If $ \boxed{X}$ is defined as $ X \minus{} \frac1X$, find $ \boxed{\boxed{X}}$. Express your answer as a common fraction.

$ X \minus{} \frac{1}{X}$ boxed is $ \left(X \minus{}\frac{1}{X}\right)\minus{}\left(\frac{1}{X\minus{}\frac{1}{X}}\right)$.

$ \boxed{\boxed{x}}\equal{}\boxed{\frac{x^2\minus{}1}{x}}\equal{}\frac{x^2\minus{}1}{x}\minus{}\frac{x}{x^2\minus{}1}\equal{}\frac{(x^2\minus{}1)^2\minus{}x^2}{x(x^2\minus{}1)}\equal{}\frac{(x^2\minus{}x\minus{}1)(x^2\plus{}x\minus{}1)}{x(x\minus{}1)(x\plus{}1)}$ As this doesn't simplify I feel I made a mistake somewhere.

You didn't make a mistake, something's wrong in the problem.

hello, i have got the same as you $ \frac{(x^2\minus{}x\minus{}1)(x^2\plus{}x\minus{}1)}{(x\minus{}1)x(x\plus{}1)}$.
Sonnhard.

the answer can't be 5/6.  It is in terms of x.  I think it is (x^4-3x^2+1)/(x)(x^2-1).

[hide="The simplest Solution"]
Take $X=2$. Therefore, $\boxed{X}=\dfrac{3}{2}$. And $\boxed{\boxed{X}}=\dfrac{3}{2}-\dfrac{2}{3}=\boxed{\dfrac{5}{6}}$.
[/hide]


From the triangle $ABC$, $(a,A,mb+nc)$ are known. Construct the triangle. ($m$ and $n$ are two given lengths.)

Merged with http://www.mathlinks.ro/Forum/viewtopic.php?t=27490 .

  darij


This one seems really ugly and it is, but I like it. Prove that in any triangle
  $\frac{m_a^2\cdot m_b^2}{m_c^2}+\frac{m_b^2\cdot m_c^2}{m_a^2}+\frac{m_c^2\cdot m_a^2}{m_b^2}\geq\frac{9}{4}\cdot\(\frac{a^2+b^2+c^2}{a+b+c})^2 $. Is it trivial? I have a very hard solution for it. I would like no computations.

Harazi,please could you clarify the notation on the right side of the inequality.
I am eager to take this one on.

Thank you

Michael Niland

Well the left_hand side is $\frac{9}{4}\cdot (\frac{a^2+b^2+c^2}{a+b+c})^2 $. The constant is chosen to have equality for equilateral triangles:


Prove that if a triangle is streched in one direction by a factor of k, its area is also multiplied by a factor of k.

I'll just post the proof for all cases:

[hide]
This involves a little bit of calculus (which the AOPS book says is dividing a weird figure up into many many small rectangles). Align these rectanges up with the direction of distortion. When the figure is distorted, one dimension of the rectangle will increase by a factor of k. Thus, the area of the whole figure, which is the sum of the areas of the rectangles, will also increase by a factor of k. The reason this proof is not so good is because for some figures they cannot be entirely broken into little rectangles.[/hide]

It's not quite calculus, but it's really really close.  :)

EDIT: OK, nevermind, it says stretch one direction...
EDIT#2: I think this works, I dreamt of it on my sleep :)

If the direction of stretching is parallel to a side, take that side as the base and it is stretched from $b$ to $kb$ (height is kept constant) and so the area of the stretched triangle is $A_k=\frac12(kb)h=kA$, where $A$ is the area of the original triangle.

If the direction of stretching is not parallel to any sides, draw a line parallel to the stretching direction that passes through the 'middle' vertex in such a way that it intersects the opposite side of the triangle, hence my strategy is to divide the triangle into 2 triangles with a common base. Apply above argument twice and add the areas and we get again $A_k=kA$

[quote="10000th User"][quote="236factorial"][quote="Bestone"]Do you want a proof for all geometric shapes alike (which is not rigorous and incomplete) or for more conventional shapes (specify)(in which the proof would be more elegant and rigorous)?[/quote]

I just want a general triangle...[/quote]
Is my proof above OK for you?[/quote]

That sounds like the aoPS solution  :D

[quote="236factorial"][quote="10000th User"][quote="236factorial"]That sounds like the aoPS solution  :D[/quote]
Sorry to disappoint you then because I do not have any AoPS books :D[/quote]

Oh well, I get it now anyway (but you should get the books  :D )[/quote]
Can you please refrain from continuing the conversation? This should be serious discussion and therefore you can PM for a thorough :D conversation if you want :D

[quote="236factorial"][quote="10000th User"]EDIT: OK, nevermind, it says stretch one direction...
[/quote]

Fine, a more serious conversation  :roll:

what if it [i]was[/i] stretched in two directions?[/quote]
Simple, then the height changes as well. But, if both directions are separated at an angle $\theta\neq 90^\circ$.... I give you some time to try it :D

[quote="10000th User"][/quote][quote="236factorial"][quote="10000th User"][quote="236factorial"]That sounds like the aoPS solution  :D[/quote]
Sorry to disappoint you then because I do not have any AoPS books :D[/quote]

Oh well, I get it now anyway (but you should get the books  :D )[/quote][quote="10000th User"]
Can you please refrain from continuing the conversation? This should be serious discussion and therefore you can PM for a thorough :D conversation if you want :D[/quote]Wow, that conversation really developed there didn't it?  :roll:


In a triangle ABC, let D be the intersections of the bisectors of ∠BAC with the side BC.
Prove that:
\[ AD^2 \plus{} BD.DC \equal{} AB.AC\]

[img]http://img11.imageshack.us/img11/2033/41732588.jpg[/img]

Let $ AD\equal{}d,AB\equal{}c,AC\equal{}b,BD\equal{}n,DC\equal{}m$.
By stewart's theorem: $ b^2m\plus{}c^2n\equal{}d^2a\plus{}amn$. Since $ AD$ bisect $ A$, $ \frac{m}{c}\equal{}\frac{n}{b}$.
Now
\[ b^2(nc/b)\plus{}c^2(mb/c)\equal{}a(d^2\plus{}mn)\Rightarrow bc(m\plus{}n)\equal{}a(d^2\plus{}mn)\Rightarrow bc \equal{} d^2\plus{}mn\]
and we are done.

It's an old, classic result. Let's call $ \{ E \}$ the second intersection with the circle $ (ABC)$; it's easy to see $ \triangle ABD \sim \triangle AEC$, hence $ AD \cdot AE \equal{} AB \cdot AC$, but $ AD \cdot AE \equal{} AD\cdot (AD\plus{}DE)\equal{}AD^2 \plus{} AD \cdot DE$, and, from power of $ D$ w.r.t. $ (ABC)$, $ AD\cdot DE \equal{} BD\cdot CD$, done.

Best regards,
sunken rock


How do you solve this..

Six people are chosen to participate in a contest to win a car. There are six keys placed in a box. The six contestants will in random order take a key from the box. If the key starts the car, they win. If the key does not, it is discarded and the next person tries. Which position (1st-6th) gives you the largest probabilty of winning?

There is an equal probability for each person.  Just imagine if each key was given out randomly, instead of people selecting them.  Same process, and then you see each person has a 1/6 probability.


function
topology
advanced fields
advanced fields unsolved

Hi, Here is a question that I couldnt solve:

Prove that If f is an upper semi continuos functions there exists decrasing continiuos functions converging to f.

thanks in advance

On a bounded metric space $f_{n}(x): =\sup_{y}f(y)-nd(x,y)$ works. One can extend this to general metric spaces using a partition of unity or something. I suppose that the proper context here is a topological space, but don't know a proof for this case.

Don't we need the compacity of our metric space, in order to make sure that our $sup$ exists ?

$\sup$ always exists; boundedness makes it finite. I do not claim that $\sup$ is attained.

f_{n}(x): =\sup_{y} [f(y)-nd(x,y)]

you mean the sup of the subtraction? or the opposite if so y depens on what for the metric.

thnks.

Yeah, that's the sup of the whole thing.


inequalities
geometry
circumcircle
geometry proposed

In a triangle ABC we have
min(h_a,h_b,h_c)<=R+r<=max(h_a,h_b,h_c)

This is the first time I see this.

Let da, db, dc be the distances from O (the circumcenter) to the sides BC, CA, AB respectively. It's well-known that R+r=\sumda. At the same time, 2S (twice the area of ABC)=ha*a=hb*b=hc*c=\sumda*a. It's easy to derive both the inequalities from here.

[quote="grobber"]It's well-known that R+r=\sum da[/quote]
   In fact, if ABC is obtuse at A you will have
     R+r = d_b+d_c-d_a. :D

Hmm. I guess you're right, but I think the solution still works, because in that case we have ha*a=hb*b=hc*c=db*b+dc*c-da*a.


I think that's right, plus it wouldn't matter what order they're in, the 8 could be anywhere.

Are you sure?  :?
\[ \begin{aligned} S & = \sum \left(3x_i^2 + 2x_i^3\right) = \sum \left(x_i^3 + (x_i+1)^3 - 3x_i - 1\right) \\ & = \sum \left(x_i^3+(x_i+1)^3\right) - \sum \left(3x_i+1\right) = \sum \left(x_i^3+(x_i+1)^3\right) - (3\cdot 2007 + 2000),\end{aligned}  \] so we need to maximize $\sum \left(x_i^3+(x_i+1)^3\right)$. But if $x_1=x_2=...=x_{1999} = 0$ and $x_{2000} = 2007$, I think we get a bigger sum... Or am I wrong? If not, who can complete the proof?  :)

0 isn't considered positive though, right?

In Belgium, 0 is positive, but in this problem it is not :)

hmm, there are not very many possibilities with this problem, one simplification though would be to have $x_i-1=y_i$, then you would not have to worry about as much, you would have

$\sum y_i=7$

the other sum becomes then

$\sum (3y_i^2+6y_i+3)+(2y_i^3+6y_i^2+6y_i+2)=2\sum y_i^3+9\sum y_i^2+12\sum y_i+5\sum 1$

you know the second 2 summations, that leaves you with the first 2 summations, and there are a limited number of cases to deal with and it is pretty clear otherwise that $y_1=7$ is the max

[quote="Altheman"]hmm, there are not very many possibilities with this problem, one simplification though would be to have $x_i-1=y_i$, then you would not have to worry about as much, you would have

$\sum y_i=7$

the other sum becomes then

$\sum (3y_i^2+6y_i+3)+(2y_i^3+6y_i^2+6y_i+2)=2\sum y_i^3+9\sum y_i^2+12\sum y_i+5\sum 1$

you know the second 2 summations, that leaves you with the first 2 summations, and [b]there are a limited number of cases to deal with[/b] and it is pretty clear otherwise that $y_1=7$ is the max[/quote]

What happens if we change the question to $x_1+x_2+...+x_{100 }= 2007$?
In this case, with your method, then we need to deal with many different cases....


I need Radon's inequality's proof, so i need a help.PLZ!
Let $ a_i,b_i,p \geq 0$ then
\[ \frac {(\sum a_i)^{p \plus{} 1}}{(\sum b_i)^p}\leq \sum \frac {a_i^{p \plus{} 1}}{b_i^p}
\]

Particular case of http://www.mathlinks.ro/Forum/viewtopic.php?t=20547 , Theorem 1 when $ r\equal{}s\plus{}1$.

  darij

[quote="darij grinberg"]Particular case of http://www.mathlinks.ro/Forum/viewtopic.php?t=20547 , Theorem 1 when $ r \equal{} s \plus{} 1$.

  darij[/quote]
Thank You


[hide]

18*2 is what Charrlote had.
So dave gave Charollote 36 pogs[/hide]

[quote="math92"]Dave gave Charlotte half of his pogs. Charlotte gave Johnnie half of the pogs she received from Dave. Johnnie kept 8 of those pogs and gave the remaining 10 to Dana. How many pogs did Dave give Charlotte?[/quote]
[hide]
Johnnie got $\frac{1}{4}$ of Dave's original pogs. She got 18 pogs, so the original number of pogs that Dave had was $18*4=72$

He gave half of those, so the answer is 36[/hide]

[quote="math92"]Dave gave Charlotte half of his pogs. Charlotte gave Johnnie half of the pogs she received from Dave. Johnnie kept 8 of those pogs and gave the remaining 10 to Dana. How many pogs did Dave give Charlotte?[/quote]
[hide]
C=0.5D
J=0.5C
J=18
C=36

the answer is 36
[/hide]


What do you think about a good teacher / a good math teacher should be? Also, what kind of teachers are considered to be bad teachers?

I major in math education in university, and in the education courses, we always focus on this question. Thus I want to know your opinion.

I guess it would depend.
I've seen a poster recently that said: "Teachers open the door. You must enter yourself."
I guess that would make sense, wouldn't it? :yup:

A good teacher, I suppose would be someone that teaches the concepts, hopefully in a way that isn't always straight from the textbook--the students can read that themselves. Maybe in a more abstract manner and one that encourages lots of questions and "What if"s. The teacher should plan ahead for what the students may ask, and ask question themselves.

Something else important: the teacher should love what they are doing and love teaching/helping others. A people personish-type of person. They should appreciate the beauty of their subject. I guess that probably wouldn't be a problem since you're on AoPS...:lol:

Hmmm. There's probably more, but that's all I can think of at the moment.

[b]Good teacher:[/b]

One that's had a lot of experience with the subject they are teaching (like my math and social studies teacher). For example, going to the historic sites of the French and Indian War, going to Philadelphia, etc. He/she should also be fairly involved with the subject, not always answering your question as, "Hmmmm...That's a good question. I'll look it up." and never does look it up. A good teacher should have straightforward methods, and never ever yell at someone for no reason. If you ask a question, they should answer it directly and never criticize you for not paying attention. Other qualities are having a sense of humor that is direct, focused, and all out grabbing your attention. They will leave you to do independent assignments but not bookwork all day long, day after day.

[b]Bad teacher:[/b]

A bad teacher would be one who constantly breaks rules of the school and feels that there is no importance of rules to teachers. Like, using cell-phones during class, listening to Ipod during teaching sessions, or harassing the students. The bad teacher would either a) hold your hand all of the time and walk you through every question or b) always typing at the computer and making you work out of the book. Dressing sloppily, is one. Always being late to class or ignorant to those who come late are some qualities of a bad teacher. Or, not even bothering to grade tests or book reports (as an example of a fourth grade teacher who was an ok teacher) to give everyone A+ and oblivious to those reports that the writers have just jotted down from watching a movie.

Well, so far that's what I think of teachers...Of course, it [b]will[/b] be much different in middle school?

This is my ranking of the teacher qualities in order of their importance (you may think that the weight on number $ n$ is $ 2^{\minus{}n}$):

1) Knowing the subject in and out (without that nothing helps; if the teacher doesn't know what he is talking about, there is no chance the students will).
2) Readiness to answer student questions and ability to adjust the material to the student level (this is basically what teaching is about; otherwise just reading a good book would suffice).
3) General organization (like preparing the lectures, timely grading, etc., it may be not so important for student's understanding the material (under condition 1, you can improvise fairly well), but it somehow raises the class moral).
4) Open-mindedness and friendliness.
5) Dressing, blackboard techniques, and everything else that comes to your mind that makes listening to a lecture a more pleasant experience.

I should confess that I'm reasonably good with 1 and 2, quite sloppy on 3, cannot really evaluate myself on 4, and mediocre on 5, which results in my evaluations varying from "The best teacher I have ever had" to "This guy should be fired immediately". :P

Also, I believe that there is no more sense in trying to discuss what is a good teacher and what is a bad teacher in education classes than in discussing what is a good proof and what is a bad proof in mathematics classes. What would be much more useful is just to spend 3/4 of the time making the students to do some presentations and pointing out the strong and the weak sides every time. (The other 1/4 of the time the teacher should present the material telling why he does it this or that way on the go). This way one can really learn something about how to teach.

Dressing is an important quality? Jeez, I live in a different society  :P

Well, read [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=195299]this[/url]. It is essentially the same story. Interestingly enough, when I was a student, the old professors always came wearing a suit and a tie but the younger ones were comfortable wearing a shirt or a sweater. My generation was probably the first to start running classes in shorts and T-shirts. I do not know what they do now :lol:.

I believe that a good math teacher must:
1. Be very good at math, not just know just enough to teach off notes. This includes being able to explain extra things that may not appear on the curriculum and are brought up.
2. Be passionate about teaching.
3. Care about students.
4. Have good presentation skills (goes with knowing subject well)
5. Have decent handwriting.
6. Not lazy (like not multiple choicing the whole test).

It is extremely important for math teachers to be able to answer any subject-related student question, no matter how roundabout or poorly phrased, as many questions are. A teacher that waffles loses credence from the students.

I think it's also important that teachers give structured notes and solutions that are thorough and easy to follow. They can believe that their students don't need the notes, but there are always several who do, and what can it hurt? Their presenting the notes will serve as aural discussion for those who don't take notes.

[quote="D3m0n Shad0w"][b]Bad teacher:[/b]

A bad teacher would be one who constantly breaks rules of the school and feels that there is no importance of rules to teachers. Like, using cell-phones during class, listening to Ipod during teaching sessions, or harassing the students.

Well, so far that's what I think of teachers...Of course, it [b]will[/b] be much different in middle school?[/quote]

@D3m0n Shad0w
Erm. Well, you [i]will[/i] continue to get teachers like that. Most likely. You will also get better teachers. Hopefully. :)

[quote="dingzhou"][quote="The_Scintillator"]
Maybe in a more abstract manner and one that encourages lots of questions and "What if"s. The teacher should plan ahead for what the students may ask, and ask question themselves.
[/quote]

hmm....i had a teacher last year who banned all what if questions.

apparently...theres no what ifs in history...[/quote]

Heh. Been there. Done that. Found a different way to phrase the question. Got yelled at for being a smart-aleck. Caused a stir in the classroom. Saw the chaos, confusion, and laughter--dusted myself off, and said: "My work here is done."

[quote="dingzhou"]hmm....i had a teacher last year who banned all what if questions.

apparently...theres no what ifs in history...[/quote]

Well, if a teacher must get through a set amount of material, discussions extraneous to the matter at hand simply detract from that objective.  While in certain subjects "what if" questions may reinforce particular concepts by applying general principles to hypothetical situations, in history (at least at a high school level) it is mostly idle speculation that is tangential to the broader point.  Then again, my history teacher is fond of discussing the latest exploits of Sarkozy's wife or spending an entire class period reading Obama's speech on race, and I certainly prefer that to an exploration of the Safavid Empire.

ah good question...You know I've grown up with good teachers throughout my schooling (I'm in eighth grade). For me a good math teacher is someone who doesnt get there material straight from a textbook. Another atribute that makes a good math teacher for me is someone who tells you about their experience learning about math and says it's OK if you don't get it right now. So really I guess I'm saying a good math teacher is someone who acts human and not like a robot and adds a personal touch.

Handwriting as someone mentioned is a big issue.

I couldn't read any of the notes my math teacher wrote on my test.
I had to ask her to translate every word.

and they have to make the class active.
Good teachers keep students in focus, and they put information into the brain in a harmless manner.

Err...well my math teacher lets the students answer questions if some are comfortable and others arent...

[quote="SimonM"][quote="The_Scintillator"]
Heh. Been there. Done that. Found a different way to phrase the question. Got yelled at for being a smart-aleck. Caused a stir in the classroom. Saw the chaos, confusion, and laughter--dusted myself off, and said: "My work here is done."[/quote]

And we wonder why people do not want to become teachers?

Its not cool or clever to be the person who picks the teacher up on things in front of the class. The teacher will dislike you, the students will eventually get annoyed with in and generally you'll be unpopular. Eventually the joke will be on you.[/quote]

I realize that. I guess I put it in a bad way.
...But to be honest, I learned absolutely nothing from that teacher. They would never teach us anything. At the time, I felt that asking that question would actually get us to learn something.

....

It didn't quite work.

I guess you're right. But now I'm making this off topic. Carry on. :D

[quote="ZhangPeijin"]Ok let's just put it bluntly.
[b]Good Teacher[/b]
Teaches without relying on textbooks/handouts that were not of his/her own creation

[b]Bad Teacher[/b]
Teaches using textbook all the time, reads the text during class, assigns homework, doesn't do anything else.[/quote]

In my opinion, you can't really dichotomize good teacher/bad teacher so easily. Many teachers carry traits that you would place all over the goodness/badness spectrum, and it is incredibly rare to find a teacher that is wholly good or (thankfully) wholly bad.

That being said, I think it really depends on the student. For instance, I think my chemistry teacher is a great teacher who just happens to be new at teaching the subject, whereas some of my friends think she "doesn't know what the f*** she's doing". We all have our own personal rubric of what constitutes a good teacher, but if I had to pick one universal quality that makes a teacher "good", it would be

[quote="ifai"]I think a good teacher can make students have interest and motivation to study outside the class since it is not enough for students to study only in class or school.
[/quote]

[quote="ZhangPeijin"][b]Bad Teacher[/b]
Teaches using textbook all the time, reads the text during class, assigns homework, doesn't do anything else.[/quote]
I have seen such teachers. One of my physics teachers just read the text in classes; and one of the physics teachers of my friend told the students to read the text...

And now I have a worse lecturer in computer: he talks about anything except computer during class. He jokes and jests almost the whole class (3 hours!), but he remembers what he has said to us, and he doesn't tell a thing twice. I think he is better to be a comedian than an assistant professor. Actually, I don't know why such person could get a PhD and become a teacher in university.

Why does dressing come into account??? :huh:

@1=2 - Dressing is important, although not [i]that[/i] important, there is still a place for it. Essentially it fits into teacher as role model.

With regards to bad teachers, I have just had a teacher set a third of the sylabus as personal study in class time using the textbooks. This is the only time in my entire life that I have actually considered something bad teaching, and I walked out (although only after trying the proper route 1) Asking him 2) Writing a letter to the school via parents )

I have seen some very poor teachers in my time, but all of them have gained my respect for managing to tolerate the classes which I am in.

A good teacher should find ways to teach concepts that allow students to explore on their own. If a new mathematical concept is best learned by trying a certain problem, the teacher should have the student explore that problem before teaching it in class. For example, a teacher might ask students to experiment with values of $ 5^n$ (for integer $ n$) mod 7 and similar examples before teaching Fermat's Little Theorem.

a good teacher by my definition would be the one who would really teach you not just do something 'by the book' and then turn around and walk away...

[quote="ZhangPeijin"]
[b]Bad Teacher[/b]
Teaches using textbook all the time, reads the text during class, assigns homework, doesn't do anything else.[/quote]
That's my geometry teacher.
Exactly why I hate her.

She assigns worksheets copied from the book company, she makes us memorize theorems (that isn't bad) and the NUBMERS THAT THEY APPEAR IN THE BOOK.
and if you don't word a theorem word by word correctly, she takes points off.

Last year, I had a fantastic Algebra teacher. Sure her handwriting was a bit hard to read, but she had a good personality. She made quizzes herself, she makes the word problems herself, and she was a student-friendly teacher.

Wow no offense but some of you guys seem to have bad school districts....Mine is awesome number 1 in the state at math...number 15 at comm arts...10 state championships in sports in 8 years. My teachers are awesome BTW :D

for me, if i get a bad grade (A- or lower  :rotfl:) the teacher is bad.
If i get an A, the teacher is good.

I really can't decide whether or not a teacher is good or not. Whenever I feel annoyed with a teacher, I think about what I would've done, and I usually can't think of anything that would be much better... I think teaching math is a pretty hard job. Especially when you don't know what you're talking about. Like most math teachers for me so far. But it's hard even if you do know what you're talking about.

My most basic qualification for a good teacher: throwing the textbook into the garbage can (or at least ignoring it at much as the school district allows him/her) 2 reasons:

1). Textbooks don't teach you how to think. (and they don't have any interesting, i.e. hard, problems either)

2). We're all capable of reading through textbooks on our own. If a teacher can't go beyond the textbook, then they shouldn't be teaching the subject in the first place.

I really don't care about dress, Ipods, cellphones, etc. The best math teacher I ever had chewed gum all the time in class. My biggest question is: am I actually going to learn something in that class? If I am, then that's what matters to me.

BTW, I tend to be fairly harsh on judging my math teachers, if you haven't already realized.

Hmm. It seems that my calculus teacher fits $ \frac {99}{100}$ of those qualities.

Your calculus teacher?!?!??!?!?!?!??! More reminders of my stupidity.  :(

[quote="piggyinc"]2). We're all capable of reading through textbooks on our own. If a teacher can't go beyond the textbook, then they shouldn't be teaching the subject in the first place.
[/quote]
that is EXACTLY the point I want to get across to my math teacher.

Oh. and She uses me. I was explaining something privately to my friend and she butts in and tells me to let her teach. I was like wt*? cuz she wasn't explaining anything else. and when she wants me to help her she expects me to go fully elaborating.

and she uses the tests from the textbook company. and she doesn't even know how to solve them herself.

so she took off some of the questions on the test. it's really stupid.


The question is 'what is 2 to the power of pi?' I know the answer, I used the calculator, but how do I solve it with paper and pencil and with no calculator?

$ \pi$ can be written as $ \frac{22}{7}$ so the question is $ 2^{\frac{22}{7}}$ which is equivelent to $ \boxed{\sqrt[7]{2}^{22}}$

It doesn't work like that. We know that $ \pi$ is irrational so this would actually be $ \lim_{x \to \pi}{2^{x}}$. Type 2^Pi [url=http://www.wolframalpha.com/]here[/url].

Thanks, I appreciate the help, but I still not sure that I understand. I am a freshman in high school, and I haven't quite learned everything in those solutions that you gave me. Could you explain please?

What [b]SuperNerd123[/b] gave you was an approximation to the real value.

$ \lim_{x\to\pi}{2^{x}}$ means the value of $ 2^x$ as $ x$ approaches $ \pi$. As $ x$ becomes closer and closer to $ \pi$, $ 2^x$ becomes closer and closer to the value of $ 2^\pi$. In this case, we have $ \lim_{x\to\pi}{2^{x}} \equal{} 2^\pi$ because $ 2^x$ is a continuous function. You will learn all about this when you study calculus. (We can never represent $ 2^\pi$ in a more exact form because it is irrational.)

For those who know Taylor Series (Calculus), we can also view $ 2^{\pi}$ as $ e^{\pi \ln 2}$, where $ \ln (2) = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots$ and $ e^y = 1 + y + \frac{y^2}{2!} + \frac{y^3}{3!} + \frac{y^4}{4!} + \cdots$, with ${ y=\pi \ln 2}$. But since this series for $ \ln (2)$ converges so slowly, it isn't very practical.

[quote="gauss202"]For those who know Taylor Series (Calculus), we can also view $ 2^{\pi}$ as $ e^{\pi \ln 2}$, where $ \ln (2) = 1 - \frac {1}{2} + \frac {1}{3} - \frac {1}{4} + \cdots$ and $ e^y = 1 + y + \frac {y^2}{2!} + \frac {y^3}{3!} + \frac {y^4}{4!} + \cdots$, with ${ y = \pi \ln 2}$. But since this series for $ \ln (2)$ converges so slowly, it isn't very practical.[/quote]

All of these formulas mentioned in this quote box are important
in more than one area of mathematics.
Suppose you want the value of $ 2^{\pi}$ correct to two decimal
places.
If a value of ln(2) has been attained to 0.693, and if a value of $ \pi$ is
known to a few decimal places, then $ \pi$ln(2) > 2, then about
10 terms need to be calculated and summed together.  As the
exponents get larger, this reasonably takes more time each time
to calculate them by hand (this thread topic).

Instead, suppose you look at $ 2^{3.141}$ for an approximation to
be correct to two decimal places (not the rounded value).

$ 2^{3.141}$

= $ 2^3 \cdot 2^{.1} \cdot 2^{.04} \cdot 2^{.001}$

= $ 8[e^{.1ln(2)}][e^{.04ln(2)}][e^{.001ln(2)}]$

Each of three of the exponents inside the square brackets
become a y-value in the $ e^y$ formula, during "its turn."

The following should cut down on the number of terms:

y = .1ln(2) ~ 0.0693 . . . . . . up to and including the $ y^4$ term in the formula
y = .04ln(2) ~ 0.02772 . . . . up to and including the $ y^3$ term in the formula
y = .001ln(2) ~ 0.000693 . . up to and including the $ y^2$ term in the formula

Then multiply 8 by these three results.


Given a right triangle with $ \angle{A}\equal{}90^\circ.$Let $ M$ be mid point of $ BC$, denote inradii of $ \triangle{ABM},\ \triangle{ACM}$ by $ r_1,\ r_2$ respectively. Find the range of $ \frac{r_2}{r_1}.$

[hide]Let $ s_1$ and $ s_2$ denote the semi-perimeters of triangles $ ABM$ and $ ACM$ respectively. Then we have from $ (ABM) \equal{} (ACM)$ that $ r_1s_1 \equal{} r_2s_2$

Or $ r_1(AB \plus{} BM \plus{} AM) \equal{} r_2(AC \plus{} CM \plus{} AM)\newline\implies r_1(AB \plus{} BC) \equal{} r_2(AC \plus{} BC)$

Hence $ \frac {r_2}{r_1} \equal{} \frac {AB \plus{} BC}{AC \plus{} BC} \equal{} \frac {\cos B \plus{} 1}{\sin B \plus{} 1}$.

Now, as $ 0 < B < \frac {\pi}{2}$, $ 0 < \cos B < 1$. Also note that as $ \cos B$ increases, $ \sin B$ decreases and vice versa. Thus

$ \frac {0 \plus{} 1}{1 \plus{} 1} < \frac {\cos B \plus{} 1}{\sin B \plus{} 1} < \frac {1 \plus{} 1}{0 \plus{} 1}$

$ \implies \frac 12 < \frac {r_2}{r_1} < 2$[/hide]

pls tell me how does AB+BC divided by AC+BC becomes cosB+1 divided by sinB+1???

sorry...i dunno that....

$ \frac {r_2}{r_1} \equal{} \frac {AB \plus{} BC}{AC \plus{} BC} \equal{} \frac{\frac{AB}{BC}\plus{}1}{\frac{AC}{BC}\plus{}1}\equal{}\frac {\cos B \plus{} 1}{\sin B \plus{} 1}$.


A $ 4 \times 3 \times 2$ solid is painted green on all 6 faces and then cut into
24 unit cubes. One unit cube is randomly selected and rolled. Its
top face is green. What is the probability that its bottom face is also
green? Express your answer as a percent.

If we think about the question, we realize that no cubes have green on the top and bottom face, so our answer is $ \boxed{0\%}$. WOW......

I put 0% and it said that I was wrong. Then they put 0 and they were right even though it asked for percent.

If we think, then no green cubes have green on the top and bottom face, so our answer is 0%.


Hey guys, I got a sudden urge to try for USNCO next year. How would one go about practicing for it? Is there a difference between competitive chemistry and school chemistry?

Try the [url=http://www.artofproblemsolving.com/Forum/index.php?f=330]chemistry forum[/url] ;)

Oh, hehe, forgot about that....I just remembered seeing things like USNCO and USACO in this forum once upon a time...


Prove that the number $99999+111111\sqrt{3}$ cannot be written in the form $(A+B\sqrt{3})^2$, where $A$ and $B$ are integers.

$ 99999\plus{}111111\sqrt{3}\equal{}(A\plus{}\sqrt{3}B)^2$
$ \Rightarrow 99999\minus{}111111\sqrt{3}\equal{}(A\minus{}\sqrt{3}B)^2$
But $ 99999\minus{}111111\sqrt{3}<0$ tradition.
So hasn't $ (A,B)$ satisfy the equality.

[hide="Another One-Line Solution"]
Suppose that $99999+111111\sqrt 3$ can be written in the form $(A+B\sqrt 3)^2$.  Then \[99999+111111\sqrt 3=A^2+2AB\sqrt 3+3B^2\implies 111111=2AB,\] contradiction.[/hide]


geometry
inequalities
inequalities unsolved

errr... what is $E$? the area of the triangle? :D  :D

[quote="nickolas"]Let  x,y,z>0 .In any  triangle ABC  show  that $\frac{xa^2}{y+z}+\frac{yb^2}{z+x}+\frac{zc^2}{x+y}\geq2E\sqrt{4-\frac{2r}{R}}  $[/quote]

To solve this problem we must use the lemma:
$ (\sum\ tan\frac{A}{2})^2 \geq\ 4-\frac{2r}{R} $
This lemma was posted in mathlinks recently.

Apply this lemma and the identity $\sum\ tan\frac{A}{2} = \frac{4R+r}{p} $, we have to prove the inequality :
$\frac{xa^2}{y+z}+\frac{yb^2}{z+x}+\frac{zc^2}{x+y}\geq\ 2r(4R+r) $
Use the identity $ ab+bc+ca=p^2+4Rr+r^2$, our inequality is equivalent to :
$\frac{xa^2}{y+z}+\frac{yb^2}{z+x}+\frac{zc^2}{x+y}\geq\frac{2(ab+bc+ca) -a^2-b^2-c^2}{2} $
$\frac{xa^2}{y+z}+a^2+\frac{yb^2}{z+x}+b^2+\frac{zc^2}{x+y}+c^2\geq\frac{(a+b+c)^2}{2} $
$(x+y+z)(\frac{a^2}{y+z}+\frac{b^2}{z+x}+\frac{c^2}{x+y})\geq\frac{(a+b+c)^2}{2} $
The last inequality is clearly true based on BCS's inequality.


Hi, I dont own a credit card..However I happen to have gift certificates at amazon and barnes and noble.. can i buy books at aops directly from this site?

Also, I have observed that the AOPS books in amazon are ridiculously overpriced.Is it possible for you to sell the books on amazon directly?

This [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=244861]topic[/url] might help.


superior algebra
superior algebra theorems

a first course in abstract algebra John B. Fraliegh 6th edition

has anyone read this book? if so what do you think of it? im trying to figure out if this is a good book to begin with.. any comments appreciated.

The book is great for an introduction (and by that I mean one of favorites), but doesn't cover some of the deeper concepts. The homework problems are good for getting a grasp of the ideas, but not too good for developing more sophisticated understanding. I would recommend this book if this is your first exposure to algebra, then after you get the basics, you should pick up a different book. I would recommend Hungerford. (And skip the chapter on homology)

perfect, thats what i was hoping to hear. i am a beginner. thank you.


vector
trigonometry
calculus
derivative
calculus computations

The position vector of a particle is r(t). Find the requested vector.

The velocity at t =45 (pi/4)  for r(t) = 4sec2(t)i - 6tan(t)j + 7t2k

do I need to calculuate the velocity first?

So $ \textbf{r}(t)\equal{}4\sec^{2}{t}\textbf{i}\minus{}6\tan{t}\textbf{j}\plus{}7t^{2}\textbf{k}$?  And you need the velocity at $ t\equal{}\frac{45\pi}{4}$?  Then yes, you need to find the velocity vector first, which is just the derivative of the position vector.

Well the question is to find the requested vector. How do we found that, what's the start ? and sorry t= pi/4

[hide="solution"]

$ v(t) 4(2\sec(t) \cdot \sec(t)\tan(t))i \minus{} 6\sec^{2}(t)j \plus{} 14tk$

With $ t \equal{} \frac{\pi}{4}$ we have

$ 4 \left(2 \left(\frac{2}{\sqrt{2}}\right)^{2} \right) i \minus{} 6\left(\frac{2}{\sqrt{2}}\right)^{2}j \plus{} \frac{14\pi}{4}k$
$ \equal{} 4(4)i \minus{} 6(2)j \plus{} \frac{7\pi}{2} k$
$ \boxed{ 16i \minus{}12j \plus{} \frac{7\pi}{2} k}$
[/hide]


for any of u chess phenoms out there....i need a decent solid(maybe unorthodox) opening for black against e4...any suggestions?? (list ur USCF or FIDE rating if u have one)

I dont know how efficient/useful it is these days, but when I used to play (represented my county at states, won some tourneys, and such), I used to love the black on white matchup using the Vienna. Its like a hybrid of defense and offense. A guy down here (I think hes a master, he'll probably be a gm in a few years if he isnt already) who has been playing for quite a long time and really really knows the game tells me that Vienna is useless nowadays, but eh. I guess it depends on what type of level you are playing at.

Hm. So many e5 players. How dull. I play the Dragon, which is about as far from a solid opening as you can possibly get. :) If you like boring games and want to play solidly, play the Caro-Kann. If you like interesting games and want to play solidly, play the French.

yea the dragon can get very dangerous for either side...one theoretical mistake and ur done....one of my chess phenom friends in ky knows the dragon 20 moves deep..pretty amazing...that's y he's rated 2100  :lol:

I know theory to nearly 20 moves (actually more than 20 in some lines), but I'm only 1700.

My teacher would teach me common openings and then teach tricks to trap my opponent.

I prefer French Defense. It is a very dynamic defense in a very French way...
On the other hand, Caro-Kann does not faire very well in tournaments(just my opinion) as the result of Leko-Kramnik match shows.  ;)

the copycat strategy
typically, a few moves into the game and he will realize you're copying him and he will make a move that, if you copied, would be horrible for you but if you did something else, he would be wasting tempo
it starts with the common e4 knights move, d4, and then he starts seeing that ur copying him
i won against a really high rated guy using that

[url=http://turbulence.org/spotlight/thinking/chess.html]One answer[/url]

(Make e2-e4 moe to find out)

I play the French, but I used to play the Scilian Najdorf/Schievengan(i think that's how you spell it). The French is a pretty solid opening, with queenside expansion and sometimes even an attack on the kingside. For a good example of a French attacking game, look at GM Akobian's 2004 World Open game against Wojetickew (i think, it was one of his later games). By the way, my USCF rating is about 1800.  :)

nice...mine is 1400 but according to GM Goldin....i play like a 1800  :lol: ...i keep on losing to these 1200 freaks that play these random moves that are actually good  :)

i used to play najdorf too...a6 can be pretty effective..then switched to dragon..ughh too much theory...then caro-kann ...not a good reputation and kinda slow...for french btw...what line do you play..and how do you usually develop the queen bishop??

copycat strategy sucks....Scrambled...when u say u've beaten high rated ppl, how high?? a decent player wouldn't fall for that

i [u]always[/u] play e5.  i don't know why. i like e5.  my ratings 1380 something.  i used to be the top 100 in the country...not anymore...i was when i was younger.  not now...yeah, there's a tournament next week. are you guys going?

I'm a boring e5 player. I should probably switch to something like the French, but I never get around to it. And as for losing to low rated players - I'm the same, I've lost to people rated 900 before, while I've gotten draws and one win against players rating in the 2100s and 2200s..

tru e5 is friggin boring...but it is solid and leaves room for tactical play and less theory...king's gambit can get pretty exciting and ruy lopez is also interesting...four knights ugghhhh boring crap

if any of you guys know, i went to a chess tournament 2 or 3 weeks ago. i got first place in a quad :)  and you guys could check my games online somewhere. i don't know which internet website though.

[color=green]When I first learnt how to play chess, I would copy whatever the other person did, but now my favourite openings to use are either one of the Sicillian openings or Caro-Kann.

Something else that's always really fun, but you can usually only do this on a beginner, is to try and pull a Scholar's Mate on them. :D[/color]

I play Reti for white and Alekhin for black. My rating is ~1850.

[quote="JasonWu"]
copycat strategy sucks....Scrambled...when u say u've beaten high rated ppl, how high?? a decent player wouldn't fall for that[/quote]
1500 i think

[quote="Scrambled"]i dont know how high but hes about as good as the 2nd highest rated guy in California[/quote]

Hmm...isn't that Simon?...I believe he was at the top of the CA list...

If it IS, and even if it ISNT...
It's extremely difficult to believe you could get anywhere copying them...I would expect the 2nd place of CA (among students) to definitely have 1700+

And no 1700 player would let you win by copying...even if your rating is like 2300 or something, copying is still copying...

The only way is if he/she purposely let you win...

[quote="ComplexZeta"]Hm. I was never anywhere near top in California. I'm just a lowly 1708 who might be about 2000 (almost certainly not better than that) if I were to play more often.[/quote]

Okay well then I'm getting incorrect info  from somewhere...but it doesn't matter.

The point is i dont think 2nd place in CA would fall for that.

OOooooooooooooooooooppppppppppsssssssssss!!!
i meant kid still in 8thgrade-

[quote="dts"]Okay, I suppose I'll go ahead and make a fool of myself here.

Usually, when I play as white, I open with e4.  The other day, someone who was about as good as I was countered with c5.  This made me nervous, because I know no Sicilian theory, and I wondered if my opponent had been studying.  So I tried something unusual:

1. e4      e5
2. b4?!

I dub this the "insanity gambit".  Although I actually won that game...
Anyone heard of this opening before?  Has it been analyzed?[/quote]

After fixing that little typo there where you put e5.. :) It does have a name, of course, you have to go a bit further into a game than 1.5 moves to find something which hasn't been analysed. Its called the Sicilian wing gambit. Doubt its any good; but on the other hand, playing a move like this can often cause the other player a lot more troubles than you, especially in a timed game.

i play in the uscf and my rating is 1500 but i rarely get to play in it.  However online i'm about 2100.  I love using the french defense

i was 2200...my rating is out of date now but against e4 i always play gambits...Icelandic Gambit if they take it...e4 d5 exd5 e6...if u have any questions email me or pm me

if anyone wants to play a game or two i'd be happy to play online someplace...i've been out of the chess world for too long

i don't think e5 is boring.... it fights back for space in centre directly!

[quote=hannahptl]Why does this poll have only 2 votes?[/quote]

The question should be:why are you bumping this?

The question should be: Why did you notice?


Given $ a, b, c > 0$. Prove that: $ \frac{a}{\sqrt{ab\plus{}8c^2}}\plus{}\frac{b}{\sqrt{bc\plus{}8a^2}}\plus{}\frac{c}{\sqrt{ca\plus{}8b^2}} \ge 1$  :)

[quote="nguoivn"]Given $ a, b, c > 0$. Prove that: $ \frac {a}{\sqrt {ab \plus{} 8c^2}} \plus{} \frac {b}{\sqrt {bc \plus{} 8a^2}} \plus{} \frac {c}{\sqrt {ca \plus{} 8b^2}} \ge 1$  :)[/quote]
By Holder we obtain: $ \left(\sum_{cyc}\frac {a}{\sqrt {ab \plus{} 8c^2}}\right)^2\sum_{cyc}a(a \plus{} 2b)^3(ab \plus{} 8c^2)\geq(a \plus{} b \plus{} c)^6.$
Hence, it remains to prove that
$ (a \plus{} b \plus{} c)^6\geq\sum_{cyc}a(a \plus{} 2b)^3(ab \plus{} 8c^2),$ which is equivalent to $ \sum_{cyc}(a^6 \plus{} 5a^5b \plus{} 6a^5c \plus{}$
$ \plus{} 9a^4b^2 \minus{} a^4c^2 \plus{} 30a^4bc \plus{} 8a^3b^3 \minus{} 4a^3b^2c \plus{} 12a^3c^2b \minus{} 66a^2b^2c^2)\geq0,$ which is obviously true.

[quote="nguoivn"]Given $ a, b, c > 0$. Prove that: $ \frac {a}{\sqrt {ab \plus{} 8c^2}} \plus{} \frac {b}{\sqrt {bc \plus{} 8a^2}} \plus{} \frac {c}{\sqrt {ca \plus{} 8b^2}} \ge 1$  :)[/quote]
My hint is to apply Holder's Inequality as follows:
$ \left( \sum \frac{a}{\sqrt{ab\plus{}8c^2}}\right)^2 \left[ \sum a(a\plus{}b)(ab\plus{}8c^2)\right] \ge \left( \sum a\sqrt[3]{a\plus{}b}\right)^3.$


Can I post here asking for clarification on a problem? If not, how do I get it?

You cannot ask for a "clarification" of a question while the round is in progress.

If you feel that a question is ambiguous, then (in your solution) you should state any assumption that you are making to rectify your perceived ambiguity.

However, I should state that all of the problems on the round were read by at least 6 different mathematicians, and none felt that any of the questions were ambiguous, so please read the problems again and ask yourself what the most reasonable interpretation is.

i am not doubting that they are very thoroughly read over nor am i doubting that i think that any of them are ambiguous, i just have a general question.

What if a problem is ambiguous to you and you write your assumption down in your proof and you have perfect solution but your assumption is wrong. can you still get a 5 in this case?

i don't know if that can be answered during the test.

[quote="peregrinefalcon88"]i do not answer the meanig in Q5 "the expected value" and i dont understand why definition cannot be clarified for its not hinting us towards the answer in an illegal way we should have a right to know the definitions :)  :mad:[/quote]You are allowed to look things up!

[quote="cognos599"]What if a problem is ambiguous to you and you write your assumption down in your proof and you have perfect solution but your assumption is wrong. can you still get a 5 in this case?[/quote]Read [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=233027]this thread[/url].

thats a good idea to look things up well if anyone sees  this posts tell me any math dictionaries u no about and wut not if ur allowed to if not dont worry im still looking on my own  :)

Heh...these threads refer to each other...

Anyway, I had a clarification question and I sent it through e-mail. But then I realized that...it's pretty obvious.......so.....disregard what I sent.

[quote="peregrinefalcon88"]is question 5 in radians or degrees? and if u cant tell me its ok i have a way of working around it but itll just save me some trouble if i dont have to.

plz go to the other topics section of the forum to vote for president in a poll to see the preference of math nerds.:D[/quote]

Let me just say this. If you had actually looked into the problem and thought about it for a while, you would not have asked this question.


Let's define sequence of positive numbers
\[ a_{i} : \equal{} x_{1} \cdots x_{i} \textrm{, \ \ } i \equal{} 1, 2, \ldots, n \plus{} 1 \textrm{.}
\]
Given inequality becomes
\[ \frac {1}{a_{1}} \plus{} \frac {a_{1}^{2}}{a_{2}} \plus{} \frac {a_{2}^{2}}{a_{3}} \plus{} \ldots \plus{} \frac {a_{n}^{2}}{a_{n \plus{} 1}} \geqslant 4 \left( 1 \minus{} a_{n \plus{} 1} \right) \textrm{,}
\]
or equivalently
\[ \frac {1}{a_{1}} \plus{} \frac {a_{1}^{2}}{a_{2}} \plus{} \frac {a_{2}^{2}}{a_{3}} \plus{} \ldots \plus{} \frac {a_{n}^{2}}{a_{n \plus{} 1}} \plus{} 4a_{n \plus{} 1} \geqslant 4 \textrm{.}
\]
Applying AM-GM inequality gives
\[ \frac {a_{i}^{2}}{a_{i \plus{} 1}} \plus{} 4a_{i \plus{} 1} \overset{AM\minus{}GM}{\geqslant} 4a_{i} \textrm{, \ \ } i \equal{} 1, 2, \ldots, n \textrm{,}
\]
so remains to prove
\[ \frac {1}{a_{1}} \plus{} 4a_{1} \geqslant 4 \textrm{,}
\]
which is valid according to AM-GM inequality again.

Equality holds if and only if
\[ x_{i} \equal{} \frac {1}{2} \textrm{, \ \ } i \equal{} 1, 2, \ldots, n \plus{} 1 \textrm{.}
\]
Q. E. D.

Why you post inequality problem at this forum?


AMC
AIME
AMC 10
AMC 10 A
AMC 12
geometry
number theory

Intro to Number Theory would be probably too easy for you if you have done AoPS Vol. 1. But AoPs Vol. 1 doesn't cover linear congruences, but that is covered in AoPS Vol. 2 along with other important number theory topics. AoPS Vol 2 is better if you are aiming to get better on the AIME.

By the way have you done all or about 95% of the problems in AoPS Vol 1? Reading is not sufficient to master the material.

I'm not currently done with the book yet. But yes, I don't leave a single problem undone before I go on to the next chapter. Would volume 2 be a bit hard for me? Should I just get intermediate algbera? Or would it be fine?

If you've done all the problems in Volume 1, then you should be ready for Vol. 2.

go with Vol.2, but remember to review AoPS vol.1 every once in a while or you'll find AoPS vol.2 quite a hassle. I know this from experience... :oops:

[quote="bulutcocuk"]I M looking for a good followup book for art of problem solving volume 1. My goal is to pass amc 10 and get a decent score on the aime. So should I go with volume 2, get intro to number theory, or first steps for mathematics olympians?[/quote]

First Steps for Math Olympians is a great book. They give great explanations and have all the topics needed to prepare. It is a high level, more AMC 12/AIME level.
It depends on how you are in the AMC 10. Vol.2 does not help that much with AMC 10/MATHCOUNTs, but does with AMC 12/AIME.
So, i reccomend the First Steps, and Vol.2 and previous tests(mock). Mastering those will give you a good chance to do well at the AIME!

It may be ideal to start practicing with real AMC 10 questions though... Volume II helps little bit for AMC 10, but it's more for AMC 12 and beyond.

Honestly, my 144 on AMC10A this year was solely from Volume I practice. If you've mastered every concept in Volume I, you should be set regarding the AMC10. Now, I'd recommend you move on to AMC12 and AIME (the last few problems on AMC12 and easy-mid level AIME questions). Vol. 2 is probably the best choice for this, although the problems in Volume II can get pretty tough.

[quote="asethi01"][quote="bulutcocuk"]Well, the thing is, I totally epic phailed on the AMC 10, because I had like no idea it was going to be hard. I thought it was like curriculum math. So I didn't get ready at all. So I got a 94.5. But my goal for next year is to pass AMC 10 and get 4-5 on AIME. The AIME part isn't really important, but it would be awesome. :D[/quote]

You have a 94.5 on the AMC 10, try the AMC 10 Problem Series class, check out the link Online Classes on the left. That'll help your score go up to the 125/130s. If you dedicate to that, you'll do much better.
THe [i]First Steps for Math Olympians[/i] is great. Keep doing that, then try to do vol.1 when you get experience with those problems.
  :)[/quote]

i am already doing volume 1, and its going pretty good. (except for geometry, which is kinda hard.)


There were two threads in which we were spamming even though they are irrelevant now. So please post/repost your experiences here :)
Me-
[hide]I know I created some noise here before it, but when I went there(extremely ill prepared as I was) I was hardly nervous. First they made us sit at the ground floor with all the guardians for an hour or two. Then I saw my friends who had their interview in the morning session coming out. Thats when I understood why I didn't get any replies to my frantic calls and smses about the interview.  :rotfl: I never imagined that the interviews would be taking soooo long.
Anyways, we went upstairs. There were 3 of my friends from CMS along with me in that interview slot. I thought we'd be taken to some sort of spooky hall, but surprisingly it was just a bedroom. And there was a young madam there. She told us to sit and wait for our turn. It was 1:30-2 then. She told us all to relax and cheer up. Then she gave us the transportation reimbursements. After that she opened her laptop and started chatting on gtalk and opened a RNBDJ video on youtube(Hell! There LAN connection was FAST) with intermittent comments like 'arrey yahan bolney ki manahi nahi hai! Thodha tez bolo, hum bhi suney....' She was really the cheerful types and greatly diffused the atmosphere and all of us sorta intermingled and started having fun cracking jokes and discussing all kinds of stupid things like the roadies interviewers and what not! All expect Ritwik...(ULTIMATE PADHAKOO! Never done anything in life except studying literally...hardly any friends or anything)
Anyways, my turn was the third. After another fellow CMSian. There were two rooms. One with bio and other w/o bio.
So the first two guys went for their interviewers while I waited for my turn. Oh, btw I was the only one in full formal dress there  :rotfl: Anyways, finally the call came about 20-30 mins later probably. I was hardly nervous then...I was actually feeling really funny and giggly. So the peon led me to the room and turned away instantly. I got confused coz there were two doors there.  :ninja: For a moment I wondered if it was a part of the test lol, then ran back and cought the peon on his way downstairs and asked which door. So I opened the door and asked 'may I come in'.  I went in gracefully and closed the door. So far so good. Then, though I'm not too sure of it, I believe I said ''Good morning ma'am, good morning sirs" at 3 pm!  :oops_sign:   Anyways, I didn't realise it at the moment. I went and stood behind the chair with a big folder of certificates and all. They asked me to relax and sit down....[/hide]
(To be continued....Watch this space for edits around 2 1/2 hrs from now)   :icecream:

(Contd....)
[hide]So then the bio sir asked for my Self Appraisal(3 pages printouts) teacher recommendation form(I wrote it up for my teacher, my mom copied it down and sir signed it at school :p )and study certificate. Thankfully not the board marksheet  :blush: But they refused to have a look at my PILE of certificates and all.   :wallbash_red: After that the physics ma'am asked me where I was from. I told her city and school. She said, 'Oh! CMS, they have a robotics lab and stuff na?' I accidentally said, 'Yes ma'am, in LDA branch' though it is in Gomti Nagar one. Side effect of our INMO training camp in LDA branch I guess :p
Then she asked me about the subject which interests me most. I said physics, especially nuclear physics. And I am aspiring to become a great nuclear scientist and help in the development of nuclear fusion as a clean inexhaustible and economicaly viable energy sources. Then a small discussion ensued on the topic and I commented on its necessity in the wake of environmental problems and energy crisis. After that they asked me what was nuclear fusion. I told them about both nuclear fusion and fission in a lil detail. Though I accidentally said 'evolution' of energy instead of 'release'. Maths sir also initially objected when I mentioned the role of mass defect in fusion as well, but then I explained a bit more and cleared up. Somehow the word nucleon was uttered by me and physics ma'am asked me what a nucleon was. I said, nucleon refers to the particles that exist inside the nucleus of an atom. ie protons and neutrons. And basically there are twelve types of particles and anti particles. Among the particles, 6 leptons and 6 quarks- up down strange charm bottom top in increasing order of their masses. They have charges plus or minus 2/3. Then ma'am quickly smiled and said 2/3? And I said no ma'am I meant +2/3 and -1/3 And then I went on a bit about hadrons and fermions and how the nucleons came about. At that moment, the interviewers were VERY impressed. Then they asked me if I read any ther books outside the school curriculum. I said yes ma'am, I'm very fond of reading scientific boks and magazines like science reporter and am currently reading a brief history of time and origins and gave a brief description of the authors of origins. I also added I'm very lucky to have great friends and we are always discussing complex scientific questions and I'm a member of online scientific communities like art of problem solving etc.  :rotfl: [/hide]
(More later :p )

You are a real braino :omighty:

If only my interview could have been perfect like yours. :(

P.S.  your post really deppressed me.

Lol, don't be depressed. You haven't heard all of it yet. Till now its only been the sundarkand....its here on things begin to go awry.  :rotfl:
Wait for the next episode when I have time ;-)

welll,thats toooo toooo similar to mine..

just that we were placed in a library instead of bedroom :rotfl: and sth similar happened at the entry to inteerview room....

the physics is almost all same..only that it was first 2 b asked
and i didnt took ne certificate...and a single page write up...which i had REALLY not read :P

I posted my interview on my blog http://www.sidbee.blogspot.com/ and I have got selected for the scholarship in SB+2 stream.

@nuclear
as gr8 as b4 :P :omighty:

@mittal
kya yaar, itna depress na hua kar............

@sid
congo dude...... :)

@all
mine is at http://bestapplication.wetpaint.com/page/My+KVPY+interview
:)

@sid
read ur interview
really well described :)
and i noticed that in delhi the biscuit ritual continues for those selected :D :D :D


induction
ratio
inequalities
limit
algebra unsolved
algebra

[quote="shobber"]Two sequences $\{ a_n \}$ and $\{ b_n \}$ satisfy:
(a): $a_0=1 \geq a_1$, $a_n(b_{n+1}+b_{n-1})=a_{n-1}b_{n-1}+a_{n+1}b_{n+1}$, $n \geq 1$.

(b): $\sum_{i=1}^{n} b_i \leq n^{\frac{3}{2}}$, $n \geq 1$.

Find the general term of $\{ a_n \}$.[/quote]
It is incorrect problem. Let $a_n$ any strongly  decrease sequence. From a) we have $b_{n+1}=\frac{b_{n-1}(a_{n-1}-a_n)}{a_n-a_{n+1}}$.
If $\frac{a_{n-1}-a_n}{a_n-a_{n+1}}\le 1,b_0=0,b_1\le 1$ we have $0\le b_n\le 1$, therefore (b) is true.
General term of $a_n$ not defined.

[quote="shobber"]Two sequences $\{ a_{n}\}$ and $\{ b_{n}\}$ satisfy:
(a): $a_{0}=1 \geq a_{1}$, $a_{n}(b_{n+1}+b_{n-1})=a_{n-1}b_{n-1}+a_{n+1}b_{n+1}$, $n \geq 1$.

(b): $\sum_{i=1}^{n}b_{i}\leq n^{\frac{3}{2}}$, $n \geq 1$.

Find the general term of $\{ a_{n}\}$.[/quote]

Actually shobber had made a mistake during the translation. The original question required the two sequences be positive, not arbitrary, and in this case, $a_{n}= 1$ is the only solution

Thanks for pointing out the mistake. I have editted my post.  :blush:

Time for a solution now?
Claim: $a_{n}=1$ for all $n$ integer.
Proof by contradiction: Note ${a_{i}}$ is a decreasing sequence, since:
$b_{n-1}(a_{n}-a_{n-1}) = b_{n+1}(a_{n+1}-a_{n})$
note $a_{0}> a_{1}$ and so by induction since $b_{i}$ are positive, it follows $a_{n}>a_{n+1}$. Next, in that equality, if $a_{n}=a_{n-1}$, then $a_{n+1) = a_{n}}$, and backwards so by another induction and contrapositive if $a_{n}<1$ for some $n$ then $a_{1}<1$. It remains to show this is impossible, ie it doesn't meet the second criteria.
Consider positive sequence {c_i} with $c_{0}= 1-a_{1}$, $c_{i}= a_{i}-a_{i+1}$. note $c_{1}+c_{2}+\cdots+c_{n}< 1$ for all integers $n$.
Now it is possible to see, after finding out the horrendous formulaes for $b_{i}$ in terms of $b_{0}, b_{1}$ and $a_{i}$ with induction that:
$b_{2n}b_{2n+1}= b_{0}b_{1}\frac{c_{0}}{c_{2n}}$
$b_{2n-1}b_{2n}= b_{0}b_{1}\frac{c_{0}}{c_{2n-1}}$
therefore:
$b_{1}+b_{2}+\cdots+b_{2n}\geq \sqrt (b_{1}b_{2})+\sqrt (b_{2}b_{3})+\cdots \sqrt (b_{2n-1}b_{2n})$. But note that we can vary $b_{i}$ by a linear factor and not change much, so the question actually boils down to proving that $\frac{b_{1}+b_{2}+\cdots+b_{n}}{n^{(}\frac{3}{2})}$ is unbounded as $n$ tends to infinity. So ignoring the constant stuff in the last expression: $\sqrt (b_{1}b_{2})+\sqrt (b_{2}b_{3})+\cdots \sqrt (b_{2n-1}b_{2n})$ it reduces to proving, using the previous inducted formulas, that the ratio $\frac{c_{1}^{(}\frac{-1}{2})+\cdots+c_{n}^{(}\frac{-1}{2})}{n^{(}\frac{3}{2})}$ is unbounded. assume a counterexample exists.  now we can assume that the sum $c_{1}+c_{2}+\cdots+c_{n}$ tends to $1$, because if it doesn't and limits to something smaller we can multiply it by a certain factor and then we will have a satisfactory sequence.
Next we note the very easy inequality: $(\sqrt(c_{1}+c_{2}+\cdots+c_{n}))(c_{1}^{(}\frac{-1}{2})+\cdots+c_{n}^{(}\frac{-1}{2}) \geq n^{(}\frac{3}{2})$.
I need to go now, but it's not hard to finish off from here, i'll finish posting later, basically you just use the inequality and say for arbitrarily large $m$ that if it holds, you extend it to $n$ as $n$ approaches infinity and find using easy limits that the sum $c_{1}+..+c_{m}$ is bounded by constant $1-C^{(}-2)$, which isn't true since it tends to $1$ .. i'll post the rest sometime .. anyway very nice pb.

Let $c_i=a_i-a_{i+1}$ for all $i\ge0$, so $c_0\ge 0$, $S=\sum_{i\ge 0} c_i \le 1$, and $b_{n+1}c_n = b_{n-1}c_{n-1}$ for all $n\ge1$. Then $c_i=0\Longleftrightarrow c_{i+1}=0$, so if $c_j=0$ for some $j\ge0$ then $c_i=0$ for all $i\ge0$ and so $a_i=1$ for all $i\ge0$.

Otherwise, $c_i\ne0\implies c_i>0$ for all $i\ge0$, so by induction on $n\ge0$, $b_{2n}=\frac{b_0c_0c_2\cdots c_{2n-2}}{c_1c_3\cdots c_{2n-1}}$ and $b_{2n+1}=\frac{b_1c_1c_3\cdots c_{2n-1}}{c_2c_4\cdots c_{2n}}$, whence by AM-GM, $b_n+b_{n+1}\ge 2\sqrt{\frac{b_0b_1c_0}{c_n}}$ for all $n\ge0$. Summing up, we get
\[2n^{3/2}\ge b_1+2b_2+\cdots+2b_{n-1}+b_n\ge 2\sqrt{b_0b_1c_0}\sum_{k=1}^{n-1}\frac{1}{\sqrt{c_k}},\]so by Hölder's inequality,
\[\frac{n^3}{b_0b_1c_0}\sum_{k\ge m}c_k \ge \left(\sum_{i=m}^{n-1}\frac{1}{\sqrt{c_k}}\right)^2\left(\sum_{k=m}^{n-1}c_k\right) \ge (n-m)^3\]for $n>m\ge1$. Taking $n\to\infty$ for fixed $m$ yields $S-\sum_{k=0}^{m-1}c_k = \sum_{k\ge m}c_k\ge b_0b_1c_0$ for all $m\ge1$, which is impossible since $\lim_{m\to\infty}S-\sum_{k=0}^{m-1}c_k = S-S=0$ (which is valid since $S\le 1$).

If $a_1=a_0=1$ we can easily show with induction that $a_i=1$ for all $i$ and that many choices for $\{b_i\}$ satisfy condition (b), so this is a valid sequence. Now assume $a_1<1$. I'll show $\displaystyle\sum_{i=1}^n b_i$ is eventually greater than $cn\sqrt{n}$ for any constant $c$.

Note the first condition rewrites as $b_{n+1} = b_{n-1} \cdot \dfrac{a_{n-1}-a_n}{a_n-a_{n+1}}$. Letting $c_{n-1} = a_{n-1}-a_n$ this is just $b_{n+1} = b_{n-1} \dfrac{c_{n-1}}{c_n}$, and since $c_0>0$ an easy induction shows $c_i>0$ for all $i$. We also trivially have $\sum_{i=1}^n c_i <1$ for all $n$.

Now let $b = \min (b_0, b_1, b_2)$. We can easily verify that $b_{2n} =b_2 \dfrac{c_2c_4...c_{2n-2}}{c_3c_5...c_{2n-1}}, b_{2n-1} = b_1\dfrac{c_1c_3...c_{2n-3}}{c_2c_4...c_{2n-2}}$ so by AM-GM $b_{2n-1}+b_{2n} \ge 2b\sqrt{\dfrac{c_1}{c_{2n-1}}}$. Similarly we can show $b_{2n}+b_{2n+1} \ge 2b\sqrt{\dfrac{c_1}{c_{2n}}}$. Therefore, after clearing some constant terms, it's enough to show that for any constant $c$, we eventually have $\displaystyle\sum_{i=1}^n \dfrac{1}{\sqrt{c_i}} \ge cn\sqrt{n}$.

Let $A_k = \displaystyle\sum_{i=1}^{2^k} \dfrac{1}{\sqrt{c_i}}$ and $B_k = \displaystyle\sum_{i=1}^{2^k} c_i$. By Jensen on the convex function $f(x) = x^{-\frac{1}{2}}$, we have $A_{k+1} - A_k = \displaystyle\sum_{i=2^k+1}^{2^{k+1}} \dfrac{1}{\sqrt{c_i}} \ge \dfrac{2^k}{\sqrt{\frac{B_{k+1}-B_k}{2^k}}} = \dfrac{2^k\sqrt{2}^k}{\sqrt{B_{k+1}-B_k}}$. Since $\displaystyle\sum_{k\ge 0} (B_{k+1}-B_k) \le 1$ and each of the terms in parentheses is positive, there is some $k>0$ for which $B_{k+1}-B_k \le \dfrac{1}{8c^2}$, and for this choice of $n=2^{k+1}$ we have $\displaystyle\sum_{i=1}^n \dfrac{1}{\sqrt{c_i}} > A_{k+1}-A_k = \dfrac{2^k\sqrt{2}^k}{\sqrt{B_{k+1}-B_k}} \ge 2^{k+1}\sqrt{2}^{k+1} c= cn\sqrt{n}$, as desired. Hence the only valid sequence is $a_i\equiv 1$ for all $i$.


probability
calculus
integration
geometry
rectangle
function
probability and stats

1. If the random variables X and Y have the joint pdf f(x,y) = 2e^(-x-y), 0<x<y, zero elsewhere, show that X and Y are not independent.

2. Let X and Y have the joint pdf f(x,y) = 3x, 0<y<x<1, zero elsewhere. Are X and Y independent? If not, find E(X|y).

3. Suppose a man leaves for work between 8 and 8.30 am and takes between 40 and 50 minutes to get to the office. Let X denote the time of departure and Y denote the time of travel. If we assume that these random variables are independent and uniformly distributed, fin dthe probability that he arrives at the office before 9 am.

Here are my hints for you :

1/ show that $ P(X<a\cap Y<b)\not\equal{}P(X<a).P( Y<b)$ for $ a>0,b>0$ that is sufficient
2/obviously the problem is not well posed beacause f(x,y) the integral over the rectangle  [0,1]x[0,3] is not 1 but 3/2.Ignoring this and supposing we have the right constant in f, then obviously X and Y are not independant ( you can reproduce the calculation of 1/ for example to prove it ) , to express the conditional expectation write x as a function of y with a little help of geometry ( draw the subset of the rectangle  [0,1]x[0,3]  where the f is non zero) .

3/it should easy using the same trick as in 2/, once you have noticed that you are trying to calculate $ P(X\plus{}Y<a)\equal{}normalistation cte \int_{40,50}P(X<9\minus{}y)dy$

hope it helps


Prove that the equation $k^{4}+(k+1)^{4}+(k+2)^{4}=n^{2}+(n+1)^{2}$ has no integer solutions.

[hide]
mod 9.

sum of two consecutive squares = 1, 4, 5, 7 (mod 9)

sum of three consecutive fourth powers = 2, 8 (mod 9)([/hide]


Find two least right triangles whose areas are in a given ratio and such that the two legs of the larger triangle differ
by 1.

I'm confused about what you want here...

Are you saying that we want to find the two smallest right triangles (smallest by area?) such that the ratio of their areas is some constant, $ r$ and the larger triangle has legs $ a$ and $ a\plus{}1$?  This does not quite make sense because we are simply trying to find $ b$ and $ c$ that satisfy $ a(a\plus{}1)\equal{}r\cdot b\cdot c$, which has many, many solutions all of which have the same area.

Could you clarify your question?

That's right [b]bFlat[/b], plenty of possible answers, what i am looking for is the smallest two triangles with same proportions as stated above. That too bothers me alot, until now.

In that case, I'm pretty sure there is no "smallest" set of two triangles since for any set area, $ \frac{A}2$, there will be a solution to $ a(a\plus{}1)\equal{}A$, given by $ a\equal{}\frac{\minus{}1\plus{}\sqrt{1\plus{}4A}}{2}$, which is clearly always a positive real and once we have the solution for the "larger" triangle, we can easily find a "smaller" triangle by letting $ b$ be whatever we want and $ c\equal{}\frac{A\cdot r}{b}$.  Hence, there cannot be any triangle to satisfy this with "smallest area" (if we did have one, we could always just choose a smaller value of $ A$ and then we would find a new, smaller triangle)...

Perhaps licu-an means an integer right triangle.

[hide="hint, maybe"]If the two legs of a pythagorean triangle differ by 1, then it must be primitive.  We can thus write the two legs as $ m^2\minus{}n^2$ and $ 2mn$.  These differ by 1:

$ m^2\minus{}2mn\minus{}n^2\equal{}\pm 1$

$ m^2\minus{}2mn\plus{}n^2\minus{}2n^2\equal{}\pm 1$

$ (m\minus{}n)^2\minus{}2n^2\equal{}\pm 1$

The solutions to this are given by Pell's equation:
$ (m\minus{}n)\equal{}1, 3, 7, 17, 41\ldots$ and $ n\equal{}1, 2, 5, 12, 29\ldots$

Thus $ (m, n)\equal{}(2, 1) (5, 2) (12, 5) (29, 12), (70, 29)\ldots$

The triangles associated with these five pairs are the 3-4-5, 20-21-29, 119-120-169, 696-697-985, 4059-4060-5741 triangles.  You can create more triangles this way and perhaps solve your own problem. :wink: [/hide]

I hope this is a start to what your question requires.

[b]tenniskidperson3, [/b]you are right. What i want was an integer sided right triangle ([b]PRIMITIVE ONE[/b]). THanks for the big hint, it is almost what i need . . .


[quote="The question"]The great rhombicuboctahedron....has 26 faces: 12 square faces, 6 regular octagonal faces, and 8 regular hexagonal faces. At each vertex of the great rhombicuboctahedron, one square, one hexagon, and one octagon meet. How many of the segments formed by connecting two of the vertices are inside the polyhedron? [/quote]

[hide="My apparently faulty solution..."]Faces: 26 (given)
Vertices: $ \frac {12(4) \plus{} 8(6) \plus{} 6(8)}{3} \equal{} \frac {144}{3} \equal{} 48$ (overcounting and dividing method)
Since faces plus vertices minus edges is 2 for any polyhedron (that's right, right?), there are $ 72$ edges.
There are $ \binom{48}{2} \equal{} 1128$ different segments with endpoints as vertices of the rhombicuboctahedron, so there are $ 1128 \minus{} 72 \equal{} 1056$ internal segments.

[hide="But the correct answer apparently is..."]$ \fbox{840}$

What's wrong? [/hide][/hide]

Compare your solution to the one on the AoPSWiki:
[url]http://www.artofproblemsolving.com/Wiki/index.php/1988_AIME_Problems/Problem_10[/url]

Oh, I see. Thanks a lot for the solution. Just two questions.

In this case, is every AIME/AHSME problem available on AoPSWiki?

Also,
[quote="The solution"]The number of diagonals of a square is $ \frac {n(n \plus{} 3)}{2} \equal{} 2$, ...[/quote]

I have always known the formula as $ \frac {n(n \minus{} 3)}{2}$. Is there something I'm not aware of?

It's probably a typo, as $ \frac{4(1)}{2}\equal{}2$, and $ \frac{4(7)}{2}\equal{}14$.

Um.. yeah the formula for the number of diagonals for a convex polygon is n(n-3)/2
n vertices to choose from
n-3 choices for the other point (one vertex is taken, and two adjacent vertices are excluded from possible vertices)
and divide by 2 because we overcounted :D


quadratics
number theory unsolved
number theory

Let $ n$ be an integer. Prove that if the equation $ x^2\plus{}xy\plus{}y^2\equal{}n$ has a rational solution, then it also has an integer solution.

It is enough to prove the following lemma:  if $ m^2 n$ has the above form, so does $ n$.  Can you adapt the proof given [url=http://forums.xkcd.com/viewtopic.php?f=17&t=30566]here[/url]?  (You will want to think about the primes that do or do not have $ \minus{}3$ as a quadratic residue.)


vector
linear algebra
matrix
real analysis
real analysis unsolved

In order to find streamlines for vector field, we need to solve this system of differential equations:

$\frac{dx}{dt} = -x+y+z$
$\frac{dy}{dt} = x-y+z$
$\frac{dz}{dt} = x+y-z$
where
$x(0) = a$
$y(0) = b$
$z(0) = c$

I have used Mathematica in order to find solutions for these equations and it calculated the following:
$x = \frac{2a-b-c+ae^{3t}+be^{3t}+ce^{3t}}{3e^{2t}}$
$y = \frac{-a+2b-c+ae^{3t}+be^{3t}+ce^{3t}}{3e^{2t}}$
$z = \frac{-a-b+2c+ae^{3t}+be^{3t}+ce^{3t}}{3e^{2t}}$

it is, of course, right - I have check parametric plot and it is streamline, but I think how it came to that solution? Does anybody can explain how I can solve these equations (step-by-step would be very very helpful  :roll: ) without using of Mathematica? I do not understand why there is $3e^{2t}$?

Thank you

First, diagonalize the matrix of coefficients $\begin{bmatrix}-1&1&1\\1&-1&1\\1&1&-1\end{bmatrix}$. The matrix has eigenvalues $1,-2,-2$ with associated eigenvectors $\begin{bmatrix}1\\1\\1\end{bmatrix}, \begin{bmatrix}1\\-1\\0\end{bmatrix}, \begin{bmatrix}1\\1\\-2\end{bmatrix}$ (These were chosen to be orthogonal for convenience).

This gives us three linearly independent solutions: $v_1(t)=\begin{bmatrix}e^t\\e^t\\e^t\end{bmatrix}, v_2(t)=\begin{bmatrix}e^{-2t}\\-e^{-2t}\\0\end{bmatrix}, v_3(t)=\begin{bmatrix}e^{-2t}\\e^{-2t}\\-2e^{-2t}\end{bmatrix}$.

Now, our desired solution is a linear combination $v=\alpha v_1+\beta v_2+\gamma v_3$. To match the initial condition $v(0)=\begin{bmatrix}a\\b\\c\end{bmatrix}$, we need (multiplying by the inverse of the similarity matrix $\begin{bmatrix}1&1&1\\1&-1&1\\1&0&-2\end{bmatrix}$), $\alpha=\frac{a+b+c}{3},\beta=\frac{a-b}{2},\gamma=\frac{a+b-2c}{6}$. This should be equivalent to the solution Mathematica gave you.

Thank you for reply.

Is that technique is calling Laplace transform? or I am wrong?

As I understand we derive the matrix from the system, then we calculate det(A-xE) and we get cubic equation (which is it by the way?) and then roots: 1, -2, -2 which are eigenvalues.

But then how we get eigenvectors???

I got an equation: $-x^3-3x^2+4 = 0$ so roots are indeed: 1, -2, -2.

I got eigenvectors:
$\begin{bmatrix}1\\1\\1\end{bmatrix}, \begin{bmatrix}-1\\0\\1\end{bmatrix}, \begin{bmatrix}-1\\1\\0\end{bmatrix}$
but I guess orthogonal matrix is indeed:
$\begin{bmatrix}1\\1\\1\end{bmatrix}, \begin{bmatrix}1\\-1\\0\end{bmatrix}, \begin{bmatrix}1\\1\\-2\end{bmatrix}$

Could you please, clarify a little bit, how we get further solutions by using eigenvectors?
and I have difficulies in understanding, how to substitute $\alpha=\frac{a+b+c}{3},\beta=\frac{a-b}{2},\gamma=\frac{a+b-2c}{6}$ to $v=\alpha v_1 + ...$ equation, so that I will get Mathematica solution (some brief explanation would be very helpful). In my calc. I do not get $3e^{2t}$...

The Laplace transform is something entirely different.

I didn't use any systematic method to get the eigenvalues and eigenvectors- I read them off from familiarity with matrices like this.

The space of eigenvectors associated with the $-2$ eigenvalue is 2-dimensional; I chose a pair of orthogonal vectors in that space as the basis because it made computing coefficients easier (This isn't exactly an orthogonal matrix, by the way). Your choice of basis would also work.

The general equation looks like $v'=Av$. If $v=c(t)v_0$, where $v_0$ is an eigenvector of $A$ with eigenvalue $\lambda$, the equation becomes $c'(t)v_0=c(t)\lambda v_0$, so $c'(t)=\lambda c(t)$ and $v=e^{\lambda t}v_0$ is a solution.

I'd write that Mathematica solution as $x=\frac{2a-b-c}{3}e^{-2t}+\frac{a+b+c}{3}e^t$, $y=\frac{-a+2b-c}{3}e^{-2t}+\frac{a+b+c}{3}e^t$, $z=\frac{-a-b+2c}{3}e^{-2t}+\frac{a+b+c}{3}e^t$.
Now, look at $\beta\begin{bmatrix}1\\-1\\0\end{bmatrix}+ \gamma\begin{bmatrix}1\\1\\-2\end{bmatrix}$. That should get you the first term.

Trying to put everything over a common denominator is just silly for this problem.

[quote="jmerry"]The Laplace transform is something entirely different.[/quote]
Different, yes. But maybe not all [i]that[/i] different.

Let $X(s),Y(s),Z(s)$ be the Laplace transforms of $x,y,z,$ respectively.

Apply the transform to the differential equation to get:

$sX-a=-X+Y+Z$
$sY-b=X-Y+Z$
$sZ-c=X+Y-Z$

Moving all of the unknowns to the left hand side and putting this in matrix form:

$\begin{bmatrix}s+1&-1&-1\\-1&s+1&-1\\-1&-1&s+1\end{bmatrix} \begin{bmatrix}X\\Y\\Z\end{bmatrix}= \begin{bmatrix}a\\b\\c\end{bmatrix}$

Suppose you try solving this by Cramer's rule, using determinants. The determinant of that coefficient matrix will turn out to be $s^3+3s^2-4.$ Anyone reading Alteran's post shouldn't be the slightest bit surprised to see this polynomial. In fact, there will be many connections between this method and the one outlined variously by Alteran and jmerry.

Thanks,

seems that I solved that problem completely and, Yes, it is very easy... now..  :oops:  I got exactly the same solutions as Mathematica gave me.
I have refreshed that method in my memory.


This just seems impossible to me... i have no idea where to even begin with...

James is producing and selling CD's for his rock band. When he had sold 10 CD's, his net profit was $\$$ 6. When he had sold 20 CD's, he hit his low net profit of $\$$ 4 due to increased production expenses. Once he sold more than 20 CD's, however, his profit began to grow again. If James' profit based on number of CD's sold can be modeled by a quadratic equation, what is the maximnum number of CD's James can sell to have an average profit per CD of less than $\$$ 1.20?

You're allowed a calculator for this.

Anyone knows how to do this???
Because I have no idea.

the way to go about this would be to find the quadratic equation modeling his net profit, which would be of the form $p(x) = ax^2+bx+c$, where you would need to solve for a, b, and c. they help you out by saying that the minimum profit is at $x=20$, which means the vertex of the parabola given by $p(x)$ occurs at $x=20$. knowing that the vertex of a parabola occurs at $x=\frac{-b}{2a}$, you can solve for $a$ or $b$ in terms of the other. you can then substitute this into the conditions you've been given, where you'll have two equations and have to solve for two unknowns. this will yield you the quadratic that models the profit...from there i think you can get the rest, but if any of this isnt clear, feel free to ask...

o, cool. i didn't know about that equation for the vertex of a proabala.  :(
thanks, everything is clear now.  :)


Let $ {\mathbb T}^2 : \equal{} [0,1]/\{0,1\}\times [0,1]/\{0,1\}$ be the $ 2$-torus, let $ \alpha: [0,1]\to[0,1]/\{0,1\}\stackrel{[x]\mapsto ([x],[0])}{\longrightarrow} [0,1]/\{0,1\}\times [0,1]/\{0,1\} \equal{} {\mathbb T}^2$ be the canonical loop in the first factor, and $ \beta$ be the analogously defined loop in the second factor. Further, let $ D : \equal{} [1/3,2/3]\times [1/3,2/3]$ be a disc inside $ {\mathbb T}^2$.

It is clear that $ H_1({\mathbb T}^2\setminus D^{\circ},\partial D)$ is free abelian on the basis $ \{\alpha,\beta\}$ (here we identify the loops $ \alpha$ and $ \beta$ with the homology classes the represent), and that $ H_2({\mathbb T}^2\setminus D^{\circ},\partial D)\stackrel{\partial}{\longrightarrow} H_1(\partial D)$ is an isomorphism. Therefore, we have $ H_2({\mathbb T}^2\setminus D^{\circ},\partial D)\equal{}{\mathbb Z}\{\varsigma\}$, where $ \varsigma$ is the unique element such that $ \partial\varsigma$ gives the counter clockwise orientation on $ H_1(\partial D)$.

Now, we dualize the situation to cohomology: we get dual cohomology classes $ \alpha^*,\beta^*\in H^1({\mathbb T}^2\setminus D^{\circ},\partial D)$ and $ \varsigma^*\in H^2({\mathbb T}^2\setminus D^{\circ},\partial D)$.

The problem is now to show that $ \alpha^*\cup\beta^* \equal{} \minus{}\varsigma^*$.

From $ H_*({\mathbb T}^2\setminus D^{\circ},\partial D)\cong H_*({\mathbb T}^2/D,D)\cong \tilde{H}_*({\mathbb T}^2)$ it is clear that $ \alpha^*\cup\beta*$ is a generator of $ H^2({\mathbb T}^2\setminus D^{\circ},\partial D)$, but the problem is to find the correct sign.

The only way I could solve this problem to give a very explicit representative of $ \varsigma$ and to compute $ (\alpha^*\cup\beta^*)(\varsigma) \equal{} \minus{}1$ by hand (you can find the details of this calculationon [url]http://www.uni-bonn.de/~habecker/dokumente/tutorium/ExTopIII/Sheet08.pdf[/url]) This is quite tedious, though.

Is there an easier way to determine the correct sign?

Thanks,
Hanno

[disclaimer: everything i'm saying may be pure non-sense]

i would be tempted to poincarè-dualize everything, and i'd be also tempted to say that $ PD(\alpha)=\beta^*$ and vice-versa.
now, cup product translates to intersection product (since dimension=codimension) and everything should be kind of fine, assuming that $ \varsigma^*$ is the poincarè dual of the positive generator of ${ H_0(\mathbb{T}^2\setminus Int(D)})$.


function
induction
algebra unsolved
algebra

find all function with $x,y \geq0$ reel numbers wich satisfy
$f(x+y)+f(x)f(y)=f(xy)+f(x)+f(y)$. :lol:

I'll find all $f: [0,\infty)\to\mathbb{R}$, which satisfy
[b](*)[/b] $f(x+y)+f(x)f(y)=f(xy)+f(x)+f(y)$ for all $x,y\geq 0$.

$f(1)\in\{0,1,2,3\}$:
$x=y=2$ in (*) gives $f(2)^{2}=2f(2)$ and $f(2)\in\{0,2\}$.
$x=y=1$ in (*) gives $f(1)^{2}-3f(1)+f(2)=0$.
So $f(2)=0\Rightarrow 0=f(1)(f(1)-3)\Rightarrow f(1)\in\{0,3\}$
and $f(2)=2\Rightarrow 0=(f(1)-1)(f(1)-2)\Rightarrow f(1)\in\{1,2\}$.

[u]Case 1:[/u] $f(1)=0$
Then $y=1$ in (*) gives $f(x+1)=2f(x)$ for all $x\geq 0$.
This gives $f(x+n)=2^{n}f(x)$ for all $x\geq 0, n\in\mathbb{N}$ by induction,
and then $f(1)=0$ gives $f(n)=0$ for all $n\in\mathbb{N}$.
Then $y=n\in\mathbb{N}$ in (*) gives $f(nx)=(2^{n}-1)f(x)$ for all $x\geq 0,n\in\mathbb{N}$,
and then $(2^{4}-1)f(x)=f(4x)=(2^{2}-1)f(2x)=(2^{2}-1)^{2}f(x)$ for all $x\geq 0$.
So $15f(x)=9f(x)$ and $f(x)=0$ for all $x\geq 0$.
This gives solution $f\equiv 0$ of (*).

[u]Case 2:[/u] $f(1)=2$
Then $y=1$ in (*) gives $f(x+1)=2$ for all $x\geq 0$. So $f(x)=2$ for all $x\geq 1$.
For $0<x<1$ we have $1\leq{1\over x}\leq x+{1\over x}$ and $y={1\over x}$ in (*) gives $2+f(x)2=2+f(x)+2$ and $f(x)=2$.
$x=y=0$ in (*) gives $f(0)^{2}=2f(0)$ and $f(0)\in\{0,2\}$.
This gives solutions $f\equiv 2$ and $f(x)=\{\begin{array}{cc}0&\text{for }x=0\\2&\text{for }x>0\end{array}$ of (*).

[u]Case 3:[/u] $f(1)=3$
Then $y=1$ in (*) gives $f(x+1)=3-f(x)$ for all $x\geq 0$.
Then $f(2)=3-f(1)=0$ and $f(x+2)=3-f(x+1)=f(x)$ for all $x\geq 0$.
But then $y=2$ in (*) gives $f(2x)=0$ for all $x\geq 0$, which contradicts $3=f(1)=f(2\cdot{1\over 2})$.

[u]Case 4:[/u] $f(1)=1$
$y=1$ in (*) gives $f(x+1)=f(x)+1$ for all $x\geq 0$.
Then $f(x+n)=f(x)+n$ for all $x\geq 0$ and $n\in\mathbb{N}_{0}$ by induction,
and then $f(1)=1$ gives $f(n)=n$ for all $n\in\mathbb{N}_{0}$.
Then $y=n\in\mathbb{N}_{0}$ in (*) gives $f(nx)=nf(x)$ for all $x\geq 0$ and $n\in\mathbb{N}_{0}$,
and then also $f(rx)=rf(x)$ and $f(r)=r$ for all $x\geq 0$ and $r\in\mathbb{Q},r\geq 0$.
Then $y=r\in\mathbb{Q},r\geq 0$ in (*) gives $f(x+r)=f(x)+r$ for all $x\geq 0$ and $r\in\mathbb{Q},r\geq 0$.
$x=y$ in (*) gives $f(2x)+f(x)^{2}=f(x^{2})+2f(x)$
and $f(x^{2})=f(x)^{2}\geq 0$ for all $x\geq 0$. So $f(x)\geq 0$ for all $x\geq 0$.
And now for all $0\leq r\leq x,\ r\in\mathbb{Q}$ we get $f(x)=f((x-r)+r)=f(x-r)+r\geq r$.
So $f(x)\geq \sup\{r\in\mathbb{Q}: 0\leq r\leq x\}=x$ for all $x\geq 0$.
If $x<f(x)$ for some $x\geq 0$, then $x<r<f(x)$ and ${x\over r}<1<f({x\over r})$ for some $r\in\mathbb{Q}_{+}$.
So assume $x<1<f(x)$ for some $x\geq 0$.
Then $x>0$ and $f({1\over x})\geq{1\over x}>1$.
So $y={1\over x}$ in (*) gives $f(x+{1\over x})+f(x)f({1\over x})=1+f(x)+f({1\over x})$
and $0<(f(x)-1)(f({1\over x})-1)=2-f(x+{1\over x})\leq 2-x-{1\over x}=-{(1-x)^{2}\over x}\leq 0$. Contradiction!
So we get the solution $f(x)=x$ for all $x\geq 0$ of (*).

There are $4$ solutions of (*) $f\equiv 0$, $f\equiv 2$, $f(x)=\{\begin{array}{cc}0&\text{for }x=0\\2&\text{for }x>0\end{array}$ and $f(x)=x$ for all $x\geq 0$.

verry nice ;thinks again olorin  :thumbup:


group theory
abstract algebra
linear algebra
matrix
vector
Ring Theory
superior algebra

Let $K$ be a commutative field.
Determine the unitary ring morphisms from $M_n(K)$ to $M_m(K)$.
----------------------------------------------------------------------------------
Perhaps there should be some tough restrictions for m and n, so that the problem is solvable. Let's find them.

Problem (*):
Let $A=\{X^n| X\mbox{ is nilpotent in }M_m(K)\}$ for $m>n$.
Then the additive subgroup generated by $A$ is  contained in $Tr0=\{X \in M_m(K)| Tr(X)=0\}$ and contains invertilbe matrices.

I'll come back on this (with a proof). Lets see how this helps the initial problem:
For $m>n$ there is no unitary morphism $M_m(K)\to M_n(K)$.

Suppose there is such a morphism $f$.

Let $X$ be a nilpotent matrix in $M_m(K)$. Then $X^m=0$, so $f(X)^m=0$, therefore $f(X)$ is nilpotent and then $f(X)^n=0\Rightarrow f(X^n)=0$, so $X^n\in Ker(f)$ for all nilpotent matrices. Therefore $X\in Ker(f)$ for all $X$ in $A$ ($A$ is the one in problem (*)).

If $X$ is invertible then $f(X)$ is invertible. So $f(J)$ is invertible for all invertible $J$ in $Tr0$, so $J$ cannot be in $Ker(f)$ (a contradiction).

For problem (*)
-----------------------------------------------------------------------------------
Let Eij be the matrix that has an 1 on the position (i,j) and 0 in the rest.
Let Fij be the matrix that has 1 on the postitions (i,i) and (i,j), -1 on the positions (j,i) and (j,j) and 0 in the rest.
It is easy to see that Eij 2 =Fij 2 =0, for i<>j.
-----------------------------------------------------------------------------------

Let G be the subgroup generated by A (the set defined in the problem's  text).
The G \leq Tr0 is trivial since every element of G is a finite sum of nilpotent matrices. Every nilpotent matrix has trace 0, so every element in G has trace 0, hence the conclusion.

To prove that it contains invertible elements, I will use the following:
Lema 1: {x*Eij | x \in K, i<>j} and {x*Fij| x\in K, i<>j} generate Tr0.
-----------------------------------------------------------------------------------
Proof of lemma 1: Let A=(ai,j)1 \leq i,j \leq n be a matrix such that TrA=0.
It is easy to see that A-a1,1F12-(a1,1+a2,2)F23-...-(a1,1+a2,2+...+an-1,n-1)Fn-1n is a matrix A' that has 0 on the main diagonal, because TrA=0.
A' is a sum of matrices from the set {xEij| x\in K, i<>j}.
The conclusion of the lemma follows.
------------------------------------------------------------------------------------

I will prove that Eij and Fij are in G. I will actually prove that they are in A.
This is equivalent to proving that the equations Xn=Eij and Xn=Fij have solutions for i<>j.

--------------------------------------------------------------------------------
For Eij:
Consider the matrix
Q=E1 2+E2 3+...+En n+1. It is easy to see that Qn=E1 n+1.
So the equation Xn=E1 n+1 has the solution Q.

I will prove that for i<>j, Ei j~E1 n+1 (i.e. there exists an invertible matrix U such as U*E1 n+1*U-1=Ei j. If we prove this, it is obvious that the equation Xn=Ei j  has the solution U*Q*U-1.

Denote by Pi j the matrix that has 1 on the positions (l,l) for all l<>i or j, on (i,j) and (j,i) and 0 in the rest.
It is easy to see that Pi j 2=I.
Pi j*Y is the matrix obtained from Y by permuting the lines i,j. Y*Pi j is the matrix obtained from Y by permuting the columns i,j.

For i<>n+1 or 1 and i<>j,
P1 i*E1 n+1*P1 i=Ei n+1
P1 j*Ei n+1*P 1 j=Ei j.
So (P1 j*P1 i)*E1 n+1*(P1 j*P1 i)-1=Ei j.

There can be found similar writings for the rest.

It can be proved that Fij~E12.

So Fij and Eij are in A for i<>j.
------------------------------------------------------------------------------------
The matrix Em 1+E1 2+E2 3+...+Em-1 m is invertible and it is in G(A).

For problem (*). It can be proved that $A$ generates $Tr0$ over $K$ (as a system of generators in the vector space $M_m(K)$ over $K$, for $Tr0$).

This is because of the lemma and the fact that $E_{ij}$ and $F_{ij}$ lie in $A$ for all $n<m$.

I wanted to prove that $G(A)=Tr0$ but got stuck when realized that if $X$ is in $G(A)$ it doesn't follow that $k\cdot X$ is in $G(A)$ for all $k \in K$, but that $k^n\cdot X$ is in $G(A)$.

So the natural problem arises:
Let $K$ be a commutative field, $n$ a positive integer. Let $N=\{x^n| x \in K\}$. When does $N$ generate $(K,+)$ as a group?

It has been a while since I have thought about this problem.

[b]Proposition[/b]: Let $f: M_n(K)\to M_m(K)$ be a nonzero ring homomorphism. Then $f$ is injective.
[b]Proof:[/b] $\ker f$ must be a two sided ideal of $M_n(K)$, and we know that there are no proper bilateral ideals of $M_n(K)$. For example by
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=15595[/url]
Therefore $\ker f=0$ or $\ker f=M_n(K)$. The second possibility fails because $f$ is nonzero. Therefore $f$ is injective.

[b]Proposition[/b]: If $n|m$, then there exists a nontrivial ring homomorphism $M_n(K)\to M_m(K)$.
[b]Proof:[/b] Let $m=kn$. Regard $M_m(K)$ as $M_k(M_n(K))$ and consider the diagonal inclusion $M_n(K)\to M_k(M_n(K))$ i.e a matrix $A$ is sent to the matrix that has $k$ blocks of dimension $n\times n$ each equal to $A$ on the main diagonal, and has $0$ everywhere else. This generalizes the canonical inclusion $K\to M_n(K): x\to xI_n$.

[b]Remark:[/b]: $\varphi_T: M_n(K)\to M_n(K)$ given by $\varphi_T(X)=TXT^{-1}$ is a ring homomorphism for every $T\in GL_n(K)$.

[b]Remark:[/b] It may be a very difficult task to find all the ring homomorphsims $M_n(K)\to M_m(K)$. Even for $m=n=1$ and $K=\mathbb C$.
Perhaps a more human chalange would be to find the $K$-algebra homomorphisms $M_n(K)\to M_m(K)$.

For the problem of finding the $K$-algebra endomorphisms of $M_n(K)$.

Let $f\in End_{K-algebra}M_n(K)$. If $f$ is nonzero, then we have proved that $f$ is injective. Being a $K$-algebra endomorphsims, $f$ is also an application of $K$-vector spaces. An injective linear application between vector spaces of the same dimension is a vector space isomorphism, therefore $f$ is a bijection. It easily follows that $f$ is a $K$-algebra isomorphism, hence an automorphism of $M_n(K)$.
By [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=40753[/url], such an automorphism is a conjugation $\varphi_T$ (i.e. $\varphi_T(X)=TXT^{-1}$) for some $T\in GL_n(K)$.

[b]Proposition[/b] Let $f: M_n(K)\to M_m(K)$ be a nonzero $K$-algebra homomorphism. Then $n|m$.

[b]Proof:[/b] This problem goes a bit into repesentation theory. Let $R=M_n(K)$.

To give a $K$-algebra homomorphism $f: R\to M_m(K)$ is the same as giving a left $R$-module structure to $K^m$. This structure is given by $r\cdot x=f(r)x$ for every $r\in R$ and $x\in K^m$. $f(r)x$ is the multiplication of the matrix $f(r)$ and $x$ thought as a column $m\times 1$ vector. Or equivalently you can think of $M_m(K)$ as the set of $K$-vector space endomorphisms of $K^m$ and then $f(r)x=f(r)(x)$.

[b]Lemma[/b] $K^n$ is a left $R$-module with the most natural action:
\[ A\cdot (x_1,\ldots, x_n)=A\cdot\left(\begin{array}{c}x_1\\\vdots\\x_n\end{array}\right), \] $A$ is a $n\times n$ matrix.

$K^n$ is a simple (has no proper submodules) left $R$-module with this action and up to ismomorphism it is the only nonzero finitely dimesnional (as $K$-vector space) simple left $R$-module.

[b]Proof of Lemma [/b] Checking that it is a left $R$-module with the given action is obvious.
For every nonzero $x$ and $y$ in $K^n$ there exists an invertible $A\in R$ such that $Ax=y$. This means that $Rx=K^n$ for every $x\neq 0$ in $K^n$. It is easy to conlcude that $K^n$ is simple.

Let $V$ be a simple nonzero $R$-module that has finite dimension $e$ as $K$-vector space. There is a morphism $f: R\to M_e(K)$ given by $f(r)(v)=rv$ for every $r\in R$ and $v\in V$. We have identified $M_e(K)$ and $=\mathcal L(V)=End_K(V)$. Because it is a morphism between rings of matrices, $f$ is injective or $0$. It is easy to see that it is not $0$ because $V$ is simple and nozero.
We define a nonzero $R$-module homomorphism $\varphi: K^n\to V$. By Schur's Lemma, it follows that $K^n$ are $V$ are isomorphic. To construct $\varphi$, notice that $K^n$ embeds in $R$ as the left sub-module of matrices that have $0$ on all the columns $2,\ldots, n$. Take $v\in V$ such that with the previous embedding there exists $x\in K^n$ such that $xv=f(x)(v)\neq 0$ ($xv$ makes sense because $x$ is now thought as a $n\times n$ matrix, not as a vector). Such a $v$ exists because $f$ is injective. Define $\varphi(x)=xv$. It is not hard to check that $\varphi$ is a left $R$-modules homomorphism and it is nonzero because $xv\neq 0$. By Schur's lemma we conclude that $\varphi$ is an isomorphism.

[b]Back to the proof of the proposition[/b]: It is known that $R$ is semisimple (every $R$-module decomposes as a direct sum of simple $R$-modules).
By assuming that $f: R\to M_m(K)$ is a nonzero $K$-algebra homomorphism, we give $K^m$ a left $R$-module structure. Because $R$ is semisimple, $K^m$ decomposes as a direct sum $V_1\oplus\ldots\oplus V_k$ of simple left $R$-modules. By the lemma, $V_i\simeq K^n$ for all $i=\overline{1,k}$, so $dim_K(V_i)=n\ \forall i$, hence $m=nk$.


Can you tell me what is (POI's site)'s language??I come from macao(a part of china,city)~~but I think many people unkown...
and then,My english is too bad,hope you can fortive~~
 :blush:

u mean at polish informatics olympiad site? I can see english there.


I understand that the formula for the sum of squares function is $r_k(n)$, and that this is the number of ways to represent n by k squares, so for example $r_2(5) = 8$, because there are 8 ways to represent 5 as the sum of two squares. However, what I don't understand is this, the formula for lattice points on a circle, given a radius R:

$N(R) = r_2(R^2)$

What is this formula trying to say? From MathWorld's description, R = radius of the circle. So for example, if we had a radius of 30, $N(30) = r_2(30^2)$, so the number of ways to represent 900 as the sum of two squares? Is this what this is asking? If so, how would we calculate that number as opposed to listing them all out.

you need to post the link to what you are asking about

Here's the sum of squares function he's talking about [url]http://mathworld.wolfram.com/SumofSquaresFunction.html[/url]

But isn't it obvious? If the coordinates of a point ON a circle (not INSIDE the circle) is $(x,y)$, then by Pythagoras we must have $x^2+y^2=R^2$, and the conclusion follows.


Show that any integer of the form $ 6k\plus{}5$ is alos of the form $ 3k\plus{}2$ but not conversely.


[quote="DaReaper"]FInd the exterior angle of regular polygon PICBAP in the polygon ABCDEFGHIJKLMNOPQRSTUVWXYZ[/quote]
What exactly are you trying to find? which exterior angle are you finding, and is it the measure that you're looking for?

[quote="DaReaper"]to be honest, i dont know, that was the question verbatim, however, if it will help you, the choices are:
[hide]52,72,60,and 108[/hide][/quote]
Oh, it's asking for a measure.

[hide="solution"]
The measure of an interior angle of a regular hexagon is $ 120$ degrees. Since the exterior angle is the supplement of the interior angle, the measure of the exterior angle is $ 60$.
[/hide]


calculus
trigonometry
real analysis
real analysis theorems

Hi, in my calculus book, for {n}={1,2,3,...}, it says that a lower bound for the sequence is 1 or any smaller number. Can u explain how any smaller number(<1) can be suitable for it?i think a lower bound may not be in the range. thanks

Its just a matter of definition.

A lower bound for a sequence {$ a_i$} is (by definition) any number $ x$ such that $ x \le a_i$ for all $ i$.
In this case $ a_i \equal{} i$ so $ x$ is a lower bound for any $ x \le 1$.

You are perhaps confusing the term "lower bound" with "infinum" or "greatest lower bound" which would be 1 in this case.

A lower bound for a sequence $ x_n$ is [i]any[/i] number $ a$ such that $ x_n\ge a$ for all $ n$. It doesn't need to be an element of the sequence and it doesn't need to be sharp (i.e., the fact that you may find a much larger number than $ a$ that is still a lower bound doesn't disqualify $ a$ as a lower bound).

Here's an example: let $ a_n\equal{}\cos 1\cdot \cos\frac12\cdot\cos\frac13\cdots\cos\frac1n.$ That is, $ a_n\equal{}\prod_{k\equal{}1}^n\cos\frac1k.$

Can you think of a lower bound for this sequence? Not necessarily the limit, not necessarily the greatest lower bound - just a lower bound that you could convince yourself of without making the proof difficult.

Because if you can think of that, that would be a major item: a decreasing sequence (which this is) which is bounded below must have a limit - even if the limit is some other number which you haven't stated yet.


What is the largest possible value of A in the cryptarithm shown, where each instance of a particular letter represents the same digit (0-9) and different letters represent different digits?

  ABC
[u]+CAB[/u]
BBDA

[hide="Hint #1"]What must B be?[/hide]

[hide="Hint #2"]Once you know B, you can set up a system of equations to find A.[/hide]

[hide="Full solution"]A+C (in the hundreds column) gives a number (B) in the thousands column. That number can only be 1 (do you see why?).  Knowing B, we can use a system of equations to find what A must be.

From the hundreds column, we know that A+C=11*.
From the ones column, we know that C+1=A.
Subtracting 1 from both sides of the second equation gives C=A-1.
Substuting the second equation into the first gives A+A-1=11
We solve the equation for A and find that A=6.

*Notice that there was a possibility that A+C in the hundreds column could have included an extra 1 from the tens column.  This would mean that A would be 9, D would be 0, and C would be 8.  However, 9+8+1 is not 11, so this solution is not valid.[/hide]


We are given twenty distinct integers b_1 ,b_2,......,b_20. The set of their pairwise sums { b_i + b_j I 1<=i<=j <=20 } has 201 elements. What is the shortest number of elements in the set of their pairwise differences
  { b_i - b_j I 1<=i<=j <=20 }?

I'm sorry, I really don't get this one. The maximum number of distinct sums is ${20 \choose 19}=190$ !

Oops, there I go assuming $i \neq j$.

The maximum number of distinct pairwise sums is actually $210$.

We want to create as many equations of the form $b_i-b_j=b_k-b_l$ as possible. An equation of the form $b_i+b_j=b_k+b_l$ creates 2 of these. Since we are missing 9 sums, we may have $b_1+b_2=b_3+b_4=...=b_{19}+b_{20}$. There are ${10 \choose 2} = 45$ "sum" equations here, creating $90$ missing differences. So we have at least $190-90+1=101$ differences.

Am I correct here? I can't help but feel slightly uneasy about the above.

i thnk: the shortest number = the least number..
i think the least numbers of piarwise distinct sum is 100
now i'm trying to complete my proof


Suppose $n \geq 4$ is an integer. Let $b_1, b_2, ..., b_n$ be non-negative real numbers such that $b_1 + b_2 + ... + b_n = 1$. Prove that

$b_1 b_2 + b_2 b_3 + ... + b_{n-1} b_n + b_n b_1 \leq \frac{1}{4}$.


Suppose that $ X$ is the union od open sets $ U_0,\cdots, U_n$ such that all homology groups $ H_k(Y)$ vanish for any intersection $ Y\equal{}U_{i_0}\cap\cdots\cap U_{i_r}$ of these open sets and all $ k>0$ (We call the open cover $ \{U_0,\cdots, U_n\}$ an acyclic cover in this case).

1. Show that $ H_k(X)\equal{}0$ for $ k>n$.
2.  If, in addition, each intersection $ Y$ is path-connected or empty, and $ n\geq 1$, show that $ H_n(X)\equal{}0$.

Thanks in advance.

I haven't worked it out in awhile, but I recall this being a straightforward induction using Mayer-Vietoris.

Yes, I see. I prove by induction and found that $ H_k(X)\cong H_k(U_1)\oplus\cdots\oplus H_k(U_n)$. But can not get 0.

Thanks.

But $ H_k(U_i) \equal{} 0$ for $ k > 0$.

That means we can apply $ H_k(Y)\equal{}0$ when $ Y\equal{}U_{i_{0}}\cap\cdots\cap U_{i_{r}}$, and $ r\equal{}0$???.

Thanks a lot.

Presumably... it should really specify in the problem, though.

EDIT: In fact, yes, that is necessary, otherwise you could take $ U_0 \equal{} X$ and satisfy the hypotheses if you are not going to allow intersections of only one set.