inequalities
Ring Theory
inequalities proposed
109
Belarus
algebra

Counterexample: let $b$ be very very big or very very small in comparison to $a$ and $c$.

I think you need $a \le b \le c$ or $c \le b \le a$...

[quote="K81o7"]Counterexample: let $b$ be very very big or very very small in comparison to $a$ and $c$.[/quote]

I don't see how that's a counterexample.

First case: $LHS$ tends to $\infty$, whilst $RHS$ tends to $3$.

Second case: $LHS$ tends to $\infty$, whilst $RHS$ tends to $\frac{a}{c}+\frac{c}{a}+1$.

Oh...it's $\ge$...my bad...i thought it was the other way  :D

[quote="Gibbenergy"][quote="Leon"]Let $a$, $b$, $c$ be real positive numbers. Show that \[\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq \frac{a+b}{b+c}+\frac{b+c}{a+b}+1\] [/quote]

Expanding all, we have to prove:
$-a^{2}b^{2}c+a^{3}c^{2}+a^{2}b^{3}+ab^{4}-2ab^{3}c-2ab^{2}c^{2}+b^{3}c^{2}+c^{3}b^{2}\geq 0$
$\leftrightarrow$$(a^{2}b^{3}+b^{3}c^{2}-2ab^{3}c^{2})+(a^{3}c^{2}+ab^{4}-2a^{2}b^{2}c)+(a^{2}b^{2}c+b^{2}c^{3}-2ab^{2}c^{2}) \geq 0$[/quote]

I have proved the inequality exactly in the same way  :lol:
Other solutions without expanding ?

This problems was used in Korea Mathematical Olympiad Winter School.

Expanding was the shortest and the simplest solution in there.

It's glad to me showing another nice proof using other way.

 :D

[b]Solution 1(of Nguyen Bao Khanh)[/b]
$ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \geq \frac {a \plus{} b}{b \plus{} c} \plus{} \frac {b \plus{} c}{c \plus{} a} \plus{} 1$
<=> $ M(b \minus{} c)^2 \plus{} N(a \minus{} b)(a \minus{} c)\ge 0$
In there ($ a\ge bc,$)
$ M \equal{} \frac {1}{bc}$
$ N \equal{} \frac {1}{ab} \minus{} \frac {1}{(a \plus{} c)(b \plus{} c)}\ge 0$
So ineq was proved
[b]Solution 2(of Vu Minh Thang)[/b]
We need prove that:
$ \sum \frac {a}{b} \minus{} 3 \geq \frac {a \plus{} b}{b \plus{} c} \plus{} \frac {b \plus{} c}{a \plus{} b} \minus{} 2$
or $ \sum \frac {a \minus{} b}{b} \geq \frac {(a \plus{} b)^2 \plus{} (b \plus{} c)^2 \minus{} 2(b \plus{} c)(a \plus{} b)}{a \plus{} b)(b \plus{} c)}$
<=>$ (a \minus{} b)(\frac {1}{b} \minus{} \frac {1}{c}) \plus{} (a \minus{} c)(\frac {1}{c} \minus{} \frac {1}{a}) \geq \frac {(a \minus{} c)^2}{(a \plus{} b)(b \plus{} c)}$
<=.$ \frac {(a \minus{} c)^2b(a \plus{} b \plus{} c)}{ac(a \plus{} b)(b \plus{} c)} \geq \frac {(a \minus{} b)(b \minus{} c)}{bc}$
Because $ \frac {a \plus{} b \plus{} c}{a} > 1$,we need prove that:
$ \frac {(a \minus{} c)^2b}{c(a \plus{} b)(b \plus{} c)} \geq \frac {(a \minus{} b)(b \minus{} c)}{bc}$
<=>$ (b^2 \minus{} ac)^2 \geq 0$
Done!

[quote="FOURRIER"][quote="quangpbc"]I think he want to say that "In there $ a\ge b,c$ "  :)[/quote]
Yes, I also think so

But why  :lol: ?[/quote]
Oh,sorry.My English is poor  :blush:
$ a\ge b,c$ mean suppose $ a\ge b,c$
[color=red][size=200]Sorry[/size][/color]  :blush:

[quote="chien than"][quote="FOURRIER"][quote="chien than"]
In there ($ a\ge bc,$)
[/quote]

Why?[/quote]
Sorry,it is $ a \geq b;c$ or $ a \equal{} max\{a;b;c\}$[/quote]

[size=150]But the Inequalitie is not:
[b][color=red]-Symetric
-Cyclic[/color][/b][/size]

[quote="Leon"]Let $ a, b, c$ be positive real numbers. Show that

$ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}\geq \frac {a \plus{} b}{b \plus{} c} \plus{} \frac {b \plus{} c}{a \plus{} b} \plus{} 1.$[/quote]$ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \minus{} \frac {a \plus{} b}{b \plus{} c} \minus{} \frac {b \plus{} c}{a \plus{} b} \minus{} 1 \equal{} \frac {(b \minus{} c)^2}{c(b \plus{} c)} \plus{} \frac {c(a \minus{} b)^2}{ab(b \plus{} c)} \plus{} \frac {b(c \minus{} a)^2}{a(a \plus{} b)(b \plus{} c)}\geq0.$

This problem is introduced as Example 2.2.9 on page 44 in new book :

Ji Chen, Chao-Cheng Ji, [url=http://www.2688.com/Shop/dProduct.aspx?Pid=2532746]Algebraic Inequalities (Mathematical Olympiad Proposition people Lecture)[/url], Shanghai Scientific & Technological Educational Publishing House, 2009, 225p. (ISBN 978-7-5428-4848-2/O·613)

Another related problem was proposed in Comments on page 45 :

$ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} > \frac {a \plus{} b}{b \plus{} c} \plus{} \frac {b \plus{} c}{c \plus{} a} \plus{} \frac {10}{11}.$

See also : http://www.mathlinks.ro/viewtopic.php?&t=216132

[quote=Leon]Let $a$, $b$, $c$ be real positive numbers. Show that \[\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq \frac{a+b}{b+c}+\frac{b+c}{a+b}+1\][/quote]
$\frac{ac}{b}+\frac{bc}{a}\geq 2c ,\frac{b^{2}}{c}+c\geq 2b ,b+\frac{c^{2}}{a}\geq \frac{(b+c)^{2}}{a+b}$
$\implies \frac{ac}{b}+\frac{bc}{a}+\frac{c^{2}}{a}+\frac{b^{2}}{c}\geq b+c+\frac{(b+c)^{2}}{a+b}$
$\Leftrightarrow \frac{a(b+c)}{b}+\frac{b(b+c)}{c}+\frac{c(b+c)}{a}\geq a+2b+c+\frac{(b+c)^{2}}{a+b}$
$\Leftrightarrow
\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq \frac{a+b}{b+c}+\frac{b+c}{a+b}+1$

$\frac{a}{b} + \frac{b}{c} + \frac{c}{a} \geq \frac{a+b}{b+c} + \frac{b+c}{a+b} + 1\iff b^2(b+c)(c-a)^2+a(ac-b^2)^2\ge 0$ which is true because $a,b,c>0$. Equality iff $a=b=c$

[quote=Leon]Let $a$, $b$, $c$ be real positive numbers. Show that \[\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq \frac{a+b}{b+c}+\frac{b+c}{a+b}+1\][/quote]
Let $a$, $b$, $c$ be real positive numbers. Show that \[\frac{c+a}{b}+\frac{b^2}{ca}\geq \frac{a+b}{b+c}+\frac{b+c}{a+b}+1\]
[size=50](Hung Nguyen Viet)[/size]
Solution of Hung Nguyen Viet:
$$\frac{c+a}{b}+\frac{b^2}{ca}=\frac{c^2}{bc}+\frac{a^2}{ab}+\frac{b^2}{ca}+\frac{b^2}{b^2}-1\geq \frac{(c+a+b+b)^2}{bc+ab+ca+b^2}-1=\frac{a+b}{b+c}+\frac{b+c}{a+b}+1$$

[quote=apollo][quote="Leon"]Let $ a$, $ b$, $ c$ be real positive numbers. Show that
\[ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}\geq \frac {a \plus{} b}{b \plus{} c} \plus{} \frac {b \plus{} c}{a \plus{} b} \plus{} 1
\]
[/quote]
we make the following substitution : $ x\equal{}\frac{a}{b},y\equal{}\frac{c}{b}$,we get $ \frac{c}{a}\equal{}\frac{y}{x}
,\frac{a\plus{}b}{b\plus{}c}\equal{}\frac{x\plus{}1}{y\plus{}1},\frac{b\plus{}c}{c\plus{}a}\equal{}\frac{y\plus{}z}{x\plus{}y}$
One may rewrite the inequality as$ x^{3}y^{2}\plus{}x^{2}\plus{}x\plus{}y^{3}\plus{}y^{2} \geq x^{2}y\plus{}2xy\plus{}2xy^{2}$
Apply AM-GM ; $ \frac{x^{3}y^{2}\plus{}x}{2} \geq x^2y,\frac{x^{3}y^{2}\plus{}x\plus{}y^3\plus{}y^3}{2} \geq 2xy^2,
x^2\plus{}y^2 \geq 2xy$ adding these three inequalities,we get the result[/quote]

What is the $z$?


algebra
polynomial
integration
function
quadratics
superior algebra
superior algebra unsolved

(1)
a)We consider the polynomial P(x)=x 5 . show that for every a in R* the polynomial P(x+a)-P(x) has no real roots.
b)Let P in R[x] be a polynomial of degree n>=2 with real and distinct roots. prove that there exists a in Q* s.t. the polynomial P(x+a)-P(x) has only real roots.

[i]Radu Gologan[/i]

(2)
Let A be a finite ring. Show that there exist 2 natural numbers m>p>=1 s.t. a m = a p for all a in A.

[i]Ion Savu[/i]

(3)
Let f:[-1,1]->R be a continuous function. Show that:

a)if  \int {0->1}f(sin(x+a))dx=0 for every real a then f(x)=0 for all x.
b)if  \int {0-.1}f(sin(nx))dx=0 for every integer n then f(x)=0 for all x.

[i]Dorin Andrica & Mihai Piticari[/i]

(4)
Let f:[0,00)->R be a periodic function with period 1, integrable on [0,1]. For a strictly increasing and unbounded sequence (x_n)n>=0, with x_0=0 and lim{n->00}(x_(n+1)-x_n)=0 we denote r(n)=max{k | x_k<=n}. Show that:

a)lim{n->00}(1/n)* \sum {k=1->r(n)}(x_k-x_(k-1))*f(x_k) =
\int {0->1}f(x)dx
b)lim{n-.00}(1/lnn)* \sum {k=1->n}f(lnk)/k =  \int {0->1}f(x)dx.

[i]Cristinel Mortici[/i]

[quote="grobber"](1)
a)We consider the polynomial P(x)=x 5 . show that for every a in R* the polynomial P(x+a)-P(x) has no real roots.
b)Let P in R[x] be a polynomial of degree n>=2 with real and distinct roots. prove that there exists a in Q* s.t. the polynomial P(x+a)-P(x) has only real roots.

[i]Radu Gologan[/i]
[/quote]

At this moment I only have a solution for a)
Suppose there exist a \in R* such that P(x+a)-P(x) has a real root
P(x+a)-P(x)=5ax^4+10a^2x^3+10a^3x^2+5a^4x+a^5
since a=/=0
(*) x^4+2ax^3+2a^2x^2+a^3x+a^4/5=0 has a real root
Technic for solving quadratic equation invite me to do change variable
x=z-2a/4 in (*) to eliminate term of degree 3, after computation
z^4 + a^2/2.z^2 +a^4/80=0
(z^2 +a^2/4 + a^2/(2 \sqrt 5))(z^2+a^2/4 - a^2/(2 \sqrt 5))=0

z^2= -a^2/4 - a^2/(2 \sqrt 5) impossible
z^2 = -a^2/4 + a^2/(2 \sqrt 5) ~ -0.026a^2 < 0 impossible

Do we really need something so complicated for (a)? We know that
P(x)=x 5 is a strictly increasing function, so it's injective, so for a=/=0 there is no real x s.t. P(x+a)=P(x), thus P(x+a)-P(x) has no real roots.

Often I really don't see the elementary way :D

(3) has been solved before at

[url]http://www.mathlinks.ro/phpBB/viewtopic.php?t=2545[/url]

Well, I did mention some of them might have been solved b4 on this forum. I think I've posted 2 proofs for (2) as well.


geometry
inequalities
trigonometry
ratio
limit
geometric sequence

$\Delta oa_1b_1$ is isosceles with $\angle a_1ob_1 = 36^\circ$. Construct $a_2,b_2,a_3,b_3,...$ as below, with $|oa_{i+1}| = |a_ib_i|$ and $\angle a_iob_i = 36^\circ$, Call the summed area of the first $k$ triangles $A_k$.
Let $S$ be the area of the isocseles triangle, drawn in - - -, with top angle $108^\circ$ and $|oc|=|od|=|oa_1|$,  going through the points $b_2$ and $a_2$ as shown on the picture.
(yes, $cd$ is parallel to $a_1b_1$ there)

Show $A_k < S$ for every positive integer $k$.

[img]http://www.mathlinks.ro/Forum/album_pic.php?pic_id=284[/img]

Denote the areas of the triangles $\triangle oa_1b_1, \triangle oa_2b_2, ...$ as $S_1, S_2, ...$. Then $A_k = S_1 + S_2 + ... S_k$. The base angles of the isosceles triangle $\triangle ocd$ are

$\angle ocd = \angle odc = \dfrac{180^\circ - \angle cod}{2} = \dfrac{72^\circ}{2} = 36^\circ$

Denote the intersection $e \equiv oa_1 \cap cd$. The isosceles triangles $\triangle oea_2 \cong \triangle oa_2b_2$ are congruent, having the same internal angles and the common side $oa_2$. The triangles $\triangle oa_1b_1 \cong \triangle ca_2o \cong \triangle deo$ are also congruent, having the equal sides $a_1b_1 = oa_2 = oe$ and the same angles $\angle a_1ob_1 = \angle oca_2 = \angle ode = 36^\circ$ and $\angle oa_1b_1 = \angle ca_2o = \angle deo = 72^\circ$. Therefore,

$S = S(\triangle ocd) = S(\triangle ca_2o) + S(\triangle deo) - S(\triangle oea_2) =$

$= 2 S(\triangle oa_1b_1) - S(\triangle oa_2b_2) = 2S_1 - S_2$

Using the sine theorem for the triangle $\triangle oa_1b_1$,

$\dfrac{a_1b_1}{oa_1} = \dfrac{\sin 36^\circ}{\sin 72^\circ} = \dfrac{\sin 36^\circ}{2\sin 36^\circ \cos 36^\circ} = \dfrac{1}{2 \cos 36^\circ} = \frac 1 G$

where $G = \dfrac{1 + \sqrt 5}{2}$ is the golden ratio and $\cos 36^\circ = \frac G 2$. Thus the similarity coefficient of the triangles $\triangle oa_2b_2, \triangle oa_1b_1$ is $\frac 1 G$ and their areas are in the square of this ratio, $\dfrac{S_2}{S_1} = \dfrac{1}{G^2}$. Therefore,

$S = S(\triangle ocd) = 2S_1 - S_2 = 2S_1 - \dfrac{S_1}{G^2} = S_1 \dfrac{2G^2 - 1}{G^2} =$

$= S_1 \dfrac{G^2 + (G + 1) - 1}{G^2} = S_1 \dfrac{G(G + 1)}{G + 1} = S_1 G$

where we repeatedly used $G^2 = G + 1$. Denote $A = \lim_{k \rightarrow \infty} A_k = S_1 + S_2 + S_3 + ...$ the limit of the infinite sequence $\{A_k\}$, which is increasing, as all the areas are $S_k > 0$. Earlier, we found that $\dfrac{S_2}{S_1} = \dfrac{1}{G^2}$ and similarly, we could deduce $\dfrac{S_3}{S_2} = ... = \dfrac{S_{k+1}}{S_k} = ... = \dfrac{1}{G^2}$ for all k. Thus the limit A is the sum of an infinite geometric sequence with the 1st term $S_1$ and with the quotient $q = \dfrac{1}{G^2} < 1$:

$A = \dfrac{S_1}{1 - q} = \dfrac{S_1}{1 - \dfrac{1}{G^2}} = \dfrac{S_1 G^2}{G^2 - 1} = \dfrac{S_1 G^2}{G + 1 - 1} = S_1 G$

Hence, we found that $S = A$. Since the sequence $\{A_k\}$ is increasing, $A_k < A = S$ for any k.

$\Delta oa_1b_1$ is isosceles with $\angle a_1ob_1 = 36^\circ$. Construct $a_2,b_2,a_3,b_3,...$ as below, with $|oa_{i+1}| = |a_ib_i|$ and $\angle a_iob_i = 36^\circ$, Call the summed area of the first $k$ triangles $A_k$. Let $S$ be the area of the isocseles triangle, drawn in - - -, with top angle $108^\circ$ and $|oc|=|od|=|oa_1|$,  going through the points $b_2$ and $a_2$ as shown on the picture. (yes, $cd$ is parallel to $a_1b_1$ there)
Show $A_k < S$ for every positive integer $k$.
[img]https://1.bp.blogspot.com/-Wi2fEsdckDE/XWuwIdw6hqI/AAAAAAAAKpc/_quN1EHOxYURpBfEgc8HiUN4bC-ZvD7pwCK4BGAYYCw/s400/1997%2Bflanders%2Bp3.png[/img]


Let $T$ be a triangle with side lengths $26$, $51$, and $73$. Let $S$ be the set of points inside $T$ which do not lie within $4$ units of any side of $T$. Compute the area of $S$.

if r is the inradius, A is the area, then we want

$\frac{(r-5)^{2}}{r^{2}}*A$

because we have a smaller similar triangle, and the ratio of the inradii squared is the area ratio

You mean 4 instead of 5 :wink:

I was coming to the same conclusion but couldn't decide if a better point to consider was the centroid, since it is unchanged in the transformation I believe. Could you offer proof?


Hey guys, this is just sort of a random thread where everyone can say how fast they run the mile, or some other distance.

I am on xc, but I havn't timed myself in a long time (for flat surfaces)

yesterday i did run a really uphill mile in 5:58...

oh and does anyone know if there's anything you can do about getting a little bit of chest pain when you run? it's rather annoying after a few miles on some days

It's a buildup of lactic acid that causes the pain.  Really, the only way to get rid of it is to train.

The last time I ran the mile was 2 years ago, when I was really out of shape.  I ran a 6:45.  When I was in 8th/9th grade, I was at my peak and I could probably run closer to 6.

ah, that explains the warning my coach gave not to drink milk right before or after running

*looks down*

My best mile is a little under 10 minutes. I'm more of a short distance sprinter/hurdler...

Our school's idea of a mile run is almost 2 miles, so my mile time is a bit more than 16 minutes.

Because I am athletically challenged, my mile time is 11 minutes or something.

I hate the mile. I HATE it. I have no idea why I allow people to make me run a mile. Especially since after you run around a track, you usually end up back where you started.

[quote="herefishyfishy1"] I hate the mile. I HATE it. I have no idea why I allow people to make me run a mile. Especially since after you run around a track, you usually end up back where you started.[/quote]

Exactly! I get bored with running around and around and around again and I end up counting (which isn't all that bad).

Edit: Then, I get yelled at by coaches for not being focused and for slacking.  :D

I find sprinting much more enjoyable than distance running.  I love doing 50-55m sprints, and I love doing the 200m.  In my opinion, one of the hardest runs since you need to sprint consistently for 200m which is pretty draining.

Im more of a marathoner because its a match of pacing yourself. Last time I ran a mile I did it in 5:35.

Is an 8 minute mile okay for a 12 year old girl? I've never been in a real race, I just measured it out and have run it a couple times. So I can probably go a little under 8 minutes if there was some competition. I ran a 5k in spring 07 and got 31 minutes... I was really disappointed that I didn't even get under half an hour. Maybe I'll try again sometime. :)

not great, but it is good - the freshman girls at the high school (the ones on xc anyways) run 6-7 minute miles (usually)

10 mins... I expend all my energy doing the first half of the mile in 3:45.

well, a rule of thumb is that for every second you run faster in the beginning, you lose 4 or so seconds on average (usually) from your overall time

Haven't done a mile for a while. (that rhymes xD, like 2 years, BTW)  I think my last mile time was 9-10 minutes.

About a year ago mine was 6:15, but I don't know if it is longer than a mile or shorter because it was around a parking lot.

Apparently, an 8-minute mile is average.  WR is 3:43:13 (Hicham El-Guerrouj).

Yeah, amazingly both the men's and women's miles times have nearly reached a limit...they've held their place for nearly 10 years now (99/96 respectively)...and the best part was that the men's holder is from Morocco

it's weird for me seeing all these times over 7 since I run track and xc and I'm always hearing numbers between 4 (we had a kid almost break 4 last year) and 6:30

Wow I'm bad at running. 7th Grade, best time = 6:42

i really suk at running. last mile time, a few months ago, was 9:35. best yet. at begining of last year, time was 11:14. i'm better at the shuttle run/agility...

[quote="JRav"][quote]it's weird for me seeing all these times over 7 since I run track and xc and I'm always hearing numbers between 4 (we had a kid almost break 4 last year) and 6:30[/quote]

Indeed, most top XC runners in high school should be able to run a 4:20.  Only 5 high schoolers have ever run a sub-4, with Alan Webb the American record holder.[/quote]

Yeah, that kid at my school ran a 4:06 and then ended up finishing a disappointing (at least for him he said) 8th in the nationals for xc...especially after he ran a 3200 in 15:09

But I really do think anyone who can run even under 9 minute miles should consider xc...it's real easy to train for and there's usually no try-outs.

A 15 minute 3200 m run is not that good.

sorry, I definitely forgot to edit that...

it was a 5000  :blush:

[quote="jhredsox93"]
But I really do think anyone who can run even under 9 minute miles should consider xc...it's real easy to train for and there's usually no try-outs.[/quote]

Really? ... Something I qualify for????!!!

My best mile was 5:13, and yet I still am pathetic compared to my school  :oops:
I'm more of a distance runner anyway (18:04 5k), and yes XC is definitely one of the most fun and simple sports out there. You don't need any equipment (or skill really), just training.

I am the slowest person ever, more than 13 min... :(  :(  :(

[quote="jhredsox93"]yeah as for competitive running goes...that's about in the top quarter of mile times for 8th graders (though overall, it's much much higher)

edit: actually, it's at the regional class level...so you could compete in the new england region for example and do pretty well[/quote]

where could you find this information?

well, personal experience...

and my coach also uses this thing...

google vdot...the first link that looks like a running website => click on that...then there should be something on that page (left side) that says something about ages...

btw, I screwed up earlier with that 5k time the kid from my school ran...it was actually a 14:38 or so

Just ran it today at school. Got  a 7:09. At least I improved....

I got a 7:37 at school, but it was very unoffical. hmmmmm i guess this is close. maybe 4 1/2 laps? no 4 1/4! and thsi trash can here will slow you down so well take off 10 seconds....hmmmm why dont we make u guys run all the way up to the field right before we start :rotfl:  :rotfl:  :|

My teachers did that too! Oh wait, we go to the same school.


calculus
integration
geometry
trapezoid
function
calculus computations

No, I'm not a college student, but I've been trying to figure out this integral.

$\int_{a}^{b}\left(y=mx+c\right)$

[hide="my solution"]If I'm not mistaken,

$\int_{a}^{b}\left(y=mx+c\right)=\int_{0}^{b}\left(y=mx+c\right)-\int_{0}^{a}\left(y=mx+c\right)$

and

$\int_{0}^{a}\left(y=mx+c\right)= \int_{0}^{a}(y=c)+\int_{0}^{a}(y=mx)=ac+\frac{ma^{2}}{2}$.

So

$\int_{a}^{b}\left(y=mx+c\right) =bc+\frac{mb^{2}}{2}-ac-\frac{ma^{2}}{2}=c(b-a)+\frac{m(b^{2}-a^{2})}{2}=(b-a)(c+\frac{m(b+a)}{2})$[/hide]

Thanks!

The notation is funny. I would have written this:

$\int_{a}^{b}(mx+c)\,dx$

Now, what is this? If $mx+b>0$ on $[a,b],$ then this integral is the area of a trapezoid, and we can compute it as that area. Alternatively: the integral is $(b-a)$ times the average value of the function. For a linear function, the average value is the average of the endpoint values (this is the same principle as the area of the trapezoid). The value at $a$ is $ma+c.$ The value of $b$ is $mb+c$. The average of those is $\frac{m(a+b)}2+c.$

Multiply that by $(b-a)$ to get that

$\int_{a}^{b}(mx+c)\,dx=(b-a)\left(\frac{m(a+b)}2+c\right).$

So yes, you're correct. I'm not a big fan of using antiderivatives when you can use geometry or something similar instead.

Thanks. I'm new to integrals, but, since I was done with school, I just wanted to find the integral of that function.


Feipang is very heavy. He is standing a 13 foot pole in the air but the place it was planted in was very unstable so it leans and rocks around in a perfect circle. (Chinese people should know why) However, the pole is very sturdy so it remains straight the whole time. If the farthest distance between two spots Feipang was in is 10 ft. What is the number of complete revolutions he will make before traveling 600ft?

[quote="mtms5467"][hide]The diameter of the circle is 10, so the circumference is $10\pi$. $\frac{600}{10\pi}$ simplifies to 19.0985931710274, which, to the next largest integer, is $20$.[/hide][/quote]
Close, but incorrect.
[quote][b]complete[/b] revolutions[/quote]

Yeah, you were close.

[hide]If the diameter of the circle is $10$, then the circumference is $10\pi$.  Now he will make $\frac{600}{10\pi}\approx19.099$ revolutions, so he will make $\left\lfloor19.099\right\rfloor=19$ complete revolutions.[/hide]

[quote="bpms"][quote="Ignite168"]Umm.. why was this moved to Getting Started? It's formula for diameter of a circle and division. Calculators are permitted on this btw.[/quote]
Where was this originally?[/quote]

Classroom

[hide]10pi is about 31.4, and 31.4 * 19 is about 600. so i guess it's 19.[/hide]

Once again, I must ask, what's the secret Chinese meaning?

[quote="lotrgreengrapes7926"]Doesn't fei pang mean fat and heavy, or something?[/quote]

Yeah, it means fat but they're two words that mean different kind of fat. It's hard to explain, maybe Ignite can do it himself.


[quote="hwenterprise"]Bob says that a 55 percent chance exists that he will go to the library tomorrow if it is raining at noon and a 30 percent chance if it is not raining at noon. The weather person on TV forcasts a 40 percent chance of rain at noon. Based on this information, what is the probability that Bob would go to the library?[/quote]

[hide]
0.40x0.55=0.22
0.60x0.30=0.18

0.22+0.18=0.40

44%[/hide]


function
vector
limit
real analysis
real analysis unsolved

If $ f: \mathbb{R}^n\rightarrow \mathbb{R}$  and $ |f(x) \minus{} f(y)|\leq |x \minus{} y|$ for every $ x,y\in\mathbb{R}^n$ then from Rademacher's theorem we know that $ f$ is differentiable almost everywhere.
Is it true that $ |Df(x)|\leq 1$ for almost all $ x\in\mathbb{R}^n$ ?

Straight from the limit definition, it is impossible to have $ |Df|>1$ anywhere (assuming that we use the operator norm corresponding to the metrics in each space). If $ |Df(x)|\ge 1\plus{}\epsilon$, we can find some nearby $ y$ such that $ |f(y)\minus{}f(x)|>(1\plus{}\frac{\epsilon}{2})|y\minus{}x|$.

Oh, right :)
Or like this:

$ \frac{|f(x\plus{}tv)\minus{}f(x)|}{t}\leq|v|$

$ Df(x)\cdot v\leq|v|$

and take $ v\equal{}Df(x)$, hm ?

In essence, yes. I'm thinking slightly more generally, in the context that $ f$ is a map between two normed vector spaces. In that case, that $ |Df(x)(v)|\equal{}\left|\lim_{t\to 0}\frac{f(x\plus{}tv)\minus{}f(x)}{t}\right|\le |v|$ for all $ v$ is precisely the statement that $ |Df(x)|\le 1$ in the operator norm. In the $ (n,1)$ case you initially stated, we know that $ |Df(x)(v)|$ is maximized at $ v\equal{}Df(x)^T$, and can look only at that case.


A cylindrical bar of gold that is 1.5 inches high and 0.25 inches in diameter has a mass of 23.1984 grams, as determined on an analytical balance.  An empty graduated cylinder is weighed on a triple-beam balance and has a mass of 73.47 grams.  After pouring a small amount of a liquid into the graduated cylinder, the mass is 79.16 grams.  When the gold cylinder is placed in the graduated cylinder (the liquid covers the top of the gold cylinder), the volume indicated on the graduated cylinder is 8.5 mL.  Assume that the temperature of the gold bar and the liquid are 86 degrees Farhenheit.  If the density of the liquid decreases by 1.0% for each 10. degrees Celsius rise in temperature (over the range of 0 to 50 degrees Celsius), determine
a) the density of the gold at 86 degrees Farhenheit
b) the density of the liquid at 40. degrees Farhenheit.

a)I found the volume of the gold:
(.125 in)(2.54 cm/1 in) = 0.3175 cm = 0.318 cm
(1.5 in)(2.54 cm/1 in) = 3.81 cm = 3.8 cm
V = (pi(0.318)^2)(3.8) = 1.206597185 cm^3 = 1.2 cc.

Then I divided the mass of the gold (23.1984 grams) by the volume of the gold (1.2 cc) to get 19.22630046 g/mL, or 19 g/mL in significant figures.

b)I found the mass and volume of the liquid first:
mass of liquid = 79.16 g - 73.47 g = 5.69 g.
volume of liquid = 8.5 mL - 1.2066 mL = 7.293 mL = 7.3 mL

Then, I found the density of the liquid at 86 degress Farhenheit, or 30 degrees Celsius ((86 - 32)/1.8 = 30 degress Celsius):
Density = mass/volume = 5.69 g/7.3 mL = 0.78 g/mL.

Finally, I had to adjust the density to agree with the temperature.  According the problem, "The density of the liquid decreases by 1.0% for each 10. degrees Celsius rise in temperature."  I have to find the density of the liquid at 40. degrees Farhenheit.  This is where I get lost.
But, I'll attempt it:
40. degress Farhenheit = (40. - 32)/1.8 = 4.4 degrees Celsius.
1.0% of the density (0.78 g/mL) is 0.0078 g/mL.
4.4 degress Celsius - 30 degrees Celsius = -25.6 degrees Celsius. Does that mean that there is a decrease in 25.6 degrees Celsius?
Then, I set up a proportion:
(0.0078g/mL/10 degrees Celsius) = (x/25.6 degrees Celsius).
I solved to get x = 0.019968 g/mL = 0.020 g/mL
Now, should I add that to the density?
0.78 g/mL + 0.020 g/mL = 0.80 g/mL.

I really don't know if this is correct?  Please work the problem out and see if I made any mistakes.  I'd appreciate any help!

There is a chemistry forum available for less math-specific questions.

http://www.artofproblemsolving.com/Forum/index.php?f=330


Just wondering, but what are your real names and where do you live? :lol:
How long are your names?

Not telling my name.  Nothing personal, just a precaution.  I will say, however, that I live in Virginia.

It's okay if you want to keep anything secret... I was just wondering at the variety of names at AOPS.

This is just our first name right? I live on the middle-eastern coast of the US.

That probably explains why you chose that avatar ;)

My last name alone is eighteen characters.

I'll say that my first name is Katherine and that I live in North Texas...where all the tornados are....lol

My first name is Carlos and I live in north-eastern Ohio.  Also, I highly doubt there is anyone who's first and last names are only one letter. :D

My last name rhymes with Yao Ming, the basketball player. And, my first name is on this website somewhere, and I am the only person with that first name, according to google.

My name's Ralph Nguyen. I don't really care who knows my name, because if you google it, you'll get a ton of results with it, but only one is me. ;) I live in Florida.

Wow! Florida? are you in the midst of the hurricanes? Just wondering...


$ x^2\plus{}y^2\plus{}8x\minus{}10y\plus{}40\equal{}0$
$ a\equal{}max[(x\plus{}2)^2\plus{}(y\minus{}2)^2]$
$ b\equal{}min [(x\plus{}2)^2\plus{}(y\minus{}2)^2]$

Prove that $ a\plus{}b\equal{}18$

rearranging the first equation gives $ x^2\plus{}8x\plus{}16\plus{}y^2\minus{}10y\plus{}25\equal{}0 \implies (x\plus{}4)^2\plus{}(y\minus{}5)^2\equal{}0 \implies x\equal{}\minus{}4, y\equal{}5$

i dont get what you mean by max and min though

ehh

$ x^2\plus{}8x\plus{}16\plus{}y^2\minus{}10x\plus{}25\equal{}1$

[quote="Euclid_great"]$ x^2 \plus{} y^2 \plus{} 8x \minus{} 10y \plus{} 40 \equal{} 0$
$ a \equal{} max[(x \plus{} 2)^2 \plus{} (y \minus{} 2)^2]$
$ b \equal{} min [(x \plus{} 2)^2 \plus{} (y \minus{} 2)^2]$

Prove that $ a \plus{} b \equal{} 18$[/quote]

Make the first equation into a circle form:
\[ (x\plus{}4)^2\plus{}(y\minus{}5)^2\equal{}1
\]
It is asking for the circles radius tangent internally and externally.  The distances are $ \sqrt{13}\plus{}1$ and $ \sqrt{13}\minus{}1$ so square them and add gives $ \boxed{28}$.  :maybe:


Solutions to problems in this issue should arrive no later than 29 th October, 2009. Solutions received after this date will only be considered if there is time before the publication of the solutions. Include your name, affiliation for each submission.

Each problem is given in English and Vietnamese to serve international readership.

Email: mathvn2008@gmail.com

Read more: http://mathvn.org/viewpage.php?page_id=4

How the hell am I expected to read this vietnamese website?
                                         Is it only Olympiad math or beyond?

@scary
Try reading all that stuff aloud  :rotfl:

Chi hua hua!
Hope game theory wont mind!I permit him to make fun of my Bengali,if he knows any!

I guess it's called Cantonese,eh,I dunno a bit of it,like yat,yee................... :w00tb:


calculus
integration
real analysis
real analysis unsolved

Calculate:
$ \int \frac{x^{2}(lnx\minus{}1)}{x^{4}\minus{}ln^{4}x}\,dx$

Make the substitution $ u \equal{} \frac{\ln(x)}{x}$ (!)
Thus $ \minus{}x^2du \equal{} (\ln(x) \minus{} 1)dx$, so the integral becomes
$ \int \frac{\minus{}1}{1 \minus{} u^4} du$
which is straightforward:
$ \frac{1}{u^4 \minus{} 1} \equal{} \frac{1}{2(u^2\plus{}1)} \plus{} \frac{1}{4(u\plus{}1)} \minus{} \frac{1}{4(u\minus{}1)}$
so the integral is
$ \frac{1}{2}\arctan(u) \plus{} \frac{1}{4} \ln(u \plus{} 1) \minus{} \frac{1}{4}\ln(u\minus{}1)$
Changing back to $ x$ is a mess so I'm not going to do it.

I hope it's right...  :blush:

:) You are right. The only thing you forgot is + C...


How many ways can 137 indistinguishable pieces of candy be distributed to Joe,Moe,Chloe, and Poe?

Hint: $ \dbinom{hockey stick}{identity}$

Isn't it Balls and Urns problem?

Balls = Candies and Urns = Boys

Answer: (137+4-1)C(4) = 140C4

Nope BobbyM is right. Think of it as a row of symbols.

There are 137 indistinguishable C's (candys), and 3 indistinguishable dividers D.

The number of way we can arrange these in a word is 140C3 or 140C137


[quote="drunner2007"]Compute $\sum_{n=0}^{\infty}\lfloor \frac{10000+2^n}{2^{n+1}} \rfloor$
Hint: Find a nice greatest integer function identity and telescope.[/quote]

[hide]$\sum_{n=0}^{\infty}\lfloor \frac{10000+2^n}{2^{n+1}} \rfloor=\sum_{n=0}^{\infty}\lfloor \frac{5^4\cdot 2^4+2^n}{2^{n+1}} \rfloor=\sum_{n=0}^{\infty}\lfloor \frac{1}{2}+5^4\cdot 2^{3-n} \rfloor$

$\sum_{n=0}^{\infty}2^{3-n} \rfloor=16$, so we have $\lfloor \frac{1}{2}+625\cdot 16\rfloor =\boxed{10000}$? can you do that?[/hide]

The argument is not valid but your answer is correct. That's really strange that you could do that and get the answer. Maybe there's something I'm missing...

Here's a hint. There is an identity that really simplifies it
[hide]$\lfloor x+\frac{1}{2} \rfloor=\ldots$[/hide]

another good problem that i thought i should bump  :)


geometry
geometric transformation
reflection
circumcircle
perpendicular bisector
geometry proposed

On the circumcircle $ \omega$ of triangle $ ABC$, two points $ D,E$ are situated. $ AD$ and $ AE$ intersect $ BC$ at $ X$ and $ Y$, respectively. Let $ D',E'$ be the reflections of $ D,E$ across the perpendicular bisector of $ BC$. Prove that $ D'Y, E'X$ intersect on $ \omega$.

Anybody?  :)  (My solution to this one is quite ugly...)

It is a nice consequence of the Pascal theorem.

Denote $ P \in E'X \cap \omega$, and by Pascal in the cyclic hexagon $ ADD'PE'E$, we deduce that the points $ X \in AD \cap PE'$, $ \infty \in DD' \cap EE'$ and $ Y' \in D'P \cap EA$ are collinear, i.e. $ XY' \| EE' \| DD'$.

Thus $ Y' \in BC$, therefore $ Y' \equiv Y \in EA \cap BC$, so $ D'Y$ and $ E'X$ intersect on $ \omega$.

$ YD'$ meets $ w$ at $ P'$. $ P'X$ meets $ w$ at $ E''$
$ 180=\angle DAP+\angle DD'P'=\angle DAP+\angle XYP$ (because $ DD'//XY$)
then $ A,P',Y,X$ are cyclic
then $ \angle DAE=\angle D'P'E''$
so $ EE''//DD'$
 then $ E''\equiv\ E'$ and $ P'\equiv\ P$ $ q.e.d$


I read in a textbook that the number e was found by doing a compound interest problem.

Using the equation A= P(1+R/N)NT
where:
A = total profit
P = intial amount of money
R = # of times money was compounded per year
T = time interval between when the initial amount was deposited and the total profit was withdrawn

By setting the value of P, R and T all to 1, the person found that no matter how high the value of n was increased, the amount of money earned would never exceed 2.73...... dollars (this irrational value being the number e of course).

My question is, how can the number e be solved using a method that doesn't involve trial and error or experimentation? Can the value of e be derived from some other known information or proven?

This is definetly the wrong section.
I guess your formula is $A=P\cdot \left (1+\frac{R}{n}\right )^{nT}$.
It is a known thing that $\left ( 1+\frac{1}{n}\right )^{n}\to e$.
The number $e$ can be computed through several ways but since $e$ is irrational you will never know it's value, you can only be accurate about decimals of some order of $e$. The above limit is the definition of $e$.
Also, $e=1+\frac{1}{1!}+\frac{1}{2!}+....$, from where you can compute it up to any order.
I hope I haven'nt misunderstood your question.

$e$ has also been proven to be transcendental, meaning it can not be expressed in a finite number of algebraic operations ie Its value as a decimal expansion can not be written exactly. But as stated before, there are many series and limits which evaluate to e, such as the 2 spix gave. By the way its around 2.718281828, i remember it because the 2.7 is easy, then it repeats 1828 twice :)


inequalities
calculus
derivative
induction
inequalities proposed

Let $ a_1,a_2,\ldots, a_n$ $ (n\ge 2)$ be positive real numbers. Prove that for all $ k >0$, the following inequality holds
\[ \frac{a_1}{a_2} \plus{}\frac{a_2}{a_3} \plus{}\cdots \plus{}\frac{a_n}{a_1} \ge \frac{a_1\plus{}k}{a_2\plus{}k} \plus{}\frac{a_2\plus{}k}{a_3\plus{}k} \plus{}\cdots \plus{}\frac{a_n\plus{}k}{a_1\plus{}k} .\]
:)

Each sum is over all indices.
Clearly equiv to \[ \sum \frac{a_1\minus{}a_2}{a_2} \geq \sum \frac{a_1\minus{}a_2}{a_2\plus{}k}\]
Let \[ f(k) : \equal{} \sum \frac{a_1\minus{}a_2}{a_2\plus{}k}\]
Then, \[ \minus{}f'(k)\equal{}\sum \frac{a_1\minus{}a_2}{(a_2\plus{}k)^2} \\ \equal{} \sum \frac{a_1\plus{}k}{(a_2\plus{}k)^2} \minus{} \sum \frac{1}{a_1\plus{}k} \geq 0\]
which follows after applying Cauchy Schwarz in Engel form to the sequences $ \left\{ \frac{1}{(a_2\plus{}k)^2}\right\},\left\{\frac{1}{a_1\plus{}k}\right\}$

[quote="me@home"]Each sum is over all indices.
Clearly equiv to
\[ \sum \frac {a_1 \minus{} a_2}{a_2} \geq \sum \frac {a_1 \minus{} a_2}{a_2 \plus{} k}
\]
Let
\[ f(k) : \equal{} \sum \frac {a_1 \minus{} a_2}{a_2 \plus{} k}
\]
Then,
\[ \minus{} f'(k) \equal{} \sum \frac {a_1 \minus{} a_2}{(a_2 \plus{} k)^2} \\
\equal{} \sum \frac {a_1 \plus{} k}{(a_2 \plus{} k)^2} \minus{} \sum \frac {1}{a_1 \plus{} k} \geq 0
\]
which follows after applying Cauchy Schwarz in Engel form to the sequences $ \left\{\frac {1}{(a_2 \plus{} k)^2}\right\},\left\{\frac {1}{a_1 \plus{} k}\right\}$[/quote]
That's it  :lol: The inequality is very easy but I really like it. :)

[quote="shfdfzhjj"][quote="me@home"]Each sum is over all indices.
Clearly equiv to
\[ \sum \frac {a_1 \minus{} a_2}{a_2} \geq \sum \frac {a_1 \minus{} a_2}{a_2 \plus{} k}
\]
Let
\[ f(k) : \equal{} \sum \frac {a_1 \minus{} a_2}{a_2 \plus{} k}
\]
Then,
\[ \minus{} f'(k) \equal{} \sum \frac {a_1 \minus{} a_2}{(a_2 \plus{} k)^2} \\
\equal{} \sum \frac {a_1 \plus{} k}{(a_2 \plus{} k)^2} \minus{} \sum \frac {1}{a_1 \plus{} k} \geq 0
\]
which follows after applying Cauchy Schwarz in Engel form to the sequences $ \left\{\frac {1}{(a_2 \plus{} k)^2}\right\},\left\{\frac {1}{a_1 \plus{} k}\right\}$[/quote]
It is also my proof. :)[/quote]
Anyways, If we dont want to use derivatives, the induction method will work here and the case we only need to prove is just $ n\equal{}2$. ;) :)

You really had the exact same proof? What a coincidence!

Also, I was thinking more about the problem $ \sum (a_1\minus{}a_2)f(a_2)\geq 0$ - the conditions on $ f$. My conjecture is that it should be true as long as $ f'(x)<0$. Can anyone prove it? It's a massive generalization and probably false.

[quote="can_hang2007"]Let $ a_1,a_2,\ldots, a_n$ $ (n\ge 2)$ be positive real numbers. Prove that for all $ k > 0$, the following inequality holds
\[ \frac {a_1}{a_2} \plus{} \frac {a_2}{a_3} \plus{} \cdots \plus{} \frac {a_n}{a_1} \ge \frac {a_1 \plus{} k}{a_2 \plus{} k} \plus{} \frac {a_2 \plus{} k}{a_3 \plus{} k} \plus{} \cdots \plus{} \frac {a_n \plus{} k}{a_1 \plus{} k} .
\]
:)[/quote]

Let $ a_1,a_2,\cdots ,a_n,b_1,b_2,\cdots ,b_n$ be positive real numbers with $ a_1\ge a_2\ge \cdots \ge a_n$ and

$ \sum_{i\equal{}1}^k a_i \ge \sum_{i\equal{}1}^k b_i (\forall 1\le k< n\minus{}1)$ and

$ \sum_{i\equal{}1}^n a_i \equal{}\sum_{i\equal{}1}^n b_i$.

Prove that for all $ k\ge 0$,
$ \sum_{i\equal{}1}^n\frac{b_i}{a_i} \ge \sum_{i\equal{}1}^n \frac{b_i \plus{}k}{a_i\plus{}k}$


abc is a triangle.Gis the centroid.O is any point inside triangle.prove GA^2+GB^2+GC^2=OA^2+OB^2+OC^2-2OG^2

[color=darkred][quote="mathsrox"]Let $ABC$ is a triangle. $G$ is the centroid. $M$ is any point inside triangle. Prove: \[GA^{2}+GB^{2}+GC^{2}=MA^{2}+MB^{2}+MC^{2}-3MG^{2}\][/quote][/color]
[color=indigo][b]Proof.[/b] We have:
$MA^{2}+MB^{2}+MC^{2}=\left(\overrightarrow{MG}+\overrightarrow{GA}\right)^{2}+\left(\overrightarrow{MG}+\overrightarrow{GB}\right)^{2}+\left(\overrightarrow{MG}+\overrightarrow{GC}\right)^{2}\\ =3MG^{2}+2\overrightarrow{MG}\left(\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}\right)+GA^{2}+GB^{2}+GC^{2}\\ =3MG^{2}+GA^{2}+GB^{2}+GC^{2}$[/color]

So it's also possible that the point is located outside the triangle :?:

[color=indigo]Yes, that right. And from this problem we have:

If $G$ be the centroid of triangle $ABC.\,M$ be an arbitrary point in plane, we have: $MA^{2}+MB^{2}+MC^{2}\ge GA^{2}+GB^{2}+GC^{2}$

And if $(O,\,R)$ be the circumcircle of triangle $ABC,$ we have: $3\left(R^{2}-OG^{2}\right)=GA^{2}+GB^{2}+GC^{2}$ [/color]


Παιδιά  γειά  εδώ  είμαι  και  εγώ  μετά  από  πολύ  καιρό ............... οι  πανελλήνιες  δεν  με  άφηναν  να  ασχοληθώ  πολύ  με  το  φόρουμ  και  τελευταία  καθόλου . :D
  Βάζω  μια  Γεωμετρία  που  την  θεωρώ  καλή ,  οχι  ιδιαιέτερα  δύσκολη    για  τους  μαθλινκερς ....... αλλά  λίγο  παράξενη .

Να  εξετάσετε  αν  μπορούμε  να  κατασκευάσουμε  ( γεωμετρικά )    το  ευθύγραμμο  τμήμα  a  από  δεδομένα
 τμήματα    $a_1 , a_2 ,...,a_n$   ωστε  να  ισχύει  οτι   $\frac{ 1}{a} = \frac{1}{a_1} + \frac{1}{a_2} + ... +\frac{1}{a_n}$

Εξαιρετικη η άσκηση φίλε μου . :D   .Λοιπόν θα δούμε πώς από δύο δεδομένα  τμήματα
$a,b$  μπορούμε να κατασκευάσουμε τον αρμονικό τους μέσο .
Παίρνουμε τα δύο τμήματα και τα θέτουμε επί ευθείας και παίρνουμε τμήμα $AB=a+b$ . Στη συνέχεια γράφουμε κύκλο κέντρου $O$ από το μέσον του τμήματος $AB$  ακτίνας $\frac{a+b}{2}$ .Στο σημείο διαχωρισμού των τμημάτων $a,b$  έστω $C$ (δηλαδή  $AC=a$ και $CB=b$) αυτό το σημείο φέρουμε κάθετη η οποία τέμνει τον κύκλο στο σημείο $D$ .Τότε $CD=\sqrt{ab}$ δηλαδή ο γεωμετρικός μέσος των $a,b$ .Στη συνέχεια φέρουμε την $DO$ και έστω $DM$ η προβολή του $CD$ πάνω στην $DO$ .Tότε $DM=\frac{2}{\frac{1}{a}+\frac{1}{b}}$  δηλαδή ο αρμονικός μέσος των $a,b$ .
Κάνουμε αυτή τη διαδικασία ανά δύο .Αν $n$ άρτιος δεν υπάρχει πρόβλημα.Αν $n$ περιττός τότε φτάνουμε μέχρι το  $n-1$ κάνοντας αυτή τη διαδικασία και στη συνέχεια το $\frac{1}{a_n}$ είναι γνωστό . ;) . Άρα προσθέτουμε γνωστά τμήματα άρα μπορούμε να κατασκευάσουμε το άλλο . :D


Congrats guys!

I spent my time at the Atlanta Heart Walk.

I heard Vestavia came. :maybe:

And that Eric won.

And that Walton won.

:\

gj to everyone, though.

If someone really wants the rankings, Mr. Fulton has the list and he happens to be online right now....

must get post numbers up....

Varsity Results:
Team:
1. Walton - 129
2. Northview - 120
3. Vestavia Hills - 112 (Nice! :D )
4. Chamblee - 104
5. Lassiter - 98
6. Woodward - 94
7. Brookwood - 87
8. Soli Deo Gloria - 74

Individual: (Ties broken by difficulty or tiebreaker)
1. Eric - 8
2. Allen - 8
3. Ben - 8
4. Henry - 8
5. Billy - 7
6. Wayne Zhu (VH) - 7
7. Jackson Bewley (VH) - 7
8. Maja - 7
9. Arvind - 6
10. Robert - 6
11. Abubakar (Walton for those that haven't met him yet) - 6
12. Heng - 6
13. Miles - 6
14. Jimmy - 6
15. Eric - 6

JV Results:
Team:
1. Vestavia A - 82
2. Vestavia C - 62
3. Vestavia B - 60
4. Walton - 51
5. Chamblee - 50
6. Westminster - 47
7. Northview - 44
8. Alpharetta - 38

Individual:
Non-Vestavia folks:
5. Grant Carlson (Westminster) - 5
7. Brian Basco (Westminster) - 5
8. Jonathan Austin (Northview) - 5
9. Hyun Kim (Walton) - 4
10. Murtaza Husain (Lakeside-Atl) - 4
12. David Fridorich-Keil (Westminster) - 4
14. Peter Woolfitt (Chamblee) - 4

lol, Northview JV lawst.  Enough said.

As for the individual results, who's the Eric in 15th place?  Was Eric so 1337 that he got 1st and 15th place with one test paper cuz his score had to carry over? :P


algebra
polynomial
AMC
USA(J)MO
USAMO
number theory
greatest common divisor

How do I show that for every polynomial f(x) in Z[x], there exist infinitely many primes such that there is an x such that f(x) is divisible by the prime? I'd like some hints on how to approach the problem.

See the solutions to [url=http://www.unl.edu/amc/a-activities/a7-problems/USAMO-IMO/q-usamo/-pdf/usamo2008-s.pdf]USAMO 2008 #1[/url].  The ideas can be generalized to polynomials with arbitrary constant coefficient; alternately, for arbitrary $ P(x)$ use the above lemma on $ Q(x) \equiv \frac{P(P(0) x)}{P(0)}$.

another solution:
$ deg(f) > 0$
let $ p_i \equal{} i^{th}\ prime\ number$ and $ A\equal{}max(d: \ |f(d)|\equal{}1)$
$ \forall n\in\mathbb{N}: \ gcd(\prod_{k \equal{} 1}^{|P(0)| \plus{} A \plus{} n \plus{} 1}p_i,P(\prod_{k \equal{} 1}^{|P(0)| \plus{} A \plus{} n \plus{} 1}p_i) \equal{} 1$
then $ \exists p\ prime$ such that $ p > p_{|P(0)| \plus{} n \plus{} 1}\ and\ p|P(\prod_{k \equal{} 1}^{|P(0)| \plus{} n \plus{} 1}p_i)$
so $ sup(\{p\ prime: \ p|f(x) ,\ for\ some\ x\in\mathbb{Z}\}) \equal{} \plus{} \infty$ gives $ \{p\ prime: \ p|f(x) ,\ for\ some\ x\in\mathbb{Z}\}$ is infini.


Prove that for all positive reals x, y and z we have

(x + (y + z^(1/4))^(1/3))^(1/2) :ge: (xyz)^(1/32).

This is another oneliner :mrgreen:

here you can find a solution!!!

http://www.mathlinks.ro/phpBB/viewtopic.php?t=748

cheers! :D :D

Yeah I know Lagrangia but I posted them such that people who want to see & solve some nice inequalities can work on it ... I don't know whether it will help to post just a link to the solution ... :? ?


analytic geometry
graphing lines
slope
arithmetic sequence

Ok i forgot to ask on forum before i closed the window but the question was something like this

The 3rd term of an arithmetic sequence is 13 and the eighth term is 38, find the product of the 1st and tenth terms.

I used slope and junk to figure out it was 144, but it ended up taking me 25 sec.

FASTER WAY NEEDED

Thanks in Advance  :D

here's how i do the problem

8th term, 38
3rd term, 13

difference of 25 and there are 5 terms in between
therefore, the common difference is +5
the first term would be 13-10 and the 10th term is 38+10
therefore, the answer would be $ 3 \times 48 \equal{} 144$
believe me, it's pretty fast

$ 3\cdot 43$? I believe you made a mistake and should be $ 3\cdot 48$.

oops
really srry
false information =)
i'll change that
but now that the question has been answered, plz drop this thread to prevent spam from coming in

[quote="isabella2296"]You could just find the common difference and proceed from there. It took me about 12 seconds.[/quote]
What do you mean by common diifference

In an arithmetic sequence, you add the same number over and over again, so two consecutive numbers has a common difference.

For example, 1, 3, 5, 7, 9, etc., has a common difference of 2.


search
inequalities
inequalities unsolved

We want to prove $(x+y+z)[\frac{1}{(x+y)^{2}}+\frac{1}{(y+z)^{2}}+\frac{1}{(z+x)^{2}}]\geq 9/4$
and why we only need to prove $\sum\frac{yz}{x(y+z)^{2}}\geq \frac{9}{4(x+y+z)}$?

[quote="Twins"]We want to prove $(x+y+z)[\frac{1}{(x+y)^{2}}+\frac{1}{(y+z)^{2}}+\frac{1}{(z+x)^{2}}]\geq 9/4$
[/quote]
Are you sure? Please search for : Iran 1996  :wink:

i think that $(x+y+z)$ should be replaced by $(xy+yz+zx)$ to Iran 96.
N.T.TUAN [b]see ur pm please[/b]. :D

Don't know what the initial one is supposed to mean, but note that at least $\sum\frac{yz}{x(y+z)^{2}}\geq \frac{9}{4(x+y+z)}$ is equivalent to Iran96 - just substitute $a^{2}=\frac{yz}x$ etc. which yields $x=bc$ etc.  :)

[hide="hopefully this works"]
\[(ab+bc+ca)\sum_\text{cyc}\frac{1}{(a+b)^{2}}=\sum_\text{cyc}\left( \frac{c}{a+b}+\frac{ab}{(a+b)^{2}}\right)\]
Multiplying by $4(a+b)(b+c)(c+a)$, we get
\[\sum_\text{cyc}\left( 4c(c+a)(c+b)+\frac{4ab(c+a)(c+b)}{a+b}\right)\]
\[=\sum_\text{cyc}\left( 4c^{3}+4c^{2}a+4c^{2}b+4abc+\frac{4a^{2}b^{2}}{a+b}+4abc(a+b+c)\cdot \frac{1}{a+b}\right)\]
But AM-HM gives $2(a+b+c)\sum_\text{cyc}\frac{1}{a+b}\geq 9$. Also $\frac{4a^{2}b^{2}}{a+b}=ab(a+b)-\frac{ab}{a+b}(a-b)^{2}$, so the above is at least
\[\sum_\text{cyc}\left( 4c^{3}+5c^{2}a+5c^{2}b+10abc-\frac{ab}{a+b}(a-b)^{2}\right)\]
which we must show is at least $9(a+b)(b+c)(c+a)=\sum_\text{cyc}(9c^{2}a+9c^{2}b+6abc)$. So our inequality is reduced to
\[\sum_\text{cyc}\left( 4c(c-a)(c-b)-\frac{ab}{a+b}(a-b)^{2}\right) \geq 0\]
The left hand side of the above can be expressed as
\[\sum_\text{cyc}\left( 2a+2b-2c-\frac{ab}{a+b}\right) (a-b)^{2}\]
To show that this is nonnegative, we can WLOG set $a\geq b\geq c$ and then expand $(a-c)^{2}$ as $(a-b)^{2}+(b-c)^{2}+2(a-b)(b-c)$, so it turns out to equal
\[a\left( 2+\frac{a}{a+b}+\frac{a}{a+c}\right) (a-b)^{2}+c\left( 2+\frac{c}{b+c}+\frac{c}{a+c}\right) (b-c)^{2}\]
\[+2\left( 2(a-b)+c+\frac{c^{2}}{a+c}\right) (a-b)(b-c)\]
which is nonnegative... assuming I didn't mess up on the algebra anywhere... :|
[/hide]


What are some good topics of research in mathematics for an amateur (the type of topics that might be researched by students in a program such as RSI)?

do I have to be a math genius to get into RSI?? is just being a USAMO qualifier who does many other activities enough??

[quote="t0rajir0u"]Those are two different questions :P  The research conducted at RSI is done under talented mentors and many of the students there have some research experience, so I don't know if it's right to call them amateurs.  They tell me one of the '08 kids was published in AMM :)  So which question do you want to emphasize?[/quote]
When I say amateurs, I mean in the realm of research itself, comparing the brilliantly talented students at a program like RSI to those professors that have been researching for years.
What type of topics to people research in pure mathematics in RSI?


search
linear algebra
matrix
function
geometry
inequalities
complex numbers

Let $n$ be given. Find the maximum value of $\prod_{1\leq i>j\leq n}{|z_i-z_j|}$ when all these complex numbers are on the unit circle.

You have to pose $z_i=e^{i\theta_i}$ then if $f(\theta_1,...,\theta_n)=\prod_{1\leq i>j\leq n}{|z_i-z_j|}$ search how to have $(\frac{\partial f}{\partial \theta_1},...,\frac{\partial f}{\partial \theta_n})=(0,...0)$ .You can have it for $z_i$ forming a regular polygon.
Now the difficult point is to show there is no other way of having this egality.
In fact if you look to the Hessian matrix you can easily prove it is definite negative in the points forming a regular polygon.
I had difficulties to conclude from here, I supposed I found an other point where the gradient was 0 and by studying the function on a straight line from this point to a regular polygon one I drew a contradiction (showing the first point couldn't be a maximum).
I do not exactly remember it but I think it worked.

So... any other thoughts on this problem?

[quote]Now the difficult point is to show there is no other way of having this egality. [/quote]

You should be able to prove this directly with geometry. Unfortunately I suck at geometry so I don't know how.

[i]Sometimes, knowing who posted a problem can be a hint. :)[/i]

Consider the Vandermonde matrix $ V \equal{} (z_j^{i\minus{}1})_{1 \leq i, j \leq n}$
Then notice that the product we're trying to maximize is nothing but $ |\det V|.$
Finally, remember [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=78303]Hadamard's inequality[/url] to get that $ |\det V| \leq n^{n/2},$ where the equality is attained for the Vandermonde matrix of the $ n\minus{}$th roots of unity.
Therefore,
\[ \max_{(z_1, \ldots, z_n) \in \mathbb{U}^n} \prod_{1\leq i>j\leq n}{|z_i\minus{}z_j|} \equal{} n^{n/2}.\]


[quote="perfect628"]Let $ \parallel{x}\parallel$ be the distance from $ x$ to the nearest integer to $ x$. Evaluate $ {\parallel\pi\parallel}^{2}\plus{}\frac{{\parallel2\pi\parallel}^{2}}{2}\plus{}\frac{{\parallel4\pi\parallel}^{2}}{4}...$.[/quote]

Dude this problem is actually pretty tough if you don't have those steps given in that team round

[hide="hint"]for 0<x<1/2, biject the ||x|| function to a more convenient one, by finding relations that the ||x|| function satisfies.[/hide]

Hmm.  Could someone post the relevant team questions?  This seems like one of those "tricky non-intuitive lemma" questions.

Here it is, but it is quite easy with it, as perfect628 mentioned. These are reworded to avoid copyright issues.

Define $ f(x)\equal{}\sum_{k\equal{}0}^{\infty}\frac{\parallel 2^{k}x\parallel ^{2}}{2^{k}}$.

(i,ii,iii) Compute $ f(x)$ for a few values of $ x$ between 0 and 1 inclusive, and sketch a graph. From this, conjecture a formula for $ f(x)$ valid between 0 and 1 inclusive, and compute $ f(1998/1999)$.
(iv) Show that $ f(1\minus{}x)\equal{}f(x)\forall x\in\mathbb{R}$ and $ f(2x)\equal{}2f(x)\minus{}2x^{2}\forall x\in[0,\frac{1}{2}]$.
(v) Show that for rational values of $ x$, the two equations uniquely determine $ f(x)$. (Harder: solve them over the reals)
(vi) Show that the conjectured formula for $ f(x)$ satisfies the two equations, show that $ f(x)$ is periodic with period 1, and compute $ f(\pi)$.

I vote we slaughter perfect628 for calling this trivial  :lol:


group theory
abstract algebra
superior algebra
superior algebra unsolved

Suppose that $G$ is a group of order $24$ such that none of its $p$-Sylow for all $p$ is normal. Prove that $G$ is isomorphic to $S_{4}$. Does this work for other values as well (I mean, can we find such a characterization of $S_{n}$ for larger $n$? I don't know that).

I've struggled to find at least one such larger $n$, but I couldn't  :D  Well, I haven't looked up groups tables though ...

The original question is easier: since $G$ has no normal $3$-Sylow, it must have exactly $4$ of them (by Sylow's theorem); now $G$ acts by inner automorphisms on the set of its $3$-Sylows. The induced homomorphism $\rho: G \rightarrow S_{4}$ must be injective by Sylow's theorem (all $3$-Sylows being conjugated). Whence $\rho$ is bijective.

@julien_santini:
Perhaps you should explain a little more explicit, why this homomorphism has to be bijective: Since all 3-sylows are conjugated, the image must be a transitive subgroup of $S_{4}$.
There is no such transitive subgroup with two elements, so $A_{4}$ can't be the kernel. There is no transitive subgroup with 6 elements, so $V_{4}$ can't be the kernel either. Hence the kernel is trivial.

Yes, that's the idea. However, I tried the same approach for $S_{5}$, but it becomes immediately very cumbersome.

This can't work for any $n>4$.
If $n>4$, both $S_{n}$ and $A_{n}\times \mathbb{Z}_{2}$ have this property. The number of groups with this property proliferates rapidly as $n$ increases.

I was wondering, in order to prove that $S_{4}$ has no transitive subgroup of order $6$, is there a quicker way than saying that it must contain a cycle and a transposition, and dealing manually with the two cases ?


Let ABC be a triangle such that
            \sum  m_a = ( \sqrt 3)/2*(a+b+c).
   Prove that prod(m_a -  (\sqrt 3)/2*a)=0.

In fact,we have [tex] \Delta ABC [/tex] is equilateral :)
We can caculate [tex] m_a+m_b+m_c, m_am_b+m_bm_c+m_cm_a [/tex] and[tex] m_am_bm_c [/tex],so we have done.

I don't think it's right.
 If a<=b<=c then the triangle just needs 2b 2=c 2 +a 2 .


Let n be a positive integer.
Consider T(n) is the set of all positive divisors of n.
S(n) is a subset of T(n) satisfied : For any distinct couple (m,n) in S(n) then max(n,m) is not divided by min(m,n).
Find the largest of |S(n)|. :D
Note:I want to find another solution not used De bruijn's theorem.Does it exit? :maybe:

maybe you can consider the poset of the divisors of n and we are loocking for the maximun size of an antichain=M, by dilworths theorem M=the minimun number of chains wich are disjoint and their union is all the poset.

But does that help? I mean, is there any reason why finding the minimal number of chains which cover $T(n)$ is easier?

This has a nice geometric representation, but or the case when $n$ has more than three prime factors you can't imagine it any more :).

If $n=p_1^{a_1}\ldots p_k^{a_k}$, then $T(n)$ is the set of all lattice points in $k$ dimensions which are contained in the parallelepiped of corners $(0,0,\ldots 0)$ and $(a_1,a_2,\ldots, a_k)$. We need to find the maximal set of points s.t. no point has all components greater than those of another point. Intuitively, they should lie somewhere close (as close as possible) to the hyperplane passing through $(a_1,0,0,\ldots,0,0),\ (0,a_2,0,\ldots,0,0),\ldots,(0,0,0,\ldots,0,a_k)$.

I admit, this is not really helpful :).


[quote="shencaili"]Let a=b=1,c=0[/quote]
$a,b,c\in R^{+}$

I remember that (as a case in my inequality posted before)

\[\sum a^{4}bc-\sum a^{3}b^{3}=(a^{2}-bc)(b^{2}-ca)(c^{2}-ab).\]

Therefore, if $a\ge b\ge c$ then

\[\sum a^{4}bc \ge \sum a^{3}b^{2}\Leftrightarrow b^{2}\le ca.\]

\[\sum a^{4}bc \le \sum a^{3}b^{2}\Leftrightarrow b^{2}\ge ca.\]

[quote="me@home"]I don't get it, its true by Muirhead. THere shouldn't be an issue with negatives...[/quote]

...how?  Neither sequence majorizes the other (unless what I know as Muirhead is a weaker version...)


calculus
integration
logarithms
trigonometry
calculus computations

Compute $ \int_{0}^{\infty}\frac{\ln^{n}(t)}{1\plus{}t^{2}}\, dt$ for all $ n\in\Bbb{N}$ .

[hide="Possible approach"]Start by making $ t =\tan x$, and so $ I =\int^{\infty}_{0}\frac{\ln^{n}(t)}{1+t^{2}}\,dt =\int^{\pi/2}_{0}[\ln(\tan x)]^{n}\,dx =\int^{\pi/2}_{0}[\ln(\sin x)-\ln(\cos x)]^{n}\,dx$

$ =\int^{\pi/2}_{0}\left[\sum_{k=0}^{n}(-1){}^{k}{}^{n}_{k}C(\ln\sin x)^{n-k}(\ln\cos x)^{k}\right]\,dx$

For $ n$ odd, this integral is then zero, because $ \int^{\pi/2}_{0}(\ln\sin x)^{k}\,dx =\int^{\pi/2}_{0}(\ln\cos x)^{k}\,dx$ and $ \int^{\pi/2}_{0}(\ln\sin x)^{a}(\ln\cos x)^{b}\,dx =\int^{\pi/2}_{0}(\ln\cos x)^{a}(\ln\sin x)^{b}\,dx$. For $ n$ even things get more complicated and the integral equals something like

$ I = 2\int^{\pi/2}_{0}(\ln\sin x)^{n}\,dx-2^{n}_{1}C\int^{\pi/2}_{0}(\ln\sin x)^{n-1}(\ln\cos x)\,dx+...$.

[/hide]

Nice trial, Carcul.  :wink:

The solution can be a bit simpler.

I only seem to be able to work out an answer for odd $ n$. Substitute $ t\equal{}\frac{1}{x}$.
You get that the integral is equal to $ \int_{0}^{\infty}\frac{(\minus{}\ln(t))^{n}}{1\plus{}t^{2}}$ which in the case of odd $ n$ implies that the integral is $ 0$ as it is equal to negative one times itself.

Yes, for odd $ n$ the integral is zero. But what's up if $ n$ is even?

Hi there!

That's my first problem. So don't be too rude  :wink:  I couldn't solve it either, but maybe the following could help
We substitute $ u \equal{}\ln t$ and so we get using $ du \equal{}\frac{1}{t}dt$

$ \int_{0}^{\infty}\frac{\ln^{n}t}{1\plus{}t^{2}}\, dt \equal{}\int_{\minus{}\infty}^{\infty}\frac{u^{n}}{1\plus{}t^{2}}\, t\, du \equal{}\int_{\minus{}\infty}^{\infty}u^{n}\,\frac{e^{u}}{1\plus{}e^{2u}}\, du \equal{}\frac{1}{2}\int_{\minus{}\infty}^{\infty}u^{n}\,\frac{2}{e^{u}\plus{}e^{\minus{}u}}\, du \equal{}$

$ \frac{1}{2}\int_{\minus{}\infty}^{\infty}\frac{u^{n}}{\cosh u}\, du$

As above for odd $ n$ we get

$ \frac{1}{2}\int_{\minus{}\infty}^{\infty}\frac{u^{n}}{\cosh u}\, du \equal{}\frac{1}{2}\int_{\minus{}\infty}^{0}\frac{u^{n}}{\cosh u}\, du\plus{}\frac{1}{2}\int_{0}^{\infty}\frac{u^{n}}{\cosh u}\, du \equal{}$

$ \minus{}\frac{1}{2}\int_{0}^{\infty}\frac{u^{n}}{\cosh u}\, du\plus{}\frac{1}{2}\int_{0}^{\infty}\frac{u^{n}}{\cosh u}\, du \equal{} 0$

For even $ n$ we get $ \int_{0}^{\infty}\frac{u^{n}}{\cosh u}\, du$ which I cannot compute either  :|

Yes, for even $ n$ the integral $ I_{n}\equal{}\int_{0}^{\infty}\frac{\ln^{n}t}{1\plus{}t^{2}}dt$ agrees with $ \int_{0}^{\infty}\frac{t^{n}}{\cosh t}dt$ .

It is always possible to write $ I_{n}$ as rational number times power of $ \pi$ .

Via residue theorem, using the contour $ \Omega\equal{}\{z: r<|z|<R, Im(z)>0\}$ we get
$ I_{n}\equal{}\minus{}\frac{1}{2}\sum_{k\equal{}1}^{n}{n\choose k}(i\pi)^{n\minus{}k}I_{n\minus{}k}\plus{}\frac{i^{n}\pi^{n\plus{}1}}{2^{n\plus{}1}}$,
$ I_{0}\equal{}\frac{\pi}{2}$.

According to this recursion formula $ I_{1}$ would be $ \minus{}\frac{\pi}{4}\plus{}i\frac{\pi^{2}}{4}$ . But actually it's $ 0$ .

Sorry, the recursion is
$ I_{n}\equal{}\minus{}\frac{1}{2}\sum_{k \equal{} 1}^{n}{n\choose k}(i\pi)^{k}I_{n\minus{}k}\plus{}\frac{i^{n}\pi^{n\plus{}1}}{2^{n\plus{}1}}$,
$ I_{0}\equal{}\frac{\pi}{2}$.

This recursion formula is correct.  :lol:

Can you give the explicit formula $ I_{n}\equal{} |E_{n}|\left(\frac{\pi}{2}\right)^{n\plus{}1}$ , too?

Whereas $ E_{n}$ are the so called Eulerian numbers.

Your recursion would show $ E_{2n}\equal{} 1\minus{}\frac{1}{2}\sum_{k \equal{} 0}^{n\minus{}1}{2n\choose 2k}\frac{2^{2n}}{2^{2k}}E_{2k}$ .

$ \int_{0}^{\infty}\frac{t^{\alpha\minus{}1}}{1\plus{}t^{\beta}}dt \equal{}\frac{\pi}{\beta}\csc\frac{\alpha\pi}{\beta}$

Set $ \beta \equal{} 2$ and differentiate $ n$ times with respect to $ \alpha$ . After that set $ \alpha \equal{} 1$ .

$ \int_{0}^{\infty}\frac{\ln^{n}t}{1\plus{}t^{2}}dt \equal{}\frac{\pi}{2}D_{\alpha}^{n}\sec\left(\frac{(\alpha\minus{}1)\pi}{2}\right)\Big|_{\alpha \equal{} 1}\equal{}\frac{\pi}{2}|E_{n}|\left(\frac{\pi}{2}\right)^{n}$

[quote="Dr. Doe"]Compute $ \int_{0}^{\infty}\frac{\ln^{n}(t)}{1\plus{}t^{2}}\, dt$ for all $ n\in\Bbb{N}$[/quote]
and so, is it also possible to do :|

$ \int_{0}^{\infty}\frac{\ln^{n}\; (1\plus{}t^{2})}{1\plus{}t^{2}}\;dt$

for all $ n\in\mathbb{N^{\plus{}}}$


i used red and green in my solution to number 5 (the colors are essential to my solution), but i forgot faxes arent colored :blush:
so do I have to resend it?

Better hope the words in your solutions are sufficiently clear.  Send a copy by snail mail tomorrow that we can check against the fax in case it's not clear.  (Make a copy to keep for yourself, first, of course.)


The greatest sum that can't be achieved with an unlimited amount of a's and b's when a and b are relatively prime is ab-a-b.  Can anyone explain to me why this works?  Also is there any rule that you know  of for 3 numbers a, b, and c? (Greatest sum that can't be achieved)

Some people call this the "chicken mcnuggest theorem".  Showing why this works involves setting up a table of remainders when all the natural numbers are divided by 'a' and another table of remainders for all the natural number divided by 'b'.

The Intro to Number Theory class here covers this theorem as well as broader material related to Diophantine equations.

The problem for three numbers is very difficult (i.e. the formula is not known).

I feel like some of these questions get asked so often they should end up in the FAQ ...

Intro to Number Theory covers quite a bit of diophantine if my memory isn't failing me.

Argh...must review transcripts this summer...

[quote="TripleM"][quote="MCrawford"]Some people call this the "chicken mcnuggest theorem".  [/quote]

Quite a few more call it the "chicken mcnugget theorem".

;)[/quote]

And quite a few more call it "what is this strange fact on middle school math competitions that none of my teachers know how to prove?"

And they usually spell it correctly.

heck i just called it spiffin-thing-to-know-for-MC theorem...nobody had a problem with it except the coaches who spent forever on guess and check when I just plugged in our nice lil integers and got an answer... :-D


integration
calculus
calculus computations


[quote="risinglegend"]#1. What are the four consecutive numbers whose sum is 50?

#2. If a number is trebled, itr gives the same result as when 28 is added to it. What is the number?

#3. When a number is added to another number 5 times as large, the result is 24. What is the first number? :?:[/quote]

[hide="Number One"]

Formula: x + (x + 1) + (x + 2) + (x + 3) = 50

= 4x + 6 = 50

4x = 44

x = $\frac{44}{4}$ = 11

x = 11

x + 1 = 12

x + 2 = 13

x + 3 = 14

[/hide]

#3

x + 5(x) = 24
6x = 24
x = 4

What's trebled mean?

i oso don't noe wats trebled : ( :blush:

Okay, I got it. Trebled = tripled. I thought you made a typo, but i just learned a new vocabulary.

3x = x + 28
2x = 28
x = 14

you're so cool,do you know that?
i think you are really fall in love with math(?) don't you?
it is very beautiful.because i see you enjoy any problems even an obvious problems and even very hard problems,that's good.
 :D

[quote="risinglegend"]#1. What are the four consecutive numbers whose sum is 50?

#2. If a number is trebled, itr gives the same result as when 28 is added to it. What is the number?

#3. When a number is added to another number 5 times as large, the result is 24. What is the first number? :?:[/quote]

Click Mr. Green to reveal hidden content!

[hide=" :D "]1.)
x + (x+1) + (x+2) + (x+3) = 50
4x = 44
x = 11
The consecutive numbers are 11, 12, 13, and 14.

2.)
3x = x + 28
2x = 28
x = 14

3.)
x + 5x = 24
6x = 24
x = 6[/hide]

[quote="Toby"]#1.  11,12,13,14[/quote]

Solutions are good. tHat's why you should provide one.

[quote="risinglegend"]#1. What are the four consecutive numbers whose sum is 50?

#2. If a number is trebled, itr gives the same result as when 28 is added to it. What is the number?

#3. When a number is added to another number 5 times as large, the result is 24. What is the first number? :?:[/quote]
[hide="1"]
1+2+3+4=10, and 10*4=40, so the answer is 11,12,13, and 14[/hide]
[hide="2"]
$3x=x+28$ Simplify and you get $2x=28$, so the number is $\framebox{14}$[/hide]
[hide="3"]
$x+5x=6x=24$ therefore, x is 4[/hide]


The non-negative reals $a, b, c, d$ satisfy $a^5 + b^5 \leq 1, c^5 + d^5 \leq 1$. Show that $a^2c^3 + b^2d^3 \leq 1$


analytic geometry
graphing lines
slope
conics
parabola
calculus
derivative

Can anyone derive a formula that gives the slope of the tangent, for arbitrary values of x, of a parabola of the form Ax^2 + Bx + C = 0,  without calculus?  Show your work.

Well without calculus, it's not going to be precise.  You'll have to approximate it.  Slope: $ m\equal{}\frac{f(b)\minus{}f(a)}{b\minus{}a}$.

couldnt you do something with the directrix/focus?

Well...

[hide]Let $ A\equal{}x$ and $ B\equal{}x\plus{}\epsilon$.  So we have:
$ \frac{a(x\plus{}\epsilon)^2\plus{}b(x\plus{}\epsilon)\plus{}c\minus{}(ax^2\plus{}bx\plus{}c)}{(x\plus{}\epsilon)\minus{}x}$
$ \frac{ax^2\plus{}2ax\epsilon\plus{}a\epsilon^2\plus{}bx\plus{}b\epsilon\plus{}c\minus{}ax^2\minus{}bx\minus{}c}{\epsilon}$
$ \frac{2ax\epsilon\plus{}a\epsilon^2\plus{}b\epsilon}{\epsilon}$
$ 2ax\plus{}a\epsilon\plus{}b$
Taking the limit of this as $ \epsilon\to 0$, we get $ \boxed{2ax\plus{}b}$.[/hide]

[quote="zolojetto"]derivatives are calculus right???[/quote]
yes. with dy/dx, etc...

Let me put this in math: $ \mathrm{Derivatives}\in\mathrm{Calculus}$ but $ \mathrm{Calculus}\notin\mathrm{Derivatives}$.  Usually, we split calculus into differential and integral calculus.  Differential calculus is concerned with finding derivatives, that is, the slope of the tangent line.  We can use derivatives to approximate change in many different ways.  Integrals are a bit tricky but equally important.

Working out the slope of a secant line can be non-calculus, but

once any poster shows work with the limit in it for the slope

of the tangent line, then it is in the calculus realm.

** EDIT **

Working out the slope requires calculus to make it rigorous, I think

If you throw a ball straight down into a parabola, it will always bounce through the same point, maybe that works?

You can't. Without calculus... -_-... That's the point. lol.


mipos kseri kanis pu mporo na vro 8emata paleoteron diagonismon tis ma8imatikis eterias gia ma8ites dimotiku?????

thanks....

 :D

Ράλλη δες στη διεύθυνση
 http://www.de.sch.gr/~amakrid/emethessalonikis/uemata%20prvtobaumias/uemata%20diagvnismvn.htm

και

http://www.emepne.gr

και αυτή

 http://users.chan.sch.gr/atzanidaki/

eyxaristo poli niko!!!!! :thumbup: ayto akrivos eyaxna........

 :D

Δές και εδώ....
http://users.sch.gr/atmatzidis/


There are n people in a room, and n sheets of paper, each with the name of a distinct person on it. The papers are then distributed to the n people (one per person) such that no one receives the paper with his or her name on it. How many ways are there to do this?

Try using the principle of inclusion-exclusion.

First count the number of derrangements for a single person.  Multiply this by n and adjust for the overlap.  You'll arrive at the formula given in

http://mathworld.wolfram.com/Derangement.html


quadratics
conics
parabola
function
geometry
geometric transformation

[quote="happyme"]Suppose all the three points $(-2,10)\text{,}(1,-8), \text{and}(4,10)$ lie on the graph of $y=ax^{2}+bx+c$. What is $abc$?[/quote]

[hide="a common way when random 3 points are given"]From substitution,

$10=4a-2b+c$ ...(1)
$-8=a+b+c$ ...(2)
$10=16a+4b+c$ ...(3)

$(1)-(2) \rightarrow 18=3a-3b$ ...(4)
$(3)-(4) \rightarrow-18=-15a-3b$ ...(5)

Substitute $-3b=18-3a$ from (4) to (5) and obtain
$-18=-15a+18-3a$, and therefore $-36=-18a$ and $a=2$.

It follows that $b=\frac{1}{3}\cdot (6-18)=-4$

Then it follows that $c=-8-2+4=-6$.

Therefore $abc=\boxed{48}$.[/hide]

[hide="a quicker way in this case"]Quickly notice that it is a parabola with vertex at $(1,-8)$ (it is because the function values at -2 and 4 are equal - then the vertex is in the between, which is $(1,-8)$).

Translate vertex to the origin - then the points are $(-3,18)$, $(0,0)$, and $(3,18)$
This translated parabola is thus $y=kx^{2}$, while we find that $k=2$.

Therefore the original function before the translation is
$y=2(x-1)^{2}-8$
$y=2x^{2}-4x+2-8$
$y=2x^{2}-4x-6=ax^{2}+bx+c$

Therefore $abc=2 \cdot-4 \cdot-6=\boxed{48}$.[/hide]


Hi!
Prove that any linear operater of a rank m is the summ of the m operaters each of rank 1.

Do you assume that both relevant vectorspaces have a finite dimension?

And are you familiar with this theorem : if $T$ is a linear operator from $A$ to $B$ then $dim( A)= dim(Im(T))+dim(ker(T))$

Let $T$ map $V$ to $W$, and suppose $T$ has rank $m$. Let $w_{1},w_{2},\dots,w_{m}$ be a basis for the image of $T$. Extend these $w_{i}$ to a basis for $W$, and let $\pi_{i}$ be projection onto the $i$-th basis element in $W$ ($\pi_{i}(\sum_{j}c_{j}w_{j})=c_{i}w_{i}$). These $\pi_{i}$ are linear maps of rank 1, and $\sum_{i=1}^{m}\pi_{i}$ restricts to the identity on $T(V)$.
Now $T=\sum_{i=1}^{n}\pi_{i}\circ T$, and we are done.

This argument doesn't need finite-dimensionality, only finiteness of rank.


$ a+b=ab \leftrightarrow a+b-1=ab-1\leftrightarrow (a-1)(b-1)=1$.
If $ b=1 \rightarrow a+1=a$ -false$ \rightarrow b!=1$. So, $ a=\frac{b}{b-1}$.
So, $ ab=\frac{b^{2}}{b-1}$ But $ ab$ has to be an integer so $ b-1|b^{2}\rightarrow b-1|1$. As$ b!=0$ $ b=2 \rightarrow a=2$.
$ 4$ is the only integer value that $ ab$can assume, for any $ a, b$ real numbers.

LaTeX hint: \neq gives $ \neq$.  We can get double bars on the arrows by capitalizing the first letter, e.g. \Leftrightarrow $ A \Leftrightarrow B$, \Longrightarrow $ A \Longrightarrow B$, etc., instead of \longrightarrow $ \longrightarrow$.

[quote="Fabiomr1"][quote="t0rajir0u"]dina013:  $ b$ is not necessarily integer.

[hide="Solution"] In other words, what is the smallest integer $ n$ such that $ x^{2}-nx+n = 0$ has two distinct real roots.  Well, the discriminant is just $ n^{2}-4n$, which we want to be positive, which occurs when $ n < 0, n > 4$.  So depending on what you mean by "smallest," we can have $ n = \boxed{-1, 5}$. [/hide][/quote]

hey you forgot that a.b = a + b[/quote]

[hide="No,"] he's saying $ ab=a+b=n$ and using the polynomial that has $ a,b$ as its roots (with Vieta's formula).[/hide]

[quote="Fabiomr1"]Be $ a$ and $ b$ two real and distinct numbers. $ a+b = ab$ and $ ab \neq 0$. What is the smaller integer value that $ ab$ can assume?[/quote]
Okay .. may be I am missing something very trivial but ...
Assume zillion   is the largest  integer you know .. so $ -zillion$ is the  smallest integer  (at least of the numbers you know ..)

Now  $ a=(1/2)(-zillion+(\sqrt{zillion^{2}+4 \ zillion})$
and $ b=(1/2)(-zillion-(\sqrt{zillion^{2}+4 \ zillion})$

Now, (a,b)  are real  , distinct, ... and a+b = ab , and ab = - zillion  which is not anything like zero .... etc... :ninja:

[quote="t0rajir0u"]I think he means smallest [i]positive[/i] integer value, which would just be $ 5$.[/quote]
:) But ...are you really sure he is [i]positive[/i] about that ..:)


who you think is gonna win?
i say Rafeal Nadal or Coria
and for the women, Clijsters or Justine H.H.
but i'm cheering for Agassi, Sharapova, Nadal,Vaidisova, and Myskina

i say :
mens:Gaudio, Coria, or Nadal
womens: Henin-hardene, Clijesters, or maybe Sharapova
Nadal's on fire coming off a win at Monte Carlo which is clay and he has like a 20 match win streak.
well we'll see who wins! :lol:
Does anyone play tennis?
I do 5 days a week and i'm in the top 20 in my state.

i cheer for roddick cause he can't win anything :rotfl: and he's american :rotfl: and i like cheering for underdogs. same with agassi, but he's also really old and fit so i cheer for him also.

i'd like to see a nadal vs. federer final again cause the last time they met in the finals, it was a great match (nadal won first 2 sets, federer stormed for win). nadal might be able to beat federer on clay.

wow math92, thats very impressive. I play 3 times a week and I have no idea what my rank in my state is. :D

Men: Federer Def. Nadal/Safin(Latter would be a revenge  :lol: )
Woman: Justin Henin-Herdenne Def. Sharapova/S. Williams

Junior: Kim(KOR) def. Young(USA) (I watch a lot of Junior)

People are just not focusing on Federer at all ever since the Austrailian Open loss to Safin. I expect him to come and win his first grand slam title this year and look on towards Wimbledon.

Arg, French Open finals are right when I have exams. Awesome.

What about Lleyton? I heard he wasnt gonna play or something..but im not sure

[quote="yif man12"]Hewitt is out.[/quote]

 :censored:

i don't think dent is playing either (for refrence, dent is the 3rd best american player) hewitt is out as well with an injury he suffered at home

i wonder why the french open and wimbledon are so close together? :?:  :?:  :?:

[quote="juicybooty911"]i don't think Federer or Sharapova or A-Rod can win, maybe QF or SF
i play 7 times a wk, live in a apt, so got my own court :D(i'm the only one who plays tennis in the apt)
last yr i was like 400 in the state, but only played 3 tournaments. and this yr i decided not to play any tournament :(
math92 what state are u in? i'm in GA[/quote]
i'm in Maryland.  I play about 2-3 tournaments a week.   I don't have a tennis court at my house, but i play on the school tennis team and i also go to a clinic.  You find out your state ranking on http://www.usta.com but you have to be registered and play in a tournament.  I don't think Federer will have that much of a chance, maybe he'll get to the QF.  He's not that good on clay, same with all the Americans.

[quote="tetrahedr0n"]I seriously hope Federer wins. I mean, the man started his season 40-2. It only makes sense for him to win the Roland Garros and complete the career slam.[/quote]

it'd make sense, but it rarely happens. federer would've gotten the grand slam last year but he failed on the french. clay is less about winners, powerful serves, etc. and more about cutting down on mistakes and hustle, hustle, hustle.

[quote="juicybooty911"]how can you play 2-3 tournament a wk? :huh:   do you mean you play tournament everyday?
i had a 14's ranking, but now i'm 15 and played no tournament, so i don't have one.[/quote]sorry i meant 2-3 tournaments a month

nadal will break guga's record for sure.. he is 19!!

Being young has nothing to do with seeting record.

[quote="juicybooty911"]O yeah,
Naaaadaaaaaaaaaalllllllllllllllllll
 :winner_first:
i predict that Nadal gonnna win the Big W w/o dropping a set. :D  :D  :D[/quote]

Somehow, I doubt that. Especially if he faces Federer again.

[quote="manuel"]I predict nadal is going to lose around the third round in wimbledon..
federer all the way without droping a set....(maybe tim henman can make him a good match)[/quote]

I could see that happening. Henman's actually had a pretty good career record against Federer. Must be that serve+volley thing...

Henman always seems like he loses in the early rounds in all four majors though.

he got to the semis of last year french open though. and it is his weakest surface..

And lost in the first round this year. It'd be cool to see him win at Wimbledon, since englishmen haven't had a winner there in a loooong time.

Yeah, but I'm pretty sure Federer can make pretty quick work of anyone he faces; only if he has some catastrophic game like against Nadal will the possibility of losing even exist. I'd like to see him do it without dropping a set - that'd be quite fantastic indeed. But I'd also like to see Roddick DO SOMETHING for once. Oh well I guess I can wait for hard court season to come around...

Out of the people that have had World #1 ranking in their career, Roddick is having the worst downfall right now. He went all the way from #1 to #5, and still somehow managed to get #2 seed (because Lleyton was absent, Nadal was too young, and French just hate Safin :lol: ), then he loses in the 3rd round. If he doenst do well in Wimbledon, I bet he will face another slidedown in the ranking.

Lleyton made the finals of the last 2 majors he played, he will do the same in Wimbledon.

See.. Nadal just loss with an unseeded player at Queens.. thats a grasscourt tournament...
 i think he wont get to the quaterfinals in wimbledon.

[quote="manuel"]I predict nadal is going to lose around the third round in wimbledon..
federer all the way without droping a set....(maybe tim henman can make him a good match)[/quote]

Henman WILL lose, no matter what, he just can't win the Big W
i say henman in the QF,(wont be surprised if he lose in the 3rd)


complex numbers
superior algebra
superior algebra unsolved

In a ring of Quaternion where all coefficient is from ring of integer modulo p prove that there is no proper ideal.

In general, the quaternion algebra over a field $ k$ is central and simple, the latter being what you want. As finite fields have trivial Brauer group, you can prove that these quaternions are isomorphic to the ring of $ 2 \times 2$ matrices.
For a more elementary proof, show that ideals must be inveriant under conjugation. Thus if an element is in an ideal, so is it's "real" part (exceptions at $ p\equal{}2$). But that easily gives the result.


$a,b,c,m,n>0$ prove :
\[\frac{a}{mb+nc}+\frac{b}{mc+na}+\frac{a}{ma+nb}\geq\frac{3}{m+n}\]

See http://www.mathlinks.ro/Forum/viewtopic.php?t=1565 .

  darij

Mmmm , it as simple and good problem. A Hint:

  Multiply the terms of the  first quotient by $ a $ , the second with $  b $ , the third with  $ c $ and use a special form of  $BCS$ ( Engels form : [b]a^2/x + b^2/y +c^2/z >= (a+b+c)^2/(x+y+z)[/b] )

But $\ (a + b+ c )^2 $  $\geq $ $3(ab + bc+ca)^2$

I hope that someone will coplete this easy solution. Of course you can use BCS from the beginnig , but it needs some techniq with roots.

[u]Babis[/u]

Darij , Sorry!

   You sent your post 2 seconds before me and I did not see it !!!


logarithms
number theory open
number theory

Prove that $lim_{n \to \infty}S(A^{n}) = \infty$, where $S(A^{n})$ is the sum of the digits of $A^{n}$ and $gcd(A,5)=1$

It is true forall $A\not =10^{n},n=0,1,..$. But my proofe is long.
If A=1, then $(A,5)=1$, but these statement is not true.

[quote="harazi"]Rust, is your proof elementary, I mean without use of linear forms in logarithms estimations? For even $A$ there is a short and beautiful proof, but for $A$ odd...[/quote]
It is elementary. I proof it a few years ago. Can you post short and beautiful proof?

I will post it tonight, now I'm quite busy. But I'm very interested in your elementary approach.

Ok, here it goes:
  Theorem: for any even $A$ which is not multiple by $10$, there exists a constant $C$ (actually, $C$ does not depend on $A$) such that $s(A^{n})\geq C\ln n$ for all $n$.

    The proof is due to Schinzel. Let $a_{0}=0$ and $a_{n+1}=1+[a_{n}\log_{2}(10)]$. It is clear that there exists $C$ not depending on $A$ such that $a_{n}<C^{n}$ for all $n$ and also that this sequence is increasing (strictly). Suppose that $k\geq a_{n}$ and write $A^{k}$ in base 10 as $c_{0}+c_{1}\cdot 10+...$. Clearly, for all $j\leq n-1$ we have $2^{a_{j}}|A^{k}$. Also, we clearly have $2^{a_{j}}$ divides $c_{0}+c_{1}\cdot 10+...+c_{a_{j}-1}10^{a_{j}-1}$. The definition of $a_{i}$ implies therefore that at least one of the digits $c_{a_{i-1}},...,c_{a_{i}-1}$ is nonzero. Thus at least $n$ digits of $A^{k}$ are nonzero and thus $s(A^{k})\geq n\geq C'\ln k$ for some constant $C'$.

It is very nice! you can see for A=2 in mathlinks contest

Let $A=A_{1}10^{k}, \ 10\not |A_{1}.$ Because $S(A^{k})=S(A_{1}^{k})$ sufficiently consider only case k=0 ($10\not |A$). Method harasy works, when $2|A,5\not |A$ or $5|A, 2\not |A$  ($10\not |A$).


Find $ y\in \mathbb{Q}$ such that $ [x\plus{}y]\equal{}\left[\frac{x}{y}\right]\minus{}\left[\frac{x}{2y}\right]\ ,\ (\forall)x\in \mathbb{R}$

[hide]Since the identity must hold for any real $ x$, it must hold for $ x \equal{} (2n \minus{} 1)y$ where $ n\in\mathbb{N}$:

$ [2ny] \equal{} 2n \minus{} 1 \minus{} n \plus{} 1\iff n\leqslant 2ny < n \plus{} 1\iff {1\over 2}\leqslant y < {1\over 2} \plus{} {1\over 2n}$

But as this must hold for arbitrarily large $ n$, we conclude that the only possible value for $ y$ is $ y \equal{} {1\over 2}$

Now we must prove $ \left[x \plus{} {1\over 2}\right] \equal{} [2x] \minus{} [x]$ for all real $ x$, but this is easily shown by splitting the cases $ n\leqslant x < n \plus{} {1\over 2}$ and $ n \plus{} {1\over 2}\leqslant x < n \plus{} 1$

Therefore $ y \equal{} {1\over 2}$.[/hide]


I want to see different solutions. If it was posted again please give a link

$a_1,a_2,a_3 >1$ and $a_1+a_2+a_3=S$ and $\dfrac{a_i^2}{a_i-1} >S$

Prove that $\dfrac{1}{a_1+a_2}+\dfrac{1}{a_2+a_3}+\dfrac{1}{a_1+a_3}> 1$


IMO Shortlist
factorial
Inequality
IMO Longlist

Prove that
\[\frac{1}{3}n^2 + \frac{1}{2}n + \frac{1}{6} \geq (n!)^{\frac{2}{n}},\]
and let $n \geq 1$ be an integer. Prove that this inequality is only possible in the case $n = 1.$

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]  :)

TRIVIAL.

A formula, more than well-known and more than straightforward to prove, states that

$1^2+2^2+...+n^2 = \frac13 n^3 + \frac12 n^2 + \frac16 n$.

Hence,

$\frac13 n^2 + \frac12 n + \frac16 = \frac{\frac13 n^3 + \frac12 n^2 + \frac16 n}{n} = \frac{1^2+2^2+...+n^2}{n}$.

On the other hand,

$\left(n!\right)^{\frac{2}{n}} = \sqrt[n]{\left(n!\right)^2} = \sqrt[n]{\left(1\cdot 2\cdot ...\cdot n\right)^2} = \sqrt[n]{1^2\cdot 2^2\cdot ...\cdot n^2}$.

Hence, the inequality we have to prove transforms into

$\frac{1^2+2^2+...+n^2}{n} \geq \sqrt[n]{1^2\cdot 2^2\cdot ...\cdot n^2}$,

what is clear from the AM-GM inequality. Equality holds only if the numbers $1^2$, $2^2$, ..., $n^2$ are all equal, what is only possible for n = 1.

  Darij

Sorry for bumping an old thread, but I wanted to post another solution :D
[Hide=similar solution] Observe that
$\frac{1}{3}n^2+\frac{1}{2}n+\frac{1}{6}$
$\geq\frac{1}{4}n^2+\frac{1}{2}n+\frac{1}{4}$
$=\left(\frac{n+1}{2}\right)^2$
$=\left(\frac{1+2+\ldots+n}{n}\right)^2$
By AM-GM on the part inside the brackets, $=\left(\frac{1+2+\ldots+n}{n}\right)^2\geq\left((n!)^{\frac{1}{n}}\right)^2=(n!)^{\frac{2}{n}}$
Equality holds when
$\frac{1}{3}n^2+\frac{1}{2}n+\frac{1}{6}=\frac{1}{4}n^2+\frac{1}{2}n+\frac{1}{4}$
Simplifying we get $n=1,-1$ but as $n\in\mathbb{N}$, equality holds when $n=1$

Thus Proved :)

We have that $1^2+2^2+...+n^2 = \frac13 n^3 + \frac12 n^2 + \frac16 n$, which can be proved [url=https://math.stackexchange.com/questions/1166027/prove-that-1222-cdotsn2-fracnn12n16-for-n-in-mathbbn]here.[/url] Now $\frac13 n^2 + \frac12 n + \frac16 = \frac{1^2+2^2+...+n^2}{n}$, now by AM-GM (which can be proved [url=https://en.wikipedia.org/wiki/AM-GM_Inequality]here[/url]) we get that $\frac{1^2+2^2+...+n^2}{n} \ge \sqrt[n]{1^2\cdot 2^2\cdot ...\cdot n^2}.$
again sorry for bumping


algebra
polynomial
trigonometry
probability
conditional probability
AMC

But are both book one and book two accessible to people studying only for the amc?

Both books should be accessible to someone preparing for the AMC (if Vol 2 isn't immediately accessible, it will be upon going through Vol 1).  The information in Vol 2 is necessary to success on the AMC - Vol 1 information alone won't be enough (for example, polynomials, trigonometry, and conditional probability, which are all typical AMC->12 subjects, are covered in volume 2).

And if you ever get confused about anything, you always have the community to help you...


calculus
integration
calculus computations

Compute the improper integral $\displaystyle\int^{\infty}_0 \frac{1}{e^x+e^{-x}}\,dx$

[quote="Kalle"]Just change variable $t = e^x$.[/quote]
After first multiplying the integrand by $\frac{e^x}{e^x}.$


Same here riddler ;)

anirudh's 4th n 5th step look wrong to me :?:

[quote="riddler"][quote="anirudh"]$\displaystyle x^2=3(x+1)^2-x^2(x+1)^2$
Sqrt both sides:
$x=\sqrt{3}(x+1)-x(x+1)$[/quote]
how can you do that?[/quote]
I don't think that's right...

Well, actually you can do something similar to that, though.
$x^2=(3-x^2)(x+1)^2\Longrightarrow x=\sqrt{3-x^2}(x+1)$

But I don't think that helps much...

Riddlr u know prerak right,he said he got it!!!! :(
i'll ask him tomrrow,i kept buggun himm but he doesnt tell :rotfl:  :rotfl:

well I doubt that but did you get it? more trouble for me. :?  :(

[quote="shadysaysurspammed"]no i didnt get ti,stuck on the same part as u.

Anyway,i would like to notify everyone that this sum was given to me by eminem(he hasnt figured it as well)[/quote]
how can you say then that it doesnt reqire any binomial theorem?

i JUST SAID THE SUM IS EMINEM'S SO THAT HE WOULDNT SAY I STOLE HIS SUM :rotfl:  :rotfl:  :rotfl:

i wont a simple method.ofcourse u can do it by bintheorem,but i want a simpler sol.

If you're going to make private conversations, then pm each other, don't do it in the middle of a thread ;)

[quote="riddler"]I dont seem to be able to factorise this...
$x^4+2x^3-x^2-6x-3=0$[/quote]

Instead of x^2, I'm getting 2(x^2).  :?

Edit: Oh, nevermind. I understand now.

ARE you sure? I think probably one of us has done a calculation mistake. :?

I've got the solution to this.Its easy.If no one gets it i'll pos it :lol:

mATHMAN,thats one set of the values of x :)

I plotted the biquadratic eqn. and this was the only set of solutions I obtained!!!

what's your wondeful soltion!!! I think I did this once but don't really remember

Let the others try Rushil,i'm  sure someone will get it.comeon guys,this sum shouldnt take so long(unless my solution is wrong somewhwre :oops: )

Should i give a hint if no ones gettin this problem ;)  ;)  :huh:  :huh:  :huh:  :)  :)

[quote="shadysaysurspammed"]I've got those 2 equations in a really simple way.all u do is write a^2+b^2 as (a-b)^2+2ab.substitute the values and u'll get the 2 equations.but i cant understand why Rushil and 100idunno how many zero's have neglected or rejected 1 of the equations.could someone pleaseee tell me why.
ps-10000 user,can ya explain ur mehod?is it a method,as riddler said,to use in typical questions? :D[/quote]

Look at your question, you did not say what type of x you wanted

[quote="shadysaysurspammed"]I posted this simple question in frt;s evaluate question but no one saw it-

x^2+x^2/{(x+1)^2}=3

find x.(only simple 9th grde fomulae please,no binomail theorem or any high suff :D )[/quote]

[quote="shadysaysurspammed"]But $x^2+3x+3=(x+\frac32)^2+\frac34>0$

1000th user,what do ya mean by this?i didnt understand,cant the value of x be complex or somethin?because this is my second equaton. :?[/quote]

Complete square and this shows two complex roots

[quote="riddler"]10000th user how did you do it? is there a specifc way of getting answers like that?[/quote]

That is my solution, I multiply equations, do not add like terms yet because maybe I can find pattern to factor and I did. It works sometimes and it doesn't sometimes. No, I don't know of [i]specific way[/i] that you want. If you don't like my solution, post yours!!!

[EDIT]About shadysayurspammed suggestion: $a^2+b^2=(a-b)^2+2ab=\left(x-\frac{x}{x+1}\right)^2+\frac{2x^2}{x+1}=\left(\frac{x^2}{x+1}\right)^2+\frac{2x^2}{x+1}=3$

Let $y=\frac{x^2}{x+1}$ and you get $y^2+2y=3$. Solve and you get $y=\frac{x^2}{x+1}=-3\,\text{or}\,1$

But this is same question, how did you know $a^2+b^2=(a-b)^2+2ab$ this works? Is there a specific way to know which identity to use? You see? So you can only answer your own question, not me.


vector
group theory
abstract algebra
superior algebra
superior algebra unsolved

The symmetric group $ S_n$ acts on the set of $ k$-element subsets of an $ n$-element set in the obvious way.  This becomes a representation of $ S_n$ on a vector space of dimension $ {n \choose k}$ when extended by linearity.  How does this representation decompose into irreducible representations?

Representations of the symmetric group can be approached in many ways, so it is difficult to tell what sort of answer you expect.  I'll describe a bunch of things that are true without worrying too much about whether it forms a complete answer (since it might be way more than complete, or completely deficient depending on your point of view).

The stabilizer of {1,...,k} in the action of G=Sym(n) on k-subsets of {1..n} is H=Sym({1..k}) x Sym({k+1..n}) =~ Sym(k) x Sym(n).  This is a "parabolic subgroup" in the language of Coxeter groups, and a "Young subgroup" in the language of representations of the symmetric group.  The linear representation is the induced module $ 1_H^G$, sometimes denoted $ M^\lambda$ where λ = (k,n-k).  It contains the Specht module $ S^\lambda$ which is irreducible over fields of characteristic 0 like $ \mathbb{C}$.  For any partition μ of n, the Specht module appears as a composition factor $ K_{\lambda,\mu}$ times, where $ K_{\lambda,\mu}$ is a so-called "Kostka number".   Now $ K_{\lambda,\mu}$ is usually 0, and is 1 for λ=μ.  In this case λ=(k,n-k) has a particularly simple form, and the only μs with $ K_{\lambda,\mu}$ are those of the form μ=(j,n-j) for 0 ≤ j ≤ k, assuming 2k ≤ n and that you like partitions to be non-decreasing.

The Kostka numbers also measure a combinatorial problem: $ K_{ (k,n\minus{}k), (j,n\minus{}j) }$ is the number of ways of placing j 1s and n-j 2s in two rows, the first of length k, and the second of the length n-k, such that each row is non-decreasing and each column is increasing.  It is not too hard to see there is exactly one such way.  That is, $ K_{\lambda,\mu}$ is 1 for μ=(j,n-j) for 0 ≤ j ≤ k, and 0 otherwise.

In other words, if you work from k=0 to floor(n/2), then at each step you only need to find one new irreducible representation, and it is the Specht module $ S^{(k,n\minus{}k)}$.  The permutation representation of Sym(n) on k-subsets over a field of characteristic 0 is then the direct sum of one copy each of those irreducible representations $ S^{(j,n\minus{}j)}$ you found for 0 ≤ j ≤ k.

I don't recall a particularly nice description of $ S^{(j,n\minus{}j)}$ other than the typical descriptions of $ S^\lambda$ with the idea that things can't be too bad for just 2 rows.  Its dimension is typical denoted $ f^\lambda$ and has a few zillion combinatorial expressions.  The previous discussion gives a pretty simple description: $ f^{(k,n\minus{}k)} \equal{} \binom{n}{k} \minus{} \binom{n}{k\minus{}1}$ for 2k ≤ n.

To make sure our conventions agree, let me specify that $ S^{(0,n)}$ is the trivial module, $ S^{(1,n\minus{}1)}$ is the natural module modulo the centralizer of Sym(n) inside it, etc.

Thanks!  It turns out I have a paper that describes the decomposition of the modules $ M^{\lambda}$, but I didn't realize that this representation was a representation of that form.


So there is a company that gives a rating to each employee each year. The possible ratings are 3.0, 3.5, and 4.0. The percentage of people that get a 3.0 is 25%, 3.5 is 40%, and 4.0 is 35%. Calculate the percentage of people that will average over 3.75 over seven years.

I had to use a computer program to solve this and I don't know if I did it right. If anyone could solve this and confirm my answer that'd be great. BTW my answer is

[hide] 8.2% [/hide]

First of all, how can you get over 3.75?

You need at least 4 4.0's (this is easy to check), and the rest 3.5/4
Otherwise, you need 5 4.0s, 1 3.0, and 1 3.5/4

So lets calculate probabilities now:

Probability of getting 4 4.0's --> (.35)^4
Probability of getting 3 3.5/4's -->(.75)^3

Probability of getting 5 4.0's --> (.35)^5
Probability of getting 1 3.0's --> (.25)
Probability of getting 1 3.5/40. --> (.75)

We're not overcounting, so this is good.

Now we need to calculate all the different arrangements, which is tedious, but straightforward.  Can you finish it off from here?  (remember to count distinct permutations)

One has to be [b]VERY careful [/b]to draw any staitistical conclusion form this kind of problem. ( Unfortunately, this is a very common  mistake  that many make)

If the problem was worded as, [b]probablity of a given employee gettting a   rank of 4.0 is 35%.. etc[/b] the above answer would be statistically correct.

BUT, [b]It is very different than saying 35% of the employees get  a rank of 4.0 .[/b]..

For example,  say,   1% of people who take AMC qualify for AIME in any given year - To draw a conclusion that  percentage  of students who  qualify for AIME in all four years is   .000001% (or   1 in 100 million)  ( rules  out ANY one from US, or for that matter in the whole  world,  doing that !!!  - as there are  not that many students in the whole world taking AIME) .. Yet we all know MANY people  who qualify for AIME (and USAMO) year after year.

We all know someone who had an A average in all (say  16 years)  of   school - imagine  how rare that would be if  we applied that logic..(Certainly less than 1 in a billion if we take 20% of students getting A average in any given year) .

Point is,  Most likely, if  you really looked  at any normal company where this type of distribution exists  you would find the actual value is  MUCH higher than  the calculated value... and you are even  likely to find an employee rated 4.0 in ALL seven years (the theoratical value for that to happen  would be about .06%)..

You have to be very carreful when you apply this kind of statistics blindly ..

Hope this helps.

(PS -  In a normal world - If an employee is ranked high, he is more likely to be ranlked high  again - assuming the ranking porcess is not  decided by a random event - such as lottery)


Prove that the set $ \left \{ 1,2,\ldots,2007  \right \}$ can be expressed as the union of disjoint subsets $ A_{i}$ with $ i\equal{} 1,2,\ldots,223$ such that
(a) each $ A_{i}$ contains 9 elements, and
(b) the sum of all the elements in each $ A_{i}$ is the same.

[quote="james digol"]Prove that the set $ \left \{ 1,2,\ldots,2007 \right \}$ can be expressed as the union of disjoint subsets $ A_{i}$ with $ i \equal{} 1,2,\ldots,223$ such that
(a) each $ A_{i}$ contains 9 elements, and
(b) the sum of all the elements in each $ A_{i}$ is the same.[/quote]

Here comes one example:
Lets divide in two each $ A_i$

The first part with $ 6$ elements and equal sum of $ 8031$
$ B_1\equal{}\{2007,1562,1561,1116,1115,670\}$
$ B_2\equal{}\{2006,1563,1560,1117,1114,671\}$
$ \cdots$
$ B_{223}\equal{}\{1785,1784,1339,1338,893,892\}$

The second part with the other $ 3$ elements and equal sum of $ 1005$:
$ C_1\equal{}\{669,225,111\}$
$ C_2\equal{}\{668,227,110\}$
$ \cdots$
$ C_{111}\equal{}\{559,445,1\}$
and the subsets
$ C_{112}\equal{}\{558,224,223\}$
$ C_{113}\equal{}\{557,226,222\}$
$ \cdots$
$ C_{223}\equal{}\{447,446,112\}$