calculus
derivative
function
absolute value
topology
real analysis
real analysis unsolved

$f'$ is defined on $[a,b]$. Prove that its all continuous points is dense on $[a,b]$.

This doesn't help to solve the problem, but it helps me :):

Are you trying to say that the set of points where the derivative of $f:[a,b]\to \mathbb R$ is continuous must be dense in $[a,b]$?

I saw in a book that if a sequence of functions for which the set of continuity points is dense in a Baire space converges to a function f, then also the set of continuity points for f is dense. The proof was however very very difficult. Note that ypur problem follows from that one. But there must be an easier way!

[quote="grobber"]Once somebody solves point (b) [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=Rudin&t=39000]here[/url], we're done with this one as well. Let's prove a particular case of the result stated by harazi above (using the problem you'll find by following the link :)):

Given real continuous functions $f_n:[a,b]\to\mathbb R$ s.t. $f_n(x)\to f(x),\ \forall x\in[a,b]$, the set of continuity points of $f$ is dense in $[a,b]$.

Since what we'll do is applicable to any subinterval of $[a,b]$, it suffices to show that $f$ has one continuity point. Use point (b) from that problem in order to prove that there is an open subset $V$ of $[a,b]$ and some natural $N$ s.t. $|f(x)-f_n(x)|<\frac\varepsilon 6$ forall $x\in V$ and for all $n\ge N$, which also means that $|f_m(x)-f_n(x)|\le \frac\varepsilon 3,\ \forall x\in V,\ \forall m,n\ge N$. There's also a compact subinterval $I_1$ of $V$ on which the oscillation of $f_N$ (the absolute value of the difference between the supremum and the infimum of $f_N$ on that interval) is smaller than $\frac\varepsilon 3$. Now, given $m,n\ge N$ and $x,y\in I_1$, we have $|f_m(x)-f_n(y)|\le |f_m(x)-f_N(x)|+|f_n(y)-f_N(y)|+|f_N(x)-f_N(y)|\le\varepsilon$, which means, in particular, that $|f(x)-f(y)|\le\varepsilon,\ \forall x,y\in I_1$.

We repeat the procedure, finding a compact subinterval $I_2$ of $I_1$ on which the oscillation of $f$ is $\le\frac\varepsilon 2$, then a compact subinterval $I_3$ of $I_2$ on which the oscillation of $f$ is $\le\frac\varepsilon 3$, and so on. If $x$ belongs to the intersection of all these intervals, then the oscillation of $f$ in $x$ is $0$, which is another way of saying that $x$ is a continuity point of $f$.

All we need now is for someone to solve that problem I posted :).[/quote]

 This is a good problem to practice using Baire's theorem
 Let's $V_m$ be the set of point $x$ such that : $|f_{n'}(x)-f_n(x)| \leq \epsilon$ for all $n,n' \geq m$.  Then each $x$ belongs to some $V_m$ . Those are closed sets , whose union is $\mathbb{R}$, so one of them must contain an open set $V$...


inequalities
function
parameterization
algebra
polynomial
Vieta
Cauchy Inequality

Let $ a,b,c$ be positive real numbers such that $ ab\plus{}bc\plus{}ca\equal{}3$. Prove that
\[ \frac 1{1\plus{}a^2(b\plus{}c)} \plus{} \frac 1{1\plus{}b^2(c\plus{}a)} \plus{} \frac 1 {1\plus{}c^2(a\plus{}b) } \leq \frac 3 {1\plus{}2abc} .\]

$ \frac 1{1\plus{}a^2(b\plus{}c)} \plus{} \frac 1{1\plus{}b^2(c\plus{}a)} \plus{} \frac 1 {1\plus{}c^2(a\plus{}b)} \leq \frac 3 {1\plus{}2abc}$
$ \iff  \frac 1{1\plus{}a^2(b\plus{}c)}\minus{}\frac{1}{1\plus{}2abc} \plus{}\frac 1{1\plus{}b^2(c\plus{}a)}\minus{}\frac{1}{1\plus{}2abc} \plus{} \frac 1 {1\plus{}c^2(a\plus{}b)}\minus{}\frac{1}{1\plus{}2abc} \leq 0$
Note that :
$ \frac 1{1\plus{}a^2(b\plus{}c)}\minus{} \frac{1}{1\plus{}2abc}\equal{}\frac{ab(c\minus{}a)\plus{}ac(b\minus{}a)}{1\plus{}a^2(b\plus{}c)(1\plus{}2abc)}$
Hence  $ LHS\equal{} \frac{abc\minus{}1}{2abc\plus{}1}[(c\minus{}a)^2S_b\plus{}(a\minus{}b)^2S_c\plus{}(b\minus{}c)^2S_a)$
where :$ S_b\equal{}\frac{b}{(1\plus{}a^2(b\plus{}c))(1\plus{}c^2(a\plus{}b))}$ and  similar $ S_a\equal{}\frac{a}{(1\plus{}b^2(a\plus{}c))(1\plus{}c^2(a\plus{}b))}$ and $ S_c\equal{}\frac{c}{(1\plus{}a^2(b\plus{}c))(1\plus{}b^2(a\plus{}c))}$ with $ S_a,S_b,S_c \geq 0$(*)
By ineq AM-GM : $ ab\plus{}bc\plus{}ca \geq 3\sqrt[3]{a^2b^2c^2} \iff abc \leq 1$(**)
 From (*) and (**), we have : $ LHS \leq 0$
 Occur  equal if and only if :$ a\equal{}b\equal{}c\equal{}1$

From $ ab \plus{} bc \plus{} ca \equal{} 3$(multiply by $ a$) we have: $ a^2(b \plus{} c) \equal{} 3a \minus{} abc$ ,  so it's enough to prove the following inequality:
\[ \sum\frac {1}{1 \minus{} abc \plus{} 3a} \leq \frac {3}{1 \plus{} 2abc}
\]
Let $ p \equal{} 1 \minus{} abc$ ,  obviously $ p \geq 0$ because with AM-GM inequality we have: $ 3 \equal{} ab \plus{} bc \plus{} ca \geq 3\sqrt [3]{a^2b^2c^2}$ so we have $ abc \leq 1$ ,  with equality if and only if $ a \equal{} b \equal{} c$.
With our new marks, we have to prove the following inequality:
\[ \sum\frac {1}{p \plus{} 3a} \leq \frac {3}{3 \minus{} 2p}
\]
or, after expanding:
\[ (3 \minus{} 2p)(p^2 \plus{} 2p(a \plus{} b \plus{} c) \plus{} 27) \leq (p \plus{} 3a)(p \plus{} 3b)(p \plus{} 3c)
\]

\[ \leftrightarrow 3p^2 \plus{} 6p(a \plus{} b \plus{} c) \plus{} 27(1 \minus{} abc) \minus{} 2p^3 \minus{} 4p^2(a \plus{} b \plus{} c) \minus{} 18p \leq p^3 \plus{} p^2(3a \plus{} 3b \plus{} 3c) \plus{} 27p
\]

\[ \leftrightarrow p(3p^2 \plus{} p(7a \plus{} 7b \plus{} 7c) \plus{} 18 \minus{} 6a \minus{} 6b \minus{} 6c) \geq 0
\]
Since $ p \geq 0$ ,  it's enough to prove the thing inside the brackets is $ \geq 0$. Now if we return $ p \equal{} 1 \minus{} abc$ and expand:
\[ p(3a^2b^2c^2 \plus{} 3 \minus{} 6abc \plus{} 18 \plus{} a \plus{} b \plus{} c \minus{} 7abc( a \plus{} b \plus{} c )) \geq 0
\]
$ 3a^2b^2c^2 \plus{} 3 \geq 6abc$ is due to AM-GM;
$ 18 \geq 6abc(a\plus{}b\plus{}c)$ is the same as :
$ T[2,2,0]\plus{}2T[2,1,1] \geq 3T[2,1,1]$ which is Muirhead ,  and:
$ a\plus{}b\plus{}c \geq abc(a\plus{}b\plus{}c)$ because $ abc \leq 1$
Summing up last inequalities we get the wanted one. Equality if and only if $ a\equal{}b\equal{}c$.

My solution is the following
The inequality is equivalent to
\[ \frac{a^2(b\plus{}c)}{1\plus{}a^2(b\plus{}c)} \plus{}\frac{b^2(c\plus{}a)}{1\plus{}b^2(c\plus{}a)} \plus{}\frac{c^2(a\plus{}b)}{1\plus{}c^2(a\plus{}b)} \plus{}\frac{3}{1\plus{}2abc} \ge 3\]
By the Cauchy Schwarz Inequality, we get
\[ \frac{a^2(b\plus{}c)}{1\plus{}a^2(b\plus{}c)} \plus{}\frac{b^2(c\plus{}a)}{1\plus{}b^2(c\plus{}a)} \plus{}\frac{c^2(a\plus{}b)}{1\plus{}c^2(a\plus{}b)} \ge \frac{36}{(b\plus{}c)[1\plus{}a^2(b\plus{}c)]\plus{}(c\plus{}a)[1\plus{}b^2(c\plus{}a)]\plus{}(a\plus{}b)[1\plus{}c^2(a\plus{}b)]} \equal{}\frac{18}{(ab\plus{}bc\plus{}ca)^2 \plus{}(a\plus{}b\plus{}c)(1\minus{}abc)} \equal{}\frac{18}{9\plus{}(a\plus{}b\plus{}c)(1\minus{}abc)}\]
Hence, it suffices to prove that
\[ \frac{6}{9\plus{}(a\plus{}b\plus{}c)(1\minus{}abc)} \plus{}\frac{1}{2abc\plus{}1} \ge 1\]
Setting $ p\equal{}a\plus{}b\plus{}c,r\equal{}abc$, then the above inequality is equivalent to
\[ \frac{6}{9\plus{}p(1\minus{}r)} \plus{}\frac{1}{2r\plus{}1} \ge 1\]
\[ \Leftrightarrow \frac{6}{9\plus{}p(1\minus{}r)} \ge \frac{2r}{1\plus{}2r}\]
\[ \Leftrightarrow 3(2r\plus{}1) \ge r[9\plus{}p(1\minus{}r)]\]
\[ \Leftrightarrow (1\minus{}r)(3\minus{}pr) \ge 0\]
Which is obviously true since
\[ 27\equal{}(ab\plus{}bc\plus{}ca)^3 \ge 27a^2b^2c^2\equal{}27r^2 \Rightarrow 1 \ge r\]
\[ 9\equal{}(ab\plus{}bc\plus{}ca)^2 \ge 3abc(a\plus{}b\plus{}c)\equal{}3pr \Rightarrow 3 \ge pr\]
We have done.
Equality holds if and only if $ a\equal{}b\equal{}c\equal{}1$.

@Moderators: How can I propose a problem to the contest? Please reply me as soon as possible. Thank you very much.

We have $ \frac{1}{{1 \plus{} a^2 \left( {b \plus{} c} \right)}} \equal{} 1 \minus{} \frac{{a^2 \left( {b \plus{} c} \right)}}{{1 \plus{} a^2 \left( {b \plus{} c} \right)}}$
So by Cauchy Schwarz we have
\[ \sum {\frac{1}{{1 \plus{} a^2 \left( {b \plus{} c} \right)}}}  \equal{} 3 \minus{} \sum {\frac{{a^2 \left( {b \plus{} c} \right)}}{{1 \plus{} a^2 \left( {b \plus{} c} \right)}}}  \le 3 \minus{} \frac{{4\left( {ab \plus{} bc \plus{} ca} \right)^2 }}{{2\left( {a \plus{} b \plus{} c} \right) \plus{} 2\sum {a^2 b^2 }  \plus{} 2abc\left( {a \plus{} b \plus{} c} \right)}}\]
Assume $ p\equal{}a\plus{}b\plus{}c,r\equal{}abc$,it suffices us to show that
\[ 3 \minus{} \frac{{36}}{{2p \plus{} 18 \minus{} 2rp}} \le \frac{3}{{1 \plus{} 2r}}\]
\[ \Leftrightarrow pr \le 3\]
which is clearly true because we have $ \left( {ab \plus{} bc \plus{} ca} \right)^2  \ge 3abc\left( {a \plus{} b \plus{} c} \right) \Leftrightarrow pr \le 3$.
We have done, equality holds if and only if $ a\equal{}b\equal{}c\equal{}1$.

We have to prove that
$ \sum\frac {1}{1 \plus{} a^2(b \plus{} c)}\leq\frac {3}{1 \plus{} 2abc}$ or equivalently

$ \sum\left(\frac {1}{1 \plus{} a^2(b \plus{} c)} \minus{} \frac {1}{1 \plus{} 2abc}\right)\leq 0$ or

$ \sum\left(\frac {a(b(a \minus{} c) \plus{} c(a \minus{} b))}{(1 \plus{} a^2(b \plus{} c))(1 \plus{} 2abc)}\right)\geq 0$  or

$ \sum\left(\frac {ac(a \minus{} b)}{(1 \plus{} a^2(b \plus{} c))(1 \plus{} 2abc)} \plus{} \frac {bc(b \minus{} a)}{(1 \plus{} b^2(a \plus{} c))(1 \plus{} 2abc)}\right)\geq 0$

or

$ \sum\left(\frac {c(a \minus{} b)^2(1 \minus{} abc)}{(1 \plus{} a^2(b \plus{} c))(1 \plus{} b^2(c \plus{} a))(1 \plus{} 2abc)}\right)\geq 0$

which holds since $ ab \plus{} bc \plus{} ca\geq 3\sqrt [3]{a^2b^2c^2}$ which means that $ abc\leq 1$ and we are done .

$ RHS\minus{}LHS\equal{}\sum_{cyc}(\frac{1}{1\plus{}2abc}\minus{}\frac{1}{1\plus{}a^{2}(b\plus{}c)})\equal{}\sum_{cyc}\frac{ab(a\minus{}c)\plus{}ac(a\minus{}b)}{(1\plus{}2abc)(1\plus{}a^{2}(b\plus{}c))}\equal{}\sum_{cyc}\frac{a(b\minus{}c)}{1\plus{}2abc}(\frac{b}{1\plus{}b^{2}(c\plus{}a)}\minus{}\frac{c}{1\plus{}c^{2}(a\plus{}b)})\equal{}\sum_{cyc}\frac{a(1\minus{}abc)(b\minus{}c)^{2}}{(1\plus{}2abc)(1\plus{}b^{2}(c\plus{}a))(1\plus{}c^{2}(a\plus{}b))}$.
Because $ ab\plus{}bc\plus{}ca\equal{}3$ , $ abc\le1$. (AM-GM) Therefore $ RHS\ge LHS$. Equality occurs iff $ a\equal{}b\equal{}c\equal{}1$.

$ \sum\frac {1}{1 \plus{} a^2(b \plus{} c)}\le \frac {3}{1 \plus{} 2abc}\Leftrightarrow \sum(1 \minus{} \frac {1}{1 \plus{} a^2(b \plus{} c)})\ge 3 \minus{} \frac {3}{1 \plus{} 2abc}\Leftrightarrow \sum\frac {a^2(b \plus{} c)}{1 \plus{} a^2(b \plus{} c)}\ge \frac {6abc}{1 \plus{} 2abc}$

By Titu inequality(or cauchy or T'2 Lemma)we have

$ \sum\frac {a^2(b \plus{} c)}{1 \plus{} a^2(b \plus{} c)} \equal{} \sum\frac {a^2(b \plus{} c)^2}{b \plus{} c \plus{} a^2(b \plus{} c)^2}\ge \frac {(2(ab \plus{} bc \plus{} ca))^2}{2(\sum a) \plus{} \sum(ab \plus{} ac)^2}$

by the hypothesis we must prove that
\[ \frac {36}{2(\sum a) \plus{} \sum(3 \minus{} bc)^2}\ge \frac {6abc}{1 \plus{} 2abc}\Leftrightarrow \frac {6}{2(\sum a) \plus{} (\sum (bc)^2) \minus{} 6(\sum bc) \plus{} 27}\ge \frac {abc}{1 \plus{} 2abc}
\]
Or
\[ \frac {6}{2(\sum a) \plus{} (\sum (bc)^2) \plus{} 9}\ge \frac {abc}{1 \plus{} 2abc}
\]
Now let $ abc \equal{} r$ and $ ab \plus{} bc \plus{} ca \equal{} q \equal{} 3$ and $ a \plus{} b \plus{} c \equal{} p$ so the last inequality is equivalent to
$ \frac {6}{2p \plus{} (q^2 \minus{} 2pr) \plus{} 9}\ge \frac {r}{1 \plus{} 2r} \Leftrightarrow 6(1 \plus{} 2r)\ge r(2p \minus{} 2pr \plus{} 18)\Leftrightarrow pr^2 \minus{} r(p \plus{} 3) \plus{} 3\ge 0 \Leftrightarrow p(r \minus{} 1)(r \minus{} \frac {3}{p})\ge 0 \Leftrightarrow (r \minus{} 1)(r \minus{} \frac {3}{p})\ge 0$ this inequality is true if we have $ r\le Min(1,\frac {3}{p})$
but by two simple inequality we have $ q^2\ge 3pr$ so $ 3\ge pr$ and $ p^2\ge 3q$ so $ p\ge 3$
So

$ \frac {3}{p}\le 1$
$ r\le \frac {3}{p}$
so the inequality is proved and the quality hols for $ a \equal{} b \equal{} c \equal{} 1$

- $ \frac {1}{1 \plus{} c^2(a \plus{} b)} \plus{} \frac {1}{1 \plus{} b^2(c \plus{} a)} \plus{} \frac {1}{1 \plus{} a^2(b \plus{} c)}\leq \frac {3}{1 \plus{} 2abc}$
- denote $ f(x) \equal{} \frac {x}{x \plus{} k}$
- function $ f(x)$ is concave for all positive real $ x$ and for a positive real parameter $ k$ because $ 0 > f''(x) \equal{} \frac { \minus{} 2k}{(x \plus{} k)^3}$
- if $ ab \plus{} bc \plus{} ca \equal{} 3$ than by AMGM $ 1\geq abc$. if $ abc \equal{} 1$ than $ a \equal{} b \equal{} c \equal{} 1$ and inequality holds for this case. so assume without a loss of generality that $ 1 > abc$
- since $ \frac {ab \plus{} bc \plus{} ca}{3} \equal{} 1$ we know that
$ \frac {1}{1 \plus{} c^2(a \plus{} b)} \plus{} \frac {1}{1 \plus{} b^2(c \plus{} a)} \plus{} \frac {1}{1 \plus{} a^2(b \plus{} c)} \equal{} \frac {c(a \plus{} b) \plus{} ab}{3(1 \plus{} c^2(a \plus{} b))} \plus{} \frac {b(c \plus{} a) \plus{} ca}{3(1 \plus{} b^2(c \plus{} a))} \plus{} \frac {a(b \plus{} c) \plus{} bc}{3(1 \plus{} a^2(b \plus{} c))}$
- now by multiplying both sides of the inequality by $ abc$ we get equivalent
$ \frac {ab}{3}\frac {c^2(a \plus{} b) \plus{} abc}{1 \plus{} c^2(a \plus{} b)} \plus{} \frac {ca}{3}\frac {b^2(c \plus{} a) \plus{} abc}{1 \plus{} b^2(c \plus{} a)} \plus{} \frac {ab}{3}\frac {c^2(a \plus{} b) \plus{} abc}{1 \plus{} c^2(a \plus{} b)}\leq\frac {3abc}{1 \plus{} 2abc}$
- denote $ k \equal{} 1 \minus{} abc$. we know $ k > 0$
- left hand side is therefore equivalent to:
$ \frac {ab}{3}\frac {c^2(a \plus{} b) \plus{} abc}{abc \plus{} c^2(a \plus{} b) \plus{} k} \plus{} \frac {ca}{3}\frac {b^2(c \plus{} a) \plus{} abc}{abc \plus{} b^2(c \plus{} a) \plus{} k} \plus{} \frac {ab}{3}\frac {c^2(a \plus{} b) \plus{} abc}{abc \plus{} c^2(a \plus{} b) \plus{} k}$
- denote $ t_1 \equal{} \frac {ab}{3}$,,$ t_2 \equal{} \frac {ca}{3}$,,$ t_3 \equal{} \frac {bc}{3}$. note that $ t_1 \plus{} t_2 \plus{} t_3 \equal{} 1$
- denote $ x \equal{} c^2(a \plus{} b) \plus{} abc$,,$ y \equal{} b^2(c \plus{} a) \plus{} abc$,,$ z \equal{} a^2(b \plus{} c) \plus{} abc$
- left hand side is further equivalent to $ t_1 \frac {x}{x \plus{} k} \plus{} t_2 \frac {y}{y \plus{} k} \plus{} t_3 \frac {z}{z \plus{} k}$ further equivalent to $ t_1f(x) \plus{} t_2f(y) \plus{} t_3f(z)$ which by Jensen inequality for concave functions $ t_1f(x) \plus{} t_2f(y) \plus{} t_3f(z)\leq f(t_1x \plus{} t_2y \plus{} t_3z)$ since $ x,y,z > 0$ and $ f$ is a concave function for positive reals
- $ f(t_1x \plus{} t_2y \plus{} t_3z) \equal{} \frac {\frac {ab}{3}\cdot(c^2(a \plus{} b) \plus{} abc) \plus{} \frac {ca}{3}\cdot(b^2(a \plus{} c) \plus{} abc) \plus{} \frac {bc}{3}\cdot(a^2(b \plus{} c) \plus{} abc)}{\frac {ab}{3}\cdot(c^2(a \plus{} b) \plus{} abc) \plus{} \frac {ca}{3}\cdot(b^2(a \plus{} c) \plus{} abc) \plus{} \frac {bc}{3}\cdot(a^2(b \plus{} c) \plus{} abc) \plus{} k}$
- since $ c^2(a \plus{} b) \plus{} abc \equal{} c(ab \plus{} bc \plus{} ca) \equal{} 3c$ and analogously for $ b,a$ we get that
$ f(t_1x \plus{} t_2y \plus{} t_3z) \equal{} \frac {3abc}{3abc \plus{} k} \equal{} \frac {3abc}{1 \plus{} 2abc}$

$ \frac {ab}{3}\frac {c^2(a \plus{} b) \plus{} abc}{1 \plus{} c^2(a \plus{} b} \plus{} \frac {ca}{3}\frac {b^2(c \plus{} a) \plus{} abc}{1 \plus{} b^2(c \plus{} a)} \plus{} \frac {ab}{3}\frac {c^2(a \plus{} b) \plus{} abc}{1 \plus{} c^2(a \plus{} b)}\leq\frac {3abc}{1 \plus{} 2abc}$
now obviously holds for all positive real $ a,b,c$ such that $ ab \plus{} bc \plus{} ca \equal{} 3$. therefore also the equivalent inequality must hold
$ \frac {1}{1 \plus{} c^2(a \plus{} b)} \plus{} \frac {1}{1 \plus{} b^2(c \plus{} a)} \plus{} \frac {1}{1 \plus{} a^2(b \plus{} c)}\leq \frac {3}{1 \plus{} 2abc}$
Q.E.D

Lest find (a+b) from the ab+bc+ac=3 :
c(a+b)=3-ab
a+b=(3-ab)/c
Lets do it for a+c and b+c and put it into our inequality
1/(1+a * (3-bc))  +  1/(1+b * (3-ac))  +  1/(1+c * (3-ba)) <= 3/2+abc
1/(1+3a-abc)  +  1/(1+3b-abc))  +  1/(1+3c-abc)) <= 3/2+abc
Lets put abc=x
1/(1+3a-x)  +  1/(1+3b-x))  +  1/(1+3c-x)) <= 3/2+x
9+2(a+b+c)(1-x)+(1-x)^2   /  9x + 9(1-x)+(a+b+c)(1-x)^2+(1-x)^3    <=   3/2+x
After entreating this inequality we will achive :
4x*x*x-9*x*x+15*x-10 <= (a+b+c)(5x*x-4*x-1)
If we do the Cauchi theorem to the ab+bc+ca=3 we will achive that abc<=1 (x <= 1). If abc<=1 we can easily prove the inequality 1/a+1/b+1/c >= a+b+c
ab+bc+ac >= abc*(a+b+c)
a+b+c <= 3/abc
a+b+c <= 3/x
If we put it to our inequality we will achive

3<= 4*(x^4)-9*(x^3)+15*(x^2)-10*x /  5*(x^2)-4*x-1
4*(x^4)-9*(x^3)+2x+3 >= 0
We can easily prove this inequality, which is right, so the starting inequality was also right.

[b]$ \frac 1{1\plus{}a^2(b\plus{}c)} \plus{} \frac 1{1\plus{}b^2(c\plus{}a)} \plus{} \frac 1 {1\plus{}c^2(a\plus{}b) } \leq \frac 3 {1\plus{}2abc}$[/b]
------
[b]Lemma: $ ab\plus{}bc\plus{}ca\equal{}3$, $ a,b,c \in R^\plus{}$
.$ \equal{}> \frac {6abc}{1\plus{}2abc}\leq \frac {((a(b\plus{}c)\plus{}b(a\plus{}c)\plus{}c(a\plus{}b))^2}{2(a\plus{}b\plus{}c)\plus{}a^2(b\plus{}c)^2\plus{}b^2(a\plus{}c)^2\plus{}c^2(a\plus{}b)^2}$ .[/b]
Dem:
$ <\equal{}> 6abc(2(a\plus{}b\plus{}c)\plus{}a^2(b\plus{}c)^2\plus{}b^2(a\plus{}c)^2\plus{}c^2(a\plus{}b)^2) \leq 36(1\plus{}2abc)$,
*$ a^2b^2\plus{}b^2c^2\plus{}c^2a^2\equal{}(ab\plus{}bc\plus{}ca)^2\minus{}2(abc)(a\plus{}b\plus{}c)\equal{}9\minus{}2(abc)(a\plus{}b\plus{}c)$
$ <\equal{}> abc(2(a\plus{}b\plus{}c)\plus{}18\minus{}2(abc)(a\plus{}b\plus{}c) \leq 6(1\plus{}2abc)$
$ <\equal{}> abc(a\plus{}b\plus{}c)\plus{}3abc\minus{}(abc)^2(a\plus{}b\plus{}c) \leq 3$
$ <\equal{}> 0 \leq (3\minus{}abc(a\plus{}b\plus{}c))(abc\minus{}1)$..(1)
* $ (abc)^{2/3}\leq (ab\plus{}bc\plus{}ca)/3\equal{}1$ $ \equal{}> abc \leq 1$ => $ abc\minus{}1 \leq 0$..(2)
* $ xy\plus{}yz\plus{}zx \leq x^2\plus{}y^2\plus{}z^2 \equal{}> 3(xy\plus{}yz\plus{}zx)\leq (x\plus{}y\plus{}z)^2$, if $ x\equal{}ab, y\equal{}bc, z\equal{}ca$
=>$ 3(abc)(a\plus{}b\plus{}c)\leq (ab\plus{}bc\plus{}ca)^2\equal{}9 \equal{}> (abc)(a\plus{}b\plus{}c)\leq 3  \equal{}> abc(a\plus{}b\plus{}c)\minus{}3 \leq 0$..(3)
**(3)and (2):
$ 0\leq (abc\minus{}1)(abc(a\plus{}b\plus{}c)\minus{}3)$=> (1) is true!. $ Q.E.D.$
---
$ S_1\equal{}\frac 1{1\plus{}a^2(b\plus{}c)} \plus{} \frac 1{1\plus{}b^2(c\plus{}a)} \plus{} \frac 1 {1\plus{}c^2(a\plus{}b) }$
$ S_2\equal{}\frac 3 {1\plus{}2abc}$
$ S_3\equal{}\frac {((a(b\plus{}c)\plus{}b(a\plus{}c)\plus{}c(a\plus{}b))^2}{2(a\plus{}b\plus{}c)\plus{}a^2(b\plus{}c)^2\plus{}b^2(a\plus{}c)^2\plus{}c^2(a\plus{}b)^2}$
$ 3\minus{}S_1\equal{}\frac {a^2(b\plus{}c)^2}{b\plus{}c\plus{}a^2(b\plus{}c)^2} \plus{} \frac {b^2(a\plus{}c)^2}{c\plus{}a\plus{}b^2(c\plus{}a)^2} \plus{} \frac {c^2(a\plus{}b)^2}{a\plus{}b\plus{}c^2(a\plus{}b)^2 }$
$ 3\minus{}S_2\equal{}\frac {6abc}{1\plus{}2abc}$
For Cauchy.
$ \frac {(a(b\plus{}c)\plus{}b(a\plus{}c)\plus{}c(a\plus{}b)^2}{2(a\plus{}b\plus{}c)\plus{}a^2(b\plus{}c)^2\plus{}b^2(a\plus{}c)^2\plus{}c^2(a\plus{}b)^2} \leq \frac {a^2(b\plus{}c)^2}{b\plus{}c\plus{}a^2(b\plus{}c)^2} \plus{} \frac {b^2(a\plus{}c)^2}{c\plus{}a\plus{}b^2(c\plus{}a)^2} \plus{} \frac {c^2(a\plus{}b)^2}{a\plus{}b\plus{}c^2(a\plus{}b)^2 }$
For the Lemma.
$ 3\minus{}S_2 \leq S_3$, but, $ S_3\leq 3\minus{}S_1$=> $ 3\minus{}S_2\leq 3\minus{}S_1$ =>$ S_1\leq S_2$
                                $ Q.E.D.$

Adding 3 to each side the inequality becomes: $ \sum_{cyc}\frac {a^2(b \plus{} c)}{1 \plus{} a^2(b \plus{} c)} > \equal{} \frac {6abc}{1 \plus{} 2abc}$, if and only if:
$ \sum_{cyc}\frac {(ab \plus{} ac)^2}{(b \plus{} c) \plus{} a^2(b \plus{} c)^2} > \equal{} \frac {6abc}{1 \plus{} 2abc}$, now in order to prove this i am going to use the cauchy inequality giving LHS>=R
$ R \equal{} \frac {(2ab \plus{} 2bc \plus{} 2ca)^2}{\sum_{cyc}(b \plus{} c) \plus{} (ab \plus{} ac)^2} \equal{} \frac {36}{2(a \plus{} b \plus{} c) \plus{} (3 \minus{} bc)^2 \plus{} (3 \minus{} ca)^2 \plus{} (3 \minus{} ab)^2}$
$ \equal{} \frac {36}{2(a \plus{} b \plus{} c) \plus{} 9 \minus{} 6bc \plus{} b^2c^2 \plus{} 9 \minus{} 6ca \plus{} c^2a^2 \plus{} 9 \minus{} 6ab \plus{} a^2b^2}$
$ \equal{} \frac {36}{2(a \plus{} b \plus{} c) \plus{} 27 \minus{} 6(ab \plus{} bc \plus{} ca) \plus{} (a^2b^2 \plus{} b^2c^2 \plus{} c^2a^2)} \equal{} \frac {36}{2(a \plus{} b \plus{} c) \plus{} 9 \plus{} (ab \plus{} bc \plus{} ca)^2 \minus{} 2abc(a \plus{} b \plus{} c)}$
$ \equal{} \frac {18}{a \plus{} b \plus{} c \plus{} 9 \minus{} abc(a \plus{} b \plus{} c)}$, and $ R > \equal{} \frac {6abc}{1 \plus{} 2abc}$, if and only if:
$ 3(1 \plus{} 2abc) > \equal{} abc(a \plus{} b \plus{} c \plus{} 9 \minus{} abc(a \plus{} b \plus{} c))$, letting $ s \equal{} a \plus{} b \plus{} c, p \equal{} abc$
$ 3(1 \plus{} 2p) > \equal{} p(s \plus{} 9 \minus{} sp)$ if and only if $ 3 \minus{} 3p > \equal{} ps \minus{} sp^2 \equal{} ps(1 \minus{} p)$ iff $ (1 \minus{} p)(3 \minus{} ps) > \equal{} 0$, this can be easily done by multiplying 2 inequalities:

$ \frac {\sqrt{(ab \plus{} bc \plus{} ca)^3}}{\sqrt{3^3}} > \equal{} abc$ by AM-GM inequality, this gives: $ 1 \minus{} p > \equal{} 0$
$ \frac {(ab \plus{} bc \plus{} ca)^2}{3} > \equal{} (ab)(bc) \plus{} (bc)(ca) \plus{} (ca)(ab) \equal{} abc(a \plus{} b \plus{} c)$, this is for the well known inequality: $ (p \plus{} q \plus{} r)^2 > \equal{} 3(pq \plus{} qr \plus{} rp)$, the last inequality is equivalent to $ 3 \minus{} ps > \equal{} 0$, the problem is done

Let $ P \equal{} abc$, and $ S \equal{} a \plus{} b \plus{} c$. Notice $ a^2(b \plus{} c) \equal{} a(ab \plus{} ac) \plus{} abc \minus{} abc \equal{} a(ab \plus{} bc \plus{} ac) \minus{} P \equal{} 3a \minus{} P$. Similarly,
\[ \frac1{1 \plus{} a^2(b \plus{} c)} \plus{} \frac 1{1 \plus{} b^2(a \plus{} c)} \plus{} \frac1{1 \plus{} c^2(a \plus{} b)} \equal{} \frac1{1 \plus{} 3a \minus{} P} \plus{} \frac {1}{1 \plus{} 3b \minus{} P} \plus{} \frac1{1 \plus{} 3c \minus{} P}
\]
Lemma A: $ P\leq1$. Consider the given expression $ ab \plus{} bc \plus{} ac \equal{} 3$. It states the arithmetic mean of $ ab$, $ bc$, and $ ac$ is 1. Since $ AM\geq GM$, $ 1\geq (xyz)^{\frac23}$, so $ P\leq1$, equality holding when $ a \equal{} b \equal{} c \equal{} 1$.

Lemma B: $ S\geq3$. Using Cauchy on $ < a,b,c >$ and $ < b,c,a >$, $ ab \plus{} bc \plus{} ac \leq a^2 \plus{} b^2 \plus{} c^2 \Rightarrow 3(ab \plus{} bc \plus{} ac) \leq a^2 \plus{} b^2 \plus{} c^2 \plus{} 2(ab \plus{} bc \plus{} ac) \Rightarrow 9\leq (a \plus{} b \plus{} c)^2 \Rightarrow 3 \leq S$.

Lemma C: $ SP \leq 3$. Let $ p(x) \equal{} x^3 \minus{} 3x^2 \plus{} kx \plus{} m$, and let $ a' \equal{} ab$, $ b' \equal{} bc$, and $ c' \equal{} ac$. Let these be roots of the polynomial $ p(x)$. Note $ 3 \equal{} a' \plus{} b' \plus{} c'$, which corresponds to the coefficient of $ x^2$ by Vieta's theorem. Also note $ k \equal{} a'b' \plus{} b'c' \plus{} a'c'$, and that this is equal to $ PS$ (they both reduce to $ a^2bc \plus{} ab^2c \plus{} abc^2$). $ m \equal{} \minus{} (a^2b^2c^2)$, so since $ P \leq 1$, $ \minus{} 1 \leq m \leq 0$.

Now suppose $ k > 3$. Then $ p(x)$ is always increasing, for $ p'(x) > 3x^2 \minus{} 6x \plus{} 3 \equal{} 3(x \minus{} 1)^2 \geq0$. But a cubic polynomial (or any function) that is always increasing can only have one real root of multiplicity 1. The other two roots are imaginary, but $ a',b',c' \in \Re$, so we have reached a contradiction. $ k\leq3$, equality holding at $ k \equal{} 3$ because $ a^2bc \plus{} ab^2c \plus{} abc^2 \equal{} 3$ when $ a \equal{} b \equal{} c \equal{} 1$, so $ SP \leq 3$.

Now we will prove the following inequality: $ 3(P \minus{} P^2) \plus{} (7P \minus{} 1)S \leq 18$

Let's suppose the contrary, that $ 3(P \minus{} P^2) \plus{} (7P \minus{} 1)S > 18$. Then the following lines are implied:
\[ 3(P \minus{} P^2) \plus{} 7PS \minus{} S > 18
\]

\[ 3(P \minus{} P^2) \plus{} 7PS \minus{} 18 > S
\]

\[ 3(P \minus{} P^2) \plus{} PS \plus{} 6PS \minus{} 18 > S
\]

\[ 3(P \minus{} P^2) \plus{} PS > S \quad \quad \quad (PS \leq 3)
\]

\[ 3P(1 \minus{} P) > S(1 \minus{} P)
\]

\[ 3P > S \quad \quad \quad (1 \minus{} P \,\texttt{is positive})
\]

\[ 3 > S
\]
We have reached a contradiction, because $ S \geq 3$ because of lemma B.

The rest is simply algebra (it may be a little easier to follow this logic backwards; I present it in this order for rigor):
\[ 3(P \minus{} P^2) \plus{} (7P \minus{} 1)S \leq 18
\]

\[ 9(P \minus{} P^2) \plus{} 21PS \minus{} 3S \leq 54
\]
Letting, $ \rho \equal{} 1 \minus{} P$, and noticing that remarkably $ P \minus{} P^2 \equal{} \rho \minus{} \rho^2$,
\[ 9(\rho \minus{} \rho^2) \plus{} 21S \minus{} 21 \rho S \minus{} 3S \leq 54
\]

\[ 9\rho \plus{} 18S \leq 9 \rho ^2 \plus{} 21\rho S \plus{} 54
\]

\[ 9\rho \plus{} 18S \minus{} 6\rho^2 \minus{} 12\rho S \minus{} 54 \leq 3\rho^2 \plus{} 9\rho S
\]

\[ 9\rho^2 \plus{} 18\rho S \minus{} 6\rho^3 \minus{} 12\rho^2S \minus{} 54\rho \plus{} 81\leq 3\rho^3 \plus{} 9\rho^2S \plus{} 81
\]

\[ (3\rho^2 \plus{} 6\rho S \plus{} 27)(3 \minus{} 2\rho) \leq 3(\rho^3 \plus{} 3\rho^2S \plus{} 27\rho \plus{} 27(1 \minus{} \rho))
\]

\[ (3\rho^2 \plus{} 6\rho S \plus{} 27)(3 \minus{} 2\rho) \leq 3(\rho \plus{} 3a)(\rho \plus{} 3b)(\rho \plus{} 3c)
\]

\[ \frac {(\rho \plus{} 3a)(\rho \plus{} 3b) \plus{} (\rho \plus{} 3a)(\rho \plus{} 3c) \plus{} (\rho \plus{} 3b)(\rho \plus{} 3c)}{(\rho \plus{} 3a)(\rho \plus{} 3b)(\rho \plus{} 3c)} \leq \frac3{3 \minus{} 2\rho}
\]

\[ \frac1{\rho \plus{} 3a} \plus{} \frac1{\rho \plus{} 3b} \plus{} \frac1{\rho \plus{} 3c} \leq \frac 3{3 \minus{} 2\rho}
\]
Substituting back $ 1 \minus{} \rho \equal{} P$, we get the desired inequality.
\[ \frac1{1 \plus{} 3a \minus{} P} \plus{} \frac 1{1 \plus{} 3b \minus{} P} \plus{} \frac1{1 \plus{} 3c \minus{} P}\leq \frac 3{1 \plus{} 2P}
\]


calculus
integration
trigonometry
calculus computations

Compute:

$ \int_{0}^{1} \sqrt{\frac{1\minus{}x}{1\plus{}x}}dx$

[hide="HINT:"]Let $ x\equal{} sin(t)$[/hide]

better     let  $ x \equal{} cost$
    maybe   concluse   that   $ A \equal{}1  .$

[b]Hong Quy[/b] is right, better get $ x \equal{} \cos t$ than $ x \equal{} \sin t$.

then we'll have $ \text{d}x \equal{} \minus{} \sin t\text{d}t$ and the integral will be:

$ I \equal{} \int^{0}_{\frac {\pi}{2}}\sqrt {\frac {1 \minus{} \cos t}{1 \plus{} \cos t}}( \minus{} \sin t)\text{d}t$

$ \equal{} \int^{\frac {\pi}{2}}_{0}\sqrt {\frac {2\sin^{2}\left(\frac {t}{2}\right)}{2\cos^{2}\left(\frac {t}{2}\right)}}\times\sin t\text{d}t$

$ \equal{} \int^{\frac {\pi}{2}}_{0}\frac {\sin\left(\frac {t}{2}\right)}{\cos\left(\frac {t}{2}\right)}\times2\sin \left(\frac {t}{2}\right)\cos \left(\frac {t}{2}\right)\text{d}t$

$ \equal{} 2\int^{\frac {\pi}{2}}_{0}\sin^{2}\left(\frac {t}{2}\right)\text{d}t$

$ \equal{} \int^{\frac {\pi}{2}}_{0}\left(1 \minus{} \cos t\right)\text{d}t \equal{} \left[t \minus{} \sin t\right]^{\frac {\pi}{2}}_{0} \equal{} \frac {\pi}{2}\minus{}1$

 :)

edit: sorry for my miscalculation too! :oops:

first multiply it by $ \sqrt {\frac {1 \minus{} x}{1 \minus{} x}}$

so the integral would be:
$ \int_0^1 \frac {1 \minus{} x}{\sqrt {1 \minus{} x^2}}dx$
let $ x \equal{} \sin{\theta} \rightarrow dx \equal{} \cos{\theta}d{\theta}$

so the integral would be:
$ \int_0^{\frac {\pi}{2}} \frac {1}{\cos{\theta}} {\cos{\theta}} d{\theta} \minus{} \int_0^{\frac {\pi}{2}} \tan{\theta} \cos{\theta}d{\theta}$
$ \int_0^{\frac {\pi}{2}} {\theta} d{\theta} \minus{} \int_0^{\frac {\pi}{2}} \sin{\theta}d{\theta}$
$ \equal{} \frac {\pi}{2} \minus{} 1$

[edit:sorry for my mistake :blush: ]

[quote="mhar_teens"]first multiply it by $ \sqrt {\frac {1 \minus{} x}{1 \minus{} x}}$

so the integral would be:
$ \int_0^1 \frac {1 \minus{} x}{\sqrt {1 \minus{} x^2}}dx$
let $ x \equal{} \sin{\theta} \rightarrow dx \equal{} \cos{\theta}d{\theta}$

so the integral would be:
$ \int_0^{\frac {\pi}{2}} \cos^2{\theta}d{\theta} \minus{} \int_0^{\frac {\pi}{2}} \tan{\theta} \cos{\theta}d{\theta}$
$ \int_0^{\frac {\pi}{2}} \frac {\cos{2\theta} \plus{} 1}{2} d{\theta} \minus{} \int_0^{\frac {\pi}{2}} \sin{\theta}d{\theta}$
$ \equal{} \frac {\pi}{4} \minus{} 1$[/quote]

theres a little problem in your solution, after taking $ x \equal{} \sin{\theta} \rightarrow dx \equal{} \cos{\theta}d{\theta}$

we'll have:

$ \int^{1}_{0}\frac{1\minus{}x}{\sqrt{1\minus{}x^{2}}}dx\equal{}\int^{\frac{\pi}{2}}_{0}\frac{1\minus{}\sin\theta}{\sin\theta}\times\cos\theta d\theta$
$ \equal{}\int^{\frac{\pi}{2}}_{0}(1\minus{}\sin\theta)\cot\theta d\theta\equal{}\int^{\frac{\pi}{2}}_{0}\cot\theta d\theta\plus{}\int^{\frac{\pi}{2}}_{0}\sin\theta\cot\theta d\theta\equal{}\cdots$

(after putting $ x\equal{}\cos\,2\theta$)
$ \int_0^1\sqrt{\;\dfrac{1\minus{}x}{1\plus{}x}\;}\;\textbf dx\equal{}2\int_0^{\dfrac{\pi}{4}}\dfrac{\sin\,\theta}{\cos\,\theta}\sin\,2\theta\;\textbf d\theta$

$ \equal{}2\int_0^{\dfrac{\pi}{4}}\left(1\minus{}\cos\,2\theta\right)\;\textbf d\theta\equal{}2\left(\theta\minus{}\dfrac{1}{2}\sin\,2\theta\right)\;\bigg|_0^{\dfrac{\pi}{4}}\equal{}\boxed{\dfrac{\pi}{2}\minus{}1}$

Lots of good ideas here, but lots of mistakes. I like mhar_teens' idea the best, though Misan is the only one with the right answer.

Pretty much the same solution as mhar_teens.

Multiply the integrand by $ \sqrt {\frac {1 - x}{1 - x}}$ to obtain:
$ \int_0^1 \frac {1 - x}{\sqrt {1 - x^2}} \; dx = \int_0^1 \left(\frac {1}{\sqrt {1 - x^2}} - \frac {x}{\sqrt {1 - x^2}}\right) dx$

$ = (\arcsin {x} + \sqrt {1 - x^2}) \Big|_0^1$

$ = \dfrac{\pi}{2} - 1$

[quote="gauss202"]Lots of good ideas here, but lots of mistakes. I like mhar_teens' idea the best, though Misan is the only one with the right answer.
[/quote]

thanks..
sorry for my little mistake


linear algebra
matrix
invariant
algebra
polynomial
linear algebra unsolved

$M \in M_n(K)$ a matrix in an arbitrary field. $C$ the subspace of matrix that commute with $M$.
Prove that $dim(C) \leq (n-1)^2+1$ if $dim(C) < n^2$.

The fact that $dimC(A)$ is the sum of squares of some natural numbers whose sum is $n$ solves the problem very rapidly. Do you have a nicer solution?

I don't remember this proof with the sum of squares.
Anyway, I think that if $P_1,P_2,..,P_r$ are the invariant polynomial of $u$, then the dimension of the commutant is exactly $\sum_1^r dim(Ker(P_i(u)))$. This solves quite quickly the pb.

i dont think that Dim(C) is always sum  of squars ... that's only when the matrix is diagonalzable !!! in general we dont have any expression of the dimenssion ? , if what say Harazi is always true so it will be clear why dim(C) >=n  witch is not easy to prove !! please alekk, why the dimension is that sum ?

I have two reasons to contradict you: any natural number is a sum of 4 squares, so your first remark has some serious mistakes. Secondly, the proposition remains true even for non-diagonalisable matrices.

:lol:  i know also that the set of positive integers which cannot be represented as a sum of distinct squars is finite ( try to prove it  :P )... sorry ,i wasn't concentrated , i see why it's always a some of squars ... so it's easy to show that dim(C(A))>=n  for every A  ( my teacher told me that not so easy to prove it , that's why i didn't think about  :D )
But alekk seems forget to put the squars (Sum(Dim( Ker(P_i(u)))²),i=1...r) ! so are meaning by P_i the factors of the minimal polynome of A which are a power of an irredictible polynome ?

no, I do not forget to put the squares, and the $P_i$ are the invariant factors (http://mathworld.wolfram.com/RationalCanonicalForm.html).

think you alekk but can you gave me more resource about this ...i want to see why the dimension of the commutant is that sum !! i have made mistakes when i thought to prove it  :oops:


This exam was much like IITJEE in terms of pattern, matks etc but not difficulty level. Use this topic only to declare AIEEE marks (expectations).

[hide="And..."]Can anyone tell me where I can find [b]reliable[/b] solutions?[/hide]

went well except for damn silly errrors and wrong qs in maths (AR screwed me )

chem-124
maths-109
phy-144 (yippy me gettin full :D  )

total -377       wat rank can i expect ??

chem - 134
math - 123
phys - 124

t- 381

why do i fk up physics always even though I am far far better in it than the other two :(

[quote]
Can anyone tell me where I can find reliable solutions?
[/quote]

I think resonance would be most reliable.
[hide]
and i am scoring most acc to resonance :D
[/hide]

maths 110
phy 118
chem 132
total 360

no solution is perfectly reliable ( aakash is better )

im gettin xtra 10 in maths so make it to 387 :D

i didnt knew aakash iitjee is so famous....

u kno frm delhi branch my score is 3rd highest among aakash regulars (i m in medical batch :rotfl: )

and also,i have been giving tests here 4 past 3 years and the errors in answer sheets have increased over the years.... :o  :rotfl:

Aakash is getting more and more famous each year. Fees increased by over 50% this year in Lucknow. Expected to increase much much more after my batch passes out :P There are some serious brainos who get 75-90% on tests  at >IIT-JEE level
Plus there are 2 geniuses in FIIT-JEE classroom who are sure to be in top 100.


Give an example of a problem that you can use Cauchy's Inequality in.

[hide]$\left(\sum\limits_{k=1}^{n}a_{k} b_{k}\right)^{2}\le \left(\sum\limits_{k=1}^{n}a_{k}^{2}\right)\left(\limits\sum_{k=1}^{n}b_{k}^{2}\right)$[/hide]

Prove:

$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \ge \frac{3}{2}$.

It's an extension of Cauchy, but its basically the same thing.


Let $ \omega$ is the circle. Let $ p_n$ is the perimeter of inscribed n-gon of $ \omega$, $ P_n$ is the perimeter of the circumscribed n-gon of $ \omega$. Prove that $ \frac{2}{3} p_n \plus{} \frac{1}{3} P_n$ is greater than lenght of the circle($ 2\pi R$)

I think the problem is wrong. Counterexample: Let $ w$ be a unit circle, $ n\equal{}3$, the circumscribed triangle is regular, and the inscribed triangle is very small. So $ P_n\equal{}6\sqrt3$ and $ p_n$ is arbitrarily small. Hence $ \frac23pn\plus{}\frac13P_n$ approaches $ 2\sqrt3\approx 3,4$. On the other hand, the circumference of the circle is $ 2\pi\approx 6,28$.


Can the number A consisting of 600 sixes and some zeros be a square?

No.

If the number ends in $ 2k+1$ zeros, then it cannot be square (you'll need an additional factor of 5, and you won't be getting it when the last nonzero digit's a 6).

If the number ends in $ 2k$ zeros, then you can effectively factor out $ (10^{k})^{2}$. However, the remaining number will either end in 06 or 66. This means it'll be equivalent to 2 mod 4, so it cannot be square either.


No, It's:
$0<x^2y+y^2z+z^2x<\frac{4}{27}$

I asked about it few months ago [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=x%5E2y%2By%5E2z%2Bz%5E2x&t=24113]here[/url] ;)


search
function
algebra
polynomial
number theory open
number theory

A friend and I were working on a project when a question arose and we have spent the last 2 weeks trying to figure out if it is possible. Given A, B, C, and D which are all integers 0-1000 is there an equation that will result in A, Z, and D where Z is also a integer 0 to 1000. But the tough part is, is there an equation that can take A,Z, and D and return the original values of A,B,C, and D. It is driving me crazy and any help would be appreciated. Thanks

You search if you can find an equation giving Z in function of A, B, C and D and an other equation giving A, B, C, D knowing A, Z and D ?
A and D are fixed, Z can take $1000$ values but (B,C) can take $1000^2$ values so it is not possible to find $(B,C)$ with $Z<1000$ I think

Hum... I had a feeling that was the case I just wasn't sure if there was something I was missing. So basically, for any sequence of numbers that is N long, there is no way that it can be reduced to another sequence of numbers that is n-1 long if all the values are in the same range. Correct?

Yes I guess
You could find something which works if you can have some properties on numbers A, B, ...
For an example if you know that A, B, C, D are (it's just an example) are $f(1), f(2), f(3), f(4)$ with $f$ a polynomial of degree $2$ you can find A, B, C, D from the three coefficient of $f$ so you can compress your $4$ numbers in $3$ numbers but you see it's very restrictive, have you some properties on A, B, C, D ?

Unfortunatly A,B,C, and D can be any possible combination of the numbers 0-1000


geometry calculus vector Stanford college Support MATHCOUNTS

I would like to address and discuss the fact that many talented students who get ahead of their math program end up being forced to take classes that do not challenge them, or especially, teach them any new material, whatsoever.

So, how does your school and situation fit into this? Have you been bored in math class all your life? Does your school actually offer good options? Have they let you get ahead and do advanced/college/independent study? Or, have you had some good and some bad?

I started thinking about this today while sitting in class "learning" about Taylor Series. I realize that, in fact, I have never actually learned much in a math class (except maybe Geometry), and I am wondering how others' experiences have been.

I've learned absolutely nothing. I have to take math classes that I took 3 years ago and have complete mastery over. It's a complete waste of time of a period in school and the homework hours at home.

[quote="ZetaX"]You can learn a lot in math classes reading Neukirch or Hatcher...[/quote]
 :wink:

This is a good example of how public school sucks.

My math class doesn't challenge me at all, and I scarcely learn anything, but I can't say that I never learn anything, because sometimes a small hole will get filled. My main disapprovement with my class is simply the level of difficult of stuff we do; most of the problems are brute force-like. I would learn more if we focused more on how to [i]think[/i] about problems and find elegant solutions, though I don't think that is really a skill that can be easily taught.

[quote="mathnerd314"]I am taking Precalculus, and it's basically been a review of Algebra II. The really interesting topics like vectors are relegated to next year, and we've skipped over a lot of the book.

What disappoints me is that there is virtually no element of proof in my class. The teacher just gives you a formula and tells you to memorize it, without showing where it came from. Instead, the teacher throws a bunch of examples at you, and you have to copy every single one down in your notes or you lose points on the notebook check.[/quote]
You have one of those method nazi people also?

I hate those. Especially when they teach the wrong method.

But seriously guys, go steal a car and use that to pay for Exeter or Andover or something.

[quote="calc rulz"]Right now I decided to start reading [i]Modern Higher Algebra[/i] by E. Artin (which I had been reading at home) during BC Calculus class  :lol:[/quote]

Might wanna go direct with Lang, if you can handle it :D

[quote="calc rulz"]
So, how does your school and situation fit into this? Have you been bored in math class all your life? Does your school actually offer good options? Have they let you get ahead and do advanced/college/independent study? Or, have you had some good and some bad?
[/quote]

My school (Henry M. Gunn in Palo Alto, California) has in my opinion  one of the better math curriculum in the country (add in public school to be more accurate). I'm only going to talk about the highest course levels. Algebra II/Geometry is taught in one year in ninth grade, with ample elements of proof I believe. The next year is spent on Trig/Analytical Geometry (this is the only class I took at my high school). Then next in the sequence is this course called "Analysis". It has nothing to do with the higher level real/complex analysis courses. Rather, combinatorics, introduction to linear algebra, and calculus A is taught. Then it goes on to BC Calculus. One thing of note here is that I think LOTS of students are encouraged to skip ahead and take classes 2 to 3 years beyond their age level.
Currently, I am a junior and I am taking the 50H sequence at Stanford with support from my math department. I am getting independent study credits. Most kids who are ahead in math go on to take either the 50 or 50H sequence at Stanford once they are done with BC Calc.

Also, next year is going to be the first year that we open an after school problem solving class. Students enrolled DO get credits. I'll be the teaching assistant next year, and no one really knows exactly what it will be like yet, but.... :D

The class is not challenging at all. Since my teacher doesn't care, I just study AoPS books, and do practice AMC's in class (we have blocked schedule, so I can finish a whole test in one class period).

I hate the homework (btw, I have a 76% in hw  :P ), because we learn like one thing per class, and the hw is just the same exercises 50 times over.

I also don't think my school district encourages people to advance in math classes. I've tried to skip precalc next year and take calc bc, but apparently precalc is a district requirement, and it falls under a honors class, so there is no possibility of testing out. Out of about 380+ freshman at my school, only about half a dozen are a year advanced in math, and I only know of one freshman in the entire district taking precalc this year (not to say there aren't more, but I don't know of any). (He was also the state mathcounts winner last year).

Not to say my school sucks, it just does in math.

I'm in grade 7, in a grade 10 academic class, and half the kids can't do the opposite theroum :mad:  :o . Its also anying that we have pretty much done the same stuff all semester long. in other words: every thing I learn, I learn at home.

I've learned a lot...I must be the only one in this thread that is still learning stuff in Precal or AP Calc..

[quote="mgao"]Like I ever learn anything in Geometry... I just sit there.  I'm sure all of you guys took "geometry" in 6th grade and found that even "INTRO" to geometry at AoPS was harder...

[hide="Anecdote"]
Once we were having a class on 3d solids.  The teacher "taught" us the formula for a pyramid/cone.  I decided to spice up the lesson a little bit.

ME:"Prove the formula!"
TEACHER:"Oh sure I know how to prove it!"

By then I was actually pretty surprised that the teacher knew how to... hopes quickly went down.

"I've done it empirically but I'm sure there's a more mathematical way..."

ME: *disappointed*
[/hide][/quote]

small side note... isn't the formula for a 3D solid(tetrahedron) one of hilbert's problems (ie, proving the formula independent of integral calculus)

[quote="infinity4ever"]Well, compared to a lot of other countries who still don't have flushable toilets in nearly ever house, I guess you could say America is a pretty rich country.  I agree though, America isn't spending nearly enough money educating its young scholars.  In my own opinion, America is spending way too much money on war. War gets you nowhere, yet America still plunders away on it![/quote]

america is doing a lot better job in educational money than other countries, i mean look at china, their money in the bank are higher than the U.S., they just keep saving it and saving it, and some of it disappear mysteriously into government officials bank accounts, and you have to pay several hundreds dollars, and several thousand for middle school, and tens of thousands for high school a year for education where some people can barely afford to eat meat once every half years.

Just a reality check: the standard (if somewhat flawed) measure of national wealth is the GDP or GDP per capita.  The GDP of the United States is about 13 trillion USD, while the (adjusted; goods and services are valued at US rates) GDP of China is around 10 trillion USD.  This puts the Chinese per-capita GDP at just under 8000 USD.  The US per-capita GDP is 44000 USD, or more that 5 times larger.  (The American national debt is a shade over three times larger than the Chinese national debt; unlike for households, debt is not the opposite of wealth.)

yes i know debt does not equal to wealth, but since china has all the money stored in the bank, why don't you put it to use for the people? it is people's "republic" of china.

By "people" they mean the government. And the government officials are fairly well off there, I'd say?

The public school mathematics education in the US is horrendous, simply put. Personally, I think this is due to a shortage of good teachers and good books.

[hide="teachers"]For me, I like to make sure I get the core, basic concepts down, and from there it is easy for me to develop new concepts, since they extend and re-enforce the basic concepts. However, the skill distribution of teachers is exactly the opposite: the more basic the principle, the worse the teacher that teaches it. By worse, I don't mean math PhD's in college vs. Masters in HS; I mean compared to that level, as in an algebra ii teacher who understands quadratic equations only at a mechanical level. Money alone isn't going to fix it.[/hide]
[hide="books"]And then there is the fact that math textbooks are progressively getting worse and worse, and that attempting to interpret them without a competent teacher becomes harder and harder. I recently helped a middle school kid who was having trouble with Algebra i. He came to me with his textbook on the first day and handed it to me. Integrated Elementary Mathematics, or something along those lines, and started flipping through it out of curiosity. It was full of crap.

Half of it was statistics. [i]Why [/i]would someone want to know about Pearson correlation coefficients and paired T tests in middle school? And it didn't even give the full picture, for Pearson it was just like, 0 if no trend, 1 if strong positive trend, -1 if strong negative trend. And after all that getting confused over out-of-place statistics, they don't even know the formula for calculating the coefficients. Then they have this thing about "the power function". And the "exponential function" (arbitrary base) and the "logarithmic function". The poor little critters don't even understand exponents at this point in time because they've just been bashed with all this bogus statistics stuff. What's worse is that they failed to properly define a function before going about this. Long story short, it was horrible.

Needless to say, I pushed it away and pulled out my AoPS 1 book, sans 1/2-page pictures and pages of calculator directions, and just used that, and it worked pretty well. I don't get what's with this new trend of terrible integrated math books; rather than alleviating the root problem (teachers), they manage to create a second part to the problem.[/hide]
Fortunately, America has many opportunities that are not as readily available to students in other countries, such as math clubs and teams, math contests, summer programs, programs run by colleges (for those prepared to pay) and such. There are also great resources on the internet  :) As a result, those who manage to maintain an interest in mathematics and the sciences are still able to develop their skills. However, this still leaves the uninspired to bear the loss.

As for me, I am going to be a 10th grader next year and will take Calc BC, after skipping alg 2 and precalc (the geometry teacher was better than the precalc teacher, so I took geometry in 9th grade and reviewed precalc on my own). I haven't decided where I'll go from there, probably AP Statistics in junior year, and figure something out for senior year.

Also about the teachers: they should not only delve into the textbook more, they should expand and appy it in real life scenerios. For example, in sixth grade (at the end of the year), my teacher gave us stuff that we'd see if Alg I and II and taught us basic concepts (two equation systems etc) and how to solve them. He also gave us AoPS-difficulty questions about our past math year.

My 7th and 8th grade math teachers just stuck to the curriculm. The 8th grade math teacher crammed everything in at the last minute (like the day before finals). We were supposed to learn the binomial theorem but we never got to that. And on the last day of school we just did some origami stuff  :mad: .

[quote="7h3.D3m0n.117"]And on the last day of school we just did some origami stuff  :mad: .[/quote]

Oragami in math class, dude you gotta be kidding me.

Um, it was the last day of school.  Last days of school are [i]never[/i] for work.  More importantly, there's lots of mathematics behind origami.  The [url=http://mathworld.wolfram.com/Origami.html]mathworld article on origami[/url] has some nice pictures and also an excellent bibliography.

[quote="JBL"]Last days of school are [i]never[/i] for work.[/quote]

Then why do we even bother to go?

[quote="JBL"]More importantly, there's lots of mathematics behind origami.[/quote]

Now that I think about it you're right, but still, all the teacher ever did was hand out instruction sheets and say "get to work"; it'd be more math-related if she explained the math connections.

My original school math program was wonderful for K-5. But as soon as I went to 6th, it was terrible. Our teacher never really explained anything. She never let us do anything fun, like origami. That is why I am home-schooled. (It was the same for my brother)

actually, i found that it was the other way around. For K - 5, im all like math is kinda easy but kinda boring... but then in 6th grade i happened to meet a good math teacher and now i consider myself to have an extreme interest in math

[quote="7h3.D3m0n.117"][quote="JBL"]Last days of school are [i]never[/i] for work.[/quote]

Then why do we even bother to go?

[/quote]
You know, then they would call the 2nd to last school day "last day", and we wouldn't bother to go there on that day either.  Nor the 3rd.  Nor the 179th.

[quote="7h3.D3m0n.117"][quote="JBL"]Last days of school are [i]never[/i] for work.[/quote]

Then why do we even bother to go?
[/quote]

if you dont want to go, then don't. what's stopping you? as if you're going to get in trouble for one little absence.

A friend of mine made this point the other day regarding investing in education. The people who have the power to do this mostly fall in the categories of

(1) politicians, or
(2) businesses.

For the former, they'll most likely be out of office in 20 years when the investment will actually pay off, so their interest in this is out of the goodness of their hearts. For businesses, the same situation occurs, as in 20 years the business will probably be gone. In the rare case of a very successful business, they do invest in education and probably do the most out of anyone in this country (my opinion, I don't have the facts). So you can see why the education system might not get the attention it deserves.

The good thing is that after middle/high school you can go to private colleges who can actually fund good programs.

Learn?  In school?  Where I come from, that seems to present an oxymoron.

Actually, I don't credit the online high school I'm going to enough.  I learned things from Apex Learning through Geometry and Algebra 2 that Mathcounts, Aops, my mom, and myself failed to teach me.  Middle school and under...nope can't think of anything I learned there.

I personally hated math until 6th grade.  Then I met Mathcounts.


in the iteval $[0 , 1 ]$ we have $n+1$ points such that $ x_0=0 , x_1 , x_2 , ... , x_n=1 $ and let $A$ be the squence of these points distences . each number in $A$ beside of $1$ has come at least two times.
show that all of the $x_i$s are rational.

Here it is:

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=algebra&t=4821[/url]

and here's another one, with a similar setting:

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=fantastic+algebra&t=4845[/url]


quadratics
probability
USAMTS
algebra
quadratic formula

Given the sum and product of two numbers, find a formula for these numbers.  Then, prove that it works.

By "prove that it works", I mean to plug it back in and test it.  For example, to prove that the quadratic formula is correct, you'd show that $a\left(\frac{-b\pm\sqrt{b^2-4ac}}{2a}\right)^2+b\left(\frac{-b\pm\sqrt{b^2-4ac}}{2a}\right)+c=0$.

[quote="José"]Well, this problem is well-known if I'm not mistaken... for example, try with $35$ and $300$[/quote]

[hide]Is it 15 and 20?[/hide]

I must've been unclear.  What I meant was this:

If the sum of two numbers is $S$ their product is $P$, find a formula for those two numbers in terms of $S$ and $P$.  Then, prove that it works.

to prove that something works, you don't need to check it back in, you just have to make sure that all your steps are correct, like in a probability problem for the usamts, it says to "prove your answer" you cannot check it, but try to have all correct steps

[hide]
$x+y = S$
$xy = P$

$x = S - y$

Put this into the second equation

$(S-y)y = P$

$y = \displaystyle \frac{S\pm \sqrt{S^2 -4P}}{2}$

$x$ is equal to $S - y$.
[/hide]


superior algebra
superior algebra solved

1)Let A,B be M_n(R) so that A^2B^2=B^2-A^2. Prove that (A+In)B^2A=A(A+In)B^2.
(GM 7/8-2003)

10x!

It is easy to see that this staytment we can prove with assumption, that
A,B,A-I are invertible matrices. If staytment holds for all invertible matrices than it holds for all matrices (this is a system of polynomyal equations on coefficients of the matrices).

In our assumption
A^2B^2=B^2-A^2  <=>  (A+I)B^2=-(A-I)^{-1}A^2. So
(A+I)B^2A=A(A+I)B^2 <=> -(A-I)^{-1}A^3=-A(A-I)^{-1}A^2 <=>
(A-I)^{-1}A=A(A-I)^{-1} <=> (A^{-1}(A-I))^{-1}=((A-I)A^{-1})^{-1}
<=>(I-A^{-1})^{-1}=(I-A^{-1})^{-1}.

I don't understand why it can be assumed that A,B,A+I_n are invertible.

I proved that A+I_n is invertible. The condition in the problem rewrites:
(A 2-I_n)(B 2 +I_n)=I_n, so A 2 -I_2 is invertible, so A \pm I_n are invertible.

[quote="amfulger"]I don't understand why it can be assumed that A,B,A+I_n are invertible.
[/quote]

Because the set GL_n(K) of invertible matrices is dense in the set of matrices M_n(K) with K= R or C


[quote="myc"]I was working on proportions and I got stuck on a problem.
   "Given that x is directly proportional to y and to z and is inversly proportional to w, and that x=4 when (w,y,z) = (6,8,5), what is x when (w,y,z) = (4,10,9)?"

the solution the book came up with is:
xw/yz = 4*6/8*5 =3/5    x=3yz/5w= 27/2
   i don't understand why they did that.  if x and w are inversly proportional, and x=4 when w=6 , then xw=24   and if w=4   then 4x=24  x=6.  I figured i must be wrong, because then i wouldn't need y and z at all.[/quote]
[hide]
Since x and w are inversely proportional, wx is a constant. So they're together. Since y and z are directly proportional to x, x/y and x/z are constants, so then xw/yz is a constant. So then we have if (w,x,y,z)=(6,4,8,5), xw/yz=3/5. So then since it is constant, we have the three values, so 3/5=4x/90, 3/5=2x/45, $\boxed{x=\frac{27}{2}}$.[/hide]

[hide]I just used some variation and got $\frac{27}{2}$.[/hide]

[quote="myc"]why did multiply y and z?  x/y * x/z = x^2/yz , so shouldn't it be xw/x^2/yz = wyz/x^2?    :?

sry, I haven't mastered latex yet, though I'm working on it.
Thanks
 :P[/quote]

You multiply y and z because they are both directly proportional to x. So you have x/y and x/z as constants, so x/yz is also constant. I don't think you necessarily need $x^2$..

[hide]
$x = k \frac{yz}{w} \\\frac{xw}{yz} = k \\\frac{4*6}{8*5} = k \\\frac35 = k \\x=\frac{3*10*9}{4*5} \\x=\frac{27}2$
Where $k$ is the constant of proportionality.
[/hide]


counting
distinguishability
algebra
binomial theorem

Find S if :

S = 1 + 3 * (1/4) + 6 * (1/4)^2 + 10 * (1/4)^3 + .......... + ((n(n+1))/2)* (1/4)^(n-1) + ..........

I have  been trying to solve this problem for a while, but have made no progress.  Any help would be appreciated.

I have the solution, I just do not know how to get it.
[hide="the solution"]64/27[/hide]

Apply the [url=http://en.wikipedia.org/wiki/Binomial_theorem]generalized binomial theorem[/url] to $ \left( 1 \minus{} \frac{1}{4} \right)^{\minus{}3}$.  Equivalently, if you have a good intuitive grasp of series multiplication, what is

$ \left( 1 \plus{} x \plus{} x^2 \plus{} ... \right)^3$?

Let $ f(n) \equal{} (an^2 \plus{} bn \plus{} c)\left(\frac {1}{4}\right)^{n \minus{} 1}$, determine $ a,\ b,\ c$ such that $ f(k) \minus{} f(k \plus{} 1) \equal{} \frac {k(k \plus{} 1)}{2}\left(\frac {1}{4}\right)^{k \minus{} 1}$,

yielding $ a \equal{} \frac {2}{3},\ b \equal{} \frac {10}{9},\ c \equal{} \frac {16}{27}$.

Use Telescoping Method,

$ \sum_{k \equal{} 1}^n \frac {k(k \plus{} 1)}{2}\left(\frac {1}{4}\right)^{k \minus{} 1}$

$ \equal{} \sum_{k \equal{} 1}^n \{f(k) \minus{} f(k \plus{} 1)\} \equal{} f(1) \minus{} f(n \plus{} 1)\longrightarrow a \plus{} b \plus{} c \equal{} \boxed{\frac {64}{27}}\ (n\to\infty)$.

[quote="t0rajir0u"]Equivalently, if you have a good intuitive grasp of series multiplication, what is

$ \left( 1 \plus{} x \plus{} x^2 \plus{} ... \right)^3$?[/quote]

$ \left( 1 \plus{} x \plus{} x^2 \plus{} ... \right)\left( 1 \plus{} x \plus{} x^2 \plus{} ... \right)\left( 1 \plus{} x \plus{} x^2 \plus{} ... \right) \implies$ For any $ x^n$, the coefficient is determined by the solutions to $ a \plus{} b \plus{} c \equal{} n$ (do you see why?), where $ a, b, c$ are nonnegative integers. This is just $ \binom{n \plus{} 2}{2}$, so $ \left( 1 \plus{} x \plus{} x^2 \plus{} ... \right)^3 \equal{} 1 \plus{} 3x \plus{} 6x^2 \plus{} 10x^3 \plus{} \dots$.

Letting $ x \equal{} \frac 14$, we have $ 1 \plus{} 3 \cdot \frac 14 \plus{} 6 \cdot \frac 1{4^2} \plus{} \dots \equal{} \left(1 \plus{} \frac 14 \plus{} \frac 1{4^2} \plus{} \dots\right)^3 \equal{} \left(\frac 43\right)^3 \equal{} \frac {64}{27}$.

[color=darkred]Edit: Whoops, confused indistinguishable a,b,c with distinguishable :blush: [/color]

I understand why [quote]For any x^n, the coefficient is determined by the solutions to a+b+c = n[/quote]
but i do not understand how to determine the number of solutions.  Of course, it's easy enough to find with small values of n, but why does the number of solutions equal (x-1      n).......and yes i know what the binomial theorem is, just do not understand [i]how[/i] coefficients are determined.

Sorry....I need to learn how to format soon....and thanks to all the repliers.

Yongyi doesn't have quite the right expression.  The solutions to $ a \plus{} b \plus{} c \equal{} n$ in non-negative integers can be put in bijection with the number of ways to place $ 2$ dividers among $ n$ objects.  Those dividers partition the $ n$ objects into three sets with $ a, b$, and $ c$ elements respectively.  And the number of ways to place the dividers is just the number of ways to pick $ 2$ elements out of $ n\plus{}2$ elements to be the dividers, which is just $ {n\plus{}2 \choose 2}$.  (We index starting from $ 0$.)

Generally, there are $ {n\plus{}k\minus{}1 \choose k\minus{}1}$ solutions to $ a_1 \plus{} a_2 \plus{} ... \plus{} a_k \equal{} n$ in non-negative integers.  This is reflected algebraically in an identity you can prove about negative binomial coefficients.

Here is an interesting article.[url]http://www.mathlinks.ro/viewtopic.php?t=213408[/url]


AMC USA(J)MO USAMO calculus integration function factorial

Show that for any non-negative reals x, y, [5x] + [5y] ≥ [3x+y] + [x+3y]. Hence or otherwise show that (5a)! (5b)!/( a! b! (3a+b)! (a+3b)! ) is integral for any positive integers a, b.

Note : [x] is the step function.

By "step function," do you mean "floor" a.k.a. greatest integer :le: x?

Clearly, if we increase x or y by 1, we increase the left side by 5 and the right side by 4.  Thus, our worst-case scenario is when
0 :le: x, y < 1.

Assume wlog that x :ge: y.  Then we can do a quick case analysis (I couldn't think of a better way to do this, but when you have so much time, it shouldn't matter):

x and y fall into intervals of the type m/5 :le: x < (m + 1)/5,
n/5 :le: y < (n + 1)/5, where m, n are members of {0, 1, 2, 3, 4}.  So then we just have to check the various cases.  It's most easily done graphically, and is left as an exercize for the reader ;).  Basically, you just show that the minimum for the left side in any given interval space is :ge: m + n.  Then you show the maximum for the right side is
< m + n + 1 or :le: m + n.

yeah i found that, and after that we let x=a/p^r and y=b/p^r, where p is any prime and r any natural number to see that the factorial thing is an integer. ugh i didn't like case analysis though. there must be a better way!


find the number of binary n word with exactly m 01 blocks. there is a nice solution at engel, any other way?

[hide]
Case 1: the sequence starts with 0.

There are $n-1$ spaces between each digit. Consider the number of "transtions" which is either where a sequence of one or more ones ends and a sequence of one or more zeroes begin, OR where a sequence of one or more zeroes ends and a sequence of one or more ones begin. Every other transition will make a "01" block, so if there are $2m$ or $2m-1$ transitions there will be $m$ "01" blocks. The total number of binary sequences is $\binom{n-1}{2m}+\binom{n-1}{2m-1}=\binom{n}{2m}$

Case 2: the sequence starts with 1.

If there are $2m$ or $2m+1$ transitions there will be $m$ "01" blocks. That makes $\binom{n-1}{2m}+\binom{n-1}{2m+1}=\binom{n}{2m+1}$

The total of both cases is $\binom{n}{2m}+\binom{n}{2m+1}=\boxed{\binom{n+1}{2m+1}}$
[/hide]


For every integer $ n\geq 1$, let $ a_n\equal{}\frac{(\minus{}1)^{n\plus{}1}}{n}$. Prove that there exists a permutation $ (b_i)$ of the $ (a_i)$ such that the infinite series $ b_1\plus{}b_2\plus{}...\plus{}b_n$ converges to $ \pi$.

it's Riemann series theorem
http://en.wikipedia.org/wiki/Riemann_series_theorem


Hai folks, I would like to ask you people some advice regarding -- concentration.

Can you suggest some advice/tips to improve my concentration drastically (or at least to some extent), say- [I mean a good amount of change is visible over a period of a month]? I can concentrate well, but sometimes I tend to lose concentration/attention; I don't want to lose my concentration/attention, and want to have  sustained focus on the task at hand.

Since I believe many people in this community can concentrate much better than me, I want suggestions/tips/advice to improve concentration over a period of time.

Is it possible to concentrate on a task for more than 3 hours (Assuming we like and also dislike certain aspects of the task)?

I thank you all in advance.

I can only recommend three things:

Make sure there are no distractions around you and everything is extremely quiet (this means no "on" TVs, siblings, computers)

Time yourself. Make a goal as to how much time you'll take to finish a worksheet. Try to achieve it! This may seem ridiculous but trust me, it really helps.

Self-control: don't purposely let yourself be distracted. It might be tempting but every bit of practice you can get is necessary.

[u]Throwing out a few things, in no particular order:[/u]

Everyone works best at a certain time of day. My best concentration comes in the dead of night, as does my dad's. My mom, on the other hand, works best really early when other people have not yet woken up. Unfortunately, I'm not a morning person, but when I do wake up early, I find myself very concentrated, and draw satisfaction from the fact that I get so much done by late morning. Find your time.

There are also different places where you might work best. Try working in different rooms of your house with different lighting. I personally like working at the head of a long table (with noone else there), and prefer a view that looks outside. Some people, on the other hand, feel distracted when there is a lot of space or when they are looking outside. Find your place.

If TV distracts you, don't wach TV within 30 minutes of working.

Higher concentration can result from better health. Try regularizing your sleeping and eating cycles if they are not already, eating a balanced diet, and implementing a regular exercise regimen.

You will also find that if you feel organized (in terms of spatial organization, time management, etc), you will also think in a more organized fashion. If you are a mess, waking up late, eating randomly, wasting time, not taking showers, etc, your thinking will also be less organized.

[b] @ TZF: [/b]  Thanks a lot dude for your suggestions.

Can anyone give some more suggestions/advice?

I'm not sure that distractions (like tv, computer, .etc) are as important as time of day, location. Even things like the day of the week can make a noticeable difference.

  I myself tend to have problems concentrating in quiet, still areas because my mind wanders off of what I am trying to concentrate on. I do notice that if there is some kind of  light  pointed directly at the area I am working on I concentrate a lot better.

  It also helps if you aren't multitasking. If you try to do two things at once you [b]might[/b] be able to them 50% as well as you would do them seperately, and beyond that, it is almost impossible to do them.

  To sum it up I would say spend a week or so finding your best location and time, have good lighting, and don't multitask.

I used to take a supplement advertised as improving memory. I remember the brand name but that would not help you because it was in another country and in another century.
I also found that if I was not eating well I could not think straight. Switching from sandwiches to fish helped.
Finding a "study buddy" helps too. You do not need someone studying the same subject, but you would want someone who stays on task for a long time. It worked for me; I studied organic chemistry while my buddy did descriptive geometry assignments.

I can study with most concentration when I am in a really nice coffee shop such as Seattle's Best, or Java House sipping a cup of my favorite coffee (sugar free of course). I would prefer the coffee place that is inside a book store such as Borders or Barnes and Noble.

[u]Here’s a tip for concentration:[/u] Water your eyes before you sit for study. Let the eyes dry naturally. Our eyes are the only part of our body that does not get into contact with water as often as others. So you need to clean the eyes with water especially if you want to concentrate for hours together.

[u]Here’s a tip for making your learning more permanent:[/u] Your goal should be to make the contents of the study to become part of your mind and nature and not just be part of the book. For this you will have to make yourself as comfortable with your surroundings as you possibly can while you are studying. This means you should be willing to accept your surroundings the way they are. Let the people walk past you as they want, let them talk loud enough that you can hear their conversations, let the kids laugh and be hushed by their parents, let the things be as natural as they can be. It is how the life really is. This is the kind of world that you live in and you need to make this be part of your study too. It may seee that you will be contradicting yourself by doing so, but you have to first make sure that you did a thourough search for a quiter and/or acceptable surrounding and are satisfied with it.

The most important rule to follow is that, you and [u]only you[/u] must decide who you like to go out to study with, which place to go (coffee shop, library, school, or home) and which corner of the chosen place should you sit. This is to make sure your learning experience is not hindered by unwanted disturbances.

For better concentration, always imagine everything fitting together in a cosmic way. It is the whole big picture that you must try to see and not the inconsequential details.


Putnam
function
calculus
algebra
difference of squares
special factorizations
college contests

Find the minimum value of
\[ \frac{(x\plus{}1/x)^6\minus{}(x^6\plus{}1/x^6)\minus{}2}{(x\plus{}1/x)^3\plus{}(x^3\plus{}1/x^3)}\]
for $ x>0$.

I'm posting this for the benefit of my class. We were discussing this problem today, and worked through it with some pretty ugly brute force, although we got it right.

[hide="A more elegant method."]Call the function to be minimized $ f(x).$

Note that $ x^6\plus{}2\plus{}\frac1{x^6}\equal{}\left(x^3\plus{}\frac1{x^3}\right)^2.$ Write $ f(x)$ using that and then do a difference-of-squares factorization in the numerator:

$ f(x)\equal{}\frac{\left(x\plus{}\frac1x\right)^6\minus{}\left(x^3\plus{}\frac1{x^3}\right)^2} {\left(x\plus{}\frac1x\right)^3\plus{}x^3\plus{}\frac1{x^3}}$

$ \equal{}\frac{\left[\left(x\plus{}\frac1x\right)^3\plus{}x^3\plus{}\frac1{x^3}\right] \left[\left(x\plus{}\frac1x\right)^3\minus{}x^3\minus{}\frac1{x^3}\right]}{\left(x\plus{}\frac1x\right)^3\plus{}x^3\plus{}\frac1{x^3}}$

$ \equal{}\left(x\plus{}\frac1x\right)^3\minus{}x^3\minus{}\frac1{x^3}\equal{}3x\plus{}\frac3x\equal{}3\left(x\plus{}\frac1x\right)$

Then by either AM-GM or by simple calculus, we see that for $ x>0,$ $ x\plus{}\frac1x\ge2,$ with the minimum value occurring when $ x\equal{}1.$ Hence the minimum value of $ f(x)\equal{}6,$ when $ x\equal{}1.$[/hide]

Hey,
[quote="Kent Merryfield"]
I'm posting this for the benefit of my class. We were discussing this problem today, and worked through it with some pretty ugly brute force, although we got it right.
[/quote]
What was the brute force method?

Thanks,

-[url=http://www.artofproblemsolving.com/Forum/profile.php?mode=viewprofile&u=33715]APSStudent[/url]

I've only jst glanced at this problem but isn't it really trivial, as the expression drops down to $ (x\plus{}1/x)^3\minus{}(x^3\plus{}1/x^3)$

Kent Merryfield already have pointed that. You should see other posts, me@home. :wink:

[quote="APSStudent"]What was the brute force method?[/quote]
At one intermediate stage, we had the expression
\[ \frac{6x^{10}\plus{}15x^8\plus{}18x^6\plus{}15x^4\plus{}6x^2} {2x^7\plus{}3x^5\plus{}3x^3\plus{}2x}.\]
Only then did we start looking for common factors.


n is a positive integer.Prove that
$\sum^{n}_{k=0}(-1)^{k}{C_{n}}^{k}{(n-k+1)}^{n}=n!$

[hide="hint"]Use Lagrange Interpolating Polynomial for polynomial $f(x)=(x+1)^{n}$ and points are $x_{i}=i(i=0,1,2,...,n)$.[/hide]

My experienced and problem's form , that is all :P


Galois Theory
superior algebra
superior algebra theorems

If $E$ is finite Galois over $F$ with Galois group $G$, then $H^{1}(G,F^{*})=\{0\}$. In my opinion, it is enough to have $E$ any extension of $F$ such that $|Aut(E/F)| < +\infty$, and to take $G=Aut(E/F)$.

Also, in Hilbert's theorem 90, we would consider any extension $E$ of $F$ such that $Aut(E/F)$ is finite and cyclic, which is different from considering $E$ a finite cyclic extension of $F$.

Does it work ?

Can you please post Hilbert 90 theorem? I've heard so much about it. ;)

Sure, it is the one saying that in a finite cyclic extension $E$ of $F$ with Galois group $<\sigma>$, every element with norm $1$ can be written as $\frac{\alpha}{\sigma (\alpha)}$ for some $\alpha \in E$.

Ad Hilbert 90 : I think it works, but: You have to assume a galois extension, so that the norm maps to $K=Fix(G)$. Anyway, this is not needed in the theorem. But like with your question to characters: Generalizations are only accepted, when they are somewhat useful :P

[quote="-oo-"]but: You have to assume a galois extension, so that the norm maps to $K=Fix(G)$.[/quote]

True ... actually $K$ does not need to play any role in my generalization; also, as you said, this generalization may be useless  :D


Hi, im having a little problem with this equation
The answer must be in days but my solution involves zeros that arent natural numbers...
Can you help me?
The equation is
* means exponent
-d*3+4d*2 - 4d + 3 < 2

The zeros i found were 1 and two unnatural numbers.  :?:


AMC
AMC 8
MATHCOUNTS
AIME
ARML
geometry
Alcumus

10 Rounds, 10 questions each. Simple, no?
[b]Sign-ups start NOW up until July 13th, 2009, at 11:59:59 PST (forum time)[/b]
PLEASE DO NOT PM ME IF YOU WISH TO SIGN UP!

Please include the following:
Grade (2009-2010 year)
Math-ish grade (if you know it. if you do not know it, say "The math I took last year...")
Contests you've participated in, if any.

This will help me determine the level of difficulty the problems will be.
I will probably post a sampling of problems on...Wednesday? (my birthday. hahahahaha.)

Oh, please tell me if you'd like to be a helper! :)
Thanks!
-Ducky

I will join. I am going to go to 8th grade. I got 5th at state in MC, 20 on AMC8. I have participated in Math Kangaroo (11th in US) and Purple Comet (fail).I took Pre-Algebra last year. I have taken the Algebra 1 test and done pretty good. I would say I am Algebra 2/Geometry level.

About the "forum time", where does it say it?

I will join.
I got like 25th state mathcounts.  I am like Algebra 1/2 Level.

I am going to the eighth grade

Yay I will join.

going to 8th grade. Next year Im going to be in honors algebra 2 and pre. calc. I got a 24/25 on amc8, 120/150 on amc 10a, 25/25 on math olympiad, 29/30 CML, 39/40 on mathleague, and failed state mathcounts and AIME (very low score on both).

i'm not in middle school anymore, but i was wondering if i could join because my level of math is comparable if not lower than most of these middleschoolers.

@rd5493: Sure!

Also, can everyone please say what AoPS classes they have taken? Thanks again!! :)

11th Grade
Pre-Calc
2007 National MATHCOUNTS (172nd Place)
2009 WV ARML Team (33rd Place)
AoPS Online Classes: N/A
AoPS Books: Intermediate Algebra and Intermediate C & P

EDIT: Never had chance to take AMC or AIME, but on practice tests I've taken:

AMC 10: Average score around 120
AMC 12: Average score around 105
AIME: Average score ranging from 3-5

Depending on the level of the other users' math skills, you might have to be a helper, if that's okay. :wink:

I'll join.

9th Grade.
Geometry and Algebra II
AoPS Classes: None
8th state MC
HM on Purple Comet
24 on AMC 8
126 on AMC 10
40/40 Math League

I will join:

8th grade
6th chapter MATHCOUNTS
55th State MATHCOUNTS
Took ALG 1
Geometry level

[hide="Sign up List"]
1. PowerOfPi
2. venkataraman
3. Thunder365
4. rd5493
5. PerpetualMotion
6. SuperNerd123
[/hide]
Also, I have the whole Introduction series, Intermediate Algebra, and AoPS Volume 1, so I will pull problems from there. (as well as some from Alcumus, too)

AoPS Classes: None
AoPS Books: this is the 11th business day I have been waiting for it, and I checked 3-7 day delivery.

What level will the questions be at? Like State MathCounts or harder?

I'm not sure yet. I'll have to look online.
The first round will be pretty easy, but over time it will get harder. (Also, I may need someone to take over for rounds 9 and 10 since I may start school around that time...)

Will you post the questions here or pm or what?
Srry but i have never done a 10 round tourny

I will post the questions here. (PM might take too long.)

Er...

[quote="myyellowducky82"]
Sign up List

1. PowerOfPi
2. venkataraman
3. Thunder365
4. rd5493
5. PerpetualMotion
6. SuperNerd123
[b]7. batteredbutnotdefeated[/b]
8. cgyao15
9. yaofan
10. athunder
11. srfan[/quote]

[quote="batteredbutnotdefeated"]agh... I am forced to withdraw from this tournament also... Oh well... Ducky can I be a helper/problem maker? I love making problems! :lol:[/quote]

Also, when will the results be posted?

after helpers send me solutons! (wink, wink)

eh, I think I need one more participant...they'll just have to join round 2. (elimination issue.)

Um...is this how you're supposed to post? D: I'm such a newbie...

Well I wanna join. :D
I'll be going into eighth grade for 2009-2010,
and I'll be taking Geometry. Though with the brainwashing
of summer, I'm probably not up to that level....
No contests before...=/

Heh, lots of eighth graders.
Sure you can join for round 2. I'm going to post the problems/Round 1 solutions tomorrow.

[hide="Round 1 solutions"]
1. 0

2. ONE MILE.

3. 0

4. Two kids row across. One brings the boat back. Then Mom rows across, and teh other kid brings the boat back. Both kids row across. One brings the boat back. Then Dad rows across, and finally the second kid takes the boat back to collect her sibling.

5.  1960, 441

6. 1215

7. 5

8.  the number having no 1's among its digits [/hide]

If you find any errors, please let me know (I never got answers for 2 problems)

[hide="I provided #5 and #6, Solutions hidden"]
5. 1960, 441 (It's a 9-40-41 triangle multiplied by 49)
6 is correct.
[/hide]

[hide="Results..."]
1. srfan--8/8
2. Thunder365--6/8
3. venkataraman--5/8
4. PowerOfPi--6/8
5. yaofan--5/8
6. SuperNerd123--6/8
[/hide]
[b]Before I post the problems, please post if your username is NOT up there.[/b]

I will no longer have time to run this tournament.
Please, someone, feel free to take over. :P[/hide]

Could someone explain number 2? I don't get it

Here see if this makes sense, if u need further explanation tell me

[hide="solution to number two"]When the boats first meet, the total distance they have traveled is equal to the width of the river. By the time they meet again, the total distance traveled equals three times the length of the river. The boats are traveling at a constant speed, so they each will have traveled three times as far by the second meeting as teh distance they'd traveled by the first time they meet. Because the boat starting in New York had traveled 720 yards at the first meeting, it must have traveled 2,160 yards at the time of the second meeting. BUt this distance is 400 yards from teh other shore, so the width of the river equals 2,160 - 400 = 1760 yards which is exactly  $ \fbox{one mile}$[/hide]

[quote="inoit"]So i guess this is over, I could continue it if anyone wanted me 2

btw ducky u stole my signature :P[/quote]
yeah, you should continue for me.
i didn't steal it, i always say it =]


Weird.  Just so I have this straight - you're already logged in, then you click My Classes and it will log you out, then you click the login button on the side and re-login, and you're ok?

Can you try deleting your cookies and see if the problem persists?

To delete the cookies on your computer I would just search for the temporary internet files folder and delete everything in there.  Or you could try finding something under internet options on your browser.  This works for me.  However I use IE so if you use something else I am not sure.

Pretty much the same - tools, options, its under privacy.


induction
algebra
function
domain
number theory unsolved
number theory

Determine sum of maximal odd divisors of $n+1,n+2,...,2n$

This is a very simple exercise on induction!  :P
We claim that the required sum is $n^2$. This is obviously true for $n=1$, since $\{n+1, \ldots , 2n \}$ would then just be the single element set $\{ 2 \}$ and the corresponding sum would be the greatest odd divisor of $2$, which is $1$.

We denote $f(x)$ to be the maximum odd divisor of $x$, where the domain of $f$ is over the natural numbers. By Induction Hypothesis, $f(n+1) + \ldots + f(2n) = n^2$. Note that $f((n+1)+1) + \ldots + f(2(n+1))$ $= [f(n+1) + \ldots + f(2n) ] + f(2n+1) + f(2n+2) - f(n+1)$ $= n^2 + f(2n+1) + f(2(n+1)) - f(n+1)$. Finally, note that $f(2n+1) = 2n+1$, while $f(2(n+1)) = f(n+1)$, so $f((n+1)+1) + \ldots + f(2(n+1)) = n^2 + 2n + 1 = {(n+1)}^2$, thus completing the induction step.


Prove that $X^{n}-2$ is irreducible over $Q[X]$
(directly,without using Schonemann/Eisenstain,etc

Please refer here:
http://www.mathlinks.ro/Forum/viewtopic.php?t=68010

Oh, I see.  It's clearly irreducible when $n = 1$, and for $n > 1$ it's irreducible $\bmod 4$.

Hold on, is my logic valid?  $4$ isn't prime, so I don't think irreducibility $\bmod 4$ implies irreducibility over $\mathbb{Q}$.


[quote="mathdog15"]For a particular peculiar pair of dice, the probabilities of rolling 1, 2, 3, 4, 5, and 6 on each die are in the ratio 1:2:3:4:5:6.  What is the probability of rolling a total of 7 on the two dice? (source: AMC 10B 2006)

Thanks[/quote]

[hide]One way to do this is by using casework.  There are three cases that add up to 7.  6 and 1, 5 and 2, and 4 and 3.  Instead of looking at the dice as 6 sided dice, look at them as 1+2+3...6=21 sided dice.  Therefore, there are $ 21^2 \equal{} 441$ possibilities total.  Now for the successful outcomes.

Case 1 - 6 and 1:  6 cases.  This is clear.

Case 2 - 5 and 2:  10 cases.  This is also clear.

Case 3 - 4 and 3: 12 cases.  Clearly.

Also, since 6 and 1 can also be 1 and 6, there are $ 28\cdot2$ successful possibilities.
Therefore, the probability is 56/441=8/63.[/hide]

[hide="Alternative Solution"]
As AdithyaGanesh pointed out $ 7\equal{}6\plus{}1\equal{}5\plus{}2\equal{}4\plus{}3$.

So, to find the probability of getting a sum of $ 7$, we simply add up the probability of getting a $ 6$ and $ 1$, a $ 5$ and $ 2$, or a $ 4$ and $ 3$.

The probability of getting a $ 6$ and $ 1$ is $ 2\left (\frac{6}{21} \right )\left ( \frac{1}{21} \right )$. (I multiplied the probability of getting a $ 6$ with the probability of getting a $ 2$.  Then, I added a factor of $ 2$ because we can get a $ 6$ then a $ 1$, or a $ 1$ then a $ 6$.

Similarly,
The probability of getting a $ 5$ and $ 2$ is $ 2\left (\frac{5}{21} \right )\left ( \frac{2}{21} \right )$.
The probability of getting a $ 4$ and $ 3$ is $ 2\left (\frac{4}{21} \right )\left ( \frac{3}{21} \right )$.

Adding the probabilities, we get $ 2\left (\frac{6}{21} \right )\left ( \frac{1}{21} \right )\plus{}2\left (\frac{5}{21} \right )\left ( \frac{2}{21} \right )\plus{}2\left (\frac{4}{21} \right )\left ( \frac{3}{21} \right )\equal{}\frac{8}{63}$[/hide]

[hide="Perhaps even more straightforward"]
Clearly the probability of rolling $ x$ is $ \frac{x}{21}$. We notice that for every available $ x$ there is a counterpart $ 7\minus{}x$ that must be rolled on the other die.

The probability, then, is $ \sum_{k\equal{}1}^{6}\frac{x}{21}\cdot \frac{7\minus{}x}{21}\equal{}\frac{1}{21^2}\left(\frac{7\cdot 6\cdot 7}{2}\minus{}\frac{6\cdot 7\cdot 13}{6}\right)\equal{}\frac{7(21\minus{}13)}{21^2}\equal{}\boxed{\frac{8}{63}}$.
[/hide]

These are all pretty much the same solution with different computation methods...


inequalities
LaTeX
geometry
perimeter
inequalities unsolved

I found this problem at a math olympiads training book. Ít´s an attachment, because I still don´t have latex, an i can´t post hard formulas... sorry  :blush:

I will write the statement here:
Let $ a,b,c$ be the sides and $ \alpha,\beta,\gamma$ be the angles of the triangle.Also let $ s$ be its semi-perimeter,i.e $ s\equal{}\frac{a\plus{}b\plus{}c}{2}$.
Prove that:
$ \frac{bc}{\alpha(s\minus{}a)}\plus{}\frac{ac}{\beta(s\minus{}b)}\plus{}\frac{ab}{\gamma(s\minus{}c)}\geq\frac{12}{\pi}$.


Find, with proof, the greatest $\alpha \in \mathbb{R}$ s.t.
$\sum \sqrt{x(x+y)(x+z)}> \alpha \sqrt{(x+y+z)^{3}}$
takes place for any $x, y, z > 0$.

[hide="Hint"]The maximum value is $\alpha = 1$.
Can anyone prove the inequality for this constant?[/hide]

[quote="PhilAndrew"]The maximum value is $\alpha = 1$.
Can anyone prove the inequality for this constant?[/quote]
See here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=145158


[quote="ckck"]If the smallest value of y satisfying teh equation $y=3x^{2}+6x+k$ is 4, find the value of k.[/quote]
[hide]The x value of the vertex is $\frac{-b}{2a}=\frac{-6}{2\time3}=-1$ and plugging infor y, you get the vertex is $(-1,-3)$ you want the y value to be 4 so you must add 7 to get $y=3x^{2}+6x+7$ so $\boxed{k=7}$[/hide]

Just take the derivative out of laziness.  :D


Prove that $ (x\plus{}y\minus{}z)^8\plus{}(y\plus{}z\minus{}x)^8\plus{}(z\plus{}x\minus{}y)^8\plus{}(x\plus{}y\plus{}z)^8\le 2188(x^8\plus{}y^8\plus{}z^8)$ for every $ x,y,z\in \mathbb R$

[quote="mr.danh"]Prove that $ (x \plus{} y \minus{} z)^8 \plus{} (y \plus{} z \minus{} x)^8 \plus{} (z \plus{} x \minus{} y)^8 \plus{} (x \plus{} y \plus{} z)^8\le 2188(x^8 \plus{} y^8 \plus{} z^8)$ for every $ x,y,z\in \mathbb R$[/quote]
It 's here
http://www.mathlinks.ro/viewtopic.php?t=232107


If $ A$ is an infinite set and $ N$ is the set of natural numbers, prove that $ A$ and $ A X N$ have the same cardinality(, where $ X$ denotes the Cartesian product).

Using the Axiom of Choice, any infinite set $ A$ can be written as $ A \equal{} \bigcup_{i \in I} B_i$, with $ B_i$ countable and pairwise disjoint. Therefore $ C_i \equal{} B_i \times \mathbb{N}$ is countable, hence of same cardinality with $ B_i$, written $ C_i \sim B_i$.
Then $ A \times \mathbb{N} \equal{} \left ( \bigcup_{i \in I} B_i \right ) \times \mathbb{N} \equal{} \bigcup_{i \in I} C_i  \sim \bigcup_{i \in I} B_i \equal{} A$.


algebra
polynomial
function
calculus
derivative
analytic geometry
graphing lines

\[ \begin{gathered}
  {\text{Let }}\alpha ,b,\gamma  > 0{\text{ and }}2\alpha \gamma  \leqslant {\beta ^2} \hfill \\
  P(t) \equal{} \alpha  \minus{} \beta t \plus{} \frac{\gamma }
{2}{t^2} \hfill \\
  {t_{k \plus{} 1}} \equal{} {t_k} \minus{} {\left( {P'\left( {{t_k}} \right)} \right)^{ \minus{} 1}}P({t_k}) \hfill \\
  {\text{I need to show that: }} \hfill \\
  a){t_{k \plus{} 1}} \equal{} {t_k} \plus{} \frac{\gamma }
{2}\frac{{{{\left( {{t_k} \minus{} {t_{k \minus{} 1}}} \right)}^2}}}
{{\beta  \minus{} \gamma {t_k}}}{\text{ and }}{\left\{ {{t_k}} \right\}_k}{\text{ is monotonically increasing}} \hfill \\
  b){\text{If }}2\alpha \gamma  < {\beta ^2}{\text{, then }}{t_{k \plus{} 1}}{\text{ converges to the smallest root of P: }} \hfill \\
  {t_*} \equal{} \frac{1}
{\gamma }\left( {\beta  \minus{} \sqrt {{\beta ^2} \minus{} 2\alpha \gamma } } \right) \hfill \\
   \hfill \\
  a){t_{k \plus{} 1}} \equal{} {t_k} \minus{} \frac{{P({t_k})}}
{{\left( {P'\left( {{t_k}} \right)} \right)}} \equal{} {t_k} \minus{} \frac{{\alpha  \minus{} \beta {t_k} \plus{} \frac{\gamma }
{2}{t_k}^2}}
{{ \minus{} \beta  \plus{} \gamma {t_k}}} \equal{} {t_k} \plus{} \frac{1}
{2}\frac{{2\alpha  \minus{} 2\beta {t_k} \plus{} \gamma {t_k}^2}}
{{\beta  \minus{} \gamma {t_k}}} \hfill \\
  {\text{Now I}}{\text{ have to show that: }} \hfill \\
  \gamma {\left( {{t_k} \minus{} {t_{k \minus{} 1}}} \right)^2} \equal{} 2\alpha  \minus{} 2\beta {t_k} \plus{} \gamma {t_k}^2 \hfill \\
   \Leftrightarrow \gamma {t_{k \minus{} 1}}^2 \minus{} 2{t_k}{t_{k \minus{} 1}}\gamma  \plus{} 2\beta {t_k} \minus{} 2\alpha  \equal{} 0 \hfill \\
\end{gathered} \]

How do I do this?

b) How do I approach this one?

Have you ever seen the picture of the Newton method iterations for convex functions? (If not, I'll draw one for you; this is the case when a picture is worth a thousand words). By the way, the statement is correct only if $ t_0$ is to the left of $ t_*$.

[quote="fedja"] the Newton method iterations for convex functions? [/quote]

What is it?

Exactly what the above formula is. If you want to solve the equation $ f(x)\equal{}0$ with smooth $ f$, you start with some initial guess $ x_0$ and then use the approximate formula $ f(x)\approx f(x_0)\plus{}f'(x_0)(x\minus{}x_0)$ to improve it to $ x_1\equal{}x_0\minus{}\frac{f(x_0)}{f'(x_0)}$, which is the solution of the linearized equation $ f(x_0)\plus{}f'(x_0)(x\minus{}x_0)\equal{}0$. Then you repeat this procedure using $ x_1$ in place of $ x_0$, and so on. If the root is simple and you started close enough to it, the sequence converges extremely fast.

The picture is here:
[asy]unitsize(1cm);
D((-5,0)--(5,0),black);
real f(real x){return x^2/2-1;}
path P=graph(f,-3.2,3.2);
D(P);
real a=-3;
for(int i=0;i<3;++i)
{
real b=a-f(a)/a;
D(D("t_"+string(i),7,(a,0))--(a,f(a))--(b,0),red);
a=b;
}[/asy]

fedja : if the root is not simple,  can the Newton method be applied/?

Under normal circumstances ($ C^\infty$-function of one variable with at least one non-zero derivative at the root) the answer is yes, but the convergence speed is much worse, so other methods like bisection may be preferable in this case. Indeed, it you do the Newton method for $ f(x)\equal{}x^3$ starting with $ x_0\equal{}1$, you get $ x_n\equal{}\left(\frac 23\right)^n$ while the bisection always guarantees you the $ 2^{\minus{}n}$ precision after $ n$ steps. Besides, it'll require division by numbers close to $ 0$, so you may end up with the rounding error problems (especially if the computation of the derivative involves subtracting 2 large values to get that almost zero). I don't remember now exactly what happens when the number of variables is 2 or more (the Newton method works then too if the differential at the root is invertible) but I suspect that there you may encounter even more problems with it (one indication of trouble is that in that case there may be many other points in an arbitrarily small neighborhood of the root where the differential is not invertible, so the next step is just undefined). So, as a rule of thumb, it is better to use Newton only when you believe that your root is simple. Fortunately, this is a "generic" case and (unless your function has some special structure) the multiple root case is an exception.

[quote="PiDeltaPhi"]I think I need a little bit more help on this one...[/quote]
OK, what do we see on the picture?

If $ t_0 < t_*$ and the function $ f$ is convex, then $ t_1$ lies on the tangent line to the graph at the point $ (t_0,f(t_0))$. By convexity, this tangent line lies below the graph and its slope is negative. Thus, $ t_1 > t_0$ but still $ t_1 < t_*$.

Thus $ t_n$ is a bounded increasing sequence, so it has a limit. Also, if $ f'$ is bounded on any finite interval, then $ t_{n \plus{} 1} \minus{} t_n\ge cf(t_n)$ with some $ c > 0$. Since $ t_{n \plus{} 1} \minus{} t_n\to 0$ and $ f(t_n)\ge 0$ for all $ n$, we see that $ f(t_n)\to 0$, so $ f(\lim_n t_n) \equal{} 0$. But $ t_*$ is the only point with this property that is not greater than $ t_*$. Thus, $ t_n\to t_*$ as $ n\to\infty$.

The exact shape of the graph is completely irrelevant here as long as the convexity and the boundedness of the derivative properties hold. Each statement above can be rewritten as an ugly inequality for the particular polynomial in question but to do it will be just a total waste of time.

Ahhh, I understand now. You were right, a graph says sometimes more than 1000 words.
Thank you!


function
trigonometry
quadratics
algebra
polynomial
algebra open

My teacher tells me to make a paper about this type of Trigonometric Equations: Change an trigonometric Function into an Algebraic Function. But I don't have much information, exercises about this topic, and I don't know that there are Olympiad problems related with this topic or not  :what?: ? Or Are there any other interesting topics that related with this topic?  :maybe:
I think that your help will be some thing so great with my paper.

[b]timlaiminh
CTptnk[/b]

[quote="t0rajir0u"]But trigonometric functions aren't algebraic... I'm not quite sure what you're asking.

Do you want infinite algebraic expansions?  Because we have the well-known series expansions

$\sin x = \sum_{k=0}^{\infty}(-1)^{k}\frac{x^{2k+1}}{(2k+1)!}$
$\cos x = \sum_{k=0}^{\infty}(-1)^{k}\frac{x^{2k}}{(2k)!}$

And the perhaps less well-known product expansions

$\sin x = x \prod_{k=1}^{\infty}\left( 1-\frac{x^{2}}{\pi^{2}k^{2}}\right)$
$\cos x = \prod_{k=1}^{\infty}\left( 1-\frac{4x^{2}}{ \pi^{2}(2k-1)^{2}}\right)$[/quote]

Thanks for your help.

What I want is problem about trigonometric equations solved by turn it into algebraic equation, for example this prolem:
$sinx+cosx+sinx.cosx=1$
We can have: $sinx+cosx=t$ so that $sinx.cosx=\frac{(t^{2}-1)}{2}$
We obtain a new equation:
$\frac{(t^{2}-1)}{2}+t = 1$
The equation above is algebraic problem.

I hope that there are some more difficult trigonometric equations that use the idea above to solve, especially Olympiad problems.

Hope all your help.
[b]
timlaiminh[/b]

[quote="Svejk"]An equation of the form
$a_{0}\cdot sin^{n}(x)+a_{1}\cdot sin^{n-1}(x)\cdot cos(x)+...+a_{n-2}\cdot sin^{2}cos^{n-2}(x)+a_{n-1}\cdot sin(x)\cdot cos^{n-1}(x)+a_{n}\cdot cos^{n}(x)=0$
It's easy to prove that $cos(x)\neq 0$ ,so by dividing the equiation by $cos^{n}(x)$ and denoting $tg(x)=t$ the equation is equivalent to:
$a_{0}\cdot t^{n}+a_{1}\cdot t^{n-1}+...+a_{n-1}\cdot t+a_{n}=0$ ,equation which it is considered to be algebraic.
Suppose that $t_{i}, 1\leq i \leq n$ are the real roots of that polynomial then $x=arctg(t_{i})+k\pi$ , $k\in \mathbb Z$.

Is this what you are searching for timlaiminh ?[/quote]

Yah, it 's right, but how can we solved this equation: $a_{0}\cdot t^{n}+a_{1}\cdot t^{n-1}+...+a_{n-1}\cdot t+a_{n}=0$ , i think it 's rather difficult if the degree is so high. And are there any other type of transform an trigonometric equation into an algebraic one?
However, your help give me some idea and I will use it.

Thank you.

[quote="t0rajir0u"]One example is the transformation

$\sin x = \frac{u-\frac{1}{u}}{2}$
$\cos x = \frac{u+\frac{1}{u}}{2}$

Where $u = e^{ix}$.  Then your problem becomes a quadratic in $u$.  In fact, any general quadratic in $\sin x, \cos x$ becomes a quadratic in $u$.

Another is the substitution $t = \tan \frac{x}{2}$.  Then

$\sin x = \frac{2t}{1+t^{2}}$
$\cos x = \frac{1-t^{2}}{1+t^{2}}$

This transformation has more specific uses, since only in certain problems will the transformed equation be easy to solve.  I think your equation, in particular, is transformed into a quartic.[/quote]

Yup, but can you give me some exercises about that?

One example: $\frac{3}{4}-2sin2x-2(sinx+cosx)+\frac{cos4x}{2}= 0$
We can easily transform it into algebraic equation by the subsitution: $sinx+cosx=t$. But after that, we obtain a quartic equation and roots is irrational number. We cannot find out exactly how much $x$ is.

Sorry, that first transformation should be $\sin x = \frac{u-\frac{1}{u}}{2i}$.

Anyway, among other things that second transformation can be used to solve

$A \sin x+B \cos x = C$

Which becomes quadratic in $t$.  In particular, it becomes

$(B+C) t^{2}-(2A) t+(C-B) = 0$

Of course this is a rather simple example.  I don't currently have any better examples...

Make everything become difficult is not difficult as making easy thing become interesting. That 's the problem of this topic, it is not difficult and doesn't have attraction to everyone.
And that what I need all your helps, in the mathlinks.ro forum . I have found out some more interesting exercises and might be some Olympiad problems are enough. Are there any ones?


function
limit
real analysis
real analysis unsolved

Does pointwise convergence of a sequence of continuous functions $ f_n: [0,1] \minus{} \minus{} > R$ to the constant zero function $ g: [0,1] \minus{} \minus{} > R, g(x) \equal{} 0$, necessarily imply uniform convergence?

Without additional assumptions it is not true( it is not difficult to find an example).

My favorite example, on $ [0,1]:$

For $ n\ge 2,$ let $ f_n(x)$ be the piecewise linear continuous function whose graphs have nodes at $ (0,0),\left(\frac1n,n\right),\left(\frac2n,0\right),(1,0).$

You can make the height $ 1$ instead of $ n$ and it will still serve as a counterexample to this particular problem; making the height $ n$ also messes up the interchange of limit and integral.

Here's an entirely different question:

Suppose $ f_n(x)\ge f_{n\plus{}1}(x)\ge 0$ for all $ n$ and all $ x\in [0,1]$ and suppose that $ \lim_{n\to\infty} f_n(x)\equal{}0$ for all $ x.$ Prove that $ f_n\to0$ uniformly on $ [0,1].$

Would this still be true if we made that $ [0,\infty)$ instead of $ [0,1]?$

[quote="Kent Merryfield"]Would this still be true if we made that $ [0,\infty)$ instead of $ [0,1]?$[/quote]  No: consider $ f_n(x) \equal{} \frac{\pi}{2} \plus{} \arctan(x \minus{} n)$.

My example (which, like you, I wanted to make strictly decreasing and uniformly bounded) was $ f_n(x)\equal{}\max\left(\frac{x\plus{}1}n,1\plus{}\frac1n\right).$

Pretty close to the same idea.

All of which suggests that if you're going to try proving my claim, you might want to consider that $ [0,1]$ is compact.

[quote="Kent Merryfield"]My favorite example, on $ [0,1]$:...[/quote]
Is this one of those "bump-like" functions that, as $ n\rightarrow\infty$, the bump gets closer and closer to the origin but taller and taller (or the other way around)?


As you may know, scoring highly on the USAMTS is a qualification method for the American Invitational Mathematics Exam (AIME).  Because of the grading irregularities with Round 2 this year, we have decided, in fairness to all participants (and in particular those whose Round 2 papers were lost), to only use scores from Rounds 1 and 3 to determine AIME qualification.

[b]If you have submitted solutions for all of the first 3 rounds of the 2008-09 USAMTS, and scored a total of 45 points or more on Rounds 1 and 3, then you have qualified for the AIME.[/b]  In order for the USAMTS to submit your name to the American Mathematics Competition (AMC) for AIME qualification, you must log on to [url]http://www.usamts.org[/url] and enter the following information on your Profile page:

1) Your School's CEEB code (if you are homeschooled, just enter 'Homeschool' here).
2) The name of the teacher in your school who will administer the AIME to you.

[b]You must complete this information by March 1.[/b]  If you have qualified for the AIME via the usual process of a high score on the AMC 10 or AMC 12, then you may elect to ignore this.  If you do not know what the AIME is, visit [url]http://www.unl.edu/amc[/url] for more details.

[quote="Brut3Forc3"]Will a similar process be used to determine prizes after round 4?[/quote]We have not even thought about prize distribution yet.


geometry
geometric transformation
rotation

does anyone know how to calculate the amount of heat energy generated after impact?

what kind of impact? a meteorite impact?

Are you talking about heat generated after impact of like say a ball on a wall or even water molecules on a glass jar?In that case  i think,not too sure that friction comes into play...Anyway even I want to know about this. :roll:

Just consider the difference in energy between the initial state and the final state.  Most of the deficit is in the form of heat.  (Note that some energy will be lost to sound, light, ad nauseum.)

I agree with inteluster, except one has to take into account the destruction, reaction, change in rotation and deformation (elastic) of both objects, as well as any mass transfer from one impacting object to another.

Typically, one sees these problems stated in tests where one assumes no special effects for either object and a rebound (or sticking) and asking for a temperature increase.  The change in free energy is used with the mass and specific heat of the objects to calculate the temperature increase.

As was said earlier, it is very small, absent a chemical reaction, such as an explosion.


let p be an odd prime . Prove that exist unique numbers $m,n$ such that $m^2=n(n+p)$.

There $infinitely$ many solutions for $p$ $prime$ when $m$ and $n$ are equal to $0$...  :lol:  :lol:  :lol:

Davron