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Have you taken the SAT's (1 or 2) before? if so, state which one you did, your age when you took it, and what was your score?

i've taken the SAT I twice but never the SAT II... even the math SAT II is hard :( anyway, when i took it in 7th grade it was bad but last year i took it (i was 13) in 8th grade and got a 1450, 800 math 650 verbal (a phenomenal improvement over one year, thankfully)... anyway, i'm sure i'll be taking many SAT II's over the next few years... *sigh*

I'm only taking the ACT this year in 8th.  No SAT, maybe next year.

I did astronomically badly (1280 on sat 1) when i was in 6th grade(this year)
I also took a book SAT from [i]real sat II from people who bring you sats[/i] or something like that.  The math section is easy (coming from a moron like me) but i didnt finish it.  I also tried the world history section and got a mere 680.

I took the SAT 1 a couple days ago...so I won't find out my score for a while yet...

I took it a few days ago too. The verbal was more challenging than the math - no duh. I hope to have gotten at least a 1450. Yo. Scrambled. Don't be so harsh -  I didn't even take them in 6th, and I only got a 1370 in 7th(hoping for better). You can improve alot 1280 is great for a 6th grader(people who are hard working and smart I find improve an average of 100 per year up to 8th grade).
By the time you're in 8th, your scores will be real high.

I've never taken the SATs at all...some of my friends have taken tem but I just don't have the chance to take a whack at it I guess and I don't feel like researching how to take the SAT in my area and stuff...oh well.

Just wondering do all SAT score have a 0 in the first digit or is it just an astronomically rare coincidence?

Btw are any of you going to watch The Perfect Score?

P.S. I'm wondering if the title is correct...mmm...

i took it in 7th grade and got a 1440. ok i guess. i dunno.

um whitehorse, that wasnt an official SAT.  it was one of the book ones.  My brother said the real one is more challenging.

By the way, he got 1530 (8th grade).
[quote]Just wondering do all SAT score have a 0 in the first digit or is it just an astronomically rare coincidence?[/quote]

umm... By first digit do you mean units or thousands
[quote]You can improve alot 1280 is great for a 6th grader(people who are hard working and smart I find improve an average of 100 per year up to 8th grade). [/quote]  Im neither smart nor hard working :mrgreen::mrgreen::mrgreen::mrgreen::mrgreen::mrgreen:

SAT I: 1490 (780 math, 710 verbal)
SAT II: Math IIC: 800 Physics: 750 Verbal: 750

I took 'em last year.

[quote]um whitehorse, that wasnt an official SAT. it was one of the book ones. My brother said the real one is more challenging.

By the way, he got 1580 (8th grade).
[/quote]

by any chance, do you live in washington (state)? and the real test isn't necessarily harder... i get distracted when its not the real one and do not quite as well but during the real one its full concentration so i do well.

and Tare, yes, all SAT scores end in a 0. Everything is a multiple of 10, which i guess is kinda strange since you might as well divide everything by 10, but i mean i like 1000 sounds better than 100 in my opinion at least...

[quote]You can improve alot 1280 is great for a 6th grader(people who are hard working and smart I find improve an average of 100 per year up to 8th grade).
[/quote]

heh, i only got 1200 in 7th grade but i had a 250 improvement (i was amazed) which is uncharacteristic, you could say, but very possible

No i dont live in washington.  I live in california.

What should I expect to get on the ACT?  Anyone have any clue?

i took the SAT II math 1 and 2 C...  somehow their grading system was wrong because i got 800 on both when I didn't answer 1

The SAT's are a set of tests used by colleges in their admission's processes.  The word "SAT" used to stand for "Scholastic Aptitude Test" or "Scholastic Achievement Test," but it is no longer an acronym for anything.  (That is, the full names of the test is SAT.)

The scoring system is a scaled system: the scores received by students (from 200 to 800, in multiples of 10) do not correspond directly to the percentage of answers the student gets correct.

Also, I want to remind people to not brag and not complain about high scores.  Thanks.

[quote="plokoon51"]could somebody tell me how they factor in the essay score into your overall writing portion of the SAT?[/quote]

The essay is scaled so that the 12 possible points count for 25% of the total essay + multiple choice.  Those numbers (raw multiple choice score and scaled essay score) are added to give you a total raw score, which is then scaled to give you your score between 200 and 800.

(I'm guessing here somewhat, but it's the only process that really makes sense.  Obviously, the last scaling means that a more exact relationship between essay and outcome on a 200-800 scale will not be more quantifiable until the College Board releases statistics on that portion of the test which should be sometime during next school year, after they've administered it for a year.)

[quote="LynnelleYe"][quote="left_unattended"][quote="LynnelleYe"]
Oh yeah, I saw your name on the program, if you are who I think you are anyway. I didn't hear them call your name though...oh well.[/quote]

really?? I'm "MM" lol. I was definitely there, haha.[/quote]
Oh, then I guessed right. Yeah, I wasn't paying much attention to everyone else so I probably missed you.[/quote]

did they call you? 1580=top score.

I took the ACT last year as a seventh grader, but I've never taken the SAT. I still have lots of time to take it though. And plus, I'm glad I took the ACT. I got a scolarship for 1 free class at the university here where I live.

[quote="xtrememath8"][quote="mathbomberII"]i've heard of SAT's before but don't know what they stand for :mad:[/quote]

Standard Admission Test[/quote]
no, it used be an acronym for Scholastic Aptitude Test, then was changed to Standardized Assessment Test, and as JBL once said, it doesnt stand for anything.

http://www.sparknotes.com/testprep/books/newsat/chapter1.html

I took it in 7th grade and got a 1260 if I remember correctly.  I took the ACT this year (freshman) and got a 30.  Hopefully I will improve by the time I'm a junior.  I don't generally do well on tests like this; I dunno I guess I just get nervous or something.

What's with so many people taking the SAT's in middle school? I know this is an advanced board and all, but I've never heard of anyone doing this before. *shrugs*

I'll be taking my NMSQT(PSAT) and SAT and perhaps SATIC or SATIIC or both or whatever this year. I gotta research a few more SATII's to figure out what other ones I'd feel comfortable with taking.

[quote="WindSlicer"]What's with so many people taking the SAT's in middle school? I know this is an advanced board and all, but I've never heard of anyone doing this before. *shrugs*[/quote]
There may be people doing it just for fun/practice/experience, but I think most of us took it for some talent search or other. (The only search I know uses it is Johns Hopkins, but there may be others.)

[quote="h_s_potter2002"]Yeah, there's one by Northwestern, Duke, Colorado, and maybe more.[/quote]

..dont forget JHUCTY = John Hopkins


geometry
geometric transformation
rotation
analytic geometry
vector
LaTeX

Consider a small ring of mass m fitted on a frictionless rod of length l.Initially the rod is rotating with a constant angular velocity about one of its end in the horizontal plane  with the ring at its centre and given an initial velocity v towards the other end of the rod.Will its velocity increase with time ?Can you explain the answer in inertial frame?[/img]

Let particle be at centre of rod at time t=0.

At a later time t suppose the particle is distant r from the centre and that the rod has turned through an angle $\theta$. Assume rotation is anti-clockwise.

We know that $\dot{\theta}$  is constant =$\omega$ say .

Using  polar  coordinates,  the  velocity  component  of the particle  along the rod $\dot{r}$ is v since there are no forces acting in this direction.

The velocity component of the particle in a direction at right angles to the rod in the direction of increasing $\theta$ is r$\dot{\theta}$ or r$\omega$

Thus the resultant velocity is $\sqrt (v^{2}+(r\omega)^{2})$which increases with r.

.[quote]Using polar coordinates, the velocity component of the particle along the rod  is v since there are no forces acting in this direction
Although there is NO FORCE in inertial frame along radial direction still its velocity in this direction seems to change when we consider a centrifugal force in rotating frame.In my question I was chiefly intrested about its velocity in radial direction.
What is your opinion on this?[/quote]

Actually, it needs a force of magnitude mv^2/r, directed inward, just to stay in place.

Let + denote outward direction and - denote inward direction.

Then, -ma = -F - mv^2/r.

Therefore, ma = F + mv^2/r.

But, since F = 0, ma = mv^2/r,
and therefore, a = v^2/r.

So, now we have dv/dt = v^2/r,
therefore, d^2(r)/dt^2 = (dr/dt)^2/r.

Then, switching to primes for aesthetic reasons,
r'' = (r')^2/r, so r''/r' = r'/r.

Integrating both sides, ln(r') = ln(r) + C.

Then, let k = e^C, so r' = k*r.

Clearly, the solution is of the form e^(kt),
so r' = v*e^(kt), which satisfies the given initial velocity.

Since C was real, k > 0, and therefore the velocity is increasing as it moves outwards.

very simple.  Since the ring is moving out the rod will slow in rotational velocity so it will lose kinetic energy.  That energy must go somewhere and since there is no friction it goes into the ring which increases its velocity.

[quote]demji wrote:
very simple. Since the ring is moving out the rod will slow in rotational velocity so it will lose kinetic energy. That energy must go somewhere and since there is no friction it goes into the ring which increases its velocity[/quote]
will its radial speed increase?
try to solve the problem by assuming that the rod's angular velocity is maintained constant.

[quote]demji wrote:
the radial speed will increase thats what i said by the ring's velocity will increase. Its a similiar effect as when a ice skater is spinning and pulls in their arms, when they pull them in they spin faster which gives them a higher KE. This increase in KE is caused by the work required to pull the arms in. In this case the situation is run in reverse. Since the ring is moving out the rod spins slower and this slowing, loss of KE, means that a force must be put somewhere. The only place it can go is into the ring in a radial direction. Since there is no friction to overcome this directly leads to an acceleration in the ring's radial velocity.
[/quote]
As Michael Niland wrote,its the normal component of velocity which increases.Even if you assume that the rod's K.E. decreases(actually I intended to maintain its angular velocity constant),the ring's K.E. increases due to an increase in normal component of its velocity as the rod doesn't provides it any force in radial direction but with only a normal force. :!:

[quote="Anurag Agrawal"][quote]PTynan89 wrote:
Actually, it needs a force of magnitude mv^2/r, directed inward, just to stay in place.

Let + denote outward direction and - denote inward direction.

Then, -ma = -F - mv^2/r.

[/quote]What is the force which provides mv^2/r ?It is not a force by itself;rather it is the resultant of the forces acting on the ring.
         -ma=-mv^2/r  just because a=v^2/r and not because latter is a force. :)[/quote]

I was using the general form of the equation.  It could also be stated as F = ma - mv^2/r.

But, there is obviously work done on it to make it accelerate outwards.  Normally, the rod would slow down as energy is transferred to the ring as it moves outwards.  However, there must be an additional source doign work on the rod to keep it rotating at the same rate.  This is adding energy to the system, thus accounting for the added energy of the ring as it moves outwards.

but for the angular velocity to remain constant a force must be applied since the moment of inertia is increasing, actually a torque, and that torque then translates into a radial force of the ring which then causes an acceleration.  Either way you look at it you get the right result.

how can its radial speed increase whedn the rod can only provide it with a NORMAL force.Actually their is NO FORCE in the inertial frame and thats what my ques. is all about.

Can someone help me?I'm still not able to get this problem.

now if there is like a ditch or something that prevents the ring from falling, then there is a tangental force acting on it (the side will be "pushing" the ring). the angular momentum must increase:
$\tau=\frac{dL}{dt}$
well if$\omega$stays constant:
$rF=\omega 2mr\frac{dr}{dt}$
$\frac{dr}{dt}=kF$

well... is the force from the side increasing? in a way, the further the ring is from the center... the greater the momentum the side has and the greater the impact...(need to think about it for some time for a better explaination)

[quote]
Tim_Lou wrote:
i dont think this problem makes sense... there is no force but normal and gravity acting on the ring, what is keeping it from falling off the rod????

it should fall off the rod and continue moving in constant velocity.
[/quote]
Now observe the problem in a frame rotating with the rod. :maybe:
you will find that the centrifugal force is increasing its velocity. :o

in a rotating frame there are more forces than the centrifugal force, known as the Coriolis force.
http://scienceworld.wolfram.com/physics/CoriolisForce.html

notice the cross product of v and omega, the object will go side way in the rotating frame. (and fall off the rod)

the full equation for newton's second law in rotating frame is:
$m\vec{a}=\vec{F}+2m\vec{v}\times{\vec{\omega}}+m(\vec{\omega}\times{\vec{r}})\times{\vec{\omega}}$

the first term involving omega is the coriolis force, the second term involving the two cross product is the centrifugal forces.

I think you have not understood the question.There is a small circular ring an a rod is fitted well inside it and the ring is at its centre;how can it fly off the rod???
Now the coriolis force is  2mv cross omega which is along normal to the rod and is compensated by the normal reaction from the rod hence only a centrifugal force of magnitude mw^2r
acts outwards as vectors w and r are perpendicular and their cross product with vector w is parallel to vector r.
Retry this if you think you had understood it wrongly or please give reasons if you still think that you were correct.

the thing is, coriolis force is not a real force, it only exists in the [b]rotating frame[/b]. in the rotating frame, you can say coriolis force cancels out with the normal force thus centrifugal force is making it fly!

but the thing is, in an [b]inertial frame[/b], there is no coriolis force, (in the context of the equation above). [b]Nothing[/b] cancels the normal force! in fact, it is the only force acting on the ring, hence making it fly!

these forces and acceleration are all related in some mathematical way. the normal can be derived using polar coordinate and the fact that the force does not result in angular acceleration. (in an inertial frame)

Conclusion: Coriolis force and Centrifugal only exist in Non-inertial frames. They are not "real" forces.

Conclusively,do you want to say that it is the normal force which is increasing the ring's radial speed until it fall's off the rod(viewed from inertial frame)?because we cannot deny the fact that its radial speed is increasing when we view it in rotating frame.
If this is true,tell me how the normal force will increase its radial speed(viewed from inertial frame).
If you think that its radial speed will be constant(as Micheal Niland wrote)please explain the case in rotating frame of reference(as we have got a net centrifugal force which says that its radial speed should increase)
My concepts about the fact that coriolis and centrifugal are not 'real' forces were clear.

in polar coordinates, we know that (in inertial system):
$\vec{F_{r}}=m[\frac{d^{2}r}{dt^{2}}-r(\frac{d\theta}{dt})^{2}]*\vec{r}$
 (*sry, i dont know how to do dot notation in latex)

we know that this radial force is zero. $\frac{d^{2}r}{dt^{2}}=r(\frac{d\theta}{dt})^{2}$

and
$\frac{d^{2}r}{dt^{2}}$ must be positive.

also:
$\vec{F_{\theta}}=m[r\frac{d^{2}\theta}{dt^{2}}+2(\frac{dr}{dt})(\frac{d\theta}{dt})]*\vec{\theta}$

angular acceleration is zero, hence, the normal force:
$\vec{F_{\theta}}=2m(\frac{dr}{dt})(\frac{d\theta}{dt})*\vec{\theta}$

omega is constant, dr/dt is positive.
if calculated in the rotate frame, the result will be the same.

Conclusion, yes the increasing normal force accelerates the ring, making it to fly outward.

edit: you can't integrate the equations directly...they turned out to be some second order differiential equations. they aint too hard to solve though.

But looking at one instant of time we see that normal force is perpendicular to the radial speed.
So how could you explain the result that normal force is increasing its radial speed? :D

well... the results follow from mathematical derivations... angular momentum has to increase... the mathematical relationship implies that the radial velocity has to be increasing if angular velocity stays the same... I guess you just have to work it out mathematically.
 :wink:

*the equations i use can be easily derived by differientiating the position vectors in polar coordinate form twice.


geometry
vector
trigonometry
linear algebra

I havnt done linear algebra in a while and would like someone to push me in the right direction to this previous problem i encountered. Its very basic but i would like to see how others do it since i ended up using geometry and not vectors for my solution.

Question:
A swimmer wishes to swim across a river which is flowing at 4miles/h. The swimmer can swim at 5miles/h. Find the direction the swimmer must swim to go straight across the river. If the river is 0.5 miles wide at the point of crossing, how long will it take the swimmer to get across the river?

Any suggestions? This is preperation for a test

not sure that i would call this linear algebra, but here is how you would normally go about solving this problem

if we are looking at the river flowing from top to bottom, and the swimmer must swim from left to right. the question you need to ask is, at what angle above horizontal does he need to swim so that his vertical speed is exactly canceled out by the downward flow of the water, so his effective direction of motion is straight across.

you'll need to use some trig here.


All the digits used to number the pages of a mathematics magazine were placed in a sack. If the sack contained 297 digits, how many
numbered pages did the magazine have?

of the 297, $ 9$ are 1 digit, $ 90*2\equal{}180$ are 2 digit, leaving 108 digits for 3 digit numbers. $ \frac{108}{3}\equal{}36\plus{}9\plus{}90\equal{}\boxed{135}$


function
complex analysis
complex analysis unsolved

Consider the function $ G(x,y(x)) \equal{} 0$ and differentiate both sides with respect to $ x$.  Then rearrange.  This requires one application of the chain rule.  Also, this does not belong to the subject of complex analysis for it to be posted here.


PROVE $ \cos 3a\equal{}4\cdot \cos^{3}a\minus{}3\cos a$
$ f\sqrt[u]{n}!^{!}_{!}!\longrightarrow M\alpha T\sqrt[h]{eM\sqrt{a}t_{i}^{c^{S}}}$$ F^{T}W$
...yildobekir...

prove $ \cos a\plus{}\cos b\equal{}2\cdot \cos \frac{a\plus{}b}{2}\cdot \cos \frac{a\minus{}b}{2}$

$ f\sqrt[u]{n}!^{!}_{!}!\longrightarrow M\alpha T\sqrt[h]{eM\sqrt{a}t_{i}^{c^{S}}}$$ F^{T}W$
...yildobekir...

Another approach would be to expand $ (\cos a\plus{}i\sin a)^3$ and then take the real part.

For the second one, Expand $ \cos(a\plus{}b)\plus{}\cos(a\minus{}b)$, simplify, and then make a substitution to get rid of the $ a\plus{}b$ and $ a\minus{}b$ parameters.

[quote="yildobekir"]PROVE $ \cos 3a \equal{} 4\cdot \cos^{3}a \minus{} 3\cos a$
[/quote]
[hide="solution"]
$ \cos 3a$

$ \equal{} \cos(a\plus{}2a)$

$ \equal{} \cos a \cdot \cos 2a \minus{} \sin a \cdot \sin 2a$

$ \equal{} \cos a ( \cos^2 a \minus{} \sin^2 a ) \minus{} 2\sin^2 a \cos a$

$ \equal{} \cos^3 a \minus{}3(1\minus{}\cos^2 a ) \cos a$

$ \equal{} 4 \cos^3 a \minus{} 3 \cos a$ [/hide]
Have a look at these similar [hide="identities"]

\[ 1. \sin 3a \equal{} 3 \sin a \minus{} 4 \sin^3 a\]
\[ 2. \tan 3a\equal{} \frac{3\tan a\minus{}\tan^3 a}{1\minus{}3\tan^2 a}\]
\[ 3. \cot 3a\equal{}\frac{\cot^3 a\minus{}3\cot a}{3\cot^2 a\minus{}1}\]
[/hide] which solutions are similar  :)

[quote="yildobekir"]prove $ \cos a \plus{} \cos b \equal{} 2\cdot \cos \frac {a \plus{} b}{2}\cdot \cos \frac {a \minus{} b}{2}$
[/quote]

[hide="solution"]
Since the following identities are true
\[ (1) \ \sin(a\plus{}b) \plus{}\sin(a\minus{}b)\equal{}2\sin a \cos b\]
\[ (2) \ \sin(a\plus{}b) \minus{} \sin(a\minus{}b)\equal{}2\sin b \cos a\]
\[ (3) \ \cos(a\plus{}b) \plus{}\cos(a\minus{}b)\equal{}2\cos a \cos b\]
\[ (4) \ \cos(a\plus{}b) \minus{} \cos(a\minus{}b)\equal{}\minus{}2\sin a \sin b\]
If we note $ a\plus{}b\equal{}\gamma$  and  $ a\minus{}b\equal{}\beta$
therefore
\[ a\equal{}\frac{\gamma\plus{}\beta}{2} \ \cdots (*)\] and
\[ b\equal{}\frac{\gamma\minus{}\beta}{2} \ \cdots (**)\]
If we substitute $ (*),(**)$ in $ (1),(2),(3)$ and $ (4)$ we obtain the following identities:
\[ \sin\gamma \plus{}\sin\beta\equal{}2\sin (\frac{\gamma\plus{}\beta}{2}) \cos( \frac{\gamma\minus{}\beta}{2} )\]
\[ \sin\gamma \minus{} \sin\beta\equal{}2\cos (\frac{\gamma\plus{}\beta}{2}) \sin( \frac{\gamma\minus{}\beta}{2} )\]
\[ \cos\gamma\plus{}\cos\beta\equal{}2\cos (\frac{\gamma\plus{}\beta}{2}) \cos( \frac{\gamma\minus{}\beta}{2} )\]
\[ \cos\gamma \minus{}\cos\beta\equal{}\minus{}2\sin (\frac{\gamma\plus{}\beta}{2}) \sin( \frac{\gamma\minus{}\beta}{2} )\]

$ (3)$ denotes your problem  :)
[/hide]

Here's an elaboration on that complex number idea.

By DeMoivre, $ \cos 3a+i\sin 3a=(\cos a+i\sin a)^3$

Therefore:

\begin{eqnarray*}
\cos 3a&=&\text{Re}\;\{(\cos a + i\sin a)^3\}\\
&=&\text{Re}\;\{\cos^3a-3\cos a\sin^2 a\pm\text{(stuff involving }i)\}\\
&=&\cos^3 a-3\cos a(1-\cos^2 a)\\
&=&\boxed{4\cos^3a-3\cos a}
\end{eqnarray*}


If $a,b,c$ are positive numbers such that $ab+bc+ca=3$, then

$\frac{\sqrt a}{b+c}+\frac{\sqrt b}{c+a}+\frac{\sqrt c}{a+b}\geq \frac{3}{2}$.

It's not hard, although not very easy.

By Holder Inequality \[\left(\sum_{sym}\frac{\sqrt{a}}{b+c}\right)^{2}\left(\sum_{sym}a^{2}(b+c)^{2}\right)\ge (a+b+c)^{3}\] So it's enough to prove that \[\frac{4}{9}(a+b+c)^{3}\ge \sum_{sym}a^{2}(b+c)^{2}=9-abc(a+b+c) \ \ \ (*)\] By Schur Inequality, \[a^{3}+b^{3}+c^{3}+3abc\ge ab(a+b)+bc(b+c)+ca(c+a)\] which yields that \[(a+b+c)^{3}\ge 4(a+b+c)(ab+bc+ca)-9abc\] Substituting it to (*), we get \[abc(a+b+c)+\frac{8}{3}(a+b+c)-2abc \ge 9\] which is exactly the sum of \[abc(a+b+c) \ge 3abc\]  \[\frac{8}{3}(a+b+c)+abc \ge 9\] (the second is easy by mixing variables, or something else, the general problem was posted here and not hard, the first is easy).

 :P We are done.

There is also a nice solution by Cauchy-Schwarz Inequality. :P

The Cauchy-Schwarz way permits us to find the smallest $p$ such that the inequality below holds for $ab+bc+ca=3$:

$\frac{a^{p}}{b+c}+\frac{b^{p}}{c+a}+\frac{c^{p}}{a+b}\geq \frac 3{2}$.

Ok, it's also not hard I think

By C-S inequality \[\left(\frac{a^{p}}{b+c}+\frac{b^{p}}{c+a}+\frac{c^{p}}{a+b}\right)\left(a(b+c)+b(c+a)+c(a+b)\right) \ge \left(a^{\frac{p+1}{2}}+b^{\frac{p+1}{2}}+c^{\frac{p+1}{2}}\right)^{2}.\] Now we must find the best constant $p$ for which \[a^{\frac{p+1}{2}}+b^{\frac{p+1}{2}}+c^{\frac{p+1}{2}}\ge 3\] which is not hard because I remember that (by mixing variables) \[a^{\frac{p+1}{2}}+b^{\frac{p+1}{2}}+c^{\frac{p+1}{2}}\ge \min \left(3,2.3^{\frac{p+1}{4}\right).}\] So the best constant p is \[3^{\frac{p+1}{4}}=\frac{3}{2}\leftrightarrow p\approx 0.4763.\]


given the sequence ${x_{n}}$ satisfies:
              $x_1=1,x_2=0,x_3=2,x_{n+1}=2*x_{n-1}+x_{n-2}   (n \geq 3)$
prove that for all $m \in N$ there exist infinitely two consecutive terms of the sequence such that two terms are divisible to m.
 :D  :D

We see that $x_0=x_{-1}=0$. And we can reconstruct in reverse order all terms if we know $x_n,x_{n+1},x_{n+2}$. Since number of triple $(x_n,x_{n+1},x_{n+2})$ is finite modulo $m$, we conclude that sequence is periodic modulo $m$. Hence the result.

yes,you are right.Your solution is similar my solution. :)


The number of circular pipes with an inside diameter of 1 inch which will carry the same amount of water as a pipe with an inside diameter of 6 inches is:
\[ \text{(A)}\ 6 \pi \qquad \text{(B)}\ 6 \qquad \text{(C)}\ 12 \qquad \text{(D)}\ 36 \qquad \text{(E)}\ 36 \pi \]

[hide="click to reveal answer"]D, 36 assuming they are the same height.  Compute the volume by finding the radius and using bh.  Then divide the two volumes (9/(1/4), ignoring pi) to get the answer.[/hide]

[hide] assuming they have the same height, the base area of the first one is (1/2)^2=1/4 and the base area of the second one is 3^2=9.  9/(1/4)= 36.  Thus, the answer is D. [/hide]

[hide](D)
[hide]Assuming that the pipes are of same length, l,

the ratio of the volume =( Pi(6)^2l)/(Pi(1)^2l)
                                      = 36  (D)[/hide][/hide]


LaTeX
function
number theory
modular arithmetic

$ 21^1 \equal{} 21 mod 100.... 21^2 \equal{} 41 mod 100..... 21^3 \equal{} 61 mod 100 .....21^4 \equal{} 81 mod 100 ....21^5 \equal{} 1 mod 100$
so $ 21^705 \equal{} (21^5)^141 \equal{} (1 mod 100)^141 \equal{} 1 mod 100$
therefore the answer of the tens digit is $ 0$

Btw, how do you latex the hide thinggy?

EDIT: cant seem to seperate the 100 and 21

Note that $ \lambda(100)\equal{}20$ and $ 705\equiv 5\bmod 20$, so $ 21^{705}\equiv 21^5\equiv 21\cdot 41^2\equiv 21\cdot (\minus{}19)\equiv \boxed{0}1\bmod 100$. (Where $ \lambda(n)$ denotes Carmichael's Lambda Function)

If you really insist on not using modular arithmetic:

$ (20\plus{}1)^{705}\equal{}\sum_{k\equal{}0}^{705}\binom{705}{k}2^k\cdot 10^k$

If we only want the tens digit, we only need $ k\leq 1$. So we sum $ 1\plus{}705\cdot 20\equal{}141\boxed{0}1$


inequalities
calculus
inequalities solved

Sorry,it is wrong. I have made a terrible error.

actually a nice Cauchy does the trick:

S.\sum (a4 + a2b2 + a2c2 + 2a2bc ) \geq ( \sum a2 )2  which means that

S \geq (\sum a2)2 / (\sum a4+2\sum a2b2+2abc \sum a) \geq 3/5 that is

5 \sum a4 +10 \sum a2b2 \geq  3 \sum a4 + 6 \sum a2b2 + 6abc \sum a

and this follows easily from \sum a4  \geq \sum a2b2 \geq abc \sum a (multiply by two, bring everything in the LHS and everything will group into squares of differences).

Here's what I did:

We amplify each fraction to get S= \sum a^3/(a+2abc) >= 3/5. I used that famous lemma (used like a hundred times here):  \sum a^2/x >= ( \sum a)^2/( \sum x). Here's what we get: S >= ( \sum a^(3/2))^2/[(a+b+c)(a^2+b^2+c^2)+6abc], so it wouyld be enough to prove
( \sum a^(3/2))^2/[(a+b+c)(a^2+b^2+c^2)+6abc] >= 3/5, but this is pretty easy (we just expand everything and then use some 'smart' applications of AM-GM. Maybe I'll post it all tomorrow; I'm too tired for that right now...

Here is my approach:
We have  a^2/(1+2bc) =  a^2/(a^2 + b^2 + c^2 +2bc) >= a^2/(a^2 + 2b^2 + 2c^2). We will show that P = Sum a^2/(a^2+2b^2+2c^2) >=3/5 (1). Putting  A = a^2+2b^2+2c^2, B = 2a^2+b^2+2c^2, C = 2a^2+2b^2+c^2, we get a^2 = (2B+2C-3A)/5 ... and  (1) becomes
(2B+2C-3A)/5 + (2A-3B+2C)/5 + (2A+2B-3C)/5 >= 3/5 <=> (B/A + A/B)+(A/C+C/A)+(B/C+C/A) >= 6 which is trivial.

This approach comes from famous proof of Shapiro inequality for n=3: a/(b+c)+ b/(c+a) + c/(a+b) >= 3/2 and can be useful in some cases.

Namdung

Sum a^2/(a^2+2b^2+2c^2) >=3/5 can also be proved like this (so can the Shapiro ineq for n=3):

let a^2=x, b^2=y, c^2=z. We want
 \sum x/(x+2y+2z) >= 3/5. We have x+y+z=1 (if we didn't have this we could consider we did). Let f(x)=1/x. f is convex, so we can apply Jensen:
 \sum xf(x+2y+2z) >= f(x^2+y^2+z^2+4(xy+yz+zx)), which is, pretty obviously, >=3/5 (we use xy+yz+zx<=x^2+y^2+z^2 and x+y+z=1).

Well, well, ... this standard inequality gives a lot of interesting solutions :D

In fact, this inequality is extremely simple. It is trivial to prove that 1+2bc>=0, etc. So, by Cauchy we have:
   sum a^2/(1+2bc)=sum a^4/(a^2+2bca^2)>=(sum a^2)^2/(sum a^2+2abc(a+b+c))=1/(1+2abc(a+b+c))>=3/5 since abc(a+b+c)<=sum a^2b^2<=(sum a^2)^2/3=1/3.

Well, I think the fact that it's so simple is one of the main reasons why it has received so many solutions... :D


geometry
calculus
integration
area of a triangle
Heron\u0027s formula

The smallest(in area) triangle with integer area and consecutive integer sides has sides $3,4,5$. The next larger such triangle has sides $13,14,15$. Compute the three sides of the next larger such triangle, given that its "middle" side is a multiple of $4$.

[hide="Solution"]
Let $x$ be the length of the middle side, then the other two sides of the triangle are $x+1$ and $x-1$. Now using the Heron's Formula, u get the area of the triangle equals to $\frac{x\sqrt{3(x^2-4)}}{4}$. Note that x is divisible by 4 and the area of the trianlge should be an integer. Considering these things, u will get $x=52$[/hide]

i gott the sides as 51,52 and 53..
Let them be (n-1),n,(n+1)

oops!!!msg posted by accident..
let the sides be (n-1) n and (n+1)
then area=(n/4)*sqrt[3*(n^2-4)]
thus (n^2-4) is of the form 3*x^2..where n=4k
thus 16k^2-4=3x^2
put k^2=169 we get x=30 and area=90*13
n=4*k=52..thus gettin the sides as 51,52 and 53..

Agolsme: here's my solution from a while ago. Hope it makes sense.  :)

[hide]Let the middle side has length $j$.

By Heron's formula, we have:
$A=\sqrt{\left(\frac{3j}{2}\right)\left(\frac{j}{2}\right)\left(\frac{j-2}{2}\right)\left(\frac{j+2}{2}\right)}$

$\Rightarrow \left(\frac{3j}{2}\right)\left(\frac{j}{2}\right)\left(\frac{j-2}{2}\right)\left(\frac{j+2}{2}\right)=m^2$

Rearranging, we have: $\frac{(3j^2)(j^2-4)}{16}=m^2$.

In order for $m^2$ to be integral, $j$ must be even.

As a result, we let the middle side $j$ of the triangle be $2n$, and the others be $2n-1$ and $2n+1$. We then have the area, by Heron's Formula, as: $A= \sqrt{(3n)(n)(n-1)(n+1)}$.
So we want $(3n)(n)(n-1)(n+1)=m^2$ for some integers $m,n$

Now $n^2$ on the LHS is already a perfect square, and the RHS becomes $3k^2$ after getting rid of the $n^2$ on the LHS, and realizing multiples of $3$ must come in odd powers. So we have:
$n^2-1=3k^2$
$n^2-3k^2=1$
$(n-k\sqrt{3})(n-k\sqrt{3})=1$

Now the first solution to this is $(n,k)=(2,1)$, so $(2-\sqrt{3})(2+\sqrt{3})=1$. This yields our $3-4-5$ triangle, since the middle side is $2n$. Its area is $6$

Squaring both sides of this equation, as before, we have:
$(7-4\sqrt{3})(7+4\sqrt{3})=1$. This leads to the solution $(n,k)=(7,4)$, which yields our $13-14-15$ triangle, with area $84$.

Cubing both sides of this equation, as before, we have:
$(26-15\sqrt{3})(26+15\sqrt{3})=1$. This leads to the solution $(n,k)=(26,15)$, which implies the middle side is $52$, and we have a $\boxed{51-52-53}$ triangle and the area of our triangle is: $A=\sqrt{(78)(26)(25)(27)}=1170$.

Just for good measure, we'll see that our method works for the fourth power. We raise our original expression to the fourth power, and we have: $(97-56\sqrt{3})(97+56\sqrt{3})=1$.   This leads to the solution $(n,k)=(97,56)$, which implies the middle side is $112$, and we have a $193-194-195$ triangle and the area of our triangle is: $A=\sqrt{(168)(56)(55)(57)}=16296$.

So we have not only found the third triangle, but a general way to find the $n^{th}$ triangle of this form: find the rational part of $((2-\sqrt{3})(2+\sqrt{3}))^n$, and double it to get the middle side, and then use Heron's to solve for the area.[/hide]


If $ x,y,z,w \in R^ \plus{}$, where $ 0 < A \leq Min${$ x/z , z/x$},
Then:
$ 2 \leq (y^2)/(x^2 \plus{} Ay^2) \plus{} (z^2)/(y^2 \plus{} Az^2) \plus{} (w^2)/(z^2 \plus{} Aw^2) \plus{} (x^2)/(w^2 \plus{} Ax^2)$

---
Gyu Lim and Sunkyu Lim(sleeper :p).jeje friends .Bye

Hi!!!I'm Sunkyu Lim~

Wow it's been very long time to see you. :lol:

I'm really sorry to see this post after it has been posted 6 months ago  :blush:

Thanks for your posting and not forgetting my name.

Let's see you in Germany this year :lol:

But i'm doubt if i can go there--;;;;;

Bye and take care~


In a set of four consecutive positive integers, the smallest integer is less than half the largest integer. What is the largest integer that could be in the set?

If we let $ x$ be the largest integer in the set, we get \[ x\minus{}3<\frac{x}{2} \Rightarrow x<6\], so the largest possible value of $ x$ is $ \boxed{5}$


trigonometry algebra polynomial function logarithms

cos(x) + x - 21 = 0

How would one go about solving the above equation in an algebraic method (not using approximation)

Thanks.

Most "transcendental" equations of this nature do not have algebraic solutions.

The only thing I can think of is an approxiamtion using Taylor Series, but that involves calculus.

One could also use $ x\equal{}21\minus{}\cos x$ and use recursion (i.e. $ x_{n\plus{}1}\equal{}21\minus{}\cos x_n$). As this is a contraction mapping (I think), it will converge. ;)

Is it proven that there exists no algebraic solution, or do you think that some time into the future one will be found?

I think it's proven... $ \cos a$ is transcendental for algebraic $ a$ which implies that if $ a$ with $ \cos a \minus{} a \plus{} 21$ is algebraic, then $ a\minus{}21$ is transcendental which is impossible. :)

The discussion here conflates two meanings of the word "algebraic" which need to be separated.

bertram is using the word "algebraic" to mean "is the root of a polynomial with rational (equivalently, integer) coefficients."  In other words, solving for $ x$ reduces to solving a polynomial.  This is a [b]very tight[/b] condition.

Karan Bhangui is using the word "algebraic" to mean "has a closed form in terms of elementary functions" (at least, if I interpret the question correctly), which is both a much looser condition and does not contain the above condition (algebraic numbers of degree five and higher cannot, in general, be expressed in terms of elementary functions (unless you consider elliptic functions elementary)).  For example, the equation $ e^x \equal{} 3$ has a solution that has a closed form in terms of elementary functions ($ x \equal{} \ln 3$) because we defined $ \ln x$ to be an "elementary function" (whatever that means).

The real question you're asking, then, is whether the inverse of the function $ \cos x \plus{} x$ is an "elementary" function.  A cheap solution is simply to define a new function $ g(x)$ to be the inverse of $ \cos x \plus{} x$ (on the appropriate interval), call it an elementary function, and then the solution is $ a \equal{} g(21)$.

In fact, such an approach has been used to "solve" a related problem:  to solve an equation like $ xe^x \equal{} C$, the Lambert W-function $ W(x)$ has been defined to be the inverse of $ xe^x$, so the solution is $ x \equal{} W(C)$.

Hmmm I never thought of it like that, cool stuff. Defining inverses is a pretty lazy thing to do I guess, but at least you can manipulate them to yield some good properties :)


[quote="Treething"]1) tayl0r has eight fingernails 1 mm. in length and the average length of all of his fingernails is 2 mm. What is the average length of his other two fingernails?

2) Timmy is two and a half times older than his son. If Timmy had his son at age 33, how old is his son?

3) One can of Hufable's Spray can paint 5 square feet of wall. How many cans of Hufable's Spray need to be bought if Ahmed wants to paint a mural 38 feet long and 9 feet high?

4) A can of strawberry yogurt is 8 ounces and consists of 5% strawberries, 35% milk, and 60% sugar. How many ounces of sugar need to be mixed in if Nutz wants his strawberry yogurt to be 80% sugar?

special question: Treething takes six seconds to say "negative two" while Ardito takes one second to say "negative two". Given that it takes both Treething and Ardito half a second to say "two", how many times longer does Treething spend saying "negative" than Ardito, assuming that the pause between the two words is negligable?[/quote]

[hide]
problem 2:
$33+x=2.5x$
$33=1.5x$
$x=22$
His son is 22 yrs old[/hide]

[hide]1: $\frac{8+2x}{10}= 2.$ Average of other 2 is 6 mm.  :w00t:

Special: 11 times longer.[/hide]

[hide]3. $\frac{38 \cdot 9}{5}= 68.4.$ Ahmed needs 69 cans to paint the mural.

4. To make it 80%, the amount of sugar has to be 4 times the amount of everything else added together. That's 12.8, or 8 more ounces.[/hide]

[quote="bpms"][quote="Treething"]1) tayl0r has eight fingernails 1 mm. in length and the average length of all of his fingernails is 2 mm. What is the average length of his other two fingernails?

2) Timmy is two and a half times older than his son. If Timmy had his son at age 33, how old is his son?

3) One can of Hufable's Spray can paint 5 square feet of wall. How many cans of Hufable's Spray need to be bought if Ahmed wants to paint a mural 38 feet long and 9 feet high?

4) A can of strawberry yogurt is 8 ounces and consists of 5% strawberries, 35% milk, and 60% sugar. How many ounces of sugar need to be mixed in if Nutz wants his strawberry yogurt to be 80% sugar?

special question: Treething takes six seconds to say "negative two" while Ardito takes one second to say "negative two". Given that it takes both Treething and Ardito half a second to say "two", how many times longer does Treething spend saying "negative" than Ardito, assuming that the pause between the two words is negligable?[/quote]
I remember my coach saying that he was rooting for ardito in the finals because he saw that you were using cheap tricks :D[/quote]

ahaha. the reactions towards my cheap tricks was more mixed than i thought it'd be, there were actually people congratulating me on my tactics instead of 100% "boo you suck give ardito the trophy"... honestly i'm not all that great at countdown, I just have the "mental dexterity" to think of "varied" "tactics" on the spot

and yeah, 3 and 4 are correct, 3 is correct without any doubt :)

[quote="b-flat"]How is number 1 possible?  You only have 8 fingers to have nails on them...  if you include thumbnails, however

[hide="thumbnails included"]The total must be $2\times10=20$ millimeters, and of that, we know there are $8$ on 8 fingers, leaving 2 fingers for the other $12$ mm, making an average of $6$.[/hide][/quote]

Thumbnails are included, of course, and you are correct.

i obviously have two more fingers than b-flat :o

yeah your answers seem correct if i didn't say it yet.

[quote="Treething"]1) tayl0r has eight fingernails 1 mm. in length and the average length of all of his fingernails is 2 mm. What is the average length of his other two fingernails?

2) Timmy is two and a half times older than his son. If Timmy had his son at age 33, how old is his son?

3) One can of Hufable's Spray can paint 5 square feet of wall. How many cans of Hufable's Spray need to be bought if Ahmed wants to paint a mural 38 feet long and 9 feet high?

4) A can of strawberry yogurt is 8 ounces and consists of 5% strawberries, 35% milk, and 60% sugar. How many ounces of sugar need to be mixed in if Nutz wants his strawberry yogurt to be 80% sugar?

special question: Treething takes six seconds to say "negative two" while Ardito takes one second to say "negative two". Given that it takes both Treething and Ardito half a second to say "two", how many times longer does Treething spend saying "negative" than Ardito, assuming that the pause between the two words is negligable?[/quote]

[hide="question 3"]
$\frac{38\cdot 9}{5}=68.4$
since you can't have 2/5 of a can, you will need 69 cans[/hide]


If you went to AMP 2007 and are going to Iowa ARML, I'm arranging a reunion as several people have asked me to do. As of right now, I'm pretty sure it'll be at Whitey's (the ice cream place) on Friday night, with more details to come later.

I am going to ARML first time this year to Iowa (don't know the ice cream place)

I went to AMP 2007 so see you there.

My name is quite revealing on my nickname or ID or whatever

Aha! The internet has come through!

Whitey's is at 112 East Washington

Here's a map:
http://www.mapquest.com/maps/112+East+Washington++Iowa+City++IA+52240/

I might come, but Wisconsin is practicing Friday night


yes isnt that obvous for the object to float....

thinking abt ur second question...

The answers are Yes, and upwards. It might help to think about the reaction to the bouyant force on each liquid.

[quote="hash_include"]a doubt regaurding buoyant force...

suppose a solid sphere is placed in a container that is filled with mercury and water, such that the sphere is EXACTLY half immersed in each..

the water layer is obviously above the mercury layer..

my doubts are:

1. will the water layer provide a buoyant force to the sphere??

2. If so, in what direction will it act?? upwards or downwards??[/quote]

Archimedes’ Principle states that an object is buoyed up by a force equal to the weight of the fluid it displaces.

It doesn’t matter whether some of this fluid covers the object or not. Think of a beach ball. You can force it to the bottom of a swimming pool, but if you let it go it will rise to the surface, rather abruptly too.

Your sphere immersed in the mercury-water bath will be buoyed-up by a force equal to:

	\[ F_B \equal{} m_{Hg}g \plus{} m_{H_2O}g \equal{}\rho_{Hg}\left(\frac{2\pi}{3}R^3\right)g \plus{}\rho_{H_2O}\left(\frac{2\pi}{3}R^3\right)g
\]

	\[ F_B \equal{} m_{Hg}g \plus{} m_{H_2O}g \equal{}\left(\rho_{Hg} \plus{}\rho_{H_2O}\right)\left(\frac{2\pi}{3}R^3\right)g
\]

@ shreyas and Dr No. :

But isn't buoyant force just a contact force, which results due to pressure differnces on different portions of an object??

So in that case, can you tell me exactly on which portion of the sphere will it act??
As far as i can visualise it, contact forces due to the water layer can only act downward  :maybe:
It's not just some magic force which suddenly just puses up a body right??
 :maybe:

bouyant force is nothinng but the force that teh surrounding partciles would have applied on a liquid that would have been there instead of the ball :)

ok.. so is it a contact force or not? :)

then your answer still doesn't convince me  :maybe:

in case of it being a contact force, as i've asked earlier, on which portion of the sphere does/can it act??

my main doubt lies there only  :(

see the condition is same as we replace the portion of sphere dissolved in water by water and that dissolved in mercury by mercury now whatever forces acting on this part keep it in equilibriumw ill act on the sphere also :)

[quote="pardesi"] now whatever forces acting on this part keep it in equilibriumw ill act on the sphere also :)[/quote]
i didn't understand this part. :(

and you still havent answered me da.. In the case of the mercury, we can say that buoyant force acts at the bottom most point of the sphere. But what about the water layer?? where does it act??

 :huh:

that's what i replied for...my go would be cannot be predicted all u can tell is that the forces acting on the sphere are same as that would have acted upon water+mercury had they been there in their respective portions :)
also the net bouyant force is equal to it's weight

ok ok fine, i kind of understood now... thanks a lot :)

actually if u have H.C.Verma do refer it...i think one of the very few books which explains this intricate fact carefully :)

The buoyant force comes from the difference of the pressure. That is the source of the archimedes principle.
Maybe we could do this problem by calculating the total difference of the pressure suffered by the ball.

[quote="hash_include"]@ shreyas and Dr No. :

But isn't buoyant force just a contact force, which results due to pressure differnces on different portions of an object??[/quote]

Static fluid pressure depends only on depth, i.e. $ \rho g h$.

Consequently there is a higher pressure on the bottom of a submerged object than on the top. This pressure difference results in an unbalanced upward force on the object. The unbalanced upward force is called buoyancy.

that's what i said u can't predict that nor can u say that the mercury layer will give an force only in th upward direction THAT'S WRONG.... :)

I don't know if my explanation can convince you. When you calculate the differece of the pressue on the surface of the ball, you will calculate the downward pressue on the top half of the ball exerted by water, and the upward pressue on the bottom half of the ball exerted by the mercury. But when you calculate the pressue exerted by the mercury, you will have to add the water's pressue on the mercury.So there is still a two-side pressue exerted by water, and you can consider that force the buoyant force of the water.

[quote="hash_include"]All that said and done, no one still has properly answered this question of mine, wherein my major lies  :huh:

[quote="hash_include"]
In the case of the mercury, we can say that buoyant force acts at the bottom most point of the sphere. But what about the water layer?? where does it act??

 :huh:[/quote][/quote]

There is no one specific point on a submerged object where pressure acts in order to create the unbalanced force. This is why Archimedes Principle says that the buoyant force is equal to the weight of the fluid displaced.

Consider the static pressure of some arbitrary fluid: $ P \equal{}\rho\, g y$, where $ y$ is positive downward.

Then $ \nabla P \equal{} \rho g\,\, \hat{u}_y$. This means pressure increases continuously with depth, and any two slices of an object at different depths will experience different forces.

i am still not convinced of ur explanation....

CAn u explain it properly???

i think maomao has given me an idea which may lead to the answer.
plz correct me if i'm wrong:

originally if the sphere is ONLY in mercury, then in equilibrium, it will not be exactly half inside the mercury. (it will be less than half inside the mercury).
Now, when the water is added, due to the pressure exerted by it(the water) on the sphere, it pushes the sphere down. so after it attains eqbm again, it is half immersed in both liquids.
Since the water pushes the sphere down of sorts, there is an extra buoyant force offered by the mercury which is caused due to the water. and we can say that force is the buoyant force  due to water..

hope every1 understood it :)  :maybe:


calculus
derivative
function
calculus computations

Find the critical points of p(p-1)^4.
Classify the critical points as local maxima, local minima, or neither.
What are the maximum and minimum values of this fuction on 0<x<1?

Don't post homework problems. What is it you're having trouble with?

alright.
after the first differentiation i set the equation equal to 0 and get p = 0.2 as a local minimum, but the answer said 0.2 is minimum and 1 is maximum.

You just wrote that you agreed with the book, and it's the wrong answer.

I agree with you. For $ f(p)\equal{}p(p\minus{}1)^4$, the sign of $ f'(p)$ is coincides with one of $ 1\minus{}5p\ (0<p<1)$.


algebra
polynomial
function
complex analysis
complex analysis unsolved

OK, so here are my thoughts, not sure if this is on the way to a correct solution...

$ f(z)$ can have at most finitely many zeros in a given ball (or else be identically zero by the identity theorem and the fact that it is entire). So let $ a_1,...,a_n$ be the zeros with respective multiplicities $ k_1,...,k_n$ and write $ f(z)\equal{}(z\minus{}a_1)^{k_1}...(z\minus{}a_n)^{k_n}g(z)$, where $ g(z)$ is entire and nonzero in $ D(0,1)$ and thus in all of $ \mathbb{C}$.

Now I may not be thinking hard at all, but the only entire functions I can think of with no zeros are either constant or of the form $ e^{h(z)}$; but $ e^z$ has an essential singularity at $ \infty$ where as our function does not (I think?), some kind of contradiction?

Sorry, it's late, if this is way off or makes no sense I apologize...

[quote="santug"]Suppose f(z) is an entire function, and  |f(z) | > 1  for |z|>1. Prove that f(z) is a polynomial[/quote]

Note $ |f(1/z)| > 1$ if $ 0 < |z| < 1$, hence $ |f(1/z)|$ cannot have an essential singularity at $ 0$; proceed from there.


induction
inequalities
combinatorics unsolved
combinatorics

Let $U, V$ be two subsets of $P(X)$ , $Card(X)=n$, such that
$a\in U, b\subseteq a$ implies $b\in U$
$a\in V, b\subseteq a$ implies $b\in V$
Prove that $Card(U). Card(V)\leq 2^{n}Card(U\cap V).$

We start with the following lemma.
[b]Lemma[/b]. For sets $U_{0}\subset U_{1}$ and $V_{0}\subset V_{1}$, \[|U_{0}\cap V_{1}|+|U_{1}\cap V_{0}| \leq |U_{0}\cap V_{0}|+|U_{1}\cap V_{1}|.\] [b]Proof[/b]. Follows from \[0 \leq |(U_{1}\setminus U_{0})\cap(V_{1}\setminus V_{0})| = |U_{0}\cap V_{0}|+|U_{1}\cap V_{1}|-|U_{0}\cap V_{1}|-|U_{1}\cap V_{0}|.\] $.$

We will prove the original statement by induction on $|X|$. For $|X|=0$, the statement is trivial. Suppose that it holds for $|X|=n$, we will prove it for $|X|=n+1.$

Let $X=X'\cup \{x\}$ where $|X'|=n$. Let $U, V$ be subsets of the power set $P(X)$ satisfying the given property (i.e., each of them is closed under taking a subset of its element) which we denote (*). Then \[U = \left\{ u\cup\{x\}: u\in U_{0}\right\}\cup U_{1}\] and \[V = \left\{ v\cup\{x\}: v\in V_{0}\right\}\cup V_{1}\] where $U_{0}, U_{1}, V_{0}, V_{1}\subset P(X').$
It follows from the property (*) of $U$ and $V$ that $U_{0}\subset U_{1}$ and $V_{0}\subset V_{1}$. Moreover, the sets $U_{0}, U_{1}, V_{0}, V_{1}$ also satisfy the property (*).

It is clear that \[U\cap V = \left\{ w\cup\{x\}: w\in U_{0}\cap V_{0}\right\}\cup \left(U_{1}\cap V_{1}\right)\] implying \[|U\cap V| = |U_{0}\cap V_{0}|+|U_{1}\cap V_{1}|.\] Using Lemma we have \[|U|\cdot |V|=(|U_{0}|+|U_{1}|)\cdot (|V_{0}|+|V_{1}) \leq 2\left(|U_{0}|\cdot |V_{0}|+|U_{1}|\cdot |V_{1}|\right).\] Then by induction, \[|U_{0}|\cdot |V_{0}|+|U_{1}|\cdot |V_{1}| \leq 2^{n}\cdot\left(|U_{0}\cap V_{0}|+|U_{1}\cap V_{1}|\right) = 2^{n}|U\cap V|.\] Plugging all together we have \[|U|\cdot |V| \leq 2\left(|U_{0}|\cdot |V_{0}|+|U_{1}|\cdot |V_{1}|\right) \leq 2^{n+1}|U\cap V|\] that finish the proof.

Sorry, it was a silly mistake of mine. :blush:

I think he proved the right result (notice that the inequality holds for $|X|=n+1$).


Help me solve this problem.
You are given a weighted tree(every edge has her weight). Problem is to compute for every vertex the distance from the furthest vertex.

What do you mean by furthest vertex?  The vertex with the longest shortest path to it?  In that case I guess a naive solution would be to use Floyd-Warshall ($n^3$) and then check each vertex for the furthest vertex ($n^2$)

Yes. The furthest vertex means the vertex with the longest shortest path to it. But $O(n^3)$ is slow because the time limit is 0.3sec and 1<=n<=10000

I think that the problem can be solved like Zorro solved Centroid. We assign value $f[xy]$ to every edge $xy$ and let $f[xy]$ be maximal distance $x$ from vertex $v$ witch is connected with $y$(path from $v$ to $y$ do not contain $x$). Then $f[xy]=a[xy] + max\{ f[y, v]_{x<>v \in adj(y)} \}$. From this easily compute what we need in complexity $O(n)$.


Pirate Pete shares his treasure with Pirate Paul in an interesting way. Pete first says, ``One for me, one for you,'' giving himself one coin and starting Paul's pile with one coin. Then Pete says, ``Two for me, and two for you,'' giving himself two more coins but making Paul's pile two coins in total. Next Pete says, ``Three for me, three for you'' while giving himself three more coins and making Paul's pile three coins in total. This pattern continues until Pete gives himself $ x$ more coins but makes Paul's pile $ x$ coins in total. At this point all of the coins have been distributed, and Pirate Pete has exactly four times as many coins as Pirate Paul. How many gold coins do they have in total?

If this takes $ n$ turns, then there are $ \frac{n(n\plus{}1)}2\plus{}n$ coins in total.

By the given, we must also have $ \frac{n(n\plus{}1)}2\equal{}4n\implies n\plus{}1\equal{}8\implies n\equal{}7$

$ \implies\frac{n(n\plus{}1)}2\plus{}n\equal{}\boxed{35}$.

After n distributions, Pete has $ 1\plus{}2\plus{} ... \plus{} n\equal{}\frac{n(n\plus{}1)}{2}$ coins. Paul clearly just has $ n$

$ \frac{n(n\plus{}1)}{2}\equal{}4n$

$ n(n\plus{}1)\equal{}8n$

$ n^2\plus{}n\equal{}8n$

$ n^2\minus{}7n\equal{}0$

$ n\equal{}0,7$

Obviously n is 7, meaning the total is 7+28=35.


Consider the equation \[\frac{xy-C}{x+y}= k ,\] where all symbols used are positive integers.

1. Show that, for any (fixed) values $C, k$ this equation has at least a solution $x, y$;

2. Show that, for any (fixed) values $C, k$ this equation has at most a finite number of solutions $x, y$;

3. Show that, for any $C, n$ there exists $k = k(C,n)$ such that the equation has more than $n$ solutions $x, y$.

Equation equavalent to
(1) $(x-k)(y-k)=C+k^{2}$.
Therefore, I think that $x,y,k,C$ are integers (else it had infinity solutions).
For integer solutions your statements is true and obviosly.


algebra
polynomial
function
algebra proposed

[quote="matex (L)(L)(L)"]Find all polynomials such that if for &a& $ p(a)$ is integer then $ a$ is integer.

thanks for help[/quote]

WLOG, we consider the case that the leading coefficient is positive first. (Obviously $ P(x)$ will not be a constant.)

Using the facts that $ P(x)$ is strictly increasing and that $ P(x)$ is a continuous function , we know the number of integer between the $ P(a\plus{}x)$ and $ P(a)$ $ \le x\plus{}1$ for all $ a,x$ so as to satisfy the condition.

But we know if $ P(x)$ is of degree $ \ge 2$ or of the form $ kx\plus{}c$, where $ k>1$, then $ P(a\plus{}x)\minus{}P(a)\minus{}(x\plus{}1)$ can be made arbitrarily large by choosing artirarily large $ x$, contradiction.

So $ P(x)\equal{}mx\plus{}b$, where $ 1 \ge m>0$. We have now $ P(x)$ is an integer $ \Leftrightarrow P(x\plus{}\frac{1}{m})$ is an integer also. So $ m\equal{}\frac{1}{n}$ for $ n \in N$

Now let $ P(x)\equal{}\frac{x\plus{}b}{n}$ then $ P(x)$ is an integer gives $ x\plus{}b$ is an integer , with $ x$ must be an integer gives $ b$ is an integer. And it's easy to check that $ P(x)\equal{}\frac{x\plus{}b}{n}$ or $ P(x)\equal{}\minus{}\frac{x\plus{}b}{n}$ , where $ n$ is natural, $ b$ is integer, satisfies the condition.


geometry
parallelogram
trigonometry
vector
geometric transformation
rotation
analytic geometry

What is the area of the triangle with vertices: (2, 1, 3) (-1,0,4) and (4,2,2)

This is to be done without a calculator. Thanks in advance!

$ A(2, 1,3),\ B(-1,0,4),\ C(4,2,2)\Longrightarrow [ABC]=\frac{1}{2}\sqrt{|\overrightarrow {AB}|^2{|\overrightarrow {AC}|^2}-(\overrightarrow {AB}\cdot \overrightarrow {AC})^2}$.

Would you mind elaborating as to how you derived that formula?

(If I'm guessing right, it has something to do with the area of the parallelogram?

$ [ABC] \equal{} \frac {1}{2}|\overrightarrow {AB}\parallel{}\overrightarrow {AC}|\sin \angle{BAC}$

$ \equal{}\frac {1}{2}|\overrightarrow {AB}\parallel{}\overrightarrow {AC}|\sqrt{1\minus{}\cos ^ 2 \angle{BAC}}$

The rest is left for U. :D

[hide="my way"]Let's find the side lengths! :D

$ AB\equal{}\sqrt{3^2\plus{}1^2\plus{}1^2}\equal{}\sqrt{11}$
$ BC\equal{}\sqrt{5^2\plus{}2^2\plus{}2^2}\equal{}\sqrt{33}$
$ CA\equal{}\sqrt{2^2\plus{}1^2\plus{}1^2}\equal{}\sqrt{6}$

Now we can use the [[Law of Cosines]]:

$ AB^2\equal{}BC^2\plus{}CA^2\minus{}2CA\cdot BC\cos{C}$.

$ 11\equal{}33\plus{}6\minus{}6\sqrt{22}\cos{C}$

$ \cos{C}\equal{}\frac{14}{3\sqrt{22}}$.

Now note that the area of $ ABC$ is $ \frac{AC*BC*\sin{C}}{2}$. So

$ \sin{C}\equal{}\sqrt{1\minus{}\cos^2{C}}\equal{}\sqrt{1\minus{}\frac{196}{198}}\equal{}\sqrt{\frac{1}{49}}\equal{}\frac{1}{7}$

Thus $ [ABC]\equal{}\frac{3\sqrt{22}}{14}$[/hide]

I hope I calculated right...

[quote="1=2"]$ \cdots\equal{}\sqrt{1\minus{}\frac{196}{198}}\equal{}\sqrt{\frac1{49}}\equal{}\cdots$[/quote]

It should be $ \sqrt{\frac1{99}}$.

Ko-n-ni-chi-wa.

[hide="Answer"]$ \frac{1}{\sqrt{2}}$[/hide]

We can use the cross-product in the general case.
Here is a simple approach that works for this particular problem:

[hide]We have the three points: $ (2, 1, 3); \; ( \minus{} 1,0,4); \; (4,2,2)$

Translate all points by vector (1,0, - 4). We get: $ (3, 1, \minus{} 1); \; (0,0,0) ; \; (5,2, \minus{} 2)$

Note that these all lie on the plane $ y \equal{} \minus{} z$ ! So let's do what is essentially a rotation transformation:
$ (s,t, \minus{} t) \to (s, t \sqrt {2}, 0) \to (s, t \sqrt {2})$, which preserves area.*

We get: $ (0,0); \; (3, \sqrt {2}); \; (5, 2\sqrt {2})$

There are easy methods for finding the area of this, such as the shoelace algorithm. We get $ A \equal{} \frac {1}{\sqrt {2}}$

[b](*)[/b] [i]The first step is a 45-degree rotation around the x-axis.
The second step drops the $ z$ coordinate, which is the same for all points.[/i][/hide]

[b]@ 1=2:[/b] If you are familiar with [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=236044&start=5]other forms of Heron's Formula[/url], you could just use the most appropriate one after finding the three side-lengths.

To be honest, in the first my solution was exatly same as yours.

By the way, when do you study vector, say, scalar product in your country?

In Japan, we study averagely vector in senior high school 2 at the age of 15~16, although we don't study cross product.

I know they're the same at the core, but I wanted to show how one could go about the same problem without vectors. Vectors (and matrices) are usually not treated in High School in the US. When they are, they are usually not treated properly.

Some learn elementary properties of vectors through problem solving, but most learn vectors in college in their first course in multivariable calculus or linear algebra. That's another quirk in the US system, too... many learn multivariable calculus before taking a solid first course in linear algebra.


[hide]This problem can be done without a calculator, and easily!  Because $55x-15x=40x$ we can simply multiply $40*1212=48480$[/hide]

[quote="Arvind_sn"]Still one of my first posts... This site is really helpful for me!!!

[hide]I just factored out and I got 48480[/hide][/quote]Welcome to AoPS/ML! Here you can improve your problem solving skills and develop your math abilities.

 :welcomeani:

[hide]To do this problem mentally, use distributive property to get $1212(55-15$ or $1212*40$. Now, just do $1212*4$ which is $4848$ and add the 0 at the end for multiplying by 10 by on the end to get $\boxed {48480}$ [/hide]


(1) Paul made a round trip between two points $ A$ and $ B$ in $ 10$ hours.  His speed going up is $ 6$ mph.  His speed going down is $ 12$ mph.  His speed going on a level surface is $ 8$ mph.  What is the distance in miles between $ A$ and $ B$?

[hide="ANSWER"]$ 40$[/hide]

-------------------------------------------------------------------------------

(2) For how many single-digit positive integers $ d$ are there an integer $ n$ for which the units digit of $ d^{n}\equal{}1$?

[hide="ANSWER"]$ 4 \Rightarrow$ FROM ANSWER KEY

My interpretation of this question is that $ n$ does not have to be positive.  If this is the case, the answer would be $ 9$.  Could someone please clarify this for me?  Thanks.[/hide]

[hide="#2"] Clearly, d must be odd. 5 doesn't work, but others do, so 4.[/hide]

For problems like number one, the speed uphill and down hill have to average out to be the speed on a flat surface so the problem has only one answer, therefore, $ \frac{10\times8}{2}\equal{}40$

Number 2 is $ \boxed{9}$, let $ n\equal{}0$

I think he left out the part where it says $ n > 0$.
It can't be that easy.

"read the question" type problem, maybe?

[hide="#2"]As xpath said, all even numbers are out and the only odd number that doesn't work is 5, so its just 5-1=4, i dont think theres a way to do it besides testing the odd numbers out[/hide]

[hide="Probably the hardest way to do #1"]Well the average speed for his trip*10/2 is just the harmonic mean of 6 and 12 times 10/2. So its

$ \frac{2}{\frac{1}{6} \plus{} \frac{1}{12}}$ $ \times$ $ 5$ $ \Longrightarrow$ $ \frac{2}{\frac{1}{4}}$ $ \times$ $ 5$ $ \Longrightarrow$ $ 8$ $ \times$ $ 5$ $ \Longrightarrow$ $ \fbox{40}$[/hide]

[quote="ZhangPeijin"]For problems like number one, the speed uphill and down hill have to average out to be the speed on a flat surface so the problem has only one answer, therefore, $ \frac {10\times8}{2} \equal{} 40$[/quote]

[quote="cognos599"][hide]Probably the hardest way to do #1
Well the average speed for his trip*10/2 is just the harmonic mean of 6 and 12 times 10/2. So its

$ \frac {2}{\frac {1}{6} \plus{} \frac {1}{12}} \times 5 \Longrightarrow \frac {2}{\frac {1}{4}} \times 5 \Longrightarrow 8 \times 5 \Longrightarrow \fbox{40}$[/hide][/quote]

I'm confused by both of the above solutions.  Could some please clarify?  By the way, what is a harmonic mean, and how does it help for problem 1?  Thanks.

Those were for problem 1 :wink:

The harmonic mean of 2 numbers $ a$ and $ b$ is the reciprocal of the mean of the reciprocals of $ a$ and $ b$.

[quote="Cube"][quote="ZhangPeijin"]For problems like number one, the speed uphill and down hill have to average out to be the speed on a flat surface so the problem has only one answer, therefore, $ \frac {10\times8}{2} \equal{} 40$[/quote]

[quote="cognos599"][hide]Probably the hardest way to do #1
Well the average speed for his trip*10/2 is just the harmonic mean of 6 and 12 times 10/2. So its

$ \frac {2}{\frac {1}{6} \plus{} \frac {1}{12}} \times 5 \Longrightarrow \frac {2}{\frac {1}{4}} \times 5 \Longrightarrow 8 \times 5 \Longrightarrow \fbox{40}$[/hide][/quote]

I'm confused by both of the above solutions.  Could some please clarify?  By the way, what is a harmonic mean, and how does it help for problem 2?  Thanks.[/quote]
For my solution, if you were to travel uphill one way, you need to travel downhill the other way, so THEREFORE, the average speed while going on an incline both ways, (up and down) must be equal to the speed of someone going on a flat surface or else the question will have more than one answer.
Example:
This kid goes 1mph uphill, and 3mph uphill, and 2mph on flat surface, he goes from point A to B and back in 4 hours, how many miles are between A and B?
Now, if you were to say the entire thing was flat, the distance from A to B and back would be 8 miles, so 4 miles between A and B.
However, if you were to say the entire thing was an uphill incline, then you would take 3 hours to to from A to B and one hour to go from B to A, therefore saying that the total distance front and back is 6 miles, so the distance between A and B is 3 miles.
See?
So the speed uphill and the speed downhill have to average out to the speed on flat surface in order for that problem to have only one answer.

[quote="Cube"](1) Paul made a round trip between two points $ A$ and $ B$ in $ 10$ hours.  His speed going up is $ 6$ mph.  His speed going down is $ 12$ mph.  His speed going on a level surface is $ 8$ mph.  What is the distance in miles between $ A$ and $ B$?

-------------------------------------------------------------------------------

(2) For how many single-digit positive integers $ d$ are there an integer $ n$ for which the units digit of $ d^{n} \equal{} 1$?

[/quote]

[hide]

1.Well since it didn't say whether you are going uphill,downhill, or level when you are going from point A to point B, I assume it doesn't matter.

So to simplify things, I will say that it is level ground the entire way. Since he is going at an equal speed the entire time, I can split the 10 hour round trip in half and make it into a 5-hour trip to get from A to B.

so he was going 8mph for 5 hours

or 40 miles

2. Since an even * an even number never equals an odd number, I will try all of the odd numbers.

1^1,^2,^3,etc equals 1, so 1 works.
3^4 = 81 so 3 works.
5^1,^2,^3,etc. always ends in 5 so it doesn't work.
7^4=2401 so 7 works
9^2=81 so 9 works

So there are 4 numbers.
[/hide]

So, for problem 2, can $ n$ be equal to $ 0$?

It would seem like it could be now that I look at it.
So the answer key is probably wrong.


What does mod mean? Or modulo? And how is it linked with bases, and logs? Can anyone breifly sum up these?

I don't understand them at all, and keep seeing them used in proofs. I don't think they're on our syllabus even.!

thanks for any help with this weird subject
mike


An island consists of three types of inhabitants, namely – truthers, liars and alternators. It is known that  a truther always tell the truth; a liar always speak falsely and an alternator always alternate between a true statement and a false statement

A visitor from a nearby town  approaches three inhabitants  A, B and C ( of the island )   and ask them whether they are truthers, liars or alternators.  He knows that precisely one among A, B and C is a truther; exactly one of the three persons is  a liar and the remaining person is the alternator.
The following are the response made by A, B and C.

A’s response:
 1. C is the truther.
 2. B is the alternator..

B’s response:
 1.  A is the liar. .
 2. C is the alternator.

C ‘s response:
1. A is the alternator. .
2. I am the truther.

Determine the identity of each of A, B and C form the abovementioned  statements.

[hide]A alternator
B liar
C truther[/hide]

[hide="Full Solution"]Assume that A is the truther. This means that his sentence 1 is true, which leads to contradiction. So A is not a truther.

Assume that A is the liar. This would mean that C is not the truther, and B is not the alternator. Since the liar is already "taken up", this means that B is the truther and C is the alternator. If C is the alternator, then since his first sentence is false, his second sentence must be true. This leads to a contradiction, so A is not the liar.

Thus, by process of elimination. [b]A is the alternator.[/b]

If A's second statement is true, that leads to B being an alternator as well, so that means A's first statement is true. This means that [b]C is the truther[/b]. by process of eliminations,  [b] B is the liar.[/b][/hide]

[quote="turboturtle"][hide]
You can go with letter C as truther, since it says that, and if you take that as true, the problem gets easy, which ends up yielding the result :
A alternator
B Liar
C truther.
[/hide][/quote]I think after that you should go with C as liar, alternator too to see all possibilities.

does an alternator has to speak one false and one true statement ? Or he can speak both truth / both false / one truth one false which there are 3 options .

One false and one truth because he always alternates between false and truth.


analytic geometry
vector
combinatorics unsolved
combinatorics

Find all positive integer numbers [tex]n\ge 3[/tex] such that there [tex]n-gon[/tex] [tex]A_1A_2...A_n[/tex] with two propertis:
a)[tex]A_i[/tex] has cordinatis integer
b)Sides are [tex]\sqrt{4+i^2}[/tex],[tex]i=1,2,...,n[/tex]

First of all $n=4k+3$ or $n=4k$. Indeed, there must even number of odd numbers $4+i^2$, since parity of $4+i^2$ coincides with the parity of sum of coordinates of the correcponding vector, and we know that sum of all vectors is 0.

It is easy to construct an example for $n=4k$. But I don't know how to tackle $n=4k+3$ :(


search
function
real analysis
real analysis unsolved

just click on the image and get a clear image. can someone help me. thax.

It is reallly a difficult question, since so many people have read it, but no one solved it.

i made a clear image of my question, i keenly need your help.

The more you bump it up, the less your chances to get an answer are. This is [b]not[/b] a math. help hotline. Besides, your constant cries for immediate help make everybody suspect that it is either a contest problem or a homework due tomorrow that you are assumed to do all by yourself... As to the problem itself, it is nothing special and similar things have been discussed on the forum a few times; that's why nobody answered. I'm too lazy/busy right away to search for an appropriate link but the general idea is that if you start above $ 0$ and have $ f'\ge f^4/2$, you'll get a blow up in finite time, so the function cannot stay large positive. On the other hand, when it is negative, and $ f'\ge f^4/2$, it goes up, so it cannot stay large negative either. I leave it to you (or someone else) to turn this into a formal reasoning.


function
ratio
induction
limit
logarithms
number theory
prime numbers

Let’s start off by discussing recursion.  Defining a sequence recursively requires 1) a set of initial values and 2) a function that tells you how to use previous term(s) of the sequence to produce the next term.  Below are some  familiar examples that are built using recursion.

-The recursive sequence that started it all, I suppose, can be defined as such:
X0       = 1
Xn+1 = 1+Xn

This sequence gives the natural numbers.  It is divergent and unbounded.  The prime numbers are a subset of this set.

	-Speaking of the primes, a sequence that took me by surprise is defined as such:
W0    = 2
    		Wn+1 = Wn ^ (Wn /(Wn-1))

The nth term of this sequence puts you very close to the nth prime.

For reference, the 64,667th prime =  809,797
  and W64,667 = 808,677.02
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
-A related sequence is defined in this way:
Y0       = 1
Yn+1 = 1/(n+1) + Yn

This is the harmonic series.  It is divergent but bounded.

-A sequence that makes a pretty graph can be defined in terms of the natural numbers and the harmonic series (using the definitions of Xn and Yn defined above):
Hn = mod(Xn ,Yn)
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
-Let’s define another sequence.
Z0       = 1
Zn+1 = 1+1/Zn

Please observe that as n goes to ∞, then
{Zn} converges to 1.61803398874989…
aka, the Golden Ratio; (1+√5)/2

Also, observe that this number is a solution to the equation
(1/(z -1)) – z = 0
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Please observe the following limits for e (from http://mathworld.wolfram.com/e.html).
e = 2.71828182845905…

As n goes to ∞
lim (1+1/n)n = e
lim ( (n+1)(n+1)/nn – nn/(n-1)(n-1) ) = e

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Now, let’s define a sequence that we’re not so familiar with.
E0       = 1
E1       = 2
En+1 = En-1 ^ (1-1/ En) + En/En-1

Please observe that as n goes to ∞, then

{En} converges to 3.29316628741186…

Also, observe that this number is a solution to the equation

(x/(x-1))x – x = 0
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Let’s define another sequence.

D0       = 1
Dn+1 = (1+1/Dn) ^ Dn

Please observe that as n goes to ∞, then

{Dn} converges to 2.29316628741186…

Also, observe that this number is a solution to the equation

((y+1)/y)y – y = 0

Finally, observe that

(2.29316628741186…)^3.29316628741186… = (3.29316628741186…)^2.29316628741186…
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Questions

-How do I go about analyzing the irrationality of a number like 2.29316628741186…?
-How do I find out if a number like 2.29316628741186… and how it might be related to another irrational number, perhaps in an algebraic way?  Like, say, (1+√5)/2 and the Golden ratio?
-How are the recursive equations of the sequences related to the equations for which their limits are solutions?  How is Dn+1 = (1+1/Dn) ^ Dn  related to ((y+1)/y)y – y = 0?  Because the limit of the sequence was known first.  Once the value of the limit was established, only then was it evident what equation it was a solution for.  In other words, the solution was found first, and the equation was deduced afterwards.
-Can you draw any conclusions about either recursion and induction when you compare the two limits, as n goes to ∞,
lim (1+1/n) ^ n = 2.71828182845905…
lim (1+1/Dn) ^ Dn = 2.29316628741186…

For example:
$ D_1 \equal{} 1$ and
$ D_{n \plus{} 1} \equal{} \left(1 \plus{} \frac {1}{D_n}\right)^{D_n}$

Function $ f(x) \equal{} \left(1 \plus{} \frac {1}{x}\right)^{x}$ is monothonous and is increacing.
That means: if $ a < b$ then $ f(a) < f(b)$ and back: if $ f(a) < f(b)$ then $ a < b$.

Now let $ A$ is a solution of the equation: $ \left(1 \plus{} \frac {1}{A}\right)^{A} \equal{} A$.

We know, that $ D_1 < A$.
Let`s prove that if $ D_n < A$ then $ D_{n \plus{} 1} < A$. We have:
$ D_{n \plus{} 1} \equal{} \left(1 \plus{} \frac {1}{D_n}\right)^{D_n} < \left(1 \plus{} \frac {1}{A}\right)^{A} \equal{} A$
We used, that $ f(D_n) < f(A)$.

Now, we know that sequence $ D_n$ is increacing and all $ D_n < A$. This means that exist the limit of this sequence. And we can find it if we put $ D_{n \plus{} 1} \equal{} D_n$. Easy to show, that:

$ \lim_{n\to \infty}{D_n} \equal{} A$

----------------------------------------------

Now let: $ f(n)\equal{}\left(1 \plus{} \frac {1}{n}\right)^{n}$.
We have:
$ \ln{f(n)}\equal{}n\cdot\ln{\left(1 \plus{} \frac {1}{n}\right)}$
For big $ n$ we can write: $ \ln{\left(1 \plus{} \frac {1}{n}\right)}\equal{}\frac{1}{n}\plus{}O{\left(\frac{1}{n^2}\right)}$
And finally:
$ \ln{f(n)}\equal{}1\plus{}O\left(\frac{1}{n}\right)$

So:
$ \lim_{n\to\infty}{\ln{f(n)}}\equal{}1$

This means, that:
$ \lim_{n\to\infty}{f(n)}\equal{}e$


Let a,b,c be triangle's sides. We know that: ab+bc+ca=27.
Prove that: a+b+c < 11

we have: $ (a \plus{} b \plus{} c)^2 \equal{} a^2 \plus{} b^2 \plus{} c^2 \plus{} 2(ab \plus{} bc \plus{} ca)$
$ < a(b \plus{} c) \plus{} b(c \plus{} a) \plus{} c(a \plus{} b) \plus{} 2(ab \plus{} bc \plus{} ca) \equal{} 4.27 < 121$


How do I get the rating not to show up under my username on the LHS of posts?

[quote="pacman2812"]It doesn't show to other members. It only shows to you.[/quote]

However, moderators can see a user's rating.

[quote="hurdler"]I have wondered about this too.

what is a good rating? What is the max rating?

Does rating tell you the average of how other users rate your posts?[/quote]A good rating is between 4 and 6. An average rating is between 2.5 and 4. A bad rating is anything lower than 2.5 (which means that your posts are crap and you should probably be more careful with what you write.)

It does tell you the average of how other users rate your posts. But bare in mind different users have different powers of rating. If you with a small rating rate someone a 1 and someone else rates him a 5, his rating on that post won't be 1+5/2 = 3, but rather something closer to 4.5 (due to the fact that you, having a small rating, are supposed to not have a pertinent opinion on the matter.)

Oh ok. Is there a thread/ or site on this website that explains what the characteristics of a good post are?

General common sense. Have a meaningful post, that is not just a mere repetition of what someone just said above you, etc. Also note that in some forums (GFF for example) your posts are not counted. This means that they are also not rated.

Many fora have such threads as a permanent announcement -- for example, the relevant thread in the Round Table is [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=214779]this one[/url], and in the Inequalities Unsolved Problems forum it's [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=21455]this one[/url].  Darij had some relevant musings, too:
[quote="darij grinberg"]To be honest, I have been wondering why people spam, and I haven't found a good answer. But there are things that provoke people to spam more and there are things that keep them from spamming. Many new members come here and, being confused at the first moment, start imitating others; this is nothing bad as a behaviour, but when the person has not really much to say and he sees lots of spam, then he often starts spamming too. And I am not using the word "spam" for [i]off-topic[/i] discussions (non-mathematical or whatever), but for [i]nonsense[/i] discussions, may they be related to mathematical topics or not. A good example of what I mean is [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=31812]this wonderful conversation on whether something is a lemma or a problem[/url], of course, mainly led by our dear Ashegh. (I have not deleted it basically because it is such a clear example of spam.)

José [i]is[/i] a spammer, as are Abdurashidjon, Davron (to a smaller extent), as Ashegh was (he apparently got better), but I am pretty sure that these people would also spam less if there were less spam around. It's all those discussions which really have no sense, where people just post some messages and don't even understand what the others write etc.. Probably the most active, and, to me, annoying discussions of this sort in the olympiad section are the ones that come every year, each time in several versions: The "what will be the problems of the next IMO" ones. I mean, it was a nice joke in 2004, people even posted funny comments. Since 2005, it starts getting boring. Is it so hard to imagine that you [i]cannot[/i] predict what the problems of the next IMO will be? Or do some people really believe that by bugging Arne with stupid questions for a long enough time, they will make him reveal his IMO proposal? I think a part of these discussions, as well as lots of useless mono- and dialogues in mathematical threads ("why does nobody solve my problem?" - "is it so hard?" - "i have a soln, wait for me a week" - "two weeks passed, where's your f'in soln?") should be simply deleted or moved to the game and fun factory if they are in some way funny (what seldomly is the case). New users should not get the idea that useless posts will stay forever.

Valentin, I don't even think there is much of a difference between spam in maths forums and spam in the games&fun factory. As for me, spam = useless posts, that's all. As a rule of thumb, before making a post, one should ask oneself the following questions:

[quote][u][b]1.[/b] Does my post contain informative value?[/u]
This needs not be mathematical. There are informative things to be said about World Cup 2006 too. On the other hand, few informative things can be said about problem 5 of the IMO 2006 before the IMO has started.
[u][b]2.[/b] Could my post interest or be useful to anyone?[/u]
This is closely related to the question before. If a post doesn't interest anyone, why should it be posted?
A few tautologies: A solution of a problem surely interests people, at least the person who posted the problem. A problem, in most cases, also interests people. On the other hand, a problem from an ongoing competition may interest somebody a bit, but it is by no means useful; instead, it damages the competition.
By the way, you can well assume that a good joke will interest people. On the other hand, a statement of the kind "I think so too" is less likely to do that.
[u][b]3.[/b] Will anyone understand this post?[/u]
Yes, this is something many people just don't think about. If you know very well that your English sucks, try to write your problem in two different versions, or just explain it with more details, or use formula language instead of words where you can... And ffs check your questions for typos before posting them! Don't expect us to both correctly interpret and solve your problems!

If statement [b]3.[/b] is not fulfilled, or none of [b]1.[/b] and [b]2.[/b] is fulfilled, then the post is spam.[/quote]

These are, in my opinion, the basic rules of useful posting. What spammers like José and the others I mentioned do is breaking these rules. Of course, there are more things inappropriate to ML that don't fall under these rules, but I think the above is most important. On another level, there are problems like misplaced topics (sorry for telling you this, Valentin, but this has much to do with the structure of ML which is way too confusing in my opinion - half of people on the Olympiad Section don't understand the exact meaning of Unsolved, Proposed&Own, Solved and Open subsections...) and flames. But flames are, fortunately, seldom, and misplaced topics can be moved. Talking about spammers, I think promoting the above rules is most important - and, of course, consequent deletion of nonsense conversations in all fores.

  darij[/quote]

However, if people only rate bad posts, wouldn't that deflate the meaning of a rating?

Yeah most people only rate bad posts bad and not as much with good posts. However, I think it should be fine because there really aren't as many bad post ratings either.

And also I noticed. On all threads about rating, people rate the posts. lol. :)

How accurate would you say post rating is? Most people only rate bad posts. I hardly ever see posts with good ratings - maybe 1 good post rating for every 3 bad post ratings.

It seems that most of the people I classified as spammers in that post have got significantly better during the last years...

As for the post rating system, I had a better opinion of it before, but now as I noticed that it only works if one has activated scripts from googleapis.com, what several people don't really want to (already for performance reasons), I have stopped using them. What I would rather propose now is a kind of "yellow/red card" system against cheaters: If some person posts at least two current contest problems within one week, he gets a yellow card which is visible on every posts he makes, so people will check first before answering his posts whether the problem is from an ongoing contest.

  darij

In conformity with the thread's name,I would suggest to change few options.Namely why don't you add an option which could hide and show the rating in accordance to preferences of each person.I wonder what is the rating of our respected moderators and administrators?
I think I'm not the only one who wants to know it,just out of curiosity.
Also in my opinion it makes sense to increase the number of moderators,especially in olympiad section,due to increasing of senseless posts, and posts containing problems from ongoing contests.
I'm curious if there is someone who has more than 50 posts and has rating approximately 5.8-6.0...what is the highest rating you've ever seen?
What is the point of adding someone into the list of friends or foes?
Valentin Vornicu have also mentioned that in some forums some posts are not counted, but it is not completely true.I noted that once you post anything in National Forum your posts number temporarily increases,but after several days it usually returns to the initial value.
I wonder if I am the only one who has met such strangenesses.

Highest rating I've ever seen? 5.9, and the person having it has substantially more than 50 posts (not far form 5-digit number of posts, actually).

Your idea of increasing the number of mods in Olympiad section is IMHO very good indeed. Removing not-Olympiad-level problems from the forums, locking up problems which have appeared before, M&Y problems, NZ problems, Polish MO's... There's a lot of work to be done every day.

Well, as long as we're on the subject: the top 14 overall posters on the fora (which at the moment is all posters with 5000 or more posts) have collectively the following multiset of ratings: 5.9, 5.8, 5.6, 5.5, 5.5, 5.3, 5.2, 5.1, 5, 4.8, 4.8, 4.4, 3.9, 3.7.  The college playground moderator group collectively has the following ratings: 5.8, 5.8, 5.6, 5.5, 5.3, 5.1, 5, 4.6, 4.6.

How come my rating is as low as it is (2.2)? I haven't made that much spam.

[quote="alexhhmun"]lotsofmath, how ironic.  That post is spam.

Is this topic the reason that ratings have become visible to everyone?[/quote]
Ratings are not visible to everyone...

[quote="JBL"]The college playground moderator group collectively has the following ratings: 5.8, 5.8, 5.6, 5.5, 5.3, 5.1, 5, 4.6, 4.6.[/quote]

my rating only tanked once i started posting in the round table regularly :(

[quote="Erken"]In conformity with the thread's name,I would suggest to change few options.Namely why don't you add an option which could hide and show the rating in accordance to preferences of each person.I wonder what is the rating of our respected moderators and administrators?
Also in my opinion it makes sense to increase the number of moderators,especially in olympiad section,due to increasing of senseless posts, and posts containing problems from ongoing contests.
What is the point of adding someone into the list of friends or foes?
Valentin Vornicu have also mentioned that in some forums some posts are not counted, but it is not completely true.I noted that once you post anything in National Forum your posts number temporarily increases,but after several days it usually returns to the initial value.
I wonder if I am the only one who has met such strangenesses.[/quote]

1st paragraph: There is an option under consideration: hide and show messages written by people with a 2 rating or under...not sure how far people are along on this.
2nd: More than 5 isn't enough for you? ;)
3rd: This is so you know which members to *stay away* from and/or associate with...like you wouldn't really want to deal with spammers (unless you are also a spammer). (But that's my opinion)
4th: At one point I believe that was a glitch...that's why the Games and Fun Factory was split into two sections: Mafia and Fun Factory. I THINK in the Mafia section, your posts are counted??

No, they aren't because they belong in the Games forum. Games posts don't count to a rating (I think, but I'm not sure).

[quote="cf249"]Posts in the Games forum and Mafia forum count toward post total and rating, while posts in the Fun Factory forum do not.[/quote]
Oh...so now it got changed...wow.

[quote="hsiljak"][quote="Erken"]
I'm curious if there is someone who has more than 50 posts and has rating approximately 5.8-6.0...what is the highest rating you've ever seen?
[/quote]

Just for the sake of completeness (and curiosity): I found a user with $ \approx 400$ posts and rating $ 6$.[/quote]
Who is that guy?for the sake of secretness,private message would be the most suitable variant.
well...let me guess,is it...galois?

[quote="hsiljak"][quote="Erken"]
I'm curious if there is someone who has more than 50 posts and has rating approximately 5.8-6.0...what is the highest rating you've ever seen?
[/quote]

Just for the sake of completeness (and curiosity): I found a user with $ \approx 400$ posts and rating $ 6$.[/quote]

So this user may very well only have one rating associated with his account, i.e. one user rated 6.

I think so too. ;)

P. S. is there any way we could find which of our posts have been rated? egosearch isn't really an option for a poster with more that 50 posts (for instance, I've seen only two of my posts rated-and it's quite obvious to me, looking at the fluctuation of the rating, that more than few posts were rated).

P. P. S. Is it possible we don't see a rating on a post that has in fact been rated?

i have a question(sorry for reviving an old thread.) How good are post ratings. What if someone is mad at a person and starts rating all his posts 1. Also what if a person jsut goes around rating bad posts.

Well, I believe the current rule in effect is that a user cannot rate more than 3 posts an hour. So I doubt someone will sit there for the entire day just to hold a grudge against another person.

Second, the admins will usually catch his fairly quickly, and the result would be that the abusing user gets his rating rights stripped.

Post ratings don't mean anything, but if yours is under 2, then you should consider making your posts more helpful and less frequent  :wink:

[quote="ernie"][quote="alexhhmun"]lotsofmath, how ironic.  That post is spam.

Is this topic the reason that ratings have become visible to everyone?[/quote]

Ratings are only visible to moderators and administrators.[/quote]

What about the stars?  Doesn't that mean ratings are available to all, if not exactly?

The stars are something else entirely; they signify the number of posts a user makes.

[quote="manjil"]I just wondered if it was possible to know which user has rated you how much to the admin and mods?[/quote]

I believe this was suggested a while back.

I believe administrators can see which person rated whom, but moderators can't.

[quote="Valentin Vornicu"][quote="hsiljak"]P. P. S. Is it possible we don't see a rating on a post that has in fact been rated?[/quote]No, in order to count you have to see it (you need a few people to vote first before their average is considered).

At this moment there is no page for you to see which posts of yours are voted. There will be one in the future.[/quote]

Well... has this useful page been created? And if it has, where is it then? :?

2 Important questions about RATING.

1) I always post relevant posts (questions, problem, math discussion). Never once have I posted silly post. Never but once someone ranked a solution I wrote up to a problem a two. That was my only rating for a while.  But don't know why I got 2 for that post. It was regular solution to regular problem that I had worked a while on. I'm confused could someone explian?

2) When I rated (just now) a over-all RATED-6 post a 6 (full score), the over-all score went down to 5.5 Why? Doesnt make mathetical sense. Average rating should still be 6. Then  seconds after I rate a rate-5 post a 6 (FULL SCORE) goes up to 5.5. That makes sense so its not a glitch. Thanks if you can help.

[quote="zhuzhu88"]But don't know why I got 2 for that post. It was regular solution to regular problem that I had worked a while on. I'm confused could someone explian?
[/quote]

You might not have explained it very thoroughly, or maybe the answer you got was incorrect. If neither of those is the case, someone was probably just abusing the post rating system. Just continue making productive posts and your rating will inevitably increase.

[quote="zhuzhu88"]When I rated (just now) a over-all RATED-6 post a 6 (full score), the over-all score went down to 5.5[/quote]

This might have been due to rounding errors. The overall rating of the post might have been 5.5 or something, and when you rated it, the forum recalculated the rating and changed it to 5.5. I say this because once I saw a post rated 6, then when I refreshed the page, it went down to 5.5.

If you had actually read this thread, you would've seen that the ratings were weighted.

amparvardi's rating has more weight than shazi's, for example.

The post ratings are weighted differently, but how does that make a 6 rated post go down to 5.5 when someone rates it a 6? I'm sure even shazi doesn't have a negative post rating weight.

Hmmm... That's weird. I saw my brother rate a post a 1. It went down a half cube. Once I canceled it, it went up a full cube. I tried this on another rated post to test, and it worked again.

My solution post under topic "7-10 double" was rated 2. Why is that what do I do to get rated higher. Did I do something wrong? Thanks!

Note: I think if you search members then click on the posts you should find it. It was a recent post. thanks.

[quote="zhuzhu88"]My solution post under topic "7-10 double" was rated 2. Why is that what do I do to get rated higher. Did I do something wrong?[/quote]
Can you give us a link?

Fixing the rating issue of a rating seemingly going down when you think it shouldn't is a relatively low priority as it does not affect the usage of the site. It also may not be a bug as per the case listed previously in this topic (someone else rated it lower just beforehand). However, when reporting the issue, please include the post number, otherwise we are unable to look in to it. To find the post number, click the little number at the top right of each message then copy the URL. The post number can be found at the end of the URL.

For example, the URL of this post is http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2368232#p2368232.

The post number is 2368232.


geometry
3D geometry
dodecahedron
rotation
ratio
geometric transformation
trigonometry

I was supposed to do this for class, it's past due, don't worry.
Problem is the teacher won't tell me how it was supposed to be done, and nobody else in class has a clue.
I know you have to create a circle and two points on it to rotate to change perspective, but I get lost in mapping my points and creating my matrices.

It was not that difficult and I have never tried anything like this with the Sketchpad before. The thing can be rotated around the vertical and horizontal axes, separately or simultaneously, no fault. I did not need any matrices, just 3 ratios: the golden ratio G (construction of a regular pentagon / decagon), the insphere / circumsphere radii ratio $ \frac {r}{R} \equal{} \frac {G}{2}$ and the distance ratio of the top face plane and of the plane containing the uppermost vertices of the lower faces from the horizontal plane through the dodecahedron center, equal to $ G^3.$ The rest was just very simple rotation of 2 pairs of equivalent planes and parallel projection.

Draw arbitrary circumsphere centered at O, fixed x-y axes, and an arbitrary large concentric circle outside. The y-axis intersects (O) at A, A'. Pick 2 arbitrary points V, H on the large circle for animation. Start with small angle from horizontal for rotation around the vertical axis (V) and a small angle from vertical for rotation around the the horizontal axis (H). Draw a line from V to O and a perpendicular to it at O, intersecting (O) at P, P'. On this cross, construct a regular decagon inscribed in the circumsphere (O), just rays to points on (O), and color them alternately red /green.

Construct the insphere with radius in the required ratio - the insphere radius is average of the circumsphere radius and decagon side. The top / bottom faces are initially degenereted, just circumsphere chords perpendicular to PP' tangent to the insphere at T, T', yielding circumradii of the top / bottom faces.  Using double the insphere radius (equal to G) and geometric division, construct planes of the uppermost vertices of the bottom faces / lowermost vertices of the top faces, i.e., circumsphere chords perpedicular to PP' with midpoints M, M', such that $ OT / OM \equal{} G^3.$

Draw a line from H to O. Using $ \cos \widehat{HOA}$ on the y-axis insphere radius, rotate the top / bottom face chords down / up, keeping their lengths constant. Do the same with the other 2 planes, on small spheres concentric to (O) and tangent to their chord, again keeping their lengths constant. Construct circles on the 4 rotated chords with diameters equal to their lengths. In the 2 top circles, construct regular pentagons oriented one way, on the bottom 2 circles, construct regular pentagons oriented the other way. This has to be done carefully, not using any central similarity with the 2 pentagons inscribed in (O), because Sketchpad switches around intersections of a line and a circle, when passing throu a tangency point - the animation would fail. Therefore, use parallels to the alternate rays of the regular decagon inscribed in (O).

These pentagons now have to be projected to the inclined planes, i.e., to ellipses with these pedal circles, 1 vertex at a time. All ellipses have the ratio of the minor / major axis equal to $ \sin \widehat{HOA}.$ No ellipses are drawn, just 5 points of each are constructed. I hope you can do this parallel projection, if not, I will describe it for one pentagon vertex.  Projected pentagons of the top / bottom faces are obvious. Again, it has to be done carefully for animation. A line through the regular pentagon vertex and center cuts its circumcircle horizontal tangents at 2 point. This line has to be drawn connecting the 2 points, not one point and the pentagon circumcenter. Likewise, a line through the projected pentagon vertex cuts horizontal tangents of a smaller circle, circumradius times $ \sin \widehat{HOA},$ at 2 points and the line also has to be drawn connecting these 2 points, not one point and the circle center. Otherwise, the projected pentagon vertices would disappear for some angles during animation, because the sketch would not know intersection of the projecting line with one tangent. Draw the appropriate pentagon sides downward from the top face and upward from the bottom face, while testing the figure with the animation points V, H, preferrably very small $ \angle HOA.$ Mutually connect the free hanging downward / upward pentagon legs. Clean up and animate.


In the figure to the left, ABCD, AEHI, and EFGH are all congruent squares.  Points D, J, C,  and F are all collinear.  How many degrees are in the measure of angle DJE?

[hide]
Because $ AF \equal{} 2AD$, Triangle $ ADF$ is a 30-60-90 one.
Looking at the quadrilateral $ ADJE$
we know that
$ \angle{ADJ} \equal{} 90$ degrees
$ \angle{DAE} \equal{} 60$ degrees
and
$ \angle{AEJ} \equal{} 90$ degrees

Therefore $ \angle{DJE} \equal{} 360 \minus{} (90 \plus{} 60 \plus{} 90) \equal{} 360 \minus{} 240 \equal{} \boxed{120}$
[/hide]

[hide]
Or alternately, we can note that triangle ADF is 30-60-90, then triangle EJF is similar because EJ is perpendicular to AF and the two triangles share an angle, so angle EJF is 60 and the answer is 180-60=$ 120$
[/hide]

[quote="CatalystOfNostalgia"]See 11th and 12th posts here: [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=168612&start=60[/url][/quote]

My bad Catalyst, hadn't seen that you had used the problem and I was cleaning off my desktop and wanted to make sure it got used somewhere.

Both answers above are right by the way!