This is directly copied from one of local competitions around my area.

The Protagoras Paradox

Perhaps one of the earliest known paradoxes is about the Greek law teacher Protagoras, who took a poor but talented student and agreed to teach him out a fee on condition that after the student completed his studies and won his first law case, he would pay Protagoras a certain sum. The student agreed to do this. Well, the student completed his studies but did not take any law cases. Some time elasped and Protagoras sued the student for the sum. Here are the arguments they gave in court.

Student: If I win the case, then by definition, I don't have to pay. If I lose the case, then I will not yet have won my first case, and I have not contracted to pay Protagoras until after I have won my first case. So whether I win the case or lose the case, I don't have to pay.

Protagoras: If he loses the case, then by definition he has to pay me (after all, this is what the case is about). If he wins the case, then he will have won his first case, hence he has to pay me. In either case, he has to pay me.

[From Raymond Smullyan's [i]What is the Name of this Book[/i]]

Thanks.

P.S. It seems like every time I post, you have something to teach me.  :lol:

In my opinion, Protagoras was right.

Regarding the student's first statement: "If I win the case, then by definition, I don't have to pay"
Winning the case obviosly does not grant the student exemption to any payment to Protagoras in the future, but it is after he wins the case that he should pay the fee.

Regarding the student's first statement: "If I lose the case, then I will not yet have won my first case, and I have not contracted to pay Protagoras until after I have won my first case."
Note that losing the case => paying the fee, but the opposite is not true.
If the court decided that the student should pay, then although it was unreasonable, the student must pay.

[quote="Nightmare124"]the student is right because he only has to pay his teacher if he doesnt win  his first case[/quote]
It is if, not only if, as the problem states. Signing a contract that says "I cannot give money to somebody if I have not win any cases" is completely nonsense. He may give the money to Protagoras if he wants to, or someone force him to.

In fact Protagoras could just sue the student for murder :D

Maybe you guys are both right. Or, both wrong. That's why it's paradox.  :)

I think they're both right, and they should just forget about the whole thing because they both owe each other. :D

[quote="Nightmare124"]double post!!! XD[/quote]
Beyond unnecessary. :|

I cannot see where the paradox is
What the student says makes no sense at all :mad:
I have explained it in the forth post
Can someone please explain why this is a paradox?

[quote="erabel"]Okay, let's see...

Student: Sued for money owed. If he wins, he does not have to pay the fine he is being sued for. If he loses, since he hasn't won his first law case, he does not have to pay. Either way, he doesn't pay. [b]Both arguements are valid.[/b]

Protagoras: Suing for money owed. If he wins, he gets the money he sued for. If he loses, his student has won his first law case, and must pay. [b]Both arguements are valid.[/b]

Now you get it?[/quote]

The first statement of the student is invalid. Of course he need not pay the sum Protagoras sue him for, but his victory triggers another payment, which does not exist at the time he is sued. Unless the court says, "The contract is annulled. You don't need to pay Protagoras any more." which sounds really weird.

The court should let the student win, then the student still have to pay Protagoras the sum. The student may sue Protagoras back, but this is another case, nothing related to the first case. What effect will a verdict bring should not be a factor affecting the court's decision.

[quote="tjhance"]Okay basically we have:

If the student loses the case then he needs to pay Protagoras. But the student claims that he won't have to pay Protagoras because he hasn't won a case yet. But winning a case is not the only reason he would need to pay Portagoras. (aka "student winning case $ \implies$ he pays" but NOT "student winning case $ \leftrightarrow$ he pays"). In this case, he would have to pay as a result of the court commanding him to pay.

Now if the student wins, then he will NOT have to pay as a result of the court commanding him to. But he WILL have to pay as a result of the original contract.

The key is noticing the difference between paying money as a result of losing the case and paying money as a result of the original contract.[/quote]

Totally agree :D
The student should hire another lawyer :P

I think I should sue them both on the grounds of mental harrasment
:D lol


How many lines of symmetry exist for a regular heptagon (a regular seven-sided polygon)?

This is simply one for each vertex, since the same lines go through the opposing side, thus $ \fbox{7}$.

there will be a symmetry for each side of the vertex so the answer is 7

so the answer is 7

rate my post thanks

There will be 1 line of symmetry for each vertex. So the answer would be 7


for each integer n>2 find the smallest natural number f(n) such that for any subset of A={1,2,...,n} like B with f(n) elements , there is 3 elements of B like

x,y,z that are pairwise coprime.

This won't be a very useful post...

The problem was posted here: http://www.mathlinks.ro/viewtopic.php?t=85233

And the solution from Kalva is here:

http://web.archive.org/web/20040810234501/www.kalva.demon.co.uk/balkan/bsoln/bsol924.html


analytic geometry
geometry
quadratics
graphing lines
slope
perpendicular bisector
geometry solved

The problem is:-

Find the coordinates of point $P$ lying inside triangle $ABC$ such that $\angle APB = \angle BPC = \angle CPA$,

[b]where A =(0,0) B =(8,0) and C =(5,4).[/b]

Please give the whole numerical soln. and not just the method.I know the method but want a sol. which doesn't involve solving Complicated eqns.

[hide]NOTE:- The answer is quite complicated involving surds. An elegant soln with minimum of algebra will be preferred.[/hide]

The 1st Fermat point P is the intersection of 3 circles centered at the orthocenters of the equilateral triangles constructed on the triangle sides c = AB, a = BC, b = CA and passing through the adjacent triangle vertices. We could get the equations of 2 of these circles and find their intersection other than the common vertex. On the other hand, the 1st Fermat point P is also the intersection of the lines connecting the outward vertices of the equilateral triangles with the opposite vertices of the triangle $\triangle ABC$. Using the circles would lead to quadratic equations, but using these lines would lead to linear equations, hence, it is easier. We need 2 of these lines to get the common intersection of the 3. The triangle side lengths are

$c = AB = 8$

$a = BC = \sqrt{(x_C - x_B)^2 + (y_C - y_B)^2} = \sqrt{9 + 16} = 5$

$b = CA = \sqrt{(x_A - x_C)^2 + (y_A - y_C)^2} = \sqrt{25 + 16} = \sqrt{41}$

Let D, E be the midpoints of the sides AB, BC and G, H the outward vertices of the equilateral triangles $\triangle ABG, \triangle BCH$. The coordinates of the midpoints D, E are

$D = (x_D, y_D) = \left(\frac{x_A + x_B}{2}, \frac{y_A + y_B}{2}\right) = (4, 0)$

$E = (x_E, y_E) = \left(\frac{x_B + x_C}{2}, \frac{y_B + y_C}{2}\right) = \left(\frac{13}{2}, 2\right)$

The coordinates of the vertex G of the equilateral triangle $\triangle ABG$ is

$G = (x_G, y_G) = \left(x_E, -\frac{c \sqrt 3}{2}\right) = \left(4, -4 \sqrt 3\right)$

The equation of the line BC is

$BC:\ \ y - y_B = \frac{y_C - y_B}{x_C - x_B}\ (x - x_B),\ \ y = -\frac 4 3\ (x - 8) = -\frac 4 3\ x + \frac{32}{3}$

The slope of this line is $m = -\frac 4 3$. The vertex H of the equilateral triangle $\triangle BCH$ lies on the perpendicular bisector of the segment BC. The perpendicular bisector of the segment BC passes through the midpoint E and it has the slope $k = -\frac 1 m = \frac 3 4$, because the product of slopes of 2 perpendicular lines is equal to -1. Hence, the equation of this perpendicular bisector is

$EH:\ \ y - y_E = k (x - x_E),\ \ y = \frac 3 4 \left(x - \frac{13}{2}\right) + 2 = \frac 3 4\ x - \frac{23}{8}$

The distance EH is equal to

$EH = \sqrt{(x_H - x_E)^2 + (y_H - y_E)^2} = \frac{a \sqrt 3}{2} = \frac{5 \sqrt 3}{2}$

Since the point H lies on the line EH, $y_H = \frac 3 4\ x_H - \frac{23}{8}$ and

${\left(x_H - \frac{13}{2}\right)^2 + \left(\frac 3 4\ x_H - \frac{39}{8}\right)^2} = \frac{75}{4}$

$x_H^2 - 13x_H + \frac{169}{4} + \frac{9}{16}\ x_H^2 - \frac{117}{16}\ x_H + \frac{1521}{64} = \frac{75}{4}$

$64x_H^2 - 832x_H + 2704 + 36x_H^2 - 468x_H + 1521 = 1200$

$100x_H^2 - 1300x_H + 3025 = 0$

$4x_H^2 - 52x_H + 121 = 0$

$x_H = \frac{52 + \sqrt{2704 - 1936}}{8} = \frac{13}{2} + 2 \sqrt 3$

$y_H = \frac 3 4\ x_H - \frac{23}{8} = 2 + \frac{3 \sqrt 3}{2}$

For the outward vertex H of the equilateral triangle $\triangle BCH$, we had to take the root with the plus sign. Having the coordinates of the points G, H, we can find the equations of the lines AH, CG.

$AH:\ \ y - y_A = \frac{y_H - y_A}{x_H - x_A}\ (x - x_A)$

$y = \frac{2 + \frac{3 \sqrt 3}{2}}{\frac{13}{2} + 2 \sqrt 3}\ x = \frac{4 + 3 \sqrt 3}{13 + 4 \sqrt 3}\ x = \frac{16 + 23 \sqrt 3}{121}\ x$

$CG:\ \ y - y_C = \frac{y_C - y_G}{x_C - x_G}\ (x - x_C)$

${y = \frac{4 + 4 \sqrt 3}{5 - 4}\ (x - 5)} + 4 = 4(1 + \sqrt 3)x - 4(4 + 5 \sqrt 3)$

Solving the set of 2 equations for the lines AH, CG, we get their intersection P

$\frac{16 + 23 \sqrt 3}{121}\ x = 4(1 + \sqrt 3)x - 4(4 + 5 \sqrt 3)$

$(16 + 23 \sqrt 3) x = 484(1 + \sqrt 3)x - 484(4 + 5 \sqrt 3)$

$(468 x + 461 \sqrt 3) x = 484(4 + 5 \sqrt 3)$

$x_P = \frac{484(4 + 5 \sqrt 3)}{468 + 461 \sqrt 3} = \frac{2^2 \cdot 11^2 (3 \cdot 41^2 - 2^4 \cdot 31 \sqrt 3)}{3 \cdot 11^2 \cdot 1153} = \frac{2^2 (3 \cdot 41^2 - 2^4 \cdot 31 \sqrt 3)}{3 \cdot 1153}$

$y_P = \frac{16 + 23 \sqrt 3}{121}\ x_P = \frac{16 + 23 \sqrt 3}{121} \cdot \frac{2^2 (3 \cdot 41^2 - 2^4 \cdot 31 \sqrt 3)}{3 \cdot 1153} =$

$= \frac{2^2 (2^7 \cdot 3 \cdot 11^2 + 11^2 \cdot 19 \cdot 47 \sqrt 3)}{3 \cdot 11^2 \cdot 1153} = \frac{2^2 (2^7 \cdot 3 + 19 \cdot 47 \sqrt 3)}{3 \cdot 1153}$

Yetti


conics
hyperbola
geometry
geometric transformation
reflection
Asymptote
geometry unsolved

[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14640[/img]

 :(

This problem is similar to the problem [url]http://www.mathlinks.ro/viewtopic.php?t=209728[/url], click the link for details. $ J$ is the intersection of the directrix $ d_1$ with the line $ a_1 \parallel s_1,$ where $ s_1$ is the unknown hyperbola asymptote. Perpendicular to the directrix $ d_1$ from $ P$ cuts $ d_1, a_1$ at $ D, A,$ respectively. Perpendicular to the directrix $ d_1$ from $ Q$ cuts $ d_1, a_1$ at $ E, B,$ respectively. $ \frac{AJ}{AD} \equal{}\frac{BJ}{BE} \equal{} \epsilon$ is the hyperbola excentricity. Reflect $ a_1$ into $ a_1'$ in the line $ PDA.$ Parallels to $ a_1'$ through $ P, Q$ cut the directrix $ d_1$ at $ S, T$ and $ \frac{PD}{PS} \equal{} \frac{QE}{QT} \equal{} \epsilon$ $ \Longrightarrow$ the hyperbola focus $ F_1$ correcponding to the directrix $ d_1$ is intersection of circles $ (P), (Q)$ with centers $ P, Q$ and radii $ PS, QT,$ respectively. If the 2 circles intersect at all, we get 2 possible foci $ F_1, F_1'$ and again, 2 different hyperbolas exist with the given directrix $ d_1,$ the given asymptote direction $ a_1,$ and passing through the 2 given points $ P, Q.$

Proceeding with one possible focus $ F_1$: Perpendicular $ p$ to $ d_1$ through $ F_1$ is the hyperbola major axis. Perpendicular to $ a_1$ through $ F_1$ is also perpendicular to the unknown asymptote $ s_1 \parallel a_1,$ it cuts $ d_1$ at $ K.$ This point lies on the hyperbola pedal circle and perpendicular to $ F_1K$ through $ K$ (parallel to $ a_1$) cuts the hyperbola major axis $ p$ at the hyperbola center $ C,$ identical with the center of the pedal circle $ (C)$ with radius $ CK.$ The other hyperbola focus is the reflection $ F_2$ of $ F_1$ in $ C,$ one hyperbola asymptote is $ s_1 \equiv CK$ and the other asymptote is reflection $ s_2$ of $ s_1$ in the major axis $ p \equiv F_1CF_2.$ Reflections $ P', Q'$ of $ P, Q$ in $ C$ also lie on the hyperbola. The pedal circle $ (C)$ cuts the hyperbola major axis $ p$ at the hyperbola vertices $ V_1, V_2,$ so that we have 6 points $ P, Q, P', Q', V_1, V_2.$ See the above link on how to construct more hyperbola points.


function
absolute value
algebra proposed
algebra

once I've heard a question :

Is it possible that $(a_1+a_2+...+a_n) -> g \in R$ and
$b_n=a_1^3+a_2^3+...+a_n^3$ is not congruent ?

Just a silly question...

Is it congruent or .... convergent?

cheers! :D :D

in fact you can prove the following statement. If $f$ is a function such that for every series $\sum_{n=1}^{\infty}u_n$ converges then $\sum_{n=1}^{\infty}f(u_n)$ also converges then
$f(x)=kx$ for every $x$ in $[a,b]$ where $a<0<b$.

To give an example and get rid of it, :):

Take first $2^{-\frac 13}$, and then two times $-\frac 12\cdot 2^{-\frac 13}$, then $3^{-\frac 13}$, and after that three times $-\frac 13\cdot 3^{-\frac 13}$, then $4^{-\frac 13}$, and afterwards four times $-\frac 14\cdot 4^{-\frac 13}$, and so on (I'm sure you see the pattern). This sequence works, I think. The series converges to $0$, while the series generated by $a_n^3$ diverges to $\infty$ (because of the coefficient $\frac 1n$ in front of the $-n^{-\frac 13}$, the negative terms become too small in absolute value to be able to stop the positive terms from going to $\infty$).


In this topic, [url]http://www.mathlinks.ro/viewtopic.php?p=993910#993910[/url]

when I tried to post my image id=11744 with [ img.]11744[ /img ] (spaces added to disable image) it didn't show the image, like not recognizing the tag. But when I introduced the 'url' tags, the image showed just fine... Is this some type of problem in MATHCOUNTS forum? This problem doesn't happen when it is in other forums (I tested on some but not all) though.

testing...

[img]11744[/img]

testing again...
[img]11744[/img]
Now notice that there are no 'url' tags in this post.

"Consecutive posts not allowed for less than 10 minutes"  :D


Find the number of ordered pairs of real numbers (a,b) such that (a + bi)^2002 = a - bi

(i being the   :sqrt: -1 and not another variable of course)

Please include an explanation, if you would be so kind.

Here's a hint (in spoiler):

[hide]If z=a+bi, then z*conj(z)=a2+b2.  Try taking (a+bi)2002=a-bi and multiplying by a+bi on both sides and see what happens.[/hide]

it says find the number of ordered pairs, and 1 is not the answer... there are more.

[hide]

if (a+bi)^2002 = a-bi, we know |(a+bi)^2002| = |a-bi| =>. But |a+bi| = |a-bi|, so |a+bi| = 1. Therefore we can write z = a + bi = cis  :theta:. Furthermore, a-bi = cis (- :theta: ). cis :theta: ^ 2002 = cis (2002 * :theta: ). Therefore 2002 * :theta: = - :theta: (mod 2pi) or 2003 * :theta: = 0 (mod 2pi). So we should get 2003 unique solutions for :theta: ( 2pi * i / 2003 for 0  :le: i :le: 2002 ). and therefore 2003 ordered pairs (a,b). (I might be wrong, just guessing)[/hide]

[hide]I get 2004 since we have (0,0) as well. But other than that small detail, I agree with everything Nukular did.[/hide]

[hide]

yeah i forgot (0,0)...

always forget the trivial solution

[/hide][/hide]

[quote]So we should get 2003 unique solutions for  ( 2pi * i / 2003 for 0  i  2002 ). and therefore 2003 ordered pairs (a,b).[/quote]

Could you explain that?

Well, if we have 2003 :theta:  =  0 (mod 2pi), then

2003 :theta: = 2pi * k for some integer k right?. Since 0 <= :theta: < 2*pi, (thats the standard for dealing with unique angles), any integer k from 0 to 2002 will give us a unique :theta:. Therefore we will have 2003 unique :theta: and therefore 2003 unique pairs (a,b). Add in the (0,0) case and you get 2004 solutions.


Hmm. I went to see this with some of my friends and wasted a good $5.75 on a very dumb movie. There was no plot, and it wasn't even that scary, because when the monsters came you knew it was all fake. I was actually laughing really hard about the movie.

Don't spoil the movie man.
The beginning and middle were good, but I hated the ending.

I have no idea what this is but I'm the only person who voted in the poll

I also have no idea what this is about so I chose randomly


hi there
prove this plz
i need hlp
tan(a)+tan(b)+tan(c)=tan(z)tan(b)tan(c)
a,b,c <>90 :oops:
also
a,b,c are the angel of a triangle

Simple use $c = \pi-a-b$ and trig rules for tangent and a little bit of algebra.

try it


limit
logarithms
calculus
calculus computations

we all know

$\lim_{p\rightarrow\infty}\;\sum_{n=1}^{p}\;\frac{1}{n}\quad\;\;\;$ diverges.

what about

$\lim_{p\rightarrow\infty}\;\sum_{n=1}^{p}\;\frac{1}{n^{1+\imath}}$ ?

does it also diverge ? or converge ?  :D

note: $\;\;\;\imath\;\;=\;\;\sqrt{-1}$

Since $n^{i}=e^{i\log n}$ is essentially constant on any interval of the form $N\le n\le 1.01N$, the sum of $\frac{1}{n^{1+i}}$ over such an interval is about $0.01$ in absolute value. The series diverges.


Let be given a right triangle $ABC$ at $A$. Denote: $O$ be the midpoint of the side $BC;\,M\in AB,\,N\in AC: \,BM^{2}+CN^{2}=MN^{2}$
Prove that: $\angle{MON}=90^{\circ}$

$ON^{2}={a^{2}\over 4}+CN^{2}-aCN\cos\gamma={a^{2}\over 4}+CN^{2}-bCN$

$OM^{2}={a^{2}\over 4}+BM^{2}-cBM$

Adding those two we get

$OM^{2}+ON^{2}={a^{2}\over 2}+MN^{2}-(bCN+cBM)\qquad (*)$

From $(b-CN)^{2}+(c-BM)^{2}=MN^{2}$, after simplification, we get $bCN+cBM={a^{2}\over 2}$, hence $(*)$ becomes $OM^{2}+ON^{2}=MN^{2}$. QED

Through vertex $B$ of $\bigtriangleup ABC,$ we draw a line perpendicular to $AB$ and we denote as $L,$ it’s intersection point, from the segment line $NO.$ So, we have $BL = CN$ $($ because of $OB = OC$ $).$

From the right triangle $\bigtriangleup BLM$ $($ $\angle LBM = 90^{o}$ $),$ $\Longrightarrow$ $(BM)^{2}+(BL)^{2}= (ML)^{2}$ and so, we have that $ML= MN.$

Hence, because of the point $O,$ is also the midpoint of the segment $LN,$ we conclude that the segment line $MO,$ is midperpenticular of the segment $LN$ and the proof is completed.

Kostas Vittas.


Let $ABC$ be an equilateral triangle and $\Gamma$ its incircle. If $D$ and $E$ are points on the segments $AB$ and $AC$ such that $DE$ is tangent to $\Gamma$, show that $\frac{AD}{DB}+\frac{AE}{EC}=1$.

Let $M$ be the tangency point of the circle with $AB$, we define the same way $N(N \in BC)$,$P(P \in AC)$,$Q(Q \in DE)$. Also $MB=BN=NC=CP=c$, $MD=DQ=a$, $QE=EP=b$, $AD=c-a$, $AE=c-b$.

Then our statement is equivalent to show:$3ab=c^{2}-ac-bc$

By cosine law in triangle $AED$: $(b+c)^{2}=(c-a)^{2}+(c-b)^{2}-2cos60^\circ(c-a)(c-b)$, and the result follows.

With the last notations. $BE$ cut $CD$ at $J$ and $AJ$ cut $BC$ at $K$, then: $\frac{AD}{DB}+\frac{AE}{EC}=\frac{AJ}{JK}$
But $BE, CD, MP$ concur at $J$, now $J$ is the midpoint of $AK$,i.e. $AJ=JK$, therefore: $\frac{AD}{DB}+\frac{AE}{EC}=1$

Dear Mathlinkers,
revisiting the last proof, we have
1. the relation of van Aubel
2. the Newton theorem
3. the midline.

Sincerely
Jean-Louis

This problem is a special case of the problem [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=615952]Modern Sangaku[/url] ;) :)


Ampere's Law applies when we have a contour around a wire with a current on it. My question is, does this current have to be infinitely long?

Because we can do it for an infinitely long wire. If $ B$ is the magnetic field at radius $ R$, then we have $ 2 \pi R B \equal{} \mu_0 i$ or $ B \equal{} \frac{\mu_0 i}{2 \pi R}$, which is the correct formula.

However, what if we said that Ampere's Law applied for a small segment of wire, like the one in the diagram? The magnetic field should still be symmetric around the wire, and so we should get $ 2 \pi R B \equal{} \mu_0 i$ and arrive at the same formula. But by the Biot-Savart law, the field strength should be $ \frac{\mu_0 i dl}{4 \pi R^2}$. It thus seems to me like Ampere's Law only applies when the wire is infinitely long.

However, Ampere's Law also works for a solenoid. So this doesn't make sense. (Do the solenoid and the infinite wire supposed be similar because they have the same topology or something?? That is, if we embed the infinite wire in a projective space...)

So where did I go wrong?

Ampere's law applies only when you have steady state currents, and these occur only in closed loops.

If the current is infinitely long, you can put the rest of the loop infinitely far away so that its effect on the magnetic field is negligible. The circuit can still be closed, so Ampere's Law still applies.

On the other hand, if the current is finitely long, either (a) you do not have a closed loop or (b) the rest of the loop has a non-negligible effect on the magnetic field.
In case (a) Ampere's Law no longer applies. You need Maxwell's correction, which deals with time rates of change of electric flux caused by the accumulation of charge at one end of the wire.
In case (b) the symmetry of the magnetic field is broken. $ B$ is no longer constant over the line integral.


Fro two prob. measures $ P$ and $ Q$ on $ (\mathbb{R}, \mathcal{B})$, the total variation distance is defined as $ \parallel P\minus{}Q\parallel \equal{}\sup_{B\subset\mathcal{B}}|P(B)\minus{}Q(B)|$
The sup metric is defines as $ d(P,Q) \equal{}\sup_{x\in \mathbb{R}}|P((\minus{}\infty, x])\minus{}Q((\minus{}\infty, x])|$
Aren't the topologies generated by these two the same? Isn't there any connection between these two?

What about $ P_n\equal{}\sum_1^n \delta_{\frac{k}{n}}$ ?

I don't understand your point alekk. Could you please be more precise?

$ P_n \equal{} \sum_1^n \frac{1}{n} \delta_{\frac {k}{n}}$ converges towards the uniform distribution on $ [0;1]$ for the distance $ d$ but not for $ \| \cdot \|$, isn't it?

Well that's true, because $ P_n$ converges in distribution to the uniform distribution on $ [0,1]$.
We have $ d(P,Q)\le \parallel P\minus{}Q\parallel$. So we maynot expect convergence with respect to $ \|\cdot\|$.
Can we say that the topology generated by $ d$ is finer than that by $ \|\cdot\|$?

In fact, if $ P_{n}\equal{}\sum_{1}^{n}\frac{1}{n}\delta_{\frac{k}{n}}$ and $ P$ corresponds to the uniform distribution on $ [0,1]$, it can be shown that $ \parallel P_n\minus{}P\parallel \equal{}1$ by choosing $ B_n\equal{} \displaystyle\cup_{i\equal{}0}^{n\minus{}1} \left(\frac{i}{n}, \frac{i\plus{}1}{n}\right)$. Am I right?

yes, this is what I meant in the previous post  :lol:


quadratics
complex numbers
algebra
algebra solved

Hey, Ive just been introduced to complex numbers and I'm having real trouble working out the question below....

The question is solve:

(1 + i)z2  + (3 - 2i)z - (21 - 7i) = 0

answer.  z = 3 - i OR -7/2 + 7/2i

Can someone please explain how to work out the roots for the equation, I have been looking through textbooks with no luck, any help is greatly appreciated!

Cheers,
Roo


vector
linear algebra
linear algebra unsolved

Let $n\geq 2$ be an integer and denote by $H_{n}$ the set of column vectors $^{T}(x_{1},\ x_{2},\ \ldots, x_{n})\in\mathbb{R}^{n}$, such that $\sum |x_{i}|=1$.

Prove that there exist only a finite number of matrices $A\in\mathcal{M}_{n}(\mathbb{R})$ such that the linear map $f: \mathbb{R}^{n}\rightarrow\mathbb{R}^{n}$ given by $f(x)=Ax$ has the property $f(H_{n})=H_{n}$.

[hide="Comment"]In the contest, the problem was given with a) and b):

a) Prove the above for $n=2$;

b) Prove the above for $n\geq 3$ as well.[/hide]

The standard basis vectors $e_{n}$ (and also $-e_{n}$) are the [url=http://en.wikipedia.org/wiki/Extreme_point]extreme points[/url] of $H_{n}$. It follows that $A$ permutes $\{\pm e_{n}\}$. Therefore, there are exactly $2^{n}n!$ such matrices.


limit algebra polynomial function calculus

What is the value of $ 0^0$?

Google results in $ 1$. [url=http://www.google.com/search?client=opera&rls=en&q=0%5E0&sourceid=opera&ie=utf-8&oe=utf-8]link[/url]

Wolfram Alpha results in "indeterminate". [url=http://www51.wolframalpha.com/input/?i=0%5E0]link[/url]

$ 0^0$ means $ \lim_{(x, y) \to (0, 0)} x^y$.
$ 0^0$ [b]don't[/b] mean $ \lim_{x \to 0} x^x$.

Can you image $ z\equal{}x^y$?

In some cases it can be defined as 1 or 0 for some purpose, but generally it is indeterminate.

What's the difference?

$ \lim_{x\to0^\plus{}}x^x\equal{}1.$

$ \lim_{x\to0^\plus{}}\left(e^{\minus{}1/x}\right)^x\equal{}e^{\minus{}1},$ which is neither one nor zero.

$ \lim_{x\to0^\plus{}}\left(e^{\minus{}1/x}\right)^{\sqrt{x}}\equal{}0.$

Actually the main reason I posted was to mention another interpretation:

When using summation notation to represent polynomials or power series, [b]by convention,[/b] the symbol $ x^0$ stands for the the constant everywhere-defined function $ 1.$ Hence within that particular convention, $ 0^0\equal{}1.$ In other words:

$ f(x)\equal{}\sum_{k\equal{}0}^na_kx^k\equal{}a_0\plus{}a_1x\plus{}a_2x^2\plus{}\cdots\plus{}a_nx^n$ and by convention, $ f(0)\equal{}a_0.$

Most calculators are programmed to spit out "undefined" on $ 0^0$.

[b]Proposition:[/b]  Let $ A, B$ be finite sets, and let $ A^B$ denote the set of functions from $ B$ to $ A$.  Then $ |A^B| \equal{} |A|^{|B|}$.

There is one function from the empty set to itself:  it does nothing.  Therefore $ 0^0 \equal{} 1$ in [b]combinatorics.[/b]  (This is the same function that makes $ 0! \equal{} 1$, the empty permutation.)

What t0rajir0u just said is also essentially a convention. If you didn't accept what he said, mathematics would not be destroyed - but it's mildly more convenient and consistent to accept it.

[quote="FantasyLover"][quote="Kouichi Nakagawa"]$ 0^0$ means $ \lim_{(x, y) \to (0, 0)} x^y$.
$ 0^0$ [b]don't[/b] mean $ \lim_{x \to 0} x^x$.

Can you image $ z \equal{} x^y$?[/quote]Um, may I ask what is the difference? :blush:[/quote]
[b]Remark 1[/b]
$ \lim_{(x, y) \to (0, 0)} x^y \neq \lim_{x \to 0} \left(\lim_{y \to 0} x^y \right)\equal{}1$.
$ \lim_{(x, y) \to (0, 0)} x^y \neq \lim_{y \to 0} \left(\lim_{x \to 0} x^y \right)\equal{}0$.
$ \lim_{(x, y) \to (0, 0)} x^y \neq \lim_{y \to 0} \left(\lim_{x \to (1/2)^{1/y}} x^y \right)\equal{}\frac{1}{2}$.
and more...

[b]Remark 2[/b]
$ \lim_{x \to 0} \left(\lim_{y \to 0} f(x, y) \right)\equal{}0, \ \lim_{y \to 0} \left(\lim_{x \to 0} f(x, y) \right)\equal{}0 \nRightarrow \lim_{(x, y) \to (0, 0)} f(x, y) \ \text{is undefined}$.
For example, $ f(x, y)\equal{}\frac{x^2y^2}{x^2y^2\plus{}(x\minus{}y)^2}, \ x^2y^2\plus{}(x\minus{}y)^2\neq0$.

[quote="Kent Merryfield"]it's mildly more convenient and consistent to accept it.[/quote]
Right.  In combinatorics one often finds that it is possible to extrapolate the value of a sequence defined on the positive integers to a value at $ 0$, and usually (but not always!) this value is consistent with a certain way of thinking about functions on the empty set, or empty permutations, or empty products.  For example, if you define the Fibonacci numbers $ F_n$ as the number of ways to tile a board of size $ n$ with tiles of size $ 1$ and $ 2$, then you get $ F_1 \equal{} 1, F_2 \equal{} 2$, etc. (note that this is the "combinatorialist's indexing" instead of the "number theorist's indexing," which is the one consistent with $ F_n | F_{mn}$).  Interpolating the Fibonacci recursion backwards gives $ F_0 \equal{} 1$, which suggests that the number of ways to tile the empty board is $ 1$ - the empty tiling.  The reason this makes sense is that the Fibonacci recursion is based on deleting the last tile, but if a tiling only has one tile then deleting the last tile gives an empty tiling.

(The really curious thing is that often it is possible to extrapolate the values of a sequence defined on the positive integers to the [b]negative[/b] integers, and moreover that these numbers have combinatorial significance; write down the negative Fibonacci numbers, for example.)

Note that $ 0^0 \equal{} 1$ is also the most convenient convention for writing down power series, which is not an accident; power series and combinatorics are closely related.

0^0=0^x/0^x=0/0=indeterminate.
But on one of the tests i took the choices for 0^0 were 1)complex 2)irrational 3)rational 4)real 5) undefined.
So I think it should be undefined even though when you write it it is indeterminate.

[quote="t0rajir0u"][b]Everything depends on context.[/b]  In a calculus class $ 0^0$ is a classic example of an "indeterminate form," and what this means is essentially what Kent Merryfield outlined:  it means it's an example of a situation in which a standard arithmetic operation isn't continuous.  In a combinatorial context, these concerns are irrelevant.[/quote]

So, we can generally say that in DISCREET contexts, $ 0^0$ is often taken to be $ 1$. In CONTINUOUS contexts, though, it's not determined.

This is rather simple compared to the rigorous reasons involving limits given above, but I'll post it anyway because it's kind of interesting.  :)

I have read that mathematicians originally defined the zeroth power of a number because of the rules they had proved for dividing powers with like bases.  They had come up with and proved rules for simplifying $ \frac{x^a}{x^b}=x^{a-b}$ when $ a>b$ and $ a<b$.  For example, $ \frac{x^5}{x^3}=x^2$ and $ \frac{x^3}{x^5}=\frac{1}{x^2}=x^{-2}$, where the penultimate expression is obtained by cancellation and the last expression is obtained by the rule.

But what about $ \frac{x^a}{x^a}$?  By simple cancellation, we get $ 1$.  But by the rule, we get $ x^{a-a}=x^0$.  So that is how mathematicians defined $ x^0$--as $ 1$.  But there's one kink in this derivation.  For $ \frac{0^a}{0^a}$, we cannot cancel the numerator with the denominator since ${ lim_{x\to0}\frac{a}{x}=\pm\infty}$.  So in this framework, $ 0^0$ is undefined.

wow very interesting posts here, thanks you all for your responses, some of it is over my head...

This is not High School Basics, especially when it's Calculus-related now. I'm moving to other forum.

[quote="JBL"][quote="Farenhajt"]... in DISCREET contexts ... In CONTINUOUS contexts ...[/quote]  Just a minor language note: the alternative to "continuous" that you were looking for is "discrete."  To be "discreet" is to be prudent, especially in not sharing information.[/quote]

Indeed it is  :| And the worst is that I knew that. Just didn't pay attention.

I never know the discrete arguments, for why \[0^0]\ was $1$ and not indeterminable (continuous argument), have there been any math problems where it mattered? For now it just seems to be a friendly convention, and usually a trivial case with no ambiguity. Thoughts?


Show that a square with side $1$ cannot be covered by five squares with side less than $\frac{1}{2}$.

This was posted at http://www.mathlinks.ro/Forum/viewtopic.php?t=34087 , but no solution was given. The source is Iberoamerican Mathematical Olympiad 2001 problem 6.

  Darij

hmm, doesn't this work?

Let square be ABCD with centre O, and sides AB, BC, CD, DA have midpoints a,b,c,d, respectively. First look at side AB, since each square is less than 1/2 in side length, there must be at least 3 small squares that lie inwardly on AB. Similar situation applies to side BC. However, these 6 squares can have one in common (the one at vertex B), so at least 5 squares covering those two sides. However, square DdOc is not covered.  :)

In fact, it isn't hard to prove using similar reasoning, I think, that the smallest number of squares with side length less than 1/2 that can cover an unit square is 9.

The small squares do not necessarily have sides parallel to the sides of the large square.

err...I can't even prove the impossibility of 4 smaller squares now... :(

Actually, now I kinda see it (at least for 4). Note that all 4 edges have to be covered by the squares. Suppose a square covers "a" of one side, then "b" of one of its adjacent sides, s.t. a^2 + b^2 < 1/2. Using this, we can quite easily show that 4 cannot cover all.  :)


[quote="Victory.US"]the equation $ x^3 \plus{} ax^2 \plus{} bx \plus{} c \equal{} 0$ has three roots :$ x_1 < x_2 < x_3$
Prove that $ (x_2 \minus{} x_1)^3 \ge \frac {2}{3.\sqrt {3}}(2a^3 \minus{} 9ab \plus{} 27c)$[/quote]

$ p(x)\equal{}x^3 \plus{} ax^2 \plus{} bx \plus{} c \equal{}(x\minus{}x_1)(x\minus{}x_2)(x\minus{}x_3)$

$ 2a^3 \minus{} 9ab \plus{} 27c\equal{}27p(\frac{x_1\plus{}x_2\plus{}x_3}{3})$

$ \Leftrightarrow (x_2\minus{}x_1)^3 \ge \frac{2}{3\sqrt 3}(x_1\plus{}x_2\minus{}2x_3)(x_1\minus{}2x_2\plus{}x_3)(\minus{}2x_1\plus{}x_2\plus{}x_3)$ ; ($ x1<x_2<x_3$)

We can assume : $ x_1\equal{}\minus{}1$ and $ x_2\equal{}0$ , $ x_3\equal{}t >0$

$ \Leftrightarrow 2t^3\plus{}3t^2\minus{}3t\minus{}2\plus{}\frac{3\sqrt 3}{2} \ge 0$ , ($ t >0$ ) - which is true .

Equality holdes for : $ x_1\equal{} \minus{}1$ , $ x_2\equal{}0$ , $ x_3 \equal{} \frac{\minus{}1\plus{}\sqrt 3}{2}$ ;

[quote="modularmarc101"]Why is it correct to assume values for $ x_1$ and $ x_2$ ?[/quote]

$ \Leftrightarrow (x_2 \minus{} x_1)^3 \ge \frac {2}{3\sqrt 3}(x_1 \plus{} x_2 \minus{} 2x_3)(x_1 \minus{} 2x_2 \plus{} x_3)( \minus{} 2x_1 \plus{} x_2 \plus{} x_3)$ ; ($ x1 < x_2 < x_3$)

1. Substitution : $ x_1\plus{}a$ , $ x_2\plus{}a$ , $ x_3\plus{}a$ don"t change this inecuality , so we can assume : $ x_2\equal{}0$

2. It is homogenous inequality => we can assume : $ x_1\equal{}\minus{}1$ , ($ x_1<x_2$)

[quote="Sergic Primazon"][quote="modularmarc101"]Why is it correct to assume values for $ x_1$ and $ x_2$ ?[/quote]

$ \Leftrightarrow (x_2 \minus{} x_1)^3 \ge \frac {2}{3\sqrt 3}(x_1 \plus{} x_2 \minus{} 2x_3)(x_1 \minus{} 2x_2 \plus{} x_3)( \minus{} 2x_1 \plus{} x_2 \plus{} x_3)$ ; ($ x1 < x_2 < x_3$)

1. Substitution : $ x_1 \plus{} a$ , $ x_2 \plus{} a$ , $ x_3 \plus{} a$ don"t change this inecuality , so we can assume : $ x_2 \equal{} 0$

2. It is homogenous inequality => we can assume : $ x_1 \equal{} \minus{} 1$ , ($ x_1 < x_2$)[/quote]

who can help me with other solution. :)


[quote="ffdbzathf"]What are the differences between incandescence, fluorescence, and phosphorescence?  How do these relate to quantum mechanics and wave-particle duality?[/quote]

Thing one: A photon is a quantum of light. Its energy is given by the expression $h\nu$. Quantum mechanics says that physical things, energy, momentum and so on come in discrete bits; i.e. particles, called quanta. So then a particle of light is a photon. Since a photon travels at speed $c=\lambda\nu$, light - a group of photons - can be viewed as a wave having wavelength, $\lambda$.

Incandescence: the emission of light (energy); i.e. photons, due to a rise in the temperature of the source of the energy.

Phosphoresence: luminescence which persists after the energy source which produced it is removed.

Fluorescence: luminescence which is not phosphoresence.

Luminesence: the emission of light other than by incandescence.


[quote="NguyenDungTN"]Let $a,b,c$ be the positive real numbers.Prove that:
$\sum a(a+2b+2)^{2}\geq (a+b+c)(a+b+c+2)^{2}$[/quote]
The inequality is equivalent to
\[\sum a(a+2b)^{2}+4\sum a(a+2b)+4\sum a \ge (a+b+c)^{3}+4(a+b+c)^{2}+4(a+b+c) \]
or
\[\sum a(a+2b)^{2}\ge (a+b+c)^{3}\]
or
\[\sum c(a-b)^{2}\ge 0 \]
which is trivial. We have done. :)


i want to talk to ppl but i can't in 'for the win'

I think you need to play around 40 games to chat on FTW.

I thinks that what you asked, right?


(x,y) is either (20188/27,1), (196,2), (123200,3), or (29645/216).

The maximum of the left side occurs at x=0 and it is always decreasing. The only integers it passes through are 1, 2, 3, and 4. Setting y equal to these values and solving for x produces those four points.


The length of each side of the larger square is 12 cm. The remaining figures are an inscribed square and two inscribed circles.What is the number of square centimeters in the area of the smaller circle? Express your answer in terms of $ \pi$.

[asy]size(100); import graph;
draw(Circle((0,0),1)); draw((sqrt(2),0)--(0,sqrt(2))--(-sqrt(2),0)--(0,-sqrt(2))--cycle); draw(Circle((0,0),sqrt(2))); draw((sqrt(2),sqrt(2))--(-sqrt(2),sqrt(2))--(-sqrt(2),-sqrt(2))--(sqrt(2),-sqrt(2))--cycle);[/asy]

Since the side length of the large square is 12, the diameter of the large circle is 12. 12 is also the diagonal of the small square, so the side length of the small square is $ 6\sqrt{2}$. That means the radius of the small circle is $ 3\sqrt{2}$, so the area of the sqmall circle is $ 18\pi$.


At a turnament between $n$ persons, everyone playes exactly one time against everyone else, and at one game there is everytime a winner and a looser.
Prove that one can arrange the participants in a chain$P_1 \to P_2 \to ... \to P_n$ such that the $i$-th person has won against the $(i+1)$-th person.

Hmm maybe I'm missing something, but I this problem seems rather trivial...

Using induction,

with only 2 players, if player 1 beats player 2, then we have P1->P2 , otherwise we have P2->P1.

Now, our induction hypothesis will be that for n players, there is a way of ordering them P1->P2->...Pn-1->Pn .

So, now we add 1 more player to these n players, producing an additional n matches.
We start at the beginning of our ordering. If Pn+1 beat P1, then he will go to the front and the ordering will be Pn+1->P1->P2...->Pn. If he lost to P1, then we see how he did against P2. If he beat P2, then the ordering will be P1->Pn+1->P2->...->Pn. We keep on going in this manner until we find a player that he lost to who is pointing to a player that he beat, and we can put Pn+1 exactly in between these 2 players. If it turns out that he didn't beat anyone, then he just goes to the end, leaving a final ordering of P1->P2->...Pn->Pn+1.

This feels a little bit too easy for this forum, maybe I missed something ? :o


Νομίζω πως έχω την εικόνα αλλά η αλήθεια είναι πως δυσκολεύομαι στη διατύπωση λόγω 3 διαστάσεων.

Το πρόβλημα το δημιουργεί το μικρότερο κυβάκι που δε γίνεται να το καλύψεις με κύβους (αναγκαστικά μεγαλύτερους) και από τις 6 μεριές. Θα μείνει κενό.
Όταν κολλήσεις στο μικρότερο κυβάκι ένα πιο μεγάλο τότε αν το μικρό τετραγωνάκι περιέχεται μέσα στο μεγάλο τότε το μεγάλο ξεχειλίζει και από τις 4 ακμές. Αν βάλεις και ένα απέναντι δε χωράει μετά να μπει άλλο στο κενό.

Το καλύτερο που μπορείς να κάνεις για να αυξήσεις το πλήθος των γειτόνων είναι να κολλήσεις τις γωνίες (μεγάλου-μικρού κύβου) έτσι ώστε το μεγάλο τετράγωνο να ξεχειλίζει το μεγάλο κυβάκι μόνο από 2 ακμές. Και πάλι όμως ότι και να κάνεις σχηματίζονται κενά που δεν χωράει μεγαλύτερο κυβάκι.

Δεν ξέρω αν βγάζεις το νόημα σε αυτά που γράφω. Είναι μπέρδεμα χωρίς σχήμα, καλύτερα να πάρει κανείς ένα ζάρι και να σημειώνει πάνω του.

Ενδιαφέρον είναι ότι στο αντίστοιχο πρόβλημα στις 2 διαστάσεις το αντίστοιχο επιχείρημα με τετράγωνο δεν εφαρμόζεται και μπορείς ένα τετράγωνο να το περιλυκλώσεις από μεγαλύτερα έτσι ώστε το καθένα να ξεχειλίζει από μια μόνο πλευρά

Nomizo pos katalaba tin perigrafi sou. Alla einai entiposiako oti auto pou perigrafeis mporei na ginei. Diladi, mporeis na kalipseis ena mikro kivaki me mi-temnomenous kivous olous diaforetikou megethous kai megaliterous apo to arxiko kivaki, xoris na afinoun kanena keno. Einai diskolo na to ferei kapoios sta matia tou os eikona, alla einai mia poli oraia askisi xeirotexnias (Ego basika to dokimasa me xarti :-) )

Episeis sosti i paratirisi sou me ta tetragona, gia tin akriveia o antistoixos diamerismos einai dinatos  se 2 diastaseis. Paratheto to epomeno link : http://www.srcf.ucam.org/tms/sqsq.php

Stergios

Μου δώσατε μια ιδέα, ελπίζω να είναι σωστή...
[hide]
Έστω ότι γίνεται ο διαμερισμός του κύβου που περιγράφει ο Anto.

Πάρτε την κάτω έδρα του κύβου κ' έστω $ K_1$ το μικρότερο κυβάκι της έδρας.

Το κυβάκι αναγκαστικά δε θα είναι στην άκρη γιατί η τετράγωνη έδρα του είναι η μικρότερη δυνατή (σκεφτείτε ένα τετράγωνο διαμερισμένο a la Anto: το μικρότερο τετράγωνο δεν μπορεί να είναι στην άκρη). Άρα, το $ K_1$ θα είναι οπωσδήποτε περικυκλωμένο από αυστηρά μεγαλύτερους κύβους που θα σχηματίζουν ένα "τείχος" γύρω του. Έτσι, πάνω από το $ K_1$ θα μπαίνουν κυβάκια μόνο αυστηρά μικρότερα και θα συναρμολογούν ένα τέλεια διαμερισμένο τετράγωνο στην πάνω έδρα του $ K_1$.

Έστω λοιπόν $ K_2$ το μικρότερο κυβάκι που στέκεται πάνω στο $ K_1$. Όπως και πριν, η πάνω έδρα του $ K_2$ θα πρέπει να είναι ένα τέλεια διαμερισμένο τετράγωνο.

Proceed ad nauseum :D
[/hide]
Cheerio,

Durandal 1707

Poli oraia :-)

Auti einai i idea !

Stergios

Sorry για την καθυστέρηση.
Στέργιο έχεις δίκιο για τη διόρθωση. Να ένας τρόπος να δει κανείς πως ένας κύβος περικυκλώνεται από μεγαλύτερους κύβους :
Βάζουμε στο πάτωμα το μικρό κύβο και μετά τον καλύπτουμε γύρω γύρω από πιο μεγάλους έτσι ώστε ο καθένας να ξεχειλίζει μόνο από μια ακμή.
Βλέποντας το από πάνω είναι σαν να έχεις ένα χαμηλό κτίριο που περικυκλώνεται από ψηλότερες πολυκατοικίες.

Ζουμπάμε μέσα στη γη 2 διαδοχικές από τις πολυκατοικίες έτσι ώστε να φτάσουν να έχουν ταράτσα στο ίδιο ύψος με το κτίριό μας. Τώρα υπάρχει χώρος για νακολλήσει από πάνω και ένας ακόμα κύβος και το ίδιο γίνεται και στο υπόγειο που ξεχωρίζουν προς τα κάτω οι βυθισμένοι κύβοι.

Του Παναγιώτη είναι ΟΚ. Ίσως πρέπει να προσθέσουμε ότι αφού αυτό γίνεται συνέχεια και κάθε φορά κατεβαίνουμε μια τάξη μεγέθους επ’ άπειρο, τότε μας βάζει φράγμα το μικρότερο κυβάκι από όλα αφού δεν μπορούμε να πέσουμε κάτω από αυτό. Νομίζω πως αν οι διαμερίσεις είναι πεπερασμένες τότε όντως τα πράγματα είναι έτσι. Αν επιτρέπεται οι διαμερίσεις να είναι άπειρες τότε νομίζω πως δεν ισχύει το επιχείρημα.

[quote="gpapargi"]Στέργιο έχεις δίκιο για τη διόρθωση. Να ένας τρόπος να δει κανείς πως ένας κύβος περικυκλώνεται από μεγαλύτερους κύβους :
Βάζουμε στο πάτωμα το μικρό κύβο και μετά τον καλύπτουμε γύρω γύρω από πιο μεγάλους έτσι ώστε ο καθένας να ξεχειλίζει μόνο από μια ακμή.
Βλέποντας το από πάνω είναι σαν να έχεις ένα χαμηλό κτίριο που περικυκλώνεται από ψηλότερες πολυκατοικίες.

Ζουμπάμε μέσα στη γη 2 διαδοχικές από τις πολυκατοικίες έτσι ώστε να φτάσουν να έχουν ταράτσα στο ίδιο ύψος με το κτίριό μας. Τώρα υπάρχει χώρος για νακολλήσει από πάνω και ένας ακόμα κύβος και το ίδιο γίνεται και στο υπόγειο που ξεχωρίζουν προς τα κάτω οι βυθισμένοι κύβοι.
[/quote]

Ωραία περιγραφή. Λίγο μπέρδεμα, eh? ;)

[quote]
Του Παναγιώτη είναι ΟΚ. Ίσως πρέπει να προσθέσουμε ότι αφού αυτό γίνεται συνέχεια και κάθε φορά κατεβαίνουμε μια τάξη μεγέθους επ’ άπειρο, τότε μας βάζει φράγμα το μικρότερο κυβάκι από όλα αφού δεν μπορούμε να πέσουμε κάτω από αυτό. Νομίζω πως αν οι διαμερίσεις είναι πεπερασμένες τότε όντως τα πράγματα είναι έτσι. Αν επιτρέπεται οι διαμερίσεις να είναι άπειρες τότε νομίζω πως δεν ισχύει το επιχείρημα.[/quote]

Προφανώς σε πεπερασμένη διαμέριση αναφερόμουν, το ad nauseum εκεί πήγαινε :P

That said... think you can do an infinite partition? :)

Cheerio,

Durandal 1707

Νομίζω ότι γίνεται διαμέριση με άπειρους κύβους.
Φαντάζομαι κάτι τέτοιο:
Παίρνεις το μεγάλο κύβο και χωρίζεις κάθε ακμή σε 2 ίσα μέρη. Έτσι βάζεις μέσα στο μεγάλο κύβο $ 2^3$ κυβάκια. Το πρώτο το κρατάς. Τα άλλα 7 πρέπει να τα αλλάξεις γιατί θέλεις κύβους διαφορετικού μεγέθους.

Τον επόμενο θα τον αντικαταστήσεις με άλλα κυβάκια όλα ίσα μεταξύ τους και στη συνέχεια το καθένα από αυτά θα αντικατασταθεί με άλλα μικρότερα όλα ίσα μεταξύ τους. Κάθε φορά τελειώνεις ένα επίπεδο (δηλαδή όλα τα κυβάκια της διαμέρισης του συγκεκριμένου επιπέδου) και μετά κατεβαίνεις επίπεδο (δηλαδή πρώτα η αρχική 7άδα και μετά διαμερίζεις την κάθε μια από αυτές). Αυτό για να περάσεις από όλα και να μην ξεμείνει κάτι.
Κάθε φορά χωρίζεις την ακμή σε p ίσα κομμάτια και παίρνεις $ p^3$ κυβάκια.
Για να μην έχουμε ίδιο μέγεθος κάποιου κύβου με κάποιον άλλο (έστω και στο μέλλον μετά από τις διαμερίσεις τους), κάθε φορά παίρνουμε p πρώτο. Αφού υπάρχουν άπειροι γίνεται.

Ελπίζω να μη μου ξέφυγε κάτι

[quote="Demetres"]gpapargi, φοβάμαι πως δεν δουλεύει η απόδειξή σου. Το πρόβλημα είναι ότι αυτά τα κυβάκια δεν αρκούν για να διαμερίσεις όλο τον κύβο.

Αν ο αρχικός κύβος έχει μήκος 1, τότε το κυβάκια σου έχουν μήκη 1/2,1/3,1/4,.... Αλλά τότε, καλύπτουν συνολικό όγκο $ (1/2)^3 \plus{} (1/3)^3 \plus{} \ldots \equal{} 0.202... < 1$.[/quote]

H αλήθεια είναι ότι κι εγώ δεν κατάλαβα πώς γεμίζει όλος ο κύβος...

Cheerio,

Durandal 1707

Πρέπει να έχεις δίκιο Demetres. Όφειλα να δω το άθροισμα αν φτάνει στο 1. Ήταν τόσο απλό τεστ και από βιασύνη το παράλειψα. Δυστυχώς το μέγεθος του κύβου ελαττώνεται πάρα πολύ γρήγορα. Θυμίζει κάτι από Ζήνωνα.

Εδώ που τα λέμε, αν περιγραφεί μια διαμέριση που δίνει άθροισμα πάνω από 1 αυτό μας κάνει ή θέλουμε να δίνει και άθροισμα ακριβώς 1; Μάλλον το δεύτερο. Έχει κανείς υπόψη του αν λύνεται το πρόβλημα δηλαδή αν υπάρχει τέτοια διαμέριση ή αποδεικνύεται ότι δεν υπάρχει (λχ μια απόδειξη με άτοπο πχ ότι αν υπάρχει άπειρη διαμέριση τότε θα υπάρχει και πεπερασμένη που αποδείχτηκε ότι δεν ισχύει); Ρωτάω για να ξέρουμε με τι έχουμε μπλέξει.  :)

Σίγουρα θες το άθροισμα να ισούται με 1. (Ακόμη και αυτό δεν είναι αρκετό. Πώς θα είμαστε σίγουροι για παράδειγμα ότι δεν έχουμε παραλείψει να καλύψουμε κάποιο (μή κενό) σύνολο σημείων με συνολικό όγκο 0; )

Μετά από αρκετή προσπάθεια έχω λύσει το πρόβλημα. Δεν θα πω προς το παρόν αν υπάρχει τέτοια διαμέριση ή όχι, διότι πιστεύω ότι είναι αρκετά μεγάλη βοήθεια. Θα αρκεστώ να πω ότι η λύση είναι πολύ απλή.

------------------------

Απορία formatting: Στο τέλος της πρώτης παραγράφου, θέλω να κλείσω την παρένθεση αμέσως μετά το ερωτηματικό, αλλά μου βγάζει ;) . Πώς το αποφεύγω αυτό;

[quote="Demetres"]Σίγουρα θες το άθροισμα να ισούται με 1. (Ακόμη και αυτό δεν είναι αρκετό. Πώς θα είμαστε σίγουροι για παράδειγμα ότι δεν έχουμε παραλείψει να καλύψουμε κάποιο (μή κενό) σύνολο σημείων με συνολικό όγκο 0; )

Μετά από αρκετή προσπάθεια έχω λύσει το πρόβλημα. Δεν θα πω προς το παρόν αν υπάρχει τέτοια διαμέριση ή όχι, διότι πιστεύω ότι είναι αρκετά μεγάλη βοήθεια. Θα αρκεστώ να πω ότι η λύση είναι πολύ απλή.[/quote]
[hide="Hint"]Στο παραπάνω quote υπάρχει ήδη ένα hint ;)[/hide]
Cheerio,

Durandal 1707

[quote="Demetres"]Σίγουρα θες το άθροισμα να ισούται με 1. (Ακόμη και αυτό δεν είναι αρκετό. Πώς θα είμαστε σίγουροι για παράδειγμα ότι δεν έχουμε παραλείψει να καλύψουμε κάποιο (μή κενό) σύνολο σημείων με συνολικό όγκο 0; )[/quote]

Κανείς βρε guys?

Toυλάχιστον για το ευκολότερο πρόβλημα που θίγει ο Demetres, τι λέτε? Έχει βρει κανείς αν υπάρχει μία μη πεπερασμένη διαμέριση του κύβου σε κυβάκια διαφορετικών μεγεθών που να καλύπτουν όλον τον [i]όγκο[/i] του αρχικού κύβου?

Demetres, αν δεις ότι πάει να θαφτεί το thread, βάλε τη λύση σου αν είναι :)

Cheerio,

Durandal 1707

Edit: να βάλω ένα [hide="hint."]Just do it :)[/hide]

Καλησπερα και απο εμενα μετα απο αρκετο καιρο. Δυστυχως οι σχολικες υποχρεωσεις δεν μου επιτρεπουν να επισκεπτομαι την μικρη μας κοινοτητα οσο συχνα θα ηθελα.
Τεσπα ειδα το προβλημα και σκεφτηκα ενα ισοδυναμο του, που δεν εχω προσπαθησει ακομη να λυσω. Το γραφω μπας και θελετε να το προσπαθησετε και ετσι.

Να δειξετε οτι αν παρουμε κυβους διαφορετικου μεγεθους μπορουμε να τους ενωσουμε με τετοιο τροπο ωστε να δημιουργησουμε ενα νεο κυβο οποιονδηποτε διατασεων.

Υ.Γ. Οι κυβοι που θα παρουμε μπορουν να ειναι οσοι θελουμε και οποιου μεγεθους θελουμε(αρκει να ειναι διαφορετικου μεγεθους μεταξυ τους οπως λεει και ο Ντουρανταλ).

Κύριοι, προσοχή:

[b]@basted[/b]
[quote="basted"][quote="Anto"]Na deiksete oti den einai dinaton na diameriseis ena kivo se mikra kivakia ola ek ton opoio exoun diaforetiko megethos.

Stergios[/quote]

Sygnwmh pou grafw se greeklish alla o sygkekrimenos ypologisths de mou parexei ellhnika. Den exw diabasei thn ypoloiph koubenta, alla me mia grhgorh matia, den einai to teleftaio 8ewrhma tou Ferma gia n=3?[/quote]

1. Δε γράφουμε έναν κύβο ως το άθροισμα δύο κύβων αλλά περισσοτέρων.
2. Πιο σημαντικό: κανείς δε μας περιορίζει σε ρητά μήκη πλευρών.

[b]@filosofimenos[/b]
[quote="filosofimenos"]Να δειξετε οτι αν παρουμε κυβους διαφορετικου μεγεθους μπορουμε να τους ενωσουμε με τετοιο τροπο ωστε να δημιουργησουμε ενα νεο κυβο οποιονδηποτε διατασεων. [/quote]

Τα προβλήματα δεν είναι ισοδύναμα και επίσης αυτό που λες δεν ισχύει (αφού ένας κύβος δε διαμερίζεται σε πεπερασμένο αριθμό κύβων). Π.χ. αν σου δώσω 7 κύβους με μήκος $ 1,2\ldots 7$ δεν μπορείς να φτιάξεις έναν καινούριο κύβο. Επίσης, αν σου δοθούν άπειροι κύβοι $ C_n$ με πλευρές π.χ. $ \frac {1}{\sqrt [3]{n}}$, και πάλι δεν μπορείς να συναρμολογήσεις έναν κύβο χρησιμοποιώντας τους όλους.

Cheerio,

Durandal 1707

[quote="Durandal"]
Demetres, αν δεις ότι πάει να θαφτεί το thread, βάλε τη λύση σου αν είναι :)
[/quote]

Θα βάλω την λύση μου σε 2-3 μέρες. Για όσους θέλουν να ξαναδοκιμάσουν, να πω ότι δεν μπορεί να διαμεριστεί ένας κυβος σε (άπειρα) κυβάκια όλα εκ των οποίων να έχουν διαφορετικές διαστάσεις.

Η αρχική μου λύση είχε ένα λάθος. Η καινούργια μου λύση (ελπίζω πως δεν έχει λάθος) θέλει λίγη ανάλυση. (Χρησιμοποιώ συμπάγεια.) Όσοι δεν γνωρίζουν τι είναι η συμπάγεια (compactness) μπορούν να χρησιμοποιήσουν...

[hide]
Υπάρχει τουλάχιστον ένα κυβάκι με διαστάσεις $ 0 \times 0 \times 0$. (Φυσικά, από την εκφώνηση του προβλήματος, συμπεραίνουμε ότι υπάρχει ακριβώς ένα τέτοιο κυβάκι.)
[/hide]

[quote="Demetres"][quote="Durandal"]
Demetres, αν δεις ότι πάει να θαφτεί το thread, βάλε τη λύση σου αν είναι :)
[/quote]

Θα βάλω την λύση μου σε 2-3 μέρες. Για όσους θέλουν να ξαναδοκιμάσουν, να πω ότι δεν μπορεί να διαμεριστεί ένας κυβος σε (άπειρα) κυβάκια όλα εκ των οποίων να έχουν διαφορετικές διαστάσεις.

Η αρχική μου λύση είχε ένα λάθος. Η καινούργια μου λύση (ελπίζω πως δεν έχει λάθος) θέλει λίγη ανάλυση. (Χρησιμοποιώ συμπάγεια.)[/quote]

Μάλλον την ίδια λύση έχουμε στο μυαλό μας :)

Cheerio,

Durandal 1707

Δεν μπορούμε να διαμερίσουμε ένα κύβο σε άπειρα κυβάκια διαφορετικών διαστάσεων.

Όπως είπα και σε προηγούμενο post, η αρχική μου λύση (που ισχυρίστηκα ότι είναι αρκετά απλή) κατέρρευσε. Ιδού μια άλλη λυση που θέλει λίγη ανάλυση:

Ας υποθέσουμε ότι ο αρχικός κύβος είναι ο $ A \equal{} [0,1] \times [0,1] \times [0,1]$. Μπορώ να υποθέσω ότι αν υπάρχει κυβάκι διαστάσεων $ 0 \times 0 \times 0$, τότε αυτό βρίσκεται στο $ B \equal{} [0,1] \times [0,1] \times [1/2,1]$. Κοιτάζω την κάτω έδρα του κύβου. Είναι ένα τετράγωνο διαμερισμένο σε τετραγωνάκια. Ισχυρίζομαι ότι δεν υπάρχουν άπειρα τέτοια τετραγωνάκια. Πράγματι, αν υπαρχουν άπειρα τέτοια, τότε από την συμπάγεια του $ [0,1] \times [0,1]$, υπάρχει τετραγωνάκι διαστάσεων $ 0 \times 0$, άτοπο. Συνεχίζουμε όπως στην λύση του Durandal για πεπερασμένα κυβάκια: Ας είναι $ K_1$ το μικρότερο κυβάκι που εφάπτεται στην κάτω έδρα, $ K_2$ το μικρότερο κυβάκι που εφάπτεται στο $ K_1$ κ.τ.λ. Μέχρι πιο ύψος μπορούμε να φτάσουμε; Ισχύρίζομαι ότι δεν μπορούμε να ξεπεράσουμε το 1/2. Για να το δείξω αυτό, αρκεί να δείξω ότι το μήκος του $ K_1$ είναι μικρότερο του 1/3. (Επειδή 1/3 + 1/9 + ... = 1/2.)  Αυτό αφήνεται σαν άσκηση στον αναγνώστη  :D .

Για να είμαι τελείως σωστός, νομίζω πως πρέπει να δείξω ότι τα κυβάκια είναι αριθμήσιμα αλλά αυτό είναι γνωστή άσκηση στην αριθμησιμότητα.

[quote="Demetres"]Ας υποθέσουμε ότι ο αρχικός κύβος είναι ο $ A \equal{} [0,1] \times [0,1] \times [0,1]$. Μπορώ να υποθέσω ότι αν υπάρχει κυβάκι διαστάσεων $ 0 \times 0 \times 0$, τότε αυτό βρίσκεται στο $ B \equal{} [0,1] \times [0,1] \times [1/2,1]$. Κοιτάζω την κάτω έδρα του κύβου. Είναι ένα τετράγωνο διαμερισμένο σε τετραγωνάκια. Ισχυρίζομαι ότι δεν υπάρχουν άπειρα τέτοια τετραγωνάκια. Πράγματι, αν υπαρχουν άπειρα τέτοια, τότε από την συμπάγεια του $ [0,1] \times [0,1]$, υπάρχει τετραγωνάκι διαστάσεων $ 0 \times 0$, άτοπο. Συνεχίζουμε όπως στην λύση του Durandal για πεπερασμένα κυβάκια: Ας είναι $ K_1$ το μικρότερο κυβάκι που εφάπτεται στην κάτω έδρα, $ K_2$ το μικρότερο κυβάκι που εφάπτεται στο $ K_1$ κ.τ.λ. Μέχρι πιο ύψος μπορούμε να φτάσουμε; Ισχύρίζομαι ότι δεν μπορούμε να ξεπεράσουμε το 1/2. Για να το δείξω αυτό, αρκεί να δείξω ότι το μήκος του $ K_1$ είναι μικρότερο του 1/3. (Επειδή 1/3 + 1/9 + ... = 1/2.)  Αυτό αφήνεται σαν άσκηση στον αναγνώστη  :D.[/quote]

Έξοχα, well done! :clap2:

Αυτή ήταν η ιδέα μου, ότι πηγαίνοντας από το μικρότερο κυβάκι στο μικρότερο κυβάκι δε φτάνεις ποτέ στο ήμισυ του ύψους του κύβου :)

Ωστόσο, θα επιμείνω στο προηγούμενό μου ερώτημα που είναι πολύ πιο σχολικό (και εύκολο):

[b]Υπάρχει μία διαμέριση του κύβου σε (άπειρους) μικρότερους κύβους διαφορετικού μεγέθους που να γεμίζουν τουλάχιστον όλον τον όγκο του αρχικού κύβου?[/b]

Cheerio,

Durandal 1707


i've a nice question for all...
write a program in c++ such that when a name (with multiple constituents) for ex:JOHN SMITH,GEORGE THOMAS,etc., the output shud be J.SMITH, G.THOMAS,......
ALSO PLZ RATE THIS QUESTION AND SHARE YOUR FEELINGS ABOU IT..........

[b] @ shibin.george [/b]: Okay, I have developed the code for it, and here it is:

[code]

#include <iostream.h>
#include <conio.h>
#include <string.h>
using namespace std;
void main()
{
string a= "John";
string b= "Smith";
a.insert(4,"."); // This statement inserts "." at the 4th position in string a //
a.erase(1,3);  // This erases three characters of "string a" starting from 'O'//
string c=a+b; // concatenation of Strings 'a' and 'b' //
puts(c);        // Printing the final string, you can also use "cout" too //
getch();
}

[/code]

[b] Output [/b]: J.Smith

This is a pretty easy programming problem, but you need to be aware of "insert","erase" operations that are part of the C++ string class (Which is a  part of the std namespace that allows you to manipulate strings safely)

If you intend to enter the strings at run-time, you need to be careful while using the "insert" and "erase" operations. (For example , if you enter the strings as-- Michael , and Slater, and you desire an output as -- M.Slater, you need to use insert and erase operation as follows:

[code]

a.insert(7,".");
a.erase(1,6);
string c=a+b;
cout<<c;
getch();
}

[/code]

[b] Output [/b]: M.Slater

No good. This is hard-coding the string sizes and does not work for strings of arbitrary length. You don't even need to know the length. Just print the first character of the first name, followed by ".", followed by the entire last name. Simple as that.

(Actually, I lied. You do need to know the length to determine the size of the second substring.)

[code]
#include <iostream>
#include <string>

using namespace std;

void printShortName(string firstName, string lastName)
{
    cout << firstName[0] << "." << lastName << endl;
}

void printShortName(string fullName)
{
    int index = fullName.find(' ');
    printShortName(
        fullName.substr(0, index - 1),
        fullName.substr(index + 1, fullName.length() - index));
}

int main()
{
    printShortName("John Smith");
    printShortName("Leonhard Euler");
    // etc...
    return 0;
}[/code]

[b] @Yongyi781 [/b] : Actually, I was developing the C++ code for the given problem, so that when we enter a string at run-time , say, something like -- David West, the output would be -- D.West, before you came up with your idea.

I really can't understand how you could so [b] rudely [/b] (I'm construing your remark -- "No Good" to be unintentional though) dismiss my program as being "No Good", I know for sure , my code works really fine if the length of the string is static. But since the OP didn't explicitly mention whether the string should be entered before run-time or after, I thought of first posting the code assuming that the "string" is mentioned before run-time.

I think you had developed the code for "strings of arbitrary size" which I feel is good, but it would have been much better had you desisted from using that obnoxious, snooty remark.(i.e.-- regarding my program)

sorry @gen8, but i just gave JOHN SMITH as an example....
my intention is that this code should sould work for all types of names of any length and any characters......
pleeaase,make the necessary corrections......

[quote="Gen8"](I'm construing your remark -- "No Good" to be unintentional though)[/quote]

No problem. It was unintentional that it would be construed as rude :)

But I believe that a good program should be generalizable, like a good math problem should be, unless there is a compelling reason not to (such as performance, etc).

shibin.george: I have posted my code. Does it work?


Harvard
college
Yale
Princeton
geometry
geometric transformation
reflection

[quote="qweretyq"]I do not have too much money but I don't think I will recieve all that much financial aid plus I am looking to go through grad school so I was wondering what the best colleges for your money were?
I am looking for great academics as well as the money...
so colleges like Harvard and Yale are out because of the money[/quote]

I have been said (by Harvard Alumni) that if you happen to be admited in Harvard, the money is not really a problem, that a big percentage (greater than 90%) receive BIG waivers, otherwise they couldn't study there. I haven't checked the facts, but otherwise, I doubt that a person I know, with a "Descent Latin American Salary" (~$30K a year) could have afford to have both of his sons at Harvard at once!

qweretyq says, "...but I don't think I will recieve all that much financial aid plus I am looking to go through grad school..."

If you're planning to go to grad school in mathematics or almost any of the sciences, then grad school (Ph.D. programs and a few types of masters' programs) is essentially free.  Usually a program which admits you will supply a tuition waiver and some sort of stipend; the stipend may be in exchange for part-time teaching/grading/etc. or for doing research, depending on whether you're a US citizen or not and how much money the department has.  Stuff like health insurance is also often covered, or at least subsidized.

On the other hand, if you're talking about professional (business, law, med) school, you're going to end up paying a bundle.

One thing to point out, though: if you don't have much money, you'd almost certainly be [i]better[/i] off going to Harvard or Yale than to schools that cost, say, USD25,000 instead of USD40,000 per year.  This is because they actually have the endowments to give financial aid packages to many, many students that other schools can't necessarily afford.  (For Harvard, at least, all financial aid is need-based rather than academic scholarships, so you don't have to compete with the rest of the student body for the money.)

That said, of course, not everyone who needs a good financial aid package will get into Harvard or Yale.  Depending where in the country you are (or, I suppose, what country you are in), local state schools often have programs that make them very economically attractive to bright students, so you can get simultaneously a really cheap education (some will pay you to come if you are a nat'l merit finalist, for example) and a collection of bright kids in the same program as you.

I agree that you should not allow financial aid concerns to deter you from applying to top colleges.  Many offer generous packages.  Harvard, Princeton and Yale, for example, all recently eliminated the parental contribution for families earning less than $40,000 a year, and greatly reduced the contribution for families earning between $40,000 and $60,000.

Financial aid offices are usually flexible; if the financial aid application does not accurately reflect your family's situation, write them a letter.  If you're accepted into your dream school but they offer less aid than another school, call up your dream school.  They might be able to match a competing offer.

You should consider your state school, especially if it offers an honors program.  Many selective colleges offer merit scholarships to top applicants.  Harvey Mudd and University of Chicago come to mind.  Rice University's base tuition is significantly lower than comparable schools; they're also generous with financial aid.  There are also many private scholarships you can apply for.

Pulling out my 2005 Fiske guide to colleges, it lists the following schools as "Bangs for the buck":

Among private colleges and universities:
Antioch
Bard
BYU
Centre College
Clark U
Cooper Union (well, duh, it's free)
Deep Springs
Earlham
Elon U
Grinnell
Hendrix
Kenyon
Lawrence U
Macalester
U. of Redlands
Rhodes College
Rice U
St. Olaf
University of the South (Sewanee)
Trinity U
Wake Forest U
Warren Wilson College
Washington and Lee U

and among public colleges and universities:
U. of British Columbia *
U. of Florida
U. of Kansas
Mary Washington U.
UMass
McGill *
UNC Asheville
UNC Chapel Hill
U. of Oregon
Queens University *
SUNY Bing +
SUNY Albany +
SUNY Geneseo +
SUNY Stony Brook +
U. of Texas
Texas A&M
U. of Toronto *
College of William and Mary
U. of Wisconsin - Madison

(*: Canadian schools)
(+: SUNY = State University of New York)

Notably missing from the public colleges are U. Michigan - Ann Arbor and U. California - many, which provide really really good educations and fall in a range of very cheap by private school standards but very pricy by public school standards.

cost don't matter bc when you get a nice job you can pay back the debt
if you don't go to a good school (you can save money) you will not get bigger salaries

[quote="qweretyq"]I do not have too much money but I don't think I will recieve all that much financial aid.[/quote]

If you don't have much money, you should get good financial aid. As JBL and others in the thread have pointed out, sometimes the very top schools, if you can get into them, actually have the best resources for providing financial aid. You should definitely apply to the schools you strongly desire to attend, after preparing well to be a strong candidate for admission, and then let the school show you what it can offer in a financial aid package. Don't ASSUME you won't get enough financial aid; apply to various schools and find out what you'll get.

[quote="tokenadult"][quote="DPatrick"][quote="juicybooty911"]if you don't go to a good school (you can save money) you will not get bigger salaries[/quote]This is most assuredly [b]not[/b] true.[/quote]

I suppose you have counterexamples in mind, but what are they?[/quote]
It's not that I have specific counterexamples in mind, it's more the certitude of juicybooty911's statement that I objected to: "If you don't go to a good school...you [b]will not[/b] get bigger salaries" (emphasis added).

My main point is that (in my opinion, of course) the difference between attending a so-called "top-tier" school (like Harvard/MIT/Caltech) and a so-called "second-tier" school (like many in the list that JBL posted) is much, much smaller than many people think.  I'm not saying it's nonexistant, but you can go to a state school and be rich and/or successful and/or happy (whichever of these is most important to you), and you can go to Harvard and get nothing out of it.

If you want a counterexample, how about Paul Allen?

How did Paul Allen meet Bill Gates?

Bill Gates dropped out of Harvard because he had a one-million-dollar trust fund from his grandfather already (from birth), so he had enough capital to found a business. He had also attended the top private school in the Seattle area, which at that time had a higher annual tuition bill than Harvard. So he is adequately prepared to self-educate as an adult, without a need for a college degree. But I don't know as many biographical details about Paul Allen.

You might want to take a look at the article [url=http://www.theatlantic.com/doc/200410/easterbrook]Who Needs Harvard?[/url] in the [i]Atlantic Monthly[/i].  It discusses a study that tracked students who were accepted into Ivy League schools but chose to attend less prestigious colleges.  Twenty years after graduation, these students were earning the same salaries as Ivy League grads.

I've heard of that conclusion, but I think that study badly needs replication before I'd bet my children's future on its conclusion.

New HP CEO Mark Hurd had a BS degree from Baylor. Stanford must be steaming mad that one of his graduates was not selected for a job in its own backyard.  :D

ok, here's a kicker, see if anyone can solve THIS one:
1. your parents make upper-middle class amounts of money.
2. they are only going to pay four years of state school tuition for you.
3. you would LIKE to go to a top-tier school.
4. you have potential to be accepted at said school.
5. however, because of #1, your financial aid is squat.
6. you also have SOME money saved up, which you are required to report when applying for financial aid, but have plans for this money to be spent on a trip you have been saving for for four years.
7. oh yes, you also want to attend a professonal school.
8. and your job will not pay large sums of money after you get out. so loans are not a good idea.
9. but you REALLY need/want to go to said professional school, because therein lies your deepest passion.

btw, tokenadult- i recognize your argument but i really don't think its valid at all. well, maybe it is, but i don't think it matters. do you think  your children are really going to be ALL THAT much happier if they are making, oh, i dunno, $10k more than "very-well-off" amounts? because i think regardless of where your children (or anyones) go to school, as long as it is a decent, respectable school, they are going to be making enough money to be happy. i just beg you not to make them think that ivies/top-tiers are the ONLY good schools out there. because [b]that[/b] is not true.  :)

[ [i]edited by moderator because double dollar sign was read as LaTeX code--no changes in substance of post[/i] ]

Well, always the toughest situation is to have a higher EFC (estimated family contribution) than your parents are willing to pay at a suitable school. Young people in that situation get my greatest sympathy, because I had to deal with that situation when I was college age. I couldn't obtain financial aid based on my own (meager) income, because my parents' income had to be counted, by law, to determine what *I* could afford to pay for college, but my parents weren't actually making their resources available to me for college expenses after a certain age.

Maybe in your situation you need to work backwards from what you will do when you have your professional school degree to pursue your passion. Is there any kind of loan forgiveness program or other way to erase student loan debt for someone who pursues that occupation? Does it matter, in that occupation, where a person attended college before starting professional school? Maybe you do need to minimize expense (and debt) up-front in your case to make achieving your goal possible. I would interview people who pursue that occupation now (something, by the way, that I stupidly DIDN'T do when I was your age, thinking about my dream occupation at the time) and find out what they would suggest.

Do any of the other participants here have suggestions for this situation?

I understand the desire to attend an elite college like H/Y/P/S/M/C.  Ears do perk up when I mention I attended one of these schools but ultimately you will be judged by who you are and what you accomplish.  Admittance to one of these schools is not an automatic ticket to anywhere.  At best they offer opportunities.  Such opportunities are also available at less selective schools.

Take a look at the most recent US Presidents.  Six of the last eight (the GB's being the exceptions) did not attend elite colleges as undergraduates : [list]
Bill Clinton (Georgetown U)
Ronald Reagan (Eureka College)
Jimmy Carter (US Naval Academy)
Gerald Ford (U Michigan)
Richard Nixon (Whittier College)
Lyndon Johnson (Southwest Texas State Teachers College)
[/list]

Take a look at the Forbes Top 10 list of richest Americans.  Yes, Bill Gates, the Harvard dropout, is at the top of the list but the rest did not attend elite colleges:
[list]
Warren Buffett (U Nebraska)
Paul Allen (Washington St U dropout)
Michael Dell (U Texas dropout)
Lawrence Ellison (U Illinois dropout)
[/list]
(The remaining slots are filled by five non-Ivy heirs of WalMart founder Sam Walton, who graduated from the University of Missouri.)

Even in academia, if you examine the CV's of math professors at Princeton and MIT, many attended Ivies as undergraduates, and many graduated from foreign universities.  The rest attended schools like U Chicago, U Maryland, U Colorado, U Arizona, Brandeis, and Purdue.  For an academic career, it's more important which graduate school you attend than which college.

It is a misconception to think that your Ivy League roommate's megarich dad will offer you a high-paying job upon graduation.  Today's Ivy admissions process is much more of a meritocracy than it was in the old days when prep schools fed their graduates straight into the elite schools. This is especially true if you hang out with math and science majors who typically come from middle class backgrounds.

I met intelligent, creative people at both the liberal arts college and big name university I attended.  The teaching was definitely superior at the small college.  Professors who excel at research do not necessarily excel at teaching.  In fact there's often an inverse correlation.

Elite schools have a lot to offer but there are many other excellent schools where you can get a quality education.  It's up to you to make the most of your environment wherever you are.

Bleumoose, if you end up attending your state school, it's not the end of the world.  If you want to shoot for your dream school, there are private scholarships available.  Books like [url=http://www.amazon.com/exec/obidos/ASIN/0060937653/qid=1112478928/sr=2-1/ref=pd_bbs_b_2_1/104-3755691-1763125]How to Go to College Almost for Free[/url] give tips on how to win merit-based scholarships.  I wouldn't worry about graduate/professional school for now.  Your interests may very well change.

So it would probably be best for me to excel as an honors student at the state school, which also offers a very good professional school in that field, instead of (maybe) entering an ivy league school which also offers a great professional school in that field, but costs 10x as much (no exaggeration) and is harder to get into (it is easier for a state resident to get into a professional school in THIS field)? because that is what all of the people that i have talked to did.

i guess i halfway know that's the right thing to do but when people you have topped academically and extracurricularly for so long are getting into top universities, and you are going to State U., it's still a little discouraging. one feels that one must be at the top of the game to "succeed."

on another note: does anyone know of any cheap, small-medium (15,000-) liberal arts school that have good science/biology programs? esp in the southeast? (sorry, off-topic?)

[quote="bleumoose"]ok, here's a kicker, see if anyone can solve THIS one:
1. your parents make upper-middle class amounts of money.
2. they are only going to pay four years of state school tuition for you.
3. you would LIKE to go to a top-tier school.
4. you have potential to be accepted at said school.
5. however, because of #1, your financial aid is squat.
6. you also have SOME money saved up, which you are required to report when applying for financial aid, but have plans for this money to be spent on a trip you have been saving for for four years.
7. oh yes, you also want to attend a professonal school.
8. and your job will not pay large sums of money after you get out. so loans are not a good idea.
9. but you REALLY need/want to go to said professional school, because therein lies your deepest passion.
[/quote]

My family was upper-middle class (roughly six figures annual income in canadian dollars) and I still got about 32000 in aid from princeton per year.  Trust me, (4) is by far the biggest hurdle on your list.  Of the people I know who got into Ivy League caliber universities last year (mainly the big three and one at MIT and one at Caltech), almost all got truly massive amounts of financial aid.  Mine was the smallest!

Wow.  Financial aid is more generous than I thought.  I recommend checking the college discussion websites to get a feel for which schools are generous with aid (both merit-based and need-based).  Many students share information about their awards.

Bleumoose,

I sense that the college decision choice for you is a bit more complicated than the money.  You mentioned science/biology... that sounds like you'd like to do pre-med?  And you also mentioned that your eventual job wouldn't make tons of money.  Just as an example, let's suppose you wanted to get an M.D. and then work in rural areas (which doesn't make much money).  You could go to an undergraduate institution which has an excellent acceptance rate for med school (there are lots, and they don't all appear to be top-tier), and then do a program (such as at Michigan State, or at least they used to have this) where your tuition etc. are cut if you do a special program which concentrates on doing medicine in rural area.

If you're willing to disclose a few more of the details of your dream (pm me if you'd like), we might be able to give you some more specific ideas on what to do.

But very generally, my philosophy is that you should always apply to the places you really want to go, and see what happens.  Maybe you'll get in, and maybe they'll give you lots of financial aid!  But if you don't apply, then they can't possibly give you financial aid.  Best not to cut off your options if you can avoid it.

it's veternary medicine. esp at a zoo/circus (exotics). they have programs for vets that cut tuition if you work in rural areas but i am unaware of any for zoo/circus vets. either way, however, one is still looking at a salary of <$50k, which wouldnt pay off large school loans. it is complicated and scary. i appreciate EVERYONE for your help and advice!

(People in the math department here sometimes have very strange conversations, so that's where I got this information.) If you get a student loan, you will have to pay back a certain amount of it each year until either it is paid off or a certain amount of time has expired (I think it's 20 years, but I'm not sure). The amount you pay each year depends on your salary. Therefore if you take out a big loan and get a low-paying job, you won't end up paying back the full loan. Furthermore, the loan will be interest-free (I think), and so it won't adjust for inflation.

In the US at least I don't think you will be able to find loans that are interest-free or forgivable after a certain number of years.  Student loans are generally interest-free during the time you're a student and then for a grace period after graduation.  After that the interest starts to accrue.

During the repayment period you can claim hardship (forbearance) to temporarily postpone or lessen payments but you still have to repay the full amount in the future.

Sometimes loans (up to a certain amount) will be forgiven if you choose to enter jobs like teaching, medicine or law in underserved areas.

Wow -- veterinary medicine with a zoo/exotic specialization is pretty difficult.  While in graduate school, I had a housemate for a year who was applying to veterinary schools; in fact, she was partially considering a zoo specialization!  As I recall, not all states have veterinary schools, and few veterinary schools have zoo/exotic specializations.  My impression at the time was that veterinary schools didn't seem to care much about the prestige of one's undergraduate institution.  They were much more interested in whether you were a resident of the state they were in, and in how well you'd done in your undergrad program, and in how much vet-tech volunteer experience you'd had.  Things might be different now, of course.  But to deal with the money factor, start looking at state residency requirements for the vet schools that have the zoo/exotic specialization; the way to make this affordable might be to move to such a state after college and work there as a vet assistant until you achieve residency.  (In-state tuition is usually significantly cheaper.)

[quote="ComplexZeta"]People in the math department here sometimes have very strange conversations, so that's where I got this information. . . . [ details omitted]  [/quote]

Most federal student loans in the United States have to be paid back in full, and I'm sure the same is true of most state loans and private college loans. Nor is low income in itself a basis for loan extension. The best reason to borrow money for higher education is to obtain work that allows a higher income, income sufficient to pay off the loan. Otherwise it is probably more shrewd to find a less expensive way to go to college.


inequalities
combinatorics unsolved
combinatorics

How many subsets of $\{1, 2, ... , 2n\}$ do not contain two numbers with sum $2n+1$?

We have $n$ pairs of the form $(k,2n + 1 - k)$ with $k \in \{1,2,\dots,n\}$. For each of these pairs, either $k$ belongs to the "chosen" subset, or $2n + 1 - K$ does, or neither $k$ nor $2n + 1 - k$ do. So, there are 3 "possibilities" for each pair, giving $3^n$ subsets :)

^very nice, but since I am bad at combo I use algebra. Partition $\{1, 2, ... , 2n\}$ into $A=\{1, 2, ... , n\}$ and $B=\{n+1, n+2, ... , 2n\}$ and let $S$ be a subset of $\{1, 2, ... , 2n\}$. There are $\dbinom{n}{k}$ ways to pick $k$ elements from $A$ to be contained in $S$, where $0 \le k \le n$. Let them be $a_1,a_2,...,a_k$. Now we cannot have in $S$ the elements $(2n+1)-a_1,(2n+1)-a_2,...,(2n+1)-a_k$, which are all contained in $B$. Choose to include the remaining $n-k$ elements in $B$ in $2^{n-k}$ ways. Thus there are $\dbinom{n}{k}2^{n-k}$ ways total and summing over all $k$ gives

\[\sum_{k=0}^{n}\dbinom{n}{k}2^{n-k}=\sum_{i=0}^{n}\dbinom{n}{i}2^i=(1+2)^n=3^n\]

where we used the substitutions $n-k=i$ and $\dbinom{n}{k}=\dbinom{n}{n-k}$, and in the second to last inequality we used the binomial theorem.


Call a positive real number special if it has a decimal representation that consists entirely of 0 and 7. For example, $\frac{700}{99}$=7.070707070707... and 77.007 are special numbers. What is the smallest n such that 1 can be written as a sum of n special numbers?

[hide]since 7 is not as easy to work with, you can divide everything by 7, making the problem with 0's and 1's summing up to 1/7.

1/7 = $0.\overline{142857}$

then we can subtract repeating sequences of 0's and 1's until we get 0. you want to subtract sequences with as many 1's as possible to minimize the n.

first, subtract $0.\overline{111111}$ (that makes 1 number) to get $0.\overline{031746}$.

then subtract $0.\overline{011111}$ (that makes 2 numbers) to get $0.\overline{020635}$.

then subtract $0.\overline{010111}$ twice (4 numbers so far) to get $0.\overline{000413}$.

then subtract $0.\overline{000111}$ (5 numbers so far) to get $0.\overline{000302}$.

then subtract $0.\overline{000101}$ twice (7 numbers) to get $0.\overline{000100}$.

finally, subtract $0.\overline{000100}$ to get 0.

we subtracted 8 numbers (meaning the sum of the 8 numbers = 1/7), so the answer is [b]8[/b]?[/hide]

That's a very interesting way to do it :)


algebra
polynomial
LaTeX
real analysis
real analysis unsolved

In the following  $T_0,T_1,...,T_n,...$  are [b] Chebychev polynomials[/b] of first kind, that is $T_0(x)=1 , T_1(x)=x$  and
 \[ T_{n+1}(x)=2xT_n(x)-T_{n-1}(x) \; \; \; \mbox{for}\; \; n\in {\mathbb N}^* .  \]
$l_0(x),l_1(x),...,l_n(x),...$ are [b] Lucas polynomials [/b] defined as  $l_0(x)=2 , l_1(x)=x ,$  and for $n \ge 1$
\[ l_{n+1}(x)=x l_n(x)+l_{n-1}(x) \; .  \]
It is supposed that a polynomial   $f$  [b] is given [/b], and  moreover      $\displaystyle f(x) =\displaystyle \sum\limits_{k=0}^{n} c_kT_k(x) \; .$
[b] Find a closed form for the sum [/b]    $(Tf)(x,y): = \sum\limits_{k=0}^{n} c_k l_k(x)l_k(y) .$

[quote="cimabue"]$Tf(x,y)=....$[/quote] Please take care how you  use LaTex: I do'nt understand !


[b][size=100][color=DarkBlue]Δίνεται τρίγωνο $ \bigtriangleup ABC$ και έστω $ D,$ του σημείο τομής του περιγεγραμμένου κύκλου του $ (O),$ από τη συμμετροδιάμεσο δια της κορυφής $ A.$ Εάν $ D'$ είναι το αντιδιαμετρικό σημείο του $ D,$ αποδείξτε ότι  η σημειοσειρά $ K\equiv BC\cap AD',\ B,\ L\equiv BC\cap DD',\ C,$ είναι αρμονική.[/color][/size][/b]

Κώστας Βήττας.

Λοιπόν το τετράπλευρο $ ABDC$ είναι αρμονικό επομένως $ (D'A,D'B,D'D,D'C)\equal{}\minus{}1$ πάρουμε την τομή αυτής της δέσμης με την $ BC$ παίρνουμε το ζητούμενο .

**Συγχωρέστε με για τυχόν λάθη γιατίείμαι σε internet cafe και δεν έχω σχήμα


group theory
abstract algebra
superior algebra
superior algebra unsolved

hi

what's the dual of $ T$, the order 12 rotational subgroup of $ T_d$, the tetrahedral group? dual group is defined as the group of all homomorphisms from $ T$ to $ \mu(\mathbb{C})$.

thanks

It's $ Hom(T,\mathbb{C}^x)\equal{}Hom(T/T',\mathbb{C}^x)\cong T/T'\cong \mathbb{Z}/3$.

i'm really sorry i cannot understand. what's $ T/T'$? and why is it isomorphic to $ \mathbb{Z}/3\mathbb{Z}$?


[quote="AndrewTom"]Consider the pair of four-digit positive integers

$ (M, N) \equal{} (3600, 2500)$.

Notice that $ M$ and $ N$ are both perfect squares, with equal digits in two places, and differing digits in the remaining two places. Moreover, when the digits differ, the digit in $ M$ is exactly one greater than the corresponding digit in $ N$.

Find all pairs of four-digit positive integers $ (M, N)$ with these properties.[/quote]
So $ x^2\minus{}y^2\in\{11,101,110,1001,1010,1100\}$. from which we'll only keep $ \{11,101,1001,1100\}$ since $ x^2\minus{}y^2$ cant be $ 2$ mod 4.

$ x^2\minus{}y^2\equal{}11$ $ \implies$ $ (x\plus{}y,x\minus{}y)\in\{(11,1)\}$ and so $ (x,y)\in\{(6,5)\}$ so no solution (no 4 digits squares)
$ x^2\minus{}y^2\equal{}101$ $ \implies$ $ (x\plus{}y,x\minus{}y)\in\{(101,1)\}$ and so $ (x,y)\in\{(51,50)\}$ which is a solution
$ x^2\minus{}y^2\equal{}1001$ $ \implies$ $ (x\plus{}y,x\minus{}y)\in\{(1001,1),(143,7),(91,11),(77,13)\}$ and so $ (x,y)\in\{(501,500),(75,68),(51,40),(45,32)\}$ from which only $ \{(75,68),(51,40),(45,32)\}$ are solutions.
$ x^2\minus{}y^2\equal{}1100$ $ \implies$ $ (x\plus{}y,x\minus{}y)\in\{(550,2),(110,10),(50,22)\}$ (I kept only couple where elements are both odd or both even) and so $ (x,y)\in\{(276,274),(60,50),(36,14)\}$ from which only $ \{(60,50)\}$ is solution.

Hence the answer : $ \boxed{(M,N)\in\{2601,2500),(5625,4624),(2601,1600),(2025,1024),(3600,2500)\}}$


(Me) For some positive real $ x,k$,
\[ x^5 \plus{} \frac {1}{x^5} \equal{} k^5 \minus{} 5k^3 \plus{} 5k \equal{} \frac {33}{4 \sqrt {2}}
\]
If $ k \equal{} x \plus{} \frac {1}{x}$ and $ x \equal{} \sqrt {a}$ where $ a$ is a positive integer not divisible by square of any prime, find $ a^{a^a}$.

Level: Medium Hidden Discoveries Mathematics Challenge (equivalent to AIME)

Erm, it looks like this problem is missing something.  Like a way to determine $ a$. :P

When I plug in $ k\equal{}x\plus{}\frac{1}{x}$ on the RHS, I get $ x^5\plus{}\frac{1}{x^5}\equal{}x^5\plus{}\frac{1}{x^5}$, so the equation works for all values of $ x$...?

Ah, I forgot to put the last part to this question (I had it but I forgot).

I fixed it now.

[hide="Easy bounding solution"]
We can calculate that $ \frac {33}{4 \sqrt {2}} \approx 5.8 < 6$.  So
\[ a^2\sqrt {a} \plus{} \frac {1}{a^2 \sqrt {a}} < 6
\]
But the only integers $ a$ such that $ a^2 < 6$ are $ a \equal{} 1$ and $ 2$.  We check and see that $ (\sqrt {2})^5 \plus{} (1/\sqrt {2})^5 \equal{} \frac {33}{4 \sqrt {2}}$.  So 2^(2^2) = 16.[/hide]

[hide="Even easier"]
3^(3^3) is too big to calculate without a calculator, so 2^(2^2) = 16.  QED :P[/hide]

[hide="Or since a is squarefree..."]
\[ a^2 \sqrt {a} \plus{} \frac {\sqrt {a}}{a^3} \equal{} \frac {33}{8} \cdot \sqrt {2}
\]

\[ \frac {a^2 \plus{} 1/a^3}{33/8} \equal{} \sqrt {2/a}
\]
The left hand side is rational.  Therefore $ 2/a$ must be the square of a rational number, so $ a \equal{} 2\cdot b^2$ where $ b$ is an integer.  However, $ a$ is squarefree, and so $ a \equal{} 2$.[/hide]

[quote="Altheman"]What is the purpose of $ k$ in this problem?[/quote]

Red herring, I guess?  Though it's pretty hard to get led off track by it; probably why he's not using the question! :)

\[ \left(x^5 \plus{} x^{ \minus{} 5}\right)^2 \equal{} a^5 \plus{} a^{ \minus{} 5} \plus{} 2 \equal{} \frac {33^2}{32}
\]

\[ a^{10} \minus{} \frac {1025}{32}\cdot a^5 \plus{} 1 \equal{} 0
\]

\[ a^5(32a^5 \minus{} 1025) \equal{} \minus{} 32
\]
Since $ a$ is an integer, a fifth power of some integer divides the LHS, so it must divide the RHS. It is clear then that, necessarily, $ a\in \{ \minus{} 2, \minus{} 1,1,2\}$. Checking gives $ a \equal{} 2$.

[quote="Altheman"]What is the purpose of $ k$ in this problem?[/quote]

Well, I pondered on whether to put $ x^5 \plus{} \frac{1}{x^5}$ or not. Obviously, having that made the problem very easy.

But, if you were to just solve the problem like this:

$ k^5 \minus{} 5k^3 \plus{} 5k \equal{} \frac {33}{4 \sqrt {2}}$

It is not possible to "get lucky" on this. But then, only with this form, it is practically impossible to solve this problem. The only way is if you're familiar with going up higher exponents from $ x \plus{} \frac{1}{x}$ (which is how I came up with $ k^5 \minus{} 5k^3 \plus{} 5k$) but still, this is not what AMC/AIME will likely to test people. It is not just a very approachable problem as most AMC problems are so I decided to not put on the contest.


What is the thousands digit of $ 9\cdot37^{2}\cdot77\cdot91\cdot143$?

Yes, there's an easier way to do this.

EDIT: I added my signature

[hide="Umm...lets see"]

Answer: 4

Yeah, all i did was type the numbers into my TI-89... :rotfl:  :blush:

[/hide]

Note that that whole thing is equal to $ (3\cdot37\cdot7\cdot11\cdot13)^{2}$. Then you pull out your 1337 bag and notice that $ 3\cdot37=111$ and $ 7\cdot11\cdot13=1001$ like we all have memorized, then we reach further into our 1337 bag, and remember that when we multiply a three digit number $ \overline{abc}$ by 1001 the product is $ \overline{abcabc}$, which we also all have memorized. Then we find find that the quantity we want is just $ (111111)^{2}$, which, by another part of our 1337 bag, is just 12345654321, which we, also have memorized. But if we don't have that memorized, we note that if we were to long multiply $ 111111\cdot111111$ the traditional way, when we add at the end, we would have one 1, then two one's, then a stack of three one's, then four, then five, then six, then five, back down to one 1.

Hence the thousands digit is 4.

As you can see, then 1337 bag was very helpful in this problem.


calculus
derivative
analytic geometry
graphing lines
slope
trigonometry
calculus computations

an equation of the line tangent to the graph of y=cos(2x) at x=pi/4 is?

What are you having trouble with? Use a derivative to get your slope, then you have a slope and a point, so you can find the equation.

$ y'\equal{}\minus{}2\sin{2x}$.  Plug in $ x\equal{}\frac{\pi}{4}$ to get $ m\equal{}\minus{}2\sin{\pi}/{2}\equal{}\minus{}2$.  Then find the equation from
$ y_{1}\minus{}y_{0}\equal{}m(x_{1}\minus{}x_{0})$
Take $ (x_{1},y_{1})\equal{}(\pi/4,0)$ so we get
$ \minus{}y\equal{}\minus{}2(\pi/4\minus{}x)\Rightarrow y\equal{}\frac{\pi}{2}\minus{}2x$


function algebra polynomial integration calculus derivative quadratics

$f(x)=x$ and $f(x)=0$ work. not sure about proving these are the only two functions.

[quote="franck"]My be there are general method to solve this equation , but l got simple idea ,
 if we write $f(x) = \sum\limits_{i = 0}^n {a_i x^i }$   as a polynomial , n must be  $\le 1$ cause if  n > 1  $\deg (F(f(x)) +(f(F(x)) \ge 6$   contradict   $\deg (x^2 ) = 2$
there are tow cases :
(1) $f(x) = ax + b$ substitute in the equation ,we get   a = 1  and  b = o and one of solution is  $f(x) = x$
(2) $f(x) = m$  and with substituting we get m= 0  and the other solution is  $f(x) = 0$[/quote]
Ok.If f is not polynomials function :D ??

csabafarkas, can you post the solution please?

[quote="franck"]ok!                  :D[/quote]

What is OK!! :D

[quote="franck"]What about this trying ,  :D

With checking we found that $f(x)=0$ is a solution  , otherwise ,

Let $g(x)=F(f(x))+f(F(x))=x^2$ ( * ) g is even , yielding $g(x)=g(-x)$

$F(f(x))+f(F(x))=F(-f(x))+f(-F(x))$

$F(f(x)) - F(-f(x)) = - (f(F(x))-f(-F(x)))$  --- ( ** )

Any function , so f ,  is odd or even or ( neither odd or even )

From ( ** ) We conclude that  ( 1 ) : $F(f(x)) = f(F(x)$  or  ( 2 ) :  $F(f(x)) = - f(F(x)$ ( the last case doesn't exist ( * )

Sub ( 1 ) in ( * ) we get  $F(f(x)) = \frac{{x^2 }}{2}$  or

$\int\limits_0^{f(x)} {f(t)dt = \frac{{x^2 }}{2}} = \int\limits_0^x {tdt}$

$\int\limits_0^0 {f(t)dt} = \int\limits_{f(x)}^x {tdt} \Rightarrow 0 = \frac{{x^2 }}{2} - \frac{{f^2 (x)}}{2}$

$f^2 (x) = x^2 \Rightarrow f(x) = \pm x$  ( by checking , we found the other solution is $f(x) = x$ ).[/quote]

I think is not ok!My opinion is $g(-x)=F(f(-x))+f(F(-x))$ :)

[quote="JBL"][quote="franck"]From ( ** ) We conclude that  ( 1 ) : $F(f(x)) = f(F(x)$  or  ( 2 ) :  $F(f(x)) = -f(F(x)$ ( the last case doesn't exist ( * )[/quote]
Where does this conclusion come from?

[quote="csabafarkas"]$F'(-x)=F'(x)$ and $F(x)=F(-x)+c ,F(0)=0$ so $F(-x)=F(x)$[/quote]
This is wrong -- if $F'(-x)=F'(x)$ then $F(x)=-F(-x)+c ,F(0)=0$ so $F(-x)=-F(x)$ -- the derivative of an odd function is even, and vice versa.[/quote]
yes Thanks JBL
 :blush:  :D  :!:

[quote="JBL"][quote="franck"]From ( ** ) We conclude that  ( 1 ) : $F(f(x)) = f(F(x)$  or  ( 2 ) :  $F(f(x)) = -f(F(x)$ ( the last case doesn't exist ( * )[/quote]
Where does this conclusion come from?[/quote]
I have conclude the last two solutions by notice  symmetrical between both sides of the equation .

 $F(f(x)) - F(-f(x)) = - (f(F(x))-f(-F(x)))$ --- ( ** )

                                            $F(f(x)) = f(F(x))$   exist (**)

 sub in both sides $f(F(x))-F(-f(x))= - ( F(f(x)-f(-F( x))$

                                       so $F(f(x))-F(-f(x)) = -f(F(x)+f(-F(x))$

                                   $F(f(x))-F(-f(x))=-(f(F(x)-f(-F(x))$ as required

, another symmetric $F(f(x)) = - f(F(x))$  doesn't exist ( ** )

Jbl , l appreciate your opinion that you mentioned above ,Please tell me if there is falses , May be there are another solutions through another ideas but i achieved this .  :)

[quote="franck"] $F(f(x)) - F(-f(x)) = - (f(F(x))-f(-F(x)))$ --- ( ** )

                                            $F(f(x)) = f(F(x))$   exist (**)[/quote]

I am confused about this step.  Perhaps I am missing something obvious.  Also, it's not clear to me: you are saying that there exists some value $x_0$ such that $f(F(x_0)) = F(f(x_0))$, right?  Then, you make a substitution into the equation (**) and end up pulling out the equation (**) again ...  I don't think I know what you're proving.  Perhaps you can add more explanation to your steps?

A suggestion that this question may be very very hard:
If we drop the condition $F(0) = 0$ and try to solve instead the equation $f(F(x)) + F(f(x)) = x^2 + (3 + 3\sqrt5)x -\frac{15+21\sqrt5}2$, we would find not only a quadratic solution for $F$ but also a cubic.  Now, it is possible that dropping $F(0) = 0$ increases the wildness of the solution, and it's possible that changing the quadratic we're trying to get increases the wildness of the solution set, but I think this suggests that this question could have a great many solutions we won't find.

Franck teh solution is not correct... :(

[quote="franck"][quote="JBL"]This problem looks really hard to me :)
I don't know why $f(x) = 0$ has been mentioned, though -- it's clearly not a solution of the original equation.[/quote]
As JBL said : $f(x) = 0$ doesn't exist the equation  because let $f(x)=0 \Rightarrow F(x) = c \,\, \forall x \in R$  , But  $F(0)=0$ , So $F(x)=0 \,\, \forall x \in R$ , substitute in ( * ) we find that  $0=x^2 \,\,\forall x \in R$  which isn't true .[/quote]
From this result , we see that  $f(x) = 0 \Rightarrow x = 0 , f(0)=0$
I will (try) to proof that $f(x)=x$ is the only solution on $R$ ,
Let ${f(x)}>x$ in some interval $( a , b ) \subset R^+$  ,  $\frac{{f(x)}}{x} = \frac{{f(x) - f(0)}}{{x - 0}} > 1$ on ( a,b) , take the limit when  $x \to 0$ , we obtain  $f'(x) > 1 > 0$ , So $f$ is strictly increasing on ( a, b) .
Now  \[ f(x) > x \Rightarrow f(f(x)) > f(x) \Rightarrow f(f(x))f'(x) > f(x)f'(x)  \]  on ( a,b)
Integrate both sides on any interval  $\subset ( a,b)$ , then :

\[ F(f(x)) > \frac{{f^2 (x)}}{2} > \frac{{x^2 }}{2} ........ ( 1 )  \]

Integrate both sides in $f(x)>x$ on ( a,b) we get  $F(x) > \frac{{x^2 }}{2}$  , since $f$is strictly increasing on ( a , b ) , then  \[ f(F(x)) > f(\frac{{x^2 }}{2}) > \frac{{x^2 }}{2} ......... ( 2 )  \]
Plug  ( 1 ) , ( 2 ) on the equation , we get $f(F(x)) + F(f(x)) > x^2$  , a contradiction ,
Same with $R^-$ , and the other case $f(x)<x$ .

Franck is not correct :(
1) we can take only one x where f(x)>x
2) f is not certanly derivable
 ;)

[quote="csabafarkas"]Determine all continous functions $f: R\to\mathbb R$ such that
  $F(f(x))+f(F(x))=x^2$, $\forall$ $x$ $\in$ $R$, where $F'(x)=f(x)$ ,$F(0)=0$.[/quote]

[quote="csabafarkas"]I have a problem:
1)f is not certanly dervable
2) we can take only one x where f(x)>x :)[/quote]

1)
we have given  $F'(x) = f(x) \forall x\in R$  , that is $F(x)$ is differentiable $\forall x\in R$  , now suppose for contrary that $F(f(x))$ isn't differentiable at $x = x_0 \in R$ , since $f(x) : R \to R , \exists y_0 \in R .s.t . f(x_0 ) = y_0$  implies that $F(x)$ isn't differentiable at $y_0=f(x_0)$ ( contradict that  $F(x)$ is differentiable $\forall x\in R$  )
we deduce that $F(f(x))$ is differentiable $\forall x\in R$  , so  $\left( {F(f(x))} \right)^\prime$ is defined as well  $\forall x\in R$  , in other words , $f(f(x)).f'(x)$ is exist  $\forall x\in R .....(I)$.
now , if we say that (  f is not certanly direvable ) then there are $x_0 \in R$ .s.t. $f'(x_0 )$ doesn't exist  ,which contradict .(I).

2) if $f(x) \ne x$ , let $f(x_0 ) > x_0$ , for one point $x = x_0 \in R$ , we claim that $f(x) > x$ in some neighborhood around $x_0$ , say  $(x_0 - \varepsilon ,x_0 + \varepsilon ) = (a,b)$
If isn't then $f(x_0 ) =y_0> x_0$ , while $\mathop {\lim }\limits_{x \to x_0 } f(x)\ne y_0$  , which contradict that $f$ is continuous on R as we search for .

[quote="csabafarkas"][quote="csabafarkas"][quote="franck"]csabafarkas , ?  :lol:[/quote]
:P[/quote] is not correct
by taking the limit with x tends to 0, we obtain the derivative in 0, and this doen't imply the monotonicity on the interval (a,b) doesn't[/quote]

franck? :D

[quote="csabafarkas"][/quote][quote="csabafarkas"][quote="csabafarkas"][quote="franck"]csabafarkas , ?  :lol:[/quote]
:P[/quote] is not correct
by taking the limit with x tends to 0, we obtain the derivative in 0, and this doen't imply the monotonicity on the interval (a,b) doesn't[/quote][quote="csabafarkas"]

franck? :D[/quote]

by taking the limit with x tends to 0 ( taking limit for what , what is the function you have limit for ? )
and i dont know what are you  talking about  :evilgrin:
can you give more details ? :roll:

[quote="franck"][/quote][quote="csabafarkas"][/quote][quote="csabafarkas"][quote="csabafarkas"][quote="franck"]csabafarkas , ?  :lol:[/quote]
:P[/quote] is not correct
by taking the limit with x tends to 0, we obtain the derivative in 0, and this doen't imply the monotonicity on the interval (a,b) doesn't[/quote][quote="csabafarkas"]

franck? :D[/quote][quote="franck"]

by taking the limit with x tends to 0 ( taking limit for what , what is the function you have limit for ? )
and i dont know what are you  talking about  :evilgrin:
can you give more details ? :roll:[/quote]
Sorry is not correct :!:


[quote="filomen"]If $ a,b,c > 0$, then

$ \frac {a^2}{a^2 \plus{} b^2} \plus{} \frac {b^2}{b^2 \plus{} c^2} \plus{} \frac {c^2}{c^2 \plus{} a^2}\leq\frac {1}{2}(\frac {a^2}{b^2} \plus{} \frac {b^2}{c^2} \plus{} \frac {c^2}{a^2})$[/quote]
Using the AM-GM Ineq, we have:
\[ \sum \frac{a^2}{a^2\plus{}b^2} \le \frac{1}{2}(\sum \frac{a}{b})\]
So that we need to prove:
\[ \sum \frac{a}{b} \le \sum \frac{a^2}{b^2}\]
This's a known result with $ a,b,c \in R$
 :lol:


Find a countable normal T1- Spape but not metrizable.

thanks in advance

Does countable mean countably infinite? If not, then the standard example of a $ T_1$ but not $ T_2$ will work, since every metrizable space is $ T_2$.

$ X\equal{}\{a, b\}$ and $ \mathcal{T}\equal{}\{\emptyset, \{a\}, \{a, b\}\}$.

it need to be [b]NORMAL[/b], your example is not

Oh sorry,  I misread that. I'd usually call normal + T_1 the $ T_4$ axiom.

This countable condition makes this quite hard. I have tons of good uncountable examples.

thanks for your attention

My idea is : make a normal space but not metrizable , and then take a countable subspace but I have a difficulty is : subspace of normal space [b]may not[/b] be normal .  I am still getting stuck

I got one!

My first instinct was to try the "Fort space." Unfortunately, the uncountable one works but the countable one does not. Here is a variant on that space that I think works. It is sort of complicated, though.

Let $ X\equal{}\mathbb{Z}^\plus{}$ (positive integers). Let $ N(n, E)$ be the number of integers in a set $ E\subset X$ that are less than or equal to $ n$. So for $ E\equal{}\{1, 3, 7, 27\}$, we have $ N(8, E)\equal{}3$.

Define a topology by declaring a set $ U$ open if it excludes 1, or if $ \lim_{n\to\infty}N(n, U)/n\equal{}1$.

So note that $ E$ is closed iff $ 1\in E$ or $ 1\notin E$ and $ \lim_{n\to\infty}N(n, E)/n\equal{}0$.

To get normal, break it into cases. If $ A$ and $ B$ are closed disjoint sets, and $ 1$ is not in either set, then both are open so we are done. So assume $ 1\in A$. Then since $ 1\notin B$, we have $ \lim_{n\to\infty} N(n, B)/n\equal{}0$, so $ B\supset B$ and $ X\setminus B\supset A$ are both open and we have normal.

$ T_1$ is trivial since if neither of your points are 1, the are both open sets. And if one of the points is 1, the other is not and it is open.

The tricky part is that $ X$ is not first countable, and hence not metrizable. So the claim is that $ 1$ has no countable local basis. Suppose it does, say $ \{B_n\}$. Since each $ B_n$ contains 1, it must be infinite. Choose $ x_n\in B_n$ such that $ x_n>e^n$. Then define $ U\equal{}X\setminus \{x_n\}$. Clearly $ U$ does not contain any $ B_n$. but it is a neighborhood of $ 1$ since $ N(n, U)>n\minus{}\ln n$, so $ \lim_{n\to\infty}N(n, U)/n\equal{}1$.

Hopefully that works :)

thanks, I am trying to understand your example