solve for x:
$ x \plus{} \sqrt {x \plus{} 1}  \equal{} \frac{1}{{\sqrt x }}$

DON'T POST HOMEWORK PROBLEMS HERE...AoPS NOT FOR EASY ANSWERS BUT AS FOR LEARNING!!

But if you show the steps you took, and then say why you don;t understand it after some specific step, that is fine.

[hide="A very general hint."]
What thing(s) in the equation are making it hard to isolate x, and how can you get rid of it/them?[/hide]

Does anyone have a nice solution for this problem ? :|

[hide="Brute Force"]
Let $ \sqrt{x}\equal{}a$, then $ x\equal{}a^2$
Then $ x\plus{}\sqrt{x\plus{}1}\equal{}\frac 1a$
$ \sqrt{x\plus{}1}\equal{}\frac{1\minus{}xa}{a}$
$ x\plus{}1\equal{}\frac{(1\minus{}xa)^2}{a^2}$
$ x(x\plus{}1)\equal{}1\plus{}x^2a^2\minus{}2ax$
$ a^4\plus{}a^2\equal{}a^6\minus{}2a^3\plus{}1$
$ a^6\minus{}a^4\minus{}2a^3\minus{}a^2\plus{}1\equal{}0$[/hide]


The chess club at Madison Middle School has 12 members. The club must select from its members a president, a vice president and a treasurer. In how many ways can this be done if no member may hold more than one office?

From the 12 people, you have 12 choices for the first office, 11 for the second, and 10 for the third, giving you $ 12*11*10\equal{}\boxed{1320}$


Let$n\ge 2$ and $f(x)=max(|sinx|,|sin(x+2)|,...,|sin(x+n)|)\forall x\in R$
Prove that $f(x)\ge \sqrt{(\frac{2n}{2n+1})^{2n+1}}$

[quote="Rust"]You are wrong, because $f(k\pi )=0$.[/quote]
oh,sorry.I edited my problem. :oops:

[quote="Rust"]I think $f(x)=max(|sin(x+1)|,|sin(x+2)|,...,|sin(x+n)|)\forall x\in R$.
Exist k: $|\frac{x+k}{\pi}-\frac 12|<\frac{1}{n+1}$. Therefore
$f(x)>\cos{\frac{\pi}{n+1}}$. This approach is better your.[/quote]
I think you need write:
$|$ {$\frac{x+k}{\pi}$} $-\frac{1}{2}$ $|<\frac{1}{n+1}$ :)


Let x,y,z be positive real numbers.

Prove or disprove:

$1+xyz \ \geq \ \min\{\ x(1+y)\ ,\ y(1+z)\ ,\ z(1+x)\ \}$

This is true: For any three nonnegative reals x, y, z, we have

$1+xyz\geq\min\left\{x\left(1+y\right);\;y\left(1+z\right);\;z\left(1+x\right)\right\}$.

In order to prove this inequality, assume the contrary: Assume that

$1+xyz<\min\left\{x\left(1+y\right);\;y\left(1+z\right);\;z\left(1+x\right)\right\}$.

Then, for instance, $1+xyz<x\left(1+y\right)$, what transforms to $1+xyz<x+xy$, or, equivalently, $1-x<xy-xyz$, what rewrites as $1-x<xy\left(1-z\right)$. Similarly, $1-y<yz\left(1-x\right)$ and $1-z<zx\left(1-y\right)$. Now, two cases are possible:

[b]Case 1.[/b] We have $1-z\leq 0$. Then, $1-x<xy\left(1-z\right)$ yields 1 - x < 0. Thus, $1-y<yz\left(1-x\right)$ yields 1 - y < 0. Finally, $1-z<zx\left(1-y\right)$ yields 1 - z < 0. Hence, all three numbers 1 - x, 1 - y, 1 - z are negative. Therefore, all sides of the three inequalities $1-x<xy\left(1-z\right)$, $1-y<yz\left(1-x\right)$ and $1-z<zx\left(1-y\right)$ are $\leq 0$; thus, we can multiply these three inequalities, but we must keep in mind that the sign gets switched two times (so it finally is the same as before); we obtain

$\left(1-x\right)\cdot\left(1-y\right)\cdot\left(1-z\right)<xy\left(1-z\right)\cdot yz\left(1-x\right)\cdot zx\left(1-y\right)$,  i. e.
$\left(1-x\right)\left(1-y\right)\left(1-z\right)<x^2y^2z^2\left(1-x\right)\left(1-y\right)\left(1-z\right)$.

Now, we divide this inequality by the negative number (1 - x) (1 - y) (1 - z) (negative because 1 - x, 1 - y, 1 - z are all negative), and it becomes $1>x^2y^2z^2$. But this cannot be true, since the numbers x, y, z are all > 1 (because 1 - x, 1 - y, 1 - z are negative). Contradiction!

[b]Case 2.[/b] We have 1 - z > 0. Then, $1-z<zx\left(1-y\right)$ yields 1 - y > 0, and $1-y<yz\left(1-x\right)$ yields 1 - x > 0. Hence, all three numbers 1 - x, 1 - y, 1 - z are positive.  Therefore, all sides of the three inequalities $1-x<xy\left(1-z\right)$, $1-y<yz\left(1-x\right)$ and $1-z<zx\left(1-y\right)$ are $\geq 0$; thus, we can multiply these three inequalities, and obtain

$\left(1-x\right)\cdot\left(1-y\right)\cdot\left(1-z\right)<xy\left(1-z\right)\cdot yz\left(1-x\right)\cdot zx\left(1-y\right)$,  i. e.
$\left(1-x\right)\left(1-y\right)\left(1-z\right)<x^2y^2z^2\left(1-x\right)\left(1-y\right)\left(1-z\right)$.

Dividing this inequality by the positive number (1 - x) (1 - y) (1 - z) (positive because 1 - x, 1 - y, 1 - z are all positive), we obtain $1<x^2y^2z^2$. But this cannot be true, since the numbers x, y, z are all < 1 (because 1 - x, 1 - y, 1 - z are positive). Contradiction!

As we obtained a contradiction in both cases, our assumption that $1+xyz<\min\left\{x\left(1+y\right);\;y\left(1+z\right);\;z\left(1+x\right)\right\}$ cannot be valid, and the inequality is proven.

  Darij


I can only take the B date (at a college/university) because my school isn't doing the amc10/12 this year.  Is there any way for me to still take the A date? I really want to so I can take both amc 10 and 12 or take the amc 10 twice depending on how I do.  :)

Thanks,
Truffles

Frequently Asked Question, #4
http://www.unl.edu/amc/f-miscellaneous/faq.shtml

Steve Dunbar
MAA Director, Mathematics Competitions

[quote="Truffles"]I can only take the B date (at a college/university) because my school isn't doing the amc10/12 this year.  Is there any way for me to still take the A date?[/quote]

You can also go to other schools in your local area to take AMC, as long as they let you. I know some students in our area did this.


The lengths of the sides of a triangle are $ x$, 13 and 37 units.
How many integer values of $ x$ are possible?

$ 37\minus{}13<x<37\plus{}13$  (by triangle inequality)
i.e.
$ 24<x<40$

Therefore, $ x\equal{}25, 26, \cdots 38, 39$.

Thus, $ \boxed{15}$.

$ 37 \minus{} 13 < x < 37 \plus{} 13$  (by triangle inequality)
i.e.
$ 24 < x < 50$

Therefore, $ x \equal{} 25, 26, \cdots 48, 49$.

Thus, $ \boxed{25}$.

 :P

OOPS, I have made a mistake in addition :o

Thank you for mewto55555 :)


Find $ p$ and $ q$ in terms of $ a$ , given that $ x^{2} \plus{}px\plus{}a$ is a factor of $ x^{4} \plus{} qx^{2} \plus{}p$.

[hide="Hint"]$ x^4 \plus{} qx^2 \plus{} p \equal{} (x^2 \plus{} px \plus{} a)(x^2 \plus{} bx \plus{} c)$

Then expand RHS and match terms.[/hide]

Thanks modularmarc101. I had

[hide=" Click to reveal hidden solution"]

$ x^{4} \plus{}qx^{2} \plus{}p \equal{} (x^{2} \plus{}px \plus{} a)(x^{2} \plus{}bx \plus{} \frac{p}{a})$, so that $ b\equal{}\minus{}p$ and

$ x^{4} \plus{} qx^{2} \plus{}p \equal{} (x^{2} \plus{} px \plus{}a)(x^{2} \minus{}px \plus{} \frac{p}{a})$, giving two possibilities: $ p\equal{}0, q \equal{} a$ and $ p\equal{}a^{2}, q \equal{} 2a\minus{}a^{4}$.

[/hide]


In http://www.mathlinks.ro/Forum/viewtopic.php?t=65049 , I wrote:

[quote="darij grinberg"]In http://www.mathlinks.ro/Forum/viewtopic.php?t=50773 , I wrote:

[quote="darij grinberg"]See http://www.mathlinks.ro/Forum/viewtopic.php?t=17207 and http://www.mathlinks.ro/Forum/viewtopic.php?t=34563 and the generalizations at http://www.mathlinks.ro/Forum/viewtopic.php?t=17230 , http://www.mathlinks.ro/Forum/viewtopic.php?t=23416 , http://www.mathlinks.ro/Forum/viewtopic.php?t=23752 .[/quote]

  darij[/quote]

;)

  darij

Wow!!! Thank you very much, darij grinberg  :wink:


[quote="arqady"][quote="Inequalities Master"]Let $ a,b,c > 0$, s.t. $ a \plus{} b \plus{} c \equal{} 3$. Prove that:
$ 6(\frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}) \plus{} 3\geq 7(a^2 \plus{} b^2 \plus{} c^2)$[/quote]
Here is my proof.
Let $ a \plus{} b \plus{} c \equal{} 3u,$ $ ab \plus{} ac \plus{} bc \equal{} 3v^2,$ $ abc \equal{} w^3$ and $ u^2 \equal{} tv^2.$ Hence, $ t\geq1$ and
$ 6\left(\frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}\right) \plus{} 3\geq 7(a^2 \plus{} b^2 \plus{} c^2)\Leftrightarrow$
$ \Leftrightarrow2\left(\frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}\right) \plus{} 1\geq \frac {21(a^2 \plus{} b^2 \plus{} c^2)}{(a \plus{} b \plus{} c)^2}\Leftrightarrow$
$ \Leftrightarrow2\sum_{cyc}a^2c \plus{} w^3\geq\frac {7(3u^2 \minus{} 2v^2)w^3}{u^2}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(a^2b \plus{} a^2c) \plus{} w^3 \minus{} \frac {7(3u^2 \minus{} 2v^2)w^3}{u^2}\geq\sum_{cyc}(a^2b \minus{} a^2c)\Leftrightarrow$
$ \Leftrightarrow9uv^2 \minus{} 2w^3 \minus{} \frac {7(3u^2 \minus{} 2v^2)w^3}{u^2}\geq(a \minus{} b)(a \minus{} c)(b \minus{} c)\Leftrightarrow$
$ \Leftrightarrow9u^3v^2 \minus{} (23u^2 \minus{} 14v^2)w^3\geq u^2(a \minus{} b)(a \minus{} c)(b \minus{} c).$
Now we'll prove that $ 9u^3v^2 \minus{} (23u^2 \minus{} 14v^2)w^3\geq0.$
$ (a \minus{} b)^2(a \minus{} c)^2(b \minus{} c)^2\geq0$ gives $ w^3\leq3uv^2 \minus{} 2u^3 \plus{} 2\sqrt {(u^2 \minus{} v^2)^3}.$
Hence, here it remains to prove that
$ 9u^3v^2\geq(23u^2 \minus{} 14v^2)\left(3uv^2 \minus{} 2u^3 \plus{} 2\sqrt {(u^2 \minus{} v^2)^3}\right).$
But $ 9u^3v^2\geq(23u^2 \minus{} 14v^2)\left(3uv^2 \minus{} 2u^3 \plus{} 2\sqrt {(u^2 \minus{} v^2)^3}\right)\Leftrightarrow$
$ \Leftrightarrow9t\geq(23t \minus{} 14)\left(3 \minus{} 2t \plus{} 2\sqrt {\frac {(t \minus{} 1)^3}{t}}\right)\Leftrightarrow$
$ \Leftrightarrow23t^2 \minus{} 44t \plus{} 21\geq(23t \minus{} 14)\sqrt {\frac {(t \minus{} 1)^3}{t}}\Leftrightarrow$
$ \Leftrightarrow23t \minus{} 21\geq(23t \minus{} 14)\sqrt {\frac {t \minus{} 1}{t}}\Leftrightarrow207t^2 \minus{} 399t \plus{} 196\geq0,$ which is true.
Thus, we need to prove that
$ \left(9u^3v^2 \minus{} (23u^2 \minus{} 14v^2)w^3\right)^2\geq u^4(a \minus{} b)^2(a \minus{} c)^2(b \minus{} c)^2,$
which is equivalent to the following inequality:
$ (139u^4 \minus{} 161u^2v^2 \plus{} 49v^4)w^6 \plus{} 9(3u^4 \minus{} 16u^2v^2 \plus{} 7v^4)u^3w^3 \plus{} 27u^4v^6\geq0$
because $ (a \minus{} b)^2(a \minus{} c)^2(b \minus{} c)^2 \equal{} 27(3u^2v^4 \minus{} 4v^6 \plus{} 6uv^2w^3 \minus{} 4u^3w^3 \minus{} w^6).$
If $ 3u^4 \minus{} 16u^2v^2 \plus{} 7v^4\geq0$ then our inequality is proved.
But $ 3u^4 \minus{} 16u^2v^2 \plus{} 7v^4 < 0$ gives $ 1\leq t < \frac {8 \plus{} \sqrt {43}}{3}.$
Id est, it remains to prove that
$ 81(3u^4 \minus{} 16u^2v^2 \plus{} 7v^4)u^6 \minus{} 108(139u^4 \minus{} 161u^2v^2 \plus{} 49v^4)u^4v^6\leq0.$
But $ 81(3u^4 \minus{} 16u^2v^2 \plus{} 7v^4)u^6 \minus{} 108(139u^4 \minus{} 161u^2v^2 \plus{} 49v^4)u^4v^6\leq0\Leftrightarrow$
$ \Leftrightarrow3(3t^2 \minus{} 16t \plus{} 7)^2t \minus{} 4(139t^2 \minus{} 161t \plus{} 49)\leq0\Leftrightarrow$
$ \Leftrightarrow(t \minus{} 1)^2(27t^3 \minus{} 234t^2 \plus{} 399t \minus{} 196)\leq0,$ which is true for all $ 1\leq t < \frac {8 \plus{} \sqrt {43}}{3}.$
Done!  :)[/quote]

Mr. Rozenberg, I admire your patience, but maybe there is a shorter way to prove this?
For example, after expanding, it gives that we need to prove only this (if I'm not wrong in the calculations):

$ \sum_{cyc}{a^4c} \plus{} \sum_{cyc}{a^3c^2} \plus{} \sum{a^2b^2(a \plus{} b)} \plus{} 2\sum{abc(ab \plus{} bc)}\geq 8abc(a^2 \plus{} b^2 \plus{} c^2)$
But I can't prove the last one. :(

[quote="Inequalities Master"]Let $ a,b,c > 0$, s.t. $ a \plus{} b \plus{} c \equal{} 3$. Prove that:
$ 6(\frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}) \plus{} 3\geq 7(a^2 \plus{} b^2 \plus{} c^2)$[/quote]
Assum $ c\equal{}min${$ a;b;c$}

$ 6(\frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}) \plus{} 3\geq 7(a^2 \plus{} b^2 \plus{} c^2)$

$ \leftrightarrow 6(\frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}\minus{}3)\minus{}7(a^2 \plus{} b^2 \plus{} c^2\minus{}\frac{(a\plus{}b\plus{}c)^2}{3}) \ge\ 0$

$ \leftrightarrow (\frac{6}{ab}\minus{}\frac{7}{3})(a\minus{}b)^2 \plus{} (\frac{6}{ac}\minus{}\frac{7}{3})(a\minus{}c)(b\minus{}c) \ge\ 0$

We have

$ M \equal{} \frac{6}{ab}\minus{}\frac{7}{3}\equal{} \frac{18\minus{}7ab}{3ab} \ge\ \frac{18\minus{}7(\frac{a\plus{}b}{2})^2}{3ab}$

$ M \ge\ \frac{18\minus{}7.\frac{9}{4}}{3ab}  \ge\ 0$

Similar to :

$ N \equal{} \frac{6}{ac}\minus{}\frac{7}{3} \ge\ 0$

We have done! :)

[quote="onlylove_math"][quote="Inequalities Master"]Let $ a,b,c > 0$, s.t. $ a \plus{} b \plus{} c \equal{} 3$. Prove that:
$ 6(\frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}) \plus{} 3\geq 7(a^2 \plus{} b^2 \plus{} c^2)$[/quote]
Assum $ c \equal{} min${$ a;b;c$}

$ 6(\frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}) \plus{} 3\geq 7(a^2 \plus{} b^2 \plus{} c^2)$

$ \leftrightarrow 6(\frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \minus{} 3) \minus{} 7(a^2 \plus{} b^2 \plus{} c^2 \minus{} \frac {(a \plus{} b \plus{} c)^2}{3}) \ge\ 0$

$ \leftrightarrow (\frac {6}{ab} \minus{} \frac {7}{3})(a \minus{} b)^2 \plus{} (\frac {6}{ac} \minus{} \frac {7}{3})(a \minus{} c)(b \minus{} c) \ge\ 0$

[/quote]
It must be
\[ \Leftrightarrow (\frac {6}{ab} \minus{} \frac {14}{3})(a \minus{} b)^2 \plus{} (\frac {6}{ca} \minus{} \frac {14}{3})(a \minus{} c)(b \minus{} c) \ge 0
\]
;)

Inequalities Master, if I am not wrong, can_hang2007 proves it by the $ b\equal{}a\plus{}x,$ $ c\equal{}a\plus{}y...$ method. :wink:

[quote="Inequalities Master"]Can we solve it by SOS, arqady?[/quote]
I don't think so.

[quote="can_hang2007"][quote="arqady"]Inequalities Master, if I am not wrong, can_hang2007 proves it by the $ b \equal{} a \plus{} x,$ $ c \equal{} a \plus{} y...$ method. :wink:[/quote]
No, I am not, arqady. I proved it by SOS ;) :). See here: http://www.mathlinks.ro/viewtopic.php?t=120859 :)
And here for the full proof of a weaker one: http://www.mathlinks.ro/viewtopic.php?t=145431
(I proceeded my proof similar to this weaker one:) )[/quote]
Thank you Can!
I have meant $ 2\left(\frac{a^2}{b}\plus{}\frac{b^2}{c}\plus{}\frac{c^2}{a}\right)\plus{}3(a\plus{}b\plus{}c)\geq\frac{15(a^2\plus{}b^2\plus{}c^2)}{a\plus{}b\plus{}c}.$
It something another. :D

[quote="Inequalities Master"]Let $ a,b,c > 0$, s.t. $ a \plus{} b \plus{} c \equal{} 3$. Prove that:
$ 6(\frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}) \plus{} 3\geq 7(a^2 \plus{} b^2 \plus{} c^2)$[/quote]
Here is the full proof :)

The stronger seems still trues:
$ \frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a}\plus{}\frac{24(ab\plus{}bc\plus{}ca)}{(a\plus{}b\plus{}c)^2} \geq\ 11$
 :wink:

[quote="nguoivn"]The stronger seems still trues:
$ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \plus{} \frac {24(ab \plus{} bc \plus{} ca)}{(a \plus{} b \plus{} c)^2} \geq\ 11$
 :wink:[/quote]
Try $ a \equal{} 9,$ $ b \equal{} 3$ and $ c \equal{} 2.$ :wink:
But the following inequality is true.
Let $ a,$ $ b$ and $ c$ are positive numbers. Prove that
\[ \frac {a\plus{}b}{c} \plus{} \frac {b\plus{}c}{a} \plus{} \frac {c\plus{}a}{b} \plus{} \frac {48(ab \plus{} bc \plus{} ca)}{(a \plus{} b \plus{} c)^2} \geq\ 22\]

[quote="Inequalities Master"]Let $ a,b,c > 0$, s.t. $ a \plus{} b \plus{} c \equal{} 3$. Prove that:
$ 6(\frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}) \plus{} 3\geq 7(a^2 \plus{} b^2 \plus{} c^2)$[/quote]
An interesting proof for you, [i] manilo[/i]!  :lol:
\[{ 6\left(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\right)+3-7(a^2+b^2+c^2)=\dfrac{2}{3} \cdot \left[\dfrac{(2b-c)^2(a-b)^2}{bc}+\dfrac{(2c-a)^2(b-c)^2}{ca}+\dfrac{(2a-b)^2(c-a)^2}{ab} \right].}\]

Wonderful !!!

Thank you very much.

I checked it.

$\[{ 6\left(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\right)+3-7(a^2+b^2+c^2)=\dfrac{2}{3} \cdot \left[\dfrac{(b-2c)^2(a-b)^2}{bc}+\dfrac{(c-2a)^2(b-c)^2}{ca}+\dfrac{(a-2b)^2(c-a)^2}{ab} \right].}$


The question is

If $\sum_{n \geq 0}^\infty c_n$ converges, and $\sum_{n \geq 0}^\infty c_n (-1)^n$ diverges for $\sum_{n \geq 0}^\infty c_n (x+1)^n$, what is the interval of convergence for this series?

I'm not even sure that I understand the question....
I understand the absolute convergence theorem, but don't know if that applies here....I am confused...and would appreciate a point in the direction of the solution....

Thanks

For $\ x\in(-1,0)$ the sequence  $\ (x+1)^n$ is decreasing and converges to $\ 0$, so from Abel's criterion we have that
$\ \sum_{n \geq 0}^\infty c_n (x+1)^n$ converges.
Are the $\ c_n$ pozitive ?

You have a power series centered at $-1.$ The sum $\sum_{n=0}^{\infty}c_n$ is that power series at $x=0$ and $\sum_{n=0}^{\infty}(-1)^nc_n$ is that power series at $x=-2.$

Here's the key theorem:

If the series $\sum_{x=0}^{\infty}c_n(x-x_0)^n$ converges for $x=x_1,$ then the series converges for all $x$ such that $|x-x_0|<x_1,$ and

If the series $\sum_{x=0}^{\infty}c_n(x-x_0)^n$ diverges for $x=x_2,$ then the series diverges for all $x$ such that $|x-x_0|>x_2.$

The proof of the first part of this theorem follows from the fact that the terms of a convergent series are bounded, followed by an explicit comparison to a geometric series. The second part is a direct consequence of the first part. The alternating series test (or related items, such as Abel's criterion) is irrelevant; this is an argment about absolute convergence.

Appling that, if we restrict ourselves to real values of $x,$ we see that this power series converges for $-2<x\le 0.$

The coefficients $c_n$ are most assuredly not positive. It's likely that $c_n$ has an alternating sign.


[quote="abacadaea"]if x percent of y is 125, then what percent of x is 2y?

i get 250, the answer is 160[/quote]

it can be anything, are u sure that you didn't mistype the question?

So say x is 25. so x percent or 25% of y is 125, so y is 125*4 or 500

so 2y=1000 so 2y is 4000% of x

say x=100

so x percent or 100% of y is 125, so y is 125*1 or 125.

so 2y=250 so 2y is 250% of x.

As you see, it can be anything.


group theory
abstract algebra
superior algebra
superior algebra unsolved

Prove that if $ k$ is a field and $ k^{*}$ is cyclic, then $ k$ is finite!

Let x be the generator of the multiplicative group of k.

If x=-1, then the field is order 2 or 3. (Set n=1 here)

Otherwise, 1+x=x^n for some n, and x is algebraic.

In any case, if F is the prime field of k, then k=F[x].

If F=Q, then the subgroup Q* of k* is not cyclic, a contradiction.

If F=Z/pZ, then F[x] is finite of order at most p^|n| (where n is defined above).

Obviously we may assume $ k^{*}$ is isomorphic to $ \mathbb{Z}$ as otherwise $ k$ will be finite. Assume $ k$ is of characteristic $ 0$, then $ k^{*}$ will contain $ \mathbb{Q}^{*}$ but this is impossible (check this). Otherwise we have that the characteristic of $ k$ is $ p$ for some prime $ p$. Then we have $ k^{*}$ contains $ \mathbb{Z}_{p}^{*}$ which is cyclic of order $ p \minus{} 1$ (I assume you know the converse of the result you posted) however this also contradicts the assumption that $ k^{*}$ is isomorphic to $ \mathbb{Z}$.

Edit - didn't see the other post here, I'm too slow...

Alternative proof: Let $ v$ generate the multiplicative group, then we have $ \minus{} 1 \equal{} v^k$ for some $ k\neq 0$, hence $ v^{2k} \equal{} 1$, so the multiplicative group is finite.

EDIT: I guess we do need to separately handle the case $ \minus{}1\equal{}1$, where we still need to write $ \mathbb{F}_2 [v]$ as a quotient of $ \mathbb{F}_2 [x]$ in much the same way as the other solution.


function
real analysis
calculus
calculus computations

For what class of functions  the two notions 'sup norm' $ sup_{x\in X}|f(x)|$ and the 'ess. sup norm' $ \|.\|_{\infty}$ coincide? where $ f$ is a complex valued measurable function on a measure space $ X$.

The continuous functions, and a bunch of stuff that doesn't come in any useful form. There are a few spaces of almost continuous functions that work (one case: left and right limits exist everywhere, function value is the average of them), but you can't go very far with that.

You use the sup norm on spaces of continuous functions, and the essential supremum on spaces of functions mod "equal almost everywhere" such as $ L^p$. The two norms are equal on the first class of spaces, and the first type of norm doesn't even make sense on the second class of spaces.

[quote="jmerry"] and the first type of norm doesn't even make sense on the second class of spaces.[/quote]
Can you explain why? Thanks.

The key is that the elements of $ L^p$ aren't actually functions - they're equivalence classes of functions.

Let $ f(x) \equal{} 1$ everywhere and $ g(x) \equal{} \begin{cases}2, & x \equal{} 0 \\
1 & \text{otherwise}\end{cases}.$

Then the supremum of $ f$ is $ 1$ and the supremum of $ g$ is $ 2.$ They are not, literally speaking, the same function.

But the Lebesgue attitude intervenes, and says, "I don't care that they're not literally the same function. They might as well be the same, since I don't recognize any difference between them."

Or to say that more precisely: these two functions both belong to the same equivalence class under the equivalence relation $ f\sim g$ iff $ f(x) \equal{} g(x)$ almost everywhere. We may loosely refer to a member of $ L^p$ as a function but it's not a function - it's an equivalence class of functions. And both $ f$ and $ g$ are members of the same equivalence class; they're both realizations of the same element of $ L^{\infty}.$ And $ \vert|f\vert|_{\infty} \equal{} \vert|g\vert|_{\infty} \equal{} 1.$

jmerry's "doesn't even make sense" applies to the notion that $ f(0) \equal{} 2$ makes no sense; the notion that $ f(0)$ equals any particular value at all makes no sense.

[quote="Kent Merryfield"]jmerry's "doesn't even make sense" applies to the notion that $ f(0) \equal{} 2$ makes no sense; the notion that $ f(0)$ equals any particular value at all makes no sense.[/quote]
Thanks jmerry, Kent for your replies. But I couldn't understand the above statement!


Mafia Game 13 has started!

noneoftheabove
Sunny
Peter VDD
LadyKn1ght
jli
Klebian
falconwing64
S2
MithsApprentice
samath
yif man12
slamchess
zeb
la mafioso
apusey
Melissa

Those who have * by their name have not confirmed.

The roles for this game are:

[b]4 mafia.[/b] No day communication outside of this thread, but you may communicate before the first day. One of the mafia has a criminal background. When he is killed, it will not be revealed that he is mafia, only that he has a criminal background. When inspected, he comes up guilty. He knows that he has a criminal background.

[b]1 inspector.[/b]

[b]2 angels.[/b] May communicate at night only.

[b]1 devil.[/b] The devil gets to inspect one person each night to find out if they are the inspector, angel, or priest (and which one). The devil is guilty but may not communicate with the mafia at any time outside of this thread.

[b]1 priest.[/b] The priest is an innocent role who inspects one person each night for whether they are the inspector, angel, or devil (and which one).

[b]1 bomber.[/b] Every night, the bomber chooses a player. The bomber plants a bomb outside the house of that player. Anyone who comes to that player's house in the night - the mafia, inspector, angel, devil, priest - is killed, along with the person whose house it is. If no one visits that person's house during the night, nothing  happens. The bomber is innocent, but must play carefully so he does not harm the citizens instead of the mafia. [i]Note that this requires the mafia and the angels to choose one person to send to the chosen person's house, so that person will die if there is a bomb.[/i]

[b]3 masons.[/b] As with mafia, night communication is allowed, day communication is not.

[b]3 citizens.[/b] One of the citizens has a criminal background. When he dies, it is not revealed that he is a citizen, only that he has a criminal background. When inspected, he comes up innocent. This citizen does not know that he has a criminal background.

I have confirmed!

Hopefully there's not random lynching like game 12 has been going so far lol :roll:

I think it can be confirmed here.

EDIT: Yes, I am 100% sure that we can confirm here.

Come on peeps confirm already! There will never *EVER* be a mafia thirteen again!

You are about to enter the most legendary mafia game that will ever occur (except for 1 and 10, and anything else that gets past 10's page limit without meaninless psots)

[quote="la mafioso"]Come on peeps confirm already! There will never *EVER* be a mafia thirteen again![/quote]

Unless one of the mods doesn't know how to count, or uses a different base besides 10, or we go [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=61057]back in time[/url].

There are really a lot of ways it could happen if you think about it.

Lol ya you're right...but some peeps need to get a push to get themselves goiubng.

Hopefully it's ready by the time i get home from school to morrow.

Yes, you can sign up on the mafia sign up thread.

I think it's a little late for that. Ask on the other forum and see if you can get put in.

If you couldn't tell, I'm here.

Ready for 13. Darn this minimum length of post.

HI, I'm ready... sort of. At least, I will be ready around 8:30... I have to go bye!

YYYYYYYYAAAAAAAAAAAAAAH! Now it's a party when zeb gets here  :P

[quote="G-UNIT"]Can I be a sub for anyone who hasn't confirmed.... I didin't know the sing-up process had begun.... :([/quote]Luckily there hasn't been a [i]sing-up[/i] process. :D

The roles were:
[code]noneoftheabove -	inspector
Sunny -			mason
Peter VDD -		mafia
LadyKn1ght -		mafia
jli -			priest
Klebian -		bomber
falconwing64 -		mason
S2 -			citizen with criminal backround
MithsApprentice -	mason
samath -		citizen
yif man12 -		angel
slamchess -		angel
zeb -			devil
la mafioso -		mafia
apusey -		citizen
Melissa -		mafia with criminal background[/code]

FOR THE LAST TIME PETER I'M A HER NOT A HIM!!!

anyways, good job... i feel stupid now.
heh.. my goal in life had been to stay alive... didnt help me win...

having a criminal record is cool though...  :)

Sunny was lynched. That's 5 ayes...

2 nays...

Sunny was a...

(jmadsen will reveal, i just wanted to put it lolol)

OMG WE WON! WOOOOOOHOOOOOOOOOOOOOOO!

EDIT: Notice that it was game 13 that the mafia won...

kinda unlucky isn't it? =p

I don't know about the very first games, but from all the ones I know this is the only one won by mafia?

HA! See? I was innocent.  Why do you have a grudge against me, Peter?

What's a grudge?

No, I don't have anything against you :) but you just helped us a lot without realising it.

Ya lol klebian got it from the start!

I don't wanna put guilty on anyone at all...but if you would've accuseed peter like klebian said, the mafia would've spared sunny cuz they thought they would be killed next.

We prolly would've killed devil on that last night and sealed our fate.

But, you didn't...hahahha :D

Actually, I was inactive for other reasons...

I thought you would realize I was devil once I pointed out the flaw with what I thought was your plan... because I would be the person who would think that you were planning something...

And, strangely enough, I forgot that la mafioso was a mafia. I almost changed my vote at the end to a nay... :blush:

But why didn't anyone suspect me? (In the end, that is.)

I shouldn't have died. I submitted my angel message an hour and a half after the deadline given, so it shouldn't have counted, but whatever.

[quote="Klebian"][quote]DANG IT!!!!!!!!
sorry guys....
My parting words:
Peter's in on it with Melissa, kill him!
[/quote]
But no one listens to the bomber.[/quote]
You weren't the only one who was on to them, Noneoftheabove had Peter and LadyKn1ght next in his inspection queue.

Killing noneoftheabove just was the crucial point in the game. I knew we were 'in the spot', and an inspector's word would have us killed regardless the defense. Even if we killed the angel, that would not have happened the next turn, so we would both have been revealed and killed. By contrary, without an inspector in the game, all we needed was some good talking. ;)

yes precisely :mad: . oh well i didnt have enough time after the speed lynch because i inspected sunny and melissa, so i had no time to tell you guys. Jeez, i think Klebian should have told the angel to save him, and then he bomb someone else.


geometry
Gauss
ratio
function
probability
geometric transformation
articles

[quote="pardesi"]because i love it and ad been reading from it then suddenly i remembered that indians understood a completely diff organic chemistry :rotfl:[/quote]

The focus in 99% of the nation (you know where all) is on IIT oriented practise and not real learning. Here i'd say its somewhat inbetween, so I am happy.

A guy from Chennai who got 729th rank in JEE was the first year topper last time in IIT-M. He got a phenomenal GPA of 9.9. Exceptional eh? As I said, JEE is more about hardwork and a little bit of talent.

Personally I feel that maths olympiad is also about loads of hardwork though talent is a must. Ofcourse with many of the Indian questions from intl sources, luck factor is very high.

BTW @ Soumyashant: I have your photo man!!! HBCSE sent me a pamphlet and the last page had the photos of olympiad winners and you were there in astronomy :D

[b]MO[/b]
i would disagree a  bit on this point.yes maths is the only olympiad where a guy of every class has equal chance(nearly).But i feel though clearing it requires very little knowledge of the requisite)if any.But brings with it a huge amount of luck factor carrying.It "does not" depend how good u r in "maths" or has in some sense nothing to do with what "maths" has in store for u...but rather it's just a question of "striking" th eright thing at right time.
No i don't mean to insult it any way but what i am stressing is it's almost a question of hide and seek .u just need to know the approach who cares how much u know of maths or even had a taste of it and such questions are being increasingly used in INMO
See for example
INMO 2007(2,1(almost))
INMO 2006(2,6(ii))
INMO 2005(1,)
INMO 2004(5) and the list goes on...
see these questions  required that "one thing" striking u or otherwise u were dead...
see for example the famous Engel problem IMO 1988 .if u went through engel u will see how he says none of the proffesors to whome these questions were given could do it though 12 students came up with perfect solutions and there started the "VIETE JUMPING" if u don't know what that was the question is half dead for u .
Ok now how do we solve such problems...
well the obvious answer is experience.and how does experience come???by solving problems&solving problems&solving problems&solving problems....
and this is Mathematical HARD WORK
[b]Rest Olympiads[/b]
Although the Physics and Chemistry olympiad questions are particularly for those who know it..but one thing is guaranteed there is very very very little luck factor involved(till the copies get checked well  :D ).if u r good at physics and chemistry then u r bound to get through them(if u r copies are checked well  :rotfl: ),there is almost nothink like "striking the right path"
and how do i know everything...the only naswer read read read read....
hence work hard work hard and work hard.
Further Physics and chemistry olympiads have lot more do with what actual physics and chemistry is than what actual maths is that is taught at olypiad level.Though only "maths" questions are the nly tough ones among the internationals.
But part of the reason being the very limited nature of our course of study about physics and chemistry in our 11th and 12th.and by limited i am not referring at the amount but on the variuos nuances of the subject.
[b]Conclusion[/b]
Comparisons of either is not valid :P

Comparison is invalid, no doubt. Its like comparing eminem and linkin park. Both are good in their own genres. :D

Anyways pardesi has a really valid point. No Chappli, I didn't say that to infuriate you.. As you know very well, that has always been my opinion (it has also been suresh sirs :P remember).

I am NOT denying the role that talent has to play in the olympiads. Obviously talent HAS, and WILL continue to play a major role in any olympiad, maths more so than physics and chem. But yes as pardesi said, Maths olympiad has luck factor associated with it. If you have tried "ONE" ||| problem to it before, you strike gold, after all the cut off is usually less than 40% of the total (correct me, if i'm wrong)

I do not say more hardwork than talent but definitely hardwork is required. I know what you are trying to imply, that you are filled with 20000% of natural talent and that you got in through sheer talent and you didn't have any extra knowledge other than the NCERT textbook and you got in with just school knowledge. You yourself know that you are wrong. I agree clearing RMO used only ur talent but you did some work for INMO surely..... You may not have done as much work as some of the other 10th'ers who cleared the round but defn you did do hard work and knew SOME STUFF (Much more than the above average guy who attempts the INMO) to clear it. You did not know TOO MUCH STUFF, you did KNOW some amt of it. I have heard from some of your friends who were with in till class X :D that you worked pretty hard for INMO. correct me if I;m wrong. I think anybody who clears the INMO does some hard work for it.

Today the scenario has changed to more practise than hard work. All your natural talent can do is clear RMO for you. (At the most) ........ Consider the case of 2 9th'ers in our school you know them. They cleared the RMO but they definitely did enough hardwork though they were extremely talented.
The tide is shifting today dear friend, as it is shifting in the IIT-JEE scenario as well.

P.S: Happy Mattu Pongal

i really would like people to define the word 'talent'.i beleive no one in thie world is talented except "Ramanujan" if what he used to say about his ideas was true.
all have worked hard(not necessarily physical may be mental also) hence suceeded

i am also of the same opinion as chappli (dont think i am putting jalra i am going to show dissent somewhere else  :wink: ).
I have always felt that i lack somethin to clear the RMO ( dunno what? ). I just happen to do well at the unrequired times . Its just that all talented people (in math) are attracted towards maths and start working at a young age ( even though if they hadn't worked they would have still got thro  :P ).
i strongly feel that natural talent is very much a necessary and sufficient requirement for clearing the inmo ( even though certain basics are reqd. ).
Also this topic has been weather beaten plenty of times in AOPS also.
I will post one such link ( even though i disagree with many of these views )
P.S. Even Gauss was much talented and so were many other mathematicians  :D

[quote="madness"]. Its just that all talented people (in math) are attracted towards maths and start working at a young age ( even though if they hadn't worked they would have still got thro  :P ).[/quote]
and this attraction u refer to is sharpening ur mind and developing skilss and working hard to sharen them.
i don't think so they wold have got through .Borrow the book men of mathematics from chappli(in case u don't have it) and read about Galois ...he never qualified the entrance to Ecole Poythechnique :o
the reason being the questions he was tested on belonged to topics he didn't like nor had any intesret explaining what he thought of them...
[quote="madness"]
 i strongly feel that natural talent is very much a necessary and sufficient requirement for clearing the inmo ( even though certain basics are reqd. ).
Also this topic has been weather beaten plenty of times in AOPS also.
I will post one such link ( even though i disagree with many of these views )
P.S. Even Gauss was much talented and so were many other mathematicians  :D[/quote]
taht's why i asked u to define natural talent.Any thing if only sharpened u get to knoe that guy has "talent".Even ramanujan would have been nothing without the book by *****(i forgot the name)
u need something to spark out interset and love for the subject.and maths olympiad though develops ur interest in maths but it never teelss u how good ur in mathematics because it is more interested in testing ur skills than knowing what u can do ?
the real test is what u can do what others couldn't???
Gauss was talented but something was there that showed him the way and he followe it and i bet Rohit u lck nothing ...the only probable reason u couldnb't clear it because the questions weren't good eneough to show other what talent u have

thank you for that but still i feel i lack somethin' .
I could do well but @ the wrong times only  :(
maybe pressure was the thing that made it so.
And btw the proof reading part should be scrapped according to me.
If one has a correct method to solve a sum and one has made valid points and got the ans , I personally feel one should be given the full credit  :)

yes u said it 'pressure' why should one check a person's 'talent' puting him ona hot stove rather make him sit calm and check his 'talent'

Math olympiad is the best of all olympiads in terms of necessity of talent...
I don't think physics, astro and others require much talent
(loads of hard work and problem solving mostly conventional)

But I will disagree with G. Vignesh...

Math olympiad requires more hardwork than talent...(although the ratio is less than other olympiads)
and still I don't think the best 30 clear the INMO...
I think the best 10-15 of the country (not necessarily the top ranked 10-15) and the rest are partial flukes ...

In IMOTC... where if a guy has solved 1000 problems from good books he definitely has a better chance than the person who clearly does not take interest in olympiad-guessing kind of problem ...
[b]And training is a must...[/b] and you can't afford to think about the philosophical aspects of the subject when you are ground to compete for the best spot ...
(btw, some problems are amazing but most are copied from other sources and it seems people have come across these problems in AOPS themselves)

G.V., you seem to idolise MO but later in your life you will realise that MO is never a measure of the best people of the country in mathematics...
(only a general idea of the best people in the country)
But MO is a very good forum for exposure to various people and knowing profs of mathematics and sharing ideas, various institutes, getting good books, scholarships  :lol: , money...

You yourself might come across sb in IIT who never did olympiad mathematics or qualified any MO and is better than you...
and it often happens...

yes that puts it all ...thats what we think here :)

Yes that's how the system is in india but i think other countries might have a system where more natural talent is encouraged.
And by natural talent i mean that which makes a student inquisitive abt maths (and probably makes him discover things on his own) without being subjected to any external influence and without any necessity at all to do maths (he could have done any other thing in life but was attracted by math )

[quote="Chapplikka"]I tell u shre, if an olympiad depends most on talent, it is the maths.. I got thru inmo in 10th.. N i tell u i din knoe even half of wat u wud hav expected from an imotc guy... there was a geometry problem which asked me to prove
 3cos x + 4sin x <= 5...
And u say math olympiad needs more hardwork than talent.. I think that would b the case fer physics, n more so in chem.. def not math...
 And as far as i know, the only place where classes are taken in full flow for olympiad stuffs is in bhaskaracharya prathistan, n now that is also changin.. Last yr only 4 made it to the tc frm pune.. dat tells u how the story is changin, n it is made of more TALENTED GUYS than more NV GUYS..
And i know shre y u made this statement..
U JUS WANTED TO INFURIATE ME...  AND IM NOT INFURIATED...  :lol:[/quote]
Last year only 4 made it to the TC from Pune coz I believe that the josh in BP is no longer there ...

Prof M. Prakash of BP started the 'project gold' in 1999 to groom 8th class students to get gold in IMO and He achieved the results in the next few years...

West bengal had a comprehensive training programme but now it doesn't
(in 2000, ten cleared INMO and Four went to IMO - 2001) :(

I tell you, if they start the programme with the same seriousness again,many others might have to say good-bye...

And the story actually keeps on changing constantly.. :oops:
So you never know when they will come back...

Why have you to go berserk over the idea that math olympiad needs lots of hard work...?? :rotfl:

Atleast pardesi has a moustache and a beard. :)

summa sorrakka nu solli cycle gap la insult pannara pathiya. Dei onne apparum pathukkarein da ippo shreyassa than gavanikkanum apparum chappliya galli pannanum  :D
Dei Shreyas vendam naan daily shave pannarein athu onnakku theriyum illaya??? :P

Pinna yen da ippadi abhandamma poi sollare?  :mad:

[quote="shruthibaskaran"]Shreyasa paakanum, chappliya gaali pananuma? Lol... Rohit a idhu? Apparently, my opinions are disillusioned! Lol

But I have to say this... Semma chinna vayasu, semma talent rohit![/quote]

Vandham naan yaarayum veena vambhukku izhukkalai. Ellarum ennai kindal seidhal naan onnum sollanena eppadi?  :D  Naanum koncham sollanum illaya  :D . Enna pannardhu indha ollagathil ippadi irrundhal thaan pozhakka mudiyum polairrukku  :(

Btw adhu rohit illa MADNESS :D .

[quote="shruthibaskaran"]Lol.. Rohit illa Madness, you sound like one of the stereotyped tamil cinema villains :D

Again, chinna vayasu semma talent da! :D Semma thathuvam la pesara!

Shre, you still alive here?[/quote]

Ok let it be so .  :no: and btw i also like this website bcos there are smileys available. No need to put lol when there is  :lol:
And Shreyas is offline

dei vayavetchikittu summa irruda  :D   :wink:  :P

[hide="again the same kadi"]vayi ellai kayu :P [/hide]

Ahem this topic till today was on Olympiads and talent if i am not wrong  :D  :D   :P

out of topic: @ pardesi r u coming for a round of that game???

hey dudes i am new so plz tell me about talent and skill in the astronomy olympiad
thanx!
 :roll:

talent .....[hide="click u r bound to do so"] [size=200][color=red]0[/color][/size][/hide]
skill.........[hide="click u r again bound to do so"]not required if u have teh required talent :P  :P  :rotfl:  :rotfl: [/hide]

hey i mean resorces where i can practice or u giove me some questions

hey look u all are genuises since i am in neither of the olymiads but i still would like to tell u all that pardesi is right coz in mathematics u need to do a lot of problems no matter how much good u r at the conceptual .this not exactly holds good in physics or chemistry and we were even told this at our camp of intel international science and engineering fair coz in these competitons pure sciences revolve around our mind which we instill from square 1 of r studies .

            so i think i agree wit pardesi's point


counting
distinguishability
geometric series

[b]In how many ways can we distribute 7 pieces of taffy and 8 pieces of licorice to 5 kids, such that each kid receives exactly 3 pieces of candy.[/b]

Let me know if I posted it in the right place. Thanks :)

Since the kids don't care what type of candy they want. You just give each of them $3$ pieces of candy... what kind of question is that?

Oh... do you mean how many different [b]ways[/b] you can distribute the candy to these 5 kids?

I got a pretty large answer... so I'm not sure if it's right.

[hide]$4!\displaystyle\left(\frac{15!}{8!7!}\right)=\boxed{154440}?$[/hide]

[quote="surge"]Is each candy distinct? As in, if someone has 3 taffy candies (say $T_1, T_2, T_3$) and later has 3 other taffy candies (say $T_4, T_5, T_6$) then these would be counted as two seperate "ways".[/quote]

I don't think that would count as two seperate ways.  Otherwise, what would be the point of some of the candy being licorice and some being taffy?

[hide="Solution"]We can give each person at most three pieces of taffy, and then give the people whatever number of pieces of licorice necessary so that each person has three pieces of candy.

Ignoring the restriction that each person gets at most three pieces of taffy, the number of ways to distribute the taffy is equivalent to the number of rearrangements of TTTTTTT////, which is $\binom{11}{4}=330$.

Now, let's count the number of ways to distribute the taffy so that the first person gets more than three pieces.  We give the first person four pieces, and then we distribute the remaining three pieces in any way.  The number of ways to distribute the remaining pieces is equivalent to the number of rearrangements of TTT////, which is $\binom{7}{3}=35$.  There are also $35$ ways to distribute the taffy so that the second person gets more than three pieces, and so on.

The answer is $330-5\cdot 35=\boxed{155}$.[/hide]

[quote="aidan"]I suppose order doesn't matter in this problem? We allocate the taffy first and the licorice can occupy the spaces between the taffy later. Every three adjacent spaces represents one student. Then it would be just $15C7$ = 6435 ways right?[/quote]

You're correct in that order doesn't matter in this problem, but your method doesn't work because if you did that then you'd be taking into account the order at which candies are given.

Consider the 15 taffy's allocated to 5 kids:

TTT|TTT|TTT|TTT|TTT

Using your method of adding licorise later would produce

TLL|TLT|TLT|LLL|TTT

and

LTL|LTT|TTL|LLL|TTT

and many others as different solutions, even though each kid is getting the same number of each type of candy

[quote="surge"][quote]Ignoring the restriction that each person gets at most three pieces of taffy, the number of ways to distribute the taffy is equivalent to the number of rearrangements of TTTTTTT////, which is $\binom{11}{4}=330$. [/quote]

Huh?  :huh:[/quote]

I think matt276eagles meant the number of ways to place 4 dividers among the 7 pieces of taffy (ball and urn problem), not distinguishable permutations of TTTTTTT////.

[quote="surge"][quote]Ignoring the restriction that each person gets at most three pieces of taffy, the number of ways to distribute the taffy is equivalent to the number of rearrangements of TTTTTTT////, which is $\binom{11}{4}=330$. [/quote]

Huh?  :huh:[/quote]

Each rearrangement of TTTTTTT//// corresponds to a different way to distribute taffy.  The number of T's to the left of the first slash is the number of pieces of taffy that go to the first person.  The number of T's between the first and second slashes is the number of pieces of taffy that go to the second person, and so on.

For example, one rearrangement is TT/TT/T/T/T.  It represents giving two pieces of taffy to the first and second person, and one piece of taffy to the third, fourth, and fifth people.

As I said in my solution, this method ignores the restriction that no person can get more than three pieces of taffy.  For example, one rearrangement is TTTT/T/T/T/.  It represents giving four pieces of taffy to the first person, one piece of taffy to the second, third, and fourth people, and no taffy to the fifth person.  But no person can get four or more pieces of taffy!  That's why I subtracted certain possibilities later in my solution.

This method is very useful in certain problems.  For example, find the number of solutions in nonnegative integers to $x+y+z=10$.

[quote="Kalle"]This problem is like distributing 7 pieces of taffy to 5 children so that no children gets more than 3 taffys. This is the coefficient of $x^7$ in the series expansion of $(1+x+x^2+x^3)^5 = \frac{(1-x^4)^5}{(1-x)^5} = (1-5x^4+10x^8 +\text{higher order terms})\sum_{k=0}^\infty \binom{4+k}{k}x^k$
The coefficient of $x^7$ is $\binom{11}{7} - 5\binom{7}{3} = 155$[/quote]

I understand the method you used, but I don't see how you found the answer.  Could you explain?

Yes, the method I used is usually called using '(ordinary) generating functions'. It can be used to solve distribution problems where there is no reference of order. Where order is important consider 'exponential generating function'. I am going to explain my solution without formalism - in a stringent presentation you would need to discuss what is meant by a formal power series where we are only concerned with the coefficients of powers of x, not convergence when substituted for a real number. Either way, all multiplication/addition between series occurs as you would imagine.

I am sure you have seen that $(1+x+x^2+ \cdots + x^n) = \frac{1-x^{n+1}}{1-x}$. This is what happens in the first step. Then, it is possible to show that for every $n$ that $\frac{1}{(1-x)^n} = \sum_{k=0}^\infty \binom{(n-1) + k}{k}x^k$. This is in fact the taylor series of the left function. This second formula and expansion of $(1-x^4)^5$ (note that the series formula is also true here with n=-5) gives the last equality. After that I just take out all terms that give a coefficient of $x^7$.

$1 = x^0$ has to be combined with $k=7$ in the series to give a coefficient for $x^7$. Similarly $x^4$ needs to be combined with $k=3$ and $x^8$ is a too large an exponent. If you feel unsure about this you can just write out the multiplications/additions.

Here is a more advanced example: Find the number of non-negative integer solutions to $x + 2y + 4z =50$. The answer to this question is the coefficient of $x^{50}$ in the series expansion of  $(1+x+x^2+x^3+ \cdots)(1+x^2 + x^4 + \cdots)(1+x^4+x^8+ \cdots)$. The first paranthesis represents the value of $x$ (1=x^0 for x=0, x^1 for x=1 ..), the second for $y$ and the third for $z$. According to the series formula above with n=1 (or indeed just the geometric series) we have that this expression is equal to
$\frac{1}{(1-x)(1-x^2)(1-x^4)} = \frac{(1+x)}{(1-x^2)^2(1-x^4)} = \frac{(1+x)(1+x^2)^2}{(1-x^4)^3}$
$= (1+x)(1+x^2)^2 \sum_{k=0}^\infty \binom{2+k}{k}x^{4k}$

From here it is easy to get the coefficient of $x^{50}$ as I did in the first example.

Try to use this method to find the number of ways to choose 30 balls among a large number of red,green and blue balls where you can choose at most two red balls, an odd number of green balls and at least 6 blue balls.

[quote="matt276eagles"]This method is very useful in certain problems. For example, find the number of solutions in nonnegative integers to x+y+z=10.[/quote]

Oh thanks, I see now. Very efficient.

So this would be... ///1111111111

$\frac{13!}{3!10!}=\boxed{286}$

I'm getting something else for that one.

There are 9 places to place the first divider, and 8 places to place the second divider, and there's a duplicate of everything. So theres 9 * 8 / 2 = 36 ways?

Alternatively, if x = 1, theres 8 ways to choose y and z, if x = 2, theres 7 and so on.. so 1+2+3+...+8 = 36?

Hmm...

$C(12,2)-(3+9\cdot3)=\boxed{36}$.

I see what my previous solution does. It overcounts the 10-0-0 solutions when subtracting since I permute by 3 and there's two slashes. So it already counts 10-0-0 and 0-10-0 and then when I multiply by three it counts its 6 more times (when it only needs to count it 3 times).

[quote="matt276eagles"]Close!  You should only have two slashes, so that the 1's are divided into three groups.  Three slashes divide the 1's into four groups.

Another extension:  Find the number of solutions in [i]positive integers[/i] to $x+y+z=10$.  :)[/quote]

heh, I think I get it... so instead of making a list, I'd do this:

$(1 + x + x^2 + x^3 + ...)(1 + x + x^2 + x^3 + ...)(1 + x + x^2 + x^3 + ...) = cx^{10}$ and find c?


Can somebody give me  some links about all russian mathematical Olympiad ...
Thanks !

[quote="Russian_Man"]I know sait with ARO,but russian version, you can read russian words?:)[/quote]

Well, I can read. Can you give the link?


Let D, E, F be the midpoints of AB, BC and CA in triangle ABC. Draw tangent lines from D, E, F to the inscribed circle of triangle ABC.  Let the tangent lines intersect EF, FD, and DE at I, L, M respectively. Prove I, L, M are colinear.

but it is not true in case of an equilateral triangle.....either that or i'm misunderstanding ur question

well, equilateral triangle is a special case. I guess we shouldn't consider it here. Just consider acute triangle. I am quite sure u need to apply Menelaus theorem here in solving it. But too many lines, so I just gave up. ==T

thxxx!


In an isosceles triangle $ABC$ with the base $AB$, $O$ is the circumcenter and
$S$ the incenter. Point $M$is chosen on side $BC$. Prove that $SM \parallel AC$ if
and only if $OM$ ⊥ $BS$.

[color=darkred][b][u]Very nice and ... easy problem ![/u][/b]

In an isosceles triangle $ABC$ with the base $BC$, $O$ is the circumcenter and $I$ is the incenter. Point $M$ is chosen on side $AC$. Prove that $IM \parallel AB\Longleftrightarrow OM\perp CI$.[/color]

[color=darkblue][b]Proof.[/b] Denote the middlepoint $D$ of the base $BC$ and the projection $P$ of the point $O$ to the line $CI\ .$ Thus, $\boxed{\ OM\perp CI\ }$ $\Longleftrightarrow$ the quadrilateral $OPDC$ is cyclically $\Longleftrightarrow$ $m(\widehat{POI})=\frac{C}{2}$ $\Longleftrightarrow$ $m(\widehat{AOM})=\frac{C}{2}$ $\Longleftrightarrow$ the quadrilateral $OICM$ is cyclically $\Longleftrightarrow$ $\widehat{IMC}\equiv\widehat{IOC}$ $\Longleftrightarrow$ $m(\widehat{IMC})=A$ $\Longleftrightarrow$ $\boxed{\ IM\parallel AB\ }\ .$[/color]


A fair six-sided number cube with the digits 1-6 on its faces is rolled three times. The positive difference between the values on the first two rolls is equal to the value of the third roll. What is the probability that at least one 3 was rolled? Express your answer as a common fraction.

We notice that there are $ 15$ possibilities. Quickly listing the possibilities, we find that there are $ 7$ cases in which at least one $ 3$ was rolled. (If we think of the differences first, we can stop after going through the differences that result in $ 3$). So the probability is $ \boxed{\frac{7}{15}}$.


function
real analysis
real analysis unsolved

A sequence {f suff n} convergs to f with respect to the metric of c(X)

if and only if {f suff n} convergs to f uniformly on X

plz send related artcls,
by
guna,

I suppose $C(X)$ is the space of all bounded* continuous functions on $X$ with a uniform norm $\parallel f\parallel =\sup_{x\in X}|f(x)|$. The claim to prove is: $\parallel f_{n}-f\parallel \to 0$ if and only if $f_{n}\to f$ uniformly on $X$. The proof is just a restatement of the definition of uniform convergence.

(*) I don't know if $X$ is compact; if it is, then "bounded" is not necessary.


Does anyone have any suggestions on any books or websites that can help me prepare for the Avogadro chemistry contest?

Thanks

Avogadro is very easy if you know your chem.  Just a grade 11 chem textbook should be fine.  There might be some previous contests posted online.

If you're serious about chem contests, you should look at the CHEM 13 News Exam.  It's a bit tougher (though still quite easy) but covers harder topics (mostly grade 12).


[quote="aadil"]this thread is for the discussion of the undergoing ashes.who will win this time?[/quote]

How many people here know what the Ashes are? And what are the Ashes?This is not my question, but it may be asked by people?

[quote="aadil"]this thread is for the discussion of the undergoing ashes.who will win this time?[/quote]
I think it would have been far better had you provided some link which explained what "The Ashes" was all about; I think you have presumed that many people here (on AoPS) know about "The Ashes" series. I wholly second muks's implied suggestion. Here's a link explaining what "The Ashes" is all about.
[url]http://en.wikipedia.org/wiki/The_Ashes[/url]
Now,coming back to the original question.I think,this time around it's going to be a fairly even contest-- which is in stark contrast to the drubbing England had suffered in the last series. (that was in 2006-2007)
[url]http://en.wikipedia.org/wiki/2006%E2%80%9307_Ashes_series[/url]

Actually,I am not watching the Ashes.Still,Australia may get the better of England.


quadratics
analytic geometry
graphing lines
slope
LaTeX

Given a line: y = mx + b
And a circle: (x-h)**2 + (y-k)**2 = r**2   where (h,k) is the circle center and r is the radius

I get this far, substituting mx+b for y:

    (x-h)**2 + (mx + b - k)**2 = r**2

But how to solve this monster for x?

Or is there a better/easier approach?

It's possible to expand everything out, rearrange, and use the quadratic forumla to get \[ x\equal{}\frac{2h\minus{}2m(b\minus{}k)\pm\sqrt{(2m(b\minus{}k)^2\minus{}2h)^2\minus{}4(m^2\plus{}1)((b\minus{}k)^2\plus{}h^2\minus{}r^2)}}{2(m^2\plus{}1)}.\] I didn't bother trying to simplify this further :maybe: .

If you're trying to obtain a quick formula to memorize, I wouldn't--it is much simpler to solve this kind of problem when specific coefficients are given :) .

Isn't it amazing that something as simple conceptually as a line and circle interaction could require so many operations to calculate its motion?

Using your solution, I tried some real world circles and lines, and I wasn't getting the results I expected.  Correct me if I'm wrong, but is it possible that there's an extra 2 exponent being applied to (b-k) in the second occurrence of (b-k)?  If minus b is 2h-2m(b-k), then b squared should be (2m(b-k)-2h) squared, rather than (2m(b-k)squared-2h) squared.  Tomorrow I will try it again.  Sorry I have not learned how to do proper notation.

[u]Just a general note:[/u]

You don't always have to use "standard form" to get a general solution. It's perfectly fine to use alternate variables.

In this case, let's instead write the line in the form $ (y\minus{}k) \equal{} m(x\minus{}h)\plus{}a$. If you want to go back into slope-intercept form, we can use the substitution $ a \equal{} b\minus{}k\plus{}mh$.

Now, for ease of writing, let $ X \equal{} x\minus{}h$ and $ Y \equal{} y\minus{}k$.

Then, $ Y\equal{}mX\plus{}a$ and $ X^2\plus{}Y^2 \equal{} r^2$, which is much easier to keep track of.

When you are done with that, you can go ahead and substitute in for $ X,Y,a$ if you really want to, but chances are that for all practical purposes, you won't need to (especially for $ a$).

[quote="WillDotson"]Correct me if I'm wrong, but is it possible that there's an extra 2 exponent being applied to (b-k) in the second occurrence of (b-k)?[/quote]

Sorry Will--I just noticed that too :| . Here's what it should be then:
\[ x\equal{}\frac{2h\minus{}2m(b\minus{}k)\pm\sqrt{(2m(b\minus{}k)\minus{}2h)^2\minus{}4(m^2\plus{}1)((b\minus{}k)^2\plus{}h^2\minus{}r^2)}}{2(m^2\plus{}1)}.\]

[quote="WillDotson"]Sorry I have not learned how to do proper notation.[/quote]

Yeah, you would need to type in LaTeX code for the math stuff to show up nicely. It's actually not too hard to learn, IMO, and if you need to type math a lot, it comes in really handy.

If you just want to check whether the line intersects with the circle, there is a much easier way (besides the determinant of the quadratic)

The distance from $ (h,k)$ (your circle's centre) to the line $ mx\minus{}y\plus{}b\equal{}0$
is given by

$ \frac{|mh\minus{}k\plus{}b|}{\sqrt{m^{2}\plus{}1}}$

In general, the distance from a point $ (x_{0},y_{0})$ to a line $ Ax\plus{}By\plus{}C\equal{}0$ is

$ \frac{|Ax_{0}\plus{}By_{0}\plus{}C|}{\sqrt{A^{2}\plus{}B^{2}}}$


function
combinatorics unsolved
combinatorics

Let $ 0\leq k\leq n$ and $ m\geq 2$ be integers. Consider the $ k$-element subsets of $ \{1,2,...,n\}$; in every such subset, compute the residue of the sum of elements, modulo $ m$. Prove that if the $ m$ residues are distributed uniformly - i.e. every residue occurs exactly $ \frac {\binom{n}{k}}{m}$ times - then $ n\geq m$.


When we use the angle formula $\tan \theta = \frac{m_{1}-m_{2}}{1+m_{1}m_{2}}$, what's the difference between using $m_{1}-m_{2}$ and $m_{2}-m_{1}$?

For example when you solve the following problem, you are to consider two cases.

Find the value of $a$  for which two lines $y=\frac{1}{2}x$ and $y=ax$ make the angle of $\frac{\pi}{4}$.

Solution:

Let $m_{1}=\frac{1}{2},\ m_{2}=a$, we have $\left|\frac{m_{1}-m_{2}}{1+m\1m_{2}}\right|=\tan \frac{\pi}{4}$.

I think there's some meaning to it (but I can't remember). Something like if you use $m_{1}-m_{2}$ it means the angle going counterclockwise from $m_{1}$ to $m_{2}$...  :maybe:


$a,b,c,d$ are positive prove that :
$\frac{a}{b+2c+3d}+\frac{b}{c+2d+3a}+\frac{c}{d+2a+3b}+\frac{d}{a+2b+3c}\ge\frac{2}{3}$

It's quite easy,too.

Use Cauchy-Swacter:

$\sum\frac{a}{b+2c+3d}\sum{a(b+2c+3d)} \ge (a+b+c+d)^2$

$\leftrightarrow4\sum\frac{a}{b+2c+3d}(ab+bc+cd+da+ac+bd) \ge (a+b+c+d)^2$

Because $\frac{8}{3}\sum{ab} \le(\sum{a})^2$

We are done!


Let $ p$ and $ q$ be twin primes.  Also,

\[ a\cdot p \equiv1 (q)\\
b\cdot q \equiv 1(p)
\]

where $ a$ and $ b$ are positive and the smallest possible.  Show that $ a\equal{}b$

If by smallest possible, you mean that $ 0 < a,b < p, q$ and by $ (n)$, you mean $ \mod n$, then:

[hide="Boop"]WLOG, let $ q \equal{} p \plus{} 2$. Reducing both congruences, we have

$ a\cdot p \equiv \minus{} 2a \equiv 1 \mod p \plus{} 2 \Longrightarrow$
$ 2a \equiv \minus{} 1 \mod p \plus{} 2$,

and

$ b\cdot q \equiv 2b \equiv 1 \mod p$. Now, because $ a,b < p$,

$ 2a \plus{} 1 \equal{} p \plus{} 2$,
$ 2b \minus{} 1 \equal{} p$.

Subtracting,

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

$ \therefore a \equal{} b$.[/hide]

I'm not sure why you specified twin primes since this is true of any consecutive odd integers.

oh yea good point.  i just noticed it for twin primes because i was doing a problem


Hello. This is a topic for a desperate student. I have taken AMC A and B at two DIFFERENT test places. And then I took AIME at the school where I took AMC A. My AMC A score+AIME score is not enough for making the USAMO; However, AMC B+AIME is above the USAMO index.
Because of the school issue, my name is not on the USAMO qualifier list. So how should I report the issue??
Please help!! It is very unlucky to be in my circumstance.

The contact information for the AMC is listed on their website, http://www.unl.edu/amc.  I suggest you contact them directly.  Have all the contact information about the schools with you when you contact them (school name, teacher name, CEEB code, etc).

You might also have the test supervisor from one of the schools involved contact the AMC office on your behalf. I've done that, and it does work.

Thanks for all your suggestions. I will contact the AMC office now, see what I should do next. Hope everything turns out well.
This kind of suspense is awful.


If theta is angle in standard position and its terminal side passes through the point (-1/2, sqrt{3}/2) on a unit circle, then find a possible value of theta.

Have you even tried this problem yet? Just make a rough sketch and I think the answer will be really, really obvious. We will help you, but not to the point of writing out solutions that you can copy.

If you're posting these to challenge people, they are definitely below the level of Intermediate.

he is posting homework...it is so obvious, this is just like my algebra 2/trig book, have you noticed his average post number is almost 11 posts per day

[quote="Bruno Bonagura"]" Moderators:  beta, joml88, tetrahedr0n, white_horse_king88 "

Where are they?[/quote]

I'm right here.  I'll take care of it as best I can.

I certainly do not want to start trouble here.

If you can help, fine.  If not, no problem.

hey its $\frac{2 \pi}{3}$.

We're here to do problem solving, not answer your homework questions.


[u][b]Problem 1. (Hungkhtn)[/b][/u] [i](a). Let $a,b,c$ be three positive real numbers and $a+b+c=3$. Find the maximum value of \[P=(a^{2}+b^{2})(b^{2}+c^{2})(c^{2}+a^{2}).\] (b). Let $a,b,c,d$ be four positive real numbers and $a+b+c+d=4$. Find the maximum value of \[P=(a^{2}+b^{2}+c^{2})(b^{2}+c^{2}+d^{2})(c^{2}+d^{2}+a^{2})(d^{2}+a^{2}+b^{2}).\] (c). Solve the general problem.[/i]

PS. [b]Flove= Hungkhtn[/b], but Hungkhtn is returning.  :lol:

By EV-Theorem, we get that the maximum value in the case $n=4$ is $4^{5}$, for $a=b=0, c=d=2$.

Yes, I know this problem ca be solved by EV, but I hope if others can soved it without EV-method? After that, I will post som problems which are more difficult and non-EV.  :lol:

Dear VASC, could you check the general case? I just prove that the maximum is only attacked when there're $k$ numbers equal together, $n-1-k$ numbers equal $0$.

[u][b]Problem 2 (Hungkhtn)[/b][/u]. [i]Let $a,b,c,d$ be four non-negative real numbers and $a+b+c+d=3$.
(i). Find the maximum value of \[(a^{2}+b^{2})(b^{2}+c^{2})(c^{2}+a^{2})(a^{2}+d^{2})(b^{2}+d^{2})(c^{2}+d^{2}).\] (ii). Find the maximum value of[/i] \[\frac{1}{a^{2}+b^{2}}+\frac{1}{b^{2}+c^{2}}+\frac{1}{c^{2}+a^{2}}+\frac{1}{a^{2}+d^{2}}+\frac{1}{b^{2}+d^{2}}+\frac{1}{c^{2}+d^{2}}.\] This is the very main result I want to post here, and it's suitable with the name "for the honor". It's unable to use EV-theorem, my solution uses SMV (nSMV) theorem, and it's able to solve the general problems.

Good luck, I will return in one week.

[quote="ehku"]The result of general case is $\frac{n^{2n}}{2^{n+2}}$, using mixing variables.[/quote]

It's very amazing that mixing variables can be used here. Can you show it, my friend?

sorry for my bad knowledge, but what is EV method? :?:

I proposed the general problem, in addition to this particular case, the following result (in the second volume)

$\bigstar$
Given non-negative real numbers $a_{1},a_{2},...,a_{n}\ (n\ge 4)$ with sum $2$. Prove that
\[(a_{2}^{2}+a_{3}^{2}+...+a_{n}^{2})(a_{1}^{2}+a_{3}^{2}+...+a_{n}^{2})...(a_{1}^{2}+a_{2}^{2}+...+a_{n-1}^{2}) \le \frac{4^{n}(n-2)^{n-2}}{n^{n}}.\]
For $n=3$, the maximal value is $2$, with equality for $a_{1}=a_{2}=1,a_{3}=0.$


At the MATHCOUNTS practice, every team member hi-fives each of the other members. Altogether there are 45 hi-fives. How many team members attended the practice?

We have $ \frac{n(n\minus{}1)}{2}\equal{}45 \implies n(n\minus{}1)\equal{}90$, easily notice that $ n\equal{}10$ is the positive root, so answer is $ \boxed{10}$.

AIME15 technically is right, but I like to use this method:
45=9+8+7+6+5+4+3+2+1.
This method implies that there are 9 members, but we are forgetting 1 member, so the answer is 10. QED  :D


Can anyone give an elementary solution of the equation $x^2=2y^4-1$?

What do you mean by "elementary solution"? Solving $x$ or $y$ in the equation is very easy for example if $x,y\in \mathbb{C}$. Can you please write more formally what do you want to solve.

Well it seems to be meant in integers ;)
(and I think it appeared already)

[quote="{x}"]To make things easier, can you tell us if this equation has any non-trival solutions (i.e $x,y>1$) ?[/quote]
The only solutions of this equations are $\left(x,y\right)=(1,1),(239,13)$.
Naturally we're searching the positive integers which are solutions. ;)


geometry
3D geometry
Support
combinatorics unsolved
combinatorics

If a cube is cut into finite number of smaller cubes, prove that at least two of them must be of same size

consider all 6 faces of this cube and between the cubes who have a face sharing with one of the 6 faces consider the smallest one and name it $ a$ ,
then try to arrive to a contradiction by assumming that all cubes have different sides,
(find a cube sharing a face with one of the faces which is smaller than $ a$)

Maybe there is more then one cube in the partition  :wink: .
Find the smallest cube(and name him $ a_1$) sharing the edge with big cube lower edge. Then the top edge of $ a_1$ has the walls around it(so it is not sharing edge with big cube). From thus all cubes contacting with $ a_1$ top edge has got the whole edge sharing with $ a_1$ top edge. Among them is the smallest one named $ a_2$. Simmilary we get cubes $ a_3, a_4,...$. So we described an infinite set of cubes from our partition.
Contradiction.
P.S. Sorry for my English  :blush:

Can you go in a bit more detail with your proof?  What is the different about cubes (where there is no dissection) than squares (where there is one)?

The difference is that the smallest cube touching a face is surrounded by larger cubes, so its opposite face is the support for another partition of cubes, and so on ... It is not easy to describe in one sentence, but easy to understand once you focus on the process.

Martin Gardner, when describing it, made reference to the rhyme of fleas having smaller fleas, and so on, at nauseating  :)


calculus
integration
limit
complex numbers
complex analysis

Not very familiar with complex integration. I would like to know how to show this:

$\lim_{k \to \infty} \int_0^{-ik} v^m e^{-v} dv = \lim_{k \to \infty} \int_0^{k} u^m e^{-u} du$

where $m < 0$ and $i$ is imaginary unit.

If $m$ is an integer, then both integrals diverge (this happens for all $m\le-1$). Suppose that $-1<m<0$. The difference of the two integrals is a path integral from $k$ to $-ik$ taken along the real and imaginary axes. Suppose we add a path $\Gamma$ from $-ik$ to $k$ to complete the contour of integration. By Cauchy's theorem the integral over the closed contour is zero, so we must prove that the integral of $z^{m}e^{-z}$ over $\Gamma$ goes to zero as $k\to\infty$. And... I don't see how this could work.  :|

It's the pole (or branch point) at zero which is a problem- $\infty$ is fine. We can show that $\int_\epsilon^\infty z^{-1}e^{-z}\,dz-\int_{i\epsilon}^{i\infty}z^{-1}e^{-z}\,dz=\frac{\pi\cdot i}{2}$, but any exponent less than $-1$ breaks this.

mlok's idea does work for $-1<m<0$, provided that choose a definition for $z^{m}$ which applies on the entire first quadrant. The easiest path to use is a square- the exponential decay as the imaginary part goes to $\infty$ is the key to the estimate.


what is usually covered in the classes? i am thinking about skipping the subject next year and plan on starting to learn the material.

i have intro to geometry. will that cover most of the geometry part?

are there any good books on trigonometry? i have one but it stinks.

Intro to Geo should cover the areas. Trig, on the other hand, might be more difficult to cover. However, I hear a new Intermediate to Trig book is coming out soon  :) . If you find trig as one of your weaker areas (like me  :oops: ), then I suggest taking the course.

i was, but its on thursday, which is when my piano lessons are, so i cant take it.


For history class, we have to be a country and the countries are:
U.S., Great Britain, France, Austria-Hungary, Germany, and Russia

Our group is Germany and we have to come up with a plan that tries to avoid war from breaking out, while giving us the maximum possible benefits.
We have to discuss with other groups our proposals. (we want this while we'll give you that...etc)

Any history expert here have any ideas?

The USA joined the World War 1 later than Britain, France and Russia. So I think the best way is to have more trade communications with the USA so if the war breaks out, you won't fear the US will join the other side.

To prevent war from breaking out, avoid giving too much to Austria-Hungary.  Then they won't go begging to you for help when Franz Ferdinand gets killed, and they would be forced to seek a less violent way to deal with Serbia.  Alternatively, you could suck up to Russia so that they wouldn't declare war on you when A-H invades Serbia.  And also try to be friends with the US.

You could *secretly* build a larger navy, and hide the ships under large mats or disguise them as ocean liners.  Germany had major issues with blockades during the war, like there were starvation problems.  If war *does* break out, there wouldn't be a blockade situation.

this indeed seems like homework but I am interested

first of all, where is belgium in all of this, sure we are not 'a big player'
but what was the first country to be attacked by german troops
why did england jump in?
where did the weapon come from that gavrillo princip used? (okay this is a bit lame i admit)

i would advise to try to cut some ties with austria, or at least advise them to keep their arch duke ferdinand home

Mmm well, then were is Mexico? We didn't were a big player but we were one of the causes that the US enter in the war: The Zimberman (not sure how spell it) telegram, were Germany was telling to Mexico that if we join forces whit them, attaking the US then they'll help us to take back the territories of California and Texas which were lost in the XIX century. But the US intercepted that telegram and Mexico stayed neutral all the war

Germany tried to be independent from alliances shortly before the war, but when "everyone else" had linked into alliances, they felt exposed and fell into what "everyone else was doing."

In these days, war was considered a reasonable way for a country to expand its borders and reap the benefits of expenditures on armies, navies and the like.  Our current reluctance of major wars was not a feeling shared by leaders in Europe at the time.

Interestingly, if the Spanish flue (which started in the US and was largely initially distributed by US soldiers, many of whom died in carrying the disease to Europe) stopped what was generally thought to be the winning push by the German army in France.  This was one of those things that the best laid plans of men can never include.

Good luck, I think you need to think things through and have a reasonable rationale for what you suggest.  This is a good time to think outside of the box.

ok UPDATE on what happened:

As it was suggested, we're trading with U.S. and even have a port set up in Boston, while they access us through our northwestern border.
However, stupid FRANCE gave Austria-Hungary full access (with tarrifs though) to all their African ports. Plus Aus-Hung has 15 battleships stationed in each port. We got Morocco from France while they got Alsance (sp?) and we got Lorainne.

In case war breaks out, we hav aus-hung protecting morrocco because of their troops so I don't know if that was good.

Thanks for the advice. We still have till Wednesday to add/modify anything. So all input is appreciated.

(I think the biggest winner was Austria-Hungary)

[quote="somerandomguy"](I think the biggest winner was Austria-Hungary)[/quote]

It's hard to imagine that A-H could come out [i]worse[/i] than they did in WWI.

When do we assume this situation is?
Immediately after Franz Ferdinand's assassination?

how about some discussion on World War 2 ? i know it was disastrous for the world and the poor guys who died, but i find it interesting , speaking historically.

what i didnt understand was why did hitler attack USSR in the first place , couldnt he take them as his allies, cos stalin was totally with hitler before the attack

Do you think the americans were right in bombing japan with the nuclear bomb, as a result of anger over pearl harbor?
one might argue that it was excessive , i mean you cant compare a Nuclear strike with an ordinary bombing , but of course "The japs started it" , so what do u think?

I think their alliance was just for the time they divided poland under them.
Also their political points of view are the most opposites possible.
And Hitler wanted to battle all his (future) enemies in a blitzkrieg (in the "normal" way of war he wouldn't have any chance even against UdSSR alone, the german army wasn't that good).

The bombing...
advantage: their impact was one of the reasons why they weren't used again.
But I thinky they were an overkill of revenge (revenge is  what produces a lot of wars), the bombs produced most of their destruction in civilian areas, not military ones (in total, the cruel destruction of everything, especially civilian targets, were together with the fights the reason why the war was so bloody).


Find the solutions of the following equations:

a) $\sin x \sin 2x \sin 3x =1$
b) $\sin x \cos 4x \sin 5x = - \frac{1}{2}$

[i]Moderator Note: Please read the instructions before posting.

Trigonometry should not be posted in Getting Started.

This thread has been moved.[/i]

Well a is obviously

[hide]
$0 +2d\pi$ for any integer $d$.

[/hide]

I'm REALLY bad with trigonometry, so this might be wrong.


function
search
algebra proposed
algebra

The function $ f(n)$ is defined on the positive integers and takes non-negative integer values. $ f(2) \equal{} 0, f(3) > 0, f(9999) \equal{} 3333$ and for all $ m, n$: $ f(m\plus{}n)\minus{}f(m)\minus{}f(n)\equal{}0$ or $ 1$. Determine $ f(1982)$

and here:

[url]http://www.mathlinks.ro/viewtopic.php?t=150769[/url]


is there a (relatively) simple formula in terms of a,b,c,d and mod bd for an x satisfying

x=a mod b
x=c mod d

simultaneously? in other words a formula f(a,b,c,d)+qbd for some integer q.

also, is that fancy looking Z the set of all integers?

First part: Yes and no: we have the [url=http://en.wikipedia.org/wiki/Chinese_remainder_theorem]Chinese Remainder Theorem[/url]. The theorem states that we can find one and only one
solution modulo $ bd$ if $ (b,d) \equal{} 1$.There is an algorithm to find the solution, but no real formula. If the moduli are small, trying a few values will usually work faster than the algorithm.

Second part: $ \mathbb{Z}$ denotes the set of integers. $ \mathbb{Z}/n\mathbb{Z}$ denotes the set of integers modulo $ n$.

One "formula" is as follows:  $ x \equal{} au \plus{} cv$ where $ u$ is the unique residue class such that

$ u \equiv 1 \bmod b, u \equiv 0 \bmod d$

and $ v$ is the unique residue class such that

$ v \equiv 0 \bmod b, v \equiv 1 \bmod d$.

A "simple formula" for $ x$ therefore implies a simple formula for $ u$ and $ v$.  Can you see why it might be difficult to find such a formula (without using, say, modular inverses)?


probability
inequalities
triangle inequality

A line segment of length 6 is divided at random into 3 parts with the restriction that the longest part is at least 0.5 more than the next longest part, and that the next longest part is twice the shortest part. Find the probability that the three parts can serve as the 3 sides of a triangle.

[hide]I get $\frac{1}{11}$

If the shortest part has length $x$ the next largest must be $2x$ and the largest must be $6-3x$.

Knowing that the largest part must be at least $.5$ more than the medium side, $6-3x > 2x+.5 \rightarrow 5x < 5.5 \rightarrow x < 1.1$.

Applying the triangle inequality, $x+2x > 6-3x \rightarrow x > 1$.

$\therefore x \in (0, 1.1)$ and in order to make a triangle $x \in (1, 1.1) \Rightarrow P = \frac{1}{11}$[/hide]


Instead of a "Currently" bar, could there just be a subject description bar, sort of like the one on posts in the forum?

Can't you just use the Currently bar as a description...?


[size=150]Be  a1,a2….an >0

Be  S=∑ 1/ai

Prove that:

 S+2√S<∑ (i√bi)/bi
                      [/size]

Sorry,i mean in the sum ai instead of bi

[quote="riemann abdellah"]how do you write like that,i don't know?[/quote]
see http://www.mathlinks.ro/Forum/index.php?f=123

can anyone solve this probelm,i'm curious to see if someone can solve it

there is no bi here,it's ai,it was a mistake :lol:

but let $a_i=1$$i=1,2,\cdots,,n$ then you will see it is false....


[quote="José"]Related topic: http://www.mathlinks.ro/Forum/viewtopic.php?t=104023[/quote]
Thanks José, would you like to try this problem? :)
You seem to know this kind.

It was taken in an Argentinian Olympiad, and I didn't know how to solve it. That's why I posted it.
I asked Rust to solve it, let's see if he post a solution.

[quote="computer_nuke"]For the problem posted, I don't understand why the larges solution is n=176 (I'm refering to the problem in the link that Jose gave). Can anyone explain it?[/quote]

I asked there. Let's see the answer there, but here let's talk about this problem.

[quote="Inspired By Nature"]Find the maximum value of x such that the expression $4^{27}+4^{1000}+4^{x}$ is a perfect square.[/quote]

[hide]This can be written as $2^{54}+2^{2000}+2^{2x}$.

is this a good way to start doing the problem.[/hide]


geometry
rectangle
function
projective geometry
combinatorics unsolved
combinatorics

Consider a K by K square array of dots.  For each row and column, there are exactly N marked dots.  Furthermore, no four of those marked dots form a rectangle with sides parallel to the original square array of dots.
It is obvious that the lower bound for K is N(N-1)+1 For example, for N=3 and K=7 we have

xxxaaaa
xaaxxaa
xaaaaxx
axaxaxa
axaaxax
aaxxaax
aaxaxxa

where x is marked and 'a' is unmarked
does K=N(N-1)+1 always work?

[/tex]

Well, I'm not sure the following is exactly what you are asking for because it is easier to express upper bound for $n$ as a function of $k$ and that I do not assume there are exactly $n$ marked dots in each lines and each column.

It is not difficult to prove that if we mark more than $\frac k 2 (1+\sqrt {4k-3})$ then there is a rectangle whose vertices are marked dots and whose sides are parallel to the side of the given square.
Thus, to avoid such a rectangle, you have to mark at most $\frac k 2 (1+\sqrt {4k-3})$ dots.
This number is an integer if and only if $k = q^2+q + 1$ for some integer $q$, and in that case its value is $k(q+1)$.
If $q$ is the power of a prime, it may be proved that this bound is sharp : It comes from the existence of a projective plan on the field with $q$ elements. The $k$ lines of the plane represent the subscript for the lines of the square, and the $k$ points represent the subscripts for the columns. Then, you mark a dot if and only if the point belongs to the line. The non-existence of a rectangle is now equivalent to the following theorem of plane projective geometry :
"Two distinct lines of the plane have a unique common point".

The existennce of projective plane for the other value of $q$ is still an open question, thus the reasoning cannot be generalized.
As far as I know, for $k=43,q=6$, it is knwon that tha maximal 'theoretical'value 301 is not reached (I do not remember any reason why). I don't know if 300 can be reached.

In tha case where the above bound is not an integer, I have no idea if the integer part of it is the desired optimal value.

Pierre.


$3^{\pi}$,but my proof isn`t non-computational[actually I just checked it through computer :blush: ].Is there a nice way to prove this?

$f(x)=\frac{ln x}{x}$, then $f'(x)=\frac{1-ln x}{x^{2}}$,
which becomes zero when $x=e$ and becomes negative $x>e$.
As $\pi>3>e$, $f(\pi)<f(3)$, which means
$\frac{ln3}{3}>\frac{ln\pi}{\pi}$.
So $\pi ln3 > 3 ln \pi$, which means $3^{\pi}>\pi^{3}$.

[quote="Hawk Tiger"]use calculator.[/quote]
also ask teacher? :rotfl:


Use each of the digits 2, 3, 4, 5, and 6 exactly once and the following clues to find the number. The number is less than 50,000 and the digits are such that the 3 is next to the 2, the 3 is not next to the 4 or 5, the 6 is not next to the 4 or 5, and the 2 is not next to the 4 or 6.

$ \boxed{45236}$ is the answer, I believe. Take the clues one by one to figure out where the digits can go, and find the overlap of the valid regions.

Focus on the number with the most restrictions.  Note that 6 is not next to the 4, 5, or 2.  So 6 must be next to the 3 alone, putting 6 on one end.  It must be the ones place because the number is less than 50,000.  So we have:
_ _ _ 3 6

3 is next to 2 giving:
_ _ 2 3 6

2 is not next to 4 producing:
_ 5 2 3 6

Thus:
45236


geometry
geometric transformation
reflection
complex analysis

Let $ f$ be analytic on $ \{z: 0 < \Im z < 1\}$ and continuous on $ \{z: 0 \leq \Im z \leq 1\}$.

Suppose that $ f(z) \equal{} 0$ if $ z \in \mathbb{R}$. Show that $ f$ is identically zero.

What happens when you try reflecting this to the strip below the real axis?

Of course - the Schwarz reflection principle.


Find all the solutions of the equation \[\left(x+1\right)^y-x^z=1\] For $x,y,z$ integers greater than 1.

a solution is [url=http://www.kalva.demon.co.uk/ibero/isoln/isol003.html]here[/url]
 :)

I think it's number theory problem. :roll:

Would it be valid to invoke Mihailescu's Theorem?  It wasn't a proven result in 2000, but it is now.   :P

When I tried to see the solution, I found the link dead, so I post a new one:

Clearly $y$ divides $x$, since $z\geq2$ hence $x^2$ divides $x^z=(x+1)^y-1=Sx^2+yx$, where $Sx^2$ is the sum of all terms in $(x+1)^y$ that have exponent $2$ or larger in $x$.

Assume that $x>2$.  Therefore, either $x$ has an odd prime factor $p$, or $x$ is divisible by $2$ with multiplicity at least $2$.  Call $u\leq v$ the exponent of $p$ respectively in $x,y$ if $x$ has an odd prime factor, or $2\leqq\leqq$ the exponent of $2$ if $x$ has no odd prime factor.  Now, we may write

\[x^z=xy+\sum_{k=2}^y\binom{y}{k}x^k=xy+\sum_{k=2}\binom{y-1}{k-1}\frac{yx^k}{k}.\]

For any $k$, let $w$ be the multiplicity with which $p$ divides $k$.  Clearly the numerator of $\frac{yx^k}{k}$ expressed as an irreducible fraction is divisible by $p$ with multiplicity $ku+v-w$, and since $\binom{y-1}{k-1}$ and $\binom{y-1}{k-1}\frac{yx^k}{k}$ are integers, then $\binom{y-1}{k-1}\frac{yx^k}{k}$ will be divisible by $p$ with multiplicity at least $ku+v-w$.  Now, if $w\geq2$, then $k\geq p^w\geq2^w\geq w+2$, or $ku+v-w\geq k(u-1)+v+2\geq u+v+1$, because $k(u-1)+2\geq2(u-1)+2=2u\geq u+1$.  If $w=1$ and $p$ is an odd prime, $k\geq3$ and $ku+v-w=u+v+(2u-1)>u+v+1$.  If $w=1$ and $p=2$, then $k\geq2$, $u\geq2$, and $ku+v-w=u+v+(u-1)\geq u+v+1$ with equality iff $u=2$.  Finally, if $w=0$, then $ku+v-w\geq2u+v\geq u+v+1$.  Therefore, the sum on the RHS is divisible by $p$ with multiplicity at least $u+v+1$, but the first term is divisible with multiplicity $u+v$, hence the LHS is divisible by $p$ with multiplicity exactly equal to $u+v$, or $u+v=zu$.  Now $y\geq2^v\geq v+1$ with equality iff $v=1$, hence $v\leq y-1$, and $zu=u+v\leq u+y-1$, ie $z=1+\frac{y-1}{u}\leq y$, or $1=(x+1)^y-x^z\geq(x+1)^y-x^y\geq yx^{y-1}\geq4$, absurd.  We conclude that $x=2$.

The equation then becomes $2^z=3^y-1=(3^t+1)(3^t-1)$, where $y=2t$ is even because $x$ is even and $x$ divides $y$.  We therefore need two consecutive even numbers that are both powers of $2$, which means that $3^t-1=2$ and $3^t+1=4$, or $t=1$, $y=2$, and $2^z=3^2-1=8=2^3$, and the only possible solution is $(x,y,z)=(2,2,3)$.

This problem today is really simple considering lifting:
$v_p(a)=\alpha$ implies that $p^\alpha \mid a$ and $p^{\alpha+1} \nmid a$
It's easy to see that $z>y$
$x^z=(x+1)^y-1^y \Rightarrow$ Since $(x,x+1)=1$ and $x\mid (x+1)-1$

Then $zv_p(x)=v_p(x^z)=v_p((x+1)^y-1^y)=v_p((x+1)-1) + v_p(y)$
$\Rightarrow v_p(y)=(z-1)v_p(x)$

If $4\mid x$ then $x^{z-1}\mid y$
$\Rightarrow z>y\ge x^{z-1}$ No solutions here.

Then $v_2(x)=1 \Rightarrow z>y\ge (\frac{x}{2})^{z-1}\Rightarrow x=2$ and we solve the equation like @daniel73, so the only solution is $(x,y,z)=(2,2,3)$
$Q.E.D$

How did no one came up with this  :P.
Assume that $(x,z) \ne (2,3)$ then by Zsigmondy there exists a prime $p$ such that $p \mid x^z+1$ and $p \not \; \mid x+1$.
Thus we have that $p \mid x^z+1=(x+1)^y$ thus $p \mid x+1$ giving a contradiction. Thus $x=2$ and $z=3$.
By replacing we have: $3^y=9$ thus $y=2$ meaning that the unique sol is $(x,y,z)=(2,2,3)$
Thus we are done  :blush:

[color=purple][b]Solution 1:[/b][/color] This solution is really similar to @MathLuis's solution only because he hinted me at Zsigmondy's.

We claim that the only solution is $(x, y, z)=(2, 2, 3)$. Now for the rest of the solution assume that $(x, z) \neq (2, 3)$.
By Zsigmondy's, we have $p\mid x^z+1$. Since it $p$ is a primitive prime factor, we have $p\nmid x+1$. Now because $x^z+1=(x+1)^y$, we have $p\mid (x+1)^y$ which is a contradiction.

[color=purple][b]Solution 2:[/b][/color] Just use Catalan's conjecture and you are done.


Let $ f(x)\equal{}x^2\plus{}mx\plus{}2005$ $ (m\in R)$
and $ a,b,c\in Z$ .Prove
$ max\{\ f(a),f(b),f(c)\}\geq\frac{1}{2}$

counterexample for $ m\equal{}2n$ with $ n\in\mathbb{Z}$ and $ n^2\geq 2006$:

$ \max\{f(\minus{}n),f(\minus{}n\plus{}1),f(\minus{}n\minus{}1)\}\equal{}\max\{2005\minus{}n^2,2006\minus{}n^2\}\leq 0$


There is a square $ ABCD$, $ x$ on a side.
Put point $ P$ in the inside of the square $ ABCD$.
And, $ \overline{AP}\equal{}17$, $ \overline{DP}\equal{}20$, $ \overline{CP}\equal{}13$.
What is one side of length of the square?
[asy]pair A=(0, 3sqrt(41)), B=(0, 0), C=(3sqrt(41), 0), D=(3sqrt(41), 3sqrt(41)), P=(43/sqrt(41), 23/sqrt(41));
draw(A--B--C--D--cycle);
draw("$17$", A--P);
draw("$20$", D--P);
draw("$13$", C--P);
label("$A$", A, NW);
label("$B$", B, SW);
label("$C$", C, SE);
label("$D$", D, NE);
label("$P$", P, SW);[/asy]

[hide="Answe"]$ 3\sqrt{41}$[/hide]
I hope as possible elementary geometric solution.

hello,let $ \angle DPA\equal{}\alpha$, $ \angle CPD\equal{}\beta$, applying the theorem of cosine and solving the equation system
$ x^2\equal{}17^2\plus{}20^2\minus{}2\cdot17\cdot20\cos(\alpha)$
$ x^2\equal{}13^2\plus{}20^2\minus{}2\cdot13\cdot20\cos(\beta)$
$ 2x^2\equal{}17^2\plus{}13^2\minus{}2\cdot17\cdot13\cos(\alpha\plus{}\beta)$
we get $ x\equal{}3\cdot\sqrt{41}$.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello,let $ \angle DPA \equal{} \alpha$, $ \angle CPD \equal{} \beta$, applying the theorem of cosine and solving the equation system
$ x^2 \equal{} 17^2 \plus{} 20^2 \minus{} 2\cdot17\cdot20\cos(\alpha)$
$ x^2 \equal{} 13^2 \plus{} 20^2 \minus{} 2\cdot13\cdot20\cos(\beta)$
$ 2x^2 \equal{} 17^2 \plus{} 13^2 \minus{} 2\cdot17\cdot13\cos(\alpha \plus{} \beta)$
we get $ x \equal{} 3\cdot\sqrt {41}$.
Sonnhard.[/quote]

Yes.
But your solution has much computational complexity.

[quote="earldbest"]drop perpendiculars from P to the four sides of the square, then the result follows from the Pythagorean Theorem. :idea:[/quote]

$ \left\{\begin{array}{l}
x^2\plus{}(z\minus{}y)^2\equal{}17^2 \\
(z\minus{}x)^2\plus{}(z\minus{}y)^2\equal{}20^2 \\
(z\minus{}x)^2\plus{}y^2\equal{}13^2
\end{array}\right.
\iff x^2\plus{}y^2\equal{}17^2\plus{}13^2\minus{}20^2\equal{}58$

So,
$ \left\{\begin{array}{l}
z^2\minus{}2yz\equal{}17^2\minus{}58 \\
2z^2\minus{}2xz\minus{}2yz\equal{}20^2\minus{}58 \\
z^2\minus{}2xz\equal{}13^2\minus{}58
\end{array}\right.
\iff
\left\{\begin{array}{l}
x\equal{}\frac{z^2\minus{}111}{2z} \\
y\equal{}\frac{z^2\minus{}231}{2z}
\end{array}\right.$

So, $ \left(\frac{z^2\minus{}111}{2z}\right)^2\plus{}\left(y\equal{}\frac{z^2\minus{}231}{2z}\right)^2\equal{}58 \iff z^4\minus{}458z^2\plus{}32841\equal{}0$.
Therefore $ z\equal{}\pm\sqrt{89}, \pm3\sqrt{41}$.

Thus, $ \boxed{3\sqrt{41}}$.

Umm, I think your solution has much computational complexity, too... :(

[quote="broniran"][quote="Kouichi Nakagawa"]Yes.
But your solution has much computational complexity.[/quote] So what?[/quote]

I hope more elementary solution.
Elementary geometric solution is better :)

The simple solution that I know is as follows.

Let $ \angle PDA \equal{} \theta$.
So,
$ \cos \angle PDA \equal{} \cos \theta \equal{} \frac{x^2\plus{}111}{40x}$  by $ \triangle APD$
$ \cos \angle PDC \equal{} \cos \left(\frac{\pi}{2} \minus{} \theta\right) \equal{} \sin \theta \equal{} \frac{x^2\plus{}231}{40x}$  by $ \triangle CDP$

Hence, $ \left(\frac{x^2\plus{}111}{40x}\right)^2 \plus{} \left(\frac{x^2\plus{}231}{40x}\right)^2 \equal{} 1$  by $ \cos^2 \theta \plus{} \sin^2 \theta \equal{} 1$
Therefore $ x\equal{}\pm\sqrt{89}, \pm3\sqrt{41}$.

Thus, $ \boxed{3\sqrt{41}}$.

But, I hope the solution that is simpler than this.


limit
trigonometry
algebra
polynomial
calculus
calculus computations

\[ \mathop {\lim }\limits_{x \to 0} \frac{{1 \minus{} \sqrt {\cos x} }}
{{x^2 }}\]

Halp!ª

Just use L'Hôpital:

$ \lim_{x\to0} \frac{1\minus{}\sqrt{\cos x}}{x^2} \equal{} \lim_{x\to0} \frac{\sin x}{4x \sqrt{\cos x}}$, since $ \frac{\sin x}{x} \to 1$ and $ \cos(0)\equal{}1$, the answer is $ \boxed{\frac{1}{4}}$.

By L'Hopital, we have:

$ \lim_{x\to 0}\frac{1\minus{}\sqrt{\cos x}}{x^2} \equal{} \lim_{x\to 0}\frac{\left(1\minus{}\sqrt{\cos x}\right)'}{\left(x^2\right)'} \equal{} \frac{1}{4}\lim_{x\to 0}\left(\frac{1}{\sqrt{\cos x}} \cdot \frac{\sin x}{x}\right) \equal{} \frac{1}{4}$

[quote="uglysolutions"]Just use L'Hôpital:

$ \lim_{x\to0} \frac {1 \minus{} \sqrt {\cos x}}{x^2} \equal{} \lim_{x\to0} \frac {\sin x}{4x \sqrt {\cos x}}$, since $ \frac {\sin x}{x} \to 1$ and $ \cos(0) \equal{} 1$, the answer is $ \boxed{\frac {1}{4}}$.[/quote]

Without l'hopital please, Idk how to use it atm.
Thanks

Multiply and divide by $ 1 \plus{} \sqrt {\cos x}$ to get $ \lim_{x \to 0} \frac {1 \minus{} \cos x}{x^2} \cdot \frac {1}{1 \plus{} \sqrt {\cos x}} \equal{} \frac {1}{2} \cdot \frac {1}{2}$.

Or use $ \sqrt {1 \plus{} t} \equal{} 1 \plus{} \frac {t}{2} \plus{} O(t^2)$ to arrive at $ \sqrt {\cos x} \equal{} 1 \plus{} \frac {\cos x \minus{} 1}{2} \plus{} O((\cos x \minus{} 1)^2)$.  Then with $ \cos x \equal{} 1 \minus{} \frac {x^2}{2} \plus{} O(x^3)$ we have $ \sqrt {\cos x} \equal{} 1 \minus{} \frac {x^2}{4} \plus{} O(x^3)$.

[Edit: I was adding the Taylor approach while vazk and Kent posted.]

$ \cos x \equal{} 1 \minus{} \frac {x^2}2 \plus{} O(x^4)$ as $ x\to 0$ and $ \sqrt {1 \plus{} u} \equal{} 1 \plus{} \frac12u \plus{} O(u^2)$ as $ u\to 0.$

Hence, $ \sqrt {\cos x} \equal{} 1 \minus{} \frac {x^2}4 \plus{} O(x^4).$ This gives us that

$ \frac {1 \minus{} \sqrt {\cos x}}{x^2} \equal{} \frac14 \plus{} O(x^2)\to \frac14$ as $ x\to 0.$

Give me a choice between L'Hôpital and Taylor polynomials, and I'll go with Taylor polynomials 95% of the time. I claim that it should be possible to do this one without writing anything down - such can be the power of the Taylor polynomial perspective.

Edit: concerning that "power of the Taylor polynomial perspective," it seems that several of us can do it quickly.


Here are the official solutions of the six IMO problems from the shortlist.

Actually, there is something interesting in seeing official solutions. As you maybe know, the jury gets the shortlist and has some 24 hours (sometime less, and in Mexico definitely less) to solve ( or, at least, to value) the problems (usually 30 or more...). Therefore, solutions provided by the problem selection committee are very important. They clearly influence the vote. For instance, problem G1, was considered (according to the booklet solution) a very easy one. The IMO contestants can speak about this ;)

[quote="silouan"]Enescu do you have IMO shortlist 2005 ?From where?[/quote]He was observer for the Pakistan at IMO 2005. I also have the ISL 2005, however there will be no disclosal of this until July 2006, as the IMO regulations require.


Let $BC$ be a fixed segment. Find the locus of $A$ such that the median $AM$ of $\triangle ABC$ is the geometric mean of $AB,AC$.

P.S. Particular case at [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=54821]54821[/url] and [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=64205]64205[/url]

In more general case, M is an arbitrary point between B, C and it is wanted the locus of A, such that the segment AM to be the geometric mean of the AB, AC.

     A solution of this general problem, has already been posted in Hyacinthos forum.

http://groups.yahoo.com/group/Hyacinthos/message/13956 Please see also:

http://groups.yahoo.com/group/Hyacinthos/message/13958

    Kostas Vittas.


did neone of u guys read todays TIMES. it sez last years cut off was-
math-1
phy-4
chem-3
 :o

ada paavingala  :o  apadina kavalayeillai intha vatiyum apadi irrundhal  :D  I thought i would never clear the cut off in phy  :(  But hopefully, this wont make me complacent

what?! impossible!
anyway, times of india hasn't come to chennai (yet)

btw varun, how did u do 2day? :D

Whoa...Thats stupid!! Considering last years JEE paper was acutally easy. Can you please provide me an online link to that Varun. We dont get that paper here. :(

@hashinclude- not very bad, not very good either
but i hardly give a flying fu** abt boards :P
but im serious abt the iitjee cut off, it seems this year also we will cut offs like these. :D
@ shre - hers the link http://timesofindia.indiatimes.com/IIT-JEE_cutoffs_Single_digits_in_2007/articleshow/2872487.cms

For all the chennai people here is what Varun said
[url]http://timesofindia.indiatimes.com/IIT-JEE_cutoffs_Single_digits_in_2007/articleshow/2872487.cms[/url]

[url=http://timesofindia.indiatimes.com/IIT-JEE_cutoffs_Single_digits_in_2007/articleshow/2872487.cms]here[/url] it is  :o

shrids this link has already been posted twice

[quote="hash_include"]yeah i know.  :D  didnt see it  :blush:[/quote]

its impossible coz from what i heard from my seniors the cut off last time was 238 and the highest was 414 and how can one go down to single digits that too in an obj paper!!! unbelievalble!!! :o

why the hell was the cutoff dat low? it was also mentioned in the newspaper that the cut off wil be similar this year too, tho its not mentioned in the link. albert we're talking abt the subject cutoff

didnt you know that before? I thought we all heard of it on the day of the results itself   :o   :(

[quote="Nitishexpress"]I cant believe it.....
For Chennai stds
 [hide]That boy in DAV who got all India CBSE topper and was also one of TRS sir's Intgn test topper, He didnt get a rank.[/hide][/quote]

Oh you mean vijaykumar Hes in NUS now!!!! :maybe:

That is what i meant ....
He has missed out the cutoff in something

[quote="madness"]does clearing the cutoof mean a confirmed seat for you in iit?  :maybe:[/quote]]]
ofcourse but your rank will mateer more for the group you take!!! :!:

[quote="madness"]does clearing the cutoof mean a confirmed seat for you in iit?  :maybe:[/quote]
Comedy pannatha rohit. You need to get a decent rank.
anyway einstein alias srini karaichu kudichified maths eh?? :rotfl:

no i was saying that as a doubt to nitish's remark bcos he said vijaykumar didnt clear some cutoff  :maybe:
Maybe he cleared everything but got a huge rank   :maybe:

Well, I believe he was also the pet student of DPS sir. (Not sure) and cleared NSEC (not sure).

Oh! Well, I can't believe that so many people can get negative marks. Isn't basic school knowledge enough to get 10-20 in last years paper.

And it seems 390 might be all India 46 (Rohith Sing).... Dunno about A-C-hin Ban-sal.

Its also stupid to say that out of 3laks who wrote JEE< only 10000 odd actually got more than this. :(

The news I heard was that Vijaykumar didnt get a rank......

I feel The pattern might have been the trouble


algebra
polynomial
quadratics
algebra proposed

Find all polynomial $ P(x)\in  R$ such that:$ 1\plus{}P(x)\equal{}\frac{1}{2}.(P(x\plus{}1)\plus{}P(x\minus{}1))$

[quote="Xantos C. Guin"][hide]$ 1 \plus{} P(x) \equal{} \frac {1}{2}\cdot(P(x \plus{} 1) \plus{} P(x \minus{} 1)) \Longleftrightarrow [P(x \plus{} 1) \minus{} P(x)] \minus{} [P(x) \minus{} P(x \minus{} 1)] \equal{} 2$.

The second order finite difference is a constant: $ 2$.

Thus, $ P(x)$ is a quadratic with a leading coefficient of $ \dfrac{2}{2!} \equal{} 1$.

Therefore, $ P(x) \equal{} x^2 \plus{} ax \plus{} b$ for some constants $ a,b \in \mathbb{R}$.

Plugging in this into the original equation shows that any $ a,b \in \mathbb{R}$ yield solutions. [/hide][/quote]
I agree with you :)


A conical tank with radius 4 and height 12 is filled with liquid. How much liquid has poured from the tip of the cone if the water level is 9 feet from the tip?

[hide]Well since the new height is $\frac{3}{4}$ of the original cone each  new dimensions is $\frac{3}{4}$ of the original too.  So the volume of the new cone is  ${\frac{3}{4}}^{3}$ or $\frac{27}{64}$ the volume of the original. The volume of the original $64\pi$ so the new one is $27\pi$ thus the amout poured out is $64\pi-27\pi=\boxed{37\pi}$[/hide]