modular arithmetic
arithmetic sequence
number theory unsolved
number theory

[quote="AndrewTom"]What is the least possible value of the greatest term of an arithmetic sequence of 7 primes?[/quote]

It's $ 2$ : $ \{2,2,2,2,2,2,2\}$

If you modify the problem : "What is the least possible value of the greatest term of an arithmetic sequence of 7 [u][b]different [/b][/u]primes?", then :

WLOG consider $ r>0$ (because the problem is about the greatest, not the last) and let then the sequence be $ \{p,p\plus{}r,p\plus{}2r,p\plus{}3r,p\plus{}4r,p\plus{}5r,p\plus{}6r\}$

If $ r\neq 0\pmod{2}$, one of $ (p,p\plus{}r)$ is even, so $ p\equal{}2$ and $ p\plus{}2r$ is not prime; So $ r\equal{}0\pmod{2}$

If $ r\neq 0\pmod{3}$, one of $ (p,p\plus{}r,p\plus{}2r)$ is divisible by $ 3$, so $ p\equal{}3$ and $ p\plus{}3r$ is not prime; So $ r\equal{}0\pmod{3}$

If $ r\neq 0\pmod{5}$, one of $ (p,p\plus{}r,p\plus{}2r, p\plus{}3r, p\plus{}4r)$ is divisible by $ 5$, so $ p\equal{}5$ and $ p\plus{}5r$ is not prime; So $ r\equal{}0\pmod{5}$

If $ r\neq 0\pmod{7}$, one of $ (p,p\plus{}r,p\plus{}2r, p\plus{}3r, p\plus{}4r, p\plus{}5r, p\plus{}6r)$ is divisible by $ 7$, so $ p\equal{}7$ and we have a sequence
$ \{7,7\plus{}30a,7\plus{}60a,7\plus{}90a,7\plus{}120a,7\plus{}150a,7\plus{}180a\}$
$ a\equal{}1$ then  the sequence is $ \{7,37,67,97,127,157,187\}$ and $ 187$ is not prime.
$ a\equal{}2$ then  the sequence is $ \{7,67,127,187,247,307,367\}$ and $ 187$ is not prime.
$ a\equal{}3$ then  the sequence is $ \{7,97,187,277,367,457,547\}$ and $ 187$ is not prime.
$ a\equal{}4$ then  the sequence is $ \{7,127,247,367,487,607,727\}$ and $ 247$ is not prime.
$ a\equal{}5$ then  the sequence is $ \{7,157,307,457,607,757,907\}$ and all numbers are primes.

If $ r\equal{}0\pmod{7}$, the sequence is $ \{p,p\plus{}210b,p\plus{}420b,p\plus{}630b,p\plus{}840b,p\plus{}1050b,p\plus{}1260b\}$ and the last (greatest) is $ >1260$

So the answer is $ 907$


[quote="Otávio_Menezes"]Let $ABC$ be a triangle with $AB=AC$. Let $D$ be a point on $BC$ such that $BD=2CD$ and $P$ a point on $AD$ such that $A\hat{B}P=B\hat{A}C$. Prove that $B\hat{A}C=2D\hat{P}C$.[/quote]
What do you mean by $2D\hat{P}C$? do you mean angle?

[quote="Otávio_Menezes"]Let $ABC$ be a triangle with $AB=AC$. Let $D$ be a point on $BC$ such that $BD=2CD$ and $P$ a point on $AD$ such that $A\hat{B}P=B\hat{A}C$. Prove that $B\hat{A}C=2D\hat{P}C$.[/quote]
I have done this problem before, I believe it's supposed to be $\angle BPD=\angle BAC$


trigonometry
geometry
circumcircle
cyclic quadrilateral
geometry unsolved

From point P of arc BC of circumscribed circle of triangle ABC perpendiculars PX, PY and PZ are dropped to BC, CA and AB, respectively. Prove that:   [b]BC/PX = AC/PY + AB/PZ[/b]

I know trigonometry solution, but i cannot find "normal" solution. Can anybody help me?

Theorem 2 in [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=157240#157240]http://www.mathlinks.ro/Forum/viewtopic.php?t=24945 post #3[/url].

  Darij

first of all by $ S_{MNL}$ we mean the area of $ \triangle MNL$and also let $ AB=c,BC=a,CA=b,\angle BAC=\angle A,\angle CBA=\angle B,\angle ACB=\angle C$

now we know that:

$ 2S_{APB}=PZ.c=AP.BP.\sin APB$

$ \Rightarrow PZ.c=AP.BP.\sin C$

$ \Rightarrow \frac{PZ}{c}=\frac{AP.BP.\sin C}{c^{2}}$

now by $ \sin$ law in $ \triangle ABC$ we know that $ \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R$ where $ R$ is the circumradius...
so we get that:

$ \frac{PZ}{c}=\frac{AP.BP}{c.2R}$

and in a similar way we get that:

$ \frac{PY}{b}=\frac{CP.AP}{b.2R}$

$ \frac{PX}{a}=\frac{AP.BP}{a.2R}$

so we have to prove that:

$ \frac{a.2R}{BP.CP}=\frac{b.2R}{CP.AP}+\frac{c.2R}{AP.BP}$

thus its sufficient to show that:

$ \frac{a}{BP.CP}=\frac{b}{CP.AP}+\frac{c}{AP.BP}$

now multiply both sides by $ AP.BP.CP$,so it remains to prove that:

$ a.AP=b.BP+c.CP$

which is true because of [b]ptolemy`s theorem[/b] in cyclic quadrilateral $ ABPC$...  :wink:

Lema.
in triangle ABC, Let H be the point on BCsuch AH is perpendicular to BC, s is a diameter of triangle ABC. Then AB*AC=AH*s

Let $ k$ be a point on $ ZY$ such $ PK$ is perpendicular to $ ZY$. By Lema, $ PY*PZ=PK*PA, PX*PZ=PK*PB,PX*PY=PK*PC$
By ptolemy theorem,  $ BC*PA=AC*PB+AB*PC$. multiply $ PK$, t $ BC*PY*PZ=AC*PX*PZ+AB*PX*PY$. Divde $ PX*PY*PZ$ then proven.


f(x) is differentiable
f(x+y)=f(x-y)+2f(y)+xy
f'(0)=1

find f(x)

what I tried
[hide]
let x=0
f(y)=f(-y)+2f(y)
f(y)=-f(-y)

let y=0
f(x)=f(x)+2f(0)
f(0)=0

let x=y
f(2x)=2f(x)+x^2

let f(x)=x
xy=0

f(x)=akx^k + a(k-1)x^(k-1) + ... + x

stuck here[/hide]

[quote="max1546"]f(x) is differentiable
f(x+y)=f(x-y)+2f(y)+xy
f'(0)=1

find f(x)[/quote]

Let $ y\ne 0$ and divide both sides by $ 2y$ and you get $ \frac{f(x\plus{}y)\minus{}f(x\minus{}y)}{2y}\equal{}\frac{f(y)}{y}\plus{}\frac x2$

Set $ y\to 0$ and this becomes $ f'(x)\equal{}f'(0)\plus{}\frac x2\equal{}\frac x2\plus{}1$ and so $ f(x)\equal{}\frac{x^2}4\plus{}x\plus{}a$

Plugging this back in the original equation, we get $ \frac {y^2}2\plus{}2a\equal{}0$ and so no solution.


number theory
greatest common divisor
number theory proposed

Find all natural numbers $ n>1$ such that the sum $ 2^2 \plus{} 3^2 \plus{} 4^2.... \plus{} n^2$ is a power of a prime number.

In closed form, the sum is $ \frac{1}{6}n(n+1)(2n+1)-1$.  Expanding this out and factoring, we get $ \frac{1}{6}(n-1)(2n^2+5n+6)$.  Since $ (n-1)(2n+7)=2n^2+5n-7$, the two factors must be coprime or have a gcd of 13.  We can now examine cases.

[u]Case 1[/u]:  One of the factors equals 1, the other is of the form $ 6\cdot p^m$.

\begin{align*}
n-1&=1\\
n&=2
\end{align*}
When $ n=2$, the sum is clearly $ 4$, which is a power of a prime number.

$ 2n^2+5n+6=1$
$ n$ is not a real number

[u]Case 2[/u]:  One of the factors equals 2, the other is of the form $ 3\cdot p^m$.

\begin{align*}
n-1&=2\\
n&=3
\end{align*}
When $ n=3$, the sum is 13, which is a power of a prime number.

$ 2n^2+5n+6=2$
$ n$ is not a real number.

[u]Case 3[/u]:  One of the factors is 3, the other is of the form $ 2\cdot p^m$.

\begin{align*}
n-1&=3\\
n&=4
\end{align*}
When $ n=4$, the sum is 29, which is a power of a prime number.

$ 2n^2+5n+6=3$
$ n$ is not a positive integer.

[u]Case 4[/u]:  One of the factors is 6, the other is of the form $ p^n$.

\begin{align*}
n-1&=6\\
n&=7
\end{align*}
$ 2n^2+5n+6=139$, which is a power of a prime number.

$ 2n^2+5n+6=6$
$ n$ is not a positive integer.

[u]Case 5[/u]:  The two factors are of the form $ a\cdot 13^x$ and $ b\cdot 13^y$, such that $ ab=6$.

There are 8 separate sub-cases to check (setting the two factors equal to 13, 26, 39, and 78) in this case.  One of these sub-cases gives the repeated solution of $ n=3$ and none of the other sub-cases give solutions.

The solutions are therefore $ n=2, 3, 4, 7$.

[hide]$ 2^{2}\plus{}3^{2}\plus{}4^{2}....\plus{}n^{2} \equal{} \dfrac{n(n\plus{}1)(2n\plus{}1)}{6} \minus{} 1 \equal{} \dfrac{(n\minus{}1)(2n^2\plus{}5n\plus{}6)}{6} \equal{} p^k$

Thus, $ (n\minus{}1)(2n^2\plus{}5n\plus{}6) \equal{} 6p^k$.

Note that $ gcd(2n^2\plus{}5n\plus{}6 , n\minus{}1) \equal{} gcd(13 , n\minus{}1)$, and that $ 2n^2\plus{}5n\plus{}6 > n\minus{}1$ for all $ n \in \mathbb{N}$.

So, if $ 13 \nmid n\minus{}1$, then we must have $ n\minus{}1 \equal{} \dfrac{6}{q}$ AND $ 2n^2\plus{}5n\plus{}6 \equal{} q p^k$ where $ q \mid 6$.

If $ 13 \nmid n\minus{}1$, then $ n\minus{}1 \equal{} 1,2,3,6 \Rightarrow n \equal{} 2,3,4,7$. Which yield sums of $ 4,13,29,139$ which are all primes raised to a power.

If $ 13 \mid n\minus{}1$ then we must have $ n \equal{} 13m\plus{}1$ for some $ m \in \mathbb{N}$ AND $ p\equal{}13$.

Then, the equation becomes: $ (13m)(338m^2\plus{}117m\plus{}13) \equal{} 6\cdot13^k \Rightarrow m(26m^2\plus{}9m\plus{}1) \equal{} 6 \cdot 13^{k\minus{}2}$.

However, $ gcd(26m^2\plus{}9m\plus{}1,m) \equal{} 1$, $ 26m^2\plus{}9m\plus{}1 > m$, and $ 26m^2\plus{}9m\plus{}1 \ge 36$ for all $ m \in \mathbb{N}$.

Thus, we must have $ m \equal{} \dfrac{6}{q}$ AND $ 26m^2\plus{}9m\plus{}1 \equal{} q 13^{k\minus{}2}$ where $ q \mid 6$.

Then, $ m \equal{} 1,2,3,6 \Rightarrow 26m^2\plus{}9m\plus{}1 \equal{} 36, 123, 262, 991$ none of which are divisible by $ 13$.

Therefore, the only such $ n$ are $ n \equal{} 2,3,4,7$.[/hide]


geometry
3D geometry
sphere
prism
combinatorics proposed
combinatorics

There is a lamp in space.(Consider lamp a point)
Do there exist finite number of equal sphers in space that the light of the lamp can not go to the infinite?(If a ray crash in a sphere it stops)

i post 4 month ago but no one gave a soln

Maybe it's time to post a solution?
In plane, I think it's not true, but it space it seems true.
What about other dimensions?
I mean , I know a problem like this:
Given a system of
a) Circles
b) Spheres
is it always possible to move far away from the system one of the pieces ?

Well, it's clear that we can block a lamp in the plane with circles, but the spacial analog seems really tricky.

Well consider all lattice points in $2n*2n*2n$ cube with center the lamp.
Now consider a sphere of radius $\frac13$ of center of each lattice point.
Now use Minkowski theorem for big $n$

If $A$ is a convex set of points in space with volume more than 8 and if $(x,y,z) \in A$ then $(-x,-y,-z) \in A$ then $A$ has a lattice point except $(0,0,0)$

Now that you said, I remember a solution....
The great idea is to take all the lattice points, then build circles of equal radiuses $R$ in those latice points and with Minkowski you show that for a quite big radius , at the distance $\frac{1}{R}$ there will be no light. Am I right? Or maybe left? Just kidding .... :D

Sorry to bump such an old problem but I have a comment here. I encountered this problem(labeled as Iran 2003) in the form:

Prove that there does not exist a finite set of spheres of equal radius such that no two intersect and that the ray never goes to infinity.

I have been trying to solve this problem for a while with no success. Does anyone have a solution?

Without using Minkowski, it's enough to use compactness. The claim can be also generalized to $\mathbb R^d$, for any finite $d$.

Place the light in the origin, and place equal open $d$-discs of any radius $r< 1/2$ centered at every lattice point except the origin. Then put a $(d-1)$-sphere $S$ of the same radius $r$ centered at the origin, and consider the projections of all discs on $S$. Every disc projects to an open subset of $S$, while the projections of all lattice points on $S$ form a dense subset of $S$. Therefore the projections form a cover of $S$ and, by compactness of $S$, there's a finite subset of projections which still covers it. Keep the corresponding discs, and remove all the others. Now you're blocking every ray of light with a finite number of arbitrarily small equal discs in $\mathbb R^d$.

@RaduSorici: either the above proof is flawed, or you misread the problem. Maybe there were more constraints on the placement of your spheres?

[quote="MindFlyer"]... the projections of all lattice points on $S$ form a dense subset of $S$. Therefore the projections form a cover of $S$ and ...[/quote]
Take the interval $[0,1]$; enumerate the rationals in it by $(r_n)_{n\geq 1}$. Choose $\varepsilon > 0$ and open intervals $I_n = (r_n - \varepsilon/2^{n+1}, r_n + \varepsilon/2^{n+1})$. Then the union of the $I_n$ has measure at most $\varepsilon$, so it does not cover $[0,1]$.

Thus just qualitative density and compactness arguments are not enough.

Oh right, I don't know why I said that. However the proof can be fixed quite straightforwardly. Any straight line through the origin passes arbitrarly close to some other lattice point, so the projections of the disks indeed cover the sphere.

A similar problem appears in Engel and he claims six sphere is sufficient.

Consider a rectangular prism with three different sized faces and the lamp is the center point of the prism. Then consider 6 infinite cones that orgininate at the center of the prism and each contain one face of the prism. Then these cones intersect and cover the light source. We can inscribe spheres into these cones. If the spheres intersect then we can vary the dimensions of the prism so that they no longer intersect.

Any complete solution to this problem? I am unable to understand the application of Minkowski's theorem mentioned in one of the posts above.Also, the initial constraint includes the fact that no two spheres touch each other. I am pretty sure that the problem can be translated in terms of circles instead of spheres and the plane $\mathbb{R}^2$. Any complete solutions using Minkowski's theorem??


when is it?
how much time do we have?
are calculators allowed?

[quote="CatalystOfNostalgia"]you get a calculator.[/quote]

No 89s or HP50s as far as I can remember.


ratio
trigonometry
function
linear algebra
matrix
arithmetic sequence
geometric sequence

:arrow: Consider that $ \{a_1, a_2, a_3,$ $ \ldots\}$ is an arithmetic progression in wich the first term $ a_1$ is $ \alpha \neq 0$.
Let $ \{b_1, b_2, b_3,$ $ \ldots\}$ be a sequence in such way that $ \{b_1 \minus{} a_1, b_2 \minus{} a_2, b_3 \minus{} a_3,$ $ \ldots\}$ forms, in the order the terms are presented, a geometric progression.
Let $ \beta$ be the ratio of the geometric progression with $ 0 < \beta < 1$. Let $ b_1 \equal{} \cos(\phi)$ with $ \frac {\pi}{4} < \phi < \frac {\pi}{2}$, $ b_k \equal{} \gamma$ and $ k > 1$. Determine the ratio of the arithmetic progression in function of the variables $ k$, $ \alpha$, $ \beta$ and $ \gamma$ and the term $ b_1$ such as the matrix $ A \equal{} \left[\begin{array}{cc} \tan(\phi) & 1 \minus{} \sin(\phi) \\
1 & 1 \end{array} \right]$ do not admit the submatrix that has the largest determinant.

[i][Problem revised and presented exactly as proposed][/i]

Explaining what [i]"do not admit the submatrix that has the largest determinant"[/i] means would be quite enough. I think it means the we shoud determine the submatrix that has the greatest value and then take the elements of this submatrix and take them out from $ A$. What is the mathematical meaning of this? How do I proceed?


limit
logarithms
floor function
ceiling function
number theory unsolved
number theory

Statement of the problem:\\
Calculate the sum of the series :
\[L=\lim_{n\to + \infty}\sum_{d|n}\frac{1}{d!+1}\]
\\

what i've done in the following lines is only establishing some
lower and upper bounds.
i don't know if i'ts correct,and i would like to ask for suggestions.
also if someone know a solution you are welcome to post it.
the problem is not from any contest currently open(as i've seen happens
these days on mathlinks) so it's perfectly "legal" to solve :)
thank you again for any suggestions or solutions.

altough the problem asks for more i couldn't
manage to solve it.
	So the inferior bound can be established if we consider
\[p_{n}\]
,instead of n, the prime number sequence wich should have the same limit
as the initial series.
	So the sum becomes:
$L=\lim_{n\to + \infty}\sum_{d|p_{n}}\frac{1}{d!+1}=\frac{1}{0!+1}+\frac{1}{1!+1}=1$

So the limit should be 1 but that does not make for a proof.
We will consider 1 as lower bound because it can be easily
seen that any n has 2 divisors and the sum has always at least
two ters adding up to 1.

	For a simple upper bound :
\[lim_{n\to + \infty}\sum_{d|n}\frac{1}{d!+1}< \lim_{n\to + \infty}\sum_{d|n}\frac{1}{d!}< \sum_{i=1}^{\infty}\frac{1}{i!}=e<2.72 \]
So we have an upper bound namely e\\
	For a better upper bound we will see in the following lines what we can do.
	To proove this new upper bound we will first proove:
\[{(2) \sum_{i=1}^{2^{k}}\log_{2}(i) \geq \{4k^{3}+3k^{2}-k}\]
proof:
	we consider the sequence
\[1,2,2^{2},2^{3},...,2^{k-1},2^{k}\]
These are some of the values that i takes in sum (2)
	wich are also powers of 2 so the logarithm will be a nice integer
	(they will have to be multiplied by a constant to be brought to ln)

	So the sum of the logarithms will be
\[\frac{k(k+1)}{2}\]
.
	We can extract from the interval
\[[1,2^{k}] \]
numbers whose logarithm
	can be minorated.
	We choose the following numbers:
\[2+1,2^{2}+1,2^{3}+1,...,2^{k-1}+1 \]
,so we can say that the sum of the logarithms
	of these numbers is for sure bigger than
\[1+2+3+...+(k-1) \]

\[\]
We can use the same thing for the following numbers
\[2^{2}+2,2^{3}+2,...,2^{k-1}+2 \]
And we obtain a lower bound wich is
\[2+3+...+(k-1) \]
.
	If we generalise we find that at step
\[i \]
the sum wich is a lower bound:
\[\frac{k(k+1)}{2}-\frac{i(i+1)}{2}\]
.
	If we add all the sums we get:
	we obtain
\[4k^{3}+3k^{2}-k \]
,
	so :
\[\sum_{i=1}^{2^{k}}\log_{2}(i) \geq 4k^{3}+3k^{2}-k \]

\[4k^{3}+3k^{2}-k \geq 5k^{3}\]
for k sufficiently large.
	inverting we obtain:
\[\frac{1}{\sum_{i=1}^{2^{k}}\log_{2}(i)}\leq \frac{1}{4k^{3}+3k^{2}-k}\leq \frac{1}{5k^{3}}\leq \frac{1}{k^{3}}\]
we are interested only in the first and last members:
\[\frac{1}{\sum_{i=1}^{2^{k}}\log_{2}(i)}\leq \frac{1}{k^{3}}\]
we write it this way too:
$\frac{1}{\sum_{i=1}^{2^{\lfloor \log_{2}{d}\rfloor }}\log_{2}(i)}\leq \frac{1}{ \lceil\log_{2}{d}\rceil^{3}}$
	we multiply by the constant we talked about at the beggining so that we get the
	logarithms in base e
	the constant will be
\[c=\frac{1}{\log_{2}{e}}\]
and making the notation
\[t= \lfloor \log_{2}{d}\rfloor \]
also the notation
\[h= \lfloor \ln{d}\rfloor \]
$\frac{1}{\sum_{i=1}^{d+1}\ln(i) }\leq \frac{1}{\sum_{i=1}^{2^{t}}\ln(i) }\leq \frac{1}{ct^{3}}\leq \frac{1}{h^{3}}$

now summing up to infinity both first and last terms we obtain we get

$\sum_{d=1}^{+\infty}\frac{1}{\sum_{i=1}^{d+1}\ln(i) }\leq \sum_{h=1}^{+\infty}\frac{1}{h^{3}}=\zeta(3)\thickapprox 1.202 \\ \\ \sum_{d=1}^{+\infty}\frac{1}{\sum_{i=1}^{d+1}\ln(i)}\geq \sum_{d|n}\frac{1}{\ln((d+1)!)}\geq \sum_{d|n}\frac{1}{d!+1}$

so:
\[\sum_{d|n}\frac{1}{d!+1}< 1.202 \]

i am sorry ... someone suggested to me that the series is not at all convergent.
:(
i believe him because he came up with 2 subseries with different limit points.
i found this also but i didn't wanted to believe it,i thought the series had a limit so the problem must certainly be wrong.
does someone on the forum know of any problem similar to this ?
if so please give me a link to it.
thank you

The first counterexample I could come up with is this one:

Let $f(n) = \sum_{d | n}\frac1{d!+1}$. Then consider $L_{1}= \lim_{k \to \infty}f \left( 2^{k}\right) = \sum_{k=0}^\infty \frac1{\left( 2^{k}\right)!+1}$ and $L_{2}= \lim_{k \to \infty}f \left( 3^{k}\right) = \sum_{k=0}^\infty \frac1{\left( 3^{k}\right)!+1}$.
$L_{1}$ and $L_{2}$ obviously exist and are finite, but $L_{1}-L_{2}= \sum_{k=0}^\infty \left( \frac1{\left( 2^{k}\right)!+1}-\frac1{\left( 3^{k}\right)!+1}\right)$ is surely $> 0$, so $L_{1}\neq L_{2}$.

But there are more...

Note that $\sup \left\{ f(n) : n \in \mathbb N \right\}= \sum_{n=1}^\infty \frac1{n!+1}$ ($=S$). That's true because:
(i) $f(n) \leq S$;
(ii) $f(n!) > \sum_{k=1}^{n}\frac1{k!+1}$, from where, with the help of (i), we get that $\lim_{n \to \infty}f \left( n! \right) = S$.

We have that $\sum_{k=1}^{7}\frac1{k!+1}\approx 1.0258 \ldots$, so by [url=http://mathworld.wolfram.com/LagrangeRemainder.html]this[/url] we must have $1.0258 < S < 1.0258+5 \cdot 10^{-4}$.

PS: $\inf \left\{ f(n) : n \in \mathbb N \right\}= \frac12$ since $f(p) \to \frac12$ for large primes $p$.

very good perfect_radio
it seems the range is small
what is it with this series ?
what can one study about it ?
i find it quite interesting and it seems it does not converge
because if one takes a random number in N ,it cannot be
said by the size of that number how much divisors it has and
also what are the divisors.


The $ \emph{Catch The Spirit}$ group is conducting a raffle. Each ticket
costs $ \$$2, and the total expenses are $ \$$500. What is the
minimum number of tickets that must be sold to yield a profit
of $ \$$2000?

We know that each ticket we sell, we earn $ 2$ dollars.  However, our total expense is $ 500$ dollars, so we need to subtract $ 500$ from our earnings to calculate our profit.  Call the number of tickets that you have to sell to get a profit of $ 2000$ dollars $ n$.  We can then set up the equation $ 2n \minus{} 500 \equal{} 2000\implies2n \equal{} 2500\implies n \equal{} \boxed{1250}$


inequalities
quadratics
geometry unsolved
geometry

Let be given a convex polygon $ M_0M_1\ldots M_{2n}$ ($ n\ge 1$), where $ 2n \plus{} 1$ points $ M_0$, $ M_1$, $ \ldots$, $ M_{2n}$ lie on a circle $ (C)$ with diameter $ R$ in an anticlockwise direction. Suppose that there is a point $ A$ inside this convex polygon such that $ \angle M_0AM_1$, $ \angle M_1AM_2$, $ \ldots$, $ \angle M_{2n \minus{} 1}AM_{2n}$, $ \angle M_{2n}AM_0$ are equal. Assume that $ A$ is not coincide with the center of the circle $ (C)$ and $ B$ be a point lies on $ (C)$ such that $ AB$ is perpendicular to the diameter of $ (C)$ passes through $ A$. Prove that
\[ \frac {2n \plus{} 1}{\frac {1}{AM_0} \plus{} \frac {1}{AM_1} \plus{} \cdots \plus{} \frac {1}{AM_{2n}}} < AB < \frac {AM_0 \plus{} AM_1 \plus{} \cdots \plus{} AM_{2n}}{2n \plus{} 1} < R
\]

[quote="hungkg"]Who can help me?Thank?[/quote]

I only can prove the last inequality (by using vector). The first and the second ones are harder and I haven't gotten any idea for them.

do someone have an idea for this problem ? i think it looks like some kind of means inequalities. that is : harmonic mean < geometric mean < arithmetic mean < quadratic mean. in fact the first two inequalities are equivalent ;)


ratio
arithmetic sequence
geometric sequence
number theory proposed
number theory

Let the sequence $\left(a, a + d, a + 2d, \ldots, a + nd, \ldots\right)$ be an infinite arithmetic progression, where $d \neq 0$. Prove that this sequence contains an infinite subsequence which is a geometric progression if and only if the ratio $\frac{a}{d}$ is rational.

  Darij

- If $\displaystyle \frac a d = \frac p q$ for some integers $p,q$ and $q > 0$ :
Note that for all integers $k \geq 0$, we have $(1+q)^k = 1 + m_k \cdot q$ for some nonegaitive integer $m_k$.
Let $n_k = p\cdot m_k$. Then :
$\displaystyle a+n_kd = \frac {dp} q (1+n_kq) = \frac {dp} q (1+q)^k$. Thus all terms of the geometrical sequence $\displaystyle \{ \frac {dp} q (1+q)^k \}$ are terms from the original arithmetical sequence.

- Conversely, assume that the arithmetical sequence contains a subsequence which is a geometrical sequence with ratio $q$. That is, there is a increasing sequence $\{n_i\}$ of nonnegative integers and real number $\alpha$ such that $a+n_id = \alpha \cdot q^i$ for all $i$. Let $x = \frac a d$.
Since $d \neq 0$ we must have $r \neq 1$ and $\alpha \neq 0$.
Now, for all $i$, we have :
$\displaystyle q = \frac {a+n_{i+1}d} {a+n_id} = \frac {x + n_{i+1}} {x + n_i}$.
Thus :
$\displaystyle x = \frac {n_iq - n_{i+1}} {1-q}$. (1)

Moreover :
$\frac {n_iq - n_{i+1}} {1-q} = \frac {n_{i+1}q - n_{i+2}} {1-q}$ so that $\displaystyle q = \frac {n_{i+1} - n_{i+2}} {n_i - n_{i+1}}$ is rational.
Thus, from (1), $x$ is rational too, and we are done.

Pierre.

(Note mostly for students in my 235 class:)

[quote=pbornsztein]- If $\displaystyle \frac a d = \frac p q$ for some integers $p,q$ and $q > 0$ :
Note that for all integers $k \geq 0$, we have $(1+q)^k = 1 + m_k \cdot q$ for some nonegaitive integer $m_k$.
Let $n_k = p\cdot m_k$. Then :
$\displaystyle a+n_kd = \frac {dp} q (1+n_kq) = \frac {dp} q (1+q)^k$. Thus all terms of the geometrical sequence $\displaystyle \{ \frac {dp} q (1+q)^k \}$ are terms from the original arithmetical sequence.[/quote]

Careful: $n_k$ will be negative if $p$ is negative. This needs a little bit of yak shaving (by throwing away the first few entries of our sequence, we can ensure that $a$ and $d$ have the same sign, and then $p$ will be positive).

[[b]EDIT:[/b] A full detailed solution can now be found in [url=http://www.cip.ifi.lmu.de/~grinberg/t/20f/mps.pdf]my [i]Mathematical Problem Solving[/i] notes[/url] (Exercise 4.10.10 in the current numbering; if the numbering changes, search for "infinite geometric progression").]


Demonstrati ca pentru orice n natural exista n nr naturale a caror suma este si si cel mai mic multiplu comun al numerelor.

Exista si pentru $n=2$?

Daca avem $a,b > 0$ astfel incat $\textrm{cmmmc}\left( a, b \right) = a+b$, atunci $a | a+b, b | a+b$. Cum $a,b < a+b$, rezulta ca $a,b \leq \frac{a+b}2$, i.e. $a = b$. Dar in acest caz avem $\textrm{cmmmc}\left( a, a \right) = a \neq a+a$  :(

[hide="parere"]pt $n\geq 3$ merge sa consider (?) :
$x,y,2,1,1,1,\ldots,1,1$.
unde $x$ impar si $y\in\left\{1,2\right\}$.
cel mai mic multiplu comun al numerelor este $2x$.
ramane sa rezolv ecuatiile in $x$ si $y$ :
$2x=x+y+2+n-3$, sau $x=n-1+y$.
acum aleg $y=1$ daca $n$ este impar si $y=2$ daca $n$ este par.
[/hide]


In following ,construction means using compass and ruler.
Let $s$ be a given circle & $AB$ & $CD$ two segments s.t they dont intersect inside the circle .
[color=blue]Problem1[/color]find the point $X$ on the premeter of $s$ such that $XA $ and $BX$ cut $CD$ in $E$,$F$ and $EF$= a given constant.
2.[color=blue]Problem2[/color] let $\mathcal E$ be a given point in $CD$ .find $X$ on the premeter of $s$ such that $XA$&$XB$ cut $CD$ in $E$&$F$ and $\mathcal E$ be the midpoint of $EF$.

guys ,my bad ,$S$ is a circle  :blush: (What were i thinking about?)(grobber remark,Thanks)


Suppose $ x_n>0$ and $ y_n\equal{}\sqrt{x_n\plus{}\sqrt{x_{n\minus{}1}\plus{}\cdots\plus{}\sqrt{x_2\plus{}\sqrt{x_1}}}}$ for all $ n\in\mathbb N$. Prove that $ \{x_n\}$ converges iff $ \{y_n\}$ converges.

The "if part" $ (\Leftarrow)$ is easy. We can see that $ x_n\equal{}y_n^2\minus{}y_{n\minus{}1}$. Take the limits for both sides we get

$ \lim_{n\to\infty} x_n \equal{}\lim_{n\to\infty}(y_n^2\minus{}y_n).$

Since the limit of $ y_n^2\minus{}y_n$ exists, the sequence $ \{x_n\}$ converges.

I also try the "only if part" $ (\Rightarrow)$. Originally, I thought about the Weierstrass Theorem that "monotonic & bounded $ \Rightarrow$ convergent". However, I found some mistake in my "proof". Now I am trying, but I would like to know how do you prove it.

Suppose that $ \{x_{n}\}$ converges, and set $ \varepsilon > 0$.
We have:
$ |x_{n} \minus{} x_{m}| \equal{} |y_{n}^{2} \minus{} y_{m}^{2} \plus{} y_{m} \minus{} y_{n}| \equal{} |y_{n} \plus{} y_{m} \minus{} 1|\cdot|y_{n} \minus{} y_{m}|$
Since $ x_{k} > 0$ it follows $ y_{k} > x_{1}^{\frac{1}{2^{k}}}\ \stackrel{k\to\infty}{\longrightarrow} 1> \tfrac{3}{4}$, and it is easy to conclude that for sufficiently large $ n, m$ there is $ y_{n} \plus{} y_{m} \minus{} 1 > \tfrac{1}{2}$.
We have also $ |x_{n} \minus{} x_{m}|<\frac{\varepsilon}{2}$ for almost all $ n,m$, due to convergence of $ \{x_{n}\}$, so all but finite number of pairs $ (n,m)\in \mathbb{N}\times \mathbb{N}$ satisfy:
$ |y_{n} \minus{} y_{m}| < 2|x_{n} \minus{} x_{m}| < \varepsilon$, hence $ \{y_{n}\}$ is Cauchy Sequence, and we are done.


trigonometry
inequalities unsolved
inequalities

Prove that:
  $ \frac{1}{1\plus{}cos4a}\plus{}\frac{1}{1\plus{}cos8a}\plus{}\frac{1}{1\minus{}cos12a}>2$

[quote="karis"]Prove that:
  $ \frac {1}{1 \plus{} cos4a} \plus{} \frac {1}{1 \plus{} cos8a} \plus{} \frac {1}{1 \minus{} cos12a} > 2$[/quote]
Try Cauchy-Schwartz and $ 4\cos^22a \equal{} t.$ :wink:


calculus
integration
trigonometry
function
real analysis
real analysis unsolved

Find all continuous $ f$ such that

$ \int_{a}^{b}f(t) \sin t \, dt = \int_{a}^{b}f(t) \cos t \, dt = 0$.

I was only able to do this (I think) for $ a=-\pi$ and $ b=\pi$  :maybe:

Consider $ S=\{\sin x,\cos x, \sin 2x, \cos 2x, ... \}$.

Now,
1)$ \int_{-\pi}^{\pi}\sin nx \cos mx dx = 0$ for all $ n,m$.
2)$ \int_{-\pi}^{\pi}\sin nx \sin mx dx = \pi \delta_{nm}$
3)$ \int_{-\pi}^{\pi}\cos nx \cos mx dx = \pi \delta_{nm}$

This shows that $ S$ is an [i]orthogonal[/i] set of functions on $ [-\pi,\pi]$.

Now [url=http://en.wikipedia.org/wiki/Convergence_of_Fourier_series]Dirichlet's Fourier Convergence[/url] theorem shows that $ S$ is a [b]closed[/b] set of orthogonal functions on $ [-\pi,\pi]$. Using a result from harmonic analysis, which states: "a system of orthogonal functions is closed if and only if it is complete".

The completeness property states that given an orthogonal system of functions $ \{\phi_{n}(x)\}$ and $ \int_{a}^{b}f(x) \phi_{n}(x) dx = 0$ then $ \int_{a}^{b}f^{2}(x) dx = 0$ if $ f(x)\not \in \{\phi_{n}(x)\}$.

So if $ \int_{a}^{b}f(x)\sin x dx = 0$ and $ \int_{a}^{b}f(x)\cos x dx =0$ then either $ \int_{a}^{b}f^{2}(x) dx = 0$ or $ f(x) = \sin nx$ or $ f(x)=\cos nx$ for $ n\geq 2$ by completeness. Since $ f(x)$ is continous then $ f(x)\equiv 0$ on $ [-\pi,\pi]$.

So,
$ 0,\sin 2x,\cos 2x, \sin 3x, \cos 3x , ...$
Are all such functions.

Uh, no. Linear combinations work, too. The set of solutions is very large; it's a subspace of an infinite-dimensional space with codimension 2.

That assumes that $ a$ and $ b$ are fixed; if we require that the condition hold for all pairs $ a,b$, $ f$ clearly must be zero.

[quote="jmerry"]Uh, no. Linear combinations work, too. The set of solutions is very large; it's a subspace of an infinite-dimensional space with codimension 2.
[/quote]
So in what I did above, the answer would be all the finite linear combinations of the functions? For the specific example.

Can it also be an [i]infinite[/i] linear combination? (As in seperation of variables when solving a partial differential equation).

I do mean infinite linear combinations; if you want to be technical, it's the closure of the set of finite linear combinations. If we're on an interval of length $ 2\pi$, the condition is precisely that the $ \pm 1$ terms of the Fourier series are zero.


I am pretty sure you can just take a/b the x distance and a/b the y distance and apply the Pythagorean Theorem.

Use the formula that I wrote [url=http://www.artofproblemsolving.com/wforum/viewtopic.php?t=703]here[/url]
[size=75]I knew it. I should've kept it a sticky[/size] :) </jk>

And yes, this is MATHCOUNTS level. (or slightly below)

Well, the x distance is 2-(-1) = 3, and the y distance is 4 - 1 = 3. The x coordinate of the point we want to find is -1 + (1/3)*3 = 0, while the y coordinate is 1 + (1/3)*3 = 2, yielding (0, 2). Note how I used 1/3 and not 2/3 in my equation, since you wanted the point 2/3 of the way from (2,4) to (-1,1), and not from from (-1,1) to (2,4).

The m:n is the ratio of AM and BM, so in this case it'll be m=2, n=1 (2:1)


Xan, Yan, and Treething are playing a game. Two dice are rolled, and the numbers are added together. If a 1, 2, 3, or 4 shows up, Xan wins. If a 5, 6, 7, or 8 shows up, Yan wins. If anything else shows up, Treething wins. Treething, being a Mathcounts national champion, realizes that this game is unfair, Yin having the advantage. How many ways are there to distribute the sums to Xan, Yan, and Treething so that the game is fair? Two ways are not considered different if two players’ winning sums are just switched. (For example, Xan having a, b, and c, and Yin having d, e, f, g, h is the same as Xan having d, e, f, g, h, and Yin having a, b, c.)

I was going to put this in my competition, but I realized I didn't know how to solve it.  :P


What is the digit in the tens place when $ 7^{2005}$ is expressed in decimal notation?

let's list out the exponential values of 7

7^1 = 07
7^2 = 49
7^3 = 343
7^4 = 2401
7^5 = 16807
7^6 = 117649
7^7 = 823543
7^8 = 5764801

as you can see, in the tens digit the numbers go back and forth between $ 0$ and $ 4$
with some figuring we find out that $ 7^{2005}$ has a tens value of $ 0$

edit: beaten to it....

edit2: turns out that when using latex, u can't really control what lines the numbers are on...

You could also note that the last 2 digits repeat every 20 (more specifically, for all integers $ a$, $ a^{22}\equiv a^2\text{ (mod 100)}$)  Thus, we just look at the tens digit of $ 7^5$.  I happen to know that $ 7^4\equal{}2401$...

[hide=sol]By Euler's totient theorem, $7^{40}$ is equivalent to 1 modulo 100, so $7^{2005}\equiv 7^{(2005-40\cdot50)}=7^{2005-2000}=7^5=16807\equiv 07$ modulo 100, so the answer is 0[/hide]


If $\tan{\alpha}$ and $\tan{\beta}$ are the roots of $x^2 - px + q = 0$, and $\cot{\alpha}$ and $\cot{\beta}$ are the roots of $x^2 - rx + s = 0$, then $rs$ is necessarily
$\text{(A)} \ pq \qquad \text{(B)} \ \frac{1}{pq} \qquad \text{(C)} \ \frac{p}{q^2} \qquad \text{(D)} \ \frac{q}{p^2} \qquad \text{(E)} \ \frac{p}{q}$

[hide]
$\cot\alpha$ is the reciprocal of $\tan\alpha$, and $\cot\beta$ is the reciprocal of $\tan\beta$.

Therefore, if $\tan\alpha$ and $\tan\beta$ are the roots of $x^2-px+q=0$, then $qx^2-px+1=0$ (the coefficients reversed) has roots $\cot\alpha$ and $\cot\beta$.

Dividing the equation by $q$, we get $x^2-\frac{p}{q}x+\frac{1}{q}=0$, so $rs=\frac{p}{q^2}$.

$\boxed{\text{(C)}}$

[/hide]

[hide="Answer"]$r=\cot \alpha +\cot \beta$, $s=\cot \alpha \cot \beta$, $rs=(\cot \alpha +\cot \beta)(\cot \alpha \cot \beta)\Rightarrow$
$rs=\left(\frac{1}{\tan \alpha}+\frac{1}{\tan \beta}\right)\left(\frac{1}{\tan \alpha \tan \beta}\right)=\frac{\tan \alpha +\tan \beta}{(\tan \alpha \tan \beta)^2}$
$p=\tan \alpha +\tan \beta$, $q=\tan \alpha \tan \beta$, $rs=\frac{p}{q^2}\Rightarrow \boxed{C}$[/hide]

[hide]$\cot {\alpha}$ and $\cot {\beta}$ are the roots of $qx^2-px+1=0$*, or $x^2-p/qx+1/q=0$, so $(p/q)(1/q)=p/q^2\implies C$.

*if you don't understand this, read AoPS 2: polynomials :D[/hide]


With the Physics Olympiad preliminary test coming soon, where can I go to cram in a very rigorous and comprehensive review of the physics required? I don't believe the physics problems presented to me in school are creative enough to prepare for the olympiad.

You can't cram for physics, you have to take the time to understand it.  [url=http://www.amazon.com/Problems-Solutions-Mechanics-Universities-Qualifying/dp/9810212984/ref=sr_1_3?ie=UTF8&s=books&qid=1261661087&sr=8-3-spell]This[/url] series of books has hundreds of Ph.D. level problems on mechanics, E&M, quantum,...

Yea, I know I have to understand it. I've never been the one to just take concepts and memorize them. I don't really accept things unless I can prove it or understand it. Thanks for the link, but is PhD level physics necessary for knowing physics well enough to do well on preliminary/semifinal of Physics Olympiad?

The problems aren't really that hard, although they are PhD problems, them emphasize a lot more on the concepts than complicated math (if you know calculus, and are willing to learn vector calc, you should be fine).  Don't be intimidated by the "PhD" label.

Just have a solid mechanics textbook to back this up (Kleppner&Kolenkow is good, Goldstein is excellent).


Let $ a_1,a_2,...,a_n$ with aritmetic mean equals zero; what is the value of:
$ \sum_{j=1}^n{\frac{1}{a_j(a_j+a_{j+1})(a_j+a_{j+1}+a_{j+2})...(a_j+a_{j+1}+...+a_{j+n-2})}}$ , where $ a_{n+k}=a_k$ ?

ps. the real [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=242972]final round[/url] is still open..

Hint: You can always find numbers $ u_1$, $ u_2$, ..., $ u_n$ such that $ u_i \minus{} u_{i \plus{} 1} \equal{} a_i$ for every $ i\in\left\{1,2,...,n\right\}$. Rewrite your sum. Then think of Lagrange interpolation.

I remember seeing this problem in an old AMM (elementary problems, I think).
[b]EDIT:[/b] Ah, it was Crux Mathematicorum problem #2004 in Crux 1995/12 by Waldemar Pompe.

  darij

[quote="darij grinberg"]Then think of Lagrange interpolation.[/quote]

you can also get a simpler solution only observing one addend  :D


[quote="MIA6"]How do you convert 6x^4-3x^2+1 to (3x^2 / 2 - 1)^2 + 15x^4 / 4?  I think we normally divide 6x by 6, but here it's not the case. Then how can I figure out? thanks for help.[/quote]

You'll have to “complete the square”:

$ 6x^4\minus{}3x^2\plus{}1 \equal{}$
$ \equal{} \frac{15}{4}x^4 \minus{} \frac{15}{4}x^4 \plus{} 6x^4\minus{}3x^2\plus{}1 \equal{}$
$ \equal{} \frac{15}{4}x^4 \plus{} \frac{9}{4}x^4 \minus{}3x^2\plus{}1 \equal{}$
$ \equal{} \frac{15}{4}x^4 \plus{} \left(\frac{3}{2}x^2 \minus{} 1\right)^2$

Hope this helps.


Hello everyone,

If my answer is wrong, could anyone please show me how to get the answer as stated in the book?

p(x) = 2x^2 - 1, x < 0

y = 2x^2 - 1

x = 2y^2 - 1, y < 0

2y^2 = x + 1

y^2 = (x + 1)/2

y = sqrt[(x + 1)/2]

y = - sqrt[(x + 1)/2] = p-1(x)

However, the answer in the book is p-1(x) = -0.5 sqrt(2x + 2).

Thank you.

They are both correct:
$ -\sqrt{\frac{x+1}{2}}=-\sqrt{\frac{2x+2}{4}}=-\frac{\sqrt{2x+2}}{\sqrt{4}}=\frac{\sqrt{2x+2}}{2}=-0.5\sqrt{2x+2}$.
You should rationalize the denominators in you fractions.

It would be better to leave it as

$ -\frac{\sqrt{2x+2}}{2}$

, that's simplest radical form, while the other is not.

Thank you very much, archimedes1 and Temperal1, for your help!


function
integration
probability
induction
symmetry
calculus
calculus computations

Let $ I_{n}$ denote $ \int_{ \minus{} \infty}^{\infty} x^{n} \phi(x)\ dx$, where $ \phi(x)$ is the normal probability function given by $ \phi(x) \equal{} \frac {1}{\sqrt{2\pi}}e^{\frac { \minus{} 1}{2}x^{2}}$.

By splitting up the  integrand as the product of $ x^{n \minus{} 1}$ and $ x\phi(x)$, establish the reduction formula $ I_{n} \equal{} (n \minus{} 1)I_{n \minus{} 2}$ provided that $ n > 1$.

Given that $ I_{0} \equal{} 1$, find an expression for $ I_{n}$ for any positive integer $ n$.

$ \phi'(x)\equal{}\minus{}x\phi(x)$ so

$ I_{n\plus{}2}\equal{}\int_{\minus{}\infty}^\infty x^{n\plus{}1}\cdot x\phi(x)dx \equal{}$
$ \left[\minus{}x^{n\plus{}1}\phi(x)\right]_{\minus{}\infty}^\infty\plus{}\int_{\minus{}\infty}^\infty (n\plus{}1)x^n \phi(x)dx\equal{}(n\plus{}1)I_n$

and $ I_0\equal{}1, I_1\equal{}0$ since it is odd,

so $ I_{2n}\equal{}(2n\minus{}1)I_{2(n\minus{}1)}\equal{}(2n\minus{}1)(2n\minus{}3)\cdots 1\cdot I_0\equal{}\frac{2n\minus{}1!}{n\minus{}1!}2^{1\minus{}n}$
while $ I_{2n\plus{}1}\equal{}0$ since its always odd

One note: the correct definition is $ \phi(x)\equal{}\frac1{\sqrt{2\pi}}e^{\minus{}\frac{x^2}2}.$ You left out the square root.

An alternative (with the induction hiding in a different place):

It is clear by symmetry that $ \int_{\minus{}\infty}^{\infty}x^n\phi(x)\,dx\equal{}0$ for odd $ n.$ Hence we will consider only $ I_{2k}.$

$ I_{2k}\equal{}2\int_0^{\infty}\frac{x^{2k}}{\sqrt{2\pi}}e^{\minus{}x^2/2}\,dx.$

Now make the substitution $ t\equal{}\frac{x^2}2$ or $ x\equal{}\sqrt{2t}.$ That gives us $ x^{2k}\equal{}2^kt^k$ and $ dx\equal{}\frac1{\sqrt{2t}}\,dt.$ The integral becomes:

$ I_{2k}\equal{}\frac{2^{k\plus{}1/2}}{\sqrt{2\pi}}\int_0^{\infty}t^{k\minus{}1/2}e^{\minus{}t}\,dt$

$ \equal{}\frac{2^k}{\sqrt{\pi}}\,\Gamma\left(k\plus{}\frac12\right).$

Now, we have that $ \Gamma\left(k\plus{}\frac12\right)\equal{}\left(k\minus{}\frac12\right)\Gamma\left(k\minus{}\frac12\right).$ (That's where the induction is hiding.)

So $ I_{2k}\equal{}2\left(k\minus{}\frac12\right)I_{2k\minus{}2}\equal{}(2k\minus{}1)I_{2k\minus{}2}.$

We have $ I_0\equal{}\frac{\Gamma\left(\frac12\right)}{\sqrt{\pi}}\equal{}1.$ Which had better be the case.

$ I_2\equal{}\frac2{\sqrt{\pi}}\,\Gamma\left(\frac32\right)\equal{}\frac2{\sqrt{\pi}}\cdot\sqrt{\pi}\cdot\frac12\equal{}1$ Yes, that's the variance of the standard normal.

$ I_4\equal{}3I_2\equal{}3.$ $ I_6\equal{}5I_4\equal{}15.$ And so on.

Thanks, Kent. I've edited it; I typed \sqrrt instead of \sqrt.


linear algebra
matrix
linear algebra unsolved

Let $\{S_{n},\, n\ge1\}$ be a sequence of $m\times m$ matrices such that $S_{n}S_{n}^\mathrm{T}$ tends to identity matrix. Prove that there exist a sequence $\{U_{n},\, n\ge1\}$ of orthogonal matrices such that $S_{n}-U_{n}\to0,\, n\to\infty.$

For large $n,\ S_{n}$ will be invertible. Let $S_{n}=P_{n}U_{n}$ be the polar decomposition of $S_{n}$, i.e. $P_{n}$ is the unique symmetric positive definite matrix such that $P_{n}^{2}=S_{n}S_{n}^{t}$, and $U_{n}$ is orthogonal.

Since $S_{n}S_{n}^{t}\to 1,\ P_{n}\to 1$ as well (the operation of taking square roots is continuous on the set of real symmetric matrices). We have $S_{n}-U_{n}=(P_{n}-1)U_{n}$. Now, $(U_{n})_{n}$ is bounded, while $P_{n}-1\to 0$, so $S_{n}-U_{n}\to 0$, as desired.

Or if $S_{n}=(v_{1}, ..,v_{n})$ one easily see that $U_{n}=(w_{1},..,w_{n})$ works where $(w_{1},..,w_{n})$ is obtained from $(v_{1},..,v_{n})$ by the Gram-Schmidt process, I think.


Given that $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}=n$ and $\frac{a}{c}+\frac{b}{a}+\frac{c}{b}=m$ both m,n are natural numbers so prove that : $|a|=|b|=|c|$.

davron latipov

Let $x_{1}=\frac{a}{b},x_{2}=\frac{b}{c},x_{3}=\frac{c}{a}$, then
$\sigma_{1}=x_{1}+x_{2}+x_{3}=n,\sigma_{2}=x_{1}x_{2}+x_{2}x_{3}+x_{3}x_{1}=m,\sigma_{3}=x_{1}x_{2}x_{3}=1$, therefore $x_{i}$ are roots $x^{3}-nx^{2}+mx-1$. This equation had only rational roots if and only if m=n=3. But you don't say, that $a,b,c$ are rational. Therefore it had many solutions: $a,ax_{2}x_{3},ax_{3}$, were $x_{i}$ are solutions $x^{3}-nx^{2}+mx-1=0$.

Yes the numbers are rational. I forgot to write.

davron latipov


geometry
3D geometry
probability
Alcumus
MATHCOUNTS
algebra
polynomial

Here are a couple of questions I need help with:

1. A large cube is dipped into red paint and divided into 125 smaller cubes. What is the probability that one  randomly selected cube will have at least 25% of its surface area painted red?

[hide="My answer"]I got 71/125 because 54 cubes will have 1 side painted red (9x6), but the answer is 44/125. How?[/hide]

2. In a game, you can earn 0,1,3,7,or 10 points with each shot of an arrow. How many scores less than 30 cannot be made in 3 shots?

3. I just realized that i stink at counting problems. What should I do?

1. If you have the Counting and Probability Books, go over the lessons that you are unsure about.
2. Go on Alcumus often, and watch the videos.
3. I don't know any other tips.

[hide="1"][hide="First way"]If $ V \equal{} 125$, then $ s \equal{} 5$. $ 3$ cubes in every edge have $ 2$ sides ($ 25$%) painted red and there are $ 12$ edges. Hence there are $ 36$ cubes with $ 2$ sides painted red. Now, we notice that only the corner cubes have $ 3$ sides painted red and there are $ 8$ of them. $ \frac {36 \plus{} 8}{125} \equal{} \boxed{\frac {44}{125}}$.[/hide]
[hide="Second way"]The reason you got it wrong is because you forgot there are cubes with no sides painted. There are $ (5 \minus{} 1 \minus{} 1)^3 \equal{} 3^3$ (why?) of such cubes. And as you have already stated, there are $ 54$ cubes with only one side painted. Hence our answer is $ \frac {125 \minus{} (54 \plus{} 27)}{125} \equal{} \boxed{\frac {44}{125}}$.[/hide][/hide]

[hide="2"]The best solution I can think of is casework albeit very systematically.

[hide="Cases"]
10+10+10=30
10+10+7=27
10+10+3=23
10+10+1=21
10+10+0=20
10+7+7=24
10+7+3=20
10+7+1=18
10+7+0=17
10+3+3=16
10+3+1=14
10+3+0=13
10+1+1=12
10+1+0=11
10+0+0=10
7+7+7=21
7+7+3=17
7+7+1=15
7+7+0=14
7+3+3=13
7+3+1=11
7+3+0=10
7+1+1=9
7+1+0=8
7+0+0=7
3+3+3=9
3+3+1=7
3+3+0=6
3+1+1=5
3+1+0=4
3+0+0=3
1+1+1=3
1+1+0=2
1+0+0=1
0+0+0=0
[/hide]

Analyzing the data, we find the missing numbers to be $ 19, 22, 25, 26, 28,$ and $ 29$. Hence, our answer is $ \boxed{6}$.[/hide]

[quote="pytheagle"]
The reason you got it wrong is because you forgot there are cubes with no sides painted.[/quote]

Dangit, how did I miss that??

Also, is there an AoPS book that only focuses on counting?

EDIT: Wait I just found one called intro to counting and probability. Is that a good book to use, because in the description of it there doesnt seem to be a lot of counting involved. Also, are there any other books that would help with mathcounts counting?

Yeah. I have both the Introduction and the Intermediate. They're worded really well, and teach you ez ways to do things.

For #2, I think you can try generating functions. In \[ (1\plus{}x^3\plus{}x^7\plus{}x^{10})^3,\] find the missing terms (powers of $ x$ that aren't in the expansion in the range from $ 1$ to $ x^{29}$). The ones that don't appear should be the ones that you can't make.

I'm not exactly sure about this since I didn't multiply it out, so could someone confirm it? If it works, it would be a LOT easier than casework.

Yes, that does work.  However, raising a polynomial with 4 terms to the third power takes a little time.

Anyway, you should buy the Intro CP book and (if you can) take the class.  Alcumus is also a good place to get practice problems.


Maybe this isn't a technical problem, but... what steps can be taken to lower spam levels on AoPS? (MathLinks has far less, probably because there are more math forums on the ML part and more "spammable" forums on the AoPS side.)

The ratings system is a good idea, but has generated spam itself... for example,
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=147632[/url]. It also does nothing to prevent spam; it discourages it, but may not be a strong enough incentive to stop spamming.

I thought ML and AoPS were exactly the same except for the styles? :ninja:

Yeah, but like the ML "section" is the olympiad forums, and the AoPS "section" has Middle School, High School, Contests, Other Topics, Regional+Local (I think...)

There's no way to lessen human spam except by more active moderators, and any method (i.e. post ratings) to discourage it aren't always effective.

One way would be to delete all non-math forums, but I don't think many people would like that.  :P

We can start enforcing a no spam policy in math forums? This might be too drastic, but the idea would be to delete all spam found in math forums, while letting the other forums go without such strict control. This would obviously be a very controversial proposal, so I'm not sure it would get anywhere.

Yep. I agree with the reporting policy. I've seen those kinds of things on other boards as well.

[quote="#H34N1"]We could have a "Report Post" thing, where a post is taken into consideration, and maybe even put in a state of pending if a responsible person reports it.

The person who reported it will be logged, and privileges will be curtailed if they are wrongly doing so[/quote]We have in mind such a feature for the forums, however that will take some coding time, and right now we're focusing all our time into more important projects like developing new books for students, and new classes.

[quote="7h3.D3m0n.117"]Haha I have been thinking of the spammers on AoPS for a while. My random ideas came to the "Janus List." I believe it was first mentioned on NUMB3RS as a list of international double/triple agents. I thought it would work to compile a list of spammers and publicly post it (blogs maybe) and try to use "positive peer pressure"  to discourage these people. But of course, that would be mean  :D .[/quote]

You know, this might be a good idea. I actually have several people in mind... :wink:


email
geometry
geometric transformation
reflection
AMC
USA(J)MO
USAMO

Today is the postmark date for admission tests. I don't think we are supposed to talk about it until much later, so please refrain from talking about scores until like Wednesday, just to be safe.

I did all of mine yesterday, so they are some pretty shoddy proofs, but I really want to get in.

The concept of this thread is very simple. If you receive your admission letter, you click Accepted. If you don't and it seems everyone else had their admission letters, you click the Rejected button.

I'm guessing that my recommendations are what got me in 0.o

You already received decisions with Admission Test B?

I sent in my application just before mail pickup on 4/9..and still havent turned in my recs yet, so it may be a while before I find out

Isn't tomorrow the day they send out the letters?

Has anyone gotten their admission letters yet.

 :maybe:

Dr. Andreescu called me and told me that it looked positive for me, but he still needed my recommendation letters. :) By the way, I got six correct, two small mistakes, and two unanswered.

I got my acceptance call from Dr. A a few minutes ago. I nailed 6 on the test, missed one because I read it incorrectly, and had one fairly decent proof (10), but I made a small assumption that I know is true but I don't know how to prove.

Penguin, I got a call and then I got and email 30 seconds later. But I'm in Dallas so I dunno. Long distance calls may be different.

[quote="kohjhsd"]email for me. I am disturbed because I got two questions wrong, where both were casework counting.  :([/quote]
Well, I'm disturbed because I suck at math and may not get in *at all*. So pick your battles. :wink:

[quote="roadnottaken"][quote="PenguinIntegral"]Well, I'm disturbed because I suck at math and may not get in *at all*. So pick your battles. :wink:[/quote]
You didn't answer my question.  How many did you answer/get correct/partial credit on the admissions test, if you don't mind my curiosity?[/quote]
Let's see, I wrote up very crappy proofs to 1,2,3,5,6,10.

I would say my chances are not good, but I'd at least like a rejection letter.

On reflection, I'd say I probably should have used my brute forcish methods for 2 other problems, because that's what the recommended solution is. :(

[quote="contradictory"]Accepted...[/quote]
It's good to know they are still sending letters.

*Really nervous now*

Does anyone know if they send rejection letters????

 :blush:  :blush:  :blush:
**SUPER NERVOUS**
- Nawahd

I'm not too sure. Good luck, though.  :)

(I did Test A and got in...)

wait, if they call they don't email, or the do both? Cuz I didn't receive any emails, but i got a call which nobody replied to on the on the other hand of the line, so i don't know if it was from them or  not :maybe:  :huh:  :maybe:

Do you really think that they would call long distance to Toronto? I figured that they qould just send an email...?

 :maybe:  :maybe:

I know. But to call is much easier than to write an email (at least for me). *Extremely Nervous*
Until what time do they send out all responses?

the third choice is so appealing. anyone who made it should vote it too...

Do the admissions/rejections always get sent out in one day, or they can be a bit extended towards the next day and beyond?

I got my email on a different day than others did.

Perhaps a day or two?

I can't exactly remember. I do remember not being able to function at school for a couple of days...

I'm in.
Advice for those who haven't received any letter from ASMP: call or Email Dr. Andreescu about it because for example my application and quiz were lost and never reviewed until today.

I'm in, and I just got it today.

So that means I'm a border choice. PHEW.

i got in...the test wasnt really that hard.  they called my mom yesterday

I got an official acceptance letter nine minutes ago because I finally got my recommendation letters in.

[quote="Aryth"]He called and said I got 8.5/10. Dunno what that's supposed to mean.[/quote]

ummmm...that means you got 8.5 questions out of 10

Specifically, it probably means that you answered 8 questions correctly, and got partial credit on a ninth (or maybe you answered 7 questions correctly and got partial credit on the other three).

[quote="Phelpedo"]@roadnottaken - you know that poem is supposed to be sarcastic, right? Frost was mocking some arrogant guy.[/quote]

the common interpretation is that frost is saying something to the effect of: choose the path of an individual

@ phelpedo: i don't know where you got your idea from...what details suggest that?

my school's english department gives the interpretation that: frost is divided between both choices...and the details in the poem suggest that both paths have their merits...

the real problem here is that you don't have the full poem so you are loosing out on the details in the other stanzas

I'm In YAY!!!!! Dr. Andreescu just called -- he said it was soo late because of my letters... lazy math teacher!

@ lena_siz: Since you're from Toronto, and you're going, wanna PM me and go on the same flight or sumthing? Cuz, I guess we're both going through Pearson.. no?

- Nawahd Werdna

I think that just about all of us got in (of the five rejection votes on the test A thread, I am almost certain that at least two were false alarms [late acceptance letters]).  And that is why AoPS rules! :yup:


function
integration
real analysis
real analysis unsolved

Let $f,g\ : \ [-a,a]\rightarrow R$ be two functions so that $a>0$, $f,g\in C^1$ and
$\left(\forall\right) x\in [-a,a],\ g(x)\ne 0,\ g(-x)=-g(x)$. Prove that
exists $d\in (-a,a)$ so that $\int^a_{-a} f(x)g(x) \ dx=2f'(d)\int^a_0 xg(x)\ dx.$

Do you mean that $g(x) \neq 0$ for $x\neq 0$ ? Then
\[ \int_{-a}^a f(x) g(x) dx = \int_{-a}^a (f(x)- f(0)) g(x) dx + \underbrace{\int_{-a}^a f(0) g(x) dx}_{{}= 0} = \int_{-a}^a \frac{f(x) -f(0)}{x} x g(x) dx = \int_{-a}^a f'(\xi_x) x g(x) dx. \]
where $0\leq |\xi_x| \leq |x|$. Since $x g(x)$ is either always positive or always negative, $x g(x) dx$ (or  $-x g(x) dx$) is a measure, therefore
\[ \int_{-a}^a f'(\xi_x) x g(x) dx = (\mathrm{average\ value\ of\ } f'(\xi_x)) \cdot{(\mathrm{measure\ of\ } [-a,a]\mathrm{\ w.r.t.\ }xg(x) dx) = f'(d) \int_{-a}^a x g(x) dx = 2 f'(d) \int_{0}^a x g(x) dx. }\]


[quote="Orif"]Does anybody have the PDF file "PROBLEMS IN LINEAR ALGEBRA" by Hojoo Lee. I'll be very thankful if you could help me. If anybody has the PDf file mail it to me please !
My e-mail: oorif14@mail.ru[/quote]

Tomorrow, I'll send it to you via email. (Now, I'm not at home.)


Find the bounds of
(a) $\sin{A}\sin{B}\sin{C}$
(b) $\sin{A}+\sin{B}+\sin{C}$
where $A,B,C$ are angles in a triangle.

I did (b) first.
(b)Say: $S = \sin A + \sin B + \sin C$
Sine is positive on $(0,180)$ so we know $0< S \leq 3$. We know $A$ and $B$ can become arbitrarily close to $0$ while $C$ becomes arbitrarily close to $180$. This tells us $S$ can approach, but never reach $\sin 0 + \sin 0 + \sin 180 = 0 + 0 + 0 = 0$. We know sine is concave down on $(0,180)$, so by Jensen's $\displaystyle\frac{\sin A + \sin B + \sin C}{3} \leq \sin\frac{A+B+C}{3} = \sin 60 = \frac{\sqrt{3}}{2}$. So, combining, we get $\boxed{0 < \sin A + \sin B + \sin C \leq \frac{3\sqrt{3}}{2}}$.
(a)Say: $P = \sin A \sin B \sin C$
Again, since sine is positive on the given interval we know $0 < \sin j \leq 1$ for all $j$ in it. So, since $A$, $B$, and $C$, can approach but not reach $0$, $0$, and $180$, respectively, $P$ can approach but not reach $\sin 0 \sin 0 \sin 180 = 0 \cdot 0 \cdot 0 = 0$. Now, by AM-GM, we know:

$\displaystyle\frac{\sin A + \sin B + \sin C}{3} \geq \sqrt[3]{P}$

And, just like in (b), Jensen's tells us:

$\displaystyle\frac{\sin A + \sin B + \sin C}{3} \leq \frac{\sqrt{3}}{2}$.

So, putting the two together, we get:

$\sqrt[3]{P} \leq \displaystyle\frac{\sin A + \sin B + \sin C}{3} \leq \frac{\sqrt{3}}{2}$

$\Rightarrow P \leq \left(\frac{\sqrt{3}}{2}\right)^3 = \frac{3\sqrt{3}}{8}$.

And we get:

$\boxed{0 < \sin A \sin B \sin C \leq \frac{3\sqrt{3}}{8}}$

**For both, equality occurs on the RHS when $A = B = C = 60^o$ due to Jensen.


when i go on peoples profiles, i see that theres no email provided in the email blank.  How can that happen?

There's an option to hide it because we don't want the little people asking us for help with mathcounts problems?


algebra
polynomial
probability
superior algebra
superior algebra solved

Let I an ideal in the set of complex polynomials in n variables and let P be a polynomial with the property that $ P(x_1,...,x_n)=0$ whenever $(x_1,...,x_n)$ is a common root of all polynomials of I. Prove that there exists $k$ such that $P^k\in I$. I think this one is one of the most difficult problems given in an oral examination at ENS.

This is Hilbert's Nullstellensatz for $\mathbb{C}[x_1,\dotsc,x_n]$.

Boris

En français le théorème des zéros de Hilbert

Thanks a lot, but it does not help me very much. This shows just the probability to solve this problem in one hour is well reduced to 0.


does the equation

$e^{x}=x$

have eny complex roots

These equation had not real solution, but had infinetely many complex solutions in any interval $2k\pi \le Im x<2(k+1)\pi .$


I looked on the Mathcounts website and they said:

February 1-26, 2006: Chapter competitions
Date & Location info. is being updated as it becomes available. (Please confirm this info. with your local MATHCOUNTS coordinator.)

Yes, yesterday was the last day to take the chapter competition.  What are you asking?  The ban on discussing scores, difficutly, problems, and answers ends today.

Here is a link to when you can post scores:
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=73354]www.artofproblemsolving.com/Forum/viewtopic.php?t=73354[/url]
It says do not discuss scores until after February 26.  Today is February 26, so I think you have to wait until 12:01 am tomorrow.

Doesn't it say on that forum 'announcement end March 11'? I'm assuming thats when we're allowed to start talking.
Edit: never mind, the competition window's over.

Although hypothetically, if my chapter's competition, which had already been delayed to 2/25 due to snow, was delayed again, they would have had to administer it next weedend.

Exactly why i said you shouldn't post questions yet.


calculus
integration
trigonometry
real analysis
real analysis solved

an easy one:
$I_n=\int_{0}^{+\infty} \frac{dx}{1+x^n}$
study the series : $\sum_1^{+\infty} u_n$ where $u_n=I_n-1$

We'd better make that $\sum_{n=2}^{\infty}u_n$ since $u_1=\infty.$

If we do that, then I know the series converges since I know a closed form expression for $I_n$. I'm working on doing it without that closed form expression.

Ah, here it is: By making the substitution $y=\frac1x$ (and then turning $y$ back into $x$), we get

$\int_1^{\infty}\frac{dx}{1+x^n}=\int_0^1\frac{x^{n-2}dx}{1+x^n}$

Hence $I_n=\int_0^1\frac{1+x^{n-2}}{1+x^n}dx.$

Subtract $\int_0^11\,dx$ to get

${u_n=\int_0^1\frac{x^{n-2}-x^n}{1+x^n}\,dx< \int_0^1(x^{n-2}-x^n})dx$

$ =\frac1{n-1}-\frac1{n+1}=\frac2{n^2-1}$

and the series converges.

Just for reference, the closed form expression I was talking about is

$I_n=\frac{\frac{\pi}n}{\sin\left(\frac{\pi}n\right)}.$

From that, $u_n=\frac1{6n^2}+O\left(\frac1{n^4}\right).$

That suggests the sum may be considerably smaller than the direct simple estimate above.


Let 0=x0<x1<...<xn=1 are n+1 real numbers. Each point xi (except x0 and xn) is a midpoint of segment [xi1,xi2] for some i1 and i2. Prove that all xi are rational numbers.

This problem continues "fantastic algebra" theme. In my opinion, described problems have more elegant solution than pbornsztein's one for "fantastic algebra".

:(  snif.... I don't deserve that.... snif...

Pierre.

Solution is based on the following statement:
Let y1,...,yn - real numbers, N is an arbitrary positive integer, N>1. Then there are fractions ai/a such that |yi-ai/a| < N.a for all i=1,2,...,n.

From the $x_i$'s, choose a basis for the vector space spanned by the $x_i$'s over the rationals. Assume both $1$ and the smallest irrational $x_i$ (call it $x_{i_0}$) belong to the basis.

Let $x_{j_0}$ be one of our numbers which have the greatest rational coefficient (call it $a$) multiplied by $x_{i_0}$ in the decomposition according to our basis. There must then be some $j_1>j_0$ s.t. $x_{j_1}$ also has $x_{i_0}$ multiplied by $a$ in its decomposition. Then we find some $j_2>j_1$ s.t. $x_{j_2}$ has the same property, and so on. We can only stop when we reach our largest number, which is $1$, but $1$ does not have $x_{i_0}$ in its decomposition, meaning that we have a contradiction. This means that there were no irrational $x_i$'s to begin with, Q.E.D.

I hope it is correct. I've been trying to post it for quite some time now (10-15 minutes), and this is the third version (the first two had fatal flaws :)).

It is the same solution than the one I proposed but that Myth considered as not elegant enough  :(

Pierre.

I remember that other problem, in the thread named "Fantastic Algebra". I never read that, but maybe it just seemed too long to Myth. I myself remember it was a significantly longer argument. Maybe I'm wrong though.

An aside: I think this is also an Iranian olympiad problem.


A ball is thrown upward from a second floor window of a building.  It takes 3 seconds for it to reach the ground below.  It strikes the ground with a velocity of -57 ft/sec.  Find the ball's initial velocity and the height from which it was thrown.  Acceleration due to gravity is -32 ft/sec^2

An entertaining possible approach: reverse time.

If we threw the ball upward from the ground at (a vertical component of) 57 ft/sec, where would it be and how fast would it be moving 3 seconds later?


[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14196[/img]

 :(

WLOG we assume $ P\not\equal{}O$.
Let $ A,X$ be the intersections of the circunference and $ OP$, with $ AP>PX$. With the compass we make a circle with center $ X$ and radius $ XP$ that intersect the first circle in $ B,C$. Because of the equalities $ BX\equal{}XC\equal{}XP$ we get that $ P$ is the incenter and $ O$ is the circuncenter of $ \triangle ABC$.
If $ P\equal{}O$ then we make the same steps than before but the line $ OP$ can be any diameter, and the triangle will be equilateral.

[img]http://img214.imageshack.us/img214/1060/constructionle0.png[/img]

[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14204[/img]

 :thumbup:


Hi this is a problem I have but I do not understand the solution please help!

[i]The resistances are $ R_1 = 2.0 \Omega$, $ R_2 = 4.0 \Omega$, $ R_3 = 6.0 \Omega$, and the indicated current is $ i = 6.0A$. The electric potential difference between points A and B that connect the section to the rest of the circuit is $ V_A - V_B = 78V$
(a) Is the device represented by "Box" absorbing or providing energy to the circuit, and (b) at what rate?[/i]

[b]THE SOLUTION[/b]
[i](a) The voltage across $ R_3 = 6.0 \Omega$ is $ V_3 = iR_3 = (6.0 A) (6.0 \Omega) = 36 V$.  Now, the voltage across $ R_1 = 2.0 \Omega$ is
 $ (V_A - V_B) - V_3 = 78 - 36 = 42 V$,
which implies the current is $ i_1 = (42 V)/(2.0 \Omega) = 21 A$.  By the junction rule, then, the current in $ R_2 = 4.0 A$ is $ i_2 = i_1 - i = 21 \times 6.0 = 15 A$.  The total power dissipated by the resistors is (using Eq. 26-27)
  $ (2.0 i_1^2) + (4.0 i_2^2) + (6.0 i^2) = 1998 W \approx 2.0 kW.$
By contrast, the power supplied (externally) to this section is $ P_A = i_A (V_A - V_B)$ where $ i_A = i_1 = 21 A$.  Thus, $ P_A = 1638 W$.  Therefore, the "Box" must be providing energy.

(b) The rate of supplying energy is $ (1998 - 1638 )W = 3.6\times 10^2 W$.[/i]

How is it that ${ V_1 = (V_A - V_B) - V_3}$. What about $ V_2$???
From My reasoning, the ratio of current through $ R_2$ and $ R_3$ is 6:4, hence $ i_2 = 9A, \ i_1 = 9 + 6 + 15A!$

Please help!

It's a parallel circuit.  They work differently than series circuits.  Recall your rules and go through the solution carefully.  It makes sense and appears correct.

I've looked through it again and answered my first question. But for my second, about the slitting of current, I always thought that at a junction the current will always split in ratios inversely according to the amount of resistance in the path. This gives the current answers I got. Is this not correct? Or does it only hold sometimes?

I'm not sure how much you know about circuitry but since we have no reason to believe that this is a non-conservative field, consult Kirchhoff's Rules.


Solve the following equation in the set of the real numbers :

$ \prod_{k\equal{}1}^n \left(x_{k}^{k}\plus{}k\minus{}1\right)\equal{}n!\prod_{k\equal{}1}^nx_k$

$ \prod_{k\equal{}1}^{n} (x_{k} ^k\plus{}k\minus{}1)\equal{}n!\prod_{k\equal{}1}^{n}x_{k}$
$ x_{k} ^k\plus{}k\minus{}1\equal{}x_{k} ^k\plus{}1\plus{}1\plus{}...\plus{}1\ge k x_{k}$   with $ AM\minus{}GM$  inequality
So $ \prod_{k\equal{}1}^{n}(x_{k}^k\plus{}k\minus{}1)\ge2*3*4*...*n \prod_{k\equal{}1}^{n}x_{k}\equal{}n!\prod_{k\equal{}1}^{n}x_{k}$,so
$ x_{1}\equal{}x_{2}\equal{}...\equal{}x_{n}\equal{}1$


calculus
integration
email
real analysis
real analysis theorems

Yesterday I had an exam in integration theory which I did quite poorly on (I believe I passed, but not with good grade). Anyway, I felt that the exam was unfair, particularly because I think the first problem is not correct. Before I send an email to my professor however I would like someone to check so that I am not mistaken.

The (first page of the) exam is here: http://lutfisk.se/Integration1.jpg

I believe the following defines a counterexample:
Set $X = \{1,2,3,\ldots\}$, define $\nu(\emptyset) = 0$, $\nu(\{1\}) = \nu(\{2\}) = 1$, $\nu(\{1,2\}) = \frac{3}{2}$. Now for any $A \subset X$ let $\mu(A) = \nu(A \cap \{1,2\})$.

Is this not a finite outer measure with the given property that is not additive?

Since $\mu$ is finite, monotone and countably subadditive, it is indeed a counterexample. Both questions 1a and 1b seem strange to me.

I think that there is no problem with question 1b. What would the problem be?

I think I had a very bad exam strategy, because I kept thinking about 1a) for way too long, not realizing that it is probably not worth that many points. I didn't even look at problem 6, and I couldn't do 3b). If I had skipped 1a) and went for 6 I would have scored a lot more points.

The link is down, could someone post the exam/problems ?
Thanks


geometry
cyclic quadrilateral
geometry proposed

Let $P$ a point out of the circle $\Gamma$. Draw two tangent lines $PA$ and $PB$ and a secant line $PCD$ such that $A, B, C, D$ are on the circle. $Q$ is a point on chord $CD$ such that $\angle{DAQ}=\angle{PBC}$.
Prove that $\angle{DBQ}=\angle{PAC}$.

[hide="Here is a solution which uses the properties of a harmonical cyclic quadrilateral."] Denote $C\in (PD)$ and $T\in CD\cap AB$. From the well-known property (see http://www.mathlinks.ro/Forum/viewtopic.php?t=43202 and
http://www.mathlinks.ro/Forum/viewtopic.php?t=43652 ),
the cyclic quadrilateral $ACBD$ is harmonically, i.e. $AD\cdot BC=AC\cdot BD$.

Therefore, $\boxed{\ \widehat{DAQ}\equiv \widehat{PBC}\ }\Longleftrightarrow \widehat{QAD}\equiv \widehat{CAQ}\Longleftrightarrow$

the rays $[AQ$ and $[AT$ are isogonals in the angle $\widehat{CAD}\Longleftrightarrow$

$\frac{QD}{QC}\cdot \frac{TD}{TC}=\left(\frac{AD}{AC}\right)^{2}$ $\Longleftrightarrow\frac{QD}{QC}\cdot \frac{TD}{TC}=\left(\frac{BD}{BC}\right)^{2}$ $\Longleftrightarrow$

the rays $[BQ$ and $[BC$ are isogonals in the angle $\widehat{CBD}\Longleftrightarrow$

$\widehat{DBQ}\equiv \widehat{CBT}$ $\Longleftrightarrow$ $\boxed{\ \widehat{DBQ}\equiv \widehat{PAC}\ }\ .$ [/hide]

Q  is the midpoint of CD.

   Thank you.

   M.T.


What would be the best EPGY courses to take in mathematics if i'm in Cal BC.

I'm planning to take Number Theory, but does it worth it to take it in EPGY. should i just take the Number Theory course through online at  http://www.utexas.edu/cee/dec and save money.

Is introductory physics equivalent to AP Physics?

anyone?
should i take Number THeory course in texas or EPGY. the price difference is about 200 dollars EPGY being more expensive...

EPGY physics fits the AP syllabus, with one of the courses being like the "B" course and one being like the "C" course. I have no opinion about the number theory issue, other than thinking that number theory on that level is a great subject for any math-liking high school student to study.

EPGY's number theory is definitely a very good course.  Whether it's worth the extra money, I don't know, as I've never even heard of the other course.


real analysis
calculus
integration
function
real analysis unsolved

Prove that id $f$ is integrable on an interval, then so is $f^{2}$.

Clearly, "integrable" in this question is intended in the Riemann sense.

For Lebesgue integrals (or for improper integrals), this is false: $\frac1{\sqrt{x}}$ is integrable on $[0,1];$ its square is not.

The cheap answer is to quote a famous theorem: a bounded function on a bounded interval is Riemann integrable iff the set of its discontinuities has measure zero. Since $f$ is integrable, it is continuous almost everywhere, and $f^{2}$ is continuous at (at least) every point that $f$ is, hence almost everywhere.


A line which passes through $O$, the point which lies at the intersection of the diagonals of the convex quadrilateral $ABCD$ :maybe: , cuts $[AB]$ and $[CD]$ in $E$ and $F$, respectively. Knowing that $\frac{AE}{AB}= \frac{CF}{CD}= r \in (0,1)$, prove that $AB,CD$ are parallel.

If, furthermore, $r = \frac12$ and $EF = \frac{AD+BC}2$, then prove that $ABCD$ is actually a parallelogram.

[i]Mihai Boloş[/i]

$\frac{AE}{AB}\cdot \frac{CD}{CF}= \frac{\triangle{AOE}}{\triangle{AOB}}\cdot \frac{\triangle{COD}}{\triangle{COF}}=\frac{\triangle{AOE}}{\triangle{COF}}\cdot \frac{\triangle{COD}}{\triangle{AOB}}$

Since $EF$, $AC$ and $BD$ are concurrent at $O$, so we can perform area ratios:

$\frac{\triangle{AOE}}{\triangle{COF}}\cdot \frac{\triangle{COD}}{\triangle{AOB}}= \frac{AO \cdot EO}{CO \cdot FO}\cdot \frac{CO \cdot DO}{BO \cdot AO}=\frac{EO}{FO}\cdot \frac{DO}{BO}=1$

Therefore $\frac{EO}{FO}=\frac{BO}{DO}$, thus $AB \parallel CD$.

If $r=\frac{1}{2}$, then $E$ and $F$ are the midpoints of $AB$ and $CD$, hence: \[\overrightarrow{EF}= \frac{\overrightarrow{AD}+\overrightarrow{BC}}{2}\] Thus: \[EF^{2}=\frac{AD^{2}+BC^{2}+2AD \cdot BC \cdot \cos{\angle{(AD,BC)}}}{4}\] But since $EF=\frac{AD+BC}{2}$, so we also have: \[EF^{2}=\frac{AD^{2}+BC^{2}+2AD \cdot BC}{4}\] Hence we get $\angle{(AD,BC)}=0^{o}$, thus $AD \parallel BC$. Therefore $ABCD$ is a parallelogram.


Each point in the plane is colored in red or blue. Prove that there exist two blue points a distance $ 1$ unit apart, or there exist four distinct collinear red points $ A_1,A_2,A_3,A_4$ such that $ A_1A_2\equal{}A_2A_3\equal{}A_3A_4\equal{}1$.