calculus
integration
trigonometry
calculus computations

[u][b]without[/b][/u] using contour integration  :whistling: , show

$ \int _0^\infty \; \dfrac { x \minus{} \sin \, x} { (\pi^2\plus{}x^2) \; x^3 } \;\;\textbf dx \; \equal{} \boxed {\dfrac{1}{2\pi^3} \; \left( 1 \plus{} 3\zeta(2) \minus{} \pi \minus{} \dfrac{1}{e^\pi} \right)  }$

$ I = -\int_0^{\infty} \frac{\sin\left( x\right) }{\left( 2\,{x}^{5}+2\,{\pi }^{2}\,{x}^{3}\right) \,{\sin\left( x\right) }^{2}+\left( 2\,{x}^{5}+2\,{\pi }^{2}\,{x}^{3}\right) \,{\cos\left( x\right) }^{2}}dx-\int_0^{\infty} \frac{\sin\left( x\right) }{2\,{x}^{5}+2\,{\pi }^{2}\,{x}^{3}}dx-\frac{%Error. "atan" is a bad command.
\left( \frac{x}{\pi }\right) }{{\pi }^{3}}-\frac{1}{{\pi }^{2}\,x}$

I'm not sure if that helps. I'm unfamiliar with how to do these in general.


function
number theory unsolved
number theory

For $ n\in\mathbb{N}$ prove that:

$ \sum_{d|n}\frac{\mu^{2}(d)}{\phi(d)}\equal{}\frac{n}{\phi(n)}$

[hide="Hint"]define a function $ f: \mathbb{N}\rightarrow\mathbb{R}$ as $ f(n)\equal{}\frac{\mu^{2}(n)}{\phi(n)}$ and then discuss on $ F(n)\equal{}\sum_{d|n}f(d)$.[/hide]

\[ \sum_{d|n}\frac {\mu^2(d)}{\phi(d)} \equal{} \prod_{p|n}(1 \plus{} \frac {1}{p \minus{} 1}) \equal{} \frac {rad(n)}{\phi(rad(n)} \equal{} \frac {n}{\phi(n)}.
\]

[quote="sinajackson"]For $ n\in\mathbb{N}$ prove that:

$ \sum_{d|n}\frac {\mu^{2}(d)}{\phi(d)} \equal{} \frac {n}{\phi(n)}$

[hide="Hint"]define a function $ f: \mathbb{N}\rightarrow\mathbb{R}$ as $ f(n) \equal{} \frac {\mu^{2}(n)}{\phi(n)}$ and then discuss on $ F(n) \equal{} \sum_{d|n}f(d)$.[/hide][/quote]
$ F(n) \equal{} \frac {\mu(n)^2}{\varphi (n)}$ is a multiple function so
$ G(n) \equal{} \sum_{d|n} F(d)$ is all so multiple function .
So enough to prove $ G(p^a) \equal{} \frac {p^a}{\varphi (p^a)} \equal{} \frac {p}{p \minus{} 1}$
Easy to check that it is true.


ratio
geometric series
arithmetic sequence
geometric sequence

Hi all,

I am looking for the equations for solving arithmetic & geometric series and the equations for solving arithmetic & geometric sequences.

Cheers.

nth term of an arithmetic sequence: $a_{n}=a_{1}+(n-1)d$
sum of n terms of an arithmetic sequence: $S_{n}=n(\frac{a_{1}+a_{n}}{2})$

($a_{1}$=1st term, $d$=common difference)

nth term of a geo sequence: $g_{n}=g_{1}(r^{n-1})$
sum of n terms of a geo sequence: $S_{n}=g_{1}(\frac{1-r^{n}}{1-r})$
sum of an infinite geo sequence: $S_{\infty}=\frac{g_{1}}{1-r}$

($g_{1}$=1st term, $r$=common ratio)

Here is all you really need to know (no, they aren't formulas) about them.  The rest can easily be derived.

An arithmetic sequence is a sequence where the difference between consecutive terms is constant, such as -2,1,4,7,10...

To find the sum, simply take the average of the terms and multiply by the number of terms.

A geometric sequence is a sequence where the ratio between any two consecutive terms is constant.

To find the sum, divide the sum by $r$ and see what you get  :D


abstract algebra
Ring Theory
superior algebra
superior algebra unsolved

Prove that if $ M$ is a torsion free module on a PID, then $ M$ is flat.

tensor product commutes with colimits, and $ M$ is a colimit of finite free, thus flat modules.


Find angle BAC if A = (2, 2, 2), B = (4, 2, 1), and C = (2, 3, 1).

how do I solve this?

[quote="kunny"]You can use inner product or scalar product.[/quote]

how do you use scalar product? I assume this is a vector question related

AB = 2i - k
AC = j - k

so is it:

$ AB . AC \equal{} sqrt(5) * sqrt(2) * cos(x)$
-2 = sqrt(5) * sqrt(2) * cos(x)

and I got the angle to be 129.23 degrees

$ cos \angle{BAC}\equal{}\frac{1}{\sqrt{10}}\equal{}\frac{1}{3.1622\cdots}$.

Maybe your answer is right. :)

I guess your answer is right, where did you get the 1 in the denominator from?

$ \overrightarrow{AB}\equal{}(2,\ 0,\ \minus{}1),\ \overrightarrow{AC}\equal{}(0,\ 1,\ \minus{}1)$.

$ \therefore \cos \angle{BAC}\equal{}\frac{\overrightarrow{AB}\cdot \overrightarrow{AC}}{|\overrightarrow{AB}\parallel{}\overrightarrow{AC}|}$

$ \equal{}\frac{2*0\plus{}0*1\plus{}(\minus{}1)*(\minus{}1)}{\sqrt{2^2\plus{}0^2\plus{}(\minus{}1)^2}\sqrt{0^2\plus{}1^2\plus{}(\minus{}1)^2}}\equal{}\frac{1}{\sqrt{10}}$


geometry
circumcircle
inradius
trigonometry
area of a triangle
Heron\u0027s formula

Is there a formula for circumradius and inradius if you're given sides of a triangle?

Well for inradius the best I can think of is $A=rs$ where A is the area, r is the inradius and s is the semiperimeter. So applying Heron's formula, the inradius is $\frac{\sqrt{s(s-a)(s-b)(s-c)}}{s}$ where a,b,c are sides and s is semiperimeter

[quote="sugarwergoingdown"]wats inradius... :blush:[/quote]

It's the radius of the circle that is inscribed in a triangle

Any ideas for circumradius?  :)  :juggle:

well i know it's more than half the longest side ._.

[quote="sugarwergoingdown"]o...now i get it, thanks![/quote]
:lol: yep

o.0 Wikipedia

http://en.wikipedia.org/wiki/Circumradius#Circumcircles_of_triangles

The diameter is the length of any side $\div$ the sine of the angle opposite of that side, so the radius is just that $\div$ 2. Is that correct..?  :|  :|

Looks fine to me. I didn't know that before, but it's entirely possible.

I seem to remember there being another way, though...

There's an easier formula for circumradius: $R={abc\over 4A}$, where $A$ is the area of the triangle (can be calculated via Heron's).

And the statement ${a\over\sin\alpha}={b\over\sin\beta}={c\over\sin\gamma}=2R$ is better known as [b]Sine Law[/b] for triangle.

Ah, right, that's what the other way was. Possible to do without a calculator if you don't have nice angles. ;)


Once again, thanks again for all for your help in future advance.

I have one troublesome problem - The question reads [b]"Derive an expression for the potential energy V between two parallel dipoles seperated by distance r."[/b] There is a picture beneath the problem which simply shows the two parallel dipoles, one with charge +q1 and -q1 on the sides, and the other with +q2 and -q2 on the sides. The length of the dipole is l, and the distance between the two dipoles, as described in the problem, is r. (the picture sort of looks like parallel plate capacitors, if that helps - eventhough capacitance has nothing to do with the problem.)

I attempted solving the problem, and I believe that the pythagorean theorem is necessary to be used.

If you all could help me out I once again would really appreciate it. Thanks!


geometry
incenter
advanced fields
advanced fields unsolved

Hi,

I have a computer program that ends up defining three lines that intersect to form a triangle. I am now trying to find the point within this arbitary tringle which minimises the perpendicular distance from each of the three lines which make up the triangle. Is there a sensible way to do this without just using a Monte Carlo method of trial and error for random points?

Thanks for your help

James Atkinson

James,

What exactly do you want to minimize?

say for example that the three distances are $a,b,c$

do you want to minimize the sum $a+b+c$ or you want to make the three distances equal (incenter) ?


geometry
rhombus
inequalities
geometry proposed

Let $ ABCD$ be a rhombus with $ \angle BAD\equal{}60^{\circ}$. Points $ S,R$ are chosen in triangles $ ABD,DBC$ respectively such that $ \angle SBR\equal{}\angle RDS\equal{}60^{\circ}$. Prove that $ SR^2\geq AS\cdot CR$.

http://www.mathlinks.ro/viewtopic.php?t=49088

  darij


http://www.artofproblemsolving.com/Forum/resources.php?c=182&cid=44&year=2008&p=1038484

In the solution, Altheman assumes that A'B'O is equilateral. How is this assumption justified? (I'm assuming there's some theorem here that I haven't encountered before)

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1038484]Original thread[/url]

In the future, click on the problem's number to go to the original thread. Feel free to revive old mathematical threads.


At Penn State, they were both site winners of Penn State but in the website's results: http://www.arml.com/score2008/divisionb_view.php

It says they are both tied for 6th nationally. So which is it? Will CTA and TJB have to compete in A division next year? And also, will Nspires and Mathematica be going to the 30 team members of CTA and TJB?

Yeah, I was curious too...

I feel so... undefined.

It may have to do with a question of whether or not Hong Kong counts in the official division B rankings.  Up through last year, international teams competed in their own division.  I believe that's why Hong Kong and Vermont are both listed as rank N1 in Division B.

I hope that we were a national winner. I could do with an Nspire...
Would we have to give back those ripped books?

I think the decision was to list the international teams in divisions A or B as appropriate (and Paradise Mall Hong Kong will surely be listed as Division A next year, assuming they return) but that they are not eligible for prizes. Hence, Vermont and Hong Kong are both listed as "N1" in Division B and Taiwan B and San Diego Surf are  both listed as "SL" or Las Vegas site winner in Division A, but only Vermont and San Diego get the official prizes for that.

Found it.

NYC Y was supposed to be in division A but were placed in division B (did well in 2006). They got 6th in division B but really didn't belong there, so they were moved to division A. Thus, TJ B and CT A are moved up and get national awards. Yay!

From the website: http://arml.com/arml_registration/index.php

The tiebreaker format for teams receiving prizes would be as follows:

    * Tiebreaker #1 is the sum of the Team Round and Power Question.
    * Tiebreaker #2 is the sum of the individual scores.

If this system was used, CTA is ahead of TJB leaving TJB as Site Winner.


2 points A and B move in the same plane with velocities $ u$ and $ v$ respectively, constant in magnitude.($ v>u$). A follows a rectilinear path independent of B. (It moves linearly with constant velocity). B moves such that the velocity vector $ \vec v$ is always "aimed" at A. Initially, the particles are such that $ \vec u \perp \vec v$ and the particles are separated by a distance $ d$. After what time will the points converge?

[hide="hint"]
if l be the distance between two points then we have $ \frac{dl}{dt}\equal{}$their relative speed along the line joining them
[/hide]

Never

If " A looks to B it will see that B's velocity points always behind him".

this is an easy one

let the distance between the two particles be d

let the time taken to converge be t

then first
let l be the distance between them and a be the angle made by B's velocity with y axis

dl/dt=usina-v

dl=(usinadt)(t=0to t)-(vdt)(t=0 to t)

but we know that the toal distance travelled along the x axis = vsinadt(t=0 to t)=ut
hence sinadt=ut/v
or usinadt=u^2t/v

replacing the values in the original equation

we get time=vd/(v^2-u^2)
[/img][/code]

It should be better if you use Latex, but seems you are mixing (summing)  finite with infinitesimal expressions.


Two players alternately replace the stars in the expression
	\[*x^{2000}+*x^{1999}+...+*x+1 \]
by real numbers. The player who makes the last move loses if the resulting polynomial has a real root $t$ with $|t| < 1$, and wins otherwise. Give a winning strategy for one of the players.

This problem is very interesting because it combines combinatorics with polynomials and algebra in an elegant way!

Actually Bolzano's Theorem seems to be usefull because then equivellantly we have to find a way such that $f(1)\cdot f(-1)$ is either negative and so the first player wins or nonnegative and so the second player wins.

[i]This is the first though!I' ll think about it! [/i]

I know this problem for 10 not for 2000.But solution is similar.In both variants the player who makes last move loses.The other player by each move before his 1000's move put 0 (or something else) near some $x^{m}$.Where m is even if it posible
Let P(x) is our polynomial after 1998 moves where * is near $x^{m}$ and $x^{2n+1}$ first player put $a$ near $x^{m}$ and second near $x^{2n+1}$ .Let other part of polinomial is $G(x)$.
So $P(x)=ax^{m}+bx^{2n+1}+G(x)$.Consider $P(1)$ and $P(\frac{-1}{2})$.
We have $a+b+G(1)$ and $a*(\frac{-1}{2})^{m}-b(\frac{1}{2})^{2n+1}+G(\frac{-1}{2})$.Consider $P(1)+2^{2n+1}P({\frac{-1}{2}})=a(1+(-1)^{m}*(\frac{-1}{2})^{2n+1-m})+2^{2n+1}G(\frac{-1}{2})+G(1)$
By puting 'good' $a$ we get that  $P(1)+2^{2n+1}P({\frac{-1}{2}})=0$ for all $b$.Using Bolzano's Theorem finish solution.

This is quite similar to USSR math olympiad 1977

A polynomial f (x) = x^4 +∗x^3 +∗x^2 +∗x +1 has three undetermined coefficients
denoted by stars. The players A and B move alternately, replacing a star by a real
number until all stars are replaced. A wins if all zeros of the polynomial are complex.
B wins if at least one zero is real. Show that B can win in spite of his only second
move.

Given a polynomial with (a) natural (b) integral coefficients, let an be the digital sum
in the decimal representation of f (n). Show that there is a number, which occurs in
a1, a2, a3, . . . infinitely often.

If $n=2k$ then $A$ will have the wimming strategy, otherwise $B$ has the winning strategy, where $A$ and $B$ are first and second player respectively

Indeed, denote the asterisk before $x^k$ as the $k$ index asterisk. Then, if $n=2k$, the number of odd index asterisk equal to the number of even index asterisk.Hence, $A$ can always perform the replacement such that after each replacement, the number of even index asterisk does not exceed the number of odd index asterisk. Thus, $A$ can replace such that after the $2k-2$ replacement, there are only two asterisks left, let it be $t$ index asterisk and $p$ index asterisk and $p$ is odd. We will prove that there exist $a_t$ such that $P(x)$ has real root $\alpha$ such that $\vert\alpha\vert<1$ for all $a_i$ such that $i\ne t$. Meaning after the $2k-1$ replacement, $A$ will replace the $t$ asterisk and win the game. Indeed, set $a,b>0$ such that $a\ne b$, $[-a,b]\subset (-1,1)$. Let $\gamma=\left(\frac{b}{a}\right)^p>0$. Set $a_t=\frac{-\gamma Q(-a)-Q(b)}{\gamma (-a)^t+b^t}$ (since $a\ne b, t\ne t$ so $\gamma (-a)^t+b^p\ne 0$) where $Q(x)=\sum_{i\ne t,p}a_ix^i$. Then we have \[\begin{aligned} \gamma P(-a)+P(b)&=\gamma[Q(-a)+a_t(-a)^t+a_p(-a)^p]+[Q(b)+a_tb^t+a_pb^p]\\  &=\gamma Q(-a)+Q(b)+a_t[\gamma (-a)^t+b^t]\\ &=0 \end{aligned}\] Hence $P(-a)P(-b)\leq 0$ (since $\gamma >0$). Hence $P(x)$ always has real root in $(-1,1)$. So $A$ is winner. If $n=2k+1$, then player $B$ can always replace such that the number of even index does not exceed the number of odd index and similarly, we have $B$ is the winner

@boldoo60: Which book is that? Is that Arthur Engel's PSS


function calculus derivative algebra polynomial trigonometry

Not quite sure is this the right place for the question. Probably not. Let me just edit your statement:

The function is $ f_n(x) \equal{} x^n \plus{} x^{n \minus{} 1} \plus{} ... \plus{} x \minus{} 1$.

Now, the answer: Derive $ f$ with respect to $ x$. It is obvious that $ f'(x) > 0$ for each positive $ x$ (so $ f$ is rising for $ x > 0$). Since $ f(0) \equal{} \minus{} 1$, $ f(1) \equal{} nx$, $ f$ crosses the x-axis somewhere between 0 and 1, and due to continuousness of $ f$ and $ f'$, that is the only solution, as $ f'$ cannot change its sign, hence the function cannot drop (and cross $ x$ axis for another time).

Edit: procedure above is valid for $ n\geq 2$. Specially, the case $ n\equal{}1$ is trivial, and for $ n\equal{}0$, the statement isn't true.

[quote="*210*"]Hello
$ n \in \mathbb{N}^{\large\ast}$ and $ f_n$  the function defined on $ \mathbb{R}_ \plus{}$ by $ f_n(x) \equal{} x^n \plus{} x^{n \minus{} 1} \plus{} ... \plus{} x \minus{} 1$
to show  the equation $ f_n( x) \equal{} 0$ admits one only one solution in $ \mathbb{R}_ \plus{}$
thank you !![/quote]
hi
$ f' >0$  and $ f(0)*f(1) <0$ ==> $ \ni! u  \in \mathbb{R}_ \plus{}$  that  $ f(u)\equal{}0$

or we can simplify$ f$

$ f(x)\plus{}2 \equal{}  \frac{x^{n\plus{}1} \minus{}1} {x\minus{}1}$ and  we solve the equation

thank you  hsiljak and Ibn Rochde
can you  continue your method,?? Ibn Rochde
it is very interesting  , $ D_f \neq D_{f\plus{}2}$

[quote="*210*"]thank you  hsiljak and Ibn Rochde
can you  continue your method,?? Ibn Rochde
it is very interesting  , $ D_f \neq D_{f \plus{} 2}$[/quote]

sorry i am late i had some work to do ........ :blush:

$ f(x)\equal{}0$ <=> $ x^{n\plus{}1}\minus{}2x\plus{}1\equal{}0$ by using Descartes' Rule of Signs like  azn6021023 had said

$ p(x)\equal{}x^{n\plus{}1}\minus{}2x\plus{}1\equal{}0$ <==> $ \ni ! (u.v)$   $ p(u)\equal{}p(v)\equal{}0$

and we have $ p(1)\equal{}0$ and $ f(1)$ different to $ 0$  so $ \ni ! u$ $ p(u)\equal{}0$

[quote="azn6021023"]Or we could cheat and not even realize the nice factoring of the original function because of Descartes. :D[/quote]

$ f(x)\equal{}0$ ==> $ x^{n\plus{}1}\minus{}2x\plus{}1 \equal{}0$ supposing $ g(x)\equal{}x^{n\plus{}1}\minus{}2x\plus{}1$

so $ g'(x)  \equal{} (n\plus{}1)x^n \minus{} 2$ so  $ g'(x)\equal{}0$ ==> $ x\equal{}(\frac{2} {n\plus{}1})^{n^{\minus{}1} }$ =$ a$

$ 0<x<a$  ==>$ g'<0$

$ a<x$ ==> $ g'>0$  and $ g(0)\equal{}1$ , $ g(1)\equal{}0$ , $ a<1$   lim g(x) = +00

so there is one $ u$ that  $ g(u) \equal{}0$

without using any descatres rules  :roll:

[quote="JBL"][quote="hsiljak"]Derive $ f$ with respect to $ x$.[/quote]  You needed to take a derivative to argue that a polynomial with only positive coefficients on its nonconstant terms was increasing?  :roll:  :wink:[/quote]

Did it automatically, not thinking at all... I guess that my exams made me do even the most elementary things using calculus :rotfl:

[quote="azn6021023"]We can also apply Descartes' Rule of Signs to solve for positive reals.[/quote]

by using descatrtes rules you prouve the exicting of double solution of $ f(x)\equal{}0$ moreover you can't prouve that one of the

solution is in $ R\minus{}$ and other in $ R\plus{}$ :noo:

Use roots of unity (yes, this is definitely overkill)!

Noting that 1 is not a root if $ n>1$, then $ f(x)\equal{}\frac{x^{n\plus{}1}\minus{}1}{x\minus{}1}\equal{}0\Rightarrow x^{n\plus{}1}\equal{}1$

The solutions to this are of the form $ e^{\frac{\imath k\pi}{n\plus{}1}}$, where $ k\in\{2,4,...,2n\}$.  Let $ \theta\equal{}\frac{k\pi}{n\plus{}1}$.  Also, $ e^{\imath\theta}\equal{}\cos(\theta)\plus{}\imath\sin(\theta)$, so for this to be real, $ \sin(\theta)\equal{}0$, which implies that $ \theta\equal{}\pi$ given the restrictions $ 0<\theta<2\pi$.  However, even if this actually can be a solution (it only is half the time), then we have $ e^{\imath\pi}\equal{}\cos(\pi)\plus{}\imath\sin(\pi)\equal{}\minus{}1$, which is not positive.


Let $n\in \mathcal N$.We write a line containing  $n$ numbers. Below each number $a$ we write $f(a)$ which stands for the number of apparitions of $a$ on the previous line. Prove that eventually we get two consecutive lines that conicide.

I think this problem comes from All Russian Math Olympiad.


We know that $0<a,b,c,d<1$ sice all terms are positive and the sum of the squares is 1. Now consider $f(x)=\sqrt{1-x}$. It is convex for $x\in [0,1]$. So by Jensen we have that:
$\sum \sqrt{1-ab}\ge 4\sqrt{\frac{4 -ab-bc-cd-da}{4}}\ge 4\sqrt{\frac{4-a^2-b^2-c^2-d^2}{4}}= 2\sqrt 3$

[quote="pvthuan"]Let $a,b,c,d$ be four positive real numbers satisfying the restriction $a^2+b^2+c^2+d^2=1$, prove that
\[ \sqrt{1-ab}+\sqrt{1-bc}+\sqrt{1-cd}+\sqrt{1-da}\geq2\sqrt3.  \][/quote]

Let $\sqrt{a}=x,\sqrt{b}=y,\dots$ and so on. So we have $x+y+z+t=1$, and let $F(x,y,z,t)=-(\sqrt{1-\sqrt{xy}}+\sqrt{1-\sqrt{yz}}+\sqrt{1-\sqrt{zt}}+\sqrt{1-\sqrt{tx}}$, by the AC method we have $F(x)\leq max{F(1,0,0,0),F(\frac12,\frac12,0,0),F(\frac13,\frac13,\frac13,0),F(\frac14,\frac14,\frac14,\frac14)}=-2\sqrt3$, which gives us the deisired inequality. We just have to prove that $F(x,y,z,t)\leq max{F(\frac{x+y}2,\frac{x+y}2,z,t),F(0,x+y,z,t)}$, and we're done.


logarithms
integration
calculus
Euler
function
real analysis
real analysis unsolved

Examine the behavior of the expression
$ \sum_{\nu\equal{}1}^{n\minus{}1}\frac{\log(n\minus{}\nu)}{\nu}\minus{}\log^2 n$
as $ n\rightarrow \infty$

The passage from the sum to the integral comes from the Euler Maclaurin summation formula or by another way ?

This follows from the definition of Rieman integral:
Riemann integral is the limit of the Riemann sums of a function as the partitions [k/n,(k+1)/n] get finer.


Geia sas eiani to prwto mou post an kai parakolou8w to forum arketo kairo. Exw brei mia endiaferousa exiswsh:
$ 8x^3 \minus{} 1 \equal{}6x$

[hide="Let me try..."]
$ 8x^3 \minus{} 1 \equal{} 6x \Leftrightarrow$

$ 8x^3 \minus{}6x \equal{} 1 \Leftrightarrow$

$ 4x^3 \minus{}3x \equal{} \frac{1}{2}\ (1)$

Η εξίσωση έχει το πολύ 3 πραγματικές ρίζες. Αρχικά θα την εξετάσουμε στο $ [\minus{}1,1]$. Αν βρούμε 3 ρίζες στο διάστημα αυτό δε θα χρειαστεί να ψάξουμε για ρίζες αλλού.

Έστω $ x\in[\minus{}1,1]$. Θέτω $ x\equal{}\cos\varphi$ με $ \varphi \in [0,\pi]$

Η $ (1)$ γράφεται $ 4\cos^3\varphi \minus{}3\cos\varphi\equal{}\frac{1}{2} \Leftrightarrow \cos 3\varphi \equal{} \frac{1}{2} \Leftrightarrow \cos 3\varphi \equal{} \cos \frac{\pi}{3} \Leftrightarrow$

$ 3\varphi \equal{} 2k\pi \pm \frac{\pi}{3}, \ k\in \mathbb{Z}\Leftrightarrow$

$ \varphi \equal{} \frac{6k\pi \pm \pi}{9}$ και επειδή πρέπει $ \varphi \in [0,\pi]$ παίρνουμε $ \varphi \equal{} \frac{\pi}{9}$ ή $ \frac{5\pi}{9}$ ή $ \frac{7\pi}{9}$

έτσι οι λύσεις της αρχικής είναι $ \cos\frac{\pi}{9}$ , $ \cos\frac{5\pi}{9}$ και $ \cos\frac{7\pi}{9}$

Έχουμε βρει 3 διακεκριμένες ρίζες επομένως δε χρειάζεται να ψάξουμε εκτός του $ [\minus{}1,1]$

Λουκάς[/hide]


Mary is 21 years older than her son.
In 6 years, she'll be 5 times older then him.
Where is the father?

Though not necessarily in [i]a[/i] bed  :lol:

So the only thing we actually know for definite about the postion of the father is that hes continous in $M$ Where $M=\{\text{space that mary occupies}\}$

Therefore the correct question is "Where is the father's?"  ;)

Or the correct answer is "The father is inside" ;)

Aw man this topic is so locked!  :rotfl:

[quote]Mary is 21 years older than her [b]son[/b]. [/quote]
Incorrect problem - must be improved.
How Mary knew that there would be son?  :roll:

Or the problem is from "99 Vatican-approved Mathematical Problems for Budding Christians", and we're supposed to know what Annunciation is?  :P

If I've understood correctly the set of fathers is empty now and so ... what we are looking for?

It seems to me I've caught. This is a great amount problem.  :D
There exist very many Mary's in the World who has at the moment 21 years of age and are waiting a kid to the waiting time. In this great amount there would be at least one desired Mary.

Unless of course the person in question is Jesus, in which case his father could have been anywhere except inside.

I'm quite a dork when it comes to Bible - where Joseph was during Annunciation?

They just [i]to be married.[/i] Joseph felt he shouldn't marry Mary until an angel came to him in a dream.
Is that what you are asking?
And why is this considered math? Is there some inside joke I don't get?

Oh okay...hahahahahahaha! :blush:
Well if that's all see you guys on math problems...


group theory
abstract algebra
superior algebra
superior algebra unsolved

Find all types of isomorphism of groups of order $13\cdot 27$ which have a subgroup isomorphic to $(Z/3Z)^{3}$.

There is obviously only one abelian group satisfying the conditions. As for the rest, you've got to dig into semi-products, since (exactly) one Sylow is normal in each of the non-commutative cases.


Triangle $ABC$ is inscribed in circle $w$. Line $l_{1}$ bisects $\angle BAC$ and meets segments $BC$ and $w$ in $D$ and $M$,respectively. Let $y$ denote the circle centered at $M$ with radius $BM$. Line $l_{2}$ passes through $D$ and meets circle $y$ at $X$ and $Y$. Prove that line $l_{1}$ also bisects $\angle XAY$

$1) DX.DY=DB.DC$
$2) DA.DM=DB.DC$
$1),2)-\to  DA.DM=DX.DY , \longrightarrow$ , $AMXY$ is cyclic $\longrightarrow$ $\angle XAM$=$\angle XYM$ , $\angle YAM$=$\angle YXM$ , $MX=MY$ $\longrightarrow$ $\angle XAM$=$\angle XYM$=$\angle YAM$=$\angle YXM$ $\longrightarrow$ $l_{1}$ bisects $\angle XAY$

$X$, $Y$, $M$, $A$ cyclic (POP) and $XM = $MY$ (Center of circle).


function
inequalities
inequalities proposed

Let $ f$ be a convex function and $ a_{1},a_{2}\cdots,a_{n}$ be arbitrary real numbers. Prove that:

$ f(a_{1})+f(a_{2})+\cdots+f(a_{n})\leq f(a_{1}+a_{2}+\cdots+a_{n})+(n-1)f(0)$

Can someone please provide the proof of this, this is helping me prove a lot of other ineqalities, after it was mentioned here:

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=902359#902359

I just cant find the proof.


How can I see if a number is the sum of a series of consecutive integers?

[hide]It will be a sum of consecutive integers if it has the form $\frac{m(m+1)}{2}-\frac{n(n+1)}{2}$, where $0\leq n<m-1$. Here the smallest integer is $n+1$ and the largest is $m$, which makes the number is a sum of $m-n$ consecutive integers.
Note that if $x=\frac{m(m+1)}{2}-\frac{n(n+1)}{2}$, then $2x=\left(m+\frac{1}{2}\right)^2-\left(n+\frac{1}{2}\right)^2=(m+n+1)(m-n)$, so factoring $x$ can provide a clue to how it is a sum of consecutive integers. (if $x$ is a power of 2, it can't be written as a sum of more than 1 consecutive integer.)[/hide]

When writing the number $x$ as a sum of consecutive positive integers, $m$ will be the largest and $n+1$ will be the smallest. You can find possible values for $m$ and $n$ by factoring as I showed above. (i.e. 34=2*17, so $m+n+1=17$, $m-n=4$ giving $m=10$, $n=6$, which checks because 34=10+9+8+7.)

Why is the largest minus smallest the number of consecutive integers? Wouldn't that just give you the sum of the consecutive terms minus the first term?

Why do you multiply by $\frac{m}{2}$ and $\frac{n}{2}$? Is it suppose to be a series sum of the $a_1$ term and of the difference?

[quote="breez"]Nvm, I understand this now. Good method. However, scorpius, why can't a power of 2 ever be a sum of a consecutive series?[/quote]

[hide]If $x$ is a power of 2, then so is $2x$. The factors of $2x$ differ by $2n+1$, so one factor is odd, the other even. The only way odd*even=power of 2 is if the odd factor is 1. 1 will be less than $2x$, so it have to be the smaller of the two factors, $m-n$, so that $x$ can only be expressed as a "sum" of 1 consecutive integer.[/hide]

You should be more careful with your terms.  We can represent [i]any[/i] integer as a sum of consecutive integers.  For intance, if we want a positive number $x$, we take the sum $-(x-1)+\cdots x$.  I believe that you all ment "positive integers" or "non-negative integers" rather than "integers."  (Sorry; I couldn't resist.)


calculus
integration
function
real analysis
real analysis unsolved

Let $f_n: [0..1]\to R$ be a sequence of functions with finite number of discontinuites.
1°.Show that we can have  $\forall x ,lim_{n\to \infty} f_n(x)=\infty$ and $lim_{n\to \infty}\int_0^1f_n(t)dt=-\infty$.
2°.If there exists $M>0$ such that $\forall x ,lim_{n\to \infty} f_n(x)=\infty$ and $\forall x,\forall n,f_n(x)\geq M$ then $lim_{n\to \infty}\int_0^1f_n(t)dt=\infty$.

[quote="Killer"]Let $f_n: [0..1]\to R$ be a sequence of functions with finite number of discontinuites.
1°.Show that we can have  $\forall x ,lim_{n\to \infty} f_n(x)=\infty$ and $lim_{n\to \infty}\int_0^1f_n(t)dt=-\infty$.
2°.If there exists $M>0$ such that $\forall x ,lim_{n\to \infty} f_n(x)=\infty$ and $\forall x,\forall n,|f_n(x)|\geq M$ then $lim_{n\to \infty}\int_0^1f_n(t)dt=\infty$.[/quote]
Set $f_n(x) = n$ for $0 \leq x \leq 1-1/n$ and $f_n(x) = -n^3$ for $1-1/n <x < 1, \, f_n(1) = n$. This is an example for 1/ and appears to be a counterexample for 2/ (with $M = 1$).


Παρακινούμενος από το [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=117768](Επιστροφή στο Δημοτικό !!!)[/url] θα ήθελα να δημοσιεύσω ένα δικό μου πρόβλημα, σχετικό με ρολόγια.

Έχουμε ένα κλασσικό ρολόι, με δείκτες. Από λάθος του εργοστασίου ο ωροδείκτης και ο λεπτοδείκτης είναι πανομοιότυποι (ίδιο μήκος  κλπ).

[color=brown]Ο προβληματισμός μου προέκυψε από την εξής σκέψη:
Θα μπορούμε να ξέρουμε [u]σε οποιαδήποτε[/u] στιγμή της ημέρας με σιγουριά τι ώρα είναι;
Το λέω γιατί υπάρχουν ώρες, όπως 1:20 και 4:05 περίπου, οι οποίες είναι "συμμετρικές", δηλαδή οι δύο δείκτες έχουν ανταλλάξει τη θέση τους.
Είναι όμως "ακριβώς" συμμετρικές;

Πιο συγκεκριμένα:[/color]
$\hline$

Αν $\phi(x)$ και $\theta(x)$ είναι η θέση του ωροδείκτη και του λεπτοδείκτη αντίστοιχα κατά τη χρονική στιγμή $x$
και $A = \{(\phi(x), \theta(x))\ |\ x \in[0,12)\}$ , τότε:

Να βρεθούν όλες οι τιμές του $x\in[0,12)$ έτσι ώστε $(\theta(x),\phi(x)) \in A$.

Γι αυτές τις τιμές να βρεθεί το $y\in[0,12)$ έτσι ώστε $\phi(y)=\theta(x), \theta(y)= \phi(x)$. Δηλαδή να βρεθεί ποια ώρα είναι "συμμετρική" με τη $x$.

Ελπίζω να σας αρέσει  :)
Εμένα μου άρεσε σαν προβληματισμός, αλλά όσο προχωρούσα στη διαδικασία της λύσης άρχισε να με συναρπάζει!

αδικημενη ασκηση,με εχει μπερδεψει,την ελυσε κανεις ?

Να κάνω μία διευκρίνιση...

Προφανώς εννοώ ότι οι δείκτες κινούνται σε συνεχή χρόνο. Δηλαδή με σταθερή γωνιακή ταχύτητα ([i]δεν πρόκειται για εκείνα τα ρολόγια στα οποία ο λεπτοδείκτης κάνει ένα "τίναγμα" κάθε λεπτό[/i])

Έτσι, [u]και μόνο[/u] η θέση του ωροδείκτη είναι αρκετή για να ξέρουμε με ακρίβεια τι ώρα είναι. Επομένως ξέρουμε και τη θέση του λεπτοδείκτη.

Έστω $A,B$ οι δύο δείκτες.
Αν υποθέσουμε ότι ο $A$ είναι ο ωροδείκτης, βρίσκουμε τη θέση $f(A)$ που θα επρεπε να έχει ο λεπτοδείκτης.
Αντίστοιχα, αν $B$ ο ωροδείκτης, βρίσκουμε το $f(B)$

Αν [u]μόνο[/u] $f(A)=B$, τότε ωροδείκτης είναι ο $A$.
Αν [u]μόνο[/u] $f(B)=A$, τότε ωροδείκτης είναι ο $B$.

Μία από τις παραπάνω ισχύει πάντα.
Το πρόβλημα παρουσιάζεται όταν ισχύουν και οι δύο σχέσεις ταυτόχρονα $f(A)=B$ και $f(B)=A$
Με άλλα λόγια όταν $f(f(A))=A\ \ (1)$.

Ζητούνται οι λύσεις της $(1)$ (και είναι αρκετές)
Εξαιρούνται βέβαια αυτές που δίνουν $f(A)=A$, αφού τότε οι δύο δείκτες ταυτίζονται.

[color=brown]Υ.Γ.
Το πρόβλημα αυτό το είχα λύσει παλιότερα με ugly τρόπο, αλλά τώρα έχω βρει ένα τύπο που μου δίνει όλες τις λύσεις[/color]

Λουκάς

Λουκά θεωρώντας ως δεδομένο ότι $\phi(0)=\theta(0)$ (...γιατί ανέφερες ότι έχουμε ένα κλασσικό ρολόι) βρήκα ότι η ώρα $x=\frac{12(k+12r)}{143}$ είναι "συμμετρική" με την ώρα $y=\frac{12(r+12k)}{143}$ για κάθε $k,r \in \{0,1,\cdots,11\}$ με $k \neq r$ και επίσης ότι αυτές είναι και οι μόνες "συμμετρικές" ώρες.

Αν πράγματι είναι αυτή η απάντηση πες μου για να γράψω και το σκεπτικό μου, γιατί μπορεί και να το έχω προσεγγίσει εντελώς λάθος το θέμα. :whistling:

Χαίρομαι που τελικά ασχολήθηκε κάποιος.
Μπράβο, Στέλιο! Αυτή είναι η λύση :roll:
Γράψε το σκεπτικό σου..  :wink:

Λοιπόν, μετά από $x$ ώρες η γωνία που θα έχει διαγράψει ο ωροδείκτης θα είναι $v(x)=\frac{\pi}{6}\cdot x$ , ενώ η γωνία που θα έχει διαγράψει ο λεπτοδείκτης θα είναι $2\pi x$, επειδή όμως μας ενδιαφέρει η θέση του λεπτοδείκτη και όχι πόσες στροφές έχει κάνει, αρκεί να πάρουμε $l(x)=2\pi\{x\}$ (...όπου $\{x\}$ το δεκαδικό μέρος του $x$), που είναι η γωνία που σχηματίζει ο λεπτοδείκτης με τον κατακόρυφο άξονα, καθώς περιστρέφεται κατά τη φορά του ρολογιού ( :P ).

Εμείς θέλουμε να βρούμε τις λύσεις $x,y \in [0,12)$ (με $x \neq y$) του συστήματος $v(x)=l(y)$ και $v(y)=l(x)$, δηλαδή του συστήματος

$\frac{\pi}{6}\cdot x=2\pi\{y\}$
$\frac{\pi}{6}\cdot y=2\pi\{x\}$

ή αλλιώς

$\{y\}=\frac{x}{12}$ $(1)$
$\{x\}=\frac{y}{12}$ $(2)$

Αντικαθιστώντας το $y$ από την δεύτερη  στην πρώτη πέρνουμε

$\{12\{x\}\}=\frac{x}{12}$ $(3)$

Επειδή τώρα $x \in [0,12)$ ο $x$ θα ανήκει σε ένα διάστημα της μορφής $\left[\frac{r}{12},\frac{r+1}{12}\right)$ ,όπου $r=0,1,\cdots,11$.

Οπότε θα ισχύει $\{12\{x\}\}=12\{x\}-r$ και αντικαθιστώντας στην $(3)$ έχουμε $12\{x\}-r=\frac{x}{12}$ ή αλλιώς

$\{x\}=\frac{[x]+12r}{143}$

(...όπου $[x]$ είναι το ακέραιο μέρος του $x$)

Το $[x]$ όμως μπορεί να πάρει τις τιμές $0,1,\cdots,11$ οπότε θέτωντας $[x]=k$ με $k=0,1,\cdots,11$ προκύπτει

$x=[x]+\{x\}=k+\frac{k+12r}{143}=\frac{12(12k+r)}{143}$

και αντικαθιστώντας στην $(2)$ βγαίνει

$y=\frac{12(12r+k)}{143}$.

Τέλος, επειδή θέλουμε $x \neq y$ θα ισχύει και $k \neq r$.

Στέλιος

Άψογα!  :)

Η δική μου λύση δε διαφέρει και πολύ:

Χωρίζουμε τον κύκλο σε ώρες αντί για μοίρες.
Έστω $x\in[0,12)$ ο χρόνος σε ώρες.
Γράφουμε $x=[x]+k(x)$, όπου $[x]$ είναι το ακέραιο μέρος του $x$ και $k(x)$ το δεκαδικό.

Μπορούμε να βρούμε τα λεπτά πολλαπλασιάζοντας το δεκαδικό μέρος του $x$ με $60$. Δηλαδή $m(x) = 60k(x)$, όπου $m(x)$ είναι η συναρτήση που δείχνει τα λεπτά. Θα είναι $0\leq m(x)<60$

Επομένως, η θέση του λεπτοδείκτη (όταν μετρηθεί σε ώρες) θα είναι $f(x) = \frac{m(x)}{5}= 12k(x) \in [0,12)$

Έστω $x=x_{0}.x_{1}x_{2}x_{3}...$ η ανάλυση του $x$ στο 12-αδικό σύστημα.

Τότε $k(x)=0.x_{1}x_{2}x_{3}...$
$\Rightarrow f(x) = 12k(x)=x_{1}.x_{2}x_{3}x_{4}...$

$\Rightarrow  k(f(x))=0.x_{2}x_{3}x_{4}...$
$\Rightarrow  f(f(x)) = 12k(f(x)) = x_{2}.x_{3}x_{4}...$

Επομένως $f(f(x))=x \Leftrightarrow$

$x_{0}=x_{2}=x_{4}=...=a$ και
$x_{1}=x_{3}=x_{5}=...=b$ και

Άρα το $x$ είναι της μορφής $x=a.bababab...$

δηλαδή $\frac{x}{12}= \frac{12a+b}{12^{2}-1}\Rightarrow x=12\frac{12a+b}{143}$, όπου $a\neq b$

και η συμμετρική θα είναι $f(x)=12\frac{12b+a}{143}$

Α! Το πλήθος των λύσεων
Έχουμε $a,b \in\{0,1,2,...,11\}$
Συνολικές περιπτώσεις $12^{2}=144$
Εξαιρείται η περίπτωση $a=b=11$ γιατί τότε προκύπτει $x=12$

Αφαιρούμε και όλες τις άλλες περιπτώσεις $a=b=0,1,2,...,10$

Μένουν $144-1-11 = 132$, δηλαδή $66$ ζεύγη λύσεων


[b]I recently got this e-mail.  I wonder if anyone is attending it.  I'll probably will.[/b]

Hello,

My name is Keigo Kawaji, and I am an instructor of mathematics problem-solving at the 2003 and 2004 Candian Mathematics Society (CMS)/ESSO National Mathematics Camps.  I was also the assistant to the director of the University of Toronto's Math Club from 2001 to 2003 and helped Dr. Olga Ivrii in running both the Tournament of Towns and the MathBattle programs; in which many of you are already involved.

First of all, I'd like to thank many of you for participating in either or both the 2003-2004 Tournament of Towns and MathBattle.  I am hoping that you had a great time participating in these competitions this year, and that you will continue to do so next year!

The purpose of this e-mail is to let you know about a wonderful opportunity during this summer in July and August.  I would like to welcome all of you to the Toronto Summer Mathematics Enrichment Program, (T-SMEP) which will be held at the University of Toronto on the following dates and times:

Sunday July 18, from 10:30 a.m. to 2:00 p.m.
Sunday August 1, from 10:30 a.m. to 2:00 p.m.
Sunday August 15, from 10:30 a.m. to 2:00 p.m.

The program is intended to help students develop both good problem-solving and solution-writing skills.  These skills will be very benecifial for students who are intending to write any future math contests that require full-solutions.
In particular, significant emphasis will be placed on problem-solving strategies for performing well in the Canadian Open Mathematics Challenge contest in November.  This contest is the qualifier to the Canadian Mathematics Olympiad, and the Canadian Mathematics Society strongly encourages students in Grades 9 or 10 to write the COMC, which is also used to select the top 23 students each year for their ESSO Mathematics Camp program.

Here are some more details about the program:

--------------------------------------------------------------------
Location:  University of Toronto Sidney Smith Building Room 5017A

A Typical Day's Timetable:
10:30-11:00  Introduction and discussion of topic.
                      A short talk is presented to introduce a new topic.

11:00-12:00  Problem-solving session.
                      Students get some time to work on solutions and get some feedbacks.

12:00-12:15 or 12:30  Lunch Break
                      Pizza lunch!  (Note: we may charge a few dollars for lunch)

12:15-12:45  Option 1:  Guest speaker session
                      There will be someone presenting a fun talk.
                   Option 2:  Problem-solving session
                      We will continue with the problem-solving session.

12:45-1:30    Option 1:  Problem-solving session.
                      This takes place right after the guest speaker session.
                   Option 2:  Contest
                      This will be a 75-minute contest that ends at 2:00.

1:30-2:00      Option 1:  Contest
                      This will be a 30-minute contest that ends at 2:00.
                   Option 2:  Contest to wrap up the day.
                      This is a continuation of the contest from 12:45.

All students are welcome to attend any of the three sessions.  Different topics will be presented each day, so there would be no disadvantage in missing any of the sessions.  The first topic on July 18 will be on Game Theory.
The other two sessions will be on Geometry and Number Theory.
There will be guest speakers for some of the sessions--many of whom are youths excelled in problem-solving still in their late teens or early 20's.

Fees:  $5 or less.
  Although we have tried to make this program free, it is impossible for us to run this program without asking every student for a nominal fee of $5 per session to cover your food, photocopy, and miscellaneous expenses.  This is a non-profit program with minimal costs:  none of the staff members are paid, and they must provide for their own costs as well.  But all the money will be spent in a great way!  We will have a wonderful pizza lunch for everyone!
-------------------------------------------------------------------------

Finally, we would like to get a rough idea of how many people will be at each session.
If you are interested, please write to torontomath2004@yahoo.com
and say that you're interested.
(Write your full name and which session(s) you will be able to attend.)
A new mailing list will be created from the replies, and all detailed correspondence regarding the program will be made through this list.

I am very excited about this summer program, and am hoping that you will enjoy a lot of really fun math through this amazing opportunity!

Hope to see you there!

Keigo Kawaji
Program Organizer, T-SMEP[/b]

I got the same email today, but I can't go since I'll in Croatia from July 15th to August 27th. Otherwise I probably would go to one and decide whether I wanted to go back.

I went to Today's session.  The material was very interesting but I found some of the later questions in the contest very hard.

I thought that this session would cover Number Theory and Geometry but I was wrong, it was just geometry.  I should have went to the session on August 1st.  Heh.  Maybe next year.

If anyone else went, post your comments.

[quote]Dr. Olga Ivrii[/quote]

Just curious... Is this Dr. Ivrii in some way related to Oleg Ivrii??

The names seem frighteningly close...

Jenn

Your suspicions are probably correct.

Dr. Ivrii is Oleg's mom.

Oddly enough, I coincidenally happened to mention Oleg to my math teacher just earlier this evening.  Apparently, his mother and father are both heavily involved in mathematics, and (I'm not quite as sure about this one) his brother has earned a reputation in mathematics as well.

That's interesting...
Although I can't really picture anyone being related to Oleg!!

Jenn


Define $ a \star b$ to be the sum of the digits of the product of $ a$ and $ b$. Evaluate $ 9 \star 3$.


Ah hunter is back, perhaps the spam content can increase.

[quote="buzzer11"]haah i don't think it'll ever take off like it did before... good times.[/quote]

Who are you


What is the frequency of the A two octaves below A-440?


$ a,b,c$ are non-negative real numbers such that $ a^{2}\plus{}b^{2}\plus{}c^{2}\equal{}1$,
prove $ a\plus{}b\plus{}c\minus{}abc\le\sqrt{2}$

Note: This problem is like the previous USMO problem:
$ a^{2}\plus{}b^{2}\plus{}c^{2}\plus{}4abc\equal{}1$, prove $ 0\le ab\plus{}bc\plus{}ca\minus{}abc\le 2$, but compared to the one I posted, it seems much more simple. I wait for a nice solution. :)

[quote="hanschinaman"]$ a,b,c$ are non-negative real numbers such that $ a^{2}\plus{}b^{2}\plus{}c^{2}\equal{} 1$,
prove $ a\plus{}b\plus{}c\minus{}abc\le\sqrt{2}$
[/quote]

What if $ a\equal{}b\equal{}c\equal{}\frac{1}{\sqrt{3}}$?

Let  $ f(a,b,c)\equal{}a\plus{}b\plus{}c\minus{}abc$

$ f(a,b,c)\minus{}f(a,t,t)\equal{}\minus{}\frac{(b\minus{}c)^{2}}{b\plus{}c\plus{}\sqrt{2(b^{2}\plus{}c^{2})}}\plus{}\frac{a}{2}(b\minus{}c)^{2}\equal{}\left(\minus{}\frac{1}{b\plus{}c\plus{}\sqrt{2(b^{2}\plus{}c^{2})}}\plus{}\frac{a}{2}\right)*\\(b\minus{}c)^{2},t\equal{}\frac{\sqrt{b^{2}\plus{}c^{2}}}{2}$

We have :

$ ab\plus{}ac\plus{}a\sqrt{2(b^{2}\plus{}c^{2})}\le 2a\sqrt{2(b^{2}\plus{}c^{2})}\le\sqrt{2}<2$ (by AM-GM)

Hence :

$ f(a,b,c)\le f(a,t,t)$

Easy to continued ...  :pilot:

I have seen this for at least 3 times in this forum. :wink:
And i think this is the most beautiful solution :
By Cauchy :
$ a(1\minus{}bc)\plus{}(b\plus{}c).1\leq\sqrt{(1\plus{}2bc)(2\plus{}b^{2}c^{2}\minus{}2bc)}$
Now let $ bc\equal{}x$,we must prove
$ (2x\plus{}1)(x^{2}\minus{}2x\plus{}2)\leq 2$
$ \Leftrightarrow 2x^{3}\minus{}3x^{2}\plus{}2x\leq 0$
$ \Leftrightarrow x(2x^{2}\minus{}3x\plus{}2)\leq 0$,which true,since
$ b^{2}\plus{}c^{2}\leq 1$ :D

[quote="stephen38"]I have seen this for at least 3 times in this forum. :wink:
And i think this is the most beautiful solution :
By Cauchy :
$ a(1\minus{}bc)\plus{}(b\plus{}c).1\leq\sqrt{(1\plus{}2bc)(2\plus{}b^{2}c^{2}\minus{}2bc)}$
Now let $ bc \equal{} x$,we must prove
$ (2x\plus{}1)(x^{2}\minus{}2x\plus{}2)\leq 2$
$ \Leftrightarrow 2x^{3}\minus{}3x^{2}\plus{}2x\leq 0$
$ \Leftrightarrow x(2x^{2}\minus{}3x\plus{}2)\leq 0$,which true,since
$ b^{2}\plus{}c^{2}\leq 1$ :D[/quote]
Actually this solution looks like mine,that i posted one month ago here http://www.mathlinks.ro/viewtopic.php?t=163928 ,isn't it?

[quote="hanschinaman"]$ a,b,c$ are non-negative real numbers such that $ a^{2}\plus{}b^{2}\plus{}c^{2}\equal{} 1$,
prove $ a\plus{}b\plus{}c\minus{}abc\le\sqrt{2}$
  [/quote]
Maybe you mean $ a\plus{}b\plus{}c\minus{}2abc\leq\sqrt{2}$,if so,then your inequality is  [url=http://www.mathlinks.ro/viewtopic.php?t=163928]Longlist 1987[/url]  :wink:

[quote="Erken"]
Actually this solution looks like mine,that i posted one month ago here http://www.mathlinks.ro/viewtopic.php?t=163928 ,isn't it?[/quote]
Yes,it does.But the first time i saw this solution was in http://www.mathlinks.ro/Forum/viewtopic.php?t=32399.
Anyway,your solution is very nice too,erken. :)
Btw,i also think it should be $ 2abc$
 :wink:


a, b, c and d are reals such that a+b+c+d=0
Show that (ab+ac+ad+bc+bd+cd)^2+12>=6(abc+abd+acd+bcd)

[quote="kaydika"]$a, b, c$ and $d$ are reals such that $a+b+c+d=0$
Show that $(ab+ac+ad+bc+bd+cd)^{2}+12\geq6(abc+abd+acd+bcd)$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=71504


This forum is for the discussion and creation of non-frivolous games.  Moderators will delete any nonsense posts, and repeat offenders will be denied access to this forum.  That sort of silliness belongs in the [url=http://www.artofproblemsolving.com/Forum/index.php?f=139]Fun Factory[/url].  If you have any doubt as to whether your post belongs here or in the Fun Factory, post in the Fun Factory.

Cool. Finally G&FF won't be as messy as it used to be :)

Can things like Big Brother, The Mole, and Elect the Executioner be moved back?

So, could you give me an example of an appropriate topic for this forum?
Like would my "Smash Discussion Thread" be considered a "serious discussion of games?" Also, by games, does that include video games?

To Brut3Forc3: yes. I will work on moving threads back to this forum today. NotA: yes it would. To quote rrusczyk: Things like "what video game do you like" are OK, but "Guess who will make the next post" are not.

Finally. No need to weed through a half page of BS before finding my threads, which I posted in 6 hours before...

so now posts here count but posts in fun factory don't?

This is completely irrelevant. Post count is irrelevant. And don't ever bother again.

Should Math & Strategy Games
http://www.artofproblemsolving.com/Forum/index.php?f=360

and Puzzles & Brainteasers
http://www.artofproblemsolving.com/Forum/index.php?f=337

be merged with this serious Games forum? As currently, both subforums receive very little traffic and it would probably get more people to discuss Rubik's Cubes and play 24 and the like while removing a bit of redundancy. Two forums on the same topic competing for one's attention is a bit unnecessary.

[quote="mathelete"]Quick! Lock the DHMO Topic![/quote]

Uh that was total spam, pm the mods for stuff like that. No one reads the stickies.

I do read them, sometimes ;)
But I don't see a reason to lock the DHMO thread. An unbiased discussion from our usage of dangerous DHMO is really necessary! Thousends of scientists and industries lie to the world about DHMO being not dangerous; this MUST be stopped!

I think he's trying to lock the topic due to a false statement, and he doesn't want us having fun with it.
I am against locking the topic, and most stickies are [b]locked[/b] by now because people don't [i]want[/i] these sort of discussions. Take it up with the forum mods if you must.

Strangely this didn't happen when there was none... *sigh*

Games Forum = Relaxation for some brain-weary math users, especially those who are younger.

[quote="jeremylu"][quote="math26"]Why is there a Games forum?
It has nothing to do with math.[/quote]
If you think that this Games forum is useless, i suggest you stop posting in this forum.[/quote]

Yobuddyyoumightwanttonotbesoaddictedtothegamesforumandwhendidhesaythatitwasuseless.

Games forum for me is just a place to pass the time.

[quote="EggyLv.999"]Stop these city games! This is almost as bad as the FF flooding with hurt and heals![/quote]

Sorry! The moderators didn't say anything against that for a long time! :stretcher:

Maybe cause [i]that[/i] post was posted a long time ago?

Sorry for this revive, but somebody please clarify the following scenario.

I want to ask how to download a game on my computer. It's complicated and difficult to download the game on my computer. Should I post said question in the Games Forum or the Computer Science forum?

[s]Computer Science, as the "Games" Forum is about discussing games and playing games, and not troubleshooting software.[/s]
What chaotic_iak said. Yeah.

No, for how to download a game, it's in Games forum, the same with the discussion of said game. Computer Science is not for general computing topics (where "downloading software" is included).

What if you want to ask for advice on how to sell a gaming device or videogame? Should you post your question in the Games Forum or somewhere else?

Try Fun Factory. Although your topic is clearly not frivolous, it doesn't fit anywhere else either. State clearly that you're looking for serious answers to avoid trolls.

[quote="MathMan1234"]Question: What about discussing games such as the Game of Life (by John Horton Conway, not the board game)? Like, what if you discovered some cool pattern you want to share?[/quote]
It's perfectly fine to discuss games rather than playing them. Some people do both. :yup:

Err, I don't think the games forum is a good place to discuss cellular automata.

It is, for the lack of any better forum. It's not frivolous enough for FF but it doesn't fit anywhere else.


I have a GUI for a program I'm writing in Java. When the user clicks a button or if 30 seconds elapses, I want the main program to check the clicked button and update the GUI. Is it better to have the main program in another thread and have it sleep until 30 seconds elapses or a click occurs, or is it better to make ActionListener call it in a function and have it create a global timer that executes it in 30 seconds and is cleared if a button is clicked?

I don't know too much about Java but if I wanted to check something after 30 seconds, I'd definitely use timer rather than create a new thread. That's exactly what timers are designed to do, threads should be used if you need to do two different things at the same time, if the only thing your thread does is waiting for 30 seconds and then checking some condition it is definitely wasting resources. That's my point of view :).


geometry
3D geometry
LaTeX
algebra
difference of squares
special factorizations

Write with a rational denominator:

$\frac{1}{\sqrt{2+\sqrt[7]{3}}}$

[i]$\text{\TeX}$ed up by moderator so that it will be easier to read.[/i]

You really should start with a simpler problem.

For example:   $\displaystyle{\frac{1}{\sqrt{2-\sqrt[3]{7}} }}$[hide]
Multiply top and bottom by $\sqrt{2-\sqrt[3]{7}}:$

$\displaystyle{.\quad \frac{1}{\sqrt{2-\sqrt[3]{7}}}\cdot\frac{\sqrt{2-\sqrt[3]{7}}} {\sqrt{2-\sqrt[3]{7}}}  \;= \;\frac{\sqrt{2 - \sqrt[3]{7}}} {2 - \sqrt[3]{7}}}$ . . . but we're not done yet!

Multiply top and bottom by $4 + 2\sqrt[3]{7} + \sqrt[3]{49}:$

$\displaystyle{.\qquad\frac{\sqrt{2 - \sqrt[3]{7}}}{2 - \sqrt[3]{7}}\cdot\frac{4 + 2\sqrt[3]{7} + \sqrt[3]{49}}{4 + 2\sqrt[3]{7} +  \sqrt[3]{49}} }$

$\displaystyle{.\quad= \;\frac{(4 + 2\sqrt[3]{7} + \sqrt[3]{49})\;\sqrt{2 - \sqrt[3]{7}}}{8 + 4\sqrt[3]{7} + 2\sqrt[3]{49} - 4\sqrt[3]{7} - 2\sqrt[3]{49} - \sqrt[3]{343}} }$

$\.\quad= \;(4 + 2\sqrt[3]{7} + \sqrt[3]{49})\;\sqrt{2 - \sqrt[3]{7}} $

Your problem, with a 7th root under the radical, will be much harder.
[/hide]

Yes, i know it will be harder with a 7th root.  But it's still not much more difficult than the example you showed with a 3rd root.

I'll give a hint (I don't know how to make those hint things, sorry)
.........

figure out how to factor a difference of nth powers...

[ spoiler]

Place your hint here

[ /hide]

Minus the space after the [ .

Like this:

[hide]

Remember, x^3 - y^3 = (x - y)(x^2 + xy + 1). That should help in rationalizing the cube root part.

[/hide]

You can also modify the text this way:

[code][hide=text to replace Click to reveal hidden content]
Text that's hidden
[/hide][/code]

Yes, hidden text is now preferred to spoiler text because it is easier for newbies to reveal.

I think that when you put the [ spoiler] tags around the text, it hides them the same way the [ hide] tags do.

Code:[hide]$x^2+2x^2$[/hide]

This is shown as:

Hidden Text

Code:[spoiler]$x^2+2x^2$[/spoiler]

This is shown as:

[hide][/hide]

Currently, they're same. The reason admins added this was so that the Latex can be hidden in the spoiler text.

And for these problems, please use Latex so it's easily readable.

It's hard to see without Latex. If you don't know the Latex, learn it. It doesn't take much time to do it.

you guys need to focus... leave Latex discussion in another thread, the current problem is still unsolved. Perhaps you might want to solve the problem now. :)

haha well isn't anybody going to at least try?

Well, I don't really enjoy seventh rooted fractions but I think I'm close to finish it.

$\frac {1}{\sqrt {2 + \sqrt [7] {3}}}$

Multiply by $\frac {\sqrt {2 + \sqrt [7] {3}}}{\sqrt {2 + \sqrt [7] {3}}}$ and erase the square root.

Then, you'll get:

$\frac {\sqrt {2 + \sqrt [7] {3}}}{2+ \sqrt [7] {3}}$ and multiply by conjugate but instead of $\sqrt [7] 3$, but $\sqrt [7] {729}$ so that the result will be 2^2-3 or 1 on the bottom. Resulting the answer:

${(\sqrt {2+ \sqrt [7] {3}})(2- \sqrt [7] {729}})$

I'm not so confident whether it can be simpler than this or not but this is close to an answer.

Well since no one seems to be solving this i'll expand on my previous hint...

[hide="hint"]Take, for example, an easier problem with only a square root in the denominator:
1/(1+sqrt(2))

To rationalize the denominator we must multiply by the conjugate of 1+sqrt(2)

1/(1+sqrt(2))*(1-sqrt(2))/(1+sqrt(2))= -1+sqrt(2)

What we have just done to the denominator is basically made it a difference of squares, since (a-b)(a+b)=a^2-b^2, and thus removed the radical

Thus, to solve our problem we must change the denominator into a sum of seventh powers.  To do this we must first know how to factor (a+b) from a^7+b^7....

I think someone can solve this from here[/hide]

Well, first post on these boards, but hopefully this will be a good start. I'm fairly sure I've got this one now...

As done previously in this thread, start by going from

to

by multiplying through by

.

If I'm right, I'd rather keep from spoiling it for others, so:

[hide]

A sum of seventh powers factors to:

.

Therefore,

.

Using this, the fraction rationalizes to:

.

In an alternate form:

.

[/hide]

Any comments, including those on mistakes, are more than welcome!

Yes that's correct, EPAstor.  Very nice. :P


geometry perimeter SFFT circumcircle ratio

[quote="marsupilami"]Please help me

Find the triangle with integer sides has perimeter = area

thank you :lol:[/quote]
6-8-10 works as well.

[quote="Ubemaya"]Those are, in fact, the only solutions.

Here's an extension:
Prove that there are infinite many right triangles with real sides such that its perimeter equals its area.[/quote]

[hide]
a, b are the legs
the perimeter is $ a \plus{} b \plus{} \sqrt {a^2 \plus{} b^2}$
area is $ \frac {ab}{2}$

$ a \plus{} b \minus{} \frac {ab}{2} \equal{} \sqrt {a^2 \plus{} b^2}$
$ a^2 \plus{} b^2 \plus{} 2ab \minus{} \frac {a^2b}{2} \minus{} \frac {ab^2}{2} \plus{} \frac {a^2b^2}{4} \equal{} a^2 \plus{} b^2$
$ \minus{}8ab \plus{} 2a^2b \plus{} 2ab^2 \equal{} a^2b^2$
$ \minus{}8 \plus{} 2a \plus{} 2b \equal{} ab$

uhh this is not looking good.  by that i mean i must be wrong.

unfortunately, even with the edit, this isn't looking happy
sfft says there's only a few solns.
[/hide]

edit: thanks ghjk

[quote="CircleSquared"][quote="Ubemaya"]Those are, in fact, the only solutions.

Here's an extension:
Prove that there are infinite many right triangles with real sides such that its perimeter equals its area.[/quote]

[hide]
a, b are the legs
the perimeter is $ a \plus{} b \plus{} \sqrt {a^2 \plus{} b^2}$
area is $ \frac {ab}{2}$

$ a \plus{} b \minus{} \frac {ab}{2} \equal{} \sqrt {a^2 \plus{} b^2}$
$ a^2 \plus{} b^2 \minus{} 2ab \minus{} \frac {a^2b}{2} \minus{} \frac {ab^2}{2} \plus{} \frac {a^2b^2}{4} \equal{} a^2 \plus{} b^2$
$ 8ab \plus{} 2a^2b \plus{} 2ab^2 \equal{} a^2b^2$
$ 8 \plus{} 2a \plus{} 2b \equal{} ab$
Your mistake is that in the expansion of (a+b-ab/2)^2, you will have the factor 2ab not -2ab :wink:

uhh this is not looking good.  by that i mean i must be wrong.

[/hide][/quote]

After the algebra is corrected, that factors as $ (a\minus{}4)(b\minus{}4)\equal{}8$, which clearly has an infinite number of real solutions.

[quote="azjps"]After the algebra is corrected, that factors as $ (a \minus{} 4)(b \minus{} 4) \equal{} 8$, which clearly has an infinite number of real solutions.[/quote]

REAL SOLUTIONS? i thought it said integer  :(

[quote="Ubemaya"]Those are, in fact, the only solutions.

Here's an extension:
Prove that there are infinite many right triangles with real sides such that its perimeter equals its area.[/quote]

[hide="Solution to Extension"]
Let r be half the hypotenuse. Then, the circumcircle of the triangle has radius r, and the point at which we have the right angle is somewhere on the semicircle.

Now, let the right angle be at C, and the endpoints of the hypotenuse be A and B. Let t be the measure of angle CAB, and let P(r,t) and A(r,t) be the perimeter and area, respectively, of the triangle with circumradius r and angle CAB equal to t.

P(r,0) = 4r and A(r,0)=0, so P(r,0)>A(r,0).
P(r,45) = (2+sqrt2)r and A(r,45)=r^2, so, provided r>2+sqrt2, P(r,45)<A(r,45)

Hence, so long as r>= 2+sqrt2, for some t between 0 and 45, P(r,t)=A(r,t) by the intermediate value theorem.

Since there are infinitely many values of r, there are infinitely many triangles with A=P
[/hide]

[b]@CircleSquared:[/b] The extension posed by Ubemaya dealt with right triangles and real sides. The original problem dealt with all triangles and integer sides. I personally think the original problem is more difficult.

[quote="SimonM"]Another extension. So that for all triangles, there exists a similar triangle such that

$ A(\bigtriangleup ABC ) \equal{} 1 \times ( |AB| \plus{} |BC| \plus{} |CA| )$

(The 1 is to maintain dimensional correctness)

[hide="Small hint"]Consider the ratio between the area and perimeter[/hide]

[hide="Massive hint"]Triangles ABC and A'B'C' are similar and AB=kA'B'
Prove that
Area ( ABC ) = $ k^2$ Area ( A'B'C' )
[/hide][/quote]
[hide="Solution"]
[b]Lemma:[/b] If $ \triangle ABC\sim \triangle A_1B_1C_1$ such that $ \frac {AB}{A_1B_1} \equal{} k$, then $ [ABC] \equal{} k^2[A_1B_1C_1]$.
[i]Proof: [/i] Let $ AX$ be an altitude of $ \triangle ABC$ and let $ A_1X_1$ be an altitude of $ \triangle A_1B_1C_1$.  Thus, $ \angle ABX \equal{} \angle A_1B_1X_1$ and $ \angle AXB \equal{} 90 \equal{} \angle A_1X_1B_1$, so AA similarity gives us that $ \triangle ABX\sim \triangle A_1B_1X_1\implies \frac {AX}{A_1X_1} \equal{} \frac {AB}{A_1B_1} \equal{} k\implies [ABC] \equal{} \frac {AX\cdot BC}{2} \equal{} \frac {k\cdot A_1X_1\cdot B_1C_1\cdot k}{2} \equal{} k^2[A_1B_1C_1]$, as desired.

Let $ [ABC] \equal{} kp$, where $ k$ is a positive real, $ [ABC]$ denotes the area of $ \triangle ABC$, and $ p$ is the perimeter of $ \triangle ABC$.  Now, consider $ \triangle A_1B_1C_1$ such that $ \triangle ABC\sim \triangle A_1B_1C_1$ with $ \frac {AB}{A_1B_1} \equal{} k$.  Hence, $ p \equal{} AB \plus{} BC \plus{} AC \equal{} k(A_1B_1 \plus{} A_1C_1 \plus{} B_1C_1)$ and $ [ABC] \equal{} k^2[A_1B_1C_1]$, so $ [A_1B_1C_1] \equal{} \frac {[ABC]}{k^2} \equal{} \frac {p}{k} \equal{} A_1B_1 \plus{} A_1C_1 \plus{} B_1C_1$, so $ \triangle A_1B_1C_1$ is the desired triangle.  [/hide]


geometry
circumcircle
cyclic quadrilateral
geometry proposed

Let $ ABCD$ be a cyclic quadrilateral and $ P$ be a variable point on it's plane. Show that independent on the position of $ P$, the common chord of the circumcircles of triangles $ ABP$ and $ CDP$ passes through a fixed point.

That line passes through the radical center of the three circles, which is the intersection of $ AB$ and $ CD$.


algorithm
probability
blogs
geometry
analytic geometry
real analysis
number theory

I am studying a major where I will take those courses :
C/C++/Java/J2EE/VB.NET/ASP.NET/PHP, Software engineering, Sql/Oracle, System Analysis and design, software Project management, E-commerce, Information systems for commerce and industry.
The problem is that we won't study any course about [b]Math[/b], [b]Algorithms [/b]and [b]data structure[/b], so I wonder what will be the benefit of studying those 3 subjects especially Math for someone who wanna be one of these : developer, software engineer, Project manager or DataBase Administrator ?
If Math is important, then could u specify which are the Topics in which I need to concentrate more.
Thanks in Advance  :)

Yes it is
My mom and dad are both software engineers and both use it a lot
i am not exactly sure for what, though

Is studying computer science necessary to be a software engineer?

If you think so, then the answer is yes because computer science is essentially a branch of mathematics.

The real answer is a little more involved because there is a big difference between studying theoretical computer science and actually being a software developer, but clearly understanding software from a theoretical perspective will help you get a handle on it in practical situations.  It's like the distinction between basic research and medicine.

how is computer science a branch of mathematics?

Short answer:  theoretical computer science can roughly be described as the study of algorithms, and algorithms can be defined mathematically.  One can prove theorems about algorithms the same as any other mathematical object.  And perhaps the strongest validation of this viewpoint is the fact that [url=http://en.wikipedia.org/wiki/P_%3D_NP_problem]P vs. NP[/url], a problem in computer science, is one of the Millennium problems.

[quote]so computer science can be understood with the help of mathematics.but computer science is in no way a branch of mathematics[/quote]

How do you then define 'a branch of mathematics'?

@aadil:

Computer science is, [i]by definition[/i], a branch of mathematics. t0rajir0u has given a good idea of what this business is all about. Ironically, computer science has little to do with computers, and isn't really scientific ("Computer Science is no more about computers than astronomy is about telescopes").

I think you had the preconceived notion that "computer engineering" and "computer science" were equivalent.

@yfehehey:

Very few programming jobs these days required knowledge of heavy math. There are many absolutely [url=http://www.loudthinking.com/]stellar[/url] [url=http://www.codinghorror.com/blog/]hackers[/url] out there who aren't great at math.

But I do encourage you to learn some CS stuff just to get a perspective on the subject. And it seems to me that you are focusing on specific technologies rather than learning programming in general. [url=http://stackoverflow.com/questions/1711/what-is-the-single-most-influential-book-every-programmer-should-read]Here[/url] is a pretty comprehensive list of books that will help you. It's just a start, though.

[quote="Myself"][quote="yfehehey"]
The problem is that we won't study any course about [b]Math[/b], [b]Algorithms [/b]and [b]data structure[/b][/quote]
That is surprising for me. How are people expected to write good software if they do not know algorithms and data structures, and at least enough math to understand/analyze them?[/quote]

Because commonly used algorithms and data structures already exist in the language's standard library?

Most software, good or not, doesn't aim at solving highly algorithmic problems.

At the university where I earned my undergraduate degree Computer Science is hosted as a separate major within the Math Department. Nevertheless, I once took an introduction to programming course there which was taught by a CS graduate student, who at times appeared to have an inadequate understanding of the mathematical foundations of the algorithms he was teaching us to program.

What maths one ought to study as preparation for computer science depend completely upon what sub-field of computer science one intends to specialize in. For example, symbolic logic, which is the backbone of the study of algorithms in my opinion, was a topic studied as part of my real analysis course. If you intend to do any simulation work you'll have to study numerical analysis at some point. Number theory is necessary for many cryptographic applications.

The final answer is that it depends upon what you're going to do in computer science, but in my experience the difference between a computer scientist and a code monkey is an understanding of math.

[quote="Dr. No"]
in my experience the difference between a computer scientist and a code monkey is an understanding of math.[/quote]

ESR says "computer science education cannot make anybody an expert programmer any more than studying brushes and pigment can make somebody an expert painter."

I think your choice of words was kind of odd. You made what is pretty much an apples to oranges comparison: "computer scientists" are hardcore academics who rarely write code, while "code monkeys" are incompetent programmers who sit in cubicles.

Great programmers generally tend to be good at math, but I think that's not really the differentiating factor between the good and the bad. There are so many computer science graduates who can analyze esoteric algorithms in a matter of seconds, but cannot make a simple Ruby on Rails application if their lives depended on it.

Even most top universities don't teach source control management and test driven development; things programmers use everyday. Instead, they teach compiler design and other stuff programmers will never use professionally.

I'm not arguing that a standard CS education is useless. It's just that as a programmer, you are expected to learn 99% of this subject yourself.

[quote="mamama"][quote="Dr. No"]
in my experience the difference between a computer scientist and a code monkey is an understanding of math.[/quote]

I think your choice of words was kind of odd. You made what is pretty much an apples to oranges comparison: "computer scientists" are hardcore academics who rarely write code, while "code monkeys" are incompetent programmers who sit in cubicles.[/quote]

There are mathematicians who can't program and there are programmers who can't do math. Provided that one intends to make one's living at a keyboard, it is best to find one's self in neither of those categories.

[quote="mamama"]Great programmers generally tend to be good at math, but I think that's not really the differentiating factor between the good and the bad.[/quote]
Agreed, however, the answer to the thesis question of this thread is still "yes" in my opinion.

[quote="mamama"]Even most top universities don't teach source control management and test driven development; things programmers use everyday. Instead, they teach compiler design and other stuff programmers will never use professionally.[/quote]
Perhaps, but that brings to mind a remark from one of my favorite professors, an electrical engineer, "Be an expert technician, but don't become one."

[quote="mamama"]I'm not arguing that a standard CS education is useless. It's just that as a programmer, you are expected to learn 99% of this subject yourself.[/quote]
That is also quite true, however, since it is necessary to be self-taught  there is no reason to remain willfully ignorant of the use of one's greatest tool for said self-education.

Thx all for replying, so I understood that it's important to have a background in Math. What are the topics in which I need to spend more time studying and solving problems ? The only important topics that i know i have to study them so much are  [b]Discrete Math[/b], [b]Probability & statistics[/b] Are there any others ?
I wish I could see a work done by a software engineer using Math, just to know whether I need to know only the basics of each lesson in math (by solving only simple exercises to help me understand better the lesson and move to the next) or if I really need to master the lesson by doing many complex exercises. ?

It seems that I need to reiterate the fact that [i]most software engineers [b]never[/b] use any math[/i].

You will find a much better discussion on this [url=http://stackoverflow.com/questions/157354/is-mathematics-necessary-for-programming]at StackOverflow[/url]. AoPS doesn't really have many programmers.

If you decide to learn some math, add [i]linear algebra[/i] and [i]algorithms[/i] to your topics list. A good way to learn about algorithms is to participate in programming contests. Steve Yegge shows you a good way to learn math in [url=http://steve-yegge.blogspot.com/2006/03/math-for-programmers.html]this blog post[/url] of his.

[quote="mamama"]It seems that I need to reiterate the fact that [i]most software engineers [b]never[/b] use any math[/i].
[/quote]
I agree most business application developers can avoid math. But even in any large business application, discrete math and other areas of math are going to rear their beautiful heads. For example of some elements you find in a modern enterprise application: rules engines (predicate logic), calendars & schedulers (date arithmetic),  flow charting (graph theory, (coordinate geometry and numerical analysis for rendering)), etc. If lucky, an average developer finds third party components that deal with this stuff. Unless you are that third party components developer.

Vertical fields such as 3d game programming get an extreme exposure to math.

So while it is possible to avoid, not being strong in math puts a drag or limitation on the endeavors you may participate. Coming out of school, you want as many options as possible before you settle on a niche.


analytic geometry
graphing lines
slope
trigonometry
calculus
calculus computations

How can I put the Slopefield program into TI-83 Calculator, any ideas?

Where is the lowercase x in the TI-83 calculator??

Somewhere in vars I think. Consider drawing your slope line at some point $ (x_p,y_p)$.

$ \theta \equal{} \arctan{(f^{\prime}(x,y))}$
$ x_{1,2} \equal{} x_p\pm r\cos{\theta}$
$ y_{1,2} \equal{} y_p\pm r\sin{\theta}$
$ \text{line(}x_1,y_1,x_2,y_2\text{)}$

Just throw that inside a loop to refresh your x and y values.
 :wink:


MATHCOUNTS
LaTeX
FTW
AMC
AMC 12
blogs
AMC 10

It's that time of year again!
Mathcounts is coming up in a few months and now you get a chance to test your skills in this mock mathcounts.

If you are interested, please sign up here.

The test will take place in a few months.
Exact dates will be posted as we near the time and we finish the test.

Any questions, please post here.

Good Luck to all that participate!

What level is it going to be at and what round (Sprint, Target, Team, CD)? Do you need help making the test?

I will sign up for problem-maker if it is available, and participant if not.

Oringinally, pytheagle created this in High School Basics and wanted the level to be about national.
It is likely that the level will not change too much.

However, it might be possible to create an easier test available to anyone who wishes to take that one.
The only problem is that it will take a longer time.
If very little people want to take this test, then we will not be very motivated to make it.
On the other hand, if many people respond saying they wish to take this test, then it might be possible to create it.

The test that is currently being created is a sprint round.
Other rounds might be made if time allows.

I will get back to you on whether or not we need another problem-maker.
Seeing as how only you responded to this thread, I'm kind of uncertain as to how successful this will be, and thus am uncertain whether we will need more help with the questions.

If you wish to sign up for this, please post here
If you wish to take the lower level test (probably gonna be a mix of chapter and state), then please post saying you wish to take the easier test

Any questions, please post here

[quote="Thunder365"]Wait isnt chapters in like end of february? Thats almost half a year. I got scared for a second when you said "few months" lol.

I will join.[/quote]

haha
i'm sorry, i don't exactly remember when it is =)

do you want to take the national level test or the chapter/state test?

I would like to join as well.  I would like to try the national level test if that's possible  :)

I join.
I'll take the higher level test, and, if possible, i'll take the chapter/state test for fun also.  :lol:

I'll join and take chapter, state, nats if possible :)

I'll take all if possible, but just the national one if not possible

Geez, I really hope people wont cheat... otherwise the qualifying score would be 46...

Are we going to get answers or solutions tomorrow?

Is it too late to sign up?  I just found out.  :l

No, but you only have until tomorrow to submit your answers. For every day after Monday late, you would lose 2 points, but since you just signed up, I will add 4 points. In other words, if you submit today, you get no points off. If you submit tomorrow, you get 2 points off. I can't accept anything after that.

The problems can be found [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=310266]here[/url]. Please read the rules and make sure you download the latest version of the test. Also, be sure to read everything that is in bold.

Finally, submit just answers by PMing me.

[hide="Signups (49)"]
jxl28
athunder
Thunder365
mathboy101
AIME15USAMO
cgyao15
the cliu
TheMan1998
jerry125
SuperNerd123
\$LaTeX\$
Canton
algebra1337
AwesomeToad
bulldog23
ButteredButNotEaten
KingJamesMvP
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Maybach
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yaofan
alexautin
King!
SonyWii
isabella2296
BOGTRO
dafjbomb
bbgun34
basketball9
pi31415926
BuckeyeFan100
cheeseyicecream
syoo
snowpaw
mathdriven
sandy999
AIME15
math.is.intense
15!!!
2lnx
The Hobbit
MathisGreat
txguest
xxrxxhxx
math rulz
superpi83
Just_Beginner
jgao1995
AdithyaGanesh[/hide]

If 50 people sign up and turn in their test, could the number of people qualifying for chapter be increased to 25?

[quote="PowerOfPi"]I doubt 33 people would turn in their test in 6 hours. The qualifying number of participants 20, but since we have less than 20 submissions, everyone qualifies![/quote]

I had a similar problem with 5 Round Tournament, except with less people :(

In general, I think we should really specify to stay COMITTED unless they say "may drop out"

so everyone qualified for chapter :?:
This is the first time I have been excited to qualify for chapter :P

So I quallified with my ...... score??

Awesome

[quote="PowerOfPi"]Answers and results the day after tomorrow, which is Friday, in the afternoon.[/quote]
That probably answers your question, AwesomeToad

I have the results from all the graders but one, which will be posted tomorrow (hopefully). This adds to the submissions number, making 26-27. I am not sure about who will qualify yet (that will be said at a later time).

I will still post answers and *current* results (more to come, so don't panic if it isn't there) on the other thread (now).


(1) Miku and John began biking from the same place on the same road, heading in the same direction.  Each day, Miku biked $ 25$ miles.  John biked $ 5$ miles the first day, $ 8$ miles the second day, and so on, each day biking $ 3$ miles more than he biked the previous day.  If they began on July $ 1^{st}$, on what date did John pass Miku?

[hide]Answer: $ \boxed{July\ 15^{th}}$[/hide]

***Is there another way to this problem without guessing and checking.***

(2) Julian tiled a $ 15$ feet by $ 21$ feet rectangular ballroom with one-foot square tiles.  When he finished, he drew both diagonals on the floor connecting the opposite corners of the room.  What is the total number of tiles that the diagonals pass through?

[hide]Answer: $ \boxed{63}$[/hide]

[hide="1"]We want for $ 5\plus{}8\plus{}\hdots\plus{}(3k\plus{}2)>25k$
Solve for $ k$
On the left hand side, there are $ k$ terms, giving us the sum $ \frac{(3k\plus{}7)(k)}{2}$
$ 3k^2\plus{}7k>50k$
$ 3k^2\minus{}43k>0$
$ k>\frac{43}{3}\approx14$

He passes him between 14 and 15 days.

So the earliest date: July 15th[/hide]

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=178840[/url]

This has been stated many times already...

Hmmm... on #2, I did $ m\plus{}n\minus{}(GCF(m,n))$ and got $ \boxed{33}$.

But I think I have the formula right..

[quote="Quickster94"]well, u used the formula to get the number of tiles one passes through, but he drew both diagonals[/quote]
Oh..Okay. But how do you get the overlapping ones? Like the ones with 2 lines.

[quote="mathcrazed"]On the left hand side, there are $ k$ terms, giving us the sum $ \frac{(3k\plus{}7)(k)}{2}$[/quote]

mathcrazed - How did you get $ \frac{(3k\plus{}7)(k)}{2}$?

Thanks.

[quote="Cube"][quote="DonkeyKong"]Hmmm... on #2, I did $ m \plus{} n \minus{} (GCF(m,n))$ and got $ \boxed{33}$.

But I think I have the formula right..[/quote]

This does [b]one[/b] diagonal, but what about [b]two[/b] diagonals?[/quote]

That's the one that I don't know. I know that there will be some tiles that overlap.. So don't know..

I would guess that you do this

(m+n)-GCF(m,n))2-GCF(m,n)

so u would get

(15+21-3)2-3

which simplifies to 63

but I can't be sure

okay, after looking over it, I realized my previous equation was wrong but I found out the equation for squares covered in two diagonals when both sides are even is (m+n-GCF(m,n))2 because they don't overlap.

[quote="Cube"][quote="mathcrazed"]On the left hand side, there are $ k$ terms, giving us the sum $ \frac {(3k \plus{} 7)(k)}{2}$[/quote]

mathcrazed - How did you get $ \frac {(3k \plus{} 7)(k)}{2}$?

Thanks.[/quote]
Sum of an arithmetic series ($ a\plus{}(a\plus{}d)\plus{}...\plus{}a(n\minus{}1)d$ is sum of the first and last term times the #of terms/2.


geometry
geometric transformation
reflection
floor function
AMC

A ray of light originates from point $A$ and and travels in a plane, being reflected $n$ times between lines $AD$ and $CD$, before striking a point $B$  (which may be on $AD$ or $CD$) perpendicularly and retracing its path to $A$. (At each point of reflection the light makes two equal angles as indicated in the adjoining figure. The figure shows the light path for $n = 3.$) If $\measuredangle CDA = 8^\circ$, what is the largest value $n$ can have?

$\text{(A)} \ 6 \qquad \text{(B)} \ 10 \qquad \text{(C)} \ 38 \qquad \text{(D)} \ 98 \qquad \text{(E)} \ \text{There is no largest value.}$

[hide="Answer"]Every time the beam is reflected, the angle of reflection increases by $8^\circ$. Or, to look at it a different way, if we begin with the perpendicular, our first reflected angle must be $82^\circ$, and it loses $8^\circ$ with each reflection. The original angle of the beam of light cannot be $<0^\circ$, so our maximum value of $n$ is $n=10\Rightarrow \boxed{B}$.[/hide]

Is it

[hide]B?[/hide]

If it is, I'll post my solution. Solution:

[hide]A particle bouncing off the wall is equivalent to the plane being flipped over and the path continuing in a straight line due to equivalent angles (angle of inflection = angle of reflection). So you can quickly see that it has to be perpendicular before the table has flipped more htan total of , and can quickly see that .[/hide]

Elemennop wrote:Is it

[hide]B?[/hide]

If it is, I'll post my solution.

Yes it is. So post your solution.

omg...a physics problem!

[hide="hint"]the idea is that once the angles formed are 90, then it will go back the way it came[/hide]

[quote="JesusFreak197"][hide="Answer"]Every time the beam is reflected, the angle of reflection increases by $ 8^\circ$. Or, to look at it a different way, if we begin with the perpendicular, our first reflected angle must be $ 82^\circ$, and it loses $ 8^\circ$ with each reflection. The original angle of the beam of light cannot be $ < 0^\circ$, so our maximum value of $ n$ is $ n \equal{} 10\Rightarrow \boxed{B}$.[/hide][/quote]
Shouldn't it be 11 then (since $ \lfloor{90/8}\rfloor\equal{}11$)?

[quote="mathwizarddude"][quote="JesusFreak197"][hide="Answer"]Every time the beam is reflected, the angle of reflection increases by $ 8^\circ$. Or, to look at it a different way, if we begin with the perpendicular, our first reflected angle must be $ 82^\circ$, and it loses $ 8^\circ$ with each reflection. The original angle of the beam of light cannot be $ < 0^\circ$, so our maximum value of $ n$ is $ n \equal{} 10\Rightarrow \boxed{B}$.[/hide][/quote]
Shouldn't it be 11 then (since $ \lfloor{90/8}\rfloor \equal{} 11$)?[/quote]

No, because n is determined by the number of times the light beam hits the walls BEFORE it hits B.  So it is floor(82/8) or 10.


Find the number of subsets $ S$ of $ \{1,2, \dots 63\}$ the sum of whose elements is $ 2008$.

[hide="Solution"]
Note that $ 1\plus{}2\plus{} \ldots \plus{} 63 \equal{} \frac{63 \cdot 64}{2} \equal{} 2016 \equal{} 2008\plus{}8$. Thus, picking a subset that sums to 2008 is equivalent to leaving out a complementary subset that sums to 8.

8 = 7+1 = 6+2 = 5+3 = 5+2+1 = 4+3+1, so there are 6 ways to pick complementary subsets that sum to 8, and the answer is $ \boxed{6}$[/hide]


In certain country there are $2n$ airports. Between some airports there are direct connections, but for any three of them there are at least one pair of airports between which there isn't a direct connection. What is the greatest number of direct connections in the whole country??

This is just Turan's Theorem. Check out the Theorem's Section.

So can you show how to solve it with Turan's Theorem ? Because I'm still not used to it and I'd like to see it in work  :D

Sure: it means that there are at most $\frac {(2n)^2}4=n^2$ connections.

Due to statement corresponding graph doesn't contain 3-clique. In this case Turan's theorem states that there are at most $\frac{3-2}{3-1}\cdot\frac{(2n)^2}{2}=n^2$ edges. And we can easily construct appropriate example: just divide $2n$ airports into two equal groups and connect all airports of first group with all airports of second group.

.... thanks but I don't know from where do you get this $\frac{(2n)^2}{4}$  :? ...
I probably don't understand Turan's Theorem: r are edges, n - vertices, k excluded subgraph,  and what is t ?

Well, I for one don't even know the full version of Turan's theorem. However, a restricted statement is this: if a graph with $n$ vertices has no triangles, then there are at most $\frac {n^2}4$ edges.

@Megus: Did you read Turan's theorem or Harry Potter? :)

Definition.
Complete subgraph with $q$ vetrices is called $q$-clique.

Turan theorem.
Suppose graph $G$ with $n$ vertices doesn't contains $q$-clique. Then there are at most $\frac{q-2}{q-1}\cdot\frac{n^2}{2}$ edges in $G$.

I don't know nothing about that Turan's theorem but I saw a very nice solution to this problem:

Let [b]A[/b] be the airport connected with [b]m[/b] airports, where [b]m[/b] is maximal (other airports are not connected with more airports).Then these [b]m[/b] airports cannot be connected together. They can be connected just to [b]A[/b] and [b]2n-m-1[/b] airports we haven't used yet (together [b]2n-m[/b]). Each of that [b]2n-m[/b] airports is connected not more than to [b]m[/b] airports. So there's not more than $ m(2n-m) = (n-(n-m))\cdot (n+(n-m)) = n^2-(n-m)^2 \leq n^2$ connections.

This problem is not hard.. can be solved also by induction.

thanks I wasn't sure of my solution, besides I solved it without Turan's theorem and could learn something new:)

[quote="Myth"]@Megus: Did you read Turan's theorem or Harry Potter? :)
[/quote]
both   ;)

[quote="Myth"]
Turan theorem.
Suppose graph $G$ with $n$ vertices doesn't contains $q$-clique. Then there are at most $\frac{q-2}{q-1}\cdot\frac{n^2}{2}$ edges in $G$.[/quote]

In topic called Turan's theorem it is a bit different so that's why I didn't get it...
Thanks again


Integers $ d_1, d_2, \ldots, d_n$ satisfy that $ d_i \geqslant 1$ and at least three of them are greater than or equal to $ 2$. If $ d_1 \plus{} d_2 \plus{} \ldots \plus{} d_n \equal{} 2n$, find maximum and minimum value of
\[ (n \minus{} 1) \cdot \sum_{k \equal{} 1}^n d_k^2 \minus{} \frac {1}{3} \sum_{k \equal{} 1}^n d_k^3.
\]

Maximum is achieved for sequence $ (2, 2, 2, \ldots, 2)$ and minimum for $ (n \minus{} 1, 2, 2, 1, \ldots, 1)$.


The vertical width of the United States flag is called its "hoist" and the horizontal length is called its "fly." The "union" is the upper-left portion of the flag where the stars of the $ 50$ states are displayed. Suppose a flag has a fly of $ 9 \frac{1}{2}$ feet and a hoist of $ 5$ feet. The dimensions of the flag's union are $ \frac{2}{5}$ of the fly and $ \frac{7}{13}$ of the hoist. How many square feet are in the area occupied by the flag's stripes? Express your answer as a decimal to the nearest hundredth.

However, we need a decimal.

$ \frac{6147}{30} \equal{} (6147 \div 3) \div 10 \equal{} 2049 \div 10 \equal{} \boxed{204.9}$

Except its 6147/130=47.3. Doesnt that make more sense than the stripes having area 205 on a flag with area 47.5. Wait I think we messed up somewhere.

Hm, I got it to be $ 49.5\minus{}\frac{133}{13} \equal{} \frac{1287\minus{}266}{26} \equal{} \frac{1021}{26}$

Also it says to nearest HUNDREDTH, so I get $ \boxed{ 39.27}$...hm someone should check....I might have computed the decimal wrong (too lazy to go get calculator)

The dimensions of the flag are $ 5$ by $ \frac{19}2$, and the dimensions of the union are $ 5\cdot\frac7{13}\equal{}\frac{35}{13}$ by $ \frac{19}2\cdot\frac25\equal{}\frac{19}5$, and the answer is $ 5\cdot\frac{19}2\minus{}\frac{35}{13}\cdot\frac{19}5\equal{}47.5\minus{}10.23076923\approx\boxed{37.27}$.

hurdler was almost correct, but he put $ 49.5$ instead of $ 47.5$ as the area of the flag.


$ Given:$

$ \frac {\sqrt {w^4 \plus{} 4w^3 \plus{} 4w^2} \plus{} 8}{\sqrt {y^4 \plus{} 12y^3 \plus{} 36y^2} \plus{} 13} \equal{} \frac {7}{4}$

 and

$ \frac {\sqrt {x^4 \plus{} 24x^3 \plus{} 144x^2} \plus{} 43}{\sqrt {z^4 \plus{} 8z^3 \plus{} 16z^2} \plus{} 8} \equal{} \frac {7}{4}$

$ Evaluate:$$ \frac {wx}{zy \plus{} 2x}$

[hide="Note"]
Consider $ \sqrt{a^2}\equal{}\minus{}a$
[/hide]

[hide="Partial Solution."]

$ \sqrt {w^4 \plus{} 4w^3 \plus{} 4w^2} \plus{} 8 \equal{} 7$
$ \Longrightarrow \sqrt {(w^2)(w \plus{} 2)^2} \plus{} 1 \equal{} 0$
$ w^2 \plus{} 2w \plus{} 1 \equal{} 0$
$ (w \plus{} 1)^2 \equal{} 0\Longrightarrow\fbox{w \equal{} \minus{} 1}$

$ \sqrt {y^4 \plus{} 12y^3 \plus{} 36y^2} \plus{} 13 \equal{} 4$
$ \Longrightarrow \sqrt {(y^2)(y \plus{} 6)^2} \plus{} 9 \equal{} 0$
$ y^2 \plus{} 6y \plus{} 9 \equal{} 0$
$ (y \plus{} 3)^2 \equal{} 0\Longrightarrow\fbox{y \equal{} \minus{} 3}$

$ \sqrt {x^4 \plus{} 24x^3 \plus{} 144x^2} \plus{} 43 \equal{} 7$
$ \Longrightarrow \sqrt {(x^2)(x \plus{} 12)^2} \plus{} 36 \equal{} 0$
$ x^2 \plus{} 12y \plus{} 36 \equal{} 0$
$ (x \plus{} 6)^2 \equal{} 0\Longrightarrow\fbox{x \equal{} \minus{} 6}$

$ \sqrt {z^4 \plus{} 8z^3 \plus{} 16z^2} \plus{} 8 \equal{} 4$
$ \Longrightarrow \sqrt {(z^2)(z \plus{} 4)^2} \plus{} 4 \equal{} 0$
$ z^2 \plus{} 4z \plus{} 4 \equal{} 0$
$ (z \plus{} 2)^2 \equal{} 0\Longrightarrow\fbox{z \equal{} \minus{} 2}$

From here, You can now evaluatE:$ \frac {wx}{zy \plus{} 2x}$

[/hide]


(1) Find $p$ such that $p,2p+1,4p+1$ are all prime numbers.

(2) Find $q$ such that $q,2q+1,4q-1,6q-1,8q+1$ are all prime numbers.

1)if $p=1$(mod 3) then $3|2p+1$ ,if $p=2$(mod 3) then $3|4p+1$
So $p=3$
2)Now we consider mod 5
if it is $1$ so $5|6q-1$, f it is $2$ so $5|2q+1$, f it is $3$ so $5|8q+1$,f it is $4$ so $5|4q-1$
So $q=5$.
I think it was easy ;)

a_vakilian, the answer for (1) is right, but for (2) you will get another $q$.


I am using an Internet Explorer browser and whenever I tried to access AoPSWiki, the following message shows:

The website cannot display the page
 HTTP 500
   Most likely causes:
•The website is under maintenance.
•The website has a programming error.

Is the wiki having issues or something?

Perhaps it was taken down like Alcumus? I dunno.