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number theory
prime numbers
number theory unsolved

A New Way to Generate Prime Number Lists
by Zhou, Shiyang

<Note: this article is translated from the Chinese version by my newphew. If you have any
comments, please email to wrglin <at> gmail.com>

This article present a new way in the category of sifting methods to make a prime number list.

Definition: sub-natural numbers
A natural number that is not divisible by none of 2, 3, and 5 is defined as a "sub natural
number".

Define R as the set of all possible remainders of a sub-natural number n divided by 90,
then R is the a set consisting of the following 24 numbers:

1, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 49, 53, 59, 61, 67, 71, 73, 77, 79, 83,
89

Obviously any sub-natural number n can be represented as:

n = 90*m + r

where m is a non-negative integer and r is a member of R.

Definition: number axles
A number axle denoted as f(r), is the set of all sub-natural numbers that can be
represented as 90*m +r, where m is a non-negative integer and r is a member of R. A number
in axel f(r) can be denoted as f(r, m).

Obviously, for two different numbers r1 and r2 from R, f(r1) and f(r2) are distinctive
sets without any overlapping members. Also obviously, the combinations of f(r) for every r
in R is the whole set of sub-natural numbers.

Definition: factorization of a number axle.
f(r2) and f(r3) is defined as a factorization of f(r1) and denoted as f(r1) := f(r2)*f(r3),
if the result of any number from f(r2) multipled by any number from f(r3) is a member of
f(r1).

It is easy to find out all the possible factorizations for all number axles, as we can
see:

if a = 90*m + x, b = 90*n + y, then

a*b = (90*m + x) * (90*n + y) = 90*(90*m*n + m*y +n*x) + x*y
                                = 90*(90*m*n + m*y +n*x ) + 90*k + z
                                = 90*(90*m*n + m*y +n*x + k) + z
                                where r < 90

Apparently, a*b is a member of number axle f(z).

Suppose we have the full list of factorization for f(r), as represented as:

f(r) := f(x_1) * f(y_1)
f(r) := f(x_2) * f(y_2)
...
f(r) := f(x_k) * f(y_k)

If we sift out all the numbers in f(r) that can be represented by one of its
factorizations, the remaining numbers in f(r) are prime numbers. The proof is as
following:

Suppose we have a sub-natural number x in f(r) that is a composite number, then it can be
factored as:

x = y*z

Obviously, y and z must be sub-natural numbers too, otherwise x won't be a sub-natural
number, hence y and z must belong to one of the number axles, and therefore x must be
sifted out by one of the number axle factorizations. So all composite numbers in f(r) will
be sifted out by the its factorizations.

<END>

Appendix: A new approach to solve the "1+1 = 2" problem

1. Break down an even number into combination of number axles

If we divide an even number n by 9, and denote the remainder as r(n), then all even
numbers can be classified into 45 categories depending on r(n) and its last decimal
digit. We can denote a category as C(r, d), where r is the remainder from division of 9,
and d is the last decimal digit.

Further, each category can be expressed as a combination of number axels, namely:
f(r1, k - i) + f(r2, i) = 2m, where 0<=i<=k, k = floor(2m/90)

Let's pick the category C(0, 0) as an example, it can be expresses as the following
combination:
C(8,9) + C(1,1): f(89) + f(91)
C(1,9) + C(8,1): f(19) + f(71)
C(8,7) + C(1,3): f(17) + f(73)
C(1,7) + C(8,3): f(37) + f(53)
C(5,9) + C(4,1): f(59) + f(31)
C(4,9) + C(5,1): f(49) + f(41)
C(5,7) + C(4,3): f(77) + f(13)
C(7,9) + C(2,1): f(79) + f(11)
C(2,9) + C(7,1): f(29) + f(61)
C(7,7) + C(2,3): f(7) + f(83)
C(2,7) + C(7,3): F(47) + f(43)
C(4,7) + C(5,3): f(67) + f(23)

The combination above can be prime number pair, composite number pair, prime-composite
number pair.

Given an number axle combination:
f(r1) + f(r2) = 2m

In all the number pairs, if we denote the prime numbers in f(r1) and f(r2) as p(r1) and
p(r2), the composite numbers as c(r1) and c(r2), all the composite numbers in the
composite number pair as c(r1) and c(r2), the remaining composite numbers as r(r1) and
r(r2). If we can prove that in a number axle combination, the size of p(r1) is bigger than
the size of r(r2), then there must be at least one prime number pairs. If we can prove
that is true for at least one of the axle combinations for an even number, then 1+1 = 2 is
proved.

[quote="wanrong.lin"]If we sift out all the numbers in f(r) that can be represented by one of its
factorizations, the remaining numbers in f(r) are prime numbers. [/quote]

This is the only step that would make this paper meaningful, and you haven't given a sieving algorithm.  The rest of the first part of the paper is just a discussion of the multiplication table of $ U_{90}$, and I can't make any sense of the appendix.  Are you trying to prove the Goldbach Conjecture?


Euler
logarithms
quadratics
algebra
quadratic formula

[hide="Solution"]
Note that $\cosh x = \frac{e^x + e^{-x}}{2}$.  Rewriting, we see that we want to solve $\cosh x = 3 \implies x = \cosh^{-1} 3$.  Using the formula $\cosh^{-1} z = \ln (z + \sqrt{z^2 - 1})$, we see that our answer is $\ln (3 + \sqrt{9 - 1}) = \boxed{\ln (3 + 2\sqrt{2})}$.
[/hide]

[hide="Edit: Easier solution"]
Substitute $u = e^x$.  Then we see that we have $u + \frac{1}{u} = 6 \implies u^2 - 6u + 1 = 0$.  Using quadratic formula gives us $u = \frac{6 \pm \sqrt{36 - 4}}{2} = 3 \pm 2\sqrt{2} \implies x = \ln(3 \pm 2 \sqrt{2})$, a more complete solution.
[/hide]

[quote="tiredepartment"]Solve for x, where e represents the Euler number (commonly associated with natural logarithms).

$\frac{e^x + e^{-x}}{3} = 2$[/quote]

[hide="Solution"]
$\frac{e^x + e^{-x}}{3} = 2$
$(e^x)+\frac{1}{e^x}=6$
$(e^x)=y$  :?
$y+\frac{1}{y}=6$
$y^2 + 1 = 6y$
$y^2 - 6y + 1 = 0$
solving the quadratic formula we get
$y = 3\pm2\sqrt{2}$
and solving $(e^x)=y$ we have
$x = ln(3+2\sqrt{2})$
$x = ln(3-2\sqrt{2})$[/hide]

Note from mod: Please hide your solution next time. Thanks.  :)

[hide]
e^(-x)+e^x=6, now multiply both sides by e^
e^2x+1-6e^x=0 rearrange
Set e^x=a

we have a^2-6a+1=0
Quadratic Equation for a: a=3+2sqrt(2), 3-2sqrt(2)
so e^x=3+2sqrt(2), 3-2sqrt(2)

x= ln(3+2sqrt(2)) and ln(3-2sqrt(2))

And we are done
[/hide]


Prove if a,b,c>0  such that $ \sqrt[30]{a}\plus{}\sqrt[4]{b}\plus{}\sqrt[1991]{c}\equal{}3$ then
$ \sum\sqrt{a^2\plus{}3}>\equal{}\frac{a\plus{}b\plus{}c\plus{}9}{2}$

By Am-Gm we can prove :
$ \sqrt{a^2\plus{}3} \ge  \frac{a\plus{}3}{2}$
Done :D

[quote="mitdac123"]By Am-Gm we can prove :
$ \sqrt {a^2 \plus{} 3} \ge \frac {a \plus{} 3}{2}$
Done :D[/quote]
$ \sqrt{a^2\plus{}3}\geq\frac{a\plus{}3}{\sqrt{2}}\geq\frac{a\plus{}3}{2}$

[quote="Sudoku"]$ \sqrt {a^2 \plus{} 3}\geq\frac {a \plus{} 3}{\sqrt {2}}$[/quote]

Well, [i]this[/i] isn't true (take eg. $ a\equal{}1$).

[quote="Sudoku"]$ \frac {a \plus{} \sqrt3}{\sqrt {2}}\geq\frac {a \plus{} 3}{2}$[/quote]

Then this is the part which is not true.. (take $ a\equal{}1$)

However $ \sqrt{\frac{a^2\plus{}3*1^2}4}\ge \frac{a\plus{}3*1}4$ [i]is[/i] true.


Sal the Magician asks you to pick any five cards from a standard deck.  You do so, and then show them to Sal's assistant Pat, who places one of the five cards back in the deck and then puts the remaining four cards into a pile.  Sal is blindfolded, and does not witness any of this.  Then Sal takes off the blindfold, takes the pile of four cards, reads the four cards that Pat has arranged, and is able to find the fifth card in the deck (even if you shuffle the deck after Pat puts the card in the deck).  Assume that neither Sal nor Pat has supernatural powers, and that the deck of cards is not marked.  How is the trick done?  Harder version: you pick which of the five cards goes back into the deck (instead of Pat).

I understand how to do the first part:
[hide]Pat selects one of the two cards of the same suit (there must be, by the Pigeonhole Principle), the one whose number is 6 or less above the other, and puts that one in the deck.  He then puts the second card of that suit on top of the pile and arranges the 3 other cards to signify the number 1-6.  Sal then adds this number 1-6 to the top card and takes the suit of the top card, and finds that card in the deck.[/hide]

But I have no idea how to do the second part, and do not believe it is possible.  This is because there are only 24 arrangements of the 4 remaining cards, and 48 cards in the rest of the deck.  So no matter what, there are 24 cards Pat cannot encode.

Anyone know how it's done?

[quote="tenniskidperson3"]I understand how to do the first part:
[hide]Pat selects one of the two cards of the same suit (there must be, by the Pigeonhole Principle), the one whose number is 6 or less below the other, and puts that one in the deck.  He then puts the second card of that suit on top of the pile and arranges the 3 other cards to signify the number 1-6.  Sal then adds this number 1-6 to the top card and takes the suit of the top card, and finds that card in the deck.[/hide][/quote]

1-6? Don't you mean 1-12? I don't get how you would have Pat do a [size=167]A♠3♣[color=red]8♥A♦K♦[/color][/size], for example (assuming king-high).

My proposal: (I haven't thought this through, so it might not work)
[hide]
He places the only card of a suit if you chose 4 different suits (2-1-1-1), leaving 3 suits left and indicating the suit already with 2-1-1, leaving 4 cards to indicate rank. If you chose 3 different suits in a 2-2-1 or a 3-1-1, then he puts an "only card" in. I think he's still free to indicate both suit and rank with 2-2 or 3-1. If you did 2 suits in a 3-2 or 4-1, reduce it to a 3-1 (or a 2-2?), and with a 5 (one suit), he just puts any card in to reduce to 4-0.[/hide]

Maybe the trick involves something discreet, like putting a hidden card face-up instead of face-down.

You don't really need 1 through 12, because Pat is free to choose either card of the suit.  And if one card is more than 6 above the other, like the K and A you have, then the A is only one more than the K, going in a circle.  So Pat would put the A back, put the K on top, and arrange the other 3 cards in a predetermined way signifying the number 1.  Sal would then look at the K and add 1 to get the A.  It is also the same suit as the K, so Sal knows that it is the A of diamonds.

If this doesn't really make sense, I learned about this trick from here:
[url]http://www.qbyte.org/puzzles/puzzle02.html[/url]
Number 19.


number theory
greatest common divisor
number theory proposed

1.: Why is this in Solved problems
2.: what means $pgcd$ ¿ (if its standard $gcd$ its trivial)

[quote="bénabar"]Let $a$, $b$ and $c$ in $\mathbb{N}$ with $ab=c^2$ and $pgcd(a,b)=1$.
Prove that: $\exists (c,f) \in \mathbb{N}^2 / e^2=a$ and $f^2=b$.[/quote]
zetax a raison .Pour moi il s'agit juste d'utiliser la decomposition en produits de facteurs premier pour le voir exactement comment ça fonctionne

Someone please give me a dictionary it starts to be interesting in here :D  :D


[hide]Hint

You can factor the $ y^{2}\minus{}4xy \plus{} 4x^{2}$ to $ (y\minus{}2x)^{2}$

then you have $ (y\minus{}2x)^{2} \plus{} y \minus{} 2x \equal{} 6$

From there factor out a $ y\minus{}2x$ to get

$ (y \minus{} 2x)(y \minus{} 2x \plus{} 1) \equal{} 6$[/hide]

Thanks for your reply, SnipedYou.

Could you offer another tip as to how I can set the quadratic equation to $ ... \equal{} 0$; $ 6$ needs to be incorporated into the quadratic equation so it can be solved?

[quote="SnipedYou"][hide]Hint

You can factor the $ y^{2} \minus{} 4xy \plus{} 4x^{2}$ to $ (y \minus{} 2x)^{2}$

then you have $ (y \minus{} 2x)^{2} \plus{} y \minus{} 2x \equal{} 6$

From there factor out a $ y \minus{} 2x$ to get

$ (y \minus{} 2x)(y \minus{} 2x \plus{} 1) \equal{} 6$[/hide][/quote]

I would go from

[hide]
$ (y \minus{} 2x)^{2} \plus{} y \minus{} 2x \equal{} 6$
Let $ y \minus{} 2x \equal{} z$
$ z^2 \plus{} z \minus{} 6 \equal{} 0$
Solve for $ z$
[/hide]

[hide="Complete Solution"]
$ z^2 \plus{} z \minus{} 6 \equal{} 0$
$ (z\plus{}6)(z\minus{}1)\equal{}0$
Hence $ z\plus{}6\equal{}0$ or $ z\minus{}1\equal{}0$
So
$ y\minus{}2x\equal{}\minus{}6$ or $ y\minus{}2x\equal{}1$
[/hide]

[quote="SimonM"]

I would go from

[hide]
$ (y \minus{} 2x)^{2} \plus{} y \minus{} 2x \equal{} 6$
Let $ y \minus{} 2x \equal{} z$
$ z^2 \plus{} z \minus{} 6 \equal{} 0$
Solve for $ z$
[/hide]

[/hide][/quote]

Thanks for your help, SimonM. This was the bridge which I needed to complete the question.

Its a useful technique that, make a note of it. I find it comes up all time (thats in normal school maths (sorry math))

Watch your factorization though. Shouldn't it be

[hide]$ (z \minus{} 2)(z \plus{} 3)$

and from there you get

$ y \minus{} 2x \equal{} 2 , \minus{} 3$

Yea I lost the $ \minus{} 3$ the way I factored thanks.
[/hide]

[quote="catalan"]Hello,

I am having trouble finding the corresponding [i]a[/i], [i]b[/i], [i]c[/i] values for the following quadratic equation.

Could someone please offer a hint/tip (as opposed to the full answer)?

Thank you very much!

---

1. Express [i]y[/i] in terms of [i]x[/i] for the relation $ 4x^2 \minus{} 4xy \plus{} y^2 \minus{} 2x \plus{} y \minus{} 6 \equal{} 0$.

I am trying to find the [i]a[/i], [i]b[/i], [i]c[/i] values so I can solve this equation with the quadratic formula. The problem is: I do not know how to factor and group these terms appropriately.[/quote]

The equation is equivalent to $ 4x^2\minus{}(4y\plus{}2)x\plus{}y^2\plus{}y\minus{}6\equal{}0$ and thus $ a\equal{}4, b\equal{}\minus{}4y\minus{}2, c\equal{}y^2\plus{}y\minus{}6$. While the above solution is obviously faster, it cannot always be applied. This one can. :)

[quote="SimonM"][quote="bertram"][quote="catalan"]Hello,

I am having trouble finding the corresponding [i]a[/i], [i]b[/i], [i]c[/i] values for the following quadratic equation.

Could someone please offer a hint/tip (as opposed to the full answer)?

Thank you very much!

---

1. Express [i]y[/i] in terms of [i]x[/i] for the relation $ 4x^2 \minus{} 4xy \plus{} y^2 \minus{} 2x \plus{} y \minus{} 6 \equal{} 0$.

I am trying to find the [i]a[/i], [i]b[/i], [i]c[/i] values so I can solve this equation with the quadratic formula. The problem is: I do not know how to factor and group these terms appropriately.[/quote]

The equation is equivalent to $ 4x^2 \minus{} (4y \plus{} 2)x \plus{} y^2 \plus{} y \minus{} 6 \equal{} 0$ and thus $ a \equal{} 4, b \equal{} \minus{} 4y \minus{} 2, c \equal{} y^2 \plus{} y \minus{} 6$. While the above solution is obviously faster, it cannot always be applied. This one can. :)[/quote]

That method can get nasty very quickly if you're not careful though[/quote]
But the approach usually works somehow if any other approach using factorization works, it's only a bit more complicated. In the end, you will always get a solution for $ x$ but the terms can be very nasty  :maybe:


analytic geometry
graphing lines
slope
geometry unsolved
geometry

[quote="Wull"]Does Anyone know what  (15 over 2, 4)  Slope of 2 over 7?[/quote]

Whatever this means, it doesn't seem to be olympiad geometry...

  darij


function
real analysis
real analysis unsolved

The Borel sets are generated from the open intervals in R. How can Q be generated from open intervals?
And is this functions measurable \[f_{n}(x)=\left\{\begin{array}{ll}e^{-\frac{x^{2}}{n}}&x\textrm{ irrational}\\ (-1)^{n}&x\textrm{ rational}\end{array}\right.\]

For example $\mathbb{Q}= \bigcup_{q \in \mathbb{Q}}\bigcap_{n \in \mathbb{N}}(q-1/n,q+1/n)$.

And $f_{n}= e^{-x^{2}/ n}$ a.e., $e^{-x^{2}/ n}$ is continous (so measurable), so $f_{n}$ is measurable.

Ok, thanks. Can you give me an example of a subset of $\mathbb{R}$ which is not a Borel set.

We need some facts:

(1) Let $C$ be the Cantor set. For $x \in C$, $x = \sum_{n \geq 1}\frac{2a_{n}}{3^{n}}$, where $a_{n}\in \{0,1\}$ define $\phi(x) = \sum_{n \geq 1}\frac{a_{n}}{2^{n}}$. Now extend $\phi$ to $[0,1]$ by $\phi(x) = \sup \{\phi(y) : y \in C, \, y < x\}$. One can show that $\phi$ is increasing, continous and maps $C$ onto $[0,1]$.

(2) If $f: \mathbb{R}\to \mathbb{R}$ is continous and $f(\mathbb{R}) = \mathbb{R}$ then $f(B)$ is Borel for every Borel set $B$.

Let us extend $\phi$ to $\mathbb{R}$ by setting $\phi(x) = 0$ for $x < 0$ and $\phi(x) = 1$ for $x > 1$. Now let $f(x) = x+\phi(x)$. One can see that $f$ is continous, $f \nearrow$ and $f(\mathbb{R}) = \mathbb{R}$.

We heve
\begin{eqnarray*}\mu(f([0,1] \setminus C)) &=& \mu f \left ( \bigcup_{n \geq 1}(x_{n},y_{n}) \right) \\ &=& \sum_{n \geq 1}\mu (f((x_{n},y_{n}))) = \sum_{n \geq 1}\mu (x_{n}+\phi(x_{n}) , y_{n}+\phi (y_{n})). \end{eqnarray*}
But $\phi(x_{n}) = \phi(y_{n})$ because $x_{n}, y_{n}\in C$. So $\mu(f([0,1] \setminus C)) = 1$.

Hence $\mu (f(C)) = 1$, so there is some nonmeasurable set $V \subset f(C)$. But $f^{-1}(V) \subset C$ so is measurable and not Borel (by (2)).

For more details see for example "Problems in Mathematical Analysis III" by W. Kaczor, M. Nowak, AMS.


Plz help me doing one question... :?:

[u]Create a user define file and write a program to read date from this file.[/u]

Plz Help Soon!!

(I recommend moving this to the Computer Science and Informatics forum)

Here is a program that reads an integer an integer from a file called "yourfile.in".

[code]

#include <fstream>

using namespace std;

int a;

int  main()
{
ifstream infile("yourfile.in");

infile >> a;

infile.close();
}[/code]


There is possible to insert pictures in latex ? ".jpg files" ? If yes , how ?


inequalities
function
algebra unsolved
algebra

Find all $ \ f(t): (0,\pi)\to R$ such that :
$ \ f(A)\plus{}cot(\frac{B}{2})\plus{}cot(\frac{C}{2})\le 3\sqrt{3}$
for every triangle $ ABC$

By Jensen's inequality,
$ f(A)\le cot(\frac {A}2)$. so infinitely many f exists.

[quote="Mashimaru"]ith_power, I think you have just pointed out a set of infinitely many funtions $ f$ satisfies the condition. But our mission is to [b]find every funtion[/b] so there should be a lot of other work in this problem... :([/quote]
But the point is: $ f\in A$ where $ A\equal{}\{f|f(x)\le cot\frac{x}2\}$ is set of all f, isn't it? :)

If we fix $ A$ and let $ B\rightarrow0$, we will have $ f(A)\le \minus{}\infty$.

Hence, there is no such function.


group theory
abstract algebra
superior algebra
superior algebra solved

Let (G,.) be a group and let e be its unit element. There exist exactly 2 elements in G\{e} s.t. xy=yx. Prove that G is isomorphic with Z3 or S3 (S3 is the group of the permutations of order 3).

It's from the Romanian NMO 1996 (12'th grade).

[quote="grobber"] There exist exactly 2 elements in G\{e} s.t. xy=yx. [/quote]

What does this mean?
x and y are the only 2 distinct elements in G\{e} such that xy=xy?

If yes, here is the solution:

Prop1. There is no element x in G such that |<x>|=ord(x)>3. If else then x, x 2  and x 3 commute.

Lemma2. If for all x in G we have x 2=e, then G is commutative.
xyxy=e, so xxyxyy=xy, so xy=yx

Prop3. If G is commutative, then G[u]~[/u] Z/3Z.
This easily follows from the condition in the problem text.

Prop4. If G is not commutative then there is exactly one element y in G such that |<y>|=3.
Apply Prop1, Lemma2, Prop3 and the condition in the problem text.

We now consider only the case when G is not commutative.

Prop5.For all x in G\<y> we have xy=y 2x (because xyxy=e).
Prop6. If x is in G\<y> we have <x,y>[u]~[/u]S3.

If there is some other element x' in G\<x,y> we have by prop4 x'x'=e and (xx')(xx')=e. It is easy to see that <x,x'> is a commutative subgroup of G and x,x' and xx' commute. This is false, so G is S3

Yes, that's what it means. I thought it was pretty clear... :D And the problem is pretty easy. Now, why did I post that? :D


The fact that

$\Bigg[ \, \frac{x \: \pm \: \sqrt{x^{2}-4}}{2}\, \Bigg]^{3}\;=\; \frac{x^{3}\:-\: 3x \: \pm \: (x^{2}-4)\sqrt{x^{2}-1}}{2}$

simplifies this equation to

$(1) \;\;\; x \;=\; \sqrt{x+4022030}$

which implies that

$x^{2}-x-4022030 \;=\; 0,$

i.e.

$(x+2005)(x-2006) \;=\; 0.$

So the only possible solutions are $x =-2005$ and $x = 2006$. But it is just the last solution that satisfies equation (1).  Hence the solution of the equation is $x = 2006$


ARML
conics
ellipse
geometry
perpendicular bisector

$ \boxed{1}$ Find the equation of the circle passing through 3 points $ ( \minus{} 3,\ 4), (4,\ 5), (1, \minus{} 4).$

$ \boxed{2}$ Find the equation of the circle passing through 2 points $ (0,\ 2),\ ( \minus{} 1,\ 1)$ and the center is on the line $ y \equal{} 2x \minus{} 8.$

$ \boxed{3}$ Find the equation of the tangent line which touches to the circle $ x^2 \minus{} 2x \plus{} y^2 \plus{} 6y \equal{} 0$ and passes through the point $ (3,\ 1).$

1.

Using the fact that the perpendicular bisector of a chord passes through the centre, I get

$ (x\minus{}1)^{2} \plus{} (y\minus{}1)^{2} \equal{} 25$.

2.

Same thing, then solving with $ y\equal{} 2x\minus{}8$, I get $ (x\minus{}3)^{2} \plus{} (y\plus{}2)^{2} \equal{} 25$.

3.

The line is $ y\equal{} mx \plus{}(1\minus{}3m)$, since it passes through $ (3, 1)$. The circle is $ (x\minus{} 1)^{2} \plus{} (y\plus{}3)^{2} \equal{} 10$.

Solving between them and using the fact that the discriminant, $ b^{2} \minus{} 4ac$ must be zero, I get $ m \equal{} 2$ and so the tangent is  $ y \equal{} 2x \minus{} 5$.

Your answer is correct for #1,2, but incorrect for #3.

[b]@ AndrewTom:[/b] [i]Your approach is perfect, you just didn't calculate $ m$ correctly.[/i]

[b]Note:[/b] The purpose of my solution is to publicize a relevant cute trick for finding the equation of a tangent to a circle at the point.

It's called the "stretch method", first introduced to me by an ARML Power Contest problem, and it generalizes to ellipses! Look it up. For circles, it works as follows:

[b]Step 1:[/b] Write the equation of the circle in standard form:
$ (x \minus{} 1)^2 \plus{} (y \plus{} 3)^2 \equal{} 10$

[b]Step 2:[/b] Replace one factor of $ (x \minus{} 1)$ and one factor of $ (y \plus{} 3)$ with $ (x_0 \minus{} 1)$ and $ (y_0 \plus{} 3)$, respectively, where $ (x_0, y_0)$ is the point of tangency, to get the equation of the tangent.
$ (x_0 \minus{} 1)(x \minus{} 1) \plus{} (y_0 \plus{} 3)(y \plus{} 3) \equal{} 10$

I'll hide the rest of the solution:

[hide="SOLUTION"]Well now that we have the generalized tangent line, substitute in the given point for $ (x,y)$.
$ (3 \minus{} 1)(x_0 \minus{} 1) \plus{} (1 \plus{} 3)(y_0 \plus{} 3) \equal{} 10$
$ \iff x_0 \plus{} 2y_0 \equal{} 0$

Stick this back into the original circle equation to get:
$ ( \minus{} 2y_0 \minus{} 1)^2 \plus{} (y_0 \plus{} 3)^2 \equal{} 10$
$ y_0 \equal{} \minus{} 2, \; 0$
$ x_0 \equal{} 4, \; 0$

Plugging these back into our generalized tangent, we get two possible lines:
$ 3x \plus{} y \equal{} 10$
$ \minus{} x \plus{} 3y \equal{} 0$[/hide]

Thanks, TZF. I now see that I should have had

$ (8m \minus{} 6m^{2} \minus{} 2)^{2} \equal{} 4(m^{2} \plus{} 1)(7 \minus{} 24m \equal{} 9m^{2})$

$ 24m^{2} \plus{} 64m \minus{} 24 \equal{} 0$

$ (3m \minus{} 1)(m \plus{} 3)$

$ m \equal{} \minus{} 3$ so $ y \equal{} \minus{} 3x \plus{} 10$

or $ x \equal{} \frac {1}{3}$ so $ 3y \equal{} x$.

I should also have realised, despite the singular "tangent line" in the question, that there would be two tangents since $ (3, 1)$ is outside the circle.

As for your "stretch method", it reminds me of finding a tangent at $ (x_0, y_0)$ by differentiating $ x^{2} \plus{} y^{2} \plus{} 2gx \plus{} 2fy \plus{} c \equal{} 0$, giving a result which is equivalent to replacing $ 2x$ by $ (x \plus{} x_0)$ and $ 2y$ by $ (y \plus{} y_0)$; $ x^{2}$ by $ x_0x$ and $ y^{2}$ by $ y_0y$. I'll work through it. Thanks again.


trigonometry
geometry
calculus
derivative
function
analytic geometry
graphing lines

I have lots of trouble with trig and geo so I decided to start a marathon for all the people that do!

1st problem(lets start off simple):

Simplify $ sqrt{\frac{2 sin x\minus{}cos x sin 2x}{sec x sin 2x}}$

[hide="There's no hint in here!"]$ sin 2x\equal{}2 sin x cos x$[/hide][/hide]

[hide="nicer looking form"]
We want to find \[ \sqrt{\frac{2\sin x-\cos x\sin2x}{\sec x\sin2x}.}\][/hide]

[hide="Solution"]
Using some trig identities, \[ \sqrt{\frac{2\sin x-\cos x\cdot2\sin x\cos x}{\frac{1}{\cos x}2\sin x\cos x}}\newline\sqrt{\frac{2\sin x(1-\cos^2x)}{2\sin x}}=\sqrt{1-\cos^2x}=\boxed{\sin x}.\] I'm not extremely good with trig so I might have messed up with some of the identities...[/hide]

[hide="New problem (if solution is correct)"]
Evaluate \[ \csc\left(\text{Arcsin }\frac{1}{2}-\text{Arccos }\frac{1}{2}\right).\][/hide]

$ \csc{\frac{\pi}{6}\minus{}\frac{\pi}{3}}\equal{}\csc{\minus{}\frac{\pi}{6}}\equal{}\minus{}\frac{1}{\frac{1}{2}}\equal{}\boxed{\minus{}2}$

Hope it's right...

Good timing, I'm just starting to learn actual trig now.

Find the amplitude of

$ f(x) \equal{} 5\sin x \plus{} 2\cos x$

the  derivative of  f(x)  is  5cosx-2sinx

                  5cosx-2sinx=0  this  implies   tanx=5/2

                                               so  sinx =5/sqr root(29)   and  cosx =2/sqr root (29)

                                     then  substituting  these  in the   given function  , the value of the  function  becomes  sqr root  (29)

method 2:

              asinx +b cosx        has  range  of   [- sqr root  (a^2+b^2),  +  sqr root  (a^2+b^2) ]

[quote="modularmarc101"]Yes you're solution is correct. I'm also horrible at trig, its one of my biggest weaknesses. Should I memorize lots of identities? Where can I find some?[/quote]
No matter what you do, don't memorize identities. Get Volume 2, take the Precalculus class, or get Plane Trignometry by S.L. Loney.

EDIT: Please don't use calculus. We need a new problem...

I believe the amplitude of a function \[ f(x)\equal{}a\sin x\plus{}b\cos x\] is $ \sqrt{a^2\plus{}b^2}.$ In that problem, $ a\equal{}5$ and $ b\equal{}2,$ so the amplitude is \[ \sqrt{5^2\plus{}2^2}\equal{}\sqrt{25\plus{}4}\equal{}\boxed{\sqrt{29}}.\]

[hide="New problem"]
Prove the amplitude of the function \[ f(x)\equal{}a\sin x\plus{}b\cos x\] is $ \sqrt{a^2\plus{}b^2}.$[/hide]

$ A, B,C$ are the angles of a triangle.
Prove that:
$ \tan \frac{A}{2} \plus{} \tan \frac{B\plus{}C}{2} \equal{} \sec A . \sec \frac{B\plus{}C}{2}$

[quote="communist"]$ A, B,C$ are the angles of a triangle.
Prove that:
$ \tan \frac {A}{2} \plus{} \tan \frac {B \plus{} C}{2} \equal{} \sec A . \sec \frac {B \plus{} C}{2}$[/quote]

  LHS=  tanA/2  + cotA/2  = secA/2 cosecA/2        (  since  tan (90-A/2)=cot  A/2)

     RHS  = secA/2  cosecA/2         (since   sec (90-A/2= cosecA/2 )

Okay. If you're going to post a solution, post a problem with it (or find someone that will), or don't post the solution at all.

OH!  IS  THAT  A RULE  , I AM SORRY THEN .!

HERE  GOES  A NEW QUESTION:

prove  that   cos(60-@)cos@cos(60+@) = 1/4.cos3@

              here  `@`  means  `theta`

Use $ 2\cos A.\cos B$=$ \cos (A\plus{}B)\plus{}cos(A\minus{}B)$

Next Problem,
Evaluate:
$ \tan^{2} \frac{\pi}{16}\plus{}\tan^{2} \frac{2\pi}{16}\plus{}\tan^{2} \frac{3\pi}{16}\plus{}....\plus{}\tan^{2} \frac{7\pi}{16}$

now  raghav   i dont think  the  question  u gave  comes  under  high school intemediates  section.  :?:

[quote="RAGHAV GROVER"][b]New Problem[/b]

Number of real values of $ x$ which satisfies $ 2\sin x \equal{} \sqrt {1 \minus{} 4x^{2}}$

I think it is an easy one!  :wink:[/quote]

Solve it graphically? I only saw [b]one [/b]real value of x in my graph.

New Problem (I hope it's not too hard. The solution is simple, so try it):

Let $ x\equal{}\frac{\displaystyle\sum_{n\equal{}1}^{44} \cos n^\circ}{\displaystyle \sum_{n\equal{}1}^{44} \sin n^\circ}$

What is the greatest integer that does not exceed $ 100x$?

[quote]
Let $ x \equal{} \frac {\displaystyle\sum_{n \equal{} 1}^{44} \cos n^\circ}{\displaystyle \sum_{n \equal{} 1}^{44} \sin n^\circ}$

What is the greatest integer that does not exceed $ 100x$?[/quote]

$ x$ comes out to be $ \cot 22.5^\circ$,using telescopic formulas, which equals $ \frac{1}{\sqrt{2}\minus{}1}$ now the problem is trivial... :roll:

EDITED POST

[b]New Problem[/b]
Find the average of $ n\sin n^\circ$ for $ n\equal{}2,4,6,8...180$

Based on the concept of the previous problem so it should not be that much difficult.. :)

This is more or less [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=353049#353049]USAMO 1996 #1[/url], except you don't even know the value...are you sure it belongs in Intermediate?

The last one was #11 on AIME from some year, can't remember.

The value of $ \cot 22.5^\circ$can be derived easily ..

EDIT:Sorry mistake in #42 $ x$ should be $ \cot 22.5^\circ$ so now the the value will be reciprocal of the above value...

I have edited that one

I guess by telescoping formula he means $ \frac {\cos x \plus{} \cos(45 \minus{} x)}{\sin x \plus{} \sin(45 \minus{} x)}$. Try expanding that - it equals to a constant.

Next problem: Let $ ABC$ be a triangle, $ M$ is the midpoint of $ AB$, $ N$ is the midpoint of $ AC$, $ BN$ meets $ CM$ at $ O$. If $ \angle{BOC} \equal{} 90^o$, find the minimum value of $ \cot B \plus{} \cot C$

If we want to find
$ \cos (\alpha)\plus{}\cos (\alpha\plus{}\beta)\plus{}\cos (\alpha\plus{}2\beta)$....upto $ n$
then it equals $ \frac{\cos(\alpha\plus{}\frac{(n\minus{}1)\beta}{2})\sin \frac{n\beta}{2}}{\sin\frac{\beta}{2}}$

and if u want to do the same for $ \sin$ then it equals $ \frac{\sin(\alpha\plus{}\frac{(n\minus{}1)\beta}{2})\sin \frac{n\beta}{2}}{\sin\frac{\beta}{2}}$

Shall i post the solution to the problem!? :wink:

Sorry for posting so late,but i have found solution to Raghav's problem
[hide="Solution"]
Multiply the given expression by $ \cos 1^\circ$ and use $ \sin(A\plus{}B)\minus{}\sin(A\minus{}B)\equal{}2\cos A\sin B$ and then apply $ \frac{\sin(\alpha\plus{}\frac{(n\minus{}1)\beta}{2})\sin\frac{n\beta}{2}}{\sin\frac{\beta}{2}}$
[/hide]

[hide="Answer"]$ \cot 1^\circ$ [/hide]

Someone plz post new problem

New Problem: Find the period and amplitude of: $ f(x) \equal{} 3\sin(3x\plus{}\frac{\pi}{2}) \plus{} \cos x$. (I'm pretty sure this is solvable.)

[hide]
The period of the first part of the function is 2pi/3  and the period of cos(x) is 2pi. In one full period T both functions must go through an integral number of periods, so

for some x and y,    2xpi/3 =  y2pi

so y = x/3  and the smallest values of (x,y) are (3,1)

so T = y2pi = 2pi

and the amplitude is sqrt(9+1) = sqrt(10)
[/hide]

The period is correct, but not the amplitude (assuming it's defined as $ \frac{\max{f(x)}\minus{}\min{f(x)}}2$ here). Take $ f(0)\equal{}4$, $ f(\pi)\equal{}\minus{}4$...now just prove $ |f(x)|\le4$. :wink:

Since no one is posting I'll post the next problem:

Let $ ABC$ be a triangle, $ M$ is the midpoint of $ AB$, $ N$ is the midpoint of $ AC$, $ BN$ meets $ CM$ at $ O$. If $ \angle{BOC} \equal{} 90^o$, find the minimum value of $ \cot B \plus{} \cot C$.

[quote="RoFlLoLcOpT"][hide="nicer looking form"]
We want to find
\[ \sqrt {\frac {2\sin x - \cos x\sin2x}{\sec x\sin2x}.}\]
[/hide]

[hide="Solution"]
Using some trig identities,
\[ \sqrt {\frac {2\sin x - \cos x\cdot2\sin x\cos x}{\frac {1}{\cos x}2\sin x\cos x}}\newline\sqrt {\frac {2\sin x(1 - \cos^2x)}{2\sin x}} = \sqrt {1 - \cos^2x} = \boxed{\sin x}.\]
I'm not extremely good with trig so I might have messed up with some of the identities...[/hide]

[hide="New problem (if solution is correct)"]
Evaluate
\[ \csc\left(\text{Arcsin }\frac {1}{2} - \text{Arccos }\frac {1}{2}\right).\]
[/hide][/quote]

hello.  there's a minor mistake in your simplification. $ \sqrt {1 - \cos^2x} = |\sin x|$


When is it valid to use the "By symmetry, ..." argument in geometry problems? I want to make sure I am using it right before I actually use it.

Whenever you can write down what the symmetry is and prove that it does what you think it does.


A rectangular volleyball court measures 20 meters by 10 meters. There is one line judge on each side of the volleyball net, standing on opposite corners of the court. If each judge has accurate vision up to 10 meters from his spot, what percent of the total court is accurately seen by at least one of the two line judges? Express your answer to the nearest whole number.

So....It's basically a 20 by 10 rectangle filled with 2 10-inch radius quarter circles.

$ 2[(10^2)\pi/4]\equal{}50\pi$

$ 20*10\equal{}200$

$ \frac{50\pi}{200}\sim 0.785\sim \boxed{79}percent....$


algebra
polynomial
superior algebra
superior algebra unsolved

How to find:

1. degree of splitting field of X^3 - 2  over Q
2. degree of splitting field of X^3 - 2 over Z/7Z

1. $ 6$. In general, the splitting field of $ x^k-2$ has degree $ k\phi(k)$ over $ \mathbb{Q}$ when $ k$ is not div. by $ 8$ and $ \frac{k\phi(k)}{2}$ when it is.

2. If $ \alpha$ is a root, the other two are $ 2\alpha$ and $ 4\alpha$. Hence, the degree is $ 3$.

For $ 1$:

From http://www.mathlinks.ro/Forum/viewtopic.php?t=209243 we have that the splitting field extension of a third degree polynomial is one of $ 1,2,3,6$. But adjoining $ \sqrt[3]{2}$ does not give the whole splitting field and hence it must be of degree $ 6$.

For an irreducible cubic, the degree of the splitting field is 3 if the discriminant is a perfect square and 6 otherwise.

[quote="t0rajir0u"]Over $ \mathbb{Z}/7\mathbb{Z}$ you also need to verify that $ \alpha \not \in \mathbb{Z}/7\mathbb{Z}$ (in other words, that the degree of the extension is not $ 1$), but this is straightforward either computationally or by the observation that $ (a^3)^2 \equiv 0, 1 \bmod 7 \implies a^3 \equiv 0, 1, \minus{} 1 \bmod 7$.

[b]Abel:[/b]  Could you explain your comment on 1?  It's not obvious to me why $ 8 | k$ is special.[/quote]My guess would be that to obtain the splitting field of $ x^k\minus{}2$, you need to throw in $ \sqrt[k]{2}$ and the $ k^{th}$ roots of unity.  When k is divisible by 8, $ \frac{\sqrt{2}}{2} (1\plus{}i)$ is a $ k^{th}$ root of unity, but it's an extension of degree 2, not 4 now, since $ \sqrt{2} \in \mathbb{Q}(\sqrt[k]{2})$

Thank you; I arrived at the same conclusion after I posted  :oops:


counting
distinguishability
geometry
3D geometry
MATHCOUNTS
search
function

7.  In how many different ways may 15 indistinguishable dimes be distributed among Susan, Frank and Harold if each must receive at least one dime?

7.  Each face of a cube is assigned a different integer.  Then each vertex is assigned the sum of the integer values on the faces that meet at the vertex.  Finally,  the vertex numbers are added.  What is the largest number that must divide the final sum for everypossible numbering of the faces?

3.  A 10 by 10 square is composed of 100 unit squares.  How many unit squares are entirely contained within the inscribed circle of diameter 10?

Sorry, danb80 if I stole your job.... :lol:

[hide]
For #1, the answer is the same as the number of positive integer solutions to $ a \plus{} b \plus{} c \equal{} 15$.

And here's a harder version: how about the number of ways to distribute the 15 dimes if some people can receive nothing at all and not all 15 dimes must be used?  Like, (0,5,10) and (0,0,0) both work.
--------------------------------------------------------------------------------------------------------
#2 can be done with algebra by using 6 variables and figuring stuff out from there
--------------------------------------------------------------------------------------------------------
#3 is in the Mathcounts forum
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1828283922&t=179936[/url]

[/hide]

Yeah, I believe #1 is a distribution problem.

[hide="overkill"]
Use generating functions somehow
[/hide]

[quote="numberdance"]You could also use combinatorics for the first problem.[/quote]
Yeah, it would be $ \binom{14}{2}$, I think...

[quote="numberdance"]That's right. You would have 12 dimes and two "dividers." (I don't know what else to call them...)[/quote]
Little typo.. but I understood.

I was taught to call it "The Donuts and Dividers Method".. :)

names don't matter, though

As for #2, I believe the answer is [hide="Click"]$ 4$[/hide].  Kinda harder to explain

For #2,

If you label each edge $ x_1, x_2, \cdots, x_8$, Then the desired sum would be $ 3\cdot\displaystyle\sum_{i \equal{} 1}^{8} x_i$. So wouldn't the answer be [hide="click"]$ \boxed{3}$[/hide]?

They're faces, not edges.

For edges, your answer is correct though  :)

lol, DK, I think I just thought of that name one day...

And did you get 12 slots and 2 dividers after you took out the 1 coin each person had to have?

That works too.

(DK, you did 14 slots for 2 dividers right?)

[quote="Chinaboy"]
(DK, you did 14 slots for 2 dividers right?)[/quote]

Yes. That's what I did.


I'm an Irish student hoping to study maths in England. I was hoping someone who has gone through the system over there (or who is going through it now) could explain some things to me.

- I'm told that most people need Maths and Further Maths (this a seperate subject right?) I'm doing Irish maths, which is equivalent to A - Level Maths but I'm not doing anything like further maths. Can anyone recomend a book (or even better a website) that covers this course?

- I've been asked to sit STEP exams. What are these supposed to test? Where can i find information about these?

- Finally, is there anywhere I can get STEP past papers?

I realise thats a lot of questions but if anyone can even answer one of them that would be a great help.

Thanks!

(I posted this in another area as well, I think it may be more relevant here)

I just answered on Other Problem Solving Topics. Only the moderator of that forum can move the thread, but I agree it should really move over here.


I'll post the solution to make sure you guys get the right answer.

Do not look until you solved it.

[hide="View"]

Rewrite the logarithrm into exponent form.

$(2x+1)^2 = 3x+1\\
4x^2 + 4x + 1 = 3x+1\\
4x^2 + x = 0\\
x(4x + 1) = 0\\
x = 0$

$4x = -1 \Rightarrow x = \frac{-1}{4}$.

$\therefore x = 0, \frac{-1}{4}$.[/hide]

Silverfalcon, you made a little mistake in your solution:

[hide]
When we see $\log_{f(x)}{g(x)}$, we are given that $f(x) \in (0,1) \cup (1, +\infty)$ and $g(x)>0$

So $2x+1 \neq 1$.
$x \neq 0$[/hide]

Yeah I think he is misinterpreting the base or something, I dunno.

But I agree with SilverFalcon.

Function $\log_ax$ is defined as: $a \neq 1$

[quote="WindSlicer"]Hm, I wasn't taught that.

Why?[/quote]
I think it is because: $1^x=1$....

Exactly, just because $1 \cdot x = x$ doesn't mean we can't do it.

Note: if there seems like there's an aggressive tone to this post - there's not. I just reread it and it could be misinterpreted.. just trying to get answers, because I've had three teachers teach me logs and no one's mentioned that, but if it is true and widely accepted it'd be a good thing to know.

[quote="Silverfalcon"]Wait I'm confused.

Putting zero in for $x$ gives:

$\log_{2 \cdot 0 + 1} 3 \cdot 0 + 1 = 2$
$\log_1 1 = 2$
$1^2 = 1$
$1 = 1$

So, this is true. What's wrong again?

Please explain without using those symbols because that's confusing me.[/quote]

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

If $a=1$, we then have $\frac{\ln b}{0}$ which is undefined.

Ahh.

Noted, and thanks for explaining it.

Since no one has posted the correct [hide="Solution"]$3x+1=(2x+1)^2 \\ 4x^2 + x = 0\\ x(4x + 1) = 0\\ x = 0$
or $4x = -1 \Rightarrow x = -\frac{1}{4}$

But if $x=0$, $2x+1=1$ which makes the equation undefined

$\therefore x = - \frac{1}{4}$[/hide]

:oops:

I apologise to Silverfalcon. When I posted this problem I wrote $\log_{2x+1}{3x+1}=2$ instead $\log_{2x+1}(3x+1)=2$. It wont happen again... ;)

[quote="prototypex24"]Er, shobber, does
$f(x) \in (0,1) \cup (1, +\infty)$ mean that $f(x) >0, f(x) \neq 1$[/quote]
Yes. It is the opening interval.

[quote="Jimmy"][quote="Iversonfan2005"][quote="Slizzel"]Solve the following ecuation $\log_{2x+1}(3x+1)=2$.

$ \text{(A)}\ 1\ \ \text{(B)}\ 0\ \ \text{(C)}\ 0\ \text{and} -\frac{1}{4}\ \ \text{(D)}\ -\frac{1}{4}\ \ \text{(E)}\ \text{no solution}$[/quote]

okay, then you know that (2x+1)^2=3x+1, so then when you solve it out you get either 0 or -1/4. But I think you want[b] -1/4[/b][/quote]

$x=0$ is an extraneous solution, because $\log_{1} a$ is undefined.  Read what I wrote above.[/quote]

oh yeah, you're right, thanks :)


Let $ a,b,c$ be positive real numbers such that $ abc\equal{}1$. Prove that $ \sqrt{\frac{2}{1\plus{}a}}\plus{}\sqrt{\frac{2}{1\plus{}b}}\plus{}\sqrt{\frac{2}{1\plus{}c}}\leq 3$.
I have not had a solution for this. Please help me to solve it or show me some books concerning with this problem.

[quote="chinhlh"]Let $ a,b,c$ be positive real numbers such that $ abc \equal{} 1$. Prove that $ \sqrt {\frac {2}{1 \plus{} a}} \plus{} \sqrt {\frac {2}{1 \plus{} b}} \plus{} \sqrt {\frac {2}{1 \plus{} c}}\leq 3$.
I have not had a solution for this. Please help me to solve it or show me some books concerning with this problem.[/quote]
Assume  $ a \equal{} \frac {y}{x},$ $ b \equal{} \frac {z}{y}$ and $ c\equal{}\frac {x}{z},$ where $ x,$ $ y$ and $ z$ are positive and see here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=17549


If $ a,b,c\ge0$ such that $ max\{a,b,c\}\le3min\{a,b,c\}$, then

$ 3abc(a\plus{}b\plus{}c)(a^2\plus{}b^2\plus{}c^2)\ge(ab\plus{}bc\plus{}ca)^3$

[quote="dduclam"]If $ a,b,c\ge0$ such that $ max\{a,b,c\}\le3min\{a,b,c\}$, then

$ 3abc(a \plus{} b \plus{} c)(a^2 \plus{} b^2 \plus{} c^2)\ge(ab \plus{} bc \plus{} ca)^3$[/quote]
See here: http://canhang2007.wordpress.com/2008/12/09/3-problems-duong-duc-lam-and-me/ :)


parameterization
inequalities
combinatorics unsolved
combinatorics

Let a;b are integer numbers.There are $a$ sets:$A_i$ satisfy $|A_i|=b$ and $|A_{i} \cap A_{j}|\leq 1$
Find $min\cup A_i$.

I assume you meant minimum cardinality of the union of all $A_i$.

It is easy to give an answer in the case when $1+4ab(b-1)$ is a square, i.e., $1+4ab(b-1)=t^2$ for some integer $t$. In that case $A_i$ must form a [url=http://mathworld.wolfram.com/BlockDesign.html]block design[/url] with parameters $(\frac{1+t}{2},b,1,\frac{1+t}{2(b-1)},a)$, and the required minimum is $\frac{1+t}{2}$.

Sorry!I didn't understand what you said :huh: ( ...)!
But I know your answer in that case is WRONG.
I have just known  an ineq about $n,a,b$
(Call n be minimum of $|\cup_{i=1}^a A_i|$)
$n\geq \frac{2}{n+1}ab-\frac{2}{r(r+1)}(_2^a)$

[quote="anhminh"]But I know your answer in that case is WRONG.
I have just known  an ineq about $n,a,b$
(Call n be minimum of $|\cup_{i=1}^a A_i|$)
$n\geq \frac{2}{n+1}ab-\frac{2}{r(r+1)}(_2^a)$[/quote]
What is $r$ here? And how this inequality contradicts to what I said?

Oh!$r$ is a real number .The ineq is true with all $r$.
With $a=5;b=7$.Your $n$ doesn't satisfy my ineq( Please trying! :D )
I will prove that ineq next time.

Oh, I think it should be also the case that $\frac{1+t}{2(b-1)}$ is integer.


trigonometry function geometry calculus ratio articles Gauss

Hey everybody!
   How do you RIGOROUSLY prove that if angle A is greater than angle B (assuming both are acute), then sin A is greater than sin B? Please help!

[hide]\[ \mbox{Let }\begin{cases} r\sin A &= y_1 \\ r\sin B &= y_2\end{cases}\]

\begin{eqnarray*} A & > & B  \\ \sin^{-1}\left(\frac{y_1}{r}\right) & >&\sin^{-1}\left(\frac{y_2}{r}\right) \\ \sin\left(\sin^{-1}\left(\frac{y_1}{r}\right)\right) & >&\sin\left(\sin^{-1}\left(\frac{y_2}{r}\right)\right) \\ \frac{y_1}{r} & >& \frac{y_2}{r} \\ \sin A &>& \sin B\quad \mbox{Q.E.D.}\end{eqnarray*}[/hide]

The key here is the fact that $ \sin$ is an increasing function on the interval $ \left(0,\frac {\pi}2\right)$.

Or go back to your geometry days, and recall the theorem that in a triangle, the side opposite the larger angle is larger. Then apply this to a right triangle.

what's sin to the power of -1? Is it cosecant?

1. Ignore what Dr. No wrote, as it's circular reasoning.  (The step from line 2 to line 3 follows from the result you're trying to prove.)
2. $ \sin^{ \minus{} 1}$ means the inverse function of sine, not the multiplicative inverse.  (The multiplicative inverse is indeed cosecant.)
3. The relevant geometric result, as Yongyi noted, is the "hinge theorem."

[quote="JBL"]1. Ignore what Dr. No wrote, as it's circular reasoning.  (The step from line 2 to line 3 follows from the result you're trying to prove.)[/quote]

I can certainly see how you would think so, however, the definitions of $ A$ and $ B$ are their respective arcsines, and it should be self-evident, if one is able to draw the figure -- I'm not able to, at least for inclusion here, that $ y_1 > y_2$, from which it is immediately apparent that $ \frac {y_1}{r} > \frac {y_2}{r}$, which is itself what was to be proven. The last line is merely redundant, but I included it for the sake of clarity.

[hide="Rigorous proof (with calculus)"]
$ \frac{d}{dx}\sin x \equal{} \cos x$, and in the interval $ \left(0, \frac{\pi}2\right)$, $ \cos x > 0$. Therefore, $ \sin x$ is increasing in the interval $ \left(0, \frac{\pi}2\right)$, and therefore, for any $ A>B$, $ \sin A > \sin B$.
[/hide]

Dr. No, references to diagrams (especially those you cannot draw) are not particularly useful for proving something like this.

[quote="Yongyi781"][hide="Rigorous proof (with calculus)"]
$ \frac {d}{dx}\sin x \equal{} \cos x$, and in the interval $ \left(0, \frac {\pi}2\right)$, $ \cos x > 0$. Therefore, $ \sin x$ is increasing in the interval $ \left(0, \frac {\pi}2\right)$, and therefore, for any $ A > B$, $ \sin A > \sin B$.
[/hide]

Dr. No, references to diagrams (especially those you cannot draw) are not particularly useful for proving something like this.[/quote]

That, sir, is a truism. Nevertheless, my argument is the same as your appeal to the theorem that angle with the larger opposite side is itself larger. (I had the law of sines in mind, but didn't mention it because that would essentially make this problem a one liner, and I was uncertain as to whether the trig students here would have seen that yet.)

Also proofs using calculus, as rigorous as they might be, are unacceptable for the purposes of high school trigonometry.

Thanks for the responses! :)

But what I was trying to say was that where in the AOPS Volume 1 problem 196 in the solution manual, it says, "... Since angle A > angle B, we have sin A > sin B. (Why?)" we did not even learn what sine graphs were yet (they are covered in AOPS Volume 2) - and definitely not calculus yet! I'm sorry if I made anything confusing.  :blush: We have only learned the basic (very basic) trig. identities and what the trig. ratios meant. Based on that, how would you prove it???????

[quote="JBL"]3. The relevant geometric result, as Yongyi noted, is the "hinge theorem."[/quote]

sorry, but I don't get how that helps?

Wait, this is the hinge theorem, right? - http://en.wikipedia.org/wiki/Hinge_theorem

Thanks very much!

[quote="theax"]Well, in the AOPS Volume 1 problem 196 in the solution manual, it says, "... Since angle A > angle B, we have sin A > sin B. (Why?)"[/quote]

However well or ill I might have defended my reasoning in my previous attempt at proof, what JBL said is true -- it's circular reasoning, in as much as I used the sine function -- what I intended to prove -- to undo the arcsine function I had used to define the angles.

In order to redeem myself I'll try again in a way that I hope is more clear.

Consider a right triangle $ \triangle ABC$ such that $ A$ and $ B$ are the acute angles thereof, and $ A > B$ then by the [url=http://en.wikipedia.org/wiki/Law_of_sines]law of sines[/url],

\[ \frac{\sin A}{\sin B} \equal{}\frac{a}{b}\]

\[ \text{By the Hinge Theorem,}\: \frac{a}{b} > 1\]

\[ \text{Therefore,}\: \sin A > \sin B\]

[quote="theax"]Thanks very much! :)

But I'm afraid I still don't get it - are you guys using the hinge theorem on wikipedia - http://en.wikipedia.org/wiki/Hinge_theorem ??????????

Because I don't really see how the "two sides of one triangle are congruent to two sides of another triangle"...

Please help![/quote]

Not in the same sense as the wiki article explains it.

Wikipedia explains the Hinge theorem in terms of comparing angles of two different triangles having each having two sides that are congruent to each other. In such a case the third side of the triangle having the larger angle is larger than the third side of the triangle having the smaller angle.

As we have referred to it here, we apply the hinge theorem to a single triangle, to compare two of its internal angles with one another. This is legitimate because a given side of any figure is congruent with itself. Thus a single triangle has two included angles with two congruent sides, and by the Hinge theorem then this implies that the side opposite the larger angle is longer than the side opposite the smaller angle, which is exactly the relationship needed to prove your original question by the law of sines.

Edit: In order to see that this is so you can redraw the triangle in question, label it differently if it helps, and put one angle in one copy, and the other angle in the second copy, then you'll have the same situation the Wikipedia article describes.

[quote="theax"]in order to make it rigorous, don't you also have to consider the cases where measure of angle A + measure of angle B is not = 90? (because aren't you assuming A + B = 90)?[/quote]  Well, I wouldn't say that "rigor" is what's lacking, just that the result Dr. No proved is not quite as general as the result you want.

Consider three triangles, $ \triangle ABC$, $ \triangle DEF$ and $ \triangle GHI$, with $ m\angle C = \m angle I = 90^\circ$, $ AB = DE$, $ BC = EF = HI$, $ m\angle B > m\angle E$, $ DF = GI$.  Apply the Hinge Theorem twice to arrive at $ BC = HI$, $ AB < GH$ while $ \m\angle B < m\angle G$ and so $ m\angle A > m\angle H$.  The result follows immediately.

[quote="theax"]ahh... I see. Thanks very much! :)

But in your proof with the law of sines. Dr. No, I still don't get it - in order to make it rigorous, don't you also have to consider the cases where measure of angle A + measure of angle B is not = 90? (because aren't you assuming A + B = 90)?[/quote]

I assumed a right triangle because $ m\angle A > m\angle B \implies \sin A > \sin B\; \mbox{ only if }\; m\angle B \le m\angle A \le 90^{\circ}$. Certainly for a right triangle $ m\angle A \plus{} m\angle B \equal{} 90^{\circ}$, but this guarantees that the necessary condition is satisfied.

[quote="JBL"]But what you've done doesn't, for example, prove that $ \sin 30^\circ < \sin 45^\circ$.[/quote]

Why not? I could have specified that $ A$ and $ B$ are the acute interior angles of an arbitrary triangle just as easily, and obtained the same result for exactly the same reason. I chose a right triangle because I wanted more generality, not less. Using a right triangle, it doesn't matter which two angles one picks, the pivotal relation, $ \frac {a}{b} > 1$, still holds.

In other words the key point is that both $ \angle A$ and $ \angle B$ are acute, not that they total $ 90^{\circ}$.

Edit: For example, consider a 30 - 60 - 90 triangle. Let $ m\angle A \equal{} 90^{\circ}$ and let $ m\angle B \equal{} 30^{\circ}$.

Is $ m\angle A \plus{} m\angle B \equal{} 90^{\circ}$ ?

Nevertheless, $ \frac{a}{b} \equal{} \frac{1}{\frac12} \equal{} 2 > 1$

[quote="theax"]i don't quite explain your proof...
Could you please re-explain it?[/quote]

This is becoming tedious. If you still don't understand why the proofs you've been given do in fact prove what you asked us to prove then perhaps you should just accept the statement of your text without proof ... for now.

However not being one willing to leave others confused, I shall make one more attempt to dispel your confusion.

1. The law of sines is valid for [b]any[/b] single triangle: equilateral, isosceles, or scalene; it doesn't matter what kind.

2. The Hinge theorem applies to any pair of triangles such that each has a pair sides that are congruent with a pair of sides of the other. When this is the case, the triangle having the larger included angle has the longer side opposite that angle. (As I explained earlier, you can extend the theorem to apply to a single triangle by drawing an exact copy of one triangle and placing one of the angles you wish to compare in the original triangle and the other angle in the copy.) How you work this out in your own head is of no consequence. What matters is that it entails $ \;m\angle A > m\angle B \implies a > b$, where $ a$ and $ b$ are the lengths of the sides opposite $ A$ and $ B$ respectively.

3. Since $ a > b$ then $ \;\frac{a}{b} > 1$.

4. From the law of sines we are guaranteed that $ \: \frac{\sin A}{\sin B} \equal{}\frac{a}{b}$.

5. From step (3) then $ \;\frac{\sin A}{\sin B} > 1$.

6. It follows immediately from (5) that $ \;\sin A > \sin B\;$ and with that, that which was to be demonstrated is demonstrated!

How about this:

Let triangle $ ABC$ be a triangle with right angle at $ C$. Draw an overlapping triangle $ ABD$ with a right angle at $ D$ such that $ \angle BAD > \angle BAC$, and let $ E$ be the intersection of $ AC$ and $ BD$.

Since $ BD > BE > BC$,
$ \frac {BD}{AC} > \frac {BC}{AC}$

So $ \sin {\angle BAD} > \sin{\angle BAC}$

[quote="Dr. No"]This is becoming tedious.[/quote]  I know, life is so hard when you try to prove things to someone just beginning trigonometry using the Law of Sines (but without actually invoking it at the necessary moment!) and s/he doesn't understand  :roll:

gauss: I think you intended to divide by AB (the length of the common hypotenuse), not AC.  And, while I'm here -- how would you prove that segments AC and BD must really intersect?

Imagine you have two angles from 0 to 90 degrees (Such as two in a right triangle). Put them both on a unit circle so that their vertexes are on the center, and one side is at 0 degrees. The other sides should be intersecting the circle at some point in the first quadrant. Drop lines straight down and you have right triangles. As the lines come out from the circle, the line that is part of the larger angle will be above the line of the bottom angle. It will hit the circle at a higher point because of this, and have a larger sine.


function
Gauss
geometry
3D geometry
sphere
ceiling function
floor function

In class we are doing discussions on dating. I was wondering what people oon aops thought of it.
For instance, we discussed how old you should be, who should do the asking, who should pay, ect.
another thing is should your parents meet the person your going to date?

1.) If my parents didn't freak out about me dating, then yes I would've gone out on one.

Anyway,

[hide="Who Pays?"]
Although probably not good for my wallet, I believe the guys should pay. Seems like the right thing to do.
[/hide]

[hide="How Old?"]
Around 13-14 years old. Even though you might be "ready" to date earlier, I think you really shouldn't start until you hit 8th or 9th grade.
[/hide]

[hide="Who Should Ask?"]
Err, mostly the guy? I've never heard a situation where the girl asked out the guy  :huh: .
[/hide]

[hide="Parent Meeting"]
This is a bit iffy for me. Parents should definately meet the one you're going to marry, but I don't think it's a big deal if it's a date.
[/hide]

[quote="Anaxerzia"]dude this is clearly the wrong forum to get any good answers  :P[/quote]

Heh. Partially true. :lol:

...I haven't even really tried to broach the topic of dating with my parents. I have a feeling it wouldn't end nicely. For me. :wink:

But I can answer some questions:

[hide="Parents Meeting"]
Erm. It depends on your parental units. They might want to, but it really isn't incredibly necessary/important. It would be nice for them to meet face-to-face, although not necessary.
[/hide]

[hide="Oldness"]
Probably middle-school to high-school to start. Elem. is just too early.
[/hide]

[hide="Asking..."]
Doesn't particularly matter. Depends on what gender you are and what the personality of person you're asking out is like. If she's an outgoing girl, then possible she might ask. Society seems to prefer the males, however.
[/hide]
@7h3.D3m0n.117
[quote="7h3.D3m0n.117"]
Err, mostly the guy? I've never heard a situation where the girl asked out the guy :huh:.
[/quote]
Meh. There are plently of situations. I just don't know for sure how often this happens to math nerds. :P :wink: :roll:

[hide="Shelling out the bills..."]
Again, depends on the girl. Most girls would prefer if the guy paid, some are more....equality-rebelling-against-stereotypical-society-oriented. And they would probably opt to split unless they're comfortable with that guy.
[/hide]

....and what class was this in?

And if she tells you she cares about you're beasting MOP-ness, then she is lying :P

So yeah my experience is you should play dumb. You are most likely to score that way.

Unless you're actually looking for an intelligent girl [sic], then you should flaunt all you want :)

[quote="7h3.D3m0n.117"]
I've never heard a situation where the girl asked out the guy  :huh: .
[/quote]

Seriously?  :lol:

I shall disprove that by saying a girl just asked me to prom a couple weeks ago. (Yes I am a freshman, this is for "in two years," still counts).

Girls...should not pay.

I believe that parents meeting is a bit overboard, especially for these teen "Relationships."

[quote="The_Scintillator"]Meh. There are plently of situations. I just don't know for sure how often this happens to math nerds. :P :wink: :roll:[/quote]

I should clarify. I don't know of any situation [b]personally[/b]. I do agree that society seems to dictate the male asking the female out. That, or the male friends will constantly badger the individual in question about being a coward.

cough arnav is needed cough.

Methinks high school should be young enough. Middle schoolers are too young and not ripe.

It doesn't matter who starts it but it must be started to go on. Sometimes you need a push or a shove (And in special cases a punch)

Everything else can be decided later.

I [b]really[/b] should bring this topic out in class. It would prove interesting...We once talked about refrigerators in the middle of math. Just today we talked about good deals and bad deals in stores, etc.

[b]Oldness:[/b] I guess 6th, 7th, 8th grade you should start. My friends have already gone on some.

[b]Parents meeting:[/b] Scary thought. No. Not because it's for your safety, I think, just that you're making a good choice. I mean, if the person you're dating is a drug dealer I'm sure you'd take one good whiff and know that right away, wouldn't you?

[b]Who asks?:[/b] I usually think it's the boys who ask. Seems right. In some dances, I heard the girls ask the boys...

[b]Who should pay?:[/b] A date doesn't have to be at the movies or a fancy restaurant...Could be at the library, the park, and places like that. No one pays for stuff there. Perfectly affordable.

Age: Middle schoolers dating, eh, not really the real thing. High school is where it's at.
Paying: I usually offer to pay, but if she wants to split, it's cool.
Asking: I've always asked, until this year when a junior girl asked me to prom.
Parents meeting: Are you kidding? Hell no.

I don't believe that old-fashioned gender roles should still apply. I think either gender could ask (and the gender of the person being asked needn't be a function of the gender of the person asking, either). I do think it's polite if the person who asks also pays.

As for age: I imagine it's natural for people to start getting interested in dating around 13-15 or so. I don't think there's anything wrong with dating that young, but if you're 13 and you're asking a girl out on a date, don't pretend to yourself that it's something serious. You're just exploring. I don't think people are generally ready for a "serious" relationship until at least the end of high school.

Meeting parents: Why wouldn't you want your parents to meet your date? If it's because you're afraid of what your parents would say, either you shouldn't be dating that person or there's something wrong with your parents.

age: definitely 13+. The "relationships" before that age is never serious dating and it's more like you get a brownie from a guy in the lunch line.

ask: well.. i've asked guys out, but mostly it should be the guy asking. unless it's a sadie hawkins (girls ask guys)

parent meeting: umm.. well. it depends on parents.

pay: well.. i prefer to pay my part of the date, but if the guy really insists, then i'd let him pay. it's because i'm sometimes way too polite.

I lack experience in dating, since it's hard to meet girls when studying mathematics. If you take a course like intermediate algebra you might come across one or two, but they are usually fat germans. Nevertheless I shall share my views.

[hide="Who Pays?"]
I've never heard of anything but the man paying. Usually you pay about 10 minutes in, at the part where you share some wine and just chit-chat to get comfortable. This has always made me uneasy though, because I feel unsure of when I should tip. Should I wait until the end, as if I'm judging her during the act?
[/hide]

[hide="How Old?"]
Although our law allows 15 year olds to have consensual sex (with anyone of any age > 15), I feel in this case definitely 18. For me personally, I like 24-28 or so.
[/hide]

[hide="Who Should Ask?"]
Err, mostly the guy? I've never heard a situation where the girl asked out the guy  :huh: .
[/hide]

[hide="Parent Meeting"]
This is a bit iffy for me. Parents should definately meet the one you're going to marry, but I don't think it's a big deal.
[/hide]

i am a firm believer that the guy should definitely ask the girl, having had negative experiences with the other way around.

however, I would feel incredibly weird if my date always paid, or vice versa. split it!

and as to playing dumb, never! she won't fall for it and unless you tell her what's unique about you you'll be boring! if she doesn't care you're too good for her anyway! (however, all things in moderation, you might not want to go on about it for too long)

I mean it's your opinion. If you don't believe it, it is up to you. I don't have time to argue all day.
Have a nice day!

[quote="The Visionary"]Who says having a relationship in elementary school is too early? I did it and I am (and was) a normal human being.[/quote]

My friend [had] a boyfriend but then she decided to wait until sixth grade.

About this whole elementary school relationship thing...

It could happen, but not most of the time. [i]Most[/i] of the time.  :wink:

[quote="7h3.D3m0n.117"]I haven't heard of a legit elementary school relationship either. To me, I really doubt the kids there are mature enough understand dating and relationships.[/quote] Bad analogy: Which elementary school kid is mature enough to understand mathematics?

Must it be serious? If it's an elementary school relationship, then it is probably an innocent relationship. The kids are mote friends than "couples". Nothing scary with that...

[quote="daboa"]Must it be serious? If it's an elementary school relationship, then it is probably an innocent relationship. The kids are mote friends than "couples". Nothing scary with that...[/quote]

Some serious, some just to get talked about. One specific couple I know are really good friends. The teachers know about this too, and constantly pester him and her about it. A bit funny. I doubt they have anything serious going on.

Ok, my dad doesn't want me dating until 18. I think 15 is good enough for me.

I will always let my parents know who i'm dating, and what shes like, untill i'm an adult.

The guy should pay in the begining, to show generosity. If the girl chooses to, then she should in the later part of the relationship.

If the girl asks, it would be great. As long as you like the girl. one girl I know wouldn't think of asking a boy out.

So this has gotten way off topic, but I do have a serious question. I'm going to grad school in fall and all I've ever been told is that you study and work all the time. I haven't been able to experience whether this is true or not (I assume it is), but if it is it sounds like one of two things can happen. You don't have time to date, or if you do date, it will end soon because you will be spending all your time studying and not with the person you are dating. I was wondering if anyone who has been through this could comment. I guess a third option is that you try to spend enough time with the person to make it work, and you end up not doing as well as you should in grad school.

Well, if you're in college and doing well then you should wait until everything hard in schoolwork dies down. That way less stress is on your shoulders; or you could always start dating a month or two before a huge test and study with them.

Relationships in graduate school? One of the people I've shared an office with the last few years is getting married this summer. It can work.

I started "dating" in 6th grade, although it wasnt really dating because all we did was talk and go to dances and the movies like twice. But I do think its good for experience, just so it wont be so "traumatic(as my mom says :P)" when you want to date seriously. However, I think it mostly depends on personal preference.

elementary school is sorta creepy :ninja:
our school's first relationship was probably like 7th grade
do hormones act that quickly?

[quote="zapi2007"]do hormones act that quickly?[/quote]

I wouldn't say 10 or 11+ years "quickly." That's extremely long in terms of biochemical reaction times.

[quote="12markkram34"]Earlier in the topic, people were talking about early dating for "experience". This doesn't make very much sense to me...why would you go through so many broken relationships and cause you and others so much misery only to do it over again, perhaps less tragically. To me, it would make sense to wait until you are older (late high school, college) to start dating, when you are old and mature enough to run things properly and not do the wrong things.[/quote]

Maturity comes with experience

middle school- there are some relationships in our school but none too serious, nothing past kissing, but there are some... difficulties, and from experience i know this is way too early. i was in one in 5th grade started ok, ended HORRIBLY, it was over in just a few months and now i realize learn about the person u fall for... she was just a terrible person... [size=75]stupid violent hippy... dumb annoying ****faced ****head *****[/size]


Given the homogenous transportation equation $\frac{du}{dt}(t,x)+\zeta.\triangledown u(t,x)=0$ and $u(0,x)=g(x)$, where $u: \mathbb{R}^{n+1}\rightarrow \mathbb{C}$ and $\zeta \in \mathbb{R}^{n+1}$, and the gradient operator $\triangledown$ operates on $x=x_{1},\dots,x_{n}$, show that the function $g(x-t\zeta)$ is a solution. Here $g$ is assumed to be continuous, and $u$ is $C^{1}$.

Here's the thing: it is easy if we differentiate with respect to $t$. But how can we do it by differentiating by $x$ and just verifying the inital equation (brute-force like ?)?

I don't really understand the question. To show that $u(x,t): =g(x-t\zeta)$ is a solution, one plugs it in and finds from the Chain Rule that $u_{t}=-\zeta\cdot \nabla g$ and $\nabla u=\nabla g$. Why avoid a computation of $u_{t}$ when your goal is to show that it equals $\zeta\cdot\nabla u$?

Thanks. The course I [i]have to[/i] follow is "engineer-style" writing. The notations are also approximate and vague, so it takes me ages to go through it. In this case, there's a definition I hadn't understood well (!!)


Solve in prime numbers $ 2p^q \minus{} q^p\equal{}7$

It's been posted here  http://www.mathlinks.ro/Forum/viewtopic.php?p=1154604#p1154604


You can say the first thing that comes to your mind or think about it for a while, but don't look up properties of numbers after reading the question before answering just to impress me that you knew that 8833 was 88^2+33^2 for example.

Mine was 216, because of Pythagoras mainly, but also because it looks nice.

Thanks.

I like a lot of numbers.  I think $ \frac{\pi^{2}}{6}$ is a pretty interesting number.  It's the solution to $ \zeta(2)$ and couldn't be solved by many of the world's great mathematicians until Euler cracked it.  I think it's neat that
$ 1\plus{}\frac{1}{4}\plus{}\frac{1}{9}\plus{}\frac{1}{36}\plus{}\dots$
sums to a number such as $ \frac{\pi^{2}}{6}$.

1729 is also an interesting number. It is the first natural number that can be expressed as the sum of two cubes in two different ways.


$ x,y,z\succ0$  prove that :

$ \frac{1}{x^2\plus{}yz}\plus{}\frac{1}{y^2\plus{}xz}\plus{}\frac{1}{z^2\plus{}xy}\leq\,\frac{1}{2}(\frac{1}{xy}\plus{}\frac{1}{yz}\plus{}\frac{1}{xz})$

[quote="mehdi cherif"]$ x,y,z\succ0$  prove that :

$ \frac {1}{x^2 \plus{} yz} \plus{} \frac {1}{y^2 \plus{} xz} \plus{} \frac {1}{z^2 \plus{} xy}\leq\,\frac {1}{2}(\frac {1}{xy} \plus{} \frac {1}{yz} \plus{} \frac {1}{xz})$[/quote]
$ \sum_{cyc}\frac {1}{x^2 \plus{} yz}\leq\sum_{cyc}\frac {1}{2\sqrt{x^2yz}}\leq\sum_{cyc}\frac {1}{2xy}.$

[quote="mehdi cherif"]$ x,y,z\succ0$  prove that :

$ \frac {1}{x^2 \plus{} yz} \plus{} \frac {1}{y^2 \plus{} xz} \plus{} \frac {1}{z^2 \plus{} xy}\leq\,\frac {1}{2}(\frac {1}{xy} \plus{} \frac {1}{yz} \plus{} \frac {1}{xz})$[/quote]

Just notice that:

$ \frac {1}{x^2 \plus{} yz} \le \frac {1}{4} (\frac {1}{xy} \plus{} \frac {1}{xz})$

and then sum cycly.


factorial
AMC
AIME
floor function
function

Define $ n!!$ to be $ n(n\minus{}2)(n\minus{}4)\ldots3\cdot1$ for $ n$ odd and $ n(n\minus{}2)(n\minus{}4)\ldots4\cdot2$ for $ n$ even. When $ \displaystyle \sum_{i\equal{}1}^{2009} \frac{(2i\minus{}1)!!}{(2i)!!}$ is expressed as a fraction in lowest terms, its denominator is $ 2^ab$ with $ b$ odd. Find $ \displaystyle \frac{ab}{10}$.

[hide]
After playing around with the partial sums, notice that all of the odd factors in the denominator cancel with the numerator (skipping rigorous proof because its the AIME, maybe I'll add one later).

Given that, we can then notice that the numerator of any partial sum will be odd because the numerator of the last term is odd, and the numerators from the other terms are all multiplied by a power of 2 to get a common denominator.

Therefore, the denominator of the sum will just be 2 raised to the exponent of 2 in the prime factorization of 2(2009) = 4018. This can be calculated as $ \lfloor 4018/2 \rfloor \plus{}\lfloor 4018/4 \rfloor\plus{}\lfloor 4018/8 \rfloor\plus{}\lfloor 4018/16 \rfloor\plus{}\cdots$. Grinding that out, we get $ 4010$. So, $ a \equal{} 4010$ and $ b \equal{} 1$ which makes the answer $ \boxed{401}$.
[/hide]

[hide]
Note that $ (2i)!! = 2^i \times i!$ and $ (2i - 1)!! \times (2i)!! = (2i)!$. Thus
\[ \sum _{i = 1}^{2009} \frac {(2i - 1)!! \times (2i)!!}{2^{2i} (i!)^2} = \sum_{i = 1}^{2009} \frac {1}{2^{2i}} {2i \choose i} =  (\sum_{i = 1}^{2009} 2^{4018 - 2i} {2i \choose i} ) / 2^{4018 } .
\] As $ {2i \choose i}$ is an integer, the odd part of the denominator is $ 1$. It thus suffices to find the minimal value of the power $ p$ of $ 2$ in an addend of the numerator, and then subtract from $ 4018$. By Legendre's Theorem, this is \begin{align*} p = 4018 - 2i + \sum_{k = 1} \lfloor \frac {2i}{2^k} \rfloor - 2\sum_{k = 1} \lfloor \frac {i}{2^k} \rfloor & = 4018 - i + \sum_{k = 1} \lfloor \frac {2i}{2^{k + 1}} \rfloor - 2\sum_{k = 1} \lfloor \frac {i}{2^k} \rfloor \\
& = 4018 - i - \sum_{k = 1} \lfloor \frac {i}{2^k}. \rfloor
\end{align*} Evidentally this quantity is minimized when $ i = 2009$, which gives $ p = 2009 - \sum_{k = 1} \lfloor \frac {2009}{2^k} \rfloor = 8$. The answer is $ \frac {(4018 - 8) \cdot 1}{10} = \boxed{401}$.  [/hide]

Here's how I did it.
[hide]
Notice that the power of 2 in the denominator is the highest power of two that didives 4018!. This is easy to find and it is 4010. Thus ab/10=401a. But a is odd so if a>=3, ab/10>=1000 which is not possible (since all answers are less than 1000). Thus a=1 and ab/10=401.
[/hide]

[quote="dgreenb801"]Anyone have a rigorous proof?[/quote]

um, azjps gave a nice proof.

[hide="More elementary"]Consider an odd prime power $ p^n$. It will appear in the numerator of $ \frac {(2i \minus{} 1)!!}{(2i)!!}$ at least as many times as it will in the denominator - in the top, $ p^n, 3p^n, 5p^n,\ldots$ and in the bottom $ 2p^n, 4p^n, 6p^n, \ldots$. If $ 2kp^n$ is in the denominator, $ (2k \minus{} 1)p^n$ is in the numerator. So all powers of odd primes cancel in every term of the sum. Then the denominator of the $ 2009$th sum is a power of two as well; the top will be odd since it is the sum of an odd number and many even numbers, so the denominator is simply the largest power of two in $ (4018)!!$.[/hide]

This was one of the few I actually got right. I kept checking to see if there was no odd number (besides 1) in the denominator. I was convinced at the eight number  :P

From there, I just found how many factors of 2 were present in 2009. Basically, its just:

2009/2 + 2009/4 + 2009/8 + 2009/16...2009/1024 and applying the greatest integer function to each of these of course.

One thing that I'm seeing is that people are showing that the denominator of the reduced $ \frac{(2n\minus{}1)!!}{(2n)!!}$ is equal to the largest power of two that will divide $ (2n)!!$, but people are not being very rigorous in showing that this is also equal to the reduced denominator of $ \sum_{i\equal{}1}^n \frac{(2i\minus{}1)!!}{(2i)!!}$.  Richard Rusczyk even seemed to gloss over this in his AIME math jams solution session (I think he would have been docked some USAMO points for that :)).  While the proof is pretty easy to do by induction -- the meat of which roughly goes like this:

\[ S_{i\plus{}1} \equal{} S_i \plus{} a_{i\plus{}1} \equal{} \frac{\mbox{odd}}{2^m} \plus{} \frac{\mbox{odd}}{2^{m\plus{}n}} \equal{} \frac{\mbox{odd}\cdot2^n\plus{}\mbox{odd}}{2^{m\plus{}n}} \equal{} \frac{\mbox{odd}}{2^{m\plus{}n}}\]

it still needs to be done for full rigor.

It is trivial if you express each term as $ \frac1{2^{2i}}\binom{2i}i$, since the $ \binom{2i}i$ clearly must be an integer.

Consider the two series:

\[ \frac{1}{4}\plus{}\frac{2}{8}\plus{}\frac{4}{16}\plus{}\frac{8}{32} \equal{} \frac{1}{1}\]
\[ \frac{1}{4}\plus{}\frac{1}{8}\plus{}\frac{1}{16}\plus{}\frac{1}{16} \equal{} \frac{1}{2}\]

Note that the reduced denominator in both of these cases is not the same as the reduced denominator of any term in the series.  It is even the case that the numerator is odd for every term in the 2nd example.

I'm not seeing how observing that each term is $ \frac{1}{2^{2i}}\binom{2i}{i}$ (which I'll grant is a true observation) helps to show ('trivially', mind you) that the sum of 2009 such terms has a denominator equal to the denominator of the 2009th term.  Somehow I have to get confidence that I'm not going to create extra powers of 2 in the numerator when combining the terms.

The fact is that you don't get extra powers of 2 in the numerator is provable mind you, and it isn't really hard to prove.  I just don't think that this is such a trivial observation that one should gloss over it in a rigorous proof.  Perhaps I'm missing something here that makes this more trivial than I am seeing it to be.

My solution has quite a simple explanation.

[quote="m1sterzer0"]I'm not seeing how observing that each term is $ \frac {1}{2^{2i}}\binom{2i}{i}$ (which I'll grant is a true observation) helps to show ('trivially', mind you) that the sum of 2009 such terms has a denominator equal to the denominator of the 2009th term.  Somehow I have to get confidence that I'm not going to create extra powers of 2 in the numerator when combining the terms.[/quote]

After you simplify each fraction, you get $ \frac {\text{something odd}}{2^\text{natural number}}$. Since the last term $ \frac1{2^{2\cdot2009}}\binom{2\cdot2009}{2009}$ has the most twos in the denominator, we can express the whole sum as $ \frac1{2^{2\cdot1}}\binom{2\cdot1}1 + \frac1{2^{2\cdot2}}\binom{2\cdot2}2 + \cdots + \frac1{2^{2\cdot2008}}\binom{2\cdot2008}{2008} + \frac1{2^{2\cdot2009}}\binom{2\cdot2009}{2009}$
$ = \frac {\text{something odd}}{2^\text{natural number less than 4010}} + \frac {\text{something odd}}{2^\text{natural number less than 4010}} + \cdots$
${ + \frac {\text{something odd}}{2^\text{natural number less than 4010}} + \frac {\text{something odd}}{2^{4010}}}$
$ = \frac {\text{something odd}\cdot2^\text{natural number} + \cdots + \text{something odd}\cdot2^\text{natural number} + \text{something odd}}{2^{4010}}$
$ = \frac {\text{something odd}}{2^{4010}}$.

[hide=Solution]

Note that $(2i)!! = i! \cdot 2^i$, and that $(2i)!! \cdot (2i-1)!! = (2i)! \implies (2i-1)!! = \dfrac{(2i)!}{i! \cdot 2^i}.$

Thus, the summed terms can be rewritten as $\dfrac{(2i-1)!!}{(2i)!!} = \dfrac{(2i-1)!!}{i! \cdot 2^i} = \dfrac{(2i)!}{(i!)^2 \cdot 2^{2i}}. $

I claim that all the terms (and thus the sum) have a denominator of the form $2^n$.

For the proof, use the form $\dfrac{(2i)!}{(i!)^2 \cdot 2^{2i}}$. We can rewrite this as $\dfrac{(2i)!}{(i!)^2} \cdot \dfrac{1}{2^{2i}} = \binom{2i}{i} \cdot \dfrac{1}{2^{2i}}$.

$\binom{2i}{i}$ is equal to the number of ways to choose $i$ items out of a total of $2i$ items so it is always an integer, and so clearly the only prime factor in the denominator is $2$.

Also, the denominator of the final result is the denominator of the final term. To prove this, use the original form $\dfrac{(2i-1)!!}{(2i)!!}.$ Note that the expression for $i+1$ has more $2$s in the denominator than $i$.

This is because $\dfrac{(2i+1)!!}{(2i+2)!!} = \dfrac{(2i-1)!!}{(2i)!!} \cdot \dfrac{2i+1}{2i+2}$. The $2i+2$ introduces at least one more factor of $2$ in the denominator while the numerator has only odd numbers so the powers of $2$ in the denominator increases as $i$ increases.'

The sum has the denominator of the final term because writing everything under a common denominator means multiplying the numerators of all the terms (except the last) by a nonzero power of 2 and summing all of those (even numbers) with the odd numerator of the last term, meaning the numerator is odd and the denominator is that of the last term.

So now it remains to calculate the powers of $2$ in $\dfrac{(2\cdot 2009 -1)!!}{(2 \cdot 2009)!!}$. This is equal to the number of $2$s in $2009!$ plus those in $2^{2009}$, which is just $2009$.

Using [url=https://artofproblemsolving.com/wiki/index.php/Legendre%27s_Formula]Legendre formula[/url] we find that $2009$ has $2001$ $2s$ so the denominator is $2^{4010}$.

Thus the answer is $\dfrac{4010 \cdot 1}{10} = \boxed{410}$.

[/hide]


abstract algebra
vector
superior algebra
superior algebra unsolved

let O be a nonzero divisible Z-module. Prove that O is not a projective Z-module. Deduce that the rational numbers Q is not a projective Z-module.

You can't divide a basis element by anything.

You can embed a projective module in a free module; that means that for any $x$, the set $\{n: ny=x\text{ for some }y\}$ is the set of divisors of some element of the ring (or an intersection of such sets). Unless every nonzero element is invertible (certainly not true for $\mathbb{Z}$), no projective module can be divisible.
If everything is invertible, every module is projective and every module is injective; we're essentially in the vector space case then.


Hi AoPSers,

A couple weeks ago, some kids in my high school were assigned a project for their photography class.  One of the requirements was that the project had to be displayed somewhere visible on the school campus.  One group of students decided to do their project on the Pakistani side of the Middle East conflict.  He obtained a picture of a dead Pakistani child, killed by Israeli rockets, and surrounding the picture were rocks with factoids like the number of Pakistani children killed so far.  In addition, he printed out statistics from an unbiased, world-renouned organization and posted them on a bullitin board nearby.  He also recieved permission from the principle, vice-principle, dean, photography teacher, and many more.

Less than a week later, one of our teachers apparently found this offensive and called a Jewish activist group to look at the project.  They threatened the school with negative publicity so our school administration made the student take down the project.

This issue brought to my attention how sheltered Americans are from learning about the Pakistani's side of the conflict.  Whenever we turn on the news or read the paper, we mostly see stories about Pakistani suicide bombers or rockets, but we see much less about Israeli violence relative to the Pakistani's.  One possible reason for this one-sided propaganda is that the US is currently allied with Israel.

Thus, because I feel that Americans are missing a large part of the story, here is quick summary of the whole conflict:

Thousands of years ago, there was a group of people in the Middle East called the Caninites.  Some converted to Judaism and when the Roman Empire invaded, they forced the Jews out.  Jews spread out around the world, most going to Europe while others went to Russia and Africa.  Up to the late 1800's Jews had been oppressed constantly and blamed for things like the Black Plague and bad harvests.  Finally Theodore Hertzl created a movement called Zionism, which encouraged his fellow Jews to find a new homeland for themselves in Palestine and Jews began emigrating there.  After World War II, the UN decided to partition land in the Middle East to give to the Jews.  Pakistanis living in the Israeli part were evicted and those who refused were killed.  Finally, the Jews had their homeland.

One would think the Jews would be satisfied but they began expanding.  The Middle Eastern countries watched as Israel kept gobbling up their territory.  Finally, they banded together against Israel in the Six Days War.

Here is a link that shows a map of Israel:
http://en.wikipedia.org/wiki/File:1947-UN-Partition-Plan-1949-Armistice-Comparison.png
The blue is what was originally given to the Jews in 1947 by the UN and the pink is the territory that they took over by 1949.

The point of this story is that the Jews got their land AND their countries but by expanding beyond their borders, they violated the United Nation's Partition Plan.  Of course, the Jews have gone through thousands of years of oppression and endured the Holocaust, but that should not be an excuse for their current actions.

-Johnny

Did you write to newspapers about it? Lobby groups getting involved in censoring projects for high school. Yeah, nothing funny going on there! Being controlled by weakminded morons (in this case the school administration) is the worst feeling in the world. I would be very angry if I was involved in the project; in fact I am getting worked up just reading about it.

Also, as an ACLU member I feel obligated to inform you that they would probably love to get involved. There are many instances of students being censored in this way, and the ACLU will essentially always win. Look up your [url=http://www.aclu.org/affiliates/index.html]local chapter[/url].

[quote="jli"]
The point of this story is that the Jews got their land AND their countries but by expanding beyond their borders, they violated the United Nation's Partition Plan.  Of course, the Jews have gone through thousands of years of oppression and endured the Holocaust, but that should not be an excuse for their current actions.

-Johnny[/quote]

I tolally agree with this , and i dont want to ofend jews . They were victims and now they are making Pakistanies victims, The saddest part of all of these situation is that the majority of people are clueless about the real facts.

[quote="jli"]... Pakistani side of the Middle East conflict.  ... Pakistani child, killed by Israeli rockets...[/quote]  Have you examined a map of the world recently?   :|  (Also, "principle" vs. "principal," "Pakistanis" vs. "Pakistani's.")  Your history of Israel also [url=http://en.wikipedia.org/wiki/1948_Arab-Israeli_War]leaves something to be desired[/url].  (Though why I'm treating as serious a one-paragraph history lesson covering 2000+ years from someone who doesn't know the difference between Pakistan and Palestine, I don't know.  I really like the "one would think" line, though -- it's a good capstone to the ridiculousness of the whole thing.)

To the extent that the underlying issue is not just a bad joke, I'd like to second HilbertThm's suggestion that the ACLU is a good group to call about this sort of thing.

Dear JBL,

Originally, I wrote a really nasty reply to yours, but then I slapped myself across the head and told myself "You are 16 now and you need to act like a big boy."  Thus, I decided to be mature because I remembered that violence (including web violence), never solves anything. (a bit ironic that we're on the subject of the Middl East Conflict :D ).  Anyways, without further ado, here is my post to you.

I should definitely watch my grammar mistakes.  You have my sincerest apologies for casting those ugly typos on your magnificent forum.  Hopefully my horrendous errors will not completely mar the message I was trying to get across: that the non-Israeli side is being grossly under-represented.  Also, thank you for posting the link; I have learned something that I did not know before.  What I posted was basically what I learned in school and obviously my teacher has not been telling me the whole story.  I will have to scold him when school starts again.  I just hope that one day, we will be able to put this fiasco behind us.

Thanks again for opening my eyes.
-Johnny

Have you really not figured out that you are referring to 'Palestine' and not 'Pakistan'?  Do you realize that this incredibly egregious error makes it completely impossible to take you seriously?

Well of course I did, blahblahblah, which is why I posted my dearest apology before (and if you look closely, I carefully avoided saying those two words so I couldn't possibly be wrong again :D).  For everyone reading, I am referring to 'Palestine', not 'Pakistan'.  And please everyone, completely disregard my initial post above because it is full of despicable mistakes and untrue facts.  If someone who is more educated in this topic could post a better version of the story, that would be greatly appreciated. :)

-Johnny


AMC
AIME
MATHCOUNTS
USA(J)MO
USAMO
AMC 10

Did you make nationals?

What was your rank?

I had a bad day. 34th and no, I didn't make it :P .

Dan made nats, and our team came in 3rd.

I'm a little angry.

-Jorian

[quote="jhredsox"]
I had a bad day. 34th and no, I didn't make it :P .
[/quote]
Didn't you make nats last year though?

[quote="PenguinIntegral"]
Didn't you make nats last year though?[/quote]

I wish I did

a kid who got into nationals last year from CT didn't this year...

I can't believe I got 34th!!!!!!!!!!!!!!!!

it's no fair

i can never predict my placement right!

-jorian

[quote="jhredsox"]I didn't make it :P .
-Jorian[/quote]

I guess that was because you wasted too much of your time in this forum

Your team came in third? No hope for our team

[quote="solafidefarms"]Please stick to discussing _your_ placement, _your school's_ placement, and who made Nats, not other people's placements, even if they're from your school.

Also, @jhredsox, it's hard to improve if you blame your performance on a bad day and being unable to predict your placement. Accept your score and move on and try to make it so that even on a bad day you could make Nats.[/quote]

I wasn't blaming it on having a bad day, just that I can never guess my score.

I just...messed up, that's all (not a bad day, I just meant it wasn't a day where I did good, not blaming it on the day)

"who made nats, not other people's placements, even if they're from your school."

is Dan allright for this?

on my team (check), mad nats (check)

sorry about the kid who finished 18th. just trying to say that this is a oneshot thing and if you don't do good on that test, you could get placed a lot lower than your expectancy.

-Jorian

Yes, I made Nationals, that's all I can say.

Note to ANYONE FROM INDIANA: We are t3h sup3r s3cr3t t3am [/nerdy spy voice].

DO NOT POST ANY INDIANA SCORES. NOT UNTIL [b]NATIONALS[/b] is over. You heard me, Nationals. We aren't even allowed to tell anyone from indiana's scores after everyone else has taken state.

lucky...

I'm going to fail, as usual :( . Probably 5th, again. Or maybe I'll be like this guy who was 5th in 7th grade and 12th in 8th grade.

Our team second in CT (after a private school), so first among public schools!

Guys who did not make it to Nationals, forget it and start preparing for AIME!

Mom-coach

i got 8th. i think the pressure got to me during the sprint.  I WISH CD counted in IN!!!! :D (@fanatic, hopefully, i don't sound like i'm showing off and rubbing it in. i'm just shocked). however, i would certainly trade 1st in CD for 4th Written. I am MAD. :mad:

Not even close to showing off, don't worry. Also if you never saw my reply in the other thread, I said you shouldn't underestimate yourself. not that i thought you were bragging.

Yeah you pwnd me in countdown. My sugar high was way early because my parents told me I made 1st right before lunch  :mad:.

Well ok I got first. I guess I can say more than just "I made nats". Seeing as how slamchess has already said it in another thread....

remember how i didn't even get through the first round at chapter in the CD? well we practiced the 2006 state CD at school practice and somehow it liked turned on my brain.

I'M GOING TO NATS FOR THE 3RD YEAR IN A ROW!!! w00t!!! 1st place written and CD (official) My team got 5th out of 26 or so, which surprised me. What really scared me is at my state they post the rankings after the target round (which is first) and I was in 15th! The same thing happened last year. But it all turned out good anyway :lol:

Just watch, everyone's going to be like "UBEMAYA'S GOING TO GET 1ST IN CALIFORNIA WHAT A GENIUS" and then I'm going to get 86th with a score of under 10 and everyone's going to finally realize that I'm stupid and suck at math.

kstan013, what state are you from?

OK this is going to be a rant!

I missed nats by [color=red]XX[/color] points because I forgot [color=red]xxx xxxxx xxxxxxx[/color] (which I shall not disclose). I made seventh written and bombed CD. 2 brothers are on the team, one girl who wouldn't talk at all, and the guy who beat me on CD. My team took fifth only because of an amazing team round score. Since when did they award fifth place a trophy?
[size=75]
[color=red]This may not have crossed the line, but let's take no chances, please.  -- rcv>[/color][/size]

why do you complain? others might feel worse

Meh.  I was too nervous during CountDown so I dropped down to 4th from 3rd...  In WI CountDown is official...   :mad:  :mad:

Ugh, I have a bad feeling about states. I only got 1 44 on last years just now, which I've already done twice. At least  got the same on the 05 last week, which I'd somehow never taken.

Well that's why they regulate official cds.

[quote="jhredsox"]did yo have to face kstan in the third round?

i guess in the states where cd is official, they can't put 1 with 2, 3 with 4, 5 with 6, etc.!

-jorian[/quote]

No, kstan got [color=red]xxxx[/color] written and [color=red]xxxx[/color] CountDown, I think.  I think for WI they do it ladder-style, 10 vs 9, winner vs 8, winner of that vs 7, etc.  It's best out of 3 until top 4...  Then it's first to get 3 right.

[size=75][color=red]<You may list your own rank, but please don't disclose other people's ranks -- rcv>[/color][/size]

Just as a note to jhredsox, it's pretty annoying to see your real name popping up all over the place because of you, apparently it is to Ubemaya, I'm sure it is also to many others that you use (bpms for example).

By the way, 8th graders who did not make nationals. No exceptions.

Move on from MathCounts! Don't sit around waiting for high school. Give yourself that advantage, start doing high school math! Study AoPS 2! Prepare for next year's AMC's (or the AIME if you made the AIME this year)! If you do more MathCounts, then you're going to get your brains sucked out. No joke. It's a waste of your time to be doing it at the end of middle school when you don't even need it.

7th graders...

Take a break. Yes, if you didn't make it, study hard next year. But don't spend the whole year preparing for last year's states, unless you got like 5! Again, give yourself that advantage, work on high school math. There are bigger and better things. Of course, return to doing MathCounts when it's important.

And people going to nationals...

Two things. Yes, study for nationals, but then do the other things. Study high school math and stuff.
Do not become complacent, and do not inflate your head. Yes, you made it. Congratulations. Don't start making predictions about your placing. Yes, make goals. But shoot for them, don't chase them.

Trust me on all of this, personal experiences here (except the first part)

[quote="jhredsox"]Are you saying I shouldn't do "-jorian"? Until then I wil do it.

What if you're in 8th grade, and aren't good enough for AoPS 2?

I don't get the point of amc 10 or 12 if you're going to get killed by the aime or for others the usamo until maybe, maybe, maybe end of high school for some. Could someone explain?

Thanks,
Jorian[/quote]
No, he's saying that you shouldn't use other people's names in your posts and mention anything about who their aops identity is without their permission.
Even if they've posted it themselves.

Didn't I edit those recent posts?

-Jorian

our state is on March 24th...

i hope I make it to Nationals

really? our state is on March 17 (california)

getting in to the nationals?
are you kidding? that's beyond hope for me... :(

[quote="13375P34K43V312"]By the way, 8th graders who did not make nationals. No exceptions.

Move on from MathCounts! Don't sit around waiting for high school. Give yourself that advantage, start doing high school math! Study AoPS 2! Prepare for next year's AMC's (or the AIME if you made the AIME this year)! If you do more MathCounts, then you're going to get your brains sucked out. No joke. It's a waste of your time to be doing it at the end of middle school when you don't even need it.[/quote]

Thanks for the advice. :)

I think people who rant/complain are getting too much flak. Yeah ok so maybe you're going to make someone else feel bad because you did better than them and are still complaining. Still, even though it may be "just junior high", that doesn't mean people haven't invested a lot of time, energy, and emotion into their preparation.

The world can be a cruel place where days, hours, minutes, or even seconds can make or break what took years to build. While it's true there are probably other good ways to let out some frustration/disappointment/anger, a few lines on a forum aren't going to end the world either.

I'm not condoning the public complaints. However, it's been almost 5 years since I last did MathCounts, and when I look back on it now I wonder how I was able to invest myself so heavily into something knowing that a matter of minutes (sprint/target rounds) and then later on literally a matter of seconds (countdown) would define two years of blood, sweat, and tears (not literally, well maybe a little). I honestly don't know if I can even do that anymore.

I salute everybody that gives their all for things like MathCounts, and for those of you who are disappointed since things just didn't work out in the end (after all, only one person can win each year...), far be it from me or anyone else here to condemn you for a few lines expressing your feelings.


My school got a ddr game (supernova 2) at the end of last year. I'm just wondering, does anyone here play ddr? you guys are all probably better than me... i'm good compared to most of the people at school, but i can't pass anything on heavy/expert...

 :ddr:  :ddr:  :ddr:  :ddr:

i. suck. at. that. game. a. lot. I. cant. beat. easy. our. school. has. the. game. with. no. memory. card. i. am. failiure. and. green. guy. -->:ddr:  :ddr:  :ddr:

Played it today at school. Got honey punch difficult A, take on me difficult A, and felm beginner A. And I STILL can't beat anything on expert  :mad:  :mad:  :mad:

I like the song "Keep smiling". The cool part is being chased by a bullet.

I play FFR but mostly StepMania.

StepMania is beastly.

EDIT:
On actual DDR I play standard and heavy. I suck at heavy tho. I haven't played DDR in a long time.

[quote="lotsofmath"]Isn't that "Always Smiling"?[/quote]
Oh yeah.

I kept my felm school record from being broken... i didn't even break it  :oops: hopefully my sister's friend won't keep saying "Your record is going down!" for another week...  :ddr: must beat pink girl...

Wasn't there a thread like this before?

Our school doesn't have DDR, at home I have DDR Max, Max 2, and In the Groove (another dancing game made by an American company).

Favorite songs in max would be:
Max 300 (WAY better than Maxx Unlimited)
Insertion (not that hard, but fun)
So Deep (probably the hardest song in max, really fun)
Tsugaru (love the part in the middle)


function
algebra
domain
calculus
derivative
inequalities
calculus computations

Hi there im new here
I was wondering if anyone had any ideas on how to prove that a function is convex.
we have that f(x)<f(a)-[f(b)-f(a)/b-a](x-a) and that f''(x)>0
we need to show that f is a convex function using the mean value theorem (possibly more than once)
Please HELP!  Im desperate!
Grinny

you mean to show that if $f(x)<f(a)-\frac{f(b)-f(a)}{b-a}(x-a)$ and $f$ is twice derivable, then $f'' > 0$?
i have a proof.. nothing difficult, just a bit longer...
if none posts it, i'll be writing it later... give me 2 hours, and i'll be back

Assume that the domain is an interval- we're not interested in funny things happening over holes.

A function is convex if it lies below its secant lines; $f(tx+(1-t)y)\le tf(x)+(1-t)f(y)$ for $0\le t\le 1$. This is the definition.
A differentiable function is convex iff it lies above its tangent lines; $f(x+a)\ge f(x)+af'(x)$ for all $a$.
A twice differentiable function is convex iff its second derivative is nonnegative.

We have three statements to relate:
(1) $f(tx+(1-t)y)\le tf(x)+(1-t)f(y)$ for $0\le t\le 1$ and all $x,y$.
(2) $f(x+a)\ge f(x)+af'(x)$ for all $a,x$.
(3) $f''(x)\ge0$ for all $x$.

(1)+(f is diff.) implies (2): Set $y=x+a$. We have $f(x+(1-t)a)-f(x)\le (1-t)(f(x+a)-f(x))$, so $a\frac{f(x+(1-t)a)-f(x)}{(1-t)a}\lef(x+a)-f(x)$; taking the limit as $t\to 1^-$, $af'(x)\le\frac{f(x+a)-f(x)}{a}$.

(2) implies (1): Suppose $f$ is differentiable and (1) is violated; we have $x,y,z$ such that $x<y<z$ and $f(y)>\frac{z-y}{z-x}f(x)+\frac{y-x}{z-x}f(z)$. Let $w$ maximize $f(w)-\frac{z-w}{z-x}f(x)-\frac{w-x}{z-x}f(z)$; by Fermat's theorem, $f'(w)=\frac{f(z)-f(x)}{z-x}$. Then $f(x)$ and $f(z)$ lie below the tangent line at $w$, and (2) is violated, proving the contrapositive.

(2)+ (f is twice diff.) implies (3): Since $f(x+a)-f(x)\ge af'(x)$ and $f(x)-f(x+a)\ge -af'(x+a)$, we have $af'(x)\le af'(x+a)$ and $f'$ is increasing. Therefore $f''\ge0$.

(3) implies (2): Taylor's theorem (a refined form of the mean value theorem) tells us that $f(x+a)=f(x)+af'(x)+\frac{a^2}{2}f''(y)$ for some $y$. Since $f''(y)$ is nonnegative, we have the inequality in (2).