Find all positive integers $n \ge 3$ such that there exists a positive constant $M_n$ satisfying the following inequality for any $n$ positive reals $a_1, a_2,\dots\>,a_n$: \[\displaystyle \frac{a_1+a_2+\cdots\>+a_n}{\sqrt[n]{a_1a_2\cdots\>a_n}} \le M_n \biggl( \frac{a_2}{a_1} + \frac{a_3}{a_2} +\cdots\>+ \frac{a_n}{a_{n-1}} + \frac {a_1}{a_n} \biggr).\] Moreover, find the minimum value of $M_n$ for such $n$.

The difficulty is finding $M_n$...

The only possible value of $n$ is $3$. It is easy to see taking $a_i = m^{i-1}$ for large enough $m$.

For $n=3$ the inequality is
\[\frac{a_1 + a_2 + a_3}{\sqrt[3]{a_1 a_2 a_3}} \leq M \left( \frac{a_2}{a_1}+\frac{a_3}{a_2}+\frac{a_1}{a_3}\right).\]

Let $x=a_1/\sqrt[3]{a_1 a_2 a_3}$, $y=a_2/\sqrt[3]{a_1 a_2 a_3}$, $z=a_3/\sqrt[3]{a_1 a_2 a_3}$. Then $xyz=1$ and
\[x+y+z \leq M \left(  \frac{y}{x}+\frac{z}{y}+\frac{x}{z}\right).\]

It is clear that $M\geq 1$ (take $x=y=z=1$). Prove that $\min M=1$. It is enough to show that for positive $x, y, z$ with $xyz=1$,
\[x+y+z \leq \frac{y}{x}+\frac{z}{y}+\frac{x}{z}.\]
or
\[y\left(\frac{1}{x}-1\right) + z\left(\frac{1}{y}-1\right) + x\left(\frac{1}{z}-1\right) \geq 0.\]
Now if we take $a=1/x, b=1/y, c=1/z$, then $abc=1$ and the required inequality takes form
\[\frac{a-1}{b} + \frac{b-1}{c} + \frac{c-1}{a} \geq 0\]
which is a known beast.

I believe you are right! Also I was wrong; I meant that the difficulty was finding $n$ (i.e. proving $n=3$). Many of us conjectured that $n=3$ and found the minimal value of $M_n$, but failed to prove that $n=3$. (Actually I was one of them)


vector
analytic geometry
integration
calculus
trigonometry
derivative
function

I'm taking a class in math (vector calc and complex analysis) after graduating 2 years ago with an engineering degree. It's already difficult to me...

Anyway, here's my problem. We're asked to solve for the arc length of a curve using Cartesian coordinates, with the formula:
\[ {S} \equal{} \int_{} \left| \left| \frac {d\vec{{r}}}{dt} \right| \right|\,dt
\]
The particular problem I'm looking at is an astroid (4 point hypocycloid):
\[ \vec{r} \equal{} cos^3 {t} ~ \hat{i} \plus{} sin^3 {t} ~ \hat{j}
\]
on the interval
\[ t \equal{} 0 , t \equal{} 2\pi
\]
So I go through the steps of finding the magnitude of dr/dt the way the professor wants us to (by taking the root of the sum of the squares). That yields the following:
\[ {S} \equal{} 3 \int^{2\pi}_0 \sqrt {sin^2t ~ cos^2t} \,dt
\]
I know the danger of doing this is that you have be careful how you apply the square root.

I'm having two problems with this:

1) Ignoring the square root dilemma, what's the integral of cos(t)*sin(t)? Wouldn't sin(t)/2 AND -cos(t)/2 both work? The problem with using one or the other kind of ties into the next question below.

2) Now for the square root dilemma. Suppose for the integrand I use the half angle formula and get sin(t)/2 for the integral. How would you evaluate this from 0 to 2pi? Obviously if you just plugged in 0 and 2pi you'd get zero. Using my intuition I could just integrate from 0 to pi/2 and multiply by four to get the right answers, but this isn't very satisfying. How do you integrate this thing?

I'm having similar difficulty on another problem - a cardioid (doing it in a polar basis, but have the same problem with integration).

Thanks for your help!

You have to break it up- you're integrating $ |\sin t\cos t|,$ which has corners at each multiple of $ \frac{\pi}{2}$. There is no nice formula for an antiderivative that works across those corners.
Whenever you see an absolute value in the integrand, the same applies.
[quote="kcharlie"]I could just integrate from $ 0$ to $ \frac{\pi}2$ and multiply by four[/quote]
This is exactly the way to do it.

$ S\equal{}\int_\alpha^\beta\;\bigg|\;\frac{d\textbf r}{dt}\;\bigg|\;dt$
$ \equal{}\int_\alpha^\beta\;\sqrt{\;\left(\dfrac{dx}{dt}\right)^2\plus{}\left(\dfrac{dy}{dt}\right)^2\;}\;dt$

for a generic astroid, the parametric equations are $ x\equal{}a\;\cos^3\,t,\;\;\;y\equal{}b\;\sin^3\,t$

so, $ S\equal{}4\int_0^{\frac{\pi}{2}}\;\sqrt{\;9a^2\cos^4\,t\;\sin^2\,t\plus{}9b^2\sin^4\,t\;\cos^2\,t\;}\;dt$
$ \equal{}12\int_0^{\frac{\pi}{2}}\;\sqrt{\;a^2\,\cos^2\,t\plus{}b^2\,\sin^2\,t\;}\;\sin\,t\;\cos\,t\;dt$
$ \equal{}6\int_0^{\frac{\pi}{2}}\;\sqrt{\;a^2\plus{}(b^2\minus{}a^2)\,\sin^2\,t\;}\;(2\,\sin\,t\;\cos\,t)\;dt$
$ \equal{}6\int_0^1\;\sqrt{\;a^2\plus{}(b^2\minus{}a^2)\cdot u\;}\;du$

where the substitution $ u\;\equal{}\;\sin^2\,t$ has been used

$ S\equal{}6\left(\frac{2}{3}\;\frac{{\;(a^2\plus{}(b^2\minus{}a^2)\cdot u)\;}^{\frac{3}{2}}}{b^2\minus{}a^2}\right)\;\bigg|_0^1$
$ \equal{}4\;\dfrac{\left[\;(a^2\plus{}(b^2\minus{}a^2)\cdot 1)^{\frac{3}{2}}\;\minus{}\;(a^2)^{\frac{3}{2}}\;\right]}{b^2\minus{}a^2}$
$ \equal{}4\;\dfrac{a^2\plus{}ab\plus{}b^2}{a\plus{}b}$

put $ a\equal{}b\equal{}1$ to get $ \boxed{6}$. this is the curve length that you want.

I'm still a bit unsure how to handle these. So for the astroid, it reduces to sqrt(sin^2 + cos^2), which reduces to sin*cos only in the first and second quadrant (since it has to be positive). What about the other two quadrants? Do you just take the absolute value?

Another example I can think of is a cardiod with r=1-cos(t)
$ {S} \equal{} \int_{} \sqrt {dx^2 \plus{} dy^2}$

$ {S} \equal{} \int^{2\pi}_0 \sqrt {\dot{r}^2 \plus{} (r\dot{\theta})^2}dt$

$ {S} \equal{} \int^{2\pi}_0 \sqrt {\sin^2{t} \plus{} (1 \minus{} \cos{t} )^2}dt$

$ {S} \equal{} \int^{2\pi}_0 \sqrt {2(1 \minus{} \cos{t} ) }dt$

$ {S} \equal{} \int^{2\pi}_0 \sqrt {2(2\sin^2{\frac {t}{2} ) }}dt$

$ {S} \equal{} 2 \int^{2\pi}_0 \sqrt {\sin^2{\frac {t}{2} }}dt$

Okay... so how do we deal with this square root problem in this case? We could say that one solution to the integral is

$ \minus{} 2 \cos{\frac {t}{2}}$

However, this is only positive in the 3rd and 4th quadrants. What about the other two? In the previous problem (astroid), I arm-waved my way into reversing the sign in the negative quadrant. This doesn't work here - if I leave the sign of the integral as is and integrate from 0 to pi then pi to 2pi, I get the correct answer.

I guess I'm just confused over how to deal with the different quadrants - how do you break up the integral and what integral do you use over those sections?

Thanks!

There are enough inconsistencies in your derivative calculations that I'd like to redo them.
$ x'(t)\equal{}3\cos^2 t\cdot (\minus{}\sin t)$
$ y'(t)\equal{}3\sin^2 t\cdot (\minus{}\cos t)$
$ (x'(t)^2\plus{}y'(t)^2)\equal{}3^2\cdot \sin^2 t\cdot \cos^2 t\cdot (\sin^2 t\plus{}\cos^2 t)\equal{}9\sin^2 t\cos^2 t$.

All right, it is multiplication rather than addition or the ~ you originally wrote.

The general rule on dealing with absolute values: break up the integral at every point where it changes sign. Remember the piecewise definition: $ |u|\equal{}\begin{cases}u&u\ge 0\\\minus{}u&u<0\end{cases}$

$ S\equal{}2\int_0^{2\pi}\sqrt{\sin^2\left(\frac{t}{2}\right)}\,dt.$

(Note: the exact same integral occurs in the calculation of the arclength of one arch of a cycloid.)

Note that that's not $ \sin^2t;$ rather it is $ \sin^2\left(\frac{t}{2}\right).$ For $ 0\le t\le2\pi,$ the only quadrants that $ \frac{t}{2}$ belongs to are the first and second quadrants. In those quadrants, the sine is positive. That justifies our writing

$ S\equal{}2\int_0^{2\pi}\sin\left(\frac{t}{2}\right)\,dt\equal{}
\left.\minus{}4\cos\left(\frac{t}{2}\right)\right|_0^{2\pi}\equal{}\minus{}(\minus{}4)\minus{}(\minus{}4)\equal{}8.$

Note that the antiderivative is $ \minus{}4\cos\left(\frac{t}{2}\right)\plus{}C$ for some completely arbitrary constant $ C.$ It makes absolutely no sense to talk about where or whether this [b]function[/b] is positive. The thing that must be positive here is the definite integral, and we only know what that is after doing a suitable evaluation of the antiderivative at the appropriate endpoints.

[quote="Kent Merryfield"]
 The thing that must be positive here is the definite integral, and we only know what that is after doing a suitable evaluation of the antiderivative at the appropriate endpoints.[/quote]

That is exactly what I wanted to hear.


analytic geometry
calculus
integration
search
combinatorics unsolved
combinatorics

n points in the plane are such that for any three points we can find a cartesian coordinate system in which the points have integral coordinates. Show that there is a cartesian coordinate system in which all n points have integral coordinates.

Have a look here :
http://www.mathlinks.ro/Forum/viewtopic.php?search_id=393236182&t=201137

Pierre.


Problem:
Let S be the set of real numbers that can be represented as repeating decimals of the form (abc)/999 where a,b,c are distinct digits. Find the sum of the elements of  S

Solution:
Numbers of the form  can be written as (abc)/999. There are 10*9*8 = 720 such numbers. Each digit will appear in each place value 720/10 = 72 times, and the sum of the digits, 0 through 9, is 45. So the sum of all the numbers is (45*72*111)/999=360.

http://www.artofproblemsolving.com/Wiki/index.php/2006_AIME_I_Problems/Problem_6

Where did the number 111 come from? :huh:


algebra
function
domain
integration
inequalities
real analysis
real analysis unsolved

Let $ \{w_k\}_{k\equal{}1}^{\infty}$ be the complete set of (smooth) eigenfunctions for $ L\equal{}\minus{}\nabla$ defined in a bounded smooth domain $ U \subset R^n$.
Let $ u_m$ be a function in the subspace spanned by $ w_1,...,w_m$. Is it true that:
$ \|u_m\|_{H^2(U)}^2<\equal{}C\|\nabla u_m\|_{L^2(U)}^2$ for some constant independent of $ m$?

($ \|u_m\|_{H^2(U)}^2\equal{}\|u_m\|_{L^2(U)}^2\plus{}\|Du_m\|_{L^2(U)}^2
\plus{}\|D^2u_m\|_{L^2(U)}^2$, the first term is easy to prove: since $ 1/\lambda_m$ is bounded. But how to bound the other terms?)

[quote="kalazealot"]the first term is easy to prove: [b]since $ 1/\lambda_m$ is bounded[/b]. But how to bound the other terms?)[/quote]

How do you mean? It is known that for a smooth bounded domain, spectrum of $ \minus{}\Delta$ consits of
\[ 0 < \lambda _1  < \lambda _2  \leqslant \lambda _3  \leqslant \cdots.
\]This means that
\[ 0 \leqslant \frac{1}
{{\lambda _m }} \leqslant \frac{1}
{{\lambda _1 }},\quad \forall m.
\]

[quote="bookworm_vn"][quote="kalazealot"]the first term is easy to prove: [b]since $ 1/\lambda_m$ is bounded[/b]. But how to bound the other terms?)[/quote]

How do you mean? It is known that for a smooth bounded domain, spectrum of $ \minus{} \Delta$ consits of
\[ 0 < \lambda _1 < \lambda _2 \leqslant \lambda _3 \leqslant \cdots.
\]
This means that
\[ 0 \leqslant \frac {1} {{\lambda _m }} \leqslant \frac {1} {{\lambda _1 }},\quad \forall m.
\]
[/quote]
yes, this is what I meant. (the posed statement is in Evans "PDE" p363 step 5, he state without proof).

after I reviewed Evans's "PDE" book, I found Thm in p317 is useful here:

therefore, we have $ \|u\|_{H^2} < \equal{} C (\|\nabla u\|_{L^2}\plus{}\|u\|_{L^2})$ and then use what you said.

my proof was incomplete: above method only proved that $ w_k$ satisfy that inequality. However, for
$ u_m \in span\{w_1,...w_m \}$, we still don't know.

maybe the following help: (but we need to assume $ n>2$)

 First still use Thm in p317 :

therefore, we have $ \|u_m\|_{H^2} < \equal{} C (\|\nabla u_m\|_{L^2} \plus{} \|u_m\|_{L^2})$

Since $ u_m$ vanish at boundary, we have, by intgration by part, $ \int |D^2u_m|^2 \equal{}\int (\nabla u_m)^2$. i.e. $ \|D^2 u_m\|_{L^2} \equal{}\|\nabla u_m\|_{L^2}$.

however, by an inequality (Evans, p265, Thm3), since $ n>2$, $ \|u_m\|_{L^2}<\equal{} C\|D u_m\|_{L^2}<\equal{}C\|D^2 u_m\|_{L^2}$. And we are done.

For $ n\equal{}2$, any ideas?

[quote="kalazealot"]however, by an inequality (Evans, p265, Thm3), since $ n > 2$, $ \|u_m\|_{L^2} < \equal{} C\|D u_m\|_{L^2} < \equal{} C\|D^2 u_m\|_{L^2}$. And we are done.

For $ n \equal{} 2$, any ideas?[/quote]

I found no restriction for $ n\equal{}2$ in Theorem 3, p265.

[quote="bookworm_vn"][quote="kalazealot"]however, by an inequality (Evans, p265, Thm3), since $ n > 2$, $ \|u_m\|_{L^2} < \equal{} C\|D u_m\|_{L^2} < \equal{} C\|D^2 u_m\|_{L^2}$. And we are done.

For $ n \equal{} 2$, any ideas?[/quote]

I found no restriction for $ n \equal{} 2$ in Theorem 3, p265.[/quote]

THEOREM 3  Assume U is a bounded,  open subset of $ R^n$. Suppose $ u\in W_0^{1,p}(U)$,  for
some $ 1 < \equal{} p < n$  Then we have the estimate
$ \|u\|_{L^2} < \equal{} C\|D u\|_{L^2}$

Here p=2, so n>2.

This theorem is from "Gagliardo-Nirenberg-Sobolev inequality" , so n>p is needed.

Thm1 in p275 ((Poincare inequality) remove this restriction, but need subtract an average of $ u$ over $ U$, I wanted to use this but can't work through....


rotation
combinatorics proposed
combinatorics

Let $ A$ be a set of $ 2n$ points on the plane such that no three points are collinear. Prove that for any distinct two points $ a,b\in A$ there exists a line that partitions $ A$ into two subsets each containing $ n$ points and such that $ a,b$ lie on different sides of the line.

[hide="Please check this solution"]Lemma. Given points A, B, C, D, no three are collinear. If there exist a line between A and B such that C and D are on the same side, there is a line such that C and D are on the different side.

Proof: Consider the triangle ACD. Firstly, the line intersects side AC. We move the line such that it intersects CD, but does not go too far s.t. A and B are on the same side. Since A, C, D are not collinear, this can always be done.

Now, assume we have an arbitrary line between A and B, there is $ n\plus{}k$ points on the right side and $ n\minus{}k$ points on the left side, where $ k\in\mathbb{N}$. By our lemma, we can move the line such that the number of points on the left increases to $ n\minus{}k\plus{}1$, while the number of points on the right reduces to $ n\plus{}k\minus{}1$. If we repeat this process, we will have $ n$ points on each side.[/hide]

[quote=JBL]Pick a point on segment $ ab$ that is not on the line joining any two of the other points.  Now choose a line through this point and rotate it slowly, keeping track of the number of points to its "left."  If this value is initially $ k$, after rotating 180 degrees it becomes $ 2n \minus{} k$.  Since this value changes by at most 1 at a time, at some intermediate stage it must be equal to exactly $ n$.  Also, any such line separates $ a$ from $ b$.[/quote]

nice solution to a beautiful problem  :D  :thumbup:


What is the sum of the coordinates of the midpoint of the line segment whose endpoints are $ (3,9)$ and $ (19,21)$?

The midpoint formula is $ (\frac{X_{1}\plus{}X_{2}}{2},\frac{Y_{1}\plus{}Y_{2}}{2})$.  So we have $ (\frac{3\plus{}19}{2},\frac{9\plus{}21}{2})$.  So then the points are $ (11,15)$, so the sum of the coordinates is $ 11\plus{}15\equal{}\boxed{26}$


well, if you read the what is your beest MC score topic, youll see some people have gotten perfect scores. but again, i dont know if you are talking about the real thing or just during liketeam practices and stuff

[quote="gauss1181"]Has anyone ever gotten a perfect score on a MATHCOUNTS Chapter Competition? (For real!) :?:[/quote]
Yes.
Numerous people do it each year.
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1773197730&t=136574

In fact, people full score nationals (cheers to Mr. Crawford)

see other thread please.


Find three perfect fourth power numbers whose sum is a perfect fourth power.

One of my books: [i]Recreations in the Theory of Numbers[/i], Albert H. Beiler, Dover,
states: "It has been proved that $x^4 + y^4 + z^4 = t^4$ is impossible."
It also states that it is unknown if four quartics can equal another quartic.
(The book is a 1964 edition.)

It then shows that [b]five[/b] quartics can equal a quartic
 and provides two generating formulas:

$(8a^2+40ab-24b^2)^4 + (6a^2 - 44ab - 18b^2)^4 + (14a^2-4ab-42b^2)^4$
$.\qquad +(9a^2+27b^2)^4 + (4a^2+12b^2)^4 \;= \;(15a^2 + 45b^2)^4$

$(4a^2-12b^2)^4 + (2a^2-12ab-6b^2)^4 + (4a^2+12b^2)^4$
$.\qquad + (2a^2+12ab-6b^2)^4 + (3a^2+9b^2)^4 \;= \;(5a^2 + 15b^2)^4$

I recommend the book as a cure for insomnia . . .


find all functions from set of naturals N to itself (N duznt contain 0)
satisfying
h(h(n))+h(n+1)=n+2

find all functions from the set of integers Z to itself satisfying
f(m+n)+f(m)f(n)=f(mn+1)

The first one is difficult to understand because a right parenthesis is missing.

Pierre.


probability
function
modular arithmetic
algebra
binomial theorem
binomial coefficients
number theory

please can you post some problems of the binomial theorem :?  :(  I have just started it and I need help.
Find the coefficient of $x^2$ in the expansion of $(1-3x)^5(2+x)^4$
Is there a special method for this apart from multiplying out the brackets? :?  :(
NEED SOLUTION :?  :(  can somebody help, please.

Of course you don't have to multiply everything out...you can use the binomial theorem for that problem :!: You won't be able to get the answer on just one application as there are two binomials raised to powers.  So think about how to get around this.

[hide="Dont look until youve tried the problem again"]
The binomial theorem states that

\begin{eqnarray*}
(x+y)^n &=& \sum_{k=0}^n{\binom{n}{k}x^{n-k}y^{k}}\\
&=& x^n+\binom{n}{1}x^{n-1}y+\cdots+\binom{n}{n-1}xy^{n-1}+y^n
\end{eqnarray*}

Let $a_i$ be the coefficient of $x^i$ in the first binomial and $b_i$ be the coefficient of $x^i$ in the second.  The cofficient of $x^2$ in the total product will be $a_0b_2+a_1b_1+a_2b_0$.  Now we just compute each term.  It comes down to

\[(1)\left(\binom{4}{2}2^2\right)+\left(-3\binom{5}{1}\right)\left(2^3\binom{4}{1}\right)+\left((-3)^2\binom{5}{2}\right)(2^4).\]

(sry, I don't feel like doing that arithmetic :P )[/hide]

Additional Problems

1.  Find the value of:

\[\binom{n}{0}+\binom{n}{1}+\cdots+\binom{n}{n}.\]

2.  There are 8 different toppings that can go on a pizza.  How many different pizzas can be made (don't count the pizza with no toppings).

3.  Find the last three digits of $7^{200}.$

4.  Find the value of

\[\binom{n}{0}-\binom{n}{1}+\cdots+(-1)^n\binom{n}{n}.\]

Try to find as many solutions to these problems as possible.

If you want to find the coefficient of $x^2$, only worry about terms with degree 2 or lower, because any term with degree higher than two, when multiplied with anything (since the lowest degree is 0) will have degree higher than 2.  So instead of using the full binomial theorem, only use it to find the ones we need.  For first relation,
$-(3x-1)^5$ (I just like to organize it with x first)
$\binom 5 3 (3x)^2 (-1)^3 + \binom 5 4 (3x)^1 (-1)^4 + \binom 5 5 (3x)^0 (-1)^5$
$-(-90x^2 + 15x - 1)$
For the second,
$(x+2)^4$
$\binom 4 2 x^2 \cdot 2^2 + \binom 4 3 x^1 \cdot 2^3 + \binom 4 4 x^0 \cdot 2^4$
$24x^2 + 32x + 16$

So we have
$(... + 90x^2 - 15x + 1)(24x^2 + 32x + 16)$
Here of course we only multiply the terms that will yield a term with degree 2.
$90x^2 \cdot 16 + -15x \cdot 32x + 1 \cdot 24x^2$
$1440x^2 - 480x^2 + 24x^2$
$984x^2$

[hide]Isn't number one to joml88's additional problems $2^n$?

I think number two is $2^8-1=255$.

[/hide]

I have a question. When is the binomial theorem taught?

I had to learn the binomial theorem by myself.

It was not taught to me in algebra 2.

[quote="deej21"]236factorial, in my school, and probably most of USA, it is taught in Algebra II (although our Algebra II is split up into Alg II and Alg III, and binomial theorem taught in the latter), so usually the junior year (11th grade).  However, here it's only taught in the advanced classes, and even there merely glossed over as unimportant.  My school is not so good.  It might be taught earlier in magnet schools and such.[/quote]

I guess I don't have to worry about it too much since i'm only entering 8th grade  :P

[quote="236factorial"]
I guess I don't have to worry about it too much since i'm only entering 8th grade  :P[/quote]

*gasp* Learn it early!  You want to do well in the AMC's in 8th grade don't you?  :)

[quote="236factorial"]What's AMC? American Mathmatics Competition?[/quote]

Yes.

Yes, do learn it as soon as possible. :( I did NOT get it covered in Saxon's Alg. II... It wasn't till AoPS's Intro to Counting and Probability that I learned it.

can somebody post a solution to the $1st$ problem of Joml88?
Or wait is my solution right?
$=nC_0+nC_1+......+nC_n$
$=(1+1)^n$
So , $=2^n$?

Try plugging in 1 and 1 into the binomial theorem.

I can get the last digit of (7 ^ 200), but I can't get the last three.

Since it was posted in a binomial coefficients thread, does (50 - 1) ^ (100) help?

Are my answers correct??

Also , I am not able to get the third didit in the 3 question . Just the last two. Any ideas

(the 2nd last digit comes from looking at $ 7^4 = 2401 $ )

[quote="deej21"][quote="Iversonfan2005"]For number $2$, it should be $8C1+8C2....+8C7+8C8$.[/quote]

this is the same as saying $( \binom 8 0 + \binom 8 1 + \binom 8 2 + ... + \binom 8 7 + \binom 8 8 ) - 1$
Using binomial theorem "backwards", it becomes $(1+1)^8 - 1$, or $2^8 - 1$ as 236factorial said.

236factorial, in my school, and probably most of USA, it is taught in Algebra II (although our Algebra II is split up into Alg II and Alg III, and binomial theorem taught in the latter), so usually the junior year (11th grade).  However, here it's only taught in the advanced classes, and even there merely glossed over as unimportant.  My school is not so good.  It might be taught earlier in magnet schools and such.[/quote]
can you explain why you can use it like that?

For number 4 you can use this property by Pascal I think
that the sum of odd terms equal the sum of evens

so for 4 its the sum of odds - sum of evens = x - x = 0

you can also use (1-1)^n.

[quote="Iversonfan2005"]Use Pascal's triangle also to figure out coefficients easily.[/quote]

Or you can just use the formula:
\[\text{(coefficient of next term)}=\frac{\text{(coefficient of current term)}\text{(power of current x)}}{\text{(number of current term)}}\]

This will help you expand expressions that have huge powers and that require a huge pascal triangle.

Masoud Zargar

you could consider the number mod 1000
$7^{200} \mod{1000}$
$(10-3)^{200} \mod{1000}$
if you think of the binomial expansion,

$\sum_{k=0}^{200}\binom{200}{k}10^{n-k}*(-3)^k$
you see that everything is divisible by 1000, except where $k =198,199,200$, but this is not that simple to do either

the only thing that i see might help would be using euler's generalization of fermat's little theorem
in this case, $\phi(1000) = 400$ so $7^{400} \mod{1000} \equiv (7^{200})^2 \equiv 1$, but i don't think that one can take square roots in mods so i am at a loss

[quote="erik-the-red"]How do you determine the last three digits of 7 ^ 200 ?[/quote]

similar to http://www.artofproblemsolving.com/Forum/viewtopic.php?t=52301

but harder :roll:

Altheman, here's how you finish it:
Note: this procedure can be used with 7 instead of 3. Do you see why?

[hide="My solution"]$3^{200} = ((((3^5)^2)^2)^2)^5$

$(((243^2)^2)^2)^5$

Now it suffices to consider each thing $\mod 1000$

So rather than write out the multipliocation and the mod signs cumbersomely, everything is in $\mod 1000$.

$((049^2)^2)^5$

$(401^2)^5$

$801^5$

$801^2 * 801^2 * 801$

$601 * 601 * 801$

$201 * 801$

$001$

So the last three digits are $\fbox{001}$[/hide]

Hopefully, I helped. Only problem is that each of the multiplication steps takes a minute to do. Took me 5 minutes to do this problem.

yes, i do know that can be done, but i rather try and use more clever techniques before using brute force, but that does work from the fact using the toitient function and doing the square root, but i could not assure that the other digitis were one as well

[quote="1234567890"]

$(401^2)^5$

$801^5$

$801^2 * 801^2 * 801$

$601 * 601 * 801$

$201 * 801$

$001$
 Took me 5 minutes to do this problem.[/quote]
When you get to $(401^2)^5$ I'm pretty sure you can just look at the last two digits squared to the 5th for mod 1000. I don't exactly know how to explain this but if you have any number $((x...01)^2)^5$ they will all be the same mod 1000. Can anyone explain this better?


abstract algebra
vector
inequalities
quadratics
linear algebra

If $ A\in R^{n\times n}$ is symmertric positive definite, then entry with the largest module must be a diagonal entry.

$ A$ is positive definite iff $ x^TAx>0$ for all $ x\in\mathbb{R}^n\setminus\{0\}.$

Hint: by choosing vectors $ x$ with all but two entries equal to zero, we can reduce this to the case of a $ 2\times 2$ matrix.

Hi miwalin,
Your result is valid if $ A$ is symmetric and non negative because the Cauchy Schwarz inequality is valid for a non negative quadratic form. You can also extend your result with a density argument.

I have not correctly read the question. If you want that necessarily an entry with the largest module must be a diagonal entry, then you can't do better than the hypothesis $ A$ positive definite.


$god=gcd/c \cdot 0 = 0$  Thus, since anything times 0 is 0, and $0<\frac{23}{33}$, so this statement is always true?

i do not believe you understand why $\frac{23}{33}> \text{God}$


function
inequalities
search
real analysis
real analysis unsolved

$ \{ a_n \}_{n\equal{}1}^{\infty}$ is a sequence satisfying $ 0 < a_n \leq a_{n\plus{}1} \plus{} a_{n^2}$ for all natural numbers n. Is $ \sum_{n\equal{}1}^{\infty} a_n$ necessarily divergent?

I guess it was here long enough to post a solution.

Let us assume that the series converges and $ A$ is its sum. Fix any non-increasing function $ g: \mathbb N\to (0,1)$ such that $ ng(n)$ is increasing. Now call $ n$ good if $ \min_{n^2\le m< (n\plus{}1)^2}a_k\ge g(n)a_{n^2}$ and bad otherwise. Note that if $ n$ is good, then $ a_{n^2}\le \frac 1{ng(n)}\sum_{n^2\le m<(n\plus{}1)^2}a_m$.

Adding the inequalities $ a_n\minus{}a_{n\plus{}1}\le a_{n^2}$ and taking into account that $ a_n\to 0$, we get
\[ a_n\le \sum_{m\ge n}a_{n^2}\equal{}\sum_{m\ge n, m\text{ good}}a_{m^2}
\plus{}\sum_{m\ge n, m\text{ bad}}a_{m^2}\,.\]
The first sum does not exceed $ \frac A{ng(n)}$ (see the remark at the end of the previous paragraph). To estimate the second sum, we observe that for every $ m$ and every $ k<2m\plus{}1$, we have
\[ a_{m^2}\le a_{m^2\plus{}k}\plus{}\sum_{m^2\le \ell< (m\plus{}1)^2}a_{\ell^2}\]
By the definition of badness, we can choose $ k$ so that $ a_{m^2\plus{}k}<g(m)a_{m^2}$. Hence, we can conclude that
\[ a_{m^2}\le (1\minus{}g(m))^{\minus{}1}\sum_{m^2\le \ell< (m\plus{}1)^2}a_{\ell^2}\]
and, finally,
\[ \sum_{m\ge n, m\text{ bad}}a_{m^2}\le (1\minus{}g(n))^{\minus{}1}\sum_{\ell\ge n^2}a_{\ell^2}\]
so
\[ a_n\le \frac A{ng(n)}\plus{}(1\minus{}g(n))^{\minus{}1}\sum_{\ell\ge n^2}a_{\ell^2}\]
Now we can split the sum on the right into the good and bad parts as before and, repeating the above argument, we'll get
\[ a_n\le \frac A{ng(n)}\plus{}(1\minus{}g(n))^{\minus{}1}\frac{A}{n^2g(n^2)}\plus{}(1\minus{}g(n))^{\minus{}1}(1\minus{}g(n^2))^{\minus{}1}\sum_{\ell\ge n^4}a_{\ell^2}\,.\]
Repeating this trick again and again, we get the general inequality
\[ a_n\le \sum_{k\equal{}0}^{K\minus{}1}\frac A{n^{2^k}g(n^{2^k})}\prod_{s\equal{}0}^{k\minus{}1}(1\minus{}g(n^{2^s}))^{\minus{}1}
\
\plus{}\ \prod_{s\equal{}0}^{K\minus{}1}(1\minus{}g(n^{2^s}))^{\minus{}1}\sum_{\ell\ge n^{2^K}}a_{\ell^2}\,.\]
At this stage we only know that $ \sum_{\ell\ge n^{2^K}}a_{\ell^2}\to 0$ as $ K\to\infty$ (provided that $ n\ge 2$, of course). Still, we can choose $ g(n)\equal{}n^{\minus{}1/3}$ and insure that the factors $ \prod_{s\equal{}0}^{k\minus{}1}(1\minus{}g(n^{2^s}))^{\minus{}1}$ are bounded by some constant independent of $ k\ge 0$ and $ n\ge 2$. This immediately yields the bound
\[ a_n\le \text{Const\,}\sum_{k\equal{}0}^{\infty}\frac A{n^{2^k}g(n^{2^k})}   \equal{}O(n^{\minus{}2/3})\].

Now, knowing that, we have more freedom because this inequality implies $ \sum_{\ell\ge n^{2^K}}a_{\ell^2}\equal{}O(n^{\minus{}2^K/3})$, which tends to $ 0$ even if multiplied by $ 2^K$. So, we can choose $ g(n)\equal{}\frac 12$ and, passing to the limit as $ K\to\infty$ get
\[ a_n\le \sum_{k\equal{}0}^{\infty}\frac {2^{k\plus{}1}A}{n^{2^k}}\equal{}O(n^{\minus{}1})\].

Our next task will be to improve that to $ a_n\equal{}o(n^{\minus{}1})$. Now it is easy: since $ a_m\le Cn^{\minus{}2}$ for $ m\ge n^2$, we see that every $ a_n$ is followed by at least $ (2C)^{\minus{}1}n^2a_n\minus{}1$ terms that are at least $ a_n/2$. Hence, $ \sum_{m\ge n}a_m \ge (4C)^{\minus{}1}n^2a_n^2$. Since this sum tends to $ 0$ as $ n\to\infty$, the conclusion follows.

Now we can be sure that the product $ (n\minus{}1)a_n$ attains its maximum $ M$ for some $ n\ge 2$. But then
\[ a_n\equal{}\frac{M}{n\minus{}1}>\frac{M}n\plus{}\frac M{n^2\minus{}1}\ge a_{n\plus{}1}\plus{}a_{n^2}\]
and we get the desired contradiction.

Wow, nice solution. Could you please explain in little more detail this implication, i don't quite understand it:

[quote="fedja"] since $ a_m\le Cn^{ \minus{} 2}$ for $ m\ge n^2$, we see that every $ a_n$ is followed by at least $ (2C)^{ \minus{} 1}n^2a_n \minus{} 1$ terms that are at least $ a_n/2$. [/quote]

Also, how did you come up withe the solution?  :lol: This is brilliant

Your first question is easy. Just note that $ a_{n\plus{}k}\ge a_n\minus{}a_{n^2}\minus{}a_{(n\plus{}1)^2}\minus{}\dots\minus{}a_{(n\plus{}k\minus{}1)^2}\ge a_n\minus{}Cn^{\minus{}2}k$. Now solve the inequality $ a_n\minus{}Cn^{\minus{}2}k\ge a_n/2$ for $ k$.

Your second question is more difficult to answer. I'll try to describe the way I followed but, since I'll omit quite a few thought chains that led to dead ends, the description will be somewhat misleading anyway.

I started with a classical similarly formulated problem that many mathlinkers may know well (it even appeared here but I'm too lazy to search for the link now). The problem asked to show that a series of positive terms satisfying $ a_n\le a_{2n}\plus{}a_{2n\plus{}1}$ must diverge. The solution to that one is just to push the sum forward using this recurrence and to obtain the general inequality $ a_n\le \sum_{2^k n\le m<2^k(n\plus{}1)}a_m$. If the series had converged, the right hand side would tend to $ 0$, so we would get $ a_n\equal{}0$.

The main difficulty was that all my attempts to push the sum forward in the original problem resulted in a mess. What we would need to do to repeat the solution verbatim is to show that for every $ q$, the $ 2^q$ numbers each of which is obtained from a fixed number $ n>1$ by applying $ q$ operations of the type $ n\mapsto n\plus{}1$ and $ n\mapsto n^2$ are all different.

[color=green]
Today I see that this is actually true and not that hard: start with the end of two equally long chains of operations resulting in the same number, remove the common $ \plus{}1$'s and look at the last squaring in one chain. Suppose it goes from $ M$ to $ M^2$. If there is no squaring in the other chain at the same place, we should have at least $ 2M\minus{}2$ operations preceding this squaring (otherwise the other chain would not be able to cover the distance from its last squaring to $ M^2$). But after $ 2M\minus{}2$ operations we would certainly end up with a number not less than $ 2M\minus{}2\plus{}n>M$.
[/color]

Yesterday I just failed to see this alternative solution, so I decided to abandon this route as "not promising". The next thing I did was to try to construct a counterexample. I realized that you cannot find a decreasing sequence with the desired property rather quickly (the terms in the decreasing series decay faster than $ 1/n$). So, I would need a series with terms going up and down. Probably, the abnormally large terms should occur in the positions that are perfect squares. So, my idea was to consider a sequence that has large terms at exact squares followed by very short tails of comparable terms not reaching the next square. It took me some time to realize that this is impossible because in this case we would have a perfect and easy opportunity to push forward (this is the case when all $ n$ are bad and $ g(n)\equal{}0$).

This gave me the idea to consider bad and good indexes separately and push forward those I could. At this point I kept $ g$ free and just looked at what I could gain from any particular choice. It turned out that I could get $ a_n\equal{}O(n^{\minus{}1\plus{}\delta})$ very easily (see the first part). It took a few more minutes to realize that once I knew that, I had a much better bound for the pushed forward tails, so $ g(n)\equal{}\frac 12$ was now possible to use. It gave $ a_n\equal{}O(n^{\minus{}1})$. I still needed $ o()$, not $ O()$, but to find the last missing trick (the one you asked to clarify) was not really that hard.

Here is another solution,suppose the sum is finite.

Let $ A \subset \mathbb{N}$ finite.

Let's note $ C = \{n^2| n \in \mathbb{N}\}$

let's note
\[ \delta(A) = min\{r \in \N | A \subset C \cup (C + 1) \cup (C + 2) \cup ... \cup(C + r)\}
\]
and
\[ min(A) = min\{p | p \in A\}
\]
For the sake of clarty we wil note $ A^2 = \{p^2|p \in A\}$ ; $ A + 1 = \{p + 1|p \in A\}$ ; $ T(A) = A^2 \cup (A + 1)$ and $ \mathcal{P}$ the property that $ \delta(A) \leq min(A)$.

We are going to prove that if $ A$ satisfies $ \mathcal{P}$ then :

1) $ T(A)$ also satisfies $ \mathcal{P}$

2)$ A^2$ and $ A + 1$ are disjoint.

--------------------------------------
1)

$ min(T(A)) = min(A) + 1$   :D  and $ \delta(T(A))\leq \delta(A) + 1$.
Indeed $ A^2 \subset C$ and $ A + 1 \subset (C + 1)\cup...\cup (C + \delta(A) + 1)$ so $ A^2 \cup (A + 1) \subset C \cup (C + 1)\cup ... \cup (C + \delta(A) + 1)$

2)Suppose that $ n \in A^2 \cap (A + 1)$, so there exists $ p \in A$ such that $ p^2 = n$ and $ p^2 - 1 \in A$

so $ n - 1 = j^2 + q$ with $ 0 < q \leq \delta(A)$; we have $ j \leq p - 1$ because $ p^2 > n - 1$.

Now
\[ 2min(A) - 2 \leq 2p - 2 = (p^2 - 1) - (p - 1)^2 \leq q \leq \delta (A) \leq min(A) \Rightarrow min(A) \leq 2
\]
Thus if $ min(A) > 2$ you can iterate : $ T^{n}(A)$ is a subset of $ \mathbb{N}$ with no repetitions and then
\[ \sum_{A} u_k \leq \sum _{T^{p}(A)} u_k \leq \sum_{k \geq min(A) + p} u_k \longrightarrow 0,p\to \infty
\]

What's the difference with my comment in green? :? Anyway, nice to have you back! :lol:

Yes, sorry, i didn't read your second post. Indeed, this is rather a detailed version of your second solution. And nice to have you back too  :P


[b]3 cards are chosen from a standard 52-card deck. What is the probability that the form a pair (e.g. 2 cards match in rank and 1 is different)?[/b]

I'm not a fan of these problems either, and I'm not that sure of my answer, so somebody correct me if they see something wrong.
[hide]
For now, we will assume that the first two cards make a pair and the third card is different.
The first card can be anything.
The second card has to match the first card, so the probability for that is $ \frac{3}{51}\equal{}\frac{1}{17}$
The third card just doesn't have to match the first two cards, so the probability for that is $ \frac{48}{50}\equal{}\frac{24}{25}$
Now, we have to multiply them together, and also multiply that by 3, because cards 1 and 2, 1 and 3, or 2 and 3 could be the pair.
$ 1\times\frac{1}{17}\times\frac{24}{25}\times3\equal{}\boxed{\frac{72}{425}}$
[/hide]

it is to long to do... but ill try  :arrow:  is it with or wiithout

[quote="mathcrazed"]Are they choosing the cards with or without replacement?[/quote]
without replacement. imagine a deck of cards and everytime you choose one you put it on the side.
@anderson. that's wrong.

[quote="andersonw"]I'm not a fan of these problems either, and I'm not that sure of my answer, so somebody correct me if they see something wrong.
[hide]
For now, we will assume that the first two cards make a pair and the third card is different.
The first card can be anything.
The second card has to match the first card, so the probability for that is $ \frac {3}{51} \equal{} \frac {1}{17}$
The third card just doesn't have to match the first two cards, so the probability for that is $ \frac {48}{50} \equal{} \frac {24}{25}$
Now, we have to multiply them together, and also multiply that by 3, because cards 1 and 2, 1 and 3, or 2 and 3 could be the pair.
$ 1\times\frac {1}{17}\times\frac {24}{25}\times3 \equal{} \boxed{\frac {72}{425}}$
[/hide][/quote]
yea.. thats probilly it...
theres a formula for it... I forgot it though...

[hide="Solution 2"]$ \text{Total Possibilities } \equal{} \binom{52}{3}$

This is because we have $ 52$ cards and are choosing $ 3$.

$ \text{Possibilities for a Pair} \equal{} \binom{13}{1}\binom{4}{2}\binom{48}{1}$

This is because first we have $ 13$ ways to choose one rank.  After that, there are $ 4$ denominations (H, D, S, C) to choose from for each type of card, and we are picking two cards for one pair.  Then there are $ 48$ cards left (the two we picked and we cannot pick the other two cards matching those two) and we must choose one.

Thus:
$ P(1\text{ Pair}) \equal{} \boxed{ \frac{\binom{13}{1}\binom{4}{2}\binom{48}{1}}{\binom{52}{3}}}$
$ \implies P(1\text{ Pair}) \equal{} \frac{72}{425}$

Which agrees with andersonw's answer.  I generally try to approach these problems in this manner.  Check out the wikipedia article "Poker Probabilities."  It has a nice layout for you to figure out how to do it -- which are problems analogous to this, only 5 card hands.[/hide]


geometry
circumcircle
inequalities
geometry open

My friend  found out something interesting in geometry (this work qualified to the national, final stage of the contest for "young scientists" - as they call it, quite popular in Europe).

His discoveries are about three geometrical problems:

(1). Given a natural number $n$ and an acute triangle $ABC$ find a point $P$ inside it's plane
such that $|AP|^n+|BP|^n+|CP|^n$ is minimal (the first Euler's point is for $n=1$, for $n=2$ the point has something to do - I dont know what - with the Appolonius circle). He found out a method which finds this point using a geometrical construction.

(2). Given a triangle $ABC$ and a natural number $n$, find a triangle $KLM$ inscribed in this triangle for
which $|KL|^n+|LM|^n+|MK|^n$ is minimal, he also found a method of finding $KLM$ (the Malafaty's triangle is for $n=1$).

(3). And the last one: where does point $P$ - from point (1) - lie when $n->\infty$?
He obtained the answer that it's going to oscilate near the center of the circumcircle of $ABC$ or near the middlepoint of one of the triangle's sides - however the radius of this oscilasion is strictly bounded.

In addition, while working on that task, he received some interesting triangle inequalities with powers such that no standard inequalities and tricks can solve it. I will present them soon.

He was actively searching with his professor whether no-one has written anything including these problems, it seems that no-one has indeed thought about it, so it looks like an interesting discovery.

What do you think about it ?

Results look nice and surely nontrivial - he has written a paper about it hasn't he? Is it available in the Internet then? I'd like to read it.

I'm especially looking to these:

[quote="Rafal"] he received some interesting triangle inequalities with powers such that no standard inequalities and tricks can solve[/quote]

 :D  :D  :D


inequalities
Euler
inequalities proposed

I don't understand why anyone deleted the topic "abc=1" in the box "Inequalities Unsolved". Because there were 4 nice results, then I will repost them here.

If $ a, b, c$ be positives real numbers such that $ abc\equal{}1$, then

$ 1) \ \frac{1}{(1\plus{}a)^2}\plus{}\frac{1}{(1\plus{}b)^2}\plus{}\frac{1}{(1\plus{}c)^2}\plus{}\frac{1}{1\plus{}a\plus{}b\plus{}c} \ge 1$
(posted by pvthuan)

$ 2) \ \frac{1}{(1\plus{}a)^2}\plus{}\frac{1}{(1\plus{}b)^2}\plus{}\frac{1}{(1\plus{}c)^2}\plus{}\frac{3}{4(a\plus{}b\plus{}c)} \ge 1$
(arqady)

$ 3) \ \frac{1}{(1\plus{}a)^2}\plus{}\frac{1}{(1\plus{}b)^2}\plus{}\frac{1}{(1\plus{}c)^2}\plus{}\frac{3}{5(a\plus{}b\plus{}c)\minus{}3} \ge 1$
(ehku)

$ 4) \ \frac{1}{(1\plus{}a)^2}\plus{}\frac{1}{(1\plus{}b)^2}\plus{}\frac{1}{(1\plus{}c)^2}\plus{}\frac{1}{2(a\plus{}b\plus{}c\minus{}1)} \ge 1$
(ehku)

By the way of my solution to arqady's one, we can be done easily 3) and 4), where the 4's is nicest. Notice that from this solution, we have the following nice identity
$ \frac{(u\plus{}v\plus{}w)^2}{\sum (u\plus{}1)^2}\plus{}\frac{1}{2(uv\plus{}vw\plus{}wu\minus{}1)}\equal{}$
$ \equal{}1\plus{}\frac{(2(uv\plus{}vw\plus{}wu)\minus{}u\minus{}v\minus{}w\minus{}3)^2}{2(uv\plus{}vw\plus{}wu\minus{}1)(\sum (u\plus{}1)^2)}$
where $ uvw\equal{}1$.

It was moved into a hidden forum because of being in a running contest. I will move it back in March unless Leonhard Euler (the first poster) tells me when the real deadline will be.

  darij

[quote="ehku"]
$ \frac {1}{(1 \plus{} a)^2} \plus{} \frac {1}{(1 \plus{} b)^2} \plus{} \frac {1}{(1 \plus{} c)^2} \plus{} \frac {1}{2(a \plus{} b \plus{} c \minus{} 1)} \ge 1$
[/quote]

The following inequality is very nice, but very hard:

$ \frac {1}{(1 \plus{} a)^2} \plus{} \frac {1}{(1 \plus{} b)^2} \plus{} \frac {1}{(1 \plus{} c)^2} \plus{} \frac {4}{(1\plus{}a \plus{} b \plus{} c )^2} \ge 1$,

where $ a,b,c$ are positive numbers such that $ abc \equal{} 1$.

For $ a \equal{} b$, it reduces to $ (a \minus{} 1)^4(2a^2 \plus{} 1)^2 \ge 0$.

[quote="Vasc"][quote="ehku"]
$ \frac {1}{(1 \plus{} a)^2} \plus{} \frac {1}{(1 \plus{} b)^2} \plus{} \frac {1}{(1 \plus{} c)^2} \plus{} \frac {1}{2(a \plus{} b \plus{} c \minus{} 1)} \ge 1$
[/quote]

The following inequality is very nice, but very hard:

$ \frac {1}{(1 \plus{} a)^2} \plus{} \frac {1}{(1 \plus{} b)^2} \plus{} \frac {1}{(1 \plus{} c)^2} \plus{} \frac {4}{(1 \plus{} a \plus{} b \plus{} c )^2} \ge 1$,

where $ a,b,c$ are positive numbers such that $ abc \equal{} 1$.

For $ a \equal{} b$, it reduces to $ (a \minus{} 1)^4(2a^2 \plus{} 1)^2 \ge 0$.[/quote]
Which is true because
$ \frac {1}{(1 \plus{} a)^2} \plus{} \frac {1}{(1 \plus{} b)^2} \plus{} \frac {1}{(1 \plus{} c)^2} \plus{} \frac {4}{(1 \plus{} a \plus{} b \plus{} c )^2} \equal{} 1\plus{}\frac{(a^2\plus{}b^2\plus{}c^2\minus{}3)^2}{(a\plus{}b\plus{}c\plus{}1)^2(a\plus{}1)^2(b\plus{}1)^2(c\plus{}1)^2}$

(The my first solution for Arqady's one is by the same way).

[quote="Vasc"][quote="ehku"]
$ \frac {1}{(1 \plus{} a)^2} \plus{} \frac {1}{(1 \plus{} b)^2} \plus{} \frac {1}{(1 \plus{} c)^2} \plus{} \frac {4}{(1 \plus{} a \plus{} b \plus{} c )^2} \equal{} 1 \plus{} \frac {(a^2 \plus{} b^2 \plus{} c^2 \minus{} 3)^2}{(a \plus{} b \plus{} c \plus{} 1)^2(a \plus{} 1)^2(b \plus{} 1)^2(c \plus{} 1)^2}$
[/quote]

How can you prove this nice identity ?  :!: :([/quote]
Put $ p\equal{}a\plus{}b\plus{}c,q\equal{}ab\plus{}bc\plus{}ca$, then
\[ \sum (1\plus{}a)^2(1\plus{}b)^2\equal{}2p^2\plus{}q^2\plus{}2pq\plus{}2p\minus{}3
\]
and
\[ (1\plus{}a)(1\plus{}b)(1\plus{}c)\equal{}p\plus{}q\plus{}2
\]
Your inequality becomes
\[ \frac{2p^2\plus{}q^2\plus{}2pq\plus{}2p\minus{}3}{(p\plus{}q\plus{}2)^2}\plus{}\frac{4}{(q\plus{}1)^2} \ge 1
\]
or
\[ \frac{4}{(q\plus{}1)^2} \ge \frac{\minus{}p^2\plus{}4q\plus{}2p\plus{}7}{(p\plus{}q\plus{}2)^2}
\]
Notice that
\[ \frac{4}{(q\plus{}1)^2} \minus{} \frac{\minus{}p^2\plus{}4q\plus{}2p\plus{}7}{(p\plus{}q\plus{}2)^2} \equal{} \frac{(p^2\minus{}2q\minus{}3)^2}{(q\plus{}1)^2(p\plus{}q\plus{}2)^2} \ge 0
\]
we have the conclusion.

From this solution, we have the above identity.

Nice, ehku! :lol:
I created and proved this inequality using EV-Theorem.


calculus
derivative
function
calculus computations

This does not belong in the intermediate forums.  Most people here do no know derivatives or even more, partial derivatives.  ;) Would a mod please move this somewhere in the college or olympiad sections?

Masoud Zargar

A better section would be in Other Problem Solving Topics.

[quote="Azat123"]Show that F_xz=F_zx and  F_xzz=F_zxz=F_zzx for the function given by

              F(x,y,z)=ye^x+x Ln z

        ;)[/quote]

$f_x = ye^x+\lnz \Rightarrow f_{xz} = \frac{1}{z}$

$f_z = \frac{x}{z} \Rightarrow f_{zx} = \frac{1}{z}$

and

$f_{xzz} = -\frac{1}{z^2}$

$f_{zxz} = -\frac{1}{z^2}$

$f_{zz} = -\frac{x}{z^2} \Rightarrow f_{zzx} = -\frac{1}{z^2}$


my amc12b score is slightly below 100

and because amc12a cutoff was 97.5
i think i have slight chance to take aime

if i can take it
ill totally take Special AIME Problem Seminar

but how am i supposed to know
i made to aime and enroll to the seminar ??????

b/c if i don get to take aime
there is no reason to take the seminar

Chill...the floor will be out soon.  :wink:

[quote="indianamath"]when will they have the state stats on the AMC website?[/quote]As soon as they are compiled, and no sooner.

Please, do not ask questions like this.  The AMC staff is working as fast as they can, and questions like this are somewhat annoying.  In particular, the time it takes the AMC staff to respond to such inquiries takes away from the time that they can spend on more useful tasks, such as compiling statistics. :)

And what does this have to do with the Special AIME Problem Seminar?  Please keep threads on-topic.


geometric series
number theory unsolved
number theory

any one show me the solution about this problem:
proof $\prod\limits_{p=4k-1}(1-\frac1p)$ tend to $0$

I assume the sum goes over all primes¿
Assume the contrary. Since $1-\frac{1}{p+1} > 1-\frac{1}{p}$, also the product $\frac{1}{\prod_{p\equiv 3 \mod 4} \left( 1-\frac{1}{p+1} \right)} = \prod_{p\equiv 3 \mod 4} \left( 1+\frac{1}{p} \right)$ converges to some nonzero constant.
Then for $s > 1$ we get (by geometric series expansion and unique prime decomposition):
$\prod_{p\equiv 1 \mod 4} \left( 1-\frac{1}{p^s} \right) \cdot \prod_{p\equiv 3 \mod 4} \left( 1+\frac{1}{p^s} \right) = \left( \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^s} \right) ^{-1}$
Since the right factor on the LHS and the whole RHS converge for $s=1$ to some nonzero constant (the RHS tends to $\frac{4}{\pi}$), also the left factor does so.
But then $\prod_{p\equiv 1 \mod 4} \left( 1-\frac{1}{p} \right) \cdot \prod_{p\equiv 3 \mod 4} \left( 1-\frac{1}{p} \right) = \prod_{p} \left( 1-\frac{1}{p} \right)$ would converge to some nonzero constant, which is wrong.


algebra
function
domain
superior algebra
superior algebra unsolved

Prove that the quotient ring $ C[X, Y ]\equal{}(X^2 \plus{}Y^2 \minus{} 1)$ is a unique fac-
torization domain but the ring $ R[X, Y ]\equal{}(X^2 \plus{} Y^2 \minus{} 1)$ is not a unique
factorization domain.

I'm not sure I understand your notation.  Do you mean, e.g. $ \mathbb{C}[X,Y] / (X^2 \plus{}Y^2 \minus{} 1)$?

oh. yes. never noticed. sorry.
replace $ "\equal{}"$ with $ "/"$


:D M and N are the midpoints of sides BC and AD,respectively, of a square ABCD. K is an arbritrary point on the extensions of diagonal AC beyond A. The segment KM intersects the side AB on some point L. Prove that the angle KNA equals the angle LNA. :D (this was on the Junior and Senior O-level test)

Extend LA to cut KN at E. Let the centre of the square be O. Then line LAE parallel line MON. So $\frac{AE}{ON}=\frac{KA}{KO}=\frac{LA}{MO}$. As ON=OM, we have AE=LA and so triangles ANE and ANL are congruent. Thus angle KNA=angle ENA=angle LNA.

Dear Mathlinkers,
an article concerning "Geometric miniatures on a square" with original proofs has been put on my website
http://perso.orange.fr/jl.ayme    vol. 7 Miniatures géométriques sur un carré  p. 35
Sincerely
Jean-Louis

Actually, the problem is true for any rectangle.

Best regards,
sunken rock


limit
real analysis
real analysis theorems

I am sure this is an easy question, but I just can't get around it, can someone please help.

Prove that $\lim_{n\to \infty}\frac{a^{n}}{n!}=0$

Please prove it step by step, because I am very stupid. Thank you in advance for helping.

I mean that $a\in [0,\infty)$.

If $0\leq a\leq 1$ then $0\leq \frac{a^{n}}{n!}\leq \frac{1}{n!}$, note that $\lim_{n\to\infty}\frac{1}{n!}=0$ we are done.

If $a>1$ then $0<\frac{a^{n}}{n!}<\frac{([a]+1)^{n}}{n!}=C\cdot \frac{([a]+1)([a]+1)...([a]+1)}{([a]+2)([a]+3])...n}\leq\frac{C}{n}\forall n>[a]+3$. And we are done.

Denote $u_{n}=\frac{a^{n}}{n!}$. Observe that $\frac{u_{n+1}}{u_{n}}=\frac{a}{n+1}\rightarrow 0<1$. In conclusion, $u_{n}\rightarrow 0$.

Otherwise, exists $n_{1}\in\mathcal N^{*}$ so that for any $n\in N\ ,\ n>n_{1}$, $u_{n}>u_{n+1}$ because $\frac{u_{n+1}}{u_{n}}=\frac{a}{n+1}<1$ for any $n\in N^{*}$ greater than a certain $n_{1}$ (or $\frac{a}{n+1}\searrow 0<1$). Thus, the sequence $u_{n}$ is decreasing with positive terms, therefore is convergent, i.e. exists $u\in [0,a_{1})$ so that $u_{n}\rightarrow u$. But $u_{2n}=u_{n}\cdot\frac{a^{n}}{(n+1)(n+2)\ldots (n+n)}$ and prove more simply that $\frac{a^{n}}{(n+1)(n+2)\ldots (n+n)}\rightarrow 0$. For $n\rightarrow\infty$ obtain $u=u\cdot 0$, i.e. $u=0$.


induction
modular arithmetic
combinatorics unsolved
combinatorics

Prove that, for each integer $\large n\geq 2$, there is a set $S$ of $n$ integers such that $ab$ is divisible by $(a-b)^{2}$ for all distinct $\large a,b \in S$.

Use induction on $n:$
$n=2$ take $S_{2}=\{0,1\}.$
Suppose that the statement is true for $n=k,$ i.e. there exists set $S_{k}$ such that $(a-b)^{2}/ab$ for all $a,b\in S_{k}.$ Take $S_{k+1}=\{L+a|a\in S_{k}\}\cup \{0\}$ where $L=lcm(a,(a-b)^{2}|a,b\in S_{k}).$ Thus if $a,b\in S_{k+1}$ and $a=0$ or $b=0$ it is obvious. Otherwise, $a=L+a',b=L+b'$ for some $a',b'\in S_{K}.$ Thus $a-b=a'-b'$ and \[(L+a')(L+b')=L^{2}+L(a'+b')+a'b'\equiv a'b'\pmod{(a'-b')^{2}}\] which is true by the property of $S_{k}.$


factorial
number theory proposed
number theory

are there exist  k number everyone greater than k,and the product of every (k-1)tuple of them is divisible by k!(1*2*...*k=k!)

[b]i dont understand what will happen if all of them be multiples of k!  :!:  [/b]

i ment does there exist  k consequative number such a.a+1,...,a+k-1

and  pruduct of any k-1  tupel of them is devisible by k!

Let $l = [log_{2}k]$  Now, let $f(n)$ denote the number of multiples of $2^{n}$ amongst $a, a+1,..., a+k-1$.  We have that $f(n)-[k/2^{n}]=0 or 1$ for all n.  Now, let $j$ be the largest integer so that $2^{j}$ divides some of the $a, a+1,..., a+k-1$  If j>k, then $2^j$ divides only one of these, and if we remove the one divisible by $2^j$, the product of the other k-1 is not divisible by k! (because the largest power of two which divides this product is $(f(1)-1) + (f(2)-1) +...+ (f(j)-1) < [x/2] + [x/4] +...+ [x/2^{k}]$ and the RHS is the largest power of two which divides k!.

The case where j=k is much the same, .


[quote="#H34N1"][img]http://i268.photobucket.com/albums/jj11/Mersenne3731/1996_Nationals_Target_5.gif[/img][/quote]

[hide]
the distance from the midpoint of MN to S is $ 16\sqrt{2}$
the distance from the midpoint of MN to U is $ 48\sqrt{2}$
drop the perpendiculars from the center of the circle to UQ and UA
then you can see that $ r(1\plus{}\sqrt{2})\equal{}48\sqrt{2}$
so $ r\equal{}96\minus{}48\sqrt{2}$

something like that

[/hide]

dont you mean midpoint of MN to S is 2rt2 and midpoint of MN to U is 6rt2....

[hide]Let X be the midpoint of MN. Do everything that circle did, except you get $ \tex{SX} = 2\sqrt {2}$ so $ \text{UX} = 6\sqrt {2}$. Now, let the radius be r. By dropping perpendiculars to sides, we get a square with side r.
$ r = \frac {6\sqrt {2}}{1 + \sqrt {2}}$, and we use a calculator to get 39.[/hide]

Atleast, I think so :maybe:

This problem was to confusing and not enough info.

well its nationals its supposed to be hard


Is there any value of n other than 1, 2, and 4, such that n^n+1 is a prime?

I think you have posted in a wrong forum.

Anyway, note that if $ k$ is an odd integer $ \geq 3$, then $ x^k \plus{} 1$ has a factor $ x \plus{} 1$, so $ n$ is of the form $ 2^m$. Also, $ n^n \equal{} 2^{m2^m}$ shows that $ m$ itself is again of the form $ 2^l$. Then $ n \equal{} 2^{2^l}$ for some $ l$, giving $ n^n \plus{} 1 \equal{} 2^{2^{l \plus{} 2^l}} \plus{} 1$.

Prime number of the form $ 2^{2^n} \plus{} 1$ is called Fermat's prime, and there is little known fact about this kind of numbers.


vector
linear algebra
matrix
linear algebra unsolved

Let $A$ be nonsingular matrix. Prove that if $\mathrm{rank}A=\mathrm{rank}\left(\begin{smallmatrix}A&B\\C&D\end{smallmatrix}\right)$ then $D=CA^{-1}B.$

$M=\left(\begin{smallmatrix}A&B\\C&D\end{smallmatrix}\right)$ with

$A\in M_{p}(K)$ with $A$ invertible

In fact we have $rank(M)=rank(A)+rank(D-CA^{-1}B)$

We note that the vector [B D] can be written as linear combination of [A C], since [A C] is of rank n already. This guarantees the existence of a matrix S satisfying $D = CS$, $B=AS$. The result now follows.


rotation
geometry
geometric transformation
calculus

Hello everyone,

I would appreciate any hints or tips for the following two word problems. Thank you!

---

1. Two pulleys, one with radius 2 inches and the other with radius 8 inches, are connected by a belt. If the 2-inch pulley is caused to rotate at 3 revolutions per minute, determine the revolutions per minute of the 8-inch pulley. (Hint: The linear speeds of the pulley, that is, the speed of the belt, are the same.)

I calculated the circumferences of both circles but I do not know what to do from there.

C of 2" = 12.5664 inches

C of 8" = 50.2655"

2. To approximate the speed of a river's current, a circular wheel with radius 4 feet is lowered into the water. If the current causes the wheel to rotate a speed of 10 RPM, what is the speed of the current?

Again, I know the wheel's circumference is 25.13274 feet.

[hide="#1"]I [i]think[/i] this problem is easier than it appears.
note that the pulleys must rotate together, and the same amount of belt is covered for both wheels, irregardless of their individual rpm.  The circumference of the smaller wheel is $ 4\pi$ (inches). It rotated 3 times.
So, $ 3(4\pi) = 12\pi$.
Now, find the circumference of the larger wheel ($ 16\pi$). divide the belt distance covered in a minute by the circ. of the larger wheel to find its rpm.
$ %Error. "Largerightarrow" is a bad command.
\frac{3}{4}$ rpm.
[/hide]
EDIT----
[hide="hint #2"]$ r\omega=V$[/hide]

[quote="catalan"]Thanks, igiul, for replying.

I got question #2, but could you please tell me how I can find the belt distance covered in a minute by the larger wheel?[/quote]
 :jump: I feel appreciated.
[hide="The belt has an RPM too!"]It's easy enough to realize that the belt is spinning over its track at a constant rate, or it would be ripped apart or... something. Anyways, we had to find the rpm of the larger wheel yes?

RPM is the number of full rotations any circle object undergoes in a minute.
One way to measure distance traveled in circular motions is $ r\theta \equal{} d$. ($ \theta$ being radians of course)
Remember that $ 2\pi$ is one full revolution.

The little one revolved 3 times in one minute. So, the belt distance it traveled is $ 2(3)(2\pi) \equal{} 12\pi$, which is actually 12 [i]inches[/i] times Pi.

They move together.

That means the Large wheel spun through $ 12\pi$ of its circumference too. But when we actually do the calculation, this is only $ \frac{3}{4}$ of a revolution for the much larger wheel (whose circumference is $ 16\pi$).

This happened in one minute. Therefore, the RPM is $ .75$ revolutions per minute.
[/hide]
Hope that helps.

Are these problems taken directly from a Pre-Calculus textbook? (Specifically one by Sullivan and Sullivan)

Thank you very much for your response igiul; I hope the two modifiers make you even happier!

Archimedes: Yes, I am studying the 1999 edition of Precalculus authored by Sullivan. I hope that his books aren't bad or anything...


1)We've known that $ sin18^0 \equal{} \frac {\sqrt {5} \minus{} 1}{4}$.Prove that:
$ sin2^osin18^osin22^osin38^osin42^osin58^osin62^osin78^osin82^o \equal{} \frac {\sqrt {5} \minus{} 1}{1024}$
2) We've known that $ \cos36^o \equal{} \frac {\sqrt {5} \plus{} 1}{4}$ and with $ 0^o < \alpha < 45^o$ we have $ \cot\alpha \equal{} \cot2\alpha \plus{} \sqrt {1 \plus{} \cot^22\alpha}$. Prove that:
$ 4\cos36^o \plus{} \cos7^o30' \equal{} \sqrt {1} \plus{} \sqrt {2} \plus{} \sqrt {3} \plus{} \sqrt {4} \plus{} \sqrt {5} \plus{} \sqrt {6}$

Do not double post:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=245491[/url]


USAMTS
MATHCOUNTS
trigonometry
induction
arithmetic sequence

[quote="knexpert"]What do you think you got on the USAMTS Round 2?[/quote]
Can you please edit your post so that there is a 26 option to filter out liars.
I voted 22-24, though I may get a 21. I strongly doubt anything lower.

I think I got between 24 and 21, probably a 3 5 4 5 5 I might have made a small mistake on number 3, and I somehow missed a number on number 1. Much better than round 1 :D

I think 5 5 5 1 5, depending on how much they hate my brute force-ish solution for #1. Maybe I'll get 2 on #4 for my special case, but I doubt it. I voted for 21, but I could get 22. 22 is pretty much the maximum I can get, but I'll be happy with that.

Probably 5 5 5 [b]5[/b] 5. Possible commendation on 4 and slightly possible on 2.

Most likely my score will be 5 5 5 [b]5[/b] 5 with the same likelihood of commendation as lotsofgreengrapes for number 4 because I got the same generalization.

I think [b]5[/b] [b]5[/b] 5 [b]5[/b] 5...

If I'm lucky anyway

I think I got them all, though.

Actually, I changed my mind.

5 [b][color=red]5[/color][/b] 5 2 [b][color=red]5[/color][/b]

I think I got a 25, with a commendation possible on 3, 4, or 5.
[quote="bpms"][quote="Iversonfan2007"]any combination of 6 6 6 - - if I'm lucky[/quote]
16 25s so far :roll:[/quote]On the first round most of the people who thought they got 25 were right.

[quote="bookaholic"][quote="renatzu"]On the first round most of the people who thought they got 25 were right.[/quote]
Yup--18 people voted that they thought they had a 25, and when the scores came out 22 people actually did. People, stop underestimating your fellow AoPSers.

I think I got all the problems right, but I have no hope of a commendation. Which makes me sad. Last year I got commended twice on the second round and never again pinned down just what I did right that one time. Sigh.[/quote]
What's the chance that 18 of those 22 are AoPSers?
So far 18 and it's only been a couple days. This is possible I suppose.

[quote="bookaholic"][quote="bpms"]What's the chance that 18 of those 22 are AoPSers?
So far 18 and it's only been a couple days. This is possible I suppose.[/quote]
You misunderstood me. In the AoPS USAMTS forum thread titled along the lines of "What do you think you got?", 18 people voted 25. In the AoPS USAMTS forum thread titled along the lines of "So what did you get?", 22 people voted 25. i.e. There are at least 22 AoPSers who actually did get a 25. Far more than 22 people got a 25 in the whole country.[/quote]
Oh sorry, my mistake. Just ignore me, I'm an idiot who has no clue what he's saying.

No, it wasn't your fault--my first post was pretty unclear. Sorry. Anyway...

[quote="mustafa"]But still, some people might just lie and say they got 25s. Remember when 15 people voted that they got in the top 12 at mathcounts last year? :rotfl:[/quote]
3 people said they had 46s at states, when none of them really did as well.

Yes, people tend to lie in polls. You can't stop that. Unless it's a subjective poll that everyone knows about.

I'm hoping for ~20... something like... 5 4 4 4 3 (Though I believe I'm underestimating)

22-24 due to my extreme sloppiness THAT I REFUSE TO FIX.  Bleh.

[quote="13375P34K43V312"]Someone comment on this please:

My solution to #1 was brute force-ish. Here's what I did:

1. Proved that it must have four digits
2. Proved that it must start with a 2, the next digit must be 0, 1, or 2, or it will be too big
3. Used upper/lower bounds to prove that 22ab doesn't work
4. Used mod 2 and mod 4 to narrow down 21ab into like 4 possibilities, tested all and none worked
5. Used mod 2 and mod 4 to narrow down 20ab into like 20 possibilities, tested and all but 2023 and 2083 didn't work.

Will this get full credit, if I did all the steps correctly?

Also, how much credit do you think I will get on #4 for doing the special case k=2?[/quote]
You can expect 0-2 points for number four for the special case k=2, most likely 1.

[quote="13375P34K43V312"]Someone comment on this please:

My solution to #1 was brute force-ish. Here's what I did:

1. Proved that it must have four digits
2. Proved that it must start with a 2, the next digit must be 0, 1, or 2, or it will be too big
3. Used upper/lower bounds to prove that 22ab doesn't work
4. Used mod 2 and mod 4 to narrow down 21ab into like 4 possibilities, tested all and none worked
5. Used mod 2 and mod 4 to narrow down 20ab into like 20 possibilities, tested and all but 2023 and 2083 didn't work.

Will this get full credit, if I did all the steps correctly?
[/quote]

Hmm... should get points. Steps 4, 5 are... well, brute-force-ish.

[quote="mustafa"][quote="13375P34K43V312"]I somehow got 2 on #4 last time for my load of garbage....it was a special special special case...hoping for 2 but whatever[/quote]

I only got 1 for my garbage load :( ...... What was in yours?[/quote]
Best case, I get 4 points for being ridiculously close to the right answer. Worst case, I get one for nonrigorous incorrect answer. Probably 2/3 points for it.

1. On number 2, for some reason I boxed my $\theta$, which turned out to be 66, but then I had $\angle MNA = 180-\theta = 113$ right below it >_< ... do you think they'll dock me a point for that one?

2. On number 5, I solved some trig equations and got $x = \frac{13}{2}$ (I had $AF = 8-x$ written somewhere earlier) and then for some crazy reason, I thought we had to find $EF$ and so I plugged in to get it and got the right length.  I boxed this and called it $AF$ by mistake!  I had all the stuff there in the right answer... think they'll understand what I meant or be like, "this kid's an idiot, -1 lol"

i'll probably get 25, unless i didn't explain #4 well enough.

Do you think not rigorously proving the patterns in #3, just looking at them at trivial would be a minor shortcoming or a larger hole?

[quote="13375P34K43V312"][quote="lotrgreengrapes7926"][quote="13375P34K43V312"]Also, how much credit do you think I will get on #4 for doing the special case k=2?[/quote]
It depends how you solved it. If it was just brute force, like testing n=2, n=3, n=4, n=5, probably 1 point. If you were slightly on your way to a full solution, it could get 2 points.
[quote="USMATS Site"][b][color=darkred]5[/color][/b]/5 = commended solution;
5/5 = clear, concise, correct solution;
4/5 = minor shortcomings in the solution;
3/5 = larger holes in a promising solution;
2/5 = some good ideas but not much more;
1/5 = evidence of understanding the problem;
0/5 = nothing presented that is credible. [/quote][/quote]

I did solve k=2 (or n=2, whatever it was), in the way that was supposed to be applied to the general case, the way they discussed in the math jam. Hoping for 2. It has some good ideas, it uses the way you're supposed to solve for the general case. Is induction possible for that problem?[/quote]
That doesn't mean much though because in the first round, problem 4 I used the method that was supposed to be applied to the general case the way they discussed in the math jam to prove that the vertices couldn't all have the same color and I didn't get any credit whatsoever. Also, on the third problem I solved part a with the method that was supposed to be applied to the general case, the way they discussed in the math jam and I still got only one point.
Sorry if I sound negative,

-bpms

[quote="13375P34K43V312"]Well the third problem specifically said 'one point for part a, four for part b,' and if you didn't answer part b then I would assume they graded it correctly.[/quote]
Well, I solved it with the formula for the sum of an arithmetic sequence, so bleh. I was hoping for two points since it wasn't completely brute forced, but whatever.


geometry
3D geometry
algebra unsolved
algebra

Consider cubes of edge length 5 composed of 125 cubes of edge length 1 where each of the 125 cubes is either coloured black or white. A cube of edge length 5 is called "big", a cube od edge length is called "small". A posititve integer $ n$ is called "representable" if there is a big cube with exactly $ n$ small cubes where each row of five small cubes has an even number of black cubes whose centres lie on a line with distances $ 1,2,3,4$ (zero counts as even number).

(a) What is the smallest and biggest representable number?
(b) Construct 45 representable numbers.

I don't really understand the problem:
[quote]a cube od edge length is called "small"[/quote]
Do you mean: a cube of edge length 1 is called "small"?
[quote]if there is a big cube with exactly n small cubes[/quote]
Doesn't a big cube always have 25 small cubes? How many big cubes do we have?I thought only one.
[quote]whose centres lie on a line with distances 1,2,3,4 [/quote]
If you have five squares in a row, isn't the distances of their centers always 1,2,3 or 4?


logarithms
function
inequalities
inequalities unsolved

Let $ a,b,c>0$ satisfy $ abc\equal{}1$.Prove that
$ \frac{a}{a^2\plus{}3}\plus{}\frac{b}{b^2\plus{}3}\plus{}\frac{c}{c^2\plus{}3}$≤$ \frac{3}{4}$

[quote="SUPERMAN2"]Let $ a,b,c > 0$ satisfy $ abc \equal{} 1$.Prove that
$ \frac {a}{a^2 \plus{} 3} \plus{} \frac {b}{b^2 \plus{} 3} \plus{} \frac {c}{c^2 \plus{} 3}$≤$ \frac {3}{4}$[/quote]
Try to prove that $ \sum_{cyc}\left(\frac{1}{4}\minus{}\frac{a}{a^2\plus{}3}\plus{}\frac{\ln a}{8}\right)\geq0.$ :wink:

For you, micf.
Let $ f(x) \equal{} \frac {1}{4} \minus{} \frac {x}{x^2 \plus{} 3} \plus{} \frac {\ln x}{8}.$
Hence, $ f'(x) \equal{} \frac {(x \minus{} 1)(x \plus{} 3)(x^2 \plus{} 6x \minus{} 3)}{8x(x^2 \plus{} 3)^2}.$
Thus, $ f$ - increasing function on $ (0, 2\sqrt3 \minus{} 3].$
Hence, it exists only one $ x_1\in(0, 2\sqrt3 \minus{} 3],$ for which $ f(x_1) \equal{} 0.$
We obtain, $ x_1 \equal{} 0.280...$
Id est, for $ \min\{a,b,c\}\geq x_1$ our inequality is proven.
Let $ 0 < a\leq x_1$ and $ g(x) \equal{} \frac {x}{x^2 \plus{} 3}.$ Hence, $ g$ - increasing function on $ (0, \sqrt3].$
Hence, $ \sum_{cyc}\frac {a}{a^2 \plus{} 3}\leq g(x_1) \plus{} 2\max_{x > 0} g \equal{} g(x_1) \plus{} 2g(\sqrt3) < \frac {3}{4}.$ Done! :)


$ Q \to R$:
  $ f(x \plus{} y) \plus{} f(xy \minus{} 1) \equal{} \left( {f(x) \plus{} 1} \right)\left( {f(y) \plus{} 1} \right)$

Let $P(x,y)$ denote the assertion. It is easy to see that constants do not work, so $P(x,0)$ gives that \[f(x)f(0)=f(-1)-f(0)-1\] so $f(0)=0$ and therefore $f(-1)=1$. Notice that $P(-1,-1)$ now gives $f(-2)=(1+f(-1))^2=4$. $P(-1,1)$ then gives $f(-2)=2(1+f(1))$ so $f(1)=1$.

Thus, $P(x,1)$ gives \[f(x+1)+f(x-1)=2(f(x)+1)\] and inductively we can figure out $f(x)=x^2$ for all integers $x$. I don't quite sure how to do the rational

Probably a combination of P(m, n/m) and P(x,1) should do the trick, but I am struggling with the big expression


logarithms
LaTeX
number theory
number theory proposed

[quote="Matharith"]Hi everybody,

While working on discrete logarithm I have found this.
It is like Syracuse conjecture.
Something very strange is that GCD(U(n),m) is always equal to 1.
Maybe someone can prove it if the statement holds for any m. I checked it for a lot of values it holds.
Here is the statement.

Thank you for any comment.[/quote]

I think I misunderstood your conjecture since it seems very very obvious.

If I understand well

Let the sequence :
$ u_0 \equal{} 1$
If $ u_n$ is even, $ u_{n \plus{} 1} \equal{} \frac {u_n}{2}$
If $ u_n$ is odd, $ u_{n \plus{} 1} \equal{} u_n \plus{} m$

$ \gcd(u_0, m) \equal{} 1$
if $ u_n$ is odd and $ \gcd(u_n,m) \equal{} 1$, then obviously $ \gcd(u_{n \plus{} 1},m) \equal{} \gcd(\frac {u_n}{2},m) \equal{} 1$
if $ u_n$ is even and $ \gcd(u_n,m) \equal{} 1$, then obviously $ \gcd(u_{n \plus{} 1},m) \equal{} \gcd(u_n \plus{} m,m) \equal{} 1$

What problem did you encounter when you tried to demonstrate this ?

It seems that you have inverted odd and even in your proof.

The statement seems to be true then

let $ m=3\cdot 5\cdot7\cdot11\cdot13\cdots P(k)$

let us generate $ U(n)$ and select only odd numbers.

let us call $ M$ the set of those odd numbers

Can we state that if $ P(k)$ is [u]bigger enough  [/u] $ M$ will be composed with only primes?

[color=red]Moderator edit: $ \text{\LaTeX}$ed[/color]

[quote="Matharith"]It seems that you have inverted odd and even in your proof.[/quote]

Just a typo error :

Let the sequence :
$ u_0 \equal{} 1$
If $ u_n$ is even, $ u_{n \plus{} 1} \equal{} \frac {u_n}{2}$
If $ u_n$ is odd, $ u_{n \plus{} 1} \equal{} u_n \plus{} m$

$ \gcd(u_0, m) \equal{} 1$
if $ u_n$ is even and $ \gcd(u_n,m) \equal{} 1$, then obviously $ \gcd(u_{n \plus{} 1},m) \equal{} \gcd(\frac {u_n}{2},m) \equal{} 1$
if $ u_n$ is odd and $ \gcd(u_n,m) \equal{} 1$, then obviously $ \gcd(u_{n \plus{} 1},m) \equal{} \gcd(u_n \plus{} m,m) \equal{} 1$

What problem did you encounter when you tried to demonstrate this ?

Are you only kidding when asking this in an olympiad forum or is there something I misunderstood ?

The sequence above U(n) has some link to the equation 2^y mod n = a

[quote="Differ"]@Matharith:

I think that before attempting to solve difficult problems, it is better if one gains a basic grasp of number theory first.

A resource I would recommend is Introduction to Number Theory at http://www.artofproblemsolving.com/Books/AoPS_B_Item.php?item_id=206 . I myself have not read it, but from reviews that other people have given me, it should be an excellent starting point.[/quote]

Thank you for advise.
So your thinking is that if we solve easy problems then we will be able to solve hard one.
I do not agree with you.
Only ideas solve the problems not the tools.
I'm not going to read your book. I have a lot of books downloaded from other sites.
I'm going to stop posting. It is better for me and for you.
It is the right solution.
So have a good night.
Good bye.

[quote="Matharith"]Only ideas solve the problems not the tools.[/quote]

In case you still read this:
Just because the ideas are an essential part doesn't say that you don't need tools. Most tools were developed in more than 200 years of mathematics and contain lots of ideas themself. Ignoring them and thinking they are useless or that you can solve serious problems without them is arrogant.

Hi,

Who is arrogant?
The one who reject at the starting any idea coming from an amateur or the one who is trying to come with ideas?
Here is the link between U(n) ant the problem of "2^y mod n = a"
Look at this picture you have the solution if you are smart enough to find it.

Thank you.

My last post.

Ps : le raisonnement flou vous savez ce que c'est? Je dis bien flou pas fou. Je ne parle pas de logique floue.
Vous avez dans cette image de quoi solutionner de manière définitive le problème de la factorisation des semi-premiers.

[quote="ZetaX"][quote="Matharith"]Who is arrogant?
The one who reject at the starting any idea coming from an amateur or the one who is trying to come with ideas?[/quote]

The one thinking he has new ideas that in fact have been brought up a hundred times before, but who can't stand that and ignores tons of helpful advices.[/quote]

The answer's right even if I said yesterday that I'm leaving.

If nothing is new with my ideas :
- why you did not give a reference (only one I'm not looking for humdreds)
- why my 2 posts are so popular with hundreds of visits??? look at the icone.

Here is another picture
Take look at it.

[b]And now it will be my last post. I will not answer to anyone.[/b]
I swear to cut my internet line forever.
Sorry to tell you that you have big trouble when an amateur come with a[u] NEW IDEAS[/u] it makes you angry.
Here you have a way to compute big 2^y mod n without storing huge numbers or even using fast exponentiation.

Here is edited picture (some mistakes corrected)
Sorry for the mistakes


On game theory problems that can end in draws like this one:
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=340040#340040
And you want to show that one player has the winning strategy, is it enough to show that the other player doesnt have the winning strategy?

Your question practically answers itself: if player 1 doesn't have a winning strategy, this means that no matter what player 2 can force the game to end in a position in which player 1 has not won.  If there are draws, this does not say whether or not player 2 can force the game to end in a position in which player 2 wins.  (Think, for example, of Tic-Tac-Toe.)


what is $\dfrac{7^{2005}-1}{6} \mod{1000}$? (also show that we are dealing with an integer)

It is $1 + 7 + 7^2 + \dots + 7^{2004} \mod{1000}$, so its integer.

Now by (http://mathworld.wolfram.com/EulersTotientTheorem.html) we can find 7^2005 mod 6000.  Its the same as 7^(405) mod 6000.  We can find 7^405 mod 16, 3, and 125, and chinese-remainder-theorem them together.  The values are 7 mod 16, 1 mod 3, and 57 mod 125.    It gives us 4807 (mod 6000).  Now we easily get the result of 801.

that is what i got too, i just was not sure about some of the steps that i did, like making it mod 6000, (eventhough it makes sense to do so) and the dividing it back to mod 1000


Να ενα πρόβλημα που έφτιαξα  σήμερα αλλά ίσως είναι και ήδη γνωστό .
Το αφιερώνω εξαιρετικά στα παιδιά που γράφουν πανελλήνιες . :D

Έστω $a_1,...,a_n>0$  πραγματικοί αριθμοί ώστε  $a_1+a_2+....+a_n=1$

Δείξτε ότι  $\sum_{i=1}^n {\sqrt a_i}+\sum_{i=1}^n{\frac{1}{\sqrt a_i}}\geq (n+1)\sqrt n$

[quote="kostas"]Μου φαινεται πολυ καλη.μπορει κανεις με γεωμετρια ή με διανυσματα?[/quote]

Η λύση  μου δεν περιέχει διανύσματα ή γεωμετρία.... Αν όμως βρει κάποιος μια τέτοια λύση θα είναι εξαιρετικά ενδιαφέρουσα περίπτωση :lol:

Συγγνώμη, αλλά επειδή βαριέμαι το πολύ latex, $\sum \gamma_i$ θα πει $\sum_{i=1}^{n} \gamma_i$

Το πρόβλημα είναι ισοδύναμο με:

$a_i \in (0, 1)$ τέτοια ώστε $\sum a_i^2 = 1$

νδο

$\sum a_i + \sum \frac{1}{a_i} \geq (n+1) \: \sqrt{n}$

Η τετριμμένη περίπτωση για $n = 1$ παραλείπεται... Θεωρούμε $n > 1$

Θεωρώντας την $f(x) = x + \frac{1}{x}$, που είναι γν. φθίνουσα στο (0,1) και κυρτή σε αυτό, έχουμε

$\sum a_i + \sum \frac{1}{a_i} = \sum{a_i + \frac{1}{a_i}} = \sum f(a_i)$

από Jensen:

$\sum f(a_i) \geq n \cdot f \left( \frac{\sum a_i}{n} \right)$

Επίσης, λόγω της: $n \: \sum x_i^2 \geq \left( \sum x_i \right) ^2$ και από το ότι η $f$ είναι γν. φθίνουσα στο (0, 1) και τα $\frac{\sqrt{n}}{n}$ και $\frac{\sum a_i}{n}$ είναι $< 1$, έχουμε:

$n \cdot f \left( \frac{\sum a_i}{n} \right) \geq n \cdot f \left( \frac{\sqrt{n}}{n} \right) = n \cdot \left( \frac{\sqrt{n}}{n} + \frac{n}{\sqrt{n}} \right) = \ldots = (n+1)\sqrt{n}$

Σωστά...... :D . Η ιδέα ήταν ο Jensen τίποτα ιδιαίτερο,,,,,, ;)


There are 360 people in my school.  15 take calculus, physics, and chemistry, and 15 don't take any of them.  180 take calculus.  Twice as many students take chemistry as take physics.  75 take both calculus and chemistry, and 75 take both physics and chemistry.  Only 30 take both physics and calculus.  How many students take physics?

I got  a different answer.  Let $x$ = The number of people taking physics and let $2x$ = the number of people taking chemistry.  Then using the Principle of Inclusion-Exclusion, we get:

\[180+2x+x-75-75-30+15+15=360\]
\[3x+30=360\]
\[3x=330\]
\[x=\boxed{110}\]

Therefore, $\boxed{110}$ people take physics.

[hide=sol]Let number of students taking physics be $n$
by PIE
$180+n+2n-75-75-30+15=360-15$
solve for $n$ to get $\boxed{110}$[/hide]

how do you guys envision this without drawing a venn diagram

[quote=V_V]what is PIE??[/quote]

[url]https://artofproblemsolving.com/wiki/index.php/Principle_of_Inclusion-Exclusion[/url]

Just look it up if you don't know what it is.

[quote=ksdicecream]lol this thread was made before I was born[/quote]

wow you're young!

[quote=Possible][quote=V_V]But I don't think it is worth memorizing formulas right?[/quote]

Its not really hard to mem.
Total number of terms for 3 items is a+b+c-ab-bc-ac+abc[/quote]

I think the point is to learn and understand before memorizing

[quote=peace09][quote=ksdicecream]lol this thread was made before I was born[/quote]

um henlo i was too and same with many other people so pwease don't post that[/quote]

How you have so much posts if your like only 10 - 11???????????
And you only joined in January 2020. HOW???

Also the thread was older than me lol.


find all surjective functions f such that f(m) divides f(n) if and only if m divides n

1. For all n f(1)|f(n). Because f is surjective f(1)=1.
2. If f(n)=1, then n|m for all m. It give n=1.
3. For all n f(p)|f(pn). Let prime q|f(p). Because f surjective, exist x, suth that f(x)=q. It give x|p. Therefore f(p) prime if and only if p is prime. If $f(x_{1})=f(x_{2})=q$ then $x_{1}=x_{2}.$
4. Because f is surjective $f(p^{n})=f(p)^{n}$, and $f(p_{1}^{n_{1}}p_{2}^{n_{2}}...p_{k}^{n_{k}})=f(p_{1})^{n_{1}}f(p_{2})^{n_{2}}...f(p_{k})^{n^{k}}$.
All functions defined by bijective functions $g: P\to P$, were P - is set of primes.


inequalities
integration
function
limit
probability and stats

Let $ X\geq 0$ be a random variable. If $ P(X\geq n)\geq \frac{1}n$ prove that $ EX\equal{}\infty$.

On one hand, $ E(X;\{X>n\})\geq 1$.  On the other hand, if $ EX < \infty$, for every $ \varepsilon > 0$ there is $ \delta > 0$ such that $ P(\Omega) < \delta$ implies $ E(X;\Omega) < \varepsilon$ (why?)  So (since $ X\geq 0$) $ EX \equal{} \infty$.

I'm sorry but I don't seem to understand what you mean by $ E(A;B)$ (I just never encountred this notation before).
Also I'm confused about your $ \Omega$ (usually by $ \Omega$ one denotes the space of all elementary outcomes). Is it an event?

$ E(X;A)\triangleq \int_A X(w)dP(w)$, $ \Omega$ is just an event.

Let $ E_n$ be the event $ X\ge n.$ Define the random variable $ Y_n$ to be $ n$ on $ E_n$ and zero otherwise.

$ E(Y_n) \equal{} nP(E_n) \equal{} nP(X\ge n).$ Hence by hypothesis, $ E(Y_n)\ge 1.$

However, note that $ Y_n\le X$ always. Note also that if we assume $ X$ is integrable ($ E(X) < \infty),$ then $ X$ is finite almost surely and $ P(\cap E_n) \equal{} 0.$ Hence the sequence of random variables $ Y_n$ tend to zero pointwise almost surely.

If $ X$ is integrable, then it is a dominating function for the Dominated Convergence Theorem. By the DCT, we would have $ \lim E(Y_n) \equal{} E(\lim Y_n) \equal{} E(0) \equal{} 0,$ contradicting our hypothesis that $ E(Y_n)\ge 1.$

Hence the DCT must fail, and for the DCT to fail, we must have $ E(X) \equal{} \infty.$


For AMSP 2006 Students that wish to apply to AMSP 2007, was the policy that you just must do one of the admission tests?

I am also interested in knowing the answer. I know titu sent us an email some time ago, but i deleted it :(


Given $ a, b, c \geq\ 0$ satisfy $ a\plus{}b\plus{}c\equal{}6.$
Prove that: $ (11\plus{}a^2)(11\plus{}b^2)(11\plus{}c^2)\plus{}120abc \geq\ 4320$
Equality occurs when $ (a, b, c) \equal{} (1, 2, 3).$
My dear friend, I hope you'll like it :)

let $ b \geq 1$
then $ f(a,b,c) \geq f(a,1,c)$
but:$ f(a,1,c)\minus{}2989\geq 0$is true because $ (a\minus{}2)^2(a\minus{}3)^2 \geq 0$

$ LHS\minus{}RHS\equal{} 11(ab\plus{}bc\plus{}ca\minus{}11)^2\plus{}(abc\minus{}6)^2\plus{}121(a\plus{}b\plus{}c)^2\minus{}36 \geq 4356\minus{}36\equal{}4320$
Equality occurs iff $ a\plus{}b\plus{}c\equal{}6, ab\plus{}bc\plus{}ca\equal{}11$ and$ abc\equal{}6$  or $ a\equal{}1,b\equal{}2,c\equal{}3$ or any symmetric permutation.

[quote="Hong Quy"]$ LHS \minus{} RHS \equal{} 11(ab \plus{} bc \plus{} ca \minus{} 11)^2 \plus{} (abc \minus{} 6)^2 \plus{} 121(a \plus{} b \plus{} c)^2 \minus{} 36 \geq 4356 \minus{} 36 \equal{} 4320$
Equality occurs iff $ a \plus{} b \plus{} c \equal{} 6, ab \plus{} bc \plus{} ca \equal{} 11$ and$ abc \equal{} 6$  or $ a \equal{} 1,b \equal{} 2,c \equal{} 3$ or any symmetric permutation.[/quote]
Yes, that's it, my friend.  :lol:
[quote="nguoivn"]
And a similar + harder is one of problems of MIC Round 3

[b]Problem 3 (Round 3)[/b]
Given the reals $ a, b, c$ satisfy $ ab \plus{} bc \plus{} ca \geq\ 11.$
Prove that: $ \sqrt [3]{a^2 \plus{} 3} \plus{} \frac {7}{5\sqrt [3]{14}} .\sqrt [3]{b^2 \plus{} 3} \plus{} \frac {\sqrt [3]{9}}{5}. \sqrt [3]{c^2 \plus{} 3} \geq\ \frac {23}{5\sqrt [3]{2}}$
Equality also occurs when $ (a, b, c) \equal{} (1, 2, 3)$  :)[/quote]
 :)


In a convex heptagon, the degree measures of the interior angles are $ x, \sim x, \sim x\minus{}2, \sim x\minus{}2,
\sim x \plus{} 2, \sim x \plus{} 2$ and $ x \plus{} 4$. What is the degree measure of the largest interior angle?

[quote="GameBot"]In a convex heptagon, the degree measures of the interior angles are $ x, \sim x, \sim x \minus{} 2, \sim x \minus{} 2, \sim x \plus{} 2, \sim x \plus{} 2$ and $ x \plus{} 4$. What is the degree measure of the largest interior angle?[/quote]

sum of angles in 7-gon is $ (7\minus{}2)\cdot 180 \equal{} 900$ The angles are $ x, \sim x, \sim x \minus{} 2, \sim x \minus{} 2, \sim x \plus{} 2, \sim x \plus{} 2$ and $ x \plus{} 4$. you can simplify that as $ 7x \plus{} 4$.
$ 7x \plus{} 4 \equal{} 900$
$ 7x\equal{}896$
$ x\equal{}128$
so largest angle is 128+4 = $ \boxed{132}$


Let $ K\equal{}(2,4)$
$ f: K\to K$ and $ g: K\to K$ satisfy
a. $ f(g(x))\equal{}g(f(x))\equal{}x$   $ \forall x\in K$
b.$ f(x)g(x)\equal{}x^{2}$   $ \forall x\in K$
Prove that $ f(3)\equal{}g(3)$

Choose $ x_{0}\in(2,4)$ and let $ a =\frac{f(x_{0})}{x_{0}}$ then $ a > 0$.

By intruction we have :

$ f(a^{n}x_{0}) = a^{n+1}x_{0},\forall n\in%Error. "mathh" is a bad command.
{N}$

If $ a > 1$ then $ a^{n}x_{0}\to+\infty$ contracdiction .

If $ 0 < a < 1$ then $ a^{n}x_{0}\to 0$ contracdiction ,too .

Then $ f(x_{0}) = x_{0}$ hence $ f(x) = x,\forall x\in (2,4)$

Simirlary $ g(x) = x,\forall x\in (2,4)$

We're done .  :)


function
complex analysis
complex analysis unsolved

A set $S \subset \mathbb{C}$ is called a URSEF (unique range set for entire function) if it has the following property: if $f,g$ are entire functions with the property that for every $b \in S$, every $k$-order zero of $f(z)-b$ is also a $k$-order zero of $g(z)-b$, and vice versa, then $f \equiv g$.

[Equivalently: $f(z) \in S$ iff $g(z) \in S$, and with multiplicities counted, then $f \equiv g$. ]

Determine whether or not there exists a finite URSEF.

Has someone come up with a construction that does not require Nevanlinna theory? If you actually have a construction that is not too involved I'd really really like to see it.

[quote="Kalle"]If you actually have a construction that is not too involved I'd really really like to see it.[/quote]

Even if it requires Nevanlinna theory to prove that it works? Because the construction in itself (as a statement) is not at all involved :)

Well, what I'd like to see is of course a proof I can understand. So I guess I'd like to see the simplest proof of a set being a finite URS :) I'd like to study Nevanlinna theory, but I wonder when I am going to find time for that.


the problem must be moved to "Inequalities Proposed & Own Problems".
I'm so sorry :blush:

[quote="vietnam2007"]let $a,b$ and $c$ be non-negative real numbers

prove that :

$\frac{a+b}{c^{3}+2abc}+\frac{b+c}{a^{3}+2abc}+\frac{c+a}{b^{3}+2abc}\geq \frac{6}{a^{2}+b^{2}+c^{2}}$[/quote]
The following inequality is true too:

$\frac{a+b}{c^{3}+2abc}+\frac{b+c}{a^{3}+2abc}+\frac{c+a}{b^{3}+2abc}\geq \frac{36}{a^{2}+b^{2}+c^{2}+5(ab+ac+bc)}.$ :wink:

$\sum \frac{a+b}{c^{3}+2abc}= \sum \frac{(a+b)^{2}}{ac^{3}+bc^{3}+2a^{2}bc+2ab^{2}c}\geq \frac{4(a+b+c)^{2}}{\sum (ac^{3}+bc^{3}+2a^{2}bc+2ab^{2}c)}\geq \frac{12}{\sum a^{2}+\sum ab}$ \[\leftrightarrow\]
\[(\sum a^{2}+2 \sum ab)(\sum a^{2}+\sum ab)\geq 3 \sum (ac^{3}+bc^{3}+2a^{2}bc+2ab^{2}c)\] \[\leftrightarrow\]
\[\sum a^{2}b^{2}\geq \sum a^{2}bc\] which is rearrangements

Is there a solution without all that computations ?


Can anyone show me how to do the following problem?

Find the largest $ a$ for which these matrices are stable or neutrally stable.

The first 2 x 2 matrix has in its first row:   ($ a$ and $ \minus{}0.8$) and second row: ($ 0.8$ and $ 0.2$).

The second matrix (2x2) has in its first row ($ a$ and $ 0.8$) and in the second row: ($ 0.2$ and $ a$).

I know that the stability of the matrix depends on the eigenvalues and it is stable if all eigenvalues satisfy $ |\lambda|<1$ and neutrally stable if some $ |\lambda_i|\equal{} 1$ and all others have absolute value < 1. But I don't know hot to solve for the variables themselves?

Stable. What is your definition ?
You give a condition about the $ A$-eigenvalues which is near the caracterization of the matrices $ A$ s.t. the sequence $ \{A^n\}$ is bounded.


function
complex numbers
complex analysis

Well...I don't really get one thing.
Zeta function is...
(1/1)^s+(1/2)^s+...
right?
Then how come s=-2,-4,... makes this function a zero???
for example,
(1/1)^-2+(1/2)^-2+...=1+4+...

Somebody explain to me...plz.

The series definition for $\zeta(s)$ is only valid when the series converges, which is when the real part of $s$ is greater than 1. The definition elsewhere only makes sense with complex analysis, and I'm not going to try to explain that here.

I'd like to know why negative even integers are trivial zeros of the Riemann  Zeta function too.

Can anyone shed some light on this?


Suppose that:
$\{\begin{array}{l}cosx=cosy.cos\alpha=cosz.cos\beta\neq0\\sinx=2sin\frac{\alpha}{2}.sin\frac{\beta}{2}\end{array}$
Prove that: $tan^{2}\frac{x}{2}=tan^{2}\frac{y}{2}.tan^{2}\frac{z}{2}$

We have: $\sin^{2}x=4\sin^{2}\alpha\sin^{2}\beta=(1-\cos\alpha)(1-\cos\beta)=\left(1-\frac{\cos x}{\cos y}\right)\left(1-\frac{\cos x}{\cos z}\right)$
Therefore:
$\cos x=\frac{\cos y+\cos z}{1+\cos y.\cos z}$
From that implying:
$\tan^{2}\frac{x}{2}=\frac{1-\cos x}{1+\cos x}=\tan^{2}\frac{y}{2}.\tan^{2}\frac{z}{2}$


The sequence $a_{1},a_{2}, ...$ of natural numbers satisfies
\[\gcd(a_{i}, a_{j}) = \gcd(i,j) \]
for all $i \neq j$. Prove that $a_{i}=i$ for all $i$.

that is so easy..put j=i  :arrow: a[i]/i is integer.such that{gcd(i,q)=1},but a[i]/q is integer. put j=q so gcd(a[i],a[q])=1 but a[q]/q is integer so gcd (a[i],a[j])>=q.so a[i]/q isnot integer.

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=125053

PS. To the above poster, I'm having a lot of trouble following your solution...


algebra
polynomial
function
algebra proposed

Prove that the polinomial equation P(x) = t.(a^k) - k.(a^t) + (k-t) = 0 have unique real positive solution, where k,t are distinct positive integers.

Of course $x=1$ is this unique solution

Suppose there is some other $a \not =1$, then $a^k-1$ and $a^t-1$ are not equal zero. Then:

$\frac{k}{t}=\frac{a^k-1}{a^t-1}$

We may suppose $k>t$. Now looking at the graphs of the functions $f(x)=x$ and $h(x)=a^x-1$ we see that equality $\frac{f(k)}{f(t)}=\frac{h(k)}{h(t)}$ is not possible. So there exist only one solution.

my solution is with MA >= MG.

(t.(a^k) + (k - t))/k  >=   (a^kt)^(1/k) = a^t

when equality holds if a^k = 1 => a = 1.


m equispaced horizontal lines are intersected by n equally spaced vertical lines.If m<n and the distance between 2 successive horizontal lines is same as that between 2 successive vertical lines,show that the number of squares formed by lines is $ \frac {(m\minus{}1)(3n\minus{}m\minus{}1)}{6}$

i'm think your formula may be incorrect, for $ n \equal{} 3$ and $ m \equal{} 2$, i count $ 2$ squares. whereas the formula gives $ 1$.

edit:
i got the following, if you can check if it matches yours (assuming the original was a typo) i can give some kind of proof/explanation

$ \frac{m(m\minus{}1)(3n\minus{}m\minus{}1)}{6}$

[quote="jackie-chan01"]Ok,
then prove the formula you think should be[/quote]

the way i did it was this:

on the grid of $ n$ by $ m$ points, draw $ 45^\circ$ diagonal lines through the points, you need $ n\plus{}m\minus{}1$ lines to cover all the points.

going from the top most diagonal line to the bottom:
the first crosses a single point, the second crosses two, up to the $ m\minus{}1$th one which crosses $ m\minus{}1$ points
similarly the bottom $ m\minus{}1$ lines do the same
the remaining $ n\minus{}m\plus{}1$ diagonals each cross $ m$ points

to create a square, you need to pick two points that are on a diagonal from each other to be two of the opposite corners of a square, as long as you're consistent about picking the same corners (i.e. top right and bottom left) each time, you wont repeat any squares.

so for each diagonal line you can make $ {k \choose 2}$ squares from those points, where $ k$ is the number of points that diagonal crosses.

which is how i got

$ (n\minus{}m\plus{}1){m \choose 2} \plus{} 2\sum_{k\equal{}1}^{m\minus{}1} {k\choose 2}$

$ \equal{}(n\minus{}m\plus{}1)\left( \frac{m(m\minus{}1)}{2}\right) \plus{} \frac{m(m\plus{}1)(2m\plus{}1)}{6} \minus{} \frac{m(m\plus{}1)}{2}$

after simplification

$ \equal{}\frac{m(m\minus{}1)(3n\minus{}m\minus{}1)}{6}$