trigonometry
function
geometry
area of a triangle
Heron\u0027s formula
algebra
system of equations

tan of both sides + tan double angle --> tan(4/3)=tan(4/3)

[hide="A little bit more detailed (i.e. important step left out)"] \[\tan \left( LHS \right)=\tan \arcsin\frac45 = \frac{\sin \arcsin \frac45 }{\cos \arcsin \frac45}=\frac{\frac45}{\sqrt{1-\frac45^{2}}}=\frac43 \\ \tan \left( RHS \right)=\tan 2\arctan \frac12 = \frac{2\tan \arctan \frac12}{1-\left(\tan \arctan \frac12\right)^{2}}=\frac43\] Therefore the two are equal :)
Thats a nice identity lingomaniac, it looks like fun to derive a general version:
[b]Problem:[/b] \[\text{Find nice rationals }p,q{ s.t. }\frac{2p}{1-p^{2}}=\frac{q}{\sqrt{1-q^{2}}}\] [/hide]

[quote="me@home"][hide="A little bit more detailed (i.e. important step left out)"] \[\tan \left( LHS \right)=\tan \arcsin\frac45 = \frac{\sin \arcsin \frac45 }{\cos \arcsin \frac45}=\frac{\frac45}{\sqrt{1-\frac45^{2}}}=\frac43 \\ \tan \left( RHS \right)=\tan 2\arctan \frac12 = \frac{2\tan \arctan \frac12}{1-\left(\tan \arctan \frac12\right)^{2}}=\frac43\] Therefore the two are equal :)[/hide][/quote]

$\tan\alpha =\tan\beta\not\Rightarrow \alpha =\beta$.

For example, $\tan \frac{\pi}{4}=\tan \frac{5\pi}{4}$, but $\frac{\pi}{4}\ne\frac{5\pi}{4}$.

yea, just then just consider all the cases i guess (where sin is 4/5 or w/e)

[hide="Best proof"] $(2+i)^{2}= 3+4i \implies 2 \arg(2+i) = \arg(3+4i)$ [/hide]

Edit:  So, me@home, here is the solution to the [hide="General case ;)"] $(a+bi)^{2}= (a^{2}-b^{2})+(2ab) i \implies 2 \arg(a+bi) = \arg( (a^{2}-b^{2})+(2ab) i ) \Leftrightarrow 2 \arctan \frac{b}{a}= \arcsin \frac{2ab}{a^{2}+b^{2}}$ [/hide]

Use $\sin 2x = \frac{2\tan x}{1+\tan^{2}x}$ to obtain that $\sin(2\arctan x) = \frac{2x}{1+x^{2}}$.
Note that $\arctan \frac12< \arctan 1=\frac{\pi}4$ so $0<2\arctan \frac12<\frac{\pi}2$. And we can  apply $\sin$ to both sides of the equality we want to prove because $\sin$ is monotonous for $x \in [0,\frac{\pi}2]$

The proof I gave [i]is[/i] a geometric proof.  ;)

Did you come up with that method on your own T0r? I've seen you use it in these kind of problems always under the label of best solution.
Its very nice

Well, I was taught the method formally at PROMYS.  The fact that $\arg$ is a multiplicative function is really a consequence of the tangent addition formula in disguise, so the substance of my proof doesn't differ significantly from the others, although the connections it suggests are very elegant :)

I think I first saw a solution of that type in response to a question about the Machin formula:

$\arctan 1 = 4 \arctan \frac{1}{5}-\arctan \frac{1}{239}$

Which similarly also has an interpretation in the complex numbers.

Edit:  Specifically,

$\frac{ (5+i)^{4}}{ 239+i}= 2+2i$

Yes thats were I remember you first posting this method, i assumed you did it more often

[quote="lingomaniac88"]Prove that $2\arctan\frac12=\arcsin\frac45 \ .$[/quote]
[color=darkblue]In the my opinion, the [u]Delta's trigonometrical proof[/u] is the best. I"ll present a geometrical proof of this problem.

Let an $A$- isosceles triangle $ABC$ for which the length $h_{a}$ of the $A$- altitude is equally with the length $a$ of the side $[BC]\ .$ Suppose w.l.o.g. $a=2\ .$

Denote : the projection $D\in BC$ , $E\in AC$ of the vertices $A$ , $B$ respectively to the opposite sides ; $m(\widehat{BAD})=\phi\ .$

Observe that : $m(\widehat{BAE})=2\phi$ ; $DB=DC=1$ ;  $AD=2$ ; $AB=2\sqrt 5$ ; $BE=\frac{8\sqrt 5}{5}\ .$ Therefore,

$\{\begin{array}{c}\tan \phi =\tan \widehat{BAD}=\frac{DB}{AD}\Longrightarrow \tan\phi=\frac{1}{2}\Longrightarrow\arctan\frac{1}{2}=\phi\\\\ \sin 2\phi =\sin \widehat{BAE}=\frac{BE}{AB}\Longrightarrow \sin 2\phi=\frac{4}{5}\Longrightarrow \arcsin \frac{4}{5}=2\phi\end{array}\Longrightarrow$ $\boxed{\ 2\arctan \frac{1}{2}=\arcsin \frac{4}{5}\ }\ .$

[b]Remark.[/b] Generally, for $BD=DC=1$ , $AD=x>1$ obtain $AB=\sqrt{1+x^{2}}$ and $BE=\frac{2x}{\sqrt{1+x^{2}}}\ .$ Therefore,

$\{\begin{array}{c}\tan\phi =\frac{BD}{AD}=\frac{1}{x}\Longrightarrow\phi =\arctan\frac{1}{x}\\\\ \sin 2\phi =\frac{BE}{AB}=\frac{2x}{1+x^{2}}\Longrightarrow 2\phi =\arcsin\frac{2x}{1+x^{2}}\end{array}$ $\Longrightarrow$ $\boxed{\ 2\arctan \frac{1}{x}=\arcsin\frac{2x}{1+x^{2}}\ }\ .$

If $f(x)=\frac{2x}{1+x^{2}}$ then $f(\frac{1}{x})=f(x)\ !$[/color]

Very nice!!
However, I am wondering how you determine the length of BE (thus AE) so easily - the way I do it would either use Heron's formula or solve a system of equations in the two right triangles (I am almost certain you have a nicer way).
Here is a diagram:

[hide="For Me@home."][b][u]With gladly and ... with areas ![/u][/b] $2[ABC]=BE\cdot AC=AD\cdot BC\Longrightarrow BE\cdot \sqrt{1+x^{2}}=x\cdot 2\Longrightarrow BE=\frac{2x}{\sqrt{1+x^{2}}}\ .$[/hide]

[hide="Virgil Nicula"]
I feel embarassed :oops: that is very elementary, I guess I didn't realize AB=AC...! Thank you for posting that.
By the way, a while ago you mentioned something about how we are almost neighbors. I'm wondering what part of washington Yakima is in, also are you affiliated with a college like University of Washington?
[/hide]


geometry
geometric transformation
reflection

This is a topic which I mean as an announcement. Please let us stop discussing communism for at least two weeks. I am horribly tired saying the same things over and over again. After two weeks, I may come back with renewed vigour, or I may call it a day. My intentions were not to create bitterness or spread hatred. Many people are taking it in a wrong way. I am also snapping rather nastily at people against whom I really have no enmity. Please forget temporarily what I have said over the past few days in these threads. If you need to carry on conversation with me, you are delighted to mail me at extreme.atheist@gmail.com.

Thanking you,

Yours truly,
Comrade bubka

EDIT: The last line will be: "... me, [b]I'll be delighted if you mail me[/b] at ..." or " ... you are [b]welcome[/b] to ...". What I wrote just reflects my mental conditon!

-----------------------------------------------------------------------------------

NOTE: This is an exact copy of the post in The Round Table. I thought I might put that here as well for not everybody goes to TRT everyday.

[quote="mathnerd314"]:P :) :spam:

What was the point of the last two posts?![/quote]

what? a picture says a thousand words, and i posted three pictures..so that's like...more than a thousand words.

I believe that communism is not dead in the world - communism occurs in online communities as well to a certain extent. Bubka, what is your opinion on this?

[quote="ccy"]You know what, Bubka?  Your topic made my access to AoPS drop for an hour (im in china right now).

I'm serious.

Next time I will now use caution before reading your topics.[/quote]

Oh... wow. :o

[quote="ShoeFactory"]When your access to aops drops like that... does it mean they are reviewing what you were looking at ?[/quote]

I think ccy means that she only has a limited time on the internet and she wasted it on bubka's topics.

[quote="ShoeFactory"]Are you sure?

Why did she mention that she was in china then.

Plus this thread happens to have the word communist and democracy.[/quote]
Well, I thought people in China and other countries in that region had a certain amount of time for each month, or something like that.

I would like to shower bubka with Independency.

Ugh bubka, this topic made my internet drop for a minute and AoPS 404d for about 5 minutes while other sites worked. Please put a more telling title next time without actually saying the topic, and yes, I'm in China.


function
inequalities
algebra unsolved
algebra

Find all function $f: \mathbb{R}\rightarrow \mathbb{R}$ s.t. it's continuouns on $0$ and for every $x\in \mathbb{R} , f(3x)-f(x)\geq 8x^2+2x$ and $f(2x)-f(x)\leq 3x^2+x$

Replace in the first inequality $x$ by $\frac{x}{3}$, $\frac{x}{3^2}$, $\dots$ $\frac{x}{3^n}$, $\dots$ and sum up. You'll find that
\[
f(x)-f(0)\geq 8x^2\sum_{n\geq 1}\frac{1}{9^n}+2x\sum_{n\geq 1}\frac{1}{3^n}=x^2+x\,.
\]
Working similarly in the second inequality, you'll have
\[
f(x)-f(0)\leq 3x^2\sum_{n\geq 1}\frac{1}{4^n}+x\sum_{n\geq 1}\frac{1}{2^n}=x^2+x\,.
\]
So, $f\in\int(2x+1)\mbox{d}x$. ;)


calculus integration trigonometry logarithms derivative function Putnam

Compute $\int_{0}^{\pi/4} \tan x (x^3+x) dx$

Is it any help that $\int_{-k}^{k} \tan x (x^3+x) dx = 0$?

My gut feeling says that this is going to be ugly.

If it was $\int_{0}^{\pi/4} (\cos x + \sin x) (x^3+x) dx$, the question would be easier.  Sorry.

You're right.  And yes that is the actual problem.

I have recently started serious study of integrations.  What I have here may be of some help.
$\int_{0}^{\frac{\pi}{4}}\tan x(x^3+x)\text{ d}x$

Let $u=\tan x=\frac{\sin x}{\cos x}$ and $\text{d}u=\frac{\text dx}{\cos^2 x}\rightarrow \text{d}x=\cos^2 x\text{ d}u$ and $x=\tan^{-1} u$
$u(0)=0$ and $u(\frac{\pi}{4})=1$.

$\\\int_{0}^{1}u[(\tan^{-1}u)^3+\tan^{-1}u]\cos^2(\tan^{-1}u)\text{ d}u$

Notify me if this helps, and if it does, please continue it so I'll see what happens.

Masoud Zargar

I also did that, but I didn't like the $(\tan^{-1} (u))^3$ so I quit with that approach.

Note also that $\cos (\tan^{-1}u) = \frac1{1+u^2}$, which looked disgusting also.

Let me try something completely random.

$\int_0^{\pi/4} (x^3+x) \int_0^{x} \ln (\sec t)\, dt \, dx = \int_0^{\pi/4} \int_0^{x} (x^3+x) \ln (\sec t) \, dt \, dx$

Then maybe I could do something with jacobians, because the integration region is triangular.

I'm not sure right now.

Can someone please help me finish this?  I think I wrote down most of the steps, but I just need to think of an appropriate substitution to make the double integral look nice.

I could apply fubini to it also:

$\int_0^{\pi/4} (x^3+x) \int_0^{x} \ln (\sec t)\, dt \, dx = \int_0^{\pi/4} \int_0^{x} (x^3+x) \ln (\sec t) \, dt \, dx =$

$\int_0^{\pi/4} \int_t^{\pi/4} (x^3+x) \ln (\sec t) \, dx \, dt$

I'm not sure where to go from there either.

I could also try to apply partial differentiation:

$F(k) = \int_{0}^{\frac{\pi}{4}}\tan (x)(x^3+kx) \, dx$

$\frac{d}{dk} F(k) = \int_0^{\frac{\pi}{4}} \tan (x) x \, dx$

I'm also having difficulty figuring this out, so maybe I put the made a poor $F(k)$.

Any help?

[quote="Hokkage"]fnInt(tan(X)(X^3+X),X,0,$\frac{\pi}{4}$) = .2564541108
Looks like nobody has tried integration by parts yet so I'll give it a shot and get back to you guys if it works...[/quote](tan(X)(X^3+X) is an odd function. so Int(tan(X)(X^3+X),X,-k,k)=0 is apparent.

I'm currently using the integration by parts method, but it seems that I am suck in a loop of continuous trigonometric integrals. :(  I'll try to get this done as soon as possible.  Until then, I'll think and wait for replies. :)

Masoud Zargar

$x^3+x$ is odd.
$\tan x$ is odd.
Thus the product is even.

may Tabulation method can help. Let $f(x) = x^3 + x$ and $g(x) = \tan{x}$. but i get $catalan{}number$ and $Polylog{}function$ . that was terrible integral. :o l

So....Anyone have any solutions yet?  I'm still stuck :(

[quote="boxedexe"][quote="BaBaK Ghalebi"]I think this might help you
$f(x)=x^3+x+1$
$f^{-1}(x)=?$[/quote]
Would you please continue this so I'll see where you're trying to get?  Thanks in advance. daste shoma dard nakone. :)

Masoud Zargar[/quote]
sorry but I have to write this in persian:bebin be tore ejab avari makoose in tabe be soorate avabe mosalasati dar miad age rqahe halle dorosti beri ide esbate oon kelide masale toe
trust me ;)
sorry again for the language thing

Do you have solution Babak Ghalebi?

The only method i tried (before went to internet cafe )  :) ) is something similar with FMako's . I still don't finish my calculation.

Let $I(\alpha) = \int_{0}^{\frac{\pi}{4}} (\alpha x^3 + x) \tan x \, dx$ then differentiating with respect to $\alpha$ will yield $I'(\alpha) = \int_{0}^{\frac{\pi}{4}} x^3 \tan x \, dx$ .So
\[ I(\alpha) = \alpha \, I' (\alpha) + \int_{0}^{\frac{\pi}{4}} x \tan x \, dx  \]

If we can compute $\int_{0}^{\frac{\pi}{4}} x \tan x \, dx$ , then we have first-order differential equation. For  $\int_{0}^{\frac{\pi}{4}} x \tan x \, dx$, it becomes $\int_{0}^{1} \frac{u \, \arctan u}{u^2 + 1} \, du$ or  $\frac{\pi}{8} \ln 2 - \int_{0}^{\frac{\pi}{4}} \ln \sec u \, du$. Can someone  solve this? :?

Just want to post what i fot during this 10 minutes. Somewhat, I proceed like blahblahblah did to Putnam A5 problem.
Suppose $\Phi (a) = \int_{0}^1 \frac{u \arctan au}{u^2+1} \, du$, then
\begin{eqnarray*} \Phi '(a) &=& \int_{0}^{1} \frac{u^2}{(u^2+1)(a^2 u^2 + 1)} \, du \\ &=& \frac{1}{a^2 -1} \left( \int_{0}^1 \frac{1}{u^2+1} \, du - \int_{0}^1 \frac{1}{(au)^2+1} \, du \right) \\ &=& \frac{1}{a^2 -1} \left( \frac{\pi}{4} - \frac{\arctan a}{a} \right) \\ \end{eqnarray*}
$\Phi(0) = 0$ and $\int_{0}^1 \Phi(a) \, da = \Phi(1) - \Phi(0) = \Phi(1) = \int_{0}^1 \frac{u \arctan u}{u^2+1} \, du$.

If that doesn't work then  :ninja: .
Maybe I should try residue theorem.

OK my post again.
Now i thinks my method doesn't work. That ODE need initial condition which is more hard to find.
I found this in my Library, from "Joseph W. Kitchen, Jr . Calculus of one variable".
In Russia it is known as [i]Lobachevsky's integral[/i] that is
\[ \varphi(x) = \int_{0}^{x} \ln \sec t \, dt  \]
It is appears as exercise to prove that
$\varphi(x) = 2 \varphi \left( \frac{\pi}{4} + \frac{x}{2} \right) - 2 \varphi \left(\frac{\pi}{4} - \frac{x}{2} \right) - x \ln 2$ .
Althought I can prove that using subtitution and integration by part, but i'm just can use that to find the value of $\varphi \left(\frac{\pi}{2} \right)$.

Edit: I misstypo the formula sorry:

[quote="boxedexe"]I have an asnwer, but no solution.  The answer is
\[ \frac{1}{1024}\tan\pi^4+\frac{1}{32}\tan\pi^2  \].

can anyone get something out of this?

Masoud Zargar[/quote]

How did you arrive at that answer?

[quote="Beginner"]$\frac{\pi}{8} \ln 2 - \int_{0}^{\frac{\pi}{4}} \ln \sec u \, du$. Can someone  solve this? :?[/quote]

[quote="Beginner"]OK my post again.
Now i thinks my method doesn't work. That ODE need initial condition which is more hard to find.
I found this in my Library, from "Joseph W. Kitchen, Jr . Calculus of one variable".
In Russia it is known as [i]Lobachevsky's integral[/i] that is
\[ \varphi(x) = \int_{0}^{x} \ln \sec t \, dt  \]
It is appears as exercise to prove that
$\varphi(x) = 2 \varphi \left( \frac{\pi}{4} + \frac{x}{2} \right) - 2 \varphi \left(\frac{x}{2}- \frac{\pi}{4} \right) - x \ln 2$ .
Althought I can prove that using subtitution and integration by part, but i'm just can use that to find the value of $\varphi \left(\frac{\pi}{2} \right)$.

Edit: I misstypo the formula sorry:[/quote]

Beginner, don't you want to calculate $\varphi (\frac{\pi}{4})$ or are you trying to say you can only use that formula to calculate $\varphi (\frac{\pi}{2})$?

[quote="Kalle"]http://www.mathlinks.ro/Forum/viewtopic.php?t=67095

This deals with calculating a very similar integral. Maybe it can be of help.[/quote]

I was able to prove the lobachesvsky integral formula using a similar method, thanks.

Darn, I didn't think this would be such a tough problem.

To FMAKO:
I meant I can't use the formula to calculate $\varphi \left( \frac{\pi}{4}\right)$.
 :P  Again I misstypo the formula, i've edit it sorry.

[quote="blahblahblah"]boxedexe's 'answer' doesn't come close to agreeing with the numerical results.  Unless he explains why he believes it is the case, I'm inclined to ignore his answer.[/quote]
Maple gave this answer.

I have Maple 10, and that's not the answer I get.

I uploaded an image on the internet.  Take a look and see if you see any errors.

[url=http://img484.imageshack.us/my.php?image=integral3jb.gif]Maple-integral[/url]

Masoud Zargar

It is treating "tanx" as a constant. You must write tan(x)

lol, idiot me.  You're right. Well, I see that even the computer took a pretty long time (30 secs) to evaluate that.

Does it now evaluate to a nice form or is it a purely numerical solution?


let a,b,c, > 0 that : a>c,b>d
prove that :
\[ (a\plus{}b\plus{}c\plus{}d)^2 > 8(ad\plus{}bc)\]

From rearrangement/Chebyshev, $ 8(ad\plus{}bc)<4(a\plus{}c)(b\plus{}d)\le (a\plus{}b\plus{}c\plus{}d)^2$

where the last inequality follows from AM-GM.


best by CS.
left and right Equivalent f(x,y,z) and f(x,z,y),
is Interesting.
BQ

[quote="xzlbq"]Let $ x,y,z > 0$,prove that

$ \frac {x}{y} \plus{} \frac {y}{z} \plus{} \frac {z}{x} \geq \frac {((y \plus{} z)(z \plus{} x)(x \plus{} y) \plus{} yzx)^2}{(x \plus{} y \plus{} z)^3xyz}\geq \frac {(y \plus{} z)x}{(z \plus{} x)y} \plus{} \frac {(z \plus{} x)y}{(x \plus{} y)z} \plus{} \frac {(x \plus{} y)z}{(y \plus{} z)x}.$

BQ[/quote]

$ \frac {x}{y} \plus{} \frac {y}{z} \plus{} \frac {z}{x} \geq \frac {((y \plus{} z)(z \plus{} x)(x \plus{} y) \plus{} yzx)^2}{(x \plus{} y \plus{} z)^3xyz}$

$ \Longleftrightarrow (y^3x \minus{} zxy^2 \plus{} z^3y \minus{} zx^2y \minus{} z^2xy \plus{} zx^3)(x \plus{} y \plus{} z)^2\geq 0.$

$ y^3x \minus{} zxy^2 \plus{} z^3y \minus{} zx^2y \minus{} z^2xy \plus{} zx^3\geq 0$

true due to AM-GM Inequality.

$ \frac {((y \plus{} z)(z \plus{} x)(x \plus{} y) \plus{} yzx)^2}{(x \plus{} y \plus{} z)^3xyz}\geq \frac {(y \plus{} z)x}{(z \plus{} x)y} \plus{} \frac {(z \plus{} x)y}{(x \plus{} y)z} \plus{} \frac {(x \plus{} y)z}{(y \plus{} z)x}.$

$ \Longleftrightarrow xyz(zy^3\minus{}zxy^2\plus{}x^3y\minus{}zx^2y\minus{}z^2xy\plus{}xz^3)(x\plus{}y\plus{}z)^2\geq 0.$

$ zy^3\minus{}zxy^2\plus{}x^3y\minus{}zx^2y\minus{}z^2xy\plus{}xz^3\geq 0$

true due to AM-GM Inequality.


So who got in? Don't vote if you don't know yet, just post in the comment chain.

i sent my reply today and  they said it would take 3 days it to reach them

I got in! :D

I guess this is kind of subdues, considering that I got in for the early application, but w/e. Any other sixth/seventh graders???? I want to know who I'll be sharing a room with.

i stayed up till 2am finishing up the application

 i answered a decent amount of question but im not sure.......

just very impatient.

[quote="AIME15"]When are we notified (I applied regular)?[/quote]

April 22nd.

I also applied regular.

Same here...but some Test A applicants say they got their results earlier or later...

Also, a question: do they tell you if you DON'T get in?

[quote="AIME15"]When are we notified (I applied regular)?

Also, do they email you back when they receive your application?  I keep hoping they don't lose mine.[/quote]

I don't think so, they just emailed me back saying that I got in. I'm pretty sure they tell you if you don't get in, too.

For those of you who applied and got in, which site (CA, TX) are you going to?

if i get in im going to texas. so nervous  :maybe: i did eight problems on the admission test, so maybe, but i did two some really dumb way.

Bleh, I did 9 problems, but I somehow misinterpreted number 10 into something that's even more trivial than what it asks for right now.

They still haven't emailed me yet, but that's probably because the post office was closed Saturday, so I just mailed the application fee today. Going to CA if I get in.

i got my email and got accepted! AIME15, you'll probably make it, dont worry, you seem smarter than me

i got in. Somehow I doubt that camp admission will be highly competitive given the $ \$$3500 admission cost; presumably you would be accepted as long as you did not completely fail the admission test.
[quote="lukezli"]dangit i meant i would go to california. i live in california, so  :D im finishing eighth grade this year.[/quote]
D4mn, I am going to be one of the oldest there; finishing 11th grade.

YES! Made it!

Lol, I was at SJMC (math circle) and was biting my nails the whole time...:D

YAY!!  congrats

which campus will you go to?

I'm gonna being going to CA. See you all!

Well, in response to that, I say:

I might try to sneak in my drum set. Heh. Who else will be bringing their drum sets?

I also might bring my tap dancing shoes. Yay tap dancing. xD

This is a bit late, but I got in too!!!!! I'm going to TX. I had complete solutions to all the problems, but I probably didn't get many right. BTW, I'm finishing 7th grade, and I'm a guy........yep.........

I got in to!!!!!!!!!!!!!!!!! :lol: , I skipped a few problems... mostly the algebra ones.
Going to CA

I failed the test. But oh well, still got in  :lol: .

All that one of my friends that got in will do is complain that its going to be hot there.

[quote="mathelete9999"].

All that one of my friends that got in will do is complain that its going to be hot there.[/quote]

it is.
esp in texas.
but its so much fun. the heat wont matter.

You're going to be inside for the hottest parts of the day.

Out at 6PM-10PM for some frisbee action :D


A small square $PQRS$ is contained in a big square. Produce $PQ$ to $A$, $QR$ to $B$, $RS$ to $C$ and $SP$ to $D$ so that $A$, $B$, $C$ and $D$ lie on the four sides of the large square in order, produced if necessary. Prove that $AC = BD$ and $AC \perp BD$.

[i]I have just come back from a journey and I ' m tired. So I write a hint to the solution that I was thinking while I was driving !!![/i]

[b] Hint [/b]

  Produce both pairs of the the oposite sides of $PQRS$ to meet the sides of the big square . We take then two parallelogrammes which are equal.( We are based on a well problem according to which:if two perpedicular lines  intesect the oposite sides of a square , then the coresponding  segments are equal). In these parallelogrammes segments  $AC$ and  $DB$ are diagonals , so they are equal  , too. A little playing  with angles gives us that  $AC$ and $DB$ are perpendicular.

   [u]Babis[/u]

  [i]Does anyone have time to write down this solution with it's details?Thanks![/i]


Find the value of $ 1\minus{}\frac{1}{1996}$, expressed as a common fraction.


[quote="cowboy"]Let A, B, C is 3 angles of an acute triangle ABC. Prove that:
sqrt(1+k(cosA)^2)/sinB + sqrt(1+k(cosB)^2)/sinC + sqrt(1+k(cosC)^2)/sinA>=sqrt((4+k)/3)
k is a positive integer.

(Sorry because I don't know how to make the problem nice.)[/quote]
Here it is nicer:
Prove $\frac{\sqrt{1+k\cos^2{{A}}}}{\sin{{B}}}+\frac{\sqrt{1+k\cos^2{{B}}}}{\sin{{C}}}+\frac{\sqrt{1+k\cos^2{{C}}}}{\sin{{A}}}\geq\sqrt{\frac{4+k}3}$
use this:
[code]\frac{\sqrt{1+k\cos^2{{A}}}}{\sin{{B}}}+\frac{\sqrt{1+k\cos^2{{B}}}}{\sin{{C}}}+\frac{\sqrt{1+k\cos^2{{C}}}}{\sin{{A}}}\geq\sqrt{\frac{4+k}3}
[/code]
surrounded by $ \$'s

Are you saying that I can use Tex code in a post and then it will appear as a Tex format in the forum?
 :D

I couldn't make out anything. Anyone else???


trigonometry
inequalities unsolved
inequalities

For any real number $ x$ and any positive integer $ n$ prove that \[ \left| {\sum\limits_{k \equal{} 1}^n {\frac{{\sin kx}}
{k}} } \right| \leqslant 2\pi
\]

[quote="Xaenir"]For any real number $ x$ and any positive integer $ n$ prove that
\[ \left| {\sum\limits_{k \equal{} 1}^n {\frac {{\sin kx}} {k}} } \right| \leqslant 2\pi
\]
[/quote]
I think it must be $ \left| \sum_{k\equal{}1}^n{\frac{sinkx}{k}} \right|$≤$ 2\sqrt{\pi}$

[/quote]
I think it must be $ \left| \sum_{k \equal{} 1}^n{\frac {sinkx}{k}} \right|$≤$ 2\sqrt {\pi}$[/quote]

Yes,I think so.

Actually a friend gave the problem. Later when I asked him to verify the problem , he says that it should be this:

It should be $ \left| \sum_{k \equal{} 1}^n{\frac {sinkx}{k}} \right|$≤$ 2\sqrt {\pi}$

What's the source of the problem? :wink:


Greetings,

Given a finite-dimensional inner product space $ V$ and a linear operator $ T$, is there another way of finding $ T$'s adjoint $ T^*$? other than to: find an ordered basis of $ V$, use Gram-Schmidt's orthorgonalization process and normalize the vectors to get an orthonormal basis $ \beta$, compute $ [T]_\beta$, then find $ T^*$ using $ [T^*]_\beta\equal{}[T]^*_\beta$.

Sometimes $ T^*$ can be calculated directly in the inner product, for example, if $ T\equal{}L_A$ then $ \langle T(x),y\rangle\equal{}\langle Ax,y\rangle\equal{}\langle x,A^*y\rangle$. But I don't know how it can be done in general.

Denote the Gram matrix by $ G$. If the matrix of the linear operator is $ A$, then the matrix of the adjoint operator is
\[ G^{\minus{}1} A^T G.\]
So first you need to invert $ G$, and then multiply 3 matrices.


calculus
integration
trigonometry
ratio
calculus computations

$\int_{0}^{\infty}\frac{\sin\sqrt{x}}{1+x}\; dx$

(Hint. The value of this integral is the ratio of two important constants in math)

If $x=t^{2}$ then your integral equals  the famous Laplace integral which ,for this particular case, is equal to:

$\int_{0}^{\infty}\frac{2t\sin t}{1+t^{2}}dt=\frac{\pi}{e}$.


pigeonhole principle
number theory unsolved
number theory

A set $S$ of $n-1$ natural numbers is given ($n\ge 3$). There exist at least at least two elements in this set whose difference is not divisible by $n$. Prove that it is possible to choose a non-empty subset of $S$ so that the sum of its elements is divisible by $n$.

here's a generalization:

lemma: let $T$ be a set of $k$ natural numbers. if there is no way to choose a non-empty subset of $T$ such that the sum of elements is divisible by $n$, then there are at least $k$ different sums of elements of non-empty subsets of $T$ modulo $n$ and it are $k$ iff all elements of $T$ are equal modulo $n$.

proof: reduce everything modulo $n$. we don't get a zero by the condition; let the elements be $a_1,...,a_k$. if $a_1+...+a_i=a_1+...+a_j$ for some $i\neq j$, we would have a zero sum. thus, all the sums $a_1+...+a_i$ are different giving $k$ different values. now assume it are only $k$. if $a_2\neq a_1$ we have $a_2=a_1+...+a_i$ for some $i\geq 2$ giving a zero sum, contradiction; thus $a_2=a_1$. by reordering the numbers, we get all numbers to be equal.

now, let $T=S$ where $k=n-1$. the lemma implies that there are at least $n$ different sums of non-empty subsets of $S$; by the pigeonhole principle, we get a zero sum.

[hide="I proved it in my mind but it might be jumbled coming out"]
Assume there is a set of naturals so that the special subset doesn't exist.
Let $ a_1,...,a_{n\minus{}1}$ be the given naturals reduced mod $ n$, with $ a_3\not\equal{}a_2$ (guaranteed to exist by the statement). Then each is distinct and nonzero (otherwise the subset would be that element). Thus $ \{0,a_1,a_1\plus{}a_2,\cdots,a_1\plus{}\cdots\plus{}a_{n\minus{}1}\}\equal{}\{0,1,2,\cdots,n\minus{}1\}$.
Let $ f(\textbf{a}): \equal{}\{0,a_1,a_1\plus{}a_2,...,a_1\plus{}\cdots\plus{}a_{n\minus{}1}\}$
Consider the sets $ A: \equal{}f(a_1,a_2,a_3,\cdots,a_n)$ and $ B: \equal{}f(a_1,a_3,a_2,a_4,\cdots,a_n)$. Among each, no two elements are equal, because then their difference is the sum of a subset and therefore nonzero.
Thus $ A\equal{}B\equal{}\{0,1,2,\cdots,n\minus{}1\}$, and every element of $ A$ besides the second is virtually identical to the corresponding element in $ B$; thus the second must also be equal, so $ a_1\plus{}a_2\equal{}a_1\plus{}a_3$.
This contradicts our assumption.
[/hide]

Let $S=\{a_1,\dots a_{n-1}\}$.We consider following $n$ numbers.

$a_1$
$a_2$
$a_1+a_2$
$a_1+a_2+a_3$
$\dots$
$a_1+\dots +a_{n-1}$

If there is $\equiv 0 \mod{n}$, we can prove.If not, by PHP, there are two numbers which are congruent to each other in $\mod{n}$.These are not $(a_1,a_2)$.Consider the difference between two numbers,then we can prove.$\blacksquare$

:coolspeak:


Let $ (I) $ be the incircle of the triangle $ ABC $ . $ M \in (I) $ . $ d $ is the tangent at $ M $ of $ (I) $ . The perpendicular with $ AI $ at $ I $ intersects $ d $ at $ A_1 $ . The perpendicular with $ BI $ at $ I $ intersects $ d $ at $ B_1 $ . The perpendicular with $ CI $ at $ I $ intersects $ d $ at $ C_1 $ . Prove that : $ AA_1 , BB_1 $ and $ CC_1 $ are concurrent .

I posted this a while back: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=18907]www.mathlinks.ro/Forum/viewtopic.php?t=18907[/url]


If $ a,b,c$ are positive reals and $ a\plus{}b\plus{}c\equal{}1$,then prove that
$ \frac{1}{a^{2}\plus{}b}\plus{}\frac{1}{b^{2}\plus{}c}\plus{}\frac{1}{c^{2}\plus{}a}\ge\frac{27}{4}$

[quote="RAGHAV GROVER"]If $ a,b,c$ are positive reals and $ a \plus{} b \plus{} c \equal{} 1$,then prove that
$ \frac {1}{a^{2} \plus{} b} \plus{} \frac {1}{b^{2} \plus{} c} \plus{} \frac {1}{c^{2} \plus{} a}\ge\frac {27}{4}$[/quote]
It's wrong. See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=160017


Prove that among $ 1996$ different integers, there are at least two of whose difference is divisible by $ 1995.$

There are only $ 1995$ possible remainders when dividing by $ 1995$. So, at least two numbers out of the $ 1996$ integers must leave the same remainder when divided by $ 1995$; The difference of one such pair will be divisible by $ 1995$.


integration trigonometry limit calculus function geometry rectangle

Show that
\[ \int_{0}^{\infty}\frac{\sin{x}}{x}\,dx=\frac{\pi}{2}. \]

For some real analysis methods, look [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=41288]here.[/url]

In particular, note posts 7,8, and 9, which connect that problem to this one.

The standard complex analysis method:
Rewrite as $\frac12\int_{-\infty}^\infty \frac{\sin x}x\,dx$.
Rewrite as $\frac12\lim_{\epsilon\to0,R\to\infty}\text{Im}\left(\int_{-R}^{-\epsilon}\frac{e^{ix}}x\,dx+ \int_{\epsilon}^{R}\frac{e^{ix}}x\,dx\right)$.
This is a principal value integral, for a function with its only pole at zero. Set up an appropriate contour (probably a large rectangle with a small circle removed), and argue that the outer pieces go to zero.

So, to state it explicitly, the real analysis methods involve writing
\begin{eqnarray*} \int^{\infty}_0 \frac{\sin x}{x} dx &=& \int^{\infty}_0 \sin x \int^{\infty}_0 \exp(-tx)dtdx \\ &=&\int^{\infty}_0\int^{\infty}_0 \sin x\exp(-tx)dxdt \\ &=&\int^{\infty}_0\frac{dt}{t^2+1} \\ &=&\frac{\pi}{2} \end{eqnarray*}
The middle equality is Fubini.

[quote]The middle equality is Fubini.[/quote]
Without either of the justifications for Fubini - the functions are not nonnegative, and the double integral does not converge absolutely. You've got to do some very delicate analysis to justify this.

That's why I like to integrate by parts first and turn it into

$\int_0^{\infty}\frac{1-\cos x}{x^2}\,dx$

before introducing $e^{-tx}$ and setting up the use of Fubini.

I'm going to pretend that I have not written those exact computations on work that I have submitted (and a final!) on at least three occasions.


limit
trigonometry
logarithms
calculus
derivative
calculus computations

We want to find the   $ \lim_{x\rightarrow0}\frac{3^{x^2}\minus{}\cos {5x}}{4x^2}$

Applying l'Hopital's rule twice gives $ \frac{2\ln 3 \plus{} 25}{8}$.

You're right of course, Carcul, it is better to use power series.

nice solutions.  thanks.    :)

For me, I call L'Hopital's rule a "cheater's method" since it takes the fun out of solving these problems, haha.
But of course, I use the rule myself, lol.   :)

Rewrite as $ (1/4)[\frac {3^{x^2} \minus{} 1}{x^2} \plus{} \frac {1 \minus{} \cos {5x}}{x^2}]$. The first term in brackets $ \rightarrow \ln3$,  by the definition of the derivative of $ 3^u$ at $ u \equal{} 0$. For the second term, recall $ \frac {1 \minus{} cos(u)}{u^2} \equal{} \frac {1}{1 \plus{} cos(u)}\frac {sin^2(u)}{u^2}\rightarrow \frac {1}{2}$. So the answer is  $ (1/4)[\ln3 \plus{} \frac {25}{2}]$.


function
integration
calculus
complex numbers
algebra
functional equation
calculus computations

Define $\Gamma (x)$ as:

$\Gamma (x)=\int_{0}^{\infty}e^{t}t^{x-1}dt$

Then, $\Gamma (x) = \int_{0}^{\infty}e^{t}t^{x-1}dt > \int_{0}^{1}e^{t}t^{x-1}dt > \frac{1}{e}\int_{0}^{1}t^{x-1}dt$, which diverges positively for all negative $x$.

It would seem that $\Gamma (x)$ diverges to positive infinity for all $x<0$, but plotting on Mathematica (which gives the definition I used) suggests otherwise.

Can someone please tell me where I went wrong?

Thanks!

~ tZf

The integral definition does diverge, but there is an alternate definition of the Gamma function for $x$ with negative real part: we use the functional equation $\Gamma(x+1)=x\Gamma(x)$ to walk backward. This gives meaningful values everywhere except the nonpositive integers, at which there are poles.

The [url=http://functions.wolfram.com/]Wolfram website[/url] also offers this representation, which works for $x\in\mathbb R\setminus\{0,-1,-2,\dots\}$:
\[\Gamma(x)=\int_{1}^{\infty}t^{x-1}e^{-t}dt+\sum_{k=0}^{\infty}\frac{(-1)^{k}}{k!(x+k)}\]


I'll make this plain and simple: we have many subject areas supporting different science olympiads such as Chemistry, Biology, Physics, CS and Info. We even have Music and Graphics Art. After I saw all those different International Science Olympiads, [b]Linguistics[/b] and [b]Philosophy[/b] are not established in AoPS/ML. If anybody is interested please reply. Administrators: what do you think of two new forums for these subjects? Or is it not enough interested in it? Although I am neither linguistic person nor a philosopher, I would like to see them bloom in our forums :)

I would say [i]definitely[/i] make a philosophy forum (I'm surprised they don't have one already!), and a linguistics forum might be a good idea too.

I don't know if you've wandered into the Round Table forum, but we have a wide range of discussions there, a number of which could be classed as either Linguistic or Philosophical.  Not every subject needs its own forum ...

I have wondered into The Round Table a few times and read a couple of the active discussions which I found interesting. Perhaps The Round Table has more uses than just serious discussions which I am not aware of. In any case, I'm glad there is a forum for general discussions in AoPS. But, I wonder how would the Linguistics category be fit into that forum... or would Linguistics fall into [b]Computer Science and Informatics[/b] forum due to Chomsky's works? For example, let's say I am interesting in ILO ([url=http://olympiads.win.tue.nl/ilo/index.html]International Linguistics Olympiad[/url]) and I want to post a problem in this site. Can there be a place for it in AoPS? It's just a suggestion I thought about, you can say [b]NO[/b] if that's the appropriate answer :)

I didn't even know such an olympiad existed.  That would be an interesting addition to the site.

Does the US participate in the ILO?


integration
logarithms
calculus
calculus computations

Substituting from the previous line, we get

$ \minus{} \frac {1}{2^{2}} \plus{} \frac {1}{5^{2}} \minus{} \frac {1}{8^{2}} \plus{} ... \plus{} \frac {1}{1^{2}} \minus{} \frac {1} {4^{2}} \plus{} \frac {1}{7^{2}} \minus{} ...$.

This is $ \frac {1}{1^{2}} \minus{} \frac {1}{2^{2}} \minus{} \frac {1}{4^{2}} \plus{} \frac{1}{5^{2}} \plus{} \frac{1}{7^{2}} ...$.

If we then add $ \frac {1}{3^{2}} \minus{} \frac {1}{6^{2}} \plus{} ...$, we must also subtract it and this gives the expression inside the square brackets in your last line.


Λίγο στα γρήγορα

1) Η τρίτη γραμμή είναι λάθος. Από $ A \Rightarrow B$ και $ \neg A$ δεν μπορούμε να συμπεράνουμε το Β

2) Η πρώτη γραμμή είναι λάθος αφού 2+2=4 αλλά $ 2\plus{}3 \neq 6$

3) Ορθό

4) Η πρώτη γραμμή είναι λάθος αφού 2+2=4 αλλά $ 2\plus{}3 \neq 6$

[quote="Demetres"]Λίγο στα γρήγορα

1) Η τρίτη γραμμή είναι λάθος. Από $ A \Rightarrow B$ και $ \neg A$ δεν μπορούμε να συμπεράνουμε το Β

2) Η πρώτη γραμμή είναι λάθος αφού 2+2=4 αλλά $ 2 \plus{} 3 \neq 6$

3) Ορθό

4) Η πρώτη γραμμή είναι λάθος αφού 2+2=4 αλλά $ 2 \plus{} 3 \neq 6$[/quote]

Aλλα ,Δημητρη, εαν βαλουμε:

 2)(2+2=4)=p kai  .....( 2+3=6)=q ,τοτε το επιχειρημα : [(p----->q) & p]------->q ,δεν ειναι σωστο??( M.Ponens)

4) Επισης (2+2=4)=p kai.........(2+3=6)=q, τοτε το επιχειρημα : [(p------>q) &~q]-------->~p, δεν ειναι σωστο??( M.Tolens)

[quote="τρικλινο"]
Aλλα ,Δημητρη, εαν βαλουμε:

 2)(2+2=4)=p kai  .....( 2+3=6)=q ,τοτε το επιχειρημα : [(p----->q) & p]------->q ,δεν ειναι σωστο??( M.Ponens)
[/quote]

Το [(p----->q) & p]------->q είναι σωστό. (Δεν είπα πως είναι λάθος.)

Πρόσεξε όμως αυτό δεν είναι modus ponens. Το modus ponens λέει πως αν έχεις στην απόδειξη σου τις γραμμές (p----->q) και την γραμμή p, τότε μπορείς στην απόδειξή σου να γράψεις και την γραμμή q.

Εγώ έγραψα πως η γραμμή (p----->q) είναι λανθασμένη, άρα και το όλο επιχείρημα είναι λανθασμένο.

Ομοίως και για το (4)

[quote="τρικλινο"]Δημητρη, συμφωνεις οτι η φορμουλα :[(p-------.>q) & p]------->q (M.Ponens) ειναι παντοτε Αληθης.........για οποιαδηποτε τιμη(αληθης,ψευδης) και ΑΝ παρουν τα p and q??[/quote]

Νομίζω ότι ήμουν αρκετά σαφής, αλλά να το ξαναπώ. Η απόδειξη

1) [(p-------.>q) & p]------->q

είναι πάντοτε ορθή. Εσύ όμως έγραψες την απόδειξη

1) p --> q
2) p
3) q

Νομίζω ότι έχουν κάποια διαφορά. Η ορθότητα της 3 γραμμής εξασφαλίζεται από την ορθότητα της πρώτης και της δεύτερης γραμμής. Με τα p και q που διάλεξες, η δεύτερη γραμμή είναι ορθή αλλά η πρώτη γραμμή είναι λανθασμένη. Άρα το επιχείρημα είναι λάθος. Θα μπορούσα βέβαια να ήμουν πιο σύντομος (όπως έκανα στο πρώτο μου post) και να έλεγα: "Η πρώτη γραμμή είναι λάθος".

[quote="τρικλινο"]Δηλαδη εννοεις,αν καταλαβα καλα , το επιχειρημα:

[(εαν 2+2=4 ,τοτε 2+3=6),αλλα 2+2=4] αρα 2+3=6............ειναι διαφορετικο απο το επιχειρημα:

1) εαν 2+2=4 ,τοτε 2+3=6

2)...................αλλα 2+2=4

3) ...................αρα 2+3=6   ??[/quote]

Εννοώ ότι το επιχείρημα

1) [(p-------.>q) & p]------->q

είναι διαφορετικό από το επιχείρημα.

1) p --> q
2) p
3) q

Στο πρώτο επιχείρημα υπάρχει μια αληθής πρόταση. Στο δεύτερο επιχείρημα, από που συμπεραίνουμε ότι οι προτάσεις (1) και (2) είναι αληθείς; (Φυσικά αν είναι τότε και η πρόταση (3) είναι από modus ponens.)

[quote="Demetres"]

Εννοώ ότι το επιχείρημα

1) [(p-------.>q) & p]------->q

είναι διαφορετικό από το επιχείρημα.

1) p --> q
2) p
3) q

Στο πρώτο επιχείρημα υπάρχει μια αληθής πρόταση. Στο δεύτερο επιχείρημα, από που συμπεραίνουμε ότι οι προτάσεις (1) και (2) είναι αληθείς; (Φυσικά αν είναι τότε και η πρόταση (3) είναι από modus ponens.)[/quote]

Μα και τα δυο επιχειρηματα ειναι M.Ponens.

Το ενα ειναι γραμμενο οριζοντια και το αλλο καθετα.

Που ειναι η αληθης προταση στο πρωτο επιχειρημα

Σε μια συντακτικη αποδειξη δεν μας ενδιαφερουν οι τιμες των προτασεων,μονο σε semantical αποδειξη μας ενδιαφερουν οι τιμες των προτασεων ,αληθης η ψευδης

Νομίζω πως δεν οφελεί σε τίποτα πια αυτή η συζήτηση. Αντιλαμβανόμαστε διαφορετικά ορισμένα πράγματα. Αν καταλαβαίνω καλά και οι δύο δεχόμαστε ως ορθό ότι

[((2+2 = 4) ==> (2+3=6)) & (2+2=4)] ==> (2+3=6)

Άρα ας μείνουμε ως εδώ


Let $ C_1$ and $ C_2$ be circles defined by
\[ (x \minus{} 10)^2 \plus{} y^2 \equal{} 36\]and
\[ (x \plus{} 15)^2 \plus{} y^2 \equal{} 81,\]respectively. What is the length of the shortest line segment $ \overline{PQ}$ that is tangent to $ C_1$ at $ P$ and to $ C_2$ at $ Q$?

$ \textbf{(A)}\ 15 \qquad \textbf{(B)}\ 18 \qquad \textbf{(C)}\ 20 \qquad \textbf{(D)}\ 21 \qquad \textbf{(E)}\ 24$

[hide]The centers of the two circles are at $ (10, 0)$ and $ ( \minus{} 15, 0)$, and their respective radii are $ 6$ and $ 9$. The shortest tangent is the internal tangent, so we find that. Let the point of the intersection of the tangent and the $ x$-axis be $ I$. Then, two similar triangles are formed, and we can determine that the point of intersection is $ (0,0)$. The two triangles are standard Pythagorean triangles, and we determine that the length of the tangent is $ 8 \plus{} 12 \equal{} 20 \implies \boxed{C}$.[/hide]


Hi everyone.

I'm trying to make a symbol that looks like this: [img]http://img215.imageshack.us/img215/5594/bootstrapmeanko0.png[/img]

I've tried using $\bar{{\bar{x}^{*}}}$ and $\bar{x}^{\bar{*}}$ but as you can see, in both cases, the alignment isn't right. Does anyone have any idea how I can do this?

This might not be quite what you want, but it's very close: $\bar{\bar{x}}^{*}$.

Even closer: $a\,\,\bar{{\!\!\bar x}}^{*}b$. ($a$ and $b$ are there just to check spacing). You may also prefer $\overline{\bar x^{*}}$.

$\bar{{\!\bar x}}^{*}$ is close enough for my purposes.

Thanks for the help and quick replies!


So I've set up asymptote as presented on the Wiki, using Ctrl-Shift-A as a shortcut for running the EPS creating script from an ASY file. The problem is that, unlike running pdftex or any of the other set up commands, this one doesn't save the file before it runs it. So I sometimes forget to press Ctrl-S before running Asymptote, which can become annoying. Is there any (quick) solution to that?

I can't see a way round that. There doesn't seem to be a way of programming TeXnicCenter so that it saves a file - the DDE commands that it supports are only for opening and printing files not for saving them (otherwise a batch file could have saved the file then run asymptote). I expect various things could be done via the build profiles but there's nothing that I can think of there that wouldn't be more trouble than it's worth.
The only suggestions that I can think of is:
1. setting auto-save to be the minimum of 1 minute;
2. if you are running Asymptote via a DOS file then putting on the first line
echo HAVE YOU SAVED THE FILE?
It wouldn't stop Asymptote from running but might remind you.

It's merely a matter of convenience. Whenever I make changes to an already build asy picture, I tend to preview a lot to see what's happening, but sometimes I forget to save, thus I belive my new code isn't working, etc.

This is another suggestion - a new profile LaTeX=>ASY which I am attaching*. Please change any paths for asy.exe and Acrobat Reader.
Advantages
[list][*]auto saves the file before running Asymptote
[*]shows errors in the TexnicCenter window at the bottom (no need for a batch file)  :)
[*]runs Ghostscript automatically to show the resulting eps file
[*]can be run in a separate copy of TexnicCenter[/list]
Disadvantages
[list][*]C
can only be used on .asy files not .tex files (these would require latex to be run first to produce a new asy file)
[*]need to change the profile between LaTeX=>ASY and LaTeX=>PDF (though this can be overcome by editing asy files in a separate instance of TexnicCenter)[/list]
Change the command line arguments to -tex "pdflatex" %bm.asy if you'd prefer a pdf rather than eps to be created; I have put in the viewer tab the assumption that example.asy is compiled to example_1.pdf but that is only for when you ask for a view - an automatic viewing follows a compilation anyway.

You may want to play with this some more.

Hope this is useful.

* Unzip to a .tco file then in TexniCenter, Build, Define Output Profiles, Import


The sum of an arithmetic sequence with difference 1, 100 terms, and first term x is $ \dfrac{100}{2} (2x \plus{} (99)(1)) \equal{} 100x \plus{} 4950$ Thus the only viable answer is A.

WE know that sum of 100 consequtive integers is expressed as 50(2x+99).let it be =p.
i.e p/50-99 must be even.
so p/50 must be odd and GREATER THAN 99.only option 1 satisfies the conditions.
so 1 option jscorrect.


I have noticed that by default, the blog "post your comment" button is actually named "%s comments" and the "view comments" button is named "post your comment."

A lot of users don't change that setting, so it's a bit silly. I didn't notice on the Princeton Math Club officer blog until today, when I realized that comments had been posted, but there was no indication of that.

The strange thing is that on the blog options page the 2 comments are exactly flip-flopped from the order they actually appear. I think that that is the problem. I think Valentin will have to re-format something


I know the strategy for simplifying congruences like      7^7386(mod 7387)   the method of successive squares but I was wondering besides going through the whole process again whether there was software out there that would let me check my answer rather than going through the whole mundane process?

Maple, Mathematica work.
Probably Octave too (never used it, but it's freeware).

By the way, if $7387$ is prime, Fermats little theorem gives you that this number is $1 \mod 7387$.

I don't check that 7387 is prime. For all prime p $a^{p-1}=1(mod \ p), p\not |a$.

Thanks for the replies guys, I appreciate it, last question what about        28^749 (mod 1147), I did it two different ways and got two different answers, I thought were right.


Let $x$ be a real number such that $x^3+4x=8$. Determione the value of $x^7+64x^2$.

[hide="solution"]$x^3+4x=8 \implies x^3=8-4x$.
$N=x^7+64x^2=x(x^3)^2+64x$.
Putting in the value for $x^3$:
$x(8-4x)^2+64x^2=(64x-64x^2+16x^3)+64x^2=16(x^3+4x)=16\cdot 8=\boxed{128}$. [/hide]
Looked challenging but actually wasnt that bad  :D


Prove that if y>eq 0 and y(y+1)<eq (x+1) 2  then y(y-1)<eq x 2

Clearly, we may assume that $y>1$ otherwise the conclusion holds trivially.
From $y(y+1) \leq (x+1)^2$ and since $\sqrt {(a+1)(b+1)} \geq \sqrt {ab} + 1$ for $a,b \geq 0$, it follows that
$|x| + 1 \geq |x+1| \geq \sqrt {y(y+1)} = \sqrt {[(y-1)+1][y+1]} \geq \sqrt {y(y-1)} + 1$, hence result.

Pierre.

Very nice solution.There was a monstruous,extremly long one in my book and i couldn't understand.


find all function $ f$ defined and continious on $ R$ such that:
$ (\forall x\in R) f(x)\equal{}f(x^2\plus{}\frac{1}{4})$
from moroccan TST 2007

General problem : Let $ k > 0$. Find all continuous functions $ f: \mathbb{R}\to\mathbb{R}$ such that $ f(x) \equal{} f(x^2 \plus{} k)\forall x\in\mathbb{R}$.

$ \longrightarrow f(x)\equal{}\text{constant}$

$ f(x)\equal{}f(\minus{}x)$

[color=green]Try[/color] $ x_1\equal{}a \;\le \frac{1}{2}, x_{n\plus{}1}\equal{}x_n^2\plus{}\frac{1}{4}$

[color=green]or[/color] $ x_{n\plus{}1}\equal{}\sqrt{x_n\minus{}\frac{1}{4}}$ [color=green]other wise[/color]


hi,
Let $ f: \mathbb{R} \rightarrow (a,a\plus{}1)$ be a increasing function and two times differentiable.
Prove:
\[ f''(x)f(x) \minus{} af''(x) \minus{} (f'(x))^{2} < 0, \forall x \in \mathbb{R}.\]

This doesn't seem true.  Let $ f(x) \equal{} \frac {2\arctan x \plus{} \pi}{2\pi}$; it is smooth, strictly increasing, and $ f\left(\mathbb{R}\right) \equal{} (0,1)$.  But, unless I've made an algebraic error,
\[ f''f \minus{} (f')^2 \equal{} \minus{} \frac {2x\arctan x \plus{} \pi x \plus{} 1}{\pi^2 (1 \plus{} x^2)^2}
\]
which at $ x \equal{} \minus{} 1$ is $ \frac {\pi \minus{} 2}{8\pi^2} > 0$.

yes i see! my first sighn was i substitute $ g(x) \equal{} f(x) \minus{} a$ and this function is increasing and two times differentiable and $ g: \mathbb{R} \rightarrow (0,1)$. and i tried to prove that $ \frac{g(x)}{g'(x)}$ is a increazing function. do you have an exemple form which $ \frac{g(x)}{g'(x)}$ is not increasing ?


Hello, MCrawford!

A delightful problem . . .

Quote:A right circular cylinder has the same volume as a rectangular box with square bases.
If both figures have the same lateral area,
compute the ratio of the diameter of the circular base to the side of the square base.[hide]

Let R = radius of the cylinder, H = height of the cylinder.

Let x = side of square base of the box, y = height of the box.

Since their volumes are equal: [pi]R2H = x2y . . . [1]

Since their lateral areas are equal: 2[pi]RH = 4xy . . . [2]

Divide [1] by [2]: ([pi]R2H) / (2[pi]RH) = (x2y) / (4xy) . . . which simplifies to: 2R = x

Since 2R is the diameter of the circular base and x is the side of the square base,

they are equal . . . their ratio is 1:1.

[/hide]


A line that passes through $ (\minus{}5, \minus{}8)$ and $ (\minus{}3, \minus{}4)$ will cross the $ y$-axis at what $ y$-coordinate?

Slope- -4+8/-3+5=2 Slope intercept form: y+8=2(x+5) When x=0, y+8=10 so 2.

[hide=Sol]Cross the y axis at x = 0.
For every 1 increase in x, there is a 2 increase in y.
Thus, the answer is 3*2+-4 = 2[/hide]


inequalities
limit
algebra unsolved
algebra

1.Let f(x)=x-e^(-x). Prove equation f(x)=0 have a unique root x=a satisfaction  :1/e <a<1
2. Find m max satisfaction inequality: [1+1/n]^(n+m) < 1/e with n positive integer

1) $ f(x) \equal{} x \minus{} e^{ \minus{} x}$
$ f'(x) \equal{} 1 \plus{} e^{ \minus{} x}$  This is always positive, so $ f(x)$ is always increasing.

To show there is a $ \text{root }a$ such that $ \frac {1}{e} < a < 1$, just test the end points.  So $ f(\frac {1}{e}) \equal{} \frac {1}{e} \minus{} e^{ \minus{} \frac {1}{e}} < 0$
and $ f(1) \equal{} 1 \minus{} e^{ \minus{} 1} > 0$
Therefore with the increasingness and the change in signs between the bounds, I guess by the Intermediate Value Theorem, there is a root.

2) I will leave this one up to you. But as a hint notice the definition of
$ e \equal{} \lim_{n\rightarrow\infty} (1 \plus{} \frac {1}{n})^n$ anywhere? Might help.

Edit: Btw, to show the inequalities in 1)$ e^{\minus{}1} \equal{} \frac{1}{e} < 1$ since $ e>1$
and for the other, $ \frac{1}{e} \minus{} e^{\minus{}\frac{1}{e}} \equal{} \frac{1}{e} \minus{} (\frac{1}{e})^{\frac{1}{e}}$ and since $ 0< \frac{1}{e} <1$,
taking any root will make it closer to 1, thus $ )\frac{1}{e})^{\frac{1}{e}} > \frac{1}{e}$ therefore the difference is negative.


floor function
modular arithmetic
function
number theory
number theory proposed

Let $ a>1$ be a positive integer. Prove that every non-zero positive integer $ N$ has a multiple in the sequence $ (a_n)_{n\ge1}$, $ a_n\equal{}\left\lfloor\frac{a^n}n\right\rfloor$.

Strange problem. Let $ N\equal{}N_1N_2$, where $ N_2$ and $ a$ are coprime, and $ N_1$ divides $ a^C$ for suitable positive integer $ C$. Without loss of generality, $ CN_1$ divides $ a^{C}$ (we may take $ C$ being large power of $ a$). Take large prime $ p$ such that $ p\minus{}1$ is divisible by $ \varphi(n)$. The existence of such $ p$ is well-known elementary case of the Dirichlet theorem, which may be obtained using the cyclotomic polynomials. Then take $ n\equal{}pC$. Then by Fermat's little theorem we have $ [a^{Cp}/(Cp)]\equal{}(a^{Cp}\minus{}a^C)/(Cp)$, which is divisible by $ N\equal{}N_1N_2$.

We can solve this problem without the Dirichlet theorem.

There are positive integers $m$ and $M$ such that $GCD(m,M)=1, N\mid a^{m}M$.
We consider $M>1$(it is obvious when $M=1$)
Let $R$ is an integer such that $M^{R}>a^{m+\varphi(M)}$ and
$a^{R} mod \varphi(M), a^{R+1} mod \varphi(M), a^{R+2} mod \varphi(M), ...$
is a periodical sequence(the period is not more than $\varphi(M)$).
Then there exists an integer $0\le i\le\varphi(M)$ and
$a^{a^{R}M^{R}-m-i}\equiv a^{R}\pmod{\varphi(M)}$.
Let $k=a^{R}M^{R}-m-i$ and $n=a^{k}M^{R}$, then

$
\varphi(M^{R+1})=M^{R}\varphi(M)\mid n-k-m-i\\
\Rightarrow a^{n-k-m-i}\equiv 1\pmod{M^{R+1}}\\
\Rightarrow a^{n-k}\equiv a^{m+i}\pmod{a^{m}M^{R+1}}\\
\Rightarrow \lfloor\dfrac{a^{n-k}}{M^{R}}\rfloor=\lfloor\dfrac{a^{n}}{n}\rfloor\\
=\lfloor\dfrac{Ia^{m}M^{R+1}+a^{m+i}}{M^{R}}\rfloor (I\in\mathbb{Z})\\
=Ia^{m}M+\lfloor\dfrac{a^{m+i}}{M^{R}}\rfloor=Ia^{m}M   ( a^{m+i}\le a^{m+\varphi(M)}<M^{R})\\
\Rightarrow N\mid \lfloor\dfrac{a^{n}}{n}\rfloor\\
$

This solution was found in collaboration with [b]proglote.[/b] it is quite similar to [b]Fedor Petrov's[/b] solution actually.

Write $N = bc$ where $\gcd(c,a) = \gcd(b,c) = 1$. Let $b|a^k$ and $k$ clearly exists since $b$'s prime factors are purely those of $a$ as well. Then consider $n = a^kp$ for a prime $p \equiv 1 \pmod{\text{ord}_c(a)}$ and $p > \max(c,a^k - k)$ which exists by Dirichlet. It is easy to see that: \[\frac{a^{a^kp}}{a^kp} = \frac{a^{a^kp-k}}{p}\]
Then remark that $a^{a^kp-k} \equiv a^{a^k-k} \pmod{p}$, so it follows \[a_{a^kp} = \frac{a^{a^kp - k} - a^{a^k-k}}{p}\]

It is clear that $a^{a^k(p-1)}-1$ divides the numerator. It follows that as $c|(a^{a^k(p-1)}-1)$ due to the order condition and $\gcd(c,p) = 1$ due to $p > c$ that $c$ divides $a_{a^kp}$. It is also clear that $b$ must divide $a_{a^kp}$ as $a^{a^k-k}$ divides the numerator. Thus we are done as we have shown an arbitrary $N$ divides some term of the sequence.

I solved it pretty ugly, no Dirichlet. Sorry I used different values. I want $k | floor(a^r/r)$

Instead of $r$ I used $ra^x$, and then I define $P$ very big and then $Q$ such that $ord_Q a=P$ and then $r$ such that $ord_r a= Q$ using two Zsigmondys.

Then I can control $x$ in mod $d=ord_{k_1} a$ (where $k_1$ is biggest divisor of $k$ coprime to $a$), mod $ORD_d a$ (where $ORD$ is like the $ord$ function but without caring about $a$ and the modulus coprime), mod $P$ and mod $Q$ and use Chinese Theorem to finish, choosing $a^{ra^x-x} \equiv a^{c+i}$ (mod $kr$), where $c$ is such that $g | a^c$ ($g=k/k_1$) and $i<d$).

$i$ is useful since $ORD_da$ and $d$ aren't necessarily coprime.

Probably i am mistaken or this is on the easier end of RMM. Indeed let $p$ be a prime such that $p>a$ and $p$ is congruent to 1 modulo $\phi(N)$.(Clearly by the properties of cyclotomic polynomials infinitely many such $p$ exist or just use Weak Dirichlet's theorem) Now clearly $N$ divides $a_p$

[hide=Solution]Note that $a^p \equiv a \pmod p$, so for all primes $p>a$, $\lfloor \frac{a^p}{n} \rfloor = \frac{a^p-a}{n} $. Now, as long $p$ is coprime to $N$ and $p \equiv 1 \pmod \phi{N}$, we have a solution. By Dirichlet's theorem, we can always find such a $p$.[/hide]

I'll just post this cuz it's different from the above solutions.
(Actually, #6 and #7 are really nice, but proving that there are infinitely many primes in the $nk+1$ takes a lot of time.)

My main idea is [hide='Power Towers']
Define a sequence $(b_n)_{n=1}^{\infty}$
$$b_1=a, ~ b_{n+1}=b^{b_n}~(n \ge 1)$$
, where $a$ is a positive integer. It is well known that for each $k$, there exists $N_k$ such that
$$n>N_k \Longrightarrow b_n\equiv b_{n-1}~(\text{mod }k)$$
[/hide].

We'll prove that $n=b_m-1$ works for sufficiently large $m>2$. One can easily check that
$$a_n=\left\lfloor\frac{a^n}n\right\rfloor=\left\lfloor\frac{a^{a^{b_{m-1}}-1}}{a^{a^{b_{m-2}}}-1}\right\rfloor=\frac{a^{a^{b_{m-1}}-1}-a^{a^{b_{m-2}}-1}}{a^{a^{b_{m-2}}}-1}=a^{b_{m-1}-1}\cdot\frac{a^{b_m-b_{m-1}}-1}{a^{b_{m-1}}-1}$$
Now let $N=N_1N_2$, where $N_2$ is the largest divisor of $N_2$ such that $(a,N_2)=1$. We'll choose $m$ large enough so that $N_1|a^{b_{m-1}-1}$. Observe that for each $p|N_2$, the sequence $v_p(a^{b_m}-1)$ is bounded, so let $x_p$ be the upper bound. Using the well-known property of power-towers, find some $M$ such that
$$m>M\Longrightarrow \phi(N_2\cdot\prod_{p|N_2}p^{x_p})|{b_m-b_{m-1}}$$, and we're done. :)

Same as above solutions, posting for storage.
Solution. Take a prime $p$, such that $p>a$ (1) and $p=k.\phi{N}$ (*) and $(p,N)=1$ (**)(existence is guaranteed by Dirichlet`s theorem). By FLT, $a^p-a$ is divisible by $p$ (2).From (1) and (2), we have that $a_p=\lfloor\frac{a^p}{p}\rfloor=\frac {a^p-a}{p}$. Using (*), (**) and Euler`s theorem, we immediately see that  $a_p$ is divisible by $N$, which is what we need.  $\blacksquare $


[hide="Answer"][hide="Synthesis I"]1) Aniline + DMDO -----> Nitrobenzene

2) 2 Nitrobenzene + sodium arsenite -----> Azoxybenzene[/hide]

[hide="Synthesis II"]1) Aniline + Caro's acid -----> Nitrosobenzene

2) Nitrosobenzene + Aniline -----> Azobenzene

3) Azobenzene + Peracetic acid -----> Azoxybenzene[/hide]
[/hide]

Ok abt synthesis II
but what is DMDO????


function
vector
real analysis
real analysis unsolved

Let I be a real interval. Is there an $ f: I\rightarrow\mathbf{R}$ function such that the following properties hold?
(1) There is a $ 0<p<1$, $ p\neq 1/2$ real number, that for all $ x,y\in I$: $ f(x)\equal{}f(y)$ or $ f(px\plus{}(1\minus{}p)y)\equal{}f((1\minus{}p)x\plus{}py)$.
(2) $ f$ takes uncountably many values.

Cute problem :).

Let $ F$ be the subfield of $ \mathbb R$ generated by $ 1$ and $ p$. The real line can be viewed as the continuum-dimensional vector space over $ F$. Choose some basis $ X\subset I$ of this space and consider the subspaces $ S_x\equal{}\text{span\,}\{y\in X, y\le x\}$, $ x\in X$. Now define $ f(z)$ to be the set of all $ x\in X$ such that $ z\in S_x$.

Now let's see if we can avoid the axiom of choice here. Can we prove that there is no such function that is Lebesgue measurable? ;)

Thank you, beautiful solution!

And is there an $ f: I\rightarrow \mathbf{R}$ such that (1) holds and $ f$ takes countably infinite many values?


superior algebra
superior algebra unsolved

Show that the center $ Z$ of the general linear group $ GL_2 (\mathbb{C})$ consists of all scalar matrices. Identify the center of the special linear group $ SL_2(\mathbb{C})$.

[quote="JRav"]This question has been posted on the forums several times before.  Multiply by elementary matrices to show that all matrices in the center must be diagonal, and then conclude that all of them must be the same (since the order can be reversed).[/quote]

My bad.. but why do we multiply only by elementary matrices? Do elementary matrices refer to permutation matrices?

Multiplying by elementary matrices corresponds to elementary row and column operations on matrices (on the left is row and the right is column).


vector
function
linear algebra
matrix
trigonometry
limit
absolute value

let $V$ be the real vector space of functions
$f:E_p*E_q -> R$ where $E_k={1,2,..,k}$.
Let $L$ be the endomorphism of V defined by:
$L(f)(x,y)=\frac{f(x-1,y)+f(x+1,y)+f(x,y-1)+f(x,y+1)}{4}$.
(we suppose that $f(0,y)=f(y,0)=f(p+1,y)=f(x,q+1)=0$)
Find the eigenvalues and eigenvectors of $L$.

One simple question: do you really manage to solve alone all these problems you post? Boy, you really are incredible! I passed quite a long time trying the problem with the exponential matrix and nothing. And now, this one.  :(

of course not ! there are often hints with pb . But i prefer not to post hints to see other solutions, that are often more natural than those proposed !

I'm released, for some moments I really thought I will never have ENS being like this.  :blush:

In this case, it helps a great deal if you understand where the problem is coming from, look at $L$ as an approximation to a partial differential operator, and and make a reasonable guess as to the solutions.

If alekk will forgive me for this, I'd like to tweak the notation slightly. Replace his $p+1$ by $p$ and his $q+1$ by $q.$ Think of $f(x,y)$ being defined for $0\le x\le p$ and $0\le y \le q,$ except that $f(x,y)=0$ when $x=0$ or $x=p$ or $y=0$ or $y=q.$ Then $L$ is a linear mapping from a real vector space of dimension $(p-1)(q-1)$ into itself.

Assume a form for the eigenvectors. Assume that for $1\le j \le p-1,\,1\le k\le q-1,$ we have

$g_{j,k}(x,y)=\sin\left(\frac{j\pi x}p\right)\sin\left(\frac{k\pi y}q\right).$

This satisfies the boundary conditions, and it follows by direct calculation that

$L(g_{j,k})=\left[\frac12\cos\left(\frac{j\pi}p\right)+ \frac12\cos\left(\frac{k\pi}q\right)\right]g_{j,k}.$

The eigenvalues may not all be distinct, but it shouldn't be too hard to show that this set of eigenvectors is independent. Since there are the required $(p-1)(q-1)$ of them, they diagonalize the operator. Actually, $L$ is self-adjoint and this set of eigenvectors is orthogonal.

Note also that all of the eigenvalues have absolute value less than 1. Hence, for all $u$,

$\lim_{n\to\infty}L^n(u)=0.$


Are the flags going to be put again or not?

Thanks.

And what about gender? Will it appear any time soon? :maybe:


if (1/a+1/b+1/c+1/d+1/e) is an integer.Find a,b,c,d,e.

Probably under the condition of $a,b,c,d,e$ being (positive) integers ;)

Here is my reasoning for three integers:
WLOG a<=b<=c
then a<=3
1/a +1/b +1/c= k=  1 or 2 or 3
k=3 -> (1,1,1)
k=2 -> (1,2,2)
k=1 ->
a=2
1/b + 1/c = ½ -> c= 1/ (1/2- 1/b) -> c= 2 b / (b-2)
      b=2 no solution
      b=3 (2,3,6)
      b=4 (2,4,4)
      b>=5 -> c <= 10 / 8 no solution
a=3 -> (3,3,3)

You can do the similar for five integers.

Except five integers requires a LOT more casework. For example, $(2,3,7,43,1806)$ is a solution :o


A positive integer $ d$ is said to be strictly ascending if in its decimal representation $ d_{m}d_{m-1}\ldots d_{2}d_{1}$ we have $ 0<d_{m}<d_{m-1}<\ldots<d_{2}<d_{1}$. For instance, 146 and 23456 are strictly ascending integers. Find the number of strictly ascending integers which are less than (a) $ 10^{9}$ (b) $ 10^{5}$.

[hide="just a hint"]
not that for every number $ n$ with distinct digits,we can get a number $ f(n)$ which the digits of $ f(n)$ are strickly increasing...
and also for every number $ d$ with strickly increasing digits,there are $ k_{d}$ numbers which are maped to $ d$ (for example for number $ n=x_{1}x_{2}...x_{t}$ where $ x_{i}\not=0$ and for every $ i<j: x_{i}<x_{j}$ we have $ k_{n}=t!$)...
now the rest is some calculations...  :wink: [/hide]


Let a,b,c > 0.Prove that
 $ \frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a}\plus{}1\geq\ 4\sqrt{\frac{a^2\plus{}b^2\plus{}c^2}{ab\plus{}bc\plus{}ca}}$

The inequality sholud be $ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \plus{} 1\geq\ 4\sqrt{\frac{ab\plus{}bc\plus{}ac}{a^2\plus{}b^2\plus{}c^2}}$
Using [b]AM-GM[/b] we have:
$ \sum_{cyc} \frac {a}{b}\geq 3\sqrt {\frac {a}{b}.\frac {b}{c}.\frac {c}{a}} \equal{} 1\Rightarrow \sum_{cyc} \frac {a}{b} \plus{} 1 \geq4$
  It's sufficient to prove that $ \sum_{cyc} a^2 \geq \sum_{cyc} ab \Rightarrow \sum_{cyc} (a \minus{} b)^2 \geq 0$ and we are done!
  Equality holds if and only if $ a \equal{} b \equal{} c$

[b]Lemma:[/b] Let $ a \plus{} b \plus{} c \equal{} 1$ and $ t \equal{} \sqrt {\frac {\sum{(a \minus{} b)^2}}{\sum(ab)}}\ge 0$, then
\[ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}\geq \frac {2 \plus{} (1 \plus{} t)^3}{1 \plus{} t}
\]

[quote="vanhoadh"]Lemma: Let $ a \plus{} b \plus{} c \equal{} 1\le \frac {1}{3}$ and $ t \equal{} \sqrt {\frac {\sum{(a \minus{} b)^2}}{\sum(ab)}}\ge 0$, then
$ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}\geq \frac {2 \plus{} (1 \plus{} t)^3}{1 \plus{} t}$
[/quote] Since when $ 1\le \frac{1}{3}$ ?

[quote="broniran"]The inequality sholud be $ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \plus{} 1\geq\ 4\sqrt {\frac {ab \plus{} bc \plus{} ac}{a^2 \plus{} b^2 \plus{} c^2}}$
Using [b]AM-GM[/b] we have:
$ \sum_{cyc} \frac {a}{b}\geq 3\sqrt {\frac {a}{b}.\frac {b}{c}.\frac {c}{a}} \equal{} 1\Rightarrow \sum_{cyc} \frac {a}{b} \plus{} 1 \geq4$
  It's sufficient to prove that $ \sum_{cyc} a^2 \geq \sum_{cyc} ab \Rightarrow \sum_{cyc} (a \minus{} b)^2 \geq 0$ and we are done!
  Equality holds if and only if $ a \equal{} b \equal{} c$[/quote]
Are you kidding? :rotfl:
let's use your method on smth else,for example let's prove that $ ab \plus{} bc \plus{} ca\geq a^2 \plus{} b^2 \plus{} c^2$:
$ ab \plus{} bc \plus{} ca\geq 3\sqrt {a^2b^2c^2}$ and $ a^2 \plus{} b^2 \plus{} c^2\geq 3\sqrt {a^2b^2c^2}$.
Q.E.D
But as we know $ a^2 \plus{} b^2 \plus{} c^2\geq ab \plus{} bc \plus{} ca$.So where did we make a mistake? :huh:
To PHONGVAN:
Very meaningful title,congratulations! :rotfl:

[quote="broniran"][quote="vanhoadh"]Lemma: Let $ a \plus{} b \plus{} c \equal{} 1\le \frac {1}{3}$ and $ t \equal{} \sqrt {\frac {\sum{(a \minus{} b)^2}}{\sum(ab)}}\ge 0$, then
$ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}\geq \frac {2 \plus{} (1 \plus{} t)^3}{1 \plus{} t}$
[/quote] Since when $ 1\le \frac {1}{3}$ ?[/quote]

Sorry, At first I want to write $ a \plus{} b \plus{} c \equal{} 1 \Rightarrow ab\plus{}bc\plus{}ca \le \frac {1}{3}$, but then I deleted.

[quote="Erken"][quote="broniran"]Can you explain what is wrong with mine solution? I prove that Left side is always $ \geq 4$ and the Right side is $ \le 4$  :blush:[/quote]
You've proved another inequality,not the one we need to prove...[/quote] I've written it  wrong in my notebook. Sorry :blush:

[quote="arqady"][quote="PHONGVAN"]Let a,b,c > 0.Prove that
 $ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \plus{} 1\geq\ 4\sqrt {\frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca}}$[/quote]
Why not
Let $ a,b,c > 0.$ Prove that
 $ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \plus{} 2\geq\ 5\sqrt {\frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca}}$ $ ?$  :wink:[/quote]
 It's wrong. :(
try $ a\equal{}2 ,b\equal{}\frac{3}{2} , c\equal{}6$

[quote="broniran"]
  It's sufficient to prove that $ \sum_{cyc} a^2 \geq \sum_{cyc} ab$ [/quote]

 It's suffices to prove that $ \sum {a^2}\le \sum {ab}$ :!:

Sorry, the lemma I showed is wrong (there were some erros in my calculation), here is the right one:

Let $ a\plus{}b\plus{}c\equal{}1$ and $ t \equal{} \sqrt {\frac {\sum{(a \minus{} b)^2}}{\sum(ab)}}\ge 0$, then
\[ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}\geq \frac{t^5\minus{}6t^3\plus{}6\sqrt{2}t^2\plus{}36\sqrt{2}}{2(t^3\minus{}2\sqrt{2}t^2\plus{}6\sqrt{2})}\]

And the largest value of k which the inequality still true is approximately 1.701132746583...

[quote="PHONGVAN"][quote="arqady"][quote="PHONGVAN"]Let a,b,c > 0.Prove that
 $ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \plus{} 1\geq\ 4\sqrt {\frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca}}$[/quote]
Why not
Let $ a,b,c > 0.$ Prove that
 $ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \plus{} 2\geq\ 5\sqrt {\frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca}}$ $ ?$  :wink:[/quote]
 It's wrong. :(
try $ a \equal{} 2 ,b \equal{} \frac {3}{2} , c \equal{} 6$[/quote]
Thanks!
If so here is my ugly proof:
Let $ a \plus{} b \plus{} c \equal{} 3u,$ $ ab \plus{} ac \plus{} bc \equal{} 3v^2,$ $ abc \equal{} w^3$ and $ u^2 \equal{} tv^2.$
Then $ t\geq1$ and $ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \plus{} 1\geq\ 4\sqrt {\frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca}}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(a^2b \plus{} a^2c) \plus{} 2abc\geq8abc\sqrt {\frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca}} \plus{} \sum_{cyc}(a^2b \minus{} a^2c)\Leftrightarrow$
$ \Leftrightarrow9uv^2 \minus{} w^3\geq8w^3\sqrt {3t \minus{} 2} \plus{} (a \minus{} b)(a \minus{} c)(b \minus{} c).$
Easy to check that $ 9uv^2 \minus{} w^3\geq8w^3\sqrt {3t \minus{} 2}$ is true. Hence, it remains to prove that
$ 9uv^2 \minus{} w^3\geq8w^3\sqrt {3t \minus{} 2} \plus{} \sqrt {(a \minus{} b)^2(a \minus{} c)^2(b \minus{} c)^2}.$
But $ 9uv^2 \minus{} w^3\geq8w^3\sqrt {3t \minus{} 2} \plus{} \sqrt {(a \minus{} b)^2(a \minus{} c)^2(b \minus{} c)^2}\Leftrightarrow$
$ \Leftrightarrow9uv^2 \minus{} w^3\geq8w^3\sqrt {3t \minus{} 2} \plus{} \sqrt {27(3u^2v^4 \plus{} 6uv^2w^3 \minus{} 4u^3w^3 \minus{} 4v^6 \minus{} w^6)}\Leftrightarrow$
$ \Leftrightarrow\left(9uv^2 \minus{} (1 \plus{} 8\sqrt {3t \minus{} 2})w^3\right)^2\geq27(3u^2v^4 \plus{} 6uv^2w^3 \minus{} 4u^3w^3 \minus{} 4v^6 \minus{} w^6)\Leftrightarrow$
$ \Leftrightarrow(48t \minus{} 25 \plus{} 4\sqrt {3t \minus{} 2})w^6 \minus{} 9\sqrt t(5 \minus{} 3t \plus{} 4\sqrt {3t \minus{} 2})v^3w^3 \plus{} 27v^6\geq0.$
For $ t\geq1$ we obtain: $ 5 \minus{} 3t \plus{} 4\sqrt {3t \minus{} 2}\leq0\Leftrightarrow t\geq\frac {13 \plus{} \sqrt {112}}{3}.$
Id est, for $ t\geq\frac {13 \plus{} \sqrt {112}}{3}$ your inequality is proved.
Let $ 1\leq t < \frac {13 \plus{} \sqrt {112}}{3}.$ Hence, it remains to prove that:
$ 3t(5 \minus{} 3t \plus{} 4\sqrt {3t \minus{} 2})^2 \minus{} 4(48t \minus{} 25 \plus{} 4\sqrt {3t \minus{} 2})\leq0,$ which is easy to sheck.
What is your proof, PHONGVAN?

The following inequality easier to prove.
Let $ a,$ $ b$ and $ c$ are positive numbers. Prove that:
\[ 2\left(\frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}\right) \plus{} 3\geq\ 9\sqrt{\frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca}}\]

[quote="broniran"]
Using [b]AM-GM[/b] we have:
$ \sum_{cyc} \frac {a}{b}\geq 3\sqrt {\frac {a}{b}.\frac {b}{c}.\frac {c}{a}} \equal{} 1\Rightarrow \sum_{cyc} \frac {a}{b} \plus{} 1 \geq4$[/quote]
[b]Broniran[/b], since when $ 3\sqrt {\frac {a}{b}.\frac {b}{c}.\frac {c}{a}} \equal{} 1$  :wink:  :noo:

[quote="Math pro"]raelly ugly  :D[/quote]
Prove better! :P
[quote="vanhoadh"]
And the largest value of k which the inequality still true is approximately 1.701132746583...[/quote]
I proved this inequality for $ k\equal{}1.8.$ :maybe:

Can you post it? :roll:

Edit:
Ah i see the problem  :blush:
My CAS have some troubles with very small value calculating...
k should be somewhere between 1.7 and 1.82...

[quote="PHONGVAN"]Let a,b,c > 0.Prove that
 $ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \plus{} 1\geq\ 4\sqrt {\frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca}}$[/quote]
Nice inequality, Thach. :)
Here is the lemma I just used to prove this inequality :)
Assume that $ a\plus{}b\plus{}c\equal{}1,q\equal{}ab\plus{}bc\plus{}ca,r\equal{}abc$, then we have
\[ \frac{a}{b}\plus{}\frac{b}{c}\plus{}\frac{c}{a} \ge \frac{1\plus{}q\minus{}6q^2}{q(1\minus{}q)}\]
Now, using this inequality, it suffices for us to prove that
\[ \frac{1\plus{}q\minus{}6q^2}{q(1\minus{}q)} \plus{}1 \ge 4\sqrt{\frac{1\minus{}2q}{q}}\]
which is just equivalent to
\[ (1\minus{}3q)^4 \ge 0\]
it is trivial and we have done. :)
There is nothing special at this solution, but I want us to discuss how to figure out the above lemma. Was it a magic? :)
Regards

[quote]There is nothing special at this solution, but I want us to discuss how to figure out the above lemma. Was it a magic?[/quote]
Really? I think it is very special  (so nice and so strange) :) I will think about this.

Can anyone tell me how \[ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \ge \frac {1 \plus{} q \minus{} 6q^2}{q(1 \minus{} q)}\]

[quote="can_hang2007"]
Here is the lemma I just used to prove this inequality :)
Assume that $ a \plus{} b \plus{} c \equal{} 1,q \equal{} ab \plus{} bc \plus{} ca,r \equal{} abc$, then we have
\[ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \ge \frac {1 \plus{} q \minus{} 6q^2}{q(1 \minus{} q)}\][/quote]
good lemma

[quote="ye109"][quote="can_hang2007"]
Here is the lemma I just used to prove this inequality :)
Assume that $ a \plus{} b \plus{} c \equal{} 1,q \equal{} ab \plus{} bc \plus{} ca,r \equal{} abc$, then we have
\[ \frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a} \ge \frac {1 \plus{} q \minus{} 6q^2}{q(1 \minus{} q)}\]
[/quote]
good lemma[/quote]
I agree with you, but it does not help for the following
$ 2\left(\frac {a}{b} \plus{} \frac {b}{c} \plus{} \frac {c}{a}\right) \plus{} 3\geq\ 9\sqrt {\frac {a^2 \plus{} b^2 \plus{} c^2}{ab \plus{} bc \plus{} ca}}.$ :wink:

My solution is based on two known inequalities:
(1)   ${ a,b,c>0\Rightarrow\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge\sqrt{15\frac{a^2+b^2+c^2}{ab+bc+ca}-6}} $
 see the book "567 Nice and hard inequalities ",pb. 547
and
(2)   $ a,b,c>0\Rightarrow\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge\frac{(a+b+c)^2}{ab+bc+ca} $
The last inequality is an immediate consequence of inequality Cauchy-Scwarz.
We note $ t=\sqrt{\frac{a^2+b^2+c^2}{ab+bc+ca}},t\ge 1 $
From
$ \frac{(a+b+c)^2}{ab+bc+ca}\ge 4\sqrt\frac{a^2+b^2+c^2}{ab+bc+ca}-1\Leftrightarrow $
$ \Leftrightarrow t^2+2\ge 4t-1 $
result that demonstrated  inequality is true if $ t\ge 3 $
From
${ \sqrt{15\frac{a^2+b^2+c^2}{ab+bc+ca}-6}}\ge\sqrt{4\frac{a^2+b^2+c^2}{ab+bc+ca}}-1\Leftrightarrow $
 $ \Leftrightarrow\sqrt{15t^2-6}\ge 4t-1\Leftrightarrow t^2-8t+7\le 0 $
result  that demonstrated inequality is true if $ 1\le t\le 7 $
In conclusion, the inequality $ \frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge 4\sqrt{\frac{a^2+b^2+c^2}{ab+bc+ca}}-1 $ is true for any $ t\ge 1 $
_____
Marin Sandu


Let $ a_1,a_2,...,a_n,x_1,x_2,...,x_n$ are positive real number such that :
  $ \sum_{k=1}^{n}{\frac{a_k}{a_k+x_k}=1}$
Prove that : $ \prod_{k=1}^{n}{x_k} \geq (n-1)^n\prod_{k=1}^{n}{a_k}$

put that $ A_i\equal{}\frac{a_i}{a_i\plus{}x_i}$  then $ \sum_{k\equal{}1}^n A_k\equal{}1$ and $ x_i\equal{}\frac{a_i(1\minus{}A_i)}{A_i}$
then $ \prod_{k\equal{}1}^n x_k \equal{} \prod_{k\equal{}1}^n a_k \times \prod_{k\equal{}1}^n \frac{1\minus{}A_i}{A_i}$
we have by AM-GM $ 1\minus{}A_i \geq (n\minus{}1)\sqrt[n\minus{}1]{\frac{\prod_{k\equal{}1}^n A_k}{A_i}}$
then $ \prod_{k\equal{}1}^n (1\minus{}A_k) \geq (n\minus{}1)^n \prod_{k\equal{}1}^n A_k$
from where we deduct that $ \prod_{k\equal{}1}^n x_k \geq (n\minus{}1)^n\prod_{k\equal{}1}^n a_k$
is my solution nice? rate it.

$ \sum_{k \equal{} 1}^n \frac {x_k}{(a_k \plus{} x_k)} \equal{} n \minus{} 1$  ===> $ \sum_{k \equal{} 1}^n \frac {1}{1 \plus{} \frac {a_k}{x_k}} \equal{} n \minus{} 1$

  by tchbechyv ==>$ \frac {n^2}{n \minus{} 1} \geq n \plus{} \sum_{k \equal{} 1}^n \frac {a_k}{x_k}$ ==>

$ (\frac {n}{n \minus{} 1} \geq \sum_{k \equal{} 1}^n\frac {a_k}{x_k}$  by IAG ==>

 $ (n \minus{} 1)^n \geq \frac {(\sum_{k \equal{} 1}^n \frac {a_k}{x_k})^n}{n^n} \geq prod_{k \equal{} 1}^n(\frac {a_k}{x_k})$

so $ \prod_{k \equal{} 1}^n x_k \geq (n \minus{} 1)^n\prod_{k \equal{} 1}^n a_k$


The matrix $ A$ is given by $ A\equal{} \begin {bmatrix} 2 & \minus{}1 & 1 \\ 0 & 3 & 1 \\ 1 & 1 & a \end {bmatrix}$, where $ a \neq 1$.

(i) Find $ A^{\minus{}1}$.

(ii) Hence, or otherwise, solve the equations

$ 2x\minus{}y\plus{}z\equal{}1,$

$ \    3y\plus{}z\equal{}2,$

$ x\plus{}y\plus{}az\equal{}2.$

I get the following:

(i) $ A^{\minus{}1} \equal{} \frac{1}{6(a\minus{}1)} \begin {bmatrix} 3a\minus{}1 & a\plus{}1 & \minus{}4 \\ 1 & 2a\minus{}1 & \minus{}2 \\ \minus{}3 & \minus{}3 & 6 \end {bmatrix}$.

(ii) $ x\equal{} \frac{5a\minus{}7}{6a\minus{}6}, y \equal{} \frac{4a\minus{}5}{6a\minus{}6}, z\equal{} \frac{3}{6a\minus{}6} \equal{} \frac{1}{2a\minus{}2}$.


we got some light matter and dark matter, about 1/10 light matter and 9/10 dark.  i agree that the light matter is speeding up, but for as much as light matter speeds up, dark matter will slow.  all the gravity pulling directly out is negated but the gravity pulling directly in.  but there's unbalanced forces for the diagonal forces, its hard to explain, but like a when a completely circlular outstretched rubber band is released it'll go in.

i dont know why the light matter would speed up though, hypothetically all the matter would slow proportionally to r squared.  but the dark matter has to slow down as the light matter speeds up.

any thoughts?

Who says the universe is going to behave like a rubber band?  Who says we know all the forces that exist?

Why would dark matter slow down as light matter speeds up?

i dont know how to explain the dark matter will slow down, there's newtons laws.  if there is a gravitational pull on the light matter out, than there will be a pull on the dark matter in.  also conservation of energy.

we dont know all the forces, i have to concede that, but anything out there that isnt fixed will be pulled in as the light matter is pulled out.

i'll abandon the rubber band idea.  if you look at only matter in the universe, than if there is more matter to one side of a body, than there's more gravitational pull to that side of the body.  so because the distribution of matter is most dense in the center and creshedos out, than that would cause unbalanced forces towards the center.  also if you look at the outskirts of the universe there is gravity pulling in but not so much pulling out.

i could put up a diagram of what i was trying to say about the diagonal forces if anyone would like.


Call a positive real number [i]special[/i] if it has a decimal representation that consists entirely of digits $0$ and $7$. For example, $700/99$ and $77.007$ are special numbers. What is the smallest $n$ such that $1$ can be written as the sum of $n$ special numbers?

Hint:

[hide]

Consider instead if you divided everything through by seven. What does that change the problem to instead?

[/hide]

Ahh, Thanks, so you get the repeating decimal 0.142857... and 8 is the largest of those so n is at least 8.


calculus
inequalities unsolved
inequalities

[color=darkblue]Let $ x,y>0$ and $ x\plus{}y \ge 6$. Find minimum value of $ P\equal{}5x\plus{}3y\plus{}\frac{12}{x}\plus{}\frac{16}{y}$.[/color]

[quote="thanhnam2902"][color=darkblue]Let $ x,y > 0$ and $ x \plus{} y \ge 6$. Find minimum value of $ P \equal{} 5x \plus{} 3y \plus{} \frac {12}{x} \plus{} \frac {16}{y}$.[/color][/quote]
We have : $ P\equal{}2(x\plus{}y)\plus{}(3x\plus{}\frac{12}{x})\plus{}(y\plus{}\frac{16}{y}) \ge 2.6\plus{}2.6\plus{}2.4\equal{}32$
So $ P_{min}\equal{}32$ when $ x\equal{}2,,y\equal{}4$

hello, with the help of calculus we can prove that $ 5x\plus{}3y\plus{}\frac{12}{x}\plus{}\frac{16}{y}\geq32$, the equal sign holds iff $ x\equal{}2$ and $ y\equal{}4$.

[quote="thanhnam2902"][color=darkblue]Let $ x,y > 0$ and $ x \plus{} y \ge 6$. Find minimum value of $ P \equal{} 5x \plus{} 3y \plus{} \frac {12}{x} \plus{} \frac {16}{y}$.[/color][/quote]

Eg: $ x\plus{}y \ge 4$, find the min of $ S\equal{}2x\plus{}3y\plus{}6/x\plus{}10/y$

P/S: $ x\plus{}y \ge 6$ is a special condition. Can we find the result of problem if $ x\plus{}y \ge a$  $ (a \ne 6)$! ;)


also, I was in an non-existing game the same time i was in the lobby

Meanwhile, in Canada, someone was walking.


A list of five positive integers has mean $12$ and range $18$. The mode and median are both $8$. How many different values are possible for the second largest element of the list?

$\textbf{(A)}\ 4\qquad
\textbf{(B)}\ 6 \qquad
\textbf{(C)}\ 8\qquad
\textbf{(D)}\ 10\qquad
\textbf{(E)}\ 12$

[hide="Answer"]The integers add to $60$ and there is a difference of $18$ between the largest and smallest. The mode is $8$, so we can have either two or three $8$'s. We cannot have four because the last number must be $28$ to keep the correct mean, but that would not keep the correct range.

With two $8$'s, we must have one as the median, so we have one number smaller and two larger, since two smaller would make the largest too large for the range. Taking $1$ as the smaller, the larger is $19$ and the last one must be $24$, which is then the largest, so that is impossible. We move up to $3$ as the smallest, and we now have $21$ as the largest, so the last number is $20$. We continue this and find $(4,18,22)$, $(5,16,23)$, $(6,14,24)$, and $(7,12,25)$, so there are five possibilities here.

Next, we move to three $8$'s. We can first have them as the three smallest, in which case we have the largest as $26$ and the remaining number as $10$. If the three $8$'s are in the middle, then the range will be too large. Therefore, we have $6$ possible values. $\boxed{B}$[/hide]

If [b]a[/b] is the first (smallest), the last one (bigger) will be [b]a[/b]+18, and assuming (at least) two [b]8[/b]'s, the sum should be

$5\cdot12\,=\,60\,=a+2\cdot8+b+a+18\,\,\,\Rightarrow \,\,2a+b\,=\,26\,\,$  (note that [b]b[/b] is even).

As $\,a\,<\,b\,<\,a+18\,\,\Rightarrow \,\,3\,\le\,a\,\,\le\,8\,\,$ that makes the answer [b]8 - 3 +1 = 6.[/b]


If $ 2x \plus{} y \equal{} 8$, what is the value of $ 4x \plus{} 2y$?

Simply multiply by 2 to get 4x+2y and get 16.


Let $ a,b,c \in R$. Prove that:
$ a^2\plus{}c^2\plus{}4ac\plus{}b^2\plus{}\frac{a^2c^2}{b^2} \ge 2(a\plus{}c)(b\plus{}\frac{ac}{b})$

The inequality is equivalent to $ (a\plus{}c)^2\plus{}\left(b\plus{}\displaystyle\frac{ac}{b}\right)^2\geq 2(a\plus{}c)\left(b\plus{}\frac{ac}{b}\right)$
or $ \displaystyle\left[(a\plus{}c)\minus{}\left(b\plus{}\frac{ac}{b}\right)\right]^2\geq 0$


[hide="Is it"]4. The probability that it will be a 6 is $\frac{1}{6}$ the first roll. The probability that it won't is $\frac{5}{6}$. The second time around, the probability that it will be a 6 is $\frac{1}{6}\frac{5}{6}=\frac{5}{36}$. The probability that we have a 6 by then is $\frac{5}{36}+\frac{1}{6}=\frac{11}{36}$ If we continue this, we find by the 4th roll, there's $\frac{671}{1296}$ which is greater than $\frac{1}{2}$. [/hide]

[quote="8parks11"]this is basically $6$.

$\frac{1}{p}$

$\frac{1}{\frac{1}{6}}= 6$

unless im not getting something in the problem[/quote]

Could you please explain your reasoning?

Thanks.

well
$\mu=1(p)+(2(pq)+3(pq^{2})+4(pq^{3})+...$
$=p(1+2q+3q^{2}+4q^{3}+...)$
$=p(\frac{1}{1-2q+q^{2})}$
$=p(\frac{1}{1-q^{2}}$
$=p(1/p^{2}) =1/p$

If $X$ is a geometric random variable with probability of success $p$ on each trial, then the mean, or expected value of the random variable, that is, the expected number of trials required to get the first success, is $\mu=\frac{1}{p}$

[quote="mathnerd314"]Mathnerd314 rolls a six-sided die repeatedly until a 6 comes up. What is the expected number of times he will have to roll the die?[/quote]
Intuition tells me 3, but intuition often fails. Look at cantor...

Followup: What is the expected number of times he has to roll to get every number at least once?

suppose x is the probability, then

$x=\frac{1}{6}+\frac{5}{6}*(x+1)$

since the first roll can be
1) a 6, where the game ends
2) something other than a 6, and the game resets, just there has already been one roll

so x=6

shouldnt this be

$geometcdf(\frac{1}{6},x)$

now, we can plug in values for x until we get a probability close enough to $1$ - which is around $26$

it doesn't really make sense - expected number he has to roll to get every number at least once : $6$?

im not sure.... it could just be $6$...

weak at probablities  :oops:


link: [url=http://www.artofproblemsolving.com/Wiki/index.php/2004_AMC_10A_Problems/Problem_20]clickie[/url]

why is EA=FC?

o dang i need to do intro to geo again! thank you so much!

EDIT: wait i don't get how HL congruence applies...

EDIT2: Nvm...

BE=BF since triangle BEF is equilateral. AB=BC since they are sides of the square. Angle A=Angle C since they're both right. Using HL congruency, we have triangles BAE and BCF are congruent, so EA=FC.

Wait... HL is the equivalent of SSA in a right triangle, right?

Yes. It works because of the fact that the A is a right angle.


percent
function
calculus
calculus computations

An automobile gets 28 miles per gallon of gasoline for speeds up to 50 miles per hour. Over 50 miles per hour, the number of miles per gallon drops at the rate of 12 percent for each 10 miles per hour.
(a) s is the speed and y is the number of miles per gallon. Find y as a function of by solving the differential equation
$ \frac{dy}{dx}\equal{}\minus{}0.012y, s>50$
(b) Use the function in part(a) to find y when s=50 and s=60.

$ \frac{dy}{ds} \equal{} \minus{}ky \to \frac{dy}{y}\equal{}kds \to ln |y| \equal{} ks\plus{}C' \to y\equal{}Ce^{ks}$

so when s>50, $ y\equal{}Ce^{\minus{}.012s}$,

and $ y(50)\equal{}28$ is given

so $ y\equal{}28e^{\minus{}.012(s\minus{}50)}$

and $ y(60)\equal{}28e^{\minus{}.12}$

Question: if fuel economy goes down by 12% every 10 mph, why is the claimed number $ y(60) \equal{} 28e^{ \minus{} .12}$ not 12% less than the number $ y(50) \equal{} 28$?


geometry
rectangle
calculus
integration
function
rotation
calculus computations

[hide="Click for problem"]

1. Let $R$ be the shaded region bounded by the graphs of $y = \sqrt{x}$ and $y = e^{-3x}$ and the vertical line $x = 1.$
(a) Find the area of $R$.
(b) Find the volume of the solid generated when $R$ is revolved about the horizontal line $y = 1$.
(c) The region $R$ is the base of a solid. For this solid, each cross section perpendicular to the x-axis is a rectangle whose height is 5 times the length of its base in region $R$. [/hide]

I got (a). That one was regular integral and piece of cake.  :D But I really don't understand how I use these washer, disc, and shell methods. I've taken test on this early this year but even back then, I didn't know what I was doing  :( . Now that AP's approaching, I need to know them well since many questions in Free Response use this idea.

Can anyone explain this to me? Like I can't even visualizing this whole "rotation."

Let's start with c, which should be a bit easier to visualize.

I may explain this a bit differently than you have learned it, but I find that this is a very effective (though not entirely precise) way to describe it.
[hide="c"]
You are creating infinitely thin boxes, each with width $\sqrt{x}-e^{-3x}$ (simply the distance between the two curves), height $5(\sqrt{x}-e^{-3x})$, and depth $dx$. So, adding together the volumes of the infinitely many boxes, or integrating their volumes, we find that the volume of the solid is

$\int_{.2387}^{1}(\sqrt{x}-e^{-3x})(5(\sqrt{x}-e^{-3x})) dx$.[/hide]

[hide="b"]
This one is probably hard to visualize. Instead of doing the given problem, we'll just revolve a simple region (between some function $y=f(x)$ and the x-axis) around the x-axis, and you should be able to generalize to other lines and regions pretty easily. I believe that this is known as the washer method:

Imagine that you're slicing up your region with vertical lines infinitely close together, and we'll call the width $dx$. Now, take one of those lines (ignoring parts outside of the region) and rotate it around the x-axis. You just made a cylinder with radius $f(x)$ and height $dx$. I hope you see why... So, the volume of that cylinder is $\pi (f(x))^{2}dx$. Now, adding together the infinitely many cylinders (integrating) shows that the volume of the whole solid is $\int_{a}^{b}\pi (f(x))^{2}dx$.

Again, the idea of adding up infinitely many infinitely thin objects is an oversimplification and very imprecise, but I found that it helped me understand the concepts more.

If you want to a region between two curves, simply do the process for both curves, and subtract appropriately.[/hide]


probability
geometry
rectangle
quadratics
circumcircle
Pythagorean Theorem
power of a point

I was ordered to post more problems from the Polya competition, so here it goes:

(From Polya 1996)

1.8  The five-legged Basafroz owns only white and black socks.  It wakes each morning, goes to its sock drawer, and pulls out five socks at random.  If the probability that all five socks are white is 1/2, what is the smallest possible number of socks in the drawer?    [i](Hint: The number of socks the Basafroz owns need not be a multiple of 5.) [/i]

2.6 Circle A and Circle B are concentric.  Circle A has radius 12; circle B has radius 16.  A rectangle is drawn, with two adjacent vertices one one circle and the other two vertices on the other circle.  Find the dimensions for a rectangle that maximizes the area.

3.3 Determine the smallest positive multiple of 17 that has only 1s in its decimal expansion.

3.6 A right triangle has a circumscribed circle of radius R and an inscribed circle of radius r.  Find the area of the triangle in terms of R and r.

Please include solutions.  Exceptions may be made for people who cannot spell, because it is painful to watch the English language being butchered.

1.8[hide] With w white socks and h sock total, the probability is [w!/(w-5)!]/[h!/(h-5)!]=1/2. Solve. The bottom part has 1 factor of 2 more than the top. It also contains the same factors as the top. The question is, what are 2 sets of 5 positive, consecutive integers such that the products of one are equal to twice the products of the other? One answer is that one of the sets starts with number n and the other ends with 2n. The other 4 numbers are shared. The first and last number are given by x+5=2x. x=5. The first sets that work are {5,6,7,8,9} and {6,7,8,9,10}. (Note how this telescopes). So, 9 whites and 10 in all. [/hide]

2.6 [hide]

If we were to maximize the area, then the two verticies on the smaller circle would be opposite each other and so one dimension would be 24 and by pythagorean theorem with 16 and 12 we get the other side to be 4rt7 so the dimensions are 24 by 4rt7[/hide]

Dark_Shadow wrote:2.6 [hide]
If we were to maximize the area, then the two verticies on the smaller circle would be opposite each other[/hide]

Can you prove this?  What about a longer, skinnier rectangle -- could such a rectangle have larger area?

Okay, so lets do this:  (not in spoiler, so don't read if you don't want)

2.6.  See the diagram below for points of reference.  We have rectangle ABCD, two circles O with radii 12 and 16, OG and OF perpendicular to AB and BC respectively, angle OBF equal to :theta:.  Then BF = 12cos :theta:, so BC = 24cos :theta:.  BG = 12sin :theta:, so BH = 24sin :theta:.

Use power of a point on A with the smaller circle: one one hand, use AHB, and on the other, use the line from A through O.  We get:
AH*AB = (AB - 24sin :theta:)*AB = 4*28 = 112.
24sin :theta: = AB - 112/AB.
Now we have 24sin :theta: and 24cos :theta:.  So square them and add to get
242 = BC :^2: + AB :^2: - 224 + (112/AB) :^2:.
800 = BC :^2: + AB :^2: + (112/AB) :^2:, or, multiplying through by AB :^2:, 800AB :^2: = (AB * BC) :^2:  + AB4 + 112 :^2: .  Now, notice that we have the expression (AB * BC), which is exactly what we want to solve for.
So (AB * BC) :^2: = -AB4 + 800AB :^2: - 112 :^2:.  This is a quadratic which has its maximum at AB :^2: = 400, or AB = 20.  At this value, BC = 96/5 and the area is equal to 384.

As an interesting note, at this particular instant, we also have that AO is perpendicular to BO.  I suspect that this is always the maximizing case, but I haven't done the work to be sure.


calculus
integration
function
complex analysis

Integrate $ f(z)\equal{}|z|^2$ over the polygon $ [1,1\plus{}i,i]$.

It seems I don't get zero. So I must be doing something stupid over and over again  :blush:

$ f(z) \equal{} |z|^2$ is not holomorphic.

[quote="Jacobson"]So I must be doing something stupid over and over again  :blush:[/quote] Yes -- expecting the integral of a positive real function to give you 0 :P

[quote="JBL"]Yes -- expecting the integral of a positive real function to give you 0 :P[/quote]

I don't think [i]this[/i] is the problem. Remember what integrating over a curve means; in an approximation, you multiply the integrand with a complex number (corresponding to the tangent direction to the curve at the point). The integrand being positive doesn't make the product positive (it is not even real). And $ g\left(z\right) \equal{} 2$ is also positive and holomorphic (so that the integral is zero)... It is $ f$ not being holomorphic that makes the integral nonzero (though it still could be zero by a random coincidence).

  darij

Yes, you're absolutely right ... I should go rate my own post a 1, if I could.  (It's obviously been too long since I've done any complex analysis.)


biologists are firly certain that oxygen was built up in the atmosphere by the invention of photosynthesis.The appereance of oxygen and therefore the beginning of extensive aerobic life was made possible by hte appereance of

fermentation
photosystem I
photosystem II
glycolisis
the nucleus

i have a clue but im not really sure about it...

what does "the nucleus " has to do with it??

i agree with other parts...

:arrow: The fhotosynthesis is a great work of the nature, beginning that is and has been a marivillosa delivers to the world, according to the scientists (and it is ratified for my :P ) thanks to this phenomenon the life has been achieved, now scientists of the FISHNET infestigan since there can realize photosynthesis by means of a machine, which to my to seem sera very difficultly.
Someone knows on this investigacióon, if it knows it make it come to the Community.
Américo Arone  ;)
Peru

photosystem I was present in cynobacteria, which produced the atomospheric oxygen necessary for aerobic organisms to exist. so I think that's the answer.

Photosystem I is associated with cyclic photophosphorylation. This evolved prior to noncyclic photophosphorylation involving both photosystems I and II. Some cyanobacteria are photosynthesizers and most of them utilize photosystem I, cyclic photophosphorylation.


[quote="Quevvy"]One of the terms of an expanded binomial is $78732x^{7}y^{2}$. The binomial is $(3x+y)^{z}$. What is $z$?[/quote]

[hide]Hint:
You don't even have to do any computation[/hide]
[hide]Just note that z is just the degree of any term, so it is 9

$9$[/hide]

[quote="mathgeniuse^ln(x)"]
[hide]Just note that z is just the degree of any term, so it is 9

$9$[/hide][/quote]
What does that mean, "the degree of any term"?

The degree of any term is just the sum of the exponents of unlike bases.

So that is why that is the answer.

[hide="I agree"] A term of a polynomial $(ax+by)^{z}$ is in form of

$(ax)^{w}(by)^{z-w}$.

So we sum 7 and 2 to get z:

9[/hide]


Show that, for each prime p, there exist positive integers k, x and y such that
(1) x2 + y2 = kp + 3
(2) 0 < 2k < p.

Hello, Arne.

Since there are exactly (p+1)/2 distinct residues of x2 (mod p) for any s equation x2+y2 = s (mod p) have solution with 0 \leq x,y \leq p-1.


Prove that there are no perfect squares in the form n2^k+2005.

Do you means $n*2^{k}+2005$?
but it is not true for $k\leq 2$
and very triavial when $k\geq 3$(just $mod 8$)

Yes, for k>2. How do you prove it using mod8?

[quote="Hawk Tiger"]You should post the problem in the pre-olympaid section.
A square can not $\equiv 5(mod 8)$[/quote]
I don't think this belongs in pre olympiad because it isn't that complicated  :huh: . Then again, maybe I don't know what I'm talking about.
Anyways, to extend on what hawk tiger said, when you square a number in mod 8 the only possible value in mod 8 is 1, 0, or 4. The reason is any number above 8 can be written as $8x+y$. When you square this and put that in mod 8, you find that this is equal to $y^{2}$. So, the value of any square in mod 8 is going to be equal to the remainders of the squares from 0-7. Squaring these we get $0^{2}=0$
$1^{2}=1$
$2^{2}=4$
$3^{2}=1$
$4^{2}=0$
$5^{2}=(-3)^{2}=1$
and so on, so the only squares are 0, 1, 4 in mod 8.


linear algebra
matrix
geometry
geometric transformation
rotation
reflection

Let A be an [b]orthogonal[/b] matrix. Show that all its eigenvalues satisfy:
l lambda l = 1

I won't attempt to formally prove this, but to see that it is true, think of the matrix as a linear transformation $T$. Then the eigenvalues satisfy: $Tv=\lambda v$, and since $T$ performs a rotation or reflection of $v$, the length of $v$ hasn't changed and thus $|\lambda|=1$. Of course this is generalizing and making some assumptions, so it may not make physical sense in some cases.

It does make sense if you go complex (think of a unitary matrix rather than just an orthogonal matrix). Besides, how do non-real eigenvalues make sense if you don't do things over $\mathbb{C}$?


group theory
abstract algebra
superior algebra
superior algebra unsolved

Let $ (G,\cdot )$ be a finite group.Prove that $ G$ is cyclic if and only if $ \forall d\in \mathbb{N^{*}}$, there is at most a subgroup $ H$ of $ G$ such that $ |H| \equal{}d$.

The interesting direction is showing that if there is at most one subgroup of order $ d$ for all $ d$ then $ G$ must be cyclic.

If $ x$ is an element of order $ d$, the subgroup $ \langle x \rangle$ has order $ d$ and is thus the only group of this order in $ G$. We then have exactly $ \varphi(d)$ elements of order $ d$ in $ G$, since they all belong to $ \langle x \rangle$ and they are all of the form $ x^k$ with $ (k,d)\equal{}1$. Thus the number $ o_d$ of elements of order $ d$ is either $ 0$ (when no such element $ x$ exists) or $ \varphi(d)$.

Now $ n \equal{} \sum_{d|n} o_d$, since every element in $ G$ has [i]some [/i]order $ d$ dividing $ n$. But $ n \equal{} \sum_{d|n} \varphi(d)$, so we necessarily have $ o_d \equal{} \varphi(d)$ for every $ d$ dividing $ n$. In particular, $ o_n \equal{} \varphi(n) > 0$, so we have elements of order $ n$ and the group is cyclic.


Prove that series \[\sum_{n=1}^{\infty}(-1)^{n}\frac{x^{2}+n}{n^{2}}\]
convegers uniformly in every bounded interval, but does not conveger absolutely for any value of $x$.

[quote="N.T.TUAN"]
 but does not conveger absolutely for any value of $x$.[/quote]

This one is easy,just separate it and see that your series is $>\sum_{k=1}^{\infty}\frac{1}{k}$ ,which diverges (harmonic series).

I know , but with first part, can you ? :D