inequalities
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IMO 2006

Determine the least real number $M$ such that the inequality \[|ab(a^{2}-b^{2})+bc(b^{2}-c^{2})+ca(c^{2}-a^{2})| \leq M(a^{2}+b^{2}+c^{2})^{2}\] holds for all real numbers $a$, $b$ and $c$.

after checking through computer , the least value of such $M$ is $\frac{9}{16\sqrt{2}}$ with equality case $a: b: c=\frac{3\sqrt{2}+2}{2}: \frac{2-3\sqrt{2}}{2}: 1$ and its permutation . So who can prove this ?

My friend Arman's solution (Mathlinks ID : armanf)
The left hand side factors like this: \[(a-b)(b-c)(c-a)(a+b+c)\] So just let $x=a-b, y=b-c, z=c-a, s=a+b+c$ then the inequality becomes \[|xyzs|\leq \frac{M}{9}(x^{2}+y^{2}+z^{2}+s^{2})^{2}\] With the property that $x+y+z=0$
Two of $x,y,z$ have the same sign. Let $x,y$ be that two. Now putting their arithmetic mean instead of them makes the inequality stricter (l.h.s. gets bigger and r.h.s. gets smaller)
So assume that $x=y$. Now $z=-2x$ and the inequality becomes: \[|2x^{3}s|\leq \frac{M}{9}(s^{2}+6x^{2})^{2}\] Now using the AM-GM inequality for $s^{2},2x^{2},2x^{2},2x^{2}$ we get \[(s^{2}+6x^{2})^{2}\geq 16\sqrt{8s^{2}x^{6}}=32\sqrt 2sx^{3}\] So $M=\frac 9{16\sqrt 2}$ works.

Hi gays,I solved this problem during the competition,and my solution is similiar to Nimas' solution,and shyong  the equality holds not only in that case :wink: .

I guess this might be a more direct way of seeing it (essentially the same argument):

Define $x,y,z,s$ as before and suppose $x,y$ have the same sign.  WLOG, they're both positive (otherwise we can replace $a,b,c$ with $-a,-b,-c$.)  Let $m = \frac{x+y}{2}$, so $z =-2m$. Now,

$xy \leq m^{2}\leq \frac{x^{2}+y^{2}}{2}$ so

$|(a-b)(b-c)(c-a)(a+b+c)|$

$= |xyzs| \leq |2m^{3}s|$

$\leq \frac{1}{16\sqrt{2}}\cdot (2m^{2}+2m^{2}+2m^{2}+s^{2})^{2}$

$= \frac{1}{16\sqrt{2}}\cdot (2m^{2}+z^{2}+s^{2})^{2}$

$\leq \frac{1}{16\sqrt{2}}\cdot (x^{2}+y^{2}+z^{2}+s^{2})^{2}$

$= \frac{9}{16\sqrt{2}}\cdot (a^{2}+b^{2}+c^{2})^{2}$

(and as before, we can check the equality case)

[quote="kakalotta123"][quote="Nima Ahmadi Pour"]My friend Arman's solution (Mathlinks ID : armanf)
Two of $x,y,z$ have the same sign. Let $x,y$ be that two. Now putting their arithmetic mean instead of them makes the inequality stricter (l.h.s. gets bigger and r.h.s. gets smaller)
[/quote]
What is it?
With $x+y=const$ ?[/quote]
I didn't get it exactly, just put their arithmetic mean instead of them, I didn't say to continously change them or something like that.

Let me help [i]Nima Ahmadi Pour[/i] explain a little more clearly.

He is replacing $x$ and $y$ with $x^{\prime}= \frac{x+y}{2}$ and $x^{\prime}= \frac{x+y}{2}$ respectively. So looking at our inequality \[\mid xyzs \mid \leq \frac{M}{9}(x^{2}+y^{2}+z^{2}+s^{2})^{2}\] after our substitution, if $x \neq y$, we get
$\mid xyzs \mid < \mid x^{\prime}y^{\prime}zs \mid$
and
$\frac{M}{9}(x^{2}+y^{2}+z^{2}+s^{2})^{2}> \frac{M}{9}(x^{\prime}^{2}+y^{\prime}^{2}+z^{2}+s^{2})^{2}$
by checking. (Remember that $x+y+z = 0$ and that $x,y,z$ are defined in terms of $a,b,c$.)

Of course, when $x=y$, we have $x = x^{\prime}$ and $y = y^{\prime}$, and there's no change.

Argh!! I don't like the keyboards at Slovenia. The labels on the keyboards are very misleading. The '&' key is actually '^', the z and y are interchanged, and  many more symbols are interchanged. I have to not look at the keyboard to type this.

It seems to be pretty doable using spherical coordinates, as well.

What do you mean by spherical coordinates?

Spherical coordinates: one radius coordinate, two angle coordinates.

I did this problem by first noting that the value we seek is the maximum of $ab(a^{2}-b^{2})+bc(b^{2}-c^{2})+ca(c^{2}-a^{2})$ on the unit sphere, and using Lagrange multipliers. After a great deal of algebra, I found the critical points; most led to zero values, leaving only one possibility (and permutations) for the maximum.
It was really pretty routine, except for the many guessed factorizations while finding critical points and a mistake in calculation.

I really should have used the coordinate system in Nima's post- I probably would have thought of them in my high school days, since I remember using them on a MOP homework problem.

According to the UK Leader, an Iranian observer found a solution to this question using Lagrange multiplier bashing.

This problem can be rephrased as follows.

Let $x,y,z$ be reals such that $x^{2}+y^{2}+z^{2}=1.$

Find the minimum value $M$ such that $|xy(x^{2}-y^{2})+yz(y^{2}-z^{2})+zx(z^{2}-x^{2})|\leq M.$

Hi ;

It Can Be Solved By $uvw$ Method :)

Best Regard

Firstly we have to note the well known factorization:$|ab(a^2-b^2)+bc(b^2-c^2)+ca(c^2-a^2)|=|(a-b)(b-c)(a-c)(a+b+c)|$.On the other hand one can think of(even though its IMO problem) to write RHS in some terms of factors that are in LHS,and then use AM-GM.Indeed it wont take long to find the following equality:$$(3(a^2+b^2+c^2))^2=(2(a-b)^2+2(a-c)(b-c)+(a+b+c)^2)^2\geq (4^2\sqrt{2(a-b)(a-c)(b-c)(a+b+c)})=16\sqrt{2}|(a-b)(b-c)(a-c)(a+b+c)|$$ hence we have$(a^2+b^2+c^2)^2\frac{9}{16\sqrt{2}}\geq |ab(a^2-b^2)+bc(b^2-c^2)+ca(c^2-a^2)|$ and we know this is the smallest $M$ because we can find equliaty case.
Note that without checking equality case one can easily find the wrong constant,for example if we rewrite as $3(a^2+b^2+c^2)=(a+b+c)^2+(a-b)^2+(b-c)^2+(c-a)^2$ with simple AM-GM we will only get $\frac{9}{16}$ which is by far to big.

[quote=blackbird97]Firstly we have to note the well known factorization:$|ab(a^2-b^2)+bc(b^2-c^2)+ca(c^2-a^2)|=|(a-b)(b-c)(a-c)(a+b+c)|$.On the other hand one can think of(even though its IMO problem) to write RHS in some terms of factors that are in LHS,and then use AM-GM.Indeed it wont take long to find the following equality:$$(3(a^2+b^2+c^2))^2=(2(a-b)^2+2(a-b)(b-c)+(a+b+c)^2)^2\geq (4^2\sqrt{2(a-b)(a-c)(b-c)(a+b+c)})=16\sqrt{2}|(a-b)(b-c)(a-c)(a+b+c)|$$ hence we have$(a^2+b^2+c^2)^2\frac{9}{16\sqrt{2}}\geq |ab(a^2-b^2)+bc(b^2-c^2)+ca(c^2-a^2)|$ and we know this is the smallest $M$ because we can find equliaty case.
Note that without checking equality case one can easily find the wrong constant,for example if we rewrite as $3(a^2+b^2+c^2)=(a+b+c)^2+(a-b)^2+(b-c)^2+(c-a)^2$ with simple AM-GM we will only get $\frac{9}{16}$ which is by far to big.[/quote]

It should be
$(3(a^2+b^2+c^2))^2=(2(a-b)^2+2(a-c)(b-c)+(a+b+c)^2)^2$

And you should suppose $min\{a,b,c\}=c$ to use that and also you forgot the square after $AM-GM$ you did applied. But this is most beautiful solution I ever seen to this problem it's just awesome congratulation.

An interesting fact is that if we include the condition $a,b,c \ge 0, M=\frac{1}{4}$ and is possible for
 $a : b : c = 0 : 1 : (1+\sqrt {2})$.

The answer is $\tfrac{9\sqrt{2}}{32}$, achieved at $(a,b,c)=(\sqrt{2}-3, \sqrt{2}, \sqrt{2}+3)$. To prove the bound, WLOG $a\geq b\geq c$ and note the following bounds.
\begin{align*}
(a-b)(b-c) &\leq (a-b)(b-c) + \frac 38 (a-2b+c)^2 = X\\
\frac{1}{2\sqrt{2}} (a-c)(a+b+c) &\leq \frac 18 (a-c)^2 + \frac 14 (a+b+c)^2 =Y\\
\end{align*}
However, after a full expansion,
$$X+Y = \frac 34(a^2+b^2+c^2).$$
Hence,
\begin{align*}\left\lvert\sum_{\mathrm{cyc}} ab(a^2-b^2)\right\rvert &= (a+b+c)(a-b)(a-c)(b-c) \\
&\leq 2\sqrt{2}\cdot XY \\
&\leq 2\sqrt{2}\left(\frac{X+Y}{2}\right)^2 \\
&= \frac{9\sqrt{2}}{32}(a^2+b^2+c^2)^2
\end{align*}
as desired.

Note that the inequality always holds for $a=b=c$, with any value of $M$. Thus, the answer to the problem is
\[M = \max_{\substack{(a,b,c)\in\mathbb{R}^3\\(a,b,c)\ne(a,a,a)}}\frac{|ab(a^{2}-b^{2})+bc(b^{2}-c^{2})+ca(c^{2}-a^{2})|}{(a^2+b^2+c^2)^2}.\]
Note that
\[|ab(a^{2}-b^{2})+bc(b^{2}-c^{2})+ca(c^{2}-a^{2})| = |(a-b)(b-c)(c-a)(a+b+c)|.\]
This motivates reparameterizing with $x=b-c$, $y=c-a$, $z=a-b$, and $t=a+b+c$, where now the only restriction is that $x+y+z=0$ and $x,y,z$ are not all equal. It is easy to check that
\[(a^2+b^2+c^2)^2 = \frac{1}{81}(3t^2+\sum(x-y)^2)^2,\]
so we have
\[M = 81\cdot\max_{\substack{(x,y,z,t)\in\mathbb{R}^4\\x+y+z=0\\(x,y,z)\ne(x,x,x)}}\frac{|xyzt|}{[3t^2+\sum(x-y)^2]^2}.\]
Fixing $(x,y,z)$, we see that it suffices to look at $t\ge 0$ since the thing we are maximizing stays the same if we negate $t$. Let $a=\sum(x-y)^2$, and let
\[f(t) = \frac{t}{(3t^2+a)^2}.\]
Note that $f(0)=0$ and $f(t)\to 0$ as $t\to\infty$, so the maximum of $f(t)$ is at a critical point of $f$ somewhere in $(0,\infty)$. It is easy to see that
\[f'(t)=0 \iff (3t^2+a)^2 = t\cdot(2(3t^2+a))\cdot(6t)\iff 3t^2+a =  12t^2\iff t=\sqrt{a}/3.\]
Thus, the maximum of $f(t)$ is
\[f(\sqrt{a}/3) = \frac{\sqrt{a}/3}{16a^2/9} = \frac{3}{16}a^{-3/2},\]
so the answer to the problem is
\[M = \frac{243}{16}\cdot\max_{\substack{(x,y,z)\in\mathbb{R}^3\\x+y+z=0\\(x,y,z)\ne(x,x,x)}}\frac{|xyz|}{\left[\sum(x-y)^2\right]^{3/2}}.\]
Note that replacinig $(x,y,z)$ to $(-x,-y,-z)$ does not change the expression within the maximum, so we can WLOG assume $x,y\ge 0$. In this case, we see that
\[M = \frac{243}{16}\cdot\max_{x,y\ge 0}\frac{xy(x+y)}{[6x^2+6xy+6y^2]^{3/2}}.\]
By de-homogenizing, we see that
\[M = \frac{243}{16}\cdot\max_{x\ge 0}\frac{x+1/x}{[6(x^2+1/x^2-1)]^{3/2}} = \frac{243}{16}\cdot\max_{\alpha\ge 2}\frac{\alpha}{[6(\alpha^2-1)]^{3/2}},\]
and it is easy to check that
\[g(\alpha)=\frac{\alpha}{(\alpha^2-1)^{3/2}}\]
is maximized at $\alpha=2$ if we restrict to $\alpha\in[2,\infty)$. Therefore,
\[M = \frac{243}{16}\cdot\frac{2}{6^{3/2}\cdot 3^{3/2}} = \frac{243}{16}\cdot\frac{1}{27\sqrt{2}} = \boxed{\frac{9}{16\sqrt{2}}}.\]

Here is [i][b]COMPLEX VERSION[/b][/i] of this problem.

Determine the least real number $M$ such that the inequality

$$|xy(x^2-y^2)+yz(y^2-z^2)+zx(z^2-x^2)|\leq M(|x|^2+|y|^2+|z|^2)$$

holds for all [b][i]COMPLEX[/i][/b] numbers $x,y,z$.

[i](Used for CMO training?)[/i]

Nothing new but I'll post for storage. Solved with [b]andyxpandy99[/b].

Notice that this the desired inequality is equivalent to proving $$|ab(a^2-b^2)+bc(b^2-c^2)+ca(c^2-a^2)| = |(a-b)(b-c)(c-a)(a+b+c)| \le M(a^2+b^2+c^2)^2.$$ WLOG let $a-b, b-c \ge 0$. Now, substituting $p = a-b, q = b-c, -2r = c-a, s = a+b+c$, we get $$|pq(-2r)s| \le \frac{M}{9}(p^2+q^2+(-2r)^2+s^2)^2.$$ Notice that $p+q-2r = 0 \implies p+q = 2r$, so fixing the value of $c-a$, we see that we can increase the LHS and decrease the RHS by making $p = q = r$. We then have $$2r^3s \le \frac{M}{9}(6r^2+s^2)^2 = \frac{M}{9}(2r^2+2r^2+2r^2+s^2)^2$$ which by AM-GM on the RHS gives $$\frac{M}{9}(2r^2+2r^2+2r^2+s^2)^2 \ge \frac{4M}{9}\left(\sqrt[4]{8r^6s^2}\right)^2 = \frac{8M\sqrt{2}}{9}r^3s \ge 2r^3s$$ $\implies M \ge \frac{9\sqrt2}{32}$ with equality when $(a, b, c) = (\sqrt{2}-3, \sqrt{2}, \sqrt{2}+3)$. $\blacksquare$

The answer is $\boxed{\frac{9\sqrt{2}}{32}}$.

This is equivalent to $|(a-b)(b-c)(c-a)(a+b+c)|\le M(a^2+b^2+c^2)^2$.

In fact, setting $(a,b,c)=(\sqrt{2}+3,\sqrt{2},\sqrt{2}-3)$ gives $162\sqrt{2}\le M\cdot 576\implies \frac{9\sqrt{2}}{32}\le M$. Also, equality holds if $M=\frac{9\sqrt{2}}{32}$. All we have to do now is show that the inequality is true.

Let $w=a-b, x=b-c, -2y=c-a, z=a+b+c$.

The inequality becomes \[|wxyz|\le \frac{M}{9}(w^2+x^2+4y^2+z^2)^2.\]

If one of $w,x,y,z$ are $0$, then the inequality obviously holds true. So assume they are all nonzero. WLOG $w,x,y,z>0$ (if this is not true, then changing $x$ to $-x$ keeps the inequality exactly the same).

Claim: If $w\ne x$, changing $w$ and $x$ both to their arithmetic mean makes the inequality stricter (as in, the LHS increases but the RHS decreases).
Proof: Note that $w+x=2y$. By AM-GM, we have \[w(2y-w)\le y^2\] so the LHS increases.
To show that the RHS decreases, we have \begin{align*}
w^2+(2y-w)^2 \\
=2w^2-4wy+4y^2 \\
=2(w^2-2wy+2y^2) \\
=2((w-y)^2+y^2) \\
\ge 2y^2 \\
\end{align*}

Equality holds iff $w=x=y$, so the RHS decreases also. $\blacksquare$

Now we can set $w=x=y$. We get \[2y^3z\le \frac{M}{9}(2y^2+2y^2+2y^2+z^2)^2\]

We have \begin{align*}
2y^2+2y^2+2y^2+z^2 \\
\ge 4\sqrt[4]{8y^6z^2} \\
\implies (y^2+y^2+4y^2+z^2)\ge 16\sqrt{8y^6z^2} \\
=32\sqrt{2}y^3z
\end{align*}

Multiplying by $\frac{M}{9}=\frac{\sqrt{2}}{32}$ gives $2y^3z$, as desired.

Let $x=a-b, y=b-c, z=c-a,$ and $u=a+b+c.$ So the inequality is: $$9|xyz| \leq M(x^2+y^2+z^2+u^2)^2.$$ Assume $x,y \geq 0.$ By AM-GM: $$|uxyz|=|uxy(x+y)| \leq |u| \frac{(x+y)^3}{4},~~~~~(\alpha)$$ equality when $x=y.$
Let $v=x+y,$ by AM-GM: \begin{align*} 2u^2v^6=2u^2\cdot v^2\cdot v^2\cdot v^2 \leq \frac{(2u^2+3v^2)^4}{4^4} \\ \implies 4\sqrt{2}|u|v^3\leq (u^2+\frac{3}{2}v^2)^2 \leq (u^2+x^2+y^2+z^2)^2.~~~~~(\beta)\end{align*} Combination of $(\alpha)$ and $(\beta)$ imply, $$|uxyz| \leq \frac{1}{16\sqrt{2}} (u^2+x^2+y^2+z^2)^2 \implies M \geq \frac{9\sqrt{2}}{32}.$$ $\boxed{M=\frac{9\sqrt{2}}{32}},$ i.e. the least value, when $(a,b,c)=(1-\frac{3}{\sqrt{2}}, 1, 1+\frac{3}{\sqrt{2}}).$ The opposite case, that is $x,y \leq 0,$ can be shown similar. $\square$

First observe that the LHS factors as $(a-b)(b-c)(a-c)(a+b+c)$. WLOG $a>b>c$. Let $f(a,b,c)$ = $M(a^2+b^2+c^2)^2-(a-b)(b-c)(a-c)(a+b+c)$. I claim the answer is $M=\frac{9\sqrt2}{32}$, which is achievable with $(a,b,c)=(\sqrt2-3,\sqrt2,\sqrt2+3)$. Note that if we increase all of $(a,b,c)$ at the same rate, the only terms that change are the $(a+b+c)$ term and the $(a^2+b^2+c^2)^2$ term. Now take the derivative with respect to $a+b+c$ (more precisely, take the limit as $h$ approaches $0$ of $\frac{f(a+h,b+h,c+h)-f(a,b,c)}{h}$. This is equal to $$A=4M(a+b+c)(a^2+b^2+c^2)-3(a-b)(b-c)(a-c)$$. In the equality case, this should be equal to zero (We know there has to be an equality case because the inequality is homogenous and the equality will not be reached by a limit, because we can simply rescale downwards. Sorry if this is a little handwavy). Thus, we substitute A=0 into our inequality and eliminate $M$, to get that, in the equality case, $4(a+b+c)^2=3(a^2+b^2+c^2)$. Thus it is sufficient to maximize $$\frac{(a-b)(b-c)(a-c)(a+b+c)}{(a^2+b^2+c^2)^2}$$ subject to the constraint $4(a+b+c)^2-3(a^2+b^2+c^2)=0$. This is equivalent to optimizing $\frac{(a-b)(a-c)(b-c)}{(a+b+c)^3}$. I claim that this expression is maximized at the equality case I described above, which is equivalent to showing that (after plugging it in and verifying that it satisfies the constraint), showing that $\frac{\sqrt2}2(a+b+c)^3-(a-b)(b-c)(a-c)\ge 0$ subject to the constraint. We now apply Lagrange multipliers on $(a,b,c)$ to get $$\frac{3\sqrt{2}}{2}(a+b+c)^2-(b-c)(2a-b-c)=\lambda(a+4b+4c)$$. Cyclically summing this and rearranging gives that $\lambda=\frac{\sqrt{2}}{2}(a+b+c)$.. Plugging this value of $\lambda$ back in and subtracting it from both sides, we get that $$\frac{\sqrt{2}}{2}(a+b+c)(3a+3b+3c-a-4b-4c)-(b-c)(2a-b-c)=0$$, which rearranges to $$(\frac{\sqrt{2}}{2}(a+b+c)(2a-b-c))=(b-c)(2a-b-c)$$. If it is not the case that $(a,b,c)$ are an arithmetic progression in some order, we can divide by $(2a-b-c)$ and cyclically sum to get a contradiction. If they are in arithmetic progression, WLOG $b$ is the middle term and some simple algebra gets that this is optimized at the equality case I gave above.

The left hand side factors to $|(a-b)(b-c)(c-a)(a+b+c)|$. Let $p$, $q$, $r$, $s$ be $a-b$, $b-c$, $c-a$, and $a+b+c$, respectively. Without loss of generality, let $s\ge 0$. Then, since
\[p^2+q^2+r^2+s^2=3(a^2+b^2+c^2)\]
we have the right-hand side as $\tfrac{M}{9} (p^2+q^2+r^2+s^2)^2$. We claim that the minimum value of $M$ is $\tfrac{9\sqrt2}{32}$. Suppose that the following is a local minimum with respect to $s$:
\[f(s)=\frac{(p^2+q^2+r^2+s^2)^2}{9s|pqr|}=\frac{(x+s^2)^2}{ys}\]
Then, we have
\[f(s)=\frac{s^4+2s^2y+x^2}{ys}=\frac1y(s^3+2sx+x^2s^-1)\]
\[f'(s)=\frac1y(3s^2+2x-x^2s^-2)=\frac1{s^2y}(3s^2-x)(s^2+x)\]
Clearly, $s^2+x>0$ we when $f$ is at a local minimum, $3s^2=x=p^2+q^2+r^2$.
\\$~$\\
We claim that given $p+q+r=0$ and $p^2+q^2+r^2=3s^2$, we have $|pqr|\le \frac{s^3}{\sqrt2}$. Let $P(x)$ be a monic cubic with roots $p$, $q$, $r$. Evidently, $P(x)$ has three roots and $P(x)=x^3-\tfrac32 s^2x-pqr$, so if $(x_1,y_1)$ and $(x_2,y_2)$ are $P$'s two local extrema, then $y_1y_2<0$. Since $P'(x)=3x^2-\tfrac32 s^2$, we have $x_{1,2}=\pm \tfrac{s}{\sqrt2}$ so
\[y_{1,2}=\pm \frac{s^3}{\sqrt2}-pqr\] Therefore, $(pqr)^2-(\tfrac{s^3}{\sqrt2})^2\le 0$, which implies $|pqr|\le \tfrac{s^3}{\sqrt2}$ as desired. Thus, $f(s)\ge \tfrac{16s^4}{9s^4/\sqrt2}=\tfrac{32}{9\sqrt2}$. Thus, $M\le \tfrac{9\sqrt{2}}{32}$. Our construction: \[(a,b,c)=(\tfrac{\sqrt{2}}{3}+1,\frac{\sqrt{2}}{3}),\frac{\sqrt{2}}{3})+1\]
Yields $(p,q,r,s)=(1,1,-2,\sqrt2)$ which yields the desired equality.


geometry
parallelogram
geometry unsolved

The quadrilateral $ ABCD$ is inscribed in the circle $ (O)$ and $ AB$ intersects $ CD$ at some point, let $ I$ be the point of intersection of the two diagonals. Let $ M$ and $ N$ be the midpoints of $ BC$ and $ CD$. Prove that if $ NI$ is perpendicular to $ AB$ then $ MI$ is perpendicular to $ AD$.

Denote X the foot of projection of I to BC  S=IX∩AD   R the midpoint of AB
1.∠BIX=∠AIP=∠SID=∠MIC
△SDI∽△MCI   =>   S is the midpoint of AD
2.SRMN is a parallelogram  O1=SM∩RN   SO1=MO1   RO1=NO1
ISOM is a parallelogram   =>  IO∩RN=O1
=> INOR is a parallelogram  =>  IN∥OR  => IN⊥AB

PS:  IM⊥AD  〈=〉 IN⊥AB

[hide="Solution"]I'll use a
[b]Lemma[/b]
If M,N,O,P are the midpoints of sides AB,BC,CD,DA of inscribed quadrilateral ABCD, then perpendiculars from M,N,O,P to CD,DA,AB,BC intersect in one point.
[b]Proof[/b]
Let Q be the intersection point of diagonals in MNOP. It's enough to consider a quadrilateral A'B'C'D' symmetric to ABCD through Q, because then our lines become midperpendiculars of sides of A'B'C'D', which obviously intersect in one point.

By lemma- intersection point of theese lines lies on perpendicular from M to AD, and assume that it's X (X and I are distinct). Let K be the midpoint of AB, k the perpendicular from I to CD and E, F the intersection points of k with CD and AB. We see that $ \angle BAC=\angle DIE=\angle CIE=\angle FIA$ and $ \angle DBA=\angle DCA=\angle IDE=\angle FIB$. Hence FB=FI=FA, which means that F=K - contradiction. Thus we see that the intersection point of our lines is I. [/hide]


A polygon has n sides and n diagonals. What is n?

[hide]N=5. The figure is a pentagon.[/hide]

I posted the following :

n = n(n - 3)/2 <=> n = 5

since a convex polygon with n sides has n(n - 3)/2 diagonals.
Why was it deleted ? Isn't it correct ?

Correcto MathFiend.

:?: I didn't delete that, nor have I seen it before. Where and when did you post it?
As for the equation it is correct I believe.

Well, I'm pretty sure I posted it ... never mind.

 :?


Given that $ I \minus{} AB$ is invertible for complex matrices $ A$ and $ B$,
Prove that $ I \minus{} BA$ is invertible.
Hence find its inverse.

Suppose that $ A$ is invertable,then we have:

 $ \det(I_n\plus{}AB)\equal{}\det(A)\cdot \det(A^{\minus{}1}\plus{}B)\equal{}\det(A^{\minus{}1}\plus{}B)\cdot \det(A)\equal{}\det(I_n\plus{}BA)$. $ (*)$

 So it is enough to prove that $ \det(I\minus{}AB)\equal{}\det(I\minus{}BA)$ for $ A,B$ noninvertable.Let $ T\in\mathcal{M}_n(\mathbb{R})$ and $ \lambda_0\in\mathbb{R}$ such that $ A\equal{}T\minus{}\lambda_0 I_n$.
 Let $ P(x)\equal{}\det(I_n\plus{}(T\minus{}xI_n)B)$ and $ Q(x)\equal{}(I_n\plus{}B(T\minus{}xI_n))$ two polynomilas and also consider $ Sp(T)\equal{}\{\lambda \in\mathbb{R}| \det(T\minus{}\lambda I_n)\equal{}0\}$ (the eigenvalues of  $ T$).
From $ (*)$ we have that $ P(x)\equal{}Q(x)$, $ \forall x\in\mathbb{R}\minus{}Sp(T)$.But the set $ \mathbb{R}\minus{}Sp(T)$ has infinetly many points, so $ P(x)\equal{}Q(x)$ is true for infinetly many values of $ x$.Therefore $ P(x)\equal{}Q(x)$, $ \forall x\in\mathbb{R}$, and in our case $ P(\lambda_0)\equal{}Q(\lambda_0)$.

This is true in any ring with identity.   One also has the identity

$ (1 \plus{} X)(1 \minus{} YX)^{\minus{}1}(1 \plus{} Y) \equal{} (1 \plus{} Y)(1 \minus{} XY)^{\minus{}1}(1 \plus{} X)$.

to t0rajir0u : why the result is true in any ring ? your identity don't explain the reason

This is true for non-square matrices $ A$ and $ B$ also. Take a look at here
http://www.mathlinks.ro/viewtopic.php?t=234376

I wonder how we could proceed to find the inverse?

$ I \minus{} AB$ being invertible means that $ det(I \minus{} AB) \neq 0$, which means that 1 is not an eigenvalue of $ AB$. Since the eigenvalues of $ AB$ and $ BA$ are the same, it turns out that 1 is also not an eigenvalue of $ BA$, and so $ det(I \minus{} BA) \neq 0$, which means that $ I \minus{} BA$ is invertible.

[quote]I wonder how we could proceed to find the inverse?[/quote]
Precisely. I like both Carcul's and Svejk ' s Methods for proving existance of inverse. Could someone please throw some light upon how one can find the inverse?

[quote="t0rajir0u"][quote="othman"]to t0rajir0u : why the result is true in any ring ? [/quote]
I don't actually know, which is why I'm wondering if anyone has a nice proof.  This problem was given to me by Kiran Kedlaya.  According to him, the motivation is that both the result and the identity are obvious if $ AB$ is nilpotent, but that doesn't provide a means to a proof at all.[/quote]
I hope [url=http://www.mathlinks.ro/weblog_entry.php?t=241701]this[/url]helps.


Consider a rainstorm with no wind and a uniform rainfall. In order to stay the driest, is it better to run through the rain from point $C$ to point $P$, walk slowly, or does it even make a difference?

isnt it like if you were to map the points of a raindrops on a cartesian plane..there are infinitely many lines that you can draw such that no more than a drop touches it...hmm i cant remember but i have seen it before and i think i still have the proof..

It depends: on whether you have infintesimal thickness. Also, I think for arbitrarily chosen points P and C, there is probability       .0000something that a drop will touch you, but if you walk slowly, I think that more drops will hit you...

Is it raining on a cartesian coordinate plane that has nothing on it except you?

in the real world with the rain going straight down, 4km/h is roughly the best speed, apparently...


Fie \[l=\lim_{n\to\infty}\frac{2n}{2n-1}\cdot\frac{2n+2}{2n+1}\cdot\frac{2n+4}{2n+3}\cdots\frac{4n}{4n-1}\] atunci l este egal cu:
a) 2
b) $+\infty$
c) $\sqrt{2}$
d) 1

could you explain a little bit how did you get that inequality pls.?
or maybe there's a  link where i could see the solution?tks.


This is just like an equation; $ \ 3x \plus{} 2 \equal{} 14$ where $ \ x$ is the amount of marbles in a bag.
Solve for $ \ x \equal{} \boxed {4}$


inequalities
Cauchy Inequality
inequalities proposed

Prove that: $ \forall a;b;c>0:$
    $ \sum\frac {1}{b(c \plus{} 1)}\geq\frac {3}{abc \plus{} 1}$

[quote="karis"]Prove that: $ \forall a;b;c > 0:$
    $ \sum\frac {1}{b(c \plus{} 1)}\geq\frac {3}{abc \plus{} 1}$[/quote]
It's ture.And we have the stronger one：
Let $ a,b,c>0$,then $ \sum\frac{1}{b(1\plus{}c)}\geq\frac{3}{\sqrt[3]{abc}(1\plus{}\sqrt[3]{abc})}$

Can you post your solution ,Jianxing113725?

Let $ k=\sqrt[3]{abc},k>0$,then $ abc=k^3$,denote $ a=\frac{ka_2}{a_1},b=\frac{ka_3}{a_2},c=\frac{ka_1}{a_3},a_1,a_2,a_3>0$
So our inequality becomes
\[ \sum\frac{1}{\frac{ka_2}{a_1}+\frac{k^2a_3}{a_1}}\geq\frac{3}{k(1+k)}\]
\[ <==>\sum\frac{a_1}{a_2+ka_3}\geq\frac{3}{1+k}\]
Applying Cauchy inequality and notice that $ (a_1+a_2+a_3)^2\geq3(a_1a_2+a_2a_3+a_3a_1)$,
\[ \sum\frac{a_1}{a_2+ka_3}=\sum\frac{a_1^2}{a_1a_2+ka_1a_3}\]
\[ \geq\frac{(a_1+a_2+a_3)^2}{(1+k)(a_1a_2+a_2a_3+a_3a_1)}\geq\frac{3}{1+k}\]
so the stronger is ture.
As $ k>0$,$ 1+k^3\geqq(k+1)$ is obvious ture,so the original is ture.


I roll a fair dice with the numbers 1,2,3,4,5,6.  What is the probability that I will roll five 5s before three 3s?

[hide="hint"]Only consider the 5s and 3s because everything else wont matter [/hide]

[quote="abacadaea"]wouldnt it simply just be 1/2[/quote]

no because there are 5 fives and only 3 threes, they arent equal.

[quote="xpmath"][hide] There are a few ways:
roll all 5s and then  three 3s; roll 5 fives, and a 3 in between, and then 2 more 3s; roll 5 fives and 2 threes and finish with 1 more 3.
Since each has a $ \frac{1}{2}$ chance of being rolled, it's:
$ \frac{1}{32}+\frac{4}{64}+\frac{6}{128}=\frac{9}{64}$?
I'm really tired so I probably made a mistake, but oh well.[/hide][/quote]

Umm actually not roll all 5s and three 3s.  Its roll all 5s only, because once you rolled all the 5s, you are done, you already rolled the five 5s before three 3s.  Same with other cases, you dont need to tack on the extra 3s at the end, you are done with rolling that final 5.  So basically

[hide="giveaway"]P(rolling five 5s in a row)+P(rolling 5 fives and 1 three, with the three not coming last-because theres no need to tact on extra 3s)+P(rolling 5 fives and 2 threes without a three coming in last)[/hide]

[quote="abacadaea"]using the hint, ignore everything else. roll a 5 or a three both one half. using logic, the probability of rolling a three is the same as a five, so rolling a three before a five has the same probability as rolling a five before a three. so ignoring everything else, we have 1/2. it doesnt say i will rol a five before a 3 on the first try. it just says i will.[/quote]

thats rolling 1 five before 1 three, not rolling [b]5[/b] fives before [b]3[/b] threes.  Its dependent on the other rolls, not independent.  Its not 1 five and 1 three.

Thats like say "what are the chances I roll 2 strait 6s on a dice?".  You would say 1/6 because there is a 1/6 chance of rolling a 6 on an individual throw.  However, thats not right because your second toss is dependent on your first toss.

Just like here, your fives that you roll are dependent if you have rolled fives before inorder to get 5 fives.


Jučer, u neobaveznom razgovoru sa profesorom Diskretne matematike (na temu 10. Hilbertovog problema-algoritmizacije rješavanja diofantskih jednačina), doznao sam da su svojevremeno dvojica matematičara iz Makedonije (inače Albanaca) predstavili tzv. makedonski algoritam koji pobija Matiyasevichevu tvrdnju o nepostojanju algoritamskog metoda rješavanja općih diofantskih jednačina. Dakako, poslije par godina, niko više nije spominjao (bez sumnje, pronađene su rupe u algoritmu), no zanimljivo je da nigdje na internetu nema pomena istog. Zna li iko išta više?

Bodane, Makaveli?

Oprosti na neznanju, što su opće diofantske jednadžbe?

Dobro pitanje, izostavio sam riječ polinomne (multinomne). Opće sam napisao jer su linearne algoritmizirane, one drugog stepena poprilično algoritmizirane, za treći stepen se vjeruje da algoritam postoji, ali nije nađen, a za stepen veći od tri gore pomenuti Matiyasevich je pokazao nemogućnost postojanja takvih algoritama.

Dakle, u ovom kontekstu, opća (loš izraz, istina, postoje i općenitije) polinomna (pa i nije baš polinomna, više multinomna) diofantska jednačina bi bila jednačina oblika $ p(x_1,\ldots,x_n)\equal{}0$ složena od stepenih i multivarijabilnih stepenih funkcija sa cjelobrojnim koeficijentima.

Kako sam ovo uspio zakomplikovati...

Baš me zanima Matijasevičev dokaz (kako bi izgledao dokaz da neki algoritam NE postoji?). Mogu li negdje vidjeti ideju kako je on to napravio?

[quote="Željko Jurić"]Među najvažnijim algoritamski nerješivim problemima nalazi se deseti Hilbertov problem, odnosno problem utvrđivanja da li zadana Diofantova jednačina ima rješenja ili ne. Pokazano je da je ovaj problem algoritamski nerješiv već za Diofantove jednačine četvrtog stepena, što je izazvalo veliko razočarenje. Nerješivost desetog Hilbertovog problema neposredna je posljedica jednog opštijeg rezultata koji je izveo Yuri Matiyasevich. Da bismo rekli nešto više o tome, preciziraćemo neke pojmove koje smo ranije intuitivno pomenuli prilikom izlaganja osnovnih pojmova teorije skupova. Za skup kažemo da je rekurzivno nabrojiv ukoliko postoji algoritam koji generira njegove elemente, preciznije algoritam koji za ma kakav zadani prirodan broj n terminira dajući na izlazu nekih n elemenata tog skupa. S obzirom da ima neprebrojivo mnogo podskupova skupa N (koji su svi prebrojivi), a samo prebrojivo mnogo algoritama, jasno je da postoje prebrojivi skupovi koji nisu rekurzivno nabrojivi. Dalje, za skup kažemo da je rekurzivan ukoliko postoji algoritam za utvrđivanje pripadnosti tom skupu, odnosno algoritam koji za svaki ulazni podatak x terminira dajući na izlazu odgovor da li x pripada tom skupu ili ne. Značajno je da postoje rekurzivno nabrojivi skupovi koji nisu rekurzivni. Na primjer, skup svih Turingovih mašina je prebrojiv, tako da sve Turingove mašine možemo indeksirati brojevima, i govoriti o prvoj, drugoj, trećoj i, općenito, n-toj Turingovom mašini. Posmatrajmo sada podskup H skupa dat kao
H = {n | n-ta Turingova mašina terminira} Prilično je lako pokazati da je H rekurzivno nabrojiv skup, ali nije rekurzivan (da jeste, Halting problem bio bi rješiv). Sada, Matiyasevitchev rezultat (poznat kao Matiyasevicheva teorema) tvrdi da je neki podskup A skupa rekurzivno nabrojiv ako i samo ako postoji Diofantova jednačina sa jednim parametrom a takva da je A jednak skupu svih vrijednosti a za koje ta Diofantova jednačina ima rješenja u skupu N (na primjer, skup složenih brojeva je skup svih brojeva a za koje Diofantova jednačina (x + 2) (y + 2) = a ima rješenja u skupu ). Iz Matiyasevicheve teoreme neposredno slijedi da bi, kada bi deseti Hilbertov problem bio algoritamski rješiv, tada bi se za svaki rekurzivno nabrojivi
skup moglo algoritamski utvrditi da li mu neki element pripada ili ne, odnosno svaki rekurzivno nabrojivi skup bio bi rekurzivan, a vidjeli smo da nije tako. Iz toga slijedi da deseti Hilbertov problem nije algoritamski rješiv.[/quote]

Eh, kako je do toga Yuri došao, kako su tu pomogli Fibonaccijevi brojevi, predstavlja već višu višu matematiku. Na netu ima resursa za dalje čitanje, a evo jedan ne pretjerano detaljan:

http://www.goldenmuseum.com/1612Hilbert_engl.html

[quote]Makedonski algoritam[/quote]
Dva članka o pomenutim autorima objavljeni su u časopisu "Računari u vašoj kući" broj 18 i broj 21. Pomenuti autori su navodno očas rešavali jednačine poput $ x^4 \plus{} 7x^3 \minus{} 2x^2 \plus{} F*x \plus{} 2F \minus{} 344 \equal{} 0$. Ovaj primer se može rešiti raznim metodama tako što se F prikaže kao kompleksni polinom. Priča se po Beogradu da su vodili pregovore sa nekim profesorima sa Beogradskog matematičkog fakulteta radi kandidovanja za neke nagrade. Pri tome su odbili da objave detalje o svom algoritmu. To se navodno dogodilo pre nekoliko godina. Ne želim da pominjem imena jer sam informaciju dobio od treće strane pa nemam potvrde o svemu tome.

U svakom slučaju reč je o dogadjanjima koja traju poduže verme. Kako je Matiyasevich davno pre pojave "makedonskog algoritma" objavio svoj dokaz, sumnjam da autori u to vreme nisu znali za to. Ostaje nejasno zašto su i dalje se držali svojih tvrdnji tim pre što do danas (prošlo je gotovo četvrt veka) nema nikakvih potvrda u njihovu korist.

Ako pretpostavimo da je njihov algoiritam zaista nešto novo, zanivljiv bi bio odgovor na pitanje da li je poznat primer jednačine koji je rešen "Makedonskim algoritmom" a ne može biti rešen drugim metodama. Takodje, interesantno je pitanje da li se "Makedonskim algoritmom" brže stiže do rešenja. Ko je uopšte, osim autora, video taj algoritam ?

Hvala na novim vijestima vezanim za naše 'južnjake'  :lol: Sad se i ja vraćam temi, nakon čitanja ovoga:

http://www.mkd.com.mk/science/algorith.htm

Ko ima vremena neka pročita ovaj review, ko ima knjigu neka pročita knjigu, i neka mi neko konačno objasni o čemu se ovdje radi  :lol: Svaka čast gospodinu inženjeru G., ali meni sve ovo visi u zraku...

Imam jedno pitanje za sve vas: Zasto nazivate algoritham "Makedonski"! ..

"Kvazi naucnice" i nesta slicno ima svugda u svetu (ili mozda u Bosni ne?), tako da mislim da nije u redu da "dajete ime" na ovo po celu jednu drzavu (kao i "ove nase juznjake").. So to vredzate puno ljudi koji stvarno nesta rade (mozda ne toliko bombasticno da pobiju Matijasevic ili rese P=NP, ali ipak stvarno rade)

Jos vedna stvar ovo je kako ste rekli neki albanac ili g-din Trpe. Posto se ne uklapa.. Ovaj prvi je mislim "dokazivao" velika Fermat teorema, dopustili su mu celi amfiteatar na fakultetu ali jedan profesor ga ja na samu prezentaciju pobio!

Stojan

P.S. Ja se izvinjavam za moji Srpski/Hrvatski/Bosanski ili kako zelite da go nazovete, u skoli ni smo ga ucili.

Stojane, nije mi bila namjera ikoga da vrijeđam - naime, za ovaj algoritam sam čuo pod imenom Makedonski algoritam, te sam mislio da, ako iko može baciti malo više svjetla na ovo, vjerovatno je čuo za taj algoritam pod ovim imenom.
Zašto mene zanima taj algoritam? Jasno da ne pobija Matijasevičevo rješenje, ali možda može biti koristan za rješavanje određenog tipa diofantskih jednačina - to bi moglo biti interesantno.

'naše južnjake' sam upotrijebio jer je Makedonija višenacionalna zemlja - pa kako više ne znam da li su makedonski Albanci ili Makedonci iza tog pokušaja, ne mogu napisati ni jedno ni drugo.

Primjetiti ovaj citat kod dr. Gruevskog:

[quote]Kako {to e poznato 10-tiot Hilbertov problem go privlekuvalo vnimanieto na mnogu istaknati matemati~ari vo svetot, no sepak vo op{t slu~aj (koga e dadena ravenkata so dve ili pove}e nepoznati) baraniot algoritam denes ne e pronajden. 1969 godina, Ruskiot matemati~ar M.Matija{evi} doka`al deka takov algoritam nikoga{ nema ni da se pronajde ( vo idnina ). Za `al, tvrdeweto ne e to~no ![/quote]

Zbog ovog paragrafa sam povezao Gruevskog sa pričom o ovom algoritmu.

Eto, ni sad ne znam detalje o ovom algoritmu, ali to više nije ni bitno. Samo mi je žao što je ovakav naslov shvaćen kao uvreda - bio je to prosto izraz koji sam ja čuo za taj algoritam - pa sam mislio da se taj naslov udomaćio.

Još jednom se izvinjavam - predloži kako da nazovem temu, nazvaću je tako.

Ma za ime sta znam.. ja mislim da i ne zasluzuje toliko paznje koliko mi dajemu  :)

Znas kako je bilo.. onako prelistujes lagano i vidis 'Makedonski algoritam' -> ajmo i to da vidimo i vidis nesta sta cesto cujes za svoju drzavu, ali to je

Ja mislim da slicna nesta ce da nadzes i u Makedoniju i u Bosni i u Americi, kao i pozitivne. Ono sta ja znam:

1) g-din Trpe vs. Matijasevic: hvala na linku koji si posocio, ni sam pre toga vidio tekst, mada nesto cuo. Primetio sam da je to (ili bar dio) objavljen u casopis iz Matematike 'Numerus' koji citaju skolarce u osnovne skole u 1996. Tad sam bio 6-ti razred i probacu da nadzem kot kuci (ama kat cu biti u Mkd) pa da skeniram i da ti posaljem nekako ako zelis.. I to je nazalost bio bas u periodu kat sam redovno citao (jer nije bilo neke knige  :P ), ali sva sreca ni sam previse citao statije, nego vise trazio neke zadace :) .

2) ovo za albanca je bio nesta slicno za velika Fermat-ova teorema. Sta ne znaci da mislim da su svi Albance takvi. Ja znam PhD student, albanac iz Makedonija, koji radi algebru sa prof. Mazur. Tako da uvek ima i iz jedne i iz druge svugda :)

Pozdrav

Ma naravno, gdje ima naučnika, ima i crackpot-naučnika, ne sumnjam - svagdje je tako (upoznao sam i previše takvih - i to samo u par godina, sve Bosanci koji mogu izvršiti trisekciju ugla ili oboriti teoriju relativiteta :) )

Da ti pravo kažem, ne moraš to ni skenirati - ovu sam temu (poslije godinu ipol) već uspio i zaboraviti, da ti danas nisi napisao odgovor, ne bih je se ni sjetio :D

Pozdrav!


Let $ a,b,c$ be positive real numbers, such that $ abc\equal{}1$.  Show that $ a^2\plus{}b^2\plus{}c^2\le a^3\plus{}b^3\plus{}c^3$, with equality if and only if $ a\equal{}b\equal{}c$.

From Cauchy, we have $ (a \plus{} b \plus{} c)(a^3 \plus{} b^3 \plus{} c^3)\geq(a^2 \plus{} b^2 \plus{} c^2)^2$

$ \implies a^3 \plus{} b^3 \plus{} c^3\geq\frac {(a^2 \plus{} b^2 \plus{} c^2)^2}{a \plus{} b \plus{} c}$.

Again, by Cauchy, $ (1 \plus{} 1 \plus{} 1)(a^2 \plus{} b^2 \plus{} c^2)\geq(a \plus{} b \plus{} c)^2$

$ \implies\frac {a^2 \plus{} b^2 \plus{} c^2}{a \plus{} b \plus{} c}\geq\frac {a \plus{} b \plus{} c}3\geq\sqrt [3]{abc} \equal{} 1$.

Combining this with the first result, we have $ a^3 \plus{} b^3 \plus{} c^3\geq(a^2 \plus{} b^2 \plus{} c^2)\cdot\frac {a^2 \plus{} b^2 \plus{} c^2}{a \plus{} b \plus{} c}\geq a^2 \plus{} b^2 \plus{} c^2$, as desired.

Equality occurs when $ a^2 \equal{} b^2 \equal{} c^2\implies a \equal{} b \equal{} c \equal{} 1$.


trigonometry
calculus
derivative
algebra proposed
algebra

Why it is said that no of roots of x3=0 is 3 repeated roots...

Then why sin(x)=1 has only one root in (0,pi)

An equation $ (x\minus{}r)^n \equal{} 0$ is said to have root $ r$ with multiplicity $ n$, so $ x^3 \equal{} 0$ has root $ 0$ with multiplicity $ 3$.
The equation $ \sin x \minus{} 1 \equal{} 0$ has one single root in $ (0, \pi)$, namely $ \pi/2$. But equation $ (\sin x \minus{} 1)^7 \equal{} 0$ will have root $ \pi/2$ with multiplicity $ 7$.

but sir,,

x^3 has degree of tangency 3 at x=0;
as up to second derivative it is zero at x=0;

this is true for sinx also at pi/2 cosx=0...
so should be two solns.??

where am i going wrong in this concept of mine?

Well, let's take $ f(x) \equal{} x^3$, with $ f'(x) \equal{} 3x^2$ and $ f''(x) \equal{} 6x$. Indeed $ f(0) \equal{} f'(0) \equal{} f''(0) \equal{} 0$.
Now let's take $ g(x) \equal{} \sin x$, with $ g'(x) \equal{} \cos x$ and $ g''(x) \equal{} \minus{}\sin x$. We have $ g(\frac {\pi} {2}) \equal{} 1$, $ g'(\frac {\pi} {2}) \equal{} 0$, $ g''(\frac {\pi} {2}) \equal{} \minus{}1$, so as you see the situation is quite different.


calculus
derivative
inequalities theorems
inequalities

Where can I get information about Lagrange multipliers? (Russian or english please)

Wikipedia should be sufficient, as should any textbook on multivariable calculus.

I used a calculus book for my first year called
$ \text{Calculus \minus{} Early Transcendentals 6th edition}$ by $ \text{James Stewart}$. I think it has a very good section on Lagrange Multipliers.

Really any calc 3 book will do.  If you know how to take partial derivatives, the method is pretty simple.


I came across this problem while preparing for a test: $ \prod_{j \equal{} 2}^{n}\left(1 \minus{} \frac {1}{j^{2}}\right)$
After working with it for about 4 days now without coming to a conclusion, I would like some help with it.  I know that this infinite product converges on $ \frac {n \plus{} 1}{2n}$
and I do see that it does by plugging in a few values, but I would like to know how to derive that answer.

[quote="mathisalanguage"]
After working with it for about 4 days now without coming to a conclusion, I would like some help with it.  I know that this infinite product converges on $ \frac {n \plus{} 1}{2n}$
[/quote]
Go ahead and prove it! If you don't see how you get it, use induction. I assume what you are saying is
$ \prod_{k \equal{} 2}^{n}\left(1 \minus{} \frac {1}{k^2}\right) \equal{} \frac {n \plus{} 1}{2n}$ which is correct.

If you write out the product, writing each term as a fraction, it is quite clear this is the value.

I don't quite understand what you said there Kalle. Could you (or anyone else) explain a bit more?

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

Kalle, how did you know to express it as $ \prod_{k \equal{} 2}^{n}\frac {(k \minus{} 1)(k \plus{} 1)}{k^2}$ and where did the $ \frac {1}{2}$ come from in this: $ \frac {(n \minus{} 1)!\frac {(n \plus{} 1)!}{2}}{(n!)^2}$ ?

[quote="mathisalanguage"]how did you know to express it as $ \prod_{k \equal{} 2}^{n}\frac {(k \minus{} 1)(k \plus{} 1)}{k^2}$ and not $ \prod_{k \equal{} 2}^{n}\frac {(k^{2}) \minus{} (k \plus{} 1)}{k^2}$[/quote]

$ 1 \minus{} \frac{1}{k^2} \equal{} \frac{k^2 \minus{} 1}{k^2} \equal{} \frac{(k \plus{} 1)(k \minus{} 1)}{k^2}$.  This is a pretty standard example of a telescoping product.  Also, your original product is finite, not infinite -- there are exactly $ n \minus{} 1$ terms.

Thanks, JBL. Now I see that the $ (k\minus{}1)(k\plus{}1)$ is just a factorization of $ k^{2}\minus{}1$. I'm still curious as to where the $ \frac{1}{2}$ came from.

Oh alright dividing by 2 to get rid of the term that wasn't included in the series.  Thanks to everyone that helped resolve this problem.


I was doing 2004 Nats round the other day and got 24 wrong.
I got 2/5 and so did everyone from here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1837764692&t=134020

yet, the answer key says 5/16. What is the real answer?

Henri stands in the square on the bottom left corner marked "start." Tresa flips a fair coin. If the coin lands Heads, Henri moves one square upward: if it lands tails, henri moves one square to the right? This process is continued until henri moves into a square with a star, at which time the game ends. If this final square witht he star is shaded, Teresa wins; if it is not shaded, Henri wins. What's the probablility that Teresa wins? Express as a common fraction.

For the picture, search the other topic.

I got 5/16 on that one, so I'll explain.
The only ways she can get to the stars on the right are right-right-right, up-right-right-right, right-up-right-right, and right-right-up-right.

So the answer is

$ \frac {1}{8} \plus{} \frac {1}{16} \plus{} \frac {1}{16} \plus{} \frac {1}{16} \equal{} \boxed{\frac {5}{16}}$

Edit: said left instead of right

Yes. $ \frac {5}{16}$ IS the right answer (doing the problem the opposite way as andersonw). Find the number of ways Henri can win, and then subtract from 1 to get Teresa's chance. OK. Call the three unshaded squares that Henri will win if he lands on them 1, 2, and 3 in order from left  to right. The chance he goes to 1 is $ \left(\frac {1}{2}\right)^2 \equal{} \frac {1}{4}$. There are two paths to 2, so $ 2\left(\frac {1}{2}\right)^3 \equal{} \frac {1}{4}$. There are three paths to 3, so $ 3\left(\frac {1}{2}\right)^4 \equal{} \frac {3}{16}$. $ 1 \minus{} \frac {1}{4} \minus{} \frac {1}{4} \minus{} \frac {3}{16} \equal{} \boxed{\frac {5}{16}}$

Yeah, I got 5/16 too.
I have no idea how to use LATEX.
Haha.

Ok. So the topmost shaded square is inaccessible if Henri must stop when he reaches a starred squared. That's out.
Then, there's the middle shaded square. Well, so reach that in a series of rights(R's) and ups(U's), there must be three R's and one U. There are 4 ways to arrange the letters URRR, but RRRU passes through a starred square, so there are only three possibilities there.
The bottommost shaded square is only accesible one way (since Henri can't move down) : RRR.
So since the probability of a R or a U is equal, the probability of getting to the middle shaded square is 3 x (1/2)^4 because there are four moves, and each has the proability of 1/2 of happening. That's 3/16. Then RRR has a probability of (1/2)^3 because there are three moves and each has the probability of 1/2 of happening. That's 1/8.
3/16 + 1/8 = 3/16 + 2/16 =[b] 5/16[/b]


Let $a, b, c$ three positive real numbers. Prove the following inequality

                         \[ 2(a^{3}+b^{3}+c^{3})\geq 3(a^{2}+b^{2}+c^{2}-1).  \]

 [i] Cezar Lupu [/i]

[hide=Very nice. Thank MathsZ.][quote=MathsZ][quote=sqing]Let $a, b, c$ be positive real numbers. Prove that
$$a^3+b^3+c^3\geq \frac{3\sqrt 2}{2}(a^2+b^2+c^2-2)$$
$$a^3+b^3+c^3\geq \frac{3\sqrt  3}{2}(a^2+b^2+c^2-3)$$
$$a^3+b^3+c^3\geq 3(a^2+b^2+c^2-4)$$[/quote]

[u][b]Generalization :[/b][/u] Let $a, b, c$ be positive real numbers. We have $$a^3+b^3+c^3\geq\frac{3\sqrt k}{2}(a^2+b^2+c^2-k)$$

[i]Proof :[/i] By $AM\geq GM$, $$2\left(a^3+b^3+c^3\right)+3k\sqrt{k}=a^3+a^3+k\sqrt{k}+b^3+b^3+k\sqrt{k}+c^3+c^3+k\sqrt{k}\geq3\sqrt{k}a^2+\sqrt{k}b^2+\sqrt{k}c^2=3\sqrt{k}\left(a^2+b^2+c^2\right)$$$$\Leftrightarrow a^3+b^3+c^3\geq\frac{3\sqrt k}{2}(a^2+b^2+c^2-k)$$

$\square$.

Thanks sqing ![/quote]

[/hide]
Let $a, b, c$ be positive real numbers. Prove that$$a^3+b^3+c^3\geq\frac{3\sqrt k}{2}(a^2+b^2+c^2-k)$$
Where $k>0.$

It suffices to show that $\sum_{\text{cyc}}(2a^3-3a^2+1)\ge0$, which is true since $2a^3-3a^2+1=(2a+1)(a-1)^2\ge0$.


number theory
least common multiple
geometry
3D geometry

I remember seeing this problem.. I don't know how to prove it.

It says:

For $x,y,z>0$, prove that there are infinitely many pairs in which $x^3+y^5 = z^7$. I think the question is right unless I remember it wrong of course.

Can anyone show proof on this? (This seems bit obvious but I can't find a way to hit this problem)

[quote="blahblahblah"]Here's one way.

Step $1$

Find a solution $(a,b,c)$

Step $2$

$(at^{35},bt^{21},ct^{15})$ is also a solution.[/quote]

Ok few little problems..

I can't do step 1...  :( It seems easy to just guess the numbers but none of my guess (I tried for some with calculator) works.. Maybe more algebraic way is better.

For step 2, can you tell me how you came up with those?

Thanks.

For step 2, he just found $LCM(3,5,7) = 105$, so if you multiply through by any number to the power of $105$, the equality remains true. We need the LCM so that it can be distributed into the cube, 5th power, and 7th power... namely, $(t^3)^{35} = (t^5)^{21} = (t^7)^{15} = t^{105}$.

Step 1 is very difficult, unfortunately.  There is probably a better way.

Try finding solutions of the form $(x,y,z) = (2^a, 2^b, 2^c)$.

[quote="blahblahblah"]Wow, I'm pretty stupid sometimes.  So we just need to solve the equation $30t+1=7c$ in positive integers.  $7^{-1}_{15}=13$, so $c=13, t=6$ works.  This inevitably leads us to the solution $(2^{30},2^{18},2^{13})$, and the rest follows from what I previously noted.[/quote]

How did you get your answer? What do you mean by solving $30t+1 =7c$.

Sorry, it takes sometime for me to understand these things. I got step 2 though. :lol:

Silverfalcon,

I condensed my work a lot there.  We want to find powers of $2$ such that $2^{3p}+2^{5q}=2^{7r}$.

Step 1:  If $3p=5q$, then the LHS becomes $2^{3p+1}=2^{7r}$.  We can write $p=5t$, $q=3t$, and equating exponents in the previous equation means we just need to solve $15t+1=7r$ (there's a typo in my last post, but the answer is correct).  We want to find solutions $(t,r)$ where both are positive integers.

We want to find the inverse of $7$ modulo $15$.  This is $13$, as I stated earlier.  So setting $r=13$, we see $t=6$

Hence $p=30$, $q=18$, $r=13$.


geometry
geometric transformation
reflection
percent
calculus
AMC
AIME

As a student entering college next year, I face one unsolved dilemma.

During years in high school, I participated in a lot of math cotests but never did well enough to be nationally recognized as many others do.

I do indeed enjoy math but I often question my ability or "worthiness" to pursue mathematics since there are so many nationally or internationally recognized geniuses out there.

Should I be opmitstic and pursue mathematics anyway or should I be "realistic" and pursue other areas that may involve some math?? How good is performance on such math contests an indicator of success as a mathematician??

Help!

I don't think you should bother yourself with the question of being 'worthy' to pursue Mathematics. The best Mathematicians aren't the ones who do best at Mathematical Olympiads but the ones who pursue a problem until they reach a solution. To a Mathematician, it does not matter if it takes you 5 hours or 5 years (or never) to reach a solution, it is the fact that a solution exists that matters. High school competitions are not a good reflection of how 'worthy' you are for a Mathematical career they are merely a way for students to push themselves beyond the school curriculum. Olympiads (and competitions) often require 'tricks' while real Mathematics rely more upon theorems (of course, a trick now and then does help) but the point is these competitions aren't really a good indicator of what Mathematics as a whole is about.

So I think you should just go with what you are passionate about. Go ahead, study Mathematics.

You should also change "Worthy enough to be a Mathematician?" to "Love math enough to be a Mathematician?". If you love math, then you are born to do math :D

[quote]How good is performance on such math contests an indicator of success as a mathematician??[/quote]
No good at all. Contest conditions could be a downright bugger, but you could (mathematically) thrive in a different environment. Don't be put off from what you really want to do because you don't think you'll be good enough for it - there are a heck of a lot of worse people than you that will have gone off to 'do mathematics' and ultimately do really well.

I know exactly the way you feel, because once I felt like you. I have also participated in math contest, but have not been "enough good" to get to national contests, i had only one chace because of my age. when i didnt go to the national contest, i felt bad, mostly because i REALLY wanted to be recognized, you know, its good!!

I also wanted to study math, but i thougth i wasnt "that" good at it. Look, being a mathamathician is cool, but being a "not-so-good mathemathician" or "not-good-enough mathemathician" it's really VERY VERY BAD, also in terms of money and personal success. I wanted to be a good one, because if you want to study something you obviously want to be the best. But when you know that you are goig to be "more for the stock" the illusion disappears. I recommend you to study engineering, even when you are not going to have the same emotions you would experiment by being a mthemathician, you will be a person that can travel into both worlds, the imaginary maths and the real things.

Also, by havent gone to a national olympiad it is not the end of world. It only infdicates that you were not yet prepared for that. I still feel nostalgic for having not gone to national contest. But i make my try, I still can do something, propose a problem, help teaching the smaller ones, theaching your friends, I dont know, its a frustratig experience but life continues and the point is not to get stuck on math, you know, there is a wide world outside waiting for you , people of talent.

Dont feel like you are dumb or something, if you want to be a matemathician, go on it and dont expect to be a bad mathematician, some people like us have to invest more effort, and I even think is an advantage, no one can tell us that "we have cheated". No one can tell us that we didnt have efforts. No one can.

I'll tell you what i done. I'm studying engineering, the idea of investing all time in solving math problems is good, but is good for me at the age of 17, not by all life. I can take as a creative, beautiful, overhelming, precious entertainment, for all my life, but when my kiddies starve for being "not-a good-enough mathemathician" i would regret. I also like engineering, because we have maths and we have real applications.

At last, I want to say that school are not only schools. Schools are magic schools. They teach us magic, maths teach us about thinking, chemistry teaches us about matter manipulation , physics tells us the cause of things being that way, medicine gives us the power to heal human beings,..., I dont know, every power is good, you have to decide what power do you want to give you the bread at home.

Att. Ernesto H. Escobedo.

GOOD LUCK

Let us take an example.

Andrew John Wiles

Famous for proving Fermat's Last Theorem at the age of 41

What did he do during his high school years? Who knows! Who cares!

The important thing is that he resolved an old riddle.

Did his high school achievements really matter compared to his success in proving Fermat's Last Theorem? Probably not.

And also, if you dont think you are willing to be a mathemathician, you can take it as a hobby and have MUCH, MUCH MORE FUN!!!!!! :lol:  smile!!!

If you really wanna be a mathematician, then tell yoursef, don't mistake absolute advantages for comparative advantages.
But I don't really believe in that  :P
I  think it's better to take it as a hobby

Andrew Wiles heard about Fermat's last theorem in [i]high school[/i].

Listen man, I'm majoring in math too. Fall semester of 2008. I also, I'm no olympiad, in fact, I've never even qualified for AIME. Here's my take on things.

Many of the other responses hit on the heart of the subject. I think how you proceed here really has no bearing on your innate abilities. I will say however, that if you have a talent for math, that is a very significant factor for consideration. The ability to define and solve problems mathematically can put food on the table. I do not mean in front of a classroom if you want to live comfortably.

For the people who have no real love for math, nor a dislike for it, but they posses talent: engineering. A good 40 percent of the kids in my calculus class are intending on majoring in some type of engineering.

I am the only math major.

[i]Why?[/i] comes to mind. Well, one sign is I can do problems that they can't. If you are even considering a math major I'll bet you can too.  Sure, maybe you can't do it in three minutes (the average time alotted for each problem on the AMC if you were to complete all 25), but you can probably reach the solution.  What really drives me though, isn't that I'm good solving hard problems.

It's that I [i]have[/i] to solve that problem. I will take as long as I need.

Ultimately, who gives a flying **** if you don't have a good score on those tests. It takes so much time to become knowledgable enough to do that. Most the kids who are perfect scorers have been studying comprehensively for years. If you know you should be a math major, be a math major. Go in general or something to get a feel for it. You don't have to declare for a while. But I must reiterate that a mathematics major isn't really something you should dedicate yourself to if you just want to be nationally reknowned, or desire a very comfortable life.

Whatever choice you settle on, good luck and don't second-guess your abilities. At all times, have faith in yourself.

Sincerely,
Grant Holzhauer

[quote="igiul"] Well, one sign is I can do problems that they can't. If you are even considering a math major I'll bet you can too.  Sure, maybe you can't do it in three minutes (the average time alotted for each problem on the AMC if you were to complete all 25), but you can probably reach the solution.  What really drives me though, isn't that I'm good solving hard problems.

It's that I [i]have[/i] to solve that problem. I will take as long as I need.
[/quote]

After all, you get satisfcation from doing problems others can't. What if somebody in your class is interested in math, can't do many problems, and get frustrated?

[quote="igiul"]Andrew Wiles heard about Fermat's last theorem in [i]high school[/i]. [/quote]

Actually, he heard about it when he was ten years old.  But that in no way detracts from the validity of your point.  :)

[quote="t0rajir0u"]Actually, he heard about it when he was ten years old.  But that in no way detracts from the validity of your point.  :)[/quote]

 :rotfl:
Only you. (That's in a good way <3)

And in response to soarer, there are many problems I can't do. Snag an AIME, any AIME; I'm sure one will get me. I'm still trying to solve [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=182703]this[/url] without becoming a danger to myself an others (I'm succeeding at the latter, but regrettably not the former) and I got so angry the other day because I couldn't solve a problem after 3 hours of intensive labor. Eventually I found out what I needed to learn though, so I'm calm now but my point still stands:

We all have our difficulties. Read [url=http://meganmcardle.theatlantic.com/archives/2007/12/a_for_effort.php]this.[/url]

Well, what do you mean by "be a mathematician?"  This is a different from "pursuing mathematics."  Here are some different answers to different questions:

Majoring in math in college:  Anyone who enjoys math can major in math in college.  Majoring in something doesn't necessarily mean you're exceptional at it.  You don't have to be the best in your department or take the hardest classes.  You would never hear a chemistry major say, "Gee, I like chemistry, but am I really good enough to study chemistry?"  (This is the unfortunate mentality that comes from contest math culture.)  Moreover, a math degree leaves doors to a variety of careers open, so I say: If you want to learn math, go learn math.

Getting a PhD in math:  To get a PhD in math, you really have to *love* math.  It will be a big chunk of your life for many years, and you have to love math in order to be able to produce a thesis.  So-called "ability" is less important than you might think.  One reason is that there are plenty of 3rd rate PhD programs out there.  Getting a PhD from a top school requires more "ability" but it also means that you had to get in to the school first, and most admits have the ability to finish.  *Much* of the attrition in grad school comes from a lack of drive or a lack of interest.  However, just because you love math, that doesn't mean you should get a PhD in math, because there's the thorny issue of what to do with your life after the PhD.  (See below.)  There's also the risk that maybe you don't love math as much as you think you do, and you lose 3 years of your life figuring that out.

Being a research mathematician:  I won't lie to you.  This is hard.  There are tons of freshly-minted math PhD's every year, and only a small fraction will go on to long careers in research.  Even with a PhD from a top graduate school, landing a permanent research job is a challenge.  One must have both passion and talent for research to succeed.  However, talent for research is NOT even remotely the same thing as contest math ability.  It isn't even the same thing as "classroom math" ability.  It's a totally different beast.  Are Putnam champs likely to be good at research?  Yes, but most working research mathematicians were not contest math superstars.

Teaching math at the college level:  This is also pretty hard.  Again, there are many more PhDs produced than there are permanent teaching positions.  The "ability" thing isn't a big deal, and contest math is 100% irrelevant.  And if you really really want to teach math, there's always community college and high school.

Becoming a world-famous, Fields Medal-winning, Millennium problem-solving mathematician:  If this were even a remote possibility, it should already be obvious to those around you that you are a bona fide math genius.  (This comment is meant to rebut the notion that maybe you're the next Andrew Wiles, and you don't even realize it.  This scenario, while theoretically possible, seems highly unlikely to me.)

Btw how can people find if they really really like math anyways??
People usually "like" things they are good...

what about age ?
Is it too late to start, suppose, at 25-30-40 etc ?

There's a reason the Fields medal is only awarded to mathematicians under 40.  To [i]enjoy[/i] mathematics, age is irrelevant.  To [i]be accomplished[/i] in mathematics is a different matter; I believe it's very difficult to start if you haven't been doing mathematics for a long time already.

i don't think olympiads are a good test of ur mathematics ...or ur luv for it...it's a test of ur skills trick and knowledge....
it's not the point whether Andrew wiles knew Feremat's theorem when he was a child or not....and by some **** law there exists another person in this world who knew about the Fermat's theorem at an age when he was samller than wiles when wiles came to know about it and and when the theorem was unsolved...
obviously he didn't suceed or may be he didn't try...
not all mathematicians proved Fermat's theorem...nor can all Prove Riemann hypothesis...but who cares...u do what u want nad in that process if u prove Riemann hypothesis it's done but if u don't who cares again....
atleast u did what  luv and u  rnever bored of it...
every year since USAMO,Putnam started there have been toppers ...say $ x$ number of them ...but surely not all $ x$ of them have won the Field's Medal...
ok i am not pushing u towards something but just telling u what is the truth and u need to beleive it...
love for mathematics is enuf and u need to be good at it and that follows if u luv it...
don't listen to all those who wrote here(including me) listen to what ur heart says and where u would like to see ur self 10 years hence.
Best of luck

You are worthy if you are interested in math.

I am a lousy problem solver in high school. I never have a chance to take part in IMO. Even if I have a chance, I think that I'll score less than 5 points.

But now I have solved several tough IMO problems, those that are solved by less than 29 contestants. I am confident about olympiad problems and confident about math research.

It took about 5 years for me to improve my math ability from a high school student who most likely score less than 5 points in IMO to a good problem solver who can solve tough IMO problems and confident about math research.

What you need is time. Your math ability will increase as you grow older. If you keep doing math passionately for 10 years, your math ability will definately improve a lot.

u can improve ur skills if u want and hence ur supposed 'talent' but what u can't is increase is ur instinctve love...yes ome argue u enjoy...but u only enjoy because u like it :)

True, but not quite: sometimes you start like doing something because you discover that you can be really good at doing it. So, in some cases your "instinctive love" arises or increases because of your skills. It is a chain reaction really: you (do not)enjoy something -> you spend (less)more time doing it -> you do it (worse)better -> you enjoy it (less)more ->...

But the answer to the main question of this thread is "yes, if you derive pleasure from doing mathematics, by all means go ahead and try it!". Just stop worrying about comparing yourself to other people and start comparing yourself to yourself a month ago. If you lose this comparison, then you are in trouble, indeed.

I think that the main dificult in the way to become a good mathematician,
(indeed if you are talented and enjoy math) is good supervisor, becouse I think
that one can really go on in math if he has for solving a real problem, which can be stated
by man who is truly established as a scientist.


just wondering who got in it?

I did, so did Rking

Livingston beat millburn by one point ><


A box contains exactly five chips, three red and two white. Chips
are randomly removed one at a time without replacement until all the
red chips are drawn or all the white chips are drawn. What is the
probability that the last chip drawn is white?

ends with white
WW
RWW, WRW
RRWW, RWRW, WRRW

ends with red
RRR
WRRR, RWRR, RRWR

$ \frac{6}{10}\equal{}\frac{3}{5}$

i got $ \frac{3}{5}$.

How do you know that all of the outcomes are equally likely?

That is an interesting question, because it *does* seem likely that three red balls would make WW very unlikely... but check it out, if you compute the probability for each one, you'll get 1/10 every time:
A few diverse examples:
WW=(2/5)(1/4)=1/10
RRWW=(3/5)(2/4)(2/3)(1/2)=1/10

Here's the easiest way I can think of to see how they're equally likely.

Consider ALL the patterns of length 5 possible in a list called T.  These are certainly all equally likely, right?  After all, it's much easier to see that all of their probabilities are $ \frac {3\cdot 2\cdot 1\cdot 2\cdot 1}{5\cdot 4\cdot 3\cdot 2\cdot 1} \equal{} \frac {1}{10}$.  Whoa... that means there are only 10 possibilities?  Of course!  BECAUSE our list WW, RWW WRW...etc [i]is just the list T where we ignore anything after the third R or the second W![/i]

Just think of it:  If we have pulled out all the reds, is there any doubt what the color of the next two balls is? No! Probability 1 that the next one (or two) drawn will be white.  Same thing goes for white, if we pull out all the whites.  The next ball(s) are guaranteed to be red, so it doesn't affect the probability.

So what I'm saying is, don't think of it as WW, think of it as WWRRR in that big list T.  We just ignore the RRR part because we know that's what HAS to come up next.

Hope this helps... or at least gets you thinking!

Homework:  Consider the experiment where you only pull three balls.  Will the outcomes be equally likely?  What is the probability now that the third ball is white?


If $ a,b,c$ are positive real numbers then prove the following inequality:
$ \frac{a}{\sqrt[3]{a+b}}+\frac{b}{\sqrt[3]{b+c}}+\frac{c}{\sqrt[3]{c+a}}\ge\frac{1}{\sqrt[3]{2}}(\sqrt[3]{a^{2}}+\sqrt[3]{b^{2}}+\sqrt[3]{c^{2}})$

this inequality was proposed in Crux Magazine  :) It itn't hard
You can see here
http://www.ch06k.com/forum/showthread.php?t=125

OK Thanks for link But I didn't understand what was written there
Please can anybody shortly explain me if there is any solution in your (chinese or vietnamese) forum...

P.S. I found this problem somewhere Maybe it was given in training camp.

[b]canada[/b] see also this topic: [url]http://www.mathlinks.ro/viewtopic.php?p=879005#879005[/url]

It is something like generalization of that solution.


$ \sum \equal{} \sum_{cyc}$.

$ a,b,c \in \mathbb{R^ \plus{} }$,

Prove that

$ 45\sum{x^4y^2} \minus{} 6\sum{x^3y^3} \plus{} 9\sum{x^4z^2} \plus{} 30x^2y^2z^2 \minus{} 10\sum{x^3y^2z} \minus{} 42\sum{x^3yz^2} \minus{} 6\sum{x^4yz} \ge 0$.

[hide="Postscript."]

I want prove it by AM-GM and power mean inequlity.

[/hide]

Could anybody give me the answer or a cue?

[quote="can_hang2007"]You should try first $ a \equal{} 1,$ $ b \equal{} 1000,$ $ c \equal{} 0.001.$[/quote]

I've found that I've copied wrong from a contest and I've edited.


function
real analysis
real analysis unsolved

$f(x,y)$ is defined on $(0,1]\times(0,1]$. $f(x,y)$ is measurable for a fixed $y$ and it is increasing for a fixed $x$. Prove that $f$ is measurable on $(0,1]\times(0,1]$.

Actually, we need to prove $\{(x,y)\,|\,f(x,y)<c\}$ is a mesuarable set.
It follows from the fact that:
\[\{(x,y)\,|\,f(x,y)<c\}=\bigcup_{y\in[0,1]} (0,y]\times\{x\,|\,f(x,y)<c\}=\]
\[=\bigcup_{q\in Q} (0,q]\times\{x\,|\,f(x,q)<c\}+\Omega_c,\]
where $Q=(0,1]\cap\mathbb{Q}$ and $\Omega_c$ is a zero-measure set. Oops... Now I think I don't know why $\Omega_c$ is a zero-measure set.


Prove that when breaking a candy with n-square shapes that after n-1 breaking it is completley divided

If the candy has irregular shape, then it doesn't always take $ n \minus{} 1$ breaks so i assume that the candy in question is rectangular.

[hide="Solution"]If it is rectangular, then it has $ a$ columns, and $ b$ rows, where $ a \cdot b \equal{} n$

The way we break it up is by separating the rows from each other, and then breaking up each row into unit squares individually. The number of breaks required to separate the rows is $ b \minus{} 1$, and the number of breaks for each row individually is $ a \minus{} 1$. There are $ b$ rows, so the number of breaks needed to completely separate the candy is $ b \minus{} 1 \plus{} b(a \minus{} 1)$ which multiplies out to $ b \minus{} 1 \plus{} n \minus{} b$.

This simplifies to $ n \minus{} 1$[/hide]


Bobby Shen is the best! He and his team has won nationals 5 years running! Hope his team can do it again.  :)

I didn't know that he's been on the Texas, Virginia, and Illinois teams for 5 years.  :wink:

His [b]school[/b] won TEXAS Mathcounts 5 years in a row. His state has only won 2 years in a row.

First Colony (Bobby's team) has won state the last 4 years.  Fort Settlement won state in 2005.  What they have done in the past means nothing.  The only thing that counts is how the teams do on the day of nationals.  That will be determined by ability and practice.  Championships are won in the practice room.

The best person at mathcounts at my school is abacadaea

I'm probably not the best, but I won our state, though.

Either I'm the best, I'm 2nd, or I'm tied with someone else for 1st. Because that person pretty much tied with me on school round, but on chapters, he got 3rd and i got 9th, and in states, he got 8th and i got 2nd. so... which one really counts, school, chapter, or state...?

depends on the difficulty levels.. I mean last year, the chapter meant virtually nothing more than a dumb mistake test to state champ-level ppl.


group theory
abstract algebra
topology
superior algebra
superior algebra solved

prove that the commutator subgroup of free group with rank $2$ is free of rank infinity.

More generally, a non-trivial normal subgroup of a finitely generated free group with infinite index cannot be finitely generated. In our case, the finitely generated free group is $F_2$, the free group on $2$ generators, and the normal subgroup is $H$, the derived subgroup, whose corresponding quotient group $F_2/H$ is the free abelian group on two generators, which is infinite.

In order to prove the first assertion, we could do this:

Let $F$ be the free group on $n$ generators, and let $N\ne 1$ be a normal subgroup of $F$ of infinite index. We want to show that $N$ is not free on finitely many generators (this looks awfully familiar, by the way; I might have posted something very similar before).

Suppose the contrary, i.e. that $N$ is a free group on finitely many generators. Let $X$ be a wedge sum of $n$ circles, which is a graph with fundamental group $F$, and let $\tilde X$ be a covering space of $X$ such that the homomorphism $p_*$ from the fundamental group of $\tilde X$ to that of $X$ induced by the covering map $p: \tilde X\to X$ is a monomorphism of $\pi_1(\tilde X)$ onto $N$. Since $N$ is normal in $F$, $\tilde X$ is a normal covering space, i.e. if $x$ is the unique vertex of $X$ and $\tilde x_1,\tilde x_2$ are vertices of $\tilde X$, then there is a homeomorphism $\varphi: \tilde X\to\tilde X$ commuting with $p$ which sends $x_1$ onto $x_2$.

We assumed that $N$ is a finitely generated free group, so $\tilde X$ is obtained from a tree by adding finitely many edges $e_1,\ldots,e_s$. Consider a point $\tilde x_1\in\tilde X$, and then another point $\tilde x_2\in\tilde X$ such that the distance from $\tilde x_2$ to $e_j$ is larger than that from $\tilde x_1$ to $e_i$ for all $i,j\in\overline{1,s}$. Let $\varphi$ be a homeomorphism of $\tilde X$ which commutes with $p$, and maps $x_1$ onto $x_2$, and let $\gamma$ be a closed path through $\tilde x_1$ which follows the shortest way to an endpoint of $e_1$, for example, traverses $e_1$, and then goes back to $\tilde x_1$ along the shortest path. $\gamma$ is mapped by $p$ onto a closed path in $X$ which represents a non-trivial element of $\pi_1(X)$. Because of the choice of $\tilde x_2$, however, $\varphi(\gamma)$ is a closed path in $\tilde X$ which represents the identity in $\pi_1(\tilde X)$, so it’s mapped by $p$ onto a path with the same property in $X$. This contradicts the fact that $p=p\circ\varphi$.

Think it might be Ok?

That's nice, but we want a stronger result in this case- that the group is actually free on some infinite set of generators.

I assumed the fact that a subgroup of a free group is free.

@grobber: your other post is here ;) : http://www.mathlinks.ro/Forum/viewtopic.php?t=56412


Just by AM-GM inequality
[hide]$abc(\frac1{a^{2}}+\frac1{b^{2}}+\frac1{c^{2}})$
$=\frac{ab}c+\frac{ac}b+\frac{bc}a$
$\ge a+b+c$
$\ge \sqrt{ab}+\sqrt{bc}+\sqrt{ac}$
$\ge \sqrt{a\sqrt{bc}}+\sqrt{b\sqrt{ac}}+\sqrt{c\sqrt{ab}}$
$\ge \frac{2a \sqrt{bc}}{\sqrt{(a+b)(a+c)}}$+$\frac{2b\sqrt{ca}}{\sqrt{(b+c)(b+a)}}+\frac{2c\sqrt{ab}}{\sqrt{(c+a)(c+b)}}$
[/hide]

[quote="white_horse_king88"]
So $\frac{\sqrt{bc}}{\sqrt{(a+b)(a+c)}}\leq \frac{1}{2}$
$\frac{bc}{(a+b)(a+c)}\leq \frac{1}{4}$
$a^{2}+a(b+c) \geq 3bc$, which if we sum cyclically gives $a^{2}+b^{2}+c^{2}+2(ab+bc+ca) \geq 3(ab+bc+ca)$, which is true by AM-GM.[/quote]

I might be wrong but I don't think this is a valid method of proof.

Yeah, you're right Splash.
Hikaru, how does the last line follow by AM-GM. I can't see very well. ... Probably very obvious.

[quote="white_horse_king88"]
Hikaru, how does the last line follow by AM-GM. I can't see very well. ... Probably very obvious.[/quote]
by AM-GM we have
$(a+b)(a+c)\ge 4a\sqrt{bc}$ :P
OK?

[quote="Hamster1800"]The last line could also be proved by rearrangement since $a^{2}+b^{2}+c^{2}\geq ab+bc+ca$.

Or by the trivial inequality
$(a-b)^{2}\geq 0$
$a^{2}-2ab+b^{2}\geq 0$
$a^{2}+b^{2}\geq 2ab$
Repeat for $(b-c)^{2}$ and $(c-a)^{2}$ and add to get

$2a^{2}+2b^{2}+2c^{2}\geq 2ab+2bc+2ca$
$a^{2}+b^{2}+c^{2}\geq ab+bc+ca$[/quote]
What do you mean by 'the last line'? the last line of my solution or white_house's. If that is white_house's i think this proof cant work.

That was white_horse's but I didn't check that his full proof was valid. I was just pointing out that the last line of his solution was indeed true.

[hide]
By Cauchy, $(a+\sqrt{bc})^{2}\leq (a+b)(a+c)$ etc., so that
\[\sum_\text{cyc}\frac{a\sqrt{bc}}{\sqrt{(a+b)(a+c)}}\leq \sum_\text{cyc}\frac{a\sqrt{bc}}{a+\sqrt{bc}}\]
But also $\frac{xy}{x+y}\leq \frac{x+y}{4}$ for all positive $x,y$ by AM-HM:
\[\sum_\text{cyc}\frac{a\sqrt{bc}}{a+\sqrt{bc}}\leq \sum_\text{cyc}\frac{a+\sqrt{bc}}{4}\]
Then by AM-GM, $\sqrt{bc}\leq \frac{b+c}{2}$ so the right hand side of above is less than or equal to $\frac{a+b+c}{2}$. It only remains to prove
\[abc\left( \frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}\right) \geq a+b+c\]
After clearing denominators, it's equivalent to
\[a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}\geq a^{2}bc+ab^{2}c+abc^{2}\]
which can be proven by AM-GM.
[/hide]

[quote="April"]Let $a,\,b,\,c>0$, prove that:
\[\frac{a\sqrt{bc}}{\sqrt{(a+b)(a+c)}}+\frac{b\sqrt{ca}}{\sqrt{(b+c)(b+a)}}+\frac{c\sqrt{ab}}{\sqrt{(c+a)(c+b)}}\le\frac{abc}{2}\left(\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}\right) \]
[/quote]
It's an easy problem and it's have many solutions.
[hide="Another way"] We have: \[\sum\frac{a\sqrt{bc}}{\sqrt{(a+b)(a+c)}}\le\sum\frac{a\sqrt{bc}}{\sqrt{4\sqrt{a^{2}bc}}}=\sum\frac{\sqrt{a\sqrt{bc}}}{2}\le\sum\frac{a+\sqrt{bc}}{4}\le\sum\frac{2a+b+c}{8}=\sum\frac{a}{2}\le\sum\frac{bc}{2a}=\frac{abc}{2}\sum\frac{1}{a^{2}}\][/hide]

After seeing the solutions, I was amazed at how weak the inequality was. So I tried to find a solution with a lot of simplification (weakening between steps)

First I threw a Cauchy:

\[\frac{a\sqrt{bc}}{\sqrt{(a+b)(a+c)}}+\frac{b\sqrt{ca}}{\sqrt{(b+c)(b+a)}}+\frac{c\sqrt{ab}}{\sqrt{(c+a)(c+b)}}\le \sqrt{\left(\sum a^{2}\right) \left(\frac{bc}{(a+b)(a+c)}\right)}\]

Now we hope that \[\sqrt{\left(\sum a^{2}\right) \left(\frac{bc}{(a+b)(a+c)}\right)}\le\frac{abc}{2}\left(\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}\right) \]
This is $\iff$: \[\left( \sum a^{2}\right)\left( \sum_{sym}a^{2}b \right) \le \left( \prod (a+b) \right) \frac{(abc)^{2}}{4}\left( \sum \frac{1}{a^{2}}\right)^{2}\]

We weaken even more, by throwing in a $\prod (a+b) \geq  8abc$ (weak am gm's multiplied) (oh yes, and raise the lower bound with a muirhead): \[\left( \sum a^{2}\right)\left( \sum_{sym}a^{2}b \right) \le 2\left(\sum a^{3}\right)\left( \sum a^{2}\right) \le \text{?? to be proved}\le 2(abc)^{3}\left(\sum \frac{1}{a^{2}}\right)^{2}\le \left( \prod (a+b) \right) \frac{(abc)^{2}}{4}\left( \sum \frac{1}{a^{2}}\right)^{2}\]

The ?? inequality to be proven is just \[\left(\sum b^{2}c^{2}\right)^{2}\geq  \left(\sum a^{3}\right) \left(\sum a^{2}\right) abc\]
which $\iff$ \[2\sum a^{2}b^{2}c^{4}+\sum b^{4}c^{4}\geq  \sum a^{6}bc+2\sum a^{4}b^{3}c\]
which is completely false by muirhead applied to cyclic sums, so we basically went way too far in the weakening process.

Just thought I'd post that little tidbit


i am new to latex anyone know anything i can use to learn latex?

Read the stickies and announcements in this forum. They contain many useful tips.

[quote="hasan4444"]Have a look at this:
http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php[/quote]

That's not up to date anymore. When you get there click on the AoPS Wiki Page.


function
combinatorics unsolved
combinatorics

The function $f : (0,...2005) \rightarrow N$ has the properties that $f(2x+1)=f(2x)$, $f(3x+1)=f(3x)$ and $f(5x+1)=f(5x)$ with $x \in (0,1,2,...,2005)$. How many different values can the function assume?

We must count how many numbers $n$ have their own value $f(n)$

For example, $f(9) = f(2*4+1) = f(2*4) = f(8)$
So the number 9 has not its own value because $9-1=8$ is divisible by 2

$0$ has its own value (because $0-1=-1$ is not a natural number)

A number n in $\{1,...,2005\}$ has NOT its own value iff n-1 is divisible by 2 either 3 or 5

But  $ n-1 \in \{0,...,2004\} .$
Since $f(1)=f(0), \ 1$ has not its own value, so we don't care about the case n=1,
hence we may count how many multiples of 2,3 and 5 are there in $\{1,...,2004\}$

In $\{1,...,2004\}$ there are

$[\frac{2004}{2}] \ = 1002 $ multiples of $2$

$[\frac{2004}{3}] \ = 668 $ multiples of $3$

$[\frac{2004}{5}] \ = 400 $ multiples of $5$

$Total=2070$

But we count two times the number 6 (one as a multiple of 2, one more as a multiple of 3)

So, we have to substract all multiples of 6.
Also, all the multiples of $10=2*5$ and the multiples of $15=3*5$

In $\{1,...,2004\}$ there are

$[\frac{2004}{6}] \ = 334 $ multiples of $6$

$[\frac{2004}{10}] \ = 200 $ multiples of $10$

$[\frac{2004}{15}] \ = 133 $ multiples of $15$

$Total=2070-(334+200+133)=1403$

We didn't finish yet!

Let's think how many times did we count the number $30=2*3*5$
-one time as a multiple of 2
-one time as a multiple of 3
-one time as a multiple of 5
Total: 3 times

Then we substracted:
-one time as a multiple of 6
-one time as a multiple of 10
-one time as a multiple of 15

Total 3-3=0 times

We lost at all number 30. Now we shall count once again the number 30 and all its multiples

In $\{1,...,2004\}$ there are

$[\frac{2004}{30}] \ = 66 $ multiples of $30$

$Total = 1403+66 = 1469$

There are $1469$ numbers in $\{1,...,2004\}$ which are divisible by at least one of 2,3 and 5

So, there are 2004-1469 = 535 numbers which are not divisible by either. We also marked number 0 as one of them.

Finally, the function can assume at most 536 different values

I'll post a generalization of this question

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=42897]Here![/url]


LaTeX
complex analysis
complex analysis unsolved

I am asked to find $ (i\minus{}1)^{\frac{1}{17}}$. This is my first class in complex analysis. I used this formula $ z^{\frac{1}{n}}\equal{} ^n\sqrt{r} \cdot exp [i(\frac{\theta}{n}\plus{}\frac{2k\pi}{n})]$ where $ k\equal{}0, \ldots, n\minus{}1$ and I got $ ^{34}\sqrt{2} \cdot exp [i(\frac{\minus{}\pi}{68}\plus{}\frac{2k\pi}{17})] k\equal{}0, \ldots, 16$. That's what I got after fully simplfying everything. Is this correct? I just want to make sure I did right.

it's the correct way to go except i think $ i\minus{}1\equal{}\sqrt{2}e^{\frac{3\pi i}{4}}$ while $ \sqrt{2}e^{\minus{}\frac{\pi i}{4}}\equal{}1\minus{}i$  :)

[hide="LaTeX tip..."]$ \sqrt[n]{r}$ looks better than $ ^n\sqrt{r}$  :) [/hide]


Can you suggest any viola pieces that has the melody, i've been working myself off just to find those type of pieces, all i know is that harry potter has the melody, but i need something harder!

Well, if there's another violist to share with, there's always the 6th Brandenburg Concerto. And if there's a violinist to share with, the Mozart Sinfonia Concertante.

Many pieces are borrowed from other instruments - for instance, the C minor concerto by J.C. Bach which was written for cello but works well on viola.

In orchestral pieces, we have to savor the moments we get. For example, the community orchestra I'm in is doing the Tchaikovsky [i]Romeo and Juliet[/i] fantasy overture. It's just one very brief moment in the piece, but the violas do get the very first appearance of the "love theme" all to ourselves. With mutes on. In a key signature with an absurdly large number of flats.

[quote="Kent Merryfield"]In a key signature with an absurdly large number of flats.[/quote]
To be specific, five flats. The lovely and unexpected key of D flat major. (In a piece for which perhaps the most common key is B minor - and which ends in B major. Tchaikowsky uses many different key signatures.) The melody is just 10 measures long; the accidentals in the melody are an A natural, a C flat, and a B double flat.


Let $\{x_n\}$ be a sequence are defined by:
\[x_{n+1}=x_n^2-2,x_1=5\]
Find :
\[\lim_{n\to \infty}\frac{x_{n+1}}{x_1 \cdot x_2 \cdots x_n}\]

I think we discussed similar problem sometime ago. We are to put $x=z+\frac{1}{z}$, then $x_n=z^{2^n}+\frac{1}{z^{2n}}$, therefore \[\frac{1}{x_1x_2...x_n}=\frac{z-1/z}{(z-1/z)x_1...x_n}=\frac{z-1/z}{z^{2^{n+1}}-1/z^{2^{n+1}}}.\] Thus the required limit is $z-1/z$ (since $z>1$ for the case $x_1=5$).


I think it is called [b]Vecten's point[/b] :
[color=darkblue][b]Let $ ABC$ be a triangle, and on the sides $ BC,AC,AB$ contruct squares in exterior with centers $ A_1,B_1,C_1$.Prove that $ AA_1,BB_1,CC_1$ are intersectating in the same point $ V$[/b][/color]

[b]Proof[/b]:
$ \Delta ACB_1 \sim \Delta ABC_1$ $ \Rightarrow \frac {AB}{AC} \equal{} \frac {AC_1}{AB_1}$ $ \Rightarrow$
\[ AB\cdot AB_1 \equal{} AC \cdot AC_1 \clubsuit
\]
But $ \angle BAB_1 \equal{} \angle CAC_1$ $ \spadesuit$
From $ \clubsuit$ ^ $ \spadesuit$ $ \Rightarrow AB \cdot AB_1 \cdot \sin BAB_1 \equal{} AC\cdot AC_1 \cdot \sin CAC_1$
$ \Rightarrow [ABB_1] \equal{} [ACC_1] (1).$
In the same way we have $ [BCC_1] \equal{} [BAA_1] (2)$ and $ [CAA_1] \equal{} [CBB_1] (3)$
$ \frac {[ABA_1]}{[ACA_1]} \equal{} \displaystyle\frac {\frac {1}{2} \cdot h_1 \cdot AA_1}{\frac {1}{2} \cdot h_2 \cdot AA_1} \equal{} \frac {h_1}{h_2}$ where $ h_1$ is the projection of point $ B$ in $ AA_1$ and $ h_2$ is the projection of point $ C$ on $ AA_1$.
But $ h_1 \parallel{} h_2$ $ \Rightarrow \frac {h_1}{h_2} \equal{} \frac {BA'}{A'C}$
In the same way
\[ \frac {[BB_1C]}{[BB_1A]} \equal{} \frac {B'C}{B'A}
\]

\[ \frac {[AC_1C]}{[BC_1C]} \equal{} \frac {C'A}{C'B}
\]
Observe that
\[ \frac {BA'}{A'C} \cdot \frac {B'C}{B'A} \cdot \frac {C'A}{C'B} \equal{} \frac {[ABA_1]}{[ACA_1]} \cdot \frac {[BB_1C]}{[BB_1A]} \cdot \frac {[AC_1C]}{[BC_1C]} \equal{} 1
\]
that is Ceva , therefore $ AA_1,BB_1,CC_1$ are ntersectating in the same point $ V$

[color=red][b]It is called Vecten's point?[/b][/color]

Dear Mathlinkers,
you are right, it is the first Vecten's point wrt ABC
The second is obtains by considering the three inner squares...
A synthetic proof is possible and Known.
Sincerely
Jean-Louis


superior algebra
superior algebra unsolved

i read from my book that $\sum_{d|n} \phi (d) =n$ (*)
they say that one can consider $M_d=\left[x\in G , |x|=d \right ]$
and the justification of * is true ,they say because of the fact
that : $|G|=\sum_{d|n} |M_d|$
but there they say that all $M_d$ are disjoint ,why is this ?!?!
because for example if we take $M_d$ and $M_{kd}$ ,we have$M_d\leq M_{kd}$ and
they are not disjoint altough $d|n,kd|n$ for example.
i dont understand,this book is really really annoying
i'm reading from nita,nastasescu "bazele algebrei",please help me!

Well it would be good to define everything ;)
I assume that $n=|G|$ and that $|x|$ means the order of $x$ (I prefer $ord(x)$).
The sets are disjoint because every element has just one order (the order of $x$ is the smallest postive integer $k$ with $x^k=e$; there can't be two different smallest inetgers for some property).


Prove that no positive integer power of 2 ends with four equal digits.

[hide="Solution"]
An integer is divisible by 16 iff the last four digits are divisible by 16. Therefore, any positive integer power of 2 (that is large enough to have four digits) must be a multiple of 16 and hence have the last four digits as a multiple of $ 2^4 \equal{}16$.

If we require the four digits to be equal (as in $ 0000, 1111, \ldots, 9999$), we can conclude that they must be 0000. Hence, any such power of 2 would end in four zeros, which would give it a factor of $ 5^4$, meaning that it could not be a power of 2. Hence such a number cannot exist; there can be no positive integer power of 2 that ends with four equal digits.
[/hide]

Shouldn't you also consider 4444 and 8888?


Given $n\in\mathbb{N}$ and positive reals $a_1, a_2, ..., a_n$ that satisfy $a_1 \cdot a_2 \cdot ... \cdot a_n=1$, we have that:

\[ \boxed{a_1^n+a_2^n+...+a_n^n\geq a_1^{n-2}+a_2^{n-2}+...+a_n^{n-2}}  \]

[hide="Prove"]
By nM-GM, we have that:

\begin{eqnarray*} \sqrt[n]{\frac{(n^2-2n+2)\cdot a_1^n+2a_2^n+...+2a_n^n}{n^2}} &\geq & \sqrt[n^2]{a_1^{n^2-2n+2}\cdot a_2^2 \cdot ... \cdot a_n^2} \\&=& \sqrt[n^2]{a_1^{n^2-2n}} \\&=& \sqrt[n]{a_1^{n-2}}\end{eqnarray*}

So, $(n^2-2n+2)\cdot a_1^n+2a_2^n+...+2a_n^n \geq n^2\cdot a_1^{n-2}$ and so on. Note that each number appears $n$ times and the sum of all its coefficients is $n^2-2(n-1)+2(n-1)=n^2$.

Summing all those inequalities, we get that:

$n^2\cdot(a_1^n+a_2^n+...+a_n^n)\geq n^2\cdot (a_1^{n-2}+a_2^{n-2}+...+a_n^{n-2})$

Finally:

$a_1^n+a_2^n+...+a_n^2\geq a_1^{n-2}+a_2^{n-2}+...+a_n^{n-2}$ $\blacksquare$
[/hide]

[b]Comment:[/b]

This is a generalization of the technique used by [b]darij grinberg[/b] to prove that $a^3+b^3+c^3\geq a+b+c$.

[quote="blahblahblah"]Stop posting the answers to your own questions.  One, the questions are excessively simple (this is just an application of Muirhead's inequality, after all) and two, this forum is for posing questions for others to solve, not for posting your solutions and having others critique them.[/quote]

I am sorry, but I thought that in this forum ([b]Inequalities Theorems and Formulas[/b]), I could post my own answers. :(

[quote="Ezbakhe Yassin"][quote="zhaobin"]it is ture that :
$\boxed{a_1^k+a_2^k+...+a_n^k\geq a_1^{m}+a_2^{m}+...+a_n^{m}}$
$k \ge m\ge 0$[/quote]

Without any condition?[/quote]
NO，
Sure the condition is
$\prod_{i=1}^na_i=1,a_i >0$

[quote="Ezbakhe Yassin"][quote="zhaobin"]it is ture that :
$\boxed{a_1^k+a_2^k+...+a_n^k\geq a_1^{m}+a_2^{m}+...+a_n^{m}}$
$k \ge m\ge 0$[/quote]

Could you prove it?[/quote]
use power mean inequality ,we get :
$\boxed{\frac{a_1^k+a_2^k+...+a_n^k}{n}\geq ( \frac{a_1^{m}+a_2^{m}+...+a_n^{m}}n)^{k/m}}$
is it obvious now ;)


If we write out this sum, we will group every term
2 ^ k /  (2 ^ k + 2 ^ l), where k isn't equal to l, with the term 2  ^ l ( 2 ^ l + 2 ^ k ).  However, when k = l, there won't exist a term to be grouped with this term. There  are 2006 terms that have k = l. This means that the number of pairs is 2006 ^ 2 - 2006 -- because 2006^2 is the total number of terms.
The sum of terms in each pair is 1, the value of each one of the remaining terms is 1/2. So the total sum will be:
2006 ^ 2 - 2006 + 2006/2 = 2006 ^ 2 - 1003
= 4023033.

I get the different result:

$\sum_{n=1}^{2006}\sum_{m=1}^{2006}\frac{2^n}{2^m+2^n}=\sum_{n=1}^{2006}\sum_{m=1}^{2006}\left(1-\frac{2^m}{2^m+2^n}\right)=2006^2-\sum_{n=1}^{2006}\sum_{m=1}^{2006}\frac{2^m}{2^m+2^n}$

In the second sum we can just exchange the letters $m$ and $n$ (write $m$ instead of $n$ and vice versa) to obtain the initial sum. Therefore we get $S=2006^2-S$ or $S=\frac{2006^2}{2}$

That would be correct, Farenhajt. My solution was to arrange the terms in a square matrix, where the $(m,n)$th sum is the entry $a_{mn}$, and then just pair terms on opposite symmetric diagonals, so they each sum to $1$. Thus, each term has a unique pair that sums to one, and so the total sum is half of the number of elements, which is $\frac{2006^2}{2}$.

Nice problem, but shouldn't the topic be called "Finite Sum?"

Masoud Zargar

Your answer is right, Farenhajt. I made a mistake in my solution -- I got the number of pairs wrong. It's actually (2006^2 - 1003)/2.Then, the total sum is
2006 ^ 2 / 2  - 1003 / 2 + 1003 / 2 = 2006 ^ 2 / 2.


Points A and B are located at $ \frac{1}{8}$ and $ \frac{1}{4}$, respectively, on a number line. What is the sum of the coordinates of the two points that trisect segment AB? Express your answer as a common fraction.

its $ \frac{3}{24}$ and $ \frac{6}{24}$.

two points that trisect AB are $ \frac{4}{24}$ and $ \frac{5}{24}$.

sum = $ \frac{9}{24}\equal{}\frac{3}{8}$.

ansewr : $ \frac{3}{8}$

How to find those points:

When two points trisect a segment, that means they split the segment evenly into 3 parts.

So first we find the distance between $ \frac{1}{8}$ and $ \frac{1}{4}$. It's $ \frac{1}{8}$. Then we divide this distance by 3, yielding $ \frac{1}{24}$.

We add this to $ \frac{1}{8}$ to get $ \frac{1}{6}$ and subtract this from $ \frac{1}{4}$ to get $ \frac{5}{24}$. (Do you see why?)

Better yet, just do $ \frac18 \plus{} \frac14 \equal{} \frac38$.

The average of the two points trisecting is equal to the average of the original two points.  Therefore, if you want to add the two trisecting points, simply multiply the average by 2, which equals the sum of the two original points.


geometry parallelogram perimeter trigonometry

The perimeter of a parallelogram is $40$, and its altitudes are $4$ and $7$. Compute $\sin{A}$.

-1991 NYSML

Let $x$ be one side, then the other side is $20-x$.  You have $x\sin A=7$, so then

$\left(20-x\right)\sin A=4=20\sin A-7\implies \sin A=\boxed{\frac{11}{20}}$

Let , x and y be the length of adjacent sides ,

 then x+y=20
also , 4x=7y

from above eqns , x =140/11

Hence sinA = 7 /x =7/140 *11=11/20

$P=40cm$
$AB=DC=a$
$AD=BC=20-a$
$A(ABCD)=(20-a)asinA=4a=7(20-a)$
$sinA=\frac{4}{20-a}$
$sinA=\frac{7}{a}$
By solving this system we get
$sinA=\frac{11}{20}$


calculus
integration
trigonometry
function
limit
logarithms
calculus computations

[quote="Fermat -Euler"]Find
\[ I = \displaystyle \int_{0}^{\displaystyle \pi/4} \left( \frac{1+\sqrt{\tan \theta}}{2} \right) \left( \frac{1+ \root \scriptstyle 4 \of { \tan \theta}}{2} \right) \left( \frac{1+ \root \scriptstyle 8 \of { \tan \theta}}{2} \right) \left( \frac{1+ \root \scriptstyle 16 \of { \tan \theta}}{2} \right)\, \, ...\, \, d\theta  \]
 ;)[/quote]

Are the dots suppose to mean that the function will go on?

like this for example:
$\displaystyle \int_{0}^{\displaystyle \pi/4} \left( \frac{1+ \root \scriptstyle 2n \of { \tan \theta}}{2} \right) ?$

[quote="Mutant"]
Are the dots suppose to mean that the function will go on?

like this for example:
$\displaystyle \int_{0}^{\displaystyle \pi/4} \left( \frac{1+ \root \scriptstyle 2n \of { \tan \theta}}{2} \right) ?$[/quote]

No, he means $\int_0^{\frac{\pi}{4}} \prod_{n=1}^\infty \frac{1+ (\tan \theta)^{\frac{1}{2^n}}}{2} \ d\theta$.

I think if you just examine $\tan \theta$ within the integrating interval, you'll see that the integral is zero.

[quote="Altheman"]say there are a finite number of terms, and we then compute the limit as the number of terms approaches infinity

let $\tan\theta=y$
$dy=\sec^2\theta d\theta=\sqrt{1+y^2}\, d\theta$
and we have $k$ terms, then

$\frac{1}{2^k}\int (1+y^{\frac{1}{2^1}})*(1+y^{\frac{1}{2^2}})*(1+y^{\frac{1}{2^3}})*...*(1+y^{\frac{1}{2^{k-1}}})*(1+y^{\frac{1}{2^k}})$

then multiply top and bottom by $(1-y^{\frac{1}{2^k}})$, then it becomes

$\frac{1}{2^k}\int(1+y^{\frac{1}{2^1}})*(1+y^{\frac{1}{2^2}})*(1+y^{\frac{1}{2^3}})*...*(1+y^{\frac{1}{2^{k-1}}})*(1-y^{\frac{1}{2^{k-1}}})*\frac{1}{ (1-y^{\frac{1}{2^k}})}$
this keeps on going to give

$\lim_{k\rightarrow\infty} \frac{1}{2^k}\int_{0}^{1} \frac{ 1-y }{ 1-y^{\frac{1}{2^k}} }*\frac{dy}{\sqrt{1+y^2}}$

sadly, i don't know much about integrals, so somebody else should finish[/quote]

You just answered the question  :P .

The answer will become zero because of $\lim_{k\rightarrow\infty} \frac{1}{2^k}\int_{0}^{1} \frac{ 1-y }{ 1-y^{\frac{1}{2^k}} }*\frac{dy}{\sqrt{1+y^2}}$

Notice that when $\lim_{k\rightarrow\infty} \frac{1}{2^k}$...............$\frac{1}{2^k}$ becomes zero when $k$ reaches $\infty$, so you multiply the integral with zero...

In the end...your answer is zero.

A correction: $dt=\frac{dy}{1+y^2}$, not $\frac{dy}{\sqrt{1+y^2}}$

No, it's not zero. We can interchange limit and integral; since $\lim_{k\to\infty}\frac{1-y^{\frac1{2^k}}}{\frac1{2^k}}=\lim_{t\to0}\frac{y^0-y^t}t=-\ln y$, the limit is
$\int_0^1\frac{1-y}{-\ln y\cdot(1+y^2)}\,dy$

That's the integral of a positive function, so it's positive.

That logarithm in the denominator is rather difficult to work with; I'm not sure there is a reasonable solution.

I guess I'll work with my last post a little more now...  If you expand the product for finite $k$, you'll get a geometric series.  The numerators of the exponents of the terms will be all nonnegative integers less than $2^k$ that can be written as the sum of distinct nonnegative integral powers of two, as well as zero; but all nonnegative integers can be written this way (think about binary), so you get a geometric series.

We then just have to take the limit of our last expression as $k$ goes to infinity.  Using some L'Hopital, we see that the integral is just ${\int_{0}^{\frac{\pi}{4}}\frac{\tan\theta-1}{\ln(\tan\theta)}}d\theta$ (don't worry too much about the endpoints, they're easy to resolve).  I don't think this integral is reducible, but you can use it anyway to get some good approximations for the original product integral: my calculator says the integral is approximately .512376629648.

Can anyone figure out how to do it for $\int_0^{\frac{\pi}{4}}\prod_{n=1}^\infty\frac{1+(\tan \theta)^{\frac{1}{3^n}}}{2}d\theta$?


An open paper tray is made of a rectangular piece of cardboard by cutting a square from each corner and turning up the flaps. Each corner is a 4-inch square. The length of the tray bottom is three times its width and the volume of the tray is 1452 cubic inches. What is the number of square inches in the surface area of the interior of the tray?
[asy]draw((0,0)--(11,0)--(11,5)--(0,5)--cycle);
draw((1,0)--(1,1)--(0,1));
draw((10,0)--(10,1)--(11,1));
draw((11,4)--(10,4)--(10,5));
draw((0,4)--(1,4)--(1,5));[/asy]

We have $ V\equal{}LWH\equal{}3W\cdot W\cdot 4\equal{}12W^{2}\equal{}1452\Rightarrow W\equal{}11$.  The surface area of the interior is given by $ 2\times (4W)\plus{}2\times (4\cdot 3W)\plus{}W\cdot 3W\equal{}32W\plus{}3W^{2}\Rightarrow S\equal{}715$.

Looks like Kouichi beat me to it.


See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=1041


algebra
polynomial
superior algebra
superior algebra unsolved

I have to prove that the number of element of a finite field is of the form $ p^n$ where $ p$ is a prime and $ n\in\mathbb{Z}_\plus{}$. I took the polynomial $ f(x)\equal{}x^{p^n}\minus{}x\in\mathbb{F}_p[x]$ and let $ F$ be its splitting field over $ \mathbb{F}$. Let $ S: \equal{}\{a\in F: a^{p^n}\equal{}a\}$. Then I showed that $ S$ is a subfield of $ F$. Now I have to show that $ F\equal{}S$ but how to do it? My idea was to show that every $ x\in \mathbb{F}_{p^n}$ satisfies $ x^{p^n}\equal{}x$ but the I realized that I don't know if $ \mathbb{F}_{p^n}$ exists.

[i]A[/i] finite field with $ p^n$ elements can be readily constructed.  You need to start off with a generic finite field of characteristic $ p$ and show that its cardinality is not divisible by any other prime; in other words, you're approaching the problem backwards.  Pick $ n$ so that $ x \mapsto x^{p^n}$ is the identity map and see what happens from there.

Okay. But doesn't this prove only that the number of elements in a finite field is of the form $ p^n$, not that for every $ n$ there exists a field with $ p^n$ elements?

Yes, which is what you asked in your question ;)

[quote="MeKnowsNothing"]Let $ S: \equal{} \{a\in F: a^{p^n} \equal{} a\}$. Then I showed that $ S$ is a subfield of $ F$. Now I have to show that $ F \equal{} S$ but how to do it? My idea was to show that every $ x\in \mathbb{F}_{p^n}$ satisfies $ x^{p^n} \equal{} x$ but the I realized that I don't know if $ \mathbb{F}_{p^n}$ exists.[/quote]

The way to do this is to show that $ S$ is a field. (The only part of this that is interesting is showing that it is closed under addition.)  Since a splitting field for $ f$ is [i]defined[/i] as a minimal field containing $ S$, the proof of existence is quite surely complete.

Like the previous posters, I not really sure whether you are trying to prove that every finite field has order $ p^n$, or that for every number $ p^n$, there exists a finite field of that order.


Source: AIME 1990 #9

A fair coin is to be tossed ten times. Let $i/j$, in lowest terms, be the probability that heads never occur on consecutive tosses. Find $i+j$.

[hide]
hmm..this calls for recursion.

Let $P(n)$ be the probability that a series of $n$ tosses have no consecutive heads.  Then If we threw a tail, we need the rest of the sequence to have no consecutive heads.  However, If we threw a head, we must throw a tail next, and a sequence that has no consecutive heads.  As the result, we get $P(n) = \frac{1}{2}(P(n-1)) + \frac{1}{2}(\frac{1}{2}P(n-2))$.  Obviously, $P(1) = 1$, and $P(2) = \frac{3}{4}$ (I'm not brave enough to assume $P(0) = 1$...), now work our way back up we get $P(10) = \frac{281}{512}$, so the answer is $793$
[/hide]

hope I didn't do anything wrong...If not, I just solved a number 9 in 5 min.!!! :first:


LaTeX
geometry
3D geometry
algebra
polynomial

If X-1/X=3+2*(root 2)
what is (X cubed)+(1/X cubed)
sorry, cannot make real question. do the best you can. please reply :D

[quote="Woohoo"]If $ X - \frac {1}{X} = 3 + 2 \sqrt {2}$
what is $ X^3 - \frac {1}{X^3}$[/quote]

$ \text{\LaTeX}$'ed.  As for the question, think complex numbers.

I posted it wrong. It's supposed to be Xcubed-1/Xcubed

Ah, then an elementary solution is much easier.  Try the difference-of-cubes formula to start with.

Could you please just solve it with the steps!?
Thanks

[hide="Solution"] Let $ a \equal{} x \minus{} \frac{1}{x}$.  Then

$ x^3 \minus{} \frac{1}{x^3} \equal{} \left( x \minus{} \frac{1}{x} \right) \left( x^2 \plus{} 1 \plus{} \frac{1}{x^2} \right) \equal{} a \left( \left( x^2 \minus{} 2 \plus{} \frac{1}{x^2} \right) \plus{} 3 \right) \equal{} a(a^2 \plus{} 3)$

Now, $ a^2 \plus{} 3 \equal{} (3 \plus{} 2 \sqrt{2})^2 \plus{} 3 \equal{} 20 \plus{} 12 \sqrt{2}$, so finally

$ x^3 \minus{} \frac{1}{x^3} \equal{} (3 \plus{} 2 \sqrt{2})(20 \plus{} 12 \sqrt{2}) \equal{} \boxed{ 108 \plus{} 76 \sqrt{2} }$

Generalization:  $ x^{2n\plus{}1} \minus{} \frac{1}{x^{2n\plus{}1}}$ can be written as a polynomial in $ a$ with integer coefficients. [/hide]


calculus
integration
calculus computations

[hide]Change variables according to x = -u:

$ I \equal{} \int_{\minus{}1}^1\frac{1}{(1\plus{}e^x)(1\plus{}x^2)}\,dx \equal{} \int_{\minus{}1}^1\frac{e^u}{(1\plus{}e^u)(1\plus{}u^2)}\,du$

Therefore,

$ \int_{\minus{}1}^1\frac{1}{(1\plus{}e^x)(1\plus{}x^2)}\,dx \plus{} \int_{\minus{}1}^1\frac{e^x}{(1\plus{}e^x)(1\plus{}x^2)}\,dx \equal{} 2I \equal{}$

$ \equal{} \int_{\minus{}1}^1\frac{1\plus{}e^x}{(1\plus{}e^x)(1\plus{}x^2)}\,dx \equal{} \int_{\minus{}1}^1\frac{1}{1\plus{}x^2}\,dx \equal{} 2 \int_0^1 \frac{1}{1 \plus{} x^2}\,dx \equal{} 2[\arctan x]_0^1 \equal{} 2 \cdot \frac{\pi}{4}$

and so $ I \equal{} \frac{\pi}{4}$.[/hide]

I can answer that.

I've seen many posts of kunny saying that japanese students don't learn somewhat advanced integration techniques, that's why kunny gives solutions like this one.