Helo!!
Proof that:
$ \forall(a;b;n)\in \mathbb{N}^3: (\frac{a\plus{}b}{2})^n\le a^n\plus{}b^n$

[quote="sami.d"]Helo!!
Proof that:
$ \forall(a;b;n)\in \mathbb{N}^3: (\frac {a \plus{} b}{2})^n\le a^n \plus{} b^n$[/quote]
[hide]
We apply induction on n. It is obvious that n works for the base case (n=1).
\[ (\frac {a \plus{} b}{2})^n\le a^n \plus{} b^n
\]
and we wish to prove that
\[ (\frac {a \plus{} b}{2})^{n \plus{} 1}\le a^{n \plus{} 1} \plus{} b^{n \plus{} 1}
\]
So, if we prove that $ \frac {a \plus{} b}{2}\le \frac {a^{n \plus{} 1} \plus{} b^{n \plus{} 1}}{a^n \plus{} b^n}$, we are done.  So, we want to show that $ 2a^{n \plus{} 1} \plus{} 2b^{n \plus{} 1}\ge a^{n \plus{} 1} \plus{} b^{n \plus{} 1} \plus{} ab(a^{n \minus{} 1} \plus{} b^{n \minus{} 1})$.  So, we wish to prove that $ a^{n \plus{} 1} \plus{} b^{n \plus{} 1} \minus{} a^nb \minus{} ab^n\ge 0$.
\[ a^n(a \minus{} b) \minus{} b^n(a \minus{} b)\ge 0
\]

\[ (a^n \minus{} b^n)(a \minus{} b)\ge 0
\]

\[ (a \minus{} b)^2(a^{n \minus{} 1} \plus{} a^{n \minus{} 2}b \plus{} a^{n \minus{} 3}b^2... \plus{} b^{n \minus{} 1})\ge 0
\]
which is obviously true.
[/hide]

You can use induction , Jensen's inequality.

Helo
$ \frac{a^{n\plus{}1}\plus{}b^{n\plus{}1}\plus{}a^nb\plus{}b^na}{2}\equal{}(\frac{a\plus{}b}{2})^{n\plus{}1}$
It's true??
Thx

[quote="sami.d"]Helo
nice answer,me  I have not used the recurrence,but I can't accomplish my reasoning:
$ (\frac {a \plus{} b}{2})^2\le a^n \plus{} b^n\Longleftrightarrow (\frac {1}{2})^n\le (\frac {a}{a \plus{} b})^n \plus{} (\frac {b}{a \plus{} b})^n$
put $ x \equal{} \frac {a}{a \plus{} b}$ and $ y \equal{} \frac {b}{a \plus{} b}$:
$ (\frac {1}{2})^n\le x^n \plus{} y^n$

From there, I do not know what to do.
 :blush:[/quote]

Remark $ x\plus{}y\equal{}1\Longrightarrow y\equal{}1\minus{}x.$

Helo
if I suppose:
$ x\ge y\Longrightarrow x^n\ge y^n\Longrightarrow x^n\plus{}y^n\ge 2y^n\ge(\frac{1}{2})\Longrightarrow x^n\plus{}y^n \ge (\frac{1}{2})^n$
and if we suppose:
$ y\ge x \Longrightarrow y^n\ge x^n\Longrightarrow x^n\plus{}y^n\ge (\frac{1}{2})^n$
it's true?


One ticket to a show costs $ \$20$ at full price.  Susan buys 4 tickets
using a coupon that gives her a $ 25\%$ discount.  Pam buys 5 tickets
using a coupon that gives her a $ 30\%$ discount.  How many more
dollars does Pam pay than Susan?

Susan pays $ \$20 \times 4 \times 75\% \equal{} \$60$.
Pam pays $ \$20 \times 5 \times 70\% \equal{} \$70$.
Thus, Pam pays $ \$10$ more.


[b] Problem 1[/b]. Find all function $f: \mathbb{R}\to\mathbb{R}$ such that
a) $f(x)+f(y)\neq 0$  $\forall x,y\in\mathbb{R}$
b) $\frac{f(x)-f(x-y)}{f(x)+f(x+y)}+\frac{f(x)-f(x+y)}{f(x)+f(x-y)}=0$ $\forall x,y\in\mathbb{R}$

[b] Problem 2[/b]. Find all function $f: \mathbb{R}\to\mathbb{R}^{+}$ such that: $(f(x))^{2}=1+xf(x+1)$  $\forall x\in\mathbb{R}$

[hide="2"]
$(f(x))^{2}-xf(x)-1=0$
$f(x)$ is the positive root of the above polynomial, where $a=1,b=-x,c=-1$
$f(x)=\frac{x+\sqrt{x^{2}+4}}{2}$
The other root is indeed negative because $\sqrt{x^{2}+4}>x$
[/hide]

Hmm?

$(x+1)^{2}= x^{2}+2x+1$
$1+x(x+1) = x^{2}+x+1$

Edit:  tjhance, you haven't proven that $f(x)$ is always positive.

Yes, this is my mistake typing. Part 2 of problem i have edited!

[hide="Hint for 1"]Simplify condition (b) and substitute $b_{n}=[f(n)]^{2}$.  Then use characteristic equations.[/hide]
[hide="Hint for 2"]Induction[/hide]

[hide]
Let $f(x)=x+1+h(x)$

We then get $2xh(x)+2h(x)+h(x)^{2}=h(x+1)$
$h(x+1)=h(x)[2x+2+h(x)]$

From the original equation with $f$, we know that $f(0)=1 \implies h(0)=0$.

From here we can induct to show that $h(x)=0$ for all positive integer x values...

... but that doesn't help too much since this has to be proven for all real input ... boxedexe, what do you mean by use induction?  :huh:

Does it help to write this in the form:
$f(x+1)=\frac{f(x)^{2}-f(0)^{2}}{x-0}$
[/hide]

I don't see how induction or characteristic equations can work over all reals... Anyway, here's my solution:

[hide="1"]We get a common denominator and then have numerator $2*f(x)^{2}-f(x-y)^{2}-f(x+y)^{2}=0$, or $\frac{f(y)^{2}-f(0)^{2}}y=\frac{f(0)^{2}-f(-y)^{2}}y=\frac{f(-y)^{2}-f(-2y)^{2}}y=...$, and so on over all multiples of y (that is, we have a line that passes through all $f(yn)^{2}$). As we let y approach zero, we see that $f(x)^{2}$ is actually just a line mx+b. But then mx+b>0 by the first condition and since $f(x)^{2}>0$, which is only possible if m=0. In that case, $\boxed{f(x)=c}$ for constant c.[/hide]


calculus
integration
trigonometry
logarithms
function
calculus computations

Evaluate: \[ I\equal{}\int_{\minus{}\frac{\pi}{3}}^{\frac{\pi}{3}}\frac{|x\sin x|}{\cos^2x}dx\]

[hide="Just Answer"]$ \frac{4}{3}\pi \plus{}2\ln (1\plus{}\sqrt{2})$.[/hide]

[quote="April"]Evaluate:
\[ I \equal{} \int_{ \minus{} \frac {\pi}{3}}^{\frac {\pi}{3}}\frac {|x\sin x|}{\cos^2x}dx
\]
[/quote]

ummm... why do you have the absolute sign ? it is not necessary since the "even" function is [i]always[/i] $ \geq 0$ within the specified limits.

$ I\equal{}2\int_0^{\frac {\pi}{3}}\frac {x\sin x}{\cos^2x}dx\equal{}2\int_0^{\frac {\pi}{3}}x\sec x\tan x\;dx$

$ \equal{}2\left\{ \left(x\int\sec x\tan x\;dx\right)\bigg|_0^{\frac {\pi}{3}}\minus{}\int_0^{\frac {\pi}{3}}1\; \int \sec x\tan x\; dx\;\;\;dx \right\}$

$ \equal{}2\left\{ \left(x\;\sec x\right)\bigg|_0^{\frac {\pi}{3}}\minus{}\left(\ln\,\tan\left(\frac{\pi}{4}\plus{}\frac{x}{2}\right) \right) \bigg|_0^{\frac {\pi}{3}}  \right\}$

$ \equal{}\frac{4\pi}{3}\minus{}2\ln\tan\left(\frac{5\pi}{12}\right)\equal{}\boxed{\frac{4\pi}{3}\minus{}2\ln(2\plus{}\sqrt3)}$


Find all real values of $ x$ that satisfy
$ x^4 \minus{} 4x^3 \plus{} 5x^2 \minus{} 4x \plus{} 1 \equal{} 0$.

$ x^4 \minus{} 4x^3 \plus{} 5x^2 \minus{} 4x \plus{} 1 \equal{} 0 \equal{} (x \minus{} 1)^4 \minus{} x^2 \equal{} (x^2 \minus{} 3x \plus{} 1)(x^2 \minus{} x \plus{} 1) \equal{} 0$

$ \implies x \equal{} \boxed{\left\{\frac{3\pm\sqrt5}2\right\}}$ (the latter doesn't have real roots).


function
integration
limit
real analysis
calculus
derivative
real analysis unsolved

Let the functions $ u:\mathbb{R}\times [0,\ \infty)\rightarrow\mathbb{C}$ be $ C^{2}$ and $ f:\ \mathbb{R}\rightarrow\mathbb{C}$ be continuous. Let $ u(x,\ t),\ f(x)$ be the functions with periods of $ 2\pi$ such that $ \frac{\partial u}{\partial t}(x,\ t)\minus{}\frac{\partial^{2}u}{\partial x^{2}}\equal{} f(x)\ (t > 0,\ \ x\in\mathbb{R}),\ \int_{\minus{}\pi}^{\pi}f(x)\ dx \equal{} 0$. Prove that $ \lim_{t\rightarrow\infty}u(x,\ t)$ converges to some function and converges uniformly for $ x\in\mathbb{R}$ then express the Fourier expansion of $ u(x,\ t)$ in terms of $ f(x)$ and $ u(x,\ 0)$.

Express $ u(x,t)\equal{}\sum_{n\in\mathbb{Z}}\widehat{u}(n,t)e^{inx}$ and $ f(x)\equal{}\sum_{n\in\mathbb{Z}}\widehat{f}(n)e^{inx}$

Applying the differential equation (the nonhomogeneous heat equation) to this, we get two cases:

Case 1: $ n\ne0.$

$ \frac{\partial}{\partial t}\widehat{u}(n,t)\plus{}n^{2}\widehat{u}(n,t)\equal{}\widehat{f}(n)$

$ \frac{\partial}{\partial t}\left(e^{n^{2}t}\widehat{u}(n,t)\right)\equal{}\widehat{f}(n)e^{n^{2}t}$

$ e^{n^{2}t}\widehat{u}(n,t)\equal{}\frac{\widehat{f}(n)}{n^{2}}\left(e^{n^{2}t}\minus{}1\right)\plus{}\widehat{u}(n,0)$

$ \widehat{u}(n,t)\equal{}\frac{\widehat{f}(n)}{n^{2}}\plus{}e^{\minus{}n^{2}t}C_{n}$ where $ C_{n}\equal{}\widehat{u}(n,0)\minus{}\frac{\widehat{f}(n)}{n^{2}}.$

Case 2: $ n\equal{}0.$ In this case, note that $ \widehat{f}(0)\equal{}0.$ Hence we have

$ \frac{\partial}{\partial t}\widehat{u}(0,t)\equal{}0$ or $ \widehat{u}(0,t)\equal{}\widehat{u}(0,0).$

Then the solution is this:

$ u(x,t)\equal{}\widehat{u}(0,0)\plus{}\sum_{n\in\mathbb{Z}\setminus\{0\}}\left[\frac{\widehat{f}(n)}{n^{2}}\plus{}C_{n}e^{\minus{}n^{2}t}\right]e^{inx}.$

Since the sequence $ C_{n}$ is bounded (at the very least - it's better than that), then we have that

$ \lim_{t\to\infty}\sum_{n\in\mathbb{Z}\setminus\{0\}}C_{n}e^{\minus{}n^{2}t}e^{inx}$ tends uniformly to zero.

Given that, we have

$ \lim_{t\to\infty}u(x,t)\equal{}\widehat{u}(0,0)\plus{}\sum_{n\in\mathbb{Z}\setminus\{0\}}\frac{\widehat{f}(n)}{n^{2}}\,e^{inx}.$

Since the coefficients of this series are $ O(n^{\minus{}2})$ (actually, by Riemann-Lebesgue, $ o(n^{\minus{}2})$) and hence absolutely convergent, this is a continuous function.

We can rewrite this as

$ \lim_{t\to\infty}u(x,t)\equal{}\frac{1}{2\pi}\int_{\minus{}\pi}^{\pi}u(x,0)\,dx\plus{}\frac{1}{2\pi}\int_{\minus{}\pi}^{\pi}K(x\minus{}y)f(y)\,dy$ where

$ K(x)\equal{}\sum_{n\in\mathbb{Z}\setminus\{0\}}\frac{1}{n^{2}}\,e^{inx}.$

It should be possible to give an explicit calculation of $ K(x).$

[quote="Kent Merryfield"]

$ K(x) \equal{}\sum_{n\in\mathbb{Z}\setminus\{0\}}\frac{1}{n^{2}}\,e^{inx}.$

It should be possible to give an explicit calculation of $ K(x).$[/quote]

I did that recently in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=162601]this[/url] thread (it's near the bottom).

I'll do my own calculation, making full use of distributions and weak derivatives.

For functions periodic of period $ 2\pi,$

$ \widehat{\delta}(n)\equal{}\frac{1}{2\pi}$ and $ \widehat{1}(n)\equal{}\begin{cases}1,&n\equal{}0\\0,&n\ne0\end{cases}.$

We have that $ \widehat{K}(n)\equal{}\frac{1}{n^{2}}$ except that $ \widehat{K}(0)\equal{}0.$ Take two derivatives:

$ \widehat{K''}(n)\equal{}\begin{cases}0,&n\equal{}0\\\minus{}1&n\ne0\end{cases}.$

From this, $ K''(x)\equal{}1\minus{}2\pi\delta(x).$

So $ K'(x)\equal{}x\plus{}C$ on $ (0,2\pi)$ with $ C$ chosen so that $ K'$ has mean value zero. This leads to

$ K'(x)\equal{}x\minus{}\pi$ on $ (0,2\pi).$ The size of the jump discontinuity at zero is correct.

Integrate again: $ K(x)\equal{}\frac{1}2x^{2}\minus{}\pi x\plus{}C$ on $ (0,2\pi).$ Again, the constant must be chosen so the function has mean value zero. That leads to

$ K(x)\equal{}\frac{1}2x^{2}\minus{}\pi x\plus{}\frac{1}3\pi^{2}$ on $ [0,2\pi]$

(Note that simple evaluation of $ K(0)$ gives a computation of $ \zeta(2)$ and computation of $ \|K\|_{2}$ gives a computation of $ \zeta(4).$)


calculus
integration
real analysis
real analysis unsolved

Suppose that 1 < q < infinity and let

F = {F in C[0, 1] : F(x) = integral from 0 to x of f where f is in Lq and q-norm(f) less than or equal to 1.}

Show that F is compact in C[0, 1]

Toad said:

Suppose that $1<q<\infty.$

$\mathfrak{F}=\{F\in C[0,1] : F(x)=\int_0^xf\,\text{where}\,f\in L^q,\,||f||_q \le1\}$

Prove that $\mathfrak{F}$ is compact in $C[0,1].$

--

(It won't do to have the plain letter "F" meaning two different things; I got that version in Tex by using \mathfrak{F}.)


there are four knights on a 3x3 chessboard.  On the upper left and lower left corners there are white knights and on the upper right and lower right corners there are black knights.  Prove that it it is impossible, using legal moves, to make it so that there are white knights in the upper left and lower right corners and black knights are on the upper right and lower left corners.
[hide="hint"]use graph theory[/hide]
[hide="another hint"]what is the path a knight can take isomorphic to?[/hide]

okay, since nobody is getting this here is another hint:
[hide="hint"]what possible paths can the knights take?[/hide]

[hide="Solution"]Numbering the 'spaces' in a 3X3 chess board with the numbers 1 through 9 as seen below, we see that a knight in position 1 can only move to position 6 or 8.  By drawing the 'spaces' as vertices and connecting vertices with edges when it is possible to make a legal move we get the following graph.  Notice that a black piece can not get in between the white pieces without occupying the same vertex at the same instant, and vice versa.  Therefore the black knights can not be diagonal from one another and the white knights can not either.[/hide]


Ross Mathematics Program
abstract algebra
topology
advanced fields
advanced fields unsolved

The book Abstract Harmonic Analysis by Hewitt and Ross states that every infinite abelian topological group admits a non-discrete Hausdorff topology. Is there an example of a non-abelian infinite topological group that only admits the discrete topology?

do you mean not discrete and indiscrete? Since every group admits an indiscrete topology.

That's the Hausdorff condition. A fixed statement: does there exist an infinite nonabelian group such that its only Hausdorff topology (as a top. group) is the discrete topology?


Find all $f: N \rightarrow N$ satisfy the conditions A and B
$A. f(n+1)>f(n)$ $\forall n \in Z^{+}$
$B. f(f(n))= n+2006$ $\forall n \in Z^{+}$
 :blush:  :D

From B $f(n+2006)=f(n)+2006$. Therefore if $f(n)>n+1003$ then $f(n+1003)\ge f(n)+1003$. It give f(f(n)>n+2006 (contradition). If f(n)<n+1003 then f(f(n))>f(n)+1003 give contradition. Therefore $f(n)=n+1003 \ \forall n\in N$.


Again, I think this goes here...

Does anyone have an easy way to convert numbers from base 10 to whatever base?
Or vice versa?

Thanks!

Hi MathIsGreat.

How about here: http://en.wikipedia.org/wiki/Positional_notation

Hope this helps.

thanks!
However, I was looking for a more basic explanation...

I assume you already know what a number in base $ n$ means. (For example, $ (123)_4$ equals $ 1\cdot4^2\plus{}2\cdot4^1\plus{}3\cdot4^0\equal{}27_{10}$)
Converting from base 10 to another base is not too difficult, since it only involves division that you are familiar with. For example, let's try converting $ (1000)_{10}$ into base 2. (Note, in the calculations below, I will omit the subscript for base 10 because we are converting from base 10).
$ 1000/2\equal{}500\plus{}0$
$ 500/2\equal{}250\plus{}0$
$ 250/2\equal{}125\plus{}0$
$ 125/2\equal{}62\plus{}1$
$ 62/2\equal{}31\plus{}0$
$ 31/2\equal{}15\plus{}1$
$ 15/2\equal{}7\plus{}1$
$ 7/2\equal{}3\plus{}1$
$ 3/2\equal{}1\plus{}1$
$ 1/2\equal{}0\plus{}1$
We stop here because furthering this process will just create more zeroes. If you think about it, the last remainder has gone through the most divisions by 2, so it's the highest power of 2, so we take the remainders in reverse order. Therefore, $ 1000\equal{}(1111101000)_2$.

Hi MathIsGreat.

Is the following site any better?

http://www.mathpath.org/concepts/Num/conv.htm


[quote]Hola a todos :

Estaba pensando que seria interesante crear una seccion para matematicas universitarias aqui en la comunidad hispanohablante . Y poder debatir topicos interesantes ya sea en los campos de algebra y analisis y porque no olvidar el calculo .

Este seccion seria un punto para postear topicos acerca de la olimpiada iberoamericana universitaria de matematicas .

Bueno no tengo mas nada que decir solo que Espero tener sus opiniones y sugerencias . y que es lo que se decidira en la comunidad .[/quote]

Bueno, compañeros ...
en vista que aun no se ha tenido una respuesta favorable de parte de los Administradores,...
esta podria ser nuestra Seccion U: Universitaria ... (se podria entender como una secuencia plana de topicos y comentarios)
espero opiniones en PM evitar colgar comentarios aki...

Comentarios aki :
[url=http://www.mathlinks.ro/Forum/posting.php?mode=editpost&p=447548]NIVEL UNIVERSSITARIO - COMENTARIOS[/url]

Los Moderadores

Aki cuelgo dos preguntas de nivel Universitario,
espero que  sean de su agrado...
Se sugiere colgar las soluciones y comentarios en este mismo Topic..
--------------------------------------------------------------
[color=blue][size=150]PROBLEMAS - NIVEL UNIVERSITARIO.[/size][/color]

[color=blue]Enunciado 01.[/color] [Problemas de Entrenamiento]
Sea $f: [0,1] \rightarrow [0,1]$ una función continua y
diferenciable tal que $f(0)=0$ y $f(1)=1$. Demostrar que existen
$a, b \in (0,1)$, con $a \neq b$, tales que

$f' (a) . f'(b) = 1.$

[color=blue]Enunciado 02.[/color]  [OMC 2005]
En el espacio lineal de todas las matrices reales $n \times n$,
econtrar la m\'axima dimension de un subespacio lineal $V$, tal
que:
$\forall X,Y \in V, traza (XY) = 0.$

Carlos Bravo ;)

Amigos,
 Aki la version en pdf  y tex. ..

 espero les sirva

Carlos Bravo ;)

Para mayor informacion sobre las olimpiadas de
Nivel Universitario
revisar esta: [url=http://imc-math.org/]OMC -web site[/url]

La cual es la pag oficial de estos eventos :
aki encuentras:
- el link a todos los host, con info de cada evento,
- los examenes y sus soluciones,
- los resultsos,
- e informacion fotografica.

Saludos,

Carlos Bravo ;)
Lima - Peru

Hola estoy estudiando y he encontrado este magnifico foro... ¿podéis ayudarme con estos términos y hacer algún ejemplo?

Muchas gracias

Si sabés inglés, buscá el foro apropiado para estudiantes universitarios, ahí te pueden ayudar........

Muchas Gracias José

Es que soy nuevo, y como este foro es TAN GRANDE!!! :huh:  pues ando perdido la verdad, por eso al ver que es seccion universitaria... y yo estudio Admón y Direccion de Empresas

Un saludo


When I try to just draw a single line I get:
[asy]draw((0,0)--(10,0));[/asy]

Also when I try to link to the Asymptote page, I get:
http://www.artofproblemsolving.com/Wiki/index.php/Asymptote_(Vector_Graphics_Language)
Which doesnt work.
Am I doing something wrong?


Two mathematicians, Kelly and Jason, play a cooperative game. The computer selects some secret positive integer $ n < 60$ (both Kelly and Jason know that $ n < 60$, but that they don't know what the value of $ n$ is). The computer tells Kelly the unit digit of $ n$, and it tells Jason the number of divisors of $ n$. Then, Kelly and Jason have the following dialogue:

Kelly: I don't know what $ n$ is, and I'm sure that you don't know either. However, I know that $ n$ is divisible by at least two different primes.

Jason: Oh, then I know what the value of $ n$ is.

Kelly: Now I also know what $ n$ is.

Assuming that both Kelly and Jason speak truthfully and to the best of their knowledge, what are all the possible values of $ n$?

Kelly's first statement implies that $ n$ has a last digit of 0, and must be divisible by 10. Since 10, 20, 30, 40, and 50 have 4, 6, 8, 8, and 6 factors, respectively, Jason can only be sure of the value of $ n$ if he knows $ n$ has 4 factors. Therefore, the only possible value of $ n$ is 10.


how would you do number 28 with a quick and accurate method?

please include an explantion

 :spam:  :clap2:

I'm assuming you mean the chapter round.
[hide="solution"]Call the number of pages Charlie reads every day x, because he reads the same number of pages each day. Therefore, the page he finishes on each day would be x, 2x,3x,..8x (these are the numbers he writes down.) Charlie's mother added all these to get 432 pages, so x+2x+3x+4x+5x+6x+7x+8x=36x=432. Solve for x and you get 12, so by the eighth day Charlie read 96 pages.[/hide]

You should post the question so I know which one your talking about, :roll:

Here's the question:
When reading a book, Charlie made a list by writing down the page number of the last page he finished reading at the end of each day. (He always finished reading a page that he started.) His mom thought his list indicated the amount of pages he had read on each day. At the end of the 8th day of reading, she added the numbers on his list and thought Charlie had read 432 pages. If Charlie started reading the book on page one, and he read the same amount of pages each day of this eight-day period, how many pages did he actually read by the end of the 8th day?


The figure shown consists of a regular hexagon and two squares. What is the number of degrees in the sum $ m\angle1 \plus{} m\angle2 \plus{} m\angle3$?

[asy]pair A,B,C,D,E;
A = (1,0); B = dir(60); C = dir(120); D = (0,0); E = (0,-1);
draw(D--dir(210)--dir(210)+C--C--C+B--C+B+A--B+A--A--D--C);
draw(C+B--E--A+E--A);
label("1",C,NW); label("2",.1E,SW); label("3",C+B+.3E,SW);[/asy]

Angle 1 is 150, Angle 3 is 30, and Angle 2 is 60. Thus our desired sum is $ \boxed{240}$.

Sorry, that was a mistake.

Angle 1 is 150, not 30...unless my eyes are bad.

Thus, the sum would be $ \boxed{240}$.


geometry
trapezoid
trigonometry
geometry proposed

Let four mobile points $A,B,C,D$ which belong to the fixed circlr $w=C(O,r)$
so that $AB=r$, $AD\parallel BC$, $x=AD$, $y=BC$. Prove that:
$AB\cap CD=\emptyset\Longrightarrow x^2-xy+y^2=3r^2$ (constant);
$AB\cap CD\ne \emptyset\Longrightarrow x^2+xy+y^2=3r^2$ (constant).

Assume that the points C, D are both on the major arc AB of the circle C(O, r). Every cyclic trapezoid is isosceles, hence, CD = AB = r. The triangles $\triangle OAB \cong \triangle OCD$ are both equilateral, hence, $\angle AOB = \angle COD = 60^\circ$. Denote $\beta = \angle BOC,\ \delta = \angle DOA$. Obviously, $\beta + \delta = 360^\circ - (\angle AOB + \angle COD) = 240^\circ$. Using the cosine theorem for the isosceles triangles $\triangle ODA, \triangle OBC$ with the bases x = AD, y = BC and shoulders OD = OA = OB = OC = r,

$x^2 = 2r^2 - 2r^2 \cos \delta,\ \ y^2 = 2r^2 - 2r^2 \cos \beta$

Using the sine theorem for the triangles $\triangle ABD, \triangle BCD$

$\frac x r = \frac{AD}{AB} = \frac{\sin \widehat{ABD}}{\sin \widehat{ADB}} = \frac{\sin \frac \delta 2}{\sin 30^\circ} = 2 \sin \frac \delta 2$

$\frac y r = \frac{BC}{CD} = \frac{\sin \widehat{BDC}}{\sin \widehat{CDB}} = \frac{\sin \frac \beta 2}{\sin 30^\circ} = 2 \sin \frac \beta 2$

$xy = 4r^2 \sin \frac \delta 2 \sin \frac \beta 2$

Adding $x^2, y^2$ and subtracting $xy$,

$x^2 + y^2 - xy = 4r^2 - 2r^2 (\cos \delta + \cos \beta) - 4r^2\sin \frac \delta 2 \sin \frac \beta 2 =$

$= 4r^2 - 4r^2 \cos \frac{\delta + \beta}{2} \cos \frac{\delta - \beta}{2} - 2r^2 \left(\cos \frac{\delta - \beta}{2} - \cos \frac{\delta + \beta}{2}\right)$

Since $\frac{\beta + \delta}{2} = 120^\circ$, $\cos \frac{\beta + \delta}{2} = -\frac 1 2$. Then

$x^2 + y^2 - xy = 4r^2 + 2r^2 \cos \frac{\delta - \beta}{2} - 2r^2 \left(\cos \frac{\delta - \beta}{2} + \frac 1 2\right) = 3r^2$

Assume that one of the points C, D is on the minor arc AB of the circle C(O, r), say the point D. The cyclic isosceles trapezoid ADBC has equal diagonals AB = CD = r, hence, the point C is on the major arc AB of the circle C(O, r). Again, the triangles $\triangle OAB \cong \triangle OCD$ are both equilateral, hence, $\angle AOB = \angle COD = 60^\circ$. Denote $\beta = \angle BOC,\ \delta = \angle DOA$ as before. Obviously, $\beta + \delta = \angle AOB + \angle COD = 120^\circ$. Using the cosine theorem for the isosceles triangles $\triangle ODA, \triangle OBC$ with the bases x = AD, y = BC and shoulders OD = OA = OB = OC = r,

$x^2 = 2r^2 - 2r^2 \cos \delta,\ \ y^2 = 2r^2 - 2r^2 \cos \beta$

Using the sine theorem for the triangles $\triangle ABD, \triangle BCD$

$\frac x r = \frac{AD}{AB} = \frac{\sin \widehat{ABD}}{\sin \widehat{ADB}} = \frac{\sin \frac \delta 2}{\sin(180^\circ - 30^\circ)} = 2 \sin \frac \delta 2$

$\frac y r = \frac{BC}{CD} = \frac{\sin \widehat{BDC}}{\sin \widehat{CDB}} = \frac{\sin \frac \beta 2}{\sin(180^\circ - 30^\circ)} = 2 \sin \frac \beta 2$

$xy = 4r^2 \sin \frac \delta 2 \sin \frac \beta 2$

Adding $x^2, y^2$ and $xy$,

$x^2 + y^2 + xy = 4r^2 - 2r^2 (\cos \delta + \cos \beta) + 4r^2\sin \frac \delta 2 \sin \frac \beta 2 =$

$= 4r^2 - 4r^2 \cos \frac{\delta + \beta}{2} \cos \frac{\delta - \beta}{2} + 2r^2 \left(\cos \frac{\delta - \beta}{2} - \cos \frac{\delta + \beta}{2}\right)$

Since $\frac{\beta + \delta}{2} = 60^\circ$, $\cos \frac{\beta + \delta}{2} = +\frac 1 2$. Then

$x^2 + y^2 + xy = 4r^2 - 2r^2 \cos \frac{\delta - \beta}{2} + 2r^2 \left(\cos \frac{\delta - \beta}{2} - \frac 1 2\right) = 3r^2$


Find all pairs of positive integers $ (x,y)$ and y is a prime number such that
$ y^x\minus{}1\equal{}(y\minus{}1)!$

[b]Liouville's Theorem[/b]: If $ p>5$ is a prime then the equality $ (p\minus{}1)!\plus{}1\equal{}p^m$ is impossible for any natural $ m\geq 2$.

Hello, [b]ehsan2004[/b]
Please share to me the proving of the [i]Liouville's Theorem[/i].
I searched in google, and i found some following links:
http://en.wikipedia.org/wiki/Liouville's_theorem_(Hamiltonian)
http://planetmath.org/encyclopedia/LiouvillesTheorem2.html
http://www.mathreference.com/cx,lioub.html
http://math.fullerton.edu/mathews/c2003/LiouvilleTheoremBib/Links/LiouvilleTheoremBib_lnk_1.html
...
But i can't look forward any page of Liouville's Theorem in number theory!
Help me, please!

it's easy , we have $ p^{m \minus{} 1} \plus{} p^{m \minus{} 2} \plus{} \cdots \plus{} p \plus{} 1 \equal{} (p \minus{} 2)!$ , $ p \minus{} 1|(p \minus{} 2)!(p\minus{}1\geq6$ and is composite) . but the left side is $ m (mod p \minus{} 1)$ . so we have $ p \minus{} 1|m$ , now u can prove that if $ p > 5$ then $ p^{p \minus{} 1} > (p \minus{} 1)! \plus{} 1$ and proof is complete :)

http://www.mathlinks.ro/Forum/viewtopic.php?t=3452

  darij


[quote="shyong"]For the second one , we write it as $111(1000a+b)$ . Hence $1000a+b$ must be divisible by $37$ or $a+b\equiv 0\mod 37$ . But $0\leq a,b\leq 9$ so $a=b=0$ ???[/quote]

means there are no solutions?
i couldn't get one either.
i got $aaabbb \equiv 0\mod9$, and $b$ must be $4$.
so trying the few possibilities, i got no answer

No solutions, although there are 3 such numbers which are exactly one less than a perfect square.

I couldn't get one, neither.. I spent like an hour.. :(

BTW, can anyone check my proof for part i, thanks.

If i'm not wrong, we can use Fermat's for part 1. but it might not be essential

[quote="kimby_102"][quote="violetcraze"]i.) Prove that  $2^{37} -1 = 137438953471$ is not a prime.

ii.) There exixt a six digit number $aaabbb$ which is a complete square. Find it[/quote]
[hide="Part i"]
We want to prove that $2^{37}-1\equiv0\mod(m)$, for some positive integer $m$.  In other words, we have to prove that $2^{37{n}}\equiv1\mod(m)$ for a positive integer $n$. This means that $\phi(m)=37n$. Assuming that $m$ is a prime number, then $\phi(m)=m-1=37n$. And when $n=6$, $m$ is a prime number, namely $223$. Hence, we found that $223$ divides $2^{37}-1$.
[/hide][/quote]

That's definitely a start; it proves that any prime divisor of $2^{37}-1$ would have to be congruent to 1 mod 74. But it doesn't guarantee that such a prime will divide $2^{37}-1$. (For example, 53 is a prime less than $\sqrt{2^{13}-1}$ and congruent to 1 mod 26, but $2^{13}-1$ is prime anyway.) It does turn out to be divisible by 223, though; this can be shown through modular arithmetic:

For $2^8\equiv 256\equiv 33$, so squaring, $2^{16}\equiv 33^2\equiv 3\cdot 363\equiv 3\cdot 140\equiv 420\equiv 197\equiv -26$, and squaring again, $2^{32}\equiv 26^2\equiv 2\cdot 338\equiv 2\cdot 115\equiv 230\equiv 7$. Multiplying by 32 gives $2^{37}\equiv 1$.

Yes (for the above) but interestingly,

if $p$  is an odd prime of the type $4k-1$ and $2p+1$ is also prime then $2^p - 1$ is divisible by $2p+1$  :)
(For example you can say that $2^{11} -1$ is divisible by 23.)


The letters in the word SLEEP are rearranged. How many different arrangements begin with the letter E?

[hide]Put an E at the beginning. Now we have S,L,E, and P left. Arrange these for 4!=24 arrangements[/hide]

[hide]Actually, 24 is correct.
Even though the two E's are indistinguishable, this is accounted for by not multiplying by two to place the other E into the spot where the first E already was.[/hide]

[quote=PeterF]I think it should be 12 instead of 24 since there is a double counting for two letter Es![/quote]

We are putting an E at the beginning to begin with so the problem becomes how many ways can you arrange the letters S, L, E, P which is $4!=\boxed{24}$.

Alternatively There are 5!/2 = 60 possibilities and 2/5 of these start with E -> 2/5*60=24


Does there exist a sequence of positive integers $ a_{2003} > a_{2002} > ... > a_2 > a_1$ with $ a_1 \equal{} 2003$ such that the following two conditions are satisfied:
1) all integers in the interval $ (2003;a_{2003})$ are either a member of this sequence or a non-prime;
2) $ A \equal{} \frac {2004}{a_1} \plus{} \frac {2004}{a_2} \plus{} ... \plus{} \frac {2004}{a_{2003}}$ is an integer?

[color=red][[b]Moderator edit:[/b] This problem was from an ongoing M&Y contest ( http://www.artofproblemsolving.com/viewtopic.php?p=1352325#1352325 ) but now it can be discussed freely again.][/color]

For example $ a_i\equal{}\frac{2003i(i\plus{}1)}{2}$, then
$ A\equal{}\frac{2004*2}{2003}\sum_{i\equal{}1}^{2003}(\frac 1i \minus{}\frac{1}{i\plus{}1})\equal{}\frac{2004*2}{2003}(1\minus{}\frac{1}{2004})\equal{}2$.


calculus
integration
logarithms
LaTeX
real analysis
real analysis unsolved

hello, for $ (i)$ is the answer $ \frac{1}{8}\pi\ln(2)$.
Sonnhard.

I think that the first integral can be generalized as follows:

Let $ 0<a<4$ be a real number. Calculate

$ \int\limits_{0}^{\infty}\frac{\ln x}{x^{2}\plus{}ax\plus{}a}dx$.

A possible argument: Using the substitution $ x\equal{}\sqrt{a}t$ we get that

$ I\equal{}\frac{1}{\sqrt{a}}\int\limits_{0}^{\infty}\frac{\ln \sqrt{a}\plus{}\ln t}{t^{2}\plus{}\sqrt{a}t\plus{}1}dt$

$ \equal{}\frac{\ln a}{2\sqrt{a}}\int\limits_{0}^{\infty}\frac{dt}{t^{2}\plus{}\sqrt{a}t\plus{}1}dt$ since the second integral is 0.

When $ a\equal{}2$ we get, as Dr. Sonnhard observed, that the value of your integral is $ \frac{\pi\ln 2}{8}$.

[quote="zhongym8903"]Though your proof is pretty good, I cannot write anything about it for I am new here and do ont know how to edit my formula on this forum.could you tell me how to edit the formulas such as integrals,is it necessary to use Word 2003 to give out my proof?
RVSP. :blush:[/quote]
Actually, this board uses a technology called "LaTeX" to typeset all the math you see here, not Microsoft Word. The AoPS site itself has a good primer to LaTeX, available here: http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php

The main plus of the LaTeX on the board is that it is generated server-side, so you do not need to have anything installed on your computer to post LaTeX; if you can access this forum, you can post nicely-formatted formulas on it :)

$ I\equal{}\int_1^\infty \frac{\ln(1\plus{}x\plus{}x^2)}{1\plus{}x^2}\, dx$ is after substitution $ x\mapsto \frac{1}{x}$

$ \tilde{I}\equal{}\int_0^1 \frac{\ln(1\plus{}x\plus{}x^2)\minus{}\ln(x^2)}{1\plus{}x^2}\, dx$

Hence $ 2I\equal{}I\plus{}\tilde{I}\equal{}\underbrace{\int_0^\infty \frac{\ln(1\plus{}x\plus{}x^2)}{1\plus{}x^2}\, dx}_{\frac13 \left(\pi \ln(2\plus{}\sqrt{3})\plus{}4K\right)}\plus{}\underbrace{\int_0^1 \frac{\minus{}2 \ln x}{1\plus{}x^2}\, dx}_{\equal{}2K}$

$ \Rightarrow I\equal{}\frac16 \left(\pi \ln(2\plus{}\sqrt{3})\plus{}10K\right)$ .

For the calculation of $ \int_0^\infty \frac{\ln(1\plus{}x\plus{}x^2)}{1\plus{}x^2}\, dx$ see this [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=1045396#1045396]link[/url].

Also on the $ \text{\LaTeX}$: There is a subforum devoted to it, and several guides. If you want to see the codes someone else used, quote their post.


function
calculus
derivative
calculus computations

How do we show that, for a twice-differentiable in $ x$ and $ y$ function $ f: \mathbb{R}^{2}\rightarrow\mathbb{R}$, we can approximate $ f\left(x,y\right)$ near the point $ \left(x_{0},y_{0}\right)$ by the following?:

\[ f\left(x,y\right)\cong f\left(x_{0},y_{0}\right)\plus{}f_{x}\left(x_{0},y_{0}\right)\left(x\minus{}x_{0}\right)\plus{}f_{y}\left(x_{0},y_{0}\right)\left(y\minus{}y_{0}\right)\plus{}\frac{1}{2}f_{xx}\left(x_{0},y_{0}\right)\left(x\minus{}x_{0}\right)^{2}\plus{}f_{xy}\left(x_{0},y_{0}\right)\left(x\minus{}x_{0}\right)\left(y\minus{}y_{0}\right)\plus{}\frac{1}{2}f_{yy}\left(x_{0},y_{0}\right)\left(y\minus{}y_{0}\right)^{2}\]

What are you allowed to assume? Its a second order Taylor series approximation, if your allowed that.

I'm trying to show it without using a Taylor series expansion.  Is there a concise way to do so?

Hi, I'm a bit confused. What do you mean by differentiable? In my understanding differentiability means existence of linear approximation, and twice differentiable here means existence of bilinear approximation and the coefficients can be easily computed. Are you actually assuming the partial derivatives exists and are continuous up to second order?


$ \text{Let }  \theta \text{ is an acute angle.}$

$ \text{Find the min. value  of}$

$ 9\sec^2 \theta\plus{} 4\csc^2 \theta\plus{}12\sec \theta\csc \theta  \text{ , without using Calculus.}$

[hide]
$ 9\sec^2 \theta \plus{} 4\csc^2 \theta \plus{} 12\sec \theta\csc \theta \equal{} (3 \sec \theta \plus{} 2 \csc \theta)^2 \ge 0$[/hide]

[quote="vinskman"][hide]$ \text{ Since the angle }\theta \text{ is acute ,}$

$ 9\sec^2 \theta \plus{} 4\csc^2 \theta \plus{} 12\sec \theta\csc \theta \equal{} (3 \sec \theta \plus{} 2 \csc \theta)^2 > 0$

$ \text {How about its min. value?}$[/hide][/quote]

We can transform the expression $ \frac 3{\cos \theta} \plus{} \frac 2{\sin \theta}\equal{}0$ to \[ \frac {2\cos \theta \plus{} 3\sin \theta}{\sin \theta \cos \theta} \equal{} 0\] So we must have $ 2\cos \theta \plus{} 3\sin \theta \equal{} 0$ and neither $ \sin \theta$ or $ \cos \theta$ equal 0.

So we have $ 2\cos \theta \equal{} \minus{}3\sin \theta$, or $ \tan \theta \equal{} \minus{}\frac 23$. But since $ \theta$ is acute, by definition, $ \tan \theta$ is positive, so the expression can never equal 0.


logarithms Putnam ratio limit function calculus derivative

How does one proceed to solve the following problem?

Find the sum of the series

$\frac{\ln 2}{2}-\frac{\ln 3}{3}+\frac{\ln 4}{4}-\frac{\ln 5}{5}+\dots$

I would proceed as follows:
(1) type sum((-1)^n*log(n)/n,n=2..infinity) into Maple;
(2) get no answer;
(3) forget about the problem.

well seeing as how this problem came from a set of problems used to help students prepare for the putnam I am sure there is an answer that does not require any computer work.
 :wink:

Use summation to write it as mlok has it and then use ratio test to conclude that it diverges. It diverges not to $\pm \infty$. I think it oscillates.

That's wrong, akech. The ratio test is useless, and you could forsee that it would be useless. The series does converge, conditionally, as an alternating series. The conditional convergence hangs a cloud of doubt over any strenuous manipulations we might attempt to compute the sum - but it is a convergent series.

As for what manipulations there might be - I tried a couple of things that aren't even worth reporting. I don't see anything mlok didn't see.

[quote="akech"]sorry, I just realized that ;-[/quote]
simpl
$\sum_{0<n\leq 2N}(-1)^{n}\frac{\ln{n}}{n}=\ln{2}\sum_{1<n\leq N}\frac{1}{n}-\sum_{N<n\leq N}\frac{\ln{n}}{n}$
=$\frac{2\gamma\ln{2}-\ln^{2}{2}}{2}+o(1)$

[quote="sos440"]Let

$S_{n}= \sum_{k=1}^{n}\frac{(-1)^{k}\ln k}{k}$
$H_{n}= \sum_{k=1}^{n}\frac{1}{k}$,
$T_{n}= H_{2n}-H_{n}= \frac{1}{n}\sum_{k=1}^{n}\frac{1}{1+(k/n)}$

Then as Extremal showed,

\begin{eqnarray*}S_{2n}& = & H_{n}\ln 2-\sum_{k = 1}^{n}\frac{\ln(n+k)}{n+k}\\ & = & H_{n}\ln 2-T_{n}\ln n-\frac{1}{n}\sum_{k=1}^{n}\frac{\ln(1+k/n)}{1+(k/n)}\\ & = & \left( (H_{n}-\ln n) \ln 2-\frac{1}{n}\sum_{k=1}^{n}\frac{\ln(1+k/n)}{1+(k/n)}\right)+(\ln2-T_{n})\ln n \end{eqnarray*}

But one can show that $\lim_{n \to \infty}n(\ln2-T_{n}) = \frac{1}{2}$, which implies that $\lim_{n \to \infty}(\ln2-T_{n})\ln n = 0$. Therefore, we have

$\lim_{n \to \infty}S_{2n}= \gamma \ln2-\frac{(\ln 2)^{2}}{2}= 0.1598689037424309717...$

Numerical calculation shows that

$S_{10^{5}}= 0.1599264681069326862...$[/quote]
In my sulution I use $\sum_{1<k\leq n}\frac{\ln{k}}{k}$~$\frac{\ln^{2}{n}}{2}+0(1)$
I think there must be $\sum_{1<k\leq n}\frac{\ln{k}}{k}=\frac{\ln^{2}{n}}{2}+C+0(1)$ where $C$ is constant. I can prove it. And therefore we can take more exact result:
$\lim_{n \to \infty}S_{2n}= \gamma \ln2-\frac{(\ln 2)^{2}}{2}+C'+o(1)$ where $C'$ is some constant.
sos440 can you calculate this constant by computer?

because of its poor convergency rate, i was able to achieve only 4-degree accuracy of approximation of

$C = \lim_{n \to \infty}\left( \sum_{k=1}^{n}\frac{\ln k}{k}-\frac{\ln^{2}n}{2}\right)$.

the estimated value is $-0.07281 \pm 10^{-5}$

Is it permissible to take the exponential of the sum?

\[\sum_{n}(-1)^{n}\frac{\ln n}{n}=\ln \prod_{n}e^{\frac{1}{n}\ln n}=\ln \prod_{n}\sqrt[n(-1)^{n}]{n}\]

and perhaps the latter is convenient

That's not very useful- the sum is actually a better form, since it's related to the zeta function and its derivative.
In fact, $\sum_{n=1}^{\infty}(-1)^{n}\frac{\ln n}{n^{s}}=\frac{d}{ds}\sum_{n=1}^{\infty}(-1)^{n}n^{-s}= \frac{d}{ds}\left(2^{1-s}\zeta(s)-\zeta(s)\right)$

Evaluating the limit of that derivative as $s\to 1$ is equivalent to finding the first two terms of the Laurent expansion of $\zeta$ at 1.

By the way, the title of this thread has been bugging me- that's a really hideous misspelling of "natural".

[color=green][Is that better? - K.M.][/color]
OK.

It works because of a relative of Abel's theorem. Try summation by parts, summing the $(-1)^{n}$ and differencing the rest; this makes our sum absolutely convergent for $s\ge 1$, without changing the values.

[quote="jmerry"]

Evaluating the limit of that derivative as $ s\to 1$ is equivalent to finding the first two terms of the Laurent expansion of $ \zeta$ at 1.

[/quote]
Why? I don't know about Laurent expansion, How to calculate $ \lim_{s \minus{} > 1}\frac {d}{ds}(2^{1 \minus{} s}\zeta{(s)} \minus{} \zeta{(s)})$

wait a minute !!! My solution is correct :
$ \sum_{1}^{2N}\frac{(\minus{}1)^{n}\ln n}{n}\equal{}\ln 2 \sum_{1}^{N}\frac{1}{k}\minus{}\sum_{N}^{2N}\frac{\ln k}{k}\equal{}\ln 2 (ln N\plus{}\gamma \plus{}o(1)
)\minus{}\frac{1}{2}(\ln^{2} (2N)\plus{}c\minus{}\ln^{2}N \minus{} c \plus{}o(1)) \minus{}>\gamma \ln 2 \minus{}\frac{\ln^{2} 2}{2}$ this is true .
jmerry you can check it whith your answer by using Laurent expansion of \zeta at 1.  (http://en.wikipedia.org/wiki/Laurent_expansion) you will takke analogisly answer.

Here's the calculation:
$ \lim_{s\to 1^\plus{}}\frac{d}{ds}\left(2^{1\minus{}s}\zeta(s)\minus{}\zeta(s)\right)\equal{} \lim_{s\to 1^\plus{}}\frac{d}{ds}\left[\left((1\minus{}s)\ln 2\plus{}\frac{(1\minus{}s)^2\ln^2 2}{2}\plus{}O(s^3)\right)\zeta(s)\right].$ If $ \zeta(s)\equal{}\frac{a_{\minus{}1}}{s\minus{}1}\plus{}a_0\plus{}a_1(s\minus{}1)\plus{}\cdots$ near $ s\equal{}1$, this becomes
$ \frac{d}{ds}\left(\minus{}a_{\minus{}1}\ln 2\plus{}(s\minus{}1)\left(\minus{}a_{0}\ln 2\plus{}\frac{a_{\minus{}1}\ln^2 2}{2}\right)\plus{}O((s\minus{}1)^2)\right)$
$ \equal{}\minus{}a_{0}\ln 2\plus{}\frac{a_{\minus{}1}\ln^2 2}{2}$ at $ s\equal{}1$.

At this point, it's clear that there was a sign mistake in my original calculation; the sum we want is actually the negative of this derivative.

It's pretty easy to show that $ a_{\minus{}1}\equal{}1$; a simple integral estimate for $ \sum_{n\equal{}1}^{\infty}n^{\minus{}s}$ suffices. Getting at $ a_0$ is a lot harder; I can show that it's $ \gamma$, but I don't want to put in all of the detail right now.

yes, i agree with you.
here [quote="jmerry"]$ \sum_{n \equal{} 1}^{\infty}( \minus{} 1)^{n}\frac {\ln n}{n^{s}} \equal{} \frac {d}{ds}\sum_{n \equal{} 1}^{\infty}( \minus{} 1)^{n}n^{ \minus{} s}$ [/quote]
must be
$ \sum_{n \equal{} 1}^{\infty}( \minus{} 1)^{n}\frac {\ln n}{n^{s}} \equal{} \minus{} \frac {d}{ds}\sum_{n \equal{} 1}^{\infty}( \minus{} 1)^{n}n^{ \minus{} s}$
but it's no so important.
More interesting is : we can generalizate this problem like this :
$ \sum_{k \equal{} 1}^{ \plus{} oo}\frac {( \minus{} 1)^{k}\ln k }{k^{s}}$

[quote="jmerry"]No. You can't calculate $ \zeta(s)$ in general- what makes you think you can calculate its derivative?[/quote]
you are right. I wanted try by asymptoic but I can't..

One direction that does make sense is to increase to power of $ \ln n$: we get $ \sum_n(\minus{}1)^n \frac{\ln^2 n}{n}\equal{}2\left(\minus{}\frac{\ln^3 2}{6}\plus{}\frac{\ln^2 2}{2}\gamma\plus{}\ln 2\cdot c\right)$, where $ c\equal{}\lim_{N\to\infty}\sum_{n\equal{}1}^N\frac{\ln n}{n}\minus{}\frac{\ln^2 N}{2}$.

In general, set $ \gamma_m\equal{}\lim_{N\to\infty}\sum_{n\equal{}1}^N\frac{\ln^m n}{\cdot n}\minus{}\frac{\ln^{m\plus{}1}N}{m\plus{}1}$. We get $ \zeta(s)\equal{}\frac1{1\minus{}s}\plus{}\sum_{m\equal{}0}^{k\minus{}1} (\minus{}1)^m\gamma_m\frac{(s\minus{}1)^m}{m!}\plus{}O(s\minus{}1)^{k}$ near $ s\equal{}1$, and the sum $ \sum_n(\minus{}1)^n \frac{\ln^k n}{n}$ is $ k!\left(\minus{}\frac{\ln^{k\plus{}1} 2}{(k\plus{}1)!}\plus{}\sum_{m\equal{}0}^{k\minus{}1}\gamma_m\frac{\ln^{k\minus{}m} 2}{m!(k\minus{}m)!}\right) \equal{}\minus{}\frac{(\ln 2)^{k\plus{}1}}{k\plus{}1}\plus{}\sum_{m\equal{}0}^{k\minus{}1}\binom{k}{m}\gamma_m(\ln 2)^{k\minus{}m}$

I imagine that all of the $ \gamma_m$ are just as intractable as $ \gamma$.

[quote="mlok"](1) type sum((-1)^n*log(n)/n,n=2..infinity) into Maple;[/quote]

LOL, you shoud type as following

[size=150]sum((-1)^n*log(n)/n,n=2..infinity);[/size]


Any of you know how to construct a unique sudoku? Please tell me.

[quote="i_like_pie"]The least complex (but not necessarily easiest) method would be to start out with a complete grid and erase numbers, making sure they can be logically deduced from the information left on the grid. I think this is what the [url=http://en.wikipedia.org/wiki/Sudoku]Sudoku[/url] article on Wikipedia refers to:

[quote="Wikipeida"]Building a Sudoku puzzle can be performed by predetermining the locations of the givens and assigning them values only as needed to make deductive progress. This technique gives the constructor greater control over the flow of puzzle solving, leading the solver along the same path the compiler used in building the puzzle. Great caution is required, however, as failing to recognize where a number can be logically deduced at any point in construction—regardless of how tortuous that logic may be—can result in an unsolvable puzzle when defining a future given contradicts what has already been built. Building a Sudoku with symmetrical givens is a simple matter of placing the undefined givens in a symmetrical pattern to begin with.[/quote][/quote]

I ACTUALLY got one that was impossible (typo) I looked in the asnwer key and the 2 numbers were different

[quote="7h3.D3m0n.117"]About numbers being erased: Many sudokus are in shapes, for example, I did a sudoku with the starting numbers in a shape of a smiley face. (Haha I did try to creat a sudoku in a Nazi sign just for fun to see if I could solve it again but it didn't work  :(  - no offense to any Jews out there).[/quote]

I've done one shaped kind of like an arrow but with some added numbers. It was expert though so I couldn't finish it. :(


Suppose you have $ L$ meters to go in straith line with your car.
The maximum acceleration of the motor is $ Am$ the maximum acceleration of the breaks are $ \minus{}Ab$.
The weigth of the car is $ m$ the gravity is $ g$ the coefficient of friction is $ \mu$ (both).
The force due air friction $ F_{air} \equal{} \minus{}k v$.
You wand to go from $ A$ to $ B$ and $ AB\equal{}L$
the initial conditions are:
In $ A$: $ x,v \equal{} 0$
In $ B$: $ x\equal{}L,v\equal{}0$

What is the minimum time that this car can race in this conditions?

Set up the equation of motion:

$ x''(t) + \frac {k}{m} x'(t) = a(t) - \mu g$, where $ a(t)$ is the acceleration delivered by the car.

You want $ a(t)$ to be some function of the form $ a(t) = \left \{\begin{array}{cc} + A_m , & t \leq \tau & - A_b , & t > \tau \end{array} \right \;$.

Solve the equation for the first piece using the known conditions at $ x = 0$.

At $ t = \tau$, find the velocity and position from the first segment, and use these as initial conditions to solve the second piece.

Now, force $ x' = 0$ at $ x = L$, which will involve solving for $ \tau$ in terms of the other parameters, and then you are basically done.

Yes. using Heaviside theta and Laplace Transform. Or a less sofisticaded method.
but the the equation for the solution of the time wen it start to break is non-linear.
You never tried to solve, only put the equation, ne?


geometry
3D geometry
sphere
conics
ellipse
analytic geometry
graphing lines

How many long questions did u guys solve?

Does ne1 know last years inao jr cutoff or how much would be enough 2 qualify this year.I might have got around 55 to 65 this year.By the way injso physics was to easy!

[quote="adityadegala"][quote="Aravind Srinivas L"]Aditya, which class are you from?
I did not appear for Astro {even 1st level}. I wrote INJSO. How much are you expecting in it? Me around 65-70.
Astro seemed to be very easy when I saw the paper. I think Astro cutoff will be 75 atleast.[/quote]
Im in 10th.
Im expecting about 60 in INJSO
I lost many marks in biology[/quote]

What was last years cutoff in INAO?

can ne1 post the answers of the copper wood question,the gravitation one(where distance=r)for 2 spheres and the minimum potential energy question

Gravitation answer is- C)
For the copper and wood cube question, I wrote answer D)
For the potential question, I wrote answers B) and D)

How did you approach the estimation question. How many subjective questions did you get?

Does anyone has the idea about INAO(Sr) cutoffs??I thnk this year's paper was tougher than last year's..

I got all subjectives except the solar eclipse 1.
I got 10 raise to 14 as the answer for estimation

[quote="Aravind Srinivas L"]Aashay Rock, I think the subjectives  are mostly common in Junior as well as Senior. So, when I saw the junior paper, it ppeared quite easy only man. I think cutoffs in Astro last year was something like 85 out of 100. Maybe this year it will be 75 since u people say it was tougher than last year.[/quote]

Not true..

Last year's cutoff was around 50/100...
The highest marks maybe arnd 85 or 90 or whatever u say... But the cutoff will never be that high

Dont worry guys, you will get a lot less marks than you expect...  :lol:

And is that Soumyashant Nayak? Why this change in ur username?
Anyway, I told 85 coz pardesi told me so. I dunno.
If the cutoff in astro itself is 50 or 55, then junior science would be much lesser coz of Biology. But as Soumya says, we will get lot leeser marks than waht we expect.

hey!guys, i was in contact with aniket sir in prev year, the highest in INAO Jr. prev year was 66/100!!!!

and this year, i will be much lesser!!!!!!!!

And i made many silly mistakes in both astro and junior!(SO NOT EXPECTING IN ANY)

BTW, i got 4 subjectives(didn't get solar one!!!!!!!)
got 2*r*r in 1st one
in aryabhatta one i got:_ a=3,b=7,c=16,d=11!!
in estimation one, i got 12*10^12 tonnes!!!!!!!!

anybody check if these are right!!!!!!!!!!
(BTW,expecting 50-55 in INJSO!)

ur aryabhatta one is correct
in sr. we didnt have the estimation one so cant verify
first one is 2r*r thats correct
even i din do the solar one

dear all................
                           congrats on being selected.....this camp would be really exciting....

and fbaggins-am i supposed to assume it stands for friedo baggins(lord of rings)

well-wish to see you all soon...

@ tiger prowl, congrats for being selected!!!!!

yes fbaggins is for frodo baggins of LOTR Trilogy!!!

did any one get the intimation letter??

if so,please inform as early as possible......
thanx.
 :coolspeak:

ya i got the letter 3 days back but im not attending the camp  :(

well well,its worth a great experiance..

y r u not attending..??

I thought you were joking on gtalk!!  :o I'd attend the camp even if I had to walk all the way from Lucknow!

seriously my main goal is JEE... :) this only prevents me.. actually me an my friend (who got INMO) had decided to go ... but then we both decided not to go... (INAO -- 24 days + KVPY 15 - days or so) ... 39 days gone... and i dont know much math and chem.. .(nuclear knows that :P) ... thats y btw if i get selected for any of the Inpho/cho/ao then i will go NEXT yr.. this yr no... :(

Firstly, if you cleared INAO, you KNOW physics! And in any case, you aren't going for a 24 day picnic. You're gonna learn physics which even your coaching teachers can't....
KVPY camp might still be a bit of a waste I dunno? But INAO....I'm pretty shocked. How I wish I could have qualified :(

kvpy camp was for just 5 days,and in no way a waste (if u r serious...i was not)

and the camp didnt make me loose anything in my studies,my rank in exams wasnt affected :)

so do attend the camps,
more than nething,it gives u fun and experiance(of compatriots) which i think is a must for JEE

PS.not only me,even the bansal topper shubham tulsiani, and the bests from VMC attended the camp,so u can guess.....

and i think u r frm Orissa,rite??ask our mod,aditya acharya,etc.....no 1 missed the camp yet managed a 3 figure rank in JEE....

Looks like they give some super JEE food at the camp :P I'll definitely go!  :D

If there is good food I will really love to attend the camp :blush:

the food was reallyyy great  :D (excpt the south indian breakfast :(  :rotfl: )

dear friends,
                   i need a small information........i am from hyderabad....
i just want to know whether anybody else from hyderabad got selected for the astronomy camp this year...

thanks a lot...!!


salam doostan  :|  ye masaleye jaleb!!  :idea:
rabeteye bazgashtiye benevisid ta koliyeye tarkibat arghame $ 1$ ta $ n$ ra toolid namayad  :!: tavajo konid ke koliyeye tarkibat az $ 1$ to $ n\minus{}1$ ra dashte bashim. sepas jomleye omoomiye an ra bedast avarid :?:

$ f(n)\equal{}nf(n \minus{} 1)$? :maybe:
shayadam man manzooreto doros nafahmide basham

$ f(n) \equal{} 9^{n \minus{} 1}...$ baghiye be ohdeye shoma  :D

pas man aslan manzoore soaleto nafahmidam,mitooni vazehtar tozih bedi? :maybe:

be onvane mesal mikhahim tamame aadade $ 3$ ragaghami ra peyda konim tarkibate an besorate $ 9*10*10$ khob hala ma bayad 2 raghamiha va yek raghami ha ra ham bedast avarim va be tor kolli agar $ n$ teadade raghamha bashad bayad $ 1,2,...,n\minus{}1$ ra ham peyda konim


What is the second number in the row of Pascal's triangle that has 43 numbers?

Let's find a pattern.

1
11
121
1331
14641

The number of numbers in the first row is $ 1$, second row is $ 2$.  Thus the number of numbers in the $ n^{th}$ row is $ n$.  Thus the row number with $ 43$ numbers is $ 43$.  We also see a pattern in the second number in each row.  The second number is always $ n\minus{}1$ where $ n$ is the row number.  $ 43$ is our row number, so the answer is $ 43\minus{} 1 \equal{} \boxed{42}$


[img]http://img696.imageshack.us/img696/8386/sgia3b.jpg[/img]

I say the answer is 3, my buddy says it's the empty set.

Who's right?

Have you tried evaluating the left and right sides at the suggested solution $ m\equal{}3$?

That's the main crux of our argument, He says because if you put 3 into the initial equation, it doesn't work out that it is the empty set.

On the other hand I say since this could be simplified to:

m-4(m-3)=3

That putting in 3 would then work and reduce to

3=3

So you're saying that $ \frac{3}{0}\minus{}4\equal{}\frac{3}{0}$? That's not true in the standard number system, since neither side is a number.

Yeah but then isn't that adding extra garbage to a number?

I look at it from the point of view, that this equation can be simplified. and that m-3 denominator is just white noise.

Okay, here's a simplification:
Rearrange to $ \frac{m}{m\minus{}3}\minus{}\frac{3}{m\minus{}3} ?\equal{} 4$.
Simplify the left-hand side: $ \frac{m}{m\minus{}3}\minus{}\frac{3}{m\minus{}3}\equal{}\frac{m\minus{}3}{m\minus{}3}\equal{}1\neq 4$. Never true, as the difference between the two sides is a constant $ 3$.

There's a reason we don't allow all operations.

[quote="Anderson2006"]That's the main crux of our argument, He says because if you put 3 into the initial equation, it doesn't work out that it is the empty set.
[/quote]

If you put substitute 3 into the original equation, you get a result that is not algebraically valid, therefore 3 is not a solution.  How does that invalidate the fact that there is no solution?

Suppose you tell me that the solution to $ \frac{1}{x} \equal{} 0$ is $ x \equal{} 0$.  I plug it in and it gives me $ \frac{1}{0} \equal{} 0$.  That is a false statement.  So $ x \equal{} 0$ is not a solution.

[quote]On the other hand I say since this could be simplified to:

m-4(m-3)=3

That putting in 3 would then work and reduce to

3=3[/quote]

Your simplification is only valid if $ m \ne 3$.  Suppose we have the equation

$ x\plus{}2 \equal{} 3$.

Clearly $ x \equal{} 1$ is a solution.  But now I multiply both sides by $ (x\plus{}1)$ and get

$ (x\plus{}1)(x\plus{}2) \equal{} 3(x\plus{}1)$.

Expanding,

$ x^2 \plus{} 3x \plus{} 2 \equal{} 3x \plus{} 3$.

Now subtracting from both sides gives

$ x^2  \equal{} 1$.

So we have $ x \equal{} 1$ as before, but now $ x \equal{} \minus{}1$ is also a solution?  This is incorrect because $ \minus{}1 \plus{} 2 \ne 3$ in the original equation.  Where did this new "solution" come from?  It came from the fact that we multiplied by $ (x\plus{}1)$.  When $ x \equal{} \minus{}1$, the value we multiplied by is equal to zero.  So you have introduced a spurious solution because now your equation says $ 0 \equal{} 0$.  And notice that in this example, there is no strange issues with division by zero.  Everything is always well-defined, so you can't blame it on division by zero.

In a similar vein, when you multiplied both sides by the factor (m-3), and in doing so you discovered that $ m \equal{} 3$ is a "solution," you basically multiplied both sides of the equation by zero.  The situation is perfectly analogous to my example.


A ``size 2 square of squares'' is a 2 by 2 grid of digits where the
digits in each row, from left to right, form a two-digit perfect
square, and the digits in each column, from top to bottom, also
form a two-digit perfect square. How many ``size 2 squares of
squares'' are there?

16
64

36
64

64
49

81
16

all work

so total is [size=200][color=darkred][i][b]4 [/b][/i][/color][/size]2 by 2 grids


A solid rectangular block is formed by gluing together $N$ congruent 1-cm cubes face to face. When the block is viewed so that three of its faces are visible, exactly 231 of the 1-cm cubes cannot be seen. Find the smallest possible value of $N$.

[hide="Answer"]Essentially, if we have an $A\text{x}B\text{x}C$ block, then $231=(A-1)\text{x}(B-1)\text{x}(C-1)$. $231=3\cdot 7\cdot 11$, so we have $A=4$, $B=8$, $C=12$, and $N=4\cdot 8\cdot 12=\boxed{384}$, since the smallest possible value is when $A,B,C$ are closest together.[/hide]

[quote="mna851"]How did you get $\ 231=(A-1)\text{x}(B-1)\text{x}(C-1)$ ?[/quote]
essentially, you cannot see what is under the first layer, i.e. a rectangular solid with dimensions that are each one less than the dimensions of the one you see.

[hide="Solution"]
Let the dimensions of the box be $x, y,$ and $z$.  Now consider taking away all of the bricks that we can see.  We will be left with a box of dimensions $x-1, y-1,$ and $z-1$.  Thus \[ (x-1)(y-1)(z-1) = 231. \]

We factor $231=3\cdot7\cdot11$, and note that we wish to minimize $xyz$.  We know that the product of three numbers whose sum is constant is smallest when they are equal.  We can't make them equal, so we make them as close to equal as possible.  So we take the dimensions to be $3+1, 7+1,$ and $11+1$.  Thus, our answer is $4\cdot 8\cdot 12= \boxed{384}.$[/hide]

Here is alternative solution using PIE.

[hide]Let $l, w, h$ be the length, width, and height, respectively, of the rectangular block. We will first figure out the number of cubes that [i]can[/i] be seen, and then subtract that from the total number of cubes $lwh$.

To get the number of cubes that can be seen could be determined using the Principle of Inclusion and Exclusion, namely
Cube can be seen = $\sum$ cubes that occupy each of the three faces - $\sum$ cubes that occupy each of the edges + $\sum$ cube(s) that occupy both the three faces AND three edges, or
$lw + lh + wh - l - w - h + 1$

Thus
$lwh - lw - lh - wh + l + w + h - 1 = 231 = 3 \cdot 7 \cdot 11$
$(l-1)(w-1)(h-1) = 3 \cdot 7 \cdot 11$

And the solution follows as above. The answer turned out to be $4 \cdot 8 \cdot 12 = \boxed{384}$[/hide]


here's a question i've been trying for quite some time but unable to get a breakthrough
prove $ a_{2n}\equal{}a_nb_n$ where $ a_i$ belongs to the series of fibonacci numbers and $ b_i$ belongs to the series of lucas numbers
for defenition of lucas numbers you can look here http://en.wikipedia.org/wiki/Lucas_number

[quote="rahul1993"]here's a question i've been trying for quite some time but unable to get a breakthrough
prove $ a_{2n} \equal{} a_nb_n$ where $ a_i$ belongs to the series of fibonacci numbers and $ b_i$ belongs to the series of lucas numbers
for defenition of lucas numbers you can look here http://en.wikipedia.org/wiki/Lucas_number[/quote]

I think everyone will think of this immediately.

[hide]
What are their closed formulas? (General formulas)
[/hide]

@stephencheng
their closed formula is $ b_n\equal{}(\frac{1\plus{}\sqrt{5}}{2})^n \plus{} (\frac{1\minus{}\sqrt{5}}{2})^n$

Try plugging in the closed formula, and substituting $ \alpha$ as $ \frac{1\plus{}\sqrt{5}}{2}$ and $ \beta$ as $ \frac{1\minus{}\sqrt{5}}{2}$.

[quote="rahul1993"]@AIME 15 could you please elaborate on that. i've tried that approach :maybe:[/quote]

General formula of lucas number $ b_n\equal{}(\frac{1\plus{}\sqrt{5}}{2})^{n}\plus{}(\frac{1\minus{}\sqrt{5}}{2})^{n}$

General formula of fibonacci number $ a_n\equal{}\frac{1}{\sqrt{5}}((\frac{1\plus{}\sqrt{5}}{2})^{n}\minus{}(\frac{1\minus{}\sqrt{5}}{2})^{n})$

So $ a_nb_n\equal{}\frac{1}{\sqrt{5}}((\frac{1\plus{}\sqrt{5}}{2})^{2n}\minus{}(\frac{1\minus{}\sqrt{5}}{2})^{2n})\equal{}a_{2n}$


Let $f\in\mathbb{R}[X],\ f=aX^{n+2}+bX^{n}+2$ where $n\in\mathbb{N}^{*}$ and $a,b\in\mathbb{R}$. If we know that $(X-1)^2|f$, evaluate $\lambda=q(1)$ where $q\in\mathbb{R}[X]$ is the quotient of dividing $f$ by $(X-1)^2$.

slizzel do you have a solution?  why do you define lambda, instead of just saying q(1)?

We know that 1 is a root, so $a+b=-2$.  Since 1 is a double root, $f ' (1)=0$, which gives us $a=n$, so the polynomial is $f(x)=nx^{n+2}-(2+n)x^n+2$, and we want to find $q(1)$... maybe long division will show some pattern and then we can plug in $x=1$ but i dont see how else to do it.  the answer appears to be $n^2+2n$.  The pattern  is $q(x)$=$nx^n+2nx^{n-1} +2(n-1)x^{n-2} +2(n-2)x^{n-3}+ \ldots + 2$, so $q(1)=n+2(n+\ldots 1)=n^2+2n$.

i agree with answer.  i see now, i didn't think to get Q(x) using product rule for derivatives, thats good solution.  Why the $ax+b$ though, isnt it just 0

Probably should be. I started out with the generic remainder, instead of throwing it out.

Well, in the book was written $\lambda=q(1)$ :). Your solutions are right (are same as in my book) But I haven't learned about $f'(x)$ or $f''(x)$. I have some small knowledge about it, but still not enough :(. Isn't there any other way?

slizzel, i suppose you could do out the long division and look for patterns.  But that's tedious to do.  If you know the basics of differentiation, then you can understand the solution.  it's a good tip whenever you see multiple roots to differentiate and plug in the root and set it equal to 0.  if you take anything away from the problem, remember that.

if arent able to follow the differentiation, then forget about the problem for now and come back to it later i guess.  Or try and follow through with the long division, but i am not really sure that it'll be easy.

I think I'll come back when later :) I'm very grateful for your solution ;)


limit calculus integration ratio logarithms search function

Today's calculation of Limit 1

Without resorting to the use of Stirling's formula, calculate the value of the limit $\lim_{n\rightarrow\infty}\frac{(n!)^\frac{1}{n}}{n}$.

Notice: I decided to start a new series of exercises (inspired by Kunny's Today's calculation of Integral ### series), namely at least challenging if not down right difficult limit calculations. Note: this series of exercises is authored by you :yup: (so that limit your thinking of, post it!).

[quote="benorin"]Today's calculation of Limit 1

 $\lim_{n\rightarrow\infty}\frac{(n!)^\frac{1}{n}}{n}=L\implies lnL=\lim_{n\rightarrow\infty}ln(\frac{(n!)^\frac{1}{n}}{n})$. Results $lnL=\lim_{n\rightarrow\infty}\frac{lnn!-nlnn}{n}$ Aplying Cesaro-Stolz method we got:
$lnL=-\lim_{n\rightarrow\infty}ln\left(1+\frac{1}{n}\right)^{n}\implies L=\frac{1}{e}$

Lemma: if $\lim_{n\to\infty}\frac{a_{n+1}}{a_{n}}$ and $\lim_{n\to\infty}(a_{n})^{1/n}$ both exist, then

$\lim_{n\to\infty}\frac{a_{n+1}}{a_{n}}= \lim_{n\to\infty}(a_{n})^{1/n}$

Proof: consider the power series $\sum_{n}a_{n}x^{n}.$ It has radius of convergence $R.$ By the ratio test and the root test (each of which has its own proof), if both of the limits above exist, then they both equal $\frac1R.$

--

Now, for this problem, let $a_{n}=\frac{n!}{n^{n}}.$ The $n$-th root is the sequence benorin wishes to compute the limit of, so we consider the ratio:

$\lim_{n\to\infty}\frac{(n+1)!}{(n+1)^{n+1}}\cdot \frac{n^{n}}{n!}= \lim_{n\to\infty}\left(\frac{n}{n+1}\right)^{n}$

$=\lim_{n\to\infty}\frac1{\left(1+\frac1n\right)^{n}}= \frac1e$ (Well-known limit.)

[quote="cckek"]Today's calculation of Limit 1

 $\lim_{n\rightarrow\infty}\frac{(n!)^\frac{1}{n}}{n}=L\implies lnL=\lim_{n\rightarrow\infty}ln(\frac{(n!)^\frac{1}{n}}{n})$. Results $lnL=\lim_{n\rightarrow\infty}\frac{lnn!-nlnn}{n}$ Aplying Cesaro-Stolz method we got:
$lnL=-\lim_{n\rightarrow\infty}ln\left(1+\frac{1}{n}\right)^{n}\implies L=\frac{1}{e}$[/quote]

Effective choice of method, well done cckek. The [url=http://planetmath.org/encyclopedia/StolzCesaroTheorem.html]Cesaro-Stolz Theorem[/url] equates the limits of the ratios of the zeroth and first order forward differences of two real sequences provided that the sequence forming the denominator is positive, strictly increasing, and unbounded, (for reference: the zeroth and first order forward differences of $a_{n}$ are $\Delta^{0}a_{n}=a_{n}$ and $\Delta a_{n}=a_{n+1}-a_{n}$, and higher orders by the relation $\Delta^{k+1}a_{n}=\Delta^{k}a_{n+1}-\Delta^{k}a_{n}$), does anyone know of any higher order relations in this context?

[quote="Kent Merryfield"]Lemma: if $\lim_{n\to\infty}\frac{a_{n+1}}{a_{n}}$ and $\lim_{n\to\infty}(a_{n})^{1/n}$ both exist, then

$\lim_{n\to\infty}\frac{a_{n+1}}{a_{n}}= \lim_{n\to\infty}(a_{n})^{1/n}$

Proof: consider the power series $\sum_{n}a_{n}x^{n}.$ It has radius of convergence $R.$ By the ratio test and the root test (each of which has its own proof), if both of the limits above exist, then they both equal $\frac1R.$[/quote]

This is quite an excellent proof of this remarkable formula, professor Merryfield. Thank you.

[quote="Kent Merryfield"]Now, for this problem, let $a_{n}=\frac{n!}{n^{n}}.$ The $n$-th root is the sequence benorin wishes to compute the limit of, so we consider the ratio:

$\lim_{n\to\infty}\frac{(n+1)!}{(n+1)^{n+1}}\cdot \frac{n^{n}}{n!}= \lim_{n\to\infty}\left(\frac{n}{n+1}\right)^{n}$

$=\lim_{n\to\infty}\frac1{\left(1+\frac1n\right)^{n}}= \frac1e$ (Well-known limit.)[/quote]

It is really quite odd that both of these seemly disparate approaches are almost exactly the same on the next to last step:

Now cckek took $a_{n}= \ln (n!)-n\ln(n)$ and $b_{n}=n$ so the hypotheses of the [url=http://planetmath.org/encyclopedia/StolzCesaroTheorem.html]Cesaro-Stolz theorem[/url] are satisfied, and we may now apply the conclusion that
\[\lim_{n\rightarrow\infty}\frac{a_{n}}{b_{n}}=\lim_{n\rightarrow\infty}\frac{\Delta a_{n}}{\Delta b_{n}}: = \ell \]
for which calculation I had noted that
\[\frac{\Delta a_{n}}{\Delta b_{n}}=\frac{\ln ((n+1)!)-(n+1)\ln(n+1)-\ln (n!)+n\ln(n)}{(n+1)-n}= \ln\left[ \frac{(n+1)!}{(n+1)^{n+1}}\cdot \frac{n^{n}}{n!}\right] \]
but this is precisely what prof. Merryfield had, except that he did not use the lograthmic evalution technique to comput his limit. Tell me professor, why does this coincidence occur, was it dumb luck that this particular limit happened to do that, or are the two techniques in use implicitly linked?

--Ben

[color=red]An addition[/color]: Can't post back-to-back, so I'll put it here:

[quote="perfect_radio"][quote="Kent Merryfield"]Lemma: if $\lim_{n\to\infty}\frac{a_{n+1}}{a_{n}}$ and $\lim_{n\to\infty}(a_{n})^{1/n}$ both exist, then

$\lim_{n\to\infty}\frac{a_{n+1}}{a_{n}}= \lim_{n\to\infty}(a_{n})^{1/n}$[/quote]
I don't think that we need to know that $\lim_{n \to \infty}\left( a_{n}\right)^{1/n}$ exists. The proof that we did in class last year involved taking logarithms and then applying Cesaro-Stolz (exactly as $\textrm{cckek}$ did).

@benorin: Higher order relations? Well, you can do the obvious (apply Cesaro-Stolz repeatedly). I don't have a nice example in mind right now, but I remember that there were quite a few problems which could be solved by applying C-S twice. I'll let you search "cesaro stolz" on the forum.[/quote]

I should have known about using repeated applications of it, I have done this very compound operation to sum divergent series: but for some reason computing limits seemed rather unrelated to me at the time. On this business of existence: we reason that since we must have $\limsup_{n\rightarrow\infty}\frac{a_{n+1}}{a_{n}}<1$ in order that the series converge according to the [url=http://planetmath.org/encyclopedia/RatioTest.html]Ratio test[/url], and said convergence is absolute, which, by the [url=http://en.wikipedia.org/wiki/Root_test]Root test[/url], implies that $\limsup_{n\rightarrow\infty}\sqrt[n]{\left| a_{n}\right|}<1$. Are we done, or just nearly?

--Ben

This is how I had done it ([url=http://www.physicsforums.com/showthread.php?t=113698&page=2]original posting[/url]):

Let the limit be denoted Y so that
\[Y= \lim_{n\rightarrow\infty}\frac{(n!)^{\frac{1}{n}}}{n}= \lim_{n\rightarrow\infty}e^{{\log \left( \frac{(n!)^{\frac{1}{n}}}{n}\right)}}= \lim_{n\rightarrow\infty}e^{{\frac{1}{n}\log \left( \frac{n!}{n^{n}}\right)}}\]

\[= \lim_{n\rightarrow\infty}e^{{\frac{1}{n}\log \left( \prod_{k=1}^{n}\frac{k}{n}\right)}}= \lim_{n\rightarrow\infty}e^{{\frac{1}{n}\sum_{k=1}^{n}\log \left( \frac{k}{n}\right)}}\]

\[= e^{{\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{k=1}^{n}\log \left( \frac{k}{n}\right)}}={e^{{\lim_{r\rightarrow 0^{+}}\int_{r}^{1}\log (x)\,dx}}}, \]
where the limit was passed through to the arguement of the exponential function by reason of continuity and the sum was seen to be a Riemann sum of the form $\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{k=1}^{n}f \left( \frac{k}{n}\right) = \int_{0}^{1}f(x)\, dx,$ which is improper integral in this case becase of the discontinuity of $\log x$ at $x=0$; the limit becomes
\[Y = e^{{\lim_{r\rightarrow 0^{+}}\int_{r}^{1}\log (x)\, dx }= e^{{\lim_{r\rightarrow 0^{+}}\left[ x(\log (x)-1)\right]_{x=r}^{1}}}}\]

\[= e^{{\lim_{r\rightarrow 0^{+}}\left[-1-r(\log (r)-1)\right]}}= e^{-1}e^{{\lim_{r\rightarrow 0^{+}}\frac{1-\log r}{r^{-1}}}}\]

\[{=^{H}e^{-1}e^{{\lim_{r\rightarrow 0^{+}}\frac{-r^{-1}}{-r^{-2}}}}=e^{-1}e^{{\lim_{r\rightarrow 0^{+}}r}}= e^{-1}}\]
where $=^{H}$ denoted the use of l'Hospital's rule and whereby we may conclude that ${ \lim_{n\rightarrow\infty}\frac{(n!)^{1/n}}{n}= \frac{1}{e}}$.

PS How do I make certian lines of TeX display in a larger font size? On http://www.physicsforums.com I could use \displaystyle{}, what works here?


geometry
circumcircle
trigonometry
angle bisector
geometry proposed

Let a quadrilateral $ ABCD$ with $ AD\equal{}CD$ and $ \angle DAB\equal{}\angle ABC<90^o$. Denote M be the midpoint of BC. The line DM meets AB at E. Prove that $ \angle BEC\equal{}\angle DAC$

Let $ E$ be, the point on the sideline $ AB$ of the given quadrilateral $ ABCD$ $ ($ $ B,$ between $ A,$ $ E$ $ ),$ such that $ \angle BEC \equal{} \angle DAC$ and it is enough to prove that $ MB \equal{} MC,$ where $ M\equiv BC\cap DE.$

Let $ \bigtriangleup PAB,\ \bigtriangleup QAE$ be, the isosceles triangles, where $ P\equiv AD\cap BC$ and $ Q$ is the intersection point of the line segment $ AP,$ from the line through the point $ E$ and parallel to $ BC.$

It is easy to show that the circumcircle $ (K)$ of the triangle $ \bigtriangleup ACE,$ tangents to $ AD,\ CD,\ QE,$ at points $ A,\ C,\ E,$ respectively and so we have that the points $ F,\ C,\ D,\ S,$ where $ F\equiv QE\cap CD$ and $ S\equiv AE\cap CD,$ are in harmonic conjugation $ ($ easy to prove from the concurrency of the line segments $ QC,\ DE,\ FA,$ applying the [b][size=100]Ceva theorem[/size][/b], in the triangle $ \bigtriangleup QDE,$ taken the circle $ (K),$ as its $ Q$-excircle $ ).$

So, we have that the pencil $ E.FCDS$ is also in harmonic conjugation and hence, because of $ BC\parallel EF,$ we conclude that $ MB \equal{} MC,$ where $ M\equiv BC\cap DE$ and the proof is completed.

Kostas Vittas.

Here is my solution:

Let $ F\equal{}DA \cap BC$. The parallel from $ E$ to $ FB$ intersects $ FA$ in $ G$. Since $ MA\equal{}MB$ and $ BC \parallel GE$ we deduce that $ EB,EM,EC$ and $ EG$ are the rays of an harmonic pencil.

Now, the parallel from $ D$ to $ AB$ intersects $ CE$ in $ P$ and $ GE$ in $ R$. The division $ (R,D,P,\infty)$ is harmonic $ \Rightarrow$ $ PD\equal{}PR$.

$ \frac {PD} {\angle DCP}\equal{}\frac {DC} {\angle CPD}\equal{}\frac {RE} {\angle RPE}\equal{}\frac {PR} {\angle PER}$

But $ \angle GEC\equal{}\angle BCE \Rightarrow \angle DCP \plus{} \angle CER \equal{} 180^{\circ}$

$ \Rightarrow \angle DCB \plus{} 2\angle PER\equal{}180^{\circ},(1)$

Let $ \angle DAB\equal{}A$ and $ \angle DAC\equal{}X \Rightarrow \angle DCB\equal{}360^{\circ}\minus{}2A\minus{}(180^{\circ}\minus{}2X)\equal{}180^{\circ}\minus{}2A\plus{}2X ,(2)$

$ (1),(2) \Rightarrow 180^{\circ}\minus{}2A\plus{}2X\plus{}2(A\minus{}\angle AEC)\equal{}180^{\circ} \Rightarrow \angle AEC\equal{}X.$

Seems to be a giant  step forward from  Prop. 2 in Archimedes' Book of Lemmas.

   M.T.

I was a participant in this olympiad ... I found a solution using Sine law ... I think it is not a difficult problem. Official solution is based on menelaus and angle bisector theorem.


topology
real analysis
calculus
integration
limit
geometry
rectangle

Hmmm....just wondering - How do mathematically define infinty ? "A number that is greater than every real number" doesnt exactly seem satisfactory...

http://mathworld.wolfram.com/Infinity.html

[quote="Mathworld"]
Infinity, most often denoted as $\infty$, is an unbounded quantity that is greater than every real number.[/quote]

[quote="h78reg"]infiniti isn't really a number because its undefined. 1/0 is infinity, just as tan 90 is infinity. anything undefined is infiniti.[/quote]

Wrong. 1/0 does not equal infinity. Suppose 1/0 does equal infinity. Then 2/0 would also equal infinity, because twice of infinity is still infinity (consider the Infinity Hotel), and then multiplying by 0 on both sides, 1=2. Contradiction. And since tan 90 = 1/0, the same argument holds.

Infinity is neither real nor complex, and it belongs in its own category. There are different types of infinity: one is the cardinality of the naturals/integers/rationals (they all have the same cardinality) which is aleph-null (countable infinity) and another is the cardinality of the reals (uncountable infinity). But there are infinitely many classes of infinity, including aleph-one (the set of subsets of aleph-null), aleph-two, and so on.

Hope that helps. But correct me if I'm wrong.

I think that this discussion is nonsense, because the definition of Infinity depends on the System you use it.

So if you look at the real numbers, you will never have to calculate with $\infty$ like $\infty+1$ a.s.o., you only use it i.e. for Limes etc. But everywhere you use it, you use it (if only indirectly) as a further Element of the relation $<$ as the "Number", that is greater than every real number.

In $\mathbb{C}$ you define it by the Riemann-Sphere.

In general topology (espc. Hausdorff-Topology ;) ) you also only have Relations of Order, you can do the same.

In ZFC, you can define: A Set contents of infinitely many Elements iff there is a bijection from this Set to a Subset of this Set.

Actually, you can do arithmetic quite well in $[0,\infty]$, as long as you stay away from $\infty-\infty$.  But you can make expressions like $0\cdot\infty=0$ work quite well (you use this convention in Lebesgue integration)

[quote="Magnara"]Infinity is not a real or complex number.  It cannot be added, multiplied, subtracted, or divided in any sensible way, so it isn't a member of the fields.  There isn't one simple notion of infinity either.  And, in most cases, mathematicians don't really talk about infinity.  Or, rather, I should say that when they do, they use incredibly precise definitions.

For example, lim x_n=L as n goes to infinity is defined as for every epsilon greater than zero there exists an N such that n>N implies that |x_n-L|<epsilon.  This matches up with intuitive ideas about infinity, but never actually mentions infinity.[/quote]Wasnt this almost exactly the same I said?

[quote="leepakhin"][quote="mathnerd314"][quote="h78reg"]infiniti isn't really a number because its undefined. 1/0 is infinity, just as tan 90 is infinity. anything undefined is infiniti.[/quote]

Wrong. 1/0 does not equal infinity. Suppose 1/0 does equal infinity. Then 2/0 would also equal infinity, because twice of infinity is still infinity (consider the Infinity Hotel), and then multiplying by 0 on both sides, 1=2. Contradiction. And since tan 90 = 1/0, the same argument holds.

[/quote]

$\frac{1}{0}$ is infinity. You can't multiply $\frac{1}{0}$ with $0$ since this is an indeterminate form.[/quote]

to be exact , I would say $\frac 10$ is [b]undefined[/b] rather than [b]infinity[/b] :)

[quote="karthik_k"]Well , that's the mathworld definition - But then it implies infinity cannot be a real number ...... but looking at it , I doubt infinity can be defined as a complex number either..... What do you regard infinity as then ? { Real ? Complex ? An exception to both ?}[/quote]

Infinity is not a complex number because it is greater than every real number implying that infinity belongs to an ordered field of numbers. The complex numbers are not an ordered field, and therefore infinity is not a complex number.

I think it would be safe to view infinity as an extension of the real numbers. The mindblowing thing is that merely because one finds a number greater than every real number does not imply that it is the only such number. In fact there are provably an infinite number of "infinities."

And just to clarify the difference between approaching and equaling...

Consider a rectangle with dimensions a, b. Draw "stairs" from a corner to the opposite corner. Obviously, the length of this segment is equal to a+b. This can be easily proven with "opposite sides of rectangle are congruent". But let the stairs become smaller and smaller until they approach a straight line. The right triangle formed with two sides and a diagonal would have sides of a, b, and a+b, right? WRONG, as we can see.

[quote="Magnara"][quote="leepakhin"][quote="shyong"]to be exact , I would say $\frac 10$ is [b]undefined[/b] rather than [b]infinity[/b] :)[/quote]
Okay, then $\displaystyle\lim_{x\to 0}\frac{1}{x}=\infty$[/quote]

This isn't true either.  It is a common mistake though.  This, however, is true: $\displaystyle\lim_{x\to 0^+}\frac{1}{x}=\infty$.[/quote]

This depends on $x$ being a Real or a complex number. If $x$ is complex, than there ist only one $\infty$, i.e. $\overline{\mathbb{R}} = \mathbb{R} \cup \{\infty, -\infty\}$ and $\overline{\mathbb{C}}=\mathbb{C} \cup \{\infty\}$.

[quote="Schoppenhauer"][/quote][quote="Magnara"][quote="leepakhin"][quote="shyong"]to be exact , I would say $\frac 10$ is [b]undefined[/b] rather than [b]infinity[/b] :)[/quote]
Okay, then $\displaystyle\lim_{x\to 0}\frac{1}{x}=\infty$[/quote]

This isn't true either.  It is a common mistake though.  This, however, is true: $\displaystyle\lim_{x\to 0^+}\frac{1}{x}=\infty$.[/quote][quote="Schoppenhauer"]

This depends on $x$ being a Real or a complex number. If $x$ is complex, than there ist only one $\infty$, i.e. $\overline{\mathbb{R}} = \mathbb{R} \cup \{\infty, -\infty\}$ and $\overline{\mathbb{C}}=\mathbb{C} \cup \{\infty\}$.[/quote]Everyone seems to be missing the point. The idea of $\displaystyle \infty$ as an element of any field of numbers is nonsense. $\displaystyle \infty$ is a symbol indicating, and only indicating, that limits such as $\displaystyle\lim_{x\to 0^+}\frac{1}{x}=\infty$ don't exist.

There is a whole section in Rudin's Real and Complex Analysis called "Arithmetic in $[0,\infty]$".  Now, I'm not going to say that $[0,\infty]$ is a field, or even that $[-\infty,\infty]$ is a field, but it is nontheless a structure that we can do a great deal of effective analysis in.

Similarly, when you work in $S^2$ (the Riemann sphere) in complex analysis, it is useful to refer to the value of a function at $\infty$.  And I'm not even going to talk about the projective plane :).

The algebraic structures here may be a bit less than satisfying (actually, $S^2$ shouldn't, because it's just a one-point compactification of $\mathbb{R}^2$, but they are very useful in practice.  But the usefulness of being able to talk about $\infty$ and the degree to which it simplifies notation makes it a very useful construct.

[quote="Schoppenhauer"][/quote][quote="Magnara"][quote="leepakhin"][quote="shyong"]to be exact , I would say $\frac 10$ is [b]undefined[/b] rather than [b]infinity[/b] :)[/quote]
Okay, then $\displaystyle\lim_{x\to 0}\frac{1}{x}=\infty$[/quote]

This isn't true either.  It is a common mistake though.  This, however, is true: $\displaystyle\lim_{x\to 0^+}\frac{1}{x}=\infty$.[/quote][quote="Schoppenhauer"]

This depends on $x$ being a Real or a complex number. If $x$ is complex, than there ist only one $\infty$, i.e. $\overline{\mathbb{R}} = \mathbb{R} \cup \{\infty, -\infty\}$ and $\overline{\mathbb{C}}=\mathbb{C} \cup \{\infty\}$.[/quote]

But all real numbers are complex, since they fit the form of a+bi with b=0.

He is pointing out that the most commonly used compactification of the real line is a two point compactification, whereas the most common compactification of the complex plane is the Riemann sphere.  We may use a one point compactification of the real line (the projectively extended real line), but I know no use for it in, say, analysis.

[quote="Bictor717"]I want to emphasize something Dr. No said, even though I don't entirely agree with his statement.
It's wrong to say $\lim_{x\to 0^+}\frac 1x=\infty$. This misleading notation is often abused when you really mean that the limit diverges. We established that $\infty$ is not a number in the everyday sense, so how can a sequence approach a non-number?

And to mathnerd314, the staircase does not "approach" the diagonal, in the literal sense, at least. The behavior of the staircase never appraoches the behavior of the line.[/quote]

why is it wrong to say that $\lim_{x\to 0^+}\frac 1x=\infty$? after all, it's just a convention... $\infty$ is just a symbol for the diverging limit, right?

It was OK the last time I checked, at any rate. It was OK in AoPS volume 2, the last time I checked.

Although infinity is not a number, we can use that notation. Saying that the limit of f(x) as it approaches blah is infinity doesn't imply that infinity is a real number; rather it implies that f(x) would approach infinity if we could carry out the calculation at blah; and the closer x gets to blah, the larger f(x) becomes.

Hm.. I was taught in the following way:

$\lim_{x\to0^+}\frac{1}{x}=\infty$

$\lim_{x\to0^-}\frac{1}{x}=-\infty$

$\lim_{x\to0}\frac{1}{x}\ \text{is not defined.}$

and

$\lim_{x\to0^+}\frac{1}{x^2}=\infty$

$\lim_{x\to0^-}\frac{1}{x^2}=\infty$

$\lim_{x\to0}\frac{1}{x^2}=\infty$

and

$\lim_{x\to\infty}x=\infty$


search
inequalities
induction
inequalities unsolved

[quote="CDP100"]If $ a, b, c > 0$, prove that

1.

$ 4(a^3 \plus{} b^3)\ge (a \plus{} b)^3$

2.

$ 9(a^3 \plus{} b^3 \plus{} c^3) \ge (a \plus{} b \plus{} c)^3$[/quote]
Only [url=http://www.mathlinks.ro/viewtopic.php?search_id=1574573256&t=167515]Holder inequality[/url] :wink:
$ (1\plus{}1\plus{}1)(1\plus{}1\plus{}1)(a^3\plus{}b^3\plus{}c^3)\geq (a\plus{}b\plus{}c)^3$.
$ (1\plus{}1)(1\plus{}1)(a^3\plus{}b^3)\geq (a\plus{}b)^3$.

[quote="CDP100"]If $ a, b, c > 0$, prove that

1.

$ 4(a^3 \plus{} b^3)\ge (a \plus{} b)^3$

2.

$ 9(a^3 \plus{} b^3 \plus{} c^3) \ge (a \plus{} b \plus{} c)^3$[/quote]
1.$ 4(a^3 \plus{} b^3)\ge (a \plus{} b)^3 \Leftrightarrow a^3\plus{}b^3 \ge ab^2\plus{}a^2b$
Which is obviously true by AM-GM:
$ a^3\plus{}a^3\plus{}b^3 \ge 3a^2b$
$ a^3\plus{}b^3\plus{}b^3 \ge 3ab^2$
2.$ 9(a^3 \plus{} b^3 \plus{} c^3) \ge (a \plus{} b \plus{} c)^3 \Leftrightarrow 8(a^3\plus{}b^3\plus{}c^3) \ge 3(a\plus{}b)(b\plus{}c)(c\plus{}a)$
$ \Leftrightarrow 8(a^3\plus{}b^3\plus{}c^3) \ge 6abc\plus{}3\sum a^2b\plus{}3\sum ab^2$
By AM-GM inequality, we have:
$ 2(a^3\plus{}b^3\plus{}c^3) \ge 6abc$
and using 1:
$ 2(a^3\plus{}b^3\plus{}c^3 \ge \sum a^2b\plus{}\sum ab^2$
So we are done!

Jensen's inequality kills them.

1. ${ 4(a^3 + b^3)\geq (a + b)^3\Longleftrightarrow \frac{a^3+b^3}{2}\geq \left(\frac{a+b}{2}\right)^3}$.

2.$ 9(a^3 + b^3 + c^3) \geq (a + b + c)^3\Longleftrightarrow \frac{a^3+b^3+c^3}{3}\geq \left(\frac{a+b+c}{3}\right)^3$.

[quote="kunny"]Jensen's inequality kills them.

1. ${ 4(a^3 + b^3)\geq (a + b)^3\Longleftrightarrow \frac {a^3 + b^3}{2}\geq \left(\frac {a + b}{2}\right)^3}$.

2.$ 9(a^3 + b^3 + c^3) \geq (a + b + c)^3\Longleftrightarrow \frac {a^3 + b^3 + c^3}{3}\geq \left(\frac {a + b + c}{3}\right)^3$.[/quote]

I agree  :lol:

Generally we have:

$ (\frac {1}{n} \sum x_i )^n \leq \frac {1}{n} \sum x_{i} ^n$

For nonegative reals $ x_i$
 :)

[quote="FOURRIER"] I agree  :lol:

Generally we have:

$ (\frac {1}{n} \sum x_i )^n \leq \frac {1}{n} \sum x_{i} ^n$

For nonegative reals $ x_i$
 :)[/quote]
We can prove that $ \frac {1}{n} \sum_{i \equal{} 1}^n a_i^m \geq (\frac {1}{n} \sum^{n}_{i \equal{} 1} a_i)^m$
where $ a_i$ are positive real numbers and $ n \in N^*$
[hide]Induction[/hide]


In the binomial expansion of $(1+1.2)^{2007}$, what is the coefficient of $1.2$ in the largest term?

hint: consider the ratio between two terms

More generally, for $x > 0$, among the terms $\{x^{n}, \binom{n}{1}x^{n-1},\binom{n}{2}x^{n-2}, \ldots, \binom{n}{n-1}x,1\}$, of what set (possibly empty) must $x$ be a member if the $i$-th term is maximal?


Let $ x>0, y>0$ and $ x^2\plus{}y^2\equal{}4$. Find maximum of $ P\equal{}\frac{xy}{x\plus{}y\plus{}2}$

[quote="kunny"]$ P \equal{} \frac {xy}{x \plus{} y \plus{} 2} \equal{} \frac {2xy}{2(x \plus{} y \plus{} 2)} \equal{} \frac {(x \plus{} y)^2 \minus{} 4}{2(x \plus{} y \plus{} 2)} \equal{} \frac {1}{2}(x \plus{} y) \minus{} 1$.

For $ x > 0,\ y > 0,\ x^2 \plus{} y^2 \equal{} 4$, we have $ x \plus{} y\leq 2\sqrt {2}$, the equality holds when $ x \equal{} y \equal{} \sqrt {2}$.

$ \therefore P_{max} \equal{} \sqrt {2} \minus{} 1$.[/quote]

My solution get  $ P_{max} \equal{} \sqrt {2} \minus{} 1$ as your  solution.

I have done this problem by let $ x\equal{}2sin\alpha, y \equal{} 2cos\alpha \Longrightarrow P\equal{}\frac{sin2\alpha}{sin\alpha\plus{}cos\alpha\plus{}1}$, with $ 0<\alpha <\frac{\pi}{2}$.

Let $ t\equal{}sin\alpha\plus{}cos\alpha$, we get $ P\equal{}\frac{t^2\minus{}1}{t\plus{}1}\equal{}t\minus{}1$, with $ 1<t \le \sqrt{2}$.

$ \Longrightarrow P_{max} \equal{} \sqrt {2} \minus{} 1$ when $ \alpha \equal{}\frac{\pi}{4} \Longrightarrow x\equal{}y \equal{} \sqrt{2}$


algebra
polynomial
calculus
derivative
calculus computations

Show that the polynomial equation with real coefficients

$a_{n}x^{n}+a_{n-1}x^{n-1}+\dots+a_{3}x^{3}+x^{2}+x+1=0$

has fewer than $n$ real roots.

If an $n$th degree polynomial has $n$ real roots, then its derivative has $(n-1)$ real roots, and so on (by Rolle's theorem). So you want to differentiate until you reach a contradiction. Unfortunately, the few known coefficients will disappear pretty soon, taking any hope of contradiction with them. A suggestion: consider $x^{n}p(1/x)$ instead of $p(x)$.


algebra
polynomial
linear algebra
linear algebra unsolved

If $ d,n \in \mathbb{N}$, and $ dn \not\equal{} 0$

show that: $ x^{d}\minus{}1|x^{n}\minus{}1$ if and only if $ d|n$

where $ x^d\minus{}1, x^n\minus{}1 \in K[x]$

and $ K$ is a field of numbers

the if part is obvious.

assume that $ x^d \minus{} 1|x^n \minus{} 1$. if $ d\not|n$, we may write $ n \equal{} dq \plus{} r$ where $ 0 < r < d$. then $ x^d \minus{} 1|x^n \minus{} 1 \equal{} x^{dq}x^r \minus{} 1 \equal{} (x^{dq} \minus{} 1)x^r \plus{} x^r \minus{} 1$, from which $ x^d \minus{} 1|x^r \minus{} 1$, which is impossible (by comparing degrees). so we must have $ d|n$.


um

1.5 * 5-across is not equal to (2/3) * 5-across.

[hide="An answer?"]
Try...
P l ~
~ 2 l
S l X
Where the l's function as both I's and 1's, and are used to spell out roman numerals, or words.
[/hide]

Yep, that's correct.

And tjhance, 2/3 of SIX is IX. Get it? And 1.5 times 6 is 9.


Se considera un cub de latura $\ 1$ si in interiorul sau o multime finita de puncte $\mathcal{A}$ cu proprietatile:
$\ (a)\mathcal{A}$ are cel putin trei elemente.
$\ (b)$Distanta dintre oricare doua puncte din $\mathcal{A}$ este cel mult egala cu $\ d\in\ (0,1)$.
Pentru fiecare punct $\ M\in\mathcal{A}$ se noteaza cu $\ x_{M}$ cea mai mica distanta de la $\ M$ la celelalte puncte din $\mathcal{A}$.Demonstrati ca \[\sum_{M\in\mathcal{A}}x_{M}^{3}<(d+\sqrt 3)^{3}\] [i]Marius Ghergu, Slatina[/i]

Poate sa dea cineva o solutie corecta?Ca eu am incercat ceva, dar nu  am reusit mai nimik.

Da, ca de obicei, cand vine vorba de dl Marius Ghergu, se ascund toti. :)


The maximum value of the function $ f: [\minus{}1,1] \rightarrow \mathbb{R}, f(x)\equal{}\minus{}2x\plus{}3$ is:
$ (A)$ $ 5$
$ (B)$ $ 6$
$ (C)$ $ 7$
$ (D)$ $ 10$
$ (E)$ $ 8$

$ \minus{}2x\plus{}3$ is greatest when $ x\equal{}\minus{}1$ (in the given interval). So the answer is $ (A)$  $ 5$


geometry
rectangle
inequalities
geometry unsolved

Given an isoscles triangle. Inscribe in it a rectangle with minimal diagonal.

We were given such a problem at the geometry lesson (about affine transformations). It's quite easy - I solved it without affine transformation.

At first, let $MNPR$ be the rectangle so that $M\in AB$, $N\in AC$, $\{P,R\}\subset BC$. Denote $MN=PR=u$ and $MR=NP=v$. From $MN\parallel BC$ and $MR\parallel AD$, where $D\in BC$, $AD\perp BC$ $(AD=h_a)$ obtain : $1=\frac{AM}{AB}+\frac{MB}{AB}=$ $\frac ua+\frac{v}{h_a}$ $\Longrightarrow$ $\frac ua+\frac{v}{h_a}=1$ and the length of the diagonal $MP$ is $d=\sqrt{u^2+v^2}$. The proposed problem becomes : $\min \left( u^2+v^2\right)$, where $0\le u\le a$, $0\le v\le h_a$ and $\frac ua+\frac{v}{h_a}=1$.  From the remarkable C.B.S.'s inequality obtain : $1=\left( u\cdot \frac 1a+v\cdot \frac{1}{h_a}\right)^2\le$ $\left( u^2+v^2\right)\left(\frac{1}{a^2}+\frac{1}{h_a^2}\right)$ $\Longleftrightarrow$ $\frac{ah_a}{\sqrt{a^2+h_a^2}}\le d$ with equality if and only if $\frac{u}{h_a}=\frac va=\frac{ah_a}{a^2+h^2_a}$ and in this case $\boxed {d^{(a)}_{\mathrm {min}}=\frac{ah_a}{\sqrt{a^2+h^2_a}}}$. Denote $l=\max \{a,b,c\}$. Prove easily that  $\boxed {\min_{x\in \{a,b,c\}} d^{(x)}_{\math {min}} =\frac{2S}{\sqrt{l^2+h^2_l}}}$. Indeed, for $x\in \{a,b,c\}$ we have $d^{(x)}_{\mathrm {min}}\ -\ \min$ $\Longleftrightarrow$ $\left(x^2+h_x^2\right)\ -\ \max$ $\Longleftrightarrow$ $x^2+\frac{4S^2}{x^2}\ -\ \max$ $\Longleftrightarrow$ $x\ -\ \max$.


(a) We have 2 matrixes $A,B\in\mathcal M_{3}(R)$. Show that if we have $BA=O_{3}$ then $(AB)^k=O_{3}$, for any natural $k\geq 2$. Is the reciprocal true? Justify the answer.

(b) Let $A\in\mathcal M_{3}(R)$ a matrix with the propriety that exist $n\in\mathbb{N}$ such that $A^n=O_{3}$. Show that the matrix $I_{3}-A$ is inversible and find $(I_{3}-A)^{-1}$

[quote="Slizzel"](a) We have 2 matrixes $A,B\in\mathcal M_3 \left( \mathbb R \right)$. Show that if we have $BA=O_3$ then $(AB)^k=O_3$, for any natural $k\geq 2$. Is the reciprocal true? Justify the answer.[/quote]

We have $(AB)^k = A (BA) B \ldots = O_3$.

The reciprocal ain't true, as proven by $\displaystyle A = B = \left( \begin{array}{*{20}c} 0 & 1 & 1 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{array} \right)$.

[quote="Slizzel"](b) Let $A\in\mathcal M_3 \left( \mathbb R \right)$ a matrix with the propriety that exist $n\in\mathbb{N}$ such that $A^n=O_{3}$. Show that the matrix $I_{3}-A$ is inversible and find $(I_{3}-A)^{-1}$[/quote]

$(I-A)(I+A+A^2)=I-A^3=I$, because $A^3=O_3$. The easiest way to see this is Cayley-Hamilton + eigenvalues (all must be $0$).


geometry
pigeonhole principle
geometric transformation
homothety
combinatorics open
combinatorics

Given mn points in the plane with no two points further than k distance apart, where m,n are positive integers and k a positive real number. Show that there always exists a circle with radius at most k/ :sqrt: m that covers at least n points.

Hmm...my approach fails for $m=4$  :(  but it works for all other values... :)

Let $A_1, \cdots , A_{mn}$ be the points.
First, wlog we may assume that $k=1$. Now, a well-known theorem of Jung's states that any finite set of points in plane with mutual distances not greater than 1 can be covered by a disk, say $D$, with radius $\frac 1 {\sqrt 3}.$
Therefore, if $m \leq 3$ we are done.

Now, we consider the disks $D_i$ with center $A_i$ and radius $\frac 1 {\sqrt m }$.
The total sum of the areas of these disks is $S = mn \frac {\pi} {m} = n \pi$.
In another hand each of these disks is contained in the disk $D'$ with the same center as $D$ and with radius $\frac 1 {\sqrt m} + \frac 1 {\sqrt 3}.$
Thus, if $S > (n-1) \pi (\frac 1 {\sqrt m} + \frac 1 {\sqrt 3})^2$ (1) this would mean (using pigeonhole principle) that there is point, say $O$, in the disk $D'$ which is covered at least $n$ times by the $D_i$'s. In that case, the disk with center $O$ and radius $\frac 1 {\sqrt m} $ gives the desired result.

But, it is straightforward to see that (1) is satisfied for $m \geq 5$ for all $n$.
I didn't check to see whether this can be improved to get $m \geq 4$ for all $n$ except some first values, which could be handled directly...

Pierre.

In fact, it is easier when $m = 4$ :
Just divide the Jung disk into $4$ quarters. From the pigeonhole principle, at least $n$ of the $4n$ points belong to a same quarter.
Now, this quarter can easily been covered by a disk with radius $\frac {\sqrt 2} {2\sqrt 3} = \frac 1 {\sqrt 6} < \frac 1 {\sqrt 4}.$

Pierre.

Pierre, a promising solution, but I have two questions:

1) why is it wlog that k=1, when i thought of the problem, i thought the generalization would be intuitive, but difficult to prove (showing that indeed we can reduce every k value to just 1)

2) why is S greater than that expression sufficient to show that a point is covered n times? I don't really see the pigeonhole's principle...

Thanks in advance.

[quote="al.M.V."]
1) why is it wlog that k=1, when i thought of the problem, i thought the generalization would be intuitive, but difficult to prove (showing that indeed we can reduce every k value to just 1)[/quote]

Maybe am I misunderstanding the problem, but you are asking for $AB \leq k$ for all points $A,B$ in the given set, am I wrong?
So, just use an appropriate scaling (an homothety) to change it to $1$. The radius of every circle will be changed in the same way, with the same change of scaling.

[quote]

2) why is S greater than that expression sufficient to show that a point is covered n times? I don't really see the pigeonhole's principle...[/quote]

The expression is $n$ times the area of the 'big' disk. If none of the points is covered at least $n$ times then the total area cannot be greater than this expression.

Pierre.

Sorry Pierre, I don't know what I was thinking, but we can indeed assume k=1 by an appropriate scaling factor of the plane....silly me  :blush:

But I am still not certain about the pigeonhole thing. You have shown that S is mn times the area of the small disk. Now if S is greater than (n-1) times the area of the large disk, at least one point is covered at least n times? Why? What are the pigeons and what are the holes? I can't seem to get that point.

Apologize for being so slow (again!)  :(  :)

[quote="al.M.V."]You have shown that S is mn times the area of the small disk. Now if S is greater than (n-1) times the area of the large disk, at least one point is covered at least n times? Why? What are the pigeons and what are the holes? I can't seem to get that point.
[/quote]

If none of the points is covered at least $n$ times, then it is covered at most $n-1$ times, so that the total area cannot be greater than $n-1$ times the area of the greater disk.
This is a continuous version of the pigeonhole principle. In that case, the sum $S$ of the areas of the small disks is the sum of the areas of the regions (in the big disk) covered exactly one time + two times the sum of the areas covered exactly two times + ... + $(n-1)$ times the sum of the areas covered exactly $n-1$ times (and not more since we assume that there is no point covered at least $n$ times). This last sum is clearly less than or equal to $n-1$ times the area of the big disk (since we first sum over the areas covered exactly $k$times, so that these partial sums are pairwise 'disjoints').

Pierre.

Thanks Pierre, I finally got this... :)

Thanks a lot for your patience.  :P


We assume $ a^2  \plus{} b^2  \plus{} c^2  \equal{} 3$ then the inequality becomes

$ \frac{{a^2 }}{{b^3  \plus{} c^3 }} \plus{} \frac{{b^2 }}{{c^3  \plus{} a^3 }} \plus{} \frac{{c^2 }}{{a^3  \plus{} b^3 }} \ge \frac{3}{2}$

Note that for $ a,b,c\ge0$ and $ a\plus{}b\plus{}c\equal{}3$ then $ a^{\frac{3}{2}} b \plus{} b^{\frac{3}{2}} c \plus{} c^{\frac{3}{2}} a \le 3$

By the Cauchy Schwarz we get

$ \frac{{a^2 }}{{b^3  \plus{} c^3 }} \plus{} \frac{{b^2 }}{{c^3  \plus{} a^3 }} \plus{} \frac{{c^2 }}{{a^3  \plus{} b^3 }} \ge \frac{{\left( {a^2  \plus{} b^2  \plus{} c^2 } \right)^2 }}{{\sum {a^2 b^3 }  \plus{} \sum {a^3 b^2 } }} \ge \frac{9}{6} \equal{} \frac{3}{2}$

We have done.

[quote="tranquoc"]Prove If a,b,c>0  then
$ \frac {a^2}{b^3 \plus{} c^3} \plus{} \frac {b^2}{c^3 \plus{} a^3} \plus{} \frac {c^2}{a^3 \plus{} b^3}$>=$ \frac {3\sqrt [2]{3}}{2\sqrt [2]{(a^2 \plus{} b^2 \plus{} c^2}}$[/quote]
Let $ a^2 \plus{} b^2 \plus{} c^2 \equal{} 3.$ Then we need to prove that $ \sum_{cyc}\frac {a^2}{b^3 \plus{} c^3}\geq\frac {3}{2}.$
But $ \sum_{cyc}\frac {a^2}{b^3 \plus{} c^3} \equal{} \sum_{cyc}\frac {a^4}{b^3a^2 \plus{} c^3a^2}\geq\frac {9}{\sum(a^3b^2 \plus{} a^3c^2)}.$
Id est, it remains to prove that $ \sum_{cyc}(a^3b^2 \plus{} a^3c^2)\leq6.$
But $ \sum_{cyc}(a^3b^2 \plus{} a^3c^2)\leq6\Leftrightarrow\sum_{cyc}a^3(3 \minus{} a^2)\leq6\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(a^5 \minus{} 3a^3 \plus{} 2)\geq0\Leftrightarrow\sum_{cyc}(a^5 \minus{} 3a^3 \plus{} 2 \plus{} 2(a^2 \minus{} 1))\geq0\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}a^2(a \plus{} 2)(a \minus{} 1)^2\geq0.$


Two bodies of masses $m_{1}, m_{2}$ (the first body is above the second) are connected using a string.
We have two situations:
(a) A force $F$ acts upwards on $m_{1}$;
(b) A force $F^\prime$ acts downwards on $m_{2}$.
Find the acceleration of the system and tension in the string in both situations.

The acceleration part is pretty obvious: $\left( m_{1}+m_{2}\right) a = F-g \left( m_{1}+m_{2}\right)$ and $\left( m_{1}+m_{2}\right) a^\prime = F^\prime+g \left( m_{1}+m_{2}\right)$.
But I can't compute the tension... Can you help me?

If we accept your answer as correct, that implies that you are considering the string to be massless and you are ignoring any elastic stretching of the string. The two masses move together in your model.

Accept that, and consider the first situation. Look at the lower mass: it is experiencing an acceleration of

$a=\frac{F-g(m_{1}+m_{2})}{m_{1}+m_{2}}= \frac{F}{m_{1}+m_{2}}-g.$

The force required to do that to this mass would be

$\frac{m_{2}F}{m_{1}+m_{2}}-m_{2}g$

Take gravity away from that, and what could possibly be delivering the rest of this force? The tension on the string. Hence, the tension is

$T=\frac{m_{2}F}{m_{1}+m_{2}}.$

Now, let's check this answer by considering the forces on the first mass. Those should be

$F-m_{1}g-T=F-m_{1}g-\frac{m_{2}F}{m_{1}+m_{2}}$

$=\frac{m_{1}F+m_{2}F-m_{2}F}{m_{1}+m2}-m_{1}g= \left(\frac{F}{m_{1}+m_{2}}-g\right)m_{1}$

And that is exactly the right force for the acceleration you claim.

You can work the second part similarly.


The sequence $ \{a_{n}\}$ is defined as:

$ \{\begin{array}{c}a_{1}=6\\\\ a_{n+1}=\lfloor\frac{5}{4}a_{n}+\frac{3}{4}\sqrt{a^{2}_{n}-12}\rfloor\end{array}$

Prove that for all $ n\geq2$ we have: $ a_{n}\equiv1\text{ }(\text{mod 10})$

I think that could be easy if we proof $ \forall_{n\in\mathbb{N}\wedge n\geq 2}\ a_{n}=2^{n-2}\cdot 10+1$, then $ a_{2}=11$, so it's o.k., thereafter:
$ a_{n+1}=\left\lfloor\frac{5}{4}\left(10\cdot 2^{n-2}+1\right)+\frac{3}{4}\sqrt{\left(2^{n-2}\cdot 10+1\right)^{2}-12}\right\rfloor$, and lets $ a_{n+1}=\Big\lfloor A\Big\rfloor$, we should proof now:
$ 10\cdot 2^{n-1}+1\leq A<10\cdot 2^{n-1}+2$, and it isn't so hard.


solve for x:    arcsin(3x)+arcsin(5x)=arccos(3x)+arccos(5x)

another one......no clue

As written, that's nonsense. Do you mean $\cos^{-1} 3x$?

Use the fact that $\cos^{-1} a+\sin^{-1}a=\frac\pi2$; this gives you that each side of your equation adds to a right angle. Now you have a right triangle with sides 3 and 5 and hypotenuse $\frac1x$. (This only gives you the positive solution; the negative of this root is another root).

This is really very easy.

so we can ignore the negative answer

cause I put these in my calculator, the graph only intersect when x=0.17149859

That's not what I meant; add the two sides instead of subtracting them, or replace arcsin by $\frac\pi2$-arccos. Don't try to isolate one of the arguments.

yeah..I got that....just never knew that arccos(x)+arcsin(x)=pi/2

so what do we do about the negative root?

I was wrong about the negative root- it's only there if you interpret the equation mod $2\pi$. At that point, one side is $-\frac\pi2$ and the other is $\frac{3\pi}2$.

I didn't put the number in my post, but you shouldn't settle for an approximation on this one; the exact answer is right there.

That identity is really very obvious from the geometric definition of $\sin$ and $\cos$.


Hi,

Just wanted to know - how does this admission thing work, anyway?  Who will be reading my personal statement?

Thanks.


LaTeX
inequalities unsolved
inequalities

$ \left(8\plus{}\frac{2\sqrt{ab}}{a\plus{}b}\right)\left(a(a\plus{}b)\frac{a\plus{}b\plus{}c}{6}\right)^{\frac{1}{3}}\geq3\left(a\plus{}\sqrt{ab}(abc)^{\frac{1}{3}}\right)$
where a,b,c are positive.

sorry but i don't know how to use the math type

hello, why don't you use $ \text{\LaTeX}$? Is this your problem
$ \left(8+\frac{2\sqrt{ab}}{a+b}\right)\left(a(a+b)\frac{a+b+c}{6}\right)^{\frac{1}{3}}\geq3\left(a+\sqrt{ab}(abc)^{\frac{1}{3}}\right)$? and $ a,b,c \in ...?$
Sonnhard.

hello, and are $ a,b,c$ positive real numbers?
Sonnhard.