Fifty watches, all showing correct time, are on a table.  Prove that at a certain moment the sum of the distances from the center $O$ of the table to the endpoints of the minute hands is greater than the sums of the distances from $O$ to the centers of the watches.

This proof probably isn't nearly rigorous enough, but whatever.

[hide]Let the sum of the distances from the tip of the minute hands to the center of the table at any one point be $f(\theta)$, where $\theta$ is the angle formed by the 12 o'clock mark and the minute hand, going clockwise. Let $f(m)$ denote the minimum value of $f(\theta)$, and let $f(M)$ denote the maximum value of $f(\theta)$. Clearly, $f(m-180) = f(M)$, as they are on opposite sides of the clock. Drawing a line through $m$ and $n$ of each clock and the center of the table makes 60 straight lines. Denoting the center of a clock $c$ and the center of the table $C$, we see that $\overline{Cm} < \overline{Cc} < \overline{CM}$,  $\overline{Cc} < \overline{CM}$ being what we intended to prove.

$Q.E.D.$ (Sort of)
edit:fixed spelling mistake
[/hide]

[quote="t0rajir0u"]But none of the watches are guaranteed to be similarly oriented with respect to $O$ - they won't necessarily take maxima and minima at the same time.  Your ideas only work for one watch (trivial anyway), I'm afraid.[/quote]
[quote="LordoftheMorons"]Let the [b]sum[/b] of the distances from the tip of the minute hands to the center of the table at any one point be $f(\theta)$, where $\theta$ is the angle formed by the 12 o'clock mark and the minute hand, going clockwise.[/quote]
Note bolded word  :|
Though there  very well may be something wrong with my proof.

I saw that.  I still don't think the proof works.  Among other things, you haven't shown $f(M) > f(m)$.

What a delightful problem! Would be much better explained with a picture, but I'm not in the mood to draw one.

[color=#BF00BF]Let our desired inequality be $\mathcal{A}\ge \mathcal{B}$. So, for the time being, consider just one watch. Let its center be $C$. Let the endpoint of the minute hand lie on the boundary of the watch (oh well, or consider the circle traced by the endpoint). Let this varying endpoint be $E$. Now draw a line perpendicular to $OC$ through $O$. Call this line $l$. Let $F$ be the varying foot of perpendicular from $E$ to $l$. Note that $OE\ge EF$. Denote by $avg(Q)$ the average of the quantity $Q$ as $E$ completes one rotation. So we have $avg(OE)\ge avg(EF)=CO$. Hence the average distance from the table-center to the endpoint is at least as much as the distance from the table-center to the watch-center. Summing up for all watches and noting that they are similarly oriented and paced, we get $avg(\mathcal A)\ge avg(\mathcal B)$. So there will be a moment in time when the required is fulfilled.[/color]

Note that this "moment in time" comes periodically every one hour.


algebra
polynomial
inequalities
complex numbers
algebra unsolved

Let $n$ be a positive integer and let $a_0, a_1, \dots a_n$ be complex numbers with
\[ |a_0|^2 + |a_1|^2 + \cdots + |a_{n-1}|^2 \leq 1.  \]
Let $z_1,z_2,\cdots z_n$ be the (complex) roots of the polynomial $f(z) = z^n+a_{n-1}z^{n-1}+\cdots + a_0$.  Prove that
\[ |z_1|^2+|z_2|^2 +\cdots + |z_n|^2 \leq n.  \]

Try here: http://www.mathlinks.ro/Forum/viewtopic.php?highlight=calculus&t=27903 .


geometry
incenter
circumcircle
geometry proposed

[quote="Jutaro"]In a triangle $ABC$ the incircle is tangent to $AB$ and $AC$ in $N$ and $M$, respectively. $P$ and $Q$ are the points of intersection of $MN$ and the circumcircle. If $I$ is the incenter, $R$ is the circumradius and $S$ is the circumradius of $PQI$, show that $S=2R$.[/quote]

This is wrong, but I guess you meant http://www.mathlinks.ro/Forum/viewtopic.php?t=86561 .

  darij


Hi,

In my school, physics is only offered to the seniors, and I'm hoping to learn it by myself over the summer. Is it possible? :? I have so far took biology and I'll be done with trigonometry and geometry before the summer. If it's possible for me to learn the physics, what textbooks do you recommend?

Also, I was hoping if I can attend some kind of physics competition like AMC for math. But I don't know any of them. Can anyone give me some information on this issue? Thanks. Plus, if there is, do I have to get school's permission? :? You see, my school's very picky about letting me do these stuff because I'm only student on our school who is trying to active on the rural area. :(

Physics really should be learned once you know calculus (calculus invented to do physics). If you were to take physics w/o calculus, then you really would miss some of the more theoretical aspect of it.

However, dont let me discourage you, mayhap others have a different opinion

Well, high school physics is a breeze.  And some concepts in calculus you really should already know (like instantaneous slope).  So I think you will have no problem.  I think I could have gone through in under a week if I had 3+ hours per day for it.  Some concepts however might not sink in until a couple months later, but you don't have to make any effort in between.

I use Physics (Fifth Edition) by Giancoli, based on non-calculus physics for the AP B test. I've only seen 2 physics textbooks, so I wouldn't know, but this one seems to be the better one.. It's also recommended by College Board, EPGY uses it, too.

Calculus does help the theoretical part, so if you're only going to do physics once, I'd agree with mcalderbank to wait until you have calculus. But if you're planning to take physics twice, there's still a lot of cool non-calculus physics you can do over the summer.

I'm taking AP Physics C this year, and the teacher uses calculus to make the concepts more confusing. It's more important to understand the concepts than to just apply the formula. The test is centered on computations.

There is a competition in physics analogous to the AMC/AIME/USAMO series, but it's not as competitive (I think) or widely taken, so you'll probably have more trouble convincing your school to pay for it.

[url=http://aapt.org/Contests/olympiad.cfm]Here's a link[/url]

I also had Giancoli 5th edition last year. It was a pretty good book, a lot of interesting problems.

we began to study physics in the 8th year. And i prefer pure physics stuff.
Though Calculus helps a lot when solve a physics problem (especially a very difficult one), i still love the solutions only with elementary math and excellent analytics!

[quote="paladin8"][quote="ankur87"]I use the same book as Sky

However, up to this point, our class has covered only about 5 chapters

Time to cram for the test....[/quote]

Wow, cram indeed. We've covered up to Ch 17 and we're way behind. We also did Ch 13/14/15 in one day so I don't really remember anything from it.[/quote]

At least you guys are... moving along (well, sort of) :)
The hopeless regular physics class I'm taking has just started work & energy...
Well, self-study last-minute cramming, I guess.

The official text for the Texas State Physics competition, (which I won in 1995, woohoo!) is Tipler.


geometry
circumcircle
geometric transformation
reflection

ABC is a triangle with orthocenter H. The median from A meets the circumcircle again at A1, and A2 is the reflection of A1 in the midpoint of BC. The points B2 and C2 are defined similarly. Show that H, A2, B2 and C2 lie on a circle.


function
real analysis
real analysis unsolved

We spoke about it long time ago, but noone posted a solution: if $f:[a,b]->R$ has finite limit in every point of the interval then it is Riemann integrable.

But very similar problems have been discussed numerous times.

The set $D$ of discontinuities of $f$ is countable, since it's the union of the sets $A_n$ consisting of those points $x$ where the oscillation of $f$ is at least $\frac 1n$, and every such set is isolated and thus countable. This means that, in particular, $D$ has null measure, so the function is Riemann integrable, satisfying Lebesgue's criterion.

Of course that this is the most natural way to do it, but I would like to see a solution without Lebesgue's theorem.

[b]Sketch of Proof[/b]. Suppose the interval is $[a,b]$ and the length $L=b-a$. For any given $\epsilon$, since the limit exists everywhere, by Cauchy's Criterion, we have $\delta_x$ for each $x$ such that $|f(x)-f(y)|<\epsilon/2L$ if $0<|x-y|<\delta_x$. In this way, we have a family of open intervals $(x-\delta_x,x+\delta_x)$ covering $[a,b]$. Apply finite covering theorem, we can choose $(x_i-\delta_i,x_i+\delta_i) (i=1,\dots,m)$ covering $[a,b]$.

Suppose the limit of $f$ at point $x=x_0$ is $A_{x_0}$. We can prove $f$ is bounded on $[a,b]$: $|f|\leq \Omega = \max\{|A_{x_i}|+\frac{\epsilon}{L}, |f(x_i)|\}$

We can make a partition of $[a,b]$. Divide $[x_i,x_{i+1}]$ into two groups:
(1) the $x_i$ are contained in the first group of intervals and the total length of intervals in this group is less than $\epsilon/2\Omega$;
(2) Each interval in the second group is contained completely in one of $(x_i-\delta_i,x_i+\delta_i)\setminus\{x_i\}$. The amplitude of $f$ on each interval is less than $\epsilon/2L$; the total length of intervals in this group is less than $L$.

Thus $\sum \omega_i \Delta x_i \leq \Omega\cdot\frac{\epsilon}{2\Omega} + L\cdot \frac{\epsilon}{2L}<\epsilon$.

Hence $f$ is Riemann-integrable on $[a,b]$.


The quadrilateral $A_1A_2A_3A_4$ is cyclic, and its sides are $a_1 = A_1A_2, a_2 = A_2A_3, a_3 = A_3A_4$ and $a_4 = A_4A_1.$ The respective circles with centres $I_i$ and radii $r_i$ are tangent externally to each side $a_i$ and to the sides $a_{i+1}$ and $a_{i-1}$ extended. ($a_0 = a_4$). Show that \[ \prod^4_{i=1} \frac{a_i}{r_i} = 4 \cdot (\csc (A_1) + \csc (A_2) )^2. \]

$ABCD$ is cyclic with exradii $ r_a,r_b,r_c,r_d.$ We denote  $ AB = a,$ $BC = b,$ $CD = c,$ $DA = d.$ Let $ X$ be the tangency point of $ (I_a,r_a)$ with $ AB.$ Then we have

$ \tan \frac {A}{2} = \frac {AX}{r_a} \ , \ \tan \frac {B}{2} = \frac {BX}{r_a} \Longrightarrow$

$ \frac {a}{r_a} = \tan \frac {A}{2}+ \tan \frac {B}{2} = \csc A + \csc B - \cot A - \cot B$

Since the cosecant of opposite angles are equal and the cotangent of opposite angles have different signs, the cyclic product yields

$ \frac {abcd}{r_ar_br_cr_d} = [(\csc A + \csc B)^2 - (\cot A + \cot B)^2][ (\csc A + \csc B)^2 - (\cot A - \cot B)^2]$

$ \frac {abcd}{r_ar_br_cr_d} = 4[1 + \csc A %Error. "cscB" is a bad command.
(1 - \cos A \cos B)][1 + \csc A\csc B(1 + \cos A \cos B)]$

$ \frac {abcd}{r_ar_br_cr_d} = 4[1 + 2\csc A \csc B + \csc^2 A \csc^2 B - (\csc^2 A - 1)(\csc^2 B - 1)]$

$ \frac {abcd}{r_ar_br_cr_d} = 4(\csc^2 A + \csc^2 B + 2\csc A \csc B)= 4(\csc A + \csc B)^2.$


The median of the list
\[ n, n \plus{} 3, n \plus{} 4, n \plus{} 5, n \plus{} 6, n \plus{} 8, n \plus{} 10, n \plus{} 12, n \plus{} 15
\]is $ 10$. What is the mean?

$ \textbf{(A) }4\qquad\textbf{(B) }6\qquad\textbf{(C) }7\qquad\textbf{(D) }10\qquad\textbf{(E) }11$

[hide="for the last time, hide it.  u know the drill"]the middle number would be:
$n+6=10, n=4$

The mean is: $\frac{99}{9}=\boxed{11}\Rightarrow\boxed{E}$[/hide]

[hide=Solution]Since $n+6$ is the median, $n+6 = 10$, or $n = 4$. Substituting this value of $n$ into each of the terms in the list then calculating the mean gives $11$. [/hide]


Please use the search function. [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=282533]Thanks.[/url]

that questions a little different..
Usually, for ftw you do NOT use calculators, because it's called cheating. For sure, in countdowns you do not use a calculator..

[quote="k00lperson"]that questions a little different..
Usually, for ftw you do NOT use calculators, because[b] it's called cheating[/b]. For sure, in countdowns you do not use a calculator..[/quote]

You may not like it when someone uses a calculator, but in no way is it against the rules.

[hide="rules"][quote="TOS"]No inappropriate language should be used (this includes acronyms and substituting letters to avoid language filter)
No multiple accounts
No using proxy servers
No excessive spam or usage of all CAPS
No discussion of any type of illegal activity
Do not reveal your or others' name or location
Do not harrass other users
Do not concern yourself with the actions of others or why they were muted/banned
Act responsibly and be respectful at all times
Follow all guidelines as outlined in the Community Terms and Conditions[/quote][/hide]

i never said it was in the rules, but people will start to hate you.

lol, you use calculators??ok..well people aren't gonna really like you.because to the people in ftw, it's sort of counted as cheating..that's what I mean by cheating.

how is it cheating then i geuss problems that tell you to round to the nearest thousandth is cheating?


We need prove that $ \left(\frac {2n}{2n \plus{} 1}\right)^n \plus{} \left(\frac {2n \minus{} 1}{2n \plus{} 1}\right)^n < 1$

 But ,if $ x_n \equal{} \left(\frac {2n}{2n \plus{} 1}\right)^n$ then $ x_n > x_{n \plus{} 1}$ and     $ y_n \equal{} \left(\frac {2n \minus{} 1}{2n \plus{} 1}\right)^n$  then $ y_n > y_{n \plus{} 1}$

 So     $ x_n \plus{} y_n < x_1 \plus{} y_1 \equal{} \frac {2}{3} \plus{} \frac {1}{3} \equal{} 1$

 PS) For proving $ x_n>x_{n\plus{}1}$  and $ y_n>y_{n\plus{}1}$, follows Bernoulli Inequality,$ (1\plus{}x)^n\geq1\plus{}nx$.

[quote="ghostchungho"]prove that
$ (2n \plus{} 1)^n > (2n)^n \plus{} (2n \minus{} 1)^n$[/quote]
I think you meant $ n\geq{2}$, otherwise the inequality doesn't hold. :wink:

I edit my post.....cause I found that fixing $ n\equal{}2$ we get, considering the inequality to be true,
$ (2n\plus{}1)^2>(2n)^2\plus{}(2n\minus{}1)^2\Leftrightarrow{4n^2\plus{}4n\plus{}1>8n^2\minus{}4n\plus{}1}\Leftrightarrow{8n>4n^2}\Leftrightarrow{n<2}$
..... a contradiction!
So I think $ n>2$.


28. Two concentric circles are drawn so that the tangent segment (shown) to the smaller circle is a chord of the larger circle. If the area of the annulus (region outside the smaller circle and inside the larger one) is 100 , find the length of the chord shown ( ).

[hide]Call the radius of the inner circle $r$, and the radius of the outer circle $s$. The area of the inner circle is then $\pi r^{2}$ and the area of the outer circle is $\pi s^{2}$. The problem tells us that $\pi s^{2}-\pi r^{2}=100$. We manipulate the equation as shown below:

$\pi s^{2}-\pi r^{2}=100$
$\pi (s^{2}-r^{2})=100$
$s^{2}-r^{2}=\frac{100}{\pi}$

We then need to find the length of the chord. Draw in the radius of the smaller circle perpendicular to the chord, and the radius of the larger circle so that it touches one endpoint of the chord. We have created a right triangle. By the Pythagorean theorem, the length of the "non-radius side" of this triangle is:
${{\sqrt{s^{2}-r^{2}}=\sqrt{\frac{100}{\pi}}=\frac{\sqrt{100}\sqrt{\pi}}{\pi}}=\frac{10\sqrt{\pi}}{\pi}}$

We notice that this leg is half of of the chord. We multiply by 2 to get our final chord length of ${\boxed{\frac{20\sqrt{\pi}}{\pi}}}$[/hide]


calculus
integration
logarithms
function
calculus computations

Guys,

I don't have any math softaware in my home PC and I cannot do this by hand:

it's an  integral from 0 to 1 of log of base 2 (3-2(square root of x)) dx

and same interval

log of base 2 (5-3(square root of x)) dx

Thanks!

PS

I just need the answers.

$\int_0^1 \log_2(3-2\sqrt{x})dx$
$\int_0^1 \log_2(5-3\sqrt{x})dx$

[quote="nicknitro71"]PS

I just need the answers.[/quote]

It's nice to know that you're being driven by the pursuit of knowledge, rather than something less wholesome.

Thanks for the re-written formulas.

This is by the way to find out the U-uncertainty of a fuzzy interval of a infinite set, if curious.

You can always use wolfram. http://integrals.wolfram.com/index.jsp

I tried, but it gives nonesense for both x=0 and 1.

I know the solution for the second equation should be 1.546 but I don't know the one for the first equation.

 :(

$\int_0^1 \ln(a-b\sqrt{x})dx = -\frac12 - \frac ab + \frac{a^2 \ln a}{b^2} + (1-\frac{a^2}{b^2})\ln(a-b)$, or so says Mathematica.

It's still off!

int_0^1 \log2(5-3\sqrt{x})dx should be equal to 1.546

And I still don't know the first one.

By hand I got on the first one 0.7925 but I have no clue how accurate my calculations are.

No, it's not off: you just haven't translated the integral I gave you into the correct form.  I gave you an integral with $\ln$ -- that is, log base $e$.  To convert it to a log base 2, you need to divide by $\ln 2$.

[quote]I get the feeling that you have no idea what you are talking about nitro
[/quote]

That might be the case but explain this:

use that equation,

plug in a = 5 and b = 3.

Convert the ln into log2 by log2 x = lnx/ln2

And then tell me is equal to 1.546! I got 2.500053.

Well yea if you done everything correctly.

[quote="nicknitro71"]Ok,

use that equation,

plug in a = 5 and b = 3.

Convert the ln into log2 by log2 x = lnx/ln2

And then tell me is equal to 1.546![/quote]

Yes.  I'll also note that while the integrator ozo suggested picks a bad branch of the Log function in order to solve this problem, giving you complex values for the integral at both endpoints of your interval, their difference is real and is the correct answer, which is totally appropriate since it shouldn't be branch-dependent.

[quote="JBL"]Oh, heh.  You have to change the $\ln$ [i]in the original integral[/i] to a $\log_2$ by dividing everything by $\ln 2$.[/quote]

Got it, 0.6807.

By the way by doing the whole integral by hand (took me 5 hours) I got .7925.

Thanks.


[quote="cezar lupu"]Prove that for any $a, b\in\mathbb{Q^{*}_{+}}$ we have the following inequality:

                                    \[ a^{\frac{a}{b}}+b^{\frac{b}{a}}\geq a+b.  \][/quote]

I will prove the problem for $a,b\in \Re$
Define $f(a,b)=a^{\frac{a}{b}}+b^{\frac{b}{a}}-(a+b)$ then $f(a,b)=f(ka,kb)$ for any $k>0$, so we can take $a\ge b=1$ Hence $f(a,1)=a^a-a=a(a^{a-1}-1)\ge 0$ with equality iff $a=b=1.$ So in the original inequality equality occurs iff $a=b$


graph theory
combinatorics unsolved
combinatorics

Greeting and thank you for helping.  I am working on a project and I need to show that R(2K3 , 2K3) =10.  My advisor say I should get monochrome bow first.  I seem to be doing too much work for this proof I think there is smarter way.  K3 is complete graph 3 vertices. Thank you

[quote="AidanKitty"]K3 is complete graph 3 vertices.[/quote]

Ok, but what is 2K3 ?

Pierre.

2 disjoint copies of a complete graph with 3 vertices, sorry


Mechanism:

NO2 + O3 <--> NO3 + O2   fast

NO3 + NO2 --> N2O5          slow

I am having trouble and got the rate law to be = [NO2][O3]/ [O2] but I can't get rid of the [O2] term for some reason.

I believe for pre-equilibrium conditions you generally assume that the concentrations of all intermediates are constant (which makes sense, if you have an equilibrium in the first place, all the participants better be constant).  So [O2] is just another constant and can be incorporated into the rate constant (which it looks like you accidentally forgot).  Also, in general there is nothing wrong with having a rate law depend on the products.  I've seen a couple of examples that do that, but can't think of any off the top of my head.  Hope that helps.

Thanks a lot! The thing that got me was that O2 was a product and the answer key did not have it in the rate law


If $x,y,z$ are real numbers such that $xyz=1$, evaluate \[ \frac{x+1}{xy+x+1}+\frac{y+1}{yz+y+1}+\frac{z+1}{xz+z+1}. \]

You are right, deej21.

As for me, plugg $z=\frac{1}{xy}$ into the given equation.

Also you can make the substitution $\ x= \frac{a}{b}$ $\ y=\frac{c}{a}$ and $\ z=\frac{b}{c}$.  Things simplify nicely.


[quote="arqady"]Let $a,$ $b$ and $c$ are positive numbers such that $a^{6}+b^{6}+c^{6}=3.$ Prove that:
$(ab+ac+bc)\left(\frac{a^{2}}{b^{2}}+\frac{b^{2}}{c^{2}}+\frac{c^{2}}{a^{2}}\right)\geq9.$[/quote]
Are you sure the above holds? Please check this counter example for me: $(a,b,c)=(5,8,6)$.

[quote="arqady"] $a^{6}+b^{6}+c^{6}=3.$[/quote]
i always forget the conditions too...;)

The counter example is for the homogenizied version.

You are right, ductrung. This consequence of a mistake here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=155488


Let $f$ be a function that satisfies :
\[ \displaystyle f(x)+2f\left(\frac{x+\frac{2001}2}{x-1}\right) = 4014-x. \]
Find $f(2004)$.

Sorry to revive this one.This one is pretty easy.We can actually find $f$.Consider $g(x)=\frac{x+\frac{2001}{2}}{x-1}$ considering its orbit we get $g(g(x))=\text{id}_x$.Now rewrite the equation as $f(x)+2f(g(x))=4014-x$ replace $x\mapsto g(x)$ and the equation becomes $f(g(x))+2f(x)=4014-g(x)$ now we can solve for $f$ and get $3f(x)=x+4014-\frac{2x+2001}{x-1}$ and putting $x=2004$ indeed gives $f(2004)=2005$


calculus
integration
trigonometry
induction
inequalities
calculus computations

Prove the following inequality.

$ \frac {1}{n} < \int_0^{\frac {\pi}{2}} \frac {1}{(1 + \cos x)^n}\ dx < \frac {n + 5}{n(n + 1)}\ (n =2,3,\  \cdots)$.

I bet the reason you haven't received solutions yet is because there doesn't appear to be any 'nice' way to solve this - just tedious brute force. I fiddled around for a while but didn't find any clean expression for the integral so I took a nap and give you this mess. :p

$ \cos ^2 {\textstyle{x \over 2}} \equal{} \frac {{1 \plus{} \cos x}}{2}$; $ \sec ^2 {\textstyle{x \over 2}} \equal{} 1 \plus{} \tan ^2 {\textstyle{x \over 2}}$

U-substitutions:
$ B_n \equal{} \int\limits_0^{{\textstyle{\pi \over 2}}} {\frac {1}{{(1 \plus{} \cos x)^n }}dx} \equal{} \frac {1}{{2^{n \minus{} 1} }}\int\limits_0^1 {(1 \plus{} x^2 )^{n \minus{} 1} dx}$

Integration by parts:
$ B_{n \plus{} 1} \equal{} \frac {{1 \plus{} nB_n }}{{1 \plus{} 2n}}$

Induction:
$ B_1 \equal{} 1$

P.S. Strict inequality does not hold in this case ;)

[quote="§outh§tar"]

P.S. Strict inequality does not hold in this case ;)[/quote]

Yeh, you are right.

I will edit the condition.

Thank you! :oops:

kunny


geometry
geometric transformation
rotation
LaTeX
vector
calculus
integration

A homogeneous circular disk lies on a wall to wall carpet exerting a uniform pressure on it.By pulling the string attached to a circumferential point and pointing in the radial direction, a force Fo is needed  to move it.What force F can move the disk , if  the string is pulled in the tangential direction?(The string is horizontal in both cases)

The problem is quite easy , but i have to check the answer i got .There's no answer given !

sorry

can u describe the first line in ur problem statement...that is the disc lies on a wall to wall carpet...

i am not able to understand it

It probably means that the disk lies on a carpet which(the carpet) extends between two walls

is it touching the vertical wall to its left...
suppose if it is just resting
write the force equation..........assuming there is friction..ok

so we get ${ \{F}{ - }{f}{ = }{Ma}$ where a is the acceleration ok

now to get the discs into motion the min force required is f...only after applying a force which is equal to the max force of friction ...will the object move

It is not touching any wall (the disk)

Your equation is correct but due to tendency of rotation in second case the direction of friction wil be different at different points so we have to find f

and moving is equivalent to any kind of motion
- translational and rotational ..

further i think answer will be in terms of Fo

[quote="rohan2007"]
...disk lies on a wall to wall carpet...

.By pulling the string attached to a circumferential point and pointing in the radial direction....

....the string is pulled in the tangential direction?....

[/quote]

I did not quite appreciate the setup, probably since the description is too hazy.

[quote="rohan2007"]
The problem is quite easy , but i have to check the answer i got .There's no answer given !

[/quote]

You could post your version of the solution or answer, that might help clear things up for us in terms of understanding.

Seems we could just take a disk resting on an horizontal rough surface, and find

\[ F \equal{} {2\over 3} F_0\]

[quote="Immanuel Bonfils"]
Seems we could just take a disk resting on an horizontal rough surface, and find
\[ F \equal{} {2\over 3} F_0
\]

[/quote]

You mean the circular surface of the disc touching the floor? That could be a possibility.

Yet I didn't understand what is meant by radial direction, tangential direction, etc.

Can anyone explain?

Actually as it is mentioned that the disk exerts a uniform pressure on the surface  - > It should lie flat on the surface..

I will post my working on it a bit later ..

[quote="rohan2007"]radial direction points to the direction normal to the disk at the outer edge ..

tangential direction indicates that a torque FR is given along with F ...[/quote]

Yep! I understood the arrangement. What I can't understand is what will be the IAR when the force is applied tangentially. Is it the centre, or the diametrically opposite point of point of application of force. I strongly its the latter.

Please, what is IAR?

rohan2007 how about posting your answer? Even hiding it...

[quote="Immanuel Bonfils"]
What is IAR?

[/quote]

Instantaneous Axis of Rotation.

[quote="Immanuel"]

rohan2007 how about posting your answer? Even hiding it...

[/quote]

I too suggested the same a while ago.

Since the disk doesn't move, any axis rigidly "tied" to him can be the "IAR" (in that case PAR, hehehe...)

Ok actually i dont know how to use latex              ..

so i will just give the approach

consider a ring of thckness dr at a dstance r from the centre

take an  line passing through the centre of the ring and paralell to force applied .Take it as the x- axis .

let us take an element making an angle d(a)  and the line joining the centre of the ring to the element make angle 'a' with x-axis

then ,  as friction oppose motion or tedency of motion .. immediately after applying the force F there will be rotationl and translational motion tendencies of that element .

As friction acts opposing the net velocity friction will act at a direction which represents the opposite direction of the vector sum of the rotational and translatonal accelerations which are

F/m and 2F/m(r/R) respectively

friction will be udmg u = coefficient of friction

so integrating frictional torque from 0 to 2pi and again in terms of r from 0 to R we get to know the total opposing torque

then again integrating the friction 's component along negative x axis we get the total opposed translational tendency

(actually from the first situation we know that F=umg
as there is no tendency of rotation there
)

now equating FR to the frictional torque and F to friction's component along negative x axis , we will get two values of F

the smaller one is the required force to make it move -  translation or only rotation

Now  have enountered another problem  -> the integral is coming a bit too big ..

earlier i had obtained the value F0/4

Seems you're making it in the difficult way: the force resultant due to the elementary tangential components cancels out, remaining only that due to translational acceleration. So, you only have to calculate the the resultant moment of the formers and compare it with that of the force F.
Did you see the  value I pointed out before, above?

Well friction at ech point will be directed exactly opposite to the net resultnt of the translational and rotational accerelations
It will have a component along the axis i pointed out opposite to F

and it will also have a torque pointing opposte to direction of torque if only F would have been there .

And the value of the net torque by friction will wont be zero ! It is because when you solve the integral  .THERE WILL BE a non zero expres

Please, read more carefully; I didn't say (rather write, he he he ) that the torque cancels out, but [b]the force resultant due to the elementary tangential components[/b], do.


robbin monro process
limit
probability
induction
convex function
algorithm
rate problems

The problem is a little lengthy to state
Suppose $g:\mathbb{R}^K \mapsto \mathbb{R}$ is a convex function. And for any given $\boldsymbol{\mu} \in \mathbb{R}^K$, there exists a unique point denoted by $\boldsymbol{R}(\boldsymbol{\mu})\in \mathbb{R}^K$ such that $g$ has a  subgradient $\boldsymbol{\mu}$ at $\boldsymbol{R}(\boldsymbol{\mu})$. However both $g$ and $\boldsymbol{R}(\boldsymbol{\mu})$ are unknown. We can only observe a random variable $\boldsymbol{Y}(\boldsymbol{\mu})$ with $\mathbb{E}[\boldsymbol{Y}(\boldsymbol{\mu})]=\boldsymbol{R}(\boldsymbol{\mu})$ and $\sup_{\boldsymbol{\mu}}\mathbb{E}[\boldsymbol{Y}(\boldsymbol{\mu})^2]<\infty$.
Now we want to find a $\boldsymbol{\mu}^*$ such that $\boldsymbol{R}(\boldsymbol{\mu}^*)=0$.
Note that $\boldsymbol{\mu}^*$ might not be unique and constitutes a convex set $\mathcal{S} \subset \mathbb{R}^K$.
Consider the following Robbin-Monro (RM) process begins with an inital estimate $\boldsymbol{\mu}_1$, and $\boldsymbol{\mu}_n$ is updated by the recursion

(1) $\boldsymbol{\mu}_{n+1}=\boldsymbol{\mu}_{n}-a_n \cdot \boldsymbol{Y}_n$

where $\sum_{n=1}^{\infty} a_n^2<\infty$ and $\sum_{n=1}^{\infty} a_n=\infty$.
The claim is that $\boldsymbol{\mu}_{n} \to \boldsymbol{\mu}^*$ wp 1, where $\boldsymbol{\mu}^* \in \mathcal{S}$.

********************************************************************

I can only prove the case when $\mathcal{S}$ is a singleton. Here is the proof:

Take $\boldsymbol{\mu}^* \in \mathcal{S}$, From (1) and  $\mathbb{E}[\boldsymbol{Y}(\boldsymbol{\mu})]=\boldsymbol{R}(\boldsymbol{\mu})$, we have

(2) $\mathbb{E}[\| \boldsymbol{\mu}_{n+1}-\boldsymbol{\mu}^*\|^2 | \boldsymbol{\mu}_n ] = \| \boldsymbol{\mu}_n-\boldsymbol{\mu}^*\|^2 - 2a_n(\boldsymbol{\mu}_n - \boldsymbol{\mu}^*)^T \boldsymbol{R}(\boldsymbol{\mu}_n)
+a_n^2 \mathbb{E}[\boldsymbol{Y}(\boldsymbol{\mu}_n)^2] $

Convexity implies $(\boldsymbol{\mu}_n - \boldsymbol{\mu}^*)^T \boldsymbol{R}(\boldsymbol{\mu}_n) \geq 0$.
Note also $\sum a_n^2 \mathbb{E}[\boldsymbol{Y}(\boldsymbol{\mu}_n)^2] <\infty$, therefore,
$\| \boldsymbol{\mu}_n-\boldsymbol{\mu}^*\|^2$ is a [i]nonnegative almost supermartingale[/i]. From a theorem by Robbins and Siegmund (1971), we have

(3) $\| \boldsymbol{\mu}_n-\boldsymbol{\mu}^*\|^2$ converges a.s. and

(4) $\sum a_n (\boldsymbol{\mu}_n - \boldsymbol{\mu}^*)^T \boldsymbol{R}(\boldsymbol{\mu}_n)$ converges a.s., which implies

(5) $\liminf_n (\boldsymbol{\mu}_n - \boldsymbol{\mu}^*)^T \boldsymbol{R}(\boldsymbol{\mu}_n)=0$

If $\mathcal{S}$ is a singleton, we can show that for any $\epsilon>0$

(6) $\inf\{(\boldsymbol{\mu} - \boldsymbol{\mu}^*)^T \boldsymbol{R}(\boldsymbol{\mu}):\|\boldsymbol{\mu}-\boldsymbol{\mu}^*\|>\epsilon\}>0$

This combining (3) and (5) implies $\boldsymbol{\mu}_n \to \boldsymbol{\mu}^*$ a.s.
However, if $\mathcal{S}$ is a not singleton, but a convex set, I still haven't figured out how to prove it (Since (6) doesn't hold).
Intuitively, it should still converge. But anyone has ideals about the proof?

Essentially you've done all the hard work. It follows from $(3)$ that almost surely $\|\mu_n-\mu^*\|$ converges for all $\mu^*\in S$ (check it on a dense countable subset of $S$ first). WLOG, we may assume that $g(0)=0$ and $0\in S$ (just subtract a linear function from $g$). Since $\|\mu_n\|$ converges and  $\liminf_{n\to\infty} \mu_n^T R(\mu_n)=0$, we may choose a subsequence $\mu_{n_k}$ such that $\mu_{n_k}$ converges to some $\mu_*$ and $\mu_{n_k}^TR(\mu_{n_k})$ converges to $0$. If we show that $\mu_*\in S$, we are done because then $\|\mu_n-\mu_*\|$ must converge, but the limit can be only $0$. Now, for each subgradient $\mu_{n_k}$ and the corresponding point $R(\mu_{n_k})$, we can write
$g(x)\geq g(R(\mu_{n_k}))+\mu_{n_k}^T(x-R(\mu_{n_k}))\geq \mu_{n_k}^T(x-R(\mu_{n_k}))$.
Passing to the limit as $k\to\infty$, we obtain $g(x)\geq \mu_*^T x$, i.e., $\mu_*$ is a subgradient of $g$ at $0$.
Thus, the limit $\lim_{n\to\infty}\mu_n$ exists almost surely and belongs to $S$. Of course the limit of $\mu_n$ will be a non-constant random variable with values in $S$ in general...

Constructing a convergent subsequence is really a powerful tool.
Thanks! fedja.
You've solved all my "suppose to be hard" problems in minuates!  Amazing...

[quote="fedja"][quote="remgenesis"]A quick question: How do you guarantee that we can always find a subsequence $\mu_{n_k}$ such that $\mu_{n_k}$ converges to some $\mu_*$?[/quote]
Since $\|\mu_n\|$ converges, $\mu_n$ is a bounded sequence and any bounded sequence in $\mathbb R^n$ contains a convergent subsequence.[/quote]

I am just a little confused about the boundness of a sequence of random variables.

$\|\mu_n\|$ converges means  for each $\omega$, $\|\mu_n(\omega)\|$ converges. But the convergence may not be uniform in $\omega$. So how can you conclude  $\|\mu_n\|<C$ for some constant $C$. or maybe this is not the boundness you mean? How do you define a bounded sequence?

All my considerations were made for fixed $\omega$ satisfying various properties each of which holds with probability $1$. So, the entire reasoning works for "good" $\omega$, which are almost all $\omega$. Of course, the bound on $\|\mu_n(\omega)\|$ and the limit $\mu_*(\omega)$ vary with $\omega$.

But for diffrent $\omega$, you would find different convergent subsequence. How could you finally get a same subsequence regardless $\omega$? Still confused...  :(

Yes, for different $\omega$ there will be different subsequences. What's wrong with that?  :? I have never claimed that I chose the same subsequence for all $\omega$. My point is that, despite you start with a convergent subsequence depending on $\omega$, you end up with the result that the entire sequence converges to $\mu_*$.

Oh, I see. So all your reasoning is for each good $\omega$, not for the whole random variable.
so rigorous speaking, we should say for almost every $\omega$, we can find $\mu_{n_k}(\omega)$ converges to $\mu^*(\omega)$. But we might not be able to find a convergent subsequence of random variables.

Is my understanding correct?

Sorry for bothering with such basic questions.

[quote="remgenesis"]
so rigorous speaking, we should say for almost every $\omega$, we can find $\mu_{n_k}(\omega)$ converges to $\mu^*(\omega)$. But we might not be able to find a convergent subsequence of random variables.
[/quote]
Exactly.

Just get a little confused about the underlying $\sigma$ fields of this stochastic process.

Should $\mathcal{F}_n=\sigma(\mu_1,\mu_2,....,\mu_n)$, which due to markov property of the process can be simplied to $\mathcal{F}_n=\sigma(\mu_n)$.

or

$\mathcal{F}_n=\sigma(\mu_1,Y_1,\mu_2, Y_2....,\mu_n, Y_n)$?

I used the first one but my professor said it is not correct and must define  $\mathcal{F}_n$ using the underlining process ($\{Y_n\}$) and then prove $\mu_n$ is $\mathcal{F}_n$ measurable...really confused. Anything wrong with the first definiation?

This makes me confused. My understanding was that we are able to take any $\mu$ and observe $Y(\mu)$ once, different observations being independent of each other (so $Y(\mu_n)$ is actually $Y(\mu_n,\omega_n)$ and, by induction, $\mu_n=\mu_n(\omega_1,\dots,\omega_{n-1})$ (otherwise I don't understand how (2) in your first post was derived). So, in my opinion, $\mathcal F_n$ is just the $\sigma$-field of events depending on the first $n-1$ omegas only in the product probability space.

Using the first $n-1$ $Y$'s to define $\mathcal F_n$ seems also possible in this case, though a bit confusing. It'll result in somewhat smaller $\sigma$-fields. Using first $n$ $\mu$'s seems also OK and gives exactly the same result as using $Y$'s if all $a_k\neq 0$ because then the smallest $\sigma$-field with respect to which $\mu_1,\dots,\mu_n$ are measurable is the same as the smallest $\sigma$-field with respect to which $Y_1,\dots,Y_{n-1}$ are measurable. The only problem is that, firstly, you add randomness to $\mu$'s using $Y$'s, not vice versa, so, in a sense, the $\sigma$-field generated by $Y$'s is primary and that generated by $\mu$'s is secondary and, secondly, some $a_k$ may be $0$, in which case you can lose measurability of the corresponding $Y$.

The remark that the definition using $\mu$'s can be simplified to $\sigma(\mu_n)$ due to the Markov property is incorrect. Consider two random variables on the standard 4-element probability space given by (-1,-1,1,1) and $(-2,0,0,2)$. The second one is obtained from the first one by adding (-1,1,-1,1), which is independent of the first random variable. Nevertheless, the $\sigma$-field generated by the second variable alone is not the same as the $\sigma$-field generated by both.

Hope I haven't said something stupid (alas, probability is not my field and I remember now only about half of what I've been taught during my student years). So you may want to check my remarks with your professor and see what (s)he has to say before taking them too seriously...  :)

Your first explanation is very similar to my professors's.
So you mean we can take $\mathcal{F}_n=\mathcal{B}(\Omega_1 \times \Omega_2,...., \times \Omega_{n-1})$.
I am just a little confusing about using "product probability space", since the dimension of $\mathcal{F}_n$ will ever increase with $n$.
To avoid using the product probability space, suppose we change the underlying probability space $\{\omega_n\}$ to a sequence of i.i.d. random variables $\{X_n\}$. That is, instead of writting $Y_n=Y(\mu_n, \omega_n)$, we use $Y_n=Y(\mu_n, X_n)$.
Then we can define $\mathcal{F}_n=\sigma(X_1, X_2,....,X_{n-1})$. Since now $Y_{n-1}$ and $\mu_n$ are functions of $X_1,X_2...,X_{n-1}$, $\sigma(Y_1,...,Y_{n-1})$ and $\sigma(\mu_1,...,\mu_n)$ will be smaller $\sigma$-fields than $\mathcal{F}_n$.

But what I am still confusing is do we always have
$E(f(\mu_{n+1})| \mu_n) = E(f(\mu_{n+1})| \mu_n,...,\mu_1)=E(f(\mu_{n+1})| Y_{n-1},...,Y_1)=E(f(\mu_{n+1})| X_{n-1},...,X_1)$? since they generate different $\sigma$-sigma fields.

BTW: What is your main field of interests. Seems to me you do know a lot! :)

[quote="remgenesis"]
But what I am still confusing is do we always have
$E(f(\mu_{n+1})| \mu_n) = E(f(\mu_{n+1})| \mu_n,...,\mu_1)=E(f(\mu_{n+1})| Y_{n-1},...,Y_1)=E(f(\mu_{n+1})| X_{n-1},...,X_1)$? since they generate different $\sigma$-sigma fields.
[/quote] In this particular case the conditional expectations coincide because, in general, if you have a $\sigma$-field $\mathcal F$, a random variable $\mu$ measurable with respect to $\mathcal F$ and a random variable $X$ independent with $\mathcal F$, and construct some new random variable $\xi=\Xi(\mu,X)$ where $\Xi$ is some fixed function, then $E(\xi|\mathcal F)=E(\xi|\mathcal \mu)$. Note that it is important here that $X$ is independent with the entire $\mathcal F$, not just with $\mu$.

[quote="remgenesis"]BTW: What is your main field of interest? [/quote] Harmonic analysis. I also like combinatorial geometry and a few other things.

I am considering a relaxation of the condition of the problem in the orginal post.
Now the convex function $g$ is not strict convex. So it may has linear part.
Since now the function $\boldsymbol{R}(\boldsymbol{\mu})$ is not uniquely defined,
$\boldsymbol{R}(\boldsymbol{\mu})$ denote the set of points at which $g$ has a  subgradient $\boldsymbol{\mu}$

Both $g$ and $\boldsymbol{R}(\boldsymbol{\mu})$ are unknown. We can only observe a random variable $\boldsymbol{Y}(\boldsymbol{\mu})$ with $\mathbb{E}[\boldsymbol{Y}(\boldsymbol{\mu})] \in \boldsymbol{R}(\boldsymbol{\mu})$ and $\sup_{\boldsymbol{\mu}}\mathbb{E}[\boldsymbol{Y}(\boldsymbol{\mu})^2]<\infty$.

We still consider the following Robbin-Monro (RM) process begins with an inital estimate $\boldsymbol{\mu}_1$, and $\boldsymbol{\mu}_n$ is updated by the recursion

$\boldsymbol{\mu}_{n+1}=\boldsymbol{\mu}_{n}-a_n \cdot \boldsymbol{Y}_n$

Claim 1:We still have $\boldsymbol{\mu}_{n} \to \boldsymbol{\mu}^*$ wp 1, with here $0
\in \boldsymbol{R}(\boldsymbol{\mu}^*)$.
Claim 2: Moreover,  $\frac{\sum_i^n \boldsymbol{Y}_n}{n} \to 0$ wp1.

For the claim 1, I think the oringinal proof can apply to this case with little modifcations.
For the claim 2, it seems easy to prove in the orginal setting. But  I don't know if it can be proved in this settings.

[quote="remgenesis"]
Claim 1:We still have $\boldsymbol{\mu}_{n} \to \boldsymbol{\mu}^*$ wp 1, with here $0
\in \boldsymbol{R}(\boldsymbol{\mu}^*)$.
Claim 2: Moreover,  $\frac{\sum_i^n \boldsymbol{Y}_n}{n} \to 0$ wp1.

For the claim 1, I think the oringinal proof can apply to this case with little modifcations.
For the claim 2, it seems easy to prove in the orginal setting. But  I don't know if it can be proved in this settings.[/quote]

Indeed, the original proof remains valid for claim 1.

Claim 2 is false unless you impose some extra assumptions. Indeed, suppose that some non-zero $x_0\in R(\mu^*)$. Let $\mu_1=\mu^*$. Then we can take a sequence $a_n$ with $a_n=0$ for even $n$ and choose $Y_n=0$ for odd $n$ and $Y_n=x_0$ for even $n$. Our sequence then stays at $\mu^*$ forever but $\frac 1n\sum_{i=1}^n Y_i\to \frac12 x_0$.

Thanks.
Then what if I assume $Y_n=Y(\mu,X_n)$, where $X_n$ are i.i.d random variables and $Y$ is a fixed function.
So for the same $\mu$, $Y_n$ have the same statistics.

I think now I can prove it if $a_n=\frac{b}{n+c}$ for any fixed positive $b$, $c$.
For simplicity, just let $a_n=1/n$ and consider the scalar case.

Since we have proved that $\mu_n$ converges, $\mu_n$ is Cauchy.

Given $\epsilon>0$, there exits $N$ such that $|\mu_n-\mu_m|<\epsilon$ for $m,n \geq N$.

By the algorithm we have $n\mu_{n+1}-n \mu_n=Y_n$

So $\sum_{n=N}^M (n\mu_{n+1}-n \mu_n)=\sum_{n=N}^M Y_n$

So $|\sum_{n=N}^M Y_n|=|M\mu_{M+1}-\sum_{n=N+1}^M \mu_n-N \mu_N|
\leq N |\mu_{M+1}-\mu_N |+\sum_{n=N+1}^M |\mu_{M+1}-\mu_n|<M\epsilon$

Let $C=|\sum_{n=1}^{N-1} Y_n|$. Then $|\sum_{n=1}^M Y_n|<M\epsilon+C$

So for $M>C/\epsilon$ we have $\frac{1}{M}|\sum_{n=1}^M Y_n|<2 \epsilon$.

The proof seems so straightfoward in this case. Just want to make sure that I didn't make any stupid mistakes. :blush:

[quote="remgenesis"]
The proof seems so straightfoward in this case. Just want to make sure that I didn't make any stupid mistakes. :blush:[/quote]
No mistakes (as far as I can see  :P ). BTW, why is it so important to have $\lim_{n\to\infty}\frac1n\sum_{k=1}^n Y_k=0$?

The above proof can be easily extended to the case where we only require $\frac{1}{a_n}=O(n)$ and $a_n$ is nonincreasing. I think monotonicity might be not needed. But donno how to eliminate this condition.

Another question is I don't know if the oringinal conditon $\sum a_n=\infty$ plus $a_n$ is nonincreasing implies $\frac{1}{a_n}=O(n)$?
Update: This is not true.

In my problem, $R^*+Y_n$ here is some random transmit rate due to random noise in a  wireless channel at time $n$. I want the long term average rate converge to $R^*$, which is the target rate we want to achieve.


superior algebra
superior algebra unsolved

Let $ a_{1}, a_{2}, ..., a_{\phi(n)}$ be a reduced residue system modulo n and let N be the number of solutions to $ x^2 \equiv 1 \mod n$. Show that $ \prod_{i\equal{}1}^{\phi(n)}a_{i} \equiv (\minus{}1)^{\frac{N}{2}} \mod n$

HINT Pair elements using the involutions $ x^{\minus{}1}$, then $ \minus{}x$. Note: must assume $ n>2$.

Thank you for the hint, but I am not able to make any headway with it;-)

The map $ x \mapsto x^{\minus{}1}$ is a permutation on the units with cyles of length 2 and 1. The 2-cycles have form $ (x \; x^{\minus{}1})$ with product $ 1$. Thus the product of all units reduces to the product of all 1-cycles, i.e. those $ x$  with $ x^{\minus{}1} \equal{} x \iff x^2 \equal{} 1$. Now similarly consider the map $ x \mapsto \minus{}x$ on this set.


http://www.mathlinks.ro/Forum/viewtopic.php?t=12554
http://www.mathlinks.ro/Forum/viewtopic.php?t=14014

  darij

Actualy, if $A=\{(\sqrt{a} - \sqrt{b})^2, (\sqrt{b} - \sqrt{c})^2, (\sqrt{c} - \sqrt{a})^2\}$ and $a\geq0,b\geq0,c\geq0$ then
  $\frac{1}{2}minA\leq\frac{a+b+c}{3}-\sqrt[3]{abc}\leq maxA.$  :)

What about the genaralisation  ?
If $a_1,...,a_n\geq 0$  then prove that  $\frac{m}{2}\leq\frac{a_1+a_2+.....+a_n}{n}-\sqrt[n]{a_1a_2.....a_n}\leq\frac{(n-1)M}{2}$

where  $m=min_{1\leq i<j\leq n}\left\{ (\sqrt{a_i}-\sqrt{a_j})^2\right\}$

and $M=max_{1\leq i<j\leq n}\left\{ (\sqrt{a_i}-\sqrt{a_j})^2\right\}$

we should prove that LHS is smaller than the average and the problem is solved.
$ a \plus{} b \plus{} c \minus{} 3\sqrt [3]{abc}\le{2(a \plus{} b \plus{} c) \minus{} 2(\sqrt {ab} \plus{} \sqrt {bc} \plus{} \sqrt {ca})}$
$ a \plus{} b \plus{} c \plus{} 3\sqrt [3]{abc}\ge{2(\sqrt {ab} \plus{} \sqrt {bc} \plus{} \sqrt {ca})}$
$ a \equal{} x^6,b \equal{} y^6,c \equal{} z^6$
by schur inequality:
$ x^6 \plus{} y^6 \plus{} z^6 \plus{} 3x^2y^2z^2\ge{\sum{x^2y^4} \plus{} \sum{x^4y^2}}$
and by AM-GM:
$ \sum{x^4y^2} \plus{} \sum{x^2y^4}\ge{2\sum{x^3y^3}}$
proved!

Since $\max\{(\sqrt{a} - \sqrt{b})^2, (\sqrt{b} - \sqrt{c})^2, (\sqrt{c} - \sqrt{a})^2\}\ge \dfrac{(\sqrt{a} - \sqrt{b})^2+(\sqrt{b} - \sqrt{c})^2+ (\sqrt{c} - \sqrt{a})^2}{3}=\dfrac{2}{3}(a+b+c-\sqrt{ab}-\sqrt{bc}-\sqrt{ca})$ so it remains to prove after some algebra $$2(\sqrt{ab}+\sqrt{bc}+\sqrt{ca})\le a+b+c+3\sqrt[3]{abc}$$
However, note that $2(\sqrt{ab}+\sqrt{bc}+\sqrt{ca})\le \sum_{sym}\sqrt[3]{a^2b}$ by AM-GM and the problem reduces to Schur.

Observe that \[ \max\{(\sqrt{a} - \sqrt{b})^2, (\sqrt{b} - \sqrt{c})^2, (\sqrt{c} - \sqrt{a})^2\} \ge \frac{(\sqrt{a}-\sqrt{b})^2+(\sqrt{b}-\sqrt{c})^2+(\sqrt{c}-\sqrt{a})^2}{3} \]
We wish to show that, upon some expanding, that \[ \frac{a+b+c}{3} + \sqrt[3]{abc} \ge \frac{2}{3} (\sqrt{ab}+\sqrt{bc}+\sqrt{ca}) \]
Let $a=x^6, b=y^6, c=z^6$. Then by degree three Schur on $x^2, y^2, z^2$ and Muirhead we have
\[ x^6+y^6+z^6+3x^2y^2z^2 \ge \sum_{sym} x^4y^2 \ge 2(x^3y^3+y^3z^3+z^3x^3) \]
as requested.

WLOG, $a \le b \le c$. Fix $a,c$, the function $f(b) = \[ \frac{a+b+c}{3} - \sqrt[3]{abc}$ is convex and thus achieves a maximum when $b= a$ or $b=c$. Let $b =a$, then it suffices to prove $2(c/a) -6(c/a)^{1/2} + 3(c/a)^{1/3} +1 \ge 0$. If $b= c$ then it suffices to prove $2(a/c) - 6(a/c)^{1/2} + 3(a/c)^{1/3} +1 \ge 0$. Note that the polynomial $2x^6 + 3x^2 -6x^3 +1 = (x-1)^2(2x^4 + 4 x^3 + 6x^2+2x +1)$ which is nonnegative for all $x>0$. The inequality is proved.

[quote=MithsApprentice]Let $a, b, c$ be nonnegative real numbers.  Prove that
\[ \frac{a+b+c}{3} - \sqrt[3]{abc} \leq \max\{(\sqrt{a} - \sqrt{b})^2, (\sqrt{b} - \sqrt{c})^2, (\sqrt{c} - \sqrt{a})^2\}.  \][/quote]

[b]Claim : [/b] $x^6+y^6+z^6+3x^2y^2z^2\ge 2(x^3y^3+y^3z^3+z^3x^3)$
[i]Proof : [/i] Since $x^2+y^2\ge 2xy$ and By Schur's Inequality we get $$x^6+y^6+z^6+3x^2y^2z^2\ge x^2y^2(x^2+y^2)+y^2z^2(y^2+z^2)+z^2x^2(z^2+x^2)\ge 2(x^3y^3+y^3z^3+z^3x^3)$$ so the claim is proven.$\square$

                 -----------------------------------
Back to the original problem FTSOC assume \[ \frac{a+b+c}{3} - \sqrt[3]{abc} \geq \max\{(\sqrt{a} - \sqrt{b})^2, (\sqrt{b} - \sqrt{c})^2, (\sqrt{c} - \sqrt{a})^2\}. \].Adding the cyclic variants we get $a+b+c-3\sqrt[3]{abc}\geq (\sqrt{a} - \sqrt{b})^2+(\sqrt{b} - \sqrt{c})^2+(\sqrt{c} - \sqrt{a})^2 \Rightarrow 2(\sqrt{ab}+\sqrt{bc}+\sqrt{ca})\geq a+b+c+3\sqrt[3]{abc}$.

Plugging $x=\sqrt[6]{a}$,$y=\sqrt[6]{b}$,$z=\sqrt[6]{c}$ we get $2(x^3y^3+y^3z^3+z^3x^3)\geq x^6+y^6+z^6+3x^2y^2z^2$ a contradiction to our claim so we are done.$\blacksquare$

Since $\max\{(\sqrt{a} - \sqrt{b})^2, (\sqrt{b} - \sqrt{c})^2, (\sqrt{c} - \sqrt{a})^2\} \ge \frac{(\sqrt a-\sqrt b)^2+(\sqrt b-\sqrt c)^2+(\sqrt c-\sqrt a)^2}{3}=\frac{2(a+b+c)}{3}-\frac{2(\sqrt{ab}+\sqrt{bc}+\sqrt{ca})}{3}$, we ought to prove that \[a+b+c+3\sqrt[3]{abc} \ge 2(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}).\] By using Schur Inequality, we have \[a+b+c+3\sqrt[3]{abc}\ge (\sqrt[3]{a^2b}+\sqrt[3]{ab^2})+(\sqrt[3]{b^2c}+\sqrt[3]{bc^2})+(\sqrt[3]{c^2a}+\sqrt[3]{ca^2})\ge 2(\sqrt{ab}+\sqrt{bc}+\sqrt{ca})\] with the use of AM-GM at last. $\blacksquare$

Suppose otherwise, so
$$\frac{a+b+c}{3}-\sqrt[3]{abc}>\max\{(\sqrt{a} - \sqrt{b})^2, (\sqrt{b} - \sqrt{c})^2, (\sqrt{c} - \sqrt{a})^2\} \implies \frac{a+b+c}{3}-\sqrt[3]{abc}>(\sqrt{a}-\sqrt{b})^2$$
and all cyclic variations. So summing cyclically:
$$a+b+c-3\sqrt[3]{abc}>\sum_{\mathrm{cyc}} (\sqrt{a}-\sqrt{b})^2=2(a+b+c)-2\sum_\mathrm{cyc} \sqrt{ab} \implies a+b+c+3\sqrt[3]{abc} \leq 2\sum_\mathrm{cyc} \sqrt{ab}.$$
But by Schur's and Muirhead, as $(\tfrac{2}{3},\tfrac{1}{3},0) \succ (\tfrac{1}{2},\tfrac{1}{2},0)$:
$$a+b+c+3\sqrt[3]{abc} \geq \sum_\mathrm{sym} \sqrt[3]{a^2b}\geq 2\sum_\mathrm{cyc} \sqrt{ab},$$
which is a contradiction, so the desired inequality is true. $\blacksquare$

WLOG $a\geq b\geq c$. Then, this becomes $$\frac{a+b+c}{3}-\sqrt[3]{abc}\leq (\sqrt{a}-\sqrt{c})^2$$ $$a+b+c-3\sqrt[3]{abc}=3a+3c-6\sqrt{ac}$$ $$2a+2c+3\sqrt[3]{abc}\geq b+6\sqrt{ac}.$$ This is a homogenous inequality, so WLOG $ac=1$. Then, this is $$2a+2c+3\sqrt[3]{b}-b\geq 6.$$ Note that now, all of the terms involving $b$ are by themselves. This means that we only have to check the "worst case", which is when $3\sqrt[3]{b}-b$ is at the minimum, since this is the only part of the equation that $b$ affects. The derivative of this is $\frac{1}{b^{2/3}}-1$. which means that the function $3\sqrt[3]{b}-b$ increases on $(0,1]$ and decreases on $[1,\infty)$. Thus, given any interval that contains only nonnegatives, the minimum value in that interval has to occur at either the left or right endpoint. Thus, it suffices to show the inequality for the cases $b=a$ and $b=c$.\\

When $b=a$, we have $$a+2c+3\sqrt[3]{a}\geq 6$$ $$a+\frac{2}{a}+3\sqrt[3]{a}\geq 6.$$ The derivative of the left side is $$\frac{a^2+a^{4/3}-2}{a^2},$$ so it will decrease up until $a=1$ and then increase past that, so the minimum value occurs at $a=1$ which gives 6, which shows this case.\\

When $b=c$ we have $$2a+c+3\sqrt[3]{c}\geq 6$$ $$\frac{2}{c}+c+3\sqrt[3]{c}\geq 6,$$ which is the exact same thing, hence done.

Refinement and generalization, here
[url]https://artofproblemsolving.com/community/c6t243f6h3117508_httpsartofproblemsolvingcomcommunityc3339224[/url]


For a positive number, prove that $ (\cos 2^n)(\cos 2^{n \minus{} 1})\cdots (\cos 2)(\cos 1) < \frac {\sqrt {2}}{2^{n \plus{} 1}}$.

Note that a unit of an angle is in radian.

I'm assuming this problem is in radians. And that I'm not wrong. :P

.... Yes, I did mean radians.  :huh:
[hide]$ (\cos 2^n)(\cos 2^{n \minus{} 1})\cdots(\cos 2)(\cos 1) \equal{} \frac {\sin(2^{n \plus{} 1})}{2^{n \plus{} 1}\sin 1}$, so it remains to prove $ \frac {\sin(2^{n \plus{} 1})}{2^{n \plus{} 1}\sin 1} < \frac {\sqrt2}{2^{n \plus{} 1}} \iff \sin(2^{n \plus{} 1}) < \sqrt {2}\sin1$.
$ 1 < \sqrt2 \sin 1$, so this is true.[/hide]

But $ \dfrac{1}{\sqrt{2}}\equal{}sin45>sin1$.  Therefore, it has to be radians.


factorial
number theory solved
number theory

prove that 10^(-n) <> 1/n1!+1/n2!+...+1/nk! (! means factorial in this problem)
n1<n2<...<nk

The following solution was given by one of my friends, Adrian Iuga:
    Suppose we can find them and take [tex] P=\prod^k_{i=1} n_i! [/tex]. Put also [tex] 1+n_k\cdot (n_k-1)\cdot...\cdot (n_k-n_{k-1}+1)+...+n_k\cdot(n_k-1)\cdot...\cdot(n_k-n_1+1) [/tex]. Since [tex] n_k!=10^n\cdot C [/tex] and [tex] (n_k,C)=1 [/tex] we must have [tex] n_k=2^x\cdot 5^y [/tex] for some x and y. Let [tex] v_2(u), v_5(u) [/tex] the exponent of 2 and 5 in u. We have three cases:
 1. if xy is not 0. Then C is 1 mod 10. and it follows that [tex] v_2(n_k!)=v_5(n_k!)=n [/tex] and this is impossible since [tex] v_2(x)>v_5(x) [/tex].
 2. if y=0. Then [tex] n_k=2^x [/tex] and so C is odd. It follows that [tex] v_2(n_k!)=n+v_2(C)=n, v_5(n_k!)=n+v_5(C)\geqn=v_2(n_k!) [/tex], false.
  3. [tex] n_k=5^y [/tex]. If [tex] n_k>n_{k-1}+1 [/tex] then C is odd and we treat this as in the case 2. So, we may take [tex]n_{k-1}=5^y-1 [/tex]. It follows that
 C is 2 mod 4 and thus [tex] v_2(n_k!)=1+v_5(n_k!) [/tex] , from where we find that [tex] n_k [/tex] is very small and we can check that by hand.


What is the measure of the acute angle formed by the hands
of a clock at 4:20 a.m.?  Express your answer as number of degrees.

Dang for ftw everyone mems answers so I will, against the ideas of AoPS, post a formula.

$ |5.5m\minus{}30h|$

At 4:20, they would be together, making it $ 0$, except it has been $ \frac {1}{3}$ of an hour, meaning the hour hand has moved $ \frac {1}{3}$ of $ 30^\circ$ , meaning the answer is $ 10^\circ$.

EDIT: TY math154  :D  :D

[code]x^\circ[/code]

And dude actually look I posted the formula.


Let n be a product of any amount of numbers in the set {1,2,3,4,5,6,7}. Find the sum of all possible values of 1/n.

[hide="Are the n distinct?"]
If so, the factorization of $ \left(1 \plus{} \frac 12\right)\left(1 \plus{} \frac 13\right)\cdots\left(1 \plus{} \frac 17\right) \equal{} \frac{3}{2} \cdot \frac{4}{3} \cdots \frac{8}{7} \equal{} \boxed{4}$, upon expansion, yields the sum of all the desired values.
[/hide]

Yes, that is the answer. All n are distinct, I was gonna edit it but I was too lazy :lol: .

That is obvious...

Also the empty product shouldn't count. So the answer should be 3 or whatever

It is a nice problem..

Actually $ 1$ was specifically in the set  :wink: Though I agree its nice.

[quote="Altheman"]
It is a nice problem..[/quote]
I thought so too. The original Mathcounts problem just asked for the product azjps gave, which was very easy.

Why shouldn't the empty product count?  (Although ideally, the problem would be written to avoid it being an issue.)

[quote="Altheman"]I feel like the answer should be:

$ \left(1 \plus{} \frac {1}{1}\right)\left(1 \plus{} \frac {1}{2}\right)\cdots \left(1 \plus{} \frac {1}{7}\right) \minus{} 1 \equal{} 7$[/quote]

This is true if the problem says the sum of products of any distinct combination of the numbers, instead of distinct products.

We can get 6, 12, 24 and several other products in two different ways without using the 1.  dgreen's claim that s/he wants distinct products (i.e., no multiplicities) is wrong if s/he wants the solution to be this nice.  (If we do solve that way, the answer turns out to be $ \frac{18}{5}$.)

:oops: How does the expression yield the sum of all the values of $ \frac{1}{n}$?

[quote="AstroPhys"]:oops: How does the expression yield the sum of all the values of $ n$?[/quote]

The expression yields the sum of the values of $ \frac {1}{n}$ -- multiply it out and see! It's similar to the argument used to show that the number of divisors of some number $ n \equal{} \prod_{i \equal{} 1}^{k} p_ie^i$, where $ p_i$ is the $ i^{th}$ prime, is $ d(n) \equal{} \prod_{i \equal{} 1}^{k}(e_i \plus{} 1)$.

I see it now.  :lol: I should have just multiplied it out in the first place, I just thought there was a specific formula or method to getting the expression.

Fixed from $ n$ to $ \frac {1}{n}$ in my first post, which is what I intended.


[quote="Redriver"]Let $a_0=4,a_1=22$ and define $a_n=6a_{n-1}-a_{n-2}.$
Prove that there exit sequence $(x_n),(y_n)$ of positive integers so that
$a_n=\frac{y_n^2+7}{y_n-x_n}$[/quote]

Is it from Bulgaria 2001?
If is, it then i think $a_n$ must be $\frac{y_n^2+7}{x_n-y_n}$  :D

I have a solution:
we know $a_n=Ax_1^n+Bx_2^n$
where $x_1,x_2$is the solution of the equation $x^2-6x-1=0$
let$y_n=\frac{a_n-a_{n-1}}{2}, x_n=\frac{a_n+a_{n-1}}{2}$
then one can solve the coeffient $A,B$and $x_1,x_2$and calculate $x_n,y_n$
then you will find $x_n,y_n,a_n$satisfy the required equality.


These are speed questions and should be done in under a minute each.

1.Find the probability of rolling at leat one 6 in 4 rolls of a fair six sided dice.:P

2. How many ordered triples (x,y,z) of reals satisfy
$ 3x^2\plus{}6y^2\plus{}4z^2\minus{}4xy\plus{}4yz\equal{}0$ :)

3. The third one is not as easy as the first two if u get it with a correct proof.(take like 2.5 minutes for this problem) :rotfl:

$ x^3 \plus{} y^4 \equal{} 2009$

Find number of ordered pair of integer solutions.

I am not posting solution to 1 as itis too easy.Here go the solutions for 2 and 3.
2. Write the given equality as $ 2(x \minus{} y)^2 \plus{} (2y \plus{} z)^2 \plus{} x^2 \plus{} 3z^2 \equal{} 0$.{TIme Taken 8 seconds}
3. Check Mod 13. $ x^3 \ equiv\minus{}(0,1,5,8,12) (mod 13)$
                          $ y^4 \ equiv\minus{} (0,1,3,9) (mod 13)$
                           $ 2009 \ equiv\minus{} 7(mod 13)$. No solutions for postive integers $ x,y,z$. TIme taken{7 seconds-remebered the mods 13 very well coz already solved a similar problem :rotfl:  :rotfl: }.
Thanks Krishna {befuddlers} for warm up.

Answer for the first one is

$ 1 \minus{} 5^4/6^4$


doubt
consider a capillary tube -dipped in water .its curved surface is (concave up) .now this because of the surface tension of the water.but isnt it also due to reason that
there is a large adhesive force between the glass and water?

so when we consider the height of water in tube shouldnt we also take in the adhesive attraciton  to find height?
is there a way to include it?

i think the adhesive force thing DETERMINES the ANGLE OF CONTACT. From there the ST of the liquid handles the rest of the phenomena...

what say?

ya- adhesive force determines angle of conatct  but is that all?
while calculating the height of water in tube,
we take the component of ST acting upwards and balance it with weight of the water.sholudnt the adhesive force also be counted?

[quote="pradeepmathias"]ok... But I still say that the surface tension concepts of excess pressure handles the rest...

Lets assume there is no Surface Tension... but there are adhesive forces. In this case, isn't it likely that the liquid will climb up the sides of the capillary tube to the top atleast (if not overflow, dunno abt that)? Whats keeping it is the surface tension between the liquid molecules, which i presume balances out the adhesive force factor except for the angle of contact thing...

Just a way of looking at it... what dyou think?[/quote]

Hmm.....Valid Point..

1) Concave surface has excess pressure than convex surface, you can derive that.
2) And so to make the pressure uniform at all pts of the same height there is a capillary rise which makes up for the 2T/R difference.
3) However the bending of the surface and creating of angle of contact is due to adhesive forces, which inturn results in 1 & 2.

So they are kinda interrelated but Adhesive force I think is not responsible for Excess Pressure. Anyways in case of pure water and pure glass, the angle of contact the angle of contact is zero (or was it 90, heck it doesn't matter), and since it is that way, there is no cap rise.

Disclaimer:
I would like to add that Surface Tension is not one concept in which I am conceptually thorough cause all the explanations I have heard or seen or read so far are kinda vague and mechanical *ripped off from some foreign author probably*. And its not pretty easy to visualize it though the books say elastic membrane, blah!

[quote="pardesi"]actually we consider only the vertical components while finding the excess pressure...the horizontal component is balance by the adhesive forces...which is like the gravitational force along the line joining the tube and the liquid that is horizontal...so[/quote]

Is Adhesive force only horizontal???

[quote="pardesi"]yes thats because it's an intermolecuular attraction and has to act along the line joining the particles[/quote]

Yes, you are right but you forget one thing. There is force in the vertical direction due to glass molecules above and below it but they cancel due to symmetry.


inequalities
geometry
rectangle
integration
inequalities proposed

Let $ n$ be a positive integer. Prove that for all $ k \ge 1$, we have the following inequality
\[ \frac{(n\plus{}1)^{k\plus{}1}}{k\plus{}1} \minus{}\frac{(n\plus{}1)^k}{2} \le \sum^{n}_{i\equal{}1}i^k < \frac{(n\plus{}1)^{k\plus{}1}}{k\plus{}1}.\]
:)

Consider the rectangles with height $i^k$ and base $A_iA_{i+1}$ where $A_i=(i,0)$ and $A_{i+1}=(i+1,0)$.
Now $i$ varies from $1$ to $n$
Clearly the sum of areas of all rectangles is $\sum_{i=1}^n i^k$
All the rectangles lie under the curve $f(x)=x^k$
So,
\[\sum_{i=1}^n i^k < \int_{1}^{n+1} x^k\, dx =\frac{(n+1)^{k+1}}{k+1}-\frac1{k+1}\]
Thus we got a better bound for right hand inequality.

Note that $i^k$ is convex. Hence the area above each rectangle but below $f(x)$ is less than the triangle formed there.
Area of each triangle is $\frac12((i+1)^k-i^k)$.
So we have
\[\sum_{i=1}^n i^k \ge \int_0^{n+1} x^k \, dx -\frac12\sum_{i=0}^n ((i+1)^k-i^k)= \frac{(n+1)^{k+1}}{k+1}-\frac{(n+1)^k}{2}\]
Here equality holds when $x^k$ is linear i.e. $k=1$
$\blacksquare$
Edit: the left inequality can also be sharpened as $\frac{(n+1)^{k+1}}{k+1}-\frac{(n+1)^k}{2}+\frac12-\frac1{k+1}$
I would attach a diagram later if possible.


If $ x\equal{}log_{x}21$, which of these expresses $ log_{21}147$ in terms of $ x$?

I only have a very rudimentary understanding of logarithms; if someone can explain, I would be eternally greatful.

[hide="Another Hint"]$ x\equal{}log_{x}21\text{ can be rewritten as }x\equal{}\frac{log_{10}21}{log_{10}x}$ (which yields buzzer11's hint)

And $ log_{21}147\equal{}\frac{log_{10}147}{log_{10}21}$[/hide]

If you understand the basics, then don't go to wikipedia and search it. Because, wiki has the basic relationships of logs.


$ E$ is a point on $ (O)$.$ CE$ is a tangent of $ (O)$
$ DF \perp BC$
Prove that $ K$ is a fixed point

[quote="feliz"]Is $ CD$ parallel to $ AB$?

Else $ D$ is not completely determinated, right? (I am supposing you want to prove $ K$ does not depend on $ E$)[/quote]
$ CD$ is a diameter of circle $ (I)$

I saw it... But any point on the line $ OE$ would have the property that $ CD$ is the diameter of $ CDE$.

Why is $ D$ determinated? (or why is $ I$ determinated?)

If we call $ d$ be a straight line which touches (O) at A.Choose E is a point on the circle (O),different from A and B .The tagent line with (O) through E meets $ d$ at C.(I) is the circle through E and touches $ d$ at C.

[quote="vittasko"]Let us to try another formulation of the proposed problem and I hope that I am not mistaken.

[b][size=100][color=DarkBlue]A circle $ (O)$ is given with diameter $ AB$ and let $ E$ be, a mobile point on it. We draw the tangent lines of $ (O)$ at points $ A,\ E$ and we denote as $ C$ their intersection point. We draw the circle $ (O'),$ taken as cord the segment $ CE$ and tangents to the line segment $ AC,$ at point $ C.$ We denote the point $ D\equiv (O')\cap OE,$ where $ O$ is the center of the given circle $ (O).$ Prove that the line through the point $ D$ and perpendicular to $ BC,$ passes through a fixed point.[/color][/size][/b]

In this case, we denote as $ F$ the orthogonal projection of $ D$ on $ BC$ and we can prove that $ K\equiv AB\cap DF$ $ ($ not $ K\equiv AC\cap DF$ $ ),$ is a fixed point, based on the below Lemma :

[b][size=100][color=DarkBlue]LEMMA. – Let $ M,\ N$ be, the midpoints of the bases $ AB,\ CD$ respectively, of a given orthogonal trapezium $ ABCD$ with $ AD\perp AB\parallel CD$ and suppose that $ \angle ANB \equal{} 90^{o}.$ Prove that the line through the point $ M$ and perpendicular to $ AC,$ bisects the segment $ DN.$[/color][/size][/b]

Kostas Vittas.

PS. I will post here later more details, if I am not mistaken as for the proposed problem.[/quote]
Yes,you are true

[quote="vittasko"][b][size=100][color=DarkBlue]A circle $ (O)$ is given with diameter $ AB$ and let $ E$ be, a mobile point on it. We draw the tangent lines of $ (O)$ at points $ A,\ E$ and we denote as $ C$ their intersection point. We draw the circle $ (O'),$ taken as cord the segment $ CE$ and tangents to the line segment $ AC,$ at point $ C.$ We denote the point $ D\equiv (O')\cap OE,$ where $ O$ is the center of the given circle $ (O).$ Prove that the line through the point $ D$ and perpendicular to $ BC,$ passes through a fixed
point.[/color][/size][/b][/quote]
[b][size=100]PROOF.[/size][/b] - Because of the circle $ (I)$ tangents to $ AC$ at point $ C,$ we have that $ IC\perp AC,$ where $ I$ is the center of $ (I).$

Also, because of $ OE\perp CE,$ we conclude that the point $ D\equiv (I)\cap OE,$ is the [b][size=100]antidiametric point[/size][/b] (= diametric opposite) of $ C$ and so, we have $ CD\parallel AB$ $ ,(1)$

From $ (1)$ and because of $ \angle EOC \equal{} \angle AOC,$ we have $ \angle DCO \equal{} \angle DOC$ $ \Longrightarrow$ $ DC \equal{} DO$ $ ,(2)$

From $ (2),$ we conclude that the line through $ I$ and perpendicular to $ CE,$ bisects the segment $ CO$ and let $ P$ be, its midpoint.

From $ IP\parallel DO$ $ \Longrightarrow$ $ \frac {IP}{DO} \equal{} \frac {CI}{CD}$ $ ,(3)$

From $ (3),$ we conclude that the point $ P$ lies on the circle $ (I)$ and so, we have $ CP\perp PD.$

$ \bullet$ The line through $ O$ and parallel to $ PD,$ intersects the line segment $ CD,$ at one point so be it $ Q$ and because of $ CP \equal{} PO,$ we conclude that $ CD \equal{} DQ$ $ ,(4)$

We have now, the configuration of the orthogonal trapezium $ ACQB,$ with $ AC\perp CQ\parallel AB,$ and based on the [b][size=100]Lemma[/size][/b] I mentioned above, we conclude that the line segment $ DF\perp BC,$ where $ F\in BC,$ bisects the segment $ AO$ and the solution is completed.

Kostas Vittas.

[quote="vittasko"][b][size=100][color=DarkBlue]LEMMA. – Let $ M,\ N$ be, the midpoints of the bases $ AB,\ CD$ respectively, of a given orthogonal trapezium $ ABCD$ with $ AD\perp AB\parallel CD$ and suppose that $ \angle ANB \equal{} 90^{o}.$ Prove that the line through the point $ M$ and perpendicular to $ AC,$ bisects the segment $ DN.$[/color][/size][/b][/quote]

[b][size=100]PROOF.[/size][/b] - Let be the point $ P\equiv AD\cap BN.$

From the right triangle $ \bigtriangleup ABP$ with $ AN\perp BP,$ we have $ \frac {PA}{AB} \equal{} \frac {NA}{NB}$ $ \Longrightarrow$ $ \frac {PA}{2(AM)} \equal{} \frac {AN}{NB}$ $ \Longrightarrow$ $ PA \equal{} 2(AM)\cdot \frac {AN}{NB}$ $ ,(1)$

Let be the point $ P'\equiv AD\cap MF,$ where $ MF\perp AC$ and $ F\in AC.$

We will prove that $ P'\equiv P.$

From the right triangle $ \bigtriangleup AMP'$ with $ AF\perp MP',$ we have $ \frac {P'A}{AM} \equal{} \frac {AF}{FM}$ $ \Longrightarrow$  $ P'A \equal{} (AM)\cdot \frac {AF}{FM}$ $ ,(2)$

From $ (1),$ $ (2),$ in order to be $ P'\equiv P,$ it is enough to prove that $ (AM)\cdot \frac {AF}{FM} \equal{} 2(AM)\cdot \frac {AN}{NB}$ $ \Longrightarrow$ $ \frac {AF}{FM} \equal{} 2\cdot \frac {AN}{NB}$ $ ,(3)$

$ \bullet$ From the similar right triangles $ \bigtriangleup DAN,\ \bigtriangleup NBA,$ we have that $ \frac {AN}{ND} \equal{} \frac {NB}{AD}$ $ \Longrightarrow$ $ \frac {AN}{NB} \equal{} \frac {ND}{AD}$ $ ,(4)$

From $ (3),$ $ (4),$ it is enough to prove that $ \frac {AF}{FM} \equal{} 2\cdot \frac {ND}{AD}$ $ ,(5)$

But, the relation $ (5)$ is true, from the similar right triangles $ \bigtriangleup FAM,\ \bigtriangleup DCA$ which give us $ \frac {AF}{CD} \equal{} \frac {FM}{AD} \equal{} 2\cdot \frac {ND}{AD}$

So, we conclude that $ P'\equiv P$ and then, from the trapezium $ ABND,$ we have that the line segment $ PM\equiv FM,$ bisects its base $ ND$ and let $ E$ be, its midpoint.

Hence, the line segment $ FM\perp AC,$ passes through the constant point $ E,$ as the midpoint of the segment $ ND$ and the proof of the [b][size=100]Lemma[/size][/b] is completed.

Kostas Vittas.


linear algebra
matrix
linear algebra unsolved

Let $A = (a_{ij}) \in \mathcal M_p \left( \mathbb C \right)$ s.t. $a_{12} = a_{23} = \ldots = a_{p-1,p} = 1$ and $a_{ij} = 0$ for the other $(i,j)$s.

(a) Prove that $A^{p-1} \neq 0$ and $A^p = 0$.

(b) Let $X \in \mathcal M_p \left( \mathbb C \right)$ s.t. $AX=XA$. Then there are $a_1,\ldots,a_p \in \mathbb C$ s.t. \[ \begin{pmatrix} a_1 & a_2 & a_3 & \ldots & a_p \\ 0 & a_1 & a_2 & \ldots & a_{p-1} \\ 0 & 0 & a_1 & \ldots & a_{p-2} \\ \ldots & \ldots & \ldots & \ldots & \ldots \\ 0 & 0 & 0 & \ldots & a_1 \end{pmatrix} .  \] (a particular case of some theorem which I posted recently)

(c) If $B,C \in \mathcal M_p \left( \mathbb C \right)$ are $2$ matrices s.t. $\left( I_p + A \right)^n = B^n+C^n$ for all $n \in \mathbb N^{\ast}$, then $B=0_p$ or $C=0_p$.

At first we prove that $B^2(I+A)=(I+A)B^2$ and same goes for $C$ thus we have $B^2A=AB^2$ and $C^2A=AC^2$ according to the question $(b)$ we have the form of the matrices $B^2$ and $C^2$ .
for $n \in N^*$:$(I+A)^{2n}=B^{2n}+C^{2n}$ let's consider $b_1$and $c_1$ the element of the diagonals of $B^2$ and $C^2$ respectively we have $b_1+c_1=1$ and $b_1^2+c_1^2=1$  thus $b_1=0$ or $c_1=0$ let's suppose $b_1=0$ then $B$ is a nilpotent matrix $\exists m \in N*$ such that for $n\geq m B^m=0_{M_n(C)}$ so for $n\geq m$:$(I+A)^{2n}=C^{2n}$ moreover $C^2=I+\sum_{i=1}^{n-1} \alpha_i A^i$ then we deduce that $\sum_{i=1}^{n-1} \alpha_i A^i-A^2-2A=0_{M_n(C)}$ so $C^2=(I+A)^2$ and this way we've proven that $B^2=0_{M_n(C)}$ thus $(I+A)^3=C^3\to (I+A)^3=(I+A)^2C$ so $C=I+A$ and this time it's proved that $B=0_{M_n(C)}$ :)

[quote="MoHaMeD"]At first we prove that $B^2(I+A)=(I+A)B^2$ [/quote]

I obtain $B^2 (I+A) = - (I+A) B^2$. How  did you get your identity?

because $BC=-CB$ and $B^2C=-BCB=CB^2$ ;)

[quote="MoHaMeD"]... $n\geq m$:$(I+A)^n=C^n$ moreover $C=I+\sum_{i=1}^{n-1} \alpha_i A^i$...[/quote]

Could you explain how do you deduce  that $C=I+\sum_{i=1}^{n-1} \alpha_i A^i$ ?
thank's

Thx Moubinool i've edited my solution it's $C^2$ instead of $C$ and the conclusion as harazi said is from $(b)$

[quote="MoHaMeD"]then we deduce that $\sum_{i=1}^{n-1} \alpha_i A^i-A^2-2A=0_{M_n(C)}$[/quote]

How?

factorise $C^{2n}-(A+I)^{2n}=(C^2-(A+I)^2)D$ and you can prove that $D$ is an invertible matrix.


modular arithmetic
Diophantine Equations

Find all positive integers $m$ and $n$ for which \[1!+2!+3!+\cdots+n!=m^{2}\]

$1!=1^2$ giving the solution $(1,1)$

$1!+2!=3$ which is not a square.

$1!+2!+3!=3^2$ giving the solution $(3,3)$

$1!+2!+3!+4!=33$ which is not a square.

Since $10|k!$ for all $k \geq 5$, $1!+2!+ \dots n! \equiv 3 \mod 10$ for all $n \geq 4$. But there are no perfect squares congruent to 3 modulo 10: hence the only solutions are those already found.

We have that any square is congruent to $ 0,1,4,5,6$ or $ 9$ modulo $ 10$. But in the expression given for $ n$ graeter than $ 3$ we get the expression to be congruent to $ 3$ modulo $ 10$.
So, the possible values for $ n$ and $ m$ are $ 1$ and $ 3$ only!

How is that a different solution? That's exactly the same, or am I missing something?

Sorry, I did not see the prevoius post. Anyway i was thinking can we replace $ m^{2}$ with $ m^{k}$ where $ k$ is greater than $ 2$ ?

For k>2, we can look at the sum modulo 27 for n=8:
$ 1\plus{}2\plus{}6\minus{}3\plus{}12\minus{}9\plus{}18\plus{}9\equiv9\pmod{27}$
Since for all $ n>8$ $ 27|n!$, we have that $ S_n\equiv9\pmod{27}\forall n>7$. Only for $ n\equal{}7$ $ 27|S_7$, but $ S_7\equal{}5913\equal{}3^4\cdot73$, so there are no solution for $ k>2$


$ a,$ $ b$ and $ c$ are positive numbers. Prove that:
$ 2\sqrt{a^{2}\plus{}b^{2}\plus{}c^{2}\minus{}ab\minus{}ac\minus{}bc}\geq\frac{(a\minus{}b)^{2}}{a\plus{}b}\plus{}\frac{(a\minus{}c)^{2}}{a\plus{}c}\plus{}\frac{(b\minus{}c)^{2}}{b\plus{}c}.$

I did it will full expansion but it is very ugly  :P . Arqady could you give a hint for a nice solution?

[quote="silouan"] Arqady could you give a hint for a nice solution?[/quote]
The hint: it's very easy! :wink:

[hide]
WLOG let $ c\equal{}\min\{ a,b,c\}$ and $ x\equal{}a\minus{}c,y\equal{}b\minus{}c$. Then we have
\[ L(a,b,c)\equal{}L(x,y,0),\ R(a,b,c)\leq R(x,y,0)\]
(where $ L$ and $ R$ denote the left and right hand sides of the inequality). So it suffices to prove
\[ 2\sqrt{x^{2}\minus{}xy\plus{}y^{2}}\geq x\plus{}y\plus{}\frac{(x\minus{}y)^{2}}{x\plus{}y}\]
But we have
\[ 2\sqrt{x^{2}\minus{}xy\plus{}y^{2}}\minus{}(x\plus{}y)\equal{}\frac{3(x\minus{}y)^{2}}{x\plus{}y\plus{}2\sqrt{x^{2}\minus{}xy\plus{}y^{2}}}\]
This reduces the problem to $ \sqrt{x^{2}\minus{}xy\plus{}y^{2}}\leq x\plus{}y$, which is clearly true.
[/hide]


1) prove that the series :
x_n=prod(1-1/(4k^2)), k=1,n converges and compute lim(n-->oo) (x_n/l)^n if we know that l=lim n->oo(x_n), l<>0

2) Let f:R--> be a function with the property that lim (x-->0) f(x)/x=1. Prove that lim (n-->00) sum f(p/(n+k))=pln2, k=1,n where p>0

thank you very much!!

Since nobody seems to be much interested, a few hints for the 2nd one :

Lemma :
lim (n-->00) sum 1/(n+k) = ln(2)

There are a lot of ways to show this well known identity. The easiest is probably to compare with an integral : 1/(n+k) <= integral(n+k-1),(n+k)) dt/t <= 1/(n+k-1) and then sum + integral(dt/t) = log

lim (x-->0) f(x)/x = 1 can be rewritten more conveniently  f(x) = x + x*e(x) with lim (x-->0) e(x) = 0.

sum f(p/(n+k))= sum p/(n+k) + sum e(p/(n+k))p/(n+k)

By lemma we just have to show that sum e(p/(n+k))p/(n+k) -> 0 when n ->+oo

For n > n0, e(p/n) < epsilon.

sum e(p/(n+k))p/(n+k) < epsilon * p * sum (1/(n+k))

sum (1/(n+k)) is convergent then majored : sum (1/(n+k)) < M.

sum e(p/(n+k))p/(n+k) < epsilon * p * M

Then set epsilon' = epsilon / (p * M) and so on

HTH

I think I have a soln for the second one:

Let's take p=1, it's the same for the general case. Just use the fact that min (f(1/n+k)/(1/n+k)) <= sum (f(1/n+k))/sum (1/n+k) <= max (f(1/n+k)/(1/n+k)) for a certain n. Then we use the fact that min blablabla and max blablabla both ->1 as n->infinity, and the fact that sum (1/n+k) ->ln2 as n->infinity.

Hope I made myself understood..

For the first one maybe Stirling's formula could help: n! is approximately n^n*e^(-n)*(2Pin)^1/2. Check it on the Internet for a proof. I don't know if it gets us anywhere with the actual problem, but it helps you calculate l=2/Pi (I think).


function
limit
inequalities
real analysis
real analysis unsolved

Consider $ (a_n)_{n \geq 1}$ a sequence of real numbers, with $ 0 < a_n \leq 1 , \forall n \geq 1$ and $ \lim_{n \to \infty} \left(a_1 \plus{} a_2 \plus{} \dots \plus{} a_n \right) \equal{} \infty$.

Prove that for every $ l \in [1, \infty) \cup \{\infty \}$ we can find a function $ L: \mathbb{N}^* \to \mathbb{N}^{*}$, strictly increasing, such that:
\[ \lim_{n \to \infty} \frac {a_1 \plus{} a_2 \plus{} \dots \plus{} a_{L(n \plus{} 1)}}{a_1 \plus{} a_2 \plus{} \dots \plus{} a_{L(n)}}\equal{}l.
\]

For a fixed $ L$ denote $ R_n\equal{}\frac{\sum_{i\equal{}1}^{L(n\plus{}1)}a_i}{\sum_{i\equal{}1}^{L(n)}a_i}\equal{}1\plus{}\frac{\sum_{i\equal{}L(n)\plus{}1}^{L(n\plus{}1)}a_i}{\sum_{i\equal{}1}^{L(n)}a_i}$. We solve the problem via a case distinction:

[b]Case 1:[/b] For $ l\equal{}1$ set $ L(n)\equal{}n$ strictly increasing and $ 1\leq R_n\equal{}1\plus{}\frac{a_{n\plus{}1}}{\sum_{i\equal{}1}^na_i}\leq 1\plus{}\frac{1}{\sum_{i\equal{}1}^na_i} \rightarrow 1$

[b]Case 2:[/b] For $ 1< l<\infty$ set $ L(1)\equal{}\min\left\{M\in\mathbb{N}\,\Big |\,(l\minus{}1)\sum_{i\equal{}1}^Ma_i>1 \right\}$ and if $ L(1),\ldots,L(n)$ have been determined, choose $ L(n\plus{}1)\equal{}\min\left\{M\in\mathbb{N}\,\Big |\,M>L(n)\,\mbox{and}\,\sum_{i\equal{}L(n)\plus{}1}^Ma_i\geq (l\minus{}1)\sum_{i\equal{}1}^{L(n)}a_i\right\}$. Both minimum definitions are valid, since $ \forall A\in\mathbb{N}: \,\sum_{i\equal{}A}^{\infty}a_i\equal{}\infty$ holds. We see that $ L$ is strictly increasing. To be more precise, $ L(n\plus{}1)>L(n)\plus{}1$ holds, since $ a_{L(n)\plus{}1}\leq 1<(l\minus{}1)\sum_{i\equal{}1}^{L(1)}a_i\leq (l\minus{}1)\sum_{i\equal{}1}^{L(n)}a_i$. This implies $ (l\minus{}1)\sum_{i\equal{}1}^{L(n)}a_i\leq \sum_{i\equal{}L(n)\plus{}1}^{L(n\plus{}1)}a_i\equal{}a_{L(n\plus{}1)}\plus{}\sum_{i\equal{}L(n)\plus{}1}^{L(n\plus{}1)\minus{}1}a_i\leq 1\plus{}(l\minus{}1)\sum_{i\equal{}1}^{L(n)}a_i$. Applying this inequality to the computation of $ R_n$ we get:

$ l\leq R_n\leq  l\plus{}\frac{1}{\sum_{i\equal{}1}^na_i}\rightarrow l$

[b]Case 3:[/b] For $ l\equal{}\infty$ define $ L(1)\equal{}1$ and if $ L(1),\ldots,L(n)$ have been determined, choose $ L(n\plus{}1)\equal{}\min\left\{M\in\mathbb{N}\,\Big |\,M>L(n)\,\mbox{and}\,\sum_{i\equal{}L(n)\plus{}1}^Ma_i\geq (n\plus{}1)\sum_{i\equal{}1}^{L(n)}a_i\right\}$. This yields a valid and strictly increasing function $ L$ and we find $ R_n\geq n\plus{}2\rightarrow\infty$


trigonometry
arithmetic sequence
inequalities unsolved
inequalities

Find the smallest value of the following expression :
\[ P \equal{} \frac{1}{{\cos ^6 a}} \plus{} \frac{1}{{\cos ^6 b}} \plus{} \frac{1}{{\cos ^6 c}}
\]
 Where a, b, and c form an arithmetic sequence whose commom difference is \[ \frac{\pi }{3}
\]

This is problem in M&Y,and it have posted before but it have been locked up.


im writing a multiple choice test, and for the answer choices, i currently use qquad, like
\[ \text{(A) }22\qquad\text{(B) }23\qquad\text{(C) }24\qquad\text{(D) }25\qquad\text{(E) }NOTA
\]
however, the spacing is Dependant on the answer choices and my school doesn't want that
so how can i make it so that the answer choices are spread out evenly, and not centered

To make it left-justified, use dollar signs instead of square brackets:

$ \text{(A)}\ 22\qquad\text{(B)}\ 23\qquad\text{(C)}\ 24\qquad\text{(D)}\ 25\qquad\text{(E)}\ \text{NOTA}$

\qquad should work like a set number of spaces instead of like a tab in some programs (until the next pre-defined column), and it looks evenly spaced to me...

i mean on the pdf, i want it spread out so that the choices are set. if you used qquad, you'll notice it's dependent on the length of the answer choices

I'm still not quite sure if I understand what exactly you want but try putting \hfill between the choices. If you write something like
[code]
Here are the choices:

1\hfill 2\hfill 3\hfill 444444\hfill 5

whatever follows
[/code]
the result will be
Here are the choices:
1*****2*****3*****4444444*****5
whatever follows
with the distances between choices spread evenly on the entire page width.

I hope that is what you are after, but if you just want
1*****2*****3**4444444**5
instead, you can do something like:
$ \mbox{
\begin{tabular}[t]{ccccc}
\phantom{44444444}&\phantom{44444444}&
\phantom{44444444}&\phantom{44444444}&
\phantom{44444444}
\\[-10pt]
1 & 2 & 3 & 4444444 &5
\end{tabular}
}$
(of course, you'll not need the \mbox in dollar signs in your TeX-file). I think stevem can offer a better solution but this works :)

[url=http://www.math.uakron.edu/~dpstory/eqexam.html]The eqexam packege[/url] do this very well.
This is a simple file [url=http://www.math.uakron.edu/~dpstory/eqExam/test01.pdf]test01[/url]


Functional Analysis
topology
advanced fields
advanced fields unsolved

It is well known that if $V$ is a Banach whose dual is separable, $V$ is separable, and the converse is not true. Prove that if $V$ is separable, $V^{*}$ is weak* separable.

Let $A\subset V$ be a dense countable set. For any finite linearly independent subset $B=\{a_{1},\dots,a_{n}\}\subset A$ choose a countable set $B'\subset V^{*}$ such that $\{(\varphi(a_{1}),\dots,\varphi(a_{n}))\colon \varphi\in B'\}$ is dense in $\mathbb R^{n}$. (This uses the Hahn-Banach theorem.) The union of all such $B'$ is countable and weak* dense in $V^{*}$. Indeed, for any weak* open set $U$ containing $\varphi\in V^{*}$ there is $B$ as above and $\epsilon>0$ such that $\{\psi\in V^{*}\colon |\psi(a)-\varphi(a)|<\epsilon\ \forall a\in B\}\subset U$.

A follow-up: find a nonseparable Banach space $V$ such that $V^{*}$ is weak* separable.


trigonometry
function
algebra unsolved
algebra

[quote="ali"]very simple.
First I shall show that f(x) is onto. for any x in R : f(0)=2x+f(f(-f(x)) -x)=2x+f(u)
so f(u)=f(0)-2x and thus f(x) is onto. Now let f(x0)=0.
then f(f(x)-x0)=f(x)-2x0 (replace x with x0 and y with x in original equation)
So f(z-x0)= z - 2x0 were z belongs to range of f(x). but since this range is R, we get
f(x)=x-x0 for all x.
-Ali[/quote]

Dont understand why f is unto?

[quote="romano"]Find f:R-->R satisfying :
 $ f(f(x)+y) = 2x+ f(f(y)-x) $ for all $ x,y \in R $[/quote]
do you think Ali's answer right?

An other one  :D  Find $f,g:R\rightarrow R $ such that $f(f(x)+y) = x.sin(x) + f(f(y)+g(x)) \ , \forall_{x,y\in R}$

 :cool: See you ..  :cool:

[quote="Diogene"]An other one  :D  Find $f,g:R\rightarrow R $ such that $f(f(x)+y) = x.sin(x) + f(f(y)+g(x)) \ , \forall_{x,y\in R}$  [/quote]
Nobody answers ! Too hard for this site  :D
 :cool:

[quote="Armo"]Can you please post your solution?[/quote]
Yes I can, of course ! All the solutions are $f(x)=x-c \ ,\  g(x)=x(1-\sin x)$ ,  where  $c$  is a constant .
 But could post a proof ? Need you a hint ?  :D Could you generalize this problem ?

 :cool:

Heh, I even can't solve this one and you are asking if I can generalize it? :D
Of course it will be very nice if you post your full solution.

[quote="Armo"]Heh, I even can't solve this one and you are asking if I can generalize it? [/quote]
In fact the problem is as easy(or hard) than the initial problem of Romano :lol:
One : Put $y=-f(x)$ and you see that  $f$ is a surjective function , so there is $x_0$ such that $f(x_0)=0$
Two :  Put $x=x_0$  and because of surjective $f$, you obtain $f(x)= x - c$ , were $c$ is a constant
Three : Replace $f$ , and you  find $g(x)= x - x.sin(x)$   :cool: Do you follow me ? :cool:

 :cool:  Do you know Charles Aznavour(ian) ? :cool:

Thank you for your solution, Diogene. Let's speak with pm.,ok?


how can you express the positive number x whose value is approx 1.3...... such that x^3=2, without writing cubert(2)?

the problem with cubert(2) is that there are 3 cuberoots including the complex value. how can i express such a number x without the ambiguity of whether i am talking about the complex values or teh real value.

and hopefully answers will not be as follows-"just call it the real cubert of 2, and the reader will know what u are talking about"

the problem is t hat i would like to have a succint mathematical notation so i dont have to explain which cube root i am talking about....

note:for those that are unconvinced that 2 has 3 cuberoots, take the cube root of 1 for example.

the cube roots of 1 are:1, cos 120+isin120, and cos240+i sin240

What about let $ x\in\mathbb{R}$ and $ x^3\equal{}2$?

"The cube roots of $ 2$" does not have the same meaning as the expression $ \sqrt [3]{2}$.  It is understood that the radical function is a function $ \mathbb{R} \to \mathbb{R}$ (for odd radicals; for even radicals, we take $ \mathbb{R}^{ \plus{} } \to \mathbb{R}^{ \plus{} }$); for example, when we take square roots of non-negative reals we understand them to be non-negative and therefore write $ \sqrt {2} \equal{} 1.414...$.

Edit:  So I guess a short answer is that it is largely a matter of convention.  If one wants to distinguish the complex roots from the real ones, usually one refers to "the primitive third roots of unity" or something along those lines.

ok, so saying $ \sqrt[3]{2}\equal{}x$, and saying $ x^3\equal{}2$ have different solutions, and the cuberoot function is defined (when written like that) to be from R to R? is that correct?

if so, thankyou t0rajir0u!(hope i spelled that right!)

just out of curiosity, are you in high school?

He just graduated this year and will be starting his first year in college.

Funny how that became part of the discussion.  :P  Just read the information he posted and do not let age be a factor in his validity.

Indeed, I am pretty sure many university students couldn't follow most of his blogs :P


Prove that

$ \sum_{k \equal{} 1}^{n}( \frac {k}{a_1 \plus{} a_2 \plus{} .... \plus{} a_k}) \leq s (\frac {1}{a_1} \plus{} ... \plus{} \frac {1}{a_n})$

is true for $ s \ge 2$, not true for $ s < 2$.

($ a_i > 0$)

You can find a proof for $ s\equal{}2$,here:
http://www.mathlinks.ro/viewtopic.php?t=169630

Thank you very much!!

It's wonderful solution!!!

But how can we know s=2 is the best constant.

You know proof maybe??

You have to distinguish between two things: the minimal $ s$ for a certain fixed $ n$, and the minimal $ s$ that makes the inequality hold for [i]all[/i] $ n$.

As for fixed $ n$, we have $ s_{\text{min}} \equal{} 1$ for $ n \equal{} 1$ and $ s_{\text{min}} \equal{} \frac98$ for $ n \equal{} 2$ (assumed you replace $ <$ by $ \leq$ in your inequality). Already for $ n \equal{} 3$, I can only obtain a numerical answer ($ s_{\text{min}}\approx 1.204692...$) which does not seem to correspond to any nice number.

The minimal $ s$ for which the inequality holds for all $ n$ is $ s \equal{} 2$. This easily follows from Erken's (resp. Sedrakyan's) proof: Take $ a_i \equal{} i$ for every $ i\in\mathbb{N}$. Then, the inequality

$ \frac {1}{a_1} \plus{} \frac {2}{a_1 \plus{} a_2} \plus{} \frac {3}{a_1 \plus{} a_2 \plus{} a_3} \plus{} ... \plus{} \frac {n}{a_1 \plus{} a_2 \plus{} ... \plus{} a_n} < s\left(\frac {1}{a_1} \plus{} \frac {1}{a_2} \plus{} ... \plus{} \frac {1}{a_n}\right)$

becomes

$ 2\left(\frac12 \plus{} \frac13 \plus{} ... \plus{} \frac {1}{n \plus{} 1}\right) < s\left(\frac11 \plus{} \frac12 \plus{} ... \plus{} \frac {1}{n}\right)$,

so that

$ \frac {s}{2} > \frac {\frac12 \plus{} \frac13 \plus{} ... \plus{} \frac {1}{n \plus{} 1}}{\frac11 \plus{} \frac12 \plus{} ... \plus{} \frac {1}{n}}$.

But $ \frac {\frac12 \plus{} \frac13 \plus{} ... \plus{} \frac {1}{n \plus{} 1}}{\frac11 \plus{} \frac12 \plus{} ... \plus{} \frac {1}{n}}\to 1$ as $ n\to\infty$ (you can prove this by [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=25862]the Stolz (aka Cesaro-Stolz) theorem[/url]), so that $ \frac {s}{2}\geq 1$ and thus $ s\geq 2$.

  darij


(x+y+z)^5?

More generally, what's (x+y+z)^n?
Please show how you solved the problem.  I think that for the general case that the number of distinct terms is equal to (n+1)th trianglar number (n is the exponent) and that the coefficients add up to 3^n.

The number of distinct terms is $\binom{n+2}{2}$, which is the $n+1$th triangular number as you stated.

And the coefficients do add up to $3^n$, since the sum of the coefficients is just the value of the F(x,y,z) at F(1,1,1).

How about the number of distinct terms and the sum of the coefficients for
$(a_1+a_2+a_3+...+a_k)^n$

In retrospect, I can see that the sum of the coefficients is $k^n$

But how about the number of distinct terms  :huh:

The number of distinct terms in $(a_1+a_2+\cdot\cdot\cdot+a_m)^n$ is $\binom{n+m-1}{n}$.

If I am not mistaken, I think you're searching for the [url=http://mathworld.wolfram.com/MultinomialSeries.html]multinomial theorem[/url]. :)

:D yep, that's what I was looking for.

Thanks a lot varun & mathmanman   :)

hmm... can anyone prove the multinomial theorem?

actually now that i think about it... im not quite sure about the proof of the binomial theorem either... although I have a slight idea..

Well, I'm going to put a sketch proof ;)

Starting from the example :
$(x+y+z)^5 = (x+y+z)(x+y+z)(x+y+z)(x+y+z)(x+y+z)$

The product of the 5 terms is the sum of all the obtained terms taking one term in each parenthese and multiplying them.
All the terms of the developement are consequently under the form : $x^{k_1}y^{k_2}z^{k_3}$ with $k_1 + k_2 + k_3 = 5$.

So, to get, for example, for $(x+y+z)^7$, $x^4y^2z$, we choose 4 parentheses among the 7 where we take tha a (${7} \choose {4}$ choices).
Once we did it, there is ${3} \choose {2}$ choices for the 2 parentheses among 3 where we take the b, the c is taken in the last parenthese.
So, total, we get : ${7} \choose {4}$ $\cdot$ ${3} \choose {2}$ $\cdot$ ${1} \choose {1}$ = $\frac{7!}{4!3!} \cdot \frac{3!}{2!1!} \cdot \frac{1!}{1!}$ = $\frac{7!}{4!2!1!}$

Therefore, in a general way, if we want to develop : $(x_1 + x_2 + ... + x_p)^n$, we get the desired formula ;)

I hope I made it clear... :)

or you could just use an extension of the binomial theroem

$\sum_{k=0}^n\binom{n}{k}x^k\left(\sum_{i=0}^{n-k}\binom{n-k}{i}y^iz^{n-k-i}\right)^{n-k}$


Prove that if y>=(y)^3+(x)^2+mod(x)+1,then (x)^2+(y)^2>=1. :lol:
Find all pairs of (x,y),such that the first inequality holds while equality in the second one attains. :|

Please use latex and ps this should be in the other forum

[b]is my post unredable without latex..............[/b]

spraking truthfully yes
but it is still one ofd your early posts so it is OK but ps try latexing

and please post htis in Maths olym forum :rotfl:  :)  :rotfl:


[hide="Solution"]By power of a point, $ CM \times CA \equal{} CN \times CB$. $ CM \equal{} \frac{CN \times CB}{CA} \equal{} \frac{8\times 2}{6} \equal{} \boxed{\frac{8}{3}}$[/hide]


Find all functions $ f: R\minus{}>R$ with the proprety that \[ \frac{1}{2}f(xy)\plus{}\frac{1}{2}f(xz)\minus{} f(x)f(yz) \geq \frac{1}{4} , x,y,z \in R\]

-) $ x\equal{}y\equal{}z\equal{}1: f(1)\minus{}f(1)^2\geq \frac14\Leftrightarrow (f(1)\minus{}\frac12)^2\leq 0\Leftrightarrow f(1)\equal{}\frac12$
-) $ y\equal{}z\equal{}1: f(x)\minus{}f(x).f(1)\geq \frac14\Leftrightarrow f(x)\geq \frac12$
-) $ x\equal{}y\equal{}z\equal{}0: f(0)\minus{}f(0)^2\geq \frac14\Leftrightarrow (f(0)\minus{}\frac12)^2\leq 0\Leftrightarrow f(0)\equal{}\frac12$
-) $ x\equal{}0,y\equal{}x,z\equal{}1: f(0)\minus{}f(0)f(x)\geq \frac14\Leftrightarrow f(x)\leq \frac12$
Then $ f(x)\equal{}\frac12$

[quote="quangchuc"]This is problem 1 in VietNamese Mathematical Olimpiad 1991.[/quote]
Do you have more problems for the vietNamese math olympiad and other problems just below this level?I need them  :D

Educated Pulishers have published a book:The Viet Namese Mathematical Olimpiad(1990-2006). Everyone that participates IMO 2007 has it. I only have the Vietnamese these. If you need it, you can contact to pvthuan moderator. I think he has ebook!


Show that every inifinite connected graph contains an infinite path or an infinite star.

yes, can you prove konig's infinty lemma?

Assume every vertex has finite degree (so that there are no infinite stars). Also notice that we can assume the graph is a tree: the set of subgraphs which are trees is inductively ordered when we partially order it by inclusion, and the union of a chain of trees is again a tree, so we can apply Zorn's lemma to conclude that there is a maximal subgraph of our graph which is a tree. It's clear now that this maximal tree must be infinite.

Now start from the vertex of your choice, and construct all the finite paths starting from this vertex. There is an edge starting from our vertex which appears on infinitely many paths, so choose this edge. Now repeat the procedure for the second endpoint of the chosen edge, choosing another edge adjacent to it, and so on. We cannot return to a previously-chosen vertex, because the graph we're working on is a tree, so the only possibility is that we're actualy constructing an infinite path, as desired.


limit
integration
function
calculus
derivative
logarithms
real analysis

Here is 3-d problem from NMSO 2005:

Let $f:{[0,1]\to [0,1]}$ with the properties: $f(0)=1$ and $0\leq f(x)<1$  $ \forall x  \n (0,1]$. Prove that $\displaystyle \lim_{n \to \infty}n \int_0^1f^{n}(x)= \frac{1}{f^{'}(0)}$

If anybody want , after some time I'll post another problems from NMSO

This problem can't be right :?. Think about the function $f(x)=1-x$. That integral is positive for all $n$, while $f'(0)=-1$.

It should be $-\frac 1{f'(0)}$, of course. Another thing is that you need some more assumptions about $f$. It should be differentiable at $0$ and bounded away from $1$ on any interval $[\delta,1]$ with $\delta>0$. Then it is just the classical Laplace asymptotic formula. The standard proof goes like this. Let $f'(0)=-a<0$ (the case $a=0$ is similar but simpler; the limit is $+\infty$ then). Let $\alpha<a<\beta$ be any two positive numbers. Using the definition of the derivative for the function $g=\ln f$ (which is also differentiable at $0$), we obtain $e^{-\beta x}\leqslant f(x)\leqslant e^{-\alpha x}$ on $[0,\delta]$ with some $\delta>0$. Also, $f\leqslant e^{-\varepsilon}$ on $[\delta, 1]$ with some $\varepsilon>0$. Therefore, $\frac 1{\beta n}(1-e^{-\beta\delta n})\leqslant \int_0^1 f(x)^n\,dx\leqslant \frac 1{\alpha n}+e^{-\varepsilon n}$, whence $\frac 1\beta \leqslant\liminf_{n\to\infty}n\int_0^1 f(x)^n\,dx \leqslant\limsup_{n\to\infty}n\int_0^1 f(x)^n\,dx\leqslant\frac 1\alpha$.  Since $\alpha<a<\beta$ were arbitrary, the statement of the problem follows.

:blush: I am really sory!! I wrote this problem very quickly  and have lose many properties.I'm sorry!
Yes it's right: $f(x)$ is differentiable and $\displaystyle \lim_{n \to \infty}n \int_0^1f^{n}(x)=- \frac{1}{f^{'}(0)} $

fedja, you solution is very nice and it is more simple  than the solution of judges of the competition.Thank you


I'm locking this thread this time.

From now on, 4everwise will delete all such advertisements.  ;)

[quote="ebookmania"]visit this link and learn for free

[url]http://www.seopromotionlinks.com/database_integration.asp[/url][/quote]

this is an advertisement, don't go there


geometry
circumcircle
incenter
geometry unsolved

I searched it but didn´t find it.

In the rays AB and CB of triangle ABC points M and N are marked, respectively, such that segments AM = CN = s, where s is the semiperimeter of ABC. K is the point diametrally opposite to B on the circumcenter of ABC. Prove that KI, where I es the incenter of ABC, is perpendicular to MN.

We had this at http://www.mathlinks.ro/Forum/viewtopic.php?t=6472 . ;)

  darij


Given $ a,b,c,d$ be none-gative numbers such that:$ a\plus{}b\plus{}c\plus{}d \equal{} abc\plus{}bcd\plus{}cda\plus{}dab$.Prove that:
$ (a^2\plus{}2)(b^2\plus{}2)(c^2\plus{}2)(d^2\plus{}2) \ge 81$

[quote="quykhtn-qa1"]Given $ a,b,c,d$ be none-gative numbers such that:$ a \plus{} b \plus{} c \plus{} d \equal{} abc \plus{} bcd \plus{} cda \plus{} dab$.Prove that:
$ (a^2 \plus{} 2)(b^2 \plus{} 2)(c^2 \plus{} 2)(d^2 \plus{} 2) \ge 81$[/quote]
Try $ a\equal{}b\equal{}c\equal{}d\equal{}0.$ :wink:

[quote="arqady"][quote="quykhtn-qa1"]Given $ a,b,c,d$ be none-gative numbers such that:$ a \plus{} b \plus{} c \plus{} d \equal{} abc \plus{} bcd \plus{} cda \plus{} dab$.Prove that:
$ (a^2 \plus{} 2)(b^2 \plus{} 2)(c^2 \plus{} 2)(d^2 \plus{} 2) \ge 81$[/quote]
Try $ a \equal{} b \equal{} c \equal{} d \equal{} 0.$ :wink:[/quote]
Sorry, must be $ a,b,c,d>0$ :)


algebra
polynomial
conics
function
abstract algebra

I've been put into a group of 4 and we have to choose a topic to present a seminar to the class on.  There's a big list of topics but we've narrowed it down to a few here:
[b]
Mechanics of Bungee Jumping[/b] - Hooke's Law etc
[b]Roots of Cubic & Quartic Polynomials[/b] - investigating solutions
[b]Rainbows[/b] - how they occur, Snell's Law... Fermat... Descartes etc
[b]Factorising Large Numbers[/b] - Exploring various methods and implications to computer security.
[b]Harmonic Series[/b] - How to estimate partial sums, Euler's constant etc.
[b]Conic Sections[/b]
[b]Lambert-W Function[/b]

That's a quick summary I can give more detail if it's of interest.
I want a topic that isn't going to be too hard but is large enough to make a decent length presentation on.  Some of the topics are very much physics based and I think it would be preferable if the topic we choose was something that could be useful when we go further with mathematics.  Any suggestions?

Conic Sections look to be not that difficult, but is still large enough to make a good presentation.

[quote="mdk"]Roots of Cubic & Quartic Poloynomials - investigating solutions  looks very interesting and there is both a math and a history
aspect to it.[/quote]
That one's actually my personal favourite as it's the most interesting to me.  Not too sure how easy a presentation would be though.  The rest of the group seem to be leaning towards the bungee jumping problem which does tie in fairly nicely with the Newtonian Mechanics module we're doing.

the bungee jumping problem is also very interesting it deals with Hooke's law, figuring out the spring constant etc...
Just make sure you're okay with this. You might actually like it because it is a lot of fun.

We put our first choice as Harmonic Series, which doesn't really interest me but it has without a doubt the best tutor.  We just need to wait and see which one we get.

I'm sure I'll be asking everyone here for help once we get our topic!  :)

It amuses me to see that I will be the first person with an avatar to post in this thread.

But, yea, why not bungee jumping? That would be interesting and easy.


A plane makes a trip of 5040 km in 7 h, flying with the wind. Returning against the wind, the plane makes the trip in 9 h. What is the speed of the wind?

 :)

check out the post on your other question and try to see if you can figure this one yourself :P

Well, let's give it a try,

d=5040 km
t=7 h
s= 720 km/h

d=5040 km
t= 9 h
s= 560 km/h

Let x represent the speed of the wind.
Let y represent the speed of the plane.

x+y=720
y-x=560
2y=1280
y=640
x+640=720
x=80

Thus, the speed of the wind is 80 km/hour...right?  :?

[quote="demonface"]well, seeing how its right, you really don't need to ask, just figure it out yourself.[/quote]

I asked the question because I didn't know how to do it at first. I posted two questions on linear systems. Since someone solved the other, I got and idea about how to solve this one. What is wrong with asking if it is right? You know...I'm not the kind of person who would just ask a question and then solve it myself just to kill time! I don't waste time like that.

[quote=DuoDuoling0]Bump. xDDD[/quote]

I can't tell if you bumped this as a joke or if you want another solution (which probably doesn't exist)