Princeton
college
calculus
integration
function
complex analysis
complex analysis theorems

Which books do you recommend in complex analysis.
Please mention good introductory books and others.

[hide="Some examples, but please mention others (and if possible comment these)"]
Anyone read Fundamentals of Complex Analysis with Applications to Engineering, Science, and Mathematics (3rd Edition) ?
by Edward Saff , Arthur D. Snider

Anyone read Basic Complex Analysis (Hardcover) by Jerrold E. Marsden , Michael J. Hoffman  ?

[/hide]

i've heard from my friend that marsden's book is good. i follow complex analysis by elias stein and rami shakarchi ( vol 2 of the princeton lectures in analysis), this book is very good but its meant to give a quick introduction to the most important tools  of complex analysis so it might miss some things. the excersises are excellent. another good book is complex analysis by joseph bak and donald newman.

If you want one that is insanely cheap (it is Dover) then Complex Variables by Stephen D. Fisher is quite good. Lots of cookbook problems to get you a good feel for what to do, but also some good standard problems that require thought and teach you things. The reading is pretty straight forward. Any details left out are easy to fill in yourself.

Down-side: Not the most standard presentation order. Line integrals come before Cauchy-Riemann Eqns. This didn't throw me off though.

Thanks  :) .
Please post more comments!
Perhaps this forum part is not that visited
as other.

[quote="color"]Which books do you recommend in complex analysis.
Please mention good introductory books and others. [/quote]
B. V. Sabat (orig. Szabat), Introduction to complex analysis.

The classic Ahlfor's "Complex Analysis", another is Stein and Shakarchi's "Complex Analysis".  Finally, Complex Variables: Harmonic and Analytic Functions by Flanagan.

I agreed for gka and randomizer but complete : Function Theory of Several Complex Variables by Steven G. Krantz :)

[b]You 've definitely got to check out [color=red]Tristan Needham[/color]'s one...

Quite possibly, [color=red]Roger Penrose[/color]'s praise for the book might provide you with further motivation to do so:[/b]

[b][color=green]"...By his innovative and exclusive use of the geometrical perspective,[/color] [color=red]Tristan Needham[/color] [color=green]uncovers many surprising and largely unappreciated aspects of the beauty of Complex Analysis."[/color][/b]


[quote="cincodemayo5590"]I'm not sure if this is possible, but...

Solve for $x \geq 0$:  $x= \frac{1}{x-1} + \frac{1}{x-2} + ... +1=\sum_{k=1}^{x-1} (\frac{1}{x-k})$[/quote]

note $x\in\mathbb{Z}^+$, and $x>1$ for the summation to be possible

this is taking a partial sum of the harmonic series up to $x-1$, that sum being equal to x

let $S_n=\sum_{1}^{n} \frac{1}{i}$ (this is your summation, just a bit clearer)

we want $S_{n-1}=n$

it is pretty easy to show that $S_{n}-1<\log_2 n$, so we want $n$ such that

$n-1<\log_2(n-1)$
$2^n<2(n-1)$

which is true for no positive integers $>1$...(easily shown by induction)

thus there are no solutions


yo tengo una solucion pero usted derrepente tiene otra solucion; compartamos soluciones!!

EN EL PROBLEMA 1.

                     X= medida del angulo en C

Estos dos problemas son variaciones de este:
http://www.mathlinks.ro/Forum/viewtopic.php?t=84696


function
induction
algebra proposed
algebra

I got it from my brother and i cant do it -*-

Find all function f:N->N such that
1.f(1)=1
2.f(2n)=2f(n)+1 for all x in N
3.f(f(n))=4n-3 for all x in N
....

I dont have the way to overcome that f(n)=2n-1 ...

We can prove it by induction!
Assume that $ f(n)\equal{}2n\minus{}1$ for all $ 1\leq n\leq k$, we have to prove that $ f(k\plus{}1)\equal{}2k\plus{}1$
In fact:
-) $ k\plus{}1\equal{}2m$ then $ f(k\plus{}1)\equal{}f(2m)\equal{}2f(m)\plus{}1\equal{}2(2m\minus{}1)\plus{}1\equal{}4m\minus{}1$ (Because $ m\leq k$)
-) $ k\plus{}1\equal{}2m\minus{}1$ then $ f(k\plus{}1)\equal{}f(2m\minus{}1)\equal{}f(f(m))\equal{}4m\minus{}3$ (Because $ m\leq k$)
Then OK!


[hide="My sorution"]
Use change of base formula to narrow down the equation to \[{\frac{\text{log}6}{\text{log}2}}\]
Then simplifying even further we get
\[{{\frac{\text{log}6}{\text{log}2}}=\frac{\text{log}2+\text{log}3}{\text{log}2}}=1+\frac{\text{log}3}{\text{log}2}=\boxed{1+\text{log}_{2}3}\]
Well that's the most I can simplify it without a calculator  :wink:

[/hide]

[quote="math92"]Evaluate the product:

$(log_{2}{3})(log_{3}{4})(log_{4}{5})(log_{5}{6})$[/quote]

*sigh* It looks like you took this from AoPS Volume 2 -_-

[hide]The chain collapses to $log_{2}{6}$

This can be proven, and it is proven in AoPS Volume 2[/hide]

[quote="math92"]Evaluate the product:

$(log_{2}{3})(log_{3}{4})(log_{4}{5})(log_{5}{6})$[/quote]
[hide="solution with lazy jim log property"]The lazy jim is a generalization to the chain rule. It says that $(\log_{a}b)(\log_{c}d=(\log_{a}d)(\log_{c}b)$
Applying it this simplifies to $\log_{2}6$[/hide]

drunkenmath and bpms u guys are wrong neveroddoreven simlified it more :D

It's hard to measure "simplified" and I'd definitely not say they're wrong :)

[hide="hint"]Try $2^{\text{Whole thing..}}$[/hide]

[hide="solution"]We switch the 3 and the 4 in the first two terms, and then switch the 5 and the 6 in the third and fourth terms and get thus:

$2\log_{4}{6}=\log_{4}{36}=\log_{2}{6}=1+\log_{2}{3}$

It is impossible to simplify any more.[/hide]

[hide][quote="math92"]Evaluate the product:

$(log_{2}{3})(log_{3}{4})(log_{4}{5})(log_{5}{6})$[/quote]

that equals log3/log2*log4/log 3*log5/log4*log6/log5 using change of base
cancel out log3, log4, and log5
and you get log6/log2
log 6= log2+log3
so you get 1+log3/log2
simplifies into 1+log base 2 3
(ok I fail in latex)[/hide]


integration
LaTeX
algebra proposed
algebra

If for each $x\geq 0$ holds that $ax^2+bx+c\geq 0$  and  $a,b,c\in R$  prove that

$\frac{a}{2008}+\frac{b}{2007}+\frac{c}{2006}\geq 0$

What do you mean by saying "and $a,b,c$"?

It was a latex problem. Now i fix it. Try to prove it

[quote="silouan"]This is also my solution but i am trying to find a solution without using integrals. Is it possible ?Please help[/quote]
Of cours that is possible :lol:.
When $x=0$=>$c\geq 0$
If $a<0$=>the ineq. $ax^2+bx+c\geq 0$ have only finity solution(contradiction)=>
$a\geq 0$.If $b>0$ the problem is done.
Let $b=-d$ wher $d>0$.
If $D=b^2-4ac>0$=> $ax^2+bx+c\geq 0$ only if $x\leq \frac{1}{2a}(d-\sqrt{d^2-4ac})$
or $x\geq \frac{1}{2a}(d+\sqrt{d^2-4ac})$=>$\frac{1}{2a}(d+\sqrt{d^2-4ac})\leq 0$(contradiction again)=>$D=b^2-4ac<0$.
$\frac{a}{2008}+\frac{b}{2007}+\frac{c}{2006}\geq 0$<=>
$(\frac{a}{2008}+\frac{c}{2006})^2\geq(\frac{d}{2007})^2$but
$(\frac{d}{2007})^2\leq \frac{4ac}{2007^2}\leq 4\frac{a}{2006}\frac{c}{2008}\leq  (\frac{a}{2008}+\frac{c}{2006})^2$ ;)

Thank you very much tiks nice solution .

Thank you too Solouan for interesting problem ;).


6. Let $a, b, c, d$ be positive integers such that $ab = cd$. Prove that $a + b + c + d$ is not prime.

8. Consider the sequence
$x_{1} = 19$, $x_{2} = 95$, $x_{n+2} = lcm(x_{n+1}, x_{n}) + x_{n}$
for $n > 1$.  Find the greatest common divisor of $x_{1995}$ and $x_{1996}$.

[quote="Andreas"]6. Let $a, b, c, d$ be positive integers such that $ab = cd$. Prove that $a + b + c + d$ is not prime.
[/quote]
Posted quite a lot of time at ML.. :) :
http://www.mathlinks.ro/Forum/viewtopic.php?t=38348 for example

I think this is $19$, furthermore this is $19$ in every cases !

First $19 \mid x_n$

If $d \mid x_n$ and $d \mid x_{n+1}$ then $d \mid x_{n-1}$ so $d \mid gcd (x_1,x_2) = 19$


logarithms
inequalities
algebra unsolved
algebra

We notice $ x\in(0,3]$. The inequality can be rewritten as
$ (x\log_2^x\minus{}5)(\sqrt{6\plus{}x\minus{}x^2}\minus{}(x\minus{}1))>0$.
Since $ 3\log_2^3<5$, we have only
$ \sqrt{6\plus{}x\minus{}x^2}<x\minus{}1$.

Hence, $ x\in (2.5,3]$.


induction
pigeonhole principle
combinatorics
Probabilistic Method

For a set $S$, let $|S|$ denote the number of elements in $S$. Let $A$ be a set of positive integers with $|A| = 2001$. Prove that there exists a set $B$ such that

(i) $B \subseteq A$;
(ii) $|B| \ge 668$;
(iii) for any $u, v \in B$ (not necessarily distinct), $u+v \not\in B$.

Also MathLinks Contest, 1st Edition, 7th Round :P :)

Btw, will the solutions to the problems from the contest be on the site?
Because in the contest section, they are missing :) :D

I have a solution myself but I am wondering it is true or not.Please check it and tell me if my solution is wrong.Thanks.
[hide="my solution"]We gerneralize the problem :
Let A be a set of positive integers such that $ |A| \equal{} 3n(n \in N*)$.Then there exists a subset B of A such that
i)$ |B| > n$
ii)$ \forall u,v \in B,u \plus{} v \not\in B$
 Proof:
We call a set S "good" if it satisfies $ \forall u,v \in S,u \plus{} v \not\in S$
We use induction to solve this porblem.
For the base case n=1,it is easy to see.
Assuming that it is true for $ n$,we will prove it is true for $ n \plus{} 1$.
Let A be a set such that $ |A| \equal{} 3n \plus{} 3$.Let $ A \equal{} T \cup \{a_1,a_2,a_3\}$ where $ a_1$ be the smallest number among the elements in set A and $ |T| \equal{} 3n$.
We see that $ |T| \equal{} 3n$.From the assumption above,we can choose a "good" subset $ B_1$ satisfying$ |B_1| > n$.Let $ B \equal{} B_1 \cup \{a_1\}$,we'll have a good subset $ B$ satisfying $ |B| > n \plus{} 1$,$ Q.E.D$[/hide]

Hmm...Why do you think that $ B$ is a good subset? What it has to do with that $ a_1$ is the smallest element.

[quote="iura"]For a set $ S$, let $ |S|$ denote the number of elements in $ S$. Let $ A$ be
a set of positive integers with $ |A| = 2001$. Prove that there exists a
set $ B$ such that

(i) $ B \subseteq A$;
(ii) $ \dsp |B| \ge 668$;
(iii) for any $ u, v \in B$ (not necessarily distinct), $ u + v \not\in B$.[/quote]
I saw this problem in a chinese book, but with $ 2001$ changed to $ n$ and $ \ge 668$ changed to $ > \frac{n}{3}$   :D
Here is the solution, given by a Israel mathematician N.Alon.

Let the elements in $ A$ be $ a_1,a_2,...,a_n$, and choose a prime of the form $ 3k-1$ and is greater than all the elements in $ A$.
Then consider $ a_i,2a_i,...,pa_i$ where $ i$ is an integer such that $ 1\le i\le n$. They form a complete set of residues of $ p$. Therefore there are $ k$ of them congruent to $ k,k+1,k+2,...2k-1$ modulo $ p$.

Let $ x_j$ be the number of numbers among $ ja_1,ja_2,...,ja_n$ that congruent to $ k,k+1,k+2,...2k-1$ modulo $ p$. Now $ x_1+x_2+x_3+...+x_p=kn$
So there exists a $ x_t$ such that $ x_t\ge\frac{kn}{p}=\frac{kn}{3k-1}\ge\frac{n}{3}$

Then $ B={a\in A}$ such that $ ta$ congruent to $ k,k+1,k+2,...2k-1$ modulo ${ p}$  :)

Let $ A \equal{} \{a_1,a_2,...,a_{2001}\}$.Let $ p$ be a prime number such that $ p \equiv 2 (mod 3)$ and $ p > a_i \forall i \equal{} 1,2,...,2001$.
Such a prime p exists by Dirichlet's Theorem,althought the result can also be easily proven directly.There is at least 1 prime congruent to 2 modulo 3(namely,2).Suppose there were only finitely many primes congruent to 2 modulo 3,and let their product be $ P$.Then $ 3P \minus{} 1$,which is larger than $ P$ and congruent to 2 modulo 3,must have another prime divisor congruent to 2 modulo 3,contradiction.Thus,the orginal assumption was wrong,and there are infinitely many odd primes that are congruent to 2 modulo 3.Specifically,one such prime is larger than all $ a_i$.
All elements of $ S$ are distinct and non-zero modulo $ p$.
Call a number $ n$ "mediocre" is the least positive residue of $ n$ mod $ p$ lies in $ [ \frac {p \plus{} 1}{3},\frac {2p \minus{} 1}{3}]$.For any $ 1$≤$ i$≤$ 2001$,there are exactly $ \frac {p \plus{} 1}{3}$ integer values of $ k \in [1,p \minus{} 1]$ such that $ ka_i$ is mediocre.Thus,there are $ \frac {2001(p \plus{} 1)}{3} \equal{} 667(p \plus{} 1)$ pairs of $ (k,i)$ such that $ ka_i$ is mediocre.
By Pigeonhole Principle,there exists some $ k$ for which the set $ B \equal{} \{a_i|ka_i$ is medicre$ \}$ has it least $ 668$ elements.
We now claim that this $ B$ satisfies the desired properties.It suffices to show that $ k$ times the sum of any 2 elements of $ B$  cannot be mediocre and hence connot equal $ k$ times any elements of $ B$.To that end,note that  $ k$ times the sum of any 2 elements of $ B$ cannot be medicre because it is congruent mod $ p$ to some number in $ [\frac{2(p\plus{}1)}{3},\frac{2(2p\minus{}1)}{3}]$ or, eqivalently,to some number in $ [0,\frac{(p\minus{}1)}{3}] \cup [\frac{(2p\plus{}2)}{3},p\minus{}1]$,which is a set containing no mediocre numbers.Thus,the set $ B$ satisfies the desired properties.

It is a special case of a theorem due to Paul.Erdos about sum-free set ( not have elements $u$ and $v$ such that $u+v$ belong also to it.)
every set of positive integers A always have a sub set S with $|S|>|A|/3$ which is sum free.you can see it in many combinatoric book.
The same as VMO 2006 pro 5 :rotfl: . :oops:
and probabilistic method will work properly.

This solution seems different from everyone else's (though it uses the same $(\tfrac{1}{3},\tfrac{2}{3})$ interval modulo $1$ idea).

Let $n=2001$, and note that $668=\tfrac{n}{3}+1$. For a positive real number $r$, define $S_r=\{rx \colon x \in \mathbb{R}_{>0},\tfrac{1}{3}<x-\lfloor x \rfloor<\tfrac{2}{3}\}$. Notice that $a,b \in S_r$ implies $a+b \notin S_r$. Let $m$ be the largest element of $A$, and let $N$ be an arbitrarily large number. We claim that
\[\int_\frac{m}{(N+1)m-\frac{1}{3}}^\frac{m}{Nm+\frac{1}{3}} |S_r \cap A| \,\mathrm{d}r>\frac{n}{3}\left(\frac{m}{Nm+\frac{1}{3}}-\frac{m}{(N+1)m-\frac{1}{3}}\right),\]
which shows that there exists a real number $r \in [\tfrac{m}{(N+1)m-\frac{1}{3}},\tfrac{m}{Nm+\frac{1}{3}}]$ such that if $B=S_r \cap A$, then $|B|>\tfrac{n}{3}$.

It suffices to prove that
\[\int_\frac{m}{(N+1)m-\frac{1}{3}}^\frac{m}{Nm+\frac{1}{3}} |S_r \cap \{k\}| \,\mathrm{d}r>\frac{1}{3}\left(\frac{m}{Nm+\frac{1}{3}}-\frac{m}{(N+1)m-\frac{1}{3}}\right)\]
for all $k \in A$. But we have
\begin{align*}
\int_\frac{m}{(N+1)m-\frac{1}{3}}^\frac{m}{Nm+\frac{1}{3}} |S_r \cap \{k\}| \,\mathrm{d}r&=\frac{k}{Nk+\frac{1}{3}}-\frac{k}{Nk+\frac{2}{3}}+\frac{k}{Nk+\frac{4}{3}}-\frac{k}{Nk+\frac{5}{3}}+\cdots+\frac{k}{(N+1)k-\frac{2}{3}}-\frac{k}{(N+1)k-\frac{1}{3}} \\
&\ge \frac{1}{3}\left(\frac{k}{Nk+\frac{1}{3}}-\frac{k}{Nk+\frac{2}{3}}+\frac{k}{Nk+\frac{2}{3}}-\frac{k}{Nk+\frac{3}{3}}+\cdots+\frac{k}{(N+1)k}-\frac{k}{(N+1)k+\frac{1}{3}}\right) \\
&=\frac{1}{3}\left(\frac{k}{Nk+\frac{1}{3}}-\frac{k}{(N+1)k+\frac{1}{3}}\right).
\end{align*}
Thus, it suffices to prove that
\begin{align*}
\frac{k}{Nk+\frac{1}{3}}-\frac{k}{(N+1)k+\frac{1}{3}}&>\frac{m}{Nm+\frac{1}{3}}-\frac{m}{(N+1)m-\frac{1}{3}} \\
\Longleftrightarrow \frac{k^2}{(Nk+\frac{1}{3})((N+1)k+\frac{1}{3})}&>\frac{m^2-\frac{2}{3}m}{(Nm+\frac{1}{3})((N+1)m-\frac{1}{3})}.
\end{align*}
Indeed, notice that the left-hand side tends to $\tfrac{1}{N(N+1)}$ while the right-hand side tends to $\tfrac{1-\frac{2}{3m}}{N(N+1)}$ as $N$ goes to infinity, as desired. $\square$

[b][color=#9a00ff]Remark:[/color][/b] This solution can be motivated, despite having a bunch of random constructions that appear seemingly out of nowhere. At first, I tried to make an argument using $\nu_2$: if there are more than $\tfrac{n}{3}$ odd numbers in $A$, choose those; otherwise, look at the subset of even numbers and try to choose something. This argument isn't strong enough, but it uses a prototypical version of $S_r$: the set $\{rx \colon x \in \mathbb{N},x \equiv 1 \pmod{2}\}$ where $r$ is a power of $2$. Eventually, I realized two things: we could set $r$ to be any real number, and we could relax $x \equiv 1 \pmod{2}$ to $x \approx 1 \pmod{2}$. This gave me the version of $S_r$, the set of all positive real numbers less than $\tfrac{1}{3}r$ away from the nearest odd multiple of $r$, which I used to solve the problem. I later scaled the definition of $S_r$ by $2$ for writeup purposes.

The biggest conceptual jump in this problem for me was seriously thinking about using small $r$, as you might not expect small $r$ to give anything useful, and indeed, most of the solutions in this thread use a large prime $p$, which can be thought of as a large $r$ except you can transform $A$ modulo $p$ with multiplication. My first try was with $r \in [\tfrac{1}{N+1},\tfrac{1}{N}]$ (using the definition of $S_r$ used in this writeup), but this wasn't quite enough because it gave a density slightly less than $\tfrac{1}{3}$. After flailing with different constructions, I realized that you could cut off very small bits of the interval $[\tfrac{1}{N+1},\tfrac{1}{N}]$ to create $[\tfrac{m}{(N+1)m-\frac{1}{3}},\tfrac{m}{Nm+\frac{1}{3}}]$ while not changing the sum. I was reluctant to try this because I thought it would almost certainly be insufficient, but it turned out to be enough.


In cate feluri putem acoperi un dreptunhi 1Xn cu k dominouri.
Obs: un domino reprezinta un dreptunghi 1X2.

[hide="parere"]
hmm...daca nu ma insel eu, raspunsul este $\binom{n-k}{k}$ (evident problema nu are sens pentru $2k>n$).
consider patratelele dreptunghilui ca fiind un sir de $k$ $1$-uri si $n-k$ $0$-uri, unde $1$ reprezinta patratelul din stanga al unui domino.
atunci, din ipoteza, nu exista doi de $1$ alaturati. acum, din sir, tai cele $k$ zerouri situate imediat in dreapta unui $1$, si se obtine un sir de $n-k$ cifre, dintre care $k$ sunt $1$.
reciproc, dintr`un sir de $n-k$ cifre, din care $k$ sunt $1$, adaug cate un $0$ dupa fiecare $1$ si obtin un sir "bun" de $n$ cifre.
acum, dintr`un sir de $n-k$ cifre unde am $1$ pe pozitiile $x_{1},x_{2},\ldots,x_{k}$, sirul de $n$ cifre obtinut va avea $1$ pe pozitiile $x_{1},x_{2}+1,x_{3}+2,\ldots,x_{k}+k-1$.
deci corespondenta obtinuta este bijectiva, asadar numarul de siruri "bune" de lungime $n$ este chiar numarul sirurilor cu $k$ de $1$ de lungime $n-k$, si de aici vine raspunsul.
[/hide]

La solutia ta nu vad nimic sa fie gresit...insa pe un caz particular nu imi da ....de ex : fie n=8 si k=3...dupa cum spui tu ar trebui sa dea 10....fie a,b,c cele 3 dominouri...fixam primele 2 dominouri( adica pe pozitiile 1 si 2 asezam 2 dominouri a)...pentru pozitia 3 avem 3 modalitati de asezare (a,b sau c),dar pentru fiecare dintre acestea mai avem 3 modalitati in care putem sa asezam dominourile pentru pozitia 4....deci 9 asezari...pana acum ( am considerat fixe doar primele doua) ...


i was wondering if anyone could help me out with some problems on messenger i am having troubles with? mainly slope problems, linear equations, and inequalities.

i have all messengers but ICQ, AIM =eazee ryd3r(or iamalexwayne)
MSN=amentopringles@hotmail.com
yahoo=fuzzy000wuzzy00

thanks  :)

[quote="Valentin Vornicu"]Welcome!

I moved your post from the Olympiad section because it doesn't sound that you need help with olmypiad stuff. I would recommend taking a look on the [url=http://www.mathlinks.ro/Classes/AoPS_C_About.php]AoPS Classes curriculum[/url] and see if any of the intro classes help you with this.

Or try to mingle in the forums, by posting problems on the forum (read the descriptions first) and trying to solve problems from your level.[/quote]

i'm sorry, i'm new here, i don't even know what olympiad means, i was just searching math forum on google in hopes someone who enjoys or is knowledgeable about algebra would help me out. truthfully i just want someone to work some problems out for me.. and show me how they did some parts i didnt understand, since i cant do that on exams with the teacher of my homeschool course. i didnt want to post the exact problems on the board itself, i was hoping for private, direct communication via messenger, so was just advertising for a helper basically.

its algrebra 2, but this lesson is basically a review of algebra one, i imagine its very easy for some of the people on here, and would just like some help to speed the class up.

You are encouraged to post your problems in the appropriate forums: either [url=http://www.mathlinks.ro/Forum/index.php?f=298]middle school forum[/url] if you're a middle schooler (or equivalent of it) or [url=http://www.mathlinks.ro/Forum/index.php?f=214]high school forum[/url].

For private lessons I think you can check out [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=23348]Richard Rusczyk's offer[/url] or any of the [url=http://www.mathlinks.ro/Classes/AoPS_C_About.php]AoPS classes[/url] I mentioned.

i'm gonna post an example here, and maybe you can help or redirect me to where i would best be helped? not looking for private lessons really, just someone in private direct communication(via instant message) to help me with who knows this stuff algrebra/algrebra 2 stuff.

wat kind of system uv equationz is      2x-3y=5 and -2x+5y=-1
a. k(c)onsistent and e(i)ndependent
b. Konsistent and Dependent
c. InKonsistent
d. Inkonsistent and depindent

sorry for the intentional typos, wasdone for seurity purposes, still readable.

:?: Security purposes :?:

Please explain

I think that you should find a tutor near you. None of the Art of Problem Solving classes would help you with this.

I would recommend using the Middle School or Getting Started forums to ask for help ( I know you said that you don't want to use the boards but I still recommend it...).  As for someone working problems out for you do you not have a textbook?  Every textbook works out problems.  There are probably sites online that solve problems.

[quote="tokenadult"]Are you trying to get someone to help you do a take-home test?[/quote]
pretty much. :blush:
i've already did a good % of it, i'd just like some people to confirm the answers i got and help with the ones i couldnt get.

so what forum would be appropiate for

wat kind of system uv equationz is 2x-3y=5 and -2x+5y=-1
a. k(c)onsistent and e(i)ndependent
b. Konsistent and Dependent
c. InKonsistent
d. Inkonsistent and depindent

and

wat is the slope intercept form of the equation of a line that goes through (1, 4) and is perepndicular to a line whose equation is   y=2/3x+5 ?

a. y=2/3x+5
b. y=-3/2x+7
c. y=-3/2x+11/2
d.y=-3/2x+5/2

those 3/2's are fractions just in case ya didnt know.

also another one i had trouble with:
what is the standard form of an equation for a line that passes through (-1, -2) and has slope -3?

a. 3x+y=-5
b. 3x+y=-1
c. 3x+y=-1
c.3x+y=5
d. 3x+y=-7

and 2 more but ill have to make a picture of graphs to show the problem. surely someone can help me with these, i bet these are cake to some of you on this site.

Come on...we're not doing your homework/take home tests!!!

And plus, I already told you which forums to use.


Let p; q and r be prime numbers. It is given that p divides qr - 1, q divides rp - 1, and r divides pq - 1.
Determine all possible values of pqr.

We will use the following fact: $a \mid x, b \mid y \Rightarrow ab \mid xy$.

Since $p \mid qr-1, q \mid pr-1, r \mid pq-1$ we have $pqr \mid (pq-1)(qr-1)(pr-1)$. But now expanding and substracting multiples of $pqr$ we get $pqr \mid pq+qr+pr-1$. Suppose WLOG $p \leq q \leq r$. Then $pqr \leq pq+qr+pr-1 < 3qr$, so $p<3$. Then $p=2$. Now we have $2 \mid qr-1$ so $qr$ is odd, which means $2<q \leq r$.

Now $2qr \mid qr+2q+2r-1$, so $2qr \leq qr+2q+2r-1$, $qr \leq 2q+2r-1<4r$. So $q<4$. Then since $2<q, q=3$. Now we have $6r \mid 3r+6+2r-1=5r+5$. Then $6r \leq 5r+5, r \leq 5$. We see $r=3$ doesn't work and $r=5$ does, so $pqr=30$.


let $ (u_{n})_{n\in N}$ a sequence such that
$ \forall (m;n) \in N$*$ ^2  u_{mn}\equal{} u_{m}$x$ u_{n}$
$ u_{10}\equal{}0$
$ u(k)\equal{}0$ if $ k\equiv 3$[$ 10$]

find $ u_{2007}$


Ok we know that the majory of spammers are new users,  so what if we try to teach them how to be a good aops members. With....
1. New user forum with stickies like what is spamming how to rate posts, etc. all in one forum.
or 2. New User Groups with mods, who they can ask questions, and to look at their posts and see what good posts look like.
or 3. New User Oriention a math Jam like setting where they can ask question met admins/mods and meet other users.

I know that this my seem like a lot of work but look at the benefits.
1.  New user learn what a good post is and the benfits of having good posts.
2. The forums will have less spam meaning less work for mods.
3. The site becomes a more happy and fun environment.

I and many other members wish they knew this infromation before they spammed.

Those are good ideas monkeygirl112

I think we should have a "New Users Forum" so to speak, where all information in this site about posting, spam, etc. is listed.  New users could ask the questions they want.

Additionally, we could possibly have a video too.  I have a feeling new users will be engaged by a video and will take lessons from it and they would be more likely to pay attention.

When I was a new user, I didn't know the definition of spam, and my post rating is really hurting(it's only 2.4).  If I learned a little bit more about the standards before having to learn my lesson, my post rating would probably be 3+.

That's a good idea, new users [i]should[/i] have there own forum where they can learn the effect of spam.
They can learn there instead of learning the consequences on their own...

Oh yeah ditto maybach... your reasons are very reasonable. :)

You want to teach them not to spam? Make it compulsive.

Putting a spam policy in the Terms and Services area is good.

I believe I mentioned this in another forum (FTW, likely, but the same gist applies):

A flash applet. The TaS scrolls at a -readable- pace. After everything is read, the "OK" button is enabled. if no movement is registered in 1-2 seconds, then the "OK" button is disabled - preventing people who just wait it out and don't actually read it.

If many people don't read stickies, what makes you think they'll read a particular forum? I like james's idea because it's compulsory. Note that for the greatest effect the TaS could be a tad less dry, or maybe just let them have a shortened version that takes 30 seconds max to read (and is in a friendly tone, written somehow in an interesting way - that worked for the rules in my mock MATHCOUNTS last year. :P )

Or you could have someone read (audio) it so they [i]have[/i] to listen to it. Just a note: This forum has lots of topics about new user spam, and nothing has really happened. We might have to wait for "AoPS V2".

I (and many people I know) keep our computers muted, so that wouldn't be truly "compulsory".

How about this idea:  In order for someone to start posting on AoPS, they have to read everything with the Flash Applet as suggested by James41, and then have to take a quiz.  If they get a perfect on the quiz, they can start posting.  And we need to make sure the answers aren't so obvious so people may actually have to refer to the Instructions on Posting to answer the questions.   Additionally, we can make it so you have to pass the quiz before you can even play FTW or do Alcumus.  This way, people will be STRONGLY encouraged to read a little about proper posting.

As for me, at least my post rating is improving(now it's 2.6.  I learned posting productively in the FTW and this forum really helps your rating)

[quote="maybach"]How about this idea:  In order for someone to start posting on AoPS, they have to read everything with the Flash Applet as suggested by James41, and then have to take a quiz.  If they get a perfect on the quiz, they can start posting.  And we need to make sure the answers aren't so obvious so people may actually have to refer to the Instructions on Posting to answer the questions.   Additionally, we can make it so you have to pass the quiz before you can even play FTW or do Alcumus.  This way, people will be STRONGLY encouraged to read a little about proper posting.
[/quote]

That is pretty overkill, and would probably discourage many new users. Most of the TAS is common sense.

How about this idea: when new users spam, PM them and let them know in a nice way. Problem solved. :)

I think anyone who has been on this site less than 6 months and is posting in this thread should take immediate action to improve the situation of new users making low-quality posts by deleting their AoPS account.

For those of you (representing the majority of posts on this thread) whose ratings are in the range [1, 3] (but even more so [1, 2.5]), see also this: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1565092#1565092

Changing the Terms of Service or having a forum for new users wouldn't make much of a difference.  People joining this forum have started from other forums, where the expectations of the post quality is lower. So if you see someone spam, just PM them and tell them how to improve their posts. If not, then this user is fully aware that he/she is spamming. A forum or extra long terms of service will not help much here.


[quote="jayme"]Dear Mathlinkers,
see for example: http://perso.orange.fr/jl.ayme/ vol. 4 An unlikely concurrence revisited and generalized p. 5-7
Sincerely
Jean-Louis[/quote]
Dear [b]jayme[/b] is your notes in english too???

Dear Dimitris and Mathlinkers,
for the present they are not in English, but I think they will be in the future.
In this time, there is so a lot of figure that you can with a little effort understand the proof.
Sincerely
Jean-Louis

[quote="CCMath1"]Circle I is inscribed in triangle ABC and the circle touches BC , AC at points M ,N. E ,F are the mid points of segment AB,AC;let $ {BI \cap EF \equal{} D}$  SHOW THAT $ {D,M,N }$ are collinear.[/quote]
Assume BI intersects MN at D. we will prove that E. D. F are collinear.
Denote AD cuts BC at G
Easily, we have $ \widehat {BDA} \equal{} 90^o$, and BI is the bicsector of angle A. So, D is the midpoint of
AG. (QED)


[quote="admire9898"]the problem is:

a(n)=2a(n-1)+3a(n-2)+3^n
(n>1)  a(0)=-1,a(1)=1
find  a(n)
thanks!! :lol:[/quote]

Let $ a_n \equal{} b_n \plus{} \frac {3^{n \plus{} 1}}{4}n$. The sequence becomes $ b_n \equal{} 2b_{n \minus{} 1} \plus{} 3b_{n \minus{} 2}$ whose characteristic equation is $ x^2\minus{}2x\minus{}3\equal{}0$, or also $ (x\minus{}3)(x\plus{}1)\equal{}0$

So $ b_n \equal{} \alpha 3^n \plus{} \beta( \minus{} 1)^n$

Starting conditions $ b_0 \equal{} \minus{} 1$ and $ b_1 \equal{} \minus{} \frac 54$ imply $ \alpha \equal{} \minus{} \frac {9}{16}$ and $ \beta \equal{} \minus{} \frac {7}{16}$

And so $ a_n \equal{} \minus{} \frac {9}{16}3^n \minus{} \frac {7}{16}( \minus{} 1)^n \plus{} \frac {3^{n \plus{} 1}}{4}n$


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $ G$ be a group and $ H$ be a subgroup of $ G$. Suppose that $ aH \equal{} bH$ if and only if $ Ha \equal{} Hb$, where $ a, b\in G$. Show that $ H$ is a normal subgroup of $ G$.

$ aH\equal{}bH$ $ \Leftrightarrow$ $ Ha\equal{}Hb$ is equivalent to $ b^{\minus{}1}a \in H$ $ \Leftrightarrow$ $ ab^{\minus{}1} \in H$.

$ \forall h \in H$ and $ \forall a \in G$ there is $ b \in G$ such that $ ab^{\minus{}1}\equal{}h$.
$ ab^{\minus{}1} \in H$ $ \Leftrightarrow$ $ b^{\minus{}1}a \equal{}a^{\minus{}1}(ab^{\minus{}1})a\equal{}a^{\minus{}1}ha \in H$ , therefore $ H \triangleleft G$ .


$\angle BPT= \angle TPD$

Prove that $B,Z,L$ are collinear.

[b]Lemma[/b]. [color=blue]agar shekle $(1)$ ra dashte bahim .angah:$\angle BAD=\angle DAC$[/color](ba enekas be markaze $A$ mitooonid sabe konid)

[color=red](rahehal shekle aghaye madani)[/color]
hal ba borhan kholf mitavanid sabet konid dayere mohitye mosalase$ZTQ$  bar khatol markazein va do dayere momas ast.

baraye inke sabet konim $B,Z,L$  hamkhatand.bayad sabet konim ke cheharzeliye $BZQL$ mohati ke ba enekas be markaze $P$ hal mishe.

mibakhsid ke hal nadashtam ziad toozih bedam.

baayad saabet konim chehar zelie $BZQD$ mohaatie. pas baayad saabet konim ke agar $S$ markaze tajaanose daakhelie do daayer baashad; aangaah $Q,Z,S$ ham khatand.
dar mosalase $\triangle NTM$baraaye noghaate $Q,Z,S$ menelaaoos minevisim; tavajoh konid:
$\frac{MS}{SN}=\frac{R}{r}$
$TZ*TM=TO^2-R^2$
$TQ*TN=-TO'^2+r^2$
$MZ*MT=2R^2$
$NQ*NT=2r^2$

age dorost mohaasebe konid; hamechi montaghel mishe be khatol markazeyn; va kaafie tool haaye rooye khatol markazeyn ro mohaasebe konid ;)


group theory
abstract algebra
superior algebra
superior algebra solved

Since this one seems really easy (and since sometimes I do some pretty obvious mistakes :(), I just want to make sure I got it right:

$G$ is a group generated by two elements of order $2$. Show that $G$ contains a subgroup of index $2$.

I don't see why it needs to be finite- index 2 makes perfectly good sense for infinite groups.

I think something like this should work (regardless of the order of the group $G$):

Let $H$ be the subset of $G$ (obviously a subgroup) consisting of those words in $a,b$ (the mentioned generators) with an even number of letters. A word having an even number of letters is equivalent to its reduced form, with alternating $a$'s and $b$'s having an even number of leters.

What we should prove now is that no two words, one in $H$ and one not in $H$, can ever be equal. It's easy to see that this means proving that no reduced word with an odd number of letters, like $abab\ldots ba$ can be equal to the identity. This is easy: assuming WLOG that $abab\ldots ba=1$, multiply this relation both to the left and to the right with $a$ to get $bab\ldots ab=1$, then multiply this both to the left and right with $b$ to get an even smaller word on the left side and so on, until we get either $a=1$ or $b=1$, which contradicts the fact that the orders of $a,b$ are both $2$.

I specialize in short sentences that make people doubt their own "proofs"  :D

Thanks, I got the same Grobber ( equivalently the subgroup $<ab>$ generated by the product $ab$, which will also contain $(ab)^{-1}=ba$ ).


Source: Ireland 1993
1. The real numbers $\alpha$ and $\beta$ satisfy the equations
$\alpha^{3}-3\alpha^{2}+5\alpha-17=0$,
$\beta^{3}-3\beta^{2}+5\beta+11=0$.
Compute $\alpha+\beta$.

any cubic of the form $x^{3}-3x^{2}+5x+c$, where $c$ is constant, is symmetric about its inflection point, which is at $x=1$.  Pick $c=-3$ (chosen because it's the average of $-17$ and $11$.  It has a real root at $x=1$, and so the sum of the two values we want is $1+1=2$.

is this the only way to do it? sorry i dont understand your solution. :blush:

i dont know that it's the only solution, but it's the only one i found.  The idea to use cubic equation's symmetry about the inflection point is pretty useful though.

You can do it by factoring.

See, $x^{3}-3x^{2}+5x = (x-1)^{3}+2x+1$. So, adding the two equalities we get $(\alpha-1)^{3}+2\alpha+1+(\beta-1)^{3}+2\beta+1 = 17-11 = 6 \iff (\alpha-1)^{3}+(\beta-1)^{3}+2(\alpha+\beta-2) = 0$. Factoring $(\alpha-1)^{3}+(\beta-1)^{3}$ we obtain $\alpha+\beta-2$ as a factor, so the equation factors and we're close home.


A sequence of real numbers x_n is defined recursively as follows: x_0, x_1 are arbitrary positive real numbers, and
x_(n+2)=[1+x_(n+1)]/(x_n).

Find x_1998.

Nice problem. Where did it come from? It looks so familiar...

Do you mean find x_1998 in terms of x_0 and x_1, or is there a numerical value no matter what numbers x_0 and x_1 are?

Im pretty sure its in terms of x_0 and x_1.

[color=cyan]Definitely in terms of x_0 and x_1.  As for the other bit, it might be an old AIME question.  I don't really know, though.  Sorry.[/color]

#12 on AIME 2002 was similar but not identical. If you have a copy of the AIME 2002, please post #12 here. I forget what it says exactly.

Is it #12 at http://www.kalva.demon.co.uk/aime/aime02.html ?

Yeah, that will do. It's in different words, but the problem is the same.

So how about we do this problem??
Any takers??

Joel, Simon, u guys wait awhile...

I shall stand aside as well...not because Im as smart as Simon and Joel, but because I want to take the 2002 AIME as a real test.

but this question isn't the same one on the AIME

Oh, you were refering to your original question?

Well, at least Ill have the solutions to the questions after I finish the AIME.  8-)

I solved this problem on a piece of paper. I'm waiting for others to solve it. I'm already too old to take the AIME again.

C'mon
someones gotta answer dis

no fun if simon or I post the solution...

Well, you could probably calculate x_2, x_3, x_4, and x_5, and a simple pattern will emerge. I am too lazy to do right now though, I have summer reading to finish.  8-)

Why don't you try it? Maybe something nice will happen. Maybe it won't. I shall say no more (for now).

yea
don't be lazy like me
thats how u do bad on the AIME

remember, u need that 150 so ripply can see ur not-so-handsome picture on the perfect scores site

[color=cyan]ouch, that was harsh Long . . .

LOL[/color]

Ok, so will someone actually post the ENTIRE  solution?????

I agree. If you solve a problem, post the full solution. If you're giving a hint, that's ok too, but if you're a solver for a particular problem, post the full solution.


analytic geometry
number theory
relatively prime

In a rectangular coordinate system, a lattice point $ (a,b)$ is defined as a point $ (a,b)$ whose coordinates $ a$ and $ b$ are integers. If a dot is placed in the diagram on every lattice point whose coordinates are relatively prime, how many dots will there be?
[asy]draw((2,2)--(2,5)--(5,5));
draw((3,2)--(3,4)--(5,4));
draw((4,2)--(4,3)--(5,3));
draw((2,3)--(4,3)--(4,5));
draw((2,4)--(3,4)--(3,5));
draw((2,2)--(5,2)--(5,5));
label("$(2,2)$",(2,2),SW);
label("$(2,5)$",(2,5),NW);
label("$(5,5)$",(5,5),NE);
label("$(5,2)$",(5,2),SE);[/asy]


I have a question.

To prove with complex numbers that a triangle ABC is equilateral it's sufficient to show that a,b,c are distinct roots of a complex number.

   where A(a), B(b), C(c) .

[quote="tarta"] a,b,c are distinct roots of a complex number.
[/quote]

What is this? i.e?  :maybe:

Yes , if a,b,c are the roots of the equiation $z^{3}=r(cos\phi+i\cdot sin\phi)$ $\rightarrow$ A,B,C are points on the circle C(o,r) and ABC is equilateral.


I am using Asymptote in MiKTex + TeXnicCenter, details can be found the next pages:

http://www.mathlinks.ro/Wiki/index.php/Asymptote:_Getting_Started/Windows/Interactive_Mode

http://www.mathlinks.ro/Wiki/index.php/Asymptote:_Advanced_Configuration

Some time ago, I wrote the asymptote code in TeXnicCenter, and save the file as
filename.asy, then the pdf document with the image show up automatically. But now , on contrary, the pdf didn't open up automatically, though the pdf with the asympyote picture can generate.

I didn't konw what's wrong. How to get the pdf show up automatically ?
Many thanks

I use windows at the moment and created a batch file. I know this is a late post, but I figure its as good a place as any to share.
The shell variables should be pretty self explanatory.

1) Copy the code into a text file.
2) Change the ASYMPTOTE_EXE and PDFLATEX_EXE to represent the location of your executables.
3) Change the file extension from *.txt to *.bat.
4) Use customize in TeXnicCenter to run the batch file (directions in file).

[code]
rem Purpose: This batch file is for creating pdf output from TeXnicCenter using Asymptote.
rem Input: Takes a *.tex latex file that presumably has an Asymptote section in it.
rem Output: outputs a *.pdf file in the same directory as the input file.
rem Modification History: Created on 4/22/2010
rem
rem ------------------------------------
rem  SETUP INSTRUCTIONS FOR USE WITH TEXNICCENTER
rem In TeXniCCenter menubar, Tools->Customize, Tools tab, New Command (box looking icon)
rem 	Command: point to this batch file e.g. "C:\your_favorite_file.bat"
rem 	Arguments: the *.tex file. Use "%tc"
rem	Initial directory: directory the *.tex file was in. Use "%dc"
rem ------------------------------------

rem set the location of your asymptote executable
set ASYMPTOTE_EXE="C:\Program Files (x86)\Asymptote\asy.exe"
echo ASYMPTOTE_EXE: %ASYMPTOTE_EXE%

rem set the location of your pdflatex executable
set PDFLATEX_EXE="C:\Program Files (x86)\MiKTeX 2.8\miktex\bin\pdflatex.exe"
echo PDFLATEX_EXE: %PDFLATEX_EXE%

rem convert the input file to long name
set TEX_FILE_IN=%~f1%
echo TEX_FILE_IN: %TEX_FILE_IN%

rem get the drivename
set DRIVENAME=%~d1%

rem get the basename
set BASENAME=%~n1%

rem get the directory name (aka path)
set DIRNAME=%~p1%

rem set the asymptote filename that will be generated by pdflatex
set ASYMPTOTE_FILE_GENERATED=%DRIVENAME%%DIRNAME%%BASENAME%.asy
echo ASYMPTOTE_FILE_GENERATED: %ASYMPTOTE_FILE_GENERATED%

rem generate *.asy file
%PDFLATEX_EXE% "%TEX_FILE_IN%"

rem generate pdf from *.asy file
%ASYMPTOTE_EXE% "%ASYMPTOTE_FILE_GENERATED%"

rem generate pdf from *.tex file and now includes Asymptote info
%PDFLATEX_EXE% "%TEX_FILE_IN%"

rem cleanup shell variables
set "TEX_FILE_IN="
set "DRIVENAME="
set "BASENAME="
set "DIRNAME="
set "PDFLATEX_EXE="
set "ASYMPTOTE_EXE="
set "ASYMPTOTE_FILE_GENERATED="

rem uncomment last line to troubleshoot
rem pause[/code]


Let $ (f_n) \in L_p(X)$ be a sequence that converges almost everywhere to $ f$ and there exist $ g\in L_p(X)$ such that $ |f_n| \le g a.e.$, for all $ n \in \mathbb{N}$.

Prove that $ (f_n)$ converges almost uniformly to $ f$.

I let $ E \subseteq X$ such that $ m(E) \equal{} 0.$
So, for every $ x\in X\cap E^c$,and for all $ \epsilon >0,$ there exist an $ N(\epsilon,x)\in \mathbb{N}$ such that for all $ n>N(\epsilon,x),$ $ |f_n(x)\minus{}f(x)| < \epsilon$

In order to prove almost uniform convergence,I try to construct $ (f_n)$ so that $ f_n \rightarrow f$uniformly on $ X\cap E_\delta$ where $ m(E_\delta)< \delta$ for $ \delta >0$.

But how can I get rid of the dependence of the natural number  on x?(so that it will become a uniform convergence on $ X\cap E_\delta)$

Hint: The set where $ g>\epsilon$ has finite measure, and on the set where $ g\leq\epsilon$, how big can $ |f_n\minus{}f|$ be?


It's given a set $A$ of infinitely many distinct positive integers. Prove that there exists a prime $p>C$ ($C$ is an arbitrary constant) which devides $a+b$ where $a$ and $b$ are different elements of $A.$

Where this problem are from?
It seems very nice! I don't have nice results, at this moment.

You know any generalization?

Maybe this problem relation for sumsets theory, I thinks we can use following theorem

[b]Theorem (Alon, Nathanson, Ruzsa)[/b]. Let $\mathcal{A},\mathcal{B}$ be nonempty subsets of $\mathbb{Z}_{p}$ with $|\mathcal{A}|\not =|\mathcal{B}|$, and let $\mathcal{C}: =\{a+b|a\in\mathcal{A},b\in\mathcal{B}, a\not =b\}$. Then $|\mathcal{C}|\geq \min (p,|\mathcal{A}|+|\mathcal{B}|-2)$.

I think that in this case you cannot find the prime ...you can only prove that there is some because we can choose the elements of $A$ to give some recidue mod prime and so no sum would be devicible by this prime.

Здравей  :) ,

Yes, Alon's theorem stated above has no relevance to the problem. This problem (and many similar ones) are instantly killed by a deep theorem in diophantine approximations, the finiteness of the nondegenerate solution to an [i]$S$-unit equation[/i]. This theorem states that if $S = \{p_{1},\ldots,p_{r}\}$ is any given finite set of primes and $n$ is fixed, then an equation $x_{1}+\cdots+x_{n}= 1$ has only finitely many solutions $x_{i}$ in rational numbers of the shape $\pm p_{1}^{k_{1}}\cdots p_{r}^{k_{r}}$, with $k_{j}\in \mathbb{Z}$ (that is to say, only primes in $S$ appear in the numerators and denominators of the $x_{i}$), and such that no proper subsum of $x_{1}+\cdots+x_{n}$ vanishes. To instantly kill your problem using this [extraordinarily deep!] result, assume to the contrary that all prime factors of the numbers $a+b$, $a,b \in A, a \neq b$, form a finite set $S$, and consider all various quadruples $a,b,c,d$ of pairwise distinct elements in $A$. Then by assumption, $X : = a+b, Y: = c+d, Z: = a+c, T : = b+d$ are all "$S$-units" (have all their prime factors in $S$), and moreover satisfy $X+Y = Z+T = a+b+c+d$, whereas $X/T+Y/T-Z/T = 1$, and all $X/T, Y/T, Z/T$ are $S$-units, and moreover it is self-evident that no subsum of $X/T+Y/T-Z/T$ vanishes. Applying the above-stated theorem with $n=3$, we conclude that $X/T, Y/T, Z/T$ can take on only finitely many values; but $X/T = (a+b)/(b+d)$ and $a,b,d \in A$ were chosen perfectly arbitrary distinct elements of $A$, whence $A$ is finite, contradiction.

But of course, this problem doesn't require one to recourse to such a formidable result as the $3$-variable $S$-unit equation (which was only proven in the 1970s, I believe, and after previous futile efforts!) :) In fact, there is a simpler solution -- but I wouldn't like to spoil the fun, so I won't write it... yet :)

This is a bit ugly but it seems to work.

Suppose we have a counterexample: every element of $\{a+b | a\neq b\in A\}$ is divisible only by the primes in $S=\{2,p_{1},\ldots,p_{k}\}$.  Let $S$ be minimal among all such counterexamples.

We must have $k>0$.  Otherwise, fixing $a,b\in A$, we would have arbitrarily large pairs of powers of 2 that differ by the constant $b-a$, which is absurd.

Suppose $p\in S$ is an odd prime.  For $i=0,\ldots,p-1$, let $A_{i}=\{a\in A | a\equiv i \mod p\}$.  If $A_{i}$ is infinite, with $i\neq 0$, then every sum of two elements of $A_{i}$ is $\equiv 2i \mod p$, hence divisible only by primes in $S\setminus\{p\}$, contradicting minimality of $S$.  Thus all but finitely many elements of $A$ are divisible by $p$.  Passing to this cofinite set, and dividing out by $p$, we may similarly conclude that all but finitely many elements of $A$ are divisible by $p^{2}$.  In fact, for all $n>0$, all but finitely many elements of $A$ are divisible by $Q^{n}$, where $Q=p_{1}\ldots p_{k}$ is the product of the odd primes in $S$.

Dividing $A$ by a power of 2 if necessary, fix $x\in A$ odd, and write $x=p_{1}^{\alpha_{1}}\ldots p_{k}^{\alpha_{k}}u$, where $(u,Q)=1$.  Fix $n>\max\{\alpha_{i}\}$.

Case I: There are infinitely many even elements of $A$.  By the above, we may pick $y=2Q^{n}v \in A$.  Now $x+y$ is odd, so by assumption, we can write
$x+y = p_{1}^{\alpha_{1}}\ldots p_{k}^{\alpha_{k}}u+2Q^{n}v = p_{1}^{\beta_{1}}\ldots p_{k}^{\beta_{k}}$
But $\alpha_{i}<n$ forces $\beta_{i}=\alpha_{i}$, and this gives a contradiction.

Case II: There are finitely many even elements of $A$.  Then for some $r\in\{1,3\}$, we may pass to an infinite subset of $A$ all of whose elements are $\equiv r \mod 4$.  Now picking $y=Q^{n}v\in A$, we must have
$x+y = p_{1}^{\alpha_{1}}\ldots p_{k}^{\alpha_{k}}u+Q^{n}v = 2p_{1}^{\beta_{1}}\ldots p_{k}^{\beta_{k}}$,
and again $\beta_{i}=\alpha_{i}$, which is impossible.

So there can be no such counterexample.


[quote="mr. math"]How do you do this problem?

A cone is formed by roatating the region described by y is larger then or equal to 2x+1, and x is greater than or equal to 0, and y is less than or equal to 7 about the y-axis. What is the volume of this figure?

Please include an explantion

 :w00tb:  :agent:  :bomb:[/quote]
[hide]*Graphs*
The y-intercept of the first inequality is 1. The length of the y-axis between this and the last equation is a length 6 segment. The cone has height 6. The point the first and third inequality cross is (3,7) and point where the third inequality and the y intercept cross is (0,7) so the radius is 3. APplying the formula, $\frac{1}{3}\pi r^2 h$, the volume is $\boxed{18\pi}$[/hide]

[quote="kidwithshirt"]hmm isnt the height 7? cuz y=7 is the last one[/quote]

No, because the Y-intercept of the graph y=2x+1 is (0,1) and the distance between (0,7) and (0,1) is 6.


That looks right for the most part. Except

[hide]1+2+3+...+10 is actually 55 so its E[/hide]

I think E should be the answer.  Ryan22 can you explain your how you got D?


For a set $ P$ of five points in the plane, no three of them being collinear, let $ s(P)$ be the numbers of acute triangles formed by vertices in $ P$.
Find the maximum value of $ s(P)$ over all such sets $ P$.

The regular pentagon has $ 5$ acute triangles and
there are $ C_{5,3}\equal{}10$ triangles
but the sum of all the interior angles of any convex pentagon is $ 540$ so at least one angle is bigger than $ 90$ same for a non-convex pentagon thus:
$ 5\leq s(P)\leq 9$

Is there actually a set $ P$ having $ s(P) \equal{} 9$ ?
I myself feel that maximum value of $ s(P)$ is between $ 6$ and $ 7$ (although I have no proof).

In the following link, I showed that $ s(P)\le 7$ for all sets $ P$ of 5 points:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=14796[/url].
Actually, looking back, all I did was link to someone else's solution to Problem 6 from IMO 1970.

It's not too hard to come up with a convex pentagon showing that $ s(P) \equal{} 7$ can be achieved.

Can you post the figure for $ s(P)\equal{}7$ ?
The maximum I have got is $ 6$.


inequalities
linear algebra
linear algebra unsolved

Let $A,B \in {\mathbb C}^{ n \times n}$,invertibles and let $rank(A+B)=m$. Show that $rank(AB^{-1}+BA^{-1}+2I_n) \geq 2m-n$.  :ninja:


Two boys and four girls are officers of the Math Club. When the photographer takes a picture for the school yearbook, she asks the club's six officers and the faculty sponsor to sit in a row with the faculty sponsoR in the middle and the two boys not next to one another. How many different arrangements are possible?

By the way, the answer is 528, I have the answer key but do not know how to do it, so ya!

You have 3 seats either side of the faculty sponsor, and we need
consider only the six seats of the officers.  The number of
arrangements of these six people with no restrictions on where they
sit is 6! = 720.

If the two boys are a and b, then if we group them as (ab) and move
them about as a single entity, we have only 5 objects to arrange, and
this can be done in 5! ways = 120 ways.  However the boys could be
paired as (ba) and again moved around as a single entity, giving a
further 120 arrangements. So the total number of arrangements when the
boys ARE together is 240.

If now we subtract 240 from 720 we shall get the number of
arrangements when they are NOT together.

Number of arrangements with boys not together is 480.

[b]But wait theres more[/b]

There is a case where they sit like this  * * O | O * *
That is 48 more. the answer is then $ 48\plus{}480$ which is $ 528$

Nice copy-pasting from [url=http://mathforum.org/library/drmath/view/56160.html]here[/url] :wink:

But anyway, even if you're copying a solution, you should try to rewrite it on your own, so you might actually learn from it.

Well you know how people search. Like they say: Please search before posting. With the highlighted posting link to a site.

You should at least give credit to who wrote the solution...

Complementary counting, just subtract: 720-528=[b]192[/b]

[quote="tiger21"]Is there an easy way to  count how many ways the boys DO sit next to each other?[/quote]
Or you could think of it like this if you didn't have the 528:

You take the number of places where two people can sit together, which is 4, and multiply it by 2 because there are 2 ways the boys can sit next to eachother (ab and ba). That wil give you 8. Then you multiply that by 4! for the other 4 people. This gives you:
$ 8*4!\equal{}8*24\equal{}\boxed{192}$


topology
real analysis
real analysis unsolved

Prove that any perfect compact metric space which is totally disconnected is homeomorphic to the Cantor set. Also, any compact metric space is a continuous image of the Cantor set. Finally, a perfect nowhere dense subset of the real numbers is homeomorphic to the Cantor set.

[quote="harazi"]any compact metric space is a continuous image of the Cantor set.[/quote]

What happens if we try to generalize this?

A [i]Cantor space[/i] is a topological space of the form $ \{0,1\}^{A}$ (where $ A$ is some set), with the product topology induced by the discrete topology on $ \{0,1\}$ (as is well-known, the Cantor set is homeomorphic to $ \{0,1\}^{\mathbb N}$, i.e. a product of countably many copies of $ \{0,1\}$).

Is every compact Hausdorff topological space a continuous image of some Cantor space?


Is there an easy way to figure this out:

Dan is given a 20 question multiple choice test with four possible answers for every question and only on of these is right.  If he randomly guesses what will his expected score (average score, at least I think these two things are the same) be?

call me crazy, but is it.. 5?  expected gain of .25 each question, .25*20 = 5.

Or is there something I'm overlooking..

I agree, $\frac 14$ change of being right. $\frac 14 \times 20 = 5$

Interpreting this as a binomial setting.
Dan has .25 prob of success=p.
Expected/average value for binomial setting=np
.25*20=5
You are correct

Another easy way to think about it is this

Since there are 4 answers for each question, he can expect to get one out of every four questions correct. One out of four multiplied by 20 would give the same answer: 5


The ODE is:         m*(d^2x/dt^2)=-c(Vw+dx/dt)^2,
                                 dx/dt(t=0)=v0*cosA
                                          x(0)=0
let x1=x,x2=dx1/dt,then the systematic form is：
dx2/dt=-(c/m)*(Vw+x2)^2，x1（0）=0， x2（0）=vocosA
I let the matlab solve it and get a strange looking solution

x(t)=Int(Int(-c/m*(Vw+diff(y1(_z2),_z2))^2,_z2=0.._z1),_z1=0..t)+Vo*cosA*t

I just wonder how to translate those codes to hand written version?

It's kind of hard to decipher what you meant without latex

did you mean $ \frac {dX_2}{dt} \equal{} \minus{} \frac {c}{m}(V_w \plus{} X_2)^2$?
because if so, it's trivial to solve by hand, (assuming $ V_w$ is a constant)

$ X_2$ is just a run off the mill logistic equation


modular arithmetic
pigeonhole principle
number theory unsolved
number theory

Prove that for every prime $p>3$ there exist integers $x,y,k$ with $0<2k<p$ and $kp+3=x^2+y^2$

I think this problem is equivalent to the problem that you can find two squares $x$ and $y$ which satisfie $x^2+y^2\equiv 3 \pmod{p}$ because you can choose any $k$ if there is a congruence and $x,y>p$ has no sense so $x^2+y^2<2p^2$. And $k=0$ can't be OK because there aren't any squares whose sum is 3.

Naphthalin

Yes, and this was posted before:
http://www.mathlinks.ro/Forum/viewtopic.php?t=76830 post #4

To show it for the prime fields in another way:
Lemma: every residue $\mod p$ is sum of two squares.

Proof: Look at the numbers of type $x^2+1 \mod p$.
If all of them would be squares, we would get that $1,2,3,...$ all are squares solving our problem (but this can only happen for $p=2$ indeed).
So there is at least one non-square of type $s=x^2+1 \mod p$ otherwise.
But when $t,s$ are non-squares $\mod p$, then $ts^{-1} \mod p$ is a square. Using this for any given nonsquare $t$ and the $s$ above we get an $y$ with $y^2 \equiv ts^{-1} \mod p$, thus $t \equiv tss^{-1} \equiv y^2 (x^2+1) = (xy)^2+y^2 \mod p$.

More generally we have:

- Every prime residue $\mod p^n$ is sum of two squares as long as $p \neq 2$.
Also the residues $ap^k$ with $a$ coprime to $p$ and $k$ even and $<n$ and $p^n$ itself are sum of two squares, but the other ones only when $p \equiv 1 \mod 4$.

- As shown in the link this also holds in any finite field.

[quote="Naphthalin"]I think this problem is equivalent to the problem that you can find two squares $x$ and $y$ which satisfy $x^2+y^2\equiv 3 \pmod{p}$ because you can choose any $k$ if there is a congruence and $x,y>p$ has no sense so $x^2+y^2<2p^2$. And $k=0$ can't be OK because there aren't any squares whose sum is 3.

Naphthalin[/quote]
You are not right. You proved, that there is $k<2p$, but not $2k<p$ or $k<\frac{p}{2}$. For proving it we:
1) can find $x,y$ such, that $x^2+y^2\equiv 3\pmod p$(use Dirichlet Principle: from $p+1$ remainders of numbers $0^2, 1^2,...(\frac{p-1}{2})^2, 3-0^2, 3-1^2,..., 3-(\frac{p-1}{2})^2$ modulo $p$ there is two same);
2) can consider, that $0\leq x,y<p$(simply we take remainders of $x,y$ modulo $p$;
3) can think, that $0\leq x,y<\frac{p}{2}$(we take the least of numbers $x$ and $p-x$, $y$ and $p-y$).
Then there is an integer $k$, that $kp=x^2+y^2-3<\frac{p^2}{2}$ :arrow: $k<\frac{p}{2}$ and if $p>3$ we can see that $k>0$(neither $k\leq -1$ nor $k=0$($x^2+y^2\not=3$).


trigonometry
function
real analysis
real analysis unsolved

Is there some way to show that

for x = $ k\pi/2\sqrt2$,  k a (edit: nonzero) integer,

cos(x) is never of the form $ a/2^{b}$ with a,b integers?

To generalize - given that cos(x) is rational, can we say anything about x?

I've been playing with convergence arguments on the sum / product formulas for the function, but I'm kind of stuck...

Thanks very much for your help.

Take $ k=0$. Then $ \cos 0=1$. Then take $ a=4,b=2$.

Well, the Gelfond-Schneider theorem says that $ \alpha^\beta$ is transcendental if $ \alpha$ is algebraic and different from $ 0$, $ 1$; and $ \beta$ is algebraic and irrational. This takes care of your question (just take $ \alpha=e^{i\pi/4}$, $ \beta=\sqrt 2$) but it is not an easy theorem. On the other hand, I doubt some nice elementary approach is possible (as usual, I'll be excited to be proved wrong on this account :)).


30% of bottles produced in a certain plant are defective. If a bottle is defective, the probability that an inspector will notice and remove it from the filling line is 0.9. If the bottle is not defective, the probability that the inspector will think it is defective and remove it from the filling line is 0.2.
If a bottle is removed, what is the probability that it is defective?

since there is a .9 probability that a bottle will be removed a worker, which is defective, that means 90 out of 100 bottles will be thought to be defective and removed. Also, since there is a .2 probability that there is a normal bottle that might be removed by a worker, that means 20 bottles out of 100 will be removed, but is not defective.  So the probability that it is defective is 70%.


let $ f(x,y,z)\equal{}xy^2\plus{}yz^2\plus{}zx^2$

given $ c>0$ (const.) find
\[ \sup_{x,y,z>0\land x\plus{}y\plus{}z\equal{}c}f(x,y,z)\]
you mustn't use Lagrange multipliers


function
calculus
integration
trigonometry
algebra
polynomial
calculus computations

Hi! I just fished a math project and got the grade on it but the one question I got wrong is bothering me I must know the answer can anyone help me? here is that question.
[img]http://img.photobucket.com/albums/v705/pantha2/bipartisan-problem.gif[/img]
Also Could you show me how you got that answer? Thank you.

I am assuming $y$ is a function of $t.$ This is an example of a first order linear ODE, and the method of solving it is a standard one that can be found in most Calculus textbooks.

The general form of such an ODE is, $\frac{dy}{dt}+P(t)\cdot y = Q(t),$ where $P(t), Q(t)$ and $y$ are functions of $t.$

First, we need to find an integrating factor (I.F.), which in the general case is $e^{\int P(t)\, dt}.$ In this problem, the I.F. is $e^{\int-\frac12 \, dt}= e^{-\frac{t}{2}}.$ We multiply both sides of the equation $\frac{dy}{dt}-\frac12 y = 2\cos t$ with the above I.F to get

$e^{-\frac{t}{2}}\cdot \frac{dy}{dt}-y \cdot \frac12 e^{-\frac{t}{2}}= 2\cos t \cdot e^{-\frac{t}{2}}$

$\Rightarrow \frac{d}{dt}\left( y\cdot e^{-\frac{t}{2}}\right) = 2\cos t \cdot e^{-\frac{t}{2}}$

$\Rightarrow \int d(y\cdot e^{-\frac{t}{2}}) = \int 2\cos t \cdot e^{-\frac{t}{2}}\, dt$

$\Rightarrow y\cdot e^{-\frac{t}{2}}= 2\int \cos t \cdot e^{-\frac{t}{2}}\, dt+C$

Just use integration by parts to evaluate the integral on the RHS  of the equation above. I will that as an exercise for you.

Another method is to break the problem into its homogenous and particular equations.  The homogeneous equation is $y'-\frac{1}{2}y = 0$, which is straight-forward to solve.  Then you just need one particular solution to the equation.  If you just guess $y = a\cos t+b\sin t$ you get $y' =-a \sin t+b\cos t$ and so $y'-\frac{y}{2}= (-a-\frac{b}{2})\sin t+(b-\frac{a}{2})\cos t = 2\cos t$ so $-a-\frac{b}{2}= 0$ and $b-\frac{a}{2}= 2$, an system which is easy enough to solve.  Then add the two solutions together to get the solution to the original equation.

We spend more energy classifying problems in a differential equations course than in most other courses I know. Why? Because a succesful classification leads us to a method.

FieryHydra classified this as a first order linear ODE. That leads to the method of integrating factors.

JBL classified this as a constant coefficient linear ODE, nonhomogeneous, but with the nonhomegenity lying within the "exponential-polynomial" class of functions generated by sums of products of polynomials, exponentials, sines, and cosines. That classification leads to the method of undetermined coefficients.

The useful classifications are not disjoint, and this particular problem does lie within the appropriate intersection. JBL's classification is usually seen in the context of second order or higher equations, but there's no reason not to include first order equations as part of the same classification, as appropriate.


modular arithmetic
number theory proposed
number theory

Let $ x,y\in N$ satisfy :
$ x|y^{x\minus{}1}\plus{}1$ where $ x>1$
Prove that $ y\equiv 1 (\mod 2)$

[quote="TTsphn"]Let $ x,y\in N$ satisfy :
$ x|y^{x \minus{} 1} \plus{} 1$ where $ x > 1$
Prove that $ y\equiv 1 (\mod 2)$[/quote]

If $ y$ is even, then $ x$ is odd and all prime divisors of $ x$ are equal to 1 modulus $ 2^1$

Assume now, that all prime divisors of $ x$ are equal to 1 modulus $ 2^p$ for some positive integer $ p$.
Then $ x\equal{}1\pmod{2^p}$ and then $ x\minus{}1\equal{}k2^p$ for some k.
Then, for any prime divisor $ q2^p\plus{}1$ of x, we have $ \minus{}1\equal{}A^{2^p}\pmod{q2^p\plus{}1}$ where $ A\equal{}y^k$
Raising this equality at power $ q$, we have $ (\minus{}1)^q\equal{}A^{q2^p}\pmod{q2^p\plus{}1}$; But, since $ q2^p\plus{}1$ divides x, it does not divide A and so $ A^{q2^p}\equal{}1\pmod{q2^p\plus{}1}$ (fermat) and so $ q$ is even.

So, if all prime divisors of $ x$ are equal to 1 modulus $ 2^p$ for some positive integer $ p$, all prime divisors of $ x$ are equal to 1 modulus $ 2^{p\plus{}1}$.

And, since $ y$ even implies all prime divisors of $ x$ are equal to 1 modulus $ 2^1$, all prime divisors of $ x$ are equal to 1 modulus $ 2^p$ for any positive integer $ p$. And this is clearly impossible if $ x\neq 1$

So $ x>1$ implies $ y$ odd.

[quote="pco"][quote="TTsphn"]Let $ x,y\in N$ satisfy :
$ x|y^{x \minus{} 1} \plus{} 1$ where $ x > 1$
Prove that $ y\equiv 1 (\mod 2)$[/quote]

If $ y$ is even, then $ x$ is odd and all prime divisors of $ x$ are equal to 1 modulus $ 2^1$

Assume now, that all prime divisors of $ x$ are equal to 1 modulus $ 2^p$ for some positive integer $ p$.
Then $ x \equal{} 1\pmod{2^p}$ and then $ x \minus{} 1 \equal{} k2^p$ for some k.
Then, for any prime divisor $ q2^p \plus{} 1$ of x, we have $ \minus{} 1 \equal{} A^{2^p}\pmod{q2^p \plus{} 1}$ where $ A \equal{} y^k$
Raising this equality at power $ q$, we have $ ( \minus{} 1)^q \equal{} A^{q2^p}\pmod{q2^p \plus{} 1}$; But, since $ q2^p \plus{} 1$ divides x, it does not divide A and so $ A^{q2^p} \equal{} 1\pmod{q2^p \plus{} 1}$ (fermat) and so $ q$ is even.

So, if all prime divisors of $ x$ are equal to 1 modulus $ 2^p$ for some positive integer $ p$, all prime divisors of $ x$ are equal to 1 modulus $ 2^{p \plus{} 1}$.

And, since $ y$ even implies all prime divisors of $ x$ are equal to 1 modulus $ 2^1$, all prime divisors of $ x$ are equal to 1 modulus $ 2^p$ for any positive integer $ p$. And this is clearly impossible if $ x\neq 1$

So $ x > 1$ implies $ y$ odd.[/quote]
Good solution.I have other method but I will post after you solve same problem:
Prove that $ x\equiv 0 (\mod 2)$
I think here is an hard problem : Does $ (x,y)$ infinite?

[quote="TTsphn"]
Prove that $ x\equiv 0 (\mod 2)$
I think here is an hard problem : Does $ (x,y)$ infinite?[/quote]

Let $ p>2$ be an odd prime. Then $ (2p\minus{}1)^{2p\minus{}1}\plus{}1\equal{}(\minus{}1)^{2p\minus{}1}\plus{}1\equal{}0\pmod{2p}$

And so $ x\equal{}2p$ and $ y\equal{}2p\minus{}1$ are a solution to the equation $ x|y^{x\minus{}1}\plus{}1$

And there are an infinity of such pairs $ (x,y)$

[quote="TTsphn"]Good solution. But can you prove $ x\equiv 0(\mod 2)$[/quote]
Hello TTsphn,

It's in my first post in this thread :

$ x$ odd implies all prime divisors of $ x$ are equal to 1 modulus $ 2^p$ for any positive integer p.

So $ x$ odd implies $ x\equal{}1$

So $ x>1$ implies $ x$ even (and so $ y$ odd, which was the target in the first post).

Hello pco. Here is my solution.
Call p is a prime divisor of n such that $ v_2(p\minus{}1)$ is min.
Imply that $ v_2(p\minus{}1)\leq v_2(n\minus{}1)$
But $ p|\gcd(2^{p\minus{}1}\minus{}1,2^{2(n\minus{}1)}\minus{}1)\equal{}2^{\gcd(p\minus{}1,2(n\minus{}1)}\minus{}1|2^{n\minus{}1}\minus{}1$
So $ p|2\Longleftrightarrow p\equal{}2$
So $ x\equiv 0(\mod 2)$
From this $ y\equiv 1(\mod 2)$

[quote="TTsphn"]Hello pco. Here is my solution.
Call p is a prime divisor of n such that $ v_2(p \minus{} 1)$ is min.
Imply that $ v_2(p \minus{} 1)\leq v_2(n \minus{} 1)$
But $ p|\gcd(2^{p \minus{} 1} \minus{} 1,2^{2(n \minus{} 1)} \minus{} 1) \equal{} 2^{\gcd(p \minus{} 1,2(n \minus{} 1)} \minus{} 1|2^{n \minus{} 1} \minus{} 1$
So $ p|2\Longleftrightarrow p \equal{} 2$
So $ x\equiv 0(\mod 2)$
From this $ y\equiv 1(\mod 2)$[/quote]

Euhhh, I'm sorry.

I don't understand why $ p|\gcd(2^{p \minus{} 1} \minus{} 1,2^{2(n \minus{} 1)} \minus{} 1)$. Sure, $ p|2^{p \minus{} 1} \minus{} 1$ but there is no reason for $ p|2^{2(n \minus{} 1)} \minus{} 1$.

For example, with $ n\equal{}33$ and $ p\equal{}11$, $ p\equal{}11$ divides $ 2^{p\minus{}1}\minus{}1\equal{}1023\equal{}11\times 93$ but $ p$ does not divide  $ 2^{2(n\minus{}1)}\minus{}1\equal{}2^{64}\minus{}1\equal{}4\pmod{11}$

Here is real hard problem .
For each $ y\in N$ if there exist  infinite $ x$ such that $ x|y^{x\minus{}1}\plus{}1$
For the case $ y\equiv 0(\mod 2)$ then it has no solution $ x>1$
But in other case?

It is not hard. Let $ x|y^{x-1}+1$ and x>1 is odd, $ p|x$ is prime divisor. $ x-1$ is even, therefore period y mod p - $ T{}_y{}_p$ must be multiple 4 ($ 4|T{}_y{}_p$). If for any prime divisor $ p=1\mod 2^k$, we get $ 2^k|x-1$, therefore for period $ T{}_y{}_p$ must be $ 2^{k+1}|T{}_y{}_p\Longrightarrow p=1\mod 2^{k+1}$. Therefore no odd solution x>1.
If $ y$ is even $ 2|y^{x-1}+1$ for $ x>1$, therefore no solution x>1.
If y>1 is odd, we consider odd primes $ p_i$, suth that $ m_p=\frac{T{}_y{}_p}{2}$ is odd and we can find solution $ x= 2^kp_1...p_m$. It is solution if and only if $ m{}_{p{}_i}|2^{k}p_1...p_m-1$.
For any prime p, suth that $ m_p$ is odd $ x=2^kp$ solution if $ 2^k|y+1$ and $ m_p|2^k-1$.


lol, i don't even know how to begin.

There are several values for a prime, p, with the property that any five-digit multiple of p remains a multiple of p under "cyclic permutation." Once such value is 41 (for example, since 50635 is a multiple of 41, so are 55063, 35506, 63550, and 6355); another such value of p is 3. Computer the value of p greater than 41.


function
inequalities
AMC
USA(J)MO
USAMO
floor function
algebra unsolved

Prove that [nx] => [x]+[2x]/2+...+[nx]/n where n is a natural number and x any real number

I think should be $ x>0$
Let
$ f(x)\equal{}[nx]\minus{}(\frac{[x]}{1}\plus{}\frac{[2x]}{2}\plus{}...\plus{}\frac{[nx]}{n}$
$ f(x\plus{}1)\equal{}f(x)$
So enough to prove that $ f(x)\geq 0 ,\forall x\in [0,1)$
We will prove the equalities:
$ [x\plus{}\frac{n\minus{}k}{n}]\geq \frac{[kx]}{k},\forall 0<k<n$
$ g(x)\equal{}[x\plus{}\frac{n\minus{}k}{n}]\minus{}\frac{[kx]}{k}$
And $ g(x)\equal{}g(x\plus{}1)$
More than $ g(x)\geq 0,\forall x\in [0,1)$ so the innequlities claim.
But from $ [nx]\equal{}[x\plus{}\frac{1}{n}]\plus{}...\plus{}[x\plus{}\frac{n\minus{}1}{n}]$
Apply two result this problem claim.

But g(x)>0 doesn't hold when k=5, n=9, x=4/9

This is an old USAMO problem (1981/5?).  A generalization was proposed as APMO 1999/2 (?), which was the following (paraphrased): If $ \{a_i\}$ is a sequence of real numbers satisfying $ a_i\plus{}a_j\geq a_{i\plus{}j}\ \forall i,j\in [1,n]\cap\mathbb{N}$, then $ \sum_{i\equal{}1}^n \frac{a_i}{i}\geq a_n$.  Taking $ a_i\equal{}\minus{}\lfloor ix\rfloor$, the result is obtained.  The generalization can be solved using a simple induction.

Can someone write the solution using Hermite's Id. ( manuscript is OK )!?
tnx a lot


Let $ a,b,c\geq 0$, s.t. $ a^3\plus{}b^3\plus{}c^3\equal{}3$.Prove that:
$ a^4b^3\plus{}b^4c^3\plus{}c^4a^3\leq 3$

We have: $ \sum\ a^6b^3.\sum\ a^3b^3. \sum\ a^3b^3 \geq\ (\sum\ a^4b^3)^3$
We only need to prove: $ \sum\ a^6b^3.\sum\ a^3b^3. \sum\ a^3b^3 \leq\ 27$
It can rewrite: Given $ a, b, c$ satisfy $ a \plus{} b \plus{} c \equal{} 3.$
Prove that: $ (a^2b \plus{} b^2 \plus{} c^2a)(ab \plus{} bc \plus{} ca)^2 \leq\ 27$
It's obviously trues because: $ (a^2b \plus{} b^2c \plus{} c^2a)(ab \plus{} bc \plus{} ca) \leq\ 9$
and $ ab \plus{} bc \plus{} ca \leq\ 3$
 :)


Find all pairs of integers $ (x,y)$ satisfying $ 1\plus{}1996x\plus{}1998y\equal{}xy.$


Marcus and Al are playing Rock/Paper/Scissors. The first person to win 10 times does not have to do the dishes. The score is now 9 to 7 in favor of Marcus. If ties are ignored and not counted, what is the probability Marcus will not have to do the dishes? Express your answers as a common fraction.

[hide]i got 1/2 + 1/4 + 1/8, so 7/8.  there is a 1/2 chance of him winning the first game, a 1/4 of losing the first but winning the second, and a 1/8 of losing the first 2 but winning the last one.[/hide]

[hide]Well, the prob that he loses is 1/2 * 1/2 * 1/2, because to lose he has to lose 3 games in a row.

So the prob he wins is 7/8.[/hide]

[hide]The probability of him winning one game is $\frac12$. The probability of him losing then winning is $\frac12\cdot\frac12=\frac14$.
The probability of him losing two then winning is $\frac18$. $\frac12+\frac14+\frac18=\frac78$.[/hide]


number theory
least common multiple
inequalities
superior algebra
superior algebra unsolved

$4$

$6$

$2$ (note that $\frac{2}{\sqrt{3}+\sqrt{5}}=-\sqrt{3}+\sqrt{5}$, thus $\sqrt{3}, \sqrt{5}$ are contained in the right one)

All by using $\mathrm{lcm}([L: K],[M: K]) \leq [LM : K] \leq [L: K] \cdot [M: K]$.

But, it would be better if you could post full solutions. I have some friends who are watching the topic and they don't understand all you have said. In fact, they asked me to post this kind of exercices. :)

The inequality $\mathrm{lcm}([L: K],[M: K]) \leq [LM : K] \leq [L: K] \cdot [M: K]$ comes from:
- $[LM : K]$ must be a multiple of $[L: K]$ and $[M: K]$, thus the LHS.
- when $a_1 , ... , a_{[L: K]}$ is a basis of $L|K$ and $b_1 , ... , b_{[M: K]}$ for $M|K$, then $LM|K$ is generated by the products $a_i b_j$, thus the RHS.

To most ugly part indeed is to show that some of the degrees, e.g. that one of $\sqrt{1+\sqrt{3}}$ is $4$ and not $2$, which can be done by calculation:
assume that $\sqrt{1+\sqrt{3}}$ has just degree $2$, then $\sqrt{1+\sqrt{3}}$ must be a square $a^2$ in $\mathbb{Q}[\sqrt{3}]$. Then also $\overline{a}^2 = \overline{1+\sqrt{3}} = 1-\sqrt{3}$ (overline=conjugation). But $\overline{a}$ is real, thus $\overline{a}^2 \geq 0$, which contradicts $1-\sqrt{3}<0$.

Others are similar, something that works everytime is just ugly calculation that often gets simpler by conjugates, norms etc.


http://www.sporcle.com/games/world.php

Post your results!

Mine:
[hide="Spoiler"]
[img]http://i42.tinypic.com/5vskfd.png[/img]
I MISSED EGYPT. Argh.
[/hide]

Argh.

Missed Hungary, Honduras, Cote d'Ivoire, Republic of the Congo, and Boliva...for a total of 190/195.  ARGH.

I got to about 60 before I resorted to [url]http://www.youtube.com/watch?v=IDtdQ8bTvRc[/url].

I got 190...because I can't spell; I memorized all the countries in 5th grade, but I just can't spell some of them properly. :blush:  I did pretty well on the fifty states "typing test," though. (Yes, since it's so easy, it's practically a typing test.) I typed fifty states in 94 seconds, which is 1.88 seconds per state.

BTW...animaniacs = very unreliable source. According to them, the following are countries:
"Republic Dominican" (should be Dominican Republic)
Caribbean (region with many countries)
Greenland (owned by Denmark)
Puerto Rico (owned by United States)
Bermuda (owned by United Kingdom
"Tobago" (should be Trinidad and Tobago)
San Juan (capital of Puerto Rico)
French Guyana (possibly still a country when Animaniacs made the song, but is now "reclaimed" property of France)
Guam (owned by United States)
Czechoslovakia (split into Czech Republic and Slovakia)
Scotland (one of the divisions of United Kingdom)
"Both Yemens" (I believe that Animaniacs isn't so old that the Yemens haven't joined yet)
England (one of the divisions of United Kingdom)
Burma (the proper name is Myanmar)
Kampuchea (the proper name is Cambodia)
Asia (a continent; wow, did they actually think people won't notice?)
Korea (Korea is split into North Korea and South Korea)
Tibet (owned by China)
"New Guinea" (should be Papua New Guinea; New Guinea [or the alternate Indonesian name of Irian Jaya] refers to the island)
Borneo,Sumatra (islands, part of Indonesia, can't believe they actually put that)
Okay, they said "the Spanish Sahara is gone"...I guess I'll assume that they know it's gone...
"Algier" (should be Algeria)
Dahomey (ancient empire centered in Benin)
Zaire (forgivable; when Animaniacs came out it was probably still called Zaire; now it's the Democratic Republic of Congo)
Mahore (owned by France)
"Cayman" (if they're referring to "Cayman Islands," that's owned by the United Kingdom, if they're referring to something else, they're referring to something inexistant)
Hong Kong (still owned by China)
Abu Dhabi (capital of the United Arab Emirates)
Yugoslavia (split into Slovenia, Croatia, Bosnia and Herzegovina, Serbia, Montenegro, Kosovo, and Macedonia)
Crete (island belonging to Greece)
Transylvania (regoin on Romania
Palestine (still considered part of Israel)

That's a lot; do you think they thought they'd get away with all that crap? Most of us would definitely be able to identify most of that by like 5th grade. They also left a whole bunch out. Seriously, bad, bad, bad. They also did a similar thing with making up chemical compounds in an extremely blatant fashion in the Ice Cream Song; seriously, who would fall for "deoxylite trisilicon"?

Saddest Quote Ever: "It took them 40 minutes to locate Canada on the map."

[quote="r15s11z55y89w21"]"Both Yemens" (I believe that Animaniacs isn't so old that the Yemens haven't joined yet)

Zaire (forgivable; when Animaniacs came out it was probably still called Zaire; now it's the Democratic Republic of Congo)

Hong Kong (still owned by China)

Yugoslavia (split into Slovenia, Croatia, Bosnia and Herzegovina, Serbia, Montenegro, Kosovo, and Macedonia)[/quote]

Airdate of said animaniacs song: 1993

The two Yemens joined cerca 1990, so that one is a mistake.

Zaire still existed up until 1997.

Hong Kong was still a dependent territory of the UK until its return to China in 1997. Even then, it's largely autonomous.

Federal Republic of Yugoslavia existed up until 2003 (albeit only established in 1992).

Otherwise... yeah. XD

Epically failed the 50 states typing test (07:22).

I got all of the Africa countries though cause I can draw a mental map of all 53.

I got $ 88$.

For U.S states, I got $ 2: 42$. :lol:

I got 106.

I was going to type dominican republic and then it put it as Dominica. That is one bad thing about it.

112, I am really slow at typing.  :blush:

88.  I missed like everything in africa except the really obvious ones  :oops:

46...for some reason it wouldn't let me do pretty easy countries, like eqypt.  :(

132... failed all of the western african countries and the pacific islands...

It took me 2 minutes 4 seconds to name all the states. :D

101 countries...got only 14 from Africa and like 3 from Oceania/Austrailia...also I can't remember all the -istan countries :P.

Pakistan
Afghanistan
Uzbekistan
Turkmenistan
Kazakhstan
Kyrgyzstan
I know I'm forgetting one...

Tajikastan (something tells me I misspelled that)

I got 125 countries. I used to go on geography sprees, thinking I could bang out every single country in about 5 continents but the rest threw me off. Need more practice, especially in Central America and the Pacific nations.

[quote="r15s11z55y89w21"]Tajikastan[/quote]
THAT'S it!
And it's Tajikistan

122.. I don't know if it's good... OR BAD O:

109.  I can't believe I missed Belgium  :wallbash_red:

113.  But I forgot a bunch, including:
North Korea
South Korea
Kenya
Belgium
Guatemala
Phillippines
Macedonia
and a bunch of other obvious ones........................ :censored:

138
I used to know all the countries in 4th grade, but I forgot some of the African countries. :o  :o  :o

44

Dang it. I used to know much more than that.  :(

I improved to 154.

The trick is to tackle everything continent by continent..............


Here's the ques

The major product will be (2)

The question is why??

plz answer fast

Not sure but, Since 2 is a conjugate system Resonance comes into play thereby stabilizing the system. In 1 the double bond is quite away from the site of reaction and is less stable as campared to 2. :maybe:

[quote="murgi"]Here allylic addition of bromine has taken place in 2 which stabilises 2 :)[/quote]
if u see carefully,it is vinylic addition rather than allylic

@pulkit-Do u mean the resonance betw lp of Br and double bond in 2 :)

-acc to resonance,dounle bond will be in resonance with intermediate carbocation only in 1 :P

an edit... (not my mistake-my teacher wrote it wrong :mad: )
Now addition does takes place at allylic carbon
But still carbocation intermediate of 1 is more stabilised by resonance.... :?:

I think that alkene in first is a terminal alkene thereby is more reactive than 2, Hence less stable.

But I also really feel that 1 will be more stable due to tertiary carbocation,more hyperconjugative pairs :fool:

Guess I would have to wait till my tutor teaches me alkene  :blush:

The most stable ressonance structure was the first one on my mechanism, not the second. However, one reason the major product is the 1-bromo is that its double bond is more stable, since it is more substituted.

But still how is the third figure in your post the most stable one? :o

The carbocation in carcul's second figure is a tertiary one and should be most stable. :(

carcul sir, we were able to guess out this mechanism...ques is just that  why the  primary c+ is more stable resonance structure :(  :)

[quote="Conjurer"]I actually think, this reaction is done in present of peroxide. First addition then elimination gives you product 2 which is also the major product.[/quote]
I agree................

Peroxide is completely absent,i m sure :(  :maybe:

got the answer...Themodyanamically stable product :mad:  :maybe: but how??? :(

More substituted double bond?

Nice question btw.

Case of Kinetics versus thermodynamics...

Would you like to go for a kinetically stable product (which proceed through a more stable intermediate) or would you like to go for a thermodynamically stable product (which in a majority of the cases, means a more stable product)?

The choice is yours... :P

And Conjurer, what makes you think of peroxide?

I THINK THE QUESTION IS WRONG. THE DOUBLE BOND IN THE OPTION 2 WILL SHIFT TOWARDS LEFT FOR ONE BECAUSE OF HYPERCONJUGATION OF POSITIVE CHARGE. :D  :D  :D

It is true that the double bond in choice 2 is misplaced, and the question is basically asking if it is better to 1) have a more stable positive charge on the tertiary center or 2) whether it is better to have a double bond with more substituents. Kinetically, the first product will likely be favored (as hell_ever noted above) since the positive charge is in the more stable place, while thermodynamically, the second will be favored. This is because the final products do not have a positive charge anywhere, so the main factor is the number of substituents on the double bond

Not only in thermo aspect but you could determine stability based on arylic and vinylic position of $Br$


Hi. I'll like to see if there wich are the differences between the mafia games that are played in your country, for example, here in Mexico (and in most of latinamerica):
The only roles are civilians, mafia (also called assasins), inspector and angel (mostly called doctor).

There are other games like one called Vampires, th diferences are:
1) Instead of mafia, there is a Vampire who instead of killing other people, makes them vampires (one person who has another role, might be also converted, in which case, will have both roles).
2) Inspector and angel roles are similars (searchs for vampires and heals from being converted to a vampire)
3) There is another character, The "santo" who select a person each turn, if that person is a vampire (or is turned into a vampire in that turn) dies.

Also, Does anybody knows were the mafia started? I've heared that it was in Argentina, but i'm not sure

For United States
well if u re not playing online, and ure just playing with cards, there are usually citizens, mafia, inspector, and doctor...
Sometimes there are serial killers and magicians


factorial
function
calculus
integration
algebra
functional equation
calculus computations

What is the process for proving the relationship of the gamma function and the factorial, starting with the factorial?

Why would you start with the factorial?  The thing to be proven is about the Gamma function, and it is usually done by integration by parts.  Once you've established the functional equation $ \Gamma(z \plus{} 1) \equal{} z \Gamma (z)$, you're pretty much done.

show that 'at integers', the gamma function (shifted by one) and the factorial satisfy the same recursion, and are equal at $ 0$.  depending on your definition, the recursion may actually define the factorial.

[quote="0714446459923"]How do you prove that the Gamma function is the only (log convex) function with this property?[/quote]
That is the so-called [url=http://en.wikipedia.org/wiki/Bohr-Mollerup_theorem]Bohr-Mollerup Theorem[/url] (proofs can be found using links on that page).


Find all natural numbers $n$ such that

(a) $n$ is not a power of two

(b) $2^n+1=p^m$ for some prime number $p$ and natural number $m$

Only $2^3+1=3^2$ works.

This is what is needed for http://www.mathlinks.ro/Forum/viewtopic.php?t=82146 and can e.g. be proven by seeing that $p=2^k+1$, setting in and looking at the $p$-adic valuations.


abstract algebra
Ring Theory
superior algebra
superior algebra unsolved

Let $ M$ be the $ \mathbb{Z}[i]$-module generated by the elements $ v_1$, $ v_2$ such that $ (1\plus{}i)v_1\plus{}(2\minus{}i)v_2\equal{}0$ and $ 3v_1\plus{}5iv_2\equal{}0$. Find an integer $ r \geq 0$ and a torsion $ \mathbb{Z}[i]$-module $ T$ such that $ M \cong \mathbb{Z}[i]^r \times T$.

Are you familiar with how to compute [url=http://en.wikipedia.org/wiki/Smith_normal_form]Smith normal forms[/url] over a generic PID?

Yes I am.  However, I do not know how to use it in this problem.  How do I use that here?  If not that way, if you have any other recommendations for solving the problem, that would be great too.

What is the Smith normal form of $ \left[ \begin{array}{cc} 1 \plus{} i & 2 \minus{} i \\ 3 & 5i \end{array} \right]$?  What does that tell you about a good choice for $ T$?


Let x,y,z>0,s.t.x+y+z=1,and k is a specified nonnegative integer
prove that:
$ \frac{x^{k\plus{}2}}{{x^{k\plus{}1}}\plus{}{y^k}\plus{}{z^k}}\plus{}\frac {y^{k\plus{}2}}{{y^{k\plus{}1}}\plus{}{x^k}\plus{}{z^k}}\plus{}\frac {z^{k\plus{}2}}{{z^{k\plus{}1}}\plus{}{x^k}\plus{}{y^k}}\geq \frac {1}{7}$

posted before, It's serbia 2007
see here : http://www.mathlinks.ro/viewtopic.php?p=860869#p860869


geometry
3D geometry
sphere
projective geometry
advanced fields
advanced fields unsolved

Hello,

I am looking for explicit formula for canonical (Riemannian) metric on the real projective space $ \mathbb{RP}^n$. Could anyone give me some references or put this formula here?

There is well known Fubini-Study (Kahler) metric for complex projective space and I could not find something similar for real space.

The same as the sphere since the covering map is a local isometry.


[quote="Interval"]If a < b, show that a < (a + b)/2 < b. The number (a + b)/2 is called the ARITHMETIC MEAN of a and b.[/quote]

OK, it is easy to show that the midpoint of two numbers is betweeen the two numbers.

So, this should be in Getting Started.

Just multiply through by 2...yeah someone should move this.

[quote="Interval"]Can you provide me with the steps for proving this question.[/quote]Multiply through by 2:

$2a < a+b < 2b$

Then look at the two inequalities seperately:

$2a < a+b \Rightarrow a <b$ (I subtracted $a$ from both sides), which we assumed true.

$a+b < 2b \Rightarrow a < b$, which is true.

both parts of the inequality are true, so the whole thing is true.

Absolutely excellent replies as usual.

Thanks.


A protractor is made so that the edge of its scale is 7.5 cm from the center point. If the scale is marked in degrees, how far apart arethe marks along the edge?

If the protractor has a 7.5 cm radius, you can look at it as a semi-cirlce with a 7.5 cm radius.
We can make a nice sector of the full circle.

The circumference of the circle is $15\pi$. If we look at the markings from 0 degrees to 90 degrees, you can see they mark out a quarter of the equivalent circle. So we know the distance from 0 degrees to 90 degrees is 1/4th that of the whole cirlce's circumference, or $\frac{1}{4}\cdot 15\pi$, or $3.75\pi$ cm.

Because there are 90 degrees in this arc, we can see that each degree is $\frac{1}{90}$ of $3.75\pi$ cm, or $\frac{\pi}{24}$ cm between each degree mark.

is it the strait distance between the points or the distance between the points on the arc?

The problem says "along the edge", and since the edge is curved....

No, it's correct, I was replying to jli.