calculus
integration
real analysis
Functional Analysis
complex analysis
real analysis theorems

Please tell me about good PDE books.
Which one is the best?

I have come across these ones, what do you think about them?
How about Asmar ?:
Partial Differential Equations and Boundary Value Problems with Fourier Series (2nd Edition) (Hardcover)
by Nakhle H. Asmar (Author)

or Haberman?:
Applied Partial Differential Equations, Fourth Edition (Hardcover)
by Richard Haberman (Author)

Please tell me about others too.
Please comment.

As far as I understand it you have just completed Matematik 2 in Lund. Is this correct? If so, let me tell you that it's highly unlikely that you are mature enough to read about PDEs. Most likely you will not get anywhere if you borrow a book. It seems wisest to take integration theory and functional analysis before trying to deal with PDEs, but before that you should probably take a couple of Matematik 3-courses for "maturity". There is a reason why the PDE courses are for PhD students...

This is not meant to sound elitist or anything, just advice. It is quite accurate; it would be extremely rare if someone managed to read such a book with the background of matematik 2 only. However, my guess on your background is only from the problems you've posted, so I may be wrong. If I am, then ignore!

Lund, that must be your location? How come your sure that I'm from Lund  :D

Well, I cant see why the books I mentioned are considered so difficult. Perhaps they are
difficult, but are you sure one has to be acquainted with integration theory and fuctional
analysis (at the PhD-level)?

As far as I have understood there are books that contain a mixture of Fourier analysis, Sturm-Liouville and a little bit (if anything) spectral theory among with the equations solving methods themselves. And I assumed that these books above are more or less of this kind.
Otherwise there are books like [url=http://www.amazon.com/Partial-Differential-Equations-Graduate-Mathematics/dp/0821807722/ref=pd_bbs_2/103-0881056-6028629?ie=UTF8&s=books&qid=1182443920&sr=1-2]this[/url] one for which you definitely need some more background.

Have you read these books? Have you read anything on PDE?

[quote="color"]Please tell me about good PDE books.
Which one is the best?
[/quote]
There are two ways to look at PDE's the applied way and the pure way. The pure way concerns the theory of PDE's and existence of solutions along with well-posed problems. This is far to difficult to learn as an introduction. I recommend to first learn the applied side first. The applied side is concerned with actually solving PDE's, different techiques to yield solutions. After you get good with solving PDE's you might want to consider learning the theory.

I think that the book [u]Analytic Methods in Partial Differetial Equations[/u], released by Springer series is a good way to start learning PDE's.

[quote]
Partial Differential Equations and Boundary Value Problems with Fourier Series (2nd Edition) (Hardcover)
by Nakhle H. Asmar (Author)
[/quote]
That book looks okay, the one I posed above is a little more advanced. But its good as a first reading.

And please avoid the following book at all costs if your a beginnger:
[url=http://www.amazon.com/Partial-Differential-Equations-Graduate-Mathematics/dp/0821807722/ref=pd_bbs_2/103-0881056-6028629?ie=UTF8&s=books&qid=1182443920&sr=1-2]Link[/url]

[quote="color"]Lund, that must be your location? How come your sure that I'm from Lund  :D
[/quote]
I  have deduced it using detective skills on par with Sherlock Holmes. So am I not right? Maybe we've even seen each othr.
[quote]
Well, I cant see why the books I mentioned are considered so difficult. Perhaps they are
difficult, but are you sure one has to be acquainted with integration theory and fuctional
analysis (at the PhD-level)?
[/quote]
You still haven't confirmed that I am correct about your level of education. If so, I am fairly certain study of PDEs will be over your head. I don't know, however, what the "applied way" of studying PDEs is. Is it learning methods without proof to solve some very particular types of equations? if you're a mathematician, how is this particularly useful if you can't even show that you've found all solutions? If you're an engineer, how is this useful when you have computers? Of course there are self-contained books that don't require prerequisite knowledge of anything, but functional analysis is definitely the language of modern analysis.

[quote]
As far as I have understood there are books that contain a mixture of Fourier analysis, Sturm-Liouville and a little bit (if anything) spectral theory among with the equations solving methods themselves. And I assumed that these books above are more or less of this kind.
Otherwise there are books like [url=http://www.amazon.com/Partial-Differential-Equations-Graduate-Mathematics/dp/0821807722/ref=pd_bbs_2/103-0881056-6028629?ie=UTF8&s=books&qid=1182443920&sr=1-2]this[/url] one for which you definitely need some more background.
[/quote]
As to what these terms mean to me understanding fourier analysis requires integration, sturm-liouville theory complex analysis and spectral theory functional analysis. Of course I can't definitely say whether or not you will have an exchange from reading the books, but I am pretty sceptical (again I am only guessing as to your background).

[quote]Have you read these books? Have you read anything on PDE?[/quote]
"No" and "just bits and pieces". But I have used Hörmanders' books for certain things.

These are just my opinions, and may not even be relevant. If you are desperately curious about PDEs I see no reason to not get a book to look in. You could just borrow some from the library. I am just saying it's an advanced topic, and serious study is best left for later.

Thank you for you replies!  :lol:
[b]
blahblahblah[/b]: Yes I've been looking in that book, it had some intro too, and as you say
it didn't seem that hard enough to read, except for the part that requires more background perhaps.

[b]sylow_theory[/b]: Thanks, please give me a full reference to the book you mentioned
"Analytic Methods in Partial Differential Equations"

[b]Kalle[/b]:
My "level of education" should be clear from the suggestions I made, or perhaps
your inferior role as Sherlock Holmes  :lol: .
[i]I am open for all kind of suggestions and I am aware of the distinction between
applied books and pure such. I've read applied books on PDE's, but I just wanted to
hear from others experience and perhaps catch up some names on, later on, useful books.
[/i]
[quote="Kalle"]
 "applied way" of studying PDEs is. Is it learning methods without proof to solve some very particular types of equations? if you're a mathematician, how is this particularly useful if you can't even show that you've found all solutions?
[/quote]
Its good to be critical on this issue and especially
when it comes to "bad" applied books, but I think you're being too harsh here. Perhaps this example is too but I'm tired of the discussion applied/pure.
[i]Example: Even if I had the ability to read Rudin calculus books, I would probably still study calculus books too, since it has some benefits (of course it should be a decent calculus book)[/i]

Thanks for your reply anyway  :) . Hörmander is a good suggestion for advanced level
isn't it?

[quote="blahblahblah"]Despite the advice in this thread, Evans (the link is in sylow_theory's post) is the bible and it's easy enough to read.  If you just want to get familiar with the mechanics of solving PDE (heat/wave/laplace equations, seperation of variables, characteristics, maybe some fourier transform) and are less interested in the theory, then there are plenty of texts available - the one I used was Strauss (but I didn't like it very much).[/quote]
I took it out from the library and I looked out it. I immediately realized this book requires a very very solid knowledge of analysis. So I didnt take it out.

[quote]
Thanks, please give me a full reference to the book you mentioned
"Analytic Methods in Partial Differential Equations"
[/quote]
Just use amazon.
[url=http://www.amazon.com/Analytic-Methods-Partial-Differential-Equations/dp/3540761241/ref=sr_1_1/104-0362355-0628725?ie=UTF8&s=books&qid=1182464430&sr=8-1]Link.[/url]

Use the [url] tags or http://www.tinyurl.com to avoid stretching my screen with long urls, please.  (To use the url tags in this way, write [url=http://www.whatevergoeshere.com]whatever text you want[/url].)

I'll just mention that I taught a PDE course a dozen or so years ago, and used Haberman as the text. The course had only a first ODE course as prerequisite, and explicitly did not have a "real analysis" or "advanced calculus" course as prerequisite. It was a classical approach, focusing mostly on linear constant coefficient equations, especially the "big three" from elementary physics: the wave equation, the heat equation, and Laplace's equation. (The "type specimens" for the classification of differential operators into hyperbolic, parabolic, or elliptic.) Haberman's book fitted that kind of course well. Of course I skipped some things in the book and supplemented others. That's just me - I can never follow any textbook exactly.

Asmar is a somewhat newer book. I know Dr. Asmar personally and even have a joint paper with him. His book is also pitched at the same level - no explicit analysis prerequisite. I haven't had a chance to teach from it. I also haven't taught from Strauss, but I know that some other faculty members in my department have used it for the same course I was talking about above.

[color=green][I just used my moderator editing powers to apply one of JBL's suggestions to sylow_theory's posts.][/color]

I just realized that the book I suggested and the book you suggested (blahblahblah) are written both by Evans!  :rotfl:


Looks like there will be a 30,000th user soon. Time to party!!!

oh dudes i am so getting the 30000th user

woohoo
dude a new user's rating is 4.2?
i wonder why?

[quote="bubka"]oh yes
last time solafidefarms got it if i amn't mistaken[/quote]

I did? Time to go look it up...

EDIT: You are/were mistaken, I'm innocent of the 20000th User.

Why is there a 30000th User already...? According to the memberlist it's still only at 28224 users...

The count on the front page I think includes accounts deleted... like spam accounts
btw, i know who created 20000th user

[quote="vishalarul"]I just noticed that only the first 285 pages of the member list include people that have at least one post.[/quote]
[quote="tjhance"]I find it interesting that, out of 565 pages on the memberlist, 345 of them are filled completely by users with 0 posts.[/quote]
565-345=220, not 285.

@ tjhance - Some of those people come to AoPS only to read the topics, but do not actively (if ever) post.


Putnam
logarithms
binomial coefficients
Pascal\u0027s Triangle

Is there an easy way to find the sum of the digits in a number? Like an explicit formula or anything like that?

i think it depends on what number you have.
for instance, you can never find the sum of the digits of $\pi$.

Not much is known about the mathematical properties of the sum of digits function. One well-known and easy to prove property is: (b is the base)

SODb(n) = n mod b-1.

∑n≥1 SODb(n) / (n(n+1)) = b/(b-1)ln(b).
The case b=2 was presented in the Putnam competition 1981 and generalized by Shallit 1985 to arbitrary bases b.

∑n≥2 SOD2(n) (2n+1) / (n2(n+1)2) =  :pi:2 / 9,
∑n≥1 SOD2(n)2 (8n3+4n2+n-1) / (4n(n2-1)(4n2-1)) = 17/24 +ln2.
(Allouche and Shallit 1992)

∑n<m SODb(n) = [(b-1)/2] · m · logbm +O(m).
(Bush 1940, Mirsky 1949)

Let e(n) be the exponent of the highest power of 2 dividing n!. Then n = SOD2(n) + e(n).
(Legendre)

The number of odd binomial coefficients in the n-th row of Pascal's triangle is exactly 2^SOD2(n).
(Fine 1957)

[quote="eryaman"]If you know the number...don't you know the digits?  Or do you mean scientific notation, like how many digits are in $50^{70}$.  If so there's a method involving $\log$ to find that.[/quote]

i dont think i totally understand the question but if u wanna know how many digits in a number its just the log of the number (round down) plus 1 i think...
e.g. 50^70= 70log50=118.927 ------> 119 digits

sorry i thought you wanted to know the sum of the digits, (why is your topic called sums then?)

I'm think he does want the sum and accidentally wrote "num" instead of "sum".

yeah. i answered for the sum. because the number of integers is pretty easy to find! it's just the $log$ of the number $+1$


prove that $ f(x,y)$ is irreducible in $ z[x,y]$
where
$ f(x,y) \equal{} x^{200}y^{5} \plus{} x^{51}y^{100} \plus{} x^{106} \minus{} 4x^{100}y^{5} \plus{} x^{100} \minus{} 2y^{100} \minus{} 2x^{6} \plus{} 4y^{5} \minus{} 2$

i have a short and nice soloution for that.

this is related to mathemathical subjects...
so we can find the problems and the solutions for those problems..
==========================================

Lenin
[url=http://www.globalinternetmarketing.net]WideCircles[/url]


So lil' bro Danny has joined AoPS and is signed in. I was showing him how to rate a post and I was told that he had to wait 31071 hours before he could vote again. I don't know if this is a function of him being a very new user or what, but I don't really think he need wait till I'm off at college to rate posts :).

That's a bug. It goes down dramatically soon. My friend, Mersenne, had joined too. Once he posted 25 times (or 35 or something like that) it goes away.

I'm now looking at this and I have no idea how it could have told your brother that. You cannot vote unless you have 25 posts on the board (and there is no hourly limit for that).

Ok, bug solved. As suspected it was a forgotten division by 3600 :D


Find all integers $ n\geq 3$ such that for any two positive integers $ m, r < n\minus{}1$ there exist $ m$ distinct elements of the set $ \{1, 2, . . .n \minus{} 1\}$ whose sum is
congruent to $ r$ modulo $ n$.

The desired integers are the odd integers.

- If $ n\equal{}2k >3$ is even.
Note that $ \sum_{i\equal{}1}^{2k\minus{}1} i \equal{} k(2k\minus{}1)\equal{}k \mod[2k].$
If $ n$ has the desired property then, for $ m\equal{}2k\minus{}2$ and $ r\equal{}k$, you may select $ 2k\minus{}2$ from the $ 2k\minus{}1$ given numbers which add up to $ k$. It follows that the remaining number (the only one which has not been selected) must be $ 0 \mod [2k]$, which is impossible.
Thus, if $ n$ is a solution then $ n$ is odd.

- If $ n\equal{}2k\plus{}1$ where $ k \geq 3$ (the case $ n\equal{}3$ is obvious).
Note that we may pair the given numbers as $ (1,2k),(2,2k\minus{}1), \cdots, (k,k\plus{}1)$ so that the sum of each pair is $ 0 \mod[2k\plus{}1]$.
Now, it is obvious that for $ m\equal{}1$, we have the desired property for any positive $ r < n\minus{}1$ (just choose $ r$).
Adding numbers by pair (there are $ k\minus{}1$ admissible pairs because $ (r,2k\minus{}r)$ is not allowed), it follows that, for each positive $ r < n\minus{}1$ and any odd positive $ m<n\minus{}1$, we may find $ m$ of the given numbers which add up to $ r$ modulo $ 2k\plus{}1$.
In another hand, considering the numbers $ i\plus{}1$ for $ i\equal{}2,\cdots, 2k\minus{}2$, and $ (2k)\plus{}2$ and $ (2k)\plus{}3$, we deduce that for any positive $ r<n\minus{}1$, we may select two of the given integers which add up to $ r$. Thus, as above, adding numbers by pairs (there are $ k\minus{}2$ admissible pairs), we deduce that for any positive $ r,m<n\minus{}1$ with $ m$ even, we may select $ m$ of the given numebrs which add up to $ r$ modulo $ 2k\plus{}1$.
Therefore, $ n$ is a solution.

Pierre.


demetine all pair(a,b) of integer with the property that number
      2a 2 +5b and 2b 2 +5a are both perfect squares  :(

locking modulo 5 there are to options, a is divisible by 5 or 2 is a square modulo 5, like 2 doesnt holds, we conclude a is divisible by 5, but then so does b, say a=5A, b=5B, after expansion and cancelation we have
2A 2 +B and 2B 2 +A are squares, now it is obvious that A and B have the same parity, so if both are even we have: A=2m B=2n, and the condition becomes:
4m 2 +n and 4n 2 +m are squares, but look they are almost a square, now supose |m|>=|n|, then we have:

(2m+1) 2 >4m 2 +n>(2m-1) 2 , so we can only have n=0 and then m is a perfect square, and this gives tne only solution in this case.

Now we just have to analise what happens when they are both odd, in fact they ill be 1 mod 4, but i dont know how to do this part!

Pascual

Your solution  is very nice but can you solve more beautiful ? thank

[quote="pascual2004"]locking modulo 5 there are to options, a is divisible by 5 or 2 is a square modulo 5, like 2 doesnt holds, we conclude a is divisible by 5, but then so does b, say a=5A, b=5B, after expansion and cancelation we have
2A 2 +B and 2B 2 +A are squares,
[/quote]
Until here it is correct, but from here it goes wrong:
[quote="pascual2004"] now it is obvious that A and B have the same parity, so if both are even we have: A=2m B=2n, and the condition becomes:
4m 2 +n and 4n 2 +m are squares,
Pascual[/quote]
firstly, don't need to have the same parity. For example A=1; B=2 is a solution. And even if they were both even then the contition what you get, is incorrect.
I honestly have no clue how to solve this problem, it's clear that when A = B, then it has infinitely many solutions but the other cases seem more difficult.
I hope that somebody (pascual2004 maybe you ?) comes up with a solution.


[quote="plane geometry"]ABCD is a cyclic quadrilateral. Lines AB CD intersect at E, lines AD BC intersect at F, and EM and FN are tangents to the circumcircle of ABCD. Two circles are constructed with E,F, their centers and EM,FN their radii, respectively. K is one of their intersections. Prove that EK is perpendicular to FK .[/quote]
Nice problem!
[hide="Solution"]
Notice that $ FA\cdot FD \equal{} FB\cdot FC \equal{} FN^2 \equal{} FK^2$, which implies that $ FK$ is tangent to the circumcircle of $ \triangle CKB$ and $ FK$ is tangent to the circumcircle of $ \triangle AKD$.  Yet, this implies that $ FK$ is tangent to both the circumcircles of $ \triangle AKD$ and $ \triangle CKB$ at $ K$, so the circumcircles of $ \triangle AKD$ and $ \triangle BKC$ are tangent at $ K$.  Similarly, the circumcircles of $ \triangle AKB$ and $ \triangle CKD$ are tangent at $ K$.

Now, invert around $ K$, which maps any point $ P$ to $ P'$.  Since the circumcircles of $ \triangle AKB$ and $ \triangle CDK$ are tangent at $ K$, we see that $ A'B'\parallel C'D$.  Similarly, $ A'D'\parallel B'C'$, so $ A'B'C'D'$ is a parallelogram.  Yet, $ ABCD$ is cyclic, so $ A'B'C'D'$ is cyclic parallelogram.  Hence, $ A'B'C'D'$ is a rectangle.  Now, this means that $ \angle KA'D' \plus{} \angle KA'B' \equal{} \angle B'A'D'\equal{}90$.  Note that $ \angle KA'D' \equal{} \angle KDA$ and $ \angle KA'B' \equal{} \angle KBA$, so $ \angle KBA \plus{} \angle KDA \equal{} 90$.  Yet, since $ FK$ is tangent to the circumcircle of $ \triangle AKD$, we have that $ \angle FKA \equal{} \angle FDK\equal{}\angle KDA$.  Similarly, $ \angle EKA \equal{} \angle KBA$, so \[ \angle FKE \equal{} \angle FKA \plus{} \angle EKA \equal{} \angle KDA \plus{} \angle KBA \equal{} 90\] [/hide]

thank The QuattoMaster 6000 a lot! :lol:
small type mistake $ A'B'\parallel C'D'$


[color=darkblue]Find $ (x; y)$ of this system: $ \begin {cases}\log_{2}{(x \plus{} y)} \equal{} x \plus{} y \minus{} 1 \\
\log_{x \plus{} y \plus{} 2}{(xy \plus{} 1)} \equal{} x \plus{} y \minus{} 1 \end {cases}$[/color]

[quote="quark1005"]what does 'root' of the system mean? Do you mean solutions for x and y ?[/quote]

Yes, solutions for x and y! :lol:  :P

[quote="seriousstone89"]you can get the first equations as follows
 let x+y = n
 n = 2^(n-1)
 n = (2^n)/2
 2n = 2^n which impiles that n =1,2
 so x + y = 1 or x + y = 2
for the second equation
use this result and proceed[/quote]

[color=darkblue]Thank seriousstone89. Your solution is nice.[/color]


[quote="Rushil"]Solve,

$ (x)^2  $ + $ (x+1)^2 $ = 25 , where $ (x)$ is the least integer greater than or equal to $x$[/quote]

The soln. set should be [-4,-3) U [3,4) ....

[quote="Andreas"]The only ways to obtain $25$ as a sum of two integer square $a^2 + b^2 = 25$ is $a = 5, 0, 3, 4$
and $b = 0, 5, 4, 3$ respectively.
So it's just to look every single case that in total are: $4$ for $a = 0, 5$ and $8$ for $a = 4, 3$.
It results that solutions are $-5 < x \leq -4 \vee 2 < x \leq 3$[/quote]

Ooopsss!!! I thought (x) to be the greatest integer <= x


calculus
integration
function
trigonometry
induction
calculus computations

A function $f(\theta)$ satisfies the following conditions $(a),(b)$.

$(a)\ f(\theta)\geq 0$

$(b)\ \int_0^{\pi} f(\theta)\sin \theta d\theta =1$

Prove the following inequality.

\[\int_0^{\pi} f(\theta)\sin n\theta \ d\theta \leq n\ (n=1,2,\cdots)\]

$|\sin(n+1)\theta | = |\sin n\theta \cos \theta + \sin \theta \cos n\theta | \leq  |\sin n\theta | + |\sin \theta |$

The induction works verry well : $\int_0^{\pi} f(\theta)|\sin (n+1)\theta| \ d\theta \leq \int_0^{\pi} f(\theta)|\sin n\theta| \ d\theta + \int_0^{\pi} f(\theta)|\sin \theta| \ d\theta \leq n+1 $
:cool:

That's right,Deogene.The point is Induction.


calculus
integration
real analysis
real analysis solved

u(n) a sequence given by
u(0)>0 and u(n+1) = 2u(n) + 2 \sqrt u(n) \sqrt(1+u(n))

i) Show that int_{u(n)^(+oo)} dx/(x \sqrt (1+x)) =

2int_{u(n+1)^+oo}dx/(x \sqrt (x+1))

ii) Calculate the limite of the sequence of u(n)/4^n when n-->+oo

Oral Polytechnique


probability
geometry
geometric transformation
reflection

[quote="mdk"]A computer is programmed to randomly generate a string of six symbols using only the letters A,B,C. What is the probability that the string will not contain three consecutive A’s?[/quote]

[hide]Let's do complementary counting.
We want to find how many ways we CAN have 3 consecutive A's.

CASE 1: Only 3 consecutive As.
Let $ AAA$ be a "super A" block, denoted by $ \mathbf{A}$.

We have four cases:
$ \mathbf{A}ijk$
$ i\mathbf{A}jk$
$ ij\mathbf{A}k$
$ ijk\mathbf{A}$

In the first case, $ i$ can be $ B,C$, and $ j,k$ can be $ A, B, C$. So we have $ 2\cdot 3^{2}$ ways.
In the second case, $ i,j$ can be $ B,C$, and $ k$ can be $ A, B, C$. So we have $ 2^{2}\cdot 3$ ways.
The third case is similar to the second case, so we have $ 2^{2}\cdot 3$ ways.
The fourth case is similar to the first, so we have $ 2\cdot 3^{2}$ ways.

Adding up, we get $ 60$ ways.

CASE 2: 4 consecutive As.
Let $ AAAA$ be a "super A" block, denoted by $ \mathbf{A}$.

We have three cases:
$ \mathbf{A}ij$
$ i\mathbf{A}j$
$ ij\mathbf{A}$

In the first, $ i$ can be $ B,C$, and $ j$ can be $ A, B, C$. So we have $ 6$ ways.
In the second, $ i,j$ can be $ B, C$. So we have $ 4$ ways.
The third case is similar to the first, so we have $ 6$ ways.

Adding up, we get $ 16$ ways.

CASE 3: 5 consecutive As.
Let $ AAAAA$ be a "super A" block , denoted by $ \mathbf{A}$.

We have two cases:
$ \mathbf{A}i$
$ i\mathbf{A}$

In the first, $ i$ can be $ B,C$. So we have $ 2$ ways.
The second case is similar to the first, so we have $ 2$ ways.

Adding up, we get $ 4$ ways.

CASE 3: 6 consecutive As.
This case is obvious, we only have $ 1$ way.

Hence, our probability is:
$ 1-\frac{60+16+4+1}{3^{6}}= 1-\frac{3^{4}}{3^{6}}= \boxed{\frac{8}{9}}$[/hide]

EDIT: I think I have the right answer. Gr... I hate casework. How would you use PIE?

hint

[hide]Let $ s_{n}$ denote the number of strings of length n with no three consecutive
A's. Thus $ s_{1}= 3, s_{2}= 9$ and $ s_{3}= 26$[/hide]

One possibility is [hide="a method"]a solution by recursion[/hide] perhaps suggested by mdk's hint.

Vishalarul, there's a problem with your solution: if the string were of length 29, you would have been arguing that the probability of getting AAA is 1, which is clearly not the case.  (If the string was of length 30, you'd be arguing that the probability of getting AAA is bigger than 1 :wink:.)  You're overcounting, and you need some [hide="food"]PIE[/hide] to make it work out properly.

Theres a 1/9 chance it will have atleast 3 consecutive As.  1-1/9=$ \frac{8}{9}$ it wont have 3 consecutive As.

vis, your way is correct except you have overcounted some things...

however, we do not need to use PIE to subtract overcounts..

we just need to use logic.

da ban man, can you show us how you get your chance of $ \frac{1}{9}$?

I tried to use recursion (albiet, my abilities in recursion are poor), and I reached the probability $ \frac{2}{9}$:

[hide]
Let $ A_{n}$ be the number of ways a string of $ n$ characters can attain 3 consecutive A's.

Trivially, $ A_{1}=0; A_{2}=0; A_{3}= 1$.

For $ A_{4}$, we can draw a diagram:
_ _ _ _
and this string of four spaces can be divided into two parts: the string containing the first 3 spaces, and the string containing the last three spaces.

For the string containing the first three spaces (the red line divides the first three space from the extraneous last space),
_ _ _ [color=red]|[/color] _
Notice that the first three spaces (_ _ _ [color=red]|[/color]) is equivalent to $ A_{3}$, which is 1. We have to factor in the last space ([color=red]|[/color] _), which can be either $ A, B, or C$, so we multiply $ A_{3}$ by $ 3$. The same goes for the string containing the last three spaces (it is just a reflection of this case); there are $ 3*A_{3}$ cases of three consecutive A's.

Then we have to subtract overcounting of 3 consecutive A's, which is either the union or the intersection of the two split cases. If it is the union, then all four spaces has to be A's, so that is $ 1$ case.

If it is the intersection, then the intersection of both strings are:
_ [color=white]|[/color] _ _ [color=red]|[/color] _
_ [color=red]|[/color] _ _ [color=white]|[/color] _
[color=white]----[/color]$ \leftrightarrow$
_ [color=red]|[/color] _ _ [color=red]|[/color] _
So any consecutive-3-A's in the middle two spaces, which is $ A_{2}$, must be subtracted. Since $ A_{2}=0$, 0 cases must be subtracted.

So $ A_{4}= 6A_{3}-1-A_{2}= 5$
Which is how I came up with the recursion $ A_{n}= 6A_{n-1}-1-A_{n-2}$, giving $ A_{6}= 162 = 3^{4}*2$.

Since a string of 6 letters can have $ 3^{6}$ possible cases, the probability is
$ \frac{3^{4}*2}{3^{6}}=\frac{2}{9}$
and the complement of this is $ \frac{7}{9}$

Sorry for the lengthy solution![/hide]

[quote="JBL"]Here is how I would solve it with recursion:

[hide]Let $ a_{n}$ be the number of sequences of length $ n$ with no AAA which end in no As.  Let $ b_{n}$ be the number with no AAA which end in a single A, and $ c_{n}$ be the number with no AAA which end in a double A.  Then we have $ a_{0}= 1, b_{0}= c_{0}= 0$ (or, if you'd rather start at $ n = 1$, $ a_{1}= 2$, $ b_{1}= 1$, $ c_{1}= 0$) and $ a_{n+1}= 2a_{n}+2b_{n}+2c_{n}$, $ b_{n+1}= a_{n}$ and $ c_{n+1}= b_{n}$.  Then it's easy to make a table to calculate than $ a_{6}= 444$, $ b_{6}= 152$ and $ c_{6}= 52$, so our answer is $ \frac{a_{6}+b_{6}+c_{6}}{3^{6}}= \frac{648}{729}= \frac{8}{9}$.[/hide]

da ban man, you may have a correct solution (although I suspect that what you call "logic" is, in fact, an application of PIE or possibly just casework, which is easy enough in this example), but it's rather difficult to tell since you gave no indication whatsoever of how you reached your solution.[/quote]

well i guess it was like PIE
from what vis already did
ijkA
ijAk
iAjk
Aijk
only the last one has 3^3 solutions.  The rest have only 3^2*2 (so there are 3 of them) solutions  (this preventing repeats).

Thus the probability is 1/9 thenn 1-1/9=8/9
although recursion is alot cooler..

[quote="JBL"]Here is how I would solve it with recursion:

[hide]Let $ a_{n}$ be the number of sequences of length $ n$ with no AAA which end in no As.  Let $ b_{n}$ be the number with no AAA which end in a single A, and $ c_{n}$ be the number with no AAA which end in a double A.  Then we have $ a_{0}= 1, b_{0}= c_{0}= 0$ (or, if you'd rather start at $ n = 1$, $ a_{1}= 2$, $ b_{1}= 1$, $ c_{1}= 0$) and $ a_{n+1}= 2a_{n}+2b_{n}+2c_{n}$, $ b_{n+1}= a_{n}$ and $ c_{n+1}= b_{n}$.  Then it's easy to make a table to calculate than $ a_{6}= 444$, $ b_{6}= 152$ and $ c_{6}= 52$, so our answer is $ \frac{a_{6}+b_{6}+c_{6}}{3^{6}}= \frac{648}{729}= \frac{8}{9}$.[/hide]

da ban man, you may have a correct solution (although I suspect that what you call "logic" is, in fact, an application of PIE or possibly just casework, which is easy enough in this example), but it's rather difficult to tell since you gave no indication whatsoever of how you reached your solution.[/quote]

nice answer. 8/9 is indeed the answer.

[quote="JBL"]$ a_{n+1}= 2a_{n}+2b_{n}+2c_{n}$[/quote]

How did you derive that?

I thought that it would be:
$ a_{n+1}= 3(a_{n}+b_{n}+c_{n})$,
but this is wrong since you got the right answer.

A string in $ a_{n+1}$ must not end in any As.  You can get such a string by adding B or C to the end of any string of length $ n$, and in no other way.  Thus, $ a_{n+1}= 2(a_{n}+b_{n}+c_{n})$.


integration
limit
logarithms
calculus
calculus computations

[hide="And so"]$L = \lim_{a\to1}\left[\frac{\frac{\pi}{2a}+2a\ln a}{\frac{\pi a^{2}}{2}-2\ln a}\right]^{\frac{1}{a-1}}$. As this is an indetermination $1^{\infty}$, lets make $L = e^{L'}$ where

$L' = \lim_{a\to1}\frac{1}{a-1}\left(\frac{\frac{\pi}{2a}+2a\ln a}{\frac{\pi a^{2}}{2}-2\ln a}-1\right) = \lim_{a\to1}\left(\frac{2\frac{a+1}{a-1}\ln a-\frac{\pi}{2}\cdot \frac{a^{2}+a+1}{a}}{\frac{\pi a^{2}}{2}-2\ln a}\right)$
$= \frac{2\lim_{a\to1}\frac{\ln a}{\frac{a-1}{a+1}}}{\pi/2}-\frac{3\pi/2}{\pi/2}= \frac{4}{\pi}\cdot 2-3 = \frac{8}{\pi}-3$

Finally, $L = e^{\frac{8}{\pi}-3}$.
[/hide]

Just a question: what is the meaning of "CMJ"?

I think it means [color=indigo]College Mathematics Journal[/color]


Should we assume that rocket after launch behaves as free particle :?:

I am not very sure. Assume and see if it works, I think we should

i think the ans is

[hide="answer"]time=T(.5+(1/pi)*cos(theta/2))
max distance=Rcos(theta/2)
as theta ->0
time=T(.5+1/pi) dist=radius of the earth[/hide]

Yep! The answer is correct.... Post your method, and hide it...

yes u need to hide it to give others equal chance...i did this one for u  :)


in fact.if $k=1$
we can prove $(x^k - \frac{1}{{x^k }})(y^k - \frac{1}{{y^k }}) \le (2^k - \frac{1}{{2^k }})^2$
so we can let $k \to 1^+$ and the inequality will certainly wrong.


Prove that there exists  constants  $a , b , c \; ,\; \; 0\le a< b\le 1\le c \; ,$ such that
\[ (*)\; \; \; \; \; \; \frac{\pi^2}{6}-\frac{1}{n^c+b}<\sum\limits_{k=1}^{n}\frac{1}{k^2}<\frac{\pi^2}{6}-\frac{1}{n^c +a}\; \; ,\; \; \forall n\in \{1,2,...\}.  \]
Which are the best  (admisible)  constants $a,b,c$  in (*)  ?

[quote="siuhochung"]It's possible to take c=1 and a to be any number in [0.4,1). I haven't worked out the value for b yet.[/quote]
Right, $c=1$  !


What is the $ y$-coordinate of the $ y$-intercept of the line that passes through the point $ (2,\minus{}3)$ and has slope $ \frac12$?

The equation for a line that has slope of $ 1/2$ is $ y\equal{}\frac {x}{2}\plus{}b$. $ (2,\minus{}3)$ should be true in this equation so plugging this in gives $ \minus{}3\equal{}1\plus{}b$, and $ b\equal{}\minus{}4$. $ b$ is equal to the $ y$-coordinate of the $ y$-intercept, so the answer is $ \minus{}4$.


integration
parameterization
analytic geometry
calculus
function
derivative
vector

Hey guys =) I am just wondering what this equation is and would love any information about it =)
$ F_8 = \underbrace{\int\int\int}_V \beta_p dv + \underbrace{\int\int\int}_V [\"{R} + [2 \omega x\vee_{xyz}] + [ \dot{\omega} xr] + \omega x[\omega xr]]\rho dv = \oint\oint \vee_{xyz}[\rho \vee_{xyz} \cdot dA] + \frac {\partial}{\partial t_{xyz}}\int\int\int \vee_{xyz}[\rho dv]$

Thanks a lot for any information.
By the way the $ \.{omega} xr$ is because I could not figure out how to dot an omega....i tried \.{\omega} but it keeps cutting off the code from there :( I don't know how else.

Edit: Followed JRav's suggestion. Thanks a bunch =)
Could you explain to me the difference between \.{} and \dot{}? =)

Also how can I have the long equation all show up in one line? Even if its advised not to, just to know =)

Is this what you want: $ \dot{\omega}$?  Just use \dot{\omega}.

I think the forum automatically moves the syntax to the next line after some point.  I'm not sure if you can get it all on one line.

As for your syntax question, here is \.{x}: $ \.{x}$, and here is \dot{x}: $ \dot{x}$.  It's hard to see but the second $ x$ is a little bigger.  The second one is more useful (I think) because as an extension, you can use \ddot{x}, which gives $ \ddot{x}$.

And I would answer your actual question if I knew the answer. :wink: Where did you get it from?

Can you tell us what the parameters mean? And where did you get that equation from?

It looks like some sort of coordinate system, and angular quantities.

The second integral is very interesting:
\[ \underbrace{\int\int\int}_{V}[\"{R}+[2\omega x\vee_{xyz}]+[\dot{\omega}xr]+\omega x[\omega xr]]\rho dv\]
This looks like a sum of fictional accelerations in a noninertial reference frame.
The first term is linear force due to linear acceleration.
The second term is Coriolis, with $ \vee_{xyz}$ playing the role of velocity, and $ x$ being some sort of dimensionless(?) factor.
The third seems to be azimuthal force, which I'm not too familiar with, but I know goes with $ \dot{\omega}$.
The fourth is centrifugal force, I think, with some x's floating around.

And the $ \rho$ reminds me of fluid flow. That's about all I can offer.

I can't give you any more than you already know :P
My friend's parents visited her in Waterloo from China and they toured around a bit. THey went to Boston and MIT/Harvard, etc.
She bought me a sweater with Maxwell's 4 equations, and a white T-shirt with this equation.
It says : What part of $ bla$ don't you understand? =)

Also what is the significance of the [] brackets? The shirt uses them...as opposed to ().

Hmm, i didnt notice this thread until now.., its interesting..

So basically its the equation of motion of some continuum in a rotating and accelerating frame.

lets clear your notation a bit. Under $ \underbrace{\int\int\int}_{V}dv$ you mean integration over the volume $ V$. I will denote this as $ \int_V dV$. Under $ \oint\oint d\vec A$ you mean surface integral over a closed surface. I will write this as $ \oint d%Error. "vecA" is a bad command.
$. Under $ V_{xyz}$ you mean the velocity field of the continuum. i will write this as $ \vec v (\vec r,t)$. Under $ \rho$ you mean the mass density, which is a function of position and time, i will write this as $ \rho( \vec r,t)$. by $ t_{xyz}$ you mean time.. i will write this as simply $ t$.

The other notation a bit later.

so Newton says:

\[ \vec F_s= \frac{d\vec p}{dt}\]

$ \vec p$ is the linear momentum.

so lets observe the term on the RHS:

the momentum is:

\[ \vec p = \int_V \rho \vec v dV\]

Now when calculating the time derivative we have to be careful, since the volume $ V$ we are integrating on, changes too. Hence the total derivative will look like this:

\[ \frac{d}{dt}\int_V \rho \vec v dV= \frac{\partial}{\partial t}\int_V \rho \vec v dV + \oint (\rho \vec v) \vec v \vec dA\]

So as you see we have the RHS of your equation. And basically it is just the change of momentum in time.

Now if we are in a rotating frame we have to add fictious forces, so the LHS will contain these forces too.

In your notation the $ x$ represents the vector product, of two vectors.
So we have term by term:

\[ \int_V\rho\ddot\vec R dV\]

this represents the linear acceleration of the rotating frame. $ \vec R$ is the coordinate pointing to the origin of the rotating frame.

\[ \int_V \rho 2\vec\omega \times \vec v dV\]

is the coriolis force.

\[ \int_V \rho \dot\vec\omega\times\vec r dV\]

Is the euler force.

\[ \int \rho \vec\omega\times(\vec\omega\times\vec r)dV\]

Is the centrifugal force.

And we have one more term left.

this is the force resulting from some outer force field, and the inner stress of the continuum.

So the $ \vec \beta$ term is basically:

\[ \vec\beta = \vec f + \text{Div}\sigma\]

where $ \sigma$ is the stress tensor.

Hence we have all the terms of your equation. Hope it was clear... :D


In an $ ABC$ triangle such that $ m(\angle B)>m(\angle C)$, the internal and external bisectors of vertice $ A$ intersects $ BC$ respectively at points $ D$ and $ E$. $ P$ is a variable point on $ EA$ such that $ A$ is on $ [EP]$. $ DP$ intersects $ AC$ at $ M$ and $ ME$ intersects $ AD$ at $ Q$. Prove that all $ PQ$ lines have a common point as $ P$ varies.

We will use with unoriented segments. According to Menelaus' theorem, applied in triangle $ PED$ for the transversal $ \overline{AMC}$,
\[ \frac {PA}{AE} \cdot \frac {EC}{CD} \cdot \frac {DM}{MP} \equal{} 1,
\]
and since the quadruple $ (E, B, D, C)$ is harmonic, $ EC/CD \equal{} EB/BD$. Therefore,
\[ \frac {PA}{AE} \cdot \frac {EB}{BD} \cdot \frac {DM}{MP} \equal{} 1,
\]
and thus, by Ceva's theorem, we conclude that the lines $ EM$, $ DA$, and $ PB$ are concurrent at $ Q$ (i.e. $ PQ$ passes through $ B$).

What means Q1 ?! This problem is very easy !

[quote="anonymous1173"][color=darkred]Consider in a triangle $ ABC$ two fixed  points $ \{D,E\}\subset BC$ so that $ D\in (BC)$ and $ B\in (DE)$ . For a [u]mobile point[/u]

$ P\in AE$ such that $ A\in (EP)$ define the points  $ \{\begin{array}{c} M\in DP\cap AC \\
\ Q\in ME\cap AD\end{array}$ . Prove that $ PQ$ passes through a [u]fixed point[/u].[/color][/quote]
[color=darkblue][b][u]Proof.[/u][/b] Denote $ F\in PQ\cap BC$ . Since the lines$ PF$ , $ AD$ , $ EM$ are concurrently ( in a point $ Q$ ), then

the point $ F$ is conjugate of the point $ C\in BC\cap AM$ w.r.t the points $ D$ , $ E$ , i.e. the point $ F$ is fixed.

[b][u]Another particular cases.[/u][/b]

[b]1.[/b]  $ E\in AA\cap BC$ and $ D$ - the foot of the $ A$ - symmedian $ \Longrightarrow$ $ F = B$ .

[b]2.[/b] $ AP\parallel ME\parallel BC$ and $ D$ - the middlepoint of the side $ [BC]$ $ \implies$ $ F = B$ .[/color]

[quote="Virgil Nicula"]What means Q1 ?! This problem is very easy ![/quote]Q1 means Question 1 (first problem in the test). The easiness of the problem is relative to its solvers.

[quote="Valentin Vornicu"]The easiness of the problem is relative to its solvers.[/quote]
 ... and to ... ? !

P can move on line AE ,the problem is also true.


Compute:

lim (n-->oo)  ((n+1)^p-n^p)/n!, where p is natural<>0.

cheers!

The solution I have is:

let b_n=n^p/n!, n>=1. Because the series b_(n+1)/b_n=1/(n+1)*(1+1/n)^p which converges to 0 we have that b_n -->0. And we have that (n+1)^p/n!=(n+1)^p/n^p*b_n=(1+1/n)^p*b_n so we get that  lim (n-->oo) ((n+1)^p-n^p)/n!=0 ...

does anyone have a much more direct solution?

cheers!

It looks pretty direct to me. Anyway, the fact that the sequence n^p/n! -> 0 is well known, so it's even more obvious that the initial sequence tends to 0. I saw this problem in Gazeta Matematica (last year I think) and I keep wondering why they proposed it there.

We can also use Stirling formula

n! ~ sqrt(2.pi.n).n^n.exp(-n)

Bn = ((n+1)^p - n^p)/n! = n^p/n! ( ( 1+ 1/n)^p -1))

= n^ p/n! ( p/n + O(1/n^2)), where O is big O of Ladau notation

Bn = n^(p-1)/n! (p +O(1/n))

n^(p-1)/sqrt(2.pi.n).n^n.exp(-n) ----> 0 when n -->+oo

lim Bn =  0

 :D

Thank you moubi.. that is  I was looking for... :D :D .. I love this solution...

cheers!!!


Let a_1,...,a_5>0 such that sum a_1/(1+4a_1)=4. Prove that:
  25 prod a_i>=5+4 sum a_i.

[quote="harazi"]I'm an idiot. It is $\frac{a_1}{1+4a_1}=1$[/quote]
If you were an idiot, so I would be a very very stupid boy, Harazi. :) Your inequality is really nice.Let
[tex] a_i=\frac{\sum x_i -x_i}{4 x_i} [/tex] and we must prove that:
[tex] 25\prod(x_1+x_2+x_3+x_4+x_5) \geq 4^5(x_1+x_2+x_3+x_4+x_5)(\sum x_1x_2x_3x_4) [/tex]
But this is clearly because number of sets $ (1,1,1,1,1) $ and $ (2,1,1,1,0) $ on the left hand side are less than the right hand side. :)
Somewhere in this forum,I have seen you post a similar inequality,and you call it Popoviciu inequality,can you post the general one, Harazi.

No, I cannot. I don't know the general form.


i'm looking for some training material on symmetric polynomials.

This is a file by Eugene Mukhin of BMC. It might help.


by $AM \geq GM$:
$\frac{a^{3}}{bc}+\frac{b^{3}}{ca}\geq 2\sqrt{\frac{a^{2}b^{2}}{c^{2}}}=2\frac{ab}{c}$
so we have :
$2(\sum \frac{a^{3}}{bc}) \geq 2(\sum \frac{ab}{c})$
again by $AM \geq GM$:
$\frac{ab}{c}+\frac{bc}{a}\geq 2b$
so we have:
$2(\sum \frac{ab}{c}) \geq 2(\sum a)$ , and we're done..... :)

[quote="VANCHANH123"]have a,b,c show that : a,b,c>0 .
:$\to$ $\sum\frac{a^{3}}{bc}\geq \sum a$[/quote]

$a^{4}+b^{4}+c^{4}\geq abc(a+b+c)$
use :
 $x^{2}+y^{2}+z^{2}\geq xy+yz+xz$

i think this ineq shouldn`t have been posted here at unsolved problems.


I need help on this problem.

A circle is circumscribed about an isosceles right triangle whose short side has a length of 3. What is the circumference of the circle?

Since the triangle is isoceles and right, the long side is the diameter of the circle. Using the formula $h=s \sqrt{2}$ where $h$ is the hypotenuse or long side and $s$ is the measure of the short side (leg). So the long side of the triangle is $3 \sqrt{2}$, so is the diameter of the circle. Using the formula $C= \pi d$, the circumference is $3 \sqrt{2} \pi$

[hide]A right isosceles triangle means the ratio of the legs to the hypotenuse (which is the diameter because it is inscribed in a circle) is $1: \sqrt2$. That means the diameter of the circle is $3\sqrt2$. and the circumference is $3\sqrt2\pi$ [/hide]


Find all primes p such that
P(x)=x(x-a)(x-b)(x-c)+p is reducible.

What are a,b,c  :huh: arbitary integers? :?:


function
algebra
floor function
superior algebra
superior algebra solved

Prove that :
      There exists a realnumber [b]a[/b] such that
        For all n>=1 :
           x_n = [[b]a[/b]^(3^n)]  is a prime number

                  {[x] is integer part of x}

surprisingly, it is true and it is known as Mills' theorem.

for a proof of the generalized form of Mills' theorem see http://www.utm.edu/research/primes/notes/proofs/A3n.html

and there's another thorem by julia robinson (i guess) about a some tower of power of 2's raised to a real number  always being a prime (of course the greatest integer function attached to it)
mill's theorem's a "thing of beauty" ,a real gem

such a real number hasnt been found yet (maybe you could find it someday :D ,ya who knows??!!) but it's existence is proved
cheers

[quote="grobber"]Can such a number be found or do we just know it exists?[/quote]

I believe such a number can be calculuated to arbitrarily fine precision, but no closed form is known (or may even exist) for it.

Another sequence gives primes numbers
[L.10^(n^2)] - [L.10^(n-1)^2].10^(2n-1) gives p_n
p_1=2, p_2=3, p_3=5, p_4=7 ...where L is a constant in wich primes
numbers appear
L=0,2003000050000007000000011....
where p_n  is in n^2 position

[x] is the integer part of x
---------------------------------------------------------
Another one Minac and Willians
http://www.cnrs.fr/Cnrspresse/math2000/html/math10.htm


Finally a thread to talk about tennis rules, strategy and top tennis players.
I am a undefeated single player in the 7th grade and I will start playing tourneys in December.

[quote="PhireKaLk6781"]Badminton pwns tennis.[/quote]

Does not...

Hey who do you think is the most promising player worse than 10th in the world ATP...

I think Marin Cilic...

I fail at tennis.
Our high school does not have a tennis team.
Not that it affects me anyways.

My school has tennis, but unfortunately its the same time as track. I'm quite a bit better at track than tennis, so I think I'll do track.

[quote="tennis123"]lucky you. We don't have a tennis team until high school.  :(   :mad:[/quote]

Me neither

Yay there's a thread on aops that talks about tennis :D

[quote="dragon96"]Baseball obviously pwns tennis.

Tennis is fine, but I'm not good at it so I don't like it as much as baseball/softball. :P[/quote]

Baseball dnt pwn tennis, duyy........

[quote="spaceguy524"][quote="jjfun1"]I fail at tennis.
[/quote]

So do I.[/quote]

Me too! Make a club? (jk)

Har har!

I suck too, but still at it.
Taking lessons.
I'm having trouble getting that topspin.

[quote="AwesomeToad"][quote="Scamper"]This forum is definitely for me.
Even though I'm only in sixth grade.
What really annoys me is when people
describe tennis as "...hitting the ball over
a piece of string..."[/quote]

Me too.

[quote="Jock"]What's so great about a sport you just hit a ball to each other?[/quote]

Fottball playing nontennis people...[/quote]
if tennis isn't "...hitting the ball over
a piece of string...", what is it?
Unlike soccer and basketball, tennis lacks teamwork and strategy. It's also less beneficial exercise.
To some degree, tennis also requires less sportsmanship.
I suspect a lot of my friends play tennis because they want the excuse "I play a sport" to put on their college resume, however they lack the dedication to play a team sport. It is an attempt to be "well-rounded", but an overdone attempt, and one that colleges have probably caught onto.
Of course, my outlook as a die hard soccer player could be somewhat biased, but if you want to play sports for entertainment (not merit), I would recommend a team sport (it also encourages proper development of social skills - 99% of social skills are developed from one's sports team).
Come to think of it, playing tennis for pure merit may die out soon as well... then team sports would be the obvious choice.

Tennis isn't a single-player sport...sometimes its doubles.  Thats two-player and therefore has teamwork, and all that other stuff.  Also, pro tennis players cough cough ROGER FEDERER....BOO NADAL cough cough..thats all i have to say.

I also suck at tennis, at least my serves anyway, well I guess my forehands aren't that great either...and neither are my backhands...or overhands...or slices...or any other move.

MY 1337TH POST!.

Baseball is WAY better than tennis. Not that I'm particularly good at either.

I spent a lot of time practicing tennis, yet I barely go anywhere. Baseball took me about two units (6th and 7th grade) to get the hang of. And you know how shortly schools go over baseball/softball.

[quote]<what funtwo said above>[/quote]

And some top colleges are looking for college applications that do not consist of the following:

- qualified AIME
- plays tennis
- plays piano/violin
- other stereotypes that are actually true

I still don't see how tennis can give you more exercise than baseball/basketball/poker.

[quote="dragon96"]
And some top colleges are looking for college applications that do not consist of the following:

- qualified AIME
- plays tennis
- plays piano/violin
- other stereotypes that are actually true

I still don't see how tennis can give you more exercise than baseball/basketball/poker.[/quote]
Exactly. Which is why it's so hard for those people to get into college. So instead of changing their focus, the college-obsessed community decides they need to work harder toward their steroetype.
The point of life is enjoyment, and therefore so is the point of sports. People make it way too complicated when they turn life into a strategy game.

@Scamper
I think topspin is hard to, but its great for making you looki better then you actually are!
@all
If only if shots where as easy as the pros make them look.
Serving is way to hard! I get 5+ double faults every match.

[quote="dragon96"]

And some top colleges are looking for college applications that do not consist of the following:

- qualified AIME
- plays tennis
- plays piano/violin
- other stereotypes that are actually true

[/quote]

Dang it, I'll probably fall under all those categories by the time I go to college (playing violin now, will join tennis, want to qualify for AIME).

Will a group of 100,000 Indians act like gangsters so our stereotype changes?  Thanks.

Maybe I should join the varsity football team, that will make me not fall under all those stereotypes (but there could be a chance that I won't be alive to go to college then)

[quote="funtwo"]The point of life is enjoyment, and therefore so is the point of sports. People make it way too complicated when they turn life into a strategy game.[/quote]

Strategy games are fun. :D

By the way, about what I said, I'm not saying qualifying AIME and playing an instrument is bad.


$\mathbb{Z}\ni a,b,c$, solve equation:

\[ a^{100}+b^{100}=c^{101}.  \]

I have no idea =)


Does anyone know any good physics books which aren't so complicated and explain theories and stuff very well?  I've read A Brief History of Time and thought it was a bit difficult for me, so something easier than that.

there is a topic about the physics resourses in this forum.

i recommend "The Elegant Universe" by Brian Greene.  it discusses special relativity, general relativity, quantum mechanics, and string theory in a very fun and interesting way.  it also does a good job of explaining where conflicts arise and why new theories need(ed) be created.

I really liked [u]Hyperspace[/u] by Michio Kaku

It is a bit difficult, but he explains concepts extremely well. And if you like physics (which I assume you do), it is a fun read.


[url=http://www.amazon.com/exec/obidos/ASIN/B0006L7WKO/qid=1116540399/sr=2-2/ref=pd_bbs_b_2_2/103-0789830-4636609]What is the point of this???!!!![/url]

I find it quite funny. Perhaps they forgot to mention that it's Yen instead of USD.

BTW , there just might be someone foolish enogh to buy that stuff...As far as I'm concerned , I'm ready to download that book if THEY pay me 1500 dollars.... :D

gotta be some mess up and if it's really worth $1500, someone would have pirated it.