is computerscience related physics  or computer science related mathmatics research similar to the stuff you do in AP computer science or is  it very different?  for example: biology research is very different from the stuff you do in a biology class.

thanks

I'm going to say that it's not at all similar (although I have no experience with research in CS) just because of the nature of AP and knowing what a typical CS1 and CS2 sequence is like. That being said, CS at a research level is a very broad field, one that is arguably a branch of mathematics in certain spectra. What are you particularly interested in?

When I say CS1 and CS2 sequence, I mean the most typical sequence followed by a university for a first year CS class (and this is what AP CS AB is modeled on). As for things you could look at to pursue your interest, there are people on the forum that could do a better job than me of leading you in the right direction, but you could look at some number theoretic topics in computer science, like the factoring algorithms, primality testing, in fact, cryptography in general is a good field to look at.

what do you mean by typical CS1 and CS2 sequence? I'm interested in number theory, I don't know if that's relevant, and I really enjoyed my AP Computer Science class. I'm also interested in physics.
[url=http://www.math123xyz.com]Interactive Math Resources[/url]


Happy Valentine's day. :)
We may have forgotten Saint Valentine and his great courage, but, we may still have the joy in writing love letters and hiding them into others' lockers. :lol:

Plus, I wish all of us may have fun in doing math today. ;)

At the time you posted... it WASN'T valentine's day yet!

It's 12:15 here, so at this point I can officially wish everyone a happy valentine's day (of course, especially to Jenn and Sarah!  :blush:  :blush:)

Don't overdose on the chocolate!

Whoops, sorry, I messed up my agenda planner. But anyways.
Valentine's day, happy chocolate factories!
I've heard that Toronto-er can eat up to 100 tons of chocolate each valentine's day, whoa! :o

(Oh yes, I've just gotten dumped weeks ago, so I can eat all my chocolate. No sharing with anybody else. :D )

Hey is it hard for math geeks to look for girlfriends?   :P  Any bright ideas will be very appriciated  :D

[quote="Kalimdor"]Well, in my case, math is not the point (the reason I got dumped)
I hate her music, and the stupid band System of a Down. And she's a german, really serious about every single tiny problem.  :D
No prejudice, but I'm just wondering why that girl loves those kinda garbage-like metal rocks :?[/quote]

??????????? speak english man

Hehehe yay!  Happy Valentine's day!  Wishes for lots of flowers, chocolates and love! Hehehe.  I know I got some.    :D

Valentines problem!

Let a "heart" of "heartradius" r and "heartlength" l be a shape defined as follows:

- Two circles of radius r are tangent (externally)
- A point P is located on the line of tangency at a distance l from the point of contact between the two circles.  To maximize heart-like-ness, we'll assume that l > r.
- All of the tangent line segments from P to the circles are drawn.
- The area enclosed by the two circles and the tangent line segments is the "heart".

Here is a "rough idea" of the shape of the "heart".  The bulges at the top are a bit less circular than in my description, but you can ignore its bloatedness.  P would be the point at the bottom:

[img]http://www.schools.pinellas.k12.fl.us/gallery/variety/Heart.gif[/img]

The problem:
a) Calculate the area of a heart of heartradius r and heartlength l.
b) The heart is inscribed in a circular target.  Cupid (working for Elyot of course) is practicing shooting his arrows at the target (so that he can hit Sarah later).  If Cupid's arrow strikes a random point on the circle, what's the chance that it hits the heart (ie what's the ratio of the heart's area to the circle's area) in terms of l and r?

EDIT: OK... the general solution is not that pretty.  So let's try the specific solution to part b if the ratio l:r is 12:5.  First person to get it done gets... ummm... to be my valentine?  LOL

[quote="Elyot"][quote="Mathgod"]Lol I like that question! :D[/quote]

Then solve it!  Feels a lot like about Euclid 8b) difficulty level.[/quote]

I don't even understand the question.... I just thought it's a nice question on valentines day...  :roll:

[quote="Elyot"]Don't overdose on the chocolate![/quote]

Oops.. too late!

[quote="Mathgod"][quote="Kalimdor"]Well, in my case, math is not the point (the reason I got dumped)
I hate her music, and the stupid band System of a Down. And she's a german, really serious about every single tiny problem.  :D
No prejudice, but I'm just wondering why that girl loves those kinda garbage-like metal rocks :?[/quote]

er... is that suppose to answer my question? anyway... valentine is almost over and i still didn't hook up with a girl....  :([/quote]
Danny from Windermere? Oh well aren't you very young like Grade 9.... should a 9th grader really complain about not having a girlfriend? :P

Random thoughts. No offense. :blush:

[quote="Elyot"][quote="Singular"]elyot you should work with the euclid committee[/quote]

BTW... the arrow Cupid shoots at Sarah in my problem is of course with you in the room, so you two can learn to get along and be nice loving friends.[/quote]

gangsters dont do math

[quote="Singular"]
gangsters dont do math[/quote]
:huh: I'm lost... :huh:

The point being...?  Are you saying I'm a gangster?   :ninja:


Where's the button that tells me what classes I enrolled for?

i think I enrolled for AIME problem series B but I'm not sure

Doesn't it show up on the "My Classes" section of the site?

Click My Classes above when you are logged in.  If you don't see the AIME class, then you're not enrolled.


A sequence of functions $\, \{f_n(x) \} \,$ is defined recursively as follows: \begin{align*}f_1(x) &= \sqrt{x^2 + 48}, \quad \mbox{and} \\ f_{n+1}(x) &= \sqrt{x^2 + 6f_n(x)} \quad \mbox{for } n \geq 1.\end{align*}
(Recall that $\sqrt{\makebox[5mm]{}}$ is understood to represent the positive square root.) For each positive integer $n$, find all real solutions of the equation $\, f_n(x) = 2x \,$.

[hide]
$f_1(x) = \sqrt{x^2+48} = 2x \Rightarrow 48 = 3x^2 \Rightarrow x = 4$ (negative answer is thrown out because of the definition of square root).

Note that $f_{n+1}(4) = \sqrt{4^2+6f_n(4)}$ so if $f_n(4) = 8$, then $f_{n+1}(4) = \sqrt{16+48} = 8$, which means $x = 4$ is a solution for all $n$.

Suppose $x < 4$. We know $x$ must be non-negative since we only use the positive square root.

Then $f_1(x) = \sqrt{x^2+48}> 2x$. But if $f_n(x) > 2x$, we have $f_{n+1}(x) = \sqrt{x^2+6f_n(x)} > \sqrt{x^2+12x} > 2x$ also (the last step since $x<4$).

Now suppose $x > 4$.

Then $f_1(x) = \sqrt{x^2+48} < 2x$. But if $f_n(x) < 2x$, we have $f_{n+1}(x) = \sqrt{x^2+6f_n(x)} < \sqrt{x^2+12x} < 2x$ also (the last step since $x>4$).

Therefore $x = 4$ is the only solution for all $n$.[/hide]

[quote="PhireKaLk6781"]Sorry for reviving, but I don't quite understand the question. Are we supposed to be solving for n, x, or a function?[/quote]
You should find a real $x$ that is a solution for all $n \in \mathbb N$.


The problem is: "Let S a linear transformation from R^3 to R^2 and T a linear transformation from R^2 to R^3. Show that TS is not invertible." Have anyone can help me to the above problem. Thanks anyone in advance.

In other words, a $ 3\times 2$ matrix times a $ 2\times 3$ matrix is not invertible.

One way is to use the concept of rank: $ \text{rank}\,AB\le \min(\text{rank}\,A,\text{rank}\,B).$

add $ \text{rank}\,AB\le \min(\text{rank}\,A,\text{rank}\,B) \leq min(2,3) \equal{} 2$
$ TS$ is a 3x3 matrix have $ rank(TS) \leq 2$ and we have $ TS$ not invertible.
 :spider:


integration
function
inequalities
Functional Analysis
advanced fields
advanced fields unsolved

This is a part of a proof of (a special case of) the Radon-Nikodym Theorem using the Riesz Representation Theorem in the Hilbert space $ L^{2}$.

Let $ \mu$ and $ \nu$ be finite measures on a measurable space and set $ \lambda\equal{}\mu\plus{}\nu$. Define $ F(f)\equal{}\int f \; d\mu$. Then $ F$ is a bounded linear functional on $ L^{2}(\lambda)$. Thus there is a unique function $ g \in L^{2}(\lambda)$ such that $ F(f)\equal{}\langle f,g \rangle$.

The function $ g$ is supposed to have the property that $ 0 \leq g \leq 1$. How is that so?

Suppose that $ g\ge a>1$ on $ E$. Take $ f\equal{}\chi_E$ and get $ \mu(E)\equal{}\int_E g\,d(\mu\plus{}\nu)\ge a(\mu(E)\plus{}\nu(E))$ whence $ (a\minus{}1)\mu(E)\plus{}\nu(E)\le 0$, so $ (\mu\plus{}\nu)(E)\equal{}0$. Now just take a decreasing to $ 1$ sequence of $ a$ to conclude that $ g\le 1$ almost everywhere with respect to $ \mu\plus{}\nu$. The other inequality is similar.


conics
analytic geometry
hyperbola
quadratics
Gauss
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[quote="harazi"]Let $ p$ be a prime number.

How many elements does $ SO_{2}\left(\mathbb{Z}/p\mathbb{Z}\right)$ have? What about $ SO_{3}\left(\mathbb{Z}/p\mathbb{Z}\right)$? And the general case (this seems terribly difficult)?

Now, how many solutions does the equation $ x_{1}^{2} \plus{} ... \plus{} x_{n}^{2} \equal{} 1$ in $ \mathbb{Z}/p\mathbb{Z}$ have? (very difficult)[/quote]

The last question is the easiest not the hardest. The conic has a rational point and therefore (by projection from that point) the number of points is the same as for $ P^{n \minus{} 1}$. The number of points over $ F_{q}$ is $ 1 \plus{} q \plus{} q^{2} \plus{} \dots \plus{} q^{n \minus{} 1}$, here $ q \equal{} p$.

(Edit:  the answer should be $ 1 \plus{} q^{n \minus{} 1}$ because the center of projection corresponds to a whole $ P^{n \minus{} 2}$ of lines tangent to the conic, not a single line.)

You're wrong. The question is absolutely non trivial and this is not the good answer.

I don't know such advanced arguments, all I know is that even for n=2 your answer is not always correct.

easy number theory leads that for $n=2$ if $p=4k+1$ the solution is $p-1$ and $p+1$ otherwise but for$n>2$ seems we should use another machinary :(

I saw a solution in an article and it was extremely technical. The answer is also quite ugly and this is why I said fleeting guest's approach cannot work.

strange ,today i come across a book due to Micheal Rosen ,I saw in application of Jacobian sum the solution of harazi's problem  :)  it is $p^{n-1}+(-1)^{\frac{n-1}{2}\cdot \frac{p-1}{2}}\cdot p^{\frac{n-1}{2}}$ if $n$ is odd and $p^{n-1}-(-1)^{\frac{n}{2}\cdot \frac{p-1}{2}}\cdot p^{\frac{n-1}{2}-1}$ otherwise .

sam-n,
Did you also see in that book how many elements there are in $\mbox{SO}_{n}(\mathbb{Z}/p\mathbb{Z})$, which was one of harazi's questions?
If yes, could you please post a short sketch of the proof?

[quote="fleeting_guest"]harazi,    what do you claim is the problem with the following argument, which is just a precis of the statement above?

$x_{1}^{2}+... x_{n}^{2}= 1$  is a smooth quadric over the field $F_{p}$ with $p$ elements.   It has a rational point $Q = (1,0,0,...)$.  To any point $R$ of the conic with coordinates in $F_{p}$, associate the line $QR$.   This is a bijection between points on the conic (different than $Q$) and rational lines through $Q$ (not in the tangent space to the conic at $Q$).  The set of non-tangent lines can be parametrized by the intersection of the lines with the plane $x_{1}= 0$, i.e. the $(n-1)$ dimensional affine space, which has $p^{n-1}$ points over $F$.  There is one additional point $Q$,  so the number of solutions is $1+p^{n-1}$.[/quote]

Be careful, as you are working in the [i]affine[/i] space, and some lines through $Q$ which don't lie in the tangent space to the quadric may still intersect the quadric at the point $Q$ only. These are those lines which, in the projective extension of our affine space, would pass through an infinite point of the quadric (e. g., if our quadric is a hyperbola, these are parallels to its asymptotes). You can count the number of such lines - this comes down to counting the infinite points of the quadric, what, in turn, is equivalent to solving the original problem for $n-1$ instead of $n$ - and thus you get a recurrent formula for the required number. However I am far from willing to make this computation now.

  darij

BTW: you can deduce the quadratic reciprocity law from this (take $ n\equal{}q$, the formula can be proved without using the reciprocity law)

I believe the book to which sam-n refers is [url=http://www.amazon.com/Classical-Introduction-Modern-Graduate-Mathematics/dp/038797329X]Ireland and Rosen[/url], in which the third problem is attacked using the technique of [url=http://en.wikipedia.org/wiki/Jacobi_sum]Jacobi sums[/url] (Chapter 8).  It is indeed possible to prove quadratic reciprocity along these lines.  It is also possible to define Gauss and Jacobi sums over arbitrary finite fields.

[url=http://en.wikipedia.org/wiki/Orthogonal_group#Over_finite_fields]Wikipedia[/url] gives the number of elements of $ O_n(\mathbb{F}_q)$.

my proof was also by reduction, but $ n \to n\plus{}2$ (variant: $ 1,2 \to n$). of course, Jacobi sums are more conceptual.

how is the quadratic reciprocity law derived from this in Ireland and Rosen (sketch)? the relevant pages are not available in google books. I did this by counting fixed points similar to that proof of Cauchy's theorem about elements of order $ p$ in a group.

(determining the number of solutions to $ \sum x_i^2 \equal{} x$ and deriving qrg from this was an exercise on an exercise set I can't find anymore)

Fix a prime $ p$.  Let $ \chi$ denote a character on $ \mathbb{F}_p$, let $ g(\chi) \equal{} \sum_{t \in \mathbb{F}_p} \chi(t) \zeta_p^{t}$ be a Gauss sum on $ \mathbb{F}_p$, and let

$ J(\chi_1, ... \chi_n) \equal{} \sum_{t_1 \plus{} ... \plus{} t_n \equal{} 1} \chi_1(t_1) ... \chi_n(t_n)$

be a Jacobi sum on $ \mathbb{F}_p$.

[b]Theorem:[/b]  If $ \chi_1, ... \chi_r$ are nontrivial characters with nontrivial product, then $ g(\chi_1) ... g(\chi_r) \equal{} J(\chi_1, ... \chi_r) g(\chi_1 ... \chi_r)$.

[b]Corollary:[/b]  If $ \chi_1, ... \chi_r$ are nontrivial characters with trivial product, then $ g(\chi_1) ... g(\chi_r) \equal{} \chi_r( \minus{} 1) J(\chi_1, ... \chi_{r \minus{} 1})$.

Let $ q$ be an odd prime not equal to $ p$ and let $ \chi$ denote the nontrivial character of order $ 2$ (that is, the Legendre symbol).  Then

$ g(\chi)^{q \plus{} 1} \equal{} ( \minus{} 1)^{\frac {p \minus{} 1}{2} } p J(\chi, \chi, ... \chi)$

where the Jacobi sum has $ q$ components.

[b]Lemma:[/b]  $ g(\chi)^2 \equal{} ( \minus{} 1)^{\frac {p \minus{} 1}{2} } p$.

It follows that $ g(\chi)^{q \plus{} 1} \equal{} ( \minus{} 1)^{ (p \minus{} 1)(q \plus{} 1)/4 } p^{(q \plus{} 1)/2}$.  Now, the terms in the sum $ J(\chi, \chi, ... \chi)$ can be grouped into two groups.  If $ t_1 \equal{} ... \equal{} t_r$, then $ t_i \equiv \frac {1}{q} \bmod p$ and the corresponding term has value $ \chi \left( \frac {1}{q} \right)^q \equal{} \chi(q)^{ \minus{} q} \equal{} \chi(q) \equal{} \left( \frac {q}{p} \right)$.  If the $ t_i$ are not all equal, then the corresponding terms in the sum can be grouped into groups of $ q$ by cyclic permutation, and since the corresponding terms have the same value they cancel $ \bmod q$.  Substituting, it follows that

$ ( \minus{} 1)^{ (p \minus{} 1)(q \minus{} 1)/4} p^{ (q \minus{} 1)/2 } \equiv \left( \frac {q}{p} \right) \bmod q$.

Since $ p^{ (q \minus{} 1)/2 } \equiv \left( \frac {p}{q} \right) \bmod q$, the conclusion follows.

OK, so the derivation of the qrg from this in Ireland and Rosen is in principle the same as I did (all $ t_i$ the same corresponds to "no fixed points")

[strange, now I can access the pages in google books---do the pages I can access change when re-dialing up?]

BTW, the calculations for
[quote]Theorem: If  are nontrivial characters with nontrivial product, then g(\chi_1) ... g(\chi_r) = J(\chi_1, ... \chi_r) g(\chi_1 ... \chi_r).

Corollary: If are nontrivial characters with trivial product, then g(\chi_1) ... g(\chi_r) = \chi_r( - 1) J(\chi_1, ... \chi_{r - 1}). [/quote]
in Ireland and Rosen can be made more conceptual and consise by using the Fourier transform and its basic properties (proof formally the same as usual---like Laumon did on some derived category of sheaves).

In my experience the pages you can access depend on the search terms you use.

I was wondering about that - the issue I have with Ireland and Rosen's presentation of Gauss and Jacobi sums is that the "classical" presentation comes at the cost of removing the motivation from many of the theorems.

How do you show that $ SO_2(\mathbb{Z}/p\mathbb{Z})$ is isomorphic to the dihedral group of order $ 2(p\minus{}1)$ or $ 2(p\plus{}1)$ depending if $ \minus{}1$ is a square in $ \mathbb{Z}/p\mathbb{Z}$ or not?

Is it? :? To me, it looks like it's cyclic of order $ p \minus{} 1$ or $ p \plus{} 1$ (according, respectively, to whether or not $ \minus{} 1$ is quadratic). Where did you see that statement, JC_math?

Notice that $ SO_2$ is the group consisting of matrices of the form $ \begin{pmatrix}a & b \\
\minus{} b & a\end{pmatrix}$ with $ a,b\in\mathbb F_p,\ a^2 \plus{} b^2 \equal{} 1$. Now just count the solutions $ (a,b)$ to $ a^2 \plus{} b^2 \equal{} 1$. In order to see that the groups are cyclic, we can argue like this:

When $ \left(\frac { \minus{} 1}p\right) \equal{} 1$, another simple counting argument shows that every non-zero element of $ \mathbb F_p$ is a root of an equation of the form $ x^2 \minus{} 2ax \plus{} 1 \equal{} 0$ for some $ a^2 \plus{} b^2 \equal{} 1$, so every element of $ \mathbb F_p^*$ is an eigenvalue of some special orthogonal matrix. In particular, there is a special orthogonal matrix which has a primitive root of unity as an eigenvalue, and this matrix can only have order $ p \minus{} 1$. When $ \left(\frac { \minus{} 1}p\right) \equal{} \minus{} 1$, on the other hand, append a square root $ i$ of $ \minus{} 1$ to $ \mathbb F_p$ to get the field $ \mathbb F_{p^2}$. Then, for $ a,b\in\mathbb F_p$, multiplication by $ a \plus{} bi$ is an $ \mathbb F_p$-linear map whose matrix with respect to the basis $ (1,i)$ is the one depicted above. The special orthogonal group corresponds in this way to the group of elements in $ \mathbb F_{p^2}$ having norm $ 1$, and this group is indeed cyclic (of order $ p \plus{} 1$).

It sounds like JC_math is referring to $ O_2(\mathbb{Z}/p\mathbb{Z})$.

Ah.. Well, that should be clear now: conjugation by $ \begin{pmatrix}1&0\\0&\minus{}1\end{pmatrix}$ acts on the special orthogonal group by sending each element to its inverse.

Indeed, I meant $ O_2(\mathbb{Z}/p\mathbb{Z})$.

What do we know about  $ O_2(\mathbb{Z}/p^k\mathbb{Z})$ and  $ SO_2(\mathbb{Z}/p^k\mathbb{Z})$?

use "smooth => formally smooth" and determine the kernel and multiplicativity of cardinality in exact sequences

myan,
I'm not familiar with "smooth => formally smooth", could you detail a bit more?


factorial
ceiling function
function
floor function

Let $ p$ be a prime number. Determine a real number $ \alpha$ such that
i) There exist infinitely many positive integers $ n$ such that $ p^{\lceil \alpha n \rceil \minus{} 1} \mid n!$.
ii) There does not exist a positive integer $ n$ such that $ p^{\lceil \alpha n \rceil} \mid n!$.

( $ \lceil x \rceil$ is the ceiling function, equal to the least integer greater than or equal to $ x$ )

[hide="Hint"] Let $ n \equal{} p^k$. [/hide]

Okay, not surprised I didn't get a response. When I actually wrote out the solution I found that this problem is far harder than I thought, even after I cut the challenge by not asking for all values $ \alpha$, which requires additional finesse. This topic should probably moved to either Pre-Olympiad or NT Proposed & Own.

[hide="More hints"] Let $ v_p(n)$ denote the greatest $ e$ such that $ p^e | n$.  It's well-known that

$ v_p(n!) \equal{} \sum_{i\equal{}1} \left\lfloor \frac{n}{p^i} \right\rfloor$

In particular, let $ n \equal{} p^k$.  Then

$ v_p((p^k)!) \equal{} p^k \plus{} p^{k\minus{}1} \plus{} p^{k\minus{}2} \plus{} ... \plus{} 1 \equal{} \frac{p^{k\plus{}1} \minus{} 1}{p \minus{} 1} \equal{} \frac{pn \minus{} 1}{p \minus{} 1}$

[b]Lemma:[/b]  $ \forall n \neq p^k$ we have $ v_p(n!) < \frac{pn \minus{} 1}{p \minus{} 1}$.  [/hide]


function
vector
linear algebra
matrix
complex numbers
linear algebra unsolved

Let $ a_1,...,a_n$ be real numbers satisfying $ 0 < a_1 < a_2 < \cdots < a_n$. Prove that the functions $ \sin(a_1 x),...,\sin(a_n x)$ are linearly independent (as elements in the vector space of functions $ \mathbb{R} \to \mathbb{R}$ over the field $ \mathbb{R}$).

One can prove a more general result that implies this one:  for any $ n$ distinct complex numbers $ \alpha_1, \alpha_2, ... \alpha_n$ the functions

$ e^{\alpha_1 z}, e^{\alpha_2 z}, ... e^{\alpha_n z}$

are linearly independent over $ \mathbb{C}$ because their (edited: thanks!) [url=http://en.wikipedia.org/wiki/Wronskian]Wronskian[/url], evaluated at $ z \equal{} 0$, is [url=http://en.wikipedia.org/wiki/Vandermonde_matrix]Vandermonde[/url].

I got it. Thank you very much.

(BTW, you probably meant the Wronskian, not the Hessian)


Let be given a positive real sequence:$(x_{n})$
Prove that : $limsup \frac{x_{1}+x_{2}+...+x_{n+1}}{x_{n}} \geq 4$

Let $y_n=\frac{x_{1}+x_{2}+...+x_{n+1}}{x_{n}}, \ z_n=\frac{x_{n+1}}{x_n} \ X_a=\{i|z_i \ge a\}.$ If capasity: $\#|X_3|=\infty, a_0=3$, then $limsup \frac{x_{1}+x_{2}+...+x_{n+1}}{x_{n}} \ge 4.$
If $X_{a_0}$ have finite terms, then for big n $\frac{x_1+x_2+ \dots+x_{n-1}}{x_n}>\frac{1}{a_0-1}-eps$ It give prove for case capasity $\#|X_{2.5}|=\infty, a_1=2.5$. This way we prove for cases $\#|X_{a_k}|=infty,$ were $a_{k+1}=3-\frac{1}{a_k-1}, \ lim_k a_k=2.$ Therefore we find prove for case when ${\{i|z_i>2}$ is infty. Other cases it easy.

This is not too hard,assume that $lim \frac{a_{n+1}}{a_n}=a<\infty$(if a is infinite the result is obvious)
For a given $t>0$ there are exist k such that $\frac{a_{n+1}}{a_n}<a+t$ for all $n\geq\ k$
then $\frac{a_1+a_2+...a_{n+1}}{a_n}\geq\ \sum_{i=1}^{n-k}\frac{1}{(a+t)^i}+a+t$
If a<1 then it's obvious
If a>1 $LHS>a+t+\frac{a+t}{a+t-1}$ then the result is true by AM-GM


Evaluate the infinite series

$S = \sum_{n = 1}^{\infty}\arctan \frac{2}{n^{2}}$.

Elemennop wrote:[hide][/hide]

How did you think of that?

The motive is similar to http://www.artofproblemsolving.com/Forum/viewtopic.php?t=144287 .

Yeah...tricky telescoping with the arctan formula...I think there was a similar problem from a Mildorf problem set or something (posted before).

EDIT: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=128400


My teacher wants to update our old library with new books. So in his Friday class, he mentions it to all of us. I have no idea which books are good for college level. I would like to hear from all of Math Linkers.

  Thank you very much.

[quote]We have 26180 registered users[/quote] Do you really want to hear from all of them?  :)

I remember seeing a neat list of math books on Xevarion's blog. However, different books are good for different colleges. Why not just browse through the [url=http://www.ams.org/bookstore]AMS bookstore[/url] and find what best suits your college's mission?


geometry
inequalities unsolved
inequalities

For triangle of sides $a,b,c $ area $A$ and semiperimeter $p$

let  $ Q =(b-c)^2 + (c-a)^2 + (a-b)^2 $ and $ q=\sqrt \frac{Q}{2} $. Prove

$ \frac{p(p-q)^2(p+2q)}{27} \geq A^2 \geq \frac{p(p+q)^2(p-2q)}{27} $

This is a very hard problem.

I have a solution but I would like to know if someone else has any idea.


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I was thinknig we should have all CHuck Norris jokes somewhere. Post if you know a good one!

Chuck Norris shot down a plane by pointing his finger at it and saying BOOM!

God said, "Let there be light!" Chuck Norris said, "Say please".

Why does Chuck Norris sleep with a nightlight on?
The dark is afraid of him.

There is no 'ctrl' button on Chuck Norris's computer. Chuck Norris is always in control
If you have five dollars and Chuck Norris has five dollars, Chuck Norris has more money than you
When the Boogeyman goes to sleep every night he checks his closet for Chuck Norris
Q: What’s 30 times Chuck Norris?
[hide]A: Oblivion.[/hide]
Chuck Norris can kill two stones with one bird.

Chuck Norris can delete the Recycle Bin.

Chuck Norris uppercutted a horse and turned it into a giraffe.

Chuck Norris can win the game (if you know it, you just lost :) )

until few months ago, if you type chuck norris on google and press "i'm feeling lucky", it would say something like "you can't search for chuck norris. only chuck norris can search you"

When the Boogeyman goes to sleep every night, he checks his closet for Chuck Norris.

Chuck Norris doesn't read books. He stares them down until he gets the information he wants.

There is no theory of evolution. Just a list of creatures Chuck Norris has allowed to live.

Outer space exists because it's afraid to be on the same planet with Chuck Norris.

Chuck Norris does not sleep. He waits.

Chuck Norris is currently suing NBC, claiming Law and Order are trademarked names for his left and right legs.

Chuck Norris is the reason why Waldo is hiding.

Chuck Norris counted to infinity - twice.

There is no chin behind Chuck Norris’ beard. There is only another fist.

When Chuck Norris does a pushup, he isn’t lifting himself up, he’s pushing the Earth down.

Chuck Norris is so fast, he can run around the world and punch himself in the back of the head.

Chuck Norris’ hand is the only hand that can beat a Royal Flush.

Chuck Norris can lead a horse to water AND make it drink.

Chuck Norris doesn’t wear a watch, HE decides what time it is.

Chuck Norris can slam a revolving door.

Chuck Norris does not get frostbite. Chuck Norris bites frost

Remember the Soviet Union? They decided to quit after watching a DeltaForce marathon on Satellite TV.

Contrary to popular belief, America is not a democracy, it is a Chucktatorship.

When Chuck Norris calls 1-900 numbers, he doesnt get charged. He holds up the phone and money falls out.

Chuck Norris once ate a whole cake before his friends could tell him there was a stripper in it.

Some people like to eat frogs' legs. Chuck Norris likes to eat lizard legs. Hence, snakes.

There are no races, only countries of people Chuck Norris has beaten to different shades of black and blue.

When Chuck Norris was denied an Egg McMuffin at McDonald's because it was 10:35, he roundhouse kicked the store so hard it became a Wendy's.

Chuck Norris can't finish a "color by numbers" because his markers are filled with the blood of his victims. Unfortunately, all blood is dark red.

A Chuck Norris-delivered Roundhouse Kick is the preferred method of execution in 16 states.

When Chuck Norris falls in water, Chuck Norris doesn't get wet. Water gets Chuck Norris.

Scientists have estimated that the energy given off during the Big Bang is roughly equal to 1CNRhK (Chuck Norris Roundhouse Kick)

Chuck Norris’ house has no doors, only walls that he walks through.

How much wood would a woodchuck chuck if a woodchuck could Chuck Norris? ...All of it.

Chuck Norris doesn't actually write books, the words assemble themselves out of fear.

In honor of Chuck Norris, all McDonald's in Texas have an even larger size than the super-size. When ordering, just ask to be Chucksized.

Chuck Norris CAN believe it's not butter.

If tapped, a Chuck Norris roundhouse kick could power the country of Australia for 44 minutes.

Chuck Norris can divide by zero.

The grass is always greener on the other side, unless Chuck Norris has been there. In that case the grass is most likely soaked in blood and tears.

A picture is worth a thousand words. A Chuck Norris is worth 1 billion words.

Newton's Third Law is wrong: Although it states that for each action, there is an equal and opposite reaction, there is no force equal in reaction to a Chuck Norris roundhouse kick.

Chuck Norris invented his own type of karate. It's called Chuck-Will-Kill.

When an episode of Walker Texas Ranger was aired in France, the French surrendered to Chuck Norris just to be on the safe side.

While urinating, Chuck Norris is easily capable of welding titanium.

Chuck Norris once sued the Houghton-Mifflin textbook company when it became apparent that their account of the war of 1812 was plagiarized from his autobiography.

When Chuck Norris talks, everybody listens. And dies.

When Steven Seagal kills a ninja, he only takes its hide. When Chuck Norris kills a ninja, he uses every part.

Contrary to popular belief, there is indeed enough Chuck Norris to go around.

Chuck Norris doesnt shave; he kicks himself in the face. The only thing that can cut Chuck Norris is Chuck Norris.

For some, the left testicle is larger than the right one. For Chuck Norris, each testicle is larger than the other one.

Chuck Norris always knows the EXACT location of Carmen SanDiego.

When taking the SAT, write "Chuck Norris" for every answer. You will score over 8000.

Chuck Norris invented black. In fact, he invented the entire spectrum of visible light. Except pink. Tom Cruise invented pink.

When you're Chuck Norris, anything + anything is equal to 1. One roundhouse kick to the face.

Chuck Norris has the greatest Poker-Face of all time. He won the 1983 World Series of Poker, despite holding only a Joker, a Get out of Jail Free Monopoloy card, a 2 of clubs, 7 of spades and a green #4 card from the game UNO.

On his birthday, Chuck Norris randomly selects one lucky child to be thrown into the sun.

Nobody doesn't like Sara Lee. Except Chuck Norris.

Chuck Norris doesn't throw up if he drinks too much. Chuck Norris throws down!

In the beginning there was nothing...then Chuck Norris Roundhouse kicked that nothing in the face and said "Get a job". That is the story of the universe.

Chuck Norris has 12 moons. One of those moons is the Earth.

Chuck Norris grinds his coffee with his teeth and boils the water with his own rage.

Archeologists unearthed an old english dictionary dating back to the year 1236. It defined "victim" as "one who has encountered Chuck Norris"

Chuck Norris ordered a Big Mac at Burger King, and got one.

Chuck Norris and Mr. T walked into a bar. The bar was instantly destroyed, as that level of awesome cannot be contained in one building.

If you Google search "Chuck Norris getting his ass kicked" you will generate zero results. It just doesn't happen.

Chuck Norris can drink an entire gallon of milk in thirty-seven seconds.

Little known medical fact: Chuck Norris invented the Caesarean section when he roundhouse-kicked his way out of his monther's womb.

Chuck Norris doesn't bowl strikes, he just knocks down one pin and the other nine faint.

The show Survivor had the original premise of putting people on an island with Chuck Norris. There were no survivors, and nobody is brave enough to go to the island to retrieve the footage.

It takes Chuck Norris 20 minutes to watch 60 Minutes.

You know how they say if you die in your dream then you will die in real life? In actuality, if you dream of death then Chuck Norris will find you and kill you.

Chuck Norris has a deep and abiding respect for human life... unless it gets in his way.

The Bermuda Triangle used to be the Bermuda Square, until Chuck Norris Roundhouse kicked one of the corners off.

There are no weapons of mass destruction in Iraq, Chuck Norris lives in Oklahoma.

Chuck Norris doesn't believe in Germany.

When Chuck Norris is in a crowded area, he doesn't walk around people. He walks through them.

Chuck Norris once ate an entire bottle of sleeping pills. They made him blink.

James Cameron wanted Chuck Norris to play the Terminator. However, upon reflection, he realized that would have turned his movie into a documentary, so he went with Arnold Schwarzenegger.

Chuck Norris can touch MC Hammer.

Thousands of years ago Chuck Norris came across a bear. It was so terrified that it fled north into the arctic. It was also so terrified that all of its decendents now have white hair.

Chuck Norris played Russian Roulette with a fully loaded gun and won.

It takes 14 puppeteers to make Chuck Norris smile, but only 2 to make him destroy an orphanage

[quote]Chuck Norris has the greatest Poker-Face of all time. He won the 1983 World Series of Poker, despite holding only a Joker, a Get out of Jail Free Monopoloy card, a 2 of clubs, 7 of spades and a green #4 card from the game UNO. [/quote]

:rotfl:

THE BEST!

:rotfl:

Apple pays Chuck Norris 99 cents every time he listens to a song.
Chuck Norris can sneeze with his eyes open.
Chuck Norris can eat just one Lay's potato chip.
Chuck Norris is suing Myspace for taking the name of what he calls everything around you.
Chuck Norris destroyed the periodic table, because he only recognizes the element of surprise.

[quote="jjx1"]geez everyone, stop posting jokes that people have already posted! :mad:[/quote]

chuck norri likes posting old jokes

[quote="dannyhamtx"]Chuck Norris can win the game (if you know it, you just lost :) )[/quote]I was gonna do that! :mad:

[quote="(^_^)"]When taking the SAT, write "Chuck Norris" for every answer. You will score OVER NINE THOUSAND!!![/quote]Fixed. :D

Chuck Norris can score 100% on Through the Fire and Flames on Expert... using only his tongue.

[quote="Nerd_of_the_Ages"][quote="DiLo"]Who is Chuck Norris?[/quote]
Yeah, who's chuck norris, I only know michael phelps.  :P
Dude, chuck norris jokes are outdated. Michael phelps is the new chuck norris.[/quote]

[img]http://www.pokemon-pearl-diamond.com/characters/pikachu.jpg[/img]

PIKAHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHH!!!!!!!!!

[quote="Nerd_of_the_Ages"]Michael phelps is the new chuck norris.[/quote]
no. just no.

Dude how do you think Kim Jong Il could lose? He's like Chuck Norris of the east!

Bonus cookies to anyone who can tell the person who originally said that.

[quote="1=2"]

Bonus cookies to anyone who can tell the person who originally said that.[/quote]

michael phelps???

oh no, can't decide between two people. I'm gonna PM one of them to see [b]WHO DUNNIT[/b]

ok the person who did it said that Treething quoted that person in his sig.

Jesus can walk on water, but Chuck Norris can swim through land.

Chuck Norris can judge a book by its cover.

chuck norris doesnt have to give peas a chance
chuck norris's age doesnt have to always go up
ocean exists because chuck norris gathered the tears of all his victims (and thats why they are salty)

Chuck Norris eats beef jerky, craps out bullets, and uses it to shoot a cow to make more beef jerky. This is known as the cycle of life.

Some people can pee their name in the snow. Chuck Norris pees his name into cement.
A variation:
Most people pee into a toilet. Chuck Norris craps on the ceiling.

Chuck Norris once visited the Virgin Islands. They are now simply known as the Islands.


[hide]
Expanding and using Muirhead

$ (3,0, \minus{} 1) \plus{} 3(2,1, \minus{} 1) \succ 4(2,0,0)$
[/hide]

Let a+b+c=1
$ \frac{(a\plus{}b)^3}{c}\plus{}\frac{(b\plus{}c)^3}{a}\plus{}\frac{(c\plus{}a)^3}{b} \geq 8(a^2\plus{}b^2\plus{}c^2$
$ \Leftrightarrow \frac{{(1 \minus{} c)^3 }}{c} \plus{} \frac{{(1 \minus{} b)^3 }}{b} \plus{} \frac{{(1 \minus{} a){}^3}}{a} \ge 8(a^2  \plus{} b^2  \plus{} c^2 )$
$ \Leftrightarrow \frac{1}{a} \plus{} \frac{1}{b} \plus{} \frac{1}{c} \minus{} 9 \plus{} 3(a \plus{} b \plus{} c) \ge 9(a^2  \plus{} b^2  \plus{} c^2 )$
$ \Leftrightarrow \left( {\frac{1}{{ab}} \minus{} 3} \right)(a \minus{} b)^2  \plus{} (\frac{1}{{bc}} \minus{} 3)(b \minus{} c)^2  \plus{} (\frac{1}{{ac}} \minus{} 3)(a \minus{} c)^2  \ge 0$
Let $ a \ge b \ge c$
Thus,
 $ \begin{array}{l}
 a \le 1,c \le \frac{1}{3} \Rightarrow ac \le \frac{1}{3} \\
 \frac{1}{{ab}} \plus{} \frac{1}{{bc}} \minus{} 6 \equal{} \frac{{b \plus{} c}}{{abc}} \ge \frac{2}{{a\sqrt {bc} }} \ge \frac{2}{{\sqrt {abc} }} \ge 6 \\
 \end{array}$
(Q.E.D) :) Is that right?

This one is easy.
The inequality to prove is equivalent to

ab(a + b)^3 + bc(b + c)^3 + ca(c + a)^3 => 8abc(a^2 + b^2 + c^2)

LHS = ab(a+b)^3 + bc(b+c)^3 + ca(c+a)^3

        = ab(a+b)(a+b)^2 + bc(b+c)(b+c)^2 + ca(c+a)(c+a)^2

        => 4(ab)^2(a + b) + 4(bc)^2(b + c) + 4(ca)^2(c + a).

thus, it is sufficient to prove that

        [(ab)^2(a + b) + (bc)^2(b + c) + (ca)^2(c + a)] => 2abc(a^2 + b^2 + c^2)

or    [a^3 b(b-c) + b^3 c(c-a) + c^3 a(a-b)] => a^3 c(b-c) + c^3 b(c - b) + b^3 a(b - a)

or   a^3 (b - c)^2 + b^3(c - a)^2 + c^3(a - b)^2 => 0

which is true since a, b, c > 0.

Equality occurs when a = b = c. :jump:

[quote="Attila the Hun"]@TRAN THAI HUNG : how a + b + c = 1?[/quote]
Let $ x\equal{}\frac{a}{a\plus{}b\plus{}c}$ and define y,z similar. Now plug in $ a\equal{}x(a\plus{}b\plus{}c)$ and b,c in inequality and you will see that the inequality remains the same if we change a,b and c with x,y and z. But x+y+z=1. This is normailzation.

[quote="Bugi"][quote="Attila the Hun"]@TRAN THAI HUNG : how a + b + c = 1?[/quote]
Let $ x \equal{} \frac {a}{a \plus{} b \plus{} c}$ and define y,z similar. Now plug in $ a \equal{} x(a \plus{} b \plus{} c)$ and b,c in inequality and you will see that the inequality remains the same if we change a,b and c with x,y and z. But x+y+z=1. This is normailzation.[/quote]

thanks a lot. I didn't know about normalisation.

[quote="Bugi"][quote="Attila the Hun"]@TRAN THAI HUNG : how a + b + c = 1?[/quote]
Let $ x \equal{} \frac {a}{a \plus{} b \plus{} c}$ and define y,z similar. Now plug in $ a \equal{} x(a \plus{} b \plus{} c)$ and b,c in inequality and you will see that the inequality remains the same if we change a,b and c with x,y and z. But x+y+z=1. This is normailzation.[/quote]
you are right, thanks.

Maybe helpful about normalizations: http://www.mathlinks.ro/viewtopic.php?search_id=1691612910&t=192080

[quote="TRAN THAI HUNG"]
$ \begin{array}{l} a \le 1,c \le \frac {1}{3} \Rightarrow ac \le \frac {1}{3} \\
\frac {1}{{ab}} \plus{} \frac {1}{{bc}} \minus{} 6 \equal{} \frac {{b \plus{} c}}{{abc}} \ge \frac {2}{{a\sqrt {bc} }} \ge \frac {2}{{\sqrt {abc} }} \ge 6 \\
\end{array}$[/quote] in post #3 .
If I am not mistaken, how come $ \frac {2}{a\sqrt {bc}}\geq \frac {2}{\sqrt {abc}}$  ???? :huh:
As $ a\sqrt {bc}\geq\sqrt {abc}$, hence $ \frac {2}{a\sqrt {bc}}\leq\frac {2}{\sqrt {abc}}$, isn't it so?

If I am mistaken, please forgive me......

$ a\sqrt{bc}\le \sqrt{abc}$, because $ a\le 1$ :wink:

oooooooooops! :blush:
Now it has been really proved that I am a dummy-head..... :(
Sorry for the inconvenience and thanks to Bugi.....


[quote="arash nk"]$ a,b,c$ are real and Positive numbers and $ abc\ge 1$.prove that:
[size=150]$ \frac {a \plus{} 1}{a^2 \plus{} a \plus{} 1} \plus{} \frac {b \plus{} 1}{b^2 \plus{} b \plus{} 1} \plus{} \frac {c \plus{} 1}{c^2 \plus{} c \plus{} 1}\leq2$[/size][/quote]
Let $ a\equal{}kx,b\equal{}ky,c\equal{}kz$ with $ a,b,c,k>0 such that xyz\equal{}1$
$ \equal{}>\sum{\frac{a\plus{}1}{a^2\plus{}a\plus{}1}} \leq \sum{\frac{x\plus{}1}{x^2\plus{}x\plus{}1}} \leq2$


function
calculus
derivative
trigonometry
calculus computations

the above question has multiple answers
according to me the answer is abcd
is option a correct?

Indeed it seems that all four options are correct.

Use [url=http://www.walterzorn.com/grapher/grapher_e.htm]this[/url] applet to graph the functions and see.

j rav :)

thanks a lot for the applet
this question was part of an indian entrance examination and the answer key released shows it as (a) as a wrong option. how :o

"Strictly increasing" means that for all $ x < y$, $ f(x) < f(y)$.  There are many strictly increasing functions that have derivative 0 at some points, such as the function in question and $ g(x) \equal{} x^3$.  Positive derivative on an interval implies strictly increasing on that interval, but these are not equivalent.

a is definitely false.  $ f'(x) \equal{} 1\plus{}\cos x \ge 0$.  Since the derivative is never negative, the function is never decreasing.  The only way for the function to be constant on an interval is for the derivative to be 0 for the whole interval, and it is easy to show that the derivative is only zero when $ x$ is an odd multiple of $ \pi$.  Thus since the function is nondecreasing and never constant, it must be strictly increasing.

A function being increasing or strictly increasing is not associated with points; it's associated with intervals. The phrase "strictly increasing at every point" is thus nonsense. It's a very popular misconception, which is why this is a tricky question.


vector
real analysis
real analysis unsolved

$X$ is a subset of $R^n$ such that if $x$ and $y$ are two arbitrary points in $X$, then there exists an open ball included in $X$ that contains $x$ and $y$. Prove that $X$ is an open ball.

I'll assume that $X$ is bounded.

We can easily prove that a similar statement takes place in $\overline X$, where "open ball" is replaced with "closed ball": we take sequences of interior points points $x_n\to x,y_n\to y$. For every pair $(x_n,y_n)$ there is a ball $B(a_n,R_n)$ which contains both. From the sequence $(a_n,R_n)_n$ we can extract a convergent subsequence, and if the limit is $(a,R)$, then the ball $B(a,R)$ will be contained in $\overline X$ and will contain $x,y$.

Now, since we're working with the compact set $\overline X$, things are much easier: consider the (closed) ball $B(a,R)$ of largest radius contaned in $\overline X$. If there were a point $x\in\overline X$ outside $B$, then $\overline X$ would contain a ball in which both $x$ and the point in $B$ farthest from $x$ lie, and this ball would have a radius $>R$, contradiction. This means that, in fact, $\overline X=B$, and, since $X$  is clearly open, it must be the interior of $\overline X=B$, so $X$ is the interior of the ball $B$.

[quote="grobber"]I'll assume that $X$ is bounded.[/quote]
You need this assumption- otherwise a half-space would work.

[quote="grobber"]Consider the (closed) ball $B(a,R)$ of largest radius contained in $\overline X.$ If there were a point $x\in\overline X$ outside $B,$ then $\overline X$ would contain a ball in which both $x$ and the point in $B$ farthest from $x$ lie, and this ball would have a radius $>R$[/quote]
There's a geometric property of the space used here; the diameter of the union of a closed ball and another point is greater than twice the radius of the ball. This would also work in other normed vector spaces, but not general metric spaces.

An odd counterexample, in the world of metric spaces.
Let $Y=\{a,b,c,e,f,g,x\}$ with distances $d(a,b)=d(a,c)=d(b,c)=d(a,e)=d(b,f)=d(c,g)=2$, $d(p,x)=1.5$ for $p\in Y/x$, and $d(p,q)=1$ for the other nine pairs $p\neq q$. Let $X=\{a,b,c,e,f,g\}$. Any two points of $X$ ar contained in one of the (closed) balls of radius 1 centered at $e,f,g$, but any ball containing all of $X$ also contains $x$.


There's pop. There's more traditional music...

But is it just me, or is there usually also a little Asian kid across the street who plays the violin 6 hours a day? Not trying to be facetious or anything... I've just noticed alot of... well...

i love chinese traditional music, I picked up the guzheng when i was in china 4 yrs ago, and have been haphazardly learning and taking lessons since.  It's a great instrument, but I wish more people played traditional chinese music in my area, I'm almost the only 1

erhu rocks!!! my chinese school started offering lessons like 2 years ago. it's amazing!! poor quickstar. i played in a group so i know like 6 people that play traditional chinese music in my area

I don't know any.

I can play a little tradition Korean music,
but no lessons. Have to use intuition.  :wink:

oh, wow, that's great

I wish we had that here.  In Montreal, where I drive for lessons, there are a number of ppl who play other instruments, but it's mostly through private teachers, not through a culture center.  There is one, but they don't have much music there.  Here in Vermont, there's a chinese school, but unfortunately we don't have many ppl who can do music and arts.

I don't get this topic... how are any of the poll choices "oriental music"?

[quote]But is it just me, or is there usually also a little Asian kid across the street who plays the violin 6 hours a day?[/quote]

Asians are like ants. 1) They are everywhere and very numerous. 2) They work too hard for their own good. (See blog for a deeper analysis of this analogy.)

[quote="guineapiggies"][quote="istos"]:rotfl:

You forgot something.

3) They tend to be smaller than average.

Just kidding.[/quote]

lol actually all the "superstars" in Taiwan are around 6ft tall :rotfl:  but yea i'm mini. everyone uses my head as a armrest :D[/quote]

Same here! ;D

[quote="dreambold22"]i'm not mini! (11-year-old at 5'1" and 84 pounds) yay!

i seriously know someone (this little 6-year-old chinese boy) who plays more than 8 hrs a day. and man, is he good. he just spends his entire day practicing  :o . it must get boring... (he's my cousin, btw.)[/quote]
It actually isn't boring.

[quote="moogra"][quote="serialk11r"]I really don't understand the topic...
But 6 hours a day is cool :) I wish I had time to play that much.[/quote]

You almost play 6. Close enough.[/quote]
Not quite, 3-4 is far from 6.

Hahaha.  Is 4-5 hours a day considered close?  :)
I'm oriental btw.   :P


A small body $ A$ is fixed to the inside of a thin rigid hoop of radius $ R$ and mass equal to that of the body $ A$. The hoop rolls without slipping over a horizontal plane; at the moments when the body $ A$ gets into the lower position, the centre of the hoop moves with velocity $ v_{0}$. At what values of $ v_{0}$ will the hoop move without bouncing?

OK, so the hoop probably bounces when the body $ A$ is at the top? What then? The normal force becomes zero at the bottom. I'm not getting enough equations. (I don't even think the ones I've got are right so I'm not posting them)

If the diagram I've attached is confusing, then heres the clarification, the dark rectangular 'thing' attached to the hoop (the ring) is the body given as $ A$, and the line that goes from the boundary of the hoop to the centre is obviously indicating the radius $ R$. Moreover, the centre of the hoop has a horizontal velocity of $ v_{0}$ which is also given.

[hide="outline of solution"]
First note that if there is no friction on the ground the ball cannot pure roll
assume the ring moves without bouncing.
let my frame be attached to the centre of the ring.This is an accelarated frame .
from this frame we have there are three forces acting on the mass  at the upper most point
1.pseudo force along horizontal
2.Normal($ N$) due to hoop downward
3.$ mg$ downward
Let $ N^{'}$ be the normal force from ground to hoop
now since from this frame t moves in a circle let it's speed be $ v$
thus $ N+mg =\frac{mv^{2}}{r}$
also at lower most point $ v_{o}=\omega r$
at $ v_{o,max}\rightarrow N^{'}\approx 0\rightarrow N=mg\rightarrow v =\sqrt{2rg}$
also see that the net work is done only by gravity so conserving energy from lowermost to upper most point
$ \frac{mv_{0}^{2}}{2}$ + $ \frac{Iv_{o}^{2}}{2r^{2}}$ $ =$ $ mg(2r)$ + $ \frac{mv^{2}}{2}$ $ +$ $ {}frac{mv^{2}{2\times 4}}$ $ +$ $ {}frac{I v^{2}{2\times 4r^{2}}}$
putting the values we get $ v_{o,max}=\sqrt{8gr}$ [/hide]

I'm not sure I understand your outline too well.

First of all, are you sure the horizontal pseudo force does have any importance at all, I think only the vertical equilibrium should be sufficient. In which case the centrifugal force should have a say. Ofcourse thats pseudo.  :roll:
"now since from this frame t moves in a circle let it's speed be $ v$" If you meant the body $ A$ to be moving in a circle in the frame of the centre of the hoop, then you have later said that the lower part is having a velocity $ v_{0}\equal{}\omega R$. That couldn't be happening at the same time since its a rigid hoop and its undergoing pure rolling. Or maybe I didn't understand what your driving at, so there might be a misunderstanding from my part.  :maybe: Either case please do elaborate on what you intended to say. Also the energy equation, couldn't figure out whats happening there... The initial conditions (left hand side is pretty self evident), but when the body $ A$ rises to the top, are we supposed to consider the rolling kinetic energy of $ A$ aswell as the hoop, and also, what is its velocity(translational) considering the fact that the centre of the hoop is also moving (at a velocity $ v$)

ok let's revisit what i have done
first
[quote="cyprusad"]First of all, are you sure the horizontal pseudo force does have any importance at all, I think only the vertical equilibrium should be sufficient. In which case the centrifugal force should have a say. Ofcourse thats pseudo[/quote]

see my frame is the centre of mass frame which is moving  towards the rightwith say accelaration $ a(t)$ hence the any object of mass $ m$ in my frame doesn't matter whether that is moving in a circle or translationally is supposed to experience a force of magnitude $ ma(t)$ towards left so as to apply Newton's laws the way they are.

secondly 'centrifugal' force is only necessary if my frame were rotating with respect to some inertial frame which here is not it is moving translationally with respect to the ground.
also note that my equation
[quote="pardesi"]
$ N+mg =\frac{mv^{2}}{r}$[/quote]

only consider sthe vertical component of the forces acting on the body and equates it with the vertical accelaration.
[quote="cyprusad"]"now since from this frame t moves in a circle let it's speed be v" If you meant the body A to be moving in a circle in the frame of the centre of the hoop, then you have later said that the lower part is having a velocity v_{0}=\omega R. That couldn't be happening at the same time since its a rigid hoop and its undergoing pure rolling. Or maybe I didn't understand what your driving at, so there might be a misunderstanding from my part. maybe Either case please do elaborate on what you intended to say[/quote]
well now as the hoop pure rolls the lower most point from ground(earth) frame has $ 0$ speed hence if $ \omega$ is it's angular speed as seen from the centre frame then clearly $ v_{0}=\omega r$
now as the body is small we can safely claim that at the time when it is just below it's speed is also $ 0$
but the hoop's speed is $ v_{0}$
.now consider the moment the body is at the [b]top[/b] of the hoop again it's speed from centre frame is $ \omega^{'}r = v$ towards right .but again the centre has to move with the same speed $ v^{'}$ toward right otherwise the [b]lowermost[/b] point slips.
hence the speed of the body as seen from earth frame is $ 2v$
also note that $ \omega^{'}=\frac{v}{r}$
[quote="cyprusad"]Also the energy equation, couldn't figure out whats happening there... The initial conditions (left hand side is pretty self evident), but when the body A rises to the top, are we supposed to consider the rolling kinetic energy of A aswell as the hoop, and also, what is its velocity(translational) considering the fact that the centre of the hoop is also moving (at a velocity v)
[/quote]
the most important part of the proof and probably the wonderful.
let my frame of refrence be earth.
now what are the forces acting on the mass at any instant
1.the normal force which is radial due to the hoop
2.gravity
now consider the work done by normal force on the hoop + the work done by gravity on the mass
let the sped of the hoop at any instant is $ \vec u$ then speed of mass is $ \vec u+\vec\omega\times\vec r$
now the normal force on hoop is anti-radial so
net work done is $ -(\vec N .\vec u)dt$ $ +$ $ \vec N.(\vec u+\vec\omega\times\vec r)$
note that noraml is perpendicular to $ \vec\omega\times\vec r$
and the other parts cancel out so normal does no net work.
also note that this normal force provides no torque about the axis through centre.
now again there is a frictional force acting on the hoop but since the body is pure rolling work done by frictional force and work done by the torque due to frictional force cancel each other
now let's have the energy equations
[u][b]for the mass[/b][/u]
$ \Delta K = W_{N}+W_{g}$ where i beleivce subscripts are clear

[u][b]work done by forces on hoop[/b][/u]

$ \Delta K_{trans}= W_{frictional force}+W^{,}_{N}$

[u][b]work done by torque[/b][/u]
$ \Delta K_{rot}= W_{frictional torque}$

adding all note that $ \boxed{W_{N}=-W_{N}^{,}}$

$ \boxed{W_{frictional force}=-W_{frictional torque}}$

so what remains is $ \boxed{\Delta K+\Delta_{trans}+\Delta_{rot}= W_{g}}$

so thsi is the reason why there is a rotational energy term and no other.
also we can epress all the varibles that is $ \omega$ in terms of $ v$ as done above.
hence solve for $ v_{o max}$ knowing that at $ v_{omax}$
as i had derived earlier
[quote="pardesi"]Let $ N^{'}$ be the normal force from ground to hoop
now since from this frame t moves in a circle let it's speed be $ v$
thus
$ N+mg =\frac{mv^{2}}{r}$
also at lower most point $ v_{o}=\omega r$
$ v_{o,max}\rightarrow N^{'}\approx 0\rightarrow N = mg\rightarrow v =\sqrt{2rg}$
[/quote]
 :)


[quote="leroyjenkens"]I don't understand how to get the reference angle. I'm not even really sure exactly what a reference angle is. The book doesn't explain it very well at all.
For example, in this book, it has a couple of examples of how they got a reference angle:

Q. Find the reference angle for 300 degrees.

A. The measure of 300 degrees lies between 270 and 360. Subtract 360 - 300 = 60 degrees.

Where does the 270 and 360 angles come from? They're not mentioned in the question.

Thanks.[/quote]
The reference angle of any angle is the distance of that angle from the x-axis. So to find the reference angle for 300 degrees, you would see that(by drawing the unit circle), you would subtract 300 from 360 to obtain 60 as the reference angle for 300. We are trying to find its "distance" in terms of an angle from the x-axis. So as another example, the reference angle of 240 would be 240-180=60.
They mention 270 to note that 300 is in the 4th quadrant.


inequalities
geometry
3D geometry
tetrahedron
complex numbers
geometry proposed

For a given tetrahedron $ ABCD,$ prove that $ \overline{AB}\cdot \overline{CD}<\overline{AC}\cdot \overline{BD}\plus{}\overline{AD}\cdot \overline{BC}.$

to: kunny

Attached, 2 ways of solving.

Best regards,

sunken rock

A blind man sees better the details

This is Ptolemy's inequality extended to three dimensions.

First, note that $ (a\minus{}b)(c\minus{}d)\equal{}(a\minus{}c)(b\minus{}d)\plus{}(a\minus{}d)(c\minus{}b)$ for any numbers $ a,b,c,d$, or even anything with commutative multiplication. In two dimensions, we can apply this to get $ (A\minus{}B)(C\minus{}D)\equal{}(A\minus{}C)(B\minus{}D)\plus{}(A\minus{}D)(C\minus{}B)$ as complex numbers, and apply the triangle inequality. In three dimensions, there's nothing quite so convenient.
What we do instead is project onto planes. Let $ P$ be a plane spanned by the origin spanned by $ A\minus{}B$ and $ C\minus{}D$, and let $ A',B',C',D'$ be the projections of $ A,B,C,D$ onto $ P$ respectively. We have $ A'B'\equal{}AB,C'D'\equal{}CD$, $ A'C'\le AC,A'D'\le AD,B'C'\le BC,$ and $ B'D'\le BD$. Apply Ptolemy's inequality to the projection, and we are done.
The equality case is the same as in two dimensions: all four points lie on a circle.


For the egg question, what does it mean to have worst possible luck? Does it mean that the eggs will break on the first try? That seems strange. Sorry If I am not allowed to do this, but I beleive the question is harmless.

Also, for the last one, can you have (18, 4) and then have the next two numbers (10,4)? i.e. 18 = 2*4 + 10.

Yes, I believe you're right about #10 (i.e. if you start out with the pair (18,4), you can create the pair (10,4)).  For further clarification though, I suggest that you email quiz@mathcamp.org.


geometry
geometric transformation
reflection
combinatorics unsolved
combinatorics

Points $A_1,...,A_n, B, C$ satisfy
$\sum_{i=1}^n A_iB = \sum_{i=1}^n A_iC = d$
Moreover, $\sum_{i=1}^n A_iP \geq d$ holds for all points $P$ in the plane.
Prove that $A_1,...,A_n$ are collinear

Take $P$ the midpoint of $BC$. Then $A_iP$ is the median of a triangle with vertices $A_i, B, C$, so it's $< \frac{A_iB+A_iC}{2}$ (to prove this just take $A_i'$ the reflection of $A$ through $P$ and look at $A_iA_i', A_iBA_i', A_iCA_i'$), unless $A_i, B, C$ are collinear, which gives equality. If some $A_i$ were not collinear with $BC$, we would have $2 \sum A_iP< \sum A_iB + \sum A_iC=2d$, absurd. So $A_1, \ldots, A_n$ must all lie on $BC$.


Very nice inequality. But, this is old inequality. VIETNAM National Olympiad 1991.

[quote="caubetoanhoc94"][quote="kuing"][quote="airrafer"]x,y,z>0

(x^2y)/z + (y^2z)/x + (z^2x)/y >= x^2 + y^2 + z^2[/quote]

try $ x \equal{} 4,y \equal{} 9,z \equal{} 36$[/quote]

It's true with $ x \ge y \ge z$  :)[/quote]
Yes, It's true with $ x \ge y \ge z$
Because by Chebyshev and AM-GM inequality,we have :
$ \frac {x^2y}{z} \plus{} \frac {y^2z}{x} \plus{} \frac {z^2x}{y}
\equal{} \frac {x^3y^2 \plus{} y^3z^2 \plus{} z^3x^2}{xyz}
\geq \frac {(x^3 \plus{} y^3 \plus{} z^3)(x^2 \plus{} y^2 \plus{} z^2)}{3xyz}
\geq  \frac {3xyz(x^2 \plus{} y^2 \plus{} z^2)}{3xyz} \equal{} (x^2 \plus{} y^2 \plus{} z^2)$
[hide] :!: good bye mathlinks... :( [/hide]


The figures are missing in the Contests>>USA>>AIME, such as 2006 AIME I.  I wonder if some one may help to post them?  Thank you very much!

Try looking here:
[url]http://www.artofproblemsolving.com/Forum/album.php?t=126317[/url]

[quote="Karth"]Go to kalva.[/quote]
Kalva stops at 2003.


function
real analysis
real analysis unsolved

What function is equal to the taylor series
$1+\frac{x^{3}}{3!}+\frac{x^{6}}{6!}+\frac{x^{9}}{9!}+...+\frac{x^{(3n)}}{(3n)!}+...$?

[hide="hint"]The solution I found was a sum of 3 other functions.[/hide]

It has actually been discussed before, [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=86275&start=62]here[/url].  It's a nice question, though.

i've taken a look at that topic, and the approach i have in my mind doesn't seem to appear there, so i post it here...

so.. call your sum $S(x)$. just consider $\zeta \in \mathbb{C}$, such that $\zeta^{3}= 1$ (and, of course, $\zeta \neq 1$.
then you have $3S(x) = e^{x}+e^{\zeta x}+e^{\zeta^{2}x}$, and you're done...


Let $ ABC$ be an acute triangle. Denote by $ \omega$ and $ \Omega$ the incircle and the circumcircle, respectively, of $ ABC$. Consider the circle that is internally tangent in $ A$ to $ \Omega$ and externally tangent to $ \omega$ in $ A_1$. Define $ B_1$ and $ C_1$ in a similar way. Prove that $ AA_1$, $ BB_1$ and $ CC_1$ are concurrent.

[hide][i]Edited by pohoatza[/i][/hide]

The external similarity center $ X_{56}$ of $ \Omega, \omega$ lies on $ AA_1, BB_1, CC_1.$

[u][b]Lemma[/b][/u]: Let ABC a triangle and M the midpoint of AB, O the center of the circumcircle $ \Gamma$ and H the orhocenter. Let $ \Gamma'$ the circle tangent to the feuerbach circle of ABC at M and to the circumcircle and P its center. Call $ K: CM \cap \Gamma$, so K stay on $ \Gamma'$.

[u][b]Proof of the Lemma[/b][/u]: Call N the midpoint of CH, so cause $ CO \parallel NM$, $ \angle NMC \equal{} \angle MCO \equal{} \angle OKC$. Call $ J: NM \cap KO$, so we have that J stay on OM, K, J and O are alight and FJ=JM. So J is the center of $ \Gamma'$.

[u][b]Proof[/b][/u]: invert all w.r.t. the incircle. Call D,E,F the touch point of the incircle with the side a,b,c. A go on A'', B on B'' and C on C''.
We have that A'' is the midpoint of EF, B'' of DF and C'' of DE. So the circumcircle go in the feuerbach circle of $ \triangle DEF$ and the circle tangent to the circumcircle in A and to the incircle is the circle tangent to the circumcircle of $ \triangle DEF$ and to its feuerbach circle on the midpoint of the side EF. So for the [b]Lemma[/b] the point of tangency with the incircle and the inverted circle is the point where DA'' meet EF and equally for the others circle.
So the A'A'', B'B'' and C'C'' concur on the centroid of $ \triangle DEF$.

This problem, in the posted form, appears in a preprint of A. Hatzipolakis and P. Yiu, called "Triads of circles", 2001. In fact, the result here is more general, and I assume that it dates from Apollonius himself:

[color=darkblue][b]Theorem[/b]: Given $ \omega_{1}$, $ \omega_{2}$, $ \omega_{3}$ three non-intersecting circles. Consider $ \Delta_{1}$, $ \Delta_{2}$ the circles simultanously tangent to $ \omega_{1}$, $ \omega_{2}$, $ \omega_{3}$ at $ A_{1}$, $ B_{1}$, $ C_{1}$, and $ A_{2}$, $ B_{2}$, $ C_{2}$, respectively. Then, the triangles $ A_{1}B_{1}C_{1}$ and $ A_{2}B_{2}C_{2}$ are perspective.[/color]

This is a simple consequence of Monge's circle theorem (See here: [url]http://mathworld.wolfram.com/MongesCircleTheorem.html[/url]).


an equilateral triangle of side $ n$ ($ n >\equal{} 2$) is divided into $ n^2$
equilateral triangles of side 1 with the lines parallel to the sides of the
triangle. Can one number these unit-sided triangles with the numbers from 1
up to $ n^2$ so that both the following conditions hold?

1. The triangles with the numbers $ i$ and $ i \plus{} 1$ have at least one common
vertex, $ 1 <\equal{} i <\equal{} n^2 \minus{} 1$.

2. the triangles numbered $ i$ and $ i \plus{} 2$ have at least one common vertex, $ 1
<\equal{} i <\equal{} n^2 \minus{} 2$.

once more, the deadline for the olympiad problems are really over. you can check with anyone you know.

Consider the extreme corner triangles. Think about what numbers THEY can occupy. Can they occupy any number between $ 3$ and $ n^2 \minus{} 2$? If they did, then they would necessarily have to have four numbers surrounding them in the adjacent triangles (Because for any $ i$ such that $ 3 <\equal{} i <\equal{} n^2 \minus{} 2$, and $ i$ is a natural number, $ i \minus{} 2, i \minus{} 1, i \plus{} 1, i \plus{} 2$ also have to occupy adjacent triangles). But each corner triangle has only THREE adjacent triangles. So we conclude that the corner triangles can be occupied only by numbers from the set $ {1, 2, n^2 \minus{} 1, n^2}$.

Now define a "region" as the set of four triangles at the corner (i.e. the corner triangle itself and its three adjacent triangles). There are THREE regions in each triangle for $ n >\equal{} 3$. Note that if a "region" contains $ 1$, it must also contain $ 2$, and if a "region" contains $ n^2$, it must also contain $ n^2 \minus{} 1$. So, for large $ n$ (in fact, $ n >\equal{} 3$), it can be easily shown that two of the regions take up all numbers from the set $ {1, 2, n^2 \minus{} 1, n^2}$. This leaves the third corner triangle unaccounted for, since it can no longer occupy any of the admissible numbers.

The cases $ n \equal{} 1$ and $ n \equal{} 2$ are trivial, and are left as an easy exercise.

So we can conclude that it is impossible to number the triangles in the way shown for any $ n$ except the trivial cases $ n \equal{} 1$ and $ n \equal{} 2$.


I dont to be risrespectful but this is how I feel...

I dont think people on this forum are prone to being Myspace additcs, but I wanted to say that I hate it. I dont have one, I dont plan to have one. It a place for people who have nothing do with their lives. It creates an "addiction" and they are all saying "oh, I need to check my mspace today ... "

Now, my point is that internet can be very destructive. It depends how you spend you time with it. For example, me and a lot of people here use the internet for doing math, i.e. AOPS. That's a good way to spend your time. I go to Wikipedia and read math and other stuff, ect .... I dont waste me time. And neither do you (I hope).

You agree?

I agree that internet can be distructive. But, I personally actually use MySpace. I get on about once or twice a week. I hope that doesn't count as "destructive"...ehmmm. :mad:

Yes, the internet can indeed be destructive. But I guess it's okay to have a little enjoyment besides math, as long as you don't spend so much time that it conflicts with other activities. I do not have a MySpace, and don't plan on getting one either.

sylow_theory, stop knocking off myspace. In fact, myspace can be quite interesting and can also be a good way to spend your time and meet new people. Just because you don't like it, doesn't mean its bad. There are some people who think that aaops is a waste of time. "waste of time" is a relative concept.

True, the internet can be destructive, usually making you twisted.

Myspace is good website but what makes it bad is what people put on it.

Myspace used to be good, but then people came and they ruined it all, so it's dangerous.

I never used Myspace.

I prefer facebook for several reasons: standard layout of pages, clearly defined networks, real names of users, no autoplay for music. Both facebook and myspace can be a waste of time or a useful tool, depending on what you do with it.

Myspace is very, very, slow. Which is why I never use it. That may be why it wastes a lot of people's time. :)

Can you specify what you hate so much about MySpace?

Personally, I used to go onto the science forum there and it was hilarious how much people argued over the simplest of things. In fact, one time they had a heated debate over proving that $ a^{0}= 1$  :rotfl:

I suppose MySpace is a social internet gathering for all the mindless idiots that are into MSN and humiliating people with youtube videos and all that frivolous stupidity.

myspace and facebook are a waste of time... unless you desperately need friends

I use myspace to keep in touch with my friends from my old town where I used to live; it's really not a waste of time if you use it for a certain purpose.

I like how he says that people who are on Myspace don't have lives and then proceeds to talk about how he reads math things on Wikipedia.

[quote="Lazarus"]I'm a Facebook addict. Whatever.[/quote]

DUDE I SEEM TO RECALL YOU CALLING ME EMO FOR HAVING A FACEBOOK AROUND 9 MONTHS AGO.

[quote="Lazarus"]four CHAN[/quote]

Someone broke Rules 1 and 2.

Haha... facebook ppl dont have lives and then "i read math on wikipedia" lmao

and emo for having a facebook???

[quote="bubka"]myspace and orkut are okay, as the first of their kind.
But dash it - facebook addict??
Next people will be bebo and fixster addicts. :roll:[/quote]

LOLWUT

WTF IS ORKUT

Hey don't you go rubbishing orkut! I have 350 friends and 63 fans there. And last time I checked, it was as about twice as better as myspace would be in five more years.

[quote="Anaxerzia"][quote="Lazarus"][quote="hunter34"]True, the internet can be destructive, usually making you twisted.[/quote]

four CHAN[/quote]

HAHA so true[/quote]

LOL 4chan and Ebaumsworld :rotfl:

[quote="azhang"][/quote][quote="Anaxerzia"][quote="Lazarus"][quote="hunter34"]True, the internet can be destructive, usually making you twisted.[/quote]

four CHAN[/quote]

HAHA so true[/quote][quote="azhang"]

LOL 4chan and Ebaumsworld :rotfl:[/quote]

Dude 4chan is on a completely different dimension from eBaumsworld.. seriously lol

thats true. but i have one question for you

so i herd u liek mudkipz?


function
inequalities
quadratics
algebra
polynomial
inequalities unsolved

Suppose $x$ and $y$ satisfy \[xy=\sqrt{1-x^2}+\sqrt{1-y^2}.\]
Find the maximum of one of the following expressions:
a) $\frac{2xy}{x+y}$;

b) $\frac{x+y}{2}$;

c) $\sqrt\frac{x^2+y^2}{2}$.

Mostly I am interested in a), but I guess they are all the same.

a) The maximum value of $ \frac{2xy}{x+y} $ occurs when $ x=y=\sqrt{\sqrt3-1} $
 b,c) the same result
  My result is OK ?

No, it is not. I can write many similar results.
I need the proof, not the final answer.

Thank you.

I have found something  :lol:  :
We have $\frac{2xy}{x+y} \leq \frac{x+y}{2} \leq \sqrt{\frac{x^2+y^2}{2}}$ with equality if and only if $x=y$. So if we find the maximum of $\sqrt{\frac{x^2+y^2}{2}}$, It's going to be the maximum of the two others. Is that wrong ?  :?

You are right. That's why I gave all three problems. I hope that one of them is solvable.

Ok. but until now I didn't find the maximum of $\sqrt{\frac{x^2+y^2}{2}}$  :(
Any hints please ??  :?

b) is easily solvable with Lagrange multipliers (probably, there is an elementary solution too). Clearly, the maximum is attained when $x,y>0$. The equation you get is $y+\frac x{\sqrt{1-x^2}}=x+\frac y{\sqrt{1-y^2}}$, which is equivalent to $x\left(\frac1{\sqrt{1-x^2}}-1\right)=y\left(\frac1{\sqrt{1-y^2}}-1\right)$. But the function $F(z)=z\left(\frac1{\sqrt{1-z^2}}-1\right)$ is increasing on $(0,1]$ as a product of two positive increasing functions. So, the maximum is attained either on the boundary ($x$ or $y=1$), or for $x=y$. But it is not hard to see that the case $x=y=\sqrt{2\sqrt 2-2}$ gives a larger value for the sum than the boundary case $x=1$, $y=\frac1{\sqrt 2}$. Not that I am very fond of this solution, but, at least, it gives the desired result  :)

Thank you for your efforts. Though I don't know whether this solution is applicable for the initial problem http://www.mathlinks.ro/Forum/viewtopic.php?t=3744

It looks like

$cosx * cosy = sinx + siny$

Part c is equivalent to find the maximal of

$cos^{2}x + cos^{2}y$

Sorry for my result !
  I checked it  . The maximum value of $ \frac{2xy}{x+y} $ occurs when $ x=y= \sqrt{2\sqrt2-2} $
  I don't have a nice solution but I have an ugly solution  ;)
   Consider $ x,y > 0 $
  From the equation , we can easily show $ x= \frac{y.(\sqrt2 +(-) \sqrt{1-y^2})}{y^2+1} $
  We'll find the minimum value of $ f(y) = \frac{1}{y} + \frac{y^2+1}{y.(\sqrt2 +(-) \sqrt{1-y^2})} $
   $ f'(y) = 0 \iff (\sqrt2+1)t^3 + (3+2\sqrt2)t^2 + 2t -2 =0 $ , where $ t = \sqrt{1-y^2} $
   So , we can easily show the minimum of f(y) occurs when $ y = \sqrt{2\sqrt2-2} $

I think your approach is fine. Don't say it's ugly ;)

P.S. You can use \pm command to produce $\pm$.

These are simpler than one might think. First I will show:

[color=blue][b]Lemma 1.[/b] For any two real numbers x and y from the interval [-1; 1] satisfying $xy=\sqrt{1-x^2}+\sqrt{1-y^2}$, we have $x^2+y^2\leq 4\left(\sqrt2-1\right)$.[/color]

[i]Proof of Lemma 1.[/i] Let $X=x^2$ and $Y=y^2$. Since the numbers x and y are from the interval [-1; 1], their squares X and Y are from the interval [0; 1].

Since $xy=\sqrt{1-x^2}+\sqrt{1-y^2}$, we have $xy\geq 0$, so that $xy=\left|xy\right|=\left|x\right|\cdot\left|y\right|=\sqrt{x^2}\cdot\sqrt{y^2}=\sqrt{X}\cdot\sqrt{Y}=\sqrt{XY}$. Thus, the equation $xy=\sqrt{1-x^2}+\sqrt{1-y^2}$ becomes $\sqrt{XY}=\sqrt{1-X}+\sqrt{1-Y}$. On the other hand, the inequality in question, $x^2+y^2\leq 4\left(\sqrt2-1\right)$, rewrites as $X+Y\leq 4\left(\sqrt2-1\right)$. Now, we will prove this inequality by contradiction: We assume that $X+Y>4\left(\sqrt2-1\right)$ and look for a contradiction. Once we have found one, Lemma 1 is proven.

Since $X+Y>4\left(\sqrt2-1\right)$, we have $\frac{X+Y}{2}>2\left(\sqrt2-1\right)$.

The function $f\left(t\right)=t^2-4\left(1-t\right)$ is a quadratic polynomial with zeros $t=-2\left(\sqrt2+1\right)$ and $t=2\left(\sqrt2-1\right)$; thus, since its leading coefficient is positive, it is positive for all $t>2\left(\sqrt2-1\right)$. Hence, since $\frac{X+Y}{2}>2\left(\sqrt2-1\right)$, we get $f\left(\frac{X+Y}{2}\right)>0$. In other words, $\left(\frac{X+Y}{2}\right)^2-4\left(1-\frac{X+Y}{2}\right)>0$. Hence, $\left(\frac{X+Y}{2}\right)^2>4\left(1-\frac{X+Y}{2}\right)$. This rewrites as

$\left(1-\frac{\left(1-X\right)+\left(1-Y\right)}{2}\right)^2>2\left(\left(1-X\right)+\left(1-Y\right)\right)$.

Now, since the numbers X and Y are both from the interval [0; 1], the numbers 1 - X and 1 - Y are positive, so that we can apply AM-GM to get $\frac{\left(1-X\right)+\left(1-Y\right)}{2}\geq\sqrt{\left(1-X\right)\left(1-Y\right)}$. Thus, $1-\frac{\left(1-X\right)+\left(1-Y\right)}{2}\leq 1-\sqrt{\left(1-X\right)\left(1-Y\right)}$. Both sides of this inequality are nonnegative (since the numbers X and Y are from the interval [0; 1], and thus the numbers 1 - X and 1 - Y are < 1, so that their arithmetic and geometric means $\frac{\left(1-X\right)+\left(1-Y\right)}{2}$ and $\sqrt{\left(1-X\right)\left(1-Y\right)}$ are < 1 as well); hence, we can square it to get

$\left(1-\frac{\left(1-X\right)+\left(1-Y\right)}{2}\right)^2\leq\left(1-\sqrt{\left(1-X\right)\left(1-Y\right)}\right)^2$.

Combining this with the inequality $\left(1-\frac{\left(1-X\right)+\left(1-Y\right)}{2}\right)^2>2\left(\left(1-X\right)+\left(1-Y\right)\right)$ from the above, we get

$\left(1-\sqrt{\left(1-X\right)\left(1-Y\right)}\right)^2>2\left(\left(1-X\right)+\left(1-Y\right)\right)$.

In other words,

$1-2\sqrt{\left(1-X\right)\left(1-Y\right)}+\left(1-X\right)\left(1-Y\right)>2\left(\left(1-X\right)+\left(1-Y\right)\right)$.

Hence,

$1-\left(\left(1-X\right)+\left(1-Y\right)\right)+\left(1-X\right)\left(1-Y\right)>\left(1-X\right)+\left(1-Y\right)+2\sqrt{\left(1-X\right)\left(1-Y\right)}$.

This simplifies to

$XY>\left(\sqrt{1-X}+\sqrt{1-Y}\right)^2$.

Since X and Y are nonnegative, we can take the square root of this inequality, so that we get $\sqrt{XY}>\sqrt{1-X}+\sqrt{1-Y}$, what contradicts our condition $\sqrt{XY}=\sqrt{1-X}+\sqrt{1-Y}$. So Lemma 1 is proven.

Lemma 1 yields $\sqrt{\frac{x^2+y^2}{2}}\leq\sqrt{\frac{4\left(\sqrt2-1\right)}{2}}=\sqrt{2\left(\sqrt2-1\right)}$. It is easy to see that this inequality becomes an equality if we take $x=y=\sqrt{2\left(\sqrt2-1\right)}$, so it follows that $\sqrt{2\left(\sqrt2-1\right)}$ is the maximum of the expression $\sqrt{\frac{x^2+y^2}{2}}$. Hence, problem [b]c)[/b] is solved. Now, problems [b]a)[/b] and [b]b)[/b] are almost trivial: Again, the maximum is $\sqrt{2\left(\sqrt2-1\right)}$. In fact, as we showed above, $xy\geq 0$, so that the numbers x and y are either both $\geq 0$, and thus we can apply HM-AM-QM to see that $\frac{2xy}{x+y}\leq\frac{x+y}{2}\leq\sqrt{\frac{x^2+y^2}{2}}\leq\sqrt{2\left(\sqrt2-1\right)}$, or the numbers x and y are both $\leq 0$, what yields $x+y\leq 0$, and thus the expressions $\frac{2xy}{x+y}$ and $\frac{x+y}{2}$ are both $\leq 0$ and thus also $\leq\sqrt{2\left(\sqrt2-1\right)}$. So problems [b]a)[/b] and [b]b)[/b] are both solved.

Nice problem, Myth!

  darij

I see many solutions here, but i'll post mine for (a) because i think it needs not that much calculations.
Let $x \in (0,1)$ and $f(x)=\sqrt{1-x^2}$, which is concave. So by considering the tangent of $f$ in the point $(\sqrt{2\sqrt{2}-2},f(\sqrt{2\sqrt{2}-2}))$ we get that :
$\sqrt{2} +1 -\sqrt{2} \sqrt{\sqrt{2}+1}\cdot x \geq f(x)$.
Let $a=\frac{2xy}{x+y}$ then $xy\geq a^2$ and $x+y\geq2a$. So by the condition above we get :
$2(\sqrt{2}+1) \geq xy +\sqrt{2} \sqrt{\sqrt{2}+1}\cdot (x+y) \geq a^2 + 2\sqrt{2} \sqrt{\sqrt{2}+1}\cdot a $ Whence, the result.

[quote="darij grinberg"]These are simpler than one might think. [/quote]
well, if the question was to prove Lemma 1, maybe it isn't very difficult. But I can't see how you approach this problem (eg. you invented function $f(t)$) Can you explain briefly?

[quote="indybar"]But I can't see how you approach this problem (eg. you invented function $f(t)$) Can you explain briefly?[/quote]

I just conjectured that all three terms $\frac{2xy}{x+y}$, $\frac{x+y}{2}$, $\sqrt{\frac{x^2+y^2}{2}}$ have their maximum for x = y; then, computing this maximum was easy (in fact, for x = y, the condition $xy=\sqrt{1-x^2}+\sqrt{1-y^2}$ takes the form $x^2=\sqrt{1-x^2}+\sqrt{1-x^2}$, what is a quadratic equation in $x^2$, so we can find x, and thus we can find the maximum, which is $\sqrt{2\left(\sqrt2-1\right)}$). Okay, now it remains to prove that it is really the maximum, but this is equivalent to proving Lemma 1. The proof of Lemma 1 is more straightforward than one might think. At first, we note that the squares of the variables x and y occur in our inequalities more often than x and y themselves, so we substitute $x^2=X$ and $y^2=Y$. Now we use a well-known trick from Harazi: If you are given that A = B and want to show that $C\leq D$, assume for the sake of contradiction that C > D and show that this implies either A < B or A > B (so you have a contradiction). This is especially useful if the condition A = B is more complicated than the required inequality $C\leq D$ (in fact, $C\leq D$ is, in our case, just a linear inequality: $X+Y\leq 4\left(\sqrt2-1\right)$, while A = B is the strange condition $\sqrt{XY}=\sqrt{1-X}+\sqrt{1-Y}$). So now, we assume that $X+Y>4\left(\sqrt2-1\right)$, and have to show that either $\sqrt{XY}>\sqrt{1-X}+\sqrt{1-Y}$ or $\sqrt{XY}<\sqrt{1-X}+\sqrt{1-Y}$; by checking a numerical example (X = 1 and Y = 1, which yields $\sqrt{XY}=1$ and $\sqrt{1-X}+\sqrt{1-Y}=0$), we see that we should try the former of these aims. Now, the first thing one would try if one sees an inequality like $\sqrt{XY}>\sqrt{1-X}+\sqrt{1-Y}$ is squaring it. After some transformations (that are more obvious if you play around with substitutions like a = 1 - X and b = 1 - Y), the inequality in question takes the form $\left(1-\sqrt{\left(1-X\right)\left(1-Y\right)}\right)^2>2\left(2-\left(X+Y\right)\right)$. Now, we would finally like to use the assumption $X+Y>4\left(\sqrt2-1\right)$, but the left hand side contains a strange geometric mean which hinders us. Anyway, at least, we can majorate a geometric mean by an arithmetic mean. So, if the sharper inequality $\left(1-\frac{\left(1-X\right)+\left(1-Y\right)}{2}\right)^2>2\left(2-\left(X+Y\right)\right)$ would be true, we would be happy. In fact, this inequality is a quadratic inequality in X + Y, or, if you wish, in $\frac{X+Y}{2}$, and rewrites as $f\left(\frac{X+Y}{2}\right)>0$, where $f\left(t\right)=t^2-4\left(1-t\right)$. But now we are almost done - in fact, in order to prove that $f\left(\frac{X+Y}{2}\right)>0$, we have to show that the number $\frac{X+Y}{2}$ is greater than the greatest zero or smaller than the least zero of the quadratic function f(t); but actually, the assumption $X+Y\leq 4\left(\sqrt2-1\right)$ yields that it is greater than the greatest zero, and thus the inequality in question is proven. The function f(t) here naturally arose by writing out $\left(1-\frac{\left(1-X\right)+\left(1-Y\right)}{2}\right)^2-2\left(2-\left(X+Y\right)\right)$ as a function of $\frac{X+Y}{2}$.

Hope this was more or less understandable (I have never been good in explaining heuristics).

  Darij

Ah yes, yes, yes! Conjecture $x=y$. I see now!


topology
advanced fields
advanced fields unsolved

I'm looking for an example of subspace of semiregular space which is not semiregular.

I found this link for Willard's book pages on google books :

[url=http://books.google.jo/books?id=-o8xJQ7Ag2cC&pg=PA98&lpg=PA98&dq=subspace+of+semi+regular+is+not+necessary+semiregular+space&source=web&ots=TQhhHiUzFy&sig=6OeAvYzhJOdhOaF0CJVHUsyZPos&hl=en&sa=X&oi=book_result&resnum=1&ct=result]Semiregular space[/url]

But Willard proved that every space $ X$( which is not semiregular) can be embedded in a semiregular space.

I couldn't understand his nhds for the topology he constructed .

Also , I have a counterexamples books for seen and seebach , I searched inside it for example . But , I couldn't find one . it doesn't talk about subspace of semiregular space.

Is there anyone here knows an example for that, or could he/she explains the nhds of Willard's proof i.e. to write them in other way?

Willard say a semiregular , $ T_1$ space need not be $ T_2$ . I think we could adjust on example 3 in 14F, pg 99 (look at the link above) to have what we need . if so, how we can do it ? or give an ex if u know.BTW, I also searched in Counterexamples books , but , it hasn't an example.

I appreciate ur help

 :)

The counterexample is to just choose $ X$ to be any non-semiregular space. Build $ Z\equal{}X\times I$ in the way described (and attached in picture form). Then $ Z$ is semiregular, but has $ X$ as a non-semiregular subspace.

O.k that's nice HilbertThm90

Did u see part 3 of 14F,pg99 ? Look at link please

Give me your opinion :

What if we make the nhds base for $ \plus{}\infty$ and $ \minus{}\infty$ as :

nhd of $ \plus{}\infty$ is of the form $ \{\plus{}\infty\}\cup (a,\infty) ,a<0$ and
nhd of $ \minus{}\infty$ is of the form $ \{\minus{}\infty\}\cup (\minus{}\infty,a),a>0$

Do we get a semiregular , $ T_1$ space which is not $ T_2$?

 :)

I don't think so, because take any $ 0 < x < \infty$ and $ \infty$ as distinct points. Of course, you have $ \{x\}$ to separate $ x$ from $ \infty$, but there is no neighborhood of $ \infty$ that excludes $ x$.

So I'd say you have $ T_0$ but not $ T_1$.

you're right dude.

I still think about it and I can't figure out of good choice. do you think we can find an easy example for that?

 :maybe:

Ah. I don't know why I didn't think of this immediately. Take any common $ T_1$ but not $ T_2$ space. Say $ X \equal{} \mathbb{Q}$ with finite complement topology. Now construct $ Z \equal{} X\times I$ as given before. It has already been shown that $ Z$ is semiregular. $ Z$ is $ T_1$ since given $ (x,y)$ such that $ y\neq 0$ we can separate in a $ T_2$ manner by small enough $ \varepsilon$'s. Now given $ (x_1,0)$ and $ (x_2, 0)$, choose $ U_1$ so that $ x_1\in U_1$ and $ x_2\notin U_1$ (possible by $ X$ being $ T_1$). Building the $ \varepsilon_x$ type neighborhoods won't cause intersection since it is strictly "vertical." Do this symmetrically the other way.

We get that $ Z$ is $ T_1$. We definitely don't have $ T_2$, though, because $ X$ is a subspace of $ Z$ and is not $ T_2$.


conics
ellipse
invariant
geometry solved
geometry

Kedlaya's booklet states that you can construct the centre of an ellipse using only straightedge.. And it seems quite hard, does anyone know how to do this?

[quote="georgesand"]Kedlaya's booklet states that you can construct the centre of an ellipse using only straightedge..[/quote]

Well, at first, everything you can construct with a straightedge alone is projectively invariant, while the notion of the "center" of an ellipse is not. So you cannot construct the center of an ellipse.

However, this will change if you have some additional data given. The most classical setup is: You are given a straightedge, and two pairs of parallel lines (but not all four lines parallel to each other) are drawn in the plane. Then, you can construct quite a lot of things, particularly

- the midpoint of a given segment (see http://www.mathlinks.ro/Forum/viewtopic.php?t=16966 post #5 tool 1);
- the parallel to a given line through a given point (see http://www.mathlinks.ro/Forum/viewtopic.php?t=16966 post #5 tool 2).

However, before we can talk about the construction of the center of an ellipse, I must know an important thing: What exactly do we know about the ellipse? I see a number of possible cases:

[b]Case 1:[/b] The ellipse is drawn on paper such that we can take any line and intersect it with the ellipse.
[b]Case 2:[/b] The focii of the ellipse are given.
[b]Case 3:[/b] Five distinct points A, B, C, D, E on the ellipse are given.

Well, in Case 2, the problem is trivial (the center of the ellipse is the midpoint between the focii, so we can construct it as we can construct the midpoint of any segment).

Now, let's consider Cases 1 and 3. For Case 3, we need an auxiliary construction:

[color=blue][b]Auxiliary construction.[/b] Given five distinct points A, B, C, D, E and a line g through one of these points - say, through A. Construct the point of intersection of this line g with the conic through the points A, B, C, D, E different from the point A.[/color]

This construction is easy using the Pascal theorem: If we denote by F the point of intersection of the line g with the conic through the points A, B, C, D, E different from the point A, then the hexagon ABCDEF is inscribed into a conic; thus, by the Pascal theorem, the points $AB\cap DE$, $BC\cap EF$, $CD\cap FA$ lie on line. The points $AB\cap DE$ and $CD\cap FA$ can be easily constructed (since the points A, B, C, D, E are known, and the line FA is simply the line g); thus, this line can be constructed, and the point $BC\cap EF$ is then located as the point of intersection of this line with the line BC. Thus, we can construct the line EF as the line passing through the two known points E and $BC\cap EF$. Now, we can locate the point F as the point where this line EF meets the line g. So the auxiliary construction is done.

Now, consider five points A, B, C, D, E on our ellipse (in Case 3, these points are given; in Case 1, they can be just arbitrarily picked - the only requirement is that they are distinct). Draw an arbitrary line g through the point A, and draw the parallel h to the line g through the point B. We can construct the point of intersection A' of the line g with the ellipse different from the point A, and the point of intersection B' of the line h with the ellipse different from the point B (in fact, in Case 1, these constructions are obvious; in Case 3, they require the auxiliary construction). Now, the segments AA' and BB' are two parallel chords in the ellipse; thus, the line joining their midpoints is a diameter of this ellipse, i. e. it passes through its center. Needless to say, we can construct the midpoints of these segments AA' and BB' and join them by a line. Thus we obtain a line through the center of the ellipse. Now, we can change the direction of the line g, and we obtain another line through the center of the ellipse. The intersection of these two lines is the center of the ellipse. Done!

By the way, Case 1 was solved in http://www.mathlinks.ro/Forum/viewtopic.php?t=16966 for the case when the ellipse is a circle. However, by an affine transformation, we can map any ellipse to a circle, and hence, of course, the same constructions must work equally well for generic ellipses.

  Darij


If I type the following $ \text{\LaTeX}$ code on a forum:
[code]$(r-s)|2$[/code]
it displays as $ (r-s)|2$.

However, if I make any change at all to that code, including changing one of the variable names to another letter, changing the 2 to another number, or putting curly braces around the 2, it displays normally, like $ (r-s)|{2}$. Obviously, this is a very minor and fixable error, but it seems very unusual. Does anyone know what's causing it?

hello, do you mean that $ (r\minus{}s) | 2$?
Sonnhard.

$ (r\minus{}s)|2$
$ (r\minus{}s) |2$

Even a single space fixes it. It's also definitely specific to the forum; I tested it in a document, and it works fine there.

I'm not sure why it went wrong in the first place, but images are cached for speed so once there is an error (r-s)|2 will always produce the same wrong image until Valentin deletes it. Whatever the problem was, it looks like it has been fixed because new images, created when you change something, produce the correct result.


calculus
integration
limit
calculus computations

Let $a$ be a real constant number. Evaluate $\lim_{n\to\infty}n\int_{-1}^{1}e^{-n|a-x|}\ dx.$

Thank you for your reply, Ottem.
Your answer is correct. :)


I think you should note that this only holds for acute angles...... if B or C are large obtuse angles the inequality doesn't hold

of course, i forgot to mention that A,B,C are angles of a triangle. I have found this inequality in a book and although it seems an easy one I can't figure out how to prove it. It maybe a mistake... try to get to a contradiction.

[hide]
$\sin A=\sin (180^\circ-B-C)=\sin (B+C)=\sin B\cos C+\cos B\sin C$. Since $\sin B$ and $\sin C$ are less than or equal to 1, $\sin A\leq \cos C+\cos B$. There would be equality if $\sin B=\sin C=1$, but there would be two right angles which can't happen in a triangle. Therefore the inequality is strict and $\sin A<\cos B+\cos C$.
[/hide]


The first ARML practice (like those I mentioned before) will happen at my house (2628 31st Ave W) this Saturday, 3:00-7:00. The email from my mom about it:

As you recall, last June we talked a bit about gathering on a regular basis in Seattle to prepare for ARML Contests.  Our first session will be on Saturday, September 24th from 3 - 7pm.  I'll ask everyone to chip in $ \$ $5 and we'll do a pizza run or something for dinner.  You guys who were seniors last year, I've kept you on the list for now, just in case you're at the UW or SU and want to help mentor the proceedings.  At least to start we'll be meeting at my house.  If our group outgrows the space, I'll look for an alternative setting.  Please reply and let me know whether or not you can make it.  Also, reply and let me know if you're interested in coming, but just can't make it on the 24th.

Tom and Mary and Elissa, these sessions are open to anyone who'd like to come, not just the kids who went to San Jose last year.  Please put a bug in the ear of any student who you think might be interested, and pass the word on to other math coaches in the area, since I only have email addressess for you three.

Do you want us to bring anything? (besides the money for pizza)And is it just problem solving or lecturing or something?

>Do you want us to bring anything? (besides the money for pizza)

Scratch paper, pencil(s), and a calculator.

>And is it just problem solving or lecturing or something?

We'll do some special practice problems from ARML, and there'll be mini-lectures on better ways to do certain types of problems.

-Adam Hesterberg


[b]Fifth International Internet Mathematical Olympiad for Students[/b]

First Competition of the Season 2009-2010
will be held on [b]December 17, 2009[/b] at 10:00 (GSM + 2)

[quote]We invite students to take part in our International Internet Mathematical Olympiad. Last year students from 17countries participated. There were more than 45000 visitors of the website of International Internet Olympiad.

The Olympiad problems’ website  http://www.ariel.ac.il/cs/projects/dom/itpm will be opened on [b]December 17, 2009 at 10:00[/b] (GSM + 2).  Participants have to send their solutions until [b]December 17,  2 p.m[/b] via e-mail  olymp@ariel.ac.il

For registration students have to send their Registration form (see http://www.ariel.ac.il/cs/projects/dom/itpm/reg.doc ) to the secretary of Faculty of Natural Sciences Ms Julia Smolensky  olymp@ariel.ac.il   before [b]December 10, 2009[/b].

Enjoy mathematics!

Prof. Alexander Domoshnitsky
Dean of Faculty of Natural Sciences
Ariel University Center- Israel[/quote]

http://www.ariel.ac.il/cs/projects/dom/itpm/
Invitation in PDF format: http://www.ariel.ac.il/cs/projects/dom/itpm/i9.pdf
Problems of Internet Olympiad 2008: http://www.ariel.ac.il/cs/projects/dom/itpm/i09.asp

Which book do you recommend in order to prepare for this contest?:) Or some types of problems...

I have easy solutions problem6 and problem8 :) :rotfl:


hi guys me appearin fr kvpy exam dis november..
i wanna now wich are d concepts asked in d exam wich are nt taught in ssc portion.
i heard calculas is also included ..
so shall i do somethin abt it???

stream sa science.
rply fast and wish u  a happy diwali..

(a) Please only post your question once.

(b) Please post in English.

Lol. It is in english, just shorthanded with many abbreviations.

Happy diwali back to you.

Anyways, why don't you google search this stuff?

Well, at least English that we can comprehend without having to think too hard. :P

I hope any of the Indian students who know the general answer to the question will post it, with links to further sources of information. I'd Google it up myself (I'm generally curious about examinations in India) but I've been busy looking up information about A-level exams and admission requirements for Cambridge.


A circular ceiling fan rotates at a constant speed of 80 revolutions per minute. A point halfway between the center of the fan and the outer edge of the fan travels 97,968 inches in 15 minutes. How far does a point on the outer edge of the fan travel in 30 minutes?

Firstly, this should be posted in the mathcounts section.
secondly, the answer is 4*97,968 which is.. 391,872 i think.

The distance travelled is proportional to the circumference which is also proportional to the radius. Therefore, a radius twice as much will travel twice as far. And twice the time, multiply by 2 again.

Therefore, $ 97968\cdot4\equal{}\boxed{391872}$.

Edit: Dang, beat


A scout troop buys $ 1000$ candy bars at a price of five for $ \$2$. They sell all the candy bars at a price of two for $ \$1$. What was their proﬁt, in dollars?

$ \textbf{(A)}\ 100 \qquad
\textbf{(B)}\ 200 \qquad
\textbf{(C)}\ 300 \qquad
\textbf{(D)}\ 400 \qquad
\textbf{(E)}\ 500$

[hide="Click for solution"]
They bought $ 200$ groups of five candy bars for a cost of $ 200 \times 2\equal{}\$400$.  They sold $ 500$ groups of $ 2$ candy bars for $ 500$ dollars.  They thus made $ 500\minus{}400\equal{}\$100$, or $ \boxed{\textbf{(A)}}$.
[/hide]

Easier way to do this in your head is simply calculate cost of an individual bar, $5$ bars for $2$ means they cost $40$ cents and selling them $2$ for $1$, means they got $50$ cents gross, hence $1000(50-40)$ or $100$


advanced fields
advanced fields unsolved

Prove that $S^\infty$ is a contractible space.

Not only has this been posted before, but it's been posted before, on this forum, sufficiently recently that it's still among the top 15 topics.

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=73270

Thanks you but can you find for me the link had solution?, thanks!

If you're not satisfied with what's in that thread, why not use that thread to ask for more? There was no reason to make a new topic on the issue.

I only think there are many topics about it, can moderator delete this my topic or move all for the main topic ? :arrow:


integration
calculus
calculus computations

OK, so if you guys could gimme some feedback on this I would appreciate it.

Denote the volume generated when the region bounded by the curve $ y \equal{} x^2 \plus{} 1$, the y-axis, and the line $ y \equal{} k$ is revolved about the y-axis as $ V$. If $ k$ is given by $ k \equal{} 1 \plus{} e^t$, find the change in $ V$ when $ t \equal{} 7$.

[b]Edit:[/b] Not quite related rates, now that I think about it.

$ V\equal{}\int_1^{1\plus{}e^t} \pi x^2 dy\equal{}\pi \int_1^{1\plus{}e^t} (y\minus{}1)\ dy$.
Use the Fundamental theorem of Calculus, $ \frac{dV}{dt}\equal{}\pi (1\plus{}e^t\minus{}1)(1\plus{}e^t)'\equal{}\pi e^{2t}$.

$ \therefore \left[\frac{dV}{dt}\right]_{t\equal{}7} \equal{}\pi e^{14}$.

Of course.  :)

I did
$ V\equal{}\pi \int^k_1 (y\minus{}2)dy\equal{}\frac{\pi}{2}(k\minus{}1)^2$
$ \Leftrightarrow \frac{dV}{dt}\equal{}\pi e^{2t}\Rightarrow \frac{dV}{dt}\equal{}\pi e^{14}$


If the solutions of x^2 + bx + 36 = 0 are integers, then the number of integer values b can have is?

and also     12^2 + 5^2 = 13^2
                  12^2 + 9^2 =  15^2
                  12^2 + 16^2 = 20^2
                  12^2 +........=.........  HOW do u find how many similar such equations are there?

Comment deleted by author -  was  in error.

About your second question, your basically asking how to figure out how many pairs of perfect squares differ by a given quantity, 12^2 in the example you wrote?

[hide]If so, I think the way to solve the problem is to look at it in cases based on how many perfect squares are between the two in question. Since 12^2 is even, there must be an odd # of perfect squares between the two we want:

If there is one in between:

a^2+(2x+1)+(2x+3)=b^2, so if the equation 2x+1+2x+3=4x+4=12^2 has an integer solution, we add one to our tally of "similar such equations there are". There is, x+1=36, x=35.

If there are 3 in between, we get 2x+1+2x+3+2x+5+2x+7=8x+16=12^2

If 5, 12x+36=12^2. There is, x+3=12, x=8.

etc.

Notice that all equations that are solutions are such because the coefficient of x is a factor of 12^2. Notice also that coefficients of x in the equations include all and only multiples of 4. Therefore, the number of equations that will work is equal to the number of factors of 144 that are also multiples of 4. This is the same as the # of factors of 36 that are multiples of 1, which, since 36=2^2*3^2, is 9.

However, this includes 4,8,12,16,24,36,48...

but all of these that are greater than 24 have a negative solution for x, which will turn out to be the same as another solution given the way that the orignal problem was stated, so 5 is the answer.[/hide]

I'm pretty sure you could do any variation similar to this method, if it is correct.

LynnelleYe wrote:I think...
[hide]If x^2+bx+36=0 has integer solutions, then it can be factored as
(x+r)(x+s)=0
Where r and s are integers. If this is multiplied out, you get
x^2+xs+rx+rs=0
x^2+(s+r)x+rs=0
So we want all pairs of integers that multiply to 36, since each pair will give a distinct sum, which is the value of b. There are 5 pairs of factors that multiply to 36-
1*36
2*18
3*12
4*9
6*6
And multiplying the additive inverses of the factors in each pait work as well. That's 10 values b can be.[/hide]

Follow LynnelleYe's solution. Few things I want to add:

[hide]let b=-(r+s). Since rs=36, r=36/s, and substituting into the first equation,

b=-(36/s+s). What makes 'b' an integer? As you can see, only those integers 's' that divide 36. 36=(2^2)(3^2) and has 3*3=9 divisors. So possible values for s and the respective values of b are noted below:

s      1     2      3      4      6      9      12      18      36

b   -37  -20    -15   -13   -12   -13   -15     -20     -37

We got 4 that are duplicates, so there are only 5 values here. But if s<0, it works as well, s being the additive inverse of those 9 divisors above. Those 10 pairs by LynnelleYe yield the integer values for 'b'[/hide]

BTW, nice try using pythagorean triples but you forgot a very important thing, what is x? Not only that, 36=6^2 can only be in a pythagorean triple finitely many times. Unfortunately, this fact has nothing to do with this problem.

Hello, mohd_ahsan!

I have too much written in LaTeX ... I won't try to hide it . . .

[quote]If the solutions of $x^2 + bx + 36 = 0$ are integers,
then the number of integer values $b$ can have is __?[/quote]
From the Quadratic Equation: $\displaystyle{x \;= \;\frac{-b \pm \sqrt{b^2 - 144}}{2}}$
we see that: $b^2 - 144 = k^2$, for some integer $k\;\;\Longrightarrow\;\;b^2 - k^2 = 144$

Then: $(b + k)(b - k) = P\cdot Q$, where $P\cdot Q = 144.\;$  (Assume $P > Q$.)

Then we have the system: $\begin{array}{cc}b + k = P\\b - k = Q\end{array}$ ... with solutions: $b = \frac{P+Q}{2},\;\;k = \frac{P-Q}{2} $

We note that $P$ and $Q$ must be of the same parity (both even or both odd).

Then we have: $(P,Q) \;= \;(72,2),\;(36,4),\;(24,6),\;(18,8)$
  which gives us: $(b,k) \;= \;(37,35),\;(20,16),\;(15,9),\;(13,5)$

Therefore, there are $4$ possible values for $b$.

The resulting Pythagorean triples containing a "12" are:
$(5,12,13),\;(9,12,15),\;(12,16,20),\;(12,35,37)$

[quote="LynnelleYe"]Weren't those counted in the last sentence? (Though I seem to have written 'pait' instead of 'pair' for some reason)[/quote]

Sorry my error. I have deleted the original message to remove the confusion. Thanks.


Find all n such that :
 (x+y+z) 3 = nxyz has integer root

what knd of numbers are x,y,z? are they real or integer numbers? :?

I assume "integer roots" means that they are integers.  :D

I guess

elseway this would be extremely complex to prove :?

hi!
you're right. x,y,z are integer
This problem's not difficult .Try! :)