. Let $ (a_n)$ be a cauchy sequence wrt $ v$. Show that $ \lim_n v(a_n)$ exists.

Thanks.

$ v$ is a real-valued uniformly continuous function. The image of a Cauchy sequence under a uniformly continuous map is a Cauchy sequence. A Cauchy sequence in $ \mathbb{R}$ has a limit.

This argument has almost nothing to do with algebra; it works in any space with a norm-type matric, and the only place the underlying algebraic structure is involved is in noting that the metric $ d(a,b)\equal{}v(a\minus{}b)$ makes sense.

I thought of something on the same line. I was worried about the fact that $ v$ is a map from a field ONTO $ \mathbb{Z}$. And I didn't know that $ v$ is also uniformly continuous.

Stronger; it's Lipschitz with constant 1, which lets you explicitly use the same $ \epsilon$ at the same $ n$ in the image.
What does discreteness add? This sequence of $ v$ values, which has a limit, is eventually constant.

On second thought, we don't really want the integer-valued version of $ v$. We really want $ v(0)\equal{}0$, and if it's not, the case in which $ a_n$ converges to zero involves $ v(a_n)$ converging to $ \minus{}\infty$.

I don't see how $ d(a,b) \equal{} v(a \minus{} b)$ makes sense. The only metric I know induces by $ v$ is $ d(a,b) \equal{} c^{v(a \minus{} b)}$, where $ 0 < c < 1$.

And I don't see what constant $ K$ would work for

$ |v(a) \minus{} v(b)| \leq Kd(a,b) \equal{} Kc^{v(a\minus{}b)}$.

Sorry for being a bit slow, but I don't quite follow with some of your statements.

Everything I said was about the metric, and I was calling the metric $ v$. That's a conflict of notation with what you were used to. Saying that it's a Cauchy sequence "with respect to $ v$" only makes sense if $ v$ is a metric, so I took that as the definition.

There is no Lipschitz relation between the integer-valued and metric versions of what we're looking at. That's merely a continuous map.


Hey I was just wondering if anyone had a spare Drivers & Utilities CD for a Dell Dimension 4600 (the cd ships with the system). I think the same cd ships with all 4600's. If not, please tell me. I'm asking because I plan on redoing my XP computer - and I have all the discs I need and have backed up my system on second hard drive, but I am only missing this. I don't think this cd is illegal to transfer online as the individual downloads of drivers is freely available on the site (but the site confuses me). Does anyone have the CD and can send it to me (image of the cd (.iso format for example, among others)).

Thanks.

Hm, I have a CD for my Dell Inspiron but I would (would = I've actually done this several times) download all the drivers from Dell's website anyway, they're probably more up to date and you don't have to instal any additional Dell software. Regarding the website, I think it's quite straight forward - just go to http://support.dell.com/support/downloads/index.aspx?~ck=mn and choose the right computer / operating system.


calculus
integration
trigonometry
inequalities
calculus computations

[hide="Answer"]$ I \equal{} \int_0^{2 \pi}e^{\cos x} \cos(\sin x) \cos x \,dx \equal{} \int_0^{2 \pi} \cos x \cdot \sum_{n \equal{} 0}^{\infty} \frac {\cos(nx)}{n!}$

$ \equal{} \int_0^{2 \pi} \left(\sum_{n \equal{} 0}^{\infty} \frac {\cos(nx) \cos x}{n!}\right)\,dx \equal{} \sum_{n \equal{} 0}^{\infty} \frac {1}{n!} \left(\int_0^{2 \pi} \cos(nx) \cos x \,dx\right)$

$ \equal{} \sum_{n \equal{} 0}^{\infty} \frac {1}{n!} \left(\frac {1}{2} \int_0^{2 \pi} [\cos[(n \plus{} 1)x] \plus{} \cos[(n \minus{} 1)x]]\,dx\right)$.

$ \equal{} 1 \cdot \left(0 \plus{} \frac{1}{2}\int_0^{2 \pi} \cos(0)\,dx\right) \equal{} \pi$.[/hide]

I think the result should be $ \pi$.
$ \cos[(n\minus{}1)x]\equal{}1$ for $ n\equal{}1$.

Yes, that's right, thanks. I forgot about the case n = 1 (one more time).

Why does [quote="Carcul"]$ \int_0^{2 \pi}e^{\cos x} \cos(\sin x) \cos x \,dx \equal{} \int_0^{2 \pi} \cos x \cdot \sum_{n \equal{} 0}^{\infty} \frac {\cos(nx)}{n!}$
[/quote]
Sorry if I ask a silly question

[hide="Because"]$ \sum_{n\equal{}0}^{\infty} \frac{\cos(nx)}{n!} \equal{} e^{\cos x} \cos(\sin x)$[/hide]

$ \sum_0^\infty \frac {e^{inx}}{n!} \equal{} \exp(e^{ix}) \equal{} e^{\cos x}(\cos(\sin x) \plus{} i \sin(\sin x))$

I doubt I could have seen this approach personally. I am not good at hard integrals

That's right: now take the real part of it and you get the sum above.

I tried proving it but I'm getting an incorrect result. Can anyone tell me what's wrong with this:

$ 0 \le \int\limits_0^{2\pi } {e^{\cos x} \cos (\sin x)\cos xdx}  \le e\int\limits_0^{2\pi } {\cos (\sin x)\cos xdx}  \equal{} e*0 \equal{} 0$
$ \to \int\limits_0^{2\pi } {e^{\cos x} \cos (\sin x)\cos xdx \equal{} 0}$

That inequality only applies to absolute values. The integrand is not always positive.


For $ a,b,c,d \ge 0$, show that
\[ a(a\minus{}b)(a\minus{}c)(a\minus{}d)\plus{}b(b\minus{}c)(b\minus{}d)(b\minus{}a)\plus{}c(c\minus{}d)(c\minus{}a)(c\minus{}b)\plus{}d(d\minus{}a)(d\minus{}b)(d\minus{}c)\plus{}abcd \ge 0
\]
:)

The inequality is symmetric so we may assume that $a\le b\le c\le d$. Thus there exist nonnegative $x, y, z$ such that $b=a+x, c=a+x+y, d=a+x+y+z$. Then the inequality is equivalent to the following which is obvious:
$$a^4+(3x+2y+z)a^3+(3x^2+y^2+4xy+2zx+yz)a^2+$$
$$+(x(x+y)(x+y+z)+xy^2+xyz+xz^2+z^3+2yz^2)a+$$
$$+x^2y^2+x^2yz+3x^2z^2+y^2z^2+3yz^3+z^4+4xyz^2+2xz^3\ge 0.$$


OK,

Whenever I make a title, name, etc... it shows all that information right in the middle of the page. Is there a way to make it so it's at the top of the page?

I can't seem to find this information in the LaTeX section of artofproblemsolving...

so can someone help me?

I'm not exactly sure what you mean, but I assume that you're talking about the output from the \maketitle command.  Basically, its effects are determined by whatever documentclass you're using, for example if you have \documentclass{article} at the top of your document, the article class defines the effects of \maketitle.  You can't directly change it short of redefining the \maketitle command.

One option is to just do that stuff by hand.  For example, instead of using \maketitle, just put something like

\begin{center}
\Large Math is Fun! \\[0.5in]
\large David Patrick \\[0.25in]
September 13, 2005
\end{center}

at the top of your document.  Alternatively, I like the way that the amsart documentclass does titles, so change the top line of your document to \documentclass{amsart} and see what it looks like.


I'm back, after a long time... I'll try to be more active, but I can't make any promises  :maybe:

So, I'm dealing with this problem:

A small, lone (as in isolated from any other object) sphere is positioned on a large distance $d$ from the Sun. If the temperature of the surface of the Sun is $T_{0}$ and its radius is $R$, find the equilibrium temperature of the sphere, considering it is the same in every point of the sphere. Both the Sun and the sphere radiate as blackbodies and heat is lost (or gained) solely through radiation.

I have two ways to solve this one and they lead to different solutions. How would you handle this?

Also on this subject, does anybody know how to prove the relation between the emission power $E$ and the energy density (energy per unit volume) $u$ of a radiating blackbody:
\[E = \frac{1}{ 4}cu, \]
where $c$ is the speed of light. I don't know where the 4 comes from...

this seems to be the simplest way:

equilibrium implies that the intensity of radiation received by the planet is equal to the intensity of radiation emitted by the planet.  stefan's law tells us $I_{0}= \sigma T_{0}^{4}$.  at a distance $d-R$ from the surface of the sun, energy conservation implies that, at the surface of the planet, $I = \frac{I_{0}R^{2}}{(d-R)^{2}}$.  Setting this equal to the stefan's law equation for the planet, we have $\sigma T^{4}= \frac{I_{0}R^{2}}{(d-R)^{2}}$ or $T = T_{0}\sqrt{\frac{R}{d-R}}$.

as for where the 4 comes from, I do not know, but I'll hunt around.

The equation of radiation black body can be proved by statistic mechanics, which always gives us strange coefficient( such as 1/4)
other question is from the mean free path equation.
Why is the term $\sqrt{2}$ added to get the true eQuation?
it is just a little trick in side. :lol:

long time ,did'nt see you Djole .
Long time Disappear in this forum. :P

I'm afraid you've made a mistake, [b]iostream.h[/b]. You equated the total energy emitted from the Sun with the energy emitted by the sphere; the problem is that not all of the Sun's radiated energy falls on the sphere.

In addition, I can't see why is $I = I_{0}R^{2}/(d-R)^{2}$. I would've expected $I = I_{0}R^{2}/d^{2}$, by comparing areas of two spheres (you know what I mean...) This point is not very relevant as $d\gg R \Rightarrow d-R \approx d$, though.

What I constantly get wrong is the part of the Sun's radiation which falls on the sphere  :mad:

@[b]cuong[/b]: Yeah, I've been busy ever since IPhO, applying to universities and such... I've started training for the next IPhO, I hope I'll be there!

the prob
Refer IPHO problem 1992 Findland, it is harder than yours


logarithms
calculus
complex analysis
calculus computations

Hello friends,
plz differentiate: d(log(y))/dx  :!:

Hmm indeed. :D
Shouldnt it be ?
\[\frac{d(log^{y}_{a})}{dx}= \frac{y'}{(\ln{a}){y}}\]
hammoda:
A simple google seach can help you much more.
And also this section is for [i]Complex Analysis[/i]. Calculus problems should be posted here http://www.mathlinks.ro/Forum/index.php?f=296


modular arithmetic
number theory
relatively prime
number theory unsolved

For which integers k is 32k^4 - 8r^2 + 1 a perfect square?

Yes, it should be 32k^4 - 8k^2 + 1. Sorry about that.

I recognise a problem from Bulgarian MO 2003, is that right, Cauchyguy?

Is this problem really Bulgaria MO 2003? Actually, I have been trying to use Pell's equation, but I haven't been successful? Can anyone help me?

Well, tt's Bulgaria 2003/3.
I also tried to solve this equation then, by using Pell's equations but after one day I got nothing!
Maybe this equation is just unsolvable!
As I remember the problem, it involved a linear reccurence whose terms were all perfect squares.
this expression is just x_2 or x_3.
I remember that I took x_4 or x_5 which was much easier to solve then your problem, and I was told this was the author's solution too..

Yes, iura, I think I know which problem you are talking about. That is the very problem I was working on when I came up with this. However, I did not know it was a Bulgarian problem. Thank you for the tip on considering x_4 or x_5. Still, I would like to know if there is a solution for the problem I posted.

[quote="kennyshovalto"]Actually, it's not that hard!
Suppose that 32k^4 - 8k^2 + 1 = r² with r is an integer then:
64k^4 – 16k² +2 = 2r²
=>(8k² - 1) ² + 1 = 2r².
This is kind of Fermat equation so:
8k² - 1 = m² - n² + 2mn;    (1)
1 = n² - m²  + 2mn;           (2)
r = n² + m².                        (3)
With m, n are integers; m, n are relative primes; and one of them is even, one is odd.By solving (2), we got m, n and then can count k and r
 :ddr:[/quote]

For example $ m\equal{}12$ and $ n\equal{}5$ is a solution of $ (2)$ but then $ (1)$ becomes $ 8k^2\minus{}1\equal{}239$ and so $ k^2\equal{}30$ oooops  :maybe:

So you just transformed the problem, not solved it.
Sorry.

Here is my solution. It's so long. but I think, I finally solved it (6 roots).
U can download my solution here:
[url]http://www.mediafire.com/?mcy02dhnf4w[/url] (.prc)
or pdf:
[url]http://www.mediafire.com/?awwmmizhjlf[/url]

In your second case (the nontrivial one), the problem is more or less equivalent to finding the solutions to $y^2=8x^4+1$. This means that $y=2k+1$, so
\[x^4=\frac{k(k+1)}{2}.\]Depending on the parity of $k$, we either have $k=m^4$ or $k=2m^4$, yielding $2r^4=m^4+1$ and $r^4=2m^4+1$ respectively. In the second case, we have
\[4\left(\frac{m}{2}\right)^4=\left(\frac{r-1}{2}\right)\left(\frac{r+1}{2}\right)\left(\frac{r^2+1}{2}\right).\]Since $r$ is odd, $r^2+1\equiv2\pmod4$, and $2=\gcd(r-1,r+1)$. The three RHS factors are clearly pairwise coprime, so switching the sign of $r$ if necessary, WLOG take $(r-1)/2=4u^4$ and $(r+1)/2=v^4$ so that $1=v^4-4u^4$, a contradiction (in fact, it's well known that $x^4-y^4=2z^2$ has no solutions; this can be proven by infinite descent easily).

In the first case, it suffices to solve the more general equation $m^4+n^4=2t^2$ with pairwise relatively prime $m,n,t$. There must exist $u,v$, then, such that $m^2=u^2-v^2+2uv$ and $n^2=v^2-u^2+2uv$. Since $2|m-n$, we can take $m=a+b$ and $n=a-b$ to get $a^2+b^2=2uv$ and $2ab=u^2-v^2$. From here, we can use an infinite descent argument to get that $m=n$ or note that
\[(a^2+b^2)^2+(2ab)^2=(2uv)^2+(u^2-v^2)^2=(u^2+v^2)^2\]while
\[(a^2+b^2)^2-(2ab)^2=(a^2-b^2)^2,\]meaning that $2ab=0$ (the proof of this fact also follows easily by infinite descent).


Browsing through the LACC competition's website( http://www.lacitycollege.edu/academic/departments/math/mathdept/contest/ ), I decided to check their previous problems/solutions.

I am confused with one of the solution:
http://www.lacitycollege.edu/academic/departments/math/mathdept/contest/samplequestions/2005prob4.html

It basically says that in the equation log (base 2x-1) (x^3 + 3x^2 - 13x + 10) = 2, you solve for x and get -3, 1, and 3. But x cannot be 1 nor -3 because that will make the base (2x-1) be -7 (if x is -3) or 1 (if x is 1).

However, I plugged in x as -3, and I get log (base -7) (49) = 2, and when I plug in x as 1, I get log (base 1) (1) = 2. Aren't these statements true; doesnt (-7)^2 = 49 and 1^2 = 1? Why did they exclude solutions x = -3 and x = 1?

[quote="p4fn2w"]It basically says that in the equation log (base 2x-1) (x^3 + 3x^2 - 13x + 10) = 2, you solve for x and get -3, 1, and 3. But x cannot be 1 nor -3 because that will make the base (2x-1) be -7 (if x is -3) or 1 (if x is 1).

However, I plugged in x as -3, and I get log (base -7) (49) = 2, and when I plug in x as 1, I get log (base 1) (1) = 2. Aren't these statements true; doesnt (-7)^2 = 49 and 1^2 = 1? Why did they exclude solutions x = -3 and x = 1?[/quote]
$\log_{1}a$ is not well defined for any number $a$ :  if $a=1$, then 1 raised to any power equals $a$; if $a\neq 1$, then you cannot raise 1 to any power and get $a$.

At first, $\log_{-7}x$ does indeed seem to be working smoothly.  But if you tried setting $x$ to some number that is not a perfect square power of 7, you would have a non-real answer, which is problematic, because when you stop restricting yourself to real numbers, logs are no longer well-defined.  (The reason for this is that $w^{z}= w^{z+2\pi i}$, for any complex numbers $w, z$.)

Furthermore, having a negative number or the number 1 as the base of the log ruins the convenient identity/definition $\log_{a}b = \frac{\ln a}{\ln b}$, as the natural logarithm is not well defined for negative numbers and $\ln 1 = 0$, which accurately encodes the difficulties we encounter.

[quote="Boy Soprano II"]Furthermore, having a negative number or the number 1 as the base of the log ruins the convenient identity/definition $\log_{a}b = \frac{\ln a}{\ln b}$, as the natural logarithm is not well defined for negative numbers and $\ln 1 = 0$, which accurately encodes the difficulties we encounter.[/quote]

Ah. I see... That makes sense. I didn't realize such restrictions exists for logarithms. Thanks for clearing it up.

[quote="MathChauffeur"][quote="p4fn2w"]Browsing through the LACC competition's website ... [/quote] It's on this Saturday! Are you going?  :)[/quote]
Yes I am. However, some of the questions seem tough to me, so I dont expect to do too well...

[quote="p4fn2w"]Yes I am. However, some of the questions seem tough to me, so I dont expect to do too well...[/quote]
Don't worry, solving only 3 of the 10 problems fully will likely be a top 20 score already! They really do not expect any participants to solve all (or even half) of the problems in 2 hours. Besides, friendly host and, free food!  :D

Try to talk a couple more of your schoolmates to go with you so you can have a full school score. Which school are you from (only if you don't mind disclosing)?

[quote="MathChauffeur"] Which school are you from (only if you don't mind disclosing)?[/quote]
Mark Keppel. But thats all I'm going to say, though.  :)

[quote="p4fn2w"]Mark Keppel. But thats all I'm going to say, though.  :)[/quote]
Fair enough.  :wink:

By the way, do you know students from other schools to whom you might want to introduce LACC? The organizers love to see more new faces.

[quote="MathChauffeur"]By the way, do you know students from other schools to whom you might want to introduce LACC? The organizers love to see more new faces.[/quote]
None from other schools, no, but that is because I dont know too many people outside of my school.

I can try talking to a few people that *might* be interested within my own school. Not sure how they will respond, so I cant really promise anything...

[quote="p4fn2w"]... I can try talking to a few people that *might* be interested within my own school ...[/quote]
They will surely appreciate it.  :)


vector
combinatorial geometry
combinatorics proposed
combinatorics

1) Suppose that there are $ n$ identical balls on plane. At time $ t\equal{}0$, each ball moves in an arbitrary direction (with the same speed), and balls that collide will obey Newton rules. Prove that after a finite time there will be no collision.
2) Prove the same thing for balls in space

Please, Omid Hatami, post the solution for this nice question!

Hi (1st post).

If the balls have same speed and are all identical, can you tell the difference between a collision, and balls that simply cross without contact ?

The same question might be more difficult if speeds are different. Does someone know an answer ?

Let us have points $ p_1,p_2,\dots,p_n$ spread in two dimensional space. We associate a vector (a arrow like line segment) with each point. Since the speeds are same , the vectors just provide the information of angle made by a vector (corresponding to every point) with respect to a co ordinate system.By writing to two conservation equations for momentum and two equations for restitution we can prove that collision of two points (point masses) can be treated as two particles just passing through each other without altering their speeds. Two vectors corresponding to two different points either are parallel(hence don't meet), not parallel but do not meet, or meet. Let $ i_1,i_2,\dots,i_r$ be the intersection points of the vectors that meet where $ r\leq {n\choose 2}$. We can find a convex hull of these points. Let $ d$ be the maximum among the lengths of the vectors (a vector starts at a point $ p_k$ and ends at $ i_l$). There would be no collisions after $ \frac{d}{s}$ where $ s$ is the speed of every point.

We can extend the same argument to $ \mathbf{R^n}$ where we would need to solve $ 2n$ equations.

@sclormu: In a general case where the speed of every point is different, we may not have the same result.

Instead of colliding, just imagine te balls passing through each other (it really doesn't matter).

Then each ball can only intersect every other ball at least once, and if tt will happen, it will happen in a finite amount of time.

After the last possible collision, there can be no more. Yay.

Edit: NVM, I thought that they were points, lol.


[quote="hzbrl"]ABCD be any quadrilateral. O is the point of intersection of the two diagonals AC and BD.
P.Q,R,S are the circumcentres of triangle OAB,OBC,OCD,ODA.
Prove that PQRS is a parallelogram[/quote][hide][asy]pair A=(0,0), B=(0,10), C=(10,13), D=(8,0), O, P, Q, R, S; draw(A--B--C--D--cycle); O=intersectionpoint(A--C,B--D); P=circumcenter(O, A, B); Q=circumcenter(O, B, C); R=circumcenter(O, C, D); S=circumcenter(O, D, A); draw(A--C..B--D); draw(P--Q--R--S--cycle); dot(A^^B^^C^^D^^O^^P^^Q^^R^^S); label("A",A,SW); label("B",B,NW); label("C",C,NE); label("D",D,SE); label("O",O,E); label("P",P,W); label("Q",Q,SW); label("R",R,E); label("S",S,N);[/asy]
$ P$ is the intersection of the perp. bisectors of $ OA$ and $ OB$.
$ Q$ is the intersection of the perp. bisectors of $ OB$ and $ OC$.
$ R$ is the intersection of the perp. bisectors of $ OC$ and $ OD$.
$ S$ is the intersection of the perp. bisectors of $ OD$ and $ OA$.
Because $ P$ and $ Q$ are both on the perp. bisector of $ OB$ and $ R$ and $ S$ are both on the perp. bisector of $ OD$, $ PQ\parallel RS$. Similarly, $ PS\parallel QR$, so $ PQRS$ is a parallelogram.[/hide]


geometry
circumcircle
ratio
incenter
inradius
radical axis
geometry unsolved

Identify triangle center where the radical lines of three mixtilinear circles (tangent to 2 sides of triangle ABC
   and internally to it's circumcircle) concur.

   Maj. Pestich

[url=http://www.math.fau.edu/yiu/mixtilinear2.pdf]Paul Yiu, [i]Mixtilinear Incircles, II[/i][/url], paragraph 1, Theorem 1:

The radical center of the three mixtilinear incircles of a triangle is the point P dividing the segment OI externally in the ratio $OP:PI=R:\left(-\frac{r}{2}\right)$.

Hereby, of course, O is the circumcenter, I is the incenter, R is the circumradius and r is the inradius of triangle ABC.

See also [url=http://www.math.fau.edu/yiu/mixtilinear3.pdf]Paul Yiu, [i]Mixtilinear Incircles, III[/i][/url].

  Darij


Two players play a game:
A large circular paper is provided and there are numerous thin identical rectangular strips of paper.The players alternately cover up the circle with the strips in a way that the strips don't overlap.The one who first finds that he cannot place the strip inside the circle such that the whole strip remains inside the circle(and no part remains outside the circle) looses.If you had had the first move , what would be your strategy to ensure a win? :rotfl:

How long/thin is said strip?

I would just place the first one over a diameter.

the strip's length is shorter than the diameter and so also is its width

Okay, so you can legally place the first strip. (Whew!)

Well, obviously the first player can place his strip at the center of the circle and copy the other guy's moves by a 180 degree rotation. This guarantees a win.


Let $ A\in M_{m,n}(R)$, where $ R$ is a ring with identity. Define $ S \equal{} \{X|\text{X is obtained from A by multiplying some rows and columns by \minus{} 1 in an arbitrary sequence}\}$. Find an upper bound for the size of $ S$. Give an example of an $ A$ such that this bound for $ |S|$ is obtained.

Well, here is a not so good upper bound: $ 2^{mn}$

but looking for a sharp upper bound will result in this:
choose $ i$ rows and $ j$ columns of $ A$ and multiply them by $ \minus{} 1$. this is easy to check that every sequence of multiplying rows and columns will result in such a matrix. (But I can not say it in exact words! :blush: )
now some over all such choices of $ i$ and $ j$, it means that:
$ \sum_{i,j}^{}{\binom{n}{i}\binom{m}{j}}$

As an example consider a matrix whose $ mn$ entiries have different absolute values. for instance $ 1, 2, ... , mn$ in real numbers field.

That's an overly complicated way of looking at it. We can go directly to a closed form.

If $ R$ has characteristic 2, the sharp upper bound is 1. Otherwise, it's $ 2^{m\plus{}n\minus{}1}$, achieved when all entries $ a_{ij}$ of $ A$ satisfy $ 2a_{ij}\neq 0$.

Choose which rows are flipped and which columns are flipped. Order is irrelevant. That's $ 2^{m\plus{}n}$. Since flipping all rows and all columns changes nothing, divide by 2.

Dear jmerry

I had seen your approach in Russian Olympiad book a long time ago, but recently a friend told me that there exists a linear algebraic solution to this simple problem. I failed to find that kind of solution. That's why I sent it in this section and I have considered the problem over a ring with identity. (Actually, we don't need the identity; we are just replacing entries with their additive inverses.)


have you got the book Mathematical Olympiads 1999-2000: Problems and Solutions from around the World by Titu Andreescu, Z. W. Feng ?
The turkish olympiads are needed.
Page 176 - 183 and page 290 - 292
can anyone who has the book,   -send me these pages.

You might want to check on the copyright of that first...


Three fourths of a truck is covered in mud. If half of the mud gets washed off, then what fraction of the truck has no mud of it?
[hide]
1/2*3/4=3/8
5/8 has no more mud[/hide]

[hide]So far $\frac{1}{4}$ of the truck isn't coverd in mud. The other $\frac{3}{4}$ is. So half of the mud gets off or $\frac{3}{4}\cdot\frac{1}{2}=\frac{3}{8}$. So $\frac{1}{4}+\frac{3}{8}=\frac{5}{8}$.[/hide]


which is this language--can ne1 translate it 4 me??
Vindem baza de date cu 1.000.000 adrese de e-mail, cuprinzand atat adrese ale persoanelor juridice cat si ale persoanelor fizice cu domiciliul in Romania, la cel mai mic pret de pe piata!
Va puteti promova acum afacerea sau ideile dumneavoastra, prin mail, in cel mai ieftin mod cu putinta, cu impact maxim.
La achizitionarea pachetului veti primi in mod GRATUIT si un program de trimitere a e-mailurilor care poate trimite pana la 50.000 de mailuri pe ora, program care in mod normal ar costa 100 Euro.
Aveti asadar toate premizele unei promovari IEFTINE si de succes, tinand cont cati bani ar trebui sa investiti in publicitatea clasica, asa incat reclama dumneavoastra sa fie vazuta de 1 milion de oameni.

Pretul bazei de date care include cele 1 milion de mailuri si al programului de trimitere oferit gratuit este 150 RON.

Pentru comenzi si detalii ne puteti scrie la adresa de mail: echipa.promovare@yahoo.com

Telefon : 0751.709848

My guess is Romanian.  Googling "scrie la adresa" gets a lot of hits from Romanian sites (URLs ending in .ro).

It looks like Romance language, in any case.

Definitely Romanian, going by "150 RON", "Romania", and all the Romanian diphthongs and u's :). I don't know any Romanian at all though.


$n > 1$ is any integer not divisible by $1997$. Put $a_{m}= m+\frac{mn}{1997}$ for $m = 1, 2, ... , 1996$ and $b_{m}= m+\frac{1997m}{n}$ for $m = 1, 2, ... , n-1$. Arrange all the terms $a_{i}, b_{j}$ in a single sequence in ascending order. Show that the difference between any two consecutive terms is less than $2.$

[quote="Goblin"]First assume that $n> 1997$. From $b_{m+1}-b_{m}=1+\frac{1997}{n}< 2$,  $b_{1}\leq a_{1}$ and $b_{n-1}\geq a_{1996}$ we see that the difference between every two consecutive terms is less than $2$. Simularly we can examen the case $n<1997$. For $n=1997$ we have $a_{m}=b_{m}$ and therefore the proof is comlete.[/quote]
Please read carefully problem!

Well I don't know what is wrong... The sequence  $a_{m}$ is "inside" the sequence $b_{m}$ ($b_{1}\leq\ldots\leq a_{1}\leq\ldots\leq a_{1996}\leq\ldots\leq b_{n-1}$) and since the difference between any two consecutive terms of the sequence $b_{m}$ is less than 2, so is the difference between any two consecutive terms of new sequence. We only don't need to examen the case $n=1997$.

But $1$ and $-1$ are not consecutive in new sequance ($-1$ and $0$; $1$ and $0$ are, but their difference is 1)

Yes, I am wrong!  :oops: But why $b_{n-1}\geq a_{1996}$?

[quote="Goblin"] Simularly we can examen the case $n<1997$. [/quote]
Please post concrete!  :oops:


geometry
circumcircle
ratio
perpendicular bisector
angle bisector
geometry unsolved

Denote M, N, P the middles of AI, BI, CI; obviously, an omothethy of centre I and ratio 1/2 matches A to M, etc.
The circumcenter of AIA_1 is the intersection of NP and perpendicular bisector of AI, that is the perpendicular onto AI through M, so it is the external angle bisector of angle NMP.
See that the external angle bisectors cut the opposite sides of triangle at 3 collinear points.

Note: few months ago Pohoatza solved here a similar problem, using the inversion.

Thank [color=darkblue][b]sunken rockvery [/b][/color]much.

[quote="apollo"][quote="Erken"][quote="apollo"]Let $ I$ be the center of the inscribed circle of the non-isosceles $ \triangle{ABC}$,and let the circle touch the sides $ BC,CA,AB$ at the points $ A_1,B_1,C_1$ respectively.Prove that the centers of the circumcircles of $ \triangle{AIA_1},\triangle{BIB_1},\triangle{CIC_1}$are collinear[/quote]
Actually,as noticed by sunken rock,this problem was posted by me before  and pohoatza posted there a proof with inversion...[/quote]
Can you post link ?  :)
Thank you[/quote]
Of course:
http://www.mathlinks.ro/viewtopic.php?t=181827


Hello,

I’m trying to use a (very interesting) package pdfscreen.sty to format a presentation but still get some problems. I was wondering if anyone maybe could help me?

First of all : I’m working under Windows XP with WinEdt and the Miktek distribution. Du to the amount of figures I use (created under XFig and imported with .pstex and pstex_t) I need to go through the .dvi  then .ps (with dvips) before using ps2pdf to obtain the pdf file. So I do not use the pdflatex to convert from tex 2 pdf (but as seen from the documentation, this doesn’t seem to be a problem : I specify the option dvips)

This all works very well as long as :

-          I insert \usepackage[dvips]{hyperref} before the \usepackage{pdfscreen} (strange but this works)

-          The option screen is chosen.

If a want to have the print layout (option print) I got some errors, I realy can’t fix ! :

If I just change screen to print : I’ve got an option clash for package hyperref 1.1142 \def\pagedissolve#1{}

If I change the screen to print and I remove the hyperref package inserted before : I’ve got hyperref using driver hpdftex* !undefined control sequence 1.320 \pdfoutput = 1

Does anyone know what the problem could be  ?
Thank you

Stephane

If you want to get pdf format from ps, the best result give Ghostscript8.5&GhostView4.7. It has an option convert to pdf.

Thank you for your answer but I use both of them.
Nevertheless, I found a solution by modifying some lines in the macro itself.

Thanks anyway !

and by the way, I've got an other question. I would like to manage bilingual versions of a text :
Say I have a "slave" file containing text in both french and dutch but would like to pass a sort of command in the master file to only include one or the other version - how could I do that ?

Thanks

Stephane


Hello,

We have $ 2n\plus{}1$ pieces of metal which have integer weight. If we remove any one of those, the other $ 2n$ remaining can be arranged so both piles consisting of $ n$ pieces of metal are having the same weight. Prove that all $ 2n\plus{}1$ pieces of metal have the same weight.

Here's a recent one, with links to versions having solutions:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=312178


ratio
quadratics
number theory proposed
number theory

This might be totally wrong, but if $ n_k$ are too big, it means there are a lot of 0s, which means that there are a bunch in a row, so $ \phi$ would have rather good rational approximations (take the number given by the binary digits of $ \phi$ up to a long sequence of 0). Then there are theorems which say that algebraic numbers can't have too good rational approximations.

First of all this is not the golden ratio exactly, but ok whatever. Denote $\phi = \frac{\sqrt{5} - 1}{2}$.
We prove this by induction.
Let us say we know all the $n_k$ for $k \le m$, and these satisfy $n_k \le 2^{k-1} - 2$ for $k \ge 4$. Let us suppose $n_{m+1} > 2^{m} - 2 \implies n_{m+1} \ge 2^{m} - 1$. Let us denote the rational expression using the first $m$ $1$'s be $\frac{a}{b}$.
Then obviously because the next $1$ occurs later than the $2^{m} - 2$ spot:
$\left | \phi - \frac{a}{b} \right | \le \frac{1}{2^{2^{m} - 2}}$
However as $\phi$ is a quadratic integer, it is badly approximable. Thus:
$\left |\phi - \frac{a}{b} \right | \ge \frac{1}{Cb^2}$

But $b = 2^{n_m}$. By our inductive hypothesis $n_m \le 2^{m-1} - 2$, we have:
$\left |\phi - \frac{a}{b} \right | \ge \frac{1}{C(2^{2^{m-1}-2})^2}$
$ = \frac{1}{C 2^{2^{m}-4}}$

$ \ge \frac{1}{2^{2^{m} - 2}}$
A clear contradiction, and thus we are done.
(Note that it can be easily shown $C \le 4$ by showing $C$ is actually quite close to $\sqrt{5}$)


Let be $ a,b,c > 0$ such that $ abc \equal{} 1$ . Then prove that :

$ \frac {a}{bc^2} \plus{} \frac {b}{ca^2} \plus{} \frac {c}{ab^2}\ge \frac {1}{ab} \plus{} \frac {1}{bc} \plus{} \frac {1}{ca}$

[quote="alex2008"]
$ ...\plus{} \frac {b}{ab^2} \plus{} ...\ge ...$[/quote]
is that supposed to be $ \frac{b}{ac^2}$?

Using $ abc\equal{}1$ gives

$ \frac{a^2}{c}\plus{}\frac{b^2}{a}\plus{}\frac{c^2}{b} \ge a\plus{}b\plus{}c$

From Cauchy-Schwarz

$ (a\plus{}b\plus{}c)\left(\frac{a^2}{c}\plus{}\frac{b^2}{a}\plus{}\frac{c^2}{b}\right) \ge (a\plus{}b\plus{}c)^2$

Done

[quote="alex2008"]Let be $ a,b,c > 0$ such that $ abc \equal{} 1$ . Then prove that :

$ \frac {a}{bc^2} \plus{} \frac {b}{ac^2} \plus{} \frac {c}{ab^2}\ge \frac {1}{ab} \plus{} \frac {1}{bc} \plus{} \frac {1}{ca}$[/quote]
It must be $ \frac {b}{ca^2}$, not $ \frac {b}{ac^2}$.
And we don't need $ abc\equal{}1$
$ (\frac {1}{ab} \plus{} \frac {1}{bc} \plus{} \frac {1}{ca})(\frac {a}{bc^2} \plus{} \frac {b}{ca^2} \plus{} \frac {c}{ab^2})\ge (\frac {1}{bc} \plus{} \frac {1}{ca} \plus{} \frac {1}{ab})^2$

[quote="ocha"]Using $ abc \equal{} 1$ gives

$ \frac {a^2}{c} \plus{} \frac {b^2}{a} \plus{} \frac {c^2}{b} \ge a \plus{} b \plus{} c$

From Cauchy-Schwarz

$ (a \plus{} b \plus{} c)\left(\frac {a^2}{c} \plus{} \frac {b^2}{a} \plus{} \frac {c^2}{b}\right) \ge (a \plus{} b \plus{} c)^2$

Done[/quote]

nice solution   :lol:


Suppose that $A$ is a $2 \times 2$ matrix with nonreal eigenvalues $\lambda = \alpha \pm \beta i$.

(a) Show that one of the nondiagonal elements, but not both, must be negative in order for the eigenvalues to be nonreal.

(b) Show that the eigenvalues are imaginary ($\alpha = 0$) if and only if $\text{Tr}A = 0$ and $|A| > 0$.

For (a)  I tried a proof by contradiction (i.e. assumed that both were positive, or both were negative). So I did the following:

$\begin{bmatrix}a & b \\ c & d \end{bmatrix}$

$\left| \begin{matrix}a-\lambda & b \\ c & d-\lambda \end{matrix}\right| = (a-\lambda)(d-\lambda)-bc = 0$

$\lambda(\lambda-a-d)+(ad-bc) = 0$

I dont know where to go from here.

You have a quadratic equation. What is its discriminant?


[quote="midorya"]$\sum_{k=0}^{n}\binom{k}{n}\cos((k+1)x)$[/quote]

[hide="Improve"]$\binom{k}{n}$ by $\binom{n}{k}$[/hide]

[hide="hint"]Apply
$\cos (k+1)x = \cos kx \cos x-...$

$\cos x+i sin x = e^{ix}$

$(1+x)^{n}=\sum_{k=0}^{n}\binom{n}{k}x^{k}$

$1+e^{ix}=2\cos \frac{x}{2}e^{i\frac{x}{2}}$

to evaluate both $\sum_{k=0}^{n}\binom{n}{k}\cos kx$ and  $\sum_{k=0}^{n}\binom{n}{k}\sin kx$[/hide]


geometry
3D geometry
algebra
polynomial
superior algebra
superior algebra unsolved

[b][u]The author of this posting is: sam-n.[/u][/b]
_____________________________________________________________

let $H_1, \ldots, H_m$ be a family of hyperplane in $\mathbb{R}^n$ that cover all vertices of unit cube $\{0, 1\}^n$ but one .then $m \geq n$.

[b][u]The author of this posting is: vess.[/u][/b]
_____________________________________________________________

Combinatorial Nullstellensatz ;) . If $a_i \cdot x = b_i$ are the hyperplanes and $(0, \ldots, 0)$ be W.L.G. the uncovered vertex, then apply CN for the polynomial

\[\prod_{1 \leq i \leq m}{a_ix-b_i} + \delta \prod_{1 \leq i \leq n}(1-x_j),\]

where $x=(x_1, \ldots, x_n)$, and $\delta \neq 0$ is chosen so that the polynomial vanishes in $(0, \ldots, 0)$. Now, the assumption that $m \leq n$ would imply that our polynomial has degree $n$ and the coefficient of $x_1 \cdots x_n$ is $\delta(-1)^n \neq 0$, but this polynomial is identically zero in $\{0, 1\}^n$ --- contradicting CN.

--Vesselin


inequalities
real analysis
real analysis unsolved

this is a norm property i need to finish a problem but i cant see right away why is true. it probably stupid.

if  $ x$ is an element of an inner product space, show that
$ \parallel{}x\parallel{} \equal{} sup_{\parallel{}y\parallel{} \equal{} 1}| < x,y > |$

where $ \parallel{} . \parallel{}$ is the norm defined by the inner product: $ \parallel{}x\parallel{} \equal{} \sqrt { < x,x > }$

thanks  :blush:

oh. it is just cauchy scharz, sorry for my ignorance.

 :wink:


Nα λυθει στους ακεραιους το συστημα
$x^{3}+y^{3}+z^{3}=x+y+z=8$

 :D  :D  :D

Βρήκα τη βασική ιδέα νομίζω
Είναι $(x+y+z)^{3}-(x^{3}+y^{3}+z^{3})=3(x+y)(y+z)(z+x)$
άρα $(8-x)(8-y)(8-z)=168$
και $(8-x)+(8-y)+(8-z)=16$ (1)
Από την (1) έχουμε ότι οι και οι τρεις είναι άρτιοι ή μόνο ένας είναι άρτιος .
Αν και οι τρεις είναι άρτιοι τότε αφού
168=3*8*7  θα είναι
$\left|8-x\right|=6$
$\left|8-y\right|=14$
$\left|8-z\right|=2$
 που αν δεν κάνω λάθος δεν ικανοποιεί την(1).
Αν ένας μόνο είναι άρτιος τότε
$\left|8-x\right|=8$
$\left|8-y\right|=3$
$\left|8-z\right|=7$
και τότε δεν έχουμε λύση αν δεν κάνω λάθος .
Μπορεί όμως να είναι και
$\left|8-x\right|=8$
$\left|8-y\right|=21$
$\left|8-z\right|=1$ και πάλι όμως δεν έχουμε λύση
Μπορεί τέλος να είναι
$\left|8-x\right|=168$
$\left|8-y\right|=1$
$\left|8-z\right|=1$

ή
$\left|8-x\right|=24$
$\left|8-y\right|=7$
$\left|8-z\right|=1$
από όπου προκύπτει λύση

καλο σκεπτικο.... :wink:
Εγω βεβαια την δουλεψα λιγο διαφορετικα(+λιγο διαφορετικα σε εκταση....)
+τελικα παλι δεν κατεληξα σε λυση....
Παντως η λυση σου φενετε σωστη :first:  :clap:

 :D

Ευχαριστώ Ράλλη .Πάντως πολύ ωραία άσκηση.......
Πάντως προβληματίστηκα στο εξής
Είναι δυνατόν άραγε να βρούμε τις ακέραιες λύσεις της εξίσωσης
$x^{3}+y^{3}+z^{3}=3$ στους ακεραίους ? :maybe:

[quote="silouan"]
Πάντως προβληματίστηκα στο εξής
Είναι δυνατόν άραγε να βρούμε τις ακέραιες λύσεις της εξίσωσης
$x^{3}+y^{3}+z^{3}=3$ στους ακεραίους ? :maybe:[/quote]
(x,y,z)=(1,1,1)
mipos =8?
I mipos enois oles tis lisis autis tis eksisosis.....?

 :D

Εννοώ όλες τις λύσεις της εξίσωσης .
Μπορούμε να τις βρούμε?

Μια αλλη προσεγγιση ειναι και η παρακατω:ισχυει οτι 3(x^3+y^3+z^3)>=(x+y+z)(x^2+y^2+z^2) αρα χ^2+y^2+z^2<=3 αρα x=y=z=+-1 ομως δεν επαληθευονται αρα δεν εχει λυσεις.Οσο για την ερωτηση σου Σιλουαν αν η εξισωση x^3+y^3+z^3=3 εχει λυσεις στους ακεραιους αρκει να ερευνησουμε αν οι εξισωσεις χ^3+y^3=z^3+3,x^3+y^3=z^3-3 εχουν λυσεις στους θετικους ακεραιους.Θα τις κοιταξω και θα σου πω.

σχετικα με αυτο που ρωτα ο Σιλουαν η απαντηση ειναι μαλλον αρνητικη  :)

Έχεις δίκαιο Κώστα .Είναι ανοιχτή η άσκηση .
Μια όχι και τόσο προφανής λύση είναι η $x=y=4$  , $z=-5$   :wink:

[quote="RAKAR"]Nα λυθει στους ακεραιους το συστημα
$x^{3}+y^{3}+z^{3}=x+y+z=8$[/quote][url=http://www.mathematica.gr/forum/viewtopic.php?f=109&t=37106]εδώ[/url]


probability
conditional probability
probability and stats

Any help will be appreciated!

[i]n[/i] people are to be seated randomly around a circular table. Find the probability that two particular people (Call them person A and person B) will be seated next to each other.

Fix the position of one person, say A.
1) What is the number of all possible positions for B?
2) What is the number of the positions when B sits next to A
3) What is the probability that B sits next to A with the position of A fixed (the so-called conditional probability)
4) Does it depend on the position of A?
5) Can you answer your question now?

Once you've followed fedja's hints and solved the original problem, then what about this one?

The same $ n$ people are seated at random in a straight row of $ n$ seats. What is the probability that A and B are seated next to each other?

Should the answer to that be the same probability as for the circular table problem? Should it be a higher probability? Should it be a lower probability?


modular arithmetic
number theory unsolved
number theory

prove or disprove the following propositions.

problem 1. someone gave me this problem: prove that $\forall$ positive integers $a\ge3$,  $\frac{a!-2}{2}$ is a prime number.

ps. this is not for you kathrin. :wink:

problem 2. Prove that $\prod_{i=1}^{n}{p_{i}}+1$ is prime where $p_{k}$ is the k-th prime.

Anyway the second problem must be $S = \prod_{i=1}^{k}p_{i}+1$  :)
And I'm sorry that I've already had a big mistake since if $p_{k}= 13$, then $S = 59 \cdot 509$. There's more counter examples.. ($p_{k}= 17, 19$ also makes $S$ to be composite)

[quote="RDeepMath91"]Anyway the second problem must be $S = \prod_{i=1}^{k}p_{i}+1$  :)
And I'm sorry that I've already had a big mistake since if $k = 13$, then $S = 59 \cdot 509$. There's more counter examples.. ($k = 17, 19$ also makes $S$ to be composite)[/quote]

sorry my mistake! but i've corrected it. :lol:

this is what i've been talking about. these expressions can have prime divisors >a or $p_{k}$! ain't i right?  :D

[quote="nayel"]prove or disprove the following propositions.

problem 1. someone gave me this problem: prove that $\forall$ positive integers $a\ge3$,  $\frac{a!-2}{2}$ is a prime number.[/quote]

$11|\frac{7!-2}{2}$ because $5!*42-2=120*42-2\equiv-1*42-2=-44\equiv 0 \pmod{11}$


function
inequalities unsolved
inequalities

Suppose I have a convex function which satisfies for some two points $ (x_{i},y_{i}),i\equal{}1,2$
$ f(\lambda x_{1}\plus{}(1\minus{}\lambda)x_{2})\equal{}\lambda f(x_{1})\plus{} (1\minus{}\lambda) f(x_{2})$
and $ f(x_{1}) \neq f(x_{2})$
What does it tell about the behavior of the function at these points

You can, of course,  deduce from this conditon that our function is linear on the segment $ [x_1,x_2]$ :wink:

Sure. You can just note that every point $ x\in [x_1,x_2]$ can be uniquely represented in the form $ x\equal{}\lambda x_1 \plus{}(1\minus{}\lambda )x_2$ for some $ \lambda\in [0,1]$ . From this condition we can deduce that $ \lambda \equal{}\frac {x \minus{} x_2}{x_1 \minus{}x_2}$ and substituting this to the condition of the problem we obtain:
$ f(x)\equal{}\lambda f(x_1) \plus{} (1\minus{}\lambda) f(x_2)\equal{}\frac {x \minus{} x_2}{x_1 \minus{}x_2} (f(x_1)\minus{}f(x_2)) \plus{}f(x_2)$. This provides linearity. Another way to prove this fact is based on the geometrical arguments :wink:

Thanks!
suppose the same condition holds true in three dimensional space that is for some two points $ \vec{r_{1}}$ and $ \vec{r_{2}}$ we have the same relation.What would be the meaning of linearity here?


[quote="anirudh"]Prove that the perpendicular bisector of a segment consists of all the points which are equidistant from the two endpoints of a segment.

Mods-please move this if this is not the right level[/quote]

[hide="I just learned proofs last week"]Point C is the midpoint of line segment AB. Point D is not on segment AB and CD forms the perpendicular bisector for segment AB. AB has length 2a, so AC=CB and let's call their length a. Let's call CD's length b. $\angle DCA= \angle DCB = 90^\circ$ By SAS, $\triangle DCA \equiv \triangle DCB$. By CPCTC, DA=DB and let's call their length c. Point D is c away from A and B, Point D is equidistiant from A and B, Point D is a point on the perpendicular bisector, so the perpendicular bisector of a segment consists of all the points which are equidistant from the two endpoints of a segment.[/hide]

This is ridiculously coincidential. Last night, I just read the AoPS 2 chapter about locuses, and that was the example.

[quote="Totally Zealous"]Maybe that's where anirudth got it from.[/quote]
If it was exactly the same, then I wouldn't be surprised.

[quote="Totally Zealous"]Maybe that's where anirudth got it from.[/quote]

your right


There is a list of $n$ football matches. Determine how many possible columns, with an even number of draws, there are.

I think that you mean this problem:

Find the number of words of $n$ letters from the alphabet $\{a, b, c\}$ which contain an even number of $a$’s.

Solution: If we denote as $a_n$ number of words which satisfy given condition then we can get easily recurrence relation: $a_n=a_{n-2}+2a_{n-1}$ and by standard methods we will find explicitly $a_n$.


A fair die is tossed $100$ times. What is the most likely number of $6$'s we would obtain?

[hide="Hint"] Binomial Probability[/hide]

[hide]
would this not just be $\frac{100}{6}= 16.666 \implies 17$ ?
[/hide]

Yes, chickendude.
[hide]Let $X$ represent the number of 6's tossed out of 100 tosses of the dice.  Then $X$ has a binomial distribtution with probability of success $\frac{1}{6}$.  The expected value of a binomial distribution is given by $np = \frac{100}{6}\approx 17$.[/hide]


function
trigonometry
calculus
integration
probability and stats

Dear friends,

I have a small doubt like, can characteristic function of a r.v. have real roots?

For example let $ X$ be r.v. with PDF $ f_X(x) \equal{} \frac {1}{2}e^{\minus{}|x|}$. Its characteristic function is given by $ \phi_X(\omega) \equal{} \dfrac{1}{1 \plus{} \omega^2}$. Its poles are at $ r_1,r_2 \equal{} \pm i$. Can i have any r.v. say $ Y$ with characteristic function $ \phi_Y(\omega)$ whose poles are on real axis? Can you please provide me with an example?

thank you for the help!

bhupala!

uniform random variables, e.g., the zero set of $ U[0,1]$ is pure real $ \pi \plus{}2\pi\mathbb{Z}$

[quote]say $ Y$ with characteristic function $ \phi_Y(\omega)$ whose poles are on real axis? Can you please provide me with an example?[/quote]
To talk about poles is the language of analytic functions: there is no reason that the characteristic function of a random variable must be analytic. However, a characteristic function of a random variable cannot have a pole - or anything remotely like a pole - on the real axis.

$ \phi_X(\omega)\equal{}E\left(e^{i\omega X}\right),$ so for $ \omega\in\mathbb{R},$

$ |\phi_X(\omega)\equal{}\left|E\left(e^{i\omega X}\right)\right|
\le E\left|e^{i\omega X}\right|\equal{}E(1)\equal{}1.$

A characteristic function must be bounded by $ 1$ on the real line. For an absolutely continuous random variable it must be continuous and tend to zero as $ |\omega|\to\infty.$ (That's a property of the Fourier transform of an $ L^1$ function.) Drop "absolutely continuous" and we'll drop the conclusion about the characteristic function tending to zero at infinity.

For a gentle look at what could happen on the line, let's flip your example around.

Define $ Y$ with PDF $ f_Y(y)\equal{}\frac1{\pi(1\plus{}y^2)}.$ Then $ \phi_Y(\omega)\equal{}e^{\minus{}|\omega|}.$

That's a bounded continuous function, but it's not even differentiable at zero. (Which corresponds rather neatly to the random variable $ Y$ not having an expected value.)

Then what do you say about the example given by hjbrasch?

regards

bhupala

Roughly the same example, but a little more symmetric and thus prettier: let $ X$ be uniformly distributed on $ [ \minus{} 1,1].$ The characteristic function is $ \phi_X(\omega) \equal{} \frac {\sin\omega}{\omega}.$

$ \frac {\sin\omega}{\omega}$ does have zeros whenever $ \omega \equal{} 2k\pi$ for $ k$ an integer other than zero. And that's OK. I never said that a characteristic function couldn't have zeros. I said that a characteristic function had to be bounded and couldn't go to infinity.

You seem to have wrapped "pole" and "zero" together in your mind into one bundle and are acting as if they're the same concept. They aren't.

We can even have a characteristic function which is zero on an interval. In fact, there is a random variable $ U$ such that

$ \phi_U(\omega) \equal{} \begin{cases}1 \plus{} \omega, & \minus{} 1\le\omega\le0 \\
1 \minus{} \omega, & 0\le\omega\le1 \\
0, & \text{otherwise}\end{cases}.$

Oh yeah,

sorry for misunderstanding. By the way one more question can a characteristic function be discontinuous? And also how do you sovle the improper integral like
\[ \int_{ \minus{} \infty}^{\infty}\dfrac{\phi_X(\omega)}{\omega}d\omega
\]
should we take cauchy principal value? Will c.p.v. exists for all $ \phi_X(\omega)$?

regards
bhupala


Prove that M={f:N->N / for any n natural f(n)>=f(n+1)} is countable.

For all decreasing $f : \mathbb{N} \longrightarrow \mathbb{N}$ there is an $x_f \in \mathbb{N}$ such that for all postive integers $m, n \geq x_f$ we have $f (m ) = f ( n )$. (Intuitively we observe that $f ( n )$ "gets constant" as $n \longrightarrow \infty$) Having this in mind, suppose that $x_f$ is the minimal integer with the proprety that $m,n \geq x_f$ implies $f(m)=f(n)$. Then we construct the bijection $\Phi : M \longleftrightarrow \mathbb{N}$ by definining
\[ \Phi ( f ) = \prod_{j = 1}^{x_f} p_j^{f ( j )}  \]
where $p_k$ is the $k_{\mbox{th}}$ prime. I leave to you the verification that this is indeed a bijection.


Can you describe it?  If you have a drawing program (even just Microsoft Paint), you can make an image and upload it.

So save the picture as a .jpg or .gif instead.  (When you save the file, it has a "save as type" option.  Change that.)

Well, for number 21 you can use something like pascal's triangle with each intersection. it's pretty hard to explain, but you give point A a value of 1, and then for each intersection, it's the sum of the point to its left and the point below it. It's kinda like figuring out how many possible routes there are to get to every single point.
as for number 19, i also didn't think there was enough info, but i didn't make any typing mistakes.

Rep -- you know how you can drag the edges of the image around on Paint?  If you could do that (to shrink the picture down somewhat), it would be really nice because it wouldn't stretch the screen like it's doing.  Then you can use the re-upload button after resaving the file.

And there is definitely enough information in that second question.  Write it out!  Do what you always have to do for distance-rate-time questions!

You observed that Alyssa's hilly problem lacks sufficient information.  Your observation leads to a simple solution.

My own solutions to the 2000, 2001, and 2002 National Competition can be found on my Web Site at [url]http://home.earthlink.net/~rcv007[/url].

BTW, my copy of the 2000 National Competition has gone missing.  [I lost some MATHCOUNTS material in a water-heater flood in my home about a year ago.]  Anybody have a good, clean copy they'd be willing to loan or donate?  Please message me.

Rep123max wrote:Oh and also, how does this question have enough information? Alyssa rode her bike for 7 hours. She started on level ground and rode at 8mph. She came to a hill and slowed down to 5mph. Upon reaching the top of the hill, she turned around and descended at a rate of 20mph. Finally, she returned home on the flat portion at 8mph. What is the total number of miles she rode?

[hide]x=distance from home to hill.  y=length of the hill.

x/8 time from home to hill

y/5 time going up the hill

y/20 time going down the hill

x/8 time from hill to home

2*x/8+y/5+y/20=7

x/4+y/5+y/20=7 multiply both sides by 20

5x+4y+y=140

5x+5y=140 divide by 5

x+y=28 (one way trip)

2x+2y=56 miles (two way trip)[/hide]

hey rcv your mathcounts timer program doesn't work. i downloaded it and it says beta version has expired.

where does alyssa's problem come from, i mean is it nationals? and which one?

2000 nationals. if you want, i can send it too you through email.

Finding her avg. speed on the hill will yield a quick solution.

Rebel


p(x) is a polynom with coefficients that belong to Z
and p(-2006)<p(2006)=2005
prove that
p(-2006)<=-2007

I'm sorry if the wording isn't so good. It's in hebrew and I don't know all the mathematical terms in English.
Anyway, can anyone help me solve this?


Does anyone know the exact scoring system used to grade the Chemistry AP test?

I know that the free response is really funky, because each section is given a weight, so for any given section, there are score multipliers. Also, on the multiple choice, -1/4 a point for a wrong answer, 0 for blank, 1 for right answer.

I do not know if this a general case, but out of 160 possible points, 100-160 was a 5 for one 'mock' AP I took in school.

i think this was the way the 2003 test was graded
it is pretty obvious how the MC is graded.
for FR:
#1:(your grade out of 8) * 1.88
#2/3:(your grade out of 8)*1.65
#4: (your grade out of 8)*0.88
#5:(your grade out of 8 )*1.65
#6:(grade out of 8)*1.65
#7/8:(grade out of 8)*1.65
add this score to the score of the MC.
if you get between: $100-160$ you'll get a $5$
my guess for my score this year is:$129.92$ :)

[quote="pkothari13"]wait are you sure all of them are out of 8? i thought 4 (the reaction predictions) were out of 15.[/quote]
yeah,,,, i guess you're right...
but i'm positive that the equations are worth less (*0.88).
anyways, i know this:
if you get a 48-75 on the MC, you have a 75-83 percent chance of getting a 5...
so don't worry much... :lol:
i can dig the exact numbers for all grades, if anyone interested...

[quote="pkothari13"]oh sweet, woo hoo, that boosted my confidence a bunch! i was just digging through some papers and found this:

Section 1 (mult. choice) = [# correct (out of 75) - .25(wrong)]*.96 = weighted Sect. 1 score

Section 2 (free response)

Question 1: Points out of 10 * 1.76
Question 2 or 3: Points out of 10 * 1.76
Question 4: Points out of 15 * .88
Question 5: Points out of 10 * 1.32
Question 6: Points out of 8 * 1.65
Question 7 or 8: Points out of 8 * 1.65

Then 107-160 = 5
85-106 = 4
61-84 = 3
42-60 = 2
0-41 = 1

I dunno how accurate this is, but yeah.[/quote]
yeah, this makes more sense now...

oh geez, you guys better not be getting my hopes up..  i'm   [b]really[/b] nervous about my grade.


[quote="mr. math"]i had trouble solving this mental math question:

A cube with the volume of eight is combined with 7 others to form a larger cube. What is th surface are of the new larger cube?[/quote]
[hide]
well the new volume will be 64. you take the cube root of 64 and that's $4$. 16 times 6 is $96$
[/hide]


Were you suprised at the outcome of the NCAA Basketball finals  (Florida over UCLA)?

[quote="jhollenbeck"]Personally, I was UCLA all the way :lol:[/quote]

Me too......


solve
$ 2l \plus{}1\equal{}3k^2 \minus{}8k \minus{} 15$
where $ l,k$ are whole numbers.

is this solvable?pls forgive if it is dumb.

You can write it as:

$ l \equal{} \frac {3k^2 \minus{} 8k \minus{} 16}{2}$,

So, since $ l$ is integer, $ 3k^2 \minus{} 8k \minus{} 16$ must be even.

You have to find for what values

$ 3k^2 \minus{} 8k \minus{} 16 \equiv 0 (mod 2)$

which lead,

$ x^2 \equiv 0 (mod 2)$

which is given by $ x \equiv 0 (mod 2)$

You can take a $ z \in \mathbb{Z}$, and the solution is:

$ k \equal{} 2z$ and
$ l \equal{} 6z^2 \minus{} 8z \minus{} 8$

Gabriele  :)

well i think you forgot to specify that $ z\ge\ 2$
as for $ z\equal{}1$ i think you get $ l<0$

Sorry,
you're right!

I thought you meant the solutions in $ \mathbb{Z}$ writing "whole numbers"...

Bye,
Gabriele :)


calculus
integration
analytic geometry
trigonometry
function
algebra
domain

Hey.
I had this double integral that when u map it, has a shifted center at (0.5,0) and a radious of 0.5. After some thinking about the problem i figured that my bounds should have a variable in them, however i am not sure how to find the bounds for teh integral. I subbed in the polar formula, and converted the equation, but the bounds of a shifter circle are not correct. Any ideas on how to find the bounds?
Thx in advance.

The boundary has equation $ (x-1/2)^{2}+y^{2}=1/4$. Which, when you expand and simplify it, becomes $ x^{2}+y^{2}=x$. In polar coordinates this is $ r^{2}=r\cos\theta$. Cancel $ r$ and observe that $ \cos\theta$ must be nonnegative, which limits the interval for $ \theta$ to $ [-\pi/2,\pi/2]$.

hmm.....
seems that i have made a mistake at some point:
f(x)= \int_{0}^1  \int_{0}^{(x-x^2)^{1/2}} x/(x^2 + y^2)^{1/2} dydx
y^2 + x^2 = x
x = r*cos(\theta) + 0.5
y = r*sin(\theta)

so naturally i sub in for x and y, and then multiply by r to convert to polar, however that creates a function that looks awefully incorrect.
Is this the right way to do it?
Thx

First, you are trying to use shifted polar coordinates, with $ x=1/2+r\cos\theta$. On one hand, this is a natural thing to try, considering that we have a shifted circle. However, the function you are to integrate has $ \sqrt{x^{2}+y^{2}}$ in the denominator. With your substitution this will become $ \sqrt{1/4+r\cos\theta+r^{2}}$, which is no fun to integrate. So, I recommend using the standard polar coordinates with $ x=r\cos\theta$ and $ y=r\sin\theta$.

Second observation: apparently, you first put limits for $ x$ and $ y$ and then tried a substitution into the limits. This is a completely wrong approach. Even when an integral is already given to you with limits for $ x$ and $ y$, you should go back to a sketch of the domain of integration and find polar limits from it. Substitution into limits simply does not work.

The correct polar limits are $ 0\le r\le \cos\theta$ and $ -\pi/2 \le\theta\le\pi/2$. Basically, you start integrating from the origin $ r=0$ and continue increasing $ r$ (going straight away from the origin) until you hit the boundary $ r=\cos\theta$.