function
calculus
derivative
algebra unsolved
algebra

[quote="greatestmaths"][color=darkred]Prove that $ \left\{\begin{array}{c}\{x,y\}\subset\mathbb R\\ \\ 1\le x^{2}-xy+y^{2}\le 2\end{array}\right\|$ $ \implies$ $ \frac{2}{9}\le x^{4}+y^{4}\le 8$.[/color][/quote]
[color=darkblue][b]Proof.[/b] Consider the function $ f:\mathbb R\rightarrow\mathbb R$, where $ f(t) =\frac{t^{4}+1}{(t^{2}-t+1)^{2}}$. Show easily that $ Im(f)\equiv f(\mathbb R) =\left[\frac{2}{9},2\right]$, i.e. for any $ t\in\mathbb R$, $ \frac{2}{9}\le f(t)\le 2$..

Using the substitution $ y = tx\,\t\in\mathbb R$ obtain $ \left\{\begin{array}{c}\frac{1}{t^{2}-t+1}\le x^{2}\le\frac{2}{t^{2}-t+1}\\ \\ \frac{t^{2}}{t^{2}-t+1}\le y^{2}\le\frac{2t^{2}}{t^{2}-t+1}\end{array}\right\|$ $ \implies$ $ \frac{2}{9}\le f(t)\le x^{4}+y^{4}\le 4\cdot f(t)\le 8$ $ \implies$ $ \frac{2}{9}\le x^{4}+y^{4}\le 8$.[/color]

[quote="Virgil Nicula"][quote="greatestmaths"][color=darkred]Prove that $ \{\begin{array}{c}\{x,y\}\subset\mathbb R\\ \\ 1\le x^{2}-xy+y^{2}\le 2\end{array}\|$ $ \implies$ $ \frac{2}{9}\le x^{4}+y^{4}\le 8$.[/color][/quote]
[color=darkblue][b]Proof.[/b] Consider the function $ f:\mathbb R\rightarrow\mathbb R$, where $ f(t) =\frac{t^{4}+1}{(t^{2}-t+1)^{2}}$.

Show easily that ([b]with derivatives, only if need be[/b] !) $ Im(f)\equiv f(\mathbb R) =[\frac{2}{9},2]$, i.e. for any $ t\in\mathbb R$, $ \frac{2}{9}\le f(t)\le 2$.

Using the substitution $ y = tx\,\t\in\mathbb R$ obtain $ \{\begin{array}{c}\frac{1}{t^{2}-t+1}\le x^{2}\le\frac{2}{t^{2}-t+1}\\ \\ \frac{t^{2}}{t^{2}-t+1}\le y^{2}\le\frac{2t^{2}}{t^{2}-t+1}\end{array}\|$ $ \implies$ $ \frac{2}{9}\le f(t)\le x^{4}+y^{4}\le 4\cdot f(t)\le 8$ $ \implies$ $ \frac{2}{9}\le x^{4}+y^{4}\le 8$.[/color][/quote]
[color=darkblue][b]Remark.[/b] Observe that $ f(t) =\frac{t^{2}+\frac{1}{t^{2}}}{(t-1+\frac{1}{t})^{2}}=$ $ \frac{(t+\frac{1}{t})^{2}-2}{[(t+\frac{1}{t})-1]^{2}}=$ $ \frac{[(t+\frac{1}{t})-1]^{2}+2\cdot[(t+\frac{1}{t})-1]-1}{[(t+\frac{1}{t})-1]^{2}}=$ $ -[\frac{1}{(t+\frac{1}{t})-1}]^{2}+2\cdot[\frac{1}{(t+\frac{1}{t})-1}]+1$.

This upper expression of $ f$ allows us to write it as a composition of more simple surjective functions and to ascertain more easily the image $ f(\mathbb R)$ of the $ f$.

I"ll try to prove [b]without derivatives[/b] that $ Im(f)\equiv f(\mathbb R) =[\frac{2}{9},2]$. Indeed, $ f(0) = 1$ and for any $ t\ne 0$, $ f(t) = (\alpha\circ\beta\circ\gamma\circ\delta)(t) =\alpha (\beta (\gamma (\delta (t))))$,

where the functions $ \{\begin{array}{cccccc}\alpha\ : &[-\frac{1}{3},1]^{*}&\rightarrow &[\frac{2}{9},2] & , &\alpha (x) =-x^{2}+2x+1\\ \\ \beta\ : & (-\infty ,-3]\cup [1,\infty ) &\rightarrow &[-\frac{1}{3},1]^{*}& , &\beta (x) =\frac{1}{x}\\ \\ \gamma\ : & (-\infty ,-2]\cup [2,\infty ) &\rightarrow & (-\infty ,-3]\cup [1,\infty ) & , &\gamma(x) = x-1\\ \\ \delta\ :\ &\mathbb R^{*}&\rightarrow & (-\infty ,-2]\cup [2,\infty ) & , &\delta (x) = x+\frac{1}{x}\end{array}$ are surjective (show easily !).

So, the function $ \delta$ is odd, decreasing on $ (0,1]$ and increasing on $ [1,\infty )$, the function $ \alpha$ is increasing on $ [-\frac{1}{3},1]^{*}$ a.s.o.

Thus, the function $ f: R^{*}\rightarrow[\frac{2}{9},2]$ is surjective, i.e. $ f(\mathbb R^{*}) =[\frac{2}{9},2]$. In conclusion, $ f(\mathbb R) =\{1\}\cup[\frac{2}{9},2]\implies\boxed{\ f(\mathbb R) =[\frac{2}{9},2]\ }$.[/color]

For $ x,\ y\in\mathbb{R}$, let $ X \equal{} x\plus{}y,\ Y \equal{} xy$ such that $ 1\leq x^{2}\minus{}xy\plus{}y^{2}\leq 2$ we have $ \left\{\begin{array}{ll}1\leq X^{2}\minus{}3Y\leq 2 &\quad\\ X^{2}\geq 4Y &\quad\end{array}\right.\Longleftrightarrow D:\ \left\{\begin{array}{ll}\frac{X^{2}\minus{}2}{3}\leq Y\leq\frac{X^{2}\minus{}1}{3}&\quad\\ Y\leq\frac{1}{4}X^{2}&\quad\end{array}\right.$

The region of $ D$ shows as below, including boader. Let $ F(X,\ Y) \equal{} x^{4}\plus{}y^{4}\equal{} (x^{2}\plus{}y^{2})^{2}\minus{}2 \equal{} (X^{2}\minus{}2Y)^{2}\minus{}2Y^{2}$.

Now $ F(X,\ Y)$ is maximized when $ Y \equal{}\frac{1}{4}X^{2},\ P \equal{} (4Y\minus{}2Y)^{2}\minus{}2Y^{2}\equal{} 2Y^{2}$, since $ \minus{}\frac{2}{3}\leq Y\leq 2$, we have maximum value of $ F(X,\ Y)$ is $ F(\pm 2\sqrt{2},\ 2) \equal{} 8$.

[color=darkred][size=134][b][u]An easy and nice extension.[/u][/b][/size] $ \left\{\begin{array}{c}\{a,b,m,x,y\}\subset\mathbb R\ ,\ 1\le |m| < 2\\ \\ 0 < a\le x^{2}\plus{}mxy\plus{}y^{2}\le b\end{array}\right\|$ $ \implies$ $ \frac{a\sqrt 2}{2\plus{}|m|}\le\sqrt{x^{4}\plus{}y^{4}}\le\frac{b\sqrt 2}{2\minus{}|m|}$.[/color]

Let $ g\equal{}x^{2}\minus{}xy\plus{}y^{2}\minus{}t$ where $ 1\le t\le 2$ and $ f\equal{}x^{4}\plus{}y^{4}$. Since $ f$ and $ g$ are both continuous and differentiable functions, we can apply the theory of lagrange multipliers and weisstrass' theorem to say that the extrema of $ f$ occur when

$ <4x^{3},4y^{3}>\, \equal{}\alpha *<2x\minus{}y, 2y\minus{}x>$

If $ 2y\equal{} x$, then $ 4y^{3}\equal{}0$ so $ y\equal{}z\equal{}0$, but that contradicts the bound on $ t$.

Solving $ \alpha\equal{}\frac{4x^{3}}{2x\minus{}y}\equal{}\frac{4y^{3}}{2y\minus{}x}$, we have $ x\equal{}\pm y$.

If $ x\equal{}y$, then $ x\equal{}\sqrt{t}$, and $ 1\le 2t^{2}\le 8$.

If $ x\equal{}\minus{}y$, then $ x\equal{}\sqrt{\frac{t}{3}}$, and $ \frac{2}{9}\le\frac{2}{9}t^{2}\le\frac{8}{9}$.

I guess I am not sure how to show that these are indeed relative maximums and minimums...I will get back to this later...The problem should be almost done.

[Summarizing our results, $ \frac{2}{9}\le f\le 8$, which is the desired result.]

[quote="greatestmaths"]let xand y real numbers such that:

1=<x^2-xy+y^2=<2

1) prove that 2/9=< x^4+y^4=<8[/quote]

I have just found a basic solution only using a qudratic function.

Let $ x^{2}\plus{}y^{2}\equal{} a,\ xy \equal{} b$, this problem can be written as $ 1\leq a\minus{}b\leq 2$ and $ a\geq 2|b|\Longrightarrow\frac{2}{9}\leq a^{2}\minus{}2b\leq 8$.

Let fix $ a\minus{}b \equal{} k\ (1\leq k\leq 2)$, we have $ a\geq 2|b|\Longleftrightarrow a\geq 2|a\minus{}k|$.

Case 1: When $ a\geq k$,  $ a\geq 2|a\minus{}k|\Longleftrightarrow a\geq 2(a\minus{}k)$, yielding $ a\leq 2k$, since $ a\geq k$ we have $ k\leq a\leq 2k$.

Denote $ f(a,\ b) \equal{} a^{2}\minus{}2b \equal{} a^{2}\minus{}2(a\minus{}k)^{2}\equal{}\minus{}a^{2}\plus{}4ka\minus{}2k^{2}\equal{}\minus{}(a\minus{}2k)^{2}\plus{}2k^{2}$,

we have $ k^{2}\leq f(a,\ b)\leq 2k^{2}$. Let $ k$ vary in $ 1\leq k\leq 2$, we have $ 1\leq f(a,\ b)\leq 8$.

Case 2: When $ a<k$, $ a\geq 2|a\minus{}k|\Longleftrightarrow a\geq\minus{}2(a\minus{}k)$, yielding $ a\geq\frac{2}{3}k$, since $ a\geq k$ we have $ \frac{2}{3}\leq a<k$,

we have $ \frac{2}{9}k^{2}\leq f(a,\ b)<k^{2}$, doing same as case 1, we obtain $ \frac{2}{9}\leq f(a,\ b)<8$.

Therefore we have $ \frac{2}{9}\leq f(a,\ b)\leq 8\ Q.E.D.$

[color=darkblue][b]Remark.[/b] $ x^{2}\minus{}xy\plus{}y^{2}\equal{}\lambda\in [1,2]$ $ \implies$ $ x^{2}\plus{}y^{2}\equal{}\lambda\plus{}xy$ $ \implies$ $ x^{4}\plus{}y^{4}\equal{}\lambda^{2}\plus{}2xy\cdot\lambda\minus{}(xy)^{2}\le 2\lambda^{2}\le 8$ $ \implies$

$ \boxed{\ x^{4}\plus{}y^{4}\le 8\ }$. We"ll have the equality if and only if $ xy \equal{}\lambda \equal{} 2$, i.e. $ x \equal{} y\in\{\pm\sqrt 2\}$.[/color]

[quote="Virgil Nicula"]Remark $ x^{2}\minus{}xy\plus{}y^{2}\equal{}\lambda\in [1,2]$ $ \implies$ $ x^{2}\plus{}y^{2}\equal{}\lambda\plus{}xy$ $ \implies$ $ x^{4}\plus{}y^{4}\equal{}\lambda^{2}\plus{}2xy\cdot\lambda\minus{}(xy)^{2}\le 2\lambda^{2}$.[/quote]

What is the range of $ xy$?

That should be $ \minus{}\frac{\lambda }{3}\leq xy\leq\lambda$.

[quote="kunny"][quote="Virgil Nicula"]Remark $ x^{2}\minus{}xy\plus{}y^{2}\equal{}\lambda\in [1,2]$ $ \implies$ $ x^{2}\plus{}y^{2}\equal{}\lambda\plus{}xy$ $ \implies$ $ x^{4}\plus{}y^{4}\equal{}\lambda^{2}\plus{}2xy\cdot\lambda\minus{}(xy)^{2}\le 2\lambda^{2}$.[/quote]What is the range of $ xy$? That should be $ \minus{}\frac{\lambda }{3}\leq xy\leq\lambda$.[/quote]

[color=darkred][b]$ \lambda^{2}\plus{}2xy\cdot\lambda\minus{}(xy)^{2}\le 2\lambda^{2}$ $ \Longleftrightarrow$ $ \left(\lambda\minus{}xy\right)^{2}\ge 0$. Is it O.K. ? The range of $ xy$ isn't interesting for me. Here is the complete proof.[/b][/color]

[color=darkblue][b]Another shorter proof.[/b]

$ 1\blacktriangleright$ Prove easily that $ \frac{xy}{x^{2}\minus{}xy\plus{}y^{2}}\ge\minus{}\frac{1}{3}$, with equality iff $ x \equal{}\minus{}y$. Thus, $ x^{4}\plus{}y^{4}\ge\frac{x^{4}\plus{}y^{4}}{(x^{2}\minus{}xy\plus{}y^{2})^{2}}\ge$ $ \frac{1}{2}\cdot\left(\frac{x^{2}\plus{}y^{2}}{x^{2}\minus{}xy\plus{}y^{2}}\right)^{2}\equal{}$

$ \frac{1}{2}\cdot\left(1\plus{}\frac{xy}{x^{2}\minus{}xy\plus{}y^{2}}\right)^{2}\ge$ $ \frac{1}{2}\cdot\left(1\minus{}\frac{1}{3}\right)^{2}\equal{}\frac{2}{9}$ $ \implies$ $ \boxed{\ x^{4}\plus{}y^{4}\ge\frac{2}{9}\ }$. We"ll have the equality if and only if $ x \equal{}\minus{}y\in\left\{\pm\frac{\sqrt 3}{3}\right\}$.

$ 2\blacktriangleright$ $ x^{2}\minus{}xy\plus{}y^{2}\equal{}\lambda\in [1,2]$ $ \implies$ $ x^{2}\plus{}y^{2}\equal{}\lambda\plus{}xy$ $ \implies$ $ x^{4}\plus{}y^{4}\equal{}\lambda^{2}\plus{}2xy\cdot\lambda\minus{}(xy)^{2}\le 2\lambda^{2}\le 8$ $ \implies$

$ \boxed{\ x^{4}\plus{}y^{4}\le 8\ }$. We"ll have the equality if and only if $ xy \equal{}\lambda \equal{} 2$, i.e. $ x \equal{} y\in\{\pm\sqrt 2\}$.[/color]

I  have noticed that, in my first post, I couldn't solve for minimum value.


consider the set of natural nos(S:1,2,3,4,.......)

two consecutive nos are randomly removed from a subset of this set and the average of the remaining nos of the subset is equal to  105/4........

then  find the subset of S wich contains the nos  (1,2,3,....n) i.e. find n and the two nos that have been removed from this set......

You've already posted this, only 12 hours ago: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=275183  It's quite rude to double-post, especially after such a short period of time when discussion has already begun on the first topic!

No there is a difference in both the problems,here two consecutive numbers are involved and in that post it was two random no.s...


here goes the solution
note that
(a+b)/c >= 2
0<= (b+c)/a <= 2
(a+c)/b >= 1
now
(a 2 - b 2 )/c  >= 2(a-b)
(c 2  - b 2 )/a  >= 2(c-b)
(a 2  - c 2 )/b   >= a-c
add these and u get wot u want :D
cheers

Isn't this solution wrong, or am I making a mistake myself ?

No I think this solution is corect.Why do you say it's wrong?

Because (b - c)/a \leq 2(b - c), not (b - c)/a \geq 2(b - c) I think.

Yes, now I think you are wright cause the problem has in the left member:

(a^2-b^2)/c+(b^2-c^2)/a+(c^2-a^2)/b
not
(a^2-b^2)/c+(c^2-b^2)/a+(a^2-c^2)/b

The idea that galois had but a bit diferent should work.

oops ,i shud've suggested the correction.the problem is incorrectly stated ,i've solved it b4.see u can get a contradiction with(a,b,c) = (4,3,2)
cheers

Yes galois you are richt. I'm sorry! the problem is:

(a^2-b^2)/c+(c^2-b^2)/a+(a^2-c^2)/b \geq ...and so on
your solution is corect. Sorry again!

[quote=Maverick]a \geq b \geq c \geq 0 Prove that

(a 2 -b 2 )/c+ (b 2 -c 2 )/a+ (c 2 -a 2 )/b \geq 3a-4b+c[/quote]
Let $a \geq b \geq c \geq 0$. Prove that$$\frac{a^2-b^2}{c}+\frac{b^2-c^2}{a}+\frac{c^2-a^2}{b} \geq 3a-4b+c.$$
[url=http://www.artofproblemsolving.com/community/c4h1095742p4909285]here[/url]

Can anybody latex this? I didn’t understand.


I emailed Mr.Rusczyk about this idea, and he said to post it on a forum.  As I am a middle-schooler, I have decided to post it in this forum.  My brother (maxmay221) and I have come up with an idea-Fantasy Math.  Similar to Fantasy Football or Fantasy Baseball, the organizers can compile the statistics of the best mathematicians country-wide==>how they do on competitions such as the Mandelbrot, ARML, AMC, HMMT (and its similar correspondents), and such.  I do not know what this idea will result in, but I feel it has potential (if somebody else would like to donate time to do it =) )..

Not a bad idea.

It will be [i]very[/i] time consuming, but it'll be a nice way for people to learn about history in mathematics. I guess one could compile mathematicians throughout the ages and have them compete in IMO or something  :lol: .

This might be better off in Questions/Suggestions than Classroom Math, so can a mod move this?

Huh?

I'm not really getting this fantasy math thing... :huh:  :?:

It seems like the idea is you would find all of the people who (say) were in 11th grade or younger in 06-07 and did well on national math competitions. Then you could create some sort of total score which is a weighted combination of all of the major national competitions. Each of those people would be given a value according to their performance last year.

Then players (who could be anyone) would have a fixed amount of money and they would be able to make a roster of people (from the list I mentioned at the beginning) (the total of the "salaries" could not exceed the fixed amount that each player has). The players' teams' performances would be a function of the total score of their team members on this year's national math competitions.

It's exactly like fantasy sports, only with math competitions.


Find distance of point (-2,3,4) from the line (x+2)/3=(2y+3)/4=(3z+4)/5 measured parallel to plane 4x+12y-3z+1=0.
 I have some doubt about this question .	there are infinite line parallel to plane

1.Place a plane through the given point parallel to the given one
2.Find the point in this plane where the given line cut it
3.The distance from this point to the given one is the answer.The line that go through these points is parallel to the given plane


calculus
number theory
real analysis
University of Chicago
college

What are some good courses to prepare for college math? I'm asking because several me and several of my friends will be doing an independent study program at my school next year (senior year for all of us). We have all finished calculus (including multivariable) and abstract algebra and by next year we will have finished an introductory course in differential equations, a number theory course, and linear algebra. So again, what would you say are some good courses for people that like math? We can pick two classes and are currently considering

1) Real analysis (heard this is very challenging)
2) Combinatorics (good for me since I am pursuing research at the University of Chicago in this topic and I find it quite fascinating :lol: )

All of our schedules should be pretty easy next year (except for the math) since we all enrolled in the most rigorous classes this year (the school's equivalent of Chemistry as well as a Physics C course), so feel free to make any reasonable suggestions.

Thanks,

lu

In several of these cases - especially in the case of the real analysis course - assume that you will very likely wind up taking the course over again once you're officially a college student. And there's nothing at all wrong with that.

The key thing is to keep your mind engaged. The right professor is more important that the right course. Get advice, find out who offers a stimulating course. At this point don't worry about whether you'll ulitmately get credit for it or not.

Take real analysis without question.  I think that of all the 'introductory' math courses, it's the most important, and the most foundational.  Combinatorics lends itself well to self-study, in any event.

Yes, me and my friends are doing these classes just to learn some math and have some fun together before college, not to get out of courses. I should also mention that, although they are called "independent studies," they are actually teacher guided.

Yes, combinatorics is a fine course, and arguably more important for a future computer science major, but real analysis ties into most of what a pure math major will study, so a well-taught real analysis course is definitely good preparation for other courses taken by a math major. There is a LOT of supplementary reading to be found about real analysis, so make use of that as you study.


If we have n positive numbers , how to arrange them like power tower to make it the biggest

I mean if we have 4 numbers 2,3,4,5 then $\square^{\square^{\square^{\square}}}$ , how to arrange 2,3,4,5 then fill in the squares can make the result become the biggest number of it's all possible combination results

my question is that generalization

Even the case $n = 2$ is not so simple.

[hide="Discussion"] Given two positive reals $a > b$, we want to discern whether $a^{b}> b^{a}$ or whether $b^{a}> a^{b}$.  Clearly the prudent thing to do is to investigate the equality case $a^{b}= b^{a}$.

This is equivalent to

$b \ln a = a \ln b \Leftrightarrow$
$\frac{\ln a}{a}= \frac{\ln b}{b}$

The function $f(x) = \frac{\ln x}{x}$ attains its maximum when $x = e$, and decreases on both sides.  So, two subcases:

- If $a, b > e$, then $f(a) < f(b) \implies a^{b}< b^{a}$.  For example, $4^{3}= 64, 3^{4}= 81$.
- If $a, b < e$, then $f(a) > f(b) \implies a^{b}> b^{a}$.  For example, $2^{1}= 2, 1^{2}= 1$.

The final case is unsatisfying.

- If $a \ge e, b \le e$, then it could be the case that either $f(a) < f(b)$ or $f(a) > f(b)$.  [/hide]

[quote="t0rajir0u"]... $a \ge e, b \ge e$ ...[/quote]
This should, of course, read "... $a$ and $b$ are on opposite sides of $e$ ...."  Exact solutions for equality can be expressed in this case using the Lambert W-function,* and if one can compute the values of the W-function then one can compare $a^{b}$ and $b^{a}$ in this case.  (Of course, if one can compute the W-function, one can probably also just compute $a^{b}$ and $b^{a}$ directly.)

For the general case, you most likely will want to restrict yourself to the case where all numbers are greater than or equal to $e$ -- hopefully, that should make the answer "nice."  A little experimentation suggests that under these conditions, you want the smallest number on the bottom.  (Inductively, finding out which number goes on the bottom gives you the entire solution.)

* This is the inverse function of $f(W) = W e^{W}$.


:?: My karma won't go up or down, and I am pretty sure it is changing. I can't tell what my karma is.  Why is this?

I don't know, but I don't think it is updated in live time. (Or am I wrong??)

Try refreshing the page. It should be updated. If it still doesn't work, contact an administrator.

1.  Its been like this for several months, so it is not updating slowly, but thanks for trying to help.

2.  How should I contact an admin?

You are still in the diagnostic phase, which lasts for 60 problems. When you have completed the diagnostic phase your karma will begin to adjust.


1. How many positive integers not exceeding $2001$ are mutiples of $3$ or $4$ but not $5$?

2. Let \[x=0.123456789101112\ldots 998999\] where the digits are obtained by writing the integers $1$ through $999$ in order. Find the $1983^{rd}$ digit to the right of the decimal point.

[hide="1"]
Multiples of 3 $\leq 2001$ = 667
Multiples of 4 $\leq 2001$ =500

Multiples of 12 $\leq 2001$=166

Hence multples of 3 or 4 $\leq 2001$=667+500-166=1001 by inclusion/exclusion.

But we must take away multiples of 15 (of which there are 133) and multiples of 20 (of which there are 100).

This gives us 768.

But we have removed multiples of 60 (of which there are 33) twice, so we must add them back in and get a total of $\boxed{801}$.[/hide]

[quote="rnwang2"]1. How many positive integers not exceeding $2001$ are mutiples of $3$ or $4$ but not $5$?

2. Let
\[x=0.123456789101112\ldots 998999 \]
where the digits are obtained by writing the integers $1$ through $999$ in order. Find the $1983^{rd}$ digit to the right of the decimal point.[/quote]

[hide="1"]
Using PIE:
 Let $X$ be the set of integers in the given range that are divisible by $3$, $Y$ those by $4$, and $Z$ those by $5$. Then, $X \cap Y$ are those divisible by 12; $X \cup Y$ are those divisible by either 3 or 4; etc. (Some notation, as I've been taught: $|S|$ is the cardinality of set $S$ and is equal to the number of terms in that set.)

Then, $| \ (X \cup Y)-Z \ | = (|X|+|Y|-| X \cap Y |)-\left( |X \cap Z|+|Y \cap Z|-|X \cap Y \cap Z|\right) = (667+500-166)-(133+100-33) = \boxed{801}$

(Those numbers are just the integer just less than $\lfloor{ \frac{2001}{\text{whatever combination of factors}}}$)[/hide]

[hide="2"]
Counting digits:
Numbers $1 \rightarrow 9$ give $9(1)=9$ digits.
Numbers $10 \rightarrow 99$ give $90(2)=180$ digits.
Numbers $100 \rightarrow 100+x$ give $x(3)=3x$ digits, as the number $100+x$ begins. (i.e. for $x=1$, there are the three digits $100$ as the number $101$ begins.)

Then, $9+180+3x=1983 \Longrightarrow x=598$. That means that there are $1983$ digits as the number $698$ begins; thus the $1983^{\text{th}}$ digit is the last of the number before, or $\boxed{7}$.[/hide]

[quote="rnwang2"]1. How many positive integers not exceeding $2001$ are mutiples of $3$ or $4$ but not $5$?[/quote]

[hide]

The multiples of three less than or equal to 2001 are 667.
The multiples of four less than or equal to 2001 are 500.
The multiples of 12 (the LCM of 3 and 4) less than or equal to 12 are 2001.
The multiples of 15 less than or equal to 2001 are 133.
The multiples of 20 less than or equal to 2001 are 100.

Therefore, the answer is $667+500-166-133-100\implies \boxed{801}$.

[/hide]


geometry
geometric transformation
reflection

[quote="mathwizarddude"]Solar heating takes advantage of solar collectors.  About 50% of the solar radiation received at the top of the atmosphere reaches the Earth during daylight hours.  How much energy would be collected on average by the cylindrical collector shown in the figure during 10 h of collections?

So is the answer just the 50% * average intensity of solar radiation multiplied by 10 h and then multiplied by whatever the exposed surface area is for the solar collectors?  (That is, $ 50\% *1400J/(m^2s) * 10h * A$ where 1400J/(m²s) = average intensity of solar radiation and A is the surface area.)[/quote]
[hide="My thoughts"]
First, the [i]albedo[/i] of the Earth is approximately $ 0.37$. The albedo of a body is its average reflection coefficient. Thus the average transmission coefficient, $ \tau \equal{} 1 \minus{}0.37 \approx 0.63$. There is some further attenuation due to atmospheric absorption, which is why your text gives a figure of 50% for the transmission coefficient.

Of the power radiated from the Sun only a fraction shines on the Earth,

\[ f \equal{} \frac{4\pi R_{S}^{2}}{4\pi R_{S\to E}^{2}}\]

\[ \begin{array}{cc} R_{S} \equal{}695, 500\quad km \\ R_{S\to E} \equal{} 149, 598, 000\quad km\end{array}\]

\[ \therefore f \equal{} 2.161\times 10^{\minus{}5}\]

The power density @ the Earth is $ \mathcal{P}_E \equal{} f \sigma T_{S}^{4}$

$ \sigma$ is the Stefan-Boltzmann constant, $ \sigma \equal{} 5.67\times 10^{\minus{}8}\quad W/{m^2 K^4}$.

The surface temperature of the sun is $ T_S \equal{} 5, 780 \quad K$.

Therefore, $ \mathcal{P}_E \equal{} (2.161\times 10^{\minus{}5})(5.67\times 10^{\minus{}8})(5.780\times 10^{3})^4 \equal{} 1, 370\quad W/m^2$.

The incident power at the solar array is then $ \mathcal{P}_{in} \equal{}\tau\mathcal{P}_E \equal{}(0.63)(1,370) \equal{} 860\quad W/m^2$.

Multiply this figure by the exposed surface area of the solar collector and then by time to calculate the energy collected in the ten hour time period.[/hide]


The positive five-digit integers that use each of the digits 1, 2, 3, 4 and 5 exactly once are ordered from least to greatest. What is the $ 50^{\text{th}}$ integer in the list?

If we order these from least to greatest, we find that obviously, the first number begins with a 1.

There are $ 4!$ or 24 ways to arrange the other digits.

Then, there are $ 4!$ or 24 more ways the arrange the other digits when the first digit is 2.

Thus, the 50th integer is the second smallest number that we can make that begins with a 3, which is $ \boxed{31254}$.


I once noticed that the fraction 1/89 = 0.01123595..., and then read something that 1/9899 was equal to 0.0001010203050813... etc.

If it is indeed the case that $ \frac{1}{n^2-n-1} = \frac{1}{n^2} + \frac{1}{n^3} + \frac{2}{n^4} + \frac{3}{n^5} + \ldots$, my question is: why?

If there's some resource I should be looking at, could somebody please point me to it?

The generating function for the Fibonacci numbers is
$ \sum_{n \equal{} 1}^\infty F_n x^n \equal{} \frac {x}{1 \minus{} x \minus{} x^2}$.
By plugging in x=1/10, this gives the curious addition tree illustrated above,
$ \sum_{n \equal{} 0}^\infty \frac {F_n}{10^n} \equal{} \frac {10}{89}$,
so
$ \sum_{n \equal{} 0}^\infty \frac {F_n}{10^{n \plus{} 1}} \equal{} \frac {1}{89}$.

Try the first chapter or two of generatingfunctionology (free online) to establish the identity Kouichi is using.


Euler
ratio
analytic geometry
geometry
circumcircle
geometry proposed

Let $a$, $b$, $c$ be the sides of triangle $ABC$.  Prove that the distances from the vertices $A$, $B$, $C$ to the Euler line, up to sign, are in the ratio
\[(b^2 - c^2)(b^2 + c^2 - a^2) : (c^2 - a^2)(a^2 + c^2 - b^2) : (a^2 - b^2)(b^2 + a^2 - c^2).\]

Actually, you don't even need the "up to sign", if you assume the distances from the vertices A, B, C to the Euler line to be directed distances (hereby, which of the two halfplanes with respect to the Euler line exactly is taken to be the positive halfplane is not important, since we consider the ratios of the distances only).

More generally, for an arbitrary line g in the plane of a triangle ABC, the directed distances from the vertices A, B, C to the line g are in the ratio l : m : n, where ( l : m : n ) are the homogeneous barycentric coordinates of the line g with respect to triangle ABC. Hence, in order to solve your problem, one has to prove that the Euler line of triangle ABC has the homogeneous barycentric coordinates

$\left( \left(b^2-c^2\right) \left(b^2+c^2-a^2\right) : \left(c^2-a^2\right) \left(c^2+a^2-b^2\right) : \left(a^2-b^2\right) \left(a^2+b^2-c^2\right) \right)$.

But this is easy: It is well-known that if two points $P_1$ and $P_2$ have the homogeneous barycentric coordinates $P_1 \left( x_1:y_1:z_1\right)$ and $P_2 \left( x_2:y_2:z_2\right)$, respectively, with respect to the triangle ABC, then the line $P_1P_2$ has the homogeneous barycentric coordinates $\left( y_1z_2-z_1y_2:z_1x_2-x_1z_2:x_1y_2-y_1x_2\right)$. Now, the centroid G of triangle ABC has the homogeneous barycentric coordinates G ( 1 : 1 : 1 ), and the circumcenter O of triangle ABC has the homogeneous barycentric coordinates $O \left( a^2 \left(b^2+c^2-a^2\right) : b^2 \left(c^2+a^2-b^2\right) : c^2 \left(a^2+b^2-c^2\right) \right)$. Thus, the line GO, which is actually the Euler line of triangle ABC, has the homogeneous barycentric coordinates

$\left( 1\cdot c^2 \left(a^2+b^2-c^2\right) - 1 \cdot b^2 \left(c^2+a^2-b^2\right) : 1 \cdot a^2 \left(b^2+c^2-a^2\right) - 1 \cdot c^2 \left(a^2+b^2-c^2\right) \right.$
$\left. : 1 \cdot b^2 \left(c^2+a^2-b^2\right) - 1 \cdot a^2 \left(b^2+c^2-a^2\right) \right)$.

We will transform the first coordinate:

$1\cdot c^2 \left(a^2+b^2-c^2\right) - 1 \cdot b^2 \left(c^2+a^2-b^2\right) = c^2 \left(a^2+b^2-c^2\right) - b^2 \left(c^2+a^2-b^2\right) = c^2a^2 - c^4 - b^2a^2 + b^4 = \left(c^2a^2-b^2a^2\right) - \left(c^4-b^4\right) = \left(c^2-b^2\right) a^2 - \left(c^2-b^2\right) \left(c^2+b^2\right) = \left(c^2-b^2\right) \left(a^2-\left(c^2+b^2\right)\right) = \left(b^2-c^2\right) \left(b^2+c^2-a^2\right)$.

Similarly we can transform the other two coordinates, and see that the homogeneous barycentric coordinates of the Euler line are

$\left( \left(b^2-c^2\right) \left(b^2+c^2-a^2\right) : \left(c^2-a^2\right) \left(c^2+a^2-b^2\right) : \left(a^2-b^2\right) \left(a^2+b^2-c^2\right) \right) $.

Proof complete.

  Darij


search
function
ratio
number theory
least common multiple
number theory unsolved

Prove or disprove the following statement :
Each positive rational number can be written in the form
         $\large\frac{a^{2}+b^{3}}{c^{5}+d^{7}}$
where $a,b,c,d$ are positive integers.

Pretty sure I've seen this before, but the Search function is not proving fruitful.

The statement is true. Notice that if $a=x^{3},b=x^{2}, c=y^{7}, d=y^{5}$ we get $\frac{x^{6}}{y^{35}}$. Now it remains to prove that each $\frac{p}{q}$ can be expressed as a ratio of 6-th and 35-th power.
$\frac{p}{q}= \frac{p^{36}q^{174}}{p^{35}q^{175}}=\frac{(p^{6}q^{29})^{6}}{(p^{1}q^{5})^{35}}$

More general statement is true we need only lcm(2,3) and lcm(5,7) to be coprime.

They need to be _co_prime. The reason is that you need to represent $1$ as linear combination of the $\mathrm{lcm}$ (or the product) of the coefficients.


Note:  7+241+83+569 is a sum of prime numbers that uses each of the digits 1 through 9 exactly once.

What is the least possible sum of primes that uses each digit 1 through 9 exactly once?

A.   207	B.   218	C.   225	D.   228	E.   252

I began with the fact that 1, 4, 6, 8, 9 must be paired up with another digit (because they are not prime, they cannot stand alone.) From there it was relatively easy guess and check. With the sum 2+3+5+47+61+89=207, I ended with a final answer of $ \boxed{\mbox{A}}$

Thats a pretty good solution...or you could place the possible units digits first and go from there


limit
abstract algebra
calculus
integration
algebra
polynomial

Find $\lim_{a \to 0} \Bigg[\frac{\pi \coth{(a \pi)}}{2a} - \frac{1}{2a^2}\Bigg]$

I used the identity $\coth(z) = \frac{e^{2z}+1}{e^{2z}-1}$ and L'Hospital a few times to get
\[\lim_{a\to 0} \left(\frac{\pi\coth {(a\pi)}}{2a}-\frac{1}{2a^2}\right) = \frac{\pi^2}{6}\]
How did you come up with this limit?

One of the things I always say I turn to before L'Hôpital's rule is power series. I can't read gauss202's mind, but I would guess that he was thinking in terms of power series and Taylor polynomials.

Here's a look:

$\cosh(x)=1+\frac{x^2}2+O(x^4)$

$\sinh(x)=x+\frac{x^3}6+O(x^5)$

$\coth(x)=\frac{1+\frac{x^2}2+O(x^4)}{x+\frac{x^3}6+O(x^5)}$

(by long division) $=\frac1x+\frac x3+O(x^3)$

So $\frac{\pi\coth(a\pi)}{2a}-\frac1{2a^2}=\frac{\pi}{2a}\left( \frac1{a\pi}+\frac {a\pi}3\right)-\frac1{2a^2}+O(a^2)$

$=\frac1{2a^2}+\frac{\pi^2}6-\frac1{2a^2}+O(a^2)=\frac{\pi^2}6+O(a^2)$

[quote="Jimmy"]How did you come up with this limit?[/quote]

I was looking for a closed form for the series $\displaystyle \sum_{n=1}^\infty \frac{1}{n^2 + a^2}$ for any value a. That was the expression that I came up with, and to check its reasonableness I looked to see what the limit was as $a \to 0$.

To prove the expression I used some things that I'm not sure belong in this forum (namely fourier series).

But Kent is correct, I solved the limit using Taylor Series as he did it. But L'Hopital's rule works too.

If $a>0$, then

$\sum_{-\infty}^{\infty} \frac {1}{k^2+a^2}=\frac {\pi}{a}\left(\frac {1+e^{-2\pi a}}{1-e^{-2\pi a}}\right)$

You can use the Poisson kernel to prove this.

Actually your formula is true for $a \neq 0$, and it reduces to mine for the sum over positive k.

But I don't think discussion of this part of the problem belongs in this forum. Maybe we can discuss it more in the Proposed Problems section. It has interesting connections to some other things.

Your method for solving it might even be worth putting in the Theorems and Formulas section.

[quote="gauss202"]... might even be worth putting in the Theorems and Formulas section.[/quote]
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=327396#p327396]I just did.[/url]

The most "elementary" method I know involves a change of variables in a double integral (and thus a Jacobian). Other methods: infinite products and the infinite analogue or Viete's relations (that was Euler's method); Fourier series; the residue theorem from complex analysis.

The method implied by this topic belongs somewhere in Fourier analysis and/or complex analysis.

I don't think any of those qualify as "high school level."

[quote="Ravi B"]This question might be a little off topic, but what's the most elementary way to evaluate the sum of $\frac{1}{n^2}$?  Can we show that the sum is $\frac{\pi^2}{6}$ with just high-school math?[/quote]

There are 14 methods here, none of which are particularly (IMHO) elementary.

http://www.maths.ex.ac.uk/~rjc/etc/zeta2.pdf

[quote]The most "elementary" method I know involves a change of variables in a double integral (and thus a Jacobian).[/quote]
I was referring to Proof 2 in the article blahblahblah found.

Probably 8 or 10 of the 14 methods could be adjusted to the computation of $\zeta(4), \zeta(6),$ and so on. There are many other piecewise-polynomial functions whose Fourier series could be used as in Proof 4 and Proof 5, so you could vary the details on those quite a bit.

But, as blahblahblah said, none of these methods is "high school" material.

Thanks, Kent and blahblahblah, for the info and reference.

I would say that Proof 9 from blahblahblah's reference is high-school level.  Maybe the average high-school student wouldn't follow it, but arguably it uses nothing beyond precalculus.  What do you think?

Well, neither 9 nor 13 use any calculus, although 13 uses Taylor series twice (although it doesn't expand anything as a taylor series; it just uses the series for certain constants.

My personal favorites are 6 and the second proof in 7, as you only need basic analysis for 6, and 7 is little more than summation of geometric series and uniqueness of Taylor/Laurent series.

Could you tell me what is O (capital o) here
Like in O(x^4)

An elementary method exists using the power of complex numbers, finite vieta relations, and some elementary inequalities.

Firstly, establish the inequality $\cot{x} < \frac{1}{x} < \csc{x}$

Secondly, use the binomial theorem to determine the expansion of $(z+i)^n +(z-i)^n = 0$ in terms of $z$.

Thirdly, find the roots of the polynomial and use them to evaluate: $\sum_{k=1}^n \cot^2{\left( \frac{2k-1}{4n} \right)}$ and hence evaluate

$\sum_{k=1}^n \csc^2{\left( \frac{2k-1}{4n} \right)}$ in terms of $n$.

Fourthly, use the inequality obtained to bound both sides of the truncated odd part of the Basel's Sum, and squeeze to evaluate the odd part of Basel's Sum.

Finally, divide the odd part by 4 and add to obtain the entire Basel's Sum.


I saw a question that forall $a,b>0$ such that $a+b=s$ the minimum of
$\left(a+\frac{1}{a}\right)\left(b+\frac{1}{b}\right)$
is
$\left(\frac{s}{2}+\frac{2}{s}\right)^{2},\ 0<s\leqslant 2\sqrt{2+\sqrt{5}}\\ 2\sqrt{s^{2}+1}-2,\ s>2\sqrt{2+\sqrt{5}}$
It's so complex and now I consider about the minimum of
$\prod_{i=1}^{n}\left(a_{i}+\frac{1}{a_{i}}\right)$
where $a_{1},a_{2},\ldots,a_{n}$ is positive and
$\sum_{i=1}^{n}a_{i}=s$
Is it possible to solve it? I'm not very confident. (Or is it an old problem and I didn't find it? If so, please point it out.)

By LCRCF-Theorem (Algebraic Inequalities, page 148), it follows that the product
$P=\prod_{i=1}^{n}(x_{i}+\frac{1}{x_{i}})$
is minimal when $x_{1}=...=x_{n-1}\le x_{n}$.
Problem 8 from page 151 establishes that for $s\le n+\sqrt{2n-2}$, P is minimal when  $x_{1}=...=x_{n-1}=x_{n}$.
Actually, P is minimal for either  $x_{1}=...=x_{n-1}=x_{n}$ or $x_{1}=...=x_{n-1}=p$ and $x_{n}=q$, with $1<p\le q$ and $p^{2}q^{2}=(p+q)^{2}+1$.


geometry
quadrilateral
geometric inequality
inequalities
IMO Shortlist
similar triangles

[color=blue][b]Problem.[/b] Let ABCD be a convex quadrilateral and O a point inside it. Let the parallels to the lines BC, AB, DA, CD through the point O meet the sides AB, BC, CD, DA of the quadrilateral ABCD at the points E, F, G, H, respectively. Then, prove that $ \sqrt {\left|AHOE\right|} \plus{} \sqrt {\left|CFOG\right|}\leq\sqrt {\left|ABCD\right|}$, where $ \left|P_1P_2...P_n\right|$ is an abbreviation for the non-directed area of an arbitrary polygon $ P_1P_2...P_n$.[/color]

This problem is, indeed, problem G7 from the IMO Shortlist 1995, and respawned as problem 3 in the 2nd round of the German competition BWM (Bundeswettbewerb Mathematik) 1999.

[i]Solution.[/i] Let g be the parallel to the line BD through the point O. Let this line g meet the segment AC at a point O'. The problem will be solved once we show that $ \sqrt {\left|AHOE\right|}\leq\frac {AO^{\prime}}{AC}\cdot\sqrt {\left|ABCD\right|}$ and $ \sqrt {\left|CFOG\right|}\leq\frac {CO^{\prime}}{AC}\cdot\sqrt {\left|ABCD\right|}$; in fact, these relations will yield

$ \sqrt {\left|AHOE\right|} \plus{} \sqrt {\left|CFOG\right|}\leq\frac {AO^{\prime}}{AC}\cdot\sqrt {\left|ABCD\right|} \plus{} \frac {CO^{\prime}}{AC}\cdot\sqrt {\left|ABCD\right|}$
$ \equal{} \frac {AO^{\prime} \plus{} CO^{\prime}}{AC}\cdot\sqrt {\left|ABCD\right|} \equal{} \frac {AC}{AC}\cdot\sqrt {\left|ABCD\right|} \equal{} \sqrt {\left|ABCD\right|}$,

and this is what we have to prove.

We will only show that $ \sqrt {\left|AHOE\right|}\leq\frac {AO^{\prime}}{AC}\cdot\sqrt {\left|ABCD\right|}$; the relation $ \sqrt {\left|CFOG\right|}\leq\frac {CO^{\prime}}{AC}\cdot\sqrt {\left|ABCD\right|}$ will then follow by analogy.

Let the parallels to the lines BC and CD through the point O' meet the lines AB and DA at the points E' and H', respectively. Then, by Thales, $ \frac {AH^{\prime}}{AD} \equal{} \frac {AO^{\prime}}{AC}$ and $ \frac {AE^{\prime}}{AB} \equal{} \frac {AO^{\prime}}{AC}$, so that the homothety with center A and factor $ \frac {AO^{\prime}}{AC}$ maps the points C, D and B to the points O', H' and E', respectively. This has two important consequences: first, E'H' || BD (since homotheties map lines to parallel lines), and then, $ \frac {\left|AE^{\prime}O^{\prime}H^{\prime}\right|}{\left|ABCD\right|} \equal{} \left(\frac {AO^{\prime}}{AC}\right)^2$ (since homotheties multiply areas with the squared homothety factor). The latter equation yields $ \left|AE^{\prime}O^{\prime}H^{\prime}\right| \equal{} \left(\frac {AO^{\prime}}{AC}\right)^2\cdot\left|ABCD\right|$, so that $ \sqrt {\left|AE^{\prime}O^{\prime}H^{\prime}\right|} \equal{} \frac {AO^{\prime}}{AC}\cdot\sqrt {\left|ABCD\right|}$; therefore, in order to prove $ \sqrt {\left|AHOE\right|}\leq\frac {AO^{\prime}}{AC}\cdot\sqrt {\left|ABCD\right|}$, it is enough to show that $ \left|AHOE\right|\leq\left|AE^{\prime}O^{\prime}H^{\prime}\right|$.

Let the line g meet the lines AB and DA at the points U and V, respectively. Since $ \left|AHOE\right| \equal{} \left|AUV\right| \minus{} \left(\left|UOE\right| \plus{} \left|VOH\right|\right)$ and $ \left|AE^{\prime}O^{\prime}H^{\prime}\right| \equal{} \left|AUV\right| \minus{} \left(\left|UO^{\prime}E^{\prime}\right| \plus{} \left|VO^{\prime}H^{\prime}\right|\right)$, proving that $ \left|AHOE\right|\leq\left|AE^{\prime}O^{\prime}H^{\prime}\right|$ is equivalent to proving that $ \left|UOE\right| \plus{} \left|VOH\right|\geq\left|UO^{\prime}E^{\prime}\right| \plus{} \left|VO^{\prime}H^{\prime}\right|$. This can be done as follows:

Let e be the distance from the point E to the line g, or, equivalently, the length of the E-altitude in triangle UOE. Then, the area of triangle UOE equals $ \left|UOE\right| \equal{} \frac12\cdot UO\cdot e$ (in fact, the area of a triangle equals $ \frac12\cdot$ sidelength $ \cdot$ corresponding altitude). Similarly, if h is the distance from the point H to the line g, then $ \left|VOH\right| \equal{} \frac12\cdot VO\cdot h$. Thus,

$ \left|UOE\right| \plus{} \left|VOH\right| \equal{} \frac12\cdot UO\cdot e \plus{} \frac12\cdot VO\cdot h \equal{} \frac12\cdot\left(UO\cdot e \plus{} VO\cdot h\right)$.

Similarly, working with the point O' instead of O, we get

$ \left|UO^{\prime}E^{\prime}\right| \plus{} \left|VO^{\prime}H^{\prime}\right| \equal{} \frac12\cdot\left(UO^{\prime}\cdot e^{\prime} \plus{} VO^{\prime}\cdot h^{\prime}\right)$,

where e' and h' are the distances from the points E' and H' to the line g, respectively. Thus, in order to prove $ \left|UOE\right| \plus{} \left|VOH\right|\geq\left|UO^{\prime}E^{\prime}\right| \plus{} \left|VO^{\prime}H^{\prime}\right|$, it is enough to show that

$ UO\cdot e \plus{} VO\cdot h\geq UO^{\prime}\cdot e^{\prime} \plus{} VO^{\prime}\cdot h^{\prime}$.

Since OE || BC and O'E' || BC, we have OE || O'E', and thus, there is a homothety with center U which maps the points O and E to the points O' and E'. This homothety must obviously leave the point U fixed (since U is the center of this homothety). Thus, the triangles UOE and UO'E' are similar. In similar triangles, the ratio of a sidelength to the corresponding altitude is equal; thus, in the triangles UOE and UO'E', we have $ \frac {UO}{e} \equal{} \frac {UO^{\prime}}{e^{\prime}}$ (in fact, e and e' are the altitudes of these triangles corresponding to the sidelengths UO and UO'). Similarly, $ \frac {VO}{h} \equal{} \frac {VO^{\prime}}{h^{\prime}}$.

Now, the points E' and H' have a property which the points E and H don't have (at least not necessarily): Since E'H' || BD and g || BD, we have E'H' || g, so that the distances e' and h' from the points E' and H' to the line g are equal. Thus,

$ UO^{\prime}\cdot e^{\prime} \plus{} VO^{\prime}\cdot h^{\prime} \equal{} UO^{\prime}\cdot e^{\prime} \plus{} VO^{\prime}\cdot e^{\prime}$
$ \equal{} \left(UO^{\prime} \plus{} VO^{\prime}\right)\cdot e^{\prime} \equal{} UV\cdot e^{\prime} \equal{} \frac {UV^2}{\left(\frac {UV}{e^{\prime}}\right)} \equal{} \frac {\left(UO \plus{} VO\right)^2}{\left(\frac {UO^{\prime} \plus{} VO^{\prime}}{e^{\prime}}\right)}$
$ \equal{} \frac {\left(UO \plus{} VO\right)^2}{\frac {UO^{\prime}}{e^{\prime}} \plus{} \frac {VO^{\prime}}{e^{\prime}}} \equal{} \frac {\left(UO \plus{} VO\right)^2}{\frac {UO^{\prime}}{e^{\prime}} \plus{} \frac {VO^{\prime}}{h^{\prime}}} \equal{} \frac {\left(UO \plus{} VO\right)^2}{\frac {UO}{e} \plus{} \frac {VO}{h}}$.

So the inequality in question, $ UO\cdot e \plus{} VO\cdot h\geq UO^{\prime}\cdot e^{\prime} \plus{} VO^{\prime}\cdot h^{\prime}$, becomes $ UO\cdot e \plus{} VO\cdot h\geq\frac {\left(UO \plus{} VO\right)^2}{\frac {UO}{e} \plus{} \frac {VO}{h}}$, or, equivalently, $ \left(UO\cdot e \plus{} VO\cdot h\right)\left(\frac {UO}{e} \plus{} \frac {VO}{h}\right)\geq\left(UO \plus{} VO\right)^2$, what is trivial from Cauchy-Schwarz:

$ \left(UO\cdot e \plus{} VO\cdot h\right)\left(\frac {UO}{e} \plus{} \frac {VO}{h}\right)\geq\left(\sqrt {\left(UO\cdot e\right)\cdot\frac {UO}{e}} \plus{} \sqrt {\left(VO\cdot h\right)\cdot\frac {VO}{h}}\right)^2$
$ \equal{} \left(UO \plus{} VO\right)^2$.

This completes the solution of the problem.

  Darij

This problem is simply following from a Brunn-Minkovsky inequality. Am I right?


Let X be a random variable such that P(X<=0) = 0 and let E(X) exist. Show that P(X>=2E(X)) <= 0.5.

I think this is an application of Chebychev but I can't seem to figure it out... Am I missing something dreadfully simple?

As a matter of fact, it is a straightforward application of [url=http://mathworld.wolfram.com/MarkovsInequality.html]Markov's inequality:[/url]

$ P(X \geq a) \leq \frac {E(X)}{a}.$

[hide]Hint: Use the substitution $ a \equal{} 2E(X).$[/hide]

Math-ly,

Oh yes... I feel like an idiot now haha... Thanks so much :)


Let $ABC$ be a triangle and let $S$ be the foot of the A-symedian, i.e. $S\in (BC)$ and $\frac{SB}{SC}=\left(\frac cb\right)^2$. Denote the points: $E\in (AC)$, $SE\parallel AB$; $F\in (AB)$, $SF\parallel AC$; $P\in BE$, $CP\parallel AB$; $R\in CF$, $BR\parallel AC$. Prove that $A\in PR$, $\frac{AP}{AR}=\left(\frac bc\right)^2$ and the quadrilateral $BCPR$ is cyclically.

Just notice that $BFEC$ is cyclic, consequently $PCBR$ is cyclic too.
Moreover, $\frac{CF}{RF}=\frac{AF}{BF}=\frac{SC}{BS}=\frac{EC}{AE}$, thus $AR||FE$.
Analogously $AP||FE$, hence $A\in{PR}$.
Since $\frac{AR}{FE}=\frac{a}{SC}$ and $\frac{FE}{AP}=\frac{a}{BS}$ we get the remaining relation!


Hi,

I am a fairly logical person and programmer.

In my new job I am touching upon algrebra, I would like to know for personal knowledge a reasonable amount,  I am not going to be problem solving, just breaking down algrebra to code.

Can anyone recommend a range of books from beginner to above high school I can use to learn an reference.

Thanks

T

A beginner book would be Introduction to Algebra which is on this website, and an advanced book would be Intermediate Algebra, which is also on this website.


wth?

where did you guys take the test?  Like did you guys have chairs if you guys didnt have desks?

we were in a classroom on a normal size desk with a proctor and everyone was spaced out atleast 1 desk.

Whoa I am so sorry for your troubles. We did it in the library, nice large desks but shared so not so large and nice, a little more space than the classroom desks.

[quote="mchoi815"]At my school, one would bring his or her own writing surface, typically a textbook, a binder or something sort of small.[/quote]

Couldn't someone cheat using a binder/textbook?

We did it in the library. Only two qualifiers, so we got our own tables.

[quote="diophantient"]I was the only taker, so I had a 20 sq. ft desk all to myself :)[/quote]

Me too, the table was really bumpy, though, so the bubbling in and writing of the answers was kinda sloppy...

We did it in the cafeteria, except we had like 30+ or something like that qualifiers. We each had about half of a large table, which was enough room, I guess.

Public School  :) But how does that apply to the type of tables we get to use? Couldn't you ask your teachers for a desk or something? That's pretty ridiculous, considering making it to AIME should be considered a pretty big achievement.

[quote="firecricket91"]Do you guys go to private or public schools?[/quote]
Huh I don't think that's very good to ask because it would create a lot of extra posts...
I think the vast majority of people go to public. I think we got nice big tables because school math departments generally don't really care that much about math, and because we made AIME, we got special treatment because its so "hard" and all.

How can you take it in the cafeteria or library?  Wouldn't it be too noisy?  The ~15 people who took it at my school were crammed into a small classroom (the only room which could be reserved for 4 consecutive periods), but at least it was quiet.

I took it in this student conference center, and it was pretty open.  I basically was given the test and told I could sit anywhere in the room.

[quote="Carbon57"]It wouldn't really be noisy...since nobody's talking while taking the test... We took in from 7:45-10:45 AM, so we finished right before brunch.[/quote]

No, I mean the other people in the cafeteria/library who aren't taking the test; at my school, the sound level is pretty much around 70 dB constantly in either place.

We had six people in a conference room, but the tables were big. I had plenty of room for myself...probably about three-quarters to a whole table to myself. And it was quiet, except for announcements every hour or so, and people walking by when they change periods.

We were in a lecture hall, which, as far as I know, is only used for detentions now. Hence, the desks, which are about 10*12 inches big, have some very distracting material on them, most of which isn't appropriate to discuss here.

A 10" by 15" blue clipboard. In a poorly lighted auditorium.

We had a 3 by 5(or maybe 6?) ft conference table in a conference room. But we had to share it with 3 other people. And then, we had to use these dividers that always make be feel claustrophobic. ugh. I got to a public school, I think around 30 people qualified (not as many as last year, hmm)

Last year, we had the library and so only 2 ppl were at at table--so we didn't need dividers.

[quote="mchoi815"]At my school, one would bring his or her own writing surface, typically a textbook, a binder or something sort of small.  As for me, I used the top of one of those boxes that hold reams of paper.  It was relatively large, but in my opinion not large enough.  I'm wondering what treatment you got while taking the AIME.[/quote]
I remember in a school I used to go to where we used book sox to cover all our books. Whenever I use book sox as a writing surface, the pencil would rip through the paper...
Anyways I had a nice large table.

What're a book sox?
I used a nice large table shared with 2 people

Yeah, I hate those things.  They're pretty much just stretchable book covers.  Luckily, we don't need them on our math books :lol:

I was the only qualifier at my school, so I had the whole Math Lab (just a room with a few computers and some math books) to myself, which featured a nice large table. Sadly, since a lot of the teachers are completely oblivious to the AIMEs, there was a lot of bureaucratic mess to deal with; if I wasn't dressed up so nice for the test (and of course the proctor showing up to explain things may also have helped), they may not have believed that the whole Math Lab was reserved for just one student for three hours.

[quote="mchoi815"]lol, you dressed up for the aime?

wow...maybe i should pay more respect for it.[/quote]

I did, but it didn't work.  :P

[quote="tjhance"]I took it on a table near the cafeteria. The conditions were nice, and the table was large enough, though not as large as I would have liked. (There was plenty of room for everything, but I couldn't "spread out").

Though I once had to take a test in the in-school suspension room.[/quote]

Man you got it good in the ISS room. Our ISS room has desks and is basically consists of black-felted cubicles. No distractions at ALL xD.

3 hours in the morning would conflict with some lunches, so cafeteria was not feasible for me. Instead I used a normal sized desk (I had two chairs next to me to place papers) in a place very close to the size of a walk-in closet (which was for some reason my teacher's office). I was the only one taking it, so it wasn't too bad. After about 15 minutes of getting into to the test there isn't very much that will keep you from biting into the test... Of course, I didn't have to use a binder...  :huh: Seriously tell your teacher to get you a desk or use the library...

My bell rang and I could hear loud kids in the hall way pretty well, but only when changing classes so it wasnt a big deal. Next time I will try the media center for sure (if I make USAMO that us..) :D

[quote="Anaxerzia"]Man you got it good in the ISS room. Our ISS room has desks and is basically consists of black-felted cubicles. No distractions at ALL xD.[/quote]

Yeah it was very nice in there. The problem was when the test was over and I found that my math teacher had told everybody I was at ISS without telling them why.

a problem? Dude, that would be so fraking cool.

oh wow.. we somehow had 9 people.. 7 from our school
and the desks were decent, except we were all about a foot away from each other....
ABSOLUTELY NO CHEATING RISK AT ALL :roll:


Since many people seem to be wanting to know what sort of references are available for programming lately, would it be possible to start a thread gathering a list of them? Currently I know of only three major ones, if you have any others please add them, and perhaps a mod could compile what we have into another thread and sticky it if it proves useful:

[url=http://www.cppreference.com]C++ Reference[/url]
Contains information about all standard C/C++ classes/functions.
[url=http://java.sun.com/reference/api/]JAVA API[/url]
Contains information about all standard JAVA classes/functions.
[url=http://www.nr.com/]Numerical Recipes in C[/url]
Contains information about numerical methods. There is a forum as well as a free online copy of the book [i]Numerical Recipes in C[/i]. You can also order the book, as well as [i]Numerical Recpies in Pascal[/i], [i]C++[/i], [i]JAVA[/i], etc.

Well, there's always the standard reference...

[url]http://msdn2.microsoft.com/en-us/library/aa388745.aspx[/url]

for .NET languages and

[url]http://www.sgi.com/tech/stl/[/url]

for C++ Standard Template Library.


Let $ 0<\alpha <\pi$ ,then prove that $ ctg \frac{\alpha}{2} \geq 1\plus{}ctg\alpha$

[hide="Solution"]Let $ x=\alpha: 2$, $ \ctg x=m$

$ \ctg x\ge1+\ctg 2x$
$ m\ge1+\frac{m^2-1}{2m}$
$ 2m^2\ge(m^2+2m-1)$
$ m^2\ge2m-1$
This is true by AM-GM [/hide]


algebra
function
domain
Ring Theory
superior algebra
superior algebra solved

Show that for $D<-1$ and $D=3(mod 4)$(and squarefree)(i.e $D=-5,-13.-17,-21...$),  the ring of integers is not euclidean by finding an ideal which is not principal

i guess you mean to say that $\mathbb{Z}\left[i\sqrt{m}\right]$ is not an euclidean domain for any $m>0, m\equiv 1 \mod 4$ is not an euclidean domain, right?
well, if i recall correctly, it just sufficies to prove that it isn't a unique factorization domain (UFD). (and therefore you should be able to find two non-principal ideals).
so, consider the $(1-i\sqrt{m})(1+i\sqrt{m}) = 1+m = 2\cdot\frac{1+m}2$.
you can easily show these are two distinct factorizations.. and i'm not wrong, $(2)\cap(1+i\sqrt{m})$ is not a principal ideal...


Let $a,b,c,r$ be positive real numbers such that $r\geq \frac{2}{3}$, prove that
$(\frac{a}{b+c})^r+(\frac{b}{c+a})^r+(\frac{c}{a+b})^r\geq \frac{3}{2^r}$

http://www.mathlinks.ro/Forum/topic-12763.html
http://www.mathlinks.ro/Forum/topic-73225.html
.............
  It has been discussed already.


Well some people seem to think #3 was very hard.... :?
But if you realized that how the different circles are dependent on each other doesn't matter, quite easy  :lol:

So did you think it was easy or hard?

Personally I fail to see how someone could have read the article on Expected Value and not have thought this problem is easy, unless they didn't really understand the article in the first place. So, I say easy.

I didn't realize that expected values summed even if they were dependant, so at the time of the contest, I thought that it was hard, but now I think that it's extremely easy.

It's [b]expected value[/b].  Consider an easy example: what is the expected value of the number shown facing up on a standard, 6-sided die when you roll it?  By [b]expected value[/b] we mean average value you expect to get if you roll the value many, many times...like expected payout in Vegas.

So, for the die problem.  You would have that the expected value equals $\frac 16(1)+\frac 16(2)+\frac 16(3)+\frac 16(4)+\frac 16 (5)+\frac 16(6)=3.5.$

I thought.  Then I realized that that would have to be the case (based on what I have learned about expected value so far), that the circles can be counted independent of each other.  So then I proved it.  But I had a hunch that I was missing something somewhere, because that made the problem kinda trivial...

[quote="Altheman"]the one thing that i don't understand is how the different triangles are independant, if you have 1 circle, the chances of the triangles around it forming it are 0 since the corner is already used for the initial circle[/quote]
They're dependent only if you look at each circle in the context of other circles, which is unnecessary.  Expected values sum for random events, even dependent random events, so simply calculating the expectected value of an individual circle forming without regarding what it does to other circles (which is easy to do) is all you have to do to solve this problem.

[quote="Altheman"]the one thing that i don't understand is how the different triangles are independant, if you have 1 circle, the chances of the triangles around it forming it are 0 since the corner is already used for the initial circle[/quote]
The different triangles, or to be more precise, the chances of the triangles forming circles at different points, are [b]not[/b] independent. The trick is that expected value is not affected by the dependence.

In a posting above, joml88 shows that the expected value of a die roll is 3.5. Let's perform that calculation another way. There are 1+2+3+4+5+6 = 21 dots on a die. Given a particular dot, for example, the upper left dot on the 3-side, the chance of that dot showing up on top to be counted is 1/6. For every dot, the chance is 1/6. So the expected number of dots to be rolled is $21\times\frac{1}{6} = 3.5$. Are the dots independent? No, not at all. The dot on the upper left corner of the 3-side always shows up with two other particular dots. It never shows up with any of the dots on the 6-side. But for expected value, that dependence has no effect on the calculation. If we could magically move the dots around before rolling the die, altering the dependencies, it would not change the expected value.

Erin Schram

Well I said, let's say there are two events, A and B, with equal probability.
If A and B are independent, ie, they don't affect each other, then we have
Expected value= P(A and ~B) + P(B and ~A) + P(B and A) = P(A)+P(B).
If they're dependent, ie, one cannot happen while the other happens, then
P(A and ~B) = P(A), P(B and ~A)=P(B), and P(A and B)=0.  So we still have P(A)+P(B).
So it works either way.  Comes out nicely.

I thought problem 3 was relatively easy. Expected value problems just don't seem to give me any trouble for some reason.


How to prove DE=2BE

[img]http://agutie.homestead.com/files/problem/problem018.gif[/img]

[hide]This Problem is pretty simple if you think about it long enough

1)Draw segment [b]AF[/b] that bisects angle [b]EAD[/b]
2)This creates two angles that equal angle [b]BAE[/b]
3) Using the theorem (If two angles are congruent then the sides opposite the angles are congruent) you make [b]BE=EF=DF[/b]
4)By segment addition [b]DF+EF=DE[/b]
5)Through the substitution property [b]BE+BE=DE[/b] or [b]2BE=DE[/b] [/hide]

[quote="fireforce"]3) Using the theorem (If two angles are congruent then the sides opposite the angles are congruent) you make [b]BE=EF=DF[/b][/quote]

Wrong.

[quote="fireforce"]3) Using the theorem (If two angles are congruent then the sides opposite the angles are congruent) you make [b]BE=EF=DF[/b][/quote]
This is only true if the 2 angles are part of the same triangle... (isoceles triangles)  ;)

[quote="Altheman"][hide="hint"]find $\alpha$ in terms of $\theta$ and use the law of sines[/hide][/quote]
I think the answer like this :
from triangle $ABE$ we get $AE = \frac{BE}{\sin{\alpha}}$
and by doing the hint from Altheman i get $\angle ADE = 90 + \alpha$
finally, using sine law at triangle $ADE$
$\frac{sin(2\alpha)}{DE}$=$\frac{sin(90+\alpha)}{AE}$=$\frac{cos(\alpha)}{AE}$
by substitution $AE = \frac{BE}{\sin{\alpha}}$ and sine's identity $sin(2\alpha)=2sin(\alpha)cos(\alpha)$. we get $DE=2BE$
thanks to Altheman :P

Can't see your picture.
Can you please describe the problem a bit?

Ok, I'll do it for you :P

We are given right triangle $ABC$ with $\angle B=90^\circ$. Point $E$ is on $BC$ such that $\angle BAE=\alpha$ and $\angle EAC=2\alpha$. Point $D$ is on $AC$ such that $\angle AED=\angle BCA=\theta$. Prove that $DE=2BE$.


number theory
prime numbers
number theory unsolved

Hi
n , k are natural numbers and k>2.
prove that    x^n - y^n =2^k   has no natural answers.

i couldn't solve it but i think i find somthing
i think we should just solve it for prime  n. (if x-y>4)
(if that you can prove that for pime n tell me that tell you why prime numbers are enough)

thanks and sorry for my bad english

Right now, i cant solve that one, but i can solve this one that you shuld try replacing the $+$ with a $-$.

First it is easy to prove $n$ is a power of $2$, or otherwise, lhs would have an odd factor. Now, if n>1, the lhs is $2$ mod $4$, so we get the result.

We set $q=\max(i,2^i|x\textnormal{ and }2^i|y)$ and $x'=\frac{x}{2^q}$, $y'=\frac{y}{2^q}$
Now $x$ or $y$ is odd so they are two of them odd since $x-y|x^n-y^n\Leftrightarrow x-y|2^k$
So we have
\[x'^n-y'^n=2^k'\textnormal{ with }k'=k-qn\]
$x'-y'|2^k'\Rightarrow \exists r<k', x'=y'+2^r$
Now $x'^n-y'^n=2^k'\Leftrightarrow (y'+2{}^r){}^n-y'^n=2^k'$ and develop, divide by $2^r$ to obtain $y'$ even

hi
but why $ y' $ must  be odd ? it can be $ x=2^a(2t+1)$  and $y = 2^{a+1}(2t'+1) $
then $q=2^a$ and $y'=2(2t'+1)$ that it is even
please tell me if i didn't understand your soulution :(

$x$ even $\Leftrightarrow y$ even
If we take $x=2^a(2t+1)$ and $y = 2^{a+1}(2t'+1)$ and $x'=2t+1$, $y'=2(2t'+1)$ but $x-y|2^k'\Leftrightarrow 2(t-2t'+1)+1|2^k'$ absurd


search
polynomial
Ring Theory
conics
ellipse
analytic geometry
graphing lines

Are there any rational numbers $ a$ and $ b$ other than the obvious $ a\equal{}\pm1,b\equal{}0$ satisfying the equation $ a^2\plus{}3b^2\equal{}1$?

http://en.wikipedia.org/wiki/Pell%27s_equation

Let $ a,b$ be the solution for this equation. So, by Euler.

$ \frac {a}{b}\approx \sqrt { \minus{} 3} \equal{} i \sqrt {3} \Longleftrightarrow \frac {a}{b} \propto i$

Wich is a contradiction, since $ a,b$ are supposed to be real numbers.

I will think that you made a mistake here. but see wiki or google pell or search here. it's not a new thing

[quote="Jane_Bennet"]Are there any rational numbers $ a$ and $ b$ other than the obvious $ a \equal{} \pm1,b \equal{} 0$ satisfying the equation $ a^2 \plus{} 3b^2 \equal{} 1$?[/quote]

There are other solutions (not sure how many though, perhaps an infinitude?)

My reasoning.

Let $ a\equal{}b$ (purely to help find a solution)

Then $ 4a^2\equal{}1$

So therefore $ a^2\equal{}\frac{1}{4}$ Therefore $ a \equal{} \pm \frac{1}{2}$ is another solution set

For any rational number $ r$, you can set $ a \equal{} \pm \frac{r^2 \minus{} 3}{r^2 \plus{} 3}$ and $ b \equal{} \frac{2r}{r^2 \plus{} 3}$, Not sure about other infinite families.

See http://www.mathlinks.ro/viewtopic.php?t=87739 for how to deal with such equations in general.

[hide="Some algebraic number theory"] Let $ a \equal{} \frac{x}{z}, b \equal{} \frac{y}{z}$ where $ x, y, z$ are integers with $ \gcd(x, y, z) \equal{} 1$.  We want to solve the Diophantine equation

$ x^2 \plus{} 3y^2 \equal{} z^2$.

Let $ x \equal{} u \minus{} \frac{v}{2}, y \equal{} \frac{v}{2}$.  Our equation becomes

$ u^2 \minus{} uv \plus{} v^2 \equal{} z^2$;

in other words, $ z^2 \equal{} N(u \plus{} v \omega)$ where $ \omega$ is a primitive third root of unity.  Since $ \mathbb{Z}[\omega]$ (the Eisenstein integers) are a UFD, it follows that $ z \equal{} N(p \plus{} q \omega)$ for some integers $ p, q$, hence we have $ u \plus{} v \omega \equal{} (p \plus{} q \omega)^2 \equal{} (p^2 \minus{} q^2) \plus{} (2pq \minus{} q^2) \omega$ and

$ z \equal{} p^2 \minus{} pq \plus{} q^2$
$ x \equal{} p^2 \minus{} pq \minus{} \frac{1}{2} q^2$
$ y \equal{} pq \minus{} \frac{1}{2} q^2$

which gives (letting $ t \equal{} \frac{p}{q}$)

$ a \equal{} \frac{x}{z} \equal{} \frac{2t^2 \minus{} 2t \minus{} 1}{2(t^2 \minus{} t \plus{} 1)}$
$ b \equal{} \frac{y}{z} \equal{} \frac{2t \minus{} 1}{2(t^2 \minus{} t \plus{} 1)}$. [/hide]

I believe that the method that ZetaX mentions is one of the gems of mathematics! Using that idea, we can clearly generate an infinite set of solutions if we just find one solution. Clearly, you have given us a set of solutions with which to start. Call $ A: (1,0)$. Now I think that we can easily characterize all of the solutions in this manner.

I claim that we can generate all possible rational points on this ellipse by taking a line of slope $ m$ that passes through $ A$ where $ m$ is any rational number.

Suppose that $ (p,q)$ is a rational point on this ellipse. Then we have that the slope is rational. Going the other direction, if $ m$ is a rational number, then we can use the root flipping argument or whatever to show that the second point of intersection is rational (there is always a second point of intersection because the vertical tangent corresponds to a slope of infinity).

That is to say, there is a bijection between the set of rational numbers and the rational points on this ellipse (except $ A$).

To be even more terse, the ellipse is isomorphic (as a [url=http://mathworld.wolfram.com/AlgebraicVariety.html]variety[/url]) to the rational [url=http://en.wikipedia.org/wiki/Projective_line]projective line[/url], as is any conic section with rational coefficients.  This is because all conic sections (and the projective line) have [url=http://mathworld.wolfram.com/CurveGenus.html]genus[/url] zero.

Edit:  I like this line of inquiry.  It suggests the question, "what is the simplest kind of curve whose set of rational points does not admit a rational parameterization?"  The answer is an [url=http://mathworld.wolfram.com/EllipticCurve.html]elliptic curve[/url].  :)

[quote="ZetaX"]By the same reasoning, $ a^2 \plus{} b^2 \equal{} 1$ has no solution. So obviously, your reasoning is wrong  :wink:[/quote]

I assumed $ a,b$ are integers, since it is a pell equation, by the definition the ratio $ a/b$ is approximately the square root of $ n$, what don't make sense if $ n$ is negative.

To be able to be rational we don't use Euler, the case is more simple and is trivialy trivial.

$ a, b$ are rational and small.  The "approximation" that you claim is only valid if $ a, b$ are large, which is obviously not the case here.  Have you read the other posts in this thread?

have you read my post?

What I said is $ \frac {a}{b} \propto i$  with is valid for large, small, even, odd, happy numbers, unhappy numbers and even prime numbers.

What do you mean by $ \frac {a}{b} \propto i$?

Er... no, it's not.  $ a \equal{} \frac{3}{5}, b \equal{} \frac{4}{5}$ is a perfectly legitimate solution to $ a^2 \plus{} b^2 \equal{} 1$, for example.  In fact, the set of rational solutions can be parameterized $ a \equal{} \frac{2t}{1 \plus{} t^2}, b \equal{} \frac{1 \minus{} t^2}{1 \plus{} t^2}$ for any rational $ t$.


Four distinct digits are used to make 2 two-digit numbers. What is the greatest possible product of the two numbers formed?


calculus
derivative
real analysis
real analysis unsolved

Let $ f: [0, 1] \to \mathbb{R}$, continuous on [a, b], differentiable on (a, b) and f(0) = 0. Furthermore, $ 0 \le f'(x) \le \alpha f(x), \alpha > 0.$ Prove that f is identically zero.

I'm stumped on this one. We were given a hint to consider the derivative of $ e^{\minus{} \alpha x}f(x)$, which is $ \minus{} \alpha e^{\minus{} \alpha x}f(x) \plus{} e^{\minus{} \alpha x}f'(x) \equal{} e^{\minus{} \alpha x}(f'(x) \minus{} \alpha f(x)) < 0$, by our assumption that $ f'(x) \le \alpha f(x)$. Somehow this is supposed to lead me to the fact that $ f(x) \le 0$, but from our hypothesis $ f(x) \ge 0$, so $ f(x) \equal{} 0$. I just can't see how to arrive at f(x) being less than zero unless I can also show that f'(x) < 0, so f(x) is decreasing, but even that is eluding me.

Let

$ g(x)\equal{}e^{\minus{}\alpha x}f(x)$. Then

$ g'(x)\equal{}e^{\minus{}\alpha x}\left(\minus{}\alpha f(x)\plus{}f'(x)\right)\leq 0$, and hence, $ g$ decreases. Now,

this implies that for all $ x\geq 0$ we have

$ g(x)\leq g(0)$, from which it follows that $ e^{\minus{}\alpha x} f(x)\leq e^{\minus{}\alpha \cdot 0}f(0)\equal{}0$.

Thus $ f(x)\leq 0$ and your problem is solved.

f should be continuous on $ [0,1]$ and not on $ [a,b]$.

Thanks, can't believe I didn't think of that. Hopefully my brain is allowed a slip-up or two like this  :blush:


I need some advice. i started my own forum and was wondering what the best way is for getting new members to join. Thanks for your ideas.

Be interesting. Be devoted. Have things worth of discussion- LOTS of discussion.

Get some activity started on your forum - you might need friends, and a crazy game called Blamo. :P

Get a cool avatar, the kind of avatars only master creators like Valentin Vornicu use! :D


group theory
abstract algebra
induction
superior algebra
superior algebra unsolved

Suppose $ G$ is a finite  $ p$-group with a unique subgroup $ H$ of index $ p$. Show $ G$ is cyclic.

Induction--for $ |G|\equal{}p^2$, $ G$ is abelian and it's clear.

Suppose $ |G|\equal{}p^n$.  Now, choose some $ a\in H\cap Z(G)$ with order $ p$. (we can do this because $ H$ is normal)  $ G/\langle a\rangle$ is cyclic by induction, so there is some $ b\in G$ such that $ \overline{b}$ has order $ p^{n\minus{}1}$ in $ G/\langle a\rangle$.

If $ b$ has order $ p^n$ in $ G$, we're done, otherwise $ b$ has order $ p^{n\minus{}1}$ in $ G$ and $ \langle b\rangle$ has trivial intersection with $ \langle a\rangle$.  So $ G\cong \langle a\rangle \times\langle b\rangle\cong Z_p \times Z_{p^{n\minus{}1}}$.  But this group clearly has more than one subgroup of order $ p^{n\minus{}1}$, contradiction.

Here's a non-inductive proof:

Pick an element $ g \notin H$. Assume this element has order $ p^i$ with $ i < n$. Because every subgroup of order $ p^i$ is contained in a subgroup of order $ p^{i\plus{}1}$, we can find in any case (also if $ i\equal{}n\plus{}1$ of course) a subgroup of index $ p$ different from $ H$. So $ g$ has order $ p^n$.

LydianRain, I'm having difficulty in understanding your proof, it's very condensed and compact.
Why is $ H \cap Z(G)$ not trivial? And why are we sure that $ \langle b \rangle$ and $ \langle a \rangle$ have trivial intersection?

Both proofs seem correct, but I'm having the same difficulty as Jan. Can you please explain? Thanks.

Yes, it seems I was a little terse, sorry. :)

Since $ H$ is normal, it contains only entire conjugacy classes.  Any non-central element of $ G$ has a conjugacy class of order $ p^i$, $ i\geq 1$, but also $ 1\in H$, so $ H\cap Z(G)$ cannot be trivial.

For the second question, we're assuming that $ b$ has the same order in $ G$ as $ \overline{b}$ does in $ G/\langle a\rangle$.  So if $ b^i\in \langle a\rangle$, then $ \overline{b}^i \equal{} \overline{1}$, hence $ b^i \equal{} 1$.

Why is a subgroup of order $ p^i$ contained in a subgroup of order $ p^{i \plus{} 1}$?

Ah, now I see LydianRain, thanks!  :)

This is the theorem I used:
Let $ G$ be a $ p$-group. Every subgroup of $ G$ with order $ p^i$ is contained in a subgroup of order $ p^{i\plus{}1}$ in $ G$. ($ i< n$)

You can prove this by a clever induction, if you want I can post the proof later, it's quite not that short, and I will need some lemma's. Or you could try to prove this yourself! If you do, I'm very interested in seeing your proof.

This theorem is very important when investigating the subgroups of a finite group $ G$. Lagrange's theorem says that every subgroup has an order which divides the order of $ G$, but what about the converse? Thanks to Sylow's theorem and the above theorem, we know that for every prime power which is a divisor of $ \left|G\right|$, there exists a subgroup of that order.


parameterization
vector
geometry proposed
geometry

In a convex quadrilateral $ABCD$ : ${AC}\cap{BD}={O}$. If $X$ and $Y$ are the midpoints of AB and CD and
${XO}\perp{CD}$, ${YO}\perp{AB}$ then $ABCD$ is cyclic.

If $S=OY\cap AB,\ T=OX\cap CD$, from the points $O,T,Y$ and the angle $\angle COD$ we can reconstruct the triangle $OCD$ (there is only one such triangle having these parameters), and from $O,S,X$ and the angle $\angle BOA$ we can reconstruct the triangle $OBA$. Since $OTY,OSX$ are similar (they are right triangles with one more pair of equal angles, $\angle YOT=\angle XOS$) and $\angle COD=\angle BOA$, we notice that a triangle similar to $COD$ has the parameters $S,X,\angle \angle BOA$ of $OBA$, which means that $OCD,OBA$ are similar, which is what we want.

I think we also have $AC\perp BD$.

Clearly, we need the following lemma.

[b]Lemma.[/b] Let $ABC$ and $A'B'C'$ be triangles, $M$ and $M'$ be the midpoints of the sides $BC$ and $B'C'$ respectively. It is known that $\angle{BAC}=\angle{B'A'C'}$ and $\angle{AMB}=\angle{A'M'B'}$. Then the triangles $ABC$ and $A'B'C'$ are similar.

Indeed, suppose the Lemma is proven. Let $XO$ meets $CD$ at $P$ and $YO$ meets $AB$ at $Q$. Then the quadrilateral $XQPY$ is cyclic, since $\angle{XPY}=\angle{YQX}=90^{\circ}$. But then $\angle{BXO}=\angle{CYO}$. And now it is easy to see that it is sufficient to apply our Lemma to the triangles $OBA$ and $OCD$. Then we will have $\angle{ABO}=\angle{DCO}$, the quadrilateral $ABCD$ is cyclic, and we are done.

So, it is sufficient to prove the Lemma.

[i]Proof of the Lemma.[/i] Without loss of generality, we may assume that $A=A'$, $B=B'$ and the points $A$, $C$ and $C'$ are collinear (and $C$ is between $A$ and $C'$). Since $\angle{AMB}=\angle{AM'B}$, the quadrilateral $AMM'B$ is cyclic. Then $\angle{BAM'}=\angle{BMM'}$. And since $MM' \parallel AC$, we have $\angle{BAM'}=\angle{BCC'}$. Therefore, $\angle{BAM'}=\angle{BCC'}$. But the angle $\angle{BCC'}$ is external for the triangle $ABC$. So, $\angle{BCC'}=\angle{BAC}+\angle{ABC}>\angle{BAC}>\angle{BAM'}$. But we have just claimed that $\angle{BCC'}=\angle{BAM'}$. Contradiction. The Lemma is completely proven. $\square$.

The problem is solved.

Ok, my solution uses vectors :D
Denote by $a=\overline{OA},b=\overline{OB},\ldots$. Now, the conditions are $(a+b)(c-d)=0$ and $(c+d)(a-b)=0$. By opening the brackets, we will get that $ac=bd$. Now, by the power of the points, it is cyclic  :lol:


Well, I usually get stuck in solving olympiad geometry, so how can I do this section successfully?
Er, what are the way to prove olympiad geometry such as 4 points on the same circle, 4 points are concurrent and what is the best way to compare angle in olympiad geometry?

For "4 points are concurrent", the best tactic is to read the problem again.
For the rest, I would begin with books like the ones listed on
http://www.cip.ifi.lmu.de/~grinberg/faq.html#books
and internet sources like the ones listed in
http://www.cip.ifi.lmu.de/~grinberg/Links.html
Particularly, http://cut-the-knot.com/geometry.shtml is recommended. Also, some topics in the Geometry Theorems & Formulas section of this forum are pretty instructive. Try solving some MathLinks problems, this always helps (both you and the forum).

  darij


induction
pigeonhole principle
MIT
college

Looking for source of modern Olympiad level induction and  pigeonhole problems. Not the easy exercises in ACOPS  Does anyone know of any good sources? :)

Kind of late to reply. But Problem Solving Strategies is pretty good in both of those areas.

mit ocw's seminar on problem solving.
http://ocw.mit.edu/OcwWeb/Mathematics/18-S34Fall-2007/Assignments/index.htm

and arthur engel's book, of course :)


topology
real analysis
real analysis unsolved

If x_n is a sequence in a metric space X such that d(x_n, x_n+1) <a_n for all n in N. If the sum of a_n is finite, show that x_n is Cauchy.

If $ \sum_{n\equal{}1}^{\infty}a_{n}<\infty$, then $ a_{n}\rightarrow 0$.  That is, for all $ \epsilon>0$, there is some $ N$ such that $ a_{n}<\epsilon$ for all $ n\geq N$.  So for any $ \epsilon>0$, if we choose $ n$ sufficiently large, $ d(x_{n},x_{n\plus{}1})<a_{n}<\epsilon$.

That is not a proof.

Moving to the analysis forum; Cauchy sequences fit better there than as general topology.

By the way, this is true. Its [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=305134]converse[/url] is false.

JRav: how can you estimate $ d(x_n,x_m)?$ And while you're at it, look at that link to the converse.

WLOG $ m<n$.
$ d(x_m,x_n)\leq d(x_m,x_{m\plus{}1})\plus{}\ldots\plus{}d(x_{n\minus{}1},x_n)\leq a_m\plus{}a_{m\plus{}1}\plus{}\ldots\plus{}a_{n\minus{}1}\leq\sum_{i\equal{}m}^\infty a_i$ where the first step is the triangle inequality.
Since $ \sum_{i\equal{}0}^\infty a_i$ convgerges, its tail sums $ \sum_{i\equal{}m}^\infty a_i$ must go to 0, so we are done.


calculus
integration
trigonometry
function
complex analysis
calculus computations

Calculate:$ \int_{\minus{}\pi}^{\pi} \frac{1}{2\pi} \frac{\sinh a}{\cosh a \minus{} \cos x} e^{\minus{} i n x} dx$
$ a$ is a positive real number, $ n$ is an integer.

It can't solve this integral, hopefully you can. But I do know the answer: [hide="answer"]\[ e^{\minus{}|n|a}\][/hide]

I always check some things that predict the behavior of Fourier coefficients. The function we are dealing with is $ \frac{c}{d\minus{}\cos x}$ for $ d>|c|.$

(1) The function is real and even. Hence the complex Fourier coefficients will be real and even. (If we were doing this is real form, all of the sine coefficients $ b_n$ would be zero.)

(2) The function is $ C^{\infty}$ on $ \mathbb{T}.$ Hence I expect the Fourier coefficients to tend to zero as $ |n|\to\infty$ faster than any power of $ n.$

(3) Complex analysis contains tools for computing this integral.

(4) But you can use Fourier inversion. You have an unsubstantiated claim that $ \widehat{f}(n)\equal{}e^{\minus{}|n|a}.$ Try verifying that via the "synthesis equation":

$ \sum_{n\equal{}\minus{}\infty}^{\infty}e^{\minus{}|n|a}e^{inx}\equal{}?$ In particular, is it $ \frac{\sinh a}{\cosh a\minus{}\cos x}\,?$

This involves the summing of geometric series.


We will remove all posts that discuss the test before this time, which is the time set by the AMC after which it is OK to discuss the tests.

Please note, this includes comments regarding the difficulty of the test or how you think you did.  Please wait until Thursday to discuss the test.

We will also be hosting a Math Jam on Saturday to discuss the test.

Would one of the instructors mind posting the answers to the AMC B after the set time has passed?


counting
distinguishability
combinatorics unsolved
combinatorics

[quote="elchip"]I have no background in combinatorics, but a problem of this nature has come up in my workplace.  So, this may be a very simple question.  (It probably is.)

I have X distinct objects and I want to put them into N indistinct buckets.

The order in which I put objects in a particular bucket doesn't matter.

But putting all of the objects into bucket 0 is exactly the same as putting them all into bucket 1... so those two situations shouldn't be counted individually.

It's okay for buckets to be empty so long all the objects are in some bucket or another.

How many ways can I put the objects into the buckets?

Let me know if I'm not clear enough.

Thanks in advance.

EDIT:

I guess this isn't as easy as I assumed?  :)

If you want to see a discrete amount of numbers, we're talking about 512 different objects and 4 indistinguishable buckets.[/quote]
$\sum_{j=1}^N \sum_{i=0}^j \frac{(-1)^i}{j!}{i \choose j}(j-i)^X$

Thanks... but in your formula, what are N and X?  Which is the buckets, and which is the objects?
And is there a source where you got that formula?

Do you have a source where I can find that equation?

do you want a proof or a book that these are writen in it??

Actually, don't worry about it.  I just read about Stirling numbers and now I see what you are doing.

Thanks for pointing me in the right direction!


Prove that there exists a real number such that the terms of the infinite sequence cos x, cos 2x, cos 4x, cos 8x, cos 16x, ... are all negative.

Simple, take $x = 120^\circ$.  Then each term is -1/2.

i used search and found there were 2 more same topics about this problem! hi welcome to MathLinks/AoPS  :welcome:  :welcomeani:
since you are new, you may not be familiar with forum guidelines. whenever you can, please use the Search feature because more than likely most competition problems/hard exercises are already discussed in the forum instead of creating a new topic to reduce SPAM. thank you and enjoy your problem solving discussion with us


function
integration
probability
logarithms
calculus
real analysis
real analysis unsolved

If $X,Y,Z$ are random variables uniformly distributed in [0,1] then so is $(XY)^Z$.

Enjoy, its not too hard.

I dont know if its necessary, but here's a hint:[hide]
IF $W$ is a uniform random variable on $[0,1]$, you might prove that $E[f(W)] = E[f((XY)^Z)]$ for a certain collection of functions and prove that its enough.[/hide]

Well, that was rather a large hint.

Let $W=(XY)^Z.$ For each nonnegative integer $k,$

$E(W^k)=\int_0^1\left(\int_0^1x^{kz}dx\right)\left( \int_0^1y^{kz}dy\right) dz$

$=\int_0^1\frac{dz}{(kz+1)^2}=-\frac1{k(k+1)}+\frac1k=\frac1{k+1}.$

Thus all of the moments of this distribution are the same as the moments of the uniform (0,1) distribution. For well-behaved distributions - and this one is bounded, so it is exceedingly well-behaved - the moments do determine the distribution.

While Kent's solution is, probably, the most elegant possible, the problem allows a direct approach too. We need to show that $P\{W<t\}=t$ for $t\in (0,1)$. This probability can be written as $\int_0^1 P\{XY<t^{1/z}\}\,dz$. Now, it is not hard to see that $P\{XY<\delta\}=\delta(1-\ln\delta)$ ($0<\delta<1$) whence the integral becomes $\int_0^1 t^{1/z}\left(1-\frac 1z \ln t\right)\,dz=\int_0^1 \left(\frac d{dz}\, zt^{1/z}\right)\,dz =t$. This is not harder than some of kunny's integrals, so it is strange that nobody solved it. Probably the probabilistic language was scaring everyone away  :? .


[color=red]Αγαπητοί φίλοι , [/color]

 με χαρά σας ανακοινώνω ότι στο φόρουμ θα δημιουργηθούν μερικά stickies- είδiκά topic - στα οποία θα βάζουμε συγκεκριμένα θέματα. Ήδη έφτιαξα τρία για τις τρεις τάξεις του Λυκείου. Με αυτόν τον τρόπο θα μπορούμε  , μπαίνοντας στην κοινότητά μας  , να βλέπουμε μπροστά μας τα πιο σημαντικά από τα πεδία που μας ενδιαφέρουν.
 Θέλω να ακούσω την άποψή σας : Ποια άλλα πεδία νομίζετε ότι χρειάζονται αυτή τη στιγμή ειδικό χώρο ; Για τις ολυμπιάδες , είπαμε , θα προσπαθήσω να πείσω το Vornicu να μας ανοίξει υποφόρουμ. ΑΝ αυτό δεν είναι εύκολο , τότε θα κάνω το ίδιο ''κόλπο'' με τα stickies.
   Βλέπω ότι το topic με τα '' αστειάκια'' έχει αρέσει. Το ίδιο και αυτό με τα ''μέλη ''. Σκέφτομαι λοιπόν να έχουμε 10 περίπου μόνιμα  topic στην αρχή της ιστοσιλίδας μας - όχι παραπάνω - για να επικοινωνούμε εύκολα και να μην ψάχνουμε να βρούμε που ...χάθηκε κάποιο μήνυμα που είχαμε δει... κάποτε.
 Στο χώρο κάτω από τα stickies θα μπορούμε έτσι να στέλνουμε ερωτήματα με άμεσο ή εντελώς επίκαιρο περιεχόμενο , ώστε να είναι εύκολα ορατά και να απαντώνται αμέσως.
 Δε γράφω περισσότερα. Ακούω τις σκέψεις σας.

  Μπάμπης

η γνώμη μου είναι ότι τα stickies ειναι πολύ καλή ιδέα μόνο που προτείνω να διατηρήσουμε την μορφή ώς έχει εκτός από μια ταξινόμηση των τόπικ αστειακια κι μέλη του φόρουμ το οποί να είναι μια ανανεώσιμη λίστα κι όχι δισεκατομμύρια πόστ
αυτά

[quote="Protonios"]Γιατί να μην έχουμε ένα για απορίες ώστε να μην είναι όλες διασκορπισμένες στο φόρουμ και κατά τα άλλα ένα για γεωμετρία, ένα για άλγεβρα, ένα για θεωρία αριθμών, ένα για συνδυαστική και ένα ξεχωριστό για Calculus? :wink:[/quote]

 Θα περιμένουμε μήπως μιας φτιάξει ο Βορνίκου τα υποφόρουμ. ΑΝ όχι , αυτό θα κάνουμε !
 Συνεχίστε με σκέψεις και διάλογο. Στο τέλος θα φανεί τι ακριβώς θέλουμε !

                      Μπάμπης


Determine if the functions $Q,P: R\rightarrow{R}$ exist such that $(xy)^{2005}+1=Q(x)P(y)$

Ok, so obviousally no constant function satisfies it. Therefore, one of $P(x),Q(x)$ has a root in $\mathbb{C}$ call it $\alpha$

Therfore, subbing this into, WLOG say $Q(x)$, we get $(\alpha y)^{2005}=-1$

This must hold for all $y$.

$y=1$ therefore $\alpha ^{2005}=-1$
$y=2$ therefore $2^{2005} {\alpha ^{2005}}=-1
$ subbing for alpha we get $2^{2005}=1$

Contradiction!!! Therefore no polynomials exist.


geometry
perimeter
number theory solved
number theory

Let n>=2 a natural number.Find all the triangles with integer sides and of perimeter n.

just how many there are, or also some method to write which ones?

I think by "Find all triangles..." means to [u]count[/u] how many triangles with perimeter n have integer sides, with n arbitrarily fixed. Because if we let n=1000000 (one million), seriously this page wouldn't even fit 10% of the solution set. I think it is counting, not the actual enumeration of all the triangles with such conditions.

Im not sure if I'm entirely right, but here it goes:

for a give n, we seek 3 positive integers a,b,c s.t. a+b+c=n and a+b>c, a+c>b, and b+c>a.

WLOG, have a \geq b \geq c

now, a+b+c=n and b+c-a = n-2a, but since b+c>a, n-2a>0 or a < floor(n/2), also a \geq ceiling(n/3) for otherwise a+b+c<n.

then b \leq a, and b \geq ceiling((n-a)/2) for otherwise b+c<n-a

finally, c is completely determined by a and b, c = n-a-b.  And by the way we selected a, the sides will be lengths of a triangle and satisfy the triangle inequality.

so for a given n:

ceiling(n/3) \leq a < floor(n/2)
ceiling((n-a)/2) \leq b \leq a
c = n-a-b

gives all possible degenerate triangles.


floor function
logarithms
combinatorics unsolved
combinatorics

Let $S$ be a set with $n$ elements. Take a positive integer $k$. Let $A_1, A_2, \ldots, A_k$ be any distinct subsets of $S$. For each $i$ take $B_i = A_i$ or $B_i = S - A_i$. Find the smallest $k$ such that we can always choose $B_i$ so that $\bigcup_{i=1}^k B_i = S$, no matter what the subsets $A_i$ are.

this is rather straightforward. for each $x\in S$ label $x$ by the label of the 'region' it lies in, i.e. label $x$ by a $k-\text{tuple}\ (B_1,B_2,\ldots,B_k)$ where $B_i=A_i$ if $x\in A_i$ and $B_i=A_i^C$ otherwise. writing in terms of complements what the problem asks for is if some 'region'(think venn-diagramatically) is empty. from the above it follows that for there are $2^k$ labels possible and so we have $2^k\geq n$ so that $k\geq\lfloor\log_2(n)+1\rfloor$.

conversely it is easy to construct such sets where this bound is achieved(throw in one point in each of the regions created by a venn-diagram).


trigonometry
algebra
polynomial
geometry unsolved
geometry

See http://www.mathlinks.ro/Forum/viewtopic.php?t=59437

$(3)\ \ 8\cos ^3 u-4\cos ^2 u -4\cos u +1=0\Longleftrightarrow$
$\sin u(8\cos ^3 u-4\cos ^2 u-4\cos u+1)=0\Longrightarrow$
$4\cos ^2 u\sin 2u-2\cos u\sin 2u-2\sin 2u+\sin u=0\Longleftrightarrow$
$2\cos u(\sin 3u+\sin u)-\sin 3u-\sin u-2\sin 2u+\sin u=0\Longleftrightarrow$
$\sin 4u+\sin 2u+\sin 2u-\sin 3u-\sin u-2\sin 2u+\sin u=0\Longleftrightarrow$
$\sin 4u=\sin 3u\Longleftrightarrow u\in \left \{ (2k+1)\cdot \frac{\pi}{7} | k\in Z\right\}\Longleftrightarrow$
$u\in \{ \frac{\pi}{7},\frac{3\pi}{7}\}\Longleftrightarrow u=\frac{3\pi}{7}\Longrightarrow A=\frac{\pi}{7}.$
[u]Remark.[/u] $BC=1=AD<AB\Longrightarrow BC<AB\Longrightarrow$
$A<C\Longrightarrow C>\frac{\pi}{3}\Longrightarrow u>\frac{\pi}{3}$.

$z^7 + 1 = (z + 1)(z^6 - z^5 + z^4 - z^3 + z^2 - z + 1) = 0$

has 7 roots $z = \cos \frac{(2k + 1) \pi}{7} + \text i \sin \frac{(2k + 1) \pi}{7},\ \ k = 1, 2, ..., 7$

(vertices of a regular heptagon on a unit circle). Factoring out the root z = -1 and substituting $v = z + \frac 1 z$ into the remaining 6th degree polynomial divided by $z^3 \neq 0$,

$v^2 = z^2 + 2 + \frac{1}{z^2}$

$v^3 = z^3 + 3z + \frac 3 z + \frac{1}{z^3} = z^3 + 3v + \frac{1}{z^3}$

$z^3 - z^2 + z - 1 + \frac 1 z -\frac{1}{z^2} + \frac{1}{z^3} =$

$= v^3 - 3v - v^2 + 2 + v - 1 = v^3 - v^2 - 2v + 1 = 0$

This polynomial has 3 different real roots (k = 1, 2, 3):

$v = \cos \frac{(2k + 1) \pi}{7} + \text i \sin \frac{(2k + 1) \pi}{7} + \frac{1}{\cos \frac{(2k + 1) \pi}{7} + \text i \sin \frac{(2k + 1) \pi}{7}} =$

${= \cos \frac{(2k + 1) \pi}{7} + \text i \sin \frac{(2k + 1)\pi}{7} + \cos \frac{(2k + 1)\pi}{7} - \text i \sin \frac{(2k + 1) \pi}{7}} =$

$= 2 \cos \frac{(2k + 1) \pi}{7}$

and consequently, if we put $v = 2 \cos u$, the polynomial

$8 \cos^3 u - 4 \cos^2 u - 4 \cos u + 1 = 0$

has 3 different roots in $(0, \pi)$ equal to $u = \frac{(2k + 1) \pi}{7}$, k = 1, 2, 3.


invariant
group theory
advanced fields
advanced fields unsolved

let $ \Gamma$ be a graph and $ G$ a group which acts on $ \Gamma$, i.e. we have a group homomorphism $ G \to Aut(\Gamma)$. an inversion consists of a $ g \in G$ and an edge $ y$ in $ \Gamma$ such that $ gy \equal{} \overline{y}$, where $ \overline{y}$ denotes the "reversed" edge of $ y$ ($ \overline{y}$ is included in serre's definition of a graph). clearly if $ G$ preserves an orientation of $ \Gamma$, this cannot exist. but in serre, trees, 3.1. the converse is claimed: if there are no inversion, then there is an orientation of $ \Gamma$ which is preserved by $ G$. I have no idea why this should be true.

I remember wondering about that too. This argument seems to work:

A [i]partial orientation[/i] will be a set of edges which contains at most one of each pair $ y,\bar y$. We say that a set of partial orientations is compatible if the union of the partial orientations in the set is a partial orientation. We can order the collection of all partial orientations by inclusion. Now look at those partial orientations $ P$ having the property (call it property $ (*)$) that $ \{g(P)\ |\ g\in G\}$ is compatible. Clearly, this collection satisfies the hypotheses of Zorn's Lemma, so there is a maximal partial orientation $ P$ with the desired property. We want to prove that $ P$ contains precisely one edge from each pair of opposite edges.

Assume there is an edge $ y$ such that neither $ y$ nor $ \bar y$ belong to $ P$. Let $ g_1,g_2$ be two elements of $ G$. If $ g_1y\in P$ and $ \overline{g_2y}\in P$, then $ g\equal{}g_1g_2^{\minus{}1}$ takes an edge in $ P$ onto the opposite of an edge in $ P$, contradicting the fact that $ P,g(P)$ are compatible. It follows that either all $ gy,\ g\in G$ are compatible with $ P$, or no edges of the form $ gy$ belong to $ P$. In the first case $ y$ can be added to $ P$, whereas in the second case $ \bar y$ can be added to $ P$ so as to preserve $ (*)$. In both cases the maximality of $ P$ is contradicted, and we're done.

thank you  :). to show that $ P \cup \{y\}$ resp. $ P \cup \{\overline{y}\}$ indeed satisfies $ (*)$, we need that there are no inversions.

I'm wondering why serre states this right after the definition of orientation preserving group actions, without any hint that it could be nontrivial and zorn's lemma is needed ..

aaah! (thx clara)

define a graph $ G \backslash \Gamma$ in the obvious way: there are no edges $ [y]$ in $ G \backslash \Gamma$ with $ \overline{[y]} \equal{} [y]$ exactly says that $ G$ acts without inversions on $ \Gamma$, and the other graph properties are clear. now every graph has an orientation (choose a set of representatives for the equivalence relation "equal or opposite"). so pick one for $ G \backslash \Gamma$; this clearly induces a $ G$-invariant orientation on $ \Gamma$.