complex analysis
function
real analysis
Functional Analysis
advanced fields
advanced fields unsolved

Characterize all compact operators from $ H(U)$ to $ H(U)$ (holomorphic functions on unit disc with standard norm being supremum).

I'm afraid you are asking too much. They are plentiful and I don't even have the slightest idea in what terms you want your characterization :( If you just want something to play with, here is a couple of reasonable problems:

a) Let $ \psi: U\to U$ be an analytic map. When is the operator $ f\mapsto f\circ\psi$ compact as an operator from $ H(U)$ to itself?

b) Suppose that $ m_n$ is a decreasing to $ 0$ convex sequence of positive numbers (say, $ m_n\equal{}n^{\minus{}\alpha}$, $ \alpha>0$). Show that the operator $ T$ defined by $ Tz^n\equal{}m_nz^n$ can be extended to a compact linear operator in $ H(U)$.

What if we demand also operators to be self-adjont/unitary - is then characterization similar to that of theorem of Hilbert-Schimdt in Hilbert spaces is possible or is $ H(U)$ with those operators still "too big"? And how does $ H(U)^*$ "look like"? (those still are probably quite general questions - but I'm just wondering :) )

Self-adjoint in the space with supremum norm  :what?: I start fearing for the outcome of your exam :roll:

$ H(U)^*$ (the standard notation for $ H(U)$ is $ H^\infty(U)$, by the way) looks hardly any better than $ (L^\infty)*$ with all those crazy Banach limits and other similar functionals constituting most of it. It is not easy to describe in any reasonable terms :(

[quote="fedja"]Self-adjoint in the space with supremum norm  :what?: I start fearing for the outcome of your exam :roll: [/quote]

I'm probably blind but still can't see what's wrong with self-adjontness: $ T\phi (x) \equal{} \phi (Tx)$  for all $ \phi \in H^{\infty} (U)$, I guess some estimates on norms show that $ T$ must be trivial or otherwise noninteresting but I miss those   :(  Fortunately, I had my Functional Analysis exam few hours ago and it went quite well so you don't have to fear no more  :D

By the way, as I see my questions are more or less dumb, can you point me to some articles/books in which they look on complex analysis of one variable from "functional analysis point of view"? (besides Rudin, who anyway doesn't have a lot on that) - I'd prefer more "fast-paced" style of writing (Lang-like, Rudin-like, Serre-like) if possible.

[quote="Megus"]
I'm probably blind but still can't see what's wrong with self-adjontness: $ T\phi (x) \equal{} \phi (Tx)$  for all $ \phi \in H^{\infty} (U)$,
[/quote]
What is $ x$ and what does $ Tx$ mean?
[quote]
Fortunately, I had my Functional Analysis exam few hours ago and it went quite well so you don't have to fear no more  :D
[/quote]
Then I fear for the qualification of your teachers :lol:
[quote]
By the way, as I see my questions are more or less dumb, can you point me to some articles/books in which they look on complex analysis of one variable from "functional analysis point of view"? [/quote]
Try to google something like "Hilbert (or Banach) spaces of analytic functions" to start with. It has been quite a while since I taught anything like that, so my memory is blank now :(.

[quote="fedja"]
What is $ x$ and what does $ Tx$ mean?
[/quote]
$ x \in H(U)$ and $ Tx$ is the same as $ T(x)$ - standard definiition of self-adjont operator  :huh: So can you explain what makes self-adjoint operators so uninteresting?
[quote]
Then I fear for the qualification of your teachers :lol:
[/quote]
As I find it a little offending for my teachers  (:) ) I want to clarify - it's basic Functional Analysis I (there's also part II) for undergraduates where no complex analysis is involved, and you do pretty much standard things like counting norms, checking continouity, compactness of operators, finding adjont operators on some standard spaces ($ c_0$,$ c$, $ L_p$, etc.), basic properties of spectrum, and such.  :)

Wait a minite: $ x\in H(U)$, $ \phi\in H(U)$, $ T\phi\in H(U)$, $ Tx\in H(U)$, what is $ \phi(Tx)$ then? :?

I mean, in order to talk of self-adjointness, we should at least have the adjoint operator acting on the same space as the operator itself and $ H(U)^*$ is so different from $ H(U)$ that there is no meaningful way to identify them. You may, of course, say that you consider $ H(U)$ as a subspace of some Hilbert space and consider the bounded operators in $ H(U)$ that can be extended to bounded self-adjoint operators in that Hilbert space but then, first, you should start with specifying which Hilbert space you mean (there is no canonical choice) and secondly, there will be no good description even then.

I had no intention to offend either you or your teachers :blush: I apologize if you consider what I said offensive. All I meant was that your understanding of functional analysis doesn't seem very impressive if one just looks at what you are writing. :P

I just realized what stupidities I wrote above  :blush:  :blush:  :blush:  Will think twice when considering to post something again. Sorry for wasting your time  :blush:


geometry
circumcircle
ratio
rectangle
parallelogram
perpendicular bisector
similar triangles

In $\triangle ABC$, let the tangents to the circumcircle at $B$ and $C$ meet at point $D$, and let $E$ be the midpoint of $BC$. Let $F$ be the midpoint of the altitude $AG$. Show that $AD$ and $EF$ intersect at the symmedian point.

A pretty synthetic proof appears in Chapter 7 of Honsbereger's Episodes, but it is not easy to see. With basic stuff only:

AD is the A-symmedian of the triangle $ \triangle ABC$. Let EF meet AD at a point K. DE is the perpendicular bisector of BC, hence $ AF \parallel DE.$ Triangles $ \triangle AFK \sim \triangle DEK$ are similar, having equal angles. Their similarity coefficient is $ k \equal{} \frac {\overline{AF}}{\overline{DE}} \equal{} \frac {\overline{AK}}{\overline{DK}} \equal{} \frac {\overline{FK}}{\overline{EK}}$. Let AD meet BC at M and let a parallel to BC through F meet AD at L. The triangles $ \triangle LKF \sim \triangle MKE$ are also similar, with the same similarity coefficient $ k \equal{} \frac {\overline{FK}}{\overline{EK}} \equal{} \frac {\overline{LK}}{\overline{MK}}$. Since L is the midpoint of AM, $ \overline{AL} \equal{} \minus{}\overline{ML}.$ Consequently,

$ \frac {\overline{AK}}{\overline{MK}} \equal{} \frac {\overline{AL} \plus{} \overline{LK}}{\overline{MK}} \equal{} \minus{}\frac {\overline{ML} \plus{} \overline{KL}}{\overline{MK}} \equal{} \minus{}\frac {\overline{MK} \plus{} 2 \overline{KL}}{\overline{MK}} \equal{} 2k \minus{} 1$

If O is circumcenter and H orthocenter and if AG cuts the circumcircle again at X, then $ \overline{AH} \equal{} 2 \overline{OE},$ $ \overline{XG} \equal{} \overline{GH}.$ From cyclic ABXC, OBDC, $ \overline{OE} \cdot \overline{DE} \equal{} \overline{BE} \cdot \overline{CE}$ and $ \overline{BG} \cdot \overline{CG} \equal{} \overline{AG} \cdot \overline{GH}.$ Then

$ k \equal{} \frac {\overline{AF}}{\overline{DE}} \equal{} \minus{}\frac {\overline{AG} \cdot \overline{AH}}{BC^2} \equal{} \minus{}\frac {AG^2 \minus{} \overline{AG} \cdot \overline{HG}}{BC^2} \equal{} \minus{}\frac {AG^2 \minus{} \overline{BG} \cdot \overline{GC}}{BC^2} < 0.$

$ \frac {\overline{AK}}{\overline{MK}} \equal{} \minus{}\frac {2(AC^2 \minus{} \overline{BG} \cdot \overline{GC}) \plus{} BC^2}{BC^2} \equal{}$ $ \minus{}\frac {2 AG^2 \plus{} BG^2 \plus{} CG^2}{BC^2} \equal{} \minus{}\frac {AB^2 \plus{} AC^2}{BC^2} < 0$

which is exactly the ratio, in which the symmedian point K divides the A-symmedian AM, so that $ K \equiv AD \cap EF$ is the symmedian point.

Well, $AD$ is the $A$-symmedian, so it suffices to prove that $EF$ passes through the symmedian point $K$.

A well-known characterization of $EF$ says that it's precisely the locus of the centers of the rectangles inscribed in $ABC$, with one side on $BC$, so all we need to do is exhibit such a rectangle with $K$ as a center. Let $B_aB_c$ and $C_aC_b$ be the antiparallel lines to $AC,AB$ respectively which pass through $K$. That is, $B_a\in AB,B_c\in BC$, $B_aB_c$ passes through $K$, and $AB_aB_cC$ is cyclic (and similarly for $C_a,C_b$). $K$ is the midpoint of both $B_aB_c,C_aC_b$, so $B_aC_bB_cC_a$ is a parallelogram with side $C_bB_c$ on $BC$ and center $K$. It is, in fact, a rectangle because $\angle KB_cC_b=\angle BAC=\angle KC_bB_c$.

[quote="yetti"]

AD is the A-symmedian of the triangle $\triangle ABC$. [/quote]

It is well-known, I suppose, but how to prove it?

The problem assumes that AD is the A-symmedian, otherwise the symmedian point could not be on this cevian. Let a circle (D) with radius DB = DC meet AB, AC extended beyond the vertices B, C at points P, Q, respectively. The triangles $\triangle s DBP, \triangle DCQ$ are both isosceles with DB = DP = DC = DQ, hence

$\angle DPB = \angle DBP = 180^\circ - \angle B - \angle A = \angle C$

$\angle DQC = \angle DCQ = 180^\circ - \angle C - \angle A = \angle B$

From the quadrilateral APDQ,

$\angle PDQ = 360^\circ - (\angle DPA + \angle DQA + \angle PAQ) =$

$= 360^\circ - (\angle C + \angle B + \angle A) = 180^\circ$

so that the points P, D, Q are collinear, the segment PQ is a diameter of the circle (D) and the circle center D is the midpoint of this diameter. In addition, the triangles $\triangle AQP \sim \triangle ABC$ with the common angle $\angle A$ and equal angles $\angle Q = \angle B,\ \angle P = \angle C$ are oppositely similar. Since D is the midpoint of PQ, AD is the A-median of the triangle $\triangle AQP$. The oppositely similar triangles $\triangle AQP \sim \triangle ABC$ have a common bisector of the common angle $\angle A$, hence the median AD of the first triangle is the symmedian of the other one.

Dear Mathlinkers,

http://jl.ayme.pagesperso-orange.fr/Docs/01.%20Problemes%20sur%20le%20point%20de%20Lemoine.pdf  p. 7…

Sincerely
Jean-Louis


Each of the four digits 1, 2, 3 and 4 is used exactly once to fill in the four blanks in the expression below, and then the expression is simplified to a single value. How many different values are possible?
\[ \_ \plus{} \_ \plus{} \_ \minus{} \_\]

The last blank can be filled with 1 digit, and we have 4.
So there are 4 possibilities.
Values:2,4,6,8

Note that by that one property, it doesn't matter what order the 3 ones in the first 3 places are. Thus it's $ \frac{4!}{3!}\equal{}4$

lolz, mewto55555, i believe that property you were talking about is called the commutative property: property where the multiplication or addition of numbers can be interchanged and the value will not chage.


calculus
integration
inequalities
geometry
calculus computations

Let $ a,\ b$ be positive real numbers.

Prove that

$ \int_{a \minus{} 2b}^{2a \minus{} b} \left|\sqrt {3b(2a \minus{} b) \plus{} 2(a \minus{} 2b)x \minus{} x^2} \minus{} \sqrt {3a(2b \minus{} a) \plus{} 2(2a \minus{} b)x \minus{} x^2}\right|dx$

$ \leq \frac {\pi}3 (a^2 \plus{} b^2).$

[color=green]Edited by moderator.[/color]

Consider $ a\equal{}2,b\equal{}1$

$ \int_{0}^{3}\left| \sqrt{9\minus{}x^{2}}\minus{}\sqrt{6x\minus{}x^{2}}\right| dx\equal{}\allowbreak \frac{9}{2}\sqrt{3}\minus{}\frac{3}{2}\pi <2\sqrt{3}$

Thank you for pointing out my mistake.

I have just edited it. How about this time?

That's correct answer.

Sorry, I made a typo.

To Moderators,

Could you change the right side : $ \frac 23\pi (a^2\plus{}b^2)$ into $ \frac{\pi}{3}(a^2\plus{}b^2)$?

Thank you in advance.

kunny

[color=green]Redundant quote removed.[/color]

thx kunny
is there any shorter approach please reply

As the result, my solutiuon would be  same as yours.

Here is a sharper inequality than what's been to show:-

$ \int_{a\minus{}2b}^{2a\minus{}b}\left| \sqrt{3b(2a\minus{}b)\plus{}2(a\minus{}2b)x\minus{}x^{2}}\minus{}\sqrt{3a(2b\minus{}a)\plus{}2(2a\minus{}b)x\minus{}x^{2}}\right| dx\leq \frac{\left( \root{4}\of{8}\right) ^{3}}{6}\left( a^{2}\plus{}b^{2}\right)$

@Indochina: Well done again Indochina, that is in fact a very sharp (possibly sharpest) inequality in this case!

@Kunny: Did you manage to achieve this? Although you have earlier claimed that your answer is same as makar's.


conics parabola algebra polynomial trigonometry

Find all real solutions of the system
$\{ \begin{array}{ll} a^2 + b = 7 \\ a + b^2 = 11 \end{array}$

[hide](2,3) just by observation :D
Don't know if there are any others though.[/hide]

Graphically, it's the intersection of two perpendicular parabolas, so there are gonna be at least two solutions and probably four... it seems like finding the others would require solving a general quartic though  :(

:rotfl:

Well Hexaditidom, there are a few more  :gleam:

That is the problem t0rajir0u. But at least we can reduce for a polynomial of degree 3.

So I'm not sure what you mean by reducing to a polynomial of degree $3$, but I'll show you how I did it.

[hide="solution"]
Well darn boxedexe started my solution.

Their sum yields: $(a+1/2)^2+(b+1/2)^2 = \frac{37}{2}$

Therefore $a+1/2 = \sqrt{\frac{37}{2}}\cos \theta$ and $b+1/2 = \sqrt{\frac{37}{2}} \sin \theta$

Their difference yields: $(b-a)(b+a-1)=4$

Thus $\frac{37}{2}(\sin^2 \theta - \cos^2 \theta) = 4$

Then $\sin \theta = \pm \sqrt{\frac{45}{74}}$ and $\cos \theta = \pm \sqrt{\frac{29}{74}}$

$a = \frac{\pm \sqrt{29}-1}{2}$

$b = \frac{\pm 3\sqrt{5} - 1}{2}$

The set of solutions are then all combinations of $(a,b)$, there are $4$.

Happy Holidays.

[/hide]

Nevermind I erred in there.

I should have written

[hide="should have"]
$\sqrt{\frac{37}{2}}(\sin \theta - \cos \theta)(\sqrt{\frac{37}{2}}(\sin \theta + \cos \theta) - 2) = 4$

Oh well I had a good start, you might be able to use the $(2,3)$ obvious solution to your advantage.

[/hide]

Good luck.

However, $(2,3)$ is also a solution, making the number of solutions greater than 4.

I think by reducing degrees he means decaying degrees using $x+\frac{1}{x}$.

I had -4 on my paper, but I thought it was wrong for some reason.  I'll try again rigth now.

Some progress:
$\\a^2+b=7\\ a+b^2=11\\ a^2+b-a-b^2=-4\\ a(a-b)=-4\\ a^2-ab=-4\\ a^2-ab+a+b^2=7\\ 7-b-a(7-a^2)+a+11-a-7=0\\ -b-7a+a^3+11=0\\ a^3+ab=7a\implies -7a=-a^3-ab\\a^2-7-a^3-ab+a^3+11=0\\a^2-ab+4=0$
However, $a^2-ab+2=0$ if we regard $(2,3)$.

Masoud Zargar

No, if you understand the graphs of the system, then it is clear that there are only $4$ real solutions.

[hide="continuation"]

$\sqrt{\frac{37}{2}}(\sin \theta - \cos \theta)(\sqrt{\frac{37}{2}}(\sin \theta + \cos \theta) - 2) = 4$

$\frac{37}{2}(2\sin^2 \theta - 1) - \sqrt{74}(\sin \theta - \cos \theta) = 4$

$\sqrt{74} \sqrt{1-\sin^2 \theta} = \frac{45}{2}+\sqrt{74} \sin \theta - 37\sin^2 \theta$

From the $(2,3)$ solution you know one of the values for $\sin \theta$ is $\frac{7}{\sqrt{74}}$.  $\cos \theta = \frac{5}{\sqrt{74}}$

Now I understand where the cubic comes from.

$74-74\sin^2 \theta = (\frac{45}{2}+\sqrt{74} \sin \theta - 37\sin^2 \theta)^2$

You can then factor out a $(\sin \theta - \frac{7}{\sqrt{74}})$

I think there might be some argument that I'm missing that will make this problem easier, because I don't want to mess around witha cubic unless I have to.  Maybe for instance, the intersections occur on some significant place on the circle, such that knowing that $(2,3)$ is a solution yields all the solutions by a simple transformation.

The rest of the problem is left as an exercise for the reader.  Hopefully you will not encounter a mean cubic.
[/hide]

Happy Holidays and Good Luck.

[quote="boxedexe"]I think this may be correct.
Add teh two equations.
$\\a^2+a+\frac{1}{4}+b^2+b+\frac{1}{4}=18+0.5=18.5\\ (a+1/2)^2+(b+1/2)^2=18.5$
This is a circle with radius $\sqrt{18.5}$.
As a result, there are infinitely many solutions, where they are selected from $a,b\in\left[-\sqrt{18.5}+\frac{1}{2},\sqrt{18.5}+\frac{1}{2}\right]$.  We conclude that there are infinitely many real values of matching $(a,b)$ satisfying that system of equations.

EDIT: The odd thing about this is that we can have an equation $a^4-14a^2+a+38=0$, which I believe it does not have infinitely many solutions.

Masoud Zargar[/quote]
Not all points on the circle satisfy the original system.

For the solution, substitute the first equation into the second one,
$(11-b^2)^2+b=7$
$\Rightarrow (b-3)(b^3+3b^2-13b-38)=0$
$\Rightarrow b=3\text{ or }b^3+3b^2-13b-38=0$
We can solve the latter cubic equation by Fragnano's method, which gives us three ugly real solutions.

[quote="FMako"]Hopefully you will not encounter a mean cubic.[/quote]
...

Yeah.  ;) I realized that after my post.  It is imposisble for it to have infinite solutions.

Okay, the solutions are: $(2,3),(a,3.58),(a_2,-3.78),(a_3,-2.81)$.
Just plug in the b values and solve for a in the original system.

Masoud Zargar

[quote="leepakhin"]For the solution, substitute the first equation into the second one,
$(11-b^2)^2+b=7$
$\Rightarrow (b-3)(b^3+3b^2-13b-38)=0$
$\Rightarrow b=3\text{ or }b^3+3b^2-13b-38=0$
We can solve the latter cubic equation by Fragnano's method, which gives us three ugly real solutions.[/quote]

Could you please explain what is Fragnano's method? One link about that would be great also. Thanks.

Well, I don't know who's Fragnano, but, you may want to look at here ;) :
http://mathworld.wolfram.com/CubicFormula.html

[b]mathmanman[/b] thanks alot for that  :winner_first:

Well i will post only a bit of my work, the rest you can deduce using the link given by [b]mathmanman[/b].

$a^2+b=7 \Rightarrow b=7-a^2$

Plugging that into second equation we get

$a^4-14a^2+a+38=0 \Leftrightarrow (a-2)(a^3+2a^2-10a-19)=0$

Hence $a=2$ is the rational root of $a^4-14a^2+a+38=0$ which gives $b=3$.

After a messy work we find the three real solutions of the equation $a^3+2a^2-10a-19=0$:

$a_1=\frac{2\sqrt{34}}{3} \cdot \cos\left(\displaystyle \frac{\arccos\frac{317\sqrt{34}}{2312}}{3}\right)-\frac{2}{3}$

$a_2=\frac{2\sqrt{34}}{3} \cdot \cos\left(\displaystyle \frac{\arccos\frac{317\sqrt{34}}{2312}+2\pi}{3}\right)-\frac{2}{3}$

$a_3=\frac{2\sqrt{34}}{3} \cdot \cos\left(\displaystyle \frac{\arccos\frac{317\sqrt{34}}{2312}+4\pi}{3}\right)-\frac{2}{3}$

With $a_1,a_2$ and $a_3$ you can get  $b_1,b_2$ and $b_3$ easier. :D

I used Matlab to calculate that roots directly. And best part is that it's correct. :winner_second:

Oh, well, you didn't need to thank me for that you know... :)
But, anyway, you're welcome!

Ah, and : [b]LoboSolitario[/b] == :first:


geometry
algebra
function
domain
integration
analytic geometry
complex analysis

I'm just having a bit of problem solving this problem, but this problem might sound easy.

Let $\Omega$ be a bounded domain in $C$ whose boundary is a curve $z=z(t), a \leqslant t \leqslant b$, and let $A(\Omega)$ be the area of $\Omega$.

Prove that
$A(\Omega)=\frac{1}{2}\int_{a}^{b}|z(t)|^{2}Im(\frac{z'(t)}{z(t)})dt$

if I'm right, it suffices to prove that if $t \to (r(t),\theta(t))$ is a closed curve in polar coordinate, then the area inside the curve is precisely
$\frac{1}{2}\int_{a}^{b}r^{2}(t) \theta'(t) dt$
To see this, write $z(t)=r(t)e^{i \theta(t)}$ and you will obtain this expression. Then you can prove thanks to the divergence theorem that this formula is correct.


function
trigonometry
algebra
functional equation
algebra unsolved

Find all $ f,g: \mathbb{R}\rightarrow\mathbb{R}$ such that for all real numbers $ a,b$,

$ f(a \plus{} b) \equal{} f(a)g(b) \plus{} g(a)f(b)$ and $ g(a \plus{} b) \equal{} g(a)g(b) \minus{} f(a)f(b)$

It's easy to see the solution pairs $ (f(x);g(x))\equal{}(sinx;cosx)$ and $ (f(x);g(x))\equal{}(\minus{}sinx;\minus{}cosx)$ but as it seems this euqation can have too much roots. Do you have solution for this one or do you know the correct answer?

lasha:  that second pair of functions doesn't work.

Write $ h(x) \equal{} g(x) \plus{} i f(x)$; then $ h(a \plus{} b) \equal{} h(a) h(b)$.  Without some kind of continuity assumption, the pathological solutions to [url=http://en.wikipedia.org/wiki/Cauchy_functional_equation]the Cauchy functional equation[/url] creep in.

With the appropriate continuity assumptions it's not hard to see that $ h(x) \equal{} e^{\alpha x}, \alpha \in \mathbb{C}$ are the only solutions.  This gives $ g(x) \equal{} e^{ax} \cos bx, f(x) \equal{} e^{ax} \sin bx, a, b \in \mathbb{R}$.


Let $ a, b, c$, be distinct positive real numbers such that  $ a \plus{} b\plus{} c \equal{} 1$. Prove that

$ \frac{bc(bc\minus{}a^3\minus{}a^2bc)}{a^2(a\minus{}b)(a\minus{}c)}\plus{}\frac{ca(ca\minus{}b^3\minus{}b^2ca)}{b^2(b\minus{}c)(b\minus{}a)}\plus{}\frac{ab(ab\minus{}c^3\minus{}c^2ab)}{c^2(c\minus{}a)(c\minus{}b)}\geq 9$

[quote="Ligouras"]Let $ a, b, c$, be distinct positive real numbers such that  $ a \plus{} b \plus{} c \equal{} 1$. Prove that

$ \frac {bc(bc \minus{} a^3 \minus{} a^2bc)}{a^2(a \minus{} b)(a \minus{} c)} \plus{} \frac {ca(ca \minus{} b^3 \minus{} b^2ca)}{b^2(b \minus{} c)(b \minus{} a)} \plus{} \frac {ab(ab \minus{} c^3 \minus{} c^2ab)}{c^2(c \minus{} a)(c \minus{} b)}\geq 9$[/quote]

substitution

$ a\equal{}\frac{x}{x\plus{}y\plus{}z},b\equal{}\frac{y}{x\plus{}y\plus{}z},c\equal{}\frac{z}{x\plus{}y\plus{}z},(x,y,z>0)$

$ \frac {bc(bc \minus{} a^3 \minus{} a^2bc)}{a^2(a \minus{} b)(a \minus{} c)} \plus{} \frac {ca(ca \minus{} b^3 \minus{} b^2ca)}{b^2(b \minus{} c)(b \minus{} a)} \plus{} \frac {ab(ab \minus{} c^3 \minus{} c^2ab)}{c^2(c \minus{} a)(c \minus{} b)}\geq 9$

 $ \Longleftrightarrow  [5,3,0]\plus{}[5,2,1]\plus{}[4,4,0]\plus{}5[4,3,1]\minus{}2[4,2,2]\minus{}5[3,3,2]\geq 0$

true due to Muirhead theorem.


inequalities
induction
triangle inequality
calculus
calculus computations

Show that the limit of the sequence $ \frac{n^2}{2^n}$ approaches zero as $ n$ approaces infinity by epsilon delta definition.

Because $ 2^n>n^3\forall n>10$ we are done.

For each $ \varepsilon>0$, $ \exists N$ such that $ \left|a_{m}\minus{}a_{n}\right|<\varepsilon$ $ \forall m,n\geq N$.  In this case, choose $ N\equal{}5$.  Then $ (a_{n})$ is a Cauchy sequence, which means it converges to some $ L$.  Then for every $ \varepsilon>0$, there is an $ N$ such that $ |a_{n}\minus{}L|<\varepsilon/2$ for all $ n$ sufficiently large.  We want $ L\equal{}0$, so for each $ \varepsilon>0$, we can find some $ n$ sufficiently large such that $ a_{n}<\varepsilon/2$.  Then by the triangle inequality,
$ \left|a_{m}\minus{}a_{n}\right|\equal{}|a_{m}\minus{}0\plus{}0\minus{}a_{n}|\leq|a_{m}|\plus{}|a_{n}|<2\left(\frac{\varepsilon}{2}\right)\equal{}\varepsilon$

I'm sorry JRav but could you explain that a bit more?

First of all, I'm confused as to why you put in "Choose $ N\equal{}5$", was it to show the sequence is Cauchy? How does that show the sequence is Cauchy?

Then after that point, it looks to me like you assumed the existence of the limit to show it was Cauchy (as in, you assume you could find a tail of the sequence arbitrarily close to $ L$ and then use that to show that any two terms are arbitrarily close in that tail...) which to me looks reaaaally circular.

JRav, it looks like you've just stated definitions without actually saying anything about the given sequence.  In particular, you haven't proven that the sequence is Cauchy.

N.T.TUAN's idea is nice; it's easy to prove it by induction and it gives the result by the squeeze theorem.  If you want a "direct" proof, consider what happens when you double $ n$.

Actually, I should elaborate:  given $ \epsilon > 0$ we wish to find $ N$ such that for all $ n > N$ we have $ \frac {n^2}{2^n} < \epsilon$.  It suffices to prove for $ n \equal{} N$ since the sequence is clearly monotonically decreasing after some point (prove this!).  Let $ k$ be the smallest positive integer such that $ \epsilon > \frac {9}{8} \cdot \frac {1}{2^k}$.  The substitution $ n \mapsto 2n$ multiplies the numerator by $ 4$ and the denominator by $ 2^n$, so for $ n \ge 3$ the substitution $ n \mapsto 2^k \cdot n$ multiplies the sequence by at most $ \frac {1}{2^k}$.  Since $ \frac {3^2}{2^3} \equal{} \frac {9}{8}$, we can take $ N \equal{} 3 \cdot 2^k$.

For a quick high-level proof, I recommend the [url=http://en.wikipedia.org/wiki/Stolz-Ces%C3%A0ro_theorem]Stolz-Cesaro theorem[/url].

Actually, I understand the epsilon-delta definition, its just that I find it hard to prove this type of problem. I have  several questions.

1.) It can be proven by induction that $ n^2 < 2^n$ for  all $ n > 3$ so we can conclude that $ \frac {n^2}{2^n} < \equal{} 1$ which implies that $ \frac {n}{2^n} < \equal{} \frac {1}{n}$. From the book, the hint says choose $ N \equal{} max(3,\frac {1}{\epsilon})$ Can you explain this part?

2.) Why can you assume that n = N if the sequence is monotonically increasing?

1)  That's just meant to exclude trivial values of $ \epsilon$.  You will really only be concerned with small values of $ \epsilon$, but to satisfy the definition (for [b]all[/b] epsilon...) it is sometimes necessary to pick $ N \equal{} 3$ instead.

2)  $ \frac {n^2}{2^n} \le \frac {N^2}{2^N}$, so if the RHS is less than $ \epsilon$ the LHS will be as well.


floor function
quadratics
function
combinatorics proposed
combinatorics

[quote="kunny"]Let [b]S[/b]$=\{1,\ 2,\ \cdots \cdots\ ,\ n\}\ (n\geq 2).$In how many ways can we take $k$ subsets of [b]S[/b] such that all of them have only two elements, and such that any two subsets have no element in common?
Next, denote the number of this choice by $f(n,\ k)$. Find all $n$ and $k\ (k\geq 2)$ which satisfies $f(n,\ k) = f(n,\ 1)$.[/quote]

A partial answer: $f(n,k)=C_{n}^{k}\sum_{1\le i_{1}<i_{2}<...<i_{k}\le{n-1}}{(n-i_{1})(n-i_{2})...(n-i_{k})}$

For $k \leq \left\lfloor \frac{n}2 \right\rfloor$ we have that $f(n,k) = \frac{\binom{n}2}{k}\cdot f(n-2,k-1)$. To see this, pick a random pair from the $\binom{n}2$ possible. Now, the rest of $k-1$ numbers can be taken from the remaining $n-2$ numbers. Each collection of subsets clearly appears $k$ times, hence the formula.
Therefore, $f(n,k) = \frac{n (n-1) \ldots (n-2k+1)}{2^{k}\cdot k!}$.

We have that $\ell (k) = \frac{f(n,k+1)}{f(n,k)}= \frac{(n-2k)(n-2k-1)}{2(k+1)}$. $\ell (k) > 1$ iff $4 k^{2}-4nk+\left( n^{2}-n+2 \right) > 0$. Consider this as a a quadratic in $k$. Then, we have that $\Delta = 16(n+2)$, which yields $r_{1,2}= 2n \pm 2 \sqrt{n+2}$. We know that $k \leq \left\lfloor \frac{n}2 \right\rfloor$, so it suffices to prove that for $n \geq 4$ we have that $\frac{n}2 < 2n-2 \sqrt{n+2}$. That's equivalent to $9 n^{2}-16 n-32 > 0$, which is obviously true for $n \geq 4$. Thus, since $k$ does not lie between the roots $r_{1,2}$, we have that $\ell (k) > 1$, i.e. $f(n,k)$ is strictly increasing when viewed as a function of $k$ (for $n \geq 3$).

I hope I didn't mess it up :huh:


A track meet awards 6, 5, 4, 3, 2 and 1 points for 1st, 2nd, 3rd, 4th, 5th and 6th places, respectively. Team A’s only two runners originally earned 6 points in the first race with both runners finishing in the top 6, but then the winner of the first race was disqualified. How many points did Team A have after the ranks were adjusted up?

[hide="Would it be"]
would it be 8, because each person would move up one place and therefore one point, for a total or two more points? [/hide]


What's the difference between Graduate Math Courses and Undergraduate Math Courses of the same title? For example graduate complex analysis course and undergraduate complex analysis course. What's the difference between the two versions?

The graduate one is much more advanced.  The graduate course builds on the undergraduate one.  Suppose you want to take graduate algebra.  Typically, you'll need to be proficient in the undergraduate course first.


Let $ a,b,c \ge 0$ . Prove that

$ \frac {a^3 \plus{} b^3 \plus{} c^3}{(a \plus{} b)(b \plus{} c)(c \plus{} a)} \plus{} \frac {135abc}{8(a \plus{} b \plus{} c)^3} \ge 1$

@mitdac123 and Honey_S: Thank you,i edited my post.

[quote="dduclam"]Let $ a,b,c \ge 0$ . Prove that

$ \frac {a^3 \plus{} b^3 \plus{} c^3}{(a \plus{} b)(b \plus{} c)(c \plus{} a)} \plus{} \frac {135abc}{8(a \plus{} b \plus{} c)^2} \ge 1$[/quote]

Your inequality is not true. For example, let try $ a \equal{} 0.1,b \equal{} 0.05,c \equal{} 0.03$ then $ LHS \equal{} 0.8$

[quote="mitdac123"]:D
Is this your ineq ?
Let $ a,b,c$ be nonegative numbers , prove that :
$ \frac {a^3 \plus{} b^3 \plus{} c^3}{(a \plus{} b)(b \plus{} c)(c \plus{} a)} \plus{} \frac {135abc}{8 (a \plus{} b \plus{} c)^3} \geq 1$
______________
Sorry for my English  :ninja:[/quote]

With this we can assume $ a \plus{} b \plus{} c \equal{} 3$ then the inequality becomes

$ \frac {{27}}{{3q \minus{} r}} \plus{} \frac {{5r}}{8} \ge 4$

$ \Leftrightarrow \left( {96 \minus{} 15r} \right)q \plus{} 5r^2 \minus{} 32r \minus{} 216 \le 0$

By Schur ineq we have $ q \le \frac {{r \plus{} 9}}{4}$ so the ineq becomes

$ r\left( {5r \minus{} 167} \right) \le 0$

The equality holds for one variable is 0.

But I think that DDLam's ineq is not $ \left( {a \plus{} b \plus{} c} \right)^3$

A similar one:

Let $ a,b,c \ge 0$ . Prove that

$ \frac {a^3 \plus{} b^3 \plus{} c^3}{(a \plus{} b)(b \plus{} c)(c \plus{} a)} \plus{} \frac {15\sqrt[3]{abc}}{8(a \plus{} b \plus{} c)} \ge 1$


calculus
integration
trigonometry
logarithms
calculus computations

can you solve this integral ? :D

$\int_{0}^{\infty}\; \left(\frac{\tan^{-1}{x}}{x}\right)^{4}\;dx$

[quote="kg2100"]really hard , I don't wanna do it , but Mathematica 6 make it  :P

[hide]$\int_{0}^{\infty}\; \left(\frac{\tan^{-1}{x}}{x}\right)^{4}\;dx=\frac{3\pi[3\zeta{(3)}+\ln 256]-\pi^{3}(1+\ln 4)}{12}$[/hide][/quote]

 :rotfl:

yes, that [b] [u]is [/u][/b]the correct answer, the one i had was

$\frac{\pi}{4}\; \left\{ \; 3\;\zeta{(3)}-2\left[ \;(1+2\;\ln 2)\cdot\zeta{(2)}-4\;\ln 2\;\right] \;\right\}$


Hi, everyone, I have got an existential proof, but fail to find a special nowhere-continuous solution now. I wonder why, could anyone help? Proof. By the axiom of choice, we can prove that every linear spaces has a base. R is a linear space over Q, so R has a base B. From the equation f(x+y)=f(x)+f(y), one can conclude that the function f of R only depends on the values of B. The cardinal number of B is c, so the cardinal number of the solution set of the equation is 2^c. It is easy to check that f is linear if f is continuous at at least one point of R, and the cardinal number of the set of linear solutions is c, so the cardinal number of the set of nowhere-continuous solutions is 2^c  . But it seems difficult to find such a special solution.

Choose some basis $ B$ of $ \mathbb R$ over $ \mathbb Q$. Assume that $ 1\in B$ and write every real $ r$ as $ a \cdot 1 \plus{} b$ for some rational $ a$ and $ b$ in the span of $ B\backslash\{1\}$. This is unique, so we can define $ f(r) \equal{} a$. This is surely not a function of type $ f(x) \equal{} cx$, so it isn't continuous.

Think you for your solution. I wonder if we could actually find a special solution that doesn't depend on the axiom of choice, Maybe it sounds fool, which is the very problem I just had failed to express correctly. I think that using the axiom of choice leads the proof somehow on a shaky ground. But it is strange to see that without the axiom we may never go any further on the equation.

No, you can't. One consequence of accepting the axiom of choice is that we must also accept the existence of strange things that can't be explicitly constructed. If the axiom of choice fails, so does existence of bases for general vector spaces- and the ability to easily construct linear maps depends on there being bases.

Thank you for you reminder. It is better to acknowledge the axiom of choice. I think it is a good idea to take it for granted that without the axiom of choice one can't prove a proposition depends on but not equal to the axiom of choice in the Z.F.S. now.

The problem that whether we can construct a nonlinear solution of the Cauchy equation in the ZFC without the axiom of choice actually has been solved several decades ago. One can proved that a solution of Cauchy equation is linear if the one is measurable, and we can't obtain a unmeasurable set in the ZFC which is not related to the axiom of choice. The result depends on the works of Solovy and Cohen.


How would explain the first, second, and third law of thermodynamics to a fifth grader (through a poster is what I was thinking, but whatever you suggest I'll appreciate)?

Please don't post twice the same thread: you have already opened this thread on the Advanced Physics Forum, so let it be there only.

Hey check out advanced section to see if you can help me out with my tri fold poster, thanks.

I am a fifth grader, and i learnt just by reading it, so just direct him/her to a website or something

use models or     ask those who framed it........... :rotfl:


quadratics
analytic geometry
conics
parabola
algebra

If the smallest value of $ y$ satisfying the equation $ y\equal{}3x^2\plus{}6x\plus{}k$ is $ 4$, find the value of $ k$.

Thanks in advance!

OK. This is a quadratic equation. The vertex for any quadratic/parabola is the maximum/minimum value (depending on whether it curves up/down has positive/negative "a" in general/standard whatever it was form). (The x-coordinate of the vertex is $ \minus{} \frac {b}{2a}$) So we would have $ \minus{} \frac {b}{2a} \equal{} \minus{} \frac {6}{3\cdot2} \equal{} \minus{} 1$. Thus the minimum (in this case) value occurs when $ x \equal{} \minus{} 1$. So plugging our known values of $ x$ and $ y$ in to the equation:
$ 4 \equal{} 3 \minus{} 6 \plus{} k\Rightarrow{k \equal{} \boxed7}$. Sorry if I didn't explain it too well. Hope it helps!

Here is another way you can do it that would arrive at the same answer
[hide]A parabola's equation may be expressed as $ y\equal{}a(x\minus{}p)^2\plus{}q$ where $ (p,q)$ is the vertex of the parabola. If y=4 is the smallest value on the parabola, then we know that it must be on the vertex. Using the equation $ y\equal{}3x^2\plus{}6x\plus{}k$, we may rewrite this as $ y\equal{}3x^2\plus{}6x\plus{}k\plus{}3\minus{}3$ now we may factor three out of the equation so
$ y\equal{}3(x^2\plus{}2x\plus{}1)\plus{}k\minus{}3 \Longrightarrow y\equal{}3(x\plus{}1)^2\plus{}k\minus{}3$ This means that the vertex of the parabola is
$ (\minus{}1,k\minus{}3)$ since we know that the y coordinate of the vertex is 4, we can write the equation $ k\minus{}3\equal{}4 \Longrightarrow k\equal{}\boxed{7}$[/hide]

We know that the general x-coordinate of the vertex of the parabola, for the general quadratic equation

$ ax^2\plus{}bx\plus{}c\equal{}0$, is $ \minus{}\frac{b}{2a}$.

Therefore, the minimum occurs at the vertex, which is $ (\minus{}1,4)$,

since the x-coordinate is as above, and the y-coordinate is given to be $ 4$.

Plugging in, we can see that $ \boxed{k\equal{}7}$.


trigonometry
number theory
least common multiple
function

Need a little help on this problem, thanks.

Find the period of
2sin(4(pi)x+(pi)/2) + 3cos(5(pi)x).

[quote="Mathematicus"]Find the period of
$2sin(4(\pi)x+(pi)/2)+3cos(5(\pi)x)$[/quote]
Well just find the period of $2sin(4(\pi)x+\frac{\pi}{2})$ and $3cos(5(\pi)x)$ and then find the lcm of the two periods.

it's what happens when a girl reaches a certain age and needs to filter out her body...

or it could be the length between repetitions of a sinusoidal function.

[quote="Treething"]it's what happens when a girl reaches a certain age and needs to filter out her body...[/quote]
Sounds like my English class... :D

[quote="Treething"]or it could be the length between repetitions of a sinusoidal function.[/quote]
It's how long it takes for a function to repeat. $\sin x$ and $\cos x$ both have periods of $2\pi$, as each segment of $2\pi$ is identical to the next segment of length $2\pi$.

[quote="Totally Zealous"]what's a period?[/quote]

A period is when the function of a trig function repeats. For example, I have marked periods in my graph by the red lines.


LaTeX
inequalities
inequalities proposed

Let a,b,c be positive numbers such that:a+b+c=1.Prove that;
4(a^2/(bc) + b^2/(ca) +c^2/(ab))+ 729 abc ≥39

hello, why don't you use $ \text{\LaTeX}$?
Let $ a,b,c$ be positive numbers such that:$ a+b+c=1$.Prove that
$ 4\left(\frac{a^2}{bc}+\frac{b^2}{ca}+\frac{c^2}{ab}\right)+729abc\geq39$.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, why don't you use $ \text{\LaTeX}$?
Let $ a,b,c$ be positive numbers such that:$ a + b + c = 1$.Prove that
$ 4\left(\frac {a^2}{bc} + \frac {b^2}{ca} + \frac {c^2}{ab}\right) + 729abc\geq39$.
Sonnhard.[/quote]
thank you. :)

Can anyone find out where my mistake is?

From $ AM\minus{}GM$, $ a^3\plus{}b^3\plus{}c^3\ge3abc \rightarrow \sum \frac{a^2}{bc}\ge 3$
So, $ 4\sum\frac{a^2}{bc}\ge 12$

So, we need to prove $ 729abc\ge 39\minus{}12\equal{}27$
$ abc\ge\frac{1}{27}$ but this is a contradiction since $ 1\equal{}a\plus{}b\plus{}c\ge 3\sqrt[3]{abc} \rightarrow abc\le \frac{1}{27}$  :(

[quote="IW@IT"]So, we need to prove $ 729abc\ge 39 \minus{} 12 \equal{} 27$[/quote]
You don't need to.

[quote="IW@IT"]Can anyone find out where my mistake is?

From $ AM \minus{} GM$, $ a^3 \plus{} b^3 \plus{} c^3\ge3abc \rightarrow \sum \frac {a^2}{bc}\ge 3$
So, $ 4\sum\frac {a^2}{bc}\ge 12$

So, we need to prove $ 729abc\ge 39 \minus{} 12 \equal{} 27$
$ abc\ge\frac {1}{27}$ but this is a contradiction since $ 1 \equal{} a \plus{} b \plus{} c\ge 3\sqrt [3]{abc} \rightarrow abc\le \frac {1}{27}$  :([/quote]
your solutions is false. :mad:  :mad:  :mad:

[quote="quykhtn-qa1"]your solutions is false. :mad:  :mad:  :mad:[/quote]
There's no reason to be angry about someone finding a bogus solution.

[hide="my progress"]Let $ a,b,c$ be positive numbers such that $ a\plus{}b\plus{}c\equal{}1$. Prove that
$ 4\left(\frac{a^2}{bc}\plus{}\frac{b^2}{ca}\plus{}\frac{c^2}{ab}\right)\plus{}729abc\geq 39$

The sum inside the parentheses can be written as $ \frac{a^3\plus{}b^3\plus{}c^3}{abc}$. Therefore we have that the original inequality is equivalent to

$ 4\left(\frac{a^3\plus{}b^3\plus{}c^3\minus{}3abc}{abc}\right)\plus{}729abc\geq 27$
or
$ 4\left(\frac{(a\plus{}b\plus{}c)((a\minus{}b)^2\plus{}(b\minus{}c)^2\plus{}(c\minus{}a)^2)}{2abc}\right)\plus{}729abc\geq 27$
or
$ \frac{2((a\minus{}b)^2\plus{}(b\minus{}c)^2\plus{}(c\minus{}a)^2)}{abc}\plus{}729abc\geq 27$[/hide]

[quote="IW@IT"]Can anyone find out where my mistake is?

From $ AM \minus{} GM$, $ a^3 \plus{} b^3 \plus{} c^3\ge3abc \rightarrow \sum \frac {a^2}{bc}\ge 3$
So, $ 4\sum\frac {a^2}{bc}\ge 12$

So, we need to prove $ 729abc\ge 39 \minus{} 12 \equal{} 27$
$ abc\ge\frac {1}{27}$ but this is a contradiction since $ 1 \equal{} a \plus{} b \plus{} c\ge 3\sqrt [3]{abc} \rightarrow abc\le \frac {1}{27}$  :([/quote]

Equality occurs in $ 4\sum\frac {a^2}{bc}\ge 12$ only when equality occurs in $ abc\le \frac {1}{27}$ - if we change $ a,b,c$ in the second inequality so it's less, then the first inequality will become less sharp. Your mistake lies in thinking that the statement "the inequality is true if $ abc\le \frac {1}{27}$" implies  "the inequality is true [b]iff[/b] $ abc\le \frac {1}{27}$"

Hmph; I'm fairly sure Jensen can crack this problem due to the "smoothing" involved which I mentioned above (though I'm equally sure that Jensen is unnecessarily complicated and that some clever AMGM can do the job equally well), but I can't seem to find the right application of it.

[hide][quote="Temperal"]
Hmph; I'm fairly sure Jensen can crack this problem due to the "smoothing" involved which I mentioned above (though I'm equally sure that Jensen is unnecessarily complicated and that some clever AMGM can do the job equally well), but I can't seem to find the right application of it.[/quote]
Using classical inequality on this one is not an easy task on this problem even when you use Weighted AM-GM.  Factorization, on the other hand, works perfectly on this kind of question!
Here is my solution:
Rewrite the inequality we need to prove:
$ 4\frac{a^3 \plus{} b^3 \plus{} c^3 \minus{} 3abc}{abc} \geq 27(1 \minus{} 27abc)$
Now factorizing both sides wisely we get:
$ 2[(a \minus{} b)^2 \plus{} (b \minus{} c)^2 \plus{} (c \minus{} a)^2]\geq 27abc\frac{(a \minus{} b)^2 \plus{} (b \minus{} c)^2 \plus{} (c \minus{} a)^2}{2} \plus{} 81abc[c(a \minus{} b)^2 \plus{} b(c \minus{} a)^2 \plus{} a(c \minus{} b)^2]$
Now grouping each $ (a \minus{} b)^2,(b \minus{} c)^2$ and $ (c \minus{} a)^2$, we get:
$ (a \minus{} b)^2(2 \minus{} 27abc \minus{} 81abc^2) \plus{} (b \minus{} c)^2(2 \minus{} 27abc \minus{} 81ab^2c) \plus{} (2 \minus{} 27abc \minus{} 81a^2bc)(c \minus{} a)^2\geq 0$
Applying the 2nd standard of S.O.S method with assumption $ a\geq b\geq c$, we're done(if you haven't known this method, find it on the "Olympiad Inequality box-Theorems and Formula").
Well, this problem is more complicated than it looks. :wink: . Hope to see a better solution![/hide]

[quote="ghjk"][hide][quote="Temperal"]
Hmph; I'm fairly sure Jensen can crack this problem due to the "smoothing" involved which I mentioned above (though I'm equally sure that Jensen is unnecessarily complicated and that some clever AMGM can do the job equally well), but I can't seem to find the right application of it.[/quote]
Using classical inequality on this one is not an easy task on this problem even when you use Weighted AM-GM.  Factorization, on the other hand, works perfectly on this kind of question!
Here is my solution:
Rewrite the inequality we need to prove:
$ 4\frac {a^3 \plus{} b^3 \plus{} c^3 \minus{} 3abc}{abc} \geq 27(1 \minus{} 27abc)$
Now factorizing both sides wisely we get:
$ 2[(a \minus{} b)^2 \plus{} (b \minus{} c)^2 \plus{} (c \minus{} a)^2]\geq 27abc\frac {(a \minus{} b)^2 \plus{} (b \minus{} c)^2 \plus{} (c \minus{} a)^2}{2} \plus{} 81abc[c(a \minus{} b)^2 \plus{} b(c \minus{} a)^2 \plus{} a(c \minus{} b)^2]$
Now grouping each $ (a \minus{} b)^2,(b \minus{} c)^2$ and $ (c \minus{} a)^2$, we get:
$ (a \minus{} b)^2(2 \minus{} 27abc \minus{} 81abc^2) \plus{} (b \minus{} c)^2(2 \minus{} 27abc \minus{} 81ab^2c) \plus{} (2 \minus{} 27abc \minus{} 81a^2bc)(c \minus{} a)^2\geq 0$
Applying the 2nd standard of S.O.S method with assumption $ a\geq b\geq c$, we're done(if you haven't known this method, find it on the "Olympiad Inequality box-Theorems and Formula").
Well, this problem is more complicated than it looks. :wink: . Hope to see a better solution![/hide][/quote]

this inequaliti uses SCHUR inequaliti. :)

[quote="nguoivn"][quote="quykhtn-qa1"]nguoivn:you can solve this inequality? :)[/quote]
Your ineq is obviously trues because: $ \frac {(a \plus{} b \plus{} c)^3}{abc} \plus{} \frac {729abc}{(a \plus{} b \plus{} c)^3} \geq\ 54$
PS: If you want your ineq is taken care of many people, you should post it in box "Inequality".
I think you'll receive many solutions for them. I rarely go to this box  :oops:[/quote]
your proof very nice  :)


Prove there is no function f: (0, :inf: )-->(0,:inf: ) such that:

f(x+y)>=f(x)+yff(x) for all x,y in  (0, :inf: )

First step: f goes to infinity for x to infinity. This is trivial.Second step: $ f(x+1)-f(x)$ goes to infinity.Obvious. Third step: f is incresing. Next: $\frac{f(x)}{x}$ goes to infinity. But  this easily implies the contradiction with the monotonicity.

To complete it:

After we prove that $\frac{f(x)}x\to\infty$ as $x\to\infty$ (also keep in mind that $f(x)-x\to\infty$), look at the relation $\frac{f(x+y)}{f(f(x))}\ge \frac{f(x)}{f(f(x))}+y$. Fix $y$ and make $x\to\infty$. The left part goes to $0$, while the right part goes to $y>0$, which yields the contradiction.


[quote="LordoftheQRings"]In Russian, when voiced consonants are at the end of a word, they become unvoiced, and when an unvoiced consonant is before a voice (except в), it becomes voiced:
б $ \leftrightarrow$ п
в $ \rightarrow$ ф
г $ \leftrightarrow$ к
д $ \leftrightarrow$ т
ж $ \leftrightarrow$ ш
з $ \leftrightarrow$ с
Three of these--б$ \rightarrow$п, г$ \rightarrow$к, д$ \rightarrow$т--correspond to Grimm's law for Germanic sound shifts. I know that Russian is not a Germanic language, but is there any sort of general law for the Russian and/or the Slavic languages?[/quote]

Early Slavic languages had only open syllables, so they never had consonants at the end of a word at all. Hence it must be a strictly Russian phenomenon. I'm not sure what you mean by an unvoiced consonant being before a voice, but if that means there's a syllable break in between, it can't be a common Slavic feature by the same token. If you mean something like tz->dz, it seems only natural since tz is very difficult to pronounce (to try to use the same rule yet again, it is impossible to start a syllable with tz-). There are certainly laws for phonetic evolution for every language, though, and this classifies as a law for Russian, but not Slavic in general.

By the way, have you looked at other Slavic languages to see if this holds? Particularly those which are not closely related to Russian, like Bulgarian, Czech, Polish, etc. I don't know anything about Russian or the alphabet or anything so I can't really check very well.

Is anyone familiar with Trubeckoy and Avanesov, the two Russian linguists?
Can anyone give me a general overview of their work?
 :lol:  :maybe:

Lol, do you think St. Cyril wanted to do this when he grew up?


[quote="kihe_freety5"]a,b,c>0
prove that:

(a+b+c)($ \frac {a^2}{b}$+$ \frac {b^2}{c}$+$ \frac {c^2}{a}$)+3(ab+bc+ac) ≥6(a²+b²+c²)[/quote]

 use the inequality in (strong inequality by M.A) :wink:

[quote="kihe_freety5"]could you please post your full solution?
help me!! :)[/quote]
[url=http://www.mathlinks.ro/viewtopic.php?t=198496]See here[/url] :)


Use Chinese remainder theorem, or otherwise, find the last four digits of 1x3x5x7x...x2007.

You've already posted this here:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=157704[/url]

Do not double post.


[b]Problem: [/b]Let  be positive real numbers such that $ a^2 \plus{} b^2 \plus{} c^2 \plus{} 2abc \equal{} 1$
Prove that:
$ 3(a \plus{} b \plus{} c) \ge 4 \plus{} 4abc$

P/s: I think we will use The mixing variables method or [b]Schur Inequality[/b] to prove it. But I haven’t succeed yet.
  I hope you can help me. Thank you!

[quote]Try a=1,b=c=0.[/quote]
what is that supposed to mean?

[quote="Dijkschneier"][quote="daigiaga1994"][b]Problem: [/b]Let  be positive real numbers such that $ a^2 \plus{} b^2 \plus{} c^2 \plus{} 2abc \equal{} 1$
Prove that:
$ 3(a \plus{} b \plus{} c) \ge 4 \plus{} 4xyz$
[/quote]
What are $ x$, $ y$ and $ z$ ?[/quote]
oh, I'm really stupid. sorry
I'll edit my post at once.
thank you!

[quote="daigiaga1994"][quote="mathstarofvn"]Try a=1,b=c=0.[/quote]
oh, I'm  sorry. "a,b,c are positive real numbers".
I'll edit my post.
Thank you![/quote]

You should realise that this won't change anything. The inequality is still wrong.

You can still make $ a\equal{}1$, $ b\equal{}c\to0$


group theory
abstract algebra
induction
superior algebra
superior algebra unsolved

So here is the following problem.

let $ G$ be a p-group of order $ p^n$. for any $ 0\leq k \leq n$ show that there exists a normal subgroup of $ G$ of order $ p^k$.

Any help is appretiated.

Here's a hint:

1) Show that there is a normal subgroup of order $ p$.

2) Use 1), and induction on $ k$.

that is a good idea. How to show that there is one normal subgroup of order $ p$?

$ p$-groups have nontrivial center, and the center is a (normal) subgroup with order a power of $ p$, hence contains a cyclic subgroup of order $ p$ (which is also normal).


Tyler has entered a buffet line in which he chooses one kind
of meat, two different vegetables and one dessert.  If the order of
food items is not important, how many different meals might he
choose? \\ \\
Meat:  beef, chicken, pork \\
Vegetables:  baked beans, corn, potatoes, tomatoes \\
Dessert:  brownies, chocolate cake, chocolate pudding, ice cream

Tyler can choose 1 type of meat out of the 3 types available.
For meat, $ 3$ ways.
In the vegetable section, there are 4 possibilities, but he must choose 2.
There are $ \frac{4!}{2!*2!} \equal{} 6$ ways he could do this.
For dessert, he can only pick 1 out of the 4 desserts.
So $ 4$ ways.
These events are dependent, so there are $ 3*6*4 \equal{} \boxed{72}$ meals he could choose from.


Compute all values of x so that the product |x-1/2||x+1/2| is prime?

Thank You.

[quote="subhash333"]Compute all values of x so that the product |x-1/2||x+1/2| is prime?[/quote]
Do you mean those to be absolute value bars or parentheses?
Either way, there's an infinite number of values for x.  Just dealing with $x>1/2$, so we don't have to worry about absolute values, for each prime number $p$, $x=\sqrt{p+1/4}$ is solution.  Since there are an infinite number of positive primes, there's an infinite number of solutions to this problem.

[hide="solution"]note that the difference of those is 1, hence the product must be even

so we either have [x-1/2]=-2, [x+1/2]=-1 or [x-1/2]=1, [x+1/2]=2

from that we get the solution set $-\frac{3}{2}\le x<-\frac{1}{2}$ and $\frac{3}{2}\le x<\frac{5}{2}$.[/hide]

[quote="junggi"]i believe it's supposed to be floors...
(it's from last year's arml)[/quote]
Oh, OK, that would make more sense.

Find all real $x$ such that
\[\left\lfloor x-\frac12\right\rfloor \left\lfloor x+\frac12\right\rfloor \]
is prime.


geometry MATHCOUNTS probability calculus derivative integration AMC

Hey I saw this game elsewhere and it turned out to be immensely popular!

Answer the previous OR question and put a new OR question of your own!

Here goes a common one -
[b]AoPS or Mathlinks?[/b]
(It sounds familiar)

Come on; get going!

father...236factorial or grobber? :rotfl:

Yes, that is dumb.

Coke

Friday or Saturday?

Math.  :D ( no seriously, food )

  Pure or Practical Applications?

Pure

Death by falling or death by poison?

poison.
Harry Potter or Hitchhiker's Guide?

urghhhhhhhhhh hard one

nowitzki... cuz he's hurt :wink:

zeitz or engel?

nope i don't know any of them  :D

Xbox 360 or PS3?

umm xbox sounds culer?

easter bunnies or easter monster?

 :clap2:
Left, where nothing is right, or right, where nothing is left

right, where nothing is left
 :wallbash: or  :wallbash_red:


If a scale model of a statue has a height of 1 foot, and the height of the actual statue is 10 feet, how many times larger is the volume of the actual statue than the volume of its scale model, assuming both are constructed in a similar manner?

We do $ 10^3$ because volume is measured in cubic units and measures a three-dimensional quantity while 1 dimension is just normal. That's why the answer is 1000.

[hide="answer"]a volume is measured in cubes and we know that 1foot=10 feet so we get the equation

10^3=1000[/hide]

[b][i][u]please rate post thanks! [=[/u][/i][/b]

please be fair though!! [=

 :weightlift:  :weightlift:  :weightlift:  :weightlift:


What is the sum of the median and the mean of the set of numbers 12, 21, 6, 11 and 30?

well, the median is the middle number, so we'll line up the numbers from smallest to largest
$ 6, 11, 12, 21, 30$
our median is 12

the mean is the average of the numbers, so we add the number together and divide by the total number of numbers
$ 6\plus{}11\plus{}12\plus{}21\plus{}30\equal{}80$
$ 80/5\equal{}16$
our mean is 16

the sum of the median and the mean would be $ 12\plus{}16\equal{}28$


If the first day of a month is Monday, what day of the week is the twenty third day?

If the first day is a Monday, then so are the 8th, 15th, and 22nd days.  So the 23rd day is a Tuesday.

Yes, I miscalculated and thought 7th,14th,21st. So I put Wednesday.

███████████████████████████████████████████████████████████████████████████████████████ The answer is Tuesday, don't be confused.                                                                                █
█                                                                                █
█           ~R@d!OAct!vÉ                                                                                █
██████████████████████████████████████████████████████████████████████████████████████
██████████████████████████████████████████████████████████████████████████████████████


Given $ a, b, c > 0$ and $ a\plus{}b\plus{}c\equal{}3$. Prove that
\[ \frac{a\sqrt{a(b\plus{}c\plus{}1)}}{2b\plus{}c}\plus{}\frac{b\sqrt{b(c\plus{}a\plus{}1)}}{2c\plus{}a}\plus{}\frac{c\sqrt{c(a\plus{}b\plus{}1)}}{2a\plus{}b} \ge \sqrt{3}\]
 :)

[quote="nguoivn"]Given $ a, b, c > 0$ and $ a \plus{} b \plus{} c \equal{} 3$. Prove that
\[ \frac {a\sqrt {a(b \plus{} c \plus{} 1)}}{2b \plus{} c} \plus{} \frac {b\sqrt {b(c \plus{} a \plus{} 1)}}{2c \plus{} a} \plus{} \frac {c\sqrt {c(a \plus{} b \plus{} 1)}}{2a \plus{} b} \ge \sqrt {3}\]
:)[/quote]
Using Holder's Inequality in combination with the known inequality $ (a \plus{} b \plus{} c)^2 \ge 3(ab \plus{} bc \plus{} ca)$, we have
$ \left[ \sum \frac {a\sqrt {a(b \plus{} c \plus{} 1)}}{2b \plus{} c}\right]^2 \left( \sum \frac {2b \plus{} c}{b \plus{} c \plus{} 1}\right) \left[ \sum a(2b \plus{} c)\right] \ge \left( \sum a\right)^4 \ge 3\left( \sum a\right)^2\left( \sum ab\right).$
It follows that
$ \left[ \sum \frac {a\sqrt {a(b \plus{} c \plus{} 1)}}{2b \plus{} c}\right]^2 \left( \sum \frac {2b \plus{} c}{b \plus{} c \plus{} 1}\right) \ge \left( \sum a\right)^2 \equal{} 9.$
From this, we see that it suffices to prove that the following inequality holds
$ \frac {2b \plus{} c}{b \plus{} c \plus{} 1} \plus{} \frac {2c \plus{} a}{c \plus{} a \plus{} 1} \plus{} \frac {2a \plus{} b}{a \plus{} b \plus{} 1} \le 3.$
Since $ \frac {2b \plus{} c}{b \plus{} c \plus{} 1} \equal{} \frac {b \minus{} c \minus{} 3}{2(b \plus{} c \plus{} 1)} \plus{} \frac {3}{2},$ this inequality can be written as
$ \frac {3 \plus{} c \minus{} b}{1 \plus{} b \plus{} c} \plus{} \frac {3 \plus{} a \minus{} c}{1 \plus{} c \plus{} a} \plus{} \frac {3 \plus{} b \minus{} a}{1 \plus{} a \plus{} b} \ge 3.$
Since $ 3 \plus{} c \minus{} b > 0,$ $ 3 \plus{} a \minus{} c > 0$ and $ 3 \plus{} b \minus{} a > 0,$ by the Cauchy-Schwarz Inequality, we have
$ \sum \frac {3 \plus{} c \minus{} b}{1 \plus{} b \plus{} c} \ge \frac {\left[ \sum (3 \plus{} c \minus{} b)\right]^2}{\sum (3 \plus{} c \minus{} b)(1 \plus{} b \plus{} c)} \equal{} 3.$


Ο κύκλος με ακτίνα ρ μπορεί να εφάπτεται και στις προεκτασεις των πλεύρων??

Ναι, γίνεται.

[quote="staretak"][quote="Nick Rapanos"][quote="maniopas"]Έστω $ ABC$ ισοσκελές τρίγωνο με βάση $ BC$. Αν $ r,R$ οι ακτίνες του εγγεγραμμένου και περιγεγραμμένου κύκλου αντίστοιχα και $ p$ η ακτίνα του κύκλου που εφάπτεται στα $ AB , AC$ και στον περιγγεγραμμένου κύκλου, μπορούμε να υπολογίσουμε το $ p$ συναρτήσει των $ r, R$ [u]μόνο[/u];

Ο τύπος που βγάζω έχει πάντα κάποιες πλευρές μέσα. :(  Και δε μπορώ να σκεφτώ τίποτα πέρα από ένα σημείο.  :mad: Όποιος μπορεί ας μου γράψει (αν βαριέται να το γράψει όλο) απλά κάποια υπόδειξη ή το αποτέλεσμα.[/quote]

Νομίζω ότι μπορείς να υπολογίσεις όλες τις πλευρές συναρτήσει των $ R,r$ οπότε όλα τα προβλήματα σου τελειώνουν.

Για ρίξε μια ματία στο σχήμα.δ[/quote]

Niko giati grafeis oti to MT einai gnwsto?? Den xreiazetai katholou...[/quote]

Eipa ego oti xreiazetai? apla ithela na deikso oti ola auta ta tmhmata einai gnosta kai epeidi akrivos den upirxe logos na grapso me poso akrivos einai iso to MT gi auto protimisa na arkesto sto gnosto...Sorry an berdepsa kapos ta pragmata..

@@@staretak: pou isouna xthes re oli mera pou se epsaxna?

Okay Nikola....

sorry, alla eida ta minimata sou afou mou ta eixes steilei. Simera pantos eimai olo online(mprosta sto pc) alla den se eida mesa. Pantos gia na apologithw tha sou po ena anekdoto.

Poia einai i diafora metaksi enos kounoupiou kai mias ginaikas?

[hide]Otan to kounoupi se roufaei den xreiazetai na tou xaideueis to kefali!..hahahaha :rotfl:  :rotfl: [/hide]

[quote="staretak"]

Pantos gia na apologithw tha sou po ena anekdoto.

Poia einai i diafora metaksi enos kounoupiou kai mias ginaikas?

[hide]Otan to kounoupi se roufaei den xreiazetai na tou xaideueis to kefali!..hahahaha :rotfl:  :rotfl: [/hide][/quote]

ok stefane...katalava gt den borousa na se vro! [b]profanos[/b] den se tsimpise kounoupi ee?


Let (a_n), n a natural number, a sequence of positive reals numbers such that
a_n+1 = sqrt(6 − 2a_n²) for all n>=1
Show that the sequence is constant.

Hmm, do we have $ a_{1}$ given?
Let us have for example $ a_{1} \equal{} 2$
Then:
$ a_{2} \equal{} \sqrt {6 \minus{} 2*2} \equal{} \sqrt {2}\neq 2 \equal{} a_{1}$ and for this $ a_{1}$ the sequence is not constant, so sth is missing...

It will be constant only for $ a_{1}\equal{}\minus{}1\plus{}\sqrt{7}$ :wink:


Find formulas for $ (f \circ g)(x)$ if

$ f(x)\equal{} 1$,if  $ x< 0$
$ f(x)\equal{}2x^2$,if  $ 0\le x \le1$
$ f(x)\equal{}1$,if    $ x>1$
and
$ g(x)\equal{}2$,if  $ x<0$
$ g(x)\equal{}3x$,if  $ 0 \le x \le1$
$ g(x)\equal{}2$,if    $ x>1$

If $ x < 0$, then $ g(x) \equal{} 2$, so $ f(g(x)) \equal{} 1$.

If $ 0 \leq x \leq \frac13$, then $ 0 \leq g(x) \leq 1$, so $ f(g(x)) \equal{} 2g(x)^2 \equal{} 2(3x)^2 \equal{} 18x^2$.

There are some other cases, but I'll leave it at that.

You only left out one case: $ x > \frac {1}{3} \Longrightarrow (f \circ g)(x) \equal{} 1$.


list good ways to find looooong digits of pi
1. use integration to to find (arctan 1)*4 in base 1000000 with a computer and so on :)

cf. [url=http://www.super-computing.org/index.html]Kanada Laboratory home page[/url]
(Kanada held the world record calculating the number of digits in the decimal expansion of pi.)


search
real analysis
real analysis unsolved

Does there exist an infinite $\sigma$-algebra which has only countably many members?

[quote="bubka"]I thought sigma algebras (borel sets) countable unions of open sets.[/quote]

What are you trying to say?

Anyways, the answer is 'no'.  Search the forum, it's been done before.

Who can help me to look for the previous link of the problem? Thank you!


modular arithmetic
number theory unsolved
number theory

Suppose that $ m\in\mathbb N$ and $ f(x_{1},\cdots, x_{n})\in\mathbb N[x_{1},\cdots,x_{n}]$. Let $ T$ be the number of solutions to \[ f(x_{1},\cdots, x_{n})\equiv 0\pmod m.\] Prove that \[ T\equal{}\frac{1}m\sum_{a\equal{}0}^{m\minus{}1}\sum_{x_{1}\equal{}0}^{m\minus{}1}\cdots\sum_{x_{n}\equal{}0}^{m\minus{}1}\exp(2\pi i\cdot af(x_{1},\cdots,x_{n})).\]

Shouldn't it be

\[ T\equal{}\frac{1}{m}\sum_{a\equal{}0}^{m\minus{}1}\sum_{x_{1}\equal{}0}^{m\minus{}1}\cdots\sum_{x_{n}\equal{}0}^{m\minus{}1}\exp\left(\frac{2\pi i}{m}\cdot af(x_{1},\cdots,x_{n})\right).\]

?

The problem becomes close to trivial if you notice that for any integer $ k$, we have $ \frac{1}{m}\sum_{a\equal{}0}^{m\minus{}1}\exp\left(\frac{2\pi i}{m}\cdot ak\right)\equal{}\left\{\begin{array}{c}1,\text{ if }m\mid k;\\0,\text{ if }m\nmid k\end{array}\right.$.

  darij


Let $ p > 3$ be a prime and $ n > p$ an integer such that the order of $ n$ modulo $ p$ is $ p \minus{} 1$. Prove that there exists a multiple of $ p$ whose digits in base $ n$ are $ 1,2,\dots,p \minus{} 1$ in some order (i.e. the multiple is of the form $ \overline{a_1a_2\dots a_{p \minus{} 1}}$ in base $ n$).

As the order of $ n$ modulo $ p$ is $ p\minus{}1$. So $ 1,n,n^2,..., n^{p\minus{}2}$ forms a complete system of residue modulo $ p$.
So for each $ i$, if $ n^i \equiv k \mod p$, $ k\equal{}1,2,3, \dots p\minus{}1$, put $ a_{p\minus{}1\minus{}i} \equal{} k$, then $ \overline{a_1a_2\dots a_{p \minus{} 1}} \equiv 1^2 \plus{} 2^2 \plus{} \dots \plus{} (p\minus{}1)^2 \equal{} \frac{(p\minus{}1)p(2p\minus{}1)}{6} \equiv 0 \mod p$.


What is the largest value of $ n$ less than 100,000 for which the expression $ 8(n\minus{}2)^5\minus{}n^2\plus{}14n\minus{}24$ is a multiple of 5?

Factoring, we eventually get $ (n \minus{} 2)((8(n \minus{} 2)^4 \minus{} (n \minus{} 12))$. Thus, we either have $ n \equal{} 7$, or the other factor a multiple of $ 5$. For the $ (n \minus{} 2)^4$ part, we check 4th power residues in modulo 5 for $ 0,1,2,3,4$= $ 0,1,1,1,0$ (8 times those mod 5 gives us $ 0,3,3,3,0$). The $ (n \minus{} 12)$ part will then have residues $ 0,1,1,1,0$. Finally, all of our possible (final) values are $ 0,2,2,2,0$. That means we either have $ n \minus{} 2 \equal{} 5k\implies{n \equal{} 5k \plus{} 2}$ or $ n \minus{} 2 \equal{} 5k \plus{} 4\implies{n \equal{} 5k \plus{} 6}$ for some integer $ k$. Since $ 5k\leq99995$, our largest possible value of $ n$ is $ \boxed{99997}$.

I would've gotten this right if not for the extra 9 ._.


[size=117][color=darkred][b]For one beginner, i.e. who don't know [u]conjugate[/u] , [u]harmonic[/u], [u]polar[/u] ! [/b][/color][/size]

[color=darkred]Let $ ABCD$ be a convex quadrilateral inscribed in the circle $ w$. The tangent $ AA$ to the circle $ w$ in the point $ A$ meet the lines $ BD$, $ CB$, $ CD$ in the points $ X$, $ Y$, $ Z$ respectively. Prove that the following three relations are equivalently :[/color]

$ 1\blacktriangleright\frac{XY}{XZ}=\frac{AY}{AZ}$.

[hide="Means that ..."]the points $ X$, $ A$ are [b]conjugate[/b] w.r.t. the points $ Y$, $ Z$ ; with another words, $ (X,Y,A,Z)$ is an [b]harmonical division[/b] on the line $ AA$.[/hide]
$ 2\blacktriangleright AB\cdot CD = AD\cdot BC$.

[hide="Means that ..."]the cyclic $ ABCD$ is [b]a harmonical quadrilateral.[/b][/hide]
$ 3\blacktriangleright X\in CC$.

[hide="Means that ..."]the line $ AC$ is the polar of the point $ X$ w.r.t. the circle $ w$.[/hide]
[hide="Proof."]
[color=darkblue]$ \left\{\begin{array}{ccccc}\triangle\ YAB\sim\triangle\ YCA &\implies &\frac{AB}{CA}=\frac{BY}{AY}&\implies & BY =\frac{AB\cdot AY}{CA}\\ \\ \triangle\ ZAD\sim\triangle\ ZCA &\implies &\frac{AD}{CA}=\frac{DZ}{AZ}&\implies & DZ =\frac{AD\cdot AZ}{CA}\end{array}\right\|$ $ \implies$ $ \frac{DZ}{BY}=\frac{AZ}{AY}\cdot\frac{AD}{AB}\ \ (*)$.

Apply the Menelaus' theorem to the transversal $ \overline{XBD}$ of $ \triangle CYZ$ and use the relation $ (*)$ :

$ \frac{XY}{XZ}\cdot\frac{DZ}{DC}\cdot\frac{BC}{BY}= 1$ $ \Longleftrightarrow$ $ \frac{XY}{XZ}\cdot\left(\frac{AZ}{AY}\cdot\frac{AD}{AB}\right)\cdot\frac{BC}{DC}= 1$ $ \Longleftrightarrow$ $ \left(\frac{XY}{XZ}\cdot\frac{AZ}{AY}\right)\cdot\left(\frac{AD}{AB}\cdot\frac{BC}{DC}\right) = 1$.

Observe that $ \boxed{\frac{XY}{XZ}=\frac{AY}{AZ}\Longleftrightarrow AB\cdot CD = AD\cdot BC}$.

Denote $ X'\in AA\cap CC$ and $ \{B',X'\}= X'D\cap w$. From the relations $ \left\{\begin{array}{c}\triangle\ AB'D\ :\ \frac{X'B'}{X'D}=\left(\frac{AB'}{AD}\right)^{2}\\ \\ \triangle\ CB'D\ :\ \frac{X'B/}{X'D}=\left(\frac{CB'}{CD}\right)^{2}\end{array}\right\|$ $ \implies$

$ \frac{CB'}{CD}=\frac{AB'}{AD}$ $ \implies$ $ \frac{B'A}{B'C}=\frac{DA}{DC}$. Therefore, $ AB\cdot CD = AD\cdot BC$ $ \Longleftrightarrow$ $ \frac{DA}{DC}=\frac{BA}{BC}$ $ \Longleftrightarrow$ $ \frac{B'A}{B'C}=\frac{BA}{BC}$ $ \Longleftrightarrow$

$ B'\equiv B$ $ \Longleftrightarrow$ $ X'\equiv X$ $ \Longleftrightarrow$ $ X\in CC$. In conclusion, $ \boxed{AB\cdot CD = AD\cdot BC\Longleftrightarrow X\in CC}$.[/color][/hide]
[color=darkblue]See for example, http://www.mathlinks.ro/Forum/viewtopic.php?t=164992 [/color]

Great work Virgil Nicula. :)
If someone want to know more about harmonic,you can read about it in Cosmin Pohoata's post: http://www.mathlinks.ro/viewtopic.php?t=161310

[quote="Virgil Nicula"][size=117][color=darkred][b]For one beginner, i.e. who don't know [u]conjugate[/u] , [u]harmonic[/u], [u]polar[/u] ! [/b][/color][/size]

[color=darkred]Let $ ABCD$ be a convex quadrilateral inscribed in the circle $ w$. The tangent $ AA$ to the circle $ w$ in the point $ A$ meet the lines $ BD$, $ CB$, $ CD$ in the points $ X$, $ Y$, $ Z$ respectively. Prove that the following three relations are equivalently :[/color]

$ 1\blacktriangleright\frac {XY}{XZ} \equal{} \frac {AY}{AZ}$.

[hide="Means that ..."]the points $ X$, $ A$ are [b]conjugate[/b] w.r.t. the points $ Y$, $ Z$ ; with another words, $ (X,Y,A,Z)$ is an [b]harmonical division[/b] on the line $ AA$.[/hide]
$ 2\blacktriangleright AB\cdot CD \equal{} AD\cdot BC$.

[hide="Means that ..."]the cyclic $ ABCD$ is [b]a harmonical quadrilateral.[/b][/hide]
$ 3\blacktriangleright X\in CC$.

[hide="Means that ..."]the line $ AC$ is the polar of the point $ X$ w.r.t. the circle $ w$.[/hide]
[hide="Proof."]
[color=darkblue]$ \left\{\begin{array}{ccccc}\triangle\ YAB\sim\triangle\ YCA & \implies & \frac {AB}{CA} \equal{} \frac {BY}{AY} & \implies & BY \equal{} \frac {AB\cdot AY}{CA} \\
 \\
\triangle\ ZAD\sim\triangle\ ZCA & \implies & \frac {AD}{CA} \equal{} \frac {DZ}{AZ} & \implies & DZ \equal{} \frac {AD\cdot AZ}{CA}\end{array}\right\|$ $ \implies$ $ \frac {DZ}{BY} \equal{} \frac {AZ}{AY}\cdot\frac {AD}{AB}\ \ (*)$.

Apply the Menelaus' theorem to the transversal $ \overline{XBD}$ of $ \triangle CYZ$ and use the relation $ (*)$ :

$ \frac {XY}{XZ}\cdot\frac {DZ}{DC}\cdot\frac {BC}{BY} \equal{} 1$ $ \Longleftrightarrow$ $ \frac {XY}{XZ}\cdot\left(\frac {AZ}{AY}\cdot\frac {AD}{AB}\right)\cdot\frac {BC}{DC} \equal{} 1$ $ \Longleftrightarrow$ $ \left(\frac {XY}{XZ}\cdot\frac {AZ}{AY}\right)\cdot\left(\frac {AD}{AB}\cdot\frac {BC}{DC}\right) \equal{} 1$.

Observe that $ \boxed{\frac {XY}{XZ} \equal{} \frac {AY}{AZ}\Longleftrightarrow AB\cdot CD \equal{} AD\cdot BC}$.

Denote $ X'\in AA\cap CC$ and $ \{B',X'\} \equal{} X'D\cap w$. From the relations $ \left\{\begin{array}{c}\triangle\ AB'D\ : \ \frac {X'B'}{X'D} \equal{} \left(\frac {AB'}{AD}\right)^{2} \\
 \\
\triangle\ CB'D\ : \ \frac {X'B/}{X'D} \equal{} \left(\frac {CB'}{CD}\right)^{2}\end{array}\right\|$ $ \implies$

$ \frac {CB'}{CD} \equal{} \frac {AB'}{AD}$ $ \implies$ $ \frac {B'A}{B'C} \equal{} \frac {DA}{DC}$. Therefore, $ AB\cdot CD \equal{} AD\cdot BC$ $ \Longleftrightarrow$ $ \frac {DA}{DC} \equal{} \frac {BA}{BC}$ $ \Longleftrightarrow$ $ \frac {B'A}{B'C} \equal{} \frac {BA}{BC}$ $ \Longleftrightarrow$

$ B'\equiv B$ $ \Longleftrightarrow$ $ X'\equiv X$ $ \Longleftrightarrow$ $ X\in CC$. In conclusion, $ \boxed{AB\cdot CD \equal{} AD\cdot BC\Longleftrightarrow X\in CC}$.[/color][/hide]
[color=darkblue]See for example, http://www.mathlinks.ro/Forum/viewtopic.php?t=164992 [/color][/quote]
I repaired the one LaTeX comand in this message !


inequalities
function
logarithms
inequalities open

The following may be trivial, but I definitely do not have a general answer:
   What are the values of n such that for all a_1,...,a_n>0 with product 1 we have sum 1/(1+a_i)<=(a_1+...+a_n+n)/4?
    I have proved that 3,4,5 are good values. I strongly believe that the answer is :all n>=2.

To prove for [tex]n \in \mathbb{Z}^{+}[/tex] and positive [tex]a_{i}[/tex] with product 1: [tex]\displaymode{\frac{1}{1 + a_{1}} + \cdots + \frac{1}{1 + a_{n}} \le \frac{a_{1} + \cdots + a_{n} + n}{4}}[/tex]

(All summation will be over [tex]i \in \{1, \dots, n\}[/tex].) Multiply by -1, add [tex]n[/tex], then multiply by [tex]\sum 1 + a_{i} = n + \sum a_{i}[/tex] as shown:

[tex]\displaymode{
\sum \frac{-1}{1 + a_{i}} \ge \frac{-n - \sum a_{i}}{4}}[/tex]
[tex]\displaymath{\sum \frac{a_{i}}{1 + a_{i}} \ge \frac{3n - \sum a_{i}}{4}}[/tex]
[tex]\displaymath{4\left(\sum \frac{a_{i}}{1 + a_{i}}\right)\left(\sum 1 + a_{i} \right) \ge \left(3n - \sum a_{i}\right)\left(n + \sum a_{i}\right)
}[/tex]

Apply Cauchy to the left, and expand the right like so:

[tex]\displaymath{
4\left(\sum \sqrt{a_{i}} \right)^2 \ge \left(2n\right)^2 - \left(n - \sum a_{i}\right)^2
}[/tex]

But by the Power Mean inequality,

[tex]\displaymath{
\left( \frac{\sum \sqrt{a_{i}}}{n} \right)^2 \ge \sqrt[n]{a_{1}a_{2} \cdots a_{n}} = 1 \iff 4\left( \sum \sqrt{a_{i}} \right)^2 \ge 4n^2}[/tex]

Substituting this into the left yields the Trivial Inequality, and we are done.

Yeah, I know this solution, but I didn't insisted on this topic because it is trivial, since it follows directly with Jensen inequality, I never see trivial things...

There's a Jensen solution!?! I thought Jensen pointed the wrong way. We have [tex]s = a_{1} + \cdots + a_{n} \ge n[/tex] by the given, but  because [tex]f(x) = \frac{1}{1+x}[/tex] is convex we have [tex]\frac{1}{1+a_{1}} + \cdots + \frac{1}{1+a_{n}} \ge \frac{n^2}{s+n}[/tex] which is, as far as I can tell, not particularly useful.

Not that trivial application! Take the function $ f(x)=\frac{e^x}{4}-\frac{1}{1+e^x} $ !

Yes, very nice idea. I am looking for more examples of this kind, Ji Chen.


I'm not sure if I am the right person to do this, but I realized that Israel doesn't have a forum... and I think we can certainly use one! Although I don't live there currently, it is important for there to be an Israeli forum, or at least some type of Hebrew speaking forum. I'll try to tell people about this... this looks wierd because this summer I started another forum, the Oregon forum... I don't know this is just an idea, I know there are some very talented Israelis here (for example see some of arqady's posts in inequalities...)

שלום! I am not from Israel, but I speak some conversational Hebrew and I think that a Hebrew speaking forum would be great!

Smiles,
lyra

Ani gamken lo chai be yisreal, aval hayom ani lamadti be kita shel "Conversational Hebrew".
Ze hiye tov meod im efshar matchil "forum" yisreali, ani rotzeh etze hamon... :)

yeah i thought about that too but didn't put it into words...
ani koreani aval garti be israel me 1993 ad 2006 achshav ani be korea menaseh lehikanes le oniversita

and by the way me@home it should be "im efshar lehatchil" and not "matchil"

lyra, ani ish lo isha, shmi Avi
yookon: ani yodea ma kulam omrim, aval ani lo yodea ledaber tov... ani gar be Oregon ve ein anashim she medabrim ivrit po... rak imi..

agreed!! Israeli forum would be awesome!


Show how to simplify the fraction (3^2008)^2/ (3^2007)^2 - (3^2006)^2/ (3^2005)^2
Note that the answer is not 0.

yes, that's what I mean, but the answer is not 0. This is a multiple choice question and none of the choices is 0.

The answer is indeed zero; no ifs, ands, or buts. :P

When three separate people get 0, and they are completely certain, I'm pretty sure that they are all right.

...or the test/book/contest/website/teacher you got it from is wrong... that's always a possibility...

Ah-ha!  Hunter, I think, is correct.

Divide everything by $ (3^{2005})^2$.

$ \frac{27}{9 \minus{} \frac{3}{1}}$

$ \frac{27}{6}$

$ \boxed{\frac{9}{2}}$

no you get $ \frac {729}{81 \minus{} \frac {9}{1}} \neq \frac {9}{2}$ or any other choice
edit: lol i trusted ernie but w/e still not 9/2

Eh... brain freeze.  IDK what I was doing.

lol...

when i saw $ \frac{9}{2}$, i was like

LET'S SIMPLIFY IT TO $ \frac{9}{2}$...

and then i used ernie's method to simplify it

it didn't work :noo:

helijet, if none of us have typed it correctly so far, this would be a good time to learn a little $ \text{\LaTeX}$ and show us what the real problem is.


calculus
integration
function
geometry
calculus computations

A  quartic funtion $ y \equal{} ax^4 \plus{} bx^3 \plus{} cx^2 \plus{} dx\plus{}e\ (a\neq 0)$ touches the line $ y \equal{} px \plus{} q$ at $ x \equal{} \alpha ,\ \beta \ (\alpha < \beta ).$ Find the area of the region bounded by these graphs in terms of $ a,\ \alpha ,\ \beta$.

how many more of these can we expect, kunny ?   :maybe: :D

[quote="kunny"][color=darkred]A  quartic polynominal function $ y = ax^4 + bx^3 + cx^2 + dx + e$, where $ a\ne 0$ touches the line $ y = px + q$ at $ x = \alpha ,\ \beta \ (\alpha < \beta ).$
Find the area of the region bounded by these graphs in terms of $ a,\ \alpha ,\ \beta$.[/color][/quote]
[b][u]Proof.[/u][/b] Suppose w.l.o.g. $ a > 0\ .$ Denote $ \|\begin{array}{c} f(x) = ax^4 + bx^3 + cx^2 + dx + e \\
\ d(x) = px + q\end{array}\ .$

Then $ |f(x) - d(x)| =$ $ a(x - \alpha )^2(x - \beta )^2$ and the required area is

$ \sigma [\Gamma_{f,d}]\equiv\int_{\alpha}^{\beta}|f(x) - d(x)|\ \mathrm {dx} =$ $ a\cdot\int_{\alpha}^{\beta}(x - \alpha )^2(x - \beta )^2\ \mathrm {dx} =$

$ a\cdot\int_{ - \frac {\beta - \alpha}{2}}^{\frac {\beta - \alpha }{2}}[x^2 - (\frac {\beta - \alpha}{2})^2]^2\ \mathrm {dx} =$ $ a\cdot(\frac {\beta - \alpha}{2})^5\cdot\int_{ - 1}^1(x^2 - 1)^2\ \mathrm {dx} =$

$ 2a\cdot(\frac {\beta - \alpha}{2})^5\cdot\int_0^1(x^2 - 1)^2\ \mathrm {dx} =$ $ 2a\cdot(\frac {\beta - \alpha}{2})^5\cdot\frac {8}{15}$ $ \implies$

$ \boxed {\sigma [\Gamma_{f,d}] = \frac {a(\beta - \alpha )^5}{30}}\ .$

That's right. :) Especially the calculation of $ \int_{\alpha}^{\beta}(x\minus{}\alpha)^2(x\minus{}\beta)^2\ dx$ is impressive to us.

You can also evaluate the integral by Integral by parts.

[b]Thank you, Kunny ![/b] Please see
http://www.mathlinks.ro/Forum/viewtopic.php?t=173349


Just to ask, how big was your chapter competition? Mine had 150+ people, and I had to battle to get 4th...

This is just so embarrasing. I get double digits on an AIME! :blush:

Probably 120-150 people, 25 or so teams plus a bunch on individuals. (I got 11th at chapter)

Our chapter had about 120-150 people
btw, what state are you from Penguin Integral?

We had a rather huge chapter competition. There were about 45 or so teams and so about 180 people plus all of the individuals. We still got 3rd place. I got 41 cuz i made a bunch of careless mistakes. :(

My chapter had about 60 people and eight schools.
I got first with a 45 on the sprint/target.
Unfortunately, Philip who got third at state last year (and got third at county on written :o ) went on a rampage and beat me 3-2 in countdown :oops: . Still, I did better than the person before me who he beat in countdown 3-0.

You guys had 45 teams? :o  That's crazy.  We had about 8 teams and 60 people at our chapter.  There were like 3 homeschools and private schools with one or two people too.  You only needed 30 to get in the top ten, and I only needed 43 to win.

Ours was around 90 people, but at states, I talked with other people and heard that they had a lot more people at their Chapter Competitions. Our state (Connecticut) only had 5 chapters and it ended up 3 people from the top 4 of our chapter ended up placing in the top 4 at states meaning we go to Nationals :D

Houston's a huge city and with its suburbs its even bigger. Especially Sugarland, a suburb. Both of their schools always gets to state but we managed to beat one of them at chapter this year.  :lol:

We had around 300-400 people in our chapter.  I don't think it is fair because it is the hardest to get to state from here...

Our chapter had the top two teams and the top 5 individuals (there may be more but I know for certain we have 1-5 and 1-2) in state

:o  :o  How do you even fit in the building?

We had about 200 people in our chapter I think... (Santa Clara Valley Chapter)

Our state is on the 18th. My goal is to get top 10

I figure I have states nailled, because everyone that has gone to states last year came from our chapter, and I beat most of them in the countdown.

we had more than like 200 ppl in our chapter....i scrued up  (didn't read the directions) so i got a crummy score of 40 and still got 2nd place in South CA indi but i got 6th cuz of stupid tiebreaker. now im just n alternate 4 state. :cursing:  :furious:  :mad:

[b]3[/b] people from my chapter made Nationals this year.

$3$ people from my Chapter (smallest in the state) made Nationals this year also.

[quote="Klebian"]3 people made nats from my chapter too! XD
4 last year though...  :P[/quote]

I think we have 0 last year, quite a difference, eh?

top five are all from our chapter, and the top two teams =P can't beat that

Vermont southwest had about 100 kids.  Vermont has 4 chapters, each about that size.

Countdown doesn't count, but they do two 12-person countdowns, so lots of kids get to participate.  They do a normal countdown for the top 12 places, plus another countdown for the top 12 6th and 7th graders who aren't in the "real" top 12.  (I'm not sure where the problems come from for that one -- they don't use the official countdown problems for it because they do it first, and need to be sure they won't run out for the official countdown.)

I'm not sure if anyone has ever gone to nationals from Hampton Roads.  Last year, someone from the Eastern Shore won, but North VA always dominates.

0 from my chapter is going to nationals.

4 for me!! (and 4 last year too and 4 the year before that and...)
yeah were DOMINANT

[quote="Klebian"]Whoa I bet your "chapter" took the top... no, all the teams!
Delaware domination![/quote]

Anyone not get the joke?

[quote="Klebian"]Whoa I bet your "chapter" took the top... no, all the teams!
Delaware domination![/quote]
I mean its AMAZING how we dominate.
Top 120 were from our chapter, as well as top 16 teams.

Were even better than greater houston chapter :o

Whatever other chapter, I bet they can't get 120 people into the state round

Delaware is the :coolest: state, exactly why 120 people from a chapter can come to states


[quote="mna851"]I've noticed that in past years there is only one floor value for sophmores and below.  Will it be like that again this year?  Also, if it will, couldn't theoretically, the juniors and seniors from the AIME II lower this years floor?[/quote]

I don't know last year something wierd happend... does anyone know someone who qualified for USAMO by getting an 8 on AIMEII? Possibly because very few people who scored in the 140's took AIME II so no one was able to set the floor?

Yea maybe.  Well also this year, I believe TJ is taking the AIME II, so there are probably some 140's over there.


modular arithmetic
induction
combinatorics proposed
combinatorics

We call an integer $ n > 1$ [i]good[/i] if, for any natural numbers $ 1 \le b_1, b_2, \ldots , b_{n\minus{}1} \le n \minus{} 1$ and any $ i \in \{0, 1, \ldots , n \minus{} 1\}$, there is a subset $ I$ of $ \{1, \ldots , n \minus{} 1\}$ such that $ \sum_{k\in I} b_k \equiv i \pmod n$. (The sum over the empty set is zero.) Find all good numbers.

Good numbers are exactly the primes.

A non-prime $ n$ has a non-trivial factorization $ n\equal{}pq$.
Then for $ b_1\equal{}b_2\equal{}\ldots\equal{}b_{n\minus{}1}\equal{}p$ all subset-sums are multiples of $ p$, and no subset-sum is $ \equiv1\pmod n$.

The reverse statement for good primes $ n$ is well-known.
Consider some fixed numbers $ 1 \le b_1, b_2, \ldots , b_{n \minus{} 1} \le n \minus{} 1$.
Let $ S_k$ denote the set of all subset-sums of the first $ k$ numbers.
We claim that $ S_{k\plus{}1}$ is either equal $ Z_n$, or strictly larger than $ S_k$.
This then implies $ S_{n\minus{}1}\equal{}Z_n$, and we are done.

Proof of the claim:
Suppose for the sake of contradiction that $ S_k\equal{}S_{k\plus{}1}\ne Z_n$.
Consider some non-element $ x$ of $ S_{k\plus{}1}$.
Then $ x\minus{}b_{k\plus{}1}$ is a non-element of $ S_k\equal{}S_{k\plus{}1}$.
By induction $ x\minus{}m\cdot b_{k\plus{}1}$ is a non-element of $ S_k\equal{}S_{k\plus{}1}$.
But the elements $ x\minus{}m\cdot b_{k\plus{}1}$ cover all of $ Z_n$, which means that $ S_k\equal{}S_{k\plus{}1}$ must be empty; contradiction.


[quote="ashpotter"]hi
I participated in the into to number therory. (it was cool!)
I didn't get how you change base 3 to base 9 the fast way.
like   $121_3$ to base nine
could someone tell me?[/quote]
There's a faster way.
We know that 2 digits in base 3 = 1 digit in base 9 (3^2=9)

So take the 21 in base 3, and you know that it's 7 in base 10, and 7 in base 9.
So the last digit is 7, and you take now 01, and that's 1 in base 9, so
$121_3 = 17_9$


For positive reals $a,b,c,d$ that $a+b+c+d=1$ Prove that:

$\sqrt{4a+1}+\sqrt{4b+1}+\sqrt{4c+1}+\sqrt{4d+1}<6$

$y=\sqrt{4x+1}$ is an increasing and concave up function.
From $\sqrt{4x+1}\leq\sqrt{2}x+\frac{3}{4}\sqrt{2}\Longleftrightarrow 2\left(x+\frac{3}{4}\right)^{2}-(4x+1)=2\left(x-\frac{1}{4}\right)^{2}\geq 0$, we have
$\sqrt{4a+1}+\sqrt{4b+1}+\sqrt{4c+1}+\sqrt{4d+1}\leq\sqrt{2}(a+b+c+d)+\frac{3}{4}\sqrt{2}\cdot 4$

$=4\sqrt{2}<6.$ Q.E.D.

Let $f(x)=\sqrt(x)$. f(x) is concave down.

By Jensen's inequality,
\[f(4a+1)+f(4b+1)+f(4c+1)+f(4d+1) \leq 4f(a+b+c+d+1)=4f(2)=4\sqrt(2)<6\]


integration
advanced fields
advanced fields unsolved

Anyone know how to do it? Im stuck on this one...

For every $x \in L^{1}(T)$ (T is torus), let $\xi_{n} = \int_{-\pi}^{\pi}x(t)\sin(nt)dt, n \in \mathbb{Z}$. Suppose that $E \subset L^{1}(T)$  is a closed subspace such that $\sum_{-\infty}^{\infty}|\xi_{n}| > \infty$ for every $x \in E$. Prove that there exist $M >0$ such that

 $\sum_{-\infty}^{\infty}|\xi_{n}| \leq M\int_{-\pi}^{\pi}|x(t)|dt$
for every $x \in E$

Thanks to all!

[quote="Xevarion"]I think it is a bijection between $l^1$ and $L^1$? Continuous map on an infinite-dimensional space is bounded, and then use the bound on the operator? This may be totally wrong; I've learned the words, but not how to put them in the winning order... :P[/quote]

I don't know. I'm trying to use Riemann lemma $\xi_{n} \to 0$ when $|n| \to  \infty$

Does it help
$\int_{-\pi}^\pi |x(t)|dt \le \sqrt{2\pi \|x\|}$
and $\|x\|^2 = \sum_n |\xi_n|^2$ ?

Then perhaps $M = \sup_n |\xi_n|$ ?


A triangle $ ABC$ is placed on a plane inclined at an angle $ \theta$ witha horizontal plane so that the height of each vertex above the horizontal plane is equal to the perpendicular from that vertex to the opposite side of the triangle.
Prove that $ cos2\theta \equal{} 4(cosA \plus{} cosB \plus{} cosC) \minus{} 5$.

So for my proceedings:
The sides of the triangle obtained by the projections on the horizontal plane are $ acos\theta, bcos\theta,ccos\theta$.
Also, we write our proof relation as  $ cos2\theta \plus{} 1 \equal{} 4(cosA \plus{} cosB \plus{} cosC \minus{} 1)$. In any triangle $ ABC$ , $ \sum cosA \equal{} 1 \plus{} \frac {r}{R}$ with $ r,R$ being the inradius and circumradious respectively. And also, $ cos2\theta \plus{} 1 \equal{} 2cos^2(\theta)$. So, we have to prove that $ cos^2(\theta) \equal{} 2\frac {r}{R}$. But so far, I have not used the projection condition effectively in the problem.

Someone please post the solution.

I didn't even read all the text, but be aware:

all the sides $ a, b, c$ can't make an angle $ \theta$ with the horizontal plane.

Is something wrong with the problem???
or is it a well known problem ??
Someone please post a solution to this nice problem.


$ a_1,a_2...a_n$ and $ b_1,b_2...b_n$ are both permutations of $ 1, \frac{1}{2},\frac{1}{3},...\frac{1}{n}$

In addition $ a_1 \plus{} b_1 \ge a_2 \plus{} b_2 \ge ...\ge a_n \plus{} b_n$

Show that $ a_m \plus{} b_m \le \frac{4}{m}$ for all $ m \in [1,n]$

[hide="Lemma"]There are at most $ m \minus{} 1$ values of $ j$ such that $ a_j \plus{} b_j\geq \frac 4m$.
[hide="Proof"]Suppose there are at least $ m$ values of $ j$ such that $ a_j \plus{} b_j \geq \frac 4m$. Then let $ A \equal{} \{a_1,a_2,\dots,a_m\},B \equal{} \{b_1,b_2,\dots,b_m\},$ and $ X \equal{} A \cup B$.
[hide="If m is even"]Let $ S \equal{} \left\{1,\frac 12,\dots,\frac 1{m/2}\right\}$ and $ s \equal{} \left\{\frac 1{m/2},\frac 1{m/2 \plus{} 1},\dots,\frac 1n\right\}$. Note that $ S\cup s \equal{} \left\{1,\frac 12,\dots,\frac 1n\right\}$ and $ S \cap s \equal{} \left\{\frac 1{m/2}\right\}$. Since $ |S| \equal{} \frac m2$, $ |X \cap S|\leq \frac m2$. Since $ |X|\geq m$, it follows that $ |X \cap s|\geq\frac m2 \plus{} 1$. Since each element from $ X\cap s$ must be paired with an element from $ X \cap S$, such a pairing cannot be achieved.
[/hide]
[hide="If m is odd"]
Let $ S \equal{} \left\{1,\frac 12,\dots,\frac 1{(m \minus{} 1)/2}\right\}$ and $ s \equal{} \left\{\frac 1{(m \plus{} 1)/2},\frac 1{(m \plus{} 3)2},\dots,\frac 1n\right\}$. By the same argument as above, there are at most $ (m \minus{} 1)/2$ elements in $ X \cap S$, so there are at least $ (m \plus{} 1)/2$ elements in $ X \cap s$, so a pairing cannot be achieved.
[/hide]
[/hide]
[/hide]
[hide="Solution"]
Suppose that $ a_m \plus{} b_m > \frac 4m$. Then $ a_1 \plus{} b_1,a_2 \plus{} b_2,\dots,a_{m \minus{} 1} \plus{} b_{m \minus{} 1} > \frac 4m$ as well, which contradicts the lemma. $ \blacksquare$[/hide]

Very Nice Solution!

I was thinking that we might be able to make the inequality strict, but I can't show it yet.

I was trying to show that $ a_n \plus{} b_n$ would be maximized when; $ a_n \plus{} b_n \equal{} \frac {2}{n} \plus{} \frac {2}{n \plus{} 2}$ for even $ n$ and $ a_n \plus{} b_n \equal{} \frac {2}{n \plus{} 1} \plus{} \frac {2}{n \plus{} 1}$ for odd $ n$

Actually this problem was from engel, but I didn't really understand the solution given.


Let $a,b,c>0$ such that $abc=1$. Show that the following inequality holds $(a+b)^n + (b+c)^n + (c+a)^n > 6\sqrt{n},\forall n \in {\mathbb N}^*$
 :ninja:

[quote="siuhochung"]just amgm on the left or am i missing sth?[/quote]

write

$\sum (a+b)^n \ge 3 \prod (a+b)^{n/3} \ge 3(2^n)$ by amgm.
$3(2^n) > 6 \sqrt{n}$ means $2^{n-1} \ge \sqrt{n}$ which is trivial isn't it.

yes u're right. my sollution is more complicated  :D  i used 1 time bernouly and 2 times am-gm  :(


Let $ x,y,z$ be positive reals satisfying $ x \plus{} y \plus{} z \equal{} 1$, then
$ x^ny \plus{} y^nz \plus{} z^nx\le \frac {n^n}{(1 \plus{} n)^{n \plus{} 1}}$,
where:
$ a)$ $ n\in \mathbb{N}, n > 1$
$ b)$ $ n\in \mathbb{R}, n > 1$

[[b]Moderator edit:[/b] Problem b) is wrong, at least for certain $ n$ between $ 1$ and $ 2$. Problem a) was solved in http://www.mathlinks.ro/Forum/viewtopic.php?t=2549 .]

assume $x\geq y \geq z$
$x^n y+y^n z+z^n x\le (x+\frac{z}{n})^n(y+\frac{(n-1)z}{n})\le \frac{n^n}{(n+1)^(n+1)}$
by $AM-GM$

This is false for $n=\frac32$, as $\frac{n^n}{(n+1)^{n+1}}=\sqrt{\frac{108}{3125}}\approx .1859$ but the equality case $x=y=z$ yields $\frac1{\sqrt{27}}\approx.19245$  :(

[quote="Centy"]$x^n y+y^n z+z^n x$ is not symmetric so I don't you think you can assume $x \geq y \geq z$[/quote]Well it is not symmetric, but $x^n y+y^n z+z^n x-(y^nx+z^ny+x^nz)=(x-y)(y-z)\sum_{i=0}^{n-1} (x^{n-1-i}y^i-y^{n-1-i}z^i)\ge0$ if $x \geq y \geq z$. This works for $n\in\mathbb N$.  :)


[quote="kunny"]Given a triangle $ABC$ such that $BC=a,\ CA=b,\ AB=c$. The incircle of the triangle is tangent at $D$ on $BC$.
Let $S$ be the area of the traingle and $T$ be the area of the rectangular having two adjacent sides $BD$ and $DC$.
Compare the area of $S$ and $T$.[/quote]

\[BD = \frac{a-b+c}2\]
\[DC = \frac{a+b-c}2\]

\[S = \frac{1}{4}\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}\]
\[T = \frac{(a-b+c)(a+b-c)}4\]

Now lets compare it. After some easy transformations we get

\[S \ge T \iff b^{2}+c^{2}\geq a^{2},\]

so depends if the angle $\angle A$ is acute, right or obtuse.


It is too much banal to solve by Cheybechev or Jensen inequality.
Sorry, I thought a proof without them.

Expanding the right side and it suffices to prove b 2004 /c+c 2004 /b>=b 2003 +c 2003 which is obviously true by multiplying bc on both sides  :D


function
integration
topology
calculus
derivative
geometry
geometric transformation

Hello, I was attempting a question in operator algebra course and having a little trouble. The question is talking about Linear transformations on X:= $ C^{\infty}[0,1]$infinitely differentiable functions . The transformation defined as $ T: X \rightarrow X\ by\ (Tf)(t) : \equal{} \int^t_0 f(s)ds$.

A very straight forward question, But the issue is when it asks to prove that the inverse transformation $ T^{ \minus{} 1}$ has an unbounded operator norm.  I can't think of the function which will help me prove $ T^{ \minus{} 1}$ is unbounded.

If I atleast knew that $ C^{\infty}[0,1]$ was a Banach space it would of great help. The Banach isomorphic theorem should give me some direction, if the space is not a Banach space I guess I will keep looking for a function.

I am not throwing out the fact that the question is wrong and we were meant to show that $ T^{ \minus{} 1}$ was bounded and not unbounded.

Thanks for reading this far.
If you remember your lecturers saying statements about $ C^{\infty}$ space, Please do reply. Any small hint would be of great help.

Have a wonderful day.
Ps.VS

Since $ \frac{d}{dx}e^{Mx}\equal{}Me^{Mx}$, differentiation is a not a bounded operator with respect to any norm or a set of seminorms.

Hmmm... I guess I managed to delude myself that since we had seminorms for every derivative, then the derivative operator would be OK. But I forgot to consider the constants involved in that. But I'm still a little confused.

Consider the operator I called $ U,$ which is a slight modification of the derivative operator. Now consider a sequence $ f_k$ which tends to $ f$ in $ C^{\infty}.$ (Since this is a metrizable space, we can discuss topological concepts using sequences.) That convergence means precisely that $ f_k^{[n]}\to f^{[n]}$ uniformly for each $ n.$ Doesn't it then follow that $ (Uf_k)^{[n]}\to (Uf)^{[n\plus{}1]}$ uniformly? Is the operator $ U$ continuous despite being unbounded in every seminorm? Or is it discontinuous and I'm missing something?

[quote="Kent Merryfield"]$ C^{\infty}$ is not a Banach space or a normed space. Its topology is given by a countable family of seminorms: $ \|f\|_k \equal{} \sup_{x\in[0,1]}|f^{[k]}(x)|.$ We can make it a metric space. However, since it is not a normed space, I am very wary of the word "bounded" (or "unbounded") as applied to an operator, and I am wary of talking about an operator norm.

The operator $ T^{ \minus{} 1},$ where it is defined, is simply the derivative operator that sends $ f$ to $ f',$ on the range of $ T.$ However, the range of $ T$ is not all of $ C^{\infty}[0,1].$ Rather, it is the subspace (of codimension 1) consisting of those functions for which $ f(0) \equal{} 0.$

It is possible to extend $ T^{ \minus{} 1}$ to a continuous operator $ U$ defined on all of $ C^{\infty}[0,1].$ Just let $ U(f)(x) \equal{} \frac {d}{dx}(f(x) \minus{} f(0)).$ Note that $ TU(f)(x) \equal{} f(x) \minus{} f(0),$ while $ UT(f) \equal{} f.$ Since $ T^{ \minus{} 1}$ can be extended to a globally continuous operator, I would not want to call it "unbounded" in any sense.

Question: is the bounded/unbounded question actually about some other topology or norm on functions on $ [0,1]?$[/quote]

Sorry for getting back late...

If an operator $ T: V\rightarrow W$ is said to be bounded then  $ \Vert T \Vert_{op} \equal{} \inf S, S: \equal{} \{C: \Vert Tv \Vert \leq C \Vert v\Vert\ \forall v \in V\}$ .

$ \Vert Tx \Vert \leq \Vert T \Vert_{op} \Vert x \Vert \ \forall x \in V$

When we apply bounded linear operators the space it dilates and scales in terms of the operator norm.

Also thank you very much for pointing out that the space maps to a subspace where functions are $ f(0) \equal{} 0$ at 0. I totally missed that point there, I was carrying the idea that there is a unique antiderivative but didn't really go the step further and see what it was. Its definitely not a Banach space.

Thank you for your reply.

[quote="mlok"]I should have said "differentiation is a not a bounded operator with respect to any norm or a seminorm". If we use a set of seminorms to construct a translation-invariant metric in the [url=http://en.wikipedia.org/wiki/Locally_convex_topological_vector_space#Further_definitions_and_properties]usual way[/url], then homogeneity is lost and the example with $ e^{Mx}$ does not tell us much. So yes, differentiation is bounded and continuous.[/quote]

hahaha =)

Thank allot man, Right under the nose.
I would like to give both of you some ratings, but unfortunately I don't have enough points to give out ratings...

Have a great day.


complex analysis
complex analysis unsolved

I need some help with the following problem please :
a) Prove that the cut plane[b] C[/b]\(-infinity,0] is polygonally path connected
b) Prove that the set [b]D[/b](0,1) U [b]D[/b](2,1), which consists of the union of open discs, is not path connected (Hint: Consider the line {z:Rez=1} and continuous paths going from one disc to another)

For a) you can just construct a straight line path from any point to say 1.
For b) you show that any path going from one disk to another has to pass the line $ \Re(z)\equal{}1$ and that no point on that line is in the set.


geometry
circumcircle
cyclic quadrilateral
geometry proposed

A cyclic quadrilateral is divided into four quadrilaterals by two lines passing through its inner point. Three of these quadrilaterals are cyclic with equal circumradii. Prove that the fourth part also is cyclic quadrilateral and its circumradius is the same.

[i]Author: A.Blinkov[/i]

Let the inner point is E and two lines IG, HF pass through E ($ I\in AD, F\in DC, G\in BC, H\in AB$). We have $ \angle IEF\equal{}180^o\minus{}\angle D\equal{}\angle B$ and HEGB is cyclic then $ \angle B\equal{}\angle D\equal{}90^o$
But $ \angle IEF\equal{}\angle A\equal{}90^o$ then $ ABCD$ is a rectangle... It's easy to show $ EGCF$ is also cyclic and its circumradius is the same.


Hi, this is a problem from a set of practise problems from a while ago.

I had a little progress on it, but not much.

Question:
Find all functions f from the integers to the integers having the property that for all integers x and y:
f(f(x) + y) = x + f(y + 2008).

Right now I'm guessing that f(x) = x + 2008 but I'm not too sure if that's the only solution, or if it is, how to prove it. Any tips? :)

Here's what I have so far:
Let y=0, then we have f(f(x))=x+f(2008)=x+k.
Let f(x)=g(x)+x, then we have f(f(x))=f(g(x)+x)=g(g(x)+x)+g(x)+x=x+k, so g(g(x)+x)+g(x)=k, or g(g(x)+x)=k-g(x).
Let x=g(z)+z. Then we have g(g(g(z)+z)+g(z)+z)+g(g(z)+z)=k, or g(k-g(z)+g(z)+z)+k-g(z)=k, or g(k+z)=z.
Have to go, ill finish later. [/hide]

[hide="Edit: Sorry, being stupid."]Since $ y \equal{} 0$ gives $ f(f(x)) \equal{} x \plus{} f(2008)$, we see that $ f(a) \equal{} f(b)\implies a \equal{} b$, so $ f$ is injective. Also $ f$ is clearly surjective since $ x \plus{} f(2008)$ takes on any value. So $ f$ is bijective and strictly monotone.

Now $ x \equal{} 0$ gives $ f(y \plus{} f(0)) \equal{} f(y \plus{} 2008)\implies y \plus{} f(0) \equal{} y \plus{} 2008\implies f(0) \equal{} 2008$.

Recall $ f(f(x)) \equal{} x \plus{} f(2008)$ so $ f(f(x)) \minus{} f(2008) \equal{} x$. But since $ f$ is bijective and monotone, and outputs only integers, $ |x| \equal{} |f(f(x)) \minus{} f(2008)|\ge|f(x) \minus{} 2008| \equal{} |f(x) \minus{} f(0)|\ge|x \minus{} 0| \equal{} |x|$, so $ f(x) \minus{} 2008 \equal{} \pm x\implies f(x) \equal{} 2008\pm x$, where the sign depends whether $ f$ is strictly increasing or decreasing.

Both $ f(x) \equal{} 2008 \plus{} x$ and $ f(x) \equal{} 2008 \minus{} x$ work after substituting.[/hide]

[quote="math154"]So $ f$ is strictly monotone.[/quote]
why is f strictly monotone ?

Otherwise it cannot be injective since for some $ a\ne b$ we have $ f(a)\equal{}f(b)$.

[quote="math154"]Otherwise it cannot be injective since for some $ a\ne b$ we have $ f(a) \equal{} f(b)$.[/quote]

I think you would need some other condition (e.g. continuity) before you can assume it's strictly monotonic.

For example, you can easily create a hybrid function that is injective but not monotonic

[hide="solution"]
Let $ P(x,y)$ be the assertion that $ f(f(x) \plus{} y) \equal{} x \plus{} f(y \plus{} 2008)$

$ P(x,0) \Longrightarrow f(f(x)) \equal{} x \plus{} f(2008) \qquad(1)$

From $ (1)$ we have
$ P(f(x),y) \Longrightarrow f(f(f(x)) \plus{} y) \equal{} f(x \plus{} f(2008) \plus{} y) \equal{} f(x) \plus{} f(y \plus{} 2008) \qquad(2)$

$ P(2008,x \plus{} y) \Longrightarrow f(f(2008) \plus{} x \plus{} y) \equal{} 2008 \plus{} f(x \plus{} y \plus{} 2008) \qquad(3)$

Subbing $ z \equal{} y \plus{} 2008$ in $ (2)$ and $ (3)$ we get

$ f(x) \plus{} f(z) \equal{} f(x \plus{} z) \plus{} 2008$

Now let $ f(x) \equal{} g(x) \plus{} 2008$, giving $ g(x) \plus{} g(z) \equal{} g(x \plus{} z)$ which is a cauchy equation with solution $ g(x) \equal{} cx$

Subbing $ f(x) \equal{} cx \plus{} 2008$ into the original equation gives $ c \equal{} \pm 1$

Hence $ \boxed{f(x) \equal{} \pm x \plus{} 2008}$[/hide]


what will be the most probable pattern of IITJEE 2008 this year comment on this!!!! what can be the newly added objective type questions like last time they introduced assertion reasoning!!!! :D

it will be the return of multiple correct answers, but this time with an evil twist!!

+ 2 for a correct answer. -6 for a wrong answer!!!

[quote="achu_182"]+ 2 for a correct answer. -6 for a wrong answer!!![/quote]

How about -5 for a right answer, +1 for a wrong answer? :rotfl: :D

Something like a flow chart completion,,, Just in order to not allow much guessingg and awarding marks for the STEPS

hello i am serious !!!! will it be the same as 2007 pattern???? coz thats what coaching institutes claim to be in mummbai and rohit if you are in doubt go to http://www.goiit.com/posts/list/community-shelf-are-you-ready-for-iit-jee-2008-49083.htm :D

it can't b such deep -ve marking!!!!!!!!!

i think roorkee is conducting jee this time. so the paper mite be slightly tougher than last year.

IN HT news paper , it is written each paper will be of 200 marks.

[quote="VARUNARASU"]yep, thats what is written in the DH newspaper too. btw, will we have multiple correct answers this time.[/quote]
are you asking us a question or stating a fact???? :lol:

I think he's asking us. And as far as the pattern goes. I think FIITJEE has done a good job of training us for everything and anything possible.

Since it is roorkee, I believe Numericals may make a comeback.  And probably so will More than 1 answer right. But I think no of matches will be reduced and assertion reasoning scrapped.

Whatever happens, I'm sure that we will all be surprised. IIT has never ever not surprised the writers. ;)


geometry
rectangle
calculus
integration
ratio
graph theory

hi guys,

why do you think of this simple pb:
a great rectangle is cut off into small rectangles so that each small rectangle have a side of integer length : prove that the great rectrangle have a side of integer length ..

bye.

Well, Alekk, It is a very very very well-known problem....
There is an old article in the AMM which compile 14 (!) proofs of that result.

Pierre.

S.Wagon, "Fourteen proofs of a result about tiling a rectangle",
A.M.M.,
1987, p.601-617.

Quadrature, n30, ex.A-2, p.43-45 (Oct/Nov 1997).

in fact I known the solution based on complex integral (it was an exercice of my techer when we studied double integral) but I was wondering if there was a simple solution ;-)

There are many, but the double-integral solution may be presented more elementary.
Look at Xavier's approach on a very recent topic on 'Maths et Dlires' :

http://boumbo.salle-s.org/mathetdelire/forum/forum.php?view=539

Pierre.

[quote="alekk"]a similar one:if a rectangle can be paved by square then the ratio of its sides is rational ..[/quote]

This was already discussed at this forum. It was posted by Omid Hatami and solved by lhvietbao:

 http://www.mathlinks.ro/viewtopic.php?t=6043

As Bao remarks, another solution can be found in Bollobas' book "Graph Theory" (Springer-Verlag New York Inc., 1979). Speaking of tiling problems, I would remind the following more difficult problem: for which $t \in \mathbb{R}^+$ can the unit square be tiled with finitely many rectangles whose side-ratio is $t$? (See http://www.mathlinks.ro/viewtopic.php?t=6055 for the answer, but a proof is missing to-date).

But for this problem cut plane into $\frac12\times\frac12$ squares and Color squares like chess. One of the sides of rectangle is integer iff area of it's black ponts is equal to area of it's white points. From this problem is obvious.