integration
real analysis
real analysis unsolved

Let $ X \subset W_0^{1,2} \left( \Omega  \right)$ with $ \dim X < \infty$. Why are the norms
\[ \begin{gathered}
  \left\| u \right\| \equal{} \left( {\int\limits_\Omega  {\left| {\nabla u\left( x \right)} \right|^2 dx} } \right)^{\frac{1}
{2}}  \hfill \\
  \left\| u \right\|_p  \equal{} \left( {\int\limits_\Omega  {\left| {u\left( x \right)} \right|^p dx} } \right)^{\frac{1}
{p}} \left( {1 < p < 2^* } \right) \hfill \\
  \left\| u \right\|_{2^* }  \equal{} \left( {\int\limits_\Omega  {\left| {u\left( x \right)} \right|^{2^* } dx} } \right)^{\frac{1}
{{2^* }}}  \hfill \\
\end{gathered}
\]
equivalent on $ X$? Why are these norms not equivalent on the whole space $ X \equal{} W_0^{1,2} \left( \Omega  \right)$.

All norms on finite-dimensional spaces are equivalent. As for examples, they are not hard to find: a localized version of $ u(x)\equal{}|x\minus{}x_0|^{\minus{}\alpha}$ belongs to some $ L^p$ spaces but not to others. It cannot tell the difference between $ W^{1,2}$ and $ L^{2^*}$ (where $ 2^*\equal{}\frac{2n}{n\minus{}2}$), but I think $ |x\minus{}x_0|^{\minus{}\alpha}|\log|x\minus{}x_0\parallel{}^{\beta}$ will do the job.


Firstly derive the Henderson-Hasselbalch (not actually part of the question).

1. Calculate the $ pH$ of a buffer solution, which contains $ 0.200M$ formic acid ($ K_a \equal{} 2.1 \times 10^{ \minus{} 4}$) and $ 0.150M$ sodium formate.

2. Calculate the change in $ pH$ of the buffer solution in question 1 when $ 0.01000M$ of sodium hydroxide is added to the solution.

3. Calculate the volume of $ 0.200M$ of sodium hydroxide which must be added to $ 100.0 cm^{3}$ of $ 0.150M$ of acetic acid ($ K_a \equal{} 1.8 \times 10^{ \minus{} 5}$) in order to prepare a buffer solution with $ pH \equal{} 5.00$.

4. The $ pH$ of a buffer solution containing $ 0.0100M$ of benzoic acid ($ K_a \equal{} 6.6 \times 10^{ \minus{} 5}$) and $ 0.0100M$ of sodium benzoate is: a) $ 5.00$ b) $ 4.18$ c) $ 9.82$, d) $ 9.00$

In the problems that follow we will mix equal volumes of $ A$ and $ B$. Find he $ pH$ of each mixture and conclude whether each solution is a) weak acid b) strong base c) buffer solution d)none of the above.

5. $ \text{A: 0.100M} CH_3COOH$, $ \text{B: 0.0500M, NaOH}$

6. $ \text{A: 0.100M} CH_3COOH$, $ \text{B: 0.150M, NaOH}$

7. $ \text{A: 0.150M} CH_3COOH$, $ \text{B: 0.100M, NaOH}$

8. $ \text{A: 0.100M} CH_3COOH$, $ \text{B: 0.100M, NaOH}$

Do you know how to answer these questions or are they just for others to practice? In the first case, where do you/others have doubts?

Fr some reason I got 4 wrong. I don't know why.

When a buffer solution is composed by equal concentrations of a weak acid/base and its conjugate base/acid, then its pH is simply pKa (or pKb). Now you are to prove this.


geometry
circumcircle
symmetry
trigonometry
cyclic quadrilateral
trig identities
Law of Cosines

Let P be any point on the circumcircle of an equilateral triangle ABC of side length $ a$.
prove that $ PA^2\plus{}PB^2\plus{}PC^2 \equal{} 2a^2$

[img]http://www.part.lt/img/476e5b7e8e887d303c19f5fe578c7d67952.JPG[/img]

We have
$ A(0,\frac{a\sqrt{3}}{3}),$
$ B(\frac{a}{2},\minus{}\frac{a\sqrt{3}}{6}),$
$ C(\minus{}\frac{a}{2},\minus{}\frac{a\sqrt{3}}{6}),$
$ P(x,y),$  $ (x^2\plus{}y^2\equal{}\frac{a^2}{3})$

$ PA^2\equal{}x^2\plus{}(\frac{a\sqrt{3}}{3}\minus{}y)^2\equal{}x^2\plus{}\frac{a^2}{3}\minus{}\frac{2ay\sqrt{3}}{3}\plus{}y^2$
$ PB^2\equal{}(\frac{a}{2}\minus{}x)^2\plus{}(\minus{}\frac{a\sqrt{3}}{6}\minus{}y)^2\equal{}\frac{a^2}{4}\minus{}ax\plus{}x^2\plus{}\frac{a^2}{12}\plus{}\frac{ay\sqrt{3}}{3}\plus{}y^2$
$ PC^2\equal{}(\minus{}\frac{a}{2}\minus{}x)^2\plus{}(\minus{}\frac{a\sqrt{3}}{6}\minus{}y)^2\equal{}\frac{a^2}{4}\plus{}ax\plus{}x^2\plus{}\frac{a^2}{12}\plus{}\frac{ay\sqrt{3}}{3}\plus{}y^2$

Adding up all equations we get $ PA^2\plus{}PB^2\plus{}PC^2\equal{}3(x^2\plus{}y^2)\plus{}a^2\equal{}2a^2$

[hide]Due to the symmetry of the equilateral triangle, we can assume that $ P$ is on minor arc $ AB$ without loss of generality. First, Ptolemy's Theorem on cyclic quadrilateral $ PABC$ gives
\[ PA \cdot BC + PB \cdot CA = PC \cdot AB \iff PA \cdot a + PB \cdot a = PC \cdot a \iff PA + PB = PC.
\]
Then, since arcs $ AC$ and $ BC$ have a measure of $ 120$, we immediately find measures of their inscribed angles: $ m \angle APC = m \angle BPC = 60$. Using law of cosines,
\begin{eqnarray*} BC^2 & = & PB^2 + PC^2 - 2 \cdot PB \cdot PC \cdot \cos 60^\circ \\
CA^2 & = & PA^2 + PC^2 - 2 \cdot PA \cdot PC \cdot \cos 60^\circ \end{eqnarray*}
and adding them,
\begin{eqnarray*} BC^2 + CA^2 & = & PA^2 + PB^2 + 2PC^2 - PA \cdot PC - PB \cdot PC \\
& = & PA^2 + PB^2 + PC^2 + PC (PC - PA - PB)\end{eqnarray*}
which, since $ BC = CA = a$ and $ PC = PA + PB$, is equivalent to
\[ 2a^2 = PA^2 + PB^2 + PC^2.
\]
[/hide]


For $x\geq 0 $ and $0\leq a,b,c \leq x$ prove that

$\frac{a+b}{x^2-ab}+\frac{b+c}{x^2-bc}+\frac{c+a}{x^2-ca} \leq \frac{2a}{x^2-a^2}+\frac{2b}{x^2-b^2}+\frac{2c}{x^2-c^2}$

Since
\[
\frac{a}{x^2 - a^2} + \frac{b}{x^2 - b^2} - \frac{a+b}{x^2 - ab} = \frac{x^2(a+b)(a-b)^2}{(x^2-a^2)(x^2-b^2)(x^2-ab)} \geq 0
\]
we have
\[
\frac{a}{x^2 - a^2} + \frac{b}{x^2 - b^2} \geq \frac{a+b}{x^2 - ab}
\]

Adding the three similar ineq yields the result.

Easier:
put x=1, A = tan :beta: , B = tan :theta: , C = tan :gamma: , then 0< :beta: , :theta: , :gamma: <pi/4
now use tan2 :beta: +tan2 :theta: >= tan((:beta: + :theta: )/2)

Ximik, it's a very nice solution. I think you meant:

tan2A+tan2B >= 2tan(A+B).

Misha


Suppose I have written code for two different diagrams in Asymptote.  Suppose I then want to place the two diagrams side-by-side, as one Asymptote picture.  I guess I could change all the coordinates of one of the diagrams, but I was hoping there was a simpler way.

This is probably a simple question, but I am an Asymptote newbie.

Do it in the same way as you would with any two pictures, put them both in a minipage and then they will be side by side.

Or look at what I did here:
[asy]size(200);
picture P,Q;
draw(P,(10,10)--(0,30)--(40,40)--cycle);
draw(Q,(10,10)--(0,30)--(40,50)--cycle);
add(P);
add(shift((50,0))*Q);[/asy]
The drawback is that this trick will require putting the explicit picture names in all drawing commands. Does anybody know if one can avoid it in some clever way?.

Thanks, steve.  There was a technical reason I wanted just one Asymptote picture, but I found a workaround.

fedja, I think you can do something like this:
[code]<draw second picture>
picture pic = shift(50, 0) * currentpicture;
currentpicture.erase();
add(pic);
<draw first picture>
[/code]

[asy]size(200);
draw((0,0)--(10,10),green);
picture pic = shift(20, 0) * currentpicture;
currentpicture.erase();
draw((0,0)--(10,10),red);
add(pic);[/asy]
Yes, Ravi, that works! Good concatenation trick. :)


calculus
integration
function
trigonometry
calculus computations

how do you do
$ \int_{ - \infty}^{\infty}\frac {cos(x)dx}{1 + x^2}$?

what I did was
$ \int_{ - \infty}^{\infty}\frac {cos(x)dx}{1 + x^2}=\int_{ - \infty}^{\infty}\frac {e^{ix}dx}{1 + x^2}$

$ \frac {1}{x - i}\frac {e^{ix}}{x + i}$

$ \frac {1}{x - i}\frac {e^{i(x - i) - 1}}{2i + x - i}$

$ \frac {e^{i(x - i)}}{e(x - i)}\frac { - .5i}{1 - .5i(x - i)}$

${ - .5ie^{ - 1}\frac {1}{x - i}\sum_{k = 0}^{\infty}\frac {i^k(x - i)^k}{k!}\sum_{k = 0}^{\infty}(.5i)^k(x - i)^k}$

if you multiply it out, the $ (x - i)^{ - 1}$ coefficient is $ \frac { - i}{2e}$
then I multiplied by $ 2i\pi$ and got $ \frac {\pi}{e}$

Is this right? It seems like this would give the integral along a contour, not along the real number axis like in the original problem.

in an even function $ f(x)$ we have $ \int_{\minus{}a}^{a} f(x) dx\equal{}0$
I think the function in this problem is even..
so the integral would be $ 0$

sorry if I'm wrong :blush:

[quote="mhar_teens"]in an even function $ f(x)$ we have $ \int_{ \minus{} a}^{a} f(x) dx \equal{} 0$[/quote]

Sorry, you are wrong. in an [color=red]odd[/color] function.

ah,,sorry for my mistake..
sorry..

the problem would be::
$ 2\int_0^{\infty} \frac {\cos{x}}{1 \plus{} x^2} dx$
maple said the the result would be:
$ \pi \cosh{1} \minus{} \pi\sinh{(1)}$

Let's find the cossinus Fourier transform of the function $ f(x) \equal{} e^{ \minus{} \beta x}$. Answer:

$ \int_0^{ \plus{} \infty}\frac {\cos(\alpha x)}{x^2 \plus{} \beta^2}\,dx \equal{} \frac {\pi e^{ \minus{} \alpha \beta}}{2 \beta}$

by residue .. consider something $ f(z) \equal{} \frac {e^{\alpha i z}}{z^{2} \plus{} a^{2}}$

That's right: we can also calculate the integral of $ f(z) \equal{} \frac{e^{iz}}{z^2 \plus{} a^2}$ around an half pie, and then consider the real part.

I think that's pretty much what I tried to do


advanced fields
advanced fields unsolved

Let $A\subset\mathbb R$ be closed with empty interior. Is it mandatory that $A$ has measure zero?

No. There is a classical counter-example for that called "the Borel set".

Let $(q_{n})_{n\ge 1}$ be an enumeration of the rationals, and take our set to be $\mathbb R\setminus\bigcup_{n\ge 1}I_{n}$, where for every $n\ge 1,\ I_{n}$ is an open interval of length $\frac 1{2^{n}}$ containing $q_{n}$.


calculus
integration
trigonometry
logarithms
calculus computations

Calculate the following indefinite integrals.

[1] $\int \sin x\sin 2x dx$

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

[3] $\int \frac{\tan ^2 x}{\cos ^2 x}dx$

[4] $\int \frac{e^x+e^{-x}}{e^x-e^{-x}}dx$

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

[quote="liyi"]Hints

[1][hide]
$\int \sin x\sin 2xdx = \int \sin x \cdot 2\sin x\cos xdx = 2\int \sin^2 x d(\sin x)$
[/hide]

[2][hide]
$\int \frac{e^{2x}}{e^x-1}dx = \int \frac{e^x}{e^x-1}d(e^x)$
[/hide]

[3][hide]
$\int \frac{\tan^2 x }{\cos^2 x}dx = \int \tan^2 xd(\tan x)$
[/hide]

[4]
[hide]$d(e^x-e^{-x}) = (e^x+e^{-x})dx$ [/hide]

[5] [hide]$ = \int \frac{d(e^x)}{e^x+1}$ [/hide][/quote]
i used the same method for $1$ and $3$.

liyi, for i shouldn't be $\cos{x} dx$?
the answer for 1 is:
$\frac{2\sin^3{x}}{3}$

[quote="amirhtlusa"]liyi, for i shouldn't be $\cos x dx$?
[/quote]
What do you mean?

nothing, your's is fine, i made a mistake :D

Here is my answer.

[hide]

[1] Using $\sin 2\theta=2\sin \theta\cos \theta$,

$\int \sin x\sin 2x dx=2\int \sin^ 2 x\cos xdx=2\int \sin ^ 2 x (\sin x)'dx=\boxed{\frac{2}{3}\sin ^ 3 x +C}$

[2] $\int \frac{e^{2x}}{e^x-1}dx=\int \frac{e^x}{e^x-1}d(e^x-1)=\int \left(1+\frac{1}{e^x-1}\right)d(e^x-1)$

$=\boxed{e^x+\ln |e^x-1|+C}$

[3] $\int \frac{\tan ^2 x}{\cos ^2 x}dx=\int \tan ^2 x\  (\tan  x)'dx=\boxed{\frac{1}{3}\tan ^ 3 x +C}$

[4] Using $\int \frac{f'(x)}{f(x)}dx=\ln |f(x)| +C$,

$\int \frac{e^x+e^{-x}}{e^x-e^{-x}}dx=\int \frac{(e^x-e^{-x})'}{e^x-e^{-x}}dx=\boxed{\ln |e^x-e^{-x}|+C}$

[5] $\int \frac{e^x}{e^x+1}dx=\int \frac{(e^x+1)'}{e^x+1}dx=\boxed{\ln (e^x+1)+C}$

[/hide]

[quote="kunny"]Calculate the following indefinite integrals.

[3] $\int \frac{\tan ^2 x}{\cos ^2 x}dx$

[/quote]
Generalization 1
Calculate
$\int \frac{\sin ^n x}{\cos ^{n+2} x}dx$


vector
linear algebra
matrix
abstract algebra
group theory
superior algebra
superior algebra unsolved

How to show $M_{n}(\mathbb{R})$ and $M_{m}(\mathbb{R})$ are/aren't isomorphic as additive groups?

What if we replace the real numbers with another field?

[quote="jmerry"]They are isomorphic, because they're isomorphic as $\mathbb{Q}$-vector spaces.

For $M_{n}(K)$ and $M_{m}(K)$, let $F$ be the smallest field with the same characteristic as $K$. If the degree of $K$ over $F$ is finite, the groups are not isomorphic. If the degree of $K$ over $F$ is infinite, they're isomorphic.[/quote]

Why do deliberatly want to see them as $\mathbb{Q}$-vectorspaces?  :huh:  I'm not following at all...

These matrix spaces are IR vector spaces by construction. If we restrict scalar multiplication to scalars from IQ, they become IQ vector spaces.
Now their dimension as IQ vector spaces equals the cardinality of IR, so the two matrix spaces have the same dimension. Therefore they are isomorphic as IQ vector spaces.

The general trick is that the K vector space M_n(K) is isomorphic to K^(n^2), and if K has infinite degree over its prime field F, then the dimensions of K and K^m (at least for any finite m) as F vector spaces are equal.

I have not yet thought about why the additive groups are not isomorphic when the degree K:F is finite.

LiMa

If that degee is finite, the two spaces have different dimension as $F$-vector spaces. A set $x_{i}$ of nonzero elements of $V$ is linearly independent over $F$ if whenever $\sum a_{i}x_{i}=0$ for integers $a_{i}$ (all but finitely many zero), $a_{i}x_{i}=0$ for each $i$. Integer multiplication belongs to the abelian group structure; every abelian group is a $\mathbb{Z}$-module.
Therefore, if two vector spaces have different dimensions over $F$, they're not isomorphic as abelian groups.

On the other hand, $M_{n}(\mathbb R)$ and $M_{m}(\mathbb R)$ are not isomorphic as additive [url=http://en.wikipedia.org/wiki/Topological_groups]topological groups[/url] for any $n\ne m$. The reason is similar: they have different dimension as vector spaces over $\mathbb R$.


function
topology
combinatorics proposed
combinatorics

If $A$ and $B$ are two subsets of $\mathbb{R}^n$ that have fix point property then what about $A+B$ where $+$ is minkowski's sum.Has $A+B$ the fix point property?

A set in $\mathbb{R}^n$ have fix point property if every continious function from this set to itself have fix point.

If $A$ and $B$ are two subsets of $\mathbb{R}^n$ then $A+B=\{x+y: x\in A,y\in B\}$.

Saman Gharib

$A+B$ does not necessarily have the fix point property; in order to find a counterexample you can use the connectivity of the closed intervals in R. Take $A = [0,1] \times \{0\}$ and $B = \{0\} \times [0,1]$.
By the intermediate value theorem it is clear that both $A$ and $B$ have the fix point property, but $A+B = [0,1] \times [0,1]$ does not (consider $f(x,y)=(1-x,y^{2})$ for instance).

$[0,1]\times[0,1]$ does not have the fixed point property? :)

First of all, [url=http://www.scs.carleton.ca/~maheshwa/MAW/MAW/node3.html]Brouwer's Fixed Point Theorem[/url] says that it does. Second of all, we can clearly find a counterexemple to that function: just take $(x,y)=\left(\frac 12,0\right)$.

Sorry if I misunderstood.

$[0,1]\times[0,1]$ does have the fixed point property (it is topologically the same as a closed unit disk). For the particular mapping in [b]julien_santini[/b]'s post, the fixed points are $(0.5, 0)$ and $(0.5,1)$.

By the way, I'm not sure that this thread belongs to combinatorics. I would place it somewhere in the College Playground (perhaps, even in Advanced Fields...)

PS. Oops, grobber was faster to reply...  :)

I think it's nice problem to think about it. taste it.... :!:

Saman Gharib

Let $A$ be an arc of a unit circle $x^2+y^2=1,\,x\le 9/10$, $B$ be a disc $x^2+y^2\le 1/4$. Then $A+B$ is an $1/2$-neighbourhood of $A$, so is homeomorphic to a ring, and has not fixed point property.


linear algebra
matrix
vector
topology
geometry
3D geometry
sphere

Note that this only makes sense as the dimension of a manifold, since $ SO_n$ is not a linear space.

Let us examine the process of assembling an orthogonal matrix column by column.

The first column may be chosen arbitrarily except for one constraint: the norm must be 1. In other words, it must be selected as a point in the unit sphere of $ \mathbb{R}^n$. That one constraint means we have $ n\minus{}1$ degrees of freedom. In selecting the second column, we have two restrictions: it must be orthogonal to the first column (which can be expressed as one linear equation), and it must have norm 1. That gives us $ n\minus{}2$ degrees of freedom for that second column. Simliarly, we will have $ n\equal{}3$ degrees of freedom for the third column, and so on. For the last column, the only choice we have is a $ \pm$ sign (and if we want the determinant to be $ 1,$ even that choice is determined.) Hence the overall dimension is the total number of degrees of freedom we had with our choices, namely
\[ (n\minus{}1)\plus{}(n\minus{}2)\plus{}\cdots\plus{}1\plus{}0\equal{}\frac{(n\minus{}1)n}2.\]

This says that $ SO_2,$ which is isomorphic to the unit circle, has dimension $ 1$ (which is correct) and that $ SO_3$ has dimension $ 3.$

$ O_n$ has the same dimension as a manifold - except that $ O_n$ is not connected. It has two connected components, one of which is $ SO_n,$ the other of which is a coset of that subgroup. Each component has dimension $ \frac{(n\minus{}1)n}2.$

Now: can you describe the tangent space at the identity for $ SO_n?$ That is, the set of all matrices $ A$ such that $ \{e^{tA},t\in\mathbb{R}\}$ is a subgroup of $ SO_n?$ This should have the same dimension that we have just computed.

See also [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=238923]here[/url] for a more algebraic argument.

The easiest way to get at the tangent space is to differentiate the defining identity $ A^TA \equal{} I$.

[Edit- link fixed. I deleted the wrong part].

[quote="jmerry"]The easiest way to get at the tangent space is to differentiate the defining identity $ A^TA \equal{} I$.[/quote]
How about I throw around symbols with reckless abandon and pretend they mean something?

$ A^TA\equal{}I$

$ (dA)^TA\plus{}A^T(dA)\equal{}0.$ Now set $ A\equal{}I.$

$ (dA)^T\plus{}dA\equal{}0,$ which implies that $ dA$ is skew-symmetric.

So, the tangent space at the identity (Lie algebra) consists precisely of the skew-symmetric matrices. Since these are completely determined by their entries in the proper upper diagonal of the matrix, the dimension of this space is $ \frac{(n\minus{}1)n}2,$ as advertised.

Note that we only used $ A^TA\equal{}I$ and not that $ \det A\equal{}1.$ But because we're following the trajectories of the exponential paths, which are continuous, we are confined to the connected component of the identity - and that's $ SO_n$ rather than all of $ O_n.$

[quote="Nemion"]What is the dimension of $ SO_n$ the space of real orthogonal matrices with determinant $ 1$.[/quote]

Aldoo, was that last p-set that bad that you even tried using ML? :D

I hated the last Pset!!! jaja Tigraaaaaaan


Euler
modular arithmetic
algorithm
number theory unsolved
number theory

[b]Reminder for EULER[/b]

If $ m$ is a positive integer and $ a$ is an integer with $ (a,m) \equal{} 1$, then $ a^{\phi(m)} \equiv 1$ $ (mod m)$

[b]my question[/b]

Let say $ m$ is prime then $ \phi(m) \equal{} m \minus{} 1$. If there is an another integer $ d$ such that $ a^d\equiv 1$ then show that $ d \mid m \minus{} 1$ [b] (d is the smallest one)[/b]

[b]Example[/b]

For $ a \equal{} 10$ and $ p \equal{} 37$ Since $ (10 , 37) \equal{} 1$ then $ 10^{36}\equiv 1 (mod 37)$. If we check for $ 10^3$ then $ 10^3 \ \equiv 1 (mod 37)$ and $ 3 \mid 36$

[quote="ertanrock"][b]Reminder for EULER[/b]

If $ m$ is a positive integer and $ a$ is an integer with $ (a,m) \equal{} 1$, then $ a^{\phi(m)} \equiv 1$ $ (mod m)$

[b]my question[/b]

Let say $ m$ is prime then $ \phi(m) \equal{} m \minus{} 1$. If there is an another integer $ d$ such that $ a^d\equiv 1$ then show that $ d \mid m \minus{} 1$

[b]Example[/b]

For $ a \equal{} 10$ and $ p \equal{} 37$ Since $ (10 , 37) \equal{} 1$ then $ 10^{36}\equiv 1 (mod 37)$. If we check for $ 10^3$ then $ 10^3 \ \equiv 1 (mod 37)$ and $ 3 \mid 36$[/quote]

That's wrong.

Choose $ a\equal{}10$, $ m\equal{}37$ and $ d\equal{}15$. Then $ 10^{15}\equiv 1\pmod{37}$ while $ 15 | 36$ is wrong.

This becomes true if $ d$ is the [u][b]smallest [/b][/u]positive integer such that $ a^d\equiv 1$. Just suppose the contrary and use euclidian division of $ m\minus{}1$ by $ d$ to get a contradiction.

[quote="pco"][quote="ertanrock"][b]Reminder for EULER[/b]

If $ m$ is a positive integer and $ a$ is an integer with $ (a,m) \equal{} 1$, then $ a^{\phi(m)} \equiv 1$ $ (mod m)$

[b]my question[/b]

Let say $ m$ is prime then $ \phi(m) \equal{} m \minus{} 1$. If there is an another integer $ d$ such that $ a^d\equiv 1$ then show that $ d \mid m \minus{} 1$

[b]Example[/b]

For $ a \equal{} 10$ and $ p \equal{} 37$ Since $ (10 , 37) \equal{} 1$ then $ 10^{36}\equiv 1 (mod 37)$. If we check for $ 10^3$ then $ 10^3 \ \equiv 1 (mod 37)$ and $ 3 \mid 36$[/quote]

That's wrong.

Choose $ a \equal{} 10$, $ m \equal{} 37$ and $ d \equal{} 15$. Then $ 10^{15}\equiv 1\pmod{37}$ while $ 15 | 36$ is wrong.

This becomes true if $ d$ is the [u][b]smallest [/b][/u]positive integer such that $ a^d\equiv 1$. Just suppose the contrary and use euclidian division of $ m \minus{} 1$ by $ d$ to get a contradiction.[/quote]
Sorry $ d$ should be smallest.


For $a,b,c$ positive real numbers prove that

$a^2+b^2+c^2+3 \geq 2(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}) $

for am-gm applied to a+1, b+1, c+1, we have
a+b+c+3 >= 2(a+b+c).
for am-gm applied to a+b, b+c, c+a, we have
2(a+b+c) >=  :sqrt: ab + :sqrt: bc + :sqrt: ca.

which is our aim :)


function
vector
calculus
calculus computations

I finally figured out how to use ode45 for a single ODE correctly in matlab, thanks for those who helped me here!

I have another question is that once I found x(t) and y(t) by using matlab ode45, how to plot(x,y)?

note:I know how to plot function y=f(x) ,but my question is if I only got the range for x(t) and the numerical solution for x(t) and y(t)  , how to plot (x,y)?

details:         1〉x(t) belongs to [0，0.4]
                    2〉 x(t)satisfies the ODE:        x1=x
                                                                    x2=dx1/dt
                                                        dx2/dt=-(0.01/0.1)*(x2-4)^2
                    3〉y(t) satisfies the ODE:      y1=y;
                                                                   y2=dy1/dt
                                                           dy2/dt=-(1/0.1)*(0.1*9.8+0.001*abs(y2)*y2)
   How to solve this type of systems of ODEs by using ode45 and plot (x，y)？

Here's an example from MatLab in which a fragment of sin function is built.

x = -pi:.1:pi;
y = sin(x);
plot(x,y)

If you have vectors $ x(t)$ and $ y(t)$ both of the same length, then you can plot the graph by using the command $ \text{plot}(x,y)$. You don't have to have $ y$ expressed explicitly in terms of $ x$ for the $ \text{plot}$ command to work. MATLAB only cares about the dimensions of the matrices (vectors).


geometry
incenter
circumcircle
trigonometry
geometry proposed

The incircle $w=C(I,r)$ of the triangle $\triangle ABC\ (AB<AC)$ touches the line $BC$ in the point $D$ and the perpendicular line from the incenter $I$ on the line $AD$ meets the line $BC$ in the point $P$. I note the point $Q\in (AC)$ for which $\widehat {ABD}\equiv \widehat {AQD}$ and the intersection $R\in AB\cap QD$. [u]Prove[/u] that the middlepoint of the side $[BC]$ belongs to the circumcircle of the triangle $PQR$.

Denote by $M$ the midpoint of $[BC]$.
$\displaystyle{PD=\frac{2(p-b)(p-c)}{b-c}}$,
$\displaystyle{DM=\frac{(b-c)}{2}}$.
Hence $\displaystyle{PD\cdot{DM}=(p-b)(p-c)}$.
From $\displaystyle{\triangle{RBD}}$ we have
$\displaystyle{RD=\frac{(p-b)\cdot{\sin{B}}}{\sin{C}}}$ and from
$\displaystyle{\triangle{DQC}}$ we get: $\displaystyle{DQ=\frac{(p-c)\cdot{\sin{C}}}{\sin{B}}}$.
Consequently: $\displaystyle{RD\cdot{DQ}=PD\cdot{DM}}$ and we are done!


number theory
greatest common divisor
modular arithmetic
relatively prime

Are the numbers $ 2^{n}$ and $ 2^{m}$ relatively prime if $ n,m$ are natural numbers and $ n > m$ ?

Two numbers are relatively prime if their common divisor is 1. 2^n and 2^m and powers of 2, so they will always have a common divisor other than 1. Is there something I'm missing?

-rts2007

No, that's right. $ 2^n$ and $ 2^m$ will never be relatively prime when $ n > m > 0$.

[quote="rts2007"]2^n and 2^m and powers of 2, so they will always have a common divisor other than 1.[/quote]

Unless $ n$ and/or $ m$ is 0, then they will have a GCD of 1.
But $ n>m>0$, so they're both divisible by 2.

What about $ 2^{n}\minus{}1$ and $ 2^{m}\minus{}1$ ?

let $ n\equal{}4$ and $ m\equal{}2$. $ 2^4\minus{}1\equal{}15$, and $ 2^2\minus{}1\equal{}3$, which clearly have a common factor greater than $ 1$.

Hi,
No, $ 2^n \minus{} 1$ and $ 2^m \minus{} 1$ are not necessarily always relatively prime. One such example following $ n>m>0$ could be that $ n \equal{} 6$ thus giving us $ 2^6 \equal{} 64$ and $ m$ could be $ 4$ giving us $ 2^4 \equal{} 16$ Both $ 16 \equiv 64 \equiv 1\pmod{3}$ Therefore, $ 2^6 \minus{} 1, 2^4 \minus{} 1$ are $ 63$ and $ 15$ respectively which both have three as their (greatest? ) common factor.
Regards,
limac


search
number theory
linear algebra
linear algebra unsolved

$x_{1},x_{2},\dots,x_{n}$ are real number such that for each $i$, the set $\{x_{1},x_{2},\dots,x_{n}\}\backslash \{x_{i}\}$ could be partitioned into two sets that sum of elements of first set is equal to the sum of the elements of the other. Prove that all of  $x_{i}$'s are zero.
[hide="Hint"]It is a number theory problem.[/hide]

This is patently not true.  Consider the set $\{3,3,3\}$.  You might prove that $x_{1}=x_{2}=..=x_{n}$, which I expect might have to hold.

No, you cannot prove that either: take $1,1,1,3,3$. However, if we assume that $n$ is odd and that the two classes of the partition have the same cardinality ($(n-1)/2$) then it becomes true. It has been discussed several times on the forum, though.

Excuse me, In this problem $n$ must be odd, which I forgot to mention it. Then the problem is correct and the solution is similar to the solution given in perfect_radio's link.

[quote=perfect_radio]I finally found [url=http://www.mathlinks.ro/Forum/viewtopic.php?search_id=1183790031&t=75852]the thread[/url] harazi was talking about.[/quote]

I cannot use the thread. It is now unreadable.


A convex pentagon $ ABCDE$ is inscribed in a circle. The distances of $ A$ from the lines $ BC,CD,DE$ are $ a,b,c,$ respectively. Compute the distance of $ A$ from the line $ BE$.

[b]Lemma:[/b] $ABCD$ is cyclic and $ P$ is a point lying on its circumcircle. Let $ X,Y,Z,V$ the orthogonal projections of $P$ on sides $AD,$ $AB,$ $BC,$ $CD.$ Then $ PX \cdot PZ \equal{} PY \cdot PV.$

$\frac {PX}{PY} \equal{} \frac {\sin \widehat{ PAX}}{\sin \widehat{PAB}} \ , \ \frac {PZ}{PV} \equal{} \frac {\sin \widehat{PCB}}{\sin \widehat{ PCD}}$

$ \angle PAB \equal{} \angle PCD$ and $\angle PAX \equal{} \angle PCD$  implies that $ PX \cdot PZ \equal{} PY \cdot PV.$


Three numbers [i]p[/i], [i]q[/i] and [i]r[/i] are all prime numbers less than 50 with the property that [i]p + q = r[/i]. How many values of [i]r[/i] are possible.

[hide]Clearly, one of $ p,q$ must be 2. So we are looking for the amount of twin prime pairs below 50. They are 3,5; 5,7; 11,13; 17,19; 29,31; 41,43. So the answer is 6. (Unless I missed a twin prime pair.)[/hide]

Yes i count six as well. Follow up: prove that for primes p,q,r,  if P + Q = R , with Q > P, then R - Q = 2

[hide]this is simple, p+q=r so, it is clear that either of p or q is 2, otherwise sum p and q is even
second condition, q>p  , 2 is the smallest prime number , so , if q>pthen q cannot be 2.
therefore from first line, we have p=2
so, 2+q=r
implies    2=r-q[/hide]


calculus
integration
trigonometry
calculus computations

[hide=" wow"]

$ \int_{\frac{\minus{}\pi}{4}}^{\frac{\pi}{4}} 7 \cos(t) \vec{i} \; dt \plus{} \int_{\frac{\minus{}\pi}{4}}^{\frac{\pi}{4}}9\sin(t)\vec{j} \; dt$

$ \equal{} 7\Big(\sin\left(\frac{\pi}{4}\right) \minus{} \sin\left(\frac{\minus{}\pi}{4}\right) \Big) \vec{i} \minus{}9\Big(\cos\left(\frac{\pi}{4}\right) \minus{} \cos\left(\frac{\minus{}\pi}{4}\right) \Big) \vec{j}$

$ \equal{} 7\sqrt{2} \vec{i}$

Thus b is correct.

[/hide]

this is solved..moderator can close my topic


Find the smallest ordered solution(a,b,c,d)(all distinct and greater than1) such that a+b+c+d is a square and the sum of any 2 pairwise also is a square

Try using $\LaTeX$. So this is what we want to find:
$a$, $b$, $c$, and $d$ such that $a+b+c+d$ is a perfect square and any $2$ pairwise is a perfect square.

The smallest solution if there are $3$ variables $(a,b,c)$ is $(2, 98, 23)$. I don't know the answer for $4$ variables.

Here's how I got my answer for $3$ variables:
[hide="Very long and messy..."]Obviously, the smallest integer is $2$. Let that be one of our integers. Now, try to find all $c$ such that $a+c$ is a perfect square. If the $2$ perfect squares are $n^{2}$ and $(n-x)^{2}$, then the difference is $x(2n-x)$.

$c=7$ is the smallest such $c$. This forces $2+b$ and $7+b$ to be a square. The difference of the squares is $5$. If you try the formula for $x=1$ or $x=5$, you get that $n=2$ and $b=2$.

Try $c=14$ now. The difference of two squares  is $12$ now.  If you try the formula for $x(2n-x)=12$, you will find that when $x=2$,$n=4$ and $b=2$. When $x=3$, there are no solutions. When $x=4$, there are no solutions. When $x=6$, $n=4$ and $b=2$. So again, this won't work.

Try $c=23$ now. The difference of two squares is $21$ now. If $x=1$, $n=11$ and $b=98$ (a possibility). If $x=3$, $n=5$ and $b=2$. Nope. If $x=7$, $n=5$ and $b=2$. Nope. If $x=21$, $n=11$ and $b=98$.

So the smallest solution is $(2, 98, 23)$. [/hide]

[quote="vishalarul"]Try using $\LaTeX$. So this is what we want to find:
$a$, $b$, $c$, and $d$ such that $a+b+c+d$ is a perfect square and any $2$ pairwise is a perfect square.

The smallest solution if there are $3$ variables $(a,b,c)$ is $(2, 98, 23)$. I don't know the answer for $4$ variables.[/quote]

can you post a solution for 3 variables? Maybe that can help us find 4..

[quote="vishalarul"]Try using $\LaTeX$. So this is what we want to find:
$a$, $b$, $c$, and $d$ such that $a+b+c+d$ is a perfect square and any $2$ pairwise is a perfect square.

The smallest solution if there are $3$ variables $(a,b,c)$ is $(2, 98, 23)$. I don't know the answer for $4$ variables.[/quote]
$a+b+c=123\neq k^{2}$ for $k\in\mathbb{Z}$.


geometry
circumcircle
cyclic quadrilateral
power of a point
radical axis
geometry proposed

Let $ABC$ be a triangle. A circle passing through $B$ and $C$ intersects the sides $AB$ and $AC$ again at $C'$ and $B'$, respectively. Prove that $BB'$, $CC'$ and $HH'$ are concurrent, where $H$ and $H'$ are the orthocentres of triangles $ABC$ and $AB'C'$ respectively.

Let (O) be the circumcircle of the cyclic quadrilateral BCB'C' and X its diagonal intersection, $XB \cdot XB' = XC \cdot XC'.$ Let E, F be the feet of the B-, C-altitudes of $\triangle ABC$ intersecting at its orthocenter H. Let E', F' be the feet of the B'-, C'-altitudes of $\triangle AB'C'$  intesecting at its orthocenter H'. Quadrilaterals BCEF, B'C'E'F' are both cyclic on account of the right angles $\angle BEC,\ \angle CFB,$ $\angle B'E'C',\ \angle C'F'B',$ hence $HB \cdot HE = HC \cdot HF$ and $H'B' \cdot H'E' = H'C' \cdot H'F'.$ Quadrilaterals BEB'E', CFC'F' are also both cyclic with circumcircles (P), (Q), on account of the right angles at their vertices E, E', F, F'. The chords BE, CF of (P), (Q) meet at H and since the power of H to both these circles is the same, H is on their radical axis. Likewise, chords B'E', C'F' of (P), (Q) meet at H' and since the power of H' to both these circles is the same, H' is on their radical axis. Thus HH' is the radical axis of (P), (Q). The chords (diameters) BB', CC' of (P), (Q) meet at X and since the power of X to both these circles is the same, X is also on their radical axis. As a result, H, X, H' are collinear and BB', CC', HH' concurrent at X.


The following statement is true.
Let $ a,b,c \ge 0 $ . Prove that :
 $ (ab+ac+bc)^4 \geq a^2b^2c^2(16(a^2+b^2+c^2)+11(ab+ac+bc)) $.
Equivalently,
$(a+b+c)^4 \ge 16(a^2b^2+b^2c^2+c^2a^2)+11abc(a+b+c)$.

Mixing variables works just fine for Vasc's problem. It suffices to put $a+b+c=1$ and to take $a$ the smallest number and to mix $b,c$ into their arithmetic mean. Vasc, do you have more intelligent solution?

The inequality can be rewritten as follows

$\sum a^2(a-b)(a-c)+5\sum bc(b-c)^2 \ge 0 $.


Use $ \sqrt 5 \equal{} 2.236$, to find $ \sqrt{\frac15}$ to the nearest hundredth.

Then, we round that to $ \boxed{0.45}$ :)

You could also write $ \sqrt{\frac{1}{5}}\equal{}\frac{\sqrt{5}}{5}$ and then perform the division to get $ \boxed{0.45}$


Why is 2.7 true? I think it must be obvious but I'm stuck with it.

[url]http://www.mathcs.emory.edu/~brussel/Scans/borevich-shafarevich.pdf[/url]

[img]http://lh6.ggpht.com/_emy8bNkpkXY/SW70WnpNt9I/AAAAAAAAABI/o7cZ2gwbvrc/s640/Borevich-Safarevich-page7.jpg[/img]

That's the definition of primitive root.


Can anyone graph the following functions
1)$ i)f_{1}(x,y)\equal{}x\minus{}y\plus{}1$
$ ii)f_{2}(x,y)\equal{}x^{2}\minus{}9y^{2}$
$ iii)f_{3}(x,y)\equal{}x^{2}\plus{}y\minus{}1$

2)$ i)f_{1}(x,y,z)\equal{}x^{2}\minus{}9y^{2}\minus{}3z^{2}$
$ ii)f_{2}(x,y,z)\equal{}2x^{2}\plus{}y^{2}$


Determine all  triples  of  positive  integers  $(x,y,z)$  with  $x \leq y \leq z$   satisfying  $xy+yz+zx-xyz=2.$

Possible algebraic hint :

[hide]Setting $f(x,y,z) = xy+yz+zx-xyz - 2$

By expanding and collecting 'a's, $f(x,x+a, x+a+b) < 0$  if $x > 2$ and $a \neq 0$ so $a=0$ or $x < 2$

Solving cases separetly gives (I think) only $1,1,1$ and $2,3,4$[/hide]

For $x,y,z$ such that $1\leq x \leq y\leq z,\ xyz=xy+yz+zx-2\Longleftrightarrow x=\frac{x}{z}+1+\frac{x}{y}\leq 1+1+1=3,$ yielding $x=1,2.$
Case 1. $x=1$ Plugg this into the equation $xyz=xy+yz+zx-2$, we have $y+z=2.\ \therefore (y,z)=(1,1).$Similarly Case 2. $x=2$, we have $yz=2y+2z-2\Longleftrightarrow y=2\cdot \frac{y}{z}+2-\frac{2}{z}<2+2=4,$ yielding $y=2,3.$ Plugg these into the equation $yz=2y+2z-2,$we have that for $y=2,$ the equation $2z=4+2z-2,$ which contradicts. Then for $y=3,$ the equation $3z=2z+4\Longleftrightarrow z=4.$ Therefore desired answer is $(x,y,z)=(1,1,1),(2,3,4).$


Umm, gold in [i]both[/i] IMO [i]and[/i] IPhO...

Yeah, that would get you into just about any college.

The number who even [i]medal[/i] in both is so small.

And much much smaller for gold in both.

I do not actually know anyone who has gotten gold in both.

Although... Arnav qualified for IPhO last year, but chose to go to IMO instead, where he got a gold medal.

However, I am fairly certain that had he chosen IPhO, he would have gotten a gold there as well.

And Sherry has gotten a silver twice at IMO so far, and a silver at IPhO once.

It depends on the college, and at almost any college an amazing accomplishment can be outweighed by other issues that would result in not getting admission. It also depends on whether or not you are an international student from the point of view of the college you are applying to.

[quote="PTynan89"]
And much much smaller for gold in both.

I do not actually know anyone who has gotten gold in both.

[/quote]

Didn't Reid Barton receive Gold Medals for IMO and IOI?

He did. But PTynan89 was talking about IMO and IPhO.

This is general commentary about the relationship between math competitions and college admissions.  Math competitions will definitely help you with college admissions, but it's nowhere as important as the usual academic stuff -- grades/class rank, standardized test scores, course selection, teacher recommendations, etc.  From my own personal knowledge, I can tell you that Harvard regularly rejects very strong math students (sometimes even American IMO team members).

On the surface, it might seem that Harvard (and probably other top schools) doesn't care very much about math competitions.  But the reality is that the vast majority of good math students are also good students all-around, so there is no incentive to admit the good math students who are not also "well-rounded."  In particular, an American applicant who had gold medals in IMO and IPhO is not going to be lock for admission without respectable grades and test scores.  The good news is that exceptional math competition performance distinguishes you from all the other kids with nearly identical transcripts; the hardest thing is to distinguish yourself from the rest of the applicant pool.

For international students, I know much less, except that everything is harder for international student admissions, and that the TOEFL is important.  I don't know much about the role of money, but I think that's been discussed in other threads.

[quote="h_s_potter2002"][quote="PTynan89"]And Sherry has gotten a silver twice at IMO so far, and a silver at IPhO once.[/quote]

Didn't she also get one like 4 years ago?  Since she lived in Puerto Rico?[/quote]

I believe that was a bronze, when she was in 8th grade. And also, are you sure she got two silvers at IMO?

Because 2003=bronze, 2004=blue MOP in USA, no IMO, 2005=silver, 2006=IPhO silver.

Meh, it's pretty hard to quantify these things, not least because most American IMOers are also generally very strong elsewhere in their resumes. I've heard that IMO participation on the American team will get a person into MIT and Caltech for sure, and will frequently be very helpful in applying to other colleges, but it's hard to measure its exact effect.

[quote="Frank"]How much does actually the IMO (and the IPHO) contribute to one's ability to join a certain college?

For example, if I were gold in both physics and mathematics, would that give me a chance to join any college in the world or it's all about the money?

Any real-life experiences?[/quote]

I really don't think thats gonna help if you are applying to West Point.


inequalities
integration
function
calculus
triangle inequality
real analysis
real analysis unsolved

The easiest way now is to write
$\begin{aligned}
\left|\int_n^{n+1}\sin(t^2)\,dt\right|&\leqslant  \left|\int_n^{n+1}\left(\sin(t^2)+\frac{\cos(t^2)}{2t^2}\right)\,dt\right|+
\left|\int_n^{n+1}\frac{\cos(t^2)}{2t^2}\,dt\right|\\
&=\left. -\frac{\cos(t^2)}{2t}\right|_n^{n+1}+\int_n^{n+1}\frac 1{2t^2}\,dt\\
&\leqslant \frac 1{2n}+\frac1{2(n+1)}+\frac 1{2n}-\frac 1{2(n+1)}=\frac 1n
\end{aligned}$

[quote="spx2"]you are probably right but i cannot understand even the first step of the proof[/quote]
I just added and subtracted $\frac{\cos(t^2)}{2t^2}$ and used the triangle inequality $\left|\int_a^b f(t)\,dt\right|\leqslant \left|\int_a^b (f(t)+g(t))\,dt\right|+\left|\int_a^b g(t)\,dt\right|$ valid for any functions $f$ and $g$ (it is a triangle inequality because $f=(f+g)-g$). After that the first integral can be computed and the second one estimated using that $|\cos|\leqslant 1$.
There is a minor typo in my post: it should be $\left|\left.-\frac{\cos(t^2)}{2t}\right|_{n}^{n+1}\right|$.

how do i proove this triangle inequality of yours?
$\left|\int_a^b f(t)\,dt\right|\leqslant \left|\int_a^b (f(t)+g(t))\,dt\right|+\left|\int_a^b g(t)\,dt\right|$

Let, for brevity, write $I(f)$ instead of $\int_a^b f(t)\,dt$. We have $f=(f+g)-g$. Therefore (linearity of integral) $I(f)=I(f+g)-I(g)$. Now we have just three real numbers $I(f)$, $I(f+g)$, $I(g)$ and the first one is the difference of the second and the third. Have you ever seen the inequality $|x-y|\leqslant |x|+|y|$ for reals?

[quote="spx2"]how do i proove the following:

$\left|\int_n^{n+1} sin(t^2) dt\right| \leq \frac{1}{n}$[/quote]

It is dubious in the sense that it may not be strong enough; the integral of $sin(t^2)$ converges (Fresnel integral).
Let's investigate it.
By a change of variable this is an integral of $\sin(u) / {2\sqrt{u}}$ on $[n^2, (n+1)^2]$ which is $O(1/{n^2})$ plus some contributions near the ends of the interval, which may be as large as $O(1/n)$.  So it seems it may not be so far from the truth, after all.

It is possible to extend this (curious) inequality further: ;)
Prove that :
\[\int_1^\infty sin(t^2) \ dt \ \leq 1\]


Harvard
college
MIT
Princeton
University of Chicago

I am interested in what others of you think about Cornell's undergraduate math program. I will be entering the School of Engineering there next year, but I am still curious about your opinions about the undergraduate math program and how it measures up to some well-known math programs like the ones at Harvard, MIT, Princeton, UChicago, etc... Perhaps some of you have some personal experiences?

Any insight would be appreciated. Thanks.

Yeah, asking zscool was going to be my suggestion too. He seemed pretty happy there the last time I saw a public post of his on the issue.

haha i didnt know u were going to cornell too.  thats awesome...  what dorm u in?  im in court hall.


On January 1, I do one pushup; January 2, I do two pushups, January 3, I do three pushups..... all through the year.  On December 31, I do 365 pushups.

How many have I done throughout the year?  How does that look as an equation?

Help!!! ...  I was thinking factorials, but that is for integers....

--  Mom of three cannot remember where she put her glasses or math books......

Hello,

Let $ S = 1 + 2 + 3 + \cdots + 365$, now we need to find a good way to calculate $ S$. Note that,
\begin{eqnarray*} S & = & 1 + 2 + \cdots + 365 \\
S & = & 365 + 364 + \cdots + 1 \\
S + S & = & (1 + 365) + (2 + 364) + \cdots(365 + 1) \\
2S & = & \underbrace{366 + 366 + \cdots + 366}_{365 \text{ times}} \\
2S & = & 366\cdot 365 \\
S & = & \dfrac{366\cdot365}{2} \\
S & = & 183\cdot 365 \\
S & = & \boxed{66795} \end{eqnarray*}
The general formula for $ 1 + 2 + 3 + \cdots + n$ is $ \frac {n(n + 1)}{2}$.

Edit:
You can even find a proof without words on the left hand side of the page, in case it doesn't show up here is the swf [url=http://www.artofproblemsolving.com/Images/Flash/Side2.swf]file[/url].

[quote="ak5528"]what the previous person said[/quote]

Obviously posting just to increase your post count. At least [i]try[/i] to make it sound like you're adding something to the discussion ;)

And just to not sound like a hypocrite by adding nothing to the discussion with this post, here's a little more info on this problem. Thinking of factorials is a good response, and is somewhat on the right track, except in this case we are dealing with addition rather than multiplication. In fact, Donald Knuth coined the term "termial" and the notation $ n?$ for $ 1\plus{}2\plus{}\cdots\plus{}n$ as an analog to the factorial $ n!$.

tonypr's method is the most widely known for the closed form of $ n?$, and was initially discovered by Gauss (I think). An alternative proof is to notice that $ n\equal{}\binom n1$, so $ \sum_{k\equal{}0}^{n}\binom{k}{1}$, by the hockey-stick identity, equals $ \binom{n\plus{}1}{2}\equal{}\frac{n(n\plus{}1)}{2}$.

Another fun fact is that these numbers are referred to as the triangular numbers and have a neat geometric interpretation.


geometry
geometric transformation
reflection
incenter
circumcircle
ratio
geometry proposed

A good one! Name the tangency of the incircle to the sides $ BC,AC,AB$,at $ E,F,G$,respectively, a briefly looking at the shape,shows that $ \triangle EFG$ and$ \triangle A_1B_1C_1$ are congruent to each other,I mean they are similar to each other with a center$ I$,and radius $ k\equal{}2$.and if u consider $ I_2$,as the incenter of $ \triangle EFG$,we have:
$ \angle FI_2G\equal{}\angle B_1A_1C_1$,then we should prove that $ FI_2IG$, is cyclic.
$ \frac{AI}{IG}\equal{}2 \Longleftrightarrow \angle GAF\equal{}\angle GEF\equal{}60^{\circ},\angle GIF\equal{}120^{\circ}$
and,$ \angle FI_2G\equal{}180^{\circ}\minus{}(\frac{180^{\circ}\minus{}60^{\circ}}{2})\equal{}120^{\circ}$
and it is proved that:$ \angle FI_2G\equal{}\angle FIG\equal{}\angle B_1I_1C_1$,and $ C_1II_1B_1$ is cyclic.

We have $ AB_1\equal{}AI\equal{}AC_1$. Moreover, since $ A$ lies on the circumcircle of $ A_1B_1C_1$ and $ I_1$ is the incenter we have that $ AB_1\equal{}AC_1\equal{}AI_1$ (since A is the mid point of the arc $ B_1C_1$). Therefore $ AB_1\equal{}AI\equal{}AI_1\equal{}AC_1$.

A solution without any contruction: By the reflection as mentioned, $ IA_1 = IB_1 = IC_1 = IA = 2r$ where $ r$ is the in radius of triangle $ ABC$.

But $ IA = r \csc (\frac {A}{2}) = 2r$ which implies $ A = 60$ degrees.  (since angle A cannot be equal to 300 degrees)

Now, if $ DEF$ is the contact triangle of triangle $ ABC$, triangles $ DEF$ and $ A{}_1B{}_1C{}_1$ are homothetic with center $ I$ and ratio $ 2$. Therefore, the angles of triangle $ A{}_1B{}_1C{}_1$ are the angles of triangle $ DEF$ respectively. This implies angles  $ B{}_1A{}_1C{}_1 = FDE$. However, the angles of the contact triangle for a given triangle $ XYZ$ is
$ \frac {180 - X}{2} , \frac {180 - Y}{2}, \frac {180 - Z}{2}$.

This implies $ B{}_1A{}_1C{}_1 = \frac {120}{2} = 60$ degrees.

Since $ I$ is the circumcenter  of triangle $ A{}_1B{}_1C{}_1$ due to the reflection, $ B{}_1IC{}_1 = 2 B{}_1A{}_1C{}_1 = 120$ degrees. Now, since $ I_1$ is the incenter of triangle $ A{}_1B{}_1C{}_1$, angle $ B{}_1I{}_1C{}_1 = 90 + \frac {60}{2}$ degrees. (Since for a triangle $ XYZ$ with incenter $ I$, angle $ ZIY = 90 + \frac {X}{2}$ in degrees angle measures).

Therefore, angles $ B{}_1I{}_1C{}_1 = B{}_1I{}_C{}_1 = 120$ degrees. Thus, $ B, I, I_1, C$ are concyclic.

by simple reflection properties $\angle C_1AB_1=2\angle BAC= 2\alpha$ (say).
so , $\angle C_1A_1B_1= \pi-2\alpha$. so , $\angle B_1I_1C_1=\pi-\alpha$
also , again just by simple angle chasing , $\angle B_1IC_1=\pi-\alpha$. hence done :D

I have no idea what you are talking about

Let $IB_1,IA_1$ meet $AC,BC$ at $K,L$,now,$LK \parallel A_1B_1$,hence $\angle IA_1B_1=\angle ILK=\angle C/2$ now easily get $\angle C_1A_1B_1=\angle (B+C)/2$,also we have $\angle C_1AB=\angle A/2$,($C_1$ is image of $I$),hence $\angle C_1AB_1=2\angle A$,but $\angle C_1AB_1+\angle A_1=180$,so $\angle A=60$.$\angle C_1IB_1=120$,again $\angle C_1I_1B_1=90+\angle (B+C)/4=120$,hence done.

Consider a homothety about $I$ with ratio $2$.
Therefore this homothety maps the incircle $\omega$ of $\bigtriangleup ABC$ to the circumcircle $\omega_{1}$ of $\bigtriangleup A_1B_1C_1$.
Now it is given to us that $\omega_{1}$ passes through $A$. Therefore we can say that $AI=2r$ where $r$ denotes the inradius of $\bigtriangleup ABC$.
Therefore $\angle BAC=60^{\circ} \implies \angle B_1IC_1=180^{\circ} - 60^{\circ} = 120^{\circ}$
Now as,  $\bigtriangleup A_1B_1C_1$ and intouch triangle $\bigtriangleup DEF$ of $\bigtriangleup ABC$ are similar to each other(where $\omega$ touches $BC$ at $D$, $AC$ at $E$ and $AB$ at $F$) $$\implies  \angle B_1I_1C_1 = 90^{\circ} + \frac{\angle B_1A_1C_1}{2}=90^{\circ} + \frac{\angle B_1IC_1}{4} = 120^{\circ}$$
Therefore, $B_1,C_1,I_1$ and $I$ are concyclic.

[quote=INMO 2008 P1]Let $ ABC$ be triangle, $ I$ its in-center; $ A_1,B_1,C_1$ be the reflections of $ I$ in $ BC, CA, AB$ respectively. Suppose the circum-circle of triangle $ A_1B_1C_1$ passes through $ A$. Prove that $ B_1,C_1,I,I_1$ are concylic, where $ I_1$ is the in-center of triangle $ A_1,B_1,C_1$.[/quote]

[list]
[*] $AB_1 = AI = AC_1$ Longrightarrow A is the midpoint of arc $B_1C_1$ not containing $A_1$.
[*] $I$ is the circumcenter of $\triangle A_1B_1C_1 \Longrightarrow IA = IC_1 = IB_1 = IA_1$
[*] By Incenter/Excenter Lemma $I_1 \in A_1A$ such that $AC_1 = AB_1 = AI_1$
[*] Since, $AI = AB_1$ ,  $AC_1 = AI_1 = AI = AB_1$
[*] Thus, $ B_1,C_1,I,I_1$ are concyclic with center $A$
[/list]

Trig bash
It is clear that $I$ is the circumcenter of $\triangle A_1B_1C_1$. Thus $AI=2r$.
$(s-a)\sec {\frac{A}{2} }=\frac{2\triangle}{s}$
From here, we get $\angle A =60^{\circ} $
Now, it is easy to obtain $\angle B_1A_1C_1=60^{\circ}$
From here, the concyclicity follows.

It is easy to note that I is the circumcenter of $\Delta A_1B_1C_1$ and that$\Delta AC_1I$ and $\Delta AB_1I$ are equilateral hence $\angle B_1AC_1 = 120^{\circ}$. since $A,B_1,A_1,C_1$ are cyclic we get $\angle B_1A_1C_1 = 60^{\circ}$. Note that $B_1IC_1 =  2\cdot\angle B_1A_1C_1 = 120^{\circ}$, $B_1I_1C_1 = 90 + \frac{\angle B_1A_1C_1}{2} = 120^{\circ}$. This implies $\angle B_1I_1C = \angle B_1IC $, Hence $B_1,I,I_1,C $ are cyclic.


function
calculus
integration
number theory
least common multiple
greatest common divisor

Find the period of sin(ax/b) + cos(cx/d),
where a, b, c, d are integers and a/b and c/d are in lowest terms.
AoPS Vol 2 gives the solution as:

\pi * [2, b, d]
----------------
     (a, c)

where [2, b, d] is least common multiple of 2, b, and d
and (a, c) is the greatest common factor of a and c.

Let a=d=2 and b=c=1. Clearly the period is 4*\pi.
But that seems to break the formula given in AoPS Vol 2.
(The formula gives an answer of 2*\pi. Am I doing something wrong?)

It looks like the AoPS Vol 2 solution is wrong  :o ,
and that the correct solution is:

2 * :pi: * lcm(b, d) / gcd(a, c)

[quote="Magnara"]Go through their proof of the formula.  Do they have an error?[/quote]

Their argument is as follows:
Since sin([i]ax[/i]/[i]b[/i]) repeats every 2[i]b[/i]:pi:/[i]a[/i] and cos([i]cx[/i]/[i]d[/i]) repeats every 2[i]d[/i]:pi:/[i]c[/i], both functions complete and [an] integral number of periods from [i]x[/i] = 0 to [i]x[/i] = [i]z[/i] if [i]z[/i]/(2[i]b[/i]:pi:/[i]a[/i]) and [i]z[/i]/(2[i]d[/i]:pi:/[i]c[/i]) are integers. Since [i]az[/i]/2[i]b[/i]:pi: and [i]cz[/i]/2[i]d[/i]:pi: are both integers, the numerator of [i]z[/i] must be divisible by 2, :pi:, [i]b[/i], and [i]d[/i]. Thus the numerator of [i]z[/i] is at least :pi: times the least common multiple of 2, [i]b[/i], and [i]d[/i]. Since [i]az[/i] and [i]cz[/i] must both be integers, the denominator of [i]z[/i] can be no greater [than] the greatest common factor of [i]a[/i] and [i]c[/i]. Putting this together, we find the minimum [i]z[/i] when the numerator is minimized and the denominator is maximized, or

      [i]z[/i] = :pi:∙[2, [i]b[/i], [i]d[/i]] / ([i]a[/i], [i]c[/i]).

[In the above formula, square brackets mean the least common multiple or lcm;
 parenthesis mean the greatest common factor,
    which is the same as greatest common denominator, or gcd.]

Note: Their formula is broken by the example in the initial post.

Their proof makes little sense.  For example, what does it mean for the numerator of z to be divisible by 2 and $\pi$?  I thought r being divisible by s means that r is an integer multiple of s.  But then no number (except 0) can be divisible by 2 and $\pi$, because $\pi$ is irrational.  They seem to have some looser definition of divisibility in mind, which has led them astray.

[quote="joml88"][quote="Myth"]What kind of problems is published in that book?[/quote]

Here is the link for the index:

[url]http://www.artofproblemsolving.com/Books/AoPS_B_V2.php[/url]

The problems are usually at about Intermediate level if you use the ranking of this site.  They have a few of the easier (would probably classify as Pre-Olympiad on this site) Olympiad problems at the end of each section however.[/quote]

I do say it again. I think the difficult level in increasing way is:

getting started, intermediate, pre-olympiad, advanced section

Why do you guys are thinking that the forum is designed in this layout ?
I really think that Valentin should confirm anyone of us, hopefully me  :P,
in order to clear this matter.  :)

[quote="orl"][quote="joml88"][quote="Myth"]What kind of problems is published in that book?[/quote]

Here is the link for the index:

[url]http://www.artofproblemsolving.com/Books/AoPS_B_V2.php[/url]

The problems are usually at about Intermediate level if you use the ranking of this site.  They have a few of the easier (would probably classify as Pre-Olympiad on this site) Olympiad problems at the end of each section however.[/quote]

I do say it again. I think the difficult level in increasing way are:

getting started, intermediate, pre-olympiad, advanced section

Why do you guys are thinking that the forum is designed in this layout ?
I really think that Valentin should confirm anyone of us, hopefully me  :P,
in order to clear this matter.  :)[/quote]

Yes thats right.. I think you misinterpreted joml as saying that the preolympiad problems are easier than intermediate, when he said that the book contains intermediate problems and easy-olympiad problems. (the easier was comparing to other olympiad problems :))

[quote="TripleM"]
Yes thats right.. I think you misinterpreted Myth as saying that the preolympiad problems are easier than intermediate, when he said that the book contains intermediate problems and easy-olympiad problems. (the easier was comparing to other olympiad problems :))[/quote]

Well, acutally I said that, but you are right.  I meant that they were easy for olympiad problems.  I also agree with Orl in that the structure of this site goes:  Getting Started<Intermediate<Pre Olympiad<Advanced.

The way I wrote that was confusing.  Sorry for the misconceptions.


Gauss
function
induction
search
inequalities unsolved
inequalities

Let n be a positive integer,and x be a positive real number. Prove that
[img]http://www.artofproblemsolving.com/Forum/latexrender/pictures/39ef05ce98cc12f269294d6d1d6854a7.gif[/img]
[x] here means Gauss function.

Is $[a]$ the floor function or the nearest integer?

[x] means the greateat integer which is smaller  than x or equal  to x.Maybe my English is poor.

[quote="Gauß"]I know a similar problem. I have solved it by induction, therefore we are ready if we had shown that
$x[\frac{k}{x}]-(x+1)[\frac{k}{x+1}]\leq 1$

I try to prove (or disprove) this inequality.[/quote]

It is not right.put x=k

I have thought about it and I want to give my results.

(i)     $k-x\leq x\left[ \frac{k}{x} \right] \leq k$

My conjecture is that we have to regard the expression:

$\sum_{k=1}^n \left[ \frac{k}{x} \right]  $

Because of the monotony we must detremine the intervall $[a,b] $ for wich $\left[ \frac{k}{x} \right]$ is constant.

$qx\leq k \leq (q+1)x \Rightarrow \left[ \frac{k}{x} \right]$ is for at most x values constant.

These are my results, so far. I'm lookink for more results yet.

These are only notes, wich I want to post.

It's seems the problem's text disappeared  :huh:  Can anyone formulate it again?

The problem statement can be found in the pdf file in the second post here: [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=681379332&t=41023[/url].

You can see the official solutions [url=http://imo.is-programmer.com/user_files/IMO/File/2005%E5%B9%B4IMO%E4%B8%AD%E5%9B%BD%E5%9B%BD%E5%AE%B6%E9%9B%86%E8%AE%AD%E9%98%9F%E9%80%89%E6%8B%94%E8%80%83%E8%AF%95.pdf]Here[/url]
But I'm not sure if it's understandable.  :noo:


How many solutions are there to the equation $ 6x^{2}\minus{}77\left[ x\right]\plus{}147\equal{}0$ where $ \left[ x\right]$ denotes the greatest integer not exceeding $ x$? :)

use tha fact $ x \minus{} 1\leq\frac {6x^{2} \plus{} 147}{77} \leq x$.and solve the two inequalities.
ssory for my incomplet solution i am busy. :oops:


Hi all! If you know your score can you please post your score and whether or not you qualified? I want to get an idea as to what score i need. Thanks

[quote="alanchou"]I think most of us don't know our scores.

And also, if it's for this year, then this should be deleted.[/quote]

Any year, any time. If you don't know post how many questions you did, how many you guessed on, what you thougth was good solution and what wasn't, etc.

I won't really say which questions,  but the questions in quizzes are pretty challenging pretty much every year. I haven't been to MathPath but I have seen tests previous years and they are pretty challenging...

BTW, Alanchou is right.... we shouldn't really discuss our scores and problems here.

We never found out our scores.

(alanchou and I went to MP last year)

Same here, I never found out my score. (I went in '06)

And here's my usual reply for how many to answer: just do your best :).

LadyKnight is right.  Just do your best.  The admissions staff have shown a keen ability to determine a student's aptitude, even if they go astray on some of the problems.  Your test doesn't have to be perfect.  They do want to see some correct solutions and/or some keen insight.  Do as many problems as you can.  Do your best on each problem.  You don't have to have the skills of the National MATHCOUNTS Champion or the skills of a college math major.  Just show them your interest, your aptitude, and your ability.  And don't sweat the rest.

If you really want more information or more guidance, come to the Math Jam, tonight.  Camp Founder, Dr. Thomas will be there, along with the Academic Director and alumni.  They should be able to answer all your questions.


Find the equation of a hyperbola with center (-2,-1) focus (-2,14) and directrix: 5y=-53.

This is to be done without a calculator. Thanks in advance!

This is a hyperbola with a horizontal directrix, so the equation is

$ \frac{(y\plus{}1)^2}{b^2}\minus{}\frac{(x\plus{}2)^2}{c^2\minus{}b^2}\equal{}1$, where

the coordinates of the vertices are $ (\minus{}2,\minus{}1\pm b)$ and the coordinates of the foci are

$ (\minus{}2,\minus{}1\pm c)\implies c\equal{}15$.

The equation of the "lower" directrix of this hyperbola is

$ y\equal{}\minus{}1\minus{}\frac{b^2}c\equal{}\frac{\minus{}b^2\minus{}15}{15}\equal{}\minus{}\frac{53}5$

$ \implies b^2\plus{}15\equal{}159\implies b^2\equal{}144$.

Finally, our equation is $ \boxed{\frac{(y\plus{}1)^2}{144}\minus{}\frac{(x\plus{}2)^2}{81}\equal{}1}$.


I am using $ \text{\LaTeX}$ to type up a problem set - I have a question about the command enumerate:

The document looks like this:

\begin{subsection}{Set 1}
\begin{enumerate}
\item <Problem 1>
\item <Problem 2>
\end{enumerate}
\begin{subsection}{Set 2}
\begin{enumerate}
\item <Problem 3>
\item <Problem 4>
\end{enumerate}
...

Is there a way to make the numbering continue, so instead of labeling Problems 3 and 4 as 1 and 2, respectively, they are labeled as 3 and 4?

Thanks in advance.

An easy way is to write
\setcounter{enumi}{2}
in the second enumerate. But that gets to be a problem if you later change things round, so the best way is to use the [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=enumitem]enumitem package[/url] and then the resume option in second and subsequent enumerates
[code]\usepackage{enumitem}
...
\subsection{Set 1}
\begin{enumerate}
\item <Problem 1>
\item <Problem 2>
\end{enumerate}
\subsection{Set 2}
\begin{enumerate}[resume]
\item <Problem 3>
\item <Problem 4>
\end{enumerate}[/code]

Why use \begin{subsection}{Set 1} which means that you have also have to use \end{subsection} as well? Just use \subsection{Set 1}.


induction
number theory
prime numbers
number theory unsolved

[quote="payman_pm"]is there any order of $1,2,3,....$ like $a_1,a_2,a_3,....$ such that for every $k\in N$ we have :
$k\mid a_1+a_2+a_3+...+a_k$[/quote]

For $n=1$ and $n=3$ there are the permutations ${1}$ and ${1,3,2}$ respectively.  Obviously there is no such permutation for $n=2$.

For a general $n>3$, there is no solution.  To show this, first note that obviously $n$ must be odd or else $n$ will not divide the sum of all the terms. So let $n=2m-1$.

Also let the last term in the permutation be $c$.  Then the sum of all the terms is $(2m-1)(m)$, so the sum of the first $2m-2$ terms is $(2m-1)(m)-c$.  Since the latter must be divisible by $2m-2$, we have $c=m\mod (2m-2)$.  Since $m>0$, this implies that $c=m$.

Now let the next-to-last term in the permutation be $d$.  Then the sum of the remaining terms (excluding $c$) is $(2m-2)(m)-d$ which must be divisible by $2m-3$.  This forces $d$ to be congruent to $m\mod(2m-3)$.

Since all the terms are $\leq 2m-1$ by definition, the only way we can have $c$ and $d$ not be the same number is to have $2m-3<m$, which implies $m<3$.  Hence the only possible solutions are $m=1$ and $m=2$ corresponding to $n=1$ and $n=3$ respectively, which are the answers given above.

I think he was asking for the set of ll positive integers  ;)
And, in that case, such a permutation does exist. It is a classical application of the chinese remainder theorem.

Pierre.

i agree but the problem he solved is better and less trivial. i love the way new problems arise so naturally :D

[quote="payman_pm"][quote="jazzmaster"]First choose [tex]a_1=1[/tex] for [tex]k=1[/tex].
So we have a sequence already for a given [tex]k[/tex].
Now we can extend it for [tex]k+1[/tex]: let's assume that [tex]a_1+...+a_k=i \mod{k+1}[/tex].
We know that the number of natural numbers n for which [tex]n=(k+1)-i \mod{k+1}[/tex] is infinite.
Let's choose one of them, and then [tex]k|a_1+...+a_{k+1}[/tex].

I hope it's correct.  ;)

[b]Edit:[/b] Actually you can prove this result in case the set you choose [tex]a_k[/tex] from is prime numbers because of the fact that any arithmetical progression has infinite number of primes in it.[/quote]
i think your solution is'nt complete,you make your sequence bu induction
but it might be there were a number like $t$ that never apear in your sequence![/quote]

Yeah, sorry. But it's easy to make sure for every [tex]k[/tex] that it will be in the sequence. Just choose [tex]a_{k+1}[/tex] such that we can make [tex]a_{k+2}=k[/tex].   :roll:

Just to clearify Pierre's idea:
 We choose such an permutation by induction:
 First,put $a_1=1$
 Assume we have choosen $a_1,...,a_k$ and $n$ is the smallest natural number not belong to $\{a_1,...,a_k\}$.
 There exists $m>\max\{a_1,...,a_k,n\}$ s.t $m\equiv -(a_1+...+a_k)(\mod k+1)$
and $m\equiv -(a_1+...+a_k+n)(\mod k+2)$.
 Now put $a_{k+1}=m,a_{k+2}=n$.
 Is it wrong? :(


Analyze...hmmm. That might be interesting. Although I think actually practicing by playing is better for increasing skill.

As a general rule, I normally analyze questionable moves in a game and how this will impact the result.

Also, when I progress through the game, and my opponent makes a crushing move, I analyze how he achieved that move and how I could have seen it before.

[quote="avemuri57"]Hello all,

I was wondering if anyone wanted to analyze a game? I have a million interesting ones to look over for scrutiny. I could post the game and we could all look at what happened for both people. I was thinking that we could analyze GM games because we can get the most out of them. In any case, please reply if you are interested.[/quote]

*sees opportunity to get better at chess*

Post some!

Hello all,

        Im actually retyping this because my computer shut off. This is not as in depth as the initial post that I had planned but here it goes:

 This is a game of Vishy versus some random dude. I liked it a lot mainly because its english (The opening I wanna learn because I HATE playing it and against it)and I felt that there were some key concepts we could all learn from it. The game goes (Vishy has black pieces)

1.Nf3 Nf6 2.c4 g6 3.b3 Bg7 4.Bb2 O-O 5.g3 d6 6.Bg2 e5 7.O-O c6
8.e3 h6 9.d4 e4 10.Nfd2 d5 11.Nc3 Be6 12.Rc1 Nbd7 13.cxd5 cxd5
14.Qe2 a6 15.Na4 b6 16.Ba3 Re8 17.Rc2 Bg4 18.Qe1 Rc8 19.Rxc8
Qxc8 20.h3 b5 21.hxg4 bxa4 22.Qd1 Qc3 23.Nb1 Qc6 24.Re1 Rb8
25.Bf1 axb3 26.axb3 Bf8 27.Bxf8 Nxf8 28.Be2 Rc8 29.Kg2 a5
30.Qd2 Qb6 31.Qa2 Ne6 32.Nd2 Ng5 33.Ra1 Kg7 34.Qb2 Qe6 35.Rxa5
Nxg4 36.Rc5 Qf5 37.f3 exf3 38.Bxf3 Nxe3+ 39.Kg1 Nxf3+ 40.Nxf3
Qxf3 41.Rxc8 Qf1+ 0-1 --Vishy wins... (not anything abnormal)

(You may need a board to read that, however if you dont have one handy check this link out;
http://www.chessgames.com/perl/chessgame?gid=1017851 )

The first thing that I see is actually a questionable move on Vishy's side. The move 8...h6 seems to have no importance to the match. I feel that a simple developing move such as 8... Bf5 or even 8...Be6 ( as he later played) would be stronger, but I may be wrong. I think the turining point of this match was turn 17 when ostermeyer played 17 Rc2. Once the black bishop goes to g4, ostermeyer is outta luck. He traps the rook on f1 and the rook on C2 is weak because the Rf1 cant protect it. I like how Vishy manages to take adavantage of this weakness and place maximum pressure on the C file. I feel that we should probably go in more  depth with his play and perhaps look more elaborately at some other variations which could have occured. These are just my thoughts on the first few moves, let me know what you guys feel. (Ill post again on the middle end ending prolly tomorrow)

Im in @serialkiller maybe for a beginner it is better, but for anyone not a complete beginner --about 2300 uscf gm games and openings and endgames are the way to go

heres another game u can analyse from my chess book. sicilian is the opening
                  Burgess-C.Jacobson

1.e4 c5 2.d4 cxd4 3.c3 dxc3 4.Nxc3 e6 5.Nf3 Nc6 6.Bc4 Bc5 7.0-0 a6 8.f4 Nge7 9. e5 0-0 10. Ne4 Ba7 11.Bg5 Qc7
12.Nf6 gxf6 13.Bxf6 Nf5 14.Nd4 h6 15.Nxe5 Nxe5 16.Qg4+ Nxg4 17. Ne7+ Kh7 18. Bd3#
[hide]
Note: all pieces are displayed with capital letters and u must follow white and blacks moves carefully.[/hide]

I find that mathblitz's chess game can't be followed :P

Maybe there is missing move...


function
geometry
perimeter
probability
ceiling function
parameterization
floor function

What is the generating function for find the number of incongruent triangles with integer sides and perimeter n?

Advance warning: there is a large positive probability of computational errors in what follows.

It's sequence [url=http://www.research.att.com/~njas/sequences/A005044]A005044[/url] in the [url=http://www.research.att.com/~njas/sequences]OEIS[/url].  We can get that result with the obvious ugly approach and a lot of arithmetic: the number of such triangles with largest side $ k$ (with $ \frac {n}{3} \leq k < \frac {n}{2}$) is $ \left \lceil \frac {3k - n + 1}{2} \right\rceil$, so we want $ \sum_{n \geq 0} \sum_{\frac {n}{3} \leq k < \frac {n}{2}} \left \lceil \frac {3k - n + 1}{2} \right\rceil x^n$.  Now split this sum up into the various options for $ n$ modulo 12 and $ k$ modulo 2 to get rid of the quasipolynomials.  For example, taking the part of the sum where $ n = 12m$ and $ k = 2j$ we have a contribution of
\begin{align*} \sum_{m \geq 0} \sum_{j = 2m}^{3m - 1} \left \lceil \frac {6j - 12m + 1}{2} \right\rceil x^{12m} & = \sum_{m \geq 0} \sum_{j = 2m}^{3m - 1} (3j - 6m + 1) x^{12m} \\
& = \sum_{m \geq 0} \frac {m(3m - 1)}{2} x^{12 m} \\
& = \frac {x^{12} + 2x^{24}}{(1 - x^{12})^3}. \end{align*}
Then you do 23 more like this (though there are obvious simplifications to be made, and they're all very close to identical with each other) and you get the answer in the link I've given.  There is, probably, a less-stupid way to go about this :)

Edit: actually, yes -- even this non-clever solution can be made more clever by rewriting our sum as $ \sum_{k \geq 0} \sum_{2k < n \leq 3k} \left \lceil \frac {3k - n + 1}{2} \right\rceil x^n$.  Now it should be enough to consider $ k$ modulo 2 and $ n$ modulo 2: the part where $ k = 2j, n = 2m$ is
\begin{align*} \sum_{j \geq 0} \sum_{m = 2j + 1}^{3j} (3j - m + 1) x^{2m} & = \sum_{j \geq 0} \frac {x^2}{(1 - x^2)^2} \cdot (j x^{4j} - (j + 1)x^{4j + 2} + x^{6j + 2} \\
& = \frac {x^6}{(1 - x^4)^2(1 - x^6)}, \end{align*}
and similarly for the other pieces.

In one of the papers by Andrews titled "MacMahon's Partition Analysis" there is an automatic derivation using the Omega method and then a second derivation using a very neat parameterization of such triangles that I've completely forgotten.  I think it's the "I" paper.  (The point is to use three variables and specialize them later.)

from an old text (in Hebrew)-
The number of such triangles, denote it by $ T_n$, is
\[ T_n \equal{} \begin{cases}\lfloor \frac {(n \plus{} 3)^2}{48} \rfloor & \mbox{if } n \equiv 1 \pmod{2} \\
\lfloor \frac {n^2}{48} \rfloor & \mbox{if } n\equiv 0 \pmod{2} \end{cases}
\]
The proof consists of 2 steps:
1. Showing that
\[ T_n \equal{} \begin{cases}p_3(\lfloor \frac {n \minus{} 3}{2} \rfloor) & \mbox{if } n \equiv 1 \pmod{2} \\
p_3(\lfloor \frac {n \minus{} 6}{2} \rfloor) & \mbox{if } n\equiv 0 \pmod{2} \end{cases}
\]
where $ p_n(m)$ is the number of partitions of $ m$ into $ n$ non-negative terms.
(I can elaborate if needed, but it is a direct argument)

2. Proving $ p_3(n) \equal{} \lfloor \frac {(n \plus{} 3)^2}{12} \rfloor$.

The generating function part is calculating $ \sum p_3(n)x^n \equal{} \frac {1}{(1 \minus{} x)(1 \minus{} x^2)(1 \minus{} x^3)}$.

How did you come up with:
\[ T_{n}\equal{}\begin{cases}\lfloor\frac{(n\plus{}3)^{2}}{48}\rfloor &\mbox{if }n\equiv 1\pmod{2}\\ \lfloor\frac{n^{2}}{48}\rfloor &\mbox{if }n\equiv 0\pmod{2}\end{cases} \]
and how does this relate to:
\[ T_{n}\equal{}\begin{cases}p_{3}(\lfloor\frac{n\minus{}3}{2}\rfloor) &\mbox{if }n\equiv 1\pmod{2}\\ p_{3}(\lfloor\frac{n\minus{}6}{2}\rfloor) &\mbox{if }n\equiv 0\pmod{2}\end{cases} \]
??????
can you clarify this more... thanks

If you can express $ T_n$ with the function $ p_3$ (your 2nd equation), and you know the explicit formula for $ p_3$ (the 2nd step of the proof I mentioned), you get your first equation.

Proof of the first step in my post:
If $ n$ is odd, and $ a,b,c$ are sides of a triangle, then $ \frac {a \plus{} b \minus{} c \minus{} 1}{2}, \frac {b \plus{} c \minus{} a \minus{} 1}{2}, \frac {c \plus{} a \minus{} b \minus{} 1}{2}$ are three non-negative integers that sum to $ \frac {n \minus{} 3}{2}$.
Also, if $ x \ge y \ge z$ are three non-negative integers that sum to $ \frac {n \minus{} 3}{2}$, then $ a \equal{} x \plus{} y \plus{} 1, b \equal{} y \plus{} z \plus{} 1, c \equal{} z \plus{} x \plus{} 1$ are sides of a triangle with $ a \ge c \ge b$.
For even $ n$, consider  $ \frac {a \plus{} b \minus{} c \minus{} 2}{2}, \frac {b \plus{} c \minus{} a \minus{} 2}{2}, \frac {c \plus{} a \minus{} b \minus{} 2}{2}$ and $ x \plus{} y \plus{} 2, \cdots$.

(this was a bijection from those triangles to the solutions to $ x \plus{} y \plus{} z \equal{} \frac {n \minus{} 3}{2}, \frac {n \minus{} 6}{2}, x \ge y \ge z \ge 0$)

Proof of the second part uses the partial fraction decomposition of $ \frac {1}{(1 \minus{} x)(1 \minus{} x^{2})(1 \minus{} x^{3})}$. To show that the generating function is right, notice that it counts the number of solutions of $ x \plus{} 2y \plus{} 3z \equal{} n$. The bijection $ (x,y,z) \to (x \plus{} y \plus{} z,y \plus{} z,z)$ shows that it is the same as the number of solutions of $ x \plus{} y \plus{} z \equal{} n, x \ge y \ge z$.

I would be quite interested to find out what the original source for this solution was, even better if there existed an ebook :)


linear algebra
matrix
inequalities
vector
superior algebra
superior algebra solved

Let $A \in Mn(R)$
$A=(a_{ij})$
$\sum_{j =/= i} a^2_{ij} < min (a^2_{ii})$
Prove that A is invertible

it reminds me of diagonally dominant matrix :lol: though they are different  :(

I find his original post

http://www.mathlinks.ro/Forum/viewtopic.php?p=92384

 :D

I forgot to say
$||X||$ is the euclidean norm of the vector $X$


The statement may be false, but how to give a counterexample.

$ \lim_{x\to 0}\ f(x) \equal{} A$ iff $ \lim_{bx\to 0}\ f(bx) \equal{} A$,for $ b \neq 0$

They are equivalent. Put $ y\equal{}bx$ and you get the first.


Hey guys...
would anyone be interested in a tournament? it could be brackets style or with teams, i'm free with anything...
i was just wondering if anyone would be interested. =)

That would be awesome!

When, though?

please tell me about the tournament if there's going to be one.

we should have it so that anyone can start a tournament, like host one at a certain time if there are enough people in the room. but yeah im in

Um, a lot of ppl mem, so it can get to be unfair.  That's why I usually do response-based games. However, a tournament does sound fun, though I have like no time.  Oh well... :(

hmmm..
Memming isnt really "cheating": it's annoying, but you can't really help it if you already know the answer (you're hardly going to let the other guy get it just because you already know it)...
So so far there are four people interested?
(I really should make a list...)

I will, but not this week, I have exams.

Sounds like fun. But make it 45 sec per problem, O.K? We don't wanna have a long tournament, but we can't have a 20 sec annoying problem.

1=2 is right. please make them 45 seconds or higher but not too long.

also, i agree with the idea of making it reponse based. If u make it as time or ratings, it'll be very unfair to ppl who aren't memming.

45 is too long.

25 is good. Response based might not be a bad idea for a tournament though

Yea... i just realized i don't really need whole 45 seconds on questions if it's based on responses. 30 seconds would be good

... I would like to, but only on Mondays 7:30-9:00 pm EST.

[quote="perfect628"]At most 30 sec, any higher is irritating. I can't stand getting the answer in like 6 seconds, then having to wait 45.[/quote]

That makes 51 seconds, not 45.

How are we going to determine a bracket? Are we doing countdowns or games? Maybe for later rounds (like semis and finals), we can convince the AoPS people to save problems that aren't yet in circumlation just for us.

If we use countdowns, My idea for the format is World Cup style. Assign like N groups of 5-6 (based on rating, and randomly generated), have them do round robins, and top 2 advance from each group. Then we can add wildcards if necessary, and throw everyone into a bracket based on some type of index (rating * # of wins in prelims or something), and go from there.

[quote="CatalystOfNostalgia"]That makes 51 seconds, not 45.[/quote]

You are truly hateful.

Anyway, I think the saving problems for a tournament would be a good idea. (If possible of course. :lol: ) It would eliminate the element of chance.

Wait so are we doing two person teams or what? And how are we going to definitively decide the format of things? Maybe 5 people or so should volunteer to organize this officially...

I have a few great NY people for a NY team. Me, Hexahedron, SmitSchu, and Alison. I haven't checked if they want to compete, though.

When will this tournament be?

Actually we should do per region, MAX of four people, as in a real competition.

Fanatic does.

I know you guys are best friends ;-)

The future does too...

and firecricket...

[quote="mathcrazed"]No one wants to be on my team T.T... No one in my state does this -_-. except xpmath. hmmmmmm xD[/quote]

Even though you are probably not in NY, I can be on your team.

[quote="mathcrazed"]No one wants to be on my team T.T... No one in my state does this -_-. except xpmath. hmmmmmm xD[/quote]
I thought I lived in Ohio...

um... this subject has been going on for a while... so are we really going to have a tournament? and i don't think it'll be easy to find out a time for us to have a tournament where everyone who wants to join will be able to.

also, i think this post is starting to come off the subject(a little) when people talk about what teams they want.

Yes, Kevin, when is this tourney going to take place...

i wanna play i wanna play i wanna play i wanna play i wanna play i wanna play!!!!

We really shouldn't talk about how we're going to organize a tournament, we should actually do it. Teams by region sounds fine, but we shouldn't make people do that. I'll be decisive and put up a poll about how many people per team we're going to have.

I think we should do the signups in this topic:     http://www.artofproblemsolving.com/Forum/viewtopic.php?t=186252

i suggest joining the march tournament 08


[quote="7h3.D3m0n.117"]If there is a rule in a game, and it simply dictated "no rules", is that rule canceled out?[/quote]
That rule will not be a rule anymore, but it won't be canceled out.
in otherwords, that "rule" will become a commonly agreed position, and won't be a rule anymore.

Similar to the barber who cannot shave himself?

You cannot make a game with the rule "no rules are allowed" for a similar reason that you can't make a game that has the rule "you lose if and only if you win"

All I know is I know nothing.
Isn't that something!!

Fraenkel,

let me guess.

You are Sgt. Schultz from "Hogan's Heroes."


[quote="riddler"]Rip Van Winkle awoke from his 20-year sleep with no idea where he was or what day it was. He approached a nearby woodsman and asked: 'What day is it?' The woodsman was naturally suspicious at the sight of this curious character with a very long beard, so he answered (truthfully): 'When the day after tomorrow is yesterday, today will be as far from Tuesday as today was from Tuesday when the day before yesterday was tomorrow.' What day is it?

Have fun! :D[/quote]

Whoa.

[hide]I got Saturday [/hide]

And the answer is...

[hide]Tuesday![/hide]I think.

Riddler, you call that easy? :what?:

tuesday is correct! :D
Easy because it was..... nice.... appears to be very easy when you have done it! :D


[quote="mahanmath"]Hello !! I really need Russia 2002 combinatorics . I couldn`t  find it on Mathlinks (Resources , ....).Please help me :blush:  !![/quote]
link: http://www.imomath.com/index.php?options=oth|Rus|Ne|Ne&ttn=Russia&p=0


Given positive reals $ a,b,c,d \in (0,\frac {1}{2})$
Prove that $ \frac{a^4\plus{}b^4\plus{}c^4\plus{}d^4}{abcd} \ge \frac {(1\minus{}a)^4\plus{}(1\minus{}b)^4\plus{}(1\minus{}c)^4\plus{}(1\minus{}d)^4}{(1\minus{}a)(1\minus{}b)(1\minus{}c)(1\minus{}d)}$

I will be very glad if I can see some alternative solutions here :)


Let be given a function f from C^2(R) satisfying condition f(x) + f"(x) >=0  for all x from R. Show that  f(x) + f(x+Pi) >=0 for all x from R.

<i>Source: VUMC 2001</i>

Namdung

It was solved
http://www.mathlinks.ro/phpBB/viewtopic.php?t=495

Normally how many hours you have for Vietnam Undergraduate Mathematics Competitions ?
How many problems you have to solve ?

Students have to solve 4 or 5 problems for 3 hours.

Namdung


[quote="SOUMYASHANT NAYAK"][quote="arqady"][quote="BaBaK Ghalebi"]$a,b,c,d$ are positive reals and such that $a^{4}+b^{4}+c^{4}+d^{4}=4$ and $S=\sqrt[4]{abcd}$,prove that
$a+b+c+d-4S\geq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}-\frac{4}{S}$
 :?:[/quote]
What if $a\rightarrow0^{+}$ $?$ :wink:[/quote]

it indicates literally nothing.
if 1/a tends to become very very large
then at the same time 4/S also becomes very very large.

(or in simple words -4/S might just cancel out the effect of 1/a.)

so we cannot conclude anything from your statement.[/quote]
Why?
$a+b+c+d-4S\geq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}-\frac{4}{S}\Leftrightarrow$
$\Leftrightarrow(a+b+c+d)abcd-4Sabcd\geq(abc+abd+acd+bcd)-4S^{3}.$
Try $a\rightarrow0^{+}.$ :wink:
You agree with me? Thank you. :)

you are right. :lol:

i thought it was some premonition from your side.(stupid of me)
i dint check it out.

i see you worked it out.

[quote="arqady"][/quote][quote="SOUMYASHANT NAYAK"][quote="arqady"][quote="BaBaK Ghalebi"]$a,b,c,d$ are positive reals and such that $a^{4}+b^{4}+c^{4}+d^{4}=4$ and $S=\sqrt[4]{abcd}$,prove that
$a+b+c+d-4S\geq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}-\frac{4}{S}$
 :?:[/quote]
What if $a\rightarrow0^{+}$ $?$ :wink:[/quote]

it indicates literally nothing.
if 1/a tends to become very very large
then at the same time 4/S also becomes very very large.

(or in simple words -4/S might just cancel out the effect of 1/a.)

so we cannot conclude anything from your statement.[/quote][quote="arqady"]
Why?
$a+b+c+d-4S\geq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}-\frac{4}{S}\Leftrightarrow$
$\Leftrightarrow(a+b+c+d)abcd-4Sabcd\geq(abc+abd+acd+bcd)-4S^{3}.$
Try $a\rightarrow0^{+}.$ :wink:
You agree with me? Thank you. :)[/quote]
c`mon guys,this is tooooo wrong,read the problem again,it has some bounds,it says that $a^{4}+b^{4}+c^{4}+d^{4}=4$ so you can`t just change one of them they are not independent,if you change $a$ then $b,c,d$ will change to...
in other words we can put $a^{4}=4-b^{4}-c^{4}-d^{4}$
and then if we put the new $a$ in the inequality,then I think it would make sense... :huh:
right? :maybe:


geometry
3D geometry
sphere
graph theory
puzzles

In a neighborhood there are three houses and three service companies: One of water, one of light and one of gas. Is it possible to connect each hose to each servise provider though lines without crossing each other?

I remember seeing this a long time ago...

[hide]If I remember properly, the answer is no, unless it's 3 dimensional.[/hide]

[quote="nr1337"][hide]
Yes, it is.
[/hide][/quote]
how? :huh:

Is it implied that you can arrange the houses/companies however you want? I don't know if that would help anyway, but...

Yes, you can put the three houses and service companies anywhere you want.  I believe, for the purposes of this problem, that you are to assume that the world is a plane.  (In fact, you could use the surface of a sphere as well, but that's not fundamentally more interesting.)

[quote="probability1.01"]The correct answer is no.  I don't know much about graph theory, so I can't give a very good proof, but essentially you must end up with something equivalent to:
----W----
--/---\---
H - L - H
--\---/---
----G----[/quote]

As someone already said, the answer is yes, there is a way (we all have telephone, water and internet service, and our lines do not cross each other). The problem is that you are thinking in two dimensions, but you only need one extra and you'll have what you need. You are right to say that $K_{3,3}$ is not planar, that is 100% true!

Haha, well I didn't think that would count :P.

[quote=SmartZX]It's not possible in 2d, but it is in 3d[/quote]

Its not even possible on a sphere. It needs to be something with a hole in it, like a donut or a mug.


analytic geometry
Support
vector
trigonometry
ratio
parameterization
parametric equation

Hello, just a quick question, There are three coordinates planes in 3-space. A line in R3 is parallel to xy-plane, but not to any of the axes. Explain what this tells you about parametric and symmetric equations in R3. Support your answer using examples,    Is this just indicating that the line is perpendicular to the xz plane and zy plane?

Saying that a line is parallel to the plane $ z\equal{}0$ is saying that it lies in some plane $ z\equal{}c$. It's not perpendicular to any particular plane.

ok but what about the axis would it be perpendicular to the z axis and therefore  making the parametric equation for z= to the point on the line since the direction vector is (a,b,0)?

What do the parametric equations for a line look like? It's not just a single equation...

[quote="mikee"]Hello, just a quick question, There are three coordinates planes in 3-space. A line in R3 is parallel to xy-plane, but not to any of the axes. [/quote]

that's easy..... do you know what are direction cosines (or direction ratios) ? you  have (if the line passes through $ (x_0,y_0)$ )

$ \frac {x \minus{} x_0}{a\;( \equal{} \;x_1 \minus{} x_0)} \equal{} \frac {y \minus{} y_0}{b\;( \equal{} \;y_1 \minus{} y_0)} \quad( \equal{} \;\frac {d}{\sqrt {a^2 \plus{} b^2}} )$

where

$ \frac {l}{m}\neq \frac {1}{0} \quad \left( or \quad\neq \frac {0}{1} \right)$

this line can't be perpendicular to either the xz plane or the zy plane...  you know  that right ? :|


$\ln{y}=x^{2}\ln{x}$, therefore $\frac{y'}{y}=2x\ln{x}+x$. Done :D


I have a problem with understanding algebra in discrete mathematics. The problem is with the following question:

P -> (Q \/ R) = ~R -> (P -> Q)

to prove that P -> (Q \/ R) = ~R -> (P -> Q)

I have manage to move to this level:

P -> (Q \/ R)

(P -> Q) \/ (P -> R)

(~P \/ Q) \/ (P -> R) - Identity Law

I am not sure what to do next to proof that

P -> (Q \/ R) = ~R -> (P -> Q)

How do I solve the problem.

Also I include an attachment of the laws of algebra.

Logically, the left side of the equation is only false if P is true, and Q V R is false, due to the imply. Therefore, the only time when The entire right side is false, is when P is true, and both Q, and R are False. Then plugging into the right side gives False as well.

Now, the only time the right hand side could be false is if R is false, P is true, and Q is false. Plugging into the left hand side gives  false as well. Then, for any other valuse, both sides will be true. I could probably write up a proof form of it if you want.


I'm not so sure, but how do you prove this?
$ f(x)\equal{} x^2$ when $ x$ is rational and $ 0$ when $ x$ is irrational. Prove that $ \lim_{x\rightarrow 0} f(x) \equal{} 0$. is it ok to have cases: like case 1) $ x$ is irrational, 2)case is rational? how do you prove it? thanks

Nothing complicated like that.  The [url=http://en.wikipedia.org/wiki/Squeeze_theorem]squeeze theorem[/url] suffices.


Is there a statistic on this site that tells how many (or what percentage ) of mathlinkers belong(that is they show they belong) to a particular country? Might I suggest one of the admins to make such a statistic if they are free! It would be fun!

[quote="Rushil"]if they are free![/quote]Unfortunatelly they are not :D

Also, what would you use to decide country? All of the profile information is self-reported, and many people choose that in a way that doesn't match their actual place of residence.

I've said above that you will obviously use the country the user chooses to show in his/her profile.

Of course, I am sure every user has a good reason to show another country, if so then that is what we want to know
It is already free in a sense, just click on the person?

So why is there no organised way of seeing it all?  Is it a technical difficulty or a privacy issue (i can understand both cases)

It will be soon. For now there are other things to worry about ;)

[quote="fredbel6"]So why is there no organised way of seeing it all?  Is it a technical difficulty or a privacy issue (i can understand both cases)[/quote]It's a in-the-queue item. :) there are just so many things that need to be done. And I guess Valentin has a life outside this site as well. :)

What I wonder: you want to know it, fredbel, but you don't indicate your country either. ;)

[quote="Peter VDD"]:) there are just so many things that need to be done.[/quote]That's right :D

[quote="Peter VDD"]And I guess Valentin has a life outside this site as well. :)[/quote]Unbelivable, isn't it? :)

Much to learn this newbie has... :oops:
Is it good now?

The closest thing we got right now is simply going to members, and sort by location, but that is sorting by their city they gave, and one guy wrote 'owl's armpit' or something so that doesn't help much

It might help to get some information and let contacts grow.  But I deduce from what Valentin said it mainly is a technical issue and it requires time to get it done, and there are no privacy obstacles apart from that?  Well then I'll just wait, anticipating, yet patiently


function
quadratics
calculus
derivative
calculus computations

$f(x)=6-\frac{1}{x}$

Solve this equation:

$f(x)=f'(x)$

What i do:

$6-\frac{1}{x}=\frac{1}{x^2}$

$0=\frac{1}{x^2}+\frac{1}{x}-6$

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

$x=2...x=-3$

Where am i doing this wrong? My calculator says i'm doing it wrong when i write the two functions on my calculator.

That's a quadratic in $\frac1x$, not $x$. The roots are $\frac1x=2,-3$, which is $x=\frac12,-\frac13$.

OK, i have never learned that, but it makes perfect sence as my calculator seems to have the same (or very simular) results. One more rule i have to keep in mind when i see $\frac{1}{x}$. Thx for the help...

[quote]One more rule i have to keep in mind when i see $\frac{1}{x}.$[/quote]
NO! Don't think that way. It's not "one more rule." In fact, there really aren't very many rules. You shouldn't be trying to add new rules; you should be trying to effectively apply the ones you already know.

What we actually have here is an instance of [i]substitution.[/i] When you see a particular algebraic expression recur in a number of places, [b]rename it.[/b]

You have $\frac1{x^2}+\frac1x-6=0.$

And I think: hmm, there are a lot of appearances of $\frac1x$ there. What if I gave that a new name: let $u=\frac1x.$ Then what does that make $\frac1{x^2}$ ? Oh, that's $u^2.$

So we have $u^2+u-6=0.$ And [i]that[/i] is familiar; I know how to solve it. (I'd just factor it by inspection; no need to pull out the quadratic formula.)

Note that the substitution didn't "do" anything - the only change was in your perception of the problem. And that's enough, right there.

Here's your quiz based on this principle. Find all solutions of the following equation:

$e^{2x}-5e^x+6=0.$

Yes,that is the idea of function.

$f(x)=y : x\Rightarrow f()\Rightarrow y$

Oh yes, that's true. Just haven't worked with this trick for quite some time and secondly i didn't know it could help me here.

$e^{2x}-5e^x+6=0$

Note: $u=e^x$

$u^x-5u+6=0$

Right?

OK, thx, i think i've got this one confirmed now. Thx to everyone. Can i ask you to plz attend to this thread i'm also having problem with (see the bottom of the thread) [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=76006]Another differentiation problem[/url]. Thx in advance.


I've heard this question asked several times but I still don't believe the answer that is typically given. The problem is:

In a family with two children, one of them is a girl. What is the probability that the other one is a girl? The typical answer that people say is 1/3 because there are 3 possibilities:

1. Boy, Girl
2. Girl, Boy
3. Girl, Girl

and only one of them has the other be a girl. But isn't this assuming that the girls are indistinguishable from girl but not from boys? Shouldn't it be

1. Boy, Girl you are given
2. Girl you are give, Boy
3. Girl you are give, Girl
4. Girl, Girl you are given

and the probability should be 1/2

Or am I missing something?

Its a well known paradox: [url=http://en.wikipedia.org/wiki/Boy_or_Girl]Read about it here[/url]

Isn't this similar to the Monty and the 3 doors question? [url]http://en.wikipedia.org/wiki/Monty_Hall_problem[/url]

Yeah I think they can by analyzed in similar ways. In essence. :)

Yeah, both of these problems involve [url=http://en.wikipedia.org/wiki/Conditional_probability]conditional probability[/url].


Euler
geometry
circumcircle
geometry unsolved

[quote="nguyentrang"]In $ \triangle ABC$, let $ B',C'$ be the midpoints of $ AC,AB$. A tangent at $ A$ to the circumcircle of $ ABC$ intersect $ B'C'$ at $ A_1$. We define $ B_1,C_1$ similarly. Prove that $ A_1,B_1,C_1$ are collinear on a line perpendicular to the Euler line of triangle $ ABC$.[/quote]
Let N is midpoint of OA so $ (O,\frac{OA}{2})$ is circumcircle of $ AB'C'$ and Its tangent at A is $ OA_1$
so $ AA_1^2\equal{}A_1B'.A_1C'$ so $ P_{A_1|(O)}\equal{}P_{A_1|(K)}$ (K is center of  Euler cycle)
Similar we have $ A_1,B_1,C_1$ are collinear on a line perpendicular to the Euler line of triangle $ ABC$


[quote="sinajackson"]how can I make a AoPS blog?
i went to the [b]Blogs[/b] and entered sth like:
http://www.artofproblemsolving.com/(here was the name that i wanted my blog to be) as the URL of my blog and it said it has to be approved within 3 buisness days....
but i did exactly this a month ago but got nothing, where am i making a mistake?[/quote]

Did you use the profile?

From profile you hit "Creat your own blog"

[url]http://www.artofproblemsolving.com/Forum/usercp.php?mode=blog&sub=create[/url]

on another note when it says "Owner" at the Blogs page how do change that to a picture?

Go to Blog Options>Customizatio

and in teh Custom CSS block copy and paste:

[code]
strong {
font-size: 0pt;
font-color: #F3F3F3;
content:url('url of picture goes here');
}[/code]

Where I put 'url of picture goes here', put your http url to that picture.
Is that what you want?

umm no not really.

when you go to

[url]http://www.artofproblemsolving.com/Forum/apsjour.php[/url]

the blog list shows owners with their avatar pictures. (if u made ur own blog then yours should be on the top). my blog under the owner column only shows my name, but i see other people with picutures of their avatar under owner...how do i change it?

Aah. yes. I see your avatar.

You don't, but I do.

[quote="anirudh"]He already has a blog :roll:[/quote]

yes Dude...I already have a blog...but i amde it 2 days after not getting reply from this post.......

thanks


algebra
function
domain
superior algebra
superior algebra solved

Show that a Dedekind domain with only finitely many primes is a PID.


This challenge is reserved for Middle School students who have never attended a State or National Competition.  Please DO NOT ANSWER if you have attended a State or National Competition.

Express $4.\overline{2}$ as a common fraction.

[quote="math92"][hide]
$4 \frac{2}{9}$
[/hide][/quote]

Where's the work?  ;)

I have never attended a state competition, let me try...

[hide]Let $n=4.\overline{2}$
$10n=42.\overline{2}$
Subtracting, we have
$9n=38$

$n=\frac{38}{9}=4\frac{2}{9}$[/hide]

[hide]well i know .222... is equal to 2/9 so 4.222... must be 4 and 2/9[/hide]

[quote="rcv"]So far, there are no correct answers.  [This challenge is still open only to those who have never attended a State or National competition.][/quote]

Exactly why I got a point off on my chapters, I didn't follow directions

[hide]$4\frac{2}{9}=\frac{38}{9}$, and 38/9 is the answer. [/hide]

[hide="Hint"]You do know what a common fraction is, right...?[/hide]

I've attended a Nationals round twice, so I'm just posting a hint :P .


Not sure if this is the right place for this, but here goes:

[b]2SinASinB = Sin[A+B] + Sin[A-B][/b]

I know that this is true, but I can't figure out how to prove it.
Help anyone?

I've tried doing it like this:

sin(A+B)=sin A cos B + cos A sin B
sin(A-B)=sin A cos B - cos A sin B
(sin A cos B + cos A sin B) + (sin A cos B - cos A sin B)

but then, I end up with:
2 sin A cos B

How do I solve this?

it doesn't seem true to my calculator
are you sure it's correct?

Well, it's not true.
Counterexample: $ 2{\sin{\frac{\pi}{2}}\sin{0} = 0 \neq \sin{\frac{\pi}{2}} + \sin{\frac{\pi}{2}} = 2}$

Related [i]correct[/i] identities:

$ 2 \cos a \cos b \equal{} \cos(a \plus{} b) \plus{} \cos(a \minus{} b)$
$ 2 \sin a \sin b \equal{} \cos (a \minus{} b) \minus{} \cos(a \plus{} b)$ (note the change in sign!)
$ 2 \sin a \cos b \equal{} \sin (a \plus{} b) \plus{} \sin (a \minus{} b)$
$ 2 \cos a \sin b \equal{} \sin (a \plus{} b) \minus{} \sin (a \minus{} b)$

After division by $ 2$, these are the [b]product-to-sum[/b] formulae.


Se tiene un triangulo cualquiera ABC, se conoce el valor de AC, AB y el angulo entre ellos "Beta" (en el caso del attachment 90 grados) que sera variable, tambien el angulo "Alfa" se sabe que es constante.

Alguien sabe identificar alguna formula que calcule el valor AD o AE o AF simplemente introduciendo en ella un nuevo valor del angulo "Beta".

Estuve buscando la solucion por varios dias y no me sale...(la vejez ??) tiro la toalla y pido ayuda.

Gracias
Diego

tres preguntas y una idea:

a) ¿Qué es D?
b) ¿Qué es E?
c) ¿Qué es F?

Consejo: Me parece que podría salir usando Stewart y algún otro teorema, el más conveniente cuando sepa el significado de esos puntos... :(

Gracias por interesarte PacMan2812 !

Los puntos D, E y F son puntos cualquiera, lo que me interesa es introducir un angulo menor que el existente (90 en este caso) y que la formula me de el valor del segmento.

En el caso D, el angulo podria ser 30 grados.
En el caso E, el angulo podria ser 45 grados.
En el caso F, el angulo podria ser 75 grados.

Puse esos puntos a modo de ejemplo.
Con la formula correcta deberia poder conocer la medida del segmento resultante, simplemente introduciendo el angulo deseado (entre 1 y 89 en este caso).

Si, Pacman, el teorema de Stewart le pega en el palo a lo que yo necesito.... pero necesita alguna variante, porque en mi caso yo no se cuanto valen los segmentos BD (o BE o BF) ni CB (o CE o CF), lo que conozco son los angulos. Me estoy rompiendo la cabeza para encontrar la relacion entre los angulos y las medidas de los segmentos...
En fin, me voy a llorar a otro lado.
Saludos

Bueno, me contestaron en otro lado en ingles, igual gracias a Pacman.
La respuesta satisfactoria la tiene el teorema de senos

El segmento en cuestion mide = AB * seno(B)
                                           -------------
                                               seno (D, E o F)


integration
calculus
calculus computations

if
\int (upper limit b, lower limit a) f(x)dx = 12.6   and
\int (upper limit b, lower limit 0) f(x)dx = 3.9, then
\int (upper limit a, lower limit 0) f(x) dx = ?

a. 0
b. 16.5
c. 8.7
d. it does not exist...

My answer is C) 8.7

^^^
Try to use latex..
$ \int^b_a f(x)dx \equal{} F(b)\minus{}F(a)\equal{}12.6$
$ \int^b_0 f(x)dx \equal{} F(b)\minus{}F(0)\equal{} 3.9$
$ \int^a_0 f(x) dx\equal{}F(a)\minus{}F(0)\equal{}\minus{}12.6\plus{}3.9\equal{}\minus{}8.7$