What is this "rating" thing? I recently see that I have been rated 3, but what does this mean?
Who rated me?
I did not understand.... :huh:

I think it's the average of your post ratings.

You see the rate post text and hexagons above the text of each post? First group of hexagons is average rating of that post, and other group of 6 hexagons is for you to rate someone's post. Mark it as 6 if you think it's excellent, mark it with 1 if you think it's very bad (spam). Of course, you don't have to rate posts. But, someone else will rate yours maybe. The average of all ratings to your posts is your rating. If it's around 3, it's average and normal.

Read the stickie that was in the forum you chose to post this thread in:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=146659


I'd like your views on which book is better for IMO and general enrichment in Geometry : -

Both by Coxeter
(1) Intro to Geometry
(2) Geometry Revisited
What about Challenging Problems in Geometry??? Is it better to get these problems Online than get the above book?

If what I gather from Amazon is correct, (1) is too advanced with undergrad topics while (2) is OK....  so which one do you recommend?? Thanks!

I recently came across Ravi Vakil's Mathematical Mosaic --- the initial impression from Amazon suggest its more elementary and for lower classes...is this correct? Thanks

Regarding the fames Honsberger books, I have the first 2 Gems ... I want to know the contents of Gems III --- does somebody have them? Is III more advanced than I and II??? Also I want to know whether you agree with this --- Honsberger's all books( Like " In Polyas footstpes" or Gems or Morsels etc) are books meant to expose great problems with great solutions to better understanding and learn new skills --- sort of like a good problem-pasttime ... however, for actual problem solving , engel etc are better//// Do you agree??

Also, How does 'Problem solving through problems' by Larson compare with Engel's book for IMO? I have Engel -- is it necessary to go into Larson?

Also, What abouut ACoPS ---- is it needed after Engel?

Finally, I shall also like your opinion on Yaglom's "Challenging Problems with Elementary Solutions Vol I and II' and 'USSR Olympiad book' .,2nd is good and I have(sort of ) it? I'd like your opinion on 1st... Thanks! Basically, is 1st also a problem with loads of solutions(like Honsberger as I have tried to understand above!)

Geometry revisetedis very basic , good however for preolympiad level. The other in an undergraduate uni level book. I have both. I think that for IMO level the second is more usefull , but there you will not find imo problems . Only good tools and nice theory!

babis

Thanks for the reply!
I once had a chance to look at both Books... In Intro , I found that coxeter covered not only basic Euclidean geometry and Projections etc. in great details, but also focused on things like 4D spaces or Braggs lattices etc --- things I believe(not sure!) belong to undergrad level and NOT IMO! I really dont know ! Maybe it helps ! But I got the impression that Intro was probably  beyond IMO --- even you say its an undergrad text! Then why are you recommending it for IMO? Please give your honest opinion Thanks!

I also say revisited --- it was quite short and the author briefly touhes on most things --- maybe thats good for brushing up things and learning basic new stuff like Projection, affine etc! What do you think abt this?

Also, can you please answer my other questions on books!
[color=darkblue][size=150]Note ACOPS - Art and Craft of PRoblem SOlving by Paul Zeitz...[/size][/color]

I got them from NBHM when I cleared INMO!!! I dont think MAA books are available in India besides in major univ libraries!

NBHM?And good luck for qualifying for IMO.(There is one more selection test after INMO,Right?)

[quote="shadysaysurspammed"]Rushil,I have ACOPS and according to the reviews i've got about Engel(PSS),You don't need ACoPS.

P.S---Rushil,Can you pleeeaaassse tell me the [size=150]EXACT [/size]place where you got the hosenberg books and Engel????!!!![size=100]Which City[/size]?I need them very desperately.[/quote]

You can get PSS from Universal Book Stall,Pune
It costs about Rs.300


inequalities
function
trigonometry
inequalities unsolved

[quote="dogcat"]Prove that if $ abc \equal{} 8$ then:
\[ \frac {1}{\sqrt {a^2 \plus{} 1}} \plus{} \frac {1}{\sqrt {b^2 \plus{} 1}} \plus{} \frac {1}{\sqrt {c^2 \plus{} 1}} > 1\text{.}
\][/quote]
Enough to prove it for positive $ a,$ $ b$ and $ c.$
Let $ a\equal{}\frac{2\sqrt{yz}}{x},$ $ b\equal{}\frac{2\sqrt{xz}}{y}$ and $ c\equal{}\frac{2\sqrt{xy}}{z},$ where $ x,$ $ y$ and $ z$ are positive numbers.
Then your inequality is equivalent to $ \sum_{cyc}\frac{x}{\sqrt{x^2\plus{}4yz}}>1.$
The last inequality is homogeneous. Hence, we can assume that $ x\plus{}y\plus{}z\equal{}1.$
$ f(x)\equal{}\frac{1}{\sqrt x}$ is a convex function.
Thus, $ \sum_{cyc}\frac{x}{\sqrt{x^2\plus{}4yz}}\geq\frac{1}{\sqrt{x^3\plus{}y^3\plus{}z^3\plus{}12xyz}}.$
Id est, it remains to prove that $ (x\plus{}y\plus{}z)^3>\sum_{cyc}(x^3\plus{}4xyz),$ which obvious.

hello arqady
what do you mean by $ \sum_{cyc}(x^3 \plus{} 4xyz)$
do you mean $ {(x \plus{} y \plus{} z)^3 \geq x^3 \plus{} y^3 \plus{} z^3 \plus{} 24 xyz}$
because it's true

[quote="y-a-s-s-i-n-e"]hello arqady
what do you mean by $ \sum_{cyc}(x^3 + 4xyz)$
do you mean $ {(x + y + z)^3 \geq x^3 + y^3 + z^3 + 24 xyz}$
because it's true[/quote]
No! ${ \sum_{cyc}(x^3 + 4xyz)=x^3 + y^3 + z^3 + 12xyz}.$  :wink:

thank s again . arqady and nice solution

if we try the same arqady did and we use imo 2001 ineq ( :P ) we obtain that

$ \sum\dfrac{x}{\sqrt{x^2\plus{}4yz}}>\sum\dfrac{x}{\sqrt{x^2\plus{}8yz}}\geq 1$  :ninja:

[quote="arqady"]Prove that if $ abc \equal{} 8$ then:
....
$ f(x) \equal{} \frac {1}{\sqrt x}$ is a convex function.
Thus, $ \sum_{cyc}\frac {x}{\sqrt {x^2 \plus{} 4yz}}\geq\frac {1}{\sqrt {x^3 \plus{} y^3 \plus{} z^3 \plus{} 12xyz}}.$
....[/quote]
Can you explain me clearly where does this come from...how can you use Jensen with the function with 3 variables?
or you do like this?
$ \frac {x}{\sqrt {x^2 \plus{} 4yz}} \equal{} \frac {1}{\sqrt {1 \plus{} \frac {4yz}{x^2}}}$ and have $ 1 \plus{} \frac {4yz}{x^2}$ for a? in $ f(a)$ ?
but the result i get in the way i think isnt like you  :blush: help me plz

[quote="Rose_joker"]
Can you explain me clearly where does this come from...how can you use Jensen with the function with 3 variables?
[/quote]
If $ f$ is convex function on interval $ I$ then for all non-negative $ \alpha_1,$ $ \alpha_2,...,\alpha_n$ such that $ \sum_{i \equal{} 1}^n\alpha_i \equal{} 1$ and for all $ \{x_1,x_2,...,x_n\}\subset I$ the following inequality holds:
\[ \alpha_1f\left(x_1\right) \plus{} \alpha_2f\left(x_2\right) \plus{} ... \plus{} \alpha_nf\left(x_n\right)\geq f\left(\alpha_1x_1 \plus{} \alpha_2x_2 \plus{} ... \plus{} \alpha_nx_n\right)
\]
In our case $ n \equal{} 3,$ $ \alpha_1 \equal{} x,$ $ \alpha_2 \equal{} y,$ $ \alpha_3 \equal{} z,$ $ f(x) \equal{} \frac {1}{\sqrt x},$ $ I \equal{} \left(0, \plus{} \infty\right)$
and $ x_1 \equal{} x^2 \plus{} 4yz,$ $ x_2 \equal{} y^2 \plus{} 4xz,$ $ x_3 \equal{} z^2 \plus{} 4xy.$

No one thought of trig subsitution?
Can we let a=tan x, b=tan y, c=tan z and instead prove
$ tanx \cdot tany \cdot tanz \equal{} 8 \implies cosx \plus{} cosy \plus{} cosz\ge1$

We can assume that $ a$ , $ b$ and $ c$ are positive . so we can take :
$ a\equal{}\frac{2x}{y}$ , $ b\equal{}\frac{2y}{z}$ and , $ c\equal{}\frac{2z}{x}$
and we have : $ (a\plus{}1)^{2} > a^{2}\plus{}1$ so :
$ \frac{1}{\sqrt{a^2\plus{}1}} > \frac{1}{a\plus{}1}$
Thus we have to show that :
$ \frac{1}{a\plus{}1}\plus{}\frac{1}{b\plus{}1}\plus{}\frac{1}{c\plus{}1} \geq 1$ . and by substitution we must show that :
$ \frac{y}{2x\plus{}y}\plus{}\frac{z}{2y\plus{}z}\plus{}\frac{x}{2z\plus{}x} \geq 1$
by Cauchy we have :
$ \frac{y}{2x\plus{}y}\plus{}\frac{z}{2y\plus{}z}\plus{}\frac{x}{2z\plus{}x}\geq \frac{(x\plus{}y\plus{}z)^2}{x^{2}\plus{}y^{2}\plus{}z^{2}\plus{}2xy\plus{}2yz\plus{}2zx}\equal{}1$
 :wink:

[quote=dogcat]Prove that if $ abc\equal{}8$ then:
\[ \frac{1}{\sqrt{a^2\plus{}1}}\plus{}\frac{1}{\sqrt{b^2\plus{}1}}\plus{}\frac{1}{\sqrt{c^2\plus{}1}}>1\text{.}\][/quote]
$ a,b,c \in \mathbb{R^{+}}, \ abc=8. $ [url=https://artofproblemsolving.com/community/c6h553533p3215844]Show that[/url]\[
1 < \frac{1}{\sqrt{a+1}}+ \frac{1}{\sqrt{b+1}}+ \frac{1}{\sqrt{c+1}} <2. \]


[quote="10000th User"]If $f(x)=\frac{x^4+x^3+1}{x+1}$, find all integers $x$ such that $f(x)$ is also an integer.[/quote]

[hide]Divide to get $x^3+\frac{1}{x+1}$ so the only values are $\boxed{0,-2}$.[/hide]

[hide]I used synthetic division to get $x^3+\frac{1}{x+1}$, so $x=0,-2$.[/hide]

Trial and error gets me the results

[hide="Answer"]
0, -2
[/hide]

your solution looks less fun

[hide]I like trial and error, that is why I got an answer of 0 and -2[/hide]


where does your username come from?

mine is from music class in like 3rd grade when we learned about rhondo or whatever form

Anaxerzia made mine for me.  I don't exactly know how he got it, though...

[url]http://www.mathlinks.ro/viewtopic.php?t=158658[/url]  :P .

If you don't know then I fear for your mental health.

nice Peijin.

With me, my last name is Kimmy, and Kimmy was already taken. (hmm i wonder why) so I decided to add something pretty. like star. and then I added 94. year of my birth.

and kimmystar94 sounds decent. it just came to me one day..

and it was actually first used when i was making a neopets account four years ago.. tee hee.

Three things for [i]distracted[/i]:

1. I can spend time on nonproductive things a lot.
2. I frequently make stupid mistakes.
3. I sometimes don't pay attention when people talk.

[i]523[/i] is something random. The number has some connections to me, but I thought of it mostly unconsciously.

Edited: I forgot to close the italic tag properly  :lol:

ra5249 is a random chain of letters i typed in when i was bored

eh well obviously mine is crazy chinese maniac...lol i have no idea where this came from, it is my yahoo email acc name but thats about it..I think I just like its originality and funniness..

I totally own at BF2142 (haven't played in a while though), and its my handle. I used to play crappy free korean games by softnyx (gunbound, rakion) that lagged like crazy. On gunbound I was youredeaddog, cuz I had a babysitter (well he was paid to drive us around and watch us at home) and he was like "DAWG" a lot. On rakion I came up with "serialkiller". I quit when it got way too laggy. Then on BF2142, serialkiller, serialk111er, blah combos of 1s and 3s in the word got taken.
BF2142 is the best game ever...

Mine is a tool of a software called Lightwave 3D. I used to be addicted to it.
In general, it can be any 3D software or someone who make 3D models.

Everything comes from the Dictionary these days.
-sigh-
Floccinaucinihilipilification!

No, hm does not stand for the H & M store nor does it stand for Hannah Montana.

Frankly,  it was the first name that came out of a username generator thing for my Yahoo account.  :D

I'm still technically using my mom's acc.

Random string of letters I typed in in like 4th grade. (Notice they're all in the home row)

[quote="perfect628"]The perfect numbers are 6, 28, 496, 8128, etc. Plus it's an ironic pun on perfect=flawless.[/quote]

Arg, your username made me think 628 was a perfect number, which messed me up on some test. Thankfully it was practice...

Huh. I love the originality. :lol:

Mine--hmm. Well, we got a bookmark in 4th grade with a bunch of words ending in the suffix "-ate". Then, at the bottom, it said "Be proud of what you ate." It was some sort of eat healthy promotion thing cycling throughout my district. I saw the word "scintillate" on there, looked it up and thought it was cool. I've always had an affinity for shiny things....

Fast forward. The word came back for me in the 7th grade during my preparation for Scripps (The Howard Scripps National Spelling Bee--yeah, [i]that[/i] Scripps). I looked it up and saw it had another definition besides "shining" and "sparkly". It also meant some form of "witty". Thought it described me fairly well. :P Adapted it for my yahoo e-mail and use on here, as well.

I don't remember if I used it anywhere else...:P

[quote="The_Scintillator"]Mine--hmm. Well, we got a bookmark in 4th grade with a bunch of words ending in the suffix "-ate". Then, at the bottom, it said "Be proud of what you ate." It was some sort of eat healthy promotion thing cycling throughout my district.[/quote]

I GOT THOSE TOO!!
i loved them. its just that i lost that bookmark that week.

reason 1:

from the english alphabet

a,b,c,d,e,f,g,h,i,j,k,l,m,n,o,p,q,r,s,t,u,v,w,x,y,z

and randomly choose 8 letters...and you have dingzhou

reason 2:

an alien came down and put a stick on the computer screen that said dingzhou, so my username is now dingzhou.

reason 3:

its my name

reason 4:

its his name

reason 5:

its not a name

reason 6:

a guy created this one ring that is above all rings, and named the ring ding cause i rhymes with ring, and then he got the very last letter of the alphabet, and then add the world hound, but he hated the letter n, so he deleted, and then zhoud just doesn't seem right, so he took away the d, so we have zhou, and he pieced it together with ding, and became dingzhou, the ring is now called dingzhou, and then a dog came and ate it, and dingzhou saw it and said, hey! why don't i have my username as dingzhou named after the ring dingzhou!

reason 7:

my name is uohzgnid

reason 8:

dingzhou just sound good

reason 9:

d stand for determination, i stand for intelligence, n stand for nuclear, g stand for gorilla, z stand for zimbabewe, h stand for horrible, o stand for odd, and u stand for uncle.

reason 10:

its better than your name

HI DING! Btw, you spelled [b]Zimbabwe[/b] wrong. :roll:

[quote="dingzhou"]

reason 9:

d stand for determination, i stand for intelligence, n stand for nuclear, g stand for gorilla, z stand for zimbabewe, h stand for horrible, o stand for odd, and u stand for uncle.[/quote]

Or it could stand for
Diareah
Idiot
Nausea
Green
Zoozoo
Huge
Odor
Underwear

:P

Mine comes from my Yahoo e-mail address, which I will not explicitly post to keep my inbox free of any additional spam.  Can't quite remember why I chose that as my e-mail address, but it sounds kind of cool... I guess.

Well, my name is Milo and my parents always tell me this story of when I was three and my sister was telling knock-knock jokes.  After she had told a bunch of jokes, I made up a knock-knock joke of my own:

Ding dong!
Who's there?
Miyo!
Milo who?
Miyo Miyo!

So, yeah.  That's how that originated.

[quote="korammath"]half korean half white
so you get korean american
shortened to koram...
added math for a little extra.[/quote]

i know a half korean and half white....i think you might know him too

[quote="kimmystar94"][quote="miyomiyo"]knock-knock jokes.
[/quote]
[quote="miyomiyo"] Ding dong![/quote]

knock-knock jokes?[/quote]Hey, I was three, okay?

"aan" is the first three letters of my first name and "johri" is my last name. together it makes "aanjohri"

actually, i didnt create this username myself, i just got it somehow

[quote="The_Scintillator"][quote="kimmystar94"]
I GOT THOSE TOO!!
i loved them. its just that i lost that bookmark that week.[/quote]

Huh. What state do you live in? PM me! :D

[quote="CircleSquared"]
Mine is from an old math problem (bonus points if you can name it)[/quote]

Erm. Is there an official name for it? Was it the Greeks? who were trying to square the circle. Or was someone trying to cricle the square?[/quote]

First one :)

I don't actually remember where mine came from  :( . It might have something to do with an Anime' I was watching at the time some guy was banished. :P

My username came from the fact that it's my name spelled backwards. :P

But seriously, my real name is Keeg Tham. :|

Do not believe what my signature tells you concerning my name!! :ninja:

PS: Yeah, I still don't understand that my username wasn't already originally taken by someone else. :huh:

i just loved the spiritual aspects development and presenting in "city of angels" movie...

Mine is supposed to mean "blown by the West wind" with an x at the end, but lots of people call me zephy  :huh: . It's also my username in several other games/forums.

[quote="33286"][quote="Math Geek"]My username came from the fact that it's my name spelled backwards. :P

But seriously, my real name is Keeg Tham. :| [/quote]
Do you mean Keeg Htam? Or is there a typo in your username?
My username spells my name in base 36.
_________________
98㆚[size=134]ɛɛ[/size][/quote]

Yeah, but if I tried to spell that backwards, it would Maht Geek, which sounds like something Indian rather than nerdy, so I decided to change it.

Okay, this joke has gone too far. :P

[quote="guineapiggies"]gee what do you think? of course it's cuz i LOVE guineapiggies!!! they are so cute and chubby!!!!! :jump:[/quote]

but careful, they bite.

and some guinea pigs are actually evil monkeys in disguise using their fuzziness to take over the world, and then they will make all other animals slave.

we might end up as slaves serving the guineapigs food...oh wait..

we do have to serve the food to guineapigs....o no!!!

they took over our world already!!!!

everyone run for their lives!!!!!!!

the guineapigs are coming! the guineapigs are coming!!!!!


modular arithmetic
inequalities unsolved
inequalities

Prove that if $ n$ is an odd positive integer, then the last two digits of $ 2^{2n}(2^{2n\plus{}1}\minus{}1)$ in base $ 10$ are $ 28$.

Wrong section.
[hide="Solution"]Let us set $ n \equal{} 2k \plus{} 1$ for simplicity. We want to show that $ 2^{4k \plus{} 2}(2^{4k \plus{} 3} \minus{} 1)\equiv 28 \pmod{100}, \forall k\in \mathbb{Z}$. We will apply the Chinese remainder theorem: modulo $ 4$, we can factor $ 4\cdot 2^{4k}(2^{4k \plus{} 3} \minus{} 1)\equiv 0 \pmod{4}$; modulo $ 25$ we have $ 2^{4(k \plus{} 5)}\equiv 2^{4k \plus{} 20}\equiv 2^{4k}$ by Euler's theorem, so we only need to consider $ k\in \{0, 1, 2, 3, 4\} \pmod{25}$. Trying these cases, we find that $ 2^{4k \plus{} 2}(2^{4k \plus{} 3} \minus{} 1) \equal{} 32\cdot 256^k \minus{} 4\cdot 16^k\equiv 3\cdot 6^k \minus{} 4\cdot ( \minus{} 9)^k\equiv 3 \pmod{25}$ for each $ k$. We conclude that the expression is congruent to $ 28$ modulo $ 100$.[/hide]


the second of two #s is 3 less than twice the first. their sum is 36 .find the #s

let two numbers be x and y.

y=2x-3

so x+2x-3=36, 3x=39. x=13

answer : 13, 23

This looks like homework...

Maybe we should just have given him ONE solution, or just hints?  Also we should have hidden answers.

No, just check out some of the other responses to his posts. He says he's making them up, but they'll all get deleted if he keeps flooding the forum.


inequalities
triangle inequality
inequalities proposed

What? Nobdy likes the problem? I think it's very interesting that $ \sum_{sum}a^3b$ is bounded so tightly, between $ 5\sum_{sym}a^2b^2$ and $ 4\sum_{sym}a^2b^2$ (Muirhead).

Hint: S.O.S. and triangle inequality written somehow so that it would relate to fourth degree terms...

[quote="rofler"]In a triangle with side lengths $ a,b,c$ prove that:

$ 5\sum_{sym}a^2b^2 > 4\sum_{sym}a^3b$

Can we replace 4 with any larger constant? (Of course, allowing if we must $ >$ to be replaced by $ \ge$)

This is my first problem I have created, hope you like it. (I know that the first part has a very nice solution without Ravi substitution).

Edit: Problem statement changed completely to symmetric sum notation[/quote]
This inequality very weak.
Put $ a\equal{}y\plus{}z;b\equal{}z\plus{}x;c\equal{}x\plus{}y$,this inequality become:
$ \sum x^4\plus{}xy^3\plus{}yz^3\plus{}zx^3\plus{}2(x^3y\plus{}y^3z\plus{}z^3x)\plus{}2\sum \ x^2y^2\plus{}5xyz(x\plus{}y\plus{}z)>0$
Which is obvious true   :wink:  :)
By your way,we can prove that: $ \frac{21}{5}(a^2b^2\plus{}b^2c^2\plus{}c^2a^2) > 4(a^3b\plus{}b^3c\plus{}c^3a)$ :)


Show that $\lim_{x \to 0} \frac{x}{a} \left [ \frac{b}{x} \right ] = \frac{b}{a} ;\quad a \ne 0$

we can get
$\frac{b}{x}-1<\left[ \frac{b}{x} \right] \leq \frac{b}{x}$

$\boxed{(i) \frac{x}{a}>0}$
$\frac{x}{a} \left( \frac{b}{x}-1 \right)<\frac{x}{a} \left[ \frac{b}{x} \right] \leq \frac{x}{a}\cdot \frac{b}{x}$
$\Longleftrightarrow \frac{b}{a}-\frac{x}{a}< \frac{x}{a} \left[ \frac{b}{x} \right] \leq \frac{b}{a}$

and $\frac{b}{a}-\frac{x}{a} \longrightarrow \frac{b}{a}$　$(x\rightarrow 0)$

therefore, $\lim_{x\to\0} \frac{x}{a} \left[ \frac{b}{x} \right] = \frac{b}{a}$

$\boxed{(ii) \frac{x}{a}<0}$
$\frac{x}{a}\cdot \frac{b}{x}\leq  \frac{x}{a} \left[ \frac{b}{x} \right] < \frac{x}{a} \left( \frac{b}{x}-1 \right)$

 (this is almost same as (i) )
$\cdots \cdots$

$\cdots$  so, $\lim_{x\to\0} \frac{x}{a} \left[ \frac{b}{x} \right] = \frac{b}{a}$


[i]Moderator's Note:  Before you even ask, this got split and churchill's comment did not, not because I have any political favoritism, but because his post was tangentially on topic and none of these were close.  --JBL[/i]

Oh GOOD GRIEF! Stop getting politics into everything.

We need new restrictions on what we post on this forum starting with politics.

No, we do not.  Ignore what you don't want to read.  There is no reason to go about limiting speech just because it's a little annoying, no matter what its political viewpoint is.  Offensive, we can talk about.  Not annoying.

You're biased.  Bush hasn't done anything bad to you.  Why are you making accusations?

[quote="churchilljrhigh"]Easy Bill Gates is a friendly Democrat who has given the most money to charity of the world's richest men.  He is very equal and not corrupt like many rich men. He has criticized Bush's extravagant tax cut on the  rich.[/quote]

Everyone is biased.

I can make a good case for what W. has done to me, by the way, using a nice bit of Republican logic, no less:  everything he does with my tax dollars, he does to me.  He has dragged my money into a war it doesn't want to be in; he has given my money as welfare to large corporations while taking it away from programs to support the poor (which is, incidentally, where my tax dollars would like to be).  He has undermined my environment and my rights.

Seriously though how come my post (which had some humor in it) was deleted and these blatantly wrong and biased political comments aren't???

I didn't delete anyone's posts, on purpose at least, except for one of Miths in a different thread which was a long quote from Lunmu followed by the words "Right on, lunmu" or some such thing.  It was in this thread or another one?


[quote="me@home"]which washington is he from?[/quote]

Washington state, where I am from!

[quote="paladin8"]Define the sequence $ (a_{n})_{n \ge 0}$ by $ a_{0}= 0$, $ a_{1}= 1$, $ a_{2}= 2$, $ a_{3}= 6$ and

$ a_{n+4}= 2a_{n+3}+a_{n+2}-2a_{n+1}-a_{n}$, $ n \ge 0$.

Prove that $ n$ divides $ a_{n}$ for all $ n > 0$.[/quote]
The characteristic equation for this is $ x^{4}-2x^{3}-x^{2}+2x+1=(x^{2}-x-1)^{2}=0$
Roots are only $ \{\phi,1-\phi\}=\frac{1\pm \sqrt{5}}{2}$
Plugging in with base cases, it follows that $ a_{n}=nF_{n}$ where $ F_{n}$ are the Fibonacci numbers.


function
algebra
functional equation
algebra unsolved

Let $f : {\Bbb R}\longrightarrow {\Bbb R}$ be a function from the set ${\Bbb R}$ of real numbers to itself. Find all such functions $f$ satisfying the two properties:

(a) $f(x+f(y)) = y+f(x)$ for all $x,y\in {\Bbb R}$,

(b) the set ${\displaystyle \left\{\frac{f(x)}{x} \ : \ \mbox{x is a nonzero real number}\right\}}$ is finite.

$x=0$  :arrow:   $f(f(y))=y+f(0)$
$f(f(x+f(y)))=f(y+f(x))=x+f(y)$
$f(f(x+f(y)))=x+f(y)+f(0)$ then $f(0)=0$ thus $f(f(y))=y$
Put $f(y)=z$ then $f(x+z)=f(x+f(y))=f(x)+y=f(x)+f(z)$
So $\forall x\in \mathbb{Q}$ $f(x)=f(1)x$
Sup $f(x)=f(y)$  :arrow:  $f(f(x))=f(f(y))$ :arrow: $x=y$ thus $f$ is one to one.
As $f(f(y))=y$ we have that $f$ is surjection then $f$ is a bijection.
Then $f$ is stricty increasing or strictly decreasing.
So $\forall x\in \mathbb{R}$ $f(x)=f(1)x$ plug this in the original equation gives $f(1)^2=1$

Amizing i don't use b) :D

[quote="lordWings"][quote="AYMANE"]$f(f(x+f(y)))=f(y+f(x))=x+f(y)$[/quote]
 :?:[/quote]
What is the problem :?:

[quote="AYMANE"]$f(f(x+f(y)))=f(y+f(x))=x+f(y)$[/quote]
I understand $f(y+f(x))=x+f(y)$ (It comes from swapping $x$ and $y$ in the original equation).
But what change or substitution do you use to get $f(f(x+f(y)))=f(y+f(x))$? I can't see it right now.

In order to get $f(0)=0$, I need property b).
[hide="Details"] $y=0$ $\to$ $f(x+f(0))=f(x)$
Case 1) $f(0)=0$
Case 2) $f$ is a periodic function with period $p=f(0)\neq 0$: $f(x+nf(0))=f(x)$.
Let's take any $x\in\mathbb{R^+}$, and we'll define
$\frac{f(x+nf(0))}{x+nf(0)}=k_n\in S$. ($S$ the set defined in b.)
Using $f(x+nf(0))=f(x)$, $\frac{k_0}{k_n}=1+nd$, with $d=\frac{f(0)}{x}$.
So, for every $n\in\mathbb{Z}$, there's a different value of $\frac{k_0}{k_n}$, and then $S$ wouldn't be finite.[/hide]

To AYMANE:

I don't understand the following conclusion:

$f$ is a bijection $\Rightarrow$ $f$ is increasing or decreasing.

Somehow I got the feeling that there are bijective solutions
of the Cauchy-functional-equation which are not monotone.
(I suppose one could construct such functions using
Hamel bases.) But of course you could complete your
solution easily using (b).

Yes, there are non-monotonous bijective solutions to Cauchy. E.g. just take two base-elements from a Hamel-base and interchange them, but leave all others unchanged.

[quote="lordWings"][quote="AYMANE"]$f(f(x+f(y)))=f(y+f(x))=x+f(y)$[/quote]
I understand $f(y+f(x))=x+f(y)$ (It comes from swapping $x$ and $y$ in the original equation).
But what change or substitution do you use to get $f(f(x+f(y)))=f(y+f(x))$? I can't see it right now.[/hide][/quote]
[b]To lordwings[/b]: we have $f(x+f(y))=y+f(x)$ then $f(f(x+f(y)))=f(y+f(x))$
[b]To solyaris[/b]:Yes i am wrong  :blush:

Let's use b) :(
Dont forget $f(1)^2=1$
Put $A=\left\{f(1)x-f(x);x\in\mathbb{R}\right\}$
Suppose that $A$ is infinite :arrow: There exist infinite reals $s\neq 1$ for which there exists $x_s$such that $f(1)x_s-f(x_s)=s$
\[ \frac{f(s-1)}{s-1}=\frac{f(f(1)x_s-f(x_s)-1)}{s-1}=\frac{f(1)f(x_s)-f(1)-x_s}{s-1}=-f(1)-\frac{2f(1)}{s-1} \] thus the original given set is infinite,contradiction thus $A$ is finite.
As $A$ is finite then there exists $u$ such that $|f(1)u-f(u)|=$Max $A$
Take $v$ such that $v=f(1)u-f(u)-1$  :arrow:  $f(v)=f(1)f(u)-u-f(1)$ :arrow: $f(1)v-f(v)=2(u-f(1)f(u))=\frac{2(f(1)u-f(u))}{f(1)}$ :arrow: $|f(1)v-f(v)|=2|f(1)u-f(u)|$ contradiction then $\forall x\in\mathbb{R}$ $f(1)x-f(x)=M$ but we have  $2M=M$  :arrow: $M=0$ :(


calculus integration function conics hyperbola vector trigonometry

Let for any time instance we assume that charged particle is making angle $\phi$ with the vertical axis.Then we write equations for this situation:
r-distance to the particle
$\gamma$-linear charge density
$rd\phi =dl$

$dx=\frac{dl}{cos\phi}=v_{x}dt$

$F=\frac{q\gamma}{2\pi e_{0}r}$

$dt=\frac{dl}{v_{x}cos\ph}$

$F_{y}=F cos\phi=\frac{q\gamma cos\phi}{2\pi e_{0}r}=m\frac{dv_{y}}{dt}$

Let for any time instance we assume that charged particle is making angle $\phi$ with the vertical axis.Then we write equations for this situation:
r-distance to the particle
$\gamma$-linear charge density
$rd\phi =dl$

$dx=\frac{dl}{cos\phi}=v_{x}dt$

$F=\frac{q\gamma}{2\pi e_{0}r}$

$dt=\frac{dl}{v_{x}cos\phi}$

$F_{y}=F cos\phi=\frac{q\gamma cos\phi}{2\pi e_{0}r}=m\frac{dv_{y}}{dt}$

We put the equation of dt in the last equation.
THen we integrate $d\phi$ from $-\frac{\pi}{2}$ to $+\frac{\pi}{2}$ and get the follofing result for $v_{y}$:

$\frac{q\gamma d\phi}{2\pi e_{0}v_{x}}=mdv_{y}$

$v_{y}=\frac{q\gamma}{2m e_{0}v_{x}}$

$tan\alpha=\alpha=\frac{v_{y}}{v_{x}}$

$\alpha=\frac{q\gamma}{2m e_{0}v_{x}^{2}}$

Is that OK?

Vx is changing too so it's not correct :)

The charge is coming from $-\inf$ to $+\inf$.So So the integration of the change of $v_{x}$ gives zero.If not please write the solution.  :maybe:

at one point you have dx=dl/cos (phi) but dx=dl*cos (phi)
   at another point you have dVy=(a constant)*d(phi)/Vx but (as you can also follow in your way of thinking the problem)Vx also varies with phi so when you integrate the last relation you cannot consider Vx a constant
   another mistake is that if you fiind a function of dVy depending on phi you see that both are positive in your solution but dVy is positive all the time while d(phi) is negative when the particle is mooving on the negative part of Ox axe and positive when x is positive.so you should choose the angle that the particle is making with Ox.
    despite all these if you consider Vx as the initial velocity then the ansqer is correct  :)
     i'll try to post a correct sollution

This is a Rutherford scattering problem, with a different central force than usual; here's one way of solving it.

First of all, the force acting on the particle is $F(r) = q\gamma/2\pi\varepsilon_{0}r$, as already noted. This is a central force, so angular momentum is conserved when calculated with respect to a fixed point. The trajectory of the particle is a hyperbola.

Second, let $z$ be the line connecting the wire and the point of the hyperbola where the particle is closest to the wire. The hyperbola will be symmetric with respect to $z$. Let $\vec r$ be the distance of the particle from the wire and let $\phi$ be the angle between $z$ and $\vec r$.

Conservation of angular momentum with respect to the wire yields
\[mbv_{0}= mr^{2}\dot \phi. \]
The force acting on the particle is (in vector form)
\[\vec F = \frac{C}{r^{2}}\hat r, \]
where $\hat r$ stands for the unit vector. The change of momentum along the $z$ line is
\[\Delta \vec p = \int_{-\infty}^{+\infty}\vec F_{z}\text dt, \]
where we have defined $t = 0$ to be when $\phi = 0$.

Since the hyperbola is symmetrical, we have
\[2mv_{0}\sin\frac{\alpha}{2}= \int_{-\infty}^{+\infty}F \cos\phi \text dt. \]
Now, we should try to solve this. I don't know how :( I thought I did, but it was a mistake.

another observation namiqfiziq....the force is not perpendicular on the velocity so you cannot consider the same angle when you calculate the components of the forces and the velocities

djole how did you eliminate r?in your last line you have 2 equalities that don't have the same dimension and you have an extra pi in your final answer

Yes, you're right, I've made a fatal mistake  :blush: I'm not sure I can solve it now... Could you post the solution?

If you saw ,I wrote on the top part $dl=rd\phi$.r is the istance between the electrified wire and the charge.If you consider this you can see that $dx=\frac{dl}{cos\phi}$And the $v_{x}$ in my answer is the initial velocity.I wrote that the change of it is negligible.

:rotfl: now i understand....you confused me with your notations...i thought dx
is dl and vice-versa and the approximation was unexpected.i thought i had to use the approximation in the final answer....which obviously made the problem harder

I want to ask you something.Will you participate in IPHO in Iran?

My national olympics phase is next week :) .what about you?


geometry
3D geometry
tetrahedron
ratio
vector
number theory
relatively prime

In a regular tetrahedron the centers of the four faces are the vertices of a smaller tetrahedron. The ratio of the volume of the smaller tetrahedron to that of the larger is $m/n$, where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

[quote="Karth"]We can use

[hide="radishes"]
So we can let the vertices of the tetrahedron be $A, B, C, D$ and denote the vectors to these vertices as $a, b, c, d.$ Note that the centroids of the vertices of each side are simply:

$\frac{a+b+c}{3}, \frac{a+b+d}{3}, \frac{b+c+d}{3}, \frac{a+c+d}{3}$

If we find a vector from one centroid to another, we get:

$\frac{a+b+c}{3}-\frac{a+b+d}{3}= \frac{c-d}{3}$

Oh my... this looks similar to a side length of the original tetrahedron, which is $\frac{c-d}.$ So, we have that the length of the original tetrahedron is 3 times the length of the inner tetrahedron, so we cube the values of the length (because it's a 3D problem) and we get the ratio to be: $1/27.$
[/hide][/quote]

it is much easier to find the ratio using a pair of simlar triangles

could someone post a soln with similar triangles?

Sorry to revive an old topic, but what is wrong with my solution?

[hide="my solution"] Let the height of the original tetrahedron be 1. Thus the base will be some determineable (but we're not gonna solve for it) are ... k. So the volume is $ \frac{k}{3}$. Since the height of the inscribed tetrahedron is 1/3 that of the original, the base of the original at that height would be $ \frac{2k}{3}$ Note that the area of the base of the inscribed tetrahedron will be a quarter of that ... $ \frac{k}{6}$. So, the volume of the inscribed tetrahedron is $ \frac{k}{54}$, and the ratio follows as $ \frac{1}{18}$, so $ 1 \plus{} 18 \equal{} 19$

[/hide]

[quote="georgiamathdude"]Sorry to revive an old topic, but what is wrong with my solution?

[hide="my solution"] Let the height of the original tetrahedron be 1. Thus the base will be some determineable (but we're not gonna solve for it) are ... k. So the volume is $ \frac {k}{3}$. Since the height of the inscribed tetrahedron is 1/3 that of the original, the base of the original at that height would be $ \frac {2k}{3}$ Note that the area of the base of the inscribed tetrahedron will be a quarter of that ... $ \frac {k}{6}$. So, the volume of the inscribed tetrahedron is $ \frac {k}{54}$, and the ratio follows as $ \frac {1}{18}$, so $ 1 \plus{} 18 \equal{} 19$

[/hide][/quote]

Mistake: "Since the height of the inscribed tetrahedron is 1/3 that of the original, the base of the original at that height would be $ \frac {2k}{3}.$" The ratio of a [i]side[/i] of the scaled original tetrahedron at a height of $ \frac{2}{3}$ from one of its vertices (measured towards the center of the opposite face) to the [i]side[/i] of the original tetrahedron is $ \frac{2}{3}$--not the ratio of the respective base areas. Your ratio, however, equals $ \left(\frac{2}{3}\right)^2\equal{}\frac{4}{9}$ and therefore the area that you were seeking is $ \frac{4k}{9}$. The volume ratio is therefore $ \frac{1}{18}\cdot\frac{2}{3}\equal{}1 : 27$ and the desired sum is $ 1 \plus{} 27 \equal{} 28$.

However, note that when we get the ratio of any length of the inscribed tetrahedron to the corresponding length of the original tetrahedron, we are done, because the tetrahedra are homothetic and thereby similar.

[hide="what am i doing wrong?"]take a face of the original tetrahedron that is not the base. Drop and altitude from the centroid of this triangle to its base. If the side has length 2, this altitude has length $ \frac{1}{3}\sqrt{3}$. Then draw the line from the centroid of the bottom face, to this point (also a truncated altitude of length $ \frac{1}{3}\sqrt{3}$. Then the angle between those two sides is $ \frac{\pi}{3}$, so they form an equilateral triangle where the third side is a side of the inscribed tetrahedron. Thus the ratio of the sides is $ \frac{\sqrt{3}}{6}$ which is wrong...[/hide]

[quote="mathemonster"]Then the angle between those two sides is $ \frac {\pi}{3}$[/quote]  The angle between two faces of a regular tetrahedron is not $ \frac{\pi}{3}$.  (Let the tetrahedron have vertices $ ABCD$, and let $ M$ be the midpoint of $ AB$.  Then the angle in question is $ \angle M$ in $ \triangle MCD$, which is not $ 60^\circ$ because it is the vertex angle of a non-equilateral isosceles triangle.)

Can somebody explain to me how to do this problem without the use of vectors? I looked at the other solutions, but I didn't understand them fully. (sorry to bring back an old post)

Where can I find a good resource to learn how to use vectors for problems like these (and 2D ones)?

For your reference.

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=150&t=381249


[quote="AceOfDiamonds"]the answer is $ 111,\!112$      where after the $ ,$ there is a \!
error?[/quote]
\! is [url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_GuideLay.php#spacing]LaTeX Layout #Spacing[/url]


In this game you asked a (mostly private)question to chosen person and ordered him/her to do something stupid(funny) :) Chosen person has a choice: answer a question or do what he/she ordered him/her. And it's a queue. Averybody is asking a question. Soon I will play this with some people in real life :P This game joins people and is really funny :P
You can post here yours ideas: questions and tasks :)

[quote="easyas3.14159..."]How will you know if they do what you ordered or not? :D[/quote]
I will execute it in school :P Of course the secrets obliges.
[quote="Chocoboom"]you mean truth or dare....?  :|[/quote]
Yes, probably original name is dare. I checked this word in a dictionary. My English isn't very well but I'm trying to improve my skills :)

So, any ideas?

How do you really mean to play this [i]online[/i]?
It's okay among friends at home or at school, though I would advise you to remain decent.
But here...I don't get it! :?

Heh, I didn't want to play it online and I didn't wrote anything unseemly I suppose :/ Maybe your coupling were different to mine or I didn't write my request clearly and plain.
However I'm very sorry. Moderators, in that case you can block this topic :)

There's no need for it to get locked. If you want ideas, just make sure it doesn't get too nasty here. :P

So you want to get hold of things to ask?

Here are some boring truth questions: Favourite colour? Favourite teacher? Which parnt loves you more? Do you like pets? and so on.

Here are some naughty truth questions: Girlfriend? When last did you get your pants wet? When did you last break wind in public? and so on.

For the dare part, ask difficult things: Stare without blinking for two minutes. Stand on one leg. and so on.

And if you get really naughty: Ring that doorbell. Whistle to that girl. and so on.

Go to your sir and say  "may i go to the toilet" adn when he says yes ,just walk back to your seat and if he asks you why u didn't go,say "I don't feel like going now"

[quote="Chocoboom"]lol when i play with friends, it gets soo stupid.

sample truth: who do you like (the person said the name of a girl in the room, then the girl ran out of the room :ninja: but came back shortly
-next round-
ask same person, whats one fantasy you've had about this girl, after he answers, girl runs out of room, this time doesn't come back :ninja:[/quote]

wow... was he saying it for real? lol.

Sample dare: Assume that person is your girlfriend, and propose to him! (Yes [b]him[/b]! it's a boy)


geometry
circumcircle
trigonometry
trig identities
Law of Cosines
Law of Sines
geometry proposed

Let $ \triangle{ABC}$ have $ \hat{A}\equal{}30$.Let $ S$ be its circumcenter and $ I$ its incenter.Choose a point $ D$ on $ BA$ and a point $ E$ on $ CA$ such that $ BD \equal{} CE \equal{} BC$.Show that $ SI$ perpendicular and equal to $ DE$

We will show that $ SE^2 \minus{} SD^2 \equal{} IE^2 \minus{} ID^2$

We use Stewart's theorem in the triangle $ SAB$

$ SA^2\cdot DB \plus{} SB^2 \cdot AD \equal{} SD^2 \cdot AB \plus{} AD \cdot DB \cdot AB \Rightarrow$

$ R^2\cdot a \plus{} R^2\cdot (c \minus{} a) \equal{} SD^2 \cdot c \plus{} (c \minus{} a)ac \Rightarrow$

$ R^2 \equal{} SD^2 \plus{} (c \minus{} a)a$

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

$ \Rightarrow SE^2 \minus{} SD^2 \equal{} (c \minus{} a)a \minus{} (b \minus{} a)a \equal{} a(c \minus{} b)$

Apply the law of cosines in the triangle $ ABC$

$ \Rightarrow a^2 \equal{} b^2 \plus{} c^2 \minus{} 2bc\cdot cosA \equal{} b^2 \plus{} c^2 \minus{} bc$

Denote by $ A' \equal{} AI \cap BC$

It is well known that $ BA' \equal{} \frac {ac} {b \plus{} c}, A'C \equal{} \frac {ab} {b \plus{} c}$ and $ IA' \equal{} \frac {a \cdot IA} {b \plus{} c}$

Now we calculate $ AA'$ using Stewart's theorem in $ ABC$ and using $ a^2 \equal{} b^2 \plus{} c^2 \minus{} bc$. We obtainn:

$ AA' \equal{} \frac {(b \plus{} c)\sqrt {3}} {b \plus{} c}, IA' \equal{} \frac {a} {b \plus{} c} \cdot IA \Rightarrow \frac {AI} {A'I} \equal{} \frac {b \plus{} c} {a} \Rightarrow \frac {AI} {AA'} \equal{} \frac {b \plus{} c} {a \plus{} b \plus{} c}, (1) \Rightarrow$

$ AI \equal{} \frac {bc\sqrt {3}} {a \plus{} b \plus{} c} \Rightarrow AI^2 \equal{} \frac {bc(b \plus{} c \minus{} a)} {a \plus{} b \plus{} c}, (2)$

We use Stewart's theorem in $ IAB$ and $ IAC$ and we obtain:

$ IA^2 \cdot \frac {a} {c} \plus{} IB^2 \cdot (1 \minus{} \frac {a} {c}) \equal{} ID^2 \plus{} a(c \minus{} a)$

$ IA^2 \cdot \frac {a} {b} \plus{} IC^2 \cdot (1 \minus{} \frac {a} {b}) \equal{} IE^2 \plus{} a(b \minus{} a)$

It is enough to prove that:

$ IA^2 \cdot \frac {a} {c} \plus{} IB^2 \cdot (1 \minus{} \frac {a} {c}) \equal{} IA^2 \cdot \frac {a} {b} \plus{} IC^2 \cdot (1 \minus{} \frac {a} {b})$

Now we use the law of cosines to determine $ IB^2$

$ IA^2 \cdot \frac {a} {c} \plus{} IB^2 \cdot (1 \minus{} \frac {a} {c}) \equal{} AI^2 \plus{} c^2 \minus{} 2\cdot AI \cdot c \cdot cos15^{\circ} \minus{} ac$

Similar for $ IC^2$, and we must prove that $ 2\cdot AI \cdot cos15^{\circ} \equal{} b \plus{} c \minus{} a$

$ \Leftrightarrow AI \cdot sin30^{\circ} \equal{} (b \plus{} c \minus{} a)\cdot sin15^{\circ}$

Using the law of sines in $ ABA'$ we get: $ \frac {\frac {ac} {b \plus{} c}} {sin15^{\circ}} \equal{} \frac {AA'} {sinB}$

Using the law of sines in $ ABC$ we get: $ \frac {b} {sinB} \equal{} \frac {a} {sin30^{\circ}} \equal{} 2a$

$ \Rightarrow sin15^{\circ} \equal{} \frac {bc} {b \plus{} c} \cdot \frac {1} {AA'} \cdot \frac {1} {2}$

So it is enough to prove that:

$ AI \cdot \frac {1} {2} \equal{} (b \plus{} c \minus{} a) \cdot \frac {bc} {b \plus{} c} \cdot \frac {1} {2AA'} \equal{} (b \plus{} c \minus{} a) \cdot \frac {bc} {b \plus{} c} \cdot \frac {b \plus{} c} {a \plus{} b \plus{} c} \cdot \frac {1} {AI}$, obviously true from $ (1)$ and $ (2)$


Find positive integers $ a, b, c, d$ such that $ a \plus{} b \plus{} c \plus{} d \minus{}3 \equal{} ab \equal{} cd$ .

[hide="My Solution"]
Without loss of generality, $ 1 \le a \le b\le c \le d$ so we have $ a \plus{} b \plus{} c \plus{} d \minus{} 3 \le 4d \minus{} 3$. We also have $ a \plus{} b \plus{} c \plus{} d \minus{} 3 \equal{} cd \le 4d \minus{} 3\implies 3 \le (4 \minus{} c)d$. The product on the RHS must be positive and it follows that each factor must be positive because $ d$ must be a positive integer. Therefore, we have $ 1\le c \le 3$. From here, we have 3 cases.

Case 1: $ c \equal{} 1$
If $ c \equal{} 1$, we must have $ a \equal{} b \equal{} 1$ from our inequality chain. The equality chain becomes $ d \equal{} 1 \equal{} d$ so the solution for this case is $ a \equal{} b \equal{} c \equal{} d \equal{} 1$. Substituting values, we find that this solution works.

Case 2: $ c \equal{} 2$
If $ c \equal{} 2$, we have $ a \plus{} b \plus{} d \minus{} 1 \equal{} 2d\implies a \plus{} b \minus{} 1 \equal{} d$. Note that $ a \plus{} b \le 4\implies a \plus{} b \minus{} 1 \equal{} d \le 3$. Now suppose that $ d \equal{} 3$. Then we have $ a \plus{} b \equal{} 4$ which is only satisfied by $ a \equal{} b \equal{} 2$. Quickly checking, we find that this does not work. If $ d \equal{} 2$, then we have $ ab \equal{} 4$, which again is satisfied by $ a \equal{} b \equal{} 2$, so there are no solutions for this case.

Case 3: $ c \equal{} 3$
If $ c \equal{} 3$, we have $ a \plus{} b \plus{} d \equal{} 3d\implies a \plus{} b \equal{} 2d$. Note that $ a \plus{} b \le 6$ so that $ d \le 3$. Using the equation $ ab \equal{} cd$ and checking $ d \equal{} 3$, we find that no $ a,b$ exist. Thus, there are no solutions for this case.

The only solution is $ \boxed{(a,b,c,d) \equal{} (1,1,1,1)}$.
[/hide]

EDIT: I just realized that my assumption was wrong, so my proof is not necessarily correct.

Another solution would be that 3 of $ a,b,c,d$ are $ 0$ and 4th one is equal to $ 3$

That wouldn't be a solution because $ a,b,c,d$ are positive integers.

On the other hand, you may assume without loss of generality that $ a \leq c \leq d \leq b$ and see what that gets you.

$ ab\equal{}cd\Longrightarrow a\equal{}ck,\ d\equal{}bk$ for some positive integer $ k$.
$ \therefore a\plus{}b\plus{}c\plus{}d\minus{}3\equal{}ad\Longleftrightarrow (bc\minus{}b\minus{}c)k\equal{}b\plus{}c\minus{}3$.

[quote="Balint Florin"]after you do what JBL said ,some case work + some mod's i got these solutions
{(3,0,0,0);(0,3,0,0);(0,0,3,0);(0,0,0,3);(1,1,1,1);(3,3,3,3)}[/quote]

Again, TachyonPulse *already* reminded a user, and consequently
others here, that the solutions are *positive* integers.

Balint Florin, please discard all of your solutions where at least
one of the values of the variables is a zero.


linear algebra
matrix
vector
algebra
polynomial

If $ \sigma$is a permutation of degree n and A is an n x n matrix over the field F with row vectors $ a_{1},...,a{}_{n}$, let $ \sigma (A)$ denote the n x n matrix with row vectors $ a_{\sigma 1},...,a{}_{\sigma n}$. Is it true that $ \sigma (A)$ is similar to A?

Here is my reasoning. I let T be a linear operator on $ F^{n}$ with matrix A in the standard basis $ \{ e_{1},...,e_{n}\}$ for  $ F^{n}$.
Then the matrix of T in $ \{ e_{\sigma 1},...,e_{\sigma n}\}$is $ \sigma (A)$. However when I tried to check this using explicit calculations of the change of basis formula, the answer does not come correct. Any help?

It's false.

Note that similar matrices have the same characteristic polynomials, including the same trace and the same determinant.

Working, over a field of characteristic zero, consider:

$ A \equal{}\begin{bmatrix}1 & 0\\ 0 & 1\end{bmatrix}$ and $ B \equal{}\begin{bmatrix}0 & 1\\ 1 & 0\end{bmatrix}$

You can convert $ A$ to $ B$ by permuting its rows. But $ A$ has trace $ 2$ and determinant $ 1.$ Its characteristic polynomial is $ (x\minus{}1)^{2}.$ $ B$ has trace $ 0$ and determinant $ \minus{}1.$ Its characteristic polynomial is $ x^{2}\minus{}1.$ $ A$ is not similar to $ B.$

Your permuation can be given as a matrix $ P,$ where $ P$ is the result of permuting the rows of the identity matrix. $ P$ is orthogonal - that is, $ P^{T}P \equal{} I.$

To permute the rows of $ A$ (and do nothing else) is to create the matrix $ PA.$ In general, $ PA$ is not similar to $ A.$

If you permute the rows, [i]and also permute the columns[/i], then that's different and does preserve similarity. That creates the matrix $ PAP^{T}.$

Do you mean permuting the rows of A involves creating the matrix PA? In that case is AP permuting the columns of A by $ \sigma^{\minus{}1}$. What this says is that $ P^{T}P$ is formed by permuting the rows of I by $ \sigma$and then permuting the columns of the resulting matrix by $ \sigma$ (technically $ \sigma^{\minus{}1}$ but the result is the same). It is then easy to see why similarity won't hold aside from trivial cases like O and I.

I also see where my error was. $ \sigma (A)$ is the matrix of T with respect to the bases $ \{ e_{1},...,e_{n}\}$and $ \{ e_{\sigma 1},...,e_{\sigma n}\}$. It took a 2 x 2 matrix for me to figure that out..


induction
number theory proposed
number theory

Consider the set $E = \{1,2,\ldots,2n\}$. Prove that an element  $c \in E$ can belong to a subset $A \subset E$, having $n$ elements, and such that any two distinct elements in $A$ do not divide one each other, if and only if \[c > n \left( \frac{2}{3}\right )^{k+1},\] where  $k$ is the exponent of $2$ in the factoring of $c$.

I'm sorry but I think I misunderstood the problem  :blush:

We can take the set $A = \{n+1,n+2,\ldots,2n\}$. If $a,b \in A$ and $a \mid b$, then $b \ge 2a \ge 2(n+1)$, impossible.
Then take $c = n+1$ of $c=n+2$, the one that is odd.

Should we prove that $c > \frac{2}{3}n$ ??  :maybe:

Yes edriv, the problem probably is wrong!

I think that the correct is $c > (\frac{2}{3})^{k}$!

Ok, I did't misunderstand the problem, I just swapped n with 2n  :D
(and e.lopes' mind is identical to mine :lol:)

Here is the proof. A comment: the problem is so strict that is not hard. But it's still very nice  :)

A definition: $\mbox{god}(n)$ is the greatest odd divisor of n
Another definition: $v_{2}(n)$ is the greatest power of 2 that divides n.
Then we have the decomposition $n = \mbox{god}(n) \cdot 2^{v_{2}(n)}$

Some facts:
a) For each odd integer $1 \le k \le 2n$, there exist a unique power of 2 such that $n < 2^{e}k \le 2n$.
Proof: take the greatest power of 2 such that $2^{i}k \le 2n$. Then $2^{i+1}> 2n$ because of maximality, and then $2^{i}> n$.
b) If two distinct integers are such that $n <a,b \le 2n$, then $a \nmid b$
Proof: if $a\mid b$ and $a \neq b$ then $b \ge 2a$.
c) If $a \mid b$ then $\mbox{god}(a) \mid \mbox{god}(b)$
d) If $\mbox{god}(a)=\mbox{god}(b)$ then $a \mid b$ or $b \mid a$.
e) For each odd integer $1 \le c \le 2n$ and for each n-antichain A, there is an unique power of 2 such that $2^{e}c \in A$.
Proof: A is an antichain, then using fact (d) we prove that the god's of its elements are distinct. There are n elements and n odd integers between 1 and 2n. Then they are all.

And now to the proof, by induction on $k = v_{2}(c)$.
$k=0$ means that c is odd.
Suppose that $c \le \frac{2}{3}n$. Then $3c\le 2n$. By fact (e), a number of the form $2^{e}\cdot 3c \in A$. But $c \in A$ and $c \mid 2^{e}\cdot 3c$.
Suppose that $c > \frac{2}{3}n$. Then, put c in A, and for each odd integer d, put the unique $n < 2^{e}d \le 2n$ in A (using fact a). A is an antichain, since if $a,b \in A, a\mid b$ then $\mbox{god}(a)=\mbox{god}(b)$, and either a,b are between n and 2n, or a = c. But then the greatest odd divisor of b is an odd multiple of c, and we obtain $b \ge 3c > 2n$.

And now, suppose the problem is always true for all c such that $v_{2}(c) < i$ and let's prove it for $v_{2}(c) = k = i$, that is, $c=2^{i}c'$ with odd c'.

Suppose $c \le \left(\frac{2}{3}\right)^{i+1}n$. Then $a = \frac{3}{2}c \le \left(\frac{2}{3}\right)^{i}n$ is odd, integer, such that $v_{2}(a) =i-1$. a, multiplied by a power of 2, must belong to A, and using the induction hypotesis, prove that this is impossible.

Now suppose $c > \left(\frac{2}{3}\right)^{i+1}n$. We do this construction:
- put c in A
- If $3^{e}c' \le a < 3^{e+1}c'$, a is an odd integer and $c' \mid a$, then put $2^{i-e}a$ in A
- if a is an odd integer and $c' \nmid a$, then put $n<2^{e}a \le 2n$ in A, as in fact (a)
The proof that this works:
- if a is an odd integer, then either $c' \mid a$ or $c' \nmid a$, therefor a number of the form $2^{e}a$ is in A. A has n elements.
- If $a,b \in A, a \mid b$ then $\mbox{god}(a) \mid \mbox{god}(b)$. Then:
* if $c' \nmid \mbox{god}(a)$, then by the construction $n < a \le 2n$ and b is too big. Impossible.
* if $c' \mid \mbox{god}(a)$ then also $c' \mid \mbox{god}(b)$. But $\mbox{god}(b) \ge 3\mbox{god}(a)$, then, $\mbox{god}(a),\mbox{god}(b)$ are in two classes of the construction, then $v_{2}(b) < v_{2}(a)$ and it's impossible that $a \mid b$.

An interesting result... Consider the following $ n \times \infty $ "matrix":

$ 2^0*1, 2^1*1, 2^2*1, \ldots $
$ 2^0*3, 2^1*3, 2^2*3, \ldots $
$ \ldots $
$ 2^0*(2n-1), 2^1*(2n-1), 2^2*(2n-1), \ldots $

Clearly any set $A$ must have exactly one number from each row. But let $c$ be in row $(k+1)/2$ consider the rows $(3k+1)/2, (9k+1)/2, \ldots$. The first one must have a number in a column before the column of $c$, the second one must have a number in a column two spaces before the column of $c$, etcetera. From this we easily get $c > n*(2/3)^k$, as desired.

Now assume $c>n(2/3)^k$ and $c$ is in column $y$. If $k | \ell$, then assume $\ell / k $ has $x$ primes (including repetition) in its factorization, and from line $(\ell +1)/2$ include the number in the column $y-x$ . If $k$ doesn't divide $\ell$, then from line $(\ell +1)/2$ include the maximum number in that line that is $\le 2n$. This works.

[hide="Remark"]An interesting "expansion" to the classical problem:
"If a subset of $[2n]$ has $n+1$ elements, then one divides another."
It uses the same idea.[/hide]


inequalities calculus derivative function Support Putnam algebra

$1.$ Using calc:
[hide]Let $f(x)=x^{8}-x^{5}+x^{2}-x+1 ~~\forall ~~x\in\mathbb{R}$

$f'(x)=8x^{7}-5x^{4}+2x-1$
$8x^{7}-5x^{4}+2x-1=0$
$x \approx .753911 \rightarrow a$

$f'(0)=-1$
$f'(1)=4$

[img]http://img522.imageshack.us/img522/184/derivrq4.jpg[/img]

Because the derivative changes from negative to positive at $x \approx .753911$ (the function goes from decreasing to increasing), by the First Derivative Test, $x \approx .753911$ produces the abscissa of an absolute mininum $\forall ~~x\in\mathbb{R}$.

$f(a) \approx .6752803 \implies f(x)>0~~\forall ~~x\in\mathbb{R}\iff \boxed{x^{8}-x^{5}+x^{2}-x+1>0~~\forall ~~x\in\mathbb{R}}$[/hide]

[hide="1, without calculus"] The inequality is obvious for $x \le 0$, since all the terms are non-negative and the last is $1$.  So we restrict ourselves to $x > 0$.

By AM-GM,

$x^{8}+x^{2}\ge 2x^{5}$

So

$x^{8}-x^{5}+x^{2}-x+1 \ge x^{5}-x+1 = x(x^{4}-1)+1$

For $0 < x < 1$ the first term has absolute value less than $1$, and for $x > 1$ the first term is positive.  QED. [/hide]

#2:
[hide]
$y(\frac{1}{x}+\frac{1}{z})+\frac{1}{y}(x+z)\le(\frac{1}{x}+\frac{1}{z})(x+z)$
Divide everything by $(\frac{1}{x}+\frac{1}{z})$ to get:
$y+\frac{\frac{1}{y}(x+z)}{(\frac{1}{x}+\frac{1}{z})}\le(x+z)$
Which simplifies to:
$y+\frac{xz}{y}\le(x+z)$

Since Z is greater than Y by a certain amount (let this amount be $a$), substitute $z = y+a$;
$y+\frac{x(y+a)}{y}\le{x+y+a}$
$\frac{xy+xa}{y}\le{x+a}$

To prove this inequality, we recall that that ${x}\le{y}$, so
$\frac{xy+xa}{y}\le\frac{xy+ya}{y}$
$\frac{xy+xa}{y}\le\frac{y(x+a)}{y}$
$\frac{xy+xa}{y}\le{x+a}$

Which supports our proof.
[/hide]

By the way, what does QED mean?

[quote="p4fn2w"]By the way, what does QED mean?[/quote]

QED means in Latin, [i]quod erat demonstrandum[/i](which was to be demonstrated) and in English [i]quite elegantly done[/i].

The most elementary solution to 2:
$y\left(\frac{1}{x}+\frac{1}{z}\right)+\frac{1}{y}(x+z)\le\left(\frac{1}{x}+\frac{1}{z}\right)(x+z)$
$\Leftrightarrow \frac{1}{x}+\frac{1}{z}-\frac{y}{xz}-\frac{1}{y}\ge 0$
$\Leftrightarrow (z-y)\left(\frac{1}{xz}-\frac{1}{zy}\right)\ge 0$

[quote="me@home"][quote="hien"]As the title i said, all of them for beginer.
The solution for the part 1:
$x^{8}-x^{5}+x^{2}-x+1=\left(x^{4}-\frac{x}{2}\right)^{2}+\left(\frac{x}{2}-1\right)^{2}+\frac{x^{2}}{2}>0$[/quote]
In fact in general I guess you can use the exact same method on any true one variable inequality of this form, right? (where signs alternate in order of degree)
First get rid of largest term in square, choose second term as necessary to get next to largest, etc. down the list.[/quote]

are you sure that's correct? How would you do this problem then?

$13x^{3}-24x+13 > 0$ (I know, I took it from the other topic)

He probably meant inequalities true for $x \in \mathbb{R}$, although even then it's doubtful.

Actually, it is Putnam 1999 A2:

[quote="Putnam Exam"]A-2 Let $p(x)$ be a polynomial that is nonnegative for all real $x$. Prove that for some $k$, there are polynomials $f_{1}(x), \ldots, f_{k}(x)$ such that $p(x) = \sum_{j = 1}^{k}\left(f_{j}(x)\right)^{2}$.[/quote](This is slightly more general than me@home's conjecture.)

Solution available (.pdf) [url=http://www.unl.edu/amc/a-activities/a7-problems/putnam/-pdf/1999s.pdf]here[/url]; the Putnam Archive is [url=http://www.unl.edu/amc/a-activities/a7-problems/putnamindex.shtml]here[/url].

I should note, though, that me@home's suggested method isn't actually easy in general -- even for, say, a fourth-degree polynomial like $x^{4}-8x^{3}+20x^{2}-16x+4$, it doesn't immediately give you the (unique, in this case) such representation -- for higher-degree situations, it will be much worse.

It is of course [i]not[/i] possible for polynomials which are negative on some subset of $\bf R$.


function
real analysis
real analysis unsolved

We say that a function $f: \mathbb R \to \mathbb R$ has the property $(P)$ if, for any real numbers $x$, \[ \sup_{t\leq x} f(x) = x.  \] a) Give an example of a function with property $(P)$ which has a discontinuity in every real point.

b) Prove that if $f$ is continuous and satisfies $(P)$ then $f(x) = x$, for all $x\in \mathbb R$.

[quote=perfect_radio](a) Official solution said

- $f(x) = x$ for $x \in \mathbb Q$;

- $f(x) = x-1$ for $x \in \mathbb R \setminus \mathbb Q$.

Mine said:

- $f(x) = x$ for $x \in \mathbb Q^{\ast}$;

- $f(x) = -1$ for $x=0$;

- $\displaystyle f(x) = \frac{x}{2}$ for $x \in \mathbb R \setminus \mathbb Q$, $x>0$;

- $\displaystyle f(x) = \frac{3x}{2}$ for $x \in \mathbb R \setminus \mathbb Q$, $x<0$.

(Way too complicated, but hopefully correct)

(b) For every $\epsilon > 0$ we must have $y \in \left( x-\epsilon, x - \frac{\epsilon}{2} \right)$ such that $f(y) > x - \epsilon$.

Take $\displaystyle \epsilon = \frac{1}{n}$ and $x_n = y$. We get $(x_n) \to x$ and $f(x_n) \to x$. Since $f$ is continuous, $f$ is the identity function.[/quote]

Where did you find the official solution


I need help. :(

Thanks and Best Regards!
[img]http://img440.imageshack.us/img440/3636/adszee1.jpg[/img]

$ \frac{x}{x\plus{}8}\plus{}\frac{4}{4\plus{}5}\plus{}\frac{6}{6\plus{}7}\equal{}1$, $ x\equal{}\frac{44}{53}$ (if the triangle exists...?)

[quote="Altheman"]
(areas of those triangles)[/quote]

Does this assume that the cevians are also altitudes, or does this hold true regardless?

EDIT: Nevermind, figured it out myself.  :P

[quote="Altheman"]MBC/ABC+MCA/ABC+MAB/ABC=1

(areas of those triangles)[/quote]
This is Gergonne's theorem for any concurent cevians in a triangle


If $ \sum_{k\equal{}0}^{n} \frac{A_k}{x\plus{}k}\ \equal{} \frac{n!}{x(x\plus{}1)(x\plus{}2).........(x\plus{}n)}$ , then $ A_k \minus{} A_{k\minus{}1} \equal{}$

Thanks for any help.

Use the technique described [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=250866]here[/url].


Hi I haven't done Pell's Equations but i did read up the method to solve them. However, I'm not sure how i can show that these 2 equations each have infinite integers solutions for x and y.
$ x^2 \minus{} 2y^2 \equal{} 1$
and
$ x^2 \minus{} 6y^2 \equal{} 1$

EDIT: I think I posted in the wrong forum. please relocate.

1st - this does not belong in HSB - Preoly would be a much better place.

I think that the best way to show that the two equations have an infinite number of solutions is to find a formula for the solutions.

I'll tackle $ x^2\minus{}2y^2\equal{}1$; the other problem can be worked similarly.

First, note that $ x/y \approx \sqrt{2}$; in fact, it turns out that the smallest solution is a convergent of the continued fraction of $ \sqrt{2}$.  The continued fraction is $ 1\plus{}\frac{1}{2\plus{}\frac{1}{2\plus{}\ldots}}$ which has the convergents $ 1$, $ 3/2$, $ 7/5$, etc.  $ x\equal{}3$, $ y\equal{}2$ is our first nontrivial solution.

Define $ N(x \pm y\sqrt{2}) \equal{} x^2\minus{}y^2$.  I turns out that, in general, $ N(a)N(b)\equal{}N(ab)$.  Our equation is $ N(x\plus{}y\sqrt{2}) \equal{} N(x\minus{}y\sqrt{2})\equal{}1$, and taking both sides to the $ n$ power, $ N((x\minus{}y\sqrt{2})^n) \equal{} 1$; that is, $ x_n \minus{} y_n \sqrt{2} \equal{} (x_0\minus{}y_0\sqrt{2})^n$ is our nth solution; in the same way, $ x_n\plus{}y_n \sqrt{2} \equal{} (x_0\plus{}y_0\sqrt{2})^n$ so we have the formulas

$ x_n \equal{} \frac{(x_0\plus{}y_0\sqrt{2})^n \plus{} (x_0\minus{}y_0\sqrt{2})^n}{2}$
$ y_n \equal{} \frac{(x_0\plus{}y_0\sqrt{2})^n \minus{}(x_0\minus{}y_0\sqrt{2})^n}{2\sqrt{2}}$

In our particular situation, $ x_0\equal{}3$ and $ y_0\equal{}2$ as found before.  All nontrivial solutions are given by these formulas. (see theorem 2: [url]http://www.imomath.com/tekstkut/pelleqn_ddj.pdf[/url])

The other equation works similarly, but with $ \sqrt{6}$ instead of $ \sqrt{2}$


Let $ a$, $ a'$, $ b$, and $ b'$ be real numbers with $ a$ and $ a'$ nonzero.  The solution to $ ax \plus{} b \equal{} 0$ is less than the solution to $ a'x \plus{} b' \equal{} 0$ if and only if

$ \textbf{(A)}\ a'b < ab' \qquad \textbf{(B)}\ ab' < a'b \qquad \textbf{(C)}\ ab < a'b' \qquad \textbf{(D)}\ \frac {b}{a} < \frac {b'}{a'}$
$ \textbf{(E)}\ \frac {b'}{a'} < \frac {b}{a}$

The solutions are $ \frac {\minus{}b}{a}$ and $ \frac{\minus{}b'}{a'}$. We have the inequality $ \frac {\minus{}b}{a} < \frac {\minus{}b'}{a'}$. Dividing by -1 gives E.

From $ ax \plus{} b \equal{} 0 \Rightarrow x \equal{} \frac { \minus{} b}{a}$ ,and from $ a'x \plus{} b' \equal{} 0 \Rightarrow x \equal{} \frac { \minus{} b'}{a'}$
So : $ \frac { \minus{} b}{a} > \frac { \minus{} b'}{a'} \Longrightarrow \frac {b}{a} < \frac {b'}{a'}$ or $ \boxed E$

$ \boxed E$ because:

$ ax\plus{}b\equal{}0 \Rightarrow x\equal{}\minus{}\frac{b}{a}$

$ a'x\plus{}b'\equal{}0 \Rightarrow x\equal{}\minus{}\frac{b'}{a'}$

Now, $ \minus{}\frac{a}{b}<\minus{}\frac{a'}{b'} \Leftrightarrow \frac{a'}{b'}<\frac{a}{b}$


function
superior algebra
superior algebra unsolved

"The homomorphism condition for an isomorphism is sometimes summarized by saying the isomorphism must commute with the binary operation(s)."...I don't understand what is it meant by the conditon "the isomorphism must commute with the binary operation(s)"..Can anyone help?? [/img][/list][/quote]

I don't think I've ever seen it worded that way (it doesn't surprise me though), but anyways it just means that if your iso in question is $ f: G\to G'$, then for all $ a,b \in G$ we have that $ f(a* b)\equal{}f(a)\cdot f(b)$, where $ *$ is the operation on $ G$ and $ \cdot$ is the operation on $ G'$.

This is a good intuition, but one must be careful to distinguish the group operation in $ G$ from the group operation in $ G'$, since they are by no means related.

It makes more sense to me to say "the function preserves the operation", or "it takes the operation from $ G$ to that of $ G'$". The word 'commute' reminds me of $ ab \equal{} ba$ or $ f(a)f(b)\equal{}f(b)f(a)$, but I don't see the connection with the homomorphism property.

The way to see the connection with commutativity is to define $ M$ by $ M(x,y)\equal{}x \cdot y$ when $ x,y\in (G,\cdot)$, and abusively define $ f(x,y)\equal{}(f(x),f(y))$.  Then the homomorphism condition literally asserts that $ M\circ f \equal{} f\circ M$.

A more general (and useful) notion of commutativity is the [url=http://en.wikipedia.org/wiki/Commutative_diagram]commutative diagram[/url], of which this is a pretty nice example.

Thanks, that makes sense (although the definition of $ f$ is indeed a bit abusive).

Somewhat in general a commutativity condition looks like $ \mu_A \circ f \equal{} (Ff) \circ \mu_B$.


algebra
polynomial
linear algebra
matrix
induction
linear algebra theorems

3 questions: ($ A,B$ matrices of same dimension over field)

1. $ AB$ and $ BA$ have the same characteristic polynomial. Do they also have the same minimal polynomial?

2. What conditions do we need in order to have $ A$ and $ B$ similar? They must have the same minimal polynomial? is this necessary and sufficient?

3. $ A$ and $ B$ are simultaneously diagonalizable if and only if $ AB\equal{}BA$? what other criteria are there for simultaneous diagonalization (or for upper triangularization?)?

[quote="Pascual2005"]1. $ AB$ and $ BA$ have the same characteristic polynomial. Do they also have the same minimal polynomial?[/quote]No. Construct $ A,B\in M_2(K)$ with $ AB\equal{}0$ and $ BA\neq 0$.
[quote]2. What conditions do we need in order to have $ A$ and $ B$ similar? They must have the same minimal polynomial? is this necessary and sufficient?[/quote]No. Necessary, but not sufficient. Using the Jordan normal form try to construct a counterexample $ A,B\in M_4(K)$ with minimal polynomial $ X^2$.
[quote]3. $ A$ and $ B$ are simultaneously diagonalizable if and only if $ AB \equal{} BA$?[/quote]Yes - if $ A$ and $ B$ are diagonalizable at all. The same holds if the word [i]diagonalizable[/i] is substituted by [i]triangularizable[/i].

Of course, I must use this to push my own posts:
http://www.mathlinks.ro/viewtopic.php?t=175802 post 7.

[b](*)[/b] If $ K$ is a field, and $ A$ and $ B$ are two matrices from $ \text{M}_n\left(K\right)$, then $ A$ and $ B$ are similar if and only if for every polynomial $ P\in K\left[X\right]$ we have $ \text{Rank }\left(P\left(A\right)\right) \equal{} \text{Rank }\left(P\left(B\right)\right)$.

If someone can prove this over any field $ K$ without the use of normal forms, we can make linear algebra a little bit nicer ;)

  darij

I do not understand why if $ A$ and $ B$ are simultaneously triangularizable then we must have $ AB \equal{} BA$? this implies that triangular matrices always commute which is not true... Also how do you prove the converses of Q3?

[quote="Pascual2005"]I do not understand why if $ A$ and $ B$ are simultaneously triangularizable then we must have $ AB \equal{} BA$? How do you prove this? And also how do you prove the converses of Q3?[/quote]

I think the triangular matrix statement is false. Just pick any two upper triangular matrices that do not commute.

For Q3
=> side is trivial by considering the common eigenbasis.
<= Take an eigenspace decomposition with respect to A. The commuting relation tells us that the each A eigenspace is B invariant. Take eigenspace decomposition of B in each A-eigenspace and choose a basis. Such a basis would be an eigenbasis for both A and B.

Does this holds:

Q. If $ AB\equal{}BA$ then they are simultaneously triangularizable?

Yes. By induction, your question is the same as whether A,B have a common eigenvector, whose answer is positive. (Same reasoning as above)

A more general statement is [url=http://en.wikipedia.org/wiki/Lie%27s_theorem]Lie's theorem[/url] for solvable Lie algebras.

[quote="Soarer"][quote="Pascual2005"]I do not understand why if $ A$ and $ B$ are simultaneously triangularizable then we must have $ AB \equal{} BA$? How do you prove this? And also how do you prove the converses of Q3?[/quote]

I think the triangular matrix statement is false. Just pick any two upper triangular matrices that do not commute.[/quote]
It is only true for $ n\le 2$. This was the case I was hastily "inspired" by. ;)

[quote="Third Edition"]
It is only true for $ n\le 2$. This was the case I was hastily "inspired" by. ;)[/quote]

In fact I think even $ n \equal{} 2$ it is false. Consider $ \left(\begin{array}{cc} 1 & 1 \\
0 & 2 \\ \end{array}\right)$ and $ \left(\begin{array}{cc} 2 & 3 \\
0 & 2 \\ \end{array}\right)$

Whow, I definitely need some holiday. ;) Thanks!


since problem 5 has two parts, how much is a) and how much is part b)?

Usually the point split is something like 2-3, since part b is definitely more involved than part a.  I'm not sure they'll decide officially though until grading begins, and we won't know until the rubric is released.

No discussion of this, please.

[size=200][b]THE ROUND IS NOT OVER YET.[/b][/size]


group theory
abstract algebra
superior algebra
superior algebra unsolved

In the questions below: $G$ is a ﬁnite group; $H \leq G$ a subgroup of $G; |G : H |$ the index of $H$ in $G; |X |$ the number of elements of $X \subseteq G; Z (G)$ the center of $G; G'$ the commutator subgroup of $G; N_{G}(H )$ the normalizer of $H$ in $G; C_{G}(H )$ the centralizer of $H$ in $G$; and $S_{n}$ the $n$-th symmetric group.

Let $a, b \in G.$ Suppose that $ab^{2}= b^{3}a$ and $ba^{2}= a^{3}b.$ Prove that $a = b = 1.$

NTTuan, do you have any general methods for these types of problems? I've seen them very few times before and don't know how to tackle... is it just playing around with the quantities?

One approach, which is neither fun nor clever, is to use the Todd-Coxeter algorithm.  If $G=\langle a,b|ab^{2}=b^{3}a,ba^{2}=a^{3}b\rangle$, and $H=\langle a\rangle\leq G$, then you can compute the action of $G$ on the cosets of $H$, and you'll get the trivial action.  Therefore $b\in H$, so it commutes with $a$, and it becomes trivial.

I haven't ever taken (formally) group theory so all I know are the stupid trick proofs for these... thanks for providing the name of the deeper theorem so at least I can research that!

Okay.. so that's like a method for trick proofs... thanks


Hi everyone,
I'd really appreciate some help with a pretty simple problem, which goes like this:

To allow timely removal and disposal of snow, a city is divided up into several sectors and snow removal operations are carried out concurrently in each sector. In Montreal, accumulated snow is loaded onto trucks and hailed away to disposal sites (e.g. rivers, quarries, sewer chutes, surface holding areas). For contractual reasons, each sector may be assigned only to a single disposal site. However, each disposal site may receive snow from multiple sectors. There are 10 sectors and 5 disposal sites. How many different assignments of sectors to disposal sites are possible?

I can't really remember how to go about this, any hints appreciated.

Thanks in advance.

Each sector could go to five different places. Since the sectors' choices are independent, then each sector could go to 5 different places.

5*5*5*5*5*5*5*5*5*5=5^10=9765625


function
geometry
calculus
calculus computations

Prove that the functions $ \frac{x^{2}}{1\minus{}x^{3}}$ and $ \frac{x^{3}}{1\minus{}x^2}$ are analytic on $ (\minus{}1;1)$ and find their Maclaurin expansion.

[quote="mathema*"]Hi :) !!

that obiously that $ \forall x \in ] \minus{} 1;1[$:

$ x \rightarrow \frac {x^2}{1 \minus{} x^3}$ and $ x \rightarrow \frac {x^3}{1 \minus{} x^2}$ are analytic fuctions

(because any fuction writed as $ x \rightarrow \frac {P(x)}{Q(x)}$ with $ P,Q \in \mathbb{K}[X]$ are analytic in here area of definition ($ i.e$ $ Q(x) \neq 0 \forall x \in \mathbb{K}$)

than:

$ \frac {x^3}{1 \minus{} x^2} \equal{} x^3 \sum_{n \geq 0} x^{2n} \equal{} \sum_{n \geq 0} x^{2n \plus{} 3}$

and the same we find that:

$ \frac {x^2}{1 \minus{} x^3} \equal{} \sum_{n \geq 0} x^{3n \plus{} 2}$

and thx[/quote] Indeed! :)


Selamlar arkadaslar, Turkiye matematik olimpiyatlari sitesi hizmetinizde:
http://www.matematikolimpiyati.net adresinden sitemize ulasabilirisiniz.

hi all, Turkish math olympiad site is avaliable...
you may visit at http://www.matematikolimpiyati.net (it is turkish for now, but we hope it will also have english one soon.)

site güzel olmuş

ama imo'nun official sitesinde sizin site gözükmüor: http://imo.math.ca/

sağ tarafta ülkelerin siteleri var ama turkey yok

[quote="Oguzhan"][quote="nighty"]Hocam site güzel olmuş ama hala geçen senenin ekim sorularıyla muhatabız.+foruma da pek ilgi yok zaten.onun yerine elemanları buraya yönlendirsek ya da burdaki elemanları oraya yönlendirsek?..[/quote]

Hocam site artık yok!... :(

Singapurun bile var. Bizim yok.

Ayrıca net ortamında türkiyedeki matematik olimpiyatı sorularına ulaşılamıyor artık.

Ne yazık ki tübitak satılığa çıkarmış...

Burdan kazanacağı parayla mı Türkiyenin bilime desteğine katkıda bulunacağız. Peki yabancılar neden yapmıyor bunu? Bir akıllı bizim tübitak heralde!...[/quote]
Efsaneler ölmez sadece iekil değiştirir. :)
Sitenin adresi değişti sadece. Bundan sonra [url]http://www.matematikolimpiyati.org[/url] adresinden ulaşabileceksiniz.
Tüm 1. aşama soruları hem bizim sitede hem de Tübitak'ın sitesinde var. Ayrıca 2. aşama, takım seçme, kamp sonu sınavlarından bazıları mevcut.

Our new site's url is [url]http://www.matematikolimpiyati.org[/url]. I'll try to translate some pages to English soon...

Gençler bazı sayfaları İngilizceye çevirdim. Sorular için destek gerek. Yardımcı olursanız sevinirim.

Some parts of the site is available in English now. Translation will continue...

matematikolimpiyatlari.com tek resmi sitemiz. Ilk acildigi gunlerden beri ilgiyle takip ediyoruz, fakat site nerdeyse bir iki kisinin cabalari ile ilerliyor. Kaysi [quote]Yardımcı olursanız sevinirim. [/quote] demis gayet hak veriyorum. Her yastan binlerce matematik olimpiyatcisi var.
matematikolimpiyatlari.com 'da birkac editor yada yazar alinacaktir ilan verseniz eminim ilgilenen bir cok olimpiyatci cikar. Gonullu yazarlarla daha verimli bir site ortaya konabilir.

Ayrica siteye girip cikan cok, fakat birsey yazan yok. Matematik olimpiyatcilarinin resmi tek sitesi oldugundan zaten temel bir popularitesi var. Fakat fotograf galerisi haric sitede liseli arkadaşlara faydası dokunacak fazla içerik yok. Eskisi gibi degil artik cok fazla ozel tesebbus hobi olarak matematikcilere yardimci olmaya calisiyor. Basit arastirmalarla
http://mathematist.wordpress.com/
http://sbelian.wordpress.com/
gibi sitelere ulasabilirsiniz demek ki hevesli insanlar var. Onemli olan ulasabilmek.


[quote="conan_naruto236"]Let $ f: N^{*2} \rightarrow N^{*}$ such that :
$ (i) f(m,1) \equal{} m$  $ \forall m \in N^{*}$.
$ (ii) f(m,n \plus{} 1) \equal{} m.\sum_{k \equal{} 1}^{m \plus{} 1}f(k,n)$  $ \forall m \in N^{*}, n \in N^{*}$.
Find $ f(m,n)$.[/quote]
Please help me  :(

It can be shown by induction,
$ f(m,n)\equal{}\frac{(m\plus{}2n\minus{}2)!}{(m\minus{}1)!(n\minus{}1)!2^{n\minus{}1}}$.

The idea is to use

$ \sum_m (m\plus{}1)\cdots (m\plus{}k)\\\equal{}\sum_m \frac1{k\plus{}1}[(m\plus{}1)\cdots (m\plus{}k)(m\plus{}k\plus{}1)\minus{}m(m\plus{}1)\cdots (m\plus{}k)]$.

[quote="xxp2000"]It can be shown by induction,
$ f(m,n) \equal{} \frac {(m \plus{} 2n \minus{} 2)!}{(m \minus{} 1)!(n \minus{} 1)!2^{n \minus{} 1}}$.

The idea is to use

$ \sum_m (m \plus{} 1)\cdots (m \plus{} k) \\
\equal{} \sum_m \frac1{k \plus{} 1}[(m \plus{} 1)\cdots (m \plus{} k)(m \plus{} k \plus{} 1) \minus{} m(m \plus{} 1)\cdots (m \plus{} k)]$.[/quote]
Why do you think $ f(m,n) \equal{} \frac {(m \plus{} 2n \minus{} 2)!}{(m \minus{} 1)!(n \minus{} 1)!2^{n \minus{} 1}}$???
Please explain to me  :( .


let $ u$ denote an $ n\times 1$ column matrix such that $ u^Tu$ = $ I_{1}$.

define $ H \equal{} I \minus{} 2uu^T$. H is called a householder matrix. prove that H is symmetric an invertible.

I'm having trouble with proving that it is invertible.

Hint: what does this matrix mean geometrically? How do you find the inverse of that geometric transformation?

  darij

Hi, i know this is a huge revive but i was searching for the answer to this
I don't really understand the above hint...can anybody help? Thanks

Try to multiple $I-2uu^{T}$ by itself, and don't forget that you know by hypothesis that $u^{T}u=I_{1}$ and you can see $I_{1}$ like a scalar.
Or instead just note that a Householder matrix define a symmetry to the orthogonal complement of vector $u$

I shoudl've clarified: I played around with it and realized its inverse was itself. I just was unsure of how to justify that idea


limit
inequalities
logarithms
abstract algebra
real analysis
real analysis solved

Prove that : If the possitive sequence $ (a_n) $ : $ \lim_{n\to \infty}a_n=a $ then
  $ \lim_{n\to \infty} {^n\sqrt{a_1a_2...a_n}} = a $

Let $ A_n, G_n, H_n $ denote the arthmetic, geometric and armonic means of n positive real numbers $a_1, a_2, ...., a_n $, then
\[
A_n\geq G_n\geq H_n
\]
Using this inequality and the Toeplitz theorem with:
\[
c_{n,k}  = \frac{1}{n}\quad for\quad k = 1,2,3....n.
\]
follows the result.

Or you can take logarithm, then apply Cesaro-Stolz Theorem ... (but it needs Heine's Theorem)

Instead of just referring to "Cesaro-Stolz" this time, let's write out - and prove - the theorem we need. This is an "abelian" theorem: the analogue for Cesaro summability of Abel's theorem for power series.

[b]Theorem:[/b]

Suppose $\lim_{n\to\infty}x_n=L$. Define $\sigma_n=\frac1n\sum_{k=1}^n x_k.$

Then $\lim_{n\to\infty}\sigma_n=L.$

Proof:

Given $\epsilon>0$, find $m$ such that if $k>m$ then $|x_k-L|<\frac{\epsilon}2.$

Let $M = \max_k|x_k-L|.$ (Convergent sequences are bounded.)

$\sigma_n-L=\frac1n\sum_{k=1}^n (x_k-L)$ so if $n>m,$

$|\sigma_n-L|\le\frac1n\sum_{k=1}^m|x_k-L|+\frac1n\sum_{k=m+1}^n|x_k-L|$

$<\frac{mM}n+\sum_{k=m+1}^n\frac{\epsilon}2<\frac{mM}n+\frac{\epsilon}2.$

Then if $n>\frac{2mM}{\epsilon},$ this is all less than $\epsilon.$

---

For the problem at hand, let $x_n=\ln(a_n)$ and we have our proof.


We are given  a  trapezoid $ABCD$  with $AB \parallel CD$, $CD=2AB$ and $DB \perp BC$. Let $E$ be  the intersection  of lines  $DA$ and $CB$, and  $M$ be  the  midpoint of  $DC$.

(a) Prove that  $ABMD$ is a  rhombus.

(b) Prove that triangle $CDE$ is isosceles.

(c) If  $AM$  and $BD$ meet  at  $O$, and $OE$ and  $AB$ meet at $N,$ prove that the line $DN$ bisects segment $EB$.

[hide](a) $DM=\frac{1}{2}DC=AB$ and $AB\parallel DM$.
Also, by the midline theorem, $AD=\frac{1}{2}DE=BM$

Since $AB=DM$ and $AB\parallel DM$, $ABMD$ is a parallelogram.
Therefore $AD\parallel BM$.

So two pairs of sides are parallel and congruent, which means $ABMD$ is a rhombus.

(b) $AB=AD$, so $\angle ABD=\angle ADB$
Also, since $AB\parallel DM$, $\angle ABD=\angle BDM$

$\therefore \angle ADB=\angle BDM$

Since the perpendicular line from $D$ to $EC$ bisects $\angle EDC$, $\triangle CDE$ is an isosceles triangle.

(c) We know that $EA=AD$ and $DO=OB$, so $N$ is the centroid of $\triangle EDB$.

So $DN$ is a median from $D$ to $EB$, and thus bisects $EB$.[/hide]


topology
real analysis
real analysis unsolved

In R^n, a homeomorphism from open unit ball to all R^n is:

f: x-->2x/(1-||x||^2)

How to show the inverse mapping is:

f^(-1): y-->y/(1+(1+||y||^2)^0.5).

thanks.

Note that $y=f(x)$ is a scalar multiple of $x$. Hence $x=f^{-1}(y)$ is a scalar multiple of $y$. Equating $\parallel f(x)\parallel $ to $\parallel y\parallel $, you will find $\parallel x\parallel $ in terms of $\parallel y\parallel $...


calculus
integration
trigonometry
real analysis
real analysis solved

Let   $T_n=\int^\pi_0\frac{x^n}{n!}(\sin x+\cos x)dx\  (n=0,1,2\cdot\cdot\cdot)$ .Evaluate $\sum_{n=0}^\infty(T_n+T_{n+2})$.

Use $\sum \frac{x^n}{n!}=e^x$ and integration by part

Bonjour,monseiur Moubinool.Merci beaucoup de votre reponse.
Could I have your solution?

kunny

$\displaystyle\sum_{n\ge0}T_n=\sum_{n\ge0}\int_0^\pi\frac{x^n}{n!}(\sin x+\cos x)dx=$ $\displaystyle\int_0^\pi\sum_{n\ge0}\frac{x^n}{n!}(\sin x+\cos x)dx=\int_0^\pi e^x(\sin x+\cos x)dx = \int_0^\pi e^x\sin x
dx+\int_0^\pi e^x\cos xdx =$ $=\displaystyle e^x\sin x|_0^\pi-\int_0^\pi e^x\cos x+\int_0^\pi e^x\cos x dx=0$.
Analoguous we find that
$\displaystyle\sum_{n\ge0}T_{n+2}=0-T_0-T_1$.
$\displaystyle T_0=\int_0^\pi (\sin x+\cos x)dx=-\cos x|_0^\pi+\sin x|_0^\pi=-2$
and
$\displaystyle T_1=\int_0^\pi x(\sin x+\cos x)=\int_0^\pi x\sin
x dx+\int_0^\pi x\cos x dx=$ $\displaystyle-x\cos x|_0^\pi+\int_0^\pi \cos x
dx+x\sin x|_0^\pi-\int_0^\pi\sin x dx=$ $\pi+0+0-2=\pi-2$.
So $\displaystyle\sum_{n\ge0}(T_n+T_{n+2})=-(-2)+2-\pi=4-\pi$.

Hello! Kuba

Thank you for your reply! Regrettably you are wrong in the calculation of $ T_0.$  Here is my solution.

By integral by parts,

$T_n=\int^\pi_0 \frac{x^n}{n!}(sinx+cosx) dx   (n=0,1,2,\cdot\cdot\cdot)$

$=\left[ \frac{x^{n+1}}{(n+1)!}(sinx+cosx)\right]_0^\pi-\int^\pi_0\frac{x^{n+1}}{(n+1)!}(cosx-sinx) dx$

$=-\frac{\pi^{n+1}}{(n+1)!}- \left[ \frac{x^{n+2}}{(n+2)!}(cosx-sinx)\right]_0^\pi-\int^\pi_0 \frac{x^{n+2}}{(n+2)!}(sinx+cosx) dx$

$=-\frac{\pi^{n+1}}{(n+1)!}+\frac{\pi^{n+2}}{(n+2)!}-T_{n+2}$

Thus we  have   $T_n+T_{n+2}=\frac{\pi^{n+2}}{(n+2)!}-\frac{\pi^{n+1}}
{(n+1)!}$

Therefore by telescoping method,we obtain the partial sum is $\sum_{n=0}^m(T_n+T_{n+2})=\frac{\pi^{m+2}}{(m+2)!}-\pi\longrightarrow-\pi (m\rightarrow\infty)$  Ans.

kunny


Here we have it, the Max Pratt game.

Two or more people go to wikipedia.org (online encyclopedia), and decide on a starting point and an ending point. They should be pretty random and aloof from one another, and if you choose something very obscure you might want to make certain that a wikipedia article exists for that topic. For example, you might choose Table Tennis to start from, and Pregnancy to end. All players go to the wikipedia article on the first topic.

The point of the game is to get from the initial article to the final the quickest, by clicking on links contained in the articles. One cannot use the keyboard or any find function (control+F, etc.). Also, one cannot click on numbers or dates.

For example, if you were going from Table Tennis to Pregnancy, you might follow the path I used:

"...is the official name for the sport in [url=http://en.wikipedia.org/wiki/China]China[/url] and [url=http://en.wikipedia.org/wiki/Taiwan]Taiwan[/url]."

click on China, go do Demographics subtitle,

"...has enacted a [url=http://en.wikipedia.org/wiki/Birth_control]birth control[/url] policy, commonly known as the [url=http://en.wikipedia.org/wiki/One-child_policy]One-child policy[/url]."

Clicking on Birth Control easily leads one to find Pregnancy.

Celebrate the awesomeness of Wikipedia by playing the Max Pratt game.

Sounds like fun; but I have much more important things to do. :)
And where DOES the name come from?

[quote="la mafioso"]Anyone wanna verses me?[/quote]

Sure, I'll play you.
I think I have some time today.

Haha, I remember something similar, with MathWorld pages.

But I know a way to always win.
Since Wikipedia is fully editable, I can create an article which is related to both the starting and ending points.  :D

Well it's named after the founder- a friend of mine named Max Pratt (that's not me, my name's Rashad). Now I did make a wikipedia article on it ("..created by child prodigy and renowned genius Max Pratt...") but it was deleted in like..two days. The game is pretty popular among my friends, but I'm interested to see if it spreads....

Whoa now...no using numbers or dates. The reason Max made that a rule is because, as you've seen, years/dates connect just about [b]everything[/b].

I remember I was playing a friend, Britney Spears to Saddam Hussein. She completely killed me, took her about ten seconds, because deep in the article....

"One of the tracks from Britney, "I'm Not a Girl, Not Yet a Woman," also won the Worst Original Song award, and in 2005 Spears was again 'honored' by the Golden Raspberry judges when she was proclaimed Worst Supporting Actress for her cameo role in the documentary [url=http://en.wikipedia.org/wiki/Fahrenheit_9/11]Fahrenheit 9/11[/url], which showed her response to a [url=http://en.wikipedia.org/wiki/CNN]CNN[/url] interview question by [url=http://en.wikipedia.org/wiki/Tucker_Carlson]Tucker Carlson[/url], regarding her political views during the then-impending [url=http://en.wikipedia.org/wiki/2003_invasion_of_Iraq][b]2003 invasion of Iraq[/b][/url]."

We play this on the school media centre computers when we're supposed to be doing...research. If you're not in the same complex, it helps a lot to get everyone playing in a chatroom together. And it is a lot of fun when you're playing with others...you weren't doing it alone, were you mathclass?

[quote="mathclass"]I went from volleyball to 1952 to nobel prizes to physics... :D pretty neat!![/quote]
I went from volleyball... to paralympics... to visual disabilities... and it was all downhill from there (I can't remember it exactly, it was a bit long).

Oh goodness I was playing and we were trying to get to 'Sod' from wherever. We wiki'd and wiki'd...passed through Grass, Lawn, Lawn Care, Astroturf, Artificial turf, Garden design, everything. Simple stuff like that (same with Stapler, Paper clip, etc.) is tough to get to. Although this article did exist, it gave us the idea to always check beforehand to see if an article on something kind of obscure/unimportant exists.


algebra
polynomial
Rational Root Theorem

[quote="Arne"]... preferrably without using Fermat's last theorem :D[/quote]

of course, but nice idea........ :D   :lol:

[hide]Prove by contradiction.  Assume $(n+1)^3=n^3+(n-1)^3$ for some positive integer $n$.  Expanding we have $n^3+3n^2+3n+1=n^3-3n^2+3n-1$, simplifying we have $n^3-6n^2+2=0$.  From the Rational Root Theorem, the only possible positive integer roots are $1$ and $2$.  Testing these out show that neither of them work.  Therefore $n$ is not a positive integer, contradicting our assumtion that it is.[/hide]

[quote="socrates"]Prove that $(n + 1)^3 \ne n^3 + (n - 1)^3$ for any positive integer n.[/quote]
[hide]
Assume the contrary.
$(n+1)^3=n^3+(n-1)^3\\n^3+3n^2+3n+1=n^3+n^3-3n^2+3n-1\\3n^2+3n+1=n^3-3n^2+3n-1\\0=n^3-6n^2-2$
integer solutions=$\pm1, \pm2$.
Neither work. Thus this is proven.
[/hide]

[quote="eryaman"]  $n^3+3n^2+3n+1=n^3-3n^2+3n-1$ and $n^3-6n^2+2=0$. [/quote]
These two equations have mistake, but your solution is OK


I don't really understand your wording...if you're just looking for $f(3)$ I guess it would be [hide]$2^{3}+3+2!=8+3+2=13.$[/hide]

[quote="13375P34K43V312"]I don't really understand your wording...if you're just looking for $f(3)$ I guess it would be [hide]$2^{3}+3-2!=8+3-2=11.$[/hide][/quote]

[hide]You mean $2^{3}+3+2!=13$ right? :wink: [/hide]

[quote="solafidefarms"]Let '$\text{your life}$' be of value 3. A function $f(x)$ changes its input $x$ to $2^{x}+x+(x-1)!$. What is $f(\text{your life})$?

(-:[/quote]

[hide]$f(3)$ = $2^{3}+3+(3-1)!$
$= 8+3+2 = 13$[/hide]

[quote="bpms"][quote="klutzycherry14"][quote="solafidefarms"]Let '$\text{your life}$' be of value 3. A function $f(x)$ changes its input $x$ to $2^{x}+x+(x-1)!$. What is $f(\text{your life})$?

(-:[/quote]

[hide]$f(3)$ = $2^{3}+3+(3-1)!$
$= 8+3+2 = 13!$
$13! = 6227020800$[/hide][/quote]
You got your factorial misplaced, you aren't taking the factorial of the entire thing ;).[/quote]

oops.  :oops_sign:  i corrected it...that was a careless error. :oops:  :blush:

[quote="bpms"][quote="MichaelFaraday"]its the function of 3, = 13[/quote]
You need to stop spamming classrom math, and now this. All you do is post a solution that has been posted multiple times with very incomplete solutions that aren't hidden.
To klutzcherry, you shouldn't edit your posts if you make a mathematical mistake. Typos are okay to edit though... Anyways, the reason you shouldn't edit careless mistakes is because newcomers who aren't as proficient as most in this forum may think that everyone here is perfect. (It is discouraging, and this was my first impression when I first joined)[/quote]
Actually, there's really nothing wrong with making a careless mistakes and then changing it. But, if it's possible, try to just make a note that you've changed it.


Find all primes of the form n^n+1 which are less than 10^19

If $ n$ is odd, then it is divisible by $ 2$. Note $ n=1$ is still a solution because $ n^{n}+1=n+1$ in that case.

$ n=2$ gives $ 5$.
$ n=4$ gives $ 257$.
$ n=6$ is divisible by $ 13$.
$ n=8$ is divisible by $ 97$.
$ n=10$ is divisible by $ 101$.
$ n=12$ is divisible by $ 89$.
$ n=14$ is divisible by $ 29$.
$ n=16$ is larger than $ 10^{19}$

Hence our primes are $ p=2,5,257$ where $ n=1,2,4$, respectively.

[hide="*cough*"] If $ n$ has an odd factor $ d$, then $ n^{\frac{n}{d}}+1 | n^{n}+1$, so the integer $ n^{n}+1$ is prime only if $ n$ is a power of $ 2$.  It's enough to check $ n = 1, 2, 4, 8$.

And if you want a better way to check $ n = 8$, note that

$ p | 8^{8}+1 \implies p | 8^{16}-1 \implies 16 | \varphi(p) = p-1$

So we only have to check $ p = 17, 97$ before we conclude that $ 8^{8}+1$ is composite. [/hide]


function
logarithms
integration
vector
real analysis
real analysis theorems

Please explain me what kind of differential system is the following one:

dx/x=dy/y=dz/z;x>0,y>0,z>0

I mean i,ve seen differential systems before,but they contained x',y',x',x'',etc.
I haven't seen such differential system with dx,dy,dz.

I think that dx, dy, and dz are TOTAL differential. This means that x, y and z are three different functions of a single variable, that we can call t. This means that it is an equation defining a differntiable curve in R^3.

\[ \begin{array}{l} r(t) = \left( \begin{array}{l} x(t) \\ y(t) \\ z(t) \\ \end{array} \right) \\ \\ \frac{{dx}}{x} = d\left[ {\ln (x)} \right],\frac{{dy}}{y} = d\left[ {\ln (y)} \right],\frac{{dz}}{z} = d\left[ {\ln (z)} \right] \\ \\ \frac{{d\left[ {\ln (x)} \right]}}{{dt}} = \frac{{d\left[ {\ln (y)} \right]}}{{dt}} = \frac{{d\left[ {\ln (z)} \right]}}{{dt}} \\ \\ r_l (t) = \left( \begin{array}{l} x_l (t) \\ y_l (t) \\ z_l (t) \\ \end{array} \right) = \left( \begin{array}{l} \ln (x(t)) \\ \ln (y(t)) \\ \ln (z(t)) \\ \end{array} \right) \\ \frac{{dr_l }}{{dt}} = \left( \begin{array}{l} \frac{{d\left[ {\ln (x(t))} \right]}}{{dt}} \\ \frac{{d\left[ {\ln (y(t))} \right]}}{{dt}} \\ \frac{{d\left[ {\ln (z(t))} \right]}}{{dt}} \\ \end{array} \right) = \left( \begin{array}{l} f(t) \\ f(t) \\ f(t) \\ \end{array} \right) = f(t)\left( \begin{array}{l} 1 \\ 1 \\ 1 \\ \end{array} \right) \\ r_l (t) = r_l (0) + \left[ {\int\limits_0^t {f(s)ds} } \right]\left( \begin{array}{l} 1 \\ 1 \\ 1 \\ \end{array} \right) = r_l (0) + g(t)\left( \begin{array}{l} 1 \\ 1 \\ 1 \\ \end{array} \right) \\ r(0) = \left( \begin{array}{l} x_0 \\ y_0 \\ z_0 \\ \end{array} \right) \\ r_l (0) = \left( \begin{array}{l} \ln (x_0 ) \\ \ln (y_0 ) \\ \ln (z_0 ) \\ \end{array} \right) \\ r_l (t) = \left( \begin{array}{l} \ln (x_0 ) + g(t) \\ \ln (y_0 ) + g(t) \\ \ln (z_0 ) + g(t) \\ \end{array} \right) = \left( \begin{array}{l} \ln (x_0 e^{g(t)} ) \\ \ln (y_0 e^{g(t)} ) \\ \ln (z_0 e^{g(t)} ) \\ \end{array} \right) \\ r(t) = \left( \begin{array}{l} x_0 e^{g(t)} \\ y_0 e^{g(t)} \\ z_0 e^{g(t)} \\ \end{array} \right) = e^{g(t)} r(0) \\ \end{array}  \]

Therefore the solution to the equation is a semirecta: it's the set of all vectors that can be expressed as:
\[ \begin{array}{l} r = \lambda r_0 \\ \lambda > 0 \\ \end{array}  \]


$ 90^\circ<x<180^\circ$

$ \frac{1\plus{}sin2x}{sinx\plus{}cosx}\plus{}\sqrt{1\minus{}sin^2x}\equal{}?$

hello, because $ \frac{\pi}{2}<x<\pi$ we have
$ \frac{1\plus{}2\sin(x)\cos(x)}{\sin(x)\plus{}\cos(x)}\minus{}\cos(x)\equal{}$
$ \frac{1\plus{}2\sin(x)\cos(x)\minus{}\cos(x)\sin(x)\minus{}\cos(x)^2}{\sin(x)\plus{}\cos(x)}\equal{}$
$ \frac{\sin(x)^2\plus{}\sin(x)\cos(x)}{\sin(x)\plus{}\cos(x)}\equal{}\sin(x)$.
Sonnhard.


algebra
polynomial
function
system of equations

Find all solutions to the system of equations $xy+yz+zx=12,$ $xyz=2+x+y+z$ in positive reals $x,y,z.$

Hmm, it's getting late so my brain is no longer operating properly. :( Here are my thoughts. (I might be overthinking this)

[hide="Partial Progress"]We can find bounds for $xyz.$ Since we have positive reals, AM-GM yields
\[4=\frac{xy+yz+xz}{3}\ge \sqrt[3]{xyz}\implies xyz\le 8\implies x+y+z\le 6.\]
Also by AM-GM,
\[\frac{x+y+z}{2}=\frac{xyz-2}{2}\ge \sqrt{xyz}\implies xyz\ge 3+\sqrt{5}.\]
So now we have less values to check, but I'm not sure how to get at those values. Only obvious solution is $(2,2,2).$

Does this have something to do with that polynomial coefficient trick? I dunno, it didn't work out for me.[/hide]

[hide="Some more partial progress"] WLOG let $x \le y \le z$.  Then $yz \ge 4$ and so $y+z \ge 4$.  But then

$xy+yz+zx = yz+(y+z)x = 12 \ge 4+4x \implies x \le 2$

Furthermore, notice that $xyz-2-x-y-z$ is an increasing function in $x$, so at some value of $x$ (for a given $y, z$) there will be a solution, and no other.

The solution $x = y = z = 2$ is now natural, but I can't quite prove the nonexistence of a solution for $x$ very small.  We do know that $yz \le 12$. [/hide]

This problem is very well known, and posted many times. I love the question a lot. It was question 4 during the second round of BrMO 1998.

4everwise, check your post, it contains some mistakes.

My solution:
Clearly $xyz \leq 8$ by AM-GM.

Now, $(xyz-2)^{2}=(x+y+z)^{2}\geq 3(xy+yz+xz)=36$
So $xyz \geq 8$
This gives us that $xyz=8$, and we can now easiley find that $(x+y+z)^{2}= 3(xy+yz+xz)$ so $x=y=z=2$.

Note that the original question was posed as follows:
Find all solutions of the simultaneous equations in which all of $x,y,z$ are positive, and prove that it is the only such solution.
Show that a solution exists in which $x,y,z$ are real and [u]distinct.[/u]


analytic geometry
calculus
calculus computations

A particle is moving on the $ x$-axis, where $ x$ is in centimeters. Its velocity, $ v$, in cm/s, when it is at the point with coordinate $ x$ is given by $ v\equal{}x^2\plus{}5x\minus{}4$.

Find the acceleration of the particle when it is at the point $ x\equal{}6$.

[hide]$ v\equal{}dx/dt$

Find $ dv/dt$ when $ x\equal{}6$.

$ \frac{dv}{dt}\equal{}\frac{d}{dx}(x^2\plus{}5x\minus{}4)*\frac{dx}{dt}\equal{}(2x\plus{}5)(x^2\plus{}5x\minus{}4)$.

$ (2(6)\plus{}5)(6^2\plus{}5*6\minus{}4)\equal{}1054cm/s^2$.[/hide]


Consider the 4th degree equation of rational non-null coefficients such that $ L,M,N,P$ are roots of this equation. Calculate the value of the following expression:

$ \frac{L}{MNP}\plus{}\frac{M}{LNP}\plus{}\frac{P}{LMN}\plus{}\frac{N}{LMP}$

[quote="triplebig"]Consider the 4th degree equation of rational non-null coefficients such that $ L,M,N,P$ are roots of this equation. Calculate the value of the following expression:

$ \frac {L}{MNP} \plus{} \frac {M}{LNP} \plus{} \frac {P}{LMN} \plus{} \frac {N}{LMP}$[/quote]
[hide]$ \frac {L}{MNP} \plus{} \frac {M}{LNP} \plus{} \frac {P}{LMN} \plus{} \frac {N}{LMP}$
$ \frac {L^{2} \plus{} M^{2} \plus{} P^{2} \plus{} N^{2}}{LMNP}$
$ \frac {(L \plus{} M \plus{} P \plus{} N)^{2} \minus{} 2(LM \plus{} LP \plus{} LN \plus{} MP \plus{} MN \plus{} PN)}{LMNP}$
Call the 4th degree equation $ ax^{4} \plus{} bx^{3} \plus{} cx^{2} \plus{} dx \plus{} e \equal{} 0$
This becomes $ \frac { \minus{} \frac {b}{a} \minus{} 2\frac {c}{a}}{\frac {e}{a}} \equal{} \frac { \minus{} b \minus{} 2c}{e}$[/hide]


Do you know is there any page where we can download shortlist and solutions to problems proposed for Balkan M.O.

Also same question for other competitions (Austrian - Polish, APMO...)

[quote="Mircea Lascu"]There is a book called "Balkan Mathematical Olympiads 1984-2006". The authors are: Mircea Becheanu and Bogdan Enescu.

It is a collection of the problems proposed at the Balkan Mathematical Olympiads (BMO), also containing some of the problems discussed by the jury. It is not only a collection of elementary problems but a page of history - the history of the BMO, one of the first regional mathematical competition for highschool students. All problems are presented with complete solutions. Many problems have several alternative solutions and we also present some extensions. An additional preparatory addendum, containing notions and classical useful results has been added to the end of the book.[/quote]

 I have this book and it is really nice. I recomend it highly to all students. As about Austrian Polish there is a book(by prof. Cucma) , but not in some site for free.

 Babis


This problem is being used in a currently-running contests, so I have temporarily deleted the problem and its solution for now.  Thanks for your understanding.


linear algebra
matrix
linear algebra unsolved

Hi, I have here an adjacency matrix A of the supposedly connected graph G

\[ \left(
\begin{array}{cccccc}
0 & 1 & 0 & 0 & 0 & 0\\
1 & 0 & 1 & 0 & 0 & 1\\
0 & 1 & 0 & 1 & 0 & 0\\
0 & 0 & 1 & 0 & 1 & 1\\
0 & 0 & 0 & 1 & 0 & 1\\
0 & 1 & 0 & 1 & 1 & 0\ \end{array}
\right)\]

I need to find $ A^2$, $ A^3$ and $ A^4$ to show that this graph G is connected.

$ A^2$

\[ \left(
\begin{array}{cccccc}
1 & 0 & 1 & 0 & 0 & 1\\
0 & 3 & 0 & 2 & 1 & 0\\
1 & 0 & 2 & 0 & 1 & 2\\
0 & 2 & 0 & 3 & 1 & 1\\
0 & 1 & 1 & 1 & 2 & 1\\
1 & 0 & 2 & 1 & 1 & 3\ \end{array}
\right)\]

$ A^3$

\[ \left(
\begin{array}{cccccc}
0 & 3 & 0 & 2 & 1 & 0\\
3 & 0 & 5 & 1 & 2 & 6\\
0 & 5 & 0 & 5 & 2 & 1\\
2 & 1 & 5 & 2 & 4 & 6\\
1 & 2 & 2 & 4 & 2 & 4\\
0 & 6 & 1 & 6 & 4 & 2\ \end{array}
\right)\]

$ A^4$

\[ \left(
\begin{array}{cccccc}
3 & 0 & 5 & 1 & 2 & 6\\
0 & 14 & 1 & 13 & 7 & 3\\
5 & 1 & 10 & 3 & 6 & 12\\
1 & 13 & 3 & 15 & 8 & 7\\
2 & 7 & 6 & 8 & 8 & 8\\
6 & 3 & 12 & 7 & 8 & 16\ \end{array}
\right)\]

Not sure how to prove that this graph is connected. I know that the $ (i,j)$ entry of $ A^k$ is a $ \plus{}ve$ integer which is number of paths of length $ k$ from $ i$ to $ j$ in the graph. So, for instance the entry $ (6,5)$ in $ A^3$ is $ 4$, which means $ 4$ paths of length $ 3$ from $ 6$ to $ 5$.

Also, I am not sure why we have to go as far as $ A^4$ to show that the graph is connected.

Any help would be most appreciated!

Thank you in advance!

Antonio

I'm not sure if it's this simple.

[quote]$ (i,j)$ entry of $ A^k$ is a positive integer which is the number of paths of length $ k$ from $ i$ to $ j$ in the graph[/quote]

Since $ (1,2)$ entry of $ A$, $ (1,3)$ and $ (1,6)$ entry of $ A^2$, $ (1,4)$ and $ (1,5)$ entry of $ A^3$ are all positive, so, $ 1$ is connected to $ 2$, $ 3$, $ 4$, $ 5$ and $ 6$. So the graph is connected...

It's that simple.

Also, you only have to go to $ A^2$, if you look at the second column.


Anyone could please show me how to solve this probability question. thanks for your help.

You are given 5 red and 8 blue discs in a box. Peter, Ron and Betty, in that order, select a disc without replacement. The first to select a red disc wins. What is the probability that Ron wins.

Charikaar

First off, note that Ron can't win on his 4th draw,since his fourth draw is the eleventh pick from the box,which means that 10 blues would have to be picked before hand,and we only have 8 blues. So Ron wins on his 1st,OR 2nd,OR 3rd draw.

Define the following events: $A=\{Ron \ wins \ on \ first \ draw\}$, $B=\{Ron \  wins\ on \ 2nd\ draw \}$, $C=\{Ron \ wins\ on\ 3rd\}$. Note that $A,\ B,\ C$ are mutually exclusive so to find the probability that Ron wins we sum the probability of events $A,\ B, $and $C$.

$P(A)= \frac{8}{13} \cdot \frac{5}{12}= .2564$

$P(B)= \frac{8}{13} \cdot \frac{7}{12} \cdot \frac{6}{11} \cdot \frac{5}{10} \cdot \frac{5}{9}=.0544$

$P(C)= \frac{8}{13} \cdot \frac{7}{12} \cdot \frac{6}{11} \cdot \frac{5}{10} \cdot \frac{4}{9} \cdot \frac{3}{8} \cdot \frac{2}{7} \cdot \frac{5}{6}=.0039$

Thus, $P(A \cup B \cup C)=P(A)+P(B)+P(C)=.2564+.0544+.0039=.3147$.

Seems correct-- but I have been known to post nonsense.

Thanks for the help. Do you know if it is possible to solve the same question using Tree Diagram?

Charikaar

It would look something like this:

Peter gets Red-5/13_____Ron gets Blue-7/12--------Betty gets Red-5/11

Peter gets Blue-8/13-------Ron gets Red-5/12______Betty gets Blue-6/11------

Where an option without a dotted line means that it's the end of a branch.

The underscores are for spacing.