I am attempting to write a basic template for a math test. I would like to set something up where I notate each problem as follows:

\begin{problem}
Problem text
\end{problem}

Or something along those lines, and have the document automatically number the questions. Is there an easy way to do this?

[quote="Eugevir"]Or something along those lines, and have the document automatically number the questions. Is there an easy way to do this?[/quote]
Yes it is. Use the enumerate environment:

\begin{enumerate}

\item Blah blah blah % problem 1

\item Blah blah % problem 2

....

\item Blah % problem n

\end{enumerate}

Another two suggestions would be:

\newcounter{[i]counter[/i]}[[i]incounter[/i]]

where the incounter sets the sounter numbers to be reset when the value of that counter changes. To call use:

\the[i]counter[/i].

Or if you want your problems to have names than use:

\newenvironment{[i]environment name[/i]}[number of options]{[i]text to be displayed at the beginning[/i]}{[i]text to be displayed at the end[/i]}

Call this with:

\begin{[i]environment name[/i]}
    Put stuff here
\end{[i]environment name[/i]}


calculus
integration
logarithms
trigonometry
calculus computations

Let $ |a|<1$. Find
\[ \int_{0}^{\pi}\frac{\ln (1\plus{}a\cdot\cos x)}{\cos x}\;dx\]

i also  have this problem,  i  can't solve it.
\[{ \int \sqrt{\frac{\sin x}{\cos x}}} dx\]

hello, substituting $ t\equal{}\sqrt{\tan(x)}$ we get $ x\equal{}\arctan(t^2)$ and our integral is now
$ 2\int\frac{t^2}{1\plus{}t^4}\,dt$, to solve this use the identity $ 1\plus{}t^4\equal{}(t^2\plus{}t\sqrt{2}\minus{}1)(t^2\minus{}t\sqrt{2}\plus{}1)$.
Sonnhard.

[b]Hong Quy[/b], how you get this integral from mine?

hello
strange it is the same question which i got yesterday in my calculas test.  :maybe:
[hide="solution"]
Let $ f(a) \equal{} \int_{0}^{\pi}\frac {\ln (1 \plus{} a\cdot\cos x)}{\cos x}\;dx$
 using Leibnitz Rule differentiate it with respect to $ a$
$ f '(a) \equal{} \int_{0}^{\pi} \frac {cos(x)}{cos(x).(1 \plus{} acos(x))} dx$
putting $ tan\frac {x}{2} \equal{} y$
$ f'(a) \equal{} \int_{0}^{\infty} \frac {2dy}{(1 \minus{} a)y^2 \plus{} (1 \plus{} a)}$
$ f ' (a) \equal{} \frac {2}{1 \minus{} a} \frac {\sqrt {1 \minus{} a}}{\sqrt {1 \plus{} a}} [ tan^{ \minus{} 1} y [ \frac {\sqrt {1 \minus{} a}}{\sqrt {1 \plus{} a}}]]^{\infty}_{0}$
$ f'(a) \equal{} \frac {\pi}{\sqrt {1 \minus{} a^2}}$
$ f(a) \equal{} \pi sin^{ \minus{} 1} (a) \plus{} C$
now putting $ a \equal{} 0, f (0)$ is coming $ 0$ thus C=0
$ f(a) \equal{} \pi sin^{ \minus{} 1} (a)$
[/hide]
thank u

Your integral equals $ \int_{0}^{\pi }{\int_{0}^{a}{\frac{dy\,dx}{1\plus{}y\cos x}}}.$


Given a positive integer $ a$, calculate the sum
\[ \sum_{n \equal{} \minus{} \infty,\ n^2 \neq a^2}^{ \plus{} \infty} \frac {1}{n^2 \minus{} a^2}
\]
using complex methods.

consider something like this $ \frac {1}{(z^{2} \minus{} a^{2})\tan \frac {\pi z}{ 2}}$ where $ a$  not  in $ \mathbb{Z}$

note that it does also work for $ a \in \mathbb{Z}$, I think, which is what the question originally asked for -- now, however, you just have to note that $ a$ is a multiple pole of the function rather than just treating it as a simple pole of the reciprocal tangent


Hey y'all,

I didn't do early for MIT, but I figured some of y'all might have.

Anybody make it to MIT???

2 people I know at Northview got deferred.

Yes, my application got deferred from MIT. So far, only one person from my school has made it to MIT, but she doesn't do math.
The rest of us (about 6 or 7, at least) have gotten deferred. :|

Hey! I'm a freshman now, and my goal is to make it to MIT. I was wondering if the following were good enough to make it to MIT:

5 AP Courses

IB DIPLOMA (Equivalent to 7 rigorous APs)

300+ Hours of Volunteering (ABout volunteering, does it matter where you volunteer or if you volunteered in many places, or rather in a couple places but consistently?)

A score of 2200+ on the SAT.

Thanks!

percy314, I think this (potential) record looks somewhat competitive, but you might want to know (if you don't already) that admissions will also take into account grades, extracurricular activities, and the required essay.
Also, in response to your question about volunteering, I'm pretty sure it doesn't matter where you volunteer, just as long as you have plenty of hours.
Hope that helps!
--Ben

Thanks for your response Ben. Yeah grades are also a critical factor in choosing students. My teacher told me that one in seven 4.0 students get accepted to an ivy league college, but that gets me wondering; If there are more than 7 ivy league colleges...then doesnt that mean all 4.0 students get into an Ivy League?

@percy314: Please tell me that you're being sarcastic about that last question.

@Math Geek: The 2 I knew of were Ben, and Dan. (Dan isn't on math team as far as I know).

[quote="percy314"]My teacher told me that one in seven 4.0 students get accepted to an ivy league college, but that gets me wondering; If there are more than 7 ivy league colleges...then doesnt that mean all 4.0 students get into an Ivy League?[/quote]
I think your teacher is trying to say 1 in 7 to the ivy league colleges as a whole.
They're not THAT easy to get into.  (although some on aops might beg to differ  :P )

What's Dan's last name?  Cuz there are some token senior white kids on our math team...  :)

Though I did not apply to MIT, I am just eager to know if anybody from Georgia or if anybody from AOPS made it to MIT.

SO yeah who made it??

Walton has one that I know of. She doesn't do too much math, but she is doing A LOT of the organizational work for Math Fest.

@1st post: though he did not reply here, but someone we all know made the early admission (I will not reveal his name unless he does not mind so.)

You know, asking Tim here is pointless, because Tim won't reveal the person's name without that person's permission.

Oh yeah, I got in early action. Thanks Tim for being such a good secret-keeper :)

Sorry I didn't reply earlier...

- Oliver

Too lazy to create a new thread for Princeton.......

Michael Lai got into Princeton!!

Congrats Michael!  Someone on this forum got into Caltech, but I don't know if he wants me to reveal his identity, and two non-ARML students I know in my school made Harvard and Princeton, respectively.


geometry
AMC
USA(J)MO
USAMO
rotation
trigonometry
complex numbers

ummm, this is my first time posting so idk how to apply the TeX. and im doing this cuz darktreb (yellow C for life!) was tlaking about #14 a lot.

ok, let
y = mx + b_4 pass through (0,12) so 12 = b_4
y = mx +b_3 pass through (8,0) so 0 = 8m+b_3
y = -x/m + b_2 pass through (10, 9) so 9 = -10/m+b_2
y = -x/m + b_1 pass through (-4, 7) so 7 = 4/m + b_1

Then: b_2 - b_1 = 2+14/m and b_4 - b_3 = 12 + 8m

Drawing in angle Q so that tan Q = m, we also see that cos Q = s / (b_4 -b_3) and sin Q = s / (b_2 - b_1). Then, you find that m = 1/3 and do more stuff to get s.

That's what I did on the contest.  If you do the algebra correctly the first time, it takes 10 minutes, which isn't bad for a #14.  If, on the other hand, you make the mistake of thinking that perpendicular lines have reciprocals whose product is 1 (rather than -1), then it takes half an hour.   :lol:

Anyway, here's a very fast way that Mr. Feng showed me.  Take the point E such that E and A are on opposite sides of BD, AE = BD, and AE $\perp$ BD.  Then E is on the square.  (Why?)  So CE gives one of the sides, and from there you should have the area within one step.

[quote](yellow C for life!) [/quote]

Hey Rich, good to hear from you. I agree, Yellow Group was easily the greatest group of all time (thank you MOP 2002).

[quote]Anyway, here's a very fast way that Mr. Feng showed me. Take the point E such that E and A are on opposite sides of BD, AE = BD, and AE \perp BD. Then E is on the square. (Why?) So CE gives one of the sides, and from there you should have the area within one step.[/quote]

Yes, that's an excellent solution. I stole the idea to write an alternate solution in my little solutions packet. It makes things VERY easy.

P.S. Tell Mr. Feng to leave geometry off the USAMO ... I think it would be good for the uh.. "greater good of everyone"....   :lol:

this is the one about the four points are on different sides of the square:
I talked to Zable: he translated it to the complex plane, rotated it, and messed with it fora while, eventually getting the right answer... but this is elegant

Just to follow up on what sameer said, here's how i solved it (it took 5 mins on the test :)):

In the complex plane, we'll rotate the square until its sides are parallel to the axes with A top, B right, C down, D left. Rotating by $\theta$ is equivalent to multiplying by $\cos\theta + i \sin\theta = x+yi$, so lets do it:

A'=(12i)(x+yi)=(-12y)+(12x)i
B'=(10x-9y)+(9x+10y)i
C'=(8x)+(8y)i
D'=(-4x-7y)+(7x-4y)i

The side length is the imaginary part of A'-C' = 12x-8y, and it is also equal to the real part of B'-D' = 14x-2y. Equating these gives x=-3y, which paired with x^2+y^2=1 gives $y=\pm 1/\sqrt{10}$ and $x=\mp 3/\sqrt{10}$. It doesn't matter which we choose, but since it looks like theta is small and negative, i chose x=positive, y=negative. This gives sidelength = 14x-2y=44/sqrt(10), area 1936/10, 936 answer.

The easiest way (I think) is to use the distance from a point to a line formula with lines going through the center of the squares:
(ax+by+c)/(a^2+b^2) = d
for Points A and C
a(x_1) + b(y_1) = a(x_2) + b(x_2)
for Points B and D (the line is perpendicular)
b(x_3) - a(x_3) = b(x_4) - a(x_4)
solve the system for a and b, etc.

[quote="darktreb"]Hey Zach, which was the problem that kept you from a 15?[/quote]
i think in some other thread it said he missed 3?

if thats true...
hehe i got a problem he missed
thats sometin to be proud of :D


geometry
3D geometry
limit
function
calculus
derivative
calculus computations

([b]5[/b]) Find $ p$ so that $ \lim_{x\rightarrow\infty}x^p\left(\sqrt[3]{x\plus{}1}\plus{}\sqrt[3]{x\minus{}1}\minus{}2\sqrt[3]{x}\right)$ is some non-zero real number.

[hide="Answer"]$ \lim x^p \left(\sqrt [3]{x \plus{} 1} \plus{} \sqrt [3]{x \minus{} 1} \minus{} 2\sqrt [3]{x}\right) \equal{} \lim x^p \left[x^{1/3}\left(\sqrt [3]{1 \plus{} \frac {1}{x}} \plus{} \sqrt [3]{1 \minus{} \frac {1}{x}} \minus{} 2\right)\right]$

$ \equal{} \lim x^{p \plus{} \frac {1}{3}} \left(1 \plus{} \frac {1}{3x} \minus{} \frac {1}{9x^2} \plus{} 1 \minus{} \frac {1}{3x} \minus{} \frac {1}{9x^2} \minus{} 2\right) \equal{} \lim x^{p \plus{} \frac {1}{3}} \left( \minus{} \frac {2}{9x^2}\right)$

$ \equal{} \lim \left( \minus{} \frac {2}{9} x^{p \minus{} \frac {5}{3}}\right)$

and for this limit to be a non-zero real number we must have $ p \equal{} \frac {5}{3}$, in which case the limit equals $ \minus{} \frac{2}{9}$.[/hide]

[hide="Contest Style Approach"]
$ f(x) : \equal{} \sqrt [3]{x}$, then we have that for large $ x$:

$ f(x \plus{} 1) \plus{} f(x \minus{} 1) \minus{} 2f(x) \equal{} \frac {\left( \frac {f(x \plus{} 1) \minus{} f(x)}{1} \right) \minus{} \left( \frac {f(x) \minus{} f(x \minus{} 1)}{1} \right)}{1} \approx f''(x)$

$ \text{deg}[x^p f''] \equal{} 0 \implies \text{deg} [x^p] \plus{} \text{deg}[f] \minus{} 2 \equal{} p \plus{} \frac {1}{3} \minus{} 2 \equal{} 0 \implies \boxed{p \equal{} \frac {5}{3}}$

[b]Note:[/b] [i]This works because $ f$ is a power function of $ x$ and thus its second derivative can be approximated by its second differences for large $ x$.[/i]
[/hide]


inequalities
geometry
special factorizations

Find the area defined by the intersection of the solutions of the two inequalities:
$ |x|\plus{}|y|<4$
$ x^{2}\minus{}8x\plus{}y^{2}>0$

If $ y\ge 0$, then $ y<\minus{}|x|\plus{}4$, the other case as $ y<|x|\minus{}4$. In fact, this inequality forms a nice little square with diagonal $ 8$ and vertices $ (4,0),(\minus{}4,0),(0,4),(0,\minus{}4)$. Completing the square in the second yields $ (x\minus{}4)^2\plus{}y^2>16$, or the area [b]outside[/b] a circle with radius $ 4$ and center $ (4,0)$. Thus the area is the area of the square $ (4\sqrt{2})^2\equal{}32$ minus the a $ 90^{\circ}$ sector of the circle, with area $ 4\pi$. Thus the area is $ 32\minus{}4\pi$.


algebra
polynomial
modular arithmetic
algebra proposed

Let $P\in Z[X]$ be an integer polynomial such that there exists $n\in Z$ such that $P(n)=0$.Prove that there does not exist any natural number $m$ such that $P(1995)\cdot P(2000)=7^{m}$.

Yes, because $(1995-n)(2000-n)\equiv (1-n)(-n)\equiv 0\pmod{2}$. This is easy problem.


If $p$, $q$ and $M$ are positive numbers and $q<100$, then the number obtained by increasing $M$ by $p\%$ and decreasing the result by $q\%$ exceeds $M$ if and only if

$\text{(A)}\ p>q ~~ \text{(B)}\ p>\frac{q}{100-q} ~~ \text{(C)}\ p>\frac{q}{1-q} ~~ \text{(D)}\ p>\frac{100q}{100+q} ~~ \text{(E)}\ p>\frac{100q}{100-q}$

[hide]I feel like posting a solution, so I will.  :D

We start with the fact that $M(1+\frac{p}{100})(1-\frac{q}{100})>M$ and isolate $p$.

$1+\frac{p}{100}>\frac{1}{1-\frac{q}{100}}>\frac{100}{100-q}$
$\frac{p}{100}>\frac{100-100+q}{100-q}$
$p>\frac{100q}{100-q}$

The answer is $\text{(E)}$.[/hide]

Nice solution.  :)

Alternatively, notice that $(1+\frac{p}{100})(1-\frac{q}{100}>1$. Solve that inequality for p. It gets messy, but knowing how to multiply binomials fast will speed up the process.


trigonometry
geometry
angle bisector
cyclic quadrilateral
trig identities
Law of Sines

[quote="Shahzada"][hide][img]http://img82.imageshack.us/img82/975/adszds9.jpg[/img][/hide][/quote]
[hide="ideas"]
Angle Bisector Theorem/Law of Cosines
Also, we can assume that BD is 1 without changing the problem.[/hide]

I generally solve them with geometric ways.I don't prefer trigonometry or law of cosines.

Do the $S$s over different sides signify congruence?

Assuming that $S$ signifies congruence, I have a hint, although Shahzada won't like it...:P

[hide="hint"]
Law of Sines, look for equal angles and equal sides :).

I think the answer is $2$, but I haven't proved every part yet; I made an assumption somewhere.

EDIT: I GOT IT.
[/hide]

[hide="SOLUTION"]
Okay, SO, look at $\triangle AFC$ and $\triangle ADF$
IF we use Law of Sines on the triangle, we get
$\frac{\sin 30}{DF}= \frac{\sin \angle ADF}{AF}$ and $\frac{\sin 30}{DF}= \frac{\sin \angle ACF}{AF}$
Thus, $\sin \angle ADF = \sin \angle ACF$. Since $\angle ACF = 40$, then either $\angle ADF = 40$ or $140$. However, if $\angle ADF = 40$, then $\angle FDB = 140$, and then $\triangle BDF$ will have a sum greater than $180$ degrees...so $\angle ADF = 140$.

NOW, do some angle chasing, and you realize that $\angle DFC=120$. SO, $ADFC$ IS A CYCLIC QUADRILATERAL because opposite angles are supplementary.

So, circumscribe a circle around ADFC. SINCE $FC=DF$, arc FC = arc DF. So, if we connect $F$ and $E$, and extend it till it hits the circle(call this point $G$). THEN, connect $D$ and $G$, and connect $A$ and $G$. NOW, CDFG IS ALSO CYCLIC! Now, we realize that $\angle FGD = \angle FGC = \angle DAF = 30$ because $\angle DAF$ and $\angle FGD$ refer to the same arc $FD$, and since arc $FD$ = arc $FC$, $\angle FGC$ must equal to both of them as well.

This implies that $\angle FCG = \angle FDG = 90!$, so $\triangle DFG$ and $\triangle CFG$ are congruent!, so, $\angle GFD = \angle GFC = 60$ and $DE = CE$

If we fill in the angles, we see that $\triangle DFB = \triangle DFE$ because of ASA, so $DB=DE$.

SO, $\frac{BD+DE}{CE}\rightarrow \boxed{2}$.[/hide]

Whew, long proof. Does anyone have something more elegant?

Thank you for solution Zellex.

Create ABC equilateral triangle.Draw perpendicular AD.|DB|=|DC| because of simetry. |CE| is also equal with them.Than create FED equilateral triangle.Extend F and C point.It will be bisector of FED triangle.İf you write angles ADCF is cyclic.Put the other angles.Than we see that the AGF is also isosceles.And point M has to be on G point.
[hide][img]http://img337.imageshack.us/img337/8415/fewfefefac0.png[/img][/hide]
If we draw DNE, angle DNG will be 40.Than we find HDA 10, and HDG=60.İf we write the other angles, SBD,SDG and GDE triangles will be similar so the |BS|=|SG|=|GE|.(|BS|+|SG|)/|GE|=2
[hide][img]http://img516.imageshack.us/img516/2182/54g54g4hx0xu7.png[/img][/hide]


A piece of string is cut in two pieces at a point selected at random. What is the probability
that the longer piece is at least 4 times the length of the shorter piece?

We need to see where it is exactly 4time as long as count everything beyond that. Make the string go from 0 to 5. A cut at 4 or 1 would make one piece 4 times as long as the other. Anything from 0 to 1 or 4 to 5 works, so the answer is $ 2/5$


A right triangle and an isosceles triangle have equal areas. The right triangle has sides of length 3, 4 and 5 units. The isosceles triangle has sides of 3, $ x$ and $ x$ units. What is the value of $ x$ ? Express your answer as a decimal to the nearest thousandth.

I suppose one way you could do it is to substitute the area 6 into the Herons formula with sides 3 x and c and solve from their, (It may be somewhat messy computation though)


Let a,b,c non negatve numbers with abc=1
Prove: [img]http://www.gnux.be/latex/data/e11e088ffc7841cb41daf802c4ed59ad.png[/img]

the inequality was discussed and solved  here by arqady :
http://www.mathlinks.ro/Forum/viewtopic.php?t=141304


ceiling function
arithmetic sequence
number theory proposed
number theory

Let $A$ be the set $A = \{ 1,2, \ldots, n\}$. Determine the maximum number of elements of a subset $B\subset A$ such that for all elements $x,y$ from $B$, $x+y$ cannot be divisible by $x-y$.

[i]Mircea Lascu, Dorel Mihet[/i]

consider the arithmetic sequence  u_p=1+3(p-1)
we have u_n - u_p = 3(n-p)
             u_n + u_p = 2 + 3(n+p-2)
We can see that the difference cannot divide the sum.
If we put any other number in the sequence of the form 3q or 3q+2 we can form a difference equal to 1 which will divide the sum.
So i think that tha max number is (n- r)/3 where r is the remainder of n mod3

senouy's argument almost hits the spot.  We partition $ A$ into subsets with three consecutive intergers: $ \{1,2,3\}$, $ \{4,5,6\}$, $ \ldots$; therefore, we get $ m: \equal{}\left\lceil\frac{n}{3}\right\rceil$ subsets.  Call this partition $ \mathcal{P}$.

If we have more than $ m$ elements in $ B$, then at least two elements of $ B$ must lie in the same subset within $ \mathcal{P}$, and this violates the requirement.  So, $ B$ has at most $ m$ elements.  An example which $ |B|\equal{}m$ is easy to find: $ B\equal{}\{1,4,7,\ldots\} \cap A$.


Find all possible values of
\[ \sqrt{a^2\plus{}a\plus{}1}\minus{}\sqrt{a^2\minus{}a\plus{}1}\]
where $ a\in\mathbb R$.

Let $ A\left( \minus{} \frac {1}{2},\ \frac {\sqrt {3}}{2}\right),\ B\left(\frac {1}{2},\ \frac {\sqrt {3}}{2}\right)$ and $ P(a,\ 0)$, we have $ \sqrt {a^2 \plus{} a \plus{} 1} \minus{} \sqrt {a^2 \minus{} a \plus{} 1} \equal{} \overline{PA} \minus{} \overline{PB}$,

$ 0\leq \sqrt {a^2 \plus{} a \plus{} 1} \minus{} \sqrt {a^2 \minus{} a \plus{} 1} < 1$.

hello, i have $ \minus{}1<\sqrt{a^2\plus{}a\plus{}1}\minus{}\sqrt{a^2\minus{}a\plus{}1}<1$
Sonnhard.

You are right. $ a$ wasn't limited $ a>0$.

hello kunny, we have $ \lim_{a \to \minus{}\infty} \sqrt{a^2\plus{}a\plus{}1}\minus{}\sqrt{a^2\minus{}a\plus{}1}\equal{}\minus{}1$
Sonnhard.

hey...if you take limit upto (+-) infinity to solve such kind of problems then does it suffices for all reals???please help

No, in some graphs, the maximum value is not the limiting value. One of the famous examples is the Bell curve, defined as \[ f(x) \equal{} \frac 1{\sqrt {2\pi}}e^{\minus{}\frac {x^2}2}\]
Here, the limit as $ x$ approaches + or - infinity is 0, but you cannot say the range is $ f(x)\equal{}0$. The maximum value is $ \frac 1{\sqrt {2\pi}}$, when $ x \equal{}0$, so the range is in fact $ 0 < f(x) \leq \frac 1{\sqrt {2\pi}}$.


Mafia
AwesomeMath
summer program
function
search
MATHCOUNTS
geometry

So, the rules are simple.The next person that will post will say something
nice about me (even that is impossible  :(  ) and the next posts something  nice about the person above him and so on!
The game starts NOW!  :lol:

I like that you started this thread finally we can be nice. :wink:

Okay this is tricky but... funcia is nice because he did something less spammier than the guy before him... :spam:

[hide]I guess I put this topic to an early end as well, who is gonna make a nice comment about me now? :w00tb:  :rotfl: [/hide]

fredbel6 is nice because he said something kind of nice about me...did i spam? :huh:

[hide]he meant that i spammed  :huh: [/hide]
Funcia is nice because he has a cool name, and he bothered to look at my topic :oops:

alex ib is nice because he actually has a rather nice topic..it reminds me of one we had on a writers forum...you answered the question above you and asked one(not unlike the slot machine..)

Funcia is nice because he currently has half the posts in the "say something nice about the person above you" topic :P

uhh.... Let me see... :rotfl:

mtms5467 is good at posting things!!!

(yeah, but I couldn't find anything better.)

nutz is nice because he's 11 and has lyme disease so all the dumb things he says are excusable.

gg to the poster after me

Treething is nice, because I get to gg.

P.S. My lyme disease is gone.

nutz fore's nice because posted in my poison topic...(lyme disease..i had that when i was little..)

never mind... gg = good game, which implies game over, which pretty much meant that whoever went after me would get owned.

And owned means?

Treethings nice because he reminded me to go do math...by saying gg

Does anyone want to like, bring this back to life?

I would rather make this into a roasting thread

[quote=nicarmt]Congrats on making the thread with most posts in FF!! #DFTW!!!![/quote]

Um, thank you?

[quote=winsuan]Oh, sorry. :([/quote]

No, it's ok. I wasn't trying to be mean or anything

New rule:

Roast the person that comments above you

Edit: but don’t get personal or overly offensive, nobody likes that.

Do yo mama jokes count?
Also after you roast someone, you need to make a sentence that the next person can roast you on.

Yo mama jokes definitely count

And yeah that’s a good rule

Um, yo mama so dumb, she tried to climb mountain dew.
Next poster:
Do a yo mama joke


Prove that for $ a,b,c$ in the interval $ [0,1]$,
$ \frac {a}{b \plus{} c \plus{} 1} \plus{} \frac {b}{c \plus{} a \plus{} 1} \plus{} \frac {c}{a \plus{} b \plus{} 1} \plus{} (1 \minus{} a)(1 \minus{} b)(1 \minus{} c)\leq 1$ :)
NB:I was inspired by the thread :
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=247194., prob#4. :D

Is this your own problem?

This is U.S.A. MO 1980.

Well, you can call it a new invention that I thought I did with the help of my teacher :) .
[hide="P/s:"]Could you please tell me where the USA Mathematical olympiad questions can be downloaded from?Please post a reply ASAP[/hide]

Sorry I have only Japanese version without solution.

Let $ LHS\equal{}f(a,b,c)$.
Note that if we fix a,b then the $ \max_{c\in[0,1]}f(c)$ is $ f(1)$ or $ f(0)$
so that $ \max_{a,b,c\in[0,1]}f(a,b,c)\equal{}1$ when $ a,b,c$ are the endpoints.

I don't know whether I'm right.

Notice that

a/(ab+c+1) + b(c+a+1) + c/(a+b+1) <= a(a+b+c) + b/(b+c+a) + c/(a+b+c) = 1.

also, (1-a)(1-b)(1-c) <= (1-0)(1-0)(1-0) = 1.

does this prove that LHS <= 1  :maybe: ?


calculus integration algebra polynomial trigonometry function geometry

I'm kind of learning calculus outside of the classroom right now and the text I'm using really briefly - and by that, I mean it goes over antiderivatives of polynomials and simple trig functions (ie $ \sin{x}$, $ \sec{x}\tan{x}$, but not $ \tan{x}$ or $ \sec{x}$, etc., no substitution rule yet - then jumps right in to an introduction of the interpretation on the integral as the area under a curve and definition of the definite integral using limits and summation. It then covers advanced techniques/methods for integration (inverse subsitution/trig substitution, partial practions, rationalization, by parts, improper integration, etc.) in a later chapter.

My question is this: would it make sense for my to go through all of the techniques for indefinite integration first (substitutions, trig substitutions, etc.) and then only after I learned those, I'd go over the definite integral and its uses. I'm not sure how this is "traditionally" taught but I figured I'd learn how to integrate all the things I can first then learn how to apply that, since I'd need to learn how to antidifferentiate before doing most area under the curve problems anyways.

Thanks.

That depends on two things.  First, it depends on how well you understand differentiation; the simple integral rules are just the inverses of their corresponding differentiation rules.  Second, it depends on why you're learning calculus.  The techniques for integration are necessary if you plan on either taking a test or learning calculus as a lead-in to applications (physics, engineering, and so forth) you'll need to be well-versed in integration techniques, but if you're just learning calculus for fun I'd advise you to have a strong understanding of the Fundamental Theorem of Calculus first.


The real sequence $ x_1\geq x_2 \geq x_3 \geq ...$ satisfies $ x_1 \plus{} \frac{x_4}{2} \plus{} \frac{x_9}{3} \plus{} ... \plus{} \frac{x_{n^2}}{n}\leq 1$ for every $ n$. Show that it also satisfies $ x_1 \plus{} \frac{x_2}{2} \plus{} \frac{x_3}{3} \plus{} ... \plus{} \frac{x_n}{n}\leq 3$.

It suffices to prove the result for all $ m$ such that $ x_m\geq 0$. For each such $ m$, let $ s\equal{}\left\lfloor \sqrt{m}\right\rfloor$. Then
\[ x_1\plus{}\frac{x_2}{2}\plus{}\cdots \plus{}\frac{x_m}{m}\leq\]
\[ x_1\left( 1\plus{}\frac{1}{2}\plus{}\frac{1}{3}\right) \plus{}x_4\left( \frac{1}{4}\plus{}\frac{1}{5}\plus{}\cdots \plus{}\frac{1}{8}\right)\plus{}\cdots\]
\[ \plus{}x_{s^2}\left( \frac{1}{s^2}\plus{}\frac{1}{s^2\plus{}1}\plus{}\cdots \plus{}\frac{1}{s^2\plus{}2s}\right)\]
So we'll be done if can show
\[ \sum_{j\equal{}0}^{2k} \frac{1}{k^2\plus{}j} \leq \frac{3}{k}\]
for all positive integers $ k$.

But this is easy enough, for there are $ 2k\plus{}1$ terms, each at most $ \frac{1}{k^2}$.


Find all polynomials $ p \in \mathbb{Z}[x]$ such that for all integers $ a$ and $ b$ such that $ a|b$, $ p(a)|p(b)$.

Just consider the polynomial $ g(x)\equal{}f(kx)$, from the condition we get that for all integer $ x$ $ f(x)|g(x)$ thus for all real $ x$
$ f|g$. From this we can deduce that if $ f(x_0)\equal{}0$ then $ f(\frac{x_0}{k})\equal{}0$.Hence $ f\equal{}ax^n$ for some integers $ a,n$

Another solution - assume that the degree of $ p$ is $ d$, and that the coefficient of $ x^d$ is $ b$. $ p(k) | p(nk)$ for any $ n, k \in \mathbb{N}$ (this solution also works when we are only given the same property for [b]positive[/b] integers $ a$ and $ b$).
Notice that in the relation $ p(k) | p(nk)$ we can replace $ p(nk)$ by any linear combination (over $ \mathbb{Z}$) of $ p(n_ik)$'s. Namely, we can replace it by the $ d$'th finite difference of $ p(nk)$ (regard $ p(nk)$ as a polynomial in n): $ \Delta^d p(nk)$ ($ \Delta P \equal{} P(n \plus{} 1) \minus{} P(n)$), which equals $ d!k^db$ (known fact, can be proved by induction).
Hence, $ p(k) | d!k^db$ for all $ k \in \mathbb{N}$. Since $ p(k) | d!p(k)$, we get: $ p(k) | d!p(k) \minus{} d!k^db$.
$ p(k)$ grows faster than $ d!p(k) \minus{} d!k^db$ (polynomial of $ k \minus{} 1$'th degree), so for this relation to hold for all natural $ k$, we must have  $ d!p(k) \minus{} d!k^db \equal{} 0$, or $ p(k) \equal{} bk^d$, identically. So the solutions are $ ax^b$, $ a \in \mathbb{Z} \setminus \{ 0 \}, b \in \mathbb{N}_{0}$.

(it also works if there's a subset $ A \subseteq \mathbb{Z}$, with infinite number of elements, such that $ a|b, a \in A \implies p(a) | p(b)$. We get like before $ p(k) | d!p(k) \minus{} d!k^db$ for $ k \in A$ and proceed in the same way)

Previous solution also  works in the same cases as yours, bambaman :)

Right.  :)
I hope I don't sound over-pedant, but in the step $ \forall x \in Z, f(x)|g(x) \implies \forall x \in R, f(x)|g(x)$ I guess the $ R$ should be $ C$ (since you want to use your argument for all the roots of $ f$, not only the real ones), and that you actually meant $ f|g$ (polynomial divides another polynomial, otherwise we have $ a|b$ when $ a,b$ are not necessarily integers). You also didn't provide proof for this step (not claiming it is hard)  :D

You are right,I meant $ \mathbb C$ instead of $ \mathbb R$ :)


If $ *$ is defined by $ a*b \equal{} \frac{1}{ab}$, what is $ a*(b*c)$?

[quote="AndrewTom"]If $ *$ is defined by $ a*b \equal{} \frac {1}{ab}$, what is $ a*(b*c)$?[/quote]

$ \equal{} \frac{bc}{a}$

$ a * (b*c) \equal{} a* \frac{1}{bc} \equal{} \boxed{\frac{bc}{a}}$


Again this year, MATHCOUNTS will have a live webcast of the National Championship countdown round.  Go to http://www.mathcounts.org at 2 PM Eastern / 11 AM Pacific on Friday May 8 to watch the contest!  Also, this year there will be [url=http://news.prnewswire.com/DisplayReleaseContent.aspx?ACCT=104&STORY=/www/story/05-01-2009/0005017285&EDATE=]a special guest host[/url] for the countdown round: Danica McKellar of [i]The Wonder Years[/i] fame (ask your parents: this was a fairly popular sitcom/drama in the early 1990s).

Also, as an experiment, I will be tweeting from National MATHCOUNTS.  Go to [url]http://www.twitter.com/aopsdave[/url] to follow.

Several AoPS staff members will be on hand in Orlando near the MATHCOUNTS registration area on Thursday and around the event on Friday and Saturday.  Stop by and say hi!

I assume we can get our books signed?  :P

Texas won as a team. Top ten teams:
1. Texas
2. Missouri
3. Washington
4. California
5. Maryland
6. Indiana
7. Massachusetts
8. North Carolina
9. New Jersey
10. Virginia.

Written top 12:
1. David Yang, CA
2. Max Schindler, MO
3. Bobby Shen, TX
4. Calvin Deng, NC
5. Alan Zhou, MA
6. Anya Katsevich, FL
7. Victor Wong, MO
8. John Zhang, WA
9. Mark Sellke, IN
10. Ray Li, PA
11. Steven Chen, TX
12. Alexander Clifton, NJ

 :lol:  :D  :lol:

You could read the other threads. (It's Bobby Shen)

To HiDN428 or a moderator: I don't mind my lists being copied, but please fix that heading to say "Top ten teams:".

[color=darkred][size=75]Done.
-Neal[/size][/color]

[quote="killertomatog"]i dont know about you guys, but i feel sorry for who ever got 50th place and below. I feel sorry for the lowest scorer personally...[/quote]

You can't feel sorry for them! That's good! Everyone's a winner.  :wink:

Well, I agree with jonathanchou711, and this is just a competition. Who knows, they may do better next year! :)  :P  :D

[quote="mathlovers1731"]do u kno what NM placed? i heard they were most improved[/quote]
last year they got 53rd. this year they got 21st


geometry
rhombus
geometric transformation
reflection
rectangle
Pythagorean Theorem

a rhombus is formed by two cords and two radii of a circle with radius R meters. what the is the number of square meters in the area of the rhombus. express your answer as a common fraction in terms of r.

how would u do this problem?

[hide="Method"]You'd have to find the length of $\overline{CF}$ so that $EF=r$.[/hide]
[hide="Answer"]$\frac{r^{2}\sqrt3}{2}$[/hide]
[hide="Solution"]Note that $A$ and $B$ are reflections of $D$ and $C$ respectively over the purple axis, so $AB=CD$.  Also note that $BCFE$ is a rectangle, so $BC=r$ and $CD=\frac{r}{2}$.
Also note that $DF=r$, also.  Apply the Pythagorean Theorem to get $CF=\frac{r\sqrt3}{2}$, which is what we've been looking for.  So the area of the rhombus is $\frac{r^{2}\sqrt3}{2}$.[/hide]

Then again, why not to draw OF and add two equilateral triangles up?

I guess that works too.

EDIT: Of course, you'd still need to realize that $\triangle ODF$ is a 30-60-90 triangle.


calculus
derivative
function
algebra unsolved
algebra

Given f(x) : R -> R such that $\exists f''(x) \forall x$ and sastisfying :
$\forall 0<x<1$ : $f(x).f'(x)\ne 0$ and $|f''(x)|\ge 1$
Prove that $\exists 0\le x\le 1$ such that $|f(x)|\ge \frac{1}{2}$

It's easy to see , because of intermediate value property, that $f''(x),f'(x),f(x)$ have constant sign.
 Two cases are possible  for $f''(x)$:
(1) $\forall x\in ]0,1[, f''(x) \geq 1$, so $f'(x)$ is increasing . Two possible cases for $f'(x)$ :
-(1.1) $f'(1)\geq f'(0)\geq 0$, so $f(x)$ is is increasing . Two possible cases for $f(x)$ :
----(1.1.1)$f(1)\geq f(0)\geq 0$ . The Taylor formula gives : $f(1)=f(0)+f'(0)+\frac{f''(a)}2 \geq \frac{1}{2}$
----(1.1.2)$f(0)\leq f(1)\leq 0$ , so  $f(1)=f(0)+f'(0)+\frac{f''(a)}2 \leq 0 \Longrightarrow f(0) \leq-f'(0)-\frac{f''(a)}2 \leq-\frac{1}{2}$
-(1.2) $f'(0)\leq f'(0)\leq 0$, so $f(x)$ is is decreasing . Two possible cases for $f(x)$ :
----(1.1.1)$f(0)\geq f(1)\geq 0$ . The Taylor formula gives : $f(0)=f(1)-f'(1)+\frac{f''(b)}2 \geq \frac{1}{2}$
----(1.1.2)$f(1)\leq f(0)\leq 0$ , so  $f(0)=f(1)-f'(1)+\frac{f''(b)}2 \leq 0 \Longrightarrow f(1) \leq f'(1)-\frac{f''(b)}2 \leq-\frac{1}{2}$
(2)$\forall x\in ]0,1[, f''(x) \leq-1$ . The function $g(x) =-f(x)$ has $g''(x) \geq 1$ and from (1) we obtain $\exists u\in\{0,1\},|f(u)|=|g(u)|\geq \frac{1}{2}$

.. Q.E.D.. :cool:

Since $ff'$ has no root on $[0,1]$, $f$ is either positive or negative on the whole interval. Without loss of generality, we can assume $f > 0$ (otherwise consider $-f$). Similarly, $f$ is either increasing or decreasing
on $[0,1]$; without loss of generality, we can assume $f$ is increasing (otherwise consider $f(1-x)$). Now we have either $f'' \geq 1$ or $f'' \leq-1$ on $[0,1]$ (this is because of Darboux's theorem).

Assume $f'' \geq 1$, then $f'(x) \geq x+f'(0)$, and $f(x) \geq f(0)+\frac{x^{2}}{2}+f'(0)x$. Taking $x=1$ yields the result.

If we assume $f'' \leq-1$, then $f'(1) \leq-1+u+f'(u)$, and since $f'(1)>0$, we must have $f'(u)>1-u$, and thus $f(u)>u-\frac{u^{2}}{2}+f(0)$, which gives $f(1)>\frac{1}{2}$.


vector
geometry
geometric transformation
rotation
analytic geometry
linear algebra
superior algebra

The first three components of a four-dimensional vector vanish in each of two reference frames. If the second reference frame is not a merely a rotation of the first about $ x_4$ axis, that is, if at least one of the coefficients $ a_{i4} \; (i\equal{}1,2,3) \neq 0$, show that the fourth component vanishes in all reference frames.
--
Let's say the vector $ \vec{v}$ is covariant. Transformation between two given reference frames is (only 4th coordinate):
\[ v_{4'} \equal{} a_{14} v_1 \plus{} a_{24} v_2 \plus{} a_{34} v_3 \plus{} a_{44} v_4\]
Because $ v_1, v_2, v_3 \equal{} 0$ we have: $ v_{4'} \equal{} a_{44} v_4 \equal{} 0$ but only if $ v_4 \equal{} 0$ - how to show that? The same is when transforming to any other reference frame:

\[ v_{4''} \equal{} b_{14} v_1 \plus{} b_{24} v_2 \plus{} b_{34} v_3 \plus{} b_{44} v_4 \equal{} b_{44} v_4 \equal{} 0\;  (?)\]
My question is - what for is the assumption that $ a_{i4} \; (i\equal{}1,2,3) \neq 0$?
If fourth component vanish in one particular reference frame then we have a zero vector and it must stay a zero vector in all frames of reference.
Is this task correctly solved?

This probably belongs in the linear algebra forum, but in any case: my preference would be to solve this in a slightly more geometrical manner–it's clearly a fact that applies in more dimensions.

Let $ W$ be the subspace generated the last basis vector, and let $ v\in W$. (we're working in the first reference frame here)  Now, what we're given is a change-of-reference linear transformation $ T$ that is "not merely a rotation about $ W$", in other words, it does not simply fix $ W$.  So there is some $ w\in W$ with $ Tw\not\in W$.

You can use this to show that either $ Tv\not\in W$, contradicting our assumption, or $ v\equal{}0$.


geometry
projective geometry
geometry proposed

Incircle of triangle $ ABC$ are tangent to sides $ BC,CA,AB$ at the points $ D,E,F$ respectively. Draw a circle $ \gamma$ passing through the points $ A,B$ and having points $ D,E$ in its interior. The line $ DE$ meets the circle $ \gamma$ at points $ P,Q$. Prove that points $ P,Q,F,M$ are cyclic, here $ M$ is mid-point of the side $ AB$.

Nice problem, but easy if someone know harmonic division  :wink: .

[b]Solution.[/b] Let $ K$ be the intersection of $ DE$ and $ AB$. We have $ (ABFK)$ forms a harmonic division, therefore $ KA\cdot KB \equal{} KF\cdot KM$. On the other hand, $ KA\cdot KB \equal{} KP\cdot KQ$. It implies that points $ P$, $ Q$, $ F$, $ M$ are cyclic, as desired.


Can anyone give me some hints on how to approach this problem?

Suppose a boat goes against the stream from point X for one mile, where it passes by a log floating with the current. The boat then went against the current (same direction as before) for another hour, before going the opposite way (with the current) to return to point X. The boat and the log reached point X at the exact same time. At what rate was the current flowing?

I tried making x=rate of the current, y=rate of the boat in still water, y-x=rate of the boat going against the stream, and y+x=rate of the boat going with the steam. But I don't know how to set up the equations...

You know, whenver I try and solve problems like this, I end up going in circles. I'm going to say the rate of the boat is $ x$, and the current is $ y$. What's causing me serious problems is that I don't know the time it took for the boat to return to X once it turned around. Everything I try is thwarted by this one missing value. I've been trying to express the total time as $ \frac{D}{R}$, but I don't know the time for the second leg of the journey. The time for the first part would have to be $ \frac{x\minus{}y\plus{}1}{x\minus{}y}$...
and for the second part $ \frac{x\minus{}y\plus{}1}{x\plus{}y}$...

Then, for the log, the total time would simply be $ \frac{1}{y}$...

So, adding the two time values for the first leg together and setting these equal to the time the log traveled...I get:$ \frac{(x\minus{}y\plus{}1)((x\minus{}y)\plus{}(x\plus{}y))}{x^{2}\minus{}y^{2}}\equal{}\frac{1}{y}$.
Um...that simplifies a bit...and I get $ 2xy(x\minus{}y\plus{}1)\equal{}x^{2}\minus{}y^{2}$

What now? Am I on a good path, or completely lost?

[hide="Explanation"] I find that the easiest way to teach organized approaches to these problems is by example.  Your suggestion is not bad; let's let $ c$ be the rate of the current and $ b$ be the rate of the boat in still water (both in miles per hour).  The most important thing is to organize all of the information you know.  In chronological order, then:

The boat takes $ \frac{1}{b\minus{}c}$ hours to pass the log, and travels $ 1$ mile.

The boat continues for $ 1$ hour, traveling $ b\minus{}c$ miles.

The boat turns back and travels back to point X, a distance of $ b\minus{}c\plus{}1$ miles which it traverses at a speed of $ b\plus{}c$ miles per hour; from the point the boat has passed the log, then, it has traveled for $ 1\plus{}\frac{b\minus{}c\plus{}1}{b\plus{}c}$ hours.

Meanwhile, the log has traveled $ 1$ mile in the same amount of time.  Since the log is traveling with the current (hence a speed of $ c$ miles per hour), we conclude that

$ \frac{1}{c}\equal{} 1\plus{}\frac{b\minus{}c\plus{}1}{b\plus{}c}\Leftrightarrow$
$ b\plus{}c \equal{} c(b\plus{}c)\plus{}c(b\minus{}c\plus{}1)\Leftrightarrow$
$ b\plus{}c \equal{} 2bc\plus{}c\Leftrightarrow$
$ b \equal{} 2bc\implies$
$ \boxed{c \equal{} 0.5\text{ mph}}$

You see now the importance of organizing what you know carefully.  :) [/hide]

Ah! I counted the time before the boat met the log as well! So close, yet so far.

I think my way would have resulted in the rate of the current if the log moved 2 miles instead of 1...which is wrong. Thanks for the explanation though :D It'll be very helpful, and it's sort of reassuring to me that I was so close to being right. (Pathetic, but whatever :D)

When I do the algebra of your last couple of steps though...I get
$ b\plus{}c\equal{}2bc\plus{}c$...so
$ c\equal{}\frac{1}{2}$

The discrepancy is between the third from bottom and 4th from bottom line.

Thank you! :lol:
I had troble expressing the time it took for the boat to return, which was why I was keep on getting weird numbers...

But shouldn't it be b+c=2bc+c and b=1/2?

[quote="Icy"]When I do the algebra of your last couple of steps though...I get
$ b\plus{}c \equal{} 2bc\plus{}c$...so
$ c \equal{}\frac{1}{2}$

The discrepancy is between the third from bottom and 4th from bottom line.[/quote]

You are correct.  My mistake  :oops:


calculus
integration
logarithms
inequalities
function
derivative
calculus computations

For $ t>0$, prove that $ \frac{1}{2t}\int_t^{3t} x\ln x\ dx\geq 2t\ln 2t$.

can we use Jensen Inequality :)
we have that $ f(x)\equal{}x\ln x$ is a convex function by second derivative: $ f'(x)\equal{}1\plus{}\ln x$, $ f''(x)\equal{}\frac{1}{x}$.

Let $ I \equal{} \int_t^{3t} x\ln x\ dx$

Then use integration by parts to get:

$ I \equal{} \frac{t^2}{2}(9\ln 3\ \plus{} 8\ln t\ \minus{} 4)$

Let $ A \equal{} 2tln(2t) \equal{} 2t[ln t \plus{} ln 2]$

$ f(x) \equal{} I \minus{} 2tA \equal{} t^2[9\ln 3\ \minus{}2(\ln 4\ \plus{}1)] >0$
And the result follows.


The proposed problem is evidently ! Here is an its proof :

[b]Proof.[/b] $m(\widehat {XPY})=360^{\circ}-m(\widehat {YPZ})-m(\widehat {XPZ})=$
$\left[180^{\circ}-m(\widehat {YPZ})\right]+\left[180^{\circ}-m(\widehat {XPZ})\right]=A+B=180^{\circ}-C\Longrightarrow$
$m(\widehat {XPY})+m(\widehat {XCY})=180^{\circ}\Longrightarrow$ the quadrilateral $CXPY$ is inscribed.

Here is [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=70672]a interesting problem[/url] : [color=darkred]in the same conditions, prove that if the quadrilaterals $AYPZ$ and $BXPZ$ are circumscribed, then and the quadrilateral $CXPY$ is circumscribed. Ascertain the triangles which respect the hypothesis of the previous implication.[/color]

[quote="easternlatincup"]Let $X$, $Y$, $Z$ be points on the sides $BC$, $CA$, $AB$ of a triangle $ABC$. Assume that the lines $AX$, $BY$, $CZ$ intersect at a point $P$, and that the quadrilaterals $AYPZ$ and $BZPX$ are cyclic. Prove that the quadrilateral $CXPY$ is cyclic, too.[/quote]

This problem is a trivial consequence of the [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=64328]the Miquel theorem for triangles[/url]:

According to the Miquel theorem for triangles, the circumcircles of triangles AYZ, BZX, CXY concur at one point. In other words, the point of intersection of the circumcircles of triangles AYZ and BZX different from Z lies on the circumcircle of triangle CXY. But we know that the point of intersection of the circumcircles of triangles AYZ and BZX different from Z is the point P (since the quadrilaterals AYPZ and BZPX are cyclic). Thus, the point P lies on the circumcircle of triangle CXY, so the quadrilateral CXPY must be cyclic.

  Darij


inequalities
calculus
calculus computations

Let $ (a_n)$ be a sequence of real numbers that converges to a non-zero real number $ a$. Let $ k \in \mathbb{R}$ be such that $ 0 < k < |a|$. Then there exists $ N \in \mathbb{N}$ such that for all $ n > N$, $ |a_n| > k$. Likewise, if $ |a| \leq k$, then there exists $ N \in \mathbb{N}$ such that for all $ n > N$, $ a_n \leq |a_n| < k$.

[i]Proof[/i]:  If $ a_n \to a$ then $ |a_n| \to |a|$. Then for each $ \epsilon > 0$ there exists a $ N \in \mathbb{N}$ so that whenever $ n > N$, $ \parallel{}a_{n}\minus{}|a\parallel{} < \epsilon$. Then $ |a| \minus{} \epsilon < |a_n| < |a| \plus{} \epsilon$ for $ n > N$. Put $ k \equal{} |a| \minus{} \epsilon$. Then we get the desired result. Likewise, if $ |a| < k$ then put $ k \equal{} |a| \plus{} \epsilon$. Thus we get the result. $ \square$

Is this correct?

You can't "put" your $ k$ as you like.
First you fix $ a \in \mathbb{R}\backslash\{0\}$, and you make some sequence $ (a_n)$ tending to $ a$. Then you fix some $ k \in \mathbb{R}$ with $ 0 < k < |a|$.

Then it is your job to prove that you can find those $ N \in \mathbb{N}$. You can't change your $ k$ at the end.


calculus
derivative
function
advanced fields
advanced fields unsolved

Ah - the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=252679]very function[/url] for which I made the computation of its Fourier transform an exercise in my Fourier analysis notes.

Note that both first-order partial derivatives of $ u$ are bounded and compactly supported - in fact, those partial derivatives are simple functions, constant on triangular wedges. Integrating these simple functions back recovers $ u,$ since $ u$ is continuous. Hence, since $ \nabla u\in L^p$ for all $ p\in [1,\infty],$ we can say that $ u\in W^{1,p}(\mathbb{R}^2)$ for all such $ p.$

(I took the liberty of defining $ u$ to be zero outside of the square $ U.$ We still have a continuous function.)


A circle meets the three sides $ BC, CA, AB$ of triangle $ ABC$ at points $ D_1, D_2; E_1, E_2$ and $ F_1, F_2$ in turn. The line segments $ D_1E_1$ and $ D_2F_2$ intersect at point $ L$, line segments $ E_1F_1$ and $ E_2D_2$ intersect at point $ M$, line segments $ F_1D_1$ and $ F_2E_2$ intersect at point $ N$. Prove that the three lines $ AL, BM$ and $ CN$ are concurrent.

Dear Mathlinkers,
I think that this problem has been presented on this site, but where?
Any reference
Sincerely
Jean-Louis

You are right Mr. jayme. It is China NMO 2005.
It was [url=http://www.mathlinks.ro/viewtopic.php?search_id=1166109447&t=24378]posted here[/url] and it also appeared in [url=http://www.math.ust.hk/excalibur/v10_n3.pdf]some edition of mathematical exaclibur[/url].


I browsed around Getting Started today and realized that the problems are once again getting too difficult. Please refer to the posting guidelines to find out which forum your problems are appropriate for.

If you're not sure whether to post your problem in Getting Started of the Intermediate forum, post in the Intermediate forum.

Is "There and Back Again" a reference to Lord of the Rings, the greatest movie trilogy ever?

Do you think I care if you don't think it matters? If it mattered so little, why did you waste energy typing in this thread?

[quote="jgao"][quote="neelnal"][quote="jgao"]Why are you wasting your energy to type up the reply :P[/quote]

Why are you wasting energy to type up a reply to the reply? :rotfl:[/quote]
Why are you wasting energy to type up a reply to the reply? :rotfl: :P[/quote]

Why are you wasting energy to type up a reply to the reply? :rotfl:

Cuz I like too waste energy. I ate 59 pounds of sugar yesterday.

JK!

[quote="neelnal"][/quote][quote="jgao"][quote="neelnal"][quote="jgao"]Why are you wasting your energy to type up the reply :P[/quote]

Why are you wasting energy to type up a reply to the reply? :rotfl:[/quote]
Why are you wasting energy to type up a reply to the reply? :rotfl: :P[/quote][quote="neelnal"]

Why are you wasting energy to type up a reply to the reply? :rotfl:

Cuz I like too waste energy. I ate 59 pounds of sugar yesterday.

JK![/quote]

Why are you wasting energy to type up a reply to the reply? :rotfl:

Cuz I like too waste energy. I ate 60 pounds of sugar yesterday.

JK! I am just coping and pasting what you wrote :rotfl:  :rotfl:

[quote="jgao"][/quote][quote="neelnal"][/quote][quote="jgao"][quote="neelnal"][quote="jgao"]Why are you wasting your energy to type up the reply :P[/quote]

Why are you wasting energy to type up a reply to the reply? :rotfl:[/quote]
Why are you wasting energy to type up a reply to the reply? :rotfl: :P[/quote][quote="neelnal"]

Why are you wasting energy to type up a reply to the reply? :rotfl:

Cuz I like too waste energy. I ate 59 pounds of sugar yesterday.

JK![/quote][quote="jgao"]

Why are you wasting energy to type up a reply to the reply? :rotfl:

Cuz I like too waste energy. I ate 60 pounds of sugar yesterday.

JK! I am just coping and pasting what you wrote :rotfl:  :rotfl:[/quote]

I think it is time to go there and back again to the purpose of this thread before we get banned.

I'll just lock this thread:P

And yes, the tital is a reference to the Lord of The Rings. I needed something that would grab peoples' attention.  :D


probability
symmetry
probability and stats

The City of Someville is planning to build a commuter train to run along Main Street.
Someville is a one-street town – all its population lives on Main street, and it is
distributed approximately according to a Normal distribution centered on the center of the
town.1 The city government has enough money to build 5 train stops. Every station will
service the people closest to it. (To clarify this point, suppose that Main Street runs in a
straight line East-West, so that the station furthest West services every person who lives
West of it, as well as those to the East of it who live closer to it than to the next station.
Similarly, the station furthest East services everyone who lives East of it, as well as
everyone who lives West of it and closer to it than to the next station. The other stations
service everyone who lives up to halfway toward the next closest station in either
direction).

If each stop is to service the same number of Someville inhabitants, where
should each station be located along main street, in relation to the center of town?

I think it has to do with using Z-scores, but can someone help me here.

My professor provided the following hint for us-

Hint: To answer this question you will need to use a table of standard normal probabilities.

But still my peers and I are unable to solve this.

The standard deviation wasn't explicitly given, so let's make it easy and say the standard deviation is 1 (in whatever unit you want to measure length in).

We first want to divide the line into five segments of equal probability. The normal table I have readily available is in "tail end" form: it gives $ P(Z>z).$ Each of our segments will have probability $ \frac15,$ so I look for $ .20$ and $ .40$ in the table. That gives me $ z\approxeq .845$ (for the 20% probability) and $ z\approxeq .253$ (for the 40% probability). The negatives of these two numbers are also boundaries.

Now what we want is to locate train stations at locations $ z_1,z_2,z_3,z_4,z_5$ in such a way that the boundary points given in the paragraph above are midpoints between stations. That is, we want to satisfy:
\begin{align*}\frac{z_1+z_2}2&=-.845\\
\frac{z_2+z_3}2&=-.253\\
\frac{z_3+z_4}2&=.253\\
\frac{z_4+z_5}2&=.845\end{align*}
Note the problem: this is four linear equations in five unknowns, and hence the solution is not unique. If we want a unique solution, we'll have to pose one more condition. There's an natural candidate for that condition: among all configurations that satisfy these four equations, choose the one that minimizes the expected distance each rider will have to walk. While that could be a messy condition to institute, there's a symmetry condition that makes it easy.

I'll let you finish from there.

thanks for starting. can someone share their answer?

Is this homework? Show a little initiative and do some of your own work.

Hi guys;

I was working on the problem when Kent beat me to the punch. He also had a cleaner method. So I rated his post excellent.

I used slightly more accurate values of -.2534 and .2534 and -.8415 and .8415.

If  you use as the 5th condition z1 + z2 + z3 + z4 + z5 = 0 and then solve that simultaneous set. You get:

$ \left (z_1 \equal{} \minus{} \frac {5881}{5000},z_2 \equal{} \minus{} \frac {1267}{2500},z_3 \equal{} 0,z_4 \equal{} \frac {5881}{5000},z_5 \equal{} \frac {1267}{2500} \right )$

Checking these points as the train stations each station handles 1 / 5 of the load with an error no greater than ABS(.000054).


Let $ x,y,z \geq 0$. Prove that:
$ \frac {x \plus{} y \plus{} z}{3} \minus{} \sqrt [3]{xyz}\leq\frac {2}{3}\max\{|x \minus{} y|,|y \minus{} z|,z \minus{} x|\}$

It's incorrect.
If $ x\equal{}1,y\equal{}z\equal{}100$
$ LHS>45>33\equal{}RHS$
But we can prove this problem.
$ \frac{x\plus{}y\plus{}z}{3} \minus{} \sqrt[3]{xyz} \le \frac{2}{3}\max\{|x\minus{}y|,|y\minus{}z|,|z\minus{}x|\}$

[quote="NextPeace"]It's incorrect.
If $ x \equal{} 1,y \equal{} z \equal{} 100$
$ LHS > 45 > 33 \equal{} RHS$
But we can prove this problem.
$ \frac {x \plus{} y \plus{} z}{3} \minus{} \sqrt [3]{xyz} \le \frac {2}{3}\max\{|x \minus{} y|,|y \minus{} z|,|z \minus{} x|\}$[/quote]

I'm sorry  :D . Can you post solution?

Without loss of generality $ x\ge y\ge z$,
We have $ \max\{|x\minus{}y|,|y\minus{}z|,|z\minus{}x|\}\equal{}x\minus{}z$
however, we have two ineq because $ x\ge y\ge z$
$ \frac{x}{3}\ge\frac{y}{3}$
$ \sqrt[3]{xyz}\ge z$
Hence, we have
$ \frac{y}{3}\plus{}z\le\sqrt[3]{xyz}\plus{}\frac{x}{3} <\equal{}> \frac{x\plus{}y\plus{}z}{3}\le \sqrt[3]{xyz}\plus{}\frac{2(x\minus{}z)}{3} <\equal{}>$Problem.

Please read the first post carefully  :mad:
[url]http://www.mathlinks.ro/viewtopic.php?t=168836&sid=4c567fd9f2d9dc0c8f8b4dcb022ffc84[/url]
It is a problem in M&Y Magazine in VietNam. Please lock the topic  :wink:


Let $ (x_n)$ be the sequence of natural number such that: $ x_1\equal{}1$ and $ x_n<x_{n\plus{}1}\leq 2n$ for $ 1\leq n$. Prove that for every natural number $ k$, there exist the subscripts $ r$ and $ s$, such that $ x_r\minus{}x_s\equal{}k$.

[quote="pco"][quote="tdl"]Let $ (x_n)$ be the sequence of natural number such that: $ x_1 \equal{} 1$ and $ x_n < x_{n \plus{} 1}\leq 2n$ for $ 1\leq n$. Prove that for every natural number $ k$, there exist the subscripts $ r$ and $ s$, such that $ x_r \minus{} x_s \equal{} k$.[/quote]

That's wrong :

Let $ x_n \equal{} 2^{n \minus{} 1}$. It does not exist any subscripts $ r$ and $ s$, such that $ x_r \minus{} x_s \equal{} 5$, for example.[/quote]
I think it will be $ x_r \plus{} x_s$(It is only my guess :maybe: )

[quote="test20"][quote="pco"]That's wrong :
Let $ x_n \equal{} 2^{n \minus{} 1}$. It does not exist any subscripts $ r$ and $ s$, such that $ x_r \minus{} x_s \equal{} 5$, for example.[/quote]
But the condition is: $ x_{n \plus{} 1}\leq 2n$
And not: $ x_{n \plus{} 1}\leq 2x_n$[/quote]

Oooops  :blush: , you're right. I should read more carefully.

Now, it's OK :

Suppose it exist some positive integer $ k$ which can never be written as a difference of two terms of the sequence. Then :

At most one of the integers $ 1$ or $ k\plus{}1$ is in the sequence $ x_n$
At most one of the integers $ 2$ or $ k\plus{}2$ is in the sequence $ x_n$
...
At most one of the integers $ k$ or $ 2k$ is in the sequence $ x_n$

So at most $ k$ numbers in $ [1,2k]$ are in the sequence $ x_n$
But $ x_{k\plus{}1}\leq 2k$ and so at least $ k\plus{}1$ numbers in $ [1,2k]$ are in the sequence $ x_n$

And so contradiction.

And so no $ k$ exists which can never be written as a difference of two terms of the sequence.


A game uses a deck of $n$ different cards, where $n$ is an integer and $n \geq 6$. The number of possible sets of $6$ cards that can be drawn from the deck is $6$ times the number of possible sets of $3$ cards that can be drawn. Find $n$.

[hide="Solution"]There are ${n \choose 6}$ was to draw $6$ cards and ${n \choose 3}$ ways to draw $3$ cards.

We get ${n \choose 6}=6{n \choose 3}$.

Simplifying, we find that $720=(n-3)(n-4)(n-5)$ and $n=13$.[/hide]

Same answer same method.
It seems people here are not seeing others soluitonbefore posting there own.

[quote="JesusFreak197"]I think that these problems are intended so that as many people as want to can solve them, and it's very nice when I can type up the solution and post it. You might be right, but I would like to hear confirmation from the admins, or at least the mods, on this issue.[/quote]

See, what I meant was you can just post that "you got the same answer by the same method".
Writing the solution again makes people read it and then they realize that method is same.

I too solved the above problem by the same method. :D
It's your wish what way yuo like to do.

However, if you look at the other person's solution first, then you aren't thinking of it on your own, and if you look at it after you've typed up your solution, you've just wasted all of your time if you don't post it. Also, some people may write easier-to-read solutions, or have some insight that another didn't.

You are both right about certain points of this issure. Generally, you want to solve it by yourself and then check the other users' solutions before posting. Of course, it doesn't really matter imo if you post your solution anyways.  :)

Now, let's please stop spamming this topic.  :P


uhh... its not like she propels herself up... its like she's climbing up at a constant rate

So the total work done is $mgh$, which is $50*6*10=3000 J$, approximating g as 10 m/s/s
and then you know that power is the rate at which that work is done.
So lets call $t$ the time it takes them to climb up.
We know that it's equal to the 200 watt lightbulb, so set up the equality
$3000J/t = 200W$, or that $t=15s$ .
So she goes up 6 meters in 15 s... you know how to calculate average velocity given a distance and a time...

the work she does is just the change in potential energy

[quote="ROLEX4life"]A 50.0 kg student climbs a 6.00 m long rope and stops at the top. (a) what must her average speed be in order to match the power output of a 200 w lightbulb? (b) How much work does she do?

This has been my approach so far but the 200 w lightbulb thing throws me off:

potential energy at bottom = 0
kinetic energy at bottom = 1/2mv^2

potential energy at top = mgh
kinetic energy at top = 0

so using the law of conservation of energy

1/2mv^2 = mgh and solving for v algebraically would result in v= (2gh)^1/2

so the power that the student actually does = Fd/t = fv = F(2gh^1/2) = ma(2gh^1/2)

i have a feeling this approach is totally wrong. please help[/quote]

a) $P = \vec{F}\cdot\vec{v} \Rightarrow v =\dfrac{P}{F} =\dfrac{200 \text{ watts}}{mg}=\dfrac{200 \text{ watts }}{\left(50\text{ kg }\right)\left(9.8\text{ m/s/s}\right)}$

b) $W = \int_{0}^{h} \vec{F}\cdot\mathrm{d}\vec{s} = mgh = \left(50 \text{ kg }\right)\left(9.8\text{ m/s/s}\right)\left(6.00\text{ m }\right)$


function
real analysis
real analysis unsolved

Find a Lipschitzian function $f: \mathbb{R}^{3}\to \mathbb{R}^{2}$ for wich the plane is contained in the set

$A=\{(x,y,z)\in \mathbb{R}^{3}|f(x,y,z)=0\}$

Which plane? There's no canonical choice in the domain.

Is there anything to stop $f=0$?

I think that $f$ is supposed to be a [i]Lipschitz quotient map[/i]. This means that there exist positive constants $l$ and $L$ such that $B(f(x),lr)\subseteq f(B(x,r))\subseteq B(f(x),Lr)$ for any $x\in\mathbb R^{3}$ and any $r>0$. ($B(\cdots)$ stands for a closed ball in an appropriate space.) Note that the second inclusion is just the Lipschitz condition. The first inclusion prevents $f$ from squashing the space: for example, it implies that $f$ is an open map. Lipschitz quotients can be thought of as nonlinear projections.

Surprisingly, there does exist a Lipschitz quotient $f\colon \mathbb{R}^{3}\to \mathbb{R}^{2}$ for which the preimage of one point contains a plane. I once tried to understand its construction, but gave up. :surrender:  Here is a link (requires subscription):
[url=http://www.springerlink.com/content/y23fpw9ylvaa1846/]Can one squash the space into the plane without squashing? (M.Csornyei)[/url]

There is an open problem along these lines: must a Lipschitz quotient map $f\colon \mathbb R^{3}\to\mathbb R^{3}$ be discrete? (A map is discrete if the preimage of any point does not have accumulation points.) I'll summarize my attempts to solve it:  :wallbash: :wallbash_red: :wallbash_red:


find all (x,y) that x,y are positive integers and:
y^2 * (x-1)= x^5 -1

[quote="xxp2000"]To solve $ y^2\equal{}x^4\plus{}x^3\plus{}x^2\plus{}x\plus{}1$[/quote]

And, do not forget $ x\equal{}1$ and $ y$ is any integer....

Pierre.


inequalities
search
absolute value
inequalities unsolved

Let $ a_{1}, a_{2}, a_{3}, ..., a_{n}$ be $ n$ real numbers whose squares sum to 1. Prove that for any integer $ k\ge 2$, there exists $ n$ integers $ x_{1}, x_{2}, x_{3}, ..., x_{n}$, each with absolute value $ \le k-1$ and not all $ 0$, such that
$ \left| \sum_{i=1}^{n}a_{i}x_{i}\right| \le \frac{(k-1)\sqrt{n}}{k^{n}-1}$

This is a problem from IMO 1987, please search on forum  :wink:


What's the flaw in the following process of solving an equation?

We will attempt to find a $ k$ such that $ ( \minus{} 2)^k \equal{} 2$. We then have

$ ( \minus{} 2)^{2k} \equal{} 2^2\implies (( \minus{} 2)^2)^k \equal{} 4\implies 4^k \implies k \equal{} 1$.

You can't... do that? There isn't a solution in the real numbers.
Hmm... I mean, it's like asking why the following isn't true:

$ \sqrt{\minus{}1}\cdot\sqrt{\minus{}1}\equal{}\sqrt{(\minus{}1)^2}\equal{}\sqrt{1}\equal{}1$

Clearly it makes sense algebraically, but because you're working with complex numbers some things simply do not work.

What you'd really need to do is:

$ k\ln(\minus{}2)\equal{}\ln 2$

$ k\equal{}\frac{\ln 2}{\ln(\minus{}2)}$.

$ 2e^{\pi i\plus{}2\pi n}\equal{}\minus{}2\implies \ln 2\plus{}\pi i\plus{}2\pi n\equal{}\ln(\minus{}2)$

$ \boxed{k\equal{}\frac{\ln 2}{\ln 2\plus{}\pi(i\plus{}2n)}, n\in\mathbb{Z}}$.

I understand that the answer is a complex number, but I still don't see what's wrong with squaring both sides. Why is it wrong? The exponent laws weren't specified for only non-negative integers, so the steps I did seem right.

Well, if you square something, you may often get extraneous (spelling?) solutions, happens a lot with those equations with square roots...

Yea but there is no square root in this equation. So why is it wrong?

It isn't wrong. You just haven't taken all the solutions into account.

[hide="something like grntrtles post"]Solutions for $ ( \minus{} 2)^k \equal{} 2$ are

$ k \equal{} \frac {2 \pi n \plus{} 2 \pi \minus{} i \ln(2)}{\pi \minus{} i \ln2}, n\in \mathbb{Z}$

And solutions for $ 4^k \equal{} 4$

$ k \equal{} \frac {2 \ln 2 \plus{} 2 i \pi n}{2 \ln 2}$[/hide]

Complex logarithm...

Still, as you already know, squaring doubles the number of solutions, and the new solutions obtained that way do not satisfy the original equation. That's why one of such solutions ($ k\equal{}1$) doesn't work.

Hm....mush better explaination than mine.  thanks!

Ok thanks guys! Complex logarithms are so cool. I was actually pondering over what would happen if we hard complex arguments/bases and then I looked it up on wikipedia to see what research has been done. Because you can write negative numbers as complex numbers, it is possible to have a negative base isn't it? Just not in the normal logarithms that are used.

As written, there's no flaw in your argument except that you didn't specify what $ k$ was allowed to be.  If $ k$ is real, then your argument works out fine; it says that no such $ k$ exists.  If $ k$ isn't necessarily real, then the last step fails.


[quote="mathnerd314"]A circle of radius 6 is inscribed in a 30-60-90 triangle. What is the length of the hypotenuse?[/quote]

[hide]I got $12\sqrt{3}+12$.
x=length of side opposite 30 degree angle
$r=A/S$, $6=\frac{\frac{x^2\sqrt{3}}{2}}{\frac{3x+x\sqrt{3}}{2}}$,$6=\frac{x\sqrt{3}}{3+\sqrt{3}}$,$x=6+6\sqrt{3}$,$\boxed{12+12\sqrt{3}}$[/hide]

[hide]You can also see that there is a little square where two radii meet the triangle: 6
Then you can see there is a similar triangle that shares a side, a vertice, and the third vertice is in the center of the circle: \sqrt[2]{3}*6
(It helps with the diagram)

$(6+6*\sqrt[2]{3})2=12+12*\sqrt[2]{3}$[/hide]

[hide]Area = Semiperimeter * Inradius
The area of a 30-60-90 triangle where 2x is the hypotenuse is $\frac{x^2\sqrt{3}}{2}$.
$\frac{x^2\sqrt{3}}{2} = 6s$
Semiperimeter of a 30-60-90 triangle is always $\frac{3x + x\sqrt{3}}{2}$.
$\frac{x^2\sqrt{3}}{2} = \frac{18x+6x\sqrt{3}}{2}$
$x^2\sqrt{3} = 18x + 6x\sqrt{3}$
$x\sqrt{3} = 18 + 6\sqrt{3}$
$x = \frac{18}{\sqrt{3}} + 6$
$x = 6\sqrt{3} + 6$
$2x = \boxed {12\sqrt{3} + 12}$
[/hide]

sorry, but what exactly is the "semiperimeter" and what is the "inradius"?

unfortunately, I haven't been able to find out what it is. If there are any links, that would be great. :D

The semi-perimeter is half of the perimeter.

The inradius is the radius of the circle that is inscribed inside a triangle (or is it any polygon?).


Theaters in Glendale, California can show horror films only on Monday, Tuesday, or Wednesday.

You can't plow a cotton field with an elephant in North Carolina.

In Lehigh, Nebraska it's against the law to sell donut holes.

Under the law of Mississippi, there’s no such thing as a female Peeping Tom.

Anti-modem laws restrict Internet access in the country of Burma. Illegal possession of a modem can lead to a prison term.

Lawn darts are illegal in Canada.

In Idaho a citizen is forbidden by law to give another citizen a box of candy that weighs more than 50 pounds.

Every citizen of Kentucky is required by law to take a bath at least once a year.

It is against the law to whale hunt in Oklahoma. (Think about it...)

A Venetian law decrees that all gondolas must be painted black. The only exceptions are gondolas belonging to high public officials.

In the state of Queensland, Australia, it is still constitutional law that all pubs (hotel/bar) must have a railing outside for patrons to tie up their horse.

According to law, no store is allowed to sell a toothbrush on the Sabbath in Providence, Rhode Island. Yet these same stores are allowed to sell toothpaste and mouthwash on Sundays.

Before the enactment of the 1978 law that made it mandatory for dog owners in New York City to clean up after their pets, approximately 40 million pounds of dog excrement were deposited on the streets every year.

Chewing gum is outlawed in Singapore because it is a means of "tainting an environment free of dirt."

The handkerchief had been used by the Romans, who ordinarily wore two handkerchiefs: one on the left wrist and one tucked in at the waist or around the neck. In the fifteenth century, the handkerchief was for a time allowed only to the nobility; special laws were made to enforce this. The classical heritage was rediscovered during the Renaissance.

For hundreds of years, the Chinese zealously guarded the secret of sericulture; imperial law decreed death by torture to those who disclosed how to make silk.

An old law in Bellingham, Washington, made it illegal for a woman to take more than 3 steps backwards while dancing.

By law, information collected in a U.S. census must remain confidential for 72 years.

Candy made from pieces of barrel cactus was outlawed in the U.S. in 1952 to protect the species.

A slander case in Thailand was once settled by a witness who said nothing at all. According to the memoirs of Justice Gerald Sparrow, a 20th century British barrister who served as a judge in Bangkok, the case involved two rival Chinese merchants. Pu Lin and Swee Ho. Pu Lin had stated sneeringly at a party that Swee Ho's new wife, Li Bua, was merely a decoration to show how rich her husband was. Swee Ho, he said, could no longer "please the ladies." Swee Ho sued for slander, claiming Li Bua was his wife in every sense - and he won his case, along with substantial damages, without a word of evidence being taken. Swee Ho's lawyer simply put the blushing bride in the witness box. She had decorative, gold-painted fingernails, to be sure, but she was also quite obviously pregnant.

In Breton, Alabama, there is a law on the town's books against riding down the street in a motorboat.

Connecticut and Rhode Island never ratified the 18th Amendment: Prohibition.

A few years back, a Chinese soap hit it big with consumers in Asia. It was claimed in ads that users would lose weight with Seaweed Defat Scented Soap simply by washing with it. The soap was sold in violation to the Japanese Pharmaceutical Affairs Law and was banned. Reportedly, the craze for the soap was so great that Japanese tourists from China and Hong Kong brought back large quantities. The product was also in violation of customs regulations. In June and July 1999 alone, over 10,000 bars were seized.

In most American states, a wedding ring is exempt by law from inclusion among the assets in a bankruptcy estate. This means that a wedding ring cannot be seized by creditors, no matter how much the bankrupt person owes.

In New York State, it is still illegal to shoot a rabbit from a moving trolley car.

Vermont, Alaska, Hawaii, and Maine are the four states in the U.S. that do not allow billboards.

Wetaskiwin, Alberta from 1917: "It's against the law to tie a male horse next to a female horse on Main Street."

Women were banned by royal decree from using hotel swimming pools in Jidda, Saudi Arabia, in 1979.

In Riverside, California, there is an old law on the city's books which makes it illegal to kiss unless both people wipe their lips with rose water.

In Saudi Arabia, a woman reportedly may divorce her husband if he does not keep her supplied with coffee.

You can get the death penalty for jumping out a high window.

You can get fined for having a weird haircut in Texas.

[quote="ln(dx/dy)"]You can get fined for having a weird haircut in Texas.[/quote]Really? Please define weird in this case :)

In some state (can't remember), you can't carry an ice cream cone in your pocket.

In another state (can't remember), pickles must bounce.

[quote="RC-7th"][quote="GoBraves"]For hundreds of years, the Chinese zealously guarded the secret of sericulture; imperial law decreed death by torture to those who disclosed how to make silk.
[/quote]

I'm not sure that's supid. I don't see the recipe for CocaCola or the source code to Microsoft Windows lying around in public.[/quote]Trust me, both can be found (to some extent) on the Internet :D

[quote="RC-7th"]But don't we still have trademark laws to prevent someone from stealing someone's else's invention?[/quote]We do, but there are crackers, and hackers, and people that just don't give a **** about those laws :) Thus almost anything can be found today on internet.

Most of the strange American Laws were passed due to some extraneous incident that caused a lot of hubub (purposely being very vague)

Then, the laws stayed because politicians do not want the removal of a lawon their agenda unless enough people care about it. If he/she were to include it, then it would alienate a small percentage of voters that still care about that certain law, whereas if he/she just left it alone, the voters would not be alienated. Thus, all the old laws about horses, and baths that never get removed. They are also never enforced, (except if the police hate you  :) )

And we know someone that tried to rip off coca cola.  PEPSI MAKERS!!! IT'S A CONSPRISY lol  Coke is better.

elvenchamp777

[quote="bleumoose"]i THINK its still illegal to wrestle a bear in the state of alabama.[/quote]

Well, and if it is on self-defense?? On the other hand, I would say that who the heck is gonna wanna wrestle a bear, but I know that in this world anyone have the right to be as stupid as he/she wants!!!

[quote="RC-7th"][quote="GoBraves"]For hundreds of years, the Chinese zealously guarded the secret of sericulture; imperial law decreed death by torture to those who disclosed how to make silk.
[/quote]

I'm not sure that's supid. I don't see the recipe for CocaCola or the source code to Microsoft Windows lying around in public.[/quote]

http://www.sodamuseum.bigstep.com/generic.jhtml?pid=10

:rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:

s*t that is funny

but not very plausible, what would you have? The [b]ugly[/b] police???

 :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:

[quote="Danbert"]It is illegal for an ugly woman to walk the streets of some California city.
Or so my friends tell me (I don't know-I live in MN)

Anybody know for sure?[/quote]

Cool, go Minnesotans.

[quote="Danbert"]It is illegal for an ugly woman to walk the streets of some California city.
Or so my friends tell me (I don't know-I live in MN)

Anybody know for sure?[/quote]

Well, I remember, like 10 years ago, I read in a Newspaper that in a Brazilian beach a lady needed a permit to walk topless. If she didn't forfit some beauty requirements, the permit was denied. That was what I read (La Nacion, Revista Dominical, some sunday quite a long time ago!!)

Best,


Can any one please check my following answers are right or not?

How many 6 character license plates start with two letters and end with 4 numbers if the letters must be different and neither is an O and the numbers form a palindrome and the digit 0 is not used?
 25*24*9*9*9*9

A committee of four students needs to be chosen to organize the school dance.  One student will be in charge of decorations, one will be in charge of the entertainment, one will be in charge of food, and the last will be in charge of advertising.  If there are 10 students willing to do any for these jobs, who many different ways could four students be chosen?
 10C4=210

The first one is incorrect, but the second one is correct
[hide="1"] You were right until the part where it said that the last 4 numbers must be a palindrome.  So there are 25 choices for the first letter, 24 choices for the second letter, 9 choices for the first number, 9 choices for the second number, but the third and fourth numbers are fixed.  Thus the numbers of ways is $24*25*9*9=\boxed{48600}$ [/hide]


integration
function
limit
calculus
inequalities
geometry
real analysis

If $\int^{\infty}_{0}f(x) \,dx <+\infty$,
Prove that $lim_{x\to  \infty}xf(x) = 0$

questions:
[hide]
i let $\int^{\infty}_{0}f(x) \,dx < \int^{\infty}_{0}\frac{1}{x}\,dx$
then compare the two functions.
but i got stuck in the middle. how do i prove that $f(x)$ is always smaller than $\frac{1}{x}$?
apart from the P-series, are there any other possibilities for the function $f(x)$? i guess it would be sufficient to prove the whole thing if i can prove that $f(x)$ is a P-series. but i think there will be other functions which satisfy the criteria. so, how should i proceed?
[/hide]

$f(x)$ need not be less than $\frac{1}{x}$ everywhere. For example, consider $f(x) = 100e^{-x^{2}}$. When $x=1$, this is significantly bigger than $\frac{1}{x}$. There are in fact quite a few possibilities for $f(x)$.

We know that $\int_{0}^\infty \frac{1}{x}$ is divergent. In fact, $\int_{M}^\infty \frac{1}{x}$ is divergent for any $M > 0$ as well. This is what you need to use.

Suppose that $\lim_{x \to \infty}xf(x) = k$ (if it diverges, that's even worse, and the same argument works). Then there exists some large number $M$ so that if $x > M$, ${|xf(x)| > |\frac{k}{2}}|$. Now what problem does this cause with $f$?

EDIT: Who can see where I went wrong? ;) I think I did at least two things wrong in fact.

This is surely false.

Here's one example. Let $g(x)$ be the piecewise linear continuous function whose graph has vertices at $(-1,0),(0,1),$ and $(1,0)$ and is zero outside the interval $[-1,1].$ (Note that the integral of $g$ is $1.$)

Now let $f(x)=\sum_{n=1}^{\infty}g(n^{2}(x-n))$

We have that $\int_{0}^{\infty}f(x)\,dx=\sum_{n=1}^{\infty}\int_{0}^{\infty}g(n^{2}(x-n))\,dx$

Apply the substitution $u=n^{2}(x-n)$ to each integral in the sum to get

$\sum_{n=1}^{\infty}\frac1{n^{2}}\int_{-1}^{1}g(u)\,du= \sum_{n=1}^{\infty}\frac1{n^{2}},$ which converges.

But we do not have $\lim_{x\to\infty}xf(x)=0.$ We do not even have that $\lim_{x\to\infty}f(x)=0,$ since $f(n)=1$ for all natural numbers $n.$

---

The next challenge: can you create a nonincreasing positive function $f$ that still serves as a counterexample to violetcraze's proposed theorem?

[quote="Kent Merryfield"]

The next challenge: can you create a nonincreasing positive function $f$ that still serves as a counterexample to violetcraze's proposed theorem?[/quote]
I think nobody can.  :D
Because the proposition will be true if we add the condition that $f$ is a nonincreasing positive function.
Notice:
$xf(x)\le 2\int_{\frac{x}{2}}^{x}f(t)dt$  :)

This statment reminds me some theorem about series. If $\sum\nolimits_{n = 0}^\infty{a_{n}}$ converges so  ${\lim }_{n \to \infty }na_{n}= 0$. And i don't see any troubles to prove it in a same way.

[quote="violetcraze"]ok, what if i pose the restrictions $f: (0,+\infty) \to (0,+\infty)$ and f is a non incresing function? does the statement holds?[/quote]
Yes, the statement holds. zhaobin has supplied the key part of the argument for it - study his post.

(And "eventually nonincreasing" is good enough.)

Inequality $xf(x)\le 2\int_{\frac{x}{2}}^{x}f(t)dt$  holds because integral is area and by f(x) properties. But $\int_{\frac{x}{2}}^{x}f(t)dt \le \int_{\frac{x}{2}}^{\infty}f(t)dt \to 0$ when $x \to \infty$ :)


$1991$ students sit around a circle and play the following game. Starting from some student $A$ and counting clockwise, each student on turn says a number. The numbers are $1,2,3,1,2,3,...$ A student who says $2$ or $3$ must leave the circle. The game is over when there is only one student left. What position was the remaining student sitting at the beginning of the game?

[hide]

Assigning numbers from $1-1991$ for the positions

After the first round of everyone saying a number, the positions of those still remaining are $1, 4, 7, ..., 1990$

Since there was only one more person left after $1990$, the $1$ position is gone first in the second round. After the second round, it's $4, 13, 22, ..., 1984$

Third, $4, 31, 58, ..., 1975$ (the intervals go up in powers of $3$)

Continuing this, the position of the last person left turns out to be $\boxed{1894}$ (I think).

[/hide]

[quote="Adunakhor"]

Assigning numbers from $ 1 \minus{} 1991$ for the positions

After the first round of everyone saying a number, the positions of those still remaining are $ 1, 4, 7, ..., 1990$

Since there was only one more person left after $ 1990$, the $ 1$ position is gone first in the second round. After the second round, it's $ 4, 13, 22, ..., 1984$

Third, $ 4, 31, 58, ..., 1975$ (the intervals go up in powers of $ 3$)

Continuing this, the position of the last person left turns out to be $ \boxed{1894}$ (I think).

[/quote]
i don't think it's right

the game finish in 7 rounds. We can easily find the connections  between every round.

$ (a_1;a_2;...;a_{1991}) \rightarrow (b_1;b_2;...;b_{663}) \rightarrow (c_1;c_2;...;c_{221}) \rightarrow (d_1;d_2;...;d_{74}) \rightarrow (e_1;e_2;...;e_{25}) \rightarrow (f_1;f_2...;f_8) \rightarrow (g_1;g_2;g_3) \rightarrow (h_1)$

$ h_i \equal{} g_{(3i)}$  ,  $ g_i \equal{} f_{(3(i \minus{} 1) \plus{} 2)}$  ,  $ f_i \equal{} e_{(3i)}$ ,  $ e_i \equal{} d_{(3(i \minus{} 2) \plus{} 1)}$  ,  $ d_i \equal{} c_{(3(i \minus{} 1) \plus{} 1}$  ,  $ c_1 \equal{} b_{3(i \minus{} 1) \plus{} 2}$ , $ b_i \equal{} a_{(3(i \minus{} 1) \plus{} 1)}$.

Substituting $ h_i \equal{} h_1$ we get $ a_{1786}$.

1894 is right.
in 2 case:
1) if there were $3^n$ students standing in a circle, after count, the number of students stand in the circle is $3^{n-1}$ and the first student who got number "1" will stay till the end.
2) with $1991$ students. because $3^6<1991<3^7$ hence we turn this case to case (1) until the number of students standing on the circle is $3^6$ and student $A$ have to be the first student who says "1" in $3^6$. Hence we need $1991-729=1262$ students get out of the circle, respectively $631$ groups of students (each group of 3 students will take out 2)., hence we need $631.3=1893$ students stand before student A, that means position of A is 1894 from the first student.

I took this solution from a Vietnamese book.


algebra
polynomial
linear algebra
linear algebra unsolved

To solve on the beach:
Let a,b,c be (2,2) matrices over C such that:
ab=c,bc=a,ca=b.
     Show that a^5=a.

Correction and additional question:

the requested result in my preceding post is still valid for (n,n) matrices and even in a ring!!

Question 2: Now n=5; a,b,c are (5,5) matrices over C which satisfy the matricial system in the precedent post and  the characteristic polynomial of a is x^5-x.
    Solve the system in this case.

$A^{5}=A(BC)A(BC)A=(AB)C(AB)(CA)=CCCB=C(AB)CB=(CA)(BC)B= BAB=BC=A$

$A$ vanishes the polynomial $x^{5}-x$ wich split with simple roots in C
$A$ is diagonalisble in $M_{n}(C)$

Same way $B^{5}=B$ and $C^{5}=C$


[quote="now a ranger"]HAPPY NEW YEAR!!!

DON't SPEND IT ON AOPS!!!

TO INFINITY AND BEYOND!! THE FUTURE!!!! :lol:[/quote]
Happy New Year 2011.


let there be a 3 digit number abc

abc>cba
abc-cba= (a 3 digit number with digits a,b,c)

what is abc?

[hide="fun problem"]I'll use $\text{abc}$ when referring to $a,b,c$ as digits and $abc$ when used in the normal sense.

We have $\text{abc}-\text{cba}=100(a-c)+(c-a)=100(a-c)-(a-c)$. Since $10+c>10>a$, $a-c<10$. So $a-c\ge1\Longrightarrow100(a-c)>100(a-c)-(a-c)>100(a-c)-10=100(a-c-1)+90$, whence the $10\text{s}$ digit of $\text{abc}-\text{cba}$ is a $9$. The $100\text{s}$ digit is $a-c-1$, so $100(a-c)-(a-c)=100(a-c-1)+90+d$, where $d$ is the $1\text{s}$ digit. Solving for $d$ we get $d=10+c-a$.

Now we'll show that $b$ is not $9$. Assume that it is. Then, equating digits, $a-c-1=c$ and $10+c-a=c$ (clearly it cannot be the other way around). It is easy to check that this leads to a contradiction (namely $c=\frac{8}{3}$). Thus $b$ is not $9$, and therefore $a=9$. Now $c$ is either $a-c-1$ or $10+c-a$, but it obviously can't be the second, so $c=a-c-1=9-c-1\Longrightarrow{c}=4.$

Finally, $\text{abc}-\text{cba}=9\text{b}4-4\text{b}9=904-409=495.$ Thus $b=5$, and so $\text{abc}$ is the number $954$.[/hide]

[hide="its not that fun"]Write the numbers vertically :
abc
[u]cba[/u]
***

where the *'s are digits a,b, and c. I left out the subtraction sign.
Look at the third column from the left, c-a=*. Since c<a, we must carry from b and add 10 to c.
(c + 10) - a = *

Since we carried, the top b in the second column turns into b - 1.

(b - 1) - b = *, we must carry again, (b+9) - b = *,
* = 9, in the second column. Obviously it can't be c.

Let b = 9.

It's either abc-cba=cba, or abc-cba=abc, but the second one is impossible, so let's look at the first one.

Since we have to carry again from the second column, the first column is (a-1) - c = *
Also in the third column, (c+10) -a  =a (we had to carry earlier, remember?).
If c=1, a=3. Does 11-3=3? no.
If c=2, a=5. 12-5 doesn't equal 5, so discard.
If c=3, a=7. That doesn't work either (13-7 doesn't equal 7)
If c=4, a=9. 14-9 isn't 5.

So b can't be 9. c can't be 9, it must be a.

Now it's 9bc - cb9 = *9*

Let the *'s first be b and c, respectively, so 9bc - cb9 = b9c.
In the third column, (c+10) - 9 = c (considering we already carried).
Can c+1=c? Obviously not, so the bottom can't be b9c.

Let the *'s be c and b, respectively.
So 9bc - cb9 = c9b.

In the first column, 9-c=c. We carried in the second column, so it's actually 8-c=c. c=4.
Look in the third column. 14 - 9 = 5. So b = 5.

abc is 954.

[/hide]

[quote="AntonioMainenti"][quote="236factorial"]its not that fun[/quote]It depends on how you solve it. :D

Actually, you're right. :P It seemed fun at the time, probably because I generally dislike questions dealing with digits in numbers.[/quote]

I like those kinds of questions. I was kidding  :P

I used to do that book called "sideways arithmetic from wayside school".


I made a generialization today but I don't know if it is correct. Help please:

For all positive reals $x_i, m$:

$\left(\frac{\sum_{i=1}^n x_i}{n}\right)^m \leq \frac{\sum_{i=1}^n x_i^m}{n}$

[quote="Soarer"]sign reversed.[/quote]
Thanks for telling me it. I've editted the inequality.
Is the inequality true?

Shobber, use \left( and  \right) to get "big" brackets, if you know what I mean :)

[quote="mecrazywong"]Your inequality is true for only $m\ge1$, actually it is just power mean.[/quote]
Thanks.  :)

[quote="Arne"]use \left( and \right) to get "big" brackets, if you know what I mean. [/quote]
But now the sigma sign becomes ugly...?  :?

Ok, then you need to use \displaystyle as well:

\left(\frac{\displaystyle{\sum_{i = 1}^nx_i}}{n}\right)^m

gives

$\left(\frac{\displaystyle{\sum_{i = 1}^nx_i}}{n}\right)^m$

Hm, that doesn't look good either. Strange.

It might look a little nicer if you didn't put the brackets around the whole thing - for example:

$\frac{\left(\displaystyle\sum_{i = 1}^nx_i\right)^m}{n^m}$

Of course, this has the disadvantage of making it less clear...


AMC
USA(J)MO
USAMO
function
search
inequalities proposed
inequalities

Prove that if two positive numbers $ p<q$ are fixed, then for every numbers $ \alpha,\beta,\gamma,\delta,\epsilon\in [p,q]$
\[ \left(\alpha\plus{}\beta\plus{}\gamma\plus{}\delta\plus{}\epsilon\right)\left(\frac{1}{\alpha}\plus{}\frac{1}{\beta}\plus{}\frac{1}{\gamma}\plus{}\frac{1}{\delta}\plus{}\frac{1}{\epsilon}\right)\leq 25 \plus{} 6\left(\sqrt{\frac{p}{q}}\minus{}\sqrt{\frac{q}{p}}\right)^2\]
and when equality holds?

[quote="Yuriy Solovyov"]Prove that if two positive numbers $ p < q$ are fixed, then for every numbers $ \alpha,\beta,\gamma,\delta,\epsilon\in [p,q]$
\[ \left(\alpha \plus{} \beta \plus{} \gamma \plus{} \delta \plus{} \epsilon\right)\left(\frac {1}{\alpha} \plus{} \frac {1}{\beta} \plus{} \frac {1}{\gamma} \plus{} \frac {1}{\delta} \plus{} \frac {1}{\epsilon}\right)\leq 25 \plus{} 6\left(\sqrt {\frac {p}{q}} \minus{} \sqrt {\frac {q}{p}}\right)^2\]
and when equality holds?[/quote]

this is from USAMO 1977.

WLOG $ \alpha \le \beta \le \gamma \le \delta \le \epsilon.$

since $ LHS$ is convex as a function of $ \alpha$,$ \beta$,$ \gamma$,$ \delta$, and $ \epsilon$, it suffices to check it for $ \alpha,\beta,\gamma,\delta,\epsilon\in \{p,q\}.$

for $ \alpha\equal{}\beta\equal{}\gamma\equal{}p$, $ \delta\equal{}\epsilon\equal{}q$ or $ \alpha\equal{}\beta\equal{}p$, $ \gamma\equal{}\delta\equal{}\epsilon\equal{}q$, we get the desired result. :football:


induction
combinatorics unsolved
combinatorics

There are $1999$ people participating in an exhibition. Two of any $50$ peaople do not know each other. Prove that there are at least $41$ people, and each of them knows at most $1958$ peaple.

Let's write the general form of this problem.

Suppose there are $m(n-1)-1$ people on the exhibition. Two of any $m$ people do not know each other.
Prove that there are at least $n$ people, and each of them knows at most $(m-1)(n-1)-2$ people.

Indeed, by induction on $m$. For $m=2$ the assertion is true.

Suppose there are $n$ people s.t. one of them knows at least $(m-1)(n-1)-1$ people. But then for this set of people we know that two of any $m-1$ people do not know each other. It means, by the induction assertion, that there are $n$ people in this subset, which know at most $(m-2)(n-1)-2$ people. Besides of this subset we leaved $n-1$ people, therefore each of selected men knows at most $(m-2)(n-1)-2+(n-1)=(m-1)(n-1)-2$ other people. QED.

[Edit: sorry, the proof is wrong] - [color=red]this edit is wrong[/color]

Sorry again. I wrote that proof is wrong, though it is correct.


[quote="danee 1nonly"]sorta off topic, but was wondering if this equality is true:

tan(a+b) = cot(a - b)[/quote]

So for $b =-a$ we would get $0 = \tan(a+b) = \cot(a-b) = \cot 2a$, for all $a\in\mathbb{R}$?  :wink:

If you'd like to know when it's true: $\tan\theta = \cot\phi$ iff $\theta = \frac{\pi}{2}-\phi+k\pi$, so we should have $a+b = \frac{\pi}{2}-(a-b)+k\pi$, i.e. $a = \frac{\pi}{4}+k\frac{\pi}{2}$. It doesn't matter what $b$ is.  :)


algebra
function
domain
calculus
derivative
logarithms
calculus computations

Some silly question:

Let f(x) = ln (1 + x). Hence $\textup{dom}(f) = (-1, +\infty)$.

Now take the derivative of f, we get $f'(x) = (1 + x)^{-1}$. Is the $\textup{dom}(f') = (-\infty, +\infty) \backslash \{-1\}$? or $\textup{dom}(f') = (-1, +\infty)$?  :huh:

Except that $f'(x)=(1+x)^{-1}$, $1+x\geq 0$.  How do you compute a derivative of an undefined function?

[quote="blahblahblah"]Except that $f'(x)=(1+x)^{-1}$, $1+x\geq 0$.[/quote]
$1+x > 0$

[quote="blahblahblah"]How do you compute a derivative of an undefined function?[/quote]
No way, I can understand that. But why the derivative is defined on some point where its original function is not?

The derivative cannot be defined on a larger domain than $f$ is, precisely because we are calculating the limit of quotients

$\frac{f(x)-f(a)}{x-a}$

as $x\to a$.  Thus we need a value for $f(a)$ to calculate the derivative there.

Remember, a function is a mapping from a set $X$ to another set $Y$ such that for every $x\in X$, there exist $f(x)\in Y$.  That is, not only does the 'rule' determine the function, but also the set which it maps from.  You are assuming that just because the rule is the same on some set, it is the same function.  But it isn't.

You probably want to be looking at something like $\log |x+1|$ to get the desired result, btw.


geometry
circumcircle
Pythagorean Theorem
geometry open

P is a point in the interior of RT△ABC satisfying AP=4,BP=2,CP=1 .Q symmetric to P with respect to AC lies on the circumcircle of △ABC,please figure out the angles of △ABC.

Here's a solution from http://www.aoshoo.com/bbs1,given by "pass":

Assume the R is symmetric to P with respect to BC. ∠ACB = 90°，so R lies on CQ.

Let RB meets ⊙O at D then RD*RB=RC*RQ so RD=1,and AD⊥ BR for AB is the diameter,so AR=AB.thus △AQR≌△APB (SSS). easily we get ∠APB=∠AQR=∠APC.

Assume ∠APB=θ and according to Pythagorean theorem AC^2+BC^2=AB^2,θ=120°(it's a long but not hard process to figure out)[/img]


modular arithmetic
number theory unsolved
number theory

Show that if $ a,b \in \mathbb{N}$  then $ (a \plus{} \frac{1}{2})^n \plus{} (b \plus{} \frac{1}{2} )^n$ is
an integer for only finite many positive integer $ n$

obviously $ n$ is odd if it increases.
Let $ v_2(a\plus{}b\plus{}1)\equal{}k$.
then we have

$ \frac{(2a\plus{}1)^n\plus{}(2b\plus{}1)^n}{2^n} \in \mathbb{N}$

$ \implies 2^n \mid (2a\plus{}1)^n\plus{}(2b\plus{}1)^n$

$ \implies 2^n \mid ((2a\plus{}1)\plus{}(2b\plus{}1))((2a\plus{}1)^{n\minus{}1}\minus{}...\plus{}(2b\plus{}1)^{n\minus{}1})$

$ (2a\plus{}1)^{n\minus{}1}\minus{}...\plus{}(2b\plus{}1)^{n\minus{}1} \equiv 1 \pmod{2} \implies 2^n \mid (2a\plus{}1)\plus{}(2b\plus{}1)$.

the rest is easy since $ v_2(a\plus{}b\plus{}1)\equal{}k$ is constant.


A variable line moves in such a way that the product of the perpendiculars from$ (a,0)$ and $ (0,0)$ is equal to a constant $ k^2$. The locus of the foot of the perpendicular from $ (0,0)$ upon the variable line is a circle the square of whose radius is ..........?

Ans $ \frac {1}{4}a^2 \plus{} k^2$

A quick glance makes this problem seem like Power of a Point.


If N(x)= 3x+8/3x-9

Find N(3)

I don't understand what you do when they say "find N(3)" what do you do with the 3?

Plug in 3 for x.
So basically N(3) = 3*3 +8/3*3 - 9.

my prof also gve us problems like F(x)=4x-10 and to find (F(x+c)-F(x))/(c) I don't understand what to do with this problem either

Just take the idea from alkjash's post and expand.


[hide="3"]$\sum_{cyc}\frac{a^{2}}{ab+2ac}\ge \frac{(a+b+c)^{2}}{3\sum ab}$   by titu's lemma (i believe this is what it is called), anyway

it suffices to prove that
$\frac{(a+b+c)^{2}}{3\sum ab}\ge 1$
$\iff \sum a^{2}-\sum ab\ge 0$
$\iff \sum \frac{1}{2}*(a-b)^{2}\ge0$ which is true by the trivial inequality[/hide]

[quote]titu's lemma (i believe this is what it is called)[/quote]
Some people like to call it the Cauchy-Schwarz inequality  :lol:

I think silouan has given this inequality at least 30 names so far.  :P

Yes  :P  . It's my favorite .In the first book I read inequalities I saw this as Andreescu's inequality . But when I wrote for first time as Andreescu's inequality here in the forum , the Romanian got angry with me .So then I call it as Cauchy . But then I saw Darij called as Cauchy in Engel form , I liked it and that's it.  :lol:


geometry
Functional Analysis
advanced fields
advanced fields unsolved

First some terminology:

A [i]basis[/i] in a Banach space is a sequence $(x_{n})_{n}$ such that every element $x$ of the space can be written uniquely as a convergent series $x=\sum c_{n}x_{n}\ (*)$. A basis $(x_{n})_{n}$ in a Banach space is said to be [i]bounded[/i] if $0<\inf_{n}\|x_{n}\|\le\sup_{n}\|x_{n}\|<\infty$. Finally, a basis is said to be [i]unconditional[/i] if for every $x$ the series $(*)$ converges no matter how we permute the terms.

Prove:

For any bounded unconditional basis $(x_{n})_{n}$ in a Hilbert space $\mathcal H$ there is a bounded invertible linear transformation $S: \mathcal H\to\mathcal H$ such that $(Sx_{n})_{n}$ is an orthonormal basis of $\mathcal H$.

And what about the standart Gram–Schmidt process? Doesn't it satisfy the conditions?

This article is a survey about questions of this type

Lior Tzafriri
Uniqueness of structure in Banach spaces. Handbook of the geometry
of Banach spaces, Vol. 2, 1635--1669, North-Holland, Amsterdam, 2003.


Let
\[ \mathbf{F} = -\frac{GmM}{|\mathbf{r}|^3} \mathbf{r}  \]
be the gravitational force field defined for $\mathbf{r} \neq \mathbf{0}$. Show that there is no continuous differentiable vector field $\mathbf{H}$ satisfying $\mathbf{F} = \nabla \times \mathbf{H}$

[quote="PTynan89"]The answer is actually amazingly simple.  Since the divergence of the curl of any vector is 0, if F were the curl of a vector, it's divergence would be 0.  However, by taking the divergence of F, and realizing that it is nonzero, we reach the conclusion that F is not a curl field.[/quote]

But isn't $\text{div } \mathbf{ F } = 0$ at all places except the origin (where it's not defined)?