algebra
polynomial
calculus
geometry
3D geometry
LaTeX
number theory

2 questions I have problems with; anyone care to help. thanks a lot!

$(1)$ Find a polynomialwith integer co-efficients one of whose roots is $x=(\frac{3}{5})^\frac{1}{6}+(\frac{5}{3})^\frac{1}{6}$

$(2)$ If $ax^{4}+4bx+c$ has a double root at $x=\alpha$ show that $ac^{3}=27b^{4}$

In reference to question 1;

this guy posted a solution here; http://www.artofproblemsolving.com/Forum/viewtopic.php?t=84947

as far as i got on number three

Before you read this, “” means brackets, and the * means to the power of whatever letter is right next to it.
I was too lazy to figure out how to actually get brackets and crap like that on this thing.   :)

So since they both equal zero:
(px*4) + qx*3 + r = (rx*4) + (qx*3) + p

“(px*4) + (qx*3) + r = (rx*4) + (qx*3) + p”/ “(qx*3)”

So then…

(px*4) + r = (rx*4) + p

(px*4) – p = (rx*4) – r

“p(x*4 – 1) = r(x*4 – 1)”/ (x*4 – 1)

So p = r

Im not sure if I missed something, but this is as far as I got, and I thought I would just post this here in case this helped you any.
sorry if that didnt help any lol

Don't post anything on q3, it's in Enrichment. (It's easy.)

Problems from first sheet on polynomials I haven't solved:

3) find polynomials f(x) and g(x) s.t. $(x^{8}-1)f(x)+(x^{5}-1)g(x)=(x-1)$

[hide]I've gotten $f(x) = 0, g(x) = \frac{1}{x^{4}+x^{3}+x^{2}+x+1}$ and $f(x) = \frac{1}{x^{7}+x^{6}+x^{5}+x^{4}+x^{3}+x^{2}+x+1}, g(x)=0$ But somehow I don't think you're allowed to do that.[/hide]

9) The equation $x^{5}=bx^{2}+c=0 (c\neq0)$ has a double root. Show that $108b^{5}=3125c^{3}$

14) Prove that the polynomial $p(x) = \sum_{k=0}^{2n}\frac{1}{k!}x^{k}$ has no real roots.

ey havent been on here in awhile, hows everyone goin with the amoc prep papers?

btw u can do 9 on the polynomials using calculus pretty easily:

x^5-bx^2+c=0

so 5x^4 - 2bx = 0 when x is at that double root

so the double root is at (2b/5)^1/3

then u sub that into ur original equation and do sum algebra to get rid of the cube root

lol sorry i hav no idea how to make the actual symbols and stuff come up

I need to do a lot in the next few days of holidays. Anyone got past papers for AMOC Seniors? (Not AIMO) Could you MSN/email them to me please..?

Sean, LaTeX guides are on this site. And just see what other people use and go on that. :)

for 3 you need to find the gcd of $x^{5}-1$ and $x^{8}-1$ and work backwards

for 9, use vietas if you dont want to use calculus and it will come out nicely.

[quote="vulgarfraction"]14) Prove that the polynomial $p(x) = \sum_{k=0}^{2n}\frac{1}{k!}x^{k}$ has no real roots.[/quote]

Let's rewrite the problem:

[color=red][b]Bloody[/b][/color] [color=blue][b]problem.[/b] Let $p_{n}(x) = \sum_{i=0}^{n}\frac{x^{i}}{i!}$. Prove that for all $k \geq 0$, $p_{2k}$ has no real roots.[/color]

[i]Solution.[/i]We begin with a very important observation: $p_{n}^\prime (x) = \sum_{i=0}^{n}\left( \frac{x^{i}}{i!}\right)^\prime = \sum_{i=1}^{n}\frac{x^{i-1}}{(i-1)!}= p_{n-1}, \, \forall n \geq 1$.
Hence, $p_{n}= p_{n-1}+\frac{x^{n}}{n!}= p_{n}^\prime+\frac{x^{n}}{n!}, \, \forall n \geq 1. \, \, \, \, \, \left( \star \right)$
The case $k = 0$ is obvious, $p_{0}\equiv 1$, and this clearly has no roots.
Now suppose the opposite, i.e. for some $k \geq 1$, the polynomial $p_{2k}$ has a real root.
Let $u$ be the smallest root of $p_{2k}$. We have two cases:

(i) $p_{2k}(x) \geq 0$ for all $x \in \left[ u, u+\epsilon \right)$, where $\epsilon$ is some small positive number. Since $\lim_{x \to-\infty}p_{2k}(x) =-\infty$ and $u$ is the smallest root of $p_{2k}(x)$, it follows that $x \in \left( u-\epsilon, u \right] \geq 0$. Thus, $u$ is a local minimum point, so $p_{2k}^\prime (u) = 0$. By $\left( \star \right)$ we have $p_{2k}(u) = p_{2k}^\prime (u)+\frac{u^{2k}}{(2k)!}= \frac{u^{2k}}{(2k)!}$.
But we also have that $p_{2k}(u) = 0$, so $\frac{u^{2k}}{(2k)!}= 0$ i.e. $u = 0$. This is a contradiction, since $p_{2k}(0) = 1$.

(ii) $p_{2k}(x) \leq 0$ for all $x \in \left[ u, u+\epsilon \right)$, where $\epsilon$ is some small positive number. Since $\lim_{x \to \infty}p_{2k}(x) = \infty$, it follows that $p_{2k}$ has roots in $\left( u, \infty \right)$. Let $v$ be the smallest of them. It's obvious that $p(x) < 0, \, \forall x \in \left( u, v \right)$. Since $p(u) = 0 = p(v)$, by Rolle's theorem there is $w \in \left( u, v \right)$ such that $p_{2k}^\prime (w) = 0$. By $\left( \star \right)$ we have that $p_{2k}(w) = p_{2k}^\prime (w)+\frac{w^{2k}}{(2k)!}= \frac{w^{2k}}{(2k)!}\geq 0$. But $w \in \left( u, v \right)$, so $p_{2k}(w) < 0$, contradiction.

Therefore, $p_{2k}$ has no real roots, for all $k \geq 0. \, \, \, \, \, \, \blacksquare$

PS: The idea of the proof is the following: if a polynomial of even degree takes positive values in all the roots of its derivatives, then the polynomial is positive everywhere.


the median $ AA^'$ of triangle $ ABC$ meets the side $ B^' C^'$ of the medial triangle $ A^' B^' C^'$
at $ P$ and $ CP$ meets $ AB$ at $ Q$.prove that $ AB=3AQ$

[hide="hint"]use menealus theorem [/hide]

Using menelaus theorem in $\triangle{ABA'}$, we have:

$\dfrac{AQ}{QB} \cdot \dfrac{BC}{CA'} \cdot \dfrac{A'P}{PA} =1$

But $\dfrac{BC}{CA'}=2$ and $\dfrac{A'P}{PA}=1 \Longrightarrow 2AQ=BQ \Longrightarrow 3AQ=BQ+AQ=AB$.


pigeonhole principle
number theory unsolved
number theory

Lesha put numbers from 1 to $22^2$ into cells of $22\times 22$ board. Can Oleg always choose two cells, adjacent by the side or by vertex, the sum of numbers in which is divisible by 4?

Yes, I think he always can:

Assume, for simplicity, that we fill the board not with those numbers, but with $11^2$ copies of each of the numbers $0,1,2,3$. Divide the board in $11^2\ 2\times 2$ squares. Each such square must contain exactly one $0$ and one $2$, and the other two squares must be occupied by the same number, either a $3$ or a $1$. In particular, this means that there's an even number of $1$'s, but this is impossible, since there are $11^2\ 1$'s.

Gyah!  I must have been high or something.

Grobber's proof is right.  The point is, that if some 2x2 block contains no 2s, then some other 2x2 block must contain a pair of 2s (by the pigeonhole principle.)

Similarly for 0s.

[quote="Saul"]How do we know that a 2x2 square isn't filled with 1s, for example?[/quote]
That was a post of stupidity. Ignore it.

Sorry for opening an old topic, but in general for $(4k+2) \times (4k+2)$, Oleg can and for $4k \times 4k$ Oleg can't.
1030
1232
1030
1232
And similary we construct for every $4k \times 4k$


TÌm số thực $\alpha$ sao cho
$n^{\alpha} $ là số nguyên với m?i số tự nhiên n>2005

[quote]Xin chào, bài toán mới đê!!!!!!!! [/quote]

Zoooooooooooooooo.............ooooooooooooo :D

 [quote]TÌm số thực  sao cho
 là số nguyên với m?i số tự nhiên n>2005[/quote]

to da lam mot bai cua bon Tau , va sau do cung che ra nhieu cau tuong tu the. ;)

Chao anh pluricomplex , sao em khong vao duoc dia chi website cua anh ?

mien to dung la mien phi nen co the ban bi buc tuong lua do mot so nha cung cap dv interenet lap ra de chan. Co le ho so nhung trang nay co chua van hoa pham khong lanh manh. :D

O day nhieu nguoi Viet Nam minh nhi?chao cac ban!

Xin chao cac ban Viet  Nam ;)

 What do we have to do with $n^{\alpha}$ and $n>2005$? :D

[quote]
Xin chao cac ban Viet Nam   ;)

What do we have to do with  $n^{\alpha}$ and $n>2005$?  [/quote]

he he, chao [b]dickchimney[/b]. He want to find all $\alpha\in\mathbb{R}$ such that $n^{\alpha}$ is integer for all $n>2005$!  ;)

[quote="pigfly"]Anh Hung a`,em nghi [b]dickchimney[/b] hieu yeu cau cua bai toan,boi vi [b]dickchimney[/b] ko phai la nguoi Ukraine dau,anh ay nguoi VietNam day(Hinh nhu anh ay ten la Nguyen Hoang Dung,dan Ams 98-01) :D  :D  :D[/quote]

 You call me huh? What's the matter? :D

Chắc chắn là đ? bài này sai rùi. Nếu n đã là số tự nhiên, thì nếu a là bất kỳ số tự nhiên nào thì a n nào chẳng là số nguyên.

I think that all of you shouldn't write vietnamese

Chào, he  he he he he he he he he he h e

TOI DA CO LOI GIAI BAI TOAN NAY >     CHAO CAC BAN

Lop hoc co 21 hs nam va 21 hs nu cung giai mot so bai toan!Biet rang moi hs trong lop chi giai duoc khong qua 6 bai toan...va cu 1 hs  nam va 1 hs nu thi cung giai duoc it nhat la 1 bai toan.CMR: ton tai mot bai toan ma co  hs nam va 3 hs nu cung giai duoc! :(

Thi QT năm 2001 mà??!! Lên kalva mà xem ;)

 Em la Viet day a? :D

viet nao vay anh dickchim :D  :D  :D  :D

em cung la nguoi viet day :P

anybody know how to find the answer for # 25 in the AMC 10 ??? i know the answer but i don't know how to do it.
 anybody??

I'm just a learner. :D

[quote="dickchimney"]Tai sao ha lang tu ?? :D :?[/qu

shouldn't write Vietnamese to learnt English better

[quote="nmtruong1986"]TÌm số thực $\alpha$ sao cho
$n^{\alpha}$ là số nguyên với mỿi số tự nhiên n>2005[/quote]
Ai có thể giải bài này?


trigonometry
geometry
angle bisector
trig identities
Law of Sines
geometry theorems

Show that $l_a=\cos\frac{A}{2}\cdot\frac{2bc}{b+c}$ where $l_a$ is the bissector of the angle $A$ of a triangle $ABC$.

Are you kiding with this one? Just draw the bisector and then apply the cosine law in  those two triangles.The formula follows immediately. ;)

There is also a synthetic proof (in the sense the word "synthetic" applies to trigonometric formulas):

Let the angle bisector of the angle CAB of a triangle ABC meet the side BC at a point D. Then, we have to prove that $AD=\cos\frac{A}{2}\cdot\frac{2bc}{b+c}$.

Well, let H be the foot of the altitude of triangle ABC from the vertex A. Then, < AHD = 90°, so that the triangle AHD is right-angled; thus, $AD=\frac{AH}{\cos\measuredangle HAD}$. Now, since < AHB = 90°, the triangle AHB is right-angled, so that < HAB = 90° - < HBA = 90° - B, and since the point D lies on the angle bisector of the angle CAB, we have $\measuredangle DAB=\frac{A}{2}$. Hence, $\measuredangle HAD=\left|\measuredangle DAB-\measuredangle HAB\right|=\left|\frac{A}{2}-\left(90^{\circ}-B\right)\right|$. Since, by the sum of angles in a triangle, A + B + C = 180°, we have $90^{\circ}=\frac{180^{\circ}}{2}=\frac{A+B+C}{2}$, and thus this becomes

$\measuredangle HAD=\left|\frac{A}{2}-\left(\frac{A+B+C}{2}-B\right)\right|=\left|\frac{B-C}{2}\right|$.

Hence, $AD=\frac{AH}{\cos\measuredangle HAD}=\frac{AH}{\cos\left|\frac{B-C}{2}\right|}=\frac{AH}{\cos\frac{B-C}{2}}$ (since $\cos\left|x\right|=\cos x$ for any angle x). But in the right-angled triangle AHB, we have $AH=AB\cdot\sin\measuredangle ABH=c\cdot\sin B$; thus, $AD=\frac{c\cdot\sin B}{\cos\frac{B-C}{2}}=\frac{c\cdot\sin B}{\sin\frac{A}{2}}\cdot\frac{\sin\frac{A}{2}}{\cos\frac{B-C}{2}}$. Now, by one of the Mollweide formulas, $\frac{\cos\frac{B-C}{2}}{\sin\frac{A}{2}}=\frac{b+c}{a}$, so that $\frac{\sin\frac{A}{2}}{\cos\frac{B-C}{2}}=\frac{a}{b+c}$. Hence,

$AD=\frac{c\cdot\sin B}{\sin\frac{A}{2}}\cdot\frac{\sin\frac{A}{2}}{\cos\frac{B-C}{2}}=\frac{c\cdot\sin B}{\sin\frac{A}{2}}\cdot\frac{a}{b+c}=\frac{c\cdot a\sin B}{\sin\frac{A}{2}\cdot\left(b+c\right)}$.

Now note that a : sin A = b : sin B by the law of sines; thus, a sin B = b sin A, and this becomes

$AD=\frac{c\cdot b\sin A}{\sin\frac{A}{2}\cdot\left(b+c\right)}=\frac{c\cdot b\cdot 2\sin\frac{A}{2}\cos\frac{A}{2}}{\sin\frac{A}{2}\cdot\left(b+c\right)}=\cos\frac{A}{2}\cdot\frac{2bc}{b+c}$.

And the proof is complete.

  Darij

your solution is very cool Darij and i know you are so good in geometry but i think you can solve this problem with an easier solution and you should not think so hard like that. :D

Well it's always cool to think of alternative solutions.

$\triangle ABC,\ i=C(I,r),\ e=C(O,R),\ D\in AI\cap BC,\ \{A,M\}\subset AI\cap e\Longrightarrow$

\[ \underline {\overline {\left|\ l_a\cdot AM=bc,\ l^2_a=bc-DB\cdot DC,\ l_a=\frac{2}{b+c}\cdot \sqrt{bcs(s-a)}=\frac{2bc}{b+c}\cdot \cos \frac A2\ \right| }}.  \] [u]A my solution:[/u] $\triangle ABM\sim\triangle ADC\Longrightarrow \frac{AB}{AD}=\frac{AM}{AC}\Longrightarrow \frac{c}{l_a}=\frac{AM}{b}\Longrightarrow$

$\overline {\underline {\left| l_a\cdot AM=bc\right| }}\Longrightarrow l_a(l_a+DM)=bc\Longrightarrow$

$l^2_a=bc-DA\cdot DM\Longrightarrow \overline {\underline {\left| l^2_a=bc-DB\cdot DC\right| }}$.

$\frac{DB}{c}=\frac{DC}{b}=\frac{a}{b+c}\Longrightarrow l^2_a=bc-\frac{ac}{b+c}\cdot \frac{ab}{b+c}\Longrightarrow$

$l^2_a=\frac{bc \left[ (b+c)^2-a^2\right] }{(b+c)^2}\Longrightarrow l^2_a=\frac{4bcs(s-a)}{(b+c)^2}\Longrightarrow$

$\underline {\overline {\left| l_a=\frac{2}{b+c}\cdot \sqrt{bcs(s-a)}\right| }}.\ \ \cos\frac A2 =\sqrt{\frac{s(s-a)}{bc}} \Longrightarrow\overline {\underline {\left| l_a=\frac{2bc}{b+c}\cdot \cos \frac A2\right| }}.$

[u]Remark.[/u] What become the up relations for $l'_a$
(the length of the exterior bisector from the angle $\widehat {BAC}\ ?.\ \underline {\overline {\left|\ Try\ !\ \right| }}$


Let recurence relation $ b_{n\plus{}2}\plus{}b_{n\plus{}1}\equal{}b_n\plus{}3k$. Prove $ b_n\equal{}C\plus{}D(\minus{}2)^n\plus{}nk$.

I think you mean $ b_{n\plus{}2} \plus{} b_{n\plus{}1} \equal{} 2b_n \plus{} 3k$.  In any case, it is straightforward.

Let $ c_n \equal{} b_n \minus{} nk$; then the given recurrence becomes $ c_{n\plus{}2} \plus{} c_{n\plus{}1} \equal{} 2c_n$, with characteristic equation $ x^2 \plus{} x \minus{} 2 \equal{} 0 \Leftrightarrow (x \plus{} 2)(x \minus{} 1) \equal{} 0$.  It is a standard result that therefore $ c_n \equal{} C \plus{} D(\minus{}2)^n$.


function
trigonometry
calculus
calculus computations

Hi, i need some help with this one:

Show that $ \ arcsinx \ \equal{} arctan(\frac {x}{\sqrt{1 \minus{} x^2}})$     if |x| < 1

thanks!

Use $ \sin( \arcsin x)) \equal{} x$ and $ \sin ( \arctan (x)) \equal{} \frac{x}{\sqrt{1\plus{}x^2}}$. The second formula follows from $ \sin x \equal{} \frac{\tan x}{\sqrt{1\plus{} \tan^2x}}$.

Draw a right triangle with catheti $ x$ and $ \sqrt{1\minus{}x^2}$. By Pythagoras, the hypotenuse is $ \sqrt{x^2\plus{}1\minus{}x^2}\equal{}\pm 1$. Then the sine is clearly $ \frac{x}{1}$, so we have $ \sin\theta \equal{} x\implies \arcsin x\equal{}\theta$ and $ \tan\theta\equal{}\frac{x}{\sqrt{1\minus{}x^2}}\implies \theta\equal{}\arctan\frac{x}{\sqrt{1\minus{}x^2}}$, and the conclusion follows.


inequalities
induction
combinatorics proposed
combinatorics

I have two proofs one of which uses Induction :)
I was actually looking for more solutions .


Let $ a,b,c$ such that $ ab\plus{}bc\plus{}ca\equal{}3$. Find the maximum value of

P=$ \sum\frac{(b\minus{}c)^2\minus{}2}{b^2\plus{}c^2}$

We just need to find the minimum value of $ \sum\dfrac{bc\plus{}1}{b^2\plus{}c^2}$


HMMT MIT college geometry trigonometry search MATHCOUNTS

all aops users going to hmmt should meet in a specific spot on saturday
that would be so cool to meet all of you

[quote="greentreeroad"]for what? just for cool?[/quote]

what else...

We could meet somewhere in the student's center during lunch... I don't know any specific places though.

i am quite unfamiliar with MIT (although I live in Boston)
perhaps we can meet in the courtyard outside lobby 8
(look at MIT campus map, go to hmmt website and go to directions page to find that link)

how bout we meet on the awards stage lol jk i suck...hmm this is sketchy but cool

we could make a team called AoPS. :P

All these AoPS meet-ups have been wishful thinking in the past, and are wishful thinking now. Blah. :(

i think meeting in the courtyard at noon is a good idea.
Just call out the codename when you are there: codename=aops

If I can find the courtyard and hear the code then I shall be there.

im going to feel stupid calling out the code...
someone has to do along with me

anyways
the courtyard is right outside lobby 10, according to the map :?:

SplashD on the other hand acts exactly like he does online

[quote="cyberspace"]Why don't we meet across the street from building 7? That whole open plaza-ish area is right near all the lunch places so we don't have to go out of our way (as I'm sure we all like food  :) ) .[/quote]

I like this idea- just remember the code, and shout it

i wonder if splashD is talking about not_trig.... hm...... :ninja:

That would be pretty tricky as !trig did not actually go to HMMT last year

[quote="SplashD"]This usually doesn't work out well because nobody actually comes. Besides, people aren't what they seem on the internet. Like, I thought I knew this one guy, and last year at HMMT, he just came up, punched me in the arm, and said something along the lines of "Whaddup foo," (This person knows who he is.)[/quote]I've gotta agree with SplashD here. These don't generally work.

That's because everyone assumes that no one else will go and so they don't show up.  Besides, we might as well go for it, as the meeting place is right on the way to lunch (most of the lunch places anyway).  The only possible disadvantage is that you might look stupid shouting "aops"  :P

i agree with cyberspace
perhaps we can write AOPS on the back of our shirts so people will identify us

[quote="SplashD"]This usually doesn't work out well because nobody actually comes. Besides, people aren't what they seem on the internet. Like, I thought I knew this one guy, and last year at HMMT, he just came up, punched me in the arm, and said something along the lines of "Whaddup foo," (This person knows who he is.)[/quote]
mystic terminator?

pakman2012?

I think i did that to someone, perhaps TONY.

[quote="greentreeroad"]Why not just choose a person everybody know (like trig) as a sign so we don't need to shout out or write anything?![/quote]

problem is i dont know anyone here

anyone who has an aops shirt wear it and half of our problem is solved cause well easily find eachother

Umm it's not like you have anything more in common with aops people at hmmt than other people at hmmt...Except that you're on aops...

so havnt u ever had the desire to meet your fellow aopsers

Not random ones whom I do not know otherwise, no I haven't.

u have a good points but common itd be epic especially if someone u found turned out to be another one of those whaddup dog people

hey sam, ive seen you at the nov. hmmt so ill know what you look like
you just have to go there and we can shout the "code"

I've seen Sam via AMC 10 perfects, so I know what he looks like.
If you know my name, search 2007 AMC 8 perfects.

lol i lost my aops shirt so im wearing my schools mathcounts shirt and i wrote aops in green marker on it XD

[quote="azjps"]I'll have on my green mathcamp 07 shirt, doubt anyone here knows who I am though.[/quote]

[i]I[/i] know who you are.  But I know pretty much everyone who has a Mathcamp T-shirt to wear ('06 '07 '08 will do that to you).

In any case, these meetings never work out, and I've never really seen the point anyway.  Meeting with people you actually know is nice (Mathcamp reunion, anyone?), but AoPSers need to get over the idea that they actually know someone based off of online interaction.

i go there and i see a kid with AOPS written in green on his mathcounts t-shirt, standing there alone yelling AOPS AOPS
lol

let's give this a try

maybe next year for hmmt, we can have an AOPS team<--thats why we have to meet and get to know each other better

[quote="not_trig"]I'll be wearing a giant puffy jacket? seeing as it'll be like sub-freezing there?[/quote]

Are you kidding? It was a beautiful day! Well, not in the morning. But the awards ceremony area was quite possibly the best outdoor environment I've  ever been in during the Boston winter. Like, it felt like October. Except without birds and leaves. But like dude, if I was an MIT kid and I need a place to lie down alone and collect my thoughts at midnight on a >20 degree night, I would totally go there.

quote:
i go there and i see a kid with AOPS written in green on his mathcounts t-shirt, standing there alone yelling AOPS AOPS
lol

thats what i did except i dont recall yelling aops aops i was talking abouts aops tho and wanted to change my team name to the aopsers.

Carl, i totally agree with you.

the aops meeting was definitely a failure

[quote="E^(pi*i)=-1"][quote="azjps"]I'll have on my green mathcamp 07 shirt, doubt anyone here knows who I am though.[/quote]

[i]I[/i] know who you are.  But I know pretty much everyone who has a Mathcamp T-shirt to wear ('06 '07 '08 will do that to you).
 [/quote]

Ah, and I know who you are.

My cell phone went off during t0rajir0u's talk.  :oops:

[quote="Zhero"]Perhaps next year, we could suggest a mini-event in which AOPSers get together?[/quote]

And do what there? Stand and look around?

And talk about stuff? Or some random activity, or a lecture, or a game, or whatever. I'm not sure. This was just an idea off the top of my head for some way more organized way for AOPSers to meet. This seemed to me to be slightly more effective than, say, wearing AOPS shirts and shouting for other AOPSers. Or maybe I'm just crazy.

We could also just tell each other which mini-events we're planning on going to or something. I dunno. I'm just somewhat interested in a somewhat more systematic way to meet up with fellow AOPSers.

[quote="laughinghead505"]lol so did anyone actually do it?[/quote]

I actually didn't and I was the one who started all of this, wasn't I?

If our team had went to lunch where I expected us to I would've been shouting the code :P  but we got pizza instead so I never went to the meeting place. Maybe next year...


Does anyone know how to sum the following sequence:

1/4 + 3/16 + 5/64 + 7/256 + ... + (2k-1)/4^k + ...

What dose the sum converge to?

is the author of this cheating?
This is from the mathpath quiz.

Well techinically its not cheating because its past the application deadline (even past the extension for all nationals people)

If Neil's right, I got that question right!! (it was prolly the easiest one)

[hide]Well, you can check by using your calculator that the answer is indeed approaching $\frac{5}{9}$[/hide]

that's not always accurate though, and sometimes hard to see.
I mean, 1/2+1/4+1/8+.... everyone can see that it approaches 1, but a summation like this, its much harder.

Not really, as that's geometric and then you can apply the formula and it's easy to see it's 1.

Go ahead, you can post your way of doing that problem.

[quote="shadysaysurspammed"]This is an arithmetico-geometric sequence.It is easy to derive the summation.I'll do so if anyone wants.[/quote]please do :)


inequalities
trigonometry
inequalities proposed

Given $ a_i > 0$ such that $ \sum_{i \equal{} 1}^n a_i \equal{} \pi$ for all $ x_i > 0$ prove that
\[ \sum_{i \equal{} 1}^n x_{i}x_{i \plus{} 1}\cos a_i\le\ (x_1^2 \plus{} x_2^2 \plus{} ... \plus{} x_n^2)\cos\frac {\pi}{n},\ n \plus{} 1\equiv 1
\]

Well, I don't recall any sort of proof (quite sure it is complicated), but from what I remember it's called the Fejes Toth inequality. :wink: Maybe you can find some links or sth :wink:
If you have a proof than go ahead and please post it :wink:

I posted it because I don't know solution, but maybe it was posted in wrong place :D, solve it or move it,  please :D


Fine all primes like $p,q$ such that:
$p+q=(p-q)^3$

[quote="Arne"]First suppose that $p,q > 3$. If $p = q$ (mod 3) then 3 divides the left-hand side but not the right-hand side, contradiction. If $p \neq q$ (mod 3) then 3 divides the right-hand side, but not the left-hand side, contradiction. Hence we must have $p = 3$ or $q = 3$. Also, we need $p > q$. Therefore, if $p = 3$, we must have $q = 2$, which is not a solution. Hence we must have $q = 3$. We then get $p + 3 = (p - 3)^3$ which reduces to $(p - 5)\left((p - 2)^2 + 2\right) = 0$. Hence the only solution is $(p,q) = (5,3)$.[/quote]
Very nice solution. ;)

[quote="Arne"]Actually, things go even quicker if you notice that $p + q = (p - q)^3 = p - q$ (mod 3) by Fermat's little theorem. Hence we must have $2q = 0$ (mod 3) and $q = 3$.[/quote]
Oh... :? I think the designer hasn't paid attention to this matter; because the original solution was something like your first solution. :lol:


geometry
circumcircle
cyclic quadrilateral
power of a point
radical axis
geometry unsolved

Let $ ABCD$ cyclic quadrilateral. The internal bisector of angles $ A$ and $ B$ meet at $ P$, points $ Q,R,S$ are similarly defined by a cyclic change of letters.
Prove that $ PQRS$ is a cyclic quadrilateral
Suppose that the circles $ ABCD$ and $ PQRS$ have centres $ O$ and $ X$ respectively. Let $ AC$ meet $ BD$ at $ E$
Prove that $ O,E,X$ are collinear and $ PR$ perpendicular to $ QS$

The fact that PQRS is cyclic follows from a short angle chase.
$ m\angle{PQR} \plus{} m\angle{PSR} \equal{} (180^{\circ} \minus{} \frac {1}{2}m\angle{A} \minus{} \frac {1}{2}m\angle{D}) \plus{} (180^{\circ} \minus{} \frac {1}{2}m\angle{B} \minus{} \frac {1}{2}m\angle{C})$
$ \equal{} 180^{\circ}$

Notice that $ P$ is equidistant from $ DA, AB,$ and $ BC$, while $ R$ is equidistant from $ BC, CD,$ and $ DA$.  Extend $ DA$ and $ CB$ to meet at a point $ M$.  Then $ P$ and $ R$ lie on the bisector of $ \angle{AMB}$.  Similarly, if we extend $ AB$ and $ DC$ to meet at $ N$, then $ Q$ and $ S$ lie on the bisector of $ \angle{BNC}$.  It's a well-known fact that these two angle bisectors are perpendicular (it follows, again, by an angle/arc chase).  Therefore, $ PQ$ and $ RS$ are perpendicular.

So notice that $ P$ is the excenter of $ MAB$ and $ R$ is the excenter of $ MCD$.  Using a boring angle chasing argument, we can prove that $ PRDA$ is cyclic.  Similarly, $ QSAB$, $ RPBC$, and $ SQCD$ are cyclic.  Therefore, $ MP \cdot MR \equal{} MA \cdot MD$, so $ M$ has the same power WRT the two circles.  Similarly, so does $ N$, so $ MN$ is the radical axis of the two circles.

Okay...so it's easy to prove that $ MN$ is the polar of $ E$ WRT to the circumcircle of $ ABCD$.  This is well-known (we can prove it via Pascal's theorem).  It then follows that $ OE \perp MN$.  So it suffices to show $ OX \perp MN$.  This follows from the fact that $ MN$ is the radical axis of the two circles.


algebra
difference of squares
special factorizations

I figured out how to do this the long....... way, but I really need an way faster way to do this:?

$ \text{Compute the value of }\frac{87^2\minus{}13^2}{87^2\minus{}2(87)(13)\plus{}13^2}\text{.  Express your answer as a common fraction.}$

[quote="mathemagician1729"]uhh unless i made a computational error, shouldnt it be 50/37 ?

Oh this is in the 06-07 handbook and my answer is right.[/quote]

I already knew the answer.....I just didn't want to post it because people just say the answer even though it is given.  I need a solution, not just an answer.

[quote="math154"][quote="math154"]Let $ (x,y) \equal{} (87,13)$. Then we have

$ \frac {87^2 \minus{} 13^2}{87^2 \minus{} 2(87)(13) \plus{} 13^2} \equal{} \frac {x^2 \minus{} y^2}{x^2 \minus{} 2xy \plus{} y^2}$

$ \equal{} \frac {(x \minus{} y)(x \plus{} y)}{(x \minus{} y)^2} \equal{} \frac {x \plus{} y}{x \minus{} y} \equal{} \frac {100}{74} \equal{} \boxed{\frac {50}{37}}$.

Edit: Yes you are correct, mathemagician. Thanks. :)[/quote]

Is this a solution or not? :wink:[/quote]

Okay...I apologize to you and mathemagician.  I though mathemagician just gave me the answer, but later after you edited your post, I looked at mathemagicains more closely and found out he was correcting you.  So i'm sorry, and that is a solution (a great one by the way, i never think of substitutung variables ever..

When you see the difference of squares, you should automatically think of $ (a\minus{}b)(a\plus{}b$.  Also, when you have two numbers squared with their product times two in the middle, automatically think of $ (a\plus{}b)^2$ and $ (a\minus{}b)^2$.

[quote="AIME15"]When you see the difference of squares, you should automatically think of $ (a \minus{} b)(a \plus{} b$.  Also, when you have two numbers squared with their product times two in the middle, automatically think of $ (a \plus{} b)^2$ and $ (a \minus{} b)^2$.[/quote]

I got the difference of squares part easily, but the square of 87 + 13 I didn't get.  I there a way to do this problem without using variables?  I just want to see the other way if there is one.

Yes, there is. math154 did it, he just used variables so it's more clear. For these kinds of problems, just remember your factoring tricks.

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

$ (a \plus{} b)^2 \equal{} a^2 \plus{} 2ab \plus{} b^2$

$ (a \minus{} b)^2 \equal{} a^2 \minus{} 2ab \plus{} b^2$

$ (a \plus{} b)^3 \equal{} (a \plus{} b)(a^2 \plus{} 2ab \plus{} b^2)$

and so on.


The standard is the mathbb font. \mathbb{R} gives $ \mathbb{R}$, \mathbb{C} gives $ \mathbb{C}$, and so on.

[quote="Z-coli"]how do you write the $ LaTeX$ for the sets of Reals, Complex,...[/quote]
[code]$\text{\LaTeX}$[/code]
$ \text{\LaTeX}$

Hi,

I prefer the dsfont package with \mathds{R} etc.
See also the [url=http://www.ctan.org/get/info/symbols/blackboard/blackboard.ps]blackboard fonts demo[/url].

Stefan

--
[url=http://texblog.net]TeXblog.net[/url]


number theory
algebra
difference of squares
special factorizations
AMC

Source: WMTS 1995 Round 2 #3

Find all positive integers x and y such that:

x:^2: + y = x + y:^2: + 10

[hide]

Combining like terms and factoring, we get X^2-y^2=(x-y) +10. Then, moving the (x-y) to the left, we get (x-y)(x+y-1)=10. SInce x, y are positive integers, we set (x-y), and (x+y-1) to be the factors of 10, namely (1,10), (2,5), (5,2), (10,1), then all the negatives. I know i'm duplicating here, and that there are more resttrictions, but for beginners, it's good not to jump too far.[/hide]

[hide]Right as rain.[/hide]

Rain falling down on me splash...

x^2 - x - y^2 + y = 10, multiply by 4 and factor to (2x-1)^2 - (2y-1)^2 = 40. Thus, we want to express 40 as a difference of squares of two odd numbers. With a touch of magic we find 40 = 19 + 21 = 7 + 9 + 11 + 13, hence the only such expressions are 11^2 - 9^2, 7^2 - 3^2; 2x-1 = 11, 2y-1 = 9 => x = 6, y = 5; 2x-1 = 7, 2y-1 = 3 => x = 4, y = 2.

Maybe a year ago I would have found the superior solution, but I've been reading too many number theory books, and as a consequence have become terrible at number theory.


Let $ M$ be the biggest distance between 6 points and $ m$ the smallest one.Prove that $ M\geq m\sqrt{3}$

[quote="The Zuton Force"]Where are the points located? We can put them into 6-dimensional Euclidean space and use (1,0,0,0,0,0) and permutations  :wink:

I assume you mean either 1D, 2D, or 3D, but please clarify[/quote]
you only have to think on one plane...

Assume what you said could be true.
WLOG, let the smallest distance between and two points be 1, between A and B.
Draw the locus of all points between 1 and $ \sqrt {3}$ away from both A and B.
I think this is made of two separate shapes that look like curved quadrilaterals.
There must exist 4 points in this locus, all between 1 and $ \sqrt {3}$ away from each other.
But one must be in a separate part than one of the others to not be 1 away, in which case
it would be at least $ \sqrt {3}$ away. It would be $ \sqrt {3}$ away at the two points formed by making the two equilateral triangles with AB as a side and then taking the vertex of each of them.


I saw this once in mafiascum, and liked it (Klebian won that particular game by the way).
Send in by PM five [i]single[/i] words you can associate with each of the following. You score a point for each hit you score with each other person (hence for each of the words you send in, the number of people other than you who selected it too is the number of points added to your score). Proper nouns are allowed.

1 Soccer
2 Keyboard
3 Sheep
4 Manifold
5 Braces
6 Cereals
7 Fan
8 Dialectics
9 Prime
10 Facebook

I'll announce tha results when I've had enough entries.

I can't call the enthusiasm is overwhelming, but at least we're off the mark.

Here are our first entrants:
dragon96
perfect628
(who right now I'm sure are tied  :wink: )

I'll be expecting more, guys.

and finally, with the joining in of pacman2812, I can assure that there will be some results declared even if we have no other participants.  :)

must I put a deadline?  :)
james4l has also joined the fray by the way.


calculus
integration
trigonometry
calculus computations

Compute $ \displaystyle{\int\limits_1^6 {\frac{2}{\sqrt{x\plus{}3}}}dx}$

[quote="PeterP"]Differentiate $ \sqrt {x + 3}$.[/quote] [hide="answer"] $ \displaystyle{\frac{1}{2}(x+3)^{-\frac{1}{2})}}$ [/hide]

You should learn this by heart, because you would frequently see this kind of integral.

$ \boxed{\frac {d}{dx}\sqrt {x} \equal{} \frac {1}{2\sqrt {x}}}$

\[ \left. {\left. {\int\limits_1^6 {\frac{2}
{{\sqrt {x \plus{} 3} }}dx \equal{} } 2\int\limits_1^6 {\frac{{d\left( {x \plus{} 3} \right)}}
{{\sqrt {x \plus{} 3} }}}  \equal{} 2\int\limits_1^6 {\left( {x \plus{} 3} \right)^{ \minus{} \frac{1}
{2}} d\left( {x \plus{} 3} \right)}  \equal{} 2\left( {2\left( {x \plus{} 3} \right)^{\frac{1}
{2}} } \right)} \right|_1^6  \equal{} 4\sqrt {x \plus{} 3} } \right|_1^6
\]

How about $ \int \frac{x}{\sqrt{x^2\plus{}1}}\ dx$?

We have:
$ \int \frac {x^{n}}{\sqrt {x^{n \plus{} 1} \plus{} 1}}\ dx \equal{} \frac {2}{n \plus{} 1}\int \left(\sqrt {x^{n \plus{} 1} \plus{} 1}\right)'\ dx \equal{} \frac {2}{n \plus{} 1}\sqrt {x^{n \plus{} 1} \plus{} 1} \plus{} C$.

And $ n \equal{} 1 \implies \int \frac{x}{\sqrt{x^2\plus{}1}}\ dx \equal{} \sqrt {x^2 \plus{} 1} \plus{} C$.

Edited: to add the general case.

That's correct. How about $ \int \frac{\cos x}{\sqrt{\sin x}}\ dx$?

We have:
$ \int \frac{\cos x}{\sqrt{\sin x}}\ dx \equal{} 2\int \left(\sqrt{\sin x}\right)'\ dx \equal{} 2\sqrt{\sin x} \plus{} C$

Yes. Last one:

$ \int \frac{f'(x)}{\sqrt{f(x)}}\ dx$

We have:
$ \int \frac{f'(x)}{\sqrt{f(x)}}\ dx \equal{} 2 \int \left(\sqrt{f(x)}\right)'\ dx \equal{} 2\sqrt{f(x)} \plus{} C$.

Generalization:

$ \boxed{\int \{f(x)\}^nf'(x)\ dx\equal{}\frac{1}{n\plus{}1}\{f(x)\}^{n\plus{}1}\plus{}C\ (n\neq \minus{}1)}$.


induction
linear algebra
linear algebra unsolved


His math teachers for the past 3 or 4 years of his life have been very distant, and they didn't seem to care about math.  They probably don't know his name, and if they do, they only know him as that quiet kid in the back somewhere who aces the quizzes.  In addition, he doesn't think his teacher has ever actually seen him or any other student work in critical thinking or problem solving in a team.  His teacher is always behind her desk on his computer while the students work on problems from the book.  Who should fill out his teacher recommendation form?

I don't believe the academic recommendation absolutely must be from a math teacher.  My recommendation letter last year was actually from my Electronics Professor who I thought knew me well enough to write a recommendation.  It must have been acceptable  because I was accepted.  So if you have a science teacher who knows you better than your math teacher or any other teacher who could better evaluate you academically, go ahead and ask for a recommendation from a non-math teacher.

In situations like this, go ahead and get a recommendation letter from another teacher.  Of course, it's a benefit to us to see some statement about your problem solving and mathematical abilities, but I definitely think it's more important to have a teacher that you know!

[quote="ztbb"][quote="archimedes1"]What would your position be on recommendation letters from online instructors (such as those from AoPS classes)?[/quote]

We're always happy to get recommendation letters from AoPS instructors -- we know them and they know our program, so their letters are more useful than most.[/quote]
Excellent!  Thank you so much!

Compared to other components such as the qualifying quiz, how important is the recommendation letter?

It's not any kind of percent weighting or anything like that.  Typically the quiz results in someone we really feel like we have to accept, someone who we don't think is ready yet for camp, and someone in the middle.  The letter helps with the third case.


Ring Theory
trigonometry
function
LaTeX
calculus
integration
algebra

Let $R$ denote the field of real numbers. Show that
  $R[X,Y]/(X^2+Y^2-1)$ is not a UFD.
I am new to this stuff.So any help how to tackle these problems will be highly appreciated.
I have tried it in the following way.
$X^2+Y^2-1$ is the eqn of a circle.Any pt on it can be represnted by $(\sin \theta ,\cos \theta )$ Now since both sine & cos are periodic functions of 2k pie. So the same pt can be represented in infinitely many no of ways.Am I in the rt dir or I became completely astray

[Moderator edit to fix bad LaTeX.]

That doesn't look like a useful approach. Also, please fix your $\LaTeX$- a smiley got into the works.

Here's an example which shows that your ring does not have unique factorization:
$2x\cdot y\equiv(x+y+1)\cdot(x+y-1)$.

Since $x^2+y^2+c$ is irreducible over $\mathbb{R}$ for all constants $c$, the ring is an integral domain, and the only units are the constants; that makes the example above meaningful.

[quote="jmerry"]That doesn't look like a useful approach. Also, please fix your $\LaTeX$- a smiley got into the works.[/quote]

I dont know how it happened. Thanks for the example.But what my question is was I ever going in the right direction.

[quote="the game"]But what my question is was I ever going in the right direction.[/quote]
No.

To disprove unique factorization, you have to represent an element more than one way[i] as a product of irreducible ring elements[/i], avoiding cases that differ by a unit.


trigonometry
calculus
calculus computations

Hi all,
I have another question on series solution of ODE.
I have this equation:

4xy"+2y'+y=0

Since the x=0 is a regular-singular point, I should use the method of Frobenius. To find the indicial equation I should equate the sum of the coefficients of x^r to zero. But I get two different coefficients a0 and a1 multiplied by some other terms. I don't know what to do in this case.
Can you help me find the indicial equation?

When I teach this, I find myself abandoning sigma notation and spreading it all out. It takes a lot of paper.

Start from $y=a_{0}x^{r}+a_{1}x^{r+1}+\cdots+a_{n}x^{r+n}+\cdots$

$2y'=2ra_{0}x^{r-1}+2(r+1)a_{1}x^{r}+\cdots+2(r+n+1)a_{n+1}x^{r+n}+\cdots$

$4xy''=4r(r-1)a_{0}x^{r-1}+4(r+1)ra_{1}x^{r}+\cdots+4(r+n+1)(r+n)a_{n+1}x^{r+n}+\cdots$

Line those up making sure to align the powers of $x$ properly. The indicial equation is going to come from the lowest power of $x,$ which is $x^{r-1}.$ Notice that that power does not appear in $y,$ just in the two other terms.

We get that $4r(r-1)a_{0}+2ra_{0}=0$. If we then assume that $a_{0}\ne 0$ (that's the assumuption that underlies the indicial equation) we get that $2r(2r-1)=0.$ That gives us $r=0$ or $r=\frac12.$ Since these do not differ by an integer, we can use both directly.

Our general index equation is this:

$4(n+r+1)(n+r)a_{n+1}+2(n+r+1)a_{n+1}+a_{n}=0.$

We can solve this as

$a_{n+1}=\frac{-a_{n}}{2(n+r+1)(2n+2r+1)}$

Take the case $r=0.$ Then we have that

$a_{n+1}=\frac{-a_{n}}{(2n+2)(2n+1)}$

This works out recursively to $a_{n}=\frac{(-1)^{n}a_{0}}{(2n)!}$ or

$y(x)=a_{0}\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{n}}{(2n)!}=a_{0}\cos\left(\sqrt{x}\right).$

The other case is $r=\frac12.$ The recursion becomes

$a_{n+1}=\frac{-a_{n}}{(2n+3)(2n+2)}$

This works out recursively to

$a_{n}=\frac{(-1)^{n}a_{0}}{(2n+1)!}$

$y(x)=a_{0}\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{n+\frac12}}{(2n+1)!}=a_{0}\sin\left(\sqrt{x}\right)$

The general solution to the differential equation is

$y(x)=c_{1}\cos\left(\sqrt{x}\right)+c_{2}\sin\left(\sqrt{x}\right)$

[quote="Kent Merryfield"]Now that we've seen what the answer is - a challenge. Given the form of the answer, this has to be correct:

Make the substitution $z=\sqrt{x}$. Show that the differential equation reduces to

$\frac{d^{2}y}{dz^{2}}+y=0.$[/quote]

You are a great teacher! Thank you so much for your help to all of us  :)


How many permutations $ (a_1,a_2,...,a_{100}) $ of $ (1,2,...,100) $ such that : " There is exactly one number $ i \in (1,2,...,99) $ such that $ a_i > a_{i+1} $

[quote="romano"][quote="maxal"]$2^{100} - 101$[/quote]
  Oh , Can you explain more ?[/quote]
The number of ways to split the set $\{ 1,2,...,100 \}$ into two subsets A and B is $2^{100}$ (since there are two options for each element: it belongs either to A or to B). Elements of A form increasing sequence $a_1 < \ldots < a_i$ (so $i=|A|$) while elemets of B form increasing sequence $a_{i+1} < \ldots < a_{100}$. This gives a solution except the following cases:
1) A or B is empty (2 cases)
2) elements of A are exacly $1,2,\ldots,i$ while elements of B are $i+1,\ldots,100$ so $a_i < a_{i+1}$ (99 cases).
Therefore, the number of solutions is $2^{100} - 101$.


MATHCOUNTS
AMC
AMC 8
AIME
calculus
derivative
geometry

[quote="mathmom"]How do those "in the know" feel that the AMC-8 compares to MathCounts?  When I looked into them, it seemed like MathCounts had much more "interesting" non-routine problems.

Also, it seems like the Math Olympiads for Elementary and Middle Schools [url]http://www.moems.org[/url] are not very popular among members here, but I find them to be very nice problems, and a great place for elementary kids to get a good start.[/quote]

In middle school, MATHCOUNTS is a much larger program with greater resources than the AMC can leverage for middle schools.  The AMC has a better focus at the high school level with the AMC 10, AMC 12, AIME, and USAMO.  At that level they are unparalleled in American education (in my opinion).  The AMC has a very good grasp on the mathematics that relates to the modern world of problem solving.

Around here there is very little discussion of elementary school math education and I must admit my own ignorance given that I wasn't exposed to any programs beyond regular textbooks until middle school and never heard of moems before working at Art of Problem Solving.  I would love to be enlightened by the responses of students who have participated.

The MOEMS offers 2 divisions, E (elementary) for grades 4-6, and M (middle school) for grades 7-8.  Younger students are allowed to participate.

It seems to be "the" contest for Elementary level kids, as there is nothing else major out there.  I think it offers a great place for kids to start getting interested in math competition, without too much pressure.  There are 5 contests, each with 5 questions, admiinistered at the school, one a month from November through March.

I am the "math contest person" at our small school (30 kids, K-8) and this year we are doing MOEMS-E and MathCounts.  Next year I expect we will add MOEMS-M as well.  Since younger kids can participate, my 8yo is getting an opportunity to do math contests, and he is enjoying it.

The MOEMS contests are easier than MathCounts -- that is, there is not a whole lot of differentiation at the top.  Most high-level MathCounts participants could probably ace MOEMS.  But I do find that the problems are interesting and motivating, and not as discouraging for kids just getting started.  (My 8yo, with very little preparation, has correctly answered half of the E-level problems in the 2 contests we have done so far.)  Everyone who participates gets a certificate, and there are additional awards for top 50%, top 10% and top 5% (I think).   There are team awards based on your school's top 10 scores as well, and you don't have to choose the team in advance.   It seems to me like the perfect way to get kids started.  One of my 5th graders is now dying to do MathCounts next year.  :D

The MOEMS problems most closely resemble MathCounts Target round problems, IMO.  There is time pressure, but nothing like the Sprint round.

But to me, it is the Target/Team type questions that are the really "interesting" part of MathCounts.  The sprint round seems like very routine application of mathematical learning.  The AMC-8 seemed more like that (routine, sprint-round type competition) to me, though I must admit that I didn't analyze it too carefully yet.

But anyhow, I'm quite interested in opinions as to what is best for a newly-emerging math team at a school small enogh that fielding a whole team for anything is a challenge.  (As I posted elsewhere, there are only 2 mathematically inclined kids on our MathCounts team -- they dragged along 2 friends who are not really into it themselves.)

Thanks!

(btw, I hope this wasn't the wrong place to post this.  If it makes sense to move it to a separate therad, feel free to move it.)

Thanks for the explanation.  It sounds like a good program.

[quote="mathmom"]The MOEMS offers 2 divisions, E (elementary) for grades 4-6, and M (middle school) for grades 7-8.  Younger students are allowed to participate.

It seems to be "the" contest for Elementary level kids, as there is nothing else major out there.  I think it offers a great place for kids to start getting interested in math competition, without too much pressure.  There are 5 contests, each with 5 questions, admiinistered at the school, one a month from November through March.
[/quote]

My little sister, who is in the fourth grade at a public elementary school in the San Diego, regularly brings home practice problems for the MOEMS.  The problems are actually quite interesting and she seems to enjoy it... MATHCOUNTS is next to nonexistent at her school.  She seems to think the problems are fun, but just another part of her surprisingly heavy load of homework for a fourth grade.  She happily explained to me that the more time you get for a problem, the harder it is supposed to be... it seems like basic pre-algebra and geometry problems though, with a few problem solving curveballs thrown in.

I agree with tokenadult.  My school doesn't offer the AMC exams, so I think there should be a way to take them without having to go to a school.  I haven't tried talking to my teachers and stuff, I'm just a freshman here and the deadline is already over, so I'll probably try next year.

My daughter (5th grade) participates in MOEMS.  It's a reasonable contest for 4th to 6th graders.  MOEMS also offers a separate contest for 7th and 8th graders, but MathCounts is a better (more challenging) program for that level.

Mathcounts is certainly more challenging than MOEMS for grades 7 and 8, but that doesn't necessarily make it "better", at least not from a school's point of view.  It is great for the top few students from a school to participate in Mathcounts, but MOEMS is more a "contest for the masses" so to speak.  Since 35 students can participate for 85 dollars (Mathcounts is 20 dollars per person, and a school is limited to 8 participants) it is a good way to encourage problem solving skills and a love of mathematics and even a "gentle" start to math competition for many more students.

IMO, MOEMS complements MathCounts at the middle school level.  It offers the chance for anyone to compete, and it offers encouragement and success (high scores, awards, etc.) even to many who aren't ready to take competition seriously yet.   Limiting math competition to the top students at the middle school level seems foolish to me, from a school's point of view.

That isn't to say that for the top mathematicians at that level MathCounts isn't the "place to be" because it certainly is.

[size=75]
[edited to fix the mess caused by using dollar signs when not wanting LaTeX -- but escaping them doesn't seem to work][/size]

[quote="Rep123max"]Well anyone in 8th grade and under should definetly take AMC 8. The hardest AMC 8 will get will be really really easy algebra. AMC 10 has algebra and geometry, and this year I think it had just a little trig (like one question). AMC 10 usually has the first 10 questions extremely easy and the last 5 more challenging and the ones inbetween are manageable (what killed me was the geometry). So if you have the option you can take both, its just that you cant take the AMC 10A and AMC 12 A or the AMC 10B and the AMC 12 B but you can take AMC 8, AMC 10A, AMC 10B, or AMC 8, AMC 10A, AMC 12B, etc.[/quote]

I'd have to disagree with you. In 8th grade, I took the AMC 8 and AMC 10, both for the first time, and did better on the 10 than the 8. If you're like me and are stimulated by a challenge, I'd certainly recommend the 10. In fact, take it if you can in any grade (well, >=6, I guess), just to know what kind of problems to look forward to.

What are you disagreeing with? You can take both the AMC-8 and 10, as there is no conflict there. Its hard to compare the difficulty of the two, given that guessing is an option on the 8 but not on the 10.

I did too.  (I got a 136 :) ) If you know the math, it's always to your advantage to take the AMC 12 level contests rather than the AMC 10 level contests.  This is because it's easier to qualify for the AIME and the USAMO by taking the AMC 12.  However, if you don't know all the math on the AMC 12, you're better off taking the AMC 10.

Yo! moderator. Could you kindly delete from this thread, or move to other threads, the posts that aren't strictly relevant to a description of the AMC contests (the thread title)? I've made one round of edits based on early replies to this thread, but the replies are still posted. Now the thread is turning into the kind of conversation thread that one can find anywhere on this forum. I'd do this myself, but I don't have moderation powers on this forum. Maybe after I edit the FAQ part one more time when the extraneous replies are cleaned up then the thread could be locked--what do you think?


limit
trigonometry
calculus
calculus computations

Find
\[\begin{gathered}\mathop{\lim }_{x \to 0}\frac{{1-\cos (x) \cdot \sqrt{\cos (2x)}\cdot \sqrt[3]{{\cos (3x)}}}}{{x^{2}}}\hfill \\ \hfill \\ \mathop{\lim }_{x \to 0}\frac{{1-\cos (x) \cdot \sqrt{\cos (2x)}\cdot \sqrt[3]{{\cos (3x)}}\cdot \sqrt[4]{{\cos (4x)}}}}{{x^{2}}}\hfill \\ \vdots \hfill \\ \mathop{\lim }_{x \to 0}\frac{{1-\cos (x) \cdot \sqrt{\cos (2x)}\cdot \sqrt[3]{{\cos (3x)}}\cdot \sqrt[4]{{\cos (4x)}}\cdots \sqrt[n]{{\cos (nx)}}}}{{x^{2}}}\hfill \\ \end{gathered}\]

sos440: How do you justify the first equality?

please explain.

there was some time this problem $\lim\frac{1-cos(x)cox(2x)...cos(nx)}{x^{2}}$. you only replay sulution in  other language :wink:


geometry
parallelogram
geometry proposed

In a parallelogram ABCD (AB is not equal to BC) AC/BD=k. Let ray AM be symmetric to ray AD according to BD, M-common point of rays AM and BM. Find AM/BM.

AM be symmetric to ray AD according to BD??

Strange, I can`t paint it :(


geometry
circumcircle
incenter
geometry unsolved

Let $w$ be a circle and $AB$ a line not intersecting $w$. Given a point $P_{0}$ on $w$, define the sequence $P_{0},P_{1},\ldots $ as follows: $P_{n\plus{}1}$ is the second intersection with $w$ of the line passing through $B$ and the second intersection of the line $AP_{n}$ with $w$. Prove that for a positive integer $k$, if $P_{0}\equal{}P_{k}$ for some choice of $P_{0}$, then $P_{0}\equal{}P_{k}$ for any choice of $P_{0}$.

[i]Gheorge Eckstein[/i]

Lets consider two circles $ w_{1}$ and $ w_{2}$ ortogonal to $ w$ with  centers in points $ A$ and $ B$ respectively.
$ I\in w_{1} \cap w_{2}$.
$ I$ is exist because line $ AB$ doesn't intersect $ w$.
Lets consider inversion in point $ I$.
We see $ P'_{n + 1}$ is getting from $ P'_{n}$ by composition of two symmetries: relative to lines $ w'_{1}$ and $ w'_{2}$.
$ w'_{1}$ and $ w'_{2}$ are passing through center $ O'$ of $ w'$.
$ \angle P'_{k}O'P'_{k + 1} = 2 \angle(w'_{1},w'_{2}) = \alpha$
So $ \angle P'_{k}O'P'_{k + 1}$ is constant.
$ \angle P'_{k}O'P'_{0} = k\alpha - 2n\pi = 0$ So $ P_{0} = P_{k}$ for any sequence $ P_{0}$, $ P_{1}$, ...,$ P_{k}$ which is built in such way.

The solution is very nice, thanks!

Could the problem G8 from the IMO 1999 also be solved using
inversion:

Points A, B, C divide the circumcircle $ w$ of the $ \triangle$ABC
into three arcs. Let X be a variable point on the arc AB and
$ O_{1}$, $ O_{2}$ be the incenters of the $ \triangle$CAX and
$ \triangle$CBX. Prove that the circumcircle of the $ \triangle
XO_{1}O_{2}$ intersects $ w$ in a fixed point.

This one(IMO)is too simple just use similarities.

We prove, more generally that the statement is true for $w$ [b]any conic[/b] and $AB$ a line not intersecting $w$.

[color=#741B47][b]Problem:[/b][/color] Let $w$ be [b]any conic[/b] and $AB$ a line not intersecting $w$. Given a point $P_{0}$ on $w$, define the sequence $P_{0},P_{1},\ldots $ as follows: $P_{n+1}$ is the second intersection with $w$ of the line passing through $B$ and the second intersection of the line $AP_{n}$ with $w$. Prove that for a positive integer $k$, if $P_{0}=P_{k}$ for some choice of $P_{0}$, then $P_{0}=P_{k}$ for any choice of $P_{0}$.
Proof: Consider a projective transformation which sends $\overleftrightarrow{AB}$ to the line at infinity. So, $w$ gets sent to a conic which does not intersect the line at infinity, therefore an [i]ellipse.[/i] Now, we can take a affine transformation which makes $w$ a circle. Thus, it suffices to prove it for $w$ a circle, and $A$ and $B$ as points at infinity. Let $\ell_A, \ell_B$ be two lines which have infinite points as $A,B$.

Now, let $Q_i$ be the point on $w$ such that $Q_iP_i \parallel \ell_A$. Then we have $Q_iP_{i+1} \parallel \ell_B$. This implies that $\angle (P_iQ_i, P_{i+1}Q_{i}) = \angle (\ell_A, \ell_B) = \alpha$ and so $P_{i+1}$ is obtained from $P_i$ by a rotation of $\beta = 2 \alpha$ at the centre of $w$.

This immediately implies what we desire, since $P_0 = P_k$ if and only if $k\beta = 2n\pi$ for some integer $n$. $\square$


conics
parabola
calculus
calculus computations

[quote="barlowbarlow1"]A car is traveling at night along a highway shaped like a parabola with its vertex at the origin. The car starts at a point 100m west and 100 m north of the origin and travels in an easterly direction. There is a statue at 100 m east and 50 m north of the origin. At what point on the highway will the car's headlights illuminate the statue?[/quote]
Solution
[hide]The car starts at $(-100,100)$ and $(0,0)$ lies on the curve.  The equation of the parabola is $y=a(x-h)^{2}+k$ where $(h,k) \implies (0,0) \iff y=ax^{2}$.  The plugging in the second point: $100=100^{2}\cdot a \iff a=\frac{1}{100}\implies y=\frac{x^{2}}{100}$.

Now that we're here, there are really two ways to do this problem..  It says it travels in an "Easterly" direction and when will the car illuminate the statue.  So we can either assume that when the car is directly facing east is when the car illuminates it or when a tangent line hits the statue.  I believe the latter to be correct.  Both solutions below.

[b]Car facing East solution:[/b]
[hide]$y$ will be $50$ since the statue is at $(100,50)$.  So $50=\frac{x^{2}}{100}\iff x=\pm 50 \sqrt{2}\Rightarrow 50 \sqrt{2}$ $\implies y=50 \iff \boxed{(50 \sqrt{2}, 50) \approx (70.7107,50)}$[/hide]

[b]Tangent line of parabola hitting statue:[/b]
[hide]$y=\frac{x^{2}}{100}\implies y'=\frac{x}{50}\implies y'|_{x=p}=\frac{p}{50}$.  So the tangent line at any point $(p,q)$ is $y-q=\frac{p}{50}(x-p) \iff y-\frac{p^{2}}{100}=\frac{p}{50}(x-p)$ and now we plug the point $(100,50)$ into this equation for $(x,y)$ because this line is supposed to contain that point.

$50-\frac{p^{2}}{100}=\frac{p}{50}(100-p) \iff p=100 \pm 50 \sqrt{2}\Rightarrow p=100-50\sqrt{2}$ $\implies q=150-100\sqrt{2}$.  So the car will illuminate the statue at $(p,q) \iff \boxed{(100-50\sqrt{2},150-100\sqrt{2}) \approx (29.2893,8.5786)}$[/hide][/hide]


I posted the wrong problem, it was too easy. The answer really is 3/4. Here's what I meant to ask:

Box A:
5 Red; 15 Blue

Box B:
8 Red; 11 Blue

Probability that the red ball came from Box A is __ .

Sorry for the confusion.

This might be wrong. [hide]The answer is (1/4)/(1/4+8/19). When you simplify that fraction, you get 19/51[/hide]

[quote="xtrememath8"]I might have made a calculation error, but I think this is correct.
[hide]I think the answer is 19/152. There 1/8 of a chance of picking Box A and then picking a red ball. There is 8/38 of a chance of picking Box B and then picking a red ball. Replacing those fraction with equivelant ones, there is a 38/304 chance of getting a red ball out of Box A and a 64/304 chance of getting a red ball out of Box B. There is 99/304 of a chance of  getting a red ball at all. Since only 38 out of those 99/304ths are from Box A, the answer is 38/304, or 19/152. [/hide][/quote]

Wouldn't you do 38/99 then? I also think there were some significant calculation errors along with your conceptual error. If you simplify (1/8)/(1/8+4/19) {I simplified for obvious reasons}, you get 19/51, if all my computation is correct.

[hide]
Since the fact that a red ball was picked elimates the possibility of picking a blue ball at all, I ignored all of the blue balls. This makes the probability that it came out of the first box 5/13.
-edit-
nvm its 19/51.
Just take total probability of picking a red ball and divide it into the probability of getting it from box a.
[/hide]

One question:
I picking the box independant from the ball picking or is it dependant?
[hide="if indep"]1/2*1/4=1/8
>>1/8<<[/hide]
[hide="if dep"]20/39*1/4=5/39
>>5/39<<[/hide]


What is the terminating decimal value of $ \left(\frac{2}{15}\right)^4\times \left(\frac{3}{2}\right)^8$ ?

[quote="GameBot"]What is the [b]terminating decimal value [/b]of $ \left(\frac {2}{15}\right)^4\times \left(\frac {3}{2}\right)^8$ ?[/quote]

$ .0081$.

[quote=Kouichi Nakagawa]$ \left(\frac {2}{15}\right)^4 \times \left(\frac {3}{2}\right)^8 \equal{} \frac {2^4 \times 3^8}{3^4 \times 5^4 \times 2^8} \equal{} \frac {3^4}{2^4 \times 5^4} \equal{} \left(\frac {3}{10}\right)^4 \equal{} \boxed{\frac {81}{10000}}$.[/quote]

or 0.0081, because it's asking for a terminating decimal.


Is there a formula saying if something is (ax+by)^n, which term will have the largest coefficient?

Like a problem $(x+0.8y)^{100}$ the largest coefficient for a term will be...?

The largest term won't always be the one in the middle of the triangle, since $b$ may be much larger or smaller than $a$.   Generally speaking, the maximum will occur at $x^{k}y^{n-k}$ for $k=\frac{an}{a+b}$, rounded either up or down, but I don't know an easy way off the top of my head to tell which way to round (I don't think it will always be rounding to the nearest integer).  Instead, I suggest you use the following technique.

By the binomial theorem, the $x^{k}y^{n-k}$ coefficient of $(ax+by)^{n}$ is $a^{k}b^{n-k}\frac{n!}{k!(n-k)!}$.  Call this $f(k)$.  To see which $k$ gives the largest $f(k)$, take a look at the ratio $\frac{f(k+1)}{f(k)}$.  If the ratio is bigger than 1, $f(k+1)>f(k)$.  Otherwise, $f(k)>f(k+1)$.  By finding where the switch occurs, you'll find where the maximum is.

[quote="CircleSquared"]The sum of the exponents of $a$ and $b$ in any given term will be the same.
So when $a$ is greater than $b$ (WLOG), for example,
a term with more $a$'s than $b$'s is greater than vice versa.
So shouldnt the greatest coefficient for a term be $\max{(a,b)}^{n}$?

That's for a and b positive though.

Let's say $a$ is negative, but $b$ is positive.
If $|a|$ is greater than $|b|$, the max coefficient is $a^{n-1}b$
If $|b|$ is greater than $|a|$, the max coefficient is $b^{n}$

So $(x+0.8y)^{1}00$, the greatest coefficient is $1^{1}00$

I'm confused, how come I have a radically different answer from everyone else?  :huh:
Did I misunderstand the problem?[/quote]

I think you misunderstood the problem
because I hvae no idea what you are trying to do sorry.

So basically when you exapnad (a+0.8b)^100, it is 100C0*a^100*(0.8b)^0+100C1*a^99*(0.8b)^1...so on the coefiicants are just the numbers (not the a and b part)

and you didnt get readically differnet answers than everyone else because no one has gotten an answer yet.

You must recall that when you have $(x+y)^{n}={n\choose 0}x^{n}y^{0}+{n\choose 1}x^{n-1}y^{1}+\cdots+{n\choose n}x^{0}y^{n}$.

If $\binom{n\choose k}$ got large enough, and $x$ and $y$ were small enough, then the smallest term could be in the middle.

For example:

$(x+1)^{4}=x^{4}+4x^{3}+6x^{2}+4x+1$; the largest coefficient is 6, not $1^{4}=1$.

Ok, I did misunderstand the problem. :oops_sign:

In that case, I think alanchou's answer seems the most right.

[quote="CircleSquared"]Ok, I did misunderstand the problem. :oops_sign:

[b]In that case, I think alanchou's answer seems the most right.[/b][/quote]

[quote="alanchou"]All of these assume n is an integer greater than 1.

If n is even, the largest coefficient if all terms are variables is $\dbinom{n}{\frac{n}{2}}$.

If n is odd, the largest coefficient if all terms are variables is $\dbinom{n-1}{\frac{n}{2}}$.

Correct me if I'm wrong.[/quote]

uh no....not to be offensive to any of you but didnt you read my post?

[quote="Art of Owna"][quote="alanchou"]All of these assume n is an integer greater than 1.

If n is even, the largest coefficient if all terms are variables is $\dbinom{n}{\frac{n}{2}}$.

If n is odd, the largest coefficient if all terms are variables is $\dbinom{n-1}{\frac{n}{2}}$.

Correct me if I'm wrong.[/quote]

Well, number 1, this has nothing to do with the problem.  Number 2, your odd thing is wrong for a binomial such as $(x+y)^{n}$ (it should be $\binom{n}{\frac{n-1}{2}}$ and $\binom{n}{\frac{n+1}{2}}$) , and in this case we arent dealing with a binomial where the coeficients of x and y are 1.

[quote="AstroPhys"](hint)
Use binomial expansion, the largest term will be the one in the center of the corresponding row of Pascal's Triangle.
[/quote]

This is right but that has nothing to do with the problem.  That will only occur in a problem such as the expansion of $(x+y)^{100}$, which the largest coefficient is $\binom{100}{50}$, but if it is something like $(x+0.8y)^{100}$, the $(0.8)^{n}$ where n is between 0 and 100 will affect how big the coeficient is, as the coefficient is no longer is $\binom{100}{k}$, it is $\binom{100}{k}\cdot (0.8)^{k}$[/quote]

[quote="AstroPhys"][hide="HINT"]
Use binomial expansion, the largest term will be the one in the center of the corresponding row of Pascal's Triangle.
[/hide][/quote]

I would really appreciate it if you, DiscreetFourierTransform, and perhaps one or two other posters (edit: I was thinking of vishalarul) would learn to read questions more carefully before regurgitating facts and techniques that are not enough to solve them.


geometry
3D geometry
number theory proposed
number theory

Find all primes $ p$ for that there is an integer $ n$ such that there are no integers $ x,y$ with $ x^3 \plus{} y^3 \equiv n \mod p$ (so not all residues are the sum of two cubes).

E.g. for $ p \equal{} 7$, one could set $ n \equal{} \pm 3$ since $ x^3,y^3 \equiv 0 , \pm 1 \mod 7$, thus $ x^3 \plus{} y^3 \equiv 0 , \pm 1 , \pm 2 \mod 7$ only.

We say that only $7$ has this property.
Suppose $p>7$ is a prime of the form $3k+1$.
So we have $k+1$ cubes.
Then if $x$ can not be written as sum of two cubes then so does $xt^3$ for every $t\neq 0$. (I'm talking in $\mathbb Z_p$)
So there are at least $k$ numbers which can not be written like this.
But by davenport's theorem at least $(k+1)+(k+1)-1=2k+1$ numbers can be written as sum of two cubes.
But we have $2k+1+k=3k+1=p$ so in the davenport theorem the equality occurs.
Hence we have all cubes are in the form $xt$ for some fixed $t\neq 0$ and all $a\leq x\leq a+k$.
But note that cubes are multiplicative so for $a\leq x,x'\leq a+k$ we have $a\leq xx't\leq a+k$.
For $xt=-1$ we have for all $a\leq y\leq a+k$ $a\leq -y\leq a+k$ so we have $a=-\frac k2$.
For $x=1$ we have $-\frac k2\leq t\leq \frac k2$. Now because $p>7$ we have $-\frac k2\leq 2\leq \frac k2$.
Now take $x=2$ and move $x'$ from $1$ to $\frac k2$ so once $xx't$ will not be in $[-\frac k2,\frac k2]$ some time and we are done.

Why do we have $k+1$ cubes? I bet you are refering to some well known lemma. Can you please state it? Thank you.

Yes, that was my original solution (just another way of showing that the equality doesn't happen).
The lemma follows directly from the existence of primitive roots $\mod p$.

The more general statement is:

Let $k$ be given and let $a_1 , a_2 , ... , a_k$ be integers, not divisible by the prime $p>2k+1$.
Then for all $n$ there are $x_1 , x_2 , ... , x_k$ with $a_1 x_1^{2k-1} + a_2 x_2^{2k-1} + ... + a_k x_k^{2k-1} \equiv n \mod p$.

$ n=x^{3}+y^{3}+pz$
Then
${ \{x^{3}+y^{3}}$ is a complete set of reside mod p.
Lemma 1: If p is a prime in the form $ 6k+5$ then
$ \{0,1^{3},..,(p-1)^{3}\}$ is a full reside .
So we can chose ${ \{x^{3}+y^{3}}$ is a complete set of reside mod p
Lemma 2:If p is the form $ 6k+1$ then cubic class reside mod p is $ \frac{p-1}{3}$
I think in this case we quite hard to prove if $ \{x^{3}+y^{3}\}$ is a complete reside mod p.
But for small prime it is not true.

It's true for all primes $ p\neq 7$ as the link shows ;)


Suppose A is a set of one hundred distinct natural numbers with the property that given a, b, c in A (distinct or not), there is a triangle with no angle greater than a right angle, and with sides of lengths a, b, c. What's the smallest possible value for the largest element of A?

So, you finally post it on the forum. But why with a new pseudo? ;)

Let $a_1 < \cdots < a_{100}$ be the integers.
The desired property can be rewritten as :
For all $1 \leq i \leq j \leq k \leq 100$ we have $a_k^2 \leq a_i^2 + a_j^2$.

This reduces to the single condition $a_{100}^2 \leq 2a_1^2.$

In another hand, since the integers are pairwise distinct, we have $a_{100} \geq a_1 + 99.$
Thus $a_1$ has to verify $(a_1+99)^2 \leq 2a_1^2$ which leads to $a_1 \geq 240.$
Then $a_{100} \geq 339.$

Conversely, it is straightforward to verify that the set $\{204,241, \cdots , 339 \}$ has the desired property.
So $339$ is the desired minimum.

Pierre.


Let $a_n$,$b_n$ be two complex and periodic sequences ($a_{n+T_1}=a_n,\forall n\in{\mathbb {Z}}$ and $b_{n+T_2}=b_n,\forall n\in{\mathbb {Z}})$. Let now define $c_n=\sum_{k\in{\mathbb{Z}}}a_kb_{n-k}, \forall n \in {\mathbb{Z}}$. Is $c_n$ periodic ? If it is , find it's period.

I think it is, with period $lcm(T_1, T_2)$, right? You just want to see how long it is till you reach the start of a new period in $a_k$ and in $b_{n-k}$ at the same time.

Well, the length of the period for $a_k$ is $T_1$ and the length of the period for $b_{n-k}$ is $T_2$. Pick any arbitrary $j$ to start from. All of the indices $j + nT_1$ ($n$ any integer) are the same with respect to the sequence $a_k$. Now, all of the indices $j + mT_2$ are the same with respect to the sequence $b_k$. If $a_kb_{n-k}$ is to be periodic, there need to be indices evenly spaced so that the sequence looks the same no matter which index you look at. If we pick some $n$ and $m$ so that $nT_1 = mT_2$, it should be obvious that both $a_k$ and $b_k$ have repeated, so this is a period for the product sequence. The smallest possible value that can satisfy that equation with $n$ and $m$ restricted to integers is $lcm(T_1, T_2)$. That's the period.

[quote="jianuovidiu"]Let $a_n$,$b_n$ be two complex and periodic sequences ($a_{n+T_1}=a_n,\forall n\in{\mathbb {Z}}$ and $b_{n+T_2}=b_n,\forall n\in{\mathbb {Z}})$. Let now define $c_n=\sum_{k\in{\mathbb{Z}}}a_kb_{n-k}, \forall n \in {\mathbb{Z}}$. [/quote]
There is one thing I don't understand here: the series in the definition of $c_n$ is almost always divergent (the sequence of its terms is $T_1T_2$ periodic and, therefore, can tend to $0$ only if it is identically $0$, which can happen for all $n$ only if all $a_k=b_k=0$).

Wow, I'm sorry, I totally misread that -- there I was like an idiot checking if $a_kb_{n-k}$ is periodic....

What fedja says is exactly right (no surprise) -- if a sequence is periodic, then when you sum it as a series, it can only converge if the sum of the terms in one period is 0. (This is because you are adding infinitely many copies of the period, and if the sum of the terms in one period is nonzero, you'll already diverge.)


From a bag of coins, $ \frac{1}{3}$ were given to Mary, $ \frac{1}{5}$ to Norm, $ \frac{1}{6}$ to Anna, and $ \frac{1}{4}$ to Bjorn. The six left were given to Troy, How many coins were originally in the bag?

First, let's find what fraction of the total coins in the bag were given to Mary, Norm, Anna, and Bjorn.  This is $ \frac{1}{3} \plus{} \frac{1}{5} \plus{} \frac{1}{6} \plus{} \frac{1}{4} \equal{} \frac{57}{60}$.  Since there were $ 6$ left, we know that $ \frac{60}{60} \minus{} \frac{57}{60} \equal{} \frac{3}{60} \equal{} \frac{1}{20}$ of the total coins is $ 6$.  Thus, if we were to find the total coins in the bag, we would multiply $ 20$ with $ 6$ to get $ \boxed{120}$ total coins in the bag.

Let $ c$ be the number of coins in total, then

$ c(1/3 \plus{} 1/5 \plus{} 1/6 \plus{} 1/4) \plus{} 6 \equal{} c$

$ c(20/60 \plus{} 12/60 \plus{} 10/60 \plus{} 15/60) \plus{} 6 \equal{} c$

$ 57c/60 \plus{} 6 \equal{} c$

$ 1/20c \equal{} 6$

$ \boxed{c \equal{} 120 \ marbles}$

EDIT: Lol, yeah sorry.

@modularmarc101:  marbles?  Isn't it supposed to be coins?


Could someone please explain the solutions to parts c, d, and e to the attached problem?  Thanks in advance!

Please start by showing some of your own work. Where are exactly your doubts?

For (c) (i), I'm pretty clueless.

For (c) (ii), I plan on using $ q\equal{}mc\Delta T$.  However, I don't know the mass nor the specific heat of the solution.

For (d) (i), I think that it could be "increase" or "stays the same."  Increase seems to obvious.  I know that decreases is probably not the answer.  I'm looking for a more rigorous argument....

For (d) (ii), is there a difference between $ \Delta H _ {\text{neut}}$ and $ q$?  If so, how?  If they're the same, the wouldn't (d) (i) and (d) (ii) be the same?

For (e), would the answer be decrease?  This is my instinct, but again, I don't have a rigorous argument.

Thanks in advance for your help!


For any positive integer $ k$, let $ f_1(k)$ denote the square of the sum of the digits of $ k$.  For $ n \ge 2$, let $ f_n(k) \equal{} f_1(f_{n \minus{} 1}(k))$.  Find, with [i]proof[/i], $ f_{1988}(11)$.

[hide="Hint"]Try to brute force it. (Find $ f_1(11)$, then $ f_2(11)$, and so on...)
Yes, I'm definitely one to advocate that. :D [/hide]


a chain of length $L$ and mass $m$ is placed on a table so that a fraction of the chain is on the table and a fraction hangs.  If linear density is constant throughout and the coefficient of friction is $\mu$ then what is the smallest fraction of the chain that can rest on the table without the chain sliding.

Let's say  $x$  to the part of chain on the table and  $L-x$  to the other part.

k-coefficient of friction

$m_{1}=\frac{mx}{L}$

$m_{2}=\frac{m(L-x)}{L}$

$km_{1}g=m_{2}g$

$\frac{kmx}{L}=\frac{m(L-x)}{L}$

$\frac{x}{L-x}=\frac{1}{k}$

$kx=L-x$                  $x=\frac{L}{(1+k)}$


[quote="neelnal"]I have seen the II, or double I, notation many times and am wondering what it is called and what it is for? Give me an example problem so I can see it in use.[/quote]
Roman numerals? (probably not)

Maybe it is $\Pi$ (Pi), which is greek for capital p, and usually stands for a product?

[quote="arsemesomething"]i did a (very thorough) search here, [url]http://www.tug.org/tex-archive/info/symbols/comprehensive/symbols-letter.pdf[/url], but couldn't find it.

you couldn't have meant $\prod$, could you?[/quote]

yes that is what i mean: $\prod$

That's like the summation operator capital sigma, except it works by multiplication and EDIT: multiplying the results of the terms.

[quote="ragnarok23"][quote="i_like_pie"]$\sum^{4}_{n=1}n=10$

$\prod^{4}_{n=1}n=?$[/quote]
$=4!=24$[/quote]
Oh, I see. I thought so, but Discreet confused me. :D

Yeah, my bad  :lol: I said adding when I meant multiplying  :P Sorry

Another notation question, what does the big f like thingy do? The f without the little slash in it?

Do you mean integral marker thingy?
It defines the limits of the integral you are evaluating, I think.

$\int^{b}_{a}f(x)\,dx$ is the area between the curve $f(x)$ and the x-axis between the lines $x=a$ and $x=b$. :D

And that's calculus, so that doesn't belong here.  While we're at it, though...

$\int^{b}_{a}f(x)dx=F(b)-F(a)$, where $F(x)=\int f(x)dx$

Actually, I'm just happy about knowing that because I learned it myself from my calculus book today, even though our class is still on related rates...

I will just lock the topic so we do not start discussing calculus and such now   :)


Find the inradius of a triangle with side length $ 5,\ 5,\ 6$.

Area of triangle = $ sqrt{8*3*3*2)}$
= 12 sq units
semiperimeter = 8

Inradius = Area / semiperimeter = 12/8 = 1.5 units

Solution :$ AK\equal{}3$ ,$ AB\equal{}BC\equal{}5$,$ OK\equal{}r$,$ BO\equal{}4\minus{}r$.Such $ AO$ he bisector.$ \frac{r}{4\minus{}r}\equal{}\frac{3}{5}$ .Hence $ r\equal{}1,5$

Yes, the answer is $ \frac 32$

Here is my solution:

Draw the perpendicular $ AH$ to $ BC$, since $ BH \equal{} CH \equal{} 3$, so we get $ AH \equal{} 4$.
Thus $ \frac {5 \plus{} 5 \plus{} 6}{2}r \equal{} \frac 12\cdot 6\cdot 4$, yielding the inradius $ r \equal{} \frac 32$


Prove that for an obtuse triangle ABC with $ A \geq 90$ we have

$ b \plus{} c \leq \sqrt {2}a$

where $ a,b,c$ are sidelengths of triangle ABC opposite at angles $ A,B,C$

$ (b\plus{}c)^2\leq 2(b^2\plus{}c^2)\equal{}2a^2\plus{}4bccosA\leq 2a^2$

We know that $ 4m_a^2 \equal{} 2\left(b^2 \plus{} c^2\right) \minus{} a^2$ . Thus,

$ A\ge 90^{\circ}{\ \Longleftrightarrow\ } 2m_a\le a\Longleftrightarrow 2\left(b^2 \plus{} c^2\right) \minus{} a^2\le a^2\Longleftrightarrow b^2 \plus{} c^2\le a^2$ .

Therefore $ \frac 12\cdot (b \plus{} c)^2\stackrel{Live}{\ \le\ } b^2 \plus{} c^2\le a^2\Longleftrightarrow b \plus{} c\le a\sqrt 2$ .

Shortly, $ (b\plus{}c)^2\le 2\left(b^2\plus{}c^2\right)\equal{}4m_a^2\plus{}a^2\le 2a^2$ .


Find all functions $ f : \mathbb{R}\mapsto\mathbb{R}$ such that \[ f((x \minus{} y)^2)\equal{}(f(x))^2 \minus{}2xf(y)\plus{}y^2\] holds for all $x,y \in \mathbb R$.

[quote="mahanmath"]Hi ! Find all function $ f : \mathbb{R}\mapsto\mathbb{R}$ such that $ f((x \minus{} y)^2) \equal{} (f(x))^2 \minus{} 2xf(y) \plus{} y^2$[/quote]

I have a lengthy solution, please check if i made an error in it. Thank you.

Let $ P(x,y)$ be the assertion for $ f((x \minus{} y)^2) \equal{} (f(x))^2 \minus{} 2xf(y) \plus{} y^2$.

$ P(0,0)$, then $ f(0) \equal{} f(0)^2$ and thus $ f(0) \equal{} 0$ or $ f(0) \equal{} 1$.

[i]Case 1.[/i] $ f(0) \equal{} 0$.

$ P(0,t)$, then $ f(t^2) \equal{} t^2$, and thus $ f(t) \equal{} t$ for all $ t \ge 0$.

Thus $ (x \minus{} y)^2 \equal{} (f(x))^2 \minus{} 2xf(y) \plus{} y^2$ for all real numbers $ x$ and $ y$ and thus $ x^2 \minus{} 2xy \equal{} (f(x))^2 \minus{} 2xf(y)$ for all real numbers $ x$ and $ y$. Here, if we put $ x \equal{} y \equal{} t$, then $ \minus{} t^2 \equal{} (f(t))^2 \minus{} 2tf(t)$ and thus $ (f(t) \minus{} t)^2 \equal{} 0$ and thus $ f(t) \equal{} t$ for all real numbers $ t$. This is function obviously satisfies the condition.

[i]Case 2.[/i] $ f(0) \equal{} 1$.

$ P(t,0)$, then $ f(t^2) \equal{} (f(t))^2 \minus{} 2t$. (*)

$ P(0,t)$, then $ f(t^2) \equal{} 1 \plus{} t^2$ and thus $ f(t) \equal{} 1 \plus{} t$ for all $ t \ge 0$. (**)

Now, from (*), we have that $ (f(t))^2 \minus{} 2t \equal{} 1 \plus{} t^2$ so $ (f(t))^2 \equal{} (t \plus{} 1)^2$ for all real numbers $ t$.

The first function we get is $ f(t) \equal{} t \plus{} 1$. Check that in this case, $ f((x \minus{} y)^2) \equal{} x^2 \minus{} 2xy \plus{} y^2 \plus{} 1$ and
\[ (f(x))^2 \minus{} 2xf(y) \plus{} y^2 \equal{} (x \plus{} 1)^2 \minus{} 2x(y \plus{} 1) \plus{} y^2 \equal{} x^2 \plus{} 2x \plus{} 1 \minus{} 2xy \minus{} 2x \plus{} y^2 \equal{} x^2 \minus{} 2xy \plus{} y^2 \plus{} 1\]
, so this function satisfies the condition.

The second is $ f(x) \equal{} \minus{} (x \plus{} 1)$ for all real numbers $ x$. But $ f(0) \equal{} 1$, contradiction.

Now, assume that $ \exists a,b \in \mathbb{R}$ such that $ f(a) \equal{} a \plus{} 1$ and $ f(b) \equal{} \minus{} (b \plus{} 1)$. Here, we may assume that $ a,b \ne \minus{} 1$.
By assertion $ P(a,b)$, then
\[ f((a \minus{} b)^2) \equal{} (f(a))^2 \minus{} 2af(b) \plus{} b^2\]

\[ f((a \minus{} b)^2) \equal{} a^2 \plus{} 2a \plus{} 1 \plus{} 2a(b \plus{} 1) \plus{} b^2 \equal{} a^2 \plus{} b^2 \plus{} 2ab \plus{} 1 \plus{} 4a.\]
By assertion $ P(b,a)$, then
\[ f((a \minus{} b)^2) \equal{} (f(b))^2 \minus{} 2bf(a) \plus{} a^2 \equal{} b^2 \plus{} 2b \plus{} 1 \minus{} 2b(a \plus{} 1) \plus{} a^2 \equal{} a^2 \plus{} b^2 \minus{} 2ab \plus{} 1.\]
Now,
\[ a^2 \plus{} b^2 \plus{} 2ab \plus{} 1 \plus{} 4a \equal{} a^2 \plus{} b^2 \minus{} 2ab \plus{} 1\]

\[ 4ab \plus{} 4a \equal{} 0\]
but since $ b \ne 1$, then $ a \equal{} 0$.
Now, we obtain that $ f(x) \equal{} 1$ for $ x \equal{} 0$ and $ f(x) \equal{} \minus{} (x \plus{} 1)$ for $ x \ne 0$. Thus $ f(1) \equal{} \minus{} 2$ but by (**), $ f(1) \equal{} 1 \plus{} 1 \equal{} 2$, contradiction.

Thus, he solutions are: (1) $ f(x)\equal{}x$ for all real numbers $ x$, and (2) $ f(x)\equal{}x\plus{}1$ for all real numbers $ x$.

[quote="mahanmath"]Hi ! Find all function $ f : \mathbb{R}\mapsto\mathbb{R}$ such that $ f((x \minus{} y)^2) \equal{} (f(x))^2 \minus{} 2xf(y) \plus{} y^2$[/quote]

$ f((x \minus{} y)^2) \equal{} (f(x))^2 \minus{} 2xf(y) \plus{} y^2$
$ f((y \minus{} x)^2) \equal{} (f(y))^2 \minus{} 2yf(x) \plus{} x^2$

Subtracting : $ 0\equal{}f(x)^2\plus{}2yf(x)\plus{}y^2\minus{}f(y)^2\minus{}2xf(y)\minus{}x^2$

$ \iff$ $ (f(x)\plus{}y)^2\equal{}(f(y)\plus{}x)^2$ $ \forall x,y$

Setting $ y\equal{}0$ in this equality and naming $ f(0)\equal{}a$, we get $ f(x)^2\equal{}(x\plus{}a)^2$

And so, $ \forall x$, either $ f(x)\equal{}x\plus{}a$, either $ f(x)\equal{}\minus{}x\minus{}a$

1) Suppose $ \exists u\ne 0$ such that $ f(u)\equal{}u\plus{}a$
Suppose then $ \exists y$ such that $ f(y)\equal{}\minus{}y\minus{}a$ :
$ (f(u)\plus{}y)^2\equal{}(f(y)\plus{}u)^2$ $ \implies$ $ (u\plus{}a\plus{}y)^2\equal{}(\minus{}y\minus{}a\plus{}u)^2$ $ \implies$ $ y\equal{}\minus{}a$ and so $ f(x)\equal{}x\plus{}a$ $ \forall x$
Putting this in the original equation, we get $ a\equal{}0$ or $ a\equal{}1$ and the solutions $ f(x)\equal{}x$ and $ f(x)\equal{}x\plus{}1$

2) Suppose $ \exists u\ne \minus{}a$ such that $ f(u)\equal{}\minus{}x\minus{}a$
Suppose then $ \exists y$ such that $ f(y)\equal{}y\plus{}a$ :
$ (f(u)\plus{}y)^2\equal{}(f(y)\plus{}u)^2$ $ \implies$ $ (\minus{}u\minus{}a\plus{}y)^2\equal{}(y\plus{}a\plus{}u)^2$ $ \implies$ $ y\equal{}0$ and so $ f(x)\equal{}\minus{}x\minus{}a$ $ \forall x\ne 0$
Using this in the original equation with $ y\equal{}0$, we see that this is not a solution.

Hence the two solutions $ \boxed{f(x)\equal{}x}$ $ \forall x$ and $ \boxed{f(x)\equal{}x\plus{}1}$ $ \forall x$

Let $ y\equal{}x$, then we have
$ f(0)\equal{}(f(x)\minus{}x)^2$
Let c be the positive square root of $ f(0)$, then $ f(x)\equal{}x \pm c$, where it is plus for some values and minus for others. Substituting into the original equation, we have
$ (x\minus{}y)^2 \pm c \equal{} (x \pm c)^2 \minus{} 2x(y \pm c)\plus{}y^2$
$ \pm c\equal{}c^2 \pm 2xc \mp 2xc$
We see here that $ c\equal{}0$ is possible, and $ f(x)\equal{}x$ is a solution. Now assume $ c$ is not $ 0$.
$ \pm 1\equal{}c \pm2x \mp 2x$
The only way this can be true for all $ x$ is if $ c\equal{}1$, since $ c$ was the positive square root of $ f(0)$ and the $ 2x$'s always have opposite signs and cancel. So we try $ f(x)\equal{}x\plus{}1$, and it works.
So the only two solutions are $ f(x)\equal{}x$ and $ f(x)\equal{}x\plus{}1$

Nice solution, dgreenb801.

Here is a mixed form between our two solutions  :

Let $ P(x,y)$ be the equation $ f((x \minus{} y)^2) \equal{} (f(x))^2 \minus{} 2xf(y) \plus{} y^2$

$ P(0,0)$ $ \implies$ $ f(0) \equal{} f(0)^2$ and so either $ f(0) \equal{} 0$, either $ f(0) \equal{} 1$

1) If $ f(0) \equal{} 0$, $ P(x,x)$ $ \implies$ $ 0 \equal{} (f(x) \minus{} x)^2$ and so $ f(x) \equal{} x$

2) If $ f(0) \equal{} 1$, subtracting then $ P(0,x)$ from $ P(x,0)$, we get $ f(x)^2 \equal{} (x \plus{} 1)^2$ and so : $ \forall x$ : either $ f(x) \equal{} x \plus{} 1$, either $ f(x) \equal{} \minus{} x \minus{} 1$
$ P(x,x)$ $ \implies$ $ 1 \equal{} (f(x) \minus{} x)^2$ and so : $ \forall x$ : either $ f(x) \equal{} x \plus{} 1$, either $ f(x) \equal{} x \minus{} 1$
And so $ f(x) \equal{} x \plus{} 1$ (else we would have for some $ x$ : $ \minus{} x \minus{} 1 \equal{} x \minus{} 1$ and so $ x \equal{} 0$ and $ f(0) \equal{} \minus{} 1\ne \plus{} 1$)

And it's easy to check back that these two funtions indeed are solutions.


algebra
polynomial
blogs
binomial theorem
linear algebra

Compute $ S_n \equal{} \sum_{k\equal{}0}^{n} F_k {n \choose k}$ where $ F_k$ is the Fibonacci sequence with $ F_0 \equal{} 0, F_1 \equal{} 1$.

[hide="Big hint"] See the technique described [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=211560]here[/url]. [/hide]

$ (x \plus{} 1)^n \equal{} \sum_{k \equal{} 0}^n \binom{n}{k}x^k$

Then use the fact that $ F_n$ is a linear combination of two geometric series.

I remember coming up with this when I was learning about the binomial theorem in precalc...

Edit: We can easily evaluate any summation in the form $ \sum_{k\equal{}0}^n\binom{n}{k}a_n$ were $ a_n$ is a linear recurrence.

A well-known application of this technique is the roots of unity filter... $ a_{n\plus{}3}\equal{}a_n$; $ a_0\equal{}1$, $ a_1\equal{}a_2\equal{}0$.

There is a much, much simpler proof using finite differences alone; all you need to do is postulate the form of $ S_n$ from computation :)

This is like a one line solution already...

$ \frac {1}{\sqrt {5}}\cdot ( (\phi \plus{} 1)^n \minus{} (\phi' \plus{} 1)^n) \equal{} \sum_{k \equal{} 0}^n \binom{n}{k}\cdot \frac {1}{\sqrt {5}}\cdot ( (\phi)^n \minus{} (\phi')^n) \equal{} \sum_{k \equal{} 0}^n \binom{n}{k}\cdot F_n$

Of course the characteristic polynomial of this sum is just $ (x \minus{} 1)^2 \minus{} (x \minus{} 1) \minus{} 1 \equal{} x^2 \minus{}3 x \plus{} 1 \equal{} 0$.

I feel that this solution is better because it proves that the recurrence holds. I'm not sure how finite differences help here. They don't nullify geometric series...

[hide="Note also..."] That $ \phi \plus{} 1 \equal{} \phi^2, \varphi \plus{} 1 \equal{} \varphi^2$ so we immediately get that $ S_n \equal{} F_{2n}$. [/hide]
[hide="However..."] Given that information, note that $ \Delta S_n \equal{} F_{2n\plus{}2} \minus{} F_{2n} \equal{} F_{2n\plus{}1}$; in other words, the finite differences of the even-indexed Fibonacci numbers are the odd-indexed Fibonacci numbers, and the reverse will hold.  We can therefore prove that $ \left( \Delta^k F_{2n} \right)(0) \equal{} F_k$ as desired. [/hide]
In any case, you'll want to read my next blog post :)

[quote="t0rajir0u"]$ S_n \equal{} F_{2n}$. [/quote]

Ha thats pretty tricky. But it is reliant on the specific numbers contained in the Fibonacci sequence (that is the equation is in the form x^2=a(x+1); in this case a=1)

It is interesting that {S_n} has the same dimension as {F_n}. For that matter $ F_n$ could be replaced by any recurrence whose characteristic polynomial that has no repeated roots...I wonder what other operations have such an effect.

Also I don't know now but It would be nice to see the effect on of the operation on $ F_k\equal{}k2^k$ or something of that nature (where the characteristic polynomial is $ (x\minus{}2)^2$ ). You can still use the trick of writing $ x\equal{}2$, but then you integrate then put it in closed form, then differentiate, etc. This behaves as we wish and the dimension doesn't change (and the fact that there is a repeated root also remains).

I'll stop my rambling but I do like the connections between linear recurrences and linear algebra, and the effect of certain operators...they lead to interesting summations. (But sadly nothing here is going to be 'new' I'm sure that somebody has fleshed out these connections)


[quote="M4RI0"]Show that for $x,y,z \in \mathbb{R}$:

$\frac{x^{2}}{\sqrt{x^{2}+y^{2}}}+\frac{y^{2}}{\sqrt{y^{2}+z^{2}}}+\frac{z^{2}}{\sqrt{z^{2}+x^{2}}}\geq \frac{x+y+z}{2}$[/quote]
Maybe $\frac{x^{2}}{\sqrt{x^{2}+y^{2}}}+\frac{y^{2}}{\sqrt{y^{2}+z^{2}}}+\frac{z^{2}}{\sqrt{z^{2}+x^{2}}}\geq \frac{x+y+z}{\sqrt2}$ $?$ :wink:
It is L.Buchovsky's problem.
Peter Scholze  has found the magnificent proof of this inequality.

[quote="M4RI0"]Show that for $x,y,z \in \mathbb{R}$:

$\frac{x^{2}}{\sqrt{x^{2}+y^{2}}}+\frac{y^{2}}{\sqrt{y^{2}+z^{2}}}+\frac{z^{2}}{\sqrt{z^{2}+x^{2}}}\geq \frac{x+y+z}{2}$[/quote]
Yeah, this is really weak, because:
$\frac{x^{2}}{x+y}+\frac{y^{2}}{y+z}+\frac{z^{2}}{z+x}\geq \frac{x+y+z}{2}$
holds, and can be proved with Muirhead after mass-expanding.


induction
ratio
geometric series
calculus
calculus computations

Hi all. I need your assistance in the proving of this:

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

When I solve it till the very end, I have:

$ x^n \plus{} (x^2 y^{n \minus{} 2} \minus{} x^{n \minus{} 2} y^2) \minus{} y^n$

How come the above can turn into $ x^n \minus{} y^n$

Thanks.

You've made a mistake.  Those two middle terms shouldn't be there.

Note: any appearance of $ \cdots$ in the middle of a statement stands for an induction - either something proved by induction or something defined by recursion. In seeking to prove things, one possible approach would be to formalize this inductin.

Dear t0rajir0u; can you show me why the two middle terms should not be there? I can't find the steps needed to cancel both out.

Dear Kent Merryfield; the question indicate us to prove and not by induction. Do you think we have to solve them using induction? Can you give me some pointers too?

Thank You for all your help.

Solution.

Well a simple expansion of the brackets, would rigorously have to be proved by induction. But if we can take the formula for the sum of a geometric series as a prerequisite (although, it itself is derived by induction), then we are fine.

Now if $ x \equal{} 0$ the result, I hope, is obvious.

So assume $ x \neq 0$.

Then $ S \equal{} x^{n \minus{} 1} \plus{} x^{n \minus{} 2}y \plus{} ... \plus{} xy^{n \minus{} 2} \plus{} y^{n \minus{} 1}$ is a geometric series with $ n$ terms, with first term $ a$ and common ratio $ r \equal{} \frac {y}{x}$. The result is trivial for $ x \equal{} y$ too, so we can ignore this case.

Thus $ S \equal{} \frac {a(1 \minus{} r^{n})}{1 \minus{} r} \equal{} \frac {x^{n \minus{} 1}(1 \minus{} \frac {y^n}{x^n})}{1 \minus{} \frac {y}{x}} \equal{} \frac {x^n(1 \minus{} \frac {y^n}{x^n})}{x \minus{} y} \equal{} \frac {x^n \minus{} y^n}{x \minus{} y}$

So $ x^n \minus{} y^n \equal{} (x \minus{} y)(x^{n \minus{} 1} \plus{} x^{n \minus{} 2}y \plus{} ... \plus{} xy^{n \minus{} 2} \plus{} y^{n \minus{} 1})$

I thought I would add a small explanation of why those middle terms don't exist:

The term containing $ x^2$ arises due to multiplying the term involving $ x$ in the first bracket by the term involving $ x$ in the second bracket AND adding to this the $ x^0$ term in the first bracket (i.e. the $ y$ term), multiplied by the $ x^2$ term in the second bracket. This is $ x \cdot xy^{n \minus{} 2} \plus{} ( \minus{} y) \cdot x^2y^{n \minus{} 3} \equal{} x^2y^{n \minus{} 2} \minus{} x^2y^{n \minus{} 2} \equal{} 0$. An identical argument can be applied to the cancellation of the $ y^2$ term, and in fact any term involving powers of $ x$ and $ y$ which are strictly between $ 0$ and $ n$.

Dear sludgethrower;

Thank you for the idea. How come I didn't think of it? Thanks indeed! Now I am curious as to why the expansion method which I used cannot prove the equation. Can you give it a try by using the expansion method? Lets see if you arrive at $ x^n \plus{} (x^2 y^{n \minus{} 2} \minus{} x^{n \minus{} 2} y^2) \minus{} y^n$ like I do. I have look through my working but I didn't see any flaw in it. Hope you can give it a try too.

Thanks once again guys. You all really bring me ideas which I have never think of. :blush:

It is not logically possible to prove this, or even state it, without induction. The only choice is whether to make the induction formal or informal. Trying to argue your way through using the $ \cdots$ in the middle is being informal about the induction. That kind of informality works for many people and many arguments - all that is required is a clear understanding of what $ \cdots$ means.

The particular difficulty we are having here is that Symbian has not shown the required clear understanding of the meaning of $ \cdots.$

Here's an exercise for Symbian: write this all out in detail, showing all terms, for $ n\equal{}5.$ Does that help you see?

Dear Kent Merryfield;

Attached forwith is my working to this question.

The moment I get this $ x^n \plus{} (x^2 y^{n \minus{} 2} \minus{} x^{n \minus{} 2} y^2) \minus{} y^n$ I dont know how to continue to prove the final answer. Do have a look at the attached.

Thank You.

I already knew that's what you'd done. I repeat: the central problem is your lack of understanding of "$ \cdots$". In particular, you are not seeing that there are specific terms included in those dots.

Again, I suggest writing it out in full for some specific value of $ n.$

Dear Kent;

May I know if you could show me what is wrong? I have check with various books and they expand them like I do. This question is a bit special with the use of ... I am running out of ideas.

The "$ \dots$" implies that there are terms inbetween: after $ x^{n \minus{} 1}y$ (to the left of the second "$ \equal{}$" sign in your attachment) comes $ x^{n \minus{} 2}y^{2}$, and before $ \minus{} xy^{n \minus{} 1}$ (below) comes $ \minus{} x^{2}y^{n \minus{} 2}$, so the terms cancel out: if you take these terms into consideration, the equation becomes

$ x^{n} \minus{} y^{n} \equal{} x^n \plus{} (\textcolor{red}{x^{n \minus{} 2}y^{2}} \plus{} x^2 y^{n \minus{} 2} \minus{} x^{n \minus{} 2} y^2\; \textcolor{red}{ \minus{} \; x^{2}y^{n \minus{} 2}}) \minus{} y^n$,

which is obviously true.

Symbian, Kent has told you exactly what you need to do in order to understand what's going on -- I don't know what more you expect him to do!  Why don't you work out the problem in the particular cases $ n \equal{} 3$, $ n \equal{} 4$ and $ n \equal{} 5$ and show us what you get?

Dear metafor; thank you. I understand your explanation. Now i see what kent is referring to. Thank you. Now I finally able to solve the problem. haha....

Kent: Sorry for being so dumb...... Now I finally get what you mean. haha..... :blush:

Dear JBL, I see that you don't understand my post. As mention, I am not very clear with Kent explanation from the very start so I had posted the steps which I had done in the attachment hoping for a simplified explanation like metafor's explanation which ring my bell. haha..... :D Anyhow, your post is not very useful in clearing my doubt, dont you think? :oops:

Thanks to everyone who had help me especially to metafor.......


geometry
3D geometry
topology
real analysis
real analysis unsolved

Given a open set $U$, prove that it could be the union of countably many [b]disjoint[/b] open balls $\{B_{i}\}$ and a zero set $Z$, i.e.,
$U = \cup_{i}B_{i}\cup Z$.

Does anyone have some ideas on this? Thanks!

We may assume that $U$ is a cube $Q^{(0)}$; otherwise cut it into disjoint open cubes. Inscribe an open ball $B^{(0)}$ into $Q^{(0)}$. Note that $|B^{(0)}|=c_{n}|Q^{(0)}|$, where $c_{n}>0$ depends only on the dimension. We can say that the volume of $Q$ is reduced by the factor of $(1-c_{n})$ once the central ball is removed. Our goal is to continue this process indefinitely so that on each step we remove at least a fixed proportion of the remaining volume.

The boundary of $B^{(0)}$ is a compact set of measure zero. We can divide $Q^{(0)}$ into smaller cubes $Q_{i}^{(1)}$ so that the total volume of the cubes $Q_{i}^{(1)}$ that intersect $\partial B^{(0)}$ is less than $\frac12 |Q^{(0)}\setminus B^{(0)}|$. Then inscribe an open ball $B_{i}^{(1)}$ into each cube $Q_{i}^{(1)}$ that is disjoint from $B^{(0)}$. The total volume of such balls is at least $\frac{c_{n}}{2}|Q^{(0)}\setminus B^{(0)}|$. Remove them and repeat.

[quote="mlok"]We may assume that $U$ is a cube $Q^{(0)}$; otherwise cut it into disjoint open cubes. [/quote]

how do you cut it into disjoint open cubes? Is it about $U$ being a disjoint union of open cubes of varying size plus their measure-zero boundaries?

Yes, that is what I meant. Dyadic cubes will do the trick.

Another proof:
WLOG we may assume $U$ is bounded. (If not, $U$ is a countable disjoint union of bounded open sets plus a null set.)
It is enough to prove that there is a constant $c_{n}> 0$ depending only on the dimension $n$ and not on $U$ such that there is a finite collection of disjoint open balls in $U$ with the measure of the union at least $c_{n}\cdot m(U)$. We'll find $c_{n}$ by pouring tiny balls of the same size into the bag $U$.

Choose $\epsilon>0$ small enough that the set $V = \{x\in U: B_\epsilon(x)\subset U\}$ satisfies $m(V) > \frac12m(U)$. There is a finite maximal subset $E$ of $V$ such that the balls $B_\epsilon(y)$, $y\in E$, are disjoint.

It is easy to see that the doubled balls cover $V$, i.e. $V \subset \cup_{y\in E}B_{2\epsilon}(y)$ hence $m(V) \le |E|\cdot m(B_{2\epsilon}(0))$.

\[m(\cup_{y\in E}B_\epsilon(y)) = |E|\cdot m(B_\epsilon(0)) \ge 2^{-n}\cdot m(V) \ge 2^{-n-1}\cdot m(U) \]

[quote="mlok"]We may assume that $ U$ is a cube $ Q^{(0)}$; otherwise cut it into disjoint open cubes. Inscribe an open ball $ B^{(0)}$ into $ Q^{(0)}$. Note that $ |B^{(0)}| \equal{} c_{n}|Q^{(0)}|$, where $ c_{n} > 0$ depends only on the dimension. We can say that the volume of $ Q$ is reduced by the factor of $ (1 \minus{} c_{n})$ once the central ball is removed. Our goal is to continue this process indefinitely so that on each step we remove at least a fixed proportion of the remaining volume.

The boundary of $ B^{(0)}$ is a compact set of measure zero. We can divide $ Q^{(0)}$ into smaller cubes $ Q_{i}^{(1)}$ so that the total volume of the cubes $ Q_{i}^{(1)}$ that intersect $ \partial B^{(0)}$ is less than $ \frac12 |Q^{(0)}\setminus B^{(0)}|$. Then inscribe an open ball $ B_{i}^{(1)}$ into each cube $ Q_{i}^{(1)}$ that is disjoint from $ B^{(0)}$. The total volume of such balls is at least $ \frac {c_{n}}{2}|Q^{(0)}\setminus B^{(0)}|$. Remove them and repeat.[/quote]

Can anyone go into detail about exactly how U is supposed to be expressed as the union of disjoint open cubes? I'm new to this  :oops:

[quote]The total volume of such balls is at least $ \frac {c_{n}}{2}|Q^{(0)}\setminus B^{(0)}|$.[/quote]

And how does this follow?


how to you write fractions like:  $\frac{2pi}{3pi}$  so that the "pi" is represented by the actual symbol " $\pi$

Simple, write
[code]\frac{2 \pi}{3 \pi}[/code]
to get $\frac{2 \pi}{3 \pi}$

Yeah , in most of the Latex based stuff , all you need to do is add a \ before the pi to get $\pi$.It's irrelevent whether you type it within a fraction , a summation , or anything for that matter.

So you can also do $\frac{2}{3}\pi$

by doing [code]\frac{2}{3}\pi[/code]


$ \frac{y^2}{x} = \frac{a}{b}$ with $ \gcd(a,b) = 1$

$ \implies x = s^b, y = s^a$ for some $ s$

$ b y^2 = a x \implies b = a s^n, n = b - 2a \in \mathbb{Z}$

$ \implies n = a s^n - 2a \implies a = \frac{n}{s^n - 2}$ (or $ s^n = 2 \&\& n = 0$ which is not possible)

case $ n = 0 \implies a = b = 0 \implies x = y = 1$
case $ n = 1 \implies s^n = 3 \implies s = 3, a = 1, b = 3 \implies y = 3, x = 27$
case $ n = 2 \implies s^n = 4 \implies s = 2, a = 1, b = 4 \implies y = 2, x = 16$
case $ n > 3 \implies s > 2 \implies s^n > n + 2 \implies$ no solutions
case $ n = -m < 0 \implies a = \frac{ms^m}{2s^m - 1} \implies 2s^m -1 \mid m$ but $ 2s^m > 1 + m$ unless $ s = 1 \implies x = 1, y = 1$

so solutions are $ (x,y) = (1,1), (16,2)$ and $ (27,3)$

[quote="Obel1x"]If $ x,y \ge 1$ and $ (x,y) \in Z$.

Solve the equation :
$ x^{y^2} \equal{} y^x$[/quote]

Let $ x\equal{}az$ and $ y\equal{}bz$ with $ \gcd(a,b)\equal{}1$. The equation becomes :

$ a^{b^2z^2}z^{b^2z^2}\equal{}b^{az}z^{az}$ which may be written $ a^{b^2z}z^{b^2z}\equal{}b^{a}z^{a}$

1) If $ b^2z\geq a$, we get $ a^{b^2z}z^{b^2z\minus{}a}\equal{}b^{a}$ and so $ a\equal{}1$ (since $ \gcd(a,b)\equal{}1$) and so $ z^{b^2z\minus{}1}\equal{}b$
If $ z\equal{}1$, we get a solution $ z\equal{}b\equal{}1$
If $ z\geq 2$, $ LHS\geq 2^{2b^2\minus{}1}>b$ and so no other solution.

2) If $ b^2z<a$, we get $ a^{b^2z}\equal{}b^{a}z^{a\minus{}b^2z}$ and so $ b\equal{}1$ (since $ \gcd(a,b)\equal{}1$) and so $ a^{z}\equal{}z^{a\minus{}z}$ with $ a>z$ ($ a\equal{}z$ would imply $ a\equal{}z\equal{}1$)
Let then $ a\equal{}ut$ and $ z\equal{}vt$ with $ u>v$ and $ \gcd(u,v)\equal{}1$. We get :
$ u^{vt}t^{vt}\equal{}v^{(u\minus{}v)t}t^{(u\minus{}v)t}$ which may be written $ u^{v}t^{v}\equal{}v^{u\minus{}v}t^{u\minus{}v}$

2.1) If $ u\minus{}v\geq v$, we get $ u^{v}\equal{}v^{u\minus{}v}t^{u\minus{}2v}$ and so $ v\equal{}1$ (since $ \gcd(u,v)\equal{}1$) and so $ u\equal{}t^{u\minus{}2}$ with $ u\geq 2$
If $ u\equal{}2$, no solution
If $ u\equal{}3$, solution with $ t\equal{}3$
If $ u\equal{}4$, solution with $ t\equal{}2$
If $ u\geq 5$, $ t\geq 2$ and $ t^{u\minus{}2}\geq 2^{u\minus{}2}>u$ and no solution

2.2) $ u\minus{}v< v$, we get $ u^{v}t^{2v\minus{}u}\equal{}v^{u\minus{}v}$ and so $ u\equal{}1$ (since $ \gcd(u,v)\equal{}1$) but then we can not have $ u>v$ so no solution.

Hence the three solutions :
$ 1^{1^2}\equal{}1^1$

$ 27^{3^2}\equal{}3^{27}$

$ 16^{2^2}\equal{}2^{16}$

Too late! Congrats Philg