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How can we prove that the sequence :$ S_{n} \equal{} (\minus{}1)^n$ diverges ?

$ S_n$ alternates between the values $ 1$ and $ \minus{}1$. If it were to converge to some value, it would have to be one of those, surely. However, if we take say $ 1$ as our proposed limit, we can for each fixed integer $ N$ find an $ n \geq N$ such that $ |S_n \minus{} 1| \equal{} 2$ (namely if $ N$ is even, we can take $ N \plus{} 1$, and if $ N$ is odd, we can take $ N$ itself).

So $ \exists \epsilon > 0 \, : \, \forall N \in \mathbb{N} \, : \, \exists n \geq N \, : \, |S_n \minus{} 1| > \epsilon$, namely $ \epsilon \equal{} 1$, say. So $ S_n$ diverges.

[quote="Tyl"]$ S_n$ alternates between the values $ 1$ and $ \minus{} 1$. If it were to converge to some value, it would have to be one of those, surely. However, if we take say $ 1$ as our proposed limit, we can for each fixed integer $ N$ find an $ n \geq N$ such that $ |S_n \minus{} 1| \equal{} 2$ (namely if $ N$ is even, we can take $ N \plus{} 1$, and if $ N$ is odd, we can take $ N$ itself).

So $ \exists \epsilon > 0 \, : \, \forall N \in \mathbb{N} \, : \, \exists n \geq N \, : \, |S_n \minus{} 1| > \epsilon$, namely $ \epsilon \equal{} 1$, say. So $ S_n$ diverges.[/quote].

Thanks .

But when we say that a sequence converges we mean that: there exists a No ,m such for all , ε>0 ...e.t.c ,e.t.c

So surely if the sequence does not converge we mean that : [b] for all real Nos ,m[/b] there exists ε>0 ....e.t.c ,e.t.c

Or perhaps there a theorem or a definition that says :

if a sequence vibrates between two Nos then the only possible limits of the sequence are those two Nos

Am I correct ??

[quote="JBL"][quote="stauros"]But when we say that a sequence converges we mean that: there exists a No ,m such for all , ε>0 ...e.t.c ,e.t.c[/quote]  This is wrong.  The condition you've written here is that of being "eventually constant."  Order of quantifiers is very important!  For convergence you choose $ \varepsilon$ first, then afterwards you choose $ N$.  (By the way, please learn a little bit of [[LaTeX]]; see also http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690 .  Also, it's "etc." (which is an abbreviation for the Latin "et cetera"), not "e.t.c.")

Tyl's proof is not very convincing because the "surely" part.  (I mean, that statement is correct, but it's not more fundamental than the result we're proving right now.)  It's really not hard to break this into three cases, though: show $ 1$ is not the limit, show $ \minus{} 1$ is not the limit, and show that any point other than $ 1$ and $ \minus{} 1$ is not the limit.  Knowing what the definition of convergence actually says will be helpful  :wink:[/quote]

Why show  first that ,1 is not the limit ,then show that, - 1 is not the limit and then show that any other point than ,1 and -1 is not the limit,[b]instead of showing that any No is  not the limit  straight away[/b].

After all by showing that the limit does not exist for any No that [b]would include the Nos 1, and -1[/b].

Any way it would be very much appreciated if you can give a complete proof ,or at least a summary of a proof.

Surely quantifiers is not mathematics,using plain English words we can express every formula in mathematics quite clearly.

[quote="stauros"]Surely quantifiers is not mathematics,using plain English words we can express every formula in mathematics quite clearly.[/quote]  I have no idea what this means, but "For all $ \epsilon > 0$, there exists $ N \in \textbf{N}$ such that blahblahblah" and "There exists $ N \in \textbf{N}$ such that for all $ \epsilon > 0$, blahblahblah" mean very different things.  You wrote one, and the definition of convergence includes the other.  If you don't understand what this means or why it's important, you have a more fundamental problem than solving this question.  (It's also not true that "plain English" can express every formula in mathematics quite clearly.  A notion like equicontinuity (which might be found in the later part of an intro analysis course that includes the question you've given here at its beginning) is just inherently complicated to express.)

The function $ a_n \equal{} (\minus{}1)^n$ is essentially a piecewise function -- it's completely natural that a proof should consider different cases.  The function has a subsequence that converges to a certain value, so it's not at all surprising that one might have to consider this value in a fundamentally different way than other values.

Why don't you describe, in plain English, what you think the proof should look like, and we can go from there?

[quote="JBL"].  (It's also not true that "plain English" can express every formula in mathematics quite clearly.   ).[/quote]

Write any formula you like and i will express it in plain English.

For example the formula : $ \forall\epsilon: \epsilon>0\Longrightarrow\exists N( N\in N\wedge\forall n( n\geq N\Longrightarrow |s_{n}|<\epsilon))$ ,can be expressed in plain English as follows:

Given ε>0 we can find a natural No N such that : if $ n\geq N$ ,then $ |s_{n}|<\epsilon$

[quote="JBL"] The function $ a_n \equal{} ( \minus{} 1)^n$ is essentially a piecewise function -- it's completely natural that a proof should consider different cases.[/quote]

What is the meaning of the word "natural" in mathematics

[quote="JBL"] Why don't you describe, in plain English, what you think the proof should look like, and we can go from there?[/quote]

That i really do not understand. You mean that if somebody suggests something which is wrong ,must i except it??

I wish  i knew what the proof should look like ,i would not have asked in the first place

Oh, so is this like the opposite of those [url=http://www.artofproblemsolving.com/Forum/search.php?search_author=%CF%84%CF%81%CE%B9%CE%BA%CE%BB%CE%B9%CE%BD%CE%BF]τρικλινο[/url] threads?

The way the definition of the limit is structured, we really can't just prove all at once that there is no limit; we have to prove that for each real $ c$, the limit is not $ c$. For this sequence, that goes in two parts: (a) If $ c\neq 1$, the limit isn't $ c$ because those values of $ 1$ are too far away. (b) If $ c\neq \minus{}1$, the limit isn't $ c$ because those values of $ \minus{}1$ are too far away.

One of the key challenges in using quantifiers, whether symbolically or as words, is knowing how to negate a statement with quantifiers in it. Can you write out a precise definition of the following by negating the $ \epsilon$-$ \delta$ definition of a limit?
The sequence $ a_n$ does not converge to $ a$.

You can argue via contradiction.Assume that the limit exists and is equal to $ l$.Then any subsequence of the sequence also converges to $ l$, but the sub-sequences $ a_{k}\equal{}1$ and $ b_{k}\equal{}\minus{}1$ converge to $ 1$ and $ \minus{}1$ so the sequence doesn't converge

[quote="jmerry"]Oh, so is this like the opposite of those [url=http://www.artofproblemsolving.com/Forum/search.php?search_author=%CF%84%CF%81%CE%B9%CE%BA%CE%BB%CE%B9%CE%BD%CE%BF]τρικλινο[/url] threads?[/quote]

I had a short run over the guy's posts and i do not blame you having nightmares in discovering the "Dr Jekyl" of the guy.

Al thought at the beginning i took your remark as an insult

[quote="jmerry"]The way the definition of the limit is structured, we really can't just prove all at once that there is no limit; we have to prove that for each real $ c$, the limit is not $ c$. For this sequence, that goes in two parts: (a) If $ c\neq 1$, the limit isn't $ c$ because those values of $ 1$ are too far away. (b) If $ c\neq \minus{} 1$, the limit isn't $ c$ because those values of $ \minus{} 1$ are too far away[/quote]

I really do not understand. What is the meaning of "too far away" in mathematics

[quote="jmerry"]One of the key challenges in using quantifiers, whether symbolically or as words, is knowing how to negate a statement with quantifiers in it. Can you write out a precise definition of the following by negating the $ \epsilon$-$ \delta$ definition of a limit?
The sequence $ a_n$ does not converge to $ a$.[/quote]

Here is the definition if your question is not rhetorical:

we can find an ε>0 such that given any natural No N we can find $ n\geq N$ such that $ |a_{n}\minus{}a|\geq\epsilon$

[quote="pardesi"]You can argue via contradiction.Assume that the limit exists and is equal to $ l$.Then any subsequence of the sequence also converges to $ l$, but the sub-sequences $ a_{k} \equal{} 1$ and $ b_{k} \equal{} \minus{} 1$ converge to $ 1$ and $ \minus{} 1$ so the sequence doesn't converge[/quote]

It is correct ,i think, but does not use the ε- definition.

[quote="stauros"]Here is the definition if your question is not rhetorical:

we can find an $ \epsilon>0$ such that given any natural No N we can find $ n\geq N$ such that $ |a_{n}\minus{}a|\geq\epsilon$[/quote]
It's not rhetorical, and that's right. Now, apply it to this problem; if $ a\neq 1$ we can choose $ \epsilon\equal{}|a\minus{}1|$, and if $ a\neq \minus{}1$ we can choose $ \epsilon\equal{}|a\plus{}1|$.

One other convenient variation; "for any $ N$, there is $ n\ge N$" is equivalent to "for infinitely many $ n$".

[quote="jmerry"][quote="stauros"]Here is the definition if your question is not rhetorical:

we can find an $ \epsilon > 0$ such that given any natural No N we can find $ n\geq N$ such that $ |a_{n} \minus{} a|\geq\epsilon$[/quote]
It's not rhetorical, and that's right. Now, apply it to this problem; if $ a\neq 1$ we can choose $ \epsilon \equal{} |a \minus{} 1|$, and if $ a\neq \minus{} 1$ we can choose $ \epsilon \equal{} |a \plus{} 1|$.

One other convenient variation; "for any $ N$, there is $ n\ge N$" is equivalent to "for infinitely many $ n$".[/quote]

If you take $ a\neq 1$ and then choose ε = |a-1| etc etc , you do not prove that $ s_{n}$ does not converge to a ,a being any No .

But if $ a\neq 1$ ,then $ a_{n}$ does not converge to a

Why insist  on the same wrong proof  ,it becomes tiresome  after a while

Every number is either not $ 1$ or not $ \minus{}1$. The full proof splits into cases based on that, as it has to.

[quote="JBL"]It's really not hard to break this into three cases, though: show 1 is not the limit, show - 1 is not the limit, and show that any point other than 1 and - 1 is not the limit[/quote]

[quote="jmerry"]Every number is either not $ 1$ or not $ \minus{} 1$. The full proof splits into cases based on that, as it has to.[/quote]

JBL dictated [b] 3 cases [/b] and you dictate [b] 2 cases [/b] .

Who is right and who is wrong .

I used the word "dictate" to indicate that none of you based their verdict on a well known fact ( law)

For example it is a well known fact that : $ a\neq 1\vee a\equal{}1$ , because it is based on the law of excluded middle

I agree -- jmerry's suggestion is more efficient than mine, but the basic principle (this sequence has two special points, $ 1$ and $ \minus{} 1$, and you therefore need some special consideration for these points) is exactly the same.  Both of us based our statements on the fact that we know how to solve the problem, are happy to give hints, but aren't going to  do your work for you.  If you want to solve the problem, spend your time thinking about the math and posting your mathematical thoughts -- complaining that others aren't doing it for you is not very interesting.

I see that my solution ignores a part of the proof by using the word "surely".

However it seems that the OP instead wants to talk about how to, in general, go about proving things using separate cases, instead of applying that to this problem. The former, however, has been discussed already so it seems we are done.

[quote="jmerry"]Who's right and who's wrong? I'm right, and JBL is also right. There is more than one way to organize the proof.[/quote]

Sorry i should have been more clear ,so

[quote="JBL"]It's really not hard to break this into three cases, though: show 1 is not the limit, show - 1 is not the limit, and show that any point other than 1 and - 1 is not the limit[/quote]

Meaning that JBL examines the cases:

 a= 1 , a= -1, $ a\neq 1$ ,$ a\neq \minus{}1$

[quote="jmerry"]Every number is either not $ 1$ or not $ \minus{} 1$. The full proof splits into cases based on that, as it has to.[/quote] .

Meaning you examine the cases :

 $ a\neq 1$ , $ a\neq \minus{}1$

So we see that JBL [b] apart from examining your cases he adds two extra ,namely a=1,a=-1[/b]

The question now is :

Since your two cases ,namely $ a\neq 1$ , $ a\neq \minus{}1$  [b]solve completely the problem[/b] the extra two of JBL what do they add to the proof??

And both jmerry and I have answered that question: there are different ways of writing the proof, and jmerry's suggestion was more efficient than mine.  Why you're wasting your time on this discussion instead of solving the problem is utterly beyond me.

One thing is for certain , that you do not know what the law proof by cases is and how it works in a proof.

So i suggest you learn it before you give confusing advise to people


What was the name of the rock band playing on the farewell party?

I don't really know.. The only thing that makes me interested is the Physikkanten  :D

hmm... I love the band.. especially the girl's voice was cool you know? hahah..
physikanten (or..?) is good too but his show was excellent just for first time only... but fantastic. When I checked in youtube, I found the same thing.

So, anyone can help? What's the band name?

BTW, in farewell party I was looking for UNK leader, Geoff Smith for photo. But he was lost! Aaaa... I'm fan.. haaha.
Disco in farewell party was pretty good.

Raja Oktovin


geometry
circumcircle
inradius
geometric transformation
trapezoid
parallelogram
rhombus

Let inversion in a circle $ (M)$ take the incircle $ (I)$ of a $ \triangle ABC$ into its circumcircle $ (O).$ It takes the triangle sidelines $ a, b, c$ into circles $ (P_a), (P_b), (P_c)$ with radii $ R_a, R_b, R_c$ intersecting at the inversion center $ M$ and pairwise on the incircle $ (I)$ and they are tangent to the circumcircle $ (O).$ $ r, R$ are the triangle inradius and circumradius. Show that

$ \frac {1}{R_a} \plus{} \frac {1}{R_b} \plus{} \frac {1}{R_c} \equal{} \frac {2}{r} \plus{} \frac {2}{R}$

Let $ k^2$ be  the power of the inversion taking  $ (I)$ into $ (O).$ It's center $ M$  is the insimilicenter $X_{55}$ of $(I) \sim (O).$ Let  $ n_a,n_b,n_c$ the distances from $ M$ to $BC,CA,AB.$ By inversion properties we get $ k^2 = 2n_a \cdot R_a=2n_b \cdot R_b=2n_c \cdot R_c$

$ \Longrightarrow \  \frac {1}{R_a} + \frac {1}{R_b} + \frac {1}{R_c} = \frac {2(n_a + n_b + n_c)}{k^2} \ \ (1)$

On the other hand, we have:

$ \frac {R}{r} = \frac {k^2}{r^2 - IM^2} \ , \   \frac {OI}{IM} = \frac {R + r}{r}  \Longrightarrow \ \frac {R}{r} = \frac {k^2}{r^2 - \frac {r^2(R^2 - 2Rr)}{(R + r)^2}}$

$ \Longrightarrow \ k^2 = \frac {Rr(r^2 + 4Rr)}{(R + r)^2} \ \ (2)$

Assume that $ \triangle ABC$ is acute, the remaining case is treated analogously with appropriate choice of signs. Let $ d_a,d_b,d_c$ be the distances from $ O$ to $ BC,CA,AB.$ If $(I)$ touches $ BC$ at $ X$ and $ M_a$ is the midpoint of $ BC,$ then $ OIXM_a$ is a right trapezoid and $ n_a$ is a parallel to its bases

$\Longrightarrow \ n_a = \frac {r \cdot OM + d_a \cdot IM}{IO}$

Adding cyclic expressions, keeping in mind that $ d_a + d_b + d_c = R + r,$ we get:

$ n_a + n_b + n_c = \frac {3r \cdot OM + (R + r)IM}{IO} = 3r + \frac {IM}{IO}(R - 2r)$

$\ n_a + n_b + n_c = 3r + r\frac {R - 2r}{R + r}=\frac {r^2 + 4Rr}{R + r} \ \ (3)$

Combining $ (1), (2), (3)$ we obtain:

$ \frac {1}{R_a} + \frac {1}{R_b} + \frac {1}{R_c} = \frac {2(r^2 + 4Rr)}{R + r}\frac {(R + r)^2}{Rr(r^2 + 4Rr)} = \frac {2(R + r)}{Rr}$

${\frac {1}{R_a} + \frac {1}{R_b} + \frac {1}{R_c} = 2 \left( \frac {1}{R} + \frac {1}{r}\right)}$

Thanks for the solution. This is what I did:

Inversion center $ M$ is similarity center of a circle and its image. As the incircle $ (I)$ is inside the circumcircle $ (O),$ $ M$ is their internal similarity center. If $ d \equal{} \overline{OI},$ then $ \overline{OM} \equal{} \frac {Rd}{R \plus{} r},\ \overline{MI} \equal{} \frac {rd}{R \plus{} r}.$ Let the ray $ (OI$ cut $ (O), (I)$ at $ U, V.$ Using $ d^2 \equal{} R^2 \minus{} 2Rr,$ power of inversion is

$ k^2 \equal{} \overline{MU} \cdot \overline{MV} \equal{} (R \minus{} \overline{OM}) (r \plus{} \overline{MI}) \equal{} Rr \minus{} \overline{OM} \cdot \overline{MI} \equal{}$ $ Rr \minus{} \frac {Rrd^2}{(R \plus{} r)^2} \equal{} \frac {Rr^2(4R \plus{} r)}{(R \plus{} r)^2}.$

Internal and external bisectors of $ \angle A, \angle B, \angle C$ cut $ BC, CA, AB$ at $ D, E, F$ and $ X, Y, Z$, respectively. Then $ XEF, YFD, ZDE, XYZ$ are collinear. Let $ p_a, p_b, p_c$ be directed distances of a point $ P$ from $ BC, CA, AB.$ $ p_a$ is positive when $ P$ is on the same side of $ BC$ as $ A$ and negative otherwise; signs of $ p_b, p_c$ are defined cyclically. When $ P \in XEF,$ then $ p_b \plus{} p_c \minus{} p_a \equal{} 0$ and cyclic exchange. This is well known, was posted before, the proof was algebraic, but I cannot find it. Anyway, let $ P \in XEF$ and let perpendicular to $ AID$ through $ P$ cut $ CA$ at $ Q.$ Let $ q_c$ be directed distance of $ Q$ from $ AB.$ Let parallel to $ BC$ through $ P$ cut $ AB$ at $ C'$ and let parallel to $ AB$ through $ Q$ cut $ BC$ at $ A',$ Let $ C'P, A'Q$ meet at $ S.$ $ \triangle ABX \sim \triangle QPS$ are centrally similar with center $ E,$ hence $ BES$ are collinear, parallelogram $ BA'SC'$ is a rhombus and $ p_b \plus{} p_c \equal{} q_c \equal{} p_a.$ On the other hand, if $ P \in XYZ,$ then $ p_a \plus{} p_b \plus{} p_c \equal{} 0$ (just a sign change) and the proof is exactly the same.

Let $ I_a, I_b, I_c$ be excenters of the $ \triangle ABC$ and $ (O_e)$ circumcircle of the excentral $ \triangle I_aI_bI_c.$ $ I$ is its orthocenter, $ (O)$ its 9-point circle, $ O_eOI$ its Euler line. $ XYZ \perp O_eOI$ is radical axis of $ (O_e), (O),$ let $ K \equiv O_eOI \cap XYZ.$ Let $ P$ be arbitrary point, let parallel $ p$ to $ XYZ$ through $ P$ cut $ O_eOI$ at $ K'.$ Let the lines $ IX, IY, IZ$ cut the parallel $ p \parallel XYZ$ at $ X', Y', Z'.$ Parallels to $ BC, CA, AB$ through $ X', Y', Z',$ respectively, pairwise intersect at $ A', B', C'.$ Triangles $ \triangle A'B'C' \sim \triangle ABC$ are centrally similar with similarity center $ I$ and coefficient $ \kappa \equal{} \frac {\overline{IK'}}{\overline{IK}}.$ $ A'X', B'Y', C'Z'$ are external bisectors of the angles $ \angle A', \angle B', \angle C'.$ Let $ p_a', p_b', p_c'$ be directed distances of $ P$ from $ B'C', C'A', A'B'.$ Since $ P \in X'Y'Z',$ $ p_a' \plus{} p_b' \plus{} p_c' \equal{} 0$ and

$ p_a \plus{} p_b \plus{} p_c \equal{} p_a' \plus{} p_b' \plus{} p_c' \plus{} 3r(1 \minus{} \kappa) \equal{} 3r(1 \minus{} \kappa).$

When $ P \equiv K' \equiv M$ (the inversion center), which is inside the incircle $ (I),$ the directed distances $ n_a, n_b, n_c$ of $ M$ from $ BC, CA, AB$ are all positive and  $ n_a \plus{} n_b \plus{} n_c \equal{} 3r(1 \minus{} \mu),$ where $ \mu \equal{} \frac {\overline{IM}}{\overline{IK}} \equal{} \minus{} \frac {\overline{MI}}{\overline{IK}}.$ Powers of $ K$ to $ (O_e), (O)$ are equal,

$ (\overline{O_eO} \plus{} \overline{OK})^2 \minus{} 4R^2 \equal{} \overline{OK}^2 \minus{} R^2,$

$ d^2 \plus{} 2 d \cdot \overline{OK} \equal{} 3R^2\ \Longrightarrow \overline{IK} \equal{} \overline{OK} \minus{} \overline{OI} \equal{} \frac {3R^2 \minus{} d^2}{2d} \minus{} d \equal{} \frac {3Rr}{d},$

$ 3r(1 \minus{} \mu) \equal{} 3r\left( 1 \plus{} \frac {rd}{R \plus{} r} \cdot \frac {d}{3Rr}\right) \equal{} \frac {r(4R \plus{} r)}{R \plus{} r}$

$ \frac {1}{R_a} \plus{} \frac {1}{R_b} \plus{} \frac {1}{R_c} \equal{} \frac {2(n_a \plus{} n_b \plus{} n_c)}{k^2} \equal{} \frac {2(R \plus{} r)}{Rr} \equal{} \frac {2}{r} \plus{} \frac {2}{R}.$

[quote="yetti"]Let $ p_a, p_b, p_c$ be directed distances of a point $ P$ from $ BC, CA, AB.$ $ p_a$ is positive when $ P$ is on the same side of $ BC$ as $ A$ and negative otherwise; signs of $ p_b, p_c$ are defined cyclically. When $ P \in XEF,$ then $ p_b \plus{} p_c \minus{} p_a \equal{} 0$ and cyclic exchange. This is well known, was posted before, the proof was algebraic, but I cannot find it.[/quote]
It also follows from the trilinear equation of $\overline{XEF},$ namely $\beta+\gamma-\alpha=0.$


linear algebra
matrix
number theory unsolved
number theory

Let $ M_n$ be the number of integers N such that
(a)$ 0\leq N < 10^n$
(b)N is divisible by 4.
(c)The sum of digits of N is also divisible by 4.
Prove that  $ M_n\not \equal{} \frac {10^n}{16}$ for all positive integers  .

Let $ S(n)$ sum of digits. Let $ M_i(n)$ is number of integers $ 0\le N<10^n$, suth that
$ 4|N, 4|S(N)\plus{}i, i\equal{}0,1,2,3.$
Obviosly $ M_0(1) \equal{}3, M_1(1)\equal{}2,M_2(1)\equal{}2,M_3(1)\equal{}3$ and
$ M_0(n)\equal{}3M_0(n\minus{}1)\plus{}3M_1(n\minus{}1)\plus{}2M_2(n\minus{}1)\plus{}2M_3(n\minus{}1)$,
$ M_1(n)\equal{}2M_0(n\minus{}1)\plus{}3M_1(n\minus{}1)\plus{}3M_2(n\minus{}1)\plus{}2M_3(n\minus{}1)$,
$ M_2(n)\equal{}2M_0(n\minus{}1)\plus{}2M_1(n\minus{}1)\plus{}3M_2(n\minus{}1)\plus{}3M_3(n\minus{}1)$,
$ M_3(n)\equal{}3M_0(n\minus{}1)\plus{}2M_1(n\minus{}1)\plus{}2M_2(n\minus{}1)\plus{}3M_3(n\minus{}1)$ or for  $ X_n\equal{}(M_0(n),M_1(n),M_2(n),M_3(n))^T$
$ X_n\equal{}AX_{n\minus{}1}.$
Therefore $ M_i(n)$ can be calculate as $ (M_0(n),M_1(n),M_2(n),M_3(n))^T\equal{}A^{n\minus{}1}(3,2,2,3)^T,$ were A is matrix 4*4.
We can consider A by mod 5 and prove, that $ 5\not|M_0(n)\equal{}M_n$.


superior algebra
superior algebra unsolved

Let $ A$ be a unitary ring with $ p^2$ elements,where $ p$ is a prime number.so that exist's maximum $ p\minus{}2$ two-sided divisors of zero.Prove that A is a corp.

This problem is from IMC Shortlist 2007, isn't it? Bogdanl_yex, where did you find this shortlist?  :?:  Say, please!

I found it in to the RMO SHORTLIST 2007....sorry if  I dissapoint you...Do you have any solution?

If $ x\neq0$ is divisor of zero, then $ x,2x,\dots,(p \minus{} 1)x$ are $ p \minus{} 1$ different numbers that are divisors of zero. So $ A$ is a finite division ring, so it is field.


$ \frac{s^2}{2p}\equal{}r$ so you have the incircle and the A-excircle...these problems are all the same...


Let $a,b,c$ be real numbers such that $\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}=|\frac{1}{a}+\frac{1}{b}+\frac{1}{c}|$.
Prove $f(x)=ax^2+bx+c$ passes through a fixed point.

Just square the both sides of the given equation. :lol:

[quote="fourierseries"]I think that you are using the equality case of the Cauchy Schwarz inequality, but a,b,c do not satisfy the first expression when they're both 0.[/quote]
Cauchy Inequality gives: $\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \leq \sqrt{3} \cdot \sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}$  :)


Could we do some national mathcounts problems? thanks

does sugarcookie perhaps live in okie-homa and perhaps is heading to the 7th grade and perhaps goes by the name amy tai?

just wondering

We have done a few Mathcounts Math Jams and the transcripts are up in the Math Jams area of the site.  Don't worry, we will frequently be doing Mathcounts problems in some Math Jams and we are discussing having a more regular course specifically taylored toward Mathcounts level material.

We also hope to offer an online course this fall that is very similar to a string of MATHCOUNTS math jams organized by subject (this will be similar to our current classes, except that in lieu of homeworks and multiple enrollment options, we'll be using the message boards to spread problems).  These will be specifically designed to prepare for MATHCOUNTS. . .

[quote="sugarcookie34"]no to syntax error i have never heard of... amy tia?

sugarcooki34[/quote]

funny how you misspelled your own name. both of em


topology
abstract algebra
geometry
3D geometry
advanced fields
advanced fields unsolved

How do I show that the higher homotopy groups $\pi_{n}(X,x_{0}), n \geq 2$ are abelian?

I can't add anything to the explanation in the [url=http://www.math.cornell.edu/~hatcher/AT/ATpage.html]Hatcher's book[/url] (which you definitely should download if you are studying algebraic topology).

could you write the solution completely?

well, let's say that if $n>1$ you have room enough in $[0,1]^{n}$ to exchange with continuity $\{x_{1}<1/2\}$ with $\{x_{1}>1/2\}$, and to turn the "cube around".
i think that if one has got the point, he/she has no difficulty to translate it in formulas..


I'm looking to buy used Statistics text books. I'm looking for any good statistics textbooks as a base of groundwork. People here probably have some leftovers from high school, since y'all are going to college now. Can anyone point me in the direction of some? :)

Fundamentals of Mathematical Statistic by Nguyen,Rogers is an excellent book, for (college) beginners. There are two part, first one covers probability, and the second covers statistic (so keep an eye on the second part). It's also recommended literature for most statistic's classes.
Other excellent book is Mathematical Statistic by Shao ( I learned statistic from this book)  but it's intended for Phd student's, and it presumes that you know measure theory. But every thing you need to know about measure theory is covered in the first chapter.
Both books have plenty of problems,examples and explanations...

For AP statistics, start with the four books from the "fab five" mentioned in this recent thread

http://mathforum.org/kb/message.jspa?messageID=5849006&tstart=0

from the AP statistics teachers electronic discussion group. Especially check the article

http://www.statland.org/MAAFIXED.PDF

for what to look for in a beginning statistics book.


For those of you, which math camps/classes are you taking or have you recently taken?? Online classes and camps are the ones i am curious about.

[quote="gauss1181"]I've never taken any online math classes or camps. :([/quote]

Same here. :(

hmmm, interesting, 50% of the people are taking WOOT....


suppose that $tan(2 \pi x ) \in \mathbb{Q} - \{ -1; 0; +1\}$
Prove that $x$ is irrational.

It is classical problem about algebraic integers. I think we have already discussed several times such problems, Alekk. ;)


rotation
combinatorics unsolved
combinatorics

n lighthouses are arbitarily placed in the plane. Each has a stationary lamp which illuminates an angle of 360/n degrees. Prove that the lamps can be rotated so that at least one lamp is visible from every point of the plane.

this problem is generalized from All Soviet Union Math Competitions of year 1967 and the original problem was n=4.

The well known generalization (in Russia) is following.

Suppose i-th lighthouse illuminates angle ai and a1+a2+...+an=360. Prove that the lamps can be rotated so that at least one lamp is visible from every point of the plane.

It is very hard problem.

has it been solved?
do you have a solution or do you know where can i find some paper about this problem?

Of cause, it was solved. Solution is long enough, so I don't want to write it. Published solution only in Russian.

There are more strong statement.

Suppose we have n lamps with common vertex which illuminate whole plane. Then it is possible to shift them by mean parallel shift these lamps to given n lighthouses such way they still illuminate whole plane.

It is clear there are n! distinct shifts. The main idea of solution is to calculate for each shift special number (we may call it as mesuare of covering of the plane). And if current shift is not required (i.e. some parts of plane are uncovered) then it is possible to construct new shift with increased mesuare of covering. Since there are finite number of distinct shifts we will find required shift in some steps.

Гальперин В., Гальперин Г., Освещение плоскости прожекторами. (Квант №11, 1981)
http://kvant.mccme.ru/1981/11/osveshchenie_ploskosti_prozhek.htm

Topic is closed (?).

I do not understand russian language  :(  :(  :(

Pierre.

It is your problem, Pierre!  :lol:  :lol:  :lol:
Welcome to Russia!

I was sure that you would laugh  :(  :(  :(

Pierre.

Since the solution is in Russian, I think few people consider the topic closed :D.

Hmmm...that you write it here?  :P

Pierre.

[quote="Omid Hatami"]Could you translate the article in english, PLEASE ? ;)  ;)  ;)[/quote]If I want to preserve a common opinion that I am an inhuman evil guy, I shouldn't do it.

[quote="Myth"]If I want to preserve a common opinion that I am an inhuman evil guy, I shouldn't do it.[/quote]

Myth, don't try to "preserve" that opinion, because none of us happen to think that way. :P

[quote] Translated and texed by Myth[/quote]

So, Myth has an heart too  :roll:

Pierre.


calculus
integration
trigonometry
logarithms
calculus computations

Calculate the following indefinite integrals.

[1] $\int \sin (\ln x)dx$

[2] $\int \frac{x+\sin ^ 2 x}{x\sin ^ 2 x}dx$

[3] $\int \frac{x^3}{x^2+1}dx$

[4] $\int \frac{x^2}{\sqrt{2x-1}}dx$

[5] $\int \frac{x+\cos 2x +1}{x\cos ^ 2 x}dx$

[quote="amirhtlusa"]1
let $u=\ln{x}$ then integrate by parts
ans:
$\frac{x(\sin{(\ln{|x|})}-\cos({\ln{|x|}}))}{2}$[/quote]

You can omitt the sign of absolute value. :)

[quote="amirhtlusa"]2
$-\cot{x}+\ln{|x|}$[/quote]
 :)

[quote="litoxe2718281828"]4.
ok, my first answer was right:
$\left(\frac{x^2}{5}+\frac{2x}{15}+\frac{2}{15}\right)\sqrt{2x-1}$[/quote]

 :)


i don't know if it's in the right forum...  :?  although i think it's doable for most people who come here... (and yes: doable is an existing word :D )

Prove that for every point which lies inside an equilateral triangle, the sum of the distances to each side is the same as the height of the triangle.

This is a lot like another problem:
http://artofproblemsolving.com/Forum/viewtopic.php?t=41331

arrgg, sorry  :blush:
i don't go to 'getting started' that much :?

a mod can lock/delete this...


Hi folks,

as you can see I am posting ISL/ILL problems in order to see your solutions. I want to indicate that a problem has at least one solution (although I always appreciate further solutions). Therefore I want to integrate a SOLVED symbol  (for an example look at the symbol below). Please provide your suggestions for such a symbol. Not matter whether "self-made" or found onto the web. In order to integrate it as emoticon on the forum (due to convenience matters)  the symbol must be sufficiently small. Thanks.

Regards,
Orlando

[quote="Peter VDD"]I don't think we should copy the one from Kalva though :maybe:

Hey you know, http://www.cooltext.com is really good for making logos like this, in any style you want. :)[/quote]

Hi Peter,

did I say I want to copy it ? It is just for you that you know what I meant.  :)
My suggestions are here:


Just came back to school.. and I realize I've forgotten so many things! (I also changed schools, so some of these things I might of not learned or it looks different to the ones I've learned). ARGG! >_< So if anyone can, I would really appreciate it you could help and show/explain me how to do these problems!

1) An element consists of 1.40% of an isotope with mass 203.973 amu, 24.10% of an isotope with mass 205.9745 amu, 22.10% of an isotope with mass 206.959 amu, and 52.40% of an isotope with mass 207.9766 amu. Calculate the average atomic mass and identify the elementy.

2) The element rhenium (Re) has two naturally occuring isotopes, 185Re and 187Re, with an average atomic mass of 186.207 amu. Rhenium is 62.60% 187Re, and the atomic mass of 187Re is 186.956 amu. Calculate the mass of 185Re.

3) A compound contains only carbon, hydrogen, and oxygen. Combustion of 10.68 mg of the compound yields 16.01 mg CO2 and 4.37 mg H2O. The molar mass of the compound is 176.1 g/mol. What are the empirical and molecular formulas of the compound.

I have a few more problems, but hopefully after I solve these I might be able to solve the others, if not I might post more up.  : :oops:

They look a bit easy. Might have made mistakes when I did them :blush: .

1. $203.973 \times 0.014+205.9745 \times 0.241+206.959 \times 0.221+207.9766 \times 0.524= 207.213 a.m.u (3d.p)$

2. $186.956 \times 0.626+m \times 0.374= 186.207 \rightarrow m= 184.953 a.m.u (3d.p)$

3. Empirical Formula: $C_{3}H_{4}O_{3}$
    Molecular Formula: $C_{6}H_{8}O_{6}$.

Might have made some error on Number 3.

[quote="Franz Joseph"]
3. Empirical Formula: $C_{3}H_{4}O_{3}$
    Molecular Formula: $C_{6}H_{8}O_{6}$.
[/quote]

Could you explain how you solved #3?

Thanks.

To solve the third problem you change the grams of carbon dioxide and water to moles of carbon and hydrogen (remember to multiply water by 2 to convert to hydrogen).  This only gives the mole ratio between carbon and hydrogen.  You have to subtract the added weight between the sample and products to remove the reacting oxygen since oxygen was added in the burning of the sample.

Since you know the molecular weight you can figure out if the simplest ratio of elements needs to be increased by some integer ratio.

This should be sufficient memory "guidance" for you to become an expert (again) on these types of problems.


inequalities
number theory
prime factorization

:maybe:  well the topic may say just for fun but this inequalities truly hold, sogive them a try.

 :)

[quote="ElChapin"]A]For all integers $ n\ge2$, let $ \sigma(n)$ be sum of the positive divisors of $ n$.

$ \sqrt {\sigma(n)\phi(n)} < n$

B]For all non prime positive integers $ n\neq1$, let $ d(n)$ be the number of positive divisors of $ n$.

$ 2\sqrt {\phi(n)d(n)} \le n$

Determine if the equality is possible.

 :)[/quote]
[hide="A"]
We take the prime factorization of $ n$, so $ n \equal{} p_1^{e_1}p_2^{e_2}...p_n^{e_n}$, where $ p_i$ is the $ i$th prime.  Thus, $ \phi (n) \equal{} n[\frac {(p_1 \minus{} 1)(p_2 \minus{} 1)...(p_n \minus{} 1)}{p_1p_2...p_n}]$ and $ \sigma (n) \equal{} \frac {(p_1^{e_1 \plus{} 1} \minus{} 1)(p_2^{e_2 \plus{} 1})...(p_n^{e_n \plus{} 1})}{(p_1 \minus{} 1)(p_2 \minus{} 1)...(p_n \minus{} 1)}$.  Hence,
\[ \sigma (n)\phi (n) \equal{} n\cdot \frac {(p_1^{e_1 \plus{} 1} \minus{} 1)(p_2^{e_2 \plus{} 1} \minus{} 1)...(p_n^{e_n \plus{} 1} \minus{} 1)}{p_1p_2...p_n} < n\cdot \frac {p_1^{e_1 \plus{} 1}p_2^{e_2 \plus{} 1}...p_n^{e_n \plus{} 1}}{p_1p_2p_3...p_n} \equal{} n^2\implies \sqrt {\sigma (n)\phi (n)} < n
\]
and we are done.
[/hide]

I think that B is false for $ n\equal{}2$ since $ \phi (2)\equal{}1$ and $ d(2)\equal{}2$, so $ 2\sqrt{\phi (n) d(n)}\equal{}2\sqrt{1\cdot 2}>2\equal{}n$.

Maybe it's supposed to be positive divisors less than $ n$.

Yes bassicaly that's the correction:

For every integer $ n > 1$ the inequality:

$ \frac {n}{2}\ge max\{{\sqrt {\phi(n)[d(n) \minus{} 1]}},\sqrt {d(n)[\phi(n) \minus{} 1]}\}$

 :)


function
analytic geometry
calculus
calculus computations

Let f be a function that is defined and twice differentiable for all real numbers x and has the following properties.

(i) f(0)=2
(ii) f'(x) > 0 for all x
(iii) The graph f is concave up for all x>0 and concave down for all x<0

Let g be the function defined by g(x)=f(x^2)

(a) Find g(0) [[[[I think its 2]]]]
(b) Find the x-coordinate of all minimum points of g. Justify your answer.
(c) Where is the graph of g concave up? Justify your answer.

(b) $ x\equal{}0$. Indeed, $ g$ is decreasing for $ x<0$ and increasing for $ x>0$. Use the chain rule to express $ g'$ in terms of $ f'$.
(c) everywhere. For example, use the chain rule to find a formula for $ g''$.

I'm sorry but I'm kind of confused. Could you be more specific?

HI
here's what mlok means to say...

1.g(x) = f(x^2) ......... (a)
put x=0
g(0)=f(0)=2

2.Differentiating eq (a) wrt x;
g'(x)= 2xf'(x^2)                                      .....(chain rule)
g''(x) = 2f'(x^2) + 4x^2f''(x^2)                  ....(product rule)
Clearly g''(x) > 0 (from definition of f(x))
Hence g'(x)=0 gives minima.
This implies the x co-ordinate of the required point is 0.

3.From 2, g''(x)>0 for all x.
Hence the  graph of g(x) is always concave up.

Hope that solves ur problem!!
Shyam


remainder of 5^17 + 7^19 when divided by 8. i wanna actually learn how to do this one.
thanks

How interesting...my problem of the week were like this: "What is 7^7^7, which is practically asking for the remainder when you divide 7^7 with 4.

Anyways, I suggest

a.) breaking it up into two parts: 5^17 and 7^19

b.) Actually try dividing the first couple of them such as 5 and 25 by 8 and see what happens or [hide]find a pattern[/hide]

c.) Think about how the remainder get's "added" when the two components (5^17 and 7^19) gets added together

Is this enough info? I'm not sure how far you got...oh well, good luck!

Addendum: Even more hints, but some of them are spoiler (and the last one is the answer)

d.) [hide]It repeats every two terms[/hide]

e.) [hide]Specifically, it goes 1,5,1,5... for the remainder of 5^x and 1,7,1,7... for the other[/hide]

d.) [hide]When adding remainders, use modular arithmetic (or just common sense  )[/hide]

e.) [hide]Remainder can't be greater than the divisor, so keep subtracting 8 from the sum until it is less than 8[/hide]

answer.) [hide]4[/hide]

OK, now try this one: What is the remainder when you divide 7^49+5^12+3^85+1^154 by 4?

Tare's whole post can be simplified to one word:

MODS

Mmm, I was trying to stay away from it in case he didn't know mods, I mean I'm guessing he would've solved it by now if she knew it...

If she doesn't know what they are... it's not that hard to explain what they are.

*sigh* OK OK, maybe I do suck at explaning things...*sigh* take it away Neal...

Allow me :)
Mod = remainder when you divide by something. So 5^17 (mod 8) is the remainder you get when you divide 5^17 by 8.

Now the nice thing about mods is that you can add or multiply. So we know that 5 = 5 (mod 8), multiply both sides by 5 and we get 5^2 = 25 = 1 (mod 8) since 25 is 1 more than a multiple of 8. Multiply by 5 again and we get 5^3 = 5 (mod 8). Etc.

By now you should be able to see what 5^17 is mod 8. Do the same with 7^19.

Good explanation, except it's :equiv: when you right modular equations. Example: 1:equiv:1 (mod 5) and read :equiv: as "congruence"...weird how this forum calls it "equiv" :arrow: "equivalence"...I mean, I learned all this from AoPS 1...hmm...

Yeah, I couldn't be bothered writing ":equiv:" all the time :-D. And its sorta read "is congruent to" not just "congruence". But anyway.. I think its equiv because that sign is also used for equivalence, like when you do logic and stuff.

need any more explaining? though i didnt do anything, but i didnt see this, but yeah

Whoops, sorry.
*awkward moment*
...
*"angels floating over the line"* (it means no one is writing anything because they're so  :shock:ed)
...
*flashback of "Speak" reading discussion group* ([url=http://www.amazon.com/exec/obidos/ASIN/014131088X/qid=1064690030/sr=2-1/ref=sr_2_1/103-6933976-0892612]Speak[/url]: A book by Laurie Halse Anderson about a girl who stops speaking, which leads to tons of trouble)
*gathers up the courage to speak again* (a phrase from the previously mentioned book)
Anyways, did you understand how to solve it?

Let me try to explain this without saying the MOD.
$5^{17} + 7^{19} = (5^2)^8*5 + (8-1)^{19}$
$=(24+1)^5*5+(8-1)(8-1)...(8-1)$
We can see the remainder is $5-1 =\boxed{4}$ now when divided by $8$.

[quote="bestwillcui1"]You do realize that this was posted ten years ago, so a solution probably wouldn't help the original poster by now.[/quote]
Only after I posted my post did I realize this is an old thread.


function
symmetry
algebra
functional equation
real analysis
real analysis unsolved

Find u : R -> R such that exists f : R -> R strictly monoton with: f(x + y) = f(x)u(y) + f(y), for each x, y in R.

$u(y) = 0\ \forall y$, $f(x) = 0$

$u(y) = 1\ \forall y$, $f(x)$ is any linear function.

:D

Now how about some $u$ that isn't flat? Consider $x = y = 0$. We see that $f(0) = f(0)u(0) + f(0)$, so either $f(0)$ or $u(0)$ is 0. We should have some symmetry in $x$ and $y$, so $f(x)u(y) + f(y) = f(y)u(x) + f(x)$, so $\frac{f(x)}{f(y)} = \frac{u(x)-1}{u(y)-1}$ whenever those expressions are meaningful (I already excluded the two cases where they never are).

I can see that $u(x) = kx + 1$ and $f(x) = cx$ satisfies these two conditions, but obviously not the original problem. What other conditions do we need? Or is this not a good approach?

f is given, u is unkown. So u should be written using f, somehow.
[u = 0 does not work, because f is strictly monoton].

[quote="mareleG"]f is given, u is unkown. So u should be written using f, somehow.[/quote]

There are clearly no such $u$ for arbitrary $f$, since for instance if $f(0)\neq 0$ it is easy to show that $f$ must be constant by making the substitution $y \to 0$.  So one must find also the limitations on $f$ or, as Xev was doing, find the pairs $(u, f)$.

Later edit:
Actually, this is not so hard at all.  By making the substitution $y\to 0$ we get $f(x) = f(x)u(0) + f(0)$ so $f(x)(1 - u(0)) = f(0)$ and either $f$ is constant or $f(0) = 0, u(0) = 1$.  By monotonicity, $f(x) = 0 \Leftrightarrow x = 0$.  Taking $(x, y) \to (-x, x)$ we get $0 = f(0) = f(-x)u(x) + f(x)$ so for $x \neq 0$, $u(x) = -\frac{f(x)}{f(-x)}$.

Of course, this now gives us a functional equation we have to solve for $f$.

Of course u and f 'work' together.

My computations give $u(x) = 1 - \frac{f(x)}{k}, k \in R$ and then $u(x) = b^x, b > 0$.
But i might be wrong :), please check this.

f(x + y) = f(y + x) means: f(x)(1 - u(y)) = f(y)(1 - u(x)) [1]

f(x + y + z) = f(x + y)u(z) + f(z) = [f(x)u(y) + f(y)]u(z) + f(z) = f(x)u(y)u(z) + f(y)u(z) + f(z)
f(x + y + z) = f(x)u(y + z) + f(y + z) = f(x)u(y + z) + f(y)u(z) + f(z)

From this, f(x) [u(y + z) - u(y)u(z)] = 0, $\forall x, y, z \in R$, f is strictly monoton, so u(y + z) = u(y)u(z) [2]

Let $A = \{x \in R | u(x) \neq 1\}$

If $A = \phi$, u = 1.
If not, from [1], $\frac{f(x)}{1 - u(x)} = k, \forall x \in A$

f(x + 0) = f(x)u(0) + f(0) [3]

From [2], $u(0) \in \{0, 1\}$
If u(0) = 0, from [3], f(x) = f(0), contradicts f strictly monoton.
u(0) = 1, from [3], f(0) = 0.

Let now $x_0 \in A, u(x_0) = 1, f(x + x_0) = f(x)u(x_0) + f(x_0) = f(x) + f(x_0), \forall x \in R$
If f(x) = 0, then $f(x + x_0) = f(x_0), x + x_0 = x_0, x = 0$
If x = 0, $f(0 + x_0) = f(0) + f(x_0), f(0) = 0$
f(x) = 0 iif x = 0.

From [1], $y = x_0, f(x_0)[1 - u(x)] = 0, \forall x \in R$
If $x_0 \neq 0$, u = 1 but we are in the case $A \neq \phi$
So $x_0 = 0$

We proved u(x) = 1 iif x = 0, ie $A = R^*, f(x) = k \cdot (1 - u(x)), \forall x \in R*$[this is true for 0 also].

$u(x) = 1 - \frac{f(x)}{k}, \forall x \in R$, u is strictly monoton.

Let $y \in R, x > 0: f(x + y) - f(y) = f(x) \cdot u(y)$
If f is increasing, $f(x) \cdot u(y) > 0$, $f(x) > f(0) = 0$, so $u(y) > 0$
The same for f decreasing.

We found $u(x) > 0, \forall x \in R$

From [2]: $log_a (u(x + y)) = log_a (u(x)) + log_a (u(y)), h(x + y) = h(x) + h(y)$[h is strictly monoton also]
It is a classical result that $h(x) = c \cdot x, u(x) = a^{c \cdot x} = b^x, b > 0$[not 1 for this case].

Finally, $u(x) = b^x, b > 0$


Slove the system of the equations: $6\left(x-\frac{1}{y}\right)=3\left(y-\frac{1}{z}\right)=2\left(z-\frac{1}{x}\right)=xyz-\frac{1}{xyz}$

[hide="Solution"]Put $a=xyz-{1\over xyz}$. Then $x-{1\over y}={a\over 6}, y-{1\over z}={a\over 3}, z-{1\over x}={a\over 2}$. Summing up those three we get

$x+y+z-{1\over x}-{1\over y}-{1\over z}=a$

Multiplying them we get

$\begin{aligned}{a^{3}\over 36}& =\left(xy-{x\over z}-1+{1\over yz}\right)\left(z-{1\over x\right)\\ & = xyz-y-x+{1\over z}-z+{1\over x}+{1\over y}-{1\over xyz}\\ & = a-a=0\end{aligned}}$

Therefore $a=0$, which gives $x={1\over y}, y={1\over z}, z={1\over x}$, i.e. $(x,y,z)\in\{(1,1,1),(-1,-1,-1)\}$[/hide]


$ p^2x^2 \plus{} q^2 y^2\minus{}qx\minus{}py\plus{} (\frac{p}{q}\minus{}\frac{q}{p})^2 \plus{}2 \equal{} 0, find (p^2x\plus{}q^2y)\minus{}p\minus{}q.$

Thank you

I guess, I posted it in the wrong forum.

This is what I have done.

$ p^2x^2 \plus{} q^2 y^2\minus{}qx\minus{}py\plus{} (\frac{p}{q}\minus{}\frac{q}{p})^2 \plus{}2 \equal{} 0$

or  $ (px)^2\plus{} (qy)^2 \minus{}qx\minus{}py \plus{}(\frac{p}{q})^2 \minus{} 2 \plus{}(\frac{q}{p})^2 \plus{} qx \plus{} py \plus{} 2$=0

or $ (px)^2 \minus{} 2qx \plus{}(\frac{q}{p})^2 \plus{} (qy)^2\minus{} 2py \plus{} (\frac{p}{q})^2 \plus{} qx \plus{} py$ = 0

or $ (px\minus{} \frac{q}{p})^2 \plus{} (qy\minus{} \frac{p}{q})^2 \plus{} qx \plus{}py \equal{}0$

or $ (\frac{p^2x \minus{} q}{p})^2 \plus{} (\frac{q^2y \minus{} p}{q})^2 \plus{} qx \plus{} py\plus{} \equal{} 0$ (I cannot proceed anymore)

I am sorry I did not post this earlier.

Thank you.

It's easy to see that we can't have all of $ x,y,p,q$ real by rewrite the equation like this :

$ (px\minus{}\frac{q}{2p})^2\plus{}(qy\minus{}\frac{p}{2q})^2\plus{}\frac{3}{4}(\frac{p^2}{q^2}\plus{}\frac{q^2}{p^2})\equal{}0$

or just simply look at the discriminant :

$ (q)^2\minus{}4(p^2)(q^2y^2\plus{}\frac{p^2}{q^2}\plus{}\frac{q^2}{p^2})$
$ \equal{}\frac{\minus{}3q^4\minus{}4p^2q^4y^2\minus{}4q^4}{q^2}$
$ <0$ for $ p,q \neq 0$ is real.

And it's easy to see that if $ p,q,x,y$ are not necessary real, then the value of $ (p^{2}x\plus{}q^{2}y)\minus{}p\minus{}q$ is not uniquely defined. To find the general term is easy (treat the equation as a quadratic equation and then solve for $ x$, then put the value of $ x$ into $ (p^{2}x\plus{}q^{2}y)\minus{}p\minus{}q$) but I think that's not what you want.

Thank you. But the sum was not really correct(I came to know this today).The original one was quite easy.Sorry for troubling you.


[quote="aditzu_k"]S-au afisat cumva rezultatele de la judet pe net?sa vad si eu cine s-a calificat![/quote]
   Incearca pe http://www.olimpiade.ro!


superior algebra
superior algebra solved

Let [b]I[/b] = <2>. Prove that $I[x]$ is not a maximal ideal of $Z[x]$ even though [b]I[/b] is a maximal ideal of [b]Z[/b]

$I[x] \subset <x,2> \subset \mathbb{Z}[x]$, and all these are not equal ($<x,2>$ means the ideal in $\mathbb{Z}[x]$ generated by $2$ and $x$).


Do you consider that very funny¿¿¿   :mad:  :spam:  :ewpu:  :chief:  :mad:

Stop:
- reposting
- attaching the problem as file

http://www.mathlinks.ro/Forum/viewtopic.php?t=90335
http://www.mathlinks.ro/Forum/viewtopic.php?t=90573 [mod edit: Topic deleted; it was a repetition of the previous one]
http://www.mathlinks.ro/Forum/viewtopic.php?t=90580 [mod edit: Topic deleted; it was a post containing "hi, can you help me?" without any problem given]
http://www.mathlinks.ro/Forum/viewtopic.php?t=90581 [mod edit: Topic deleted; it was a repetition of the first one]

And I don't think that crying for "helping [u]immdently[/u]" would help.


Find the least and greatest value of
P=sinx$ \sqrt{1\plus{}cosx}$+cosx$ \sqrt{1\plus{}sinx}$

[quote="HoTuyetNham"]Find the least and greatest value of
P=sinx$ \sqrt {1 \plus{} cosx}$+cosx$ \sqrt {1 \plus{} sinx}$[/quote]
We have:
$ P^2 \leq (sin^2x\plus{}cos^2x)(2\plus{}sinx\plus{}cosx) \leq 2\plus{}\sqrt{2}$
So $ \minus{}\sqrt{2\plus{}\sqrt{2}} \leq P \leq \sqrt{2\plus{}\sqrt{2}}$

The least value is false. You can calc where the equation


Three students, with different names, line up single file. What is
the probability that they are in alphabetical order from
front-to-back?  Express your answer as a common fraction.

the probability that the first child is in the right place is 1/3. the probability that the second is, after the first being, is 1/2. if those two are good, then the probability of the last child being in place is 1/1.

1/3*1/2=1/6

ANSWER:1/6

Well, easier way.

If you have a problem like this, the answer is always $ \frac{1}{n!}$.

It's not that all "problems like this" are answered by the formula $ \frac{1}{n!}$, but the formula certainly answers this question:  if a collection of $ n!$ events are equally likely to occur, what is the probability of one of them happening?

In this case, we knew there was only one way to line the people up one direction in alphabetical order.  However, if you were asked "what is the probability they are in alphabetical order" you'd have to recognize that there are two ways that could happen, and the solution would be $ 2\times\frac{1}{n!}$.


Am nevoie de o indicatie pentru:

Sa se verifice ca: sinxsin(x+pi/n)sin(x+2pi/n)sin(x+3pi/n)...sin[x+(n-1)pi/n] = sin(nx)/2^n-1

Multumesc

Polinomul \[ P(X)=X^n-(\cos 2nx+i\sin 2nx) \] are radacinile \[ x_k=\cos\left(2x+\frac{2k\pi}{n}\right)+i\sin\left(2x+\frac{2k\pi}{n}\right),\;\forall k=0,1,\ldots,n-1 \]
Rezulta ca
\[ X^n-(\cos 2nx+i\sin 2nx)=\prod\limits_{k=0}^{n-1}\left[X-\cos\left(2x+\frac{2k\pi}{n}\right)-i\sin\left(2x+\frac{2k\pi}{n}\right)\right] \].
Luam $X=1$ si obtinem
\[ 2\sin nx(\sin nx+i\cos nx)=2^n\prod\limits_{k=0}^{n-1}\sin\left(x+\frac{k\pi}{n}\right)\cdot \prod\limits_{k=0}^{n-1}\left[\sin\left(x+\frac{k\pi}{n}\right)+i\cos\left(x+\frac{k\pi}{n}\right)\right] \] si cum
\[ \prod\limits_{k=0}^{n-1}\left[\cos\left(x+\frac{k\pi}{n}\right)+i\sin\left(x+\frac{k\pi}{n}\right)\right]=(\cos nx+i\sin nx), \] prin simplificare obtinem \[ \sin nx=2^{n-1}\prod\limits_{k=0}^{n-1}\sin\left(x+\frac{k\pi}{n}\right), \] ceea ce trebiua demonstrat.

Este frumoasa solutia domnule Daniel, va raman profund indatorat!


$ AB = 20$ cm, $ m \angle A = 30^\circ$, and $ m \angle C = 45^\circ$. Express the number of centimeters in the length of $ \overline{BC}$ in simplest radical form.
[asy]import olympiad; size(200); import geometry_dev; import graph; defaultpen(linewidth(0.8));
pair A = origin, B = (10*sqrt(3),10), C = (10*sqrt(3) + 10,0);
draw(Label("$20$",align=NW),A--B); draw(B--C); draw(A--C);
markangle(Label("$30^{\circ}$"),B,A,C,radius=20);
markangle(Label("$45^{\circ}$"),B,C,A,radius=20);
label("$A$",A,W); label("$B$",B,N); label("$C$",C,E);[/asy]

Drop an altitude from $ B$ to $ AC$.  The length of the altitude is $ 10$ by $ 30\minus{}60\minus{}90$ triangle ratios, so $ BC\equal{}\boxed{10\sqrt{2}}$ by $ 45\minus{}45\minus{}90$ ratios.


group theory
abstract algebra
superior algebra
superior algebra unsolved

show that every group is isomorphic to a subgroup of $Gl_n(Z)$.

Do you mean a finitely generated group?  :?

oups  :blush:  .. it is: every finite group (this is why it is easy) is ...
I don't know the answer for the finitely generated group ??

Every finite group is a subgroup of some symmetric group, and the group of $n\times n$ permutation matrices is a subgroup of $GL_n(\mathbb Z)$.


geometry
3D geometry
pyramid
geometry unsolved

Given to a right pyramid of hexagonal base, we make to pass for the center of the base an Alpha plan that is parallel to the one lateral face. Which the reason enters the areas of the section and the face?
tkz : )


calculus
integration
trigonometry
algebra
polynomial
logarithms
calculus computations

Is there anyone can help me to solve this integral
$ \int\frac{3tanx\plus{}1}{2sin2x\minus{}5cos2x\plus{}1}$
pls    :lol:

hello, use the $ tan$ half angle substitution with $ t\equal{}\tan(\frac{x}{2})$ and the identities
$ \sin(x)\equal{}\frac{2t}{1\plus{}t^2}$
$ \cos(x)\equal{}\frac{1\minus{}t^2}{1\plus{}t^2}$
$ \tan(x)\equal{}\frac{2t}{1\minus{}t^2}$.
Sonnhard.

Using Sonhard's hint:

You're going to want to expand $ \sin 2x \equal{} 2 \sin x \cos x$ and $ \cos 2x \equal{} \cos^2 x \minus{} \sin^2 x$
Doing that, the integral becomes
$ \int \frac{3 \tan [x]}{1\plus{}4 \cos [x] \sin [x]\minus{}5 \left(\cos^2 [x]\minus{}\sin^2[x]\right)}$

Then, using the substitution
$ t \equal{} \tan{\frac{x}{2}}$, the integral is transformed:
$ \int \frac{3 \left(\frac{2t}{1\minus{}t^2}\right)}{1\plus{}4 \left(\frac{1\minus{}t^2}{1\plus{}t^2}\right) \left(\frac{2t}{1\plus{}t^2}\right)\minus{}5 \left(\left(\frac{1\minus{}t^2}{1\plus{}t^2}\right)^2\minus{}\left(\frac{2t}{1\plus{}t^2}\right)^2\right)}$

Wait.. I'm not actually sure if that simplifies.
Grah.

Even if it did simplify, you'd be left with a polynomial of degree six on the bottom, and one of degree five on top.

I don't think there's a general method to do those.

hello, after simplifying your integral i have got
$ \minus{}\frac{1}{4}\int\frac{(t^2\minus{}6t\minus{}1)(1\plus{}t^2)^2}{(\minus{}1\plus{}t)(1\plus{}t)(t^4\plus{}2t^3\minus{}8t^2\minus{}2t\plus{}1)}\,dt$.
Sonnhard.

$ I \equal{} \int \frac {3\tan x \plus{} 1}{2\sin 2x \minus{} 5\cos 2x \plus{} 1}dx \equal{} \int \frac {3\tan x \plus{} 1}{\sqrt {29}\sin (2x \minus{} \alpha) \plus{} 1}\ \Longleftarrow 2x \minus{} \alpha \equal{} y$

where $ \tan \alpha \equal{}\frac{5}{2}$

$ \equal{} \frac {3\tan \frac {y \plus{} \alpha}{2} \plus{} 1}{\sqrt {29}\sin y \plus{} 1}\cdot \frac {dy}{2}\ \Longleftarrow \tan \frac {\alpha}{2} \equal{} \sqrt {29} \minus{} 2$

$ \equal{} \frac {1}{2}\int \frac {1}{\sqrt {29}\sin y \plus{} 1}\cdot \frac {(5 \minus{} \sqrt {29})\tan \frac {y}{2} \plus{} 3\sqrt {29} \minus{} 5}{1 \minus{} (\sqrt {29} \minus{} 2)\tan \frac {y}{2}}dy$

Let $ \tan \frac {y}{2} \equal{} z\Longrightarrow \frac {1 \plus{} z^2}{2}dy \equal{} dz$,

$ I \equal{} \frac {1}{2}\int \frac {1}{1 \plus{} 2\sqrt {29}z \plus{} z^2}\cdot \frac {(5 \minus{} \sqrt {29})z \plus{} 3\sqrt {29} \minus{} 5}{1 \minus{} (\sqrt {29} \minus{} 2)z}dz$

The integral is a bit tricky, so lets start with some manipulation:

\[ \int\dfrac{1\plus{}3\tan x}{2\sin(2x)\minus{}5\cos(2x)\plus{}1}dx\]

Now the denominator looks ugly, so lets put that into a simpler form:

\[ 2\sin(2x)\minus{}5\cos(2x)\plus{}1\equal{}2\sin(2x)\minus{}2\cos(2x)\plus{}1\minus{}3\cos(2x)\equal{}\] \\ \[ \equal{}2\sin(2x)\minus{}2\cos(2x)\minus{}2\plus{}3(\underbrace{1\minus{}\cos(2x)}_{2\sin^2x})\equal{}2\sin(2x)\minus{}2\cos(2x)\minus{}2\plus{}6\sin^2x\equal{}\] \\ \[ \equal{}2\sin(2x)\minus{}2(\underbrace{1\plus{}\cos(2x)}_{2\cos^2x})\plus{}6\sin^2x\equal{}4\sin(x)\cos(x)\minus{}4\cos^2x\plus{}6\sin^2x\] \\ \[ \equal{}\cos^2x\left(4\dfrac{\sin x}{\cos x}\minus{}4\plus{}6\dfrac{\sin^2x}{\cos^2x}\right)\equal{}\cos^2x(6\tan^2x\plus{}4\tan x\minus{}4)\]

So now our integral will be:

\[ \int\dfrac{1\plus{}3\tan x}{2\sin(2x)\minus{}5\cos(2x)\plus{}1}dx \equal{} \int\dfrac{1\plus{}3\tan x}{\cos^2x(6\tan^2x\plus{}4\tan x\minus{}4)}dx\]

Plug: $ u\equal{}\tan x \,\, \Longrightarrow \,\, dx\equal{}\cos^2x \,du$

So:

\[ \int\dfrac{1\plus{}3\tan x}{\cos^2x(6\tan^2x\plus{}4\tan x\minus{}4)}dx\equal{}\int\dfrac{1\plus{}3u}{6u^2\plus{}4u\minus{}4}du\equal{}\] \\ \[ \equal{}\dfrac14 \int \dfrac{4\plus{}12u}{6u^2\plus{}4u\minus{}4}du\equal{}\dfrac14 \ln(6u^2\plus{}4u\minus{}4)\plus{}C_1\equal{}\] \\ \[ \equal{}\dfrac14 \ln(3u^2\plus{}2u\minus{}2)\plus{}C_2\equal{}\dfrac14 \ln(3\tan^2x\plus{}2\tan u \minus{}2) \plus{}C_2\]

So finally our integral is:

\[ \int\dfrac{1\plus{}3\tan x}{2\sin(2x)\minus{}5\cos(2x)\plus{}1}dx\equal{}\dfrac14 \ln(3\tan^2x\plus{}2\tan x \minus{}2) \plus{}C\]

That's correct answer. :)

The key point is rewriting the integrand by using $ \boxed{\tan x \equal{} \frac {\sin x}{\cos x},\ 1 \plus{} \tan ^ 2 x \equal{} \frac {1}{\cos ^ 2 x}}$

$ \int \frac {3\tan x \plus{} 1}{2\sin 2x \minus{} 5\cos 2x \plus{} 1}dx$

$ \equal{} \int \frac {3\frac {\sin x}{\cos x} \plus{} 1}{4\sin x\cos x \minus{} 5(2\cos ^ 2 x \minus{} 1) \plus{} 1}dx$

$ \equal{} \int \frac {3\sin x \plus{} \cos x}{4\sin x\cos ^ 2 x \minus{} 10\cos ^ 3 x \plus{} 6\cos x}dx$

$ \equal{} \int \frac {3\tan x \plus{} 1}{4\tan x \minus{} 10 \plus{} 6\cdot \frac {1}{\cos ^ 2 x}}\cdot \frac {1}{\cos ^ 2 x}dx$

$ \equal{} \int \frac {3\tan x \plus{} 1}{4\tan x \minus{} 10 \plus{} 6(1 \plus{} \tan ^ 2 x)}\cdot \frac {1}{\cos ^ 2 x}dx$

$ \equal{} \int \frac {3\tan x \plus{} 1}{2(3\tan ^ 2 x \plus{} 2\tan x \minus{} 2)}d(\tan x)\ \Longleftarrow u \equal{} \tan x$

$ \equal{} \int \frac {3u \plus{} 1}{2(3u^2 \plus{} 2u \minus{} 2)}du$

$ \equal{} \frac {1}{4}\ln (3u^2 \plus{} 2u \minus{} 2)$, yielding yeahnigga's result.


[quote="kihe_freety5"]I have a problem from  http://batdangthuc.net/showthread.php?t=300&page=2

x,y,z≥0 ; x²+y²+z²=2;
CMR;

a)1+(x+y+z)$ \sqrt {xyz}$ ≥ x²y²z² + $ \frac {$\{xy + yz + zx}^2${ }{x + y + z}$[/quote]
Does your problem mean
$ x;y;z \geq 0;x^2+y^2+z^2=2$
Prove that
$ 1+(x+y+z)\sqrt{xyz} \geq x^2y^2z^2+\frac{(xy+yz+xz)^2}{x+y+z}$


Let $ABC$ be acute triangle. Inscribe a rectangle $DEFG$ in this triangle such that $D\in AB,E\in AC,F\in BC,G\in BC$. Describe the locus of (i.e., the curve occupied by) the intersections of the diagonals of all possible rectangles $DEFG$.

The locus of the point $P,$ as the intersection point of the diagonals $DF,$ $EG,$ of the rectangles $DEFG,$ is the line connecting the midpoints $M,$ $N,$ of the side segment $BC$ and the altitude $AA',$ of $\bigtriangleup ABC,$ respectively.

There is a more general problem presented by [b]''Number 1''[/b] and already has been discussed at http://www.mathlinks.ro/Forum/viewtopic.php?t=80163

Kostas Vittas.

Like in previous proofs, call M and N the midpoints of BC and AA', the height from A; call also K and L the midpoints of FG and DE, respectively, and P - the intersection of GE and DF, the diagonals of the rectangle. Obviously, A, L and M are collinear, while P is the midpoint of KL, hence M, P and N are collinear as well.

Best regards,
sunken rock

You could also coordinate bash this with A' (the foot of the altitude from A) as the origin.

As $F$ moves with constant velocity on $BC$, so does $E$ and hence $D$ moves with constant velocity on $AB$. Hence their midpt also moves on a straight line. Taking the extreme cases, we see that this line joins midpt of $BC$ and $A$ altitude.


logarithms
modular arithmetic
algebra proposed
algebra

[quote="LeoYard"]Discrete logarithm problem is supposed to be very difficult to solve.

Given the primitive element 2, the prime number 331, and the number 141, solve for the number x below:

2^x = 141 mod (331)

What is x?[/quote]
$ 2^{0}\equal{}1\pmod{331}$
$ 2^{1}\equal{}2\pmod{331}$
$ 2^{2}\equal{}4\pmod{331}$
$ 2^{3}\equal{}8\pmod{331}$
$ 2^{4}\equal{}16\pmod{331}$
$ 2^{5}\equal{}32\pmod{331}$
$ 2^{6}\equal{}64\pmod{331}$
$ 2^{7}\equal{}128\pmod{331}$
$ 2^{8}\equal{}256\pmod{331}$
$ 2^{9}\equal{}181\pmod{331}$
$ 2^{10}\equal{}31\pmod{331}$
$ 2^{11}\equal{}62\pmod{331}$
$ 2^{12}\equal{}124\pmod{331}$
$ 2^{13}\equal{}248\pmod{331}$
$ 2^{14}\equal{}165\pmod{331}$
$ 2^{15}\equal{}330\equal{}\minus{}1\pmod{331}$

None of these sixteen values is $ 141$ or $ \minus{}141\equal{}190$

Since we have $ 2^{15}\equal{}\minus{}1\pmod{331}$, and so $ 2^{30}\equal{}1\pmod{331}$, we know that if $ 2^x\equal{}141\pmod{331}$, then, $ 2^{mod(x;15)}$ should be either $ 141$, either $ 190$.

So the equation $ 2^x \equal{} 141 \pmod {331}$ has no solution in $ \mathbb{N}$

Could you please check the following:

Now, let's have 3^x = 141 (mod 331)

We use Shanks' method to get the answer (which is 3^206 = 141 mod (331))

We could use another method to solve for x - a method identical to the Pohlig-Hellman algorithm.

1. Notice that the prime number 331 is very smooth (ie 331-1=330=2*3*5*11).

2. Use the following theorem (described below) to try and find congruence classes for x in various residue systems:

THEOREM: If g is primitive and p is prime and g^x=k(modp), then a divisor of p-1, say q, divides x, if and only if k^(p-1/q)=1(modp).

3. Okay, let us now try and solve for x in 3^x=141(modp).

Calculating the values of 141^(330/2)mod331, 141^(330/3)mod331, 141^(330/5)mod331, and 141^(330/11)mod331, we get the values 1, 31,64, and 74.

Since only 141^(330/2)mod331=1, this means that only 2 divides x. Hence x=0(mod 2).

Since this is all we can get using 141, let us multiply both sides of the equation 3^x=141(mod p) by 3^(-1) (the inverse of 3mod331). The inverse of 3 is 221, so this gives us the following

3^(x-1)=141*221=47(mod331).

Calculating 47^(330/2)mod331, 47^(330/3)mod331, 47^(330/5)mod331, and 47^(330/11)mod331 we find that only 47^(330/5)mod331=1. Hence this means that 5 divides x-1, which means that x=1(mod5).

Now notice that since 3 doesn't divides either x or x-1 (otherwise we would have had either 141^(330/3)mod331=1 or 47^(330/3)mod331=1), this means that 3 must divide x-2. Thus x=2(mod3).

We have now found out the following congruence classes for x:

x=0(mod2)
x=2(mod3)
x=1(mod5)

Using the Chinese Remainder theorem, the above congruences give us x=26(mod30). If we wanted, we could go on and try calculating the congruence class for 11 (ie find k such that x = k(mod 11). Indeed, this would give us even more information about x, however we have obtained enough info on x that we can solve for x just by calculating x^k(mod 331), where k takes on the values 26, 56, 86, 116, 146, 176, 206, 236 and so on until we find a value for k such that 3^k=141.

Doing the above, we find that k=206. Hence 3^206 = 141(mod 331).


John is running 30 minute,10 minute is resting.
Lily is running 40 minute,10 minute is resting.
Justin is running 50 minute,10 minute is resting.
Together are running start 1.00.What time is it they will be together rest?
A)20:30
B)20:40
C)20:50
D)21:00
E)21:10
i solved but i made mistake
40,50,60 min. later they start again.
so time 10
1+10=11 they start again...
last ten min. they rest
so i found 22:50...but this wrong...pls help me!
Thanks and best regards!

I think id say:
[hide]$ (k \equal{} 10 minutes)$
John ends his rest every $ 4k$
Lily ends her rest every $ 5k$
Justin ends his rest every$ 6k$
$ (lcm) \equal{} 2*2*3*5 \equal{} 60$
So 60*k later they are finished resting, just subtract 10 minutes from thoose 10 hours[/hide]


calculus
function
limit
number theory theorems
number theory

Hello everyone.

Absolute newbie here. I hope this is right place for a question I have, which is:

Is n% of an infinite number another infinite number?

hope someone can help me,

best regards

Infinity is not a real number. It doesn't obey some of the properties that "number" should obey. For example, if you say that $ \infty \plus{} 1 \equal{} \infty$ then subtracting it from both sides yields $ 1 \equal{} 0$, an absurdity.

Hi, and thanks for the reply gauss202, I see what you mean, but I don't grasp why you decided
that infinity + 1 = infinity, surely the equation becomes infinity + 1 = infinity + 1?

(In passing, how did you manage to print you infinity symbol?)

Did you assume that because infinity is so large then the 1 is irrelevant?

As for infinity not being a real number, surely the same can be said for zero?

What I was trying to find out is that because infinity is so large, then perhaps any fraction of infinity
must likewise be as large, thus infinite.

There is no definition of addition/multiplication on $ \mathbb R \cup \{ \infty \}$ that is consistent with associativity, commutativity, distributivity and the known laws for real numbers.
Please not again another discussion about $ 0$ or $ \infty$ or $ \dagger$ being a number.....

[quote="peethagoras"]What I was trying to find out is that because infinity is so large, then perhaps any fraction of infinity must likewise be as large, thus infinite.[/quote]
You can't deal with infinity directly as past posters pointed out. In calculus, you get somewhat close with taking limits of functions going to infinity. One rule of limits is that if $ \lim_{x\rightarrow a} f(x)$ tends to $ \infty$ (where $ a$ is some real or $ \pm \infty$), then for any positive real $ b$ (no matter how high), $ \lim_{x \rightarrow a} \frac{f(x)}{b}$ tends to infinity. This is pretty close to answering your question.


I was looking through the first book of the art of problem solving (the basics) until I got to the first few pages and saw exponents. I stink at them, so can any of you'll answer? I mainly don't understand the elimnating, like a problem:

(3^4*3^-2)/(3^5*3^-2)
I know it's hard to read, but its basically three to the fourth power times three to the negative two power divided by three to the fifth power times three to the negative power.

Can anybody help a non-math guy that is trying to be a math guy here? I know this might be rated poor (spam) but I need to learn some math....  :P

$\LaTeX$ed:

$\frac{3^{4}\cdot3^{-2}}{3^{5}\cdot3^{-2}}$

[hide="Solution"]There is a  $3^{-2}$ in the numerator and in the denominator, so they cancel.
\[\frac{3^{4}}{3^{5}}\]
In a fraction, powers of the same number or variable can be reduced by subtracting or adding a number to the power.
\[\frac{3^{4-4}}{3^{5-4}}=\frac{3^{0}}{3^{1}}=\frac{1}{3}\][/hide]

um, i_like_pie, how do you get latex, or whatever you call it? i don't like putting ^ but I must, cuz i don't know about latex stuff...  :wink:

On the sidebar thing on this website, there is a link that says LaTeX.  You can find out there.

But basically you need to surround your code in $\$$ signs before doing stuff like [code]\frac{3}{4}[/code]

All you have to do is put anything you want in $\LaTeX$ between dollar signs, and you get a nice representation of what would otherwise be ugly old text.

To learn how to use $\LaTeX$, you can check out the following pages:

http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php

http://tobi.oetiker.ch/lshort/lshort.pdf

The first one will probably be enough to begin with, and the second one is a bit more advanced.

You can practice in the [url=http://www.artofproblemsolving.com/Forum/index.php?f=224]Test Forum[/url] or in the [url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_TeXer.php]TeXer[/url].

Also, you can hold your mouse over somebody else's $\LaTeX$ (or quote the post) and view the code (this helps a lot).

Other useful information can be found in [url=http://www.artofproblemsolving.com/Forum/index.php?f=123]LaTeX and Asymptote[/url].

Note: refrain from typing text in $\LaTeX$ unless it is necessary to do so, as for some people it will take a long time to load pages that have a lot of $\LaTeX$.

Thanks, alanchou and i_like_pie, but what's that language and that line above it, i_like_pie? what does it mean? also i seen people with the line and stuff under it. how you get "that"?  :lol:

You mean my signature? It's in Latin.

Go [url=http://www.artofproblemsolving.com/Forum/usercp.php?mode=editprofile&sub=sig]here[/url] to create/change your signature.

What do you mean, latex in an exponent?

"quote" this.

$x^{5y+3}-3$

[quote="toadoncart"]Another question: how do you put LaTex on exponents?  :huh:[/quote]

use the carrot symbol located on your keyboard so the code would look like this

[code]$x^6$[/code] which would provide:

$x^{6}$

If you need to insert something longer than one character long into exponents, you need brackets.
For instance, for x^(6+1), this would be:

[code]$x^{6+1}$[/code]

Which yields $x^{6+1}$.

Lemme practice...

$x^{6}$

Did it work?  :huh:

challenge question(not really, though):

$4^{2}y+3\cdot6^{7}\cdot8^{7}y8=x$

Oh, and y is 4...

this is really easy, but me just practicing LaTex  :oops:

To check, after you hit "submit", hit "view my message" and you should spot any LaTeX anomolies if there are any.

[quote="toadoncart"]Wowzers, me already know that, but me thinks that it better with preview then submit, just in case a person looks at your work before you even view your message and it wrong!  :lol:[/quote]

If you simply use preview (I do this all the time), you can test out your latex without actually posting it.  :)


calculus
integration
logarithms
geometry
rectangle
calculus computations

Prove that $ \boxed{\sum_{k \equal{} 1}^{n \minus{} 1} \frac {1}{k \plus{} 1} < \ln n < \sum_{k \equal{} 1}^{n \minus{} 1} \frac {1}{k}\ \ (n\geq 2)}$.

In a rush, but

[hide="Solution"]

The left-hand-sum is the sum of areas of rectangles below the curve $ y \equal{} \frac {1}{x}$ while the right-hand-sum is the sum of areas of rectangles above the curve. The area under the curve is $ \int _{1} ^{n} \frac {1}{x} dx \equal{} \ln n$.[/hide]


geometry
geometric transformation
homothety
ratio

A circle with center $ O$ and a point $ A$ is given. $ P$ is an arbitrary point on the circle.

i) Prove that the locus of the centroid of $ \triangle OAP$ is a circle.

ii) Find the radius of the loci circle in terms of $ OP \equal{} R$.

iii) Find the distance from $ O$ to the center of the loci circle in terms of $ OP \equal{} R$ and $ OA \equal{} x$.

[hide]Use complex numbers, taking O in the origin of the complex plane.[/hide]

The midpoint M of OP describes a circumference O' with center O and radius 1/2R . Call G the centroid of AOP .Because of AG=2/3AM , G describes a homothetic circle to O' in the homothety with center A ans ratio 2/3 . So this circle has radius  1/3R and its center K lie on OA such that AK=2/3x

[hide]Let $ M$ be the midpoint of $ OA$. $ PG / GM \equal{} 2 / 1$. Let parallel line to $ OP$ passing through $ G$ intersect $ OA$ at $ O'$. $ O'G \equal{} R/3$ which is fixed. $ OO' / O'M \equal{} 2/1$, then $ OO' \equal{} OA / 3 \equal{} x/3$ which is fixed. So the locus is a circle with center $ O'$.[/hide]


In $ \triangle ABC$ the altitıde $ [AH]$, the internal bisector $ [BN]$, and the median $ [CV]$ are passing from a common point. If $ |BH| \equal{} x$ and $ |HC| \equal{} y$, then find
a) $ \frac {Area(\triangle HVN)} {Area(\triangle ABC)}$

b) $ \frac {\sin \angle VHN } {\sin \angle ABC}$

in terms of $ x$ and $ y$.

[hide="Solution"]By Ceva's theorem we have $ \frac xy\cdot \frac {CN}{NA}\cdot \frac {AV}{VB} \equal{} 1\ \Longleftrightarrow \frac xy \equal{} \frac {AN}{NC}\ (*)\ \Longleftrightarrow HN\parallel AB$

$ \Longrightarrow \angle HNB \equal{} \angle ABN \equal{} \frac {\angle B}{2}$ and $ \angle NBH \equal{} \frac {\angle B}{2}$, so $ \triangle BHN$ is isosceles with $ \boxed{BH \equal{} HN \equal{} x}$

Because $ [BN$ is the internal bisector in $ \triangle ABC$ $ \Longrightarrow \frac ca \equal{} \frac {AN}{NC}\stackrel{(*)}{ \equal{} } \frac xy$ $ \Longleftrightarrow \frac {c}{x \plus{} y} \equal{} \frac xy\ \Longleftrightarrow \boxed{c \equal{} \frac {x(x \plus{} y)}{y}}$

In $ \triangle AHB,\ \angle B \equal{} 90^{\circ}$, $ [HV]$ is median $ \Longrightarrow \boxed{VH \equal{} \frac c2 \equal{} \frac {x(x \plus{} y)}{2y}}$

Also we get $ \angle AHV \equal{} \angle BAH$ and because $ \angle AHN \equal{} \angle BAH\ \Longrightarrow \angle VHN \equal{} 2\angle BAH$.

$ \Longrightarrow \sin(\widehat{VHN}) \equal{} 2\sin(\widehat{BAH})\cdot\cos(\widehat{BAH}) \equal{} 2\cdot \frac xc\cdot \frac {AH}{c}$

$ \Longrightarrow \frac {\sin(\widehat{VHN})}{\sin(\widehat{ABC})} \equal{} \frac {\frac {2x\cdot AH}{c^2}}{\frac {AH}{c}} \equal{} \frac {2x}{c}\ \Longleftrightarrow \boxed{\frac {\sin(\widehat{VHN})}{\sin(\widehat{ABC})} \equal{} \frac {2y}{x \plus{} y}}$

$ \frac {Area(\triangle HVN)}{Area(\triangle ABC)} \equal{} \frac {VH\cdot HN\cdot \sin(\widehat{VHN})}{AB\cdot BC\cdot \sin(\widehat{ABC})}\ \Longleftrightarrow \boxed{\frac {Area(\triangle HVN)}{Area(\triangle ABC)} \equal{} \frac {xy}{(x \plus{} y)^2}}$[/hide]


geometry
circumcircle
perpendicular bisector
geometry unsolved

$ A$, $ B$ are points on the same side of the line $ l$ on a plane. We get $ A'$ by reflecting $ A$ across $ l$. The intersection of $ l$ and the segment $ A'B$ is $ C$. The intersection of $ l$ and the perpendicular bisector of segment $ AB$ is $ D$. Show that $ A$, $ B$, $ C$ and $ D$ are concyclic. [geogebra]f6fc9eb2d671548d34850b856032e908bc3e0d39[/geogebra]

Let $ \angle {ACl} \equal{} a$
This implies $ \angle {A'Cl} \equal{} a$

Because triangle AA'C is isosceles
$ \angle {CA'A} \equal{} 90 \minus{} a$

Now $ \angle {ACB} \equal{} 180 \minus{}2a$

$ D$ is the circumcenter of triangle AA'B.
So $ \angle {ADB} \equal{} 2\angle{AA'B} \equal{} 2(90\minus{}a) \equal{} 180 \minus{} 2a$

Since $ \angle {ACB} \equal{} \angle {ADB}$ ABCD is a cyclic quadrilateral.


calculus
integration
limit
trigonometry
real analysis
real analysis unsolved

$ \lim_{n \rightarrow \infty}\int_{1}^{\plus{}\infty}\frac{\sin nx}{x}dx\equal{}?$

Make the substitution $ u\equal{}nx.$

$ \equal{}\lim_{n\to\infty}\int_n^{\infty}\frac{\sin u}{u}\,du\equal{}0.$

It equals zero because this particular improper integral is known to converge.


Is it me or does the problem that stays up the top of the page not seem to change as you go on to the next challenge? Is this meant to be like that or do you just forget or something.

Hit refresh on your browser.  The old image is cached on your machine.

Yeah I suppose thats the problem. Although refresh doesn't work, but my computer does that to me sometimes. I'll have to clear out the cache completely one day.


function
limit
real analysis
real analysis unsolved

Let f be a real valued function, thrice continuously differentiable in a neibourhood of 'a' & let f"(a)=0.
Prove that,
lim(h->0) [f(a+h) -2f(a) +f(a-h)]/h^3 = f"'(a)/3.

plz give a detailed solution
thank you... :)

The statement is false, the limit is $ 0$.  Furthermore, if $ f$ is four times continuously differentiable, then
\[ \lim_{h\to 0}\frac{f(a\plus{}h)\minus{}2f(a)\plus{}f(a\minus{}h)}{h^4}\equal{}\frac{f^{iv}(a)}{12}\]

[hide="hint"]
Taylor expand
[/hide]


Find the complex number $ x$ such that $ (1\plus{}2i)x\plus{}5\minus{}3i\equal{}0$.

Let $ x \equal{} a \plus{} bi$, where $ a$ and $ b$ are real numbers.  Substituting, we get that
\[ (1 \plus{} 2i)(a \plus{} bi) \plus{} 5 \minus{} 3i \equal{} 0\]
Expanding the LHS, we get
\[ a \minus{} 2b \plus{} 5 \plus{} 2ai \plus{} bi \minus{} 3i \equal{} 0\]
Because $ a$ and $ b$ are real numbers, it follows that $ a$ and $ \minus{} 2b$ are real, and thus
\[ a \minus{} 2b \plus{} 5 \equal{} 0\]
Similarily, $ 2ai$ and $ bi$ are imaginary, so
\[ 2ai \plus{} bi \minus{} 3i \equal{} 0\implies 2a \plus{} b \minus{} 3 \equal{} 0\]
By solving our system of equations, we get that $ a \equal{} \frac {1}{5}$ and $ b \equal{} \frac {13}{5}$.  Thus, $ x \equal{} \boxed{\frac {1}{5} \plus{} \frac {13i}{5}}$

Very Good! :)

You can solve with respect to $ x$ directly.

$ (1\plus{}2i)x \plus{} 5 \minus{} 3i \equal{} 0$

$ (1\plus{}2i)x \equal{} \minus{}5 \plus{} 3i$

$ x \equal{} \frac{\minus{}5\plus{}3i}{1\plus{}2i}$

$ x \equal{} \frac{(\minus{}5\plus{}3i)(1\minus{}2i)}{(1\plus{}2i)(1\minus{}2i)}$

$ x \equal{} \frac{\minus{}5\plus{}10i\plus{}3i\minus{}6i^2}{5}$

$ x \equal{} \frac{1\plus{}13i}{5}$

$ x \equal{} \boxed{\frac15 \plus{} \frac{13}5i}$

[hide="Short Solution"]$ (1\plus{}2i)x\plus{}5\minus{}3i \equal{} 0 \Rightarrow x\equal{}\frac{3i\minus{}5}{1\plus{}2i}\equal{}\frac{1}{5}\plus{}\frac{13}{5} i$[/hide]

[hide="my answer"]
Move $ 5\minus{}3i$ over to the RHS:
$ (1\plus{}2i)x \equal{} \minus{}5\plus{}3i$
Dividing by $ 1\plus{}2i$ gives
$ \frac{\minus{}5\plus{}3i}{1\plus{}2i} \equal{} \frac{(\minus{}5\plus{}3i)(1\minus{}2i)}{(1\plus{}2i)(1\minus{}2i)} \equal{} \boxed{\frac{1\plus{}13i}{5}}$
[/hide]


Team members tell us, how did Canada do?   :)

Wow, you post the result here already?  :huh:

So medal-wise it's 1 Silver, 3 Bronze, 1 HM.

hey 4 medals and 1. Hon mention is great for a team with only 2 seniors...futures looking good  :)

Good job guys! This was a really cruel exam. I sincerely doubt that any previous Canadian team could have performed much better on this year's extremely hard test.

ya rly surprised china didn't get first  :o


[quote="2006"][quote="mathematica"]Prove that for all positive real numbers like $x,y,z$ we have:

$\sum_{cyc}(x^{6}+x^{4}yz+2x^{2}y^{2}z^{2}-4x^{3}yz^{2})\geq 0$.[/quote]
It is not nice !  :([/quote]

nice or not nice is subjective, don't you think?

in my opionion i think it`s nice...

now i continue with the proof:

$\Longleftrightarrow \sum{x^{6}}+\sum{x^{4}yz}+6x^{2}y^{2}z^{2}\geq 4\sum{x^{3}yz^{2}}$

now let`s apply schur : $\sum{x^{6}}+3x^{2}y^{2}z^{2}\geq \sum{x^{4}y^{2}}+\sum{x^{2}y^{4}}$

but from muirhead we have : $\sum{x^{4}y^{2}}+\sum{x^{2}y^{4}}\geq 2\sum{x^{4}yz}$ , because $(4,2)\succ(4,1,1)$.

so it remais us to prove that : $3\sum{x^{4}yz}+3x^{2}y^{2}z^{2}\geq 4\sum{x^{3}yz^{2}}$

$\Longleftrightarrow 3\sum{x^{3}}+3xyz\geq 4\sum{xy^{2}}$, and now another schur :

$\sum{x^{3}}+3xyz\geq \sum{xy^{2}}+\sum{x^{2}y}$

$\Rightarrow 2\sum{x^{3}}+\sum{x^{2}y}\geq 3\sum{xy^{2}}$, which is true.


Knowing something about the product (for example, molecular formula) would help.


Let $ a,b,c\geq 0$, no two of which are zero.Prove that:
$ \frac{a^2(b\plus{}c)}{b^2\plus{}c^2}\plus{}\frac{b^2(c\plus{}a)}{c^2\plus{}a^2}\plus{}\frac{c^2(a\plus{}b)}{a^2\plus{}b^2}\geq a\plus{}b\plus{}c$.

[quote="Inequalities Master"]Let $ a,b,c\geq 0$, no two of which are zero.Prove that:
$ \frac {a^2(b \plus{} c)}{b^2 \plus{} c^2} \plus{} \frac {b^2(c \plus{} a)}{c^2 \plus{} a^2} \plus{} \frac {c^2(a \plus{} b)}{a^2 \plus{} b^2}\geq a \plus{} b \plus{} c$.[/quote]
The inequality is equivalent to $ \sum \frac{ab(a\minus{}b)^2(a\plus{}b)}{(b^2\plus{}c^2)(a^2\plus{}c^2)} \ge 0$ which is true  :)  .


Let K be a subfield of (C,+,*). Show that the following are equivalent:

a)x2 =a2 +1 has solutions in K for all a in K;
b)x2 =a2 +a+1 ha solutions in K for all a in K.

Give an example of K=/=C for which at least one of these holds.

[i]Romanian NMO/final round/1990/12'th grade[/i]

It is obvious that any subfield of (C,+,*) must contain Q

a)=>b)
Choose y, z, t such that
  y^2=(2a+1)^2 + 1
  z^2=y^2 + 1
  t^2=z^2 + 1 = y^2 + 2 = (2a+1)^2 + 3 = 4a^2 + 4a + 4
Then take x to be t/2 and find that x^2 = a^2 + a + 1

b)=>a)
For a = 1 we have that x^2=3 has solutions in K => sqrt(3) is in K
Now take y such that y^2 = a^2 + a + 1 for (sqrt(3)/2)a-1/2 instead of a
  Then y^2 = (3/4)(a^2 + 1)
Now x=2y/sqrt(3) gives the desired result.

R is one subfield that satisfies the relations. :)


A point in the first quadrant on the circle with the equation $x^{2}+y^{2}=1$ has a tangent that makes an angle of 30 degrees with the x-axis. Find its coordinates.

[hide]Somehow I got $(\frac{1}{2},\frac{\sqrt{3}}{2})$. But that answer is wrong. Whats wrong with this:
So we have a point (x,y) that lies on the circle. When we draw this to the x-axis and make that 30 degree angle, we have a 30-60-90. We got the 60 degrees from when we drew a line from the center of the circle to the tangent. So opposite of the side of the 30 degrees is 1, because thats the radius. That means the point the line touches to the x-axis touches will have a length of 2. So the point (x,y) and the point (2,0) has a distance of $\sqrt{3}$. So now we set an equation. We have $\sqrt{(2-x)^{2}+(0-y)^{2}}=\sqrt{3}$
$(2-x)^{2}+(y^{2})=3$
$4-4x+x^{2}+y^{2}=3$ Since $x^{2}+y^{2}=1$ we can substitute.
$4-4x+1=3$
$-4x=-2$ So $x=\frac{1}{2}$. PLugging this back in to the distance formula we get that $y=\frac{\sqrt{3}}{2}$. [/hide]
Can anyone tell me what I did wrong or is the test writers wrong?

[quote="anyone"]I think you're right. The point would be at 60 degrees on the unit circle, so the coordinates would be $(cos60,sin60)=(\frac{1}{2},\frac{\sqrt{3}}{2})$. What did the testwriters say was the correct answer?[/quote]
They didn't really tell me. All they said the answer was E or N.O.T.A.

[quote="anyone"]That would be correct so long as none of the other four answers were $(\frac{1}{2},\frac{\sqrt{3}}{2})$[/quote]
Yea, there was only one choice for that. I guess I was right. Thanks for your help to those who approved of my answer.


Legolas flips a fair coin 2008 times, and Gimli flips another fair coin 2009 times. What is the probability that Gimli flips more heads than Legolas?

after 2008 flips, they can have the same amount of heads, then gimli could get a heads, or gimli could already have more heads. then the probability is p+1/2(1-2p)=1/2 where p is the probability that gimli already has more heads

I STRONGLY DISAGREE WITH YOU!. Okay after 2008 throws, lets assume they have the same amount of heads and tails. Now its 1/2 probability that Gimili gets a head. So 1/2 probability that Gimili gets more. Also you messed up. You made everyone realize that 1=1. LOL. Watch. Solving for P, P+(1-2P)/2=1/2. Multiply by two on both sides. 2P+1-2P=1. 1=1. LOL. So the answer is 1/2 from your conclusions? Read the chapter called think about it in introduction to counting and probability.
ANSWER: 1/2.

Oh... I just said for the first $2008$ flips for both of them, the best chance is having an equal amount of heads. After that, Gimli has a $\frac{1}{2}$ chance of getting the next to be heads, so the answer was $\frac{1}{2}$.


geometry
power of a point
Pythagorean Theorem
geometry unsolved

I got a different answer.

[hide][img]http://img394.imageshack.us/img394/9237/ahdfsl89r32dw8.jpg[/img][/hide]

[quote="atomicwedgie"][hide]$ \frac {49}{3}\pi$[/hide][/hide][/quote]
[hide="Solution"] atomicwedgie is correct.  Call PD=DE=x.  So, by Power of a Point, \[ AP^2\equal{}PD(PE)\] \[ (7\sqrt{6})^2\equal{}x(2x)\] \[ 6(49)\equal{}2x^2\] \[ x^2\equal{}49(3)\] \[ x\equal{}7\sqrt{3}\] Now, we will find the radius.  Call the radius r.  So, we have that (by Power of a Point), \[ AP^2\equal{}PC(PB)\] \[ (7\sqrt{6})^2\equal{}(7\sqrt{7}\minus{}r)(7\sqrt{7}\plus{}r)\] \[ 6(49)\equal{}7(49)\minus{}r^2\] \[ 49\equal{}r^2\] \[ r\equal{}7\] Thus, OE=OD=7 and DE=$ 7\sqrt{3}$.  So, we can find that $ \angle DOE\equal{}120$.  Thus, we have that the area of the sector is \[ \frac{120(\pi)(7^2)}{360}\equal{}\boxed{\frac{49\pi}{3}}\] [/hide]

My solution is quite elementary.

[hide]Let $ M$ be the midpoint of $ \overline{DE}$.  Then $ \triangle PAO$ and $ \triangle PMO$ are right triangles.  It follows by the Pythagorean Theorem that the radius of the circle is $ r \equal{} \overline{AO} \equal{} 7$, and hence $ \overline{DO} \equal{} \overline{EO} \equal{} 7$.  If we let $ \overline{DM} \equal{} \overline{EM} \equal{} x$, it follows from $ \overline{PD} \equal{} \overline{ED}$ that $ \overline{PD} \equal{} 2x$.  Hence we have two Pythagorean relationships,

$ \overline{PM}^2 \plus{} \overline{OM}^2 \equal{} \overline{PO}^2$,
$ \overline{DM}^2 \plus{} \overline{OM}^2 \equal{} \overline{DO}^2$.

Their difference yields

$ \overline{PM}^2 \minus{} \overline{DM}^2 \equal{} \overline{PO}^2 \minus{} \overline{DO}^2$,

or equivalently,

$ (3x)^2 \minus{} x^2 \equal{} (7 \sqrt{7})^2  \minus{} 7^2$,

from which it easily follows that $ x \equal{} \frac{7 \sqrt{3}}{2}$.  Therefore $ \triangle DMO$ is a 30-60-90 right triangle (since $ \overline{OM} \equal{} 7/2$), and  angle $ DOE \equal{} 2\pi/3$.  Therefore the area of sector $ DOE$ is $ \frac{1}{3}\pi(7^2) \equal{} \frac{49}{3}\pi$, as claimed.[/hide]


geometry
geometric transformation
rotation
reflection

Two successive Clockwise rotations about the origin of angles x and y (0<= x,y<= pi)
resulrts in a reflection through the origin.  Find x+y.

Clearly, x+y is in radians, as the reflection is equal to a 180degree rotation, yet the answer key insists that it is 180 degreees.

I get pi/2. This question is problem 332 from AOPS volume1. :ninja:

Maybe it got radians and degrees mixed up...

i think so. Does anyone have a better solution


Suppose a set has defined on it a binary algebraic operation. If that operation is associative, then the notation $ a^n$ is unambiguous. But what if we don't have associativity? Then $ a(aa)$ may be different than $ (aa)a.$

For such a non-associative operation, let $ g(n)$ be the number of different possible meanings for $ ``a^n\,".$ We have $ g(1)\equal{}1,g(2)\equal{}1,$ and $ g(3)\equal{}2.$

Find a general expression for $ g(n).$

[hide]
We can represent the different ways of interpreting $ a^n$ in postfix notation. For example, $ (a*a)*(a*a) \longleftrightarrow a a * a a * *$.

Each sequence of $ n$ $ a$'s and $ (n\minus{}1)$ $ *$'s represents a different possible meaning of $ a^n$. Thus $ g(n)$ is the number of valid ways to arrange $ n$ $ a$'s and $ (n\minus{}1)$ $ *$'s.

The first character in the sequence must be an $ a$. After removing that $ a$, the remaining string consists of $ (n\minus{}1)$ $ a$'s and $ (n\minus{}1)$ $ *$'s such that no substring from the beginning contains more $ *$'s than $ a$'s. (If this were not the case, then we would be calling on the operator when there is only one operand in the stack.)

From this, we can see that $ g(n) \equal{} C_{n\minus{}1}$, where $ C_{n}$ is the $ n$th Catalan number. $ \boxed{g(n) \equal{} C_{n\minus{}1} \equal{} \frac{1}{n}\binom{2(n\minus{}1)}{n\minus{}1}}$.

Nice problem. I hope my solution was clear enough.
[/hide]

This is, I think, the original problem that Catalan was trying to solve that got his name attached to this sequence.

[url=http://math.mit.edu/~rstan/ec/]Richard Stanley's [i]Enumerative Combinatorics[/i] page[/url] has 231 other combinatorial interpretations of the sequence.


algorithm
geometry
3D geometry
calculus
algebra
polynomial

[quote="Benjamin Hu"]Does anyone know any algorithms to find roots of numbers other than square roots (i.e. I am looking for algorithms to find cube roots, 5th roots, 7th roots, etc.)? I am looking for algorithms for roots that can't be rewritten as compositions of other roots (for instance, the sixth root of a number is the cube root of the square root of the number, or the square root of the cube root of that number).[/quote]

Ferrari's method can be used to find 3rd roots. The method to find 4th roots is horribly disgusting. Anything above this has been proven to be impossible.

Actually, I think the method for degree four is more elegant than the method for degree three.

Anyhow, I was under the impression that we were looking for good rational approximations of roots.  In general, if we are looking $ b^{1/t}$ for $ b,t\in \mathbb{Z}$, and $ a$ is the nearest integer to $ b^{1/t}$, then we can use the binomial theorem/Taylor series estimate
\[ [a^t \plus{} (b\minus{}a^t)]^{1/t} \equal{} \sum_{k\equal{}0}^{\infty} \binom{1/t}{k} (a^t)^{(1/t)\minus{}k}(b\minus{}a^t)^k \equal{} \sum_{k\equal{}0}^{\infty} \binom{1/t}{k} a^{1\minus{}tk}(b\minus{}a^t)^k, \]
where $ \binom{r}{k} \equal{} \frac{\prod_{j\equal{}0}^{k\minus{}1}(r\minus{}j)}{k!}$, for $ r\in\mathbb{R}, k\in\mathbb{Z}^\plus{}_0$, as usual.  I think this should converge approximately geometrically.   You could also use a method known as Newton's approximation, which is outlined in many elementary calculus textbooks.  The procedure is to find an estimate for a root of some differentiable curve.  You then find the equation of the line passing through that point.  You then let the interception of this curve with the $ x$ axis become your new estimate, and iterate the process.  In general, convergence is somewhat tricky, but with plain curves like polynomials, as long as you start with a good enough estimate, you should be fine.  In fact, I would not be surprised if you get the very same approximation.


[b]Question:[/b]
suppose X1 and X2 are independent with f(x) = 2x, if 0<x<1, 0 otherwise
Find E[max{X1,X2} and Var[max{X1,X2}]

[b]Attempt:[/b]
E[max{X1,X2}]
i understand this can be solved 2∑E[X1>k|X2=k] for discrete cases (??)
so for continuous distribution,

2*∫(k to 1)∫(0 to k) x *2x dy dx

and similarly
E[max{X1,X2}^2] with x^2 *2x
and computer Var using above

right? please can anyone provide a correct solution if above is incorrect?

what is $ f$ ? the density of $ X_1$ and $ X_2$ ?

yes, they are independent and identically distributed. thanks

for example:
\[ E[\max(X_1, X_2) ] \equal{} \int_{x \in \mathbb{R}} \int_{y \in \mathbb{R}} \max(x,y) f(x) f(y) dx dy \]
do you know why ? If not, I am sure that if you read again your textbook, you should be able to understand where it comes from. Do not hesitate to ask more questions ..


my computer has shaky internet connection and when i try to resign in the classroom, it usually says direct login failed: username in use or something like that...and i have to try many times before it works.
help someone? =[[ thanks

I think what is happening there is that when you disconnect, a 'ghost' of you is left in the classroom for a little while.  I am currently on vacation, but when I get back I will contact the software vendor that created the classroom about how to best get your ghost booted out of the classroom more quickly.  I will send you a solution when we have one.

I have that problem also. (But I think that I have already informed rrusczyk of the problem)

Do the same thing I recommend here if you can't get back in.

(General note to classroom users: this most commonly happens to wireless users - some wireless connections have lots of tiny disconnects - not such a big deal if you're surfing the web, but they can be enough to knock you out of a chat room on some computers.)


trigonometry
Putnam
search
calculus
calculus computations

gaah i thought i got somewhere with this problem and then i realized i was using the wrong sines and then my head started to explode and...yeah. help an 8th grader out ;-;

[img]http://i32.photobucket.com/albums/d6/animalabuse/1-8-06-d.gif[/img]

This problem is not intended for 8th graders. See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=64436]Putnam 2005 A5.[/url]

This is the third time since this problem appeared as A5 on this year Putnam it is posted by somebody  :rotfl:. Seems it becomes a tradition to post it once a week :spam:.  Please, search the Putnam subforum for the solutions. It is very easy to find there!

PS. Oops, Kent was faster to answer. :) Still, the point remains: please, search the Putnam subforum and the Kalva web page before posting the problems from Putnam. Most likely, you'll find the solution there.


[quote="rookworm"][quote="jimhu"]What if you (mainly your brain) was taken apart (basically you die) and reassembled in the same order (then come back to life) it was originally, when you(reassembled) wake up would you still be yourself(the person youre destined to be)?

Its hard to explain because we don't know if you wake up as the same conciousness before your brain was reassembled. (For all we know that your old self might have died).

Another question: since everything happens once, do you think we are predestined? (IF you are questioning predestination right now, that event might be predestined).

Heh just a bunch of pretty dumb questions... I hope for interesting discussion...[/quote]

I think these are great questions! I've been thinking about the former one since I first saw an episode of "Star Trek". I have thought about something similar to the latter one a great deal recently. Here's an interesting question along those lines: "If we live in a deterministic, non-branching universe", what does it mean to say something is 'possible' or 'necessary' (in the metaphysical, as opposed to epistemological sense). Moreover, what kind of difference is there (if any) between counterfactuals about the past, and those about the future?

I'm tending towards a radical position (which I would like to call 'modal nihilism') which denies that statements about metaphysical possibility are meaningful (not even logical positists went this far...). For instance, the statements, "It might rain tomorrow", or "If I had only sold it when I had the chance, I'd be rich!", would be meaningless, or at least mean something quite different from what we'd normally say they do. I may put forth some of my arguments later in this thread if there is interest. I challenge anybody to explain the concept of 'possibility' without resorting to some weird multiple universe theory.

Any thoughts?[/quote]

The question stated above is actually linked with predestination... So basically the way it has to do with your question is that if predestination was true, it would make statements like "I could have been rich" not meaningful at all...

And plus I don't really understand your vocabulary such as metahpysical lol... I am only a 15 year old with a shallow word bank...

[quote="PenguinIntegral"]If Chuck Norris is late, time better slow the f*** down.[/quote]

roundhouse kick'd for blasphemy

Why can't we just treat "possibility" (i.e. randomness) as a law of nature? I mean, we make all kinds of other assumptions like gravity and Coulomb's law. With regard to human decision making, that depends on things we can't know by the FWT!

[quote="rookworm"]I challenge anybody to explain the concept of 'possibility' without resorting to some weird multiple universe theory.[/quote]
Consider the set of worlds that would appear identical to the man who makes the statement about a past or future possibility. By this I mean that if he looked around one world for ten minutes, then was picked up and placed in the other world to look around for ten minutes, he wouldn't find any difference. (However, the atoms, bacteria, and dust particles that he can't see may be different.) Then the probability that an event will occur/could have occured is the fraction of these worlds in which the event occurs.

This idea is central to the successful theory of [url=http://en.wikipedia.org/wiki/Statistical_mechanics]statistical mechanics[/url]. I would say that any philosophical idea that produces a predictive physical theory is meaningful.

These are hard questions!
I was wondering this when watching the Fly
in this film a lonely scientist builds two machines, one scans the object, the second one completely reconstructs the object
this isn't actual teleportation but cut and paste
for some reason the scientist let the first machine destroy the object, which is completely useless to me it seems
well, i was wondering, when he tried that out on himself, why did you do that? you killed yourself!!
well he got in trouble when a fly! went along in the machine

it is also explored in an episode of earth final conflict
an alien robotic explorer scans people and then destroys them
a scientist discovers that the machine can also reconstruct and begs his coworkers to force it to rebuild his wife, although they keep telling him it won't be her
in the end, surprisingly, he goes through with it, and lives happily again with the clone

what is consciousness? i believe in science, but will science ever explain this?

[quote="ccy"]Another question: If you are "teleported" to some other place, is the teleported person still you? :D[/quote]Teleportation as you imagine it (as a process of being taken apart, transformed into mere fotons, transported and then reassambled) probably from Star Trek series will not be possible.

However, using a mainfold of the space-time-continuum, bending the subspace field between you and your desired location would allow you to instantly travel to a different place (warp jump) appearing for all others as teleporting. However the amount of energy required to bend space between two locations ... is huuuuge :)

I remember my friend had this obsession with brain transplants and whether that count keep 'you' and your memories and such, when we were in Year 8.

I think you're eventually touching on the existence of a 'soul'. As this is basically a forum of maths people the very idea would probably earn you ridicule. ;)

But such a feat would take very precise nanotechnology and such, and you need to kickstart the brain again, but I guess technically it'd work, hypothetically. It's all in there. :D

[quote="vulgarfraction"][quote="fredbel6"]for me it is not about a soul
i am a die hard atheist yet i must ask wel : being aware, what is that?[/quote]

I'm maintaining that it's all chemical and electrical messages. Nothing happens when those die.[/quote]

My definition: Being aware is being conscious of your existence.

Plus, existence of souls cannot be proven or disproven. You cannot just assume that it's not "scientifically" present, it is not really present. What's wrong about talking about the concept of souls in a math forum? Mathematicians are supposed to be open to new ideas that are don't necessarily have to make sense.

Your question is a very interesting Jimhu.
I have thinkiking about these kind of questions for a long time. :)

I just have no idea what  the  answer is. :oops:

But I will ask some few more questions.

Suppose you take matter(say water) of exactly your own mass and change its subatomic configuration so that the water ends being an exavt copy of you.And now suppose I kill you seeing that there is another 'you'.Does your life still exist?

You disintegrate and reintegrate somewhere else.Are you still yourself?

If I take the atoms from your body and turn it into a some other creature.then is it still you or some other creature?

It all comes down to what exactly is consciousness?
At present we have very little idea about what conscioussness is and thats the problem.

vulgarfraction, yes  i agree I too am an atheist and i believe that it's all chemistry and electric signals , but still how there is a me inside me? I just don't understand, it's so paradoxal, it is for me the single most difficult philosophical question

I don't agree with being aware being the same as knowing you exist
because what is knowing that
a dumb animal like a hamster will not really think about that

[quote="fredbel6"]
a dumb animal like a hamster will not really think about that[/quote]
Are you sure? How do you know this?

well what is knowing that you exist?  explicitly sitting down and pondering about it
or just thinking of your next lunch

don't get me wrong, i love all rodents!  perhaps i should have said guinea pigs, hamsters are a bit more ingenious than guinea pigs in my experience :lol:

What is consciousness?  That in itself is probably one of Godel's metastatements.
Well, as study of the brain progresses, we may gain some insight into what consciousness is...but we're nowhere close...

This is just like that book I just read, Teleportation.


Multiplying by xyz in the first we obtain
$xy+yz+zx+z^{2}=0$ so $(x+z)(y+z)=0$ so x=-z or y=-z

**for x=-z the third becomes $y^{3}=8$  so $y=2$ and from the
second we have $2x^{2}=18$  so $x=3$ or $x=-3$ so $z=-3$ or $z=3$

** For y=-z the third becomes $x^{3}=8$ so $x=2$  and from the second we have that $2y^{2}=18$ so $y=3$ or  $y=-3$ and $z=3$ or $z=-3$

So the triples are $(x,y,z)=(3,2,-3)$  , $(x,y,z)=(-3,2,3)$
$(x,y,z)=(2,3,-3)$  $(x,y,z)=(2,-3,3)$