linear algebra
matrix
geometry
projective geometry
linear algebra unsolved

Hello,

I've been working on this problem for days.  I got it from a book and it's supposed to be obvious, but I find it quite hard.

Let $ F$ be a finite field of $ q$ elements.  Let $ K$ be an extension of order $ q^{t + 1}$.
Let's write $ v = \frac {q^{t + 1} - 1}{q - 1}$.

Let ${ Tr: K\rightarrow F: y\rightarrow y + y^q + \cdots + y^{q^t}}$ be the trace function.
Let $ x$ be a primitive element of $ K$

Let $ c_k = 1 - Tr(x^{k})^{q - 1}$.  Note that this can only be $ 0$ or $ 1$.
[Moderator edit: I suppose this should mean $ c_k = 1 - \left(\text{Tr}\left(x^{k}\right)\right)^{q - 1}$, and not $ c_k = 1 - \text{Tr}\left(\left(x^{k}\right)^{q - 1}\right)$. - Darij]

Now let us consider this $ v\times v$ matrix $ N$ : $ n_{ i j} = c_{j - i}$

Now we create a Vandermonde matrix...
let $ \gamma_0,\dots,\gamma_{v - 1}$ be the different $ v$-th roots of unity in $ K$.
Let the $ v\times v$ matrix $ V$ be defined by $ V_{ i j}$ be $ \gamma_{j - 1}^{i - 1}$, or thus:
\[ V = \left(\begin{array}{c c c} 1 & \cdots & 1 \\
\vdots & \vdots & \vdots \\
\gamma_{0}^{v - 1} & \cdots & \gamma_{v - 1}^{v - 1}\end{array}\right)
\]
Now show that $ V N V$ is a diagonal matrix $ diag(d_0,d_1,\dots,d_{v - 1})$ with

$ d_u = \sum_{k = 0}^{v - 1} c_k \gamma_u^k$

I already managed to find that $ (VNV)_{ a b} = \sum_{k = 0}^{v - 1} c_k (\sum_{w = 0}^{v - 1} \gamma_w^{a - 1} \gamma_{b - 1}^{w + k})$., then I got stuck.

Can anyone give me a hint? I"m really surprised that it's so hard.. :huh:

(For those interested, the article was written by K.J.C. Smith, he is trying to find the rank of the incidence matrix of the projective geometry $ PG(t,q)$ (the projective space of geometric dimension $ t$ over a field of order $ q$)  $ N$ is the incidence matrix - that's why there are only ones and zeroes, and he knows that that diagonalmatrix $ D$ has the same rank since $ V$ is invertible, which reduces the problem to counting the non-zero elements on the diagonal of $ D$)

One would expect that
$ \sum_{w\equal{}0}^{v\minus{}1} \gamma_{w}^{a\minus{}1} \gamma_{b\minus{}1}^{w\plus{}k}$ is zero when $ a\neq b$, and $ \gamma_u^k$ when $ a\equal{}b\equal{}u$.  That would prove everything, but is this a good approach? Is that even true? Note that I did not say anything about the ordering of the roots $ \gamma_i$? :maybe:

I have taken the freedom to slightly edit your post in two places where it can be misunderstood.

Still it appears to me that the problem is wrong. I have checked on an example that $ VNV$ does not need to be diagonal, even if the roots $ \gamma_0$, $ \gamma_1$, ..., $ \gamma_v$ are written in "natural" order (so that $ \gamma_k \equal{} \gamma_1^k$ for every $ k$).

What is true, however, is that $ V^{ \minus{} 1}NV \equal{} \text{diag}\left(d_0,d_1,\dots,d_{v \minus{} 1}\right)$. Or maybe $ VNV^{ \minus{} 1} \equal{} \text{diag}\left(d_0,d_1,\dots,d_{v \minus{} 1}\right)$, since I have not explicitely done the computation. In fact, the situation is easier than you might think:

First note that $ c_k \equal{} c_l$ for any two integers $ k$ and $ l$ with $ k\equiv l\text{ mod }v$ (prove this on your own, noticing that $ x^v\in F$ yields $ \text{Tr }\left(x^vy\right) \equal{} x^{v\left(t \plus{} 1\right)}\text{Tr }y$). Hence, the matrix $ N$ is a circulant matrix. The rest follows from standard properties of circulant matrices (note that $ V$ is the matrix of the discrete Fourier transformation). Please tell me if you need some details to be made clearer.

As a matter of fact, $ x$ does not need to be a primitive element of $ K$.

  Darij


calculus
integration
function
logarithms
algebra
functional equation
complex analysis

I've been wondering what the integral form of the zeta function would look like if we chose a different branch of the logarithm to integrate over.  Here's the zeta function:

$ \zeta(s) \equal{} \Gamma(1 \minus{} s)\frac {1}{2\pi i}\mathop\int\limits_{\multimap}\frac {z^{s \minus{} 1}}{e^{ \minus{} z} \minus{} 1}dz;\quad Log[z] \equal{} ln(r) \plus{} i\Theta_z;\quad \minus{} \pi < \Theta < \pi$

So we're integrating along the branch cut of the principal branch.  Suppose I define a new function:

$ \eta_n(s) \equal{} \Gamma(1 \minus{} s)\frac {1}{2\pi i}\mathop\int\limits_{\multimap}\frac {z^{s \minus{} 1}}{e^{ \minus{} z} \minus{} 1}dz;\quad Logn[z] \equal{} ln(r) \plus{} i(\Theta_z \plus{} 2n\pi)$

This then is the zeta expression with the Hankel contour traversing the branch cut of the $ n'th$ branch.

Can these questions be answered:

1)  Will $ \eta_n(s)$ have zeros?  If so will it have any "non-trivial" zeros and will these zeros bear any functional relationship to the zeros of zeta?

2) What will $ \eta(s)$ look like for $ Re(s) > 1$?  Is is possible to "collapse" the integral expression to a simpler form like is done for the zeta function when $ Re(s) > 1$?

3)  Can a functional equation be constructed for $ \eta(s)$?

4)  How far could $ \eta(s)$ be taken through the same type of rigorous analysis $ \zeta(s)$ is taken through?  For example, suppose I substituted $ \eta(s)$ in place of $ \zeta(s)$ in the proof of the Prime Number Theorem.  Granted, $ \eta(s)$ will more than likely have nothing to do with primes but would it have some meaning for something else?

5)  I suspect there will be a functional relationship between $ \eta(s)$ and $ \zeta(s)$ like $ \eta_n(s)\equal{}F(n,\zeta(s))$.  If so, can the Prime Number Theorem be expressed in terms of $ \eta(s)$?

uh . . . I guess I've made some progress with this.  Perhaps I should have asked:  Given the definitions above, show:

$ \eta_n(s) \equal{} e^{2\pi i n s} \zeta(s)$

which means of course $ \eta_n(s)$ has the same zeros as $ \zeta(s)$ and also:

$ \eta_n(s)\equal{}\Gamma(1\minus{}s)(2\pi)^{s\minus{}1}2Sin(\frac{\pi s}{2})\eta_n(1\minus{}s)$

. . . well it was news to me at least


jw... is there an easier/faster way to doing this question than expanding & solving? especially since all the choices are squares..  :maybe:

The polynomial f(x)= x^4 + ax^3 + bx^2 +cx + d has real coefficients, and f(2i) = f(2+i) = 0. What is a+b+c+d?
(A) 0
(B) 1
(C) 4
(D) 9
(E) 16

[hide]
$f(x)=(x-2i)(x+2i)(x-2-i)(x-2+i)$
Instead of expanding that to find the polynomial, you could replace an x with 1 and expand... that's a bit easier than expanding it with the x in there

Then, since $f(1)$ is the sum of the coefficients, just subtract $1$, since you aren't counting the coefficient of $x^{4}$
[/hide]

Here's how I did it. I'm not sure if this is any easier than your way.

$f(x)=(x^{2}+4)(x^{2}-4x+5)$, the first polynomial having roots 2i and -2i and the second having roots 2+i and 2-i (where the coefficients are quickly computed as the sum and products of the conjugates). Then $a+b+c+d=f(1)-1=(1^{2}+4)(1-4+5)-1=9$.

What do you mean, "since all the choices are squares"? Is there some easy way to show that it's a square?

That seems like a rather useless generalization, since the problem is too specific to ever need to remember a formula for it (that and it's not a hard problem anyways)

I was just showing that it was always a perfect square.

What I meant was the original post made that seem obvious. I don't see why it's obvious.


Given a MPG (Maximal Planar Graph, all faces are triangles, also the infinite face) on v vertices. We can give a +1 or -1 number to the triangles. If we add mod3 the triangle numbers of the triangles adjacent to a vertex we get a vertex number of 0,1 or 2 for that vertex.
It’s always possible to give triangle numbers so that all vertices have a vertex number of 0. It’s Heawood’s equivalent formulation of the 4 colour theorem (dual of Heawood’s vertex character).

But does there exist a proof or disproof for the following statement:
[i]Given all the 2^(2v-4) combinations of triangle numbers, then any set of v-2 vertex numbers has all the 3^(v-2) different combinations  of vertex numbers if the two missing vertices are adjacent? [/i]
I tested this in excell up to 9 vertices (that's the maximum for my computer and/or my programming skills) and it was ok.
 :arrow: I'm looking especially for a disproof on more vertices.

kweenie

[quote="kweenie"]But does there exist a proof or disproof for the following statement:
[i]Given all the 2^(2v-4) combinations of triangle numbers, then any set of v-2 vertex numbers has all the 3^(v-2) different combinations  of vertex numbers if the two missing vertices are adjacent? [/i]
I tested this in excell up to 9 vertices (that's the maximum for my computer and/or my programming skills) and it was ok.
 :arrow: I'm looking especially for a disproof on more vertices. [/quote]

The statement above is in fact a special case (for a degenerated polygon on 2 vertices) of the more general conjecture below:
[i]Given a triangulated planar polygon with i vertices in it, with all the combinations of the triangle numbers, then we have all the combinations of the vertex numbers for the inner vertices.[/i]

In attachment an illustration for the principle of triangle- and vertex numbering.

kweenie


[quote="mr. math"]If (2a+2b)/c=(d^2)/bcd solve for a in terms of b,c, and d.[/quote]
Original problem in LaTeX: $\frac{2a+2b}{c}=\frac{d^{2}}{bcd}$
[hide]Multiply by c $2a+2b=\frac{cd^{2}}{bcd}$
Divide by 2 $a+b=\frac{cd^{2}}{2bcd}$
Subtract b $a=\frac{cd^{2}}{2bcd}-b=\frac{cd^{2}-2b^{2}cd}{2bcd}$
$\boxed{a=\frac{cd^{2}-2b^{2}cd}{2bcd}}$

Edit:$a=\frac{cd^{2}-2b^{2}cd}{2bcd}$ -> cancel out the cd to end up with $\boxed{a=\frac{d-2b^{2}}{2b}}$[/hide]
 :!: 100th post! :!:

[quote="ckck"][hide="Think this is right"]Multiply both sides by c to get
$2a+2b=\frac{cd^{2}}{bcd}$
Dividing by two gets
$a+b=\frac{cd^{2}}{2bcd}$
Then subtracting by $b$ is
$a=\frac{d^{2}-2b^{2}cd}{2bcd}$[/hide][/quote]
Between your second and third equation, you forgot a "c" in the $a=\frac{d^{2}-2b^{2}cd}{2bcd}$, it should be $a=\frac{cd^{2}-2b^{2}cd}{2bcd}$. Then you can cancel out a "cd" out of all the fraction. :P

[quote="mr. math"]If (2a+2b)/c=(d^2)/bcd solve for a in terms of b,c, and d.[/quote]

[hide]

$2a+2b = \frac{d^{2}}{bcd}\cdot c$

$2a+2b = \frac{d}{b}$

$2a = \frac{d}{b}-2b$

$a = \frac{(\frac{d}{b}-2b)}{2}$

[/hide]


function
probability
expected value
probability and stats

There is an auction with N players. Some painting is saling by auction. B is a bet for each player and this bet depends on player's value W, i.e. Wi ~ V[0;1] and B = f(Vi)

If player won, then he must pay arithmetic average of bets of all N players. The task is to find player optimal strategy as function of V.

Let's guess.

Maybe $ f$ is linear?
In that case $ b \equal{} c\cdot v$. (I prefer $ b$ for bid and $ v$ for value)
If $ v$ is uniform then (for linear $ f$) $ b$ is also uniform.

While searching for Nash Equilibria one fix the strategies of all other players and find the best response for one selected player. So i fix strategies of all $ (n \minus{} 1)$ players as $ b_{i} \equal{} c\cdot v_{i}$ for $ i\in\{2,...,n\}$

Let's denote $ W$ - the event "I Win the auction".

Expected profit or utility for risk neutral player equals to:
$ E(U) \equal{} P(W)\cdot (v \minus{} E(\frac {b \plus{} b_{2} \plus{} ...b_{n}}{n}|W)$
Here (i am the first player - just for convinience)
$ P($I win$ ) \equal{} P(b > b_{2},b > b_{3},...,b > b_{n}) \equal{} P(cv_{2} < b)\cdot...\cdot P(cv_{n} < b) \equal{} \left(\frac {b}{c}\right)^{n \minus{} 1}$
What do we have for $ E\left(\frac {b \plus{} b_{2} \plus{} ...b_{n}}{n}|W\right)$?
Conditionally on W i have:
$ b$ - is a constant
$ b_{2}$, ...,$ b_{n}$ are iid on $ [0;b]$ (!) and their expected value is $ b/2$.
So $ E\left(\frac {b \plus{} b_{2} \plus{} ...b_{n}}{n}|W\right) \equal{} \frac {b \plus{} (n \minus{} 1)\cdot \frac {b}{2}}{n}$

Plugging this into $ E(U)$ and maximizing by $ b$:
$ E(U) \equal{} const\cdot b^{n \minus{} 1}(v \minus{} b\frac {n \plus{} 1}{2n})$
$ (n \minus{} 1)\frac {1}{b} \minus{} \frac {n \plus{} 1}{2n}\frac {1}{v \minus{} b\frac {n \plus{} 1}{2n}} \equal{} 0$
Or
$ \frac {b}{n \minus{} 1} \equal{} v\frac {n \plus{} 1}{2n} \minus{} b$
Or
$ b \equal{} v\cdot \frac {n^2 \minus{} 1}{2n^{2}}$
And it's indeed linear!

For large $ n$ you bid about one half of the value - just like in the case of "average of the others bid" auction. Because the influence of your bid disappears.

Спасибо, Борис Борисович!) Думаю, здесь не обидятся, если я на русском напишу? Задачка действительно вызвала затруднения, но теперь все стало понятно. :D


calculus
integration
trigonometry
geometry
3D geometry
tetrahedron
calculus computations

Calculate
\[ \iiint\limits_{0 < x,y,z < \pi } {\frac{dx dy dz}{{1 \minus{} \cos x\cos y\cos z}}}.
\]

$ P\equal{}\int _0^{\pi }\int _0^{\pi }\int _0^{\pi }\frac{1}{1\minus{}\cos (x) \cos (y) \cos (z)}dxdydz$

$ P\equal{}\int _0^{\pi }\int _0^{\pi }\int _0^{\pi }\left(\sum _{i\equal{}0}^{\infty } (\cos (x) \cos (y) \cos
   (z))^i\right)dxdydz$

After expand and simplify, interchangind the sum.

$ P\equal{}\sum _{i\equal{}0}^{\infty } \left(\int_0^{\pi } \cos ^i(t) \, dt\right)^3$

Is very tought but $ \int_0^{\pi } \cos ^i(t) \, dt\equal{}\frac{(\minus{}2)^i \pi ^2}{i! \Gamma
   \left(\frac{1}{2}\minus{}\frac{i}{2}\right)^2}$

To calculate the sum of the cube of this will be very hard.

i believe what bookworm asked for was
$ \iiint_{0<x\plus{}y\plus{}z\leqslant\pi}\;\dfrac{dx\;dy\;dz}{1\minus{}\cos\,x\;\cos\,y\;\cos\,z}\;\neq\;\int_0^\pi\int_0^\pi\int_0^\pi\dfrac{dx\;dy\;dz}{1\minus{}\cos\,x\;\cos\,y\;\cos\,z}$

but
$ \iiint_{0<x\plus{}y\plus{}z\leqslant\pi}\;\dfrac{dx\;dy\;dz}{1\minus{}\cos\,x\;\cos\,y\;\cos\,z}\;\equal{}\;\frac{1}{6}\;\int_0^\pi\int_0^\pi\int_0^\pi\dfrac{dx\;dy\;dz}{1\minus{}\cos\,x\;\cos\,y\;\cos\,z}$

(since the left integral is actually taken over a tetrahedron and not a cube)

the answer i have got for
$ \iiint_{0<x\plus{}y\plus{}z\leqslant\pi}\;\dfrac{dx\;dy\;dz}{1\minus{}\cos\,x\;\cos\,y\;\cos\,z}$ is $ \boxed{\dfrac{\Gamma^4\left(\frac{1}{4}\right)}{24}}$

i did this hastily at lunch today, so i don't yet know if it is correct, but i didn't spot any mistakes...   :P

bravo misan.. my integral is
\[ \int_0^\pi  {\int_0^\pi  {\int_0^\pi  {\frac{{dxdydz}}
{{1 \minus{} \cos x\cos y\cos z}}} } }
\]
and the answer is
\[ \frac{1}
{4}\left( {\Gamma \left( {\frac{1}
{4}} \right)} \right)^4 .
\]


The Caltech and Harvey Mudd math clubs would like to invite you to the first Caltech Harvey Mudd Math Competiton on [b]Saturday, February 20, 2010[/b].

The regular registration deadline is [b]December 20, 2009[/b].

For more information, visit [url]http://chmmc.caltech.edu[/url].

This is a new entry in the calendar of on-site math contests for Southern California. There are several others, notably including:

CSULB Math Day at the Beach, which will be on March 20, 2010. We'll have registration letters sent out probably in mid November. For all information about this, contact me.

Los Angeles City College (LACC) Contest. (I don't know the 2010 date - it's usually in March.)

Some comparisons:

Both Math Day at the Beach and this new Caltech/Harvey Mudd competition are for school-based teams of 6. LACC has a team category but is really more of an individual contest.

Math Day at the Beach has results available on the same day and apparently so will Caltech/Harvey Mudd. LACC, which uses a long-answer format, does not announce results until several weeks after the contest.

LACC tends to have the best prizes (cash). Yasha doesn't seem to be making any explicit promises about prizes, and neither are we.

Our schedule for the day of the contest typically has registration from 9:00 to 10:00 with competition starting at 10:00, and we're usually all done by 2:30 or 3:00; it looks like Caltech/Harvey Mudd will start earlier in the morning and last later into the afternoon, in part because they want to copy more parts of the ARML style. We have an individual round that mixes multiple choice and free response; Caltech/Harvey Mudd plans to have all free response. Both of us will have a team round of some sort.

Caltech/Harvey Mudd will allow two teams per school to compete. We won't - our venue will only accommodate about 35 teams and we will probably have that many schools. Both of us will make some provisions for team-less students. We will not accept schools or teams from outside our area (defined as the following counties: Los Angeles, Orange, Riverside, San Bernardino, Ventura, and Santa Barbara). That's probably not a question Yasha has thought about yet.

The LACC contest has been around for a long time. There used to be a contest at Orange Coast College that had been around for perhaps 40 years, but it has not happened in the last several years. CSULB Math Day at the Beach first happened in 2000, so our 2010 contest will be our 11th. This will be the first year for Caltech/Harvey Mudd.

Another difference: CHMMC will charge a per-student fee of \$10 (\$15 for late registration), while Math Day at the Beach has never yet charged a fee. (I'll add that our exact financing for 2010 isn't yet clear, but we'll try as hard as we can to keep it the way it's been.)

[quote="Yasha"]CHMMC will accept teams from outside the Los Angeles area, and we strongly encourage such teams to attend.[/quote]
Any news on which teams are coming?

The list Yasha gave me a couple of weeks ago had four organizations from San Diego: Del Norte HS, San Diego Math Circle, The Bishop's School, and Torrey Pines Community. There was also "Sonoran Science Academy" which I think is from Arizona. The rest of the teams are from the LA/Orange County area. In particular, no teams from Northern California.

The organizers will post results, but in the meantime, I do have the awards written down:

Team:

1. Irvine University HS A (214)
2. Torrey Pines A (167)
3. Troy HS (157)
4. Orange County Math Circle (155)
5. The Bishop's School (134)
6. Torrey Pines B (112)
7. San Marino HS (103
8. Palos Verdes Peninsula HS (102)
9. North Hollywood HS (89)
10. Accord Institute (83)

Individual:

1. David Yang, Diamond Bar HS (10)
2. George Xing, Irvine University HS (6)
3. Sounya Basu, Torrey Pines (6)
4. Johnny Li, Orange County Math Circle (5)
5. Raku Watari, Orange County Math Circle (5)
6. Dai Yang, Irvine University HS (5)
7. Stephen Liu, Torrey Pines (5)
8. Kevin Yin, San Marino HS (4)
9. Tony Jin, Orange County Math Circle (4)
10. Aaron Schild, The Bishop's School (4)

[quote="haoye"]I left early and missed out on the awards - are the results going to be posted online soon (either here or at the official website)?

Also, will the problems and solutions be posted as well?[/quote]
The problems are available now, and the solutions will be available soon.
http://chmmc.caltech.edu/problems.html

Edit: Also, some results are available.
http://chmmc.caltech.edu/results.html

Can anyone say a brief intro. or format of CHmmc as well as Math Day at the Beach?


Define the sequence $a_0 = 3, a_n = 3^{a_{n-1}}$.  Prove that $a_n \equiv a_{n+1} \bmod 10^{n}$.

(I'm not actually sure that this is the question, since I don't remember it very clearly; it might've only held for $n > k$ and asked to find $k$.)

Hint :

[hide]This is true for $n > 0$

This can be done by induction noting that if $10^n \mid N$ then ${10}^{n+1} | 3^N - 1$ which can be resumed to showing that ${10}^{n+1} | 3^{10^n} -1$ (*).

(*) can be also proved by induction by writing ${3^{10}}^{10} - 1 = {(3^{10}-1) + 1}^{10} - 1$ [/hide]


There are $ 26$ choices for the first letter and $ 26$ for the second.  For the digits, there are $ 5$ choices for the odd and $ 5$ for the even, then $ 2!\equal{}2$ to arrange them.  Multiplying yields $ 26 \times 26 \times 5 \times 5 \times 2\equal{}\boxed{33800}$.

[quote="satch1031"]I am confused if 0 is odd or even[/quote]
It is. 0, 2,4,6,8 are all even.

0 is even. :lol:
All even numbers are divisible by 2. 0 is divisible by 2.

[hide="solution"] $0/2=0$[/hide]

The AOPS solution says [hide]There are 26 choices of letters for each of the first two spots and 10 choices of digits for the next spot. Once the first digit has been chosen, we know whether the second digit must be even or odd. Either way, there are 5 choices for the second digit. There are a total of  different plates.[/hide] but I got 26^2*5^2, since there are 5 choices for both even and odd.

Glossysuace, you made the mistake that I almost made: the first digit can be odd [b][u]or[/u][/b] even, but the second can only be the other one.

@above you are clearly wrong... it doesn't say one of the digits can be odd [u]or[/u] even. It says one of the digits is odd and the other is even I reported the bug

Here is my solution:
[hide=answer]For the first two letters we can just do $26^2$ because there are 26 letters and there are no restraints. For the numbers, the first number can be anything therefore giving us, $10$ because there are 10 digits. For the second one, based on the first number, it has to be either odd OR even. Thus, giving us $5$. So in total, we have: $$26^2 + 10*5$$ This gives us $\boxed{33800}$ [/hide]

Thank you welp i got it wrong?

i think thats what they mean

[quote=AwesomeDude10]What does that mean?[/quote]

it means @bsu1 needs help because he/she got it wrong
sniped

LOL i mean "Thank you, welp :C ,me got it wrong"

[quote=GameBot]How many license plates consist of 2 letters followed by 2 digits, if one of the digits must be odd and the other must be even?[/quote]

[hide=sol]There's $26$ letters in the English alphabet. There are $5$ odd digits, and there are $5$ even digits. But the order doesn't matter for the digits. This gives us $26 \cdot 26 \cdot 5 \cdot 5 \cdot 2 = \boxed{33800}$.[/hide]

[quote=ArtyA6]funny that glossy sauce did an incredibly large 9 year bump to the problem.[/quote]

According to the AoPSsheriff, gamebot discussions can be bumped [b]as long as it adds to the discussion[/b]

how is this level 16?

there are $26^2$ ways to pick the two letters. the ways to have an odd and even digit is odd - even and even - odd. each would have $5^2$ ways for a total of [hide]$26^2 \cdot 5^2 \cdot 2 =\boxed{33800}$ ways[/hide]

tripped and put $16900$ at first


Suppose ,we are to prove that : if $ \sum a_{n}$ diverges, then $ \sum(1\plus{}1/n)a_{n}$ diverges.

Is it then equivalent to proving:

if $ \sum(1\plus{}1/n)a_{n}$ converges ,then $ \sum a_{n}$ converges ??

Yes or no??

In either case why?

O.K ,thank you.

So if $ \sum a_{n}$  diverges to $ +\infty$ we have to show that ${ \sum(1+1/n)a_n}$ diverges to $ +\infty$

and by contrapositive we have to show:

if $ \sum(1+1/n)a_{n}$  [b]does not diverge[/b] to $ +\infty$, then $ \sum a_{n}$  [b]does not diverge to[/b] $ +\infty$.

correct ??

$ P \Longrightarrow Q$ is equivalent to $ \neg Q \Longrightarrow \neg P$.

In this case $ P$ is "$ \sum a_{n}$ diverges", whereas $ Q$ is "$ \sum (1\plus{}1/n)\, a_{n}$ diverges".

Their negations are $ \neg P$ i.e. "$ \sum a_{n}$ converges" and $ \neg Q$ i.e. "$ \sum (1 \plus{} 1/n) \, a_{n}$ converges".

So yes.

BUT, if a series does not diverge to infinity [b]it does not mean that converges[/b]

Yes, that's correct.

So far, you appear to be making a collection of correct and somewhat related statements -- is there some point here that I'm missing?

Then the answer to my initial post should be [b]no[/b]  instead of [b]yes[/b]

No, both your first and third posts correctly take the contrapositive of a statement, but the two statements you started with are different and so the two contrapositives are different as well.  The words "to infinity" did not appear in your first post; the negation of "the series converges" is "the series diverges," and the negation of "the series diverges to infinity" is "the series does not diverge to infinity."  Again, is there some point to this?

When a series diverges ,it either diverges to infinity or alternates round a fixed point.

Hence when  it does not diverge ,it does not diverge to infinity and does not alternate round a fixed point.

[b]Do those two imply convergence?[/b]

[quote="stauros"]
Hence when  it does not diverge ,it does not diverge to infinity and does not alternate round a fixed point.

[b]Do those two imply convergence?[/b][/quote]

no..  for a series to converge,  (with respect to given problem) the individual terms of the series must approach zero.

and the 2 conditions u gaev  need not necessarily imply this.

By definition, if a given series does not converge then it diverges.  Among series that diverge, some diverge to $ \infty$ (or $ \minus{} \infty$) and some diverge in some other way.  If "alternates around a fixed point" means by definition "diverges but does not diverge to $ \infty$ (or $ \minus{} \infty$)" then the answer is, "Yes, by definition."  If "alternates around a fixed point" means something else, then the answer depends on what precisely you think it means.  (Personally, I think "alternates around a fixed point" is poor terminology for "diverges but does not diverge to $ \infty$ (or $ \minus{}\infty$).")

Again, what's the point here?  Why can't you go look up these definitions in a textbook?


Solve the system of equations:
$ \left\{ \begin{array}{l}
4xy \plus{} 4(x^2  \plus{} y^2 ) \plus{} \frac{3}{{(x \plus{} y)^2 }} \equal{} \frac{{85}}{3} \\  2x \plus{} \frac{1}{{x \plus{} y}} \equal{} \frac{{13}}{3} \\  \end{array} \right.$

[quote="devil_maths"]Solve the system of equations:
$ \left\{\begin{array}{l} 4xy \plus{} 4(x^2 \plus{} y^2 ) \plus{} \frac {3}{{(x \plus{} y)^2 }} \equal{} \frac {{85}}{3} \\
2x \plus{} \frac {1}{{x \plus{} y}} \equal{} \frac {{13}}{3} \\
\end{array} \right.$[/quote]
[hide]$ 2x \plus{} \frac {1}{x \plus{} y} \equal{} \frac {13}{3}$
$ \frac {1}{x \plus{} y} \equal{} \frac {13}{3} \minus{} 2x$
$ x\plus{}y \equal{} \frac{1}{\frac {13}{3} \minus{} 2x}$
$ y \equal{} \frac{1}{\frac {13}{3} \minus{} 2x} \minus{} x$

Then substitute into the other equation and have fun. Some factoring might make it a bit easier because of the equation for $ x\plus{}y$:

$ 4xy \plus{} 4(x^2 \plus{} y^2) \plus{} \frac {3}{(x \plus{} y)^2 } \equal{} \frac {85}{3}$
$ 4(x^2 \plus{} 2xy \plus{} y^2) \minus{} 4xy \plus{} \frac {3}{(x \plus{} y)^2 } \equal{} \frac {85}{3}$
$ 4(x \plus{} y)^2 \minus{} 4xy \plus{} \frac {3}{(x \plus{} y)^2 } \equal{} \frac {85}{3}$[/hide]

Express the above system in term of $ x\plus{}y$ and $ xy$ And using Vietta's theorem.


Kristin has 5 different pairs of earrings. What is the probability that two randomly-selected earrings will be a matching pair? Express your answer as a common fraction.

So the answer is 1/9, but how? Explaination? Please and thank you!

we treat this as conditional probability

there are 5 pairs, which makes 10 earrings

she can pick any one of them first, but then to get a matching pair, she must pick the exact matching one left in the pile, so int eh pile of 9 earrings, only 1 will be the one she wants after her first choice, so it is 1/9

thats the quicker way

teh other way is to think. she has 5 pairs of earrings, so 5 times she wil get a (different) matching pair

To pick 2 earrings out of 10 in total is $ \binom{10}{2}$ which is 45, so our answer is 5/45=1/9

hope this helps!


[quote="quykhtn-qa1"][quote="quykhtn-qa1"]Given $ a,b,c > 0$ and $ abc \equal{} 1$. Prove that:
$ \frac {2a^2 \plus{} b^2 \plus{} 3a^2b^2}{c^2(a \plus{} b)^4} \plus{} \frac {2b^2 \plus{} c^2 \plus{} 3b^2c^2}{a^2(b \plus{} c)^4} \plus{} \frac {2c^2 \plus{} a^2 \plus{} 3c^2a^2}{b^2(c \plus{} a)^4} \ge\ \frac {81}{8(a \plus{} b \plus{} c)^2}$[/quote]
Similary, prove that :
$ \frac {4a^2b^2 \plus{} a^2 \plus{} 1}{c^2(a \plus{} b)^4} \plus{} \frac {4b^2c^2 \plus{} b^2 \plus{} 1}{a^2(b \plus{} c)^4} \plus{} \frac {4c^2a^2 \plus{} c^2 \plus{} 1}{b^2(c \plus{} a)^4} \ge\ \frac {81}{8(a \plus{} b \plus{} c)^2}$[/quote]
no one interested in this inequality?? :(

[quote="nguoivn"]Sorry, quykhtn-qa1. Indeed, it's very ugly with me  :oops: . So, I don't like and can't solve it.
But I hope that many people'll like it. Good luck  :)
Sorry for spam.[/quote]
But,this inequality become nice if you use AM-GM  :)


induction
number theory proposed
number theory

Let [tex]{u_n}[/tex] be a sequence such that:[tex]u_1=1,u_2=4[/tex] and [tex]u_{n+2}=14u_{n+1}-u_n-6[/tex]
Show that:[tex]u_n[/tex] is perfect square!

We have $U_n = a_n^2$ where $a_1=1,a_2=2$ and $a_{n+2} = 4a_{n+1}-a_n.$

To prove this, you only have to verify that $a_n = \frac {2-\sqrt 3 } 2 (2+\sqrt 3)^n + \frac {2 + \sqrt 3 } 2 (2 - \sqrt 3)^n .$
Thus direct computations leads to :
$a_{n+1}^2 + a_n^2  = 4a_{n+1}a_n - 3$.
Therefore, using induction, it leads to $U_{n+2} = a_{n+2}^2$, as desired.

Pierre.


Given n numbers $ a_1,a_2,...,a_n$ in $ [\minus{}1;2]$ such that their total sum is $ 0$.Let $ U_k\equal{}\frac{a_k\sqrt{4k\minus{}1}}{(4k\minus{}3)(4k\plus{}1)}$ for $ k\equal{}1,2,...,n$.Prove that $ |U_1\plus{}U_2\plus{}...\plus{}U_n|<\frac{\sqrt{n}}{2}$

hix,your solution is very short,I don't understand

I am sorry $ U_1\equal{}a_1\frac{\sqrt 3 }{5}$ and maximal mean $ \frac{2\sqrt 3}{5}\equal{}\sqrt{\frac{12}{25}}>\frac 12$.
Therefore your condition is not correct.

[quote="Rust"]I am sorry $ U_1 \equal{} a_1\frac {\sqrt 3 }{5}$ and maximal mean $ \frac {2\sqrt 3}{5} \equal{} \sqrt {\frac {12}{25}} > \frac 12$.
Therefore your condition is not correct.[/quote]

err.. you forgot the requirement that $ {\sum_{k\equal{}1}^n a_k \equal{} 0}$. In other words, $ U_1\equal{}a_1\frac{\sqrt 3}{5}\equal{}0$ because $ n\equal{}1$, $ a_1$ must equal 0 to sum (by itself) to 0.


Hi. I am trying to make a floating figure with a border to insert some pseudocode. But no matter what I try I can't make spaces within the text. I want some lines to be indented. In other words I cant get \hspace to work or \qquad or whatever. Thanks for any help. Here's what I have:

\floatstyle{boxed} \restylefloat{figure}
\begin{figure}
\emph{Embed( S )} \\
\{  \\
\hspace{1cm}Assign coordinates of $l_{r_{i}}$ to $l_{s_{1}}$ \\
\hspace{1cm}Assign coordinates of $l_{r_{j}}$ to $l_{s_{2}}$	\\
\hspace{1cm}$l_{r_{x}}\Leftarrow$ some landmark of $R$ where $l_{r_{x}}\neq l_{r_{i}}$ and $l_{r_{x}}\neq l_{r_{j}}$ \\
\hspace{1cm}FOR EACH landmark $l_{s_{i}}$ of $S$ where $i \neq 1, 2$ \\
\hspace{2cm}$Circle_{1}\Leftarrow$ circle centered at $l_{s_{1}}$ with radius $d(l_{s_{1}}, l_{s_{i}})$ \\
\hspace{2cm}$Circle_{2}\Leftarrow$ circle centered at $l_{s_{2}}$ with radius $d(l_{s_{2}}, l_{s_{i}})$ \\
$IntersectionPoint_{1}\Leftarrow$ one intersection point of $Circle_{1}$ and $Circle_{2}$ \\
$IntersectionPoint_{2}\Leftarrow$ second intersection point of $Circle_{1}$ and $Circle_{2}$ \\
IF orientation of \{$l_{s_{1}}, l_{s_{2}}, IntersectionPoint_{1}$\} $\neq$ orientation of \{$l_{s_{1}}, l_{s_{2}}, l_{r_{x}}$\} \\
THEN Assign coordinates of $IntersectionPoint_{1}$ to $l_{s_{i}}$ \\
ELSE Assign coordinates of $IntersectionPoint_{2}$ to $l_{s_{i}}$ \\
\}
\caption{}
\label{alg:embed}
\end{figure}
\floatstyle{plain} \restylefloat{figure}

LaTeX doesn't expect to put spaces at the beginning of a line but it will do so if you replace \hspace with \hspace*


algebra
polynomial
calculus
integration
number theory
prime numbers
absolute value

Let $\mathcal P$ be a polynomail with rational coefficients such that
\[\mathcal P^{-1}(\mathbb{Q}) \subseteq \mathbb{Q}.\]
Prove that $\deg\mathcal P \leq 1$.

Lots of stronger assertions have been proven on the forum.

For this one, you could do this:

Assume WLOG the polynomial has integral coefficients, and let it be $p(x)=a_nx^n+\ldots+a_1x+a_0$. We know that for every $m\in\mathbb Z\ p(x)-m$ has rational rots. The denominators of these roots (in their lowest terms) must divide $a_n$, so the distance between two such roots is bounded from below (it's at least $a$ for some $a>0$). However, if $n\ge 2$, we have $|p(x)-p(y)|\to\infty$ if $x,y\to\pm\infty$ and $|x-y|\ge a$, which means that $p(A)$ ($A$ is the set of rational roots of polynomials of the type $p(x)-m,m\in\mathbb Z$) cannot contain all the integers.

We already have something a little bit stronger: it suffices to know that for every $m\in\mathbb Z,\ p(x)-m$ has at least one rational root.

Actually, here's a stronger statement: given integers $a_n,a_{n-1},\ldots,a_1$, for large enough prime numbers $p$ the polynomial $Q_p(x)=a_nx^n+\ldots+a_1x+p$ is irreducible:

The absolute value of $a_nx^n+\ldots+a_1x$ has a maximum $M$ on the (closed) unit disk. If $p>M$, then the equation $Q_p(x)=0$ can have no roots in the closed unit disk. However, if $Q_p$ were reducible, it would be possible to write it as $Q_p(x)=f(x)g(x)$ where $f,g$ are polynomials with integral coefficients, and one of them (say $f$) has the free term equal to $1$ (while the other one has the free term equal to $p$). But in this case, $f$, and thus $Q_p$, has at least one root of absolute value $\le 1$, contradiction.

I think this problem has been discussed before, a long time ago.


[quote="hitek"]Prove that if $a,b,c\geq0$ then
\[\sqrt{a^{2}+b^{2}+(a+b)c}+\sqrt{b^{2}+c^{2}+(b+c)a}+\sqrt{c^{2}+a^{2}+(c+a)b}\geq2(a+b+c). \]
[/quote]
SOS kills it. :wink:

A better represent:

If $a,b,c\ge 0$ and $a+b+c=2$ then
\[\sqrt{a+b-ab}+\sqrt{b+c-bc}+\sqrt{c+a-ca}\ge 2\sqrt{2}.\]

:huh: yes,it isn't so easy...
very interesting...

But even if $a^{2}+b^{2}+(a+b)c<a^{2}+c^{2}+(a+c)b$ magorization will be there , because inequality $(a+b)^{2}>$'something' will be right in both varients.


function
Support
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Name a function $f$ which is not integrable on $[a, b]$, but $f=g'$ for some function $g$.

uhm..
i'm sorry if i'm not clear, but writing everything down is pretty boring...
without any loss of generality, you can suppose $[a,b]=[0,1]$
take a cantor set of positive measure, and construct it as a decreasing intersection of compact sets $C_{k}$, where the $k$-th one is union of $2^{k}$ closed disjoint intervals $I_{j,k}= [a_{j,k}, b_{j,k}]$ such that $a_{1,k}= 0, b_{2^{k},k}=1$. finally call $d_{k}= \min_{j}\{b_{jk}-a{jk}, a_{j+1,k}-b_{j,k}\}/3$ (which is a third of the minimum distance between two endpoints of intervals of $C_{k}$).
now, consider a smooth function $s$ with support in $[-1,1]$, and which is constantly $1$ on $[-1/2, 1/2]$, and define $s_{k}(x) = \sum_{j=1}^{2^{k}}\frac1{2^{k}}\left(s\left(\frac{x-a_{j,k}}{d_{k}}\right)\sin\left(\frac1{x-a_{j,k}}\right)+s\left(\frac{x-b_{j,k}}{d_{k}}\right)\sin\left(\frac1{x-b_{j,k}}\right)\right)$.
now define $g=\sum_{k}s_{k}$.
what you can actually prove is that both the series and the series of derivatives are convergent, so you can commute series and derivative, and you get that $g'$ is discontinuous on the whole cantor set, hence not (riemann-)integrable..


topology
real analysis
real analysis unsolved

Draw Cantor set $C$ on the circle and consider the set $A$ of all chords between points of $C$.
a)prove that $A$ is compact.(easy part).
b)is $A$ convex?

Could you please clarify some things here?

First of all, what's Cantor's set on the circle? I imagine you get it by drawing the Cantor set on a segment, and then identifying the two extremities of the segment, but I'm not sure. Second of all, you say "consider the set $A$ of all chords..". What does that mean, exactly? Is $A$ the union of the chords, considered as a subsets of the plane, or is it actually the set of chords, which we endow with some sort of topology? If so, what topology do we give it? What makes two chords close to each other? In other words, are the elements of $A$ points or chords?

Sorry, but it seems really unclear to me, and it looks like a problem I might enjoy :).

When I asked the question, I was thinking about the actual Cantor set, the one where you remove the middle thirds, but, come to think of it, you might be right. Maybe this is what sam-n meant. I still think we should wait for him to clarify this though :).

What do you think about my (probably stupid? :?) questions regarding the set $A$? What do you think sam-n meant by that?

Actually, when I was reading it for first time I read it as "the set A of [b]midpoints[/b] of all chords" (I don't know where I found the word "midpoint"  :blush: ). So it was quite surpizingly to read your tangled explanations about the nature of $A$.

Anyway, I think we should consider $A$ is a set of points on the plane.

Now I realize just how stupid those tangled explanations really are :).

If $A$ were a set of chords, what on Earth would "convex" mean? I really doubt $A$ is anything else than the union of the chords :).

For the first part of the problem (the fact that the Cantor set is closed suffices here, I think):

Assume $X$ is a point in the closure of $A$. It must, of course, belong to the closed disk. In this case, there is a sequence of points $X_n$ which are projections of $X$ on some of our chords s.t. $XX_n\to 0$. It's also clear that there is a subsequence $X_{k_n}$ s.t. the direction of $XX_{k_n}$ tends to some fixed direction $d$. Now, the extremities of the chords through $X_{k_n}$ and perpendicular to $XX_{k_n}$ tend to the extremities of the chord through $X$ and perpendicular to $d$, and, since the Cantor set is closed, those two extremities belong to it, meaning that $X\in A$.

This shows that $A$ is closed, and since it's bounded, it must be compact.

Indeed, it was so clear that $A$ is compact.

you are really quick excuse me for not viewing per day :D now should I explain it more  :)

The cantor set is not only compact but it is also a perfect set of measure 0!!! ;)


is it the original problem?
if not,can you post the original one?

It is my friend's problem.  I think that this problem is not so beautiful.
But the expansion may be the original problem.  :(
$\sum_{sym}a^6 + 10\sum_{sym}a^4 b^2 +5\sum_{sym} a^4 b c \geq 6\sum_{sym} a^5 b +5\sum_{sym}a^3 b^3 + 5\sum_{sym} a^3 b^2 c$

We will use the $\sum$ notation for cyclic sums. Then, the inequality in question becomes

$\sum\left(b^5-c^5\right)\left(b-c\right)\geq\sum 5bc\left(b^2+c^2-ba-ca\right)\left(b-c\right)^2$
$\Longleftrightarrow\sum\left(b^4+b^3c+b^2c^2+bc^3+c^4\right)\left(b-c\right)^2\geq\sum 5bc\left(b^2+c^2-ba-ca\right)\left(b-c\right)^2$
$\Longleftrightarrow\sum\left(b^4+b^3c+b^2c^2+bc^3+c^4\right)\left(b-c\right)^2$
$-\sum 5bc\left(b^2+c^2-ba-ca\right)\left(b-c\right)^2\geq 0$
$\Longleftrightarrow\sum\left(\left(b^4+b^3c+b^2c^2+bc^3+c^4\right)-5bc\left(b^2+c^2-ba-ca\right)\right)\left(b-c\right)^2\geq 0$
$\Longleftrightarrow\sum\left(\left(b-c\right)^4+5bc\left(ca+ab-bc\right)\right)\left(b-c\right)^2\geq 0$
$\Longleftrightarrow\sum\left(\left(b-c\right)^6+5bc\left(ca+ab-bc\right)\left(b-c\right)^2\right)\geq 0$.

But this follows if we add together $\frac{23}{33}$ times the inequality $\sum\left(b-c\right)^6\geq 0$ and $\frac{5}{33}$ times the inequality $\sum\left(2\left(b-c\right)^6+33bc\left(ca+ab-bc\right)\left(b-c\right)^2\right)\geq 0$. While the first of these two inequalities does not require any explanation, the second is true since

$\sum\left(2\left(b-c\right)^6+33bc\left(ca+ab-bc\right)\left(b-c\right)^2\right)$
$=\left(b+c-2a\right)^2\left(c+a-2b\right)^2\left(a+b-2c\right)^2+33abc\left(a\left(b-c\right)^2+b\left(c-a\right)^2+c\left(a-b\right)^2\right)$
$\geq 0$.

This completes the proof of the inequality.

By the way, how did your friend come up with this inequality? Did he try to find the sharpest inequality of 6th degree? ;)

  Darij

[quote="darij grinberg"]

$\sum\left(2\left(b-c\right)^6+33bc\left(ca+ab-bc\right)\left(b-c\right)^2\right)$
$=\left(b+c-2a\right)^2\left(c+a-2b\right)^2\left(a+b-2c\right)^2+33abc\left(a\left(b-c\right)^2+b\left(c-a\right)^2+c\left(a-b\right)^2\right)$
$\geq 0$.

  Darij[/quote]
Hi,Darij:can you tell me how to find this equality?

darij,how can you calculate this equation?you are something. :D

[quote="zhaobin"]Hi,Darij:can you tell me how to find this equality?[/quote]

Computing the sum for a = 0, then guessing which kind of terms will occur in the general case. The remaining terms were easy to identify then.

  Darij


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I want to hear each and everyone of you who has an opinion say what you believe.

This topic is now open for discussion.  :)

Please refrain from speaking if you intend to spam.

[quote="Go Around the Tree"]God- the character in the Bible.[/quote]

Ah, now you've made the question much more specific.  Good.

If the Bible is to be read literally, there is no evidence that this God physically exists as a corporeal entity.

This is far from answering the much more general question, "does God exist?"  That question is much more complicated, and any serious attempt to answer it should begin with the search for a more satisfying definition of what it means for God to exist.

[quote="Fanatic"]Let's just put it this way.

If I was trying to take over the world, religion would be my best friend.

Infer now plz.

Some more hints..

If I was trying to takae over the world / trying to make my race/species/gender/sexuality/job more acceptable than others, religion would be my best friend.

As for god? I don't know. Agnostic. But for those who argue "if god didn't create the world, then who did?" I still find the argument "who the f created god then?" plausible no matter what the response is.

As for the bible -- absolute BS created by those who wanted and knew how to obtain power. Religion itself, at least in most of its aspects, is excellent. People's interpretation of what to take in and what not to take in, is totally wrong. I don't blame religion... I blame religious people for being hypocrites.[/quote]
"How I envy the young. For them everything is so clear. Things seem to be either one way or another." [url=http://www.imdb.com/title/tt0108265/]Swing Kids[/url]

[quote="Imperial Effect"]"How I envy the young. For them everything is so clear. Things seem to be either one way or another." [url=http://www.imdb.com/title/tt0108265/url]Swing Kids[/url][/quote]

 :D

Which of course is entirely the distinction between arguing the truth of a mathematical proposition and the truth of a statement as complex as "God exists."  It seems to me that this forum, being primarily a mathematics forum, has members which particularly love to fall prey to this fallacy.

Well, here's something...

Let's say you're talking about a certain subject.  If you keep asking "Why is this so?", you're eventually going to get to a point where it's pretty difficult to explain why.  That's where God comes in.  But then you can ask, "How did God come to exist?"  Personally, I don't think any of us will truly know the answer to this question until we die.

[quote="lingomaniac88"]you're eventually going to get to a point where it's pretty difficult to explain why.  That's where God comes in. [/quote]

So, in other words, when you can't explain something, then it's because of God?

 :wink:

[quote="Ihatepie"]If he is though he isn't nice.

Why is there 9-11,The war in Iraq, torture, terrorism, killing, hunger, racism, or many other bad things if god is so good? or real... [/quote]

http://en.wikipedia.org/wiki/Problem_of_evil

Does God exist- leave aside the Christian/Biblical aspect of the debate(I am an Indian, and not well versed with the Bible to talk upo it), I dont believe that God exists, because of the following reasons:
where is the proof that he exists?
and unless anything is proved it is non-scientific to accept it.

And if I'm not wrong didnt Jesus perform miracles(I'm not sure), like curing an ailing person , making bread from nowhere etc. Now, all of these violate known and proved scientific laws.

[quote="t0rajir0u"][quote="Ignite168"]In the beginning, God was there.[/quote]

I feel it's mostly counterproductive to debate a true/false statement such as "Does God exist?"  Let me ask you this:

What do we know about God?  What do we [i]really[/i] know about God?[/quote]

Well what do you want me to do? Make up an argument? I'd rather use the truth.

We have much knowledge of God in the Old Testament, Genesis is a very good place to start, I suggest you read it. It is there and less so Exodus, where you'll find the majority of God's interactions with people as documented by Moses.

Most of you make most obvious logical mistakes when it comes to "thinking about god". You're mathmaticians for god's sake, think in more abstract way.

Words are limited by what human can see. The world is infinite on the contrary to history of humanity. Thus, to achieve language that would be able to desribe the world as it is we would need infinite language, so prolly, infinite time.

The paradox of the question concerning god's existence is that it can be formed in most limited language but answered by none known so far.

[quote="gollywog"]You're mathmaticians for god's sake, think in more abstract way.[/quote]

For whose sake, now? :D

Define "exist". Existing as in a material sense, possessing mass and occupying a distinct position in spacetime? Existing in some other sense?

[quote="Ignite168"][/quote][quote="Ihatepie"][quote="Ignite168"][quote="Ihatepie"]I believe God isn't real. If he is though he isn't nice.

Why is there 9-11,The war in Iraq, torture, terrorism, killing, hunger, racism, or many other bad things if god is so good? or real...

I doubt any1 will ever figure out how the universe started...[/quote]

How can you even begin to understand God?

Where were you when God laid the foundation of the earth? Do you have understanding? Who determined its measurements -- surely you know! Or who stretched the line upon it? On what were its bases sunk, or who laid its cornerstone, when the morning stars sang together, and all the sons of God shouted for joy? Or who shut in the sea with doors, when it burst forth from the womb; when God made clouds its garment, and thick darkness its swaddling band, and prescribed bounds for it, and set bars and doors, and said, `Thus far shall you come, and no farther, and here shall your proud waves be stayed'? Have you commanded the morning since your days began, and caused the dawn to know its place, that it might take hold of the skirts of the earth, and the wicked be shaken out of it?[/quote]

who made God?[/quote][quote="Ignite168"]

No one. In the beginning, God was there.[/quote]
and u know that how... and also how do u know those books r right? The bible was made by some people who wanted 2 make alot of money.

[quote="Ignite168"]We have much knowledge of God in the Old Testament, Genesis is a very good place to start, I suggest you read it. It is there and less so Exodus, where you'll find the majority of God's interactions with people as documented by Moses.[/quote]

At this point it is easy to bring up a few familiar counter-arguments which I have no doubt you have already formulated defenses against:

- The Bible itself is riddled with contradictions and otherwise with moral imperatives that most people feel no need to follow today, yet are there all the same (see, for example, http://www.skepticsannotatedbible.com/ ).  Which parts do we believe and which parts don't we believe?

- Other holy books besides the Christian Bible exist.  Your belief in your particular book is spurred by the fact that you have been raised in that belief.  What makes the belief of another denomination raised in a different culture, with a different worldview, less valid?

Beside those obvious points, here are two less obvious ones:

- The Council of Nicaea, presided over by Emperor Constantine I, selected certain holy books in circulation to be included in the "official" (as it were) Bible, and omitted others.  Shall we conclude that Constantine was holy enough to discern the true word of God from fabrications and slander?

- The King James Edition of the Bible was [b]translated[/b] over a millennia and a half after the original compilation of the text.  The King's scribes were, after all, only human.  Shall we trust their interpretation of the text?  Might they not have mistranslated the word of God?  (Plenty of evidence shows that, after a careful examination of older texts, many phrases in the King James Edition are surprisingly misleading.)

[quote="gollywog"]Words are limited by what human can see. The world is infinite on the contrary to history of humanity. Thus, to achieve language that would be able to desribe the world as it is we would need infinite language, so prolly, infinite time.

The paradox of the question concerning god's existence is that it can be formed in most limited language but answered by none known so far.[/quote]

The use of the word "infinite" is self-defeating.  This is one of those nice concise "proof-like" arguments that the mathematically inclined love to use to convince themselves of the truth of a distinctly non-mathematical proposition, and therefore to put the matter to rest, hmm?

Of course, the world is not so simple.  The same argument could be applied against, for example, the pursuit of science.  Does that mean that we do not pursue science?

[quote="Ihatepie"]and also how do u know those books r right? The bible was made by some people who wanted 2 make alot of money.[/quote]

This is an [i]unfair allegation,[/i] and you know it.  It's based on the mistaken black-and-white belief that either the writers of the Bible were transcribing the true word of God, or the writers were con artists trying to fool future generations.

Are there really no third possibilities?  Do you presume that you know so much about the nature of culture and civilization, especially two thousand years in the past, that you can rule out any other motives for compiling what is a truly inspiring religious text?

How about the primal desire to tell stories?  The desire to pass on our understanding of the world to future generations?  Isn't it possible that the people writing this book really believed in their own basic message?

[quote="t0rajir0u"][quote="kstan013"]Where would we get this idea of a God if there wasn't one? [/quote]

Where would we get the idea of unicorns if there aren't any?[/quote]

Whoa ... I was going to use the example of unicorns!  The only explanation for such a coincidence is the existence of a higher force.  Praise Jesus!

t0rajir0u, most well written forum post EVER. Much respect.

[quote="JBL"]
[quote]God does it to test our faith in Him...thats what I think.[/quote]
If your god is omnipotent and omnicient, he shouldn't need to test your faith.  I still don't get it; what's so good about faith, anyhow?  Why is believing something without evidence morally superior to believing things with evidence and withholding judgement on the rest?
[/quote] From a bar mitzvah I attended way back when, I recall that they concluded with the point that the Jewish God isn't totally omnipotent or completely flawless, but still mostly.

So Jews don't believe in the omnipotence of God? or just some Jews?

Religion is such that it was [i]made[/i], arguably, to differ from science; religion is [b]separate[/b] from science. Of course, scientists want to prove everything scientifically. I don't think atheism came from scientific evidence against God, but lack of evidence for God that certain scientists who didn't want a God manipulated into being evidence against God.

As for Islam... wasn't the Koran (or Qu'ran?) against converting people by force, but Muhammed and his followers spread throughout the Middle East, even into Spain, giving the ultimatum "Convert, leave, or die." (As far as I've heard.) Also, many Islamic prophecies appear to be obvious (there was something about an army defeating another, when the "prophesized" victor was military far superior to the loser). The Christian Bible, I admit, may have some of these too. Prophecies are a bit dubious in any religion.

[quote="not_trig"]Religion is such that it was [i]made[/i], arguably, to differ from science; religion is [b]separate[/b] from science. [/quote]

The most common religions of the world have roots that all predate the scientific era, so this isn't a valid way to deflect criticism.  In fact:

[quote="Dictionary.com"]1.  a set of beliefs concerning [b]the cause, nature, and purpose of the universe[/b], esp. when considered as the creation of a superhuman agency or agencies, usually involving devotional and ritual observances, and often containing a moral code governing the conduct of human affairs.[/quote]

I think what needs to be done here is to make a distinction between religion and [b]spirituality.[/b] Refer to

http://en.wikipedia.org/wiki/Spirituality#The_spiritual_and_the_religious

Either way spirituality is taken, it is basically some form of a philosophy of life, and in that sense, yes, it is distinct from science.

Religion proper, however, has historically been a way of explaining the world - precisely the role that science has now taken from it.  That is why many people like to think of religion and science as necessarily separate; because religion no longer has a role in people's lives as a way to explain the world, its role as a source of spiritual well-being is emphasized.

That having been said, the problem with religion comes when people continue to use it as a way to explain the world, e.g. to morally justify slavery or genocide.  In this sense religion is dangerous because it [b]leads to false conclusions in some areas where science would give a more accurate answer.[/b]

I must respectfully disagree with max_tm: while that post was indeed excellent, [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=684572#684572]this[/url] was the best forum post ever.

[quote="t0rajir0u"]In this sense religion is dangerous because it [b]leads to false conclusions in some areas where science would give a more accurate answer.[/b][/quote]

The problem is worse than that: not only does it lead to false conclusions, but it leads to absolute certainty in those conclusions.  When you claim that "faith" is important, when you raise unreasoned belief above reasoned belief, you not only open yourself to error but you also automatically reject any correction of that error.

Question: the existance of a God (or god), from the atheist perspective, was used to explain natural phenomenon or to gain power. There are so many religions in the world, that I think its safe to assume that a few came soley from natural phenomena explanation. That would explain their belief in god/God. But where would the belief, for example, of something like "Zeus is the king of the gods and he rules over twelve" come from? After all, the Greeks must've seriously believed it, or they wouldn't have made 100+ big temples honoring them. For example, if I wildly invent a theology, then even if it explains and satisfies, I would be hard pressed even to convince myself that something I wildlly invented was actually true, much less have an incentive to teach it to others. Operating under the assumption that some religions were not created simply to gain power (from the Atheistic viewpoint), then how would you explain how the non-powerhungry religions with defined mythology (like Greek or Norse myths) came to be?
Just a thought.

[quote="t0rajir0u"][quote="Scrambled"]how would you explain how the non-powerhungry religions with defined mythology (like Greek or Norse myths) came to be?[/quote]

This can all be grouped under the broad category of [b]folk religion[/b], and the short and sweet explanation is that folk religion is a direct result of the human need to explain the world in human terms, as I discussed in my lengthy post.

Added to this basic need for explanation is the human urge to [b]tell stories[/b], which, as anyone can tell you, runs deep in any culture.  From explanations of the natural world come fantastic stories - the more entertaining, the better.

Out of this [b]oral history[/b] evolves what we would refer to as a mythology.[/quote]
Sure, but I find it hard to believe that folk religion inspires people to spend their riches building big temples like the greeks did. And forcing persecution on the non-greek mythology believers like Rome did.
I dunno.

[quote="Scrambled"]I find it hard to believe that folk religion inspires people to spend their riches building big temples like the greeks did. And forcing persecution on the non-greek mythology believers like Rome did.[/quote]

You're ignoring rather a lot of the non-religious cultural forces that shape history.  You might be interested in taking a world history class.

Regarding temples:  If you actually believe that powerful, capricious Gods control your world, then of course it makes perfect sense as a community to build a temple to appease a God.  This is, I think, a fairly simple principle.

Regarding Roman persecution:  You're missing a major historical point here.

The Romans established a [b]state religion[/b].  While they maintained the basic shape and nature of the twelve Gods of the Greek pantheon, this is in no way a direct continuation of Greek folk religion.  This state religion was also influenced by Etruscan or old Latin beliefs.

If you're interested, I'm sure you can find plenty of books on the subject.  It's fairly interesting.

In light of the fact that Roman religion is sponsored by the state, it makes [b]political sense[/b] to persecute non-believers, since they disrupt the social order.

[quote="t0rajir0u"]
Regarding temples:  If you actually believe that powerful, capricious Gods control your world, then of course it makes perfect sense as a community to build a temple to appease a God.  This is, I think, a fairly simple principle.
[/quote]
Right. But my point is if a guy made up random characters and told them through stories and other people added other stuff on for dramatic effect, that doesn't really lead to any real belief, as the creators didn't believe what they were saying anyways, and didn't try (or so I believe)  to achieve one's own ends in order to believe. I've studied ancient Greece a lot (well, maybe not, but I have looked through several world history books concerning this), and as far as I have read, no one used religion as a driving force of power. So what makes folk religion believed? (The other example was simply to show the magnitude of a religion in which no one should seriously believe in, assuming that it was false.)
Just to clarify, I'm not using this to advocate any religion being true; I'm simply wondering, as this topic already deals with a lot about religion.

[quote="Scrambled"]But my point is if [b]a guy[/b] made up random characters and told them through stories and other people added other stuff on for dramatic effect, that doesn't really lead to any real belief, as the creators didn't believe what they were saying anyways, and didn't try (or so I believe)  to achieve one's own ends in order to believe. I've studied ancient Greece a lot (well, maybe not, but I have looked through several world history books concerning this), and as far as I have read, no one used religion as a driving force of power. So what makes folk religion believed?[/quote]

Whoever said it was just one guy?  Oral history is passed down and accumulated over generations by many individuals.

I also think you're missing some of the historical context here.  Ancient Greece is not exactly the age of mass media.  Stories about the past from the mouths of bards (or, failing that, your family) are often both a primary source of entertainment and of historical information.  Since there aren't exactly any other sources of historical information, true histories passed down by word-of-mouth are not easily distinguished from old legends.

Furthermore, what makes you think that people passing it down didn't believe what they were saying?  Consider, for example, the modern proliferation of [b]urban legends.[/b]  How do they start?  Perhaps there is some grain of truth that a storyteller proceeds to embellish heavy.  It spreads by word-of-mouth until pretty soon everybody has heard of it.

In the information age, it's actually feasible to investigate and debunk such legends.  In more ancient times, "urban legends" become... well, legends.  People believe them.

[quote="Frank"]Sorry for bumbing a long-time-ago-posted thread, but it just seems that it deserves a lot of attention. A lot. I'll just simply try to explain my opinion towards the "existence" or "non-existence" of God. :D

I'm Christian, raised in Catholic family. To begin with.. I don't actually think that it matters whether the God exists or not. Also, I don't think that the God exists or doesn't exist. Of course, we have no scientific data to verify the existence of God. We cannot disprove it either. But if something cannot be disproved, it doesn't mean that it actually is a true fact.

It's important to believe in something. It's what makes us humans. We have a freedom to believe in anything we want to, or eanything we're [i]made to[/i] want. Made by our enviorment, our genetic makeup? I don't know, nor I care.

What's really important is to stick to morals and respect other people's faith. By doing so, we're averting any possible crisis. What we need more of is a tolerance. The religion itself is a good example. The world is splitted over the religion.

The God is a (more or less) [i]fictional summarization[/i] of the deepest and the most beautiful human feelings -- love & compassion. Most of the believers are overwhelmed by that idea in extend of throwing away anything potentially disproving God. They're blinded. But if their blindness is turning them into better people, with stronger moral stance and some higher beliefs, if it makes them accept certain thing with ease, and if it makes them set up some higher moral values, then, to believe in God means to be a human. Then we gotta stick to the idea.  :wink:

But from the other side, some people have problems with dividing their emotional and their intellectual relation with God.
On intellectual level, I believe that God is just an idea, the fictional one. On emotional level, I believe that God [i]exists[/i] with all of my heart. But most of the Christians, Muslims and other believers cannot think in this way. It's because they find the idea of God being just an idea extremly repulsive, due to their mental bonding of "emotional" and "intellectual".

So whether you believe in God or not, I don't think it matters at all. I also don't think that God's existence/non-existence changes anything. We're here to do something. And what to do, is up to us.

I hope that you won't find yourself insulted by the content of this post.
Greetings  :wink:[/quote]

You have some thought-provoking ideas. Emotionally, I agree, but intellectually, I find it hard to believe.

But seriously, it seems as if you're defining truth in a multitude of ways. Which raises the question: is what you're saying absolutely true or not?

I read some of those post specially by religious members. I do respect their opinions.
Neither existence nor inexistence of God has been proved.But religion is the best way to win your political aim.I think that all founder of religions were politicians.They took some of scientific or non-scientific Ideas to convince their Nation to follow them.
And people started believe because God was shown as a very kind and Powerful creature.Some might have fear or some might have seen fantancy kind of miracle.Now a days we see that some religious people are used or using their religion to win a war.


How can i type the binomials using latex?

{top \choose bottom} also works:
${top \choose bottom}=something$

Just a note: In $\LaTeX$-2e, \binom is preferable over \choose.

Just another note: $\LaTeXe$ (\LaTeXe) is the command you're looking for, jhaussmann5.

Ah, thanks.  I'll try to remember that.  I was just too lazy to look it up. :P

I think the $\text{\TeX}$ community is trying to move towards the new $\text{\LaTeX}e$ commands, such as \binom rather than \choose and \frac rather than \over. I think they also want to switch to

[code]\[ \][/code]

and get rid of $\$$\$ (which I have not done myself).

[quote="Ledurt"]do you mean that instead of using $\$ i sould use \ [ and  \ ] in my documents?
and for the same reasons giving before?
ledurt[/quote]
Not exactly. Typing $BLAH$ in a document produces the in-text equation BLAH. But there are two ways to create a displayed equation (i.e. it gets its own line and is in displaystyle): \[BLAH\] or $$BLAH$$. The community is leaning towards \[BLAH\] rather than $$BLAH$$, but the in-text equations are still written using $BLAH$.

(Edited... sorry)

you said:

"\[BLAH\] or \[BLAH\]. The community is leaning towards \[BLAH\] rather than \[BLAH\]"

I'm sorry but I don't see the diffirence...

but anyway, instead of using

\begin{equation}
...
\end{equation}

I should use

\[
...
\]

and what about "align" and other equation environnement?

ledurt

Sorry, the computer edited what I wrote. It is fixed, and hopefully the comment now makes sense  :blush:

[quote="djimenez"][quote="jhaussmann5"]Just a note: In $\text{\LaTeX}$-2e, \binom is preferable over \choose.[/quote]

Well, why are you so sure? It may be that you prefer it. Actually I think, as \choose is the original comand in $\text{\TeX}$, probably it compiles more efficiently (just a guess) and it is a little more natural, you usually say $n$ [i]choose[/i] $k$, and not [i]binomial[/i] $n,k$.[/quote]

It's not just me, this was part of the upgrade from $\text{\LaTeX}$ 2.09 to $\text{\LaTeX}e$; see this website, for example:[url=http://www.agnld.uni-potsdam.de/~shw/LaTeX-html/latex2elatex209.html]www.agnld.uni-potsdam.de/~shw/LaTeX-html/latex2elatex209.html[/url]

I believe part of the rationale for this change was that when $\text{\LaTeX}$ is compiling your code, it will see \binom first and know that a binomial coefficient is coming, rather than finding a \choose in the middle of the code, then having to go back and correct the formatting.  Anyway, this is how it was explained to me.

That is weird, because the school of math here at tech has always the newest version of $LaTeXe$, and I have never had any problem with \choose or \over on them, even when the version they have is even newer than the 2.09!

Yes, well probably they think that if you use lots and lots of choose in a document like say a looong article or a book on a 486 with 16 Mb of Ram machine running Linux OS then you will get in some trouble :D In our "modern" days, these so called time saving tehniques in $\LaTeX$ are no longer necessary.


Find R2 such that the red area be half of circle 1 area $\pi{R1}^{2}$

$R_{1}=OA, \ R_{2}= AB$

Let $\phi = \angle AOB$

If my work is correct, then we have to solve the equation $\phi \cos \phi-\sin \phi = \frac{\pi}{2}(2\cos\phi-1)$, which is equivalent with the equation $\phi-2\sin\phi = (\pi-\phi)(2\cos\phi-1)$

I don't know how can we solve this equation, but the solution is about $\phi \approx 1.235896924$, which gives $R_{2}\approx 1.158728473 R_{1}$

[u][color=red]Comment[/color][/u]

This problem was posted again, [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=71024[/url]


Why China didn't participate in the IMO 1998?

The 1998 IMO was in Taiwan. China and Taiwan have very poor relations, and China refused to send a team for political reasons.

[quote="Ignite168"]China was represented by Taiwan because they are one country >_> <_<[/quote]

Lol, I don't wish to offend any Taiwanese members of the community. However, I ASSURE you that the Chinese team is not well-represented by a purely Taiwanese selection. You do have a point there about the one country business though.

And yes, Mainland China and Taiwan do not have optimal political relations, but I wouldn't say they're poor.

I noticed that too, the People's Republic of China did not send a team in 1998, thus making it possible for other teams to get a better position.

They don't recognize each other as a country.  They both consider the other part to be ruled by an illegitimate government.  It would be inconsistent with their policy to send a team to their "own country".

Well, it only made +1 advance for everyone. Tho, considering distances not places it didn't make much difference. Maybe influenced statistics a little bit, but one win more or less who cares?

Sureley the Iranian team since they placed first.

[quote="MithsApprentice"]I ASSURE you that the Chinese team is not well-represented by a purely Taiwanese selection.[/quote]You're right. When you add 1.5 billion people chances of finding 6 better increase dramatically :)

Actually looking at the past history, in the last 15 years only 3 teams managed to beat China to the first place: Romania, USA and Bulgaria (Romania came pretty close in 1995 too, but ended up on the 2nd position).

About the argument of "not sending a team to its own country": that doesn't make sense. Why would they send a team to Hong Kong in 1995? Or even allow a Hong Kong team after Hong Kong became an official province of China? ...

[quote="Valentin Vornicu"]

About the argument of "not sending a team to its own country": that doesn't make sense. Why would they send a team to Hong Kong in 1995? Or even allow a Hong Kong team after Hong Kong became an official province of China? ...[/quote]
Is that referring to me? :huh:
It's more than just "different", unlike Hong Kong, the Chinese disagree with Taiwan's "status".  Or do they also show propaganda in the streets of Beijing showing what kind of a threat Hong Kong, to urge men to enlist in the army? :|

Not saying that Iran would hav won China that year, but Iran have the potential.
Also they had the only perfect scorer.

[quote="Valentin Vornicu"][quote="color"]Also they had the only perfect scorer.[/quote]Just because Ciprian Manolescu graduated one year before that :D[/quote]

Well well  :) , but that is actually not relevant.
Also there were other perfect scorers in
the 90's who potentially could copy their results but they graduated.


geometry
geometric transformation
rotation
ratio

we have a ball linked to a spring wich slides on a stick of wood.
both the stick and spring are linked to a wall.
the ball has a rotation movement with the wall taken as the rotation axis on the
horizontal plane.
we are given:

$\omega -$ angular speed of rotation of the ball.
$k -$ elasticity coeficient of the spring wich the ball is attached to.
$m -$ mass of the ball
$g \approx 10$
$l=1m$ - lenght of the stick of wood
$f=\frac {1}{5}$ - a ratio

we have to find $\mu -$ coeficient of friction between ball and stick of wood so
that the spring is stretched to $f\cdot l$ lenght

how do i solve this?
i tried to do

$F_{cf}=F_{fr} +F_{el}$ by wich i mean
centrifugal force equals friction force added to elasticity force
and further:
$m\omega ^2 \cdot x=\mu mg + kx$
$m\cdot 4\pi ^2 v^2 x=\mu mg + kx$
$\mu=\frac{m\cdot 4\pi ^2 v^2 x - kx}{mg}$
but $x=lf$ because the spring is streched that much.
and we got
$\mu=\frac{m\cdot 4\pi ^2 v^2 lf - klf}{mg}$

is this right or not?
if not please tell me the right way
thanks

I lost your train of thought when you made the substitution $\omega = 2 \pi v$. Is v supposed to be the velocity? It's not a given. Also from a quick check of the units, the answer doesn't seem right because $\mu$ is a dimensionless constant and your final answer definitely has some units.

make your own solution post it.
maybe i can learn where i'm wrong.
thanks for looking

yes your answer is correct.
the only cool thing about this problem is that the direction of the Friction changes, with as the mass moves, therfore, then most important step is to find:
$F_c=F_f+F_s$ then one other good thing is to realize that the direction of the force exerted by the spring is always in the same direction as motion, therefore, you will have some $kx$ instead of $-kx$
you did it all correct... :)

no, you have made mistake. :(
$\omega = \frac{v}{x}$
you don't need to replace it, because $\omega$ is given.
change this back, solve again, and you'll be fine.


This particular case is easy because you just have to look for 2^(-1) in Z3, and that's 2, so your eqn is, basicly, x=2(mod 3). Are you looking for a general way of solving these? I think there is a method which helps you get more solutions once you have one solution, or something like that. I'll look it up and let you know.

I'm not sure if Rep123max would understand what 2^(-1) in Z3 may mean :)

Basically now that you are down to 2x=1 (mod 3), you want to know what x is mod 3. If x was 0, 2x=0, if x was 1, 2x=2, if x was 2, 2x=4=1. So you know x must be 2 mod 3.

So x = 3a+2. Substitute that into the original and you get 15a+10+3y=472, so 3y = 462-15a, so y=154-5a (sorry, I'm doing this in my head, so I may do an arithmetic mistake somewhere..)

So every solution is like that, for any a.

[quote="Rep123max"]how do you know x=2(mod3) and not 1(mod3)????

thanks[/quote]

you know how ab (mod c) = [a (mod c)][b (mod c)] ?then you get
[2 (mod 3)][x (mod 3)]=1 (mod 3). If x is 2, you get 2*2 (mod 3) or 1, which works. If x is 1, you get 2*1 (mod 3), which leads to 2=1 (mod 3), which is definitely false.

Basicly, since 3 is so small, you can just try all residues modulo 3 and see which one works. Since 1 and 0 don't work, you find x to be 2(mod 3).

If you're still mystified, go grab your AoPS V2, read section 23.2 & 23.3, then 24.1.  The first two talk about division in congruences & solving linear congruences.  The second about the Diophantine equations you are working with.

when you get to 2x=1 (mod 3), this is equivalent to 2x=4 (mod 3) Now you can divide by two to get x=2 (mod 3). NOTE: You cannot divide in a congruence like ax=b (mod c) under normal cirmcumstances. here it works because 2 and 3 (our a and c) are relatively prime.


Let $ a,b,c \in {R}$ s.t. $ a^2\plus{}b^2\plus{}c^2\equal{}1$.Prove that $ |a^3\plus{}b^3\plus{}c^3\minus{}3abc| \le 1$

Where have u seen this inequality because i have seen an other inequality which was just like this

let $ a\plus{}b\plus{}c\equal{}x$ , we have $ a^3\plus{}b^3\plus{}c^3\minus{}3abc\equal{}(a\plus{}b\plus{}c)(a^2\plus{}b^2\plus{}c^2\minus{}ab\minus{}bc\minus{}ca)\equal{}\frac{1}{2}x(3\minus{}x^2)$
your inequality is equivalent to $ \minus{}2\le x^3\minus{}3x \le 2$ or $ (x\plus{}1)^2(x\minus{}2) \le 0$ and $ (x\minus{}1)^2(x\plus{}2) \ge 0$, which it is true since
$ \minus{}\sqrt{3} \le x \le \sqrt{3}.$

i made it...a few minutes ago until you posted the solution ( like you )

If I don't remmember mistakenly, the general inequality for this problem is:
Given a,b,c belongs to R.
Prove that:(a^3+b^3+c^3-3abc)^2=<square(a^2+b^2+c^2)^3,which is easily proven by B.C.S inequality! :)


I am confused as to what the definition of supplementary angles are... and the question is name two angles supplementary to two...I would have gotten it if i could find just one  definition for it...but i've found multiple so i'm not sure how to go about naming said two angles... :blush: [img]http://i110.photobucket.com/albums/n119/phsviking/Anothermathpic.jpg[/img]

Supplementary means two angles having a sum of $180^\circ$, or $\pi$.  They DO NOT have to be adjacent to be supplementary.

Any angles that are together and form a line add to $180^\circ$, and are therefore supplementary.

Angles such as $\angle1$ and $\angle4$ are also supplementary.

[url=http://mathworld.wolfram.com/SupplementaryAngles.html]Here[/url] is Wolfram Research's definition of supplementary.

[url=http://en.wikipedia.org/wiki/Supplementary_angles]Here[/url] is Wikipedia's definition.

Hope this helps  :)

[quote="bpms"][quote="vishalarul"]Supplementary angles are $2$ adjacent angles whose sum of measures of $180$ degreees.
Angles 1 and 2 are supplementary
Angles 3 and 4 are supplementary
Angles 1 and 3 are NOT supplementary.[/quote]
They do NOT need to be adjacent.[/quote]

To answer the OP, supplementary angles are two angles that add up to $180^\circ$.

Supplementary angles are two angles that add up to $180^\circ$

Complementary angles are two angles that add up to $90^\circ$

Supplementary angles refers to two or more angles whose sum adds up to 180. In ur pic,
1 and 4
2 and 3
1 and 2
3 and 4
Are all supplementary.
The angles supplementary to 2 are 1 and 3
finally, sumthin which i cud answer!! :D


MATHCOUNTS
ratio
geometry
Pascal\u0027s Triangle

I think darktreb allowed me to post this thread.

Like other mock tests, please say how was the test and list your solution to it.

Thanks. I did not attend the test, however, since I'm not MC participant.

I'll start.

#30

Evaluate: $\displaystyle \displaystyle \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \ldots + \frac{1}{2004 \cdot 2005} $

[hide="Click_to_read_it"]We can rewrite this whole sequence by telescoping. This sum is equal to:

1-1/2+1/2-1/3+1/3-1/4+1/4.....-1/2004+1/2004-1/2005. So, all the terms cancel except the first and last one, leaving:

1-1/2005 = 2004/2005 as final answer.[/hide]

Very Nice Silverfalcon,

I've seen that problem a couple times, butnever thought of it that way.  Here's how I always do it :

[hide]
1/(1x2) = 1/2
1/(1x2) + 1/(2x3) = 1/2 + 1/6 = 2/3
1/(1x2) + 1/(2x3) + 1/(3X4)= 1/2 + 1/6 + 1/12 = 3/4

I notice pretty quick that the sum is always a fraction made up from the denominator of the last fraction added.  No need to know how to break them up.  Sometimes if you don't know the rule/pattern/trick, just start adding the series until you notice a pattern :)

So 1/(1x2) + 1/(2x3) + 1/(3X4) + ........ 1/(2004x2005) = 1/2 + 1/6 + 1/12+....... 1/(2004x2005)= 2004/2005

I really like Falcon's method, but if you don't know or notice the differnce of unit fractions technique he used, this works out well.

 [/hide]

My favorite problem was #20. Anyone have a solution for that?

I wasn't able to attend. Can you post it?

Scroll down this page and click on the topic "Mock Mathcounts" that is not stickied.

You should find there.

I did #30 like frost did.

How about #29:  29. Find the sum of all two-digit Fibonacci numbers that are not triangular numbers.  I did not do that one.

Will a full set of answers/solutions eventually be posted?

I am especially curious to see a solution to Target round #7, especially with clear explanations (rather than just plugging in formulas).

[hide="This is how far I got"]There are 5 different colors, so if she needs to take 6 picks to get 2 of the same color, she must have first chosen 1 of the each color fior the first 5 picks.

There are 5! different orders in which she could have done that.

There are a total of 15 marbles.  So, there are 15x14x13x12x11 different ways of drawing 5 distinct marbles, but this is where I get stuck.  I know I have to account for the fact that not all of these are distinct, because some of the marbles are the same (color).  But I'm not sure how to handle that when there are multiple sets of non-distinct objects.  [/hide]

Thanks in advance from a new MathCounts coach....

Let's give answers to some questions that are posted.

Since number 29 has been requested, I'll post here.

[hide="Solution"]

Here is how we define the Fibonacci numbers:

\[F_n = F_{n-1} + F_{n-2}\]

Using this, here are all two digit numbers that are Fibonacci numbers:

\[{13,21,34,55,89}\]

Triangular number is defined as:

$\binom {n+1}{2}$ Or $\frac {1}{2} n(n+1)$

Now, putting some numbers in and following are two digits triangular numbers:

\[{10,15,21,28,36,45,55,66,78,91}\]

13,34,89 are only numbers that are Fibonnacci but not triangular so their sum is answer, which is 136.[/hide]

I'll try to post more to help out. :)

#26

This is very nice problem. If you looked into at least one MAO questions, you should know how to do this easily.

[hide="LOOK"]When this is expanded, the coefficient is equal to:

\[\binom {5}{0}, \binom {5}{1}, \binom {5}{2}, \binom {5}{3}, \binom {5}{4}, \binom {5}{5}\]

Those are the coefficients. So add them and I belive the answer is 32.

For those who might know what those means:

\[\binom {n}{k} = \frac {n!}{k!(n-k)!}\][/hide]

Alternate Solution:

[hide="Alt_Sol"]Draw a Pascal's Triangle:

*********1
*******1**1
*****1**2**1
****1*3**3**1
***1*4**6**4*1
**1*5*10*10*5*1

The last line is $(x+y)^5$ so 1+5+10+10+5+1 = 32.[/hide]

Either way's fine.

Hope this helps!

Sprint problems 20 and 22
Target problems 5, 6, 8

Those are, in my humble opinion, the best (and with 20, 5, 6, 8, arguably hardest) problems on the test

I'm not so confident about #20 since I just did it but it looks okay to me.

[hide="20"]I must say I love these similar triangles.

Call DE = x and EF = y.

\[\frac {AB}{DE} = \frac {12}{x}\\
\frac {BC}{EF} = \frac {15}{y}\]

$x+y = 12\\
x = 12-y$

\[\frac {12}{12-y} = \frac {15}{y}\\
12y = 180-15y\\
27y = 180\\
y = \frac {20}{3}\\
x = \frac {16}{3}\]

Thus, using either x or y, we find the ratio of AB:DE or BC:EF, which is same.

\[\frac {12}{\frac {16}{3}} = \frac {9}{4}\]

Because the ratio of area is $a^2:b^2$, the ratio of area must be 81:16 in this case. And since 81 satisfies the interval of area, 81 is the area of ABC and 16 is the area of triangle DEF.[/hide]

[quote="sundma"]target #7

the box has 5 different colors and to draw 2 of the same color after 6 picks it means you have the worst luck because you have to pick one of all the other colors first.

here is the solution as I did it;

[hide](4/14*4/13*3/12**2/11*1/10)*5!= 48/1001

explaination:  4 red out of 14, 4 blue out of 13 left, 3 yellow out of 12 left, 2 white out of 11 left, and 1 green out of 10 left.  Then you have 5! ways of picking the color order.[/hide][/quote]

Thanks.  I was thinking your approach wouldn't work because....

[hide]the individual probabilities are different depending on the order you pick the colors in, for example if you picked the green first, then the white, yellow, blue, red, your individual probabilities would look like:
(1/14*2/13*3/12*4/11*1/10)  but luckily, the multiplication of fractions works out the same no matter how the numerators and denominators are distributed.  [/hide]

So that does look like the right approach to me.   Thanks again.  It is sad being so rusty that there are MathCounts problems that stump me . :blush:

[quote="mathmom"]It is sad being so rusty that there are MathCounts problems that stump me.[/quote]

Why? Not everybody's born to know everything. :) Like if this happened an year agon (when I was 8th grade), trust me, I would barely got 10 questions right, if I was lucky. But I got improved in many areas (thanks to AMC Preparing), and now, MC is easy for me in most cases.

See the whole point is to improve math on how good you are, not compare to how other people are. Currently, my goal is to qualify to AIME. Not to be the number one in my school, nor the best in my grade, but to be the best I can.

And to darktreb, when you might post some answers to these?

very nicely said Silverfalcon--  it's hard not to compare yourself with others but everyone should just try and do the best they can!

[quote="Silverfalcon"][quote="mathmom"]It is sad being so rusty that there are MathCounts problems that stump me.[/quote]

Why? Not everybody's born to know everything. :) [/quote]

There was a time when these problems would have been easier for me (though not 8th grade, when I don't think I had been exposed to many of these topics yet!)  I am an adult with high school and college-level competition experience, and it is sad for me to see that I have forgotten so much!  I wasn't trying to imply that it was sad in general for someone not to be able to do particularly challenging MATHCOUNTS problems, just that it was sad for me.  Sorry if anyone misinterpreted that!

I'll wait a little longer on the answers. That way people can pop in and take the test still. I always hated having answers available when taking a test, the lure of looking at them whenever I was stumped for more than 2 seconds made it hard.

5.

[hide="Sol"]$\displaystyle (x - 2)^2 - 3 = 13\\
(x^2-4x+4)-3 = 13\\
x^2-4x-12 = 0\\
(x-6)(x+2) = 0\\
x = 6, \text {and} x = -2$

Add them, it's 4.[/hide]

6.

[hide="Sol"]

There are several ways to do this but most basic:

Because of $12 \sqrt 2$, we find that the side is 12. Therefore, its area is 144 or 12^2.[/hide]

7.

[hide="Sol"]

Here is what this is saying:

A+B = C
C = 114
A+B = 114 by substitution.

To find the average we have to:

(A+B+C)/3 = Average
114+114 = 228/3 = 76 is the average.[/hide]

8.

[hide="Sol"]

Palindromes are the numbers in form of abba.

200-300, there are:

202, 212,222,232,242,252,262,272,282,292 so 10.[/hide]

I'll add more later.

for 8) i think you meant 3000 and not 300

[quote="Silverfalcon"]Ok.

I'd like to post some solutions in order so in case they want to have some answers to these, they can have it.

I didn't like number one. Someone can post the number one. :-)

[hide="2"]20 = d $\pi$ and then r = 10. So, area is 100 $\pi$. For square, it's 40/4 = s = 10. Area is 100. Add them. The total area is:

\[100 \pi + 100\][/hide]

[hide="3"]64 players. So, the first round has 32 matches, then 16, 8, 4, 2, 1 and add those up. You should get answer.[/hide]

[hide="4"]120/x is integer only if x is a factor of 120. Count how many factors are in 120. Those are the answers (number of factors) is the answer to the question.[/hide]

Harder ones:

[hide="27"]If you change each values into mod 2005 at beginning, you'll find that the sequence is 1,2,3,....2003,2004. Now, add them and change that in mod 2005 again since this whole thing is divided. The remainder is 0.[/hide]

Perhaps I should give better explanation about this but i"m not that expert on mod neither. But the concept is this, I think.[/quote]

......for number three, i got 63.

Thanks, hatsup123.

It is 63. My addition went wrong.


inequalities
trigonometry
inequalities solved

just a quick notice: have a look on how the sources of the problems should be written. it should be Created by treegoner not "my own problem" :D

:D  :D  :D  :D  :D  :D  :D
[quote]Let a,b,c >0. Prove that
64abc[(ab+bc+ca)^2][(a+b+c)] <=min{9(ab+bc+ca),(  a +  b+  c)^4}*([(a+b)(b+c)(c+a)]^2)[/quote]
If min{9(ab+bc+ca),(  a +  b+  c)^4}is 9(ab+bc+ca) then the problem    becomes not very difficult
 64abc[ab+bc+ca](a+b+c)<=[(a+b)(b+c)(c+a)]^2
 we notice that [ab+bc+ca](a+b+c)=[(a+b)(b+c)(c+a)]+abc
 so [ab+bc+ca](a+b+c)<= 9/8 [(a+b)(b+c)(c+a)]
but how about min{9(ab+bc+ca),(  a +  b+  c)^4}is (  a +  b+  c)^4
It's really a tough one :?  :?  :?  :?

Please, post the corect statement of the problem. As it is stated it is not valid (take a=b=c).

I 'm really sorry when I mistyped.
In LHS, ab+bc+ca not (ab+bc+ca)^2 :D  :D

This is the right problem :
   Let a,b,c>0.Prove that
     64abc(ab+bc+ca)[(a+b+c)^3]<=min{27(ab+bc+ca),( \sum  \sqrt a)^4) *([(a+b){b+c)(c+a)]^2)

I'm really proud that I can use one of my own inequalities to solve this problem. So, for (sqrt(a)+...)^4 the solution is the following:
   Put a=x^2, b=y^2, c=z^2. It becomes:
  (x+y+z)^4(x^2+y^2)^2*(y^2+z^2)^2*(z^2+x^2)^2>=64(xyz)^2( sum x^2y^2)(sum x^2)^3.  But prod (x^2+y^2)>=8/9*(sum x^2)(sum x^2y^2). So, it is enough to prove that:
   (sum x)^4*rpod (x^2+y^2)>=72(xyz)^2(sum x^2)^2. Now, using one of my problems we have prod (x^2+y^2)*(1/x+1/y+1/z)^2>=8(sum x^2)^2. So, it suffices to prove that (sum x)^4>=9(xyz)^2(sum 1/x)^2, or (sum x)^2>=3sum xy, which is true.

I think your solution is really good. But how can you prove your own problem.? :D
 Here is my solution
    For every x,y,z>0 there exist 2 triangles ABC and DEF such that
        sinA =  \sqrt (x(x+y+z)/[(x+y)(x+z)]), ......
        sinD = \ \sqrt ((xy+yz+zx)/[(x+y)(x+z)]), .........
     Noticing that
  1) sinAsinD <= sin((A+D)/2)^2,....
  2) For every m,n,p are real numbers and XYZ is a triangle , we have
            \sum(mn*(sinZ)^2) <= 1/4(m+n+p)^2.
    We have
           \sum (sinA*sinD *(\sqrty)*(\sqrtz)) <=1/4( \sum \sqrt x)^2.
    and we have done.
   I really want your solution to your own problem. Thank you. :)  :)  :)

Beautiful solution, much more interesting than mine. A solution to my problem was given by Namdung using a huge identity that I don't remember. My solution is the following:
   Write x^2+y^2=2c and so on. Then the inequality becomes sum sqrt(abc/(b+c-a))>=sum a. Use Schur to obtain that abc(a+b+c)>=sum a^3(b+c-a). Then use Holder to deduce that sum (a_i)^3/(b_i)^2>=(sum a_i)^3/(sum b_i)^2. So, abc(a+b+c)>=sum a^3/(1/sqrt(b+c-a)^2>=(sum a)^3/(sum 1/sqrt(b+c-a))^2, from where we have the desired inequality.

Harazi, I 'm very happy to solve your problem with my solution relevant to trigonometry, and it will give you a tighter inequality.
   Let x,y,z>0. We have ( \sum x^2)( \sum x^2*[(y^2+z^2)/yz])<=4
prod(x^2+y^2)*( \sum 1/x)^2.
     We know there exist a triangle ABC such that
     sinA =  \sqrt (x^2( \sum x^2)/(x^2+y^2)(x^2+z^2)),.........
   We have
      \sum 1/zy (sinA)^2 <=1/4( \sum 1/x)^2.
   And we 've done. :D  :D


you're trying to turn problem solving into a giant generalization, and its not.  i seriously doubt that would be a better approach to any problem proposed in contest math..

[quote="vishalarul"]I know this way is longer than Elemennop's, but can't use take the sigma of the quotients, then the formulas for $\sum_{i=0}^{n}i^{k}$,
where k is 1, 2, or 3?

I think this method will do much better in the general case.[/quote]

What do you mean by taking the sigma of the quotients?  What are the quotients here?

[quote="vishalarul"]I meant the quotient:
$\frac{n^{3}+4n^{2}+3n+2}{n^{2}+4n+3}$[/quote]

OK, now show me how you would take the sigmas of this.

I meant sum (the sigma symbol denotes sum).

i believe he is talking about taking each of the terms on the top and giving them their own denominator...(???)

i still fail how see how this could possibly prove fruitful when calculating sums like these by hand.

if what jhcr said is right, than that's a completely fruitless method.

[quote="vishalarul"] However, when doing the generalization, doesn't my method work?[/quote]

No, not at all.

[quote="vishalarul"]I thought that the sigma of a quotient was the quotient of the sigmas (denominator is nonzero). Is this false? If so, please provide a counterexample.[/quote]

Yes, it is false.  You can't add the numerators and denominators of a bunch of fractions like that.  It's like saying that $\frac{1}{2}+\frac{1}{2}=\frac{1+1}{2+2}$, which is obviously wrong.

[quote="vishalarul"]Still, doesn't it work? I'm more concerned about it working. Your method is very nice, Elemennop, and I would probably use that method if I were in a contest. However, when doing the generalization, doesn't my method work?[/quote]

From this point (the end of what you described), you would have to then do partial fraction decomposition on EVERY TERM.  not what a call a generalization worth using on any case to speak of.

matteagles,

what i think he meant was like

(1+2+3+4)/8=

1/8+2/8+3/8+4/8

[quote="vishalarul"]Really? Can you find me a counterexample to the statement:
If $g(i)$ is nonzero for all $i$ such that $0\leq{i}\leq{n}$, then
$\sum_{i=0}^{n}\frac{f(i)}{g(i)}=\frac{\sum_{i=0}^{n}f(i)}{\sum_{i=0}^{n}g(i)}$.[/quote]

One was already given.

$\frac{1}{2}+\frac{1}{2}\neq \frac{1+1}{2+2}$

Where did you get the idea that such a summation could possibly work?   :huh:  It's valid if and only if the LHS evaluates to zero.

[quote="vishalarul"]Really? Can you find me a counterexample to the statement:
If $g(i)$ is nonzero for all $i$ such that $0\leq{i}\leq{n}$, then
$\sum_{i=0}^{n}\frac{f(i)}{g(i)}=\frac{\sum_{i=0}^{n}f(i)}{\sum_{i=0}^{n}g(i)}$.[/quote]

rofl

l'hospitals doesnt work on summations

[quote="vishalarul"]Really? Can you find me a counterexample to the statement:
If $g(i)$ is nonzero for all $i$ such that $0\leq{i}\leq{n}$, then
$\sum_{i=0}^{n}\frac{f(i)}{g(i)}=\frac{\sum_{i=0}^{n}f(i)}{\sum_{i=0}^{n}g(i)}$.[/quote]

Or $n=1$, $f(i)=1$, $g(i)=2$, which is the example I gave in my previous post.

[quote="vishalarul"]Thanks for clearing that up for me. Can you do the same thing with products? Like, disprove:
$\sum_{i=0}^{n}f(i)g(i) = (\sum_{i=0}^{n}f(i))(\sum_{i=0}^{n}g(i))$

I guess the only thing you can do with sums is add or subtract if you can disprove this...[/quote]

$(1+2)(1+2) = 9 \neq 1 \cdot 1+2 \cdot 2 = 5$

Honestly, is it that hard to come up with numerical examples yourself?  (Edit:  In any case, the quotient one and the product one imply each other.  Just divide the product one by one of the sums.)

As a note, you can do something like $\sum_{k=0}^{100}(2k^{2}+6k-4)=2\sum_{k=0}^{100}(k^{2}+3k-2)$.  Here I am just factoring a 2 out of each term of the sum.

[quote="vishalarul"]So you can factor out constants, but not variables?[/quote]

Yes.  Think of the sigma notation as just a more convenient way to write out a sum of numbers.  This should help you figure out by yourself what you can and can't do with sigma notation.

Factoring out a variable when that variable is different for every term of the sum clearly doesn't work.


find a sequence (u_n) in R that doesnt converge such that every subsequence (u_kn) converges

For example:
if $ i$ is prime, $ u_{i}\equal{} 1$
if not, $ u_{i}\equal{} 0$


for image:
http://img692.imageshack.us/img692/9996/check.gif

Aston and Thomas are doing altitude measurements. Three of the same cone placed on the table. The size of each cone is 120 cm tall and 100 cm in diameter base. Then a sphere of radius 50 cm placed in the middle of the third cone. They want to know the high sphere from the surface of the table measured. The funny thing is, they both did not bring a ruler to measure the height of the sphere (height in question is the distance from the surface of the table to the center of the sphere). Can they get a definite answer? How 'measurement' is? (I put the three cone attached)

It's rude to double-post; it's more rude to triple-post; and it's further rude to start bumping up your own posts after only a few hours.  This is made even more obnoxious by the fact that on one of your threads, people have already started to help you.  Finally, your choice of title is totally unhelpful.  In the future, you should post problems only once, wait at least a day or two before bumping, and choose informative titles.

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=310585