This is an interesting problem.

Given that an earth satellite near the earth's surface takes about 90 minutes per revolution and that a moon satellite takes about 90 minutes per revolution, what interesting statement can you derive at the moon's composition?

The moon is made of cheese. Also, it's easy to see that $ \frac {M_e}{r_e^3} \equal{} \frac {M_m}{r_m^3}$, so the moon and the earth have the same average densities. So the earth is also made of cheese?

I've never been to the moon so I can't tell you what it's made of.  But you got the correct answer.  I'll put up some more problems soon.


Gauss
analytic geometry
geometry
vector
trapezoid
geometry proposed

Given triangle $ABC,P$ is an arbitrary point and $A',B',C'$ is cevian triangle of $P$ denote $A_{1},A_{2},A_{3}$ are midpoint of $PA,B'C',BC$ then $A_{1},A_{2},A_{3}$ are colinear (it is of Gauss :D )and more denote $d_{a}$ is line connecting $A_{1},A_{2},A_{3}$ similar we have $d_{b},d_{c}$ then  $d_{a},d_{b},d_{c}$ are concurrent (maybe my own :D) .
Now if $P(x,y,z)$ calculate coodinate of concurrent point !

I think that it is a well-known result, but I don’t remember any reference.

It is very easy to prove the concurrency of the three Gauss’s lines which the problem states, but because of I am not good in coordinates, somebody else must give you the ones, of their concurrency point.

Kostas Vittas.

Ok Kostas I agree with you it is well know result ! .But here I only want to change Gauss's line (a problem of quadrilateral) to problem of triangle. Although two problems are the same but I think in geometry away of looking at a problem is improtant.
An example is famous theorem of Ptoleme
In quadrilateral it is given $ABCD$,then $AB.CD+AD.BC \ge AC.DB$
But in triangle it is given triangle $ABC$ forall $M$ then $aMA,bMB,cMC$ are three side of a new triangle.
Actually here they are the same but in version of triangle we can get more many results than in version of quadrilateral.
All I only want to say about improtance of away of looking.
Return this problem, we can understand with a triangle $ABC$ and arbitrary point $P$ we will have there Gauss's line of triangle $ABC$ corespond to point $P$ and they are concurrent at $G*$ we can call be Gauss' point of $ABC$ corespond to $P$. Actually I got many nice results with this away of looking. I will post one of them, I found the following result but I'm not sure I'm a first and maybe they were well know that I didn't know but I feel it is nice.
Given quadrilateral $ABCD$ and arbitrary point $P$ we will call Gauss point of $ABC$ corespond to $P$ be $D*$ similar with triangle $BCD,CDA,DAB$ we will get $A*,B*,C*$ prove that four lines $AA*,BB*,CC*,DD*$ are concurrent at a point that we call Gauss point of  quadrilateral $ABCD$ corespond to $P$.
And with Gauss point of quadrilareral we can get a similar problem for pentagon $ABCDE$, with an arbitrary point $P$ we call Gauss point of $ABCD$ corespond to $P$ be $E*$ similar we have $B*,C*,D*,E*$ then
$AA*,BB*,CC*,DD*,EE*$ are concurrent at a point that we call Gauss point of pentagon $ABCDE$ corespond to $P$.
and more  we can define Gauss point of n-polygon corespond to arbitrary point $P$and they have similar properties!

Hope you can understand more about my ideal in this topic! :)

I have seen this problem about G*. I show that $P, G*, G$ are colinear and $GP=4GG*$.
I think this can be generalize to every n-polygon like your idea by my result.

That's right Little Gauss, nice result,do you have solution for it?, I'm tring with vector method! :)

I showed it by Menelaus thm.
but maybe there is vector solution...

[quote="Little Gauss"]I have seen this problem about G*. I show that $P, G*, G$ are colinear and $GP=4GG*$.[/quote]

 $\bullet$ We can prove the collinearity of $P,$ $G^{*},$ $G,$ based on the below first theorem of concurrent segments.

We say three segments as concurrent, when their segment lines are concurrent. Three parallel segments are as concurrent at infinity.

[b][size=117]FIRST THEOREM OF CONCURRENT SEGMENTS.[/size][/b] – Three given segments concurrent at a point so be it $S,$ taken per two of them as opposite sides, form three quadrilaterals. Prove that:

[b][size=117]a).[/size][/b] – The intersection points of diagonals of two quadrilaterals and the intersection point of the opposite sides, the other than $S,$ of the third quadrilateral, are collinear.

[b][size=117]b).[/size][/b] – The intersection points of the opposite sides of all the quadrilaterals, the other than $S,$ are collinear.

So, in our configuration now, we consider the parallel segments $AB\parallel A_{1}B_{1}\parallel A_{3}B_{3}$ and based on above theorem, we conclude that the points $G,$ as the intersection point of diagonals of the trapezium $ABA_{3}B_{3},$ and the point $G^{*},$ as the intersection point of diagonals of $A_{1}B_{1}A_{3}B_{3}$ and the point $P,$ as the intersection point of the non parallel sides $($ the other than the point at infinity, concurrency of $AB,$ $A_{1}B_{1}$ $A_{3}B_{3}$ $),$ are collinear.

$\bullet$ As [b][size=117]Little Gauss[/size][/b] said, we can prove the relation $PG = 4\cdot (G^{*}G),$ by Menelaus’s theorem, considering the segment line $A_{1}A{3},$ as transversal of the triangle $\bigtriangleup AA_{1}A_{3}.$

So, we have that $\frac{A_{3}G}{A_{3}A}\cdot \frac{A_{1}A}{A_{1}P}\cdot \frac{G^{*}P}{G^{*}G}=1$ $,(1)$ but $A_{3}A = 3\cdot (A_{3}G)$ and $A_{1}A = A_{1}P.$

Hence, from $(1)$ $\Longrightarrow$ $G^{*}P = 3\cdot(G^{*}G)$ and the proof is completed.

Kostas Vittas.

PS. I will post here next days, the proof of the above first theorem of concurrent segments, which is true in every configuration of the given segments and their edges and it is easy to prove by Desarques’s theorem $($ In our configuration, we can see the perspective triangles $\bigtriangleup ABG,$ $\bigtriangleup A_{1}B_{1}G^{*}$ and then the conclusion of the collinearity of $P,$ $G^{*},$ $G$ $).$

[quote="gemath"]...
Hope you can understand more about my ideal in this topic! :)[/quote]

Yes dear gemath, I can understand what you mean and thank you for all the extensions (interesting and new results to me), of the problem about the Gauss's point in a triangle. :)

Kostas Vittas.

ok thank you for undestanding me!  :)
I will give another interesting about this way to looking at Gauss's line,
If we understand a Gauss's line as line connecting mid point of $PA$ and "centroid" of $BC$ we will have an extension for quadrilateral that I call it be Gauss's line of quadrilateral:
Given quadrilateral $ABCD$ and an arbitrary point $P$ let $G_{a}$ be centroid of triangle $BCD$ and $A'$ midpoint of $PA$ the line $d_{a}$ connecting $A'G_{a}$ we called a first Gauss's line of quadrilateral $ABCD$ corespond $P,A$ thus a quadrilateral $ABCD$  has four Gauss's line and they are concurrent at $G*$-Gauss's point of $ABCD$(Gauss's point of $ABCD$ I defined above).
If we understand midpoint of $PA$ as centroid of $PA$ we will have more three Gauss's line of quadrilateral as following we call $G_{ab},G_{cd}$ be centroid of triangle $PAB,PCD$ the the line $d_{a}'$ connecting $G_{ab},G_{cd}$ we still consider as the second Gauss's line thus we will get three second Gauss's line three lines are concurrent at a point $G**$ but that pity it isn't Gauss's point but it lies on $PG*$ thus it is quite nice.
Similar we can have Gauss's line for n-polygon as I do above!
I still have quite many ideal about this version I will post later!


LaTeX
geometry
3D geometry
algebra unsolved
algebra

i don't have latex to write it like that

but still looks hard

[quote="perfect_radio"]Solve in $\mathbb R$ the equation $\sqrt[3]{3 \cdot 2^x - 6}+\sqrt[3]{2 \cdot 2^x - 4}=\sqrt[3]{3 \cdot 2^x - 4}+\sqrt[3]{2 \cdot 2^x - 6}$.

Any ideas?

[i]Edited by Myth[/i][/quote]
Well, mmm, I am sure there are a LOT of solutions close to $+\infty$, in fact in the limit are equal, but I purppose take left the $6$ factors and the $4$ factor to the right and use $x^3+y^3=(x+y)(x^2-xy+y^2)$ and some factors simplify, after then you can try to solve, in my opinion it's a really HARD equation, but good luck!!

Best regards.

RX the Crazy Math

[quote="perfect_radio"]i don't have latex to write it like that[/quote]
You don't need Latex to use it on this forum. Just learn it from examples here.

take the cube of both sides and simplify the equality ,then the equation u have becomes easier

[quote="Myth"]no, no, no

See. You have $S=\sqrt[3]a+\sqrt[3]b=\sqrt[3]c+\sqrt[3]d$ and $a+b=c+d$, so after taking the cube you will receive: $S\sqrt[3]{ab}=S\sqrt[3]{cd}$. Therefore $S=0$ or $ab=cd$. But $S=0$ can be easily analyzed and $ab=cd$ with $a+b=c+d$ implies $a=c$ and $b=d$ OR $a=d$ and $b=c$.[/quote]
Good job, I thought the same, but I think this is a high school problem  :D  :D Because my students in the university solves harder problems than this one.

RX TCM

[quote="Dr Sonnhard Graubner"]hello, i have found no solutions here.
Sonnhard.[/quote]

Because, of course, $x=1$ isn't a real number. :roll:

hello, but for $x=1$ we get
$\sqrt[3]{-2}$?
Sonnhard.

Oh dear, you really can't see anything past WolframAlpha? Odd-indexed roots are defined over all $\mathbb{R}$, in spite of WA's stubborn exclusion of the negative domain.


integration vector calculus trigonometry

A satellite falls free towards the Earth starting from rest at a distance $ R$, much larger than the Earth's radius. Treating the Earth as a point of mass $ M$, use dimensional analysis to show that the time $ T$ taken by the satellite to reach the Earth is given by
\[ T \equal{} C \left( \frac{R^3}{GM}\right)^{\frac{1}{2}},\]where $ G$ is the gravitational constant and $ C$ is a dimensionless constant.

By integrating the equation of motion of the satellite, show that $ C \equal{} \frac{\pi}{2\sqrt{2}}$.

So far I have that $ [G] \equal{} \textup{kg}^{\minus{}1}\textup{m}^{3}\textup{s}^{\minus{}2}, [R] \equal{} \textup{m}, [M] \equal{} \textup{kg}$, where I've used SI units.

So by dimensional analysis, say $ T \equal{} f(R, G, M)$ so that $ \textup{s} \equal{} [\textup{kg}^{\minus{}1}\textup{m}^{3}\textup{s}^{\minus{}2}]^{\alpha}[\textup{m}]^{\beta}, [\textup{kg}]^{gamma}$ and by comparing powers on both sides, $ \minus{}2\beta \equal{} 1, \gamma \minus{} \beta \equal{} 0, \alpha \plus{} 3\beta \equal{} 0$. Then we can the part of the formula in the brackets. The constant $ C$ should be there because $ T$ is probably dependent on something else that can be expressed in dimensionless form - I'm not sure this is the correct reason so if you know please explain.

Finally, how (and which equation) do you integrate to get the value of $ C$?

[quote="aidan"]The constant $ C$ should be there because $ T$ is probably dependent on something else that can be expressed in dimensionless form - I'm not sure this is the correct reason so if you know please explain. [/quote]
Not necessarily.. Here it is due to the choice of units.. If I'd consider " [i]'C'[/i] seconds " as a unit of time.. then the 'C' would not be there..

[quote="aidan"]Finally, how (and which equation) do you integrate to get the value of ?[/quote]
[url=http://www.mathlinks.ro/viewtopic.php?p=1755469#1755469]Refer to your other thread I've explained.[/url]
It applies here also (Actually it can be aplied anywhere).
Here find $ a(x)$ .. Use $ a\equal{}v\frac{dv}{dx}$ to get $ v(x)$ and finally use $ v\equal{}\frac{dx}{dt}$ to find $ x(t)$. Put x=0 to find the $ T$. The constant in that would be your $ C$.

[quote="savy2020"]Not necessarily.. Here it is due to the choice of units.. If I'd consider " 'C' seconds " as a unit of time.. then the 'C' would not be there.. [/quote]

But we can't choose units such that $ C$ isn't present right? This has something to do with [url=http://en.wikipedia.org/wiki/Buckingham_pi_theorem]Buckingham's pi theorem[/url].. but I don't fully understand it either even after reading several times.

The acceleration of the object is given by wikipedia to be $ \frac{GM}{r^2}$ where $ G$ is the gravitational constant, $ M$ is the mass of Earth and $ r$ the distance of the object from the center of Earth.

Thus we have $ \frac{GM}{x^2=  v \frac{dv}{dx}$. Rearranging and integrating, $ GM \int \frac{dx}{x^2} = \int v \, dv$ so $ -\frac{GM}{x} = \frac{1}{2}v^2 + c$, $ c$ a constant. Correct so far?

[quote="aidan"]
But we can't choose units such that $ C$ isn't present right? This has something to do with [url=http://en.wikipedia.org/wiki/Buckingham_pi_theorem][b]Buckingham's pi theorem[/b][/url].. but I don't fully understand it either even after reading several times.
[/quote]
That's interesting.. I've never known of it  :oops:

[quote="aidan"]

Thus we have $ \frac {GM}x^2 \equal{} v \frac {dv}{dx}$. Rearranging and integrating, $ GM \int \frac {dx}{x^2} \equal{} \int v \, dv$ so $ \minus{} \frac {GM}{x} \equal{} \frac {1}{2}v^2 \plus{} c$, $ c$ a constant. Correct so far?[/quote]
Well it has to be $ \frac {GM}x^2 \equal{} \minus{} v \frac {dv}{dx}$. Because you've to consider the directions.. The 2 vectors $ \overrightarrow{x}$ and $ \overrightarrow{dv}$ are essentially in opposite directions.

Ok I will try that :)

So: $ \frac{GM}{x} \equal{} \frac{1}{2}v^2 \plus{} c$ where $ c$ is a constant. (I'm sure whether we should be doing an indefinite or definite integral at this point, but I suspected the former).

Now if $ x \equal{} R$, $ v \equal{} 0$, so we get $ c \equal{} \frac{GM}{R}$.

Rearranging, $ v \equal{} \frac{dx}{dt} \equal{} \sqrt{2GM \left( \frac1x \minus{} \frac1R \right)}$; I get the integral $ \int_R^0 \left( \frac1x \minus{} \frac1R \right)^{\minus{}\frac12} \, dx \equal{} \int_0^T \sqrt{2GM} \, dt$. The LHS isn't easy to simplify though.. have I made a mistake?

[quote="aidan"]
So: $ \frac {GM}{x} \equal{} \frac {1}{2}v^2 \plus{} c$ where $ c$ is a constant. ([i]I'm sure whether we should be doing an indefinite or definite integral at this point, but I suspected the former[/i]).
[/quote]
Anything -- definite or indefinite -- will do.

[quote="aidan"]I get the integral $ \int_R^0 \left( \frac1x \minus{} \frac1R \right)^{ \minus{} \frac12} \, dx \equal{} \int_0^T \sqrt {2GM} \, dt$. The LHS isn't easy to simplify though.. have I made a mistake?[/quote]
Well there seems to be no mistake...

[url=http://integrals.wolfram.com/index.jsp?expr=sqrt%5B%28Rx%29%2F%28R-x%29%5D&random=false]Use the online integrator for integration[/url]

You get..
[img]http://u3pwiq.blu.livefilestore.com/y1pp55DQd8jFZ_WZV6jB7rQGZKSojk7Xj6ZaK755l9iMGP65rCW6Y49D4L4VS4TSWZDX0jI61P9jdblBMTJV5spz0DrZcKXeGbd/MSP22765357527573949052_422.gif[/img]

I've gotten  a quite simpler expression:
\[{ \int \sqrt {{rx}\over{r - x}}}\; dx = \sqrt {r^3} .\left (tan^{ - 1} \sqrt {{x \over {r - x}}} - {\sqrt {x.(r - x)} \over r} \right)\]
It's equivalent...but "with blood, sweat and tears" (remember?) no integrator.

This is how I did it in the end: the integral is $ \int_R^0 \sqrt{\frac{Rx}{R \minus{} x}} \, dx$. Let $ x \equal{} R \sin^2 \theta$, a very nice substitution. Then the integral becomes
\[ 2 R\sqrt{R}\int_{\frac{\pi}{2}}^{0} \sin^2 \theta \, d\theta \equal{} \sqrt{2GM}T \\
2 R\sqrt{R}\int_{\frac{\pi}{2}}^{0} \frac{1}{2} (1 \minus{} 2 \cos \theta) \, d\theta \equal{} \sqrt{2GM}T \\ R\sqrt{R} \left[ \theta \minus{} \frac{1}{2} \sin 2\theta \right]_{\frac{\pi}{2}}^0 \equal{} \sqrt{2GM}T \\ \Rightarrow R\sqrt{R} \cdot \frac{\pi}{2} \equal{} \sqrt{2GM}T \\ \Rightarrow T \equal{} \boxed{\frac{\pi}{2\sqrt{2}} \left( \frac{R^3}{2GM}\right)^{\frac{1}{2}}}.\]


[hide]Let's take the sum of the squares of the two membrers:$5x^2+14x+9-18-x^2+x+20=5(x+1)(x-\frac{1}{5}+2)-(x+1)(x+2)$ so $(R_1)^2\equiv -(R_2)^2$ mod (x+1)

First root=8

Divide everything by $\sqrt{x+1}$: we obtain $\sqrt{5x+9-\frac{18}{x+1}}-\sqrt{x+2+\frac{18}{x+1}}$. So $\frac{18}{x+1}=c$, then $4x-18(\frac{2}{x+1}+1)=\sqrt{20(x+2)}$, and again $18(\frac{2}{x+1}+1)=d \& 16x^2-8dx+d^2=20x+40$, solve and put a system the two solution plus $d=18(\frac{2}{x+1}+1)$[/hide]


superior algebra
superior algebra unsolved

$ R$ is commutative ring with $ 1$ .$ X$ is a finite set of prime ideals of $ R$.$ R$ has only one ideal maximal $ m$.and none of members of $ X$ is maximal.and suppose  $ m\minus{} m^2 \subseteq \cup_{p \in X}$.prove  if $ m$ be finite generated then $ X$ is emty.

I try to use this theorm:if $ I$ be an ideal of $ R$ so that $ I \subseteq  \cup_{p \in X}$ then $ \exists q \in X$ so that $ I \subseteq q$.because from $ m\minus{} m^2 \subseteq \cup_{p \in X}$ we can conclude $ m \subseteq  (\cup_{p \in X}) \cup  m^2$

So what is your problem? Your idea is right. The only problem is to argue why $ m \neq m^2$. This is because when $ m$ is finitely generated, by Nakayama lemma $ m \equal{} 0$. This forces $ R$ to be a field and then of course, if $ X$ cannot contain $ (0)$ then it has to be empty.


algebra
function
domain
number theory
prime numbers
Ring Theory
number theory unsolved

Find all natural numbers $ a, b, c$ such that:
$ a^2\plus{}1$ and $ b^2\plus{}1$ are prime numbers and $ (a^2\plus{}1)(b^2\plus{}1)\equal{}c^2\plus{}1$

$ (a^2 \plus{} 1)(b^2 \plus{} 1) \equal{} (ab \plus{} 1)^2 \plus{} (a \minus{} b)^2 \equal{} (ab \minus{} 1)^2 \plus{} (a \plus{} b)^2$.  By some lemmas in the Gaussian integers, these are the only representations of $ c^2 \plus{} 1$ as a sum of squares, hence either $ a \minus{} b \equal{} 1$ or $ a \plus{} b \equal{} 1$.  But then $ a, b$ have opposite parity, so one of $ a^2 \plus{} 1, b^2 \plus{} 1$ is even.  So $ (a, b, c) \equal{} \boxed{(1, 2, 3)}, \boxed{(2,1,3)}$ are the only solutions.

[quote="t0rajir0u"] By some lemmas in the Gaussian integers, these are the only representations of $ c^2 \plus{} 1$ as a sum of squares, hence either $ a \minus{} b \equal{} 1$ or $ a \plus{} b \equal{} 1$.  [/quote]

Hmm.. For a person like me (I do not know gaussian integers lemmas, what to do with sum of squares, etc.) this proof cannot be understood... I hope someone will show more elementary proof of my question, but I also would like you -  t0rajir0u to explain some theory you have used...

thank you

I have used two lemmas.

[b]Lemma:[/b]  If a prime $ p$ is expressible as the sum of two squares, this representation is unique.  $ 2p$ is also expressible as the sum of two squares uniquely.

[b]Lemma:[/b]  If two odd primes $ p, q$ are expressible as the sum of two squares, then $ pq$ is expressible as the sum of two squares in exactly two ways.

The underlying truth behind these lemmas is that $ \mathbb{Z}[i]$ is a unique factorization domain, but there are elementary demonstrations of them, I think.  I just don't remember them at the moment...


Let r and s be natural numbers and t = (r,s). Show that

a^r is congruent to 1 which is congruent to a^s (mod m)
implies that
a^t is congruent to 1 mod m.

[hide="Solution"] By Bezout's Lemma, $\exists x, y : rx+sy = t$.  But then

$a^{rx+sy}\equiv 1 \bmod m$ [/hide]


AMC
AIME
USA(J)MO
USAMO
email
AIME I
AMC 10

When are scores from the 12B going to be sent to schools?  I'm just somewhat curious.  Alright, thanks.

i think the results for the A test, were sent with e-mail on the 17th...
that would be 17 days after the test....

assuming it takes the same time, you figure out when you will get the results. :)

[quote="lowest_reynolds"][quote="krustyteklown"]it probably won't take the same time, a lot less students take the B test, and there's no overseas students
also you need the results by the time the AIME is done[/quote]

Isn't Melanie (mel) an overseas student?[/quote]

who?

i mostly meant those people like, in china, that's why they had to lock the discussion of AMC A for so many days, where as AMC B was free the day after

Please note that we were able to score and report on about 128,000 A-date answer sheets from about 2,500 schools in 17 days.   Scoring and reporting on approximately 100,000 B-date answer sheets from about 2,000 schools is underway as I write this.  I expect that it will take about the same amount of time.  The entire AMC staff is well aware of the AIME I dealine and the entire staff is currently working overtime to get the reports out to you as fast as possible.

Steve Dunbar
MAA Director, American Mathematics Competitions

[quote="amirhtlusa"]is this for the A test or the B test?!
also are  you talking about the e-mail or the actual sent report?[/quote]
B test - and I just meant for my school, of course ;-)  Also, I don't know whether we got the actual report; I didn't think to ask.  Probably it was just the e-mail, now that I think about it, since the names and scores of the AIME qualifiers were copy-pasted into the e-mail our teacher sent us.

[quote="h_s_potter2002"][quote="zanttrang"][quote="h_s_potter2002"]Our school just got our's today.[/quote]
Our school got our scores yesterday.  Kind of sad that I was the top scorer with a 126.5...[/quote]

I was my school's highest scorers.  But that's probably because on two people took it at our school.[/quote]

I go to a middle school, but 9 out of the 10 of us took it (the 10th one who didn't make it was ranting "my calculator lied to me!" after the test.) made AIME. The 10th made AIME through the 10A though.

Scary part is: A 6th grader got 112, a 7th grader got 114.5, and an 8th grader got 139.5 (it was higher than his 10A...)

our school got ours yesterday or the day before I believe, as my teacher gave me a copy of the results today.

Funny, we haven't got the official AMC 10B results, nor received an e-mail. Do they take longer for overseas schools?

> Funny, we haven't got the official AMC 10B results, nor received an e-mail. Do they take longer > for overseas schools?

Yes, the answer sheets from international schools get processed last because the reports for international schools get sent by DHL package service, while USA schools go by regular mail and the delivery time is longer, so they need to be processed first.

Steve Director
Director, American Mathematics Competitions

[quote]Yes, the answer sheets from international schools get processed last because the reports for international schools get sent by DHL package service, while USA schools go by regular mail and the delivery time is longer, so they need to be processed first. [/quote]

Thanks for the info.

> When will the state and overall scores be posted for the AMC A?

Probably in a couple of weeks.  Right now we are focusing all of our attention on getting the AMC 10 A , AMC 12  A, AMC 10 B and AMC 12 B school reports out to schools, and simultaneously delivering the necessary AIMEs.  The AMC staff is working overtime to get that done.  Even after AIME I is delivered, the special requests for AIME II require considerable handling and attention.  We will begin scoring the AIME  I about March 8, sending the AIME IIs and preparing for USAMO at that time.

Unbelievably, in the middle of this, we are still receiving packages of A-date and B-date answer sheets in the mail.  Scoring and adding them into the mix, together with editing and data sorting and analysis  necessary to produce comprehensive state reports, it will be at least two weeks.

Steve Dunbar
Director, American Mathematics Competitions

Hi AMC Director.  My school took the contest B, but we have still not received the results.  My teacher called the AMC, and left a message, but did not get to speak to anyone in person.  I'm worried we won't get to take the AIME.  If we cannot receive our material by March 7, would it be possible to allow us to take the AIME II?  Thanks.

> My school took the contest B, but we have still not received the results.

Probable reason:  your teacher did not provide an email address to send the electronic results.
Second reason, due to an larger than usual number of schools aking hte B-date, we have had more results to mail, and it is taking longer.  You should have all necessary materials this week.

>My teacher called the AMC, and left a message, but did not get to speak to anyone in person.

Try calling between 8 and 12 and 1 and 5 Central time.  Someone always answers the phone.  But we have been extremely busy answering phone calls and sending reports, and it possible that all the lines were occupied when your teacher called.

> If we cannot receive our material by March 7, would it be possible to allow us to take the AIME II?

If you have not received materials by March 6, have your teacher contact us.  We will make an alternate arrangement.

Steve Dunbar
Director, American Mathematics Competitions


modular arithmetic
geometric series
number theory
relatively prime
number theory unsolved

[quote="dule_00"]Is this true ${x \over y}\equiv z \Leftrightarrow x\equiv yz$[/quote]

This is the definition of modular division  :huh:

[quote="dule_00"]Find the last $ 1000$ digits of the number $ N\equal{}1 \plus{} 50^1 \plus{} 50^{2} \plus{} 50^3 \plus{} ... \plus{} 50^{999}$.[/quote]

Hey what happened with the other post  :?:

Anyway let me explain again.

You need to find,
$ N\equiv r (\mod 1000)$

Where,
$ 0\leq r\leq 999$.

(Basically the remainder upon division by 1000).

Now.
You can simplify that sum because it is a geometric series and then use some mod arithmetic to find the remainder.
You get that?

[quote="kathi;)"]but you just found the last 3 digits, ice_age wanted to have the last 1000 digits, so you shoud calculate $\mod ~10^{1000}$........?[/quote]

Yes.

Note when you add them you get,

${\frac{50^{1000}-1}{49}}\equiv x (\mod 10^{1000})$

So

$50^{1000}-1\equiv 49 x (\mod 10^{1000})$

But that huge number leaves remainder of 0.

So we just got to solve

$49x \equiv 999 (\mod 10^{1000})$

Now $\gcd(49,10)=1$ so it got a unique solution.

I do not think the problem has been solved successfully here.

[quote="dule_00"]Find the last $ 1000$ digits of the number $ N\equal{}1 \plus{} 50^1 \plus{} 50^{2} \plus{} 50^3 \plus{} ... \plus{} 50^{999}$.[/quote]

Solution:

The sum is$ N \equal{} \frac {50^{1000} \minus{} 1}{50 \minus{} 1} \equal{} \frac {50^{1000} \minus{} 1}{49}$,
Note: 1/49 has a recurring decimal of a period of 42 digits:
$ \frac {1}{49} \equal{} 0.\dot02040816326530612244897959183673469387755\dot1 \equal{} 0.\dot P$, which P denotes the above 42 digits.

The multiple of 42 nearest to 1000 is 1008=24*42.  So$ \frac {10^{1008}}{49} \equal{} \underbrace{PP...P}_{24 times}.\dot P$
and$ M \equal{} \frac {10^{1008} \minus{} 1}{49} \equal{} \underbrace{PP...P}_{24 times}$is an integer having 1008 digits.

The difference between N and M is
\[ N \minus{} M \equal{} \frac {5^{1000}*10^{1000} \minus{} 1}{49} \minus{} \frac {10^{1008} \minus{} 1}{49} \equal{} \frac {5^{1000} \minus{} 10^{8}}{49}\cdot 10^{1000}
\]
.
since$ 10^{1000}$ is relatively prime to 49, so$ \frac {5^{1000} \minus{} 10^{8}}{49}$is an integer.
therefore, the final 1000 digits of N conincides with M,i.e.,$ q \underbrace{PP...P}_{23 times}$, where q has 34 digits which are the last 34 digits of the number P.

---------------------Solved---------------------


geometry AMC AIME search calculus Stanford college

On April 22, a new math event -- I hope the start of an ongoing thing -- will take place at Google in Mountain View.  For more information, see [url]http://msri.org/specials/festival[/url] or contact me directly at joshua_zucker@castilleja.org

I hope if you're in the area, you can come enjoy with us!

And even if you're not in the area, you might enjoy the sample problem (Candy Conundrum) on the web site.

--Joshua Zucker

i want to come, but i doubt my school will do it
can i just come by myself?

sounds like it's more high school geared, by the example.  is there anything for middle-schoolers?

well AoPS middle schooler$\approx$high schooler  :P

daermon: a chaperon of some sort is required, and I do prefer it if you bring a group of people (from your school, or a local math club, or something -- mostly because I have a limited amount of space at Google, and I want it to be something like 200+ kids and 50ish adults, not 100ish kids and 100ish adults).  But registration will probably fill up in the next couple weeks, so probably the best move is to register yourself and a parent, and then see if you can form or join a group of some sort.

star99: yes, there will be plenty of middle-school-level things.  Many of them will be ones like the posted problem, where a middle schooler can understand the question and make some progress, while a high schooler would be expected to get most of the way through it.  The table leaders for that type of activity will take a look at your grade level in figuring out when to start handing out the raffle tickets for your accomplishments on the problem.

and daermon: given my experiences, many AoPS middle schoolers are closer to college level.  But here in the Bay Area we get a distorted picture because we have so many strong math students (at every level!)

--Joshua Zucker

I'd like to register a team from my school (Menlo).
However, we have several seniors in our math club (myself being a freshman) who scored 12 on the AIME.

Is this going to be too easy for them?
(I'm assuming it will be proper difficulty for me, with just a 5 on the AIME).

I'd say the AIME 12+ gang will have to shop around a bit for interesting activities, but many of the problems are sufficiently hard and open-ended to keep them busy, I'd say.

There'll be something like 20-30 different activity tables/problem sets, and each one will have somewhere around 10 problems, so the overall range of difficulty will be pretty large.  Some of the problems, at least to the best of my knowledge, are unsolved -- could be published, at least in OEIS or similar online place, if you solve them, or might make good science fair/talent search project material.

Heck, I can score at least that well on an AIME, and I wrote the problems, and I still can't do some of them!

--Joshua Zucker

Ok, I'll get a team ready, and I think our math director will come with us. (crosses fingers)  :read:

Sounds like fun!  The problems seem quite deep too.  I will probably be there.

The event is nearly full so if you're planning to come you should visit http://msri.org/specials/festival and follow the link to preregister soon.

Hope to see you there,
--Joshua Zucker

i registered as an individual :)
turns out my school heard about it and told people but i dont think anybody else wants to go
anybody else here going?

Just an update on our team situation: we're definately coming, though it may be as few as four people and as many as twenty or so.

I'm looking forward to this and the first ARML meeting definately!

How do you know if you are registered?

I think I am, but I'm not sure... :maybe:

My school did not register. They tried to, but it was too late.

So what do I do?

I want to go as an individual.

Yes, that girl is very smart, but super hyper.

[quote="star99"]jing-jing is a very smart girl.  she's just different.  she's only 13 or something, freshman and taking calculus[/quote]
Being "advanced" in high school math has othing to do with how smart a person is.
I'm sure I'm definitely capable of finishing my school's entire BC calc curriculum within a month.

The main bad thing about the festival was limited space.  All the good tables were taken.  However,  I see that they are taking measures to limit the amount of people that can go.  So I will definitely go next year (if I can get myself invited.

[quote="star99"]jing-jing is a very smart girl.  she's just different.  she's only 13 or something, freshman and taking calculus[/quote]

Stop making me feel retarded.

Anyway, I threw all my tickets in the box with my name on it.  :furious:

And were the people giving out the prizes looking at the person's ticket? It didn't look like it...  :|

yeah, nobody was checking
vishal was trying to get me to go up and pretend i had a ticket  :ninja:
but i didnt :P

i didnt win anything, but it wouldnt really have mattered because i put most of my tickets in the google gear box before realizing that there were more than one :huh:

yeah....that girl was pretty weird
i wonder why paul brought her up again?

[quote="DiscreetFourierTransform"]I think he liked the challenge.  :P
I love how he asked all the kids who had questions whether it was a smart [wagon] question before they could ask it. That's totally true. At the event, basically, it was us, the humble yet PWNAGE AoPS kids, and then there were the ARML kids (no less PWNAGE, but not quite as cool), and then my Menlo Friends (eric, you and vishal met my friend Daniel Fremont? We were the guys with the TI-89 doing the characteristic polynomial to the 100th power for that Fibonacci Flips one? Then we realized that the answer was simply twice the hundredth fibonacci number? Then we tried getting the roots for the cubic, and Vishal kept typing it in wrong?) Yeah, Daniel is pretty good at this stuff. Anyways, so then, after AoPS, ARML, and Daniel, THEN comes the noobs like that girl. Wow is all I have to say. Even if she is PWNAGE at math, I think she needs to think twice before going into a calculus class with seniors who may or may not tolerate that kind of oddity.

Not that it wasn't funny, and she did a good job as cannon fodder for Zeitz to pwn at puppies and kittens.

I just used the words Cannon Fodder and Puppies in the same sentence. Now I'm a horrible human being.
Oh well.

[/end rant][/quote]
um
wow that was a ranty rant

Was she hot?

To keep this post on topic, do you people think it is worth going to?

i'll ignore the first part of your post... :o
definitely!  the problems were great :)

[quote="i_like_pie"][quote="Ubemaya"]Was she hot?[/quote]
:rotfl:

I never would've expected that from you.[/quote]
i guess you haven't seen this....
[url]http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=132839[/url]

@star99: really, taking calculus as a freshman isn't [i]that[/i] special. I can think of at least a dozen or so people off the top of my head in this area who might be allowed to do that and at least a half dozen that actually do (or are about to) See, if they'll let [i]me[/i] into calculus as a freshman, it really isn't that special...

[quote="vishalarul"] Stop making me feel retarded. [/quote]
Wow, you're going to AP Calc next year, too...

Seriously, I put about 8 tickets in the tours box, forgot to put my name on any, was told that I had put the wrong parts of the tickets in, put in 6 more later properly, and didn't get anything... :bomb:

don't make me rant about it...

overall, it was cool, though

P.S. @ubemaya: Seriously, stop that; its really disturbing


function
geometry
3D geometry
sphere
algebra unsolved
algebra

1) Find all natural numbers $(x,y,z)$ which satissfy the following equations
$x+y+z=6$ and $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2-\frac{4}{xyz}$

2) Find all integer numbers $(x,y,z)$ which satissfy the following equations
$3(x^2+y^2+z^2)=1$ and $x^2y^2+y^2z^2+z^2x^2=xyz(x+y+z)^2$

3) Find all numbers $(x,y)$ for which $x|x|+y|y|=1$ and $[x]+[y]=1$
(|-1|=1,[1,1]=1)

The solution set for Part 1 is {(2,2,2)}.

I haven't proved this, but this is what I did:

1) I solved for x in terms of y (I have the equation, but its long; I might put it up later).
2) I tried the first 1,000,000 natural number for y.
3) Out of the results, only y=2 yielded a natural number for x (specifcally x=2).
4) I substituted back in the original equation to filter out extaneuous solutions, and (2,2,2) works.)

What is [x] in Part 3?

Is this the least integer function?

[z] denotes the greatest integer less than or equal to z.
Try to solve it in an easier way  ;)

@1: arent there only $7+6+5+4+3+2+1= 28$ solutions for $x+y+z=6$
or other question: are the natural number $\mathbb N =\{0,1,\dots\}$

...

peeta you are right. So the first sistem have errors leave it. What about the second two???
How can i solve them

The second one is probably wrong...it's the same as asking:

find the integer points on the sphere $x^2+y^2+z^2=\frac{1}{3}$, such that they satisfy the other equation...but there's no such points on that sphere. So there's no solution for this system...at least not for intergers..

:surrender:

http://www.zaba.ru/cgi-bin/tasks.cgi?tour=national.pmo.1997.3

http://www.artofproblemsolving.com/Forum/resources.php?c=137&cid=39&year=1997

In the first the degree is- 2
In the second-3

 :stink:

$...xyz(x+y+z)^2$
$...xyz(x+y+z)^3$

Solve them both in real numbers

http://www.zaba.ru/cgi-bin/tasks.cgi?tour=national.pmo.1997.1
this is in real numbers too.

Now I hope it`s clear  :roll:

[quote="Beat"]3) Find all numbers  for which
(1)   $x|x|+y|y|=1$
and
(2)   $[x]+[y]=1$.[/quote]

If $x\geq 0$ and $y\geq 0$ we have either $[x]=0$ and $[y]=1$, or $[x]=1$ and $[y]=0$. In both cases $x|x|+y|y|\geq 1$ and the equality holds if and only if  ($x=0,y=1$ or $x=1,y=0$).

If $x<0$ and $y<0$ relation (1) is impossible.

If $x\geq 0$ and $y<0$, from (1) we get
        $x\geq 1$ and $y=-\sqrt{x^2-1}$.
OTOH, $y\geq [y]=1-[x]\geq 1-x$, so
        $-\sqrt{x^2-1}\geq 1-x$,
from which we easily obtain $x\leq 1$. Since we had $x \geq 1$, we get $x=1$, hence $y=0$ (see (1) ), which is false (we assumed y<0).

The 4th case is analogous to the third.


topology
geometry
advanced fields
advanced fields theorems

Can anyone tell me the name of any book on moduli spaces.I want a easy and simple treatment on the subject.

Which kind?  Moduli spaces on elliptic curves or those appear in low-dim topology?  For the latter one, Moore's lecture notes on Seiberg-Witten theory may give you some idea, but for details I think one has to turn to Donaldson's geometry on 4-manifolds.

How about the first one i.e. treatment of moduli spaces on elliptic curves.


calculus
integration
trigonometry
real analysis
real analysis unsolved

Study the serie $\sum \frac{sin(\pi\sqrt n)}{n^\alpha}$.

Getting the cheap answers out of the way first:

Clearly, this converges absolutely for $\alpha>1,$ and for $\alpha\le0,$ it diverges becuase the terms fail to go to zero. Hence the interesting cases are for $0<\alpha\le1.$

Here's an analoguous problems for integrals; however we should not expect that this problem would have the same answers:

$\int_1^\infty\frac{\sin(\pi\sqrt{x})}{x^{\alpha}}\,dx$

Make the subsitution $w=x^2$:

$=2\int_1^{\infty}\frac{\sin \pi w}{w^{2\alpha-1}}\,dw$

Integrate by parts:

$=-\frac2{\pi}\frac{\cos \pi w}{w^{2\alpha-1}}|_1^{\infty}- \frac{2(2\alpha-1)}{\pi}\int_1^{\infty}\frac{\cos\pi w}{w^{2\alpha}}\,dw$

This converges for $\alpha>\frac12.$ It diverges for $\alpha\le\frac12.$

A not too rigourous answer but perhaps the main idea is there :

${2 \times k/2n \leq sin(\pi\sqrt{n^2+k})} \leq \pi \times k/2n$ for $0 \leq k \leq n$

So if $a \leq 1/2$, we can find arbitrarily large sub-sum of the serie thus it is not Cauchy and diverges.
Conversely if $a > 1/2$ the serie is Cauchy and thus converges.


inequalities
number theory
prime numbers
number theory unsolved

Let $ A$ be a finite set of prime numbers, and let $ a$ be a positive integer. Prove that there are only finitely many positive integers $ m$ for which all prime divisors of $ a^m\minus{}1$ are in $ A$.

suppose the inverse
let$ A$={$ p_{1},....,p_{m}$}.
we can assume that every $ p_{i}$ divide  $ a \minus{} 1$.
from Zsigmondy's theorem  we have $ \parallel{}a^n \minus{} 1\parallel{}(p_{i})\leq \parallel{}a^2 \minus{} 1\parallel{}(p_{i}) \plus{} n$ for every  $ i$.
multiplying these inequality we have $ a^n_1 \leq n(a^2 \minus{} 1)$ for every $ n$ with condition of problem.but if n is large enough we have the inverse.

[quote="sumita"]
from Zsigmondy's theorem  we have $ \parallel{}a^n \minus{} 1\parallel{}(p_{i})\leq \parallel{}a^2 \minus{} 1\parallel{}(p_{i}) \plus{} n$ for every  $ i$.
.[/quote]

Are there more elementary ways?  :)


Does anyone know which textbook math 132 uses? Also, what about 25b? Does 25b introduce differential forms? And so is 132 an introduction to calculus on manifolds? Thank you!

I am a senior, but I am curious for certain reasons.

you go to Harvard? thought u were a senior...


Let $K$ be a subfield of $\mathbb C$ such that $[\mathbb C: K]=2$.

(a) Can $K$ be isomorphic to a proper subfield of $\mathbb R$?
(b) Can $K$ be such that $\mathbb R$ cannot be embedded in $K$, and $K$ cannot be embedded in $\mathbb R$?

So I haven't had too advanced of theory in this, but I think there is a theorem that says that if $E/K$ is a field extension then $E=K$ iff $[E: K]=1$. If that is incorrect then the rest of this will be too.

Suppose $K$ is isomorphic to a proper subfield of $\mathbb{R}$. Then $[\mathbb{R}: K]=n$ for some $n>1$. Now we have $[\mathbb{C}: K]=[\mathbb{C}: \mathbb{R}]\cdot [\mathbb{R}: K]=2n$ a contradiction since $n\neq 1$. So $K$ cannot be isomorphic to a proper subfield of $\mathbb{R}$. Of course unless it is possible the order of $\mathbb{R}$ over one of its proper subfields to be $1$.

The first paragraph is correct; the second is not. If $K\subset \mathbb C$ is [b]isomorphic[/b] to a proper subfield of $\mathbb R$, it does not follow that $K\subset \mathbb R$. For example, $\mathbb Q(\pi i)$ is isomorphic to a proper subfield of $\mathbb R$ (such as $\mathbb Q(\pi)$), but it is not contained in $\mathbb R$. Also, if $K'\subset \mathbb R$ is a field isomorphic to $K$, then it's not necessarily true that $[\mathbb C: K']=[\mathbb C: K]$. (Although this is true if the isomorphism between $K$ and $K'$ extends to an automorphism of $\mathbb C$).

Are there any standard counterexamples to if $K'\subset\mathbb{R}$ and $K'\cong K$ where $K\subset\mathbb{C}$, then $[\mathbb{C}: K]=[\mathbb{C}: K']$? When I try to think of one it ends up working (probably because I keep thinking of ones where the isomorphism can be extended to an automorphism of $\mathbb{C}$).

sure there are: take any two sets of algebraically independent elements of $\mathbb{R}$ (and it's known that you can choose a continuum set with such property) $X,Y$ with the same cardinality, and consider $\mathbb{Q}(X), \mathbb{Q}(Y)$. these fields are isomorphic and they're different.

anyway, if i'm not wrong, $\mathbb{R}$ itself is isomorphic to a proper subset of itself... i'm sure $\mathbb{C}$ has this property, and i think that intersecting this proper subset with $\mathbb{R}$ gives the answer to (a).

the answer to (b) seems to be negative, but i'm not at all sure about it..

just one more thing: if $[K: k]=2$, and $K$ is algebraically closed, then $K = k(i)$... maybe this could help a little...


How many different paths are possible in moving from $ A$ to $ B$ given that you must move down to the right or down to the left?

[asy]for (int i=0; i<4 ; ++i){
draw((0+1.4i, 0-1.4i)--(-4.2+1.4i,-4.2-1.4i),linewidth(0.7));
draw((0-1.4i,0-1.4i)--(4.2-1.4i,-4.2-1.4i),linewidth(0.7));}
label("$A$",(0,0),N);
label("$B$",(0,-8.4),S);[/asy]

You want to go straight down, 6 steps. That means you must move the same amount of times right as you move left. Therefore, you want to take 3 steps left and 3 right. This can be arranged in any order, so the answer is 6C3=[b]20.[/b]

Or, you could think of it like: you have to move 3 to the right eventually, and 3 to the left, how many ways can we order that?

6!(NUMBER OF OBJECTS)/3!(3 REPEATS OF LEFT)*3!(3 REPEATS OF RIGHT)
or
6!/3!*3!
or
720/6*6
or
20

YAY!


geometry
3D geometry
quadratics
number theory
number theory theorems

I am still finding my way here
So tell me if this 'lemma' should be in another number theory topic

apparently in order to do elementary treatment of  certain equations, fermat used this lemma :
if $N^3=a^2+A b^2$ with $A\in {1,2,3}^$ and $(a,b)=1$ then you can find u and v such that :
$u^2+A v^2=N$ and
$a=u^3-3 u v^2A$ and $b=3 u^2 v-A v^3$

so basically he is saying, if you do look at it with quadratic rings, that for a proper representation, you can take a third root 'of that representation'

That surprisingly comes in handy when considering certain elliptic curves like $y^2+2=x^3$

now just for the case $A=1$ any remarks, hints, of a proof, are more than welcome, elementary or with factorisation arguments, because i am kinda stuck on this

Isn't there an A missing in your second equation¿

There is an gaussian integer $a+bi$ such that $N^3=a^2+b^2=(a+bi)(a-bi)$.
Since $\gcd(a,b)=1$, we have $\gcd(a+bi,a-bi)=\gcd(a+bi,2a)|2$, and thus (the factors $(1+i)|2$ are easy to handle) $a+bi$ is the third power of a gaussian integer $u+vi$.
Then clearly $N=u^2+v^2 = (u+iv)(u-iv)$ and $a+bi=(u+iv)^3 = (u^3-3uv^2)+i(3u^2v-v^3)$, so comparing coefficients gives the result.

The same works for the other $A$.

yes i know but than you use unique factorisation, while i heard fermat had this completely elementary in a descent fashion?

you were right about the incorrect equation,it is edited

but, even if you use factorisation, what would you do when $A+3$ some steps are now more difficult because of a more complicated unit set in that quadratic ring?

The units are not a that big problem:
for $A=1$, there are four
for $A=2$, there are two
for $A=3$, there are two

Instead of using unique factorisation, you could also find such a representation by using $(a^2+b^2)(c^2+d^2)=(ac-bd)^2+(ad+bc)^2$, finding all (theres just one) representations for primes and showing that you can reach all representations for any integer by this.

wow that's got me completely confused

the last thing you said is extremely helpful, because showing that that way of constructing makes everything comes in handy

however, when $A=3$ you have two units if zou consider $\mathbb{Z}[\sqrt{-3}]$ but that is not the ring of algebraic integers of its fraction field, i mean, that is not the right ring to use when needing unique factorisation which i think you do as you say a product of two things is third power, so they both are because they are coprime....


let x,yand z be a positive real numbers such that x+y+z=1
prove that (1+1/x)(1+1/y)(1+1/z) >= 64

Yeah!!!! Someone has readen my notes on inequalities! :cool:

Pierre.

it is sufficient to prove that
\[(x+y+z+x)(x+y+z+y)(x+y+z+z)\geq 64xyz\]

$x+y+z+x\geq 4\sqrt[4]{x^2yz}$
$x+y+z+y\geq 4\sqrt[4]{xy^2z}$
$x+y+z+z\geq 4\sqrt[4]{xyz^2}$

multiply the three inequalities above, we have the problem proved.

In fact it is pretty easy just write 1+1/x=1+1+y/x+z/x >= 4(yz/x^2)^(1/4) :lol:


Compute the sum

$\varphi \left( 0\right) +\varphi \left( \frac{1}{n}\right) +\varphi \left( \frac{2}{n}\right) +...+\varphi \left( 1\right) $,

where $\varphi \left( x\right) =\frac{4^{x}}{4^{x}+2}$.

  Darij

Thanks!

Actually the problem turned out to be much simpler than I thought. I myself had conjectured that $\varphi \left( x\right) +\varphi \left( 1-x\right) =1$, but I wanted to check it with Maple, and Maple gave out some nonsense (no idea why, but I am really not an expert in using Maple, hence I probably made something wrong), hence I withdrew my conjecture... And now I see the conjecture was true...

Actually the same is true for the more general function

$\varphi \left( x\right) =\frac{\left( a^{2}\right) ^{x}}{\left( a^{2}\right) ^{x}+a}$

with an arbitrary real a. In fact, we always have $\varphi \left( x\right) +\varphi \left( 1-x\right) =1$.

  Darij


function
parameterization
ratio
calculus
calculus computations

solve $ y" \plus{} x^2y' \plus{} xy \equal{} 0$

$ y \equal{} \displaystyle\sum_{n \equal{} 0}^{\infty} a_nx^n$
so, $ xy \equal{} \displaystyle\sum_{n \equal{} 0}^{\infty} a_nx^{n \plus{} 1}$

$ y' \equal{} \displaystyle\sum_{n \equal{} 1}^{\infty}na_nx^{n \minus{} 1}$

so $ x^2y' \equal{} \displaystyle\sum_{n \equal{} 1}^{\infty}n a_nx^{n \plus{} 1}$

$ y'' \equal{} \displaystyle\sum_{n \equal{} 2}^{\infty} n(n \minus{} 1)a_nx^{n \minus{} 2}$

=$ \displaystyle\sum_{n \equal{} 0}^{\infty}(k \plus{} 2)(k \plus{} 1)a_{k \plus{} 2}x^k$

subst. into $ y" \plus{} x^2y' \plus{} xy \equal{} 0$
$ \displaystyle\sum_{k \equal{} 0}^{\infty}(k \plus{} 2)(k \plus{} 1)a_{k \plus{} 2}x^k \plus{} \displaystyle\sum_{n \equal{} 1}^{\infty} na_nx^{n \plus{} 1} \plus{} \displaystyle\sum_{n \equal{} 0}^{\infty} a_nx^{n \plus{} 1}$

How to proceed from here?

Who gave you this problem?

Mathematica gives me the answer is $ e^{ \minus{} \frac {x^3}{6}} \sqrt {x} \left(C_1 \textbf{I}_{\frac {1}{6}} \left( \frac {x^3}{6} \right) \plus{} C_2 \textbf{I}_{ \minus{} \frac {1}{6}} \left( \frac {x^3}{6} \right) \right)$

Where $ C_1$ and $ C_2$ are arbitrary constants and $ \textbf{I}_n (x)$ gives the modified Bessel function of the first kind. The Bessel function is, in turn, [i]defined [/i]as the solution to another particular differential equation. So in that sense, it's probably not a very satisfying answer.

it was a question in my lecture notes

answer is

$ y(x)\equal{}a_0(1\minus{}\frac{x^3}{6}\plus{}...)\plus{}a_1(x\minus{}\frac{x^4}{6}\plus{}...))$

What you want to do is to line up the like powers of $ x,$ and to find a generic term with the same indicated power of $ x.$ The problem with what you've written is that you haven't equalized the powers of $ x$ so you can't yet add your sums. (I also find that summation notation gets in the way as much as it helps.)

$ y'' \equal{} 2a_2 \plus{} 3\cdot 2a_3 x \plus{} 4\cdot3a_4x^2 \plus{} \cdots \plus{} (n \plus{} 2)(n \plus{} 1)a_{n \plus{} 2}x^n \plus{} \cdots$

$ x^2y' \equal{} 0 \plus{} 0x \plus{} a_1x^2 \plus{} \cdots \plus{} (n \minus{} 1)a_{n \minus{} 1}x^n \plus{} \cdots$

$ xy \equal{} 0 \plus{} a_0x \plus{} a_1x^2 \plus{} \cdots \plus{} a_{n \minus{} 1}x^n \plus{} \cdots$

The sum has to equal zero, so these series add up to a series all of whose coefficients are zero. This gives us a infinite chain of equations for the unknown coefficients $ a_n:$

$ 2a_2 \equal{} 0$
$ 3\cdot2a_3 \plus{} a_0 \equal{} 0$
$ 4\cdot3a_4 \plus{} 2a_1 \equal{} 0$
$ \qquad\vdots$
$ (n \plus{} 2)(n \plus{} 1)a_{n \plus{} 2} \plus{} na_{n \minus{} 1} \equal{} 0$

Solve each of these equations for the highest-index coefficient. By that, we don't have any equations we can solve for $ a_0$ and $ a_1,$ so they are free parameters. (We expect two such free parameters in a second order equation.)

$ a_2 \equal{} 0$
$ a_3 \equal{} \minus{} \frac1{3\cdot 2}\,a_0$
$ a_4 \equal{} \minus{} \frac {2}{4\cdot 3}\,a_1$

The generic equation becomes $ a_{n \plus{} 2} \equal{} \minus{} \frac {n}{(n \plus{} 2)(n \plus{} 1)}\,a_{n \minus{} 1}.$

Each coefficient is a multiple of the one three before it. That gives us three chains of coefficients:

$ a_2 \equal{} a_5 \equal{} a_8 \equal{} a_{11} \equal{} \cdots \equal{} 0.$

$ a_3 \equal{} \minus{} \frac1{3\cdot 2}\,a_0,\ a_6 \equal{} \frac {4}{6\cdot5\cdot3\cdot2}\,a_0,\ a_9 \equal{} \minus{} \frac {7\cdot 4}{9\cdot8\cdot6\cdot 5\cdot3\cdot2}\,a_0,\dots$

$ a_4 \equal{} \minus{} \frac2{4\cdot 3}\,a_1,\ a_7 \equal{} \frac {5\cdot 2}{7\cdot6\cdot4\cdot3}\,a_1,\ a_{10} \equal{} \minus{} \frac {8\cdot 5\cdot 2}{10\cdot9\cdot7\cdot 6\cdot4\cdot3}\,a_1,\dots$

So we can write the general solution as $ y \equal{} a_0y_1 \plus{} a_1y_2$ where

$ y_1 \equal{} 1 \minus{} \frac {x^3}{6} \plus{} \frac {4x^6}{6\cdot5\cdot3\cdot2} \minus{} \frac {7\cdot 4x^9}{9\cdot8\cdot6\cdot5\cdot3\cdot2} \plus{} \cdots$ and

$ y_2 \equal{} x \minus{} \frac {x^4}{6} \plus{} \frac {5\cdot2x^7}{7\cdot6\cdot4\cdot4} \minus{} \frac {8\cdot 5\cdot2x^{10}}{10\cdot9\cdot7\cdot6\cdot4\cdot3} \plus{} \cdots$

For each series, it is quite easy to use the ratio test to show that each has a radius of convergence of $ \infty.$

thanks for the solution

So would you recommend using power series rather than summation notation?

[quote="bos1234"]So would you recommend using power series rather than summation notation?[/quote]
These are power series no matter how we write them. The choice is a notational one: whether to use summation notation or whether to expand the series out and use $ \dots$ in the notation. This isn't a question of what is right or wrong - this is just notation and preference. Summation notation has the advantage of being much more compact; it has the disadvantage (or at least, notable property) that you must constantly shift indices. That is, you have to be willing at any moment to rewrite $ \sum_{n\equal{}2}^{\infty}n(n\minus{}1)a_nx^{n\minus{}2}$ as $ \sum_{n\equal{}0}^{\infty}(n\plus{}2)(n\plus{}1)a_{n\plus{}2}x^n.$ Or maybe as some third possibility. In this context, you are doing that quite often. Expanded-out notation takes much more paper (or, for the lecturer, chalkboard or whiteboard space), but has the advantage of a more visual presentation.

I've taught this material and tried to explain it to classes. My preference for using expanded-out notation is a pedagogical choice. I just find it easier to explain that way.


[hide="this isnt exactly rigorous but..."]
$a=b=1 \rightarrow \boxed{2^2+2^2=2^3}$.
The graph of $y=8-2^{(2-x)^4+x^2}-2^{(2-x)^2+x^4}$ looks continuous and it looks like it has only one root in $(0,3)$ at $x=1$. And I'm going to bed... maybe i'll deal w/ logarithms in the morning :P [/hide]

[quote="socrates"][quote="shyong"]Find all positive real number $a,b$ such that it fulfills both the conditions

(i)$a+b=2$
(ii)$2^{a^4+b^2}+2^{a^2+b^4}=8$

 :)[/quote]

just AM-GM

classic one,no??[/quote]

that cool , and this solves my doubt ! thanks :lol:


Solve for a positive value of $ x$:

$ \frac{5\sqrt{2}}{x} \equal{} \frac{x}{\sqrt {8}}$

First, cross-multiply. $ 5\sqrt{2} \cdot \sqrt{8} \equal{} 5\sqrt{16} \equal{} 20$. Now we have $ x^2 \equal{} 20$, so $ x \equal{} \boxed{2\sqrt{5}}$.


Kermit the frog enjoys hopping around the inﬁnite square grid in his backyard. It takes him $ 1$ Joule of energy to hop one step north or one step south, and $ 1$ Joule of energy to hop one step east or one step west. He wakes up one morning on the grid with $ 100$ Joules of energy, and hops till he falls asleep with $ 0$ energy. How many different places could he have gone to sleep?

[hide]Label each place in the grid as an ordered pair.  WLOG Kermit starts at $ (0,0)$.  He makes 100 hops, so he ends up at a point $ (x,y)$ where $ x\plus{}y$ is even.  If $ y\equal{}100$, then $ x$ has one possible value ($ 0$).  If $ y\equal{}99$, $ x$ has two possible values ($ \minus{}1$, $ 1$).  If $ y\equal{}98$, $ x$ has three possibilities ($ \minus{}2$, $ 0$, $ 2$).  And so on.  So the answer is $ 1\plus{}2\plus{}\ldots\plus{}50\plus{}51\plus{}50\plus{}\ldots\plus{}2\plus{}1\equal{}51^2 \equal{} \boxed{2601}$.[/hide]

[quote="lingomaniac88"][hide]Label each place in the grid as an ordered pair.  WLOG Kermit starts at $ (0,0)$.  He makes 100 hops, so he ends up at a point $ (x,y)$ where $ x \plus{} y$ is even.  If $ y \equal{} 100$, then $ x$ has one possible value ($ 0$).  If $ y \equal{} 99$, $ x$ has two possible values ($ \minus{} 1$, $ 1$).  If $ y \equal{} 98$, $ x$ has three possibilities ($ \minus{} 2$, $ 0$, $ 2$).  And so on.  So the answer is $ 1 \plus{} 2 \plus{} \ldots \plus{} 100 \plus{} 101 \plus{} 100 \plus{} \ldots \plus{} 2 \plus{} 1 \equal{} 101^2 \equal{} \boxed{10201}$.[/hide][/quote]

Just a note, there's a typo fixed above (at $ y\equal{}0$ there are actually $ 101$ pairs).

This was an AMC question form the 2006 AMC 12B (number 18) test
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=433613#p433613

see the similarity?

The general formula is (n+1)^2 so we have 101^2=10201

when you practice a lot of old contest problems, you are bound to find 2 that are similar...unless its the AIME...where they always comeup with some unique question (or just add in a lot of calculation)...


with a,b,c>1 solve a^a+b^b+c^c>a^b+b^c+c^a

this inequality   solution
http://www.mathlinks.ro/viewtopic.php?t=210571


[hide]Added on paper got 27/8= 3 3/8[/hide]


function
counting
distinguishability
algebra
polynomial
geometry
binomial theorem

Could someone explain the logic of negative combinatorics to me?

I was looking at manipulations in generating functions problems, and i simplified everything down to the raw combinatorics but came up with only (and its in the answer key too) -(10C0)(-10C15) - (10C1)(-10C9)....  :?:

By definition, $ { \minus{} n \choose k} \equal{} \frac {( \minus{} n)( \minus{} n \minus{} 1)...( \minus{} n \minus{} (k \minus{} 1))}{k!} \equal{} ( \minus{} 1)^k {n \plus{} k \minus{} 1 \choose k}$.  (This is the definition relevant to the generalization of the [url=http://mathworld.wolfram.com/BinomialTheorem.html]binomial theorem[/url].)  This is $ ( \minus{} 1)^k$ times the number of [url=http://en.wikipedia.org/wiki/Multiset]multisets[/url] of $ k$ elements from a set of $ n$ elements, or the number of ways to put $ n$ indistinguishable balls into $ k$ distinguishable urns.

As for the "logic" behind why this works, classically this is the result of what's called a "combinatorial reciprocity theorem" (which, according to what I've seen, is a theorem of the form "if a polynomial has combinatorially meaningful values at positive integers, it has combinatorially meaningful values at negative integers") but to tell you the truth the only meaningful explanation I have seen is quite beyond HSB level and is currently an area of active research.  See the discussion [url=http://math.ucr.edu/home/baez/counting/]here[/url].


search
number theory unsolved
number theory

How many primes $p$ are there where
$p < x$ and ($p$/23)=-1 (not a square mod 23)

In particular, could you please provide a lower bound (and upper if you desire).  Thank you!

There exists a prime number theorem for primes in arithmetic progressions, so all you need is to classify the primes you search for in arithmetical sequences of common difference $23$ and first term the non square residues and directly apply this theorem.


calculus
integration
trigonometry
calculus computations

Let $a,b$ are real numbers.
Evalute

\[\int_0^{\pi} |a\cos nx+b\sin nx|dx\ (n=1,2,\cdots)\]

$a\cos nx+b\sin nx = \sqrt{a^2+b^2}\sin(nx + \theta)$.
Thus $\int_0^{\pi} |a\cos nx+b\sin nx|dx = \sqrt{a^2+b^2}\int_0^\pi |\sin(nx + \theta)|dx = 2\sqrt{a^2+b^2}$

[quote="liyi"]$a\cos nx+b\sin nx = \sqrt{a^2+b^2}\sin(nx + \theta)$.
[/quote]

How did you get this ?

As long as $a,b\not=0$, there exists $\theta$ such that

$\cos \theta=\frac {b}{\sqrt{a^2+b^2}}$

and

$\sin \theta=\frac {a}{\sqrt{a^2+b^2}}$

Expanding $\sin(nx+\theta)$ via angle-addition formulae yields the desired result.

Yes. It is well-known in 10th grade in China...


calculus
integration
function
real analysis
real analysis unsolved

Let $ (X, %Error. "texcal" is a bad command.
{B}, \mu)$ be a measure space and $ f : X \rightarrow \mathbb{R}\cup \{ \infty \}$ be a non-negative integrable function. Prove that $ \mu (\{ x : f(x)=\infty \}) =0$.

This seems pretty simple. However, I don't know how to prove that a set has measure zero. I know that if it is integrable then the integral is finite. However, I don't see how the set would have to have measure zero right now.

Then try the contrapositive: if that set has positive measure $ a$, the integral is infinite (because it's at least $ a\cdot \infty$.

This is going to look somewhat similar to the hint jmerry gave you on your other problem.

Let ${ E=\{x: f(x)=\infty}.$ Now let $ g_n(x)$ be the function that equals $ n$ for $ x\in E$ and zero otherwise. Since $ g_n<f$ everywhere, it follow that $ \int g_n<\int f.$

Can you finish it from there?


on #8 This is how i did it: it said south east and south west...or north east north west..something of that gooby .  So..assuming that when it says SE or NE (can't remember) it's gotta be exactly that..so between sides of length 8 and 10 is a right angle..making those three cities enclose a right triangle.  so 8^2 + 10^2=164 which is about 169 so that's why A.  Maybe that isn't the most correct way to do it..but it seems to have worked just dandy!

i am pretty sure that is how u are supposed to do it fubu... i did it the same way.

I took the 10B... score unknown... yet.

Edit:

well... now i know either i got 143 or 137.. depending on if I filled in #1 wrong. I knew it was 363... (C)... but im not sure if i filled in C or A... hehehehe

I did it the same way; you have to remember that it was number 8, so it doesn't require as much thinking as one might think.

rep123max: the way you did it, the lines have the same slope... it says SE and SW so they should be perpendicular

I took the 10B and received a 137. How I got there will be a very simple exercise for the reader. (Hehe)


The sequence $ a_N$ satisfies the recursion $ a_n\equal{}3a_{n\minus{}1}\plus{}28a_{n\minus{}2}$ for integers $ n>2$ where $ a_1\equal{}3$ and $ a_2\equal{}65$.

Find the explicit formula.

There is a rather general theory of how to deal with recurrences of this type; see, for example, [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=212490]here[/url].

A well-known way to solve recurrences like these is to assume your terms will be in a geometric series, that is $ a_n\equal{}\lambda^n$. Putting this into the recurrence, we get $ \lambda^n\equal{}3\lambda^{n\minus{}1}\plus{}28\lambda^{n\minus{}2}$, and dividing by $ \lambda^{n\minus{}2}$, we have $ \lambda^2\minus{}3\lambda\minus{}28\equal{}0$. Solving the quadratic, we get $ \lambda\equal{}\minus{}4,7$, so our series translates to $ a_n\equal{}a\cdot(\minus{}4)^n\plus{}b\cdot 7^n$. (Essentially, any sequence in that form will satisfy the given recurrence, but only one will meet the requirements given by the initial terms.)
We have $ a_1\equal{}3\equal{}\minus{}4a\plus{}7b, a_2\equal{}65\equal{}16a\plus{}49b$, which gives us the solutions $ a\equal{}b\equal{}1$. Thus, the explicit formula will be $ a_n\equal{}(\minus{}4)^n\plus{}7^n$.

Thanks! I get it now! That's a really good solution!


This is the AMC forum ... this should probably be in one of the High School Forums or something.


linear algebra
matrix
linear algebra unsolved

$ A,B \in M(n,C)$

1/ $ A \equal{} B$

2/$ e^{A} \equal{} e^{B}$

3/$ AB \equal{} BA$

4/$ Ae^{B} \equal{} e^{B}A$

5/$ e^{A}e^{B} \equal{} e^{B}e^{A}$

6/$ e^{A}e^{B} \equal{} e^{A \plus{} B}$

7/$ e^{A}e^{B} \equal{} e^{B}e^{A} \equal{} e^{A \plus{} B}$

Study => or <=> between the seventh proposition above.
Find  minimal statement to get  =>
You can also give counter example

Some cheap ones: (1) implies (2), obviously, but (2) does not imply (1).

$ A\equal{}\begin{pmatrix}0&\minus{}2\pi\\2\pi&0\end{pmatrix}$ and $ B\equal{}0.$ Then $ e^A\equal{}e^B\equal{}I.$

(3) implies (4), (5), (6), and (7). There's room for doubt for most of the converses.

For instance, let $ A$ be as above and $ B\equal{}\begin{pmatrix}2\pi i&0\\0&0\end{pmatrix}.$ Then $ AB\ne BA$, but (4) and (5) are both true since $ e^A\equal{}e^B\equal{}I.$

I need a little more thought about (6) and (7).

[quote="Kent Merryfield"]
I need a little more thought about (6) and (7).[/quote]
Of course $ (7) \implies (6)$, but in general we don't have $ (6) \implies (7)$


function
integration
geometry
3D geometry
ratio
calculus
limit

In $ \mathbb R^n$, the maximal function is defined as $ f^\star(x) \equal{} \sup |Q|^{\minus{}1}\int_Q |f|$, where the supremum is taken over cubes with center $ x$. Let $ f^{\star \star}$ be defined similarly, but with the supremum taken over all $ Q$ containing $ x$. Thus $ f^\star \leq f^{\star \star}$. Show that there is a positive constant $ c$ depending only on the dimension such that $ f^{\star \star} \leq c f^\star$.

 :(

Suppose $ a$ is any number less than $ f^{**}(x).$ Then there is a cube $ Q$ containing $ x$ such that $ \int_Q|f|>a|Q|.$ Now let $ \widetilde{Q}$ be the smallest cube centered at $ x$ which contains $ Q.$ Then $ \int_{\widetilde{Q}}|f|\ge \int_Q|f|>a|Q|$ so that $ \frac1{|\widetilde{Q}|}\int_{\widetilde(Q)}|f|>\frac{a|Q|}{|\widetilde{Q}|}.$

As for that last ratio: the worst case involves $ x$ lying in a corner of $ Q,$ in which case the side length of $ \widetilde{Q}$ would be twice the side length of $ Q.$ So we can say that $ \frac{|Q|}{|\widetilde{Q}|}\ge 2^{\minus{}n}.$ Hence we have that $ f^*(x)>2^{\minus{}n}a.$ Since $ a$ was any number less than $ f^{**}(x),$ we can take the supremum over such values of $ a$ to conclude that $ f^{**}(x)\le 2^nf^*(x).$

The point of this is that you may now use whichever of $ f^*$ or $ f^{**}$ best facilitates the proof of the result you want - it doesn't matter which.

What is the usefulness of this concept  :?:

Maximal functions appear in many forms in harmonic analysis (an area of mathematics). One of the most important of these is the Hardy-Littlewood maximal function. They play an important role in understanding, for example, the differentiability properties of functions, singular integrals and partial differential equations. They often provide a deeper and more simplified approach to understanding problems in these areas than other methods.

~~> [url]http://en.wikipedia.org/wiki/Maximal_function[/url]

If you want to prove that some sequence of operators converges [i]pointwise almost everywhere[/i], then you need for an appropriate maximal operator to satisfy an appropriate weak-type inequality. The Hardy-Littlewood maximal operator directly controls the pointwise behavior of the local averages, $ \lim_{|Q|\to 0}\frac1{|Q|}\int_Qf,$ with the result that these tend to $ f(x)$ [i]a.e.[/i], which is the Lebesgue theorem on the differentiation of the integral. But this operator in turn bounds and controls many other maximal operators for many other kinds of pointwise convergence conclusions.


How many positive integers divisible by 4 can be created by using the digits (1,2,3,4) at most once for each integer?

I got 12,32,24 can be last 2 digits.  the remaining digits can be permuted starting from the left, having 2! for a 4 digit #, 2 for a 3 digit # or one for not using the remaining digits.  That would be (2+2+1)3=15 and the answer according to game 14766 is 16.  Am I missing something or is it the problem?

I think the problem is that you got all the numbers except 4. I think thats right...

Thank you.  I got all the 2, 3, and 4 digits, but missed the onedigit answer.


Hi all.

I'm trying to include in my LaTeX2e document an old style capital T that traditionally stands for a tensor space, that is

       T^k(V)

is a k-tensor space over R where V is a vector space over R.

My problem is the T.  I wish I knew how this font was called.  It's definitely not mathcal.  mathcal{T} $\mathcal{T}$ looks similar to a regular T whereas what I need looks similar to $\mathcal{J}$ (\mathcal{J}), though not as slanted.  Does anyone know what I'm talking about?

Thanks for any help.

The letter I'm talking about is also similar to I in Schwabacher font (yswab---oldgerm package) and something between I and J in the mathfrak font.

Spivak uses this old-style T on pp.75-76 of his classic book "Calculus on Manifolds."

I'm not sure what you need. Try the package 'mathrsfs' if you haven't yet found what you were looking for.

Aurel


number theory
Diophantine equation
number theory unsolved

Solve this diophantine equation in $ N^4$:
$ t^2\plus{}y^2\equal{}2hl,3yt\equal{}2h^2\plus{}l^2$.
We know that $ y>t,\gcd(y,t)\equal{}1$ and they r odd.So are $ h,l$(odd and coprime).

Hello,shoki
i don't  get this "$ \gcd(y,t)\equal{}1$". It is same as $ (y,t)\equal{}1$ right ?(if so,they're relatively primes between them)


Hey all, I came across this site on google while I was searching for microeconomics help. There is this one question that is driving me nuts on my homework and I'm HOPING someone can help me out. Thanks !

Brenda has the utility function U(x1,x2) = x1^2 x2. Then, the MUx1 = 2x1x2 and the MUx2= x1^2 . The price of good 1 is p1= 1, the price of good 2 is p2= 3 and her income is 180. Determine the utility-maximizing bundle.

Any tips or help is greatly appreciated ! Thanks once again.

This is an interesting question, to maximize the utility bundle. We have the system of the equation:
* $ \frac{MU_{x_1}}{P_{1}}\equal{}\frac{MU_{x_2}}{P_{2}}$
* $ P_{1}x_{1}\plus{}P_{2}x_{2}\equal{}180$

hello everyone,

actually i have a very important work to do using stata 9.but unfortunately i'm new to that software.i was just wondering,if i could find anyone to help me out.

Please reply,it's very important.

With Thanks


A hexagon with two side lengths of $7$ and four side lengths of $20$ is inscribed in a circle of radius $r$. Find the radius of the circle.  I used the cosine law repeatedly and it yielded a HUGE equation (sixth degree).  The solution in the AoPs book said something about flipping which I totally didn't understand. If someone could explain that solution to me, that would be great!  (It's problem 381 in chapter 20.)

The key to this is that you can rearrange the side lengths any way you want and it doesn't change the area.

So, in the book, it's arranged the hexagon with the two $7$ sides next to each other, creating two cyclic quadrilaterals with the same area.  Then it flips the lower quadrilateral (this doesn't change the area) so that $BC$ and $FE$ are parallel, which makes the areas much easier to find.

But how do you know if the flipped quadrilateral will still be cyclic or not? Because you're sort of "replacing" a side 7 with a side length of 20, so everything just falls into place?

Draw the circumradii to each vertex.  Switching the sides just implies moving triangles around, which doesn't change cyclicity or the area.


number theory
least common multiple
number theory unsolved

This is in my school test and noone solve it,can you solve it for me
For $n\geq\ k^2;n.k\in N$.Find the maximum value of the expression
$\frac{(n+1)(n+2)...(n+k)}{lcm[n+1;n+2...n+k]}$

[quote="Rust"]It have not simple solution.
Let prime p divide only one term n+i ($1\le i\le k$),then $p\not |N$, were
$N=\frac{(n+1)(n+2)...(n+k)}{lcm[n+1;n+2...n+k]}$. Therefore
$N=\prod_{p<k} p^{s(p)}$,
were $s(p)=\sum_{i\ge 1} ([\frac{n+k}{p^i}]-[\frac{n}{p^i}])-s_0(p)$. Number $s_0(p)$ is maximal degree, suth that $p^{s_0(p)}|(n+i)$ for one i: $1\le i \le k$.[/quote]
Can you post your solution?I find the difference answer(after the test,of course) :$N=\prod_{p<k} p^{s(p)}$,
where $p^{s(p)}||M_i$(i=1->k such that the degree of p in M_i is maximum) where $M_i=|\prod_{j=1;j\neq\i}^{n} (i-j)|$


Since $n\geq 3$ is odd, none of these numbers is $\equiv -1 \mod n$ since that would mean that $2^k \equiv 0 \mod n$.
Now there are $n-1$ different residues $\mod n$ left. If the residue $0$ wouldn't occure, there are two different numbers $0<k<l<n$ with $2^k-1 \equiv 2^l-1 \mod n \iff 2^{l-k}-1 \equiv 0 \mod n$, thus there is again a number $\equiv 0 \mod n$ in that set.


[hide]The slope of the line going through points $(-1,-7)$ and $(-29,13)$ is $-\frac57,$ so the slope of the line going through $(13,-7)$ and $\perp$ to this line is $\frac75.$

Using point-slope, $y+7=\frac75(x-13),$ or $7x-5y=-56.$[/hide]

Are you sure? I got

[hide]$7x-5y=126.$[/hide]

for standard form

[hide="my comments"]Well, at a cursory glance one can tell that the coordinate (13, -7) isn't on I like pie's line.[/hide]
Good bye navier stokes equation. I'll miss you.

[quote="xxxxx"]Find the equation of the line passing through the point (13,-7) and perpendicular to the line through points (-1,-7) and (-29,13)[/quote]

[hide]the slope of ( 1, -7) and ( -29,13) is determined with this formula in case you didn't know:

${y}^{2}-{y}^{1}/{x}^{2}-{x}^{1}$ This is called the how to find the [b]slope[/b] of a line. the slope is -5/7. Perpendicular to that is 7/5. Since you know that it's passing throught point ( 13, -7), you plug it into what you already know in slope-intercept form to find the y-intercept.:

y= 7/5 X + b  {passes through 13, -7)

-7=7/5(13) +b
-7=91/5 +b

b= -25 1/5

plug it in: y=7/5 X -126/5 [/hide]

please don't revive topics older than 3 weeks

thank you,
jorian


MATHCOUNTS probability geometry rectangle ratio inequalities LaTeX

1.
Darts are thrown randomly at a board showing three concentric circles of radii 2, 3, and 4.  If a dart hits inside one of the circles, what is the probability, expressed as a common fraction, that it is in the interior of the circle of radius 3 but not in the interior of the circle of radius 2?

2.
Three tennis balls are tightly sealed in a right cylindrical can, so that the balls cannot move inside the can.  If the radius of each ball is 4 cm, find the number of cubic centimeters in the volume of the space within the cylinder, not taken up by the three tennis balls.  Express your answer in terms if $\pi$.

3.
A custom made window is a single pane of glass consisting of a semicircular sheet of glass on top of the narrow end of a rectangular sheet of glass.  The rectangle has dimensions of 4 feet by 3.5 feet.  How many feet of weather-stripping will be needed to go around the glass?  Express your answer to the nearest integer.

4.
The numbers from 1 through 8 are separated into two sets, A and B.  The numbers in A are multiplied together to get $a$.  The numbers in B are multiplied together to get $b$.  The larger of the two numbers $a$ and $b$ is written down.  What is the smallest possible number that can be written down using this procedure?

5.
Eric's Florist Shop sells roses for 20 dollars per dozen, daises for 5 dollars per dozen, and chrysanthemums for 10 dollars per dozen.  At these rates how much would an arrangement of 9 roses, 6 daises, and 18 chrysanthemums cost in dollars?

6.
Given the equation $\displaystyle \frac{1}{m} + \frac{1}{n} = \frac{2}{15}$, where $m$ and $n$ are positive integers, find the value of $m+n$ such that $m>n$ and $m$ is not a multiple of $n$.

7.
Use the pattern below to express $100^2$ in the form $a^2+b^2-c^2$.  Then find $a+b+c$.
\begin{eqnarray*}
12^2 = 8^2 + 9^2 - 1^2\\
14^2 = 10^2 + 10^2 - 2^2\\
16^2 = 12^2 + 11^2 - 3^2\\
18^2 = 14^2 + 12^2 - 4^2
\end{eqnarray*}

8.
The ratio of the length of a rectangle to its width is the same as that of the diagonal to the length.  If the width is 2, how many units are in the length of the diagonal?

9.
Suppose a number $x$ satisfies $x^2-4x-1=0$.  How much greater is $x$ than its reciprocal?

10.
What is the difference between the largest and the smallest values of $x$ satisfying the following system of inequalities?
Express your answer as a mixed number.
\begin{eqnarray*}
2x-6 \le 6\\
4x+3 \ge 13
\end{eqnarray*}

I checked your answers and they are all correct. Here are the answers using correct Latex code:

1)$ \frac{5}{9}$
2)$ 128\pi$
3)$ 17$
4)$ 210$
5)$ 32.50$
6)$ 40$
7)$ 194$
8)$ 2\plus{}2\sqrt{3}$
9)$ \frac{2\plus{}3\sqrt{2}}{2}$
10)$ \frac{7}{2}$

[hide="9 is definitely wrong"]Okay, so you want $ x\minus{}\frac{1}{x}$, which is $ \frac{x^2\minus{}1}{x}$, which is $ \frac{4x}{x}$, which gives 4.[/hide]

That makes sense, I probably fudged up quadratic formula!

EARTHQUAKE!!!

Stick to Ohio forum, xpmath! Please? After all you live in OHIO...

How bout a Kansan? I was in MO 3 days ago. Your roads suck.

#9 is pretty cool. Let y = x-1/x which is your answer, than yx = x^2-1, so y = 4.

Our roads suck? We're spending billions on Highway 40 and our roads suck?!?!?

Transportation sucks in general. Duh...we need to become primitive beings again. If you know why I say so..

When we all use Pokemon to travel from town to town...

Yet Mewto55555 still plays it as if it is the most popular game in the world. Adn talks about it.

My Dialga is pwnage!! ANd it did not get old! HAve you seen all those ads for POkemon?

I see ads for your mom...and she got--nvm, jkjkjk. sry for any insult taken.

lets not direct "your mom" jokes at each other. Its racist. :D

Its funny how off topic this threadhas become...50 posts!!!

Stop spamming just so you can get to 50 posts mewto.

Yet another spam post. Do some math problems, idiot!

We did all of these problems. And idiot is not a nice word.

I was kind of talking about those in the main MC forum, IDIOT!!

OK you guys, knock it off.
Don't ever flame at someone on this forum.
you can be banned.
And if its sarcastic well, sarcasmn doesn't travel well over the internet.

I believe you've posted the exact same thing somewhere else before? Um...why would you be in the Missouri forum, just wondering?


algebra
polynomial
modular arithmetic
number theory unsolved
number theory

Give all the polynomials W(n), every coefficient of W(n)'s monomials is an integer, and (2^n-1)/W(n) is an integer.

Assume $W$ isn't constant. Let $p$ be an odd prime which divides $W(n)$ for some $n$ (try to show that an odd $p$ always exists; it's not hard). In this case, we get $p|W(n+pt),\ \forall t\in \mathbb N$. This means that $p|2^{n+pt}-1\equiv 2^n\cdot (2^t)^p-1\equiv 2^n\cdot 2^t-1\pmod p$, and since $p|2^n-1\Rightarrow 2^n\equiv 1\pmod p$, we get $p|2^t-1,\ \forall t\in\mathbb N$, which is a contradiction.

We have thus found $W$ to be constant, and from $W(n)|2-1=1,\ \forall n$, we find $W(n)=1,\ \forall n$ or $W(n)=-1,\ \forall n$.


Prove that for every positive integer $m$ there exist $n \in \mathbb N$ such that $2000^n$ begin with $m$

Surely posted before...
we have $log_{10}(2000^n)=n.log_{10}2000$. We have to show there are ka nd n such that $m.10^k<2000^n<(m+1).10^k$ which is equivalent to
$k+log_{10}m<n.log_{10}2000<k+log_{10}(m+1)$,
$log_{10}m<n.log_{10}2000-k<log_{10}(m+1)$.
Existence of solution to the last equation follows from the density of
$\{n.log_{10}2000-k\}$ in $\mathbb R$.


can anyone please post an explanation on how to solve the following problem?

a weight of 480 is supported by 3 ropes. the weight is at S(0,2,-1). the ropes are tied to the points P(2,0,0), Q(0,4,0) and R(-2,0,0). find the tension in each rope.

please provide a step by step explanation if possible, thanks. :)

[url=http://en.wikipedia.org/wiki/Lami's_theorem]Lami's theorem[/url] would solve it.

wouldn't that require the forces to be coplanar, which isn't the case in this problem. Unless I'm overlooking something.


There are four lamps in a room, that are turned on or off using three switches located in another room in the same home, one for each lamp. Both rooms are comunicated for a unique door, and there are no windows or something similar, and there is no posibility to see what lamp is turned on by any of the switchs. The problem is to match the switches with the lamps if you can walk a unique time from one room to the other.

[quote="polymorphic"]you mean three lamps, one for each switch[/quote]Yes I do!

[quote="LynnelleYe"][hide="hmmm"]Turn on two of the switches. Wait 10 minutes or so. Turn one off and go to the room with the lamps. The lit lamp goes with the switch that's still on, the off but warm lamp goes with the switch you just turned off, and the cold lamp goes with the switch you didn't touch?[/hide][/quote]Yes, that is the basic idea. Good!

[quote="fdump"]Does it have anything to do with temparature?[/quote]Yes, of course. If you turned the two lights on for ten minutes and after that you turn one off, then, there is one on (the switch that you turned on but not off), there is one off that is hot (the one that goes with the switch you turned on and off) and one that is off and cold (corresponding to the switch you didn't tuoched)

[quote="randomdragoon"]What if the lights are LEDs?[/quote]C'mon, you're thinking in the worst case scenario. Any way, as far as I know, nobody really use LEDs as home lights! too expensive!


Given:
Acute Triangle ABC and Acute Triangle A'B'C'
AD and A'D' are attitudes
BE and B'E' are angle bisectors
AD = A'D'
BE = B'E'
BC = B'C'

Prove:
ABC = A'B'C'

Note: You are not given that angle ABE = angle A'B'E' and angle EBC = angle E'B'C'

[hide="hint"]See that $\triangle {ABC}$ and $\triangle {A'B'C'}$ have the same area...[/hide]

Where to go from there? I know that already...


[hide]$ f^{\minus{}1}(x)\equal{}\frac{x}{3}$, Hence $ f(h(g(f^{\minus{}1}(3))))\equal{}f(h(g(1)))\equal{}f(h(6))\equal{}f(217)\equal{}651$.[/hide] I better have not made a calculation error...

I think I messed up on the relationship between $ f^{\minus{}1}(3)$ and $ f(3)^{\minus{}1}$.

How did you get $ f(h(g(1)))$?

Since $ f^{\minus{}1}(x) \equal{} x/3$, $ f^{\minus{}1}(3) \equal{} 1$ and then he put that back into $ f(h(g(f^{\minus{}1}(3))))$

I'm having trouble following this particular 1st part. Can someone explain it step a little bit further on the 1st part?  :blush: I'm doing something wrong, but I can't find what. I keep getting an answer that requires tons of multiplication.

how does $ f^{\minus{}1}(x) \equal{} x/3$? I think that's what I don't get.

$ f^{\minus{}1}$ denotes the inverse of the function $ f$, given that an inverse exists. It is not the same as $ \frac{1}{f}$. For example, if $ f(b) \equal{} a$, then $ f^{\minus{}1}(a) \equal{} b$. Thus, if $ f(x) \equal{} 3x$, then $ x \equal{} f(f^{\minus{}1}(x)) \equal{} 3f^{\minus{}1}(x)\Rightarrow f^{\minus{}1}(x) \equal{}\frac{x}{3}$.

this is how I always did it:

Given a function, let's say, $ f(x)=4x+3, let$f(x)=y$ . Thus we have$y=4x+3. Now just switch $ x$ and $ y$ to get the inverse of the function, and we must solve for $ y$: $ x=4y+3\Rightarrow{}x-3=4y\Rightarrow y=\frac{x-3}{4}$. This would be the inverse of the original function.

Oh!!!

I wonder what happens when they ask for the inverse when there are multiple variables.

Wow, I finally see it. Whew.  :maybe: Gracias.

I don't believe that an inverse can be taken of any 3+ dimension function.

However, if such a thing could be done, it would probably be done in one of the two following ways.

Given $ f(x,y)\equal{}2x\plus{}3y$, find $ f^{\minus{}1}(x,y)$.

[b]Way Number 1:[/b]
Let $ f(x,y)\equal{}z$ and replace all $ x$ with $ y$, all $ y$ with $ z$, and all $ z$ with $ x$.
Thus, $ z\equal{}2x\plus{}3y\Rightarrow x\equal{}2y\plus{}3z$.
Take the new equation and solve for $ z$, $ z\equal{}\frac{x\minus{}2y}{3}$.

[b]Way Number 2:[/b]
The same as #1, except replace all $ x$ with $ z$, all $ y$ with $ x$, and all $ z$ with $ y$.
Thus, $ z\equal{}2x\plus{}3y\Rightarrow y\equal{}2z\plus{}3x$.
Solve for $ z$, $ z\equal{}\frac{y\minus{}3x}{2}$.

Just don't quote me on this, I am probably incorrect and stand by my statement that there is no $ f^{\minus{}1}(x,y,...)$.

just a quick question, what does $ f^{\minus{}1}(x)$ mean?

Doesn't F(x)-1 mean the inverse of F(x)?
It would mean x/3, wouldn't it?

@abacadaea

$ f^{\minus{}1}(x)$ is read "the inverse of $ f(x)$". $ f^{\minus{}1}(x)$ is found by treating $ f(x)$ as $ y$. Then replacing all $ x$ with $ y$, and the $ y$ with $ x$. Then solve the equation for $ y$ (if possible). Finally, replace $ y$ with $ f^{\minus{}1}(x)$.
Ex.

$ f(x)\equal{}7x\plus{}5\Rightarrow y\equal{}7x\plus{}5\Rightarrow x\equal{}7y\plus{}5\Rightarrow 7y\equal{}x\minus{}5\Rightarrow y\equal{}\frac{x\minus{}5}{7}\Rightarrow f^{\minus{}1}(x)\equal{}\frac{x\minus{}5}{7}$

Tell me if that was confusing, I hope not.


Prove that for every positive integer x the number (sqrt(2006)+sqrt(2005))^2x differs from an integer by no more than 1/(4*2005)^x.

Got this from a teacher, and it will be fun, surely.

Apparently our poster doesn't like to use LaTeX, so I've TeXed the problem:

Prove that for every positive integer $x$ the number $(\sqrt{2006}+\sqrt{2005})^{2x}$ differs from an integer by no more than $\frac{1}{(4 \cdot 2005)^x}$.


[hide="a start"]
Sub $ c=19-|a+b|$ and we get

$ ab + |a+b| = 116 \qquad(1)$

$ ab + |a+b| = 78 \quad d (2)$

From $ (1)$ we get $ (a+1)(b+1) = 117$ and $ (a-1)(b-1)=117$, since $ 117=3\cdot 3\cdot 13$ the pairs are $ (\pm 1,\pm 117),(\pm 3,\pm 39),(\pm 9,\pm 13)$

From $ (2)$ we get $ (a+1)(b+1) = 79$ and $ (a-1)(b-1) = 79$ and the we can only have $ (\pm 1,\pm79)$

Now we have to count all of these :o . At first glance it looks like there should be $ 16$ distinct ordered pairs, but this is not a solution, so the only way i think we can do it is to check every answer!!!

please tell there's a better way!
[/hide]

Checking the conditions for the modulus might help cut extraneous solutions:

[hide]
Sub $ c \equal{} 19 \minus{} |a \plus{} b|$ to get:

$ ab \plus{} |19 \minus{} |a \plus{} b\parallel{} \equal{} 97$

1) If $ 19 \minus{} |a \plus{} b| > 0$, then

$ ab \minus{} |a \plus{} b| \equal{} 78$
a)If $ 19 > a \plus{} b > 0$, then (a - 1)(b - 1) = 79. Remembering the condition $ 19 > a \plus{} b > 0$, there are no possibilities here.

b)The case $ \minus{}19 < a \plus{} b < 0$ is mostly symmetrical. (a + 1)(b + 1) = 79, there are no possibilities here.

2) If $ 19 \minus{} |a \plus{} b| < 0$, then

$ ab \plus{} |a \plus{} b| \equal{} 116$
a) If $ a \plus{} b > 19$, then $ (a \plus{} 1)(b \plus{} 1) \equal{} 117$. All three factorings work, namely $ (0, 116), (2, 38), (8, 12)$.

b) The case $ a \plus{} b < \minus{} 19$ is mostly symmetrical. $ (a \minus{} 1)(b \minus{} 1) \equal{} 117$, for which we have  $ (0, \minus{} 116), ( \minus{} 2, \minus{} 38), ( \minus{} 8, \minus{} 12)$.

There are 6 triples that satisfy the equation, or option $ \textbf{(C)}$.

P.S: "The case is symmetrical" is not really a rigorous argument, but it helps saving time.

[/hide]

[hide]Case 1: $ c\ge 0$
$ |c|\equal{}c$, so subtracting the second equation from the first gives $ ab\minus{}|a\plus{}b|\equal{}78$. Note that we must have $ ab>0$, so $ a,b$ have the same sign.
If $ a,b>0$, then the equation is $ ab\minus{}a\minus{}b\equal{}78\iff (a\minus{}1)(b\minus{}1)\equal{}79$. This corresponds to the solutions $ (a,b)\equal{}(80,2),(2,80)$. However, plugging these back in reveals that $ 82\plus{}c\equal{}19\iff c\equal{}\minus{}63$, which is a contradiction, as $ c\ge 0$. Thus, these solutions are invalid.
If $ a,b<0$, then substituting $ x\equal{}\minus{}a,y\equal{}\minus{}b$ gives the same equation as before, $ (x\minus{}1)(y\minus{}1)\equal{}79$. From the similar solutions, we find a similar contradiction, so no solutions here.

Case 2: $ c<0$.
Now we have $ |c|\equal{}\minus{}c$, so adding the two equations gives $ ab\plus{}|a\plus{}b|\equal{}116$. Suppose $ ab$ was negative. But for integers, we have $ |a\plus{}b|\le |ab|\minus{}1$ (equality when either $ a$ or $ b$ is $ \pm1$). We have $ 116\plus{}|ab|\equal{}|a\plus{}b|\le |ab|\minus{}1$, contradiction. Thus, $ ab\ge 0$.
If $ a,b\ge 0$, then we have $ (a\plus{}1)(b\plus{}1)\equal{}117\equal{}3^2\cdot13$, which corresponds $ 3\cdot2\equal{}6$ solutions. (You can check to confirm that they do in fact work.)
If $ a,b\le 0$, then the substitution $ x\equal{}\minus{}a,y\equal{}\minus{}b$ gives us $ (x\plus{}1)(y\plus{}1)\equal{}117$, which gives us 6 more solutions.

Thus, there are $ \boxed{\textbf{(E) }12}$ solutions.[/hide]

[quote=TachyonPulse]How many ordered triples of integers $ (a, b, c)$ satisfy\[|a \plus{} b| \plus{} c \equal{} 19\quad\text{and}\quad ab \plus{} |c| \equal{} 97?\]
$\textbf{(A)}\ 0\qquad
\textbf{(B)}\ 4\qquad
\textbf{(C)}\ 6\qquad
\textbf{(D)}\ 10\qquad
\textbf{(E)}\ 12$[/quote]

This is kinda annoying for casework, but just consider the cases where $c>0$ and $c<0$ and the subcases in each case $a+b>0$ and $|a+b<0|$. As we can see,

If $c<0$ then there are $6$ possibilities in each of the subcases $a+b>0$ and $a+b<0$.

If $c>0$ then there are $0$ possibilities for each of the subcases $a+b>0$ and $a+b<0$.

Thus there are $6+6=\boxed{12}$ total cases.


Let p(x) be the polynomial such that p*(cos(x)) = cos(13x)
Find the sum of the coefficient of p(x).

put x=0
we will be left with the sum of coefficients which is 1

i messed up the problem...its supposed to be p(cos(x))=cos(13x)

But, i did the problem and am not sure if its right:
To find the coefficients of any polynomial, you plug in 1 and the result is the sum.

cos(x) = 1
x = 0

p(cos(0)) = p(1)
              = cos(13x*0)
              = 1


[quote="happyme"]Six athletes, F, G, H, M, N, and O, swim in two separate 100-meter events, labeled Event 1 and Event 2. Each athlete finishes both events, and there are no ties. The same athlete who finishes second in Event 1 finished fifth in Event 2. M finished sixth in Event 1. F finishes in one of the last three places in both events. If M finishes fourth in Event 2, in which place must F finish in Event 2?[/quote]

[hide]First M finishes 6th in Event 1 and [b]4th in Event 2[/b].

F should be 4th~6th in both events.

One unnamed player finished 2nd in Egent 1 but finished [b]5th in Event 2[/b].

This unnamed player cannot be F as given.

Therefore, F is neither 4th nor 5th in event 2, so F should have finished $\boxed{6\text{th}}$ in event 2.[/hide]


function
LaTeX
real analysis
real analysis unsolved

What's the minimal number $n$ of discontinuities that a real function $f$ which satisfies $f(f(x)) =-x$ can have?

If it was discussed before, could you give me a link? Searching $\LaTeX$ code is not available :(

Over all of the reals, or just some interval?

And also, it's pretty obvious that it has to be a bijection.

One more thing, the function $f$ has order $4$, meaning that $f^{4}(x) = e(x) = x$, where $e$ is the identity function.

That's all I have so far, so it's probably not very helpful.

This only really makes sense as a question about all of $\mathbb{R}$.

It's not just that $f$ has order 4; every orbit except $\{0\}$ has exactly four elements.

Infinitely many discontinuities are required.
Proof:
First, we note that $f(0)=0$. Suppose $f$ has only $n$ discontinuities $x_{1},\dots,x_{n}$ (including zero), and let $S=\{x: x=f^{i}(x_{j})\text{ for some }i,j\}\cup \{0\}$. $S$ is still finite, and contains $4k+1$ elements for some integer $k\le n$. Also, $f(S)=S$ and $f^{-1}(S)=S$.
$\mathbb{R}\setminus S$ is the union of $4k+2$ open intervals, and $f$ is continuous on each of these intervals. Since $f$ maps $\mathbb{R}\setminus S$ to itself bijectively, these intervals must be mapped to each other by $f$. Let $A$ be the set of these intervals; we define $f$ on $A$ in the natural way. Since each element of $A$ is either entirely positive or entirely negative, $f^{2}(U)\neq U$ for each $U\in A$. On the other hand, $f^{4}$ is the identity on $U$, so each orbit in $U$ has exactly four elements. The number of elements in $U$ is not divisible by 4, and we have a contradiction.


Diophantine equation
number theory unsolved
number theory

Find all positive integer $n$ satisfies $n^2+1$ divide $n!$

please post problems correctly, how are we going to find them if there are infinitely many, for example, let $n$ be a solution of the pell equation: $n^2-5m^2=-1$, then $n^2+1=5m^2$ so the factors $5$ and $m$ and $2m$ are $<n+1$, so their product divides $n!$

($n$ large enough! :D )

What about $n^2+1$ is a prime? As far as I know it is an open question whether there are infinitely many such prime numbers.

[quote="Myth"]What about $n^2+1$ is a prime?/quote]

??? But in that case, it can't divide $n!$.

Pierre.

Indeed, that is what I am talking about.


linear algebra
matrix
algebra
polynomial
induction

Denote the 2x2 identity matrix by I.

Let $ M \equal{} \begin{pmatrix} p & q \\
r & \minus{} p \end{pmatrix}$ where p,q,r are real numbers.

Let s be a real number.

Prove that if $ p^2 \plus{} qr \equal{} 1$, then for any $ n\in \mathbb {N}$.

$ (M \plus{} sI)^{2n} \equal{} \frac { (s \plus{} 1)^{2n} \minus{} (s \minus{} 1)^{2n}}{2} M \plus{} \frac { (s \plus{} 1)^{2n} \plus{} (s \minus{} 1)^{2n}}{2} I$

[hide]
we have that the characteristics polynomial of $ M$ is $ t^2\minus{}1\equal{}0$, that means $ M$ statisfies $ M^2\equal{}I$.
We will prove by induction, for $ n\equal{}1$. The LHS will be $ M^2\plus{}2sM\plus{}s^2I\equal{} 2sM\plus{}(s\plus{}1)I$. And the RHS will be $ 2sM\plus{}(s^2\plus{}1)I$, that is true.
Then, supposed the statement is true for $ n\equal{}k$ for any natural number $ k$, we have that : $ {(M\plus{}sI)}^{2k}\equal{}\frac {{(s\plus{}1)}^{2k}\minus{} {(s\minus{}1)}^{2k}}{2}M\plus{}\frac {{(s\plus{}1)}^{2k}\plus{}{(s\minus{}1)}^{2k}}{2}I$. Then for $ n\equal{}k\plus{}1$ we have that $ LHS\equal{}{(M\plus{}sI)}^{2k\plus{}2}\equal{}{(M\plus{}sI)}^{2k} .{M\plus{}sI}^2\equal{}(\frac {{(s\plus{}1)}^{2k}\minus{} {(s\minus{}1)}^{2k}}{2}M\plus{}\frac {{(s\plus{}1)}^{2k}\plus{}{(s\minus{}1)}^{2k}}{2}I)(2sM\plus{}(s^2\plus{}1)I)\equal{}2s(\frac {{(s\plus{}1)}^{2k}\minus{}{(s\minus{}1)}^{2k}}{2}I\plus{}(s^2\plus{}1)\frac {{(s\plus{}1)}^{2k}\plus{}{(s\minus{}1)}^{2k}}{2}I)\plus{}((s^2\plus{}1)(\frac {{(s\plus{}1)}^{2k}\minus{}{(s\minus{}1)}^{2k}}{2})\plus{}2s\frac {{(s\plus{}1)}^{2k}\plus{}{(s\minus{}1)}^{2k}}{2})M \equal{} \frac {{(s\plus{}1)}^{2k\plus{}2}\minus{}{(s\minus{}1)}^{2k\plus{}2}}{2}I\plus{}\frac {{(s\plus{}1)}^{2k\plus{}2}\minus{}{(s\minus{}1)}^{2k\plus{}2}}{2}$
[/hide]

[quote="Qruni"][hide]
we have that the characteristics polynomial of $ M$ is $ t^2 \minus{} 1 \equal{} 0$, that means $ M$ statisfies $ M^2 \equal{} I$.
We will prove by induction, for $ n \equal{} 1$. The LHS will be $ M^2 \plus{} 2sM \plus{} s^2I \equal{} 2sM \plus{} (s \plus{} 1)I$. And the RHS will be $ 2sM \plus{} (s^2 \plus{} 1)I$, that is true.
Then, supposed the statement is true for $ n \equal{} k$ for any natural number $ k$, we have that : $ {(M \plus{} sI)}^{2k} \equal{} \frac {{(s \plus{} 1)}^{2k} \minus{} {(s \minus{} 1)}^{2k}}{2}M \plus{} \frac {{(s \plus{} 1)}^{2k} \plus{} {(s \minus{} 1)}^{2k}}{2}I$. Then for $ n \equal{} k \plus{} 1$ we have that $ LHS \equal{} {(M \plus{} sI)}^{2k \plus{} 2} \equal{} {(M \plus{} sI)}^{2k} .{M \plus{} sI}^2 \equal{} (\frac {{(s \plus{} 1)}^{2k} \minus{} {(s \minus{} 1)}^{2k}}{2}M \plus{} \frac {{(s \plus{} 1)}^{2k} \plus{} {(s \minus{} 1)}^{2k}}{2}I)(2sM \plus{} (s^2 \plus{} 1)I) \equal{} 2s(\frac {{(s \plus{} 1)}^{2k} \minus{} {(s \minus{} 1)}^{2k}}{2}I \plus{} (s^2 \plus{} 1)\frac {{(s \plus{} 1)}^{2k} \plus{} {(s \minus{} 1)}^{2k}}{2}I) \plus{} ((s^2 \plus{} 1)(\frac {{(s \plus{} 1)}^{2k} \minus{} {(s \minus{} 1)}^{2k}}{2}) \plus{} 2s\frac {{(s \plus{} 1)}^{2k} \plus{} {(s \minus{} 1)}^{2k}}{2})M \equal{} \frac {{(s \plus{} 1)}^{2k \plus{} 2} \minus{} {(s \minus{} 1)}^{2k \plus{} 2}}{2}I \plus{} \frac {{(s \plus{} 1)}^{2k \plus{} 2} \minus{} {(s \minus{} 1)}^{2k \plus{} 2}}{2}$
[/hide][/quote]

That's right, thanks for your demo.


yep

in baseball there r penalties i think

[quote="kheg_k"]What do u think of people using steroids in the Olympics, track meats, proffessional sports leagues (like National Basketball Association)?

What should leagues do about it?

I think that steroid-users should be banned for a period of time for first offense and banned permenantly for second offense.[/quote]

I don't know much about it, but when I took some ointment and rubbed it on my shoulder due to a complaint, I was quite surprised to learn from the box that it contained EPO as ingredient.   :ninja:

Of course, steroids make competitions unfair, but I don't think something can be done to it. Sports like cyclism are really "unclean" sports and even if they banned Ullrich, Basso, Vino etc., they'll keep doing it. I really don't know how this issue can be solved.
They could authorize it to get fair competition, but that would mean you're almost obligated to take steroids to compete, which isn't fair...

Whatever you want to put in your body, I don't care, as long as you don't start hurting other people.

If the associations make it a violation to use them, then don't use. It's pretty simple.

I think steroids should be officially allowed.

In "Asterix and the Olympic Games" Obelix got banned and Asterix could not use his potion. That is [i]very[/i] unfair if you ask me.

they want more fame than their life (they are obsessed)

nah, i'd use underhanded methods to win if i wanted to (me... countdown... unfair ownage) but messing up your body sounds like it sucks

well if everyone did i would but you get banned from swimming...

For me, if you use sterois, you have to be banned per life... so, then, anybody do it again

Using steroids is not fair to people who do not use steroids and competing in the same competition. :!:

The should have 2 leagues for every sport:
a non steroid people one and a steroid people one.

[quote="Xevarion"]We are lucky that we do not live in a time when people have discovered something to produce the same effect on our minds as steroids have on our bodies. Math contests would be a lot less fun.[/quote]

nah, i know i used 'mind performance enhancers' for mathcounts nationals... it's called mountain dew (you caffeine addicts could use jolt, monster or bawls)

I don't think caffeine makes you smarter. At least, it doesn't work for me. :(

Caffeine just makes you more jumpy. :D

Anyway, I don't think steroids should be used in sports.  I think that to get good at something (not just sports, but anything in general), you should do so by working at it.

There definitely should be penalties for using steroids (or anything like it) in sports.

The best sterois you need is love for your team, that's the one fair, and it's really good. If you do your best every practice, and you try to impove yourself, you'll be good. Steroids are for non-sure people

[quote="lingomaniac88"]
There definitely should be penalties for using steroids (or anything like it) in sports.[/quote]
By the government or the sports association?

Steroid designers have much more of a profit to be made then steroid testers.  Therfore, there willl alwyas be new steroids we can't test for.  Since it's always going to happen anyway, I might be for allowing steroids.  However, steroids are unfair.  Even if everyone is allowed to use them, those who do will greatly risk their health.  So a good player who doesn't want to risk his health will not take steroids, and thus be at a disadvantage.  If they could invent a safer steroid, I would still be against steroid use, but I would be for legalizaing steroids in sports, since you can't prevent them.  Did that make sense?

[quote="Xevarion"]We are lucky that we do not live in a time when people have discovered something to produce the same effect on our minds as steroids have on our bodies. Math contests would be a lot less fun.[/quote]

Ever heard of speed? Thankfully, most nerds don't do hard drugs, and most that do hard drugs wouldn't be using them for help in math competition. There are some anti-ADD prescription amphetamines though, that could work.  :ninja:

I know speed is widely used by college kids cramming for tests, but I don't think it actually makes you smarter.  All it does is keep you alert and awake.  Besides, it would be virtually impossible to use in a math contest.  It would have to happen something like this:

1) Math nerd works up courage to attempt to purchase an illegal drug for a math competition.
1) Math nerd finds a drug dealer.
2) Math nerd confronts dealer and asks to buy speed.
3) Dealer, for some reason, lets nerdy looking kid buy drugs, even though (from what I've seen on COPS) they usually only sell to hard drug users.
4) Math nerd snorts (or injects) speed properly, and reaches desired state.
5) Math nerd is able to sit in one place for 90 minutes, even though a side effect of speed is an inability to sit still for even a short period of time.

Yea, I don't see it happening.  And I mean "math nerd" in the nicest way.  :P

[quote="mysmartmouth"]1) Math nerd works up courage to attempt to purchase an illegal drug for a math competition.
1) Math nerd finds a drug dealer.
2) Math nerd confronts dealer and asks to buy speed.
3) Dealer, for some reason, lets nerdy looking kid buy drugs, even though (from what I've seen on COPS) they usually only sell to hard drug users.
4) Math nerd snorts (or injects) speed properly, and reaches desired state.
5) Math nerd is able to sit in one place for 90 minutes, even though a side effect of speed is an inability to sit still for even a short period of time.[/quote]

You can get drugs online. I heard that's cheaper and better quality than you usually get on the street too.

Drugs are bad.