If  $ \alpha,\beta,\gamma$  are acute angles such that $ cos\alpha\equal{}tan\beta,cos\beta\equal{}tan\gamma,cos\gamma\equal{}tan\alpha$ ,prove that $ sin\alpha\equal{}sin\beta\equal{}sin\gamma\equal{}\frac{\sqrt{5}\minus{}1}{2}$

[hide="My solution"]
$ sin\alpha \equal{} cos\alpha\times tan\alpha \equal{} cos\alpha\times cos\gamma$
Let $ sin^2\alpha \equal{} x, sin^2\beta \equal{} y$ and $ sin^2\gamma \equal{} z$.
Then $ x \equal{} (1 \minus{} x)(1 \minus{} z)$, so $ z \equal{} \frac {1 \minus{} 2x}{1 \minus{} x}$
Similarly, $ y \equal{} \frac {1 \minus{} 2z}{1 \minus{} z}$ and $ x \equal{} \frac {1 \minus{} 2y}{1 \minus{} y}$.
In those 3 expressions we perform next:
Into expression for $ y$ we plug in $ z$ from 1. expression and then we got $ y$ in terms of $ x$ and $ x$ in terms of $ y$. It's easy to plug in one into another and get quadratic equation:
$ x^2 \minus{} 3x \plus{} 1 \equal{} 0$ which solution is $ x \equal{} \frac {3 \minus{} \sqrt {5}}{2}$ and same for $ y$ and $ z$.
EDIT:Then $ \sqrt{x}\equal{}\frac{1\plus{}\sqrt{5}}{2}$

[/hide]

[quote="Bugi"]Are you sure?
I got that $ sin\alpha \equal{} sin\beta \equal{} sin\gamma \equal{} \frac {3 \minus{} \sqrt {5}}{2}$
[hide="My solution-Correct it!"]
$ sin\alpha \equal{} cos\alpha\times tan\alpha \equal{} cos\alpha\times cos\gamma$
Let $ sin^2\alpha \equal{} x, sin^2\beta \equal{} y$ and $ sin^2\gamma \equal{} z$.
Then $ x \equal{} (1 \minus{} x)(1 \minus{} z)$, so $ z \equal{} \frac {1 \minus{} 2x}{1 \minus{} x}$
Similarly, $ y \equal{} \frac {1 \minus{} 2z}{1 \minus{} z}$ and $ x \equal{} \frac {1 \minus{} 2y}{1 \minus{} y}$.
In those 3 expressions we perform next:
Into expression for $ y$ we plug in $ z$ from 1. expression and then we got $ y$ in terms of $ x$ and $ x$ in terms of $ y$. It's easy to plug in one into another and get quadratic equation:
$ x^2 \minus{} 3x \plus{} 1 \equal{} 0$ which solution is $ x \equal{} \frac {3 \minus{} \sqrt {5}}{2}$ and same for $ y$ and $ z$.

[/hide][/quote]
Looks like what you have is $ \sin^2 \alpha \equal{} \sin^2\beta \equal{} \sin^2\gamma \equal{} \frac {3 \minus{} \sqrt {5}}{2}$.

[quote="Euclid_great"][quote="Bugi"][hide="My solution"]
$ sin\alpha \equal{} cos\alpha\times tan\alpha \equal{} cos\alpha\times cos\gamma$
Let $ sin^2\alpha \equal{} x, sin^2\beta \equal{} y$ and $ sin^2\gamma \equal{} z$.
Then $ x \equal{} (1 \minus{} x)(1 \minus{} z)$, so $ z \equal{} \frac {1 \minus{} 2x}{1 \minus{} x}$
Similarly, $ y \equal{} \frac {1 \minus{} 2z}{1 \minus{} z}$ and $ x \equal{} \frac {1 \minus{} 2y}{1 \minus{} y}$.
In those 3 expressions we perform next:
Into expression for $ y$ we plug in $ z$ from 1. expression and then we got $ y$ in terms of $ x$ and $ x$ in terms of $ y$. It's easy to plug in one into another and get quadratic equation:
$ x^2 \minus{} 3x \plus{} 1 \equal{} 0$ which solution is $ x \equal{} \frac {3 \minus{} \sqrt {5}}{2}$ and same for $ y$ and $ z$.
EDIT:Then $ \sqrt {x} \equal{} \frac {1 \plus{} \sqrt {5}}{2}$

[/hide][/quote]

how can you say that The last line?[/quote]

Well $ {(\frac {1 \plus{} \sqrt {5}}{2})}^2\equal{}\frac{3\minus{}\sqrt{5}}{2}$, and the negative solution is impossible because angles are acute.


[b][u]What is the smallest positive integer greater than 1 which leaves a remainder of 1 when divided by 2,3,4,5,6 or 7?[/u][/b]

[hide]LCM of those numbers, plus 1. Which I am too lazy to figure out.[/hide]

just find the LCM of them and add one which is like 420ish

421 apparently by everyone else


John and Judith are guinea pig farmers. They need to make enclosures for their many species of guinea-pig, but unfortunately, they live in a country where there is "fence tax". Consequently, they can only afford to erect a maximum of 24 fences. The enclosures can have any number of sides and can have any shape, so long as the fences are straight, and a fence can join another fence only at end points. What is the largest number of enclosures they can make?

Source: Australian Math Competition, S29, 2000

[hide="New and Improved Guess, though still probably wrong"] 14. :D [/hide]

[hide="Explanation, just in case I'm right"] The most number of sections you can make dissecting a polygon by drawing lines to the vertices is $n$ with $2n$ lines. So the minimum answer is 12. You can make this more efficient by adding on to an existing side. The more additions, the more efficient. We can make lots of figures to make it as efficient as possible, therefore we use triangles. We can keep adding on to the side of a triangle and dissecting it. We can form 4 triangles, each including 3 sections, with 21 fences. We can finish off the 3 fences any way we want, but the maximum is always $\boxed{14}$. [/hide]

Err that's still not right (nearly there though). However you were correct about adding on to an existing side.

Consider using triangles (polygon with least sides) everywhere and you've pretty much got it.


summer program
Mathcamp
Ross Mathematics Program
PROMYS

Is there a special section in the college resume that asks what you did over the summer?  (listing names of camps, activities, etc each year)  :?

I always just listed mine under "seminars, etc"... that's where I listed AoPS class, my equine reproduction seminar, and my transportation engineering internship (which was horrible, fyi). When in doubt, put it under whatever's most general.

The [url=http://app.commonapp.org/index.cfm?APP=AppOnline&ACT=Display&DSP=Forms]Common Application[/url] has a space for asking about summer activities.

Having looked at the common application...I'm concluding that we don't list all the camps we've attended during summer?
Also, what does the resume cover?
Thank you!

[quote="thisismysn"]Having looked at the common application...I'm concluding that we don't list all the camps we've attended during summer?[/quote]

Well, you could. The first general principle to consider is that you list activities since eighth grade. Some middle school accomplishments are impressive, but generally colleges want to know what you did at high school age. So start with the summer after eighth grade, usually. If you run out of space, you can write "see attachment" and put on an extra page to list all the summer programs you did. That is an issue of setting priorities: list the summer programs that were most impressive (that is, the ones that you had to take a test to get into) and maybe don't mention unknown local programs except in such terms as "local chess champ, 2003-2006" or something like that. But you DEFINITELY want to mention something like MOP, MathCamp, Ross, or PROMYS, so make space for it if it doesn't fit in the standard blanks on the form.


Let V be a vector space over the reals R. Let F(V) be the set of all linear maps T: V ->R (linear functionals), with standard operations:
(T+S)(v) = T(v)+S(v)   and (cT)(v)=c(T(v)).

a) Check that F(V) is a vector space with these operations (this one I can do)
b) Prove that if V is finite dimensional, then F(V) is finite dimensional too, and dim F(V) = dim(V).

For part (b): Choose a basis $b_{i}$ for $V$. A basis for the dual $V^{*}$ ($F(V)$ in your problem) is $u_{i}$ so that $u_{i}(v_{i})=1$ and $u_{i}(v_{j})=0$ for $i\neq j$.


O is the orthocentre of $ \triangle ABC$. $ \angle BOC$ =$ 90^o$, find $ \angle BAC$.

Thank you.

Let X be the perpendicular from B to AC, so BX passes through O. Since $ CO\perp BX$ and $ CX \perp BX$, we must have $ X \equal{} O$ (they are the same point), because there is only one perpendicular from C to BX. Thus X is the orthocenter, so $ CX \perp AB$, so $ \angle BAC \equal{} 90$.

Another way:

The image of the orthocentre lies on the circumcircle of ABC, so $ \angle BHC\equal{}180^\circ\minus{}\alpha$


geometry AMC AIME quadratics USA(J)MO USAMO probability

Suppose that you are on the last problem of the AMC.  You have solved every other problem and are fairly certain you got them all without any mistakes etc.  The last problem is stumping you and you only have less than a minute left...not nearly enough time to solve the problem.  Do you guess and hope that you are lucky OR do you leave it blank and convince yourself that you are content with a 146.5?

I would guess.  That kind of mentality helped me go from virtually guaranteed 143 to 138 this year :).  Still, there isn't that much to lose on the AMC under those circumstances, so I say why not.

As Bob puts it, "it's a sin to get a 146.5".

Seriously, what's the difference between a 146.5 and a 144  (don't say 2.5 points)? Everyone will know that one person chose to leave a blank and the other took a guess. I'd much rather have a 20% chance of getting a perfect and my picture on the AMC website than a 100% chance of not getting a perfect and no picture on the AMC website.

For me, definately the guess. First, you can always, always narrow down answer choices, sometimes with math, sometimes with logic, sometimes with a calculator... anything!

Odd analysis: Lets say that you have a 30% of being right, which isn't unreasonable. If you are a perfect scorer on the AMC, you will be remembered - the prestige is great. You're guaranteed at the top of your nation, state, school, etc.... There's a lot in it.
If you lose - you lost 2.5 points, from 146.5 -> 144. Big deal? No.

So guess!

By the way -  On last year's AMC-10, this happened to me. (But not on 25, I'm thinking on 24.) I decided to solve a geometry problem  by measuring the diagram on the test. I got the problem right. I didn't get the 150 because of other fun mistakes, like on problem 3 or something like that, but it was worth it to guess.

Haha, I wouldn't describe a 146.5 as sinful, having had one myself once.  I did guess the next time I took it (no gain in getting two 146.5s in the same year, obviously), but sadly to no avail.  I say leave it blank -- what if you're wrong about having gotten all the others?  (And, as Kent Merryfield said in another post, guessing is a losers game -- the wrong answers are sure to look more attractive than the right one.)  Sorry for being unromantic, or whatever.

[quote="JBL"](And, as Kent Merryfield said in another post, guessing is a losers game -- the wrong answers are sure to look more attractive than the right one.)[/quote]

I have a TI-30xIIS calculator (in addition to a TI-89)... if I was randomly guessing, I'd probably use the random number generator to give me an integer from 1 to 5, inclusive (unless, of course, I could eliminate a choice or two for certain; then, I'd use the generator to make an integer from 1 to 3 or something :D).

DEFINITELY GUESS... You are only gonna get 2.5 pts lower than what you are supposed to get, and the 2.5 pts is not important at all. You can easily gain it back by getting 1 more AIME question right :) .

Also, say you leave the question blank, and you get 7 problems wrong:

You'll end up with a 104.5

But if you guess, you still have a 102, you still qualify for AIME.

!

People like me count on the AIME as opposed to the AMC anyway, so I just guess--whatever :)

[quote="ednerd"]Definitely guess.  Getting a 146.5 (I'm sad to admit I have) makes you feel very unmanly.  Sort of like a 0 on the USAMO.  But I won't go there.[/quote]

What if you're a girl?

150 ALL THE WAY!!! (I got a 144 =P, but that wasn't from a guess so yeah)

Let's see...
(Glory of 150) * (1/5) ? (Closeness of 144) * (4/5)
(Glory of 150) ? 4(Closeness of 144)

Well since the Glory of 150 includes a picture on the AMC Website, we will proceed to add 1000 points.

Hence

(Glory of 150) = 150+1000 = 1150 > 576 = 4(Closeness of 144)

Thus guessing is worth it.

This has actually happened to me.  On the 2004 10A, I couldn't solve a problem (I think it was #19), so I just picked one that looked attractive -- I think it was 8/9.  It turned out that was the correct answer!

But, of course, I missed a different problem, so I got a 144.  The moral of this story: check your work.

[quote]What if you're a girl?[/quote]

It makes you feel unwomanly?

[i]If[/i] I ever come that close, I'll mull it over on the spot.

Actually, it doesn't just make you feel unmanly.  You just find yourself regretting not solving it/yelling at yourself.  Especially if not solving it involves solving a quadratic equation wrong and wondering why none of the answer choices are right.

Because I would never, ever, ever, ever have any confidence in my answers for the other 24, I'd stay. I wouldn't be happy with a perfect due to random guessing anyway. I'd feel like I didn't deserve it.

[quote="yif man12"]Miths, if you ever get a perfect, just send in a picture of BoA. Please. It would be so sick.[/quote]

Haha...perfect score?...wow...we'll see.  ;) It would actually be pretty cool if I sent in a picture of BoA though!! :D

As far as my grade level...I'm a sophomore, which makes me pretty much a newb in terms of years of experience compared to many of the users here.

As a side note...for token and qvc and anyone else who's interested (and hasn't been keeping up with my xanga :-P ), [url=http://img.photobucket.com/albums/v452/MithsApprentice/SCCHE.jpg]here's a picture[/url] from like last October. That the picture makes my face look fatter than it should be though...so...*cough*. ^_^

[quote="tetrahedr0n"]Well... if you let your hair grow you might look like BoA.[/quote]

Lol. Nice.

[quote="MysticTerminator"]whoa. if I ever get a perfect score, I'll need to send in a picture of .... something. hmm. now I need ideas. just in case.[/quote]

whatever that is on your avatar! :D

btw, if i get one, i am sending in Homer Simpson... :lol:

[quote="MysticTerminator"]whoa. if I ever get a perfect score, I'll need to send in a picture of .... something. hmm. now I need ideas. just in case.[/quote]

[img]http://home.wanadoo.nl/r.de.sain/geitebreier/catsniper.jpg[/img]

Use that picture!

Double LOL (literally) from tokenadult and A+MATH here.

First time AMC has seen a sniper kitten who's good at math.  I would send something like that in, if only my parents would let me.

In the unlikely event that I get a 150 on the AMC-12 next year, I promise you folks to send in that picture.

All the perfect scorers should send in the SAME picture :P

If that's the case, they'd definitely have to send in pictures of BoA. Tis only fair to the viewers.

I think this has been discussed several times now... use the search or ask Miths.


well, sometimes TeXNiCcenter used to notify me when i need to install a package, but now it does not appear such notification, and it's not nice because every package i need i have to download it and install it by myself.

i'm using 2.6 version, since i couldn't install the 2.7.

1) Go to MiKTeX, Settings, Install packages on-the-fly and choose Ask me first.
Or
2) Download the complete version of MiKTeX and then you won't need to keep installng.


calculus
integration
logarithms
real analysis
calculus computations

I have a few problems, the first one, I have no idea whatsoever how to convert a Riemann sum to a definite integral:

1. lim(n->infin.) (1/n) [sum of (n, i=1) 1/(1+(i/n))^2]

The other two are integrals that I don't understand the process for:

2. int((x^5)(e^(x^2))

3. int_1^4(e^(sqrt(x)))

Help would be much appreciated!

1:

$\frac{1}{n}\sum_{i=1}^n\frac{1}{1+(i/n)^2}$

The $i/n$ in the sum goes from $1/n$ to $1$. As $n\to\infty$, this is $\int_0^1\frac{dx}{1+x^2}=\mathrm{arctan}\;1=\frac{\pi}{4}$

2:

$I=\int x^5e^{x^2} dx$

Substitute $u=x^2$. Then $x\;dx=\frac{1}{2}du$, so $I=\frac{1}{2}\int u^2e^udu$. This can be done with partial integration.

3:

$I=\int_1^4e^{\sqrt{x}}dx$

We can substitute again. $u=\sqrt{x}$ gives $dx=2u\;du$ so we have

$I=2\int_1^2ue^{u}du$

And this can be done with partial integration again.

Thanks so much for your help... one question though, in the Riemann sum, the whole 1+ i/n term is squared, not just the i/n term... how does this affect the result?

Oh, sorry. Well, we get

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

I reckon. This is not as easy to evaluate though...

That shouldn't be hard; it's $(1+x)^{-2}$, which can be done by the general power rule.

the part that confuses me is the dx being over the term... I have another similar problem that I'm not sure how to work through because of this:

int( dx/(1+e^x) )

any solutions or suggestions?

That's just notation to save space. If you feel more comfortable with it that way, write the $dx$ off to the side: $\int \frac1{1+e^x}\,dx$.

For this, try the substitution $v=1+e^x$, or the partial substitution $u=e^x$. You'll have to pull $du$ or $dv$ out of thin air, so it might be helpful to write the substitutions backward ($x=\ln(v-1)$ or $x=\ln u$).

That makes sense, I feel kinda stupid for not realizing that.... oh well... I'll work through it and see what I come up with... thanks for your help!


15. There are 100 players in a singles tennis tournament. The tournament is single elimination, meaning that a player who loses a match is eliminated. In the 1rst round, the strongest 28 players are given a bye, and the remaining 72 players are paired off to play. After each round, the remaining players play in the next round. The match continues until only one player remains unbeaten. The total number of matches played is

(A) a prime number (B) divisible by 2 (C) divisible by 5
(D) divisible by 7 (E) divisible by 11

[b]The first part of the problem about "28 strongest players are given a bye" is very confusing. I don't understand this problem.[/b]

that part doesnt matter. All you need to know is that there are 100 players

since there are 100 players, 99 have to be elliminated, so you find the letter that satisfies 99, which is E :P. I think that part was meant to confuse you

Um, hello, you know you can always search for the full solution in the resources section:

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1370108#1370108[/url]

(Or look up "2003 AMC 10B Problems/Problem 15" in the wiki)

But as to your question, "the 28 strongest players are given a bye" means that those 28 players automatically progress to the second round without playing a first round match, so it's like an "automatic" win against an imaginary player (by that, that's the definition of a bye) That means those 28 players are technically not involved in any matches for the first round. But the 72 other players are paired, so 36 matches are played between those players in Round 1. Then the losers get eliminated, and the 36 winners of those matches, along with the 28 "bye" players are paired for Round 2, and so on for the rest of the rounds, until only one player remain as the winner.

Basically a bye is that they pretty much sit out for one round.

Since it is elimintion, 1 person gets out for every match played. So it is 99 matches, which is $ E$ divisible by 11

Also, please try to keep discussion about a problem in that problem's original thread. That way, if someone else has the same question, they don't have to start a new thread about it.


Prove the following ineq:
$ (3a\plus{}2b\plus{}c)(\frac{1}{a}\plus{}\frac{1}{b}\plus{}\frac{1}{c}) \leq \frac{45}{2}$ with $ a;b;c$ satisfying $ 1 \leq a;b;c \leq 2$

http://www.mathlinks.ro/viewtopic.php?p=1066439#1066439
Reffering to pvthuan that problem is from an ongoing contest. Please lock the topic...


integration
algebra
polynomial
function
real analysis
real analysis unsolved

$ f\in R[a,b]$ and $ \int_a^b{f(x)dx} > 0$, $ P$ is a polynomial s.t. $ \int_a^b{P^2(x)f(x)dx} \equal{} 0$.
Show that $ P\equiv 0$

This doesn't appear to be true.  For example, if $ a \equal{} \minus{}b$ and $ f(x)$ is a positive even function then the first condition is clearly true but $ x f(x)$ is odd.

Sorry, I made a typing mistake.
Actually it should be $ \int_a^b{P^2(x)f(x)dx} \equal{} 0$.

It's still not true: consider $ f(x) \equal{} x$, $ [a, b] \equal{} [ \minus{} 1, 2]$ and $ P(x) \equal{} x \minus{} c$ with $ c \equal{} \frac {1 \minus{} \sqrt {6}}{2}$.  (This is what you get when you solve the equation $ \int_{ \minus{} 1}^2 x(x \minus{} c)^2 dx \equal{} 0$ for $ c$.  If I were a textbook, I would have chosen $ a$ and $ b$ so that $ c$ came out rational  :wink: )  Perhaps there's supposed to be a quantifier somewhere that makes this not false?  (The conclusion looks like the statement should begin, "Suppose $ P$ is a polynomial such that for all functions $ f$ with the property ....")


Euler
geometry
circumcircle
geometric transformation
reflection
geometry proposed

prove that there is a point $ P$ on euler line of $ \triangle ABC$ that $ AG_A\equal{}BG_B\equal{}CG_C$.where $ G_A,G_B,G_C$ are centroids of $ \triangle BPC,\triangle APC,\triangle APB$

Thank you dear [b][size=100]Amir.S[/size][/b], for a nice problem.

I can only this time to present the construction of the point $ P,$ as the problem states and I will post here later more details.

The basic idea is that for every point $ P,$ lies on the Euler line $ OH$ of a given triangle $ \bigtriangleup ABC,$ where $ O,$ $ H,$ are its circumcenter and orthocenter  respectively, the line segments $ AGa,$ $ BGb,$ $ CGc,$ as the problem states, are concurrent at one point so be it $ Q,$ which lies also on the line segment $ OH.$

So, if we consider $ Q\equiv O,$ we can define two points $ Ga,$ $ Gb,$ on the line segments $ AO,$ $ BO$ respectively, such that $ GaGb\parallel MaMb,$ where $ Ma,$ $ Mb,$ are the midpoints of the side-segments $ BC,$ $ AC$ respectively and simultaneously $ GaGb \equal{} \frac {2}{3}\cdot (MaMb)$.

The line segments $ MaGa,$ $ MbGb,$ intersect each other at one point $ P,$ lies also on the Euler line $ OH$ of $ \bigtriangleup ABC,$ which has the property as the problem states.

Kostas Vittas.

[quote="Amir.S"] [color=darkred]Prove that there is a point $ P$ on [b]Euler's line[/b] of $ \triangle ABC$ that $ AG_a \equal{} BG_b \equal{} CG_c$ ,

where $ G_a,G_b,G_c$ are the [i]centroids[/i] of $ \triangle BPC$ , $ \triangle APC$ , $ \triangle APB$ .[/color] [/quote]
[color=darkblue][b][u]Proof.[/u][/b] Denote the [i]projection[/i] $ D$ of $ A$ to $ BC$ and the [i]reflection[/i] $ D_1$ of $ D$ w.r.t.

the [i]midpoint[/i] of $ [BC]$ ( $ D$ and $ D_1$ are [i]isotomic[/i] w.r.t. $ [BC]$ ). Prove easily metrically

( [b]I"ll post if you wish ![/b] ) that the required point $ P$ has the property $ PD_1\perp BC$ .

[b]Remark.[/b] Denote the [i]projections[/i] $ D$ , $ E$ , $ F$ of the [i]orthocenter[/i] $ H$ on the sidelines $ BC$ , $ CA$ , $ AB$ respectively and the its [i]isotomic[/i] points $ D_1$ , $ E_1$ , $ F_1$ w.r.t. the sides $ [BC]$ , $ [CA]$ , $ [AB]$ respectively. Prove easily ( using the [b]Carnot's lemma[/b] ) that the perpendicular lines in the points $ D_1$ , $ E_1$ , $ F_1$ on the sidelines $ BC$ , $ CA$ , $ AB$ respectively are concurrently (let $ P$ be this point). The upper proof certifies that the point $ P$ belongs to the [b]Euler's line[/b] of the triangle $ ABC$ .[/color]


Given two convergent series of real numbers, $ \sum_{j\equal{}0}^{\infty} a_j$ and $ \sum_{k\equal{}0}^{\infty} b_k$, we define their product as $ \sum_{n\equal{}0}^{\infty} c_n$ where $ c_n\equal{}\sum_{j\plus{}k\equal{}n} a_j b_k$. The series $ \sum_{n\equal{}0}^{\infty} c_n$ may be divergent. Rudin gives this example: if $ a_j\equal{}b_j\equal{}\frac{(\minus{}1)^j}{\sqrt{j\plus{}1}}$, then each of the $ n\plus{}1$ products in the formula for $ c_n$ has the same sign and is at least $ \frac{1}{n\plus{}1}$ in absolute value. Hence $ |c_n|\ge 1$ for all $ n$.
On the other hand, the product converges if at least one of the series $ \sum_{j\equal{}0}^{\infty} a_j$ and $ \sum_{k\equal{}0}^{\infty} b_k$ is absolutely convergent.

What about the converse? Given a conditionally (=non-absolutely) convergent series $ \sum_{j\equal{}0}^{\infty} a_j$, can one find a convergent series $ \sum_{k\equal{}0}^{\infty} b_k$ for which the product diverges? To begin with, take $ a_j\equal{}\frac{(\minus{}1)^j}{j\plus{}1}$.

We can take care of your $ a_j\equal{}\frac{(\minus{}1)^j}{j\plus{}1}$ example by letting $ b_j\equal{}\frac{(\minus{}1)^j}{\ln(j\plus{}2)}.$

Slightly more generally, suppose $ a_j\equal{}(\minus{}1)^jp_j$ for decreasing $ p_j>0$ and define $ q_k\equal{}\sum_{j\equal{}0}^kp_j.$ By assumption, $ q_j\to\infty$ as $ j\to\infty.$

Let $ b_j\equal{}\frac{(\minus{}1)^j}{q_j}.$

Then as in Rudin's example, all of the signs in any sum for $ c_n$ are equal, and

$ |c_n|\equal{}\sum_{j\equal{}0}^n\frac{p_j}{q_{n\minus{}j}}\ge\frac1{q_n}\sum_{j\equal{}0}^np_j\equal{}\frac{q_n}{q_n}\equal{}1.$

So $ c_n$ doesn't tend to zero and the sum of the $ c_n$ diverges.

That's not a general proof, since it's all still tied into the specifics of the alternating series test, but it seems to bode well for the prospects of your proposed theorem being true.

Oh. I made a sloppy estimate and somehow decided that 1/log doesn't work. But it does. You don't even need $ p_j$ to be decreasing.


Find the area of a regular polygon in terms of it's side length s.

If we're just given the side length, then it is impossible. But if we're given that it's a regular $n$-gon, then a formula in terms of $n$ and $s$ would be $\frac{ns^{2}\cot\left(\frac{180}{n}\right)}{4}$.

That's the exact same formula I got! How did you get it? I just found a pattern that worked with 3-5.

We can divide a regular $n$-gon into $n$ congruent triangles. We'll find the area of one of the triangles and multiply it by $n$.

Let $R$ denote the circumradius of the polygon. Then a formula for the triangle is $\frac{1}{2}\sin\left(\frac{360}{n}\right)R^{2}$. To find $R$ in terms of $s$, we can write
$\sin\left(\frac{180}{n}\right)=\frac{s}{2R}$
$\Longrightarrow R=\frac{s}{2\sin\left(\frac{180}{n}\right)}$. Substituting this in to our expression, we get $\frac{ns^{2}\cot\left(\frac{180}{n}\right)}{4}$.


On page 224 of Volume 2 (first page of Find It and Make It), I don't get the part where is says that we cannot conclude that the desired locus is the perpendicular bisector of AB, and that only the fact that every point in the bisector is in the locus was proven.  It then goes on with a similar proof and then concludes that the locus is the perpendicular bisector of the segment.  Could someone please clarify/explain this to me?  Thanks in advance!

Here are three statements:

1) Everything that is an element of $ X$ is an element of $ Y$.
2) Everything that is not an element of $ X$ is not an element of $ Y$.
3) $ X \equal{} Y$.

What is the logical relationship between these three statements?

In your case, $ X$ is the locus of points equidistant from two given points (at least this is my guess based on what you've said) and $ Y$ is the perpendicular bisector of the segment joining two given points.

[quote="JBL"]Here are three statements:

1) Everything that is an element of $ X$ is an element of $ Y$.
2) Everything that is not an element of $ X$ is not an element of $ Y$.
3) $ X \equal{} Y$.

What is the logical relationship between these three statements?[/quote]

Oh, so Vol. 2 first proved statement one with SSS.  However, the fact that the locus is the perpendicular bisector of the segment was not proven until statement two was proven, which automatically proved statement three, which represents what we were trying to prove in the first place.  Please correct me if this is not right.

[quote="JBL"]In your case, $ X$ is the locus of points equidistant from two given points (at least this is my guess based on what you've said)[/quote]

Yes, that's the locus I was referring to. :)

Thanks a lot!

Right: the third statement is equivalent to (1) and (2) together, and you need both (1) and (2) in order to conclude (3).  In other words, the first result proved that the bisector is [i]contained in[/i] (not "equal to") the locus; you then need to show that the locus is contained in the bisector, or else the locus could consist of the bisector plus a bunch of other points somewhere.


What is the point of determining the differential rate law through experimentation when it is easier to determine the integrated rate law.

For $ A+B\rightarrow C$, you would need at least 3 plots of concentration vs. time (normal, keep A constant, and keep B constant) to determine reaction orders. Then you can determine the rate constant and the actual rate law.

However, you only need conc. vs time for 1 experiment to determine the integrated rate law.

Actually, I would not think it is so easy to get the integrated rate law out of experiment.  For a reaction $ A+B\rightarrow C$ where both $ A$ and $ B$ appear in the rate law, the integrated rate law is essentially a real-valued function of two variables.  So at best you can find the equation of some curve on the surface defined by this function, but it is not possible to replicate the entire function from there.

You could probably go through a list of all possible rate laws and see if any are consistent with the data you obtained, but that is probably more work than doing 3 experiments (plus, I am not sure if you can really do this).

And in any case, it is not always so easy to track the concentrations of all reactants and products.  Unless you have some very high-level equipment, you will probably be doing this spectroscopically, in which case you probably only want one species to be absorbing.  Keep in mind also that if you were able to keep track of all products and reactants, it is likely that intermediates would also absorb, which could throw things out of wack.

Compared to these difficulties, I think the differential rate law sounds nice and easy to measure:).

Thanks, I see what you mean.

So I guess integrated rate law is easiest to find when there is only 1 reactant, like in $ A\rightarrow B$?

Or if you know that the rate law depends only on one reactant.  You could probably do it in one experiment this way, but again you would have to do curve fitting.  If you have a computer, this might be easier, but I don't think you would have fun doing so by hand :).

[quote="gregx007"]And in any case, it is not always so easy to track the concentrations of all reactants and products.  Unless you have some very high-level equipment, you will probably be doing this spectroscopically, in which case you probably only want one species to be absorbing.[/quote]

Which reminds me, at IChO last year they had a lab problem where you did have to do spectroscopy of solutions with two dyes.  It is possible to do, but unless you can make some approximations (as we got to do there) it can be a lot of grind work.  And believe me, grind work is not fun on (competitive) labs!


inequalities
LaTeX
logarithms
function
inequalities proposed

Let $a,b,c >0$ and $a+b+c=1$

Find the maximum of
$a^k(b+c)+b^k(c+a)+c^k(a+b)$

$k$ is a positive real.

------------------------------------------------------------------------------------------------
This problem I created last night and It take me all this morning to Find the solution!
I hope It hasn't been post  before!

I think we can use this ineq:
a^{k}b+b^{k}c+c^{k}a<=  n^{n}/(n+1)^{n+1}.

I think we can use this ineq:
a^{k}b+b^{k}c+c^{k}a /lef  n^{n}/(n+1)^{n+1}.

I don't think so.
Please,If you can,complete your answer!

I think the maximum is $\frac{2}{3^k}$ - is it correct?

Sorry But my result isn't it! It's only special case.
 :(  :(

Your problem is really interesting ,Hung.

my guess of answer: $max(\frac{2}{3^k},\frac{1}{2^k})$

I think it's very similar to the fraction inequality.
I used similar approach and i got stuck at the same thing again. I proved the case $k \ge \frac{2}{3}$ for fraction inequality and I prove the case $ k \le \frac{3}{2}$ for this one.

[quote="hungkhtn"]Ok,It's my result.But I think It's more difficult than The Fraction Inequality alot.
 :?  :?
Please Explain your answer,Siuhochung.I don't think It can be solved in 2 cases of $k$.[/quote]

nono, i dun hv any result yet. I can only prove the case $k \leq \frac{3}{2}$ for this inequality. (much weaker than the result of this question)

After 2 days of hard working ,i have found the solution

In my proof , i suppose that a<=b<=c .Then let a be a fixed number .Then  we can  prove that
f ( a,b,c) <= f (a,(b+c)/2 ,(b+c)/2) = g(a).
Case k<=1 and k>=2 we only need to work on g'(a) and it's easy.
Case 1<=k<=2 , i worked on g'(a) and g''(a) .

My complete solution is long and i will write it when i have time ( and when i finish the Learning Latex course  :D ) .

Suppose that you are right.What do you continue?Because the rest is already not easy,I think.
 :)  :)
I wait you post the solution after you finish Leaning Latex.

I think I have wrong result because this ineq is not easy.

I have the following results:

(1) $\sum{a^2(b+c) \le \frac{1}{4}(a+b+c)^3}$, with equality for $(a,b,c)=(0,1/2,1/2)$;
(2) $\sum{a^3(b+c) \le \frac{1}{8}(a+b+c)^4}$, with equality for $(a,b,c)=(0,1/2,1/2)$;

Additionally, I conjecture that
(3) $\sum{a^4(b+c) \le \frac{1}{12}(a+b+c)^5}$, with equality for $(a,b,c)=(0,\frac{3-\sqrt{3}}{6}, \frac{3+\sqrt{3}}{6})$.

[quote="hungkhtn"]By next 2 weeks![/quote]

hungkhtn: so now you can post your solution  :)

Hungkhtn, I prove Siuhochung's result 2/(3^k)

1. 2/3^k is definitely wrong.
2. Please clarify your last line of work.

We have 2 cases:
(A) $0<m\le m_1=\frac{\ln 2}{\ln 3-\ln 2}$;
(B) $m \ge m_1$.
     In the case (A) we can use RCF-Theorem. The expression $E(a,b,c)=\sum a^m(b+c)=\sum{a^m(1-a)}$ is maximal for $a=b=c$.
     In the case (B), we write $E$ as $E=(a^m+b^m+c^m)(a+b+c)-(a^{m+1}+b^{m+1}+c^{m+1})$, and then apply (n-1)EV Principle: if $a+b+c=1$ and $a^m+b^m+c^m=constant $, then the expression $a^{m+1}+b^{m+1}+c^{m+1}$ is minimal for either $0=a<b\le c$ or $0<a\le b=c$.
Thus it remains to consider only these cases.
Hence
$E \le max \{\max_{b+c=1}E(0,b,c), \max_{a+2b=1}E(a,b,b)\}=$
$=max \{\max_{b+c=1}bc(b^{m-1}+c^{m-1}), \max_{1/3\le b\le 1/2}2(b(1-2b)^m+b^m(1-b))\}$.
Finally, I think that:
$E \le E(1/3,1/3,1/3)$ for $0<m \le m_1$;
$E \le E(0,1/2,1/2)$ for $m_1\le m \le 3$;
there is $b_m$, $0<b_m<1/2$ such that $E \le E(0,b_m,1-b_m)$ for $m \ge 3$.
I think we can find a nice solution to show that for any integer $m\ge 2$  the expression $E(a,b,c)$ is maximal only if $\min(a,b,c)=0$. This result is also valid for $E_1=a^mb+b^mc+c^ma$, as was shown in this forum.

Well Hungkhtn what is your solution? I am in agony to see this. Please post it.

[quote="keira_khtn"]Hungkhtn,this is my proof:Because [tex]k\geq\ -1+1=0[/tex]
[tex]a^k(b+c)+b^k(c+a)+c^k(a+b)
=\frac{a^k}{(b+c)^{-1}}+\frac{b^k}{(a+c)^{-1}}+\frac{c^k}{(a+b)^{-1}}\leq\frac{(a+b+c)^k}{3^{k+1-1}[2(a+b+c)]^{-1}}[/tex]
Hence:
[tex]L.H.S\leq\frac{2(a+b+c)^{k+1}}{3^k}=\frac{2}{3^k}[/tex][/quote]
Hungkhtn,sorry because i 've post without obviously!
I gusse that this inequal is true but i've not prove it yet.Or may it only be true for some cases.I expect it will reverse sign iff a,b,c be sidelengths of any triangle(since a Vasc's ineq):
[tex]a^rb+b^rc+c^ra\geq 3(\frac{a+b+c}{3})^{r+1}[/tex]

Hungthn you promise that you will post the solution of this problem in two weeks. But this post was a month ago. Please put your solution .is not only me who wants your solution.Please post it as soon as possible. :)

the function $f(x)=x^n-x^{n-1}$ is not convex?

[quote="Kondr"]We can rewrite the expression into form $X=\sum {\left(a^k-a^{k+1}\right)}$.  From Jensen for
$f(x)=x^k-x^{k+1} $ we get $X\leq 3 f\left( \frac 1 3 \right)=\frac 2 {3^k}$. Unfortunately this holds only for convex f :([/quote]

Just wonder that if you know that it is not a solution and probably it is useless, then why should you post it..

[quote="Kondr"]Because f(x) is convex in interval $(\frac {k-1}{k+1}, \infty)$, so this is a prove for k<1. I hoped you will be able to finish the proove for greater k.[/quote]

but then Vasc has completed a solution in this direction by his RCF theorem.

Sorry because I'm too late in this Forum.
I created this one along time ago. I write my solution in a lost book...  :(

I 'll write again by the time I can... The idea is proving 2 of 3 number equal.
It's not too difficult and the remain is find the maximum without condition a+b+c=1

You check in form: $\frac{a^k(b+c)+b^k(c+a)+c^k(a+b)}{(a+b+c)^{k+1}}$

Because it's use some long compution then I don't remember now.

Please find your solution and put it because this problem interest many people in this Forum

Hunghtn do you find your solution? Please reply.

THis's the first time this problem has been posted. So here's some idea about it. I'm forgot USB when I go to the University this morning, SO I can't show the full solution

WLOG, suppose that $a \ge b \ge c$
\[ f(a,b,c)=a^k(b+c)+b^k(c+a)+c^k(a+b)  \]
We will prove that
(i). IF $k \ge 2$ then $f(a,b,c) \le f(a,b+c,0)$.
(ii). IF $k \le 1$ then $f(a,b,c) \le f(1,1,1)=6$, Use Holder Inequality.
(iii). IF $1\le k \le 2$ then $f(a,b,c) \le f(t,t,c)$ where $t=\frac{a+b}{2},u=\frac{a-b}{2}$.

So, (i) is easy and (ii) is too. If you want, I will post more details soon. I will prove (iii).
\[ f(a,b,c)=g(u)=(t+u)^k(t-u+c)+(t-u)^k(t+u+c)  \]
\[ g'(u)=k(t+u)^{k-1}(t-u+c)-(t+u)^k-k(t-u)^{k-1}(t+u+c)+(t-u)^k  \]
\[ g'(u)=(t+u)^{k-1}\Big((k-1)t-(k+1)u+kc\Big)-(t-u)^{k-1}\Big((k-1)t+(k+1)u+kc\Big)  \]
Because $1\le k \le 2$, we have $\left(\frac{t+u}{t-u}\right)^{k-1}\le \frac{t+u}{t-u}$ and to prove $g'(u) \le 0$ we only need to prove
\[ (t+u)\Big((k-1)t-(k+1)u+kc\Big)-(t-u)\Big((k-1)t+(k+1)u+kc\Big) \le 0  \]
But, of course, It's easy.

From that, we need to devide the problem into 2 case
(i). If $k \le 2$.
Consider the function $g(t)=2t^{k}(t+c)+2c^kt$ with $2t+c=3$. Let
\[ h(t,c)=\frac{t^{k}(t+c)+c^kt}{(2t+c)^{k+1}}  \]
So we can assume that $1=t \ge c$.
\[ h(c)=\frac{c^k+c+1}{(2+c)^{k+1}}  \]
It's a strange, but finding the maximum of this function $h(c)$ is more easy and clearly than function $h(t,c)$. Easy to check that the equation $h'(c)=0$ has not more than 3 root  $\le 1$ if $k \le 2$.  So
\[ h(t) \le \max\{h(1),h(0)\}  \]
So we are done.
(ii) If $k \ge 2$. We can assume that $c=0$. Consider the function
\[ f(a,b,0)=\frac{a^kb+b^ka}{(a+b)^{k+1}}  \]
Suppose that $a \ge b=1$. Let
\[ q(a)=\frac{a^k+a}{(a+1)^k}  \]
This last step is not difficult. Then
\[ q(a) \le \max \{q(1),q(\alpha)\}  \]
which $\alpha$ is root of the equation
\[ x^{k}-kx^{k-1}+x-k=0.  \]

Hungkhtn you have some Latex errors and  I make them good .
BTW, thank you very very much for the solution

[quote="hungkhtn"]THis's the first time this problem has been posted. So here's some idea about it. I'm forgot USB when I go to the University this morning, SO I can't show the full solution

WLOG, suppose that $a \ge b \ge c$
\[ f(a,b,c)=a^k(b+c)+b^k(c+a)+c^k(a+b)  \]
We will prove that
(i). IF $k \ge 2$ then $f(a,b,c) \le f(a,b+c,0)$.
(ii). IF $k \le 1$ then $f(a,b,c) \le f(1,1,1)=6$, Use Holder Inequality.
(iii). IF $1\le k \le 2$ then $f(a,b,c) \le f(t,t,c)$ where $t=\frac{a+b}{2},u=\frac{a-b}{2}$.

So, (i) is easy and (ii) is too. If you want, I will post more details soon. I will prove (iii).
\[ f(a,b,c)=g(u)=(t+u)^k(t-u+c)+(t-u)^k(t+u+c)  \]
\[ g'(u)=k(t+u)^{k-1}(t-u+c)-(t+u)^k-k(t-u)^{k-1}(t+u+c)+(t-u)^k  \]
\[ g'(u)=(t+u)^{k-1}\Big((k-1)t-(k+1)u+kc\Big)-(t-u)^{k-1}\Big((k-1)t+(k+1)u+kc\Big)  \]
Because $1\le k \le 2$, we have $\left(\frac{t+u}{t-u}\right)^{k-1}\le \frac{t+u}{t-u}$ and to prove $g'(u) \le 0$ we only need to prove
\[ (t+u)\Big((k-1)t-(k+1)u+kc\Big)-(t-u)\Big((k-1)t+(k+1)u+kc\Big) \le 0  \]
But, of course, It's easy.

From that, we need to devide the problem into 2 case
(i). If $k \le 2$.
Consider the function $g(t)=2t^{k}(t+c)+2c^kt$ with $2t+c=3$. Let
\[ h(t,c)=\frac{t^{k}(t+c)+c^kt}{(2t+c)^{k+1}}  \]
So we can assume that $1=t \ge c$.
\[ h(c)=\frac{c^k+c+1}{(2+c)^{k+1}}  \]
It's a strange, but finding the maximum of this function $h(c)$ is more easy and clearly than function $h(t,c)$. Easy to check that the equation $h'(c)=0$ has not more than 3 root  $\le 1$ if $k \le 2$.  So
\[ h(t) \le \max(h(1),h(0))  \]
So we are done.
(ii) If $k \ge 2$. We can assume that $c=0$. Consider the function
\[ f(a,b,0)=\frac{a^kb+b^ka}{(a+b)^{k+1}})  \]
Suppose that $a \ge b=1$. Let
\[ q(a)=\frac{a^k+a}{(a+1)^k}  \]
This last step is not difficult. Then
\[ q(a) \le \max q(1),q(\alpha)  \]
\alpha is root of the equation
\[ x^{k}-kx^{k-1}+x-k=0.  \][/quote]

Is the answer $\frac{3k^k}{(k+1)^{k+1}}?$

No, kunny. The answer if $k\ge 2$ is quite complicated.

[quote="Vasc"]

Additionally, I conjecture that
(3) $ \sum{a^4(b + c) \le \frac {1}{12}(a + b + c)^5}$, with equality for $ (a,b,c) = (0,\frac {3 - \sqrt {3}}{6}, \frac {3 + \sqrt {3}}{6})$.[/quote]
The following stronger inequality is true too.
Let $ a,$ $ b$ and $ c$ are positive numbers such that $ \frac {1}{a} + \frac {1}{b} + \frac {1}{c} = 3.$ Prove that:
\[ (a + b + c)^5\geq12(a^4b + a^4c + b^4a + b^4c + c^4a + c^4b) + 171\]

If I am not mistaken, the inequality is proven by using mixing variables after the setting of

 $ \frac{1}{a}\equal{}x$

My computer got mad about this problem and actually, I spent nearly 1 hour on cleaning up the mess of mixing variable consequence. Very nice inequality though!


I have a week to solve Round 2, and I wanted to typeset them with LaTeX.  Is it simple enough to learn within a week?

Thanks.

Yes, it's very easy. Take the time to learn it; it's worth it.


[quote="claudiocamera"]In order to solve an optimization problem I came across a inequality system, I think it is not dificult, but I dont remember how to solve this kind of systems, can you help me? The system is:

0.4x -0.2y>-1
-0.2x+0.6y>-1
-0.1x+0.05y+0.25z>-0.5

Thanks[/quote]
[hide]$0.4x-0.2y>-1;\,-0.2x+0.6y>-1;\,-0.1x+0.05y+0.25z>-0.5$

First, make all coefficients integers: $2x-y>-5;\,-x+3y>-5;\,-2x+y+5z>-10$

The second inequality can be changed to $2x-y>-5+3x-4y.$ The LHS is the same as in the first inequality, so $-5=-5+3x-4y,$ or $3x=4y.$ Isolate one of the variables and substitute them into one of the first two inequalities. We find $x>-4$ and $y>-3.$

I'm not sure how to find $z.$ :|[/hide]


a) Prove that every multiple of 6 can be written as the sum of four third powers of integers.
b) Prove that every integer can be written as the sum of five third powers of integers.

it says for EVERY multiple of 6 in a), which is clear nonsense...

Sorry for my mistake. It has been corrected.


linear algebra
matrix
topology
linear algebra unsolved

[b]Source: Homework[/b]

Show that the matrices $M$ $\in$ $M(n,n)$ on the form $M=$ $\begin{pmatrix}A & B\\ C & D \end{pmatrix}$ where $A$ $\in$ $M_{k}(n,n)$ is invertible, make an open set $U$ $\subseteq$ $M(n,n).$

The determinant of a matrix is continuous ...


Use the same method as in http://www.mathlinks.ro/Forum/viewtopic.php?t=63539 to find also the sum for those with $k$ even.

Reverse the sum, put 90-theta in for theta and it is the same sum


modular arithmetic
function
search
geometry
3D geometry
combinatorics unsolved
combinatorics

Define an overpartition of $ n$ to be a partition of $ n$ in which that last occurrence of a part may be overlined. For example, the overpartitions of $ 3$ are $ 3, \overline{3}, 2 \plus{} 1, \overline{2} \plus{} 1, 2 \plus{} \overline{1}, \overline{2} \plus{} \overline{1}, 1 \plus{} 1 \plus{} 1, 1 \plus{} 1 \plus{} \overline{1}$. Define the rank of an overpartition to be the largest part minus the number of parts. Let $ p_o(n)$ denote the number of overpartitions of odd rank and $ p_e(n)$ denote the number of overpartitions of even rank. Prove that, if $ n \equiv 3 \pmod{8}$, then
\[ (\minus{}1)^n (p_e(n)\minus{}p_o(n)) \equal{} r_3(n)
\]
where $ r_3(n)$ is the number of representations of $ n$ as a sum of three squares, taking order and sign into account (so $ 1^2 \plus{} 1^2 \plus{} 1^2$ is different from $ (\minus{}1)^2 \plus{} 1^2 \plus{} 1^2$).

Any ideas for approaches are very welcome. Thanks.

Looks generating-functional to me.  The GF for $ r_3(n)$ is easy enough to write down (and is related to some known number-theoretic generating functions, I believe -- this all looks very Ramanujan-y), as is (I think) the generating function for overpartitions.  And to modify this to take rank into account shouldn't be too hard.

As is obvious, I'm not thinking very long about this, but have you tried something like this?  (What context did this come up in?  Putting "overpartition" into my search engine brings up a lot of interesting-looking stuff, though I don't know if it's directly relevant.  Putting it into google scholar might bring up more.)

The $ r_3(n)$ generating function is the cube of a certain [url=http://en.wikipedia.org/wiki/Theta_function#Jacobi_theta_function]theta function[/url], so I think the Jacobi triple product will be relevant, and I'm reasonably certain it can't be too hard to write down the generating function for the overpartitions of rank $ k$.  (Then substitute $ \minus{}1$ into the rank-weighting variable.)

This is a slightly altered form of the case $ n \equiv 3 \pmod{8}$ of a theorem that says:
\[ ( - 1)^n (p_e(n) - p_o(n)) = - 16H(n) - 1/3 r_3(n)
\]
where $ H(n)$ denotes the Hurwitz Class Number of $ n$. I'm interested in proving the whole theorem, but since this case was so nice in appearance I thought there might be a slick combinatorial method (i.e. some sort of bijection from even overpartitions to odd ones that is not defined on a subset that can be corresponded with sums of squares). I've done some work with generating functions, but it seems difficult to encode the condition $ n \equiv 3 \pmod{8}$, and the general case is difficult because I know of no generating function for $ H(n)$ (that is simple). I'll post my work so far when I can.

EDIT:

Here it is. Let $ p_{k, m}(n)$ denote the number of overpartitions of $ n$ with $ k$ parts and rank $ m$. Then I claim that
\[ \sum_{k, m, n} p_{k, m}(n) z^k r^ m q^n = \frac {(1 + 1) ( - rq; q)_{k - 1} (zq)^k}{(rq; q)_k} r^{1 - k}
\]
The RHS can be explained as follows: $ (zq)^k$ represents $ k$ ones as the starting base for the partition. Then
\[ \frac {1}{(rq; q)_k} = \frac {1}{(1 - rq)(1 - rq^2)\ldots (1 - rq^k)} = (1 + rq + r^2q^2 + ...)(1 + rq^2 + r^2q^4 + ...) ... (1 + rq^k + r^2q^{2k
}\]
If we choose $ rq^{\alpha i}$ from the $ i$th bracket in the product, then we increment the first $ i$ terms of the partition by $ 1$. It is easy to see that this increases the rank by $ \alpha$, hence the coefficient of $ r^{\alpha}$. Furthermore, this process constructs a unique partition, and any partition can be uniquely decomposed into such steps of increments.

Now, the term $ (1 + 1)( - rq;q)_{k - 1} = (1 + 1)(1 + rq)(1 + rq^2)\ldots (1 + rq^{k - 1})$ generates a "partition" with distinct parts but allowing $ 0$ as a part.

Thus, the generating function generates two partitions: one ordinary one, and one with distinct nonnegative parts (by the way, this sort of "direct product" of partitions is the substance of Frobenius partitions), whose parts together add to $ n$. We establish a bijection between these and overpartitions as follows: for each part i$ \lambda$ n the second partition (with distinct nonnegative parts), we augment the first $ \lambda$ parts of the first partition and overline the $ \lambda + 1$st part.

So, let $ p_m(n)$ denote the number of overpartitions of $ n$ with rank $ m$. Then clearly
\[ \sum_{m, n} p_m(n) = \sum_{k = 1}^{\infty} \sum_{k, m, n} p_{k, m}(n) z^k r^ m q^n = \sum_{k = 1}^{\infty} \frac {(1 + 1) ( - rq; q)_{k - 1} (zq)^k}{(rq; q)_k} r^{1 - k}
\]
If we set $ z = 1$ and $ r = - 1$, then we get the generating function for the values in question, $ p_e(n) - p_o(n)$.
The latter looks similar to the LHS of Ramanujan's $ _1 \psi_1$ formula, but the sum isn't bilateral. I'm looking for any identities that might help compute such a sum, as well as any identities relating to $ \theta(q)^3$ (the generating function for sums of three squares).

The above discussion implies that the following is a generating function for partitions:
\[ \sum_{k = 1}^{\infty} \frac {(q)^k}{(q; q)_k}
\]
but the product form is more well known:
\[ \prod_{k = 1}^{\infty} \frac {1}{1 - q^k}
\]
I suspect that there should be an alternate generating function for the rank difference. Can anybody come up with one?

Finally, there is an analogous theorem for a different definition of the rank called the M2-rank. Let $ l$ be the largest part, $ m$ be the number of parts, $ m'$ be the number of overlined odd parts, and $ p = 1$ if the largest part is odd and non-overlined and $ 0$ otherwise. The M2 rank is $ l - m + m' - p$. Then
\[ ( - 1)^n (p_e(n) - p_o(n)) = - 8H(n) + 1/3 r_3(n)
\]
A notable case is that if $ n \equiv 3 \pmod(8)$, then the number of overpartitions with even M2-rank minus the number of overpartitions with odd M2-rank is $ 0$. I wonder if there is a nice bijection that proves this.


The incircle of triangle $ ABC$ is tangent to the sides $ BC,CA,AB$ at $ D,E,F$ respectively.
Suppoe the incircle intersects $ AD$ again at $ X$ such that $ AX\equal{}XD,XB$ and $ XC$
intersects the incircle again at points $ Y$ and $ Z$,respectively.Prove that $ EY\equal{}FZ$

Poon.

It is from Iberoamerican MO 1995 . See here for a solution http://www.mathlinks.ro/viewtopic.php?search_id=952312459&t=4954


trigonometry
logarithms
floor function
calculus
calculus computations

This isn't too difficult, but someone asked me this in my office today, so why not?

Consider $\sum_{n=1}^{\infty}\frac{(-1)^{n}\sin n}n$

Show that the sum converges conditionally.

Compute its exact value.

generally :  :wink:
$\sum_{n=1}^{\infty}\frac{(-1)^{n}\sin nx}n$
easy we only find this sum easy
$\sum\frac{1}{n}z^{n}$ where $z\in C$ I mean complex number. and $|z|<1$. after use Abelia theorem

We have, by Fourier analysis, that:

$x=2\sum_{n=1}^{\infty}(-1)^{n-1}\frac{\sin nx}{n}$, $-\pi<x<\pi$,

To prove that $\sum_{n=1}^{\infty}\frac{|\sin n|}{n}$ does not converge we proceed as follows. Assume it converges.  Then we get that

$\sum\frac{\sin^{2}n}{n}=\sum\frac{1-\cos 2n}{2n}$ converges which is false since

$\sum\frac{\cos 2n}{n}=-\ln 2\sin 1$ while the first series diverges.

conditional convergence, like always, involves 2 parts:

(i) showing $\sum_{n}\; | \;\mathbb{a}_{n}\; | \;$ diverges

(ii) showing $\sum_{n}\;\mathbb{a}_{n}\;$ converges

----------

(i) this is a rather peculiar series i think, and to show
$\sum_{n}\; | \;\mathbb{a}_{n}\; | \;$ diverges will require a slight trick...
[ [b]note:[/b] a comparison test against $\sum_{n}\; \frac{1}{n}\;$ will not work,
because we have $0\; \leq\; | \; \sin{n}\; | \;\leq \;1$]

so, we approach instead as follows:

$\sum_{n}\; | \;(-1)^{n}\;\frac{\sin{n}}{n}\; | \;=\;\sum_{n}\; \frac{| \; \sin{n}\; | }{n}\\ =\;\sum_{n}\; \frac{| \; \sin{n}\; |-| \; \sin{(n-1)}\; |}{2n}\;+\; \frac{| \; \sin{n}\; |+| \; \sin{(n-1)}\; |}{2n}\\ =\;\sum_{n}\;\quad 1^{st}\;+\;2^{nd}\;$

we can show the 1st one converges whereas the 2nd one diverges.
for the 2nd one, one can easily show that it will be bounded below by $\sin{1}$... i.e.

$\lfloor \;\frac{| \; \sin{n}\; |+| \; \sin{(n-1)}\; |}{2n}\; \rfloor\;=\;\sin{1}\; \quad ( \; >\;0 \; )$

hence, the 2nd series will diverge.

now, for the 1st one, use dirichlet's test, i.e. let $a_{n}$ be terms of a sequence
that is monotonically convergent to 0, and let $b_{n}$ be terms of a sequence
whose partial sums, i.e. $\sum_{n}^{p}\;b_{n}\;( \; where\; p\; \ll \; \infty )$, are bounded above by
some constant $M\; \ll \; \infty\; ( \; \Rightarrow\;Cauchy\; series )$ ;
it then follows $\;\sum_{n}\; a_{n}\; \cdot \; b_{n}\;$ is convergent.
[ PS: one can also use abel's test here. ]

this concludes

$\sum_{n}\; | \;(-1)^{n}\;\frac{\sin{n}}{n}\; |$ diverges.

(ii) to show $\sum_{n}\;\mathbb{a}_{n}\;$ converges, we use a similar argument as above,
but this time using leibnitz's alternating series test, where
the partial sums of $\sum_{n}^{p}\;(-1)^{n}\;\sin{n}$ are bounded and $\frac{1}{n}$ monotonically approaches 0.

as for the sum, didi already showed (using fourier series):

$\sum_{n=1}^{\infty}\; (-1)^{n}\;\frac{\sin{n}}{n}\; =\;-\frac{1}{2}$

(i used a different method, but i guess there is no need to post it anymore...)

In general we have:

$\sum_{n=1}^{\infty}(-1)^{n-1}\frac{\sin(nx)}{n}= \frac{x}{2}$.


Let $ABCDE$ be a convex pentagon and let $F = BC \cap DE, G = CD \cap EA, H = DE \cap AB, I = EA \cap BC, J = AB \cap CD$. Suppose that the areas of the triangles $AHI,BIJ,CJF,DFG,EGH$ are all equal. Then the lines $AF,BG,CH,DI,EJ$ are all concurrent.


Let $a,b,c >0$ such that their sum is $3.$ Prove that the following inequality holds:

                         \[ \frac{a^{3}+abc}{(b+c)^{2}}+\frac{b^{3}+abc}{(c+a)^{2}}+\frac{c^{3}+abc}{(a+b)^{2}}\geq\frac{3}{2}. \]

Cute.  Assume $a\geq b\geq c$.

By Schur's inequality,

$\sum \left(\frac{a}{(b+c)^2}\right)(a-b)(a-c)\geq 0$

It follows that

$LHS\geq \sum \frac{a^2(b+c)}{(b+c)^2}=\sum \frac{a^2}{b+c}\geq \frac{a+b+c}{2}=\frac{3}{2}$.

Sure.  It's a nice problem, in any event.

Thx. I'm really happy you appreciate it. :lol:

Another proof:

By Chebyshev, then Iran 1996, then Schur

\begin{eqnarray*} LHS&\geq&\frac{1}{3}\left(\sum x^3+xyz\right)\left(\sum \frac{1}{(x+y)^2}\right) \\ &\geq& \frac{3}{4(xy+yz+zx)}\left(\sum x^3+xyz\right) \\ &\geq& \frac{1}{2}\frac{(xy+yz+zx)(x+y+z)}{xy+yz+zx}\\ &=&\frac{x+y+z}{2} \\ &=&\frac{3}{2} \end{eqnarray*}

which yields the desired result.

:D I found a very simple way for this one. But I am not sure it is correct or not??
We know that $a^2+b^2+c^2 \geq (ab+bc+ca)$ so $a^2+bc \geq a(b+c)+(b-c)^2$
Then we get $a^3+abc \geq a^2(b+c)+a(b-c)^2$
So LHS $\geq \sum \frac{a^2}{b+c} +\sum a{(\frac{b-c}{b+c})^2}$
 But $\sum \frac{a^2}{b+c} \geq \frac32$ so we are done! :rotfl:

[quote="Gibbenergy"]:D I found a very simple way for this one. But I am not sure it is correct or not??
We know that $a^2+b^2+c^2 \geq (ab+bc+ca)$ so $a^2+bc \geq a(b+c)+(b-c)^2$
[/quote]
THe above should be $a^2+bc \geq a(b+c)-(b-c)^2$

Terrible sorry!
  I make a very silly mistake. But anyway, after long computation, I can prove it but It is very ugly(brutal force). And that is very interesting that I have found out
$\sum \frac{a^3(2a^2+bc)}{(b+c)^2} \geq \frac94$
   But I am not sure if it is correctly or not since I sometimes make stupid mistakes!

If  $ a,b,c$ be nonnegative numbers such that $ bc + ca + ab > 0$, then when $ k\leq3$ we have

$ \frac {a^3 + kabc}{(b + c)^2} + \frac {b^3 + kabc}{(c + a)^2} + \frac {c^3 + kabc}{(a + b)^2}\geq\frac {1 + k}{4}(a + b + c),$

with equality if $ k = 3,a = 0,b = c > 0.$

[b]Proof[/b] $ \sum{\frac {a^3 + kabc}{(b + c)^2} - \frac {1 + k}{4}\sum{a}}$

$ = \sum{\frac {(b - c)^2\left[(b + c - a)^2(5a + 3b + 3c) + bc(17a + 4b + 4c)\right]}{6(c + a)^2(a + b)^2}}$

$ + \frac {3 - k}{4}\sum{\frac {a(b - c)^2}{(b + c)^2}}\geq0.$

[b]Remark[/b] When $ - 1\leq k\leq 3$, it is stronger than another Cezar Lupu's inequality, see :

http://www.mathlinks.ro/viewtopic.php?t=61944


Let $ a,b,c$ positive reals such that $ ab\plus{}bc\plus{}ca\equal{}1$. Prove that

$ \sum\frac {\sqrt {a \plus{} b \plus{} c} \plus{} \sqrt {a \plus{} b} \plus{} \sqrt {c}}{a \plus{} b}\geq\frac {2 \plus{} \sqrt {2}}{2} \plus{} \frac {3(3 \plus{} \sqrt {3})}{2\sqrt {a \plus{} b \plus{} c}}$

[quote="Ligouras"]Let $ a,b,c$ positive reals. Prove that

$ \sum\frac {\sqrt {a \plus{} b \plus{} c} \plus{} \sqrt {a \plus{} b} \plus{} \sqrt {c}}{a \plus{} b}\geq\frac {2 \plus{} \sqrt {2}}{2} \plus{} \frac {3(3 \plus{} \sqrt {3})}{2\sqrt {a \plus{} b \plus{} c}}$[/quote]
The inequality is not homogenous ! Try $ a\equal{}b\equal{}c \geq \frac{ 9(1\plus{}\sqrt{2}\plus{}\sqrt{3})^2 }{ (2\plus{}\sqrt{2})^2 }$

[quote="rachid"][quote="Ligouras"]Let $ a,b,c$ positive reals. Prove that

$ \sum\frac {\sqrt {a \plus{} b \plus{} c} \plus{} \sqrt {a \plus{} b} \plus{} \sqrt {c}}{a \plus{} b}\geq\frac {2 \plus{} \sqrt {2}}{2} \plus{} \frac {3(3 \plus{} \sqrt {3})}{2\sqrt {a \plus{} b \plus{} c}}$[/quote]
The inequality is not homogenous ! Try $ a \equal{} b \equal{} c \geq \frac { 9(1 \plus{} \sqrt {2} \plus{} \sqrt {3})^2 }{ (2 \plus{} \sqrt {2})^2 }$[/quote]

My Friend

$ ab \plus{} bc \plus{} ca \equal{} 1$

sorry


[quote="stancioiu sorin"]Let's take the trapezoid $\ ABCD$ whith $\ AB\parallel CD$, $\angle A=90, \angle C=2\angle B$ and $\ AB=3CD$.
$\ a)$ Find the points $\ M,N\in (AD)$ such that $\angle DMC=\angle AMB$ and $\angle DNC=\angle ABN$.
$\ b)$ If we know $\ MN=3 cm$, find the area of $\ ABCD$.
$\text{Constantin Apostol}$[/quote]
May you mean $\ MN=3 CM$ :maybe:

I guess it mean "centimeter". So we can use it to find the area numericaly.

:blush: Thank you,I see.
It isn't hard problem,we just need to show that $AN=DN$ and $AM=3DM$,
after this it is trivial to find the area.

Ok, let me finish this problem.

The construction of $M$: $S(AD)$, then $C \to C'$, let $M = C'B \cap AD$, then $\angle{CMD}=\angle{BMA}$.

The construction of $N$: As Tiks put it, $N$ is the midpoint of $AD$. This is because $AN=DN= \sqrt{3}DC=\sqrt{3}a$, thus $\frac{ND}{CD}=\sqrt{3}=\frac{AB}{AN}$, hence $\triangle{CND}\sim \triangle{NBA}$, hence $\angle{CND}=\angle{NBA}$.

Because $\triangle{DMC}\sim \triangle{AMB}$, so $AM=3DM$, thus $AD=4MN=12$ which implies $a=2\sqrt{3}$. The area of the trapzoid is $48\sqrt{3}$.


Let D be the decomposition of the plane into concentric circles about the origin. Prove that D is homeomorphic to ${ \{x \in \mathbb{R}: x \geq 0}$. Show also that D is upper semicontinuous.

For part 1, send each circle to its radius in $ \mathbb{R}$. For part 2, I don't know what semi continuous means

It's true you can carry the quotient topology on to the quotient $ Y$ by $ X/Y$, but in that case $ D$ would be endowed with the quotient topology and we won't get $ X\equal{}D\times Y$ that easily since the product topology is different from the multiplied quotient topology.


geometry
perpendicular bisector
angle bisector
Law of Sines

In convex quadrilateral $ ABCD$, $ CB,DA$ are external angle bisectors of $ \angle DCA,\angle CDB$, respectively. Points $ E,F$ lie on the rays $ AC,BD$ respectively such that $ CEFD$ is cyclic quadrilateral. Point $ P$ lie in the plane of quadrilateral $ ABCD$ such that $ DA,CB$ are external angle bisectors of $ \angle PDE,\angle PCF$ respectively. $ AD$ intersects $ BC$ at $ Q.$  Prove that $ P$ lies on $ AB$ if and only if $ Q$ lies on segment $ EF$.

Hi, my name is dr. Ilham Kosasih, I am from Indonesia, if i have a questions about Math, can i ask u?? :) , would u pliz help me?? :lol: btw, my email is masterpool4ever@yahoo.com, i love math.

Hello!

a) $ Q\in EF$, $ \left\{P_{1}\right\}\equal{}CP\cap AB$ and $ \left\{P_{2}\right\}\equal{}DP\cap AB$.

$ \frac{P_{1}A}{P_{1}B}\equal{}\frac{CA\cdot sin\left(\angle ACP_{1}\right)}{CB\cdot sin\left(\angle BCP_{1}\right)}\equal{}\frac{CA}{CB}\cdot\frac{sin\left(\angle DCF\right)}{sin\left(\angle QCF\right)}\equal{}$
$ \equal{}\frac{CA}{CB}\cdot\frac{DF\cdot sin\left(\angle CDF\right)}{QF\cdot sin\left(\angle CQF\right)}\equal{}\frac{CA}{CB}\cdot\frac{DF}{QF}\cdot\frac{sin\left(\angle CEF\right)}{sin\left(\angle CQE\right)}\equal{}$
$ \equal{}\frac{CA}{CB}\cdot\frac{DF}{QF}\cdot\frac{CQ}{CE}$   (1)

$ \frac{P_{2}A}{P_{2}B}\equal{}\frac{DA\cdot sin\left(\angle ADP_{2}\right)}{DB\cdot sin\left(\angle BDP_{2}\right)}\equal{}\frac{DA}{DB}\cdot\frac{sin\left(\angle QDE\right)}{sin\left(\angle CDE\right)}\equal{}$
$ \equal{}\frac{DA}{DB}\cdot\frac{QE\cdot sin\left(\angle DQE\right)}{CE\cdot sin\left(\angle DCE\right)}\equal{}\frac{DA}{DB}\cdot\frac{QE}{CE}\cdot\frac{sin\left(\angle DQF\right)}{sin\left(\angle DFE\right)}\equal{}$
$ \equal{}\frac{DA}{DB}\cdot\frac{QE}{CE}\cdot\frac{DF}{DQ}$   (2)

(1), (2) $ \Rightarrow\frac{P_{1}A}{P_{1}B}\equal{}\frac{P_{2}A}{P_{2}B}\Leftrightarrow \frac{CA}{CB}\cdot\frac{DF}{QF}\cdot\frac{CQ}{CE}\equal{}\frac{DA}{DB}\cdot\frac{QE}{CE}\cdot\frac{DF}{DQ}\Leftrightarrow$
$ \Leftrightarrow\frac{QC}{QE}\cdot\frac{QD}{QF}\equal{}\frac{CB}{DB}\cdot\frac{DA}{CA}\Leftrightarrow$
$ \Leftrightarrow\frac{sin\left(\angle QEC\right)}{sin\left(\angle QCE\right)}\cdot\frac{sin\left(\angle QFD\right)}{sin\left(\angle QDF\right)}\equal{}\frac{sin\left(\angle CDB\right)}{sin\left(\angle DCB\right)}\cdot\frac{sin\left(\angle DCA\right)}{sin\left(\angle CDA\right)}\Leftrightarrow$
$ \Leftrightarrow\frac{sin\left(\angle QCD\right)}{sin\left(\angle QCE\right)}\cdot\frac{sin\left(\angle QDC\right)}{sin\left(\angle QDF\right)}\equal{}1$.
The last relation is true.
So $ P_{1}\equal{}P_{2}\equal{}P\Rightarrow P\in AB$.

b) $ P\in AB$, $ \left\{Q_{1}\right\}\equal{}BC\cap EF$ and $ \left\{Q_{2}\right\}\equal{}AD\cap EF$.

$ \frac{Q_{1}E}{Q_{1}F}\equal{}\frac{\frac{Q_{1}C\cdot sin\left(\angle Q_{1}CE\right)}{sin\left(\angle Q_{1}EC\right)}}{\frac{Q_{1}C\cdot sin\left(\angle Q_{1}CF\right)}{sin\left(\angle Q_{1}FC\right)}}\equal{}\frac{sin\left(\angle Q_{1}CE\right)}{sin\left(\angle Q_{1}EC\right)}\cdot\frac{sin\left(\angle Q_{1}FC\right)}{sin\left(\angle Q_{1}CF\right)}\equal{}$
$ \equal{}\frac{sin\left(\angle Q_{1}CD\right)}{sin\left(\angle CDB\right)}\cdot\frac{sin\left(\angle CDE\right)}{sin\left(\angle BCP\right)}\equal{}\frac{sin\left(\angle BCD\right)}{sin\left(\angle CDB\right)}\cdot\frac{sin\left(\angle BDP\right)}{sin\left(\angle BCP\right)}\equal{}$
$ \equal{}\frac{BD}{BC}\cdot\frac{sin\left(\angle BDP\right)}{sin\left(\angle BCP\right)}\equal{}\frac{BP\cdot sin\left(\angle BPD\right)}{BP\cdot sin\left(\angle BPC\right)}\equal{}\frac{sin\left(\angle BPD\right)}{sin\left(\angle BPC\right)}$   (3)

$ \frac{Q_{2}E}{Q_{2}F}\equal{}\frac{\frac{Q_{2}D\cdot sin\left(\angle Q_{2}DE\right)}{sin\left(\angle Q_{2}ED\right)}}{\frac{Q_{2}D\cdot sin\left(\angle Q_{2}DF\right)}{sin\left(\angle Q_{2}FD\right)}}\equal{}\frac{sin\left(\angle Q_{2}DE\right)}{sin\left(\angle Q_{2}ED\right)}\cdot\frac{sin\left(\angle Q_{2}FD\right)}{sin\left(\angle Q_{2}DF\right)}\equal{}$
$ \equal{}\frac{sin\left(\angle ADP\right)}{sin\left(\angle DCF\right)}\cdot\frac{sin\left(\angle DCA\right)}{sin\left(\angle Q_{2}DC\right)}\equal{}\frac{sin\left(\angle ADP\right)}{sin\left(\angle ACP\right)}\cdot\frac{sin\left(\angle DCA\right)}{sin\left(\angle CDA\right)}\equal{}$
$ \equal{}\frac{sin\left(\angle ADP\right)}{sin\left(\angle ACP\right)}\cdot\frac{AD}{AC}\equal{}\frac{AP\cdot sin\left(\angle APD\right)}{AP\cdot sin\left(\angle APC\right)}\equal{}\frac{sin\left(\angle APD\right)}{sin\left(\angle APC\right)}$   (4)

(3), (4) $ \Rightarrow\frac{Q_{1}E}{Q_{1}F}\equal{}\frac{Q_{2}E}{Q_{2}F}\Leftrightarrow\frac{sin\left(\angle BPD\right)}{sin\left(\angle BPC\right)}\equal{}\frac{sin\left(\angle APD\right)}{sin\left(\angle APC\right)}$.
The last relation is true.
So $ Q_{1}\equal{}Q_{2}\equal{}Q\Rightarrow Q\in EF$.

Best regards, Petrisor Neagoe :)

[quote="Petry"]b) $ P\in AB$, $ \left\{Q_{1}\right\}=BC\cap EF$ and $ \left\{Q_{2}\right\}=AD\cap EF$.[/quote]Nice proof! Just wanted to point out that once you've proven (a), you can just prove that $P\in AB$ must be unique (for instance, by noting $\angle{PDC}$ and $\angle{PCD}$ increase with $EF$), so no need to go through all the law of sines again for (b). :P

---

Here's another solution. Let $T=DB\cap CA$ so that $Q$ is the $T$-excenter of $\triangle{TDC}$. The convexity of $ABCD$ yields $T>C,D$.

We are given that $P$ exists, so because $DA,CB$ are external bisectors of $\angle{PDE},\angle{PCF}$, $Q,T$ must lie on opposite sides of $DE$ and $CF$.

Let $\theta = \angle{TDE}$. Since $DCEF$ is cyclic, $\angle{PDC}=\angle{TDE}=\theta=\angle{TCF}=\angle{PCD}$. Hence $P$ varies (with $\theta<90^\circ$) along the perpendicular bisector $\ell$ of $DC$, so $P\in AB$ for exactly one value of $\theta$ (say $\alpha$). Similarly, we always have $\angle{TFE}=C$ and $\angle{TEF}=D$, so $Q\in EF$ for a unique value of $\theta$ (say $\beta$). It clearly suffices to show that $\alpha=\beta$.

First suppose $\theta=\alpha$, so $FE=r_T(\csc{C}+\csc{D}) = 2[TCD](\csc{C}+\csc{D})/(c+d-t)=t(d+c)/(c+d-t)$ (by the sine area formula). Then
\[ \frac{FE}{\sin\theta} = \frac{FE}{\sin{FDE}}=\frac{CD}{\sin{CFD}} = \frac{t}{\sin(T+\theta)} \implies \frac{\sin T}{\tan\alpha}+\cos{T} = \frac{c+d-t}{c+d}, \]where we use $\sin(T+\theta)=\sin{T}\cos\theta+\cos{T}\sin\theta$.

Now suppose $\theta=\beta$, and let $A',P',B'$ be the feet from $A,P,B$ to $DC$. Then
\[ \frac{t}{2}\tan\beta = DP'\tan\beta = PP' = \frac{2[PA'B']}{A'B'} = \frac{2[P'AB]}{A'B'} = \frac{[DAB]+[CAB]}{TA\cos{C}+TB\cos{D}}. \]By the angle bisector theorem, $\frac{TA}{TC}=\frac{TA}{CA-TA}=\frac{TD}{CD-TD}$, so $TA = \frac{dc}{t-c}$ and $CA = TA+d=\frac{td}{t-c}$. Similarly, $TB = \frac{cd}{t-d}$ and $DB = \frac{tc}{t-d}$, so
\begin{align*}
\tan\beta = \frac{2}{t}\frac{[DAB]+[CAB]}{TA\cos{C}+TB\cos{D}}
&= \frac{(t-c)(t-d)\sin{T}}{tcd}\frac{DB\cdot TA+CA\cdot TB}{TA\cos{C}+TB\cos{D}} \\
&= \frac{(t-c)(t-d)\sin{T}}{tcd}\frac{\frac{tcd}{(t-c)(t-d)}(c+d)}{(t-d)\cos{C}+(t-c)\cos{D}} \\
&= \frac{(c+d)\sin{T}}{t(\cos{C}+\cos{D}) - t}.
\end{align*}Thus
\begin{align*}
\frac{\sin{T}}{\tan\beta}+\frac{t}{c+d}
&= \frac{\sin{T}\cos{C}+\sin{T}\cos{D}}{\sin{C}+\sin{D}} \\
&= \frac{\sin(T+C)-\cos{T}\sin{C}+\sin(T+D)-\cos{T}\sin{D}}{\sin{C}+\sin{D}} = 1-\cos{T} = \frac{\sin{T}}{\tan\alpha}+\frac{t}{c+d},
\end{align*}so we're done (as $\theta$ must be acute in order for $P$ to exist).

[quote=P-H-David-Clarence]This can be done using harmonic range of points. And i wonder if the condition "$CEFD$ is concyclic" can be deleted?[/quote]

You are right, the condition "$CEFD$ is cyclic" can be deleted.
Here's a short proof using barycentric coordinates.

Let $AC \cap BD = X$, and $X=(1,0,0), C=(0,1,0), D=(0,0,1)$, $a=CD, b=DX, c=XD$
$Q$ is $X$- excenter of triangle $XCD$, so $Q=(-a:b:c)$
$E$ is on $XC$, so let $E=(t, 1-t, 0)$ for some real number $t$, and similary let $F=(u, 0, 1-u)$ for some real number $u$
$P' = CF \cap DE$, so $P'=(ut:u(1-t):t(1-u))$
$P$ is isogonal conjugate of $P'$, so $P=(a^2 : \frac{t}{1-t} b^2 : \frac{u}{1-u} c^2)$. ($u, t$ are not $0, 1$)
$A = DQ \cap XC$, so $A=(-a:b:0)$, and similary $B=(-a:0:c)$

$P$ is on $AB \iff \det\begin{pmatrix} -a & b & 0 \\ -a & 0 & c \\ a^2 & \frac{t}{1-t} b^2 & \frac{u}{1-u} c^2 \end{pmatrix}=0 \iff a+ b \frac{t}{1-t} + c \frac{u}{1-u} =0$
$Q$ is on $EF $
$\iff \det \begin{pmatrix} -a & b & c \\ t & 1-t & 0 \\ u & 0 & 1-u \end{pmatrix} =0 \iff -a(1-t)(1-u)-bt(1-u)-cu(1-t)=0 \iff  a+ b \frac{t}{1-t} + c \frac{u}{1-u} =0$

So, $P$ is on $AB$ iff $Q$ is on $EF$, whether $CDFE$ is concyclic or not.


Google Harvey Mudd.

Have you? Well now you are a geek for knowing what Harvey Mudd is. The all-Math school.

So, What are your thoughts? I really want to know. Not sure whether to apply there or not.

Whether or not to apply to Harvey Mudd depends on whether you'd like to attend there if you are admitted. If you wouldn't possibly accept an offer of admission there, there is no reason to apply. I hear that Harvey Mudd is a very good engineering school, and it's in a part of California with pleasant weather. I think my oldest son will only apply to research universities (that is, colleges with attached graduate schools), but he may still change his mind. My personal preference is strongly for research universities in large urban areas.

Harvey Mudd provides very generous merit aid for many students.  Some students turn down more "prestigious" institutions to become Mudders.

[quote]The Harvey S. Mudd Merit Award

All applicants for admission are considered for the Harvey S. Mudd Merit Award, including our foreign/international student applicants. Harvey S. Mudd Merit Award recipients are selected by the Office of Admission on the basis of their academic achievement and their leadership potential. Recipients of this academic award have the following credentials:

SAT Critical Reading Score: 700 or above

SAT Writing Score: 700 or above

SAT Math Score: 750 or above

SAT Math 2 Subject Exam Score: 750 or above

High school rank within the top 10%

Evidence of character and personal promise as reflected in admission application materials.[/quote]
See [url]http://www.hmc.edu/admin/finaid/grants.html#hmcmerit[/url]

Is that given to everyone who qualifies, or is it just the minimum requirements?

We have going to have a Math Jam with the Harvey Mudd Director of Admissions on Tuesday, October 24 at 7:30 Eastern / 4:30 Pacific.

(I'm about to post an announcement in this forum.)

[quote="numberdance"]Harvery Mudd is also in the same city as 4 other colleges (Scripps, Pomona, Claremont McKenna, and Pitzer) as well as two graduate institutions. Most of the colleges offer cross-registration. They share a library system and some athletic facilities. I can see how that would easily change some peoples' minds about going to Harvery Mudd.

[url]http://www.claremont.edu/[/url][/quote]

not to mention i think caltech is close by

It's closer to HMC than most colleges, but it isn't literaly the same campus - not just city - like the Claremonts. I think cross-registration at Caltech wouldn't be very feasible, whereas the 5Cs readily offer everything numberdance mentioned.

And no one has answered my question yet... To rephrase it: Is the Merit Award given to every qualifier or is it a competition among qualifiers?

The requirements listed above are the ONLY requirements for the merit award. Typically, 35-40% of the students here receive that award. Of course, you still have to do things like maintain a decent GPA, not flunk out, etc.

As a current HMC student, I may be somewhat biased in favor of Mudd, but I would like to point out that one of the biggest advantages of Mudd: the tight student body and student-faculty relationship. Since there are no graduate students here, undergrads are first in line when a prof is looking for a research assistant, and anyone who wants to do summer research is almost guaranteed at least 1 opportunities here in their 4 years. Furthermore, we have this thing where you and a group of friends can take a prof out to lunch/dinner, and Mudd will pay for most of it (something like 7 or 8 bucks per person, can't rememeber off the top of my head). The student body is also very friendly and warm, and it would not be uncouth to randomly go up to someone and strike up a conversation.

Of course, the best way to decide if Mudd (or any college, for that matter) is right for you is to go and VISIT. Stay overnight if you can, but really, one needs to go and experience a little of the college before one can make an informed decision. It always makes me sad when people get acceptances to several awesome schools and decide solely based on the prestige of one, without actually visiting the campuses.


If the length of a diagonal of a square is $ a \plus{} b$, then the area of the square is:

$ \textbf{(A)}\ (a \plus{} b)^2 \qquad\textbf{(B)}\ \frac {1}{2}(a \plus{} b)^2 \qquad\textbf{(C)}\ a^2 \plus{} b^2$
$ \textbf{(D)}\ \frac {1}{2}(a^2 \plus{} b^2) \qquad\textbf{(E)}\ \text{none of these}$

[hide]
The area is $ \frac{1}{2}d^2$ (because $ d\equal{}s\sqrt{2}$, and $ (s\sqrt{2}^2\equal{}2s^2\equal{}2A)$, so $ \boxed{B}$, $ \frac{1}{2}(a\plus{}b)^2$ is the correct answer.
[/hide]


This has popped up a couple of times now, sometimes when submitting posts, this most recent one happened when I was trying to view a thread in the Round Table by author.

The error message is as follows
[quote]
SQL requests not achieved

DEBUG MODE

SQL Error: 1054 Unknown column 'pt.post_id' in 'on clause'

SQL Request: SELECT p.*, pt.post_subject, pt.post_sub_title, pt.post_text, pt.bbcode_uid, u.username, u.user_level, u.user_allow_viewonline, u.user_session_time, u.user_session_logged, u.user_bot_ips, u.user_bot_agent, u.user_posts, u.user_from, u.user_interests, u.user_website, u.user_email, u.user_icq, u.user_aim, u.user_yim, u.user_regdate, u.user_msnm, u.user_viewemail, u.user_rank, u.user_sig, u.user_sig_bbcode_uid, u.user_avatar, u.user_avatar_type, u.user_allowavatar, u.user_allowsmile, u.user_weblogs, u.user_current_blog, u.user_albums, u.user_gender, u.user_post_rating, r.rating FROM phpbb_posts p, phpbb_posts_text pt, phpbb_users u LEFT JOIN phpbb_post_ratings r ON ( r.post_id = pt.post_id AND r.user_id = 17514) WHERE p.topic_id = 88374 AND pt.post_id = p.post_id AND u.user_id = p.poster_id ORDER BY u.username, p.post_id LIMIT 0, 20
Line : 397
File : class_posts.php[/quote]

[quote="i_like_pie"]That happens once in a while, but usually goes away (gets fixed?) after a while. When you post and that error comes up, your post is posted.[/quote]That's a different error (not sure what causes that random error once in a while, but the thing is that the server executes the command (placing the post in).

This one is a bug (can be reproduced by trying to sort after author). Bug fixed  :lol:


\[\triangle ABC\Longrightarrow 2r\le \sqrt {\frac {a}{s}\cdot \left[ a(s-a)+{(b-c)}^2\right]}\le 2(R-r).\] [u][i]Example:[/i][/u] $a=5,\ b=4,\ c=3\Longrightarrow s=6,\ r=1,\ R=\frac 52\Longrightarrow 2<\sqrt 5<3.$

Interesting result!

The left side hand is quite easy: It can be prove by squaring both sides and use formula  r = S/s. But the right side is quite complicated. I thought about the idea to show middle term is less or equal R <= 2(R-r) but I'm not sure about it's validity.

hello, after inserting $ R\equal{}\frac{abc}{4A}$, $ r\equal{}\frac{A}{s}$ , $ A\equal{}\sqrt{s(s\minus{}a)(s\minus{}b)(s\minus{}c)}$ and squaring $ \sqrt{\frac{a}{s}\left(a(s\minus{}a)\plus{}(b\minus{}c)^2\right)}\le 2(R\minus{}r)$ your inequality is equivalent with
$ \frac{(a^2b\plus{}a^2c\plus{}b^2c\plus{}bc^2\minus{}b^3\minus{}c^3\minus{}2abc)^2}{(b\plus{}c\minus{}a)(a\plus{}b\minus{}c)(a\minus{}b\plus{}c)(a\plus{}b\plus{}c)}\geq 0$ which is obviously true.
Sonnhard.


geometry
geometric transformation
rotation
trigonometry

Because of earth's rotation, a pendolum isn't always perfectly aligned to the force of gravity. If $\alpha$ is the angle between the gravity and the pendolum, $l$ is the latitude, $g$ the gravitational field, $R$ the radius of the Earth, $T$ the period of Earth's rotation:

- Determine all values of $l$ that make $\alpha=0$
- Demonstrate that $\alpha \simeq \frac{2 {\pi}^2 R}{gT^2} \cdot sin(2l)$
- Determine which value of $l$ maximizes $\alpha$

It isn't hard...

Hm, I remember doing something very similar to (2) in the [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=55078]Balloooooneee[/url] topic. Too bad I only put the answer there. Anyway, (1) and (3) follow directly from your equation - $\alpha=0^\circ$ only for equator and poles and is maximal for $l=45^\circ$.

EDIT:
But I remember that it wasn't as straightforward as it first looked. I'll try to do it again and post here.

It's more or less all [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=55078]there[/url], you only need to divide $a_\text{tangential}/g$:

$\alpha\approx\tan\alpha=a_\text{tangential}/g=\omega^2R\cos\phi\sin\phi/g$
$\alpha\approx(\frac{2\pi}{T})^2R\frac{1}{2}(2\cos\phi\sin\phi)\frac{1}{g}$
$\alpha\approx\frac{2\pi^2R}{gT^2}\sin{2\phi}$


in trigangle $ ABC$, the shortest side is $ BC \equal{} a$ (that is,$ b>a, c>a$). Prove or disprove that:

$ (c \minus{} a)^2.(b \minus{} a)^2 \plus{} bc.[b^2 \plus{} bc \plus{} c^2 \minus{} 2a(b \plus{} c) \plus{} a^2] > 0$

                                                                                Lokman GÖKÇE

This probem is a problem or a solution?

  (c - a)².(b - a)² + bc.[b² + bc + c² - 2a(b + c) + a²]

=(c - a)².(b - a)² + bc.[(b - a)² + c(b + c - 2a)] > 0


calculus
integration
LaTeX
function
real analysis
real analysis unsolved

I will be honest, i don't have any ideea how to do this.
But i can guarantee you that this will be the only exercices that i don't know how to solve at all that i will post on this forum.
So, this is only one time help.

I have an exam tommorow, i've only studied partial derivates, rotor and the gradient.
I didn't had any time for integrals.

There are some chances that i will have to solve some type  ( very close ) of this integral tommorow.

Thanks in advance.

Why don't you just type it in Latex? I can hardly read it. Do you mean $ \int\sqrt {\frac {3}{x^2 \plus{} 1}}dx$ ?
What methods have you seen on computing integrals? Do you know the substitution rule?

If it's what Blue Velvet typed, then a trigonometric (or possibly hyperbolic) substitution will help.

If it's $ \int\sqrt {\frac {3}{x^3 \plus{} 1}}\,dx,$ then I suspect that there is no elementary function (function that can be written in a finite number of steps starting with the usual formulas) that is the antiderivative of that.

You do know that the latter situation is common, right? You shouldn't be surprised by it.

Kent Merryfield - you are correct, that's the integral

Nobody ?
I need it in about 12h maximum :(

You don't read very well, do you? At the very least you ignored the message I was trying to deliver. But since you insist on an "answer," I'll give you one:

$ F(x) \equal{} \int_0^x\sqrt {\frac {3}{t^3 \plus{} 1}}\,dt,$ to which you may add any constant.


[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14865[/img]

 :(

$ M$ is the midpoint of $ BC.$ Internal bisector of the $ \angle A$ cuts the circumcircle $ (O)$ again at $ N.$ $ P, Q$ are feet of perpendiculars from $ N \in (O)$ to $ AB, AC,$ $ PQ$ is Simson line with the pole $ P$ passing through $ M.$ Assume $ b > c.$ The right $ \triangle NBP \cong \triangle NCQ$ are congruent, because $ NB \equal{} NC$ and $ \angle NBP \equal{} 180^\circ \minus{} \angle B \minus{} \frac{\angle A}{2} \equal{} \angle C \plus{} \frac{\angle A}{2} \equal{} \angle NCQ.$ Therefore, $ BP \equal{} CQ$ and it follows $ AP \equal{} AQ \equal{} \frac{b \plus{} c}{2}.$  Midpoint $ M$ of $ a \equal{} BC$ is on $ PQ.$

Construct $ P, Q$ on the legs of the given angle $ \angle A,$ such that $ AP \equal{} AQ \equal{} \frac {b \plus{} c}{2}.$ Draw circle $ (A)$ with center $ A$ and given radius $ m_a.$ $ M$ is also on $ (A).$ The circle cuts the segment $ PQ$ at $ M, M'$ (symmetrical WRT the internal bisector of the $ \angle A$). Pick one, say $ M.$ Double the segment $ AM$ into $ AN \equal{} 2\ AM.$ Parallel to $ AQ$ through $ N$ cuts the ray $ AP$ at $ B$ and parallel to $ AP$ through $ N$ cuts the ray $ AQ$ at $ C.$

[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14873[/img]

 :thumbup:


[quote="Allnames"]$ Let,x,yz\ge 0$,$ xy \plus{} yz \plus{} zx \plus{} xyz \equal{} 2$
Find the best $ k > 0$ to this ineq is true for all $ x,y,z$

$ x \plus{} y \plus{} z\le \frac {1}{4xyz} \plus{} (12 \minus{} 8k)xyz \plus{} k \minus{} 2$
P\s>I think it is a nice ineq,with that condition,we have many other ineq :)[/quote]
It 'll be kill by CYH Put :$ a\plus{}b\plus{}c\equal{}p;ab\plus{}bc\plus{}ca\equal{} \frac{p^2\minus{}q^2}{3} ;abc\equal{}r.$

Or put $ a\equal{}cosA;b\equal{}cosB;c\equal{}cosC$

No,no.my friend,i also think it can be killed by CYH ,but follow me,it is quite hard,can you detail it :wink:
And the second way,i dont think your put is true,check care fully.my dear

Sorry,my problem has some mistakes,and i edited
By the way,i have just found my ineq is quite stong,and usefull :lol:

[quote="onlylove_math"][quote="Allnames"]$ Let,x,yz\ge 0$,$ xy \plus{} yz \plus{} zx \plus{} xyz \equal{} 2$
Find the best $ k > 0$ to this ineq is true for all $ x,y,z$

$ x \plus{} y \plus{} z\le \frac {1}{4xyz} \plus{} (12 \minus{} 8k)xyz \plus{} k \minus{} 2$
P\s>I think it is a nice ineq,with that condition,we have many other ineq :)[/quote]
It 'll be kill by CYH Put :$ a \plus{} b \plus{} c \equal{} p;ab \plus{} bc \plus{} ca \equal{} \frac {p^2 \minus{} q^2}{3} ;abc \equal{} r.$

Or put $ a \equal{} cosA;b \equal{} cosB;c \equal{} cosC$[/quote]
I think denta is much nicer

Not comment my friend,tell me about your proof,at least,what is your result :wink:

[quote="Allnames"]$ Let,x,yz\ge 0$,$ xy \plus{} yz \plus{} zx \plus{} 2xyz \equal{} 1$
Find the best $ k > 0$ to this ineq is true for all $ x,y,z$

$ x \plus{} y \plus{} z\le \frac {1}{4xyz} \plus{} (12 \minus{} 8k)xyz \plus{} k \minus{} 2$
P\s>I think it is a nice ineq,with that condition,we have many other ineq :)[/quote]
with can_hang 's help,i have found some mistake in my problem  :blush:
but how abuot $ k\equal{}3\sqrt 3 \minus{}4$ :lol:


summer program
Mathcamp
PROMYS
HCSSiM
Ross Mathematics Program
Mafia
number theory

Hi,

I'm a senior in high school this year and am interested in doing a math program next summer (before my freshman year of college); I particularly wanted to find one though where I wouldn't be the only person that old and everyone else would be in lower grades, but instead a place where it would be totally common to have people about to start college.  Does anyone have any suggestions for what this might be?  I was most interested in CanadaUSA Mathcamp and PROMYS, but have also heard good things about HCSSiM and Texas Honors Math Camp.  Please let me know if you have any advice!

Thanks, M.

I don't think it's particularly common, at any of the programs you mentioned, for students to be about to start college.  (Someone please correct me if I'm wrong!)  Both Mathcamp and PROMYS explicitly allow graduating seniors to apply; HCSSiM almost never invites graduating seniors to participate.

I spent two summers at the Ross Math Program.  I'm surprised it wasn't on your list.  Last summer there were people in college who still came as first-years.  In general, first-years are more likely to be in high school and not about to go to college.  Having said that, there were at least a handful of people about to go to college among the first year and second year students alone.  Beyond that, there are the junior counselors and counselors who stay in the same dorms as you.  Since most of the days are spent in the dorms (you have to be pretty committed to doing a lot of math), you can choose to work/ hang out with any age group you want.  I mean all counselors are required to be at least old enough to be going to college the next year and they are very very open to just chilling or helping.  As a Rosser, I would also have to say that it seems like it is the most hardcore out of all these camps.  It is 8-weeks long and it is also where I decided to be a math major in college.  You don't have to worry about the quality of the program at all.  The only thing that doesn't work as well for some people is the expectation that people work hard.  It's up to you what you want out of your summer program.  I highly recommend it!

PROMYS is really nice, though I went as a high school freshman... there were several graduated seniors, though, and really the age difference isn't that important, as the counselors are all college students and the junior counselors are graduated seniors as well. As long as you like math, you'll find yourself in good company.

PROMYS is fantastic. We had quite a few people who were transitioning between high school and college last year. Going to a math camp is great because you get to enter college with a peer group that you already know.

But like paladin8 says, the age difference really isn't that important. Most of the people there are at different stages of their life and social development anyway.

Hi Mathilde,

Last year there were a dozen students at Mathcamp who were about to start college.  Six of those were new Mathcamp students; the other six were students who had been to Mathcamp in previous years and came back.  For applicants in grade 10 and above, we don't really look closely at exactly what grade they're in, so as far as admissions goes it doesn't help or hurt you.

I have to agree with Peter about the Ross Program.  There is just the right amount of guidance from counselors and individual creativity involved to ensure a complete understanding of what you learn.  Eight weeks does seem like long time, but, with math and frisbee, time flies.

[quote="beta"][quote="mel"][quote="beta"]PROMYS is very nice, I went there this summer and I learned a lot of number theory there.

Hehe you get to play mafia to like 4:00AM on weekends :D[/quote]

hehe....you didn't stay around for the 7:00am game, Feiqi?[/quote]

5:00AM is good enough, no? :D[/quote]

Well, I guess by then, it's hard to tell the difference between 5am and 7am anyway.   :P


trigonometry
function
limit
calculus
derivative
trig identities
calculus computations

[quote="red218"]I been having a problem with trig functions. One of which is: lim x ->0 (sin2x)/x. And also, what would I do if I had the same trig function like
lim x->0 tan5x/tan2x? And for distribution, I had this problem: lim x->0 square root(x+1) -1/ x, I tried it in all different ways but somehow I cannot compute the answer which is 1/2, plz help.[/quote]

1)$\lim_{x\to 0}\frac{\sin 2x}{x}= \frac{1}{2}\lim_{x\to 0}\frac{\sin 2x}{2x}= \frac{1}{2}$

2)$\lim_{x\to 0}\frac{\tan 5x}{\tan 2x}= \lim_{x\to 0}\frac{\cos 5x}{\cos 2x}\cdot \frac{\sin 2x}{\sin 5x}= \lim_{x\to 0}\frac{\cos 5x}{\cos 2x}\cdot \lim_{x\to 0}\frac{\sin 2x}{\sin 5x}= \lim_{x\to 0}\frac{\sin 2x}{\sin 5x}$
$=\lim_{x\to 0}\frac{2x}{\sin 5x}\cdot \frac{\sin 2x}{2x}= \lim_{x\to 0}\frac{2x}{\sin 5x}\cdot \lim_{x\to 0}\frac{\sin 2x}{2x}= \lim_{x\to 0}\frac{2x}{\sin 5x}$
$=2\lim_{x\to 0}\frac{x}{\sin 5x}= \frac{2}{5}\lim_{x\to 0}\frac{5x}{\sin 5x}= \frac{2}{5}$

3)Just rationalize.

I understand now how to get 1 and 3, but in  the back of the book for 2., somehow the answer is 5/2(I apologize I did not mention that before). I even got 2/5 by doing the same steps as #1. Was I suppose to switch one of the fractions upside down?

It's $\frac52$, and the bad step is the first one. $\tan=\frac{\sin}{\cos}$.

Using the above method (correctly this time) it is not hard to show that:
$\lim_{x\to 0}\frac{\tan nx}{\tan mx}= \frac{n}{m}$ with $m\not = 0$.

[quote="me@home"]Alternatively, use l'hospital... :P[/quote]
Uck, I hate that rule

I just want to know if I did these right...

For  2. I make it (cos2x/cos5x)*(sin5x/sin2x)=(5x/sin2x)*(sin5x/5x=1)=5 lim x->0 x/sin2x= 5/2 lim x->0= 1(basically)=5/2 ? The ans. is 5/2.

For lim x->0 (sec(x)-1)/x sec(x), I split the denominators in half (x)(sec x), =(Sec(x)/Sec(x)=1)-(1/x), make the x on the bottom like, 1-1=0 ?the ans. is 0

To be honest, I really do not know what to do with this: lim x->0 (1-cos(x))/sin(x)) . The ans. is 0.

And if I did those on top wrong, I can take criticism. On a side note, The book did not explain the material well, is there any website where I can find the tools necessary to solve these types of problems( and beyond)?

Looking at those limits in your last post:
$\frac{\sec x-1}{x\sec x}=\frac{(1-\cos x)/\cos x}{x/\cos x}=\frac{1-\cos x}{x}$.
You know this limit- the fact that $\lim_{x\to 0}\frac{1-\cos x}{x}=0$ is an essential step in calculating the derivatives of trigonometric functions.

The second can also be built from pieces you know: $\frac{1-\cos x}{\sin x}=\frac{1-\cos x}{x}\cdot \frac{x}{\sin x}$.
Alternately, we use trig identities: $\frac{1-\cos x}{\sin x}=\frac{1-\cos^{2}x}{\sin x(1+\cos x)}=\frac{\sin x}{1+\cos x}$

Based on the content of these, I assume you haven't yet learned the theorem known as L'Hopital's rule. You don't need it here, so just ignore those comments.


Hi,

I don't know if I have already asked this question in this forum before but I think it's a nice problem...

Show that for every [tex]q \in \mathbb{Q}^+[/tex] exist four positiv integers [tex]a,b,c,d[/tex] such that:

[tex]q=\frac{a^3+b^3}{c^3+d^3}[/tex]

http://www.mathlinks.ro/Forum/viewtopic.php?t=6232
http://www.mathlinks.ro/Forum/viewtopic.php?t=19763

:)


Jennifer burns calories while doing activities..:

Sleeping - 10 calories/per hour

Walking - 80 cph

Basketball - 150 cph

Bicycling - 100 cph.

She spends 5 times as much time playing basketball as she does while walking, and half as much time bicycling as playing basketball. She spends 4 hours sleeping. She burns 1250 calories that day. How much time did she spend on each activity?
Round all answers to nearest hundredth.

I don't think I should show my work because I used matrices.

[hide] walking: 1.779 hours
basketball: 8.897 hours
cycling: 4.449 hours
sleeping: given[/hide]

[hide]

basketball = 5.601 hours
walking = 1.120
biking = 2.80
[/hide]

[hide]
used a one variable equation...[/hide]

xenon_nightmare152 and jimli are correct

how do you know :?:  ToBe didn't say they were right


A cube with 3-inch edges is made using 27 cubes with 1-inch edges.
Nineteen of the smaller cubes are white and eight are black. If the
eight black cubes are placed at the corners of the larger cube, what
fraction of the surface area of the larger cube is white?

The surface area of the cube counting every side is $ 54$ square inches. Every black cube has outside surface area of $ 3$, so the surface area of the larger cube that is black is $ 8*3 \equal{} 24$, and $ 34$ square inches white. So the fraction of the surface area of the larger cube that is white is $ \frac {34}{54} \equal{} \frac {17}{27}$. I think... :maybe:

Edit: Oops, I seem to need to go back to kindergarten..  :blush:

[quote="$ LaTeX$"]The surface area of the cube counting every side is $ 54$ square inches. Every black cube has outside surface area of $ 3$, so the surface area of the larger cube that is black is $ 8*3 \equal{} 24$, and [color=red][b]34 [/b][/color] square inches white. So the fraction of the surface area of the larger cube that is white is $ \frac {34}{54} \equal{} \frac {17}{27}$. I think... :maybe:[/quote]

The logic in the above is correct, except for a small mistake:

The surface area would be 54 square inches as by the formula 6s^2. The black cubes would take away 3 square inches per each of the 8. Thus 24 square inches would be black, leaving 54-24 or [b]30 square inches[/b] as white. Thus we have-

$ \dfrac{\mbox{white area}}{\mbox{total area}} \equal{} \dfrac{30}{54} \equal{} \boxed{\dfrac{5}{9}}$

But why isn't it 29/54? 1 of the white cubes is inside the 3x3x3 cube so the surface area of that is 0.

Easy Way!
One face looks like this:
$\Box \blacksquare \Box$
$\blacksquare \Box \blacksquare$
$\Box \blacksquare \Box$
All the other faces look the same.
Thus, the ration of surface area is $5/9$


function
superior algebra
superior algebra theorems

I don't know if I remember this problem correctly but here it goes.

Let $ S_n$ be the set of all permutations of $ S \equal{} {\{0, 1, 2, 3, \ldots, n\}}$. Show that if $ A$ is a permutation of $ S$, and $ B$ is a permutation of $ S$, then $ AB$ is also a permutation of $ S$.

My proof:

Since $ A$ and $ B$ are permutations of $ S$, that means that they are one-to-one and onto with the elements of $ S$. Now $ AB$ is simply a mapping of the elements of $ S$ from $ A$ to $ B$, which means that $ AB$ is still a permutation of $ S$.

Is my proof correct? Is there a more elegant proof?

The easiest way to show that a function is one-to-one and onto is to explicitly describe its inverse.  As for terminology, usually $ S_n$ denotes the number of permutations of $ \{ 1, 2, ... n \}$.

[quote="guile"]Now $ AB$ is simply a mapping of the elements of $ S$ from $ A$ to $ B$, which means that $ AB$ is still a permutation of $ S$.[/quote]  This sentence doesn't have any discernable meaning, or at least, it could be connected in your mind with the right thing to put here, but it could equally as easily be connected with things that have nothing to do with the solution.

In fact, since we're working with a finite set it is equivalent to show that the map is injective, is surjective, or has an inverse.  Proving any one of these three things is easy in this case, but you should be clear about which one you're doing and what intermediate steps you're making.


Is there a way in $ \text{\LaTeX}$ to define terms with two words inside single one curly brackets? I have typed the definitions as
The field is said to be \operatorname{algebraic} \operatorname{closed} if ...
\operatorname{algebraic closed} won't produce the space between terms.

An operator name is supposed to be a one-off [b]mathematical[/b] function defined in math mode, not a text definition. For example, (see The LaTeX Companion 8-2-11):
$ \operatorname{seg} (a,r) \equiv \{ z \in \mathbf{C} \colon \Im z < \Im a, \ |z\minus{}a| <  r\}$
Using \operatorname prevents seg being put into italics and gets the correct spacing. Here seg is only used in this formula so doesn't have to be defined elsewhere.

Is [i]algebraic closed[/i] a mathematical function? No, then you just say:
The field is said to be algebraically closed if ...
and use \emph, \textbf, \textit, \textsl etc for the algebraically closed words.


Two cows play a game where each has one playing piece, they begin by having the two pieces on opposite vertices of an octahedron, and the two cows take turns moving their piece to an adjacent vertex. The winner is the first player who moves its piece to the vertex occupied by its opponent's piece. Because cows are not the most intelligent of creatures, they move their pieces randomly. What is the probability that the first cow to move eventually wins?

Source: Duke Math Meet 2008

I'm pretty sure this problem has been posted before, but I can't seem to find where it is...

Any solution or link to this problem will be greatly appreciated. :)

[hide]Two vertices of an octahedron are either opposite each other or adjacent each other. Let $ p$ be the probability of a cow winning if it starts on a vertex opposite of the other cow, and let $ q$ be the probability of winning if it starts on a vertex adjacent to the other cow.
We have $ p\equal{}1\minus{}q$ since the first move brings the cow adjacent to the other cow, then the other cow's probability of winning becomes $ q$.
We also have $ q\equal{}\frac{1}{4}\plus{}\frac{1}{2}(1\minus{}q)\plus{}\frac{1}{4}(1\minus{}p)$ since a move has $ \frac{1}{4}$ probability of an immediate win, $ \frac{1}{2}$ probability of landing adjacent to the other cow, and $ \frac{1}{4}$ probability of landing opposite of the other cow.
Making the substitution $ 1\minus{}p\equal{}q$ in the second equation gives $ q\equal{}\frac{1}{4}\plus{}\frac{1}{2}(1\minus{}q)\plus{}\frac{q}{4} \iff q\equal{}\frac{3}{5}$, so $ p\equal{}\boxed{\frac{2}{5}}$.[/hide]


function
integration
limit
calculus
derivative
real analysis
real analysis unsolved

Hello, I know my level of math is too low compared to most of you, but I have the following problem with a GRE math subject.

If function f is strictly increasing, which of the following is necessarily wrong?

A.$\forall x f(2x)=f(x)$
B.$\int_{0}^{1}f(x)dx=\int_{1}^{2}f(x)dx$
C.$\lim_{x\to \infty}f'(x)=0$
D. $f'(1)=-f'(2)$

I think both B and D are necessarily wrong, what do you think?

D is not necessarily false. A strictly increasing function can still have a zero derivative somewhere (example: $f(x)=x^{3}$). So we could have $f'(1)=f'(2)=0.$

B is certainly false.

And, as you've written it, A is also false.

But if function f has f'(x0)=0, it means it has a maximum or minimum value at x0. It contradicts the condition that f is strictly increasing. Besides, for f(x)=x^3, f'(1)!=0 and f'(2)!=0.

[quote="fredzhang0421"]But if function f has f'(x0)=0, it means it has a maximum or minimum value at x0. It contradicts the condition that f is strictly increasing. [/quote]
That is absolutely and totally wrong. A "critical point" is not a point at which a function [i]does[/i] have a max or a min; rather it is a point at which a function [i]might[/i] have a max or a min. The example of $f(x)=x^{3}$ was intended to be a counterexample for that. $f'(0)=0$ and yet the function is strictly increasing at zero.

I didn't intend for $x^{3}$ to itself be the example; the intent was to find a function whose local behavior around 1 and again around 2 was the same as the local behavior of $x^{3}$ around zero.

For instance, consider the primitive $\int(x-1)^{2}(x-2)^{2}\,dx.$ That should do it.

Yes, yes ,absolutely! I see! Sorry for the wrong understanding of the condition f'(x)=0


MATHCOUNTS
geometry
3D geometry
factorial
calculus
ratio
articles

This is actually a nice memory technique. I give a word, and you give another word related to that, [i]and state this relation clearly[/i]. The nest person gives a word related to your word, and the game moves on. As usual don't post after yourself. [b]Proper nouns[/b] [b]are allowed[/b].

umm... might we have a first word to start the game bubka?  ;)

I am getting forgetful nowadays. Anyway let me give it an archetypal beginning:

[b]maths[/b]

(a bit banal, but works fine!)

pencils. Not just one pencil. Not a pen either. Pencils, a essential part of every maths exam.

eraser: an essential part of every pencil, particularly (in my case,) those used in every maths exam.

calculators, specifically my TI-83 Plus which has almost no available memory because that's all being used for the list of 314-ish prime numbers

ancient greeks, because they had abacuses (abacusi?  abaci?)

philosophy

because they started all that crap

my english teacher, because he seems to like philosophy

Poems, because that's all he's been talking about all week.

"Touch It", as in that new busta rhymes song

Wow, there are so many dirty words that I can think of right now...

STDs

war, guns and all that violence that makes us cry

Cold War
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the word [size=125] Frigid[/size]

And for some reason, the wii

ugly swedish models whom you somehow have a fetish for