analytic geometry
graphing lines
slope
inequalities
trigonometry
inequalities proposed
algebra

[quote="darkmaster"]\[ x>0,y>0, x^2+y^2 \leq x+y \]. Find max of
    \[ S=x+2y\]
Enjoy it![/quote]

$ x^2+y^2 \leq x+y $ implies

$ (x - \frac 12)^2 + (y - \frac 12)^2 \leq \frac 12 $

so $ (x, y) $ must be in the interior, or on, the circle $ (x - \frac 12)^2 + (y - \frac 12)^2 = \frac 12 $.  now, for a fixed $ S $, $ S = x+2y $ is a line of slope $ -1/2 $, so all such lines are parallel, and the higher a line, the greater the value of $ S $.  hence the maximum value of $ S $ under the conditions of the problem is achieved when such a line is at its "highest" -- i.e., tangent to the circle $ (x - \frac 12)^2 + (y - \frac 12)^2 = \frac 12 $ in the first quadrant (because of the conditions $x > 0$, $ y > 0$.

substituting $ x = S - 2y $ into the equation of the circle, we have $ S^2 - 4Sy + 4y^2 + y^2 = S - 2y + y $, or $ 5y^2 + (-4S + 1)y + S^2 - S = 0 $ which should have only one solution for $ y $ to get tangency.  this occurs when

$(-4S + 1)^2 - 20(S^2 - S) = 0 $;

$ 4S^2 - 12S - 1 = 0 $;

$ S = \frac {12 \pm \sqrt {160}}{8} $.

the positive of these two values is the answer: $ \frac {12 + \sqrt {160}}{8} = \frac {3 + \sqrt {10}}{2} $

let a=x-1/2 and b=y-1/2 then we have a^2+b^2<=1/2.
So S=x+2y=a+2b+3/2
and by cauchy-shwarz inequality we get
S<=sqrt(a^2+b^2)sqrt(5)+3/2<=(sqrt(10)+3)/2.
So the maximum of S is (sqrt(10)+3)/2.

When the line $x+2y-S=0$ is perpendicular to the circle $\left(x-\frac{1}{2}\right)^2+\left(y-\frac{1}{2}\right)^2=\frac{1}{2}$ at the first quadrant,

the value of $x+2y$  is maximized,then the distance between the center of $\left(\frac{1}{2},\frac{1}{2}\right)$ and the line $x+2y-S=0$ is equal to the radius $\frac{1}{\sqrt{2}}$,

we have ${\frac{\left|\frac{1}{2}+2\cdot \frac{1}{2}-S\right|}{\sqrt{1^2+2^2}}}=\frac{1}{\sqrt{2}}\Longleftrightarrow \left|\frac{3}{2}-S\right|=\frac{\sqrt{10}}{2}\Longleftrightarrow S=\frac{3\pm\sqrt{10}}{2}$.

Sketching the graph, the desired maximum value is $\frac{3+\sqrt{10}}{2}$.

[quote="darkmaster"]Yes, $ \frac{3+\sqrt{10}}{2}$ is correct.
My proof is a lot shorter:
 \[ 2S+\sqrt{10}(x+y) = (2+\sqrt{10})x+(4+\sqrt{10})y \leq \sqrt{x^2+y^2}\sqrt{40+12\sqrt{10}}\leq 2.\sqrt{\sqrt{10}+3}.\sqrt{\sqrt{10}(x+y)} \leq \sqrt{10}+3+\sqrt{10}(x+y) \]
\[ \Longrightarrow S \leq \frac{3+\sqrt{10}}{2} \][/quote]

When does equality hold? :D

It's O.K.

Another solution.

We can set the points on the given circle as follows.

$x=\frac{1}{\sqrt{2}}\cos \theta +\frac{1}{2}, y=\frac{1}{\sqrt{2}}\sin \theta +\frac{1}{2}\ \left(0<\theta <\frac{\pi}{2}\right)$.

Then let $\vec a=(1,2), \vec b=(\cos \theta ,\ \sin \theta)$, using scalar product,

we have $x+2y=\frac{1}{\sqrt{2}}\vec a\cdot \vec b +\frac{3}{2}\leqq\frac{1}{\sqrt{2}}\cdot\sqrt{5}+\frac{3}{2}=\frac{3+\sqrt{10}}{2}$.

where the equality holds when $\vec a \parallel \vec b $(in the same direction)$\Longleftrightarrow (\cos \theta , \sin \theta)=\left(\frac{1}
{\sqrt{5}},\frac{2}{\sqrt{5}}\right)$.

Therefore the desired maximum value is $\frac{3+\sqrt{10}}{2}$

when the equality holds $(x,y)=\left(\frac{1}{2},\frac{1}{2}\right)+\frac{1}{\sqrt{2}}\cdot \frac{1}{\sqrt{5}}(1,2)$,yielding [b]darkmaster[/b]'s result.

[quote=sqing]Let $ x,y>0$ and $ x^2+y^2 \leq x+y .$ Prove that$$x+3y\leq 2+\sqrt 5$$[/quote]

[quote=sqing]Let $ x,y>0$ and $ x^2+y^2 \leq x+2y .$ Prove that
$$x+2y\leq 5$$
$$x+3y\leq \frac{7+5\sqrt{2}}{2}$$
$$x+7y\leq \frac{5(3+\sqrt{10})}{2}$$
$$2x+3y\leq\frac{8+3\sqrt{65}}{2}$$
$$3x+4y\leq  \frac{11+5\sqrt{5}}{2}$$[/quote]

[hide=Hint/sketch]Use the substitution $a=\frac{2\sqrt5}5\left(x-\frac12\right)$ and $b=\frac{2\sqrt5}5(y-1)$.[/hide]

Let $x,y$ be reals such that $x^2+y^2\le x+y.$ [Prove that
$$\frac{k+1}{2} - \sqrt{\frac{k^2+1}{2}} \le kx+y \le \frac{k+1}{2} + \sqrt{\frac{k^2+1}{2}}$$

Let $ x,y>0$ and $ x^2+y^2 \leq x+2y .$ Prove that
$$x+2y\leq 5$$
Let $ x,y>0$ and $ x^2+y^2 \leq x+y .$ Prove that
$$x+y\leq 2$$


ABC is a triangle with the area 1. Take D on BC, E on CA, F on AB, so that AFDE is cyclic.
Prove that: Area of triangle DEF=<EF^2/4AD^2.

Draw a line which passes through $ D$ and parallel to $ AB$, it cuts $ AC$ at $ P$, extend $ DP$ to $ Q$ such that $ P$ is the midpoint of $ DQ$.
It's easy to prove that $ \Delta ADP\sim \Delta FED$ then $ \frac{S_{DEF}}{S_{ADP}}\equal{}(\frac{EF}{AD})^2$
But $ PD\equal{}QP$ thus $ S_{ADP}\equal{}S_{APQ}$
$ \frac{S_{APQ}}{S_{ABC}}\equal{}\frac{AP.PQ}{AB.AC}\equal{}\frac{AP}{AC}.\frac{DP}{AB}\equal{}\frac{BD}{BC}.\frac{DC}{BC}$
$ \leq \frac{1}{4}(\frac{BD}{BC}\plus{}\frac{DC}{BC})^2\equal{}\frac{1}{4}$
So $ S_{DEF}\leq (\frac{EF}{2AD})^2$

[quote="Ulanbek_Kyzylorda KTL"] [color=darkred][b][u]An easy extension[/u].[/b] Let $ ABC$ be a triangle and $ D$ , $ E$ , $ F$ be three points on the sidelines $ BC$ , $ CA$ , $ AB$ respectively

so that the quadrilateral $ ADEF$ is cyclically (can be and nonconvex !). Prove that $ \boxed {\ \frac {[DEF]}{[ABC]}\equal{}\left(\frac {EF}{AD}\right)^2\cdot\frac {DB\cdot DC}{BC^2}\ }$ .[/color] [/quote]

[color=darkblue][b][u]Proof[/u].[/b] Denote $ \left\|\begin{array}{c}
m(\angle DAB)\equal{}x\\\\
m(\angle DAC)\equal{}y\end{array}\right\|$ . Prove easily that $ \frac {AD^2}{DB\cdot DC}\equal{}\frac {\sin B\sin C}{\sin x\sin y}\ (*)$ and $ \left\|\begin{array}{c}
m(\angle DEF)\equal{}x\\\\
m(\angle DFE)\equal{}y\end{array}\right\|$ . Therefore,

$ \frac {[DEF]}{[ABC]}\equal{}\frac {EF^2\sin x\sin y}{2\sin A}\cdot \frac {2\sin A}{BC^2\sin B\sin C}\equal{}$ $ \frac {EF^2}{BC^2}\cdot \frac {\sin x\sin y}{\sin B\sin C}\stackrel{(*)}{\ \equal{}\ }\frac {EF^2}{BC^2}\cdot\frac {DB\cdot DC}{AD^2}\ \implies\ \frac {[DEF]}{[ABC]}\equal{}\left(\frac {EF}{AD}\right)^2\cdot\frac {DB\cdot DC}{BC^2}$ .[/color]


How many integers between 2 and 1004 are multiples of 5?

well, there are no multiples of 5 below 2 and the number that is a multiple of 5 that is 1000.
100/5=200

Because in every 10 numbers there are two numbers can be divided by 5, so 1004/10=100
100*2=200


integration
induction
algebra
binomial theorem
real analysis
real analysis unsolved

Sum the following two series:

$ \sum\limits_{m\equal{}0}^{\infty}\sum\limits_{n\equal{}0}^{\infty}\frac{(m\plus{}n)!}{m!n!}\left(\frac{x}{2}\right)^{m\plus{}n}$,   $ \minus{}2<x<1$,

and

$ \sum\limits_{n\equal{}0}^{\infty}\sum\limits_{m\equal{}0}^{\infty}\sum\limits_{p\equal{}0}^{\infty}\frac{(m\plus{}n\plus{}p)!}{m!n!p!}\left(\frac{x}{3}\right)^{m\plus{}n\plus{}p}$,   $ \minus{}3<x<1$,

and extend the results to any number of indices of summation.

\begin{eqnarray*}
& & \sum \sum \frac{(m+n)!}{m!n!}\left(\frac{x}{2}\right)^{m+n}\\
& = & \sum\sum \frac{1}{m!n!}\left(\frac{x}{2}\right)^{m+n} \int_{0}^{\infty} t^{m+n} e^{-t} \, dt\\
& = & \int_{0}^{\infty} \left[ \sum \frac{1}{m!}\left(\frac{xt}{2}\right)^m \right] \left[ \sum \frac{1}{n!}\left(\frac{xt}{2}\right)^n \right] e^{-t} \, dt\\
& = & \int_{0}^{\infty} e^{-t(1-x)} \, dt\\
& = & \frac{1}{1-x}
\end{eqnarray*}

this method can be generalized to any number of indices of summation, and the result is always $ \frac{1}{1-x}$.

Now, justify that in the part of the region where we don't have absolute convergence.

We may use [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=182575]this fact[/url]. That is, if $ \sum_{n\equal{}0}^{\infty} a_n$ converges, then $ \sum_{n\equal{}0}^{\infty} \int_{0}^{\infty} \frac{a_n}{n!}t^n e^{\minus{}t} \, dt \equal{} \int_{0}^{\infty} \sum_{n\equal{}0}^{\infty} \frac{a_n}{n!}t^n e^{\minus{}t} \, dt$.

Being a combinatorialist, I tend not to worry about regions of convergence. :)  I guess that means I am missing the point of this exercise, though.  Here's a formal manipulation that verifies the equality:
\begin{align*}\frac {1}{1 - x} & = \sum_{n \geq 0} x^n \\
& = \sum_{n \geq 0} \left(\frac {x}{k} + \frac {x}{k} + \ldots + \frac {x}{k}\right)^n \\
& = \sum_{n \geq 0} \sum_{\substack{a_1 + a_2 + \ldots a_k = n \\
a_i \geq 0}} \binom{n}{a_1; a_2; \ldots; a_k} \left(\frac {x}{k}\right)^{a_1} \left(\frac {x}{k}\right)^{a_2}\cdots \left(\frac {x}{k}\right)^{a_k} \\
& = \sum_{n \geq 0} \sum_{\substack{a_1 + a_2 + \ldots a_k = n \\
a_i \geq 0}} \frac {n!}{a_1!\cdot a_2!\cdots a_k!} \left(\frac {x}{k}\right)^{n} \\
& = \sum_{a_1 \geq 0} \sum_{a_2 \geq 0} \cdots \sum_{a_k \geq 0} \frac {(a_1 + \ldots + a_k)!}{a_1!\cdot \cdots a_k!} \left(\frac {x}{k}\right)^{a_1 + \ldots + a_k}. \end{align*}
Then at the end I would say, "since this is a formal equality, it holds wherever the two sides are defined."  This happens to be the interval $ ( - k, 1)$.

(Editted to fix a minor typo.)

But JBL, I don't think you're missing the point. It's a natural idea to convert this series into power series. My solution is rather an overkill.

Hi!

Maybe I'm wrong, but isn't

$ \frac {1}{1 - x} \& = \sum_{n \geq 0} x^n$

just for $ |x|<1$? I don't see why this is the case here:

Can someone help me?

EtotheX

JBL's "solution" is not really a proof that the identity holds for $ x<\minus{}1$. To rigorously treat this case, you need to examine the repeated sum carefully like sos440 did. It is not at all obvious that the right hand side is well-defined on that interval.

Another version of the argument, at least for the first sum:

Note that $ \frac{(m\plus{}n)!}{m!n!}\left(\frac x2\right)^{m\plus{}n}\equal{}\frac{x^m}{m!}\frac{d^m}{dx^m}\left(\left(\frac x2\right)^{m\plus{}n}\right)$. Summing these over $ n$, the inner sum is $ \frac{x^m}{m!}\frac{d^m}{dx^m}\left(\sum_{n\equal{}0}^{\infty}\left(\frac x2\right)^{m\plus{}n}\right)\equal{} \frac{x^m}{m!}\frac{d^m}{dx^m}\left(\sum_{k\equal{}0}^{\infty}\left(\frac x2\right)^{k}\right)\equal{}\frac{x^m}{m!}\frac{d^m}{dx^m}\left(\frac{ 1}{1\minus{}\frac x2}\right)$ for $ \minus{}2<x<2$.

Now, sum over $ m$: we get $ \sum_{m\equal{}0}^{\infty}\frac{x^m}{m!}\frac{d^m}{dx^m}\left(\frac{ 1}{1\minus{}\frac x2}\right)$. For any nice enough $ f$ (the Taylor series converges to $ f$), $ \sum_{m\equal{}0}^{\infty}\frac{t^m}{m!}f^{(m)}(x)\equal{}f(x\plus{}t)$. This gives $ \frac1{1\minus{}\frac{x\plus{}x}{2}}\equal{}\frac1{1\minus{}x}$ as long as $ |x|<2$ (convergence of the inner sum) and $ |x|<2\minus{}x$ (convergence of the outer sum). Together, these give $ \minus{}2<x<1$.

[quote="EtotheX"]isn't
\[ \frac {1}{1 - x} \& = \sum_{n \geq 0} x^n
\]
just for $ |x| < 1$?[/quote]
Yes, but this is what I mean by the terms "formal computation" and "formal equality" -- it's true whenever it makes sense (which happens to be for $ |x| < 1$).  This is true of every step in the chain of calculations I made -- as long as both sides of any equal sign make sense, they will in fact be equal.  Then at the end, we get to conclude that the original left side and the final right side are equal wherever they both make sense.

This explanation would make most combinatorialists happy, but I'm not surprised that analysts (e.g. fedja) might not like it so much.  I think this is because of the phrase "we get to conclude" -- it covers up things that have nothing to do with the combinatorial aspect of the problem but that have everything to do with the analytic aspects (e.g., where these expressions actually converge).  This is also why I said that I thought I was "missing the point" of this problem -- I don't think didilica thought of this as a problem about the binomial theorem, but rather as a problem about regions of convergence.

$ \sum_{m \equal{} 0}^\infty \sum_{n \equal{} 0}^\infty \frac {(n \plus{} m)!}{n!\, m!} \, x^n y^m \equal{} \sum_{m \equal{} 0}^\infty \sum_{n \equal{} 0}^\infty \left(\frac {d}{dx}\right)^m x^{n \plus{} m}\, \frac {y^m}{m!}$

$ \equal{} \sum_{m \equal{} 0}^\infty \left(\frac {d}{dx}\right)^m\sum_{n \equal{} 0}^\infty x^{n \plus{} m}\, \frac {y^m}{m!} \equal{} \sum_{m \equal{} 0}^\infty \left(\frac {d}{dx}\right)^m\frac {x^m}{1 \minus{} x}\, \frac {y^m}{m!}$

$ \equal{} \sum_{m \equal{} 0}^\infty \frac {m!}{(1 \minus{} x)^{m \plus{} 1}}\, \frac {y^m}{m!}$ because $ \left[\left(\frac {d}{dx}\right)^m\frac {x^m}{1 \minus{} x} \equal{} \left(\frac {d}{dx}\right)^m\frac {1}{1 \minus{} x} \minus{} \underbrace{\left(\frac {d}{dx}\right)^m\frac {1 \minus{} x^m}{1 \minus{} x}}_{ \equal{} 0} \right]$
$ \equal{} \frac {1}{1 \minus{} x} \sum_{m \equal{} 0}^\infty \left(\frac {y}{1 \minus{} x} \right)^m \equal{} \frac {1}{1 \minus{} x} \cdot \frac {1}{1 \minus{} \frac {y}{1 \minus{} x}} \equal{} \frac {1}{1 \minus{} x \minus{} y}$

This is valid iff $ |x| < 1$ and $ |y| < |1 \minus{} x|$ .

$ \sum_{m \equal{} 0}^\infty \sum_{n \equal{} 0}^\infty \sum_{p \equal{} 0}^\infty \frac {(p \plus{} n \plus{} m)!}{p! \, n!\, m!} \, x^p y^n z^m \equal{} \sum_{m \equal{} 0}^\infty \sum_{n \equal{} 0}^\infty \sum_{p \equal{} 0}^\infty \left(\frac {d}{dx}\right)^{n \plus{} m} x^{p \plus{} n \plus{} m}\, \frac {y^n}{n!} \frac {z^m}{m!}$

$ \equal{} \sum_{m \equal{} 0}^\infty \sum_{n \equal{} 0}^\infty \left(\frac {d}{dx}\right)^{n \plus{} m} \sum_{p \equal{} 0}^\infty x^{p \plus{} n \plus{} m}\, \frac {y^n}{n!} \frac {z^m}{m!} \equal{} \sum_{m \equal{} 0}^\infty \sum_{n \equal{} 0}^\infty \left(\frac {d}{dx}\right)^{n \plus{} m} \frac {x^{n \plus{} m}}{1 \minus{} x}\, \frac {y^n}{n!} \frac {z^m}{m!}$

$ \equal{} \sum_{m \equal{} 0}^\infty \sum_{n \equal{} 0}^\infty \frac {(n \plus{} m)!}{(1 \minus{} x)^{n \plus{} m \plus{} 1}} \, \frac {y^n}{n!} \frac {z^m}{m!} \equal{} \frac {1}{1 \minus{} x} \, \sum_{m \equal{} 0}^\infty \sum_{n \equal{} 0}^\infty \frac {(n \plus{} m)!}{n!\, m!} \, \left(\frac {y}{1 \minus{} x}\right)^n \, \left(\frac {z}{1 \minus{} x}\right)^m$

$ \equal{} \frac {1}{1 \minus{} x} \cdot \frac {1}{1 \minus{} \frac {y}{1 \minus{} x} \minus{} \frac {z}{1 \minus{} x}} \equal{} \frac {1}{1 \minus{} x \minus{} y \minus{} z}$

This is valid for $ |x| < 1$ and $ |y| < |1 \minus{} x|$ and $ |z| < |1 \minus{} x \minus{} y|$ .

By a longish induction step you can show

$ \sum_{k_n \equal{} 0}^\infty \cdots \sum_{k_1 \equal{} 0}^\infty \frac {(k_1 \plus{} ... \plus{} k_n)!}{k_1!\cdots k_n!} \, x_1^{k_1}\cdots x_n^{k_n} \equal{} \frac {1}{1 \minus{} x_1 \minus{} ... \minus{} x_n}$

for $ |x_1| < 1$ and $ |x_2| < |1 \minus{} x_1|$ ... and $ |x_n| < |1 \minus{} x_1 \minus{} ... \minus{} x_{n \minus{} 1}|$ .

Should $ x_1,...,x_n$ all be equal $ x_1 \equal{} ... \equal{} x_n \equal{} x$ so the convergence conditions are

$ |x| < 1 \,\, , \,\, |x| < |1 \minus{} x| \,\, , \,\, |x| < |1 \minus{} 2x|\,\, , \,\, |x| < |1 \minus{} 3x| \,\, , ... , \,\, |x| < |1 \minus{} (n \minus{} 1)x|$

whereat $ |x| < |1 \minus{} jx|$ is equivalent to $ \left[x < \frac {1}{j \plus{} 1}\, \vee\, x > \frac {1}{j \minus{} 1} \right]$ for $ j\neq 1$

and $ |x| < |1 \minus{} x|$ is equivalent to $ x < \frac12$ .

All conditions are simultaneously fulfilled iff $ \minus{} 1 < x < \frac {1}{n}$ .

Therefore $ \sum_{k_n \equal{} 0}^\infty \cdots \sum_{k_1 \equal{} 0}^\infty \frac {(k_1 \plus{} ... \plus{} k_n)!}{k_1!\cdots k_n!} \, x^{k_1 \plus{} ... \plus{} k_n} \equal{} \frac {1}{1 \minus{} nx}$ for $ \minus{} 1 < x < \frac {1}{n}$

respectively $ \sum_{k_n \equal{} 0}^\infty \cdots \sum_{k_1 \equal{} 0}^\infty \frac {(k_1 \plus{} ... \plus{} k_n)!}{k_1!\cdots k_n!} \, \left(\frac {x}{n}\right)^{k_1 \plus{} ... \plus{} k_n} \equal{} \frac {1}{1 \minus{} x}$ for $ \minus{} n < x < 1$ .

In the absolute convergence case $ \|x\|_1 < 1 \quad , \quad x \equal{} (x_1,...,x_n)$

we can use the multi-index notation to write

$ \frac {1}{1 \minus{} |x|} \equal{} \sum_{k \equal{} 0}^\infty |x|^k \equal{} \sum_{k \equal{} 0}^\infty k! \, \sum_{| \alpha | \equal{} k} \frac {x^\alpha}{\alpha!} \equal{} \sum_{|\alpha|\ge 0} |\alpha|! \,\, \frac {x^\alpha}{\alpha!}$ .


Every day, $ 20\%$ of the fish in a fish store are sold.  There are 2000 fish left in the store at the end of the day on Tuesday.  How many were there when the store opened on Monday?

[hide="solution"]
Let $ f\equal{}\text{the number of fish starting on Monday}.$ Every day, there is only $ \frac{4}{5}$ of the fish that the day started out with. We have the equation $ \left(\frac{4}{5}\right)\left(\frac{4}{5}\right)f\equal{}2000\Rightarrow f\equal{}\frac{25\times2000}{16}\equal{}25\times125\equal{}\boxed{3125}.$[/hide]


algebra
polynomial
calculus
derivative
integration
linear algebra
matrix

Yes.
But we can restrict ourselves to the case we have no double roots.
It doesn't really matter.
(In the case of double roots there are two $\frac{1}{0}$ in our equation and so we cannot say the equation is false)

Residue theorem should work (hopefully :mrgreen:). Take $\frac{1}{P(x)}$ and integrate along a suffiently large circle around the origin. By letting your radius grow, your integral will approach $0$

Take $P(x)=c(x-a_1)(x-a_2)...(x-a_n)$.
Then in relation
(1)   $\sum_{i=1}^n\frac{1}{P'(a_i)}=0$
each denominator will be a product of factors $a_i-a_j$.

Multiply both sides by $(-1)^{n-1}$ and change, if necessary, the sign before the fractions s.t. the denominators should contain only factors $a_i-a_j$ with $i>j$. After reducing to the same denominator, we obtain at each numerator a Vandermonde determinant (see  http://en.wikipedia.org/wiki/Vandermonde_determinant ).
The first fraction, for instance,  will have as numerator
$\prod_{2\le j<i\le n}(a_i-a_j)=\det \begin{bmatrix}1 & 1 & \dots &1\\a_2 & a_3 & \dots & a_n\\ a_2^2 & a_3^2 & \dots & a_n^2\\ \vdots & \vdots & & \vdots\\a_2^{n-2} & a_3^{n-2} & \dots & a_n^{n-2}\\ \end{bmatrix}$
Eliminate the common denominator in relation (1) and get an alternating  sum of $n$ Vandermonde determinants, which actually represents the expansion of the determinant
$\det \begin{bmatrix}1 & 1 & \dots &1\\1 & 1 & \dots &1\\a_1 & a_2 & \dots & a_n\\ a_1^2 & a_2^2 & \dots & a_n^2\\ \vdots & \vdots & & \vdots\\a_1^{n-2} & a_2^{n-2} & \dots & a_n^{n-2}\\ \end{bmatrix}=0$

here is another solution:
first note that:$p'(x)=p(x)[\sum_{i=1}^n\frac{1}{x-a_i}]$
now from the above your sum will be equal to:
$\frac{1}{(a_1-a_2)...(a_1-a_n)}+...\frac{1}{(a_n-a_1)...(a_n-a_{n-1})}$
noe use lagrange thorem for writing the polynomial$f(x)$ such that $f(a_i)=1$
then  because $deg(f)$ is not greater than $n-1$ , it is obvious that  $f(x)=1$
so by comparing the coefficient the result follow.

[quote="Nima Ahmadi Pour"]I found out an amazing fact while solving a problem.
If $P(x)$ is a $n>1$ degree polynomial with roots $a_1,a_2,...,a_n$ then
<center>$\sum_{i=1}^n\frac{1}{P'(a_i)}=0$</center>
where $P'$ is the derivative of $P$.
Try to prove it! :lol:[/quote]

polish 1979 (and look at the mehdi safa's book , page 242)  :D


[quote="cefer"]I think it must be
$\sum{\frac{1}{b^{2}+c^{2}}\geq{\sum{\frac{2bc}{(a^{2}+b^{2})(a^{2}+c^{2})}}}}$
[/quote]
It 's right :)
arqady'solution is very nice .(the same with me :) )


Let $ x$ and $ y$ positive real numbers such that $ (1\plus{}x)(1\plus{}y)\equal{}2$. Show that  $ xy\plus{}\frac{1}{xy}\geq\ 6$

[hide="Proof"]By AM-HM, we have that

$ \frac{x\plus{}y}{2} \geq \frac{2}{\frac{1}{x} \plus{} \frac{1}{y}}$

$ \frac{x\plus{}y}{2} \geq \frac{2xy}{x\plus{}y}$

$ (x\plus{}y)^2 \geq 4xy$

From the given equation $ (1\plus{}x)(1\plus{}y) \equal{} 2$, we can rewrite it as $ x\plus{}y \equal{} 1\minus{}xy$, so

$ (1\minus{}xy)^2 \geq 4xy$

$ 1 \minus{} 2xy \plus{} (xy)^2 \geq 4xy$

$ (xy)^2 \plus{} 1 \geq 6xy$

$ xy \plus{} \frac{1}{xy} \geq 6$

QED[/hide]

nice solution, modularmarc101  :) mine was just the same :lol:

Alternately,
\[ 2 \equal{} (1 \plus{} xy) \plus{} (x \plus{} y)\ge1 \plus{} xy \plus{} 2\sqrt {xy} \equal{} (\sqrt {xy} \plus{} 1)^2\implies\sqrt {xy}\le\sqrt {2} \minus{} 1 \\
\implies8\le(3 \minus{} xy)^2\implies xy \plus{} \frac1{xy}\ge6.\]

By Cauchy Schwarz $ 2 \equal{} (1 \plus{} x)(1 \plus{} y) \geq\ (1 \plus{} \sqrt {xy})^{2}$
Then $ 3 \minus{} 2\sqrt {2}\geq\ xy$ $ (1)$.
We have $ xy \equal{} 1 \minus{} x \minus{} y$. Changing this on the inequality, we´ll have to prove that:
$ 1 \minus{} x \minus{} y \plus{} \frac {1}{1 \minus{} x \minus{} y}\geq\ 6$  iff $ (x \plus{} y \plus{} 2 \plus{} 2\sqrt {2})(x \plus{} y \plus{} 2 \minus{} 2\sqrt {2})\geq\ 0$ iff $ 1 \minus{} xy \plus{} 2 \minus{} 2\sqrt {2}\geq\ 0$ iff $ 3 \minus{} 2\sqrt {2}\geq\ xy$ $ (1)$ which it´s true by $ (1)$. $ QED$

AM-GM gives $ 2\sqrt{(1\plus{}x)(1\plus{}y)}\leq (1\plus{}x)\plus{}(1\plus{}y)\Longleftrightarrow 2\sqrt{2}\leq x\plus{}y\plus{}2$, yielding $ x\plus{}y\geq 2(\sqrt{2}\minus{}1)$.
Thus from $ (1\plus{}x)(1\plus{}y)\equal{}2\Longleftrightarrow xy\equal{}1\minus{}(x\plus{}y)\leq 3\minus{}2\sqrt{2}$. Now for $ 0<t\leq 1$, the function $ y\equal{}t\plus{}\frac{1}{t}$ is  decreasing function, so for $ 0<xy\leq 3\minus{}2\sqrt{2}<1$, we have $ xy\plus{}\frac{1}{xy}\geq 3\minus{}2\sqrt{2}\plus{}\frac{1}{3\minus{}2\sqrt{2}}$, yielding
$ xy\plus{}\frac{1}{xy}\geq 6.\ Q.E.D.$

[quote="trejos"]Let $ x$ and $ y$ positive real numbers such that $ (1 \plus{} x)(1 \plus{} y) \equal{} 2$. Show that  $ xy \plus{} \frac {1}{xy}\geq\ 6$[/quote]
$ (1 \plus{} x)(1 \plus{} y) \equal{} 2 \iff xy \equal{} 1 \minus{} x \minus{} y \le 1 \minus{} 2\sqrt {xy} \iff (\sqrt {xy} \plus{} 1)^2 \le 2$ $ \iff \sqrt {xy} \le \sqrt {2} \minus{} 1 \iff xy \le 3 \minus{} 2\sqrt {2}$.
The function $ f(t) \equal{} t \plus{} t^{ \minus{} 1}$ is decreasing on $ (0;1)$, so $ f(xy) \ge f(3 \minus{} \sqrt {2}) \equal{} (3 \minus{} \sqrt {2}) \plus{} \frac {1}{3 \minus{} \sqrt {2}} \equal{} 3 \minus{} 2\sqrt {2} \plus{} 3 \plus{} 2\sqrt {2} \equal{} 6$, and we are done.

[quote="kunny"][quote="Mathias_DK"][quote="trejos"]Let $ x$ and $ y$ positive real numbers such that $ (1 \plus{} x)(1 \plus{} y) \equal{} 2$. Show that  $ xy \plus{} \frac {1}{xy}\geq\ 6$[/quote]
$ (1 \plus{} x)(1 \plus{} y) \equal{} 2 \iff xy \equal{} 1 \minus{} x \minus{} y \ge 1 \minus{} 2\sqrt {xy} \iff (\sqrt {xy} \plus{} 1)^2 \ge 2 \iff \sqrt {xy} \ge \sqrt {2} \minus{} 1 \iff xy \ge 3 \minus{} 2\sqrt {2}$. On the other hand: $ (1 \plus{} x)(1 \plus{} y) \equal{} 2 \iff xy \equal{} 1 \minus{} x \minus{} y \le 1$.
The function $ f(t) \equal{} t \plus{} t^{ \minus{} 1}$ is decreasing on $ (0;1)$, so $ f(xy) \ge f(3 \minus{} \sqrt {2}) \equal{} (3 \minus{} \sqrt {2}) \plus{} \frac {1}{3 \minus{} \sqrt {2}} \equal{} 3 \minus{} \sqrt {2} \plus{} 3 \plus{} \sqrt {2} \equal{} 6$, and we are done.[/quote]

The first line in your solution is wrong and your solution has been already posted. :wink:[/quote]
Thanks for correcting me - it obviously went a bit too fast.
What is wrong with posting a solution, which is already there?

We find $ y\equal{}\frac{1\minus{}x}{x\plus{}1}$, then we have to show
$ \frac{x(1\minus{}x)}{x\plus{}1}\plus{}\frac{x\plus{}1}{x(1\minus{}x)} \ge 6 \iff (x^2\plus{}2x\minus{}1)^2 \ge 0$

Notice that 1+x=a and 1+y=b, the following inequality equivalent to                                                                  (a-1)^2 (b-1)^2+1≥6(a-1)(b-1)⇔
a^2+b^2≥4, because ab=2.
With AM-GM a^2+b^2≥2ab=4.

At least you could ask before using my problem in your National Olympiad

i was a participant in the Centroamerican competition of 2009 :D and this problem was shown to all of the contestants and it wouldnt seem fair that countries would use problems like this on their national competition as people would be aware of its solution as it has been posted a long time ago on this same website. i also back up MR.Jose Gabriel that to use a problem they should have full permision of the creator

[quote="ogpg007"]We can say that $ a=xy $ so:
                                            $ a+ \frac{1}{a} \geq 6 $
$\iff$                                     $\frac{a^2+1}{a} \geq 6 $
$\iff$                                     $a^2+1 \geq 6a $
$\iff$                                     $a^2-2a+1 \geq 4a $
$\iff$                                     $(a-1)^2 \geq 4a $
That's true because of $AM-GM$
$QED$[/quote]

I think you need to state the range of $a$.

Here is my 2nd solution.

$(1+x)(1+y)=2\Longleftrightarrow x+y+xy=1$, let $x+y=s,\ xy=t\ \cdots (*)$,  we have $s+t=1\ \cdots [1]$.
We are to find the condition for which there is some positive real number $x>0,\ y>0$.
Thus the condition can be given that $s>0,\ t>0,\ s^2\geq 4t\ \cdots[2]$. From $[1],\ [2]$, we have $s^2\geq 4(1-s)\geq 0$
$\Longleftrightarrow s^2+4s-1\geq 0$, since $s>0$, the range of $s$ is $s\geq -2+2\sqrt{2}$.
$\therefore xy+\frac{1}{xy}=t+\frac{1}{t}\geq 6\Longleftrightarrow t^2-6t+1\geq 0$, which is true by $t^2-6t+1=(t-3)^2-8=(s+2)^2-8\geq 0\ Q.E.D.$

We can easily see that $xy = 1 - x - y$.
Therefore, $xy + \frac{1}{xy} \geq 6   =   (1-x-y)(1-x-y)+1 \geq 6 (1-x-y)$

Therefore: $2 - 2x - 2y + x^2 + y^2 + 2xy \geq 6-6x-6y$

Then:

\[x^2+y^2+4x+4y+2xy  \geq 4\]

Since $x,y$ are both positive this is obviously true.

I believe that there is another solution using derivatives. (as you can tell, inequalities aren’t my cup of tea)

Note that $1 + x = \frac{2}{y+1} \implies x = \frac{1-y}{y+1}$. Therefore, we have that
$$xy + \frac{1}{xy} = \frac{1}{\left(\frac{1-y}{y+1}\right)y}+y\left(\frac{1-y}{y+1}\right) = \frac{y+1}{y(1-y)} + \frac{y(1-y)}{y+1} = \frac{(y+1)^2+(y(1-y))^2}{(y+1)(y-1)(y)}$$
Now, we want to find the maximum of this, so we can simply take the derivative! A few preliminary steps to make it easier to find the derivative:
$$\frac{(y+1)^2+(y(1-y))^2}{(y+1)(y-1)(y)} = \frac{y^2+2y+1+y^2(y^2-2y+1)}{(y+1)(y-1)(y)} = \frac{y^4-2y^3+2y^2+2y+1}{y^3-y}$$
Therefore, after some computation, we have that the derivative is:
$$\frac{y^6-5y^4-5y^2+1}{y^6-2y^4+y^2} = 0 \implies y^6-5y^4-5y^2+1 = 0$$
We can do a bit of factoring: $y^6-5y^4-5y^2+1 = (y^2+1)(y^4-y^2+1)-5y^2(y^2+1) = 0$. Since $y^2 \ne -1$ (that would imply $y$ isn’t real), we have that $y^4-6y^2+1 = 0$. Clearly, $y^2 = \frac{6 + \sqrt{32}}{4}$. Plugging that into $\frac{y^4-2y^3+2y^2+2y+1}{y^3-y}$ and after lots of computation (which I’ll leave out), we get that it evaluates to $6$. Therefore, $xy + \frac{1}{xy} \ge 6$. (we can easily check endpoints)

[hide=Good.][quote=pfanni][quote="trejos"]Let $ x$ and $ y$ positive real numbers such that $ (1 \plus{} x)(1 \plus{} y) \equal{} 2$. Show that  $ xy \plus{} \frac {1}{xy}\geq\ 6$[/quote]

The inequality is equivalent to:
$ \left( \sqrt{xy}\minus{}\frac {1}{\sqrt{xy}}  \right)^2\geq 4$
Therefore I have to prove
$ |\sqrt{xy}\minus{}\frac {1}{\sqrt{xy}}|\geq 2$
$ |xy\minus{}1|\geq 2\sqrt{xy}$

The condition  $ (1 \plus{} x)(1 \plus{} y) \equal{} 2$ gives: $ x\plus{}y\plus{}xy\equal{}1$.
So we just have to show:
$ |\minus{}x\minus{}y|\geq 2\sqrt{xy}$, which is $ AM\minus{}GM$[/quote][/hide]
By AM-GM,  $|1-xy|= |x+y|=x+y\geq 2\sqrt{xy},$
$$|1-xy|\geq 2\sqrt{xy}\iff  |\sqrt{xy}-\frac {1}{\sqrt{xy}}|\geq 2\iff \left( \sqrt{xy}-\frac {1}{\sqrt{xy}}  \right)^2\geq 4\iff  xy + \frac {1}{xy}\geq  6.$$

Calling $k=xy$:
$k + \frac{1}{k} \geq 6$
$k^2 - 6k + 1 \geq 0$
Then $k\geq k_1$ or $k\leq k_2 $
Were $k_2 = \frac{6 \;  - \sqrt{36-4}}{2}=3-2\sqrt{2}$
So, by proving $xy \leq 3-2\sqrt{2}$ we are almost done.
$(x+1)(y+1)=2$
$xy + x + y = 1$
Using $AM-GM$:
$\frac{x+1+y+1}{2}\geq \sqrt{(x+1)(y+1)} = \sqrt{2}$
$x + y \geq 2\sqrt{2}-2$
Adding $xy$ to both sides
$1 \geq 2\sqrt{2}-2 + xy$
$3 - 2\sqrt{2} \geq xy$
Made it!

Let $ x$ and $ y$ positive real numbers such that $ (1+x)(1+y)=2$. Show that $$ x^2y+\frac{1}{x^2y}\geq 5\sqrt 5$$
Let $ x$ and $ y$ positive real numbers such that $ (1+x)(2+y)=3$. Show that $$ xy+\frac{1}{xy}\geq  10$$
Let $ x$ and $ y$ positive real numbers such that $ (1+x)(2+y)=4$. Show that $$ xy+\frac{2}{xy}\geq 9-2\sqrt 2$$
Let $ x$ and $ y$ positive real numbers such that $ (1+x)(3+y)=4$. Show that $$ xy+\frac{1}{xy}\geq  14$$


group theory
abstract algebra
superior algebra
superior algebra solved

Let $G$ be a group with $2n$ elements,  $H_i\le G$ and $|H_i|=n$, $i=\overline{1,m}$, $m\ge 2$
Prove that if $H_1\cap H_2\ldots\cap H_m=\{e\}$ then $G$ is abelian.

Nice problem. It reminds me of the good old days when I was competing in contests... :)

All $H_i$ are normal (this is well known), so $|G/H_i|=2$, meaning that $G/H_i\cong \mathbb Z_2,\ \forall i$. In turn, this means that $x^2\in H_i,\ \forall i,\forall x\in G$. This means that for all $x\in G$, we have $x^2\in H_1\cap H_2\cap\ldots\cap H_m=\{e\}$, and from here it's easy.


If $ \frac {\tan(x \minus{} y)}{\tan(x)} \plus{} \frac {\sin^2(z)}{\sin^2(x)} \equal{} 1$ prove that $ \tan^2(z) \equal{} \tan(x)\cdot \tan(y)$

[hide="Solution"]From $ \tan{x\minus{}y}\equal{}\frac{\tan{x}\minus{}\tan{y}}{1\plus{}\tan{x}\tan{y}}$ and $ \sin^2x\equal{}\frac{\tan^2x}{1\plus{}\tan^2x}$ we arrive at (let $ \tan{x},\tan{y},\tan{z}$ be $ a,b,c$ respectively)
\[ a(a\minus{}b)(1\plus{}c^2)\plus{}c^2(1\plus{}a^2)(1\plus{}ab)\equal{}a^2(1\plus{}c^2)(1\plus{}ab)\\\\
\implies(1\plus{}c^2)(a^2\minus{}ab\minus{}a^2\minus{}a^3b)\plus{}c^2(1\plus{}a^2)(1\plus{}ab)\equal{}0\\\\
\implies c^2(1\plus{}a^2)(1\plus{}ab)\minus{}ab(1\plus{}c^2)(1\plus{}a^2)\equal{}0\\\\
\implies(1\plus{}a^2)(c^2\plus{}abc^2\minus{}ab\minus{}abc^2)\equal{}0\\\\
\implies c^2\equal{}ab,\]as desired.[/hide]


[quote="triplebig"]Solve the equation $ x^{10} \minus{} 2x^9 \plus{} 2x^8 \plus{} x^6 \minus{} 2x^5 \plus{} 2x^4 \minus{} 12x^2 \plus{} 24x \minus{} 24 \equal{} 0$ knowing that she admits a complex root of module $ \sqrt {2}$ and argument $ \frac {\pi}{4}$[/quote]
[hide]$ x^{8}(x^{2}\minus{}2x\plus{}2)\plus{}x^{4}(x^{2}\minus{}2x\plus{}2)\minus{}12(x^{2}\plus{}2x\plus{}2)\equal{}0$
$ (x^{2}\minus{}2x\plus{}2)(x^{8}\plus{}x^{4}\minus{}12)\equal{}0$
$ (x^{2}\minus{}2x\plus{}2)(x^{4}\plus{}4)(x^{4}\minus{}3)\equal{}0$
$ x^{2}\minus{}2x\plus{}2\equal{}0$-->$ x\equal{}1\pm i$
$ x^{4}\equal{}\minus{}4$-->$ x\equal{}\sqrt[4]{4}cis(\frac{\pi}{4}\plus{}\frac{\pi k}{2})$
$ x^{4}\equal{}3$-->$ x\equal{}\sqrt[4]{3}cis(\frac{\pi k}{2})$[/hide]

Thank you for the solution, ckck.
Is there any way I could use the information of the complex root in case I didn't observe that $ x^2\minus{}2x\plus{}2$ is a common factor?

[quote="triplebig"]Thank you for the solution, ckck.
Is there any way I could use the information of the complex root in case I didn't observe that $ x^2 \minus{} 2x \plus{} 2$ is a common factor?[/quote]
Well given the modulus $ \sqrt{2}$ and argument $ \frac{\pi}{4}$, we have that one of the roots is $ \sqrt{2}cis(\frac{\pi}{4})$, and that equals $ \frac{1}{2}\plus{}i\frac{1}{2}$. With some computation you could've factored that out :D


Dear my friend Silouan,

Yes you can try to find a solution to this second gift of me to you. It is unexpectedly nice... All other Mathlinkers are welcome  :lol:

$\bigstar$ Problem.
Let $a,b,c$ be non-negative real numbers. Prove that
\[\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}+\frac{9\sqrt{ab+bc+ca}}{a+b+c}\ge 6. \]

thank you Hunghtn for this very very nice gift  :D
My solution is similar to elopes with a little bit final .

I can't imagine that this problem can be proved by classical inequalities. In this inequalities, the equality holds for very interesting case
\[(a,b,c)~ \left(\frac{7\pm\sqrt{45}}{2},1,0\right).\]
However, I can not explain why your applying of Cauchy -Schwarz in the beginning of the solution is correct because if you do that, the case of equality does not satisfy.

the equality occurs, my friend hungkhtn!

if $a+b = 3\sqrt{ab}$, we have that $a = b\frac{(3+\sqrt{5})^{2}}{4}$ or $b\frac{(3-\sqrt{5})^{2}}{4}$.

so, $a = b\frac{7+\sqrt{45}}{2}$ or $b\frac{7-\sqrt{45}}{2}$!

[quote="e.lopes"]Nice Problem!

We easily see that no more than one of the numbers can be 0.

suppose a, b, c > 0.

By Cauchy, we have that $\sum{\sqrt{\frac{a}{b+c}}}\geq \frac{(a+b+c)^{2}}{\sum{a\sqrt{a(b+c)}}}$.

By AM-GM inequality, we have that:

$\frac{(a+b+c)^{2}}{\sum{a\sqrt{a(b+c)}}}+\frac{9\sqrt{ab+bc+ca}}{a+b+c}$

$\geq 2\sqrt{\frac{9(a+b+c)\sqrt{ab+bc+ca}}{\sum{a\sqrt{a(b+c)}}}}$

now, we have:

$2\sqrt{\frac{9(a+b+c)\sqrt{ab+bc+ca}}{\sum{a\sqrt{a(b+c)}}}}\geq 6$

$\Longleftrightarrow (a+b+c)\sqrt{ab+bc+ca}\geq \sum{a\sqrt{a(b+c)}}$

but see that $a\sqrt{ab+bc+ca}> a\sqrt{ab+bc}$, because a, b and c are non-negative!

(and the same for the other parts)

Now, suppose that c = 0.

We have to prove that $\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}}+\frac{9\sqrt{ab}}{a+b}\geq 6$

but this is the same that $(a+b)^{2}+(3\sqrt{ab})^{2}\geq 6(a+b)\sqrt{ab}$, true by AM-GM![/quote]

Yes, I understand. I can not find any miskate here. I just can not explain for myself about your first step

[quote]By Cauchy, we have that
\[\sum{\sqrt{\frac{a}{b+c}}}\geq \frac{(a+b+c)^{2}}{\sum{a\sqrt{a(b+c)}}}\]
[/quote]

Because in this applying of Cauchy, the equality only holds for
$a=b=c$ or $a=b,c=0$ (not the case of equality in the end). So I don't know why this solution can go right finally.

Above all, your proof is very nice.

[quote="e.lopes"]but my friend, cauchy is to prove that LHS > 6, when a, b, c >0.

when one of them is zero, we have the equality case!  :)

thanks :D[/quote]

My solution bases on the following simple statement:

If $a\ge b\ge c$ then

\[\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}\ge \sqrt{\frac{b+c}{a}}.\]

By this one, we may assume that $c=0$.


Many useful and important algorithms use a source of random numbers
in order to achieve robustness and speedy execution. These algorithms
are often phrased as having a coin that they can ‘flip’ when they need to
make choices. (A random number can be obtained by making a sequence of
coin flips and converting the list of resulting bits into a number.) However,
Turing machines do not have built in a notion of coin flip. Suggest a way
to modify the definition of Turing machines to allow access to a source
of coin flips. (Note that I am not interested in just encoding a pseudo-
random number generator, since that is just a sequential algorithm and
I already know how to implement those.)

Thanks.

Any suggestions on this problem? It has me stumped.

If you only want to modify a definition, there's surely a lot of ways to do it!  For example, you could redefine the decision function as follows: turing machine gets from configuration "a" with 50% chance to configuration d1(a) and with 50% chance to configuration d2(a). If you don't want to use the random generated number, you set d1(x)=d2(x). If you want to make a decision based on a random number, you put decision 1 in d1(x) and decision 2 in d2(x).


Find the least value of \[ \frac{{x(1 \plus{} y) \plus{} y(1 \plus{} z) \plus{} z(1 \plus{} x)}}
{{\sqrt {xyz} }}
\]
for $ x,y,z>0$

[hide]$ E\equal{}{x\plus{}xy\plus{}y\plus{}yz\plus{}z\plus{}zx\over\sqrt{xyz}}\stackrel{\mathrm{AM\minus{}GM}}{\geqslant}{6\sqrt[6]{x^3y^3z^3}\over\sqrt{xyz}}\equal{}6$

Hence $ E_{\min}\equal{}6$ for $ x\equal{}y\equal{}z$[/hide]


I am looking for sources(books, articles, links) which help me develop methods to help students in the topics that they mistake most.

Like...If we take up chain rule in differentiation, using pigeon hole principle and many other topics where students find it difficult to use these formulae.  So, by giving them exercises which involve the application of such tough concepts, and gradually take them from elementary applications to complex applications, I think one can make the students comfortable and confident.

Any information in this direction will be of great help.

Pls do post here.

[quote="krustyteklown"]Isn't chain rule in calculus?[/quote]

yes


function
real analysis
real analysis unsolved

Let $\{a_{n}\}_{n=1}^{\infty}$ be a convergent sequence, and $f$ a bijection from $\mathbb{N}$ to $\mathbb{N}$. Is the sequence $\{b_{n}\}_{n=1}^{\infty}=\{a_{f(n)}\}_{n=1}^{\infty}$ convergent ?
(b_ has the same elements as a_n but in a different order)
[hide]Yes the new sequence is convergent.
Provide a proof that convergence holds.[/hide]

A sequence converges to $L$ iff for any $\epsilon>0$ all but finitely many terms are within $\epsilon$ of $L$. Rearranging the terms does not change that.

$ a_n$ converges to $ L$ , so $ \forall_{\epsilon > 0},\exists_A,\forall_{n>A}, |a_n\minus{}L|< \epsilon$. $ f$ is injective , so there is $ B$ such that $ \forall_{n>B},f(n)>A$
Finally $ \forall_{\epsilon > 0},\exists_B,\forall_{n>B},f(n)>A\Longrightarrow  |a_{f(n)}\minus{}L|< \epsilon$, so $ a_{f(n)}$ converges to $ L$
 :cool:

[quote="color"]Let $ \{a_{n}\}_{n = 1}^{\infty}$ be a convergent sequence, and $ f$ a bijection from $ \mathbb{N}$ to $ \mathbb{N}$. Is the sequence $ \{b_{n}\}_{n = 1}^{\infty} = \{a_{f(n)}\}_{n = 1}^{\infty}$ convergent ?
(b_ has the same elements as a_n but in a different order)
[hide]Yes the new sequence is convergent.
Provide a proof that convergence holds.[/hide][/quote]
$ a_n\to L\in\mathbb{R}$
we suppose that $ (a_{f(n)})$ is not convergente
so $ a_{f(n)}$ don't converge to $ L$
then $ \exists e > 0,\exists \phi \mathbb{N}\to \mathbb{N}(\phi(n + 1) > \phi(n)),\forall n\in\mathbb{N}: \ |a_{f(\phi(n))} - L| > e$
and we have $ \exists N\in\mathbb{N},\forall n\ge N: \ |a_n - L| < e$
then $ \forall n\in\mathbb{N}: \ f(\phi(n)) < N$ so $ \exists i,j\in\mathbb{N}: \ f(\phi(i))=f(\phi(j))$ (impossible because $ fo\phi$ injective)
so $ (a_{f(n)})$ converge.

we just need $ f$ injective

[quote="Diogene"]Finally $ \forall_{\epsilon > 0},\exists_B,\forall_{n > B},f(n) > A\Longrightarrow |a_{f(n)} \minus{} L| < \epsilon$, so $ a_{f(n)}$ converges to $ L$
[/quote]

It doesn't necessarily converge to L. For example, take rearranging the elements of an alternating harmonic series in such a manner that you represent $ a_n$ as a sum of 3 functions of $ n$ - two will genereate even elements and one the odd ones, or the other way around. You get 2 different results. Like this:

$ \sum^{\infty}_{n \equal{} 1}\left(\frac {1}{4n \minus{} 3} \plus{} \frac {1}{4n \minus{} 1} \minus{} \frac {1}{2n}\right) \equal{} \frac {3}{2}\ln(2)$

$ \sum^{\infty}_{n \equal{} 1}\left(\frac {1}{2n \minus{} 1} \plus{} \frac {1}{4n \minus{} 2} \minus{} \frac {1}{4n}\right) \equal{} \frac {1}{2}\ln(2)$

but all the elements are there.

[b]milin[/b], we are discussing the convergence of a [u]sequence[/u].

The affirmative answer is evident from the cauchy criterion

By the way, the statement is true in an arbitrary topological space (the nature of sequence $ a_n$ was never specified). Perhaps it should be proved in such generality.

Let $ X$ be a topological space. By definition, a sequence $ (a_n)$ in $ X$ converges to a point $ p$ iff for any neighborhood $ U$ of $ p$ the set $ E(U,(a_n)): \equal{}\{n\colon a_n\notin U\}$ is finite. Let $ f\colon \mathbb N\to\mathbb N$ is a bijection. Since $ E(U,(a_{f(n)}))\equal{}\{n\colon a_{f(n)}\notin U\}\equal{}f^{\minus{}1}E(U,(a_n))$, it follows that $ E(U,(a_{f(n)}))$ has the same cardinality as $ E(U,(a_{n}))$.

[quote="milin"]It doesn't necessarily converge to $ L$.[/quote]
No it converge to $ L$.
if we suppose that $ a_n\to L$ and $ a_{f(n)}\to M\neq L$
then $ \exists N\in\mathbb{N},\forall n\ge N: \ |a_n-L|<\frac{|L-M|}{2},|a_{f(n)}-M|<\frac{|L-M|}{2}$
then $ f([N,+\infty[\cap \mathbb{N})\subset [|0,N|]$ (impossible because $ f$ is injectiv)


geometry
geometric transformation
combinatorics proposed
combinatorics

Suppose that we have $ n \ge 3$ distinct colours. Let $ f(n)$ be the greatest integer with the property that every side and every diagonal of a convex polygon with $ f(n)$ vertices can be coloured with one of $ n$ colours in the following way:

(i) At least two colours are used,

(ii) any three vertices of the polygon determine either three segments of the same colour or of three different colours.

Show that $ f(n) \le (n\minus{}1)^2$ with equality for infintely many values of $ n$.

Let $ K_n$ be a graph with $ n$ vertices and $ d_i \equal{} n \minus{} 1$ for each vertice($ d_i$ denotes the degree).
we want to color the edges of this graph using $ r \ge 2$ colors with this property:
$ i)$ At least two colours are used.
$ ii)$ any three vertices of the polygon determine either three segments of the same colour or of three different colours.
we'll prove $ r \ge \sqrt {n} \plus{} 1$.
let's suppose $ r < \sqrt {n} \plus{} 1$.then consider a vertice $ a$.
$ a$ has $ n \minus{} 1$ neighbors and there exists a color that is used in a number bigger than $ \sqrt {n} \minus{} 1$ in the edges connecting this vertice to other vertices.suppose that this color is $ c_1$ and the set of vertices connected to $ a$ with an edge of this color be $ S$.
obviously all edges between the vertices of $ S$ is $ c_1$.
it's known that there exists at least one vertice ,let's call it $ b$, which is connected to $ a$ with a different color than $ c_1$, say $ c_2$.but one can easily see that all edges between the vertice $ b$ and the set $ S$ are mutually colored differently and also different from $ c_1$ and $ c_2$.so it directly implies that $ r > 2 \plus{} \sqrt {n} \minus{} 1 \equal{} \sqrt {n} \plus{} 1$.
so we have a clear contradiction.
for the next part we choose $ n$ s.t. it is a perfect square.
then $ \sqrt {n} \plus{} 1$ color $ \equal{} \{c_1,...,c_{\sqrt {n} \plus{} 1}\}$ are chosen and used in this way:
take a vertice $ a$ and connect it with other vertices by using each color exactly $ \sqrt {n} \minus{} 1$ times.
the color of other edges are defined uniquely.

Could you clarify the last bit? I don't think it is correct.

i wasn't thinking right when i said that the other edges will be defined uniquely!sorry :blush: !
i saw this problem a long time ago in a book.but the question didn't have the second part.
at first, it's obvious that if the equality occurs then $ n$ is a perfect square.
since for each color we can't choose a vertice and color more than $ \sqrt{n}\minus{}1$ of the edges connected to it by one color and also because of having $ \sqrt{n}\plus{}1$ colors and $ (\sqrt{n}\minus{}1).(\sqrt{n}\plus{}1)\equal{}n\minus{}1$ we must use $ \sqrt{n}\minus{}1$ times each color for each vertice.
so the reason for which i said that we must color the edges in that way(in my previous post) was this.
but i can't prove that what i said is correct(i think it must be so!),though i 've found how to color edges for $ n\equal{}4,9$.
anyway,any proof for this part should go by this way.
sorry again :blush: !

I proposed this problem, but I do not know all cases in which equality occurs (if anybody does, please let me know).

But I'm sure that equality will occur if $ n \minus{} 1$ is... prime. Or a prime power. Try to prove that.

I'll quote gjw from the forum of the Austrian Mathematical Olympiad.

[quote]Das ist doch ein altes, wohlbekanntes Beispiel...
Es wurde bereits in den 1980er Jahren in Bulgarien bei einem Wettbewerb verwendet.

Vollstaendig analysiert wurde es dann 1994 von Coburn Ward und Sándor Szabó, unter dem Namen "swell-colorings of complete graphs":
On swell-colored complete graphs. Acta Mathematica Universitatis Comenianae 63 (1994), 303--308.

Ward und Szabó zeigen unter anderem, dass Gleicheit genau dann gilt, wenn es eine endliche affine Ebene der Ordnung n-1 gibt.
Fuer die MEMO-Version reichen natuerlich die Primzahlfaelle aus.[/quote]
Translation:
That's an old, well-known problem...
It has been used already in 1980 in Bulgaria for a competition.

In 1994, the problem has been fully analysed of Coburn Ward and Sándor Szabó, with the name "swell-colorings of complete graphs":
On swell colored complete graphs. Acta Mathematica Universitatis Comenianae 63 (1994), 303--308.

Ward and Szabó showed among others that equality is attained if and only if there exists a finite affine plane of order n-1.
For the MEMO, the cases where n-1 is prime, are of course sufficient.

Hi, I have tried to solve this problem (equality part), and I couldn't prove that for $ n\minus{}1$ prime, equality holds. Can anyone help me?

step (1) is obvious so i will prove the second step.
----------------------------------------------------------------
let$p$ be any prime .suppose $X=\left \{ (x,y)\mid x,y\in \mathbb{N},1\leq x,y\leq p \right \}$ in the plane.
If the line which pass from the points $A=(x_{1},y_{1}),B=(x_{2},y_{2})$ ( these points are in $X$ )is not parallel by $O_{x}$ we will color the edge between $A,B$ by $\cot _{p}\widehat{BAOx}$ (The line $AO_{x}$ is parallel to $x$exists).(and we defined $\cot _{p}\widehat{CDE}=\cot \widehat{CDE}(mod p)$.
--------------------------------------------------------------------------------------------------------------------
if the line $AB$ is parallel to  $O_{x}$ we put  color of the edge between $A,B$ by $p+1$.


One-fifth of the cars in a parking lot are blue and 1/2 of the blue cars are convertibles. If 1/4 of the convertibles in the parking lot are blue, then what percent of the cars in the lot are neither blue not convertibles?

[quote="AstroPhys"][hide]
$ \frac{4}{5}\cdot 100 \% = 80 \%$
[/hide][/quote]

I get a different answer than you...
[hide]Let there be $ b$ blue cars and $ c$ convertibles, and $ n$ total cars. Since 1/5 of the cars are blue, $ n = 5b$.

Since 1/2 of the blue cars are convertibles, there are $ b/2$ blue convertibles.
Since 1/4 of the convertibles are blue, there are $ c/4$ blue convertibles.

Since these two quantities are the same,
$ b/2 = c/4$
$ c = 2b$

So, we have:
$ 5b$ total cars
$ 2b$ convertibles
$ b$ blue cars
$ b/2$ blue convertibles

Now we may apply PIE to see that:
$ 2b+b-b/2 = 5b/2$
cars are either blue or they are convertibles.

Hence, $ 5b-5b/2 = 5b/2$ cars must be neither blue nor convertibles. Hence,
$ \frac{5b/2}{5b}= \boxed{50\%}$

of the cars are neither blue nor convertibles.[/hide]

Let the number of total cars be $ 1$.

[hide="Subtract blue cars."]$ 1-\frac{1}{5}= \frac{4}{5}$[/hide]
[hide="Subtract convertibles."]Note: DO NOT RE-SUBTRACT THE BLUE CONVERTIBLES
$  3\left(\frac{1}{5}\right)\left(\frac{1}{2}\right) = \frac{3}{10}=$The number of cars that are convertibles, excluding the stupid blue ones.

$ \frac{4}{5}-\frac{3}{10}= \frac{5}{10}= 50$percent.
[/hide]

Holla @ ya vish, but seriously, you made it way to hard.


Does anyone know where I could find solutions to some of the very beginning USACO problems (in pseudo code or something)? I do try to bash my novice head on them for a long time, but after a while, I really wanna see someone else's solution  :P

Well... you'll get to see a problem analysis and a sample solution after you solve the problem... but that's not going to help you too much, is it? :) Since I already have access to the analysis of the first couple of these problems, I could tell you what it says - if that's what you want. There's also a C source code although you'd probably prefer a Java solution... if you have some specific problem in mind, just ask someone if he/she could show you some sample Java solution or give you some pointers. (I could do this for you if you're really helpless.)

Hmm... could you help me with the Greedy Gift Givers problem?

I changed the code from what I posted earlier (I misread what form the input came in), to this

[code]
import java.io.*;
import java.util.*;

public class gift1 {
    public static void main(String[] args) throws IOException {
        // Use BufferedReader rather than RandomAccessFile; it's much faster
        BufferedReader f = new BufferedReader(new FileReader("gift1.in"));    // input file name goes here
        PrintWriter out = new PrintWriter(new BufferedWriter(new FileWriter("gift1.out")));

        String s1 = f.readLine();
        int n = Integer.parseInt(s1);

        ArrayList peopleNames = new ArrayList(n);
        int amountLoss[] = new int[n];
        int amountGain[] = new int[n];
        int amountNet[] = new int[n];

        for(int i=0; i<n; i++)
            {
                String s2 = f.readLine();
                peopleNames.add(s2);
            }

        for (int j=0; j<n; j++)
            {
                String s3 = f.readLine();
                String s4 = f.readLine();
                StringTokenizer st1 = new StringTokenizer(s4);
                int amountGiven = Integer.parseInt(st1.nextToken());
                int numOfRec = Integer.parseInt(st1.nextToken());

             if (numOfRec == 0) // division by 0
                    {
                        amountLoss[j] = 0;
                    }
                else
                    {
                        amountLoss[j] = (amountGiven/numOfRec) * numOfRec;
                    }

                for (int k=0; k<numOfRec; k++)
                    {
                        String s5 = f.readLine();
                        amountGain[peopleNames.indexOf(s5)] += amountGiven/numOfRec;
                    }
            }

        for(int l=0; l<n; l++)
            {
                amountNet[l] = amountGain[l] - amountLoss[l];
            }

 for(int m=0; m<n; m++)
            {
                Object Name = peopleNames.get(m);
                out.write((String)Name + " " + amountNet[m] + "\n");
            }

        out.close();
        System.exit(0);
    }
}
[/code]

I submitted it to the site, and got the first case wrong, here was the output

[code]
> Run 1:
      Execution error: Your program did not produce an answer that
      was judged as correct. The program stopped at 0.16 seconds.

      Here are the respective outputs:
      ----- our output ---------
      mitnik_2923
      Poulsen_557
      Tanner_128
      Stallman_-311
      Ritchie_-1777
      Baran_245
      Spafford_-1997
      Farmer_440
      Venema_391
      Linus_-599
      ---- your output ---------
      mitnik_2923
      Poulsen_557
      Tanner_-636
      Stallman_-1011
      Ritchie_-311
      Baran_-1677
      Spafford_-77
      Farmer_440
      Venema_391
      Linus_-599
      --------------------------

      ------ Data for Run 1 ------
      10
      mitnik
      Poulsen
      Tanner
      Stallman
      Ritchie
      Baran
      Spafford
      Farmer
      Venema
      Linus
      mitnik
      300 3
      Poulsen
      Tanner
      Baran
      Poulsen
      1000 1
      Tanner
      Spafford
      2000 9
      mitnik
      Poulsen
      Tanner
      Stallman
      Ritchie
      Baran
      Farmer
      Venema
      Linus
      Tanner
      1234 1
      Poulsen
      Stallman
      536 3
      Farmer
      Venema
      Linus
      Ritchie
      2000 1
      mitnik
      ... [and more] ...
      ----------------------------
[/code]

If you're having trouble with the problem, try adding debugging output in order to see where it's not working. For example, if in your loop you print out how much money each person gains/loses each time, you should probably be able to figure out what the problem is from the output. (A good idea would be to try to check the results by hand to see where the code fails.)

The problem is still in the input format :). First line is the number $n$ and then there are $n$ lines with people's names, that works all right. Then there are $n$ groups with information about the gift givers - first line contains the name of the gift giver, second line the amount of money and number of people to whom we are going to give gifts, and then there are the names of gift recipients.

You assume that the $j$th group always talks about gift giver number $j$:
[code]
amountLoss[j] = (amountGiven/numOfRec) * numOfRec;
[/code]
This is not true, the whole point of the first line in each group is to tell you which person is giving gifts now. What you need to do is:
[code]
amountLoss[peopleNames.indexOf(s3)] = (amountGiven/numOfRec) * numOfRec;
[/code]
(Remember to replace both occurences of amountLoss[j].) I hope it's going to work now :wink:.

AHHH!

Heh, I just noticed that as I tried nebula42's suggestion of printing out the intermediate results. I have to be more thorough with these things...

Thanks for all your help  :)


inequalities
geometry
inradius
inequalities proposed
acute

Prove that in any $ \triangle ABC$ :          ${ \boxed{p\sqrt{1-\frac{2r}{h_a}}\le m_a}}$ .

$ m_a$ - median on side $ BC$  $ ,h_a$ - altitude on side $ BC$ $, r$- inradius , $ p$ - semiperimeter

[quote="Sergic Primazon"]Prove that in any $ \triangle ABC$ : $ \boxed{\frac {h_a}{m_a}\ge \frac {2r}{R}}$ .
$ m_a$ - median on side $ BC$ and $ h_a$ - altitude on side $ BC$
$ r$- inradius , $ R$ - circumradius[/quote]

$ \frac {h_a}{m_a}\ge \frac {2r}{R}$

$ \Longleftrightarrow \frac {h_a^2}{m_a^2}\ge \frac {4r^2}{R^2}.............(*)$

substitution $ a \equal{} x \plus{} y,b \equal{} y \plus{} z,c \equal{} z \plus{} x$

$ (*)\Longleftrightarrow (z^2 \minus{} 2yz \plus{} y^2)x^3 \plus{} (3z^3 \minus{} 9yz^2 \plus{} 13y^2z \plus{} y^3)x^2 \plus{} (3z^4 \minus{} 8z^3y$ $ \minus{} 2y^3z \minus{} 9y^2z^2)x \plus{} 3yz^4 \plus{} z^5 \plus{} 3y^2z^3 \plus{} y^3z^2\geq 0$

Note

$ (z^2 \minus{} 2yz \plus{} y^2)x^3\ge 0$

we only prove

$ (3z^3 \minus{} 9yz^2 \plus{} 13y^2z \plus{} y^3)x^2 \plus{} (3z^4 \minus{} 8z^3y \minus{} 2y^3z \minus{} 9y^2z^2)x$ $ \plus{} 3yz^4 \plus{} z^5 \plus{} 3y^2z^3 \plus{} y^3z^2\geq 0.............(**)$

$ 3z^3 \minus{} 9yz^2 \plus{} 13y^2z \plus{} y^3>0$

$ \bigtriangleup \equal{} \minus{} z^3(3z^3 \plus{} 54yz^2 \plus{} 75y^2z \plus{} 28y^3)(y \minus{} z)^2\le 0$

so $ (**)$  holds.

 :lol:

[quote="Sergic Primazon"]Prove that in any $ \triangle ABC$ : $ \boxed{\frac {h_a}{m_a}\ge \frac {2r}{R}}$ .
$ m_a$ - median on side $ BC$ and $ h_a$ - altitude on side $ BC$
$ r$- inradius , $ R$ - circumradius[/quote]

See

http://www.mathlinks.ro/viewtopic.php?t=317101


Pascal\u0027s Triangle
algebra
binomial theorem

[hide]

if we set x=1, then the expansion gives us the sum of the coefficients, so we have $2^7=128$
$\boxed{E}$
that is a pretty cool problem in that there is such an elegant solution

[/hide]

[hide="Answer"]We take the eighth row of Pascal's Triangle, for the terms $\{ 1,-7,21,-35,35,-21,7,-1\}$. We multiply each of these by the corresponding power of $3$ to get $3^7-7\cdot 3^6+21\cdot 3^5-35\cdot 3^4+35\cdot 3^3-21\cdot 3^2+7\cdot 3-1=128\Rightarrow \boxed{E}$.[/hide]

[b]EDIT:[/b] Heheh, that way is so much easier... :P

[hide]Just use Binominal theorem with the first term is $3x$ and the second term is $-1$.

$\boxed{E}$[/hide]

[quote="Altheman"]but do you know why?[/quote]

uhh.... no  :blush:  I'm guessing you just let x=1. Hey, that's not a bad idea  :P

[quote="236factorial"]uhh.... no  :blush:  I'm guessing you just let x=1. Hey, that's not a bad idea  :P[/quote]

That's because 1 is neutral element in multiplication. Just like zero is in sum. So when we let x=1, we'll be isolating the coefficients and, hence, getting the sum of them.

[quote="236factorial"]I think that the sum of the coefficients of an equation in the form $(ax+b)^n$ is $(a+b)^n$, right?[/quote]

Since the coefficient is a constant multiplied by some power of x, just use the binomial theorem and you'll see that the coefficients of $(ax+b)^n$match perfectly with the terms of $(a+b)^n$.

If we have $(ax+b)^n$, then it is obvious that if $x=1$, all terms, excluding their coefficients, will always be $1^n$. Since $n$ is positive, $1^n$ will always be $1$, resulting in the multiplacation of $1$ by each coefficient.  This will give us the sum of all the coefficient.

Masoud Zargar


calculus
integration
function
algebra
polynomial
partial fractions
calculus computations

Evaluate:  integral upper:1 lower: 0 { [x^4 (1 - x)^4]dx / (1+x^2)

then after it's answered I have to explain why 22/7 DOES NOT EQUAL pi, by using the integral above.

If anyone could help or share. I'd really appreciate it, thanks.

Let's start with a side issue:

Since $0<\frac1{1+x^2}<1,$ we have that

$0<\int_0^1\frac{x^4(1-x)^4}{1+x^2}\,dx<\int_0^1x^4(1-x)^4\,dx.$

For nonnegative integers $n$ and $m$, define

$G(n,m)=\int_0^1x^n(1-x)^m\,dx.$

Can you find the general formula for $G(n,m)$ ? (Hint: integrate by parts to set up a useful recursion. And note that it's pretty simple when $n=0$ or $m=0.$)

Of course we're interested in $G(4,4).$

Once we get that out of the way, we'll tackle the main problem.

I have no idea what you are asking of me. I'm not super math-savvy, I just take it for Computer Science cirriculum, but if you could work this problem out and detail it for me it'd be a big help. This was just such a hard problem I wanted to know the answer. Thanks a lot though. I appreciate it.

[quote]This was just such a hard problem I wanted to know the answer.[/quote]
We're not running an answer service here. If you come here, expect to do some work and make a contribution yourself.

This is an interesting problem - too interesting to just kill off with "this is the answer." Well, I can reveal that $\pi\ne\frac{22}7,$ not exactly.

it is obvious that the integral value you are asking for is above zero (you are integrating a continuous function over an interval that is positive, and not everywhere zero)

now how to evaluate the integral?
use euclidean divison  : write x^4*(1-x)^4 as (x^2+1)*q(x) +(a*x+b)

it will turn out that a is zero and b=-4

now the integral should become easy if you know how to integrate polynomials, and that the antiderivative of 1/(x*x+1) is arctan( x)

now you should get 22/7-Pi/4

I understand this isn't an answer service. I really was just wondering if someone could work out the problem and explain how they attained it.

If not that is okay, I keep coming up with the wrong answer when I try what you guys tried

Would partial fractions and integration by parts be the route to take?

Thanks

Not integration by parts, except for the auxiliary problem I posted back in post #2. For the original problem, integration by parts is not an attractive option.

And, technically speaking, not partial fractions either. Partial fractions (which is a technique of algebra, not a technique of integration) applies when you're trying to pull apart a fraction whose denominator has at least two different factors. We've only got one factor in our denominator here (assuming we want to stay in the real numbers, which we do in this instance): $x^2+1.$

But partial fractions are one part of the family of techniques for handling rational functions. The first item in the handling of rational functions is this: if it's not a proper fraction, divide it out. (A proper fraction being one whose numerator has a lower degree than its denominator.)

$\frac{P(x)}{x^2+1}=Q(x)+\frac{ax+b}{x^2+1}$ where $Q$ is a polynomial.

In our particular case, we get $a=0,$ but that was the result of the calculation, not something you could easily see beforehand.

I don't see a pretty way to do this; you're going to have to get your hands dirty and do some grunt work. In particular, you might as well multiply out the numerator first:

$\frac{x^4(1-x)^4}{x^2+1}=\frac{x^8-4x^7+6x^6-4x^5+x^4}{x^2+1}= x^6+\cdots$ (more work to be done).


[quote="Beat"]Solve in integers : $x^{4}+4y^{4}=2(z^{4}+4u^{4})$[/quote]

RHS even $\Rightarrow $ $x$ is even $\Rightarrow $ $4$ divides LHS $\Rightarrow $ $z$ is even
$x=2x_{1}$ and $z=2z_{1}$ $\Rightarrow $ $4x_{1}^{4}+y^{4}=2(4z_{1}^{4}+u^{4})$

So, if $(x,y,z,u)$ is a solution, $x$ and $z$ are even and $(y,\frac{x}{2},u,\frac{z}{2})$ is solution. Then $y$ and $u$ must be even and $(\frac{x}{2},\frac{y}{2},\frac{z}{2},\frac{u}{2})$ is solution.

And so, the only solution is $(0,0,0,0)$

--
Patrick


search
geometry
calculus
inequalities
number theory
complex numbers
graph theory

Hi All,

Can anyone give/find out/sell :) the solution manual for
"The Art and Craft of Problem Solving" (Second Edition) by Paul Zeitz?

Looking forward to hearing...
DimaACM

It is usually not given out to students.

[quote]This website has a lovely feature called a "search." With it, you can find posts that have appeared in the past addressing subjects which interest you. Searching for "Zeitz manual" works very well.[/quote]

Yup, I did search it... however did not find the actual manual.
Well, I could ask Richard about it... but doubtful he wanna sell it :)

Phelpedo, thanks for the solutions ;)

what topics are covered in the ACOPs by Zeitz? Is it olympiad level?

It's most definitely olympiad level, and has a chapter on each of algebra, combinatorics, number theory, geometry, and calculus, as well as a few chapters on general problem solving strategies, and a chapter that covers complex numbers, graph theory and generating functions.  It's definitely worth it, if that's what you're asking about.

I agree that it's a good book, but in my opinion it's a little on the easy side if you're preparing for something like the IMO. There are a few IMO #1 sand #4s, but mostly easier olympiad problems. The geometry and inequality coverage is especially lacking (well, there isn't even a geometry section).

[quote="Phelpedo"] The geometry and inequality coverage is especially lacking (well, there isn't even a geometry section).[/quote]
I got the new version (2007 one) it got a Geometry section now. Called [b]Geometry for Americans[/b], no offense.

By the way how long it took you to finish the book? I spend hours on it everyday almost, do a lot of excerices. And still didnt finish it :(. It been like 2 months and Im only half way through.

(I wish one can go through math books succesfully as quick as novels).

[quote="sylow_theory"][quote="Phelpedo"] The geometry and inequality coverage is especially lacking (well, there isn't even a geometry section).[/quote]
I got the new version (2007 one) it got a Geometry section now. Called [b]Geometry for Americans[/b], no offense.
[/quote]
 :rotfl: Are you serious?

I read it during Christmas vacation (3 weeks), minus chapter 4 (crossover techniques) and 8 (calculus) and half of chapter 7 (number theory). And yes, "Geometry for Americans" seems a silly name  :roll:

Could Phelpedo please post the TEX file for that so I could edit it with solutions as I complete htis book?

Actually, I got the official instructor's manual for ACoPS and am working through it with both your manual and the official one, as the official one does not have solutions for all of the problems.

Hi, #H34N1, could you please tell us where did you buy the offical instructor's manual?
I am interested in the one for the second edition of ACoPS. Thanks.

I cannot give the solution manual to anyone anymore. Sorry.

Hi, I am not asking you to give me the solution manual. From the discussion,
it sounds to me like the solution manual is published officially. Correct me if
I am wrong, but, I could not find it anywhere. I just need you to point me to
the web site where you purchased the solution manual. Thanks.

You can't buy the official solutions manual.
That is because if you go to ACoPS's website, at wiley.com, you'll find out that you need to be an instructor to be able to obtain the solutions

Of course, you can ask your math teacher to create an account at wiley.com and get a solutions manual for you

OK, I understand now. Unfortunately, I am doing self study in home schooling
environment. I don't have a math teacher per se and I am not planning to get
one any time soon. Thank god AoPS books do not have such ridiculously stupid
restrictions. Is there anything we as a community can do to make the ACoPS
book more home-school friendly?

I am doing self study as well, as my school isn't that crazy about math contests. Ask one of your parents to sign up as a home school teacher.

Art and Craft of Problem Solving - Paul Zeitz. Instructor's Manual can be downloaded from:
[url]http://rapidshare.com/files/261027147/ArtCraftOfProbSolv_InstManual.PDF[/url]

[quote=Phelpedo]I made a compilation of my solutions when I went through it last year for no real reason...it's by no means complete but I'll attach it in case you want it[/quote]

Thank you for the solutions


Find all real numbers a for which the sequence $ x_n=n\cdot f(a\cdot n!)$ converges. Not difficult, but I like it very much. Additional question: find a if the sequence $x_n$ is bounded. Here $f(x)$ is the fractional part of $x$.

Really, not difficult ;)
The key idea is to consider $n!$-base expansion of $a$. But I don't know what you mean for additional question. How can we find any closed form for such $a$?

Indeed, not difficult. Too bad nobody managed to find a solution in the contest. For the second question, can I see your answer?

But that is the question. We need that $n!$-base digits are bounded. But what numbers can be represented in such way?


geometry
circumcircle
inequalities
geometry proposed

Show that for a triangle with radii of circumcircle and incircle equal to $ R$, $ r$ respectively, the inequality $ R \geq 2r$ holds.

Hello , Euler's inequality http://mathworld.wolfram.com/EulersInequality.html

$\mid OI\mid=\sqrt{R(R-2r)} \Rightarrow R\ge 2r$

By Euler's theorem,
$OI^2 = R^2 - 2Rr$
As $OI \ge 0$,
$2Rr \le R^2$
Hence $R \ge 2r$

[hide="a bad approach!"]we have $ R=abc/4\triangle $ and $ r= \triangle/s $ we are required to prove $ R \geq 2r $ or $ \frac{abc}{4} \geq 2  \frac{\triangle}{s} $ or, $ abc \geq 8(s-a)(s-b)(s-c) $ or,$ abc \geq (b+c-a)(c+a-b)(a+b-c) $ let $ b+c-a = x , c+a-b = y , a+b-c = z  $  also, $a = \frac{y+z}{2}  $ ,$ b = \frac{z+x}{2} $ and  $ c = \frac{x+y}{2}  $ now we can prove that $ \frac{y+z}{2}\frac{z+x}{2}\frac{x+y}{2} \geq xyz $ eaily by the am-gm inequality.

 :P  [/hide]

[quote=J.Y.Choi]It's not a bad approach I think, earrth. I like that.

 [quote="J.Y.Choi"]Thinking a circumcirlce of the midpoint triangle of the given one might yield the proof.[/quote]
 Could anyone give the proof via my suggestion above??[/quote]

Sorry for reviving this thread but this question is still unanswered...
Use the formula $R$$=$$\frac{abc}{4\Delta}$.
For the medial ttriangle the sides are $\frac{a}{2}$,$\frac{b}{2}$,$\frac{c}{2}$ and the area is$ \frac{1}{4\Delta}$ so $R$ becomes $\frac{abc}{2\Delta}=$$\frac{R}{2}$..
Clearly this is more than the $inradius=r$..
Equality holds when triangle is equilateeral
QED


Find the remainder when $ 41^{65}$ is divided by $ 7$.

hello, we have $ 41 \equiv \minus{}1 \mod 7$ therefore we get
$ 41^{65}\equiv \minus{}1 \mod 7$ since $ 65$ is odd.
Sonnhard.


calculus
integration
trigonometry
calculus computations

[quote="modguru"]Integrate with respect to z:

sin(x+y-z)^x + x^y/2 + 2y^z + xy(cos(2z^2))

this one is very hard if you can do it you are a master[/quote]

How many times will u post it ???   :ninja:

And yes it's hard if it don't exist...  :D

[quote="modguru"]Integrate with respect to z:

sin(x+y-z)^x + x^y/2 + 2y^z + xy(cos(2z^2))

this one is very hard if you can do it you are a master[/quote]

do you mean
$\int\left(\sin^{x}(x+y-z)+\frac{x^{y}}{2}+2y^{z}+xy\cos(2z^{2})\right)dz$?


The sequence $x_0,x_1,x_2,.....$ of real numbers is called with period $p$,with $p$ being a natural number, when for each $p\ge2$, $x_n=x_{n+p}$.
Prove that,for each $p\ge2$ there exists a sequence such that $p$ is its least period
and               $x_{n+1}=x_n-\frac{1}{x_n}$  $(n=0,1,....)$

Observe that if $x_1=\cot\alpha$ then $x_2=2\cot{2\alpha}$, therefore all we have to do is to prove that for all $p$ the equation $2^p\cot{2^p\alpha}=\cot\alpha$ has real solutions, which can be done with a little bit of calculus :)

[quote="spider_boy"]The sequence $x_0,x_1,x_2,.....$ of real numbers is called with period $p$,with $p$ being a natural number, when for each $n\ge2$, $x_n=x_{n+p}$.
Prove that,for each $p\ge2$ there exists a sequence such that $p$ is its least period
and               $x_{n+1}=x_n-\frac{1}{x_n}$  $(n=0,1,....)$[/quote]
sorry for mistake with $n\ge2$


I choose a number from 1 to 144 ,u should choose a subset from {1,..144}and then ask me if my number is in your subset or not ,if my answer is yes u get 2 $ if not u get 1$ ,what is the least money u can get?

I remember that because 144 is Fibonacci number we could easily solve the problem


Anyone have any suggestions on any chemistry books I could use to study for the chemistry olympiad?

For general chemistry these are the books that I narrowed down to.
 PLEASE HELP ME CHOSE A FIRST CHOICE AND POSSIBLY A SECOND.

Chemical principles fourth edition  by Zumdahl
General chemistry custom edition for UBC A and B  by  Petrucci
Chemistry the central science  by   Brown
Chemistry third edition  McMurry     (recommended by the university hosting chemistry olympiad training camp.)

For organic chemistry these are the books that I thought of using. Can you please help me decide on one? and also a second choice?

Organic Chemistry McMurry  (recommended by univeristy hosting chemistry olympiad training camp)
Organic chemistry  eighth edition by  Solomon
Organic chemistry bu  Wade

Please also include other recommendations. Thanks.

Books have been discussed countless times, although in terms of organic chemistry, I have Wade and find it to be very comprehensive.

I recommend the book by Silberberg! It taught me a lot, and I find it a more interesting read than the text by Zumdahl. Also, I would recommend doing every practice test on the main Chemistry.org site and also http://dbhs.wvusd.k12.ca.us/webdocs/NChO/NChO-Menu.html
For the national exam, I did most of the Multiple choice tests on those two sites and I found that a lot of the questions on the test are very repetitive or asked in a different way. I didn't make the camp this year, and I found that one of my biggest weaknesses on the test was my lack of organic chemistry and descriptive chemistry knowledge, as I felt those two were the hardest sections on the multiple choice. So anyways, good luck and have fun!  :)

I recommend for a good study the edition of McMurry is the one that use and much people me has recommended it and like second option perhaps for basics of the subjects the books of solomons and hart craine hart like support :stretcher:


function
calculus
derivative
calculus computations

The problem states:

State Rolle's Theorem, then apply to $ f(x)\equal{}(x\minus{}a)^2 (x\minus{}b)^4$ on the interval $ [a,b]$

So Rolle's Theorem is just a special case of the mean value theorem that states:
     -if a function is continuous on $ [a,b]$ AND differentiable on $ (a,b)$ AND $ f(a)\equal{}f(b)$....
     THEN there exists a point c on the interval $ (a,b)$ such that $ f'(c) \equal{} 0$.

So basically, if a function has the same x value at the endpoints, at some point in the interval, the function must have a horizontal tangent. Easy enough.

So I have my interval, and have my function and I verified that x = 0 @ f(a) and f(b).

so, find $ f'(x)$. to do this you use the product rule $ f'(x) \equal{} f(x)g'(x) \plus{} f'(x)g(x)$.

$ f(x) \equal{} (x\minus{}a)^2$
$ f'(x) \equal{} 2(x\minus{}a)$

$ g(x)\equal{} (x\minus{}b)^4$
$ g'(x) \equal{} 4(x\minus{}b)^3$

So $ f'(x)\equal{}2(x\minus{}a)(x\minus{}b)^4 \plus{} 4(x\minus{}b)^3(x\minus{}a)^2 \equal{} 0$

here is where I get stuck. If I factor out a 2 I can get $ 2[(x\minus{}a)(x\minus{}b)^4 \plus{} 2(x\minus{}b)^3(x\minus{}a)^2] \equal{} 0$, but I keep getting confused with how to solve for values of x that make the function become zero.

i've also tried substituting in numbers for a and b (such as a=1 and b=4) so that I stop getting confused, and when this happens, I find that x=0 at f(-a), but that is clearly not on my interval. Any help?

Greatly appreciated.

You need to factor further. The derivative factors as $ 2(x\minus{}a)(x\minus{}b)^3q(x)$, where $ q(x)$ is another linear polynomial. The points $ a$ and $ b$ aren't interesting, but the zero of $ q$ is.

[quote="nickluto"]...I verified that x = 0 @ f(a) and f(b).[/quote]
Your terminology is very confused here; this doesn't make sense. You mean that $ f(x)\equal{}0$ at $ x\equal{}a$ and $ x\equal{}b$, or equivalently $ f(a)\equal{}f(b)\equal{}0$. Keep the functions and variables straight.

Thanks for the help J!

Im still slipping up on the factoring somehow in one of the steps, as I cannot seem to find a way to get the same result as you did. If I subtract over the right side of the derivative, I can see:

$ 2(x\minus{}a)(x\minus{}b)^4 \equal{} \minus{}4(x\minus{}b)^3(x\minus{}a)^2$

divide by $ (x\minus{}a)$ to get:

$ 2(x\minus{}b)^4 \equal{} \minus{}4(x\minus{}b)^3(x\minus{}a)$

dividing by $ (x\minus{}b)^3$ on both sides gives $ 2(x\minus{}b) \equal{} \minus{}4(x\minus{}a)$

moving it back over gives $ 2(x\minus{}b)\plus{}4(x\minus{}a)$

and factoring out a 2 gives $ 2[(x \minus{} b) \plus{} 2(x \minus{} a)]$

it's been a long day and I feel im doing some elementarily stupid mistakes...

can you spell out how you factored it for me?

You have found the factor. Simplify it and find the root.

We have $ f(x)\equal{}(x\minus{}a)^2(x\minus{}b)^4,$ and we want its derivative.

But let's be smart about the algebra - we can look ahead to see that we're going to have common factors of $ (x\minus{}a)$ and $ (x\minus{}b)$ (which is exactly what jmerry was doing and saying - he was looking ahead, and hadn't yet performed the work.) So we resist any impulse to multiply things out unnecessarily. Start with the product rule:

$ f'(x)\equal{}(x\minus{}a)^2\cdot 4(x\minus{}b)^3\plus{}2(x\minus{}a)(x\minus{}b)^4.$

Now we look for common factors - there's a common factor of $ (x\minus{}a)(x\minus{}b)^3$ in both terms.

$ f'(x)\equal{}(x\minus{}a)(x\minus{}b)^3\cdot4(x\minus{}a)\plus{}(x\minus{}a)(x\minus{}b)^3\cdot 2(x\minus{}b)$

$ \equal{}(x\minus{}a)(x\minus{}b)^3\left[4(x\minus{}a)\plus{}2(x\minus{}b)\right]\equal{}
2(x\minus{}a)(x\minus{}b)^3(3x\minus{}2a\minus{}b).$

In fact, nickluto, you were almost there. All you had do do was to rewrite  $ (x\minus{}b)\plus{}2(x\minus{}a),$ which you had, as $ (3x\minus{}2a\minus{}b).$ [There are some odd negative signs in your intermediate steps that I don't understand, however. And I wouldn't have used the word "divide" when "factor" more clearly conveys what is being done.]

The point we're interested in is the root of that last factor: $ 3x\minus{}2a\minus{}b\equal{}0$ when $ x\equal{}\frac{2a\plus{}b}{3}.$ This is the point a third of the way across the interval $ [a,b],$ closer to $ a$ than to $ b.$

---

Generalization: for arbitrary positive $ m$ and $ n,$ the function $ g(x)\equal{}(x\minus{}a)^m(x\minus{}b)^n$ has a (unique) zero of its derivative in the open interval $ (a,b).$ Locate that point.

Awesome, thanks for your help guys.

First time here, so sorry that I didn't realize all I had to do was put a dollar sign in front of my equations to make it a lot more easily readable.

And In RE: Kent: I actually didn't factor it on accident. I set the derivative to zero, moved the right hand side over by subtracting it and divided through. For some reason, I guess because I had been looking at the same problem for about an hour, it wasnt immediately obvious that $ (x \minus{} a)(x \minus{} b)^3$ could be factored out right off the bat...I am 4 years out from taking any calculus (or any math for that matter) classes and I guess I'm way rustier than I thought :P What a silly mistake.

You know how sometimes if you are typing a paper you start to question if you spell apple (or some other trivially simple word to spell) correctly because you've been looking at it too long? used to happen to me all the time in programming class.


function
trigonometry
algebra unsolved
algebra

Find all continuous functions $ f: \mathbb{R} \to \mathbb{R}$ such that for all reals $ x$ and $ y$, the equation
\[ f\left(x \minus{} y\right)f\left(x \plus{} y\right) \equal{} \left(f\left(x\right)\right)^{2} \minus{} \left(f\left(y\right)\right)^{2}
\]
holds.

$f(x-y)f(x+y)=-f(y-x)f(x+y)$ $\to$ $f(x)=0$ or $f$ is an odd function.

$f(x-y)f(x+y)=(f(x)-f(y))(f(x)+f(y))$ $\to$ From here you can see that any solution to Cauchy's equation $f(x+y)=f(x)+f(y)$ is a solution to this problem. The only continuous ones are $f(x)=ax$.

But they are not the only solutions.
I have identified another family of continuous solutions:
$f(x)=a(e^{bx}-e^{-bx})$ (interestingly, $b=i$ leads to $f(x)=\sin x$).

lomos_lupin:
Is there a source?
Do you know the solution, even if you can't solve it?

No lordwings I dont know any solution for this problem.

Note that the functions $ f\left(x\right) \equal{} x$, $ f\left(x\right) \equal{} \sin x$ and $ f\left(x\right) \equal{} \sinh x$ all satisfy the equation $ f\left(x \minus{} y\right)f\left(x \plus{} y\right) \equal{} \left(f\left(x\right)\right)^{2} \minus{} \left(f\left(y\right)\right)^{2}$. But there are more solutions.

please Refer to
http://www.mathlinks.ro/Forum/viewtopic.php?t=116627

All the solutions are of the form (assuming you're looking for continuous functions) :
$ax, \ \ a \sin(bx), \ \ a \sinh(bx).$

Note that without the assumption that $ f$ be continuous, any additive real function would be a solution of your problem and, without continuity, many of them are monsters. So, the continuity assumption is not dispensable.

Pierre.


limit
algebra
binomial theorem
calculus
calculus computations

Need help with these limits:

a) $ \lim_{x \to \infty}\frac{(x\plus{}1)^2(3\minus{}7x)^2}{(2x\minus{}1)^4}$

b) $ \lim_{x \to \minus{}\infty}\frac{\sqrt{2x^4\plus{}1}\minus{}3x^3}{x^3\plus{}x^2\plus{}2x\minus{}1}$

Useful fact: $ \lim_{x\to\infty}\frac{a_nx^n\plus{}a_{n\minus{}1}x^{n\minus{}1}\plus{}\cdots\plus{}a_0}{b_nx^n\plus{}b_{n\minus{}1}x^{n\minus{}1}\plus{}\cdots\plus{}b_0}\equal{}\frac{a_n}{b_n}$. In this case, $ a_n$ or $ b_n$ may be $ 0$, if we are to accept $ \frac{a_n}{0}\equal{}\infty$.

Proof: Either multiply by $ \frac{\frac{1}{x^n}}{\frac{1}{x^n}}$ or apply L'Hopital's rule repeatedly.

Using this lemma and the binomial theorem, the first limit is $ \frac{\binom{2}{2}\cdot\binom{2}{2}\cdot (\minus{}7)^2}{\binom{4}{4}\cdot 2^4}\equal{}\boxed{\frac{49}{16}}$.

Using this lemma again, the second limit is simply the coefficients of $ x^3$ divided, or $ \frac{\minus{}3}{1}\equal{}\boxed{\minus{}3}$.


inequalities
trigonometry
inequalities proposed

[quote="No Reason"]Prove that for all triangle ABC we have:
$ l_al_bl_c\geq \frac {8}{9}\sqrt {3}abc$

When does the equation hold ?[/quote]
Try $ a\equal{}b\equal{}1$ and $ c\rightarrow2^\minus{}.$ :wink:

[quote="No Reason"][quote="arqady"][quote="No Reason"]Prove that for all triangle ABC we have:
$ l_al_bl_c\geq \frac {8}{9}\sqrt {3}abc$

When does the equation hold ?[/quote]
Try $ a \equal{} b \equal{} 1$ and $ c\rightarrow2^ \minus{} .$ :wink:[/quote]
I don't understand what you say.
I mean $ l_a,l_b,l_c$are three bisector of a triangle ABC and I think when a=b=1,c=2 there isn't any bisector ?[/quote]
I mean that if $ a\equal{}b\equal{}1$ and $ c\rightarrow2^\minus{}$ then $ l_c\rightarrow0.$
Thus, we obtain $ 0\geq\frac{16\sqrt3}{9},$ which wrong. :wink:

In this post I will find maximum value of product of lengths of bisectors in any triangle.

It is well known that $ l_{a} \equal{} \frac {2bc}{b \plus{} c}\cos\frac {A}{2}$ and analogous equalities also hold for lengths of other two bisectors.
Therefore:

$ l_{a}l_{b}l_{c}$=$ \frac {8(abc)^{2}}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}\cos\frac {A}{2}\cos\frac {B}{2}\cos\frac {C}{2}$ $ \leq abc (\frac {\cos\frac {A}{2} \plus{} \cos\frac {B}{2} \plus{} \cos\frac {C}{2}}{3})^{3}$ $ \leq abc (\frac {\sqrt {3}}{2})^{3}$ $ \equal{} \frac {9}{8\sqrt {3}}abc$ ,

so our proof is finished and equality is achieved for equilateral triangles.

(Notice that we used Jensen’s inequality for cosines of halves of angles $ A$, $ B$ and $ C$).


geometry
angle bisector
geometry proposed

Let $ABCD$ is convex cuadrilateral.$\omega$ is inscribed sircle of $ABCD$.
$E\in{AB}\cap{\omega}$             $O\in{AC}\cap{BD}$
$F\in{AD}\cap{\omega}$             $M\in{EF}$ and $OM\perp{EF}$

Prove that $\angle{BMO}=\angle{DMO}$

[quote="sprmnt21"][quote="Tiks"][quote="sprmnt21"]If P is the intersection between EF and BP, it results that {B,D; O,P} is an harmonic quadruple so M{B,D; O,P} is an harmonic pencil then , as MP _|_ MO, MO and MP are the angle bisector of lines DM and BM.[/quote]
What do you means????????.
1.[b]If P is the intersection between EF and BP?????????[/b]
2.it results that {B,D; O,P} is an [color=red][b]harmonic quadruple[/b][/color] so M{B,D; O,P} is an [b]harmonic pencil [/b]then
 :roll:  :huh:[/quote]

Sorry. I mistyped P for D.

My intention was this.

If P is the intersection between EF and BD, it results that {B,D; O,P} is an harmonic quadruple so M{B,D; O,P} is an harmonic pencil then , as MP _|_ MO, MO and MP are the angle bisector of lines DM and BM[/quote]
Can you prove that
[b]{B,D; O,P} is an harmonic quadruple[/b]?
I think that is not well know :) .

[quote="sprmnt21"]You are rigth. My intention was to give only a hint of a solution.

Here you can find some usefull results:

http://www.maths.gla.ac.uk/~wws/cabripages/quad.html

that togheter the Brianchon's theorem should prove the fact that {B,D; O,P} is an harmonic quadruple.

Let me know if you prefer that I write down all details of the proof.[/quote]
[b]O.K. YOU WRITE ALL DETAILS OF THE PROOF[/b] :rotfl:
 In my solution I used also Menelaus' therome.
From it we have that  $\frac{DF}{FA}\frac{AE}{EB}\frac{BP}{PD}=1$<=>
$\frac{BP}{PD}=\frac{BE}{DF}=\frac{BO}{DO}$
For leste one see this[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=70299]www.mathlinks.ro/Forum/viewtopic.php?t=70299[/url]

Tiks, w.r.t. the harmonical division mentioned by Sprmnt21 see
http://www.mathlinks.ro/Forum/viewtopic.php?t=70184
where appear some properties of a circumscribed quadrilateral.
(Translate $w.r.t.\equiv$with respect to).


calculus
integration
calculus computations

what would replace "b" for the 1st and "a" for the 2nd problem?
[img]http://img151.imageshack.us/img151/6441/calcbz0.jpg[/img]

The first step is to realize that $ u\equal{}3x^2\minus{}9x\plus{}10$.


[quote="Moss"]Let $M(n)=\{n,n+1,n+2,n+3,n+4,n+5\}$ be a set of 6 consecutive integers. Let's take all values of the form \[\frac{a}{b}+\frac{c}{d}+\frac{e}{f}\] with the set $\{a,b,c,d,e,f\}=M(n)$.
Let \[\frac{x}{u}+\frac{y}{v}+\frac{z}{w}=\frac{xvw+yuw+zuv}{uvw}\] be the greatest of all these values.

a) show: for all odd $n$ hold: $\gcd (xvw+yuw+zuv, uvw)=1$ iff $\gcd (x,u)=\gcd (y,v)=\gcd (z,w)=1$.
b) for which positive integers $n$ hold $\gcd (xvw+yuw+zuv, uvw)=1$?[/quote]
we can firstly show that $t=\max{\frac{a}{b}+\frac{c}{d}+\frac{e}{f}=\frac{n+5}{n}+\frac{n+4}{n+1}+\frac{n+3}{n+2}=\frac{3n^{3}+18n^{2}+28n+10}{n(n+1)(n+2)}}$
so $gcd(3n^{3}+18n^{2}+28n+5,n(n+1)(n+2))=1\leq >(n,10)=1,(n+1,3)=1,(n+2,2)=1$

a) we can show that  \[\frac{x}{u}+\frac{y}{v}+\frac{z}{w}=\frac{n+5}{n}+\frac{n+4}{n+1}+\frac{n+3}{n+2}=\frac{3n^{3}+18n^{2}+28n+10}{n(n+1)(n+2)}\]
$\Rightarrow \gcd(xvw+yuw+zuv,uvw)=\gcd(3n^3+18n^2+28n+10,n(n+1)(n+2))=\gcd(n+1,3)=1$
we have $5\nmid n$ because if $5\mid n$ then $5\mid \gcd(3n^3+18n^2+28n+10,n(n+1)(n+2))$ , contradiction
$\Rightarrow \gcd((n+5),n)=1$
And easy to see that $\gcd (n+2,n+3)=1$
b) from a) we have $\gcd (xvw+yuw+zuv, uvw)=\gcd(n+1,3)$
so $\gcd(n+1,3)=1\Rightarrow n=3k$ or $n=3k+1$


Prove that if $g \ge 2$ is an integer, then two series \[\sum_{n=0}^{\infty}\frac{1}{g^{n^{2}}}\;\; \text{and}\;\; \sum_{n=0}^{\infty}\frac{1}{g^{n!}}\] both converge to irrational numbers.

Sounds like a trivial consequence of Liouville's Approximation Theorem and the definition of Liouville's Numbers.
See http://mathworld.wolfram.com/LiouvillesApproximationTheorem.html and http://mathworld.wolfram.com/LiouvilleNumber.html

Just write the numbers in base $ g$. For the first sum, the base-$ g$ expansion contains a $ 1$ at each $ g^{\minus{}n^2}$ place value, and zeros elsewhere. These ones are increasingly far apart, so the digits do not repeat nor terminate; it must therefore be irrational.

The same method can be used for the second sum.

Let $f(n) : \mathbb{N} \rightarrow \mathbb{N}$ be any strictly increasing function with the property that $\limsup_n \left(f(n+1) - f(n)\right) = \infty$. Clearly this property holds for $f(n) = n^2$ and $f(n) = n!$. We claim that the sum $\displaystyle \sum_{i = 0}^{\infty} \frac{1}{g^{f(i)}}$ is irrational for any such function $f$. Assume by contradiction that there exist integers $p, q$ with $\displaystyle \frac{p}{q} = \displaystyle \sum_{i = 0}^{\infty} \frac{1}{g^{f(i)}}$ and let $p_n$ be such that $\displaystyle \frac{p_n}{g^{f(n)}} = \sum_{i = 0}^{n} \frac{1}{g^{f(i)}}$. Then choose $n$ large enough so that $f(n+1) - f(n) - 1 > \log_g{q}$. We then get the following contradiction:

\begin{align*}
\dfrac{1}{qg^{f(n)}} &\le \dfrac{p}{q} - \dfrac{p_n}{g^{f(n)}} \\
&= \displaystyle \sum_{i = n+1}^{\infty} \frac{1}{g^{f(i)}} \\
&\le \frac{1}{g^{f(n+1) - 1}} \displaystyle \sum_{i = 1}^{\infty} \frac{1}{g^i} \\
&\le \frac{1}{g^{f(n+1) - 1}}\displaystyle \sum_{i = 1}^{\infty} \frac{1}{2^i} \\
&= \frac{1}{g^{f(n+1) - f(n) - 1}g^{f(n)}} \\
&< \dfrac{1}{qg^{f(n)}}
\end{align*}


Now I feel that my birthday is cursed :(

1) who's paul erdos???

2) my sister's half birthday is on sept. 11, even though she was born in 2005.

The creator of MOEMS died on my birthday :(

Shakespeare died on my birthday!

The wreakage of a fighter jet was found on my birthday :(

See how cursed mine is?! :noo:

For anyone who hasn't caught on, search Wikipedia for the day of your birth and look at the "Deaths" column.

Hey, Hubble (and Pope John Paul I) died on my birthday! And the USSR Olympiad Problem Book was published on the exact same day and year as my birth.


function
quadratics
floor function
calculus
integration
algebra
number theory solved

[a] be the integer part of a (floor function)

Find the number of distinct value of the following sequences..
[k^2/2004] for k = 1, 2, 3, 4.... 2003...

Please show me the solution..
Thanks

The question is equivalant to the number of different quadratic residues mod 2004.
Generally i dont think there is any neat way to do this. We can count the number of pairs (x,y) s.t. x^2 = y^2 (mod 2004) and then the number of different residues is 2004 - (this number). 2004 | x^2 - y^2 is equivalant to 2^2*3*167|(x-y)(x+y). Now since x-y and x+y are of the same parity and they cant both be odd
they are both even. From now it should be straightforward but *very* messy ...

How do u find the quartic residue of x^2 = y^2 (mod 2004)??
Can show the full solution...

Vinoth, I think you read the question wrongly.  It's asking for the number of distinct values of $\lfloor \frac{k^2}{2004} \rfloor$, not $\{ \frac{k^2}{2004} \}$, where the first one refers to the integral part, and the second one refers to the fractional part.

Since $2^2 - 1^2, 3^2 - 2^2, \ldots, 1002^2 - 1001^2 < 2004$, we see that all the numbers $0, 1, 2, \ldots, 501$ can be attained.  Now since $1003^2 - 1002^2, 1004^2 - 1003^2, \ldots, 2003^2 - 2002^2 > 2005$, we see that all the numbers attained when $k = 1003, 1004, \ldots, 2003$ are distinct.  Hence the number of distinct values attained is $502 + 2003 - 1003 + 1 = 1503$.

As a side note, this was given for the first part of the 2004 SMO, which constituted 25 questions to be solved (only answers, not full solutions, were required) within 2.5 hours.  The question is simple, provided that you can view it from the correct perspective.


number theory
algebra
induction
equation

Find all pairs of positive integers $ (x,y)$ such that
\[ x^y \equal{} y^{x \minus{} y}.
\]
[i]Albania[/i]

Can someone post the original sollution please :roll:

[hide]It's clearly that $ x\ge y$ because $ y^{x \minus{} y}\in N$
Case 1: $ x \equal{} y$ then $ x \equal{} y \equal{} 1$
Case 2: $ x > y$ and $ x,y\ne 1$ then $ x \equal{} y^n(n\in N, n\ne 1)$
$ \implies y^{yn} \equal{} y^{y(y^{n \minus{} 1} \minus{} 1)}\implies n \equal{} y^{n \minus{} 1} \minus{} 1\implies n \plus{} 1 \equal{} y^{n \minus{} 1}$
It's easy to see that $ n < 4$ because $ n \plus{} 1\ge 2^{n \minus{} 1}$
If $ n \equal{} 2$ then $ y \equal{} 3,x \equal{} 9$
If $ n \equal{} 3$ then $ y \equal{} 2,x \equal{} 8$
Therefore we get the solution: $ (x,y) \equal{} (1;1),(2;8),(3;9)$[/hide]
edit: oops !:blush: , a typo!

In reference with post # 4; Kyoshiro must have made a mistake in typing the answer at the last. He should have written, as the solutions are  $ x \equal{} y \equal{} 1 ; y \equal{} 3,x \equal{} 9 ;y \equal{} 2,x \equal{} 8$ ,  is:
$ (x,y) \equal{} (1,1),(8,2),(9,3)$ :roll:

[u][b]Step 1:[/b][/u]

First, If we take the $ y$ root of both sides, we have the equation:
$ x \equal{} y^\frac{x \minus{} y}{y}$.
The left side of obviously an integer. In order for the right side to be an integer, the power of $ y$ must be a positive integer, or $ \frac{x \minus{} y}{y}$ must be a positive integer. So, $ x\minus{}y$ must be a multiple of $ y$. Therefore $ x$ has to be a multiple of $ y$. If $ x$ is a multiple of $ y$, then we have the equation "$ x$ is equal to $ y$ times a number", or $ x\equal{}ay$.

[u][b]Step 2:[/b][/u]

Now we substitute $ x$ for $ ay$ in the original equation:
$ x^y \equal{} y^{x \minus{} y}$.
$ (ay)^y \equal{} y^{ay \minus{} y}$.
$ a^yy^y \equal{} \frac{y^{ay}}{y^y}$.
$ a^yy^{2y} \equal{} y^{ay}$
$ ay^2\equal{}y^a$
$ a\equal{}\frac{y^a}{y^2}$
and it finally reduces to...
$ a\equal{}y^{a\minus{}2}$
[u]
[b]Step 3:[/b][/u]

We now have an equation that allows us to easily find the corresponding $ x$ value for a given $ y$ value. We use logic to finish the problem.

Evidently, if we plug in $ 1$ for $ y$, $ a$ equals $ 1$ also. So our $ x$ $ y$ ordered pair is $ (1,1)$.
If we plug in $ 2$ for $ y$, $ a$ equals $ 4$ also. So our $ x$ $ y$ ordered pair is $ (2,8)$.
If we plug in $ 3$ for $ y$, $ a$ equals $ 3$ also. So our $ x$ $ y$ ordered pair is $ (9,3)$.

I'm not going to explain the rest, but if you use logical reasoning, you can tell that no other value for $ y$ greater than $ 3$ will have a positive integer value for $ a$. So we have three ordered pairs, and we're done.

If you want me to explain anything, just post or pm me.

[quote="liltobe9"]

Evidently, if we plug in $ 1$ for $ y$, $ a$ equals $ 1$ also. So our $ x$ $ y$ ordered pair is $ (1,1)$.
If we plug in $ 2$ for $ y$, $ a$ equals $ 4$ also. So our $ x$ $ y$ ordered pair is $ (2,8)$.
If we plug in $ 3$ for $ y$, $ a$ equals $ 3$ also. So our $ x$ $ y$ ordered pair is $ (9,3)$.

[/quote]

liltobe9,
your second ordered pair needs to be changed to $ (8,2)$.

[quote="liltobe9"]First, If we take the $ y$ root of both sides, we have the equation:
$ x \equal{} y^\frac {x \minus{} y}{y}$.
The left side of obviously an integer. In order for the right side to be an integer, the power of $ y$ must be a positive integer, or $ \frac {x \minus{} y}{y}$ must be a positive integer. [/quote]
Why does the power of $ y$ have to be a positive integer? $ 4^{\frac{1}{2}} \equal{} 2$.

I have a different way to show that $ x\equal{}ay$ for integer $ a$. Clearly $ x,y$ must have the same primes dividing them. Let $ p$ be a prime dividiing $ x,y$, and let $ m,n$ be the exponents of $ p$ in $ x,y$ respectively. In order for the equation to hold, it must be true that the exponents of $ p$ on both sides of the equation are the same. Then we have:

$ my \equal{} n(x\minus{}y)$

$ \frac{m}{n} \equal{} \frac{x\minus{}y}{y}$

Case 1: $ x>2y$ so $ \frac{m}{n} > 1$. This tells us that for all $ p$, the exponent of $ p$ in $ x$ is greater than that in $ y$. So in this case, $ y\equal{}ax$. Substituting this and it is fairly clear there are no solutions.

Case 2: $ x\equal{}2y$ so $ \frac{m}{n} \equal{} 1$. This means $ x\equal{}y$ and it is easy to see that the only solution is $ (x,y)\equal{}(1,1)$

Case 3: $ x < 2y$ so $ \frac{m}{n} < 1$. This means that for all $ p$, the exponent of $ p$ in $ x$ is less than that in $ y$. So in this case, $ x\equal{}ay$. We then continue in the same way as litobe9 to get that:

$ (x,y) \equal{} (1,1), (8,2), (9,3)$

[quote="Zhero"][quote="liltobe9"]First, If we take the $ y$ root of both sides, we have the equation:
$ x \equal{} y^\frac {x \minus{} y}{y}$.
The left side of obviously an integer. In order for the right side to be an integer, the power of $ y$ must be a positive integer, or $ \frac {x \minus{} y}{y}$ must be a positive integer. [/quote]
Why does the power of $ y$ have to be a positive integer? $ 4^{\frac {1}{2}} \equal{} 2$.[/quote]

Zhero: you didn't look at the equation properly.

You're right if $ y$ is a perfect square and we have the square root, or if $ y$ is a perfect cube and we have the cube root. However, we have the $ y$ root of $ y$.

There is no value of $ y$ for which the $ y$ root of $ y$ is an integer, except for $ 1$.

In this instance, the power of $ y$ must be a positive integer or else $ y^\frac {x \minus{} y}{y}$ cannot be a positive integer. Note that if $ y$ is $ 1$, the power is still an integer.

We cannot use $ 4^{\frac {1}{2}}$, it would have to be $ 4^{\frac {1}{4}}$.

As for the logical reasoning described in [b]liltobe9[/b]'s answer, one could say the following:
If $a\neq2, a=y^{a-2} \Rightarrow y=\sqrt[a-2]{a}$ which also means that $a-2>0 \Rightarrow a>2$ and since $a\in Z \Rightarrow a\ge3$
However we know from the AM-GM inequality that  $ \frac{a+1+1+...+1}{a-2}\ge\sqrt[a-2]{a}$. (There are $a-3$ 1's in the LHS)
So now we have that $\frac{2a-3}{a-2}\ge\sqrt[a-2]{a}=y \Rightarrow 2+\frac{1}{a-2}\ge{y} (*)$
Since $3\le{a}<\infty \Rightarrow 1\le{a}-2<\infty \Rightarrow 1\ge\frac{1}{a-2}>0 \Rightarrow$
$3\ge2+\frac{1}{a-2}>0$ and from $(*) \Rightarrow 3\ge2+\frac{1}{a-2}\ge{y} \Rightarrow y\le{3}$

[quote="andreass"]As for the logical reasoning described in [b]liltobe9[/b]'s answer, one could say the following:
If $a\neq2, a=y^{a-2} \Rightarrow y=\sqrt[a-2]{a}$ which also means that $a-2>0 \Rightarrow a>2$[/quote]  Since $(x, y) = (1, 1)$ is a solution with $a = 1 < 2$, probably you should reconsider this argument.

Clearly $ x\ge\ 2y $ for $ x,y\ge\ 1 $. Let $ gcd(x,y)=d $ and we have $ x=dx_1, y=dy_1, gcd(x_1,y_1)=1 $. So the given equation is equivalent to $ d^yx_1^y=d^{(x-y)}y_1^{(x-y)} $ or $ x_1^y=d^{(x-2y)}y_1^{(x-y)} $. Now if $ y_1 $ has a prime factor $ p $, $ p $ must divide $ y_1^{(x-y)} $ a contradiction. So $ y_1 $ must be $ 1 $, or $ y=d $. The give equation now is equivalent with this equation $ x_1=d^{(x_1-2)} \ge\ 2^{(x_1-2)} $. By trivial induction, for $ x_1\ge\ 4, 2^{(x_1-2 })\ge\ x_1 $ and now is easy to see that the solutions are $ (x,y) = (1,1),(8,2),(9,3) $

[quote="andrejilievski"]Clearly $ x\ge\ 2y $.[/quote]  Since (1, 1) is a solution, clearly not.

brothers I am posting my solution.

it is clear that x>=y

[b]case 1:[/b] x=y,

x^x = 1

or, x = 1, so (x,y) = (1,1)

[b]case 2:[/b] x>y

[b]lemma :[/b] x = a*y where a>1 and a is positive integer.

[b]proof :[/b]

x^y = y^(x - y)

or, x = y^[(x - y)/y]

as x>y and x,y are integers,

(x - y)/y is integer

so, y|x

x = a*y where a>1 and a is positive integer.(proved)

[b]back to our main problem[/b]

(a*y)^y = y^[y*(a - 1)]

or, (a^y)*y^(2*y) = y^(a*y)

or, a*y^2 = y^a

or, y^(a - 2) = a;

it is clear that a>2

if y>4 by induction we find that y^(a - 2)>a

so, y<4 then casework

So arqady's solution can be strengthened with the following theorem:

For positive integer x,y,a,b we have that $x^a = y^b \Rightarrow $ x|y or y|x. From this we split the problem into two cases, namely x|y and y|x. The former case yields that x=y, which yields (1,1). For the latter case, we let x = ky which yields $ y = k^{\frac{1}{k-2}} $, which gives us the rest of the solutions since we only have to check k = 3,4 (which both work) and can then use x > $2^x-2$ to prove that these are the only solutions.

(k = 3 and 4 give the other two solutions and the inequality with 2's in it holds only for x>4, which essentially eliminates every other possibility for k.)

Find all pairs of positive reals $(x,y)$ satisfying the equation.
Find all pairs of positive rationals $(x,y)$ satisfying the equation.

Another way to show $x$ is a multiple of $y$ when $x>y$:

Let $x=y+a, a>0$. Then $(y+a)^y=y^a \Rightarrow y^y+\cdots +a^y=y^a$. So $a>y$ and $y | a^y$. So $y|a$, hence $y |x$.

Main equation is \begin{align} x^y = y^{x-y} \end{align}

If $x=y$, then $(x,y)=(1,1)$ is a solution. If $x<y$, then $y^{x-y} \not\in \mathbb Z^+$ and there is no solution.

Now, let's examine the case: $x>y>1$. In this case, $x^y = y^{x-y} < x^{x-y}$ and $y<x-y \implies x>2y$. If $\gcd(x,y)=d$, then $x=ad$, $y=bd$, $\gcd(a,b)=1$ such that $a,b \in \mathbb Z^+$. Rewrite these in $(1)$, $a^yd^y = b^{x-y}d^{x-y} \implies a^y = b^{x-y}d^{x-2y}$. Therefore $b \mid a^y$. On the other hand, $1=(a,b)=(a^y, b)= b$ and we yields $b=1$, $y=d$. Hence we find that $$ a=d^{a-2}, \qquad x=d^{a-1}, \qquad y=d $$

Our main idea: Because of $d^{a-2}$ will be very fast increasing, equation $a=d^{a-2}$ has very limited solutions. We have to search the solutions of $a=d^{a-2}$ for $d\geq 1$ and  $a>2$.

$\bullet$ For $d=1$, $a>2$, there is no solution.

$\bullet$ For $d=2$, $a\in \{ 4, 8, 16, 32, \dots \}$. Only $a=4$ satisfy. We find the solution $(x,y)=(8,2)$.

$\bullet$ For $d=3$, $a\in \{ 3, 9, 27, 81, \dots \}$. Only $a=3$ satisfy. We find the solution $(x,y)=(9,3)$.

Now, let's examine the case: $d\geq 4$. Then, $a=d^{a-2}\geq 4^{a-2}$ and $16a \geq 4^a$. For $a=3$, $16\cdot 3 \geq 4^3$ is false. Also, for $a\geq 4$, inequality $16a \geq 4^a$ can't satisfy. Because $4^a$ faster increase than $16a$. Therefore, there is no solution for the case $d\geq 4$.

Hence, all $(x,y)$ solutions of the main equation are $(1,1)$, $(8,2)$, $(9,3)$.


integration
trigonometry
calculus
calculus computations

$\int \frac{\cos x - \sin x +1 -x}{e^x+\sin x +x}dx$

:D  :D
 :D $\int\frac {\cos x-\sin x+1-x}{e^x+\sin x+x}=\int(-1+\frac{(e^x+\sin x+x)'}{e^x+\sin x+x})dx=-x+ln|e^x+\sin x+x|+c$


Let's say that in a group of people there is a 50% probability that two have the same birthday.

How many people are in this group?

[hide="Answer with no explanation"]
The answer is 23; this is an Academic Bowl question. Why is it 23?[/hide]

Basically just find when the probability of n people not having the same birthday falls below 50%.  Said probability would be $(356/356)*(355/356)*(354/356)...$

this can be complicated because you can have situations were there are 2,3,etc birthday's in common, but if you do the probability that nobody has the same birthday, and subtract that from one, you have the probability that at least 2 people have the same birthday