$ a,b,c \geq 0$ and $ a\plus{}b\plus{}c\equal{}3$ prove that :
$ a\plus{}ab\plus{}2abc \leq \frac{9}{2}$

[quote="dgreenb801"][hide="Solution"]
The key is to note equality holds at $ (\frac {3}{2},1,\frac {1}{2})$
The given equality is equivalent to
$ \frac {a(a \plus{} b \plus{} c)^2}{9} \plus{} \frac {ab(a \plus{} b \plus{} c)}{3} \plus{} 2abc \le \frac {(a \plus{} b \plus{} c)^3}{6} \iff (a \plus{} b \plus{} c)(a^2 \plus{} 3b^2 \plus{} 3c^2 \minus{} 2ab \plus{} 6bc \plus{} 4ca) \ge 36abc$
By AM-GM, the left factor$ \equal{} (\frac {a}{3} \plus{} \frac {a}{3} \plus{} \frac {a}{3} \plus{} \frac {b}{2} \plus{} \frac {b}{2} \plus{} c) \ge \frac {6}{\sqrt [6] {3^3 \cdot 2^2}} (a^{1/2}b^{1/3}c^{1/6})$
For the second factor, note that $ a^2 \plus{} \frac {9}{4}b^2 \ge 3ab$, so the right factor is at least
$ \frac {3}{4}b^2 \plus{} 3c^2 \plus{} ab \plus{} 6bc \plus{} 4ca \equal{} \frac {3}{4}b^2 \plus{} 3c^2 \plus{} \frac {1}{2}ab \plus{} \frac {1}{2}ab \plus{} \frac {3}{2}bc \plus{} \frac {3}{2}bc \plus{} \frac {3}{2}bc \plus{} \frac {3}{2}bc \plus{} ac \plus{} ac \plus{} ac \plus{} ac \ge 12 \sqrt [12] {\frac {3^6}{2^8}} (a^{1/2}b^{2/3}c^{5/6})$ by AM-GM.
Multiplying these gives the desired result.[/hide]
This should be in the inequalities olympiad forum, unless you want to use lagrange multipliers.[/quote]
wat an application of AM-GM! very nice dgreenb801!


How do I input LaTeX into Keynote? I've tried some programs, but all of the downloads are just a bunch of nonsense codes. Help appreciated!  :D

What are you trying to do with the $ \text{\LaTeX}$? If you want to input the actual $ \text{\LaTeX}$ code into Keynote and have Keynote process it, there's a way to do this. You could probably use $ \text{\LaTeX}$ to make an image and then put that image into Keynote (Instant Alpha will allow you to extract the white background). Would this work?


[i]Let[/i] $\large a,b,c,d>0$ [i]such that [/i]$\large a\sqrt{a}+b\sqrt{b}+c\sqrt{c}+d\sqrt{d}=4$[i].prove that[/i]
...............
$\large\frac{ \frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}+\frac{1}{d^{2}}}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}}\ge\frac{a^{2}+b^{2}+c^{2}+d^{2}}{a+b+c+d}$
[b]@hungkhtn/toanhocmuonmau: hai anh giup em bai nay nhe,em chua giai duoc  :rotfl:   [/b]

[quote="svasao_qb111"]
[b]@hungkhtn/toanhocmuonmau: hai anh giup em bai nay nhe,em chua giai duoc  :rotfl:   [/b][/quote]

If it is so funny, translate it so that every one could understand!


ARML
pigeonhole principle
algebra
polynomial
Vieta
function

Hungry this before April 18th and PM/email me the answers (if you would like to compete for eternal glory, fame and honor). It might be too easy or too hard or something i haven't seen many ARML problems.

edit: fixed typo on #5
edit2: actually i fixed a typo on #4
edit3: here's a new #8 since i'm not exactly completely positively absolutely sure about the last one:
edit4: ok use the old #8, it has been confirmed....

Sorry about all this trouble!

[hide="top scorers"]
(Out of 6 instead of 8)

Alan - 6 (nice job!)
nat mc - 6 (nice job)!
Altheman - 5
scorpius119 - 5
[/hide]

Shoot i knew I spelled it wrong :D I can only remember the first part of his name...

Hey is anyone actually taking this? Please reply if you are.

[quote="chess64"]Hungry this before April 18th...[/quote]

I don't understand...  :?

But I will take it anyway and PM you my answers.  :lol:

Edit: OK, I'm done.  That was really hard, but the problems were good.  I will PM you my answers now.

i'm too busy doing other stuff (like problems for the MIMC so I don't think i'll be able to take it :( ), but the problems look really good, although a lot harder than real ARMLs :o , like your number 1 would be like a number 4 on the real one

I didn't take it, but in my (not terribly well-informed) view, the first few start out harder than real  ARML and the last couple easier than real ARML

Some of them were too easy to be on the back.... like Arnav said earlier.

OK I should really like review the questions to make sure they're okay next time...

#3 is kind of screwed up; i won't count it
#4 is way too computation-heavy and the question is unclear; won't count it either
#8 is like way too easy but whatever

yeah :blush: and about the difficulty, sorry, i realized that #8 was too easy and I couldn't think of a really hard problem. maybe #7 should have been #8...? well thanks to whoever takes it!

i think u should count them all..... i didnt see any problems with 3 or 4. Besides, I thought 4 was well calculated.

I suspect problem 3 is much more complicated than you realized, reddevil.  How did you solve it?

I used pigeonhole since the problem is just like the one they give as example on Art of Problem Solving Book's very front, with different dimensions. But i didnt know i would get this big of an answer.

also, #2 seems too easy to believe and some of the information on the problem, were they intended to be there just unnecessarily? And #4 has an ugly answer. and it doesnt say whether to answer in minutes or hours.

heh, #1 looks the hardest now that I just glanced over it...everything else looks fairly simple once you see a twist/know a factorization/etc.

[quote="xxreddevilzxx"]I used pigeonhole since the problem is just like the one they give as example on Art of Problem Solving Book's very front, with different dimensions. But i didnt know i would get this big of an answer.[/quote]

Which AoPS book? :?

ANyway, that's what I did too but smart people like JBL and probability1.01 are saying that there's something wrong with it. I don't really understand it. Whatever..

[quote]#4 has an ugly answer. and it doesnt say whether to answer in minutes or hours.[/quote]

that's why i'm not counting it

btw, you didn't send me your answers, did you? :?

@elemennop: i thought #1 was really easy, and #7 was really hard...

#3 I'm sure most people can understand what you mean by it, but the wording is a bit off...but I was never really good at reading problems anyways, haha.

#7 is alright, it can be pretty easily brute forced, and the 'elegant' way isn't too much faster.

anybody care to post the answers up? I just did it, #2 and #4 were a bit computation heavy...I wanna see if I missed any

edit> my bad, I didn't see that it ends teusday...I guess I'll submit my answers

The Pigeonhole Principle states something of the form, "If you have at least $x$ many, then event E must happen."  It does [i]not[/i] tell you that $x$ is the least number with this property.  Indeed, most pigeonhole arguments will not give you an absolute minimum.  In order to really solve this question, I would need a pigeonhole argument for my answer $n$, [i]as well as[/i] a counter-example for $n - 1$.  Without that, you can't be sure that it isn't true for some smaller number.

[quote="xxreddevilzxx"]I used pigeonhole since the problem is just like the one they give as example on Art of Problem Solving Book's very front, with different dimensions. But i didnt know i would get this big of an answer[/quote]
I did it the same way, it was a fun problem

thanks for the test!, also i did not think that #7 was that difficult, nor time consuming, i can post my solution/thinking tomorrow after the test is done

here is my solution to #7

problem let $x^5-1=0$, if $z$ is a nonreal root, compute

$(z-1)(z^2-1)(z^3-1)(z^4-1)$

so...

first i thought, it does not matter which i choose, lets do $e^{\frac{2\pi}{5}}$

after putting it in, each factor has a different root

it is the product of the nonreal roots subtracted by one, roots? product? vieta! shift the first function over to the left making the roots 1 less, or

$(x'+1)^5-1=0$, then factor out the $x'=0$, and the product by vieta is $5$

nice solution, better than mine! :)

(you guys can start discussing now...)

actually, here's a little complicated answer to number 1: $\frac{2^2}{2^2-1} \cdot \frac{3^2}{3^2-1} \cdot....\cdot \frac{2006^2}{2006^2-1}$ :P

all problems have been posted in separate threads in HSM  , I just created the related post collections

[url=https://artofproblemsolving.com/community/c3736077_2006_mock_arml_iii_individual]here[/url]

enjoy  / start solving

more USA mocks [url=https://artofproblemsolving.com/community/c2439870_usa_mocks]here [/url] (also mentioned [url=https://artofproblemsolving.com/community/c5h3200761p29225908]here[/url])
more ARML mocks [url=https://artofproblemsolving.com/community/c2442143_arml_mocks]here[/url],
more User-created contests [url=https://artofproblemsolving.com/community/c2435719_user_created_contests]here[/url] (contains all above)


Using the digits $ 1, 2, 3, 4$ and $ 5$ without repetition, the number of $ 5$-digit numbers is $ 5! \equal{} 120$. If we list these numbers in an increasing order, what is the number that comes in the $ 60th$ place?
$ 1 \minus{}\minus{}\minus{}\minus{} 12345$
$ 2 \minus{}\minus{}\minus{}\minus{} 12354$
$ 3 \minus{}\minus{}\minus{}\minus{} 12435$
$ 60 \minus{}\minus{}\minus{}\minus{} ?$
$ 120 \minus{}\minus{}\minus{} 54321$
---------------------------------------
Is there anyway to make a table to organize it in the forum?

Hmmm, this isn't probability :)
[hide="solution"]
We have $ 2\cdot 4!\equal{}48$ numbers starting with $ 1$ or $ 2$. We have $ 48\plus{}2\cdot 3!\equal{}60$ numbers starting with $ 1$, $ 2$, $ 31$ or $ 32$ so on $ 60th$ place is the largest number starting with $ 32$ which is $ \boxed{32541}$.[/hide]


The water level would go down, because the ice is less dense than the water, and therefore, water weighing the same as the ice cube would take up less space...

The water level will stay the same. Think of it as the ice cube makes a hole in the water. When it melts, the hole will be filled.

Actually, that's not a very good way of thinking about it. :huh: :lol:

I think the water level will go down because water at normal temperature takes up less volume than it does in frozen or gas form.

I'd think it'd go down-but only slightly.   :icecream:

I believe that water expands as it freezes. Most of the ice cube will be submerged when it floats in water. Also, you have to take into account the amount of evaporation occuring while it melts. I would say that the water level would drop, but not really enough to notice. I think it would also make a difference as to how much water there is compared to how much ice is in the water. If the amount of water in the ice exceeds the amount of displacement the ice causes in the water, it is likely that it would cause a rise in the water level. The only way this could happen is if the ice was not floating in the water, but just sitting in the water.

[quote="lotrgreengrapes7926"]Call the volume of the ice cube that is underwater $a$.
And call the weight of the ice cube $b$.
By Archimedes' principle, $a$ is equal to the volume of water that weighs $b$. When the ice cube melts, its weight stays the same ($b$) and its volume is equal to $a$ (because "$a$ is equal to the volume of water that weighs $b$").

So the water level stays the same.[/quote]

WOW way smart. I was beginning to wonder if this was philosophia or maths!

[quote="myc"]An ice cube is floating in a cup of water, full to the brim. As the ice melts, what happens?  Does the cup overflow or does the water level drop, or does it remain the same? explain why.  (Archimedes' Principle: A floating object displaces a volume of water whose weight equals the weight of the object).[/quote]

it wouldn't move? water is still water?

This can't be in the right forum. :|  :|  :|

Look, before this gets moved to another forum, I just want to say, ice cubes are always more compressed than water without any outside forces affecting it.  So as it expands it takes up more space.  Think of it this way:  the mass stays the same but the volume gets bigger, because the space between the molecules gets bigger, so even though it has the same weight, it does not have the same volume.

mgao, checkout this webpage.

http://www.word-detective.com/howcome/waterexpand.html

[quote="i_like_pie"][quote="bpms"]Isn't part of the ice above the water though?[/quote]
I thought it was level with the water. :huh:[/quote]
If it isn't, I think that the water level would decrease...


Compound A has molecular formula C4H9Cl. Compound A reacts
with Y in ethanol to give 1-butene, which when reacted with
Br2/H2O gives compound B. Which of the following statements
is true?

a. B could be 1,2-dibromobutane and Y could be KOH

b. A could be 1-chlorocyclobutane and Y could be H2/Pt

c. B could be 1-bromo-2-butanol and Y could be KOH

d. A could be 1-chlorobutane and Y could be H+

e. B could be 2-bromo-1-butanol and Y could be H2/Pt

sorry that was absent-mindedness :blush:
answer is C


topology
real analysis
real analysis unsolved

assume that $A$ is subset of $[0,1]$ which is homomorphic to Cantor .we say that $x$ is initial point if there exists $\sigma$ such that $A\cap (x-\sigma,x)=0$.does there exist $A$ which its initial points would be rational.

That doesn't quite make sense as written. Is this what you mean?

Let $A\subset[0,1]$ be hom[b]e[/b]omorphic to the Cantor set. We say that $x$ is an initial point of $A$ if, for some $\sigma$, $(x-\sigma,x)\cap A$ is empty. Does there exist an $A$ for which all rational numbers in $(0,1)$ are initial points of $A$?

There are other possible interpretations, but they're all trivial.

[quote="jmerry"]Does there exist an $A$ for which all rational numbers in $(0,1)$ are initial points of $A$?

There are other possible interpretations, but they're all trivial.[/quote]

I think it's the other way around: does there exist such a set $A$ all of whose initial points are rational (I also think that an initial point of $A$ is actually supposed to belong to $A$; the name "initial [b]point of $A$[/b]" makes more sense that way)? Is it trivial? :?

That other version is trivially true if you require the initial points to be in $A$ (see the actual cantor set), and trivially false otherwise, since $x$ an initial point implies that $x-c$ is an initial point for all small enough $c$, some of which will be irrational. The set of initial points isn't empty either, since $A$ is compact and is not the whole interval.

Yes, sorry about that. I'm always in a hurry when I post these days, because I do it from Internet Cafe-s and time is money :).

Edit:

And it looks like the answer is positive now. We can find a compact subset of $[0,1]$ with positive measure which does not contain any rationals (take out an open interval around each rational s.t. the lengths of the intervals sum up to less than $1$). Since every uncountable closed set contains a set homeomorphic to the Cantor set (the set should be compact, perfect, and totally disconnected), we're done.

Did I just say something stupid again? :)

No, that's just fine. Even that version is easy.


Prove that there exist irrational numbers a and b such that ab is rational.

It looks more like Algebra to me but it's very nice nevertheless

[hide]

A try :

x = :rt2::rt2:

Either x is rational either x is not rational

1) x is rational

We have found a = b = :rt2:

2) x is not rational

Then x:rt2: = :rt2::^2: = 2

Take a=x, b=:rt2:

[/hide]

By the Gelfond-Schneider theorem, sqrt(2)sqrt(2) is irrational. I just felt like pointing that out.

Gelfond-Schneider : with google I found it with some interesting results at :

http://www.math.sc.edu/~filaseta/gradcourses/Math785/Math785Notes8.pdf


The company ONIx markets $N$ products. The company has offered the clients $M$ promotional offers. Every offer consists of 2 diffrent products, which are sold together at a price that is lower than the combined prices of the 2 products (for example juice and mineral water). The products are identified with numbers from $1$ to $N$, and the promotional offers with numbers from $1$ to $M$.Because the company has recently changed their software application which takes care of the data base of the company, the employees haven't get used with the new system and, by mistake, one of them erased all the information about the products and the promotional offers. The only remaining information are the ones from the department of statistic, which uses an own data base. This information is represented by the number $M$ of offers and all the $K$ pairs of offers which have a common product (every 2 offers can have at most one common product).

Enuntiation

Using the information of the department of statistic, determine the numbers of products and the 2 products from each offer.

INPUT

On the first line of the file promo.in there are 2 numbers $M$ and $K$. The next $K$ lines contain 2 numbers $A$ and $B$, meaning that the offer $A$ and $B$ have one common product.

OUTPUT

In the file promo.out there will be a number N, representing the number of products. The next lines
have to contain 2 numbers. The $i$-est line of these M has to contain the number of the products from the $i$-est offer.

Restrictions

$1\leq M\leq 2007$
$0\leq K\leq 100000$
The number of products that have to be determined are at most equal to $2*M$
There will always be a solution to the problem

SAMPLE INPUT

11 7
1 4
4 7
7 1
2 5
5 8
8 2
10 11

SAMPLE OUTPUT

17
1 2
3 4
5 6
1 7
3 8
9 10
1 11
3 12
13 14
15 16
15 17

Time limit: 0.4 s

Is the question asking for minimum N or just any N that works?

--Artur


A sequence of functions $ \, \{f_n(x)\} \,$  is defined recursively as follows:

$ f_1(x) \equal{} \sqrt {x^2 \plus{} 48}, \quad \mbox{and} \\
f_{n \plus{} 1}(x) \equal{} \sqrt {x^2 \plus{} 6f_n(x)} \quad \mbox{for } n \geq 1.$

(Recall that $ \sqrt {\makebox[5mm]{}}$ is understood to represent the positive square root.) For each positive integer $ n$, find all real solutions of the equation $ \, f_n(x) \equal{} 2x \,$.

[hide]
Claim: $ x\equal{}4$ is the only solution. It works for all $ n$.

Proof:

We begin by solving the case for $ n \equal{} 1$.

$ f_1(x) \equal{} \sqrt {x^2 \plus{} 48} \equal{} 2x$
$ \Leftrightarrow$ $ \sqrt {x^2 \plus{} 48} \equal{} 2x$
$ \Leftrightarrow$ $ x^2 \plus{} 48 \equal{} 4x^2$
$ \Leftrightarrow$ $ 3x^2 \equal{} 48$
$ \Leftrightarrow$ $ x \equal{} 4, \minus{} 4$.

However, -4 is an extraneous solution, as it does not satisfy $ \, f_n(x) \equal{} 2x \,$.

We can prove that $ x \equal{} 4$ is a solution for all n by induction.

We have just showed that the base case, $ n \equal{} 1$, has a solution of $ x \equal{} 4$.

Assume $ x \equal{} 4$ is a solution for some $ n \equal{} k$.

$ f_{k}(4) \equal{} 8$

For $ n \equal{} k \plus{} 1$,

$ f_{k \plus{} 1}(x) \equal{} \sqrt {x^2 \plus{} 6f_{k}(x)}$, so
$ f_{k \plus{} 1}(4) \equal{} \sqrt {4^2 \plus{} 6f_{k}(4)} \equal{} \sqrt {4^2 \plus{} 6(8)} \equal{} 8 \equal{} 2(4)$.

$ x \equal{} 4$ is a solution for all $ n$.

Now, assume there is a solution $ x < 4$
For the base case, $ n \equal{} 1$, $ f_1(x) \equal{} \sqrt {x^2 \plus{} 48}$. If $ x < 4$, then  $ f_1(x) < 8$.
But then, for each successive $ f_{n \plus{} 1}(x) \equal{} \sqrt {x^2 \plus{} 6f_n(x)}$, if we have that $ x < 4$, then $ f_{n \plus{} 1}(x) < 4$ also!
So, there are no solutions less than four.

Next, assume there is a solution $ x > 4$
For the base case, $ n \equal{} 1$, $ f_1(x) \equal{} \sqrt {x^2 \plus{} 48}$. If $ x > 4$, then  $ f_1(x) > 8$.
But then, for each successive $ f_{n \plus{} 1}(x) \equal{} \sqrt {x^2 \plus{} 6f_n(x)}$, if we have that $ x > 4$, then $ f_{n \plus{} 1}(x) > 4$ also!
So, there are no solutions greater than four, either!

$ x \equal{} 4$ is the only solution. It works for all $ n$. QED [/hide]

[quote="Arvind_sn"][hide]
Now, assume there is a solution $ x < 4$
For the base case, $ n \equal{} 1$, $ f_1(x) \equal{} \sqrt {x^2 \plus{} 48}$. If $ x < 4$, then  $ f_1(x) < 8$.
But then, for each successive $ f_{n \plus{} 1}(x) \equal{} \sqrt {x^2 \plus{} 6f_n(x)}$, if we have that $ x < 4$, then $ f_{n \plus{} 1}(x) < 4$ also!
So, there are no solutions less than four.

Next, assume there is a solution $ x > 4$
For the base case, $ n \equal{} 1$, $ f_1(x) \equal{} \sqrt {x^2 \plus{} 48}$. If $ x > 4$, then  $ f_1(x) > 8$.
But then, for each successive $ f_{n \plus{} 1}(x) \equal{} \sqrt {x^2 \plus{} 6f_n(x)}$, if we have that $ x > 4$, then $ f_{n \plus{} 1}(x) > 4$ also!
So, there are no solutions greater than four, either![/hide][/quote]
[hide="Arvind_sn:"]The way you worded it doesn't necessarily prove it, but I see what you mean. (You've proven that $ f_{n\plus{}1}(x)<4$, but isn't it possible that $ 2x<4$ also?) Instead, you might want to say $ 2x>f_1(x)>f_2(x)>\ldots>f_n(x)$ if $ x<4$, and similarly for $ x>4$.[/hide]

Why do we need to do the bit at the end? After we have shown that the only solutions to $ f_1(x)\equal{}2x$ are $ \pm 4$ then doesn't it follow that they can be the only possible solutions; since we need the expression to be true for all $ n\in \mathbb{N^\plus{}}$ and clearly $ 1\in \mathbb{N^\plus{}}$...?

Oh yeah, you're right. I forgot about that bit...

No, there could be a solution for a specific n. Well, there can't, but the problem wants you to find solutions like that if they exist.


probability
AMC
AIME
analytic geometry
function

#3 basically you have to look at the parities of numbers and there are a few cases i believe that you have work out.  i think kalva would have a solution.  http://www.kalva.demon.co.uk

Yay, questions!

1) Thats not a general notation, there is in fact no general notation that I know of, whenever a problem uses digits it usually says as such. Whereever you saw this abc, it should have a definition of what the bar represents.

2) pass :)

1) OK lets see if I can get these in the right order.
A surjective function is one where for any y in the range, you can find x where f(x)=y. In simpler terms, you can "get" any number in the range by taking the function of something. This is also called "onto".
Injective, or "one to one", means that f(x) and f(y) are always different for different x and y. In other words, each x is mapped to something different.
Bijective is both of the above together - one to one and onto.
Hope thats a bit clearer.

2) You are choosing socks like AABBCDEF. The extra twos come from the fact that theres two choices for which sock of types C, D, E, and F you choose (left sock, right sock, if you wanna say it that way.)

As for number 3), yeah, well Kalva should work. If you want a very brief outline of what I would do, then for each coordinate x, y, z, we have to make sure that either both points have an even value or both have an odd value. Then there'll probably be a little bit of checking to make sure you actually have distinct points.

[quote="mathnovice"]Okay-- here are a few miscellaneous questions that I have had in my studies.

1. Does "abc" with a bar over it mean the number 100a + 10b + c in written form?

2. Why is the word "fat" used in so many problems in Andreescu's books? The names of people are always Mr, Ms, bug, etc Fat.

These are some counting questions I have.

1. What is the difference between surjective, bijective, and injective? The definitions that I have found are difficult to understand.

2. There are nine pairs of distinct socks in Adrian's drawers. He randomly grabs eight socks without looking. What is the probability that he has exactly two matching pairs of socks among the socks he grabbed?

The answer is [(9 C 2)(7 C 4)(2^4)]/[(18 C 8)]. Why is there a 2^4 term in the solution? I came up with the same answer without the 2^4.

3. (AIME 2001) Let S be the set of points whose coordinates x, y, and z are integers that satisfy 0 :le: x :le: 2, 0 :le: y :le: 3, and 0 :le: z :le: 4. Two distinct points are randomly chosen from S. What is the probability taht the midpoint of the segment they determine also belongs to S?

The answer is 23/177. Does anyone have a good explanation why?[/quote]

1). I've seen that notation before in the USSR olympiad problem book, thats what I assume it would be if I saw it on a contest, but I don't think its a well known notation.


A dollhouse models a real house at a scale of 1:15. If it takes 20 cans to paint the walls of the real house, how many cans are needed to paint the walls of the doll house? Express your answer as a common fraction.

From what I can understand, this dollhouse is a 2D product, isn't it?

[quote="Brut3Forc3"]It should take $ \frac{20}{15^2}\equal{}\frac{4}{45}$.[/quote]
@Brut3Forc3,  why do you square the $15$ in the denominator?  This is how I did it:

Let $x$ = # of cans of paint needed to paint the doll house

$\frac{1}{15}=\frac{x}{20}$

$x=\frac{20}{15}$

$x=\boxed{\frac{4}{3}}$

[quote="JSGandora"][quote="Brut3Forc3"]It should take $ \frac{20}{15^2}\equal{}\frac{4}{45}$.[/quote]
@Brut3Forc3,  [b]why do you square the $15$ in the denominator[/b]?  This is how I did it:

Let $x$ = # of cans of paint needed to paint the doll house

$\frac{1}{15}=\frac{x}{20}$

$x=\frac{20}{15}$

$x=\boxed{\frac{4}{3}}$[/quote]

I believe that you need to square the 15 in the denominator because the ratio of the surface areas (doll house to real house) is the square of the ratio of the side lengths (1:15).

For example:

The ratio of feet to yards is $1:3$, however, the ratio of square feet to square yard is $1:9$ or $(\frac{1}{3})^2=\boxed{\frac{1}{9}}$.

The answer according to FTW's review is $\boxed{\frac{4}{45}}$.


Determine all pairs (a, b) of positive integers such that

(ab + b + 7)|(ab + a + b).

Seems not too difficult at first sight, proved to be not so easy

[hide]A try : the solutions are (7c,7c) for c > 0.

A = ab + b + 7

B = ab + a + b

bB = aA + b-7a so A divides C=b-7a or C is null.

. C= b - 7a = 0, implies b = 7c for some c, implies a = 7c. Then B = cA so (7c,7c) is a solution for all c

. A - C = (a-1)b + b + 7 + 7a > 0 so if C != 0, A can't divide C if A > C. So there are no solutions for C != 0.

[/hide]

:evil:

[hide]

Just realized (11,1) is also a solution so my solution is all wrong somewhere. Time to go to bed, see that tomorrow.

[/hide]

Its basically the computer-programming symbol for not equal.

I think you were wrong because you assumed that if C!=0 then C>0, and from this you found that if A-C>0 then A can't divide C. But if C<0 then A>0>C and A can divide C (observe that for (11,1) we have C<0).
What we have to do is take into account the case C<0.

sorry, I did not mean to double post

Here it goes:

[hide]C=7a-b^2, not b^2-7a so,

A-C=(a+1)b^2+b+7-7a=a(b^2-7)+b^2+b+7

If b^2<7, then A-C could be < 0, so A<C

So A can't divide C if A > C, but A < C so A can divide C.

I'm not sure from there.[/hide]

Thanks you grobber and :sqrt:-1, you're right  :!:
It was really a stupid mistake  :(

A second chance :

So if C < 0, so we must have -C >= A (this time it's correct).
-C >= A can be rewritten (a+1)(b-7) + b <= 0 so b-7 < 0 and b=1 or 2.

. b = 1, a+8 divides 7a-1, 7(a-1)=7(a+8) - 57 so a+8 divides 57. a = 49 and a = 11 are the only solutions

. b = 2, 4a+11 divides 7a-4.
7*(4*a+11) - 4*(7*a-4) = 93 so 4*a+11 divides 93 which gives no solutions.

So :
(7c,7c), (11,1), (49,1) are the solutions.

Hope it's correct this time.

PS And yes != means different from, I'm doing some Java programming these days  :wink:

That's cool! I think this comes from a shortlist, but I can't remember which one. I always thought of it as being hard (although I never tried to solve it!! :D  :D )


geometry
incenter
circumcircle
inradius
area of a triangle
Heron\u0027s formula

Find the distance between the incenter and the circumcenter of a 8,10,12 triangle.
(or just explain how this can be done)

Just a hint:

[hide][url]http://mathworld.wolfram.com/Incenter.html[/url]

Of course, the inradius and circumradius can be found using $[A] = rs = \frac{4R}{abc}$.[/hide]

if you have the inradius and circumradius, you can find this by euler's triangle theorem

[hide]Using Heron's formula, we see that:

$\sqrt{15(15-8)(15-10)(15-12)}=\sqrt{1575}=15\sqrt{7}$

From there, you can use the formula to get the inradius, which is $rs=A$, so $r=\sqrt{7}$, then to get the circumradius, we use the formula: $R=\frac{abc}{4A}$, so we find that the circumradius is $\frac{16\sqrt{7}}{7}$. Now, to find the distance between the incenter and circumcenter, we use the formula: $d^2=R(R-2r)$, so $\boxed{d=\frac{4\sqrt{14}}{7}}$.[/hide]


calculus
integration
logarithms
Putnam
function
parameterization
parametric equation

Im just curious if someone can give me a few examples of wen and how to use this. Ive been looking on the internet and i either get an analysis explanation which is beyond my current knowledge or no explanation at all. Also along these same lines is defining a parametric equation to help work through the integral, if someone could add to that it would be greatly appreciated.

Here's a classic example:
$\int_0^1 \frac{x-1}{\ln x} \ dx$

Set $f(t) = \int_0^1 \frac{x^t-1}{\ln x} \ dx$, then we have $f(0) = 0$ and,
$f'(t) = \frac{d}{dt}\left(\int_0^1 \frac{x^t-1}{\ln x} \ dx\right) = \int_0^1 \frac{d}{dt}\left(\frac{x^t-1}{\ln x}\right) \ dx = \int_0^1 x^t \ dx = \frac{1}{1+t}$

And thus $f(1) = \int_0^1 \frac{x-1}{\ln x} \ dx= \int_0^1 \frac{dt}{1+t} = \ln 2$

thats really interesting, is there any online resources about this topic (i mean other than AoPs) it seems to be a topic little treated in standard calculus courses.

Feynman wrote (in Surely You're Joking, Mr. Feynman) that this was one of the unusual techniques in his "different box of tools", and that he was sometimes able to use this technique to figure out integrals that other students couldn't figure out with standard techniques like contour integration (which he claimed never to have learned).  For a long time I wondered what he was talking about.

This technique is interesting, but for some reason it was never mentioned in any Calculus course I ever took.

I've also never seen this technique in a calculus course, but I have seen it in competitions.  Here's another example from the 2005 Putnam:

Evaluate $\int_0^1 \frac{\ln(1+x)}{1+x^2} \ dx$

Im just curious what the logical reasoning (as opposed to the rigorous proof) behind this technique is. What is the main goal that one establishes when defining the parametric equation. Is it neccesary that $f(0)=0$? Also what class of functions can tis technique be applied to?

It's not necessary that $f(0) = 0$. You have some integral and you change a constant in it to a variable to make it $f(y)$. Then you differentiate under the integral sign to find $f'(y)$ and integrate in $y$. But if you want to evaluate $f(k)$ for some $k$, you need to know the value of $f$ at some other point so that you can do a definite integral. $f(0)=0$ is a common one, but it could be $f(5) = \pi$ or whatever is convenient in the problem.


probability
advanced fields
advanced fields unsolved

hello all

this is the problem im having trouble with, while using all my knowledge about probability, i can not figure the answer to this problem please help

a motor vehicle insurance company divides policyholders into 3 catagories: high, medium and low risk. the high risk group contains 10% of people, and they each have a 35% chance of making a claim during a particular year. the medium risk contains 20% of people, and they each have a 15% of chancw of making a claim during a yaer. the low risk group contains the remaining 70% of people, and they each have a 5% chance of making a claim during a year. the company's general procedure is to put a new policy holder into the risk group in the first year. if the person doesnt make a claim, they move into the medium risk group in the second year, and then into the low risk group in the third year.what is the probability that this will happen if the person is actually a low risk driver? what if they are actually a high risk driver? does the companys procedure seem to be fair?

steven

If the person is actually a low-risk driver, then when he is put into the high risk category for his first year, he would not have a 35% chance of making a claim, instead he would have the same 5% chance as the other low-risk drivers, so the probability that he would make it to medium-risk is 95%, and the probability that he would make it to low low-risk is 90.25%.  If the driver was really a high risk-driver, then he would have a 65% cance of becoming a medium risk driver after a year, and then a 42.25% chance of becoming a low-risk driver after two years.  Hope this helps, let me know if I misinterpreted your question.


trigonometry
function
calculus
real analysis
real analysis unsolved

$ f(\theta)\equal{}\sin^2{\theta}$ be $ 2\pi$ periodic function on $ (\minus{}\pi,\pi)$, find the fourier series of $ f$.

[hide]
i've found the answer (by using the calculus def.) to be $ \frac{1}{2}\minus{}\frac{1}{2}\cos{2\theta}$, and because of that I think a faster way should exist... I want to see a non-calculus solution for this.[/hide]

How is the question of finding the Fourier series of a function non-calculus in any way?

Regardless, the power-reduction identity gives $ \sin^2 \theta \equal{} \frac{1 \minus{} \cos 2 \theta}{2}$ identically, and since the Fourier series is unique (if it exists) we are done.


The 1st part with Ai does that just mean a random number in the sequence? and what is its relation to An if Ai is not used anywhere in the problem. Thanks.

Allow me to re-phrase the question:

An increasing sequence $A_{1}, A_{2}, \ldots$ is given such that for all $n \geq 2$, $A_{n}+A_{n+1}= \frac{9}{A_{n}-A_{n-1}}+8$.  What is the value of $A_{70}$?

Since the sequence isn't constant, however, we do need to know where it starts -- for instance, dropping the first few terms gives another, equally valid sequence.

Can we determine a rule for what the first two values can and cannot be in order to force the sequence to be increasing? I'm trying to look for one.

In principle, it's certainly possible that the sequence is unique -- the two conditions (that it's increasing and that it satisfies the recurrence relation) certainly limit the possibilities.  However, as I noted, given any sequence that works, we can just drop the first $k$ terms for any $k$, re-number, and the resulting sequence will also be valid.  So, if such a sequence exists and is unique, to answer this question still requires at least one more piece of information.

It's also possible that there are typos in the question.  Having worked with it for a while this afternoon, that's my current hypothesis.


Pythagorean Theorem
geometry
geometry unsolved

Hello  . Once you compute the radius of r2  by using Pythagorean theorem we could use the Descartes circle formula  to find the radii of the smallest circles r3,r4,r5 :)  :blush:  :D
http://mathworld.wolfram.com/DescartesCircleTheorem.html

Nebraska boy, Could you explain how do we use Descarte Circle Theorem to solve this problem?
Thanks you.  :)


Deep in our magical fantasy world, the continent is reigned by elemental tribes of highly skilled wizards. After thousands of years of wars between the tribes, a pact has been established to keep the cease-fire and to grant the world an eternal day. Until the pact is broken, there will be no nights (and thus no night actions). When a day ends, I will announce the new day starts, unless the pact is broken.

Play wise. This game will be pretty tactical.

Players left (15) in alphabetical order:
- Ars
- alanchou
- CheeseIsGood
- DonkeyKong
- erabel
- Hamster1800
- jnub
- LevatorNasolabialis
- miyomiyo
- mnmath
- perfect628
- setheus_soul
- temperal
- xpmath
- ZhangPeijin

Players dead:
- [color=blue]woser456, Lynched day 1, Ice Wizard[/color]
- [color=red]kimmystar94, Lynched day 2, Fire Wizard[/color]
- 1=2, Lynched day 4, Last Survivor
- [color=yellow]math_explorer, Lynched day 5, Lightning Wizard[/color]

Hello... Interesting theme...

So the game has started?

[b]The Rules (PLEASE READ THESE, ESPECIALLY IF THIS IS YOUR FIRST GAME):
Votes must be in bold. If you do not bold your vote, it will not be counted.

Please be attentive and unvote, if necessary, before casting a new vote. This is not required, but I’d appreciate it.

Executions will require a simple majority of votes. Once a player has reached the necessary majority, his pleas are useless and any attempts to unvote will be unheeded. You may vote: no lynch – majority votes of this kind are necessary to end the day without a death.

The game is not to be discussed outside of the thread unless your role specifically states that you may do so.

Once your death scene has been posted, you’re dead. Stop typing.

Don’t edit/delete previously submitted posts

Don’t quote any PMs from me.

If you have a night choice to make, it is due by the posted deadline. I will not wait for you – if you do not submit a choice to me, tough! No choice will be made.

NO HIDE TAGS. Also, no changing font size, or posting in any other way that obscures something you're saying. Anything you say should be clearly visible to everyone without having to click a button or enlarge your text or anything like that.

If you anticipate being unavailable for more than a 48-hour period, please post a notice to that effect in the thread. Treat this game as a commitment. Be considerate – don’t leave us hanging.

Remember, it’s a game. It’s supposed to played. It’s supposed to be fun. Do your part.[/b]

The game has started... right? Great theme. Let's see... Well, I'm here, for one thing. Boom.

Umm... do we get any special powers depending on our tribe or anything? For that matter, what're the tribes? Can you tell us?

Frustrating...

Where's everybody else?

[quote="math_explorer"]Where's everybody else?[/quote]

Behind you.

Presents! YAY!!!

So random voting? :yankchain:

I'd say wait a day or two, for people to show up.

tribes...fantasy world...nice work:D hy guys:)

I think this game is supposed to be confusing...

[b]Vote:1=2[/b] for posting a.... strange emoticon.

[b]Vote: Temperal[/b] for posting too many dots in an ellipses.

[b]Vote![/b] (That was a command. Not a vote. I just decided to bold it.)

@Peter: I didn't have enough info to know at the time that andersonw (who I targeted) was Ice, and I expected if he was, then fire would've killed him. So my kill was unnecessary. Unfortunately, Erabel messed up for fire and screwed up my plan as well. Also, why didn't 1=2's double kill go through? Ars and Hamster [b]were[/b] on the same team.

Aha, the undeads get kills too. Well it could have ended in a draw, assuming the undeads are killable.

Yeah, jnub definitely deserved to win, but erabel's stupidity just astounds me.

Hamster if you had told me it was jnub
who I had suspected of being your helper
I would have been able to still save the game
I suspected that the undead would win if we massacred each other
but I didn't know that they had kills
which as it turned out didn't work
I just wasn't sure which of the 2 undead you got to help you

I had a plan to get erabel convinced if he didn't believe me
but he failed to send anything to me

anyways I was going to kill 1=2 again
but as I wasn't sure who helped you so I agreed to your plan
and as a result we failed to kill the undead

that is why you tell your team everything
its impossible that both jnub and 1=2 got reborn unless they are on the same team
which is why I didn't originally like the plan
but as hamster said I didn't know what jnub had said

also jnub if you had talked to me instead
you would have lost

Ars - jnub admitted that they were on the same team. He claimed that he was against another team that had not appeared yet.

However, we (fire) still would have lost due to erabel...

[quote="jnub"]Also, why didn't 1=2's double kill go through? Ars and Hamster [b]were[/b] on the same team.[/quote]Because he sent me andersonw and Hamster1800. :P

@CheeseIsGood: that's what I meant with the cooperation thing. An undead can only die when it has been target of [b]all three[/b] elements. So as soon as there is no fire anymore, they're immortal.

Hmm that's quite tricky. So I would have killed us all too :) But as it turns out - it would have been possible to win if erabel hadn't killed me -_-

So could ghosts be lynched?

Oh, also, had we lynched someone else between the 2nd and 3rd gatekeepers, would there have been 3 ghosts?

Alternatively, would there ever be >2 ghosts?

Not that it matters, but 1=2, why did you switch your targets?

oops.
 :blush:

I thought that anderson and Hamster were on the same team, and I accidentally said "ars" instead of "andersonw".

Whoop de doo!

fwahahahaha! I actually inadvertantly kept the essential bits of the team layout from the ghosts :P I knew exactly where everyone was (except that erabel actually was being quite annoying about mnmath).

[quote="Hamster1800"]Oh, also, had we lynched someone else between the 2nd and 3rd gatekeepers, would there have been 3 ghosts?[/quote]Yep. Jnub's night powers were significantly less than 1=2's night powers, because I had provided there could be multiple of them in theory. (jnub was just an 'undead wizard', while 1=2 was the Devil)

But apparently jnub's tricky plan was stronger than his night powers. :P

Yeah, okay, so I made all the right decisions until jnub started talking :(

That makes me sad...

Our conclusion then: never trust an indian when he might have an incentive to lie.

Though of course only Hamster knew that I was indian =P.


function
limit
trigonometry
real analysis
real analysis unsolved

What is the limit, as x tends to 0 -, of  x^2 e^(1/sin x)?
  Can this be accomplished by Taylor Series?
  The answer is 1/2.

Here is what I got:

$x^{2}e^{-1}\leq x^{2}e^{1\over \sin(x)}\leq x^{2}e^{1}$

Then,
$\lim_{x \to 0}x^{2}e^{-1}\leq \lim_{x \to 0}x^{2}e^{1\over \sin(x)}\leq \lim_{x \to 0}x^{2}e^{1}$

$0 \leq \lim_{x \to 0}x^{2}e^{1\over \sin(x)}\leq 0$

Therefore, $\lim_{x \to 0}x^{2}e^{1\over \sin(x)}= 0$ by squeeze theorem.

But the answer is suppose to be 1/2 so I guess I am wrong.  :(

Yes, I think you are not quite right.

Remember e is raised to 1/sin x, and 1/sin x explodes to infinity when x tends to 0. The relation -1 <= 1/sin x <= 1 isn't quite right.

So does anyone have any idea?

$e^{\frac1{\sin x}}$ might as well be $e^{\frac1x}.$ As $x\to0^{+},$ it goes to infinity faster than any power of $x.$

Hence, $\lim_{x\to 0^{+}}x^{2}e^{\frac1{\sin x}}=+\infty.$

At the same time, $\lim_{x\to 0^{-}}x^{2}e^{\frac1{\sin x}}=0.$

Yes, this sounds plausible. But how do you prove that when x goes to 0+, e^(1/x) goes to infinity faster than any power of x?

actually, i'd prefer to use de l'hopital...
i don't think there's a simple way to use its power series expansion to get the desired result (take a look at $\sin x$, for example...).
am i wrong?

Taylor series for sine is not convergent outside of $|x| \le \frac{\pi}{2}$... that's why it won't work for $\frac{1}{x}$ as $x\to 0$. But the Taylor series for $e^{x}$ is convergent for all $x$ so putting in $\frac{1}{x}$ and letting $x\to 0$ should make it obvious...  :huh: or am I oversimplifying?

uhm.. i think you're oversimplifying, actually...
anyway.. sine is entire, and its power series converges everywhere...
so, try and take $\sin(x)/x$, or, if you prefer $x\sin(1/x)$... i think your messing with taylor series doesn't work here.. am i still wrong?

Ok I have no idea what i was talking about with sine. However, if we just look at taylor series for $\frac{1}{x^{n}}e^{x}$ (as this problem can be stated), it's obvious that as $x \to \infty$ the first finite number of terms go to 0 and the rest of them grow. And all are positive. I still can't see anything wrong with that.

uhm, ok :)
sorry, i just thought you took a look at the first terms, that's all...
it seems to be more than ok :)

:D ok good, I guess I'm not totally crazy. Now hopefully we didn't confuse everyone else along with each other ... :P

$p\in N^{*}\Longrightarrow\lim_{x\nearrow 0}x^{-2p}\cdot e^{\frac{1}{\sin x}}=\lim_{x\nearrow 0}\left(\frac{x}{\sin x}\right)^{-2p}\cdot (\sin x)^{-2p}\cdot e^{\frac{1}{\sin x}}=$ $\lim_{y\nearrow 0}y^{-2p}e^{\frac{1}{y}}=\lim_{z\searrow 0}z^{-2p}\cdot e^{-\frac{1}{z}}=\lim_{t\to\infty}\frac{t^{2p}}{e^{t}}=0\Longrightarrow \boxed{\ \lim_{x\nearrow 0}x^{-2p}\cdot e^{\frac{1}{\sin x}}=0\ }\ .$
[b]Remarks.[/b]

$1.\ \ x^{2}=x^{2p+2}\cdot x^{-2p}\ !$
$2.$ For $x\nearrow 0$ I used the substitutions $\sin x=y\nearrow 0$, $-y=z\searrow 0$, $\frac{1}{z}=t\to\infty$, i.e. directly, $-\frac{1}{\sin x}=t\to\infty\ .$
$3.$ I used the remarkable limits $\lim_{x\to 0}\frac{x}{\sin x}=1$, $\lim_{x\to\infty}\frac{x^{n}}{e^{x}}=0$ for any $n\in N^{*}\ .$


http://www.apple.com/ipodnano/

check it out, it's totally sick !!! :coolspeak:

:( now that I've won a Shuffle, they come out with something better.

that thing is so cool!
and the ipod flea was funny. :D

[quote="mathclass"]:( the memory is kind of small, though. and I wouldn't want something pencil thin, what if you drop it and it snaps or something?[/quote]

You'd be branded a loser for life? Lol. I like the black one. It looks really sleek.

[quote="xxreddevilzxx"]I dont like it. Whats the point of having it too thin, when ipod mini is small enough in the first place?[/quote]

the color display and it doesn't just play music
and it looks totally awesome, especially w/ a dock and a Mac
watch the last part of this video and see why
http://stream.apple.akadns.net/

[quote="mathclass"]:( the memory is kind of small, though. and I wouldn't want something pencil thin, what if you drop it and it snaps or something?[/quote]

The ArsTechnica people [url=http://arstechnica.com/reviews/hardware/nano.ars/3]stress tested[/url] the iPod nano, and found that it is far stronger than it looks.

ipod nano looks really awesome....

if I was into portable music players I'd want it..

[quote="mathclass"][quote="Drunken_Math"][quote="mathclass"]I want a 4 GB now and it's not that expensive! :O on ebay they are going for more than the retail price. I don't get it, why don't they buy it from apple.com?[/quote]

I have a friend who bought a laptop for $\$$4000 from eBay. He didn't know he could've bought it from Dell for $\$$2000. Some people don't even know they could've bought things from other places than eBay.[/quote]

o__o are you serious? that's kind of silly.[/quote]

Yup, it's kinda funny!

[quote="aznmonkey1992"]
i would just be fine with a jump drive/mp3 since i dont have many songs...
[/quote]

But people, like me, who have a lot of songs. (I have around 200)... need a large MP3 player.

second'd

I barely have any, mainly because I reformatted my computer due to a virus right before I mass uploaded songs to my iPod...

[quote="Drunken_Math"]But people, like me, who have a lot of songs. (I have around 200)... need a large MP3 player.[/quote]
200 is a lot of songs? I know some people who filled their ipods, and they hold several thousand! But I only have 300 or so.

I have more songs than an iPod Mini 6G can hold  :lol:

I have approximately 12 GB of music, but I listen to such a small fraction of that.
An iPod nano holds more than enough space for me  :)
The only way I can think of getting one for now is through winning the iTest  :P

[quote="whoknowscolinlu"]How many songs r u going to litteraly listen to?? Is it worth it?? How bout just burned CDs?? What happened to the good old days?? Then we go all electronical[/quote]
hmm...as i am a weirdo and listen to classical...i literally listen to...say about 720 songs, or 3+ GB during any given month. i am such a freak.  :lol:

anyway...200 isn't a lot. i'm over 1000 total, which, in my mind, still isn't enough. I haven't put my entire CD collection on my iPod yet.  :D

[quote="jmadsen"]I don't have CD's, I just get my songs from iTunes.[/quote]

ouch, at a dollar per song and even 200 songs...

[quote="236factorial"][quote="jmadsen"]Yeah, and how much does a CD cost per song?[/quote]

The average CD can cost anywhere from 10 to 17 dollars (the average). Unless most of the CD has the songs you want, it's not as good as getting i-tunes.[/quote]
Meh, the classical CD's I like are somewhere from 17-20. iTunes wins again. Oh, yes, on the CD's I want, there are perhaps 2 or 3 pieces I want and the others are just useless bits. The only CD that I have gottent that was worth it was the Planets, 7.99 dollars for 3 good songs. Best CD deal ever. Now on iTunes that would be 2.97, I would have saved 5.02, or 5 good songs, and a gumball. $\text{iTunes}>\text{CD's}\\\text{QED}$

wow, look at this! http://www.walmart.com/catalog/product.gsp?dest=9999999997&product_id=3756866&sourceid=1500000000000002375350

even smaller than iPod nano. smaller than a pushpin! than a quarter! and still 1 gig of music. pretty nifty.

well,
i just bought an iPOD mini and i have only used like a tenth of the gigs on it and that is around like 40 songs which is crazy. it says it can hold like 1000 songs which is total BS

[quote="elston"]i'm probably going to get a nano for christmas, when deals come out and prices go down a bit

saw someone at school with a black one, really tight[/quote]

i don't think the price will go down, it just came out
maybe if u wait 2 yrs, it will

Yay. Mine came in the mail today. I only have 1 GB filled up on it, so I think it'll be plenty of space even if it means I need to filter out some songs I don't listen to... that's always fun.

[quote="G-UNIT"][quote="jmadsen"]Yeah, and how much does a CD cost per song?[/quote]

  And how much does it cost to download them free?[/quote]

I'm guessing a little less than a dollar, but I might be wrong.


For,like a honorable mention?

Does it help a lot when I want to transfer to schools like Brown,Dartmouth(I need at least 25000$ \$$ FA  per year and I am a intl student(from China).....lol)...High GPAs(4.0 or at least 3.9) seem very common in transfering to ivy-level schools....

It might help, it might not.  It certainly won't hurt though.  Keep in mind that transfer acceptances also have to do with what the school needs.  Suppose hypothetically a college wanted a class of 1500 people but they overenrolled and got 1550, and then, very people people left.  Then they may not accept transfers, no matter how good simply because there isn't room.


according to bohr's theory angular momentum is nh/2(pi). but i have also read that it is root(L(L+1))h/2pi where L is azimuthal quantum number

Can anyone explain this

$ L\equal{}n\hbar$ is the quantized angular momentum in Bohr's theory, where $ n$ is the principal quantum number and $ \hbar\equal{}\frac{h}{2\pi}$. His theory was based on circular orbits for the electron path, called the Bohr orbit.

$ L\equal{}\sqrt{J(J\plus{}1)}\hbar$ is the quantized angular momentum provided by Quantum Mechanics, where $ J$ is the angular momentum quantum number, which happens to be the azimuthal quantum number. In today's term, it should $ L_z\equal{}\sqrt{J(J\plus{}1)}\hbar$.

Why is there a difference? Bohr's theory needed to be improved. Bohr's theory worked for hydrogen atom but as we move to more complex elements, even helium, Bohr's theory did not predict well. The rotational energy levels of electrons at different states for a helium atom were not proportional to $ J^2$, but they were proportional to $ J(J\plus{}1)$. There were more reasons but I'll leave that for now because... I need to check my notes again :P

so there is a difference isnt there?
For a hydrogen atom, both the formulae are supposed to work but they dont.
How?

There are two contributors to angular momentum:

1.Spin angular momentum
2.Orbital angular momentum

That's why your discrepancy occurs.

could u explain taking an example of 1s orbital?

[quote="srinitrs"][quote="zimrock"]ok, whats the angular momentum of an electron in a 1s orbital.

h/2pi according to bohr model and its zero if we use root(L*(L+1))h/2pi.

So, isnt there a difference?[/quote]
it is only 0 and not h/2*3.14 for any s orbital orbital the second formula has to be applied as L=0 for  s orbital!!! :D[/quote]
for hydrogen atom it is h/2pi, not zero.

First, in a 1s orbital, the electron has zero orbital angular momentum.
Second, Bohr is successful on determining the energy levels of the different states for hydrogen atom (or hydrogen-like atoms such as $ \ce{He\plus{}},\ce{Li\plus{}\plus{}}$)


Pretty much, you pick your favorite manga and you pick your favorite character in that manga!!!!

my fav manga: tie between Fairy Tail and Mahou Sensei Negima
my fav character: Mistgun in Fairy Tail, Fate in Mahou Sensei Negima

my favorite manga ((The only manga i read)): Hikaru no go
character:Hikaru

I knew that Japanese manga and anime spread through the world.
I feel very glad as a Japanese.

Thanks for fan :lol:

:D Well you should be proud , I learnt a lot a lot about Japan out of interest just reading manga..

And oh of course my favorites:

 :roll:
1)Naruto
[hide][img]http://anime-wallpapers.com/images/800x600/sasuke-and-naruto-dark.jpg[/img]
[/hide]
2)Bleach
[hide][img]http://www.bleachwallpaper.info/images/wallpapers/ichigo-bankai-115658.jpeg[/img][/hide]

Yaaaay! Mangaaaaaaaa!!!!!  :D

1. Fullmetal Alchemist - Edward Elric

2. Rurouni Kenshin - Hitokiri Battousai (Yes, I consider Kenshin and Battousai two different people.)

I also have many more series that I love, such as D.Gray-man, Bleach, Code Geass, Deathnote, Juushin Enbu, Peacemaker Kurogane, Rave Master, Franken Fran, Nabari no Ou, and Shinshi Doumei Cross (the only shojo I like), but I figure no one wants to hear about those.  :)  (I've also read pretty much every other series on Onemanga.com, in case anyone's wondering, which I'm sure they're not.)


conics
ellipse
trigonometry
geometry
integration
function
complex numbers

We have two coaxal cyllinders, but there cross sections are ellipses(As shown on the below diagram) The semi great axises are (from inside to outside) $a_{1}$, and $a_{2}$ respectively and the semi small axises are $b_{1}$ and $b_{2}$ respectively as shown on the diagram . Find the capacity of this capacitor if there is vaccum between the plates if its length is $L$.

How would you do this one? I'm very interested to see its solution, please post it...

A very useful tool for us is now the complex-potential.
First of all lets calculate the complex potential of a ellipse formed cyllinder.
The parametric form of an ellipse:
$x=a \cos \phi$ and $y=b\sin \phi$
let: $a=c\cosh \alpha$ and $b= c\sinh \alpha$ so $a^{2}-b^{2}=c^{2}$
So we see that $c$ is the half of the focus of the ellipse.
from here we get the following equation:
$z=c\cos(\Psi+i(\Phi-\alpha))$
So from here the complex potential can be written as:
$w=A \cdot arcosh \left(\frac{z}{c}\right )$
The normal potential of the two ellipses at $x=a_{1}$ $y=0$ and $x=a_{2}$ $y=0$:
$\Phi_{1}=A \cdot arcosh \left(\frac{a_{1}}{c}\right)$ . and $\Phi_{2}=A\cdot arcosh \left(\frac{a_{2}}{c}\right)$.
SO the potential differance:
$\Delta \Phi =A\cdot ln \left (\frac{a_{2}+b_{2}}{a_{1}+b_{1}}\right)$
So the complex potential:
$w=\frac{\Delta \Phi}{ln \left (\cfrac{a_{2}+b_{2}}{a_{1}+b_{1}}\right)}\left(\frac{z}{c}\right )$.
The electric field between the plates:
$|\vec E|=\frac{\Delta \Phi}{ln \left (\cfrac{a_{2}+b_{2}}{a_{1}+b_{1}}\right)}\frac{1}{|\sqrt{z^{2}-c^{2}}|}$
The amount of charge on the area between two points of the main curve:
$q_{AB}=\int_{A}^{B}\varepsilon_{0}E_{n}ds =-\varepsilon_{0}\int_{A}^{B}\frac{\partial \Phi}{\partial n}ds =-\varepsilon_{0}\int_{A}^{B}\frac{\partial \Psi}{\partial s}ds =\varepsilon_{0}(\Psi_{A}-\Psi_{B})$
Where $\Psi = Im w$ is the function that writes the field lines.
We used here that for the $w(z)=\Phi+i \Psi$ imaginary and real part of a complex analitic function the following is true:
$\frac{\partial \Phi}{\partial b}=|w^{'}(z)| =\frac{\partial \Psi}{\partial s}$
So the charge per unit length of the capacitor plate:
$q=2\pi \frac{\Delta \Phi \varepsilon_{0}}{ln \left (\cfrac{a_{2}+b_{2}}{a_{1}+b_{1}}\right)}$
So from here the capacity:
$C=\frac{2\pi \varepsilon_{0}L}{ln \left (\cfrac{a_{2}+b_{2}}{a_{1}+b_{1}}\right)}$


Came across this problem in Zeitz. I however doubt it is correct.

There are 2000 points on a circle, and each point is given a number that is equal to the average of the numbers of its two nearest neighbors. Show that all the numbers must be equal.

My issue is that it does not mention anything about the distribution of the points. For example, I could put a put a 1000 points very close together and give them all a number i. And diametrically opposite I could place the other 1000 points and give them a different number j.

The problem does not imply that neighbors are on either side of the point. So both nearest neighbors could be on one side of a point.

So is there some information missing from the problem?

It says "nearest [b]neighbors[/b]", not "nearest distance", so I don't think the complications arise.

Oh and the proof:
Assume for contradiction there exists a point $A_{i}$ that doesn't equal one of its two neighbors. Then, one of $A_{i+1}$ or $A_{i-1}$ is larger than $A_{i}$. We prove it for the case $A_{i+1}$ is larger; then the other follows by symmetry.

Thus, we have $A_{i+1}>A_{i}>A_{i-1}$. Futhermore, $A_{i}+A_{i-2}=2A_{i-1}\implies 2A_{i-1}>A_{i-1}+A_{i-2}\iff A_{i-1}>A_{i-2}$. By symmetry, it clearly holds that $A_{i+1}>A_{i}>A_{i-1}>\cdots A_{j}$ for all $j$ since this loops around the circle (counterclockwise, say). Choosing $j=i+1$, we obtain the contradiction that $A_{i+1}>A_{i+1}$, which proves the two not equal cases by symmetry, which shows that the $A_{i}$ are equal.

Yup, If you assume nearest neighbors the problem should not arise. I have a different proof.

Let the numbers be $a_{0}$ to $a_{1999}$. Lets assert that they are not all equal. Therefore there is a maximum number and let $a_{i}= m$ be that maximum number. But since it is the maximum,  $a_{i+1}= a_{i-1}= m$ too. Now using the same argument for $a_{i+1}$ and $a_{i-1}$ we get that $a_{i-2}= a_{i+2}= m$. Continuing we see that all numbers $a_{0}$ to $a_{1999}$ are equal to $m$, Contradicting the original assertion that all the numbers are not equal.

Let me know if you find errors in the logic. :)

No errors; it works :)
Two easy solutions to an easy problem.
Its really intuitive...

Or, you could do...

[hide="Induction"]We will prove the result for $n$ points, where $n \ge 3$ using mathematical induction. Clearly, the result holds when $n=3.$

[hide="Trivial Proof of n=3"]
If the numbers are $a, b, c,$ we have the equations:

$2a = b+c$
$2b = c+a$
$2c = a+b$

Subtracting the second from the first gives:

$2(a-b) = b-a \Rightarrow 2(a-b)+(a-b) = 0 \Rightarrow a = b$

Symmetrically, $a = b = c.$
[/hide]

Assume that the result holds for some positive integer, $k.$

Given the points $V_{1}, V_{2}, V_{3}, \cdots, V_{k},$ since their values are equal by our assumption, let their values all equal $m.$ If we place any point in between any $V_{i},$ its value is going to be $(m+m)/2 = m.$ Thus, the result holds for $k+1$ points as well.

Thus, since we have proven the result for $k+1$ using the assumption of validity of $k$, by mathematical induction, the result holds for all positive integers. Clearly, as $2000$ is a positive integer, the result holds for $2000$ as well. $\blacksquare$[/hide]

Sorry but I think in this case, your proof is unnecisarilly long for such a trivial problem... :(


Real numbers $ a,b,c,d$ satisfy the inequality $ abcd>a^2\plus{}b^2\plus{}c^2\plus{}d^2$.

Prove that $ abcd>a\plus{}b\plus{}c\plus{}d\plus{}8$

[quote="cancer"]Real numbers $ a,b,c,d$ satisfy the inequality $ abcd > a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2$.

Prove that $ abcd > a \plus{} b \plus{} c \plus{} d \plus{} 8$[/quote]
With the same conditions $ abcd>4(a\plus{}b\plus{}c\plus{}d)\minus{}16$ true too.  :wink:

[quote="cancer"][quote="arqady"][quote="cancer"]Real numbers $ a,b,c,d$ satisfy the inequality $ abcd > a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2$.

Prove that $ abcd > a \plus{} b \plus{} c \plus{} d \plus{} 8$[/quote]
With the same conditions $ abcd > 4(a \plus{} b \plus{} c \plus{} d) \minus{} 16$ true too.  :wink:[/quote]

Can you show?  :D[/quote]
$ abcd>\frac{(a\plus{}b\plus{}c\plus{}d)^2}{4}.$ :wink:

[quote="cancer"]Real numbers $ a,b,c,d$ satisfy the inequality $ abcd > a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2$.

Prove that $ abcd > a \plus{} b \plus{} c \plus{} d \plus{} 8$[/quote]

cancer, where's this problem from, please?

[quote]cancer, where's this problem from, please?[/quote]

Question is not for me  :) but I think that I know answer (hope it is correct one)

Last year (2007) it appeared in Belarussian M.O. as problem for category B.


calculus
integration
logarithms
calculus computations

Use the substitution $x=at$ and you should get after some calculations that the integral equals $\frac{\pi \ln a}{2a}$.

What about $\int_{0}^\infty \frac{\log x}{(x-1)(x+a)}\, dx$, where $a > 0$?

The value of your integral is:

$\frac{\pi^{2}+(\ln(a))^{2}}{2(a+1)}$.

Here is my approach for calculating this integral:

Observe that

$\frac{1}{(x-1)(x+a)}=\frac{1}{a+1}\left(\frac{1}{x-1}-\frac{1}{x+1}\right)+\frac{1}{a+1}\left(\frac{1}{x+1}-\frac{1}{x+a}\right)$

$=\frac{2}{(a+1)(x^{2}-1)}+\frac{a-1}{a+1}\frac{1}{(x+1)(x+a)}$.

Now, integrate

$\int_{0}^{\infty}\frac{\ln x}{(x-1)(x+a)}dx=\frac{2}{a+1}\int_{0}^{\infty}\frac{\ln x}{x^{2}-1}dx+\frac{a-1}{a+1}\int_{0}^{\infty}\frac{\ln x}{(x+1)(x+a)}dx$

The first integral is  $\int_{0}^{\infty}\frac{\ln x}{x^{2}-1}dx=\frac{\pi^{2}}{4}$ and the second one can be calculated via Residue theorem by using the classical formula:

$\int_{0}^{\infty}\frac{p(x) \ln x}{q(x)}dx$ where $deg(q)-deg(p) \geq 2$ and $q(x)\neq 0$ for $x \geq 0$.

Finally the integral equals: $\frac{\pi^{2}+(\ln(a))^{2}}{2(a+1)}$.


A right prism has a hexogonal base and has a height of 10cm. The side length of the base is 4 cm. What is the volume of the prism? Express your answer in simplest radical form.

[hide="solution"]The formula for the volume of a prism is $ bh $ where $ b $ is the area of the base and $ h $ is the height. We know that $ h $ is 10, but we need to find $ b $. We can make the hexagon into six equalateral triangles all with base $ 4 $. The formula for the area of an equalateral triangle is ${ \frac {s^2 \sqrt{3}}{4}} $. Where $ s $ is the length if a side. Substitute and get $ \frac {16 \sqrt{3}}{4} $ or $ 4 \sqrt {3} $. Multiply by six for all of the equalateral triangles and get $ 24 \sqrt {3} $. Multiply by the height and get $ 240 \sqrt{3} $[/hide]

[hide="Nice formula for area of a regular hexagon"]$ \frac{3s^2\sqrt{3}}{2} $[/hide]


Let $ n$ be a given positive integer,and $ a$ be a real.

Determine the range of $ a$ such that the inequality:

$ m^2\minus{}(a\minus{}1)m\plus{}\frac{n^2}{2n\plus{}1}a>0$　holds for all integer $ m$.

[hide]This is a quadratic inequality with respect to $ m$; it will hold true for all integers $ m$ if it has no roots, i.e. if it's determinant is less than $ 0$.

$ (1\minus{}a)^2\minus{}\frac{4n^2}{2n\plus{}1}a<0$
$ a^2\minus{}\frac{4n^2\plus{}4n\plus{}2}{2n\plus{}1}a\plus{}1<0$

Which in turn is a quadratic with respect to $ a$.  Manipulating the $ a$-term,

$ a^2\minus{}[\frac{4n^2\plus{}4n\plus{}1}{2n\plus{}1}\plus{}\frac{1}{2n\plus{}1}]a\plus{}1<0$
$ a^2\minus{}[(2n\plus{}1)\plus{}\frac{1}{2n\plus{}1}]a\plus{}1<0$
$ [a\minus{}(2n\plus{}1)][a\minus{}\frac{1}{2n\plus{}1}]<0$

It is clear that $ \frac{1}{2n\plus{}1}\leq{a}\leq(2n\plus{}1)$.
[/hide]

[quote="jrshoch"][hide]This is a quadratic inequality with respect to $ m$; it will hold true for all integers $ m$ if it has no roots, i.e. if it's determinant is less than $ 0$.

$ (1 \minus{} a)^2 \minus{} \frac {4n^2}{2n \plus{} 1}a < 0$
$ a^2 \minus{} \frac {4n^2 \plus{} 4n \plus{} 2}{2n \plus{} 1}a \plus{} 1 < 0$

Which in turn is a quadratic with respect to $ a$.  Manipulating the $ a$-term,

$ a^2 \minus{} [\frac {4n^2 \plus{} 4n \plus{} 1}{2n \plus{} 1} \plus{} \frac {1}{2n \plus{} 1}]a \plus{} 1 < 0$
$ a^2 \minus{} [(2n \plus{} 1) \plus{} \frac {1}{2n \plus{} 1}]a \plus{} 1 < 0$
$ [a \minus{} (2n \plus{} 1)][a \minus{} \frac {1}{2n \plus{} 1}] < 0$

It is clear that $ \frac {1}{2n \plus{} 1}\leq{a}\leq(2n \plus{} 1)$.
[/hide][/quote]

Yes,that is a sufficient condition.How do we find the necessary and sufficient?


I used paint to make one that was pretty nice.

Our school uses a mix of symbols/pictures.

We have a TI-83, a picture of winston churchill with a carrot, and I think "$e^{i\pi}+1 = 0$".


A rhombus is formed by two radii and two chords of a circle whose radius is $ 16$ feet. The area of the rhombus in square feet is:

$ \textbf{(A)}\ 128 \qquad\textbf{(B)}\ 128\sqrt {3} \qquad\textbf{(C)}\ 256 \qquad\textbf{(D)}\ 512 \qquad\textbf{(E)}\ 512\sqrt {3}$

The shorter diagonal is just another radius, it's $ 16$ feet. This divides the rhombus into two equilateral triangles, so the length of the other diagonal is $ 16\sqrt{3}$. The area of the rhombus is then $ \frac{1}{2}(16\cdot16\sqrt{3})\equal{}128\sqrt{3}\implies \boxed{B}$.


Let $ x_0(t) \equal{} 1$, $ x_{k \plus{} 1}(t) \equal{} (1 \plus{} t^{k \plus{} 1})x_{k}(t)$, for all $ k\ge0$.

Let $ y_{n,0}(t) \equal{} 1$, $ y_{n,k}(t) \equal{} \dfrac{t^{n \minus{} k \plus{} 1} \minus{} 1}{t^{k} \minus{} 1}y_{n,k \minus{} 1}(t)$, for all $ n\ge0$, and $ 1\le$ $ k\le$ $ n$.

Show that: $ \sum_{j \equal{} 0}^{n \minus{} 1}\ {( \minus{} 1)^j}x_{n \minus{} j \minus{} 1}(t)y_{n,j}(t) \equal{} \dfrac{1 \minus{} ( \minus{} 1)^n}{2}$, for all $ n\ge1$.

Thanks "milin".

I don't know yet, but it is definitely not in my taste :roll:. Besides, it is a bit naiive to expect people to answer fast during the exam weeks, so be patient: sooner or later somebody who likes such questions more than I will figure it out :P. By the way, folks, as a moderator, I should say that I'm sort of tired of answering the "why hasn't anybody solved my problem yet?" posts. The reasons vary but the main point is always the same: [color=red]this is not a hotline[/color].


linear algebra
matrix
topology
linear algebra unsolved

$D_{n}$ denotes the family of all $n\times n$ doubly stochastic matrices.
$S_{n}=\{\lambda \in \mathbb{C}~: ~\exists P \in D_{n}~ \text{such that}~\det(\lambda I_{n}-P) = 0\}$.
Is $S_{n}$ convex?

I haven't figured out how to approach this yet, but I did notice that $e^{2\pi ij/k}\in S_{n}$ for all $0\le j<k\le n.$ (Permutations.)

A simple counterexample in the case $n=3$:
We can have eigenvalues of $-1$ from $\begin{bmatrix}1&0&0\\0&0&1\\0&1&0\end{bmatrix}$ and $-\frac12\pm\sqrt{\frac34}i$ from $\begin{bmatrix}0&1&0\\0&0&1\\1&0&0\end{bmatrix}$. We can't have nonreal eigenvalues with real parts less than $-\frac12$, since the trace of any such matrix is nonnegative and any complex eigenvalues must come in conjugate pairs.
The set of eigenvalues contains an open set, but it also has spikes.

jmerry,
You are right $S_{3}$ is not convex, in fact we can prove that $S_{3}= \bigcup_{m=2}^{3}\mbox{Conv}\left( \{ e^{2i\pi \frac{k}{m}}, k=0,\dots,m-1\}\right)$.
Also we can also show easily that $S_{2}$ is convex.

What about the general case for $S_{n}$?

Here is one idea. The outline will be a bit sketchy, but I hope it will be clear what is going on. :) I'll do everything for $S_{4}$ but you can try to do something like that for any $S_{n}$. Note that $i$ is an eigenvalue only if the matrix is $P_{0}=\left[\begin{matrix}0&1&0&0\\ 0&0&1&0\\ 0&0&0&1\\ 1&0&0&0 \end{matrix}\right]$ or some other permutation with one cycle of full length. Thus, if $\lambda\approx i$ is an eigenvalue, then the corresponding matrix is $P_{0}+P$ where $P$ is small. Using the standard perturbation theory for eigenvalues, we see that $\lambda$ up to the second order (in size of $P$) terms is $i+\langle Px,x \rangle/4$ where $x=\left[\begin{matrix}1\\i\\-1\\-i\end{matrix}\right]$ is the eigenvector of $P_{0}$ corresponding to the eigenvalue $i$. Now, $P$ has negative entries at positions $(1,2)$, $(2,3)$, $(3,4)$, and $(4,1)$, which contribute their sum to the change in $\lambda$ times $i$. All other entries in $P$ are positive and give the same sum in absolute value. From here it is easy to conclude that the affine part is contained in the square $Q$ with vertices $1,i,-1,-i$. This would be the end if we could show that the linear term is not zero. We have $\langle Px,x\rangle=\sum P_{km}i^{m-k}$. We have see that the terms with $m\equiv k+1\mod 4$ give $i$ with negative coefficients whose sum is comparable to the norm of $P$. But the only terms that can kill them are those with $m\equiv k-1\mod 4$ (everything else doesn't point in vertical direction) and these terms also give a negative multiple of $i$. Thus, if $S_{4}$ were convex, it would be contained in the square $Q$. But $S_{4}\supset S_{3}\ni e^{2\pi i/3}$.


vector
algebra
polynomial
linear algebra
linear algebra unsolved

Show that if $ T$ and $ S$ are invertible linear operators on a vector space $ V$ then $ TS$ and $ ST$ have the same eigenvalues.

[hide="Hint"]If $ v$ is an eigen vector of $ AB$ to eigen value $ \lambda$, then $ Bv$ is an eigen vector of $ BA$ to eigen value $ \lambda$.[/hide]

I think it is not necessarly to be invertible.

$ TS$ and $ ST$ (if $ S,T$ are any square matrices) have always same characteristic polynomial but are not necessarily similar.  If $ S$ or $ T$ is invertible then they are obviously similar.

[quote="msaid"]How do you use the invertibility property?[/quote]... one condition for an eigen vector is that it is not the zero vector. but I think this was discussed in the thread ehsan2004 referred to.

@ loup blanc: Another nasty thing here is that $ V$ is not explicitly assumed to be finite dimensional, such that determinants do not apply a priori.  I was running into this trap at first, too.  :o

Do you have any reduction step to finite dimensions in stock? But I doubt that it exists ...   :?:

Hi third edition,
you are right; I assumed $ V$ to be finite dimensional.
      Else assume $ S$ invertible; then $ TS\equal{}S^{\minus{}1}(ST)S$ and $ S^{\minus{}1}(ker(ST\minus{}\lambda{I}))\equal{}ker(TS\minus{}\lambda{I})$. (for the eigenvalues).
     Moreover $ ST\minus{}\lambda{I}$ is not invertible iff $ TS\minus{}\lambda{I}$ is not invertible.  (for the spectrum)
      No problem ?

[quote="loup blanc"]     Moreover $ ST \minus{} \lambda{I}$ is not invertible iff $ TS \minus{} \lambda{I}$ is not invertible.  (for the spectrum)
      No problem ?[/quote]
No, no problem: Provided that at least one of the operators $ S$ and $ T$ is invertible, then $ TS\minus{}\lambda$ is surjective iff $ ST\minus{}\lambda$ is surjective, for every $ \lambda$.

If $ S,T$ are noninvertible then the result is false ($ dim(V)\equal{}\infty$); the counterexample is easy:
$ (e_i)_{i\geq1}$ is a basis of $ V$. Let $ S: e_i\rightarrow{e_{i\plus{}1}}$ and $ T: e_i\rightarrow{e_{i\minus{}1}}$ if $ i\not\equal{}1$ and $ Te_1\equal{}0$. Then $ TS\equal{}id$ and $ STe_1\equal{}0$.


function
calculus
derivative
limit
calculus computations

For all real numbers x and y, let f be a function such that f(x+y) = f(x) + f(y) +2xy and such that lim h->0 (f(h)/h) = 7.

a) Find f(0). Justify your answer.
b) Use the definition of the derivative to find f'(x).
c) Find f(x).

a)  Let $ x \equal{} y \equal{} 0$.  Then $ f(0) \equal{} 2f(0) \plus{} 0 \implies f(0) \equal{} 0$.

b)  Let $ y \equal{} h$.  Then $ f(x \plus{} h) \equal{} f(x) \plus{} f(h) \plus{} 2xh \implies$
$ f'(x) \equal{} \lim_{h \to 0} \frac{ f(x \plus{} h) \minus{} f(x) }{h} \equal{} \lim_{h \to 0} \frac{f(h) \plus{} 2x h}{h} \equal{} 7 \plus{} 2x$.

c)  We conclude that $ f(x) \equal{} x^2 \plus{} 7x \plus{} C$ for some constant $ C$.  Since $ f(0) \equal{} 0$, $ C \equal{} 0$.


algebra
function
domain
calculus
derivative
inequalities
algebra unsolved

Let $ n\in\mathbb N$. Suppose that for every $ i \in\left\{ 1,2,...,n\right\}$, we have given a complex number $ z_i$ satisfying $ \left|z_{i}\right| \equal{} 1$. Assume that $ \sum_{i \equal{} 1}^{n}z_{i} \equal{} 0$.
If $ z\in\mathbb{C}$, prove that:
$ n\leq\sum_{i \equal{} 1}^{n}\left|z \minus{} z_{i}\right|$.

Hmm... I tried that before, but what if $|z|< 1$? then we don't have a definite lower bound.

Let $f$ be defined to do $Z \to \sum_{k=1}^n |Z-Z_k|$, with $Z \not \in \{Z_1,\ldots,Z_n\}$;
let $g(Z)=\sum_{k=1}^n \frac {Z-Z_k} {|Z-Z_k|}$.
i) Let $m= \inf\{f(Z) \mid Z \in \mathbb{C}\}$. There exists a complex number $a$ such that $f(a)=m$.  (easy).
ii) If there exists $a$ in the domain of $g$ such that $g(a)=0$, then $f(a)=m$.
Proof: Let $Z$ be in $\mathbb{C}$; $\sum_{k=1}^n ((Z-a+a-Z_k)\dfrac {\overline{(a-Zk)}} {|a-Z_k|}=f(a) \leq f(Z)$.
iii) If for all $Z$ in the domain of $g$ we have $g(Z) \neq 0$, then there exists $k \in \{1,\ldots,n\}$ such that $f(Zk)=m$.
Proof: If $a$ is not in $\{Z_1,\ldots,Z_n\}$ then $f(a)=m$ implies $Df(a)=0$ or $g(a)=0$. ($Df$ is here the derivative of a function $f$ from $\mathbb{R}^2$ to $\mathbb{R}$).
In our problem we are in the case ii) with $a=0$: $g(0)=0$ and $f(0)=n$ implies $m=n$.

I don't understand your solution,loup blanc.
can you clear it,please?
Thanks.

[quote="mavropnevma"]If $z=0$ then $\sum |z-z_i| = \sum |z_i| = n$. If $z \neq 0$, take $t = \dfrac {1} {z}$. Notice that $z_i = \dfrac {1} {\overline{z_i}}$ and $|\overline{z_i}| = 1$, since $|z_i| = 1$. Then $\sum |z-z_i| = \sum \left | \dfrac {1} {t} - \dfrac {1} {\overline{z_i}} \right | = \sum \left | \dfrac {t - \overline{z_i}} {t\overline{z_i}} \right | = \dfrac {1} {|t|} \sum |t - \overline{z_i}| \geq \dfrac {1} {|t|} |\sum (t - \overline{z_i})| = \dfrac {1} {|t|} |nt - \overline{\sum z_i}| = \dfrac {1} {|t|} n|t| = n$.

Moreover, for $z\neq 0$ and $n>2$ the inequality is strict, since it is folklore that $\sum |\alpha_i| = |\sum \alpha_i|$ if and only if there exist $\alpha$, and non-negative real $\lambda_i$, such that $\alpha_i = \lambda_i\alpha$. This however leads to $t- \overline{z_i} = \lambda_i\alpha$, so $nt = (\sum \lambda_i)\alpha$, and so $\overline{z_j} = \dfrac {\sum \lambda_i - n\lambda_j} {n} \alpha$ for all $j$, absurd. Notice that for $n=2$ we have $z_1 = -z_2 = \omega$, with $|\omega| = 1$, and for any $z = \lambda\omega$, with $\lambda \in [-1,1]$, we have $|z-z_1| + |z-z_2| = (1-\lambda) + (1+\lambda) = 2$.[/quote]
thanks mavropnevma,I see your solution is true,but i don't have intuition,can you explain why you work with $\frac{1}{z}$ and $\overline{z_i}$? how you get this solution?

To tell you the truth, by trying to deal with the case $|z| < 1$ (as seen in the above, the case $|z| \geq 1$ is easy)!
So I worked, for $z\neq 0$, with $t = \dfrac {1} {z}$, and taking into account that a modulus-$1$ number has as inverse its conjugate, it all fell into place.


floor function
function
algebra
absolute value

[quote="robinhe"][b]How many real solutions are there to the equation $4x^2-40\lfloor x\rfloor +51$.[/b]

Please supply a solution.  Thanks :)

Also, tell me if I put it in the right place.  I'm not really sure how hard this is to everyone else. :?[/quote]

...Equation? Isn't there supposed to be another side to an equation...?

I'm assuming the whole thing equals zero...

Sorry, my bad. :oops:

It equals 0.  I'll change it. ;)

Placed about right, could be getting started.

[hide="Hint"]Just take the case where $\ x \geq 0$ and where $\ x < 0$[/hide]

no, that is incorrect ;), think about what floor means and call $\lfloor x\lfloor$ as $x+r$ where $r$ is the fractional part.

I'll post a solution in a little while if no one gets it ;)

[quote="ch1n353ch3s54a1l"][quote="robinhe"][b]How many real solutions are there to the equation $4x^2-40\lfloor x\rfloor +51=0$.[/b]

Please supply a solution.  Thanks :)

Also, tell me if I put it in the right place.  I'm not really sure how hard this is to everyone else. :?[/quote]

[hide]Uhh...I don't really know..but I factored $4x^2-40x+51$ to $(2x-3)(2x-17)$ so the points are $17/2$ and $3/2$. The only way the expression could be an integer is if x was a multiple of 1/2. So I tried all these cases between 3/2 and 17/2 and none of them worked... $\boxed{0}$?[/hide][/quote]

Well first of all, you factored wrong, and then you didn't take the floor function into full consideration...

no, he did not factor wrong, but yes, he did not take the floor into consideration,
[hide="Soution"]
If we let $y\leq x<y+1$, we can factor it as $(2y-3)(2y-17)\leq 0$, and so $\frac{3}{2}\leq y\leq \frac{17}{2}$, and clearly $y=2,3,4,5,6,7,8$, but a quick check shows that only $y=2,6,7,8$ works out, thus those are our possible values, so there are $\boxed{4}$ real solutions for $x$.[/hide]

Oh, I'm retarded, I thought that was an absolute value, disregard my former statement in it's entirety.  Yes, you did post this in the right forum then. :fool:


What brand of mobile phone do you  want / have / like best?

The phone that I just got, and like the best is the Motorola Razr V3 :D

Speakerphone, Bluetooth Technology, Digital Camera and an Aluminum Case....

[hide="A picture"][img]http://www.grunderen.no/mobi/razr-v3.jpg[/img][/hide]

i have a sagem myx-2 but my dream now is to have a nokia 7710

[quote="MithsApprentice"]fanaticism, does the touchpad on the V3 get dirty as fast as some people claim it does?[/quote]

Personally, I don't think that this is the case (it hasn't happened so far), but there are cases for the phone itself which prevents the touchpad from getting dirty.

[quote="shobber"]well i don't think mobile phones are so important to my life, so i choose not to have one.[/quote]I'm quite on the opposite pole here :D Can't live without one anymore :)

mine was always an electronic leash for me. its a motorola V221. or something. the standard camera phone. i just upgraded from a mid-nineties style (digital, but it looked like an analog phone).

samsung is the best suitable... they are the smallest, and has the best camera. I like small phones, so that is just my opinion.... and I just dont like bluetooth. Also, Samsung has the best quality in ringtones, pics, and other stuff.

[quote="xxreddevilzxx"]samsung is the best suitable... they are the smallest, and has the best camera.[/quote]You are mistakening (unless you live in Korea, at which point I would agree with you :) ). The best cameras at this point in mobile phones are due to Sony Ericsson, with S700 (S710i the american version with EDGE) and k750/w800 ( 2Mpx with autofocus in 100g! You won't belive what photos that little mobile can make!)

I've a dinky little Motorola v180.  Really a horrible phone.  No camera and the WXYZ/9 button sticks, but it calls and sends messages.  I'm not too picky about my phone.  If it works, it's good enough, and no one actually calls my phone save for my parents.

I don't need a mobile phone; I have [url=http://www.skype.com]Skype![/url]

[quote="JBL"]There is no option for "none?"  :-)[/quote]Didn't actually thought about that :D :D

I currently carry a Motorola i855 (nextel service). The screen is awesome. And nobody else has one :). (OK so maybe I cheat, so maybe I am sitting inside Motorola right now typing this very message!)

Someone mentioned they didn't like bluetooth. How can you not like bluetooth, did bluetooth offend you or something? Perhaps you meant you have no need for bluetooth?

yay for my ancient nokia, i really should just get a job and buy a new one, maybe next year... meh, whom im kidding, this ones good enough

Whee I have a ... lemme check :P NEC N820... lol my mom just gave this to me one day cause I didn't want one and she thought I needed one... It's pretty cool though, camera, light, GPRS, Internet... so i'm pretty happy with mine.

[quote="math92"]i don't have a phone :([/quote] i don't, too :(  but then again what the oint of having one when yoour 11-13?

yeah i don't need a phone.  Phoneless people are cool!!! :lol:


i've proved only one part...

by the power means we have that
$\frac{1}{\sqrt n}=\sqrt{\frac{\sum_{i=1}^n x_{i}^2}{n}}\geq \frac{n}{\sum_{i=1}^n \frac{1}{x_{i}}}\\\Rightarrow a(\sum_{i=1}^n \frac{1}{x_{i}})\geq an\sqrt n$
again by the power means (or cauchy-schwarz) we have that
$\frac{1}{\sqrt n}=\sqrt{\frac{\sum_{i=1}^n x_{i}^2}{n}}\geq \frac{\sum_{i=1}^n x_{i}}{n} \\\Rightarrow -b(\sum_{i=1}^n x_{i})\geq -b\sqrt n$
adding the two inequalities gives
$a(\sum_{i=1}^n \frac{1}{x_{i}})-b(\sum_{i=1}^n x_{i})\geq (an-b)\sqrt n$


good theoretical physics programs? I'm really leaning towards going into physics, so i'm wondering if there's a school with a good theoretical physics (quantum mechanics) program. Any tips?

The same advice applies here as was given in the "Math Colleges" thread. It really depends on a lot of personal preference factors as to which school would be best for you. Although one difference is, I think there is a smaller set of schools with good physics departments than there are having good math departments. A lot of that has to do with budgets and physical resources.

For example, schools that are closer to major research labs tend to attract better faculty and have very active research. But that is not universally true. The best advice is to research each individual school, and take into account other factors about the overall school itself. But I will say that I think the "name" of a school means maybe sightly more when picking a physics program than it does math. Because money and resources can go a long way in terms of good lab and research opportunities.


Let $(O)$ be the circumcle of square $ABCD$.$M \in (O)$.Prove that:
a)$MA^{4}+MB^{4}+MC^{4}+MD^{4}=24R^{4}$
b)$MA.MB.MC.MD <6R^{4}$

a) use complex numbers
b) use a) and AM-GM inequality; you easily get that equality can not occur

In fact, using complex numbers, you get $MA \cdot MB \cdot MC \cdot MD \le 2R^{4}$.


Hello, this is an inequality which I know how to solve using Karamata. But is it possible to solve this by just using Jensen alone?

If $ f$ is convex, show that
$ f(a) \plus{} f(b) \plus{} f(c) \plus{} f(\frac{a\plus{}b\plus{}c}{3}) \geq \frac{4}{3} \left( f(\frac{a\plus{}b}{2}) \plus{} f(\frac{b\plus{}c}{2}) \plus{} f(\frac{c\plus{}a}{2}) \right)$

Will anyone post me the Jensen's inequality??? I knew it, but could not remember it right now!
Moreover, hendrata01 wrote:
[quote]Hello, this is an inequality which I know how to solve using Karamata.[/quote]
What is Karamata??? :?:

try

[url]http://www.google.com[/url]

 :wink:

or, see

[url]http://www.artofproblemsolving.com/Resources/Papers/MildorfInequalities.pdf[/url]


A number $x_n$ of the form 10101...1 has $n$ ones. Find all $n$ such that $x_n$ is prime.

These numbers are of the form $\dfrac{100^n-1}{99} = \dfrac{(10^n-1)(10^n+1)}{99}$.
Now for $n>2$ all two factors are true factors, even after dividing some divisor of $99$ out, because they would be $>99$. So it's not prime then.


Hey, it's been a while since there's been a problem on this forum, so I guess I'll go ahead with one.

You have two springs of constants $k_{1}$ and $k_{2}$, connected in series.  What is the work done when you stretch the whole spring $\triangle l$ ?

[hide="Here is my soln:"]Let's say that the strings are stretched by $\Delta l_1$ and $\Delta l_2$ ($\Delta l_1+\Delta l_2=\Delta l$). The forces at the spring boundary must be in equilibrium:

$k_1\Delta l_1=k_2\Delta l_2$

The solution of these two equations is

$\Delta l_1=\Delta l\frac{k_2}{k_1+k_2}$
$\Delta l_2=\Delta l\frac{k_1}{k_1+k_2}$

And the work is:

$\frac{1}{2}(k_1 \Delta l_1+k_2 \Delta l_2)=\frac{k_1k_2}{k_1+k_2}\Delta l$.[/hide]

Well, the work of a spring is $\frac{1}{2}k{\triangle l}^2$

but besides that, it looks good  :)

Hm, oops :). In that case,
[hide="the result is right here:"]
$\frac{k_1k_2}{k_1+k_2}\Delta l^2$.

Huh? Looks strangely familiar....[/hide]

=]  springs in a mechanical system behave like capacitors in a circuit.

Interesting. Perhaps there's a deeper analogy

[hide]hidden somewhere?[/hide]


MATHCOUNTS
quadratics
algebra
polynomial
function
quadratic formula
product of roots

(a) The sum of two numbers is 110.  Their product is 1729.  What are the two numbers?

(b) The sum of two numbers is 50.  Their product is 555.  What are the two numbers?

Is there an easier way to solve this rather than solve for 2 simultaneous equations?

The subject is probably ``Quadratic Cleverness'' for a reason. I will say no more.

Syntax Error wrote:sigh. ill just solve it by means of the simultaneous equation way. it really doesnt take that long. maybe a minute or more depending on the person. sigh. well for the first one, i gots [hide]19 and 91. basically just set up the two equations x+y=110 and xy=1729. then x=110-y. plug that in the second equation and you get 110y-y:^2:=1729. rearranging, -y:^2+110y-1729=0. quadratic formula and voila!

does the cleverness have to do with like (x+y):^2:=12100. so then x:^2:+y:^2+2xy=12100. but since we know that xy=1729. plug in that times 2 to get x:^2:+y:^2:=8642. but i dont know what to do from there. or if my line of thought is actually just doing something radically different. i would do the second problem, but im currently trying to venture my way to finally finish the aops book. sigh. but i dont think it really matters because there supposedly is some trick. so ill stay quiet unless i stumble upon somat. but sigh. i guess one could think that i showed the "old" method to the new uns[/hide]

The second would be[hide]25:pm::sqrt:70[/hide], solved by the above written way.

I have no clue with the "clever" factor either.[/hide]

usually when you do problems like those enuff times they dont require much thinking on your part. but im still wondering what this "cleverness" is

The cleverness is in seeing that the two numbers could be the roots of a quadratic equation that you can construct from knowing the sum and the product of the roots of the quadratic.

While it doesn't help you quite so much in regards to quadratics, think about how much this could help when dealing with more variables and thus higher degree polynomials.

Suppose you have x + y + z = 8, xy + yz + zx = 19, and xyz = 12.  Can you relate these to a polynomial the same way that you related the two numbers in my problem to a quadratic?

The idea is to use so-called symmetric sums to relate variables to polynomials.

I probably shouldn't have tried to go this far with this idea in the MATHCOUNTS area.  You certainly won't see something this difficult on a MATHCOUNTS exam, but many of you seem bright enough to take this idea and run with it.

Still, the cleverness will save a little time even with two variables.  You don't have to bother substituting a variable and determining the quadratic.  You can simply write the quadratic down immediately because you have all the information you need to determine its coefficients.

I'm not sure if this is what you meant

[hide]when x+y=a and xy=b, x:^2:-ax+b=0[/hide]

Yes Tare, that is exactly what I meant.  Now can you use that same idea to solve the three variable problem I posed?

I solved the generic equation for three variable question.

[hide](a,b,and c is the solution, a+b+c=d, ab+ac+bc=e, and abc=f)

(t-a)(t-b)(t-c)=0

(t:^2:-(a+b)t+ab)(t-c)=0

t:^3:-(a+b)t:^2:+abt-ct:^2:+(a+b)ct-abc=0

t:^3:-(a+b+c)t:^2:+(ab+ac+bc)t-abc=0

t:^3:-dt:^2:+et-f=0[/hide]

[quote="s0mp"]never mind, its been corrected!  :wink:[/quote]
I accidentally wrote + for all the signs but I fixed it.

[quote="s0mp"]oh i see, its the sum of roots/product of roots relationships u have to use.

so the polynomial for the x,y,z one mr crawford was talking bout would be x :^3: -8x :^2: +19x+12, i think. x, y, and z would be hte roots of teh polynomial.

darn, beat to it.  :D[/quote]
You mean x :^3: -8x :^2: +19x[b]-[/b]12, right?

err..i thought the last term's c.e. in any polynomial is just leading coefficient times product of roots, and that second term's is -1(leading coefficient)(sum of roots). but i guess ur right!

I forgot that Mr. Crawford asked for the solution to x+y+z=8, xy+yz+zx=19, and xyz=12 (or t:^3:-8t:^2:+19t-12=0), so here it is.

[hide]x:^3:-8x:^2:+19x-12 factorize to (t:^2:-7t+12)(t-1)=0 or (t-3)(t-4)(t-1)=0, so the solution is t=3,4, or 1.[/hide]

[quote="MCrawford"]First, it's probably not a good idea to write the polynomial as a function of x since x is one of the three roots you're solving for.  Use a variable like t instead.  You did find the roots correctly, so what are all the solutions for (x, y, z)?[/quote]
I changed the xs into t.
The solution is x=-4, y=-3, and z=1. (the correspondance doesn't matter;all of the equations are commutative. For example, it will still work if x=1, y=-3, and z=-4)

that polynomial discussion strikes good insight  :idea:

but at least these 2 are about the same difficulty with or without that method


integration
trigonometry
function
calculus
calculus computations

Integrate $ \int\frac {\cos 7x \minus{} \cos 8x }{ 1 \plus{} 2\cos 5x } dx$

$ \int\frac {\cos 7x - \cos 8x }{ 1 + 2\cos 5x }dx$

Multiply and divide by $ \sin 5x$

$ = \int \frac {\sin 5x (\cos 7x - \cos 8x)}{\sin 5x + \sin 10x} dx$

$ = \int \frac {2 \sin (\frac {5x}{2}) \cos (\frac {5x}{2}) (2\sin (\frac{15x}{2} \sin (\frac{x}{2})}{2 \sin (\frac {15x}{2}) \cos (\frac {5x}{2})}dx$

$ = \int (2sin (\frac{5x}{2}) \sin (\frac{x}{2})dx$

$ =\int \cos 4xdx - \int \cos 6x dx$

$ =\frac{\sin 4x}{4}-\frac{\sin 6x}{6}+C$

hello, prove that your integrand $ \frac{\cos(7x)\minus{}\cos(8x)}{1\plus{}2\cos(5x)}$ is equal to
$ \minus{}4\cos(x)^3\plus{}2\cos(x)^2\plus{}3\cos(x)\minus{}1$
Sonnhard.

At picture 74 , i think that is wrong , please recheck

Again @ picture74 in the last step it should be $ \equal{} \int\cos 2xdx \minus{} \int\cos 3x dx$

Anyways thank you for the solution and the idea which are more important.

But now i have a problem i find the i can integrate such functions only when i find a pattern that is if the sum of the coffecients of x in the cos expression equals the 3 times the coffecient of x in cos in the denominator  , i find its integrable easily and otherwise not .

For example in the above problem in the num it is 15 which is 3*5 ( denominator coffecient) .

Why is this so ?? please explain why i am finding it difficult  to integrate for example $ \int\frac{\cos 11x\minus{}\cos 12x }{ 1\plus{}2\cos 9x }dx$

Sorry for the double post .

hello, your integral
$ \int\frac{\cos(11x)\minus{}\cos(12x)}{1\plus{}2\cos(9x)}\,dx$
is not so simple to solve.
Sonnhard.

Actually my whole question is that only . why is that i am able to solve some integrals of those type and not others .

Please see my post #7 .


On a beginner level Minesweeper board (9 by 9, 10 mines), what is the probability that I will win in one click if I always click the corner square, assuming that I cannot lose* on the first click?

*aka click on a mine

Cause Shobhit wanted me to post this here, and also because I happen to be bored enough to try winning in this manner.

haha, i just read this in the chatroom. well, too bad I do not have time to attempt a solution for it... but anyways if you want to check yourself, click on this link.

[url]http://www.factspider.com/mi/minesweeper-(game).html[/url]


[quote="Peter"]Prove that if $p$ is a prime, then $p^{p}-1$ has a prime factor that is congruent to $1$ modulo $p$.[/quote]

In fact, $p^{p-1}+p^{p-2}+...+p+1>1$. Thus, the number $p^{p-1}+p^{p-2}+...+p+1$ has a prime factor $q$. Then, since $p^p-1=\left(p-1\right)\left(p^{p-1}+p^{p-2}+...+p+1\right)$, we also have $q\mid p^p-1$, so that $p^p\equiv 1\text{ mod }q$. Hence, $p$ is divisible by $\text{ord}_q p$. Since $p$ is a prime, this yields that $\text{ord}_q p=1$ or $\text{ord}_q p=p$.

If $\text{ord}_q p=1$, then $p\equiv 1\text{ mod }q$, and thus $p^{p-1}+p^{p-2}+...+p+1\equiv 1^{p-1}+1^{p-2}+...+1+1=p\equiv 1\text{ mod }q$, what contradicts $q\mid p^{p-1}+p^{p-2}+...+p+1$.

Hence, $\text{ord}_q p=1$ is not possible, so we must have $\text{ord}_q p=p$. Since $\text{ord}_q p\mid q-1$ (because $q$ is a prime), this yields $p\mid q-1$, so that $q\equiv 1\mod p$. Since $q$ is prime and divides $p^p-1$, it thus follows that $q$ is a prime factor of $p^p-1$ which is congruent to $1$ modulo $p$, and we are done.

  darij

We can generalize this problem: $ p$ can be composite, too.

Yes, this follows from Zsigmondy's theorem. For those that don't like it in it's general form: just use the same proof  :wink: