Mafia
algorithm
abstract algebra
geometry
rectangle
analytic geometry

How is final exam going? I get to write math diploma in Janaury and omit English and social written exams, because of Waterloo Seminar[hide="!"]:D  :lol:  :first:  :coolspeak:  :clap2:  :heli:  :agent:  :clap:  :rotfl:  :harhar:  :yup:  :w00tb:  :thumbup:  :whistling:[/hide]
I still have chem, bio(today!), and social muliple choice to write. :(

Does Anyone want to complain about your exams?

I wrote the exams before the seminar, but screwed up on 2 out of the 5.
Darn...

but I do have a math 12 challenge to write, so boring... 5 hours

Yeah! Finished my bio exam! It was easy but had some stupid questions. :D

My bio exam was crazy.

Powerpoint presentation and 6 diagrams, 3 binomial nomenclature charts!

I hope I did well.......

[quote="lightrhee"][quote="billzhao"]You got English omitted? Wow.

I have my exams for English, French and Math, starting tomorrow, with the last one on friday.[/quote]
Friday? I thought you would be at Washington on that day. :?[/quote]

We leave on Sunday.

[quote="Sunny"]Ahhhh I'm never going to survive the week!  If you don't see me on Saturday david blame it on my crazy teachers!!!!!!

Wrote my in class social final, and in class physics final today.  Tomorrow I have all 3 french finals.  Wednesday I have my english final (phew, easy one)  Thursday I have physics dipolma exam all day.  Then friday i have the final portion of my social final.  And this isn't including my speech for tomorrow....AHHH!H!HH!!!!!H!H! :stretcher:  :wallbash_red:  :cursing:  :censored:

I'm going to die in the next 24 hours...like I said, if you don't see me on saturday david, you know who to blame...[/quote]

Yikes!  Yeah, my schedule looks similar.  Just remember, we are all in this, and we will get through this, together.  :)

Lol, you done?   :blush:

Damn i'm so jealous :mad:

[quote="TheDreamer"]Lol, you done?   :blush:

- i'm so jealous :mad:[/quote]
No, I'm not done. I just get to omit a few stressful exams. I only have chem and social part B(multiple choice) left. :)

[quote="Sunny"][quote="lightrhee"][hide="This post may offend you"][quote="TheDreamer"] Just remember, we are all in this[/quote]
Not me :harhar: [/hide][/quote]

 :furious:  :diablo:

Oh no worries David...you know what happens to people who offend me...heh heh heh...see you on Saturday.   :bomb:[/quote]
I have warned you. You shouldn't have read it if you were going to be offended. I shouldn't have given you the idea(calling me and playing the instrument, which I can not spell(something like piccolo, which seems to be spelled wrong(maybe the spelling is correct))). I better disconnect my phone tonight... :D

[quote="lightrhee"]

I have warned you. You shouldn't have read it if you were going to be offended. I shouldn't have given you the idea(calling me and playing the instrument, which I can not spell(something like piccolo, which seems to be spelled wrong(maybe the spelling is correct))). I better disconnect my phone tonight... :D[/quote]

Hahaha I don't need to call you.  I'll see you on Saturday.  And Sunday, and Monday, and Tuesday.   :D   Aren't you excited to see me again?

BTW, it is piccolo.  Hmm I should bring it... :diablo:

Hey David,
Got any place to play Mao online?

[quote="TheDreamer"]You never cease to amaze us Elyot.  Anyways, I won't go on and on about that, I'm sure you've heard enough flattery throught your life..

Do you just get instructions online?  I have a bunch of books, and I used to get them at the library..i don't really have pics of my stuff cuz it was long ago..but i do still have a few things i made, i can show you sometime.

Lol next time I see you we should fold some paper *looks for paperfolding emoticon*

Oh, and can you teach me an algorithm for trisecting a paper?  The push two ends together way is really inaccurate, so I was wondering whether someone knew a better way

Anyways, I heard you're also good at cooking, maybe you should also post pictures of crazy foods you cooked  :P[/quote]

The big one is my own design, but it's based on a technique called "cross-box pleating".  I've never seen anybody attempt a 4x4 cross-box though.  Even 2x2 is relatively insane.  That thing took me like 18 hours at least.

Kawasaki rose is pretty famous and you can get diagrams for it online.  It takes maybe 45 minutes first time... then once you get the hang of it, you can do one in 10-15 mins.

The other two are my own designs based on pretty common modular forms.  There are many popular types of modules (pieces) that you can fold, some with as few as 2 folds, and you can put them together in all sorts of intricate designs and patterns... add colours and you can do some neat stuff.

---------------------------------------------------------------

I don't have any cooking pictures, but I can post some "pseudo-recipes".  Here's a simple one with like 3 ingredients, so it's not too hard.

Try this one...
- Take about 8cm by 8cm square of brownie.
- Cut it into 16 pieces.
- Add 2 or 3 scoops of SMORES ice cream.  Mix.
- Heat up about 40 mL or so of NUTELLA!
- Pour molten nutella onto brownie/ice cream concoction.
- Enjoy (with a special someone)!

I invented this dessert impromptu while trying to impress my girlfriend.  Apparently, it worked.  It was uncompromisingly delicious and she rewarded me graciously for it later ;)

Another trick I invented to impress people is to cut an orange in half and throw it in the freezer.  Then whenever you make a dessert, take a serrated knife and shave off a tiny, paper-thin piece of orange and use it as a garnish.  If you shave it off properly, it should curl quite nicely.  The rind also becomes quite edible when it's in a small, shaved, frozen chunk.

------------------------------------------------------------

Trisecting a sheet of paper isn't hard... you can actually trisect lengths with compass and straightedge constructions...

- Take a rectangle and fold a 4x4 grid on it.  I'm gonna use coordinates, so the bottom left is (0,0), the bottom right is (4,0), top left is (0,4), top right is (4,4).
- Fold the following:
--- The line through (2,1) and (4,4)
--- The line through (3,1) and (4,4)
- These 2 lines trisect the line from (0,0) to (4,0).  It's easy to prove using similar triangles.

[quote="lightrhee"][quote="d.lam.86"][quote="TheDreamer"]
Anyways, I heard you're also good at cooking, maybe you should also post pictures of crazy foods you cooked  :P[/quote]
He is Elyot the Multi-Talented Man!  :rotfl:  ;)[/quote]
When Dr. Wright was reading Elvot's bio today(today was the media day), it was so fun. Elyot has like a few hundreds of interests. :rotfl:[/quote]

Yeah, a year ago he used to say "I'm good at every aspect of life" and I thought he was kidding...well with every passing day I believe him more and more   :D

[quote="TheDreamer"][/quote][quote="lightrhee"][quote="d.lam.86"][quote="TheDreamer"]
Anyways, I heard you're also good at cooking, maybe you should also post pictures of crazy foods you cooked  :P[/quote]
He is Elyot the Multi-Talented Man!  :rotfl:  ;)[/quote]
When Dr. Wright was reading Elvot's bio today(today was the media day), it was so fun. Elyot has like a few hundreds of interests. :rotfl:[/quote][quote="TheDreamer"]

Yeah, a year ago he used to say "I'm good at every aspect of life" and I thought he was kidding...well with every passing day I believe him more and more   :D[/quote]

I don't think I said that... I'm not that arrogant, am I???

I probably said something more like "I try to be good at every aspect of life; not just math."  "TRY!"  I would never claim that I'm good at everything.

[quote="Elyot"][/quote][quote="TheDreamer"][/quote][quote="lightrhee"][quote="d.lam.86"][quote="TheDreamer"]
Anyways, I heard you're also good at cooking, maybe you should also post pictures of crazy foods you cooked  :P[/quote]
He is Elyot the Multi-Talented Man!  :rotfl:  ;)[/quote]
When Dr. Wright was reading Elvot's bio today(today was the media day), it was so fun. Elyot has like a few hundreds of interests. :rotfl:[/quote][quote="TheDreamer"]

Yeah, a year ago he used to say "I'm good at every aspect of life" and I thought he was kidding...well with every passing day I believe him more and more   :D[/quote][quote="Elyot"]

I don't think I said that... I'm not that arrogant, am I???

I probably said something more like "I try to be good at every aspect of life; not just math."  "TRY!"  I would never claim that I'm good at everything.[/quote]

Haha, maybe I don't remember what you said exactly, but don't worry, you don't claim to be more than you are :)

[quote="TheDreamer"]

Haha, maybe I don't remember what you said exactly, but don't worry, you don't claim to be more than you are :)[/quote]

Awww... thanks.

Funny, mostly unrelated story:
I gave a long, complicated guest lecture today about "HOW TO GET A GIRLFRIEND".  Crazy.

[quote="Elyot"][quote="Sunny"][quote="Elyot"]

Awww... thanks.

Funny, mostly unrelated story:
I gave a long, complicated guest lecture today about "HOW TO GET A GIRLFRIEND".  Crazy.[/quote]

HAHAHAHA.  I wanna hear this.[/quote]

Well... first I gave the steps... I said that before they get girlfriends, they need to get big, active social lives with girls.  Then I talked about what kind of characteristics to look for in girls, how to see if you like them, how to find one that likes you too, etc.  Then I gave them some stories about some of my old girlfriends, etc.  Then they asked questions.

It was quite interesting.  Some things I will refrain from posting ;)[/quote]
Just curious, who were the audience?  :huh:

[quote="d.lam.86"][/quote][quote="Elyot"][quote="Sunny"][quote="Elyot"]

Awww... thanks.

Funny, mostly unrelated story:
I gave a long, complicated guest lecture today about "HOW TO GET A GIRLFRIEND".  Crazy.[/quote]

HAHAHAHA.  I wanna hear this.[/quote]

Well... first I gave the steps... I said that before they get girlfriends, they need to get big, active social lives with girls.  Then I talked about what kind of characteristics to look for in girls, how to see if you like them, how to find one that likes you too, etc.  Then I gave them some stories about some of my old girlfriends, etc.  Then they asked questions.

It was quite interesting.  Some things I will refrain from posting ;)[/quote][quote="d.lam.86"]
Just curious, who were the audience?  :huh:[/quote]

Everyone in the van, including the IMO team, Felix, and Dorette.

[quote="lightrhee"][quote="Sunny"]He is refering to the IMO team on their way to Banff.  Or when they got there.[/quote]
On the way. But I was sleeping, so I missed everything. :D[/quote]

You fell asleep because you already knew everything?

[quote="TheDreamer"]
You fell asleep because you already knew everything?[/quote]

Lol, probably he fell asleep because we were up to 3 in the morning the night before, went through a day of math and then faced the reporters and media for another few hours.


Find the area of the region enclosed by the graph whose equation is $ \mid y\minus{}x\mid \plus{}\mid y\mid \equal{} 2$.

The region inside $ |x \minus{} y| \plus{} |y| \equal{} 2$ is just a skew of that inside $ |x| \plus{} |y| \equal{} 2$, so they have the same area.

[b]Problem 1:[/b] Find the volume enclosed by the condition:
\[ |x \plus{} y \minus{} z| \plus{} |x \minus{} y \plus{} z| \plus{} | \minus{} x \plus{} y \plus{} z| \leq 6\]
[b]Problem 2:[/b] Show that the volume enclosed by:
\[ |a_1 x \plus{} b_1 y \plus{} c_1 z| \plus{} |a_2 x \plus{} b_2 y \plus{} c_2 z| \plus{} |a_3 x \plus{} b_3 y \plus{} c_3 z| \leq 1\]
is the same as the volume enclosed by:
\[ |a_1 x \plus{} a_2 y \plus{} a_3 z| \plus{} |b_1 x \plus{} b_2 y \plus{} b_3 z| \plus{} |c_1 x \plus{} c_2 y \plus{} c_3 z| \leq 1\]

[u]Hint:[/u] [i]First consider the analogous case of two dimensions.[/i]


I know that for most problems, either a proof or number is required as the final solution, and that as long as it's right and obvious simplifications were made, the answer is right.

However, for number 5, how would you define as locus in the final answer? Like, by properties or...what?

I would assume something like an equation of a circle, or a line, or etc.

Usually, one describes a locus either in terms of an equation representing valid points in the cartesian coordinate system or with a simple description.  For example, I can describe the locus of points equidistant from a point $ O$ as a circle with center $ O$, or I can define $ O\equal{}(0,0)$ and the locus would then be all points $ (x,y)$ such that $ x^2\plus{}y^2\equal{}r^2$ for constant $ r$.

[quote="cosinator"]Usually, one describes a locus either in terms of an equation representing valid points in the cartesian coordinate system or with a simple description.  For example, I can describe the locus of points equidistant from a point $ O$ as a circle with center $ O$, or I can define $ O \equal{} (0,0)$ and the locus would then be all points $ (x,y)$ such that $ x^2 \plus{} y^2 \equal{} r^2$ for constant $ r$.[/quote]

What if you said "the locus is a circle with center $ (0,0)$ and radius $ r$? Would that get credit?

I'm pretty sure that for a circle, either of the following would be accepted:

$ x^2\plus{}y^2\equal{}9$

A circle centered at $ (0,0)$ with radius $ 3$.

For a line, any one of the forms should be accepted.

For most others shapes, I'd use the equation, but a description (or a picture) should be accepted.

[quote="Heero Kirashami"]So, for a line, could I say "The locus of the points is the line with this certain relation [not stated here] to line [one of the ones in the problem]"?[/quote]

I think as long as your description includes every point in the locus.

This is really giving away a lot, guys >_>

It's actually not, the locuses discussed have nothing to do with the problem.

Why not just delete all the posts in this thread as the original poster got his question answered.

Hey, it doesn't really matter. We're just naming geometrical shapes.

We probably could've gone with "if the points are in this star [or other geometrical shape]" but that's more awkward as an example. Plus, stars are hard to graph without parametrics (as far as I know...).

[quote="123456789"]But if you're giving examples of loci, that's a huge clue which aren't the answer...Think about it.[/quote]

Let me clarify. I didn't mean that the loci are not the answer to the problem. I only meant that they were arbitrarily chosen and have no purposeful relation to the problem.


Two cubes, whose faces are numbered 4, 5, 6, 7, 8, and 9, are
tossed. What is the probability that the product of the two numbers
on the top faces is between 29 and 61? Express your answer as a
common fraction.


Hello!
I've taken AIME problem series B during winter 2007-2008. Unfortunately, I forgot to keep a copy of the transcripts/hw problems. Is there any way I can get a copy of them now?

Thanks.

sorry to be redundant and reviving an old thread

but i seem to have the same problem as lena_siz, a couple days ago, my computer crashed with all my AIME problem series B homework, solutions, transcripts etc....

is there any way I can access that information to study for the AIME?

I believe I still have the receipt and confirmation stuff that I took the class.

Thanks


induction
combinatorics proposed
combinatorics

The squares of a grid m*n ( or you can assume first that the grid is n*n) are coloured arbitrary with white and black. Prove that using a king we can move from the left-most column to the right-most column only on white squares or we can move with the up-most row to the down-most row only on black squares. :)

I just loved the solution I saw today. I look forward to see yours. The problem is well known, but I hope it wasn't posted before :)
You may also see this http://www.mathlinks.ro/Forum/viewtopic.php?t=138389 and figure out what is the connection :)

Anybody? C'mon the problem is really nice :)

i have one solution to this, but i'm not certain about it  :(
i do one induction and see a lot of cases.

very, very nice problem!  :wink:

could you post the solution ciprian? i'm absolutely right that your way is more nice than mine..

(again, sorry for my bad english!)

I must say, for the begining, that is NOT my solution... :blush: However, I will give just a hint for the moment cause I don't wanna spoil the fun :P
[hide="Hint"]
Consider a (planar) graph G with black and white vertices corresponding to the squares - two vertices are connected if they are on the same row or column. Then add some additional vertices (this is the crux move and gives the beauty of the problem), let's say red vertices that simulate the diagonally moves of the king. You will get a triangulated planar graph, if I remember well...Then u add 2 black and 2 white vertices outside such the triangulated graph is in the interior of that 'quadrilateral'. (the same color vertices are opposite). Triangulate again and prove the statement now...Enough hints for the moment :wink: .  I am in a rush now so I hope I haven'n made confusions...
[/hide]

[quote="AMM, March 1999"]
A Tale of Two Integrals
by Vilmos Totik
Show that there is some interval $I \subset [0,1]$ such that the integrals of $f$ and $g$ over $I$ are both equal to $\frac{1}{2}$. This problem is equivalent to the following problem without integrals: On a blackjack machine one can win or lose one dollar at a time. Suppose two players playing once per minute during a period find that eventually both of them win exactly $2N$ dollars. Show that there was a time interval during which both of them won exactly $N$ dollars.

Six solutions are presented that are based on other combinatorial or geometrical/topological results. These are:

the Borsuk-Ulam antipodal theorem on continuous mappings of spheres;
the mountain climbing problem on two climbers staying always at the same altitude;
the chord theorem for parallel chords of planar curves;
the chess king-moving theorem on king walks on cells of the same colour;
the winding number theorem for vector fields;
the Jordan curve theorem on components of simple curves.[/quote]
This is what I found related to this problem.

Could anyone post the entire article? I'd be eternally grateful :D


Using Caushy shwarz inequality we have

$ \frac{1}{5}\plus{}\frac{1}{x\plus{}1}\plus{}\frac{1}{x\plus{}1}\plus{}\frac{1}{x\plus{}1} \ge \frac{16}{3x\plus{}8} \ge \frac{16}{3x\plus{}x\plus{}\frac{16}{x}}\equal{}\frac{4x}{4\plus{}x^2}$

Similarly we apply the same thing we get

$ \sum_{i\equal{}1}^5\left(\frac{1}{5}\plus{}\frac{1}{x_i\plus{}1}\plus{}\frac{1}{x_i\plus{}1}\plus{}\frac{1}{x_i\plus{}1}\right) \ge \sum_{i\equal{}1}^5 \frac{4x_i}{4\plus{}x_i^2}$

which is equivalent to

$ 1 \ge \sum_{i\equal{}1}^5 \frac{x_i}{4\plus{}x_i^2}$

The given hypothesis implies $ \sum_{cyc}{\frac{4-x_i}{1+x_i}} =0$ .We need to prove that :
\[ 1= \sum_{cyc}{\frac {1}{{1 + {x_i}}}  \geq \sum_{cyc}{\frac {{{x_i}}}{{4 + {x_i}^2}}  \iff \sum_{cyc}{\frac{4-x_1}{1+x_1}\cdot \frac{1}{4+x_1^2}} \geq 0}}\]
Let $ x_1 \leq x_2 \leq x_3 \leq x_4 \leq x_5$ ,then :
\[ \frac{4-x_1}{1+x_1} \geq \frac{4-x_2}{1+x_2} \geq \frac{4-x_3}{1+x_3} \geq \frac{4-x_4}{1+x_4} \geq \frac{4-x_5}{1+x_5}\]
\[ \frac{1}{4+x_1^2 } \geq \frac{1}{4+x_2 ^2} \geq \frac{1}{4+x_3 ^2} \geq \frac{1}{4+x_4 ^2} \geq \frac{1}{4+x_5 ^2}\]

Now applying Chebyshev inequality for the above sequences ,we're done .Equality holds for $ x_i =4$


function
algebra
domain
calculus
integration
geometric series
arithmetic series

I've encountered some directions for problems that say "Solve for x, state any restrictions on the variables." What does this mean, and how do you find the restrictions? Thanks in advance, dynamo729.

"Restrictions" most likely means the set of values that $ x$ can take on and still have the function produce a real result, or the domain of the function.  For the most part, you just have to worry about dividing by 0 and taking square roots of negative numbers.

Restrictions are like..I'll give a few examples.

[hide="eg1"]
Find all real numbers $ a$ such that the roots of $ x^3 \minus{} 6x^2 \plus{} 21x \plus{} a \equal{} 0$ are not all real and form an arithmetic progression.
[/hide]

[hide="eg2"]
Find the only integral solution of $ (x \minus{} 1)^3 \plus{} (x \minus{} 2)^3 \plus{} (x \minus{} 3)^3 \plus{} ... \plus{} (x \minus{} 2007)^3 \equal{} 0$
[/hide]

Equations for which you want integer solutions are called [b]Diophantine equations[/b]

[hide="eg3"]
Solve the equation $ x/(x \minus{} 2) \equal{} 3/(x \plus{} 2) \plus{} 6x/(x^2 \minus{} 4)$.
[/hide]

You can't have $ x \equal{} 2 \text{ or} \minus{} 2$. If you find that 2 or $ \minus{} 2$ turn out to be the solutions, these are called extraneous solutions. (In this case, -2 is an extraneous solution, so there are no solutions to this equation)

Why "can't we divide by 0"? Actually we can, but when we divide by 0, we can infinity. I have a truly beautiful proof of this using geometric series and arithmetic series.

Consider a geometric series $ a\plus{}ar\plus{}ar^2\plus{}ar^3\plus{}...\equal{}a/(1\minus{}r)$ by geometric series formula, where $ a$ is a real number. If $ r\equal{}1$, then the geometric series is an arithmetic series $ a\plus{}a\plus{}...\equal{}\infty$. Then $ a/(1\minus{}r)\equal{}a/0\equal{}\infty$.


Is there an equivalent to Microsoft word's spellcheck in TeXnicCenter that doesn't count all the commands and colored text?

Sometimes I copy and paste everything into microsoft word and spellcheck, ignoring what word thinks as wild punctuation mistakes. Is there a better way to do this?

If you click on [b]tools[/b] button, you'll see [b]Spelling[/b] in the menu. That's the TeXnic center built-in spell checker. Try it and see if it is satisfactory for you.


I am a senior in high school trying to decide what major to study in college. I am wondering whether there are any electrical engineers here who know about the field, since I am considering majoring in electrical engineering. Particularly, I am concerned about my impression that most (or many) electrical engineers end up doing mindless tasks for their day jobs that pay five-figure salaries and lead nowhere. I do not know whether this description is true or not, but I get the feeling that there is not much left to discover in the field of electrical engineering (as opposed to, say, the field of bioengineering). Thus, I fear that after getting an electrical engineering degree, I would just be thrown in with the masses of people who perform manual labor and textbook applications of electrical engineering principles and never go anywhere with their careers.

Any opinions or remarks would be greatly appreciated. Thanks.

Even when my research is applied to Electrical engineering, I am not an engineer, nevertheless, in [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=34592]this thread[/url], wild octagon sais that

[quote]The Bureau of Labor Statistics site lists 69,970 as the median salary for actuaries, with the top 10% earning at least 137,650. Compare to salaries for mathematicians (76,470 / 112,780), electrical engineers (69,930 / 103,860) and financial analysts (57,100 / 108,060).[/quote]
This suggest that in general, Mathematicians earn more than Electrical Engineers and that your impressions may be not that wrong. I don't know much about the market for Electrical Engineers here, but I know plenty of them in Latin America, and they claim that one of the problems in their profession is that the real market is oversaturated (in Latin America, I don't know here) and that many people with Bachelors or Masters in the Area have to take jobs as technicians, and well, you don't want to spend six years and a lot of money in the education for a great job to get at the end the same job that you will get if you go year and a half to a technical institute to earn a technical degree!

By the way, this should be in the College forum I think!

The problem with speaking in generalities is that well... it's just a bunch of generalities. It doesn't necessarily reveal anything that will be specific to YOU. There are a lot of people who get degrees in electrical engineering - some of whom do well, some of whom do not; some of whom have exciting careers, and some of whom do not; some of whom enjoy what they do, and some of whom... well, who do not.

The main differences between these types of people are a.) How lucky/talented/wise are they to end up working in an exciting place, or starting an exciting business, b.) How good are they at what they do, and c.) Their personality/temperment.

Two of my roommates from college are electrical engineers and they love what they do. One works for Intel designing better and faster processors, and the other (last time I heard) was working for Hughes doing something on Space Satelites. They both have pretty intellectually challenging jobs that I don't think are too mundane. (Although [i]every[/i] job will have some aspects that are mundane have-to-be-done type of things).

I'm sure, though, there are other people with degrees in electrical who do not enjoy what they do. Maybe because they are not so great at it, or maybe they just fundamentally don't like the subject. Perhaps for those people, an electrical engineering degree wasn't the best choice for them.

What I'm saying is, that you can't use anectdotal evidence or general statistics to decide what is best for YOU. That all depends on what fundamentally interests you, and what you feel that you are good at. There are certainly paths to sucess and happiness in electrical engineering, you just have to figure out if any of those paths are suited to you.

Hmm...What do you all think about the future of electrical engineering, in terms of new scientific discoveries and advances? I am not very knowledgable in this area, but it seems to me, as a layman, that we also know most of the things to know about electricity. Am I wrong? Is the future of electrical engineering as bright as the future for, say, nanotechnology or bioengineering?

Electrical Engineers don't just study "electricity". There are new advances being made in electrical engineering right now - and lots of them. From advances in quantum computing, to cryptography, to data compression and transmission, to highly mathematical research topics like prime number and quadratic residue distributions.

It could be said that we are living in the Electronic Age - so there are lots of new and improving technologies being made in electrical engineering fields. For example, building LCD-like displays that are as thin as a sheet of transparency paper (ever seen the portable computers displays they had in the movie Red Planet?).


function
algebra
polynomial
number theory
relatively prime
number theory unsolved

Find all $ f(x)\in Z[x]$ such that there are infinitely many pairs of coprime positive integers $ (a,b)$ such that $ a\plus{}b$ divides $ f(a)\plus{}f(b)$

Let $ n\equal{}a\plus{}b$, then for infinetely many n exist x<n, suth that $ n|f(x)\plus{}f(n\minus{}x)$ or $ n|f(x)\plus{}f(\minus{}x)$.
Obviosly it is true, if $ f(x)\plus{}f(\minus{}x)\equal{}0$ (f(x) had only odd degrees).
Let $ g(x)\equal{}(f(x)\plus{}f(\minus{}x))/2\equal{}a_0\plus{}a_2x^2\plus{}...a_{2k}x^{2k}$.
Consider distinct means $ |g(x)|, x\equal{}1,2,...$ it give distinct pairs $ x,n(x)\equal{}g(x)$, when $ n|f(x)\plus{}f(\minus{}x)$, and $ (x,n(x))\equal{}1$ if $ (x,a_0)\equal{}1$.
Therefore for any $ f(x)\in Z[x]$ we had pairs $ (a,a_0)\equal{}1\to (a,b)\equal{}1$, suth that $ a\plus{}b|f(a)\plus{}f(b)$, were b choise $ b\equal{}|f(a)\plus{}f(\minus{}a)|\minus{}a$.

[quote="Rust"]
Consider distinct means $ |g(x)|, x \equal{} 1,2,...$ it give distinct pairs $ x,n(x) \equal{} g(x)$, when $ n|f(x) \plus{} f( \minus{} x)$, and $ (x,n(x)) \equal{} 1$ if $ (x,a_0) \equal{} 1$.
[/quote]

What are you talking about in this line? Can you explain more?

If $ (x,a_0)\equal{}1$ and $ n\equal{}n(x)\equal{}g(x)$, then $ n|f(x)\plus{}f(\minus{}x)\to n|f(x)\plus{}f(n\minus{}x)$. Take $ a\equal{}x, b\equal{}n\minus{}x$, then $ a\plus{}b\equal{}n|f(a)\plus{}f(b)$.

I chose $ x\equal{}a,(a,a_0)\equal{}1,n\equal{}a\plus{}b\equal{}f(a)\plus{}f(\minus{}a)$.

I am afraid this proof has flaw. What if $ f(x)\equal{}x^2$. To make sure $ a\plus{}b|a^2\plus{}b^2$, you have to have $ a\plus{}b|2ab$, or $ a\plus{}b|2$.

I believe the right answer is either
1) $ f(x)\plus{}f(\minus{}x)\equal{}0$, or
2) $ f(x)\plus{}f(\minus{}x)$ has at least two terms.

[quote="xxp2000"]or $ a \plus{} b|2$.
[/quote]
So, $ a\equal{}b\equal{}1$ ($ a$ and $ b$ must be positive integers) and there is not infinitely many pairs of $ (a,b)$ satisfy conditions.

it's correct assuming a0 different of 0
How did you find it?

Can anyone post a complete, understandable solution?

Denote $ g(x) \equal{} \frac {1}{2x^{2k}} (f(x) \plus{} f( \minus{} x))$ where $ 2k$ is the least even degree terms of $ f(x)$. If no such exists, clearly $ f$ satisfies the condition.

For any $ a$ such that $ \gcd(g(0),a) \equal{} 1$, choose $ b \equal{} g(a) \minus{} a$. Because $ a \plus{} b | f(a) \plus{} f(b) \iff (a \plus{} b) | f(a) \plus{} f( \minus{} a) \equal{} 2g(a)$

It is clear that $ a \plus{} b \equal{} a \plus{} (g(a) \minus{} a) \equal{} g(a) | 2g(a)$. What requires is $ b > 0$ or $ g(a) > a$ and $ \gcd(b,a) \equal{} 1$. The latter holds because $ \gcd(b,a) \equal{} 1 \iff \gcd(g(a) \minus{} a,a)) \equal{} 1 \iff \gcd(g(0),a) \equal{} 1$.

If $ g(x)$ is not a constant function then its degree must be greater than or equal to 2. And so, we requires the leading coefficient to be greater than or equal to 0 so that eventually $ g(x)$ will always exceed $ x$. In case the leading coefficient fails to be, try $ b \equal{} \minus{} g(a) \minus{} a$. And the same argument holds.

If $ g(x)\not\equiv 0$ is a constant function. Clearly, it does not satisfies the condition for infinitely many $ x$. We want to prove that there exists finitely many coprime $ a,b$ such that $ a \plus{} b | a^{2k} \plus{} b^{2k}$. This is in turn equivalent to $ a \plus{} b | 2$....

In conclusion, $ f(x) \plus{} f( \minus{} x)$ must not have only one term of form $ x^{2k}$ (either zero or at least 2).

If $ f(x)\equal{}f_{odd}(x)\equal{}\sum_{i\equal{}1}^N k_ix^{2i\minus{}1}$, then it will clearly work. What about terms with even powers? Suppose that exists a function $ f_{even}(x)\equal{}\sum_{i\equal{}1}^N k_ix^{2i}$ such that $ a\plus{}b|f_{even}(a)\plus{}f_{even}(b)$ for infinite coprime pairs $ a\plus{}b$. As a matter of fact, it is possible to prove that holds $ a\plus{}b|f_{even}(a)\minus{}f_{even}(b)$. In fact, for each term of $ f_{even}(a)\minus{}f_{even}(b)$ with the same power, e.g.  $ k_ja^{2j}\minus{}k_jb^{2j}$, we have that  $ k_ja^{2j}\minus{}k_jb^{2j}\equal{}k_j(a\plus{}b)(....)$, thus  $ a\plus{}b|k_ja^{2j}\minus{}k_jb^{2j}$. But then, if $ a\plus{}b|f_{even}(a)\plus{}f_{even}(b)$ and $ a\plus{}b|f_{even}(a)\minus{}f_{even}(b)$, then $ a\plus{}b|2f_{even}(a)$ for infinitely many coprime pairs $ (a,b)$. Now, we choose a $ b$, coprime to $ a$, such that $ a\plus{}b\gg 2f_{even}(a)$, but in this case $ a\plus{}b$ can't exactly divide $ 2f_{even}(a)$. Absurd. Thus, all the $ f(x)$ are of the form $ f(x)\equal{}f_{odd}(x)\equal{}\sum_{i\equal{}1}^N k_ix^{2i\minus{}1}$.

[quote="geda"]Now, we choose a $ b$, coprime to $ a$, such that $ a \plus{} b\gg 2f_{even}(a)$, but in this case $ a \plus{} b$ can't exactly divide $ 2f_{even}(a)$. Absurd. [/quote]
You have shown that for each $ a$, there are finitely many $ b$'s that satisfy the equation.  Yet, there are infinitely many possible $ a$'s so how do you know that there are not infintely many values of $ b$?

By the way, are we assuming that $ a$ and $ b$ must be positive?  If not, then we can have that $ (a \plus{} b)|\sum_{i \equal{} 0}^n {a_ib^i}$ as long $ a_i$ are not $ 0$ for all $ i$. This is because then we may continuously divide off $ b$'s until we are left with a constant term; let the new polynomial be $ \sum _{i \equal{} 0}^m {c_ib_i} \equal{} g(b)$.  From here, we pick a $ b$ that is coprime to the constant term, so $ b$ is coprime to $ g(b)$.  Let $ g(b) \minus{} b \equal{} a$, so we have found a suitable $ a$ for infinitely many $ b$, meaning that infinitely many $ (a,b)$ exist.

Yet, if this is not the problem then we can hopefully do the following:
[hide="Solution"]
As [b]geda[/b] did, we can narrow the condition down to having that $ (a \plus{} b)|2f(a)$, where $ f$ is an even polynomial.  Notice that $ \gcd (a \plus{} b, a) \equal{} 1$, so we may divide off a power of $ a$ from $ f(a)$ until we are left with a constant.  Let the new function be $ g(a)$.  This means that we require $ (a \plus{} b)|2g(a)$.  If $ g(a)$ is a constant function, then there are only finitely many factors of $ 2g(a)$, so $ a \plus{} b$ may only take on a finite number of numbers, meaning that there are a finite number of $ (a,b)$. If $ g(a)$ has a degree of at least $ 1$, then we consider $ h(x)$ so that it is $ \minus{} g(x)$ if the leading coefficient of $ g(x)$ is negative and $ g(x)$ if the leading coefficient of $ g(x)$ is positive.  Since $ g(x)$ has a term that is at least $ x^2$ (recall that $ g(x)$ is even), we have that $ h(x)$ will have its largest term have an exponent of at least $ 2 > 1$ and it will have a positive coefficient.  Thus, there will be a point when $ h(a) > a$ for $ a > k$ for a given positive $ k$.  From here, we let $ a$ be above $ k$ and relatively prime to the constant term of $ h(x)$.  There are infinitely many values of $ a$, so we will prove that there is at least one $ b$ for every $ a$ so that $ \gcd (a,b) \equal{} 1$ and $ (a \plus{} b)|f(a)$, which will establish that there are infinitely many suitable pairs $ (a,b)$.  Pick $ b \equal{} h(a) \minus{} a > 0$, so $ (a \plus{} b)|h(a)$, so $ (a \plus{} b)|2g(a)$, so $ (a \plus{} b)|2f(a)$.  Thus, as long as $ g(a)$ is not a constant term, so as long as $ f(a)$ does not only consist of one term with an even exponent, we may find infinitely many suitable pairs $ (a,b)$.  [/hide]


When n standard 6-sided dice are rolled, the probability of obtaining a sum of 1994 is greater than zero and is the same as the probability of obtaining a sum of S. The smallest possible value of S is
(A) 333
(B) 335
(C) 337
(D) 339
(E) 341

You can roll 1994 with 332 6's and 1 2. So isn't the prob. of rolling 1994 with 333 dice the same as the prob. of rolling any 332 same numbers and 1 other number? Why isn't the answer 332*1+2 = 334?

Because you can also roll 1994 with 333 dice by rolling three hundred and thirty-one 6s, one 5 and one 3  :wink:

[hide="hint"]
The probability of getting two sums is the same if and only if they are equidistant from the max and min (for example, with two die, the min is 2 and the max is 12; 4 and 10 are both 2 away from the max and min and have the same probability.
[/hide]

[hide="solution"]
the maximum sum is  $6n$ and the minimum sum is $n$.
The probability of getting a sum of $n+a$ is the same as the probability of getting a sum of $6n-a$. All other sums would have different probabilities.

$n+(6n-1994)$ has the same probability as getting $1994$.
We need to find the minimum value of $n+(6n-1994)=7n-1994$ knowing that
$6n\ge 1994$
$n\ge 332.333...$
$n\ge 333$

Then $7(333)-1994=\boxed{337}$
[/hide]


AMC AIME USA(J)MO USAMO geometry MATHCOUNTS calculus

dude AIME was evil

I felt like firmly establishing this

rather than being uncertain

like, on par with USAMO 05

evil

if it was that evil, i should get into usamo with a 5.

I thought it was quite easy (except the geometry ones which I didn't even try), so I answered 10 & got a 9 (missed 4 stupidly).

Most of the problems seemed to have straight forward ways to do them, though there were a few tricks.

so many words in the questions

i read it

and then realized there was a second line

and then at hird line

and then i quit reading and guessed

[quote="calc rulz"]8 last year

12-13 on last couple year's practice tests (recently)

6 AIME I 2007[/quote]

DUDE THAT'S EXACTLY THE SAME AS ME!

[quote="usaha"][quote="bpms"]I thought it was MUCH easier than average.
I got an 8, one careless mistake (number 4), the mathcounts problem screwed me over for that one.[/quote]

8..thats pretty good for a guy still in middle school (by the mathcounts statement)..but you might also want to erase that mathcounts statement..seeing peopel havnet taken it yet..[/quote]
No, it was on an old states.
I still have to do state mathcounts (mine is on the 30th)
But yes, I'm still in middle school.
I stayed up until 12:30 the night before cramming in AIME prep, because I was worried that I would black out.

dude...they posted the answers on the amc website :roll: everyone took it.

If you mean mathcounts, the planet one is a classic problem, not from this year.

[quote="jb05"]I thought the problems were really easy, but they were really disgusting with lots of messy arithmetic brute force that was outrageously easy to make mistakes on. But yes I very much agree it was evil, just not hard.[/quote]

I'd agree with that. I'm not sure that it would take as much mathematical knowledge as it would rough brute-forced solutions or extended casework. Not the most representative of AIMEs and certainly not the best one, let's just hope that the AIME II is about the same or I'm sure this messageboard is going to hear a lot of complaining. I got the same score I got the last 2 years on the AIME, a 9.

I've posted comments in my [url=http://www.artofproblemsolving.com/Forum/weblog.php?w=277]blog.[/url] Feel free to discuss them here.

[quote="JSteinhardt"]I concur that the AIME was quite silly, as many problems seemed to be mostly computation.

It was sort of like, wait isn't this supposed to test our math skills, not our ability to avoid off-by-one errors?[/quote]
Wait, which were the computation heavy problems?

2
3
5
6
9
10
11

Among the ones I did, which were 1-11.

I consider it to be computationally intensive if over 75% of the most elegant solution involves manipulating numbers.

I am told by Zhai that many of the other geo problems were also very algebraic.

Also the only problem (among those) that seemed to have a particularly clever solution was 8, though according to other people, brute force is not too hard either.

[quote="JSteinhardt"]I concur that the AIME was quite silly, as many problems seemed to be mostly computation.

It was sort of like, wait isn't this supposed to test our math skills, not our ability to avoid off-by-one errors?[/quote]

That IS a math skill, though maybe not the type of skill you are refering to.

yeah

#5: case work
#6: case work
#9: I messed up my equation solving but still it was a bit messy but I guess it's still average.

#10: case work

#12: a huge radical mess(possibly the most computational problem I have done yet)

#14: guessable

#15: should've switched position with #10

Hey I'm scheduled to take the II this year - who knows where I can find the problems from the AIME I?

If they're online, a link would be great

...Shocking...

Last two years, I have succumbed to careless mistakes, calculations errors, etc., missing the cutoff by 1 point every time.

This year, I made relatively few such errors, and got a 12?

See, I've always maintained that both the AMC and AIME are very much luck-based, and it's extremely easy to slip up on them.  I learned the hard way not to be overconfident and to be extremely careful in checking...but to have been on the IMO team and now be unsure about making it...that does really suck...

Maybe they'll make an exception, considering that you've been on the team?  Or you'll probably still make it?

2 wasn't too computationally heavy. I got a non-integral value for time, so my hopes were dashed, but I kept doing the problem anyways and got the answer.

Moral: Just do it.

[quote="JSteinhardt"][quote="K81o7"]Maybe they'll make an exception, considering that you've been on the team?  Or you'll probably still make it?[/quote]

Who are you referring to?[/quote]

The person who started the thread.

it was tough. i've learned a lot this past year and still only improved by 1.

practice test: did 10, got 6 correct
2007 AIME I: did 10, got 6 correct

talk about consistency  :roll:

well, i got #1, 2, 3, 5, 7, and 14 correct. proud of that #14  :P

my index: 192. so close!  :(  will the index ever go that low? :huh:

[quote="MysticTerminator"]StellarSupremacy: dude
StellarSupremacy: it is not about making it
StellarSupremacy: it is about OWNING it
StellarSupremacy: THIS HAS NOT BEEN THE CASE
StellarSupremacy: I WILL GO BE DEPRESSED NOW[/quote]

-.- At least you made it. Others are like omg^2, i hope i made it.

[quote="MysticTerminator"]StellarSupremacy: dude
StellarSupremacy: it is not about making it
StellarSupremacy: it is about OWNING it
StellarSupremacy: THIS HAS NOT BEEN THE CASE
StellarSupremacy: I WILL GO BE DEPRESSED NOW[/quote]

Exactly. I was totally shooting for 15 in my last year, but then I got to #10 and was like crap, this problem stinks. And then #12 took me about a year to work through the radicals.

[hide="my comments on each problem"]*1.  Trivial; simple computation
*2.  you just have to read this one carefully and set up the equations
*3.  All you had to do here was expand and write a simple equation
4.  WAY too similar to an old MathCounts question; I didn't really solve it, so I missed it.  This was probably put in just to trip up MathCounts people  :ninja:
*5.  This one could be solved easily with mods or pattern recognition
*6.  Nice casework here using the multiples of 13 as "jumping" points
*7.  Not very hard, but intimidating
*8.  I really liked this one.  You had to realize that the two polynomials share a root and write out a few equations using the sum and product of roots.
9.  No idea; Geometry is my weakness.  The picture was very hard to draw
*10.  This was some casework.  You had to be careful to make sure you had the cases to cover everything.
*11.  This was a nice problem, but only if you actually solved it and didn't just get it by pattern.

I haven't looked at the rest yet.[/hide]

please guys

unless it's not allowed for those people taking the II to see the problems from the I, I'd really appreciate it if someone would post the problems.

I'd rather solve the problems myself, and I will be driving back home from the mountains from about 4 tomorrow until 7 or 8 am friday. I dont care if it's the official problem statement, but if there's a few problems without diagrams that you basically remember off the top of your head, it'd we great to see them.

What is up with this test?  No geometry until the problems start getting harder?  I suck at number theory and counting and that WAS the easy part on the test.   Also, I had never even seen a problem similar 5 or 7 before.. so I skipped them because they seemed hard..then I went back to 7 with like 5 mins to go and realized they all differ by one except for the ones that are integers..so I did it really quickly and made a computational mistake and not realizing that log 1=0

4 last year, 7 this year.
Given the fact that I'm a sophomore, I'm pretty sure I made it. Though the AIME was pretty dumb this year.

Try to compact the threads.

Especially the threads that can generally be titled under "complaining".


Let f(x)=5-4x.  Find "a" so that f(a)+f^-1(a)=a

Let g(x)=x^3-9x. Determine all values of k so htat g(2k-7)=0.

Given f(x)=(2^x)-4
Find all x such that f(x)*f^-1(x)=0
Evaluate f^-1 o f(√3x^2+5)

im having some trouble figuring out what i have to do for some of these...please and thanks for any help!! =]

1) $f^{-1}(a)=-\frac{a}{4}+\frac{5}{4}$
So then you just do $5-4a-\frac{a}{4}+\frac{5}{4}=a$

2) First factor $g(x)$ and figure out what $x$ values will make $g(x)=0$. Then you can figure what values of $k$ will give you these $x$ values.

3) You now either $f(x)$ or $f^{-1}(x)$ must equal zero, so $x$ has to be either:
- an $x$ value that makes $f(x)=0$
- the value of $f(0)$, since $f^{-1}(f(0))=0$

4) By definition, $f^{-1}(f(x))=x$


X is a variable point on the disk centered at A (with radius R_1) and Y is a variable point on the disk centered at B (with radius R_2). Find the locus of points Z such that  \triangle{XYZ} equilateral :huh: [/hide][/b]

Hello - does anybody have a solution to this problem, especially as it relates to http://www.artofproblemsolving.com/Forum/viewtopic.php?p=492881#492881?

Here, I'll retype it in a little better format:

$ X$ is a variable point on the disk centered at $ A$ (with radius $ R_1$) and $ Y$ is a variable point on the disk centered at $ B$ (with radius $ R_2$). Find the locus of points $ Z$ such that $ \triangle{XYZ}$ equilateral

Thank you


I'm a junior atm.  I recently (this year) found out about the AMC contest, and the tests following it.  So, it's my first year taking the test and I hope to advance to the AIME at least (pretty sure I won't advance to USAMO).  I've taken a couple of previously administered AMC tests and have scored above 100 on them, but I want to study and make sure I safely get at least a 100, and maybe even around 120.  Considering it's this late in the stage, should I bother doing the intro texts or should I just start with vol 1?  I would like to hear what you guys have done or are doing to prepare and your suggestions on what I should do.  Right now my plan is to do about 1 chapter in vol 1 every two days while doing two practice AMC tests a week until the actual test.  Is this a sound plan?

By the way, I'm talking about the AMC 12.

In your case, I would definitely do Volume 1.  [u]Make sure you understand the concepts though.[/u]  Also, most of the concepts are pretty straightforward, but some of the problems are difficult.  Try to solve these; those kind of problems help.

Pretty good. Don't go any faster. Know absolutely everything in terms of manipulations.

I also recommend saving a 3 or 4 more recent AMC 12s for the weekend before the actual test.
Do some of the mock AMCS starting in january, because they are often harder, and taking tests slightly harder than the actual test is a amazing way to improve.

To show you how much, in freshmen year I improved from 6-7 to consistent 9s on aimes just by doing a lot of tough mock aimes.
Also, as you get closer to the test, like end of january, I recommend you figure out where you usually stop, and work from there. For example, if you usually do the first 15 problems, and strugle with 16-18, do around 10 tests just on the 15-19 area. This will give you the boost u need without wasting an hour doing the easy first 12 questions. Of course, also have a few sessions where you sit the whole test through.

I agree with NuncChaos. Doing problems harder than the contest you are preparing for, is a great method and it works very well. I also like the idea of doing just the 15-19 range (for example), I've never tried that.

[quote="AIME15USAMO"]I agree with NuncChaos. Doing problems harder than the contest you are preparing for, is a great method and it works very well. [/quote]

Not necessarily... I did this and it kind of messed me up, but it does help somewhat. Just don't keep on doing USAMO and international questions thinking that's going to make you score a 150 on the AMC10 or 12. :blush:


Is there anyway I can buy the past years USAMTS other than ones that are available on the web?

I really want to qualify the AIME through it next year but I want to have full practices on as much as possible so I can do my best job.

If this is a pm matter, please free to pm me. :)

[quote="Silverfalcon"]Is there anyway I can buy the past years USAMTS other than ones that are available on the web?

I really want to qualify the AIME through it next year but I want to have full practices on as much as possible so I can do my best job.[/quote]
I know of no such printed materials yet, though I recall George Berzsenyi, the founder and writer of the USAMTS, once saying that he hopes to publish such a book one day.

However, seven more years of former USAMTS problems are available on the web than the six years (Years 10 through 15) shown on [url=http://www.usamts.org/Problems/U_ProblemsPast.php]the USAMTS Past Problems page[/url]. George Berzsenyi used to also run the International Mathematics Talent Search, which featured the same problems. The Canadian Mathematical Society has a web page, [url=http://www.cms.math.ca/Competitions/IMTS/]International Mathematics Talent Search[/url] that contains all the problems of all 44 rounds of the IMTS. Round 29 of the IMTS corresponds to Year 10 Round 1 of the USAMTS. Round 44 of the IMTS corresponds to Year 13 Round 4 of the USAMTS.

These IMTS problems have no solutions posted. However, I suppose that after this year's USAMTS Round 4 is over, you could organize a practice session  in this AoPS forum that goes through some old rounds of the IMTS, and compare solutions with the other participants of the practice session.

Erin Schram
USAMTS grader

Thank you very much. :)

I'll try to solve as many as can.

Just wondering, is working on AIME help on USAMTS? (I know USAMTS get students to AIME but I wonder whether working on AIME helps USAMTS also)

I think working on any of the AMC tests would definitely improve your USAMTS score.

If you practice by learning the math inside and out (which is the whole reason for contests), you should improve in any math contest you take.  Math is linked so that once you grasp one concept, it becomes easier to grasp many related concepts.  So, even if AIME has a different flavor, there is still a logical process you must continually improve for any contest, not to mention life's problems.

[quote="origamimasterjared"]I find that the USAMTS questions are similar to the AIME questions, at least in difficulty.  The differerence is that for USAMTS you get a whole month for five questions, whereas for AIME you get three hours for fifteen questions.[/quote]
The major difference between AIME problems and the USAMTS problems is that many of the USAMTS problems are the "prove . . . something something something" whereas the AIME problems are (always, by necessity) of the type "find something (that has an integer value on [0,999])." This makes many of the USAMTS problems "longer," if not more difficult. Of course, USAMTS problems tend to have (especially this year) an extreme variety of difficulty levels (at least as much as the difference between an AIME problem 1 and an AIME problem 15).

I do old AIMEs to practice for the USAMTS; they certainly can't hurt your score! :)

[quote="ZennyK"]Of course, USAMTS problems tend to have (especially this year) an extreme variety of difficulty levels (at least as much as the difference between an AIME problem 1 and an AIME problem 15).[/quote]

That's for sure--on one problem set, I solve two problems in no more than five minutes but spent my 30 days on one of the others and still didn't get it after all that time.

I also agree that the USAMTS problems are similar to those in the AIME.  But I like the amount of time you get to work on them.  It keeps you from feeling like a loser when the round is over.  I'm sure we've all said this after the AMC/AIME:  "WHY didn't I get that problem!?" :P


yes, I agree with the first man

Only way I can think is this
we have:


Persons A, B, C, D, E, and F are to be sorted into $3$ groups of $2$.  How many possible different sets of $3$ groups are there?  (The group A, B, is the same as the group B, A).

Is it[hide]$\frac{\displaystyle\binom{6}{2}\binom{4}{2}}{3!}$?[/hide]

yes it is right,$\frac{6!}{(2!)^33!}$
you can general it ab to a group of b people $\Rightarrow \frac{(ab)!}{(b!)^aa!}$ ;)

[quote="Elemennop"]Is the set of groups (a,b) (c,d) (e,f) the same as (a,b) (e,f) (c,d)?

I would think it'd be the same and the answer then just be $\binom{6}{2}\cdot\binom{4}{2}$.[/quote]But the problem with that is, that considers the order that you choose it in.  So I think you have to divide it by $3!$ to compensate for the ordering thing...since (a,b) (c,d) (e,f) can be ordered in $3!$ ways.

No, because it is not important which is first or second or third!  ;)

Yeah but any permutation of the groups will still be the same member sets in each group.

So I still don't see where the division by $3!$ comes from.

Because when you do $\binom{6}{2}$ you could, say choose (a,b). Then when you do $\binom{4}{2}$ say you choose (c,d).  Then your three pairs are (a,b) (c,d) (e,f).

You could also choose (c,d) first with $\binom{6}{2}$, then (e,f) with $\binom{4}{2}$, so you have (a,b) left.  In this case the three pairs are also (in a different order) (a,b) (c,d) (e,f), but both these orders are both counted.

So groups themselves are distinct? If that's true I understand the division by $3!$, but it seemed like groups are indiscernable and that (a,b) (c,d) is equivalent to (c,d) (a,b).

The groups are indescernable.  (ooh, good vocab word.)  But the way we count them, they aren't. So we divide to account for the fact that they are indescernable.

yes you are right .It is not diffrent between (a,b)(c,d)(e,f) &(c,d)(a,b)(e,f)&...& (e,f)(c,d)(a,b).we wnt jusy to make group ;)

Ok I see now, I was looking at it completely wrong.

Haha, woops.  :blush:


(a) Determine, with careful explanation, how many ways $ 2n$ people can be paired off to form $ n$ teams of $ 2$.

(b) Prove that $ \{(mn)!\}^{2}$ is divisible by $ (m!)^{n\plus{}1}(n!)^{m\plus{}1}$ for all positive integers $ m, n$.

(a) permute the $ 2n$ people arbitrarily, then let the first two form a team, the next two form a team, and so on. there are $ (2n)!$ ways to permute the people and then form teams in this way. but once these teams are formed there are $ 2! \equal{} 2$ ways to permute the members of each team, and $ n!$ ways to permute the $ n$ teams, so in the count $ (2n)!$, each distinct selection of teams is counted $ 2^n\cdot n!$ times. so the true count is $ \frac {(2n)!}{2^nn!}$.

(b) by the same reasoning as in (a), replacing 2 with $ m$, we have $ m!^n n!|(mn)!$. reversing the roles of $ m$ and $ n$, we also get that $ m! n!^m|(mn)!$. multiplying, we therefore get $ m!^{n \plus{} 1}n!^{m \plus{} 1}|(mn)!^2$.


[quote="Honore"]If

$x^{12}-x+1=0$

then

$x^{10}+x^{5}+1=?$

I am not sure that this problem, which has been asked at another forum, can be solved using elementary operations.

Thanks and regards.[/quote]
There are 12 complex (not real) means, suth that $x^{12}-x+1=0$. It give 12 means $x^{10}+x^{5}+1$.
But we can find polinom $P(y)=y^{12}+a_{11}y^{11}+...$, suth that $P(x^{10}+x^{5}+1)=0$ if $x^{12}-x+1=0$.


function
real analysis
real analysis unsolved

sorry to both people with such stupid questions, but i was working through some problems in royden and got stuck on something that seemed easy. namely, show that the product of two absolutely continuous functions is absolutely continuous, as is the inverse of an absolutely continuous function (provided it is zero nowhere).

I was able to prove that a sum is abs cts, but these other two dont seem to come as easily.

thanks

Use $|f(x)g(x)-f(x')g(x')| \leq |(f(x)-f(x'))g(x)|+|f(x')(g(x)-g(x'))| \leq$
$[|(f(x)-f(x')|+|(g(x)-g(x')|]*K$ where $K$ exists due to the continuity of $f$ and $g$.

Are you sure about the inverse being absolutely continuous ?
Because if you take $f(x)=x$ on $(0,1)$ then $f$ will be absolutely continuous but $\forall \epsilon >0$, just take $x'=x+\epsilon$ and you will have $\frac{1}{x}-\frac{1}{x'}=\frac{\epsilon}{x(x+\epsilon)}$. It is then clear that by picking $x$ small enough the difference will become arbitrarily large, which is not what we want.


The number of teeth in three meshed gears $A$, $B$, and $C$ are $x$, $y$, and $z$, respectively. (The teeth on all gears are the same size and regularly spaced.) The angular speeds, in revolutions per minutes of $A$, $B$, and $C$ are in the proportion

$\text{(A)} \ x: y: z ~~\text{(B)} \ z: y: x ~~ \text{(C)} \ y: z: x~~ \text{(D)} \ yz: xz: xy ~~ \text{(E)} \ xz: yx: zy$

[hide="solution"]the idea in this is that the distance that they rotate is constant, suppose they all went $n$ gear teeth in a minute, then clearly, $\frac{n}{x}$ is the number of revolutions for A, we do the same for B and C, to get the ratio

$\frac{n}{x}: \frac{n}{y}: \frac{n}{z}\implies yz: xz: xy \implies \boxed{D}$[/hide]

Cool! I got all the answer choices on one line! :thumbup:  :D


Hi,
can anyone explain me in more details the proof of Newton inequality:
http://www.artofproblemsolving.com/Wiki/index.php/Newton%27s_Inequality#Proof
because i can't understand it.
Thanks.

Certainly, but you need to tell what exactly is unclear first (otherwise you'll just get another long text you'll have hard time understanding).


probability
function
integration
calculus
calculus computations

Prove that for any distribution function and for any $a\geq 0$ ;
$\int [\mathbb{F}(x+a)-\mathbb{F}(x)]dx=a$.

[hide="reveal"] My thought was to use that $\mathbb{F}$ is a densitiy function of $U(0,1)$ , but I got stuck[/hide]

Making a guess as to what this means:

Let $F(x)$ be a cumulative distribution problem. Let's start by assuming that the random variable is (absolutely) continuous and that there is a density $f$ such that

$F(x)=\int_{-\infty}^{x}f(t)\,dt.$

Then:

$\int_{-\infty}^{\infty}F(x+a)-F(a)\,dx=\int_{-\infty}^{\infty}\int_{x}^{x+a}f(t)\,dx\,dt$

A mathematician of my training seldom writes down a double integral without intending to switch the order of integration.

$=\int_{-\infty}^{\infty}\int_{t-a}^{t}f(t)\,dx\,dt=\int_{-\infty}^{\infty}af(t)\,dt=E(a)=a.$

This would make sense even in the discontinuous case: use Stieltjes integrals in $dF(t)$ in place of $f(t)\,dt$ and the same argument still works.


It's a very well known inequality called Nesbitt's inequlity.
[hide="Solution 1"]
Let's put $x=b+c$, $y=c+a$, $z=a+b$. Then we have $a=\frac{y+z-x}{2}$, $b=\frac{x+z-y}{2}$, $c=\frac{x+y-z}{2}$ and the inequality becomes: \[\frac{y+z-x}{2x}+\frac{x+z-y}{2y}+\frac{x+y-z}{2z}\geq \frac{3}{2}\iff\]  \[\iff \frac{y+z-x}{x}+\frac{x+z-y}{y}+\frac{x+y-z}{z}\geq 3.\] After substituting $u=\frac{x}{y}$, $v=\frac{y}{z}$, $w=\frac{z}{x}$ it becomes \[\frac{1}{u}+w-1+u+\frac{1}{v}-1+\frac{1}{w}+v-1 \geq 3\iff\]  \[\iff \frac{1}{u}+u+\frac{1}{v}+v+\frac{1}{w}+w \geq 6.\] This inequality is trivial after noticing $t+\frac{1}{t}\geq 2$ for $t>0$ (it is equivalent to $t^{2}+1\geq 2t\iff (t-1)^{2}\geq 0$) which ends the proof.
[/hide]
[hide="Solution 2"]
We can use also Jensen inequality. Let's put $k=a+b+c$ (of course $a,b,c<k$). Then we have to prove that \[\frac{\frac{a}{k-a}+\frac{b}{k-b}+\frac{c}{k-c}}{3}\geq \frac{1}{2}.\] Function $f(t)=\frac{t}{k-t}$ is convex on interval $(0,k)$ so we have \[\text{LHS}\geq \frac{\frac{a+b+c}{3}}{k-\frac{a+b+c}{3}}= \frac{\frac{k}{3}}{\frac{2}{3}k}=\frac{1}{2}\] which ends the proof.
[/hide]

[hide="solution 3"]
It's homogeneous, so $a+b+c=1$. The ineq. becomes:

$\frac{a}{1-a}+\frac{b}{1-b}+\frac{c}{1-c}\ge \frac{3}{2}$

Function $\frac{x}{1-x}$ is convex in $(0,1)$, so by Jensen, $\frac{1}{3}\left( f(a)+f(b)+f(c) \right) \ge f \left( \frac{a+b+c}{3}\right)$
[/hide]

[hide="solution 4"]

$(a+b+c)^{2}= \left( \sum_{cyc}\frac{\sqrt{a}}{\sqrt{b+c}}\cdot \sqrt{a}\cdot \sqrt{b+c}\right) \le LHS \cdot 2(ab+bc+ca)$ by Cauchy-Schwarz

Now we want $\frac{(a+b+c)^{2}}{2(ab+bc+ca)}\ge \frac{3}{2}\Longleftrightarrow a^{2}+b^{2}+c^{2}\ge ab+bc+ca$, that is obviously true by rearrangement[/hide]

nice solution here http://www.answers.com/topic/nesbitt-s-inequality


The most prestigious sports club in one city has exactly N members. Each of its members is strong and beautiful. More precisely, i-th member of this club (members being numbered by the time they entered the club) has strength Si and beauty Bi . Since this is a very prestigious club, its members are very rich and therefore extraordinary people, so they often extremely hate each other. Strictly speaking, i-th member of the club Mr X hates j-th member of the club Mr Y if Si ≤ Sj and Bi ≥ Bj or if Si ≥ Sj and Bi ≤ Bj (if both properties of Mr X are greater then corresponding properties of Mr Y, he doesn't even notice him, on the other hand, if both of his properties are less, he respects Mr Y very much).

To celebrate a new 2003 year, the administration of the club is planning to organize a party. However they are afraid that if two people who hate each other would simultaneouly attend the party, after a drink or two they would start a fight. So no two people who hate each other should be invited. On the other hand, to keep the club presti≥ at the apropriate level, administration wants to invite as many people as possible.

Being the only one among administration who is not afraid of touching a computer, you are to write a program which would find out whom to invite to the party.
2 <= N <= 100,000;  1 <= Si, Bi <= 1000000000

A question keeps bugging me when seeing some informatics problems (like this one): how do we handle the big numbers involved there? (in a practical way) Dinamic allocation should be the way, but I never understood how to do use it in these cases.

What big numbers are you talking about? I don't see what big numbers have to do with this task..


[quote="GameBot"]In a patio, the pattern is determined by regular hexagonal tiles, square tiles, and equilateral triangular tiles as shown in the diagram. If the area of each hexagonal tile is 96 in$ ^2$, what is the number of square inches in the area of the square tile? Express your answer to the nearest whole number.
[asy]import olympiad; size(150); import geometry_dev; import graph; defaultpen(linewidth(0.8));
path ngon(int numSides, real side, pair blCorner,real degrees=0){
 pair[] vertices = new pair[numSides];
 real intdeg = 180*(numSides - 2)/numSides;
 real ext = 180 - intdeg;
 vertices[0] = blCorner;
 for(int i = 1; i < numSides; ++i){
  vertices[i] = vertices[i - 1] + side*dir(ext*(i - 1) + degrees);
 }
 vertices.cyclic(true);
 path a = vertices[0]--vertices[1];
 for(int i = 2; i <= numSides; ++i){
  path b = a -- vertices[i];
  a = b;
 }
 return a--cycle;
}
 draw(ngon(6,1,origin));
 draw(ngon(12,1,(0,-1)));
 pair[] vertices = new pair[6];
 vertices[0] = (0,-1);
 for(int i = 0; i < 6; ++i){
  vertices[i + 1] = vertices[i] + dir(30*2*i) + dir(30*(2*i + 1));
  draw(ngon(4,1,vertices[i],60*i));
 }[/asy][/quote]

For the side length x, $ \frac{3x^2\sqrt{3}}{2}=96 \implies x^2\sqrt{3}=64 \implies x^2=\frac{64}{\sqrt{3}}$.

So roughly $ 37$.


trigonometry
conics
ratio
geometry proposed
geometry

Let $(O)$ be a circle and $ABC$ be an arbitrary triangle (not neccessary to be inscribed in $(O)$). Prove that the triangle $A'B'C'$ formed by the poles of $BC, CA,AB$ wrt $(O)$ is perspective with $ABC$.
    Does anyone have any synthetic proof of it ? I have a solution which I don't know if it can be considered synthetic   ;)

The most beautiful problem I saw in this 2 months :)
For convenience, assume that (O) lies inside $\triangle ABC$. Let D,E,F be the inverse of A',B',C' respectively. We will prove that AA',BB',CC' are concurrent. Let $R_1,R_2,R_3$ be the circumradii of quad. B'C'FE,C'A'DF,A'B'ED.
By Ceva, $1=\frac{\sin\angle BAA'}{\sin\angle CAA'}\cdot\frac{EA'\cos\angle B'EA'}{A'D}\cdot\frac{A'D}{FA'\cos\angle A'FC'}=\frac{\sin\angle BAA'}{\sin\angle CAA'}\cdot\frac{2R_3\sin\angle OB'A'\cos\angle B'EA'}{2R_2\sin\angle OC'A'\cos\angle A'FC'}$. By the same way, we obtain other 2 equalities, multiply them together and the result follows.

Anyway, don't know why but I have a strong feeling that the point of concurrency has a close relation with the point of concurrency of the lines perp. to B'C' thru A and so on.

I think this looks better in a projective setting :).

We must show that the lines connecting the vertices of a triangle $ABC$ with the poles $A',B',C'$ of the sides $BC,CA,AB$ respectively wrt a conic $\mathcal C$ are concurrent.

Let $P=BC\cap B'C',M=BC\cap AA',M'=AA'\cap B'C'$. What we want to prove is equivalent to the fact that the perspectivity from the line $BC$ to the line $B'C'$ with center $BB'\cap CC'$ takes $M$ to $M'$, or, in other words, we want to prove that $(B,C;M,P)=(B',C';M',P)$.

We have $(B,C;M,P)=(AB,AC;AM,AP)=(*)(C',B';P,M')=(B',C';M',P)$. Only the equality marked $(*)$ needs justification: it holds because the polarity wrt $\mathcal C$ turns the points $C',B',P,M'$ into their polars wrt $\mathcal C$, which are $AB,AC,AM,AP$ respectively, and this transformation preserves cross-ratios.

Very nice proof Grobber (I like it  :roll: )
  My proof is not Euclidean. I change the problem into the meaning in Klein model and reduce it in to the two easy special cases : when $A=O$ or $BC$ passes through $O$.

I don't understand your proof at all, COul you tell me where do you apply the Ceva's theorem... Also, let's solve this theorem:

Given a triangle, to prove that the triangle formed by the polars of its vertex are in perspective with this

thanks

jose carlos

It's the first time for me to explain something I proved more than one year before :D
Anyway, basically I used the trigonometric form of Ceva and played around with some ratios, and that's all. I'm sure everyone can work out the details if he really gives a try.


Given $ a, b, c > 0$. Prove that:
$ \frac{1}{\sqrt[4]{a^2\minus{}ab\plus{}b^2}}\plus{}\frac{1}{\sqrt[4]{b^2\minus{}bc\plus{}c^2}}\plus{}\frac{1}{\sqrt[4]{c^2\minus{}ca\plus{}a^2}} \geq$ $ \frac{\sqrt{2}}{3}. \frac{(a\plus{}b\plus{}c)^3}{a^2\sqrt{a^3\plus{}b^3}\plus{}b^2\sqrt{b^3\plus{}c^3}\plus{}c^2\sqrt{c^3\plus{}a^3}}$
 :)

[quote="nguoivn"]Given $ a, b, c > 0$. Prove that:
$ \frac {1}{\sqrt [4]{a^2 \minus{} ab \plus{} b^2}} \plus{} \frac {1}{\sqrt [4]{b^2 \minus{} bc \plus{} c^2}} \plus{} \frac {1}{\sqrt [4]{c^2 \minus{} ca \plus{} a^2}} \geq$ $ \frac {\sqrt {2}}{3}. \frac {(a \plus{} b \plus{} c)^3}{a^2\sqrt {a^3 \plus{} b^3} \plus{} b^2\sqrt {b^3 \plus{} c^3} \plus{} c^2\sqrt {c^3 \plus{} a^3}}$
 :)[/quote]

[hide="Just A thought"]
Again :( , I am new with the idea of homogenisations, so, I would like to ask you whether we can say:
After inserting $ a, b, c$ with $ ka, kb, kc$ the inequality does not change for $ k\neq0$ ($ \frac{1}{\sqrt{k}}$ gets cancelled.
So can we assume $ a\plus{}b\plus{}c\equal{}1$? :blush:

Extremely sorry for my ignorance in this field of normalisations and homogenisations, so can anyone give me references? :oops: [/hide]

[quote="Potla"]

[hide="Just A thought"]
Again :( , I am new with the idea of homogenisations, so, I would like to ask you whether we can say:
After inserting $ a, b, c$ with $ ka, kb, kc$ the inequality does not change for $ k\neq0$ ($ \frac {1}{\sqrt {k}}$ gets cancelled.
So can we assume $ a \plus{} b \plus{} c \equal{} 1$? :blush:

Extremely sorry for my ignorance in this field of normalisations and homogenisations, so can anyone give me references? :oops: [/hide][/quote]
[hide="to Potla"]As you said,Potla,The inequality will not change if we replace $ a,b,c$ by $ ka, kb, kc$,so when you choose $ k \equal{} \frac {1}{1 \plus{} b \plus{} c}$ the condition $ a \plus{} b \plus{} c \equal{} 1$ will be satisfied.
Sorry for my bad English.[/hide]

[quote="nguoivn"]Given $ a, b, c > 0$. Prove that:
$ \frac {1}{\sqrt [4]{a^2 \minus{} ab \plus{} b^2}} \plus{} \frac {1}{\sqrt [4]{b^2 \minus{} bc \plus{} c^2}} \plus{} \frac {1}{\sqrt [4]{c^2 \minus{} ca \plus{} a^2}} \geq$ $ \frac {\sqrt {2}}{3}. \frac {(a \plus{} b \plus{} c)^3}{a^2\sqrt {a^3 \plus{} b^3} \plus{} b^2\sqrt {b^3 \plus{} c^3} \plus{} c^2\sqrt {c^3 \plus{} a^3}}$
 :)[/quote]
This inequality follows from the inequality:
If $ a,b,c \ge 0,a\plus{}b\plus{}c\equal{}3,$ then $ a\sqrt{a\plus{}b}\plus{}b\sqrt{b\plus{}c}\plus{}c\sqrt{c\plus{}a} \ge 3\sqrt{2}.$
;) :)


function
search
algebra
partial fractions

Hi!  At a math circle I attend, the group was talking about Fibonacci and I mentioned Binet's formula.  After the class, the instructor told me about a proof of Binet's formula using generating functions.  I think he thought I knew more about generating functions than I actually do  :D, so I am unable to finish the proof past where he showed me.  Could somebody please explain how to finish this proof?
\begin{eqnarray*}f(x)=F_{0}+F_{1}x=F_{2}x^{2}+F_{3}x^{3}...\\ x\cdot f(x)=F_{0}x+F_{1}x^{2}+F_{2}x^{3}... \end{eqnarray*}
adding the first two equations together and solving for $f(x)$ gives
\begin{eqnarray*}(1+x)f(x)=F_{1}+F_{2}x+F_{3}x^{2}+F_{4}^{3}...\\ x(1+x)\cdot f(x)=f(x)\\ f(x)=\frac{1}{1-x-x^{2}}\end{eqnarray*}
now what?  :maybe:

try using Search function here.
you will get MANY results, including generating functions and other porrofs of that formula!! :wink:

$f(x) = \frac{1}{1-x-x^{2}}= \frac{1}{\left(1-\frac{1+\sqrt{5}}{2}x\right)\left(1-\frac{1-\sqrt{5}}{2}x\right)}$.

Using partial fractions, this is

$\frac{1}{\sqrt{5}}\left(\frac{\frac{1+\sqrt{5}}{2}}{1-\frac{1+\sqrt{5}}{2}x}-\frac{\frac{1-\sqrt{5}}{2}}{1-\frac{1-\sqrt{5}}{2}x}\right)$.

Can you finish it from here?

And if you'd like to learn more about the topic, I suggest taking a look at "Generatingfunctionology" by Herbert S. Wilf. This exact question is covered in the first chapter!

http://www.math.upenn.edu/~wilf/DownldGF.html


Let be 0<a<b real numbers. Find the functions f:[a,b] -> [a,b] such that                       f(x)+f(f(x))=2x for every x of [a,b].

This isn't to say that these are the [b]only[/b] solutions, but they seem to work:

\[ f\left(x\right)=ax+b\implies \left(a^{2}+a\right)x+ab+2b=2x. \]

From this, I find that two such functions are $f\left(x\right)=-2x$ and $f\left(x\right)=x$.

But f(x)=-2x is not solution and the function is not linniar. :)

Let $g(x)$ be $g(x) = \frac{f(x)}x$, and then the initial equation becomes $g(x)=\frac 2{1+g(f(x))}$

Manifestly $g(x)$ is positive and bounded. I note $\displaystyle i=\inf_{x\in [a,b] }(g(x)) ,\ s=\sup_{x\in [a,b] }(g(x))$
Now I can write $\frac 2{1+s} \leq g(x) \leq \frac 2{1+i}$, so $\frac 2{1+s} \leq i$ and $s \leq \frac 2{1+i}$, so $s=i$, so $g(x)=k(const) > 0$
I put in the initial equation and obtain $kx +k^2x = 2x$, and finally $k=1$, so $f(x)=x$
 :cool:


algebra
polynomial
inequalities
inequalities proposed

Woops...nevermind...I typed it into mathematica incorrectly at first, it seems to be correct.

Minimize[{3 (a^4 + b^4 + c^4) + 9 (a^2 b^2 + b^2 c^2 + c^2 a^2) -
   5 (a^3 (b + c) + b^3 (c + a) + c^3 (a + b)) - 2 a b c (a + b + c),
  a >= 0, b >= 0, c >= 0, a + b + c == 1}, {a, b, c}]

{0, {a -> 1/3, b -> 1/3, c -> 1/3}}

Okay it is

$ a^2(b\minus{}c)^2\plus{}b^2(c\minus{}a)^2\plus{}c^2(a\minus{}b)^2\plus{} 3 ( (a \minus{} b)^2 (b \minus{} c)^2 \plus{} (b \minus{} c)^2 (c \minus{} a)^2 \plus{} (c \minus{} a)^2 (a \minus{} b)^2)$

which is true...

[quote="hungkhtn"]Let $ a,b,c$ be non-negative real numbers. Prove that
\[ 3\sum_{cyc} a^4 \plus{} 9\sum_{cyc} a^2b^2\ge 5\sum_{cyc} a^3(b \plus{} c) \plus{} 2abc\sum_{cyc} a.
\]
[/quote]
Yes, it's really easy:
$ \sum_{cyc}(3a^4 \minus{} 5a^3b \minus{} 5a^3c \plus{} 9a^2b^2 \minus{} 2a^2bc)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(3a^4 \minus{} 10a^3b \plus{} 14a^2b^2 \minus{} 10ab^3 \plus{} 3b^4) \plus{} 4\sum_{cyc}(a^2b^2 \minus{} a^2bc)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(a\minus{}b)^2(3a^2\minus{}4ab\plus{}3b^2)\plus{}2\sum_{cyc}c^2(a\minus{}b)^2\geq0.$

[quote="arqady"][quote="hungkhtn"]Let $ a,b,c$ be non-negative real numbers. Prove that
\[ 3\sum_{cyc} a^4 \plus{} 9\sum_{cyc} a^2b^2\ge 5\sum_{cyc} a^3(b \plus{} c) \plus{} 2abc\sum_{cyc} a.
\]
[/quote]
Yes, it's really easy:
$ \sum_{cyc}(3a^4 \minus{} 5a^3b \minus{} 5a^3c \plus{} 9a^2b^2 \minus{} 2a^2bc)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(3a^4 \minus{} 10a^3b \plus{} 14a^2b^2 \minus{} 10ab^3 \plus{} 3b^4) \plus{} 4\sum_{cyc}(a^2b^2 \minus{} a^2bc)\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(a \minus{} b)^2(3a^2 \minus{} 4ab \plus{} 3b^2) \plus{} 2\sum_{cyc}c^2(a \minus{} b)^2\geq0.$[/quote]

Actually, the following one is true and it is my implication for all real numbers $ a,b,c$

\[ 3(a^4\plus{}b^4\plus{}c^4)\plus{}9(a^2b^2\plus{}b^2c^2\plus{}c^2a^2)\ge 10(a^3b\plus{}b^3c\plus{}c^3a)\plus{}2abc(a\plus{}b\plus{}c).\]

Notice the change from symmetric sum to cyclic sum.

I suppose that it is a bit out of the ordinary in the sense that you cannot just schur/muirhead bunch. You have to extract squares as well. I don't think so much about those solutions so it didn't jump out at me immediately.

[quote="hungkhtn"] Actually, the following one is true and it is my implication for all real numbers $ a,b,c$
\[ 3(a^4 \plus{} b^4 \plus{} c^4) \plus{} 9(a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2)\ge 10(a^3b \plus{} b^3c \plus{} c^3a) \plus{} 2abc(a \plus{} b \plus{} c).
\]
[/quote]
This is sharper:
\[ 3(a^4 \plus{} b^4 \plus{} c^4) \plus{} \frac {73}{9}(a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2)\ge 10(a^3b \plus{} b^3c \plus{} c^3a) \plus{} \frac {10}{9}abc(a \plus{} b \plus{} c).
\]

Let $ a \plus{} b \plus{} c \equal{} 3 , a,b,c \equal{} R$  . Prove that
\[ \sum_{cyc} a^4 \plus{} 3\sum_{cyc} a^2b^2 \plus{} 3\ge 2\sum_{cyc} ab(a^2 \plus{} b^2 ) \plus{} 6(a \minus{} b)(a \minus{} c)(b \minus{} c)
\]

You can probably write it in the form $ \sum_{\rm cyc} [p(x,y,z)]^2$ where $ p(x,y,z)$ is a homogeneous polynomial of degree 2 in $ x,y,z$.

Can you post your solution, Altheman ?  :D  In Vasc's book, having a genelizition of above (for 4th degree :D) but how do you think about that follow:
Let $ a,b,c,d$ be real numbers such that $ a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2 \equal{} 1$ Prove that:
\[ \sum a^6 \plus{} \sum a^4b^2 \plus{} 2abcd(ab \plus{} bc \plus{} cd \plus{} da\minus{}1)\ge 2\sum a^5b \plus{} \sum(abc)^2
\]
In here $ \sum \equal{} \sum_{cyc}$
Who can write it in the form $ \sum [g(a,b,c,d)]^2\ge 0$ ?  :wink:

[quote="manlio"][quote="hungkhtn"]
Actually, the following one is true and it is my implication for all real numbers $ a,b,c$
\[ 3(a^4 \plus{} b^4 \plus{} c^4) \plus{} 9(a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2)\ge 10(a^3b \plus{} b^3c \plus{} c^3a) \plus{} 2abc(a \plus{} b \plus{} c).
\]
Notice the change from symmetric sum to cyclic sum.[/quote]

Can you please give me a hint to prove it, hungkhtn? Did you use SOS or SOS-SHUR?

Thank you very much.[/quote]
I think, he solved it by EMV method  :) . This method works very well for many three-variable cyclic inequalities (include the inequality posted by Vasc as above of course).

I don't have any of these books...I just read AoPS :D

I am not sure, but that expansion contains all 'elementary' cyclic sums of degree 4....That is what I like to call them anyway

a^4+b^4+c^3

a^3b+b^3c+c^3a

ab^3+bc^3+ca^3

abc(a+b+c)

we have enough variables that it is likely that we can find the desired coefficients (but of course I am not sure, I was merely suggesting that the method would probably work)

I like your problem and I will think about it.

It obviously cannot be written in the form that you propose, but we can augment it...


Henry's little brother has 8 identical stickers and 4 identical sheets of paper. How many ways are there for him to put the stickers on the sheets of paper, if only the number of stickers on each sheet matters?

That can be only used for if the stickers were identical BUT the pieces of paper were distinct. I'm not sure how to proceed in any way other than a list, in solving the actual question, so here goes-

8 0 0 0
7 1 0 0
6 2 0 0
6 1 1 0
5 3 0 0
5 2 1 0
5 1 1 1
4 4 0 0
4 3 1 0
4 2 2 0
4 2 1 1
3 3 2 0
3 3 1 1
3 2 2 1
2 2 2 2

This totals $ \boxed{15}$ ways.

How is this different from [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=540&t=239503]this[/url]?

With all due respect for brute forcing and counting up the ways to do it, isn't there a method where I don't have to use up half a piece of scratch paper making lists? A formula? If this problem had a very large amount of stickers and a very large number of sheets of paper, I'm pretty sure listing wouldn't be a good idea. Anybody know? Thanks.

If you don't want to list them out, the best you can do is probably casework. If I told you to sum the first 1000 prime numbers, is there necessarily an easy way to do that?

Well, there isn't a formula for the first 1000 prime numbers.

That is my point. Not every problem has an easy, casework free way of doing it.

[quote=Freddie123]wait, will wang? you go to NA?[/quote]

you could've just pm'ed him instead of reviving a 2 year old thread.......

[quote=Ridley-C]Why doesn't stars and bars work?[/quote]

because both objects are indistinguishable

[quote=MathArt4]Sorry for the bump, anyway...
Is this a coincidence that this is 5+4+3+2+1?[/quote]

Yes.

Yes, as I stated on the other page for this problem.  Which makes me wonder why this problem was posted on two separate threads by Gamebot?

EDIT: Wait they're actually two different problems nvm and actually I don't think you can use a non-casework approach for this one

[quote=Zhang2018]Is this problem referring to indistinguishable items in indistinguishable boxes? however, the problem states "each sheet," so shouldn't the sheets be distinguishable? That is what I did in my first solution, and it was incorrect. On my second solution, I changed my thinking and it was correct, but I just wanted to confirm. Someone else may be able to give a more detailed explanation.[/quote]

Not sure how relevant this is at this point, but the statement in the first sentence that there are[quote=Alcumus]$8$ identical stickers[/quote]
should definitely supersede anything else.
------------
Anyways, since the answer is $6\choose2$$^a$ I was wondering if there is any reason or pattern behind that or if that is just coincidence.
------------
$^a$ the $\LaTeX$ doesn't seem to have rendered quite right there; sorry

[quote=professordad]The $4$ sheets of paper in the other problem are different colors, while here the papers are the same color.[/quote]

It is almost the same thing as the 2006 AMC 12 problem.

omg finally mastered distinguishability yayyyyyyyyy

this one was a bad problem tho, why make us just list?

i accidentally typed `5 as the answer lol

I attempted to do this w/ generating functions for partitions but it's even more work; this problem is probably inevitably ugly.

Is there any way to solve this without counting them all? I only have so much pen ink.


This came up in a problem I was thinking about and I thought it was interesting.

Prove that for $ n\in\mathbb{N}$
\[ (\minus{}2)\binom{3n\plus{}1}{n\minus{}1}\plus{}(\minus{}1)\binom{3n}{n\minus{}1}\plus{}...\plus{}(2n)\binom{n\minus{}1}{n\minus{}1}\equal{}0\]
And more generally, for $ k\in\mathbb{N}$,
\[ (\minus{}k)\binom{kn\plus{}k\plus{}n\minus{}1}{n\minus{}1}\plus{}(\minus{}k\plus{}1)\binom{kn\plus{}k\plus{}n\minus{}2}{n\minus{}1}\plus{}...\plus{}kn\binom{n\minus{}1}{n\minus{}1}\equal{}0\]
In sigma notation:
\[ \sum_{i\equal{}0}^{kn\plus{}k}(i\minus{}k)\binom{kn\plus{}k\plus{}n\minus{}1\minus{}i}{n\minus{}1}\equal{}0\]
EDIT: On second thought, this should probably be moved to Intermediate.

[quote="JBL"]It looks like you're missing the RHS of your equations, or something.  What are we supposed to prove about these expressions?[/quote]

Oh right.  Fixed.


How many degrees are in the sum of the measures of the six numbered angles pictured?
[asy]draw((3,8)--(10,4)--(1,0)--cycle);
draw((7,8)--(9,0)--(0,4)--cycle);
label("1",(3,8),SSE);
label("2",(7,8),SSW);
label("3",(10,4),2W);
label("4",(9,0),NW+NNW);
label("5",(1,0),NE+NNE);
label("6",(0,4),2E);[/asy]

If you look carefully, the figure is an intersection of two triangles.

The first triangle has angle measures 1, 3, and 5.

The second triangle has angle measures 2, 4, 6 and 6

The sum of the interior angles of a triangle is 180.

So, we are looking for 1+3+5+2+4+6= 180+180 = 360.


quadratics
complex numbers
algebra
quadratic formula

Mathematically prove that $z\overline{z}=a^{2}+b^{2}$.

[hide]
z = a + bi
z(bar thingy) = a - bi

(a + bi)(a - bi) = a^2 + b^2[/hide]

I should have made it more clear. I want something like deriving it.

what do you mean derive? this follows directly from the definition of a conjugate

See...I want to prove it to some of my friends who think that I'm wrong, so I want a rigorous derivation

suppose z=a+bi, then we denote the conjugate of z, z (with the bar on top)=a-bi, so multiplying them obviously gives a^2+b^2...the confusion lies in defining z=a+bi probably...because z does not always have to be that, z=bob, for example, is possible

[quote="anirudh"]But isn't $z=a+bi$ by convention?

When I tell them that, they don't believe it. What says that $z=a+bi$[/quote]

one generally uses z=a+bi for a complex number...and i guess with the whole conjugate thing it is "assumed" that z is a complex number...but you could also be like let z=x+yi, then zz(bar)=x^2+y^2,

i also think the confusion might be this: you are saying

a complex number * the complex conjugate is equal to the sum of the real part squared and the imaginary part squared, this is true

however, they might be thinking something to the effect of zz(bar) is always equal to a^2+b^2, which is not really understanding your statement...

[hide="proof"]z=a+bi

$z\overline{z}=(a+bi)(a-bi)=a^{2}+b^{2}$[/hide]

@Altheman: Assume bob=a+bi, because a and b could be 0.

[quote="anirudh"]But isn't $z=a+bi$ by convention?

When I tell them that, they don't believe it. What says that $z=a+bi$[/quote]

Perhaps you could say $z = \Re(z)+i\Im(z)$, with $\Re(z),\Im(z)\in\mathbb{R}$. Every complex number can be written like that, as $\mathbb{C}$ is $\mathbb{R}[i]$, an [url=http://mathworld.wolfram.com/ExtensionField.html]extension field[/url] of $\mathbb{R}$. Then it's easy to prove, as, by definition, $\overline{z}= \Re(z)-i\Im(z)$, that $z\overline{z}= \Re(z)^{2}+\Im(z)^{2}$.

[quote="anirudh"]But isn't $z=a+bi$ by convention?

When I tell them that, they don't believe it. What says that $z=a+bi$[/quote]
Nothing says that, such $a,b$ simply exist.
To answer any more question, you (or they) have to tell how they define everything that occures (and no, the answer to this is not like "as always").
Untill then already the first answer was perfectly well.

Usually $z$ is used to represent a complex number.  I have no idea why.

All complex numbers can be expressed in the form $a+bi$, where $a,b\in\mathbb{R}$.  So let $z=a+bi$ for real $a$ and $b$.  So the conjugate of $z$ is $a-bi$ by definition.  So their product is $(a+bi)(a-bi)$, or $a^{2}+b^{2}$.

[quote="anirudh"]But isn't $z=a+bi$ by convention?

When I tell them that, they don't believe it. What says that $z=a+bi$[/quote]

Well, if you want to prove that $z\overline{z}=a^{2}+b^{2}$, it helps to use the right variables ;)

You might as well say that you can't prove that $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. What says that $ax^{2}+bx+c=0$?

mathnerd's got a point.  You need to declare something first.  For the quadratic formula, you say this:

[b]IF[/b] $ax^{2}+bx+c=0$, then $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$.

For the original problem, you'd have to say that $z=a+bi$, where $a$ and $b$ are real numbers.


conics
ellipse
hyperbola
calculus
derivative
analytic geometry
graphing lines

How do you prove that the ellipse $ 4x^2\plus{}9y^2\equal{}45$ and the hyperbola $ x^2\minus{}4y^2\equal{}5$ are orthogonal.

My plan:
Figure out where they intersect(which I don't know how to do...)
Implicit differentiation and plug in the point where the intersection is
Make sure that the slopes are negative reciprocals.

How do you find out where they intersect?

They intersect at the points where both equations are satisfied.  Subtract four times the second equation from the first to find $ 25y^2 \equal{} 25\Rightarrow y \equal{} \pm 1$ and add $ \frac {9}{4}$ times the second equation to the first to find $ \frac {25}{4}x^2 \equal{} \frac {225}{4}\Rightarrow x \equal{} \pm 3$


You are not supposed to post ANYTHING like this here.

Only "technical" questions about the (meaning of the) problems, nothing else.


Hello, I have a problem with well, a problem, from AHSME. I am just starting to learn about polynomials, and I think I made an error:

Let $ P(x)$ be a polynomial such that when $ P(x)$ is divided by $ x\minus{}19$, the remainder is 99, and when it is divided by $ x\minus{}99$, the remainder is 19. Find the remainder when $ P(x)$ is divided by $ (x\minus{}19)(x\minus{}99)$.

What I did: $ P(x)\equal{}q_1(x\minus{}19)\plus{}99$, giving $ P(19)\equal{}99$, and $ P(x)\equal{} q_2(x\minus{}99)\plus{}19$, so $ P(99)\equal{}19$. $ P(x)\equal{}q_3(x\minus{}19)(x\minus{}99)\plus{}r$, giving the equations $ P(19)\equal{}P(99)\equal{}r$. This may be trivial to you, but please help!

When you're dividing $ P(x)$ by a polynomial of degree $ 1$, the remainder is of degree $ 1 \minus{} 1 \equal{} 0$. That's why you ended up with $ 99$ and $ 19$ when you divided $ x \minus{} 19$ and $ x \minus{} 99$ respectively.

But when you're dividing $ P(x)$ by a second degree polynomial, $ (x \minus{} 19)(x \minus{} 99)$, the remainder is of degree $ 2 \minus{} 1 \equal{} 1$. So, it has the form of $ ax \plus{} b$.

Just write $ P(x) \equal{} Q(x)(x \minus{} 19)(x \minus{} 99) \plus{} ax \plus{} b$. Then plug in the values you know and solve the system for $ a$ and $ b$.

Wait, why should the solution be linear? Doesn't the 0 make the remainder a constant?

Only in special cases will the remainder be constant when dividing a quadratic by a linear term.

However, it does happen. For example, $ \frac{x^2\plus{}3x\plus{}3}{x\plus{}3}$ has a remainder of 3.

So basically when you are dividing some polynomial of degree n by x-a, the deg of the remainder can be no greater than n-1.

Other way around.  When the polynomial you're dividing [b]by[/b] has degree $ n$, the remainder has degree at most $ n \minus{} 1$.  This is the [url=http://en.wikipedia.org/wiki/Division_algorithm]division algorithm[/url] for polynomials.

Plz. disregard everything I just said (I make very stupid mistakes sometimes  :( ). I re-did the problem and got -x+118.


[quote="ice_age"]1)   y and 3y are two integers, which are formed by using each of the nine digits (1,2,3,4,5,6,7,8,9) exactly once.
      what is y?

Any hints for all the above three questions?
thanks.[/quote]

y is 5823 or 5832

I found the same answer. By the, way did you use some method to find it or just brut force?

See also [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=200713]this similar topic[/url]  :)


linear algebra
matrix
vector
complex numbers
linear algebra unsolved

Prove that a normal stochastic matrix is always doubly stochastic.

I hope this is correct, since I sort of got lost in the indices :).

I'll assume the sum on every row is $0$, and the aim is to prove that the sum on every column is also zero. The usual definition is with $1$ as the common value of the sums, but we can subtract the identity matrix from the initial matrix to reach this version.

The matrix is unitarily diagonalziable, so it has the form $U^*DU$, where $D=\mbox{diag}(\lambda_1,\ldots,\lambda_n)$. If $x_1,\ldots,x_n$ are the columns of $U$, then the condition that $A$ is stochastic translates to $\forall i,\ \langle Dx_i,v\rangle=0\ (*)$, where $v=\sum_{j=1}^nx_j$, and the conclusion we want to reach, namely that the matrix is doubly stochastic, translates to $\forall i,\ \langle D^*x_i,v\rangle=0\ (**)$.

If, in general, for all vectors $u,\ u(k)$ is the $k$'th entry of $u$, then $(*)$ can be written as $\forall i,\ \sum_k\lambda_k[x_i(k)\overline{v(k)}]=0\ (\#)$. This shows that the complex numbers $\lambda_k\cdot\overline{v(k)}$ are solutions of a homogenous linear system, having $U$ (the transpose of $U$, actually, but it doesn't really matter) as its matrix, meaning that all these complex numbers are zero. In turn, this means that the complex numbers $\overline{\lambda_k\cdot v(k)}$ are also null, which shows that $(\#)$ also holds with $\overline{\lambda_k}$ instead of $\lambda_k$. This, however, is just another way of writing $(**)$, which is what we wanted to prove.

Is it correct?

Don't imagine I checked the computations, I know it's correct since my solution was also based on the same idea and my teacher said it was correct. I really liked this problem: after so many indices and stuff you arrive at such a nice conclusion.  :D


I am not especially liking it right nw because there are too many tourneys going on so I voted no although I like it sometimes like in the year when actual problems are being posted more often....

Wow.  I thought it would be like 100%.  A lot of people must not like Aops, and yet there are 43000+ members.... :huh:

no parents made me/sort of?????

i vote yes


function
geometry
3D geometry
sphere
limit
trigonometry
complex analysis

Let's consider the recurrence $ (n+1)a_{n+1}=a_1 a_n+a_2 a_{n-1}+...+a_n a_1$ with $ a_1=1$ . Prove that the sequence $\sqrt[n]{a_n}$ converges. I tried with estimations and did not suceed. I also tried with generating functions and failed. So...

Power serie $f(x)=\sum_{n\geq 0} a_ nx^n$ with $a_0=0$

$f^2(x)$ ?

Yes, Moubi, this approach is obvious, but what we do next?  ;)

We get an equation to an explicit of $f(x)$
then expand in power serie to an explicit expression of $a_n$ and compute the limit

Can you please post a complete solution?  :)

I think we get the differential equation $x f'(x) = x + f(x)^2$.  But I don't know how to solve that...

Well, this is my problem too. How to solve the monster? Moubi?

Let me try to answer. The truth seems to be that you cannot solve the equation explicitly. Fortunately, you do not need to. What you need is to show that the differential equation allows to extend $f$ to an analytic function from $\mathbb C$ to the Riemann sphere $\mathbb S$ (or, if you prefer, meromorphic function). Then you should check that $f$ has its nearest to the origin pole on the positive real semiaxis and apply a general theorem from Complex Analysis. I think that is what Moubinool had in mind  :) .

Do you know any complex analysis, harazi? If you do, I'll try to explain better but if you do not, I'd better think of some other purely real approach. The theorem I was going to use is the following:

Let $f:\mathbb C\to \mathbb S$ be a meromorphic function with Taylor expansion at the origin $f(z)=\sum_{n\geq 0}a_n z^n$. Assume that $z_0$ is the nearest to the origin pole of $f$ and that there are no other poles on the circle $|z|=|z_0|$. Then
$\lim_{n\to\infty} |a_n|^{1/n}=1/|z_0|$.

Does this sound familiar?

No, it does not, but I can imagine it, so let's accept it as it is. And how we prove the thing with the poles for our differential equation?

This question is two-fold: how do we prove that $f$ is meromorphic and how do we prove that the closest to the origin pole is unique. I'll go over the first part now. And I'll do it for a simpler equation $f'=1+f^2$, which, of course, you can solve: $f(x)=\tan (x+c)$. I hope it won't surprize you that $\tan z$ is defined for complex $z$ as well and is a meromorphic function on the entire complex plane. Let's see how to get this right from the equation. The key is the power series method. For any initial data $f(x_0)=y_0$ ($x_0,y_0\in\mathbb C$), we can create a unique Taylor series at $x_0$ that satisfies the equation in some disk around $x_0$ and starts with $y_0$. What is more, its radius of convergence can be estimated from below if $|y_0|\leq 1$. If we could get an estimate for the radius of convergence independent of $y_0$, we would be able to analytically continue $f$ along any path in $\mathbb C$ and get an analytic function. Unfortunately, this is impossible. But let us notice that if $|y_0|>1$, then $|1/y_0|<1$ and the function $g=1/f$ satisfies almost the same differential equation $g'=-1-g^2$. This allows to use the power series method for $g$ instead of $f$ with some fixed radius of convergence. The only problem is that $g$ may vanish at some points, which will become poles of $f$, so $f$ will be not analytic but meromorphic. Does this make sense? If it does, try to apply this technique to the real equation you have yourself.

Thank you very much, fedja, I understood now. Thank you. I wonder whether there exists a purely elementary approach...


trigonometry
Gauss
complex numbers
geometric series

Here's a problem you may enjoy as much as I did.  Compute the sums
\[ \sum_{k=0}^n\sin (kx)\hspace{.5in}\sum_{k=0}^n\cos (kx) \]

[hide="HINT iff you are stuck"]
Complex numbers using $e^{ikx}$
[/hide]

Masoud Zargar

[hide="solution"]Let $e^{ikx} = \cos (kx) + i \sin (kx)$.  Then

$\sum_{k=0}^n \cos (kx) = \text{Re} ( \sum_{k=0}^n (e^{ix})^k )$

and similarly for the $\sin$ series.  Evaluating this finite geometric series gives

$\sum_{k=0}^n (e^{ix})^k ) = \frac{1 - (e^{ix})^{n+1}}{1 - e^{ix}} = \frac{1 - \cos ((n+1)x) - i \sin ((n+1)x)}{1 - \cos x - i \sin x}$

Evaluating the real and imaginary parts of this sum gives us

$\sum_{k=0}^n \cos (kx) = \frac{1 - \cos x - \cos ((n+1)x) + \cos x \cos ((n+1)x) + \sin x \sin ((n+1)x) }{2 - 2 \cos x}$

$\sum_{k=0}^n \sin (kx) = \frac{\sin x - \sin x \cos ((n+1)x) - \sin ((n+1)x) + \cos x \sin ((n+1)x) }{2 - 2 \cos x}$

I'm not sure whether there's an easy way to put this into a simpler form.[/hide]

Well, that is probably right, but there is a simpler answer.

We have $e^{ikx}=\cos (kx)+i\sin (kx)$.
$\sum_{k=0}^n (e^{ix})^k=\frac{e^{ix(n+1)}-1}{e^{ix}-1}\\ =\frac{e^{ix(n+1)/2}(e^{ix(n+1)/2}-e^{-ix(n+1)/2})}{e^{ix/2}(e^{ix/2}-e^{-ix/2})}\\ =e^{ixn/2}\cdot\frac{\sin\frac{(n+1)x}{2}}{\sin\frac{x}{2}}\\ =\left(\cos\frac{nx}{2}+i\sin\frac{nx}{2}\right)\cdot\frac{\sin\frac{(n+1)x}{2}}{\sin\frac{x}{2}}\\ =\frac{\cos\frac{nx}{2}\sin\frac{(n+1)x}{2}+i\sin\frac{nx}{2}\sin\frac{(n+1)x}{2}}{\sin\frac{x}{2}}$
$\\\sum_{k=0}^n\cos (kx)=\Re\left(\sum_{k=0}^n (e^{ix})^k\right)=\frac{\cos\frac{nx}{2}\sin\frac{(n+1)x}{2}}{\sin\frac{x}{2}}\\ \sum_{k=0}^n\sin (kx)=\Im\left(\sum_{k=0}^n (e^{ix})^k\right)=\frac{\sin\frac{nx}{2}\sin\frac{(n+1)x}{2}}{\sin\frac{x}{2}}$

Are there any proofs without the use of complex numbers.  I think there is one derived in a similar fashion as Gauss' summation formula.

Masoud Zargar

Yes, that's a much nicer version than what I got.  I've never seen any method of proof other than by complex numbers.

Thanks.  I have a solution for $\sum_{k=1}^n\cos (kx)$ by using telescoping sums, but it isn't that nice.  It's just a simple application of product-to-sum formula.  You just multiply it by $2\sin\frac{x}{2}$.

MZ

I liked this problem.

It got messy though.

How about something like:

$\sum_{i=0}^{n}\frac{1}{i+1}\dbinom{n}{i}$

How do we find the sum of this?

Any ideas?

for player's sum,

[hide]

$\frac{1}{i+1}\dbinom{n}{i}=\frac{n!}{(i+1)i!(n-i)!}=\frac{n!}{(i+1)!(n-i)!}=\frac{1}{n+1}\frac{(n+1)!}{(i+1)!(n-i)!}=\frac{1}{n+1}\dbinom{n+1}{i+1}$

So the sum is $\frac{1}{n+1}$ times the sum of $\dbinom{n+1}{i+1}$ for all the i. The sum of all the combinations in row n+1 is $2^{n+1}$ but since we skip the $\dbinom{n}{0}$ we must subtract 1.

This ends up as

$\frac{2^{n+1}-1}{n+1}$

[/hide]

Tjhance,

I see that you have $\frac{n!}{(i+1)!(n-1)!}= \frac{1}{n+1}\frac{(n+1)!}{(i+1)!(n-1)!}$

Shouldn't it be:

$\frac{n!}{(i+1)!(n-1)!}= \frac{1}{(n+1)!}\frac{(n+1)!}{(i+1)!(n-1)!}$

Or am I just not understanding something?

EDIT: Yikes...Ignore this please.

Most likely a typo :wink:
In normal language, what difference does one exlcamation mark mean anyway? :P
Seeing as there is no other way for there to be a true statement, yes you are correct.

Yikes,

I see it now.

I must be sleepy.

Why is the sume of all combinations 2^(n+1)?

[quote="player"]Yikes,

I see it now.

I must be sleepy.

Why is the sum of all combinations 2^(n+1)?[/quote]
The sum is equivalent to the number subsets of an n+1-set including the empty set.

I figured out my post above yet again...

I am too sleepy...

Sorry for doubting you so much tjhance, and thanks for your informative reply boxedexe


Use intermediate calculus level to answer:
Find the general solution of the equation
y'' - 4y' + 3y = cosh x

First of all, stop re-posting the same topic.  Second, this uses calculus so don't post it in the high school forum.

hello, at first of all you must solve the homogeneous part of your equation by the ansatz $ y(x)\equal{}e^{\lambda x}$.
Sonnhard.

Do not double post:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=268915[/url]


I have two ideas for perpetual motion (don't stop reading now):

1) I know that friction eats up all of your energy, but if there was some way to always, say, cut the amount of friction on a surface in half, then the amount of useful energy you would have would be bounded by below (so it would always be visibly moving). Not sure of the math on this one.

2) A lot of perpetual motion devices don't work because of entropy. Is there any way that you could design a box that would increase the entropy within it while decreasing the entropy around it? I heard that most humans do the opposite, which is where I got the idea. If yes, then you could put one of these boxes into a perpetual motion device, and it would work. Also, if we had enough of these, we could prolong the life of the entire universe (until it freezes over).

i think this goes under one of the sciences... especially Physics.

comment on number 2....

u r an inspired scientist and it is a very good thing...
BUT...for entropy in human is the opposite then wut it seems to be is because we r a open system...

in another word...

we get organized but involuntarily the universe get more disordered...

In the box u r talking about...i assume it will be a closed system though it is not possible to apply wut u said...

Of course everything is possible... but i personally don't want or like to go against the law of physic...