geometry
rectangle
trigonometry
area of a triangle

Here are challenges for beginners in geometry.

1. Regular hexagon $ABCDEF$ is inscribed in rectangle $PQRS$ with $AB = 1,$ and $A$ And $B$ on side $PQ,$ $C$ on side $QR,$ $D$ and $E$ on side $RS,$ and $F$ on side $SP$. What is the area of $PQRS$?

2. Isoceles triangle $ABC$ has angle $BAC = 135^\circ$ and $AB = 2.$ What is its area?

3. Equilateral triangle $ABC$ is inscribed in a circle with radius 6. Find the area of the region enclosed by $AB,AC,$ and the minor arc $BC$.

Enjoy!  :)

[hide] 2sqrt(3) ummm the four triangles outside the hexagon are 30-60-90 triangles, and the hypoteneuses are 1. soo get length of sides, area. [/hide]

3)[hide] Since ABC is equilateral, all of its angles are 60 degrees and the distance from its center point to each of its vertices is 6.  Let the center point be called D.  Our goal would then be to compute the area of ABD + area of ACD + sector DBC.  The areas of the two smaller triangles are equal so its basically finding $2[ABD]$+ sector DBC.

Sector DBC has an intercept angle of 120 degrees, since the radius of the circle is 6, then the area of the sector is $12\pi$

Triangle ABD is iscoseles, with 2 of its side lengths at 6.  Being so, and that $\angle BDA$ is 120 degrees, the remaining angles must be 30 degrees each.  Using the area of a triangle as $\frac 12ab\sin c = 9\sqrt3$

Therefore, the total area bounded by the conditions is $2(9\sqrt3)+12\pi = 18\sqrt3 + 12\pi = 6(3\sqrt3+2\pi)$[/hide]

2)[hide="hint"]Since the triangle is iscoceles, we know two lengths (AB= AC) and an included angle.  Sounds like $\frac 12ab \sin c$ to me ... [/hide]

1)[hide="hint"]Since each interior angle of the hexagon is 120 degrees, and that the hexagon is inscribed inside the rectangle, the rectangle would include the hexagon and four congruent 30-60-90 triangles.  However, the area of the rectangle can be expressed as base times height.  Since the triangles have hypoteneuse 1, we can find the dimensions of each triangle ... [/hide]


When Caitlin starts running towards Asha, Asha is 100 meters
from Caitlin.  As Caitlin runs, Asha moves directly away from
Caitlin at one-third of Caitlin's speed.  How many meters has Caitlin run when she first catches up to Asha?

However, to give what the problem asked for:

We know $ D\equal{}RT$.

We make two equations: one for Caitlin and one for Asha.

Caitlin: $ D\plus{}100\equal{}RT \implies D\equal{}RT\minus{}100$

Asha: $ D\equal{}\frac{1}{3}RT$

Substituting $ D\equal{}RT\minus{}100$ into the second equation yields $ RT\minus{}100\equal{}\frac{1}{3}RT \implies 2RT\equal{}300 \implies RT\equal{}150$, so $ D\equal{}50$.  Caitlin runs $ D\plus{}100\equal{}\boxed{150}$ meters.

Mewto55555, by your reasoning, what is $ 1\plus{}\frac{1}{2}\plus{}\frac{1}{4}\plus{}\ldots$?

Something very close to 2, yet never reaching it.

Just like Caitlin gets very close to Asha, but never reaches her.  However, the sum is actually two, and Caitlin does actually pass Asha.  Have you ever passed someone going in the same direction as you?


So, this is our new home! :lol:

For some it may need some time to adapt, especially to the cosmetic side (where I will fix the design problems as soon as Valentin gives me the necessary permissions), but I'm sure everyone will feel at home here soon. Also, the good old $ \text{\LaTeX}$ works here as it should be. :)

Please continue to use the site as you've all done so far, you're a great community! :D

Thanks Valentine, Peter and Mathlinkers! I'm gonna do more work on PEN after next week. I'm now preparing a series of lectures on number theory in the KMO summer program. See you!

Hojoo, I am liking the solutions PDF starting to come together. Is the eventual aim to have a PDF of solutions for every problem in PEN (might it be more logical to have separate PDFs for the different sections or it would be a monster file?).

Also, Peter, are we reforming the PSC, etc. I have been very busy with musical activities for the last month (I occasionally masquerade as an orchestra conductor), but after this week I will be more available to verify solutions, etc. (and Hojoo, would it be helpful if I TeX some of the posted solutions that aren't currently in your PDF?).

Tom
:)

[quote="Ilthigore"]Hojoo, I am liking the solutions PDF starting to come together. Is the eventual aim to have a PDF of solutions for every problem in PEN (might it be more logical to have separate PDFs for the different sections or it would be a monster file?).[/quote]
 Eventually, I hope that we collect all solutions in a printable format. It will be looks like a big thesaurus book :) However, if we have such a monster book, then actullay something weird is gonna happen. Simply beginners will be suffering from number theoritic shock:  "Look at this huge solutions book :maybe: . Should I attack all these problems in PEN?"

So basically, my plans is doing it by a 'comfortable' book series: PEN Solutions Book A, PEN Soltutions book B, etc and the philosophy of this Solutions Book series is that a Solution Book is more than just a collection of solutions. Each Solution Book has selected problems with various difficulty from each chapter. I hope that a Solution Book offers a 'balanced' collection of 20-30 "cool" problems with various solutions. Different Soltions Book has different authors. But, they should have a unified style.

[quote="Ilthigore"] Hojoo, would it be helpful if I TeX some of the posted solutions that aren't currently in your PDF?).[/quote]
Yeah, of course! After editing the tex code of the solutions in a neat format, send it to me. An efficient way of doing this is, of course, that you do this after proof-reading the solutions. Another way to contribute Solution Book project, is simply that you begin to write your own Solution Book!

I think the first aim now is to get more problems solved. The PEN solutions book was - if I understood correctly? - in first place a way to motivate people to write more solutions.

Currenlty less than half of the problems have been solved - even many from IMO shortlists/longlists, APMO, ... have not been solved. You may just copy official solutions or other verified solutions in a readable format, that's no problem, or link them if you have verified another solution (like I've done a few times lately). As long as we just have solutions available.

Because we need to know which ones are valid problems, and which ones are wrong/open/too hard. Also it would be nice to have an idea of the difficulty level of problems. I think I have corrected the statement of 77 problems so far in total, and more may be found needed when we actually try to solve them.

As for the PSC finally, that is still on the old site.

Can some body post problem in this forum ,they were be took  from national Olympiad near year.
I take from Crux and magazine .

[quote="Peter"]Because we need to know which ones are valid problems, and which ones are wrong/open/too hard.[/quote]
There are open problems in PEN?

Normally not, but we have discovered a few which seem unsolvable.
We want to eliminate those, as it is certainly not intended to have open problems in PEN.

Hi peter I can't post on the Other Problems subforum in PEN. It asks for special permission. Can u make it open for all?

[quote="manjil"]Can u make it open for all?[/quote]Err... I thought it was open. I'll have to ask Valentin about this.

The problems are solved already here in AoPS/Mathlinks. Just follow the tag: https://artofproblemsolving.com/community/c146_pen

(Now I am interested in post some translations , hehe!)

The old pen.problem-solving.be is officialy dead for now.

Well, now we need to repost the old files [b]*and* [/b] reformat the book!

[quote=AopsUser101]What is PEN?[/quote]

Problems in Elementary Number Theory


If a ten question True/False test is given and $ 60\%$ is a passing grade, what is your percent chance, to the nearest whole percent, of passing if you guess on each question?

By symmetry, note that getting a $ 60\%$ is just as likely as getting a $ 40\%$, $ 70\%$ as $ 30\%$, etc.

However, before utilizing this, we must subtract from $ 1$ the probability that we get $ 50\%$, which occurs with probability $ \dbinom{10}5\left(\frac12\right)^{10}\equal{}\frac{63}{256}$.

Our final answer is $ \frac{1\minus{}\frac{63}{256}}2\equal{}\frac{193}{512}\approx\boxed{38\%}$.

I am sorry, but I do not understand your final step? Is there a reason for it?

So basically, you have to subtract the probability from 50, because we dont need the 50percent in our final count. Next, 40percent is the same as 60 percent, so everything is symmetric, so you just divide by 2.


Given that $\left\{\begin{array}{c}_ a+b\le c+1 \ b+c \le a+1 \ c+a\le b+1  \end{array}\right\}$ for the numbers $a,b,c \ge 0$, show that $a^2+b^2+c^2\le 2abc+1$.


function
real analysis
real analysis unsolved

Let $(a_{k})_{k}$ and $(b_{k})_{k}$ be two sequences in $l^{1}(\mathbb{N})$ such that for every integer $p \geq 1$: $\sum a_{k}^{p}= \sum b_{k}^{p}$. Does it follow that $(a_{k})_{k}$ is a permutation of $(b_{k})_{k}$ ?

It does, except for the zeros. In other words, the non-zero terms of the two sequences are the same (even with the same multiplicities): the functions $f(z)=\prod(1-a_{k}z)$ and $g(z)=\prod(1-b_{k}z)$ (the products being taken after all $a_{k}\ne 0$ and all $b_{k}\ne 0$) are equal entire functions, and the sets of $\frac{1}{a}$'s and $\frac{1}{b}$'s respectively are their zero sets (again, counting multiplicities).


quadratics
group theory
abstract algebra
probability
superior algebra
superior algebra unsolved

Let $ p \equal{} 6l \minus{} 1$ be a prime number, where $ l$ is also prime.

Show that for every  $ y\in\mathbb{F}_p$ we can find $ x\in\mathbb{F}_p$ that $ y^2 \equal{} x^3 \plus{} 1$.

First, let's answer that $ \mathbb{F}_p$ is the same with $ \mathbb{Z}_p$, the set of residues modulo $ p$.

Then, we'll clarify the good, terse post of t0rajir0u, while relaxing the condition for $ p$ to be just $ p \equal{}3\ell \minus{}1$, for any positive integer $ \ell$. The case $ p\equal{}2$ is trivial, so assume $ p\geq 3$.

Look at the mapping $ f : \mathbb{F}_p \to \mathbb{F}_p$ given by $ f(x) \equal{} x^3$. Assume $ f(x) \equal{} f(y)$, i.e. $ (x\minus{}y)(x^2\plus{}xy\plus{}y^2) \equal{} 0$. Then either $ x\equal{}y$, or $ x^2 \plus{} xy \plus{} y^2 \equal{} 0$, so $ 4(x^2 \plus{} xy \plus{} y^2) \equal{} (2x\plus{}y)^2 \plus{} 3y^2 \equal{} 0$. Now, $ y\equal{}0$ leads to $ x\equal{}0$ (so $ x\equal{}y$), while for $ y \neq 0$ it follows $ \minus{}3 \equal{} (2xy^{\minus{}1} \plus{} 1)^2$, meaning that $ \minus{}3$ needs be a quadratic residue modulo $ p$ (notice that for $ p \equal{} 3\ell \plus{} 1$ we have quite the reverse, for example for $ p\equal{}7$ one has $ 1^3 \equal{} 2^3$, so $ f$ is not anymore injective).

Using the well-known properties of the Legendre symbols we have $ \genfrac {(} {)} {} {} {\minus{}3} {p} \equal{} \genfrac {(} {)} {} {} {\minus{}1} {p} \genfrac {(} {)} {} {} {3} {p} \equal{} (\minus{}1)^{(p \minus{}1)/2} \genfrac {(} {)} {} {} {p} {3} (\minus{}1)^{(p\minus{}1)/2} \equal{} \genfrac {(} {)} {} {} {2} {3} \equal{} \minus{}1$, so $ \minus{}3$ is not a quadratic residue modulo $ p$.

Therefore $ f$ is injective, hence surjective (since $ \mathbb{F}_p$ is finite). So, for any $ z\in\mathbb{F}_p$ we can find $ x\in\mathbb{F}_p$ such that $ z \equal{} x^3$, not just for $ z\equal{}y^2\minus{}1$.

Equivalently, $ \mathbb{F}_p^{*}$ is a cyclic group with order coprime to $ 3$.

[quote="MathJimmy"]All right! And one more thing, why if we have a group of order $ 6l$ where $ l$ is prime and we choose a random element

then it is very unlikely that its order would be in $ \{1,2,3,6\}$[/quote]
If $ a$ is an element of order $ l$ then so is $ a^k, 1 \le k \le l \minus{} 1$, so there are many more of these elements than the others.  In other words, an element of large order automatically belongs to a large subgroup, whereas the opposite is true for elements of small order.

I think all group you've mentioned make the same group of order $ l$.

We have $ l\mid 6l$ then we have an element of order $ l$ and so, $ l \minus{} 1$ elements of order $ l$ (every element (except 1) from group generated by element of order $ l$ has order $ l$).

Maybe like this: if $ k_i$ is the number of elements of order $ i\in\{1,2,3,6\}$ then we have at least $ k_i\cdot (l \minus{} 1)$ elements of order $ i\cdot l$.

By counting elements:

$ 1 \plus{} k_2 \plus{} k_3 \plus{} k_6 \plus{} (l \minus{} 1) \plus{} (l \minus{} 1)(k_2 \plus{} k_3 \plus{} k_6)\leq 6l$

so

$ k_2 \plus{} k_3 \plus{} k_6\leq 5$

[quote="mavropnevma"]while relaxing the condition for $ p$ to be just $ p \equal{} 3\ell \minus{} 1$, for any positive integer $ \ell$. The case $ p \equal{} 2$ is trivial, so assume $ p\geq 3$.[/quote]

And just how many primes larger than 2 do you know of that are -1 mod 3 but not -1 mod 6?

The relaxation had mainly to do with $ \ell$ not required to be a prime! And as we were there, $ p \equal{} 6\ell \minus{} 1$ would disallow $ p \equal{} 2$, for which it still was true, so why not go whole hog?

By the way, a comment for LydianRain's eyes only
[hide]Your signature, while sarcastically good humored, is a bit offensive, similar in mood with Mercutio's uttering "a curse on both your houses".[/hide]

Ah, hadn't noticed that "restriction" in the original problem, sorry.

And yes, it's supposed to be offensive–I feel strongly about the issues at hand, as did Mercurio. :)

[quote]
We have $ l\mid 6l$ then we have an element of order $ l$ and so, $ l \minus{} 1$ elements of order $ l$ (every element (except $ 1$) from group generated by element of order $ l$ has order $ l$).

Maybe like this: if $ k_i$ is the number of elements of order $ i\in\{1,2,3,6\}$ then we have at least $ k_i\cdot (l \minus{} 1)$ elements of order $ i\cdot l$.

By counting elements:

$ 1 \plus{} k_2 \plus{} k_3 \plus{} k_6 \plus{} (l \minus{} 1) \plus{} (l \minus{} 1)(k_2 \plus{} k_3 \plus{} k_6)\leq 6l$

so

$ k_2 \plus{} k_3 \plus{} k_6\leq 5$
[/quote]

The dihedral group $ D_{42}$ has order $ 42\equal{}6\cdot7$ and $ 24$ elements of order in $ \{1,2,3,6\}$. It has $ 21$ elements of order $ 2$, yet no elements of order $ 14$ as you predicted.

Steve

[quote="MathJimmy"]All right! And one more thing, why if we have a group of order $ 6l$ where $ l$ is prime and we choose a random element then it is very unlikely that its order would be in $ \{1,2,3,6\}$[/quote]
[quote="SD5225272"]
The dihedral group $ D_{42}$ has order $ 42 \equal{} 6\cdot7$ and $ 24$ elements of order in $ \{1,2,3,6\}$.[/quote]

The other Frobenius group of order 42 (the holomorph of the cyclic group of order 7, the normalizer in S7 of a Sylow 7-subgroup) is even worse.  It has 1+7+14+14=36 elements of orders in {1,2,3,6}, and only 6 other elements, so the probability is more than 85% that a random element has order in {1,2,3,6}.

For any prime p congruent to 1 mod 6, the holomorph of Z/pZ is pretty bad, and has a 1 - (p-1)/(6p) = 5/6 + 1/6p > 83% chance of a random element having order in {1,2,3,6}.

I forgot to type that a group has to be abelian. I think that argumentation is ok then.

Please tell me if its all right or not  :roll:

In that case, assuming $ l\neq2,3$, the probability of a randomly chosen element having order in $ \{1,2,3,6\}$ is $ \frac{1}{l}$.

Steve

Then we have '$ \equal{}$' there where I had '$ \leq$' (because the other assumption which I hadn't written was that $ \ell$ is large :) ) ?


induction
linear algebra
linear algebra unsolved

Let $ X_{1},X_{2},\dots, X_{n}$ and $ Y_{1},Y_{2},\dots ,Y_{n}$ subspaces of $ X$. Prove that
$ (X_{1}\cap Y_{1})+(X_{2}\cap Y_{2})+\dots+(X_{n}\cap Y_{n})\subseteq (X_{1}+X_{2}+\dots+X_{n})\cap (Y_{1}+Y_{2}+\dots+Y_{n})$.

I tired induction, and it didn't quite work :(

take $ \alpha \in (X_{1}\cap Y_{1})+(X_{2}\cap Y_{2})+\dots+(X_{n}\cap Y_{n})$

then $ \alpha = \alpha_{1}+...+\alpha_{n}$ with $ \alpha_{i}\in X_{i}\cap Y_{i}$, for $ i=1,..n$

so $ \alpha_{i}\in X_{i}$ and  $ \alpha_{i}\in Y_{i}$ for $ i=1,..n$

now, you should be able to finish the proof yourself by showing that $ \alpha$ is also in the right hand side  :)


help me:
solve:
coscoscoscosx=sinsinsinsinx

Try to prove that $ \cos \cos \cos \cos x > \sin \sin \sin \sin x$ (in radians; the problem is trivial in degrees), so there are no solutions.

[hide="Hint"] $ \cos x$ is between $ 1$ and $ \minus{}1$.  What values can $ \cos (\cos x)$ take?  Etc. [/hide]


limit
calculus
function
derivative
integration
Irrational numbers

We should prove, that for all sequences $ \{a_i\}, a_i\in N$, such that $ 1<a_1<a_2<...$
\[ \sum_{a_i}^{}{\frac{2^{a_i}}{a_{i}!}}\]
will be irrational number.
[b](*)[/b] Let for all sequences it will be rational number.
So:
\[ \sum_{a_i}^{}{\frac{2^{a_i}}{a_{i}!}}\equal{}\frac{p}{q}\]
Let sequence $ \{b_i\}, b_i\in N$, such that $ 1<b_1<b_2<...$ and $ b_i \ne a_j$. Using formila
\[ e^x\equal{}\sum_{i\equal{}0}^{\infty}{\frac{x^n}{n!}}\]
we will have:
\[ e^2\minus{}1\equal{}\sum_{a_i}^{}{\frac{2^{a_i}}{a_{i}!}}\plus{}\sum_{b_i}^{}{\frac{2^{b_i}}{b_{i}!}}\]
But
\[ \sum_{a_i}^{}{\frac{2^{a_i}}{a_{i}!}}\equal{}\frac{p}{q}\]
So, we have:
\[ \sum_{b_i}^{}{\frac{2^{b_i}}{b_{i}!}}\equal{}e^2\minus{}1\minus{}\frac{p}{q}\]
which is irrational, because $ e^2$ is irrational. This is contradiction with [b](*)[/b].

Unfortunately, this does not prove the thesis. The negation of the proposition "for all sequences $\{a_i\}$, $a_i \in \mathbb{N}$, with $1 < a_1 < a_2 < \cdots$, $\sum_{i=1}^{\infty} \frac {2^{a_i}} {a_i!}$ is irrational"[b] is not [/b](*) "for all sequences $\{a_i\}$, $a_i \in \mathbb{N}$, with $1 < a_1 < a_2 < \cdots$, $\sum_{i=1}^{\infty} \frac {2^{a_i}} {a_i!}$ is rational", so the above only proves that if, for such a sequence $\{a_i\}$, its associated series has a rational limit, then for the complementary sequence $\mathbb{N}^* \setminus \{a_i\}$,  its associated series has an irrational limit.

I'm confused. I see, suppose its not irrational. Then it is rational.
Bla bla.
Contradiction.
Hence its irrational. :S

Try to do this one:
Let $S_{n} = \sum^{n} \frac{2^{a_{i}}}{a_{i}!}$.
Then of course
\[
S_{n+1} = \frac{(a_{1}!a_{2}!...a_{n}! \cdot A_{n})}{a_{n+1}!}
\]
where $A_{n} \in \mathbb{N}$
It's enough to prove that, there exist sequance $p_{n}, q_{n} \in \mathbb{N}$ such that
$|S_{\infty}-p_{n}/q_{n}| = o(1/q_{n})$

But  on the other hand
\[
|S_{\infty} - S_{n}| = |S_{\infty} - \frac{(a_{1}!a_{2}!...a_{n-1}! (A_{n-1}))}{a_{n}!}| = \sum_{j=n+1}^{\infty}\frac{2^{a_{j}}}{a_{j}!}
\]
Try to prove, that
\[
\lim_{n \to \infty} \frac{a_{n}!}{a_{n-1}!}\sum_{j=n+1}^{\infty}\frac{2^{a_{j}}}{a_{j}!} \to 0
\]

Try to use Stirlnig formula.

P.S.

I'm not sure is it true or not, but it's one of the possible ways  approach to such problem.

I will borrow a quill from Liouville's quiver.

For a given sequence $0\leq a_1 < a_2 < \cdots$ of non-negative integers, let me denote by $\displaystyle E_n = \sum_{a_k \leq n} \dfrac {2^{a_k}} {a_k!}$. Since the sequence $(E_n)_{n\geq 1}$ is non-decreasing, and since $\displaystyle E_n < \sum_{k=0}^{\infty} \dfrac {2^k} {k!} = \textrm{e}^2$, it follows it is convergent to its supremum $\displaystyle E = \lim_{n\to \infty} \sum_{a_k \leq n} \dfrac {2^{a_k}} {a_k!}$.
Assume $E = \dfrac {p} {q}$, with $p,q \in \mathbb{N}^*$, $\gcd(p,q) = 1$. Let us denote by $v_2(N)$ the exponent of $2$ in the canonical factorization of $N$, and take $\alpha = v_2(q)$. It is immediate from Legendre's formula that $v_2(n!) \leq n-1$, with equality if and only if $n=2^m$ for some positive integer $m$. Take then a large enough $m$ (thus a large enough $n$), so that $q \mid \dfrac {n!} {2^{n-1}} 2^{\alpha}$, hence $\dfrac {n!} {2^{n-1}} 2^{\alpha} E = \dfrac {n!} {2^{n-1}} 2^{\alpha}\dfrac {p} {q} \in \mathbb{N}^*$. On the other hand, it is clear that $\dfrac {n!} {2^{n-1}} 2^{\alpha} \dfrac {2^k} {k!} \in \mathbb{N}^*$ for all $0\leq k\leq n$, hence $\dfrac {n!} {2^{n-1}} 2^{\alpha} E_n \in \mathbb{N}^*$.

Now, $\displaystyle 0< E-E_n \leq \sum_{k=n+1}^{\infty} \dfrac {2^k} {k!} < \sum_{k=n+1}^{\infty} \dfrac {2^k} {n!(n+1)^{k-n}}$, therefore $\displaystyle 0< \dfrac {n!} {2^{n-1}} 2^{\alpha} (E-E_n) <$ $ 2^{\alpha+1} \sum_{k=n+1}^{\infty} \dfrac {2^{k-n}} {(n+1)^{k-n}} =$ $ 2^{\alpha+1} \sum_{k=n+1}^{\infty} \dfrac {1} {((n+1)/2)^{k-n}} =$ $ 2^{\alpha + 1} \dfrac {2} {n+1} \cdot \dfrac {1} {1 - 2/(n+1)} =$ $ \dfrac {2^{\alpha+2}} {n-1} < 1$ for large enough $m$ (and so large enough $n$). On the other hand, according with the above, $\displaystyle \dfrac {n!} {2^{n-1}} 2^{\alpha} (E-E_n)$ should be an integer, thus establishing a contradiction.

Re the above post: I do not think that asymptotic formulae (like Stirling's) can decide on the rationality of a limit expression.

[quote="mavropnevma"]
Re the above post: I do not think that asymptotic formulae (like Stirling's) can decide on the rationality of a limit expression.[/quote]
I don't understand well why you think so.
In my case, you just need to prove that the limit
\[
\lim_{n\to\infty}\frac{a_{n}!}{a_{n-1}!}\sum_{j=n+1}^{\infty}\frac{2^{a_{j}}}{a_{j}!}
\]
is equal to zero. And one of the best way to show that is, to use stirling formula.

yes, you are right, as always I made mistake in my calculations :(.
Therefore Let me try to explain in more approximately language, since the sence was not in the exact calculation, it was just in approximations, but anyway my first approximation was not correct.
since $ a_{j}!/a_{i}! = 2^{a_{j}-a_{i}-100}b_{i,j}$ for $j>i$, where $b_{i,j} \in \mathbb{N}$ hence
$S_{n+1}  =\frac{2^{a_{n+1}-100}b_{n+1} }{a_{n+1}!}$ where $b_{n+1} \in \mathbb{N}$
Then, if you again repeat the same things, it's enough to prove such limit
\[
\lim_{n \to \infty} \frac{a_{n}!}{2^{a_{n}}}\sum_{j=1}^{\infty}\frac{2^{a_{n+j}}}{a_{n+j}!}=0
\]
Which again seems to be true by stirling formula :).

[quote="mavropnevma"]It's enough for what? You propose an estimate of a limit, without stating what is the use you propose to make of it. Being a little more exact, and clearer in your purposes will help everybody.[/quote]
For the irrationality of $\sum_{j=1}^{\infty} 2^{a_{j}}/a_{j}!$, it's enough to prove  that the limit
\[
 \lim_{n\to\infty}\frac{a_{n}!}{2^{a_{n}}}\sum_{j=1}^{\infty}\frac{2^{a_{n+j}}}{a_{n+j}!}
 \]
 is equal to zero.

That is a [b]statement[/b], not a [b]proof[/b]. If you will elaborate on it, you will see that it amounts to the proof I gave in full detail in a previous post.

Yes :) but your post was [b]after my[/b], therefore I was not read it. What you have done, you just reduced  the problem with the same  limit  what I have written (because we are using one and the same things to  proving irrationality). Yes I agree that you have written it more detailed proof (because Only now I read it). I just was  posting some hints (with one error calculation :) ) and that is all, but anyway the idea was right.
As you see (and have done) my limit is  equal to zero, since
\[
\frac{a_{n}!}{2^{a_{n}}}\sum_{j=1}^{\infty}\frac{2^{a_{j}}}{a_{j}!} <
\frac{a_{n}!}{2^{a_{n}}}\sum_{j=a_{n+1}}^{\infty}\frac{2^{j}}{j!} <
x_{n}:=\frac{(a_{n+1}-1)!}{2^{a_{n+1}-1}}\sum_{j=a_{n+1}}^{\infty}\frac{2^{j}}{j!} \to 0
\]
since
\[
A_{n}=\frac{n!}{2^{n}}\sum_{j=1}^{\infty}\frac{2^{j+n}}{(j+n)!} \to 0
\]
for example by stirling formula (this is about what I was talking on each my posts and was answering on your first question), or for example by the simplest estimates what already have you written like this one
\[
\frac{1}{(n+j)!} < \frac{1}{(n+1)!j!}
\]
And the sequence $x_{n}$ is the  subsequence of $A_{n}$.
But anyway, I remaind you that I just was answering on the your first question

[quote="Extremal"][quote="mavropnevma"]
Re the above post: I do not think that asymptotic formulae (like Stirling's) can decide on the rationality of a limit expression.[/quote]
I don't understand well why you think so.
[/quote]
hence this sentences
[quote="mavropnevma"]
If you will elaborate on it, you will see that it amounts to the proof I gave in full detail in a previous post.
[/quote]
is more adressed to you not to me (because your solution was written after my post)
(with one remarks:
If you will elaborate on it, you will see that your detailed proof, it is detailed explanation of  my hints I gave in  my first post
[quote="Extremal"]
It's enough to prove that, there exist sequance $p_{n}, q_{n} \in \mathbb{N}$ such that
$|S_{\infty}-p_{n}/q_{n}| = o(1/q_{n})$

[/quote]
and the  proof of the limit
[quote="Extremal"] it's enough to prove such limit
\[
\lim_{n \to \infty} \frac{a_{n}!}{2^{a_{n}}}\sum_{j=1}^{\infty}\frac{2^{a_{n+j}}}{a_{n+j}!}=0
\]
Which again seems to be true by stirling formula :).[/quote]
.)
:)
)

You missed the most important point that for infinitely many $a$, the number $\frac{a!}{b!2^{a-b}}$ is an integer for all $1\le b<a$. Your tail estimate is the same as mavropnevma's but he proves that the beginning segment is an integer and you don't.

Anyway, this discussion is sort of boring. How about the same problem with $3$ instead of $2$?

Well, Aigner and Ziegler, in their Proofs from the Book, claim that proving that $\textrm{e}^3$ is irrational (certainly a most particular case of what you asked, fedja) requires heavier machinery (bit of calculus), using the function $\frac {x^n(1-x)^n} {n!}$ with various differentiation and integration tricks. Will that idea hold for the "holey" sequence $a_i$ ?

I agree with mavropnevma and it seems to be open problem in the case 3. and not only in this case, also  for all $m>2$


need help!

Bob has a collection of 5 red marbles and 5 green marbles.  If he picks a green marble, he wins.  He also has 2 boxes.  The marbles are placed in the box such that he has the greatest probabilty of winning.  First a box is chosen, and then a marble.  What is th probability that he wins

[hide="answer"]First, fill one box so its probability is the highest it can be. So, in box 1, there is 1 green, no reds. So, in the other, there are 4 greens, 5 reds. No, our desired probability is 1/2+2/9=13/18.[/hide]


This description was a problem on the SAT:

A, B, C, and D are points on a line, with D the midpoint of segment BC. The lengths of segments AB, AC, and BC are 10, 2, and 12, respectively. What is the length of segment AD?

[hide]the order of the letters are c, a, d, and b... since ac =2 and cd = 6, then ad = 4[/hide]

[hide]Yep, the answer is 4.

I always wondered why they made problems on the SAT simple enough so that many middle school students can get a perfect score.[/hide]


f: $ N*\minus{}>N* : f(m\plus{}f(n)) \equal{} n \plus{} f(2008\plus{}m)$

See here http://www.mathlinks.ro/viewtopic.php?p=1221707#1221707 .


geometry
3D geometry
group theory
abstract algebra
topology

Let $ X$ be the space obtained by attaching  a $ 2$-cell to the torus $ T^2\equal{}S^1\times S^1$. ( the attaching map is not known)  What are the possible homology groups of $ X$.

Solution:

$ X$ is the mapping cone of $ S^1 \to T$, so we get a long exact sequence in homology. Noting that all these spaces have zero homology above dimension 2, and that $ H_2(S^1) \equal{} 0$, this gives the sequence

$ 0 \to H_2(T) \to H_2(X) \to H_1(S^1) \to H_1(T) \to H_1(X) \to 0$.

The map $ H_1(S^1) \to H_1(T)$ must be either the zero map or injective, since $ H_1(T)$ is torsion-free. If it is the zero map, then $ H_2(X) \equal{} H_2(T) \oplus \mathbb{Z}$ and $ H_1(T) \equal{} H_1(X)$. If it is injective, then $ H_2(T) \equal{} H_2(X)$ and $ H_1(X) \equal{} H_1(T) / H_1(S^1)$. These are the only two possible cases, determined by whether or not the attaching map is nullhomotopic in $ T$. Of course, the quotient $ H_1(T) / H_1(S^1)$ can be any quotient of $ H_1(T)$ by a cyclic subgroup of itself, depending on the attaching map used.


[quote="silouan"]$\sqrt{a^2 + (\frac{b-c}{2})^2}+\sqrt{b^2 + (\frac{c-a}{2})^2}+\sqrt{c^2 + (\frac{a-b}{2})^2} < 2$[/quote]
This is obviously not true for any $a,b,c$.

It looks like $a+b+c=1$ gives a true inequality.

I can only prove it when the following system of inequalities are satisfied. :P
$\begin{cases} a+5b<3 \\ b+5c<3 \\ c+5a<3 \end{cases}$.


I know this has been posted before...

But what score counts for USAMO qualification if you take both AMC 12A and 12B?

I read on an earlier post by JBL that said the higher score counted, but I just wanted to confirm this and make sure that the rules didn't change.

Thanks.

Your better score is the one that counts.


calculus
integration
trigonometry
logarithms
calculus computations

Compute the following integral $ \int \frac{dx}{\cot \frac{x}{2}\cdot \cot \frac{x}{3}\cdot \cot \frac{x}{6}}$

well, i am able  to do this,  but it is leading to very cumbersome  calculation .  anyone out with a simple method?

$ P \equal{} \int \frac{1}{\cot \frac{x}{2}\cdot \cot \frac{x}{3}\cdot \cot \frac{x}{6}}\,dx \equal{} \int \tan \frac{x}{6} \tan \frac{x}{3} \tan \frac{x}{2} \,dx$

$ \equal{} 6 \int \tan (u) \tan (2u) \tan (3u) \,du$

Observe now that

$ \tan(3u \minus{} u) \equal{} \frac{\tan(3u) \minus{} \tan(u)}{1 \plus{} \tan(3u) \tan(u)} <\equal{}> \tan(2u) [1 \plus{} \tan(3u) \tan(u)] \equal{}$

$ \equal{}  \tan(3u) \minus{} \tan(u) <\equal{}> \tan (u) \tan (2u) \tan (3u) \equal{} \tan(3u) \minus{} \tan(2u) \minus{} \tan (u)$.

Therefore,

$ P \equal{} 6 \int \left(\minus{} \frac13 \frac{\minus{}3 \sin(3u)}{\cos(3u)} \plus{} \frac12 \frac{\minus{}2 \sin(2u)}{\cos(2u)} \plus{} \frac{\minus{}\sin(u)}{\cos(u)}\right)\,du$

$ \equal{} \minus{}2 \ln|\cos(3u)| \plus{} 3 \ln|\cos(2u)| \plus{} 6 \ln |\cos(u)| \plus{} C$

$ \equal{} \minus{}2 \ln| \cos(x/2)| \plus{} 3 \ln| \cos(x/3)| \plus{} 6 \ln|\cos(x/6)| \plus{} C$.


function
integration
logarithms
inequalities
topology
complex analysis
complex analysis open

Suppose that $f:\mathbb D\to\mathbb D$ is continuous and such that $f$ is analytic on the open set $\{|f|>\delta\}$, where, as usual, $\mathbb D$ is the unit disk and $\delta>0$ is some very small number. Is it true that then there exists a function $g$ analytic in $\mathbb D$ such that $|f-g|<\delta^{\alpha}$ in $\mathbb D$ for some absolute $\alpha>0$? If the answer is "No", is it true, at least, that we can approximate $f$ in the disk $\frac12\mathbb D$?

No. We cannot make the error smaller than 1 in general, as this example shows. Let $f(z)=\frac{\delta}{z}$ for $|z|>\delta$ and $f(z)=\frac{\overline{z}}{\delta}$ for $|z|\le\delta$. Then for any $g$ analytic in $\mathbb{D}$, $\int_C f(z)-g(z)\, dz=2\pi i\delta$, where $C$ is the circle of radius $\delta$. If $\epsilon$ is the maximum error in our approximation, we have $\left|\int_C f(z)-g(z)\, dz\right|\le2\pi\delta\epsilon$, so $\epsilon\ge1$.

Of course, the zero function approximates $f$ that well. In general, we can't do better.

[quote="jmerry"]$f(z)=\frac{\overline{z}}{\delta}$ for $|z|\le\delta$. [/quote]
This is not analytic on the set where $|f|>\delta$! (the function must be  analytic where $f(z)$ is large, not where $z$ is large).

I will estimate on the circle, but it's possible that you can extract something useful from here
I have  left some details, but I'm sure that  you will complete  this all  details too.

$dist(f , H^{\infty}) = \sup_{F \in H_{0}^{1},  ||F||_{1}=1} \left| \frac{1}{2\pi} \int f F d\theta \right|$

note that
$ \nabla f  \nabla F = f_{x}F_{x}+f_{y}F_{y} = 2 \left( \frac{\partial f}{\partial \bar z } \right)  \left(\frac{\partial F}{\partial z} \right)$
and since $F(0)=0$ then by the polarization  Littlewood-Paley identity we have:
$\frac{1}{2\pi}\int f F d\theta = \frac{2}{\pi} \iint \frac{\partial f}{\partial \bar z} \frac{\partial F}{\partial z} \log \frac{1}{|z|}dxdy$
WLOG $F=G^{2}$ where $G \in H^{2}$ and $||G||_{2}^{2} \leq 2||F||_{1}$ we can do it because:
put $F = BH$ where $B$ is Blashke product and $H$ has no zeros,  then
$F = \frac{(B-1)H+(B+1)H}{2} =\frac{G_{1}^{2}+G_{2}^{2}}{2}$ where $G_{j} \in H^{2}$ and $||G_{j}||_{2}^{2} \leq 2||F||_{1}$
So our aim is to estimate :
$I = \frac{4}{\pi} \iint_{|f|>\delta} \frac{\partial f}{\partial \bar z}G(z) \frac{\partial G}{\partial z} \log \frac{1}{|z|}dxdy+\frac{4}{\pi} \iint_{|f|\leq \delta} \frac{\partial f}{\partial \bar z}G(z) \frac{\partial G}{\partial z} \log \frac{1}{|z|}dxdy= \frac{4}{\pi} \iint_{|f|\leq \delta} \frac{\partial f}{\partial \bar z}G(z) \frac{\partial G}{\partial z} \log \frac{1}{|z|}dxdy$
Now use Schwarz's Inequality  :
$|I| \leq  C \left(\iint_{|f| \leq \delta} \left| \frac{\partial f}{\partial \bar z} \right|^{2} |G|^{2} \log \frac{1}{|z|}dxdy  \right)^{1/2}\times \left(\iint \left|\frac{\partial G}{ \partial z} \right|^{2} \log \frac{1}{|z|}dxdy \right)^{1/2} \leq C (||f \chi_{|f|\leq \delta}||_{BMO} ||G||_{2}) ||G||_{2} \leq C ||f \chi_{|f|\leq \delta}||_{\infty} ||F||_{1} \leq C \delta$
by theorem about carlsons measure and  also there was used some known theorem that if $g \in L^{1}(\mathbb{T})$ and $d\lambda_{g} := |\nabla g|^{2} \log \frac{1}{|z|}dxdy$ then $g \in BMO(\mathbb{T})$ if and only if
$\lambda_{g}$ is carlsons measure, and there exist absolute constants $C_{1}$ and $C_{2}$ such that
$C_{1}||g||_{BMO}^{2} \leq N(\lambda_{g}) \leq C_{2} ||g||_{BMO}^{2}$ here $N(\lambda_{g})$ is carlsons norm.

what I have written it's of course true for harmonic functions, but what I got it's trivial things :(


[quote="can_hang2007"][quote="onlylove_math"]Let $ a;b;c;k > 0$ and $ abc \equal{} 1$. Prove that:
$ a^2b^4 \plus{} b^2c^4 \plus{} c^2a^4 \ge\ \frac {9}{3 \plus{} k(a^2b \plus{} b^2c \plus{} c^2a)} \plus{} \frac {k(a \plus{} b \plus{} c)^2}{(k \plus{} 1)(ab \plus{} bc \plus{} ca)}$[/quote]
Notice that
\[ a^2b^4 \plus{} b^2c^4 \plus{} c^2a^4 \ge \frac {3}{2}\sum \frac {a \plus{} b}{c} \minus{} 6
\]
and
\[ a^2b \plus{} b^2c \plus{} c^2a \ge 3
\]
It suffices for us to prove that
\[ \frac {3}{2}\sum \frac {b \plus{} c}{a} \minus{} 6 \ge \frac {3}{k \plus{} 1} \plus{} \frac {k(a \plus{} b \plus{} c)^2}{(k \plus{} 1)(ab \plus{} bc \plus{} ca)}
\]
for all $ a,b,c,k > 0$.
\[ \Leftrightarrow \frac {3}{2} \sum \frac {(a \minus{} b)^2}{ab} \ge \frac {k\sum (a \minus{} b)^2}{2(k \plus{} 1)\sum ab}
\]

\[ \Leftrightarrow \sum (a \minus{} b)^2\left( \frac {3}{ab} \minus{} \frac {k}{(k \plus{} 1)(ab \plus{} bc \plus{} ca)}\right) \ge 0
\]
which is obviously true because
\[ \frac {3}{ab} \minus{} \frac {k}{(k \plus{} 1)(ab \plus{} bc \plus{} ca)} \ge \frac {3}{ab} \minus{} \frac {1}{ab \plus{} bc \plus{} ca} \equal{} \frac {2ab \plus{} 3bc \plus{} 3ca}{ab(ab \plus{} bc \plus{} ca)} > 0
\]
We have done.
:)[/quote]
Very nice , Thank you very much :lol: . And with stronger ineq
Let $ a;b;c;k > 0$ and $ abc \equal{} 1$. Prove that:
$ a^2b^4 \plus{} b^2c^4 \plus{} c^2a^4 \ge\ \frac {1\plus{}kbc^2}{1\plus{}kb^2c}\plus{}\frac {1\plus{}kca^2}{1\plus{}kc^2a}\plus{}\frac {1\plus{}kab^2}{1\plus{}ka^2b}$

Setting :$ x\equal{}ab^2,y\equal{}bc^2,z\equal{}ca^2$
This ineq is equivalent to :
$ \sum \frac{a^2\minus{}1}{1\plus{}\frac{k}{a} } \ge 0$ Which is obvious true by am-gm and chebysev


[hide]The expression will always be divisible by 29. Here all congruences will be mod 29.

By Fermat's Little Theorem, $2^{28k}\equiv 1\Rightarrow 2^n\equiv 2\Rightarrow 2^{2n}\equiv 4$.

Then $(2^{2n}+1)^2+2^2\equiv (4+1)^2+2^2\equiv 0$.

Since the expression is divisible by 29 and greater than or equal to $(2^{29}+1)^2+2^2$ which is much greater than 29, it is composite.[/hide]


I was doing some precalc problems and suddenly started wondering what type of function $ f(x) \equal{} \sqrt{x}$ is, can someone please help?  :blush:

Umm... a strictly increasing function?

What exactly do you mean? You can classify functions based on any characteristic...


symmetry
geometry
incenter
geometric transformation
homothety
trigonometry
angle bisector

[hide]By symmetry, F is the midpoint of arc ED, and so <EFK = <DFK. So FK is the angle bisector or <DFE. Likewise, DN and DM bisect the other two angles of triangle DEF. So, the three angle bisectors of triangle DEF meet at the incenter of DEF.[/hide]

[quote="zabelman"]By symmetry, F is the midpoint of arc ED, [/quote]
Can you explain why?

coz $AE = AD$ and $K$ and $F$ lies on bisector of angle $\angle EAD$

[quote="Risk"]$K$ and $F$ lies on bisector of angle $\angle EAD$[/quote]
How can that be true?  :?

If you know that each of those points is a tangent, then each segment connects two points of tangency.  If each is perpendicular at the two points, then each must pass through the incenter, right?

[quote="DPopov"]I think Zach meant to say $EN$ bisects an angle of $\triangle DEF$ instead of $DN[/quote]
The point is how!

[quote="DPopov"]$A$, $K$, and $F$ are collinear since $AE$ and $AD$ are tangent to both circles and also the circles are externally tangent to one another.[/quote]
Maybe it is fool to ask such a question...but why? Can you give a proof?

[quote="DPopov"]$A$ is the center of homothecy and maps point $K$ to point $F$. Similarly $B$ is the center of homothecy and maps point $N$ to point $E$ etc. Will this suffice?[/quote].

 I still have some questions if this solution is true.POints $A , K , F$ are not homothetic since $AK \perp ED$ but $AK$ in not necessary perpendicular to $BC$.

[u] Babis[/u]

you can use this theorem:
[b]assume that the incircle of $\triangle ABC$ is tangent to the sides at points $D,E,F$ and let $X,Y,Z$ be three arbitary points on the circle,then $DX,EY,ZF$ are concurrent iff $AX,BY,CZ$ were concurrent[/b]...

[quote="silouan"]Here is my idea .
Take the small circle that is tangent to BC at S and to AB at T and to the incircle(with center I) at M and let K its center and let .Because KT=KS we have that K lies on the bisector of the angle B . So B, K , I are collinear .
But we know that when two circles are externally tangent at a point , then this point lies on the line of their centers . So B, K,M,I are collinear .But IM is perpendicular to EF so M is the midpoint of the arc EF and we are done  :wink:[/quote]

 Thank you Silouan.

 Of course , the main problem was that points $A,K,F$ are not homothetic. The first solution was correct , but it was not nicely given. After your message , the solution can be understood by anybody.(Some letters in your solution  do not refer to ath given figure.For example your $E ,M$ must become $D,N$).

 [u]Babis[/u]

[color=darkblue][b]Remark.[/b] Is evidently that the intersection is the incenter of the triangle $DEF$ ![/color]

[color=darkred][b][u]Lemma[/u][/b] (well-known "atomic bomb"). Let $ABCDEF$ be a cyclic hexagon. Then $AD\cap BE\cap CF\ne\emptyset\Longleftrightarrow AB\cdot CD\cdot EF=BC\cdot DE\cdot FA$.[/color]

[color=darkblue][b]Proof.[/b] Apply the above lemma to the cyclic hexagon $KEMFND$, where $\{\begin{array}{c}DK=KE\\\ EM=MF\\\ FN=ND\end{array}$.
The relation $KE\cdot MF\cdot ND=EM\cdot FN\cdot DK$ is evidently. Therefore, the lines $KF$, $EN$, $MD$ are concurrently.[/color]

Proof of the above 'atomic bomb':
Since $DK=KE$, a.s.o., it follows that $DM, NE, FK$ are the angle bisectors of $\Delta DEF$, which are concurrent at its incenter.

Best regards,
sunken rock

Note that $N$ is the midpoint of arc $FD$ and same for the others.

This is because $A$ is the external center of similitude of the circles, and thus takes center to center, and since the center, $A$ and $K$ are collinear, $K \to F$.

Thus, $AK$ is an angle bisector, and since $\triangle AED$ is isosceles, by symmetry $K$is the midpoint of arc $\hat{ED}$.

Looking at triangle $\triangle EDF$ we have concurrency iff:

$\prod_{cyc} \dfrac{\sin EDM}{\sin FDM} = \prod_{cyc} \dfrac{EM}{FM}$.

But, these are the midpoints, so the product is $1$ meaning they are concurrent $\blacksquare$.

But wait, Remember that the angle bisector meets at the midpoints of arcs, thus their concurrence point if the incentre.


Just a few days ago:

http://www.mathlinks.ro/viewtopic.php?t=242609


Ring Theory
superior algebra
superior algebra unsolved

I want to find prime and jacobson radical radicals in Z[T]/(T^3), here Z = integers.

Is it true that Z[T]/(T^3) is a field, because T^3 is irreducibel over Z[T]. If it is true that
Z[T]/(T^3) is a field then 0 is the prime and jacobson radical radical.

Is it true please help.

$ T^3$ is not irreducible in $ \mathbb{Z}[T]$, since you have $ T^3 \equal{} T \cdot T \cdot T$. Therefore the quotient ring $ \mathbb{Z}[T] / (T^3)$ is not a field, and so the problem is not quite as trivial as you were hoping  :)

so how does prime ideals and maximal ideals in Z[T]/(T^3) look like?

You might want to represent the prime ideals of the quotient ring as prime ideals of Z[T] (with a crtain condition) to make things a bit clearer here.

Yes, I think that the prime ideals in the ring are T and T^2.

Let P be a prime ideal in the ring so T*T^2 = 0 which implies that T is in P or T^2 is in P.

But is it true that the prime and jacobson radical is 0. Please help.

[quote="peteryellow"]Yes, I think that the prime ideals in the ring are T and T^2.[/quote] You mean the ideals generated by (the residue classes of) T or T^2 I suppose. Can you prove the first is prime and second is not?

[quote]Let P be a prime ideal in the ring so T*T^2 = 0 which implies that T is in P or T^2 is in P. [/quote] You could go on with your argument to get a pretty useful result here.

[quote]But is it true that the prime and jacobson radical is 0. Please help.[/quote]So how many prime ideals are there? And what definition of Jacobson ideal do you use (something like nilpotent elements in your definition)?

$ rad(T^3) \equal{} rad(T) \equal{} (T)$ since $ (T)$ is prime. a maximal ideal of $ Z[T]$ containing $ T^3$ also contains $ T$, i.e. corresponds to a maximal ideal of $ Z[T]/T \equal{} Z$, i.e. has the form $ (p,T)$ for some prime number $ p$. if $ f$ is in all those ideals, $ f(0)$ is divisible by every prime number, so that $ f \in (T)$. this shows $ jac(T^3)\equal{}(T)$.

thus $ nil(Z[T]/T^3)\equal{}jac(Z[T]/T^3) \equal{} (\overline{T})$.

My definition of jacobson radical is that it is intersection of maximal ideals and prime radical iks intersection of prime radical.

Can you please explain in more details, how do you get that rad(T.T^2) = rad(T) = (T). and are you using the same definitions as I am using.

[quote="peteryellow"]How can I show that the ideal generated by T i.e, (T) is prime but (T^2) is not prime.[/quote]Hmm, I would do it in the ring $ Z[T]$  first, then translate it in terms of the quotient ring - already proposed in [url]http://www.mathlinks.ro/viewtopic.php?p=1238139#1238139[/url].  :wink:

Third Edition can you please helppppppp me in detail because I am not sure about how to solve this thanks.

you should try to fill in the details in the proof. I use the same definitions and only use simple standard facts such as $ rad(a^n) \equal{} rad(a)$, ideal correspondence in factor rings and the ideal structure of Z.

$ (T^2)$ is not prime because $ T^2$ is not irreducible.

Hmm, to be honest I am not sure where your problems are. You might show whar you have done so far.

There is already a sketchy proof given by -oo-. But again, try to represent the ideals in the quotient ring as ideals in Z[T] with certain conditions. Then apply the definitions.

Sorry about not being much more specific.

[quote="-oo-"] $ nil(Z[T]/T^3) \equal{} jac(Z[T]/T^3) \equal{} (\overline{T})$.[/quote]
How do you define $ \overline{T}$?

[quote="puuhikki"]How do you define $ \overline{T}$?[/quote]I interpret it as iamge of T under $ Z[T] \rightarrow Z[T]/(T^3)$.


Daisuke Matsuzaka rolled one of his beloved Dice $k$ times. The sum of the numbers on the top face was $18$, his number. For all values of $k$ from 1-4, how many different ways could this have happened.

Bad puns ROCK!

[hide]
For k=3, obviously only 1 way.
For k=4, I would do balls and urns, but I don't know how to apply it when there's a maximum instead of a minimum.
So I bruteforced it.
2 6's (3 6's gives a value of 0 for the 4th die, 4 6's gives a value of -6 for the 4th die):
6 6 5 1: 12
6 6 4 2: 12
6 6 3 3: 12
1 6:
6 5 5 2: 12
6 5 4 3: 24
6 4 4 4: 4
no 6's:
5 5 5 3:4
5 5 4 4:6
Total is 86, if I counted right.

[/hide]

Umm I'd like to know a better way other than brute force.  Anyone have one?

Wait...how many sides are oin this die...circle sdquared I don't think that your solution works only because you assumed that it was a 6-sided die. Becuase k can be as small as 1, I am drawn to believe that the die has more than 6 sides. Could you clear this up, perfect628?

They're standard dice, with 6 sides, sorry.


geometry
calculus
modular arithmetic
number theory

I'm a high school senior going on to college for engineering, but I feel that I've been missing a lot in terms of mathematics. I went through a basic geometry class up to Pre-Calculus and I have had some Calculus.

I'm not sure what AoPS book I should get, because I have no experience with Number Theory other than what's commonly introduced in Algebra 2/ Precalc classes. I'm really interested in becoming better at general problem solving and teaching myself Number Theory and whatever else is necessary to do more difficult problems.

Any recommendations for books on AoPS and any other good ones?

Thanks.

Well the AoPS book on Number Theory should help!  AoPS Volumes 1 & 2 touch on number theory (mostly congruences and some applications).  If you're already familiar with the basics of congruences (like if you know what $ a\equiv b\pmod{c}$ means, you know some of the rules...), then just skip to volume 2.  But the book number theory should cover all of that.


If x,y,z>=0 then prove that the folowing inequality holds:
3+x+y+z+1/x+1/y+1/z+x/y+y/z+z/x>=3*(1+x)(1+y)(1+z)/(1+xyz).

[i]unknown source[/i]

Just carefully expand two sides, we get that initial ineq is equivalent to

1/x + 1/y + 1/z + x/y + y/z + z/x + x^2yz+y^2xz+z^2xy + x^2z+y^2x+z^2y >= 2(xy+yz+zx+x+y+z)

But 1/x + y^2x >= 2y, 1/y+z^2y>=2z, 1/z + x^2z >= 2x,
      x/y + z^2xy >= 2xz, y/z + x^2yz >= 2xy, z/x + y^2xz >= 2yz

And we are done.

Namdung


The altitude-side triangle is 30-60-90 so $ a\equal{}18cm$. $ P\equal{}ah\equal{}162cm^2$

We don't want the area, though, we want the perimeter, which is 4 times the side length, so $ 4\cdot18\equal{}72$


[hide="Solution"]
Well, it's just basicly $ (\minus{}1 \plus{} 2) \plus{} (\minus{}3\plus{}4) \plus{} \cdots (\minus{}2005 \plus{} 2006) \minus{} 2007$ which is the same as $ 1(1003) \minus{}2007 \equal{} 1003 \minus{} 2007 \equal{} \boxed{\minus{}1004}$
[/hide]

hello, you can use this sum
$ \sum_{k\equal{}1}^n(\minus{}1)^k \cdot k\equal{}\frac{1}{4}\left((\minus{}1)^n(2n\plus{}1)\minus{}1\right)$.
Sonnhard.


geometry
trigonometry
AMC
USA(J)MO
USAMO
Olimpiada de matematicas

[quote="conejita"](...)Sobre CB tomamos un punto D tal que el incirculo de CBD y el excirculo (...)[/quote]
No entiendo si $D$ esta en $CB$ -> $C$, $D$ y $B$ colineales??

[quote="conejita"]Una disculpa, deberia decir:
Sobre AB tomamos un punto D...

Ahora si esta bien el problema.
Grax  :oops:[/quote]
jajja,  de razon, yo estaba considerando sobre el lado, osea que en la region jajaja, bueno, entonces conejita corrige  :D çchaufas saludos,.

[quote="conejita"]Sea ABC un triangulo con CB=CA. Sobre CB tomamos un punto D tal que el incirculo de CBD y el excirculo de CAD (tangente a las prolongaciones de CD y CA) sean iguales (es decir del mismo radio). Demuestra que una cuarta parte de las alturas iguales(ya sea $Ha$ o $Hb$) es igual al radio de dichos circulos.

Por favor, este problema no me ha salido nunca!!! Necesito ayuda!![/quote]

Has intentado con trigonometria?? Para aqui para el forum esta bien intentar siempre con geometria euclideana pero en un examen, este seria el tipico problema que me parece que si no me sale lindo rapidamente, blasfemaria un poco de senos y cosenos sobre el papel.
Afortunadamente, no hay necesidad de ello..

1- La idea está basada en un hecho bastante conocido y es que la suma de las perpendiculares trazadas desde un punto de la base de un triángulo isósceles hasta los lados es igual a la longitud de la altura.  Con esto en mente

2- Sean $O_{1}$, $O_{2}$ los centros de las circunferencias, exinscrita en $CAD$ e inscrita en $CBD$ respectivamente. Sean $E$ y $F$ los puntos donde la perpendicular por $O_{1}$ corta a los rayos $CA$ y $CD$ respectivamente. Ahora $CEF$ es isosceles de base $EF$. Analogamente, sean $G$ y $H$ los puntos donde la perpendicular por $O_{2}$ corta a los rayos $CD$ y $CB$ respectivamente. $CGH$ isosceles base $GH$.

3- Ahora tenemos muchos triangulos isosceles como queremos. Vease que por (1) el diametro de los circulos es igual a las alturas con respecto a los lados iguales de los triangulos $CEF$ y $CGH$.

4- $O_{1}D = DO_{2}$. Esto es por igualdad de triangulos y utilizando el hecho de que los radios de los circulos son iguales y que ambos centros estan en la bisectriz del $\angle BDC$

5- Sea $M$ el punto donde se intersectan las rectas $EF$ y $AB$. Entonces se tiene $\angle O_{1}MD = \angle O_{2}CD$(¿por que?) luego $\triangle MDO_{1}= \triangle CDO_{2}\Rightarrow CD = DM$. Si $M'$ fuera el punto donde $GH$ intersecta a $AB$ entonces se tendría $\angle O_{2}M'D = \angle O_{1}CD$, $\angle M'DO_{2}= \angle CDO_{1}$ y por (4) $O_{1}D = DO_{2}$, luego sería $\triangle DM'O_{2}= \triangle DO_{1}C$ y $DM' = CD = DM$ y todo este lio es para esto. $GH$, $AB$ y $EF$ concurren en $M$

6- Considerese la circunferencia que circunscribe al $\triangle MGF$. Sea $O$ su centro. Entonces $\angle GOD = 2\angle GMD = \angle ACD$. Trácese la paralela por $F$ a $CA$ que corta a $AB$ en $N$. Entonces $\angle GFN = \angle ACD = \angle GOD$. Luego $OGNF$ puede inscribirse en una circunferencia. Luego $\angle NGF = \angle NOF = 2\angle DMF = \angle BCD$. Luego $GN \parallel BC$.

7- Finalmente, la distancia de $N$ a $AC$ es igual a la distancia de $F$ a $AC$(recuerdese $FN \parallel AC$) que es igual a $2r$ (3). Igualmente la distancia de $N$ a $BC$ es la distancia de $G$ a $BC$,tambien $2r$. Y la suma de las dos es la altura pues $N$ pertenece a $AB$. Luego $H = 4r$. Por cierto que $N$ es claramente el punto medio de $AB$. Una forma un poco enrevesada de llegar a el.

Si alguien encuentra una solucion mas rapida y que no utilice trigonometria me encantaria verla. Seguro que existe.

Daniel

[quote="conejita - a very nice problem "][size=117][color=darkred]Sea $ABC$ un triangulo con $AB=AC$. Sobre $BC$  tomamos un punto D tal que el incirculo de $ABD$ y el excirculo de $ACD$ (tangente a las prolongaciones de $AC$ y $AD$ ) sean iguales (es decir del mismo radio). Demuestra que una cuarta parte de las alturas iguales(ya sea $h_{b}$ o $h_{c}$) es igual al radio de dichos circulos.[/color][/size][/quote]
[color=darkblue][b]Proof 1 (metrical).[/b] Denote the incircle $w_{1}=C(I_{1},\rho )$ of the triangle $ABD$ and the its  tangent points : $P\in AB$ $\mathrm{\ ,\ }$ $X\in BD\ ,\ U\in AD\ ,\ BX=x\ .$

Denote the $A$- exincircle $w_{2}=C(I_{2},\rho )$ of the triangle $ACD$ and the its tangent points : $R\in AC\ ,\ Y\in CD\ ,\ V\in AD$ $\mathrm{\ ,\ }$  $CY=y\ .$

$1.\blacktriangleright\ \triangle BXI_{1}\sim\triangle I_{2}YC$ $\Longrightarrow$ $\frac{BX}{I_{2}Y}=\frac{XI_{1}}{YC}$ $\Longrightarrow$ $\boxed{\ xy=\rho^{2}\ }\ .$

$2.\blacktriangleright$ $UV=AV-AU=AR-AP=$ $(AC+CR)-(AB-BP)=$  $CR+BP=$ $CY+BX=x+y$ $\Longrightarrow$

$UV=x+y\ .$ But $UV=XY=BC-(BX+CY)=a-(x+y)\ .$ Thus, $x+y=a-(x+y)$, i.e. $\boxed{\ x+y=\frac{a}{2}\ }\ .$

$3.\blacktriangleright\ x=\rho\cot\frac{B}{2}\ ,\ y=\rho\tan\frac{B}{2}\Longrightarrow$ $x-y=\rho \left(\cot\frac{B}{2}-\tan\frac{B}{2}\right)$ $\Longrightarrow$ $\boxed{\ x-y=2\rho\cot B\ }\ .$

Therefore, $(x+y)^{2}-(x-y)^{2}=4xy$ $\Longrightarrow$ $\frac{a^{2}}{4}-4\rho^{2}\cot^{2}B=4\rho^{2}$ $\Longrightarrow$ $a^{2}=16\rho^{2}\left(1+\cot^{2}B\right)$ $\Longrightarrow$ $4\rho=a\sin B$ $\Longrightarrow$ $\boxed{\boxed{\ 4\rho=h_{b}\ }}\ .$

[b]Remark.[/b] From trigonometry I used only the well-known relation $\cot\frac{x}{2}-\tan\frac{x}{2}=2\cot x\ .$ Prove easily that $\frac{DB}{DC}=\frac{4b+a}{4b-a}\ .$

[b]An interesting and very nice property.[/b] Denote the incircle $w_{3}=C(I',r')$ of the triangle $ACD\ .$

From the my topic http://www.mathlinks.ro/Forum/viewtopic.php?t=50559 obtain :

$4\rho=h_{b}\ ,\ pr^{2}+a\rho r'=pr(\rho+r')\ \|\bullet 4$ $\Longrightarrow$ $bh_{b}=2S=2pr\ ,\ 4pr^{2}+ah_{b}r'=pr(h_{b}+4r')\ \ \|\bullet b$ $\Longrightarrow$

$4pr^{2}b+a\cdot 2pr\cdot r'=pr\cdot 2pr+4prr'b$ $\Longrightarrow$ $r'=\frac{r(2b-p)}{2b-a}\ ,\ 2(2b-p)=2b-a$ $\Longrightarrow$ $\boxed{\ r'=\frac{r}{2}\ }\ \mathrm{-\ great\ result\ !\ }$[/color]

[quote="conejita"]Ola Daniel, disculpa pero no entiendo la construccion de las prependiculares que pasan por O1 y O2!!
Podrias explicarlo un poco mejor xfavor
Gracias[/quote]

Bueno, los rayos $CO_{1}$ y $CO_{2}$ son bisectrices de los angulos $\angle ACD$ y $\angle DCB$ respectivamente porque pasan por el ex/incentro de los triangulos en cuestion. Si ahora trazo una perpendicular por $O_{1}$ y tomo los puntos donde esta corta a los rayos $CA$ y $CD$ entonces ahora $CO_{1}$ no solo es bisectriz de ese triangulo nuevo sino tambien altura, luego tiene que ser isosceles.

Fijate que ahora el los centros de las circunferencias son puntos que pertenecen a la base de triangulos isosceles, y los radios son las distancias desde estos puntos hasta los lados respectivos. Luego, la suma de las distancias desde $O_{1}$ hasta $CA$ y $CD$, o sea $2r$ es igual a la altura desde $F$ del triangulo $CEF$.

Dime si te ayuda ahora

Daniel

hola...en estos momentos no cuento con tiempo para hacer la solucion completa pero con esto seguro les saldra facil y sin usar trigonometria:
Como el triangulo es isosceles y las circunferencia inscrita y circuncrta tienen el mismo radio, una manera de relacionar los radios con la altura sobre los lados iguales seria utilizar areas: triangulos parciales y el total:
Siendo $r$, el radio de las circunferencias. Y $H$ pie de la perpendicular trazada de B sobre el lado AC.....$[XYZ]: area de la region XYZ$
$[CBD]=(\frac{CB+DB+CD}{2})r$ $...(i)$
$[ACD]=(\frac{AC+CD-AD}{2})r$ $...(ii)$
$[ABC]=\frac{AC}{2}BH$ $...(iii)$
y tambien con $AC=CB$ se obtiene que: $BH=4r$

Disculpa pero, no es tan obvia el resultado!!
Podrias poner completa tu solucion Americoperu??
Gracias :)

[b]Proposición: [/b]Sea $ \triangle ABC$ un triángulo isosceles con $ AB \equal{} AC \equal{} L.$ $ D$ es un punto en $ BC$ tal que el incírculo de $ \triangle ABD$ y el A-excírculo del $ \triangle ADC$ son iguales a $ r.$ Demostrar que la altura $ h$ sobre los lados iguales es igual a cuatro veces $ r.$

Trazamos las perpendiculares $ DP$ y $ DL$ desde $ D$ a $ AB$ y $ AC,$  respectivamente. Usamos las formulas conocidas que relacionan el inradio y los exinradios con alturas.

$ DP \equal{} \frac {r(L \plus{} AD \plus{} BD)}{L} \ , \ DQ \equal{} \frac {r(AD \plus{} L \minus{} DC)}{L}$

$ DP \plus{} DQ \equal{} \frac {r(2L \plus{} 2AD \plus{} BD \minus{} DC)}{L}$

Pero es claro que $ DP \plus{} DQ \equal{} h \ \Longrightarrow h \equal{} \frac {r(2L \plus{} 2AD \plus{} BD \minus{} DC)}{L} \ , (*)$

Como son iguales estos dos círculos, seran iguales los segmentos de tangente desde $ D$ a ambos, lo cual en términos de  $ \triangle ABD$ y $ \triangle ACD$ implica $ L \plus{} DC \minus{} AD \equal{} AD \plus{} BD \minus{} L,$ es decir $ 2L \equal{} 2AD \plus{} BD \minus{} DC.$

De modo que en $ (*)$ resulta $ h \equal{} \frac {r(2L \plus{} 2L)}{L} \equal{} 4r.$


A tree doubled its height every year until
it reached a height of 32 feet at the end of
6 years. What was the height of the tree,
in feet, at the end of 3 years?

Divide by $ 2^{6\minus{}3}$ to get 32/8=4


inequalities
function
floor function
inequalities unsolved

Find all positive integers $m$ and $n$ such that the inequality:
\[ [ (m+n) \alpha ] + [ (m+n) \beta ] \geq [ m \alpha ] + [n \beta] + [ n(\alpha+\beta)] \]
is true for any real numbers $\alpha$ and $\beta$. Here $[x]$ denote the largest integer no larger than real number $x$.

[quote="shobber"]Find all positive integers $m$ and $n$ such that the inequality
\[\lfloor (m+n) \alpha \rfloor+ \lfloor (m+n) \beta \rfloor \ge \lfloor m \alpha \rfloor + \lfloor n \beta \rfloor + \lfloor n(\alpha+\beta) \rfloor\]
is true for any real numbers $\alpha$ and $\beta$. Here $\lfloor x \rfloor$ denotes the largest integer less than or equal to $x$.[/quote]
Are $\alpha$ and $\beta$ really supposed to be real? Not just positive real? Since for real numbers $\alpha$ and $\beta$, the questions becomes quite easy..  :whistling:

[i]Proof.[/i] Let $\alpha$, $\beta$ be integers. Then the inequality $(m-n)\beta \ge 0$ must hold true. However, if $m>n$, we can choose a negative $\beta$, if $m<n$, we choose a positive integer $\beta$. Consequently, $m=n$. The inequality becomes \[\lfloor 2m \alpha \rfloor + \lfloor 2m \beta \rfloor \ge \lfloor m \alpha \rfloor + \lfloor m \beta \rfloor + \lfloor m(\alpha+ \beta) \rfloor .\] Define integers $A$ and $B$, such that we have $A > m \alpha \ge A-1$ and $B>m \beta \ge B-1$. Now we can split the problem up into three cases:
$\bullet$ $2A > 2m \alpha \ge 2A-1$ and $2B > 2m \beta \ge 2B-1$,
$\bullet$ $2A > 2m \alpha \ge 2A-1$ and $2B-1 > 2m \beta \ge 2B-2$,
$\bullet$ $2A-1 > 2m \alpha \ge 2A-2$ and $2B-1 > 2m \beta \ge 2B-2$.
A short computation shows that the inequality holds true in each of the three cases. Hence every pair $(m$, $n)=(t$, $t)$ for some positive integer $t$ fits. $\square$


Let $N$ be a number with 9 distinct non-zero digits, such that, for each $k$ from 1 to 9 inclusive, the first $k$ digits of $N$ form a number which is divisible by $k$. Find $N$ (there is only one answer.)

how would you start to solve this other than trial and error.
i know that the 5th digit is 5 and i know that every digit in an even spot has to be even. but that is about as much as i can do w/o resorting to number crunching.  :(

Ha. Found it :wink:. [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=divisible+digits+first&t=36686]See here[/url].


How do I align equations so that the equal signs are aligned together?

for example,
how can I change this so that the equations are aligned along the equal sign?

$ 1 \pm \sqrt{1\plus{}4k} \equal{} 2x$
$ \sqrt{1\plus{}4k} \equal{} 2x\minus{}1$
$ 1\plus{}4k \equal{} (2x\minus{}1)^2$
$ 4k \equal{} (4x^2\minus{}4x\plus{}1)\minus{}1$
$ k \equal{} x^2\minus{}x, \{x \mid x \in \mathbb{N}\}$

Use the align environment:

\begin{align} 1 \pm \sqrt {1 + 4k} & = 2x \\
\sqrt {1 + 4k} & = 2x - 1 \\
1 + 4k & = (2x - 1)^2 \\
4k & = (4x^2 - 4x + 1) - 1 \\
k & = x^2 - x, \{x \mid x \in \mathbb{N}\} \end{align}

with & before the symbol you want aligned.
If you don't want the equations numbered, then use align* instead of align.

Please do [i]not[/i] use eqnarray as it is obsolete as it gives poorer results than align. See [url=http://tug.org/pracjourn/2006-4/madsen]Avoid eqnarray![/url].

Thank you for introducing a site.
The thing which improved a fault of eqnarray is align.
(But align is necessary to use amsmath package.)
This served as a reference very much :)


Δύο διανυσματικοί χώροι L και Μ πάνω σε ένα σώμα F είναι ισόμορφοι αν υπάρχει μη γραμμική απεικόνιση$ f: L\rightarrow M$ η οποία να είναι αμφιμονοσήμαντη (1-1 και επί). Αποδείξτε ότι δύο πεπερασμένης διάστασης γραμμικοί χώροι είναι ισόμορφοι αν και μόνο αν έχουν την ίδια διάσταση.

Ποιον ορισμό της διάστασης χρησιμοποιείς? Αν χρησιμοποιείς το συνηθισμένο (αριθμός γραμμικά ανεξάρτητων διανυσμάτων που παράγουν το χώρο), τότε:

1. Διάσταση => Ισομορφισμός
Πάρε τα διανύσματα της μίας βάσης, απεικόνισέ τα στα διανύσματα της άλλης βάσης (1-1 αντιστοιχία) και κάνε γραμμική επέκταση της απεικόνισης για να πάρεις ένα γραμμικό μετ/μό. Μετά δείξε (τετριμμένο) ότι αυτός ο μετ/μός είναι ισομορφισμός.

2. Ισομορφισμός => Διάσταση
Δείξε ότι η εικόνα μιας βάσης του L είναι βάση του M.

Cheerio,

Durandal 1707


Comparing the numbers $ 10^{\minus{}49}$ and $ 2\cdot 10^{\minus{}50}$ we may say:

$ \textbf{(A)}\  \text{the first exceeds the second by }{8\cdot 10^{\minus{}1}}\qquad\\
\textbf{(B)}\ \text{the first exceeds the second by }{2\cdot 10^{\minus{}1}}\qquad \\
\textbf{(C)}\ \text{the first exceeds the second by }{8\cdot 10^{\minus{}50}}\qquad \\
\textbf{(D)}\ \text{the second is five times the first}\qquad \\
\textbf{(E)}\ \text{the first exceeds the second by }{5}$

Solution fixed. I misread the problem at first.

[hide="My Solution"]
We wish to find how much the first number exceeds the second number by. We can do so by subtracting them.

$ 10^{ \minus{} 49} \minus{} 2\cdot 10^{ \minus{} 50} \equal{} 10^{ \minus{} 50}(10 \minus{} 2) \equal{} \boxed{8\cdot 10^{ \minus{} 50}\text{ (C)}}$
[/hide]

@rds5493: I am posting these problems so that they can be added to the resources section [url=http://www.artofproblemsolving.com/Forum/resources.php]here[/url]. Unfortunately, every problem must be posted separately (that's just how it works).

maybe u can make a whole topic with all of these problems on it instead of trying to double ur previous post count or w/e ur doing.  btw, if u know the answer, y r u posting all of these problems.  only post the ones u do not know how to do, and stop bombing the forum with all these topics


[quote="monster_458"]Five different CD-Rom games, Garble, Trapster, Zoom!, Bungie, and Blast'em, are offered as a promotion by SugarRush cereals. One game is randomly included with each box of cereals.Determine the probability of getting all the 5 games if 12 boxes are purchased.

Answer is 33/182

I need the whole solution with steps.
Your help will be greatly appreciated.[/quote]
Translate this to a balls and urns problem:
You have 12 balls to place into 5 distinct urns. There must be at least one ball in each urn so, we're trying to partition:
$a + b + c + d + e = 12$ with the restriction that $a,b,c,d,e > 0$ so the solution is:

$\binom{11}{ 4}$

oh oops, it's a probability problem, well, the total number of ways to arrange the balls without the restriction is:
$\binom{16}{4}$

so the total probability is : $\frac{\binom{11}{4}}{\binom{16}{4}}$


It takes Jesse 4 days longer to prepare an order of pies that it takes James. Together they baked pies for 2 days and then James left. Jesse took 7 additional days to complete the order of pies. Working alone, how long would it have taken James to prepare the pies?

Please help me :(

Kindly show the representation, working equation and the solution.

THANKS :D

Over and out.

[hide="hint"]
If there are $ P$ pies in an order
James works at a rate of $ \frac {P}{n}$ orders per day, and Jesse works at a rate of $ \frac {P}{n \plus{} 4}$...
Try to set up an equation...[/hide]

BTW: why have you posted these problems twice under different pseudonyms!!

[hide]Well we don't know how many pies there are, but we also don't know how fast they can make pies each day.

Using ocha's hint above, we let P equal to number of pies, and we  let Jesse's rate be $ n \plus{} 4$, and James rate be $ n$.

We know that they completed the order of pies in the two days they worked together and the 7 days Jesse worked alone.

So we can write the following equation;

P = (pies made in two days) + (pies made in seven days)

Now we just need to figure out what to put inside the parenthesis.

The second part is easier because it is just Jesse working alone. It is basically 7 times the number of pies she can make in one day. The number of pies she can make in one day is $ \frac{P}{n \plus{} 4}$.

The first part is the sum of two quantities. The first quantity is the number of pies Jesse can make in two days, and the second quantity is the number of pies Jesse can make in two days. So it should be equal to;

2 x (The number of pies Jesse can make in aday) + 2 x (The number of pies James can make in a day)

Using the distributive property we get;

2 x (The number of pies Jesse can make in a day + The number of pies James can make in a day)

Or

2 x $ (\frac{P}{n \plus{} 4} \plus{} \frac{P}{n})$

So now we can write the equation.

$ P \equal{} [2 X (\frac{P}{n \plus{} 4} \plus{} \frac{P}{n})] \plus{} \frac{7P}{n \plus{} 4}$

I'll leave the rest for you marcopolo. [/hide]


he sequence of real number $ (x_n)$ is defined by $ x_1 \equal{} 0,$ $ x_2 \equal{} 2$ and $ x_{n\plus{}2} \equal{} 2^{\minus{}x_n} \plus{} \frac{1}{2}$ $ \forall n \equal{} 1,2,3 \ldots$ Prove that the sequence has a limit as $ n$ approaches $ \plus{}\infty.$ Determine the limit.

My solution is diffrent,be cause my EL is very bad so i ony can write my idel
We check $ x_{4k},x_{4k\plus{}1},x_{4k\plus{}2}.x_{4k\plus{}3}$ then by inductry we will find the lim of 4 seqence
Then the result is $ 1$

$x_{n+2} = 2^{-x_n} + \frac{1}{2} \forall n = 1,2,3 \ldots$ ,we can find that $x_0=1-log_23$
suppose that as $n\to\infty$ ,$x_n\to\infty$ ,so $x_{n+2}\rightarrow{\frac{1}{2}}$, again ,we can assume $x_{n-2}\to\infty$, then $x_{n}\rightarrow{\frac{1}{2}}$ ,which contradicts our assumtion.then as $n\to\infty$ ,$x_n$ has a limit(say $L$).
then $L=2^{-L}+\frac{1}{2}$ ,hence $L=1$

This solution is more perfect.
suppose that as $n\to\infty$ ,$x_n\to\infty$.

now $\prod_{n=0}^{\infty}x_{n+2}=\prod_{n=0}^{\infty}(2^{-x_n}+\frac{1}{2})$

since after a large value of $n$, $(2^{-x_n}+\frac{1}{2})\rightarrow{\frac{1}{2}}$ so the right part looks like that:

$\prod_{n=0}^{\infty}(2^{-x_n}+\frac{1}{2})=N\lim_{t\to\infty}\frac{1}{2^t}=0$ ,where $N$ is finite.

 But the left part:

 $\prod_{n=0}^{\infty}x_{n+2}=\infty$ ,which contradict our assumption.Hence as $n\to\infty$,$x_n$ has a limit which is $L=1$.

@Manmaid : this is wrong, because $x_{n}$ can be bounded without has limit

Easy to prove that for $x\in (0,1)$ that $2^x-1 < x$.
Assume that $|x_n-1|=\epsilon_n$.
We get that \[\epsilon_{n+2}=2^{-1}\left| 2^{\epsilon_n}-1\right| < \frac{\epsilon_n}{2}\]
Hence both the even and the odd terms of the sequence converge to $1$. Hence the result follows.


search
function
algebra
polynomial
vector
Ring Theory
number theory

Tell me which one you guys think is better, and perhaps an explanation for your choice.

I already have been acquainted with your friend.
Those posts are like a year old and discussing old software versions.
Since then new things have come out Maple 11, Mathcad 13, MATLAB 7 and I dont know about the rest.
Plus they don't really compare everything.

Mathematica 6 is 99% the same as Mathematica 5, and the same is doubtless true of the others.  If there were particular questions that were not answered by previous threads, you would be much better off posing those questions in place of such a general request.

[url=http://www.sagemath.org/]Sage math[/url]

I prefer it to any of the above.

[quote="mlok"]At my department there are recurring discussions about relative merits of Maple and Matlab in Engineering Calculus. This is what our customers say (on facebook groups):
[quote]I Hate Maple 10, and all other versions. - 38 members
I f***ing hate Maple - 12 members
I Hate Maple! (and those who love it) - 8 members
I Seriously Seriously Dislike Maple and Thats Putting it Nicely - 14 members
I dislike Maple with great intensity! - 9 members [/quote] Total: 81
[quote]Ags Who Hate Matlab With a Passion of a Thousand Burning Suns - 38 members
I Hate Matlab - 18 members
I HATE MATLAB - 28 members
elseif x(n)=[I:HATE:MATLAB] end - 9 members[/quote] Total: 93
I think it's too close to call.[/quote]

HAHAHAHAAHAA.
Im guessing that poll was unbiased  :P

@ Hamster1800
First time I've heard of Sage. Why do you like it more?

It's powerful, open source, and uses a well known scripting language as a base (python - some people hate python though...). There are things in its basic documentation that I have not seen done in any of the others, such as setting a variable to the polynomial ring of the rationals modulo x^2+1. If you can do this in mathematica, please tell me how.

I personally use Maple. And mod of a polynomial is just as simple as typing and pressing the enter key.
I guess a mathematica guru will defend it's software.

http://www.math.umbc.edu/~campbell/Computers/MAPLE/numbthy.html

Not exactly what I meant. I meant setting a variable to $ \mathbb{Q}[x]/\langle x^2\plus{}1\rangle$.

It's in the link I put up.

Look under Polynomial Operations

Right, you can reduce a polynomial modulo $ x^2\plus{}1$. I mean I'm setting a variable to be the [i]field[/i], not a specific polynomial.

Ok I guess I can't help you there. I have not studied what a field is, and have no clue how to use that in Maple.

What is a field?

Mathematica is just so much cooler because I got a free copy of it and didn't get a free copy of Maple, MATLAB, or Mathcad; that is, regardless of actual product quality.

So can you do abstract algebra in general with Sage math? Like finite groups and fields, vector spaces, polynomial rings over any field, etc...?

Yes. The main designer of SAGE is a number theory specialist, and that sort of thing is what SAGE does best. (The A in the acronym stands for "Algebra")

In other fields, SAGE tends to simply use what others did first.


number theory
prime factorization
number theory unsolved

The integer $ N$ is positive. There are exactly 2005 ordered pairs $ (x,y)$ of positive integers satisfying:
\[ \frac{1}{x}+\frac{1}{y}=\frac{1}{N}\].
Prove that $ N$ is a perfect square.

$ \frac{1}{x}+\frac{1}{y}= \frac{1}{N}\Leftrightarrow$
$ (x-N)(y-N) = N^{2}$

There are $ \tau(N^{2}) = 2005 = 5 \cdot 401$ solutions.  Every divisor of $ 2005$ is congruent to $ 1 \bmod 4$, so each exponent in the prime factorization of $ N^{2}$ must be divisible by $ 4$.  The conclusion follows.


I) x,y,z anoikoun |R $ x<y<z$ ,,   $ x\plus{}y\plus{}z\equal{}6.  .xy\plus{}yz\plus{}zw\equal{}9$ ndo ,$ 0<x<1<y<3<z<4$

II) $ x,y,z >0$,   , $ x^xy^yz^z\geq (xyz)^a$ opou a einai o ari8mitikos mesos twn x,y,z

Ωραιες ασκησεις.Ας βαλω μια λυση για την 2..λοιπον..λογαριθμιζοντας εχουμε $ xlnx\plus{}ylny\plus{}zlnz\geq \frac{(x\plus{}y\plus{}z)}{3} (lnx\plus{}lny\plus{}lnz)$.Omos i $ f(x)\equal{}xlnx$ einai kurti me deuteri paragogo $ f''(x)\equal{} \frac{1}{x}$ profanos thetiki.ara apo JENSEN $ xlnx\plus{}ylny\plus{}zlnz \geq 3 \frac{(x\plus{}y\plus{}z)}{3} ln \frac{(x\plus{}y\plus{}z)}{3} \geq \frac{(x\plus{}y\plus{}z)}{3}ln(xyz)$ apo AM-GM kai teleiosame.... :wink:
LEFTERIS

[quote="lefmast"]Ωραιες ασκησεις.Ας βαλω μια λυση για την 2..λοιπον..λογαριθμιζοντας εχουμε $ xlnx \plus{} ylny \plus{} zlnz\geq \frac {(x \plus{} y \plus{} z)}{3} (lnx \plus{} lny \plus{} lnz)$.Omos i $ f(x) \equal{} xlnx$ einai kurti me deuteri paragogo $ f''(x) \equal{} \frac {1}{x}$ profanos thetiki.ara apo JENSEN $ xlnx \plus{} ylny \plus{} zlnz \geq 3 \frac {(x \plus{} y \plus{} z)}{3} ln \frac {(x \plus{} y \plus{} z)}{3} \geq \frac {(x \plus{} y \plus{} z)}{3}ln(xyz)$ apo AM-GM kai teleiosame.... :wink:
LEFTERIS[/quote]    :wink:  pare kai thn lush mou  8etw ,,$ x\geq y \geq z$ ara $ x^xy^y \geq x^yy^x$ ara $ (\frac{x}{y})^x\geq(\frac{x}{y})^y$ omios gia to $ y^yz^z \geq y^zz^y$ kai $ z^zx^x \geq x^zz^x$ kai polaplasiazodas tis treis sxeseis kata melh bgenei $ ({x^xy^yz^z})^{2} \geq x^{y\plus{}z}y^{z\plus{}x}z^{x\plus{}y}$ kai polaplasiazodeas me to $ x^xy^yz^z$ pernoume $ ({x^xy^yz^z})^{3} \geq ({xyz})^{3a}$ afou $ x\plus{}y\plus{}z\equal{}3a$ara me kubikes riza pernoume to zhtoumeno :D

Η πρωτη ειναι ωραιο θεμα, αν οχι για λυκειο τοτε για επιπεδο Θαλη.

Ευκολα βλεπουμε οτι:

$ 9 \equal{} ab \plus{} bc \plus{} ca \equal{} a(b \plus{} c) \plus{} bc \equal{} 6a \minus{} a^2 \plus{} bc \Rightarrow abc \equal{} a(9 \plus{} a^2 \minus{} 6a) \equal{} a(a \minus{} 3)^2 \geq 0$
διοτι ενας απο τους 3 πρεπει να ειναι θετικος και χβτγ μπορουμε να υποθεσουμε οτι ειναι ο $ a$
Ακομα, επηδη η παραπανω σχεση ισχυει και για τους αλλους 2 αριθμους, συμπεραινουμε οτι και οι 3 πρεπει να ειναι μη αρνητικοι.
Επισης αν $ a$ ειναι ο μικροτερος απο τους 3 αριθμους, τοτε ειναι $ a \leq 2$ (αλλιως $ a \plus{} b \plus{} c > 6$, ατοπο.), οποτε ευκολα με παραγωγηση της $ f(x) \equal{} x(x \minus{} 3)^2$ βρησκουμε οτι αυτη παρουσιαζει μεγιστο στο [0, 2] για $ x \equal{} 1$, οποτε $ k \equal{} f(a) \leq f(1) \equal{} 4$

Η εξισωση $ x^3 \minus{} 6x^2 \plus{} 9x \minus{} k, k \equal{} abc > 0$ τωρα, εχει ριζες τους συγκεκριμενους αριθμους, οποτε απο τις σχεσεις: $ f(0) \equal{} f(3) \equal{} 4 \minus{} k, f(1) \equal{} f(4) \equal{} \minus{} k$, με Θ. Bolzano στα διαστηματα $ [0, 1], [1, 3], [3, 4]$ παιρνουμε τις ζητουμενες ανισοτικες σχεσεις για τους 3 αυτους αριθμους.

Edit: Ξεχασα να σημειωσω οτι δεν γινεται να εχουμε $ k \equal{} 0$ η $ k \equal{} 4$, διοτι τοτε ευκολα προκειπτει οτι στην πρωτη περιπτωση οι δυο αριθμοι θα επρεπε να ηταν ισοι με 3 ενω στη δευτερη περιπτωση οι δυο αριθμοι θα επρεπε να ηταν ισοι με 1 (διοτι η $ f(x) \equal{} x(x \minus{} 3)^2$ παιρνει την μεγιστη τιμη της $ f(1) \equal{} 4$ στο $ [0, 2]$ μονο για $ x \equal{} 1$, οποτε ο ελαχιστος απο τους 3 αριθμους θα επρεπε να ηταν ισος με 1), αυτο ειναι ατοπο αφου οι τρεις αριθμοι ειναι διαφορετικοι ανα 2. Οποτε αφου η $ g(x) \equal{} x^3 \minus{} 6x^2 \plus{} 9x \minus{} k$ δεν μηδενιζεται στα ακρα των διαστηματων στα οποια εφαρμοζουμε το Θεωρημα του Bolzano, οι 3 αριθμοι θα βρησκονται στα εσωτερικα των συγκεκριμενων διαστηματων.

[quote="NickNafplio"]Η πρωτη ειναι ωραιο θεμα, αν οχι για λυκειο τοτε για επιπεδο Θαλη.

Ευκολα βλεπουμε οτι:

$ 9 \equal{} ab \plus{} bc \plus{} ca \equal{} a(b \plus{} c) \plus{} bc \equal{} 6a \minus{} a^2 \plus{} bc \Rightarrow abc \equal{} a(9 \plus{} a^2 \minus{} 6a) \equal{} a(a \minus{} 3)^2 \geq 0$
διοτι ενας απο τους 3 πρεπει να ειναι θετικος και χβτγ μπορουμε να υποθεσουμε οτι ειναι ο $ a$
Ακομα, επηδη η παραπανω σχεση ισχυει και για τους αλλους 2 αριθμους, συμπεραινουμε οτι και οι 3 πρεπει να ειναι μη αρνητικοι.
Επισης αν $ a$ ειναι ο μικροτερος απο τους 3 αριθμους, τοτε ειναι $ a \leq 2$ (αλλιως $ a \plus{} b \plus{} c > 6$, ατοπο.), οποτε ευκολα με παραγωγηση της $ f(x) \equal{} x(x \minus{} 3)^2$ βρησκουμε οτι αυτη παρουσιαζει μεγιστο στο [0, 2] για $ x \equal{} 1$, οποτε $ k \equal{} f(a) \leq f(1) \equal{} 4$

Η εξισωση $ x^3 \minus{} 6x^2 \plus{} 9x \minus{} k, k \equal{} abc > 0$ τωρα, εχει ριζες τους συγκεκριμενους αριθμους, οποτε απο τις σχεσεις: $ f(0) \equal{} f(3) \equal{} 4 \minus{} k, f(1) \equal{} f(4) \equal{} \minus{} k$, με Θ. Bolzano στα διαστηματα $ [0, 1], [1, 3], [3, 4]$ παιρνουμε τις ζητουμενες ανισοτικες σχεσεις για τους 3 αυτους αριθμους.[/quote]

wraia lush   :)

[quote="Athinaios"]Νικο ωραιος. Μόνο ενα μικρο typo στο bolzano στο τελος ως προς τις τιμες της f αν θες διόρθωσε αλλα ok, μικρή σημασία έχει.
Αν τωρα θέλουμε να αποφύγουμε αναλυση και να παμε μονο με στοιχειωδης αλγεβρα μπορουμε να σκεφτουμε ως εξης:
Ισχυει:
$ x(x \minus{} 3)^2 \equal{} y(y \minus{} 3)^2 \equal{} z(z \minus{} 3)^2 \equal{} xyz$
Ευκολα δείχνουμε ότι ειναι θετικοί διαφοροι του 3 (ομοσημοι ανα δυο με αθροισμα θετικο και απο την παραπανω σχεση κανενας δεν ειναι ισος με 3 ή 0)
Αν δυο πανω απο 3,$ z \minus{} 3 \equal{} \sqrt {yx},y \minus{} 3 \equal{} \sqrt {zx}$ και με αφαιρεση,$ z \minus{} y \equal{} \sqrt {x}(\sqrt {y} \minus{} \sqrt {z})$ ατοπο αφου ειναι διακεκριμενοι θετικοι.
Αν δυο κατω απο 3 με ομοιο τροπο, $ \sqrt {y} \plus{} \sqrt {z} \equal{} \sqrt {x}$. Ομως το δεξι μελος ειναι κατω απο $ \sqrt {y}$ λογω διαταξης,ατοπο.
Αρα ενας ακριβως(δηλαδη ο $ z$) πανω απο 3.
Αν τωρα δυο κατω απο 1, τοτε $ (z \minus{} 3)^2 \equal{} xy < 1 \leftrightarrow z < 4$ και αρα $ x \plus{} y \plus{} z < 6$, ατοπο.
Αν δυο πανω απο 1 τοτε ομοια $ z > 4$ και ατοπο.
Αρα $ 0 < x < 1 < y < 3 < z$.(δεν γινεται $ x \equal{} 1$ γιατι τοτε $ y \equal{} 1$ και αντιστροφα)
Tωρα μας μενει νδο $ z < 4$.Aυτο ομως ειναι αμεσο αφου σε αντιθετη περιπτωση $ 4 > x(x \minus{} 3)^2 \equal{} y(y \minus{} 3)^2 \equal{} z(z \minus{} 3)^2 \geq4$ , κατι παλι ατοπο. :wink:
η πρωτη ανισοτητα στην τελευταια σχεση ισχυει γιατι $ 4 \minus{} x(x \minus{} 3)^2 \equal{} (x \minus{} 1)^2(4 \minus{} x) > 0$[/quote] :wink: bravo hlia afthn thn lush eixa upoyh mou

[quote="KapioPulsar"]II) $ x,y,z >0$,   , $ x^xy^yz^z\geq (xyz)^a$ opou a einai o ari8mitikos mesos twn x,y,z[/quote] πάλι [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=810053#p810053]εδώ[/url]


LaTeX
trigonometry
real analysis
real analysis unsolved

Hi everyone,

Could someone please explain to me how to find out if the following diverges or converges?

ncosnpie

Would I need to use the squeeze theorem or would I just use L'Hopital's rule?

Thank you very much

Your notation isn't very clear; you can use LaTeX on this board, you should read the guide to typesetting LaTeX on the main page. If your sequence is $ n \cos(\pi n)$ then this will diverge. You should also be aware of the actual definition of a limit: a sequence $ \{a_n\}$ converges to a number $ A$ if, for all $ \epsilon > 0$, there exists an $ N$ in the natural numbers so that $ n > N$ implies that $ |a_n \minus{} A| < 0$.

[b]Chocolatelover[/b], how about $ \sum_{n\equal{}1}^{\plus{} \infty}\frac{\cos(n \pi)}{n}$?


advanced fields
advanced fields theorems

I have to give a short lecture (30 mins) about singular homology and its link to cohomology. Now, I know what singular homology is, but I'm unfamiliar with cohomology and it's not in our course either.

Even though I can find definitions and such on the internet, I'd like some recommendations:
- What do you find the most important results/theorems about cohomology?
- Interesting basic properties about cohomology that I should explain/mention?
- Interesting ways to present cohomology intuitively?
- Other ideas you can give me?

30 min!  :what?: You could teach semester long course on this.

If the audience knows about differential forms, then I'd say this is the best intuitive way to approach cohomology. See how Bott and Tu approach it in the first four or so pages (of chapter I not the Preface) of Differential Forms in Algebraic Topology. Any math library should have it, and I couldn't imagine learning it a different way. Note that they don't even assume you know differential forms, but intuitively it could be hard if you are trying to get that concept on top of the cohomology.


Prove that

\[ \frac{1}{pq}(\frac{1}{(y\plus{}z)^2}\plus{}\frac{1}{(y\plus{}z)^2}\plus{}\frac{1}{(x\plus{}y)^2})\ge \frac{1}{(px\plus{}qy)(qx\plus{}py)}\plus{}\frac{1}{(py\plus{}qz)(qy\plus{}pz)}\plus{}\frac{1}{(pz\plus{}qx)(qz\plus{}px)}\ge\frac{9}{(p\plus{}q)^2(xy\plus{}yz\plus{}zx)}\]

For all $ p,q>0,p\equal{}q\equal{}1$ we get Iran 96.

[quote="encyclopedia"]Prove that
\[ \frac {1}{pq}(\frac {1}{(y \plus{} z)^2} \plus{} \frac {1}{(y \plus{} z)^2} \plus{} \frac {1}{(x \plus{} y)^2})\ge \frac {1}{(px \plus{} qy)(qx \plus{} py)} \plus{} \frac {1}{(py \plus{} qz)(qy \plus{} pz)} \plus{} \frac {1}{(pz \plus{} qx)(qz \plus{} px)}\ge\frac {9}{(p \plus{} q)^2(xy \plus{} yz \plus{} zx)}
\]
For all $ p,q > 0,p \equal{} q \equal{} 1$ we get Iran 96.[/quote]
This inequality follows from the following generalization of Iran Inequality, which was first proposed by Hungkhtn-I think:
\[ \frac{1}{(a\plus{}kb)^2}\plus{}\frac{1}{(b\plus{}kc)^2}\plus{}\frac{1}{(c\plus{}ka)^2} \ge \frac{9}{(k\plus{}1)^2(ab\plus{}bc\plus{}ca)}\]
for all $ a,b,c,k \ge 0.$
:)

[quote="can_hang2007"][quote="encyclopedia"]Prove that
\[ \frac {1}{pq}(\frac {1}{(y \plus{} z)^2} \plus{} \frac {1}{(y \plus{} z)^2} \plus{} \frac {1}{(x \plus{} y)^2})\ge \frac {1}{(px \plus{} qy)(qx \plus{} py)} \plus{} \frac {1}{(py \plus{} qz)(qy \plus{} pz)} \plus{} \frac {1}{(pz \plus{} qx)(qz \plus{} px)}\ge\frac {9}{(p \plus{} q)^2(xy \plus{} yz \plus{} zx)}
\]
For all $ p,q > 0,p \equal{} q \equal{} 1$ we get Iran 96.[/quote]
This inequality follows from the following generalization of Iran Inequality, which was first proposed by Hungkhtn-I think:
\[ \frac {1}{(a \plus{} kb)^2} \plus{} \frac {1}{(b \plus{} kc)^2} \plus{} \frac {1}{(c \plus{} ka)^2} \ge \frac {9}{(k \plus{} 1)^2(ab \plus{} bc \plus{} ca)}
\]
for all $ a,b,c,k \ge 0.$
:)[/quote]
But there still not exist nice proof,I think. :(


quadratics inequalities calculus conics parabola

Can you guys please give me some word problems on quadratic inequalities? I want some not-so-challenging and challenging stuff so I could practice. Can't find anything on Google so I resorted to asking here. Sorry if I kinda wasted your time.  :oops:

Yeah I saw that, but I want word problems not just simple inequalities which are directly stated.

erm ok heres one i made up

a rock is thrown vertically in the air at an initial height of 6 ft and an initial speed of 42 fps and t seconds after it is thrown, it is -16t^2+42t+6 ft in the air (if u knew calculus, you could derive the parabola with the other info.)
for how long is the rock higher than 32 ft.?

[hide="answer (try to work towards this)"]
5/8 i think
[/hide]
plz dont answer this ppl this is for ihatewatermelons

plz dont answer this!

You make up a problem, and when you say the answer you say "..., I think"?

That's pretty funny... :rotfl: Not being able to know the answer to your own problem...that's what happens to my brother when he tries to stump me.. :P


induction
number theory
greatest common divisor
number theory proposed

prove that if 3^n-2^n is a prime power then n is a prime!

good luck

Assume $3^n-3^n=p^k$. We can use induction after $k$. If $n$ isn't a prime or $pq$ with $p,q$ primes, then there is a composite divisor $a$ of $n$ which is $<n$, so $3^a-2^a=p^{k'},\ k'<k$, so by the induction hypothesis $a$ is prime, thus yielding a contradiction. This means that if $n$ isn't prime then it's $p_1p_2$, with $p_i$ being prime.

We thus have $3^{p_1p_2}-2^{p_1p_2}=p^k$. From the well-known result (I've used it in this section recently) that $(\frac {a^n-1}{a-1},a-1)=(a-1,n)$, we find that $(\frac {3^{p_1p_2}-2^{p_1p_2}}{3^{p_i}-2^{p_i}},3^{p_i}-2^{p_i})=(3^{p_i}-2^{p_i},p_{i+1})$. On the other hand, the first gcd written above has to be a power of $p$, so $p_1=p_2=p$, so $3^{p^2}-2^{p^2}=p^k$, but $3^{p^2}\equiv 3(\mod\ p),2^{p^2}\equiv 2(\mod\ p)\Rightarrow p^k=3^{p^2}-2^{p^2}\equiv 1(\mod\ p)$, a contradiction.


Would you play the super lotto?

I'm talkin about the lotto where you pick 5-6 numbers.

Knowing the chances, would you spend the dollar to play? If so, how often would you try your chances?

yes. it benefits education in the state of the lottery, regardless of the winning chances. think of it as a cheap way to help your state, with a bonus- there's an excitement that MAYBE you'll win. although you probably won't. who cares, its a dollar.

Yes, because it's only a dollar and although the chances of winning are slim, if you DO win, you get a LOT of money.  :lol:

No, because a dollar can get you a 20 oz soda.

Unfortunately, lotto tix usually cost more than a dollar.

[quote="mcalderbank"]I think the lottery is a tax for people who cannot do math.[/quote]Or for people who really don't care about that dollar and want to give it away for a good reason... and if besides, there is a posibility in about several millions to get a lot of money, well, why not (I don't know how does lotto works here, but in Costa Rica there is a chance in 100 000 to win, but it is a lot more expensive than a dollar!)

If you play the lottery to win, then you're wasting your money and time.  But sometimes it's fun to see if you're lucky.  Sometimes winning something small with one of those smaller-scale lotteries can make your day.  And it's also going towards a good cause (the part that isn't eaten up by the bureaucracy monster that is).

I posed the question because I bought a lotto ticket on my 18th birthday....just for kicks. It was either that or porno or cigarettes, so I went with the lotto, porno and cigarettes aren't my thing.

I lost, of course, and then thought how dumb of me it was to play. Now, I figure I'll only buy when the jackpot is ridiculously high and I have a spare dollar....Definitely not a weekly thing....

[quote="bleumoose"]dont you go to georgia tech, david? the lotto there is a buck a draw. and the wins pretty big (theres some billboard in downtown that says how much, i cant recall the number).[/quote]Yes I do, but I don't play lotto to win (actually, I don't play lotto at all, except by some evitable discontinuities, when I am back home) but to give some money away for a good cause. The only time I won something was a 40 to 1 (the ticket was 50 colones, and I won 2000, in that time, about 25 bucks, and it was one of those things that you scrub to remove a layer and see if there is a price).

I don't know exactly what they do here with the money of the lotto, but in Costa Rica they use it for social programs. It keeps homeless old people to die in the streets, and puts some food on their stomach, hence, even when I know I am basically giving away that money, at the end, that is the porpose (and with the rules of the lotto in Costa Rica, there is a marginal recover of 25% :P )

[quote="djimenez"]
I don't know exactly what they do here with the money of the lotto, but in Costa Rica they use it for social programs. It keeps homeless old people to die in the streets, and puts some food on their stomach, hence, even when I know I am basically giving away that money, at the end, that is the porpose (and with the rules of the lotto in Costa Rica, there is a marginal recover of 25% :P )[/quote]

I *think* that in Texas they use it to help fund public education along with property taxes.  I'm not definitely sure, so don't quote me on it.

if i played the lottery, it would be to win. but since the chances of winning are so unbelievably small, there is no point to playing.  i also have no luck, so i wouldn't even bother.

[quote="juicybooty911"]i don't like lotteries but when they make it for educational stuffs, it's ok, and it makes someone very rich[/quote]Yeah! Let's be realistic, lotto wasn't done to make YOU rich, it was done to take a buck from you once in a while, in many cases with the idea of using that money for something good. Any way, there is always someone somewhere who wins the lottery. By the way, like two years ago, a guy here in Georgia won something like $160M  :? (lucky of him, I would give me satisfied with 1% of that :P )

Not even that

I'd be satisfied if I made any profit at all....

my math teacher said that a person has a higher chance of making money by pointing at a random person on the street and suing them for a random offense :P

[quote="bleumoose"]your math teacher is jaded.

in georgia, they use the money to fund public education as well.[/quote]
erm if ur saying what i think ur saying ---- my math teacher was just trying to be funny (he's not actually "jaded" ?) and he had just been telling us a story about a guy who sued him

Actually, if a lottery has gone for several weeks without a winner, it is conceivable that the jackpot will grow large enough so that participating has a positive expected value.  The people who get scammed are those who participate when the jackpots are small.


1_ find the sum of $ f(1) \plus{} f(2).... \plus{} f(20)$ if $ f(x) \equal{} x^3 \plus{} 2x^2 \minus{} 5x \plus{} 4$

I don't know how to set this one up. =D

2) If $ f(x) \equal{} g^{ \minus{} 1}(x)$, find $ f(f^{ \minus{} 1}(f(f^{ \minus{} 1}(f(f^{ \minus{} 1}(f(x))))))) \plus{} g^{ \minus{} 1}(g(g^{ \minus{} 1}(g(g^{ \minus{} 1}(g^{ \minus{} 1}(g(x)))))))$

No idea how to do this one.

1.  Do you know the formulas for $ \sum_{k\equal{}1}^{n} k, k^2, k^3$?

2.  What is the definition of $ f^{\minus{}1}(x)$?


$A$, $B$ are two fixed points. $C$ is a mobile point. Construct two squared with $AC$ and $BC$, namely $ACPD$ and $BCQE$.
Prove that the mid-point of $PQ$ is indipendent of $C$.

But...it isn't.
Suppose you put $C$ as the midpoint of $AB$?  Then $P$ and $Q$ would overlap, and we would have that midpiont right there.  Okay.  Suppose $C$ were waaayyyyyy out there.  You'd get a $P$ and $Q$ waaay out there, and thus a midpoint waaay out there...

Ok, let's say that $C$ is not on $AB$, then it should be correct.


Connie has $ \frac{5}{3}$ the amount of money as Joe.  If Joe gives Connie $ \$3$, then Connie would have three times the money Joe has.  How many dollars did Connie have originally?

$ C\equal{}\frac{5J}{3}$.  $ 3(J\minus{}3)\equal{}C$.  Two equations, two unknowns, easy to solve.  Of course, check to make sure that I set them up correctly.

Add 3 to C because Joe is giving Connie the money. Other than that it's right.

You would use the Algebra solution for harder questions than this, but this one is relatively ez. we know that the amount of money she has has to be a multiple of 5, because in this question the amount has to be an integer, and larger than 5, because 3/5(5)-3=0, so we can just try 10,15,20.... It turns out 15 is the right answer.
Solution 2:
You could realize that Joe is giving 3 dollars and the money after Connie is given $3 is still divisible by 3.. Hence, the amount of money Connie will have before will be a multiple of 15. Because we know that the amount cannot be too large, so 15 is the lowest and correct answer.

Note- Method 2 will only work for relative primes.


function
induction
algebra unsolved
algebra

Find all pairs of functions f,g:R->R such that
(i) if x<y then f(x)<f(y)
(ii) for all x,y ,f(xy)=g(y).f(x)+f(y)

Observe that if f(x)<f(y) then f(x)<f(x^2+y^2)<f(y). Suppose f is not constant. So, we can find m and n such that f(m)<f(n). Then f(m)<f(m^2+n^2)<f(n). Put x_1=n and x_(k+1)=m^2+(x_k)^2. Then an easy induction based on the first observation shows that f(m)<f(x_k)<f(x_(k-1))<...<f(x_1)<f(n) for all k. For k large you will find an absurdity. So f is constant.

o.O

It seems Harazi solves another problem here. I understood nothing.

Indeed, I don't know how, I think I posted the solution to another problem in the wrong topic: this solution is for the problem with $f(x^{2}+y^{2})=f^{2}(x)+f^{2}(y)$. I have no idea how this happened. Sorry!

Sorry again, it was $2f(x^{2}+y^{2})=f(x)+f(y)$. There are too many equations of this type...


Sa se determine functiile $f:R\to R$ continue ,cu proprietatea:\[
f(x) + 2f(2x) + f(4x) = 25x^2  + 9x + 4,\forall x \in R
\]

$f(x)=x^2+x+1$ is good
let
$g(x)=f(x)-x^2-x-1$ we have that $g(x)$ is continous too.
we have that:
$g(x)+2g(2x)+g(4x)=0$
let $h(x)=g(x)+g(2x)$ and we have that $h(x)$ is continous too.
we have:
$h(x)=-h(2x)$
so
$h(x)=h(4x)$
$h(x)=-h(2x)$
$h(0)=0$
$h$ is continous

it is easy to verify that $h \equiv 0$ so $f(x)=x^2+x+1$


Find all prime numbers $p$ such that $ p = m^2 + n^2$ and $p\mid m^3+n^3-4$.

Since $ p|(m^2 \plus{} n^2)(m \plus{} n) \equal{} m^3 \plus{} n^3 \plus{} mn(m \plus{} n)$, we have $ p|mn(m \plus{} n) \plus{} 4 \Rightarrow p|2mn(m \plus{} n) \plus{} 8$. But $ 2mn \equiv (m \plus{} n)^2 mod \text{ } p$, therefore $ p|(m \plus{} n)^3 \plus{} 2^3 \equal{} (m \plus{} n \plus{} 2)((m \plus{} n)^2 \minus{} 2(m \plus{} n) \plus{} 4)$.

$ p|m \plus{} n \plus{} 2$ or $ p|2mn \minus{} 2(m \plus{} n) \plus{} 4$. In the first case with Cauchy ($ |m \plus{} n|^2 \le (1 \plus{} 1)(m^2 \plus{} n^2)$) we get $ \sqrt{2p} \plus{} 2 \ge p$. This limits $ p$, and some calculations suffice. In the second case $ p \equal{} 2$ or $ p|mn \minus{} m \minus{} n \plus{} 2$. Using AM-GM ($ mn \le \frac {m^2 \plus{} n^2}{2}$) we also limit $ p$.


inequalities
calculus
inequalities unsolved

For a,b,c real numbers prove that

3(a^4+b^4+c^4)^3 \geq (a^3+b^3+c^3)^4

Did I miss something or this one is very easy?
(a 3  + b 3  + c 3 ) 4    =  |(a 3  + b 3  + c 3 ) 4|  \leq  ((a 3|  + |b 3|  + |c 3| ) 4, thus we may assume that a,b,c are non-negative.
But, in that case :
3*(a 3  + b 3  + c 3 ) 4  \leq  (a 4  + b 4  + c 4 ) 3, since it is the inequality between the means of order 3 and order 4....

Pierre.

indeed it is. :) i figured they wanted a solutions without using the generalised mean value theormes :)

[quote]For a,b,c real numbers prove that $3(a^4+b^4+c^4)^3 \geq (a^3+b^3+c^3)^4$[/quote]Ok, a solution without generalized MVT. $f(x)=x^{\frac43}$ is convex. So Jensen gives us:
\[ f(a^3)+f(b^3)+f(c^3) \ge 3 \cdot f\left( \frac {a^3+b^3+c^3}3 \right) \]\[ a^4 +b^4 + c^4 \ge \frac1{\sqrt[3]{3}} \left( a^3+b^3+c^3 \right)^{\frac43}\]\[  3 \left(a^4 +b^4 + c^4 \right)^3 \ge \left( a^3+b^3+c^3 \right)^4\]
We had something alike on our calculus exam this year :)


How hard are these really? Are there any good books to use for these contests, or should I just use my old calc stuff?

I was trying to teach myself calculus, and I was wondering the same thing.  What level of calculus is required??

[quote="chi^2"]level should not go past Calculus BC (or Calc 2 if you're taking it as a college course).  try look at mualphatheta.org for some practice problems.  that will help determine what you need to study a lot better.[/quote]

Too bad there has and very often is always at least a couple calc 3 questions (double/triple integrals, etc.)

I don't think they ever venture into line/surface integral stuff tho

it goes into calc 3?  wow!!  that explains why there were a couple of calc things on the Alpha trig test last year (like Taylor series . . .)

ah, guess I have to study calc 3 over the summer then . . .

[quote="mysmartmouth"]I'm using http://tutorial.math.lamar.edu/ for a great resource to learn calculus with no textbook.[/quote]

Does this mean you've learned pre calc already? :D


So, I was thinking about a problem.

Let's say you have 5 people of different ages and you have a round table with 5 seats. What is the probability that they will be arranged in assending or descending order?

What I've thought about:

5! ways to sit around the table meaning 120 ways to sit around the table.

Now finding how many ways they can sit so that they're in ascending or descending order. This is where I'm indecisive.

In ascending: the first person sits anywhere, the other people have only one place in which they can sit.
In descending: the first person sits anywhere and it's the same thing.

so that leaves 5 ways in each case meaning 10 ways total so 10/120  1/12.

But, are there really 5 ways to sit in ascending order or descending order? Isn't it all the same. One way of sitting in ascending order is the same as another way of sitting in ascending order.

Couldn't the answer be 1 and therefore the answer 2/120 or 1/60?

The correct answer is $ \frac1{12},$ as you had it the first time.

Look at it this way: let the youngest person sit down first. Since it's a round table, it really doesn't matter where that person sits. In any case, we'll have the other four seats around the table left to fill. You can fill those other four seats in $ 4!\equal{}24$ ways, of which exactly $ 2$ ways correspond to increasing or decreasing order. Hence the probability is $ \frac{2}{24}\equal{}\frac{1}{12}.$

Mhmm for some reason as I think about it I get double that answer.
This is because considering the ascending order, it sounds like it is arranged only (WLOG) clockwise. But what about counter-clockwise?

If it's ascending counterclockwise, then it's descending clockwise, and you've already counted it.


In this forum, is it possible to write an equation array while using displaystyle?

Yes it is possible but it won't affect the output. Here is an example from Kopka & Daly's excellent A Guide to LaTeX:

Displaystyle: the default on this forum
\begin{eqnarray}
(x+y)(x-y)& = & x^2-xy+xy+y^2 \nonumber\\
& = & x^2-y^2\\
(x+y)^2 & = & x^2+2xy+y^2
\end{eqnarray}

Displaystyle and centred using  double dollars
\begin{eqnarray}
(x+y)(x-y)& = & x^2-xy+xy+y^2 \nonumber\\
& = & x^2-y^2\\
(x+y)^2 & = & x^2+2xy+y^2
\end{eqnarray}

Textstyle ie non-displaystyle
$\textstyle
\begin{eqnarray}
(x+y)(x-y)& = & x^2-xy+xy+y^2 \nonumber\\
& = & x^2-y^2\\
(x+y)^2 & = & x^2+2xy+y^2
\end{eqnarray}$


[quote="pigfly"]Calculate the sum $S=\prod_{k=0}^{2^{2003}}(4\sin^2\frac{k\pi}{2^{2004}} - 3)$
Sorry if this problem was posted before![/quote]

I guess    3
$S(n)=\prod_{k=0}^{2^{2003}}(4\sin^2\frac{k\pi}{2^{2004}} - 3)$

s(0)=-3
s(1)=3
s(2)=-3

so I guess     3

I would start form the identity : $-\frac{sin(3\alpha)}{sin\alpha}=4sin^2\alpha-3$ , for nonzero $sin\alpha$.


number theory
greatest common divisor
algebra
polynomial
number theory unsolved

Prove that if f(x) is a polynomial of degree n with integer coeff. and k is an arbitrary integer, then the GCD of the nos f(x), x running over the set of integers is equal to GCD of the following n+1 integers: f(k), f(k+1),...,f(k+n)

Give a minor hint

Also suggest a book on number theory.
I am new to this branch

minor hint: if $ a, b$ are integers and $ f(x)$ has integer coefficients, then $ a\minus{}b|f(a)\minus{}f(b)$.


:) If x,y,z are three consecutive positive integers, show that 2logy=log(1+xz)

[hide]

so

we know that 2logy = logy^2

so now all we have to prove is that

y^2 = 1+xz

we are given that x, y, z are consecutive +Z

therefore, let
x= a-1
y=a
z=a+1

xz+1 = (a-1)(a+1)+1
= a^2
= LHS

QED

[/hide]

[hide="Solution"]

Say the three consecutive positive integers are $y - 1$, $y$, and $y + 1$.

$2\log{y} = \log{1 + (y-1)(y+1)}$

$\log{y^2} = \log{(1 + y^2 - 1)}$

$\log{y^2} = \log{y^2}$

$2 = 2$[/hide]

[hide]

let $x=x\\ y=x+1\\ z=x+2$

So we have $2\log{y}=\log{(1+xz)}$

Substituting:

$2\log{(x+1)}=\log{(1+x^2+2x)}$

Using the power rule for Logs on the LHS and factoring on the RHS, you get

$\log{(x+1)^2}=\log{(x+1)^2}$

QED

[/hide]

-BLt


Prove that $ 3^{3n\plus{}3} \minus{}26n \minus{}27$ is divisible by $ 169$ for all $ n \in \mathbb{N}$.

Or you may use induction:

for $ n\equal{}1$ we have $ 169|3^6\minus{}26\minus{}27\equal{}3^3(3^3\minus{}1)\minus{}26\equal{}26^2\equal{}4 \cdot 169$

now suppose that for some $ n$ it is $ 3^{3n\plus{}3}\minus{}26n\minus{}27\equal{}169m$

then, for $ n\plus{}1$ we have that

$ 3^{3n\plus{}6}\minus{}26n\minus{}26\minus{}27\equal{}3^3 \cdot (169m\plus{}26n\plus{}27) \minus{}26n\minus{}26\minus{}27\equal{}3^3 \cdot 169m \plus{} 26^2 n \plus{}26^2 \equal{}169 \cdot (27m\plus{}4n\plus{}4)$

Q.E.D :)