Let $ a_1,a_2,\ldots,a_n$ be real constants and
\[ f(x)\equal{}\cos(a_1\plus{}x)\plus{}\frac{\cos(a_2\plus{}x)}{2}\plus{}\frac{\cos(a_2\plus{}x)}{2^2}\plus{}\cdots \plus{}\frac{\cos(a_n\plus{}x)}{2^{n\minus{}1}}\]
If $ x_1$ and $x_2$ are real numbers and $ f(x_1)\equal{}f(x_2)\equal{}0$, prove that $ x_1\minus{}x_2\equal{}m\pi$ for some integer $ m$.

[quote="Jumbler"]Let $ a_1,a_2,...,a_n$ be real constants and
\[ f(x) \equal{} \cos(a_1 \plus{} x) \plus{} \frac {\cos(a_2 \plus{} x)}{2} \plus{} \frac {\cos(a_2 \plus{} x)}{2^2} \plus{} ... \plus{} \frac {\cos(a_n \plus{} x)}{2^{n \minus{} 1}}\]
If $ x_1,x_2$ are real numbers and $ f(x_1) \equal{} f(x_2) \equal{} 0$, prove that $ x_1 \minus{} x_2 \equal{} m\pi$ for some integer $ m$.[/quote]

$ f(x)\equal{}\cos(x)\sum_{k\equal{}1}^n\frac{\cos(a_k)}{2^{k\minus{}1}}\minus{}\sin(x)\sum_{k\equal{}1}^n\frac{\sin(a_k)}{2^{k\minus{}1}}$

$ f(x)\equal{}0$ $ \iff$ $ \cos(x)\sum_{k\equal{}1}^n\frac{\cos(a_k)}{2^{k\minus{}1}}\equal{}\sin(x)\sum_{k\equal{}1}^n\frac{\sin(a_k)}{2^{k\minus{}1}}$

$ \iff$ $ u\cos(x)\equal{}v\sin(x)$ with $ u\equal{}\sum_{k\equal{}1}^n\frac{\cos(a_k)}{2^{k\minus{}1}}$ and $ v\equal{}\sum_{k\equal{}1}^n\frac{\sin(a_k)}{2^{k\minus{}1}}$

If $ u\equal{}v\equal{}0$, we have $ f(x)\equal{}0$ $ \forall x$ and then
If $ n\equal{}1$, this would mean $ \cos(a_1\plus{}x)\equal{}0$ $ \forall x$, which is impossible.
If $ n>1$, using $ x\equal{}\minus{}a_1$, we get : $ f(\minus{}a_1)\equal{}0\equal{}1\plus{}\sum_{k\equal{}2}^n\frac{\cos(a_k\minus{}a_1)}{2^{k\minus{}1}}$ and so $ \sum_{k\equal{}2}^n\frac{\cos(a_k\minus{}a_1)}{2^{k\minus{}1}}\equal{}\minus{}1$

But this is still impossible since $ |\sum_{k\equal{}2}^n\frac{\cos(a_k\minus{}a_1)}{2^{k\minus{}1}}|\le$ $ \sum_{k\equal{}2}^n\frac{1}{2^{k\minus{}1}}$ $ \equal{}1\minus{}\frac 1{2^{n\minus{}1}}<1$

So we cant have $ u\equal{}v\equal{}0$ and :

either $ u\ne 0$, and $ f(x)\equal{}0$ $ \iff$ $ \cot(x)\equal{}\frac vu$ and so $ x\equal{}\text{arccot}(\frac vu)\plus{}n\pi$. Q.E.D.

either $ v\ne 0$, and $ f(x)\equal{}0$ $ \iff$ $ \tan(x)\equal{}\frac uv$ and so $ x\equal{}\arctan(\frac uv)\plus{}n\pi$. Q.E.D.


symmetry
inequalities proposed
inequalities

Let $ a,b,c$ be different positive reals. Prove that:
$ \frac{a^2(b\plus{}c)\plus{}b^2(c\plus{}a)\plus{}c^2(a\plus{}b)\minus{}6abc}{a^2\plus{}b^2\plus{}c^2\minus{}ab\minus{}bc\minus{}ca}\geq\frac{16abc}{(a\plus{}b\plus{}c)^2}$

[quote="mathfunny"]Let $ a,b,c$ be different positive reals. Prove that:
$ \frac {a^2(b \plus{} c) \plus{} b^2(c \plus{} a) \plus{} c^2(a \plus{} b) \minus{} 6abc}{a^2 \plus{} b^2 \plus{} c^2 \minus{} ab \minus{} bc \minus{} ca}\geq\frac {16abc}{(a \plus{} b \plus{} c)^2}$[/quote]
It's equivalent to $ \sum_{cyc}c(a\minus{}b)^2(a\plus{}b\minus{}3c)^2\geq0.$

I know that it's equivalent to $ \sum_{cyc}c(a\minus{}b)^2((a\plus{}b\plus{}c)^2\minus{}8ab)\geq0$.
How do you it $ \sum_{cyc}c(a\minus{}b)^2(a\plus{}b\minus{}3c)^2\geq0$


AMC AIME USA(J)MO USAMO geometry inequalities blogs

I'm in a kind of odd position, where the AIME is easy for me, but the USAMO and IMO Olympiad contests are too hard. I still can't do most 1s and 4s on the USAMO, yet I can get double digits on the AIME. I've been told to do lots of Engel and Acops, but Acops was very boring and hard, with skimpy material coverage, and Engel covers a lot of things I don't think are relevant. I think I have a good grasp of subject material, but I can't apply it creatively enough.

Suggestions? I'm hoping for experienced people like t0r and kunny to give some kind of long post that revolutionizes the way I study math, but that's a pipe dream :(

[quote="PenguinIntegral"]Suggestions? I'm hoping for experienced people like t0r and kunny to give some kind of long post that revolutionizes the way I study math, but that's a pipe dream :([/quote]

QC, you're a celebrity!

Adam would be even more of a celebrity than he is, and certainly much more so than me, if he would just post more.   :rotfl:

[quote="shobber"]I think that you should as people such as MysticTerminator. Because he is in the USA and he knows how to learn math in the USA.[/quote]
Interesting enough, I did exactly this. Before I posted this, I AIMed him and tried to talk for him for a bit. It turns out his entire vocabulary consists of, "Go do porblems", "Engels porblems are good", "wtf?", and "Go away".
[quote="t0rajir0u"][quote="PenguinIntegral"]Engel covers a lot of things I don't think are relevant. I think I have a good grasp of subject material, but I can't apply it creatively enough.[/quote]

To the first point, I think Engel covers some interesting and really general problem-solving strategies (extremal principle, convexity, etc.) and worth paying attention to.  The key to getting a lot out of the material is, like AoPS (only more so) to really be engaged by the exercises and understand how simple principles can solve complex problems.  I think Engel incorporates enough Olympiad problems into his exercises that the relevancy of his methods are clear...?

As to the second, I think looking at creative solutions is a good way to find your own creativity :P  Who knows?

[quote="PenguinIntegral"]I'm hoping for experienced people like t0r and kunny to give some kind of long post that revolutionizes the way I study math, but that's a pipe dream :([/quote]

 :D  Very flattering, thank you!

I'm afraid I don't really have much advice to give.  The way to learn math is similar to the way to learn most other skills (and Scientific American can back me up on this :) ):  Challenge yourself with what is just beyond your level of skill.  If you do this, you'll keep improving (not too far beyond, or you'll just be lost and won't learn!).  At least, that's what I hear.  Recently I've been too lazy to challenge myself much  :P[/quote]

OK, thanks! When I refered to irrelevant portions of Engel, I was thinking of the coloring chaper and other things like that, but I guess I haven't taken a good look at the table of contents lately.

I think I need to turn down the challenge though. I got a big book of IMOs (like 86 through 99) and so far I've been able to solve...one? I guess, this was partly what inspired the Engel comment, as many of the problems in their are just extremely weird. They don't fit into many of the engel catagories and instead work on the principle that somehow you will wade through the notation and discover a simplifying observation. TO reduce the damage to my ego, I'm now working under the assumption that doing enough Engel will work on my skills enough that making those observations will become easier.

[quote="tim1234133"]Try doing some slightly easier olympiad questions than USAMO (I hear it's one of the harder national olympiads) like questions from the Australian Olympiad, Flanders Olympiad (Belgium), Canadian...
Make sure what you are doing is not too easy, because it gets rather pointless, and not too hard similarly (like t0r said)

What I do, which I find helps me (I think I'm at a fairly similar level of ability to you) is to print of lots of (for example, the last 5 years of Indian olympiad questions) contest questions, and, over a longish time (days at least) try and do [b]all[/b] of the questions. I keep all my solutions organised in folders, with the questions, and feel a kind of personal pride in having completed a whole set of problems.
Completeness is important because otherwise you can end up doing (what I used to do a lot) only the questions you like, and only get better at the type of questions you like doing best (which you probably find easier anyway). If you do all the questions from a contest (or all the questions 1 or 4, not necessarily the hard ones) then you get good practise in all areas.[/quote]
Ok, I'll try that. Indian, Flanders, and Austrailian you say?

I think I really need to work on my geometry and inequality skills also. I guess CPIG+Geometry Unbound+Geometry Revisited+ another hasitly made marathon should fix that, and I need to go see if t0j responded to my comment about a second inequality lesson thread...

[quote="PenguinIntegral"]
Interesting enough, I did exactly this. Before I posted this, I AIMed him and tried to talk for him for a bit. It turns out his entire vocabulary consists of, "Go do porblems", "Engels porblems are good", "wtf?", and "Go away".[/quote]

He's right, there isn't any miraculous way of learning other than doing problems until you can't stand looking at any more problems, and then doing some more

there's no secret, and he's probably annoyed enough already with people asking him how to be as good as he is that he wants to stop being bothered

[quote="PenguinIntegral"][quote="Elemennop"]
there's no secret, and he's probably annoyed enough already with people asking him how to be as good as he is that he wants to stop being bothered[/quote]
It wasn't exactly that. He was just crazy.[/quote]

I suppose then there's no point asking him for advice whether or not to do woot...

right. The obvious way to do math problems is to go crazy.

and I didn't tell you to go away. If you tell me that you're feeling lazy, don't bother talking to me.

[quote="MysticTerminator"]If you tell me that you're feeling lazy, don't bother talking to me.[/quote]
You could have told me that before going "wtf?" and logging out. :(

I was being sarcastic, after your suggestion to go do an IMO problem. Especially when you just did it in front of me, and I already knew how to solve it.:roll:

Anyway, sorry for any percieved rudeness. I ment the crazy comment in good humor.

I agree with the doing easier olympiads thing... for instance, on my blog, I have lots of problems from easier MO's that you can check out and probably won't be too difficult for you. Just scroll back through dates and stuff; even the first two Putnam problems on each day are a lot easier than USAMO problems.

[quote="Phelpedo"]BTW, coloring is not irrelevant to solving Olympiad problems.[/quote]
I know. It's just a lot of coloring problems aren't very interesting, and the space in the book could have been put to better use. Like more systems of equations. Or a wider variety of function problems.

[quote="Xevarion"]I've been learning inequalities, and I'll tell you what I did in case it will help you:

About 2 months or 10 weeks ago, I started to read the inequalities forum. I read every post in the forum and tried to understand the solutions. When I saw an interesting theorem that I hadn't heard of before, I wrote it down. Before I started this, all I knew were AM-GM and the analysis inequalities (but I only knew those in the context of abstract vector spaces so I had no experience applying them to olympiad kind of inequality problems).

About a month later, I started trying to make and propose problems. I posted a few, most of which weren't particularly good. Usually I couldn't solve any of the problems, but sometimes I could actually do something rather than just staring at it. I looked around the Internet and found a ton of inequality documents, such as Hojoo Lee's notes, Kiran Kedlaya's notes, Mildorf's notes, some Mathematical Reflections articles, a couple of chapters from Vasc's book, etc. You can find links ot most of these on this very forum.

Now, I can actually solve about 1/2 to 1/3 of Hojoo Lee's problems (though not quickly), and every once in a while I can do a problem that's posted on the forum (or find a counterexample). I'm hoping that if I stick with it and read all the stuff I found and keep doing problems, I'll become an active and useful contributor in that forum. (I'm inspired by [b]arqady[/b]!) It's critical to be patient and refuse to give up. So what if you would never be able to think of the kind of solutions that get posted in the olympiad forums? After you read A LOT of them, you start to see what methods people use and how to look at the problems. Don't expect to be one of "those guys" whose solutions you boggle at every time you log on, at least not soon. They didn't get that way magically after reading a chapter from a book or studying for two weeks. It takes a long time and a lot of work.[/quote]

Hmm, good advice. I like your mixing variables post, by the way.

I'm sorry, I can't really think of a good reply to that other than "wow, good post."  :wink:
EDIT: Ok, maybe I can. Where are his notes, when I google him his homepage link is broken.

[quote="PenguinIntegral"]If I could just get the names of about 3 more eas(ier) olympiads with lots of problems on ML/AoPS or kalva, that would be stellar. Thanks.[/quote]
I'm revising my criteria from "easy olympiads" to "Olympiads that aren't hard". I don't want to end up doing a copy of the Romainian TST that some country stole as thier national MO :wink:

[quote="tim1234133"]There are quite a few British Olympiads [url=http://www.bmoc.maths.org/home/bmo.shtml]here[/url]

There are some Irish olympiads on Kalva, which I have never done but apparently they're not so hard.[/quote]

Yeah, they're roughly equivalent to questions in the Pre-Olympiad forum here (although some problems are much easier). 1988-2005 can be found here: [url]http://maths.ucd.ie/~dukes/imo.html[/url] (and Kalva has the solutions to most of these between 1988 and 2003).


Provided a rectangle which is composed of three congruent squares. Prove without using trigonometry formulae that that the sum of the angles $\alpha$ and $\beta$ equals 45.

Consider in the graph we have AB=AC and[tex] \angle BAC=90^\blobcircle[/tex].


[quote="ch1n353ch3s54a1l"][quote="Silverfalcon"]How many of the numbers, 100,101,..,999, have three different digits in increasing order or in decreasing order?

$\text{(A)} \ 120 \qquad \text{(B)} \ 168 \qquad \text{(C)} \ 204 \qquad \text{(D)} \ 216 \qquad \text{(E)} \ 240$[/quote]
[hide]
(B) 168
increasing digits: 28+21+15+10+6+3+1 since there's 123,124, etc. until 129 which is 7 and you do that for each tens so you get $\frac{7\times8}{2}$ which is 28, and you do that until you get to 789. Then you multiply by two because it's the same thing decreasing and you get 168[/hide][/quote]

i dont think so ... since for the decreasing , you can include 0 while for increasing you cannot include 0 ...

[hide="my solution"]
For increasing order , the center digit can range from 2 to 8 . When center = 2 , then first digit can only be 1 while third digit can have 3 to 9 (7 possiblity) . So do case work to get

$2(1\cdot 7+2\cdot 6+3\cdot 5)+4\cdot 4=84$

For decreasing , the center can be 1 to 8 . And some case work shows that it is

$2(1\cdot 8+2\cdot 7+3\cdot 6+4\cdot 5)=120$

Hence combine both cases to get $\boxed{204}$ which is $C$
[/hide]

[hide]For increasing digits, there are $1\cdot7+2\cdot6+3\cdot5+4\cdot4+5\cdot3+6\cdot2+7\cdot1=84$ numbers. There are $1\cdot8+2\cdot7+3\cdot6+4\cdot5+5\cdot4+6\cdot3+7\cdot2+8\cdot1=120$ numbers. So there are a total of $\boxed{204}$ numbers.[/hide]


linear algebra
matrix
number theory proposed
number theory

Define the Lucas sequence as usual: $ L_0 \equal{} 2,L_1 \equal{} 1,\ L_{n \plus{} 1} \equal{} L_n \plus{} L_{n \minus{} 1},\ \forall n\ge 1$. Prove that for any prime $ p > 3$ we have $ L_p\ |\ L_{\frac {p \plus{} 1}{2}}^p \minus{} 1$.

Are ther anyone hwo knows any solution for this problem? :?
$Grobber$ can you post your solution?

Sorry, but first of all I don't like to write down solutions to the problems I post (the problem is here so that people can enjoy it, not so that they can check my proof), and second, I posted this a very long time ago; it would take some time to reconstruct my proof, and I don't want to go through that :).

P.S.

Here's what happened when I tried to type the subscript of $ L_{(p \plus{} 1)/2}$ properly: $ L_{\frac {p \plus{} 1}2}$. How can I make that $ \frac {p \plus{} 1}2$ small enough for the subscript to look reasonable? :) Thank you.

Edit:

Thank you very much for the tip, Farenhajt :).

Ok,i will post my proof to this problem. :)  :) .If i have some mistakes,Please correct.
First, i denote $\alpha=\frac{1+\sqrt{5}}{2},\beta=\frac{1-\sqrt{5}}{2}$
We have the well-known:$L_{n}=(\alpha)^n+(\beta)^n$ for all $n \in N$
obviously we have $L_{m}L_{n}=L_{m+n}+(-1)^mL_{n-m} (n \geq m)$.So we have
                          $L_{n+2p} \equiv L_{n} (mod L_{p})$ (1) ($p$ is prime number $>3$)
Put $k=\frac{p+1}{2}$.In this time:
                           $L_{k}^p=(\alpha^k+\beta^k)^p=\sum_{m=0}^p \binom {p}{m} \alpha^{(p-m)k}\beta^{mk} =\sum_{m=0}^{k-1}(-1)^{mk} \binom {p}{m}\alpha^{(p-2m)k}+\sum_{m=k}^p (-1)^{(p-m)k} \binom {p}{m}\beta^{(2m-p)k}=\sum_{m=0}^{k-1}(-1)^{mk} \binom {p}{m} L_{(p-2m)k}=\sum_{m=0}^{k-1}(-1)^mk \binom {p}{m}L{(k-m)p-m}$
Use (1) we can prove some affirmatives:
        +when $m,k$ have opposite parity we have $(-1)^mkL_{(k-m)p-m} \equiv L_{p-m} (mod L_{p})$
         +when $m,k$ have same parity we have $(-1)^mkL_{(k-m)-m} \equiv L_{m} (mod L_{p})$
So we have:
         $2L_{k}^{p} \equiv 2 \sum_{\begin{matrix}0 \leq m \leq p \cr m \equiv k (mod 2) \cr \end{matrix}} {p \choose m}L_{m} \equiv 2 \sum_{\begin{matrix} 0\leq m \leq p \cr m \equiv k (mod 2) \cr \end{matrix}} {p \choose m}{\alpha^m+\beta^m} \equiv ((\alpha+1)^p+(-1)^k(\alpha-1)^p)+((\beta+1)^p+(-1)^k(\beta-1)^p) \equiv L_{2p}+(-1)^{m+k}L_{p} \equiv 2 (mod L_{p})$ (because $L_{2p}=L_{p}^2+2 \equiv 2 (mod L_{p})$)
So we finished the proof. :lol:  :lol:

Grobber, type the expression correctly (no ommissions of braces) - L_{\dfrac{p+1}{2}} works just fine

[quote="nhat"]
Use (1) we can prove some affirmatives:
        +when $ m,k$ have opposite parity we have $ ( \minus{} 1)^mkL_{(k \minus{} m)p \minus{} m} \equiv L_{p \minus{} m} (mod L_{p})$
         +when $ m,k$ have same parity we have $ ( \minus{} 1)^mkL_{(k \minus{} m) \minus{} m} \equiv L_{m} (mod L_{p})$
[/quote]

Please, anybody show why this is.

I just don't get it  :(

$ 2L_{k}^{p} \equiv 2 \sum_{\begin{matrix}0 \leq m \leq p \cr m \equiv k (mod 2) \cr \end{matrix}} {p \choose m}L_{m} \equiv 2 \sum_{\begin{matrix} 0\leq m \leq p \cr m \equiv k (mod 2) \cr \end{matrix}} {p \choose m}{\alpha^m \plus{} \beta^m} \equiv ((\alpha \plus{} 1)^p \plus{} ( \minus{} 1)^k(\alpha \minus{} 1)^p) \plus{} ((\beta \plus{} 1)^p \plus{} ( \minus{} 1)^k(\beta \minus{} 1)^p) \equiv L_{2p} \plus{} ( \minus{} 1)^{m \plus{} k}L_{p} \equiv 2 (mod L_{p})$

If  $ \L_{p}$ is even, It is not sure that $ \L_{p}$ divides $ \L_{k}^p \minus{} 1$
And, $ p$ is not necessarily prime. In nhat's proof he doesn't use the fact that p is prime.
Since $ p \equiv 1 or 2 (mod 3 )$ so $ \L_{p}$ is odd. Since $ p$ is odd so $ ( \minus{} 1)^p \equal{} \minus{}1$ .
See Moldova TST 2006 when p=59


logarithms
calculus
calculus computations

$ u_t \equal{} 2u_{xx} \quad$

$ T'(t) \equal{} \minus{}2 \lambda T(t) \quad$

and

$ X''(x) \equal{} \minus{} \lambda X(x). \quad$

Does it make any difference if i choose constant to be $ \lambda$ and not $ \minus{}\lambda$ as above

Thanks

Certainly not.  $ \lambda$ is some as yet unknown constant to begin with, one which you need to find permissible values for.  You can make the right hand side constants $ \log \left(\frac {\lambda ^3}{37\pi e \plus{} \lambda ^2}\right)^3$ for all it matters, it'll only affect what the permissible values for $ \lambda$s look like, not the solution.

While everything JoeBlow said is completely correct, I would probably use $ \minus{}\lambda$ just so the number $ \lambda$ can be positive. (Of course, that requires that I already at least suspect something about the solutions.)

Where you have a choice, choose notation that makes your life easier.

Keep in mind that you don't want heat to increase exponentially in magnitude. The transients should die out over time, not become more pronounced.


induction
function
superior algebra
superior algebra theorems

Show that every convex closed bounded set in $R^n$, either has at least n+1 extremal points, or is contained in an affine subspace of $R^n$, with dimension n-1.

Affine subspace = hyperplane $a_1x_1+...+a_nx_n=b$?

Please post in the right place. The Advanced Section is only for High-School topics (IMO curriculum that is) ...

[quote="Myth"]Then it is standart theorem: each point of closed bounded set is a convex combination of extremal points.[/quote]
It can be easily proved by induction on dimension, using theorem about separating hyperplane.

I remembered this problem today, for some reason, and I just wanted to complete it.

The volume function defined on the set of $n+1$-uples of points of $\mathcal F$ (our set) to $\mathbb R$ reaches its maximum in some $n+1$-uple. If we assume that $\mathcal F$ is not included in some $n-1$-affine subspace, then this maximum is positive.

Let $x_1,\ldots x_{n+1}$ be $n+1$ points for which the maximum is achieved. If $x_1$ is not extremal, then all the segments on which it lies (by this I mean the segments in the interior of which it lies) belong to the unique $n-1$-affine hyperplane $\mathcal H$ through $x_1$ and $\|[x_2,\ldots,x_n]$ (this is true for all the points of $\mathcal H\cap\mathcal F$). We can thus move $x_1$ until we reach an extremal point of $\mathcal H\cap\mathcal F$. This will also be an extremal point of $\mathcal F$. After that, we fix $x_1,x_3,\ldots,x_{n+1}$, and repeat the procedure for $x_2$, and so on. In the end, we have $n+1$ distinct (because the volume of the simplex they form is positive) extremal points of $\mathcal F$.


We simply look at the column for the $ '80$'s. Since France and Spain are in Europe, we just add $ 15$ and $ 9$ together and get $ \boxed{24}$.


trigonometry
vector
number theory
relatively prime
trig identities

Given that $\sum_{k=1}^{35}\sin 5k=\tan \frac mn,$ where angles are measured in degrees, and $m$ and $n$ are relatively prime positive integers that satisfy $\frac mn<90,$ find $m+n.$

[quote="yassin88"][quote="1234567890"]Now we see the LHS[/quote]

What does LHS mean?  :?:[/quote]

Left hand side of the equation.

[quote="1234567890"][hide="Solution"]So we want meow to evaluate
$ S \equal{} \sin{5} \plus{} \sin{10} \plus{} \sin{15}.....\sin{170} \plus{} \sin{75}$
$ S\sin{5} \equal{} \sin{5}(\sin{5} \plus{} \sin{10} \plus{} \sin{15}.....\sin{170} \plus{} \sin{175})$
The LHS, by the identity $ \sin{a}\sin{b} \equal{} \frac {\cos{(a \minus{} b)} \minus{} \cos{(a \plus{} b)}}{2}$ telescopes to
$ 1 \plus{} \cos{5}$
So we have: $ S\sin{5} \equal{} 1 \plus{} \cos{5}$
$ S \equal{} \frac {1 \plus{} \cos{5}}{\sin{5}}$
$ S \equal{} \frac {1 \minus{} \cos{175}}{\sin{175}}$
Using the identity
$ \tan{x} \equal{} \frac {1 \minus{} \cos{2x}}{\sin{2x}}$, we identify $ 2x \equal{} 175$ and find
$ S \equal{} \tan{\frac {175}{2}}$
So $ m \plus{} n \equal{} 175 \plus{} 2 \equal{} \boxed{177}$[/hide][/quote]

What's the name of this identity?

Addition theorem, mainly. It appears a lot in trigonometry relations.

Is there an easy way to memorize those product-to-sum (and sum-to-product) trig identities? In this problem we must know that there's a *minus* sign between the two cosines in the product of the two sines so that we multiply by $ \sin5$ to make it telescope, but what if we couldn't memorize that identity (or thought there's a plus sign instead of minus sign so it won't telescope)? Does it mean we have to derive all the identities and check which one has a minus sign?  :o

If you remember everything except a sign, plug in some easy values of $ a$ and $ b$ (for example, setting $ b \equal{} 0$ and then setting $ a \equal{} b$ tells you the sign of both terms here).  Or, yes, derive them - it doesn't take much time if you're snappy with the angle addition formulas.  As for the angle addition formulas, memorize them like this:

$ (\cos a \plus{} i \sin a)(\cos b \plus{} i \sin b) \equal{} \cos (a \plus{} b) \plus{} i \sin (a \plus{} b)$.

It's also helpful to think of $ \cos a \cos b \plus{} \sin a \sin b$ as a dot product.

[quote="mathwizarddude"]BTW @t0rajir0u is there something more to it than just the sine addition identity could be written as a dot product?[/quote]
It makes more sense to me conceptually, anyway.  $ (\cos a, \sin a) \cdot (\cos b, \sin b)$ is precisely the cosine of the angle between the two unit vectors, which is clearly $ \cos (a \minus{} b)$.

[hide="Solution(actually well written for once)"]
Assume all angle measurements to be in degrees here...

Note that by product to sum, $\frac{\cos(a-b)-\cos(a+b)}{2}=\sin(a)\sin(b)$.  We multiply the given equation by $2\sin(\frac{5}{2})$, to give us
$\sum_{i=1}^{35}(\sin(5i^\circ))(2\sin(\frac{5}{2})=\sum_{i=1}^{35}(\cos(-\frac{5}{2}+5i)-\cos(\frac{5}{2}+5i))\equal{}$ $\cos(\frac{5}{2})-\cos(\frac{15}{2})+\cos(\frac{15}{2})-\cos(\frac{25}{2})+\cdots-\cos(180-\frac{5}{2})$.  This is the same as $2\cos(\frac{5}{2})$, therefore, $\sum_{i=1}^{35}(\sin(5i^\circ))(2\sin(\frac{5}{2})=2\cos(\frac{5}{2})$, so $\sum_{i=1}^{35}(\sin(5i^\circ)))=\cot(\frac{5}{2})=\tan(\frac{175}{2})$ so the answer is $\boxed{177}$.  [/hide]

[quote=nkim9005][quote=mathwizarddude]Do I really have such poor memory? 'cause I totally have NO memory of seeing this problem before...  :P  :o
Anyways here's what I did using complex bashing (haha I hope they will keep giving this type of problems since almost all of them could be brute-forced using the complex thingie)

$ \Im \sum_{k \equal{} 0}^{35}e^{5ki} \equal{} \Im \frac {e^{5\cdot36i} \minus{} 1}{e^{5i} \minus{} 1} \equal{} \Im \minus{} \frac2{\cos5 \plus{} i\sin5 \minus{} 1} \equal{} \Im\frac { \minus{} 2(\cos5 \minus{} 1 \minus{} i\sin5)}{(\cos5 \minus{} 1)^2 \plus{} \sin^25} \equal{} \frac {2\sin5}{1 \minus{} 2\cos5 \plus{} 1} \equal{} \frac {\sin5}{1 \minus{} \cos5} \equal{} \frac {2\sin(5/2)\cos(5/2)}{1 \minus{} (1 \minus{} 2\sin^2(5/2))} \equal{} \cot(5/2) \equal{} \tan(175/2)$

somedays I hope I would learn elegance as well... :lol:
(BTW @t0rajir0u is there something more to it than just the sine addition identity could be written as a dot product? :maybe:
'cause it's really sweet to have it written as $ (\cos\alpha,\sin\alpha)\cdot(\cos\beta,\sin\beta)$ :D )[/quote]

Wait a minute... Shouldn't the summation be from $k=1$ to $35$ rather than $k=0$, and wouldn't that change the second statement from $\Im \frac {e^{5\cdot36i} - 1}{e^{5i} - 1}$ to $\Im  (e^{5i} \frac {e^{5\cdot \mathbf{35i}} - 1}{e^{5i} - 1})$[/quote]

O wait... im really dumb.
Even if its supposed to say $\Im \sum_{k = 1}^{35}e^{5ki}$, we can "add" $e^{0 i}$ without actually changing the imaginary part... whoops.

P.S. I'm not trying to b*t*h too much or anything like that but Mathwizardd..., if you ever make a similar solution, you should mention that you added something to the summation so dumb people like me don't have to spend 20 minutes figuring that out.

p5: Let the sum be $k$. Note that $\sin 5p \sin 5 = \frac{\cos(5p-5)-\cos(5p+5)}{2}$, so multiplying both sides by $\sin 5$ everything cancels into $k \sin 5 = 1+\cos 5$, so $k = \frac{1+\cos 5}{\sin 5} = \tan \frac{175}{2} \implies \boxed{177}$.


calculus
derivative
linear algebra
matrix
calculus computations

$ f: R^m  \to R$, $ F: R^{m \plus{} 1}  \to R$ are differentiable on their domains. If $ x \in R^m$, define $ (x,f(x)) \in R^{m \plus{} 1}$

If $ \frac{{\partial _1 F(x,f(x))}}{{\partial _{n \plus{} 1} F(x,f(x))}} \equal{} \partial _1 f(x),\forall x \in R^m$, compute the second order partial derivative of f at a point x.

I understand that this is the quotient rule but the $ (x,f(x)) \in R^{m \plus{} 1}$ is throwing me off since it is not standard notation for vectors.

Then introduce the map $ g(x) \equal{} (x,f(x))$, with values in $ \mathbb R^{m \plus{} 1}$. Its derivative matrix is just the $ m\times m$ identity matrix with an extra row $ \nabla f$ at the bottom. The derivatives of $ F\circ g$ can be computed with the multivariable chain rule.


integration
function
real analysis
real analysis unsolved

For $0 < x < 1$ calculate the value of
\[ \sum_{n=1}^\infty \left(1 + \frac12 + \frac13 + \cdots + \frac1n\right)x^n \]

Hard problem :(

I don't understand your solution. Please explain some of the trickier steps, especially the second step. Your final answer is also not correct, but you might have made a mistake.

I have thought of a solution that is pretty elegant, but that I wouldn't have thought of if I hadn't seen the answer first:

Let $I_p = \{(j,k)\in \mathbb{N}\times\mathbb{Z^+} | j+k = p\}$. Then we have that $\left(1 + \frac12 + \frac13 + \cdots + \frac1p\right)x^p = \sum_{I_p} \frac{x^{j+k}}{k}$

Now then, since we have positive series everywhere in the interval considered we get:
\[ \sum_{p=1}^\infty \left(1 + \frac12 + \frac13 + \cdots + \frac1p\right)x^p = \sum_{p=1}^\infty \left(\sum_{I_p} \frac{x^{j+k}}{k}\right)= \sum_{j\ge0, k\ge 1} \frac{x^k}{k}x^j  \]

Using the rule for multiplying absolutely convergent (in this case positive) series we have
 \[ \sum_{j\ge0, k\ge 1} \frac{x^k}{k}x^j = \sum_{j=0}^\infty x^j \cdot \sum_{k=1}^\infty \frac{x^k}{k} = \frac{\ln(1-x)}{x-1}  \]

Ok. I will not learn how to integrate function series until the lecture on tuesday, but I will look at the solution then :D. However, there is one step, $\sum_{i=1}^{\infty}x^i\sum_{j=1}^{i}\frac{1}{j} = \sum_{i=1}^{\infty}\frac{1}{i}\sum_{j=i}^{\infty}x^j$, that I am wondering about now already. Can you explain it?

Series is absolutely convergent, so if $\sum_{i=1}^{\infty} a_i =L$ and $\pi : {\mathbb R} \to {\mathbb R}$ -biection, then $\sum_{i=1}^{\infty} a_{\pi(i)} =L$

What does your new order of summation look like? I still don't see what you did with the indices. Sorry. Try again?  :P

I want to say that we can permute terms of a series, and sum of series will stay the same...

A corollary:
Suppose that $\sum_{k=0}^{\infty}a_k$ and $\sum_{k=0}^{\infty}a_k$ are absolutely series with sums $A$ and $B$. Then the double Series $\sum_{k,j =0}^{\infty}a_j b_k$ converges and has sum $AB$.

Yes, I am aware of the theorems, and I used them in my solution (also specifically what Beginner said). I can also intuititively see anik_andrew's 'rearrangement', but I don't see how it works in more detail. So I was asking for a more precise explanation of what he did. I use one rearrangement in my solution to come with essentially the same rearrangement that anik_andrew did, but since he could just write equality between the two he must see some easier way to do it.

$\begin{array}{l} (1) \cdot x^1 & \left(1+\frac{1}{2}\right)x^2 & \left(1+\frac{1}{2} + \frac{1}{3}\right)x^3 & \left(1+\frac{1}{2} + \frac{1}{3} + \frac{1}{4} \right)x^4&\ldots \end{array}$
Then look at the columns...
$\begin{array}{l} \left(x+x^2+x^3+x^4+x^5+x^6+\ldots \right) 1 & \left(x^2+x^3+x^4+x^5+x^6+\ldots \right)\frac12 & \left(x^3+x^4+x^5+x^6+\ldots \right)\frac13 & \left(x^4+x^5+x^6+\ldots \right)\frac14 & \ldots \end{array}$

This theorem due to Fubini (I almost forget it).  Anik_Andrew's method usually used to compute  double series.

Ok Anik_Andrew would you please to remember me where we use Absolutely convergent condition here (I've completely forget it  :( )?

[quote="Beginner"]
Ok Anik_Andrew would you please to remember me where we use Absolutely convergent condition here (I've completely forget it  :( )?[/quote]
When we change the order of a terms

Got it now.. Its Rienmann's Rearrangement theorem. Thanks.


A committee of 3 people is to be selected randomly from 6 different people.

1. What is the probability that neither John or Sara will be on the committee?

2.  What is the probability that either John or Sara will be on the committee?

I know that you need to find the ways that you can have John and Sara not be on committee and then divide that by the total number of committees, but I can't figure out how many ways the committee can be made, so it doesn't have John or Sara.

Hi,

1) So we don't want John nor Sara on the committee. So there are now only 4 out of the 6 total people left for us to pick the 3 committee members from. Therefore, there are $ \binom{4}{3} \equal{} 4$ ways. So the probability is $ \dfrac{\binom{4}{3}}{\binom{6}{3}}.$

2) Either John or Sara can be on the committee at once. So we just have two exclusive cases.

Case 1: John is on the committee. There are 4 other members to choose the remaining 2 people from (since John is already picked, and Sara cannot be picked). Therefore, there are $ \binom{4}{2}$ ways.

Case 2: Sara is on the committee. Similarly as before there are 4 other members to choose the remaining 2 people from, which can be done is $ \binom{4}{2}$ ways.

Hence, the probability is $ \dfrac{2\binom{4}{2}}{\binom{6}{3}}.$

For #2, it means that only one of them can be on the committee, not both?

Yes, the problem says John or Sara, so that's can't mean both.


geometry
circumcircle
trigonometry
geometry proposed

Two circles $ C_1$ and $ C_2$ cut themselves at $ A$ and $ B$ the tangent of the circle $ C_2$ passing  by $ A$ cuts the circle $ C_1$ in a point $ C$ and the tangent of the circle $ C_1$ passing  by $ A$ cuts the circle $ C_2$ in a point D, the half-line $ (d)$ passing by $ A$ (internal in the angle $ \angle{DAC}$) cuts respectively the circles $ C_1$ , $ C_2$ and the circumcircle of $ ADC$ in $ M$ ,$ N$ and $ P$.
Prove that: $ AM\equal{}PN$

$ \angle CAB\equal{}\angle ADB,\angle ACB\equal{}\angle DAB \Rightarrow \angle ABC\equal{}\angle ABD\equal{}\alpha$
$ \angle CDP\equal{}\angle CAP\equal{}\angle ADN \Rightarrow \angle NDP\equal{}\angle ADC\equal{}\beta$. Similarly, $ \angle ACM\equal{}\angle DCP\equal{}\gamma$
Applying the law of sine we obtain:
$ \frac{NP}{\sin \angle NDP}\equal{}\frac{PD}{\sin \angle PND} \Rightarrow NP\equal{}PD.\frac{\sin \beta}{\sin \alpha} (1)$
$ \frac{AM}{\sin \angle ACM}\equal{}\frac{AC}{\sin \angle AMC} \Rightarrow AM\equal{}AC.\frac{\sin \gamma}{\sin \alpha} (2)$
But $ \frac{AC}{\sin \angle ADC}\equal{}\frac{PD}{\sin \angle DCP}$ or $ \frac{AC}{\sin \beta}\equal{}\frac{PD}{\sin \gamma} (3)$
From $ (1),(2)$ and $ (3)$ we are done.


How hard is it to download LaTex and learn it and do stuffs and make it to pdf documents?

I like pdf very much and want to have them if I can for my many contests.

Opinions please!

and tell how long it took.

Not very long iff you follow the instructions on this site.

It should only take a few minutes to read through the AoPS latex guide.  It usually takes a while to install, but installation is very straight-foward.

i hate to come against what others are saying but it took me a long time first to figure out what LaTeX was, what a tarball was, how to get it for free, how to use and upack a diskimage/tarball (for which i needed winsomething or whatever).  I downloaded some full version that was like 400 megs and finally got a program that i didn't know how to use :)

24 hrs of messing around (from scratch, heard about it from a friend who doesn't explain things well) but it was well worth it.

Maybe if u r more computer saavy than me it would be faster for u

[quote="luther_driggers"]i hate to come against what others are saying but it took me a long time first to figure out what LaTeX was [ ... ] Maybe if u r more computer saavy than me it would be faster for u[/quote]What do you mean by figuring out? Installing the program on your computer, or using it on the board?!

[quote="luther_driggers"]i meant installing it on my computer to creat pdf's, sorry it wasn't clear.

learning enuf code to use on this board or even do a math assignment for skul is easy thou.  I just printed a reference sheet and learned from other people's code.  doesn't really take any time, it's just slow at first.

the difficult part is installing the big package thing but u don't need that for the board, right?[/quote]You only need to know to write in $\LaTeX$ code of course.

As for the pdf creation, you should just install the minimal package. I have not yet used any packages aside from those provided in the minimal instalation.


geometry
3D geometry
sphere
trigonometry

If a sphere has uniform charge density $ \sigma$, radius $ R$, and a point lies outside the sphere a distance $ z$ (from the center so $ z > R$). What is the electric field at this point without using Gauss's Law?

This is classical. If anyone knows a book or website or even here that has a solution, that would be great. I'm getting horrid things I can't integrate:

$ dE\equal{}\frac{1}{4\pi\epsilon_{0}}\frac{dq}{r^{2}}$ where $ r^{2}\equal{}z^{2}\plus{}R^{2}$. Let $ \theta$ be the angle between $ z$ and $ R$ and $ \varphi$ be the angle between $ z$ and $ r$. We can ignore all force other than in the $ z$ direction since the others cancel.

Then since $ r^{2}\equal{}z^{2}\plus{}R^{2}\minus{}2zR\cos\theta$ and $ R^{2}\equal{}z^{2}\plus{}r^{2}\minus{}2zr\cos\varphi\Rightarrow\cos\varphi\equal{}\frac{\minus{}R^{2}\plus{}z^{2}\plus{}r^{2}}{2zr}$

so $ dE\equal{}\frac{1}{4\pi\epsilon_{0}}\frac{\sigma R\sin\theta d\theta d\phi}{z^{2}\plus{}R^{2}\minus{}2zR\cos\theta}\left(\frac{\minus{}R^{2}\plus{}z^{2}\plus{}r^{2}}{2zr}\right)$.

Now I need to integrate over $ \phi$ between $ 0$ and $ 2\pi$ quite simple, and $ \theta$ between $ 0$ and $ \pi$, not humanly possible. Something has to cancel...right?

[hide="hint"]think of the sphere as a collection of shells and assume u know the electric feld outside due to a shell[/hide]


superior algebra
superior algebra theorems

Let n > 1 be an integer; suppose x^2 congruent to 1 (mod n) for all integers x such that 0 < x < n and gcd(x,n) = 1. Show that n divides 24.
[This isn't from homework or a  test; it's exercise 9.7#2 in David Benson's _Mathematics, a Musical Offering_ and I'm stuck.]

[quote="jmerry"]Step 1: show that $ n$ is not divisible by 5 (because then some of your suares would be $ 4$ mod $ 5$.
Step 2: $ 5^{2}\equiv 1\mod n$.
It's not hard to show that all divisors of 24 work.[/quote]

Do I misunderstand? You seem to be addressing the converse: if n divides 24 then for all x s.t. gcd(x,n)=1, x^2 is congruent to 1 (mod n).

I'm not addressing the converse. What can you say about $ n$ if you know that $ 5^{2}\equiv 1$?


Can someone give me hint on this problem
http://acmicpc-live-archive.uva.es/nuevoportal/data/problem.php?p=3413
Thanx in anticipation.

[hide="first hint"]
Two SINs $s_{1}$, $s_{2}$ will be congruent mod $m$ iff $m | s_{1}-s_{2}$.
[/hide]

[hide="second hint"]
The number of factors of a number is at most logarithmic in the size of the number.
[/hide]

[hide="third hint"]
You can precompute the factors of a number using a combination of a Seive and Dynamic Programming.
[/hide]

[hide="fourth hint"]
There are only at most $\binom{G}{2}$ pairs $s_{1},s_{2}$.
[/hide]


[color=red]Seriously Wrong!!![/color]

I don't know if I am missing something...
$m+n=m+p$ so we can cancel out them from the equation.
Then it leaves: $\frac{1}{p}=\frac{1}{n}$, which is obviously true.

[quote="shobber"]I don't know if I am missing something...
$m+n=m+p$ so we can cancel out them from the equation.
Then it leaves: $\frac{1}{p}=\frac{1}{n}$, which is obviously true.[/quote]

 :huh:  How is it true that $m+n=m+p$ ... ?

I guess it has to do with something like $2S_m=S_{m+n}=S_{m+p}$ ...

Perhaps he meant sum of next n terms OR sum of the next p terms after those...

hey guys, dont get confused :D  Actually i aint too sure too , but it is something that it aint necessary p=n , ITs like in a quadratic  if you get x=7 or x= 9 that is 2 POSSIBLE values of x similarly here possible values of of the LHS are the 2 RHS's

i can give the base hint if you want . Equate individually the LHS with the  RHS''s , then ob tain the values of d and in turn equate them and do some stuff .

the question's actually weird , not even our school teacher is gettin it....

[quote="eminem"]hey folks , i am desperate for a solution , can someone please post it fast??

if the sum of m terms of an A.P. = sum of next n terms OR sum of next p terms , then prove

(m+n)(1/m - 1/p) = (m+p)(1/m - 1/n)

thx[/quote]I don't understand the question. Two questions/comments:

Are we talking about some $m$ terms of an AP or any $m$ terms?

The only difference between "sum of next $n$ terms" and "sum of next $p$ terms" is that one says $n$ and one says $p$. What differentiates the two? This makes no sense to me whatsoever.

Could you rewrite the question please?

[hide="Boring idea"]
I think this sequence m is a quadratic equation with 2 answers which are not equal if we can find the quadratic sequence maybe we can use them in the equation to prove it...
[/hide]

hey antonio, its the first m terms

hey riddler i would say you are on the right track

as for the m=p thing . , i would say its not necessary that m=p , cant it be that the sum of (p-m) terms = 0???

If m isnt = to p,then i would say its a descendin A.P(Pretty obvious)

as for the m=p thing . , i would say its not necessary that m=p , cant it be that the sum of (p-m) terms = 0???
WHY :huh: ????

like if i say.........

sum of 7 terms of an AP= sum of next 9 terms Or sum of next 13 terms

means sum of those middle 4 terms(13-9) = 0 !!

means some of these middle 4 terms are negative , some are positive , only then can they add upto be 0 , thats what makes it confusing....

[quote="eminem"]hey antonio, its the first m terms[/quote]Er.. could anyone else explain? If someone could just rewrite the question to make it coherent (to me) that would be great.

[quote="varun"]The sum of first m terms of an AP is equal to the sum of next n terms.
We can also say that sum of first m terms of an AP is equal to the next p terms.
Now prove that:     (m+n)(1/m - 1/p) = (m+p)(1/m - 1/n)

Think that will help you.  :lol:[/quote]We have an arithmetical progression $a+b,a+2b,a+3b,\ldots$, and we want to show that \[\sum_{k=1}^m({a}+kb)=\sum_{k=m+1}^{m+n}({a}+kb)=\sum_{k=m+1}^{m+p}({a}+kb)\]implies $(m+n)\left(\frac1m-\frac1p\right)=(m+p)\left(\frac1m-\frac1n\right).$

Is that right? Or is it:

\[\sum_{k=1}^m({a}+kb)=\sum_{k=m+1}^{m+n}({a}+kb)=\sum_{k=m+n+1}^{m+n+p}({a}+kb)\]

[quote="AntonioMainenti"]We have an arithmetical progression $a+b,a+2b,a+3b,\ldots$, and we want to show that \[\sum_{k=1}^m({a}+kb)=\sum_{k=m+1}^{m+n}({a}+kb)=\sum_{k=m+1}^{m+p}({a}+kb)\]implies $(m+n)\left(\frac1m-\frac1p\right)=(m+p)\left(\frac1m-\frac1n\right).$

Is that right? Or is it:

\[\sum_{k=1}^m({a}+kb)=\sum_{k=m+1}^{m+n}({a}+kb)=\sum_{k=m+n+1}^{m+n+p}({a}+kb)\][/quote]

The first one is correct.

Now I think you got the question.

[quote="Altheman"]well then, it is obvious that p = n (just look at the closed forms for this)
$(m+n)(\frac1m - \frac1n) = (m+n)(\frac1m - \frac1n)$ and that is trivial[/quote]

I must say Altheman that the question is correct. We do have n=p. But it may not always be true. I have done this question last year. Well I will try to find out its soln and post it.  At present I am  also unable to solve it.

Why not any moderator is trying this question? I think they may give some useful suggestions.

[quote="Altheman"]well then, it is obvious that p = n (just look at the closed forms for this)
$(m+n)(\frac1m - \frac1n) = (m+n)(\frac1m - \frac1n)$ and that is trivial[/quote]If $p=n$ it is trivial, as you said. Otherwise we can assume WLOG that $p>n$. The equality (between the sums) holds when the $p-n$ "extra" terms are "centered" on $0$.

Well friends now let me post the solution. I have done it today morning only.
So open the pdf file attached to see the solution.
I am very happy solving this question. :rotfl:

[quote="shadysaysurspammed"]Varun,I didnt understand the step right after equation (3).

By the way,where did u see this before?Which book?[/quote]
I am again sending the same file updating itwith some more information, hope that it will help you.

I think "$A = B$ or $A = C$" should be "$A = B$ and $A = C$".

[quote="eminem"]hey antonio  :sleeping: , isnt that just what i wrote ( read my above posts )

that sum of p-n terms gotta be 0?[/quote]Well, apparently you didn't explain it very well, since Altheman still needed clarification. Why wouldn't I post what I did?

[quote="chess64"][size=150]Question[/size]

Let $A$ be the sum of the first $m$ terms of an arithmetic sequence. Let $B$ be the sum of the next $n$ terms, after $m$ terms. Let $C$ be the sum of the next $p$ terms, after $m$ terms. Given that $A=B$ or $A=C$, prove that

\[(m+n)(\frac{1}{m} - \frac{1}{p}) = (m+p)(\frac{1}{m} - \frac{1}{n}).\]

That should help. :roll:[/quote]Not really, since we've already clarified exactly what the problem is.

[quote="dakyru"]If it was "and", the problem would just become trivial.  The original wording was "or".[/quote]If it were "or" it wouldn't make any sense. Suppose that \[\sum_{k=1}^m({a}+kb)=\sum_{k=m+1}^{m+n}({a}+kb),\] but not \[\sum_{k=1}^m({a}+kb)=\sum_{k=m+1}^{m+p}({a}+kb).\]There are infinitely many values of $p$ for which the second equality does not hold, and they cannot all satisfy $(m+n)\left(\frac1m-\frac1p\right)=(m+p)\left(\frac1m-\frac1n\right).$

[quote="varun"]I am again sending the same file updating itwith some more information, hope that it will help you.[/quote]Good job varun! I checked and it's all right, except a small typo. Just before (4), the sum should be from $m+n+1$ to $m+p$.


Find the first positive integer whose square ends in three $ 4$'s.

Find all positive integers whose squares end in three $ 4$'s.

Show that no perfect square ends with four $ 4$'s.

Find the first positive integer whose square ends in three $ 4$'s.
$ 21^2\equal{}441<444<484\equal{}22^2$ is not a square
$ 1444\equal{}38^2$

Find all positive integers whose squares end in three $ 4$'s.
[hide="Hint"]Checking mod $ 10$ several times you get:
$ 1000k\plus{}a$ with $ a\in \{38,462,538,962\}$[/hide]

Show that no perfect square ends with four $ 4$'s.
[hide="hint"]Checking mod $ 10$ with the above solutions.[/hide]


Prove that for any positive numbers $ x$ , $ y$ , $ z$ exists the inequality $ xyz\plus{}\sqrt {\sum x^2\cdot\sum x^4}\ \ge\ \frac 23\cdot\sum x^2(y\plus{}z)$ .

From Cauchy-Schwarz:

$ \sqrt {\sum x^2\cdot\sum x^4} \geq \sum x^3$

Inequality is equivalent to

$ 3xyz \plus{} 3\cdot \sum x^3 \geq 2\cdot \sum x^2(y \plus{} z)$

We have well known relation : $ 2 \cdot \sum x^3 \geq \sum x^2(y \plus{} z)$
so it is enough to prove

$ 3xyz \plus{} \sum x^3 \geq \sum x^2(y \plus{} z)$

which is just Schur


show that for positve reals $ a,b,c$
the following inequality holds,
$ \sum_{cyclic}\frac{ (a \plus{} 2b)^3 } {(a \plus{} 2c)^3} \ge 3$

p.s. i know a solution but i think its quite lengthy

someone plz post solution
thanks in advance!

Hi geniusbliss, I will just give a hint:
the inequality is equivalent  to proving that
$ \sum \frac{a\plus{}2b}{a\plus{}2c} \ge 3$ and now use the Cauchy Schwartz Inequality.

well i have decided to post my solution,
incase someone might be able to
finish it in the middle directly,
so here is my solution,
we have by power mean,
$ \sum_{cyclic}\frac { (a \plus{} 2b)^3 } {(a \plus{} 2c)^3} \ge \frac {(\sum_{cyclic} \frac {a \plus{} 2b}{a \plus{} 2c})^3}{9}$
therefore to prove,
$ (\sum_{cyclic} \frac {a \plus{} 2b}{a \plus{} 2c})^3 \ge 27$,
or to prove,
$ \sum_{cyclic} \frac {a \plus{} 2b}{a \plus{} 2c} \ge 3$
subtract $ 1$ from each term,
we get,
$ 2(\sum_{cyclic} \frac {b \minus{} c}{a \plus{} 2c}) \ge 0$
divide the whole by 2
we get,
$ \sum_{cyclic} \frac {b \minus{} c}{a \plus{} 2c} \ge 0$
now again add one to each term to get,
$ (a \plus{} b \plus{} c)(\sum_{cyclic} \frac {1}{a \plus{} 2c}) \ge 3$
now multiply by $ 3$ on both sides,
and arrange suitably,
we get,
$ (\sum_{cyclic}a \plus{} 2c)(\sum_{cyclic} \frac {1}{a \plus{} 2c}) \ge 9$
which is true by AM-HM or Cauchy-Schwarz
$ Q.E.D$

[quote="geniusbliss"]
$ \sum_{cyclic}\frac { (a \plus{} 2b)^3 } {(a \plus{} 2c)^3} \ge \frac {(\sum_{cyclic} \frac {a \plus{} 2b}{a \plus{} 2c})^3}{9}$
therefore to prove,
$ (\sum_{cyclic} \frac {a \plus{} 2b}{a \plus{} 2c})^3 \ge 27$,
or to prove,
$ \sum_{cyclic} \frac {a \plus{} 2b}{a \plus{} 2c} \ge 3$
subtract $ 1$ from each term,
we get,
$ 2(\sum_{cyclic} \frac {b \minus{} c}{a \plus{} 2c}) \ge 0$
divide the whole by 2
we get,
$ \sum_{cyclic} \frac {b \minus{} c}{a \plus{} 2c} \ge 0$
now again add one to each term to get,
$ (a \plus{} b \plus{} c)(\sum_{cyclic} \frac {1}{a \plus{} 2c}) \ge 3$
$ Q.E.D$[/quote]

well instead of subtracting 1 from each term, you can just add 1 to each term and apply AM-HM from there, which will give you the desired result.


calculus
integration
trigonometry
calculus computations

Evaluate $ \int_0^\pi e^{x\sin x}(x^2\cos x\plus{}x\sin x\plus{}1)\ dx$.

It is $ \pi$. :)

[hide="solution"]
The integrand is $ \frac{d}{dx}(e^{x \sin x}x)$.
[/hide]


number theory
greatest common divisor
number theory solved

1)For any integer n>1 the numbers1/5*(2^(4n+2)+1) are not prime

2) Prove that :

[a,b,c]=abc(a,b,c)/{(a,b)(a,c)(b,c)}

where [a,b]=ab/(a,b).

3)Prove that:

[a,b,c]^2/{[a,b][b,c][a,c]}=(a,b,c)^2/{(a,c)(b,c)(a,c)}.
(USA 1972)

hope someone will post solutions! :D

bye bye

4 the 2nd one:

I think you mean the gcd and the other things! :D :D

[b]Solution: [/b]

we use that [a,b]=ab/(a,b) and the distributivity formula that: ([a,b],c)=[(a,c),(b,c)] and we get that:

[a,b,c]=[[a,b],c]=[a,b]c/([a,b],c)=abc/{(a,b)[(a,c),(b,c)]=abc((a,c),(b,c))/{(a,b)(a,c)(b,c))
=abc(a,b,c)/{(a,b)(a,c)(b,c)}

tschuess!!!!!!!!!

for the 1st one:

[b] Solution: [/b]

For any n integer we have that:

2^(4n+2)+1=(2^(2n+1)-2^(n+1)+1)(2^(2n+1)+2^(n+1)+1)
because 5=2^2+1 | 2^(4n+2)+1 for any n >1 we have that 2^(2n+1)-2^(n+1)+1=2^(n+1)*(2^n-1)+1>=2^3*3+1=25 so the number 1/5*(2^(4n+2)+1) is the product of two factors>1 so the number is not prime! :D

tschuess

I'll give some hints for the other problems.

1) Do you know the (famous) formula a4 + 4b4 = (a2 - 2ab + 2b2)(a2 + 2ab + 2b2) ?

3) Write both sides in terms of minima and maxima.

If a = \pi(pkmk) and b = \pi(pknk) for primes pi, then

(a, b) = \pi(pkmin(mk, nk))
[a, b] = \pi(pkmax(mk, nk))

For the 3rd is done exactly like the 2nd one!

cheers! :D

My solution also works.

http://www.kalva.demon.co.uk/usa/usoln/usol721.html


algebra
polynomial
induction
real analysis
real analysis unsolved

Let $P$ be a polynomial of degree $n$ with only real zeros and real coefficients.
Prove that for every real $x$ we have $(n-1)(P'(x))^2\ge nP(x)P''(x)$. When does equality occur?

I will proceed by induction. For $n=1$ that's manifestly true. Let $P$ be  polynomial of  degree $n+1$ , so
$P(x)=u(x)Q(x)$ where $u(x)$ is a monic real polynomial of degree $1$ and $Q(x)$ is a polynomial of degree $n$ with only real roots  and real coefficients.
$P(x)=u(x)Q(x) \Longrightarrow P'= Q+uQ'\Longrightarrow P'' = 2Q' + uQ''$ , where $(n-1)(Q')^2\geq nQQ''$ , so :

$n(P')^2-(n+1)PP'' = nQ^2-2uQQ'+nu^2(Q')^2-(n+1)u^2QQ''\geq$$nQ^2-2uQQ' + \frac{u^2(Q')^2}n = (Q\sqrt n - \frac{uQ'}{\sqrt n})^2 \geq 0$... (q.e.d)
  :cool:

Sorry, I forgot the equality case ! Remember : $u(x)$ is a monic real polynomial , $u(x) = x-a$
$n(P')^2-(n+1)PP'' = 0 \Longrightarrow Q\sqrt n - \frac{uQ'}{\sqrt n} = 0 \Longrightarrow \frac{Q'}{Q} = \frac nu$$\Longrightarrow Q(x)= cu^n_{(x)}\Longrightarrow P(x)=c u^{n+1}_{(x)}$
Also, It's easy to verify that the equality holds  for  $P(x)=c u^{n+1}_{(x)}$
:cool:

without induction, by the condition we have that
\[ P \equal{} b\prod^n_{i\equal{}1} (x\minus{}a_i)\]
Suppose $ x$ is not a root of $ P$ and Let $ c_i \equal{} \frac{1}{x\minus{}a_i}$, then
\[ P' \equal{} P (c_1 \plus{} c_2 \plus{} \ldots \plus{} c_n), P'' \equal{} P (\sum_{i \neq j} 2c_i c_j)\]
The problem thus becomes
\[ (n\minus{}1)(c_1 \plus{} c_2 \plus{} \ldots \plus{} c_n)^2 \ge 2n (\sum_{i \neq j}c_i c_j)\]
factoring gives
\[ \sum_{i \neq j} (c_i \minus{} c_j)^2 \ge 0\]
Which is trivially true.  Also the equality holds iff all the $ c_i$ are equal, which means all the $ a_i$ are equal and $ P\equal{}b(x\minus{}a)^n$


Prove that if there is integers $ m,n$ such as $ 24|mn\plus{}1$ then $ 24|(m\plus{}n)$.

[hide="Solution?"]
\[ 24|mn \plus{} 1\implies mn \plus{} 1\equiv 0\pmod{24}\implies mn\equiv 23\pmod{24}\]
Therefore, either
\[ m\equiv 23\pmod{24} \\
n\equiv 1\pmod{24}\]
or
\[ m\equiv 1\pmod{24} \\
n\equiv 23\pmod{24}\]
Either way,
\[ m \plus{} n\equiv 1 \plus{} 23\equiv 24\equiv 0\pmod{24}\implies 24|(m \plus{} n)\]
Am I missing anything in my proof?
[/hide]

I'm not sure if you are missing something "jxl28" but the solution seems to be true  :D

Thanks

Beware of hasty conclusions! If $ mn \equiv 23 \pmod{24}$ it does not follow one is $ 1$ and the other $ 23$. How about $ m \equiv 5 \pmod{24}$ and $ n \equiv 19 \pmod{24}$ (or vice-versa)?
One just has to exhaust all these other cases.

Thanks "mavropnevma" for your clarification   :)

I got stuck... D:

[hide="What I have so far"]We have that $ mn\plus{}1\equal{}24k$ for some integer k. This means that $ mn\equiv \minus{}1\bmod{24}$. We want to prove that $ m\plus{}n\equal{}24j$ for some integer j. This means we are trying to prove that $ m\equiv \minus{}n\bmod{24}$. But from here I don't know what to do...am I approaching this problem the wrong way?[/hide]

From $ mn \equiv \minus{}1 \pmod{24}$ follows $ m \equiv 1 \pmod{2}$, but then $ m^2 \equiv 1 \pmod{8}$. But it also follows $ m \equiv \pm 1 \pmod{3}$, therefore $ m^2 \equiv 1 \pmod{3}$.
Together,  $ m^2 \equiv 1 \pmod{24}$ so then $ \minus{}m \equiv (mn)m \equal{} m^2n \equiv n \pmod{24}$, i.e. $ m\plus{}n  \equiv 0 \pmod{24}$, or $ 24 \mid m\plus{}n$.

[quote="mavropnevma"]From $ mn \equiv \minus{} 1 \pmod{24}$ follows $ m \equiv 1 \pmod{2}$, but then $ m^2 \equiv 1 \pmod{8}$. But it also follows $ m \equiv \pm 1 \pmod{3}$, therefore $ m^2 \equiv 1 \pmod{3}$.
Together,  $ m^2 \equiv 1 \pmod{24}$ so then $ \minus{} m \equiv (mn)m \equal{} m^2n \equiv n \pmod{24}$, i.e. $ m \plus{} n \equiv 0 \pmod{24}$, or $ 24 \mid m \plus{} n$.[/quote]
I'm fairly new to mods so here are my questions:
How do you get that $ m\equiv 1\bmod{2}$ from $ mn\equiv \minus{}1\bmod{24}$? And also $ m^{2}\equiv 1\bmod{8}$? Then I guess how did you get everything from that? D:

[quote="JBL"]If $ mn$ is one more than a multiple of 24, then it's odd, so each of its factors is odd.  The square of every odd number is congruent to 1 mod 8.[/quote]
Ok I get that part, but how did mavropnevma get that $ m^{2}\equiv \pm1\bmod{3}$?

There is never a good reason to block-quote the entire post immediately preceding your own.

Just borrow the same reasoning: if $ mn$ is one more than a multiple of 24 then $ m$ _________ $ 3$.  The square of any number ________ $ 3$ is ___________ $ 3$.  (You've also misquoted mavropnevma.)


function
search
algebra proposed
algebra

For those who think that $ f(f(x))$ is always an increasing function :

Find all functions $ \mathbb{R}\rightarrow\mathbb{R}$ such that $ f(f(x)) \equal{}\minus{}x$ $ \forall x\in\mathbb{R}$

Suppose there exists such interval $ P$, that for any reals $ (x,y)$ from $ P$, where $ x>y$, $ f(x)>f(y)$. Than, $ f(f(x))>f(f(y))$, but as $ f(f(x))\equal{}\minus{}x$,$ f(f(y))\equal{}\minus{}y$, we have $ \minus{}x>\minus{}y$, but $ x>y$-contradiction. So, $ f$ is increasing function. Again take reals $ x$ and $ y$ with $ x>y$. $ f(x)<f(y)$. So, $ f(f(x))>f(f(y))$, or $ \minus{}x>\minus{}y$-again contradiction because $ x>y$.
So, there is no solution.

[quote="lasha"]Suppose there exists such interval $ P$, that for any reals $ (x,y)$ from $ P$, where $ x > y$, $ f(x) > f(y)$. Than, $ f(f(x)) > f(f(y))$, but as $ f(f(x)) \equal{}\minus{}x$,$ f(f(y)) \equal{}\minus{}y$, we have $ \minus{}x >\minus{}y$, but $ x > y$-contradiction. So, $ f$ is increasing function. Again take reals $ x$ and $ y$ with $ x > y$. $ f(x) < f(y)$. So, $ f(f(x)) > f(f(y))$, or $ \minus{}x >\minus{}y$-again contradiction because $ x > y$.
So, there is no solution.[/quote]

You're wrong.
Solutions exist

When you write "Suppose there exists such interval $ P$, that for any reals $ (x,y)$ from $ P$, where $ x > y$, $ f(x) > f(y)$. Then, $ f(f(x)) > f(f(y))$", the conclusion "Then, $ f(f(x)) > f(f(y))$" is wrong because maybe $ f(x)\notin P$ when $ x\in P$

[quote="Rust"]$ f(0) \equal{} 0$. For all $ x\not \equal{} 0$ we can consider pairs $ (x,y), x\not \equal{} 0\not \equal{} y,\ |x|\not \equal{} |y|$ and define
$ f(x) \equal{} y,f(y) \equal{}\minus{}x,f(\minus{}x) \equal{}\minus{}y, f(\minus{}y) \equal{} x$.[/quote]

Perfecty right, Rust.

In other words :
Let $ \mathbb{A}$ and $ \mathbb{B}$ to subsets of $ \mathbb{R}^{\plus{}}$ such that :
$ \mathbb{A}\cup\mathbb{B}\equal{}\mathbb{R}^{\plus{}}$
$ \mathbb{A}\cap\mathbb{B}\equal{}\emptyset$
$ \exists$ $ h(x)$ bijective function $ \mathbb{A}\rightarrow\mathbb{B}$

$ f(x)$ may be defined as :
$ f(0)\equal{}0$
$ f(x)\equal{}h(x)$ $ \forall x\in\mathbb{A}$
$ f(x)\equal{}\minus{}h^{\minus{}1}(x)$ $ \forall x\in\mathbb{B}$
$ f(x)\equal{}\minus{}f(\minus{}x)$ $ \forall x<0$

Example :
$ f(0)\equal{}0$
$ f(x)\equal{}\frac{|x|}{x}\minus{}(\minus{}1)^{[\minus{}|x|]}x$ $ \forall x\neq 0$

But there are not continiosly function $ f(x): R^{n}\to R^{n}$ suth that $ f(f(x))\equal{}\minus{}x$ if n is odd.

You are right pco! Sorry, I made mistake. :(

For pco: http://www.mathlinks.ro/viewtopic.php?search_id=1499275189&t=113408.

[In the link it is proved that every such function has infinitely many discontinuities.]

PLease see this and the link of enescu
http://www.mathlinks.ro/viewtopic.php?search_id=831547553&t=56637


Is there a difference between the words 'justify' and 'prove' when relating to trying to solve a problem. Particularly in Discrete Mathematics if that makes any difference. Thanks.

Usually proof is supposed to be a little more rigorous.  Prove generally refers to a general result, while justify usually refers to an answer to a specific question.  I think justify is more or less like a "find with proof."

In math olympiad, and in many math courses you are going to find different resoults. It depends on how the question is asked if you should they will ask you to justify or to prove. For example

Prove that the equation $x^2+1=0$ has no real solutions.

vs

Does the equation $x^2+1=0$ has real solutions? Justify your answer.

On other places, you usually find justify and proof differenciating quantitative from qualitative questions.

I grade homeworks for a math class at my college.  Often, when students think they've proven something with words, it's not that they've done something an improper way or taken up too much space but rather that they are just wrong.  It's very easy to convince yourself of mathematical truths using words, but those aren't proofs nor mathematical justifications.  In practice, there is no difference between those last two terms -- to prove something is to provide mathematical justification for it.  Mathematical justification can (and often does) come in the form of words as much as equations, but beginning proof-writers often have difficulty discerning between things they know to be true and thins that they've proven to be true.

Now, not having seen your solution, I can't say for sure if this is what your problem was, but if I were you I wouldn't just assume that you were right and your teacher is a mean grader.

In the solutions manual for the homework. They do a lot of proofs/justifications for one-to-one functions and onto functions through words. For example, "Clearly, this function can not be onto because it can not equal a negative number" (when the universe of discourse is all real numbers). If that is how the solutions manual is teaching us how to do things, I would assume that is what we're suppose to do.  I'm not saying he's a mean grader, just that he may not have been clear about what he wanted us to do since most of the class got this question wrong.

[quote="DYE"]Thanks. The reason I asked was because I was asked to  determine whether the function is a bijection from R to R. Justify your answer.
f(x)=2x-1. I justified my answer by definition of what one-to-one is and what onto is and concluded that for everyone x, there was a unique solution therefore it made it one-to-one, and was onto because those unique solutions covered all the Real numbers as specified. I went into more detail but basically that is how I justified my answer through words and by examination. However I got the whole question wrong because we were supposedly to use mathematical symbols to prove it. The answer he wanted was 2x-1=2y-1 --> x=y for 'proof' of one-to-one and f(x) = f(y+1//2) = 2((y+1)/2)-1 =y for 'proof' of onto. Can you set of a proof like that for a one-to-one function? What if the function were 2x^2-1, twhat would your answer be? I might have been wrong, but the wording of the the problem by using "justify" was misleading. Almost every student did the same reasoning as I did and did not receive any points. Thanks anyways though.[/quote]

In my view, what you wrote is fine.  The algebra shows that for every $y\in\mathbb{R}$, there exists $x\in\mathbb{R}$ such that $f(x)=y$.  This establishes surjectivity.  The algebra also establishes injectivity (there is only one solution).  Of course, this is without seeing exactly what you wrote, but you're definitely on the right track.

For the other example, it's neither one to one nor injective.  Since $2x^2-1\geq -1$, it's not surjective (there is no $x\in\mathbb{R}$ such that $f(x)=-2$) and not injective because $f(-x)=f(x)$

[quote="DYE"]This was my exact answer "One-to-One? Yes, because for every real number of x, there is a unique answer in which no two x's produce the same outcome (outcomes of the range of the codomain). Onto? Yes, because the range of the codomain contains/covers all real numbers. Therefore Bijective? Yes, because the function is one-to-one and onto." Remember, I was interpreting the word 'Justify' (as many others would) as a way to show that this is true or not true with rationality and less intensity in terms of mathematical symbolic proving. With the example that I gave if the function were f(x) = 2x^2-1, following the "proof" method from the TA, you would set 2x^2-1 = 2y^2-1. Simplifiying, you would get x=y, therefore would prove that it is one-to-one by the TA's method. This would be untrue since if we let x=5 and in another case x=-5, they both map to the same number in the codomain which is 5, which would make it not one-to-one.  Basically, through the way that the TA proved a one-to-one function, does not work in all cases such as this function because through his method of proof, this function should be a one-to-one function. I may be wrong, I'm just curious, thats all. Have a good day.[/quote]

On the last count, you are not correct. $f(x)=f(y)\rightarrow x^2=y^2\rightarrow x=\pm y$, which contradicts injectivity.

Oh yeah, thanks, but how do you know which side to put the +/- on, or does it matter? This forum community is great, I'm glad I stumbled across it. Makes me want to change to a Math major. Not really, but I really respect math majors.

It can go on either side -- if y is -x, x is -y and if y is +x, x is +y.  Plus-or-minus signs are a little bit tricky.

As for your solution, I would probably have given you most-but-not-full credit for it -- you definitely knew what you were doing (which is the important part), but the statements themselves are a little hand-wavy, which I suppose is my way of saying I think "justify" asks for something more rigorous than you thought.  In any event, you understand the mathematics of what you're doing, and that's what actually matters :)

Maybe not when an important course grade depends on it, but...

I disagree.  I would take a B in my math classes with an understanding of the material over an A without such an understanding any day.

Hehe...now would you really? ;)  Having become rather cynical with my professors, I have all but given up learning from them.  I realize that having a good teacher helps a lot, but if I don't have one I can always learn straight from the text.  I figure that if I manage to get an A in a class without mastering the material, I can always brush up on it some other time, like over the summer.  (after all, does when I learn the material really matter?  the end result is that I learned it and got an A in the course)


vector
abstract algebra
superior algebra
superior algebra unsolved

Let $ V_{1}$ and $ V_{2}$ be vector spaces over $ F$ with subspaces $ W_{1}$ and $ W_{2}$ respectively. Find a subspace $ X \subseteq V_{1} \otimes V_{2}$ and a bilinear map

$ \phi : \frac {V_{1}}{W_{1}}\times \frac {V_{2}}{W_{2}}\ \rightarrow \frac {V_{1} \otimes V_{2}}{X}$

I know $ \frac {V_{i}}{W_{i}}$ has elements that look like $ v_{i} \plus{} W_{i}$ and so I'm guessing that
if I let $ W_{1}$ have a basis $ w_{1},...,w_{n}$ and $ W_{2}$ have basis $ w'_{1},...w'_{m}$ then

$ x \equal{} span_{F} (w_{1},...w_{n},w'_{1},...w'_{m})$ but it also includes all the product elements like $ w_{1}w'_{2}$ right? How do I express the set X?

$ \phi$ sends ($ v_{1} \plus{} W_{1}$,$ v_{2} \plus{} W_{2}$) $ \mapsto$ ($ v_{1}v_{2} \plus{} X$)

Is this correct so far?

You implicitly assume the vector spaces are finite dimensional, are you intentionally doing this?

I think there's actually a "less computational" way to do this, use the universal property of the tensor product to get a bilinear map $ \psi: V_1\times V_2\rightarrow V_1\otimes V_2$, and try using this to build your new map. Define your subspace in terms of this map also. (Also, there is an isomorphism $ V_1/W_1\times V_2/W_2\cong (V_1\times V_2)/(W_1\times W_2)$)

uhh I guess your right since the problem doesnt state that they're finitely dimensional

going with what you said in the second post
$ \psi: V_{1}\times V_{2}\rightarrow V_{1}\otimes V_{2}$
is the map that sends $ (v_{1},v_{2}) \mapsto v_{1}v_{2}$
but how do I define my subspace in terms of this map?

Should $ X\equal{}W_{1} \otimes W_{2}$

well, try defining $ X\equal{}\psi(W_1\times W_2)$ whether or not that is $ W_1\otimes W_2$ I'm too lazy to check but whatever, the second map $ (v_1,v_2)\plus{}(W_1,W_2)\mapsto \psi(v_1,v_2)\plus{}(\psi(W_1),\psi(W_2))$ is going to be bilinear

now that I think of it, $ \psi(W_1,W_2)$ is probably going to be $ W_1\otimes W_2$ but I think it'd be easier to think of it as the image.


I'm kind of hooked on these games, so well this is a discussion thread on Tycoon games;

Zoo Tycoon, Roller Coaster Tycoon, Fish Tycoon, School Tycoon, County Fair... etc. etc

Just like... discuss! :)

My sister is quite good at Zoo Tycoon, although my favorite is by far the Roller Coaster Tycoon series.  The games have certainly improved a lot since the first version came out however long ago it was.  I'm not really sure what else to add... :maybe:

I really want to get roller coaster tycoon.

maybach says its really fun.

Ah, I remember playing this. I had pretty much the original roller coaster tycoon thingy (I sound old, but I'm only 14  :) ), and remember having a blast. Too bad I lost the cd though. :roll:

[quote="JRav"]This may sound odd, but if you're going to get a Roller Coaster Tycoon game for the first time, I recommend you go with one of the old ones (Roller Coaster Coaster Tycoon-perhaps with an expansion pack, or Roller Coaster Tycoon 2).  The newer games are rather different from the old ones, and in my opinion, not as good.  Yes, you can do more stuff but it doesn't have the "character" or "heart" of the older versions.[/quote]

Well, I have RTC1 and RTC3 is quite cool with a lot of things.

I guess Roller Coaster Tycoon 1 is good if you crave something feeling like an old fashioned carnival truly feeling kinda homey.

But RCT3 in my opinion is WAY COOLER.  The graphics look really realistic, I love Coaster Cam, and the billboard design feature totally pwns.  Also, some of the more tedious work(such as building tunnels) is automated with RCT3.

Solution:  Get the Platinum Edition RCT3, the regular RCT1 and play both!  It kinda depends on what you want.  Though you'll probably get more out of RCT3 if you try RCT 1 first.

Hehe, Tycoon games.

I love making tourist traps and generating ridiculous amounts of revenue.

The premise is to create a wildly popular ride and at the exit, create a miniplaza with several rides, food+drink+bathroom, but no way to get back to the entrance  :rotfl:

IMHO, graphics make the game worse, because they involve higher system specs and thus cause my computer to crash. If your computer is better than mine, feel free to splurge on graphics. But gameplay is essential.

Oh yeah, once I made a sign right by the exit that you couldnt leave. Sweetness.

I like playing Dino Park Tycoon, but it's really old.

I remember making one of those roller-coasters where you go through a loop, then go up an incline then going back down the incline

i then increased the speed by a few more miles per hour, and needless to say, it wasn't pretty

I also enjoyed dropping people in the water.  Not good for business, but I had gotten bored with the game after a while.

Never heard of these. I'll try to beg my parents for these.
Sound Fun.

I did forget to mention that in Zoo Tycoon, I sealed off the entrance, then opened a few of the fences for some of the carnivorous animals to get out.

MWAHAHA  :rotfl:   :P

@Zapi; ditto.
My most favorite part of Zoo Tycoon... see all the Lions attack the visitors held captive :evilgrin:....

[quote="polkadotman1"]Yea,

My friend used to fling people off roller coasters for fun :(

And, he trapped tourists in.

You go in, can't go out.

He earned a TON of money that way because every ride the tourists had to pay.[/quote]

This would actually work in RCT3, provided your park has ATM's because then your customers HAVE to spend money.  But in RCT1 each customer comes in with money, and once they spend it, they want to be done.  And if they can't leave, they're satisfaction will go WAY down, your park rating will drop A TON, and you'll go bankrupt because no one wants to enter.  So I don't think that's such a great idea.  And the "fling peopleoff roller coasters" I could never do in RCT1 or 3.  In RCT1, you can drown people, but in RCT3, you can't.  I ordered the RCT3 Platinum, so there's a Waterpark expansion pack which is awesome.  I create these twisty slides with huge drops.  But the intensity level is considered "Low".  :rotfl:

I have quite a few tycoons... I always had these phases where I would play them like every day. The one that I was into the most was probably Monopoly Tycoon, which I don't think has been mentioned yet... sad.

[quote="meewhee009"]@maybach
That's wow....
To me... there's this water slide with a huge drop at the amusement park that I live close too.
Haa.... wow.
I need to get RCT3... or some other tycoon...

This thread's making me anxious...
 :lol:[/quote]

I think you can download it for free of the Internet.  But it's probably not legal though.........:(

[quote="kookamunga"]The one that I was into the most was probably Monopoly Tycoon, which I don't think has been mentioned yet... sad.[/quote]

I had that too.  It never was as fun as any of the other tycoons for me.  Maybe it's because I had trouble winning at that game.

Or maybe it was the inability to unleash terror on visitors.

By the way, did anyone here remember playing Legend of the Zoombinis?

Scratch what I said before.  I don't [i]think[/i] you can find a free download of these games on the internet.  I [b] know [/b]it.  Unfortunately, even though the websites "claim" their legal, I know they aren't.  It's like saying  "Oh I stole your TV from your house, but I think it's legal, so I shouldn't be punished. " :P

Still.......if only it was legal................:sigh:

False, not all Tycoons that require you to pay money are illegal (that is, demo versions FTW)

But look at a Google search for RCT3:

[url]http://www.google.com/search?q=free+download+of+RCT3&sourceid=ie7&rls=com.microsoft:en-us:IE-SearchBox&ie=&oe=[/url]

Those don't sound like demos to me.

First, second, third, fourth, sixth, seventh... website are demos.
Fifth website is a patch.

Nearly every single one is a demo.

In my school's computers class, we are playing roller coaster tycoon 2, and we have a huge project on it.

I want to buy it, but I don't know where.

Any help?

THANK YOU!

Now if only I could convince my parents...

Is this a trustable demo?

http://www.download-free-games.com/simulation/rollercoaster_tycoon2.htm

You can watch someone fail at Roller Coaster Tycoon 3 here
https://www.youtube.com/watch?v=fbyjNahL7Rc

Any tips to complete the build your own six flags park scenario? (We only have three days to build the park, and one day has already gone bye. I only have 694 park rating because our the vomit and no restroom. My park value is 72000 dollars.)

Any actual rct2 videos that useful??

I've already watched Magikaly's videos...

And no videos that involve hacking please


geometry
function
calculus
calculus computations

find the volume of the solid generated by revolving the area between the curve y=(cos x)/x and the x-axis for pi/6 <= x <= pi/2 about the y-axis...help me please, thanks

You need to express the function as $ x\equal{}x(y)$ since you are going around the y-axis.  Even better, look up the shell method (wikipedia explains it well).

This is a [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=192492]double post.[/url]

Please don't re-post problems you've already asked, Jakey. And do read your textbook.


Could someone please assist me with this question please.
http://members.optusnet.com.au/gujju_style/fib.jpg

I understand induction to an extent (i.e. proving the first and/or second case), but then i don't know how to go about and proving something like this for n things.

That's the part i don't quite understand. Ok if i prove it for n = 1, i do that by just substituting a fixed number and showing that it works. But i can't do that for n + 1 can I? Just put in a fixed number? otherwise it wouldn't really be induction and showing it's true for all n.

This is actually just a special case of a much more general theorem about the convergents $\frac{p_{n}}{q_{n}}$ of an arbitrary irrational $r$.

Anyway, the idea of mathematical induction is this:

[b]Assume that[/b] $f_{k}^{2}-f_{k-1}f_{k+1}= (-1)^{k}$.  Try to

[b]Prove that[/b] $f_{k+1}^{2}-f_{k}f_{k+2}= (-1)^{k+1}$

The key is that you will be using your assumption in order to prove this statement.  Once you've established that (the "inductive step"), then since you know that it's true for $k = 1$, it must be true for $k+1 = 2$.  But since it's true for $k = 2$, it must be true for $k+1 = 3$.  And so forth.

[quote="Elemennop"]Assume it's true for the $n$ case, and then try to prove that it's true for the $n+1$ case (usually using some sort of algebraic manipulation of the $n$ case).

Or you can just plug in Binet's and BASH & SMASH IT[/quote]

bash and smash? can u explain?

[quote="coolio2005"]Still having problems trying to figure it out, could someone help me please. Thanks.[/quote]

[quote="t0rajir0u"][b]Assume that[/b] $f_{k}^{2}-f_{k-1}f_{k+1}= (-1)^{k}$.  Try to

[b]Prove that[/b] $f_{k+1}^{2}-f_{k}f_{k+2}= (-1)^{k+1}$[/quote]
[hide="A sketch"] We know that $f_{k+1}= f_{k}+f_{k-1}$.  We also know that $f_{k+2}= f_{k+1}+f_{k}$.  Hence

$f_{k+1}^{2}-f_{k}f_{k+2}= f_{k+1}( f_{k}+f_{k-1})-f_{k}( f_{k+1}+f_{k})$

Expanding, this gives us

$f_{k+1}f_{k}+f_{k+1}f_{k-1}-f_{k}f_{k+1}-f_{k}^{2}= f_{k+1}f_{k-1}-f_{k}^{2}$

But remember that we [b]assumed[/b] that $f_{k}^{2}-f_{k+1}f_{k-1}= (-1)^{k}$.  Therefore,

$f_{k+1}f_{k-1}-f_{k}^{2}=-( f_{k}^{2}-f_{k+1}f_{k-1}) =-1 \cdot (-1)^{k}= (-1)^{k+1}$ [/hide] [hide="An explanation of what we have done"] Let's take a step back here.  What have we proven?

We wanted to prove that $f_{n}^{2}-f_{n-1}f_{n+1}= (-1)^{n}$ is true for all integers $n$.

[i]Let $T(a)$ denote the statement "The above equation holds true for some positive integer $a$."  [/i]

We [b]assumed[/b] $T(k)$, and we [b]proved[/b] $T(k+1)$.  This means that whenever we have $T(k)$, we must also have $T(k+1)$.  Now, you can just plug in numbers to verify $T(1)$.  What do you have?  You have the statement

$T(1)$

And you also have

$T(k) \text{ implies }T(k+1)$

So what does that mean?  Well,

$T(1) \text{ implies }T(2)$
$T(2) \text{ implies }T(3)$
$T(3) \text{ implies }T(4)$

And so forth.  Hence we have $T(n)$ for all positive integers $n$, which is what we wanted to prove!  This is the heart of mathematical induction. [/hide]

[quote="danee 1nonly"]bash and smash? can u explain?[/quote]

He means to plug in the explicit formula for $F_{n}$ to prove it by "brute force."  He's kidding.  I hope. :wink:

That technique will be less useful in helping you to solve the general identity ;)

Specifically, it is this:  Let $p_{k}, q_{k}$ be the sequences defined by

$p_{0}= 1, p_{1}= r_{0}, p_{k+1}= r_{k}p_{k}+p_{k-1}$
$q_{0}= 0, q_{1}= 1, q_{k+1}= r_{k}q_{k}+q_{k-1}$

Where $r_{k}$ is some sequence of positive integers.  Show that

$p_{k}q_{k+1}-q_{k}p_{k+1}= (-1)^{k}$

(The special case in this topic is when $r_{k}= 1 \forall k$. )

[b]Remark:[/b]  $\frac{p_{k}}{q_{k}}$ denotes the series of convergents to the continued fraction

$r = r_{0}+\frac{1}{r_{1}+\frac{1}{r_{2}+\frac{1}{...}}}$

Can you prove this by induction as well?

[quote="t0rajir0u"]He's kidding. [/quote]

I wouldn't ever dare.


algebra
polynomial
algorithm
function
Galois Theory

How to solve this equation:

$ ax^3 \plus{} bx^2 \plus{} cx \plus{} d \equal{} 0$ ?  :maybe:

MUCH HARDER than I thought :o So what is the formula? I didn't find it there  :lol:

$ x_1\equal{}?$
$ x_2\equal{}?$
$ x_3\equal{}?$

Take a look at [url=http://www.math.vanderbilt.edu/~schectex/courses/cubic/]this[/url].

I think so, I mean, unless you can't derive the formula or factor it. :)

[quote="Bachukas"]Is it true that nowadays every equation can be solved?  :)[/quote]

What type of equations are you talking about (for example, do you mean equations like $ 3^x \equal{} x^2$), and what is your definition of "solved"?

I think Bachukas is asking if there is a "formula" that will solve cubic equations, one that will solve 4th degree, 5th degree, etc.  I think someone proved that theres no formula past x^3 or x^4, but there's definitely not a formula that will solve your 29348279384 degree equation.

Galois Theory and Abel-Ruffini Theorem could be useful. Polynomial equations whose degree is grater or equal to $ 5$ cannot generally be solved (more precisely, their soulutions can't be put in an exact radical form).

http://en.wikipedia.org/wiki/Galois_theory
http://en.wikipedia.org/wiki/Abel-Ruffini_theorem

Yes. No polynomial degree five or higher can be solved by root extraction.

The word "solved" should be qualified with what mathematical tools are allowed for the solution. If only the operations of addition, subtraction, multiplication, division are allowed, then only linear equations can be "solved". If in addition to these, the operation of root extraction is allowed as a tool, then all polynomials of degree less 5 can be solved with an algorithm involving these tools.

Note, however, that:
1.) certain restricted forms of higher degree polynomials CAN be solved with algorithms involving the above tools.
2.) if other tools are allowed, such as trigonometric functions or well understood special functions, then all polynomial equations can be "solved" in terms of those tools.
3.) if you consider the ability to find the solutions of an equation to an arbitrary number of decimal places as "solving the equation", then most equations, polynomial or otherwise, are "solvable" through various numerical techniques.


No comment:

[quote="[url]http://www.mo2005.info[/url]"]In der Oberstufe gewannen der Berliner Peter Schulze ( Kl.11 ), der Sachse Mathias Ost ( Kl.11), Darij Grimberg aus Baden-Württemberg und Christian Stettle aus Hamburg (beide Kl.12/13).[/quote]

Weiß jemand zufällig, durch welches Spracherkennungsprogramm die Leute unsere Namen gejagt haben?

  dg

Was ich lustig fand, dass darunter "Die volle Punktzahl, das hei�t 40 von 40 Punkten errang Peter Scholze" bereits zu Beginn der richtige Name stand.  ;)


Por qué no usas el subjuntivo en la siguiente oración?

[i]Siempre comen en esta cafetería aunque saben que la comida no es muy buena.[/i]

It expresses uncertainty that the cafeteria will serve bad food, doesn't it?

en qué verbo iría el subjuntivo??????

(about the source: it is "la gramática es aburrida")

Oops, typo on the source (I should've read it again... )

The verb is the second to last word: [i]es[/i]. Why is it not [i]sea[/i]?

[quote]Siempre comen en esta cafetería aunque saben que la comida no es muy buena.[/quote]
You are not saying anything uncertainty. "They always eat in this cafeteria, even knowing that the food isn't good"
If you want to say uncertainty that the food might be bad it'll be something like

Siempre comen en esta cafetería aunque saben que a veces la comida no es muy buena.

Or

Siempre comen en esta cafetería aunque sepan que a veces la comida no es muy buena

The difference between the first one and the second one is that in the second you are saying that even if the subject (implicit) knows that the food is sometimes bad, they still eat there, but you aren't saying that all the subject knows it. In the first one, you are also saying that all the subject knows it.

Hope it helps you

Oh I see now. I was considering the word [i]es[/i] too much, neglecting the word [i]saben[/i]. Thanks.


rotation
real analysis
real analysis theorems

Hi..all  i've a question

For $P \in \mathbb{R}^{2}$ we define the set $\mathcal{C}_{(d,r,P)}=\{x \in \mathbb{R}^{2}| d(x,P)=r \}$ to be the [i]circle with center $P$ and radius $r$ relative to metric $d$ in $\mathbb{R}^{2}$[/i]

The metric that we are usually use is Euclidian
that is $\forall a=(x_{1},y_{1}) ,b= (x_{2},y_{2}) \in \mathbb{R}^{2}$
\[d(a,b)=\sqrt{(x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2}}\]

in this case the graph of $\mathcal{C}_{(d,r,P)}$ in euclidean plane is an usual circle.

my question: is there any way that we can define a $d$ such that  when we graph $\mathcal{C}_{(d,r,P)}$ in euclidean plane we will have a square?
thx.

Sure. For a square with sides parallel to the axis, use

$d_{\infty}(a,b)=\max(|x_{1}-x_{2}|,|y_{1}-y_{2}|).$

And for a "diamond" (a square rotated $45^{\circ}$), use

$d_{1}(a,b)=|x_{1}-x_{2}|+|y_{1}-y_{2}|.$


analytic geometry
geometry
search
number theory unsolved
number theory

Let P be such a polygon in (x,y) coordinate plane that all its vertices have integers as both its coordinates. Let I(P) be a number of points with integer coordinates INSIDE of the polygon. Let B(P) be a number of points with integer coordinates ON THE BORDER of the polygon. Prove that its area is equal to I(P) + B(P)/2 -1.


Does anyone know where I can get Mathcounts competitions earlier than 2001?

Buy 'em off [url=http://www.mathcounts.org]MathCounts.[/url]

I think any other way is illegal.

However, your school may let you borrow their copy.

It really sucks though, for people whoe go to bad schools. You can't buy competitions any earlier than 1999.

Yes, that certainly makes it harder for schools that are just getting started as well.  It would be great if they were willing to put copies of the older competitions, that they no longer sell, online (or allow others to do so).

i heard they gave out CDs in some years with all of the national problems.


ok...i'm just wondering...for you people that are national Mathcounts competitors and/or perform very well in Mathcounts/other competitions such as the AMC/ARML, how do you manage your time?....do you guys practice competition math every day and about how much?...do you guys still manage to get good grades in school and keep up with your homework?...i would like as much input as possible..thanks :)

Last year, I just did test after test and I got 205.

For nationals this year, I would do one year's sprint (without time until near the competition), target and team.  I would copy all of the problems I missed into a notebook and later would try to figure out a quick solution.  I jumped up 90 places.

I hope this information is useful.  PM me if you have any questions

[/i]

this is helpful information...but did it affect your school life/homework in anyways?...about how much time a day would you spend studying...?

I didn't do much at all, I did two target rounds total between states and nationals. I got 78th.

This year, my first year, I did a few tests and studied a few theorms for around 30 mins to an hour a day.
And I didn't make the team.  :(

The simple answer: practice, practice, practice.

If you get something wrong, look at your solving process.  Check every detail.  If you just can't figure out what's wrong then you should seek help.  The last resort is to look in the solutions manual.

a lot of people think that you need to spend like 2/3 hours a day to do really good in nationals, amc, aime, etc.

but its really about making sure you know your fundamentals. and its always a good thing if you don't make stupid mistakes or get the same concepts wrong again. if you notice any kind of pattern in your mistakes (or even problems you cant solve) focus on those things before you go to something else.

i usually do about like an hour to thirty mins. a day. but its not always about doing all the tests from previous years.

and for the record, i maintain a 4.0 gpa and still was the amc 10 and 12 school winner. i also have a bunch of other mathcounts achievements.

It really depends on how much of a social life you have and how much busy work your teachers give. Work math around those two, not the other way around.


If the middle two digits of 9567 were interchanged, how much larger would the resulting number be?

Interchange the two middle digits and get 9657.
9657-9567=90.


When will the solutions be coming in?

Also, more importantly, when will the new questions?

[quote="Peter VDD"]it's been answered yet in announcements yet, but... what do you mean by "year 16" in your topic title? :) this is only gonna be the 4th edition :P[/quote]

I think he's confusing the MathLinks contest with the USAMTS.  Since he's an intermediate-topics-poster, I'd guess he's looking for USAMTS, in which case this should be moved there.

The problems should be corrected by the end of Oct.?