lets say you have $k_1(abc) + k_2(ab + ac + bc) + k_3(a+b+c) +k_4=0$ (where $k_i$ is a given constant)
solve for a,b, and c, how would you solve for this? i know how to deal with k 1,2,3 but the $k_4$ doesn't seem to work the same way
i would do $x^3-k_3x^2+k_2x-k_1=0$, but what does one do with $k_4$
is this even correct? i am not sure, any help would be appriciated

I dont' think it's correct, cause vieta's is like

$ax^3+bx^2+cx+d=0$

$r_1 r_2 r_3=-d$

etc.

Not all in one equation like you have

$k_4$ is multiplied by something as well.

Think the sum of the roots taken 0 at a time.

Ok.

Say $a, b, c, d$ are the roots of $x^4 + k_1x^3 + k_2x^2 + k_3x + k_4$.

We then know:

$k_1 = -(a+b+c+d)$
$k_2 = ab + ac + ad + bc + bd + cd$
$k_3 = -(abc + abd + bcd)$
$k_4 = abcd$

Remember that you are taking $(x - a)(x-b)(x-c)(x-d)$, so signs alternate. If you aren't sure, multiply that out and equate.


hôm nay thi VMO rồi,có đề ở đây nè mọi người:
http://www.mathlinks.ro/viewtopic.php?p=1418636#1418636

[CENTER]ĐỀ THI HỌC SINH GIỎI QUỐC GIA THPT NĂM HỌC 2008-2009.[/CENTER]

[u][b]Bài 1. (4điểm)[/b][/u]
 Giải hệ phương trình:
$ \left\{\begin{array}{l}\frac {1}{{\sqrt {1 \plus{} 2x^2 } }} \plus{} \frac {1}{{\sqrt {1 \plus{} 2y^2 } }} \equal{} \frac {2}{{\sqrt {1 \plus{} 2xy} }} \\
\sqrt {x(1 \minus{} 2x)} \plus{} \sqrt {y(1 \minus{} 2y)} \equal{} \frac {2}{9} \\
\end{array} \right.$

[u][b]Bài 2. (5điểm) [/b][/u]
Cho dãy số (x_n) xác định như sau:
$ \left\{\begin{array}{l}x_1 \equal{} \frac {1}{2} \\
x_n \equal{} \frac {{\sqrt {x_{n \minus{} 1} ^2 \plus{} 4x_{n \minus{} 1} } \plus{} x_{n \minus{} 1} }}{2} \\
\end{array} \right.$

Xét dãy số $ {y}_{n} \equal{} \sum_{1}^{n}\frac {1}{{{x}_{i}}^{2}}$. Chứng minh dãy $ (y_n)$ có giới hạn hữu hạn và tìm giới hạn đó.

[u][b]Bài 3. (5 điểm)[/b][/u]
Cho 2 điểm cố định $ A, B$ và điểm $ C$ di động trên mặt phẳng sao cho $ \hat{ACB} \equal{} a \ (0 < a < 180)$ không đổi cho trước. Hình chiếu của tâm đường tròn nội tiếp $ I$ của tam giác $ ABC$ xuống ba cạnh $ AB,\ BC,\ CA$ lần lượt là $ D,\ E,\ F$. $ AI$ và $ BI$ cắt $ EF$ lần lượt tại $ M,N$.
a) Chứng minh độ dài $ MN$ không đổi.
b) CM đường tròn $ (DMN)$ luôn đi qua một điểm cố định.

[u][b]Bài 4. (3điểm)[/b][/u]
Cho $ a,\ b,\ c$ là các số thực. Với mỗi $ n$ nguyên dương, $ a^n \plus{} b^n \plus{} c^n$
là số nguyên. Chứng minh rằng tồn tại 3 số nguyên $ p,q,r$ sao cho $ a,b,c$ là các nghiệm của pt bậc ba $ x^3 \plus{} px^2 \plus{} qx \plus{} r \equal{} 0$.[/COLOR]

[u][b]Bài 5. (3 điểm)[/b][/u]
Cho tập hợp $ S$ gồm $ 2n$ số nguyên dương đầu tiên. Tìm số tập hợp $ T$ sao cho trong $ T$ không có 2 phần tử $ a,b$ nào thỏa mãn $ |a \minus{} b| \in \left\{ {1;n} \right\}$
(chú ý tập rỗng thỏa mãn ĐK trên)

==HẾT==

http://forum.mathscope.org/showthread.php?t=7731

Mọi người vào đây có tranh luận...............


[color=blue]
Let $x, y, z$ three positive integers such that there is $n\in\mathbb{N^{*}}$ such that

                                      \[ xy+yz+zx=n(x+y+z).  \]
Prove that \[ \max \{ x, y, z \} \leq n^{2}  \]

 [/color]


Let $ a,b,c\in [\frac{1}{3},3]$
Prove that $ \frac{a}{a\plus{}b}\plus{}\frac{b}{c\plus{}b}\plus{}\frac{c}{a\plus{}c}\ge \frac{7}{5}$

[quote="arqady"][quote="Allnames"]Let $ a,b,c\in [\frac {1}{3},3]$
Prove that $ \frac {a}{a \plus{} b} \plus{} \frac {b}{c \plus{} b} \plus{} \frac {c}{a \plus{} c}\ge \frac {7}{5}$[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=175423[/quote]
 I am very glad to see your link,but I think it is not nice   :blush:
Since the equation,I think we can put $ a\equal{}3x,b\equal{}\frac{y}{3}$,and then use MV,and you


for positive a,b,c,d,e|abcde=1 prove that
$\frac{5}{3}\ge \frac{1}{1+a+ab}+\frac{1}{1+b+bc}+\frac{1}{1+c+cd}+\frac{1}{1+d+de}+\frac{1}{1+e+ea}$


I took the GPML state Math contest and I qualified for the state, but i am puzzled on how to answer the following question (without a calculator, target round):
Cos(36)-Cos(72)=?

If you know how to solve the problem, please e-mail me at saladandbeans@hotmail.com
thank you

You might also construct a geometric argument.  What shape is mostly likely to help?

As for motivating that trig solution -- you have trig identities like sums to products and products to sums.  The goal is to find a way to use the numbers given to construct numbers that you prefer to work with (because you know or can relate their trig values).

[quote="salahodeen"]Hello,
Shobber, I have a question: How did you go from 2sin54sin18 to ((2sin18cos36cos18)/cos18)?
I am sorry if this is a really stupid question, but i just could not find out how you took that step.
Thanks a lot[/quote]
Sorry, i did not post clearly:

$2\sin 54^o \sin 18^o = 2 \cos 36^o \sin 18^o = \frac{2 \sin18^o \cos 36^o \cos 18^o}{\cos 18^o}$

In other words, that was indeed a stupid question. Congrats salahodeen, maybe you should have taken integrated 1

Were you at St. Francis Community College for a qualifying contest?

yes i was; for the Rock Bridge High School team

okay i was just curious because i think i remember the name "salahodeen."
I was from the Ladue team.


function
topology
real analysis
real analysis unsolved

Let $ f$:  $ I \rightarrow J$, convex function

Then prove that $ f$ is continue on any point in $ I$ except the boundaries.

Not true.  Let $ I \equal{} [0,1]$, $ f: [0,1]\to \mathbb{R}$ with
\[ f(x) \equal{} \begin{cases} 0 & 0\leq x < 1 \\
1 & x \equal{} 1 \end{cases}
\]
$ f$ is convex and not continuous.  You should require that $ I$ be open.

ok, you are right, do you have a proof for the statment with I open, please?

it depends what is $ I$. If it is an open subinterval of $ \mathbb{R}$, it is true. If it is an open set of a general normed vectorial space, it is false (just take a non continuous linear function).

Such an example is precluded by requiring $ I$ to be finite dimensional.  PlanetMath has a [url=http://planetmath.org/encyclopedia/ContinuityOfConvexFunctions.html]proof[/url] for the $ I$ open and finite-dimensional case, which is nearly what you want.  A more self-contained development, including the the most general form of the statement you're looking for, can be found in [url=http://books.google.com/books?id=6jjY2Vi3aDEC&printsec=frontcover&dq=Infinite+Dimensional+Analysis&source=gbs_summary_r&cad=0]Infinite Dimensional Analysis[/url] by C. D. Aliprantis and K. C. Border (pg. 177-180).


Post a proof (without words) of the Pythagoreas Formula/Theorem.
Then find the proof (with words).
I'll start:

[img]http://upload.wikimedia.org/wikipedia/commons/thumb/8/84/Pythagoras_proof.svg/350px-Pythagoras_proof.svg.png[/img]

the area of the small square is $ c^2$
And we have
$ c^2\equal{}(a\plus{}b)^2\minus{}4*\frac{1}{2} ab$
$ \longrightarrow c^2\equal{}a^2\plus{}b^2\plus{}2ab\minus{}2ab$
$ \longrightarrow c^2\equal{}a^2\plus{}b^2$


calculus
integration
trigonometry
calculus computations

How would you go about finding the indefinite integral of (sin x)^2?

Two possibilities (not an exhaustive list):

1. $\sin^2x=\frac12(1-\cos(2x))$

2. For a specific definite integral - the integral over a whole quadrant - we have the "yin-yang argument." [You [i]must[/i] draw a picture of this.]

Consider $I =\int_0^{\frac{\pi}2}\sin^2x\,dx.$ Make the substitution $u=\frac{\pi}2-x,$ getting

$I=\int_0^{\frac{\pi}2}\cos^2u\,du=\int_0^{\frac{\pi}2}\cos^2x\,dx$

So $2I=\int_0^{\frac{\pi}2}(\cos^2x+\sin^2x)\,dx=\frac{\pi}2$

Thus $I=\frac{\pi}4.$

I remember the result this way: whenver you integrate either $\sin^2$ or $\cos^2$ (of a linear expression) over an interval that starts and stops at the ends of quadrants, then the integral is "half of the box."


Find $ `a$' so that the point (3, $ a$) is on the graph of $ y\equal{}4x\minus{}14$.

If $ (3,a)$ is on $ y \equal{} 4x \minus{} 14$, then if we plug the values in ($ 3$ for $ x$ and $ a$ for $ y$ the equation must make sense.  Thus:

$ a \equal{} 4(3) \minus{} 14$
$ a \equal{} 12 \minus{} 14$
$ a \equal{} \boxed{\minus{}2}$

I entered that, but I got it wrong. The answer was $-2$. But why do they make us do that?


[quote="shobber"]$a,b,c$ are positive numbers. Prove:
\[ \frac{\sqrt{a^2+8bc}}{a}+\frac{\sqrt{b^2+8ac}}{b}+\frac{\sqrt{c^2+8ab}}{c}\geq 9  \][/quote]
$\sum \frac{\sqrt{a^2+8bc}}{a} = \sum \sqrt{1 + 8\frac{bc}{a^2}} \ge \sqrt{9 + 8\left(\sum \sqrt{\frac{bc}{a^2}}\right)^2} \ge 9$

let's change the left side:

$\sum_{cycl}\frac{\sqrt{a^2+8bc}}{a}=\sum_{cycl}\sqrt{1+8\frac{b}{a}\frac{c}{a}}$

with the sostitution $d=\frac{a}{b}$, $e=\frac{b}{c}$, $f=\frac{c}{a}$  ($def=1$)

$\sum_{cycl}\sqrt{1+8\frac{b}{a}\frac{c}{a}}=\sum_{cycl}\sqrt{1+8\frac{f}{d}}$

with the sostitution $x=\frac{d}{e}$, $y=\frac{e}{f}$, $z=\frac{f}{d}$  ($xyz=1$)

$\sum_{cycl}\sqrt{1+8\frac{f}{d}}=\sum_{cycl}\sqrt{1+8x}$

now

$\sum_{cycl}\sqrt{1+8x}\geq3\sqrt[6]{(1+8x)(1+8y)(1+8z)}$

for AM-GM

always for AM-GM,

$1+8x=1+x+x+x+x+x+x+x+x\geq9x^{\frac{8}{9}}$

using it

$3\sqrt[6]{(1+8x)(1+8y)(1+8z)}\geq3\sqrt[6]{9x^{\frac{8}{9}}\cdot9y^{\frac{8}{9}}\cdot9z^{\frac{8}{9}}}=9$

that's all I wanted to demonstrate.

bye bye

$\displaystyle \sum \frac{\sqrt{a^2+8bc}}{a} = \sum \sqrt{1+\frac{bc}{a^2} + \ldots + \frac{bc}{a^2}} \geq 3\cdot\sum \sqrt[18]{\left(\frac{bc}{a^2}\right)^8)} \geq 9$

[quote="scpajmb"]generalization:
 Let  $k > 0,\alpha > 0,a_i > 0(i = 1,2, \cdots ,n),$we  have

${\rm{ }}(\frac{{a_1^2 + ka_2 a_3 }}{{a_1^2 }})^\alpha + (\frac{{a_2^2 + ka_3 a_4 }}{{a_2^2 }})^\alpha + \cdots + (\frac{{a_n^2 + ka_1 a_2 }}{{a_n^2 }})^\alpha \ge n(1 + k)^\alpha$[/quote]
Use Minkowsky to prove this ineq, too.

[quote="hendrata01"]IMO 2001 problem 2:
$\sum \frac{a}{\sqrt{a^2+8bc}} \geq 1$
This one is a direct consequence by AM-HM[/quote]

Sorry, but I fail to see this, could you please give some details?

oh I'm sorry. I got the sign reversed  :oops:


What topics do you guys think that one should study for IOI which are not on Cormen's book?
Thank you.

more on graph and more on data structure


inequalities geometry circumcircle inradius ratio trigonometry function

[b]In a triangle, which is  the best constant $\lambda \; ,\; \lambda \in \left(\frac{3}{2},4\right] ,$ such that $\hspace*{0.5cm} \displaystyle \max\{r_a,r_b,r_c\}\le \lambda R\;\hspace*{0.5cm}$ ?   [/b]
[b]References:[/b]
 [1]H.J.Baron, Tohoku Math.J., 48(1941) 185-192.
 [2]O.Bottema ,R.Z.Djordevic ,  R.R.Janic, D.S.Mitrinovic,P.M.Vasic , ,,[i]Geometric  Inequalities[/i]" , Wolters-Noordhoff Publishing Groningen ,1969,p.58.
 [3]P.Erdos , Math.Lapok 1(1949) p.72.

[quote="flip2004"][b]In a triangle, which is  the best constant $ \lambda \; ,\; \lambda \in \left(\frac{3}{2},4\right] , $ such that $ \hspace*{0.5cm} \displaystyle \max\{r_a,r_b,r_c\}\le \lambda  R\;\hspace*{0.5cm} $ ?   [/b][/quote]

The best constant is $\lambda=4$. In fact:

[b](1)[/b] For each triangle ABC with circumradius R and exradii $r_a$, $r_b$, $r_c$, we have $\max\left\{r_a;\;r_b;\;r_c\right\}<4R$.

[i]Proof.[/i] On the one hand, $r_a+r_b+r_c-r=4R$, where r is the inradius of triangle ABC. On the other hand, $\max\left\{r_a;\;r_b;\;r_c\right\}<r_a+r_b+r_c-r$, because the number $\max\left\{r_a;\;r_b;\;r_c\right\}$ is actually one of the numbers $r_a$, $r_b$, $r_c$ - let's say, WLOG, that $\max\left\{r_a;\;r_b;\;r_c\right\}=r_a$ -, and thus, the inequality $\max\left\{r_a;\;r_b;\;r_c\right\}<r_a+r_b+r_c-r$ becomes $r_a<r_a+r_b+r_c-r$, what is trivial because $r<r_b$ (what holds since $\frac{r}{r_b}=\frac{s-b}{s}$, where $s=\frac{a+b+c}{2}$ is the semiperimeter of triangle ABC, and I think it is clear that s - b < s). Thus, $\max\left\{r_a;\;r_b;\;r_c\right\}<r_a+r_b+r_c-r=4R$, and the proof is complete.

[b](2)[/b] The constant $\lambda=4$ cannot be replaced by a smaller one; in fact, considering an isosceles triangle ABC with AB = AC and letting the point A get closer and closer to the midpoint of the segment BC, the ratio $\frac{r_a}{R}$ gets arbitrary close to 4 (in fact, we have $r_a+r_b+r_c-r=4R$, and when the point A approaches the segment BC, the values of $r_a$ and R become arbitrarily large, while those of $r_b$, $r_c$ and r are bounded).

An IMHO more interesting question is to find the best constant $\lambda$ such that the inequality $\max\left\{r_a;\;r_b;\;r_c\right\}<\lambda R$ holds for every [i]acute-angled[/i] triangle ABC. The best such $\lambda$ is $\lambda=1+\sqrt2$. The proof is not considerably harder:

We will show that for any acute-angled triangle ABC with circumradius R and exradii $r_a$, $r_b$, $r_c$, we have $\max\left\{r_a;\;r_b;\;r_c\right\}<\left(1+\sqrt2\right)R$.

After a well-known formula, $\frac{r_a}{R}=\cos B+\cos C-\cos A+1$. Since the triangle ABC is acute-angled, its angle A is acute, so that cos A > 0, and thus $\cos B+\cos C-\cos A+1<\cos B+\cos C+1$. Also, since the angle A is acute, we have A < 90°; but by the sum of angles in triangle ABC, we have A = 180° - (B + C), so that this becomes 180° - (B + C) < 90°, what simplifies to B + C > 90°, so that C > 90° - B, and thus, since the cosine function is strictly monotonically decreasing on the interval [0°; 90°] (both angles C and 90° - B lie in this interval, since B and C are acute angles), we get cos C < cos (90° - B); in other words, cos C < sin B. Thus,

$\frac{r_a}{R}=\cos B+\cos C-\cos A+1<\cos B+\cos C+1<\cos B+\sin B+1$
$=\sqrt2\left(\frac{\sqrt2}{2}\cos B+\frac{\sqrt2}{2}\sin B\right)+1=\sqrt2\left(\sin 45^{\circ}\cos B+\cos 45^{\circ}\sin B\right)+1$
$=\sqrt2\sin\left(45^{\circ}+B\right)+1$.

Since $\sin\left(45^{\circ}+B\right)\leq 1$, this becomes

$\frac{r_a}{R}<\sqrt2\cdot 1+1=1+\sqrt2$,

so that $r_a<\left(1+\sqrt2\right)R$. Similarly, $r_b<\left(1+\sqrt2\right)R$ and $r_c<\left(1+\sqrt2\right)R$, and thus, since the number $\max\left\{r_a;\;r_b;\;r_c\right\}$ is one of the numbers $r_a$, $r_b$, $r_c$, we must have $\max\left\{r_a;\;r_b;\;r_c\right\}<\left(1+\sqrt2\right)R$.

So we have shown that the constant $\lambda=1+\sqrt2$ satisfies the inequality $\max\left\{r_a;\;r_b;\;r_c\right\}<\lambda R$ for every acute-angled triangle ABC. It remains to show that it is also the best constant, but this is trivial, since equality is obtained in the limiting case of an isosceles right triangle ABC.

Sorry for my confusing writeup; I am quite busy now and don't have time to carefully write down solutions.

  Darij

[quote="darij grinberg"][quote="flip2004"][b]In a triangle, which is  the best constant $ \lambda \; ,\; \lambda \in \left(\frac{3}{2},4\right] , $ such that $ \hspace*{0.5cm} \displaystyle \max\{r_a,r_b,r_c\}\le \lambda  R\;\hspace*{0.5cm} $ ?   [/b][/quote]
The best constant is $\lambda=4$.....  Darij[/quote]
Indeed, a nice and complete response ! /Proposer


Let f,g from R[X] be polinomials, g<>0 and f=(X^2-2*sqrt(3)*X+4)*g.
Prove that if all ceoficients of f are positive then the grade of g>=4.

thank you :D

Its not too hard to bash this one out with algebra. Though there may be a nicer solution, I would probably do this in a competition as it doesn't take much time and it definitely works.
g degree 0 obviously won't work.

g degree 1: let g=ax+b and when we expand fg we get b>2:rt3:a, a>2:rt3:b and a,b>0. Multiply first and second inequalities to get ab>12ab, and there are no solutions.

g degree 2: let g=ax:^2:+bx+c, we get a,c>0, b>2:rt3:a, b>2:rt3:c, a+c>2:rt3:b. 2:rt3:b > 12a and 12c so its > 6a+6c (taking average). So a+c>6(a+c), no solutions.

g degree 3: let g=ax:^3:+bx:^2:+cx+d. Now we get a,d>0, b>2:rt3:a, c>2:rt3:d, a+c>2:rt3:b, b+d>2:rt3:c.
Adding those last two and rearranging, (a+d)>(2:rt3:-1)(b+c), which is > (2:rt3:-1)(2:rt3:)(a+d). Again a and d must be 0 so there are no solutions.

So degree of g is >=4.

As I said, not "nice" but its not difficult and would work in a competition.


Prove that $(2+\sqrt5)^{\frac{1}{3}}$ +$(2-\sqrt5)^{\frac{1}{3}}$=1

There was a mistake in my first posting of this question.My apologies for this.The mistake has now been corrected.

We use $(a+b)^{3}=a^{3}+b^{3}+3ab(a+b),\forall a,b\in \mathbb R$

[quote="Michael Niland"]Prove that $(2+\sqrt3)^{\frac{1}{3}}$ +$(2-\sqrt3)^{\frac{1}{3}}$=1[/quote]
Except for the fact that it is not true...:)


geometry
trigonometry
calculus
derivative
AMC 10
AMC

There's no need to study if you have good understandings in both geometry and basic algebra. The important part is to stay in top mental state (preferably physical too) during the test (AHHHH!!!!), which means you may want to prepare psychologically for the test (AHH!!!! again). ........ I am really mad at my performance this year... I did so bad..... the problems were really obvious and I didn't see most of the geometry problems. Can anyone suggest a way to prepare better mentally?
An interesting fact: the two times i took math contests in my school library, the two times I failed miserably. This didn't happen elsewhere. Does anyone share similar experiences?

have you tried Tai Chi  :D ?
anyway, what do i need to know in order to do well in the amc 10? like some basic stuff in algebra, geometry, precal?

not precal!!!! that's in AMC 12...

my dad knows tai chi..... I tried to learn it, but with no avail..

I am pretty nothing above the trigonometry level will ever show up on the AMC-12.  They arent testing your high level of education, they're testing the depth of it.

if you take what im learning in precalc right now... *coughnothingcough* then you could say a lot of it is on the AMC-10 even... i mean we learned "simplifying radicals" and "combinations and permutations" which is stuff i knew and LEARNED already in 7th/8th grade. the only meaningful stuff was the derivatives and induction.


inequalities
function
logarithms
inequalities proposed

Let $a,b,c$ be non-negative real numbers. Prove that

$\sum_{cyc}{(a-b)(a-c).\sum_{cyc}{a^2(a-b)(a-c) \ge (\sum_{cyc}{a(a-b)(a-c))^2 }}}$.

Indeed, the inequality

$\left(\sum_{\text{cyc}}\left(a-b\right)\left(a-c\right)\right)\cdot\left(\sum_{\text{cyc}}a^2\left(a-b\right)\left(a-c\right)\right)\geq\left(\sum_{\text{cyc}}a\left(a-b\right)\left(a-c\right)\right)^2$

holds for arbitrary reals a, b, c, because

$\left(\sum_{\text{cyc}}\left(a-b\right)\left(a-c\right)\right)\cdot\left(\sum_{\text{cyc}}a^2\left(a-b\right)\left(a-c\right)\right)-\left(\sum_{\text{cyc}}a\left(a-b\right)\left(a-c\right)\right)^2$
$=3\left(b-c\right)^2\left(c-a\right)^2\left(a-b\right)^2\geq 0$.

  darij

A little generalization now.
[i]Let $p$ be a real number, and let $a,b,c$ be non-negative real numbers. Prove that
$S_0.S_{2p} \ge S_p^2$ , where $S_k=\sum_{cyc}{a^k(a-b)(a-c)}$[/i]

Why not a big generalization?

[color=blue][b]Theorem 1.[/b] Let a, b, c, x, y, z be six real (not necessarily nonnegative) numbers. Assume that $a\geq b\geq c$. Also, assume that either $x\geq y\geq z$ or $x\leq y\leq z$. Then,

$\left(\sum_{\text{cyc}}\left(a-b\right)\left(a-c\right)\right)\cdot\left(\sum_{\text{cyc}}x^2\left(a-b\right)\left(a-c\right)\right)\geq\left(\sum_{\text{cyc}}x\left(a-b\right)\left(a-c\right)\right)^2$.[/color]

[i]Proof of Theorem 1.[/i] The inequality we have to prove rewrites as

$\left(\sum_{\text{cyc}}\left(a-b\right)\left(a-c\right)\right)\cdot\left(\sum_{\text{cyc}}x^2\left(a-b\right)\left(a-c\right)\right)-\left(\sum_{\text{cyc}}x\left(a-b\right)\left(a-c\right)\right)^2\geq 0$.

But a straightforward calculation reveals that

$\left(\sum_{\text{cyc}}\left(a-b\right)\left(a-c\right)\right)\cdot\left(\sum_{\text{cyc}}x^2\left(a-b\right)\left(a-c\right)\right)-\left(\sum_{\text{cyc}}x\left(a-b\right)\left(a-c\right)\right)^2$
$=-\left(a-b\right)\left(b-c\right)\left(a-c\right)\left(\left(b-c\right)\left(y-z\right)^2+\left(c-a\right)\left(z-x\right)^2+\left(a-b\right)\left(x-y\right)^2\right)$.

Since $a-b\geq 0$, $b-c\geq 0$ and $a-c\geq 0$ (this is all because $a\geq b\geq c$), instead of proving that this product is $\geq 0$, it will be enough to show that

$\left(b-c\right)\left(y-z\right)^2+\left(c-a\right)\left(z-x\right)^2+\left(a-b\right)\left(x-y\right)^2\leq 0$.

This is equivalent to

$\left(b-c\right)\left(y-z\right)^2+\left(a-b\right)\left(x-y\right)^2\leq -\left(c-a\right)\left(z-x\right)^2$,

what rewrites as

$\left(b-c\right)\left(y-z\right)^2+\left(a-b\right)\left(x-y\right)^2\leq \left(b-c\right)\left(z-x\right)^2+\left(a-b\right)\left(z-x\right)^2$.

However, since $b-c\geq 0$ and $a-b\geq 0$, this will become trivial once we succeed to show that $\left(y-z\right)^2\leq\left(z-x\right)^2$ and $\left(x-y\right)^2\leq\left(z-x\right)^2$. But this is equivalent to $\left|y-z\right|\leq\left|z-x\right|$ and $\left|x-y\right|\leq\left|z-x\right|$, what is actually true because we have either $x\geq y\geq z$ or $x\leq y\leq z$.

This completes the proof of Theorem 1.

As a consequence of Theorem 1, we obtain one of the facts which are known under the name Vornicu-Schur inequality:

[color=blue][b]Theorem 2.[/b] Let a, b, c be three real numbers, and X, Y, Z three nonnegative numbers. Assume that $a\geq b\geq c$. Also, assume that either $X\geq Y\geq Z$ or $X\leq Y\leq Z$. Then,

$\sum_{\text{cyc}}X\left(a-b\right)\left(a-c\right)\geq 0$.[/color]

[i]Proof of Theorem 2.[/i] Apply Theorem 1 to the real numbers a, b, c, $x=\sqrt{X}$, $y=\sqrt{Y}$, $z=\sqrt{Z}$ (of course, the condition that either $X\geq Y\geq Z$ or $X\leq Y\leq Z$ becomes either $x\geq y\geq z$ or $x\leq y\leq z$). You get

$\left(\sum_{\text{cyc}}\left(a-b\right)\left(a-c\right)\right)\cdot\left(\sum_{\text{cyc}}X\left(a-b\right)\left(a-c\right)\right)=\left(\sum_{\text{cyc}}\left(a-b\right)\left(a-c\right)\right)\cdot\left(\sum_{\text{cyc}}x^2\left(a-b\right)\left(a-c\right)\right)$
$\geq\left(\sum_{\text{cyc}}x\left(a-b\right)\left(a-c\right)\right)^2\geq 0$.

But $\sum_{\text{cyc}}\left(a-b\right)\left(a-c\right)=\frac12\left(\left(b-c\right)^2+\left(c-a\right)^2+\left(a-b\right)^2\right)\geq 0$, with equality only for a = b = c (this is a trivial case, which we leave out of consideration here) - thus, we can divide by $\sum_{\text{cyc}}\left(a-b\right)\left(a-c\right)$, and get $\sum_{\text{cyc}}X\left(a-b\right)\left(a-c\right)\geq 0$. This proves Theorem 2.

Finally, here is Schur's inequality in a slightly upgraded form:

[color=blue][b]Theorem 3.[/b] Let p be a real number.

[b](a)[/b] If a, b, c are three nonnegative numbers, then the inequality $\sum_{\text{cyc}}a^p\left(a-b\right)\left(a-c\right)\geq 0$ holds.

[b](b)[/b] If p is an even nonnegative integer, then this inequality holds for arbitary reals a, b, c.[/color]

[i]Proof of Theorem 3.[/i] [b](a)[/b] Since the inequality in question is symmetric, we can WLOG assume that $a\geq b\geq c$. If $p\geq 0$, then $a^p\geq b^p\geq c^p$; if $p\leq 0$, then $a^p\leq b^p\leq c^p$. Also, the numbers a, b, c are nonnegative, and thus the numbers $a^p$, $b^p$, $c^p$ are also nonnegative. Hence, we can apply Theorem 2 to the numbers $X=a^p$, $Y=b^p$, $Z=c^p$ and immediately get $\sum_{\text{cyc}}a^p\left(a-b\right)\left(a-c\right)\geq 0$. Thus, Theorem 3 [b](a)[/b] is proven.

[b](b)[/b] Since the inequality in question is symmetric, we can WLOG assume that $a\geq b\geq c$. Since p is an even nonnegative integer, we have p = 2n for some nonnegative integer n.

Define a function sign by $\text{sign }t=\left\{\begin{array}{cc}-1,&\mbox{ if }t<0\\0,&\mbox{ if }t=0\\1,&\mbox{ if }t>0\end{array}\right.$. Define $x=\text{sign } a\cdot\left|a^n\right|$, $y=\text{sign } b\cdot\left|b^n\right|$, $z=\text{sign } c\cdot\left|c^n\right|$. Then, $x^2=\left(\text{sign } a\cdot\left|a^n\right|\right)^2=\left(a^n\right)^2=a^{2n}=a^p$ and similarly $y^2=b^p$ and $z^2=c^p$. But $a\geq b\geq c$ yields $x\geq y\geq z$ (in fact, the function $f\left(t\right)=\text{sign } t\cdot\left|t^n\right|$ is monotonically increasing on the whole real axis). Thus, we can apply Theorem 1 to get

$\left(\sum_{\text{cyc}}\left(a-b\right)\left(a-c\right)\right)\cdot\left(\sum_{\text{cyc}}a^p\left(a-b\right)\left(a-c\right)\right)=\left(\sum_{\text{cyc}}\left(a-b\right)\left(a-c\right)\right)\cdot\left(\sum_{\text{cyc}}x^2\left(a-b\right)\left(a-c\right)\right)$
$\geq\left(\sum_{\text{cyc}}x\left(a-b\right)\left(a-c\right)\right)^2\geq 0$.

By the same argument as in the proof of Theorem 2, we see that we can divide by $\sum_{\text{cyc}}\left(a-b\right)\left(a-c\right)$, and obtain $\sum_{\text{cyc}}a^p\left(a-b\right)\left(a-c\right)\geq 0$. This proves Theorem 3 [b](b)[/b].

[b]Note.[/b] An alternative proof of Theorem 3 [b](b)[/b] can be given using the Vornicu-Schur inequality for convex functions ( http://www.mathlinks.ro/Forum/viewtopic.php?t=1166 post #11) - in fact, for an even nonnegative integer p, the function $f\left(x\right)=x^p$ is neither increasing nor decreasing over all $\mathbb{R}$, but it is convex, so we can apply Vornicu-Schur.

Beautiful result, Vasc!

  Darij

And your proof and your results are beautiful, Darij !  :lol:

A rather ugly generalization.
If $a \ge b \ge c$ and $x_i$, $y_i$ satisfy
(a) $x_i \ge x_j$, $y_i \ge y_j$
(b) $y_i \ge x_i$ and
(c) $y_i - y_j \le x_i - x_j$ where $j \ge i$ and $i,j \in {1,2,3}$, then
\[ (\sum_{cyc} x_i^2 (a-b)(a-c) )(\sum_{cyc} y_i^2 (a-b)(a-c) ) \ge (\sum_{cyc} x_i y_i (a-b)(a-c) )^2 \]
Darij's generalization is a special case with all $y_i$ are equal.

I say it's ugly because I can't find any particular use of this inequality(while its special case, vornicu-schur, has much applications. So I will call that one beautiful) yet. For those who can find it, please tell me.

And more general now.
[i]Let $p$ and $q$ be real numbers such that $pq>0$, and let $a,b,c$ be non-negative real numbers. Prove that
$S_0.S_{p+q} \ge S_p.S_q$ , where $S_k=\sum_{cyc}{a^k(a-b)(a-c)}$[/i].

[quote="Vasc"]Indeed Siuhochung, this is the solution!  :)[/quote]
I think from this result, we can find a holder-like form: $S_0^{k-1} S_k \ge S_1^{k}$
Combining with an old inequality from you $S_1 (\sum \frac{1}{a}) \ge S_0$, we immediately get $S_n (\sum \frac{1}{a}) \ge S_{n-1}$ which is interesting.
Furthermore, if we suppose $T_k=\sum x^k(a-b)(a-c)$ with $x$,$y$,$z$ defined in the way darij proposed, we can also obtain $T_0 T_{p+q} \ge T_p T_q$. But I'm not sure whether $T_1 (\sum \frac{1}{a}) \ge T_0$. Actually is this true?

[quote="siuhochung"]We can again assume $a \ge b \ge c$, then write the inequality in the form $(a-b)(b-c)(a-c)((a-b)((b^q-c^q)(a^p-b^p)+(a^q-c^q)(b^p-c^p))+(b-c)((a^q-b^q)(b^p-c^p)+(a^p-b^p)(a^q-c^q))) \ge 0$
When $pq > 0$, for example, $b^q-c^q$ and $a^p-b^p$ are of the same sign. So their product is larger than 0. Same applies for other brackets, so the last inequality is true.[/quote]

could you explain me what am i doing wrong? i obtained:

$(a-b)(b-c)(a-c) ( (a-b) \left( a^p - b^p \right) \left( b^q - c^q \right) + (a-c) \left( c^p - b^p \right) \left( c^q - a^q \right) + (b-c) \left( - \left( b^p - c^p \right) \left( b^q - c^q \right) + \left( a^p - b^p \right) \left( a^q - c^q \right) \right) ) \geq 0$

the last term i obtained is not always positive i think, or at least it's not obvious to me :(

In my solution I cancelled $a-c$ by breaking it into $a-b$ and $b-c$ and regrouping of terms.

[quote="Vasc"]Siuhochung, notice that the inequality
$(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})S_p \ge S_{p-1}$
follows directly  for $p \ge 1$ by multiplying the inequalities
$S_0S_p \ge S_1S_{p-1}$
and
$(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})S_1 \ge S_0$.[/quote]
well yes i know this. What I have problem with is $T_1 (\sum \frac{1}{a}) \ge T_0$.
I'm not even sure if it's true...

[quote="siuhochung"] Furthermore, if we suppose $T_k=\sum x(a-b)(a-c)$ with $x$,$y$,$z$ defined in the way darij proposed, we can also obtain $T_0 T_{p+q} \ge T_p T_q$. But I'm not sure whether $T_1 (\sum \frac{1}{a}) \ge T_0$. Actually is this true?[/quote]
What is $k$?

oops, I actually mean $T_k = x^k (a-b)(a-c)$

[quote="Vasc"]And more general now.
[i]Let $p$ and $q$ be real numbers such that $pq>0$, and let $a,b,c$ be non-negative real numbers. Prove that
$S_{0}.S_{p+q}\ge S_{p}.S_{q}$ , where $S_{k}=\sum_{cyc}{a^{k}(a-b)(a-c)}$[/i].[/quote]
The same way we can prove a bit more generally, $S_{t}\cdot S_{t+p+q}\ge S_{t+p}\cdot S_{t+q}$ for any $t\in\mathbb R$, otherwise stated, for reals $u<v$ and for $0\le h\le v-u$, we have $S_{u}\cdot S_{v}\ge S_{u+h}\cdot S_{v-h}$, which means the function $t\mapsto\log S_{t}$ is convex. I suppose the same holds for  $t\mapsto\log T_{t}$ :) .

[quote="Soarer"]What I have problem with is $T_{1}(\sum \frac{1}{a}) \ge T_{0}$.
I'm not even sure if it's true...[/quote]But this is not homogenous, as the degree in $x,y,z$ has nothing to do with the degree in $a,b,c$. Put e.g. $x=\frac1a$ etc, I don't think that holds. :maybe:

[quote="spanferkel"]
The same way we can prove a bit more generally, $S_{t}\cdot S_{t+p+q}\ge S_{t+p}\cdot S_{t+q}$ for any $t\in\mathbb R$, otherwise stated, for reals $u<v$ and for $0\le h\le v-u$, we have $S_{u}\cdot S_{v}\ge S_{u+h}\cdot S_{v-h}$, which means the function $t\mapsto\log S_{t}$ is convex. [/quote]
I think your assertion is not true.
For example, the inequality $S_{2}*S_{4}\geq S_{3}^{2}$ doesn't hold.


When do most people commence preparing for the Putnam examination? I am currently a junior in high school, and am taking AP Calculus BC. What books and/or math levels should we know? Next year, I'm taking Advanced Calculus at Georgia Tech University. Does anybody have any advice for me? I would really appreciate any comments  :)

Don't take the Putnam (officially) as you only have four years eligibility.


superior algebra
superior algebra theorems

G is a subgrup in Sn. If all cicles of 3 elements (ijk) are in G than G=An or G=Sn.

$ A_n$ consists of those permutations which may be written as a product of an even number of transpositions. and  $ (a,b)(c,d) \equal{} (a,b,c)(b,c,d)$ shows that $ A_n$ is generated be the 3-cycles. besides, the sign shows $ |S_n : A_n| \equal{} 2$.

so if you have (a,b)(c,d) in G doesn't mean that (a,b) and (c,d) is in G. So how does this help us?


Can there be integer solutions for the depth, width, height and diagonal though space of a cuboid and none of those dimensions irrational?

In other words, you're asking

Can there be (nonzero, I'm assuming the cuboid is not degenerate) integers $ a, b, c$ such that $ a^{2}\plus{}b^{2}\plus{}c^{2}$ is a perfect square? (WLOG $ (a, b, c)\equal{}1$)

We can easily find an infinite number of solutions, because the equation $ 2x^{2}\minus{}y^{2}\equal{}\minus{}1$ has infinitely many solutions (the simplest of which is $ x\equal{}2, y\equal{}2$ yielding $ a\equal{}b\equal{}2, c\equal{}1$).  And even these are a very specific set of solutions.  So the answer is yes.

Thanks for answering. Why c = 1, and a=b=2? We are dealing with a cube.
I got this question out of a textbook.
Can we have all the dimensions (including diagonal) of a cube rational numbers?

[quote="Christine"]Thanks for answering. Why c = 1, and a=b=2? We are dealing with a cube.
[/quote]

Er...I thought it was a cuboid...

If it's a cube, then you would have to have $ 3a^{2}$ is a perfect square for some integer $ a$.  But the number of times $ 3$ divides this expression is odd: therefore, this is impossible.

For sides of rational length, the same argument applies, because you can multiply through to clear denominators.

The problem states "cuboid" or did you mean "cube" ?
At any rate, if you are looking for a cuboid that has integer side lengths, integer face diagonals.
Check this link http://mathworld.wolfram.com/PerfectCuboid.html


Let $ n$ be a natural number equal or greater than 3 . A sequence of non-negative numbers  $ (c_0,c_1,\ldots,c_n)$ satisfies the condition: $ c_{p}c_{s}\plus{}c_{r}c_{t}\equal{} c_{p\plus{}r}c_{r\plus{}s}$ for all non-negative $ p,q,r,s$ such that $ p\plus{}q\plus{}r\plus{}s\equal{}n$. Determine all possible values of $ c_2$ when $ c_1\equal{}1$.


Euler
combinatorics unsolved
combinatorics

Find the maximum number of triangulars that can be formed by $ 8$ lines in a plane with the property that any $ 2$ lines are concurent.

That's fairly simple if you employ Euler characteristic of the plane.
http://en.wikipedia.org/wiki/Euler_Characteristic


[quote="LHM"]When I still couldn't get a question right, I clicked the post on the forum button. And it told me that my post was unauthorized. But where can I find a solution to my posts? When it said "forum", what did it mean?

 :?:

Thanks,
LHM[/quote]

Someone else has mentioned this as well.  Did the message appear to come from the AOPS website or the game itself?

Try to do another Ask on Forum for any random problem and see what it does.  If it gives you the unauthorized message again, can you take a screen shot and post it?

Thanks.

OK, I just took a screenshot of it.  Its attached in PDF format.
:)
minicon


symmetry
group theory
abstract algebra
function
algebra
polynomial
superior algebra

First, I would like to ask if any of you guys know some good algebra books? mine isn't too good at explain stuff and I would like to try out another one.

I've been staring at these questions all week and have just been stump. Maybe I should give up on math, not sure
Thanks for reading and helping :lol:
Quick question, what's the greek letter that's use to denote symmetry groups?
1) If H and K are subgroups of G, show that $ H\cap K$ is also a subgroup of G
2) Find all group morphism from ( [b]Z[/b],+) to ([b]Q[/b], +)
3) Show that the set {f:[b]R[/b]-{0,1} $ \rightarrow$ [b]R[/b]-{0,1}}, of functions under composition, is isomorphic to $ \delta _{3}$
$ f_{1} \equal{} x$
$ f_{2} \equal{} 1 \minus{} x$
$ f_{3} \equal{} \frac {1}{x}$
$ f_{4} \equal{} 1 \minus{} \frac {1}{x}$
$ f_{5} \equal{} \frac {1}{1 \minus{} x}$
$ f_{6} \equal{} \frac {x}{x \minus{} 1}$

4) Is $ \alpha_{3}$ congruent to $ D_{6}$? Give Reasons
5) Find the symmetry group of the polynomial: $ (x_{1} \minus{} x_{2})(x_{3} \minus{} x_{4})$
6) Prove that the symmetry group on n is generated by the elements $ (12)(23)...(n \minus{} 1)(n)$
7) Prove that the set of all automorphisms of H forms a group under composition.

Usually, $ S_{A}$ is the standard notation for the symmetric group of some set $ A$.

1) That statement is not completely true.  The correct statement is that $ H\cup K$ is a subgroup if and only if $ H\subseteq K$ or $ K\subseteq H$ (and $ H\leq G$ and $ K\leq G$).
If $ H\subseteq K$ or $ K\subseteq H$ (assume the former just to save space, you can easily swap names), then $ H\cup K\equal{}K$, which we already know to be a subgroup.  Now assume that $ H\cup K$ is a subgroup.  If $ K\subseteq H$, we are done so we will show that $ H\subseteq K$.  Let $ x\in H$.  Since $ K\not\subseteq H$, choose $ y\in K\backslash (H\cap K)$ and consider $ z\equal{}x\circ y\in H\cup K$.  If $ z\in H$, then $ y\equal{}x^{\minus{}1}\circ z\in H\cap K$, which is a contradiction.  Hence $ z\in K$, meaning $ x\circ y^{\minus{}1}\in K$.

[quote="polarbears"]1) If H and K are subgroups of G, show that $ H \cup K$ is also a subgroup of G[/quote]  This is wrong and probably you mean $ H \cap K$.  Since you have, basically, just posted an entire homework set here, why don't you briefly list the ideas that you've tried for each question and show your work.  I'll give you a hint for the one I've quoted: begin by listing the properties an object needs to satisfy to be a subgroup.  Then check each one.

It's possible that your text is using the symbol $ \mathfrak{S}$ for the symmetric group.  Also, in the future, please proofread your posts -- this one is full of sloppy errors and typos.  There's a preview button for a reason!

1) (JBL's version) Since $ H\leq G$ and $ K\leq G$, you know they each contain some element (in particular, a common identity element).  Therefore, $ H\cap K$...  If $ h_{1}$, $ h_{2}^{\minus{}1}\in H\cap K$, then what can you say about $ h_{1}\circ h_{2}^{\minus{}1}$?  Why is this important?

7) [b]Theorem: [/b][i]Let $ G$ and $ G_{1}$ denote groups.  Then the identity map $ 1_{G}: G\rightarrow G$ is an isomorphism for every group $ G$ and if $ \sigma: G\rightarrow G_{1}$ is an isomorphism, then so is $ \sigma^{\minus{}1}: G_{1}\rightarrow G$.[/i]

[b]Corollary[/b]: [i]If $ G$ is a group, the set of all isomorphisms $ G\rightarrow G$ form a group under composition.[/i]

These are likely standard theorems in most algebra texts.  The only "work" in the theorem is showing that $ \sigma^{\minus{}1}$ is a homomorphism, but this is straightforward as long as you "follow your nose."  The corollary is immediate from the subgroup criterion.

[quote="JRav"]1) (JBL's version) Since $ H\leq G$ and $ K\leq G$, you know they each contain some element (in particular, a common identity element).  Therefore, $ H\cap K$...  If $ h_{1}$, $ h_{2}^{ \minus{} 1}\in H\cap K$, then what can you say about $ h_{1}\circ h_{2}^{ \minus{} 1}$?  Why is this important?
[/quote]

It is $ e$? So the identity exist which is one of the 2 axioms I must show?

Can I also get some hints on the other ones?

Do you know the two parts of the subgroup criterion?  Knowing them would really help.  I know you said you don't like your algebra book but I find it really hard to believe that it doesn't contain the subgroup criterion.


vector
function
topology
analytic geometry
real analysis
real analysis solved

Consider a vector-input vector-output function $\boldsymbol{f}:\mathbb{R}^K_+ \rightarrow \mathbb{R}^K_+$.
Denote $\boldsymbol{f}(\boldsymbol{x})=[f_1(\boldsymbol{x}),....,f_K(\boldsymbol{x})]^T$, satisfying:
1) $\boldsymbol{f}(\boldsymbol{x})$ is a continuous function of $\boldsymbol{x}$;
2) $f_i(\boldsymbol{x})=0$ if $x_i=0$, $i=1,2,...,K$;
3) $\| (\boldsymbol{f}(\boldsymbol{x}))_\mathcal{I}  \|\rightarrow \infty$ when $\| \boldsymbol{x}_\mathcal{I} \|\rightarrow \infty$  , $\forall \mathcal{I} \subset \{1,2,...,K\}$. Here for a vector $\boldsymbol{x} \in \mathbb{R}^K$, $\boldsymbol{x}_\mathcal{I}$ denote the vector formed by components specified by the index set $\mathcal{I}$.

Show that given any $\boldsymbol{y} \in \mathbb{R}^K_+$, there always exists some $\boldsymbol{x}^* \in \mathbb{R}^K_+$, such that $\boldsymbol{f}(\boldsymbol{x}^*) = \boldsymbol{y}$.

For $K=1$, it becomes well-known one-dimensional intermediate value theorem.
So basically, it is a generalization of intermediate value theorem to $\mathbb{R}^K$. Any ideas of the proof?

Such statements are usually reducible to the famous theorem from topology saying that if $h$ is a continuous mapping from a ball to $R^n$ whose restriction to the boundary of the ball is the identity mapping, than the image of $h$ contains the entire ball. The statement under consideration can be reduced to this theorem as follows. Note first of all that if $\varphi+\psi=1$ is any partition of unity in $R_+^k$ and $f$ and $g$ are two mappings satisfying (1)-(3), then $h=\varphi f+\psi g$ also satisfies (1)-(3). Let now $y$ be a point with all positive coordinates (if some coordinates of $y$ are $0$, we just consider the restriction of $f$ to the set of $x$ whose corresponding coordinates are also $0$ and get the same problem in lower dimension). Choose $R>0$ such that both $R$ and the norm of $f(x)$ are much bigger than the norm of $y$ whenever $\|x\|>R$. After that choose $\delta>0$ such that both $\delta$ and $f_i(x)$ are much smaller than $y_i$ whenever $x_i<\delta$ and $\|x\|\leq R$. This is possible to do due to uniform continuity of $f$ on the set $\|x\|\leq R$. Now choose a partition of unity such that $\varphi=1$ if $\|x\|\leq R$ and $x_i\geq \delta$ for all $i$ and $\psi=1$ if $\|x\|\geq R+1$ or $x_i\leq \delta/2$ for some $i$. Let $f$ be our mapping and let $g$ be the identity mapping. Let $h=\varphi f+\psi g$. Then $h$ restricted to the boundary of the symplex $x_1+\ldots+x_k\leq t$ (which topologically is as good as a ball) is the identity mapping for sufficiently large $t$ and, therefore, the image of $h$ contains $y$. It remains to note that we cannot have $h(x)=y$ for $|x|>R$ or when $x_i<\delta$ for some $i$, so we must have $f(x)=y$.

Thanks! fedja. You have solved every problem of mine!
Sorry still got questions:
1) what is a partition of unity in $R_+^k$? are $\varphi$ and $\psi$ nonnegative real value continuous functions of $x$ with $\varphi(x)+\psi(x)=1$?

2) Can you give me a reference for the theorem you mentioned in topology?

3) What additional conditions are needed to conclude that such solution $x$ is unique? For example, what if I add that for each $i$, $f_i(x)$ is an increasing function of $x_i$ when fixing $x_j (j\neq i)$?

Thanks again. It really helps me a lot!!!

[quote="remgenesis"]
1) what is a partition of unity in $R_+^k$? are $\varphi$ and $\psi$ nonnegative real value continuous functions of $x$ with $\varphi(x)+\psi(x)=1$? [/quote] Yes, that is exactly what it is.

[quote] 2) Can you give me a reference for the theorem you mentioned in topology? [/quote] I do not remember the exact name under which it goes but any decent topology textbook would have it. Unfortunately, my topology textbooks are now far away and I cannot give you the title and the page. Maybe somebody else will do it.

[quote] 3) What additional conditions are needed to conclude that such solution $x$ is unique? For example, what if I add that for each $i$, $f_i(x)$ is an increasing function of $x_i$ when fixing $x_j (j\neq i)$? [/quote] This is a subtler question. Your condition is not sufficient. Just take any increasing strictly positive $\eta:R_+\to R$ such that $\eta(1)=1$, $\eta(2)=2$ and consider the mapping $f(x_1,x_2)=(x_1\eta(x_2),x_2\eta(x_1))$. Then $f(1,2)=f(2,1)=(2,2)$. I suspect that it is enough to check that $f$ is one-to-one locally but do not have a proof as of now.


limit
induction
logarithms
inequalities
calculus
strong induction
triangle inequality

Let a sequence $f(n)$ satisfy $f(1)=1$ and $ f(n)=n-f([\frac{n}{2}])$.
Prove that $\lim\limits_{n \to \infty}{\frac{f(n)}{n}}=\frac{2}{3}$.

For all $n$, we have :
$f(2^n) = 2^n - 2^{n-1} + 2^{n-2} - \cdots + (-1)^n = \frac {2^{n+1} + (-1)^n} 3$.
Therefore $\lim \frac {f(2^n)} {2^n} = \frac 2 3.$

Direct computations shows that $f(1)=f(2)=1,f(3)=2.$
By induction it is easy to verify that $f(n+1) - f(n) = 0$ or $1$. This proves that $f$ is non decreasing.
Thus, for all $n$, let $2^p \leq n < 2^{p+1}.$ Clearly $\lim \frac {2^p} n = 1 = \lim \frac {2^{p+1}} n.$
From above, we have $f(2^p) \leq f(n) \leq f(2^{p+1).}$
It follows that $\frac {f(2^p)} {2^p} \cdot \frac {2^p} n \leq \frac {f(n)} n \leq \frac {f(2^{p+1})} {2^{p+1}} \cdot \frac {2^{p+1}} n$, hence result.

Pierre.

I don't get it:
\[
\lim \frac {2^p}{n} = 1 = \lim \frac {2^{p+1}}{n}
\]
???

Doesn't $\frac {2^p}{n}$ decrease from 1 to 1/2 in every interval $[2^p, 2^{p + 1}]$? So I guess it shouldn't have a limit at all.

You are absolutely right : This is non-sense. ;)

Let $n= a_p2^p+ \cdots + a_12 + a_0$ be the expansion of $n$ in base $2$.
Then $[\frac n 2] = a_p2^{p-1} + \cdots +a_22 + a_1.$
It easily follows that  :
$f(n) = (a_p2^p+ \cdots + a_12 + a_0) - (a_p2^{p-1} + \cdots +a_22 + a_1) + \cdots +(-1)^pa_p.$
Thus for each $i$, the contribution of $a_i$ is $a_i(2^i-2^{i-1} + \cdots +(-1)^i2^0) = \frac 2 3 2^ia_i(1- (-\frac 1 2)^{i+1}) = \frac 2 3 a_i2^i + \frac 1 3 a_i(-1)^i.$
Summing over $i$, it leads to $f(n) = \frac 2 3 n + \frac 1 3 \sum_{i=0}^p (-1)^ia_i.$

But $a_i =0$ or $1$ so that $|\sum_{i=0}^p (-1)^ia_i | \leq \sum_{i=0}^p a_i \leq p+1 \leq 1 + \log_2 (n).$
Thus $\frac 2 3 n - \frac 1 3 (1 + \log_2 (n) ) \leq f(n) \leq \frac 2 3 n + \frac 1 3 (1 + \log_2 (n) )$ from which the desired result follows easily.

Pierre.

I have really not much time now for solving problems, but this one is just too nice. Here is my solution, which, at least, looks correct to me (thanks to Peter Scholze for a slight simplification):

In the following, "natural number" means "element of the set $\left\{1,2,3,...\right\}$".

[color=blue][b]Problem.[/b] Let a sequence $\left(f\left(n\right)\right)_{n\geq 1}$ be defined by f(1) = 1 and $f\left(n\right)=n-f\left(\left[\frac{n}{2}\right]\right)$ for any natural n > 1.

Prove that $\lim_{n\to\infty}\frac{f\left(n\right)}{n}=\frac23$.[/color]

[i]Solution.[/i] Consider the sequence $\left(Q_n\right)$ defined recursively by $Q_1=\frac13$ and $Q_{n+1}=\frac12 Q_n+\frac{1}{4n}$ for any natural n. Then, direct computation using the recurrence formula yields $Q_2=\frac{5}{12}$ and $Q_3=\frac13$. Now we prove a little lemma:

[b]Lemma 1.[/b] For any $n\geq 3$, we have $0\leq Q_n\leq\frac{1}{n}$.

[hide="Proof of Lemma 1"]
[i]Proof of Lemma 1.[/i] We prove Lemma 1 by induction over n:

For n = 3, the inequality $0\leq Q_n\leq\frac{1}{n}$ is evident, since $Q_n=Q_3=\frac13=\frac{1}{n}$.

Now assume that the inequality $0\leq Q_n\leq\frac{1}{n}$ holds for some n = k, with $k\geq 3$. We have to show that it also holds for n = k + 1.

Well, since the inequality $0\leq Q_n\leq\frac{1}{n}$ holds for n = k, we have $0\leq Q_k\leq\frac{1}{k}$. The recursive definition of the sequence $\left(Q_n\right)$ yields $Q_{k+1}=\frac12 Q_k+\frac{1}{4k}$, so that $0\leq Q_k$ immediately yields $0\leq Q_{k+1}$. Also, since $k\geq 3$, we have $k-3\geq 0$, and thus

$\frac{1}{2k}+\frac{1}{4k}-\frac{1}{k+1}=-\frac{k-3}{4k\left(k+1\right)}\leq 0$,

so that $\frac{1}{2k}+\frac{1}{4k}\leq\frac{1}{k+1}$. On the other hand, we know that $Q_k\leq\frac{1}{k}$. Hence,

$Q_{k+1}=\frac12 Q_k+\frac{1}{4k}\leq \frac12\cdot\frac{1}{k}+\frac{1}{4k}=\frac{1}{2k}+\frac{1}{4k}\leq\frac{1}{k+1}$.

Combining this with $0\leq Q_{k+1}$, we get $0\leq Q_{k+1}\leq\frac{1}{k+1}$. Thus, the inequality $0\leq Q_n\leq\frac{1}{n}$ is proven for n = k + 1. Hence, the induction step is completed, and Lemma 1 is proven.[/hide]

Now, since we have $0\leq Q_n\leq\frac{1}{n}$ for all $n\geq 3$, but $\lim_{n\to\infty} 0=0$ and $\lim_{n\to\infty}\frac{1}{n}=0$, the sandwich theorem yields $\lim_{n\to\infty} Q_n=0$.

Now define a sequence $d_n=\left|\frac{f\left(n\right)}{n}-\frac23\right|$. We will show that $\lim_{n\to\infty} d_n = 0$. Clearly, once this will be proven, it will follow that $\lim_{n\to\infty}\frac{f\left(n\right)}{n}=\frac23$, and thus, the problem will be solved.

Hence, it is enough to show that $\lim_{n\to\infty} d_n = 0$.

In order to prove this, we begin with a simple fact:

[b]Lemma 2.[/b] For any natural number n, we have $0<f\left(n\right)\leq n$.

[hide="Proof of Lemma 2"]
[i]Proof of Lemma 2.[/i] We will prove Lemma 2 by strong induction over n: First we must prove Lemma 2 for n = 1; then we must show that, if k > 1 is any natural number, and Lemma 2 holds for all natural numbers n < k, then Lemma 2 must also hold for n = k.

Well, proving Lemma 2 for n = 1 is trivial: We know that f(1) = 1, so that $0<f\left(1\right)\leq 1$, and thus, Lemma 2 holds for n = 1.

Now assume that Lemma 2 holds for all natural numbers n < k, where k > 1 is some given natural number. We will show that Lemma 2 also holds for n = k.

In fact, since k > 1, we have $f\left(k\right)=k-f\left(\left[\frac{k}{2}\right]\right)$. Since $\left[\frac{k}{2}\right]<k$ (this is because $\left[\frac{k}{2}\right]\leq\frac{k}{2}$, and $\frac{k}{2}<k$), we can apply Lemma 2 to $n=\left[\frac{k}{2}\right]$, and get $0<f\left(\left[\frac{k}{2}\right]\right)\leq \left[\frac{k}{2}\right]$. Now, since $f\left(\left[\frac{k}{2}\right]\right)\leq \left[\frac{k}{2}\right]$ and $\left[\frac{k}{2}\right]<k$, we have $f\left(\left[\frac{k}{2}\right]\right)<k$, and thus $f\left(k\right)=k-f\left(\left[\frac{k}{2}\right]\right)>0$; also, since $0<f\left(\left[\frac{k}{2}\right]\right)$, we have $f\left(k\right)=k-f\left(\left[\frac{k}{2}\right]\right)<k$. Thus, $0<f\left(k\right)<k$; consequently, $0<f\left(k\right)\leq k$, and thus, Lemma 2 holds for n = k. This completes the induction step. Hence, Lemma 2 is proven. (Note in passing that, as a consequence of this proof, we see that the inequality $f\left(n\right)\leq n$ can become an equality only for n = 1. But this won't be important in our further investigations.)[/hide]

The crucial lemma of our proof will be the following one:

[b]Lemma 3.[/b] If u is any natural number, then, for any natural number n such that $2^u\leq n<2^{u+1}$, we have $d_n\leq Q_u$.

[i]Proof of Lemma 3.[/i] We prove this lemma by induction over u:

For u = 1, proving Lemma 3 comes down to verifying that $d_2\leq Q_1$ and $d_3\leq Q_1$ (in fact, the only natural numbers n such that $2^1\leq n<2^2$ are n = 2 and n = 3). But this is straightforward, since the recursive formula $f\left(n\right)=n-f\left(\left[\frac{n}{2}\right]\right)$ for any natural n > 1 yields, with the help of f(1) = 1, the values f(2) = 1 and f(3) = 2, and being plugged into the formula $d_n=\left|\frac{f\left(n\right)}{n}-\frac23\right|$, these values yield $d_2=\frac16$ and $d_3=0$, which are both $\leq\frac13=Q_1$.

Now we come to the induction step: Assume that Lemma 3 is correct for u = k, where k is some natural number. In other words, for any natural number n such that $2^k\leq n<2^{k+1}$, we have $d_n\leq Q_k$. We have to show that Lemma 3 is also correct for u = k + 1; in other words, we have to show that, for any natural number N such that $2^{k+1}\leq N<2^{k+2}$, we have $d_N\leq Q_{k+1}$. We can show this as follows:

Let N be a natural number such that $2^{k+1}\leq N<2^{k+2}$. Let $n=\left[\frac{N}{2}\right]$. Then, $n\leq\frac{N}{2}<n+1$. Dividing the chain of inequalities $2^{k+1}\leq N<2^{k+2}$ by 2, we get $2^k\leq\frac{N}{2}<2^{k+1}$. Now, from $\frac{N}{2}<n+1$ and $2^k\leq\frac{N}{2}$, it follows that $2^k<n+1$, so that, because of the integrality of n, we get $2^k\leq n$. Moreover, from $n\leq\frac{N}{2}$ and $\frac{N}{2}<2^{k+1}$, we get $n<2^{k+1}$. Thus, $2^k\leq n<2^{k+1}$. Hence, after the induction assumption (i. e., after the assumption that Lemma 3 holds for u = k), we have $d_n\leq Q_k$.

Now, by the definition of the sequence f(n), we have $f\left(N\right)=N-f\left(\left[\frac{N}{2}\right]\right)$. Since $\left[\frac{N}{2}\right]=n$, we thus have $f\left(N\right)=N-f\left(n\right)$. Consequently, $\frac{f\left(N\right)}{N}=\frac{N-f\left(n\right)}{N}=1-\frac{f\left(n\right)}{N}$. Thus,

$d_N=\left|\frac{f\left(N\right)}{N}-\frac23\right|=\left|\left(1-\frac{f\left(n\right)}{N}\right)-\frac23\right|=\left|\frac13-\frac{f\left(n\right)}{N}\right|=\left|\frac{f\left(n\right)}{N}-\frac13\right|$
$=\frac12\cdot\left|\frac{2f\left(n\right)}{N}-\frac23\right|=\frac12\cdot\left|\frac{f\left(n\right)}{\left(\frac{N}{2}\right)}-\frac23\right|=\frac12\cdot\left|\left(\frac{f\left(n\right)}{\left(\frac{N}{2}\right)}-\frac{f\left(n\right)}{n}\right)+\left(\frac{f\left(n\right)}{n}-\frac23\right)\right|$
$\leq\frac12\cdot\left(\left|\frac{f\left(n\right)}{\left(\frac{N}{2}\right)}-\frac{f\left(n\right)}{n}\right|+\left|\frac{f\left(n\right)}{n}-\frac23\right|\right)$ (by the triangle inequality)
$=\frac12\cdot\left(\left|\frac{f\left(n\right)}{\left(\frac{N}{2}\right)}-\frac{f\left(n\right)}{n}\right|+d_n\right)=\frac12\cdot\left|f\left(n\right)\left(\frac{1}{\left(\frac{N}{2}\right)}-\frac{1}{n}\right)\right|+\frac12d_n$.

Since $n\leq\frac{N}{2}$, we have $\frac{1}{n}\geq\frac{1}{\left(\frac{N}{2}\right)}$, so that $\left|\frac{1}{\left(\frac{N}{2}\right)}-\frac{1}{n}\right|=\frac{1}{n}-\frac{1}{\left(\frac{N}{2}\right)}$. On the other hand, since 0 < f(n) after Lemma 2, we have $\left|f\left(n\right)\left(\frac{1}{\left(\frac{N}{2}\right)}-\frac{1}{n}\right)\right|=f\left(n\right)\cdot\left|\frac{1}{\left(\frac{N}{2}\right)}-\frac{1}{n}\right|$. Thus,

$d_N\leq\frac12\cdot\left|f\left(n\right)\left(\frac{1}{\left(\frac{N}{2}\right)}-\frac{1}{n}\right)\right|+\frac12d_n=\frac12\cdot f\left(n\right)\cdot\left|\frac{1}{\left(\frac{N}{2}\right)}-\frac{1}{n}\right|+\frac12d_n$
$=\frac12\cdot f\left(n\right)\cdot\left(\frac{1}{n}-\frac{1}{\left(\frac{N}{2}\right)}\right)+\frac12d_n$.

But remember that $\frac{N}{2}<n+1$; thus, $\frac{1}{\left(\frac{N}{2}\right)}>\frac{1}{n+1}$, so that

$\frac{1}{n}-\frac{1}{\left(\frac{N}{2}\right)}<\frac{1}{n}-\frac{1}{n+1}=\frac{\left(n+1\right)-n}{n\left(n+1\right)}=\frac{1}{n\left(n+1\right)}$.

Since n + 1 > n, we have $n\left(n+1\right)>n^2$, and thus

$\frac{1}{n}-\frac{1}{\left(\frac{N}{2}\right)}<\frac{1}{n\left(n+1\right)}<\frac{1}{n^2}$.

Consequently,

$d_N\leq\frac12\cdot f\left(n\right)\cdot\left(\frac{1}{n}-\frac{1}{\left(\frac{N}{2}\right)}\right)+\frac12d_n<\frac12\cdot f\left(n\right)\cdot\frac{1}{n^2}+\frac12d_n$.

But Lemma 2 yields $f\left(n\right)\leq n$, and we know that $d_n\leq Q_k$. Thus,

$d_N<\frac12\cdot f\left(n\right)\cdot\frac{1}{n^2}+\frac12d_n\leq \frac12\cdot n\cdot\frac{1}{n^2}+\frac12 Q_k=\frac{1}{2n}+\frac12 Q_k$.

But since $k\geq 1$, we have $2k\leq 2^k$. Since $2^k\leq n$, we thus have $2k\leq n$, so that $n\geq 2k$, and $\frac{1}{2n}\leq\frac{1}{4k}$. Consequently,

$d_N<\frac{1}{2n}+\frac12 Q_k\leq\frac{1}{4k}+\frac12 Q_k=\frac12 Q_k+\frac{1}{4k}$.

But by the definition of the sequence $\left(Q_n\right)$, we have $\frac12 Q_k+\frac{1}{4k}=Q_{k+1}$, so that we get $d_N<Q_{k+1}$. We don't need the sharp inequality, it's enough for us to know that $d_N\leq Q_{k+1}$. This completes the induction step, and thus, Lemma 3 is proven.

Now the final argument to solve the problem is canonical:

If we consider any natural number $n\geq 2$, and define a natural number u by $u=\left[\log_2 n\right]$, then $u\leq \log_2 n < u+1$; thus, $2^u\leq 2^{\log_2 n}<2^{u+1}$. In other words, $2^u\leq n<2^{u+1}$. Hence, we can apply Lemma 3 to the numbers u and n, and get $d_n\leq Q_u$. Since $u=\left[\log_2 n\right]$, this becomes $d_n\leq Q_{\left[\log_2 n\right]}$. On the other hand, the definition of the sequence $d_n$ by the equation $d_n=\left|\frac{f\left(n\right)}{n}-\frac23\right|$ yields that $0\leq d_n$ for any n. Thus, we get $0\leq d_n \leq Q_{\left[\log_2 n\right]}$ for any $n\geq 2$. Since $\lim_{n\to\infty} 0=0$ and $\lim_{n\to\infty} Q_{\left[\log_2 n\right]}=0$ (the latter follows from $\lim_{n\to\infty} Q_n=0$ and from the fact that the logarithm grows unboundedly), the sandwich theorem yields $\lim_{n\to\infty} d_n=0$, and the problem is solved.

Let me hope that the above solution is halfways correct - that would be very nice for the first nontrivial calculus problem I ever solved. ;)

[b]EDIT:[/b] I'm just seeing Pierre's new solution - it's really cool! However, my one is a kind of elusion for somebody who is not too secure in number theory (like me)...

  darij

[quote="kunny"]\[f(n)=n-f\left(\left[\frac{n}{2}\right]\right),\ f(1)=1\Longleftrightarrow f(n)=\left\{\begin{array}{cc} \frac{2}{3}n+\frac{1}{3}(-1)^{\ log_2 n} &\mbox n:even \\
\frac{2}{3}n+\frac{1}{6}-\frac{1}{6}(-1)^{\log_2(n+1)} &\mbox n:odd\end{array}\right.\][/quote]

???

I always thought that, for [i]most[/i] natural n, the logarithm $\log_2 n$ is transcendental, so that terms like $\left(-1\right)^{\log_2 n}$ are... hmm... well, not so appropriate here, maybe?

Anyway, I can only repeat that Pierre's solution is beautiful. Maybe I should finally start learning combinatorics and number theory?...

  dg

Thank you for your comment, darij.

Anyway, There are some solution! Needless to say, my solution is one of them. ;)

Comment: [b]Assuming that $\frac{f(n)}{n}$ converges[/b],intuitively the conclusion follows.

Case 1: $f(2n)=2n-f\left(\left[\frac{2n}{2}\right]\right)\Longleftrightarrow \frac{f(2n)}{2n}=1-\frac{1}{2}\cdot \frac{f(n)}{n}$

Case 2: $f(2n+1)=2n+1-f\left(\left[\frac{2n+1}{2}\right]\right)\Longleftrightarrow \frac{f(2n+1)}{2n+1}=1-\frac{n}{2n+1}\cdot \frac{f(n)}{n}$


inequalities
function
inequalities unsolved

let   $ a , b>0$ real , $ m<0$ integer
Prove that
  $ \left( 1\plus{} \frac{a}{b} \right)^m \plus{} \left( 1\plus{} \frac{b}{a} \right)^m \ge 2^{m\plus{}1}$

[quote="aktyw19"]let   $ a , b > 0$ real , $ m < 0$ integer
Prove that
  $ \left( 1 \plus{} \frac {a}{b} \right)^m \plus{} \left( 1 \plus{} \frac {b}{a} \right)^m \ge 2^{m \plus{} 1}$[/quote]

Using Am_GM inequality we have

$ \left( 1 \plus{} \frac {a}{b} \right)^m \plus{} \left( 1 \plus{} \frac {b}{a} \right)^m \ge 2\sqrt{\left( 1 \plus{} \frac {a}{b} \right)\left( 1 \plus{} \frac {b}{a} \right)}^m$

and using Cauchy schwarz inequality we have $ (1\plus{}\frac{a}{b})(1\plus{}\frac{b}{a})\geq (1\plus{}1)^2$ Hence

$ 2\sqrt{\left( 1 \plus{} \frac {a}{b} \right)\left( 1 \plus{} \frac {b}{a} \right)}^m \geq 2(1\plus{}1)^m\equal{}2^{m\plus{}1}$

 :)

$ f(x) \equal{} x^m$ is convex for $ m<0$, $ x>0$.

by Jensen's and AM-GM, $ LHS \ge 2\left( 2\plus{}\frac{a}{b}\plus{}\frac{b}{a}\right)^{m} \ge RHS$ :read:

@[b]Evariste-Galois[/b]: wow your solution is simpler :D

[quote="aktyw19"]let   $ a , b > 0$ real , $ m < 0$ integer
Prove that
  $ \left( 1 \plus{} \frac {a}{b} \right)^m \plus{} \left( 1 \plus{} \frac {b}{a} \right)^m \ge 2^{m \plus{} 1}$[/quote]
As noted before the two other solutions are wrong...
Let $ n \equal{} \minus{}m > 0$. Then we have:
$ \left( 1\plus{}\frac{a}{b}\right)^{m}\plus{}\left( 1\plus{}\frac{b}{a}\right)^{m}\ge 2^{m\plus{}1} \iff$
$ \frac{b^n \plus{} a^n}{(a\plus{}b)^n} \ge 2^{1\minus{}n} \iff$
$ \frac{a^n\plus{}b^n}{2} \ge \left ( \frac{a\plus{}b}{2} \right)^n$, which is the power mean inequality

[quote="gelo"]@Evariste-Galois and earldbest:
Your solutions are false and precisely the errors occur in the last line :
m<0 so the power function x^m is decreasing ...  :wink:

This is my solution :
I consider f(x) = (1+x)^m+(1+1/x)^m.
By derivation you find that the function admits a minimum at x=1.  :D

Greetings, gelo[/quote]

oops :wallbash:

[quote="gelo"]@Evariste-Galois and earldbest:
Your solutions are false and precisely the errors occur in the last line :
m<0 so the power function x^m is decreasing ...  :wink:

This is my solution :
I consider f(x) = (1+x)^m+(1+1/x)^m.
By derivation you find that the function admits a minimum at x=1.  :D

Greetings, gelo[/quote]

ohhh yes !! :blush:  it seemed to me that $ m>0$   :(


[hide="Hint"]
Use the closed form for each sum and then divide to obtain a polynomial and a constant remainder. Find the values of $ n$ such that the remainder is in integer.
[/hide]

Thanks for hint, but I solved this problem, and I am unsure if my solution is correct :). So please, can you just write final solution for me, just to verify ?

[quote="Tomekk"]Hi,

for which natural numbers "n", number 1+2+3+..+n divides 1*2 + 2*3 + 3*4 + ... + n(n+1) ?[/quote]
[hide="Solution?"]The first sum can be written as $ \frac{(n)(n\plus{}1)}{2}$. The second sum can be written as $ \frac{(n)(n\plus{}1)(2n\plus{}1)}{6}\plus{}\frac{(n)(n\plus{}1)}{2}$ Dividing the first from the second we then have $ \frac{2n\plus{}1}{3}\plus{}1$.
So we just need the fractional part to be an integer and that occurs when the top is a multiple of 3. Call this 3k, so we have that $ 2n\plus{}1\equal{}3k$ where k is an integer. Solving we see that $ n\equal{}\frac{3k\minus{}1}{2}$. Not all integer values of k will make n an integer or natural, thus the only ones that do work are when k is odd. Therefore since we're only looking for the values that k is odd, k can be written as $ 2m\minus{}1$ where m is an integer. Plugging this back in we get that $ n\equal{}3m\minus{}2$. [/hide]


vector
absolute value
linear algebra
linear algebra unsolved

Suppose V is a real inner product space and (v1...vm) is a linearly independent list of vectors in V.
Prove that there exist exactly 2^m orthonormal lists (e1...em) of vectors in V such that
span(v1...vj)=span(e1...ej) for all j contained in {1,...m}

The problem here is that I cannot see how it is true that there exist exactly 2^m orthonormal
lists when m=1. When m=1, we have span(e1) with e1 being the only orthonormal vector.
If anybody could give me a hint (not the solution itself, lol) that would be great.  :)

Oh  :D That's right! What's with me and neglecting negative numbers  :P ?
Just to make sure that I'm on the right track before I start the proof, when m=1, j can only be 1 since
orthonormal lists are linearly independent. So there are two possible lists here that are
orthonormal: (e1) and (-e1). Considering that the space V is 1 dimensional, with m=1,
the vector e1 is [1] and -e1 is [-1]. Clearly lle1ll is 1 and ll-e1ll is also 1. And <e1,e1> is 1 and
<-e1,-e1> is also 1, and e1 is orthogonal to a vector with a lone component in another plane
(so lets say a vector e2 is (a column) [0 1] and e1 is [1 0] then the dot is 0 same for -e1).
Do I have this down, don't want to start the proof until I have it down. Thanks
fedja, this problem's been buggin me.  :D

The dimension of $ V$ is not specified. The two choices for $ e_1$ are $ \pm \frac{v_1}{\lVert v_1 \rVert}$.

Yes, I know. It's just that the case where m=1 has been bugging me. Would what I have said
be true if m did=1?

$ m$ is the number of vectors $ v_i$, not the dimension of $ V$. Regardless of what $ m$ is, $ e_1$ can only be $ \pm \frac {v_1}{\lVert v_1 \rVert}$. Now $ e_2$ should be chosen so that $ \{e_1, e_2\}$ is an ON-basis for $ V_2 \equal{} \text{span} \{v_1, v_2\}$.

Note that had this been stated in a complex inner product space rather than a real inner product space, the cardinality would not have been finite. You could multiply each $ e_j$ by a complex scalar of absolute value $ 1,$ and there are infinitely many of those.

What's being asked for here: a proof that in a real inner product space, the output of the Gram-Schmidt process is unique up to $ \pm$ signs, provided the order of the $ v_j$ is specified. (Permute that order and you lose uniqueness.)

This might have something to do
with combinations.
Let {e1....em} be a list of orthonormal vectors in V. Let ej=vj/llvjll. Set j=1, and let {ej} be a basis for V1, a subspace of V. Since {ej} is a basis for
the subspace of Vm, same for {-e1}, and both are orthonormal, these are the two lists of orthonormal
vectors under V1, which turns out to be 2C1=2^1

Let j=2. Let the basis for V2 be {e2}. Now include e2 with {e1} and {-e1}. The number of
different lists have not changed, but including -e2 to {e1} and {-e1} will double the
amount of different lists (since -e2 cannot share a list with e1, since that wouldn't make the
list orthonormal) so there are 2C1*2C1 lists in the union of V1 and V2, or 2^2=4.

Since each Vj where 1<=j<=m have orthonormal bases {ej}, and {-ej}, there are
twice as many different lists in the space with {e1...ej} as a basis
 after introducing ej to a basis {e1...ej-1} in that ej and -ej are the only
orthogonal bases in Vj and that ej cannot share a list with -ej in any list,
2^j-1*2C1 doubles when a basis of Vj is added to {e1...ej-1}.

I understand the concept, but I don't think I can prove it clearly. If you
could start me off on a better route, that'll be great. Thank you!

That doesn't work at all, because the the vectors $ \frac{v_i}{\|v_i\|}$ are not orthonormal. All we were ever told about the $ v_i$ is that they were independent - not orthogonal.

You [i]must[/i] use some of the mechanics of the Gram-Schmidt process.

[quote="Kent Merryfield"]That doesn't work at all, because the the vectors $ \frac {v_i}{\|v_i\|}$ are not orthonormal. All we were ever told about the $ v_i$ is that they were independent - not orthogonal.

You [i]must[/i] use some of the mechanics of the Gram-Schmidt process.[/quote]
Alright here goes
Let U be an orthogonal subspace of V with a basis {u1...um}.
 Set j=1, and let uj=vj. Let uj=vj-proj_uj(vj). Set ej=uj/llujll {ej} be an orthonormal basis for V1, a subspace of V. Since {ej} is a basis for
the subspace of Vm, same for {-e1}, and both are orthonormal, these are the two lists of orthonormal
vectors under V1, which turns out to be 2C1=2^1

Let j=2.  Set uj=vj-proj_u1(v2)-proj_u2(v2). Set ej=uj/llujll. Let the orthonormal basis for V2 be {e2}. Now include e2 with {e1} and {-e1}. The number of
different lists have not changed, but including -e2 to {e1} and {-e1} will double the
amount of different lists (since -e2 cannot share a list with e1, since that wouldn't make the
list orthonormal) so there are 2C1*2C1 lists in the union of orthonormal lists in V1 and V2, or 2^2=4.

Since each Vj where 1<=j<=m have orthonormal bases {ej}, and {-ej}, there are
twice as many different lists in the space with {e1...ej} as a basis
after introducing ej to a basis {e1...ej-1} in that ej and -ej are the only
orthonormal bases in Vj and that ej cannot share a list with -ej in any list,
2^j-1*2C1 doubles when a basis of Vj is added to {e1...ej-1}.


the third word should read part.

I will retype, if you are having problems reading.

The gnomon (part that sticks up) of a sundial casts a shadow exactly to the edge of the dial. The base of the gnomon is 6 inches long. The sundial's face has a radius of 9 inches and the long edge of the shadow was measured to be 13 inches long, as shown. What time is it (to the nearest minute)?

Thanks again

ps: the hypothenuse of the dark grey triangle should be labeled 13 inches


A negative exponent indicates a reciprocal.

$ \left(\frac{1}2\right)^{\minus{}3} \equal{}2^3\equal{}\boxed{8}$


calculus
derivative
integration
geometry
similar triangles

Source: Mathematical Morsels

Let ABC denote an equilateral triangle inscribed in a circle. For any point P on the circle, show that the two shorter segments among PA, PB, PC add up to the third one.

What is "Mathematical Morsels"?

Hint:

[hide]Ptolemy[/hide]

[quote="Iversonfan2005"][quote="riddler"]Is Ptomley theorem GS level?
 :?[/quote]

Well, it's not very difficult to memorize or understand ;)[/quote]

Or prove since it only requires use of similar triangles.  :)

[quote="Arne"]Well, it's true that the proof is elementary in that it only uses similar triangles, but it's not an easy proof to come up with (not at all) so I don't think the proof is GS level.[/quote]

Definitely. Proving once you know what to use and where to start will be easy but "deriving" part is definitely not GS.
I never said that this is a challenging problem to prove it.  :D

Anyway.. Back to the problem..  :)

[quote="Iversonfan2005"][quote="riddler"]Is Ptomley theorem GS level?
 :?[/quote]

Well, it's not very difficult to memorize or understand ;)[/quote]

Basic derivatives and integrals aren't difficult to memorize or understand either.  Does that make them GS?

[url=http://www.amazon.com/exec/obidos/ISBN%3D0883853035/ctksoftwareincA/104-6269019-6342334]Matematical Morsels[/url] is a book.

Cut the knot even cites this exact problem from the same book [url=http://www.cut-the-knot.org/proofs/ptolemy.shtml#consequence]here[/url].


vector
analytic geometry
trigonometry
rotation
complex numbers

Hmm...

I have little experience with vectors so I applied complex numbers instead.

$ f_1\equal{}150e^{i225^\circ}$
$ f_2\equal{}220e^{i125^\circ}$

$ |f_1\plus{}f_2|\approx 243.802$
$ \arg(f_1\plus{}f_2)\approx 162.294^\circ$

So I'd agree with your answer.

The notion of 125 degrees is ambiguous because you have no idea which direction 0 degrees is.

However, backtracking, we definitely know SW to be $ 225^\circ$ (in polar coordinates, where positive x-axis is $ 0^\circ$). Thus, subtracting the vectors $ (243.8\angle 134^\circ)\minus{}(150\angle 225^\circ)$ yields a vector with magnitude 288 N and pointing $ 102.7^\circ$, which does not agree at all with the vector given in the problem.

[quote="helpmeplz"]Hey guys if someone can help me out on this I`d really appreciate it.

Two forces, one 150 N and the other 220 N are applied to the same object. The first force is oriented SW and the second at 125 degrees. What single force would have the same effect on this object? Find the force and the angle of direction.

I found 243.8 N in the direction of 162 degrees. But my book found 243.8 N in the direction of 134 degrees. Which is rightÉ[/quote]
You are right; your text is wrong.

\begin{eqnarray*}\sum F_x\colon 150\mbox{ N }\cos 225^{\circ} + 220\mbox{ N }\cos 125^{\circ} &=& -232.25\mbox{ N } \\\sum F_y\colon 150\mbox{ N }\sin 225^{\circ} + 220\mbox{ N }\sin 125^{\circ} &=& 74.147\mbox{ N }\end{eqnarray*}

\[ F_R =\sqrt{(-232.25)^2 +(74.147)^2} = 243.80\mbox{ N }\]

\[ \phi = \tan^{-1}\left(\frac{74.147}{-232.25}\right) = 162^{\circ}\]

[quote="Yongyi781"]The notion of 125 degrees is ambiguous because you have no idea which direction 0 degrees is.[/quote] I disagree. Since we have stipulated $ \mbox{Southwest (SW)} = 225^{\circ}$ that defines $ 0^{\circ} =\mbox{ due North }$ and further $ 125^{\circ} = 35^{\circ}\mbox{ South of East.}$ Both directions are completely unambiguous.

[quote="Yongyi781"]The notion of 125 degrees is ambiguous because you have no idea which direction 0 degrees is.[/quote]

That's not true; we can arbitrarily assign a degree value that represents North. If we preferred, we could call North 90 degrees and rotate the coordinate axes so that it corresponds to $ 0^\circ$, and our answer will work out the same way.

[quote="Dr. No"][quote="Yongyi781"]The notion of 125 degrees is ambiguous because you have no idea which direction 0 degrees is.[/quote] I disagree. Since we have stipulated $ \mbox{Southwest (SW)} \equal{} 225^{\circ}$ that defines $ 0^{\circ} \equal{} \mbox{ due North }$ and further $ 125^{\circ} \equal{} 35^{\circ}\mbox{ South of East.}$ Both directions are completely unambiguous.[/quote]

In fact, if $ SW \equal{} 225^\circ$, then $ 0^\circ$ is due East. :wink:

It is common in polar coordinates that we designate $ 0^\circ$ as east. However, in physics textbooks, we have no idea what reference frame we have. If we choose $ 0^\circ$ to be due East, which is the most common interpretation, we have $ 125^\circ$ is $ 35^\circ$ west of north. (Keeping in mind that it goes counterclockwise). However, if we suddenly decide that $ 0^\circ$ should be North, and also that we should go clockwise, that makes $ 125^\circ$ equal to $ 35^\circ$ south of east.

You get the same answer for both since they are symmetric with respect to SW, the direction of which we intuitively fix to be a constant angle regardless of reference frame. However, the answer will change if we decide to use $ 125^\circ$ counterclockwise from South, for example.

In short: SW is fixed, but $ 125^\circ$ is not.

[quote="Yongyi781"][quote="Dr. No"][quote="Yongyi781"]The notion of 125 degrees is ambiguous because you have no idea which direction 0 degrees is.[/quote] I disagree. Since we have stipulated $ \mbox{Southwest (SW)} \equal{} 225^{\circ}$ that defines $ 0^{\circ} \equal{} \mbox{ due North }$ and further $ 125^{\circ} \equal{} 35^{\circ}\mbox{ South of East.}$ Both directions are completely unambiguous.[/quote]

In fact, if $ SW \equal{} 225^\circ$, then $ 0^\circ$ is due East. :wink:[/quote]

You are quite wrong, my friend. If $ 0^{\circ}$ were due East, then $ 225^{\circ}$ is NW; not SW. I believe you are confusing the use of polar coordinates in standard position, what is typically done in trigonometry, with compass heading, or azimuth.

In polar coordinates, in standard position, a zero angle is represented by a point on the unit circle directly to the right of the circle's center; $ 180^{\circ}$ is then a point on the unit circle directly to the left of the circle's center.

Whereas for azimuth, $ 0^{\circ}\mbox{ is due North, }90^{\circ}\mbox{ is due East, }180^{\circ}\mbox{ is due South, and }270^{\circ}\mbox{ is due West.}$ The compass directions are defined this way to ensure alignment with the Earth's magnetic field.

These directions are well defined and well known. There is no ambiguity or confusion in any of it.

[quote="Yongyi781"]Ah-ha, but the ambiguity lies in the fact that there is no indication as to which of these two systems to prefer. By just saying $ 125^\circ$, you are leaving it to the reader to choose the system of coordinates.[/quote]

Not really. By specifying the direction of the first force as SW, compass heading is implied.

Hm, ok, I guess that makes sense. It wouldn't change the answer for this problem, though, since both systems are symmetrical with respect to SW.


Ok

What is LOL or lol or whatever!

tspa

LOL is a common internet acronym that stands for [b]Laughing Out Loud[/b].  Traditionally, it was used to denote that something was very funny - funny enough, in fact, that the reader (either on a forum or in an instant message conversation) was, in fact, laughing out loud.  People can't hear you laugh over the internet, so LOL became the common way to express merriment in the absence of actual vocal communication.

Nowadays, it tends to be used in response to something that was intended to be funny, but may or may not actually have been very funny.  Rarely do people actually use it after physically Laughing Out Loud, at least from my experience.

"dude an ambulance just pulled up to my house"
"lol"

[quote="bubka"]And in connection rotfl is rolling on the floor laughing an extreme version of lol.

What is w00t?[/quote]

http://en.wikipedia.org/wiki/W00t

"Things about aops that you want to know!?"
"lol"
 :lol:  :rotfl:

What's wrong with that? Ok, I might consider changing that...to what?

honestly, the word lol is sometimes annoying
No offense, 10000th user

Tspa

how come so many people say PWN? actually, after comming to this site, i've been using it regularily too :oops:

So that's w00t sorted out.

Good.

Now can someone lay out in simple words the rules of using LaTeX? I know some basics, but am too lazy to take the tutorials given here.

So then just Download TeXaide.

And no more bothering with manual LaTeX.

lol. i like how you assumed everyone at AoPS are like math freaks that are clueless to everyhing lol. SEE OMG I WROTE LOL WITHOUT EVEN REALIZE IT.

and you also write omg without realizing it.
Here are some useful abbreviations -
BTW - by the way
OMG - oh my god
TUATLTUAE - the ultimate answer to life, the universe, and everything
TOW - tug of war
G&FF - games and fun factory
IMO - international mathematical olympiad
(oops i am going too far :oops: )


In a league with n teams, each team plays the others m times. Every game ends in either a win or a loss (no ties allowed). What's the greatest number of teams that can end with a losing record (i.e. having lost more games than they won)?

Partial solution:

[hide]We have four cases:

1) m even

2) n even, m odd

3) n odd, m odd

I'll tackle the first two:

1) n - 1 teams can have a losing record.  Clearly, all n teams cannot have losing records since the total number of wins equals the total number of losses.  Thus, let any team beat every other team in every game they play, and the all other pairs of teams split their m games evenly.  They all end up with losing records.

2) The answer is also n - 1.  Let one team beat every other team.  There are n -1 (an odd number) of teams remaining, and they each have to play (n - 2)*m games (an even number).  I claim that they can split these games half wins, half losses, and thus all end up with a losing record.

Proof:  Let us just label the teams 1, 2, ..., n - 1.  We will say that a beats b in all their m matches if b is a member of (a + 1), (a + 2), ..., (a + (n - 2)/2), all taken mod (n - 1).  There are exactly as many teams that will beat a as will be beaten by a, so we're done.[/hide]

its impossible if both m and n are odd because there need to be as many wins as loses.  Odd *odd is odd.

No, that would give an odd number of games.  Each game still gives 2 results (a win and a loss), so it's still a viable possibility.

A conjecture: [hide] for m = 1, the answer is (n - 2), but for larger m, the answer is (n - 1) again.  I think I can prove the first part (thanks to yif on the Getting Started forum for an idea how), but I'm not so sure about the second.  I don't really feel like hashing it out, honestly.  It looks kinda messy.[/hide]


[b]Hello All :

 :D

If we open drafting compass [u]5 CM[/u]  , the question is :

How can you sketch a circle with a radius [u]4 CM[/u] without moving its legs or moving its pencil.?

Note that we wanna sketch it on surface not able to move. :o

[/b]

Er...you open a second compass to 4 cm and sketch it with that?  If you can't move the components of the compass, and you can't move the surface, you need some outside help...

[b]hello K81o7 :

you haven't to use another drafting compass , you must use the same compass.

BTW, I've two solutions to my problem. But I just like to share it with you. :)

Try another solution.

[/b]

Then do you have any other tools?  Can you move the compass at all?

[b]Yes dude.

you can use any other tool, but not tool do the same work as the compass do.

you can use for e.g's :

ruler , wood , piece of wood, piece of anything sold....etc

There is hint in my e.g.

Good luck dude.
[/b]

So then, dude, does this ruler have cm markings?  And can we stand it up straight?  Because, dude, that would be awesome, because then we can create a 3-4-5 right triangle and in this way get a 4cm marking, dude.  Then we rotate that around (assuming we're allowed to rotate the compass with its pencil off of the surface, dude).

I still feel that's not what you were looking for though, dude.  I don't think it's clearly defined what you can and can't do, dude.

[b]surely without doubt you solve the problem.

I must be more accurate in my information.

yes you can do that friend.

depending on Pyth. Thm.

so try to find another solution.

you can move the ruler in any way.

 I see you are nervous  :mad:  since you repeat the dude many times.

You can :

bring ruler in cm , dcm , ...etc

bring piece of wood , bring any thing  you would like...

rotate the compass as You want. ( how could you sketch circle without rotating )

you can't :

as I defined upon, in my post replies. ( Fine ).

anything else allowed

smile  :) plz , and I'm sorry  :blush:

[/b]

Sorry about the repeating dude so many times, I just noticed you used it a couple of times so I kinda picked up on it.  No offense meant.  :D

Hm so then if the compass cannot be rotated, and we already have a 5cm measure, what use is the compass?  I have to think about this.  Can you attach a pencil point to the ruler and rotate it?  :rotfl:

[b]Great job .

Yes you can attach pencil point  :D at ruler and we can rotate it.

so the solutions are :

1) we can bring up a piece of wood has length 3 cm , and put it perpendicular to the surface .

and put one leg of compass at the top of it and rotate the compass around to sketch circle with 4 cm. by Pyth thm.

2) Explain the 1st statement as I said upon.

the rest on you dude

Nice work  :P [/b]


number theory
greatest common divisor
number theory proposed

Given a fraction $A=\frac{n^2+16}{n+5}$ where $1\leq n \leq 1994$. How many possible values can $n$ get where $A$ is irreducible?

$(n^2 + 16, n + 5) = (n^2 + 16 - (n + 5)^2, n + 5) = (10n + 9, n + 5) = (41, n + 5)$.
$A$ must be irreducible, so the gcd must be 1.
The number of multyplies of $41$, $\leq 1999$ is $48$.
Hence there are $1994 - 48 = 1946$ values of $n$.


$x,y,z$ are postive number and $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} \leq 3$ prove that :

$\frac{x^3}{1+yz+z+y} + \frac{y^3}{1+xz+x+z} + \frac{z^3}{1+xy+x+y} \geq \frac{3}{4}$

In new year, I happy everybody fun and study good! :pilot:  :first:  :icecream:

[quote="zhaobin"][quote="robinho"]$x,y,z$ are real number [/quote]
 :?:[/quote]

$x,y,z$ are postive number :P

[quote="mateivld"]well from Cebasev we have $\displaystyle(\sum\frac{x^3}{1+yz+y+z})(\sum\frac{1}{x})\geq 3\sum \frac{x^2}{1+yz+y+z}$ so we have to prove $\displaystyle \frac{x^2}{1+yz+y+z}\geq \frac{3}{4}$.we take Cauchy and we need to prove $4(\sum x)^2\geq 9+6\sum x+3\sum yz$.now $3\sum yz\leq (\sum x)^2$ so we have to prove $(\sum x)^2\geq 2\sum x+3$.We have $\sum x \geq 3$ so the ineq is proved. :)[/quote]

This is very long :D and I have a prove very short and beautifull.
We have $\frac{x^3}{1+yz+z+y} + \frac{y^3}{1+xz+x+z} + \frac{z^3}{1+xy+x+y} \Leftrightarrow \sum \frac{x^3}{(1+y)(1+z)}$
then I use Cosi inq :

$\frac{x^3}{(1+y)(1+z)} + \frac{(1+z)}{8} \frac{1+y}{8} \geq \frac{3x}{4}$


What is the probability of throwing 15 with 3 throws with a single dice?

The total number of possibilities is clearly $ 6^3\equal{}216$.

Instead of counting the number of integer solutions to $ a\plus{}b\plus{}c\equal{}15$ with $ 1\leq a,b,c\leq 6$, we note we can just count the number of nonnegative integer solutions to $ (6\minus{}a)\plus{}(6\minus{}b)\plus{}(6\minus{}c)\equal{}18\minus{}15\equal{}3$ as each solution will correspond to a solution to the original equation.  By balls and urns there are $ \binom{5}{2}\equal{}10$ ways.

So the probability is $ \boxed{5/108}$.

very clear explaination, thanks [b]5849206328x[/b]!


What does your teacher do with the solutions to the AMC? My teacher says that the rule is you can give the sllutions out when the booklet is mailed. since i took at to the end of school, and the teacher has to stamp the school address on each one (the ta does taht). i have to wait till tomoroow.


If a,b,c are positive numbers,such that a+b+c=1,prove that
(1+a)(1+b)(1+c)>=8(1-a)(1-b)(1-c)

An easy one. $(1+a)(1+b)(1+c)\ge{8(1-a)(1-b)(1-c)}\Leftrightarrow{(2a+b+c)(a+2b+c)(a+b+2c)\ge{8(b+c)(c+a)(a+b)}}.$
Now: $2a+b+c=(a+b)+(a+c)\ge{2\sqrt{(a+b)(a+c)}}$, writing in the similar way the others and multiplying we obtain the desired result with equality when $a=b=c=\frac{1}{3}$.That's all. :)

[quote="Raz"][quote="blahblahblah"]We need to prove that

$\prod (2x+y+z)\geq 8\prod (x+y)$.  Set $x+y=a$, and so on, which turns it into

$\prod (a+b)\geq 8\prod a$.  Now, since $a+b\geq 2\sqrt{ab}$ and so on, the desired inequality follows immediately.[/quote]
Is there any difference between our solutions? :D  :D  :D[/quote]

Nothing difference ... but only blahblahblah's solution "looks shorter"  :P


Find all integer solutions of the equation
$ 3^{x\plus{}1}\minus{}x\cdot3^{x}\minus{}4x\minus{}1\equal{}0$


function
inequalities
inequalities unsolved

Consider the convex function $f(x)=\frac1x$. Jensen gives us that the expression on the left is greater than or equal to: \[f\left( \sum_{cyc} (x_1^2x_2 + x_1x_2^2 + x_1x_2x_3) \right)\]

Via the usual way we see that the function argument is clearly $\leq \frac3{n^2}$, and applying $f(x)=\frac1x$ to that, we get that it's $\geq \frac{n^2}3$.

Peter


geometry
3D geometry
combinatorics unsolved
combinatorics

Eight different positive integers are used to label the vertices of a cube. Each edge of the cube is then labeled with the unsigned difference of the labels of its endpoints.

What is the smallest number of distinct edge labels that could be produced in this way?

We can have 3 differences, as in the example here:

[img]http://i191.photobucket.com/albums/z12/adrian_satja/matis.jpg[/img]

(the differences are 1, 3, 4).

It remains to prove there cannot be 2 differences.

Which is pretty easy to do because the edges on the vertices with $ 1$ (or $ 8$) have to all be distinct, so 3 is a minimum.


A rectangle is inscribed in a circle with a radius 5 units long. The ratio of the dimensions of the rectangle is 0.75. How many units are in the perimeter of the rectangle?

Hello
Let the dimensions be $ x$ units and $ 0.75x$ units.
$ x^2 \plus{} 0.5625x^2 \equal{} 100$
$ x \equal{} 8$
Thus the perimeter is $ 2(8 \plus{} 6)$ i.e.$ 28$units
Thank u.


If Un>0, Un+1/Un<=1-1/n-a/nlogn, a>1, then from n=1 to infinity ∑Un converges.

If Un>0, Un+1/Un>=1-1/n-1/nlogn, then from n=1 to infinity ∑Un diverges.

I couldn't understand your question; could you write those expression with parenthesis, at least?


[size=200]Problem 1[/size]
(3/4)t^ -(1/4) - (1/2)t^ -(3/4) = 0

yes they are to negative powers!

[hide]t = 4/9[/hide]

Rewrite: $ \frac {3}{4}t^{\frac{\minus{}1}{4}} \minus{} \frac {1}{2}t^{\frac { \minus{} 3}{4}} \equal{} 0$

[hide="Solution"]Set $ x\equal{}t^{\frac{\minus{}1}{4}}$

Then multiply through by $ 4$.

We then have $ 2x^3\minus{}3x\equal{}0$
$ x(2x^2\minus{}3)\equal{}0$
$ x\equal{}0$ (which is obviously impossible) or $ x\equal{}\pm\sqrt{\frac{3}{2}}$

To find $ t$ we take $ x^{\minus{}4}\equal{}\frac{4}{9}\equal{}t$.  Since we are taking it to an even power, the plus or minus disappears =)[/hide]

well, if you simplify everything, you will get
[hide]16*sqrt2t/81t = t[/hide]

You can do the rest from there
I think


First Day 24.02.2006

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=439272#p439272]1.[/url] Positive integers $a,b,c,x,y,z$ satisfy:

$a^2+b^2=c^2$, $x^2+y^2=z^2$

and

$|x-a| \leq 1$ , $|y-b| \leq 1$.

Prove that sets $\{a,b\}$ and $\{x,y\}$ are equal.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=439274#p439274]2.[/url] It is given a triangle $ABC$ in which $AC+BC=3AB$. Incircle with center $I$ touches $BC$ and $CA$ at respectively $D$ and $E$. Let $K$, $L$ be symmetrical points of $D$,$E$ with respect to $I$. Prove that $A,B,K,L$ lie on one circle.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=439275#p439275]3.[/url] Positive reals $a,b,c$ satisfy $ab+bc+ca=abc$. Prove that:

$\frac{a^4+b^4}{ab(a^3+b^3)} + \frac{b^4+c^4}{bc(b^3+c^3)}+\frac{c^4+a^4}{ca(c^3+a^3)} \geq 1$

Second Day 25.02.2006

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=440029#p440029]4.[/url] Let $c$ be fixed natural number. Sequence $(a_n)$ is defined by:

$a_1=1$, $a_{n+1}=d(a_n)+c$ for $n=1,2,...$.

where $d(m)$ is number of divisors of $m$. Prove that there exist $k$ natural such that sequence $a_k,a_{k+1},...$ is periodic.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=440032#p440032]5.[/url] Point $C$ is a midpoint of $AB$. Circle $o_1$ which passes through $A$ and $C$ intersect circle $o_2$ which passes through $B$ and $C$ in two different points $C$ and $D$. Point $P$ is a midpoint of arc $AD$ of circle $o_1$ which doesn't contain $C$. Point $Q$ is a midpoint of arc $BD$ of circle $o_2$ which doesn't contain $C$. Prove that $PQ \perp CD$.

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=440036#p440036]6.[/url] Given is a prime number $p$ and natural $n$ such that $p \geq n \geq 3$. Set $A$ is made of sequences of lenght $n$ with elements from the set $\{0,1,2,...,p-1\}$ and have the following property:
  For arbitrary two sequence $(x_1,...,x_n)$ and $(y_1,...,y_n)$ from the set $A$ there exist three different numbers $k,l,m$ such that:

$x_k \not = y_k$, $x_l \not = y_l$, $x_m \not = y_m$.
Find the largest possible cardinality of $A$.

P.S. I've got all from the first day but I've got one annoying 'mistake' - in the second I wrongly drawn the picture (on my picture intersection of $AK$ with $BC$ and $BL$ with $AC$ lied farer than respectively $D$, $E$ from $C$ - I don't know how many points will they cut me out - I wrote at the end that my picture was bad-drawn but that doesn't change reasoning [for which I used nice lemma + some angle chasing] -hope that won't cost me much  :oops: ). Tommorow I'll post problems from the second day as soon as I get home from it  :)

P.S. 2  Second day was worse than first for me...I've managed to get two of them (without sixth - I'm terrible at combinatorics). After all I also think they won't give me more than two points for second from the first day so my result won't be really impressive - few points over twenty perhaps (of 36 possible)  :oops:

[deleted]

Sorry, didn't know there was another thread.

[b]fuzzylogic[/b], you should have posted this here : http://www.mathlinks.ro/Forum/viewtopic.php?t=76459
(although it has already been noticed ;))

I've just posted problems from second day - enjoy  :)


geometry
circumcircle
trigonometry
incenter
cyclic quadrilateral
similar triangles
geometry proposed

Let $ \triangle\ ABC$ with $ a\ne b$ and $ C \equal{} 2B\ .$ Denote the point $ F\in (AB)$ for which $ \widehat {FCA}\equiv\widehat {FCB}$ , the midpoint  $ E$ of $ [CF]$ ,

the point $ D\in AC\cap BE$ and the point $ M\in (BE)$ for which $ m(\angle {MCB})\ \equal{} \ m(\angle{MBA})\ .$ Prove that $ MA\perp MC\ .$

[hide]Let $ N$ be, the midpoint of the side-segment $ BC,$ of the given triangle $ \bigtriangleup ABC$ and it is easy to show that $ FB \equal{} FC,$ from $ \angle B \equal{} \frac {\angle C}{2} \equal{} \angle FCB.$

The line through the midpoint $ E$ of the segment $ CF$ and perpendicular to $ CF,$ intersects the line segment $ FN$ at point so be it $ P.$

Let $ (K)$ be, the circumcircle of the cyclic quadrilateral $ PNEC$ and let be the point $ M\equiv (K)\cap AP.$

We will prove that the points $ B,\ M,\ E,$ are collinear and also that $ \angle MCB \equal{} \angle MBA.$

$ \bullet$ From the cyclic quadrilateral $ PNEC$ and $ EN\parallel AB,$ we have $ \angle CNE \equal{} \angle B \equal{} \angle BCF \equal{} \angle ACF$ $ \Longrightarrow$ $ \angle CPE \equal{} \angle ACF$ $ ,(1)$ and so, we conclude that the circumcircle $ (K)$ of $ PNEC,$ tangents to $ AC,$ at point $ C.$

Hence, from the right triangle $ \bigtriangleup CAP$ with $ CM\perp AP,$ we have that $ (AC)^{2} \equal{} (AM)\cdot (AP)$ $ ,(2)$

It is easy to show that the quadrilateral $ PQFB,$ where $ Q\equiv BC\cap EP,$ is cyclic because of $ \angle BQP \equal{} 90^{o} \minus{} \angle FPQ \equal{} 90^{o} \minus{} \angle FCB \equal{} 90^{o} \minus{} \angle B \equal{} \angle BFP$ and let $ (L)$ be its circumcircle.

$ \bullet$ From the similar triangles $ \bigtriangleup ACF,\ \bigtriangleup ABC$ we have that $ \frac {AC}{AB} \equal{} \frac {AF}{AC}$ $ \Longrightarrow$ $ (AC)^{2} \equal{} (AF)\cdot (AB)$ $ ,(3)$

From $ (2),\ (3)$ $ \Longrightarrow$ $ (AM)\cdot (AP) \equal{} (AF)\cdot (AB$ $ ,(4)$

From $ (4)$ we conclude that the point $ M$ lies on the circle $ (L).$

From the cyclic quadrilaterals $ PMEC,\ PNEC,\ PMFB,$ we have that $ \angle AME \equal{} \angle ECP \equal{} \angle ENF \equal{} \angle BFP \equal{} \angle BMP$ $ \Longrightarrow$ $ \angle AME \equal{} \angle BMP$ $ ,(5)$ and so, we conclude that the points $ B,\ M,\ E,$ are collinear.

Also, we have that $ \angle MBA \equal{} \angle MPN \equal{} \angle MCB$ and the proof is completed.[/hide]

Kostas Vittas.

[color=darkblue][u]Thank you for your very nice proof[/u], [b]Kostas Vittas[/b] ! It is more nice than mine. I hope that you liked it.
I like it very much and I obtained it by extension of  my simple and old problem :[/color]

[quote][color=darkred]Let $ \triangle ABC$ for which $ C\equal{}90^{\circ}$ and $ CA\equal{}CB\ .$ Consider the point  $ D\in (AC)$ such that

$ DA\equal{}2\cdot DC$ and the point  $ M\in (BD)$ so that  $ \widehat {MCB}\equiv\widehat {MBA}\ .$Prove that $ MA\perp MC\ .$[/color][/quote]

Solution for Virgil Nicula's old problem:

Let $ (O)$ be the circumcircle of $ \triangle ABC$ and $ BD \cap (O) \equal{} \{B,E\}$. We have: $ \widehat{ACE} \equal{} \widehat{ABE} \equal{} \widehat{BCM}, \widehat{CAE} \equal{} \widehat{CBM},CA \equal{} CB$, i.e., $ \triangle CAE$ and $ \triangle CBM$ are congruent.

Since, $ CM \equal{} CE$ and $ \widehat{CEM} \equal{} \widehat{CAB} \equal{} 45^o$ then $ ME \equal{} CE\sqrt {2}$.

In $ \triangle ACE$, we have:
$ \frac {EA}{CE} \equal{} \frac {\sin{\widehat{ECA}}}{\sin{\widehat{EAC}}} \equal{} \frac {\sin{\widehat{ABD}}}{\sin{\widehat{CBD}}} \equal{} \frac {\sin{\widehat{ABD}}}{\sin{\widehat{CBD}}}.\frac {BA}{BC}.\frac {BC}{BA} \equal{} \frac {BC}{BA}.\frac {S_{BAD}}{S_{BCD}} \equal{} \sqrt {2}$

$ \Rightarrow AE \equal{} CE\sqrt {2} \equal{} ME$

But $ \widehat{AEM} \equal{} 90^o$ so $ \widehat{AME} \equal{} 45^o$. Thus $ \widehat{AMC} \equal{} \widehat{CME} \plus{} \widehat{AME} \equal{} 45^o \plus{} 45^o \equal{} 90^o$, and we are done.

The first problem of the topic can be solved similarly:

Let $ (O)$ be the circumcircle of $ \triangle ABC$ and $ BM \cap (O) \equal{} \{B,G\}$. We have: $ \widehat{GCA} \equal{} \widehat{GBA} \equal{} \widehat{MCB}$ and $ \widehat{GAC} \equal{} \widehat{GBC}$ so $ \triangle GCA \sim \triangle MCB$, i.e., $ \triangle GCM \sim \triangle ACB$. Therefore, $ \widehat{CMG} \equal{} \widehat{ABC}$.

On the other side, we have:

$ \frac {GA}{GC} \equal{} \frac {\sin{\widehat{GCA}}}{\sin{\widehat{GAC}}} \equal{} \frac {\sin{\widehat{ABD}}}{\sin{\widehat{CBD}}} \equal{} \frac {BC}{BF} \equal{} \frac {\sin{\widehat{BFC}}}{\sin{\widehat{BCF}}} \equal{} \frac {\sin{\widehat{ACB}}}{\sin{\widehat{ABC}}} \equal{} \frac {AB}{AC} \equal{} \frac {GM}{GC}$

$ \Rightarrow GA \equal{} GM\Rightarrow \widehat{GMA} \equal{} 90^o \minus{} \frac {\widehat{AGM}}{2} \equal{} 90^o \minus{} \frac {\widehat{ACB}}{2} \equal{} 90^o \minus{} \widehat{ABC}$.

Thus, $ \widehat{CMA} \equal{} \widehat{CMG} \plus{} \widehat{GMA} \equal{} \widehat{ABC} \plus{} 90^o \minus{} \widehat{ABC} \equal{} 90^o$, and we are done!

Hello!

$ \angle ABC\equal{}\alpha\Rightarrow\angle ACB\equal{}2\alpha$. So $ \frac{c}{sin 2\alpha}\equal{}\frac{b}{sin\alpha}\Rightarrow c\equal{}2b\cdot cos\alpha$

The parallel through the point $ A$ to $ CF$ intersects the line $ BE$ at $ T$.
Let's prove that the point $ T$ lies on the circle $ \Gamma$ with diameter $ AC$.
$ \frac{EF}{AT}\equal{}\frac{BF}{BA}\Rightarrow\frac{CF}{2\cdot AT}\equal{}\frac{CF}{AB}\Rightarrow AT\equal{}\frac{AB}{2}\Rightarrow AT\equal{}\frac{c}{2}\Rightarrow AT\equal{}b\cdot cos\alpha$   (1)

$ O$ is the midpoint of $ AC$ and $ \left\{C,S\right\}\equal{}CF\cap\Gamma$.
$ \frac{CS}{sin\left(\angle SOC\right)}\equal{}\frac{SO}{sin\left(\angle OCS\right)}\Rightarrow\frac{CS}{sin 2\alpha}\equal{}\frac{b}{2\cdot sin\alpha} \Rightarrow CS\equal{}b\cdot cos\alpha$   (2)

(1),(2) $ \Rightarrow AT\equal{}CS$
$ AC$ diameter of $ \Gamma$, $ AT\parallel{}CS$ and $ AT\equal{}CS\Rightarrow T\in\Gamma$

$ \angle MBA\equal{}180^{\circ}\minus{}\angle BAT\minus{}\angle ATM\equal{}180^{\circ}\minus{}\angle BAC\minus{}\angle TAC\minus{}\angle ATM\equal{}$
$ \equal{}3\alpha \minus{}\angle ACF\minus{}\angle ATM\equal{}2\alpha \minus{}\angle ATM\Rightarrow\angle MBA\equal{}2\alpha \minus{}\angle ATM$   (3)
$ \angle MCB\equal{}\angle ACB\minus{}\angle ACM\Rightarrow\angle MCB\equal{}2\alpha \minus{}\angle ACM$   (4)
(3),(4) and $ \angle MBA\equal{}\angle MCB\Rightarrow\angle ATM\equal{}\angle ACM\Rightarrow M\in\Gamma\Rightarrow AM\bot MC$

Best regards, Petrisor Neagoe :)

Hello!

Your problem and the problem presented here ([url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=259859[/url])
help me to formulate another problem:

Let $ ABC$ le a triangle, $ I$ is the incenter, $ J$ is the $ A$-excenter and $ M$ is the midpoint of $ AB$.
$ \left\{P\right\}\equal{}MI\cap BC$, $ \left\{Q\right\}\equal{}MJ\cap BC$, $ \left\{D\right\}\equal{}AI\cap BC$, $ E$ is the midpoint of $ AD$
and $ N\in\left(BE\right)$ such that $ \angle NAB\equal{}\angle NBC$. Prove that $ CP\equal{}CQ\Leftrightarrow AN\bot NC$.

Petrisor Neagoe :)

[color=darkblue] A similar problem ([b]Marcel Chirita[/b], [u]Romania[/u]) :[/color] $ \left\|\ \begin{array}{ccc}
\triangle\ ABC & , & AB\equal{}AC\\\\
D\in (AC) & , & DC \equal{} 2\cdot DA\\\\
M\in (BD) & , & \widehat {MCB}\equiv\widehat {MBA}\end{array}\ \right\|$ $ \Longrightarrow\ MA\perp MC\ .$

$ \frac {CD} {AD} \equal{} \frac {1}{2} \equal{} \frac {BC^2} {AB^2}$ , hence BD is a symmedian, that is, AECB is harmonic quadrilateral, where BD∩(ABC) = {B,E}. Since $ \angle BCM \equal{} \angle ABD \equal{} \angle ABE \equal{} \angle ACE$ and $ \angle CAE \equal{} \angle CBE, \triangle AEC \sim \triangle BMC$ (actually they are congruent, with AC = BC). Similarly $ \triangle AEC \sim \triangle AMB$, hence $ \angle CME \equal{} \angle AME \equal{} 45^o$.
This way we re-discovered the nice property of harmonic quads, namely “[i] a diagonal is seen from the middle of the other one at an angle twice the acute angle from one end of the latter, and this angle is bisected by this latter diagonal.[/i]”

Best regards,
sunken rock


calculus
integration
function
calculus computations

How do I simplify the following expression without resorting to special function:

$\int_{0}^{1}\int_{2y}^{2}\cos(x^{2}) \, dx \, dy$.

Can people check if my answer for this question is correct:

$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}(1+\csc^{2}x) \, dx = x-\cot x |_{-\frac{\pi}{2}}^{\frac{\pi}{2}}= \pi$,
I am not too certainly about my answer as Maple gives me $\infty$ as the solution.

Just draw the diagram.  You'll then find that you can change the order of integration so that your $y$ bounds are $0$ to $x/2$ and your $x$ bounds are $0$ to $2$.

So, the integral then becomes $\int_{0}^{2}\int_{0}^{\frac x2}\cos(x^{2}) \, dy \, dx$.

Which is $\int_{0}^{2}\cos(x^{2})*\frac{x}{2}\, dx$.

This can easily be evaluated to:
$\frac{1}{4}\sin(x^{2})|_{0}^{2}= \frac{1}{4}\sin(4)$.

Three points:
- This doesn't belong in "Open Questions". It should be in "Computations and Tutorials" instead. Putting it in "Open Questions" simply makes it less likely that anyone will find it.

- First question: It's an iterated integral. What happens if you switch the order of integration (Fubini's theorem)?

- Second question: The function goes to $\infty$ in the middle of the integral, so the integral is improper. In fact the integral is improper for two reasons- it's improper at zero from above, and also at zero from below. Split the interval, and deal with each half separately- if either diverges, the integral diverges.
The antiderivative you wrote down is not valid on the full interval; it's not continuous at zero.


Point $ P$ is the midpoint of segment $ LR$ and $ LR \equal{} 8$ cm. If $ L$ is rotated clockwise around point $ P$ until equilateral triangle $ LPR$ can be formed, through how many degrees is $ L$ rotated?
[asy]unitsize(1cm);
pair A=(0,0),B=(2,0),C=(4,0);
draw(A--B--C);
dot(A);
label("$L$",A,S);
dot(B);
label("$P$",B,S);
dot(C);
label("$R$",C,S);[/asy]

Since P is the midpoint, the equilateral triangle will have side length 4 and 60 degree angles.  Therefore, the point L must be rotated $ 180\minus{}60\equal{}120$ degrees.

Note that knowing the length is useless. The interior angle of an equilateral triangle is 60 degrees,
so angle L'PR = 60 and LPL' = 180 - 60 = 120 (L' is L once it has been rotated)


calculus
integration
function
real analysis
real analysis unsolved

I saw in a post that $\displaystyle \int e^{x^{2}} dx,$ cannot be computed.. but how about.. $\displaystyle \int e^{-x^{3}} dx$ ?

i don't think it has an elementary primitive...

If $e^{-x^3}dx$ had an elementary antiderivative, then so also would $u^{-\frac23}e^{-u}du$ and there would be an elementary expression for $\Gamma\left(\frac13\right).$

The right attitude is this: for any elementary function, once we leave familiar territory, it is unlikely that that function has an elementary antiderivative. The default assumption would be to assume that it does not, unless we have a specific reason to believe it does.


Two circles $ O_1$ and $ O_2$, with centers $ P$ and $ Q$, respectively, intersect at points $ A$ and $ B$. Let $ PQ$ intersect $ O_1$ at $ O_2$ at $ C$ and $ D$, respectively, such that $ DP<DQ$ and $ CQ<CQ$. Let $ DC\equal{}d$ and $ AB\equal{}a$. Find the radii of both circles.

My solution is very ugly, so I thought you guys may have a slicker one :)

I don't think so. The original problem mentioned something called "qual radius." Maybe that means they are equal?

The problem was in Kiran Kedlaya's Geometry Notes, btw.


A given trapezoid has perpendicular diagonals.  If one of these diagonals has a length of 30, and the altitude of the trapezoid has a length of 24, find the area of the trapezoid.

Drawing the altitude from one of the vertices on the shorter side to the larger side, we see that we have $ 2$ similar triangles.  We have that both of the right angles, of where the altitude hits the larger side, and where the two diagonals meet, are equal, and that the angle that both contain are equal. By $ AA$, we have that they are similar, and the rest follows trivially.