calculus integration function induction factorial Euler Gauss

While investigating the values of certain integrals, I evaluated the following:

$ \int_{0}^{\infty}e^{-x}\, dx = 1$

$ \int_{0}^{\infty}e^{-\sqrt{x}}\, dx = 2$

$ \int_{0}^{\infty}e^{-\sqrt[3]{x}}\, dx = 6$

So, I was interested in finding the value of

$ \int_{0}^{\infty}e^{-\sqrt[n]{x}}\, dx \, \text{, where n}\geq 1 \, \text{is an integer.}$

I'm guessing the value is $ n!$, but I can't think of a way to evaluate this without using integration by parts, which I don't know how to do when an arbitrary constant $ n$ is involved.

Substitute $ x=t^{n}$; the integral becomes $ \int_{0}^{\infty}nt^{n-1}e^{-t}\,dt=n\Gamma(n)=\Gamma(n+1)$.
At least for positive integer values, the defining integral of the gamma function is easy to evaluate by integration by parts with induction. The value is indeed $ n!$.

Ah, I see.

I wouldn't know to use the Gamma function because I don't really know about it.

But, it's nice to know that my conjecture about the value being $ n!$ is correct. :)

There's nothing really complicated about the gamma function. The definition is $ \Gamma(s)=\int_{0}^{\infty}x^{s-1}e^{-x}\,dx$ for $ s>0$. We can easily calculate $ \Gamma(1)=1$, and $ \Gamma(s+1)=\left[-x^{s}e^{-x}\right]_{0}^{\infty}+\int_{0}^{\infty}sx^{s-1}e^{-x}\,dx=s\Gamma(s)$; by induction, $ \Gamma(n+1)=n!$ for $ n$ a nonnegative integer.

The gamma function is essentially an extension of the factorial to non-integer values. The shift by one exists for historical reasons, and I don't entirely understand them.

[quote="jmerry"]
The gamma function is essentially an extension of the factorial to non-integer values. The shift by one exists for historical reasons, and I don't entirely understand them.[/quote]
Really? Can you tell me what that reason is? I always wondered about that.

[quote="zakary"]I wouldn't know to use the Gamma function because I don't really know about it. [/quote]
The gamma function is one of my favorites. It really is useful, learn it! Follow jmerry's post.

And I want to add that $ \Gamma(s)$ converges for $ s>0$ and diverges for non-positive integers $ s$.

[quote="jmerry"]
The gamma function is essentially an extension of the factorial to [b]non-integer[/b] values. [/quote]

might i add fractional as well as complex numbers... essentially, it is
a [i]meromorphic [/i]function defined over the entire complex plane
$ \mathbb C \;\;\;\text{minus}\;\;\;z=-n,\;\;\;where\;\;\;n\in\mathbb N\cup \{\;0\;\}$

[quote="jmerry"]
The shift by one exists for historical reasons, and I don't entirely understand them.[/quote]

i *[i]think[/i]* the reason might have been euler was able to popularize it
more than gauss, even though legendre is credited for introducing it first
(euler certainly did a lot of work involving both $ \Gamma(z)$ and $ \gamma$, and much later,
a lot was done by weierstrass). but then again, euler came before gauss... :|
so perhaps that was why he was able to popularize it more...

i recall (don't know from where) gauss introduced something called the
[b]pi [/b]function or some name like that (or maybe it was called the [i]big[/i]-pi function
to differentiate it from the [i]little[/i]-pi function $ \pi(n)$ that then described
the distribution of prime numbers). gauss defined $ \Pi(z)=\Gamma(1+z)$,
and so by his definition, $ \Pi(n)=n!\;\;\; (\;\forall$ natural numbers
$ n\;\;\ni\;\;n\in\mathbb N\cup \{\;0\;\}\; )$

[quote="sylow_theory"]
And I want to add that $ \Gamma(s)$ converges for $ s>0$ and diverges for non-positive integers $ s$.[/quote]

that is correct. as a matter of fact, $ \Gamma(z)$ diverges for $ z\in\mathbb C\;\;\;\text{where}\;\;\;z=a+\imath b$
$ (\;\forall\;a=-n\;\;\ni\;n\in\mathbb N\cup \{\;0\;\},\;b=0\; )$

it converges [b]absolutely [/b]otherwise. :)

To clarify the talk about $ \Gamma$ on the complex plane:
The integral definition I gave works for $ \text{Re}(s)>0$, where the integral converges absolutely. We can then extend the function to the entire complex plane excluding the nonpositive integers using the functional equation $ \Gamma(s)=\frac1s\Gamma(s+1)$. This extension is analytic; viewed as a function on the whole plane, it is meromorphic with a simple pole at each nonpositive integer and no other poles.

Wow, that's some frickin' awesome stuff.

Unfortunately, I haven't taken the classes where stuff like this has been taught and/or would be necessary... the farthest math I've been able to take so far is AP Calculus BC... I'm looking forward to continuing my math in college!

Does anyone have a graph of the gamma function in the real plane?  the complex plane?

It's hard to graph something in the complex plane; there are really four dimensions there. On the positive real line, it would mostly just look very smooth and convex; $ \log\Gamma$ is a convex function.


I uh.. what? I dont understand the problem. Please clarify.

[quote="risinglegend"]Mr Lin buys an article and sells it to Mr Chen sells the article to Mr Ang at a gain of 20%. How much money did Mr Ang at a gain of 20%. How much money did Mr Lin pay for the article, if Mr Ang pays 360 dollars for it?[/quote]

I don't understand your second sentence! I think the answer to you problem is[hide="answer"] 432dollars because 20% of 360 plus 360= 432 Mr. And made 72 dollars[/hide]


Find all positive  integers  $n$ such that the number

$A=\frac{n}{1!}+\frac{n}{2!}+.....+\frac{n}{n!}$  is an integer number

It's also Slovak! We, Slovak and Czech republic have common olympiad (only different TSTs). Just this year we will have different national rounds...

To the problem: Just make common denominator (or how you call it...) and then think about modulo $n-1$ in nominator...


[quote="Kunohoshi Hitomi"]anyone going?[/quote]
nope. nobody is going. it will be canceled

but seriously, what kind of question is that? of course people are going. perhaps, you wanted to know which aopsers are going, then you might want to be more specific in the wording of your question.

Eastern coast = penn state

There's already a thread [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=80368]here[/url].

Why?
I find no sense in it, true, but nothing disagreeeable either.


Prove in a non-right triangle
$\tan \alpha+\tan \beta+\tan \gamma = \tan\alpha*\tan\beta*\tan\gamma$
where $\alpha, \beta, \gamma$ are the angles


The discussion of Firefox surely is not representative of the populace, is it?

I voted for Opera because of all the little things it offers (like mouse gestures; I love those :)), but I actually use all of them. I have one window of each open right now (Opera, IE,Firefox) :). Really.

Firefox.

Opera seems OK, but there's that annoying banner ad at the top you have to pay to get rid of...

Microsoft products annoy me.

Firefox is free, and good enough for me.

29 dollars is too small price for excellent program, moreover, I think, 29 dollars is small sum for almost all europeans, americans and canadians. Am I right?

Opera doesn't render correctly many of my pages. To a web designer that's a BIG no-no.

[quote="Osiris"]Opera doesn't render correctly [b]many of [i]my[/i] pages[/b].[/quote]
Try to use standartized HTML.  :)

[quote="MithsApprentice"][quote="Valentin Vornicu"][quote="ComplexZeta"]Myth, you seem to be missing the point. Opera should be able to render any (reasonable) website that Osiris creates. If it cannot, that is not a problem with the website. It is a problem with the browser.[/quote]I can create websites that only IE can render correctly. Does that mean that all the other browsers just suck? No, it means that I am a bad designer.

A designer should be able to design his sites in such a way that all browsers can access and display correctly (or better said "as wanted") the site.[/quote]

Um...I believe a compromise is necessary. If a designer makes a website accessible to only one browser, of course that's a lack of ability on the part of the designer. But on the other hand if every major browser except one is capable of displaying it correctly, then that's just a bad browser.[/quote]

I don't think Microsoft is really that great, but a PC can play a lot more games than a Mac can, so I have one. You can only access several of Microsoft's pages using IE. Surely that's not a lack of the designer's skill.

[quote="RC-7th"]You can only access several of Microsoft's pages using IE. Surely that's not a lack of the designer's skill.[/quote]

That's different. It's called ulterior motives - where companies make promoting their products more important than widespread access.

[quote="Myself"]I use Firefox.  I like all the extensions.  My favorite is Adblock - It can get rid of banner ads![/quote]
Opera doesn't need extensions for this, because it is embedded feature.  :)

[quote=Kea13][quote=jerry_li]19 year bump :)[/quote]

The whole point is not to do that
@bove Google chrome didn’t exist when the poll was made[/quote]

so crazy that's how old this thread is

yea we have historical glaciers in FF lol

[quote=jerry_li]19 year bump :)[/quote]

congratulations
[quote=blackjack2010]No Google or Chrome? :([/quote]

yeah google was still like
actually in the year that this thread was made, google JUST went public
lol
chrome didn't come to exist until four years later... :whistling:

i saw this post and was like "cool"
and i didn't see chrome and i was like "huh that's weird"
then i saw it was made in 2004 and i was like "welp no surprise"

[quote=Myth]I am sure Opera is the most appropriate application for internet browsing. It has many features, which simplify whole process.[/quote]

Lets go!!!


Prove that :

$ K\equal{}\sum\limits_{t\equal{}1}^{n} \frac{(\minus{}1)^{t\plus{}1}}{t}$ is not integer.

 :)

Obviosly, consider $ t\equal{}2^{[log_2(n)]}$.

The same proof considered [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=212837]here[/url] works.


What is the sum of five consecutive positive integers which have a mean of 17?

If the mean of five numbers is $ 17$, then the sum of the $ 5$ numbers must be $ 5\cdot17\equal{}\boxed{85}$.

if the mean of five numbers is 17 that means you could assume that all the numbers are 17. Intern leads you to 17*5=85

Hope this helped


In triangle ABC, a=14, b=15, and c=16. Compute $ \sin{\frac{A}{2}}$.

[hide="hint"]Heron's formula[/hide]


Nerds are cool! If you are a nerd and proud of it, join the club.  :coolspeak:  :gleam:

hmmm... take the nerd test.

http://moppers.kaseorg.com/nerdtest/

I'm 15% nerd.

35.0

TYTIA

Would probably have been more but I don't go to school. Homeschooled.

only 10%, probably because im on the school basketball team and i've been to more bar and bat mitzvahs than most of the jewish kids in my school =]...

i'd like to be a nerd though!

22.5%...actually, it's probably lower, since on a lot that were borderline, I just checked them.

It's a self-fulfilling prophecy type of thing, or an reverse-type of self-fulfilling prophecy
because a person can either consciously or unconsciously recognize certain ones of
the statements which will place them into one of the categories.  If I want less "nerd,"
then a bias will creep in, but if I fancy I'm more "nerd," it will work the other way.

It needs to be more of a double-blind study where you don't even suspect
you're being "data mined" for your nerdness, and the researcher doesn't
have a clue as to the likelihood of the nerd prospects that he or she has on hand.

But if you must, take it with a "grain of salt" and have a modest bit of fun.

[quote="FantasyLover"]ehh only 30%??

[quote]He read books on calculus and trigonometry[/quote]

 :rotfl:[/quote]

Meh, No calculus but I do Trig.

[quote="isabella2296"]70%....  :o[/quote]

Why did I expect that?

[quote="AIME15"]Uhh...wth!?

12.5%!?!?!?!??!?!![/quote]

Why didn't I expect that?

Lol i got more nerdy...now I'm at 75%.

...I guess I'll just say that doesn't count.

[quote="eggylv999"]47.5%

Because of my relation to sports[/quote]

WHAT ABOUT ME????????????????? :mad:  :mad:  :mad:

I got 15percent. Probably cause i play a sport and I'm not too smart. Although I do get all A's :D

I am a 92.5% nerd, would have been 100% but my voice is not "nasal and loud", I don't have acne, and I don't wear braces. But I do have the title of smartest in the school!  :lol:

What actually constitutes a nerd anyway? some of the things on the quiz are a little overdone... "he played games at lunch on his TI-89...", "he laughed at his teachers' bad jokes..." I think a true nerd (someone who scored 80% or more on the quiz) is someone who is exceptionally smart and shows it.

I'M A NERD AND PROUD OF IT!!!!

 :10:

[quote="CyberGenius"]I think a true nerd (someone who scored 80% or more on the quiz) is someone who is exceptionally smart and shows it.
[/quote]

Someone who shows it.. in a way generally not accepted by society? Hah.

Some of the questions were based on stereotypes.

40% here.
I guess that's sort of average.

huh. Now I'm 10% nerd.

My nerd-ness is shrinking at a constant velocity. I don't know if that's good or bad...

5%
i have braces and unfortunately have never had a girl

I'm a freshman in high school and have never gone on a date. I don't have braces, but soon will be getting them. However, people in my current school respect me for being smart instead of prosecuting me. I guess I'm lucky that way....... :)


If I interpreted you correctly, then yes.

Think of this as expected value.

Yes, I believe that is the correct answer to the problem.


inequalities
trigonometry
geometry proposed
geometry

For any triangle ABC, prove the inequality

$\sin A \left(\cos B + \cos C - \cos \left(B-C\right)\right) + \sin B \left(\cos C + \cos A - \cos \left(C-A\right)\right)$
$+ \sin C \left(\cos A + \cos B - \cos \left(A-B\right)\right) \geq 0$.

I obtained this as a side-product in proving another inequality and I know it is true (from a boring calculation which shows that it follows from Schur). But maybe it is still simpler than I think.

  Darij

Dear Darij
obviously sin A (cos B + cos C) + cyclic = sin A + sin B + sin C
Using tan A + tan B + tan C = tan A tan B tan C, we get
sin A cos B cos C + cyclic = sin A sin B sin C; hence
sin A cos (B-C) + cyclic = 4 sin A sin B sin C and your sum is
S = sin A + sin B + sin C - 4 sin A sin B sin C, which is
S = 2 sin A sin B sin C (R/r -2) and your result.
Happy 2005 to everybody.

Thanks, Jean-Pierre! Next time I should think before posting...

  Darij


function
real analysis
real analysis unsolved

Let $r_1,r_2,...$ be all rational numbers from $[0,1]$ and define $f(x)$ for $x\in [0,1]$ as the sum of $\frac{1}{2^n}$ for all numbers $n$ such that $r_n<x$. Prove that it is continuous when restricted to the irrationals of $[0,1]$ but cannot be extended to a continuous function on $[0,1]$.

It seems similar to the famous function that is often a counterexample for Riemann integrability. If we pick some $\epsilon > 0$, there is an $n$ so that $\frac{1}{2^n} < \epsilon$, and then any rationals in the ordering past $n$ (can't include $n$, but that's a minute point) don't matter for continuity. So it's okay for irrationals -- we can pick a small enough interval around an irrational so that no rationals $r_j$ with $j \le n$ are inside the interval. However, if we try to extend it to rationals, it doesn't work because there is a jump at the rational point of $\frac{1}{2^j}$. No matter how small is our interval around the point, the function would always jump by that much within the interval. So, not continuous.


advanced fields
advanced fields unsolved

I read in Atiyah's lecture notes the following statement:

"If we take the standard skew-form (,) on $ \mathbb{C}^4 \equal{} \mathbb{H}^2$, this is related to $ j$ and the positive inner product <,> by
\[ < u,v > \equal{} (u,jv)
\]
and in particular $ j$ preserves the skew-form in the sense that
\[ (ju,jv) \equal{} \overline{(u,v)}."
\]
I think the statement has something wrong... how the 2nd line follow from the first, and also I am confused if the skew form is hermitian or just symmetric, because he frequently quoted that $ z\in\mathbb{C}^4$ is annilated by itself under the skew form "because the form is skew-symmetric". (But if it is symmetric, I don't think $ (u,ju): \equal{} < u,u >$ will actually give a real number...)

Can somebody help to clear up my confusions?

Original source:
http://books.google.com/books?id=8Js3-h0JJBUC&pg=PA125&lpg=PA125&dq=C2+%22skew+form%22&source=web&ots=k8v2thPdYE&sig=Q63QugR4svPScBZu2Gk-ZJjPSI8&hl=en#PPA133,M1

Also the first 2 lines of page 57

I know what my problem is.

Basically, if $ u\equal{}z_1\plus{}z_2j$, then $ ju \equal{} \minus{}\bar{z_2}\plus{}\bar{z_1}j$, the conjugation arises from the fact that $ i$ and $ j$ does not commute.

So there actually IS a conjugation appearing, and that resolves all the hermitian/symmetric business.


Start with a right triangle with one leg 1 unit and the other leg x > 1.
Drop the altitude of the triangle, which splits it into a big and small triangle.
Drop the altitude of the bigger triangle. We now have three.
Drop the altitude of the biggest triangle one last time.
If the foot of this altitude cuts the hypotenuse in half, find x.

I'm getting [hide="1"]$\sqrt{1+\sqrt{2}}$[/hide]  The length of the last altitude is [hide="2"]$\frac{x^3}{(x^2+1)^{3/2}}$[/hide] which means half of the hypostenuse is [hide="3"]$\frac{x^4}{(x^2+1)^{3/2}}=\frac{\sqrt{x+1}}{2}$[/hide]. Thus, [hide="4"]$2x^4=(x^2+1)^2$[/hide] and we are only dealing with positive lengths, so you then get the earlier result.


limit
logarithms
calculus
calculus computations

[hide="Answer"]$ L \equal{} \lim_{x \to 0} \left(\frac {e^{x \cdot \ln a} \plus{} e^{x \cdot \ln b}}{2}\right)^{1/x} \equal{} \lim_{x \to 0} \left(\frac {1 \plus{} x \ln a \plus{} 1 \plus{} x \ln b \plus{} O(x^2)}{2}\right)^{1/x}$

$ \equal{} \lim_{x \to 0}[1 \plus{} x (\ln(\sqrt {ab}) \plus{} O(x))]^{1/x}$

$ \equal{} \lim_{x \to 0}[[1 \plus{} x (\ln(\sqrt {ab}) \plus{} O(x))]^{1/x(\ln(\sqrt {ab}) \plus{} O(x))}]^{(\ln(\sqrt {ab}) \plus{} O(x))}$

$ \equal{} \lim_{x \to 0} e^{\ln(\sqrt {ab}) \plus{} O(x)}$

$ \equal{} \sqrt {ab}$[/hide]

$ x_i  \ge 0$, $ a_i  > 0$, $ \sum\limits_{i \equal{} 1}^n {a_i }  \equal{} 1$

Find $ \mathop {\lim }\limits_{p \to 0} \left( {\sum\limits_{i \equal{} 1}^n {a_i x_i ^p } } \right)^{\frac{1}{p}}$

[hide="Or:"]
$ y \equal{} \lim_{x \to 0} \left(\frac{a^x \plus{} b^x}{2}\right)^\frac{1}{x}$
$ \ln y \equal{} \ln \left[\lim_{x \to 0} \left(\frac{a^x \plus{} b^x}{2}\right)^\frac{1}{x}\right] \equal{} \lim_{x \to 0} \ln \left(\frac{a^x \plus{} b^x}{2}\right)^\frac{1}{x} \equal{} \lim_{x \to 0} \frac{\ln(a^x \plus{} b^x) \minus{} \ln 2}{x}$

Which by L'Hopital's Rule equals, $ \lim_{x \to 0} \frac{1}{a^x \plus{} b^x}(a^x \cdot \ln a \plus{} b^x \cdot \ln b) \equal{} \frac{1}{2}(\ln a \plus{} \ln b) \equal{} \ln \sqrt{ab}$

Therefore, $ y \equal{} \sqrt{ab}$[/hide]


trigonometry
analytic geometry
linear algebra
matrix

Two points, P and Q, are on a circle. They start from the same point and go at constant speeds in the same direction around the circle but P goes twice as fast as Q, so that the chord PQ is constantly changing length. What path is followed by the midpoint of PQ?

Letting P = (cos a, sin a) and Q be (cos b, sin b), with a = 2b,
then using sum of two cosines and sum of two sines,
I get the mid-point as (cos(3b/2)cos(b/2), sin (3b/2)cos (b/2))
so that y =  (tan(3b/2))x, but have got no further.

For Q = (cos b, sin b), I make the mid-point  cos(b/2) (cos(3b/2), sin(3b/2)) or x^2 + y^2 = (cos(b/2))^2. Is this a limacon?

We take the origin to be at the centre of the circle and its radius to be $ r$. If the midpoint is $ R$, then its parametric equations are $ R\left({r\over 2}(\cos 2\phi \plus{} \cos\phi),{r\over 2}(\sin 2\phi \plus{} \sin\phi)\right)$. Now we have to eliminate $ \phi$.

$ x^2 \plus{} y^2 \equal{} {r^2\over 4}(2 \plus{} 2\cos 2\phi\cos\phi \plus{} 2\sin 2\phi\sin\phi) \equal{} {r^2\over 2}(1 \plus{} \cos\phi) \equal{} r^2\cos^2{\phi\over 2}$

$ x \equal{} {r\over 2}\cdot 2\cos{3\phi\over 2}\cos{\phi\over 2} \equal{} r\left(4\cos^3{\phi\over 2} \minus{} 3\cos{\phi\over 2}\right)\cos{\phi\over 2} \equal{} r(4\cos^4{\phi\over 2} \minus{} 3\cos^2{\phi\over 2})$

Now substitute $ \cos^2{\phi\over 2} \equal{} {x^2 \plus{} y^2\over r^2}$ and rearrange to get

$ 4(x^2 \plus{} y^2)^2 \minus{} 3r^2(x^2 \plus{} y^2) \minus{} r^3x \equal{} 0$

which is the desired equation.

In polar coordinates, where $ \rho^2 \equal{} x^2 \plus{} y^2$ and $ x \equal{} \rho\cos t$, we get

$ 4\rho^3 \minus{} 3r^2\rho \minus{} r^3\cos t \equal{} 0$

Hvala, Farenhajt. I failed to eliminate the angle. (There is an r^2 missing from the middle term in the last line.) I am still wondering if the curve is a kind of limacon and whether it is well-known, perhaps with its own name.

P.S. Have you looked at my two problems: "sum of tans" and "matrices and APs and GPs" in the Pre-Olympiad forum? I feel the first one should be easy using the diagram I described but I just can't see it and I have not been able to improve on the given matrices in the second problem.

Thanks for the remark - edited :)

Probably simpler equation would be obtained if somewhat different coordinate system were chosen (e.g. if the circle were $ (x\minus{}r)^2\plus{}y^2\equal{}r^2$). By plotting it, it does resemble limacon.

Thanks again.

(I've now got a 2X2 matrix for APs in my Pre-Olympiad post "matrices and APs and GPs".
Using letters for the elements of the matrix gives simultaneous equations.
The first row is

(1 + d/a)     (-d^2/a^2)

the second row is

1                1.)


function
number theory proposed
number theory

[b]Problem1:[/b]
Let p be a prime number.Find all $f:\mathbb{N^*}\to\mathbb{N^*}$ satisfying the condition:
[b](1)[/b]$f(p+n)=f(n),\forall n\in\mathbb{N^*}$
[b](2)[/b]$f(m\cdot n)=f(m)\cdot f(n),\forall m,n\in\mathbb{N^*}$
[b]Problem2:[/b]
Let a,b be two integer number.Find all $f:\mathbb{N^*}\to\mathbb{N^*}$ satisfying the condition:
[b](1)[/b]$f(a+n)\leq f(n)+a,\forall n\in\mathbb{N^*}$
[b](2)[/b]$f(b+n)\geq f(m)+b,\forall m\in\mathbb{N^*}$

I don't exactly remember what $\mathbb N^*$ means. It is 1,2,3,... Right?

Yes, that's what it means.

For the first one: if $n<p$ then $f(n^{p-1})=f^{p-1}(n)=f(1)$, and since it's clear that $f(1)=1$, we get $f(n)=1,\ \forall n$ s.t. $p\not|n$. On the other hand, $f(p^2)=f^2(p)=f(p)$, so $f(n)=1,\ \forall n$ s.t. $p|n$. From these two we get $f(n)=1,\ \forall n\in\mathbb N^*$.


geometry
3D geometry
algebra
factorization
sum of cubes

I don't remember how to do those problems that are like:

1*3*5+3*5*7+5*7*9+7*9*11+9*11*13+11*13*15
Can someone help?

[quote="binonunquineist"]The most obvious method I can think of is writing each triplet product as a cube of the middle number minus quadruple the middle number. Since we should know the formula for sum of cubes and sum of odd integers, we can use a little bit of algebra to get the formula

 $(1/8)((n-1)(n+3)(n^{2}+2n+5))$

where n is the middle integer of the last triplet.[/quote]
so is 4550 the answer?

jorian

now i'm just wondering here, butg wouldn't it just be easier to do it out than come up with that fancy formula?

jorian

Probably if it was on a sprint, it would be best to know a quick formula.

although wouldn't you have to take a while deriving it?

jorian

True, but it would also minimize computation mistakes, wouldn't it?


I thought I'd go a bit simpler this tim since I have to actually type in the solutions and it might take me a very long time on some problems.  Not too hard, but no picture, sorry.

An infinitely long wire carries a current of $ i\equal{}1\text{A}$.  It is bent so as to have a semi-circular detour around the origin, with radius $ r\equal{}1\text{cm}$.  Calculate the magnetic field at the origin.

If, in the future, you would rather me just post the answer rather than the full solution, then I might be inclined to post harder/longer problems.

By Biot-Savart: $ \vec{B}\equal{}\frac{\mu_0}{4\pi}I\int\frac{d\vec{l}\times \vec{r}}{r^2}$, so

$ \vec{B}\equal{}\frac{I\mu_0}{4\pi}\int_0^\pi\frac{rd\theta}{r^2}\equal{}\frac{I\mu_0}{4r}$, so in this case the magnitude of the field is $ \frac{1(4\pi\times 10^{\minus{}7})}{4(.01)}\equal{}\pi\times 10^{\minus{}5}$ T.

The noncurved part cancels.

But it's been awhile since I've done one of these...

Indeed you are correct.  I'll post another one in a few days.

Actually the  semi infinite stright portions doesn't cancel but reinforce:

\[ B \equal{}\frac{I\mu_0}{4\pi r} (\pi \plus{}2 )\]

^^ idl X r =0 at the origin for the infinite wires .

@JRav: Keep the questions like this , maybe a bit harder but more logical than mathematical. Just a recommendation.

[quote]	^^ idl X r =0 at the origin for the infinite wires . [/quote]
What is this?

Conjurer
I guess you're taking the origin in one of the straight portions (even then it would be the field due to the other one).
But the origin lays distant r from them.

I believe the figure is something like this

Conjurer, you are indeed correct, that is the picture.

[quote]As for the level, it says it is from UC Berkeley. What does this mean? Is it a General Physics II final exam question or something? Surely this is not a physics qualifying exam problem.[/quote]

This is a Ph.D. qualifying exam problem.  Apparently one of the easier ones.

I am inclined to give a somewhat harder problem next time, but not too hard.

With that "violent" (he, he, he) detour its OK.

I was thinking of something more smooth, like the detour being tangent to the straight portions.

Oh, I finally get what you were doing Immanuel. I think that what you did is a much better problem!!  :lol:


Καλησπέρα σε όλους.

Οι λάτρεις των ανισοτήτων σήμερα θα είχαν την τιμητική τους με το θέμα της ανισότητας που έπεσε στην πρώτη μέρα της φετινής IMO στην Ισπανία.

Παραθέτω το πρόβλημα:

α) Εάν οι $ x,y,z$ είναι πραγματικοί αριθμοί διαφορετικοί του $ 1$ και τέτοιοι ώστε $ xyz \equal{} 1$,  να δείξετε ότι

$ \frac {x^{2}}{\left(x \minus{} 1\right)^{2}} \plus{} \frac {y^{2}}{\left(y \minus{} 1\right)^{2}} \plus{} \frac {z^{2}}{\left(z \minus{} 1\right)^{2}} \geq 1.$

β) Να δείξετε ότι η ισότητα ισχύει για άπειρες ρητές τιμές των $ x,y,z$.

Παραθέτω παρακάτω τη λύση μου στο συγκεκριμένο πρόβλημα:

[b][u]Λύση:[/u]
[/b]
[b]Θέτουμε $ a \equal{} \frac {x}{x \minus{} 1}, b \equal{} \frac {y}{y \minus{} 1}, c \equal{} \frac {z}{z \minus{} 1}.$ και θέλουμε να δείξουμε ότι $ a^2 \plus{} b^2 \plus{} c^2\geq 1.$

Πολλαπλασιάζοντας κατά μέλη παίρνουμε $ abc \equal{} (a \minus{} 1)(b \minus{} 1)(c \minus{} 1)$ δηλαδή ουσιαστικά τη σχέση $ xyz \equal{} 1$, την αναγάγαμε σε σχέση των $ a,b,c$.

Από την τελευταία παίρνουμε $ a \plus{} b \plus{} c \equal{} ab \plus{} bc \plus{} ca \plus{} 1$. Υψώνουμε στο τετράγωνο άρα
$ a^2 \plus{} b^2 \plus{} c^2 \plus{} 2(ab \plus{} bc \plus{} ac) \equal{} (ab \plus{} bc \plus{} ca)^2 \plus{} 2(ab \plus{} bc \plus{} ca) \plus{} 1$ άρα

$ a^2 \plus{} b^2 \plus{} c^2 \equal{} (ab \plus{} bc \plus{} ca)^2 \plus{} 1 \geq 1$ και τελειώσαμε.[/b]

Το δεύτερο ερώτημα είναι σχετικά απλό αν κάποιος έχει κάνει το πρώτο με αυτό τον τρόπο. Περιμένω τις δικές σας προσεγγίσεις.

[b][u]Ερώτηση:[/u][/b] Τί διαφορά έχει αυτό το πρόβλημα από το πρόβλημα του Θαλή της Γ' Λυκείου φέτος? Τολμώ να πω ότι του Θαλή ήταν δυσκολότερο διότι σε κάθε κλάσμα είχε δύο μεταβλητές και η λύση να βρείς τα συμμετρικά πηλίκα $ \frac {a}{b}$ κτλ. ώστε να τα εκφράσεις σα συνάρτηση μίας μεταβλητής έλειπε από αυτό το πρόβλημα, στο οποίο, ήδη από την αρχή είχαμε μόνο μία μεταβλητή σε κάθε κλάσμα. Μάλιστα κι εκεί έμμεσα υπήρχε το γινόμενο $ xyz \equal{} 1$, μόνο που ήταν στη μορφή $ \frac {a}{b}\cdot\frac {b}{c}\cdot\frac {c}{a} \equal{} 1$, που ισχύει πάντα. Ο Σιλουανός τότε είχε δώσει μία εξαιρετική λύση (η οποία κατέληγε σε ίδια σχέση με την επίσημη λύση αλλά με διαφορετικό τρόπο).  Τί λέτε κι εσείς??

Φυσικά αυτό το θέμα μας συμφέρει σαν Ελληνική Ομάδα, αφού είμαι σχεδόν βέβαιος ότι οι μαθητές μας θα το αντιμετώπισαν με απόλυτη επιτυχία καθώς είναι Γ' Λυκείου οι περισσότεροι.

Για να σας θυμίσω το πρόβλημα της Γ΄Λυκείου του φετινού Θαλή και την επίσημη λύση της ΕΜΕ, σας παραπέμπω στη σελίδα

http://www.hms.gr/eme/modules/wfsection/article.php?articleid=1165

[b][u]Παρατήρηση:[/u][/b] Αν στον Θαλή ζητούσαν να αποδειχθεί η ανισότητα

$ \left(\frac {a}{a \minus{} b}\right)^2 \plus{} \left(\frac {b}{b \minus{} c}\right)^2 \plus{} \left(\frac {c}{c \minus{} a}\right)^2\geq 1$ με $ a,b,c\in \mathbb{R}$ και $ a,b,c$ διαφορετικοί μεταξύ τους ανά $ 2$ και διαφορετικοί του μηδενός,

[b]ΤΟΤΕ [/b]θα είχαμε πιάσει [b]ΑΚΡΙΒΩΣ[/b] το θέμα και θα είχε [b]ΑΚΡΙΒΩΣ[/b] την ίδια λύση με αυτή της επίσημης λύσης της ΕΜΕ. Προσπαθήστε την!!

Περιμένω και τις δικές σας απόψεις!

Φιλικά,
Αλέξανδρος Συγκελάκης

Αλέξανδρε είμαι σίγουρος ότι ο Σιλουανός και ο Τάσος το έλυσαν . Όντως το θέμα θυμίζει Θαλή 3ης Λυκείου . Άντε καλή δεύτερη μέρα στα παιδιά και καλά αποτελέσματα . Τολμώ να πω ότι επίσης και το πρώτο θέμα ήταν εύκολο για IMO αλλά ωραίο .

Αλεξανδρε οντως εκπληκτικη ομοιοτητα ...........Λετε να την προτεινε κανενας δικος μας;
Aν δεν την ελυνες με αυτον τον τροπο (υπαρχουν πολλοι τροποι στο αντιστοιχο toric του forum) και δεν καταφερνες το τελειο τετραγωνο +1 αλλα κατελληγες σε αθροισμα τετραγωνων (χωρις μετασχηματισμο) τοτε ηταν δυσκολο το δευτερο ερωτημα γιατι δεν εβγαζες ευκολα την υπαρξη των ρητων...

Η λυση  του Sil χρησιμοποιουσε καποια ιδιοτητα του πολυωνυμου Langrance που οπτικα σε παρεπεμπε η τοτε εκφωνηση του Θαλη.
Καλη τους συνεχεια αυριο

Ο εισηγητής του συγκεκριμένου προβλήματος φαίνεται να είναι ο Walther Janous από την Αυστρία. Ένας από τους μεγαλύτερους και καλύτερους λύτες προβλημάτων του περιοδικού CRUX MATHEMATICORUM  ( http://journals.cms.math.ca/CRUX/ ) που εκδίδει η Μαθηματική Εταιρεία Καναδά.

Όσο το κοιτάω τόσο δεν μπορώ να το χωνέψω ότι έπεσε σαν μέτριας δυσκολίας πρόβλημα στην ΙΜΟ. Απίστευτο!

Αλέξανδρος

[quote="r_boris"]Λύνεται εύκολα με ανάλυση αλλά όχι τόσο εύκολα όσο η λύση του Αλέξανδρου
Αργότερα θα ποστάρω την λύση
Πάντως Και το 1 και το 2 θέμα κατά την γνώμη μου είναι πολύ εύκολα για ΙΜΟ
ΜΠΡΑΒΟ ΣΤΑ ΠΑΙΔΙΑ ΜΑΣ[/quote]
Λύση στο attacment

Για πληρότητα γράφω τη λύση και στο (β) ερώτημα σύμφωνα με τη λύση που έδωσα παραπάνω στο (α). Ας ονομάσουμε $ (1)$ τη σχέση $ ab \plus{} bc \plus{} ac \plus{} 1 \equal{} a \plus{} b \plus{} c$ που βγάλαμε στο (α) ερώτημα. Τότε η ισότητα ισχύει μόνο όταν $ ab \plus{} bc \plus{} ac \equal{} 0$ και τότε η $ (1)$ γίνεται $ a \plus{} b \plus{} c \equal{} 1$ δηλαδή αν ξαναγυρίσουμε στα $ x,y,z$ και λάβουμε υπόψιν ότι $ xyz \equal{} 1$, παίρνουμε μετά τις απλοποιήσεις των παρανομαστών ότι $ xy \plus{} yz \plus{} xz \equal{} 3$. Αντικαθιστώντας $ z \equal{} \frac {1}{xy}$ παίρνουμε το τριώνυμο ως προς $ x$ : $ y^2x^2 \minus{} (3y \minus{} 1)x \plus{} y \equal{} 0$ που έχει διακρίνουσα $ \Delta \equal{} \minus{} (4y \minus{} 1)(y \minus{} 1)^2$.  Τώρα διαλέγω αριθμό $ y$, τέτοιο ώστε $ y \equal{} \frac {1 \minus{} k^2}{4}$ για οποιοδήποτε ΑΚΕΡΑΙΟ $ k$ (Αυτά τα $ y$ είναι προφανώς άπειρα σε πλήθος).  Τότε η διακρίνουσα είναι θετική και τέλειο τετράγωνο και προφανώς $ y\in \mathbb{Q}$. Τότε οι ρίζες της εξίσωσης είναι ρητές και επειδή $ z \equal{} \frac {1}{yz}$, άρα $ z\in\mathbb{Q}$. Άρα βρήκαμε μία απειρία ρητών λύσεων για τις οποίες ισχύει η ισότητα.

Κύριε Μπόρη ευχαριστούμε για τη λύση σας στο (α). Πράγματι είναι πολύ καλή!!

Αλέξανδρος

Μετά από καιρό ένα γεια αξίζει να το πω εδώ στο ελληνικά . :)
Εκ μέρους της ελληνικής ομάδας να σας πούμε ότι ευχαριστούμε τα μέλη για τις ευχές πριν το διαγωνισμό και για τη στήριξή τους μετά  :)
Για τη δυσκολία αξίζει να πω δύο πραγματάκια εδώ :
Πράγματι τα θέματα φαίνονται εύκολα αλλά ξέρετε ότι το να είσαι μέσα σε μία αίθουσα να ξέρεις ότι ο χρόνος που έχεις είναι ιδιαίτερα περιορισμένος και να έχεις άλλους 10 φοβερούς ανθρώπους δίπλα σου που να βλέπεις ότι γράφουν χιλιάδες σκέψεις δημιουργεί ένα άλλο κλίμα . Σκεφτείτε ακόμα ότι μας παίρνει πολύ χρόνο εκεί να κατανοήσουμε τη δυσκολία των θεμάτων . Εννοώ ότι όταν περιμένςι κάτι δύσκολο στο 2 και κάτι απροσπέλαστο στο 3 τότε αυτό σου δημιουργεί σκέψεις σε δύσκολες ιδέες που αν δεν πιάσουν εγκλωβίζεσαι αγχώνεσαι κτλ ... Παίζουν πολλά πράγματα ρόλο ....
Η δεύτερη μέρα ήταν όντως πολύ δυσκολότερη από την πρώτη αλλά ακόμα και αυτό είναι σχετικό ...
Τέλος σκεφτείτε ότι το ερώτημα 2)β) που είναι πολύ απλό όντως κόστισε πολλούς πόντους και χρυσά μετάλλια . Για του λόγου το αληθές ρίξτε μια ματιά στους ρώσσους αν και υπάρχουν και τόσοι τόσοι άλλοι που την πάτησαν εκεί . :wink:  (like me too αλλά αφού έλυσα το α) χαλάλι δεν θα άλλαζε τπτ )

Οπ γεια σου Σιλουανε. ΣΥγχαρητηρια και καπως πιο αμεσα για το μεταλλιο! :)  ευχομαι την καλυτερη καριερα! μαθηματικο θα πας ετσι;

Σιλουανε συγχαρητηρια! Δεν χρειαζεται να σκεφτεσαι την ΙΜΟ πια. Εισαι πια κι επισημα ενας IMO medallist!  :D

Γεια σου Σιλουανέ και συγχαρητήρια και πάλι σε σένα και σε ολόκληρη την Ελληνική Ομάδα για τις επιτυχίες της!

 Όπως είπα και σε προηγούμενο post είναι μεγάλο πράγμα σε μία τέτοια διοργάνωση (με όλα εκείνα που τη συνοδεύουν: άγχος, πίεση χρόνου, δύσκολες/περίπλοκες ιδέες κτλ κτλ) να μπορέσεις να λύσεις έστω και ένα ολόκληρο θέμα, ακόμα και να τσιμπήσεις κάποιους πόντους απο ένα θέμα (όπως ΟΛΟΙ στην Ελληνική ομάδα κάνατε φέτος). Σας αξίζουν συγχαρητήρια γι'αυτό!! Εδώ απλά έκανα μία σύγκριση του θέματος της IMO με ένα δικό μας θέμα στον Θαλή και θετικό για την Ελληνική ομάδα!! Φυσικά όλοι εμείς πίναμε τον καφέ μας μπροστά στον υπολογιστή, άνετοι και χαλαροί και προσπαθούσαμε να λύσουμε τα θέματα!! Ασφαλώς και κατανοούμε πώς νοιώθει κάποιος διαγωνιζόμενος κατά τη διάρκεια του διαγωνισμού και γι'αυτό δεν κατηγορούμε κανένα για το αν πήγε καλά ή άσχημα για το αν έλυσε ή δεν έλυσε ένα θέμα!! Ελπίζω να μη φάνηκε κάτι τέτοιο μέσα από το μήνυμά μου ή τα μηνύματα άλλων! Αν έγινε κάτι τέτοιο ζητώ ειλικρινή συγγνώμη καθώς δεν ήταν η πρόθεσή μου!! Άλλωστε είμαι ο τελευταίος που θα κρίνει κάποιον από εσάς!

Καλή συνέχεια πλέον στο Πανεπιστήμιο και ελπίζουμε να συνεχίσεις να επισκέπτεσαι και να λύνεις με το δικό σου μοναδικό τρόπο πολλά θέματα από το "ΕΛΛΗΝΙΚΑ" που τόσο αγάπησες!

Φιλικά,
Αλέξανδρος

Αλέξανδρε σε ευχαριστώ πολύ (και τους προλαλίσαντες) :) και θα προσπαθήσω να βοηθάω και εγώ με τον τρόπο μου την μαθηματική αυτή παρέα .
Σε όλα όσα είχες πει ήσουν άψογος και προφανώς δεν έθιξες κανέναν . Απλά αυτά ήθελα να τα προσπέσω και αφού είχατε βάλλει εδώ την ανισότητα (βλέπεις οι αγάπες δεν ξεχνιούνται  :P ) είπα να σχολιάσω εδώ . Όλα αυτά τα έγραψα γιατί απλά όλη αυτή η ιστορία με το ''εύκολα-δύσκολα '' δεν έχει και νόημα γιατί αυτά είναι τελείως σχετικά πράγματα . Ας πούμε πιστεύω ότι για τους περισσότερους εδώ (εξαιρείται ο Ηλίας ) το 1ο θέμα φάνηκε πολύ πιο εύκολο από το τέταρτο . Παρόλα αυτά είδα κάποιον στο αγγλικό να λέει το εντελώς αντίθετο ... Τέλος πάντων ... Να είστε όλοι καλά και προτείντε και καμιά άσκηση που και που  :P

Σιλουανέ πολλά συγχαρητήρια για το μετάλλειο. Εύχομαι ολόψυχα καλή πρόοδο.

Σίγουρα είναι άλλο να γράφεις χαλαρά  και άλλο να γράφεις με χρονικό περιορισμό και σε ένα ιδιαίτερα αγχωτικό περιβάλλον. (όσοι έχουν δώσει απλώς πανελλήνιες εξετάσεις το ξέρουν καλά)
Να κάνω και μια ερώτηση για να δω κυρίως πόσο απέχουμε ως χώρα από το να γράφουμε συστηματικά όλα τα θέματα (άλα Ρωσία και Κίνα): τα θέματα αυτά που έπεσαν τα έλυνες (εσύ και τα άλλα παιδιά) εκτός εξετάσεων δηλαδή χωρίς χρονικό περιορισμό ή ήταν πάνω από τις δυνάμεις της ελληνικής ομάδας  σε δυσκολία;


calculus
integration
function
derivative
algebra
binomial theorem

I've been working on the attached problem for a week or so now. I really hope someone can help me out.

Thanks a lot!

i can have it solved if you really need it

Alright, so, first, you differentiate under the integral sign with respect to a, then you evaluate the integral over x just obtained and integrate with respect to a (this integral is easily found using the gamma function evaluated at 1/2).  To then find the "constant of  integration" (which is a function of b and p) differentiate under the integral sign in the original problem with respect to b, integrate with respect to b, and then the "constant of integration" thus obtained is equal to the function of a just found, while the "constant of integration" from the earlier integration is just the function of b just found plus some function of p (possibly).  Set a and b equal to zero, and the whole thing is obviously zero, and set the function of p "constant of integration" desired thus equal to zero at all points, i.e. that function of p is zero.  That's the method, I don't have time myself to go through all the steps right now however.

I just noticed another method of solution (once again, I don't have time to fill in the details).  A rigorous justification can easily be given through uniform convergence:

Expand e^(-t^2) in the exp function's usual power series, then integrate and substitute t=a/sqrt(x)+b*sqrt(x).  Then, take the coefficents of the series (these coeff. are ((-1)^n)/[n!*(2n+1)]) outside the integral sign, each one multiplied by the integral from x=0 to infinity, with respect to x, of [exp(-px)]*[(b*sqrt(x)+a/sqrt(x))^(2n+1)], and then sum from n=0 to infinity.  The integral within the series can be found using the binomial theorem and the gamma function, or simply through substitutions (I think).  This should give a really cool series expansion in a, b, and p as the answer to your problem (although I doubt it would be simple to derive any closed-form expression from such a series).

P.S. This shouldn't be in the "pre-olympiad" category.  No matter how you approach the problem, it uses some (fairly) advanced calculus machinery, and some analysis (for example, differentiation under integral sign, integration of series, etc., require justification through such means as uniform convergence).  I'm sure there are some more conventional methods of solution, but I've always had really strange ways of evaluating integrals.


symmetry
probability
logarithms
expected value
probability and stats

This was a nice problem that I recently solved:

A knight starts in the corner of a usual $ 8\times 8$ chessboard and makes a move selected from his valid moves with equal probability.  What is the expected number of steps for the knight to return to his starting place?

Is this related to this one: http://www.mathlinks.ro/viewtopic.php?t=174148
An ugly solution would be to write a linear system with $ 64$ equations, and $ 64$ unknowns, and let matlab do the job  :lol: (or a little less, with the obvious symmetry of the problem). But I guess you have a nice solution  :)

the chessboard is a finite graph, and the movements of the knight on the board are a markov chain with transition probability $ p(v_i,v_j) \equal{} \text{deg}(v_i)$.

it only remains to find the stationary distribution $ \pi$ and use $ E_{v_0}T_{v_0} \equal{} \pi(v_0)^{ \minus{} 1}$.

Of course there is a little more work, like computing how many edges are actually in the graph, but this shouldn't be that difficult :)  I'll post a solution in a day or so if no one answers before then.

Hmmm... Computing the steady state requires solving a linear system with about $ n^2/8$ variables instead of $ n^2/2$ but I do not see a great advantage in this unless there is some beautiful formula that I'm missing. :?

there was a mistake in my second post; the transition probability should obviously have been $ p(v,u) = \text{deg}(v)^{-1}$ if $ (u,v)$ is an edge in the graph and $ 0$ otherwise.

So, the stationary distribution is $ \pi(v) =\deg(v)/(2|E|)$, where $ |E|$ is the number of edges in the graph.  to see this we just need to verify for all $ u\in V$ the equalities
\begin{eqnarray*}
\sum_{v\in V} \pi(v)p(v,u) &=& \pi(u) \\
\sum_{v\in V} \pi(v) &=& 1 \\
\end{eqnarray*}
The first follows since $ \pi(v)p(v,u) = 1/(2|E|)$ if $ (u,v)$ is an edge and $ 0$ otherwise, so summing over all $ v$ we get $ |\{v\in V: (u,v)\in E\}|/(2|E|) = \text{deg}(u)/(2|E|) = \pi(u)$.  (since the chain is reversible, it actually suffices to verify $ \pi(u)p(u,v) = \pi(v)p(v,u)$ for all $ u,v\in V$.)  The second condition is automatically true by one of the most basic principles in graph theory.

Finally, we compute $ |E|$ using (again) the formula
\[ |E| = \frac{1}{2}\sum_{v\in V}\text{deg}(v)
\]
and arrive (after some tedious but straightforward work) at $ |E|=168$.  As a corner $ v_0$ of the chessboard has degree $ 2$, this gives us
\[ E_{v_0}T_{v_0} = (\pi(v_0))^{-1} = \left(\frac{2}{2\cdot 168}\right)^{-1} = 168
\]
so the expected return time is $ 168$ moves.  I don't know if we can use these tools to get a quicker proof of the similar problem that alekk mentioned.

[quote="blahblahblah"]
and arrive (after some tedious but straightforward work) at $ |E| \equal{} 168$.  [/quote]
Nice solution :blush:! The "tedious work" is actually multiplying out $ 4$, $ n\minus{}1$ and $ n\minus{}2$ (just look at how many edges of each of the 4 kinds you have on the board). I doubt it can help with the other problem though: to get $ n^2\log n$ from an explicit formula is not that easy, not to mention $ \pi$ :()

In general, on an $ n\times n$ chessboard $ G_n\equal{}(V_n,E_n)$ for $ n\geq 8$, I think there are $ (n\minus{}4)^2$ vertices of degree $ 8$, $ 4(n\minus{}4)$ vertices of degree $ 6$, $ 4(n\minus{}4)\plus{}4$ vertices of degree $ 4$, $ 8$ vertices of degree $ 3$, and $ 4$ vertices of degree $ 2$, resulting in $ |E_n|\equal{}4(n^2\minus{}3n\plus{}2)$.


[quote="airelemental135"]State the largest possible domain such that the given relation is a function.

f(x) = 4/((x^2) + (4x))

I am not sure what this question is asking but is the answer simply all real numbers excluding 0 and -4?[/quote]

Yep that's right, if $f(x)=\frac{4}{x^2 + 4x}$.

You can check [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=355010&highlight=#355010]here[/url].

I hope that helps.


Three dice are rolled at once. The 3 nos. appearing on the dice are permuted to get all [u]distinct[/u] 3-digit nos. and the sum of these 3 digit nos. is noted as the 'score'.

For eg. if the 3 dice show 1,2,3 the 3-digit nos. are -
123, 132, 213, 231, 312, 321 and the 'score' is 1332.

What is the average score after infinite no. of throws?

Let's call $S_{a,b,c}$ the score of the result $\{a,b,c\}$.

There are $6\times 6\times 6=216$ possible results.

There are three different cases:

$1)$ The values are all different.
There are $6\times 5\times 4=120$ possible results in this case.
$S_{a,b,c}=111(2a+2b+2c)$.
Every result has its complementary and $S_{a,b,c}+S_{7-a,7-b,7-c}=777\times 6$.
Therefore, the average score in this case is $S_{1}=\frac{777\times 6}{2}$.

$2)$ The values are all the same.
There are $6\times 1\times 1=6$ possible results in this case.
$S_{a,a,a}=111a$.
Every result has its complementary and $S_{a,a,a}+S_{7-a,7-a,7-a}=777$.
Therefore, the average score in this case is $S_{2}=\frac{777}{2}$.

$3)$ Two of the values are equal and the other is different.
There are $216-120-6=90$ possible results in this case.
$S_{a,a,b}=111(2a+b)$.
Every result has its complementary and $S_{a,a,b}+S_{7-a,7-a,7-b}=777\times 3$.
Therefore, the average score in this case is $S_{3}=\frac{777\times 3}{2}$.

So, the average score, considering the three different cases, is
$S=\frac{120\times S_{1}+6\times S_{2}+90\times S_{3}}{216}=\frac{777}{2}\times\frac{83}{18}=1791.4\dots$

Sorry, your answer is wrong.

Note that 216 different scores are not produced.
For eg. 223,232,322 together produce a single score.

It's true that there aren't $216$ different scores, but we have to consider if some scores are more relatively frequent than others.
If one score is more relatively frequent than another (there are more dice result combinations that lead to the first score) the first score will contribute more to the statistic mean.
Therefore, we have to count the number of dice combinations for every different score, and not only the number of different scores.

Imagine another experiment: rolling a dice with four faces with a $3$ and two faces with a $6$. There are only two possible results: $3$ and $6$, but the average is not $(3+6)/2=4.5$; it's $(4\times 3+2\times 6)/6=4$, because a $3$ will appear more frequently than a $6$.


function
inequalities unsolved
inequalities

Find the maximum value of $ A \equal{} ( a \plus{} b \plus{} c ) \left( \frac{1}{a} \plus{} \frac {1}{b} \plus{} \frac{1}{c}\right)$. Suppose that $ 1 \leq a, b ,c \leq 2$

[quote="Nam Luu"]Find the maximum value of $ A \equal{} ( a \plus{} b \plus{} c ) \left( \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}\right)$. Suppose that $ 1 \leq a, b ,c \leq 2$[/quote]

in $ [1,2]^3$, function $ ( a \plus{} b \plus{} c ) \left( \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c}\right)$ is continuous and for each variable is convex, so we just need to check that $ (a,b,c) \equal{} (1,1,1),(1,1,2),(1,2,2),(2,2,2)$.

result is $ A_{\max}10$.


Is it true that a function f:R->R with the IVP and satisfing f(x)<=f(x+q), for all real x and for all positive rational q, is an increasing function?

And if it's not true...can someone give a counterexample?(i'm asking for a counterexample because i think that it is not true...but i might be wrong...)

[edit]We can easily prove that $f$ is increasing on the set of rationals. Other than that, if $f$ should have that property it would be continous.

you said that  If $x\in(p,r)$ then $f(x) \in (f(p),f(r))$ . how do you know? of, course it would be like that if x-p would be rational but this would mean that x is rational... Am I missing something? :maybe:

You are right. sorry. easter madness.  :jump:

No answer? :(  I was stiil thinking about a counterexample, and because here the monotony is equivalent to the continuity, the Hamel functions came to my mind...so a Hamel function with IVP, and with $f(1)>0$, will be a good example, but I don't know if there exists such a function... :?:

Actually, I don't know very much about Hamel functions...just the well known fact that they are solutions to Cauchy equations, others then $ax$ and are, of course, everywhere discontinuous...can one say something more about them?


function
calculus
derivative
ratio
algebra
domain
complex analysis

Would you please clarify this for me: " ... we can write $x=\frac12(z+\bar z), y=-\frac12 i(z-\bar z)$. With this change of variable we can consider $f(x,y)$ as a function of $z$ and $\bar z$ which we will treat as [b]independent variables[/b] (forgetting that they are in fact conjugate to each other)."
and using the statement above, it follows $\frac{\partial z}{\partial \bar z}=0$

I assume that the last formula in your post should actually be $\frac{\partial f}{\partial\bar z}=0$.

In my humble opinion, this is a really bad example of mathematical writing, which is almost impossible to make much sense of. The problem with it is that while we can, of course, consider the function $F(z,w)=f\left(\frac{z+w}2, -i\frac{z-w}{2}\right)$ of two independent complex variables $z,w$ and formally use the chain rule to get $\frac{\partial F}{\partial w}=\frac{\partial f}{\partial x}\frac{dx}{dw}+\frac{\partial f}{\partial y}\frac{dy}{dw}=\frac12\left(\frac{\partial f}{\partial x}+i\frac{\partial f}{\partial y} \right)$, which is what, probably, was meant, the function $F$ is actually defined only for $w=\overline z$ and to interpret the partial derivative with respect to $w$ in the usual way, i.e., as the limit of the difference ratio as $w$ changes and $z$ stays fixed is impossible (once you change $w$ a bit without changing $z$, you are outside the domain of $F$). The best we can say here is that if $f$ could be extended from $\mathbb R^2$ to $\mathbb C^2$ as a differentiable (with respect to complex variables $x$ and $y$) function, then the  partial derivative of $F$ with respect to $w$ would be given by that formula. This is something that suggests that $\frac{\partial f}{\partial \overline z}$ should be defined the way it is, but it by no means amounts to a rigorous derivation or a proof of anything.

While I do agree with everything fedja just said, I also find $\frac{\partial f}{\partial \overline{z}}$ to be a very useful way of writing the Cauchy-Riemann equations.

[quote="fedja"]I assume that the last formula in your post should actually be $\frac{\partial f}{\partial\bar z}=0$.
. The best we can say here is that if $f$ could be extended from $\mathbb R^2$ to $\mathbb C^2$ as a differentiable (with respect to [b]complex variables [/b]$x$ and $y$) function, then the  partial derivative of $F$ with respect to $w$ would be given by that formula. This is something that suggests that $\frac{\partial f}{\partial \overline z}$ should be defined the way it is, but it by no means amounts to a rigorous [b]derivation[/b] or a proof of anything.[/quote]
Thank you both. The question arose in my mind, when we were solving one of the problems of "Complex variables and applications" by Brown & Churchill in the class: "By formally applying the chain rule in [b]calculus[/b] to a function $F(x,y)$ of two variables, derive the expression $\frac{\partial F}{\partial \bar z}=\frac{\partial f}{\partial x}\frac{\partial x}{\partial w}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial w}=\frac12\left(\frac{\partial f}{\partial x}+i\frac{\partial f}{\partial y} \right)$".
My instructor answered my question this way:If $z$ and $\bar z$ were independent, the result was clear. [b]It can be shown that [/b], the result is the same in the opposite case.
The parts in the bold face type are still confusing for me.


The number 1000000! (one million factorial) ends with lots
        of zeros.  How many zeros?  What is the rightmost
non-zero  digit?

[hide="Hint"]The number of finishing zeros in a number is the number of factors 10. Which is the least amount of factors 2 and 5.[/hide]

isnt it just

[hide]Since the number of factors of 2 is greatly larger then the number of factors of 5 we see it is..

$\lfloor{\frac{1000000}{5}}\rfloor+\lfloor{\frac{1000000}{5^{2}}}\rfloor+\lfloor{\frac{1000000}{5^{3}}}\rfloor+...+\lfloor{\frac{1000000}{5^{8}}}\rfloor=\boxed{249998}$

Correct me if i'm wrong!
[/hide]

we can write every ten number block as 10! because
$1\cdot2\cdot3\cdot...\cdot10\equiv11\cdot12\cdot13...\cdot20\bmod 10$
So, we can call all ten number block as $10!$. There are $1000000/10=100000$ blocks in $1000000!$. Then, we can write this number as
${10!}^{10^{5}}$. After we exclude 0's from 10!, we exclude zeros from the whole number. So, $10!/{100}=36288\equiv8\bmod{10}$. Then the rightmost non-zero number is $8^{10^{5}}\equiv6\bmod{10}$

More interesting:

The number $1000000_{12}!$ (that is, twelve to the sixth, factorial) ends in a large number of 0s when expressed in base 12.

Before doing the calculation, do you think it will be bigger or smaller than the number in the above problem?  Can I choose numbers for the base that make the answer smaller than either, or larger than either?

Calculation: How many 0s does it actually end in?  ($12^{6}$ is very big; one might, at this point, switch to considering $2\cdot 12^{3}$ instead.)

Questions, having done the above:
What was the constraining factor on the 0s?  What if instead of $12^{6}$ (or $2\cdot 12^{3}$), I had chosen $12^{6}-1$ (or $2\cdot 12^{3}-1$); then what would be the constraining factor?  Explain.

Extra-special bonus: everything gets even worse when I replace 12 with 720.  Why?  (You might try the calculation with 24!, 26!, 30!, 35!, 39! and 44!, and keep track of which values are preventing you from adding an extra 0.)

Here is an idea, but not a solution.
[hide] Well, for this one, you have to calculate the number of factors of 2 and 3 since they are the prime factors of 12.  Since there are 2 factors of 2 in twelve, you will have to take the number of factors of 2 in 1000000 and divide it by 2.  For example, if there were n factors of two in 1000000, then there are n/2 factors of 4.  So you would have to what SM4RT did except for 2 and 3 instead of 5.  Then see which one is smaller  after you divide the number of factors of 2 by 2.  Whichever amount is smaller is the answer (if that makes any sense at all, which it probably doesn't). [/hide]


[quote="infinity"]In a triangle ABC, angle BAC=35 degrees. D is a point on AB such that AD=BC and ADC=110 degrees. E is a point on AC such that BE=DC. What is the size of ABE?

I think it is 35 degrees, but could someone please help me out? Much appreciated![/quote]

since $\angle ACD =\angle BAC$ we know that $AD=BC=DC=BE$  so it isnt hard to deduce that $\angle ABE =40$degree

[quote="infinity"]Thankyou for answering my question, but how did you deduce that ABE=40 degrees?[/quote]

$\triangle ADC$ is isosceles at $AD$ and $CD$ which means $\angle ADC=110$ degree . So you will get $\angle BDC =\angle CDB =70$ degree . Then obtain $\angle BCD =40$ degree. $\angle AEB =180 -\angle BEC = 105$ degree . And lastly , $\angle ABE =180-\angle BAE -\angle AEB =40$ degree

[quote="infinity"]How did you get the angle BEC?[/quote]

$\triangle BCE$ is isoceles at $BC$ and $BE$ ...

yes I definately agree with Shyong the answer is $40^0$. :D

[quote="mathmanman"]:rotfl: So, the answer is 1?  j/k[/quote]
what? where did $1$ come from?

$40^0$ (forty to the zero) equals 1.

Try using \circ in $\LaTeX$ in order to avoid confusion with the degree symbol (e.g. $40^\circ$).

ohhhhh I understand what mathmanman said now :blush:  :blush:  :blush:  :idea:  Ill use the \circ symbol now.

Yup, thanks to joml88 for clearing that up :)
@riddler : no offense meant, I was just kidding!

doesnt matter... oh mannn that was soo embarassing :blush:

It really doesn't matter, I didn't want to embarass you dude, but, at least, you won't do that again  :P


For [b]all[/b] positive integral $ n$?  It's not entirely clear.

sorry, it is for all positive integral n.

all positive integral $ n$ less than $ 30$?


anyone know a good source of problems and stuff, that will help get my modular arithmetic background ( ie basics ) concrete?

Search the forums.  Search wikipedia? Google it.  Take the Intro the Number Theory class.  Lots of ways :roll:

OK, I think it would be good to search the forums as this has been posted many times.  I'm pretty sure you will find the info you were looking for. :)


geometry
circumcircle
trigonometry
perpendicular bisector
Pythagorean Theorem
geometry solved

Since $AB = AC < BC$, $PC < \frac{BC}{2}$. Let $M$ be the midpoint of the segment $BC$ and $Z$ the intersection of the perpendicular bisector of the segment $BC$ with the circumcircle $(O)$ of the isosceles triangle $\triangle ABC$ different from the vertex $A$. The right angle triangles $\triangle AZC \sim \triangle CAM$ are similar, because the angle $\angle ZAC \equiv \angle MAC$ is common. Hence,

$\frac{ZA}{CA} = \frac{CA}{MA},\ \ 2R = ZA = \frac{CA^2}{MA}$

where $R$ is the circumradius. The right angle triangles $\triangle CDA \cong \triangle AZC$ are congruent, because the angles $\angle CDA = \angle CZA$ spanning the same arc $CA$ of the circumcircle $(O)$ are equal. Hence, $DA = CZ$. Using Pythagorean theorem for the triangles $\triangle AZC, \triangle CAM$,

$DA = CZ = \sqrt{ZA^2 - CA^2} =  \sqrt{\left(\frac{CA^2}{MA}\right)^2 - CA^2} =$

$= \frac{CA}{MA} \sqrt{CA^2 - MA^2} = \frac{CA \cdot MC}{MA}$

Using the sine theorem for the triangle $\triangle APC$ and the fact that $\angle DAP = 90^0 - \angle PAC$, $\cos{\widehat{DAP}} = \sin{\widehat{PAC}}$,

$\frac{PC}{AP} = \frac{\sin{\widehat{PAC}}}{\sin{\widehat{ACP}}} =\frac{CA}{MA}\ \cos{\widehat{DAP}}$

$\cos{\widehat{DAP}} = \frac{PC}{AP} \cdot \frac{MA}{CA}$

Now we can calculate the remaining side $DP$ of the triangle $\triangle DPA$ using the cosine theorem:

$DP^2 = AP^2 + DA^2 - 2AP \cdot DA\ \cos{\widehat{DAP}} =$

$= AP^2 + DA^2 - 2 AP\ \frac{CA \cdot MC}{MA} \cdot \frac{PC}{AP} \cdot \frac{MA}{CA} = $

$= AP^2 + DA^2 - 2 MC \cdot PC = AP^2 + DA^2 - BC \cdot PC = DA^2$

$DA = DP$

[quote="mathpk"]$ABC$ is an isosceles triangle with $AB=AC<BC$. Let $P$ be a point on $BC$ such that $AP^2=BC\cdot PC$, and let $CD$ be a diameter of the circumcircle of triangle $ABC$. Prove that $DA=DP$.[/quote]

Here is my solution of this problem. It is a bit algebraic, but I don't think it is worth to look for anything more synthetic for this ugly problem...

Since CD is a diameter of the circumcircle of triangle ABC, we have < DAC = 90° and < DBC = 90°. Thus, the triangles DAC and DBC are right-angled, so that the Pythagoras theorem yields $DC^2=DA^2+AC^2$ and $DC^2=DB^2+BC^2$. Thus, $DA^2+AC^2=DB^2+BC^2$.

On the other hand, since the triangle ABC is isosceles with AB = AC, the midpoint R of its base BC is simultaneously the foot of its altitude from the apex A. Hence, the triangles ARC and ARP are right-angled, so that Pythagoras yields $AP^2=AR^2+RP^2$ and $AC^2=AR^2+RC^2$. Consequently, $AC^2-AP^2=RC^2-RP^2$. Since $AP^2=BC\cdot PC$, this becomes

$AC^2-BC\cdot PC=RC^2-RP^2=\left(RC-RP\right)\cdot\left(RC+RP\right)$.

Since R is the midpoint of the segment BC, we have RC = BR, so that this becomes

$AC^2-BC\cdot PC=\left(RC-RP\right)\cdot\left(BR+RP\right)=PC\cdot BP$.

Consequently,

$AC^2=PC\cdot BP+BC\cdot PC=PC\cdot\left(BC+BP\right)$
$=\left(BC-BP\right)\cdot\left(BC+BP\right)=BC^2-BP^2$.

Hence, the equality $DA^2+AC^2=DB^2+BC^2$ transforms into $DA^2+\left(BC^2-BP^2\right)=DB^2+BC^2$. This is equivalent to $DA^2=DB^2+BP^2$. On the other hand, since < DBC = 90°, the triangle DBP is right-angled, and thus, by Pythagoras, $DP^2=DB^2+BP^2$. Hence, $DA^2=DP^2$, so that DA = DP, and the problem is solved.

  darij


function
geometry
geometric transformation
rotation
ratio
limit
calculus

the angular velocity of uniform rotation is defined as the ratio of the angle of rotation to the corresponding time interval .give the definition for the angular velocity of nonuniform rotation.

[quote="azadeh"]:!:   ? :huh:[/quote]

If you will notice Classroom Math is in the MIDDLE SCHOOL FORUM. This question is easily harder than stuff in middle school. I'm guessing...physics?

nonuniform rotation implies $\overline{\omega}$ is not constant.  We have $\overline{\omega}= \frac{\Delta\theta}{\Delta t}$.  We can say that $\omega_{x}$ is the instantaneous angular velocity defined by $\omega_{x}= \lim_{\Delta t \rightarrow 0}\frac{\Delta \theta}{\Delta t}= \frac{d\theta}{dt}$.

But that's calculus which SHOULD NOT BE HERE!!!


Pat wants to select 8 pieces of fruit to bring in the car for the people he's driving to Montana with. He randomly chooses each piece of fruit to be an orange, and apple, or a banana. What is the probability that either exactly 3 of the pieces of fruit are oranges or exactly 6 of the pieces of fruit are apples?

Probability that $ 3$ are orange:

$ \binom{8}{3} \times \left ( \frac{1}{3} \right )^3 \times \left (\frac{2}{3} \right )^5 \equal{} \frac{1792}{6561}$

Probability that $ 6$ are apples:

$ \binom{8}{6} \times \left (\frac{1}{3} \right )^6 \times \left (\frac{2}{3} \right )^2 \equal{} \frac{112}{6561}$.

Adding yields $ \boxed{\frac{1904}{6561}}$.

[quote=Eugenis]In my solution I assumed all the fruits of one type were indistinguishable. Can someone explain where I went wrong?

[hide] First consider the case where exactly 3 pieces of fruit are oranges. There are 5 pieces of fruit left, and they can either be apple of banana. By stars and bars we have $6$ ways to distribute the rest of the fruits. Then consider when exactly 6 pieces of fruit are apples. There are 2 pieces of fruit remaining for oranges and bananas. Therefore by stars and bars $3$ ways to distribute the rest. So in conclusion we have $9$ outcomes, and in total there are $45$ by stars and bars. So $\frac{9}{45}=\frac{1}{5}.$ [/hide][/quote]

That's what I did too. I think the reason this is not correct is because the order matters?

You used stars and bars. Now you have to translate back from stars and bars.


Πότε μια συνάρτηση είναι ομοιόμορφα συνεχής? Δίνω μερικές συναρτήσεις οι οποίες πρέπει να εξεταστούν για ομοιόμορφη συνέχεια.

1) $ f(x)\equal{}x\plus{}1 \quad ,\quad x\in R$

2) $ f(x)\equal{}\frac{1}{x}\quad ,\quad x\in(0,1)$

3) $ f(x)\equal{}\sqrt{x} \quad ,\quad x\geq0$

4) $ f(x)\equal{}\frac{1}{x^2\plus{}2} \quad ,\quad x\in R$

5) $ f(x)\equal{}(tanx)^x \quad ,\quad x\in (0,\frac{\pi}{2})$

FILE MAILO, GENIKA AN THELEIS NA APOFYGEIS TON E-D ORISMO THS OMOIOMORFHS SYNEXEIAS, EINAI XRHSIMA KAPOIA THEWRHMATA. MIA SYNARTHSH EINAI OMOIOMORFA SYNEXHS AN H PARAGWGOS THS EINAI FRAGMENH, EINAI OMOIOMORFA SYNEXHS STO [A,B] AN EINAI SYNEXHS STO [A,B], H EINAI OMOIOMORFA SYNEXHS STO [0, APEIRO] AN EINAI SYNEXHS STO [0,A] KAI STO [A, APEIRO] GIA A>0.

Κωστα, πολυ χρησιμο αυτο που λες γιατι ο ε-δ ορισμος με δυσκολευει αρκετα. Ομως για τις συναρτησεις που εβαλα θα ηθελα να δω λυση με ε-δ...


On Tuesday, December 13, at 7:30 PM ET (4:30 PM PT), we will have Sasha Schwartz, a past IMO gold medalist, USAMO winner, and National MATHCOUNTS champion, discuss careers in finance at D.E. Shaw group.  Many excellent problem solvers, like Sasha, go on to exciting and lucrative careers in finance.  At this Math Jam, you can learn more about this field, and how to prepare yourself for it.

[quote="hello"]Is he going to come to this site later on to answer questions? For example, if I cant be online at the time, will he answer posts in a thread?[/quote]

That I can't be sure about, but we'll likely have more similar Math Jams in the future.

[quote="hello"]Is he going to come to this site later on to answer questions? For example, if I cant be online at the time, will he answer posts in a thread?[/quote]

Yeah. That would be cool! In the meanwhile I am waiting for the transcript!  :)


I was wondering if anyone had state competions from 1999 and before that.
If you do can you pm me the test or something??
thanks alot.

P.S. I am willing to offer to trade for some national problems and 2000-2005 states and chapters.

i might do it for the nationals answer key, but not 2001-2005 states and chapters.
THOSE ARE FREE ONLINE.

You'd have to propose a trade instead of just giving it to him in the first place?

If I had it, I'd give.

I'd give it to you if my scanner wasn't messed up.

What nats do you have?
If you have 1996,1997, or pre 1990 Ill take those.
Check your PMs

I have 1996 targets too. just not sprint

if any body has any states from before 2001, please PM me

Could anyone PM me state competitions before 2000 also?

http://www.4shared.com/dir/257226/be0f945c/Mathcounts_States.html
This is my sharing folder for state tests that I have.
I will soon be downloading tests from 95 onwards.
A "donation" of a few years of recent national tests would be greatly appreciated (like 2001-2005)

i had 1992,1993 but if i had a scanner.........

[quote="Klebian"]Oh. Yeah. I had those. Heh.
Could someone give me 04 and 05 nats then? XD[/quote]
you could buy it.....
Off to request it in the coaches forum at mathcounts.org

hey, thanks for all those state tests, the resource is awesome  :lol: !

[quote="dogseatcheese"]Is it just me....or does older competitions be a littler easier than nowadays.[/quote]

It seems just a little easier. Probably because your not nervous.

or maybe it is all on purpose...

thx to evryone who  tests here...

I think the pressure issue does have a lot to do with it but it may be that as mathcounts goes on coaches are getting better and better at coaching good kids and kids are getting better and better at getting perfet scores that mathcounts has adjusted so there aren't as many scores so close.

[quote="biffanddoc"]It is because all of the files in 1999_states.zip are for 1999

The others are for 2000 states[/quote]

Naw, I checked the 1999_states.zip you sent me, and the awnser file included is indeed the same one that I uploaded. Maybe you could check again?

Wow I just realized that the book i had was messed up.
those are the answers to the national competition :oops:

will update tomorrow :sleeping:

could someone please pm me the rite answers 4 1999????
thx

*remembers*

actually im thinking of doing solutions for that year, since it seemed easy.
 :idea:  :idea:

I have the 1996-1997 SPRINT ROUND state all scanned. If you want it, pm me any pre-1999 states or pre-2001 chapters and I will pm you the sprint(thats all I have) :( .

Where do you guys get your national competitions from?  I can't seem to find them anywhere. :read:

People please help my website project.
Once it launches we won't hav this kind of trouble.
I still need some info.

[quote="Klebian"]What I don't understand is that while I've said many times that my website has pre 2001 state tests, there have been 3 people still to ask for those tests...
No offense, obviously, but a bit of reading can get you hordes (sp?) of information[/quote]

Unless I'm missing something, theres no tests on your website from your link mentioned earlier in this thread.

Really?
It's worked for me on more than 1 computer, and at least 2 AOPSers have uploaded something on it...

And just to clarify, it has every state sprint/target from 1996 to 2005.


abstract algebra
group theory
superior algebra
superior algebra solved

would there be a ring homomorphism which is onto  $\phi:2\mathbb{Z}\rightarrow \mathbb{Z}_{2}$ ?

I don't think so.

First of all, it should induce a group homomorphism whose kernel has to have index $2$ in $2\mathbb Z$. This means that there's only one choice for the kernel, namely $4\mathbb Z$, and the homomorphism is determined: it takes multiples of $4$ to $0$ and integers of the form $4k+2$ to $1$. Now simply check that this is not compatible with the multiplication in the two rings: $f(2\cdot 2)=f(4)=0$, but, on the other hand, $f(2\cdot 2)=f(2)\cdot f(2)=1$.

But it has index $2$ in $2\mathbb Z$. You have no business with $\mathbb Z$.

[quote="grobber"]Now simply check that this is not compatible with the multiplication in the two rings:[/quote]

Hmmm. Which rings would that be?

Sorry. I forgot rings without unity exist


Hey everyone,
   How can I prove this:  absol.(b^2-2a^2) is greater or equal to 1. Given: b>a>0 and both a and b are integers.

    Thanks

This is the same as proving there are no integer solutions to $b^2=2a^2$. And then this is the same thing as proving $\sqrt2$ is not rational. Finished.


Latex is automatically right-justifying a very short line at the end of one of my solution paragraphs, and it looks gross. Does anyone know how to fix it?

Try putting a \\ at the end of the line maybe.


Circles with centers $ O$ and $ P$ have radii 2 and 4, respectively, and are externally tangent.  Points $ A$ and $ B$ are on the circle centered at $ O$, and points $ C$ and $ D$ are on the circle centered at $ P$, such that $ \overline{AD}$ and $ \overline{BC}$ are common external tangents to the circles.  What is the area of hexagon $ AOBCPD$?

[asy]
unitsize(0.4 cm); defaultpen(linewidth(0.7) + fontsize(11));
pair A, B, C, D;
pair[] O;
O[1] = (6,0);
O[2] = (12,0);
A = (32/6,8*sqrt(2)/6);
B = (32/6,-8*sqrt(2)/6);
C = 2*B;
D = 2*A;
draw(Circle(O[1],2));
draw(Circle(O[2],4));
draw((0.7*A)--(1.2*D));
draw((0.7*B)--(1.2*C));
draw(O[1]--O[2]);
draw(A--O[1]);
draw(B--O[1]);
draw(C--O[2]);
draw(D--O[2]);
label("$A$", A, NW);
label("$B$", B, SW);
label("$C$", C, SW);
label("$D$", D, NW);
dot("$O$", O[1], SE);
dot("$P$", O[2], SE);
label("$2$", (A + O[1])/2, E);
label("$4$", (D + O[2])/2, E);[/asy]

$ \textbf{(A) } 18\sqrt {3} \qquad \textbf{(B) } 24\sqrt {2} \qquad \textbf{(C) } 36 \qquad \textbf{(D) } 24\sqrt {3} \qquad \textbf{(E) } 32\sqrt {2}$

[hide="Answer"]We know that $OP=6$. Drawing $OM$ parallel to $AD$, we have $MP=DM=2$. Therefore, $OM=\sqrt{6^2-2^2}=4\sqrt{2}$, so $|AOBCPD|=2(|OMP|+|AOMD|)=2\left(\frac{1}{2}(4\sqrt{2})(2)+2(4\sqrt{2})\right)=24\sqrt{2}$. $\boxed{B}$[/hide]

[hide]Note that OPBC can be reflected along the perp. bisector of OP to form a rectangle with height 6 and length AD = x.  Draw a line from a point E (on DP) to O, where DE = 2. EO is then equal to x, then by pythagorean, $2^2 + x^2 = 6^2$, $x^2=32$, x = 4¬/2. So the total area is 6*4¬/2 = 24¬/2 -> B.[/hide]


...Breaking...Please forward to anyone who would be affected or
interested...Microsoft buys TeX for a "handsome sum", plans a "seamless
union of TeX and Office" to be offered at around $800, will phase out
all existing TeX implementations and archives within 12 months,
development moves to India, Knuth was reportedly disappointed with the
dominance and stagnation of LaTeX but declines comment...
PALO ALTO, CALIFORNIA, USA (CNEWS/MSNBC) -- In a major move into the
scientific publishing market, Microsoft Corporation announced today
that it has purchased all rights to the computer language and document
compiler known as TeX (pronounced, tech), and plans a major new
product line based on the 25-year old software.
Stanford Professor Donald Knuth (pronounced, kah-nooth), the author
of the widely-used TeX software, in a joint press conference at the
university campus with Microsoft Chairman Bill Gates, acknowledged that
the two had been negotiating for some months. I felt that a quarter
century of TeX in the public domain was enough. I am reasserting the
copyright to my original work in TeX. Microsoft will carry the ball
now, and I can get back to my computer science research.
Knuth acknowledged he was paid a handsome sum from Microsoft, which
he will use to finance his work on a project he has code-named Volume
4. He declined to comment any further or answer any question. Close
associates indicated that he was disappointed in what the TeX community
had done with his creation, citing the dominance and stagnation of
LaTeX as example. Most TeX documents now employ a flavor called LaTeX,
whose version 3 was announced about ten years ago and has never been
released.
At the press conference, Microsoft chairman Bill Gates said the
acquisition was the kind of cooperation between academia and industry
that builds prosperity for both. He added that TeX would finally give
Microsoft a foothold in mathematical desktop publishing that has
eluded the software giant since its founding. Drawing gasps of surprise
from the college audience, Gates asserted that TeX will soon be
biggest jewel in the Microsoft crown.
Apparently the jewel metaphor will include a hefty, unavoidable price
tag for future TeX users.
Gates outlined plans whereby all existing TeX compilers would be phased
out, to be replaced by a new master implementation, called Microsoft
Master TeX and written in C++, which will also incorporate key ideas
and code from Microsoft Office as well. It will be optimized for
Longhorn, Microsoft's next generation operating system, but will also
work in Windows XP; versions of Windows older than XP will not be
supported. Microsoft TeX for other platforms, such as UNIX
workstations, will follow at an as-yet unspecified date. According to
Gates, Microsoft Master TeX will be a seamless union of TeX and
Microsoft Office, incorporating the very best from each, and will
ensure that the incompatibilities across platforms are once and for all
eliminated. TeX software is widely used due its portability, although
variations among operating systems have been troublesome due to
uncoordinated development.
A beta version for public testing on Windows XP is expected soon,
issuing from an executive project center at Microsoft's Redmond campus,
but developed at and supported by a brand new 450-programmer center in
Bangalore, India.
Unlike the technical aspects of the project, Gates explained that
pricing for Microsoft TeX has already been firmly set. The single-user
retail product is expected to have a street price of about $800. When
heckled by a graduate student complaining about the high price for the
formerly free product, Gates, seemingly startled, explained that a
"student edition at $399 is likely" and that "Microsoft will use the
revenue to make TeX better."

Most current users of TeX have paid nothing for their implementations,
derived from Professor Knuth's formerly-free work. Before leaving the
podium, Gates made a final comment that "TeX hasn't changed in years.
What kind of a product can that be?" and then handed the microphone to
an assistant, introduced only as the project leader for Microsoft TeX.

The assistant displayed an overhead presentation using the current test
version of Microsoft TeX. Equations and tables could be seen dissolving
into each other in a morphing action between frames. "No one has ever
done that with TeX," Gates announced from an audience seat at one
point. "It's the kind of sizzle that can really enliven a dull academic
conference." Some onlookers were not convinced, especially when the
program crashed midway through the demonstration, resulting in a
five-minute delay while the system was restarted. Microsoft technicians
later blamed a third-party display driver.

The impact on the large base of existing TeX users was unclear. During
a question-and-answer period, Gates said that the "TeX" trademark would
be registered as the exclusive property of Microsoft, and could not
appear in any competitive or free software. "We are granting of our own
good will until the 4th quarter of this year, free use to any existing
TeX vendors or public-domain authors. That's plenty of time for an
orderly phase-out and change-over to Microsoft TeX, or no TeX at all.
After that, our legal department will be contacting them."

A Microsoft attorney added that some of the project personnel would be
dedicated to searching the Internet to find non-Microsoft TeX software.
"Archives and collections of TeX-related programs will not be permitted
beyond a 12 month grace period. The standards must be enforced, or they
become meaningless. We are rescuing a fine piece of work from being
diluted into worthlessness. You would not believe the number of
programs that have been based on TeX without any central, controlling
authority. We will stop this infringement."

Some large organizations dependent on TeX were stunned by the
announcement and had not yet formed plans for dealing with the change.
At the American Mathematical Society, whose publications largely depend
on TeX for typesetting, editor Barbara Beeton was incensed. "I can't
believe Don [Professor Donald Knuth] sold us out like this. We should
have never based a publishing enterprise of this scope on so-called
public-domain software. What were we thinking?" At many journals,
publication schedules for the rest of the year were put on hold, and
editors scrambled to reassure their authors that deadlines would not
slip more than a few months.

Certain small businesses are also expected to feel the impact of the
Microsoft ownership of TeX. Palo Alto restaurant owner Wu Chen appeared
unhappy at the news, stating that "for ten year I print new menu every
day with TeX, now I will pay big time." He displayed a crumpled,
take-out flyer, and with tears in his eyes explained how multiple
columns, exotic typefaces, and daily price changes could all be printed
by TeX in a multi-lingual format.

Commercial vendors of TeX software stand to lose everything in the face
of the new Microsoft monopoly. While most derivatives of TeX were
freely published, several companies had made a business of publishing
proprietary supported versions. One anonymous source from a leading TeX
firm said that "publishing TeX was a gold mine while it lasted, and the
Internet let us mine it deeper and deeper. Now this is a cave-in right
on our heads. TeX was a monumental work of beauty and utility, freely
given to the world by one of the finest and most generous minds of the
20th century. Now it belongs to a lucky dropout. We're finished."

It's quite breaking news. Do you believe this news? :D

Well, it's obviously a hoax. You'll convince yourself better after trying to search "Microsoft buys TeX" or something like that on Google or Yahoo news. I think that the message came from a newsgroup. Must be one of those annoying chain letters.

Nicely written though. :D

ok then, thanks for telling mee :D it came from comp.tex.tex!

Oh, Oh... it became so real at once  :D I could not find it at comp.text.tex ( http://www.1001annonces.com/1001newsgroups/lmess.php?name=comp.text.tex ) but I found an article by HJhan ( http://faq.ktug.or.kr/mywiki/HJhan ) at: http://faq.ktug.or.kr/mywiki/_c6_db_bf_c2_b1_db_b5_e9_3aLa_2fTeX_bf_a1_5f_bf_ac_b0_fc_b5_c8_5f_b3_f3_b4_e3_b5_e9

...Breaking...Please forward to anyone who would be affected or
interested...

should say enough about it's credibility :maybe:

oh my god suddenly this turned into necropost/reaper

7 minute bump :)

14 minute bump
also this is a rare post by Anonymous


Prove

$ \frac{x^{\alpha+1}}{a^\alpha} +  \frac{y^{\alpha+1}}{b^\alpha} +\frac{z^{\alpha+1}}{c^\alpha} \geq \frac{(x+y+z)^{\alpha+1}}{(a+b+c)^\alpha } $

for $a,b,c,x,y,z >0 $ and $ \alpha \geq 0$

As far as I remember Laurentiu Panaitopol proved this and an even more general one in a well-known article in GM. I say well-known since from a year or two, allmost all inequalities published in GM use this inequality. I really find this reality embarassing.

Indeed Laurentiu Panaitopol proved the more general inequality:

$\displaystyle \sum \frac{x_{1}^{a+1}}{y_{1}^{a}} \geq \frac{(\sum x_{1})^{a+1}}{(\sum y_{1})^{a}}$ which is a consequence of Holder's inequality.

This is not known also as Radon's Inequality?


Given a regular hexagon,
 two players alternatevely draw different segments from its vertices
until every intersection belongs to an odd amount of segments.

Who has a winning strategy? How about a regular polygon with more than 6 sides? (edited)

But how can each of the intersections belong to an odd number of diagonals? Suppose that $ A_1,A_2,A_3,A_4$ are four consecutive vertices. Then the intersection $ A_1A_3\cap A_2A_4$ cannot belong to any other diagonal except $ A_1A_3$ and $ A_2A_4$. Am I missing something?


What is the greatest number of regions into which five planes can divide three dimensional space?

Hint:
[hide]Thinking about 2-d, when we add in a line, look about how many regions are added, and how many times the new line is intersected by the original lines...[/hide]

Solution:
[hide]
The first few are trivial:
0 planes:1 region
1 plane: 2 regions
2 planes: 4 regions
3 planes: 8 regions
However from here we get a little tricky. If you think about How you add on a new line in the 2d version of this problem, we would go to the left of all intersections of all of the lines and add it in there.
We want to do the same sort of thing. If we go all the way outwards past the intersections of the 3 planes, then we throw our new plane in, call it A. If we consider the old plane's intersections with A, then we get one of the 2d problems. For each region that A is divided into, we add a 3-d region to one side of A.
So for the fourth plane, we add in 7 3-d regions (I assume you can do the 2-d version).
For the fifth plane, we add in 11 3-d regions.
4 planes:15 regions
5 planes: 26 regions
[/hide]


In a class of 20 students, 8 girls were absent due to a basketball team trip.  Fifty percent of the students present are girls.  Given that no boys are absent, what percent of the total class is boys?

there is 20 kids in all.
8 left so there is now 12.
50% of the 12 are girls so 6 are boys.
what percent of 20 is 6?
x/100=6/20
x/100=3/10
[b]30%[/b]

[hide][color=indigo]after 8 people leave there are 12 left. that means there are 6 boys, which is 30%[/color][/hide]

[hide]OK, so first we take out the girls; now we have 12 people. The problem says the 50% of the remaining people are guys, so we know that out of the entire class, there are 6 guys ($\frac{12}{2}$). 6 guys out of 20 total people is $\frac{30}{100}$, or [b]30%. There's your answer.[/b][/hide]


[quote="4everwise"]852 digits are used to number the pages of a book consecutively from page 1. How many pages are there in the book?[/quote]
[hide]
the first 9 numbers have one digit.
852-9=843

10 though 99 have 2 digits (thats 90 numbers)

90 x2 = 180

843-180 = 663

numbers 100 though 999 have 900 numbers with 3 digits.

663/3=221

220+100=[b]320 pages[/b]
[/hide]


If two poles $ 20''$ and $ 80''$ high are $ 100''$ apart, then the height of the intersection of the lines joining the top of each pole to the foot of the opposite pole is:

$ \textbf{(A)}\ 50'' \qquad\textbf{(B)}\ 40'' \qquad\textbf{(C)}\ 16'' \qquad\textbf{(D)}\ 60'' \qquad\textbf{(E)}\ \text{none of these}$

[hide]

We let the height of the intersection be $ x$ and the distance from the pole $ 20"$ long to the height $ y$. By similarity, we have:
\[ \frac{y}{100}\equal{}\frac{x}{80}\]
\[ \frac{100\minus{}y}{100}\equal{}\frac{x}{20}\]
Cross-multiplying gives a linear system of $ 200\minus{}20y\equal{}100x$ and $ 80y\equal{}100x$. Solving, we have $ x\equal{}16$, so the answer is $ \textbf{ (C) }$[/hide]

Note that in general, if the two poles have heights $ a$ and $ b$, then the height at the intersection point $ m$ satisfies $ \frac1m\equal{}\frac1a\plus{}\frac1b$.

It can also be solved analytically....
[hide="Solution."]
Let the Top of the $ 20"$ pole at $ (0,20)$ and its foot at $ (0,0)$
While the Top of the $ 80"$ pole $ (100,80)$ and its foot at$ (100,0)$

Finding each equation of the intersecting lines... You've got

$ x \plus{} 5y \minus{} 100 \equal{} 0$
$ 5y \minus{} 4x \equal{} 0$

Solving This System, The Point of intersection is at $ (20,16)$ where $ y_1$ here is the height from the point of intersection to the ground... So it must be $ \fbox{(C) 16"}$ :lol:
[/hide]


Let $ k$ and $ n$ be natural numbers with $ n$ being odd. Prove that $ \sum_{i\equal{}1}^k i^n$ is divisible by $ \sum_{i\equal{}1}^k i$

Let $ s_n(k)\equal{}\sum_{i\equal{}1}^{k\minus{}1}i^n\equal{}\sum_{i\equal{}1}^{k\minus{}1}(k\minus{}i)^n.$ Therefore $ 2s_n(k)\equal{}\sum_{i\equal{}1}^{k\minus{}1}(i^n\plus{}(k\minus{}i)^n)$.
It give $ k|2s_n(k)$ and $ 2s_n(k\plus{}1)\equal{}2s_n(k)\plus{}2k^n$, therefore $ k(k\plus{}1)|2s_n(k\plus{}1)\equal{}2\sum_{i\equal{}1}^ki^n$.


limit
integration
calculus
calculus computations

[quote="Lauren:)"]what is the limit as n approaches infinity of

the sum from i=1 to n of

((1 + (2i/n) )^2) (2/n)

when i=(n(n+1))/2[/code][/quote]
So if I read correctly we're trying to find $ \lim_{n \to \infty} \sum_{i \equal{} 1}^n \frac {2}{n} \cdot \left( 1 \plus{} \frac {2i}{n} \right) ^2$

It is important to note the following:
$ \int_a^b f(x) ~dx \equal{} \lim_{n\to \infty} \sum_{i \equal{} 1}^n f(x_i) ~\Delta x$
$ \int_a^b f(x) ~dx \equal{} \lim_{n\to \infty} \sum_{i \equal{} 1}^n f \left( a \plus{} i \cdot \frac {b \minus{} a}{n} \right) \cdot \frac {b \minus{} a}{n}$.

So it is obvious that $ b \minus{} a \equal{} 2$ and $ a \equal{} 1$ so $ b \equal{} 3$.  Also, $ f(x_i) \equal{} (x_i)^2$.  We then note that:

$ \lim_{n \to \infty} \sum_{i \equal{} 1}^n \frac {2}{n} \cdot \left( 1 \plus{} \frac {2i}{n} \right) ^2 \equal{} \int_1^3 x^2 ~dx$
$ \equal{} \left[ \frac {x^3}{3} \right]_1^3$
$ \equal{} \frac {1}{3} (27 \minus{} 1)$
$ \equal{} \boxed{\frac {26}{3}}$

The limit of a sum of terms is a sum of the limits of each term. (similar to integral/derivative properties of sums)

That is not true in general, especially not in this situation where the number of terms in the sum is also dependent on the variable in which we are taking our limit.


Suppose {x_i} (i=1,2,...) is a sequence of reals. x_i \geq 1. Denote b_n=x_1*x_2*...*x_n and {b_n} does not have an upper bound. Prove it holds that

1/x_1 + 1/x_2 + ... + 1/x_k < k - 1999

for k which is large enough.

===============================================

My solutions is as follows:

{b_n} is increasing.
Denote S_n = \sum {i=1}^n (1-1/x_i).
For m>n, we have
S_m - S_n
= \sum {i=n+1}^m (1-1/x_i)
= \sum {i=n+1}^m (1-b_{i-1}/b_i)
= \sum {i=n+1}^m (b_i-b_{i-1})/b_i
> \sum {i=n+1}^m (b_i-b_{i-1})/b_m
= (b_m - b_n) / b_m = 1- b_n/b_m.

Take  \varepsilon = 1/2, for an integer N, no matter how large it is, since {b_n} increases to \infty, there always exists m > n such that
b_m > 2b_n
thus, |S_m - S_n| = 1 - b_n / b_m > 1/2
From Cauchy's theorem, we know that S_n diverges.
So for any given N, there exists k such that S_k > N,
which is,
k - \sum {i=1}^k 1/x_i > N
i.e., 1/x_1 + 1/x_2 + ... + 1/x_k < k - N

===============================================

I'd like to know a solution without using Cauchy's theorem. It is not a college math problem.

Here is the same problem :
Let 0<a_1<a_2<...<a_n<...  and  Lim a_n=+ \infty .
Prove that :
a_1/a_2 + a_2/a_3 + ... +a_m/a_{m+1} < m -K
for k which is large enough


Prove that a list can be made of all the subsets of a finite set in such a way that

(i) the empty set is first in the list
(ii) each subset occurs exactly once, and
(iii) each subset in the list is obtained either by adding one element to the preceding subset or by deleting one element of the preceding subset.

We can easily do it by induction. Clearly the question is equivalent with the following: order the set of all possible n-strings consisting of zeros and ones only in such way that the first string is "00...0" and every two consecutive strings differ in one position.

We will solve this equivalent problem with the additional condition that the last string is "100...0". For $n=2$ consider such a list: "00", "01", "11", "10". To obtain a new list for the next value of $n$ precede every string in our current list $A$ by "0" and add to the obtained list $'1'+A$ written in reversed order (by $'1'+A$ I mean the ordered set $\{'1k' \ |\ k\in A\}$).


geometry
rectangle
geometric transformation
rotation
search
combinatorics unsolved
combinatorics

How many distinct tables $ m\times n$ exist filled with $ 0$ or $ 1$ such that every sub-table $ 1\times 2$ contains at least one $ 0$?

Does $ 1 \times 2$ mean what it says, or does it mean $ 1 \times 2$ or $ 2 \times 1$?  If the former, this is just $ m$ disjoint copies of the problem for a $ 1 \times n$ rectangle, and this is equivalent to the number of ways of filling a $ 1 \times n$ rectangle with 0s and 1s so that there are no two consecutive 1s, and this is just the Fibonacci number $ F_n$ (with $ F_0 \equal{} F_1 \equal{} 1$), so the answer is $ F_n^m$.

Sorry for missunderstanding. $ 1\times 2$ means all rotations and translation of it i.e. $ 1\times 2$ and $ 2\times 1$.

There doesn't seem to be a closed form for the number of such matrices.

When I computed their number for $ 1 \times 1$, $ 2 \times 2$, and $ 3 \times 3$ matrices, I was led to sequence
[url=http://www.research.att.com/~njas/sequences/?q=2%2C7%2C63&language=english&go=Search]A006506[/url] in the online encyclopedia of integer sequences.

If you're interested in how the formula behaves for large $ n$, you may want to do a search for the Hard Square Entropy Constant mentioned in Sloane.  The link there I tried is broken, but Google turned up some other neat stuff.

For fixed $ m$, one can write down a recursive system involving $ 2^m$ variables by considering the possible column adjacencies.  (I know problems of this sort have appeared on the forum before, but I haven't been able to find the smart keywords to search for.)  A nearly identical discussion of this problem took place [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=315970524&t=48673]here[/url].

Thank you for response. Case $ m\equal{}n\equal{}4$ appeared on some Korea olympiad , i tried to generalize but as you see I haven't succeeded.


function
number theory
algebra unsolved
algebra

Find all $f:\mathbb N\rightarrow\mathbb Z$ satisfying both of the following properties:
(1)If a,b are positive integers, then $f(ab)+f(a^2+b^2)=f(a)+f(b)$.
(2)If a,b are positive integers and a|b, then $f(a)\ge f(b)$.

It is from Mathlinks contest 3rd edition 7th round, and i think the answer is $f$ constant, however I think this is more a number theory problem than algebra.

I created this problem and intended to give in one of our TST. Yet, this was not considered very attractive and Valentin accepted to give it in mathlinks contest. The answer is far more general than the constant function.

sorry Harazi, I will think more carefully about it.

The answer is:   [b]f(a^a'*b^b'*...d^d'*p^p'*...r^r')[/b]
              =[b][f(p)-f(1)]+...+[f(r)-f(1)]+f(1)[/b]
where  [b]a,b,...,d[/b]are primes which are not of form [b]4k+3,k's integer
[/b]
 and [b]p,...,r[/b] are primes which congruent to 3 (mod4)

 and [b]a',b',...,r'[/b] are positive inegers.

hi, well, this is my first reply  :)

for prime p=2 and p=1(mod 4), it must satisfy f(p)=f(1) and

for prime p=3 (mod 4), it must satisfy f(p)<=f(1) (i.e. it can have any integers satisfying f(p)<=f(1) independently)

for n=p_1^e_1....p_m^e_m

define f(n)=f(p_1)+f(p_2)+...+f(p_m)-(m-1)f(1)

this function satisfies the condition which is stated in the problem.

Actually, I want to post my solution ,but I don't have any idea using LAtex.  :blush:

sorry


complex analysis
complex analysis unsolved

Hi, I'm taking my first course in complex analysis. One of the very first theorems is:

"If $ G$ is open and connected, $ f: G \to \bf C$ is differentiable with $ f'(z)\equal{}0$ for all $ z \in G$, then $ f$ is constant"

I scrutinized the proof. However, I couldn't find a counterexample if we drop connectedness.

Let $ U$ be the open ball around $ \minus{}2$ with radius $ 1$, and $ V$ the open ball around $ 2$ with radius $ 1$, too. Take $ G \equal{} U \cup V$.
Define $ f(u)\equal{}\minus{}1$ and $ f(v)\equal{}1$ for all $ u \in U, v \in V$. Then $ f'(x)\equal{}0$ for all $ x \in G$, but $ f$ is not constant.


inequalities
induction
inequalities proposed

I don't for sure,but have you tried Bernoulli's inequality? ;)

well done siuhochung . There is another way to do it without induction. hint: it is an inequality closely related to the
bernoulli inequality.

Thank you siuhochung for yet another very interesting way of proving the inequality.
My solution uses the fact that
\[ \frac{1-x^{n+1}}{1-x}= 1 + x + ... + x^n \leq 1 + {n \choose 1}x + ... + x^n = (1+x)^n\]
replacing $x = 1/2005$ will result in the inequality.  :D


If $ x,y,z \ge 0$ then:

$ a)$ $ \sqrt{(x\plus{}y)(y\plus{}z)}\plus{}\sqrt{(y\plus{}z)(z\plus{}x)}\plus{}\sqrt{(z\plus{}x)(x\plus{}y)} \le 2(x\plus{}y\plus{}z)$

$ b)$ $ x\plus{}y\plus{}z<\sqrt{(x\plus{}y)(y\plus{}z)}\plus{}\sqrt{(y\plus{}z)(z\plus{}x)}\plus{}\sqrt{(z\plus{}x)(x\plus{}y)}$.

[hide="Part A"]
By AM-GM we have the following inequalities:

$ \dfrac{(x\plus{}y)\plus{}(y\plus{}z)}{2}\ge \sqrt{(x\plus{}y)(y\plus{}z)}$

$ \dfrac{(y\plus{}z)\plus{}(z\plus{}x)}{2}\ge \sqrt{(y\plus{}z)(z\plus{}x)}$

$ \dfrac{(z\plus{}x)\plus{}(x\plus{}y)}{2}\ge \sqrt{(z\plus{}x)(x\plus{}y)}$

Adding these inequalities gives

$ \dfrac{4x\plus{}4y\plus{}4z}{2}\ge \sqrt{(x\plus{}y)(y\plus{}z)}\plus{}\sqrt{(y\plus{}z)(z\plus{}x)}\plus{}\sqrt{(z\plus{}x)(x\plus{}y)}$

Or

$ 2(x\plus{}y\plus{}z)\ge \sqrt{(x\plus{}y)(y\plus{}z)}\plus{}\sqrt{(y\plus{}z)(z\plus{}x)}\plus{}\sqrt{(z\plus{}x)(x\plus{}y)}$
[/hide]

[quote="moldovan"]If $ x,y,z \ge 0$ then:

$ a)$ $ \sqrt {(x \plus{} y)(y \plus{} z)} \plus{} \sqrt {(y \plus{} z)(z \plus{} x)} \plus{} \sqrt {(z \plus{} x)(x \plus{} y)} \le 2(x \plus{} y \plus{} z)$

$ b)$ $ x \plus{} y \plus{} z < \sqrt {(x \plus{} y)(y \plus{} z)} \plus{} \sqrt {(y \plus{} z)(z \plus{} x)} \plus{} \sqrt {(z \plus{} x)(x \plus{} y)}$.[/quote]
a,we have
$ \sum \sqrt {(x \plus{} y)(y \plus{} z)}\le \sum \frac{x\plus{}2y\plus{}z}{2}\equal{}2(x\plus{}y\plus{}z)$
b,we have $ \sqrt {(x \plus{} y)(y \plus{} z)}>y$,....
so $ \sum \sqrt {(x \plus{} y)(y \plus{} z)}>\sum x$
if x=0,i think this case is simple:D

a) Your first inequality is already proved.
b) It's easy to prove that:
$ \sqrt{(x\plus{}y)(y\plus{}z)}\geq y\plus{}\sqrt{xz},$
$ \sqrt{(y\plus{}z)(z\plus{}x)}\geq z\plus{}\sqrt{xy},$
$ \sqrt{(z\plus{}x)(x\plus{}y)}\geq x\plus{}\sqrt{yz}.$
Adding up these inequalities, we obtain:
$ \sqrt{(x\plus{}y)(y\plus{}z)}\plus{}\sqrt{(y\plus{}z)(z\plus{}x)}\plus{}\sqrt{(z\plus{}x)(x\plus{}y)}\geq$
$ \geq x\plus{}y\plus{}z\plus{}\sqrt{xy}\plus{}\sqrt{yz}\plus{}\sqrt{zx}\ \ (\ >x\plus{}y\plus{}z).$
Obviously, this one is stronger than your second inequality. Your problem is solved.


Find all natural numbers $m$ such that $\frac{m(m+1)}{3}$ is a perfect square.

Masoud Zargar

boxedexe, how did you solve it?  There's probably more than one way to do it.

[quote="kokosai"]The first few are 0, 3, 48, 675, 9408, 131043.[hide="complete solution"]There are an infinite number of solutions.  For any integer $n$, $\frac{1}{4}(-2+(7-4\sqrt{3})^n+(7+4\sqrt{3})^n)$ is a solution to the equation.[/hide][/quote]
How did you find the general solution?

[quote="boxedexe"]Easy perfect square problem[/quote]

Liar  :P  :rotfl:

[quote="kokosai"][hide]$\frac{m(m+1)}{3}=n^2$
$m(m+1)=3n^2$
$4m^2+4m=12n^2$
$4m^2+4m+1=12n^2+1$
$(2m+1)^2-3(2n)^2=1$
Let $x=2m+1$ and $y=2n$.  Then we have the Pell equation
$x^2-3y^2=1$.[/hide][/quote]
Nice, I wouldn't have known that last part.


[hide] Note that twice per hour from 12:00 to 6:00 that there will be a 90 degree angle formed.  Since there are 6 hours, there is going to be $6*2=12$ times [/hide]

Twice per hour, so its 2*number of hours, or six, which gives you 12.

guys, it might be 13 cause u have to also remember 6:00 is one (and that doesn't follow ur 2*6 thing)

jorian

6:00 won't work because 6:00 forms an angle of 180 degrees between the hour and minute hands. By the way, #H34N1, it doesn't really matter which times form the 90 degree angle, it simply matters how many there are (so you don't really need to find all of the times).

[quote="#H34N1"]But what are those 2?

It took me like about 3 minutes to get this[/quote]
One either way man. The hour hand is say between the 6 and 7. There are two places the minute hand could be to make it 90 degrees, and both of them are hit.

[quote="#H34N1"]Between $12: 00$ midnight and $6: 00$ A.M., how many times do the hour hand and the minute hand of a clock form a right angle?[/quote]
[hide]
it forms 2 times every hour. that would be 12 times.
[/hide]


with a,b,c >0 and a+b+c=3
prove that :
1)(a)^3*(b)^2+(b)^3*(c)^2+(c)^3*(a)^2>=3
 2) a^3+b^3+c^3>=(ab)^2+(bc)^2+(ca)^2;

this maybe wrong  :lol:
anyone check before solving ;
i also check  three times and the answer is  :D right
 :D  :D  :D  :lol:  ;)

sorry if posted before :blush:

I think (1) is wrong for $a=3,b=0,c=0$
although they are positive number,but it is still wrong. :D

zhaobin, i don't think 0 is positive.
Furthermore it's stated that a,b,c [b]>[/b] 0

didn't you notice my last sentence?
we let $a=3-2k,b=k,c=k,k \longrightarrow 0$
we will get $a^3b^2+b^3c^2+c^3a^2 \longrightarrow 0$
right? :)
I also guss $a^3b^2+b^3c^2+c^3a^2  \leq 3$

hei your solution is beautiful :lol:

for the first one:
zhao bin say it is wrong !
many thank

so can we prove this:
 a^3*b^2+b^3*c^2+c^ 3*a^2>=ab+bc+ca;
with a,b,c>0 and a+b+c=3 :D  :D  :D

$a^3b^2+b^3c^2+c^3a^2 \leq 3 $ is also wrong.for $a=2,b=1,c=0$
and I think this $a^3b^2+b^3c^2+c^3a^2 \leq 8 $ will be true.

hei ,it's wrong
but how about this :

let a,b,c>0 and a+b+c=3
prove that:

2((a)^3(c)^2+(b)^3(a)^2+(c)^3(b)^2)+a^3+b^3+c^3>=2(ab+bc+ca)+(ab)^2+(bc)^2+(ca)^2 :D  :D  :D

it maybe wrong :blush:  ;)  ;)

hei ,anyone agree or disagree with this ;

if this one is true there are alot of interests :D


geometry
function
geometric transformation
reflection
rectangle
absolute value
analytic geometry

I have two questions that obviously require a similar approach. Unfortunately, I do not know how to approach them? Help please!
[b]1. Find the area of the region determined by the system:
$ y\ge |x|\\
y\le \minus{}|x\plus{}1|\plus{}4$
Express your answer as a decimal[/b]

[b]2. What is the number of square units in the area of the region determined by the following system?
$ |x|\plus{}|y|\le 4y\le 0$[/b]

$ |x|\plus{}|y|\le 4y \le 0$

Absolute value is positive.
$ |x|\plus{}|y| \geq 0$

You mistake, cause there is no area.

I can hard belive that you mistoke the signal, since is quite impossible.

Actually, $ |x|$ need not be positive, it can be zero as well. In that case, the only solution for the second one would be (0,0).

[hide="hint for the first one"] Note that they intersect at right angles. Draw a good diagram, and instead of finding the points of intersection by trying to equate two absolute value functions, find the points of intersection by equating two linear functions.[/hide]

Also, there is no need to post all of your problems in bold.

[hide="Solution"]You know that $ y\ge |x|$ is just the part of the Cartesian plane that is above or equal to the graph of $ y \equal{} |x|$
Then, $ y\le \minus{} |x \plus{} 1| \plus{} 4$ is just a reflection of $ y \equal{} |x|$ in the x-axis, moved 1 unit to the left, and 4 units up.

Setting the equations equal to each other, $ |x| \equal{} \minus{} |x \plus{} 1| \plus{} 4$.
We have two cases:
(1) $ x$ is positive:
$ x \equal{} \minus{} x \minus{} 1 \plus{} 4 \\
\Rightarrow 2x \equal{} 3 \\
\Rightarrow x \equal{} \frac {3}{2} \equal{} y$

(2) $ x$ is negative:
$ \minus{} x \equal{} x \plus{} 1 \plus{} 4 \\
\Rightarrow 2x \equal{} 5 \\
\Rightarrow x \equal{} \minus{} \frac {5}{2} \equal{} \minus{} y$

The region created is just a rectangle with vertices at $ (0,0),(\frac {3}{2},\frac {3}{2}),( \minus{} 1,4),( \minus{} \frac {5}{2},\frac {5}{2})$

Using the Shoelace Formula, or the Distance Formula coupled with the formula for the area of a rectangle, we get:
$ \boxed{\frac {15}{2}}$[/hide]

The solutions is a point, so I NEVER SAID that there is no solution. I Said that there is no AREA. What is right. :)


Write $ sin^4 x$ in terms of $ cos2x$ and $ cos4x$.

detailed solution would be appreciated.

$ \sin ^ 4 x \equal{} (\sin ^ 2 x)^2 \equal{} \left(\frac {1 \minus{} \cos 2x}{2}\right)^2 \equal{} \frac {1}{4}\left(1 \minus{} 2\cos 2x \plus{} \cos ^2  2 x\right)$

$ \equal{} \frac {1}{4}\left(1 \minus{} 2\cos 2x \plus{} \frac {1 \plus{} \cos 4x}{2}\right)$

$ \equal{} \frac {3}{8} \minus{} \frac {1}{2}\cos 2x \plus{} \frac {1}{8}\cos 4x$

The last term in the first line should be $ \cos^22x$.

The basic thing to remember is that $ \cos2x\equal{}\cos^2x\minus{}\sin^2x$, so you can use the identity $ \sin^2x\plus{}\cos^2x\equal{}1$ to get $ \sin^2x$ and $ \cos^2x$ in terms of $ \cos2x$.