calculus
integration
function
quadratics
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probability
probability and stats

what is the general relation between the quadratique covariation and the covariance of two continuous process (or gadlag process) ? we know that for martingal $ X$ and $ Y$ of $ L^2$ , $ Cov(X,Y)\equal{}E(<X,Y>)$, one more question : wich are the process $ X$ and $ Y$ that have a determisitic quadratique covariation ? any  :idea: ?

It is not clear what do you mean when you write $ \langle X,Y\rangle$ for  `general' processes $ X,Y$. It seems that this makes sence only when $ X,Y$ are martingales w.r.t. the same filtrations (maybe, semi-martingales).

As for your second question, the Levy theorem gives an answer. Note that this answer depends on additional properties of the martingales: if they are continuous then the pair $ (X,Y)$ is Gaussian and is defined completely by its covariances. But there is another case $ X(t) \equal{}N(t)\minus{}t$, where $ N$ is the Poisson process with intensity $ \lambda\equal{}1$. This is the case of  'unpredictible' martingales, and in this case there exists another version of the Levy theorem ... I think this reference can help you with this subject

R. J. Elliott, Stochastic Calculus and Applications. Springer-Verlag, New. York, Heidelberg, Berlin, 1982.

yes, I meant semi-martingales, thank you for the reference, one more detail , you said : for continuous martingale $ (X,Y)$ is a gaussien process ! it seems to me that this result is  true for Wiener integrals only ( martingales of the form $ \int_0 ^t \phi_s dB_s$ with $ \phi_s$ a deterministic function of $ L^2$ )?

For one continuous martingale $ X$ with non-random $ \langle X\rangle$, one should put $ B(t)\equal{}X(f(t))$, where $ f$ is the inverse function for $ t\mapsto  \langle X\rangle_t$ (here I assume that $ \langle X\rangle_t$ is strictly increasing; this assumption can be omitted, but I do not want to go deep into details). Then $ B$ is a Brownian motion by the Levy theorem. Consequently,
$ X$ is  Gaussian. Note that $ X$ not necessarily is a Wiener integral; for
$ X\equal{}\int_0^t\phi_s\,dW_s$ quadratic characteristic $ \langle X\rangle_t\equal{}\int_0^t\phi_s^2\, ds$  has $ L_2$ derivative, while, in general,
$ \langle X\rangle_t$ can be any incrreasing function (say, the d.f. for a singular probability distribution).

In order to cover 2-dim case one should be more careful because this  `time change' trick can not be made straightforwardly: every component $ X_1,X_2$ has its own `time scale' $ \langle X_1\rangle, \langle X_2\rangle$. What one can  do is, for instance, to consider  linear combinations $ c_1X_1\plus{}c_2X_2$ and use 1-dim result plus the descripion of a 2-dim Gaussian distribution in terms of 1-dim projections

So, a 2-dim continuous martingale with non-random quadratic covariations is a Gaussian process, indeed; but it is not nesessarily a Wiener integral.

a lot of thanks to you Alex-K, I'm so surprised, I couldn't see this trick of inversion !! but what if the quatratique variation is not strictly increasing ? how would you proceed ? I dont see wath can be infered about semi martingale (or martingale) process $ (X,Y)$ such that $ <X,Y>$  is deterministic ? you used the fact that $ <X>$ and $ <Y>$ are deerministics too ! is it what you have called in your first post by additonal properties ? may be you well tell me that it's a consequence of that $ <X,Y>$ is deterministic  :lol:

Ah, of course I have assumed that all the characteristics -- $ \langle X\rangle, \langle Y\rangle, \langle X,Y\rangle$ -- are deterministic. Sorry that it was not clear from my posts. One can invent a counterexample when  $ \langle X,Y\rangle\equiv 0$ but $ X,Y$ are dependent. For instance, let $ W$ be the Wiener process, $ X(t)\equal{}\int_0^t1_{W(t)<C}dW(t), Y(t)\equal{}\int_0^t1_{W(t)>C}dW(t)$. Let me leave the proof that $ X$ and $ Y$ are not independent as an exrcise  :wink:

About strict monotonocity. In the basic argument, we build the Brownian motion $ B$ such that $ X(t)\equal{}B(\langle X\rangle _t)$. If $ \langle X\rangle$ is strictly monotonous, then $ B(t)\equal{}X(f(t))$. In general case we should separate the intervals where  $ \langle X\rangle$ is constant and put the increments of $ B$ on these intervals equal to the increments of an auxiliary Brownian motion $ \tilde W$ independent of $ X$. This is a rough idea; for the careful description of the construction see e.g.  [i]A.V.Skorokhod, Stochastic Equations for Complex Systems[/i]

okey, but you have told me in an other topic that , for martingales , quadratique covariance is meanning independency and you're given me a counterexemple for two martingales such that $ <X,Y>\equal{}0$ without having independency of $ X$ and $ Y$ ! unless $ <X,Y>\equal{}0$  and $ <X,Y> \equiv 0$ are having different meaning for you !

Okay, once more: when one  assumes   $ \langle X\rangle, \langle Y\rangle$ deterministic then $ \langle X,Y\rangle\equal{}0$ is equivalent to independence (because continuous martingale $ (X,Y)$ with deterministic $ \langle X\rangle, \langle Y\rangle$, and $ \langle X,Y\rangle$ is Gaussian). But when  $ \langle X\rangle, \langle Y\rangle$ are random, then independence does not follow immedialety, as we have seen in the example at the previous post.

In other topic I had on my mind that $ X,Y$ are Brownian motions which yields that $ \langle X\rangle, \langle Y\rangle$ deterministic. Sorry that I have not wrote this explicitly and this caused misundertstanding    :(

okey, now I got the point ! I thought that for martingales à zero quadratique covariation means independency ! so for genral matingales (with same filtration...) we dont have a relation between the quadratique covaraition and covariance ?


Hello my fellow students.  :lol: . ANyone mind discussing any worthy online free rpgs?

theres this one um... ill may get back to you.

Most of the good RPGs are not free (unless you pirate it, of course. But then you usually can't play online). And then most MMORPGs are only good in private servers where the admin can twitch the exp, drop, etc. rates. So if you REALLY dont want to pay money, I suggest

1) Find a private server in some random MMORPG with a good admin
2) [NOTE: Not recommended] Pirate some RPG.

Back like four years ago I got addicted to Runescape (I would not recommend playing it though). It does have free servers, but everything good is saved for the members.

Dude you should seriously play that game called GO DO YOUR MATH AND STOP WASTING YOUR LIFE.

na..i heard that game was over rated  :D

if you're willing to pay money (I wouldn't), play WoW.

if you aren't, play Silkroad Online cause the graphics are awesome :D

serioiusly. get a girlfriend and have some sort of normal life.

on Saturday night in my timezone,
this thread has turned into a debate of
GET A GIRLFRIEND AND LIVE NORMAL LIFE
 vs.
 DO YOUR MATH AND STOP WASTING TIME
 vs.
 PLAY FREE ONLINE RPG GAMES
 vs.
(this is my recommendation to the OP) DON'T PLAY RPG, PLAY INTELLIGENT BOARD GAMES WITH A GIRLFRIEND INSTEAD!

My solution is obviously the best because it is the combination of all of the above.

... wth? :huh:

Take my advice:

Unplug your computer and make sure it powers down,
Drop it in a 43 foot hole in the ground,
Bury it completely. Rocks and boulders should be fine,
And burn any clothes you may have worn anytime you were infected with...

Computer addiction.

[quote="paganinio"](this is my recommendation to the OP) DON'T PLAY RPG, PLAY INTELLIGENT BOARD GAMES WITH A GIRLFRIEND INSTEAD!

My solution is obviously the best because it is the combination of all of the above.[/quote]

Most girls I know would leave you if you were that geeky to make them play chess.  :D

@1=2-Why don't you do that with your computer/clothes.

computer bye-bye means more time for math

clothes bye-bye means you will not catch it again.

get more clothes.

There's a game called Ladder Slasher that I play a lot, if you want [i]completely[/i] free.  Most people get bored of it within a couple minutes.  But others have played it for years and years.

Then there's Diablo II, if you mean play online for free, but pay for the actual game.  Played it off and on since it came out.  Great game.

And I don't think "get a girlfriend" or "stop playing computer games" answers his question at all.

And there's plenty of girls that play RPGs.  Perhaps you all need to talk to more girls.

you obviously didnt read what i said carefully. I didnt say girls playing RPG's, I said girls who like chess. I remember chess nationals. I think the ratio of guys to girls was like 7:1 or something.

[quote="came"]Uh, there's one called Maplestory, but you have to download it and it takes about an hour to download if you have an ok computer, 45 min for fast, and 12 min for super fast computer. Yeah, and if you have a super slow computer, don't even bother downloading it. One of my friends took 4 days to download it. :([/quote]

Not as bad as WOW (world of warcraft). It takes 1.5 days to download on an extrememly good computer.

[quote="Nerd_of_the_Ages"][quote="came"]Uh, there's one called Maplestory, but you have to download it and it takes about an hour to download if you have an ok computer, 45 min for fast, and 12 min for super fast computer. Yeah, and if you have a super slow computer, don't even bother downloading it. One of my friends took 4 days to download it. :([/quote]
Actually, when I said about private servers, I was specifically thinking about maple (and the non free mmorpgs and maybe some other ones). Also, it doesnt take that long, that's an exaggeration. No really, only takes 10 to 15 minutes. And I haven't played WoW either, but I'm guessing that that's an exaggeration, too. Where are you guys getting these times?[/quote]

WoW is not an exaggeration. I had to keep my computer on for a day and a half for it to dl.

ZOMG!!!
You guys play Maple Story?
Me too!

Okay... maybe I should get a life...
 :(

P.S. My computer takes 2 minutes 15 seconds to download.  :)

i quit maple when i was lvl 8 :D

i downloaded it in 30 minutes on just my second fastest computer :D

Some good ones: GINORMO Sword, Battle for Wesnoth


This is one problem in the homework that I got stuck. Hope anyone can give me some hint.
Let A be an m by n matrix. Prove that there exists a sequence of elementary row operations of type 1 and 3 that transforms A into an upper triangular matrix. Note that type 1 is to interchange any two rows of A and type 3 is to add any scalar multiple of a row of A to another row.

I think I need to find an algorithm for this, but I just don't know how.

The algorithm is Gaussian elimination, essentially. Starting from the $ (1,1)$ position, and moving down and to the right ...

If the entry in the $ (j,j)$ position is not zero, then use the $ j$th row in a series of type three operations designed to create zeros everywhere below that element in the $ j$th column.

If the entry in the $ (j,j)$ position is zero but there is some nonzero element below it in the $ j$th column, then do a type one row interchange to bring the nonzero element up to the $ (j,j)$ position. Once that is done, proceed as in the previous paragraph.

If the entry in the $ (j,j)$ position is zero and every entry below that in the $ j$th column is also zero, then do nothing and move on, incrementing $ j.$

The pivot elements in our upper triangular matrix won't, in general, be $ 1.$ That's fine. Leave them that way - my habits (picked up from once teaching a numerical analysis course) are to leave them that way anyway.

Thank you very much for your clear explanation of this algorithm. I really appreciate it. I didn't expect that anyone would give such a big hint, probably an entire answer. If someone gave a hint, I think they would just say Gaussian elimination then let me find out what it is. I also searched and read some more info. about this method after reading yours. Thanks a lot.


a dairy farm had 100 gallons of milk that is 4.6% butterfat. how much skim milk(no butterfat)
should be mixed with it to make milk that is 3.2% fat?
what would the equation be to find the answer?

Let the amount of skim milk be $ x$. So far, there are $ 4.6$ gallons of butterfat. At the end of this, there will be still be $ 4.6$ gallons of butterfat, but there will be $ 100\plus{}x$ gallons of milk total. This gives us the equation $ \frac{4.6}{100\plus{}x}\equal{}\frac{3.2}{100}\implies460\equal{}320\plus{}3.2x\implies140\equal{}3.2x\implies x\equal{}\boxed{43.75}$.

would you mind helping me with a really hard one

but dont give me the answer
i would like to try it on my own to see if im doing it correctly

the question is

 a solution containing 30% insecticide is to be mixed with a solution containing 50% insecticide to make 200L of a solution containing 42% insecticide.    how much of each solution should be used?

First, let the amount of 30% insecticide be $ x$ and the amount of 50% insecticide be $ y$. Then, set up two equations.

[hide="Hint after you think about that"]
Make one equation using the fact that there is a total of 200L in the final solution.
Also, how much of the final solution should be insecticide? What two quantities can you add that should equal this amount of insecticide?
[hide="Equation 1"]
$ x \plus{} y \equal{} 200$
[/hide]
[hide="Equation 2"]
$ 0.3x \plus{} 0.5y \equal{} 0.42\cdot200$
[/hide]
[/hide]

so to find the answer the equatin would be .3x+.5y=.42 times 200?
then it would be .3x +.5y = 84
i cant add the x andy
so what do i do?

another solution

make one equation

let's take the insecticide one

define $ x$ as the amount of $ 30\%$ insecticide used
therefore, the amount of $ 50\%$ insecticide would be $ 200\minus{}x$ (the total minus the 30%)
the result has to be $ 42\%$ insecticide, so that would be $ 0.42\times200$

now, we write our equation
$ 0.3x \plus{} 0.5(200\minus{}x) \equal{} 0.42(200)$

solving from there is pretty easy and i'll leave it to you to do that


Let AB be a chord of a circle and let P be a point of the tangent to A. Show that the angle PAB equals the perimeter angle AQB, where the point Q is on the perimeter of the circle and such that P and Q lays on different sides of the line AB.

Edit: I drew a picture.

I'm sorry, but I still don't get it.  :(
BTW, is it this theorem your mean. I don't know the English name for it. And I don't know how to apply it to this problem.

I get it now!! PAB + BAQ* (that's your new point) = 90 degrees. PAB = 90-BAQ*
And in the triangle ABQ*, AQ*B + BAQ* = 90 degrees <=> AQ*B = 90-BAQ*. So PAB=AQB. Q.E.D

Eh. We had a great moment to post our fancy computer geometry system pictures at least. :)

[quote="shobber"]I also sketched a picture.
P.S. This is one of the most basic theorem in plane geometry. It is called 弦切角定理 in Chinese, but how to say it in English?[/quote]

does that mean two angles that subtend the same chord are equal_

I don't know, but this law was stated in the AoPS book. I don't remember whether it has a name or not.  :(

[hide]
 an incredibly informal and pleasantly simple proof:

angle PAB is 1/2 arc PB (An angle formed by an intersecting tangent and chord which has its vertex "on" the circle has a measure 1/2 the intercepted arc).

angle AQB is 1/2 arc PB (The measure of the intercepted arc (equal to its central angle) is exactly twice the measure of the inscribed angle.)

therefore angle PAB is congruent to angle AQB

[/hide]


Find out n natural so that there exists lim(x-->0) [(1+x)^3+(1-x)^3-2]/x^n

cheers! :D :D

I think the limit always exists. Maybe the problem was asking for n s.t. the limit is finite and/or =/=0, or something like that.

[(1+x) 3 -1]/x n + [(1-x) 3 -1]/x n = 6x 2-n (do all the calculations and you'll see). The limit is 0 for n>2, 6 for n=2 and 00 for n<2.

yes.. the limit should be finite.

cheers! :D

These exercises usually ask for finite and non zero limit because these are the 'interesting' cases. For this particular problem n must be 2 (so the limit is 6).


function
inequalities solved
inequalities

If a,b,c are in (0,1) then a+b+c<2+abc.

cheers!

[i]unknown source[/i]

Define f(a) = a(bc - 1) + 2 - b - c, with b, c fixed. This is a linear function of a.
Thus it attains it extreme values at the endpoints of the prescribed interval.
So we only need to consider the cases a = 0 and a = 1.
If a = 0 then f(a) = (1 - b) + (1 - c) > 0. If a = 1 then f(a) = (1 - b)(1 - c) > 0.
So we are done.


ceiling function
algorithm
floor function
continued fraction
Diophantine equation

$ {x^2}-{3y^2}=1$.
Find the solution set.
EXTRA HELP  (If it is related to the continued fraction expansion of $ \sqrt{3}$.
If $ {a}$ is the period of the continued fraction and $ {C_k}=\frac{P_k}{Q_k}$ is the $ {kth}$ convergent,i want to know why all solutions are of the form $ {(P{}_i{}_a,Q{}_i{}_a)}$ for positive $ {i}$.)
Any kind gesture is appreciated

All solutions $ (x,y)$ are given by:

$ (2\plus{}\sqrt{3})^n\equal{}x\plus{}y\sqrt{3}$.

The key to the second implication is that any solution to $ x^2 \minus{} 3y^2 \equal{} 1$ has the property that $ \frac {x}{y}$ is a good rational approximation to $ \sqrt {3}$, and there is a theorem which states that any sufficiently good rational approximation must in fact be a convergent of the continued fraction.  In turn, the key to this result is that the triangles formed by connecting consecutive convergents in the x-y plane have no interior lattice points.  (I don't, however, know of a particularly elegant proof of the periodicity result other than that it is the only way for $ x, y$ to satisfy the right linear recurrences.)

@krazy FK thanks.
@t0raijir0u you made my day
One more thing what is the periodicity of a continued fraction?give an example with my problem above

I'm not sure what you mean.  The result I was referring to is the fact that $ P_n^2 \minus{} dQ_n^2$ is periodic with the same period (if I recall correctly) as the continued fraction.  I don't know off the top of my head a proof of this fact.

As for all sufficiently good approximations being convergents, the threshold is that if $ \left|x-\frac pq\right|<\frac1{2q^2}$, $ \frac pq$ is a convergent for $ x$. If we're trying to approximate $ \sqrt{N}$, this becomes $ |p^2-Nq^2|=q^2\left|N-\left(\frac pq\right)^2\right|=q^2\left|\left(\sqrt{N}-\frac pq\right)\left(\sqrt{N}+\frac pq\right)\right|$
$ <q^2\cdot\frac1{2q^2}\cdot 2\left\lceil\sqrt{N}\right\rceil=\left\lceil\sqrt{N}\right\rceil$

On the periodicity , we have a proof by algorithm: see [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=178278]here[/url]. It's a shallow recursion involving arithmetic on bounded integers; since it has finitely many states, it must be eventually periodic.
We track three numbers $ a,d,m$; $ a_n$ are the entries in the continued fraction, $ d_n$ is $ |p^2-Nq^2|$ for $ \frac pq$ the $ N$th convergent (signs would alternate with | | removed), and the $ m$ are really just there to make the algorithm work better. They are bounded as follows: $ m_n<\sqrt{N}$ because $ m_n^2+a_na_{n-1}=N$, $ d_n<2\sqrt{N}$ because $ \left|\frac{p}{q}-\sqrt{N}\right|<\frac1{q^2}$ for any convergent $ \frac pq$, and $ a_n<2\sqrt{N}$ by the defining recursion $ a_n=\left\lfloor\frac{\sqrt{N}+m_n}{d_n}\right\rfloor$.
The solutions to the Pell equation $ p^2-Nq^2=\pm 1$ correspond to the triples $ (a_n,d_n,m_n)$ with $ d_n=1$. When this happens, we must also have $ m_n=\sqrt{N-a_na_{n-1})> \sqrt{N-2\sqrt{N}}>\sqrt{N}-1}$, forcing $ m_n=\lfloor\sqrt{N}\rfloor$. Then $ a_n=\left\lfloor\frac{\sqrt{N}+\lfloor\sqrt{N}\rfloor}{1}\right\rfloor=2\lfloor\sqrt{N}\rfloor$.
The whole triple is the same at each solution of the Pell equation (aside from the trivial $ p=1,q=0$), so the period of the $ d_n$ is the same as the period of the continued fraction.


Fie ABC un triunghi in care m( :ang: A)=90 :^o:  si (AB-AC) 2 =2( :rt2: -1)AB*BC.Determinati masura unghiului B si C!

Why is it still here?

We have $(\sin\beta-\sin\gamma)^2=2(\sqrt{2}-1)\sin \beta\sin\gamma$. It immediately follows that $\sin 2\beta=\frac{\sqrt{2}}{2}$, so...

Ah, I see. There is $AB\cdot BC$... Maybe it is a typo? I assumed that $(AB-AC)^2=2(\sqrt{2}-1)AB\cdot AC$.


First, find the perimeter in terms of $ p$:
\[ p\minus{}2\plus{}p\minus{}4\plus{}p\minus{}4\plus{}p\minus{}1\plus{}p\minus{}3\plus{}p\plus{}3\plus{}p\plus{}1\equal{}7p\minus{}10\]
Then, we simply substitute $ 11$ for $ p$, and get
\[ 7\cdot 11\minus{}10\equal{}\boxed{67}\]


linear algebra
matrix
linear algebra unsolved

Let $ A,B \in M_{n}(R)$ 2 invertible matrix  s.t. $ xA^2 \plus{} yB^2 \equal{} z(AB \plus{} BA)$ ,where $ x,y,z \in R$
 and $ x \geq y \geq z > 0$ .
Prove that $ (A \minus{} B)^2 \equal{} O_{n}$

[quote="posabogdan"]Let $ A,B \in M_{n}(R)$ 2 invertible matrix  s.t. $ xA^2 \plus{} yB^2 \equal{} z(AB \plus{} BA)$ ,where $ x,y,z \in R$
 and $ x \geq y \geq z \geq 0$ .
Prove that $ (A \minus{} B)^2 \equal{} O_{n}$[/quote]
You need to strengthen the condition, take $ x\equal{}y\equal{}z\equal{}0$ for instance and the conclusion doesn't always hold.

[quote="JC_math"]
You need to strengthen the condition, take $ x \equal{} y \equal{} z \equal{} 0$ for instance and the conclusion doesn't always hold.[/quote]
I edit

[quote="posabogdan"]Let $ A,B \in M_{n}(R)$ 2 invertible matrix  s.t. $ xA^2 \plus{} yB^2 \equal{} z(AB \plus{} BA)$ ,where $ x,y,z \in R$
 and $ x > y > z > 0$ .
Prove that $ (A \minus{} B)^2 \equal{} O_{n}$[/quote]

I think that assumption is $ x, y, z\in\mathbb{R} \ \text{and} \ xyz \not\equal{} 0$ :maybe:

I found a solution :
$ xA^2 \plus{} yB^2 \equal{} z(AB \plus{} BA)$ | * $ A^{ \minus{} 1}$ left and right
$ xA \plus{} yA^{ \minus{} 1}B^2 \equal{} z(B \plus{} A^{ \minus{} 1}BA)$
$ xA \plus{} yB^2A^{ \minus{} 1} \equal{} z(B \plus{} ABA^{ \minus{} 1})$

$ y(A^{ \minus{} 1}B^2 \minus{} B^2A^{ \minus{} 1}) \equal{} z(A^{ \minus{} 1}BA \minus{} ABA^{ \minus{} 1})$ | * $ A$ left and right
$ y(B^2 A \minus{} A B^2) \equal{} z(BA^2 \minus{} A^2 B)$
When * with  $ B^{ \minus{} 1}$ left and right we find :
$ x(A^2 B \minus{} BA^2) \equal{} z(A B^2 \minus{} B^2 A)$
so $ xy \equal{} z^2$ so $ x \equal{} y \equal{} z$ => $ A^2 \plus{} B^2 \equal{} AB \plus{} BA$ => $ (A \minus{} B)^2 \equal{} O_{n}$


calculus
function
integration
calculus computations

What is the solution to the differential equation $ \frac{dy}{dx} = \frac{x^3}{y^2} $ where $y(2) = 3 $

What are they asking for?  I'm I supposed to do seperable differention, integrate by parts, or what?  Please help me!  Thank you.

This is a "first order separable ordinary differential equation." Most standard calculus textbooks have at least a section, maybe even a chapter, that deals with such things. Do look for this in a textbook.

A first order differential equation is an equation of the form $\frac{dy}{dx}=f(x,y),$ meant to be solved for the unknown function $y(x).$ If $f$ satisfies some reasonable continuity-smoothness conditions, we have solutions. (See "existence-uniqueness theorem" as a topic to look up.)

If $f(x,y)=p(x)y+q(x),$ we call this a linear differential equation; with the classification comes a method. (See "integrating factor" or "multiplying factor.")

If $f(x,y)=p(x)q(y),$ we call this a separable differential equation; with the classification comes a method, called separation of variables.

Your problem is separable. As for the method I'm about to show you, don't worry that much about step-by-step justifications; the overall justification comes from the existence-uniqueness theorem. If we have a function that satisifies the differential equation and the initial conditions, then that's the only solution.

$\frac{dy}{dx}=\frac{x^3}{y^2}$

$y^2dy=x^3dx$

$\int y^2dy=\int x^3dx$

$\frac13y^3=\frac14x^4+C$

$\frac13\cdot 3^3=\frac14\cdot 2^4+C$ so $C=5$

$\frac13y^3=\frac14x^4+5$

$y=\left(\frac34\,x^4+15\right)^{\frac13}$


[quote="Lovasz"]My friend gave me this problem, I don't think it's true, but he was sure it's not wrong?!?!

Let $x,y,z\in\mathbb{R}$ satisfy $xy+yz+zx=1$, then:

$\frac{1}{1+x^2}+\frac{1}{1+y^2}+\frac{1}{1+z^2} \geqq \frac{x(1-y^2)}{(1+x^2)^2}+\frac{y(1-z^2)}{(1+y^2)^2}+\frac{z(1-x^2)}{(1+z^2)^2}$[/quote]
x=0, y=100, z=0.01

Chen241290, that's not a counterexample...

[quote="spider_boy"][quote="Lovasz"]My friend gave me this problem, I don't think it's true, but he was sure it's not wrong?!?!

Let $x,y,z\in\mathbb{R}$ satisfy $xy+yz+zx=1$, then:

$\frac{1}{1+x^2}+\frac{1}{1+y^2}+\frac{1}{1+z^2} \geqq \frac{x(1-y^2)}{(1+x^2)^2}+\frac{y(1-z^2)}{(1+y^2)^2}+\frac{z(1-x^2)}{(1+z^2)^2}$[/quote]

If i'm not wrong i have met this q at arthur engels book.Try trigonometric substitution.[/quote]

Yes, you maybe right, I thought about it, but trigonometric substitutions in this problem are quite messy, go on!, and welcome your proof if I can't try, spider_boy.


Here are some of my favorite riddles

1.What can you catch but not throw?
[hide]A Cold[/hide]

2.What goes around the world but stays in a corner?
[hide]A stamp[/hide]

3.Give me food, and I will live; give me water, and I will die. What am I?
[hide]Fire[/hide]

[b]What are your favorite riddles???[/b]

There was a round mansion. The owner returned to find his wife murdered. The gardener claimed he was raking the leaves. His child claimed he was playing ball. His maid claimed that she was sweeping the corners. Who murdered his wife?

There was a flood, and many people died. Why don't you bury the survivors?

To take the fun out of these riddles, read them twice. In English, what that means is don't read them twice. (or more)

[quote="cardwacko9999"]

There was a flood, and many people died. Why don't you bury the survivors?

quote]

[hide]
because survivors means alive. unless u are an inhumane psycho, u dont bury ppl alive[/hide]

[quote="cardwacko9999"]There was a round mansion. The owner returned to find his wife murdered. The gardener claimed he was raking the leaves. His child claimed he was playing ball. His maid claimed that she was sweeping the corners. Who murdered his wife?
[/quote]

the maid, who was sweeping corners of a round house.  :D

[quote="sky9073"]A classic:

It's greater than god, and more evil than the devil
The rich want it, the poor have it
And if you eat it, you die.

[hide]nothing[/hide]

The second line's not necessarily true, but..[/quote]

i thought it was " The rich cant get it, but the poor have it"

What has legs but can't run?

[hide="The answer"]Chairs, beds, tables, etc.[/hide]

[hide="My answer"]Right and isosceles triangles[/hide]
I'm such a dork. :)

Here's some dumb riddles:

What always ends everything?
[hide]the letter g[/hide]

What makes a man bald-headed?
[hide]lack of hair[/hide]

What animal keeps the best time?
[hide]a watchdog[/hide]

stupid, aren't they? :rotfl:

[quote="h_s_potter2002"]Here's one

WHy can't a bald man find use of his keys?
[hide]Because he lost all of his locks[/hide][/quote]:? :? ... I really don't get this one ...

A bunch of hair is called a locks of hair....

Why doesnt a tank engine like to sit down?
[hide]Because it has a tender behind[/hide]

These are kinda stupid.

What has teeth but can't bite?
[hide]a comb[/hide]

What gets bigger the more you take from it?
[hide]A hole[/hide]

Am I right?

Here's another one

What is the strongest animal?
[hide]a snail, what other animal carries its house? none[/hide]

this one is kinda lame
what kind of fish has two knees?

[hide]a 2-knee (tuna) fish
i told ya it was bad[/hide]

What is it that, after you take away the whole, some still remains?

[hide="Answer"]Wholesome[/hide]

[quote="GoBraves"][quote="white_horse_king88"]A man pushes a car past a hotel and smiles.

Why?

A man sees a rare bird. He is excited at first, then sad, and then terrified. What happened?[/quote]

[hide]1. um..he's in a happy mood
2. um..he's suddenly had a mood swing[/hide][/quote]

Nah - nice try though.  ;)

[quote="plokoon51"]this isn't related to the topic but to answer sheepwarrior's ?

[quote="sheepwarrior"]uh... demonface, next time can u just post all the jokes in one post? were you trying to get more posts or sumthin? ;) (hey where do those ppl get the smiley with the guy on the floor laughing/cryuing?)[/quote]

Under all of the emotions, click on "view more emoticons". the other smiles/etc. are at the bottom so just scroll down.[/quote]
oh, ok thx  :D
i went there b4 but i guess i didnt wait long enough :P

[quote]A man sees a rare bird. He is excited at first, then sad, and then terrified. What happened?[/quote]

[hide]Saw a phoenix reborn maybe?[/hide]

[quote="white_horse_king88"]A man pushes a car past a hotel and smiles.

Why?

A man sees a rare bird. He is excited at first, then sad, and then terrified. What happened?[/quote]

[hide]Monopoly![/hide]

Heh, I only knew it because I had heard a variation on it.

A man pushes his car to a hotel, goes inside, pays the owner some money and leaves. He then pushes his car to another hotel, again pays the owner some money and leaves. What is he doing?

[quote="white_horse_king88"]A man pushes a car past a hotel and smiles.

Why?

A man sees a rare bird. He is excited at first, then sad, and then terrified. What happened?[/quote]

1) monopoly
2) man is riding on an airplane, bird gets sucked into engine

Dang - I thought the second one was a stumper.

Where'd you hear it?

[quote="demonface"]here one I really like

The first person builds this but does not keep it. The second person buys it but doesn't know it. The third person uses it but doesn't use it. What is the object?

[hide]a coffin[/hide][/quote]

Coffin.

sure answer my joke after I tell you the answer[/quote]

anyway here's another joke,

there's a shop that has a sign saying that the fries are fat free,
someone goes in and orders some fries. The owner brought a really greasy fry.
The buyer asks " I thought you said that it was fat-free"

What did the owner say?

note only look at the answer if you are stumped

[hide]No, the fat is free[/hide]


calculus
integration
function
real analysis
real analysis unsolved

Intgrate with respect to z:

sin(x+y-z)^x + x^y/2 + 2y^z + xy(cos(2z^2))

this one is very hard if you can do it you are a master

Are x and y functions of z or are they constants?

I dont think calculus belongs is this forum.

So we'll call them functions of z then. (If we're differentiating with respect to z, they must be functions of z, like y is a function of x in dy/dx.)

To moderators, this problem is not suitable in this section.
You should move it to approriate section.

kunny

btw i have got the answer here for the ones so curious.

[hide][img]http://www.techzine.nl/f/g/19734phpbD9dbj.gif[/img][/hide]

 :D  :D   :showoff:

[quote="B23"]What's the point of this question? To see if humans are still better in integrating than computers? I'm not interested in this kind of competition.[/quote] exactly!  :D

[color=darkblue][size=134]Sorry, I arn't curiously ! [/size]Modguru, you can try through the pensioners of a lunatic asylum, maybe one of them will solve it.
At least for me, from this moment, "Modguru" means "jump over time".[/color]


Hey there, if someone can answer either one of the following problems and be able to show their work I would be much obliged. Thanks in advance and I’ll be sure to choose the best answer. Peace.

1. A grid has lines at 90 degree angles. There are 12 lines in one direction and 9 lines in the other direction. Lines that are parallel are 11 inches apart. What is the minimum number of 12-inch-by-12-inch floor tiles needed to cover all of the line intersections on the grid? The titles do not have to touch each other. You must keep the tiles intact—don’t break or cut them.

2. There are nine points on a piece of paper. No three of the points are in the same straight line. How many different triangles can be formed by using three of the nine points as vertices?

3. Dionne can run around a circular track in 120 seconds. Basha, running in the opposite direction as Dionne, meets Dionne every 48 seconds. Sandra, running in the same direction as Basha, passes Basha every 240 seconds. How often does Sandra meet Dionne?

[quote="rise89"]3. Dionne can run around a circular track in 120 seconds. Basha, running in the opposite direction as Dionne, meets Dionne every 48 seconds. Sandra, running in the same direction as Basha, passes Basha every 240 seconds. How often does Sandra meet Dionne?[/quote][hide]Let's say the length of the track is $m$ meters. Then Dionne runs at a rate of $\frac{m}{120}$ m/s. If we assume that Basha runs the track in $b$ seconds, then his (her? sorry) speed is $\frac{m}{b}$ m/s.

Running together, Basha and Dionne cover the whole length of the track, $m$ meters, in $48$ seconds. Their total speed is $\frac{m}{120}+\frac{m}{b}$ m/s, and this is equal to $\frac{m}{48}$. This is because Dionne runs a total of $(48 \text{ seconds})\left(\frac{m \text{ meters}}{120\text{ seconds}}\right)$ meters during the $48$ seconds between the time interval that they meet, and Basha runs and total of $(48 \text{ seconds})\left(\frac{m \text{ meters}}{b\text{ seconds}}\right)$ meters. Since they run a total of $m$ meters every $48$ seconds, $\frac{m}{120}+\frac{m}{b}=\frac{m}{48}$. Solving this gives you $b = 80$.

You can use the similar logic to solve for Sandra's speed, which you can compare with Dionne's speed to determine how often they meet.
[/hide]

[hide="2"]
Because no three points are collinear, any three points we select will produce a unique triangle.

Therefore, the solution is just $\binom{9}{3}$, or 84 triangles.
[/hide]

For number 1, you have to visualize the maximum amount of area.  Would the area of a series of almost linear rhombii be larger, or would a grid of squares have larger area?  What about tiling?

[hide="1."]We have a 12*9 rectangle of intersections, and a tile can cover at most 4 intersections, with a 2*2 square being covered. 12/2 is 6, and 9/2=4.5, so 5*6=30 tiles are needed.[/hide]

[hide="2."]Since no three points are collinear, we can choose a random 3 points and they will form an non-degenerate triangle. $\binom{9}{3}=84$[/hide]

[hide="3."]Let's say that Dionne can run 12 miles per second. I suggest that you don't get in her way!!!

For Basha and Dionne to meet every 48 seconds, they must run the whole track in 48 seconds. Since the whole track is 1440 miles, Basha's rate plus Dionne's rate must equal 1440/48=30 miles per second.  Therefore, Basha goes 18 miles per second.(These girls are fast!!!)

Let's assume that Sandra is going faster than Basha. Therefore, the rate of Sandra minus the rate of Basha is 1440/240=6. Therefore, Sandra's rate is 24 miles per second.(I think these girls are a little to fast!!!)

So 24+12=36, and 1440/36 is the time that Sandra takes to meet Dionne, which is 40 seconds.[/hide]


Lets see how many words we can get that start with "Fun". 1 per post. Update the list as we go.

1. Function

Fundamental

I assume it's 1 word per post?

EDIT: Lol.

1) Fundamental

2) Fundamentally

Do forms of the same word count?

1) Fundamental

2) Fundamentally

3) Fundamentalist

Of course!

Having fun with funky fungus infested funnels at funfairs results in depletion of funds.

Interpret that as you will.

1) Fundamental

2) Fundamentally

3) Fundamentalist

4) Fundamentalize

5) Funeral

6) Funding

7) Funds

weren't there more?

1) fundamental
2) fundamentally
3) fundamentalist
4) fundamentalize
5) funeral
6) funding
7) funds
8) FUN <--- obviously
9) fungus
10) funnels
11) funfairs
12) function

1) fundamental
2) fundamentally
3) fundamentalist
4) fundamentalize
5) funeral
6) funding
7) funds
8) FUN <--- obviously
9) fungus
10) funnels
11) funfairs
12) function
13)fungi

Make sure you include the rest of the list.


inequalities
quadratics
algebra
inequalities proposed

[quote="Gibbenergy"]I have generate this inequality but highly doubt its valid. Can someone check it?
Let $a,b,c \geq 0$ such that $a+b+c=3$. Prove that:
$a^{4}+b^{4}+c^{4}+ab+bc+ca \geq 6$
$a^{4}+b^{4}+c^{4}-ab-bc-ca \geq 0$
Thanks :lol:[/quote]
It's true by Muirhead only.

[quote="arqady"]
It's true by Muirhead only.[/quote]

Yeah. It is true by Muirhead. How can you do it so fast, Argady? For me, I generated it by Lagrange Mutipliers in class this morning.  :D

[quote="Gibbenergy"]I have generate this inequality but highly doubt its valid. Can someone check it?
Let $a,b,c \geq 0$ such that $a+b+c=3$. Prove that:
$a^{4}+b^{4}+c^{4}+ab+bc+ca \geq 6$
$a^{4}+b^{4}+c^{4}-ab-bc-ca \geq 0$
Thanks :lol:[/quote]

  Is there any solution without Murihead? I am very curious to see how we can handle it without much of expand.Fans of classic inequality, please show it  :D ? By the way, is the inequality still true if we omit the condition $a,b,c \geq 0$?

[quote="Gibbenergy"]I have generate this inequality but highly doubt its valid. Can someone check it?
Let $a,b,c \geq 0$ such that $a+b+c=3$. Prove that:
$a^{4}+b^{4}+c^{4}+ab+bc+ca \geq 6$
$a^{4}+b^{4}+c^{4}-ab-bc-ca \geq 0$
Thanks :lol:[/quote]
$a^{4}+b^{4}+c^{4}+ab+bc+ca \geq 6\Leftrightarrow \sum a^{4}+\frac{1}{2}\sum a(b+c)\ge 6\Leftrightarrow \sum a^{4}+\frac{1}{2}\sum a(3-a)\ge 6\Leftrightarrow \sum a^{4}\ge\frac{3}{2}+\frac{1}{2}\sum a^{2}$
Similarly,
$a^{4}+b^{4}+c^{4}-ab-bc-ca \geq 0\Leftrightarrow \sum a^{4}+\frac{1}{2}\sum a^{2}\ge \frac{9}{2}$
But Vasc's problem can't be solved by this way :blush:

[quote="Vasc"]Actually, the stronger inequality holds
$2(a^{4}+b^{4}+c^{4})+19(ab+bc+ca) \geq 63$.[/quote]

Vasc, for the general problem $a+b+c=3$ Prove that:
$a^{4}+b^{4}+c^{4}+k (ab+bc+ca) \geq 3+3k$
Your k is $k= \frac{19}{2}= 9.5$. However, after some calculations, I think the best constant is $k=12$ .
The inequality becomes
$a^{4}+b^{4}+c^{4}+12(ab+bc+ca) \geq 39$
Is it also your result? :)

[quote="arqady"][quote="Gibbenergy"] I think the best constant is $k=12$ .
The inequality becomes
$a^{4}+b^{4}+c^{4}+12(ab+bc+ca) \geq 39$
Is it also your result? :)[/quote]
No, Gibbenergy. The vasc's $9.5$ is best.
And your is wrong: $a=b=1.5$ and $c=0.$ :wink:[/quote]

Thank Arqady, I found my mistake upon expand. The fact is that we have to minimize the quadratic equation
$6b^{2}-20b+26-k \geq 0$ with the constrain is $1 \leq b \leq \frac32$
this is an easy problem, just plug in $f(\frac32)$ to find the maximum $k$.
I did wrong because I plug in $f(1)$ which is not corrected.


[quote="SimonM"]The obtuse-angled triangle $ ABC$ has sides of length $ a$, $ b$ and $ c$ opposite the angles $ A$, $ B$ and $ C$ respectively. Prove that

                $ a^3 \cos A \plus{} b^3 \cos B \plus{} c^3 \cos C < abc$.[/quote]

Use cosine law,

$ a^3 \cos A \plus{} b^3 \cos B \plus{} c^3 \cos C < abc$

$ \Longleftrightarrow a^3\cdot \frac{b^2\plus{}c^2\minus{}a^2}{2bc}\plus{}b^3\cdot \frac{c^2\plus{}a^2\minus{}b^2}{2ca}\plus{}c\cdot \frac{a^2\plus{}b^2\minus{}c^2}{2ab}<abc$

$ \Longleftrightarrow a^4(b^2\plus{}c^2\minus{}a^2)\plus{}b^4(c^2\plus{}a^2\minus{}b^2)\plus{}c^4(a^2\plus{}b^2\minus{}c^2)<2abc$

$ \Longleftrightarrow a^6\minus{}(b^2\plus{}c^2)a^4\minus{}(b^2\minus{}c^2)^2a^2\plus{}(b^2\minus{}c^2)(b^4\minus{}c^4)>0$

$ \Longleftrightarrow a^4(a^2\minus{}b^2\minus{}c^2)\plus{}(b^2\minus{}c^2)^2(b^2\plus{}c^2\minus{}a^2)>0$

$ \Longleftrightarrow (b^2\plus{}c^2\minus{}a^2)\{(b^2\minus{}c^2)^2\minus{}a^4\}>0$

$ \Longleftrightarrow (b^2\plus{}c^2\minus{}a^2)(b^2\minus{}c^2\plus{}a^2)(b^2\minus{}c^2\minus{}a^2)>0$

$ \Longleftrightarrow (a^2\plus{}b^2\minus{}c^2)(b^2\plus{}c^2\minus{}a^2)(c^2\plus{}a^2\minus{}b^2)<0$

$ \ \because a^2\plus{}b^2<c^2,\ b^2\plus{}c^2<a^2,\ c^2\plus{}a^2<b^2$ Q.E.D.

That looks alright up till the very end but to say a^2 > b^2 + c^2, b^2 > c^2 + a^2 and c^2 > a^2 + b^2 is incorrect; actually only one of these is correct, the other two need to be the other way around, but the result still stands.

You are right, but I meant to write that $ a^2\plus{}b^2<c^2$ or $ b^2 \plus{} c^2 < a^2$ or $ c^2 \plus{} a^2 < b^2$.

[quote="kunny"][quote="SimonM"]The obtuse-angled triangle $ ABC$ has sides of length $ a$, $ b$ and $ c$ opposite the angles $ A$, $ B$ and $ C$ respectively. Prove that

                $ a^3 \cos A \plus{} b^3 \cos B \plus{} c^3 \cos C < abc$.[/quote]

Use cosine law,

$ a^3 \cos A \plus{} b^3 \cos B \plus{} c^3 \cos C < abc$

$ \Longleftrightarrow a^3\cdot \frac {b^2 \plus{} c^2 \minus{} a^2}{2bc} \plus{} b^3\cdot \frac {c^2 \plus{} a^2 \minus{} b^2}{2ca} \plus{} c\cdot \frac {a^2 \plus{} b^2 \minus{} c^2}{2ab} < abc$

$ \Longleftrightarrow a^4(b^2 \plus{} c^2 \minus{} a^2) \plus{} b^4(c^2 \plus{} a^2 \minus{} b^2) \plus{} c^4(a^2 \plus{} b^2 \minus{} c^2) < 2abc$

$ \Longleftrightarrow a^6 \minus{} (b^2 \plus{} c^2)a^4 \minus{} (b^2 \minus{} c^2)^2a^2 \plus{} (b^2 \minus{} c^2)(b^4 \minus{} c^4) > 0$

$ \Longleftrightarrow a^4(a^2 \minus{} b^2 \minus{} c^2) \plus{} (b^2 \minus{} c^2)^2(b^2 \plus{} c^2 \minus{} a^2) > 0$

$ \Longleftrightarrow (b^2 \plus{} c^2 \minus{} a^2)\{(b^2 \minus{} c^2)^2 \minus{} a^4\} > 0$

$ \Longleftrightarrow (b^2 \plus{} c^2 \minus{} a^2)(b^2 \minus{} c^2 \plus{} a^2)(b^2 \minus{} c^2 \minus{} a^2) > 0$

$ \Longleftrightarrow (a^2 \plus{} b^2 \minus{} c^2)(b^2 \plus{} c^2 \minus{} a^2)(c^2 \plus{} a^2 \minus{} b^2) < 0$

$ \ \because a^2 \plus{} b^2 < c^2,\ b^2 \plus{} c^2 < a^2,\ c^2 \plus{} a^2 < b^2$ Q.E.D.[/quote]

$ \Longleftrightarrow a^3\cdot \frac {b^2 \plus{} c^2 \minus{} a^2}{2bc} \plus{} b^3\cdot \frac {c^2 \plus{} a^2 \minus{} b^2}{2ca} \plus{} \boxed{c\cdot \frac {a^2 \plus{} b^2 \minus{} c^2}{2ab}} < abc$

$ \Longleftrightarrow a^4(b^2 \plus{} c^2 \minus{} a^2) \plus{} b^4(c^2 \plus{} a^2 \minus{} b^2) \plus{} c^4(a^2 \plus{} b^2 \minus{} c^2) <\boxed{ 2abc}$

Are these typos? :lol:  :D

Thanks!

[quote="kunny"][quote="SimonM"]The obtuse-angled triangle $ ABC$ has sides of length $ a$, $ b$ and $ c$ opposite the angles $ A$, $ B$ and $ C$ respectively. Prove that

                $ a^3 \cos A \plus{} b^3 \cos B \plus{} c^3 \cos C < abc$.[/quote]

Use cosine law,

$ a^3 \cos A \plus{} b^3 \cos B \plus{} c^3 \cos C < abc$

$ \Longleftrightarrow a^3\cdot \frac {b^2 \plus{} c^2 \minus{} a^2}{2bc} \plus{} b^3\cdot \frac {c^2 \plus{} a^2 \minus{} b^2}{2ca} \plus{} c^3\cdot \frac {a^2 \plus{} b^2 \minus{} c^2}{2ab} < abc$

$ \Longleftrightarrow a^4(b^2 \plus{} c^2 \minus{} a^2) \plus{} b^4(c^2 \plus{} a^2 \minus{} b^2) \plus{} c^4(a^2 \plus{} b^2 \minus{} c^2) < 2a^2b^2c^2$

$ \Longleftrightarrow a^6 \minus{} (b^2 \plus{} c^2)a^4 \minus{} (b^2 \minus{} c^2)^2a^2 \plus{} (b^2 \minus{} c^2)(b^4 \minus{} c^4) > 0$

$ \Longleftrightarrow a^4(a^2 \minus{} b^2 \minus{} c^2) \plus{} (b^2 \minus{} c^2)^2(b^2 \plus{} c^2 \minus{} a^2) > 0$

$ \Longleftrightarrow (b^2 \plus{} c^2 \minus{} a^2)\{(b^2 \minus{} c^2)^2 \minus{} a^4\} > 0$

$ \Longleftrightarrow (b^2 \plus{} c^2 \minus{} a^2)(b^2 \minus{} c^2 \plus{} a^2)(b^2 \minus{} c^2 \minus{} a^2) > 0$

$ \Longleftrightarrow (a^2 \plus{} b^2 \minus{} c^2)(b^2 \plus{} c^2 \minus{} a^2)(c^2 \plus{} a^2 \minus{} b^2) < 0$

$ \ \because a^2 \plus{} b^2 < c^2,\ b^2 \plus{} c^2 < a^2,\ c^2 \plus{} a^2 < b^2$ Q.E.D.[/quote]


i think the answer is[hide]1[/hide]

jorian

$\frac{x}{x}=x$
$x^{2}=x$ => $x^{2}-x=0$
using the Quadratic Formula we get
 x=0 or x=1  therefore what is either 1 or 0

is this true ?

ya i knoow, but how come the quad.. formula gives 0?

i think 0/0=1  because there one 0 in 0 .

no, it's like saying there are no cookies and you take no cookes, now you have all the cookies :wink:

jorian

Isn't the question supposed to say: "For what value of "what" will what divided by what be equal to what"

[quote="binomial_4eva"]what is what divided by what equals to what ?[/quote]
isnt it just x/x=x^2 so 1?

[quote="binomial_4eva"]lol true .
how come my profile signature doesnt apear in ma thing anyone knows how to do tht ?[/quote]

You have to click the attach signature box before you post (or change your profile settings).

As for the problem...

[hide]
Isn't it $\frac{x}{x}=x$
Multiplying both side to clear the fraction:
$x=x^{2}$
Subtracting $x$ from both sides:
$0=x^{2}-x$
Factoring:
$0=x(x-1)$

From that, we can "tell" (for the lack of a better word) that $x$ can equal $0\text,1$, [i]but[/i], one of the restrictions we found from our first fraction was that $x$ cannot equal zero, so our only answer is $\boxed1$. [/hide]

Crude, but it works.

thanks hapyme 4 the signature help.

we can also say

what is what devided by what equals to what
what is what times what equals to what

$\frac{x}{x}=x=x^{2}$
so
what is what devided by what equals what is what times what.

yes, the quadratic formula gives 0 and one, but zero is an extraneous root, so we "throw it out"


Give $ a,b,c >0 : a\plus{}b\plus{}c \equal{}1$
Prove that:
 $ (ab\plus{}bc\plus{}ca)^2 \geq \frac {abc}{a^a b^b c^c}$

I think it's not easy .  :P  :D

$ \frac {abc}{a^ab^bc^c} \equal{} a^{b \plus{} c}b^{c \plus{} a}c^{a \plus{} b}\leq(\frac {\sum(b \plus{} c)a}{\sum(b \plus{} c)})^{\sum(b \plus{} c)} \equal{} (\sum ab)^2$

[quote="lia_h1"]Give $ a,b,c > 0 : a \plus{} b \plus{} c \equal{} 1$
Prove that:
 $ (ab \plus{} bc \plus{} ca)^2 \geq \frac {abc}{a^a b^b c^c}$

I think it's not easy .  :P  :D[/quote]
We will prove that $ a^a b^b c^c\ge \frac{1}{3}$ or $ alna\plus{}blnb\plus{}clnc\ge ln\frac{1}{3}$ but $ f(x)\equal{}xlnx, f"(x)\equal{}\frac{1}{x}\ge0$ so its true buy Jensens ineq,so we must prove $ (ab \plus{} bc \plus{} ca)^2 \ge 3abc$ or $ (ab \plus{} bc \plus{} ca)^2 \ge 3abc(a\plus{}b\plus{}c)$ or
$ (ab)^2\plus{}(bc)^2\plus{}(ac)^2\ge abc(a\plus{}b\plus{}c)$ its true by CS


i begin in making geometry draws by computer, but i only know about cabri and it is good but i need another one, i heard about a program that you can expand, by example teaching to the program how to make armonick points, or to make inversion.

any help about the name of some program and the place to download it

I use z.u.l; it's a good program (and also free)
http://mathsrv.ku-eichstaett.de/MGF/homes/grothmann/java/zirkel/

For just neat and tidy drawings, no geometric additions such as inversion, I use a simple, but efficient 2D CAD program, like DeltaCad.  C.a.r. [z.u.l.] is also highly recommended though.  For presentation, the drawings are easily copy/pasted into a word processor along with other files or formulas or description, where they are resizable, then all converted to a PDF for ready email transport.  I find that useful for tutoring.


geometry
geometric transformation
dilation

1. A white solid is observed to be insoluble in water, soluble in excess ammonia solution, and insoluble in excess dilute hydrochloric acid.  The white solid could be:
A. AgCl
B. NaHCO3
C. Zn(OH)2
D. BaSO4
E. CaCO3
I really don't know how to approach this problem.  Please help.

2. All of the following are better conductors of electricity than gallium-31 except
A. Ge
B. Hg
C. Pb
D. Sn
E. Zn
I'm thinking that metals and transition metals conduct electricity well.  The only one that isn't a metal or a transition metal is Germanium, which is a metalloid.  However, Mercury donesn't conduct electricity that well either.  So is the answer A (Germanium) or B (Mercury)?  Please help me.

the first one is likely to be ${Zn(OH)}_{2}$ or $AgCl$ because $Zn(OH)_{2}$ forms insolube $ZnCl_{2}$ in $HCl$ and forms the soluble complex $Zn{{(NH_{3})}_{4}}^{2+}$in excess ammonia.
similarly AgCl forms a complex ${Ag(NH_{3})}_{2}^{+}$ in ammonia (tollens reagent ). others dont form complexes in ammonia


calculus
integration
probability and stats

Anyone familiar with this result?  It's proof in discrete time is quite simple, but is sort of horrible in continuous time.

What exactly do you want to know about this decomposition ?

PS:
Yes its proof is really complex you can have a look at it in Protter's book
"Stochastic integration and Differential Equations"

i don't find protter's proof to be very illuminating in terms of developing intuition.  KS's proof is maybe a little better.  still i'd like to be able to explain in a paragraph why such a result should be true.  what technical problems arise when you naively try and bootstrap from the discrete case on a finer and finer mesh

Hi gotempleu

If we compare both theorems ( KS and Protter)

KS do not care about the form of the jumps of the process and introduce natural processes instead for uniqueness when Protter introduce predictability for uniqueness (general theorem)

KS also treat both case of DL Submartingales as well as submartingales of Class D when Protter only treat the Supermartingales of class D case.

I think that both concept of predictability and "naturality" are in some sense very close to one another. But Protter's view is more "stopping time" related when KS view is more connected with the original extension of Doob's result in discrete time.

To me, if we keep in mind the uniqueness goal, it's Protter's point of view  which is the more natural because it points out the importance of stopping times (inaccessible and accessible) and predictability of the increasing process ( very important for uniqueness results in stochastic integration for semi martingales)
when KS is more based on the natural processes which are really not really intuitive in spite of their names (why does those processes are natural ?) and as the book is only on continuous stochastic integration KS do not need to focus too muh on the different famillies of stopping times.

To get back to your original question the difficulty arise because of the different nature of jumps times in continuous time compared with discrete time. Stopping times are much more easy to treat in the discrete case
for example :
-the only inaccessible stopping time in the discrete time case is $ T\equal{}\plus{}\infty$.

Also we have the fact that the notion of predictability is very esasy in the discrete time case it's only the fact that a process is $ \mathcal{F}_{n\minus{}1}$ adapted.

In the same way I think that KS give an easy characterisation of natural processes in the discrete time case when its more involved in the continuous case.

see ya

you could have a look at "Diffusions, Markov Processes, and Martingales" (volume II) by Williams and Rogers. I am reading it, and I find the proofs well motivated .. As TheBridgeTwo said, that proof is really not at all easy nor intuitive ..

this is something along the lines of what i was hoping to see, which is a bootstrapping argument to get (a very strong) convergence to the desired FV part in the continuous case.  it looks like Rao did similar things.  this proof i find more intuitive than protter's or KS's.  i will look at the williams and rogers though.

http://www-users.mat.uni.torun.pl/~adjakubo/dm1.pdf

In the spirit it's the same as the one in KS (I say that because of the use of the convergence theorem in L^1 for $ A^n$ ) but

-avoiding natural process
-classification of stopping time that you find disturbing

to go directly to predictability of the process
thanks to :
-some kind of cesaro sums mixed with a diagonal argument for construction of the entire process.

Thanks for the paper


Hello everyone,

I am studying higher-powered equations on my own, but my roots to the following equation differ from the ones in the answer key. I would appreciate any help or hints.

Thanks!

---

[b]My Work :[/b]

$ \ 3x^{5}\plus{}4x^{4}\minus{}10x^{3}\equal{} 0$

$ \ x^{3}(3x^{2}\plus{}4x\minus{}10) \equal{} 0$

Solving for the quadratic:

$ \ x \equal{}\frac{\minus{}4\pm\sqrt{16\minus{}4(3)(\minus{}10)}}{6}$

$ \ x \equal{}\frac{\minus{}4\pm\sqrt{136}}{6}$

All three roots:

$ \ x \equal{}\minus{}2.6103$

$ \ x \equal{} 1.277$

$ \ x \equal{} 0$

Answers in book: [hide] $ \ x \equal{}\minus{}2.6998$ $ \ x \equal{}\minus{}4.0332$ $ \ x \equal{} 0$[/hide]

Your method and the answers seem right to me (substituting the values back into the equation works). I think your answer key has a mistake in it.

Hello kh-,

Thanks for your reply. I posted the answer with the hidden option so that you can see if the book's answers are incorrect.

Yeah, they've made a mistake, unless you've copied the formula from the question wrong or something there's no way those answers are right. $ x \equal{}\minus{}2.6998$ and $ x \equal{}\minus{}4.0332$ are both incorrect (they give the values $ \minus{}21.0081$ and $ \minus{}1487.13$ respectively), although obviously $ x \equal{} 0$ is correct.

I guess you already know how to do this, but you can check the answers yourself by just substituting back into the original equation, although it's pretty gruelling unless you've got a calculator/Mathematica or something handy.

[hide="Very exciting substition within"]The formula being:
$ 3x^{5}\plus{}4x^{4}\minus{}10x^{3}\equal{} 0$

And the answers you got for the formula = 0 being:
$ x \equal{}\minus{}2.6103,\ x \equal{} 1.277,\ x \equal{} 0$

$ 3(\minus{}2.6103)^{5}\plus{}4(\minus{}2.6103)^{4}\minus{}10(\minus{}2.6103)^{3}\approx 0$
$ 3(1.277)^{5}\plus{}4(1.277)^{4}\minus{}10(1.277)^{3}\approx 0$
$ 0^{5}\plus{}0^{4}\minus{}0^{3}\equal{} 0$

And their answers for the formula = 0 apparently being:
$ x \equal{}\minus{}2.6998,\ x \equal{}\minus{}4.0332,\ x \equal{} 0$

$ 3(\minus{}2.6998)^{5}\plus{}4(\minus{}2.6998)^{4}\minus{}10(\minus{}2.6998)^{3}\equal{}\minus{}21.0081\neq 0$
$ 3(\minus{}4.0332)^{5}\plus{}4(\minus{}4.0332)^{4}\minus{}10(\minus{}4.0332)^{3}\equal{}\minus{}1487.13\neq 0$
$ 0^{5}\plus{}0^{4}\minus{}0^{3}\equal{} 0$

So all your answers are correct and theirs are not (although obviously you rounded so the first two are slightly off of zero, but they're clearly right).[/hide]

All right.

Thank you very much, kh-, for your in-depth replies!

I would also like to point out the fact that the answers in the back of the book $ (x =-2.6998, x =-4.0332,$ and $ x = 0)$ contains [i]two[/i] negative answers. As I agree with the solve point of the quadratic $ (\frac{-4\pm\sqrt{136}}{6})$, this makes [i]two[/i] negative answers impossible as, obviously, $ \abs{-4}$ is less than $ +\sqrt{136}$. So, there will be one positive [i]and[/i] negative answer, forcing [i]at least[/i] one the book's answers to be wrong.

If you would be so kind as to provide me with the code for the "greater than" and "less than" signs that would be appreciated.  :lol:

greater than and less than:
[code]a>b
b<a[/code]

greater than or equal to
[code] a \geq b[/code]

less than or equal to
[code] a \leq b[/code]

Thanks, chickendude. Although I should have realized that "less than" was "<".  :blush:

[quote="sludgethrower"]I also got $ e^{\frac{2\pi}{3}i}$ and $ e^{\minus{}\frac{2\pi}{3}i}$[/quote]

How? All five answers have been accounted for ($ x\equal{}\minus{}2.6103, 1.277,\text{and }0$(0 occurs three times in multiplicity)).

I would be interested to see your work.


Four Friends A,B,C & D are playing a game of cards. Only J, Q, K & A of all the 4 suites are taken for the game.(i.e they have 12 cards to play with). The cards are distributed such that each gets 1 card of every suite and 1 card of every value. In how many ways can the cards be distributed such that A gets a black jack?

My book gives a solution of 3456

but my approach tells me something is wrong..here is my approach...

1) 4 sets of different cards of different suites can be formed as follows...
first card in 4 ways
next in 3 ways
and 3rd in 2 ways
so 4x3x2 = 24 Ways

2) They can be distributed in 4! ways = 24 Ways

3) So total number of ways is 24 * 24 = 576

Now no of ways A can get a black jack should be less than the total number of ways...isn't it? :?:
So whats the flaw??? and is ther a better approach to solving this problem?

Do you mean they have 16 cards to play with?

Are you sure that you have the question right? What book did you get it from?


floor function
combinatorics unsolved
combinatorics

$n$ natives live on an island. Every two of them are either friends or enemies. One day, their chieftain ordered all inhabitants (including himself) to make themselves a necklace of small stones, following the rule : every two friends must have at least one stone of the same kind in their necklaces, and no two enemies may have any stones of the same kind in their necklaces (a necklace may be empty, as well). Prove that the chieftain's order can be performed using $\Big[\frac{n^2}{4}\Big]$ different kinds of stones. Prove that the order can not be realized, in general, using smaller number of stones.

We will first prove a lemma: if $G$ is a graph such that every vertex has degree at least $2r-1$, then we can find $r$ disjoint pairs of vertices such that each pair is joined by an edge. Proof: it is obvious for $r=1$ (take any vertex and one of its neighbors). Assume it is true for $r-1$. Then just take a random vertex and one of its neighbours and pull them out of $G$ as a pair; the resulting graph $G'$ has minimal degree at least $2r-3$, so we can find another $r-1$ disjoint pairs.

Now for the problem: if $n=2k$, then we have $k^2$ kinds of stones; if $n=2k+1$, we have $k(k+1)$. We will look at two cases:

a) There's a guy with at most $k$ friends, where $2k \leq n<2k+2$. The chieftain tells this guy to go away to get some coconuts and has the remaining $n-1$ tribesmen make their necklaces amongst themselves (i.e., supposing the coconut guy doesn't exist), with the number of kinds of stones allowed by the inductive hypothesis. When the coconut guy comes back, there are $k$ kinds of stones that haven't been used ($k(k+1)-k^2=k^2-k(k-1)=k$), so he can have those $k$ kinds on his necklace, and each of his friends adds one of them to their respective necklaces (since they are at most $k$, the kinds of stones they take can be all different). This works.

b) The minimal number of friends is $k+r$, with $r>0$. Again, the chieftain tells a guy with minimal number of friends, $X$, to go away and get some coconuts and the remaining tribesmen make their necklaces forgetting him. When $X$ comes back, there are, again, $k$ colors left for him. He has $k+r$ friends and $k-r-1$ or $k-r$ enemies. Since $k+r$ was minimal, each of his friends has at least $k+r$ friends, so at least $k+r-(k-r)-1=2r-1$ of these friends are not $X$ or his enemies. In other words, the subgraph made up of friends of $X$ has minimal degree (of friendliness) at least $2r-1$, so we can find $r$ pairs of friends among them. Again $X$'s necklace is made up of his $k$ kinds. $r$ of these kinds go to pairs of friends (since they are friends, they can share the kind of stone), and the remaining $k-r$ go to $(k+r)-2r=k-r$ lone friends. We are done.

To see that $\lfloor \frac{n^2}{4} \rfloor$ is necessary, take a graph with $2k$ vertices divided into two groups $A, B$ with $k$ tribesmen each, such that everyone hates everyone else in his group but likes the guys in the other group. Then any two men in $A$ have disjoint necklaces, and each necklace has at least $k$ kinds of stones, since they must be friends with all the $k$ $B$-guys, which are also pairwise disjoint. Then we need $k^2$.


how do i set this problem up

  An English teacher was asked how many fish she caught. she replied:

   When I tried to place a fish upon each dish, I had a fish without a dish.
When I tried to place two fish upon each dish, I had a dish without a fish.
You'll never hear boos and hisses, if you can find how many fishes?

If f = # of fish
and d = # of dish

f = d + 1
f = 2d -1 (assuming that the second line means that the last dish has one fish)
otherwise, f = 2d - 2

@DuoCapital: I think it's f = 2d-2 because there are no fish on that dish.

so if you take the two equations and subtract them:

f = 2d-2
[u]f =  d+1[/u]
0=d-3
d=3

since f = d+1, f = 3+1 = $ \boxed{4}$


hello
i am preparing for iit jee 2008
i heard some rumours that
the paper is going to be divided
screening held on dec. and then mains
some of our lecturers told
actually where did these rumours come from?
is it true

i think they originated from kota...
the best thing is to wait for the announcement by the IIT authorities(if any  :D )

but i think screening in december is for sure

i think so because one of our lecturers have been in touch with some iit kanpur professors
and they told that screening would be there in dec.
let's wait until the notification comes


[quote="vardanch"]($FIND$ $NEW$ $PROOF$, $OR$ $ORIGINAL$ $METHOD$)

$f: R /to R$

for every $x,y /in R$
   $f(x+y)+f(x)f(y)=f(x)+f(y)+f(xy)$.[/quote]

ok, maybe you will show us your original proof :-)

davron


function
integration
real analysis
calculus
derivative
real analysis unsolved

$ f: [0,1]\to \mathbb{R}$ a infinitely differential function,can we prove that $ |f(x)|'$ exist almost everywhere?

If it is ture and $ f(0) \equal{} 0$,Prove that:
\[ \int_0^1 |f(x)|dx \equal{} \int_0^1 |f(x)|'(1 \minus{} x)dx\]
In the sense of Lebesgue Integral.

$ |f|$ is absolutely continuous on $ [0, 1]$, so its derivative exists a. e. (Actually $ |f|'(x)$ exists unless $ f(x) \equal{} 0$ and $ f'(x) \neq 0$, and that set of $ x$'s is countable.) To see the second statement, write $ |f(x)|$ as the integral of its derivative and switch the order of integration.


If $m$ and $n$ are positive integers with $m$ odd, determine

$d = gcd(2^{m}-1, 2^{n}+1)$.

[hide="Hint"] This is just

$\frac{ gcd(2^{m}-1, 2^{2n}-1) }{ gcd(2^{m}-1, 2^{n}-1) }$ [/hide]

Nobody?

[hide="Hint 2"] Recall the lemma

$gcd(2^{m}-1, 2^{n}-1) = 2^{gcd(m, n)}-1$

And use this to evaluate the expression I gave in the first hint. [/hide]

i think this is what i have..not sure if its correct

[hide]
$=\frac{ gcd(2^{m}-1, 2^{2n}-1) }{ gcd(2^{m}-1, 2^{n}-1) }=\frac{2^{gcd(m, 2n)}-1}{2^{gcd(m, n)}-1}= \frac{2^{ gcd(m, n)}-1 }{2^{gcd(m, n)}-1}=\boxed{1}$

EDIT: with the correct hint :wink:
[/hide]

Yes.  I wrote the second hint improperly, but the answer is still right.

[quote="SM4RT"]how did u come up with hint 1..is that from a theorem or something..[/quote]

It's because $2^{2n-1}= (2^{n}-1)(2^{n}+1)$ and $\gcd(a,b) = \frac{\gcd(a,bc)}{\gcd(a,c)}$ when $\gcd(b,c) = 1$.


[quote="Ignite168"]It is now 4:40AM. Ignite168 does not plan to sleep at all, however, to prevent falling asleep, he must studies have shown this impairs your response time by 30%. If normally Ignite168 can catch a yard stick falling at the rate of 5760 sixteenth inches/min. at the 3 inch mark, where to the nearest tenth of an inch, will he catch a yard stick falling at the rate of 720 foot/hour?

If my calculations are right, which they probably aren't as no sleep reduces your smartness by $-626i\%$ the number of seconds in your new response time is like phi to phi squared compared to the answer.
[size=9][hide="."]0.71428571428571428571428571428571[/hide][/size][/quote]
Huh.

i don't exactly get this question. there's something with unit conversion or something with reaction time. i think. :maybe:

I think something is wrong with the problem.  Could you please explain it more clearly?

[quote="Ignite168"]however, to prevent falling asleep, he must studies have shown this impairs your response time by 30%. [/quote]

Reword that bit, please.

[quote="lotrgreengrapes7926"]It is now 4:40AM. Ignite168 does not plan to sleep at all. Studies have shown this impairs your response time by 30%. If normally Ignite168 can catch a yard stick falling at the rate of 5760 sixteenth inches/min. at the 3 inch mark, where to the nearest tenth of an inch, will he catch a yard stick falling at the rate of 720 foot/hour?[/quote]

But then why would he have written something about "however, to  prevent falling asleep..."?

OK let me try...

[hide]so he can normally catch a yard stick falling at 30 feet/min at the 3 inch mark, and he's trying to catch one at 12 feet/min.
So... it's falling 2.5 times slower but his response time is .7 times normal..
then... do we multiply 3 by $\frac1{2.5}$ and $\frac1{.7}$ or what?
The answer that way would be... $1\frac57$
although his "answer" is $\frac57$[/hide]


∑n=1 to ∞ (2n^3/2 + 5n^1/2 - 1)/(7n^5/2 -6n61/2 +3N -2)  Which test would i use to see if this sum was convergent or not and what would its limit be if it was convergent?

$ \sum_{n\equal{}1}^{\infty}\frac{2n^{3/2} \plus{} 5n^{1/2} \minus{} 1}{7n^{5/2} \minus{}6n^{1/2} \plus{}3n \minus{}2}$ should be analyzed with the Limit Comparison Test.

Let $ a_n$ be the general term of your series. Then $ \lim_{n \to \infty} \frac{a_n}{\frac{1}{n}} \equal{} ...$


At the grocery store last week, small boxes of facial tissue were
priced at 4 boxes for $ \$5$. This week they are on sale at 5 boxes for
$ \$4$. The percent decrease in the price per box during the sale was
closest to: \\ \\
A. $ 30\%$ \\
B. $ 35\%$ \\
C. $ 40\%$ \\
D. $ 45\%$ \\
E. $ 65\%$ \\ \\
Select the most appropriate letter.

:ddr:

equation is 1.25-1.25x=.8
because you start out with 5/4 and then you had a sale and end up with 4/5 and you need to subtract 1.25 by the percent discount to end up with 4/5 or .8
x=.36

nearest is B=35%

so B

At first, each box cost $ 5/4. Later, each box caust$4/5.

5/4 - 4/5 = 9/20.

[(9/20) / (5/4)] x 100 = 36%.

The closest answer is therefore B. :starwars:


[quote="anirudh"]I play black. :) But sometimes white. :)[/quote]
what openings do you use as black?

You can follow this link to a webpage called "A basic guide to chess openings", the contents of this webpage are very useful.

[url]http://www.eudesign.com/chessops/[/url]

I hope this can help you! :P

[quote="JavaMan"]FICS automatically chooses white/black so that you've played a balanced number of games with each, so I play both about the same.[/quote]

umm not really you can choose what color u r.....

and black openings are kinda hard because it all varies on white's style

Black openings for 1. e4:

Sicilian
French
Petroff

those three are highly recommnended and can lead to an interesting game

if i play black, i usually use that one defense where black moves his c-pawn up one. it's like the kann defense or something.  Some people say it's too slow, but i usually beat them with it. so i don't know.

Yes its the caro kann opening...very good if you know how it works

[quote="7h3.D3m0n.117"][quote="JavaMan"]FICS automatically chooses white/black so that you've played a balanced number of games with each, so I play both about the same.[/quote]

umm not really you can choose what color u r.....

[/quote]

You can choose, but I don't think many people do (only time i've did is when competing in an online tourney). At some places you have to choose a colour, so that's often inconvenient and at times you realise you've played 30 games in a row with white...

I like playing black better. I don't really know why, but I just feel more comfortable that way. I don't play chess very much though.

I prefer to play as black due to the vast amount of openings one needs to prepare for as white (if you're an e4 player, but Queen's Gambit players still need to know a large collection of openings). The amount of openings you need to have in stock for black is large, yet extremely manageable.
As for my opening against e4, a Sicilian Najdorf, Sveshnikov, or Schevenigan .

I like the Queen's gambit and the Sicilian Defense. My rating is 1456 what's yours?[/hide]

[quote="Dsquared"]It doesn't really matter if you play as black or white.  They have the same amount of chance of winning.  It's the endgame that counts.[/quote]
rofl

[quote="fries4guys"]In fact, I was black twice today and I came SO close to winning when we had to pick up.[/quote]
When you play harder people it is much more important.

But I was black, and if it was important then someone would have won already.

Chess tournaments rarely allow a player to choose a color, so I play as both White and Black. As Black, I currently play the Najdorf Sicilian and the Grünfeld Defense.

I don't really care what color I play on. As black, I like sicilian and as white, I like the four knight's defense

I prefer white to have the initiative and some control over what opening follows. ~60% of the time my first move on white is 1. e4

White: Queens Gambit
Black: Sicilian Defense usu. French Variation

[quote="math92"]Do you play as white or black? I usually play as white, but i want to learn to play with black more effectively.  If you do play as black usually, could you tell me some effective openings?[/quote]

I play on FICS too! Can you give me your account?

:P People use sicilian against me. I use the Smith Morra Gambit against you.

As white, I usually use the Danish Gambit or the Smith Morra Gambit.

If I'm playing white, I try a King's Pawn opening of some sort... I basically try to control the center squares early while waiting for the right moment to bring out my queen.


calculus
derivative
trigonometry
analytic geometry
graphing lines
slope
quadratics

Find the points on the graph of y = -csc(x), where the tangent is parallel to 3y-2x=4. Use exact values.

0 is less than or equal to x which is less than or equal to 2*pi

Thanks

$ y' \equal{} \frac {cosx}{ \sin^{2}(x)}$. This slope has to be the same as the slope of $ y \equal{} \frac {2}{3}x \plus{} \frac {4}{3}$. So we have:
$ \frac {2}{3} \equal{} \frac {cosx}{ \sin^{2}(x)} \equal{} \frac {cosx}{1 \minus{} \cos^{2}(x)}$
We get: $ 2\cos^{2}(x) \plus{} 3\cos(x) \minus{} 2 \equal{} 0$
Using quadratic equation we find that $ cos(x) \equal{} \minus{} 2$ or $ \frac {1}{2}$.
Our natural choice is $ \cos(x) \equal{} \frac {1}{2}$. In the interval $ [0, 2\pi]$, we have $ x \equal{} \frac {\pi}{3}$
and plugging this into your original $ y \equal{} \minus{} \csc(x)$, we get $ y \equal{} \minus{} \frac {2\sqrt {3}}{3}$. So the point we are after is: $ (x, y) \equal{} (\frac {\pi}{3}, \minus{} \frac {2\sqrt {3}}{3})$

The equation $ \cos x\equal{}\frac12$ has two solutions in $ [0,2\pi).$ The other one is also valid.


geometry
3D geometry
sphere
trigonometry
calculus
inequalities
algebra

As one of the question writers, I would like to know, what were some of your favorite/least favorite questions this year?

My favorite question was number 10 on combinatorics A, which I sadly messed up on, because I did 6C2 instead of 6C3, and missed an obvious pattern. Oh well.

I didn't particularly like the geometry questions, but that might have been because I got stuck doing geometry, even though it was my last choice.

On a different note, I'd like to know if my challenge on the marble problem was accepted.

I was definitely not a fan of the geometry test. Not that I got a 6, but that the problems were like...not geometry. And some of the team test problems were kind of lame, in my opinion, and clearly written last night. Not that it's a bad thing, but...

As for the power test, though, it was really good. I would have liked more of a team effort (cough WENYU CAO cough), but we did decently. Ish.

So Quagga was disappointing, anyhow. 8th on power, fail on the team round? But I'm convinced we had the highest individual totals. So we are all selfish?

Geometry was pretty disappointing. But I suck at geometry anyways, so it might be me (actually, #2 was a pretty nice problem, but I forgot to add the sphere to compensate for the corners). Most of the number theory test was okay/good, I guess. At least the problems made since and didn't require plotting figures in 4 dimensions or finding cosines of angles in cuboctahedrons. Except for #10 on number theory: how exactly were you supposed to get the nine numbers that worked??
The team test was like: 1-6 easy, and we couldn't figure out anything on 7-10.
And the whole thing about it being delayed by 2 hours wasn't very fun...
The power test was pretty fun, though.

There was a small bit of ambiguity on combinatorics division B.

#3 - I didn't know that you meant a tesselation figure where the hexagon with the squares and equilateral triangle was repeated infinitely. I thought that you started with a hexagon, and you added layers to it by constructing squares and equilateral triangles forever. Another member from my school also had this interpretation of the problem.

#8 -  When you said "below and to the right" you could have meant that all the x's that were both below and to the right were greater. Or you could have meant that all the x's to the right were greater AND all the x's below it were greater. You have to specify what "right" and "below" mean also. Does it mean generally to the right or directly right?
I took the first interpretation and ended up with 8C2 * 6! = 20160.

x b x x
a x
c
x

Is b to the right of a? Is c below b?

I loved number theory #8 (div b), it allowed for much creativity.
The individual finals was a great test also. #1 I missed the trivial solution (plug in n = 1), but #2 and #3 were very well worded. #3 was my favorite problem =D

Thank you for a great competition,
Michael Tan AAST

I generally liked the questions for Geometry B, except some were too easy. For example, number 2 for the test was a freebie: No one should've missed it. It was that easy.

I generally liked the questions for Geometry, but only because I was able to understand most of them :D

Blah, I was on an unofficial team, and submitted my power test work (Yes, MY work... i.e. all the work) as "AAST Alpha", because [i]someone[/i] decided to tell me that.

BTW, AAST Alpha got 2nd Power  :lol:  :lol:  :lol:

And how was #7 on the Geo B Geometry?

I need to know: is #5 25? Every time I do it I get a different answer. I put 25 and got that 3 times in a row though, but my teammate keeps annoying me about how trig and similar triangles and cyclic quadrilaterals are invovled.

And the incircles/circumcircle one was too easy. Finding the 4 sides is trivial, and then use Bragmupahta (or something) for cyclic quads.

Bleh, I would have gotten top 5 in Geo if I was official.

#3 on geo B I failed. I forgot to multiply by 3 (stupid mem 1x1x1)

When will answers/solutions be up?

NT was decent; except for that #2 had infinite solutions in integers (n,0) works for every n an integer
Overall I didn't like it:
NT was weird and basically brute force except for the $ f(x) \equal{} x^{{x^x}^x}$ I got that one it was nice with the Chinese Remainder Theorem. Otherwise which ones weren't brute force?
Also was Geometry Geometry? What was all that four dimension junk? anyways thanks tho it was fun..
It seemed like alot of beast ppl did not do very well:
Vlad didnt make individual finals, and Catalyst did not win anything, thats kind of weird. It seems as if 2007 was written much better, I liked the questions more.
The T-shirts were awesome, and mathbowl was fun too.
Who won overall for teams I left a little early? Unfortunately I did not make individ finals as I failed Geom

North Carolina won overall for teams.

I managed to solve #3 on finals 30 seconds before time - it would have taken me 2-3 minutes to write up. I agree with everyone else about geometry. I was happy about Power, we managed to get 10th with only 4 people.

Was anyone else surprised by a mild calculus problem on the finals?

Very fun fun fun day (except for the longer than expected day).My questions ( I was in Division B and did Combinatorics and Geometry):
Combinatorics: There were a few ambiguities. Number 6 I thought was just 2008 because I thought $ \binom {m}{k} m<k$ is undefined and would make the whole thing undefined. If I knew that integers less than 2008 worked then i would have put the given answer. Additionally, as someone already pointed out, the x's question I thought it meant down AND to the right.
Geometry: The first two problems were easy points but that's all I got lol. I know what to study now for AMC  :lol: .
Team: Where are the answers? My team (and the A team from my school) got $ 10\sqrt 6$. What was the answer? As for the other questions, the first 4 were too easy for a team test. Was number 6 337? And if my friend did it correectly, 10 was 4017. If that's the correct answer then it seems too easy for 10.
Individual test: Will we get our graded answers back? Was 1 null set? Was 2 4 and 13? Was 3 not comprehendable lol
Anyways, GO CHE! Hooray personal 3rd place on an individual test!

Catalyst of Nostalgia, do you remember me? I was the one hanging out with you guys during the awards ceremony haha.

Anyways I took Combinatorics and Geometry B and I pretty much agree with everything everyone else said.  Number 3 in Combinatorics was pretty unclear.  For the number with the sum and differences of combinations up to 2008, I didn't know you could have a lower number on the top of a combination.  But now I know lol.

Also for the Individual Test Division B, i thought number 1 was way too trivial.  If you let n=1 the second inequality reduced to (greatest integer of b)<0.  Since b in a positive answer, there's obviously no solution and the first inequality doesn't even come into play.  I liked the second question though.  It was easy but it made me feel good that I could do it lol.

I also won trophies in geometry and individual finals!!!! Really wasn't expecting that lol.

BOGTRO: You didn't even need to use Brahmagupta for the incircle/circumcircle, it turns out to be an isosceles trapezoid.
What was #5 on geo B?

[quote="mathemonster"]NT was weird and basically brute force except for the I got that one it was nice with the Chinese Remainder Theorem. Otherwise which ones weren't brute force? [/quote]
Umm... let's see.
#1, 2, 3, 4 (to an extent), 6 (to an extent, you have to calculate 2008*1004), 8, and 9 were all non-brute force. And possibly 7 as well, but I didn't understand the solution. :P

[quote="crazypianist1116"]Team: Where are the answers? My team (and the A team from my school) got $ 10\sqrt{6}$. What was the answer? As for the other questions, the first 4 were too easy for a team test. Was number 6 337? And if my friend did it correectly, 10 was 4017. If that's the correct answer then it seems too easy for 10. [/quote]
Our coach was a proctor, so he had the answers. I remember 10sqrt(6) and 337 were right, but number 10 was like 10320126572910670 or something like that (we guessed 6025, heh)

My team (Fairfax Math Circle Red) seemed confused in general by a lot of the questions on the tests. The difficulty changed a [b]lot[/b] from what I could see of the previous years, to the point where I was happy that I got 2 or 3 right on each (Number Theory and Algebra A). I did not go to the Individual Finals, so I can't really comment on that.
Questions I liked:
-The Power Test was excellent. I'm not quite sure where my team placed, as we had to leave before the awards ceremony. Basically two people did this whole test (me and another), but it took the whole week, to the point that we were still solving problems on the 4-hour car ride there.
-The Team round was stimulating, and really made us think.
Dislikes:
-The Math bowl didn't really have too much math on it, I expected something based more on mental math.
-The length was somewhat disappointing. I know there appeared to be problems, but I think my team wanted more updates on what was happening.

Separately, are results posted anywhere? I'd like to see how my team(s) did.

Alex Zhu = ONE MAN ARMY. enough said.

Math Bowl is just for fun with a lot of random things in the questions. For example, my friend was jumping up and down on her seat when the anime question came up.

It's mostly based on math, but not necessarily.

[quote="crazypianist1116"]In lieu of all of this, I have to say my favorite question was the one about infinitesimal Randall Munroe.[/quote]

Exactly how do you do this one without having the temptation of guessing 3/2?

[quote="darkdieuguerre"] United placed 7th, with a 1st place power test, for example.[/quote]

No way?  :huh: That is pretty funny .. considering that I did quite a bit of our power test .. and like I thought it failed (esp 3.8, 4.3) ..

I recognized the hook length formula for the combo question with the $ x$s, and (i think) saw the stack-queue question somewhat recently in the olympiad section (but naturally forgot the result), but otherwise failed computationally [actually took the time out to multiply the denominator out for the cube problem before I realized I was multiplying the wrong numbers due to addition mistake]

[quote="rjyanco"]
[quote="darkdieuguerre"]United placed 7th, with a 1st place power test, for example.[/quote]
Given the multipliers, if they finished seventh, I think it's a fair guess that their individual and team scores were very low, at least relative to other top ten teams.  We can check when results are put online; but if I'm right, that would buttress my point.

Let me state up front that PUMaC can set whatever multipliers they want; their rules are their rules (though they need to be stated somewhere in print).

My first point, though, is that it seems disingenuous for PUMaC to state that the Power Test is graded out of 75 points when its actual value is 300 points, probably 50% or more of the total points earned by most teams at PUMaC 2009.[/quote]

I [in United] just read this right now. Indeed only 2 of us put much input into the power test, and in all likelihood I could have finished it alone, though I definetly had not expected that we would do as well as we did on the power test (eg, I wrote up most of 4 on the day of, before deciding that sleep would be a better option).

I just want to note that our team and individual scores were indeed quite low, lower then we had expected at any rate, as we were down at least several slots due to last-minute substitutions and then a member simply not showing up. I wouldn't blame it particularly on a single member carrying a weak team then; we have a decent number of USAMOers and so forth amongst us, but I guess we were just not performing to full capacity on that day.

rjyanco, thanks for your comments on Power Test grading. I wasn't really involved with the organization of PUMaC this year, but I was one of the Power Test graders. If I remember correctly, the test was actually out of 82, and the top team scored 81. I think your criticism is definitely deserved -- it didn't seem to me that the selection of scoring system was very well-thought-out this year. Of course, it was impossible for us to correct that error after seeing the effect we had created: there's no way we can modify the scoring after we've graded some of the tests. But we can certainly be more careful next year about balancing the parts of the contest.

Regarding cheating, we try to create the rules of the test to make it hard to cheat. Since all resources are allowed, and it seems not unreasonable to assume that all students know how to Google, looking stuff up isn't really cheating. Perhaps you could criticize the fact that this rewards research skills as well as math skills, but consider the following: the graders could tell which teams had found Conway's book (or other references). I recall two different teams that used Conway's book did poorly anyway. They thought that they had found all the answers and got lazy, either not writing good proofs or failing to translate Conway's notation and terminology into that of our tests (which leaves great doubt as to the team's understanding of the material). So teams can't "get away with it" that easily.

When we have our PUMaC reflection and discussion meeting, we'll definitely make sure to spend a good amount of time on the points you've brought up here. Thanks again.

[quote="FantasyLover"][quote="soltixnj"]FantasyLover, I used induction for #3 of Div B finals.

Does anyone know when (or if) results will be posted?[/quote]

Soltixnj, could you please describe what you did?

Thanks.  :)[/quote]

It's true for an n=2 hypergraph.
Assume it's true for n=k.
Then consider a hypergraph of n = k+1 points.
Let q be the added point.

There are two cases, either an edge contains q or it doesnt.
If it doesn't then it's part of n=k hypergraph which works by the induction hypothesis.

If it does then it's part of n = k+1 and theres 2 more cases. Consider all the other vertices of the edge.
Either they are all the same color or they are not.
If they are not, there is no restriction on q.
If they are, then there cannot be another edge with all the opposite color, because then the two would only share q in common, which is forbidden by the problem.

When are pumac results coming out?

I don't have the results myself, but they exist so I don't see why it should take much longer for them to be posted. I'll remind the people who have access.

[quote=darkdieuguerre][quote="Xevarion"]If I remember correctly, the test was actually out of 82, and the top team scored 81.[/quote]

The results say United scored an 87, which looks correct given the rankings.

Are the actual weightings of the power test problems ever officially announced?[/quote]
Out of 88, then. All I remembered was that it was in the 80s and they were 1 point short of perfect.

I'm not qualified to answer your question (i.e. I don't feel that I have the authority to reveal that information on my own) but I'll ask for you.

[quote="jb05"]They are posted, and I disagree that the power round is overrepresented - it's only around 10-15% of the score for most people. However, the team round is dramatically underrepresented.[/quote]
Power round total was multiplied by 4 for sweepstakes. Team round score was multiplied by 5.

Man, if only I hadn't made so many dumb mistakes on Geo, I would have been in top 5.

Oh well.

Thanks for the comments on the tests. I can't respond to all of them, but some things we're keeping in mind for next year as we continue tinkering with the tests are:

-Several people have commented that that they think all the problems on test Blah were either gimmes or impossibly hard. (Classically, of course, if you can solve a problem, you call it easy, and if not, you call it hard.) The contest draws a wide range of people, even within each division, and we try to include problems accessible to both. In particular, the distribution of the lowest scores indicates that none of the problems were too easy for everyone; other solvers can just race through them. A few of the tests could apparently have used a few more of the easiest questions.
-For similar reasons, we're happy with the difficulty of the last few problems.
-Given the steep drop-off in scores in the middle, we'll work on making the middle few problems easier for next year. This varies by test, of course.
-Higher-dimensional geometry becomes more common at higher levels, and will remain on the contest. This year's tests were a bit heavy on it, though.
-We'll ponder rjyanco's points for next year's Power Test. We're happy with that it involves research (and some of the definitions were deliberately made different from what was easily available online, to trip up blind copiers), and hesitant to put it on-site because the day is so packed as it is, but we'll consider separating it further from the onsite events.

Thanks for all your comments, and see you next year!


If anyone could explain how the following is done, it would be greatly appreciated!

Consider the surface xyz=cos(x+y+z) and the point P(0, pi/4, pi/4), which lies on the surface.

Find a normal vector to this surface at the point P.

Any surface given by a single equation can be considered as a level surface of some function. So move everything to one side of the equation. Consider this surface to be the set $ \{x: g(x)=0\}$ where $ g(x)=xyz-\cos(x+y+z).$

Once we have that formulation, a normal vector is a scalar multiple of $ \nabla g.$ In this case,

$ \nabla g=\langle yz+\sin(x+y+z),xz+\sin(x+y+z),xy+\sin(x+y+z)\rangle$

Evaluated at the point $ \left(0,\frac{\pi}4,\frac{\pi}4\right),$ this becomes

$ \left\langle\frac{\pi^{2}}{16}+1,1,1\right\rangle.$

There are still some things we could do with that, including, if we wish, dividing by its magnitude to get a unit vector. We might also decide we don't like the direction this is pointing and take its negative - your instructions did not indicate a specific preference for direction, but sometimes there is one.


Don't just post an answer -- that doesn't say anything at all.  There's nothing interesting anyone can say about or learn from an answer. Post a solution.

[quote="tetrahedr0n"]This is a cool problem:

The last 7 digits of n! are 8000000. What is n?[/quote]

Let's see...in order to have six zeroes on the end of n!, n has to be between 25 and 29.

How do I know this?  Well, the number of zeroes on the end of the expansion of n! is [n/5] + [n/25] + [n/125] + [n/625] + ..., where [x] represents the greatest integer less than or equal to n.
When n is 24, n! has 4 + 0 + 0 + ... = 4 zeros on the end.  When n is 25, n! has 5 + 1 + 0 + 0 + ... = 6 zeros on the end.  When n is 30, n! has 6 + 1 + 0 + ... = 7 which is too many zeros.  Hence n must be between 25 and 29 to achieve six zeros on the end.

I think I'm halfway done, but I'm drawing a blank on how to figure out which one (25-29) gets the 8.  Any ideas?

What I did when I was taking the contest was just to write down the numbers from 1 to 25, remove 6 factors of 5 and 6 factors of 2, and just multiply. Then I checked 26 and 27, and 27 worked.

I think it's possible to do better though. Suppose 25! ends 8000000. Then 26! also ends 8000000. Thus it's not 25. If 26! ends 8000000, then 28! ends 8000000, so it's not 28. Now we start working modulo 5 after dividing by 1000000. Let f(n)=(n!/1000000) (mod 5). If f(28)=3, then f(26)=3, so again we don't have uniqueness. This argument doesn't work for 29 though unfortunately. I think we have to be clever to show that it can't be 29, but I can't think of how. Any suggestions?

I did it the same way you did -- multiplying many things together after failing to elliminate 27, 28, and 29.


What did you think of the song contest?

I was at the Penn State one. My fav was "I do math all day". (I think it was Lehigh valley)

Anyone have videos?

The attempted Rick Roll was EPIC FAIL. Though the trenchcoat was really cool, it's very hard to make anything sound good on that sound system.

If our team had done the song contest, which GA didn't have this year, Billy Dorminy would have pwnd (solafidefarms is his username). :rotfl:

Oh yeah. Although, in liu of performance at ARML, I serenaded the cheerleaders with She Thinks My Protractor's Sexy :). (I MADE 25 CENTS!!!)

:rotfl:  :rotfl:

Wow I don't know what's more awesome, the song, or the fact that you specifically went up to the cheerleaders at the campus to sing it...

Your courage must be unbounded..

The RickRoll was much better the night before
In the arena it didn't work out so well

Nice work with the cheerleaders. How come you didn't enter the contest?

the Song Contest? UGA didn't have a song contest for me :(.

I think that, perhaps, we didn't have a song contest BECAUSE of you!  :D

I still cannot believe that Lazslo Seress won third place...  I heard people talking, saying that he was a good singer but what he was singing about was not interesting.  Yeah, so anyways, I liked the "I do math all day" song too.

Well, it's not my fault everyone else didn't want to sing for me; this year Ruby seemed receptive to the idea of singing, but then we hadn't anywhere to sing, so I had to sing on my own...

*sigh*
Being a girl on math team is sometimes tough.
My team kept hooting at the cheerleaders.  :blush:

lol, I don't think there were any girls on the SFBA(A) or (B) teams...

[quote="solafidefarms"]No, worse, I sang outside their open windows :). I actually tried to get them to come cheer for us at awards, but they were leaving at noon :(. So, instead, they cheered for us on the way over, much to the dismay of my teammates. "Yay smart math people!" "Yay cute cheerleaders!"

(And what needs so much courage? They're just fellow humans...)[/quote]

That was funny yet disturbing.  You were like "YEAH!!!!!!!!!!!!  YEEEEEEEEEEEHAW!!!!"  and the cheerleaders were like "YEAAAAAAAAAAAAAAAHHHHHHHHHHHHHHHHHH!!!!" back, and Vlad was like "Yup, that's Billy's mating call."

 :arrow:  :|  :rotfl:  :rotfl:

Um billy most of those cheerleaders were like 12, so yeah.. I hope you meant cute like a baby...

[quote="eshen"]*sigh*
Being a girl on math team is sometimes tough.
My team kept hooting at the cheerleaders.[/quote]

Meh. The just ignored me...

Actually the cheerleader team sitting beside me at breakfast was composed entirely of seniors _in high school_ so they were actually older than myself. :)

I was surprised at how many cheerleaders you managed to get to cheer for us. At the same time, I wanted to take off my ARML T-shirt and yell out "TECH PRIDE!!" (because I was wearing a GT math competition T-shirt underneath).

There were a lot of teen girls.
Not to be mean, but also some surprisingly overweight ones.

I am still convinced that the judges at Penn State chose Ontario because they wanted to Rickroll everyone. :P


Hi- I am new to the forum. I read the guidelines, so I hope I am not committing some horrible faux pas in my first post...

As the subject indicates, I am interested in strategy-based math games for gifted kids...I teach 5th grade, but games can generally be adapted/differentiated for grades 4-6.

I say "strategy-based" to differentiate these from games that are designed to promote skill development in a certain area; that is, games to improve math facts, or finding factors, reducing fractions, etc, etc.

I have collected a few of these over the years but am always, always looking for more. I have discovered that they are very difficult to find...most math "games" are designed for skill develpment, or are just based on luck.

An example of the kind of game I am looking for are the NIM-based games. I'm sure most of you  are familiar with these, but just as an example, there is a pile of, say, 20 objects: pins, pennies, toothpicks, matchsticks, etc. Each player may remove 1,2, or 3 at a time. The player who removes the last piece wins.

This is a great game to play to develop math sense...and there are many variations.

Another example would be the "100 game;" (though I'm sure there are scads of games with the same name). In this game, players start with a number and add 1-10 on each turn. The one who gets to 100 wins. Wonderful game for kids to work on and try to develop the math strategy that will guarantee a win.

Please, anyone with a similar interest, or who knows of sources for such games, or has some pet ones they've used over the years...I look forward to your response.

BJN

I have a couple ... but the strategies required are far from obvious, so it'd be your judgement whether or not to use these.

Nim - There are three piles of objects. Each turn a player can take any number of objects as long as they are all from the same pile. The person to take the last object wins.

Factors - Numbers 1 through n, where n is any positive integer, is written. Each turn players may erase any number [i]and[/i] all of its factors. The player to erase the last number wins. Good game for building the concept of factors.

24 game - A classic: Use a deck of cards. Deal out four cards face up, and then try to make the number 24 using the four basic operations and [b]all four[/b] cards. Jacks are 11, queens are 12, kings are 13, and aces are 1. For example, if the cards were J 2 A A, a possible solution would be (11 + 1) * 2 * 1.

[quote="ProtestanT"]i remember that for NIM the player that takes the last object loses. you win when you leave the opponent with one object left, because he is forced to take that and lose.
i think the player-that-takes-the-last-object-wins would be too cheap...

i don't see the difference between strategy-based math games and skill-developing math games. could someone explain?[/quote]
Well, actually both versions of Nim are extremely similar to each other. Play some games for yourself.

Hi, the difference is this: in skill development games, the purpose is to develop a specific mathematical skill, such as reducing fractions, or understanding factors, or converting decimals to fractions to per cents, etc. I have many of those, such as decimal/fraction bingo, where the students' game boards have fractions on them such as 1/2,  4/100,   2 6/10, etc. The teacher calls out ".5 "    " .04"          "2.6" etc, etc., and the students have to find them on their cards. The purpose is to increase their ability to recognize the relationship between decimals, fractions, per cents.

There are SCADS of such games of every type and variety everywhere, both in books and online.

Strategy-based games are those such as NIM, where the players are looking to work out a strategy that will enable them to win the game every time. While there may be definite math skills involved in the game, the main purpose is to develop mathematical thinking; it's not to practice factors, or per cents, etc.

These games are hard to find....that's why I'm posting here in hopes of finding games and sources. And, I am always happy to share whatever I have with anyone.

NIM: I am no expert in math, math games, or in NIM. But as I understand it, this is an ancient game and there are innumerable variations. In some, you win by taking the last object. In others, you win by forcing the other player to take the last one. All of these variations are useful, as they require the players to rethink their strategy and try to apply a new one in order to win. It's a fun, rewarding and intriguing way to develop math thinking. I hated math as a kid; it was never any fun, so I would like the experience to be different for my students.