your inequality is equivalent to
\[
0\leq \sum \frac{(b+c)^2-4bc}{a^2(b+c)}=\sum\frac{(b-c)^2}{a^2(b+c)}
\]
which is obviously true.

Let $a,b,c>0$.Prove the following inequality\[ \sum \frac{b+c}{a^2}\geq 4\sum \frac{bc}{a^2(b+c)}\geq 2\sum \frac{1}{a}.\]

[quote="Cezar Lupu"]Let $a,b,c>0$.Prove the following inequality:

  $\sum\frac{b+c}{a^2}\geq 4\sum\frac{bc}{a^2(b+c)}$ ;)[/quote]
Your inequality is equivalent to
$\frac{(b+c)^2}{a^2}\geq\frac{4bc}{a^2(b+c)}$

$(b+c)^2\geq4bc$

Which is simple AM -GM!!

[quote="rachitgoel"][quote="Cezar Lupu"]Let $a,b,c>0$.Prove the following inequality:

  $\sum\frac{b+c}{a^2}\geq 4\sum\frac{bc}{a^2(b+c)}$ ;)[/quote]
Your inequality is equivalent to
$\frac{b+c}{a^2}\geq\frac{4bc}{a^2(b+c)}\iff (b+c)^2\geq4bc$

Which is simple AM -GM!![/quote]
Nice.
Please prove $2\sum \frac{bc}{a^2(b+c)}\geq \sum \frac{1}{a}.$

[quote="rachitgoel"][quote="sqing"]
Which is simple AM -GM!!
Nice.
Please prove $2\sum \frac{bc}{a^2(b+c)}\geq \sum \frac{1}{a}.$[/quote]

$2\sum \frac{bc}{a^2(b+c)}\geq \sum \frac{1}{a}\iff 2\sum\frac{b^2c^2}{a(b+c)}\geq\sum ab$

$2\sum\frac{b^2c^2}{a(b+c)}\geq\frac{2(\sum ab)^2}{2\sum ab}=\sum ab$.[/quote]
Good.
$2\sum \frac{bc}{a^2(b+c)}= \frac{2}{abc}\sum\frac{b^2c^2}{a(b+c)}\geq\frac{1}{abc}\frac{(\sum ab)^2}{\sum ab}=\sum \frac{1}{a}$

[url=http://www.artofproblemsolving.com/community/c6h545741p3156732]Stronger than Romania 2005:[/url]
Let $a,b,c>0$.Prove the following inequality$$\frac{b+c}{a^2}+\frac{c+a}{b^2}+\frac{c+a}{c^2}\geq4\left(\frac{bc}{a^2(b+c)}+\frac{ca}{b^2(c+a)}+\frac{ab}{c^2(a+b)}\right) \ge 2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)  .$$


Asymptote AMC USA(J)MO USAMO geometry incenter

Hi. I attempted to post an asymptote diagram in the topic "asy test 2" in the test forum. It was a rather complex diagram of USAMO #6. Here was the post.

[quote="Hamster1800"]Usamo #6:

[asy]size(200); real dist(pair a, pair b) { return sqrt((a.x-b.x)^2+(a.y-b.y)^2); } pair invert(pair center, real radius, pair p) { return (center+(p-center)*radius^2/dist(center,p)^2); } pair pa, pb, pc, qa, qb, qc; pair A = origin; pair B = (3,0); pair C = 2.5*dir(77); pair O = circumcenter(A,B,C); pair I = incenter(A,B,C); label("A",A,SW); label("B",B,SE); label("C",C,NW); path in = incircle(A,B,C); path cir = circumcircle(A,B,C); draw(A--B--C--cycle,linewidth(1)); draw(in,linewidth(1)); draw(cir,linewidth(1)); pair E = intersectionpoint(in,A--B); pair ta = intersectionpoint(in,I--interp(I,(I+O-A),10)); pair tt = (-(O-A).y,(O-A).x); pair tb = intersectionpoint(A--(interp(A,O,10)),interp(ta,ta-tt,-10)--interp(ta,(ta-tt),10)); ta = invert(A,dist(A,E),ta); tb = invert(A,dist(A,E),tb); draw(circumcircle(A,ta,tb),red+linewidth(1)); pair o = circumcenter(A,ta,tb); dot(o); label("$P_A$",o,E); pa = o; ta = intersectionpoint(in,I--interp(I,(I-O+A),10)); tb = intersectionpoint(A--interp(A,O,10),interp(ta,ta-tt,-10)--interp(ta,(ta-tt),10)); ta = invert(A,dist(A,E),ta); tb = invert(A,dist(A,E),tb); draw(circumcircle(A,ta,tb),red+linewidth(1)); o = circumcenter(A,ta,tb); dot(o); label("$Q_A$",o,E); qa = o; tt = (-(O-B).y,(O-B).x); ta = intersectionpoint(in,I--interp(I,(I+O-B),10)); tb = intersectionpoint(B--interp(B,O,2),interp(ta,ta-tt,-10)--interp(ta,(ta-tt),10)); ta = invert(B,dist(B,E),ta); tb = invert(B,dist(B,E),tb); draw(circumcircle(B,ta,tb),blue+linewidth(1)); o = circumcenter(B,ta,tb); dot(o); label("$P_B$",o,E); pb = o; ta = intersectionpoint(in,I--interp(I,(I-O+B),10)); tb = intersectionpoint(B--interp(B,O,2),interp(ta,ta-tt,-10)--interp(ta,(ta-tt),10)); ta = invert(B,dist(B,E),ta); tb = invert(B,dist(B,E),tb); draw(circumcircle(B,ta,tb),blue+linewidth(1)); o = circumcenter(B,ta,tb); dot(o); label("$Q_B$",o,E); qb = o; E = intersectionpoint(in,B--C); tt = (-(O-C).y,(O-C).x); ta = intersectionpoint(in,I--interp(I,(I+O-C),10)); tb = intersectionpoint(C--interp(C,O,2),interp(ta,ta-tt,-10)--interp(ta,(ta-tt),10)); ta = invert(C,dist(C,E),ta); tb = invert(C,dist(C,E),tb); draw(circumcircle(C,ta,tb),green+linewidth(1)); o = circumcenter(C,ta,tb); dot(o); label("$P_C$",o,E); pc = o; ta = intersectionpoint(in,I--interp(I,(I-O+C),10)); tb = intersectionpoint(C--interp(C,O,2),interp(ta,ta-tt,-10)--interp(ta,(ta-tt),10)); ta = invert(C,dist(C,E),ta); tb = invert(C,dist(C,E),tb); draw(circumcircle(C,ta,tb),green+linewidth(1)); o = circumcenter(C,ta,tb); dot(o); label("$Q_C$",o,E); qc = o; dot(I); label("$I$",I,E); dot(O); label("$O$",O,S); draw(pa--qa,linewidth(1.5)); draw(pb--qb,linewidth(1.5)); draw(pc--qc,linewidth(1.5));[/asy]

It doesn't like it because the code is too long? How long can it be?

Edit: just one pair of circles:

[asy]size(200); real dist(pair a, pair b) { return sqrt((a.x-b.x)^2+(a.y-b.y)^2); } pair invert(pair center, real radius, pair p) { return (center+(p-center)*radius^2/dist(center,p)^2); } pair pa, pb, pc, qa, qb, qc; pair A = origin; pair B = (3,0); pair C = 2.5*dir(77); pair O = circumcenter(A,B,C); pair I = incenter(A,B,C); label("A",A,SW); label("B",B,SE); label("C",C,NW); path in = incircle(A,B,C); path cir = circumcircle(A,B,C); draw(A--B--C--cycle,linewidth(1)); draw(in,linewidth(1)); draw(cir,linewidth(1)); pair E = intersectionpoint(in,A--B); pair ta = intersectionpoint(in,I--interp(I,(I+O-A),10)); pair tt = (-(O-A).y,(O-A).x); pair tb = intersectionpoint(A--(interp(A,O,10)),interp(ta,ta-tt,-10)--interp(ta,(ta-tt),10)); ta = invert(A,dist(A,E),ta); tb = invert(A,dist(A,E),tb); draw(circumcircle(A,ta,tb),red+linewidth(1)); pair o = circumcenter(A,ta,tb); dot(o); label("$P_A$",o,E); pa = o; ta = intersectionpoint(in,I--interp(I,(I-O+A),10)); tb = intersectionpoint(A--interp(A,O,10),interp(ta,ta-tt,-10)--interp(ta,(ta-tt),10)); ta = invert(A,dist(A,E),ta); tb = invert(A,dist(A,E),tb); draw(circumcircle(A,ta,tb),red+linewidth(1)); o = circumcenter(A,ta,tb); dot(o); label("$Q_A$",o,E); qa = o; dot(I); label("$I$",I,E); dot(O); label("$O$",O,S); draw(pa--qa,linewidth(1.5));[/asy]

Edit 2: Can I do 2 pairs of circles?

[asy]size(200); real dist(pair a, pair b) { return sqrt((a.x-b.x)^2+(a.y-b.y)^2); } pair invert(pair center, real radius, pair p) { return (center+(p-center)*radius^2/dist(center,p)^2); } pair pa, pb, pc, qa, qb, qc; pair A = origin; pair B = (3,0); pair C = 2.5*dir(77); pair O = circumcenter(A,B,C); pair I = incenter(A,B,C); label("A",A,SW); label("B",B,SE); label("C",C,NW); path in = incircle(A,B,C); path cir = circumcircle(A,B,C); draw(A--B--C--cycle,linewidth(1)); draw(in,linewidth(1)); draw(cir,linewidth(1)); pair E = intersectionpoint(in,A--B); pair ta = intersectionpoint(in,I--interp(I,(I+O-A),10)); pair tt = (-(O-A).y,(O-A).x); pair tb = intersectionpoint(A--(interp(A,O,10)),interp(ta,ta-tt,-10)--interp(ta,(ta-tt),10)); ta = invert(A,dist(A,E),ta); tb = invert(A,dist(A,E),tb); draw(circumcircle(A,ta,tb),red+linewidth(1)); pair o = circumcenter(A,ta,tb); dot(o); label("$P_A$",o,E); pa = o; ta = intersectionpoint(in,I--interp(I,(I-O+A),10)); tb = intersectionpoint(A--interp(A,O,10),interp(ta,ta-tt,-10)--interp(ta,(ta-tt),10)); ta = invert(A,dist(A,E),ta); tb = invert(A,dist(A,E),tb); draw(circumcircle(A,ta,tb),red+linewidth(1)); o = circumcenter(A,ta,tb); dot(o); label("$Q_A$",o,E); qa = o; tt = (-(O-B).y,(O-B).x); ta = intersectionpoint(in,I--interp(I,(I+O-B),10)); tb = intersectionpoint(B--interp(B,O,2),interp(ta,ta-tt,-10)--interp(ta,(ta-tt),10)); ta = invert(B,dist(B,E),ta); tb = invert(B,dist(B,E),tb); draw(circumcircle(B,ta,tb),blue+linewidth(1)); o = circumcenter(B,ta,tb); dot(o); label("$P_B$",o,E); pb = o; ta = intersectionpoint(in,I--interp(I,(I-O+B),10)); tb = intersectionpoint(B--interp(B,O,2),interp(ta,ta-tt,-10)--interp(ta,(ta-tt),10)); ta = invert(B,dist(B,E),ta); tb = invert(B,dist(B,E),tb); draw(circumcircle(B,ta,tb),blue+linewidth(1)); o = circumcenter(B,ta,tb); dot(o); label("$Q_B$",o,E); qb = o; dot(I); label("$I$",I,E); dot(O); label("$O$",O,S); draw(pa--qa,linewidth(1.5)); draw(pb--qb,linewidth(1.5));[/asy]

Guess so.

Edit 3: Making my code fewer lines

[asy]size(200); real dist(pair a, pair b) { return sqrt((a.x-b.x)^2+(a.y-b.y)^2); } pair invert(pair center, real radius, pair p) { return (center+(p-center)*radius^2/dist(center,p)^2); } pair o,pa, pb, pc, qa, qb, qc, A = origin, B = (3,0), C = 2.5*dir(77), O = circumcenter(A,B,C), I = incenter(A,B,C); label("A",A,SW); label("B",B,SE); label("C",C,NW); path in = incircle(A,B,C), cir=circumcircle(A,B,C); draw(A--B--C--cycle,linewidth(1)); draw(in,linewidth(1)); draw(cir,linewidth(1)); pair E = intersectionpoint(in,A--B), ta = intersectionpoint(in,I--interp(I,(I+O-A),10)), tt = (-(O-A).y,(O-A).x), tb = intersectionpoint(A--(interp(A,O,10)),interp(ta,ta-tt,-10)--interp(ta,(ta-tt),10)); ta = invert(A,dist(A,E),ta); tb = invert(A,dist(A,E),tb); draw(circumcircle(A,ta,tb),red+linewidth(1)); dot(pa = o = circumcenter(A,ta,tb)); label("$P_A$",o,E); ta = intersectionpoint(in,I--interp(I,(I-O+A),10)); tb = intersectionpoint(A--interp(A,O,10),interp(ta,ta-tt,-10)--interp(ta,(ta-tt),10)); ta = invert(A,dist(A,E),ta); tb = invert(A,dist(A,E),tb); draw(circumcircle(A,ta,tb),red+linewidth(1)); dot(qa = o = circumcenter(A,ta,tb)); label("$Q_A$",o,E); tt = (-(O-B).y,(O-B).x); ta = intersectionpoint(in,I--interp(I,(I+O-B),10)); tb = intersectionpoint(B--interp(B,O,2),interp(ta,ta-tt,-10)--interp(ta,(ta-tt),10)); ta = invert(B,dist(B,E),ta); tb = invert(B,dist(B,E),tb); draw(circumcircle(B,ta,tb),blue+linewidth(1)); dot(pb = o = circumcenter(B,ta,tb)); label("$P_B$",o,E); ta = intersectionpoint(in,I--interp(I,(I-O+B),10)); tb = intersectionpoint(B--interp(B,O,2),interp(ta,ta-tt,-10)--interp(ta,(ta-tt),10)); ta = invert(B,dist(B,E),ta); tb = invert(B,dist(B,E),tb); draw(circumcircle(B,ta,tb),blue+linewidth(1)); dot(qb = o = circumcenter(B,ta,tb)); label("$Q_B$",o,E); E = intersectionpoint(in,B--C); tt = (-(O-C).y,(O-C).x); ta = intersectionpoint(in,I--interp(I,(I+O-C),10)); tb = intersectionpoint(C--interp(C,O,2),interp(ta,ta-tt,-10)--interp(ta,(ta-tt),10)); ta = invert(C,dist(C,E),ta); tb = invert(C,dist(C,E),tb); draw(circumcircle(C,ta,tb),green+linewidth(1)); dot(pc = o = circumcenter(C,ta,tb)); label("$P_C$",o,E); ta = intersectionpoint(in,I--interp(I,(I-O+C),10)); tb = intersectionpoint(C--interp(C,O,2),interp(ta,ta-tt,-10)--interp(ta,(ta-tt),10)); ta = invert(C,dist(C,E),ta); tb = invert(C,dist(C,E),tb); draw(circumcircle(C,ta,tb),green+linewidth(1)); dot(qc = o = circumcenter(C,ta,tb)); label("$Q_C$",o,E); dot(I); label("$I$",I,E); dot(O); label("$O$",O,S); draw(pa--qa,linewidth(1.5)); draw(pb--qb,linewidth(1.5)); draw(pc--qc,linewidth(1.5));[/asy]

Still too long...I could abuse syntax more to make it shorter but it becomes unreadable then. Is there a reason for the limit to the length of code?[/quote]

As you can see, the two attempts at drawing the full diagram were rejected on the ground that the code was too long. I was wondering if you could specify what the limits for "too long" are. Thanks.

If I understood Valentin right, the limit is 2000 characters plus certain constructs are prohibited but most of the language is functioning as it should. As to this particular diagram, we have already discussed it in the test forum :), so I'm posting it here just in case somebody will read it here but not there.

[asy]size(200); real dist(pair a, pair b) { return sqrt((a.x-b.x)^2+(a.y-b.y)^2); } pair invert(pair center, real radius, pair p) { return (center+(p-center)*radius^2/dist(center,p)^2); } pair A, B,C, O, I, E,ta,tt,tb; pen q=linewidth(1), r=linewidth(1.5); A = origin; B = (3,0); C = 2.5*dir(77); O = circumcenter(A,B,C); I = incenter(A,B,C); label("A",A,SW); label("B",B,SE); label("C",C,NW); path in = incircle(A,B,C); path cir = circumcircle(A,B,C); draw(A--B--C--cycle,q); draw(in,q); draw(cir,q); E = intersectionpoint(in,A--B); pair routine(pair A, real sign, pen p,string lab) { ta = intersectionpoint(in,I--interp(I,(I+sign*(O-A)),10)); tt = (0,1)*(O-A); tb = intersectionpoint(A--(interp(A,O,10)),interp(ta,ta-tt,-10)--interp(ta,(ta-tt),10)); ta = invert(A,dist(A,E),ta); tb = invert(A,dist(A,E),tb); draw(circumcircle(A,ta,tb),p+q); pair o = circumcenter(A,ta,tb); dot(o); label(lab,o,plain.E); return o; } draw(routine(A,-1,red,"$P_A$")--routine(A,1,red,"$Q_A$"), r); draw(routine(B,-1,blue,"$P_B$")--routine(B,1,blue,"$Q_B$"), r); E = intersectionpoint(in,B--C); draw(routine(C,-1,green,"$P_C$")--routine(C,1,green,"$Q_C$"), r); dot(I); label("$I$",I,E); dot(O); label("$O$",O,S); void smile(pair center, real radius) { filldraw(circle(center,radius),yellow,brown); filldraw(circle(center+0.5*radius*dir(120),0.2*radius),blue,red); filldraw(circle(center+0.5*radius*dir(60),0.2*radius),blue,red); draw( (center+0.7*radius*dir(240))--(center+0.7*radius*dir(300)), red+linewidth(2) ); } smile((5,0),0.5); smile((5,1),0.3); // And so on, and so forth.[/asy]


[quote="thetruthisfake"]Below is attached file.

[url=http://img115.imageshack.us/i/22706787.jpg/][img]http://img115.imageshack.us/img115/9599/22706787.th.jpg[/img][/url][/url][/quote]

2) Let the three numbers selected by the first player be $ \ a,\ b,\ c$ such that $ \ a^2\minus{}4bc\equal{}\lambda^2,\ b^2\minus{}4ac\equal{}\mu^2,\ c^2\minus{}4ab\equal{}\gamma^2,$ where $ \ \lambda,\ \mu,\ \gamma,\in \mathbb Q/0$


binomial coefficients
number theory unsolved
number theory

[quote="8parks11"]The middle binomial coefficient is $ 2n \choose n$

Show that there innitely many values of n so that $ 2n \choose n$ is not divisible by
3, 5 or 7.[/quote]

Sorry for my question but I'm not sure to understand "[b]not [/b]divisible by 3, 5 [b]or [/b]7"

Do you mean : "Show that there are infinitely many values of n so that $ 2n \choose n$ is divisible neither by 3, neither by 5, nor by 7" ?

Or do you mean : "Show that there are infinitely many values of n so that $ 2n \choose n$ is not divisible by 3, or not divisible by 5, or not divisible by 7" (which would be quite simple to show) ?

To the best of my knowledge, this problem is [url=http://garden.irmacs.sfu.ca/?q=op/divisibility_of_central_binomial_coefficients]open[/url].  (pco: the former.)


In your opinion, (or from what you have heard is ok too), what math or science related jobs will we need within the next few decades and why?

bonus for being specific: i.e. electrical engineer instead of engineer
also bonus for justification or facts

[hide="digression"]I got in an argument with a pal the other day. He thought nanotechnology would become really important. I think it won't necessarily be much more important than it is today.

Don't get me wrong, (ex:I love MGS4) but I just don't see it happening like the way it does in the movies or in MGS to any extent... [/hide]

k how about the ones that we might be able to predict or already exist or will probably exist?

[quote="crazychinesemaniac"]MOst likely the science/math jobs that we will need in the future don't even exist and we have no idea what they will be.[/quote]
And we'll need creative minds to be pioneers in this field and lead the future? And with math and problem solving, we'll be able to develop such minds?
(Sorry, I've listened to Richard Rusczyk present this topic exactly twice and thought I'd say it. :P )


Show that $ P(x) \equal{}n.x^{n\minus{}1} \plus{}(n\minus{}1).x^{n\minus{}2}\plus{}.....\plus{} 2x\plus{}1$ is irreducible in $ Z[x]$ for any $ n \geq 2$.

i can't solve it , who can help me?

[hide="Hint"]Compute the sum $ \sum_{k\equal{}1}^{n}kx^{k\minus{}1}$
And it is lots of ways to do that ;)[/hide]

[quote="LideR_M"][hide="Hint"]Compute the sum $ \sum_{k = 1}^{n}kx^{k - 1}$
And it is lots of ways to do that ;)[/hide][/quote]
easy to check that the sum equal $ \frac {n.x^{n + 1} - (n + 1).x^n + 1} {(x - 1)^2}$

but i can't continue it ...(because i can't find $ %Error. "exit" is a bad command.
k$(or $ \no \exists$)such that when$ x: = x + k$ in expression can use with [b]extended  EISENSTEIN 's Criterion[/b] or use Complex number ...
? CAN you post your complete solution ?.


greatest common divisor
number theory
relatively prime
AMC

Source: British Math Olympiad, Round 2 (Thursday, February 26, 1998)

Suppose that x, y, and z are positive integers such that
1/x - 1/y = 1/z.

Let g be the greatest common factor of x, y, and z.

Prove that both gxyz and g(y-x) are perfect squares.

Hmm, for this question, do you actually have to find g in terms of x,y,z?

If not, then what do you do about g?

If so, how do you find g?

Basically, I'm beggin for a hint....

Well, showing that one's a square implies the other is a square since they differ by a factor of z^2. Is that enough of a hint? [hide]because I can't think of anything else[/hide]

um, i might just be being stupid again, but

did u answer my question?

I was kidding. Examine my last post very closely and all shall be revealed.

Hmm, i understand the hint, but i still don't know what to do with the g....

perhaps a hint targeted towards the g?

You want a hint for the g? Ok..:

[hide]Assume that g=1  If you look closely, proving it for g=1 will prove it for all g.[/hide]

so all u gotta prove is that y-x is a perfect square...

assuming that [hide]x,y,z have no common factor.[/hide]

yes.

well, let x=ga, y=gb, z=gc

Then, all u gotta prove is b-a is a perfect square

I think i got it!

[hide] By clearing denominators and rearranging, we get

x ( y + z) = yz,

which implies that x | yz and y+z | yz.

Now we show that (assuming WLOG that y>=z) y=az where a is a positive integer:

Assume that y= az + b.

Then y+z = ( a + 1) z + b, and the only way this divides yz is if b=0.

so y=az.

This means that (a+1)z | a z^2, or a+1 | az. Since a and a+1 are relatively prime, we have z= a+1, or a = z-1. Thus y = z ( z-1).

From this we get x ( z^2) = z^2 ( z-1), and  so x=z-1.

Thus

x=z-1

y=z(z-1)

and g=1, since g and g-1 are relatively prime.

gxyz= z^2 * (z-1)^2

g(y-x) = (z-1)^2

[/hide]

yay! hope that's right.

y is not necessarily a multiple of z. I believe the ordered triple (x,y,z)=(12,20,30) works. I'm trying to find the flaw.

hmmm...well, assuming y:ge:z is not WLOG, first of all....

Well yeah it is, if you rearrange it as 1/x = 1/y + 1/z. y and z can be interchanged (thus 12, 30, 20  is a solution too).

Actually it's not quite WLOG because the stuff you're trying to prove isn't symmetric.

A try for this one :

We have yz = x(y+z)  (*)

x=g*x'
y=g*y'
z=g*z'

gxyz= g^4 * x'y'z' and y'z' = x'(y'+z').

So we can suppose, wolog, g = 1.

xyz = x(y+z) so it suffices to prove (y+z) is square.

Let's start slowly  :) : let p prime such that p divides (y+z). Then by (*), p divides y or z. Wolog p divides y. Then p divides y, p divides (y+z) so p divides z.
Now p^2 divides x(y+z) and p doesn't divide x (because g = 1) so p^2 divides (y+z).

In the same vein. Let suppose p^(2k+1) divides (y+z) but p^(2k+2) does not divide (y+z).
By (*), p^a divides y, p^b divides z and a+b=(2k+1). Then a != b, wolog a < b. Then p^(a+b) divides p^a (1 + p^(a-b)) which is impossible (p does not divide 1 + p^(a-b)).

So (y+z) is divisible only by even powers of primes, which proves it is a square.

I didn't look for y-x, but since z(y-x)=xy it may be something similar.


algebra
polynomial
group theory
abstract algebra
induction
geometry
geometric transformation

Hello,

I'm trying to teach myself a bit of Galois theory, and I am highly interested in the problem of solvable and non solvable polynomials (by radicals)
A  simple exercise showed up in the course, and i can't solve it

proof that Sn, the group of n! permutations on the set {1,2,....n}
is generated by one n-cycle, and one transposition.

a n-cycle is a permutation of form (a1 a2 ...  an)
so a1 is sent to a2,a2 to a3, ....a(n-1) to an and an to a1 again

a transposition is an element of form (a b), so it only switches two elements, and leaves the rest alone

by generated i mean, the smallest subgroup of Sn containing a random ncycle and an random transposition, is Sn itself!

another formulation would be (as it is a finite group), can every element of Sn be written as a product of one random cycle and one random transposition.

What I already found was that the generated group is obviously transitive (every element of {1,2....n} can be sent to any other element) , so I was thinking of provng that the fixing group of some element in {1,2,....n} is the same, and thus doing some induction, but I doesn't seem to work.  Can anyone help?  It's a major obstruction for me in understand this theory?

Thx,

Frdric

another problem is: find the minimum number of transpositions that can generate Sn ?? There is a nice solution if i remember ..

Is the answer to this last question $n$? It certainly must be $\le n$, because if we show that an $n$ cycle and a transposition generate $S_n$, then the cycle $(1\ 2\ \ldots\ n)=(1\ 2)(1\ 3)\ldots(1\ n)$, so it can be obtained from $n-1$ transpositions, and if we add the other transposition then we get $n$. This means that $n$ transpositions are enough, but maybe there could be less than $n$ (I'm not sure right now).

i think it can be less than n, take S4
and take three transpositions :(12),(34),(13)

multiplying duo's gives quickly :

(12)(34)
(132)
(123)
(143)
(134)
now if we start conjugating :

(14)(132)(14)=(1)(432)
and so its inverse is also in it (423)

again : (34)(132)(34)=(142)
and so the inverse too : (124)
together with the identity, we get a subgroup with at least 13 elements in a group of 24 elements, so we have the full group

but, does anyone know how to prove that one ncycle and one transposition generate the full group?

the statement has to be wrong for a random transposition and random n-cycle: take n=4, n-cycle (1234) and transposition (13). the distance between 1 and 3 will always be 2, so we cant obtain e.g. (1324).
anyway, there is a n-cycle and transposition such that Sn is generated by them, take e.g. (123...n) and (12). now, first identify all elements which are just a cyclic rotation of another(so, one can get to other by multiplying with (123...n) some times). we want to show that every element of that quotient group can be generated(which gives the result immediately). now, we may arbritarily exchange two neighboring elements using the transposition (12)(remember, we may arbritarily rotate our numbers). now, first exchange neighboring numbers such that the first number is correct, then exchange such that the second number is correct etc., at the end the last number must be correct since it is the only one left, so we get the result.

Peter

btw, looking at grobber's post, we immediately see that n-1 transpositions suffices: just take (12), (13), ..., (1n).  now, if we want to generate $(a_1a_2a_3...a_n)$, just use first $(1a_n)$, then $(1n)$, so $a_n$ is the last number. now, use induction :) (too lazy to write the same step a thousand times :D ).

Peter

and here is the proof that it is actually n-1: obviously, there must be a way to send every number from (1,...,n) to any other number by the transpositions. now, with every transposition, we can reach at most one new number, so to reach all the other n-1 numbers beginning at 1, we need at least n-1 transpositions.
anyway, you don't need anything of this to understand galois theory, maybe you understand the Sn better, but you will understand galois theory without this exercise i guess.

Peter

that's odd
i am using the courses of a certain professor to be found at :

http://www.math.ucsb.edu/~mckernan/Teaching/02-03/Spring/111C/

in lecture 7 : http://www.math.ucsb.edu/~mckernan/Teaching/02-03/Spring/111C/l_7.pdf

in lemma 7.12 he explictly states the theorem  :?

it must however be noted , that on other sites, they only speak of Sp , with p prime, instead of Sn in general

because u c; in later lemmas he proves that the galois group of x^5-6*x+3 contains a 5-cycle and a transposition, and thus is S5, which is not solvable, which is in fact the goal i want to reach

so does anyone know then how to prove it for Sp, with p prime?

it is actually the same as my proof for (123...n) and (12): in fact we may assume that the n-cycle is (123...n) and let the transposition be (1m). now, we do the same as before looking at that quotient group. we want to show that we can get the transposition (12), so we can use my old proof. and there we got the transposition $(1m)(m(2m-1))((2m-1)(3m-2))...((km-k+1)(km+m-k))$(k arbritary) where the numbers are mod n, which equals the transposition $(1(km+m-k))$. now, we want $km+m-k=2$ (mod n) which is equivalent to $k=\frac{2-m}{m-1}$ (mod n), which is solvable($m\neq 1$) modulo a prime.

Peter

:(  hi

thx peter for all your efforts, but i'm afraid i still don't understand completely

i already found that the case (12....n) and (12) is somewhat special, and, with p prime, every power of (12...n) (except powers divisible by p) is also a pcycle as the exponent is coprime with p, and so every two elements can be made neighbouring (which would not so surely be the case voor n not prime)

though, i don't understand what you do exactly
what do you mean by 'first identify all elements which are cyclic rotations of another' , could you give an example
next you speak of quotient group, don't you need a normal subgroup N to speak of the quotient group Sn/N ?
and why would generation of this group imply generation of the full Sn

sorry for being such a pest, I highly appreciate the efforts in the topic by everyone so far, but I don't know of any other place to get help

thx,
fred

hi,

don't be afraid of posting something here, i'm lucky to solve some problems and help someone :)
anyway, sorry for posting my proofs so short.
actually i first consider the subgroup G of $S_n$ generated by $(123...n)$ and then look at the quotient group $S_n/G$. now, i show that we can generate every element of that quotient group using the transposition. thus, we can, for every a of $S_n/G$ obtain a representant of a in $S_n$. now, by using $(123...n)$, we may obtain ANY representant of a in $S_n$ since they differ just by a power of $(123...n)$ by definition of G.
hope you understand a bit better now...

Peter

hi

excuse me but i still don't really seem to figure it out

take n=5
the subgroup generated by just (12345) has only five elements right? (cyclic right?)
but it is well known i thought, that S5 only has A5 as nontrivial normal subgroup (has to do  with A5 being simple)
so we aren't allowed to take quotient groups i'm afraid?

thx,

fred

Right: in order for you to be able to do that, $G$ has to be normal, and it never is (except maybe for small values of $n$, like $n\le 4$; I didn't check). For all values $n\ge 5$, $A_n$ is simple, so the subgroup of $G$ formed by the even permutations is not normal in $S_n$ (not even in $A_n$).

sorry...my proof is not quite correct formulated. i'm always doing proofs quite intuitive in my mind and must then translate then and sometimes, the only thing we get is bullshit :D
soo...here is a new try(it's still the same idea, but now better formulated i hope): it suffices to show that using the n-cycle (123...p) and (1k) we may obtain any order of the numbers 1,2,3,...,p. so, we may rotate them cyclic and exchange the first with the k-th. now, rotating such that the m-th becomes the 1st, exchanging the 1st and k-th and rotating such that the 1st becomes the m-th, we can exchange the m-th and (m+k-1)-th number(numbers are mod p, so p+1 is the same 1 etc.). thus, we may exchange the 1st with the k-th, the k-th with (2k-1)-th and so on, till we exchange the (1+j(k-1))-th with the ((j+1)(k-1))-th number, j being arbritary. this is the same as exchanging the 1st with the (1+(j+1)(k-1))-th number. now we want to exchange the 1st with the 2nd number, so we need $1+(j+1)(k-1)=2$ modulo p, which means $j=\frac{1}{k-1}$ mod p, this is possible since $k\neq 1$. so, we can exchange the 1st and the 2nd, thus two neighboring numbers. Thus, starting with 1,2,3,...,p and wanting to end with the permutation $a_1,a_2,...,a_p$, we first exchange neighboring elements $a_p$ and $a_p+1$, $a_p+1$ and $a_p+2$, ..., $p-1$ and $p$, so $a_p$ is the last. now, do the same to get $a_{p-1}$ to be the second last and so on. we get the desired permutation.

it is completely combinatoric, so there are no problems with wrong use of the name quotient group or something like that any more(actually i could formulate the idea another way and it would be correct, but it is easier to do it purely combinatorial).

Peter

;)  okay i  am starting to believe, however, there are things i don't seem to get( a terminology problem again)

first of all
once you have proven all possible p-cycles are in G, does it follow that G=Sp because of the number of pcycles (p-1)!, and the  fact, that one supplemental transposition makes the number of elements in G more than (p-1)! ?
that's how i read it...

next, could you give an example of your first step? exchanging  m and m+k-1
it's a bit confusing, we are talking of conjugating numbers in cycles that are themselves permutations  :maybe:
so for instance, if i let you start with (12345) and (13) , are you saying you can use your first step
to get to (15342) switching 2 and 2+3=5 ?

sorry for being so slow and much thx to everyone in the topic

frederic

yeah, i hate that terminology problem. i always think of what we get if we put our permutation onto the numbers 1,2,3,...,p(or, for terminology, let's call them $x_1,x_2,...,x_p$). i then showed that we can get every ordering(i don't want to use permutation) of these. indeed, in my mind it is easier to see what our permutation has actually done then to think of the permutation itself. so, the p-cycle (123...p) rotates our numbers $x_i$. the transposition (1k) exchanges the first and the k-th of these $x_i$. so, for example my proof is like the following for p=5 and the tranposition (13):
we start with $x_1,x_2,x_3,x_4,x_5$. we may rotate these numbers by the p-cycle and we may always exchange the first and the third. now, if we want to exchange the second and the fourth, we do the following:
$x_1,x_2,x_3,x_4,x_5$->rotation ->$x_2,x_3,x_4,x_5,x_1$->transposition ->$x_4,x_3,x_2,x_5,x_1$->rotation->$x_1,x_4,x_3,x_2,x_5$.
similarly, we may exchange the third with the fifth, the fourth with the first and the fifth with the second number. thus, we may do the following:
$x_1,x_2,x_3,x_4,x_5$->exchange 1st and 3rd ->$x_3,x_2,x_1,x_4,x_5$->exchange 3rd and 5th ->$x_3,x_2,x_5,x_4,x_1$->exchange 5th and 2nd ->$x_3,x_1,x_5,x_4,x_2$->exchange 5th and 3rd ->$x_3,x_1,x_2,x_4,x_5$ ->exchange 1st and 3rd ->$x_2,x_1,x_3,x_4,x_5$,
so we can exchange the first and the second. now we may exchange two neighboring elements. so, if we want to get e.g. $x_3,x_5,x_4,x_2,x_1$, we first carry $x_1$ to the right, then $x_2$, then $x_4$ and finally $x_5$, like this:
$x_1,x_2,x_3,x_4,x_5$-> $x_2,x_1,x_3,x_4,x_5$-> $x_2,x_3,x_1,x_4,x_5$-> $x_2,x_3,x_4,x_1,x_5$-> $x_2,x_3,x_4,x_5,x_1$-> $x_3,x_2,x_4,x_5,x_1$-> $x_3,x_4,x_2,x_5,x_1$-> $x_3,x_4,x_5,x_2,x_1$-> $x_3,x_5,x_4,x_2,x_1$.
so, with our p-cycle and our transposition, we may order the numbers $x_1,x_2,x_3,x_4,x_5$ in an arbritary way(generalizing that idea).
i remarked that my proof is slightly wrong that we have the transposition of the 1st and the 2nd, but this may be corrected by doing the transpositions $(1k)$, $(k(2k-1))$, ..., $((1+j(k-1))(1+(j+1)(k-1)))$, $((1+(j-1)(k-1))(1+j(k-1)))$, ..., $(k(2k-1))$, $(1k)$ all after another, their composition actually is the transposition $(1(1+(j+1)(k-1)))$.

i'm soo sorry if you still can't understand...

Peter

:)
hi i believe u are correct but i've been thinking about i for a long time
i myself hate it when people nag about terminology, but misunderstandings are not nice either :
i thought u meant with (1,2,3,4,5) the PERMUTATION that sends 1 to two, 2 two three, 3 to four, etc...
and i thought you were manipuling THAT PERMUTATION in orde to get something else
by now i realise (?) that that just is some status, and that you intended to make TRANSPOSITIONS
(ab)

however, u are going  a bit fast when you say i 'switch the first and third',
if you want to go from (13) to the transposition, i would suggest conjugation :

(12345)(13)(15432)=(24)
(read the LHS from right to left, as we speak of compositions of actions)

first you have to make the object '2'->1 and the object '4'->3
in the known transposition 1->3  and 3->1
and then you have to do it back

anyway : you prove u can get the transposition (12) , your work modulo p , doesn't really need that two, you can in fact reach any (1a) with a<>1 right?
but rewriting (12345)=(45312) (1 isn't that important , it's just its name) we and using (a1) we can get all transpositions (ab) with b<>a , right?
so we have all transpositions and thus the full group?

yeah, you could do that as well, but then, you have to use that if you have all transpositions, then you got the whole group. anyway, my proof that we got (12) can be easily extended.

i'm happy that you do now understand :)

in real life, i'm writing better proofs but on mathlinks i'm sometimes thinking of too much problems at one time and just give very short formulated proofs where often some parts are missing or slightly wrong.

Peter


analytic geometry
logarithms
trigonometry

[quote="PenguinIntegral"]3. Given that the second coordinates of the points A(1,a), B(3,b), C(4,c) are

$a=-\frac{\sin39^\circ+\sin13^\circ}{\sin26^\circ\cdot\cos13^\circ},\qquad b=\sqrt{10^{2+\log_{10}25}},\qquad c= \left(\frac{1}{\sqrt{5}-2}\right)^3-\left( \frac{1}{\sqrt{5}+2}\right)^3$

determine whether the three points are collinear.[/quote]

[hide]
$c = \left(\frac{\sqrt{5} + 2}{3}\right)^3 - \left(\frac{\sqrt{5}-2}{3}\right)^3 = 76/27$
$b = \sqrt{2500} = 50$
$a = -\frac{\sin{3x}+\sin{x}}{\sin{2x}\sin{x}} = -\frac{2\sin{2x}\cos{x}}{\sin{x}\cos{x}} = -2\cot{x}$
$\cot{13^{\circ}}$ is positive therefore they're not all colinear.
[/hide]

well to find $a$ i used

$\sin (\alpha) + \sin (\beta) = 2\sin\left(\frac{\alpha+\beta}{2}\right)\cos\left(\frac{\alpha-\beta}{2}\right)$

so you still get $a=-2$.


Good day.!

This is the problem. I believe this has been used already as an Olympiad Problem. I just don't know whether this one was used as a Nat'l or an Int'l Problem... Here goes...

Help please.

For which m does the inequality mx^2 - 4mx + m^2 + 2m - 3 > 0 hold for all x?

Kindly include an explanation.. :D


function
real analysis
real analysis unsolved

Let $ f: R\minus{}>R$ o continuos function s.t $ f(x)\equal{}x$ don't have real solutions .Prove that the equation $ f^n(x)\equal{}x$ don't have real solution ($ n \in N$) ,where  $ f^n(x)\equal{}f(f(f(....(x)))$

Since $ f(x) \equal{} x$ has no real solutions, we have $ f(x) > x$ or $ f(x) < x$ for all $ x$.
If $ f(x) > x$ (or $ f(x) < x$), $ f^n(x) > f^{n \minus{} 1}(x) > \cdots > x$ (or $ f^n(x) < x$).
So $ f^n(x) \equal{} x$ has no real solutions.


Here's how to square a word, let's say word:

word
odor
rode
drew

Notice word, odor, rode, and drew appear twice.

Here's how to square two:

two
win
one

You get the idea, don't you?

If you do, try squaring {nose, dog, year, home, bean, percent}

I'll have more squares later...

Wow... how creative, randomdragoon.

DOG
ODE
GET

Is that how you do it?

nose
ores
sees
esse

home
ores
mets
esse

what's esse?

bubka
unary
bales
kreat
aysty

there. that's cool. only two new words.

[code]

w o r d
o d o r
r o d e
d r e w[/code]

[code]p e n i s
e l o p e
n o o o n
i p o d s
s e n s e
[/code]

yeah ignore the fact that nooon has one too many os

[size=200]I AM SHOCKED!!![/size] :|

bowl
ogre
wren
lens

Now I shall be bold:

[code]percent
emperor
rpximni
ceiling
ermines
nonnerd
trigsdy[/code]

nonnerd - somebody who is not a nerd
trigsdy - efficient in trigonometry
rpximni - no idea yet - any suggestions?

[code]s t o p
t o p s
o p t s
p s s t
[/code]

[code]m u s i c
u n i t y
s i n c c
i t c h y
c y c l e[/code]

only a small modification of the spelling of sink


analytic geometry
graphing lines
slope
rotation
geometry
geometric transformation
reflection

I was wondering how the following problems could be solved:

1.  If each point of the circle x^2+y^2=25 is reflected in the point (4,1), the set of image points satisfies the equation x^2 + ay^2 + bx + cy + d = 0.  Find a + b + c + d.

2.  The slopes of two lines through the origin are p and q, with p > q > 0.  The line y =x bisects the angle between the lines.  If p + q = root13, compute p-q.

3.  Let L1 be the line y = 2x.  The point p1(9,1) is reflected in L1 to P2.  L1 is rotated 90 degrees to the line L2 and P2 is reflected in L2 to the point P3(a, b).  Find a+b.   (A question I have is if lines L2 and L1 are paerpendicular.)

[hide]
(a) Let $ (x,y)$ be an arbitrary point on the circle. If we reflect this arbitrary point in the point $ (4,1)$, we obtain $ (x',y')$. We have $ \frac{x \plus{} x'}{2} \equal{} 4\implies x \equal{} 8 \minus{} x'$ and $ \frac{y \plus{} y'}{2} \equal{} 1\implies y \equal{} 2 \minus{} y'$. Substituting that into the equation, we have:

$ (8 \minus{} x')^2 \plus{} (2 \minus{} y')^2 \equal{} 25\\
x^2 \minus{} 16x \plus{} 64 \plus{} y^2 \minus{} 4y \plus{} 4 \equal{} 25\\
x^2 \plus{} y^2 \minus{} 16x \minus{} 4y \plus{} 43 \equal{} 0.$

(I switched from using $ x'$ and $ y'$ to $ x$ and $ y$ to make it easier to type.)

Thus, $ a \plus{} b \plus{} c \plus{} d \equal{} 1 \plus{} \minus{}16 \plus{} \minus{}4 \plus{} 43 \equal{} \boxed{24}$.

(b) The lines $ y \equal{} px$ and $ y \equal{} qx$ are inverses of each other. Therefore, if $ (x_0,y_0)$ satisfies $ y_0 \equal{} px_0$, then it must also satisfy $ x_0 \equal{} qy_0$. Thus, $ y_0 \equal{} pqy_0\implies pq \equal{} 1.$ We have

$ (p \plus{} q)^2 \equal{} p^2 \plus{} q^2 \plus{} 2pq \equal{} p^2 \plus{} q^2 \plus{} 2 \equal{} 13\implies p^2 \plus{} q^2 \equal{} 11\\
(p \minus{} q)^2 \equal{}  p^2 \plus{} q^2 \minus{} 2pq \equal{} 11 \minus{} 2 \equal{} 9\implies p \minus{} q \equal{} \boxed{3}.$

(c) Yes, $ L_1$ and $ L_2$ are perpendicular. I think you know how to reflect points across lines, so I'll leave the solution to you.
[/hide]

[quote="TachyonPulse"]$ (8 \minus{} x')^2 \plus{} (2 \minus{} y')^2 \equal{} 25$[/quote]

You know, at this point you could have just put $ x'\equal{}y'\equal{}1$ to get $ 1\plus{}a\plus{}b\plus{}c\plus{}d\equal{}7^2\plus{}1^2\minus{}25\equal{}25$, so $ a\plus{}b\plus{}c\plus{}d\equal{}25\minus{}1\equal{}\boxed{24}$.


calculus
integration
number theory
greatest common divisor
Ring Theory
superior algebra
superior algebra unsolved

What is a method to find out if $ GCD(14,1\plus{}4i\sqrt{3})$ or $ GCD(2\minus{}2i\sqrt{3},2\plus{}2i\sqrt{3})$ exist in $ \mathbb{Z}[i\sqrt{3}]$ ?

I'm going to interpret the problem like this:

Given two elements $ a$ and $ b$ in $ \mathbb{Z}[i\sqrt{3}]$ we understand their greatest common divisor to be the minimal ideal $ D$ such that $ a\subseteq D$ and $ b\subseteq D$.

With that understanding of the problem, we have the following:

$ \mathbb{Z}[i\sqrt{3}]$ is the integral closure of $ \mathbb{Z}$ in $ \mathbb{Q}(i\sqrt{3})$; in particular, it is a Dedekind domain. Thus we have that any ideal $ I$ factors uniquely into prime ideals. Factor the principal ideals $ (a)$ and $ (b)$ into primes. Compare the two factorizations and we can just pull out all the common prime ideals. The product of all the common prime ideals will give the minimal ideal containing $ (a)$ and $ (b)$.

I think the integral closure of $ \mathbb{Z}$ in $ \mathbb{Q}(i\sqrt {3})$ is $ \mathbb{Z}[\frac {1 \plus{} i\sqrt {3}}{2}]$ :)

[quote="LkNsngth"]

$ \mathbb{Z}[i\sqrt {3}]$ is the integral closure of $ \mathbb{Z}$ in $ \mathbb{Q}(i\sqrt {3})$; in particular, it is a Dedekind domain. [/quote]

$ \rm\text{Yes, post \;\#2 is incorrect. Instead, with} \;\; w \equal{} \sqrt{\minus{}3}, \; v \equal{} 1\plus{}4w
\\ \\
\text{1. Show principal ideal} \;\; (14,v) \equal{} (14\plus{}wv,v) \equal{} (2\plus{}w,v) \equal{} \; \text{...}
\\ \\
\text{2. Exists} \;\; d : \equal{} (2\plus{}2w,2\minus{}2w) \implies d \equal{} (2\plus{}2w,4) \;\; \text{so Euclid's Lemma (below)}
\\ \\
\text{yields} \;\; 2|(1\minus{}w)(1\plus{}w) \implies 2|1\plus{}w \; \text{in} \;\; \mathbb{Z}[w] \;\; \Rightarrow\Leftarrow
\\ \\
LEMMA \;\; (a,b) \equal{} (ac,bc)/c \;\;\; \text{if} \;\;\;  (ac,bc) \;\;  \text{exists}
\\ \\
\text{Proof: }  \;\;  d|a,b \iff dc|ac,bc \iff dc|(ac,bc) \iff d|(ac,bc)/c  \;\;\;\Box
\\ \\
\text{EUCLID'S LEMMA} \;\;  a|bc  \implies a|(a,b)c \;\;\ \text{if} \;\;\; (ac,bc) \;\; \text{exists}
\\ \\
\text{Proof: } \;\;   a|ac,bc \implies a|(ac,bc) \equal{} (a,b)c \;\; \text{by Lemma} \;\;\;\Box$

$ \rm\text{See also my post}$ http://google.com/groups?selm=gparte%24si4%241%40dizzy.math.ohio-state.edu

--Bill Dubuque

Keywords: Euclid's Lemma gcd exists principal ideal


[quote="cancer"]Each of the numbers $1,2,...,N$ is colored black or white. We are allowed to simultaneously change the colors of any three numbers in arithmetic progression. For which numbers $N$ can we always make all the numbers white?[/quote]

Nice question.
The answer is $\forall N\geq 8$

1) If is is possible for $N_{0}$, it is possible for $N_{0}+1$ and then for any $N>N_{0}$
It's obvious since you just have to change the three first position if the first is black and then it remains $N_{0}$ position to make all white, which is supposed possible.

So we are looking for the minimum $N$ for which it is always possible.

2) Minimum $N$ is $\geq 8$
Let $S$ be a set of independant transformations (change of 3 positions in arithmetic progression). Independant means that none of the transformation may be obtained by application of the others.
For $f,g\in S$, we have $f\circ g=g\circ f$ and $f\circ f=Id$
So the number of effective transformations using transformations in S is $\leq 2^{|S|}$
And since we want to cover $2^{N}$ cases, we need to have $|S|\geq N$

Then, notice that the transformations $(i,i+2,i+4)$ may be obtained from $(i,i+1,i+2)\circ(i+1,i+2,i+3)\circ(i+2,i+3,i+4)$ and so transformations with delta $2$ are not independant of transformations with delta $1$.
With a given N, we have at most $N-2$ transformations of delta 1.
Since transformations, with delta 2 are dependant, we need at least 2 transformations of delta $3$ plus the $N-2$ transformations of delta $1$ (assuming that transformations with delta $3$ are independant, which is not yet proven), and so we need $N\geq 7$ in order to have one transformation with delta 3, and $N\geq 8$ in order to have 2 of them.

So minimum $N$ is $\geq 8$

3) Minimum N is $8$
Let $white=0$ and $black=1$
Let $a_{1}$ to $a_{8}$ the $N$ colors of a given configuration.

If $a_{1}\oplus a_{3}\oplus a_{4}\oplus a_{6}\oplus a_{7}=1$ ($\oplus$ is exclusive or - XOR), then change $(a_{1},a_{4},a_{7})$ and you ave a new set $b_{1}$ to $b_{8}$ in which  $b_{1}\oplus b_{3}\oplus b_{4}\oplus b_{6}\oplus b_{7}=0$

Then if  $b_{1}\oplus b_{2}\oplus b_{4}\oplus b_{5}\oplus b_{7}\oplus b_{8}=1$, then change $(b_{2},b_{5},b_{8})$ and you ave a new set $c_{1}$ to $c_{8}$ in which  $c_{1}\oplus c_{3}\oplus c_{4}\oplus c_{6}\oplus c_{7}=0$ (the last transformation changed nothing to this value)  and $c_{1}\oplus c_{2}\oplus c_{4}\oplus c_{5}\oplus c_{7}\oplus c_{8}=0$

Then, if $c_{1}=1$, change $(position_{1},position_{2},position_{3})$ and you have a new configuration $(0,c_{1 }\oplus c_{2},c_{1}\oplus c_{3},c_{4},c_{5},c_{6},c_{7},c_{8})$
Then, if $c_{1}\oplus c_{2}=1$, change $(position_{2},position_{3},position_{4})$ and you have a new configuration $(0,0,c_{2}\oplus c_{3},c_{1}\oplus c_{2}\oplus c_{4},c_{5},c_{6},c_{7},c_{8})$
Then, if $c_{2}\oplus c_{3}=1$, change $(position_{3},position_{4},position_{5})$ and you have a new configuration $(0,0,0 ,c_{1}\oplus c_{3}\oplus c_{4},c_{2}\oplus c_{3}\oplus c_{5},c_{6},c_{7},c_{8})$
Then, if $c_{1}\oplus c_{3}\oplus c_{4}=1$, change $(position_{4},position_{5},position_{6})$ and you have a new configuration $(0,0,0,0 ,c_{1}c_{2}\oplus c_{4}\oplus c_{5},c_{1}\oplus c_{3}\oplus c_{4}\oplus c_{6},c_{7},c_{8})$
Then, if $c_{1}c_{2}\oplus c_{4}\oplus c_{5}=1$, change $(position_{5},position_{6},position_{7})$ and you have a new configuration $(0,0,0,0 ,0,c_{2}\oplus c_{3}\oplus c_{5}\oplus c_{6},c_{1}\oplus c_{2}\oplus c_{4}\oplus c_{5}\oplus c_{7},c_{8})$
Then, if $c_{2}\oplus c_{3}\oplus c_{5}\oplus c_{6}=1$, change $(position_{6},position_{7},position_{8})$ and you have a new configuration $(0,0,0,0 ,0,0,c_{1}\oplus c_{3}\oplus c_{4}\oplus c_{6}\oplus c_{7},c_{2}\oplus c_{3}\oplus c_{5}\oplus c_{6}\oplus c_{8})$

But $c_{1}\oplus c_{3}\oplus c_{4}\oplus c_{6}\oplus c_{7}=0$ thru previous steps
And $c_{1}\oplus c_{2}\oplus c_{4}\oplus c_{5}\oplus c_{7}\oplus c_{8}=0$ also thru previous steps, so $(c_{1}\oplus c_{3}\oplus c_{4}\oplus c_{6}\oplus c_{7})\oplus(c_{1}\oplus c_{2}\oplus c_{4}\oplus c_{5}\oplus c_{7}\oplus c_{8})=0$, and so $c_{2}\oplus c_{3}\oplus c_{5}\oplus c_{6}\oplus c_{8}=0$ and all is white.

Q.E.D.

[quote="cancer"]Please explain, what does your $\oplus$ mean?[/quote]

As I said, $\oplus$ means "exclusive or - XOR" : binary operation such that :
$0\oplus 0=0$
$0\oplus 1=1$
$1\oplus 0=1$
$1\oplus 1=0$

Many interesting properties of XOR exist :
$a\oplus(b\oplus c)=(a\oplus b)\oplus c$
$a\oplus a=0$
Changing position $i$ is just replacing $a_{i}$ by $a_{i}\oplus 1$

So, for example, an expression like "If $a_{1}\oplus a_{3}\oplus a_{4}\oplus a_{6}\oplus a_{7}=1$, ..." means : "if number of black numbers in $\{a_{1},a_{3},a_{4},a_{6},a_{7}\}$ is odd, ..."


Prove that the equation $\sqrt{x^{3}+y^{3}+z^{3}}=1969$ has not integral solutions.

Maybe start by showing that 11^2*179^2 can't be expressed as the sum of three cubes.

Working modulo 9 you see that 1969^2 = 4 (mod 9)

The only possible cubic residues mod 9 are 0,1,8. No sum of 3 of these numbers is equal to 4 mod 9 as required.


Find the number of ways to select [i][b]n[/b][/i] objects from [i][b]3n [/b][/i]objects of which [i][b]n[/b][/i] are identical and rest are different.

You don't need to double post to get attention.
http://www.mathlinks.ro/Forum/viewtopic.php?t=154260

Yep. The second time was a mistake.  :oops:


trigonometry
LaTeX
geometry
trig identities
Law of Sines

Given a triangle with $ a \equal{} 11$, $ b \equal{} 14$ and $ \alpha \equal{} 35$ which is (are) possible length(s) of c?

this is from my review packet for trig verbatum.  which angle does $ \alpha$ refer to??

my packet only has answers, no solutions, so a solution would also be nice.

btw im NOT in school.

[hide]Since the question implies that there can be more than one answer, we will apply the Law of Sines.  Furthermore, although $ 35$ has no degree sign, $ \alpha$ probably corresponds to the degrees, since radians would be really big.  I'm also guessing that $ \alpha$ corresponds to the angle across from $ a$. So, $ \frac {sinA}{a}\equal{} \frac {sinB}{b}\equal{} \frac {sinC}{c}$.  This allows us to conclude that $ B \approx 46.89 or 133.11$.  Angle $ C$ now has two possible values, $ \approx 11.89, 98.12$.  Using the Law of Sines again, we get $ \approx 18.99, 14$.[/hide]

btw, where is the degree sign for LaTex?  :huh:

[quote="azn6021023"]btw, where is the degree sign for LaTex?  :huh:[/quote]There is none.
However you can use
[code]^{\circ}[/code]as a degree sign. For example,
[code]$1337^{\circ}$[/code]makes $ 1337^{\circ}$.

Yes, generally $ \alpha$ is across from $ a$, $ \beta$ is across from $ b$, and $ \gamma$ is across from $ c$.

If we calculate $ b\sin\alpha$, we get

$ 14\sin35^{\circ}$

$ \approx 8.03$

$ a$ is greater than this, but less than $ b$, so there are two values for $ c$.

Now just solve using Law of Sines, just make sure you get both solutions!

[hide="b sin alpha"]
Earlier I calculated $ b\sin\alpha$. Here's why. Imagine you have the $ \alpha$, $ a$, and $ b$. Now arrange these in your mind so that $ \alpha$ is in the lower left corner, $ b$ is the line segment extending diagonally northeast from $ \alpha$. From the other endpoint of $ b$, begins $ a$. $ c$ is the bottom side of this almost complete right triangle. $ b\sin\alpha$ is the exact height needed for $ a$ so that this becomes a right triangle. If $ a$ is less than this value, there is no possible triangle. Even if we drop $ a$ down vertically, it will never touch $ c$. However, if $ a$ is exactly equal to $ b\sin\alpha$, then there is one possibility for $ c$, because we have just created a right triangle. If $ a$ is greater than $ b\sin\alpha$, but less than $ b$, then there are two possibilities for $ c$. From the given endpoint of $ a$, draw the arc of the given length of $ a$. There are exactly two places where it intersects $ c$. However, if $ a$ is also greater than $ b$, then it only intersects once.[/hide]

BTW: If someone could please make a diagram to show what I have just said, that would probably help.

[quote="azn6021023"][hide]Since the question implies that there can be more than one answer, we will apply the Law of Sines.  Furthermore, although $ 35$ has no degree sign, $ \alpha$ probably corresponds to the degrees, since radians would be really big.  I'm also guessing that $ \alpha$ corresponds to the angle across from $ a$. So, $ \frac {sinA}{a} \equal{} \frac {sinB}{b} \equal{} \frac {sinC}{c}$.  This allows us to conclude that $ B \approx 46.89 or 133.11$.  Angle $ C$ now has two possible values, $ \approx 11.89, 98.12$.  Using the Law of Sines again, we get $ \approx 18.99, 14$.[/hide]

btw, where is the degree sign for LaTex?  :huh:[/quote]

i get the solution. thanks. yeah 35 is degrees. didnt know how to make the degrees sign/


quadratics function algorithm parameterization rotation vector geometry

I have 4 Points in space. I want to make this 4 points lie in a plane.
How to make this?

My thought were, If those points are $ A,B,C,D$ I make $ P1,P2,P3,P4$ that lies in the plane $ z \equal{} a x \plus{} b y \plus{} c$.
Them minimize $ dis(A,P1) \plus{} \ldots$

But to find a global minimum with 11 variables is really slow.

Any Ideas?

Look for a plane $ ax\plus{}by\plus{}cz\plus{}d\equal{}0$ that is closest to the given points (in the least squares sense). Minimum of a quadratic function of four variables isn't hard to find. Once you have the plane, use orthogonal projection to move the points to the plane.

That's were my first thought
but the plane found in my method is different than the plane in your method (try to see)
with my method the points are at a closest distance (make sense, you are minimizing 11 variables, not only 3)

The amount of error is significant, since there is a lot of points changing recursively.

With my algorithm some points do not converges well, and the error is so evident, visually.

Well, the answer will depend on what "closest" means.

But in any case it should be easier to minimize the sum of squares than the sum of absolute values. That is, $ |A\minus{}P_1|^2\plus{}\dots \plus{}|D\minus{}P_4|^2$.

I don't know what's the best way to handle the constraint of coplanarity without making one of four points somehow distinguished. Maybe add a penalty $ M|\det |^2$ with large $ M$?  :maybe:

closest means here the sum of the squares of the norm.

But I think I found a solution (numeriaclly).
We fit the 4 points in the plane $ z \equal{} a x \plus{} b y \plus{} c$
The projection of the point $ (X,Y,Z)$ in this plane is $ \left\{\frac {\left(b^2 \plus{} 1\right) X \minus{} a (b Y \plus{} c \minus{} Z)}{a^2 \plus{} b^2 \plus{} 1},\frac {\left(a^2 \plus{} 1\right) Y \minus{} b (a X \plus{} c \minus{} Z)}{a^2 \plus{} b^2 \plus{} 1},\frac {\left(a^2 \plus{} b^2\right) Z \plus{} a X \plus{} b Y \plus{} c}{a^2 \plus{} b^2 \plus{} 1}\right\}$

Since this solution is very close to the minimization of the sum of the squares of the norm, it can be used as initial solution, and any method of local minimum can be used (althougth local it will find a global one)

This way the process is more numerically stable and converges well.

Edit:Using this new points and the original points as seeds, the difference in time of calculation is so low.
An my method is about 100 times slower.

I think is the best to use Least Squares and use a more smart way to make the planification

Let's imagine that we already have a solution: points P1..P4 lie in a plane S. Then it's clear that P1..P4 are the projections of A..D onto S. Therefore, the sum $ |P1\minus{}A|^2\plus{}\dots |P4\minus{}D|^4$ is the same as the sum of squares of distances from A..D to the plane S.

So we indeed should be looking for S (described by 3 parameters) instead of points P1..P4 (11 parameters).

When you do Least Squares  for $ z \equal{} a x \plus{} b y \plus{} c$, the sum that is minimized involves differences of z-coordinates. But we want distances to the plane, not z-differences. There must be a standard way of minimizing them. Let me know what it is  :roll:

yes. you are right.
I was using least squares in the z axis not in a "projection" axis to say..

Netherless, the $ dis(A,P1)$ is the actual distance (throught this projection axis) to the plane, the 2 cases are the same, isn't it?

But one have 11 variables and other only 3.
Minimizing the distance to the plane we find the plane and them project it.

Is need to minimize $ \sum _{i \equal{} 1}^n \frac {\left(a X_i \plus{} b Y_i \plus{} c \minus{} Z_i\right){}^2}{a^2 \plus{} b^2 \plus{} 1}$

Need not to say that minimizing 3 variables instead of 11 is 3.66 faster  :D;

But the general formula for $ n$ points is not linear as in the least squares in z axis, it's really a mess.

The no-linear characterisc of this minimization make it more difficult to find the minimum.

My implementation is more instable and slow with only 3 variables than 11, but I think this is just a bad implementation

The minimizing plane must pass through the "center of mass" of the four points. (To see this, rotate the space so that the minimizing plane is horizontal $ z \equal{} k$. Which value of $ k$ minimizes $ \sum (z_j \minus{} k)^2$? The mean. )

We can translate the points so that the center of mass is at the origin. Now we are down to normal vector $ (a,b,c)$ - if it's a unit vector, then the sum of distances is just $ \sum (ax_j \plus{} by_j \plus{} cz_j)^2$. ... Only two parameters left!

But instability is inherent in this problem. If the four points are the vertices of a regular tetrahedron, then you have,... um, six minimizers? And if the tetrahedron is slightly irregular, you may have one unstable minimizer.

only 2 parameters left!

I had to abandon this approch and go back to 11 variables.
I have a mesh of semi-planar quadrilaterals.
I make 4 points in a plane. Ok.
But, if I go to a adjacent quadrilateral and minimize 4 points, the other quadrilateram is not in a plane anymore.
So I need to add a constraint to make a point or more fixed and minimize with this.

And it is faster implemented with the other minimization, just adding constraints.

Oh. I suppose there's a reason why you can't use a triangular mesh.
Fixing the quadrilaterals one at a time, you may end up with a mesh that does not stay close to the original...


[size=150](Please visit link below to know what I am talking about!!!)

I am stuck on D. So far my Area for c is (x+14) (x+12) I need to get the answer for D. I can't figure it out. How do I plug the equation (x+14) (x+12)=2208?? Someone please help me

(Please visit link below to know what I am talking about!!!)

[url]http://hotmath.com/help/solutions/cpm02/8/9/Classwork/cpm02_8_9_Classwork_AP-79_294/index.html[/url]
[/size]

Thanks in advance!!! :surf:  :starwars:  :starwars:  :trampoline:  :thumbup:  :coolspeak:  :love:  :love:


Hi;

We define the arithmetic operation $\left\langle x \right\rangle$ as truncation of $x$ to one decimal place (so if $x$=3,81, then $\left\langle x \right\rangle$=3,8. If $y$=3,8, then also $\left\langle y \right\rangle$ =3,8 ). Are there unlimited pairs of positive integers $a,b$, for which these equations hold?
(1) $\left\langle \frac{a}{b}\right\rangle = \left\langle a+\frac{b}{10^{n}}\right\rangle$
(2) $\left\langle \frac{a}{b}\right\rangle = \left\langle b+\frac{a}{10^{n}}\right\rangle$

For (1) $n$ is the number of digits of$b$.

For (2) $n$ is the number of digits of $a$.

(1) should be quite easy, but I'm not sure of (2).

The first is easy: $LHS\leq a <RHS$ => no solutions.
For the second one: $a=10^{2t}+10^{t-1}$, $b=10^{t}$ generates infinitely many solutions, but there are another ones (5,2), (26,5) etc.


ceiling function
pigeonhole principle
combinatorics proposed
combinatorics

Let $ S$ be a set of $ n$ points in space($ n\ge 3$).The segments joining these points are of distinct length,and $ r$ of these segments are colored red.Let $ m$ be the smallest integer for which $ m\ge \frac{2r}{n}$.Prove that there always exists a path of $ m$ red segments with their lengths sorted increasingly

I have the following idea which I am not able to convert into a solution.

The following statement implies the statement in the problem : For  a graph with $ n$ vertices and $ r$ segments, there exists an path of length $ \big( \lceil \frac {2r}{n} \rceil \big)^2$

As it would follow from an theorem of Erdos that says - There is a monotonic subsequence of length $ n$, in any sequence of distinct numbers of length $ n^2$

Oh,you are thinking about Dilworth theorem :)
This is indeed a nice problem,and it has a two-line proof which is hard to come up with :roll:
If anyone wants it,I will post the solution.Anyway I still want to wait for a nice solution from mathlinkers :)

Can you post your solutions?hxy09.
It seems that you have a wonderful solution.

I didn't solve this problem :blush: but official solution is really nice,here we go:

Denote by the red segments $ L_1,...,L_r$ and $ l_1<...<l_r$ ($ l_k$ is the length of $ L_k$)

Put $ i$th robot on the vertex $ i$ ($ i\equal{}1,2,...n$)

In the $ j$th step,we exchange the position of robot on $ L_j$($ j\equal{}1,...r$)

By pigeonhole principle,one robot must have walked along a red path which included at least $ m$ red segments.

And the length of segments must sort increasingly. :D

QED


inequalities calculus function algebra domain logarithms integration

very wonderful problem:
$ \large \text{prove that: } \large e^{\pi}>\pi^e$

Very easy problem:
Our inequality is equivalent to $ \frac{\pi}{ln\pi}>e$
Consider $ f(x)\equal{}\frac{x}{lnx} \ x\in <e,\pi)$  (of course we know $ \pi >e$)
Observe that $ f'(x)\equal{}\frac{lnx\minus{}1}{(lnx)^{2}}$ which is greater than zero for $ x>e$
Thus the obvious inequality follows $ e\equal{}f(e)<f(\pi)\equal{}\frac{\pi}{ln\pi}$
Q.E.D. :wink:

yes Its very easy
[hide="HINT TO SOLVE IT {CALCULUS}"]
use tailor theom
[/hide]

(e^pi)>(pi^e)

to prove this inequality ,we can divide both raise both the terms to the exponent  1/e(pi).so

it is equivalent to prove the inequality
e^(1/e)>pi^(i/pi)

now use the function f(x)=x^(1/x)

the function has maxima at e^1/e.(you can verify that)

hence it is always true.
that
 e^1/e>x^1/x for all x belonging to the domain of f(x)

[quote="mathson"]very wonderful problem:
$ \large \text{prove that: } \large e^{\pi} > \pi^e$[/quote]

Consider the funtion

$ f(x)\equal{}\frac{x^{e}}{e^{x}}$ with $ x>e$

$ f'(x)\equal{}\frac{x^{e\minus{}1}(e\minus{}x)}{e^{x}}<0$ for every $ x>e$ so we have that $ f$ is strictly  decreasing.

We know that $ e<\pi \Longleftrightarrow f(e)>f(\pi)\Longleftrightarrow\frac{e^e}{e^e}>\frac{\pi^e}{e^\pi}\Longleftrightarrow e^{\pi} > \pi^e$

Consider the function $ f(x)\equal{}\frac{\ln x}{x}$ and examine the behaviour.

I think that we can use the inequality : $ e^x>1\plus{}x$ directly.

Is there a theorem that says whether $ x^y > y^x$ based on whether $ x > y$ whenever $ x > e$ and $ y > e$?

^My uncle actually has a problem similar to that in his book. I've already posted it on this forum so here is a link - http://www.artofproblemsolving.com/Forum/viewtopic.php?t=192267

Look at the the additional problem for #1

Denote $ \boxed {\cdot}\ \in\ \{\ < \ ,\ > \ \}$ . Therefore, $ \boxed {\ \begin{array}{ccc} \\
\ e^{\pi} & \boxed {\cdot} & \pi^e \\
 \\
\pi\ln e & \boxed {\cdot} & e\ln\pi \\
 \\
\frac {\ln e}{e} & \boxed {\cdot} & \frac {\ln \pi}{\pi} \\
 \\
f(e) & \boxed {\cdot} & f(\pi ) \\
\ \end{array}}$ , where  $ f(x) = \frac {\ln x}{x}$ , $ x\in [e,\pi ]$ . Observe that

$ f'(x) = \frac {1 - \ln x}{x^2} < 0$ , i.e. the function $ f$ is decreasing. In conclusion,  $ f(e) > f(\pi )$ , i.e. $ \boxed {\ e^{\pi} > \pi^e\ }$ .

[b][u]Similar exercise.[/u][/b] $ 3^{\sqrt 5}\ \ \boxed {\cdot}\ \ 5^{\sqrt 3}$ .

[b][u]Proof.[/u][/b] $ \boxed {\ \begin{array}{ccc} \\
\ 3^{\sqrt 5} & \boxed {\cdot} & 5^{\sqrt 3} \\
 \\
\sqrt 5\ln 3 & \boxed {\cdot} & \sqrt 3\ln 5 \\
 \\
\frac {\ln 3}{\sqrt 3} & \boxed {\cdot} & \frac {\ln 5}{\sqrt 5} \\
 \\
f(3) & \boxed {\cdot} & f(5 ) \\
\ \end{array}}$ , where  $ f(x) = \frac {\ln x}{\sqrt x}$ , $ x\in [3,5]$ . Observe that $ f'(x) = \frac {2 - \ln x}{2x\sqrt x} > 0$

because $ e^2 > 5 > 3$ . Thus, the function $ f$ is increasing. In conclusion, $ f(3) < f(5 )$ , i.e. $ \boxed {\ 3^{\sqrt 5}\ \ < \ \ 5^{\sqrt 3}\ }$ .

[b][u]Another exercises.[/u][/b]  $ \|\begin{array}{ccccc}
1 & \blacktriangleright & \sqrt [3]2 + \sqrt [3]5 & \boxed {\cdot} & \sqrt [3]3 + \sqrt [3] 4\\\\
2 & \blacktriangleright & \sqrt [3]5 + \sqrt 2 & \boxed {\cdot} & \sqrt [3]4 + \sqrt 3\\\\
3 & \blacktriangleright & \sqrt [3]3 + \sqrt [5]5 & \boxed {\cdot} & \sqrt [3]5 + \sqrt [5]3\\\\
4 & \blacktriangleright & \frac {e^{\pi}-e^e}{e^{\pi}+e^e} & \boxed {\cdot} & \frac {\pi -e}{2}\\\\
5 & \blacktriangleright & \frac 1e<a<b &  \implies &  \ln^2\frac {a+b}{2}\ \boxed {\cdot}\ \ln a\ln b\end{array}$

[quote="metafor"][hide="Joke Proof"]$ e \approx 2.7, \\
\pi \approx 3, \\
2.7^{3} \equal{} \left(\frac {27}{10}\right)^{3} \equal{} 19.683 > 3^{2.7} \equal{} 9 \times 3^{0.7} \approx 19.419, \\
\therefore e^{\pi} > \pi^{e}$[/hide][/quote]
Did you know that $ e^{\pi}\equal{}20\plus{}\pi$

[color=darkred][b]The oldest aproximation :[/b][/color]  $ 3,14159\ldots\ \equal{}\boxed {\pi < \frac {22}{7}} \equal{} 3,1428\ldots\ !$  [color=darkblue][b]Wao, [/b][/color] [color=red]6666[/color] :lol:

[quote="Virgil Nicula"][color=darkred][b]The oldest aproximation :[/b][/color]  $ 3,14159\ldots\ \equal{} \boxed {\pi < \frac {22}{7}} \equal{} 3,1428\ldots\ !$  [color=darkblue][b]Wao, [/b][/color] [color=red]6666[/color] :lol:[/quote]

Here is a link related to this.

[url]http://www.mathlinks.ro/Forum/viewtopic.php?p=1137598#1137598[/url]

Thank you, [b]Kunny[/b] ! Very nice this exercise : $ \boxed {\ \frac {22}{7} \minus{} \pi \equal{} \int_0^1 \frac {x^4(1 \minus{} x)^4}{1 \plus{} x^2}\ dx>0\ }$ .

You are welcome, Virgil Nicula. :)

I have posted the definite integral before, but I can't find my post.

I remember that Kent Merryfield showed  $ \frac {22}{7} \minus{} \frac {1}{630} < \pi < \frac {22}{7} \minus{} \frac {1}{1260}$.

Let's start over for 7777 from now on.

I have just founded my old post, Virgil Nicula. :lol:

[url]http://www.mathlinks.ro/Forum/viewtopic.php?p=125923#125923[/url]

[quote="diophantient"][quote="metafor"][hide="Joke Proof"]$ e \approx 2.7, \\
\pi \approx 3, \\
2.7^{3} \equal{} \left(\frac {27}{10}\right)^{3} \equal{} 19.683 > 3^{2.7} \equal{} 9 \times 3^{0.7} \approx 19.419, \\
\therefore e^{\pi} > \pi^{e}$[/hide][/quote]
Did you know that $ e^{\pi} \equal{} 20 \plus{} \pi$[/quote]

No I didn't know that... but then again it isn't true. :S

what about this (from my teacher)
by using taylor theory:

$ \Large e^x \equal{} 1 \plus{} \frac{x}{1!} \plus{}\frac{x^2}{2!} \plus{} \frac{x^3}{3!} \plus{} ........\\
 e^x > 1 \plus{} x\\
 e^\pi > 1 \plus{} \pi \\
e^{\frac {\pi}{e}} > 1 \plus{} \frac {\pi}{e}\\
e^{ln \frac {\pi}{e}} > 1 \plus{} ln \frac {\pi}{e}\\
e^{ln \frac {\pi}{e}} > ln e \plus{} ln \frac {\pi}{e}\\
e^{ln \frac {\pi}{e}} >  {ln {e  \frac {\pi}{e}}}\\
\frac {\pi}{e} > { ln \pi}\\
e^{ \frac {\pi}{e}} >   \pi\\
e^ \pi  >   \pi ^ e$

[quote="kunny"]Consider the function $ f(x) \equal{} \frac {\ln x}{x}$ and examine the behaviour.

I think that we can use the inequality : $ e^x > 1 \plus{} x$ directly.[/quote]

Plugging $ x\equal{}\ln \frac{\pi}{e}$ into the inequality gives the result. :lol:


I just finished AP computer comp sci and am looking for ways to continue honing my computer skills.  Are there any computer classes online (like the math classes aops offers)?

thanks!

I don't exactly know of any classes similar to the AoPS classes for computer science, but there are the USACO (computer science version of USAMO) contests, which also have a set of training pages that teach you a lot about algorithms.  There's more information about that at [url]http://www.usaco.org[/url].  And then there's also TopCoder contests which are similar too, and there's information about those at [url]http://www.topcoder.com[/url]

You could also try MIT's OpenCourseWare.

There was a similiar discussion on TopCoder forums. Here is a link, enjoy:

http://forums.topcoder.com/?module=Thread&threadID=507267&start=0&mc=36

--Artur


Prove that
tan(pi/7)*tan(2pi/7)*tan(3pi/7) = sqrt(7)

Can you explain furthermore?
I've done this one already, but it's a really ugly answer.
(I used Viete Theorem)

Something with DeMoivre's Theorem, probably...

I was thinking of maybe combining them with angle addition/multiple angle formulas too...

Btw, how would you do it with Vieta's?

Consider the equation tan(7x) = 0 ... (1)
then tan(7x) = tan(4x + 3x) = ... (very ugly  :( )
I get an new equation ay^3 + by^2 + cy -7 = 0 ... (2)
      with y = (tanx)^2
(I've done this already, but I don't think I should post it here.)
:stretcher:
We know that (pi/7), (2pi/7), and (3pi/7) are the roots of (1)
So tan(pi/7), tan(2pi/7), tan(3pi/7) are the roots of (2).
Using the Viete Theorem,
we get (tan(pi/7)*tan(2pi/7)*tan(3pi/7))^2 = 7
Hence A = tan(pi/7)*tan(2pi/7)*tan(3pi/7) = sqrt(7)
(Obviously, A>0)

[hide="Hint to a nicer solution"] Since $\tan \frac{4\pi}{7}=-\tan \frac{3\pi}{7}$ and so forth, the problem is equivalent to showing

$\prod_{k=1}^{6}\tan \frac{k\pi}{7}=-7$ [/hide]

oh I cannot do it. Can you give me further explanation?

Oh, I haven't worked it out yet, but that seemed like a nice place to start  :rotfl:

[hide="Alternately..."] There are explicit formulas for $\sin x \sin 2x ... \sin nx$ and $\cos x \cos 2x ... \cos nx$.  Can you find them?  And then divide one by the other? [/hide]

Bang. I give up. I spent 4 hours for it today.
 :stretcher:
do you have any idea?


ComplexZeta kindly pointed out that what I did was rubbish so I will delete it here

I see much algebra.  Would you mind writing what exactly you're trying to prove or disprove or whatever?

Every true statement implies every other true statement, so can you be a bit more precise?

It might be very long and tedious. There's a simpler proof on my website: http://www.albanyconsort.com/simon in the theorems section.

not really sure whats going on here since most of the posts have been edited, but Simon, very nice page.

I pointed out that it was rubbish? That's news to me. Thanks Spoon. If you have any suggestions, I would be willing to try to incorporate them into my site, although I rarely update it.


complex analysis
logarithms
integration
calculus
function

Show that if $ f$ is holomorphic in the unit disc, is bounded, and uniformly converges to zero in the sector $ \theta <$argz$ <\phi$ as $ |z|\rightarrow 1$ then $ f\equal{}0$

Any input is greatly appreciated; thanks in advance!

Suppose $ f(0) \neq 0$. By Jensen's formula, for $ r < 1$

$ \log |f(0)| \equal{} \sum \log \frac {|z_n|}{r} \plus{} \int_0^{2\pi} \log|f(re^{it})| \frac {dt}{2\pi} \leq \int_0^{2\pi} \log|f(re^{it})| \frac {dt}{2\pi}$
Where the sum is taken over all zeros $ z_n$ of $ f$ with $ |z_n| < r$

But under the given hypothesis the integral on the right goes to negative infinity as $ r\to 1$ which is a contradiction

Now $ \frac{f(z)}{z}$ has the same properties as $ f$ and hence this function is also zero in 0. So $ f(z)/z^k$ is zero in 0 for all $ k$ and hence $ f \equal{} 0$

Or just consider $ \prod_{k\equal{}1}^n f(\zeta^k z)$ where $ \zeta^n\equal{}1$ and $ n\phi>2\pi$ and apply the maximum principle :). Interestingly enough, the global boundedness condition is superfluous. Can anybody see how to modify my or Kalle's proof to avoid using it?


calculus
integration
trigonometry
calculus computations


Solve that :not exist a (an angle) which  satisfies :for any integer number n>=0,we have
cotg(a/(2^n)) is a relational number .

Here's my solution:
Suppose that there exists an angle A so that $ cotang \frac{A}{2^n}$ is rational $ \forall n \in N$, we will prove this is PROBABLY NOT TRUE:
because with n is sufficient large $ \frac{A}{2^n} \in (0,\frac{\pi}{2})$,we can choose A in (0,pi/2).
we put $ cotan(A)=\frac{u_0}{v_0}$ with $ u_0,v_0 \in N$.
   Put $ x=cotan(\frac{A}{2})$ then $ cotan(A)=\frac{u_0}{v_0}=\frac{x^2-1}{2x}$
$ \Longrightarrow v_0x^2-2u_0x-v_0=0$
$ (\Delta)' = (u_0)^2+(v_0)^2 >0$
$ (\Delta)'$ must be a perfect square so that we must solve Pythagore equation $ (u_0)^2+(v_0)^2=p^2$
and got $ u_0=(u_1)^2-(v_1)^2;v_0=2(u_1)(v_1)$
$ \Longrightarrow cotan(\frac{A}{2})=x=\frac{u_0+\sqrt{{u_0}^2+{v_0}^2}}{v_0}=\frac{u_1}{v_1}$
Do it again and again with cotan(A/2),cotan(A/4),... then we have two sequences of numbers:
$ u_{n-1}={u_n}^2-{v_n}^2;v_{n-1}=2u_nv_n$ and $ cotan(\frac{A}{2^n})=\frac{u_n}{v_n}$
$ \Longrightarrow {u_n}^2+{v_n}^2=\sqrt{(u_{n-1})^2+(v_{n-1})^2}=...=\sqrt[2^n]{(u_0)^2+(v_0)^2}$
So if we get n is sufficient large, we got the contradiction.
Bye,PNT.

sorry Are you famous PNT?
can you tell me exactly whoyou are
everybody care about this (probably no true is that right?)

darkmaster,you call me "Probably Not True",ok. GO AHEAD,I like it so much.
Your solution is very interesting


Hi
You know the way most of the science books are presented, with one big margin and most of the figures in it?  Do you know how to do that in latex?

Also, do you know how to change fig to figure in the caption?

tkx

Do you mean \documentclass{book} by any chance?  :?

sorry, no...

Here's a page from a book having the layout I want


Um, yeah, what's actually going on?

Is it that the deadline has now passed and round 3 is in the process of being initiated?

No guys it was a PHP error on my side, that I haven't noticed until now (division by 0! :D). So you still have some hours to complete your solutions, sorry for the troubles caused.


An airplane has three sections: First Class (24 seats), Business Class ($ 25\%$ of the total number of seats) and Economy ($ \frac{2}{3}$ of the total number of seats). How many seats does the plane have?

$ \frac14\plus{}\frac23\equal{}\frac{11}{12}$. Therefore, 24 seats is $ \frac1{12}$, or $ 24\cdot12\equal{}\boxed{288}$ seats

[hide=solution]First we add $\frac14+\frac23$ and we then subtract it from 12/12. We eventually get 24 seats are $\frac1{12}$ of the total amount of seats. Finally we do, $24\times12$ and get $\boxed{288}$.[/hide]

[quote=awesomeonfire][hide=solution]First we add $\frac{1}{4}+\frac{2}{3}$ and we then subtract it from $1$. We eventually get $24$ seats are $\frac{1}{12}$ of the total amount of seats. Finally we do, $24*12$ and get $288$.[/hide][/quote]

FTFY

The First class have 24 seats . Now let the total number of seats $x$. Business class have  25% I.e $\frac{1}{4}$ of the total number of seats . Also the economy class has $\frac{2}{3}$ of the total number of seats .
Now setting up an equation  that satisfies the above condition is:
$\frac{1}{4}*x+\frac{2}{3}*x+24=x$
$\implies$ $\frac{11}{12}*x+24=x$
$\implies$ $\frac{11x}{12}+\frac{288}{12}=x$
$\implies$ $11x+288=12x$
$\implies$ $x=288$


calculus
derivative
trigonometry
function
calculus computations

Find $ dy/dt$ when $ y\equal{}5\plus{}4.9cos(\frac{pi}{6}t)$.

My thinking is that it should be $ \boxed{\minus{}2.56563*sin(\frac{pi*t}{6})}$ however I am wondering if there is a prettier way to report the answer.

I would just leave it as $ \frac{dy}{dt}\equal{}\minus{}\frac{4.9\pi}{6}\sin\left(\frac{\pi t}{6}\right)$.  No reason to convert it to a decimal.

Also, some syntax help: For trig functions such as sine and cosine, use the code
[code]\sin[/code]
Also, symbols such as $ \pi$ are given by
[code]\pi[/code]
You can scroll over portions of my post to see what my code looks like.


inequalities
real analysis
real analysis unsolved

If $\{\cdot\}$ denotes the fractionary part and $e$ is Napier's constant, prove $\frac{1}{n+1}\left(1+\frac{1}{n+2}\exp\left({\frac{1}{n+1}}\right)\right) \le \{en!\}\le \frac{1}{n+1}\left(1+\frac{1}{n+2}\exp\left({\frac{ e-1}{n+1}}\right)\right) .$

What we know abot $e$? We know that $e=1+1+\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{n!}+..$
therefore you must use this
Question: where ?
Answer Indeed here {$en!$}
or
for small $n$ you can check this inequality.
consider
{$en!$}={$N+\frac{1}{n+1}+\frac{1}{(n+1)(n+2)}+...+$}
or it is equal to
$\frac{1}{n+1}+\frac{1}{(n+1)(n+2)}+...$
after yourself

[quote="Extremal"]What we know abot $e$? We know that $e=1+\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{n!}+..$[/quote]
A well-known fact...[b]But where is your solution ? [/b]

hm..
after few calculations you will take:
$exp(\frac{1}{n+1})<1+\frac{1}{n+3}+\frac{1}{(n+3)(n+4)}+...<exp(\frac{e-1}{n+1})$

prove it yourself

[quote="Extremal"]hm..
after few calculations you will take:
$exp(\frac{1}{n+1})<1+\frac{1}{n+1}+\frac{1}{(n+3)(n+4)}+...<exp(\frac{e-1}{n+1})$
prove it yourself[/quote]
[b]But, ... this is the question ! [/b]

Good,
now you must prove only this :
$\frac{1}{(n+1)^{m}m!}<\frac{1}{(n+3)(n+4)...(n+m+2)}$

i dont get it

the inequality is untrue.

just test for a few numbers...


Salutari
    Ma poate ajuta cineva cu niste rezolvari?Problemele sunt in attach, iar orice informatii suplimentare sau discutii se pot face si prin mail la  <snipped>
   Va multumesc.

Nu poti sa le scrii aici pe forum sau sa mi le trimiti pe lupucezar@yahoo.com , te rog. ;)
Sa stii ca Latex-ul nu-i asa greu. :P

[quote="thelord"]iar orice informatii suplimentare sau discutii se pot face si prin mail la  <snipped>[/quote]Discutii daca vrei se pot face aici. Te sfatuiesc sa inveti $\LaTeX$ (uite-te in semnatura mea) [b]si[/b] sa postezi o problema separata in fiecare topic pentru discutie ;)

Revin cu problemele scrise in LATEX...

EX1:

Fie: $x_0$ un numar real fixat; sa se arate ca operatiile:

${x}\oplus{y} =x+y-x_0$ , $(x,y\in\cal R)$    si
$\alpha\odot{x}=\alpha{x}+\(({1}-\alpha)x_0$     , $(\alpha,x\in\cal R)$

determina pe dreapta reala $\cal R$ o structura de spatiu vectorial real. (Astfel se definesc o infinitate de structuri de spatiu vectorial pe aceeasi multime fata de acelasi corp).   :?:

I tried to open the file to get the problems but no success
Could someone translate the problems in english ?
Thank's

[quote="Moubinool"]I tried to open the file to get the problems but no success [/quote]

   Se pare ca este o problema cu arhiva, am sa creez un topic pt. fiecare problema.

Sigur a inteles, Moubi. Anyway, asteptam problemele. :)


Prove that for:
$ a_1,a_2,\dots,a_n\geq 0,\ a_1\plus{}a_2\plus{}\cdots\plus{}a_n\equal{}s,\ a_1\cdot a_2\cdot...\cdot a_n\equal{}p,$
the following inequality holds:
$ a_1^{n\plus{}1}\plus{}a_2^{n\plus{}1}\plus{}\cdots\plus{}a_n^{n\plus{}1}\geq s\cdot p$

Marin Bancoș - ROMANIA

We may assume without loss of generality that $ a_{1} \geq a_{2} \geq \ldots \geq a_{n}$, so it's a simple rearrangement......

Thanks, Martin N.
My proof for inequality MB-1 :
$ LHS\equal{}a_{1}^{n\plus{}1}\plus{}a_{2}^{n\plus{}1}\plus{}\cdots\plus{}a_{n}^{n\plus{}1}\geq \frac{a_1\plus{}a_2\plus{}\cdots\plus{}a_n}{n}\cdot\left(a_{1}^{n}\plus{}a_{2}^{n}\plus{}\cdots\plus{}a_{n}^{n}\right)\geq$
$ \geq\frac{a_1\plus{}a_2\plus{}\cdots\plus{}a_n}{n}\cdot n\cdot \sqrt[n]{a_1^n\cdot a_2^n\cdot ... \cdot a_n^n}\equal{}s\cdot p\equal{}RHS$

Marin Bancoș - ROMANIA

[quote="marin.bancos"]For: $ a_{1},a_{2},\dots,a_{n}>0,$ this inequality could be written as:
$ \frac{a_1^n}{a_2\cdot a_3\cdot...\cdot a_n}\plus{}\frac{a_2^n}{a_1\cdot a_3\cdot...\cdot a_n}\plus{}\cdots\plus{}\frac{a_n^n}{a_1\cdot a_2\cdot...\cdot a_{n\minus{}1}}\geq a_1\plus{}a_2\plus{}\cdots\plus{}a_n$
This is a [b]generalization of the problem 3, Canada National Olympiad - 2002[/b].[/quote]
From the AM-GM inequality,
\[\sum_{cyc}\left(\frac{a_1^n}{a_2a_3\cdots a_n}+a_2+a_3+\cdots+a_n\right)\geq n\sum_{cyc}a_1;\]
So that
\[\sum_{cyc}\frac{a_1^n}{a_2a_3\cdots a_n}+(n-1)\sum_{i=1}^na_i\geq n\sum_{i=1}^na_i;\]
Which rewrites into
\[ \frac{a_1^n}{a_2\cdot a_3\cdots a_n}\plus{}\frac{a_2^n}{a_1\cdot a_3\cdots a_n}\plus{}\cdots\plus{}\frac{a_n^n}{a_1\cdot a_2\cdots a_{n\minus{}1}}\geq a_1\plus{}a_2\plus{}\cdots\plus{}a_n.\]


Problems 1-3, 6, 8-10, 12, 14 were all solved correctly. Problem 4 had the right answer but with no proof or justification, which was surely necessary. I estimate we would have received a scored 30 out of 40. I thought everyone did a very good job on this round.

This power round reminded me of two articles I have on nice techniques to use in geometry proofs.

Thanks!

Also, can next power round please not be geometry? I need to redeem myself  :P

How would you do this problem using mass points?

In  triangle ABC, points D and E are on AB and AC, respectively. The angle bisector of angle A intersects DE at F and BC at T. If AD = 1, DB = 3, AE = 2, and EC = 4, compute the ratio AF : AT.

What's wrong with my reasoning:

First, assign a mass of 3 to A and a mass of 1 to B.  So D has a mass of 4.  Then I assign a weight of $ \frac{3}{2}$ to C, so E has a mass of $ \frac{9}{2}$. Also, T has a mass of $ \frac{5}{2}$.  Then AF:AT = 5/2 : 3 = 5:6.

But the answer is 5:18.  And I know I did a few things wrong, because if I keep going I get contradictions for the masses of a few different points.  Plus I never used ABT...  But what is actually incorrect in my reasoning above?

$ T$ does not have a mass of $ \frac {5}{2}$.

Let $ m_1$ be the mass of $ B$ and $ m_2$ be the mass of $ C$. The mass of $ T$ is equal to $ m_1 \plus{} m_2$ if and only if $ \frac{BT}{TC} \equal{} \frac {m_2}{m_1}$. But $ T$ does not split $ BC$ in that ratio with the masses you have assigned. By the angle bisector theorem, the ratio of $ BT$ to $ CT$ is $ 2: 3$, which is not equal to the ratio of the masses you gave to $ C$ and $ B$ ($ 3: 2$).

To solve this problem with mass points you'll have to use the technique of splitting masses described in section 5.1. [Notice also that the problem asked you to find $ AF: AT$, not $ AF: FT$].

[hide="Solution:"]
Assign a mass of 2 to $ C$ and a mass of 3 to $ B$. These masses are chosen since, by the angle bisector theorem, $ CT: TB \equal{} AC: AB\equal{} 3: 2$ and it puts $ T$ as the balancing point of $ BC$ with a mass of 5. Now we apply the technique of splitting masses to the point $ A$.

Moving along the segment $ AC$ we assign as a "first approximation" a mass of 4 to point $ A$. This would make $ E$ the balancing point of $ AC$. But moving along $ AB$ we also add a mass of 9 to $ A$ (which if it were a mass of 9 alone at $ A$, would make $ D$ the balancing point of $ AB$). This now gives point $ A$ a mass of 13. What is confusing is that by adding these two masses at $ A$ neither of the points $ E$ or $ D$ remain balancing points of their respective line segments. But that's OK because it ends up making point $ F$ the balancing point of line segment $ AT$!! (Prove this).

Therefore the ratio of $ AF: FT \equal{} 5: 13$, and so $ AF: AT \equal{} 5: 18$.[/hide]

That makes more sense.  For the proof, can you just say

[hide]
F is the balancing point (or sum) of $ mA$, $ 3B$, and $ 5C$.  $ mA \plus{} 3B \plus{} 5C \equal{} mA \plus{} 5T$.  So F is the balancing point of AT.
[/hide]

or is there more to it?

Also is there an order in which you should assign masses?  Because if I assigned A and B first, then A and C, I would be stuck with a problem making B,T, and C work out nicely.  I would have to add more weight to B (2.25), which would throw everything off...

There is an order in which you should assign masses. You need to assign B and C first and not change them so that T will be the balancing point of BC with a mass that you know. You then want to start "changing" the masses at A (in accordance with B and C) in order to get F as a balancing point of line segment AT.

As a general guideline, you want to "change" the mass at the vertex of a cevian (in accordance with the other two points in the triangle) in order to make the balancing point of the cevian the point where it meets a transversal line. (I hope that made sense).


[img]http://bbs.pep.com.cn/attachments/a_n_R3asA7TYHF26.jpg[/img]

How to write $a_{n}$ with just one formula

Do you mean the sequence of partial products?  In other words,

$a_{1}= \frac{2 \cdot 2}{1 \cdot 3}= \frac{ 2^{2}}{2^{2}-1}$
$a_{k}= \frac{(2k)^{2}}{(2k)^{2}-1}a_{k-1}$

Denote $a_{n}=\frac{u_{n}}{v_{n}}$, where $\{\begin{array}{ccc}u_{n}&{ \mathrm{\ : \ }}& 2\ ,\ 2\ ,\ 4\ ,\ 4\ ,\ 6\ ,\ 6\ ,\ 8\ \ldots\\\ v_{n}& \mathrm{\ : \ }& 1\ ,\ 3\ ,\ 3\ ,\ 5\ ,\ 5\ ,\ 7\ ,\ 7\ \ldots\end{array}$.

$1\blacktriangleright u_{n}=2p\Longleftrightarrow$ $2(p-1)<n\le 2p$ $\Longleftrightarrow$ $p-1<\frac{n}{2}\le p$ $\Longleftrightarrow$ $-p\le-\frac{n}{2}<-p+1$ $\Longleftrightarrow$

$-p=[-\frac{n}{2}]$ $\Longleftrightarrow$ $p=-[-\frac{n}{2}]$ $\Longleftrightarrow$ $\boxed{u_{n}=-2[-\frac{n}{2}]}$.

$2\blacktriangleright$ $v_{n}=2q-1$ $\Longleftrightarrow$ $2(q-1)^{2}-1<n\le 2q^{2}-1$ $\Longleftrightarrow$ $q-1<\sqrt{\frac{n+1}{2}}\le q$ $\Longleftrightarrow$

$-q\le-\sqrt{\frac{n+1}{2}}<-q+1$ $\Longleftrightarrow$ $q=-[-\sqrt{\frac{n+1}{2}}]$ $\Longleftrightarrow$ $\boxed{v_{n}=-2[-\sqrt{\frac{n+1}{2}}]-1}$.

[quote="Love U"]Nope , my mean

$a_{1}=\frac{2}{1}$,$a_{2}=\frac{2}{3}$,$a_{3}=\frac{4}{3}$,$a_{4}=\frac{4}{5}$

I want to know the general term of sequence $a_{n}$[/quote]

You just wanted the rule behind the products?  Because this is just

$a_{2k-1}= \frac{2k}{2k-1}$
$a_{2k}= \frac{2k}{2k+1}$


Find all functions $ f: \mathbb{R} \rightarrow \mathbb{R}$ such that $ f(1)\equal{}1$ and
$ f(f(x)y\plus{}\frac{x}{y})\equal{}xyf(x^{2}\plus{}y^{2})$

I found this problem somewhere, but I really puzzled about it..... is it already posted??

[quote="Little Gauss"]Find all functions $ f: \mathbb{R} \rightarrow \mathbb{R}$ such that $ f(1) \equal{} 1$ and
$ f(f(x)y \plus{} \frac {x}{y}) \equal{} xyf(x^{2} \plus{} y^{2})$

I found this problem somewhere, but I really puzzled about it..... is it already posted??[/quote]

I haven't figured out a solution yet... but here some little results:

Let $ P(x,y)$ be the statement $ f(f(x)y \plus{} \frac {x}{y}) \equal{} xyf(x^{2} \plus{} y^{2})$ and let $ g(x) \equal{} f(x)x$ be.

$ P(1,y)$: $ g(x \plus{} \frac {1}{x}) \equal{} g(1 \plus{} x^2)$  (*)

$ \displaystyle\lim_{a\to\infty}g(a)\equal{}\lim_{x\to 0}g(x\plus{}\frac{1}{x})\equal{}\lim_{x\to 0}g(1\plus{}x^2)\equal{}g(1)\equal{}1$ (**)

And now from $ P(x,x)$ and (**) it follows that $ f(2) \equal{} \frac {1}{2}$

[quote="Max(i)"]
$ P(1,y)$: $ g(x \plus{} \frac {1}{x}) \equal{} g(1 \plus{} x^2)$  (*)

$ \displaystyle\lim_{a\to\infty}g(a) \equal{} \lim_{x\to 0}g(x \plus{} \frac {1}{x}) \equal{} \lim_{x\to 0}g(1 \plus{} x^2) \equal{} g(1) \equal{} 1$ (**)
[/quote]
It is not clear that $ g$ is a continuous function.... so we can't say like that.
Anyone can post a solution????

BTW, One of most important cultural heritage in Korea, even a pride of Korean, "Sung Rye Moon(for all koreans, is it right?)" was almost entirely destroyed by a small fire raised by a crazy man......ㅠㅠ I think it would be a biggest tragedy of Korea in 21th century..

I think, proving that exists value of $ y$ for which $ f(x)y\plus{}x/y\equal{}x^2\plus{}y^2$ (easy using the fact that LHS-RHS is continous and Intermediate value theorem)
and using the fact that $ xy\equal{}1$ or$ f(x^2\plus{}y^2)\equal{}0$ in that case is useful.=>$ f(y)\equal{}1/y$ or exist $ x$ $ s.t.$f$ (x^2\plus{}y^2)\equal{}0$ could be useful.
probably we should show that for nonzero values the function does not take zero as value.
but as I am not genius as above, i cannot show it.

:) it same my solution but I can't prove that : if exits$ a , b$so that $ f(a) \equal{} f(b) \equal{}  0$ then $ a \equal{} b \equal{} 0$

To ywl.... I already thought about your idea.... but I also couldn't prove that $ f(x)\equal{}0 \Rightarrow x\equal{}0$.....

[color=green]I think the problem must be : $ f: \mathbb{R} \longmapsto\mathbb{R}$, because we cant put $ y\equal{}0$ in the condition. [/color]


let $ a\equal{}b\equal{}c\equal{}2$,  then  $ abc\equal{}8 \geq  \frac{9}{64}$

AM-GM Inequality gives

$ (1 \plus{} a) (1 \plus{} b) (1 \plus{} c) ( \frac {1} {1 \plus{} a} \plus{} \frac {1} {1 \plus{} b} \plus{} \frac {1} {1 \plus{} c}) \geq 9$

Hence$ (1 \plus{} a) (1 \plus{} b) (1 \plus{} c) \geq 9$

From Chebyshev's Inequality, $ 4(1 \plus{} abc) \geq (1 \plus{} a)(1 \plus{} b)(1 \plus{} c) \geq 9$

Hence $ abc \geq 5/4$

Anything wrong here?

edit: something is wrong here

[quote="Hong Quy"]let $ a \equal{} b \equal{} c \equal{} 2$,  then  $ abc \equal{} 8 \geq \frac {9}{64}$[/quote]

sorry I ve taken the exercice wrongly, actually

$ \frac {1}{1 \plus{} a} \plus{} \frac {1}{1 \plus{} b} \plus{} \frac {1}{1 \plus{} c} \equal{} 2$

and not =1

now its corrected

[quote="FOURRIER"]$ a,b,c > 0$ $ \frac {1}{1 \plus{} a} \plus{} \frac {1}{1 \plus{} b} \plus{} \frac {1}{1 \plus{} c} \equal{} 2$ Then prove :

$ abc \le \frac {9}{64}$[/quote]
  ineq's  not strong.
 we have :
$ \frac{1}{1\plus{}a} \equal{}\frac{b}{b\plus{}1} \plus{}\frac{c}{c\plus{}1} \geq \frac{2 \sqrt{bc}}{\sqrt{(1\plus{}b)(1\plus{}c)}}$
$ \frac{1}{1\plus{}b} \equal{}\frac{a}{a\plus{}1} \plus{}\frac{c}{c\plus{}1} \geq \frac{2 \sqrt{ac}}{\sqrt{(1\plus{}a)(1\plus{}c)}}$
$ \frac{1}{1\plus{}c} \equal{}\frac{a}{a\plus{}1} \plus{}\frac{b}{b\plus{}1} \geq \frac{2 \sqrt{ab}}{\sqrt{(1\plus{}b)(1\plus{}a)}}$
 hence : $ abc \leq \frac{1}{8}  < \frac{9}{64}$


Fie $m,n \in \mathbb{N}^{*}$ cu $m<n$. Zicem ca [b]"$M$"[/b] daca dintr-o secventza $a_{1}a_{2}...a_{n}\in \mathbb{N}$ alegem $m$ numere si le marim cu $1$.

a)Daca $(m,n)=1$, demonstratzi ca pentru orice secventza $a_{1}a_{2}...a_{n}\in \mathbb{N}$ exista un sir de [b]"$M$"[/b] (finit) dupa care toate numerele sunt egale.

b)Daca $(m,n)\neq 1$, aratati ca exista secventze pt care nu se poate.

a) este suficient sa aratam ca se poate aplica M un anumit numar de ori astfel incat un anumit element al sirului (de alegerea noastra) creste cu exact o unitate relativ celorlalte elemente, pt ca atunci repetand procedura asta putem ridica fiecare element mai mic pana la 'nivelul' celui mai mare...

sa zicem ca vrem ca $a_{1}$ sa creasca cu o unitate relativ celorlalte elemente. aplicam M la elementele $a_{1},...,a_{m}$, apoi la elementele $a_{m+1},...,a_{2m}$, apoi la elementele $a_{2m+1},...,a_{3m}$, si asa mai departe, considerand ca pt $k > n$, $a_{k}$ e de fapt $a_{k\pmod{n}}$. pt ca $(m,n)=1$, exista $a,b$ naturale astfel incat $ma=nb+1$. rezulta ca daca aplicam M de $a$ ori in felul descris, atunci $a_{1}$ creste cu $b+1$ intre timp ce toate celelalte elemente cresc cu $b$. deci $a_{1}$ intr-adevar creste cu exact o unitate fatza de celelalte elemente. (QED)

(b) fie $(m,n)>1$ si fie sirul $(1,0,0,...,0)$. presupunem ca aplicand M de mai multe ori putem nivela sirul. suma sirului de fiecare data creste cu $m$, deci sirul rezultant are o suma de forma $1+am$. in acelas timp, sirul rezultant are o suma de forma $bn$ pt ca are $n$ elemente egale. dar $(m,n)>1$, deci nu putem avea $1+am=bn$, o contradictie.


I think the [url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_Downloads.php]page[/url] stating how to download $\LaTeX$ may be a bit out of date.  You advise people to first go [url=http://miktex.org/Setup.aspx]here[/url], then "Click the link that starts 'small-miktex'."  However, no such link currently exists...

In the meantime, what should we click if we want to install $\LaTeX$?

You don't install latex, you don't install anything. Its all on the website. :)

Well, yeah, as long as you only want to use $\LaTeX$ on the forums here, you don't need to install anything.  However, if you want to typeset documents using $\LaTeX$, you have to install stuff before you can do so.


calculus integration function trigonometry algebra domain limit

In AP Calculus, I was told that a function must be continuous on the interval over which the function is being integrated in order for the definite integral of the function to exist.

This explains why something like $\int_{1}^{3}\frac{1}{(x-2)^{3}}dx$ is undefined because $f(x) = \frac{1}{(x-2)^{3}}$ is not continuous at $x = 2$.

However, how can something like $\int_{0}^{\infty}\frac{e^{-x}\sin{x}}{x}dx$ exist?  Does the integrand not need to be continuous/exist at the endpoints of the interval over which the integrated is being integrated?

[quote="Zakary"]In AP Calculus, I was told that a function must be continuous on the interval over which the function is being integrated in order for the definite integral of the function to exist.
[/quote]
This is just the $0$-th order approximation to the truth. What was, probably, meant was that you can integrate a continuous on the finite closed interval function by every conceivable method of integration but, strictly speaking, there seems to be no particular integration method that has continuity as a necessary condition for a function to be integrable by this particular method. Moreover, the reason we have so many different methods of integration is that we want to be able to integrate as large class of functions as possible and, in particular, we want to go far beyond the class of continuous functions. So, you are usually quite safe when saying that every continuous function on a closed finite interval is integrable regardless of which definition of integrability you use but the converse seems never to be true. The claim was, probably, just made to prevent the students to blindly use the formulae for the antiderivatives and the correct version of it would be [i]"You'd better check that the function is continuous before integrating it over an interval; if it isn't, you MAY BE in trouble, though it is not time for you to learn when exactly you ARE in trouble yet"[/i] (Note the difference between MAY BE and ARE).

Anyway, my advice would be to ask the person who told you this for the precise definitions of "being integrable" and of the "definite integral". Without these definitions, the statement you quoted is just empty. With them, it shouldn't be hard to see what the truth (or, at least, the $1$-st order approximation to it) is.

The first type of integral you learn about is the Riemann integral. Two necessary conditions for Riemann integrability:
The function is bounded.
The domain is a bounded interval.

Any continuous function satisfying these conditions is Riemann integrable, and we can extend that to piecewise continuous functions by splitting the interval. That's not all, of course- it's merely a sufficient condition.
Removable singularities and jump discontinuities can't break integrability of any sort.

The first extension you learn is improper Riemann integrals- that framework is good enough to answer the two questions you posed.
The first function is unbounded, and analysis near the singularity reveals that the improper integral diverges.
The second function lives on an unbounded interval; there isn't actually anything wrong at zero, and the integral converges absolutely at $\infty$ by comparison to $e^{-x}$.

Key phrase here: "improper integral." That means an integral computed by first restricting the domain of integration to exclude a "bad" point (or several), then taking the limit at the restriction is relaxed away.

One of the ways I talk about it is to ask of an integral, "for how many different reasons is this integral improper?" (I'm having this conversation in a Calculus II class; some of what was in fedja's post involves issues that won't be raised until some later class.) The conventional principles for accepting that an improper integral exists include that each reason that the integral is improper must be dealt with [i]separately.[/i]

What about $\int_{1}^{3}\frac{dx}{(x-2)^{2}}\ ?$

The fact that the function is undefined at $x=2$ and likely ill-behaved near $2$ makes that point a point of interest. Looking at the limits from each side, we see that we have infinite limits and a vertical asymptote. Therefore, I will say that this integral is improper for two reasons (or in two places) - the left side of $2$ and the right side of $2.$ To resolve this, we would need to deal, separately, with both
$\int_{1}^{2}\frac{dx}{(x-2)^{2}}$ and $\int_{2}^{3}\frac{dx}{(x-2)^{2}}.$

Both of those integrals diverge (by direct calculation), with the left side being $-\infty$ and the right side being $+\infty.$ Part of the conventions of improper integrals is that we refuse to attempt to balance these opposing infinities; we simply declare the integral divergent. [We don't usually address the topic of "principal values" in that early course; when we do, we will have to take some considerable care about precise definitions.]

Now, what about $\int_{0}^{\infty}\frac{e^{-x}\sin x}x\,dx\,?$

We have two apparent problems - the fact that the domain goes to infinity (that always makes an integral improper) and the fact that the integrand isn't defined at zero - at the very least, that makes $x=0$ a "suspect" point. In response to that, we split the integral into two pieces: $\int_{0}^{1}$ and $\int_{0}^{\infty}.$ (There's nothing magical about $1.$ We could have split this into two integrals at $4.8$ if we'd felt like it.)

So we examine $\int_{0}^{1}\frac{e^{-x}\sin x}x\,dx$ and that "suspect" point at $0.$

But it turns out that $\lim_{x\to 0}\frac{e^{-x}\sin x}x=1.$ I prefer to take the attitude that "removable singularities deserve to be removed." That is, by the simple expedient of declaring that the function $\frac{e^{-x}\sin x}x$ has the value $1$ when $x=0,$ we have a continuous function on $[0,1].$ And, as fedja said, continuous functions on closed bounded intervals are integrable. Our ultimate verdict on this: not only does the integral exist, it wasn't even really improper.

That leaves $\int_{1}^{\infty}\frac{e^{-x}\sin x}x\,dx.$

Here, we note that $\left|\frac{e^{-x}\sin x}x\right|\le e^{-x}$ for $x\ge 1$ and that

$\int_{1}^{\infty}e^{-x}\,dx$ converges. Hence, we have comparison on our side - the integral is improper, but it's also absolutely convergent, hence convergent. So if we put the two pieces together, we have a convergent integral.

[I haven't addressed the question of finding an exact value of that second integral. It is possible, but the methods for doing so may tax your ingenuity if you haven't seen them.]

Operating by my principles, I said that that second integral wasn't improper at zero. (It was improper at the other, infinite end.) If I were to encounter someone who claimed that the integral is improper at zero because the function is undefined at that point, then I wouldn't feel the need to argue - the label isn't that important, and we will both agree that the integral converges.

Ok, so it seems like I'm gonna have to learn more Calculus to be able to calculate those types of integrals... >_>

nevermind
I figured out what I was asking.


inequalities proposed
inequalities
algebra
Vietnam

[quote="pigfly"]Let $x,y$ be real numbers satisfying the condition:
\[x-3\sqrt {x+1}=3\sqrt{y+2} -y\]
Find the greatest value and the smallest value of:
\[P=x+y\][/quote]
I have solved  this problem
use geometry! :D
max if x=y+1...
min if x=-1 or y=-2 :)

An algebra solution,observe that:
[tex] x+y+3 \leq (\sqrt{x+1}+\sqrt{y+2})^2 \leq 2(x+y+3) [/tex]  :)

[quote="kunny"]
min$P=\frac{9+3\sqrt{21}}{2}\ \cdots (x,y)=\left(\frac{13+3\sqrt{21}}{2},-4\right),\left(-4,\frac{13+3\sqrt{21}}{2}\right)$[/quote]

But don't you need [tex]y\geq-2[/tex] to have the root positive?

[quote="pigfly"]Let $x,y$ be real numbers satisfying the condition:
\[x-3\sqrt {x+1}=3\sqrt{y+2} -y\]
Find the greatest value and the smallest value of:
\[P=x+y\][/quote]

Let $X=\sqrt{x+1},Y=\sqrt{y+2}\Longleftrightarrow x=X^2-1,y=Y^2-2$.We can rewrite the given conditions to $X^2+Y^2-3=3(X+Y)\ (X\geqq 0,Y\geqq 0)$ and$P=3(X+Y)$

$\Longleftrightarrow \left(X-\frac{3}{2}\right)^2+\left(Y-\frac{3}{2}\right)^2=\frac{15}{2}\ (X\geqq0,Y\geqq0)$  and $X+Y=\frac{P}{3}$

Determining the condition for which a circle with centered $\left(\frac{3}{2},\frac{3}{2}\right)$,radius $\frac{\sqrt{15}}{\sqrt{2}}$ and a line $X+Y=\frac{P}{3}$ have at least of the interception point,

Max$P$ is $9+3\sqrt{15}$, when $(X,Y)=\left(\frac{3+\sqrt{15}}{2},\frac{3+\sqrt{15}}{2}\right)$

$\Longleftrightarrow (x,y)=\left(\frac{10+3\sqrt{15}}{2},\frac{8+3\sqrt{5}}{2}\right)$, and

min$P$ is $\frac{9+3\sqrt{21}}{2}$,when $(X,Y)=\left(\frac{3+\sqrt{21}}{2},0\right),\left(0,\frac{3+\sqrt{21}}{2}\right)$

$\Longleftrightarrow (x,y)=\left(\frac{13+3\sqrt{21}}{2},-2\right),\left(-2,\frac{13+3\sqrt{21}}{2}\right)$.

[quote="Anh Cuong"][quote="Pascual2005"]Anh Cuong, sorry but what i get is: $(x+y)^2=9(\sqrt{x+1}+\sqrt{y+2})^2\leq 18(x+y+3)$ this gives us a cuadratic in $x+y$ and an interval where $x+y$ could lie.[/quote]
[tex] i) P^2 \leq 18(P+3) \Rightarrow P \leq 9+3\sqrt{15} [/tex].
[tex] ii)P^2 \geq 9(P+3)  \Rightarrow P \geq \frac{9+3\sqrt{21}}{2} [/tex]
(Observe that: [tex] x+y= 3(\sqrt{x+1}+\sqrt{y+2}) > 0 [/tex])  :)[/quote]
very nice [b]Anh Cuong;minhkhoa[/b]  who have another solution? my solution is the same [b]kunny[/b]'s solution

If  $x,y $ be real number such that $x-3\sqrt{x+1}=3\sqrt{y+2}-y$ . Then$$ \frac{9}{2}+\frac{3}{2}\sqrt{21}\leqslant x+y\leqslant 9+3\sqrt{15}$$.

I have already solved the problem, sqing

See post # 6

[quote=quangminhltv99]I don't understand why but I got $\min$ $P$ is $9-3\sqrt{15}$
My solution:
Clearly, $x+y=3(\sqrt{x+1}+\sqrt{y+2})$. By Cauchy-Schwarz, we get $\sqrt{x+1}+\sqrt{y+2}\leq\sqrt{2(x+y+3)}$. Now, by setting $t=\sqrt{x+y+3}$, we get $\frac{t^2-3}{3}\leq\sqrt{2}t\Rightarrow t^2-3\sqrt{2}t-3\leq 0\Rightarrow t\in\left[ \frac{3\sqrt{2}-\sqrt{30}}{2},\frac{3\sqrt{2}+\sqrt{30}}{2}\right]$ And you get min $P$ will be $9-3\sqrt{15}$[/quote]

For which values of $x,\ y$ is attained the minimum value ?

Let $x,y $ be real number such that $x-\sqrt{x+1}=2\sqrt{y+2}-y.$   Prove that$$ \frac{1+\sqrt{13}}{2}\leq  x+y\leq \frac{5+\sqrt{85}}{2}$$
Let $x,y $ be real number such that $x-2\sqrt{x+1}=2\sqrt{y+2}-y.$   Prove that$$ 6\leq  x+y\leq 4+2\sqrt{10}$$

Let $x,y $ be real number such that $x-3\sqrt{x+1}=-3\sqrt{y+2}-y.$ Prove that
$$\frac{3(3-\sqrt{21})}{2}\le x+y\le \frac{3(3+\sqrt{21})}{2}$$

Let $x,y $ be real number such that $\sqrt{x+1}-y=x+\sqrt{y+3}.$   [url=https://artofproblemsolving.com/community/c6h1973360p13685976]Prove that [/url]
$$\frac{1-\sqrt{17}}{2}\le x+y\le \frac{1+\sqrt{17}}{2}$$