In a circle with center O and radius 1, AB is a diameter. X and Y are on the circumference such that angles XAB = angle YBA. YB and XA intersect (inside the circle) at point T. OY and AX intersect at point Q. Assume angle AQO = 90 degrees, find the area of triangle YQT

1/(8sqrt[3])...i belive they're all 30-60-90s...
You can show that AOXY is a rhombus.  First, XY is parallel to AO.  Then you know that OY=AO=OX.  If you make the arc AYX around O, and then extend ray XY (the two pieces of information we have: OY=OA=OX and that XY is parallel to AO), there is exactly one intersection point except for X, and that is when AOXY is a rhombus with angles of 60 and 120 degrees.

After you know AOXY is a rhombus, how do you get the answer?

Draw the perpendicular from O to XY.  Now you have a whole bunch of right triangles in the ide of the semicircle with Y.  These are all similar.  You know that OAX is 30 degrees.  From this, you know that all of the triangles are 30-60-90s.  Then, you know that AO is 1.  You know that QY is 1-OQ=1/2 and that QT is 1/2sqrt[3].  So the are of the triangle is 1/8sqrt[3]

It is very easy.Because angle AQO is 90 degrees,arc AY =arc YX.
And becaouse angle ABY = angle BAX,arc AY=arc XB.So arc AY=arc
YX=arc XB.So angle AOQ=60 degrees.QY=OQ=0.5. So the area of the triangle is 1/8sqrt[3]

Honglei, I can't understand why arcAY = arcYX.
And another thing, to get QY=OQ, do you need to prove AOXY is a rhombus first?

[quote="indybar"]Honglei, I can't understand why arcAY = arcYX.
?[\quote]
because $AOXY$ is a rhombus...
take a look at the picuture,
when $AOXY$ is a rhumbus, then $\angle AOY=\angle YOX$

This is very nice problem! :)

Do you have a source to this?

I have little bit different approach without proving rhombus. :)

Note that we can prove $\triangle AOQ \cong \triangle XOQ \cong \triangle XYQ \cong \triangle AYQ$ by SAS (third angle can be shown by subtracting other angle and right angle).

So, the quadrilateral $AOXY$ has $4(\text{Area of one triangle})$ and for each triangle we have hypotenuse of $\frac{1}{2}$ and shorter leg $\frac{1}{4}$ which gives other leg $\frac{\sqrt 3}{4}$.

Thus, the are of AOXY = $4 \cdot \frac{\sqrt 3}{4} \cdot \frac{1}{4} \cdot \frac{1}{2} \Rightarrow \frac{\sqrt 3}{8}$.

QED

Well, Silverfalcon, the problem asks for the area of the small triangle, not the rhombus  :D
It's in the first round of Thailand's IMO selection last year.  :)

[hide]OY // BX since both are perpendicular to AX

It's easy to show XY // OB

So OBXY is a rhombus

Therefore OXY is an equilateral and BY is perpendicular to OX

So T is the orthocenter (and centroid) of the equilateral OXY

area(YOT) = 1/3 * area(YQX) = 1/6 * area(OXY)[/hide]

[quote="luimichael"]Hello! amirthlusa
Can you tell me which software did you use in order to generate such a nice picture ?[/quote]
Paint :D
just save the picture as JPEG, so you can post it here


trigonometry
rotation
vector
complex analysis
complex analysis theorems

I was reading a complex anlysis book by Ahlfors and there were some things I couldn't understand on page 163.
At the bottom of page 163, he wrote:
" If gamma is a regular curve with the equation z=z(t), the direction of the tangent is determined by the angle alpha=arg z'(t), and we can write dx=ldzl sin(alpha). The normal which points to the right of the tangent has the direction Beta=alpha-pi/2......."
My question is: [1]what does it mean by "direction of the               tangent is determined by the angle alpha"?
                     [2] what is the definition of ldzl?
                     [3]How could he conclude such a statement above?

Thanks.

[1] the tangent line forms angle $\alpha$ with the positive direction of the real axis

[2] if $z=x+iy$ is a complex number, then $x=|z|\sin\alpha$, where $\alpha=\arg z$. Ahlfors applies this to the differential of $z$. Don't ask me what it is - I don't know.  :(

[3] if you rotate a tangent vector to a curve by $\pi/2$ radians clockwise, it will be orthogonal to the curve. As you move along the curve, the arrowhead will be to your right.

I just wanted point out that alpha is not equal to arg z(t).
Alpha is equal to arg z'(t).  Could you explain why the tangent line forms angle arg z'(t) with the positive direction of real axis?
Thanks.

$z(t) = u(t)+iv(t) = (u(t), v(t))$.  The tangent to the curve has direction $(u'(t), v'(t))$.  This is the same direction as $z'$, if you consider $z'$ just as a vector.  Its direction is given by $\arg z'$, because that's what $\arg$ means.


[quote="Pirkuliyev Rovsen"]Solve the equation
$ 9^{\sqrt {x}} \plus{} 9^{\sqrt {y}} \plus{} 9^{\frac {1}{\sqrt {xy}}} \equal{} \frac {81}{\sqrt {x} \plus{} \sqrt {y} \plus{} \frac {1}{\sqrt {xy}}}$  :?:
[/quote]

Let $ \sqrt {x} \equal{} a \ ; \ \sqrt {y} \equal{} b \ ; \ \sqrt {\frac {1}{xy}} \equal{} c$

$ \Longrightarrow \begin{cases}a,b,c \ge 0 \\
abc \equal{} 1\end{cases}$

$ (a \plus{} b \plus{} c)(9^a \plus{} 9^b \plus{} 9^c) \ge 3\sqrt [3]{abc}.3\sqrt [3]{9^{a \plus{} b \plus{} c}}$ $ \ge 3\sqrt [3]{abc}.3\sqrt [3]{9^{3\sqrt [3]{abc}}} \equal{} 81$

$ (x,y)\equal{}(1,1)$


linear algebra
matrix
algebra
polynomial
vector

Hi, I just had a few questions regarding finding the solutions of a matrix and its basis.

Question 1:
1 0 4
0 0 38

What is a basis for the column space and the row space?
I thought the basis for the row space was the above matrix (each row has an independent variable)

But the basis for the column space is...?
1 4
0 38
??

Thus each column has a independent variable in it, but this doesn't look right to me. Can someone please explain?

Question 2:

Find all polynomials p in P^3 for which 2p(7) - 3p(6) = 14.

So what I did was write out p(x) as:
ax^3 + bx^2 + cx + d

Evaluated it at the values given (you can check my math if you'd like):

38a -10b - 4c - d = 14

So there are three free variables and one dependent variable. This makes sense to me, however, I want to write the basis of the equation and the solutions to the equation in matrix/vector form, how do I do that?

I tried solving for d and got:

d = 38a - 10b - 4c -14

Where do I go from here? Any help is greatly appreciated!

Thanks.

$\begin{bmatrix}1&0&4\\0&0&38\end{bmatrix}$ is already in row echelon form (although not reduced row echelon form.) There are two pivot elements: the 1 and the 38. Hence the rank of the matrix is 2 and the row space and the column space are both both two dimensional.

A systematic procedure for finding a basis for the column space is to take the columns of the original matrix that correspond to the locations of the pivots in the row echelon form.

That's where $\left\{\begin{bmatrix}1\\0\end{bmatrix}, \begin{bmatrix}4\\38\end{bmatrix}\right\}$ comes from.

Of course, the span of these two vectors is simply all of $\mathbb{R}^2$ and you could use
$\left\{\begin{bmatrix}1\\0\end{bmatrix}, \begin{bmatrix}0\\1\end{bmatrix}\right\}$.

Note that any collection of vectors, such as the columns of the original matrix, that includes the zero vector must be linearly dependent. Omitting the zero vector from the listing will not change the span.

Somehow this wound up being posted twice, once here, and once someplace it doesn't belong. I'll delete the other thread, which did draw one response, reproduced below:

[quote="jmerry"]Eventually, a moderator will notice this and move it.

On your first question: those are correct bases for the row and column spaces. Of course, bases are never unique, and $\begin{bmatrix}1\\0\end{bmatrix},\begin{bmatrix}0\\1\end{bmatrix}$ is also a basis for the column space. The zero column, of course, does not increase the span of the columns, so we can ignore it when finding the column space.

For your second question: pick three linearly independent possibilities for $a,b,c$ (such as (1,0,0), (0,1,0), and (0,0,1)) and plug them in to find what value of $d$ fits. That will get you a basis.[/quote]

Thanks for the replies, that makes sense, however are these basis dependent on the fact that the original matrix has column of 0's?  Say we replaced one of those 0's with another number, assuming the matrix represents a linear transformation from R^3 to R^2, then the dimension must still be two, so how would I write the basis?

1 2 4
0 0 38

Then the basis for the column space would be?
1 0
2 0
0 38

???

Note that $\begin{bmatrix}1&2&4\\0&0&38\end{bmatrix}$ is also already in row echelon form, with two pivot elements. Following the same procedure as before, using the columns corresponding to the pivot elements, we once again find that $\left\{\begin{bmatrix}1\\0\end{bmatrix}, \begin{bmatrix}4\\38\end{bmatrix}\right\}$ is a basis for the column space.

The set of all of the columns, namely $\left\{\begin{bmatrix}1\\0\end{bmatrix}, \begin{bmatrix}2\\0\end{bmatrix}, \begin{bmatrix}4\\38\end{bmatrix}\right\},$ does span the column space but it is not a basis for that space becuase it is linearly dependent. We need to omit some members of this set to get down to a linearly independent set; looking for columns corresponding to pivot elements in the row echelon form is a systematic way to do that.


symmetry
LaTeX
function
rotation
inequalities proposed
inequalities

[quote="N.T.TUAN"]This problem posted before! This is general problem :
Let $0<p<q$, if  $x_{i}\in [p,q] (i=1,2,...,n)$ then find maximum value of $(\sum x_{i})(\sum \frac{1}{x_{i}}).$[/quote]
How can I get it? :blush:
or can you say me the link to the before?

General problem unsolved on this forum. If I'm not wrong! Please try solve it!  :D

If $0<p<q$ and $x_{k}\in [p,q] ,\forall k=\overline{1,n}$ then $\left( \sum x_{k}\right)\left(\sum \frac{1}{x_{k}}\right)\leq n^{2}\frac{(p+q)^{2}}{4pq}$
Let $f: [p,q]\to \mathbb R$ ,$f(x_{1})=\left( \sum_{k=1}^{n}x_{k}\right)\left(\sum_{k=1}^{n}\frac{1}{x_{k}}\right)$ $\Leftrightarrow$
${f(x_{1})=x_{1}\left(\sum_{k=2}^{n}\frac{1}{x_{k}}\right)+\frac{1}{x_{1}}\left(\sum_{k=2}^{n}x_{k}\right)}+\left( \sum_{k=2}^{n}x_{k}\right)\left(\sum_{k=2}^{n}\frac{1}{x_{k}}\right)+1$,so $f$ is obviously convex (sum of convex functions).
Because of the convexity and the symmetry we have:
\[\left( \sum_{k=1}^{n}x_{k}\right)\left(\sum_{k=1}^{n}\frac{1}{x_{k}}\right)\leq \left( \sum_{k=1}^{n}x'_{k}\right)\left(\sum_{k=1}^{n}\frac{1}{x'_{k}}\right)\]
where $x'_{k}\in \{ p,q\}$ $\forall$ $k=\overline{1,n}$
Let $a$ be the number of $p$'s and $b$ the number of $q$'s,so we have $a+b=n$.Moreover we have:
\[\left( \sum_{k=1}^{n}x_{k}\right)\left(\sum_{k=1}^{n}\frac{1}{x_{k}}\right)\leq \left( \sum_{k=1}^{n}x'_{k}\right)\left(\sum_{k=1}^{n}\frac{1}{x'_{k}}\right)=(ap+bq)\left(\frac{a}{p}+\frac{b}{q}\right)\leq n^{2}\frac{(p+q)^{2}}{4pq}\]
And the last one is equivalent to:
\[a^{2}+b^{2}+2ab+ab\left(\frac{p}{q}+\frac{q}{p}-2\right)\leq (a+b)^{2}\frac{(p+q)^{2}}{4pq}\]
$\Leftrightarrow$
\[ab\left(\frac{p}{q}+\frac{q}{p}-2\right)\leq (a+b)^{2}\frac{(p-q)^{2}}{4pq}\]
which is obviously true since $ab\leq \frac{(a+b)^{2}}{4}$ and $\left(\frac{p}{q}+\frac{q}{p}-2\right)=\frac{(p-q)^{2}}{pq}$

[quote="N.T.TUAN"][quote="Armenia"]Could you give me some lessons to use Latex??[/quote]
See here http://www.mathlinks.ro/LaTeX/AoPS_L_About.php  :wink:[/quote]
But this site doesn't work. There isn't such a site. :?:

Right  :oops: Try to .http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php

To all: Please try solve above problem  :wink:

I don't see much difference in the two problems N.T.Tuan
both seem to be solved by the same way.
each is a convex function in each variable
so it is enough to consider the case when there are a ps and b qs
from then on it is simply easy work. :lol:

No, my friend , it is cyclic :) Please try to solve it !

Sorry N.T.Tuan I thought it was symmetric :oops:
in that case I think it can be solved this way
Let us find the maximum of the function f(a_1,..,a_n)=sigma(cyclic(a_i/a_i+1)).
put the n variables a_1 from a_n (p or q) in a long strip.
since there exists at least one p let us rotate it till it comes first in the strip.
f(p,p,...,p)is smaller than f(p,p,...,q)
starting from p we perform the following going from left to right.
if ppp or qqq appears, changing to pqp or qpq will make the value larger.
if ppq appears changing it to pqp will make the value larger.
if qqpp appears (changing it to qpqp will make the value larger.
if qqpq appears  changing it to qpqq will keep the value the same.
you will notice that using these, pps and qqs and longer strings of ps and qs will be pushed to the right.
Let us keep pushing them until n-2 variables are organized in a p-q-p-q etc. string.
1) p q p q ..... p a_n-1 a_n (n is odd)
in this case (p,q,...,p,q,q),(p,q,p,q,...,p,q,p),(p,q,p,q,...,p,p,q) maximizes the function
2) p q p q ..... q a_n-1 a_n (n is even)
in this case (p,q,p,q,...,p,q) maximizes the function

(1)and 2) results can be obtained by putting (p,p),(q,q),(p,q),(q,p)and computing)

so the answer is (p/q+q/p)(n/2) (when n is even)
and (p/q*((n-1)/2)+q/p*((n-1)/2)+1 (when n is odd)
please tell me if there are any mistakes.

Why you don't use LaTex? Please learn it here .http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php

I will check your solution later.

[quote="N.T.TUAN"]This is another problem (of Nam ,too)  :
Let $ 0<p<q$, if  $ x_{i}\in [p,q] (i=1,2,...,n)$ then find maximum value of $ \frac{x_{1}}{x_{2}}+\frac{x_{2}}{x_{3}}+...+\frac{x_{n}}{x_{1}}.$

 :wink:[/quote]

$ max=k*(\frac{p}{q}+\frac{q}{p})$  if $ n=2k$ and
$ max=k*(\frac{p}{q}+\frac{q}{p})+1$  if $ n=2k+1$  :wink:


function
ceiling function
floor function
trigonometry
complex numbers
geometric series
algebra

How many solutions are there to $z+2x+3y=100$ where x, z, and y are positive integers?

I saw this problem and came up with a quick method of finding the number of solutions, but I'm wondering if there's an explicit formula.

[quote="bpms"]I just realized that my method wouldn't work on this specific problem.
Is there no closed way of finding the number of solutions of something like this?[/quote]

Modified version of Simon's Favorite Factoring trick?

General methods for solving diophantine equations maybe?
i.e. look at the equation mod 2 & 3, see what remainders are possible, so on and so forth...

[quote="Hamster1800"]Generating functions are -the- most general way to do this. Basically we have the generating functions

$1+x+x^{2}+x^{3}+x^{4}+...+x^{n}+... = \frac{1}{1-x}$

$1+x^{2}+x^{4}+x^{6}+...+x^{2n}+... = \frac{1}{1-x^{2}}$

$1+x^{3}+x^{6}+x^{9}+...+x^{3n}+... = \frac{1}{1-x^{3}}$

So basically we can take any multiple of one from the first (x), any multiple of two from the second (2y) and any multiple of three from the third (3z) to get a new term. The coefficient of the appropirate term in the resulting power series will be the answer. So we have to find the coefficient of $x^{100}$ in $\frac{1}{(1-x)(1-x^{2})(1-x^{3})}$, which is probably not easy to find.[/quote]

just partial fraction decomposition [where you do complex too]

altheman, could you actually show a complete solution please?

thanks

maybe...i have sat prep tonight...i am just procrastinating...

step 1: partial fraction decomp

no school bs "OMG IRREDUCABLE FACTOR;" use complex numbers to break up 1/(x^2+x+1)

if it is in the form: $\frac{A}{x-B}$, then you should be able to do it

i forget how to do the A/(x-B)^2...that might be something like sum nx^n-1, or something like that

cubic...sum (n)(n-1)x^{n-2}

Okay I'll try to fill in for Altheman since he's procrastinating...

We know that $1+x+x^{2}$ has roots of
$\frac{-1\pm\sqrt{1-4}}{2}$. Let this be $\omega$ and guess what, the other one is $\omega^{2}$ (think roots of unity if you don't see this)

We want the power series for $\frac{1}{(1-x)(1-x^{2})(1-x^{3})}$. We know this factors as

$\frac{1}{(1-x)^{3}(1+x)(x-\omega)(x-\omega^{2})}$

So now we have this really annoying equation

$\frac{1}{(1-x)^{3}(1+x)(x-\omega)(x-\omega^{2})}= \frac{A}{1-x}+\frac{B}{(1-x)^{2}}+\frac{C}{(1-x)^{3}}+\frac{D}{1+x}+\frac{E}{x-\omega}+\frac{F}{x-\omega^{2}}$. Now we clear the denominators:

$1 = A(1-x)(1-x^{3})+B(1-x^{3})+C(1+x+x^{2})+D(1-x)^{2}(1-x^{3})+E(1-x)^{3}(1+x)(x-\omega)+F(1-x)^{3}(1+x)(x-\omega^{2})$.

We know tha $\omega = \overline{\omega}$, so $E=F$ by pairing the complex parts

$1 = A(1-x)(1-x^{3})+B(1-x^{3})+C(1+x+x^{2})+D(1-x)^{2}(1-x^{3})+E(1-x)^{3}(1+x)(x-\omega)+E(1-x)^{3}(1+x)(x-\omega^{2})$. Clearly, if this holds for all x, it must hold for x=1. Now watch as almost every term becomes zero

Also notice that the last two terms equal $E(1-x)^{3}(1+x)(x-\omega+x-\omega^{2}= E(1-x)^{3}(1+x)(2x+1)$, since both $\omega$ and $\omega^{2}$ have real part $-1/2$

$1 = 0+0+C(3)+0+0+0$. so $C = \frac{1}{3}$.

Now let's pair up the coefficients of x.

$0 =-A+0+C-2D-3E+E+2E =-A+\frac{1}{3}-2D$

So $A =-2D$

I'm going to rewrite the expression as I now have it in my head

$1 =-2D(1-x)(1-x^{3})+B(1-x^{3})+C(1+x+x^{2})+D(1-x)^{2}(1-x^{3})+E(1-x)^{3}(1+x)(2x+1)$

Suppose we let $x=-1$ now. We get

$1 =-2D(-2)(2)+B(2)+C(1)+D(4)(2)+0$

So:
$1 = 8D+2B+\frac{1}{3}+8D$

$\frac{2}{3}= 16D+2B$
$\frac{1}{3}= 8D+B$
$B = \frac{1}{3}-8D$

So we have

$1 =-2D(1-x)(1-x^{3})+(\frac{1}{3}-8D)(1-x^{3})+(\frac{1}{3})(1+x+x^{2})+D(1-x)^{2}(1-x^{3})+E(1-x)^{3}(1+x)(2x+1)$

Now let's let $x=-1/2$ and watch magic happen!

$1 =-2D(\frac{3}{2})(\frac{7}{8})+(\frac{1}{3}-8D)(\frac{7}{8})+\frac{1}{3}(\frac{7}{4})+D(\frac{9}{4})(\frac{7}{8})+0$

Now we have an equation only in $D$! Solving this we get...

$1 = \frac{7}{8}(-3D)+\frac{7}{24}-7D+\frac{7}{24}+D(\frac{63}{32})$

$1-\frac{7}{12}= D(-\frac{21}{8}-7+\frac{63}{32})$
$\frac{5}{12}= \frac{245}{32}D$

$D = \frac{147}{8}$.

And so we get a bunch of stuff in terms of D and we get the sum of generating functions which each are the generating functions for geometric series and we can add those really easily.

By the way, if my derivation of D is wrong, please tell me. I'll try to fix it but it should give you a good idea of how to approach it either way.

Not sure if this is right, but here you go:

[hide="Solution"]Rearrange the equation to get $z+2x=100-3y$.  Then for each possible value of $y$, we know that $2x$ can be any positive even number less than $100-3y$.  Assuming $100-3y$ is positive, there are $\left\lceil\frac{100-3y}{2}\right\rceil-1$ positive even numbers less than $100-3y$.  Now we take $y$ from 1 to 33.  The answer is

$\sum_{y=1}^{33}\left(\left\lceil\frac{100-3y}{2}\right\rceil-1\right)$
$=\sum_{y=1}^{33}\left(\left\lceil 50-2y+\frac{y}{2}\right\rceil-1\right)$
$=\sum_{y=1}^{33}\left(49-2y+\left\lceil\frac{y}{2}\right\rceil\right)$
$=33\cdot 49-2\cdot\frac{33\cdot 34}{2}+(1+1+2+2+\cdots+16+16+17)$
$=33\cdot 49-33\cdot 34+\frac{16\cdot 17}{2}+\frac{17\cdot 18}{2}$
$=784$[/hide]

Actually, it's $x^{6}$ on top.  All it means is that you need the coefficient of $x^{94}$ in the resulting expression instead of $x^{100}$.

The simplest formula to OEIS gives is $\textrm{Round}(\frac{(n+3)^{2}}{12})$, although if we restrict to positive integers it should be $n-3$ instead of $n+3$.

[quote="stupidityismygam"]i brute forced it and got $784$[/quote]

whoa that must have taken you a lloooonnnggg time  :lol:

i just realized how to expand $\frac{x^{a}}{(x-b)^{c}}$ as a generating function...it is totally trivial using maclurian series techniques...

so once you do the disgusting partial fraction decomp, it is easy

[quote="worthawholebean"]Once you do the partial fraction decomposition, can't you just use the extended Binomial Theorem?

*is happy to be using his newfound knowledge from AoPS vol. 2*[/quote]

they are the same thing...extended binomial theorem is derived from maclurian series...

[quote="Hamster1800"][quote="JBL"]The entry from the OEIS is here:
http://www.research.att.com/~njas/sequences/A001399

Actually, the generating function given above is slightly off: they're solving with nonnegative, rather than positive, integers.[/quote]

Fine :(

take out the $1$s from each of the generating series then. You get the generating function $\frac{x^{3}}{(1-x)(1-x^{2})(1-x^{3})}$ I'm not reworking the whole problem because of this though.[/quote]

Much easier not to fix the function and redo the partial frac-decomp, and just subtract the three cases where $x=0$, $y=0$, and $z=0$.

BTW, is there a nice pure combinatorical solution or something?

[hide="Speculation"]
We have:

$y+(y+x)+(y+x+z)=100$

Clearly, $y+x+z>y+x>y$

This becomes:

$a+b+c=100,a>b>c$

In other words, there is exactly one solution corresponding to each distinct partitioning of 100 into three positive integers.

$a$ may be anything from $34$ to $97$

Thus $b+c$ may be anything from $3$ to $66$

For even integers, the number of pairings is $\lfloor n/2 \rfloor-1$ and for odd integers, the number of pairings is $\lfloor n/2 \rfloor$. Summing this from $3$ to $66$ gives:

$1056$

Hmm.... that's not what matt got...
[/hide]

[quote="white_horse_king88"]BTW, is there a nice pure combinatorical solution or something?[/quote]

Good question.. I rather highly suspect that you can do recursion as well which is I think fundamentally similar to the method here. Doubt you can do much better than that.

So how did I over-count...  :maybe: was there something totally fundamentally wrong with my initial strategy?  :(

TZF: you counted $(a, b, c) = (34, 50, 16)$ and many other combinations that violate your ordering $a > b > c$.


I don't understand conditional probability at all and no book or website that I've come across has been able to explain it to me in a good way. Could someone help?

Part of the issue I'm having is if:
P (B|A) = P (A & B)/ P (A)
and P (A & B) = P(A) * P(B); then isn't P (B|A) = P (B)?

P(B|A)=P(B) if A and B are independent events.  That means the equation P(A&B)=P(A)*P(B) holds when A and B are independent.  Otherwise use the conditional probability equation.

To add to above,  a simple example, to make it obvious is to work with a simple example ...  let us throw a random 12- face dice..
and  denote:

P(A) = probability that your number is divisible by 4
P(B) = Probability that your number is divisible by 6
P(C) = Probability that your number is divisible by 3

(Easy to see, P(A)=1/4, P(B)= 1/6, P(C) = 1/3 )
A&C are independent,  and P(A&C) is indeed  1/12.
A&B are not independent and P(A&B) is also 1/12 .. work out all your
formulas ..
Hope it is helpful ...
[hide](For example, given a number is divisible by 4, it is also divisible by 6 is (1/3)
(Or (1/12)/(1/4) as given by your formula ...[/hide]


calculus
integration
trigonometry
logarithms
real analysis
real analysis unsolved

prove :D

$ \int_{0}^{1}\int_{0}^{1}\;\frac{\textbf dx\;\textbf dy}{2\cos\left(\frac{\pi}{5}\right)\minus{}xy}\;\;\equal{}\;\;\boxed{\frac{\pi^{2}}{10}\minus{}\ln^{2}\left(2\cos\left(\frac{\pi}{5}\right)\right)}$


velocity includes direction, speed does not

your equations are incorrect...

displacement=velocity*time

distance=speed*time

Speed is the absolute value of velocity.

[i]OR[/i]

Speed is a scalar while velocity is a vector.

Hence, the magnitudes are the same. The former, however, has no direction, but the latter does.

[quote="Interval"]I typed the wrong formulas?

Sorry but I got both from a math book.

Thanks for letting me know.[/quote]

maybe i should have been more clear, you typed them correctly in terms of variables, but you did not indicate the difference between distance and displacement...showing a lack of understanding...blah blah...yea...

[quote="Interval"]I know that in physics speed is called velocity or is it?

Is there a difference between speed and velocity?

In pure math: D = rt, where r = rate = speed.

In Physics: D = vt, where v = velocity = speed?????

Why?[/quote]

speed=scalar.  for example, 5 mph

velocity=vector.  ie 5 mph north

yes the equations are wrong in a way.  d=vt is displacement=rate*time.  displacement is a vectot like 5 miles north

d=rt is distance=speed*time.  distance is scalor like 5 miles

Well, both "distance" and "displacement" start with a D...

Yeah, displacement has a direction (like velocity), while distance doesn't (like speed). Or, in the style of the SAT, Displacement:Velocity::Distance:Speed...

[quote="Interval"]In pure math: D = rt, where r = rate = speed.[/quote]

This isn't pure math either.  It's just the simplified physics of motion.


modular arithmetic
quadratics
number theory unsolved
number theory

$n|a^{2}+1$ so, every prime divisor of $n$ is of the form $4k+1$ or is equal to $2$. If you use      $(a^{2}+b^{2})(c^{2}+d^{2})=(ac+db)^{2}+(ad-bc)^{2}=(ac-bd)^{2}+(ad+bc)^{2}$, and the well known fact that every prime number of the form $4k+1$ is sum of two squares, $2=1^{2}+1^{2}$ and $2\parallel a^{2}+1$ the result follows.
(if $ad=bc$ and $ac=bd$ then $a=b$ so $p=2a^{2}$ and for prime $p>2$, that's imposible. so those $x$ and $y$ are surely natural)

[quote="SpongeBob"]$n|a^{2}+1$ so, every prime divisor of $n$ is of the form $4k+1$ or is equal to $2$[/quote]
Why is that ?

It follows from quadratic reciprocity (well, a supplementary lemma), since all such $p$ must satisfy $\left( \frac{-1}{p}\right) = 1$.

Edit:  This is a well-known problem in the theory of quadratic residues.  There is a [i]very[/i] nice proof using Minkowski's Theorem.  Do you all know it?

$-1$ being not a quadratic residue $\mod p$ with $p \equiv 3 \mod 4$ is trivial and doesn't need any theorem stronger than little Fermat...

I see that there are bunch of smartmouths here, everyone knows everything and it's not
their level to write proof for something as trivial as this :)

If $a^{2}\equiv-1 \mod p$ then $(a^{2})^{\frac{p-1}2}\equiv (-1)^{\frac{p-1}2}\mod p$. From this it follows that
$\frac{p-1}2$ must be even because $(a^{2})^{\frac{p-1}2}=a^{p-1}\equiv 1 \mod p$, so $p=4k+1$.

Bye

The reason why don't proof anything is that all this stuff was already written some hundred times.


[hide="hint"]From system we have $x^{8}-1=(y^{7}+1)(x-1)$ and $y^{8}-1=(x^{7}+1)(y-1)$. Now it is easy than  :maybe: [/hide]

Since $\frac{y^{7}+1}{x-1}\geq 1$, we have the following fact :
If $x \geq-1$, then $y \geq-1$, otherwise $y <-1$.
It's also obvious that if $x > 0$, then $y^{7}+1 > 1 \Rightarrow y > 0$ (and vice-versa)

Divide into cases :
i) $x > y > 0$ (or $y > x > 0$). But we have $y^{7}+1 = (x^{4}+1)(x^{2}+1)(x+1) > x^{7}+1 \Rightarrow y > x$. A contradiction.
ii) $0 > x > y$ (or $0 > y > x$). Clearly, neither $x$ nor $y$ equals $-1$, otherwise $x = y =-1$.
But we have $x^{7}+1 = (y^{4}+1)(y^{2}+1) < y^{7}+1 \Rightarrow x < y$. A contradiction.
ii) $x = y$. Now, $x^{7}+1=(x^{4}+1)(x^{2}+1)(x+1) \Rightarrow x = y =-1$ or $0$.

Hence, we can deduce that all solutions are : $(x,y) = \{ (0,0), (-1,-1) \}$

It is equavalent to
$y^{7}=x^{7}+x^{6}+x^{5}+x^{4}+x^{3}+x^{2}+x=x^{7}+g(x),$
$x^{7}=y^{7}+g(y).$
If $x>0$, then $y>x>0$, therefore $x>y>x$ contradition.
1. $x=0=y$ solution.
If $x<-1$, then $x<y<-1$, therefore $x<y<x$ contradition.
2. $x=-1$ give $x=y=-1.$
If $-1<x<0$, then $-1<y<x<0$, therefore $-1<y<x<y$ contradion.
We use only $g(x)>0$ for $x>0$ or $x<0$ and $g(x)<0$ for $-1<x<0$.

[quote="hien"][quote="Rust"]It
If $x<-1$, then $x<y<-1$, therefore $x<y<x$ contradition.
[/quote]

Could you explain it more clearly for me?[/quote]
If $x<-1$, then $g(x)=(x+1)x(x^{4}+x^{2}+1)>0$, therefore $y^{7}=x^{7}+g(x)>x^{7}$. It give $y>x$. But $x^{7}+g(x)=x(g(x)+1)<x<-1$, therefore $x<y<-1$. Because $y<-1$ from $x^{7}=y^{7}+g(y)$ we get $y<x<-1$. contradition.


hola mi nombre es Marco Antonio Huarancca Ninanya, estudio con fabiola y con otros chicos a los cuales les voy a hablar del foro, conozco a chen, jossy, franco, fredy,..;me gusta el algebra y un poco de matematica discreta  :)

espero traer mas gente al forum :spider:

Qué bueno que puedas participar de este foro, anima a tus compañeros de aula a que participen también...a y agregate a la lista de peruanos en mathlinks:

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=56422]www.mathlinks.ro/Forum/viewtopic.php?t=56422[/url]

$Tipe$


number theory
relatively prime
greatest common divisor

Find all positive integers $ a,b$ such that $ a\plus{}b\mid ab$.

we take $ \boxed{S \equal{} \{(mfh,mf(f \minus{} h)) : \ gcd(f,h) \equal{} 1\ and\ f\ge h > 0\}}$
we have $ mfh \plus{} mf(f \minus{} h) \equal{} mf^2|mfh\times mf(f \minus{} h)$ then $ \forall (a,b)\in S: \ a \plus{} b|ab$

if $ a \plus{} b|ab$ for some $ a,b > 0$ we take a=a'd and b=b'd where $ gcd(a',b') \equal{} 1$
so $ d(a' \plus{} b')|d^2a'b'$ gives $ a' \plus{} b'|d$ then a'=h and b'=f-h for somme f and h such that $ f \equal{} a' \plus{} b'|d$ and $ gcd(h,f) \equal{} gcd(a',b') \equal{} 1$.
then $ (a,b) \equal{} (mfh,mf(f \minus{} h))\in S$ where $ m \equal{} \frac {d}{f}$

finally $ \{(a,b)\in\mathbb{N} \minus{} \{(0,0)\}: \ a \plus{} b|ab\} \equal{} S\cup \{(a,0),(0,a)|\ a\in\mathbb{N}^*\}$

[hide="Solution"]Let $ d \equal{} (a,b), a \equal{} dx, b \equal{} dy$, so $ (x,y) \equal{} 1$. Then we require $ dx \plus{} dy | d^2xy$, or equivalently $ x \plus{} y | dxy$. Since $ x$ and $ y$ are relatively prime, $ x\plus{}y$ is relatively prime to both $ x$ and $ y$. Therefore $ x \plus{} y | dxy \Leftrightarrow x \plus{} y | d$. So we pick any two positive integers with greatest common divisor 1, and multiply both by any constant divisible by the sum of the two integers. Since all steps in this proof are reversible, this is necessary and sufficient.[/hide]

Take $ a\equal{}b\equal{}2k$ with $ k\in\mathbb{Z^\plus{}}$.
$ 2a\mid a^2\Rightarrow \displaystyle\frac{a^2}{2a}\equal{}\displaystyle\frac{a}{2}\in\mathbb{Z^\plus{}}$
Because $ a\equal{}2k$. :D

That's only a tiny subset of the solutions. You need to determine the whole set with proof to solve the problem. :wink:


There are many two-digit multiples of 7, but only two of the
multiples have digit sum of 10. The sum of these two multiples of 7
is what?

The two multiples are 28 and 91, and their sum is 119.


Find all polynomials with real coefficients $ P(X),Q(x),R(x)$ satisfying $ \sqrt {P(x)} \minus{} \sqrt{Q(x)} \equal{} R(x)$ for every $ x\in \mathbb R$

We have $ P(x)\minus{}Q(x)\equal{}E(x)$ for some $ x$ so $ \sqrt{P(x)}\plus{}\sqrt{Q(x)}\equal{}\frac{E(x)}{R(x)}$.
From the hypothesis we conclude that
$ P(x)\equal{}\left(\frac{E(x)}{2R(x)}\right)^{2}$
If $ E(x)$ does not divide $ 2R(x)$ so $ E(x)^2$ does not divide $ 4R(x)^2$ neither which contradict with $ P(x)$ is a polynomial
So $ P(x)\equal{}\left(A(x)\right)^2$ with $ A\in\mathbb{R}[x]$ and similar with $ Q(x)$.


Let $p$ be a prime $=1(\mod 3)$ and $q$ be the integer part of $\frac{2p}{3}$. If

$\frac{1}{(1)(2)}+\frac{1}{(3)(4)}+...+\frac{1}{(q-1)(q)}= \frac{m}{n}$, for integers $m,n$, show that $m$ is divisible by $p$.


If $ xyz\equal{}(1\minus{}x)(1\minus{}y)(1\minus{}z)$ with $ 0 \le x,y,z \le 1$ Show that

$ x(1\minus{}z)\plus{}y(1\minus{}x)\plus{}z(1\minus{}y) \ge \frac{3}{4}$

[quote="Tiger100"]If $ xyz \equal{} (1 \minus{} x)(1 \minus{} y)(1 \minus{} z)$ with $ 0 \le x,y,z \le 1$ Show that

$ x(1 \minus{} z) \plus{} y(1 \minus{} x) \plus{} z(1 \minus{} y) \ge \frac {3}{4}$[/quote]
If $ z \equal{} 0$ your inequality is obviously true.
Let $ x,$ $ y$ and $ z$ are positive numbers and $ \frac {1 \minus{} x}{x} \equal{} a,$ $ \frac {1 \minus{} y}{y} \equal{} b$ and $ \frac {1 \minus{} z}{z} \equal{} c.$
Hence, $ abc \equal{} 1$ and your inequality is equivalent to
$ \sum_{cyc}(a \plus{} ab \minus{} 2)\geq0,$ which is true by AM-GM.


In the land of Albatri, only a special breed of albatrosses live. These albatrosses have perfectly square faces that are 5 toes tall. The mouth is located exactly .5 toe above their chins and is perfectly linear across the whole face. Their nose is randomly located between the top of their heads and their mouth, and is a point. What is the probability that the upper angle of the triangle formed by connecting the corners of the mouth and the nose of any given albatross is obtuse?

Inspired by some old topics about face painting, triangles, and albatri.

Billy

Solafidefarms, which point is the vertex? The mouth corner or the nose.

The vertex in question is the one at the nose.


videos
search
geometry
geometric transformation

any of you have a favorite band?

im not talking about the bands who have one good song, like eiffel 65 with there: im blue dabadeebadabi dabadeedabadi...

im talkin bout the real shiznit, like the band that brings to good stuff year in and year out.

ill start

oh yeah, you can have more than one

so i like trust company, goo goo dolls, matchbox 20, eminem, 50 cent, good charlotte, foo fighters

theres more, i just cant remember

How about Marilyn Manson? :twisted:

Seriously speaking, Dirty Vegas, Carly Simon (well, he's not a rock band), Lou Bega, etc.

to tell the truth, ive never listened to a song by marilyn manson

ive only heard one song by dirty vegas

Let me guess: "Days go by," right?
Ever since that Jap car-maker used it everyone know about it, especially after Chapelle (I don't know how to spell his name; he's a comedian hosting a show called "Chapelle's Show" or somat) made fun of it.

You know, I think Fiery could do the same thing the guy did [url=http://launch.yahoo.com/musicvideos/player/default.asp?videoID=1084297]here[/url]:)

i know of chapelle, hes funny. especially in half baked

days go by is correct

have you heard of darude?

wooooooooooooot!

how bout lautsprecher?

i think i spelled that right, im not sure

but listen to omnibus by lautsprecher, itll roxorz fo sure

lautsprecher is how i think its spelled. some german dude

I just realized that in order for the song to come true, ripply must first leave fiery and go marry another men...*sigh* how sad :twisted:

Oh, and I found this along the way: [url=http://24.195.131.228:1300/divx/Glowstick.avi]This[/url] is my image of you :lol:

[quote="Ragingg"]Eminem, Dave Matthews, Counting Crows, John Mayer, Red Hot Chili Peppers... Not that I listen to all of them... :P[/quote]

Whoa, sick...my two favorite bands are Dave Matthews Band and Counting Crows, and I listen to both of them.  A lot.

Just for interest, to any DMB/CC fans, what are your favorite songs?

[quote="mathfanatic"][quote="Tare"]Carly Simon (well, he's not a rock band).[/quote]

What?  If Carly Simon is a he...well, first of all, he must have an extremely high voice to sing songs like "Tapestry" and "Clouds in My Coffee"...and second of all, he must have longish hair and look like a female, and third of all, his marriage to James Taylor must have been...different.[/quote]
Oh, and you reprimand [b]me[/b] for embarrassing girls...
a.) There is nothing wrong with male alto (though male sopranos are rare)
b.) he could have [b]tried[/b] to have that voice deliberately via surgery...I shall say no more
c.) The "longish hair" and the looks are totally offensive. I mean, look at all the alienated youth in Tokyo. You can't even determine their gender at all, yet they still don't point that out to one another...
d.) I'm not sure what you're talking about, but I'm pretty sure they did it in the Netherlands or Canada, if they had one. It could've been a joke, you know, like in The Onion or somat. Even if they had one, the only difference would be that there would be no bride and it wasn't done in the US, that's all. (I could say the above because my father's classmates married each other, and so did two of my teachers. traumatic, yet full of life lessons.)

what :oops:  I don't enjoy bragging or people that do. But ok, I'm pretty good.

well sorry. I still like aim lingo cuz my rents won't let me get aim, but we can go into that later if you want. like I said before...g2g ttyl lol[b]t[/b]

agreed...

okay, now bak to the original topic...

[quote="MathFiend"]g2g ttyl lol[b]t[/b][/quote]
Translation? :) "t"??? :lol:

I sang great too until my voice changed, now it's all...excrementy :lol:

*starts searching for Magnito's helmet again* :)  :wink:

lucky for me, my voice won't change that drastically, one of the advantages of being a female. do you know how the Iroquois indians treated the women in their society. look it up sometime...

Mmm, I studied in fourth...*sigh* (note: I actually studied Sioux :) I'm just sighing about [b]the days[/b]...*sigh* I was much closer to you then, physically :lol: )

Anyways, keep reminding me of that fact until it sinks through my mind...too many names starts with "math" :)

i'd rather not...juss PM it to me if its inappropriate for the forum

and I quote from my Social Studies textbook, "The women of the Iroquois were greatly respected. When a man and a woman were married, the man moved in with the woman and all her belongings. The women of the Iroquois also chose the clan leader. If the leader wasn't doing his job properly, the women had the right to choose a new clan leader." Man, I wish our society was a matriarchy.

the posts following MathFiends quote were deleted b/c they concerned a topic that should not be discussed on this section of the forum...

if the admin has nothing against it, then you may hold the conversation in Etc...

However, let's get back on topic

Once again this thread has been edited. (this time due to request by Tare). Please check "Moderator Rights".


algebra
polynomial
limit
trigonometry
logarithms
quadratics
real analysis

Suppose a sequence $\{x_n\}$ of real numbers satisfies the condition that for any $P(x)\in\mathbb{N}_0[x]$ with $\deg P=2$ the following equality holds \[\lim_{n\to\infty}(x_n+x_{P(n)})=0.\] Is true that $\lim_{n\to\infty}x_n=0$?

[quote="heartwork"]Yes.[/quote]
Prooof, pleeease... Actually I am joking  :lol:  The answer is [hide="hidden here"]Are you sure you want to read the answer? This is a cool problem! Go and think more. Only if you become completely desperate, click [hide="here"] Are you completely sure? OK, [hide="click here"]Not necessarily[/hide][/hide] [/hide]

For the details, [hide="look here if you dare"]Nice joke, fedja. [hide="One more layer"]$x_n=\sin\frac{\pi\cdot \log\log n}{\log 2}$[/hide][/hide]

Now, how about if $P$ is allowed to be either quadratic or cubic?  ;)

[quote="fedja"]Now, how about if $ P$ is allowed to be either quadratic or cubic?  ;)[/quote]
for deg $ \equal{}m$ we need :
$ x_n\equal{}\sin\frac{\pi\cdot \log\log n}{\log m}$ :)

That misses the point; it doesn't give you both at once.

I don't understand well why?  :maybe:
if I take :
$ x_{n} \equal{} \sin(\frac {\pi \ln \ln n}{\ln m})$ then
1) it's easy to note that :$ \lim_{n \to \infty} x_{n} \neq 0$
2) if $ P(x) \equal{} a_{m}x^{m} \plus{} ... \plus{} a_{0}$ then:
$ \ln (P(n)) \approx \ln a_{m} \plus{} m\ln n \plus{} O(\frac {1}{x})$
thus :
$ \ln (\ln a_{m} \plus{} m\ln n \plus{} O(\frac {1}{x}))\approx \ln (m\ln n ) \plus{} O(\frac {1}{\ln n})$
and for big $ n$ we have :
$ x_{P(n)} \equal{} \sin(\frac {\pi \ln \ln P(n)}{\ln m})\approx \sin \pi (\frac {\ln (m\ln n ) \plus{} O(\frac {1}{\ln n}) }{\ln m} )\approx \sin ( \pi \plus{} \frac {\pi \ln \ln n}{\ln m} \plus{} O(\frac {1}{\ln n}))\approx \minus{} \sin(\frac {\pi \ln \ln n}{\ln m})$
thus :
$ x_{n} \plus{} x_{p(n)}\approx 0$ for big $ n$.

So, is $ m$ equal to $ 2$ or $ 3$? You have to choose one, and then your example doesn't work for the other. The logarithms are incommensurable, and this structure can't give you both at once for any value of $ m$.

:lol: I don't translate it right:
... is allowed to be either quadratic or cubic?...
I was thinking it's means if $ P$ has degree only $ 3$
now I agree with jmerry.
for $ deg \equal{} 3$ we have $ x_{n}^{(3)}$ sequance.
for $ deg \equal{} 2$ we have $ x_{n}^{(3)}$
and if $ P$ "is allowed to be either quadratic or cubic" then :
what about some combination of   $ x_{n}^{(2)}$ and $ x_{n}^{(3)}$?
I mean define sequance
$ y_{n}^{(2,3)} \equal{} x_{n}^{(3)}$ if $ deg P(n) \equal{} 3$ and $ \equal{} x_{n}^{(2)}$ if $ deg P(n) \equal{} 2$  :huh:

I used to know the solution to this 2 or 3 question when I made my post. The point was that you couldn't get away with any explicit example this time but would have to do some non-trivial analysis on the graph with edges given by pairs $ n$, $ P(n)$ where $ P$ is one of polynomials in a finite family. The key idea was to show that there are no short cycles of odd length if you restrict yourself to sufficiently large values of $ n$. I don't remember the details right away but if somebody is interested and nobody finds the solution in a reasonable time, I'll try to recall them and post the full argument :)

Sorry, I didn't understand from this comment: answer is yes or no?  :blush:


vector
linear algebra
linear algebra unsolved

Let's define Wi=im(T^i) where T is nilpotent, i.e. there is a k s.t. T^k=0.
a)Prove that if W^i is different from 0 then dim W^(i+1) < dim W^i.
b) if U is a vector space s.t. dim U=n and T is nilpotent, prove that T^n=0.

cheers,

a) Note, that $ W_i\equal{}Im(T^i)\equal{}\{T^iv\,|\,v\in V\}$ holds, since the RHS is a sub-vectorspace. In particular, note that $ T(W_i)\equal{}W_{i\plus{}1}$ holds. If $ \dim W_i\equal{}\dim W_{i\plus{}1}$ for some $ i$, we get $ W_i\equal{}W_{i\plus{}1}\equal{}T(W_i)$ and this implies $ W_i\equal{}T^k(W_i)\equal{}0$ or in other words for all $ W_i\not\equal{}\{0\}$ we have $ \dim W_i >\dim W_{i\plus{}1}$.

b) This is a direct consequence of a) with the identification $ U\equal{}W_0$. We step-down at least one dimension between $ W_i\not\equal{}\{0\}$ and $ W_{i\plus{}1}$. Therefore, $ W_n\equal{}\{0\}$ and hence, $ T^n\equal{}0$.


calculus
integration
geometry
parameterization
vector
function
calculus computations

Ahhh I can't figure this out.

It is $\int_{C}(1+xy)ds$ where $C$ is straight line path from $(0,2)$ to $(0,-2)$.

------

I did it two ways and got +4 one way and -4 the other way but I'm not sure what is wrong with either way.  I originally got -4 and teacher says +4 and I just want to know if I should give up the fight.

So positive or negative?

Also I tried interpretting it as lateral surface area but then I wasn't sure how to in 3D.  I mean the area is 4 and it is above the xy plane.  But it seems like it would make a difference it was going (0,2) to (0,-2) or (0,-2) to (0,2).

The basic formula you need is
\[\int_{C}(1+xy)\,ds=\int_{a}^{b}f(\vec r(t))|\vec r\,{}'(t)|\,dt\] where $\vec r(t)$, $a\le t\le b$, is some parametrization of $C$. The orientation of $C$ does not matter for integrals of scalar functions.

For a line segment $AB$ use the parametrization $\vec r(t)=\vec r_{0}+t\vec v$, $0\le t\le 1$, where $\vec r_{0}$ is the position vector of $A$ and $\vec v$ is the vector from $A$ to $B$. In this case $\vec r(t)=\langle 0,2\rangle+t\langle 0,-4\rangle=\langle 0,2-4t \rangle$, hence $\vec r\,{}'(t)=\langle 0,-4\rangle$, and the integral is equal to $\int_{0}^{1}(1+0(2-4t))4\,dt=4$

Ok.  I've never seen the $r_{0}+tv$ thing, but makes sense.  I did $r = <0,-t>\ \ t\in [-2,2]$ and got the positive 4, but then I did it this way and got negative.  What is wrong, is it the going backwards with $t$?  In every other problem I have done, going backwards has worked fine.  If that is in fact the problem, are there specific conditions for when you can take $t$ as decreasing?

Let $r(t) = <0,t>$ as $t$ goes from $2$ to $-2$.  Then $|r'(t)| = 1$, so the integral is $\int_{2}^{-2}(1+0(t))(1)dt =-4$  :(

Informally speaking, the "length element" $ds$ is equal to $|\vec r\,{}'(t)\parallel dt|$, not $|\vec r\,{}'(t)|dt$. You may want to consider the simpler integral $\int_{C}1\,ds$, which represents the length of $C$.

You have stumbled upon a fact that is not normally discussed in public: the Riemann integral $\int_{a}^{b}f(x)\,dx$ is actually the line integral of [b]vector field[/b] $\vec F(x)=f(x)\vec i$, not of scalar function $f(x)$. (But remember: you did not hear it from me.  :ninja: )


function
topology
real analysis
calculus
calculus computations

What is equicontinous? and how it is different from the continous function...i need explanation and examples...
 Also i need about uniform continous...and difference between uniform continous and continous ...and how the delta and epislon are choosen in both the cases.... :(

Let $ \mathcal{F}$ be a collection of real-valued continuous functions on a metric space $ X$.  If, given any $ \varepsilon>0$, we can choose a single $ \delta>0$ that always works for every $ f\in\mathcal{F}$ and every $ x\in X$, then $ \mathcal{F}$ is equicontinuous.  So given any $ \varepsilon>0$, there is a $ \delta>0$ such that whenever $ x,y\in X$ satisfy $ d(x,y)<\delta$, we have $ \left|f(x)\minus{}f(y)\right|<\varepsilon$ for all $ f\in\mathcal{F}$.

Equicontinuous is then the "most" continuous.  Just as uniform continuity is stronger than continuity, equicontinuity is even stronger.

Uniformly continuous means that the choice of $ \delta$ depends on the function $ f$ and $ \varepsilon$, but not $ x,y$.  A function that is uniformly continuous is continuous, but not necessarily the other way around.


I can't find AMC practice tests online. I found one but then lost it...:[

You can find some in the Contests link (look at the top of your screen).
I also randomly found this site that has many links, including 9 AMC/AHSME tests...
[url]http://www.math.uncc.edu/~hbreiter/[/url]

I tried looking... I actually found a place, but I lost it when I accidentally exited it...:[

http://www.artofproblemsolving.com/Forum/resources.php?c=182

good place


inequalities
Euler
inequalities unsolved

We have positive real numbers x, y, z. Prove that:

$ \left( {\frac {x}{y} \plus{} \frac {y}{z} \plus{} \frac {z}{x}} \right)^2 \ge \frac {3}{2}\left( {\frac {{x \plus{} y}}{z} \plus{} \frac {{y \plus{} z}}{x} \plus{} \frac {{z \plus{} x}}{y}} \right)$

Thanks very much!

[quote="L_Euler"]We have positive real numbers x, y, z. Prove that:

$ \left( {\frac {x}{y} \plus{} \frac {y}{z} \plus{} \frac {z}{x}} \right)^2 \ge \frac {3}{2}\left( {\frac {{a \plus{} b}}{c} \plus{} \frac {{b \plus{} c}}{a} \plus{} \frac {{c \plus{} a}}{b}} \right)$

Thanks very much![/quote]

It's old and well known, my friend. And it's not yours.  :wink:
See in the attachment.


AMC
AIME
USA(J)MO
USAMO
MATHCOUNTS
AMC 8

Does anyone know of good websites that give you practice problems for the AMC 8?

There are a bunch of practice problems found under the AMC 8 Archive if you look on the AMC website at [url]http://www.unl.edu/amc[/url].

Also, if you want full tests, look here:

[url]http://www.math.ksu.edu/main/handbook/ProblemSets/AMC8Problems[/url]

 :)

I don't know if this is legal, but [someone I know] has a website and has 2001-2005 online [url=brockman.com/math/amcs.php]here[/url].

AMC does allow people to distribute its tests publicly. Of all I know, any of its tests (AMC 8/10/12, AIME, USAMO, MOP, IMO, etc.) can be posted online, unlike MATHCOUNTS, which doesn't allow full sets of all of its problems to be posted online.

What years are available so far? (thanks for the links BTW)

There are also some competitions here https://www.artofproblemsolving.com/Wiki/index.php/AMC_8_Problems_and_Solutions

I looked at that a while ago, but a lot of the problems/answers aren't posted.

There are a few here ( you have to choose amc8 they have other things too
http://www.unl.edu/amc/d-publication/d1-pubarchive/2008-9pub/


Solve the following system of equations for all primes p with integer solutions x,y !

(1) p + 1   = 2x^2
(2) p^2 + 1 = 2y^2

OK, this is a German question.

You can look up the solution in the 1997 - 1998 book,

http://www.unl.edu/amc/a-activities/a4-for-students/mc97-98-01feb.pdf

at page 41.

it also appears in Po Leung Kuk invitational math competition(? or something like that, just for junior form, i.e. grade 7-9) 5 years ago i think


Let $ a,b,c\geq 0$, no two of which are zero.Prove that:
$ \frac{a^2}{(2a\plus{}b)(2a\plus{}c)}\plus{}\frac{b^2}{(2b\plus{}a)(2b\plus{}c)}\plus{}\frac{c^2}{(2c\plus{}a)(2c\plus{}b)}\leq \frac{1}{3}$.

[quote="Inequalities Master"]Let $ a,b,c\geq 0$, no two of which are zero.Prove that:
$ \frac {a^2}{(2a \plus{} b)(2a \plus{} c)} \plus{} \frac {b^2}{(2b \plus{} a)(2b \plus{} c)} \plus{} \frac {c^2}{(2c \plus{} a)(2c \plus{} b)}\leq \frac {1}{3}$.[/quote]
See here:
http://www.mathlinks.ro/viewtopic.php?t=66660


i know many of u wouldn't have deemed this exam worthy to be written.
but for those of u, who have written it, how was it??
i found it a good and tricky paper, tougher than last year, and just marginally short of the level of last year's IITJEE!!

absolutely, physics was way too trickier than last year's JEE! :ninja:
any clue about solutions???????

hell yes it was much better than last years paper. last year's paper was no better tahn any CET exam. this year's paper was much better comparatively.

[url=http://www.ankur-gupta.com/blog/aieee/aieee-2008-solutions]Solutions for AIEEE 2008[/url] here!!!!! :D

all the best! :ninja:

i actually did chem better than physics  :o though i felt the physics paper was pretty good.

and did you guys check out that math question? they had actually typed "Prove that the function is invertible."   :rotfl:
looks like too much practise giving "tough" subjective qsns  :rotfl:

i hv done awfully!
207 by narayana's solutions and 231 by Career Point's

maximum (in case, some of my remaining doubts are correct) i can get 246, not more than that.
so i consider my score to be 220!

everyone pls post their scores!!

i havent checked with the solns and dont plan on doing so either. but havent done that great. mite get around 220 too.

I think narayana solutions suck! I'm waiting for a better institute to post their solutions.

why not check this????
http://www.brilliant-tutorials.com/AIEEE%202008%20Solutions.html :D

:spam:  :spam: am i going blind or did you post nothing. why cant i do that?

i flopped and am getting only 200 courtesy fiitjee solns :( :(

[quote="isomorphism"]So what is the general consensus on the AIEEE paper, IIT paper this year? Was it hard or easy?

So are there any IIT or AIEEE top 100 aspirants lurking here? :P
[/quote]
Ofc :P When there are people like you know all of the indian forum users (except me :D ), then you do have the top few ranks here.
[quote="isomorphism"][b]P.S:[/b]And does anyone of you have a link to the IIT paper, AIEEE paper for this year?[/quote]
We have given the link to the solutions by various institutes, think the questions are there itself.  :)

[quote="isomorphism"]So what is the general consensus on the AIEEE paper, IIT paper this year? Was it hard or easy?

So are there any IIT or AIEEE top 100 aspirants lurking here? :P

[b]P.S:[/b]And does anyone of you have a link to the IIT paper, AIEEE paper for this year?[/quote]

yay!! another kannadiga!! :D  :D

[quote="isomorphism"]So what is the general consensus on the AIEEE paper, IIT paper this year? Was it hard or easy?

So are there any IIT or AIEEE top 100 aspirants lurking here? :P

[b]P.S:[/b]And does anyone of you have a link to the IIT paper, AIEEE paper for this year?[/quote]

U think there won't be anyone from top 100 here !!!! :o  :o

Result on 7th June or later as per the website! :maybe:  :o

i am not able to see any knid of that thing in that site how come you saw????

hi guys...im new here.... since u guys wanted to see some scores....heres mine in AIEEE ... im getting 240-244 not considering that crap velocity of sound question...

chennai by any chance? coz there are a lot of people (rather nerds) from chennai katchi here  :D

nerd leader hash includee speaks about this ah???

any idea when results will come out i mean the AIEEE ONE??? :D

Even one of my old friends told me that it is coming out only on the 7th :play_ball:  :lol:

I found this link somewhere, dunno about its authenticity :maybe:
http://www.exam-result.com/aieee-results-2008-on-june-7-2008/

Not that anyone cares but the [url=http://www.cbseresults.nic.in/]cbse class XII results[/url] are out on the 21st (chennai) and 23rd (delhi)  :D

[color=red][size=150][u][i][b]MADNESS ROHIT BOARD TOPPER!!![/b][/i][/u][/size][/color]

Let me clear the air..............go to the official website of AIEEE and click on Results and ........ link............look for it in the main page!

It redirects u to a page where it says "7th June or later!" :huh:


Here it is:

If a, b, c are real numbers such that a:^2: + 2b = 7, b:^2: + 4c = -7, and c:^2: + 6a = -14, find a:^2: + b:^2: + c:^2: .

I know one can solve it through brute force algebra and substitutions, but a note that accompanied this problem hinted at the method of summing equations. Does anyone know of such a more elegant solution to this problem? Thanks.

Fierytycoon

Adding all three equations give a:^2: + 6a + b:^2: + 2b + c:^2: + 4c = -14.
Completing the square fo each variable gives (a+3):^2: + (b+1):^2: + (c+2):^2: = 0, in other words the only solution is a=-3, b=-1, c=-2.

Oh...that was simple. I think I was just stuck on one line of thought--not good. Thanks.

Fierytycoon


Given two real sequence $ {a_n}$ and  $ {b_n}$, what is the necessary condition  for
$ \limsup_{n \rightarrow \infty } (a_n \plus{} b_n)\equal{} \limsup_{n \rightarrow \infty }a_n \plus{} \limsup_{n \rightarrow \infty }b_n.$

Here is a place where it doesn't work: $ \{a_n\}\equal{}\{1,2,1,2,1,2,\ldots\}$ and $ \{b_n\}\equal{}\{2,1,2,1,2,1,\ldots\}$.

Also note that you always have $ \limsup_{n\to\infty}(a_n\plus{}b_n)\leq \limsup_{n\to\infty}a_n\plus{}\limsup_{n\to\infty} b_n$.

Do these things help at all? I can think of a bunch of sufficient conditions...  :lol:

The equality holds if both the sequences are convergent. Please provide the other sufficient conditions.

i think it holds iff every increasing sequence of indices $ n_j$ for which $ b_{n_j}\to\limsup_{n\to\infty} b_n$ contains an increasing subsequence of indices $ n_{j_k}$ such that $ a_{n_{j_k}}\to \limsup_{n\to\infty} a_n$.


let a and b be positive real numbers satisfying a+b>= (a-b)^2. Prove that
          x^a (1-x)^b + x^b (1-x)^a <= 1/(2^(a+b-1))
for 0<=x<=1, with equality if and only if x=1/2

[quote="JBL"]Use [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=123690]LaTeX[/url].  aviateurpilot, \max giving $ \max$ looks nicer than $ max$.[/quote]
thanks :)


first of all, if i roll 2 6-sided dice, would there be a 2/6 chance to roll a six? :?  :?  :?

[quote="h78reg"]there is a 1/6 chance of rolling a 6 on 1 die and 1/36 of rolling a 6 on 2 die.

this post doesn't belong in pre-olympiad. it's WAY to easy.[/quote]

I think he means the probability of rolling a 6 on either die, it's 1/6+1/6.

Yeah, this should be in Getting started.....

[quote="nr1337"]I think he means the probability of rolling a 6 on either die, it's 1/6+1/6.[/quote]
If he meant the probability of rolling at least one 6, then that would be $1-\left(\frac{5}{6}\right)^2=\frac{11}{36}$.

[quote="mathbeginner"][quote="nr1337"]
I think he means the probability of rolling a 6 on either die, it's 1/6+1/6.
[/quote]

so the chance of a six coming up is 2/6 right...

but then if u were to roll 6 dice, the chance of a six coming up would be 1/6+1/6+1/6+1/6+1/6+1/6 = 6/6

this is where i get confused because 6/6=100% (correct?) and there isnt a 100% chance of a six coming up obviously

srry if this isnt in the right place  :?[/quote]

Very good analysis of why the answer can't be 1/6 + 1/6.

I will move this to Getting Started, but I'll note you should take care to be precise in your question phrasing - your question could mean 'What is the probability the sum will be 6?' or 'What is the probability at least one 6 will occur?' or 'What is the probability exactly one 6 occurs?'  These are obviously very different questions.

I think you mean 'What is the probability at least one 6 will occur?'  I'll give this hint to that question: [hide]What is the probability no 6 occurs?[/hide]

[quote="rrusczyk"]I think you mean 'What is the probability at least one 6 will occur?'  I'll give this hint to that question: [hide]What is the probability no 6 occurs?[/hide][/quote]

That's a good hint.

[hide="with that"]you can easily find that it is 5*5=25 (5 choices other than 6), and then just subtract that from the total possibilities, 36, to get $\frac{11}{16}$. [/hide]

I knew I forgot something....

[hide]
1/6+1/6 counts the possibility of getting 2 sixes twice and must be subtracted once.  So the real probability of getting at least one six is 1/6+1/6-1/36=11/36

Alternatively, the probability of not getting any six is (5/6) * (5/6) = 25/36, so the probabiliy of getting at least one six is 1-25/36 - 11/36[/hide]

I need more work on probability :P


Just to ask, how do you really become a moderator?

How many posts for each of the ranks?

For example Yang Mills theory = 500 posts (?).

I'd like to introduce you to my friend.  He's named [url=http://www.artofproblemsolving.com/Forum/search.php]Seach Tool[/url].

Try the [url=http://www.artofproblemsolving.com/Wiki/index.php/AoPS_forums_tutorial]AoPSWiki[/url] also.

[quote="José"]0 - 19 New Member (Zero Stars)
20 - 49 P versus NP (Half Star)
50 - 99 Hodge Conjecture (One Star)
100 - 249 Poincare Conjecture (Two Stars)
250 - 499 Riemann's Hypothesis (Two and Half stars)
500 - 999 Yang Mills Theory (Three Stars)
1000 - 2499 Navier-Stroke's Equation (Four Stars)
2500 - $\infty$ Birch & Swinnerton Dyer. (Five Stars)
----------------- Administrator (Six Stars)[/quote]

I think that would be navier-STokes.

Moderator powers can be signed up for, or an administrator will notify you of an open position that you could take.

Administrator can only be reached if your name is : MCrawford, RRuscyzk (sorry about the spelling!), NSato, etc.


Not that hard, but cool:

1. A point $C$ is on a chord $AB$ of circle $K$ with center at $O$.  Let $D$ be the second intersection of $K$ with the circumcircle of $AOC$.  Show that $CD = CB$.

2. Let $ABCD$ be inscribed in a circle with center $O$.  Given that $AC$ is perpendicular to $BD$, show that the altitude from $O$ to $AD$ is equal to half of $BC$.

Problem one.
Connect $OD$, $OB$ and $DA$.
$\angle{OCB}=\angle{ADO}=\angle{DAO}=\angle{DCO}$
and
$\angle{CDO}=\angle{CAO}=\angle{CBO}$
Togather with $OD=OB$, $OC=OC$, we get $\triangle{ODC} \cong %Error. "trinalge" is a bad command.
{OBC}$, thus $CD=CB$.

Connect $OB$ and $OD$, let $OG \perp BC$. Thus $\angle{BOG}=\angle{BDC}$, $\angle{DOF}=\angle{DCA}$. But $\angle{BDC}+\angle{DCA}=90^o$, so $\angle{BOG}+\angle{DOF}=90^o$. Since: $\angle{DOF}+\angle{ODF}=90^o$, so: $\angle{ODF}=\angle{BOG}$. Togather with $OD=OB$, $\angle{OFD}=\angle{OGB}=90^o$, we will have: $\triangle{OFD} \cong \triangle{OGB}$, therefore $OF=BG$. Since $O$ is the circumcentre of $\triangle{DBC}$, so $BG=CG$, which gives $OF = \frac{1}{2} BC$.

[b]The second is also good! [/b]
  Let $K$ the antidiametric to  $A$  that is $AOK$is a diameter) and M the midpoind of AD. Then $KD = 2 OM$. But  $DK = BC$ , since angles DAC and KAB are equal (  $= 90- \angle ADB$ and $= 90 - \angle AKB$respectively) and finaly  $\angle ADB = \angle AKB$

[u]  Babis[/u]


Prove that among $39$ consecutive natural numbers, it is always possible to find a number such that the digits sum to a number divisible by $11$.

[hide] Do you mean among 29 natural numbers?  :maybe:

The digital sums in $\mod 11$ are $1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9,10,2,3,4,5,6,7,8,9,10,0$. This repeats every 29 numbers. In 29 consecutive numbers, there is always a number $0\mod29$. [/hide]

EDIT: Never mind.  :blush:


modular arithmetic
number theory
number theory proposed

Let $ a\in N$
Find all $ k\in N$ such that exist infinite $ n\in N$ such that
$ n|(a\plus{}k)^n\minus{}a^n$

Obviosly if $ (a,k)\equal{}d>1$, then exist infinetely many  $ n\equal{}d^m$, suth that $ n|(a\plus{}k)^n\minus{}a^n$.
If $ (a,k)\equal{}1$, let $ (p\minus{}1)|k$, then $ n\equal{}(p\minus{}1)*p^m$ satisfyed your condition.
Therefore all k work (for even p=3, for odd p=2).

I forget, that $ p\not |a\plus{}k, p\not |a$.
If a -odd work any even k ($ n\equal{}2^m$). Any even k work if a is even too ($ 2|d\equal{}(a,k)$).
For odd k n must be odd. If $ p|k$, then $ p^m|(a\plus{}k)^{p^m}\minus{}a^{p^m}$ ($ n\equal{}p^m$) work.

[quote="Dtrung"]It's not simple!
Suppose that $ \large n > 1$, we will show that there are a contradiction...
Indeed, let $ p$ be the smallest divisor of $ n$.
Since $ \large n|(a \plus{} 1)^n \minus{} a^n$, we have $ \large p|(a \plus{} 1)^n \minus{} a^n$ or $ \large (a \plus{} 1)^n \equiv a^n \pmod{p}$ .  $ (1)$
It's easy to see that $ (a,p) \equal{} 1$.
So, let $ b$ be the unique inverse such that $ \large ab\equiv 1 \pmod{p}$.
From $ (1)$, multiplying both sides by $ b^n$, we get $ \large ((a \plus{} 1)b)^n \equiv 1 \pmod{p}$.
Let $ d$ be the smallest number ($ \large d \in N$) such that $ \large ((a \plus{} 1)b)^d \equiv 1 \pmod{p}$.
Then $ \large d\mid n$.
By Fermat's Little Theorem, $ \large ((a \plus{} 1)b)^{p \minus{} 1}\equiv 1 \pmod{p}$, and so $ \large d\mid p \minus{} 1 \Rightarrow d < p \Rightarrow \gcd (d,n) \equal{} 1$, because of the definition of $ p$ as the smallest prime divisor of $ n$. Thus we must have $ d \equal{} 1$, implying that $ \large a \plus{} 1 \equiv a \pmod{p}$, which is the contradiction we want.
So $ n \equal{} 1$. :blush:[/quote]
Use lemma
$ \gcd(a,b)\equal{}1$
$ d\equal{}\gcd(m,n)$ then
$ \gcd(a^m\minus{}b^m,a^n\minus{}b^n)\equal{}a^d\minus{}b^d$
With the same method we can find out all a such that exist infinite $ n\in N$
$ n|a^n\minus{}1$

[quote="TTsphn"]Are you sure
You can prove that $ n|(a \plus{} 1)^n \minus{} a^n\leftrightarrow n \equal{} 1$[/quote]
I only want to prove this statement :blush:

Yes use lemma then the problem quite easy
$ p$ is the least prime divisor of $ n$
then
$ p|\gcd((a\plus{}1)^n\minus{}a^n,(a\plus{}1)^{p\minus{}1}\minus{}a^{p\minus{}1})\equal{}1$ tradition
So $ n\equal{}1$

I know a problem from China TST 2006, it is solved by this method.
"Find all positive integer pairs $ (a,n)$ such that $ \large \frac{(a\plus{}1)^n\minus{}a^n}{n}$ is an integer."
It's really same your idea :blush:

Another problem : "Find all $ \large a \in N$ for which there exists a number $ \large n \in N$ such that $ \large a^n \plus{} 1$ is divisible by $ \large n^2$" :blush:

I think must be $ n>1$
If $ n\equiv 0(\mod 2)$ easy to check there aren't $ a\in N$ satisfy
Now we prove $ n\equiv 1(\mod 2)$ satisfy
Call $ p$ is the least prime divisor of n then
$ p|a^n \plus{} 1$
$ p|\gcd(a^{2n} \minus{} 1,a^{p \minus{} 1} \minus{} 1) \equal{} a^2 \minus{} 1$
Because n odd then $ p|a \plus{} 1$
p odd so $ a \plus{} 1$ is not a power of 2
If $ a \plus{} 1$ is not a power of 2 then chose $ p|a \plus{} 1$ and $ p$ odd .
Easy to check $ p$ satisfy condition $ p^2|a^n \plus{} 1$


function
integration
calculus
geometry
limit
inequalities
calculus computations

1)Can we construct a positive continuos real function of real variable $f$ such that for any $n\in\mathbb{N}$ we have $f(n)=n$ and $\int^\infty_{0}f(x)dx$ converges ?

2)Suppose that $f$ is continuos and decreasing function on $[a,\infty]$ , $a\in\mathbb{R}$ , such that $\int^\infty_{a}f(x)dx$ converges . Can we say that $\int^\infty_{a}(f(x))^{2}dx$ also converges ?

I'm not going to give detailed proofs, just brief sketches.

1) Yes. You want it to be 0 most of the time, but come up to $n$ at the integers. Make the function look like a sequence of triangles with vertices at $(n, n)$. Make the widths of the triangles decreasing and very small (like $\frac{1}{n^{3}}$). The integral of $f$ is the sum of the areas of the triangles...

2) Yes. Split $[a, \infty]$ into $[a, b]$ and $[b, \infty]$ (the one point $b$ doesn't matter in the integral, we can count it twice with impunity) where $f(b) = 1$. Bound the integrals on the two intervals separately (note that squaring has a different effect on each side...)

My first instinct would be to do #2 the same way Xevarion did it, but just for the amusement, I'll try a different way.

Let $F(x)=\int_{a}^{x}f(t)\,dt$ and $M=\lim_{x\to\infty}F(x).$

It is given that $M$ is finite, and $F(x)$ is an increasing function with $0\le F(x)\le M.$

We now do an integration by parts. Since we didn't say anything about $f$ being differentiable, this will have to be a Stieltjes integral; with $f$ being monotone, we can do that.

$\int_{a}^{\infty}f(x)f(x)\,dx=F(x)f(x)\big|_{a}^{\infty}-\int_{a}^{\infty}F(x)df(x)$

$F(x)f(x)\big|_{a}^{\infty}=M\cdot 0-0\cdot f(a)=0.$ Note that the monotone $f$ must tend to zero at infinity to be integrable. So

$\int_{a}^{\infty}[f(x)]^{2}\,dx=\int_{a}^{\infty}F(x)(-df(x))$

$\le M\int_{a}^{\infty}(-df(x))=-Mf(x)\big|_{a}^{\infty}=Mf(a).$

We have the inequality

$\int_{a}^{\infty}[f(x)]^{2}\,dx\le f(a)\int_{a}^{\infty}f(x)\,dx$

----

On second though, I didn't need the integration by parts and the Stieltjes integral to get there. It suffices to note that with $f$ decreasing, $[f(x)]^{2}\le f(a)f(x).$

but what if $f(x)<0$ ? then we cannot say that

$[f(x)]^{2}<f(a)f(x)$

Isnt it ?

The stated hypotheses for $f$ were that it was continous, decreasing, and integrable on $[a,\infty).$ If it were ever negative, it would not be integrable. To satisfy those hypotheses, $f$ must necessarily be a positive function that tends to zero at infinity (and even that isn't sufficient.)


WERE YOU AWARE THAT THE RACE IS VERY CLOSE

WHY IS THERE A RANDOM LOWER CASE "G" IN THE TITLE OH IS IT BECAUSE THEY DON'T LET YOU HAVE ALL CAPS IN TITLES THANKS

AND ALSO I AM NOT SHOUTING

YOU DEFINITELY ARE
YOU'RE GOING AT IT WITH A DECISIVE AMPLITUDE OF VOICE

[quote="1=2"]wtf are we talking about? What race?[/quote]
THE VERY CLOSE RACE OF COURSE

[quote="1=2"]wtf are we talking about? What race?[/quote]

THE ONE EVERYBODY'S SHOUTING ABOUT OF COURSE

[quote="1=2"]wtf are we talking about? What race?[/quote]DO NOT TRY TO PREDICT THE WEATHER, FIND PATTERNS IN RANDOMNESS, OR ATTEMPT TO SEE WHY HoD IS RANDOMLY POSTING THESE THINGS AND PEOPLE LIKE ME ARE ACTUALLY RESPONDING WITH WAY TOO LONG RUN-ON SENTENCES THAT EVEN NOTE INSIDE THEMSELVES THAT THEY THEMSELVES ARE RUN-ON SENTENCES AND BY THE WAY THE RACE ENDED 3 DAYS BEFORE A WEEK FROM NOW AND THEN WE WILL NOW WHO WINS.


I need help making a net with a specific volume and surface area. I don't
know how to do it.

Work Done:

Volume=80
surface area=144

Do you need the net to have all integer sides?


this has a 80 percent chance its gonna get locked. i don't gurantee appropriate stuff

[19:57] Urc: im pregnant
[19:58] jjx1: how old are you
[19:58] Urc: 12
[19:58] Urc: and im a male

well it hasn't been written in a long time. i don't want to revive a thread.

6 days does not constitute "a long time".  Generally, you can revive a thread up to a month old.  This period can be even longer under certain circumstances.

w/e fine lock this idc. you can post here or teh number 2. do number 2 i guess. i just thought no one would go on it cause well its at the bottom.

[17:22] dragon96: doot doot doot doot doot doot doot doot doot doot doot doot doot doot doot [17:21] FloodFilter: floppy, cd?
[17:22] floppy: yeah
[17:22] floppy: what?!?!?
[17:22] MysticTerminator: HMM
[17:22] MysticTerminator: WHAT CLASS IS THIS
[17:22] MysticTerminator: THIS SEEMS PRETTY RANDOM
[17:22] floppy: [17:21] FloodFilter: floppy, cd?
[17:22] MysticTerminator: ANYWAY
[17:22] floppy: [17:21] FloodFilter: floppy, cd?
[17:22] floppy: WHAT?!?!?
[17:22] hurdler: what class????what?
[17:23] floppy: Why did flood filter send me a messag?!?

I forgot to get what floppy said earlier, but he was asking for a cd.

there appearantly is a username _FloodFilter. no surprise

old one from before (can't remember when)

dannyhamtx: mewto
dannyhamtx: do you know that problem about students liking pies and angles in pie graphs?
mewto55555: hmm?
mewto55555: 50

someone is not memming......

Sorry mewto, I just saw it (not stating source) and I thought it was funny.

[16:23] mewto55555: cherry pi is not hot
[16:23] mewto55555: violin321 is
[16:23] sup3rsolv3r: ...
[16:23] mewto55555: dont copy that
[16:23] mewto55555: please
[16:24] sup3rsolv3r: i just did
[16:24] mewto55555: cherry will kill me

...

[16:43] mewto55555: ill go out with violin321 then
[16:43] mewto55555: shes semi-hot

You should come to our school. Mewto55555 is obsessed with violin321.

What is your school's name and where? Are cherry and violin and mewto in the same grade?

[quote="ernie"][18:09] guineapiggies: my wife!?!?!?[/quote]

lol ernie. btw i said this b/c ernie said something about me having a wife...idk i just looked up at screen so yeah... darn i should have copied the whole thing  :wink:

[quote="tinytim"]Ladue Middle School. Cherry is in 7th. mewto, violin and I are in 8th grade.[/quote]

In St. Louis, I think.

[17:51] * alexhhmun marries Ding! HA!! ALL MINE!!

[9:06] aw9264: hello
[9:06] aw9264: helloooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo
[9:06] aw9264: be back in 5min
[9:08] aw9264: im back
[9:08] aw9264: hello
[9:08] aw9264: ne1 there
[9:08] aw9264: ???
[9:08] aw9264: im alone again??????? grrrrrrr
[9:08] aw9264: >.<

[14:09] AIME15: im scared of you
[14:09] AIME15: your rating is too low
[14:09] FantasyLover: ...
[14:09] AIME15: o_O_O_O_O_O_O_o
[14:09] FantasyLover: i wanted to go high by beating you

[quote="jjx1"]20:43] CA Math: wait is your name DOJO ^_^?
[20:43] ghost of levans: hehe
[20:43] pacman2812: then your parents hate you
[20:43] Dojo: NO CA
[20:43] hurdler: hm
[20:43] hurdler: FAN
[20:43] ghost of levans: then i'd have to think your parents are weird[/quote]

wait. did levans change his name? or is this someone else...

levans got eliminated in the top admin competition thing. so he changed his account name to ghost of levans. later he changed it back to levans.


calculus
integration
trigonometry
logarithms
calculus computations

are the answers

(1) $ \frac {\pi}{12} \minus{} \frac {1}{8}$?

(2) [b][edit][/b]$ \minus{}\frac {73}{60}$?

.... and are there really two $ dx$'s in number (3)?  :(

[quote="kunny"]Regretabbly your answers are incorrect.

How do you find $ \int_0^{\frac {1}{2}} \frac {1}{\sqrt {1 \minus{} x^2}}\ dx$ without using $ \sin ^{ \minus{} 1}x$?[/quote]

sorry, I made a mistake... I edited it :D

[quote="kunny"]Evaluate the following definite integrals.

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

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

(3) $ \int_{ \minus{} 1}^7 \frac {dx}{1 \plus{} \sqrt [3]{1 \plus{} x}}$[/quote]

For (1)

Put $ \ x\equal{}\sin \theta\implies dx\equal{}\cos \theta\  d\theta$

$ \implies\boxed{\int_0^{\frac {1}{2}} \frac {x^2}{\sqrt {1 \minus{} x^2}}\ dx\equal{}\int_0^{\frac {\pi}{6}}  \sin^2\theta\ d\theta \equal{}\frac12 \int_0^{\frac {\pi}{6}}  1\minus{}\cos2\theta\ d\theta\equal{}\frac12\left(\theta\minus{}\frac {\sin 2\theta}{2}\right)_{0}^{\frac {\pi}{6}}\equal{}\frac {\pi}{12}\minus{}\frac {\sqrt 3}{8}}$

[quote="kunny"]Evaluate the following definite integrals.

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

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

(3) $ \int_{ \minus{} 1}^7 \frac {dx}{1 \plus{} \sqrt [3]{1 \plus{} x}}$[/quote]

For (2)

Put $ \ x\equal{}\tan \theta\implies dx\equal{}\sec^2\theta d\theta$

$ \implies \boxed{\boxed{\int_0^1 \frac {1 \minus{} x}{(1 \plus{} x^2)^2}\ dx\equal{}\int_0^{\frac {\pi}{4}} \frac {1 \minus{} \tan\theta}{\sec^2\theta}\ d\theta\equal{}\frac 12\int_0^{\frac {\pi}{4}} (\cos 2\theta \plus{}1 \minus{} \sin 2\theta)\ d\theta\equal{}\frac 12\int_0^{\frac {\pi}{8}} 2\ d\theta\equal{}\frac {\pi}{8}}}$

hello, for (3), setting $ \sqrt [3]{1 \plus{} x} \equal{} t$ we get $ dx \equal{} 3t^2\,dt$ and our integral is
$ 3\int_{0}^{2}\frac {t^2}{1 \plus{} t}\,dt \equal{} 3[\frac {1}{2}t^2 \minus{} t \plus{} \ln(1 \plus{} t)]_{0}^{2} \equal{} 3\ln(3)$
Sonnhard.

That's incorrect. :(

Now that's correct. :)

hello kunny, i think i have forgotten the factor $ 3$.
Sorry,Sonnhard.

[quote="kunny"]Evaluate the following definite integrals.

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

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

(3) $ \int_{ \minus{} 1}^7 \frac {dx}{1 \plus{} \sqrt [3]{1 \plus{} x}}$[/quote]

For (3)

Put $ \ 1 \plus{} x \equal{} t^3\implies dx \equal{} 3t^2dt$

$ \boxed{\boxed{\int_{ \minus{} 1}^7 \frac {dx}{1 \plus{} \sqrt [3]{1 \plus{} x}} \equal{} 3 \int_{ 0}^2 \frac {t^2}{1 \plus{} t}dt \equal{} 3 \int_{ 0}^2 t \minus{} 1 \plus{} \frac {1}{1 \plus{} t}dt \equal{}3 \left(\frac {t^2}{2} \minus{} t \plus{} \ln (1 \plus{} t)\right)_0^2 \equal{} 3\ln3}}$

Yes, that's right. :lol:
Now all problems have been solved!

[quote="kunny"]Regretabbly your answers are incorrect.

How do you find $ \int_0^{\frac {1}{2}} \frac {1}{\sqrt {1 \minus{} x^2}}\ dx$ without using $ \sin ^{ \minus{} 1}x$?[/quote]

Is there a way to do that?

At any rate, it's same as using $ \sin^{\minus{}1}{x}$.

[quote="makar"][quote="kunny"]Evaluate the following definite integrals.

For (2)

Put $ \ x = \tan \theta\implies dx = \sec^2\theta d\theta$

$ \implies \boxed{\boxed{\int_0^1 \frac {1 - x}{(1 + x^2)^2}\ dx = \int_0^{\frac {\pi}{4}} \frac {1 - \tan\theta}{\sec^2\theta}\ d\theta = \frac 12\int_0^{\frac {\pi}{4}} (\cos 2\theta + 1 - \sin 2\theta)\ d\theta = \frac 12\int_0^{\frac {\pi}{8}} 2\ d\theta = \frac {\pi}{8}}}$[/quote][/quote]

For this part: ${{ \frac 12\int_0^{\frac {\pi}{4}} (\cos 2\theta + 1 - \sin 2\theta)\ d\theta = \frac 12\int_0^{\frac {\pi}{8}} 2\ d\theta = \frac {\pi}{8}}}$

Could you please explain?

[quote="Srinivasa Ramanujan"]I THINK FOR FIRST ONE WITHOUT USINGARC SINX THERE IS ANOTHER METHOD[/quote]

put $ 1\minus{}x^2\equal{}t^2x^2$

U CAN ALSO USE SUBSTITUTION

  ${ \[x^2 = t\] } $

[quote="Srinivasa Ramanujan"]U CAN ALSO USE SUBSTITUTION

  $ { x^2 \equal{} t }$[/quote]

how can that really help ?

put  1- x^2 = t^2

 u lastly get integral as

http://www.sitmo.com/gg/latex/latex2png.2.php?z=100&eq=-%5Cint%5Csqrt%7B1-t%5E2%20%7D%20.dt

[quote="dragoneye776"][quote="MrMagician"]
For this part: ${{ \frac 12\int_0^{\frac {\pi}{4}} (\cos 2\theta + 1 - \sin 2\theta)\ d\theta = \frac 12\int_0^{\frac {\pi}{8}} 2\ d\theta = \frac {\pi}{8}}}$

Could you please explain?[/quote]

Yes, I second that. I am at a loss as to the steps taken here.[/quote]
$\int_0^{a}f(x)dx=\int_0^{\frac {a}{2}}(f(x)+f(a-x))dx$


A sequence $(a_n)$ is defined as follows: $a_0=a_1=1, (n+3)a_{n+1}=(2n+3)a_n+3n.a_{n-1}$. Prove that each term of the sequence is an integer

Did you by any chance mean $(n+3)a_{n+1}$ instead of $(n+3)a_n+1$? I think it's like this because otherwise you wouldn't need the first two terms, just the first one.


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

Why it sees small?

Do you want something like \frac{a}{b} = \dfrac{\dfrac{a}{b}}{1} ?

Actually, I already tried that, but it doesn't work.

Yes it does work, though \dfrac{a}{b} = \dfrac{\dfrac{a}{b}}{1} is better
$ \frac{a}{b}\equal{}\frac{\frac{a}{b}}{1}$ instead of $ \tfrac{a}{b}\equal{}\frac{\frac{a}{b}}{1}$.
Remember that the top of the fraction has to be smaller otherwise it won't fit.
You could use $ \frac{a}{b}\equal{}\frac{a/b}{1}$ or, though this may look ugly, increase the font size for a displayed formula
[code]\fontsize{12}{15}\selectfont
\[ \frac{a}{b} = \frac{\frac{a}{b}}{1} \][/code]

This is what I want

[img]http://img129.imageshack.us/img129/1016/5cd8c63b1c850c3c7d356a1yj0.gif[/img]

Could it be possible to make?

It depends of the configuration of the LaTeX (in the forum)?

That image is precisely what \dfrac{a}{b} = \dfrac{\dfrac{a}{b}}{1} looks like in a document (the forum does some extra processing so it will look different in the forum).
[img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/d/2/f/d2f750538f0bdaa4d4402d2a7652414cb5900ccd.gif[/img]
(This image was produced in the [url=http://www.artofproblemsolving.com/LaTeX/AoPS_L_TeXer.php]TeXer[/url]).

And is there a way to ask Valentin if it can be fixed to produce

[img]http://www.mathlinks.ro/Forum/latexrender/pictures/d/2/f/d2f750538f0bdaa4d4402d2a7652414cb5900ccd.gif[/img]

when we post?

It'd be useful.

You could try posting a request in the [url=http://www.artofproblemsolving.com/Forum/index.php?f=144&ppage=50&sort=lastpost&order=DESC]Questions, Suggestions, & Announcements forum[/url].

There is a workaround you can use, though you may not like it, and that is to replace the inner fraction with the more complicated \genfrac:
\frac{a}{b} = \frac{\genfrac{}{}{0.5pt}{0}{a}{b}}{1} gives
$ \frac{a}{b}\equal{}\frac{\genfrac{}{}{0.5pt}{0}{a}{b}}{1}$
See [url=http://tug.ctan.org/tex-archive/info/math/voss/mathmode/Mathmode.pdf]Math Mode[/url] if you want to know how \genfrac works.

\cfrac{1}{1+\cfrac{a}{b}} gives $ \cfrac{1}{1\plus{}\cfrac{a}{b}}$

That looks like a good idea - where \dfrac is required but shows as a small fraction then use \cfrac instead.
\frac{a}{b}=\frac{\cfrac{a}{b}}{1} gives
$ \frac{a}{b}\equal{}\frac{\cfrac{a}{b}}{1}$


inequalities solved
inequalities
algebra

Let f(x) be the above expression.

- If x  \leq  0, then clearly f(x)  \geq  x 8  + x 2  + 1 > 0.
- If x  \geq  1 then x 8   \geq  x 5  and x 2   \geq  x, thus f(x)  \geq  1 > 0.
- If 0 < x < 1 :
Then x 8  - x 5  +  x 2  = x 2 (1 + x 9 )/(1 + x 3 ) >  0, and 1 - x > 0, then f(x) > 0, and we are done.

Pierre.

Let $  x $  be real number . Prove that $$x^8-x^5+x^2-5x+4\geq 0$$

[quote=Lagrangia]Prove that $\forall x \in \mathbb{R}$ we have: $x^8-x^5+x^2-x+1>0$.[/quote]

Solution:
For $x<=0$ it is evident.
For $x>1$ we can write $x^5(x^3-1)+x(x-1)+1>0$ it is true.
For $x$ from $(0,1)$ we have $x^8+x^2(1-x^3)+(1-x)>0$ it is also true.

Let $  x $  be real number . Prove that $$ x^6 - x^4+2x^3 - x^2 -6x+5  \geq 0$$

[hide=Thanks.][quote=bachkieu][quote=sqing]Let $  x $  be real number . Prove that $$ x^5 - x^4+2x^3 - x^2 -5x +4 \geq 0$$[/quote]

wait doesn't $x = -2$ provide a contradiction?[/quote]
[/hide]

Let $  x $ be real number . Prove that$$ 64x^6 +192x^5-320 x^3+192x+61  \geq 0$$


A point by point individual score distribution... with a pretty graph :)

(Someone asked for it)

thank you very much!
did you type all the scores into excel and get this bar diagram?
its a huge work to do..

You can just copy paste it and it'll format itself nicely within excel  ;)

I'm even lazier than that..... my dad did most of it  :lol: thanks dad!

Binary scores?  Since there is a bijection between Z and B, aren't all scores "binary scores"  :roll:

[quote="randomdragoon"]The distribution is clearly skewed right. So many binary scores![/quote]
obviously its skewed right.


well i have an assignment due on Friday and I was wondering if anyone here could help me with a few problems, im sorry if they may be kind of easy to you but i really suck in math.

6. Find the exact value of the six trigonometric functions of 45 degrees
sin45degrees=
cos45degrees=
tan45degrees=
csc45degrees=
sec45degrees=
cot45degrees=

Describe each of the following graphs of y. Give (a) its amplitude (if applicable), (b) its maximum and minimum(if applicable) and (c) its period. If it has no amplitude, maximum or minimum write" none" on the appropriate blank line.
10. y= 3sin 1/3x
11. y=tan2A
i have tried reading my textbook but i learn best by example and I have no one at home who knows how to do this kind of math so any help will be greatly appreciated.

[hide] sin 45 = sqrt(2)/2 and because sin x = 1/(csc x) ---- csc x =sqrt(2)
cos 45 is the same as sin 45 and sec 45 is the same as csc 45
tan 45 = 1 and the Cotangent is the same in this case.

Are your graph questions to be graphed in radians or degrees?[/hide]


In how many ways 14 identical toys can be distributed among 3 children , so that each of them atleast gets 1 toy and no two of them have the same number of toys .

please post ur solution ....

 [hide]
actually  i get the answer to be 60 , but  i am not sure . please help me .

[/hide]

Yes, I've also got $ \binom{14\minus{}1}{3\minus{}1} \minus{} 3\times 6 \equal{} 78\minus{}18\equal{}60$.

One of many possible explanations for the answer is that it is $ 3!$ times the number of partitions of 14 into three distinct parts, [url=http://www.research.att.com/~njas/sequences/A001399]A001399[/url] in the [url=http://www.research.att.com/~njas/sequences]OEIS[/url].

[quote]zhoraster
Yes, I've also got \binom{14-1}{3-1} - 3\times 6 = 78-18=60.[/quote]

can u please explain how u did that  :( ?

$ \binom{14 \minus{} 1}{3 \minus{} 1}$ is compositions of 14 into three positive parts.  Of these, there are $ 6 \cdot 3$ that come from the six positive integer solutions of $ 2a \plus{} b \equal{} 14$ with $ a \neq b$; if 14 were a multiple of 3 then we would have another case to look at, too.

[Edit: corrected silly error.]


inequalities
calculus
derivative
inequalities proposed

if k is an integer ,the there is a simple proof .if not,Let $a_{1}=1$
then it is equivlent to $f(t)=t^{k+1}(k-1)-(k+1)t^{k}+(k+1)t-(k-1)$
we have $f^{'}(t)=(k-1)t^{k}-kt^{k-1}+1\geq 0$(by AM-GM)
then $f(t)\geq f(1)=0$

Can you prove the inequality without using differentiation?

if k is intenger ,I can.
if not,I guess I  can  prove it with only AM-GM,But  I haven't got a solution.


In day 1, I finish 3/8 of the book. On day 2, I finish 3/8 of the rest of what is remaineing. In day 3, I finish 3/8 of what is remaining. Then I have 15 pages remaining. How many pages did I start with?

[hide] $ \frac58 \cdot \frac58 \cdot \frac58 = \frac{125}{512}$ of the book is remaining.

  Thus the book had $\frac{512\cdot15}{125}=61.44$ pages, which seems quite odd since books have a whole number of pages, am I doing something inccorectly?[/hide]

i dont this is right...

[hide]

Approximately 284 pages.....?
[/hide]

[quote="236factorial"][hide]3/8 + 15/64 + 75/512 = 392/512 read, or 120/512 remaining,

$15\cdot\frac{512}{120} = 64$ pages. [/hide][/quote]

That's what i was expecting. 64 pgs remaining. Good job

Oops, I didnt forgot about the remainder part. :blush:

Well, if you check it, it does not work.
After day one, he read 64*3/8=24 pages, 64-24=40 are left
After day two, he read 40*3/8=15 pages,  40-15=25 are left
After day three, he read 25*3/8=9.375 pages, 25-9.375=15.625, not 15.

G-UNIT's solution is correct, so this problem is screwed up

That's true... and if you work it backwards, you get:

$15=\frac{5}{8}\cdot x \Rightarrow x=24$
$24=\frac{5}{8}\cdot x \Rightarrow x=38.4$

Which obviously is not possible.


algebra
polynomial
quadratics
trigonometry
difference of squares
special factorizations
algebra unsolved

Suppose a and b are real numbers such that the quadratic polynomial
f(x) = x^2 + ax + b
has no non-negative real roots. Prove that ther exist two polynomials g,h, whose coefficients are non-negative real numbers, such that
f(x)= g(x)/h(x)
for all real numbers x.

From hypothesis we have $f(x)>0\forall x\geq 0$ and therefore this problem is a corollary of following problem [hide="IMO 1997 shortlisT"] http://www.kalva.demon.co.uk/short/soln/sh9711.html[/hide]
But I don't know why it isn't lies http://www.mathlinks.ro/Forum/resources.php?c=1&cid=17&year=1997

[hide]
First, $f(x)$ gets arbitrarily large as $x\rightarrow \infty$, so $f(x)>0$ for all nonnegative $x$. In particular, $b=f(0)>0$. Now letting $f(x)=bg(x/\sqrt{b})$ and $k=-a\sqrt{b}$, it suffices to prove the statement for $g(x)=x^{2}-kx+1$.

Since $f(x)>0$ for nonnegative $x$, we have $g(x)>0$ for nonnegative $x$. In particular, $g(1)>0\Rightarrow k<2$.

Now define a sequence $t_{n}$ by $t_{0}=k$ and $t_{n+1}=t_{n}^{2}-2$. Then we have (easily verified by difference of squares)
\[(x^{2}-t_{0}x+1)(x^{2}+t_{0}x+1)(x^{4}+t_{1}x^{2}+1)(x^{8}+t_{2}x^{4}+1)\cdots (x^{2^{n}}+t_{n-1}x^{2^{n-1}}+1) \]

\[=x^{2^{n+1}}-t_{n}x^{2^{n}}+1 \]
for all $n$. So we will be done if we can show that $t_{n}$ is not positive for some $n$. This is clearly true if $k$ not positive. For positive $k$, let $k=2\cos \alpha$ where $0<\alpha <\frac{\pi}{2}$. Since
\[(2\cos \theta)^{2}-2=2\cos 2\theta \]
we obtain $t_{n}=2\cos 2^{n}\alpha$ for all $n$.

Since $2^{n}\alpha$ becomes arbitrarily large for large enough $n$, there are integers $n$ such that $2^{n}\alpha \geq \frac{\pi}{2}$. If we consider the least $m$ such that $2^{m}\alpha \geq \frac{\pi}{2}$, then $2^{m}\alpha <\pi$ so $t_{m}$ is negative, as desired.
[/hide]


I wonder if they run some statistical data about IMO problems after the contest.
Especially,I'm curius to know  parameters like:
Scored Average number of points per  problem and contestant for each problem during
last 10 Olympiads  .

:thumbup:
Wow!That is exactly what I was looking for!
Thank you Tomci0.

[quote="TomciO"]http://www.srcf.ucam.org/~jsm28/imo-scores/

Great site.[/quote]

Really cool site!!!!


algebra
linear equation
geometry theorems
geometry

[quote="dondigo"]Is there any nice condition on three lines to be concurrent in terms of complex numbers :?:[/quote]

Well, if you have six points $a$, $b$, $c$, $x$, $y$, $z$ in $\mathbb{C}$ - here, I identify every point with its affix (i. e. the complex number corresponding to this point) -, then you can find out whether the lines $ax$, $by$, $cz$ concur by writing out the equations of these lines (for instance, a point $t$ lies on the line $ax$ if and only if $\frac{a-t}{x-t}=\frac{\overline{a}-\overline{t}}{\overline{x}-\overline{t}}$; this easily becomes a linear equation in $t$ and $\overline{t}$) and checking whether the common solution of two of these equations satisfies the third one (note that you can treat $t$ and $\overline{t}$ as two independent variables). Oops, I see you said "nice" and probably a computation that pwns my CAS is not what you call nice... then I am not sure whether there is a "nice" general concurrence criterion at all. However, for concyclic points, there is the following criterion (basically a complex version of http://www.mathlinks.ro/Forum/viewtopic.php?t=102695 ):

[color=blue]If $a$, $b$, $c$, $x$, $y$, $z$ are six points on one circle, then the lines $ax$, $by$, $cz$ concur if and only if

$\frac{b-x}{c-x}\cdot\frac{c-y}{a-y}\cdot\frac{a-z}{b-z}=1$.[/color]

  Darij


find all functions $ f : N \to N$that :
\[ xf(y)\plus{}yf(x)\equal{}(x\plus{}y)f(x^2\plus{}y^2)\]

$ P(x,x): f(2x^2)\equal{}f(x)$.
Define series $ x_n$ with $ x_1\equal{}1$ and $ x_{n\plus{}1}\equal{}2x_n^2$.
We see $ f(x_n)\equal{}f(1)$.
Let $ a\equal{}f(1)$ and $ y$ fixed.
We have, for natural $ n$,
$ x_n\plus{}y|x_nf(y)\plus{}yf(x_n)\Rightarrow x_n\plus{}y|(x_n\plus{}y)f(y)\plus{}y(a\minus{}f(y))\Rightarrow x_n\plus{}y|y(f(y)\minus{}a)$
Obviously we can always find $ x_n$ such that $ x_n>y(f(y)\plus{}a)$.
Then $ x_n\plus{}y|y(f(y)\minus{}a)\Rightarrow y(f(y)\minus{}a)\equal{}0\Rightarrow f(y)\equal{}a$.
Since $ y$ is arbitrary, $ f(y)\equal{}f(1)$.

in this cases you should precize if 0 is element of N or not because it might be essential for the solution


Are you sure it is solved problem? I have a feeling I've seen it recently in "Unsolved problems in number theory" by R.K. Guy  :maybe:
Just asking to be sure  :)

ah - so it gotta be something else then  :P

Edit: probably - find all $p$ such that $p^{2}| (p-1)!+1$  :)

[quote="Andrej_pp"]Prove that if p is a prime greater than 5, then
(p-1)! + 1 is not a power of p.[/quote]

Supose that $(p-1)!+1=p^{k}$.
If $k=2m+1$ then $(p-1)!=(p-1)(1+p+...+p^{2m})$ results
$(p-2)!=1+p+...+p^{2m}$ imposible becouse  one side of equality is odd the other is even.

If $k=2m$ then $(p-1)!=(p^{2}-1)(1+p^{2}+...p^{2m-2})$ imposible becouse the right side is divisible with $p+1$ the left is not. So $(p-1)!\neq p^{k}$

[quote="cckek"]If $k=2m$ then $(p-1)!=(p^{2}-1)(1+p^{2}+...p^{2m-2})$ imposible becouse the right side is divisible with $p+1$ the left is not. So $(p-1)!\neq p^{k}$[/quote]
Let p=7, then p=1=8 and $8|6!$.  :D

ur right. sorry i was sleappy when i wrote it:)

For all prime p>5 $(p^{2}-1)|(p-1)!$. But can be $(p^{4}-1)\not |(p-1)!$.
When m is odd and p>7 it is easy to chek, that $ord_{2}((p-1)!)>ord_{2}(p^{2m}-1$. In these way can be proove $(p-1)!\not =p^{k}-1$.

By the way this fact is called Liouville's Theorem. I remember having proven it half a year ago and my solution was based on Hensel's lemma for prime $2$ (so the idea overlaps with Rust's one, I bet).

I still don't get what's wrong with my solution for the general case... :maybe:


algorithm
number theory
relatively prime
Euclidean algorithm

Can it be proven that the fraction $ \frac{x}{x\plus{}1}$ can't be simplified or reduced?

Assuming $ x\in\mathbb{N}$, then it can only be reduced when the numerator and denominator are not relatively prime, but $ gcd(x \plus{} 1,x) \equal{} gcd(x,1) \equal{} 1$ by the Euclidean Algorithm, so it can't be reduced.

Seeing as you never said x is an integer, you can!


Find the minimum of
\[ \log_{x_1}{(x_2 - 1/4)} + \log_{x_2}{(x_3 - 1/4)} + \ldots + \log_{x_n}{(x_1 - 1/4)}  \]
where $x_1, x_2, \ldots, x_n$ are real numbers on the interval (1/4, 1).

[quote="#20002 pwns"][hide=hmmm...]my answer is $\boxed{n}$ (move around terms and then mass cancellation and min; choose n 1/4's and eventually log_(1/4) of 1/4^n = n.... do you want proof?

Define $\{a_i\}=\log_{x_i}{(x_{i+1 \ \text{(mod n)}}-\frac14)}$. We want to find $\min \sum{a_i}$
We have: ${x_i}^{a_i}=(x_{i+1 \ \text{(mod n)}}-\frac14)$. Multiplying together the $n$ equalities this generates, we have:
$\prod{{x_i}^{a_i}=\prod{x_i - \frac14}}$
Now here is where I make the intuitive step I can't prove: since all $x_i$ are indistinguishable, points of interest can only occur when they are all equal. So we assume $x_i=x_j \forall j \in \{1,n\}$. This yields:
${x_1}^{\sum{a_i}}={(x_1-\frac14)}^n$. Now $\sum{a_i} \text{increases/decreases with } {x_1}^{\sum{a_i}}$. The rest follows[/quote]

Could you hide your answers please?

Heh actually its o.k. because its pretty incomprehensible anyway.


Find $ \displaystyle\sum_{n \equal{} 1}^{\infty} \frac{n}{2^n}$

Another
[hide]Evaluate $ \sum_{n\equal{}1}^\infty n x^{n\minus{}1}$, then substitute $ x\equal{}1/2$.
This series is easy because it is the derivative of $ \sum_{n\equal{}1}^\infty x^n$.
[/hide]

hello, or prove by induction
$ \sum_{n\equal{}1}^k\frac{n}{2^n}\equal{}\minus{}2\left(\frac{1}{2}\right)^{k\plus{}1}(k\plus{}1)\minus{}2\left(\frac{1}{2}\right)^{k\plus{}1}\plus{}2$
Sonnhard.


I really really really need AMC 8 2007, i am willing to trade any test of AMC 8 available upto 2005. (with solutions)

dude

come on

heres the link

and go home happy

http://www.math.ksu.edu/main/handbook/ProblemSets

u are welcome

He wanted the 2007 AMC 8 and if you actually bothered to go to where you suggested, the 07 AMC is not there.

[quote="Yankzgodzilla55"]http://www.artofproblemsolving.com/Wiki/index.php/2007_AMC_8

2007[/quote]

Only 2 problems are up.

this is a disaster

anything i post in this topic, goes wrong

i will find a website and give it to you

[quote="pythag011"]
Only 2 problems are up.[/quote]

Which brings up a really good point. EDIT IT!! Put more problems up!!!

Er...I've go the AMC 8 2007.  It's just in paper format.


Let a,b,c be positive real numbers. Prove that ab/(c(c+a)) + bc/(a(a+b)) + ca/(b(b+c)) >= a/(c+a) + b/(a+b) + c/(b+c).
Who can solved?

Denote x=a/b, y=b/c, z=c/a and the inequality becomes:
sum[ (y-1)/(z+1)] >= 0 or xy^2+yz^2+zx^2+x^2+y^2+z^2>=x+y+z+3 where xyz=1.
By applying the AM-GM inequality we have that xy^2+yz^2+zx^2>=3 and x^2+y^2+z^2>=[(x+y+z)^2]/3>=x+y+z, because x+y+z>=3.

equivalent to $\sum[\frac{ab}{c(c+a)}+\frac{b}{c+a}]\geq \sum\frac{a+b}{c+a}$, so $\sum\frac{b}{c}\geq \sum\frac{a+b}{c+a}$. This is a problem from Mathlinks contest.

[quote="hxtung"]Let $ a,b,c$ be positive real numbers. Prove that

$ \frac {ab}{c(c \plus{} a)} \plus{} \frac {bc}{a(a \plus{} b)} \plus{} \frac {ca}{b(b \plus{} c)}\geq \frac {a}{c \plus{} a} \plus{} \frac {b}{a \plus{} b} \plus{} \frac {c}{b \plus{} c}.$[/quote]Using the arithmetic-mean - geometric-mean inequality,

$ \prod_{sym}{a(b \plus{} c)}\left[\sum_{cyc}\frac {ab}{c(c \plus{} a)} \minus{} \sum_{cyc}\frac {a}{c \plus{} a}\right]$

$ \equal{} \sum_{cyc}b^2c^4 \plus{} \sum_{sym}b^3c^3 \minus{} \sum_{cyc}ab^2c^3 \minus{} 3a^2b^2c^2$

$ \equal{} \sum_{cyc}\left(\frac {4b^2c^4 \plus{} c^2a^4 \plus{} a^2b^4}{6} \minus{} ab^2c^3\right) \plus{} \sum_{sym}b^3c^3 \minus{} 3a^2b^2c^2\geq 0.$

[b]Remark[/b] $ \sum_{cyc}b^2c^4\geq\sum_{cyc}ab^2c^3$ holds for all real numbers $ a,b,c$,

$ \sum_{sym}b^3c^3 \minus{} 3a^2b^2c^2 \equal{} \sum_{sym}{bc}\left(\sum_{sym}b^2c^2 \minus{} \sum_{sym}a^2bc\right)\geq 0$ holds for $ \sum_{sym}{bc}\geq 0$,

and therefore the original inequality holds for real numbers $ a,b,c$ such that

$ bc \plus{} ca \plus{} ab\geq0$ and $ abc(b \plus{} c)(c \plus{} a)(a \plus{} b)\neq 0$.

Another proof see :  http://www.mathlinks.ro/Forum/viewtopic.php?t=102568


Θέλω δύο γραμμών λύση: :wink:

$ x\geq y \geq z > 0$ and $ x^2 \plus{} y^2 \plus{} z^2 \equal{} 3$

Prove:

$ \frac{x}{z^2} \plus{}\frac{ y}{x^2} \plus{}\frac{ z}{y^2} \geq\frac{ x^2}{z} \plus{}\frac{ y^2}{x} \plus{}\frac{ z^2}{y}$

πολλ. LHS με $ \frac{3}{3}\equal{}\frac{x^2\plus{}y^2\plus{}z^2}{3}$ και μετά Tschebytsev

[quote="maniopas"]πολλ. LHS με $ \frac {3}{3} \equal{} \frac {x^2 \plus{} y^2 \plus{} z^2}{3}$ και μετά Tschebytsev[/quote]

Το εξηγείς λίγο παραπάνω αυτό? Γιατί δεν ξέρουμε διάταξη των όρων του LHS (αλλά και να ξέραμε δεν καταλαβαίνω πως προκύπτει το ζητούμενο)... δεν μου φαίνεται να βγαίνει έτσι :huh:

Είπαμε 2 γραμμές αλλά όχι κι έτσι. :)  Δεν είναι σωστός ο Τσεμπύτσεφ δυστυχώς... Είναι πολύ πιο sharp από αυτό. :wink:

Ρε Κώστα , ρε Κώστα  :P  . Γιατί τα ταλαιπωρείς τα παιδάκια με Vietnam 1991 kai Vasc ταυτόχρονα ??  :wink:
Μιας και το φόρουμ έχει πολλούς one-liners πλέον (δείτε 2ο ποστ) ας γίνω και εγώ  :P
Τώρα ισχύει ότι $ (x^2\plus{}y^2\plus{}z^2)^2\geq 3(x^3y\plus{}y^3z\plus{}z^3x)$ και $ \frac{x^2y}{z}\plus{}\frac{y^2z}{x}\plus{}\frac{z^2x}{y}\geq x^2\plus{}y^2\plus{}z^2$

[quote="silouan"]$ \frac {x^2y}{z} \plus{} \frac {y^2z}{x} \plus{} \frac {z^2x}{y}\geq x^2 \plus{} y^2 \plus{} z^2$[/quote]

παιδια δεν ξερω αν ειναι σωστο γιατι λογο της μεγαλης ποσοτητας των πραξεων δεν εκανα επαληθευση (οποτε μπορει να μου εχει ξεφυγει κατι), αλλα νομιζω οτι η παραπανω ισοδυναμα γραφεται:

$ y^3(x \minus{} z)(y \minus{} z) \plus{} zy(xy \minus{} y^2)^2 \plus{} x(y \minus{} z)(x \minus{} y)[y(x \minus{} z) \plus{} (y \minus{} z)(y \plus{} z)] \geq 0$ η οποια ισχυει. Θα ηθελα οποιος γνωριζει πως ακριβως αποδεικνειεται η παραπανω να μου πει αν αποδεικνειεται ετσι οπως το εκανα εγω.

[quote="silouan"]Νίκο κάπως έτσι ήταν η επίσημη λύση τους . Εγώ είχα βρει μια άλλη απόδειξη
http://www.mathlinks.ro/viewtopic.php?t=137623&search_id=1754682743&start=20  (πρόβλημα 11)[/quote]

Ωραια λυση Σιλουανε :)
Και εγω πρωτου να βρω την παραπανω αποδειξη, ειχα προσπαθησει να αποδειξω αυτο που αποδεικνειεις και εσυ στην αρχη της λυσης σου (οτι δηλαδη η παρασταση του αριστερου μελους ειναι μεγαλυτερη η ιση απο την ιδια με αντιστροφους τους συντελεστες των $ x^2, y^2, z^2$) και μετα με ΑΜ-ΓΜ προκειπτει αμεσως η ζητουμενη, αλλα δεν το εκανα καλα και δεν με οδηγησε πουθενα αυτο...

H λύση του Sil είναι ιδιαίτερα διδακτική ως μεθοδολογία γιατί συμμετρικοποίησε την ανισότητα. Εξαιρετική ιδέα. :)


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Wow.. addictive game.. anyone got any hints or tips? im stuck.  I know that 9 has something to do with what JAY PACK sounds like out loud (i know what it sounds like, im just not giving that away) but i dont know what to do with that.

EDIT never mind i think i got it!! Either way, any hints?

did you really get that JAY PACK? hmm im not sure if im a good hinter in these riddles: it's related to music, it is hinted by title and somewhere in the source code tells you. also, the password hint tells you exactly which one you should get according to source code...

im stuck at level 16 and Miths is stuck at ... i dunno what level, he never said it :)

[quote="Palytoxin"]Miths is stuck at ... i dunno what level, he never said it :)[/quote]

Haha it's ridiculously low. I don't think you guys want to know.

[quote="MithsApprentice"]Haha it's ridiculously low. I don't think you guys want to know.[/quote]
well, there's nothing to brag about the levels, they are very hard, and it's all about fun. so you can tell us where and maybe we can help you with hints tips

[quote="xdl615x"]Is there any inappropriate material in that game at all? I'm afraid to access it from the library computer...[/quote]

 :rotfl: i would have been long gone banned from MathLinks if i ever do that, so no, there is no inappropriate materials on the website BUT there might be one or two 'f' or 's' words creeping around, that's pretty common :D

[quote="xdl615x"]Is there any inappropriate material in that game at all? I'm afraid to access it from the library computer...[/quote]

Not that I know of.

I'm at jay pack for now.

[quote="NoSoupForYou"]I'm stuck at level 6.  :D[/quote]
wow level 6 is tough!!!
ok did you read the title? go do that and make sure you gather all important info (clues after clues)

To Miths: i hope you find my hints helpful in post #2 :)

[quote="ToBe"]Yeah.. I'm sure i got the Jay Pack lol its easy for someone who works with.. that stuff a lot.

right now im stuck on 'stereo' with the two images and the screwed up dimensions and the source code hint "GIF him a chance to show himself" there was a video and it showed the dragon distorting but dragon isnt the answer and theres nothing to convert to a GIF...[/quote]
the text below the pic is the biggest hint and look at the pic, who/what is screwing up the dimensions :D

11. Giving up for the day. I need some down time. This junk so too...uh...time-consuming.

Oh actually. Ha. I guessed that one and now I'm on 12. Yay!

:)

Dude 11 is weird.

lol yeah i just guessed on the dimensions and i got it..a fter like a half hour of trying to figure it out.

seriously, when you guys get to level 16, please reply!!! i've been stuck for 2 days already and can't figure out what those numbers mean!!!

If I remember correctly, one of the things you mentioned above actually leads to the solution, so try to do it more carefully. Maybe I will be able to say more when I am back home where I have GIMP installed.

  Darij

Ok, I got to the Morse Code one but I don't feel like searching it up and decoding. Can someone just tell me what it is?

No.
If you don't even want to do this, abandon NotPron...
And by the way, I'm strictly against big spoilers, especially if out of fooliness!

for level 8, is the other type of jay pack, j-peg, as is jpg, as in the file type?

[hide="At abacadaea"]I think so.  However, I'm not sure of the answer.  I found this:[hide="Spoiler"]mus2.jpg[/hide]  According to mathnerd314, I'm on the right track.  I'm not quite sure how the "water became wine" thing comes into play.  A Google search of the phrase "water became wine" yields two possibilities:

1. It's a song.
2. It's a Biblical reference.

I'm not sure how these tie together.  I'll accept any hints.[/hide]

On the other hand, I could just write a program that can brute-force through every single possibility for the username and password. :P

I want to see a program checking about $ 10^{30}$ combinations in human time...

[quote="lingomaniac88"][hide="At abacadaea"]water became wine[/hide][/quote]

[hide="Well,"]Mary Poppins is not it, for Bread and Water turns into [b]Tea[/b] and cakes.

I really think that it is a biblical reference. I really need to read it a lot more carefully.[/hide]

online riddle game..
i tried it..
it's impossible for a person like me...