Go to noml.org, and it has top scorers for week 0.

hmmm...I got an 11.  One away from the All Star thingy...oh well.

I missed number 2.  I got the answer 127, and can't figure out what I did wrong.  :?

Haha I see what I did wrong now.  :blush:

I thought it said $ y $ was the total value of the ridged coins, not all the coins.  Woops.  :P

i got an 11... i completely underthought number 8... i over looked a lot in that problem for some reason  :blush:

Any idea when the vertical scores will be posted?

Usually takes a week or more! Well, this year I'm not too sure. Cody has less teams and more "administration". But there's also the CX, which in some ways doubles the grading burden.

I think he was out of town for a while. Nathan told me on Saturday that only Problem #3 remained to be graded, so  round results should be posted soon.

Indeed. We'd like to be able to finish grading for each round within a week of the conclusion of Phase 2, but sometimes issues like Dimitar's parents making him study for PSAT and Cody being out of town delay things a bit. Don't worry, the scores will be out tonight.

Yay, Week 0 box scores have been sent via e-mail. :)

Hmm...apparently, my team (Seattle) did not send a Vertical Round or to any thing on Cross-Examination.  Woops.


Hi,
where can I find a proof of the fact that the connected sum of two topological manifolds does not depend on discs and homeomorphisms used?
(for the definition, see for example http://mathworld.wolfram.com/ConnectedSum.html )

Thank you  :)

How formal of a proof are you looking for, because heuristically this is how I would explain it:

Let $ M$ and $ N$ be two manifolds. Let $ M'$ be $ M$ with a disc removed and $ M''$ be $ M$ with a different disc removed. There is a homeomorphism from $ f$ from $ M'$ to $ M''$. Using the same notation with $ N$, there is a homeomorphism $ g$ from $ N'$ to $ N''$. Now these homeomorphisms take boundaries to boundaries. Therefore, there is a natural homeomorphism from $ M'\#N'$ to $ M''\#N''$ which we can make by gluing $ f$ to $ g$ along the boundary of the removed discs.

Thanks for your answer! Now I try to prove rigorously the existence of that omeomorphism, and see why the proof is wrong when one of the two manifolds is not path-connected  :). I'm doing this as a personal exercise.

Sketch: the homeomorphism exists "locally," since it exists in $ \mathbb{R}^n$. Path connectedness, combined with compactness of the unit interval, gets you the rest of the way.


I just wanted to see what other people liked and find something new to read for the summer.

I (currently) like the authors Katy Klise, JK Rowling, Tamora Pierce, and Anne McCaffrey (though her books are harder to read).  There are several others, but I can't remember them all.

didn't one of them write 2000 leagues under the sea?

Orson Scott Card for a mix of Sci-fi and philosophy   :)
Douglas Adams for good 'ol brit humour
ive only read one book (Crime and Punishment) by Fyodor Dostoyevsky but i rather liked it.

oh and Verne wrote 20,000 Leagues Under the Sea, i dont like it; many others do.

I wouldn't call this book my favorite, but some of you might be interested. The other week I read the book [i]The Man Who Knew Infinity: A Life of Ramanujan[/i], about the Indian mathematical genius who collaborated with G.H. Hardy of the UK. This book is truly a page-turner, very entertaining, engaging, and well-researched.

The only downside: throughout the book the author digresses to show some of the mathematics that Ramanujan worked on. Unfortunately, after showing a fancy equation he will proceed to explain the meanings of the symbols in very simplistic terms. It can get very annoying.

For example, aftering showing an incredible formula for the definite integral of some infinite product, he explained what the "elongated S" means in terms of a hot dog analogy. Still, it's an excellent read.

A dream of the red mansion by Cao Xue Qin.

does anyone like mark twain? i read tom sawyer a while a go and, while it was good, it didn't seem much like a classic to me.

tom saywer was okay.  i read it in school.  I liked Huck Finn alot though.

I REALLY am not a hardcore reader, but there is a book titled "The curious incident of the dog in the nighttime" by Mark Haddon. Very interesting adventure about a kid with autism (the thing where you remember every detail in your life) who tries to solve the mystery of a murdered dog. He brings up lots of math topics. The thing is, I've had it for 6 months and I still can't tell if it was a true story or a very intricate fiction. (Rated Pg-13 for foul language.)

Also my shelf is full of comic strip treasuries, like Calvin & Hobbes.

Hey i DIG the following.........

Alistair Macleans

Agatha Christies

the Da vinci code

Lord of the rings

[quote="Hexaditidom"]I REALLY am not a hardcore reader, but there is a book titled "The curious incident of the dog in the nighttime" by Mark Haddon. Very interesting adventure about a kid with autism (the thing where you remember every detail in your life) who tries to solve the mystery of a murdered dog. He brings up lots of math topics. The thing is, I've had it for 6 months and I still can't tell if it was a true story or a very intricate fiction. (Rated Pg-13 for foul language.)[/quote]
i read that book (i own it) I thought that it was a very good book but it had some foul language.

favorite book is c.s. lewis's [u]the screwtape letters[/u].

I hate mark twain... and most classic american literature

The DaVinci Code, Angels and Demons, and Deception Point (all by dan brown) are all good w/ tons of cool fact/details

The Code Book isnt fiction... but it's very interesting. i got my copy from usamts and am still doing the cipher challenge

i don't like charles dickens too much. on the other hand, i love magazines. :D

I really like Orson Scott Card & his Ender Wiggin series, Mark Twain, Roald Dahl, Eddings (their Sparhawk series are the best, they have written a lot of books but their style and character personalities are the same in every series.. It gets really annoying.), anime books (dragonball z is great!), the Star Wars series...

basically fantasy, sci-fi, classics, girly books, I'm into everything. :) heh.

Chip Hilton is the best series ever!  Its so good is undescribable! The first 3 arn't as good as the others, but are still awesome books.  I have 1-12 and 14

Dan Brown, Asimov, Card, Tolkein, Rand, Dickens, Patterson,  Crichton, Blah Blah Blah... Those are just some authors that I've found to be worthwhile.

Boy, I was scared when i read this thread until post #17.  I thought it was kinda incredible that noone had read R.A. Salvatore, until, of course, A+ Math said so.

Anyways my favorite books (and series) and authors are:

The entire set by R.A. Salvatore
Angels and Demons
Da Vinci code
Artemis Fowl trilogy (actually, now it isn't..but yea)
Nightcrawlers (too much fowl language though)
The Prince of ______ series
J.K. Rowling is okay, but her writing style is too slow for me.

I could go on...but I'm too lazy


conics
hyperbola
calculus
calculus computations

can u give me just only one problem solving here in hyperbola,
its my assignment, thx.. :P

EDIT: Kunny has provided you one, see the post (in this forum) entitled: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=115027]Hyperbola and Triangle[/url].

Your question is not very clear, what do you want? Any problem involving hyperbolas? or something else?


Ok. So basically, you say how much money you want today at least, if you're in a good mood, you would probably want 10000 bucks or something. Ok, but..this is the challenging part..
You have to find a picture of that much money and put it on here!
I start:
I want 1000000 :
Pictures
[img]http://www.overseaspropertymall.com/wp-content/uploads/2007/10/1milliondollarbill01.gif[/img]

lol
[img]http://www.brent-krueger.com/images/2004%20Uncirc%20Nickel%20obverse-gs.jpg[/img]
5 CENTS


There are problems in MATHCOUNTS that say:
Find the number of triangles in <figure>.
I get them wrong all the time because I missed a single case.
The only slick way applies to only part of these problems:
Find the number of rectangles whose sides are parallel to the gridlines.
<grid>
then I use combinatorics.
That's the only trick, does anybody else know anything about this type of problem?

Usually after I have used combinatorics, I check for those non-obvious triangles that I'm not counting.  It only takes 5 seconds more, and it always works.


calculus
integration
real analysis
real analysis unsolved

As stated the equality is false since for $n=3$

$\int_{0}^{a}(1-x^{2})^{\frac{n-1}{2}}dx=\int_{0}^{a}(1-x^{2})dx=a-\frac{a^{3}}{3}$
On the other hand

$\int_{0}^{1}(1-x^{2})dx=\frac{2}{3}$,
thus we get that we should have

$a-\frac{a^{3}}{3}=\frac{2}{3}a^{3}$.

[quote="Kent Merryfield"]Perhaps he meant to say

$\int_{0}^{a}(a^{2}-x^{2})^{(n-1)/2}\,dx= a^{n}\int_{0}^{1}(1-x^{2})^{(n-1)/2}\,dx$

That would work (it's just a change of variable.)[/quote]

yes, that's what I meant. So that's easy.
Also my book says this integral

$\int_{0}^{1}(1-x^{2})^{(n-1)/2}\,dx$

can be expressed in terms of

$\Gamma(s)=\int_{0}^{\infty}e^{-t}t^{s-1}\,dx$

How to show that?

Now you're talking about the n-volume of the n-ball. [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=26191]Here[/url] is one recent topic that's somewhat related.

[quote="kalazealot"]Also my book says this integral

$\int_{0}^{1}(1-x^{2})^{(n-1)/2}\,dx$

can be expressed in terms of

$\Gamma(s)=\int_{0}^{\infty}e^{-t}t^{s-1}\,dx$

How to show that?[/quote]
There's formula
\[B(p,q)\overset{def}{=}\int_{0}^{1}x^{p-1}(1-x)^{q-1}dx=\frac{\Gamma(p)\Gamma(q)}{\Gamma(p+q)},\ p,q>0\]
Just use it (make change of variable $t=x^{2}$ in your integral)


For a six sided dice, how many times would we expect to roll it until all numbers appear?What if we extend this to an n sided dice?

The method is to construct the probabilities and invert them, then add them.

We first roll the die. ($ 1$)
There is a $ \frac 56$ chance that the next roll will not be the same as the first. ($ \frac 65$)
There is a $ \frac 23$ chance that the next roll will not be the same as the first two. ($ \frac 32$)
There is a $ \frac 12$ chance that the next roll will not be the same as the first three. ($ 2$)
There is a $ \frac 13$ chance that the next roll will not be the same as the first four. ($ 3$)
There is a $ \frac 16$ chance that the last roll will not be the same as the first five. ($ 6$)

Adding up, $ 1 \plus{} \frac 65 \plus{} \frac 32 \plus{} 2 \plus{} 3 \plus{} 6 \equal{} \frac {147}{10} \equal{} 14.7$ rolls.

meanwhile, for n sided die, you have $ \frac{n}{n}\plus{}\frac{n}{n\minus{}1} \ldots \plus{}\frac{n}{2}\plus{}\frac{n}{1} \equal{} n\sum^n_{k\equal{}1}\frac{1}{k}$

[quote="Wickedestjr"]Wouldn't it be infinity, because you can not guarantee it. If you have a deck of cards, you would have to pull at the most 49 cards from the deck to guarantee that one of the cards that you pull is an ace.[/quote]

well obviously we arent talking about cards; we're talking about dice

The problem isn't asking how many to guarentee getting all the numbers, it's how many times you would expect to roll to get all the numbers, it's a Theoretical Probability, sorta like expected value.

Experimentally, you could roll a die until u got all the numbers, count the number of rolls, repeat a million times, avg, and u should approx get 14.7


If we know that $(\frac{a-1}{1+a})^{n+1}\geq \prod_{i=0}^{n}\frac{x_{i}-1}{1+x_{i}}$ is the inequality
$a^{n+1}\geq \prod_{i=0}^{n}\ x_{i}$ true? Prove your answer.

oops. a is definitely greater than x since $f(x) = \frac{x-1}{1+x}$ is an increasing function.

If I find that $a \geq x_{0}$,  can I say that $a \geq x$ for all n since I can permute the highest x and set it as $x_{0}$? Anyways, I don't think my previous reply is true.


USAMTS
geometry
trigonometry
3D geometry
sphere
algebra
polynomial

How did AoPSers do on the first two rounds of the USAMTS?

well, right now i have a 42, but i argued for one of my solutions in this round so i hope it'll go to 44 *crosses fingers*

50...I'm happy, I thought I'll die from geometry, but I guess not

My cumulative score is 48. Two 24s... being that close to perfect is irritating.

48...2 points off due to laziness
on number 1, round 2 I didn't even bother to do it.. just said blahblah...16 solutions

If you used [i]any[/i] approximations, you will not get full credit (unless they change the rubric, which I don't see happening).

It also says on the USAMTS site: [quote]Remember to fully justify all your answers - don't just send numerical answers.[/quote]

Using an approximation is equivalent to sending in a numerical answer...the only difference is the approximation has some explanation behind it.  Using an approximation does not PROVE that the answer is such and such.  You can use the calculator, if you can rigorously prove that it does give the correct answer.

There was some discrepancy on the grading of this problem.  Specifically, it seems that some (all?) of the trig solutions got docked 2 points for use of a calculator even though some people did not make use of a calculator in their solution.

[quote="beta"][quote="K81o7"]43...I didn't expect points to be taken off for finally using a calculator on number 5...I'm very annoyed that they didn't tell us about that beforehand, because I used a calculator at the end, but got an EXACT answer.[/quote]

No, because if you use a calculator, HOW do you know your answer is exact? Say you get 140 exactly, how do you know the exact answer is not really 140.000001 or something?[/quote]

I checked on my calculator in several ways, for example, asking it to put the answer in fractional form, etc.  But you're right: calculators do have limits...
I'll just assume that all future problems on the USAMTS can be solved without a calculator.

[quote="jackb1115"]Hmmmm ... you might want to send in an appeal about that .. it shouldn't be a big deal if you got a correct, exact answer:

[quote]Except where otherwise noted in problems, participants may use calculators and computers to solve problems. In this case, the participant must explain the program that was used to solve the problem, and must prove that the method used in the program does solve the problem. [/quote]

Unless, maybe they felt you didn't explain your use of the calculator sufficiently??

JB[/quote]

5 points: Complete solution.

4 points: Complete solution, with calculation mistakes.

3 points: Using a calculator to provide an approximate answer.

2 points: Assuming that X is the orthocenter, and then working out the answer, or finding the spheres as the locus of Q.

1 point: Some work on the problem, but no significant progress.

So, according to the rubric, my getting 3 points is justified.  And it's true: my solution was not very insightful; I didn't show that $X$ was the orthocenter, for example.  I just wish they could have told me beforehand that a calculator solution would not allow me full credit.  Then I would looked further into the problem.
This is my first year of USAMTS, so I didn't have much of a goal to aim for.  So I'm not going to make a an appeal, but I think they should tell us beforehand whether a calculator solution is okay.  In any case, I learned the lesson, and won't make that mistake next year.

Well I ended up with a huge messy expression with sines and square roots.  I suppose I could have just worked it out...

Actually NEVER use calculator in a proof in simplifying trig expressions, square roots, logs, etc. because calculators only give approximations in these and you can never argue that your answer is exact.

Actually in most proof contests calculators are banned anyway.  :)

[quote="K81o7"]43...I didn't expect points to be taken off for finally using a calculator on number 5...I'm very annoyed that they didn't tell us about that beforehand, because I used a calculator at the end, but got an EXACT answer.[/quote]
Technically speaking, you got a rounded answer, because using a calculator to simplify trigonometric expressions can never get you an exact answer.  You should be very careful when interpreting calculator use - calculators are meant only to be a tool to simplify the performing of brute-force calculations.  You should never use them to perform a calculation you couldn't do by hand (albeit slowly), because in this case you're almost certainly taking advantage of the ability of a calculator to quickly approximate to 10 or 15 decimal places, and while this may indeed get a correct answer on a math contest with nice answers, there's no guarantee that what you have gotten is indeed the correct answer.  So, I think that the graders are justified in taking off points for calculator use in this case, because it eliminates an actual proof of the answer.  What you've actually shown is that the answer is somewhere very close to 140 degrees, which is not what was intended.

[quote="joml88"]
There was some discrepancy on the grading of this problem.  Specifically, it seems that some (all?) of the trig solutions got docked 2 points for use of a calculator even though some people did not make use of a calculator in their solution.[/quote]
Yes, this confused me a LOT.  It's like as soon as they saw a trig solution, they assumed you used approximations in some way and docked the points, even though there are perfectly acceptable trig solutions that make no use of calculators.

[quote="jackb1115"][quote="calc rulz"][quote="TheAmazingOne"]There's nothing inherently wrong about using a calculator or computer program. If you had used something like Mathematica or a TI-89 to simplify expressions, that wouldn't have lost you points. The losing of points was actually as a result of the blatant lack of rigor due to using the approximations.[/quote]

Yes, but remember, the calculator or program can't solve the problem for you. If you use a Ti-89 or Mathematica to too high an extent (though I'm not completely familiar with those), it may actually solve part of the problem for you, making the problem easier with the calculator/program. Remember that it says "Show that the program used does not solve the problem for you." In such cases where the USAMTSer only has to figure out some of the problem, points should be docked.[/quote]

Where does it say that??

JB[/quote]
I think he's confusing that with the quote "You must also prove that the process you use with the computer does in fact solve the problem."  You can devise or use a computer program to solve a problem for you - you just have to show that it does what you claim that it does.  I've used Mathematica openly once in my problems, to solve a difficult polynomial equation I got on a geometry problem, and I cited the command I used to solve the equation, and though they complained about it on my solution sheet, I did still get a 5 on that problem.

47 - one point off on round 1 because i thought that $\binom{6}{3}=30$  :D , and two off on round 2 because i just put an answer, not a proof, for problem 4.

[quote]
I think he's confusing that with the quote "You must also prove that the process you use with the computer does in fact solve the problem." You can devise or use a computer program to solve a problem for you - you just have to show that it does what you claim that it does. I've used Mathematica openly once in my problems, to solve a difficult polynomial equation I got on a geometry problem, and I cited the command I used to solve the equation, and though they complained about it on my solution sheet, I did still get a 5 on that problem.[/quote]

Yes, thankyou. I thought it said "You must also prove that the process you use with the computer does not solve the problem.", ie that you should do it yourself rather than using a program to make it trivial for you.


function
analytic geometry
graphing lines
slope

Let $a,b,c\in\mathbb{R}$ (real numbers) and the function $f\text{: }\mathbb{R}\rightarrow\mathbb{R}, \ f(x)=ax^2+bx+c$ sucht that $f(-1)<1, f(1)>-1$ and $f(3)<-4$. Show that $a$ is negative.

(-1,1), (1,-1) is a line with slope -1, while (1,-1), (3,-4) is a line with slope -3/2.  If a is positive then the function is concave up.  So going right to left, between the second and third point it must increase faster than 3/2, while between the second and first point it must increase slower than 1.  Contradiction.


[quote="alex2008"]Let be $ a,b,c > 0$ such that $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3$ . Show that :

$ \frac {a^2 \plus{} b^2}{ab(ab \plus{} 1)} \plus{} \frac {b^2 \plus{} c^2}{bc(bc \plus{} 1)} \plus{} \frac {c^2 \plus{} a^2}{ca(ca \plus{} 1)}\ge 3$[/quote]

Of course, by AM-GM
\[ \sum \frac {a^2 \plus{} b^2}{ab(ab \plus{} 1)} \ge \sum \frac2{ab\plus{}1}\ge\sum\frac{18}{ab\plus{}bc\plus{}ca\plus{}3}\ge3
\]

Here is my solution:
By [b]AM-GM[/b], we get:
$ \sum \frac {a^2 \plus{} b^2}{ab(1 \plus{} ab)} \ge \frac {2}{1 \plus{} ab}$
The inequality is equivalent to:
$ \sum \frac {1}{1 \plus{} ab} \ge \frac {2}{3}$ or $ 3 \plus{} ab \plus{} bc \plus{} ca \ge abc(a \plus{} b \plus{} c) \plus{} 3a^2b^2c^2$.
Again, by [b]AM-GM[/b], we get:
$ LHS \equal{} \frac {1}{3}(a \plus{} b \plus{} c)^2 \plus{} \frac {2}{27}(a^2 \plus{} b^2 \plus{} c^2)^3 \plus{} \frac {1}{3}(ab \plus{} bc \plus{} ca) \ge abc(a \plus{} b \plus{} c) \plus{} 2a^2b^2c^2 \plus{} (a^2b^2c^2)^\frac {1}{3}$.
So we need to prove that $ (a^2b^2c^2)^\frac {1}{3} \ge a^2b^2c^2$. But this is so obviously, because $ a^2b^2c^2 \le 1$, which end our proof  :blush: .

[quote="alex2008"]Let be $ a,b,c > 0$ such that $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3$ . Show that :

$ \frac {a^2 \plus{} b^2}{ab(ab \plus{} 1)} \plus{} \frac {b^2 \plus{} c^2}{bc(bc \plus{} 1)} \plus{} \frac {c^2 \plus{} a^2}{ca(ca \plus{} 1)}\ge 3$[/quote]

Of course, by AM-GM
\[ \sum \frac {a^2 \plus{} b^2}{ab(ab \plus{} 1)} \ge \sum \frac2{ab \plus{} 1}
\]
simply use power-AM inequality in $ \sum \frac2{ab \plus{} 1}$
so it becomes,
\[ \frac {1}{3} \sum \frac2{ab \plus{} 1} \geq\ (\frac { ab \plus{} 1 \plus{} bc \plus{} 1 \plus{} ca \plus{} 1}{3.2})^{ \minus{} 1}
\]
and the max. of $ ab \plus{} bc \plus{} ca \equal{} a^2 \plus{} b^2 \plus{} c^2 \equal{} 3$

and so the ineq. becomes
$ \frac {1}{3} \sum \frac2{ab \plus{} 1} \geq\ 1$

and so the result follows


Consider the set $S=\{1,2,...,n\}$. For every $k\in S$, define $S_{k}=\{X \subseteq S, \ k \notin X, X\neq \emptyset\}$. Determine the value of the sum \[S_{k}^{*}=\sum_{\{i_{1},i_{2},...,i_{r}\}\in S_{k}}\frac{1}{i_{1}\cdot i_{2}\cdot...\cdot i_{r}}\] [hide]in fact, this problem was taken from an austrian-polish[/hide]

[quote="campos"]Consider the set $S=\{1,2,...,n\}$. For every $k\in S$, define $S_{k}=\{X \subseteq S, \ k \notin X, X\neq \emptyset\}$. Determine the value of the sum \[S_{k}^{*}=\sum_{\{i_{1},i_{2},...,i_{r}\}\in S_{k}}\frac{1}{i_{1}\cdot i_{2}\cdot...\cdot i_{r}}\] [/quote]

[i]I hope I have interpreted the question correctly, as I understand it for $S^{*}_{k}$ we discard all the terms with k in the subset forming the denominator.[/i]

First we evaluate $S^{*}_{0}$:

For example, when n = 3: $S^{*}_{0}= \frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{1\cdot2}+\frac{1}{1\cdot3}+\frac{1}{2\cdot3}+\frac{1}{1\cdot 2\cdot 3}= 3$

We get the neat expansion
$1+S^{*}_{0}= \left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right) ... \left(1+\frac{1}{n}\right) = \frac{2}{1}\cdot \frac{3}{2}... \frac{n+1}{n}= n+1$
therefore $S^{*}_{0}= n$

Now to find $S^{*}_{k}$. If we remove the term $\frac{1}{k}$ from $S^{*}_{0}$, then for every remaining term with k its denominator subset, there is a term with the same subset except without k. So the sum of the latter group is k times larger than that of the former. But the latter group consists of all the terms without k in their denominators, therefore its sum is $S^{*}_{k}$. Therefore

$S^{*}_{0}-\frac{1}{k}= \frac{k+1}{k}S^{*}_{k}$

Giving the result
$S^{*}_{k}= \frac{kS_{0}-1}{k+1}= \frac{nk-1}{k+1}$ for $k \in S$ otherwise $S^{*}_{k}= n$


A "simple sqaure" is any integer greater than 1 that has only three positve integer factors-itself, its square root, and 1. Which of the folling is a simple square?

A)121

B)100

C)81

D)64

E)33

Can someone give me an clear explanation???

It must be a prime number squared, and the only choice that fits that is $ \textbf{(A)}$ because it is equal to $ 11^2$.


Find the smallest positive integer $ n$ such that $ n^2\minus{}26n\plus{}30$ is at least 1000.

$ n^2 \minus{} 26n \plus{} 30\ge1000\implies(n \minus{} 13)^2\ge1139 \equal{} 33.\sim ^2$
$ \implies n \minus{} 13\ge34\implies n\ge\boxed{47}$.


The Round 3 solutions are now posted on the USAMTS web site, and it is OK for you to discuss the Round 3 problems and solutions.

Round 4 will be posted by the end of January; the submission deadline is March 9.

The solution to #1 doesn't seem to account for the case where the set of odd digits does not include a zero. In this case, we don't have to worry about having a zero as the leading digit.

Edit: Never mind.

I haven't read through the solution yet but I got $ \frac {11(4\cdot 4! \plus{} 5!\cdot 5!)}{9\cdot 9!}\equal{}\frac {1661}{34020}$ for number one.

actually i got $ \frac{11(5!\cdot4\cdot4!\plus{}5!^2)}{9\cdot9!}$ so i think you made and error somewhere because my answer simplifies to $ \frac{11}{126}$ which was the irght answer

When will we get our scores and feedback for Rounds 2 and 3?


linear algebra
matrix
linear algebra unsolved

Find all positive integer numbers n,if there exists $ A$,$ B$from $ M_{n}(C)$ and  $ (AB \minus{} BA)^2 \equal{} I_{n}$.

The answer is: even positive integers.
If $ n\equal{}2p\plus{}1$, suppose $ A,B$ solutions, then $ AB\minus{}BA$ is diagonalizable with eigenvalues $ \pm 1$, but $ tr(AB\minus{}BA)\equal{}0$, contradiction.
For $ n\equal{}2p$, jkust find $ A,B$ s.t. $ AB\minus{}BA\equal{} \mbox{diag}(I_p,\minus{}I_p)$.


Consider the following two scenarios:

(A) Take a random sample of 50 students from an elementary school with children in K to 5th grade.

(B) Take a random sample of 50 1st grade students from the same elementary school.

For each sample, you are interested in the average height. Which of these has the smaller margin of error for 90% confidence?

************************************************************************

So I know margin of error = critical value * standard error of the statistics or margin of error = critcal value * standard deviation of the statistic

The heights will vary more in the scenario (A) as 5th graders are usallly taller than 1st graders, making the standard deviation in scenario (A) greater. When you multiply a constant (critical value will be the same in both scenarios) by a larger number, you will get a larger result, so thus, (B) will yield the smaller margin of error for 90% confidence.

HOW IS MY REASONING?


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $ H$ be a normal subgroup of a finite group $ G$ and $ P$ be a sylow p-subgroup of $ H$ then prove that

                                \[ G\equal{}HN_G(P)\]
where $ N_G(P)\equal{}\{g\in G | gPg^{\minus{}1}\equal{}P\}$.

$ \forall g\in G: P^g\subseteq H^g\equal{}H$, i.e.

$ P^g\in Syl_p(H)\Longrightarrow \exists h\in H: P^g\equal{}P^h\Longrightarrow gh^{\minus{}1}\in N_G(P)$ adn this implies $ G\subseteq N_G(P)H\equal{}HN_G(P)$ and the other inclusion is obvious

this has a name:
[url=http://en.wikipedia.org/wiki/Frattini%27s_argument]Frattini's argument
[/url]


trigonometry
function
group theory
abstract algebra
superior algebra
superior algebra unsolved

Prove that every set can be made a group by properly picking a binary operation.

PS: I have posted this on the Romanian sub-forum, but I didn't really understand the solution given...

If it's infinite, you can even make it a field (and was discussed before: http://www.mathlinks.ro/Forum/viewtopic.php?t=63545).
For getting a group, use cyclic groups for countable sets. For uncountable ones take $\mathbb{Z}^{(\mathrm{card})}$, meaning the "sequences" [in fact mappings $\mathrm{card}\to \mathbb{Z}$] with all but finetely many entries $=0$ and coordinatewise addition.

i think it's about (a,b) intervals where a<b...at least that's in the original problem, which was proposed for the Romanian National Olympiad :wink:

For intervals it is very simple: every open interval $(a,b)\neq\emptyset$ can be mapped bijective to $\mathbb{R}$ and with this bijection we get a group operation on (a,b). (In fact every such interval is a topological group with this operation)

yeah, well...i thought about that myself, but how do you prove the interval is stable with the operation? :oops: sorry but can you be a little bit more specific Gockel? I didn't fully understand your statement :huh: ...

Use the map $f: (-1,1) \to \mathbb{R}, \ x \mapsto \tan(\pi x)$ (all other intervals can be reduced to this one).
$f$ and it's inverse are continuous, thus if we use the operations on $\mathbb{R}$ (I say operations since we can also make it a topological field that way), we get those operations to be continuous too.

yeah, i got it myself.......another function though...thanx for th help anyway...actually you can define fields on each (a,b) interval :ninja:

[quote="ZetaX"]If it's infinite, you can even make it a field (and was discussed before: http://www.mathlinks.ro/Forum/viewtopic.php?t=63545).
For getting a group, use cyclic groups for countable sets. For uncountable ones take $ \mathbb{Z}^{(\mathrm{card})}$, meaning the "sequences" [in fact mappings $ \mathrm{card}\to \mathbb{Z}$] with all but finetely many entries $ \equal{} 0$ and coordinatewise addition.[/quote]
Sorry, I didn't get it.
Could you explain? (What' s going to be the binary operation, by which the set's going to be a group?)
Thanks

In two days I'm starting serving mother-country (army). Hope we'll talk soon.

The binary operation will be addition. So essentially, for the interval $ ( - 1,1)$, you have $ x + y = \tan^{ - 1}(tan(x) + tan(y))$. Let's call this operation $ + _0$. For the interval $ (a,b)$, you have ${{ x + y = \frac {x - (a + b)/2}{b - a}} + _0\frac {y - (a + b)/2}{b - a}}$.

Hope that helps,
~Jacob Steinhardt

I have a couple of days off (is that the word for soldiers or not?)

You nean that, if the set X has the same cardinality with the set G, where  (G, +) is a group, (let f be such a bijection from X onto G), we define, for a,b in X, a*b = f^(-1) (f(a)+f(b)), right?

[quote="ZetaX"]
For getting a group, use cyclic groups for countable sets. For uncountable ones take $ \mathbb{Z}^{(\mathrm{card})}$, meaning the "sequences" [in fact mappings $ \mathrm{card}\to \mathbb{Z}$] with all but finetely many entries $ \equal{} 0$ and coordinatewise addition.[/quote]

Sorry , could u clarify more about the mappings u took here ?

 :)

Let $ C$ be a set of cardinality $ \mathrm{card}$.

It's the sum of $ C$ many copies of $ \mathbb Z$. So you consider the set $ S\equal{}\mathbb Z ^{(C)}$ defined by:
Let $ P$ be the set of all mappings $ f: C \to \mathbb Z$. If $ f,g \in P$, their "sum" is defined by $ (f\plus{}g)(x) \equal{} f(x)\plus{}g(x)$ for all $ x \in C$. This is in fact a group, the product of $ C$ many copies of $ \mathbb Z$.
Now consider the subset $ S$ of $ P$ consisting of those $ f \in P$ such that the set $ C_f : \equal{} \{ c \in C | f(c) \neq 0 \}$ is finite. This is in fact a subgroup of $ P$ (easy to check).
Then $ S$ has the correct cardinality. Proofing this may be a little hard if one hasn't encountered cardinal arithmetic before. Take a look at http://en.wikipedia.org/wiki/Cardinal_arithmetic#Cardinal_arithmetic .
The idea is, that for $ n \in \mathbb Z$, $ n>0$, the sets $ D_n : \equal{} \{ f \in S | \ \ |C_f|\equal{}n \}$ have cardinality $ |(C\times \mathbb Z)^n| \equal{} |C^n| \equal{} C$ and that $ |S| \equal{} |\dot\bigcup_{n \in \mathbb N_0} D_n| \equal{} |\dot\bigcup_{n \in \mathbb N_0} C| \equal{} |C\times \mathbb Z| \equal{} |C|$.
In this proof, a couple of times $ |C| \geq |\mathbb Z|$ was used.

[quote="ZetaX"]Let $ C$ be a set of cardinality $ \mathrm{card}$.

It's the sum of $ C$ many copies of $ \mathbb Z$. So you consider the set $ S \equal{} \mathbb Z ^{(C)}$ defined by:
Let $ P$ be the set of all mappings $ f: C \to \mathbb Z$. If $ f,g \in P$, their "sum" is defined by $ (f \plus{} g)(x) \equal{} f(x) \plus{} g(x)$ for all $ x \in C$. This is in fact a group, the product of $ C$ many copies of $ \mathbb Z$.
Now consider the subset $ S$ of $ P$ consisting of those $ f \in P$ such that the set $ C_f : \equal{} \{ c \in C | f(c) \neq 0 \}$ is finite. This is in fact a subgroup of $ P$ (easy to check).
Then $ S$ has the correct cardinality. Proofing this may be a little hard if one hasn't encountered cardinal arithmetic before. Take a look at http://en.wikipedia.org/wiki/Cardinal_arithmetic#Cardinal_arithmetic .
The idea is, that for $ n \in \mathbb Z$, $ n > 0$, the sets $ D_n : \equal{} \{ f \in S | \ \ |C_f| \equal{} n \}$ have cardinality $ |(C\times \mathbb Z)^n| \equal{} |C^n| \equal{} C$ and that $ |S| \equal{} |\dot\bigcup_{n \in \mathbb N_0} D_n| \equal{} |\dot\bigcup_{n \in \mathbb N_0} C| \equal{} |C\times \mathbb Z| \equal{} |C|$.
In this proof, a couple of times $ |C| \geq |\mathbb Z|$ was used.[/quote]

Gr8 , Thx  :)


real analysis
inequalities
calculus
calculus computations

My question is:
How do I bound t?
For example when the red pulse moves into the blue pulse how did they obtain the inequality $ {\minus{} 3\le}t\le0$

And also the fourth figure where the red pulse is about to leave the blue pulse.
how did the obtian the inequality $ t > 1 and (t \minus{} 3) < 1$

[img]http://i25.tinypic.com/2ivekh3.jpg[/img]

[img]http://i27.tinypic.com/2ue6tfc.jpg[/img]

regards,
bos1234


a and b is an integer between 1 and 100, inclusive, which is the measure of two sides of a triangle. What is the probability that the altitude to a and b are equal?

er...wouldnt that just be asking what is the prob that they're equal?

u sure?

I thought it might be like 1/100...if my previous statement is correct...

No, both of your answers are incorrect. Mith's answer is closer though.

I hate how stupid I am. Yeah your right. careless, careless!

Come on, don't say that, it happens to everyone once in a while.

...answer?

uh...Tare...there are 100x100 ways to choose a and b right?

And there's 100 ways for a and b to be the same right?

so...shouldn't it be 100/10000=1/100?

or am i missing something?

Oh whoops! I typed it up wrong, it's supposed to be much harder than this...oh well. :(

lol...u realize its kinda hard for us to get a correct answer if we dont have the correct problem...correct?

...*sigh* correct...although I don't see you *realizes that the following phrase might be hurtful*...never mind :?

*sigh* I can't say it, not to you...maybe to terat...never mind :twisted:


analytic geometry
calculus
calculus computations

How to find a center of mass of a homogeneous solid hemisphere of radius R and mass M?

  My solution for the $z$ component is $z_{cm}= \frac{3R}{4}$ But the solution in my book yield $z_{cm}= \frac{3R}{8}$.  :|

Without even doing the calculation, one can see that $\frac{3R}4$ is not plausibly correct.

Look at it from the side. $z$ ranges from $0$ to $R,$ so the $z$ coordinate of the center of mass must be in the interval $[0,R].$ But the hemisphere is fatter at the bottom near the equator, and pinches down at the pole. With more of the mass in the lower half than the upper half, the center of mass must be in the lower half. That is, we should have $z_{cm}<\frac{R}2.$


[quote="Interval"]How does one derive the distance formula used to find the distance between points on the xy-plane?[/quote]
[hide="Hint"]You use pythag.[/hide]

${A=\{x,y\}}$
${\alpha ,\beta \}}$

By Vectors, The distance is the norm of the diference of theses 2 vectors

${\{x,y,\alpha ,\beta \}\in \mathbb{R}}$

${D=\|A-B\|}$
${D=\sqrt{|x-\alpha |^{2}+|y-\beta |^{2}}}$
${D=\sqrt{(x-\alpha )^{2}+(y-\beta )^{2}}}$

Did you want to know the distance pron a point to a function, a line, etc, aren't you!?

.

[quote="Quevvy"][quote="max_tm"]Oh my god... you said "your hint does not help" less than 10 minutes after the ckck posted! Just take a pen and paper and spend a little time THINKING about it. He gave you the key to the solution![/quote]
[quote="jli"]ok...this is just really sad. have you drawn a diagram yet???

seriously, you said you spend more that 30 minutes on each of these problems. i will not believe that you cant solve this is 30 minutes

draw horizontal and vertical lines from the two points and the line between them and if you still cant get it... um, yeah[/quote]
Why do you have to be so mean?
If would be nice if you actually explained something...[/quote]

He is being mean because the guy didn't even try to solve it.

I mean seriously, he doesn't even tell the sources to these problems.

And also, he did explain it sort of.

Consider the points (a,b) and (c,d)

Then draw verticals and horizontals from a to c and b to d, and then you have pythags, and you are done.

OMG

[quote]Why do you have to be so mean?
If would be nice if you actually explained something...[/quote]

The fact that he has claimed in previous posts (including one where he refers to the people that answer to his posts as "tutors" some of whom are "inadequate tutors") to spend a lot of time solving these problems, and then the timestamp on his post suggests that he spent maximum 5 minutes using the hint (and my suspicion is he didn't even bother trying to solve, if there is no solution, he is not satisfied; look at any of his posts) and then has the nerve to say very bluntly "your hint does not help" to anyone who takes the time to even bother posting a hint to a question like this... just gets on my nerves  :P.


I was wondering how can i get to count 100! without a calculator. If i am not mistaken , 100! = 99! multiply by 100. I get to noe about factorial using the oxford dictionary btw ^.^

Frequently, problems will not ask you what $ n!$ is; they'll ask you what the last digit is or how many times some digit appears or how many factors it has or something.  In fact, most calculators can't handle $ 100!$.

^^That's the definition of factorial.  I don't think there is any quick trick to getting the answer, so you're going to have to do it by hand if you really want the exact answer.

Or get a calculator that can handle lots and lots of digits.

A question that you might get on an easy test:
What is the units digit of $ 400!$?
Solution: There is at least 1 5 and 1 2 in $ 15!$, so the units digit must be 0.

A slightly harder question:
How many times does the digit 0 appear in $ 85!$?
Solution: Count the number of 5's, since there will always be enough 2's to pair with the 5's. There are 17 multiples of 5, and 3 multiples of 25, which each add another, so there is a total of 20 0's.

[quote="mathcountsfreak"]
A slightly harder question:
How many times does the digit 0 appear in $ 85!$?
Solution: Count the number of 5's, since there will always be enough 2's to pair with the 5's. There are 17 multiples of 5, and 3 multiples of 25, which each add another, so there is a total of 20 0's.[/quote]

how do you know the only 0s are at the end?

I think he meant to say "how many zeros does 85 factorial end with?"

OMFG!!!!!!!! THAT MUCH??????wow thanks...... i jz wanna noe some formula about factorial and what question they will ask about this topic.

Most questions are similar to mathcountsfreak's so learn how to add these number theory applications.  Unfortunately, I never really learned them. :(

alanchou did u find some insanely huge online calc because how did u have enough time to do that by hand? send the url

@fishythefish, you can use programs such as Mathematica or Maple.


group theory
abstract algebra
superior algebra
superior algebra solved

We have a finite group G such that x^2=e for every x from G.Prove that
the number of elements of G is a power of 2.

I guess this problem shoul have a solution despite being so easy.
Suppose there is a prime p>2 such that p|n (cardinal of G).
By Cauhy's Theorem there is an element of order p in G, and
this is a contradiction.

This is the obvious way. The 'standard ' solution, the one you learn in class and which doesn't use Cauchy, is this (at least the outline):

You find a group with 2 elements, {e,x1}, so 2 | |G|. If there exists another element x2 in G you find the subgroup {e,x1,x2,x1x2}, so 4 | |G|. IF you find another x3 in G you repeat the process, obtaining the fact that 8 | |G|, and so on. In the end you get that every power of 2 smaller than |G| divides |G|, and this can only be satisfied if |G| itself is a power of 2.

I knew grobber's solution but I wanted my proof to have only one idea and to finish it quickly.
I don't know what this post was doing in this section and I'm a moderator.
Perhaps our guest posted it here.


I was wondering if the AoPS website could also make a competition similar to USAMTS for students in college?

This is a question that will get more notice if you move it to the USAMTS forum or if you e-mail the creators of this site rather than if you post on the forum.


function
logarithms
algebra
domain
functional equation

Find all functional equations satisfying $ f(xy)\equal{}f(x)\plus{}f(y)$ for all real $ x$ and $ y$

Put $ y\equal{}0 \implies f(x)\equal{}0$

Maybe the function is definied in natural numbers,not all reals ?

Yes sure ,$ \log(xy) \equal{} \log x \plus{} \log y$ ,but this is just for positive numbers $ x,y$

also put $ y\equal{}1 \implies f(1)\equal{}0$.

[quote="enndb0x"]

also put $ y \equal{} 1 \implies f(1) \equal{} 0$.[/quote]

And what did you prove with it?

Again there's the caveat. It would be nice if it was bounded, or monotonic, or continuous.

[quote="t0rajir0u"][quote="keyree10"]Only such function is $ f(x) \equal{} k\log_ax$
NOTE: It is defined only for positive reals.[/quote]
No; $ f(x) \equal{} g(\log x)$ for any solution $ g$ to the [url=http://en.wikipedia.org/wiki/Cauchy_functional_equation]Cauchy functional equation[/url] works, and you can mess around with positive and negative values as well.[/quote]

But whenever you put  $ f(x) \equal{} g(\log x)$ is it not assumed  that $ x>0$

As enddb0x noted, the only solution if you allow $ 0$ in the domain is trivial.  That's why keyree10 restricted the domain.  In fact, you can extend to the negative reals: just take $ f(x) \equal{} g \log |x|$.

I got the answer, but I got it intuitively. How do I get it algebraically?


Find all pairs of functions $ f,g: \mathbb{R}\rightarrow\mathbb{R}$ such that
(i) if $ x < y$, then $ f(x) < f(y)$, and
(ii) $ f(xy) \equal{} g(y)f(x) \plus{} f(y)$ for any $ x,y\in R$.

1. $ x\equal{}1\to f(1)g(y)\equal{}0$. If $ g(y)\equal{}0 \forall y$, then $ 0<x<1,y>0$ give contradition. Therefore $ f(1)\equal{}0$ and $ f(0)\equal{}\minus{}a<0$.
2. $ x\equal{}0$ give $ f(y)\equal{}a(g(y)\minus{}1)$.
and $ f(xy)\equal{}a(g(xy)\minus{}1)\equal{}g(y)a(g(x)\minus{}1)\plus{}a(g(y)\minus{}1)\to g(xy)\equal{}g(x)g(y).$
Because increasing function $ g(x)$, satisfyed $ g(xy)\equal{}g(x)g(y)$ is $ g(x)\equal{}Asign x |x|^c, c>0,A>0$,
we get $ g(x)\equal{}Asign x |x|^c,f(x)\equal{}a(g(x)\minus{}1),a>0$.

[quote="Rust"]
Because increasing function $ g(x)$, satisfyed $ g(xy) \equal{} g(x)g(y)$ is $ g(x) \equal{} Asign x |x|^c, c > 0,A > 0$,
we get $ g(x) \equal{} Asign x |x|^c,f(x) \equal{} a(g(x) \minus{} 1),a > 0$.[/quote]

How do you prove this?

Let $ \phi(x)\equal{}ln(g(e^x))$, then $ \phi(x\plus{}y)\equal{}\phi(x)\plus{}\phi(y)$ and $ \phi(x)$ increase, therefore $ \phi(x)\equal{}cx\to g(x)\equal{}x^c,x>0,c>0$.
In these way, for $ x<0$ we get $ g(x)\equal{}sign(x)|x|^{c_1}$. Let $ x_1,x_2<0$, then $ x_1^2,x_2^2>0$, therefore $ c_1\equal{}c>0$. It give $ g(x)\equal{}sign(x)|x|^c$.

Not sure what you do for $ x<0$, Rust,
but if $ g(x)\equal{}x^c$ for all $ x>0$ and $ g$ strictly increasing,
then $ g(0)^2\equal{}g(0)$ and $ g(0)<g(1)\equal{}1$ gives $ g(0)\equal{}0$
and $ g(\minus{}1)^2\equal{}g(1)\equal{}1$ and $ g(\minus{}1)<g(0)\equal{}0$ gives $ g(\minus{}1)\equal{}\minus{}1$
and then $ g(\minus{}x)\equal{}g(\minus{}1)g(x)\equal{}\minus{}x^c$ for all $ x>0$.
So $ g(x)\equal{}\text{sgn}\, x\cdot |x|^c$ for all $ x\in\mathbb{R}$.


If you toss a coin with a radius of $ .5$ inches, what is the probability it lands entirely in a circle with a radius of $ 1$ inch, given it must land within a square containing the circle where one side of the square is $ 4$ inches?

[hide]For the coin to be entirely within the circle, its center must be within a concentric circle whose radius is $ \frac{1}{2}$. The area of this circle is $ \frac{\pi}{4}\text{ inches}^2$.

The area of the region in which the center of the coin can land is $ 2^2\equal{}4\text{ inches}^2$.

Therefore, the probability is $ \frac{\frac{\pi}{4}\text{ inches}^2}{4\text{ inches}^2}\equal{}\boxed{\frac{\pi}{16}}$.[/hide]

Thanks! I figured it was this simple :).

EDIT: I think you meant $ 16 \text{ inches}^2$ in the denominator.

No. Firstly, the $ inches^2$'s cancel out. Secondly, you do not have units of measurement in a probability.

[quote="skimnc"]EDIT: I think you meant $ 16 \text{ inches}^2$ in the denominator.[/quote]
I don't think so...

$ \frac{\frac{\pi}{4}}{4}\equal{}\frac{\pi}{16}$

Uhh shouldnt the denominator be 9 since the coin also has to land completely inside the square?

[quote="skimnc"]No, he was dealing with the center of the coin landing in the square.  That much is correct.  My argument is that the center must land inside the square, which has side length $ 4 \implies$ the area is $ 16$.

I guess I should have revised the problem to state the center must land in the square.  My bad.[/quote]

So basically, the entire coin has to be in the circle, but not the entire coin (only the center) has to be in the square?

If the coin must land completely within the square,  $ P \equal{} \frac {\frac {\pi}{4}\text{ inches}^2}{9\text{ inches}^2} \equal{} \boxed{\frac {\pi}{36}}$.

If the coin's center must land within the square, $ P \equal{} \frac {\frac {\pi}{4}\text{ inches}^2}{16\text{ inches}^2} \equal{} \boxed{\frac {\pi}{64}}$.

[quote="7h3.D3m0n.117"][quote="the future"]So basically, the entire coin has to be in the circle, but not the entire coin (only the center) has to be in the square?[/quote]

The wording indicates that the [i]entire[/i] coin must land inside the square.[/quote]

So then why is the denominator not 3^2?

It should be (4-.5-.5)^2 because in order for the entire coin to land in the square, the center can be in a 3*3 square.

As I mentioned in my last post, it was a blunder on my part not typing that the center must be within the square.  My bad.

[quote="i_like_pie"]You are correct, the future. :)

For some reason, I considered the diameter instead of the radius when calculating the area of the inner square.[/quote]

well technically what skimnc is saying (only the center has to be in the square), the answer is what you poisted above pi/64.


Let $ S$ be a set of $ n$ points in plane such that the distance between any 2 points is at least 1. Prove that there exists a subset $ T$ with at least $ n/3$ points such that the distance btween any 2 points in $ T$ is at least $ \sqrt{3}$.

The correct statement asks for $ \frac n 7$ instead of $ \frac n 3$.
A solution can be found here :
http://www.math.ca/Competitions/CMO/solutions/sol_2003.pdf

Pierre.


Untrue. Put $b,a$ very large, $b$ a bit bigger than $a$ ($b=\frac{4a}{3}$) and $c$ very small. It will lead to contradiction.


The following series
\[ \sum\limits_{n \equal{} 1}^{ \plus{} \infty } {\left( { \minus{} 1} \right)^{n \minus{} 1} \frac {{nx}} {{1 \plus{} n^{\frac {5} {2}} x^2 }}} ,\qquad \sum\limits_{n \equal{} 1}^{ \plus{} \infty } {\frac {{nx}} {{1 \plus{} n^4 x^2 }}}
\]
converge uniformly or not in $ [0, \plus{} \infty)$?  :)

[quote="Sage!"][hide="Hint"]Use Wirestrass theorem[/hide][/quote]

It seems to be not  :D


real analysis
function
limit
integration
calculus
Support
real analysis unsolved

If is an integrable function on $ \mathbb R$ , then :

$ \lim_{n \to \infty}  \int\limits_{ \minus{} \infty}^{\infty} f(x) cos(nx) dx \equal{} 0$

This can't be correct, take $ f(x)\equal{}\chi_{(\minus{}\varepsilon,\varepsilon)}(x)$ ( $ \varepsilon$ small enough ), $ f$ is integrable, and the desired integral doesn't equal $ 0$.

Sorry man , I forgot to write the limit.

I edited the question.

Thx  :)

I think this can be done with use of Fourier theory. I have done so for $ f$ periodic. Using density of step functions it should be doable.

This question is just from Royden book 3rd edition pg 94 , Q#16.

Inside the book there is a hint to use the problem 15, and it said also the problem easy if $ f(x)$ is just step function .

The Problem 15 was :

$ \forall \epsilon >0$ and for any integrable function $ f$ on $ E$ then :

a) there is a simple function $ \varphi$ such that $ \int_{E} |f\minus{}\varphi| < \epsilon$.

b) there is a step function $ \psi$ such that $ \int_{E} |f\minus{}\psi| < \epsilon$.

c) there is a continuous function $ g$ vanishing outside a finite interval such that  $ \int_{E} |f\minus{}g| < \epsilon$.

 :)

[quote="Mr.Doe"]But for Fourier series to converge ( by points ) to $ f$, $ f$ and $ f'$ need to be piecewise continuous.

Edit: I know I'm a "pain in the behind", with all these comments, so I'm going to try to contribute: problem 15, just states that simple, step and continous functions with compact support are dense in space of integrable functions on $ E$ ( although I'm not sure if (c) is true, for $ E \equal{} \mathbb{R}$ :huh: ). Proving your statement for one of the mentioned functions would solve your problem.[/quote]

True. With use of the results from problem 15 it becomes a lot easier. I assumed these things were not yet proven. I don't see why this would not be true for $ E \equal{} \mathbb{R}$. By assuming your function $ f$ is integrable on $ E$ you can just split your integral up.

While proving the theorem, don't forget to elaborate [i]why[/i] doing the abovementioned would solve your problem.

This has been posted before, of course. My favorite nice dense subspace is the $ C^1$ compactly supported functions, which allow us to integrate by parts.
See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=91119]here[/url] and [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=131611]here[/url].


Asymptote
conics
parabola
analytic geometry
\/closed

[code]
import graph;
size(200);
draw((0,0)--(1,1));
xaxis(-1,3);
yaxis(-2,1);
real f(real x)
{
return x^2-2*x;
}
draw(graph(f,0,2));
[/code]
gives
[asy]import graph;
size(200);
draw((0,0)--(1,1));
xaxis(-1,3);
yaxis(-2,1);
real f(real x)
{
return x^2-2*x;
}
draw(graph(f,0,2));[/asy]
with no trace of the axes or the parabola in the picture. Moreover, if I remove the first drawing command, it just gives an error (probably because no picture is created, but I'm not sure). The code works perfectly well on my home PC, so it must be some problem on the site.

So is this supposed to draw a parabola? ... I dont understand why a line is drawn.

It is supposed to draw the line (the first drawing command), 2 coordinate axes (xaxis() and yaxis() are drawing commands as well), and the parabola (the last drawing command), out of which only the line appears on the picture.

What is interesting is that explicitly specifying that the routines are from the graph package gives the correct picture:
[asy]import graph;
size(200);
draw((0,0)--(1,1));
graph.xaxis(-1,3);
graph.yaxis(-2,1);
real f(real x)
{
return x^2-2*x;
}
draw(graph.graph(f,0,2));[/asy]

There might be an interference with another package ...

Actually, I think I know the cause by now: it is interference of graph.asy with itself. I tried to add import olympiad; command to my code at home and it crashed. The reason is that olympiad.asy as written by Maria used include instead of import and I didn't change that either (see the first few lines) and in the latest versions of Asymptote those two do not really go well together (don't ask me why :)). When I changed the include commands to the import ones in the olympiad.asy, the problem disappeared. Try to do it on the site and   see if it helps.


Mae ran one mile in 10.5 minutes. How many minutes per mile must she average for the next four miles to run five miles in 47.5 minutes? Express your answer as a decimal to the nearest hundredth.

To total $ 47.5$ minutes, over thenext 4 laps she must run $ 37$ miles. $ \frac{37}{4}\equal{}9.25$.

[hide="EXPLANATION"]ill answer this in a series of steps.

1.all together she needs to run 47.5 minutes

2.and she can run 1 mile in 10.5 minutes so we just subtract  $ 47.5-10.5=37 $

3. so next she needs to run 4 miles so you need to divide the 37/4

4.[hide="ANSWER"]9.25[/hide][/hide]

[hide="."][size=200][color=#0040FF][b][i][u]PLEASE RATE MY POST THANK YOU SO MUCH!![/u][/i][/b][/color][/size][/hide]


#5 has a mistake anyway: it must be "Ни жив, ни мертв".


modular arithmetic
combinatorics unsolved
combinatorics

A group of $ n$ people noticed that, for some period of time, three of them might be going for a dinner together, each pair meeting at exactly one dinner. Show that $ n\equal{}1$ or $ 3$ $ (\pmod 6)$

We must have that $ 3 \mid \binom{n}{2}$, so $ n$ is equivalent to 0 or 1 mod 3 and thus to 0, 1, 3 or 4 mod 6.  However, the set of triples containing a fixed person partitions the set of other people into pairs, so $ n$ is odd and thus equivalent to 1, 3 or 5 mod 6 and we have our result.


integration
logarithms
inequalities
induction
inequalities unsolved

We take a pozitive integer number $\ n\geq\ 3$.Prove that \[\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+...+\frac{1}{2n}<\frac{3}{4}\]

it's $1-\frac12+\frac13-\cdots-\frac1{2n}=\frac1{n+1}+\cdots+\frac1{2n}$ and its true :wink:

[quote="me@home"]that's not true, it would be $\sum \frac{1}{n(n+1)}$[/quote]No, it is $\sum \frac{1}{(2n-1)2n}$, or rather $\sum_{k=1}^{n}\frac{1}{(2k-1)2k}$ :wink:

Both inequalities are true !
Don't tell me that you can't prove it ??? :P

PS: try induction :D

#1 I was talking about the identity, and I now realize my stupidity because it is in fact true... I'm sorry :oops:

Yes the identity helps this problem a lot, but anyway you look at this problem it is still pretty easy

sorry for my mistake about that identity, i will make myself forgiven by proving this ineq not with that identity  :oops:

$S = \frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n}$

Multiplying by $2$:

$2S =(\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n})+(\frac{1}{2n}+...+\frac{1}{n+1})$

$\Rightarrow 2S = (\frac{1}{n+1}+\frac{1}{2n})+...+\frac{1}{2n}+\frac{1}{n+1})$

Apply the following [b]"lemma"[/b]:

For any $k \in{1,2,...n}$ it`s true the ineq:

$\frac{1}{n+k}+\frac{1}{2n+1-k}< \frac{3}{2n}$, which is trivial.

So, applying the lemma we have:

$2S < n \frac{3}{2n}\Rightarrow S < \frac{3}{4}$

Amin!


How would you formally define "palindrome"?

We have this intuitive definition: a number that reads the same forwards as backwards.

Examples:
$9$
$1234567890987654321$
$444555444$
$2332$
$31415951413$
$1001$
$2112$
$1010101$

But how do you formally define it?

Here's an interesting fact:
If you take a number, switch its digits and add it to the original number, and keep repeating this process, you eventually get a palindrome.

Is it possible to prove the above fact? Or is it just a conjecture?

[quote="fireemblem13"][quote]If you take a number, switch its digits and add it to the original number, and keep repeating this process, you eventually get a palindrome. Is it possible to prove the above fact? Or is it just a conjecture?[/quote]
Not true. Try 5996. [url]http://mathworld.wolfram.com/PalindromicNumberConjecture.html[/url]
According to that site, your intuitive definition of palindrome is the defintion they provide.[/quote]
Good try, but no.  :P

[quote="wolfram"]Among the first 100000 numbers, [u]5996 numbers[/u] apparently never generate a 'palindromic number'[/quote]
Actually, one counterexample is 196.

(5996 generates a palindrome after 9 steps)

In the base $b$, the number $A$ can be written as $A=\sum_{i=0}^{n}a_{i}b^{i}$, where $a_{i}\in \{0,1,...,b-1\}$ for every $0\leq i\leq n$ and $a_{n}\ne 0$. $A$ is a palindrome if and only if $a_{i}=a_{n-i}$ for every $0\leq i\leq n$.

[quote="fireemblem13"][quote]If you take a number, switch its digits and add it to the original number, and keep repeating this process, you eventually get a palindrome. Is it possible to prove the above fact? Or is it just a conjecture?[/quote]
Not true. Try 5996. [url]http://mathworld.wolfram.com/PalindromicNumberConjecture.html[/url]
According to that site, your intuitive definition of palindrome is the defintion they provide.[/quote]

5996 is the total number of numbers that don't work for the 196-algorithm in the first 100000. Some numbers (from that site) that don't work are 196, 887, 1675, 7436, 13783, 52514, 94039, 187088, 1067869, 10755470


trigonometry
analytic geometry
graphing lines
slope
geometry proposed
geometry

Let $a,b,c$ be real numbers more than $0$. Three points $A(a,0),B(0,b),C(1,c)$ satisfy $\angle ABC=\frac{\pi}{6},\angle BAC=\frac{\pi}{3}$.
Find the maximum and minimum value of the length of $AB$.

Equations of the line $BC$ and $CA \perp BC$ are

$BC:\ \ y - b = (c - b)x$.

$CA:\ \ y - c = -\frac 1 {c - b}(x - 1)$

Let $\tan \alpha = c - b$ be slope of the line $BC$. Slope of the line $AB$ is then

$\tan {\left( \alpha - \frac \pi 3 \right)} = \frac{\tan \alpha - \sqrt 3}{1 + \tan \alpha \ \sqrt 3} = \frac{c - b - \sqrt 3}{1 + (c - b) \sqrt 3}$

and equation of this line is

$AB:\ \ y - b = \frac{c - b - \sqrt 3}{1 + (c - b) \sqrt 3} \ x$

Point $A$ is intersection of the lines $AB$ and $CA$. Solving equations of these two lines yields:

$x_A = 1 + \sqrt 3(c - b)$

$y_A = c - \sqrt 3$

Since $y_A = 0$ $\Longrightarrow$ $c = \sqrt 3$. Since $x_A = a$ $\Longrightarrow$ $a = 4 - b \sqrt 3$.
$AB$ is minimum or maximum iff $S = AB^2$ is minimum or maximum.

$S = AB^2 = 4BC^2 = 4[1 + (c - b)^2]  = 4[1 + (\sqrt 3 - b)^2]$

$\frac{\text dS}{\text db} = 8(b - \sqrt 3) = 0$, $b = \sqrt 3$, $AB = 2$

If $b = 0$, $AB = 4$ and if $a = 0$, $AB = \frac 4 {\sqrt 3}$. Therefore, maximum is $AB = 4$ and minimum is $AB = 2$.

[quote="yetti"]
Let $\tan \alpha = c - b$ be slope of the line $BC$. Slope of the line $AB$ is then

$\tan {\left( \alpha - \frac \pi 3 \right)} = \frac{\tan \alpha - \sqrt 3}{1 + \tan \alpha \ \sqrt 3} = \frac{c - b - \sqrt 3}{1 + (c - b) \sqrt 3}$

and equation of this line is

$AB:\ \ y - b = \frac{c - b - \sqrt 3}{1 + (c - b) \sqrt 3} \ x$
[/quote]

Thank you for your detail solution,yetti.

Regrettably you were wrong in $\angle BAx$.

kunny

Sorry, I indeed switched the size of the angles $\measuredangle ABC$ and $\measuredangle BAC$. Using $\measuredangle ABC = \frac \pi 6$ and $\measuredangle BAC = \frac \pi 3$, the only difference in the solution would be

$y_A = c - \frac{\sqrt 3}{3}$

Minimum $AB = 2 \frac {\sqrt 3}{3}$ for $b = \frac{\sqrt 3}{3}$, $AB = \frac 4 3$ for $b = 0$ and maximum $AB = 4 \frac{\sqrt 3}{3}$ for $a = 0$.


$ABC$ is equilateral triangle. $L,M,N$ are points on $BC,CA,AB$ respectively. Prove that\[MA.AN+NB.BL+LC.CM<BC^2\]

[hide]Multiply through by (1/2)(sin60). The left is the area of ABC minus the area of LMN and the right is the area of ABC.[/hide]

wow same exact explanation. word for word. Ditto on the solution... only i didnt want to do it three times.


inequalities
function
inequalities solved

It is Jensen for concave function 1/(1/x+1/y)

Initial inequality is equivalent to inequality
u2/x+v2/y \geq (u+v)2/(x+y)
which is obviously true, where u=a-b, v=c-d, x=a+b, v=c+d

That's a smart approach, Myth. I did this (not so smart  :D ) :

If we write it as ab(a+b)+cd/(c+d) <= (a+c)(b+d)/(a+b+c+d) and we simply do all the calculations (common denominator and stuff) we get, in the end, 2abcd <= (ad)^2+(bc)^2, which is obviously true.

Both my approaches allow to prove generalization
1/(1/x1+1/y1) + ... + 1/(1/xn+1/yn) \leq 1/(1/(x1+...+xn)+1/(y1+...+yn))

Myth, you said that initial inequality is equivalent to :

U^2/X +V^2/Y  \geq (U+V)^2/(X+Y)

where U=a-b, V=c-d, X=a+b, Y=c+d

How did you get this equivalence?

I think it comes from here:

We can observe that U^2/X+V^2/Y+4ab/(a+b)+4cd/(c+d)=(U+V)^2/(X+Y)+4(a+c)(b+d)/(a+b+c+d) (*)=(a+b)^2/(a+b)+(c+d)^2/(c+d)=(a+b+c+d)^2/(a+b+c+d)=a+b+c+d.
So from (*) 4ab/(a+b)+4cd/(c+d)<=4(a+c)(b+d)/(a+b+c+d) iff U^2/X+V^2/Y>=(U+V)^2/(X+Y).

I said "we can observe", but I don't think I would have observed it by myself :D

Oh, yes.
I don't believe my obvious "inside" idea has such long bases.

Thank you. All this is very interesting and original.
I had never seen it before.


[quote="Tiago Guerreiro"]Hi all  :)

Find the smallest positive integer such that the sum of its nine smaller different multiples (including itself) it's a number with equal digits.[/quote]

[hide]Is it 180?

I didnt know how to solve this but the number has to be a multiple of 5 because 1+2+3+4+5+6+7+8+9=45 so the sum contains a 5 or 0. Then I just went up from there. A two digit number doesnt work because the thousands digit is too low and the hundreds for the lower numbers is too high and at 170 the tens digit equals the thousands in 170$\times$45 so then the lowest is 180?[/hide]

[hide="hint"]
A number which has no different digits and its digits equals $d$ can be
written in the form:

$d \, \times 11 \ \ldots \ 11$.

[hide="2nd hint"]
It pretends to find the minor natural number $n$ such that:

$n+2 \times n+3 \times n+\cdots+9 \times n = d \times 11 \ldots 11$
[/hide]
[/hide]

[hide]5555...(9 times)? cause, as said above, it need to be a multiplication of 45=3*3*5. so it need to end by 0 or 5. o isn't, so 5. 5x=0 mod 9, so the smallest possible is 9 times [/hide]

[hide="That is an incomplete solution"]

Now all you have to do is to find the minor number in the form $11 \ \ldots \ 11$ which is divisible by $9$.

By the criterion of divisibiliy by $9$. the sum of the digits of the number must be a multiple of $9$. Therefore, that number is $111111111$.

Then, $n=111111111/9=12345679$

[/hide]

[quote]
 it need to be a multiplication of 45=3*3*5.
[/quote]

Can you, please, explain me how do you came with that ?

we actually found the same number-I only say what is the su of the first 9 multiples of this number, so if you divide it by 45 we get the same answer. I used dividibility rules... we have the number $54x$ right? so we factor $9\cdot5x$ so it need to be divided by 9 adn 5. is it divisble by 9, the sum of the digits is dvisible by 9. if it divisble by 5, it's end with 5 or 0. we knw that all it's digits are equal! if it end with 0, it will useless. if it end with 5, all the number is made up of 5! so we need the sum to be dvisible by 9-$5x=9k$ it's first happen when $x=9$.


Let $ a,b,c\geq 0$,no two of which are zero.Prove that:
$ \frac{a}{b\plus{}c}\plus{}\frac{b}{c\plus{}a}\plus{}\frac{c}{a\plus{}b}\geq \frac{13}{6}\minus{}\frac{2(ab\plus{}bc\plus{}ca)}{3(a^2\plus{}b^2\plus{}c^2)}$.

Yes ,here is a nice proof, there is a nicer solution, It can be used pqr there p=1
The end you can finish


[tex]a,b,c,d[/tex] is positive real number. Prove that

[tex]a/\sqrt (4c^2+5bd) + b/\sqrt (4d^2+5ca) + c/\sqrt (4a^2+5db) + d/ \sqrt(4b^2+5ac) \geq 4/3[/tex]

May I miss something or this inequality does not holds if $a=c=0$ (or are very close to 0) and $b=d$ ?

Pierre.


function
complex analysis
complex analysis unsolved

If $ f$ is an analytic function so that $ f(0) \equal{} 0$ and $ f(1) \equal{} 0$, find the maximum value of $ f(\frac {1}{2})$. Thank you for the solution before.

:what?: These two conditions are not enough to bound $ f(1/2)$ by anything: consider $ f(z) \equal{} Az(z \minus{} 1)$ with arbitrary $ A\in\mathbb C$.

By the way, a friendly remark about English. Your phrase in the end of the post should be either "Thanks for the solution [b]in advance[/b]" (meaning that you thank for the solution despite it is not there yet) or "Thanks for the solution [b]of my previous problem(s)[/b]" (meaning that you thank for the solution of the problem(s) in your previous post(s)). The way you wrote it is grammatically incorrect and it is also unclear which of the above two things you meant. :)


Matt and Claire hiked to a point $\frac{3}{4}$ the way up the moutain in 43 minutes and 12 seconds. They took the same amount of time to hike each half of the distance from bottom to top. it took them 1 minute and 36 seconds longer to hike the third quarter than the fourth. how many minutges did it take them to hike the entire mountain from bottom to top? express as decimal to nearest tenth.

it takes them 43 min and 12 sec,
which is 2592 sec.
supposing they hiked for equal time for each quater of the
mountain, they would finish the hike in 4x (x standing for the
time it takes to climb one quarter).
however, the 3rd quarter took 96 sec longer, and the first
3 quarters took 2592 sec, you get the equation
3x + 96 = 2592
solving it gives x= 832
so 4x=3328sec
which in min and sec is 55 min and 28 sec

[quote="espark52"]how many minutges did it take them to hike the entire mountain from bottom to top? express as decimal to nearest tenth.[/quote]
[quote="davidlizeng"]which in min and sec is 55 min and 28 sec[/quote]
[hide]Then, it would be $55 \frac{28}{60}$ which is $55.4\bar{6}$ so then the final answer is $55.5$ minutes[/hide]

ooops....
forgot to put ans. into decimal form :D
thanks for the correction quevvy

[hide]Let $x$ be the amount of time they took to climb the third quarter.
The first half is $\frac{216}{5}-x$. And the second is $2x-\frac{8}{5}$.
$\frac{216}{5}-x=2x-\frac{8}{5}$
Solve.... and $x=\frac{224}{15}$
And for the second half they went $\frac{448}{15}-\frac{8}{5}=\frac{424}{15}$.
But the problem wants from bottom to top, so multiply $2$, and the answer is $\frac{848}{15}\approx56.5$[/hide]

davidli, where did you get all those numbers?
I think davidlizeng explains it the best

I don't think davidlizeng's solution is correct. The third quater is not equal to 96 seconds more than the first quarter. The problem said that the third quater took 96 seconds longer than the fourth quarter.

good point davidli
then the equation would be
3x +48=2592
so x=848
and in min, the ans. is 56.5
thanks for the correction
stupid me........

[hide="probably wrong because I'm dumb"]
Let's say it took x minutes to hike the 4th quarter. Then x + (x + 1.6), the time it took to hike the 3rd and 4th quarter, is equal to the time it hiked the first half. The first half is 3/4 minus the 3rd quarter, so it's 43.2 - (x + 1.6), and this is equal to 2x + 1.6 as mentioned above. Now we have 41.6 - x = 2x + 1.6, so 3x = 40. Plug that into (43.2 - (x + 1.6)) and multiply by 2 to get 56.533333... which rounds to 56.5[/hide]


logarithms
calculus
calculus computations

[img]http://img014.photo.21cn.com/photos/user/36217461/20070418162805/m/8594059.jpg[/img]
My solution is 0
[img]http://hiphotos.baidu.com/noodle%5Fmatrix/pic/item/c0ef98cc9703e21400e92876.jpg[/img]

Note that as $u\to0,\ \ln(1+u)=u-\frac{u^{2}}2+O(u^{3}).$

(That's from one of my entries on the "seven power series to memorize" sheet.)

Hence, as $x\to\infty,\ \ln\left(1+\frac1x\right)=\frac1x-\frac1{x^{2}}+O(x^{-3}).$

$x^{2}\ln\left(1+\frac1x\right)= x-\frac12+O(x^{-1})$

$x-x^{2}\ln\left(1+\frac1x\right)= \frac12+O(x^{-1})$

The limit is $\frac12.$

Could you tell me where I could find that post? Thanks

It's in several places, but I just created a [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=145324]new topic[/url] so I'll always be able to find it. In the future, just try searching for the phrase "Seven power series."


jamie owns a flower shop. he is asked for three daisies of different color jamie reaches into his vase of daisies  in which he has one purple daisy, four yellow daisies, five blue daisies, and thirteen orange daisies. how many flowers must jamie pick out to insure that he has at least three flowers of different color?

A. 5     B. 6    C. 10    D. 19    E. None of these

Look at the most extreme case (not particularly rigorous, but it works here). We can get, at worst, get $ 13$ orange daisies and $ 5$ blue daisies, and then we are guaranteed a third color, so our answer is $ 13\plus{}5\plus{}1\equal{}\boxed{\textbf{(D)}\ 19}$.


Let $P(x)$ be a polynomial of degree $n$, so that $P(k)=\frac{k}{k+1}$ for $k=0,1,2,...,n$. Find $P(n+1)$.


A triangle is removed from the corner of a rectangle as shown.  What is the number of square units in the area of the shaded pentagon?

[asy]currentpen = fontsize(10pt); int numberoflines = 50;
for(int i = 0; i<numberoflines; ++i)
draw((0,15*i/numberoflines)--(20,15*i/numberoflines)); path p = (0,0)--(20,0)--(20,10)--(10,15)--(0,15)--cycle;
size(101); clip(p); draw(p); draw((20,10)--(20,15)--(10,15));
label("20",(10,0),S); label("15",(0,7.5),W); label("10",(5,15),N); label("10",(20,5),E);[/asy]

We know the area of the rectangle is 20 * 15 = 300. To find the area of the pentagon, we subtract the area of the triangle from the rectangle. By filling in the reamining lengths, the sides of this triangle are 20 - 10 = 10 and 15 - 10 = 5. So the area of it is (10)(5)/2 = 25.

Thus, the area of the pentagon is $ 300 \minus{} 25 \equal{} \boxed{275}$

Since we are given all the sides we need we can just find the sides of the unshaded shape by subtracting to get that the area of the smaller triangle is 10*5/2=25

The total area is 15*20=300 so the shaded area is 275.


Find the minimal value of k such follow inq true fo all x>0 : $2(x^{k}+\frac{1}{x^{k}}+3)\geq 3(x+\frac{1}{x})$

Here is general :
  Let p,q>0 .Find the minimal value of k such follow inq true for all x>o
                     $p(x^{k}+\frac{1}{x^{k}}+3)\geq q(x+\frac{1}{x})$
   Any one solve it ??


[quote="Allnames"][quote="arqady"]Hello, Inequalities Master!
I think, you rewrite the Vask's book.  :D
Why do you do it?[/quote]
Can you tell me about this book?i think it is very wonderful[/quote]
I agree with you, it's a wonderful book!
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=100399

arqady 我做过此书一些题，是我国一不等式专家发的，但不全，有些题不太会做。
我在中国，买不到，您有此书电子版？如果有，方便的话，电邮到：lbylfx@sina.com
（Some problem, I have done this book is an inequality in China expert, but not all, some questions will do.
I buy in China, less than, you have the book online? If you have, it's convenient, would email to: lbylfx@sina.com
）


Express $\frac{1}{\sqrt{2005 + \sqrt{2005^2 - 1}}}$ as the difference of two square roots.

we can generalised to

$\frac{1}{\sqrt{a+\sqrt{a^2-1}}}=\frac{\sqrt 2}{2}\left(\sqrt{a+1}-\sqrt{a-1}\right)$

for $a\geq 1$

[hide="solution"]$\frac{1}{ \sqrt{a + \sqrt{a^2 - 1}}} = \frac{ \sqrt{a - \sqrt{a^2 - 1}}}{\sqrt{a^2 - (a^2-1)}} = \sqrt{a - \sqrt{a^2 - 1}}$ ?

$\sqrt{a - \sqrt{a^2 - 1}}=\sqrt{x}-\sqrt{y}$

$\implies a-\sqrt{a^2-1}=x+y-2\sqrt{xy}$

$a=x+y,a^2-1=4xy$

Now do stuff and it comes $x=1003,y=1002$?

So the answer is

${\boxed{\boxed{\boxed{\boxed{\sqrt{1003}-\sqrt{1002}}}}}}$[/hide]


Hope this one is right:

Suppose $n \ge 25$ and $t_{1}, t_{2}, ..., t_{n}$ are positive reals such that

$n^{2}+1 > (t_{1}+t_{2}+...+t_{n})(\frac{1}{t_{1}}+\frac{1}{t_{2}}+...+\frac{1}{t_{n}})$

Show that for distinct $i, j, k$, $t_{i}, t_{j}, t_{k}$ form the sides of some [i]acute[/i] angled triangle.

(In IMO 2004, it was $n \ge 3$ and we were asked to prove that they form the sides of some triangle, not necessarily acute)

An update:

A) if $t_{i}> 0$ and $n \ge 13$ and $n^{2}+1 > (t_{1}+t_{2}+...+t_{n})(\frac{1}{t_{1}}+\frac{1}{t_{2}}+...+\frac{1}{t_{n}})$ then show that for all distinct $i,j,k \le n$, $t_{i}, t_{j}, t_{k}$ form the sides of an [i][u]acute[/u][/i] angled triangle.

B) Show that there exist $t_{i}> 0$ such that for $3 \le n \le 12$, $n^{2}+1 > (t_{1}+t_{2}+...+t_{n})(\frac{1}{t_{1}}+\frac{1}{t_{2}}+...+\frac{1}{t_{n}})$ and for some distinct $i,j,k \le n$, $t_{i}, t_{j}, t_{k}$ form the sides of a [i][u]right[/u][/i] triangle.

In short show that $12$ is the best possible upper bound on $n$ for which the triangles are not forced to be acute.


inequalities
geometry
3D geometry
tetrahedron
combinatorics unsolved
combinatorics
graph theory

All edges of a polyhedron are painted with red or yellow. For an angle of a facet, if the edges determining it are of different colors, then the angle is called [i]excentric[/i]. The[i] excentricity [/i]of a vertex $A$, namely $S_A$, is defined as the number of excentric angles it has. Prove that there exist two vertices $B$ and $C$ such that $S_B + S_C \leq 4$.

I had all the ingredients for a solution as soon as I saw the problem, except the little observation at the end, that eluded me until now! For a polyhedron (or a planar map), Euler's formula $V + F = E + 2$ yields the known inequality $E \leq 3V - 6$. For conformity, here is a proof. Let $F_k$, $k\geq 3$ be the number of facets with $k$ sides. Then $F = F_3+F_4+\cdots + F_k + \cdots$, and by a double-counting argument $2E = 3F_3 + 4F_4 + \cdots + kF_k + \cdots$.
It follows that $2E-3F = F_4 + 2F_5 + \cdots + (k-3)F_k + \cdots \geq 0$, hence $3(E+2) = 3V + 3F \leq 3V + 2E$, and so $E \leq 3V - 6$.

The next step is to notice that if we can draw an extra edge (a diagonal of a facet), keeping the graph planar, then no matter how we colour this extra edge, the excentricities of the vertices do not decrease (either stay the same, or increase by $2$). Therefore it is enough to prove the problem when all facets have been triangulated; under that construction we will in fact now have $2E=3F$, $E = 3V-6$ and thus $F = 2V-4$.

Thirdly, denote by $X_f$ the number of excentric angles of a facet $f$. Since each facet is a triangle, it follows $X_f \leq 2$, and then $\sum S_v = \sum X_f \leq 2F = 4V - 8$, meaning there exist (at least) two vertices $u$ and $w$ of excentricities $S_u, S_w < 4$.

And now, the final step. In all configurations, for any vertex, its excentricity must be an even number (trivially proved, but somehow taking me a little long to realize); this means $S_u, S_w \leq 2$, and so $S_u + S_w \leq 4$.

EDIT. In fact, we proved the stronger statement that there exist four vertices $a,b,c,d$ having $S_a + S_b + S_c + S_d \leq 8$; this is the maximum that can be said, since we can colour the edges of a tetrahedron $abcd$ such that $S_a = S_b = S_c = S_d = 2$.

nice problem!

Define the excentricity of a face $A$, namely $T_A$, to be the number of excentric angles it has. Note that the excentricity of any vertex has to be even. Let the vertex set be $V$, the edge set be $E$, and the face set be $F$.

It suffices to prove that $\sum_{v\in V}S_v\leq 4|V|-4$. Let $k=\frac{2|E|}{|F|}$. By Euler's formula, we have $|V|-|E|+|F|=2$ so $|F|=\frac{|V|-2}{\frac{k}{2}-1}$. Any triangle face must have excentricity at most $2$. Let $f_m$ be the number of faces with $m$ edges.

We will prove that \[f_3\geq (4-k)|F|.\] For the sake of contradiction, assume the contrary. We have
\begin{align*}
k|F|&=2|E|\\
&=\sum_{m=3}^\infty mf_m\\
&\geq 3f_3+4(|F|-f_3)\\
&>4|F|-(4-k)|F|=k|F|,
\end{align*}
contradiction.

Now
\begin{align*}
\sum_{v\in V}S_v&=\sum_{f\in F}T_f\\
&\leq 2f_3+\sum_{m=4}^\infty mf_m\\
&=\sum_{m=3}^\infty mf_m-f_3\\
&\leq (2k-4)|F|\\
&=4|V|-8
\end{align*}
so we are done. $\square$


Let $ a_1; a_2;...; a_{2007}$ be pairwise distinct integers, all greater than 1, such that $ \sum_{i\equal{}1}^{2007} a_i\equal{}2017035$
Could it be possible that the sum $ \sum_{i\equal{}1}^{2007} a_i^{a_i^{a_i}}$ is a perfect square?


my math teacher always says not to do

$\frac{1}{a+c}=\frac{1}{a}+\frac{1}{c}$.

but i was wondering is this ever true for any numbers?

[hide]
multiply through by $ac(a+c)$

$ac = c(a+c)+a(a+c)$
$ac = ac+c^{2}+a^{2}+ac = a^{2}+c^{2}+2ac$
$a^{2}+ac+c^{2}= 0$

$a = \frac{-c \pm \sqrt{c^{2}-4c^{2}}}{2}$

under the square root:
$-3c^{2}> 0$
$c^{2}< 0$

meaning c has no real solutions
[/hide]

[hide="A Little Note"]
$\frac{1}{a+c}=\frac{1}{a}+\frac{1}{c}$
$\frac{1}{a}+\frac{1}{c}=\frac{a+c}{ac}$
is another way to write the right side of the equation
[/hide]

so no real solutions. how about imaginary?

[hide]
infinite solutions:
let c be any complex number and

$a = \frac{-c \pm \sqrt{c^{2}-4c^{2}}}{2}$

$a = \frac{-c \pm \sqrt{-3c^{2}}}{2}$

$a = \frac{-c \pm ic\sqrt{3}}{2}$

[/hide]

[quote="chickendude"][hide]
$a = \frac{-c \pm ic\sqrt{3}}{2}$
[/hide][/quote]
[hide="something cool about that number"]
Divide it by c and cube it
Both plus and minus cubed are the same thing, and it's a very special number.[/hide]

[hide="that is clever"]
It's $1 cis \pm\frac{2\pi}{3}$
which cubes to
$1 cis \pm2\pi$
that's interesting there
[/hide]

[quote="rnwang2"][hide]
$\frac{1}{a+c}=\frac{1}{a}+\frac{1}{c}$

$\implies a^{2}+ac+c^{2}=0$

$\implies a^{3}-c^{3}=0$[/hide][/quote]
Wait, what?
\[\frac{1}{1+1}\ne \frac{1}{1}+\frac{1}{1}\]
I think you made an algebraic error somewhere. But I don't have to time to find it, as the school bell is ringing right now.  :)

[quote="rnwang2"][hide]
$\frac{1}{a+c}=\frac{1}{a}+\frac{1}{c}$

$\implies a^{2}+ac+c^{2}=0$

$\implies a^{3}-c^{3}=0$[/hide][/quote]

[hide="an error"]
in your final equation, $a \neq c$

$a^{2}+ac+c^{2}=0$

$\implies (a-c)(a^{2}+ac+c^{2})=a^{3}-c^{3}= 0$
only when $a \neq c$
[/hide]

Right...I forgot to mention that.   :blush:


I wonder if it's possible to measure the correlation of 3 timeseries? Haven't fund anything serching the net..

I want to compare 3 different timeseries, and see it's correlation. How do you measure this? If to of them is highly correlated two eachother, but the third isn't.. how will this show up?

Anyone have an idea?

/Johan

I don't see it not being possible to check the correlation of three things at once and represent it as a chart in $R^{3}$, but I doubt it would be fun (or even manageable) without the aid of a computer.[/b]

My idea (boring though) would be to correlate them pairwise and you should be able to work with that.

Thanks for your replies, i think I can use both of them.

If i correlate a-b, b-c and a-c, and then multiply them as r(a)*r(b)*r(c) I should get a decent value to work out from. I only need a "signal" when all variables are highly correlated...


geometry
conics
hyperbola
perpendicular bisector
angle bisector
geometry proposed

Let $ABCD$ ($AB\neq AD$) be a quadrilateral with incircle centered at $O$. Let $OA$ and $OC$ meet the incircle at points $E$ and $F$ respectively. Prove that $\angle DAB=\angle BCD$ iff $EF\perp BD$.

Since $OA,OC$ cut the incircle in two points (each), I'll assume $E,F$ are on the rays $(OA,(OC$ respectively.

Let's first notice that $\angle BAD=\angle BCD$ iff $BD$ is the perpendicular bisector of $AC$. On the other hand, the statement $BD\perp EF$ is equivalent to $BD$ being parallel to the angle bisector of $\angle EOF$, so it's equivalent to the fact that $AO,CO$ make equal angles with $BD$ (call this statement $(*)$).

Considering that $AO,CO$ are the tangents in $A,C$ to the hyperbola with foci $B,D$ and passing through $A$, $(*)$ is obvious.


Prove that if A is an $ n\times n$ symmetric matrix, then A has an orthonormal set of n eigenvectors.

See the Hermitian matrices section [url=http://en.wikipedia.org/wiki/Spectral_theorem]here[/url].
[url=http://www.ltcconline.net/greenl/courses/203/MatrixOnVectors/symmetricMatrices.htm]This page[/url] on symmetric matrices might answer some of your other questions :)

Every hermitian matrix is unitarily similar to the diagonal matrix of its eigenvalues. This can be proved directly, or we can observe first that every $ A \in M_{n \times n} (\mathbb{C})$ is unitarily similar to a triangular matrix and then use the definition of hermitian matrix.


projective geometry
geometry proposed
geometry

$AD$ is an altitude of  $\triangle ABC$ where $D \in BC$ but differ from $B$ and $C$.  And  $\mu $ is the circle with diameter $AD$.  $E$ and $F$ are the points where $\mu$ intersects $AC$ and $AB$ respectively. $BE$ and $CF$ intersect  the circle $\mu$ again at $K$ and $L$. Prove that $KL$ is parallel to $BC$.

(I know there is a way to describe the picture shortly.)

What is this, Xeroxia ? Do you going to toilet ? We are looking forward to come back.

apply Pascal to hexagon AFLKEA

We know that the intersection of AF and KE is B, that of FL and EA is C, and if we call X the intersection of LK and AA (AA meaning the tangent at A to the circle), Pascal's theorem tells us that B, C, and X are collinear. In other words, BC, LK, and AA are concurrent (all passing through X). However, AA is perpendicular to AD (the altitude), and thus parallel to BC, so it intersects with BC only at the point of infinity corresponding to them both, so if LK is concurrent with both these lines, it too must pass through this point at infinity, implying that it is parallel to BC, as desired. I, too, urgently await your return from the toilet (?)

Welcome back, Xeroxia ! Mystic Terminator killed a fly with the atomic bomb.  This problem is like a beauty fly.

$\widehat{AFE}\equiv\widehat{ADE}\equiv\widehat{DCA}\equiv\widehat{BCE}\Longrightarrow B,F,E,C$[b] on a circle [/b]$\Longrightarrow\widehat{CFE}\equiv\widehat{CBE}.$
$\widehat{LKE}\equiv\widehat{LFE}\equiv\widehat{CFE}\equiv\widehat{CBE}\Longrightarrow\widehat{LKE}=\widehat{CBE}\Longrightarrow LK\parallel BC.$

[b](Mathematical language)[/b]


number theory
least common multiple
relatively prime
divisibility tests

This is because 111 is divisble by 3 and it is a standard result that is
$a$ divides $N$ and $b$ divides $N$ , then $ab$ divides $N$ if and only if $a$ and $b$ are relatively prime which is obviously not the case in your question!!

Yeah, i didn't realize 111 was divisible by 3.  :blush:

[quote="236factorial"]Consider the number 666. It's divisible by 2 and 3, so it's divisible by 6.

Consider the number 222,222,222. It's divisible by 2, 111, and 9, but not divisible by 1998. Why?[/quote]

It wouldn
t be true, because if 111 is divisible by 3.  :)

[hide]
Since 111 is divisible by 3, you have an extra 3; 2 * 9 * 111 which equals 1998 is not a factor of 222,222,222.[/hide]

[quote="236factorial"]Consider the number 666. It's divisible by 2 and 3, so it's divisible by 6.

Consider the number 222,222,222. It's divisible by 2, 111, and 9, but not divisible by 1998. Why?[/quote]
You just forgot that $\text{lcm}(2,9,111)\neq1998=666$

So how do you test divisibility for 1998?

It has to be divisible by $27$ and $222$, or $27$ and $37$ and $2$, or $54$ and $37$. In other words, you can't split up the powers of 3, at least one number has to have all the powers of the prime

You can't. You just have to test for divisibility by $9$; if that works, divide the number by $9$ and check for divisibility by $3$.


inequalities
quadratics
algebra
inequalities proposed

maybe this is an old inequality
left part is exactly chauchy inequality
for right part;
let $f(x)=(\sum a_i^2)x^2+\frac{(a_nb_n+a_1b_1)}{\sqrt{4a_nb_n+a_1b_1}}\sum a_ib_i*x+\sum b_i^2$
then $f(-\sqrt{\frac{b_1b_n}{a_1a_n}})\leq 0$
so there is a real solution for the quadratic equation $f(x)=0$
the rest is easy.


consider all 12! orderings of the list (1,2,3,...,12). In  how many of these lists is the kth number in the list greater than k-3 for all k, 4 <= k <= 12?
`

I tend to think of this problem like this:

[hide="Solution"] Keep in mind that there are 12 numbers, so if you choose any 2, they can be arranged in the slots of $k$ and $k-3$ in one way that allows $k > k - 3$.

Thus, for the problem, I just did $C(12,2) * C(10,2) * C(8,2) * C(6,2) * C(4,2) * C(2,2) = \frac{12!}{(2^6)} = 7484400$.

So, tell me if I thought wrong, because I'm not 100% sure on this, but it seems logical.[/hide]

Hmm.

[hide="I did it this way"]
Take the 1st number, we realize that the 4th number has to be greater than the 1st, and the 7th > 4th, and 10th > 7th.
Take the 2nd number, similarly, 11th number > 8th > 5th > 2nd.
Take the 3rd number, again, 12th number > 9th > 6th > 3rd.
This covers all 12 numbers. We notice that these three groups of 4 numbers are separate - meaning that the 10th number can be less than the 9th, and the problem's conditions will still stand. So if we pick a group of 4 numbers from the 12, we can only arrange them in one way, but we can pick whether it is the 1st group of 4 numbers (1st number, 4th, 7th, 10th), the 2nd group (2nd number, 5th, 8th, 11th), or the 3rd group (3rd number, 6th, 9th, 12th). Then we pick another group of 4 numbers, and place it in once of the remaining 2 groups. Finally, the last group of 4 numbers should fall in to place.

This gives:
$C(12, 4) * 3 * C(8, 4) * 2 * C(4, 4) = 207900$
[/hide]

Either theres a small mistake in the wording or you two have misunderstood slightly (or I've misunderstood your solutions!):
The kth number has to be greater than k-3, not the (k-3)rd number.
(ie the 4th number > 1, the 5th number > 2...)

Ok, if that's the case, then I have no clue, but here's a guess:

[hide="Guess"] I figure, start with the 12th number. There are 3 possible choices, 10, 11, and 12. Thus, 3. Now, pick one, and move onto 11th number. There are 4 choices, 9, 10, 11, and 12, but one of them is taken, so again, 3 choices. For 10th number, 5 choices, 8, 9, 10, 11, and 12, but 2 taken, so 3 choices yet again. So this continues to the 4th number. So that's $3^9$ right there. Now, there are 3 numbers remaining to fill the first 3 spaces of the arrangement, so that's $3!$ ways. So, the answer is?

$2 * 3^{10}$???? Wow, that sounds so wrong.[/hide]

You have no clue and that was a guess? :)

[thats exactly what I get]

Heh. I see. Thanks TripleM!

Yup, now that's what I get too.


algebra
polynomial
induction
number theory proposed
number theory

Fibonacci polynomials  $ f_1(x),f_2(x),...,f_n(x),... $  are defined by $ f_1(x)=1 , f_2(x)=x $ and  $  f_{n+1}(x)=xf_n(x)+f_{n-1}(x) $  for  $  n \ge 1 $.
Let $ F_n = f_n(1) $ be the n-$th$ Fibonacci number.
I am interested to solve following questions:
 1) To find all positive integers  $ x $ such that $    F_{nx} = f_n\left(L_x\right)F_x \;  . $  It's true that $ x $ may be any odd natural number ?

 2) I have observed that \[ f_n(4)=\frac{F_{3n}}{2}\; \;  ,\; \;  f_n(11)=\frac{F_{5n}}{5} . \]
  Which is the set of all positive integers $ p $ for which there exists a pair  $ (m,s) $ of positive integers such that
\[      f_n(p)= \frac{F_{nm}}{s}\; \;                ?
\]
[i]NOTE:[/i] $  L_0=2 , L_1=1 ,\;  L_{x+1}=L_x + L_{x-1} $  when  $ x $ is a positive integer. In other words, $ L_{n-1} $ is the n-$th$  Lucas number.

I have'nt a complete solution.

Answer to part 1: Yes, it works for all odd $x$. Changing notation a bit, we have
$f_{n+1}(L_k)=L_kf_n(L_k)+f_{n-1}(L_k)$, so if we have $F_{nk}=f_n(L_k)F_k$ and $ F_{(n-1)k}=f_{n-1}(L_k)F_k$, we need $F_{(n+1)k}=f_{n+1}(L_k)F_k =(L_kf_n(L_k)+f_{n-1}(L_k))F_k=L_kF_{nk}+F_{(n-1)k}$. The base cases here are $n=0$ and $n=1$, which are both trivial.
To prove that $F_{(n+1)k}=L_kF_{nk}+F_{(n-1)k}$, we prove the more general formula $F_{m+k}=L_kF_m-(-1)^kF_{m-k}$, which takes the correct sign when $k$ is odd. This formula can be proved by induction; it is trivial for $k=0,k=1$. Beyond that, assume that $F_{m+k-1}=L_{k-1}F_m+(-1)^kF_{m-k+1}$ and $F_{m+k-2}=L_{k-2}F_m-(-1)^kF_{m-k+2}$. Then, adding these, $F_{m+k-1}+F_{m+k-2}=(L_{k-1}+L_{k-2})F_m-(-1)^k(F_{m+k-2}-F_{m+k-1})$, so by the defining identities $F_{m+k}=L_kF_m-(-1)^kF_{m-k}$.

For even $k$, we need the polynomials $g_n$ defined by $g_0=1, g_1=x,g_{n+1}(x)=xg_n(x)-g_{n-1}(x)$; then $F_nk=g_n(L_k)F_k$.

Part 2:
Since $f_1(p)=1$, we must have $\frac{F_m}{s}=1$ and $s=F_m$.
Also, $f_2(p)=p$, so $L_m=\frac{F_{2m}}{F_m}=p$. This then becomes equivalent to part 1; $p$ must be $L_m$ for some odd $m$.


function
topology
advanced fields
advanced fields unsolved

"Show that if $ (X, T)$ is connected and $ \sim $ is an equivalence relation on $ X$, then $ (X/ \sim , T')$ is connected." ($ T$ and $ T'$ are topologies upon $ X$ and $ X/ \sim $, respectively)

[hide="Would following be a correct proof?"]Let $ Y \equal{} X/\sim $
Suppose $ Y$ is disconnected. Then there are $ A,B \in T'$ such that
$ Y \equal{} A \cup B$.

Define the function $ f: X \rightarrow Y$ such that $ f(x) \equal{} [x]$, for all $ x \in X$. Clearly $ f$ is onto.

By definition of quotient space (identification space and resulting topology is special case of quotient space and topology) we know $ f^{\minus{}1}(A), f^{\minus{}1}(B) \in T$.

 Since $ f$ onto, $ X \equal{} f^{\minus{}1}(A) \cup f^{\minus{}1}(B)$. Thus $ X$ disconnected.[/hide]

I think it's right, since you suppose $ A \cap B \equal{} \varnothing$. But there is an easier proof, cause your function $ f$ is continous by definition of quotient topology, so $ X$ connected implies $ f(X)\equal{}X/ \sim$ connected, cause continous functions "carry" connectedness.


[quote="pda"]Circle K has diameter AB=25. C is a point on K such that AC=7. A smaller circle is tangent to AB, to AC, and to K. What is its radius?[/quote]

Let the center of the given circle is O and that of smaller tangent circle is S.  Let the radius of smaller circle is r.  Let the smaller circle touches AB at T. O lies on bisector of the angle OAC and $ Tan(\angle OAS) = \frac{3}{4}$
We have $ OS=12.5-r$
From the triangle OST, $ OT = \sqrt{OS^{2}-r^{2}}$
$ Tan(\angle OAS) = \frac{3}{4}= \frac{{ST}}{{AT}}= \frac{r}{{AO+OT}}$
$ \Rightarrow 3(AO+OT) = 4r$
$ \Rightarrow 3(12.5+\sqrt{OS^{2}-r^{2}}) = 4r$
Solving for $ r$, we get $ r=4.68$


prove for each $ n\in\mathbb{N}$: $ \frac{n^{7}}{7}\plus{}\frac{n^{5}}{5}\plus{}\frac{23n}{35}\in\mathbb{Z}$
good luck^^

$ \frac{n^7}{7} \plus{} \frac{n^5}{5} \plus{} \frac{23n}{35} \equal{} \frac {5n^7 \plus{} 7n^5 \plus{} 23n}{35}$.

We must prove that for all $ n \in \mathbb Z$, $ 35$ divides $ 5n^7 \plus{} 7n^5 \plus{} 23n$, that is, $ 5n^7 \plus{} 7n^5 \plus{} 23n$ is divisible by both $ 5$ and $ 7$.

$ 5 | 5n^7 \plus{} 7n^5 \plus{} 23n \iff 5|7n^5 \plus{} 23n \iff 5|2n^5 \plus{} 3n \iff 5|2n \plus{} 3n$ (by Fermat's Little Theorem) which is clearly true.

Same thing for 7:

$ 7 | 5n^7 \plus{} 7n^5 \plus{} 23n \iff 7|5n^7 \plus{} 23n \iff 7|5n^7 \plus{} 2n \iff 7|5n \plus{} 2n$.

Done.


Seeing I couldn't go to the last math jam.. would someone be able to post any questions that were brought up (not sure how it works properly, are there set questions?)

The transcript is available in the Math Jams section, so you can check out the problems (and solutions) there.

Enjoy...


The cylinder has a volume of

$ \pi r^2 h$, or $ 30\pi$

The cone has a volume of

$ \frac{\pi r^2 h}{3}$, or $ 10\pi$

Adding these, we have $ \boxed{40\pi}$


Let a quadrilateral $ ABCD$. The perpendicular from $ A$ to $ BD$ meets the perpendicular from $ B$ to $ CD$ at point

$ P$  .The perpendicular from $ C$ to $ BD$ meets the perpendicular form $ D$ to $ AB$ at $ Q$. Prove that the line $ PQ$

passes through the point of intersection of the diagonals $ AC$ and $ BD$.

Babis

Let $ E,F$ be the feet of the perpendiculars from $ D,B$ on $ AB,CD$ and $ M,N$ the feet of the perpendiculars from $ A,C$ on $ BD.$ Since $ AEDM$ and $ MFPD$ are cyclic, we have $ \angle BAM \equal{} \angle QDN$ and $ \angle BPM \equal{} \angle BDF.$ Thus $ \triangle BAM \sim \triangle QDN$ and $ \triangle BPM \sim BDF.$  Then

$ \frac {\overline{CN}}{\overline{DN}} \equal{} \frac {\overline{BM}}{\overline{MP}} \ , \ \frac {\overline{QN}}{\overline{DN}} \equal{} \frac {\overline{BM}}{\overline{AM}}  \Longrightarrow  \frac {\overline{QN}}{\overline{CN}} \equal{} \frac {\overline{MP}}{\overline{AM}}$

$ \frac {\overline{QN} \plus{} \overline{CN}}{\overline{CN}} \equal{} \frac {\overline{MP} \plus{} \overline{AM}}{\overline{AM}} \Longrightarrow  \frac {\overline{AM}}{\overline{CN}} \equal{} \frac {\overline{AP}}{\overline{QC}}$

Define the intersections $ T \equiv AC \cap BD$ and $T' \equiv PQ \cap AC.$

From  $ \triangle T'QC \sim \triangle T'PA$ we obtain $ \frac {\overline{AP}}{\overline{QC}} \equal{} \frac {\overline{AT'}}{\overline{CT'}}$

But $ \frac {\overline{AM}}{\overline{CN}} \equal{} \frac {\overline{AT}}{\overline{CT}} \Longrightarrow  \frac {\overline{AT'}}{\overline{CT'}} \equal{} \frac {\overline{AT}}{\overline{CT}}  \Longrightarrow T \equiv T'$ and the conclusion follows.


integration
function
limit
real analysis
real analysis unsolved

Im studying for my midterm, trying to complete a few problems from royden for practice (which I have found unnaturally hard for some reason).

in particular, bounded variation is giving me trouble. Suppose that f is absolutely continuous on [a,b]. then

$T = \int_a^b | f^\prime |$

where T is the total variation of f.

(1) Let $a=x_1<x_2<\cdots<x_n=b$ be a partition,
$\sum |f(x_i)-f(x_{i-1})| = \sum \left|\int_{x_{i-1}}^{x_i} f'(x)dx\right| \leq \sum \int_{x_{i-1}}^{x_i} |f'(x)| dx \leq \int_a^b |f'(x)|dx$.
hence
$\bigvee_a^b (f)\leq \int_a^b |f'(x)|dx$

(2)Now we prove that $\int_a^b |f'(x)|dx \le\bigvee_a^b (f)$.

Let $a=x_1<x_2<\cdots<x_n=b$ be a partition, consider a step function $S(x)$ which is correspondent to the partition and it takes value only in $\{-1,0,1\}$. Thus $\int_a^b S(x)f'(x)dx \leq \sum \left|\int_{x_{i-1}}^{x_i} f'(x)dx\right| = \sum |f(x_i) -f(x_{i-1})| \leq \bigvee_a^b (f)$ (here we use the absolutely continuity for Newton-Leibniz's Formula)

Now let a series of step functions, say $\{f_k\}$, approaches to $f'(x)$. Let $S_k(x) = \begin{cases} 1, & f_k(x)>0 \\0, & f_k(x) = 0 \\-1, & f_k(x)<0 \end{cases}$
then $\lim S_k(x)f'(x) = |f'(x)|$. Hence $\int_a^b |f'(x)| = \lim \int_a^b S_k(x)f'(x)dx \leq \bigvee_a^b (f)$

I think that this should go to theorems and formulas. :?

liyi,

thank you very much for taking the time to type all of that. However, I feel quite stupid for asking this, but can you explain your $\vee$ notation? I cant figure it out...

[quote="candy"]oh, perhaps you denote the total variation of f on [a,b] by $\vee_a^b (f)$?[/quote]
Yes. That is what we use in China, and I thought it was an international notation  :blush: Now it seems not...


Let f R^n -->R^
Satisfy <x,y> = <f(x),f(y)>  , Remark <,> is inner product
Show f is linear transformation

Just develop ${\parallel}f({\lambda}.x + \mu.y) -{\lambda}.f(x) -\mu.f(y) {\parallel}$ using bi-linearity


calculus
derivative
calculus computations

I TA a class in introductory mathematics, and I have forgotten one of the topics tomorrow.

Let $ f: \mathbb{R} \to \mathbb{R}$. I think we can suppose $ f$ is $ C^2$ here. If the question asks to find inflection points of $ f$, is the method to find zeros of $ f''$ and see if $ f''$ changes sign around the zeros? Or should you inspect $ f'$ directly? Are there any other methods I should teach?

Inflection points are points where the second derivative changes sign. You definitely want to look at the second derivative for that; it's easier than directly finding maxima and minima of the first derivative.

There is a general result:
If $ n$ is the smallest natural number such that $ f^{(n)}(a) \neq 0$, then $ f(a)$ is an inflection point if $ n$ is even


calculus
derivative
trigonometry
calculus computations

Say I have $ f(x,y) \equal{} e^{ \minus{} x} sin(y)$

and the question asks

Use difference quotients with $ \Delta x \equal{} 0.1$ and $ \Delta y \equal{} 0.1$ to estimate $ f_x(1, 6)$ and $ f_y(1, 6)$

What I did for $ f_x(1, 6)$ :

$ \frac {f(1.1,6) \minus{} f(1,6)}{0.1}$
$ \frac {e^{ \minus{} 1.1} \sin 6 \minus{} e^{ \minus{} 1} \sin 6}{0.1}$

which gives me $ \minus{} .0365$

What I did for $ f_y(1, 6)$ :

$ \frac {f(1,6.1) \minus{} f(1,6)}{0.1}$
$ \frac {e^{ \minus{} 1} \sin (6.1) \minus{} e^{ \minus{} 1} \sin 6}{0.1}$
which gives me $ .00638494$

why is this answer wrong?

[quote="jmerry"]Calculus is done in radians.[/quote]

Oh thanks! Didn't realize my calculator was set in degrees...

[quote="jmerry"]Calculus is done in radians.[/quote]
Perhaps my favorite post of the day :)

Nice catch, JMerry!

It comes from dealing with students. You learn to spot the standard mistakes.


what is the smallest positive real r such
that a square of side lenght 1 can be covered
by 3 disks of radius r ??

Do you have any idea how much this r might be? I have a 'feeling' it's 10^(1/2)/6, but I'm not sure. I'll try to prove this. I'll be back... :D

Sorry, I got a smaller one, 65^(1/2)/16, but I'm still not sure if it's the best :?

Here's what I did (this isn't actually a rigourous proof, just some observations):
Because we have 4 corners, one of them should be covered by a single disk, and then the whole side will be covered by the same disk. Let's assume that side BC is covered by disk 1, center O1. We get the best covering (most efficient one, with disks of the smaller radius, or, in other words, I don't know how to explain this :D ) if O1B and O1C are actually equal to r (not strictly smaller).

Now let's assume that disk 1 intersects CD and AB at X and Y respectively, and let M be the midpoint of segment AD. Triangles AMY and DMX (hehe, sounds like the rapper :D) must be covered by disk 2 and disk 3. In order to get the smallest radius, I looked for such an r that M is at the intersection of disk 2 and 3, X at the intesection of disk 3 and 1, Y at the intersection of disk 2 and 1, because I thought that if we made the disks any smaller, then these points wouldn't be covered(I know it's a bad argument, just look at it and tell me what you think). Let CX=x. Then (if you draw the picture you'll see) we have 4r^2=x^2+1; 4r^2=(1-x)^2+1/4 If we solve this we get x=1/8 and r=65^(1/2)/16, so it's just a bit 'over' 1/2.

What do you think?

yeah, I was thinking like you but i was not
able to prove that was the minimum. Anybody's
an idea ??

I don't think that anybody's interested anymore but this is a nice problem. A correct solution could be formulated as follows, draw a picture :):

Suppose that we could do this with $r<\sqrt{65}/16$. It's been already said that we may assume that points B and C are covered by the same disk - let it be disk #1. Let X and Y be points on AB and CD that are in distance 1/8 from B and C and let M be the midpoint of AD. Since $|XC|=\sqrt{65}/8$, point X can't be covered by disk #1 - therefore, let it be covered by disk #2. Since $|MX|=\sqrt{65}/8$ and |MB| is even greater, point M can't be covered by disk #2 nor by disk #1 therefore it is surely covered by disk #3. Also, $|MY|=|YB|=\sqrt{65}/8$ so point Y can't be covered by disk #1 nor disk #3 therefore it is covered by disk #2.
Now we're almost there. Let Z be a point on AX such that |ZX| = 1/8. Z and Y can't be covered by the same disk and Z, C neither, therefore Z is covered by disk #3. D, B can't be covere by the same disk, D, X can't be covere by the same disk, D is covered by disk #3. But Z, D can't be covered by the same disk! Minimal $r=\sqrt{65}/16$.