I don't really know why but I'm a bit tense and nervous about posting on AoPS forums...ha...heh...

Anyways, ok I was just having an unecessarily difficult emotional problem this summer and I think I'm ready to explode if I can't get help from you, the supportive fellow AoPSers.

Here are my symptoms:

First of all whenever I have the burning desire to do math, I also have the burning desire to stay away from it... which results in me doing nothing.

When I start working on a problem, my mind is wandering and there's like a white fog circulating my head as I stared blankly on the black-printed words and had a sinking feeling in my stomach. I can't concentrate.

Third, everytime I think or about to go doing math, a sinking feeling triggered and a heavy weight seems to drop into me... so I'm crushed.

I think I'm going insane but I can't help it. All I need to do is to get over all these heaviness and start doing what I love to do... math.

Or maybe I do need somebody to control me so I could do it... maybe I don't have that independentness. Like in school... when the teacher asked to do such and such for homework, I completed it to perfection without mental intrusions. I desperately need that kind of "willingness" not contradictory feelings in order to get on with math or else I'm going to be really miserable whem math contests came.

So please help me... I have had such a hard summer, emotionally. I think I will have to go to mental institutions to get it cured so I can do math.

Thank You for reading this and help! I think I feel much better now that I'm writing this. I think I'm going insane.

so yeah.

I've never heard anything like what you have described.  First, you should consider if you like math or if you don't.  Perhaps you trying too hard problems.  If this is the case try doing problems that you would consider easy.  It's good to do hard problems.  But if you aren't able to make progress doing some easy problems where you can fully answer them boosts your confidence.

You might also be putting too much pressure on yourself.  Take a deep breath.  If you don't know how to do a problem it's not the end of the world.

Also, if this is as really a huge problem as it sounds and it doesn't get better any time soon I would honestly consider going to see a pyschiatrist or something.

I hope it works out for you :)

whoa! i never heard of this before!
you really need to talk to  a counselor ( like in school...) they're the best at handling these problems. trust me, ive got problems too  ;)

I've had minor versions of those experiences before.  For me it was because the problems were too hard, and maybe I was a little tired as well.  I kept on getting frustrated from working for hours and getting nowhere, that is until I realized I needed a rest.  I think it's healthy to (perhaps temporarily) stop working on a problem after an hour or two, and it's ok to look at solutions when you're stuck.  You should always stop doing math as soon as you start feeling tired and bored.  If you're having these feelings before you actually delve into the math, maybe you need to think about how you feel about math to begin with.  I don't know... good luck ;-).

This seems stupid suggestion but trust me, it will help you.

1. Forget about math for about an hour. Don't worry if you think not doing math for a month will lower your problem solving skills. It might but it won't kill you.

2. After the rest, sit down and start with very easy problem. If you're in high school, start with Mathcounts Problems. If you're in middle school, start with elementary problems. Anyway, start with easy problems that don't hurt your brain much. Also, it's a good idea to have tic-tae-toe for a warm-up (if you need something fun that will help you concentrate).

3. With 2nd suggestion, get milk and cookie (or chips or whatever you want) and read the bit harder problem while eating. Eating will distract you from concentrating into the hard problem and therefore, it may cure the concentration problem.

4. If this is very serious, it might not be just for math. As given above, talk with someone who can help you.

5. Make sure you remember the reason for math or anything. You'll be happy if you're doing something you like. For me and many people in this site, math is something we do for fun. I don't think any of people here hates math and come here just to spend their time. Don't pressure yourself too much neither.

Hope this works.. I personally like 3rd one best. (and not a joke)

Is math the only thing this is happening with?

While I have never heard of this, the only reason that comes to mind would be that: classmates despise math, so an idea of math being unfun and uncool was etched in your mind, and this summer you broke out of it and discovered the fun in math, but still a remnant of the past remains and you are fighting out of it. Dunon thats all i can think of.

Good god Treething...

Anyways, I've definitely had your experiences!!!

I don't know about the others here, but I have experienced that white fog / burdening heaviness.

This comes out of trying problems that you don't have a good grasp while in the back of your head - you have that burning ambition to be good. You approach problems which you can't solve with ease, and you begin to question yourself, but you decide that math is definitely good and you want to pursue it - but that lingering thought doesn't go away. You feel kind of dazed like math problems really cause alot of confusion. Something like that... I had that in 7th grade.

I was horrid horrid ... horrid at math in 7th grade. I was not at all advanced. I knew nothing about problem solving. Then I began to pursue math. At first, I asked my mom's help. She helped me find good sites (One of which was AOPS!!), and she taught me. However, everytime she talked I just kind of blanked out and just daydreamed. I wanted to understand, but it was sort of difficult at first. So she would just leave me more problems and I would get more anxious. You see, I wanted to learn more than I could handle. But you must start from the basics and do many many problems. You have to find something in math that you really enjoy. This can't be something you want to enjoy - it must be something that already appeals and just doesn't let go. For me, that was MathCounts. So I got a bunch - AKA 10 years worth of MC problems that I would just work on. And I just did problems. No pressure. NO measuring up against others. I just ignored the competition and went right ahead and did problems. Those I didn't understand, I asked for help and did again and again. Then I started to participate in the forums - writing problems, trying other people's problems, doing problems I made for making problems. Plus, the competition just pushed me a little harder because I knew there was something that I could measure up against when I felt comfortable with how I understood the methods of basic problem solving.

When you are comfortable, you will no longer have white fog - you will enjoy it and you will constantly want to try more problems. Enter MC, or AMC, do a few competitions - that's always fun. Try the Getting Started, or MC forums, or Intermediate - whatever appeals. Don't ever be afraid to post and ask questions. Have fun! That's key I think.  :)  And have a nice day.

well, sometimes, when I see a very long problem, i don't really feel as though that i want to do it, so i just thingk that it has to be done and do it.  :)

I'd like to cite "How to Solve It" by Polya on this matter.

(Polya about an intelligent problem solver)"...He should enjoy and seek what seems to him simple or instructive or beautiful. He should solve problems, [b]choose the problems which are in his line[/b], meditate upon the solution, and invent new problems."
Thus it becomes clear that one should not try to be better, i.e. one must not keep in mind one's own success over others while solving problems. [b]Look for problems which intrigue and inspire you and your imagination, which you feel affect you emotionally, which you can put your whole personality into.[/b]

(Polya on progress and achievement)"In order to solve a problem, we [b]must have some knowledge of the subject-matter[/b] and we must select and collect the relevant items of our existing but initially dormant knowledge....however, it is not enough to recollect isolated facts, we must combine these facts, and their combination must be well adapted to the problem at hand"
Thus ingenuity without knowledge leads to nowhere. You can invent things without using some new knoledge up to some moment when you reach a dead end. In fact, without realizing it you use your previous knowledge while working on any problem. The unability to solve a problem indicates the absence of corresponding knowledge and experience required to solve the problem. So solve more easier problem first, then as you notice that none of those poses difficulties anymore you can try a more complex problem.

Especially in the forums, [b]do not try to solve a problem nobody has yet solved and if your real purpose is to show off with your genius[/b]. Because if this is the case, and the problem does not intrigue your mind, it is unlikely that you will solve it. Even if you manage to solve it, after some time you will notice that you completely forgot the problem and its solution.

[b]If a problem is inaccessible, just let it go[/b], you miss nothing by doing this. It means that your brain is yet to be prepared for problems of this sort. I personally encounter lots of situautions, where after I've learnt new material it suddenly strikes me how I can connect it to a problem I didn't manage to solve on the first go.

[b]Work directly on the "body" of mathematics[/b], i.e. by reading books or textbooks and asking "why" or "what if" questions about theorems, lemmas or points you can't grasp completely. By doing this it could happen that you generate a new problem or invent something new while reviewing the material, or you could make an interesting observation which could be a basis for you future scientific work...who knows! ;) (I think I answered my thread myself I've posted today...:D )

My last advice: [b]sleep as long as you can/need[/b]...and basically try to sleep a bit when your head goes round due to "information overrun"...it's essential!...you'll see the effect immediately when you come back to learning maths after a nap. Eat healthy food, do ragular exerscise and go out for a walk to tank up some oxygen.

And never, never dispair and underestimate yourself!!!

  :)

If easy problems just make you more impatient, sit down with a hard problem (but not ridiculously hard) having no expectations to actually solve it, but just play around.

Like try finding a pattern in the integers that can be expressed as the sum of two squares. (Don't try to prove, it's really hard). See if you can say something interesting about them.

A few more suggestions.  NB: I'm not giving you advice --- I'm suggesting you think about these topics and use them to take control of your work.

Physical condition affects mental condition.  Try to get good sleep patterns.  Don't pull too many all-nighters on maths or anything else.  Read something relaxing before going to sleep.  Get reasonably fit.  Go for walks or bike rides to clear your head.  Don't exhaust yourself physically and then expect to do math.  Eat nutritionally balanced meals and don't snack on too many sugary confections.  Get into good physical habits.  Get to know what affects your physical state.

Emotional state affects mental condition.  Practice relaxation.  Try light classical music.  Find a friend or mentor to discuss mathematics with.  Find a mentor to discuss strategies with.  Get into good emotional habits.  Get to know what affects your emotional state.

Mental condition.  Engage your unconscious mind.  Read into a problem long enough to 'embed' it into the unconscious.  Don't hop from one thing to another.  If it isn't working out after you've given it a good shot, decide whether to continue or do something different.  When you decide to do maths, don't exhaust yourself mentally playing Doom, surfing the 'net, watching PBS first.  Find somewhere quiet to do math with pencil notepad and without distractions such as radio, TV, computer, hifi.  Use the library or some What time of day is best for reading, studying, problem-solving, brain-storming ideas?  Get into good mental habits.  Get to know what affects your mental state.

Intellectual condition.  Read mathematics texts.  Why did the author say that at that point?  'Deconstruct' the argument.  Look for a 'plot' and 'characters'.  List various proof strategies (induction, contradiction) and see which ones were in use.  Try a simple numerical example and work it out in detail by hand to illustrate a general point.  Do the exercises in the book.   If you think you have the solution, write it out in detail.  Read with pencil and notepad by you and use them.  Carry a notebook for ideas.  Work a problem from both ends (data and solution) and try and connect them up.  Write down everything you know about the data and look for connexions.  Are you interested in structure, elegance, powerful techniques, brute-force computatations, head-on or sneak attacks?  What proofs and solutions enthuse and inspire you?  Why?  What do they have in common?  Do you learn by example, discussion, experiment or recipe?  Get into good intellectual habits.  Get to know what affects your intellectual condition.

Morale.  Anything worth doing is going to be hard.  You wouldn't expect to run a marathon without preparation and exercise.  You wouldn't expect to play a sport without coaching and practice.  You wouldn't expect to cook a perfect souffle first time.  Easy problems get you warmed up and encourage you.  Hard problems take time and effort.  Find a problem-solving group (real or virtual) for mutual support and encouragement.

I had problems with concentration on math in late high school and all through college.  I believe that it had to do with the fact that I wasn't doing math because I enjoyed it anymore and it was suddenly harder than ever to locate math that I enjoyed.  I largely quit working on math problems that didn't relate to something I was working on from about the time I was 15 until AoPS started up.

There were times when I found enjoyable problems and the fog lifted, but finding the right problems to work on -- those that made me excited about the work -- become more difficult.

Having been through that experience, I recommend finding another subject that generates that same excitement within you.  For me it was Economics/Finance, which I looked at more from a problem solving perspective than most people.  Fortunately those subjects kept me learning about some perspectives about mathematics which were new to me.

I'm sure everyone's experience is different.  Try to find something that excites you.

Try not to stress about it.  I believe that it's a perfectly normal psychological response to a feeling or environmental condition that you just haven't quite yet identified.  Maybe you'll only identify it after working on something else.  Maybe it'll happen next week.  Maybe I'm wrong and all you need to do is put math out of your mind for a week and read a book.  Either way, do something you enjoy to try to get it out of your system.

Good luck!

the symptoms you are mentioning actually happened to me when I started doing maths in year 10, up until then I was a straight A student in maths and everything else (except language other than english).

My solution was to force myself to study as hard as I could which often lead me to sleep deprivation and I kept on doing very badly in my tests (around the 70's). I was young and thought I was just dumb not thinking it had anything to do with my lack of sleep so I studied harder but always got around the 70's. THis one time the teacher gave us an extra hard test which just about eveveryone failed except me, I got 68%. Looking back I feel like I wasted all of year ten because of these simple problems.

My advice is to find out whats wrong wtih you emotionally and have a serious discussion with yourself about it, and if you ever find it hard to concentrate on your maths what you should do is talk the math question out loud with yourself, looking for possible methods to solving it etc. It helps with the concentration.

well i have similar problems lately i havent sat to solve problems like i used tu before, i mean before every day i would study for at least 4 hours that was about last year. see last year i was obssesed with math and the olympiad  and this resulted in me always working but at the end i did bad in my tsts because i would get so nervous when i would take the tests and then i went to nationals last year and i got so nervous i got sick and during my tes i even bled through my nose obviously i only got third place in nationals and i had so much high expectations that year. so what did i decide to do this year? i decided to stop doing math and just compete thats it. i mean i already know everything i need to know to win nationals my problem is i get nervous thats why i havent won in nationals. so do you think my plan worked?
 of course!! currently im have won every single competition ive been in!! and always the top scorer!! nice huh!?? so my advice is:

IF YOU GET TOO OBSESSED STOP DOING MATH!!!  JUST COMPETE!! results will come pouring in!!!  :D

oh yeah and get a girlfriend and go drink with your friends!! that always helps!!! :D

Did not you observe that jeez is FEMALE?! And she is far below legal drinking age.

Anyway, sometimes this happens to me; usually though it's in history :( or geography :o ...Sometimes, if it is just incredible tiredness or I'm just sick of it, I remind myself that it's not me I'm studying for, and I take a break. Eat some good food. Then do something else.

By the way, don't feel tense about posting. Even if you get it wrong, those of us who are civil won't laugh at you, but point out where you went wrong.

SORRY :blush:   anyway you should never be afraid of posting because if you never post if you never make mistakes how will you learn? also i think you should consider finding another hobby to clear your mind you should find some other things to do so when you finally decide to do math well you will be happy and excersised after doing something fun or something like that . anyway good luck  :)

when I was younger, I had the same symptoms.
When I was trying to work alone with a book in front of a desk, that's what I got.

Try working with other people, most likely your friends or any people at your age/math level. you might need an adult or an older/more experienced student to keep things on track (like getting problems or something). It sounds like a lot of work, but I think it pays off. It becomes much more enjoyable. All the pressure that makes you want to be lazy just go away, sort of. The atmosphere is most important, imo, to get rid of the symptoms.

But if you're like me and you really can't find people who are not interested in math around you, that's what AoPS is for. :)

And as for the working with a book in front of a desk thing, I found that it becomes easier to handle as you get older.

Just my thoughts.

This is what I think you're talking about.
When you see the problem, you think "Oh god, this is a ton of work, I don't feel like writing it all out and doing all of it, I'm too tired." So you skim over about 20 problems not liking any of them, thinking they're just brute force. Then you settle down on one, set it up, and find its very interesting and has a lot of ideas you didn't know before. Then once you look at the next one, you start thinking the same thing all over again. So you stop, but then you start thinking about the ideas in the problems and want to start again, until you see the problems, and you keep getting into this cycle. At least that's the way it is for me. This can haunt me at competitions, when I see the problem I just don't feel like doing it so I skip over it and when I come back to it I'm just like nah I can't do it, I'll leave it blank.
Does anyone else have this, and how can you fix it?


$n$ is a natural number bigger than one and $p$ is a prime number, $n|p-1$ and $p|n^3-1$. Prove that $4p-3$ is a square.

It has been discussed before and as i know it's for Moroccan mo 2004(and Iran mo) ;)

This problem also been in the $Armenian$ national olimpiad 2005 :lol:


Find all polynomials f with real coefficients such that for any positive integer n, f(2^n) is a natural power of 2. It's not a difficult problem, rather a nice one.

Let m be the degree of f.
If  for any  positive integer n , f(A^n) is a natural power of A (A > 1), then f(x) = (A^p)*(x^m) , where p is an integer , p >= -m .
........
It is trivial if m=0. Suppose now m > 0 : f(x) = c*(x^m) + g(x) , with c <>0 and degree(g) < m
f(A^n) = A^u(n) , where u(n) is natural number .
c = lim(x-->+oo){f(x)/(x^m)} = lim(n-->+oo){f(A^n)/(A^(m*n))} = lim(n-->+oo){A^(u(n)-m*n)}
But u(n)-m*n is allways an integer, so his  limit is an integer p, so there is an integer N such that
n > N ===> u(n) = m*n + p ===> c = A^p
f(A^n) = (A^p)*(A^(m*n)) + g(A^n) = A^u(n) = A^(m*n+p) ===> g(A^n) = 0 for all n > N ===> g(x) = 0
So f(x) = (A^p)*(x^m) , so f(A) =A^(p+m) ===> p+m >=0 ===> p >= -m
..... See you ...

Yeah, that's exactly the solution I gave. Anyway, I maintain my opinion that this is a simple, but nice problem.


There are several values for a prime $p$, with the property that any 5-digit multiple of $p$ remains a multiple of $p$ under "cyclic permutation".  One such value is 41 (for example, since 50635 is a mutiple of 41, so are 55063, 35506, 63550, and [0]6355); another such value of $p$ is 3.  Compute the value of $p$ that is greater than 41.

we can be pretty confident that p divides the sum of the digits of any multiple...then for some reason, we can be like p divides 10^5-1, and we can factor that using the 3 and the 41, to get 813=3*271, so the answer is 271...not rigirous, but this solves it fast, i think thats right anyway...

EDIT: the correct logic is posted above...but i would have gotten the answer correct on the individual anyway :)


function geometry algebra polynomial calculus derivative trigonometry

Inspired by [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=242582]this.[/url]

Suppose this is the state of your knowledge:

You know the definitions of trigonometric functions and their relations to circles and triangles, along with the usual basic identities. You know and can use the definition of radian measure and understand what it has to do with, say, the areas of sectors of circles. (Note: radians are actually an anachronism here, but go along with it.)

You know how to expand $ (1\plus{}x)^{\alpha}$ as an infinite series for arbitrary (or at least arbitrary rational) $ \alpha.$

You know how to integrate and differentiate polynomials, and accept as a general principle that you can manipulate power series as if they were polynomials.

You do [b]not[/b] know the derivatives of any trigonometric functions, and you do not know any important limits involving trigonometric functions. (And in answering the challenge below, you won't redevelop these before doing anything else.)

[b]The challenge:[/b] find the power series centered at zero for $ \arcsin x.$

[b]The bigger challenge:[/b] given the series you have, solve $ \theta\equal{}\arcsin x$ for $ x$ to find at least the first few terms of the series in $ \theta$ for $ x\equal{}\sin\theta.$

$ \int_0^x \sqrt{1 \minus{} t^2} \, dt$ gives the sum of the area of a sector of the unit circle and a triangle.  The triangle has base $ x$ and height $ \sqrt{1 \minus{} x^2}$, hence area $ \frac{1}{2} x \sqrt{1 \minus{} x^2}$.  The sector subtends an angle of $ \arcsin x$, hence has area $ \frac{1}{2} \arcsin x$.

The binomial theorem gives $ \sqrt{1 \minus{} t^2} \equal{} \sum_{k\equal{}0}^{\infty} { \frac{1}{2} \choose k} (\minus{}1)^k t^{2k}$, and integrating term-by-term we conclude

$ \sum_{k\equal{}0}^{\infty} { \frac{1}{2} \choose k} (\minus{}1)^k \frac{x^{2k\plus{}1}}{2k\plus{}1} \equal{} \sum_{k\equal{}0}^{\infty} { \frac{1}{2} \choose k} (\minus{}1)^k \frac{x^{2k\plus{}1}}{2} \plus{} \frac{1}{2} \arcsin x \implies$
$ \arcsin x \equal{} \sum_{k\equal{}0}^{\infty} { \frac{1}{2} \choose k} (\minus{}1)^{k\plus{}1} \frac{2k\minus{}1}{2k\plus{}1} x^{2k\plus{}1}$

(although I'm not fond of the idea of simplifying this expression).

So Newton didn't know all these things but he knew Abel's theorem and the ideas of uniform convergence  :huh:

t0rajir0u's series can be written as

$ \arcsin x \equal{} x \plus{} \frac {x^3}6 \plus{} \frac {3x^5}{40} \plus{} \cdots \equal{} \sum_{k \equal{} 0}^{\infty} \frac {1\cdot 3\cdots (2k \minus{} 1) x^{2k \plus{} 1}}{2^kk!(2k \plus{} 1)}$

Nothing fancy there - just writing out $ \binom{\frac12}{k}$ and noticing the patterns.

Uniform convergence is a mid-19th century notion (not even available to the critics of Fourier.) As for the dawn of calculus: power series were everywhere, and power series lay very near the heart of how both Newton and Leibniz motivated things. A generation or two later, Euler was the master and artist at using power series. It was calculate first, ask questions later (a century or two later).

Now the real challenge: [i]inverting[/i] the series. Yes, Newton really did do that. Hint: use an iterative process. This, not numerical analysis, is what "Newton's method" was about in the original.


I have a lot of tables with border lines insides. Since it is really difficult to code with table format in latex. I decided to make tables as figure and use\begin{figure} ....for these tables.

1. since I use\begin{figure}, caption will be shown with a figure order. It is just like: Figure2.2 ....
2. I also needs to label these figures since I will use a lot in the text. But I need to use table
(\ref{}) since they are tables in essence.
3. Problem of inconsistency comes out: In text, it is table(2.3). But there exists only figure(2.3)...How to make those figures show with table(2.3) in the caption..

I attached a sample problem.

\renewcommand{\figurename}{Table} will change Figure to Table in the captions.

I tried to put both in the front of figures and the preamble. But neither really works.
Putting in the preamble makes other real figure into tables and the numbering of tables disappear.

Putting in the figures does not work.

Where should I put it? Do I have to put it all the places where I need to reset figure into table?

You just put it [i]before[/i] the table environment, then change it back afterwards.
The problem is using independent counters. You could define a new environment to
[code]\newenvironment{tabtofig}%
 {\setcounter{figure}{\value{table}}\renewcommand{\figurename}{Table}\begin{figure}}%
 {\end{figure}\renewcommand{\figurename}{Figure}}[/code]do all the work for you. Then you put your pseudo tables inside \begin{tabtofig}...\end{tabtofig}.
You may need to play with the new environment to get it just as you'd like it.

Forgot to increment the table counter.
[code]\newenvironment{tabtofig}%
 {\setcounter{figure}{\value{table}}\renewcommand{\figurename}{Table}\begin{figure}}%
 {\end{figure}\renewcommand{\figurename}{Figure}\stepcounter{table}}[/code]

Thanks for solving it! I have been working on it for a long time today!

Another TWO questionS..

1. Those tables still show in The List of Figures. I try to show them in The List of Tables.
How to do with it?

2. Tableofcontents show
0.1, 0.2(0.2.1.0.2.2.....),0.3(....).0.4..
how to make it as 1,2(2.1,2.2,,....),3(...),4(...).

Thank you.

Actually I've just realised that there's a much simpler way of doing this - just stick your image in a table
[code]\begin{table}
\includegraphics{your image}
\caption {tablelabel}
\end{table}[/code]then this will be listed in the list of tables and you don't have to worry about mixing it up with figures.

As for 0.1 instead of 1.1 that depends on the document class you're using, how you number sections and chapters etc. So you need to give a minimal example to show the problem.

1. Can I use this \epsffile{filename.eps} instead of \includegraphics{}?
2. Document class is Report.
3. mini example

 \chapter{this is my paper's title}
\section{introduction}
\section{model}
\section{extension1}
\subsection{}
\subsection{}
\subsection{}
\section{end}

Contents show like this:
0.1 introduction
0.2 model
0.3 extension
0.3.1 ....
0.3.2..
0.3.3
0.4 end

Is there a way to make it like:
1 intro
2 model
3 extension
3.1...
3.2...
3.3..
4. end

Thank you.  I changed the code and it works well.
BTW, would you tell me how to make the table1,table2,table3...consistent with their position.
For example, if table1 is in first section, i would like it as table1.1..table 2 in first section as table1.2
table3 in section2 as table2.1  ??

And could you give me good reference for this kinda formatting problem?

1. \epsffile is positively ancient and was meant for an older version of LaTeX.
[quote]This package was developed as a general solution to the problem of including graphics in LaTeX 2.09; as such there are obsolete copies to be found on the web (though no longer on the archive). These old versions should not be used with current LaTeX[/quote]Use the more modern \includegraphics and create a pdf via DVI ie LaTeX => DVI => PS => PDF using latex, dvips, ps2pdf. Or you could convert the eps figures to pdf or other format first - see [url=http://www.tex.ac.uk/cgi-bin/texfaq2html?label=pdftexgraphics]Imported graphics in PDFLaTeX[/url].

2. That example gives 1.1, 1.2 etc not 0.1, 0.2 as the first number is the chapter. To get rid of that first number use
[code]\renewcommand{\thesection}{\arabic{section}}[/code]

You are right.
Graduate school gives a latex format for ETD, asking to use\epsffile..

So I think they should update their format..


calculus
integration
derivative
logarithms
real analysis
real analysis unsolved

Assume the validity of differentiation under integral sign.Prove that
$ \ \int_{0}^{a} \frac{log(1\plus{}ax)}{cos(x)}dx \equal{} \frac{1}{2}tan^{\minus{}1}(a)\cdot log(1\plus{}a^{2})$

Is that supposed to be $ \frac{1}{\tan(a)}$ or $ \arctan(a)$? Either way, this doesn't seem to be true..

It is $ arctan(a)$
How could you say, it is not true ?

Put $ a\equal{}1$ and you'll see it doesn't hold.

Ok,
It was a typo mistake.
Now, prove that
$ \ \int_{0}^{a}\frac{log(1\plus{}ax)}{1\plus{}x^{2}}dx \equal{}\frac{1}{2}tan^{\minus{}1}(a)\cdot log(1\plus{}a^{2})$

Looks like a single variable problem, but needs multi-variable techniques!

note $ \frac{d}{da}[1/2 \arctan a \log (1\plus{}a^2)] \equal{} \frac{1/2 \log(1\plus{}a^2)}{1\plus{}a^2} \plus{} \frac{1/2 \arctan(a) * 2a}{1\plus{}a^2}$
$ \equal{} \frac{1/2 \log(1\plus{}a^2) \plus{} a \arctan a}{1\plus{}a^2}$.

Define $ F(x,a) \equal{} \int \frac{\log(1\plus{}ax)}{1\plus{}x^2} dx$.  Now note $ \frac{\partial F}{\partial x} \equal{} \frac{\log(1\plus{}ax)}{1\plus{}x^2}$ (fundamental theorem of calculus) and $ \frac{\partial F}{\partial a} \equal{}  \frac{\partial}{\partial a}\int \frac{\log(1\plus{}ax)}{1\plus{}x^2} dx$
$ \equal{}  \int \frac{\frac{\partial}{\partial a}[\log(1\plus{}ax)]}{1\plus{}x^2} dx \equal{} \int \frac{x}{(1\plus{}ax)(1\plus{}x^2)} dx$
$ \equal{} \frac{1}{1\plus{}a^2}\int \frac{\minus{}a}{1\plus{}ax} \plus{} \frac{x \plus{} a}{1\plus{}x^2}dx$
$ \equal{} \frac{1}{1\plus{}a^2}(\minus{}\log(1\plus{}ax) \plus{} 1/2 \log(1\plus{}x^2) \plus{} a \arctan(x)) \plus{} C$

Then with FTC and then multivariable chain rule...
$ \frac{d}{da} \bigg[ \int_0^a \frac{\log(1\plus{}ax)}{1\plus{}x^2} dx \bigg] \equal{} \frac{d}{da} \big[ F(a,a) \minus{} F(0,a)\big]$
$ \equal{} \frac{\partial F}{\partial x}(a,a) \plus{} \frac{\partial F}{\partial a}(a,a) \minus{} \frac{\partial F}{\partial a}(0,a)$.
$ \equal{} \frac{\log(1\plus{}a^2)}{1\plus{}a^2} \plus{} \frac{1}{1\plus{}a^2}(\minus{}\log(1\plus{}a^2) \plus{} 1/2 \log(1\plus{}a^2) \plus{} a \arctan(a)) \plus{} C \minus{} \bigg(\frac{1}{1\plus{}a^2}(\minus{}\log(1\plus{}a*0) \plus{} 1/2 \log(1\plus{}0^2) \plus{} a \arctan(0)) \plus{} C\bigg)$
$ \equal{} \frac{1}{1\plus{}a^2}(1/2 \log(1\plus{}a^2) \plus{} a \arctan(a))$

This is the same as the derivative of the right side of the equation!

Since the two sides of the equation have the same derivative, they must differ by at most a constant.  Plug in $ a\equal{}0$ to see that the equation reduces to $ 0\equal{}0$.  Thus the constant the two sides differ by is $ 0$, so they are the same.  QED.


topology
calculus
articles
geometry
group theory

One day, I decided to find out in which places topics in Mathematics and Physics were interlinked or used to prove results in each other's topics. Most of Mathematics is applied everywhere in Physics -  from Calculus to Group Theory etc. I considered that possibly the only field which is not talked much about is the application of Topology to Physics and vice-versa. So what do you think about this? What are the possibilities for research? What has been done? Do you have any books or online articles to recommend??
Also, do you have any other possible topics in interconnection of Physics and Mathematics that are interesting??

Someone once told me that the hairy biliard ball theorem serves as an argument for no magnetic monopoles in some models.

Although I dont completley understand why.

How about the manifold geometry associated with General Relativity as a correlation to topolgy?

[quote="Rushil"]Dr. No, Can you plz explain what you meant by "Reimann"?? Plz expalin???[/quote]Did I misspell his name? ...oops...  :blush:

Riemann, was a mathematician the lion's share of whose work concerned the connection between physics and topology trying to answer the very question you posed. Einstein's General Relativity is based in no small measure on Riemann's work.

[quote="Rushil"]Can you quote some book or article of his or any other author available online or for purchase??? (I was not asking who he was- his hypothesis has made him famous enough!!!)[/quote]I can't make a personal recommendation since I haven't read it, but I did [url=http://www.amazon.com/exec/obidos/tg/detail/-/0817637893/104-5056499-0416761?v=glance]find this[/url] on the web.


AMC
AMC 12
AMC 12 A
AIME
AMC 10
ARML
conics

Well, I got a 144 on the AMC12A, due to a silly mistake on #24. (I think it was #24. It was the sin(nx) one.)

Did anyone else (on AoPS) just barely miss getting perfect, or just barely miss making the AIME? Let's talk.

There's a big difference between barely qualifying and barely missing perfect   :!:

I don't know how well this thread bodes with the whole "stop complaining about high scores" ideology.

eh....it is inconsiderate to complain about a high score...but besides that, whatever...

[quote="Fanatic"]Some people just want to vent.

When I got 5th at state last year, every time I attempted to vent, people would get really pissed at me and think I was jealous and I should "just get over it".

It screwed me over for a month. I couldn't touch math without feeling really crappy. Looking back, I'm over it now, but everyone has their right to vent to people that listen.

That's extreme, but I poured my soul into making nationals that year and it blew up in my face when I changed #8 on the target from right to wrong at the last second. I especially loved 2nd place girl going "I don't even want to make nationals".

[b]Regardless, a public forum isn't the place to vent.[/b]

haha, i'm still venting :p.

and whoever mentioned that "conflicts with no complaining about high scores" is sort of right. I didn't see any complaining in the 1st post, just listing of facts. But I can see how others could interpret that as complaining.[/quote]

I didn't get angry at you, I remember I was sorry but I could not really do anything.  I thought that you were better than I was and you didn't go, but I did.

At least you have another year.

@Everyone: [url=http://en.wikipedia.org/wiki/Terriermon]Momentai.[/url]

(144 for me but seeing how that is a 15 point improvement over last year I'm not complaining)

144 here too.

but i'm not sad in any way nor do i wish to vent. actually, i think i got quite lucky.

and, yes, i do think that this thread is pointless and that it violates the "no complaining about high scores" policy.

it'd definitely be more useful to use the AoPS forum to discuss math problems than it is to discuss scores.

discussing problems will improve your score while discussing scores will waste your time.

[quote="Fanatic"]ehehehehe:

I meant vent in real life. Anyone that chastised me for venting on a public forum was right.[/quote]

People around me get angry if I vent.  People want to be happy, not bogged down by some math geeks troubles.

[quote="not_trig"]I try to vent, but the *derogatory term* people at my school are like, WOW, HE ONLY MISSED 2 and I'm like, wow, I misbubbled one and was just stupid on the other and tied with a certain 7th grader that I know...

(well, there was one other AMC10 qualifier with a 120, but still, my school got like 150 people to take AMCs)

[/end babbling]

lol[/quote]

I sympathize with you. It's like the women's olympic person who messed up the skiing thing this year (as an analogy  :P )... people were like: oh my gods (or god) you just won a silver medal! and after, during the interviews, she was crying, because she lost the gold medal (she accidentally fell on her last turn).

[quote="Karth"]oh my gods (or god)[/quote]
 :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:

I wouldn't vent to my non-nerd friends anyways. I'd act stupid, play stupid, and yay I fit in with the stupid. Not difficult at all. I don't blame them.

[quote="Altheman"][/quote][quote="iamagenius"][quote="Altheman"][quote="MysticTerminator"]dude

who the - actually types out the frickin hyphens[/quote]

otherwise would be contradictory to my point...you just got ownd by professor subtle

but i am capable of typing [b]pussy[/b]...if it makes you feel better

BECAUSE I AM SURE IT DOES[/quote]
 :o  :noo:  :help:  :censored:    :police:  :stretcher:[/quote][quote="Altheman"]

oh come on...

if at any point the word was used without offensive meaning, that would be the time...[/quote]
 :P  :)

[quote="iamagenius"][quote="Altheman"][quote="MysticTerminator"]dude

who the - actually types out the frickin hyphens[/quote]

otherwise would be contradictory to my point...you just got ownd by professor subtle

but i am capable of typing [b]pussy[/b]...if it makes you feel better

BECAUSE I AM SURE IT DOES[/quote]
 :o  :noo:  :help:  :censored:    :police:  :stretcher:[/quote]

man you are such a [b][i]pussy[/i][/b]

[quote="MysticTerminator"][/quote][quote="iamagenius"][quote="Altheman"][quote="MysticTerminator"]dude

who the - actually types out the frickin hyphens[/quote]

otherwise would be contradictory to my point...you just got ownd by professor subtle

but i am capable of typing [b]pussy[/b]...if it makes you feel better

BECAUSE I AM SURE IT DOES[/quote]
 :o  :noo:  :help:  :censored:    :police:  :stretcher:[/quote][quote="MysticTerminator"]

man you are such a [b][i]pussy[/i][/b][/quote]

 :rotfl:   you're probably right

[quote="MysticTerminator"][quote="Karth"]I can't believe that this thread is still going on... Arnav, you can stop acting like a retard now... we all know that you're amazing and stuff, but that doesn't mean you can use such language in a public forum.[/quote]

dude I think the point of all the above was that there wasn't actually any inappropriate language

seriously some of you guys need to work on that reading comprehension business[/quote]

Irony and Oxymoron were both used. After all, where there is no smoke, there is no fire.

There is quite a contrariety, as Mr. Arnav uses Irony well; it makes him seem quite (uhh how should I say this) mum, though we all know that implies nothing.

[quote="iamagenius"][quote="MysticTerminator"]
[b]DUDE[/b]

it was not referring to a biological part

WTF MAN

being a pussy means cowardly and stuff

soo....on a related note, isn't it fun how many topics "almost-perfect scores" can lead to[/quote]
[quote="MysticTerminator"]

[b]dude[/b]

who the - actually types out the frickin hyphens[/quote]
[quote="MysticTerminator"]
[b]dude [/b]I think the point of all the above was that there wasn't actually any inappropriate language

seriously some of you guys need to work on that reading comprehension business[/quote]

[b]suffice it to say that I am a pussy and MysticTerminator is one cool dude[/b][/quote]

dude ok I like your style

[quote="Karth"]I can't believe that this thread is still going on... Arnav, you can stop acting like a retard now... we all know that you're amazing and stuff, but that doesn't mean you can use such language in a public forum.[/quote]

Ok negative nancy, comin in here with your whining and your hollering

how bout you get off your high horse and stop accusing people just cause of their language

[quote="Elemennop"]Ok negative nancy, comin in here with your whining and your hollering

how bout you get off your high horse and stop accusing people just cause of their language[/quote]
Very good point positive polina

This just in -- this thread has become so silly!

People, get your minds out of the gutter. Nothing to see here, absolutely nothing.

BTW: The difference between 144 and 150 is your picture on the website, a plaque, and yet another page for you to hit when your search for yourself on google

[quote="probability1.01"]I would just like to point out that I lost. Also I might have gotten a 144, so uhh... I share your pain if I indeed got 144, except that I probably didn't get a 144. Anyways, who really cares about the difference between 144 and 150?[/quote]
1. You got a 150.

2. Getting a 150 is obviously better because then you can say "yo I got a [i]perfect[/i] score what now bia***"

[quote="rnwang2"][quote="probability1.01"]I would just like to point out that I lost. Also I might have gotten a 144, so uhh... I share your pain if I indeed got 144, except that I probably didn't get a 144. Anyways, who really cares about the difference between 144 and 150?[/quote]
1. You got a 150.

2. Getting a 150 is obviously better because then you can say "yo I got a [i]perfect[/i] score what now [b]bia***[/b]"[/quote]
hey i don't appreciate that kind of language

[quote="Pakman2012"]this is pretty - up guys[/quote]

yea...i guess we are showing a lack of maturity...[b]this topic should be locked[/b]...i suppose i got goaded into being obnoxious...although i feel that my point did not get through so well...in particular, regardless of whether it is justified or not to feel disrespected when somebody uses the word -, the one using it is being inconsiderate, and the purpose of my tone was to mimick arnav...to show how that is obnoxious

@rnwang: :lol:


integration calculus derivative algorithm function trigonometry search

Evaluate $\int\sec^{3}(x)dx$

[hide="what i got"]
$\frac{\sec(x)\tan(x)}{2}+\frac{\ln|\sec(x)+\tan(x)|}{2}+C$
i just want to make sure i didn't make a mistake somewhere.
if you would like i could post my work.
[/hide]

It's correct.

The easiest way to check such answers would be to just work out the derivative of your answer and see if it works. I find that easier that going through actual integration again.

There's actually an algorithm for doing any power of the trigonometric functions. If you don't have it, I'll type it up for you. Just as a hint, pretty much it involves factoring the functions and then using basic trig identities to convert it into other stuff like $sin^{2}(x)+cos^{2}(x)=1$. Also if you're doing just one trig function like this one you can always just expand it then integrate term by term.

http://www.mathlinks.ro/Forum/viewtopic.php?t=57219
(I put sec^3 in the Search Query and randomly hit topics until I got the right one.)


Ms. Albertson is randomly selecting the order in which her 25 students will each present a report next week. Five students will present each day, Monday through Friday. What is the probability that the shortest student will present his report on Thursday? Express your answer as a common fraction.

5 students will present on thursday. The probability the shortest one is one of those 5 is $ \frac{5}{25}\equal{}\frac15$


linear algebra
matrix
group theory
abstract algebra
superior algebra
superior algebra unsolved

Prove that there are no isomorphism between $ (R,\plus{})$ and $ (R^*,\cdot)$; $ (Q^*,\cdot)$ and $ (Q_{\plus{}}^{*},\cdot)$.

[quote="ZetaX"]You just need to find a property the other one doesn't have.
Count self-inverse elements.[/quote]

$ f(0) \equal{} 1$
But $ f^{\minus{}1}(\minus{}1)^{\minus{}1} \equal{} f^{\minus{}1}(\minus{}1) \equal{} 0$ because $ 0$ is the only elements in $ (\mathbb{R},\plus{})$ that is self inverse.

The demonstration for $ f: \ (\mathbb{Q}^*,\cdot) \rightarrow  (\mathbb{Q}^*_\plus{},\cdot)$ is similar.

The order of negative one is different each members of each pair of groups.

Spix:
What do you intend to do¿ $ \minus{}1$ needs not to be mapped to $ \minus{}1$.

Thanks for the help but I really don't get it. Maybe someone could elaborate.

How many solutions to $ x\circ x\equal{}e$ are there ($ \circ$ the group operation and $ e$ the neutral element)¿
Show that this number is preserved by isomorphisms, and show that this number is not equal in your groups.

I think I said the same think as you did, that is to analyze the elements of order two: $ \minus{}1$ in this case.(at least that is what I tried)
$ f((\minus{}1)(\minus{}1))\equal{}f(\minus{}1)\plus{}f(\minus{}1)\equal{}2f(\minus{}1)\to f(1)\equal{}2f(\minus{}1)\to f(\minus{}1)\equal{}0$ but also $ f(1)\equal{}0$ which destroys the isomorphism.

Another similar problem:
Let $ A=\{X\in M_{2}(\Bbb{R})|det(X)\neq 0 \}$ and $ B=\{X\in M_{3}(\Bbb{R})|det(X)\neq 0\}$.
 Prove that there are no isomorphisms between $ (A,\cdot)$ and $ (B,\cdot)$.

I did find a solution but have a question regarding it:[hide]
Let $ X_{1}=\begin{pmatrix}1 & -2 \\
1 & -1\end{pmatrix} \in A$. We get $ X^2=-I_{2}$
We assume that there exists $ Y\in B$ so that $ Y^2=-I_{3}$, but this implies $ det^{2} (Y)=-1$, which is contradictory with $ Y$ being a real matrix.
Assuming that $ f: A\rightarrow B$ is an isomorphism, we get $ f^{2}(X_{1})=f(X_{1}^2)=f(-I_{2})$. My question is: can I say that $ f(-I_{2})=-I_{3}$? If I can then the problem is solved, if not...well then it isn't.[/hide]

no, $ \minus{} I_2 \mapsto \minus{}I_3$ is not guaranteed.

[nonsense] the center of $ GL_n(\mathbb{R})$ consists exactly of the diagonal matrices with nonzero entries, thus is isomorphic to $ {\mathbb{R}^*}^n$. its torsion subgroup is $ \{\pm 1\}^n$, so that it has $ 2^n$ elements. thus $ GL_n(\mathbb{R}) \cong GL_m(\mathbb{R})$ if and only if $ n \equal{} m$.[/nonsense]

[quote]the center of $ GL_n(\mathbb{R})$ consists exactly of the diagonal matrices with nonzero entries[/quote]
Shouldn't these entries also be equal? Or, to be more precise, isn't $ Z(GL_n(\mathbb{R}))\equal{}\left\{x\cdot I_{n}|x\in\Bbb{R}^*\right\}$?

[quote="bike_me"][quote]the center of $ GL_n(\mathbb{R})$ consists exactly of the diagonal matrices with nonzero entries[/quote]
Shouldn't these entries also be equal? Or, to be more precise, isn't $ Z(GL_n(\mathbb{R})) \equal{} \left\{x\cdot I_{n}|x\in\Bbb{R}^*\right\}$?[/quote]Yes.

A correct proof counts the number of equivalence classes of unipotent matrices (Jordan canonical form) or the number of equivalence classes of idempotent matrices.

Here is a proof:
Assume that $ GL_{2}(\mathbb{R})\cong GL_{3}(\mathbb{R})$ then $ Z(GL_{2}(\mathbb{R}))\cong Z(GL_{3}(\mathbb{R}))$, so $ \left\{aI_{2}|a\in\mathbb{R}^*\right\}\cong \left\{aI_{3}|a\in \mathbb{R}^*\right\}$.
Then there exists $ a\in \mathbb{R}^*$ so that $ f(\minus{}I_{2})\equal{}aI_{3}$. Where f is the isomorphism assumed to exist. But $ I_{3}\equal{}f(I_{2})\equal{}f(\minus{}I_{2})f(\minus{}I_{2})\equal{}a^{2}I_{3}$. Thus $ a\equal{}\minus{}1$ or $ a\equal{}1$.
For the first case let me quote myself:
[quote]$ Let X_{1}\equal{}\begin{pmatrix}1 & \minus{}2 \\
 1 & \minus{}1\end{pmatrix} \in A$. We get $ X^2\equal{}\minus{}I_{2}$
We assume that there exists $ Y\in B$ so that $ Y^2\equal{}\minus{}I_{3}$, but this implies $ det^{2} (Y)\equal{}\minus{}1$, which is contradictory with $ Y$ being a real matrix.
Assuming that $ f: A\rightarrow B$ is an isomorphism, we get $ f^{2}(X_{1})\equal{}f(X_{1}^2)\equal{}f(\minus{}I_{2})$. My question is: can I say that $ f(\minus{}I_{2})\equal{}\minus{}I_{3}$? If I can then the problem is solved, if not...well then it isn't.
[/quote]
And for the second case we get: $ f(\minus{}I_{2})\equal{}f(I_{2})$, a contradiction.


In a right triangle the length of all three sides are positive integers. If the area of the triangle is 120 ,
then what is the length of the hypotenuse ?

[hide="Solution"]It is well known that any right triangle with integer lengths has one side divisible by $ 5$. Consider the $ 5\minus{}12\minus{}13$ triangle, which has an area of $ 30$. This is a factor of $ 4$ lower than expected, so we double the dimensions. The hypotenuse is therefore $ \boxed{26}$.

[/hide]


group theory
abstract algebra
number theory
least common multiple
superior algebra
superior algebra solved

I guess this should be a piece of cake for you guys:

Find all pairs of positive integers (m,n) s.t. there are at least 2 group morphisms

a) S_m->Z_n
b) Z_n->S_m

(S_m,*) is the group of permutations of order m and (Z_n,+) needs no introduction  :D

[i]Nicolae Paun (Romanian contest), 2003, 12'th grade[/i]

a)Sm ->Zn.

[b]I'll use smaller for  \leq [/b]

If n has no prime divisors smaller that m, then Z_n has all nonzero elements of order  \geq m.
Since the image of S_m through any morphism f:S_m->Z_n is a subgroup of Z_n, and for all x in S_m the order of f(x) in Z_n (denoted ord{Z_n}(f(x)) is a divisor of ord{S_m}(x), we get that f(x)=0 for all x in S_m.
Also note that for all x in S_m, ord{S_m}(x) has all prime divisors smaller that m. That is because any permuation x can be decomposed into a product of cycles and ord(x) is the lcm of these cycles.

Now let (m,n) such that n has a prime divisor smaller than m. Let that prime be p. Then let x=(1 2 ... p) (a cycle of length p this is x(1)=2, x(2)=3,...,x(p)=1 and x(k)=k for k>p).
Then ord{S_m}(x)=p.
There can be built a morphism f:S_m->Z_n such that f(x k )=k*(n/p) (as an element of Z_n) for all k in Z, and f is 0 in the rest.
Since the identity morphism <>f, we have that if n has a prime divisor smaller that m, then there are at least 2 morphisms.

The conclusion is that there are at least 2 morphism iff n has no prime divisor smaller that m.

I think b is the same. Correct me if I'm wrong.
If n has a prime divisor smaller than m. There exists f:Z_n->S_m such that f(k*n/p)=x k, where p and x are the same as in (a), and f is e in the rest.

"The conclusion is that there are at least 2 morphism iff n HAS NO prime divisor smaller that m."

I think you meant "there are at least 2 morphisms iff n has (instead of 'has no') a prime divisor <m. I got the same thing for (b), but your morphism for (a) doesn't convince me. I got n even:

Any permutation can be seen as a product of transpositions, so a morphism is determined if we know the images of all transpositions. If f((a,b)) = 0 then f(t)=0 for all permutations t, so in order to have a different morphism from f(t)=0 for all t we nedd to have a transposition (a,b) s.t. f((a,b))=k=/=0, but from here we get 0=f(e)=f((a,b)*(a,b)) = f((a,b))+f((a,b)) = k+k, so there is an element of order 2 in Zn, so n is even. This proves the necessity.

Let's now take n=2k. This is a morphism:f(t)=0 if t is an even permutation and f(t)=k if it's odd. This shows the sufficiency of n even.

The pairs are (m,2k) with positive m and k.

Hm. It seems I should stop posting at 3:49 am.
I didn't expected to be wrong in (a).

So where's the mistake? That thing you called f isn't a morphism, right?

You are right. It is possible to have a*b=xk for some a,b not on the form x^n, and we have f(ab)=f(a)+f(b)=0+0=0 and f(x^k)<>0 for some k not divisible by p.
So f is not a morphism.


trigonometry
inequalities
LaTeX
Support
function
geometry proposed
geometry

[color=purple] Prove or disprove that :
  $ \frac{cosA}{\cos\frac{A}{2}}\plus{}\frac{cosB}{\cos\frac{B}{2}}\plus{}\frac{cosC}{\cos\frac{C}{2}}\ge\sqrt{3}$
   The following is true and it's due to [b]Tran Ha Minh [/b]. It's really nice because it's eqivalent to Schur inequality with degree 2  :) :
        $ \frac{cos^2A}{\cos^2\frac{A}{2}}\plus{}\frac{cos^2B}{\cos^2\frac{B}{2}}\plus{}\frac{cos^2C}{\cos^2\frac{C}{2}}\ge 1$
  Please help me to prove or disprove the first ! Thank you very much ! [/color]

take f(x)=cos(x)/cos(x/2)=2cos(x/2)-sec(x/2).=>f"(x)= -cos(x/2)/2-sec(x/2)(sec^2(x/2)+tan^2(x/2))/4<0 for all x in(0,pi). Now apply Johnson's inequality with x=A, B, C. we will get opposite inequality.

Sorry for the way I have written. I do not know how to use Latex and also text editor here does not support it seems

Hi evarist did you check my reply. Anyways it is a nice problem

[color=purple]Hey please edit your post and I'll can check it brightly. In your post you must use LaTex .  :) [/color]

$ Let\, f(x) \equal{} \frac {cos(x)}{cos(\frac {x}{2})} \equal{} 2cos(x) \minus{} sec(\frac {x}{2}) \\
Then\,f^"(x) \equal{} \minus{} \frac {1}{2}cos(\frac {x}{2}) \minus{} \frac {1}{4}sec(\frac {x}{2})(sec^2(\frac {x}{2}) \plus{} tan^2(\frac {x}{2}) ).$
$ \equal{} > f^"(x) < 0 for\: all\: x \: in\: (0,pi). \\
Now \,apply \,Johnson's \,inequality\, for \,f(x) \,with x\equal{}A,B,C. \,Then \,we\, get\, the \,following$

$ \frac {cosA}{\cos\frac {A}{2}} \plus{} \frac {cosB}{\cos\frac {B}{2}} \plus{} \frac {cosC}{\cos\frac {C}{2}}\le\sqrt {3}$

[color=purple] I saw it yesterday  :). I forgot using specific function  :(  Thank you but in your post[b] Jensen inequality [/b]not [b]Johnson inequality[/b] :) [/color]


calculus
algebra
polynomial
floor function
inequalities
binomial theorem

Show, without calculus, that for any polynomial $P(x)$, we have for some $x$ that $2^{x}> P(x)$.

[quote="N.T.TUAN"][hide]We need only prove for all $n>2$ there exist $x$ such that $2^{x}>x^{n}$.

Choose $x=2^{n}$.[/hide][/quote]
Nice solution, but    crossing of  all the t's may be missing above...
Choosing $x=2^{n}$ when $n=3$ does not give $2^{x}>x^{n}$.

Let $n\in\mathbb{N}$ be fixed. Since $\frac{2^{x}}{x^{n}}\to+\infty$ when $x\to+\infty$, for sufficiently large $x$ we have $\frac{2^{x}}{x^{n}}>1$.

Yes, the limit holds, but the thing is, we shouldn't use calculus.

[hide]The Binomial Theorem kills the problem. Let $n$ be degree of $P$. If $x$ integer such that $x\geq n+1$, then
\[2^{x}=\sum_{k=0}^{x}\binom{x}{k}\geq \sum_{k=0}^{n+1}\binom{x}{k}\]
Let $Q(x)$ be the right hand side of the above, so that $Q(x)$ has degree $n+1$. Now we just need to find an integer $x\geq n+1$ such that $Q(x)>P(x)$.

Now observe that $Q(x)-P(x)$ is a polynomial with degree $n+1$ and positive leading coefficient. We will be done if can show that for any polynomial $f$ of degree $n$, then $x^{n+1}>f(x)$ for some integer $x\geq n+1$. This is true because if
\[f(x)=a_{0}+a_{1}x+a_{2}x^{2}+\cdots+a_{n}x^{n}\]
then letting $x=\max \{ n+1,\left\lfloor a_{0}+a_{1}+a_{2}+\cdots+a_{n}+1\right\rfloor \}$ works, since $x^{n}>(a_{0}+a_{1}+\cdots+a_{n})x^{n}\geq f(x)$ (the right inequality holds because $x\geq 1$).
[/hide]

[quote="Karth"]Or you could just cite that solving a problem in non polynomial time (particularly np complete) is *much* slower than solving a problem in polynomial time. But that's like CS so does it count?[/quote]

This doesn't even make any sense: $\bf NP$ is not known to be equal to $\bf E$, the set of languages which can be decided in linear exponential time (that is, time $2^{O(l)}$ for inputs of length $l$).  $\bf NP$ is not short for "non-polynomial," it is short for "nondeterministic polynomial," which is totally different.  It is unknown whether $\bf NP$ is in fact equal to $\bf P$, and if it is, then all $\bf NP$ problems (and in particular all $\bf NP$-complete problems) can be solved in polynomial time.

Randomizer's solution makes sense, but it's a "solution," not a solution.  Restating the thing to prove and then asserting that it is correct is not what most people are looking for in a proof.

More precise explanation: Let $n$ be fixed and take a certain $x>2^{2n}$. Let $k$ satisfy $2^{k+1}>x\ge 2^{k}$. To show that $2^{x}>x^{n}$ it suffices to show that $2^{2^{k}}\ge (2^{k+1})^{n}$. Equivalently: $2^{k}\ge n(k+1)$; sufficiently $2^{k}\ge\frac{k(k+1)}{2}$. This one is true (why?) what ends the proof.
Similarily one can get stronger inequality which shows that $\frac{2^{x}}{x^{n}}$ can be arbitrarily large.


Mary told John her score on the American High School Mathematics Examination (AHSME), which was over 80.  From this, John was able to determine the number of problems Mary solved correctly.  If Mary's score had been any lower, but still over 80, John could not have determined this.  What was Mary's score?  (Recall that the AHSME consists of 30 multiple-choice problems and that one's score, $s$, is computed by the formula $s = 30 + 4c - w$, where $c$ is the number of correct and $w$ is the number of wrong answers; students are not penalized for problems left unanswered.)

[hide="Answer"]Every score with $w<4$ up to a certain point can be made in at least two ways, either with a certain number right and wrong, or adding one right and four wrong. However, once we have $26$ answered, we cannot add this, because it would exceed the number of problems on the test. Therefore, since we must have $w<4$ (otherwise we could subtract four wrong and add one right to make the score with a different combination), we have $w=3$, $c=23$, which gives a score of $30+4(23)-3=\boxed{119}$.[/hide]

[hide=unorganized thinking=sol]First we shift the problem down by $30$ so we are looking for a score over $50$ such that for each number below it, we are unable to guess the # of questions correctly
this means there exits $w_{a},c_{a}$ and $w_{b},c_{b}$ such that
$4w_{a}-c_{a}=4w_{b}-c_{b}$
\\
$4(w_{a}-w_{b})=c_{a}-c_{b}$

Keep in mind that also $(w_{i}+c_{i})\le{30}$ for all $i$

Now if we let $w_{a}-w_{b}=k, c_{a}-c_{b}=4k$

The question tells us to start looking at ranges above $50$

We want to minimize $c_{a}$ and maximize $w_{a}$
so if $k=1$ we can have like
$(26,4)$ but the next pair is $(25,4)$ which we drop $4$ not continuous so this doesn't help
So we  need $c$ to be flexible by $3$ since as $w$ drops $1$ expression drops $4$
this means both ways flexible.
Thus $c=4$  we choose and if $c=7$ we would need $w=23$
Thus $(23,4)$ is the optimal
Shifting back gives $118$ but we want one more so $\boxed{119}$[/hide]

[quote=PIANOer]How do we get that $w<4$?. Guess around?[/quote]

If $w \geq 4$, then we could have decreased the number of correct by 1 and the number of wrong by 4 to get a new combination $(c-1, w-4)$ that produces the same value and will not exceed the 30-question limit.

First note that we can subtract $30$ from Mary's score to get an easier AHSME score calculator: $s=4c-w$. Secondly, Mary can achieve the same score in one way if she cannot get one more question correct and four more questions wrong. That is, $w+4+c+1 > 30 \implies w + c \ge 26$. However at the same time, if she gets four or more questions wrong, then she can get one more question right and still get the same score. That means that $w \le 3$ and by the inequality $w+c \le 25$, $c \ge 22$. We know that $4c-w = 4(c+w)-5w \ge 104--15=89$. Adding back the original $30$ to Mary's score, we have that $w+c \le 119$, so $\boxed{119}$ is the first occurence where John can tell how many questions Mary answered right (and how many wrong). We can easily verify this to be correct.


function
real analysis
real analysis unsolved

Let f, g be real functions, and for each x $ \in$ Dom (f) $ \cap$ Dom (g) define
(f $ \vee$g)(x) := max{f(x), g(x)} and (f $ \wedge$g)(x) := min{f(x), g(x)}.
Prove that:
$ (f \vee g)(x) \equal{} \dfrac{(f\plus{}g)(x) \plus{} \mid(f\minus{}g)(x)\mid}{2}$
for all x $ \in$ Dom (f) $ \cap$Dom (g)

without loss of generality, assume $ f(x)\geq g(x)$ and then prove that $ (f \vee g)(x) \equal{} f(x)$.


So, I guess most of you are familiar with pole vaulter problem. For those, who are not familiar with the problem here it is:
There is this pole vaulter who has a stick of length $ 8m$ and there is a barn of length $ 4m$. He decides to fit that stick in that barn by boosting himself to a certain relativistic velocity. He boosts himself up and his stick is actually $ 3m$ long in our frame. Initially both of the doors (back and front) of  are closed. When vaulter reaches the back door, it opens, he enters the barn(the back end of stick is in barn) the back door closes, then the front end of the stick reaches the front door, the door opens, and when vaulter gets out of barn the front door closes. The actually question asks what happens in vaulter's frame. It is pretty simple: his stick is $ 8m$ long, barn is even shorter than $ 4m$ since it moves towards the vaulter at a relativistic velocity. So, what happens is that the front of rod reaches the back door it opens then it reaches the front door and it opens, too, so in this frame due to simultaneity fact both are open. Then the back of rod reaches back door, the door closes, and when it reaches the front door, it closes, too.
The interesting thing is that in one frame at one point the vaulter is entirely (with his rod) in the barn, with both doors closed, and in the other frame he never faces the situation, when he is in the barn with both doors closed.
OK, let's go to the main question: say we have a shiny day with a bright sun or if you will with three bright suns :D , and, we place a bomb on the vaulter which blows up, when it doesn't receive light. In one frame it appears that he is in the barn with both doors closed, we dies, in other frame, he doesn't blow up because at any moment when he is in the barn one of the doors is open.
Finally, does the vaulter die, does he doesn't die or death is simply relative concept... :lol:

well, the bomb will trigger when the conditions for detonation are satisfied in the bomb's frame, so no, he shouldn't die?

come on I can tell you the answer, however I want to figure it out...


Circle E is internally tangent to all four sides of square ABCD. Prove that AC goes through the center of circle E.

I haven't done many geometry proofs, so I don't know if this is enough, but here goes:
[hide]Each angle of the square is a right angle, so it has 90 degrees, so each diagonal of the square, AC and BD, is a diameter of the circle. Therefore, AC must go through the center of circle E.[/hide]

I don't think that's enough because you don't show why AC must pass through the center of circle E. Also, AC and BD are not diameters of circle E.

Oh wait, I thought it was the other way around, where square ABCD is inside the circle E. My bad.

[hide]A and E are joined.
So are E and B  and  D and E

We notice that in $ \triangle AEB$ and $ \triangle BEC$,
AB=BC,
BE is common
and $ \angle ABE$=$ \angle EBC$ [since BA and BC are tangents to the circle]

so, $ \triangle AEB$ $ \cong$ $ \triangle BEC$.

So. $ \angle AEB$= $ \angle BEC$ = x
Similarly, $ \angle AED$ = $ \angle DEC$=y

$ \angle AEC$+ $ \angle BEC$ + $ \angle AED$  +$ \angle DEC$=2(x+y)=$ 360^o$

So, $ x \plus{} y \equal{} 180^o$

Hence, it is easily seen that $ \angle DEC$ + $ \angle BEC$=$ 180^o$ i.e. BD is a straight  line.
Similarly,AC is a st. line
Therefore,AC and BD pass through E.

Hence proved.[/hide]


quadratics
number theory proposed
number theory

Find all primes $p,q,r$ such that : $p^3=p^2+q^2+r^2$

If $p = q$ (respectively $p = r$), then you need $p^2(p-2) = r^2$ (respectively $p^2(p-2) = q^2$). So a fortiori $p-2=1$ and $p = q = r = 3$ in both cases, since $p, q, r$ are supposed to be primes. Similarly, letting $q = r$, you get $p^2(p-1) = 2q^2$, which leads once more to $p-1 = 2$, i.e. $p = 3$, and $q = r = p = 3$. Then admit in the following that $p, q, r$ are pairwise distinct, making $\gcd(q,r) = 1$. You want $0 \equiv q + r\bmod 2$, so that both $q$ and $r$ has to be odd. Moreover $p = 2$ makes $2^3 - 2^2 = q^2 + r^2 > 18$, absurd! Hence it's $p \equiv 1 \bmod 2$ too. Anyway $p^3 = p^2 + q^2 + r^2$ only if $(qr^{-1})^2 \equiv -1 \bmod p$, forcing $p \equiv 1 \bmod 4$, since $-1$ is not a quadratic residue of any prime $\equiv 3 \bmod 4$. It follows $p^3 = p^2 + q^2 + r^2$ only if $1 \equiv 3 \bmod 4$, which is obviously impossible! Uh...


The polynomial $ x^n \plus{} a_{n\minus{}1}x^{n\minus{}1} \plus{} ... \plus{} a_1x \plus{} 1$ has non-negative coefficients and $ n$ real roots.
Show that its value at $ 2$ is at least $ 3^n$.

Reminds me of binomial expansion of (1+2)^n   :maybe:


Given $\triangle ABC$, points $D$, $E$ and $F$ are on $BC$, $CA$ and $AB$, respectively. If the centroid of $\triangle ABC$ is also the centroid of $\triangle DEF$, prove that $\frac{BD}{DC}=\frac{CE}{EA}=\frac{AF}{FB}$.

[hide]Vector solution:

Denote the common centroid by $G$. Then by the known property of the centroid

\begin{eqnarray}\overrightarrow{GD}+\overrightarrow{GE}+\overrightarrow{GF}=\overrightarrow 0\\ \overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}=\overrightarrow 0\end{eqnarray}

On the other hand, $\overrightarrow{GD}=\overrightarrow{GB}+\overrightarrow{BD},\overrightarrow{GE}=\overrightarrow{GC}+\overrightarrow{CE},\overrightarrow{GF}=\overrightarrow{GA}+\overrightarrow{AF}.$ Substituting these into $(1)$ and taking $(2)$ into account, we get

$\overrightarrow{AF}+\overrightarrow{BD}+\overrightarrow{CE}=\overrightarrow 0\qquad (3)$

Since $\overrightarrow{AF},\overrightarrow{BD},\overrightarrow{CE}$ are collinear with $\overrightarrow{AB},\overrightarrow{BC},\overrightarrow{CA}$ respectively, we can write

$\overrightarrow{AF}=\lambda\overrightarrow{AB}, \overrightarrow{BD}=\mu\overrightarrow{BC}, \overrightarrow{CE}=\nu\overrightarrow{CA}\qquad (4)$

where $\lambda, \mu, \nu$ are some real numbers. We also note that $\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CA}=\overrightarrow 0\implies \overrightarrow{CA}=-\overrightarrow{AB}-\overrightarrow{BC}\qquad(5)$

Substituting $(4)$ into $(3)$ and taking $(5)$ into account, we get

$(\lambda-\nu)\overrightarrow{AB}+(\mu-\nu)\overrightarrow{BC}=\overrightarrow 0$

But since $\overrightarrow{AB}$ and $\overrightarrow{BC}$ are linearly independent, this can be satisfied only if $\lambda-\nu=0\land \mu-\nu=0\implies \lambda=\mu=\nu.$

Now ${{BD}\over{DC}}={{BD}\over{BC-BD}}={{\mu BC}\over{BC-\mu BC}}={\mu\over{1-\mu}}.$ The similar goes for other two sides, and since $\lambda=\mu=\nu$, the obtained ratios are all equal.[/hide]


Let $\{a,b,c,d\}\subset\mathbb R^*_+$ so that  $2p=a+b+c+d$ and $p > \max\{a,b,c,d\}$ $\implies\boxed{\ \frac{(p+a)(p+b)(p+c)(p+d)}{(p-a)(p-b)(p-c)(p-d)}\ \ge \ 81\ }$ .


Hello,

I was looking at the classes list, and saw that the AIME Problem Series B was the next AIME Problem Series class to be offered.  Shouldn't it be the Problem Series A, because it should go in order A, B, A, B ...

Right now the B class is going on.

We flip them every once in a while, so students who can't take our classes during the summer (or during the school year) will have access to both problem series.


[quote="guile"]I have solved several problems and I already have answers. I just want to check if there are other solutions.

1.) Find the number  of ways to choose a pair {a,b} of distinct numbers from the set {1,2,3,...,50} such that

i) |a - b| = 5
ii.) |a - b| <= 5

[hide="My Answers"]i. 45, ii. 235 [/hide]

2.) There are 12 students in a party. Five of them are girls. In how many ways that the 12 students can be arranged in a row if

i.) The five girls must be put together forming a block
ii.) no two girls are adjacent
iii.) between two particular boys A and B, there are no boys, but exactly three girls.

[hide=" My Answers"]i. (7!)(5!)(8), ii. (7!)(5!)(4), iii. (7!)(5!)(8)[/hide][/quote]
Different answer for 1 (ii), 2(ii):
[hide="1 (ii)"]Two cases: $a\le 5$ ,$5 < a< 46$, $a\ge 46$

CASE 1:
Note that for each number, there are $5$ other numbers we can choose. So $25$ so far.
CASE 2:
For each number, there are $10$ other number we can choose. There are $400$ so far.
CASE 3:
Same as case 1; $25$.

Adding, $450$. But we have counted everything twice, so divide by 2 to get $\boxed{225}$.[/hide]

[hide="2 (i)"]The five girls, forming a block have $5!$ ways to put themselves together in that block. For the rest, we have 7 boys and 1 girl block, so there are $8!$ ways to arrange that. So $(5!)(8!)$[/hide]

[hide="2 (ii)"]No two girls are adjacent, so it looks like this:
B G B G B G B G B G
with one boy left over. We must pick a "leftover" boy, with $7$ ways to do that. We have $11$ ways to put him somewhere. So we have $77$ ways.
There are $77(6!)(5!)$ ways.[/hide]

[quote="me@home"]1)
i) this means $a=b\pm 5$ there are 50 options for a, and either 2 or 1 on b depending. Clearly, $5(1)+40(2)+5(1)=90$ then divide by two for repetition, so 45

What are these problems for?[/quote]

actually im pretty sure 1i is 90, not 45..it asked for ordered paris, not sets

as for 2 ii, i think the correct answer is 56*7!*5!, the boys and girls can be ordred in 7!*5! ways, while there are 8C5=56 ways the girls wont be next to eachtother.

the pair is not ordered, that is why, i used the symbo {}.

[quote="me@home"]1)
i) this means $a=b\pm 5$ there are 50 options for a, and either 2 or 1 on b depending. Clearly, $5(1)+40(2)+5(1)=90$ then divide by two for repetition, so 45

What are these problems for?[/quote]

actually, i have a subject called combinatorics and im just reviewing what i have learned in probability before. sort of practice.

[quote]Adding, . But we have counted everything twice, so divide by 2 to get . [/quote]

[hide="My Solution for 1"]i) Let the first pair {a,b} be {1,6}. If you continually add 1 to the pair, then the largest possible value for a is 45 since if a is 46, b would b 51. Hence, there are 45 pairs. Note that we are asking for the number of ways to choose a pair, hence the order does not matter. {1,6} is the same as {6,1}.

ii.) Using the argument in i, if |a - b| = 4, then our first pair will be (1, 5) and the largest possible value of a is 46. This means that the number of pairs increases by 1 as we decrease the difference. Hence for there are 45 + 46 + 47 + 48 + 59 = 235 pairs of numbers. [/hide]

[hide="My solution for 2"]
i.) The possible number of positions for girls is 5! for boys is 7! But, the girls can be arranged in 8 ways. If we arranged 12 chairs in a row and number them from 1-12, the girls can occupy positions in 8 ways. That is chairs 1-5,2-6,3-7,...,8-12. Hence, the number of arrangement is 7!*5!*8.

[/hide]

I don't know but i think that is correct.

For the ordered pairs, I don't think (1,6) is the same as (6,1), but not sure.

For 2(i), think of the girls as one block themselves. Inside the block, there are 5! ways to order. For the other 8 chairs, 7 go to boys and 1 to the girl block, so $\binom{8}{7}=8$ for choosing the boys' chairs. The boys can be ordered 8! ways, so 5!*7!*8?
Don't trust me on these.

[quote="shawn1"]For ordered paris (a,b) is not (b,a), it will have to say [i]sets [/i]for them to equal so its 90.[/quote]
Yes, I think that's what it means. For example, try graphing (3,5). It's not the same as (5,3).
However, then you have to worry about a=b, which must be counted once.

Maybe the question is a bit ambiguous, but it does not really ask for ordered pairs. Note that it asks on how many ways can you choose two numbers such that their difference is 5, etc.

Maybe the question is not clear.


algebra
polynomial
complex analysis
complex analysis unsolved

Let $ p(z)\equal{}\sum_{k\equal{}0}^{n}a_k z^k,\: a_k\in\mathbb{C},\: a_n\neq 0$, be a polynomial of degree at least 2, namely $ n\geq 2$. If all roots of the polynomial are located inside of the simple, closed loop γ, show that:

$ \oint_{\gamma}\frac{1}{p(z)}dz\equal{}0$

[u]Tip:[/u] We know that $ \exists R>0$:

$ |p(z)|\geq \frac{1}{2}|a_n\parallel{}z|^n$, $ \forall |z|\geq R$

But it is the direct conclusion from residue theorem, since $ {Res}\limits_{z=\infty}\frac{1}{p(z)}=0$.

Well i still have a problem getting it...

You said that $ {Res}\limits_{z=\infty}\frac{1}{p(z)}=0$, but how do you conclude this? Could you please be a little more specific with the solution?

Thank you!


probability
analytic geometry
MIT
college
combinatorics unsolved
combinatorics

2n- children are standing in a line outside a theater which is showing a puppet

show. n-children have a one dollar coin and the rest n-children have a two dollar

coin each.

The entry ticket costs one dollar.

If all arrangements of the children are equally likely,then find the probability that

 no child will have to wait for change, if the cashier at the ticket window has no

 change to start with.

[hide]There is an obvious bijection between good orderings and Dyck paths, so the answer is $ \frac{C_{n}}{\binom{2n}{n}}\equal{}\frac{1}{n\plus{}1}$ where $ C_{n}$ is the $ n$th Catalan number.[/hide]

I am out of depth on this reply of yours. You have the correct answer. Please explain in detail or tell me the source where i can find about good orderings and Dyck paths

Sorry, here are more details.  By "good ordering," I just meant an ordering of the children which satisfies the desired condition.  A Dyck path (of length $ 2n$) is a path from $ (0, 0)$ to $ (n, n)$ in the coordinate plane composed of $ n$ steps of unit length to the right and $ n$ steps of unit length up, such that the path never passes above the line $ y \equal{} x$.  Thus, there is one Dyck path of length 2, two Dyck paths of length 4, 5 of length 6, etc.

To biject between orderings and paths, start at the beginning of the line and put your pencil at $ (0, 0)$.  As you come upon each student, if the child holds a one dollar coin, draw a horizontal segment; if the child holds a two dollar coin, draw a vertical segment.  It's clear that the initial ordering is good if and only if the resulting path is a Dyck path.  Now, the question is one of counting Dyck paths.  But this is a famous combinatorial problem, and the answer is given by the Catalan numbers $ C_{n}$, which have the closed-form $ \frac{1}{n\plus{}1}\binom{2n}{n}$.  The total number of orderings is $ \binom{2n}{n}$, so the probability is $ \frac{1}{n\plus{}1}$.

Really, I haven't done any math here: I've just translated the language of the problem to language I prefer and then used a result I already know which makes the problem easy.  The interesting part, if you haven't seen it before, is to prove that $ C_{n}\equal{}\frac{1}{n\plus{}1}\binom{2n}{n}$.  I suspect a proof of this can be found on the forum by searching for either "Catalan" or "Dyck," and it can also doubtless be found elsewhere.  The Catalan numbers are very interesting and pop up in all sorts of places.  The mathematician Richard P. Stanley, at MIT, has on his website a .pdf file which lists 149 (and counting) combinatorial questions whose answer is the Catalan numbers.


analytic geometry
graphing lines
slope
ARML

Set up the equation for the tangents to the circle given by x^2 + y^2 - 2x + 4y = 0 which are perpendiclar to the straight line given by x-2y+9=0

[quote="The Zuton Force"][hide="Solution"]We have $(x-1)^{2}+(y+2)^{2}=5$, so the tangent is of the form $(x_{0}-1)(x-1)+(y_{0}+2)(y+2)=5$, where $(x_{0},y_{0})$ is on the circle. This line has slope $\frac{1-x_{0}}{y_{0}+2}$, which must equal $-2$ since it must be perpendicular to $x-2y+9=0$

Thus, $\frac{1-x_{0}}{y_{0}+2}=-2 \implies y_{0}=\frac{x_{0}-5}{2}$

So, we have $(x_{0}-1)^{2}+\left( \frac{x_{0}-1}{2}\right)^{2}=5$, giving:
$x_{0}=-1 \implies y_{0}=-3 \qquad \text{or}\qquad x_{0}=3 \implies \; y_{0}=-1$

This gives the equations of the two tangents to be:
$(-1-1)(x-1)+(-3+2)(y+2)=5 \implies \boxed{y=-5-2x}$
$(3-1)(x-1)+(-1+2)(y+2)=5 \implies \boxed{y=5-2x}$[/hide][/quote]

How did you know about $(x_{0}-1)(x-1)+(y_{0}+2)(y+2)=5$? I have seen it somewhere but don't remember where :blush: .

I solved it another way though.

[hide]Translate the circle to the origin and make the new equation $x^{2}+y^{2}=5$.
Let the tangent line be $y=-2x+k$.

Substitution gives
$x^{2}+(-2x+k)^{2}=5$
$5x^{2}-4kx+k^{2}-5=0$.

Since the line is tangent, the discriminant has to be zero. So

$D/4=4k^{2}-5(k^{2}-5)=-k^{2}+25=0$.

This gives $k= \pm 5$.

When we find the points of tangency, we can translate them to the original point by moving them 1 unit right and 2 units down. But these translated points are on the same line, so $k$ doesn't change.

So $y=-2x \pm 5$.[/hide]

Or there is that formula that some people memorize: [hide]$y = mx \pm r \sqrt{m^{2}+1}$ for $m=-2$ and $r=\sqrt{5}$, which also gives $y=-2x \pm 5$.[/hide]

[quote="The Zuton Force"]There was something of that nature on a recent ARML Power Round... that might've been it?[/quote]

I remember that now!  :D It was called "Stretch Method" or something similar. Thanks.


Let functions$\ f(x)$, saytisfying the condition
       $\ f(xy)+f(x-y)+f(x+y+1)= x.y+2x+1$ (for all x,y in R)
And: $\P(x)$ = $\frac{n^(f(x))}{n^(f(x))+\sqrt{n}}$
Find: A= $\P(a/n)$+$\P(b/n)$ (so that a+b=n, n other 0)

Find all polyomials $f(x)$ such that:
$f(x)+f(1/x)= x+1/x$ (for all x in R, and x other 0)

[hide]*y=-1
f(-x)+f(x+1)+f(x)=x+1
*y=0
F(0)+f(x+1)+f(x)=2x+1
$\rightarrow$ f(-x)-f(0)=-x
*t=-x
f(t)-f(0)=t
$\rightarrow$f(t)-t=f(0)-0
*g(t)=f(t)-t
$\rightarrow$g(t)=g(0)  (for all t in R)
From:$\ f(xy)+f(x-y)+f(x+y+1)= x.y+2x+1$
$\rightarrow$ f(xy)-xy+f(x-y)-(x-y)+f(x+y+1)-(x+y+1)=0
g(xy)+g(x-y)+g(x+y+1)=0
$\rightarrow$3g(0)=0
$\rightarrow$g(0)=0
$\rightarrow$g(t)=0(for all t in R)
$\rightarrow$f(t)-t=0(for all t in R)
$\rightarrow$f(x)=x
$\P(x)$=$\frac{n^x}{n^x+\sqrt{n}}$
$\P(x)$+$\P(y)$=1(for all x,y so that x+y=1)[/hide]

is that right?

find all continuous funchtions f(x), defined on R satisfying the condition:
f*(x)+f(x)=2x (for all x in R)
where f*(x)=f(f(f(x)))

Your last one was posted on the forum not long ago. :)

Find f(x):
xf(x)+f(a-x)=b(for all x on R )

[b]1[/b].  find f: $\left[ {a;b} \right] \to \left[ {a;b} \right]$
f(f(x))+x=2f(x)

[b]2.[/b]  If f continuous on R
 find f: $R \to \ R$
f(f(x))+x=2f(x)

[quote=CT_14][b]1[/b].  find $f:$ $\left[ {a;b} \right] \to \left[ {a;b} \right]$
$f(f(x))+x=2f(x)$

[b]2.[/b]  If $f$ continuous on $R$
 find $ f:$ $R \to \ R$
$f(f(x))+x=2f(x)$[/quote]

[quote=CT_14]Find $ f(x)$:
$xf(x)+f(a-x)=b$(for all $x$ on $ R $ )[/quote]

[quote=CT_14]find all continuous funchtions $f(x)$, defined on $R$ satisfying the condition:
$f*(x)+f(x)=2x $ (for all x in R)
where $f*(x)=f(f(f(x)))$ [/quote]


Find the mean of the prime numbers in this set: $ \{15, 17, 19, 20, 21\}$.

hello, the searched mean of the prime numbers is$ \frac{17\plus{}19}{2}\equal{}18$
Sonnhard.


Let the sequence $a_n$ ,$n\in N*$ with integer $a_i$ integer
We know that

$a_{n+14}=a_n^4+a_{n+1}^4+....+a_{n+13}^4$ . There exist $k$, $k >14$ such that

$a_k=2015$  ?

Here is hint: $k$ can't be greater then 14 because $a_i^4\equiv_{16} 0,1$ so
\[ a_1^4+a_2^4+...+a_{14}^4\equiv_{16} 0,1,...,14 \] but never $15$ though $2015\equiv_{16}15$ BUT when $k\leq 14$ I will have to think about it....


1) Let I= {x € R: -1<x<1} and let f(x) = x/(1-x^2) for x in I. Show that f : I-->R is one-to-one and onto using calculus.
a). calculate f ‘(x) and use the result to explain why f is one-to-one.

b). calculate lim(as x approaches 1 from the left) and lim(as x approaches 1 from the right) to show that the range of f  is neither bounded above nor bounded below.

c) Show that f is onto as follows: given Y0 € R use part b) and the intermediate value theorem to show that there is an X0 in I such that f(X0) = Y0.

2) May use the standard calculus facts concerning sine and cosine functions:

a)	Show that lim(as x approaches 0) sin(1/X) does not exist.
b)	Let g(x)=x sin(1/X) if x does not equal 0 and let g(0) = 0. Show that g is continuous at x=0 but that g is not differentiable at x=0.
c)	Let h(x) =x^2 sin (1/X) if x does not = 0 and let h(0)=0. Show that h is differentiable at all x but that h’ is not continuous at x=0.

Wrong section, and it feels a bit too much like we're doing your homework for you...

Are there any you're having problems with in particular?

This is Not Homework and I don't have any problems with any of these. In fact, I have the answers. I thought that it would be a good idea to post them and see if anyone of you will be able to solve none, one, or as many as you can.
but I do understand if they are difficult to answer.

[quote="mikejones"]but I do understand if they are difficult to answer.[/quote]

They aren't.  But it always looks suspicious if someone posts a large number of textbook-type problems in the same post.


If $a_0 =1, a_1 =3, a_2 = 7$, and $a_{n+3}=3a_{n+2}-3a_{n+1}+a_n$ for $n>0$, then what is a general formula for $a_n$?

what does $a_m$ have to do with the problem? it's not even defined! maybe you meant $a_n$?

[quote="eryaman"][hide] The problem with my solution is that it assumes the sequence continues to be quadratic...so I'll work on proving that.[/hide][/quote]

Suppose $n^2+n+1=f(n)$. Since $f(0)=1, f(1)=3, f(2)=7$, and $f(n+3)=3f(n+2)-3f(n+1)+f(n)$, you don't even have to prove that it continues to be a quadratic.

[quote="eryaman"][hide]Listing out the first few terms I get 1,3,7,13,21,31.  Because their "second" difference (forget what the term called) is $2$, it is a quadratic.

So we know the equation is in the form $a_n=an^2+bn+c$, and $c=1$, since $a_0=1$.  Plugging in $a_1=3$ and $a_2=7$ gives the simultaneous equations $a+b=2$ and $2a+b=3$.  Solving gives $a=1$, $b=1$, and $c=1$.

So the general equation is $a_n=n^2+n+1$.

The problem with my solution is that it assumes the sequence continues to be quadratic...so I'll work on proving that.[/hide][/quote]

you don't have to assume it's quadratic. since we know the "second difference" is $2$, we can take the initial term, and add the sum, like so:
$1+\sum^{n}_{k=1}{2k}$
$=1+\frac{n}{2}(2+2n)$
$=1+n+n^2$
this way we dont have to assume its quadratic, its determined by the formula for arithmetic series

But "knowing" that the second difference is two is the same as assuming its quadratic

What I meant was, how do we know the sequence continues to be have a "second difference" two?  I know that for those terms it does...but what if it was just a coincidence?  Or am I missing something...?

[hide]the general formula is n^2+n+1, i just used guess and check because i don't know that much math.[/hide]


New problem.. sorry :(

If P is a 10th degree polynomial so that P(0) = P(1) = P(2) = P(3) ... = P(9) = P(10) = 1. What is P(11)?

:maybe: No one else finds this strikingly similar to a current USAMTS problem?
EDIT Nevermind.

Consider $ Q(x) \equal{} P(x) \minus{} 1$. We have that $ 0,1,2,3,\ldots,10$ are all roots of $ Q$. This is 11 roots of a $ 10$th degree polynomial. That's a contradiction. I'm going to assume you didn't mean to say $ P(0) \equal{} 1$. In this case, we can write $ Q(x) \equal{} c(x\minus{}1)(x\minus{}2)(x\minus{}3)(x\minus{}4)(x\minus{}5)(x\minus{}6)(x\minus{}7)(x\minus{}8)(x\minus{}9)(x\minus{}10)$, where $ c$ can be any nonzero real number. Then $ P(11) \equal{} Q(11) \plus{} 1 \equal{} c(10!) \plus{} 1$. So, depends on what $ c$ is. Perhaps you meant to say $ P(0)$ was some other value?

[quote="Farenhajt"][quote="MellowMelon"]Consider $ Q(x) \equal{} P(x) \minus{} 1$. We have that $ 0,1,2,3,\ldots,10$ are all roots of $ Q$. [b]This is 11 roots of a $ 10$th degree polynomial. That's a contradiction.[/b][/quote]

Not really a contradiction - the conclusion is $ Q(x)\equiv 0$, hence $ P(x)\equiv 1$.[/quote]

If $ Q(x)\equiv 0$, $ \text{deg }Q \equal{} \minus{}\infty \neq 10$

There's a contradiction, since $ P(x) \minus{} 1$ has $ 11$ different roots when the degree of $ P$ is $ 10.$

True.  The correct problem statement would have been that $ P(x)$ is a polynomial of degree at most $ 10$.


Where could I find pdfs or other kinds of files of college applications for schools? Does the application form vary for different schools?

Usually you can find each college application on the college website, under the section of the website for undergraduate admission. Many colleges use the Common Application,

https://www.commonapp.org/CommonApp/default.aspx

so the Common Application site is a good first place to look.

It's becoming increasingly common to file applications on line. My own university now gets nearly all of its applications that way ( http://www.csumentor.edu ) and I would think that is also true of the various University of California campuses.


Given triangle $ ABC$ and the incircle $ (I). IA$ cuts $ (I)$ at $ A_1 (A_1\in [AI])$. Similarly to $ B_1,C_1$. Let $ A_2,B_2,C_2$ be the midpoints of $ BC,CA,AB$. Prove that $ A_1A_2, B_1B_2, C_1C_2$ are concurrent  :wink:

Dear friend,
This nice problem have been in our forum many times (it's the fourth I've seen it). You can made a search with the key words [i]incircle midpoints concurrent[/i]

Can you show me any reference? Or post a nice solution?  :)

See my post here: [url]http://www.mathlinks.ro/viewtopic.php?p=1122010&search_id=22528217#1122010[/url].

Dear Livetolove and Mathlinkers,
next on Revistoim (in Spain) and on my site (in French) you will see two synthetic proofs of this problem.
Sincerely
Jean-Louis

Dear Mathlinkers,
two synthetic proofs of this known result originated from Clark Kimberling has been put on my website
http://perso.orange.fr/jl.ayme  vol. 5  The second mid-arc point…
Sincerely
Jean-Louis


Consider the polynomial $P(x)=X^{2n}-X^{2n-1}+\dots-x+1$, where
$n\in{N^*}$. Find the remainder of the division of polynomial
$P(x^{2n+1})$ by $P(x)$.


geometry
projective geometry
geometry solved

Let $(I)$ be inscribed circle of triangle $A_1A_2A_3$. $P_i$ lies on side $A_{i+1}A_{i+2}$ sastifying $A_1P_1,A_2P_2,A_3P_3$ concurent. $t_i$ be tangent of $(I)$ through $P_i$ ($\neq$ $A_{i+1}A_{i+2}$) meets $P_{i+1}P_{i+2}$ at $Q_i$. Proved $Q_1,Q_2,Q_3$ coliner

First I change the notations:

[color=blue][b]Problem.[/b] Let i be the incircle of a triangle ABC, and let D, E, F be three points on the sidelines BC, CA, AB of triangle ABC such that the lines AD, BE, CF are concurrent. Let d, e, f be the tangents to the circle i through the points D, E, F (different from the lines BC, CA, AB). Prove that the points $d\cap EF$, $e\cap FD$, $f\cap DE$ are collinear.[/color]

[i]Solution.[/i] In the following solution, we will use the polar transformation with respect to circles, the Desargues theorem, and the cyclocevian conjugate theorem; the latter states:

[color=blue][b]Cyclocevian conjugate theorem.[/b] Let ABC be a triangle, let X and X' be two points on its side BC, let Y and Y' be two points on its side CA, and let Z and Z' be two points on its side AB such that the points X, X', Y, Y', Z, Z' lie on one circle. Assume that the lines AX, BY, CZ concur. Then, the lines AX', BY', CZ' concur.[/color]

This theorem was problem 1 of the MathLinks Contest, 5th edition, 6th round. Here is a quick [i]proof of this theorem[/i]:

We are going to use directed segments. Since the points Y, Y', Z and Z' lie on one circle, the intersecting chords theorem yields $YA\cdot Y^{\prime}A=AZ\cdot AZ^{\prime}$. Similarly, $ZB\cdot Z^{\prime}B=BX\cdot BX^{\prime}$ and $XC\cdot X^{\prime}C=CY\cdot CY^{\prime}$. Hence,

$\left(\frac{BX}{XC}\cdot\frac{CY}{YA}\cdot\frac{AZ}{ZB}\right)\cdot\left(\frac{BX^{\prime}}{X^{\prime}C}\cdot\frac{CY^{\prime}}{Y^{\prime}A}\cdot\frac{AZ^{\prime}}{Z^{\prime}B}\right)=\frac{BX\cdot BX^{\prime}}{XC\cdot X^{\prime}C}\cdot\frac{CY\cdot CY^{\prime}}{YA\cdot Y^{\prime}A}\cdot\frac{AZ\cdot AZ^{\prime}}{ZB\cdot Z^{\prime}B}$
$=\frac{BX\cdot BX^{\prime}}{CY\cdot CY^{\prime}}\cdot\frac{CY\cdot CY^{\prime}}{AZ\cdot AZ^{\prime}}\cdot\frac{AZ\cdot AZ^{\prime}}{BX\cdot BX^{\prime}}=1$.

Now, since the lines AX, BY, CZ concur, the Ceva theorem yields $\frac{BX}{XC}\cdot\frac{CY}{YA}\cdot\frac{AZ}{ZB}=1$. Hence, $\frac{BX^{\prime}}{X^{\prime}C}\cdot\frac{CY^{\prime}}{Y^{\prime}A}\cdot\frac{AZ^{\prime}}{Z^{\prime}B}=1$, so that, by the Ceva theorem, the lines AX', BY', CZ' also concur, and the cyclocevian conjugate theorem is proven.

Now to the solution of the problem. Let the incircle i of triangle ABC touch the sides BC, CA, AB at the points X, Y, Z and the tangents d, e, f at the points X', Y', Z'. Let $U=YY^{\prime}\cap ZZ^{\prime}$, $V=ZZ^{\prime}\cap XX^{\prime}$, $W=XX^{\prime}\cap YY^{\prime}$.

The polar of a point P outside of a circle k with respect to this circle k is the line joining the points where the two tangents from the point P to the circle k touch k. Hence, the polar of the point A with respect to the circle i is the line YZ, and the polar of the point D with respect to the circle i is the line XX'. Hence, the pole of the line AD with respect to the circle i is the point $YZ\cap XX^{\prime}$. Since the line XX' coincides with the line VW, we can thus state that the pole of the line AD with respect to the circle i is the point $YZ\cap VW$. Similarly, the poles of the lines BE and CF with respect to the circle i are the points $ZX\cap WU$ and $XY\cap UV$. Since the lines AD, BE, CF concur, their poles with respect to the circle i are collinear; in other words, the points $YZ\cap VW$, $ZX\cap WU$, $XY\cap UV$ are collinear. Hence, by the Desargues theorem, the triangles UVW and XYZ are perspective, i. e. the lines UX, VY, WZ concur. But now we have the triangle UVW and the points X and X' on its side VW, the points Y and Y' on its side WU, and the points Z and Z' on its side UV, and we know that these points X, X', Y, Y', Z, Z' lie on one circle (namely, on the circle i) and that the lines UX, VY, WZ concur. By the cyclocevian conjugate theorem, this yields that the lines UX', VY', WZ' also concur.

We proved that the polar of the point D with respect to the circle i is the line XX'. Similarly, the polars of the points E and F with respect to the circle i are the lines YY' and ZZ'. Thus, the pole of the line EF with respect to the circle i is the point $YY^{\prime}\cap ZZ^{\prime}=U$. On the other hand, the pole of the line d with respect to the circle i is the point X' (since the line d is tangent to the circle i at the point X'). Hence, the polar of the point $d\cap EF$ with respect to the circle i is the line X'U = UX'. Thus, the point $d\cap EF$ is the pole of the line UX' with respect to the circle i. Similarly, the points $e\cap FD$ and $f\cap DE$ are the poles of the lines VY' and WZ' with respect to the circle i. Now, since the lines UX', VY', WZ' concur, their poles $d\cap EF$, $e\cap FD$, $f\cap DE$ must be collinear, and the problem is solved.

  Darij


superior algebra
superior algebra unsolved

Prove:
There's no such group $ G$ that: $ G > S_{4}$ and $ G^{(1)}\equal{} S_{4}$.

Hi.

$ S_{4}$ is a complete group. That means: $ Z(S_{4})\equal{}\{1\}$ and $ Inn(S_{4})\equal{}Aut(S_{4})$. One can prove, that every extension of a complete group is a direct product, so our G has the form $ G\equal{}S_{4}\times C$ where $ C\equal{}C_{G}(S_{4})$. If this would be true, then we would have $ S_{4}\equal{}G^{(1)}\equal{}S_{4}^{(1)}\times\ C^{(1)}\equal{}A_{4}\times\ C^{(1)}\neq S_{4}$

I'm sorry I don't know anything about complete groups nowtimes.
I'll go to look up in some "higher" books.
I think it should be a strong method!


is it possible to cover three dimnetional euclidean space with line s which are pairwise skew?

any ideas ? no one couldn't help me ? I'm drowning :D

Have a look at this (you've posted it before, you know :)):

[url]http://www.mathlinks.ro/viewtopic.php?t=4343[/url]

I don't know if it helps, but it's an idea...


can any one prove using high school number theory ...that the sum of any two cubed integers cannot be written as a cubed integer ...this is fermats last theorem for n=3 case.please post a  solution or a clear cut procedure

[quote="t0rajir0u"]Euler's elementary proof is [url=http://fermatslasttheorem.blogspot.com/2005/05/fermats-last-theorem-proof-for-n3.html]here[/url].[/quote]

Thank you. I was curious about at it, too. :roll: