I use the term loosely... anything that is intriguing and informative with uncommon concepts and readable for someone who isn't already an expert. Obviously math competition books would count, but we've all seen those a hundred times around here. Math in general is okay but not necessary - other sciences, politics, economics, psychology, etc are all good. (On the other hand, I've never gotten myself to read a pure philosophy book, but maybe someone can convince me.) Be sure to tell why they're worth reading!

I'm halfway through [b]Gödel, Escher, Bach: An Eternal Golden Braid[/b] by Douglas Hofstadter. The preface talks about how no one has been able to sum it up very well after 20 years, but I'll try to cover at least some of the interesting things... The author talks about what he calls 'strange loops' -- recursive, self-referencing concepts -- and how they are found in formal systems such as in Gödel's Theorem, the music of Bach, the art of Escher, many other places. A simple example of this is the Epimenides paradox, "This statement is false." Hofstadter goes on to connect the 'strange loops' to explaining conciousness and connections to computer science and artificial intelligence. Half of the chapters present the concepts with dialogues in the style of Lewis Carroll, using his Tortoise, Achilles, and other characters. The other half is expository that pretty clearly and easily explains the ideas. The book can be a little slow, but it's filled with humor, interesting historical references, and great pictures (mostly Escher's artworks).

When I was very young, I read some of Carl Sagan's [b]Cosmos[/b]. At the time I was more interested in the great pictures that were in the huge hardcover version I borrowed from the library. However, I still remember a few bits and pieces, such as explaining bizzare occurances on Earth, introducing bizzare occurances in space (delay of light travel, ideas of aliens beyond the usual humanlike scifi style), and a lot of astronomy/astrophysics history. I'm sure if I read it again, I'd get a lot more from it.

And for fiction, Neal Stephenson's [b]Cryptonomicon[/b] made me interested in World War II, encryption, data banks, gold, and more. His [b]Snow Crash[/b] and [b]The Diamond Age[/b] made me in interested in the potential awesomeness of the future (virtual reality, mind implants, nanotechnology).

I'm sure there's a lot more out there... let me hear your suggestions.

For books, I'd have to say Hyperspace by Michio Kaku (check the spelling on the author)

There is a movie I really liked, What the Bleep Do We Know

It is an interesting movie, but you have to choose what to believe from the ideas presented

Don't miss out on Edward De Bono's Serious creativity...............or some of his other books on lateral thinking......they seriously give you a completely different way of looking at things......

The book{I'm referring to serious creativity} basically argues that creativity is not just something that occurs randomly, but is something that can be approached in a methodical manner............

[quote="sugarwergoingdown"]I like mostly sci fi classics and such (Farenheit 451, 1984, etc) and books on our ever expanding horizon of science (particle science, inventions, etc.

EDIT: oops I misread the topic. I got rid of the movies and the music i like. :blush:  :([/quote]

  Hmmm... I wonder how many different interpretations one can extract from the statement "Mind Expanding [size=150][b]BOOKS[/b][/size]".

Music and movies are fine since we don't have an overwhelming number of books yet.

Thanks for the suggestions so far. Now I'm interested specifically in books on consciousness, artificial intelligence, and economics/globalism. (And, yes, I know artificial intelligence and economics are pretty much unrelated.)

[quote="sugarwergoingdown"]your sarcasm isn't appreciated. ;)[/quote]

Yes it is :D. Oh yeah Count Down by Steve Olson (sp?).

Since we talked about Godel in GEB, I suggest "Godel's Proof" by Nagel and Newman. I loved that book, and "mind expanding" is an appropriate label for it from my perspective.


probability
geometry
3D geometry
number theory
relatively prime

Hi, a question.

A fair die is rolled four times. The probability that each of the final three rolls is at least as large as the roll preceding it may be expressed in the form m/n where m and n are relatively prime positive integers. Find m+n.

actually, the answer is 126/6^4. I just don't know how to get the 126.

There are $6^{4}$ total ways to roll the die.

For any given sequence of rolls, there is exactly one way to place the rolls such that they are in ascending order.

If the rolls all had to be different, then there would be $\binom64$ ways to roll the die.  However, the last three rolls could be the same as the one before it, so there are $\binom94=126$ ways to roll the die.

$\therefore\frac{126}{6^{4}}=\frac{7}{72}$

[quote="13375P34K43V312"][hide="Brute force"]
Case 1: first roll is a 1
$\frac{1}{6}\cdot\frac{5}{6}\cdot\frac{5}{6}\cdot\frac{5}{6}=\frac{125}{1296}$.
Case 2: first roll is a 2
$\frac{1}{6}\cdot\frac{4}{6}\cdot\frac{4}{6}\cdot\frac{4}{6}=\frac{64}{1296}$..

We can see that the other four will be similar, in the form $\frac{(6-c)^{3}}{1296}$, where $c$ is the case number. Thus the numerator is the sum of the first 5 cubes, which is the sum of the first 5 natural numbers squared, $15^{2}=225$. The probability is $\frac{225}{1296}=\frac{25}{144}$.[/hide][/quote]

What if the first roll was a 2, the second roll a 6, then the 3rd roll wouldn't have a $\frac{4}{6}$ chance.


geometry
circumcircle
incenter
Euler
geometry solved

[quote="sxcxc"]In a triangle $ABC$ with circumcenter $O$, incenter $I$ and centroid $G$, prove that $OI\geq OG\geq\frac{1}{2}IG$.[/quote]

Let H be the orthocenter of triangle ABC. It is well-known that

[b](1)[/b] the point G lies on the segment HO and we have $GH=2\cdot OG$ (Euler line theorem);
[b](2)[/b] we have $\measuredangle GIH > 90^{\circ}$ (see http://www.mathlinks.ro/Forum/viewtopic.php?t=39447 ).

Now, [b](2)[/b] tells that the triangle GIH has an obtuse angle at I; thus, it has an acute angle at G (since a triangle can have at most one non-acute angle), so that $\measuredangle HGI<90^{\circ}$. But [b](1)[/b] yields $\measuredangle OGI=180^{\circ}-\measuredangle HGI$, and thus $\measuredangle OGI>180^{\circ}-90^{\circ}=90^{\circ}$. Thus, the triangle OGI is obtuse-angled at G. Since the longest side of an obtuse-angled triangle is the side opposite to the obtuse angle, it follows that OI is the longest side of triangle OGI; thus, $OI\geq OG$. Since the problem asks us to prove that $OI\geq OG\geq\frac12\cdot IG$, it remains to prove that $OG\geq\frac12\cdot IG$. In other words, we have to show that $2\cdot OG\geq IG$. Now, [b](1)[/b] yields $2\cdot OG=GH$, so we have to show that $GH\geq IG$. Since the triangle GIH has an obtuse angle at I, and the longest side of an obtuse-angled triangle is the side opposite to the obtuse angle, it follows that GH is the longest side of triangle GIH, so that $GH\geq IG$ is shown. Proof complete.

  darij


Slips of paper, each with a distinct integer from 1 through 6 written on it, are placed in a box. If the slips of paper are drawn one at a time and are placed in the order they are drawn, what is the probability that they form the number 123,456? Express the result as a common fraction.

Answer:
There are 6 possible digits for the 1st digit, 5 for the 2nd, 4 for the 3rd, etc.
thus, there are 6!=720 possibilities.
Only 1 work, so the answer is 1/720.


Find Each quantity if vector v = 3i - 5j + 2k and vector
w = -2i + 3j - 2k.

(1) 3v - 2w

(2) ||v + w||


Find all positive integers n that are within 250 of exactly
15 perfect squares.

[hide]
We are looking for $ k$ such that $ (k \plus{} 15)^2 \minus{} k^2 \le 500 \le (k \plus{} 16)^2 \minus{} k^2$. Any $ n$ that are within $ 250$ of $ k, (k \plus{} 15)$ will work.

So, we have $ 30k \plus{} 225 \le 500 \le 32k \plus{} 256.$ From the left inequality, we get $ k \le 275/30 \equal{} 55/6$, so $ k\le 9$. From the right inequality, we get $ k \ge 244/32 \equal{} 61/8$, so $ k\ge 8$. Therefore, $ k \equal{} 8, 9$ work.

So, those numbers within $ 250$ from $ 64$ and $ 529$ are $ [279, 314]$, and those within $ 250$ from $ 81$ and $ 576$ are $ [226, 331]$.
[/hide]

Just make sure you consider the overlapping, which might result in 16th and 17th square getting into the picture (which we don't want to happen).

I think I did... :maybe: because I also had the upper bound...

But... from $ k$ to $ k\plus{}15$, both inclusive, there are $ 16$ numbers... Therefore I think you need $ (k\plus{}14)^2\minus{}k^2\leqslant 500\leqslant (k\plus{}15)^2\minus{}k^2$

Oh, you are right :oops: Thank you, Farenhajt.


inequalities
Cauchy Inequality
inequalities proposed

(MOSP 2007) Let $ a,b,c,x,y,z$ be positive real numbers such that $ ax\plus{}by\plus{}cz\equal{}xyz.$ Prove that
$ x\plus{}y\plus{}z > \sqrt{a\plus{}b}\plus{}\sqrt{b\plus{}c}\plus{}\sqrt{c\plus{}a}.$

[quote="great math"]We tackle this inequality by Cauchy inequality

[u][i][b]Solution[/b][/i][/u]

$ 1 = \frac {a}{yz} + \frac {b}{xz} + \frac {c}{xy} > \frac {1}{x}(\frac {b}{z} + \frac {c}{y}) \Leftrightarrow x > \frac {b}{z} + \frac {c}{y}$

Similarly, we apply for other terms
$ \{\begin{array}{cc} y > \frac {a}{z} + \frac {c}{x} \\
 \\
z > \frac {a}{y} + \frac {b}{x} \end{array}$

Summing up three analogous inequalities, we get

$ 2( x + y + z) > (\frac {a + b}{z} + z) + (\frac {a + c}{y} + y) + (\frac {b + c}{x} + x) \ge 2\sum_{cyc}\sqrt {a + b}$

We complete our proof here.[/quote]
Very nice your proof, dear Viet. Let us continue with another nice inequality:

Prove that
$ x + y + z \ge \sqrt{2}\min \{\sqrt {a + 2b} + \sqrt {a + 2c},\sqrt {b + 2c} + \sqrt {b + 2a},\sqrt {c + 2a} + \sqrt {c + 2b}\},$
for positive numbers $ a,b,c,x,y,z$ such that $ ax + by + cz = xyz.$

A little remark:
$ \sqrt{2} \min \{\sqrt {a + 2b} + \sqrt {a + 2c},\sqrt {b + 2c} + \sqrt {b + 2a},\sqrt {c + 2a} + \sqrt {c + 2b}\}> \sqrt {a + b} + \sqrt {b + c} + \sqrt {c + a}.$

Wow, the second inequality is really nice. Thanks for sharing Canhang2007!

[u][i][b]Solution[/b][/i][/u]

Firstly, we may assume that b is between a and c. Therefore, it comes to the following estimation

[size=150][u][b]Claim 1[/b][/u][/size]

$ (b+2a)(b+2c) \le 3(ab+bc+ca)$

[b][u]Proof[/u][/b]      It is trivial since $ (b-a)(b-c) \le 0$

[size=150][b][u]Claim 2[/u][/b][/size]

$ \min\{\sqrt{a+2b}+\sqrt{a+2c},\sqrt{b+2c}+\sqrt{b+2a},\sqrt{c+2a}+\sqrt{c+2b}\}=\sqrt{b+2c}+\sqrt{b+2a}$

[b][u]Proof[/u][/b]: We construct an inteval, let's say $ 0<c<b<a$ then

\[ \{\begin{array}{cc}
\sqrt{a+2b}+\sqrt{a+2c} \ge \sqrt{b+2c}+\sqrt{b+2a} \\ \\
\sqrt{c+2a}+\sqrt{c+2b}  \ge \sqrt{b+2c}+\sqrt{b+2a} \end{array}\]

The original inequality can be written as follows based on claim 2

$ (x+y+z)^2 \ge 4(a+b+c+\sqrt{(b+2a)(b+2c)})$

Or according to claim 1, it is enough to show that

$ (x+y+z)^2 \ge 4(a+b+c+\sqrt{3(ab+bc+ca)})$

Using the condition given, we have

$ LHS=(\frac{a}{yz}+\frac{b}{xz}+\frac{c}{xy})(x+y+z)^2$

$ = \sum_{cyc}\frac{ax^2}{yz} + \sum_{cyc}x(\frac{b}{z}+\frac{c}{y})+2(a+b+c)+2\sum_{cyc}(y+z)(\frac{b}{z}+\frac{c}{y})$

We observe that

$ LHS > \sum_{cyc}b(\frac{x}{z}+\frac{z}{x}) + 6(a+b+c) \ge 8(a+b+c)$

Finally, we show that

$ 8(a+b+c) \ge 4(a+b+c)+4\sqrt{3(ab+bc+ca)}$

or equivalently

$ a+b+c \ge \sqrt{3(ab+bc+ca)}$

The last inequality is trivial so we complete our proof here

Really nice, congratulations great math ! :)

[quote="great math"]
$ LHS = \left(\frac {a}{yz} + \frac {b}{xz} + \frac {c}{xy}\right)(x + y + z)^2$

$ = \sum_{cyc}\frac {ax^2}{yz} + \sum_{cyc}x\left(\frac {b}{z} + \frac {c}{y}\right) + 2(a + b + c) + 2\sum_{cyc}(y + z)\left(\frac {b}{z} + \frac {c}{y}\right)$
[/quote]
I am sorry, Viet. But I think this identity is not correct. I think it must be
$ \left(\frac {a}{yz} + \frac {b}{xz} + \frac {c}{xy}\right)(x + y + z)^2 = \sum_{cyc}\frac {ax^2}{yz} + \sum_{cyc}x\left(\frac {b}{z} + \frac {c}{y}\right) - 2(a + b + c) + 2\sum_{cyc}(y + z)\left(\frac {b}{z} + \frac {c}{y}\right).$
Please check it again.

[quote][size=150][b][u]Claim 2[/u][/b][/size]

$ \min\left\{\sqrt {a + 2b} + \sqrt {a + 2c},\sqrt {b + 2c} + \sqrt {b + 2a},\sqrt {c + 2a} + \sqrt {c + 2b}\right\} = \sqrt {b + 2c} + \sqrt {b + 2a}$[/quote]
For this, I think it is not correct, too. With your assumption $ b$ is the number between $ a$ and $ c,$ we cannot deduce that $ \sqrt{b+2c}+\sqrt{b+2a}$ is the smallest among three expressions above. The reason is
$ (a+2b)+(a+2c)=(b+2c)+(b+2a)=(c+2a)+(c+2b),$
and
$ (b+2c)(b+2a) -(a+2b)(a+2c) =(b-a)(a+b-2c),$
$ (b+2c)(b+2a)-(c+2a)(c+2b)=(b-c)(b+c-2a).$
We dont know if either $ (b-c)(a+b-2c),(b-c)(b+c-2a)$ are negative or not. This means that we dont know if
$ \left( \sqrt{b+2c}+\sqrt{b+2a}\right)^2 \le\left( \sqrt{a+2b}+\sqrt{a+2c}\right)^2,$
$ \left( \sqrt{b+2c}+\sqrt{b+2a}\right)^2 \le \left(\sqrt{c+2a}+\sqrt{c+2b\right)^2,}$
or not. So the claim 2 of you is not correct.

However, we can use the fact
$ \min\left\{\sqrt {a + 2b} + \sqrt {a + 2c},\sqrt {b + 2c} + \sqrt {b + 2a},\sqrt {c + 2a} + \sqrt {c + 2b}\right\} \le \sqrt {b + 2c} + \sqrt {b + 2a}$
to prove our inequality. Please try it.

Dear Canhang, I think that my claim 2 is right since I stated in my solution above that we could choose an interval so that $ 0<c<b<a$ and then

$ (b\plus{}2c)(b\plus{}2a)\minus{}(a\plus{}2b)(a\plus{}2c) \equal{}(b\minus{}a)(a\plus{}b\minus{}2c) <0$

$ (b\plus{}2c)(b\plus{}2a)\minus{}(c\plus{}2a)(c\plus{}2b)\equal{}(b\minus{}c)(b\plus{}c\minus{}2a) <0$

The above statements are still true if we change the role of $ a$ and $ c$ deliberately.

However, I apologize for my miscalculation ( something was spinning in my head and  I don't know why? :( ). Also, continue the way I have done  before, there is a little alteration in my identity

The correct identity should look like this

$ \left(\frac{a}{yz}\plus{}\frac{b}{xz}\plus{}\frac{c}{xy}\right)(x\plus{}y\plus{}z)^{2}\equal{}\sum_{cyc}\frac{ax^{2}}{yz}\plus{}\sum_{cyc}x\left(\frac{b}{z}\plus{}\frac{c}{y}\right)\plus{}2(a\plus{}b\plus{}c)\plus{}2\sum_{cyc}\left(\frac{ax}{y}\plus{}\frac{ax}{z}\right).$

Hence, we have the following estimation

$ LHS > 6(a\plus{}b\plus{}c)\plus{}4(\sqrt{ab}\plus{}\sqrt{bc}\plus{}\sqrt{ca})$

Thus, it is enough to check that

$ 6(a\plus{}b\plus{}c)\plus{}4(\sqrt{ab}\plus{}\sqrt{bc}\plus{}\sqrt{ca}) \ge 4(a\plus{}b\plus{}c)\plus{}4\sqrt{3(ab\plus{}bc\plus{}ca)}$

or

$ a\plus{}b\plus{}c\plus{}2\left(\sqrt{ab}\plus{}\sqrt{bc}\plus{}\sqrt{ca}\right) \ge 2\sqrt{3(ab\plus{}bc\plus{}ca)}$

Having observed that

$ a\plus{}b\plus{}c\plus{}2\left(\sqrt{ab}\plus{}\sqrt{bc}\plus{}\sqrt{ca}\right) \ge a\plus{}b\plus{}c\plus{}2\sqrt{ab\plus{}bc\plus{}ca} \ge 2\sqrt{2\sqrt{ab\plus{}bc\plus{}ca}(a\plus{}b\plus{}c)}$

It remains to prove that

$ 2\sqrt{ab\plus{}bc\plus{}ca}(a\plus{}b\plus{}c) \ge 3(ab\plus{}bc\plus{}ca)$ or equivalently $ a\plus{}b\plus{}c \ge \frac{3}{2}\sqrt{ab\plus{}bc\plus{}ca}$

Again, Cauchy inequality works well for the last inequality and we end our proof here.

[quote="great math"]Dear Canhang, I think that my claim 2 is right since I stated in my solution above that we could choose an interval so that $ 0 < c < b < a$ and then

$ (b \plus{} 2c)(b \plus{} 2a) \minus{} (a \plus{} 2b)(a \plus{} 2c) \equal{} (b \minus{} a)(a \plus{} b \minus{} 2c) < 0$

$ (b \plus{} 2c)(b \plus{} 2a) \minus{} (c \plus{} 2a)(c \plus{} 2b) \equal{} (b \minus{} c)(b \plus{} c \minus{} 2a) < 0$

The above statements are still true if we change the role of $ a$ and $ c$ deliberately.

However, I apologize for my miscalculation ( something was spinning in my head and  I don't know why? :( ). Also, continue the way I have done  before, there is a little alteration in my identity

The correct identity should look like this

$ \left(\frac {a}{yz} \plus{} \frac {b}{xz} \plus{} \frac {c}{xy}\right)(x \plus{} y \plus{} z)^{2} \equal{} \sum_{cyc}\frac {ax^{2}}{yz} \plus{} \sum_{cyc}x\left(\frac {b}{z} \plus{} \frac {c}{y}\right) \plus{} 2(a \plus{} b \plus{} c) \plus{} 2\sum_{cyc}\left(\frac {ax}{y} \plus{} \frac {ax}{z}\right).$

Hence, we have the following estimation

$ LHS > 6(a \plus{} b \plus{} c) \plus{} 4(\sqrt {ab} \plus{} \sqrt {bc} \plus{} \sqrt {ca})$ (*)
[/quote]
Dear Viet, I apologize for my last message about your claim 2. I am so stupid, very sorry. Now, could you please explain some more lines for the estimation (*)? I think you have proved
$ \sum x\left(\frac {b}{z} \plus{} \frac {c}{y}\right) \plus{} 2\sum \left(\frac {ax}{y} \plus{} \frac {ax}{z}\right) \ge 4(a \plus{} b \plus{} c) \plus{} 4\left( \sqrt {ab} \plus{} \sqrt {bc} \plus{} \sqrt {ca}\right).$
But I think this inequality is not true, so please explain more about your solution. Thanks.

And here is my proof:
Without loss of generality, we may assume that $ x$ is the largest number among $ x,y,z.$ Then, applying the AM-GM Inequality together the Cauchy Schwarz Inequality, we have
$ \frac {(x \plus{} y \plus{} z)^2(ax \plus{} by \plus{} cz)}{xyz} \ge\frac{4x(y \plus{} z)(ax \plus{} by \plus{} cz)}{xyz} \equal{} \frac {2(y \plus{} z)(2ax \plus{} 2by \plus{} 2cz)}{yz} \ge \frac {2(y \plus{} z)[a(y \plus{} z) \plus{} 2by \plus{} 2cz]}{yz} \equal{} \frac {2(y \plus{} z)[(a \plus{} 2c)z \plus{} (a \plus{} 2b)y]}{yz} \ge \frac {2\left[ \sqrt {(a \plus{} 2c)zy} \plus{} \sqrt {(a \plus{} 2b)yz}\right]^2}{yz} \equal{} 2\left( \sqrt {a \plus{} 2b} \plus{} \sqrt {a \plus{} 2c}\right)^2.$
Since $ \frac {ax \plus{} by \plus{} cz}{xyz} \equal{} 1,$ this inequality implies
$ (x \plus{} y \plus{} z)^2 \ge 2\left( \sqrt {a \plus{} 2b} \plus{} \sqrt {a \plus{} 2c}\right)^2,$
from which we deduce that
$ x \plus{} y \plus{} z \ge \sqrt {2}\left( \sqrt {a \plus{} 2b} \plus{} \sqrt {a \plus{} 2c}\right) \ge \sqrt {2}\min \left\{\sqrt {a \plus{} 2b} \plus{} \sqrt {a \plus{} 2c},\sqrt {b \plus{} 2c} \plus{} \sqrt {b \plus{} 2a},\sqrt {c \plus{} 2a} \plus{} \sqrt {c \plus{} 2b}\right\},$
as desired. The proof is completed.

Dear Canhang, I am sorry for not wrtiting the full proof and minding you checking my solution all the times. By the way, your solution is really touching and glamorous. Here are some few lines of my explanation

$ x\left(\frac{b}{z}\plus{}\frac{c}{y}\right) \plus{} y\left(\frac{a}{z}\plus{}\frac{c}{x}\right)\plus{}z\left(\frac{a}{y}\plus{}\frac{b}{x}\right)$

$ \equal{} b\left(\frac{x}{z}\plus{}\frac{z}{x}\right)\plus{} a\left(\frac{y}{z}\plus{}\frac{z}{y}\right)\plus{}c\left(\frac{x}{y}\plus{}\frac{y}{x}\right)$

and

$ 2\sum\left(\frac{ax}{y}\plus{}\frac{ax}{z}\right)\equal{}2\left(a\frac{x}{y}\plus{}a\frac{x}{z}\plus{}b\frac{y}{x}\plus{}b\frac{y}{z}\plus{}c\frac{z}{x}\plus{}c\frac{z}{y}\right)$

Hope you  forgive me for my laziness. Thanks a lot!

[quote="great math"]Dear Canhang, I am sorry for not wrtiting the full proof and minding you checking my solution all the times. By the way, your solution is really touching and glamorous. Here are some few lines of my explanation

$ x\left(\frac {b}{z} \plus{} \frac {c}{y}\right) \plus{} y\left(\frac {a}{z} \plus{} \frac {c}{x}\right) \plus{} z\left(\frac {a}{y} \plus{} \frac {b}{x}\right)$

$ \equal{} b\left(\frac {x}{z} \plus{} \frac {z}{x}\right) \plus{} a\left(\frac {y}{z} \plus{} \frac {z}{y}\right) \plus{} c\left(\frac {x}{y} \plus{} \frac {y}{x}\right)$

and

$ 2\sum\left(\frac {ax}{y} \plus{} \frac {ax}{z}\right) \equal{} 2\left(a\frac {x}{y} \plus{} a\frac {x}{z} \plus{} b\frac {y}{x} \plus{} b\frac {y}{z} \plus{} c\frac {z}{x} \plus{} c\frac {z}{y}\right)$

Hope you  forgive me for my laziness. Thanks a lot![/quote]
I understand your cyclic sum, dear great math. But I wonder where did the constant $ 6$ came from your estimation, as we have
$ b\left(\frac {x}{z} \plus{} \frac {z}{x}\right) \plus{} a\left(\frac {y}{z} \plus{} \frac {z}{y}\right) \plus{} c\left(\frac {x}{y} \plus{} \frac {y}{x}\right) \ge 2(a\plus{}b\plus{}c),$
and
$ 2\left(a\frac {x}{y} \plus{} a\frac {x}{z} \plus{} b\frac {y}{x} \plus{} b\frac {y}{z} \plus{} c\frac {z}{x} \plus{} c\frac {z}{y}\right) \ge 4(\sqrt{ab}\plus{}\sqrt{bc}\plus{}\sqrt{ca}),$
which both adding to the expression $ 2(a\plus{}b\plus{}c)$ from the expansion gives only $ 4(a\plus{}b\plus{}c)\plus{}4(\sqrt{ab}\plus{}\sqrt{bc}\plus{}\sqrt{ca}),$ not $ 6(a\plus{}b\plus{}c)\plus{}4(\sqrt{ab}\plus{}\sqrt{bc}\plus{}\sqrt{ca}).$ That's the reason why I asked you, dear viet.

:blush: Opps. It was my uncoverable mistake. Probably, I may start with this inequality again later on. Anyway, thanks Canhang for your feedback.

[quote="great math"]We tackle this inequality by Cauchy inequality

[u][i][b]Solution[/b][/i][/u]

$ 1 = \frac {a}{yz} + \frac {b}{xz} + \frac {c}{xy} > \frac {1}{x}(\frac {b}{z} + \frac {c}{y}) \Leftrightarrow x > \frac {b}{z} + \frac {c}{y}$

Similarly, we apply for other terms
$ \{\begin{array}{cc} y > \frac {a}{z} + \frac {c}{x} \\
 \\
z > \frac {a}{y} + \frac {b}{x} \end{array}$

Summing up three analogous inequalities, we get

$ 2( x + y + z) > (\frac {a + b}{z} + z) + (\frac {a + c}{y} + y) + (\frac {b + c}{x} + x) \ge 2\sum_{cyc}\sqrt {a + b}$

We complete our proof here.[/quote]
See also the Namdung's proof here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=1899


geometry
circumcircle
ratio
angle bisector
cyclic quadrilateral
projective geometry
geometry solved

Let SR and AC meet at E and PQ and AC meet at E'. By Menelause theorem we have:

$\frac{DS}{SA}\cdot \frac{AE}{CE}\cdot \frac{CR}{RD}=1$ and $\frac{AP}{PB}\cdot \frac{BQ}{QC}\cdot \frac{CE'}{AE'}=1$.

Since DS=DR, BL=BP, CR=CQ, AS=AP, so we have: $\frac{CE'}{AE'}=\frac{CE}{AE}$. Thus $E=E'$ so AC, SR and PQ are concurrent.

Similarly, by using the angle bisector theorem, we have NM, KL and AC being concurrent. Let their concurrent point be F. So $E=F \Longleftrightarrow \frac{CR}{AS}=\frac{CM}{AN}\Longleftrightarrow \frac{CR}{AS}=\frac{CT}{AT}$. Let DC and BA meet at U, then, since R, T and P are collinear, so apply Menelaus in triangle UAC, we have $\frac{UR}{CR}\cdot \frac{CT}{AT}\cdot \frac{AP}{PU}=1$. Because UP=UR, AP=AS, we can conclude that $\frac{CR}{AS}=\frac{CT}{AT}$. Therefore $E=F$.

We have cyclic quadrilateral PQRS. Let QR and PS meet at G, the G is on line BD. Since T is the intersection of RP and QS, so X and Y are also the intersections of GT. Because E is the intersection of SR and PQ, so we have EX and EY being the tangent lines of the circumcircle of PQRS, which completes the proof.

[quote="Bismarck"]Let quadrilateral $ABCD$ be circumscribed on circle $\omega$. Denote:

$T = AC \cap BD$

$\{X,Y\}= \omega \cap BD$

$i, j$ - tangents to $\omega$ in $X$ and $Y$ respectively.

$k, l, m, n$ - angle bisectors of: $\angle ATB, \angle BTC, \angle CTD, \angle DTA$

$K = AB \cap k$
$L = BC \cap l$
$M = CD \cap m$
$N = DA \cap n$

$P = AB \cap \omega$
$Q = BC \cap \omega$
$R = CD \cap \omega$
$S = DA \cap \omega$

Prove that the lines $AC, KL, MN, PQ, RS, i, j$ concur in one point.[/quote]

A slightly different proof:

In [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=135302#135302]http://www.mathlinks.ro/Forum/viewtopic.php?t=20538 post #2[/url], I showed that the harmonic conjugate U of the point T with respect to the segment AC lies on the lines PQ and RS and is the pole of the line BD with respect to the incircle $\omega$ of the quadrilateral ABCD.

The lines i and j are the tangents to the circle $\omega$ at the points X and Y, or, equivalently, the polars of the points X and Y with respect to $\omega$. Hence, the point of intersection of these lines i and j is the pole of the line XY with respect to the circle $\omega$; but the line XY is the line BD, and thus the point of intersection of the lines i and j must be the pole of the line BD with respect to the circle $\omega$. But this pole is U, as we know. So the point U must be the point of intersection of the lines i and j.

Since U is the harmonic conjugate of the point T with respect to the segment AC, we have $\frac{CU}{UA}=-\frac{CT}{TA}$. Since we know that the point T lies inside the segment AC, we have $\frac{CT}{TA}=\frac{\left|CT\right|}{\left|AT\right|}$, and thus $\frac{CU}{UA}=-\frac{\left|CT\right|}{\left|AT\right|}$. On the other hand, since TK is the angle bisector of the angle ATB, and since the angle bisector of an angle of a triangle divides the opposite side in the ratio of the adjacent sides, we have $\frac{AK}{KB}=\frac{\left|AT\right|}{\left|BT\right|}$, and similarly $\frac{BL}{LC}=\frac{\left|BT\right|}{\left|CT\right|}$. Hence,

$\frac{AK}{KB}\cdot\frac{BL}{LC}\cdot\frac{CU}{UA}=\frac{\left|AT\right|}{\left|BT\right|}\cdot\frac{\left|BT\right|}{\left|CT\right|}\cdot\left(-\frac{\left|CT\right|}{\left|AT\right|}\right)=-1$.

Thus, using the Menelaos theorem for the triangle ABC and the points K, L, U on its sidelines AB, BC, AC, we obtain that the points K, L, U are collinear, so the point U lies on the line KL. Similarly, the point U lies on the line MN.

Combining all what we saw above, we see that the point U lies on the lines AC, KL, MN, PQ, RS, i, j. Thus, these lines concur at one point, qed.

  Darij


limit
logarithms
inequalities
complex numbers
calculus
calculus computations

For what conditions on a sequence $ \{ a_n \}$ does the following lemma hold?

[b]Lemma:[/b]  $ \sum a_n < \infty \Leftrightarrow \prod (1 \plus{} a_n) < \infty$.

I'm pretty sure $ \lim_{n \to \infty} a_n \equal{} 0$ is necessary.  Is it necessary that $ a_n$ be real?  Non-negative?  Monotonic?

The idea:  $ \ln \prod (1 \plus{} a_n) \equal{} \sum \ln (1 \plus{} a_n) \equal{} \sum (a_n \minus{} \frac{a_n^2}{2} \pm ...)$.  The question is when the "remainder term" converges, I guess.

The convention on infinite products is to say that the product "diverges" to zero if it tends to zero without any of its factors being zero. Hence $ \prod (1\plus{}a_n)<\infty$ isn't really the right way to express what you're saying.

If $ \prod_{n\equal{}1}^{\infty}(1\plus{}a_n)$ converges, then we must have $ \lim_{n\to\infty}a_n\equal{}0.$

For that reason, let's confine ourselves to the case in which $ a_n\to0.$

Rather than looking at the full Taylor series, let's just look at simple inequalities:

$ \ln(1\plus{}x)\le x$ for all $ x>\minus{}1.$

If $ 0<x<c,$ then there exists $ b>0$ such that $ \ln(1\plus{}x)\ge bx.$

If $ \minus{}1<c<x<0,$ then there exists $ b>0$ such that $ \ln(1\plus{}x)\ge bx.$

Here are some statements we can get from this:

If $ a_n>0$ for all sufficiently large $ n,$ then $ \prod(1\plus{}a_n)$ converges iff $ \sum a_n$ converges.

If $ \minus{}1<a_n<0$ for all sufficiently large $ n,$ then $ \prod(1\plus{}a_n)$ converges iff $ \sum a_n$ converges.

If $ \sum a_n$ converges absolutely, then $ \prod(1\plus{}a_n)$ converges. (In fact we can say that the product converges unconditionally - that it is left unchanged by rearrangements of the order.)

We do run into trouble when the $ a_n$ change sign and the sum isn't absolutely convergent. For instance, we can build an example in which $ \sum a_n$ converges conditionally but $ \prod(1\plus{}a_n)$ diverges to zero.

As for complex values: that simply extends the trouble with real signed values. But it's still true that absolute convergence of the sum would imply convergence of the product.

[quote="Kent Merryfield"]The convention on infinite products is to say that the product "diverges" to zero if it tends to zero without any of its factors being zero. Hence $ \prod (1 \plus{} a_n) < \infty$ isn't really the right way to express what you're saying.[/quote]

Makes sense - in other words, the log of the product diverges to $ \minus{}\infty$, which is not particularly nice behavior.  Thanks :)

[quote="Kent Merryfield"]
We do run into trouble when the $ a_n$ change sign and the sum isn't absolutely convergent. [/quote]
The good condition that still ensures that the sum and the product converge simultaneously (under the natural assumptions that $ a_k\ne \minus{}1$ and $ a_k\to 0$) is $ \sum_k |a_k|^2<\plus{}\infty$. It is, of course, not absolutely necessary if you talk about complex numbers but it is the next step beyond the absolute convergence condition. Note that for real $ a_k$, the convergence of both the sum and the product implies this condition.

In other words, the norm is finite in $ l^2$.  Could you elaborate, please?  :)

Note that $ \lim_{z\to0}\frac {\log(1 \plus{} z) \minus{} z}{z^2} \equal{} \minus{}\frac12.$ Hence there is some $ \delta > 0$ and some $ M > 0$ such that if $ |z| < \delta,$ then $ |\log(1 \plus{} z) \minus{} z|\le M|z|^2.$

So now, suppose $ a_n\to0.$ Hence, except for the finitely many terms for which $ |a_n|\ge\delta,$ we have that the difference between $ \sum_{n}\log(1 \plus{} a_n)$ and $ \sum_{n}a_n$ is no greater than $ M\sum_{n}|a_n|^2.$


One not nice solution(!) :
Let $s=a+b$ and $p=\sqrt ab$ then we have
\[ \sqrt a-\sqrt b=1 \rightarrow a+b-2\sqrt ab=1 \rightarrow s=1+2p \]
Now writing the inequality in terms of $s$ and $p$ and replacing $s$ by $1+2p$ the L.H.S. becomes
\[ p^4+8p^3+12p^2+6p+1 \]
and the R.H.S. will be
\[ p^4+8p^3+10p^2+4p+\frac 12 \]
which is clearly smaller than L.H.S.
P.S. : The calculation I didn't type is not so much long.


So, uh, is anyone else here planning on taking/looking into the North American Computational Linguistics Olympiad? Apparently Stanford offers it. Anyone know if there's anything we have to do to register other than enter information into the website? (e.g. contacting Stanford first?)

Wait, is this the same thing that Adam Hesterburg did a session on?

If so, I might look into it. When is it held?

EDIT: HI BEST FRIEND LINEL!

Yeah, it is. The open round is Feb. 4 and the invitational is Mar. 11. Their website is http://www.naclo.cs.cmu.edu.

Yeah I'm going to do this. It sounds fun.
Don't you need a teacher/parent or something? It told me I did when I tried to register a few weeks ago.

Yeah, I guess so. I just registered and put my dad under the teacher info section.

Haha yeah, I tried googling for "naclo practice problems" and found large quantities of chemistry practice problems involving NaClO. But if it had been the Salt Olympiad, I wouldn't have used such inappropriate capitalization...I swear!

If you do find any practice problems, could you please link us here?  Thanks.
EDIT: here are some-- [url]http://www.naclo.cs.cmu.edu/practice.php[/url]

[quote="DiscreetFourierTransform"]I swear to god when I read the title of this thread, I thought it was the "salt" olympiad.[/quote]
When I read it I thought it was the bleach olympiad...(NaClO=sodium hypochlorite) :D


summer program
Mathcamp
MathPath
geometry
email
number theory

Welcome to the Canada/USA Mathcamp subforum.  This is a forum for all things Mathcamp: questions (for MC alumni or staff) from prospective students or parents, discussions among alumni, etc.

I'm David Savitt, the deputy director of Mathcamp; in my day job, I'm a postdoctoral researcher at McGill University, specializing in number theory and arithmetic algebraic geometry.  A second moderator will likely be added later this week.

For now, I should mention that the Mathcamp 2005 qualifying quiz and online application should go live by the end of this week.  Keep your eyes on the Mathcamp website, http://www.mathcamp.org , for further details.

Thanks for moderating the forum. I've got a guy on my math team too young for MathCamp this year (he is applying to MathPath), but I have heard a lot of great things about MathCamp and am encouraging him to think about MathCamp for the future.

[quote="fanaticsm"][quote="rrusczyk"]I attended Mathcamp as a guest speaker in 2004.  When I first arrived, one of the students I knew from speaking at his high school told me, 'Mathcamp makes the rest of the school year look so bad.  Even the good stuff looks bad.'  From what I saw while there, I can understand his sentiment.

Those of you who are curious about Mathcamp should ask your questions in this forum - in addition to the administrators of the camp, our site has dozens and dozens of Mathcamp alums who can answer your questions.[/quote]

Are you coming back again this year?[/quote]

Yes, he is. Check the guest speakers list for 2005 at http://mathcamp.org/academic.php

You should remove the last period from the link or it won't work.

[quote="ztbb"][quote="mathboy_cntHDVN"]how to visit mathcamp ?[/quote]

If you are a student and you are in the area where Mathcamp is taking place, it is usually possible to visit Mathcamp for one or two days.  If that's what you are interested in, you can email mc-info@mathcamp.org with your request.[/quote]
His location is vietnam...

If you mean, "Can I come to Mathcamp if I live in Vietnam?", the answer is yes.  There were quite a few international students last year.


Euler trigonometry limit integration inequalities factorial calculus

:D in how many [b]different[/b] ways, other than the one Euler used  ( using the infinite product formula for $ \sin\; \theta$ ),
can you show Wallis' formula, i.e.

$ \pi\; \equal{} \; 2\times \frac {2^2 \cdot 4^2 \cdot 6^2 \cdot 8^2 \cdot 10^2}{3^2 \cdot 5^2 \cdot 7^2 \cdot 9^2 \cdot 11^2}\cdots$

hello, from
$ \sin(x)\equal{}x\prod_{n\equal{}1}^{\infty}\left(1\minus{}\frac{x^2}{\pi^2n^2}\right)$ we get
by inserting $ x\equal{}\frac{\pi}{2}$:
$ \frac{\pi}{2}\equal{}\lim_{n \to \infty}\frac{(\left(2n)!!\right)^2}{\left((2n\minus{}1)!!\right)^2(2n\minus{}1)}\equal{}\prod_{n\equal{}1}^{\infty}\frac{(2n)^2}{(2n\minus{}1)(2n\plus{}1)}$.
Sonnhard.

do you even care to read my question ?  :(

Hmm. I seem to remember something about Stirling's approximation.
$ \lim_{k \to \infty} \prod_{n\equal{}1}^{k}\frac{(2n)^2}{(2n\minus{}1)(2n\plus{}1)}$
$ \lim_{k \to \infty} \frac{1}{2k\plus{}1} \cdot \prod_{n\equal{}1}^{k}\frac{(2n)^4}{((2n)(2n\minus{}1))^2}$
Then,
$ \lim_{k \to \infty} \frac{1}{2k\plus{}1}\frac{2^{4k}(k!)^4}{((2k)!)^2}$
Using Stirling's Approx. for k!,
$ k! \sim  \sqrt{2k\pi}\cdot\left(\frac{k}{e}\right)^k$

$ \lim_{k \to \infty} \frac{1}{2k\plus{}1}\frac{2^{4k}(\sqrt{2k\pi}\cdot\left(\frac{k}{e}\right)^k)^4}{((\sqrt{2k\pi}\cdot\left(\frac{2k}{e}\right)^{2k}))^2}$

We can pull the $ 2^{4k}$ inside $ \left(\frac{k}{e}\right)^{4k}$ and cancel it with the denominator,

Which condenses to $ \frac{k \pi}{2k\plus{}1}$
Which has the limit $ \frac{\pi}{2}$

The sequence $ \{I_{n}\}_{n \in \mathbb{N}}$, where $ I_{n} = \int_{0}^{\frac {\pi}{2}} \sin^{n} t dt$, is monotonously decreasing and satisfies

\begin{eqnarray} \frac {1}{I_{2n}I_{2n - 1}} \leq \frac {1}{I_{2n}^{2}} \leq \frac {1}{I_{2n}I_{2n + 1}}\end{eqnarray}

for every $ n \in \mathbb{N}$. Since

$ I_{n + 2} = \frac {n + 1}{n + 2}I_{n},$

we also have that

$ I_{2n} = \frac {(2n)!}{2^{2n}(n!)^{2}} \cdot \frac {\pi}{2}$

and

$ I_{2n}I_{2n + 1} = \frac {\pi}{2(2n + 1)}.$

Backwards substitution (in $ 1$) gives us now

$ \frac {4n}{\pi} \leq \frac {2^{2}(2^{2n}(n!)^{2})^{2}}{\pi^{2}((2n)!)^{2}} \leq \frac {2(2n + 1)}{\pi}$

and we're done (well, we'll need to multiply all sides of the latter inequalities by $ \frac {\pi^{2}}{4(2n - 1)}$ and take [i]limits[/i] as $ n \to \infty$ )...  :lol:

[quote="DiscreetFourierTransform"]Hmm. I seem to remember something about Stirling's approximation...
[/quote]

Hi there, [b][color=blue]DFT[/color][/b],

Your approach is very curious... Quite the more so when one recalls that [b]Wallis' Formula[/b] is a basic ingredient of some proofs of the [b]Factorial estimate[/b].  :D

[i]See ya![/i]

We did this in a Calculus BC class, and I recall that we were given Stirling's Formula without proof. I assumed they were unrelated. We were asked to find the limit of that product as $ n$ approaches $ \infty$. My teacher is normally very rigorous, so I suppose he simply missed this circular reasoning.

The simplest argument towards Stirling's formula is to estimate $ \int_1^n \frac1x\,dx$ by the trapezoid rule. If you're careful, that gets you within $ O(1)$ of $ \ln(n!)$. Getting that $ \sqrt{2\pi}$ requires a stronger argument.

[quote="jmerry"]... Getting that $ \sqrt {2\pi}$ requires a stronger argument.[/quote]

Here is where [b]Wallis's Formula[/b] usually comes into play... :P

By Monotone Convergence Theorem, we know that

$ \frac{\sqrt{\pi}}{2} = \int_{0}^{\infty} e^{-x^2} \, dx = \lim_{n\to\infty} \int_{0}^{\sqrt{n}} \left( 1 - \frac{x^2}{n}\right)^n \, dx.$

But

\begin{eqnarray*}
\int_{0}^{\sqrt{n}} \left( 1 - \frac{x^2}{n}\right)^n \, dx
& = & \sqrt{n} \int_{0}^{1} \left( 1 - t^2 \right)^n \, dt\\
& = & \frac{\sqrt{n}}{2} \int_{0}^{1} (1 + t)^n (1-t)^n \, dt\\
& = & \frac{\sqrt{n}}{2} \frac{n! n!}{(2n+1)!} 2^{2n+1}\\
& = & \sqrt{n} \frac{2 \cdot 4 \cdots (2n-2) (2n)}{1 \cdot 3 \cdots (2n-1) (2n+1)}.
\end{eqnarray*}

Combining two facts gives the Wallis formula.

If you use a general infinite product with a Hurwitz zeta parameter and get the regularized product, you can then use the infinite product of the Gamma function to find an interesting generalization of Wallis formula! (sorry, too much of a bother to type the whole thing in, there have been a couple of papers, search ArXiv for 'generalized Wallis formula' or 'zeta regularization' and you'll be able to find them)

[quote="jmerry"]The simplest argument towards Stirling's formula is to estimate $ \int_1^n \frac1x\,dx$ by the trapezoid rule. If you're careful, that gets you within $ O(1)$ of $ \ln(n!)$. Getting that $ \sqrt {2\pi}$ requires a stronger argument.[/quote]

timothy gowers discussed it in [url=http://gowers.wordpress.com/2008/02/01/removing-the-magic-from-stirlings-formula/]this[/url] blog post.


Find all positive integers n so that 89|(10ⁿ + 1).

Any clever way to do this?

[hide="Stuff"]From FLT, $ 10^{88}\equiv 1\bmod 89$. $ 10^0\equiv 1$, so every $ 88$ terms it repeats.

$ 10^n\equiv 88\bmod 89\implies 2^{n\minus{}3}5^n\equiv 11\bmod 89$.

$ 10^{n\minus{}3}\cdot 36\equiv 10\plus{}1$

$ 10(36\cdot 10^{n\minus{}4}\minus{}1)\equiv 1$

$ 36\cdot 10^{n\minus{}4}\minus{}1\equiv 9$

$ 10(4\cdot 10^{n\minus{}5}\minus{}1)\equiv 1$

$ 4\cdot 10^{n\minus{}5}\minus{}1\equiv 9$

$ 4\cdot 10^{n\minus{}6}\equiv 1$

$ 10^{n\minus{}6}\equiv 67$.

I don't know where to go with this (listing out powers of 10 mod 89 would be... bad), but I don't think there exists an $ n$.
[/hide]

o_O

[hide]If there does exist some $ k$ such that $ 10^k \equiv \minus{} 1 \mbox{ mod 89}$, it is clear that $ 2k$ is the order of 10 mod 89. Therefore, we wish to find the order of 10 mod 89. By FLT, $ 10^{88} \equiv 1 \mbox{ mod 89}$, so we only need to check the factors of 88. We verify that$ 10^{44} \equiv 1 \mbox{ mod 89}$, and that the next largest factor, $ 10^{22} \equiv \minus{} 1 \mbox{ mod 89}$. Therefore, the order of 10 is 44, and the smallest $ k$ such that $ 10^{k} \equiv \minus{} 1 \mbox{ mod 89}$ is 22. Therefore, the answer to your question is $ 22 \plus{} 44k$, where $ k$ is a nonnegative integer. (44 because 44 is the order of 10 mod 88.)
[/hide]

Wait, why is 2k order of 10 mod 89? Couldn't it be less?

And, you found order of 10 mod 89, and then again finding the same order:

...it is clear that $ 2k$ is the order of 10 mod 89. Therefore, we wish to find the order of 10 mod 89...

Sorry about the lack of clarity. If we can find the order of 10 mod 89 (it must be 2k), then $ 10^{k} \equiv \minus{} 1 \mbox{ mod 89}$; we are trying to find this k. So when we get the order, we just half it. (Of course, this is assuming that there DOES exist some $ k$ with $ 10^k \equiv \minus{} 1 \mbox{ mod 89}$, but the problem statement led me to assume there was.

And that's a fact that I learned from 104 Number Theory problems (great book =D).

[hide="Proof"]Suppose $ (a, n) \equal{} 1$ and $ a^s \equiv \minus{} 1 \mbox{ mod n}$, with $ s$ being as small as possible. We will show that $ 2s$ is the order of $ n$. Clearly, $ a^{2s} \equiv 1 \mbox{ mod n}$, so it suffices to show that there does not exist any $ t < 2s$ with $ a^t \equiv 1 \mbox{ mod n}$.

Suppose for the sake of contradiction that there does. Since $ t < 2s$, $ t \minus{} s < s$. However, $ a^{t \minus{} s} \equiv \minus{} 1 \mbox{mod n}$. But since $ t \minus{} s < s$, our assumption that $ s$ is minimal is contradicted. Therefore, $ t \equal{} 2s$. [/hide]

Divides, as in $ a | b$ means that a divides b.


Find Max $ k$ such that: for $ a,b,c\ge0$ and $ a^k\plus{}b^k\plus{}c^k\equal{}3$,we have
$ \frac {a^2}b\plus{}\frac {b^2}c\plus{}\frac {c^2}a \ge 3$

I think, it's very hard! Even for $ k\in\mathbb N.$ :wink:


induction
number theory unsolved
number theory

prove that every even integer $ 2n$ can be written in the form $ 2n\equal{}(x\plus{}y)^2\plus{}3x\plus{}y$ with $ x,y$ nonnegative integers

[hide]Let $ x_n$ and $ y_n$ represent the values of $ x$ and $ y$ for the integer $ 2n$.
We proceed by induction.
We show that the statement is true for the base case, $ n \equal{} 0$. If $ n \equal{} 0$, we can write $ 2 \cdot 0$ in the form $ 2n \equal{} (x \plus{} y)^2 \plus{} 3x \plus{} y$, if $ x \equal{} y \equal{} 0$:
$ 0 \equal{} 2\cdot 0 \equal{} (0 \plus{} 0)^2 \plus{} 3\cdot 0 \plus{} 0 \equal{} 0$.
We assume that the statement is true for $ n \equal{} k$, so $ 2k \equal{} (x_k \plus{} y_k)^2 \plus{} 3x_k \plus{} y_k$.
We show that the statement must be true for $ n \equal{} k \plus{} 1$.
If $ y_k \equal{} 0$, then
$ 2k \equal{} (x_k \plus{} 0)^2 \plus{} 3x_k \plus{} 0 \equal{} x_k^2 \plus{} 3x_k$
$ 2(k \plus{} 1) \equal{} x_k^2 \plus{} 3x_k \plus{} 2 \equal{} (x_k \plus{} 1)(x_k \plus{} 2) \equal{} (x_k \plus{} 1)^2 \plus{} (x_k \plus{} 1)$
$ 2(k \plus{} 1)$ can be written in the form $ 2n \equal{} (x \plus{} y)^2 \plus{} 3x \plus{} y$ if we let $ x_{k \plus{} 1} \equal{} 0$ and $ y_{k \plus{} 1} \equal{} x_k \plus{} 1$.
The statement is true for $ n \equal{} k \plus{} 1$ in this case.
If $ y_k \neq 0$ then
$ 2(k \plus{} 1) \equal{} 2k \plus{} 2 \equal{} (x_k \plus{} y_k)^2 \plus{} 3x_k \plus{} y_k \plus{} 2 \equal{} (x_k \plus{} y_k)^2 \plus{} 3x_k \plus{} 3 \plus{} y_k \minus{} 1 \equal{} (x_k \plus{} y_k)^2 \plus{} 3(x_k \plus{} 1) \plus{} (y_k \minus{} 1)$
$ 2(k \plus{} 1)$ can be written in the form $ 2n \equal{} (x \plus{} y)^2 \plus{} 3x \plus{} y$ if we let $ x_{k \plus{} 1} \equal{} x_k \plus{} 1$ and $ y_{k \plus{} 1} \equal{} y_k \minus{} 1$.
The statement is true for $ n \equal{} k \plus{} 1$ in this case.
Induction is complete, and we are done.[/hide]

Here is another method. Let x+y=k .S0 0≤x≤k
The problem becomes
     every even integer 2n can be written in the form k(k+1)+2x
which is equivalent to
     every even integer 2n satisfies " k(k+1)≤2n=k(k+1)+2x≤k(k+3) " for an integer k
because x can be chosen arbitrarily as long as 0≤x≤k is satisfied.

Obviously,for any even integer 2n we can find an integer k such that k(k+1)≤2n<(k+1)(k+2)=k(k+3)+2
     As both 2n and k(k+3) are even, k(k+1)≤2n≤k(k+3). So we are done.


articles
calculus
integration
Pythagorean Theorem
geometry
linear algebra

I'm supposed to do a sort of paper on a topic of my choosing for my Astrophysics class. I thought it would be sort of neat (and I'm more interested in it) to do something more mathematically related (e.g., "The mathematics of the gravitational field equations") rather than something more Astrophysics-y (e.g., "A gravitational field is..."), but I'm not really sure what there is to do. Any suggestions? I'm not looking for something simple like...the applications of the Pythagorean theorem to Astronomy...or the use of trigonometry...Anything using even beyond high-school math (differential equations, linear algebra, analysis, etc.) would be fine.
For example, I saw a few references to using quaternions and knot theory in Astrophysics, but they were basically just the titles of magazine articles, and I couldn't find much else.

Using integral calculus to adapt the Lorentz transformations of special relativity to non-uniformly moving objects?  Something in which I was always interested, but never had the time to do.


inequalities
geometry
trapezoid
inequalities theorems

With all real number $a,b>0$ we well know
Root mean square $RM=\sqrt{\frac{1}{2}(a^{2}+b^{2})}$
Arithmetic mean $AM=\frac{1}{2}(a+b)$
Geometric mean $GM=\sqrt{ab}$
Harmonic mean $HM=\frac{2ab}{a+b}$
And well know the algebraic inequalities
$RM\ge AM\ge GM\ge HM\ge$ after nice algebraic interpretations of them they still have nice geometric interpretation as following
In my figure $AN=a,BN=b$ and $RM=NC\ge AM=OC\ge GM=NP\ge HM=DP$ it is geometric interpretation of them.  :)

Nice. Have you been looking for an equivalent in three variables & dimensions?

Hello Gemath!!

It's a nice idea to prove RM-GM-AM-HM by geometric interpretation.

There's also a nice interpretation in an isosceles trapezoid:
Let $ ABCD$ be an isosceles trapezoid with sides $ AD$ and $ BC$ horizontal.
The diagonals $ AC$ and $ BD$ cross at a point $ P$; the horizontal segment through $ P$ is the harmonic mean of $ AD$ and $ BC$.
The horizontal segment that cuts $ ABCD$ into two similar trapezoids is the geometric mean.
The segment between the midpoints of $ AB$ and $ CD$ is the arithmetic mean.
The horizontal segment that bisects the area is the RMS.


Hello everybody...I am interested in proposing and helping for the proposal of problems for the USAMO math contest...

He is talking about the Mock USAMO in case anyone was confused.  I know that FermatPrime wants to have it.  So no there is at least one person to write problems.  This thread can be used to talk about the Mock USAMO.  If people are interested in writing problems please say so here.

*is happy* We'd also need more judges...

Although I almost always have interesting short-answer problems floating around, I don't have too many good Olympiad problems, but I could probably whip up a nice inequality or two for a Mock USAMO.

I am also willing to be a grader if the Mock USAMO is held far enough (at least a week) before my school year starts on Tuesday, September 7.  After that date, my forum-trolling time will be cut drastically.

[quote="Mildorf"]Although I almost always have interesting short-answer problems floating around, I don't have too many good Olympiad problems, but I could probably whip up a nice inequality or two for a Mock USAMO.
[/quote]

Of course, there's thousands of Olympiad problems on Kalva; even if you don't want to use them fresh off the site, you can hopefully get some inspiration from them.

I'd be interested, if only to see if I could get a 1/42  ;)

Would we need to qualify for this through the Mock AIME?

[quote="joml88"]Who all would want to take a Mock USAMO?  It seems that FermatPrime is the only one showing interest right now.  I would like to look at the questions but the chances of me actually solving one are near impossible(at the current time).[/quote]

a very definite yes. call it the ELMO and think of some new things it can stand for.

Im interested... either in taking it, judging it, or proposing problems. I do have a nice collection of problems that haven't been publicized too much.

I would want to take it, though I would fail horribly...

now where am I gonna get 6 hours to do problems...

This is going to be interesting. This will be my first official or nonofficial taking of any sort of USAMO. I hope I can get above 0/42.

So....now that the list of qualifiers has been established, will this contest take place? Who will be hosting it? It sounds like it will be a lot of fun!

What level of questions are you looking for?

The mid-level questions I send to our national olympiad, but the easy level ones and hard ones are still up for grabs :)

I unfortunately am not much of a problem writer.. but I would be willing to help grade as long as this takes place before the school year (Aug 26)

If anyone wants to see past USAMO questions look at this link [url]http://www.kalva.demon.co.uk/usa.html[/url]

I made this post mainly because Peter VDD asked what level these were at.  I hope this helps.


trigonometry
complex numbers
algebra unsolved
algebra

Find the value of:
sec^2 pi/9 + sec^2 5pi/9 + sec^2 7pi/9

hello, did you meant
$ sec^2\left(\frac{\pi}{9}\right)\plus{}sec^2\left(\frac{5\pi}{9}\right)\plus{}sec^2\left(\frac{7\pi}{9}\right)$?
Sonnhard.

[quote="Jane_Bennet"]Let $ \frac {x \plus{} \mathrm{i}}{\sqrt {x^2 \plus{} 1}} \equal{} \cos{\theta} \plus{} \mathrm{i}\sin{\theta}$, where $ x\in\mathbb{R}.\ \therefore\ \tan{\theta} \equal{} \frac {1}{x}$.

By De Moivre’s formula, $ \frac {(x \plus{} \mathrm{i})^9}{(\sqrt {x^2 \plus{} 1})^9} \equal{} \cos{9\theta} \plus{} \mathrm{i}\sin{9\theta}$

$ \cos{9\theta} \equal{} 0\ \Rightarrow\ 9\theta \equal{} \frac {\pi}{2},\frac {3\pi}{2},\frac {5\pi}{2},\frac {7\pi}{2}$

$ \therefore\ \theta \equal{} \tan^{ \minus{} 1}{\left(\frac {1}{x}\right)} \equal{} \frac {\pi}{18},\frac {3\pi}{18},\frac {5\pi}{18},\frac {7\pi}{18}$

$ \Rightarrow\ x \equal{} \cot{\left(\frac {\pi}{18}\right)},\cot{\left(\frac {3\pi}{18}\right)},\cot{\left(\frac {5\pi}{18}\right)},\cot{\left(\frac {7\pi}{18}\right)}$
$ \color{white}.\quad\quad\color{black} \equal{} \tan{\left(\frac {\pi}{2} \minus{} \frac {\pi}{18}\right)},\tan{\left(\frac {\pi}{2} \minus{} \frac {3\pi}{18}\right)},\tan{\left(\frac {\pi}{2} \minus{} \frac {5\pi}{18}\right)},\tan{\left(\frac {\pi}{2} \minus{} \frac {7\pi}{18}\right)}$
$ \color{white}.\quad\quad\color{black} \equal{} \tan{\left(\frac {4\pi}{9}\right)},\tan{\left(\frac {\pi}{3}\right)},\tan{\left(\frac {2\pi}{9}\right)},\tan{\left(\frac {\pi}{9}\right)}$
$ \color{white}.\quad\quad\color{black} \equal{} \minus{} \tan{\left(\frac {5\pi}{9}\right)},\sqrt {3}, \minus{} \tan{\left(\frac {7\pi}{9}\right)},\tan{\left(\frac {\pi}{9}\right)}$

Hence $ \minus{} \tan{\left(\frac {5\pi}{9}\right)},\sqrt {3}, \minus{} \tan{\left(\frac {7\pi}{9}\right)},\tan{\left(\frac {\pi}{9}\right)}$ are roots of the equation

$ x^9 \minus{} 36x^7 \plus{} 126x^5 \minus{} 84x^3 \plus{} 9x \equal{} 0$ (that’s the real part of $ (x \plus{} \mathrm{i})^9$ above)

But since they are not 0, they are roots of the equation

$ x^8 \minus{} 36x^6 \plus{} 126x^4 \minus{} 84x^2 \plus{} 9 \equal{} 0$

In other words, their squares are roots of the equation

$ x^4 \minus{} 36x^3 \plus{} 126x^2 \minus{} 84x \plus{} 9 \equal{} 0$

Since $ \tan^2{\left(\frac {5\pi}{9}\right)},3,\tan^2{\left(\frac {7\pi}{9}\right)},\tan^2{\left(\frac {\pi}{9}\right)}$ are all distinct, I conclude that they form the complete solution set to the quartic equation above. Hence, their sum is the negative of the coefficient of $ x^3$ divided by the leading coefficient of the quartic equation above – in other words! … (drum roll) …

$ \color{white}.\quad\color{black}\sec^2{\left(\frac {\pi}{9}\right)} \plus{} \sec^2{\left(\frac {5\pi}{9}\right)} \plus{} \sec^2{\left(\frac {7\pi}{9}\right)}$

$ \equal{} \ \tan^2{\left(\frac {5\pi}{9}\right)} \plus{} \tan^2{\left(\frac {7\pi}{9}\right)} \plus{} \tan^2{\left(\frac {\pi}{9}\right)} \plus{} 3$

$ \equal{} \ 36$ :clap2:

If you don’t believe me, check with a calculator. :jump:[/quote]
Your solution is nice but too long! Just using Cardano's formula to calculate sin(pi/9) and you will easily get the answer! By the way, we can compute everything with sin15 or even sin5! :)

Cardano's formula would just express $ \sin \left( \frac{\pi}{9} \right)$ in terms of itself, as I recall from my experiences with it.

Here is a quicker method.

Using $ \cos{3\theta} \equal{} 4\cos^3{\theta} \minus{} 3\cos{\theta} \equal{} \cos{\theta}\left(4\cos^2{\theta} \minus{} 3\right)$ we have

$ 4\cos^2{3\theta}\ \equal{} \ 4\cos^2{\theta}\left(4\cos^2{\theta} \minus{} 3\right)^2\ \equal{} \ x(x \minus{} 3)^2$

where $ x \equal{} 4\cos^2{\theta}.$

Hence $ x_1 \equal{} 4\cos^2\frac {\pi}9,\ x_2 \equal{} 4\cos^2\frac {5\pi}9,\ x_3 \equal{} 4\cos^2\frac {7\pi}9$ are the roots of the equation $ x(x \minus{} 3)^2 \equal{} 1;$ i.e. of the equation

$ x^3 \minus{} 6x^2 \plus{} 9x \minus{} 1\ \equal{} \ 0$

Therefore

$ \color{white}.\quad.$$ \sec^2{\frac {\pi}9} \plus{} \sec^2{\frac {5\pi}9} \plus{} \sec^2{\frac {7\pi}9}$

$ \equal{}\ \dfrac{\cos^2{\dfrac{\pi}9}\cos^2{\dfrac{5\pi}9}\plus{}\cos^2{\dfrac{5\pi}9}\cos^2{\dfrac{7\pi}9}\plus{}\cos^2{\dfrac{7\pi}9}\cos^2{\dfrac{\pi}9}}{\cos^2{\dfrac{\pi}9}\cos^2{\dfrac{5\pi}9}\cos^2{\dfrac{7\pi}9}}$

$ \equal{} \ \dfrac{\dfrac{x_1x_2 \plus{} x_2x_3 \plus{} x_3x_1}{16}}{\dfrac{x_1x_2x_3}{64}}$

$ \equal{} \ 36$

:jump:


integration
trigonometry
geometry
calculus
calculus computations

I feel so stupid for not being able to get this: I would like to find the volume of a torus with small radius $ R$, and big radius $ a$.  I decided to take the revolution around the $ x$-axis approach, so we would want to revolve a circle around the $ x$-axis.  Such a circle is given by $ x^{2}\plus{}(y\minus{}a)^{2}\equal{}R^{2}$ so we can consider $ y\equal{}\sqrt{R^{2}\minus{}x^{2}}\plus{}a$ and double it.  So the volume formula would be given by
$ V\equal{}2\pi\int_{\minus{}R}^{R}(f(x))^{2}\,dx\equal{}2\pi\int_{\minus{}R}^{R}R^{2}\minus{}x^{2}\plus{}2a\sqrt{R^{2}\minus{}x^{2}}\plus{}a^{2}\,dx$
$ \equal{}2\pi\left((R^{2}\plus{}a^{2})x\minus{}\frac{x^{3}}{3}\plus{}aR^{2}\left[\sin^{\minus{}1}\left(\frac{x}{R}\right)\plus{}\frac{x\sqrt{R^{2}\minus{}x^{2}}}{R^{2}}\right]\right)\Bigg|_{\minus{}R}^{R}$
$ \equal{}2\pi\left(2R(R^{2}\plus{}a^{2})\minus{}\frac{2R^{3}}{3}\minus{}\frac{\pi aR^{2}}{2}\right)$
I have a feeling this is wrong.

...or, you could cut the torus and it becomes a cylinder of length $ 2\pi(R\plus{}a)/2\equal{}\pi(R\plus{}a)$ and cross sectional area $ \pi\left(\frac{a\minus{}R}{2}\right)^2$, so $ V\equal{}\frac{\pi^2}{4}(R\plus{}a)(a\minus{}R)^2$.

Of course, this is just a way to figure it out and is not a proof that when you lengthen the inner part and shorten the outer part the volume stays constant.


geometry
trigonometry
number theory
relatively prime
angle bisector
trig identities
Law of Cosines

In $ \triangle ABC, D$ lies on segment $ AC$ so that $ m \angle ABD \equal{} m \angle DBC \equal{} \theta$. If $ AB \equal{} 4, BC \equal{} 16,$ and $ BD \equal{} \frac{2}{\cos \theta},$ then $ BD \equal{} \frac{a}{\sqrt{b}}$ for relatively prime integers $ a$ and $ b$. Compute the ordered pair $ (a,b)$.

[quote="Silverfalcon"]In $ \triangle ABC, D$ lies on segment $ AC$ so that $ m \angle ABD \equal{} m \angle DBC \equal{} \theta$. If $ AB \equal{} 4, BC \equal{} 16,$ and $ BD \equal{} \frac {2}{\cos \theta},$ then $ BD \equal{} \frac {a}{\sqrt {b}}$ for relatively prime integers $ a$ and $ b$. Compute the ordered pair $ (a,b)$.[/quote]

Mooooore geometry...=P
[hide]
Angle bisector theorem gives $ CD\equal{}4AD$

Law of cosines on $ \triangle BCD$ and $ \triangle BDA$ and the fact that $ BD\cos \theta \equal{} 2$ gives $ 192\plus{}BD^2\equal{}16AD^2$ and $ AD^2\equal{}BD^2$. Solve and get $ AD\equal{}BD\equal{}\sqrt{\frac{192}{15}}\equal{}\frac{8}{\sqrt{5}}$ so $ (a,b)\equal{}\boxed{(8,5)}$?[/hide]

Well done. I don't think there is any other method than what you've done (Angle Bisector and Law of Cosine). That's how I solved it too.


geometry
trigonometry
trig identities
Law of Sines

The problem is:

Draw two triangles that are not congruent, but have exactly five sets of parts (sides and angles) that are congruent. [b]Make a drawing that is to scale.[/b] Label your measurements.

The only thing I've figured out is that there has to be exactly 2pairs of congruent sides and 3pairs of congruent angles. Also the pairs of congruent sides can't be corresponding or else you would end up with SAS which would make them congruent.

Remember you have to draw it to scale so not any random angle measurements and lengths would work.

Reply ASAP!!

[quote="Zenith Xan"]I don't think this is possible, is it?

If three sides are congruent, then all angles are as well.

If three angles are congruent and two sides, according to stuff like law of sines, the last would have to be too, wouldn't it?[/quote]

From what I see, the only way to do it is to make them isosceles, which is kind of exploiting the wording of the problem.

So you make a 45-45-90 right triangle, the two congruent sides are 1, which makes the hypotenuse squareroot of 2. The second triangle is also 45-45-90, with the two congruent sides squareroot of 2, and the hypotenuse 2.

I don't really like this way of doing it, because the two pairs of congruent sides are just two sides of an isosceles triangle congruent to the SAME side on the other triangle.

Blech.

If all of the angles are congruent, then if two of the sides are, then by SAS the triangles are too.

[quote="fireforce"]Here is what i got: :ninja:

Is it right?[/quote]

Yeah its basically right but you have to find exact measurements and angles :(

And to Biffandoc, its not SAS if the sides and angles don't correspond.

There are many solutions, for example first triangle 8,12,18 and the second 12,18,27.

Actually if first triangle has sides a,aq,aq^2 second has sides aq, aq^2 and aq^3 where 0<q<(1+sqrt(5))/2


For any positive odd integer n, how many positive even integers are less than n?

A. n    B. n – 1   C. n + 1  D. n/2  E. n-1/2

It's E because you could subtract 1, then you get an even number, and then exactly one half of the remaining numbers are even in that set. Therefore the answer is E, which I presume as $ \frac{n\minus{}1}{2}$


Prove there do exist a infinite sequence $ {a_n}$ satisfies
(1)$ a_n$ is positive integer $ n \equal{} 1,2,3...$
(2)$ a_n \le n^7$   $ n \equal{} 1,2,3...$
(3)Define $ T$ to be the set of all $ a_n$ and  $ S(A)$ is the sum of all elements in $ A$
     Now for all positive integer $ k$,There [b]dosen't[/b] exist a set $ A$   $ A \subset S$ and $ A \neq \{{a_k}\}$   such that $ a_k \equal{} S(A)$

Isn't it the same as [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=15463]this?[/url]


A square computer screen measures 14 inches along its diagonal. How
many inches long is the perimeter of the screen? Round your answer
to the nearest whole number.

The perimeter is 4(14/sqrt2), which is about 56/1.414, approximately 36.6 or 37

[b] Correction: [/b] We know that $\sqrt{2}\approx 1.41$, therefore, the length of an individual side is $\frac{14}{\sqrt2}\approx 10$. Therefore, the perimeter is approximately $4\cdot 10=\boxed{40}$ inches.


Μπορειτε να με βοηθησετε στο εξης προβλημα:
Πως μπορουμε να δειξουμε οτι η διαμετρος ενος συνολου στον $ R^n$ ισουται με την
διαμετρο της κυρτης του θηκης?
Υπαρχει αντιστοιχο για την κλειστοτητα αλλα ειναι πιο ευκολο.

Δεν το σκέφτηκα σοβαρά, αλλά είμαι confident ότι θα βγει γρήγορα γνωρίζοντας ότι η κυρτή θήκη κάποιου συνόλου $ \mathcal{T} \subsetneq \mathbb{R}^n$ γράφεται και ως $ \left\{\sum_{i \equal{} 1}^k \lambda_i x_i \; : \; k \geq 1, \{x_i\}_{i \equal{} 1}^k \subseteq \mathcal{T}, \{\lambda_i\}_{i\equal{}1}^k \subseteq \mathbb{R}_\plus{}, \sum_{i \equal{} 1}^k \lambda_i \equal{} 1 \right\}$ (γιατί?). Ίσως να υπάρχει και κάποιο πιο direct/elegant proof :)

(By the way, αποδεικνύεται ότι μπορείς να περιοριστείς μόνο σε $ k \equal{} n \plus{} 1$).

Να δώσω κι εγώ μία γρήγορη ιδέα (χωρίς να έχω σκεφτεί πολύ το πρόβλημα)...

Για πιο απλό συμβολισμό, ας υποθέσουμε ότι το $ 0$ ανήκει στο σύνολό μας (έστω $ A$) και έστω $ x\in\Delta(A)$ όπου $ \Delta(A)$ η κυρτή θήκη του $ A$.

Τότε το $ x$ θα γράφεται στη μορφή $ x \equal{} \sum\nolimits_i p_i x_i$ για κάποια $ x_i \in A$ ($ \sum p_i \equal{} 1$) και από Jensen (δηλ. τριγωνική) παίρνουμε: $ |x| \leq \sum\nolimits_i p_i |x_i| \leq \max_i\{|x_i|\}$, οπότε και $ \sup_{x\in \Delta(A)} |x| \leq \sup_{x\in A} |x|$. Το $ 0$ είναι αυθαίρετο (το διάλεξα μόνο και μόνο για να μην κουβαλάω σύμβολα) οπότε με άλλο ένα $ \sup$ νομίζω τελειώσαμε...

cmad, αυτό είχες στο μυαλό σου?

Cheerio,

Durandal 1707


What is the greatest real number that is at least as large as its
square?

Just thinking about the problem gives $ 1$.

The greatest real number that is larger than its square is 1 because a square is when you multiply the number by itself. 2 is too big, because its square is 4. So that leaves one answer, 1. (1*1=1).


calculus
integration
trigonometry
geometry
3D geometry
sphere
induction

I've been working on a problem and I've gotten it down to $\int_{0}^{a}{x^{2}\sqrt{a^{2}-x^{2}}}\ dx$.  That integral is in all the integration tables but I haven't been able to prove it.  Hmmm.

Try a trig substitution: $x=a\sin\theta$.

Yeah I just got it  :oops:  Thankyou

The whole problem if anyone is interested, cuz it was kinda cool.. kinda.

Find the "volume" of the "four-dimensional sphere" $x^{2}+y^{2}+z^{2}+w^{2}=a^{2}$ by evaluating $16\int_{0}^{a}\int_{0}^{\sqrt{a^{2}-x^{2}}}\int_{0}^{\sqrt{a^{2}-x^{2}-y^{2}}}\int_{0}^{\sqrt{a^{2}-x^{2}-y^{2}-z^{2}}}\ dwdzdydx$

Some alternate [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=26191]methods[/url], which extend to balls in $n$ dimensions. I like the induction by steps of two dimensions.

I see, very nice.  What I did was
[hide] rewrote it as $16\int_{0}^{\frac{\pi}{2}}\int_{0}^{\frac{\pi}{2}}\int_{0}^{a}\int_{0}^{\sqrt{a^{2}-\rho^{2}}}\rho^{2}\sin{\phi}\ dw d\rho d\phi d\theta$ which is very simple up to that integral I posted above.[/hide]


function
real analysis
real analysis solved

Let $ f :R \to R $ satisfying 2 conditions :
 $ f(0) = f(1) = 0 $
 $ \min_{x\in [0,1]}f(x) = -1 $
 Find $ \min(\max_{x\in [0,1]}f''(x)) $

[quote="romano"]
 Find $ \min(\max_{x\in [0,1]}f''(x)) $[/quote]

You mean $\max_{[0,1]}f\"$
another post the correct statement $f$ is $C^2$ ?

It's $8$.

$f''$ is $8$ when $f(x)=4x(x-1)$, for example, and now we have to show that there's no function satisfying those conditions which has $f''(x)<8,\ \forall x\in[0,1]$.

Let $x_0$ be a point in $[0,1]$ for which $f(x_0)=-1,f'(x_0)=0$. We write $f(x)=f(x_0)+(x-x_0)f'(x_0)+\frac{(x-x_0)^2}2f''(\lambda_x)=-1+\frac{(x-x_0)^2}2f''(\lambda_x)\ (*)$, with $\lambda_x$ between $x$ and $x_0$. Now use the formula $(*)$ with $x=0$ and $x=1$, and the assumption $\max f''(x)<8$. We get $x_0>\frac 12$ and $1-x_0>\frac 12$, which is clearly a contradiction.

May be my proof is more standard.
Suppose $ f(a) = -1 $. Then let's find $ b $ such that if we define
$ g(x) := f(x) + bx(1-x) $, then $ g(a) = 0 $. Easy computation shows that
$ b = \frac{1}{a(1-a)} $. So we have $ g(0) = g(a) = g(1) = 0 $. Applying twice
Rolle theorem we get that there exists $ x_{0} \in (0,1) $, such that
$ g''(x_{0}) = 0 $, but $ g''(x_{0}) = f''(x_{0}) - 2b $ , so
$ f''(x_{0}) = 2b = \frac{2}{a(1-a)} \geqslant 8 $.


geometry
perimeter
inequalities
arithmetic sequence
triangle inequality

How many distinct, non-equilateral triangles with a perimeter of 60 units have integer side lengths $ a$, $ b$, and $ c$ such that $ a$, $ b$, $ c$ is an arithmetic sequence?

Then one of the sides is 20.

The other two sides are 20-x and 20+x.

Triangle inequality: $ 20\plus{}20\minus{}x>20\plus{}x$, so $ 10>x$.

x=0 becomes equilateral.

Therefore there are 9 ways.

That is what I came up with, however the answer is 10


There are 5 monkeys and 5 ladders and at the top of each ladder there is a banana. A number of ropes connect the ladders, each rope connects two ladders. No two ropes are attached to the same rung of the same ladder. Each monkey starts at the bottom of a different ladder. The monkeys climb up the ladders but each time they encounter a rope they climb along it to the other ladder at the end of the rope and then continue climbing upwards. Show that, no matter how many ropes there are, each monkey gets a banana.

every rope switches 2 monkeys

for example at a certain moment, say monkey 1 is on the first ladder, monkey 2 is on the second ladder ...
if a rope connects 2 ladders, say the third and fourth, monkeys 3 and 4 will switch ladders

[hide="More Formal Solution"]
Let's establish this lemma.

[i]Lemma: For $ k$ monkeys and $ k$ ladders, no two monkeys are in same ladder ($ k$ is a positive integer)[/i]

Proof:

With each rope, let $ L_L$ and $ L_R$ be two ladders at its left and right, respectively. Then, whatever monkey on $ L_L$ will move to $ L_R$ and the monkey on $ L_R$ will move to $ L_L$.
Hence, the rope will simply switch the monkeys, and since there are same number of ladders as monkeys, no two monkeys will end up in same ladder, regardless of number of ropes, because all of them start in different ladder and the switches will simply change the permutation of them, without putting two on one.
Therefore, the problem follows the result of this lemma. QED [/hide]

I'm pretty bad at explaining these but I hope that made sense.  :P

I believe both above solutions are wrong, so here is my attempt.
[b]Claim:[/b]A monkey cannot get stuck

[b]Proof:[/b]
Let all the parts of ladders  that monkey [b]ascends[/b] be good.Consider a graph whose vertices are the good parts. Now after ascending a good part of ladder he can go to only one other good part through a rope. Also a good part can be accesssed from only one good part(except for the starting part). This is because every good part corresponds to one 'out' rope(through which he goes to other good part) and one 'in' rope(through which hecomes on it). This observation lets us conclude that there will be no cycles, because if there was, then one of the good parts can be accessed through two good parts(keep drawing the monkey's journey till a good part is repeated; that good part is accessed through two good parts). Note that this uses the fact that the first good part in journey is not accessible from other good part.

Now let the bananas traverse down the ladder with the same rules, except that it must descend down the ladder. It will not get stuck, hence reach a monkey. It is easily seen that this is the exact same path taken by monkey to reach that banana. Hence two monkeys cannot reach the same banana , else the path of banana will reach to two monkeys, which is absurd. Hence each monkey gets one banana.


Hello,

Alongside the new problem rating system that I am trying to implement into the Intermediate forum (which, if successful, might spread to other forums) I am going to start the [b]I-1337 AoPSer Contest.[/b]

Basically, I will keep score.  Every week a winner will be announced for that week and then there will be a fresh contest for next weeks winner.  Monthly and (eventually) yearly high scorers will also be recognized.

Points will be given on the basis of how active a user is in the Intermediate forum.  You may receive points for the following actions:

$\bullet$ posting a correct and [b]full[/b] solution to a challenge problem.  The number of points will be decided by the difficulty level.  You get however many points the difficulty level is.

-Only the first correct solution receives points.  [b]And to receive these points the poser of the original problem must confirm if the solution is adequate or not.[/b]  However, if an alternate solution is given that is different from the original solution posted by someone else points may also be given by discretion of the problem poser.

$\bullet$ For posing a problem one will receive .5 points.  The points are low because I don't really want an excess of problems being posted so much so that they can be considered spam.  (the stickied problem contests/marathon count for this)

$\bullet$ Posting tutorials (such as those in the inequalities marathon) will receive points.  The number of points is at my discretion and depends on the quality of the tutorial.

$\otimes$ Points will be taken away from individuals who do anything that is not condoned by proper board etiquette, e.g. spam, chronic double posting, disparaging comments, etc.

Here is the scoreboard:

\[\begin{tabular}{|c|c|}\hline
Competitor & Cumulative Score \\ \hline
xxreddevilzxx & .5\\ \hline
K81o7 & 4\\ \hline
\end{tabular}\]

I think it's a good system, but with too many ways to score points, how are you going to be able to manage all the points for hundreds of users who would be eligible?

Ok, I've changed it so you get points for postin a solution, alternate solution, or tutorial.  Remember, to get points in a topic where you post a solution the original poser [b]must[/b] acknowledge that you earn the points.

I think this should be manageable (at least I hope :P ).

Also, I gave xxreddevilzxx points for posting a problem.  Points will only be given for topics that are made after his post.  That is, there should not be any digging through past topics and such to revive dead threads.

[quote="joml88"]
-Only the first correct solution receives points.  [b]And to receive these points the poser of the original problem must confirm if the solution is adequate or not.[/b]
[/quote]
great... since i live in a total different timezone, most problems are already solved when i just read them  :?

Yeah, that might be a problem...I've encountered that problem once or twice...
We don't want anyone deprived of sleep or having changed sleeping patterns because of these problems.  Thus...
Points should be given to mroe than just the first person?  Or the time of day at which the problem comes up cycles?  I don't know...

[quote="WindSlicer"]This is the only way that the system works.

Some people always will be put at a disadvantage, but, hey, it gives you a reason to recruit others from your area! :)[/quote]
yeah, lol  :D
it's not like i know dozens of kids of my age who are willing to do math in their free time  :roll:

but the rule doesn't really bother me  :|  i'll try to come up with different solutions, or try the ones which haven't been answered at night yet :)

We'll see how the competition goes.  If it's too much for me to handle or if it doesn't seem to be going well then I might stop doing it.  But also, I think there are going to be certain days/weeks where the competition isn't in effect.  For example, next week I won't have internet access at all so the contest won't be going on that week.

What if we formed teams, with people from many timezones on each team?
And of course, solved more difficult problems...
Each problem could be posted at a different time of day.  That way, things might be a little more fair overall.

Hmm, well the contest seemed like a good idea at first.  But the more I think of it, the more I question how successful it would be.  Additionally, there are already a ton of contests/marathons going on around here.  I think I'm just gonna shut the contest down now.


integration
calculus
geometry
function
complex analysis
complex analysis unsolved

How to prove\[ \frac{1}
{{2\pi }}\int_0^1 {\int_{ \minus{} \pi }^\pi  {\frac{r}
{{r{e^{i\theta }} \plus{} z}}d\theta dr} }  \equal{} \bar z\]for all $ z \in D(0,1)$?

Well, by brute force, if we use the residue theorem to evaluate the inner integral we get that the expression is equal to $ \frac{1}{z} \int_0^{|z|} r \, dr \equal{} \frac{\bar{z}}{2}$ (the area of the disc is $ \pi$ not $ 2 \pi$). I did evaluate it with Stokes also which feels a little "cleaner", but is probably more or less the same thing.

I don't quite know what is going on here though, why do we get an anti-analytic function when integrating an integrand with analytic dependence (plus a simple pole)? The thing is that we cannot differentiate under the integral sign because the differentiated integrand is not integrable on the disc. What is this from? Can someone explain.

I should provide an extra statement. It says that if $ |z| \geq 1$, then
\[ \frac{1}{{2\pi }}\int_{0}^{1}{\int_{\minus{}\pi }^\pi{\frac{r}{{r{e^{i\theta }}\plus{}z}}d\theta dr}}\equal{}\frac{1}{z}.\]

via the substitution $ (e^{i \theta}=\zeta$ and using simplest residue theorem for the unit circle  we obtain that )
$ \frac{1}{2\pi}\int_{0}^{1}\int_{-\pi}^{\pi} \frac{r}{re^{i\theta}+z} d\theta dz = \frac{1}{2\pi}\int_{0}^{1}\frac{2\pi}{z}dr = \frac{1}{z}$

[quote="Kalle"]
I don't quite know what is going on here though, why do we get an anti-analytic function when integrating an integrand with analytic dependence (plus a simple pole)? The thing is that we cannot differentiate under the integral sign because the differentiated integrand is not integrable on the disc. What is this from? Can someone explain.[/quote]
This double integral is just the fancy way to write $ \int_D\frac{dA(w)}{w\minus{}z}$. Now, we [b]can[/b] differentiate under the integral sign if we work with generalized functions and recall that $ \bar\partial\frac 1z\equal{}2\pi\delta_0$. Thus, $ \bar\partial$ of the integral is $ 1$ in the unit disk and $ 0$ outside. Also, the function is continuous and decays like $ 1/z$ at infinity. Note that $ \bar z$ in $ D$ glues with $ 1/z$ outside $ D$ to give just a function with this property. The difference between our integral and this function is analytic and tends to $ 0$ at $ \infty$. Hence it is $ 0$.


[quote="A. S. K."]$a^{b}= b^{a}$, solutions in $\mathbb{N}$.[/quote]

We always have as a solution $a=b$.
For $a\neq b$, the simpliest way in my knowledge is to study $x^{y}= y^{x}$, for $x$ and $y$ in $%Error. "mathlab" is a bad command.
{R}^{+}$, which gives $\frac{\log(x)}{x}= \frac{\log(y)}{y}$
The study of $f(x)=\frac{\log(x)}{x}$ on $%Error. "mathlab" is a bad command.
{R}^{+}$ shows that for every $x_{0}\in(1,e]$ there exist a unique $y_{0}$ (in $[e,+\infty)$) such that $\frac{\log(x_{0})}{x_{0}}= \frac{\log(y_{0})}{y_{0}}$.
Since we are looking for integers solution, only $x_{0}=2$ may be a solution and it is easy to see that $y_{0}=4$ works then, and, as a consequence, is the only solution.

So the whole set of solutions of the requested problem are $(n,n)$, $(2,4)$ and $(4,2)$

--
Patrick

For $a \neq b$ let us assume WLOG $b>a$ and it is clear that:

$a= p_{1}^{\alpha_{1}}\times p_{2}^{\alpha_{2}}\times......\times p_{n}^{\alpha_{n}}$

and $b= p_{1}^{\beta_{1}}\times p_{2}^{\beta_{2}}\times......\times p_{n}^{\beta_{n}}$

Therefore $b=ka$. Now substituting that into the original equations:

$a^{ka}=(ka)^{a}$ and thus $k=a^{k-1}$.

Now it is clear (you can do this graphically) that for $k>2$ and $a \geq 2$ then $x^{k-1}>k$.

Thus under these conditions, the solutions are $(a,b)= (2,4), (4,2)$.

[quote="BanishedTraitor"]For $a \neq b$ let us assume WLOG $b>a$ and it is clear that:

$a= p_{1}^{\alpha_{1}}\times p_{2}^{\alpha_{2}}\times......\times p_{n}^{\alpha_{n}}$

and $b= p_{1}^{\beta_{1}}\times p_{2}^{\beta_{2}}\times......\times p_{n}^{\beta_{n}}$

[u]Therefore $b=ka$[/u]....[/quote]

I don't see why. $b>a$ doesn't mean $\beta_{i}\geq \alpha_{i}$ $\forall i$. For example $2^{4}3^{1}> 2^{1}3^{2}$

--
Patrick

[quote="A. S. K."]$a^{b}= b^{a}$, solutions in $\mathbb{N}$.[/quote]
More interesting in $\mathbb{Q}^{+}$.

Hint: WLOG $a\geq b$ and define that $\gcd(a,b)=d$ now argue that $\gcd(a,b)=b$.

I think because $a^{b}=b^{a}$ and $b>a$,we can conculde $b=ka$

How about if $a,b \in Q^{+}$ ?
Thx..

[quote="sylow_theory"][quote="Yosh..."]

How about if $a,b \in Q^{+}$ ?
Thx..[/quote]

The answer is unbelievable.

[hide]
$(1+1/n)^{n}$ for all positive integers $n$.

Why is it unbelievable? Because those are the truncations for the number $e$.
[/hide][/quote]
$(1+1/n)^{n}$ for all positive integers $n$.

You mean this is a or b?
I don't understand you solution. Please explain more. Thanx

[quote="linH"]
You mean this is a or b?
I don't understand you solution. Please explain more. Thanx[/quote]

I am sorry, $a=(1+1/n)^{n}$ and $b=(1+1/n)^{n+1}$

Solving for $a^{b}=b^{a}$ is positive rationals.

Let $k=\frac{b}{a}$ then $b=ka$.

Again, we have $a^{k-1}=k$. Which would then give us $b=k^{\frac{k}{k-1}}$.

Let: $\frac{1}{k-1}=\frac{p}{q}$ such that $(p,q)=1$.

Therefore $a=(\frac{p+q}{p})^{\frac{p}{q}}$ and $b=(\frac{p+q}{p})^{\frac{p+q}{q}}$.

$a,b$ are rational then we must be able to take the $q^{th}$ root of $p$ and $p+q$.

Clearly $p=n^{q}$, if $q \geq 2$ then $n^{q}< p+q < (n+1)^{q}$.

So $q=1$.

Therefore $a=(1+\frac{1}{n})^{n}$, $b=(1+\frac{1}{n})^{n+1}$.

Do you know if it was ever proposed in an Imo?

I don't think so and I wouldn't imagine it should. It is far too famous.


calculus
integration
search
number theory theorems
number theory

Let $ p$ be an odd prime and $ g$ be the primitive root wrt that prime (a well known theorem states that it always exists).
Is it true that $ g$ is also a primitive root wrt $ p^k$ for any integral $ k$?

Remark: My semi-automatic computer search suggests a positive answer...

$ p\equal{}3,g\equal{}\minus{}1$.
But if $ g$ is a primitive root $ \mod p^2$, it is one $ \mod p^k$.

Ok, what about the roots in the interval $ [1, p\minus{}1]$ (for which I originally checked)?
Would you care to post a proof/link/name of the theorem that asserts your claim?

Nevermind, I managed to prove your claim by myself. Merely a simple extension of the existence proof for the primitive roots of $ p^k$. (but still haven't found a way to figure out if my conjecture was right for the numbers in the interval $ [1, p\minus{}1]$). Still, this gives me quite a simple way of verifying primitive roots for powers of small primes.

[quote="ZetaX"]$ p \equal{} 71,g \equal{} 11$[/quote]

Is the statement works

when I change that into "there exist a primitive root g modular p in which it is still primitive root modular p^k"

That was already mentioned  :wink:
[quote="ZetaX"]But if $ g$ is a primitive root $ \mod p^2$, it is one $ \mod p^k$.[/quote]


title

The message is too small. Please make the message longer before submitting.

I guess so(he said it and I would assume that he doesn't randomly lie and lock things(assuming he locked that thread).) Too bad we can't make this forum like that of NC, eh?

Yes. He rather rudely edited one of my posts, scarring me for life. And proving his point.

[quote="kstan013"]dude I deleted a smilie. nothing emotionally scarring[/quote]
 i think that is terribly emotionally scarring. why did you?

[quote="kstan013"]in the usergroups thing, for each group it says "subscribe", but then I found out that it's only for private forums, and BOB here was being a smart @$\$\$$ and said I'll "subscribe" :rotfl: like I'm some kinda fool. And I couldn't edit my thing then because for some reason I can only edit other people's stuff, but not mine, so I deleted the smilie which made it seem like BOB was a nice young fellow, and made me seem like I knew what I was talking about. I actually helped both of us :rotfl: ok not really.

you're still safe from stalkers BOB  :rotfl:  dude you're way too paranoid, unless you were just joking in which case send me a pm saying so :rotfl:[/quote]

I was just joking. You can tell because I was like OMGOMGOMGTHESTALKERSAREGOINGTOGETMEOMGOMG.

[quote="kstan013"]well I know how you are peter :D besides it's kinda fun calling you bob :rotfl:[/quote]

How am I?

[quote="kstan013"]I don't know. I just needed something smart to say :P[/quote]

You failed.


Οραίος. Μόνο, αν θες να βάζεις μεγάλες παρενθέσεις, πχ δίπλα σε ρίζες ή κλάσματα, γράψε \left ( ή [ και \right ) ή ] για να προσαρμοστούν στο κατάλληλο μέγεθος. Πχ:
$ [\frac{x^3}{4}(\sqrt [5]{\frac {1}{x}})]^\frac{1}{2}$                  χωρίς \left και \right

$ \left[\frac{x^3}{4}\left(\sqrt [5]{\frac {1}{x}}\right)\right]^\frac{1}{2}$      με.

Orfikoss(το μικρό σου?? ) ευχαριστώ πολύ το έψαχνα αλλα δεν το βρήκα.

Χάρηκα Δημήτρης


geometry
ratio
inequalities
AMC
AIME
ARML
triangle inequality

Triangle ABC is similar to triangle MNP.  If BC=60, AB=12, MP=8 and BC>AC>AB, compute the sum of all possible integer values for the ratio of the area of triangle ABC to the area of triangle MNP.

Also what if AC>BC>AB

Is the asnwer 20?  for the first one, i just did the triangle inequality.

i got 49 for the first one...there was only one possible area ratio...   :wink:

for the 2nd one, i got 64 and there was only one possible ratio....again....

Oh wait I misread the question.

There are $20$ possible area ratios for #1 so the sum is:

the smallest is 37, the largest is 56, they are in increasing arithmetic by 1 order so we have

$\frac{(37+56)*20}{2}=930$ which I believe is correct as the sum of all possible values for #1

Pine, you didnt consider that the sides dont hvae to be integers :wink:
only the area ratios..

Solution for #1 (answer already given):

[hide]
From the given conditions, BC>AC>AB and $\frac{AB}{MN}=\frac{BC}{NP}=\frac{CA}{PM}$.

We are given AB=12 and BC=60. Therefore, by the triangle inequality, $48<AC<72$, but by the restriction BC>AC>AB, we have $48<AC<60$.

Since the ratio of similarity is $\frac{AC}{MP}$, the ratio of the areas of the similar figures is $(\frac{AC}{MP})^{2}$.

From MP=8 and $48<AC<60$ we have $6<\frac{AC}{8}<\frac{15}{2}$ and then $36<(\frac{AC}{8})^{2}<56\frac{1}{4}$.

Thus the solution is the sum the integers from 37 to 56 inclusive, which is $\boxed{930}$.

Did this come from an AIME? It seems very AIME-ish.
[/hide]

The method for #2 will be the exact same with a slightly different application of the triangle inequality.

[quote="bthings_2000"]
Did this come from an AIME? It seems very AIME-ish.
[/quote]

No, it was one of my practice arml problems..

but if it was AIME, it would be only like number 1-3ish..


modular arithmetic
floor function
Pascal\u0027s Triangle

How many even entries are there in the 2009th row of the Pascal's triangle?

[hide="Solution"](i) First we show that if $ n$ is even and $ k$ is odd, then $ \binom{n}{k}$ is always even.

$ \binom{n}{k}=\frac{n\cdot{\binom{n-1}{k-1}}}{k}$. The numerator of the fraction is even and the denominator is odd. Thus, the quotient is even (because we are dealing with integers).

(ii) Now we show that if $ n$ and $ k$ are both even, then $ \binom{n}{k}\equiv{\binom{n/2}{k/2}} \pmod{2}$.

After a bit of algebra, we get $ \binom{n}{k}=\frac{(n-1)(n-3)\cdots(n-k+1)\binom{n/2}{k/2}}{1\cdot{3}\cdots(k-1)}$. From this, we clearly see that $ \binom{n}{k}\equiv{\binom{n/2}{k/2}} \pmod{2}$.

(iii) If $ n$ and $ k$ are both odd, then we just write $ \binom{n}{k}=\frac{n\cdot{\binom{n-1}{k-1}}}{k}$ and we find that $ \binom{n}{k}$ and $ \binom{(n-1)/2}{(k-1)/2}$ have the same parity.

(iv) Similarly, we find that if $ n$ is odd and $ k$ is even, then $ \binom{n}{k}\equiv{\binom{(n-1)/2}{k/2}}$.

From (i), (ii), (iii), and (iv), we can say that $ \binom{n}{k}$ is definitely even if $ n$ is even and $ k$ is odd. Otherwise, $ \binom{n}{k}=\binom{\lfloor{n/2}\rfloor}{\lfloor{k/2}\rfloor}$.

This fact makes us think of the binary representation of $ n$ because we can tell whether $ \lfloor{n/2}\rfloor$ is even based on its binary representation. Indeed, we find that if we express both $ n$ and $ k$ in binary, we can find the parity of $ \binom{n}{k}$ by continually truncating the last digit in the binary representation until the parity becomes trivial.

From this observation, we find that the number of odd integers in the $ n$th row of Pascal's triangle is $ 2^a$ where $ a$ is the number of $ 1$'s in the binary representation of $ n$. $ 2009$ has $ 8$ $ 1$'s in its binary representation so the $ 2009$th row has $ 256$ odd numbers. This leaves $ 2010-256=\boxed{1754}$ even numbers.

I might make the last part rigorous if I have time.
[/hide]

Hm...

[hide="Something Cool"]Note that $ (1\plus{}x)^2\equiv1\plus{}x^2\pmod2$. This results in a pretty quick proof of the binary stuff.[/hide]

The lemma in the second post along with its proof is really nice  :lol:
BTW math154 how would you use that fact to prove the lemma?

Split up $ (1 \plus{} x)^{2009}$ into factors depending on the binary representation of $ 2009$.

math154's observation is, suitably generalized, equivalent to the combinatorial proof of [url=http://en.wikipedia.org/wiki/Lucas%27_theorem]Lucas' theorem[/url] given in the Wikipedia article.  This is morally related to the relationship between actions of cyclic groups and the [url=http://en.wikipedia.org/wiki/Frobenius_endomorphism]Frobenius map[/url] I briefly discuss [url=http://qchu.wordpress.com/2009/06/09/the-magic-of-the-frobenius-map-ii/]here[/url].


function
induction
limit
real analysis
real analysis unsolved

Let $\left( x_n \right)_{n \geq 1} \subset \left[ 0,\infty \right)$ such that \[ x_{n+2} = x_{n+1}^{\sqrt[n]{x_n}}, \, n \geq 2 \, . \]

Prove that $\left( x_n \right)_{n \geq 1}$ has limit and compute it as a function of $x_3$.

Well, I was a little bit busy lately, but here is the proof for the convergence of this sequence:

  [b] Problem. [/b]

       Study the convergenceof the sequence $(x_{n})_{n\geq 1}$ defined by \[ x_{n+1}=x_{n}^{\sqrt [n] x_{n}} \], for all $n\geq 2$ and $x_{2}>0$.

  [i] Solution.[/i]

    If $x_{2}=1$ then $x_{n}=1$, for all $n\geq 2$. If $x_{2}\in (0,1)$ then by an easy induction $x_{n}\in (0,1)$, so the sequence is bounded. On the other hand because $\sqrt [n] x_{n} <1$ the sequence is increasing, so it is convergent. Now let's suppose that $x_{2}\in (1,\infty)$. By induction $x_{n}\in (1,\infty)$ for all $n\geq 2$ and we have \[ x_{n+1}=x_{n}^{\sqrt [n] x_{n}} >x_{n}^{1} \], so the sequence $(x_{n})_{n\geq 2}$ is increasing. Now, one can easily show that  \[ x_{n} >x_{2}^{\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{n-1}} \] so $\lim_{n\to\infty} x_{n}=\infty$. :)

[quote="Cezar Lupu"] $x_{n+1}=x_{n}^{\sqrt [n] x_{n}}$, for all $n\geq 2$ and $x_{2}>0$.[/quote]

[quote="perfect_radio"] $x_{n+2} = x_{n+1}^{\sqrt[n]{x_n}}, \, n \geq 2$ [/quote]

What is the correct form???

The recurrence relation I wrote is the one written in GM.

Anyway, it doesn't make any difference. The idea is the same. ;)


function
calculus
derivative
calculus computations

If anyone could explain how the following is done step by step, it would be greatly appreciated - this is an example in my book, but the solution isn't really explained clearly in the book:

Find and classify all critical points of the function
f(x,y) = x^3 - 6xy + 8y^3

A critical point of such a function is a point where all the first partial derivatives are 0.  So, take partials of $ f$ with respect to $ x$ and $ y$.  Then you have two equations and two variables and you'll have to solve.  That finds all the critical points.  To classify them, you'll have to examine the higher derivatives at those points.

Totally unrelatedly, are you, Sharkbites and meebo the same person?


Let $n \ge 3$ be a positive integer.

Prove that we can find two [u]different[/u] sequences of positive integer $(a_1;a_2;...;a_n)$ and $(b_1;b_2;...;b_n)$ satisfy the following conditions:

$i) 1<a_1<a_2<...<a_n$
$ii)1<b_1<b_2<...<b_n$
$iii)(a_1-1)(a_2-1)...(a_n-1) | a_1a_2...a_n-1$.
$iv)(b_1-1)(b_2-1)...(b_n-1) |b_1b_2...b_n-1$.

Actually, I think you've made a typo in $iii)$ or $iv)$, cos' with this conditions we only have to find one sequence (the second one will be the same )

[quote="zabelman"]There's a typo somewhere, since it doesn't work for n=2.[/quote]

Thank you, here $n\ge 3$.


Well, anyone have some nice algos for those tasks?

Btw, did anyone actually understand the last problem? Wasted 30 minutes reading it, and didn't understand it after that either. :P

well, unfortunately due to the internet failure at my house i could not participate at the online version.  :mad:   I am sad :(

Anyhow, ok let's start discussion over here.

depot and sorting is not much tough.

depot needs to find minimum number of cycles

and key sorting needs to just print different chars from nehind.

Amn't i right?

and number 3 was quite difficult. I don't think it was tough to understand the problem but it was tough to derive algo! ;)

Hm... Interesting...  Got 0 points because of "output format error" on all tasks!

What's the correct output format? No newline for last line?

oh.. there was no chance to check the first test case.. really bad..

I was planning to participate, but 2-7 AM is not when I do my best work!

Ok so here is the final comments on the problems:

Dept: Have to find minimum number of cycles in graph
Key: Have to print different numbers from behind
service: 4D DP problem with clever use of index


Prove that if $(N,10)=1$ then $N^{101}$  ends with three digits, which are also the last thee three digits of N.

[hide="Solution"] We wish to show that $N^{101}\equiv N \bmod 1000$ for $gcd(N, 10) = 1$.  Of course, this suggests the Totient Theorem.  However, $\varphi(1000) = 400$.

Note that $\varphi(125) = 100$, however.  We therefore know that $N^{100}\equiv 1 \bmod 125$.  Moreover, $\varphi(8) = 4$.  It follows that $N^{100}\equiv 1 \bmod 8$.

By CRT we know that $N^{100}\equiv 1 \bmod 1000$.  QED. [/hide]

Point to note, $N^{103}$ and $N^{3}$ will have the same last 3 digits (whatever the value of     $gcd(N, 10)$may be!)

[hide="Explanation"] $N^{103}\equiv N^{3}\bmod 2^{3}, 5^{3}$.  This is not too hard to see for the cases $gcd(N, 10) = 2, 5, 10$, which are the only possible other cases. [/hide]


What is the due time for each day's answer forms to be faxed in? The teacher's manual only says "immediately". My school's fax machines do not always work (they died yesterday when my teacher tried to fax the T-II form), so my teacher might have to go outside faxing them.

It probably doesn't matter.  I think that in this year's teacher's manual they put in a clause saying that they would let people mail it in right after the second day.

Well considering that a fax machine does not have memory, and thus only one sheet can be faxed in over a line, it doesn't seem likely that faxing the sheets immediately can even work, unless they somehow had a few hundred fax machines, one for every test center :O

I know last year my teacher was given permission to scan my solutions and email them in.


Does anyone know how to derive the theorem ab-a-b?  I believe this might require higher mathematics, I don't know really.

"ab - a - b" isn't exactly a theorem.  Do you mean proving that the largest number that can't be written as the sum of non-negative integer multiples of two relatively prime integers a and b both greater than 1 is ab - a - b?

it's very simple. someone move this thread to MC and I'll tell you. (can't right now too much bio hw)

Hey, JBL, don't you mean ab-a-b is the GREATEST number that can't be written as etc..... ?

Well, in fact I think I do.  Has been edited.


Να βρεθει ο μεγαλυτερος απο τους αριθμους $1, \sqrt{2},\sqrt[3]{3},...,\sqrt[n]{n}$ οπου $n\in N$ με $n\geq 2$

Ορίζουμε συνάρτηση $f(x)=x^{1/x}$, με πεδίο ορισμού τους πραγματικούς μεγαλύτερους του 1 και την παραγωγίζουμε, οπότε παρατηρούμε ότι εμφανίζει μέγιστο για $x=e$.

Άρα μεγαλύτερος αριθμός θα είναι ή ο $2^{1/2}$ ή ο $3^{1/3}$. Υψώνοντας και τους δυο αριθμούς στην έκτη βλέπουμε ότι μεγαλύτερος είναι ο $3^{1/3}$

Ωραιος  :clap2:
Εκτος αν $n=2$ τοτε ειναι $\sqrt{2}$  ο μεγαλυτερος...

Τι εννοείς στο τελευταίο post;

Ο $3^{1/3}$ είναι μεγαλύτερος από τον $2^{1/2}$. Ύψωσε και τους δυο αριθμους στην έκτη τους δύναμη και θα δεις ποιος είναι μεγαλύτερος

Εννοεί ότι αν $n=2$, τότε δεν ορίζεται το $\sqrt[3]{3}$, αφού $3>2$, άρα έξω από το πεδίο ορισμού


given $ n\in\mathbb N,\;\;\;n > 2$, find the volume enclosed by the following surface in the first octant

$ \left(\frac {x}{a}\right)^n \plus{} \left(\frac {y}{b}\right)^n \plus{} \left(\frac {z}{c}\right)^n\; \equal{} \;1$

PS: answer involves the gamma function...

(question has been edited, mlok, thanks for pointing out the flaw in my original question! :D )

This is a closed surface only when $ n$ is even.


Find $ \sin^3 x \plus{} \cos^3 x$ given $ \sin x \plus{} \cos x \equal{} 0.8$.

[hide="Solution"]

$ \sin^2x \plus{} \cos^2x \plus{} 2\sin x\cos x \equal{} 0.64$
$ \sin x\cos x \equal{} \minus{} 0.18$

Now,

$ \sin^3x \plus{} \cos^3x \equal{} (\sin x \plus{} \cos x)(\sin^2x \plus{} cos^2x \minus{} \sin x\cos x) \equal{} 0.8\cdot1.18 \equal{} 0.944$
[/hide]


Find all functions $f: (1,+\infty)\rightarrow(1,+\infty)$ such that for all $x,y\in(1,+\infty)$ we have:

$f\left(x^{u}\cdot y^{v}\right)\leq f(x)^{\frac{1}{4u}}\cdot f(y)^{\frac{1}{4v}}$

where $u,v>0$ are constants.

[quote="sinajackson"]Find all functions $f: (1,+\infty)\rightarrow(1,+\infty)$ such that for all $x,y\in(1,+\infty)$ we have:

$f\left(x^{u}\cdot y^{v}\right)\leq f(x)^{\frac{1}{4u}}\cdot f(y)^{\frac{1}{4v}}$

where $u,v>0$ [color=blue]are constants[/color].[/quote]
Are you sure? I know following problem : Find all functions $f: (1,+\infty)\rightarrow(1,+\infty)$ such that for all $x,y\in(1,+\infty)$ and for all $u,v>0$ we have:

$f\left(x^{u}\cdot y^{v}\right)\leq f(x)^{\frac{1}{4u}}\cdot f(y)^{\frac{1}{4v}}$.

Let $g(x)=lnf(e^{x})$, then your condition equavalent to
$g(ux+vy)\le \frac{1}{4u}g(x)+\frac{1}{4v}g(y) \ \forall x,y>0$.


Find all positive integers $ n$ for which $ \cos (\pi \sqrt{n^2\plus{}n}) \ge 0.$

[hide="Probably wrong.."]$ \cos(\pi \sqrt {n^2 \plus{} n}) \geq 0$

$ \frac {5\pi}{2} \geq \pi \sqrt {n^2 \plus{} n} \geq \frac {3\pi}{2}$

$ \frac {5}{2} \geq \sqrt {n^2 \plus{} n} \geq \frac {3}{2}$

$ \frac {25}{4} \geq n^2 \plus{} n \geq \frac {9}{4}$

$ n^2 \plus{} n \geq \frac {9}{4}$

$ (n \minus{} \frac { \minus{} 1 \pm \sqrt {10}}{2}) \geq 0$

$ n \geq  \frac { \minus{} 1 \pm \sqrt {10}}{2}$

$ \frac {25}{4} \geq n^2 \plus{} n$

$ 0 \geq (n \minus{} \left(\frac { \minus{} 1 \pm \sqrt {26}}{2}\right))$

$ n \leq \frac { \minus{} 1 \pm \sqrt {26}}{2}$

$ \frac{\minus{}1 \pm \sqrt{26}}{2} \geq n \geq \frac{\minus{}1 \pm \sqrt{10}}{2}$
[/hide]

[hide]Note that $ n < \sqrt{n^2\plus{}n} < n\plus{}\dfrac{1}{2}$.

For all $ x \in \left( n , n\plus{}\frac{1}{2} \right)$, we have:
$ \cos(\pi x) > 0$ iff $ n \in \mathbb{N}$ is even.
$ \cos(\pi x) < 0$ iff $ n \in \mathbb{N}$ is odd.

Therefore, the only positive integers such that $ \cos (\pi\sqrt{n^{2}\plus{}n})\ge 0$ are the even integers.[/hide]

$ 2k\minus{}\frac 12\le \sqrt{n^2\plus{}n}\le 2k\plus{}\frac 12$.
$ n^2\plus{}n\equal{}(n\plus{}\frac 12)^2\minus{}\frac 14$. Therefore $ n\equal{}2k, k\ge 0$ work, but $ n\equal{}2k\minus{}1\to \sqrt{n^2\plus{}n}<2k\minus{}\frac 12,k\ge 0$.
Therefore work $ n$ even $ n\ge 0$ or $ n<0$ odd.

The problem was for positive integers $ n$, but its still a good extension.


If $a$, $b$ ,$c$ are three positive real numbers such that $ab+bc+ca = 1$, prove that \[ \sqrt[3]{ \frac{1}{a} + 6b} + \sqrt[3]{\frac{1}{b} + 6c} + \sqrt[3]{\frac{1}{c} + 6a } \leq \frac{1}{abc}.  \]

This is a problem in IMO shortlist 2004  ;)

http://www.mathlinks.ro/Forum/viewtopic.php?t=30865

  darij


hi,

 I am working on producing Camera ready copies for a book.
 I use winedt and convert the latex documents to pdf. I take the printouts on a A4 sheet. But i want to draw a border of
(6 x 9 in) i.e the book paper size on the pdf outputs.

 What should i do?

Hi,
use the [url=http://www.ctan.org/tex-archive/help/Catalogue/entries/fancybox.html]fancyhdr[/url] package.

It provides many commands for this:  \fbox: \shadowbox, \doublebox, \ovalbox, \Ovalbox

 :)

Is there a way to get crop marks around the four corners instead of the full line border around the page. Making sure that the textwidth and textheight are not changed.

What should i do.


in circle O, chords AB and CD intersect at right angles at E.  If CE=8, DE=6, and AE=24, what is the radius of circle O?  I can find BE=2 by power of the point, but I have no idea what to do from there to get the radius-i can get all these power of the point equations but there are too many variables.  And then i used the fact that the radius if its the perpendicular to a chord, it is also the bisector but then my circle would be screwed up.  So can someone figure this out.  Thanks.

EDIT: uh oops..I didnt mean to put here..oh well ill leave it.  am i allowed to copy and paste this to like the high school basic section?
EDIT 2: wtf..now i just figuerd this out..i was using my power of the point wrong..ill just leave it up incase you want to do it..

yepppppp
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lol initially i used the POTP on the radius instead of the whole diameter so I had (r-11)*11=49 instead of (r+11)(r-11)=49

(8+6)/2=7, thats the perpendicular bisector (the radius perpendicular bisects the chord that has 8+6), then from power of the point (r+11)(r-11)=7*7