Problems Fibonacci:
In 1225, Emperor Fre - Topic - ric II with some mathematicians have tried financial Fibonacci with the problem: "Find rational numbers x such that x and x ² + 5 ² - 5 are average means of rational numbers. "
After a while, Fibonacci has found a number that is 41/12. Indeed:
                      (41/12) ² + 5 = (1681/144) + 5 = 2401/144 = (49/12) ²
                      (41/12) ² - 5 = (1681/144) - 5 = 961/144 = (31/12) ²
Quite many people have tried to take the solution of this problem but not. Now, after a long time to learn, you have the solution but not sure how the award is true or not. Which, by themselves, this problem has a variety of experience with the original experiment is testing the  41/12 - Fibonacci has found. Indeed, for example, take a test that is more "3344161/1494696":
         (3344161/1494696) ² + 5 = (11183412793921/2234116132416) + (11170580662080/2234116132416)
                                               = (4728001/1494696) ²
        (3344161/1494696) ² - 5 = (11183412793921/2234116132416) - (11170580662080/2234116132416)
                                              = 12832131841/2234116132416
                                              = (113279/1494696) ²
1 VITAMINS FOR USE IN PROBLEMS:
       a. Fermat equation model:
Fermat equation like is the equation of the form x ⁿ + y = 2z ⁿ ⁿ with n ≥ 2 and x, y integer, x ≠ y, but here we only consider the case n = 2 ie when:
                                   x ² + y ² = 2z ²
and we will only consider the experience of its original News of the experiment so that (x, y, z) = 1.
We can see: x, y not the same even. Suppose x odd. Inferred y odd. So two sides left the residue mod 4 and z odd. If p │ x, p │ y => p = 1. So (x, y) = 1. Suppose x> y. Put u = (x + y) / 2; v = (x - y) / 2 then we obtain: u ² + v ² = z ². From (x, y) = 1, x = u + v, y = u - v it follows that (u, v) = 1. So (u, v, z) is the three Pitago original. So:
                                u = m ² - n ² (= 2mn)
                                v = 2mn (= m ² - n ²)
                               z = m + n ² ²
where m, n are positive integers m> n, (m, n) = 1 and m, n, or other parity:
                               x = m ² - n ² + 2mn
                              z = m + n ² ²
Because the role of equality of x, y should we conclude that experience:
                             x = m ² - n ² + 2mn
                             y = n ² - ² m + 2mn
                             z = n ² + m ²
where m, n are positive integers coprime and m, n different parity.
b. Phuong Trinh PYTAGO:
  Pytago equation is the equation of the form:
                              x ² + y ² = z ² (1)
with x, y, z integer, and here we will only consider the original experience that is (x, y, z) = 1.
1. Easy to find x, y and z different parity odd.
Suppose x even, y odd. Have:
                         x ² = (z + y) (z - y)
                                    ó (x / 2) ² = [(z + y) / 2] [(z - y) / 2] (2)
By z, y are odd elements together so [(z + y) / 2]; [(z-y) / 2] are primes together (you read self-proven). Combined with (2) we have [(z + y) / 2]; [(z-y) / 2] are the main method.
Set [(z + y) / 2 = m ² [(z-y) / 2] = n ² with m, n are positive integers:
                       z = m + n ² ² ² y = m - n ²
By y, z be odd m, n of different parity.
Do (m ², n ²) = 1 so (m, n) = 1
So:
                      x = 2mn
                     y = m ² - n ²
                     z = m + n ² ²
with (m, n) = 1; m, n different parity and m> n.
Island again, the number of visible (x, y, z) satisfies the above (1)
c. VITAMIN TOPIC III:
Equation (n ^ 4) - (m ^ 4) = 5γ ²
invisible experiment with n, m different parity.
Proof:
  We have: m, n other odd parity it follows that γ and conspicuous m, n not divisible by 5.
          Consider two cases:
· School hopn even, m odd:
                                           (m ^ 4) ≡ 1 (mod 16)
(n ^ 4) ≡ 0 (mod 16) which
                                               5γ ² ≡ 5, 4, 13 (mod 16)
so (m ^ 4) + 5γ ² ≡ 6, 5, 14 (mod 16). Vô lí!
· Case n odd, m even:
                                           (n ^ 4) ≡ 1 (mod 16)
(m ^ 4) ≡ 0 (mod 16) which
                                          5γ ² ≡ 5, 4, 13 (mod 16)
so (n ^ 4) - 5γ ² ≡ 12, 13, 14 (mod 16). Vô lí!
So the invisible test equation.
d. VITAMIN TOPIC 4:
        For a = a1. A1 k ² with no estimates of the numbers (except one), find raw experience of the equation:
                                       a.x ² + y ² = z ²
Here we only consider the case a1 retail raw experience of the equation is calculated by the formula:
                                     x = mn
  y = (u.m ² - en mi) / 2
  z = (u.m en mi ² +) / 2
with uv = a1 (u, v raw), (m, n) = 1, mn odd.
2. Solution:
First, let us remind the subject mathematics: "Find some rational ф such that ф ² + 5 is the square of a rational number, ф ² - 5 is the square of the rational numbers.
Solution
Suppose ф = a ² + 5 ² ф ² - 5 ² = b with a, b rational. If so, will survive taiu, v, z, t such that:
            Ф = u / v: a = t / v b = z / v
We have: (u / v) ² + 5 = (t / v) ²
                            (z / v) ² + 5 = (u / v) ²
                               5v ² = u ² + t (2)
                              5v ² + z ² = u ² (3)
Get (2) - (3) we obtain: u ² - z ² = t ² - u ²
                                 or z = 2u ² + t ² ² (4)
World (3) to (2) we have: z ² + 10V ² = t ² (1)
(4) is the equation with m = 2 Fecma style. Applying lemma 1, z, t, u of the form:
                            z = m ² - n ² + 2mn
                            t = n ² - ² m + 2mn
                             u = n ² + m ²
     where m, n are positive integers coprime and m, n different parity.
             (1) ó ² = 10V (t + z) (t - z)
= (M ² - n ² + n ² + 2mn - m ² + 2mn) (n ² - ² m + 2mn - n ² - ² m - 2mn
= 4mn (2n ² ² m -2)
= 8mn (n ² - m ²)
Follows that (1) ó ² 5v = 4mn (n ² - m ²)
Set v = 2.j then: 5J ² = mn (n ² - m ²)
To prove m, n, n ² - ² m a pair of elements together. Therefore, if one of them divisible by five, the rest will not divisible by 5.
n ² - m ² = 5μ ² is an invisible process experience. Indeed: m, n other odd parity deduced μ. N ² - m = 5μ ² ² ² = mn is from 5v (n ² - m ²) have n, m the method. Set n = n `²` ² m = m then we obtain the equation: (n ^ 4) - (m ^ 4) ² = 5μ ² (1 *). Applying lemma (3) we have equation (1 *) and experiment.
Since n ² - m ² = 5μ ² is an invisible process n or m test should be divisible by 5. Then, from 5J ² = mn (n ² - m ²) is the n ² - m ² = α or α ² ² ² + m = n ². Applying lemma 2 we have n, m of the form:
                   n = θ + ψ ² ²
                  m = 2θψ
Here invite you to try the rest. I wish you luck to find a solution short of their better solution.

This post is even better than a Samuel Becket play! I thoroughly enjoyed its absurd appeal to the limits of human communication through language than the mathematics that may lie (deeply) buried underneath ...


modular arithmetic
number theory unsolved
number theory

[quote="qntty"]If a,b and c are positive integers, find all integer values of $ \frac {a^2 \plus{} b^2 \plus{} c^2}{abc}$[/quote]

Very nice problem. I spent a lot of time on it.

[hide="My solution"]
The set $ S$ of possible integer values for $ \frac{a^2\plus{}b^2\plus{}c^2}{abc}$ when $ a,b,c\in\mathbb N$ is$ \{1,3\}$ :

1) $ a\equal{}b\equal{}c\equal{}3$ $ \implies$ $ 1\in S$

2) $ a\equal{}b\equal{}c\equal{}1$ $ \implies$ $ 3\in S$

3) $ k\notin S$ $ \forall k>3$ :
Let $ (a,b,c)\in\mathbb N^3$ such that $ a^2\plus{}b^2\plus{}c^2\equal{}kabc$ with $ k>3$. Wlog say $ a\geq b\geq c$
If $ b\equal{}1$, then $ c\equal{}1$ and we get $ a^2\plus{}2\equal{}ka$ and so $ a|2$ and so either $ a\equal{}1$, either $ a\equal{}2$. In both cases, we get $ k\equal{}3$

So, consider now $ b\geq 2$. We have $ k\equal{}\frac{a}{bc}\plus{}\frac{b}{ac}\plus{}\frac{c}{ab}$ with $ \frac{b}{ac}\leq 1$ and $ \frac{c}{ab}\leq \frac{1}{b}\leq \frac{1}{2}$

So $ \frac{a}{bc}<k\leq \frac{a}{bc}\plus{}\frac{3}{2}$
So $ k>\frac{a}{bc}\geq k\minus{}\frac{3}{2}> \frac{k}{2}$ and so $ kbc > a > \frac{1}{2}kbc$ and so $ a>kbc\minus{}a>0$

Let then $ x\equal{}kbc \minus{} a$. We know that $ a>x>0$
We have $ (kbc\minus{}x)^2\plus{}b^2\plus{}c^2\equal{}k(kbc\minus{}x)bc$ and so $ x^2\plus{}b^2\plus{}c^2\equal{}kxbc$

And so, if $ (a,b,c)$ is a solution for $ k$, $ (x,b,c)$ (with $ a>x>0$) is also a solution.
So, thru infinite descent, we have no solution for $ k>3$
Q.E.D.

4) $ 2\notin S$ :
Let $ (a,b,c)\in\mathbb N^3$ such that $ a^2\plus{}b^2\plus{}c^2\equal{}2abc$
Since RHS is even, LHS is even and so at least one of the three integers a, b or c is even. Wlog say $ c\equal{}2c'$
Then we have $ a^2\plus{}b^2\plus{}4c'^2\equal{}4abc'$
So $ a^2\plus{}b^2\equal{}0\pmod 4$ and this is possible only if both are even.
So $ a\equal{}2a'$ and $ b\equal{}2b'$ and we get $ a'^2\plus{}b'^2\plus{}c'^2\equal{}4a'b'c'$ but we know (from point 3 above) that $ 4\notin S$. So $ 2\notin S$
Q.E.D.
[/hide]

[quote="pco"][quote="qntty"]If a,b and c are positive integers, find all integer values of $ \frac {a^2 \plus{} b^2 \plus{} c^2}{abc}$[/quote]

Very nice problem. I spent a lot of time on it.

[hide="My solution"]
The set $ S$ of possible integer values for $ \frac {a^2 \plus{} b^2 \plus{} c^2}{abc}$ when $ a,b,c\in\mathbb N$ is$ \{1,3\}$ :

1) $ a \equal{} b \equal{} c \equal{} 3$ $ \implies$ $ 1\in S$

2) $ a \equal{} b \equal{} c \equal{} 1$ $ \implies$ $ 3\in S$

3) $ k\notin S$ $ \forall k > 3$ :
Let $ (a,b,c)\in\mathbb N^3$ such that $ a^2 \plus{} b^2 \plus{} c^2 \equal{} kabc$ with $ k > 3$. Wlog say $ a\geq b\geq c$
If $ b \equal{} 1$, then $ c \equal{} 1$ and we get $ a^2 \plus{} 2 \equal{} ka$ and so $ a|2$ and so either $ a \equal{} 1$, either $ a \equal{} 2$. In both cases, we get $ k \equal{} 3$

So, consider now $ b\geq 2$. We have $ k \equal{} \frac {a}{bc} \plus{} \frac {b}{ac} \plus{} \frac {c}{ab}$ with $ \frac {b}{ac}\leq 1$ and $ \frac {c}{ab}\leq \frac {1}{b}\leq \frac {1}{2}$

So $ \frac {a}{bc} < k\leq \frac {a}{bc} \plus{} \frac {3}{2}$
So $ k > \frac {a}{bc}\geq k \minus{} \frac {3}{2} > \frac {k}{2}$ and so $ kbc > a > \frac {1}{2}kbc$ and so $ a > kbc \minus{} a > 0$

Let then $ x \equal{} kbc \minus{} a$. We know that $ a > x > 0$
We have $ (kbc \minus{} x)^2 \plus{} b^2 \plus{} c^2 \equal{} k(kbc \minus{} x)bc$ and so $ x^2 \plus{} b^2 \plus{} c^2 \equal{} kxbc$

And so, if $ (a,b,c)$ is a solution for $ k$, $ (x,b,c)$ (with $ a > x > 0$) is also a solution.
So, thru infinite descent, we have no solution for $ k > 3$
Q.E.D.

4) $ 2\notin S$ :
Let $ (a,b,c)\in\mathbb N^3$ such that $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 2abc$
Since RHS is even, LHS is even and so at least one of the three integers a, b or c is even. Wlog say $ c \equal{} 2c'$
Then we have $ a^2 \plus{} b^2 \plus{} 4c'^2 \equal{} 4abc'$
So $ a^2 \plus{} b^2 \equal{} 0\pmod 4$ and this is possible only if both are even.
So $ a \equal{} 2a'$ and $ b \equal{} 2b'$ and we get $ a'^2 \plus{} b'^2 \plus{} c'^2 \equal{} 4a'b'c'$ but we know (from point 3 above) that $ 4\notin S$. So $ 2\notin S$
Q.E.D.
[/hide][/quote]

Very nice;pco  :P


how would you do this question..iv tried everything but i just dont seem to be getting the right answer.

Calculate and enter the mass of Co(NO3)2.6H2O that will be required to prepare 50.00 mL of a 0.2500 M stock solution of Co2+.

Calculate number of moles required, then convert moles to mass.

ya...i get the number of Co moles required (0.0125 mol), but i need the mass of the whole solution. i used the mole ratio to get the moles of the rest of the elements, i add all of them up (0.1125 mol) and then convert to grams, but i dont get the right answer...what am i doing wrong? (the answer i get is 32.73 grams)

I got $3.6375 grams$, don't know if that's right or not. If it is, I'll post how I did it tomorrow, but I have to go to bed now. Once I got the moles, .0125, I multiplied that by the molar mass of the solution you were given to get my answer. I got the molar mass to be 291 grams.

okay thanks...thats the right answer. i was just making a stupid mistake...thanks for your help :)

You're very welcome :D Glad I could help.


There three natural numbers a,b,c. Any two of them are multiplied. The product is divided by the remaining number where you get a remainder of 1.
Determine all solutions (a,b,c) !

a|bc-1, b|ac-1 and c|ab-1. If we multiply them we get abc|(ab-1)(bc-1)(ca-1) and if we subtract from the right part all the obvious multiples of abc we get abc|ab+bc+ca-1. Let's assume (ab+bc+ca-1)/abc=k. Then 1/a+1/b+1/c-1/abc=k, so k<1/a+1/b+1/c<=3 so k can only be 1 or 2.

Case 1  k=2 This means that 1/a+1/b+1/c-1/abc=2. Let's assume a<=b<=c. If a>=2 then 1/a+1/b+1/c-1/abc < 1/a+1/b+1/c <= 3/2 < 2, so false. Then a=1. We now have 1/b+1/c-1/bc=1. If b>=2 then 1/b+1/c-1/bc < 1/b+1/c <= 1 so false, so b=1. Now c can be anything, so one type of soln is (1, 1, c) and the permutations. We can see that this checks a|bc-1 and the others.

Case 2 k=1 We have 1/a+1/b+1/c-1/abc=1. Again a<=b<=c. if a>=3 then 1/a+1/b+1/c-1/abc < 1/a+1/b+1/c <= 1, false, so a is 1 or 2. If a=1 then 1/b+1/c-1/bc=0, so b+c=1, false, so a=2. We now have 1/b+1/c-1/2bc = 1/2 iff 2/b+2/c-1/bc=1. If b>=4 then 2/b+2/c-1/bc < 2/b+2/c <=1, false, so b is 2 or 3 (it has to be >= a=2). If b is 2 then 2/c-1/2c=0, so c=0, false, so b=3, which leaves 2/c-1/3c=1/3 iff 6/c-1/c=1 iff c=5, so the soln is (2, 3, 5). We can see that the solns (2, 3, 5) and the perutations also check a|bc-1 and the others, so this is another type of soln.

The ordered solns: (1, 1, x) where x is a positive integer and (2, 3, 5).

[quote] The ordered solns: (1, 1, x) where x is a positive integer and (2, 3, 5). [/quote]

Sorry your proof is wrong. Only one solution is correct: (2,3,5). The others solutions are (2,5,3), (3,2,5), (3,5,2), (5,2,3) and (5,3,2). So don't forget the permuations. Grobber you could have one more try !  :D

Well, I did say ORDERED solns, by that meaning a<=b<=c. And I did mention the permutations somewhere in that post. Besides, why doesn't 1, 1, x work?

1|1*x-1; 1|1*x-1 and x|1*1-1=0 since any number divides 0. What's the problem then?  :D

[quote] The ordered solns: (1, 1, x) where x is a positive integer and (2, 3, 5). [/quote]  :)

Sorry grobber but I only considered the line above. But I still think that (1, 1, x) is wrong. You can deduce the answer from my general remark:

Assume that two numbers, e.g. a and b, have the a common divisor g \geq  2, so g would also be a remainder of the quotient (ac)/b. Hence the numbers a,b,c considered in pairs can not have a common divisor. Moreover all three numbers must be greater than 1 or they at least one of the considered quotients would have a remainder 0.  :)

Ok, I can't understand anything, but did you look at the last line in my previous post? What is wrong there? I mean this line:

1|1*x-1; 1|1*x-1 and x|1*1-1=0 since any number divides 0. What's the problem then? Bye! :D

Hi grobber,

let me try explaining it ! WLOG assume a = 1, b = 1 and c = x. Hence considering the permutations isn't necessary anymore ! So let's check the three cases (remainder is abbreviated by R) :

(1) (ab)/c = 0 R 1 (2) (ac)/b = c R 0 (3) (bc)/a = c R 0

Well, I see what you mean, but remember that a and b are 1 and (mod 1) all residues are equal, so R 0= R 1 if we divide by 1! Besides, the problem you posted is equivalent to:
c|ab-1; b|ac-1; a|bc-1, and the triplet (1, 1, x) satisfies these.

One can write the equation as you have done it:

c|ab-1; b|ac-1; a|bc-1

but only after realizing my general remark ! So take your triplet (1,1,x) and check it with the problem's statement ! It won't match for the problem's statement but for your equation. But when considering the equation we already excluded this triplet from my general remark. Slightly confusing ! But that's the point !!!


I can not upload a file. Why? This is what i saw
Sorry, you have reached your maximum Upload Quota Limit of 10 MB

And how many files did you already upload¿
The upload quota is the sum of all files!

Actually you've uploaded 69 files, totaling 10.14 MB :) Anyway, I've increased your limit :P


trigonometry
calculus
integration
function
calculus computations

One answer is correct :lol: .  However, solutions may generally be written in the form $\frac{(p+1)\pi}{4}$ where $p=0,4,8,12...$.  I think this is correct anyway  :maybe:

[quote="prithwijit"]Is it $\frac{\pi}{4}$ and $\frac{3\pi}{4}$?[/quote]
Would you care to post your solution?

$\frac{\pi}4$ is correct but $\frac{3\pi}4,\frac{5\pi}4,$ and all answers with $n>\frac{\pi}2$ are absurd. The function is not integrable on any interval that contains $\frac{\pi}2$ (or ${\frac{3\pi}2},\dots$).

Thanks for clearing that up. I didn't think about the fact that $\tan{\theta}\to\infty$ as $\theta\to\frac{\pi}{2}$.  Is this the reason it isn't integrable?

First observe that $\left( \tan x \right)^\prime = \sec^{2}x$, so an antiderivative would be $\frac{\tan x}{e^{x}}$.

If you're talking about this $\tan x$, then you're right.

[quote="coffeym"]Find all values of $n$, where $n\in\Re$, for which

$e^{n}\int^{n}_{0}{\frac{\sec^{2}{\theta}-\tan{\theta}}{e^{\theta}}\,d\theta\,=1}$[/quote]

According to Maple v10, we have

$e^{n}\int^{n}_{0}{\frac{\sec^{2}{\theta}-\tan{\theta}}{e^{\theta}}\,d\theta\,=\frac{\sin(2n)}{\cos(2n)+1}}$

for $0\leq n\leq \frac{\pi}{2}$ and it diverges otherwise.


How can I make something like:

[img]http://img376.imageshack.us/img376/5272/asdasdfp1.png[/img]

(That's all, no text inside, just the image you see.)

Either use the boxes in Tables 240 & 241 of the [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=comprehensive]The Comprehensive LaTeX Symbol List[/url]or play with \shadowbox in the [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=fancybox]fancybox package[/url] sticking phantom text in it to get it to the size you want.

Thanks.

I used the code \SquareShadowB but doesn't work. The paper talks about the "ifsym." I used \usepackage{ifsym} but it didn't work.

This is weird: I realized that \SquareCastShadowBottomRight it's a better option to yield what I want, so I put \usepackage{bbding} in the preamble but when compiling, it does appear a "k" instead the symbol I want. Why this happens?

bbding is a Metafont and so is used for text. So don't use it in maths mode (if you have to do that use \text)
[code]\documentclass{article}
\usepackage{bbding}
\begin{document}
This is a bbding symbol \SquareCastShadowBottomRight
\end{document}[/code]

Okay, thanks, that works.

Another problem: since I'm using the ntheorem package, I need to compile in LaTeX => PS => PDF, but if I want to use \SquareCastShadowBottomRight, it requires to use LaTeX => PDF, so ntheorem package ain't gonna work. How can I use \SquareCastShadowBottomRight by using LaTeX => PS => PDF?

Use the pifont version instead (Table 242)
\usepackage{pifont}
...
\ding{113}

Thanks so much stevem!!

Just one more question: the size is a bit small, how can I increase it?

It's just a normal text character so you can change its font size to anything you like, or easier,  use \large, \Large, \LARGE, \huge or \Huge as you would for any other text.

Please don't spam and revive topics a month old.


Write out a sequence of 2n+1 +'s and -'s, e.g. +-++-. Now, for any element of the sequence, the "right count" is the number of +s minus the number of -s on the right of it, and the "left count" is defined similarly. Show that the number of elements for which the "right count" equals the "left count" is odd. (For example, in our sample sequence, the first, second, and third elements satisfy this property.)

[quote="paganinio"]where's the sample sequence?[/quote]
It's the one he gave, namely + - + + -.

Cute problem! :)

Here's my approach, although I'm sure there's a nicer solution out there.

[hide]

Instead of $+$'s consider $1$'s, and instead of $-$'s consider $-1$'s. We are now interested in those elements $x$ of our sequence for which the sum of the elements to the left is equal to the sum to the right of $x$. In other words, if $S$ is the sum of all the elements, we're interested in those elements $a_k$ s.t. $\sum_{i=1}^{k-1}a_i=\frac{S-\varepsilon}2,\ \sum_{i=1}^ka_i=\frac{S+\varepsilon}2$, where $\varepsilon$ is $\pm 1$. We may also assume WLOG $S>0$.

Here's the actual solution :): At first we have sum $0$. Then we start adding up the elements of the sequence successively (we compute $a_1,a_1+a_2,a_1+a_2+a_3,\ldots$). In the end we reach $S$, and we make steps of length $1$. This means that if we pass from $\frac{S+1}2$ to $\frac{S-1}2$ $k$ times, then we have to pass from $\frac{S-1}2$ to $\frac{S+1}2$ $k+1$ times, so there are $2k+1$ elements like those we're interested in.

[/hide]


Does anyone here use IRC? I have been having trouble finding some music videos and such which are supposedly on IRC. I am a beginner at using the program and was wondering if anyone could help. (PM me if possible)
Thanks!

Searching on your own for music on IRC is difficult... I use irc for it's intended purpose as a chat program. Not as a method to illegally download music.

I suggest this post be locked if the topic of discussion is going to be where to go to find illegal material.

If you want to get free music, you should go to [url]http://www.naxos.com[/url], which is definitely legal and has a fine selection.

[quote="Spoon"]I suggest this post be locked if the topic of discussion is going to be where to go to find illegal material.[/quote]
It will be, if anyone suggests doing anything illegal. Any posts that describe how to do anything illegal will be deleted, as well, if I see them.

Whoa. I wasn't talking about illegal music. I was talking about foreign music videos (anime and the like) that have been timed and subtitled from the raw. It was more of a question of exchanging files in general, which I am having trouble with.

Naxos is the best music website on the internet. Sadly, it might not apply to a majority of the people here. Simon, what kind of music do you listen to? What performers? I like the "Historical Series" very much, especially their recordings of young Horowitz and the Busch/Serkin Kreutzer Sonata.


calculus
integration
number theory unsolved
number theory

Prove that there aren't any positive integrer numbers $x,y,z$ such that $x^2+y^2=3z^2$.

$\mod 3$ gives that $3|x,y$, giving that $3|z$, thus infinite descent gives the only integral solution $x=y=z=0$.
This works for any number $k$ that is not the sum of two squares for the equation $x^2+y^2=kz^2$.

$\mod 4$ and infinite descendant.

@ZetaX. Sorry for posting the same at the same time  :o


ARML calculus geometry AMC USA(J)MO USAMO MATHCOUNTS

Hi, everyone,

I guess this is most nearly on-topic here on the Round Table. I'm wondering what all of you think would be reasonable criteria for designating a "captain" or two for a newly formed math team. The "team" is anyone in a group of mostly homeschooled students who enter local competitions together, or who do AMC contests through the same test center registration. Some participants just do one program, and some do several. Always, the "team" has been in recruiting mode to get more participants to join in local contests that involve team rounds. A few of the older participants are probably experienced enough now to coach younger participants in beginning-level contests.

I see that TJ has posted [url=http://activities.tjhsst.edu/vmt/lettering.phtml]criteria for letter in math team[/url], and that seems suggestive of some reasonably objective criteria a coach might use to designate a "captain" for a math team. Stuy has captains on its math team too, right? What other examples do you know of math teams that are so organized that they have a system for designating captains? What is that system?

Thanks for any advice you have. Best wishes to all of you going to ARML, MOP, or one of the other summer programs.

With captaining, you should be very careful with a "point system" based on test-taking abilities. The thing about math team is that, even though it's math "team," it's not really a team sport like basketball or football. Your captain should be a team player anyways- we don't have captains, per se, but there was always that one kid on the bus that everyone else went to with questions after tournaments. You know, you get stuck on this problem and you know you can't actually do anything about it anymore, but you still want to know what to do? You always end up going to the same person. Find someone like that- someone that's smart, but more importantly, doesn't mind helping the younger students find the resources they need. (I wasn't one of those kids, fyi, but I know the type).

Don't know how helpful this will be, but here's all about the Stuyvesant math team:

The Stuyvesant captains are selected by the teacher in charge of the Stuyvesant math team.  They are selected primarily on the basis of scores on competitions, although other factors may also be relevant at the margins.  I don't know if there is a particular formula for combining scores across competitions -- if there is, it's not knowledge available to the students.  The Stuy captains are responsible for a room of 30 or so students every morning, probably not equivalent to what your captains will be doing.  At the top level (students in pre-calculus or above) there is mostly problem-solving.  At the lower levels (algebra & geometry or algebra II equivalent), there is more time spent on teaching concepts.

Thanks to bleumoose and JBL. That's interesting information about other programs.

Yeah, we should be so lucky to have multiple rooms with thirty students in each room.   ;)   That's a worth recruiting goal.

At TJ officers are selected democratically.

For the last 3 years going to MOSP seems to have had a high correlation with who was Captain. I'm not sure how good an idea that actually is though, or how feasible if you have just a group of homeschoolers instead of a high school of 2,000 students.

Personally the ideal Math Team Captain would be willing to help and encourage his fellow Mathletes, take charge of situations when the need arises, objectively determine lineups for any away competitions if necesarry, give informative and understandable lectures, and have some sort of vision as to where he or she wants to take the team. Being good at math is also of course important, though the best at math is not necesarilly the best captain (though clearly when one lectures it is helpful to be able to know enough so that other people learn from you).

My two cents.

Thanks for the description of how the TJ captains have been chosen recently, and for the ideal description of a math team captain. I appreciate the ideas posted in replies to this thread.

[quote]Being good at math is also of course important, though the best at math is not necesarilly the best captain (though clearly when one lectures it is helpful to be able to know enough so that other people learn from you).[/quote]

I agree whole-heartedly.  Though, I think it is ideal for the top math student to be the captain [i]if[/i] he has the leadership abilities.  That way, the captain has complete respect from everyone.  You also need a captain who can be there all of the time and who has experience...that's why I was never president of my high school's team though I was vice president and did perhaps the same amount as stuff for the team as the president.

Wow. That's a cool description of the election process at TJ. I guess we can just keep on growing our team here, as there is already precedent about how to select officers for a really huge team.

[quote="kimby_102"][quote="JSteinhardt"][quote="kimby_102"]Just out of curiosity: how many officials are chosen in TJ??[/quote]

Well this year we will have:

Captain
2 Co-captains
2 Statisticians
Finance Committee (people form their own finance committee and run as a group)

We're actually in the middle of elections right now, it's pretty exciting, we've already selected the Captain and a Co-captain, and next week we have a run-off for 2nd Co-captain because there was a tie, as well as elections for the 2nd statistician and Finance Committee (the first statistician was picked midway through the year to be trained by the previous 2 statisticians).

Of course, TJ has something of a huge math team, so other places probably would be smaller. After hearing many of the speeches for the captainship, I think I should add that another important thing, at least in my eyes, is that the person running have the singular goal of aiding their math team, or changing things, or in some way leaving a legacy of themselves, as opposed to the all-too-common mentality of "I need this for College" or "I want it for the prestige/bragging rights", etc., though I am happy to say that for the most part the candidates for office sincerely wished to help the team (I am just mentioning it because it was brought up briefly in a few speeches).[/quote]
Sorry, what do the Statisticians do? :blush:[/quote]
Statisticians maintain the website (I'm not clear as to their other duties)

[hide="Out of topic"]Nice signature Steinhardt![/hide]

Yes, well, this year we didn't have a captain of the math team. In fact, we only had about 5 actively involved people :?  :( . But I'm sure captain is picked by the most reliable and most dedicate math student--not simply the best student at math anyway. I'm sure that next year we'll have a captain though as our school will hopefully get more involved in math contests. :)

If we get too many people involved and have too hard of a time picking a captain, we'll just have to fight each other to see who gets the spot. You know, the strongest person survives, natural selection...etc ;) jk :rotfl:

Interesting topic.  I would like to say a few words as captain of the TJ math team lol.  The TJ math team has really grown quite a lot in recent years due to our first really awesome captain Greg Price.  He had ridiculously good abilities in math and also cared deeply about the team.  He was the one who started holding TJUSAMO (our usamo practices), since he realized that olympiad type problems are often hard to get started in.  Our team's well-being really depends heavily on the student leadership.  At our school roughly, the captains and co-captains do lecture writing, lecture giving, and organizing things, while statisticians maintain the website, do anything computer related (believe me, it's a whole lot), and do more of the data type stuff.  Finance committee people are kinda newish, it's because we don't have money since we don't know how to raise money.

But back to what I was saying before, in my eyes a good captain has to really look out for the future, it is his job to find motivated or motivate his teamates especially the younger students.  Lecturing is very important, but frankly, most of what you have to do to get better will be on your own time.  For example 2 hours a week of USAMO practice isn't really enough.  That is why people who go to MOSP and come back have a higher chance of being elected.  They are really most of the ones who actually start spending serious time studying at our school and I mean like serious amounts.  It is also important that the tradition of the leadership positions is taught to the younger students.  That way, year after year there will be people who will step up to take the spot of leadership and do something with it.  Things like writing and delivering quality lectures, hunting down misguided underclassmen and talking to them, and developing team strategies take a lot more time than the administrative tasks of deciding who will be on what team or what the scores will be for that day.

The actual tasks of captains will vary from team to team, but it is important to establish some sort of tradition for what the role of captain should be so that even people who may not be as naturally inclined to lead and take on the responsibility don't feel like they have no guidance at all.  For example, I would never had spent so much time fussing over the format of TJUSAMOs had I not always heard Eric Price and Mildorf arguing over how to teach us.  I also wouldn't be hunting down freshmen if I myself hadn't been hunted down by senior officers my freshmen year.  It is also important for captains to deal with non-math issues.  For example, I have been very concerned about some of the over-competitiveness and disrespect I have seen at math team practices.  There are issues of fairness in grading, fairness in team selection, possibility of a written constitution, how much power the officers have.  I think that these types of issus are really almost like government stuff and I don't think that many other schools have to face those issues (it's just a question of the size of the team).

I'm very glad you are so interested in these issues already Steinhardt, you'll be a real good captain in the future.  You also have some real good freshmen coming in, probably not as deep as my year, but maybe comparable to the year older than my year in terms of skills.  Good luck with that.

Wow. I really appreciate the posts about this topic. I see that part of the TJ tradition is sharing information even with other math teams, and that is high-class.

My school wasn't exactly TJ (well, technically it was, but we just happened to have the same name), but here's what our officers were:

2 captains: Pretty fuzzy job description.  Basically, be the leaders.  We didn't lecture; practices were led by the coach.  My senior year we had one student-run practice (led by me) for the people going to the state tournament.  Captains were also generally in charge of organizing things during team events.

2 treasurers: Collect money for stuff and bring food to practices.

1 secretary: I think the only thing in the actual job description is to take attendance.  My senior year, the secretary was quite involved and ended up becoming a sort of team statistician as well.

1 publicity: In charge of getting stuff about us on the school announcements.

I think that was it.  All the officer positions were pretty low-key; we only had 20 to 25 people to manage, and the coach did a lot of the administrative stuff.  All the officers were elected.  With few exceptions, captains tended to be the people with the most experience (years on the team) and interest in math/the team.

[quote="pdiao"]I also wouldn't be hunting down freshmen if I myself hadn't been hunted down by senior officers my freshmen year.[/quote]
You hunt freshmen down? You never did so to me  :P
Next year I'll make a run for cocaptain, but the success of that endeavor looks kind of hazy, as the current sophomore year is really good.

Me and Haitao are the Cocaptains. Btw who is this?

[quote="math92"]On my mathcounts team, we have 1 captain who is probably the smartest.  He scores the highest on all the tests and places higher than any of us and is overall a great math student.[/quote]

Well, that's another important thing. The captain must do what he asks the others to.... I mean, if I'm a captain and I want one to score 90 out of 100 at a test, I "must" have more than 90...

For my school's math team, there are usually four captains, all seniors.  Most years, the previous year's captains choose who next year's will be at the end-of-season dinner (the exception being this year, when only one senior was on the team, so the three junior captains stay on for another year.  At least I hope so, being one myself).

We have 2 Presidents and 3 VPs. We want at least 2 VPs who are juniors or below to take the place of the Presidents (unless they aren't seniors). Other than that, it doesn't really matter in terms of age, besides the fact that you naturally have more experience.

Criteria-wise, we basically need 2+ officers to be good at math, end of story. Or else lectures/going over problems can't happen. Since we're a mostly student-run math club (almost entirely), we also want people who are responsible and willing to put some time into the club. Of course talent is important (that's why we need to the officers who are good at math), but people who can lead and are responsible are ultimately more valuable than extra people who can do math.

EDIT: Forgot to mention, elections are run democratically. Presidents are elected first, so anyone who loses that race can run for VP. Usually, however, there isn't any competition for the President spots since people run together.

[quote="RC-7th"][quote="pdiao"]... For example, I have been very concerned about some of the over-competitiveness and disrespect I have seen at math team practices ...

I'm very glad you are so interested in these issues already Steinhardt, you'll be a real good captain in the future.  You also have some real good freshmen coming in, probably not as deep as my year, but maybe comparable to the year older than my year in terms of skills.  Good luck with that.[/quote]

The new freshman class will be good, as they demonstrated with their MathCounts win, but they'll also cause problems that will have to be dealt with. Already, tensions are picking up between them and the current freshman class. In particular, there are rumors that the new class is better, which is causing disrespect and over-competiveness. I'm very interested in how you'll fix those issues.

So who got TJ Co-captain?[/quote]

And those rumors are pretty widespread when I live in CT and hear about how this year's freshmen class doesn't live up to TJ standards and next year will be a lot better.

Anyway, I was captain of my school's mathcounts team because I was the only one with a good chance of getting more than a 12 at the chapter sprint and the only one who had ever placed into top 100 at states in the history of the school (I placed 8th in my 7th grade year).

I think that electing a captain democratically helps to ensure that the captain is well respected, a very importand quality that is difficult to control otherwise.  It is also important that the captain is good at math and great at explaining math to others.  Responsibility and devotion are 2 other factors that should be considered.


$\cos 3\theta =4\cos^{3}\theta-3\cos \theta$

[hide]
Apologies, no equation software.  P = Pi.

Letting x = cosT, use the identity above, giving:
cos3T = -sqrt(3)/2.
3T = (2n+1)P +/- arccos(-sqrt(3)/2) n an integer.

x = cos(5P/18), cos(7P/18) or cos(17P/18).
[/hide]

you should at least proove how you got to that,also that x is in [-1,1]

We can consider complex $\theta$. When we find 3 real roots in [-1,1], we prove, that $\theta$ is real.

Any equation of the form cosx = B has periodic solutions as written.  The original equation is cubic so has a maximum of three solutions and I have given three distinct solutions.  Why do I need to prove that cosine values lie in the range [-1, 1]?

you misunderstood.
your proof may be correct for that PARTICULAR CASE
where x is in [-1,1],and i agree it is.
but BEFORE considering cos and that stuff you must proof
there are no solutions outside [-1,1].
THAT is what i MEANT!

No.  As Rust pointed out, for any $x$ (real or complex), there is a $\theta$ (possibly complex) such that $\cos \theta = x$.  You don't need to assume that $x$ is between $-1$ and $1$.

In any case, why would you need to prove that real solutions are real (before you know exactly what they are) in order to find them?


If $ 2^n$ is a factor of $ 20!$, what is the largest possible value of $ n$?

Or, we can do:

$ \left \lfloor \frac{20}{2} \right \rfloor \plus{} \left \lfloor \frac{20}{2^2} \right \rfloor \plus{} \left \lfloor \frac{20}{2^3} \right \rfloor \plus{} \left \lfloor \frac{20}{2^4}\right \rfloor\plus{}\ldots\equal{}10\plus{}5\plus{}2\plus{}1\equal{}\boxed{18}$.  Much faster.


What will happen to the electric potential energy of a charged particle in an electric field when the particle is released and free to move?

a. it will increase
b. it will change into kinetic energy
c. it will remain unchaged

im thinking B, but could always use a second oppinion. THANKS

I think there are two cases to consider: 1) The field is due to a single charge, and 2) the field is due to more than one charge and our charged particle is in equilibrium in that field.

are you saying that there are more than one possible solution to this question?

In my opinion yes. However, perhaps for your question we are to assume that the electric field is created, for example, by a capacitor, in which case the correct answer is (b), since the electric field is conservative.


function
geometry
3D geometry
sphere
calculus
derivative
linear algebra

Let $ S$ be a compact connected surface in $ R^3$. Prove that if $ S$ is star shaped, i.e. 0 is not in $ S$ and one cannot draw from 0 a straight line tangent to $ S$, then it is of the form $ S(f)$, where $ f$ is a positive differentiable function from $ S^2 \to R$ (S^2 is unit sphere) and $ S(f) \equal{} \{f(p)p \in R^3 | p \in S^2\}$

$ f$ is the inverse of radial projection $ g\colon S\to S^2$, so you only need to prove that $ g$ is a diffeomorphism. The derivative matrix of $ g$, considered as a map from $ \mathbb R^3\setminus\{0\}$ onto $ S^2$, has constant rank $ 2$, and its null space is a line through the origin. The star-shaped condition ensures that the restriction of $ Dg$ to the tangent space of $ S$ has trivial null space. Thus, $ g$ is a local diffeomorphism of $ S$. It remains to prove that $ g$ is injective... but I don't see a clean proof of that.  :|

There is a general lemma to the effect that a local homeomorphism between compact spaces is a covering. owk

But that is not enough - we need a global diffeomorphism.

You've got it :). The radial projection is a covering of the sphere, as owk noticed, so it must be one-to-one because the sphere is simply connected. I believe this is what he meant.

But this wouldn't work in $ \mathbb R^2$, and the result is dimension-independent.  :|
Also, $ g$ is not necessarily onto, so we don't know if the image is simply connected.

Sure $ g$ is onto: it's a local homeomorphism so it's open; in particular its image is open. Also, its image is compact. What's the only non-empty open and compact subset of the sphere? :)

And you've really lost me with that dimension thing :). Do the comments in this topic piece together to give a valid proof for Soarer's problem or not?

Valid, yes. But not quite satisfactory for some reason. The result that Soarer stated in $ \mathbb R^3$ should be true in all dimensions, in particular in $ \mathbb R^2$ where the "spheres" are not simply connected.

Define a section of $ g$ by sending a point $ x\in S^2$ to the point in $ g^{\minus{}1}(x)$ which is closest to the origin. owk

Thanks all.

Em, this problem is taken out from a first course of differential geometry, where no algebraic topology is assumed. Is there any way to get around with covering spaces?


[quote="JasonSullivan"]A student models a stage lighting system using a circuit-drawing computer package and a spreadsheet. He starts with a power supply of e.m.f. 120 V and internal resistance 15 Ω. He assumes that each lamp has fixed resistance of 60 Ω. He is interested in the effect of turning on the lamps one at a time, so that the number of lamps switched on increases from one to six. His circuit and part of his spreadsheet are shown ....
[b]The lamp marked X is the first to be switched on. Explain how lamp X would appear as successive lamps are switched on. (3 marks)[/b][/quote]That student has already done most of the work; or maybe the spreadsheet calculations came pre-programmed, along with the circuit-drawing computer package.
You see that the "net resistance" $ (R)$ of the whole set of lamps $ (60 \Omega$ for one lamp on, $ 60 \Omega /n$ for $ n$ lamps on $ )$ plus the internal resistance of the power supply has been used to calculate the "Total current" $ (I)$ in Amps.
$ I \equal{} \frac{120V}{15 \Omega \plus{} R}$
The voltage $ (V)$ or "power drop" ("p.d.") across the whole set of lamp was calculated as:
$ V\equal{}RI$
Then the power to the whole set of lamps was calculated by multiplying $ V$ times $ I$.
Because all the lamps that are turned on have the same resistance and "see" the same voltage, they will all take the same share of the total power. To figure out how much power is applied to each lamp, we just divide (by the number of lamps turned on) the total power to the whole set.
When only one lamp is on, it's using 154 Watts of power. (I hope it is a lamp rated for at least 150 Watts at 120V, because one rated for less will probably burn out immediately).
When two lamps are on, they are sharing 213 Watts (using 106.5 Watts each). Do you think they will burn as brightly as the single lamp was burning before?
When three lamps are on, each one uses less than 80 Watts (235Watts/3). Are they looking a little dimmer?
By the end, 6 lamps are sharing 230 Watts, giving each one less than 40 Watts. Are they burning very brightly?
Should they get a better power supply?

If "better" means "stronger" it can burn some lamps at the beginning... (just kidding)

[quote="JasonSullivan"]A student models a stage lighting system using a circuit-drawing computer package and a spreadsheet. He starts with a power supply of e.m.f. 120 V and internal resistance 15 Ω. He assumes that each lamp has fixed resistance of 60 Ω. He is interested in the effect of turning on the lamps one at a time, so that the number of lamps switched on increases from one to six. His circuit and part of his spreadsheet are shown below.

[img]http://farm4.static.flickr.com/3511/3254169932_9db9ec57dd.jpg[/img]

[b]The lamp marked X is the first to be switched on. Explain how lamp X would appear as successive lamps are switched on. (3 marks)[/b]

Please help. Many thanks.[/quote][hide]

First, notice that all the lamps are in parallel with one another. This suggests that current division is a good strategy to use. The power dissipated across the n-th switch is given by: $ i_n^2\cdot 60\Omega$. Brightness is proportional to the power dissipated and thus the brightness of the n-th lamp is proportional to $ i_n^2$.

Next, in order to make the current division calculation one needs to represent the circuit in question by its Norton's equivalent circuit, where $ I_{sc} \equal{} \frac {V_{oc}}{R_{Th}} \equal{} \frac {120\,V}{15\,\Omega} \equal{} 8\, A$.

When the first switch is turned on: $ i_1 \equal{} \frac {15\Omega}{15\Omega \plus{} 60\Omega}(8\,A) \equal{} 1.6\,A\implies i_1^2 \equal{} 2.56\, A^2$.

The equivalent resistance is of these first two resistors is: $ \frac {15\cdot60}{15 \plus{} 60} \equal{} 12\Omega$.

Hence, adding the second switch $ i_2 \equal{} \frac {12\Omega}{12\Omega \plus{} 60\Omega}(8\,A) \equal{} 1.33\,A \implies i_2^2 \equal{} 1.78\, A^2$.

Since the resistance of each lamp is the same, the current across each lamp is the same.

The equivalent resistance is now: $ \frac {12\cdot60}{12 \plus{} 60} \equal{} 10\Omega$.

The current across the third lamp is $ i_3 \equal{} \frac {10}{10 \plus{} 60}(8\, A) \equal{} \frac {8}{7}\, A\implies i_3^2 \equal{} \frac {64}{49}\, A^2$.

The equivalent resistance is now $ \frac {10\cdot 60}{10 \plus{} 60} \equal{} \frac {60}{7}\Omega$.

The current across the fourth lamp is $ i_4 \equal{} \frac {\frac {60}{7}}{\frac {60}{7} \plus{} 60}(8\,A) \equal{} 1\,A \implies i_4^2 \equal{} 1\,A^2$.

The equivalent resistance is now $ \frac {\frac {60}{7}\cdot 60}{\frac {60}{7} \plus{} 60} \equal{} \frac {60}{8}\Omega$.

The current across the fifth lamp is: $ i_5 \equal{} \frac {\frac {60}{8}}{\frac {60}{8} \plus{} 60}(8\,A) \equal{} \frac {8}{9}\,A\implies i_5^2\equal{}\frac {64}{81}\, A^2$.

The equivalent resistance now becomes: $ \frac {\frac {60}{8}\cdot 60}{\frac {60}{8} \plus{} 60} \equal{} \frac {60}{9}\Omega$.

The current across the sixth lamp, i.e. the current across each lamp with all switches on, is: $ i_6 \equal{} \frac {\frac {60}{9}}{\frac {60}{9} \plus{} 60}(8\,A) \equal{} 0.8\,A \implies i_6^2 \equal{} 0.64\, A^2$.

The overall pattern is that the individual lamps dim further with the activation of each additional lamp.[/hide]


[size=150]Give three circle $ (O_1),(O_2), (O_3)$ tangent outside and same tangent inside with $ (O)$ . Give $ R_1, R_3 , R_3 , R$ are radius of  $ O_1 ,O_2 ,O_3, O$
PROVE THAT
 $ R_1 \plus{} R_2 \plus{} R_3 \leq 3(2\sqrt {3} \minus{} 3)R$[/size]

I have an idea, but I do not solve the problem for now...

Think about an inversion which sends the three internal circles to three identical circles.This is posible by Steiner's Porism. Now, I need to realize a way to use it.

I will post a solution as soon as I can.


hey ya

my friend showed me this question and i don't understand this.
This must be a very exciting question,
and i think the solution would be very funny.
Can anyone explain how to solve this?

The Rainbow Game is played by a team of seven. Each player gets a hat, which can be any one of the seven colors red, orange, yellow, green, blue, indigo, and violet. The colors of the hats are independent of each other and repetitions are allowed; for instance, it may happen that all the hats are green. Each player can see only the colors of the six hats worn by the rest of the team; no player can see the color of his or her own hat. The players are to guess the colors of their own hats, and if at least one player guesses correctly then the team as a whole wins.

The players may not communicate in any way during the game, and they must all announce their guesses simultaneously. They are, however, allowed to plan out a strategy in advance, and they hope to find a strategy which will guarantee them success for every possible arrangement of hats. Is there such a strategy? Either find one or show that it cannot exist.

Hello. You must be new to this forum.

____________________________________________________________________________________
EDITED: I have checked what kyle_k pointed out. It is the qualifying quiz after all!! So I edited my solution(?) out.
             All a sudden, there are so many Koreans here! (I'm Korean too) :D

I think the question that you posted is Mathcamp qualifying quiz #8 of this year.
I think we are not suppose to post answer or discuss answer about the qualifying quiz.

I was invited but I am not going, so I didn't even look at the quiz. Good thing someone else noticed. :)


Let $f \in \mathbb{R}[X]$ be a polynomial which has all roots real. We know that $f(X^2-1)=f(X) \cdot f(-X)$. Find all such $f$.

[quote="Rafal"]you mean that $f$ has $n$ real roots where $n=degf$ ?[/quote]

yes (not necesarilly distinct)

well I don't see much problem here,
Put $f(x)=a_n\prod_{i=1}^{n}(x-x_i)$ thus
$(-1)^n(a_n)^2\prod_{i=1}^{n}(x-x_i)(x+x_i)=a_n\prod_{i=1}^{n}(x^2-1-x_i)=$$
a_n\prod_{i=1}^{n}(x-\sqrt{1+x_i})(x+\sqrt{1+x_i})$                                                                                cause we have to have $2n$ real roots, so $a_n=(-1)^n$ and
the set $\{\sqrt{x_1+1},-\sqrt{x_1+1},,...,,...,\sqrt{x_n+1},-\sqrt{x_n+1}\}$ must be the same as $ \{ x_1,-x_1,...,x_n,-x_n \} $ , the rest is easy.

[quote="harazi"]Try to see mathlinks contest archive. I think there was even more difficult problem there.[/quote]

can you tell me what problem you are talking about?
i can't find it


[quote="Silverfalcon"]Two cubes have their faces painted either red or blue. The first cube has five red faces and one blue face. When the cubes are rolled simultaneously, the probability that the two top faces show the same color is $\frac{1}{2}$. How many red faces are there on the second cube?[/quote]

[hide]$\boxed{3}$.
$x$=# of red faces
$6-x$=# of blue faces
two red faces rolled = $\frac{5}{6}\cdot\frac{x}{6}$
two blue faces rolled = $\frac{1}{6}\cdot\frac{6-x}{6}$
$\frac{5}{6}\cdot\frac{x}{6}+\frac{1}{6}\cdot\frac{6-x}{6}=\frac{1}{2}$
solving, $x=3$[/hide]

[hide="Answer"]There's a $\frac{5}{6}$ chance that the first cube will show red and a $\frac{1}{6}$ chance that the cube will show blue. Essentially, if $p$ is the probability of the second cube showing red, then we have $\frac{5}{6}p+\frac{1}{6}(1-p)=\frac{1}{2}\Rightarrow p=\frac{1}{2}$, which means that the second cube has $\boxed{3}$ red faces.[/hide]

Hmm, I solved it the same way you guys did, but can you think of it this way?

[hide]
You know that, no matter what face the first cube rolls, the probability the second cube has the same is $\frac 1 2$, meaning there must be equal blue and red faces on it, or 3 faces each.
[/hide]

Any information about the first cube is irrelavent other than the number of colors. Since there are only two colors, then the second cube needs to be half and half, thus 3 faces are red and 3 are blue.


geometry
power of a point
geometry solved

This problem is familiar with P1 in IMO 2005
I don`t know how to translate it, but i`ll try to explain it. Hope it`s clear:
ABC is an equilateral triangle. See the picture in problem 2 here:
http://www.zaba.ru/cgi-bin/tasks.cgi?tour=russia.kvant.2001.var5

Prove that $AB_2 + CA_2 + BC_2 = AC_1 + BA_1 + CB_1$

Using the power of a point theorem, the problem reduces to computations. You have $AB_2 \cdot AB_1 = AC_1 \cdot AC_2$ etc. Combining them we obtain formulas for $AB_2,\ldots$ in terms of $\ell,AB_1,\ldots$. Summing up, we get $\sum AB_2 = \sum AB_1$. (this last part is pretty ugly, but doable)

I'm not in the mood of writing the whole thing, but if you need help, feel free to ask me.

I`m so stuck :blush: . Can you please explain. Thank you ( What is $\ell$)

Well, $\ell$ is the side of the triangle. (standard notation in Romania)

Let $x=BC_2$, $y=CA_2$, $z=AB_2$, $u=CB_1$, $v=AC_1$, $w=BA_1$.

By power of a point (see first post) we have \[ u = \frac{y(\ell-w)}{\ell-z} , \ \ v= \frac{z(\ell-u)}{\ell-x}, \ \ w= \frac{x(\ell-v)}{\ell-y} .  \]

Now, in the formula for $u$ we have a $w$. Put instead of that $w$ the third formula, and instead of the new $v$ the second formula. You'll get \[ u = \frac{\ell y (\ell - y)(\ell - x) - xy \ell (\ell -x)+xyz(\ell-u)}{\prod (\ell-x)} .  \]
Solve for $u$ and then you get big formula. The other two formulas ($v,w$) follow cyclically.
Compute $\sum u$. Most probably ( :) ), you will get $\sum x$.

Maybe someone will find a better solution.

[quote="Beat"]This problem is familiar with P1 in IMO 2005[/quote]

at the first sight. I don't think it can really be used in the IMO 2005 problem.

[quote="Beat"]I don`t know how to translate it[/quote]

Let me translate:

[color=blue][b]Problem.[/b] Given an equilateral triangle ABC, and a circle which intersects its side BC at the points $A_1$ and $A_2$, its side CA at the points $B_1$ and $B_2$, and its side AB at the points $C_1$ and $C_2$.

Prove that $AB_2 + CA_2 + BC_2 = AC_1 + BA_1 + CB_1$.[/color]

[i]Solution.[/i] By the intersecting chords theorem, $BA_1\cdot BA_2=BC_2\cdot BC_1$, what rewrites as $BA_1\cdot\left( BC-CA_2\right)=BC_2\cdot \left( AB-AC_1\right)$, or, equivalently, $BA_1\cdot BC-BA_1\cdot CA_2=BC_2\cdot AB-AC_1\cdot BC_2$. Thus,

$BA_1\cdot BC-BC_2\cdot AB=BA_1\cdot CA_2-AC_1\cdot BC_2$; similarly,
$CB_1\cdot CA-CA_2\cdot BC=CB_1\cdot AB_2-BA_1\cdot CA_2$;
$AC_1\cdot AB-AB_2\cdot CA=AC_1\cdot BC_2-CB_1\cdot AB_2$.

Hence,

$\left(BA_1-CA_2\right)\cdot BC+\left(CB_1-AB_2\right)\cdot CA+\left(AC_1-BC_2\right)\cdot AB$
$=BA_1\cdot BC-CA_2\cdot BC+CB_1\cdot CA-AB_2\cdot CA+AC_1\cdot AB-BC_2\cdot AB$
$=\left(BA_1\cdot BC-BC_2\cdot AB\right)+\left(CB_1\cdot CA-CA_2\cdot BC\right)+\left(AC_1\cdot AB-AB_2\cdot CA\right)$
$=\left(BA_1\cdot CA_2-AC_1\cdot BC_2\right)+\left(CB_1\cdot AB_2-BA_1\cdot CA_2\right)+\left(AC_1\cdot BC_2-CB_1\cdot AB_2\right)$
$=0$.

Now, since the triangle ABC is equilateral, we have BC = CA = AB; thus, dividing by BC = CA = AB, we transform this equality into

$BA_1-CA_2+CB_1-AB_2+AC_1-BC_2=0$,

or, equivalently, $AB_2 + CA_2 + BC_2 = AC_1 + BA_1 + CB_1$. And the problem is solved.

Note that http://www.mathlinks.ro/Forum/viewtopic.php?t=68862 is a generalization of this problem (the order of the points on the circle is different, but that doesn't really matter).

  Darij

Thanks both of you.
btw, i didn`t say that it can be used in the imo


geometry
incenter
projective geometry
angle bisector
geometry unsolved

Show that the external angle bisectors of a triangle meet the opposite sides of the triangle in three collinear points.

There are well-known teorems that can helps us.

Let $B'$ be the intersection of the external angle bisectors of $A$ and $C$. And let define $A'$ and $C'$ in the same way. Then look that $BB'$ is the angle bisector of $<ABC$ and because of that $BB'$, $CC'$ and $AA'$ concur.

So the triangles $ABC$ and $C'A'B'$ are in perspective from $I$ the incenter, and by Desargues the intersection of their respective sides are collinear.

Look that the points of intersection, are the points the problem asks.


algebra
polynomial
function
trigonometry
complex analysis
complex analysis unsolved

Let $ P(z)$ is a complex polynomial, which has  $ n$  real roots and let  $ a$  is a real number. Then prove that  $ Q(z)\equal{}P(z\plus{}i*a)\plus{}P(z\minus{}i*a)$  also has  $ n$  real roots.

Thank you beforehand.

This is a corollary from the following general result. Let f be a real entire function of the Laguerre-Polya class.
(This means that f is a limit of real polynomials whose all zeros are real. (Uniform limit on every compact
subset of the plane)). Then the operator f(D), where D=d/dz, maps the set of polynomials with all roots
real into itself. In your case, f(D)=2 cos(aD). It is easy to verify that Q=2 (cos(aD))P in your notation,
and that cos is in the Laguerre-Polya class.

Here is a sketch of the proof of the general result (I suppose it was found by Takagi
in the beginning of XX century). Let A be the class
of all differential operators that preserve the class of polynomials whose roots are real.
First you verify that D is in A. (Using Rolle's theorem).
Then you verify that D also preserves the Laguerre-Polya class. (By passing to the limit).
Then you verify that D-a is in A for every real A. (This is because (D-a)P(z)=exp(az)D(exp(-az)P(z))).
Thus if f is a real polynomial with all roots real then f(D) is in A. Because f(D) is a product of the operators of the form D-a.
And finally, by passing to the limit you prove that for every Laguerre-Polya function f, f(D) is in A.

Pym.

In my previous message I assumed that n is the degree of your polynomial.
(Otherwise the question makes no sense).

Pym.

I think this problem has easier solution.

Armenia,
You write:
"I think this problem has easier solution".
Then why don't you tell us this easier solution?
I will be glad to see it.

Pym.

Suppose $ P$ is a complex ( :lol: ) polynomial with real
roots $ P(x)\equal{}c(x\minus{}x_1)(x\minus{}x_2)\cdot \ldots \cdot (x\minus{}x_n)$, and $ a > 0.$
Denote by $ Q(z)\equal{}P(z\plus{}ia)\plus{}P(z\minus{}ia)$. If $ Q(z_0)\equal{}0$ then
$ \left|\frac{P(z_0\plus{}ia)}{P(z_0\minus{}ia)}\right|\equal{}1$. For $ z\equal{}x\plus{}iy,$  $ y>0,$  and for every $ j\equal{}1, 2, \ldots , n$
we have $ \frac{|z\plus{}ia \minus{}x_j|^2}{|z\minus{}ia \minus{}x_j|^2}\equal{}\frac{(x\minus{}x_j)^2\plus{}(y\plus{}a)^2}{(x\minus{}x_j)^2\plus{}(y\minus{}a)^2} > 1.$
So for $ \Im z >0$ we have $ \left|\frac{P(z\plus{}ia)}{P(z\minus{}ia)}\right| > 1$
(analogously for $ \Im z <0$ we have $ \left|\frac{P(z\plus{}ia)}{P(z\minus{}ia)}\right|< 1$).
Thus $ Q(z)$ has only real roots.

(as far as I know this idea belongs to Polya).

The following statement is true.

Let $ P(z)$ be a REAL polynomial (of arbitrary degree), which has $ m$
real zeros and let $ a$ be a real number. Then the number of real zeros of  $ Q(z)\equal{}P(z\plus{}ia)\plus{}P(z\minus{}ia)$ is not less than $ m$.

A sketch of the proof is described by PYM, and the main fact is that for
every real $ b\neq 0$ and every real polynomial $ P$ the number of real zeros
of $ S(z)\equal{}P^{'}(z) \plus{}bP(z)\equal{} e^{\minus{}bz}\left(  e^{bz}P(z)\right)^{'}$ is not
less than the number of real zeros of $ P$. (If  $ P$  has $ m$ real zeros,
then by the Rolle's theorem $ S$ has $ \geq (m\minus{}1)$ real zeros, but $ P$
and $ S$ are of the same degree).

Unfortunatelly I do not know a simple proof of this statement.

AnnA-lyticity,
Your argument is correct, but apparently you lost one root in your estimate:
the polynomial S(x)=exp(-bx)(exp(bx)P(x)' with b different from 0
has at least as many real roots as P.
Indeed, suppose first that all real roots of P are simple, and there are m
of them. Then by Rolle's theorem there are at least m-1 real roots
"between the roots" of P. But there is at least one more!
Suppose, for example that b>0. If p is the smallest real root of P, then
by the same Rolle's theorem, S also has a root on (-infty,p).
Similarly, if b<0, there will be an extra root of S on the right of the
roots of P. The case of multiple roots of P is treated by perturbation, or just by thinking
carefully for 1 min about them.

On my opinion, this is a simple proof of your statement. It uses nothing but Rolle's theorem.
Pym.

Pym,

I do not think that I lost one root in my proof.
I have mentioned that the polynomials $ P$ and
$ S$ are of the same degree, and $ P$ has exactly $ m$
real zeros, and $ S$ has $ \  \geq (m\minus{}1)$ real zeros.
Obviously this implies that  $ S$ has $ \geq m$ real zeros.

About the "simplicity" of the proof. The proof in the
case when all zeros of $ P$ are real is very simple
and I can explain it for a good schoolchild. But the
proof in the general case (not for the mentioned
simple lemma, but for the statement posted by Armenia)
is not elementary. At first we consider the operator
$ \cos (aD)$,  then the representation for the $ \cos(az)$
and so on. I think that the elementary proof for this case
exists and would like to find it.


conics
parabola
analytic geometry
geometry

Im wondering if there is another method besides the derivate to find the equation of the tangent and the normal line to the parabola $ y \equal{} x^2$ at the point $ ( \minus{} 2,4)$ . I mean a more geometrical method

This belongs in the High School Forum.

I can't think of anything. I don't think it can be represented by a shape.

now  here  comes  what  we call as  the parametric  equations of  a parabola  .

         they  are of the form  (at^2,2at)   for the parabola  y^2=4ax

          now in the parametri c  form  the equation to tangent  is   yt=x+at^2    and the equation to normal is  y+xt=2at+at^3

            but  these  are  also  obtained   by finding the derivatiove  to the equation of the parabola  and  substituting the parametric coordinates in the respective  equations.

by the way does this  belong  to this  forum  only?  my god  we  are taught this  in grade 11  that is  at the age of 17.

I think that he question of Nebraska is actually very interesting. I just remember you guys the nice reflective porperties of the conic sections:

[b]The tangent and normal lines to a parabola are the external and internal bisector of the angle formed by the line that joins the foci and the contact point P and the parallel to the focal axis passing through P[/b]

the foci is (0,0.25) the focal axis is x=0 and the contact point P (2,-4)

This has more calculations than derivating , but this is pure geometry with no Calculus.


$ ABC$ has three angles $ c$ ,$ d$ and $ s$  such that $ c$ and $ d$ are acute and $ \sin^2(c)\plus{}\sin^2(d)\equal{}\sin(s)$
prove that $ s\equal{}\pi/2$

$ \triangle ABC$ has $ \angle A,\ \angle B$ acute. $ \sin^2 A \plus{} \sin^2 B \equal{} \sin C\ \Longrightarrow \angle C \equal{} \frac {\pi}{2}.$
____________________________________________________________

$ \sin^2 A \plus{} \sin^2 B \equal{} \sin C\ \Longleftrightarrow\ a^2 \plus{} b^2 \equal{} 2Rc \ge c^2\ \Longrightarrow\ \angle C \le \frac {\pi}{2},$ $ \triangle ABC$ is not obtuse. Then $ \angle B \ge \frac {\pi}{2} \minus{} \angle A,$ $ \sin B \ge \sin \left(\frac {\pi}{2} \minus{} A\right) \equal{} \cos A > 0,$ we can square the last inequality. $ \sin C \equal{} \sin^2 A \plus{} \sin^2 B \ge \sin^2 A \plus{} \cos^2 A \equal{} 1,$ $ \sin C \equal{} 1,$ $ \angle C \equal{} \frac {\pi}{2}.$


27.) The [i]Digital Sum [/i]of a number is the sum of its digits. For how many of the positive integers 24-125 inclusive is the digital sum a multiple of 7?

[hide]For 2 digit numbers, it can be 7 or 14.

And we can't make 14 until 59.

So 1 in the 20's, 1 in the 30's, 1 in the 40's, 2 in the 50's, 2 in the 60's, 2 in the 70's.
Now at 8_, we can't make 7. So 1 in the 80's, 1 in the 90's.

In 10_, we have 1, in 11_, we have 1. and 124 is the last one. Add them up and you get (i think) 14[/hide]

[hide]
The highest digital sum in your set of numbers for two digit numbers is 18 which is formed by 99. This means that the digital sum has to be 7 or 14 for two digit numbers.
The highest digital sum for three digit numbers is 8 which is formed by 125. This means the digital sum for 3 digit numbers has to be 7.
For two digit numbers, for multiples of 7, you have
25
34
43
52
61
70
For two digit multiples of 14, you have
59
68
77
86
95
For three digit multiples of 7, you have
106
115
124

Now add them up to get $\boxed {14}$.

[hide="The way I found the numbers quickly"]
For two digit numbers, the sum is 7 so the first digit can't be more than an 8. That leaves you 1-7, 1 is too low, so you have 6 possibilities for 2-7. For when the sum is 14, the first digit must be 5-9 as with 4, the second digit can't be 10. That means you have 5 possibilities. Lastly, for three digit numbers, the only three digit numbers in the set of numbers is 1. That means the other 2 numbers have to add up to 6. That leaves you with 0-6 for the second digit, 3-6 are too large though so now you have 0-2 or 3 possibilities
[/hide][/hide]

[hide]For each with tens digit:

2-4 (3 of them), there is 1 choice for the units digit (add to 7)
5-7 (3 of them), there are 2 choices for the i=units digit (add to 7 or 14)
8-9 (2 of them), there is 1 choice for the units digit (add to 14)

100s has 1 choice for the units digit.
110s has 1 " "
120s has 1 " "

3*1+3*2+2*1+3=14[/hide]

Out of curiosity, can anyone tell me a divisibility rule for 7?

Ah, I think I would much rather just divide it out, thank you.

Their method is actually quite ugly. Another is to cancel the last digit double it and subtract.
Ex: 8549. Eliminate the 9 to get 854. Double and subtract to get 854-18=836. Repeat to get 71. 71 isn't divisible by 7 so no, 8549 isn't divisible


arithmetic sequence
algebra unsolved
algebra

(4)Consider the following array:
1    2    3    4     5
6    7    8    9    10
11  12  13  14  15
16  17  18  19  20
21  22  23  24  25
show that no matter which 5 numbers are removed from this array, there exists a 5-term Arithmetic progression among the remaining 20 numbers

Do not double post:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=316562[/url]


algebra
polynomial
real analysis
real analysis unsolved

note that if we let $ f(x) \equal{} \frac{a_0}{1\cdot 2 \cdots k}x^k \plus{} \frac{a_1}{2\cdot 3 \cdots (k\plus{}1)} x^{k\plus{}1} \plus{} \cdots \plus{} \frac{a_n}{(n\plus{}1)(n\plus{}2)\cdots (n\plus{}k)}x^{n\plus{}k}$, then we have $ f(1)\equal{}f'(1)\equal{}f''(1)\equal{}\cdots\equal{}f^{(k\minus{}1)}(1)\equal{}0$...

but what exactly do you mean "find at least roots number of $ p(x)$"?


Let $ a,b,c$ be the sidelengths of an acute triangle.Prove the ineq:
$ a^2(b\minus{}c)^2\plus{}b^2(c\minus{}a)^2\plus{}c^2(a\minus{}b)^2\plus{}(a\plus{}b\plus{}c)(a\minus{}b)(b\minus{}c)(c\minus{}a)$≥$ 0$

[quote="SUPERMAN2"]Let $ a,b,c$ be the sidelengths of an acute triangle.Prove the ineq:
$ a^2(b \minus{} c)^2 \plus{} b^2(c \minus{} a)^2 \plus{} c^2(a \minus{} b)^2 \plus{} (a \plus{} b \plus{} c)(a \minus{} b)(b \minus{} c)(c \minus{} a)$≥$ 0$[/quote]
Inequality
$ <\equal{}>(a^2\plus{}b^2)(b\minus{}c)^2\plus{}(b^2\plus{}c^2)(a\minus{}b)^2\plus{}2b^2(a\minus{}b)(b\minus{}c)\plus{}(a\plus{}b\plus{}c)(a\minus{}b)(b\minus{}c)(c\minus{}a) \ge 0$
Case 1:
$ a \le b \le c$ So $ (a\plus{}b\plus{}c)(a\minus{}b)(b\minus{}c)(c\minus{}a) \ge 0;2b^2(a\minus{}b)(b\minus{}c) \ge 0$
We have done!
Case 2:
$ a \ge b \ge c$
By Am-Gm We have:
$ (a^2\plus{}b^2)(b\minus{}c)^2\plus{}(b^2\plus{}c^2)(a\minus{}b)^2 \ge 2(a\minus{}b)(b\minus{}c)\sqrt{(a^2\plus{}b^2)(b^2\plus{}c^2)} \ge (a\plus{}b)(b\plus{}c)(a\minus{}b)(b\minus{}c)$
We need prove that:
$ (a\plus{}b)(b\plus{}c)(a\minus{}b)(b\minus{}c)\plus{}2b^2(a\minus{}b)(a\minus{}c)\minus{}(a\plus{}b\plus{}c)(a\minus{}b)(b\minus{}c)(a\minus{}c)$
$ <\equal{}>(a\minus{}b)(b\minus{}c)[(a\plus{}b)(b\plus{}c)\plus{}2b^2\minus{}(a\plus{}b\plus{}c)(a\minus{}c)] \ge 0$
$ <\equal{}>(a\minus{}b)(b\minus{}c)[(3b^2\plus{}c^2\plus{}2bc\plus{}ac\minus{}a^2] \ge 0$
We have:
$ 3b^2\plus{}c^2\plus{}2bc\plus{}ac>(b\plus{}c)^2>a^2$
We have done!  :P


geometry
incenter
circumcircle
similar triangles
angle bisector

[color=darkred]In the triangle $ABC$ denote the incenter $I$ and the point $D\in BC\cap AI$.

For a mobile point $M\in [BC]$ define the points $X\in [AB]$, $Y\in [AC]$

for which $\widehat{BMX}\equiv\widehat{CAM}$, $\widehat{CMY}\equiv\widehat{BAM}$. Prove that $XY\parallel BC$ and

ascertain (construct) the position of the point $M$ for which the area $[XAY]$
is minimum and in this case show that there is the relation $\frac{XA}{XB}+1=\left(\frac{IA}{ID}\right)^{2}$.[/color]

This is a nice problem, thank you Virgil Nicula.

1) Clearly, $\measuredangle XMY = \measuredangle XAY$, so $XAYM$ is cyclic.
Therefore, $\measuredangle BXM = \measuredangle AYM$, so it follows that \[\measuredangle ABC = \measuredangle AMY = \measuredangle AXY\]
which yields $XY \parallel BC$.

2) Call $S$ the intersection of $AI$ and the circumcircle of $\Delta AXY$.
To minimize the area of $\Delta AXY$, we might as well minimize the radius of it's circumcircle. This is equivalent to minimizing $\left|AS\right|$ because of $1)$.

But we have $d(M,XY \leq d(S,XY)$ with equality when $D=M=S$.

So the case we are looking for is $M=D$.

3) I think there is a small typo, you are giving away the answer. The relation you give is independent of $M$, so it is always true. I think you mean \[\left(\frac{IA}{IM}\right)^{2}= \frac{XA}{XB}+1\]

It's easy to prove:

Because $M=D$, the circumcircle of $\Delta AXY$ is tangent to $BC$.
So $\measuredangle XMB = \measuredangle BAM$, which gives us two similar triangles and the relation: \[\frac{AB}{BD}=\frac{BD}{XB}\]
\[\iff \left(\frac{AB}{BD}\right)^{2}=\frac{AB}{XB}\]

Using the angle bisector theorem, we know that $\left(\frac{AB}{BD}\right)^{2}=\left(\frac{AI}{ID}\right)^{2}$, and of course we know that $\frac{AB}{XB}=1+\frac{AX}{XB}$, so we are done!


Ok... so I'm trying to do a problem in Alcumulus, but it isnt showing me a box where i can type in answers and there is no submit button... help?

Can you be a bit more specific? What page are you on? Are you getting a problem?

It may be a problem with your browser or the page may not have fully loaded. Did you try refreshing?

This has happened to me a couple of times.
What it basically is, is a temporary glitch due to the lack of problems available.
After some time, try again, and chance your topic that you are working on.


function
inequalities
calculus
derivative
algebra
domain
number theory

what is the meaning for a function to be convex (to apply Jensen, for example). I mean you could always use the second derivative, but there's gotta be a non-calculus definition.

similarly, what does it mean for a function to be concave?

I think: (read from some book, not sure if I recalled it all correctly)

Take two points P and Q on the graph of the function. Join P and Q. If the graph of the function lies entirely below the line, the function is said to be convex between P and Q. If within a lower bound and an upper bound, with any two points P and Q convexity is satisfied, the function is said to be strictly convex between those bounds.

hmm...isn't there some sort of algebraic defenition though? like f is convex iff f((a+b)/2):ge:(f(a)+f(b))/2 or something like that?

ok. yay i now have 3 whole defenitions with one from each major branch of mathematics

Why do you say there are three branches of mathematics? We usually say that there are four (I think): algebra, analysis, number theory, and geometry/topology.

As I recall, a definition is this:
a function is convex iff for any numbers x and y in the domain of the function and any t in [0, 1] we have tf(x)+(1-t)f(y) >= f(tx+(1-t)y).

The ineq is reversed for concave fctions


The AMC-12 A 2003 #23

How many perfect squares are divisors of the product $1! \cdot 2! \cdot 3! \cdot \cdot \cdot 9!$?

A) 504
B) 672
C) 864
D) 936
E) 1008

[hide="Answer"]Checking for multiples of each prime, we find that the number is equal to $2^{30}\cdot 3^{13}\cdot 5^5\cdot 7^3$. For perfect squares, we must have an even power of each prime, so we can simplify it to $2^{30}\cdot 3^{12}\cdot 5^4\cdot 7^2=(2^2)^{15}\cdot (3^2)^6\cdot (5^2)^2\cdot (7^2)$. This gives us a total of $(16)(7)(3)(2)=672$ perfect squares. $\boxed{B}$[/hide]

It's a double factorial! Forgot properties of double factorials, but fobtastic anyways.

Actually [url=http://mathworld.wolfram.com/DoubleFactorial.html]this[/url] is a double factorial.

Also, Bestone, you can use $\cdots$ instead of $\cdot$ thrice...just a little $\LaTeX$ tip.


Let  $a,b,c$ are reals such that $a+b+c\neq 0$ and $\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}=0.$ Prove that  $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=1.$

$\frac{a^2}{b+c}+a=\frac{a(a+b+c)}{b+c}$

$\sum\frac{a^2}{b+c}=0$ $\implies \sum\frac{a(a+b+c)}{b+c}=a+b+c$ $\implies (a+b+c)\sum\frac{a}{b+c}=a+b+c$ . Since $a+b+c\not = 0$ hence $\sum\frac{a}{b+c}=1$

Put $p=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}.$ Then

$\frac{a}{b+c}=p-\frac{b}{c+a}-\frac{c}{a+b}\implies \frac{a^2}{b+c}=ap-\frac{ab}{c+a}-\frac{ac}{a+b}\qquad(1)$

Similarly,

$\frac{b^2}{c+a}=bp-\frac{ab}{b+c}-\frac{bc}{a+b}\qquad(2)$
$\frac{c^2}{a+b}=cp-\frac{ac}{b+c}-\frac{bc}{a+c}\qquad(3)$

Adding up $(1),(2),(3)$ we get

$0=p(a+b+c)-\frac{b(a+c)}{a+c}-\frac{a(b+c)}{b+c}-\frac{c(a+b)}{a+b}=(p-1)(a+b+c)$

Since $a+b+c\neq 0$, it follows $p=1.$


modular arithmetic
algebra
polynomial
number theory unsolved
number theory

For how many primes $ p$, there exits unique integers $ r$ and $ s$ such that for every integer $ x$, $ x^{3} \minus{} x \plus{} 2\equiv \left(x \minus{} r\right)^{2} \left(x \minus{} s\right)\pmod p$?

PS: I have already asked the problem in [url=http://www.mathlinks.ro/viewtopic.php?t=272621]Pre-olympiad[/url] section. I don't know it is correct to re-ask it in this forum, but now I think it is upper than pre-olympiad.

You want it to have a multiple root, so you want $ p$ to divide the discriminant.

The formula you mentioned is
$ \partial_{3} \equal{} G^{2} \plus{} 4H^{3}$
where $ G \equal{} a^2d \minus{} 3abc \plus{} 2b^3$ and $ H \equal{} ac \minus{} b^2$ where $ P\left( x\right) \equal{}  ax^3 \plus{} 3bx^2 \plus{} 3cx \plus{} d$.

For the polynomial $ P\left(x\right) \equal{} x^3 \minus{} x \plus{} 2$, $ G \equal{} 2$ and $ H \equal{} \frac {\minus{}1}{3}$. Thus $ \partial_3 \equal{} 4 \plus{} \frac{\minus{}4}{27} \equal{} 4\cdot \frac {26}{27}$.
For the prime numner $ p$, $ 4\cdot \frac {26}{27} \equiv 0 \pmod p$. Thus  $ p \mid 4 \cdot 26 \Rightarrow p \equal{} 2$ or $ p \equal{} 13$.
In fact, $ (p,r,s) \equal{} (2,1,0)$ and $ (13, 3, 6)$ are the solutions.

But is there a simpler way to solve the problem without knowing $ \partial_3$?


Find all functions $f: \mathbb R^{\geq 0} \to \mathbb R^{\geq 0}$ such that
\[f(y)f(xf(y))\equal{}f(x\plus{}y)\]
for any non-negative real choice of $x$ and $y$.

[quote="moriarti"]Find all functions $ f: R^ \plus{} _0 \minus{} > R^ \plus{} _0$ such that

$ f(y)f(xf(y)) \equal{} f(x \plus{} y)$

for any x, y.[/quote]

Let $ P(x,y)$ be the assertion $ f(y)f(xf(y))\equal{}f(x\plus{}y)$

$ f(x)\equal{}0$ $ \forall x$ and $ f(x)\equal{}1$ $ \forall x$ are obviously the only two constant solutions. Let's now consider non constant solutions.

1) Assume $ f(x)$ is a non constant solution and $ \exists u>0$ such that $ f(u)\equal{}0$
================================================================================
Since $ f(x)$ is non constant, $ \exists c\geq 0$ such that $ f(c)\ne 0$. Then $ P(0,c)$ $ \implies$ $ f(c)f(0)\equal{}f(c)$ and so $ f(0)\equal{}1$
Let $ x\ge u$ : $ P(x\minus{}u,u)$ $ \implies$ $ f(x)\equal{}0$ $ \forall x\ge u$

Let then $ a\equal{}\inf\{x>0$ such that $ f(x)\equal{}0\}$. We got $ f(x)\equal{}0$ $ \forall x>a$
If $ a\equal{}0$, we got a solution : $ f(0)\equal{}1$ and $ f(x)\equal{}0$ $ \forall x>0$ (easy to check back that this is a solution)
Consider now $ a>0$ (and so $ f(x)\ne 0$ $ \forall x\in[0,a)$
If $ f(a)\ne 0$, $ P(\frac a{f(a)},a)$ $ \implies$ $ f(a)^2\equal{}f(a\plus{}\frac a{f(a)})$ and we got a contradiction ($ LHS\ne 0$ while $ RHS\equal{}0$) so $ f(a)\equal{}0$.

Suppose now $ \exists y_1<y_2<a$ such that $ f(y_1)\equal{}f(y_2)$. Then :
Comparing $ P(x,y_1)$ and $ P(x,y_2)$, we get $ f(x\plus{}y_1)\equal{}f(x\plus{}y_2)$
Setting $ x\equal{}a\minus{}\frac{y_1\plus{}y_2}2$ in this equation implies $ f(a\minus{}\frac{y_2\minus{}y_1 }2)\equal{}f(a\plus{}\frac{y_2\minus{}y_1}2)$ but then $ RHS\equal{}0$ while $ LHS\ne 0$ and so $ f(x)$ is injective over $ [0,a)$

Let then $ x,y<a$ :
$ P(\frac x{f(y)},y$ $ \implies$ $ f(x)f(y)\equal{}f(\frac x{f(y)}\plus{}y)$ and so $ \frac x{f(y)}\plus{}y<a$ since $ LHS\ne 0$
$ P(\frac y{f(x)},x$ $ \implies$ $ f(y)f(x)\equal{}f(\frac y{f(x)}\plus{}x)$ and so $ \frac y{f(x)}\plus{}x<a$ since $ LHS\ne 0$
So $ f(\frac x{f(y)}\plus{}y)\equal{}f(\frac y{f(x)}\plus{}x)$ with both args $ <a$ and so (since $ f(x)$ is injective :
$ \frac x{f(y)}\plus{}y\equal{}\frac y{f(x)}\plus{}x$
$ \iff$ $ xf(x)\plus{}yf(x)f(y)\equal{}yf(y)\plus{}xf(x)f(y)$
$ \iff$ $ xf(x)(f(y)\minus{}1)\equal{}yf(y)(f(x)\minus{}1$
$ \iff$ $ xf(x)\equal{}c(f(x)\minus{}1)$ for some real $ c\ne 0$
$ \iff$ $ f(x)\equal{}\frac c{c\minus{}x}$ $ \forall x\in[0,a)$

Obviously, either $ c<0$, either $ c\ge a$ (else $ f(x)<0$ for some $ x\in[0,a)$

Plugging this back in initial equation and remembering that $ f(x)\equal{}0$ $ \forall x\ge a$ :
consider $ 0\le y< a\le x\plus{}y$, we need $ xf(y)\ge a$ (else $ LHS\ne 0$ while $ RHS\equal{} 0$) so $ \frac {cx}{c\minus{}y}\ge a$ $ \forall 0\le y< a\le x\plus{}y$ and so $ 0<c\le a$, so $ c\equal{}a$

And it's easy then to check that we got a solution.

2) Assume $ f(x)$ is a non constant solution and $ f(x)>0$ $ \forall x$
===================================================================
Since $ f(x)$ is non constant, $ \exists c\geq 0$ such that $ f(c)\ne 0$. Then $ P(0,c)$ $ \implies$ $ f(c)f(0)\equal{}f(c)$ and so $ f(0)\equal{}1$
Assume $ \exists u$ such that $ f(u)>1$
$ P(\frac u{f(u)\minus{}1},u)$ $ \implies$ $ f(u)f(\frac {uf(u)}{f(u)\minus{}1})\equal{}f(\frac {uf(u)}{f(u)\minus{}1})$ and so $ f(\frac {uf(u)}{f(u)\minus{}1})\equal{}0$, which is impossible since we assumed $ f(x)>0$ $ \forall x$, so $ f(x)\le 1$ $ \forall x$

Let then $ x<y$. Then $ P(y\minus{}x,x)$ $ \implies$ $ f(x)f((y\minus{}x)f(x))\equal{}f(y)$ and, since $ f((y\minus{}x)f(x))\le 1$, we get $ f(y)\le f(x)$ and $ f(x)$ is a non increasing function.

If $ \exists u>0$ such that $ f(u)\equal{}1$, then $ P(u,u)$ $ \implies$ $ f(2u)\equal{}1$ and so $ f(2^nu)\equal{}1$ and, since $ f(x)$ is non increasing and $ f(0)\equal{}1$ : $ f(x)\equal{}1$ $ \forall x$, which is impossible (since we assume $ f(x)$ is a non constant solution).

So $ f(x)<1$ $ \forall x>0$ and $ f(x)$ is a decreasing function and, so, is injective.
Then :
$ P(\frac x{f(y)},y$ $ \implies$ $ f(x)f(y)\equal{}f(\frac x{f(y)}\plus{}y)$
$ P(\frac y{f(x)},x$ $ \implies$ $ f(y)f(x)\equal{}f(\frac y{f(x)}\plus{}x)$
So $ f(\frac x{f(y)}\plus{}y)\equal{}f(\frac y{f(x)}\plus{}x)$ and so (since $ f(x)$ is injective) :
$ \frac x{f(y)}\plus{}y\equal{}\frac y{f(x)}\plus{}x$
$ \iff$ $ xf(x)\plus{}yf(x)f(y)\equal{}yf(y)\plus{}xf(x)f(y)$
$ \iff$ $ xf(x)(f(y)\minus{}1)\equal{}yf(y)(f(x)\minus{}1$
$ \iff$ $ xf(x)\equal{}c(f(x)\minus{}1)$ for some real $ c\ne 0$
$ \iff$ $ f(x)\equal{}\frac c{c\minus{}x}$ $ \forall x$

In order to have $ f(x)\ge 0$ $ \forall x$, we need $ c<0$ and(setting $ b\equal{}\minus{}c>0$) : $ f(x)\equal{}\frac b{x\plus{}b}$ $ \forall x$
Plugging this back in the initial equation, we check that this indeed a solution.

3) synthesis of solutions :
===========================

$ f(x)\equal{}0$ $ \forall x$

$ f(x)\equal{}1$ $ \forall x$

$ f(0)\equal{}1$ and $ f(x)\equal{}0$ $ \forall x>0$

$ f(x)\equal{}\frac a{a\minus{}x}$ $ \forall x\in[0,a)$ and $ f(x)\equal{}0$ $ \forall x\ge a$ for some real $ a>0$

$ f(x)\equal{}\frac a{x\plus{}a}$ $ \forall x$ for any real $ a>0$


algebra
polynomial
calculus
calculus computations

I think you put this in a wrong place... :?:

[quote="countdownmath"]Find a polynomial with integer coefficients that has the number $\sqrt{2}+\sqrt[3]{3}$ as one of its roots.[/quote]

Hint: $x=\sqrt{2}+\sqrt[3]{3}$ and work backwards.


real analysis
integration
real analysis unsolved

Show that if $ g \in L^1(\mathbb{T}), \mu$ is a finite measure on $ \mathbb{T}\equal{}\mathbb{R}/(2\pi \mathbb{Z})$, and $ \mu(x\plus{}\alpha \pi)\minus{}\mu(x)\equal{}g(x)dx$ for some irrational $ \alpha$ (where $ dx$ denotes the Lebesgue measure), then $ \mu$ is absolutely continuous with respect to the Lebesgue measure.

[quote="PluTo"]$ \mathbb{T} \equal{} \mathbb{R}/(2\pi \mathbb{Z})$, and $ \mu(x \plus{} \alpha \pi) \minus{} \mu(x) \equal{} g(x)dx$ for some irrational $ \alpha$ [/quote]
Can you explain what do you mean here more explicitly?

I think it means for any Lebesgue measurable set $ A$,
\[ \mu(A\plus{}\alpha \pi) \minus{} \mu(A) \equal{} \int _A g(x) dx\]


To find the length of span of a proposed ski lift from A to B, a surveyors measures the angle DAB to be 35 degrees and then walks off a distance 0f 800 feet to C and measures the angle ACB to be 25 degrees. What is the distance from A to B?

Maybe you omitted a diagram with supplementary info. (Where is C? Is the line CD intended to be parallel to the line AB?)


If the six digits 1, 2, 3, 5, 5 and 8 are randomly arranged into a six-digit positive integer, what is the probability that the integer is divisible by 15? Express your answer as a common fraction.

5 has to be at the end which will happen $ \frac{2}{6}\equal{}\frac{1}{3}$


what would you say about $p=3$ ? :)

for reference , you can look at

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=2%5E%7Bp-1%7D-1&t=83190[/url]

[quote="shyong"]what would you say about $p=3$ ? :)

yeah, i thought about that too. but from my working, i got sth like p + 2 = k^2. so it ruled that out. but i guess my working was wrong.
thanks  :)

[quote="violetcraze"]If p is an odd prime and  (2^(p-1) - 1) / p is a complete square (k^2), Find p.

I got 7, but i got another answer,  47, which is wrong. How should this question be tackled?[/quote]

factor!

okay. i got it. factored wrongly. haha. thanks.  :P


let $a,b,c$ be   positive numbers    .
 prove that:

$\large \sqrt[3]{\frac{(a+b)^2(b+c)^2(c+a)^2}{abc}}\geq{\frac{4(a+b+c)}{3}}$

It's is an old inequality.
It is equivalent to
$27(a+b)^2(b+c)^2(c+a)^2\ge 64(a+b+c)^3abc$
$\leftrightarrow 27\sum(a^4b^2+a^2b^a)+54\sum a^3b^3\ge 10\sum a^4bc+ 30\sum(a^3b^2c+a^3bc^2)+114a^2b^2c^2$
and we have
$5\sum(a^4b^2+a^2b^4)\ge 10%Error. "suma" is a bad command.
^4bc$
$3\sum(a^4b^2+a^2b^4)\ge 3\sum(a^3b^2c+a^3bc^2)$
$54\sum(a^3b^3) \ge 27\sum(a^3b^2c+a^3bc^2)$
$19\sum(a^4b^2+a^2b^4)\ge 114a^2b^2c^2$

But I think this is the worst solution :(


Is it true that for every positive reals $x_{1},x_{2},\ldots,x_{n}$, the function
$f(t) = \left(\frac{x_{1}^{t}+x_{2}^{t}+\ldots+x_{n}^{t}}n \right)^{\frac{1}{t}}$
is concave when $t>0$ and convex when $t < 0$?

This fact is very beautiful, indeed!

What a pity it's false...

Let's consider the case n=3

According to your claim, $QM+GM\ge 2AM$ but $\sqrt[2]{\frac{8^{2}+8^{2}+1}{3}}+\sqrt[3]{8\cdot 8\cdot 1}< 2\cdot\frac{8+8+1}{3}$


induction
number theory proposed
number theory

Solve in integers
\[ x\cdot \phi^{n \plus{} 1} \plus{} y\cdot\phi^n \equal{} 1\]
where $ \phi \equal{} \frac {1 \plus{} \sqrt {5}}{2}$ - golden ration.

[b]Lemma 1.[/b] If $ a,b,c,d$ are integers and $ a + b\phi = c + d\phi$ then $ a = c,b = d$.
[b]Proof.[/b] It immediately follows from the fact that $ \phi$ is irrational.

[b]Lemma 2.[/b] $ \phi^n = F_n\phi + F_{n - 1}$, where $ F_i$ is the $ i$-th Fibonacci number.
[b]Proof.[/b] Follows from induction and the identity $ \phi^2 = \phi + 1$.

Using the lemmas our equation is equivalent to the following system:
\begin{align*}xF_{n + 1} + yF_n & = 0\qquad (1) \\
xF_n + yF_{n - 1} & = 1.\qquad (2)\end{align*}
Since $ \gcd(F_n,F_{n + 1}) = 1$, from $ (1)$ it follows that $ x = kF_n$ and $ y = - kF_{n + 1}$ for some integer $ k$. This implies, from $ (2)$, $ k(F_n^2 - F_{n + 1}F_{n - 1}) = 1$. But from Cassini's identity, $ F_{n + 1}F_{n - 1} - F_n^2 = ( - 1)^n$. Hence $ k(-1)^{n+1} = 1$. Therefore $ k=(-1)^{n+1}$ and we conclude that $ (x,y)=((-1)^{n+1}F_n,(-1)^nF_{n+1})$.


Find all triples of positive integers $(a,b,c)$ such that
\[\left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)\left(1+\frac{1}{c}\right) =2 \]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=110648]Click Here to Reveal Hidden Thread with This Problem[/url]


Please stop clogging up this board with these problems! I assume that you can do the AMC 10 #1, so I'm not seeing the point of posting. If it is for other users' benefit, please post them in the Wiki!

Thanks,

~BOGTRO

No, it's fine for them to post the problems. They go in the Contests section, and the solutions link to these threads.

Well at least put them in one topic.. It's annoying.

See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=191609]here[/url], they must go in separate topics.


Here's a problem that was passed on to me from a friend. I found a solution, but in a very messy way. I wonder if anyone can find a nicer one.

In the figure below,  :ang: DBE = 30, :ang: EBC = 50, :ang: ECD = 20, and :ang: DCB = 60 (all in degrees). Find the measure of :ang: CDE and :ang: BED.

There's a very slick solution:

[hide]Draw a regular 18gon with A at the center and B and C consecutive vertices. Draw a few diagonals. Can you prove that they are concurrent? (hint, use angle form of ceva) Now can you find the angles?[/hide]

Yeah, I've seen a variety of solutions to this question, but I've never seen one that is simultaneously nice and believably discoverable by even a very observant geometer.  Is this an ARML question?  Or AIME?

When I first saw this problem, I think I solved it without an especially large amount of trouble with the law of sines. I remember that there was a nicer solution in [i]Crux Mathematicorum[/i] though, so perhaps I'll try to dig out that issue and present the solution somewhere. I don't think this question was on a contest, but correct me if I am mistaken.

It is also in Geometry Revisited.  Their solution was also very slick.

A brief background of the problem, along with eight solutions, can be found [url=http://mathcircle.berkeley.edu/BMC4/Handouts/geoprob.pdf]here[/url]. Keep in mind that the lettering is a little different on that file.


Sarah and Anita are playing a game with a standard six-sided die. Each of them take turns rolling the die. If Sarah rolls a 5 or a 6, she wins and the game ends. If Anita rolls a 1,2,3 or 4, she wins and the game ends. Sarah is the first person to roll. What are their respective probabilities of winning? Your answers should be in exact values.

Thanks for helping or trying :D

Clearly someone wins with probability $1$, so it's enough to calculate the probability that Sarah wins.

Her probability of winning is:

Pr( Sarah wins on first turn) + P( Sarah doesn't win on first turn, Anita doesn't win on second turn, Sarah wins on third turn) + P( Sarah doesn't win on first turn or third turn, Anita doesn't win on second turn or fourth turn, Sarah wins on fifth turn) + ...

This is a geometric sequence - you can sum it if you're careful.

[hide="Answer"]For Sarah ${3\over 7}$, for Anita ${4\over 7}$.[/hide]

If you want an explanation, I'll post it.

[hide="if you still cant get it..."]
For Sarah , if she won in first turn then the probability is $2/6=1/3$
If Sarah won in second turn then it must be both she and Anita lost in first round multiply by the probability Sarah won which is $4/6\cdot 2/6\cdot 2/6 = 2/9\cdot 1/3 = 2/27$

So if Sarah won in the $n^{th}$ round , then her probability would be $(2/9)^{n-1}\cdot (1/3)$

Hence just write all them out and it is sum of geometric progression to infinity with first term $1/3$ and common ratio $2/9$ and it shouldnt be hard to get the total to be $3/7$ .

Similarly work out for Anita with first term $2/3\cdot 2/3=4/9$ and common ratio $2/9$  :)
[/hide]


I'm trying to put an acute accent over a vowel with a macron in a table.

pit\'{\=a} gives $ pit\'{\=a}$, which is what I want.

However, when I use this in a table, the acute accent does not appear over the vowel.  For instance, the following code produces the table below:

[quote]\begin{tabular}[t]{|l|l|}\hline
English&father\\\hline
Latin&pater\\
Ancient Greek&pat\'{\=e}r\\
Sanskrit&pit\'{\=a}\\\hline
\end{tabular}[/quote]

\begin{tabular}[t]{|l|l|}\hline
English&father\\\hline
Latin&pater\\
Ancient Greek&pat\'{\=e}r\\
Sanskrit&pit\'{\=a}\\\hline
\end{tabular}

Is there any way to fix this?

Firstly, pit\'{\=a} doesn't give what you want as it gives $ \text{pit\'{\\equal{}a}}$ in a document. It [i]appears[/i] to work because on the forum putting it between dollar signs uses maths mode. If you try it in a document you get a load of errors which the forum has shielded from you. A table is in text mode so it is showing it correctly.

So the question is how to show the accent in text mode? The following will do it in a document (though not on the forum):
[code]\usepackage{tipa}
...
\textacutemacron{a}[/code]
and this works in a table or anywhere else in text mode.

You'll fnd more details on pages 15-16 in [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=comprehensive]The Comprehensive LaTeX Symbol List[/url] and the [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=tipa]TIPA Manual[/url].

Ah, that explains it and that works, thank you!


calculus
integration
abstract algebra
limit
trigonometry
absolute value
real analysis

How prove that for all $\epsilon>0$
\[ \lim_{n\to\infty}\int_{\mid x\mid\geq \epsilon}D_{n}(x)dx=0 \]
where $D_{n}(x)=\frac{1}{2\pi}\int_{-n}^{n}e^{ixs}ds$ is a Dirichlet Kernel (the one involved in a theory about the Fourier transform on real line)

To start with, it would help to have the explicit form of the Dirichlet kernel. This is algebraically a little nicer on the line than on the circle. By explicit calculation in the integral,

$D_n(x)=\frac1{\pi}\frac{\sin (nx)}x.$ This is an even function.

(We also have $D_n(0)=\frac{n}{\pi}$ but we won't need that.)

So $\int_{|x|>\epsilon}D_n(x)\,dx= \frac2{\pi}\int_{\epsilon}^{\infty}\frac{\sin(nx)}x\,dx$

Integrate by parts, integrating the sine.

$=\left.\frac{2(1-\cos(nx))}{n\pi x}\right|_{\epsilon}^{\infty}+ \frac2{\pi n}\int_{\epsilon}^{\infty}\frac{1-\cos(nx)}{x^2}\,dx$

In absolute value, this is bounded by

$\le \frac4{n\pi\epsilon}+\frac2{\pi n}\int_{\epsilon}^{\infty}\frac2{x^2}\,dx$

$=\frac8{n\pi\epsilon}.$

For fixed $\epsilon,$ this tends to zero as $n\to\infty.$

Note that all of these integrals were conditionally convergent improper intervals. $D_n\not\in L^1(\mathbb{R})$ and

$\int_{|x|>\epsilon}|D_n(x)|\,dx=\infty.$


complex analysis
unsolved
domain
integration
calculus

Let $f$ be a holomorphic function in some neighborhood of the origin, $f(0) = 0$. Suppose there is a domain $U \subset \bar{C}$ with $f(U) \subset U$ and $f^{n}(z) \to 0$ on compact subsets of $U$. Prove that then either $|f'(0)|<1$ or $f'(0)=1$.

*Here $f^{n}$ mean the $n$'th iterate of $f$.

It's obvious that $f^{n}$ tends to zero in some neighborhood of zero if $|f'(0)|<1$, and that $f^{n}$ can't tend to zero in any neighborhood of zero if $|f'(0)|>1$. The unit circle is the interesting case; first-order behavior isn't enough to describe the action there.

Do you have an example of such a function with $f'(0)=1$ and the $f^{n}$ converging to zero?

Can I ask this problem to be clarified a little for me? What is $C$ and what is the difference between $f^{n}(z) \to 0$ on every compact subset of $U$ and just $f^{n}(z) \to 0$ for every fixed point $z$. Are we assuming uniform convergence on every compact subset?

The answer for the uniform version: If $f$ is an analytic function on some region $R$ containing the origin with $f(0)=0$, there is $U\subset R$ such that $f^{n}(z)$ is defined for all $n$ and tends to zero uniformly on compact subsets of $U$ if and only if $|f'(0)|<1$.

Proof: The cases $|f'(0)|<1$ and $|f'(0)>1$ are left as exercises. For the case $|f'(0)=1$ we use the following lemma:
Lemma: Suppose $g$ is an injective analytic function on the closure of some disk $D$ with radius $r$ centered at zero. Then the area of $g(D)$ is at least $\pi r^{2}|g'(0)|^{2}$.
Proof: The area of $g(D)$ is $\int\int_{D}|g'(z)|^{2}dA$ by the change of variables formula in two dimensions. The absolute value $|g'(z)|^{2}$ is a subharmonic function, so its average value on any circle or any disk is at least the value at the center. This central value is $|g'(0)|^{2}$, and we have the desired inequality.

Now suppose $f$ is as in the statement, $|f(0)|=1$, and $f^{n}$ converges uniformly (on compact sets) to zero in some neighborhood $U$ of zero. There is some $C$ so that $f$ is injective on the disk of radius $C$ centered at zero, and that disk is contained in $U$. For each $r<C$, let $D_{r}$ be the disk of radius $r$ centered at zero. By the uniform convergence, there is some $N(r)$ such that $f^{n}(D_{r})\subset D_{r}$. Since $|(f^{n})'(0)|=1$, the lemma implies that $f^{n}$ is not injective on $D_{r}$. In particular, $|f^{m}(z)|>C$ for some $z\in D_{r}$ and $m<n$.
Given any $k$,  we can choose $r$ small enough so that $f^{m}(D_{r})\subset D_{C}$ for all $m<k$. Doing this repeatedly, we have a sequence $z_{j}$ tending to zero so that $|f^{m(j)}(z_{j})|>C$ for some increasing sequence $m(j)$. This contradicts uniform convergence, and we are done.

No. That's a much simpler proof of what I posted.

[quote="grobber"]What's wrong with this :? :

If a sequence of holomorphic functions converges uniformly on compact sets to a holomorphic function, then the sequence of derivatives converges (again, uniformly on compact sets) to the derivative of the limit (this follows from Cauchy's integral formula, expressing $f'(z)$ in terms of the values of $f$ on a small circle around $z$). In our case, if the sequence is $f^{n}$, the derivative of $f^{n}$ in $0$ is the $n$'th power of $f'(0)$ (because $f(0)=0$). This can only converge to $0$ if $|f'(0)|<1$.

Am I missing something obvious?[/quote]

But it wasn't said that $U$- the domain on which we have the convergence on compact subsets, contains $0$. If it doesn't contain zero, the derivatives may not converge to $0$.
For the proof of this fact you are using, i think you need this small circle around zero to be contained in the domain of the convergence.

[quote="jmerry"]With that clarification, here's a positive example: $f(z)=\frac{z}{z+1}, U=\mathbb{C}-(-\infty,0]$.

The closure of $U$ must contain zero, of course. The arguments for $|f'(0)|<1$ and for $|f'(0)|>1$ still work, of course.[/quote]

Thanks for the example. But i didn't quite understand what do you mean by you second line: i still don't know how do you deal the case $|f'(0)| = 1$ out? why do we have the only possibility $f'(0) = 1$ in this case?

That part of the problem is still unsolved here. I've thought about it some, but I don't have anything yet.

It seems to me that it is just false as stated: take $f(z)=-z+z^{3}$ and $U=\{|z|<r, \arg z\in (-r,r)\cup(\pi-r,\pi+r)\}$ where $r>0$ is sufficiently small. Am I missing something? :?

That's definitely convergent: $r=\frac{\pi}6$ ought to be enough, and I suspect you could go almost to the imaginary axis with the right shape of region.


conics
parabola
geometry
vector
Ring Theory
advanced fields
advanced fields unsolved

Some time ago I purchased David Eisenbud's encylopedic "Commutative Algebra" at my university bookstore, as it had been marked down to less than half the original price. I thought this sounded like a good deal, but to this day it remains all around the most inpenetrable book I own. I would really like to understand, for instance, in what way a parabola tangent to a given line may be represented by the ring k[x,y]/(y^2-yx^2). pg 37. Any help here would be most appreciated.

This is more the realm of algebraic geometry (though the two are very closely related). You can associate to k[x,y]/A the points in k^2 where each element of A vanishes. For a principal ideal like A = (y^2-yx^2), this is just the set of points where y^2 = yx^2, or just the two components y = 0 and y=x^2.

If $ A$ is a prime ideal, the resulting zero locus is called an "affine variety". Affine varieties form a category, and its opposite category is equivalent to the category of finitely generated k-algebra domains. So you can recover information about the ring by studying it's associated affine variety. Things get a lot deeper than this, but it's just one reason we might be interested in the geometry of rings.

Thanks for the reply. I'm not very good with the notation of algebra yet; by k^2 do you mean k[x,y], or the direct product of k[x,y] with itself, or maybe something else?

Neither. It's the affine space $ \{(a,b): a,b\in k\}$. It has a vector space structure, but not meaningful ring-type multiplication.

Ah, I see. All these things are no doubt somewhere to be found in the book, but his prose is so chatty as opposed to systematic, it may take me forever to put the pieces together. Eisenbud really does converse "in the language of algebra", it is way beyond a simple definition-theorem-proof layout.


prove that if $f(x)$ and $g(x)$ are two sinusodial functions with the same period , then their linear combination is also sinusodial

[quote="aviateurpilot"]$Acos(wx+a)+Bcos(wx+b)=Re((Ae^{a}+Be^{b})e^{iw})=(Ae^{a}+Be^{b})cos(wx)$[/quote]
it doesn't look right... minor tho


Earth is represented on a map of a portion of the solar system so that its surface is the circle with equation x^2 + y^2 + 2x + 4y - 4091 = 0.  A weather satellite circles 0.8 units above Earth with the center of its circular orbit at the center of Earth.  Find the equation for the orbit of the satellite on this map.

Use completion of a square to get a circle in the form $ (x \minus{} h)^{2} \plus{} (y \minus{} k)^{2} \equal{} r^{2}$ where $ (h,k)$ is the vertex of your circle and $ r$ is the radius. From there, just increase the radius by 0.8, unless I'm badly mis-reading the question.