integration
inequalities
parameterization
function
real analysis
real analysis solved

$a>0$ $f:[-a,a]\rightarrow R$ such that $f''$ exist and Riemann-integrable
suppose $f(a)=f(-a)$
 $ f'(-a)=f'(a)=a^2$

Prove that $6a^3\leq \int_{-a}^{a}{f''(x)}^2dx$
Study equality case ?

Radu Miculescu

Integrate by parts:

$\int_{-a}^af''(x)x\,dx= f'(x)x|_{-a}^a-\int_{-a}^af'(x)\,dx$

$= 2a^3- f(x)|_{-a}^a =2a^3-0=2a^3.$

Now apply the Schwarz inequality:

$(\int_{-a}^af''(x)x\,dx)^2\le (\int_{-a}^af''(x)^2\,dx) (\int_{-a}^ax^2\,dx)$

$(2a^3)^2\le \frac23\,a^3 \int_{-a}^af''(x)^2\,dx$

So $\int_{-a}^af''(x)^2\,dx\ge 6a^3.$

The criterion for equality in the Schwarz inequality would be that $f''(x)=cx$ for some constant $c.$ Even with the boundary conditions in the problem, this leads to a considerable family of extremal solutions.

The equality case gives a one-parameter family of functions for fixed $a$.

These are $f(x)=\frac12x^3-\frac12a^2x+b$
It's not hard to work out from the given conditions and the fact that $f''(x)$ is a multiple of $x$.

nice solution Kent...
I got this after some computations
$\int_{-a}^a(f''(x)-3x)^2dx\ge0$
which is equivalent with the initial inequality
For the case of equality $f(x)=\frac12x^3-\frac12a^2x$  which verifyes the hypothesis conditions.


linear algebra
matrix
superior algebra
superior algebra solved

[color=green]Find all the units in M[size=75]2[/size](Z[size=75]2[/size])[/color]

Thanks guys!!!  :lol:

The $2 \times 2$ matrices over $\mathbb{Z}_2$¿
Then being unit is the same as having discriminant $1$, so just casework:
Let the matrix be $\binom{a \ \ b}{c \ \ d}$.
Then necessary and sufficient is that $ad+bc=ad-bc=1$, so lets take cases:
i) $a=1,d=1 \implies bc=0 \implies$ $b=0$ or $c=0$
ii) $a=0, d=0 \implies bc=1 \implies b=c=1$
iii) $a=0, d=1 \implies$ the same
iv) $a=1, d=0 \implies$ dito


Find all the real solutions of the following equation:
\[4x^2 - 40\lfloor x\rfloor + 51 = 0\] where $\lfloor x\rfloor$ is the biggest integer $\le x$.

[i][edited by mod: added $\LaTeX$][/i]

Hello Andreas,

I'am not sure, but is this what you mean:
4x²-40[x]+51=0
x_1=8.5
x_2=1.5

So the biggest integer <x_1=8

4x²-40*8+51=0
4x²-269=0
x_1=8.2006...
x_2=-8.2006...

The same we can do using x_2,however,no real solution is found then.
I hope this helps.

I didn't fully understand what Allegro meant, but a solution is given by the following:
You have $4x^2-40x+51>0$ for all $x<1.5$ or $x>8.5$.
Since $4x^2-40x+51 \leq f(x)=4x^2- 40 \left\lfloor x \right\rfloor +51$ all zeros are in the interval $[1.5,8.5]$. So you have only to look at $n \leq x < n+1$ for $n=1,2,3,4,5,6,7,8$, thus setting $\left\lfloor x \right\rfloor =n$ for all these $n$, solving the equation $x^2-40n+51$ for that constant $n$ and looking which solutions fulfil $n \leq x < n+1$.

But you can make this work a little bit easier by using the intermediate value theorem and the fact that $f(x)$ is strictly increasing in the intervals $[n,n+1[$ for positiv integers $n$:
$x^2-40n+51$ has a solution and then only one in $[n,n+1[$ iff $f(n) \leq 0$ and $f(n+1)>-40$.


inequalities
function
inequalities proposed

[quote="SUPERMAN2"]Let $ x,y,z > 0$ satisfy $ x^2 \plus{} y^2 \plus{} z^2 \plus{} xy \plus{} yz \plus{} zx \equal{} \frac {3}{2}$.Prove that
$ \sqrt {2(xyz \plus{} 1)} \geq x \plus{} y \plus{} z$[/quote]
$ \sqrt {2(xyz \plus{} 1)} \geq x \plus{} y \plus{} z\Leftrightarrow4xyz \plus{} 1\geq2(xy \plus{} xz \plus{} yz)\geq0.$
Let $ f(x,y,z) \equal{} 4xyz \plus{} 1 \minus{} 2(xy \plus{} xz \plus{} yz),$
$ x \equal{} \min\{x,y,z\}$ and $ t \equal{} \frac {\sqrt {x^2 \plus{} 3(y^2 \plus{} z^2 \plus{} xy \plus{} xz \plus{} yz)} \minus{} x}{3}.$
Hence, $ \sqrt {x^2 \plus{} 3(y^2 \plus{} z^2 \plus{} xy \plus{} xz \plus{} yz)} \equal{} 3t \plus{} x$ and
$ f(x,y,z) \minus{} f(x,t,t) \equal{} \frac {2(y \minus{} z)^2(x \plus{} (1 \minus{} 2x)(2t \plus{} x \plus{} y \plus{} z))}{6t \plus{} 4x \plus{} 3y \plus{} 3z}\geq0.$
Thus, it remains to prove that $ f(x,t,t)\geq0.$ But $ f(x,t,t)\geq0\Leftrightarrow$
$ \Leftrightarrow f\left(x,\frac {\sqrt {\frac {9}{2} \minus{} 2x^2} \minus{} x}{3},\frac {\sqrt {\frac {9}{2} \minus{} 2x^2} \minus{} x}{3}\right)\geq0\Leftrightarrow x^2(2x \minus{} 1)^2\geq0.$
What is your proof, SUPERMAN2? Thank you.

[quote="SUPERMAN2"]Let $ x,y,z > 0$ satisfy $ x^2 \plus{} y^2 \plus{} z^2 \plus{} xy \plus{} yz \plus{} zx \equal{} \frac {3}{2}$.Prove that
$ \sqrt {2(xyz \plus{} 1)} \geq x \plus{} y \plus{} z$[/quote]

$ p^2\minus{}q\equal{} \frac{3}{2}$ , $ 2r\plus{}2 \ge p^2$

$ \sqrt{\frac{3}{2}} \le p \le \frac{3}{2}$

1.By Shur : $ p^3\plus{}9r \ge 4pq \equal{}> r\ge \frac{p(4q\minus{}p^2)}{9}\equal{}\frac{p(p^2\minus{}2)}{3}$

$ \Rightarrow 2r\plus{}2 \ge \frac{2p(p^2\minus{}2)}{3}\plus{}2\ge p^2 \Leftrightarrow (p^2\minus{}2)(2p\minus{}3) \ge 0$ - true for $ \boxed{\sqrt{\frac{3}{2}} \le p \le \sqrt{2}}$

2. By "pqr" we have : $ q^*\equal{} \sqrt{p^2\minus{}3q}\equal{}\sqrt{\frac{9}{2}\minus{}2p^2}$ and $ r \ge \frac{1}{27}(p\plus{}q^*)^2(p\minus{}2q^*)$

$ \Rightarrow 2r\plus{}2 \ge \frac{14}{27}p^3\minus{}p\plus{}2\minus{}\frac{4}{27}(\frac{9}{2}\minus{}2p^2)^{\frac{3}{2}}\ge p^2$ - true for $ \boxed{\sqrt{2} \le p \le \frac{3}{2}}$

So inequality holdes for : $ \sqrt{\frac{3}{2}} \le p \le \frac{3}{2}$

Equality holdes for : $ (p\equal{}\sqrt{2}, q\equal{} \frac{1}{2}, r\equal{}0) \Leftrightarrow (x\equal{}y\equal{}\frac{\sqrt{2}}{2} , z\equal{}0)$ or $ p\equal{}\frac{3}{2}\Leftrightarrow (x\equal{}y\equal{}z\equal{}\frac{1}{2})$

[quote="SUPERMAN2"]Let $ x,y,z > 0$ satisfy $ x^2 \plus{} y^2 \plus{} z^2 \plus{} xy \plus{} yz \plus{} zx \equal{} \frac {3}{2}$.Prove that
$ \sqrt {2(xyz \plus{} 1)} \geq x \plus{} y \plus{} z$[/quote]
Another proof.
Let $ x\plus{}y\plus{}z$ and $ xy\plus{}xz\plus{}yz$ are constants.
$ \sqrt {2(xyz \plus{} 1)} \geq x \plus{} y \plus{} z\Leftrightarrow f(x,y,z)\geq0,$
where $ f(x,y,z) \equal{} 4xyz \plus{} 1 \minus{} 2(xy \plus{} xz \plus{} yz).$
But $ f$ is a linear  function of $ xyz.$ ( the condition does not depend on $ xyz$ )
Hence, $ f$ gets a minimal value on the values boundary of $ xyz.$
Id est, it remains to check $ f(x,y,y)\geq0$ and $ f(x,y,0)\geq0,$ which is true.

It's stronger than Schur 's ineq (degree 3).
And I think the stronger also true:
$ a^2\plus{}b^2\plus{}c^2\plus{} \frac{3\sqrt{3}abc}{\sqrt{a^2\plus{}b^2\plus{}c^2}}  \geq\ 2(ab\plus{}bc\plus{}ca)$
 :maybe:

[quote="nguoivn"]It's stronger than Schur 's ineq (degree 3).
[/quote]
Can you elaborate, please?  :)

[quote="nguoivn"]
And I think the stronger also true:
$ a^2 \plus{} b^2 \plus{} c^2 \plus{} \frac {3\sqrt {3}abc}{\sqrt {a^2 \plus{} b^2 \plus{} c^2}} \geq\ 2(ab \plus{} bc \plus{} ca)$
 :maybe:[/quote]
Since, $ a^2 \plus{} b^2 \plus{} c^2 \plus{} \frac {3\sqrt {3}abc}{\sqrt {a^2 \plus{} b^2 \plus{} c^2}} \minus{} 2(ab \plus{} bc \plus{} ca)$ is a linear function of $ abc$
we need  check only two cases:
1) $ b \equal{} c \equal{} 1$
2) $ c \equal{} 0.$ :wink:
The first case is wrong.  :(  Try $ a \equal{} \frac {1}{2}$ and $ b \equal{} c \equal{} 1.$
By the way, the following inequality is true already.
Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ a^2\plus{}b^2\plus{}c^2>0.$ Prove that:
\[ a^2 \plus{} b^2 \plus{} c^2 \plus{} \frac {\sqrt3(ab\plus{}ac\plus{}bc)\sqrt[3]{abc}}{\sqrt {a^2 \plus{} b^2 \plus{} c^2}} \geq\ 2(ab \plus{} bc \plus{} ca)\]

Here is my solution:
$ \sqrt{2(xyz\plus{}1)}\geq x\plus{}y\plus{}z$
<=>$ 2xyz\plus{}2 \geq x^2\plus{}y^2\plus{}z^2\plus{}2xy\plus{}2yz\plus{}2zx$
<=>$ 2xyz\plus{}2 \geq x^2\plus{}y^2\plus{}z^2\plus{}2(\frac{3}{2}\minus{}x^2\minus{}y^2\minus{}z^2)$
<=>$ 2xyz\plus{}2 \geq 3\minus{}x^2\minus{}y^2\minus{}z^2$
<=>$ x^2\plus{}y^2\plus{}z^2\plus{}2xyz \geq 1$
We use contradicting method,the problem is equivalent to:
"If $ x,y,z>0$ satisfy $ x^2\plus{}y^2\plus{}z^2\plus{}2xyz\equal{}1$,prove that $ x^2\plus{}y^2\plus{}z^2\plus{}xy\plus{}yz\plus{}zx \leq \frac{3}{2}$"
Because $ x^2\plus{}y^2\plus{}z^2\plus{}2xyz\equal{}1$,we can let $ x\equal{}cosA,y\equal{}cosB,z\equal{}cosC$,where $ \triangle ABC$ is acute.
We need to prove $ \sum_{cyc}cos^2A\plus{}\sum_{cyc}cosAcosB \leq \frac{3}{2}$,which is equivalent to an old problem:
$ cosA\plus{}cosB\plus{}cosC \leq \sqrt{sin^2A\plus{}sin^2B\plus{}sin^2C}$
See the post [hide="here"]http://www.mathlinks.ro/Forum/viewtopic.php?t=276934[/hide] for the proof of the above ineq.

[quote="arqady"]
By the way, the following inequality is true already.
Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ a^2 \plus{} b^2 \plus{} c^2 > 0.$ Prove that:
\[ a^2 \plus{} b^2 \plus{} c^2 \plus{} \frac {\sqrt3(ab \plus{} ac \plus{} bc)\sqrt [3]{abc}}{\sqrt {a^2 \plus{} b^2 \plus{} c^2}} \geq\ 2(ab \plus{} bc \plus{} ca)\]
[/quote]
We can prove it easily by Am-Gm, my friend  :)

[quote="arqady"]
By the way, the following inequality is true already.
Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ a^2 \plus{} b^2 \plus{} c^2 > 0.$ Prove that:
\[ a^2 \plus{} b^2 \plus{} c^2 \plus{} \frac {\sqrt3(ab \plus{} ac \plus{} bc)\sqrt [3]{abc}}{\sqrt {a^2 \plus{} b^2 \plus{} c^2}} \geq\ 2(ab \plus{} bc \plus{} ca)\]
[/quote]
Very nice ineq!$ uvw$ method can also kill it easily.We only have to prove it when $ a\equal{}b$,and $ c\equal{}0$.Can you post the solution using AM_GM ineq,nguoivn?Thanks.

[quote="SUPERMAN2"][quote="arqady"]
By the way, the following inequality is true already.
Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ a^2 \plus{} b^2 \plus{} c^2 > 0.$ Prove that:
\[ a^2 \plus{} b^2 \plus{} c^2 \plus{} \frac {\sqrt3(ab \plus{} ac \plus{} bc)\sqrt [3]{abc}}{\sqrt {a^2 \plus{} b^2 \plus{} c^2}} \geq\ 2(ab \plus{} bc \plus{} ca)\]
[/quote]
Very nice ineq!$ uvw$ method can also kill it easily.We only have to prove it when $ a \equal{} b$,and $ c \equal{} 0$.Can you post the solution using AM_GM ineq,nguoivn?Thanks.[/quote]
Of course, my friend.
I am writing and collecting a chapter about a method which named BTH. This method effect with the forms strange ineq which are hard to prove by basic methods).
Here is some application of this method:
1) (Rachid Benchlikha)
Given $ a, b, c > 0$. Prove that:
$ \frac {1}{(3a \plus{} 2b \plus{} c)^2} \plus{} \frac {1}{(3b \plus{} 2c \plus{} a)^2} \plus{} \frac {1}{(3c \plus{} 2a \plus{} b)^2} \leq\ \frac {9}{4(ab \plus{} bc \plus{} ca)}$
2) (Tran Quoc Luat)
Given $ a, b, c > 0$. Prove that:
$ \frac {a \plus{} b \plus{} c}{\sqrt [3]{a^2b^2c^2}} \geq\ \frac {a \plus{} b}{ab \plus{} c^2} \plus{} \frac {b \plus{} c}{bc \plus{} a^2} \plus{} \frac {c \plus{} a}{ca \plus{} b^2}$
3) (Tran Van Luan)
Given $ a, b, c > 0$. Prove that:
$ \frac {a}{a \plus{} \sqrt {2(b^2 \plus{} c^2)}} \plus{} \frac {b}{b \plus{} \sqrt {2(c^2 \plus{} a^2)}} \plus{} \frac {c}{c \plus{} \sqrt {2(a^2 \plus{} b^2)}} \geq\ \frac {2}{1 \plus{} \sqrt {2}}$
4) (arqady)
Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ a^2 \plus{} b^2 \plus{} c^2 > 0.$ Prove that:
$ a^2 \plus{} b^2 \plus{} c^2 \plus{} \frac {\sqrt3(ab \plus{} ac \plus{} bc)\sqrt [3]{abc}}{\sqrt {a^2 \plus{} b^2 \plus{} c^2}} \geq\ 2(ab \plus{} bc \plus{} ca)$
..........
When I finish, I'll post it here. Please wait for me  :)

[quote=SUPERMAN2]Let $ x,y,z>0$ satisfy $ x^2\plus{}y^2\plus{}z^2\plus{}xy\plus{}yz\plus{}zx\equal{}\frac{3}{2}$.Prove that
$ \sqrt{2(xyz\plus{}1)} \geq x\plus{}y\plus{}z$[/quote]

By computer，we have
$$LHS^2-RHS^2=\frac{1}{9}\sum_{cyc}{(x-y)^2(x+y-z)^2}+\frac{1}{3}xyz\sum_{cyc}{(x+y-1)^2}\ge{0}$$

[quote=SUPERMAN2]Let $ x,y,z>0$ satisfy $ x^2\plus{}y^2\plus{}z^2\plus{}xy\plus{}yz\plus{}zx\equal{}\frac{3}{2}$.Prove that
$ \sqrt{2(xyz\plus{}1)} \geq x\plus{}y\plus{}z$[/quote]

Very beautiful .Solucion :

[quote=SUPERMAN2]Let $ x,y,z>0$ satisfy $ x^2\plus{}y^2\plus{}z^2\plus{}xy\plus{}yz\plus{}zx\equal{}\frac{3}{2}$.Prove that
$ \sqrt{2(xyz\plus{}1)} \geq x\plus{}y\plus{}z$[/quote]

[img]http://s10.sinaimg.cn/middle/006ptkjAzy7bV0Unbv339&690[/img]


I don't understand this problem.

What is the maximum number of points in the intersection of the graphs of two differernt fourth degree polynomial function y = p(x) and y = q(x), each with leading coefficient 1?

[hide="HOW?"]From the answer, it said that the intersection of x-axis is p(x) - q(x) but how do you know it? What about the y-intercept? And I still don't understand how you get the idea of intersection of two fourth degree equation by p(x) - q(x). Can anyone give me a drawing?[/hide]

Also, according to this idea, then the maximum number of intersection between two third-degree function a(x) and b(x) has at most 2 intersection because a(x)-b(x) yields second-degree equation and it can have at most 2 intersection with the x-axis??

:-(

Believe it or not, if $p_1(x)$ and $p_2(x)$ are both quadratic polynomials with leading coefficients 1, only one intersection is found. Try it and see if you convince yourself.

You can't really draw a diagram to help here:

Let

$f(x)=x^4+ax^3+bx^2+cx+d$
$g(x)=x^4+rx^3+sx^2+tx+u$

Then the intersection of $f,g$ corresponds to $f(x)-g(x)=0\Rightarrow (a-r)x^3+(b-s)x^2+(c-t)x+(d-u)=0$

Since $f(x)-g(x)$ is a polynomial of degree at most 3, it has at most 3 roots.


If the number $ 15!$, that is, $ 15 \cdot 14 \cdot 13 \dots 1$, ends with $ k$ zeros when given to the base $ 12$ and ends with $ h$ zeros when given to the base $ 10$, then $ k \plus{} h$ equals:

$ \textbf{(A)}\ 5 \qquad \textbf{(B)}\ 6 \qquad \textbf{(C)}\ 7 \qquad \textbf{(D)}\ 8 \qquad \textbf{(E)}\ 9$

[hide="Solution"]$ 12^k\equal{}(3\cdot 2)^k$. We can factor $ 3^6$ out of $ 15!$, so $ k\equal{}6$.
$ 10^h\equal{}(2\cdot 5)^h$. We can factor $ 5^3$ out of $ 15!$, so $ h\equal{}3$.
$ k\plus{}h\equal{}\boxed{\textbf{(E)}\ 9}$.[/hide]

grn_trtle, I don't think that's right.

$ 12^k \equal{} (2^2 * 3)^k$

While although you can factor $ 3^6$ out of $ 15!$, you can only factor out a $ 2^{11}$.

Therefore, $ k \equal{} 5$

The other part is the same, so $ \boxed{8}$


Let $A$ be a set of $16$ positive integers with the property that the product of any two distinct members of $A$ will not exceed 1994. Show that there are numbers $a$ and $b$ in the set $A$ such that the gcd of $a$ and $b$ is greater than 1.

Then if a number $x \in A$ has a prime $p | x$ then we cannot have $p|y$ for any other $y \in A$.

The first 16 primes are:

2      3      5      7     11     13     17     19     23     29
     31     37     41     43     47     53

If we use 16 numbers from A, we must use 16 primes.  The lowest primes we can use are these.  If $p|x$ clearly $x\ge p$.

So the largest 2 numbers in the set are atleast $47*53 > 1994$.

Comment: Singular's solution is proof by contrapositive, for people who are learning the techniques

Original statement: Given $A$ with 16 positive integers. If for all $a,b\in A$ such that $ab\le1994$, then there exist $x,y\in A$ such that $\gcd(x,y)>1$.

Contrapositive statement: Given $A$ with 16 positive integers. If for all $x,y\in A$ such that $\gcd(x,y)=1$, then there exist $a,b\in A$ such that $ab>1994$.

Singular, since 1 is a positive integer, therefore a can be {1 2 3 5 7 11 13 17 19 23 29 31 37 41 43 47}, but your solution still works since 43*47>1994.


Diophantine equation
number theory unsolved
number theory

Prove that for every natural $n$ there exist natural $a$ and$b$ such that  $n=(a^2+b)/(b^2+a)$

Is it really hard?
Or nobody wants to help me :(

Yes it is true. Consider Pell equation $x^2-ny^2=1-n^3.$ One solution x=1,y=n. We need solution x=1(mod 2n). It give $b=\frac{x-1}{2n},a=\frac{y+n}{2}$.

Please, can you write it indetails!
thank you!


Johnny the boy-with-no-hand-eye-coordination is cutting a 10-inch paper on a given 10-inch line. Johnny goes astray of the given line with probability 2/3 for every given inch of line he cuts through. What is the probability that he does not go astray and cuts the paper with skill?

Haha, boy with no-hand-one-eye-coordination. I like that  :lol:

Probability is just $ \left(\frac {1}{3}\right)^{10}$


Determine all functions $f: \mathbb{R}\rightarrow\mathbb{R}$ so that $\forall x: x\cdot f(\frac x2) - f(\frac2x) = 1$

Actually the answer is $ f(x)\equal{}\frac{x\plus{}2}{3x}$ but your idea is correct.

Of course $0$ is not in the domain.
Anyway, [hide=solution]Let's plug in $x=\frac{4}{x}$. We see that $\frac{4}{x}f(\frac{2}{x})-f(\frac{x}{2})=1$. Let's multiply this equation by $x$: $4f(\frac{2}{x})-xf(\frac{x}{2})=x$. Adding this to the original identity allows the $f(\frac{x}{2})$ and we end up with $f(\frac{2}{x})=\frac{x+1}{3}$. Plugging in $\frac{2}{x}$ gives $f(x)=\boxed{\frac{x+2}{3x}}$.[/hide]

Above solutions are actually incorrect.
Interpreting the problem as it is, setting $x=0$ yields a contradiction, since $\text{LHS}$ will be undefined.

If the FE is true $\forall x\ne0$, then:
Moving $x\mapsto\frac4x$, we get $\frac4xf\left(\frac2x\right)-f\left(\frac x2\right)=1$, so $4f\left(\frac2x\right)-xf\left(\frac x2\right)=x$. Adding this to the original assertion, we get $3f\left(\frac2x\right)=x+1\Rightarrow f(x)=\frac2{3x}+\frac13\forall x\ne0$. Now $f(0)$ can be any real.
Easy to verify that $\boxed{f(x)=\begin{cases}\frac{x+2}{3x}&\text{if }x\ne0\\c&\text{if }x=0\end{cases}}$, where $c$ is any real constant.


A negatively charged particle is directed between two electrically charged plates. Why does the path of the charged particle bend? As the charge on the plates is increased, would you expect the bending to increase, decrease, or stay the same? As the mass of the particle is increased, would you expect the bending to increase, decrease, or stay the same?

I posted this problem in the chemistry forum at first. It was reccomended to me that I post this in the physics forum too as the problem involves Coulumb's Law and electrostatics.   :)

I already solved this problem for you.  See the original forum.


vector
linear algebra
matrix
linear algebra unsolved

Suppose that V is finite dimensional and S,T, Prove ST IS INVERTIBLE IF ONLY IF BOTH S AND T ARE INVERTIBLE

In a finite dimesnional space, invertible is equivalent to injective or surjective. But ST invertible implies that S is surjective and T is injective, ie S and T are both invertible. The reverse way is obvious.

Here's a different argument. We say that $A$ is invertible and that $A^{-1}=B$ if and only if $AB=I$ and $BA=I.$

The one "big" theorem from finite-dimensional vector spaces to quote here is that for mappings from a finite dimensional vector space into itself (which might as well be $n\times n$ matrices), a left inverse is a right inverse and vice versa. That is $AB=I$ iff $BA=I.$

(This can be proved by factoring a matrix as a product of elementary matrices - that is, by Gaussian elimination. The theorem that julien_santini quoted, namely invertible iff injective or surjective, also has its proof in Gaussian elimination - as a matter of counting pivots in each row and column.)

So: Assume $ST$ is invertible. That is, $\exists\,B$ such that $B(ST)=I$ and $(ST)B=I.$

But then $(BS)T=I,$ so $T$ has a left inverse, hence an inverse.

And $S(TB)=I,$ so $S$ has a right inverse, hence an inverse.

That's one direction. If we assume that $S$ and $T$ are both invertible, we can see by direct computation that $T^{-1}S^{-1}$ is the inverse of $ST.$

julien_santini: [b]you said : But ST invertible implies that S is surjective and T is injective,[/b]
 I miss that part I KNOW IF ST inversible means that ST injective and ST Surjective , but I miss how you implies that ST inversible  ==>  S is surjective and T is injective

Thank u in advance

If $S$ is not surjective, how can $ST$ be surjective?  Similarly, if $T$ is not injective, how can $ST$ be injective?  Think about these in terms of the definition of surjectivity and injectivity.


I need some serious help to make a collection of ray diagrams through different type of lenses!
Assume do > C, I have to give a set of below answers for concave mirror, convex mirror, converging lens and diverging lens.

do=7.5cm
ho=1cm
di=?
hi=?
M=?

How can I get those information above and how do I draw those pictures?

Use the mirror and lens equations to get type of image and image distance etc.

More specifically, the useful equations are:

$ \frac 1{di} \plus{} \frac 1{do} \equal{} \frac 1f$  Where [i]f[/i] is the focal distance.

Also:
$ \frac {\minus{}di}{do}\equal{} \frac {hi}{ho}$


calculus
integration
trigonometry
function
complex numbers
calculus computations

Perhaps we could integrate in the complex plan, using an appropriate contour, the function $\frac{e^{e^{iz}}}{z}$ and then take the imaginary part of it.


For how many positive integers $ n$ is $ n^2\minus{}3n\plus{}2$ a prime number?

$ \textbf{(A)}\ \text{none} \qquad
\textbf{(B)}\ \text{one} \qquad
\textbf{(C)}\ \text{two} \qquad
\textbf{(D)}\ \text{more than two, but finitely many}\\
\textbf{(E)}\ \text{infinitely many}$

[hide="hinto"]
Factor it.

Also, remember what a prime is.
[/hide]

[hide]$ n^2 \minus{} 3n \plus{} 2 \equal{} (n \minus{} 1)(n \minus{} 2)$ for all positive integer $ n$. But the only prime number that can be generated is by letting $ n \equal{} 3 \implies (3 \minus{} 1)(3 \minus{} 2) \equal{} 2(1) \equal{} 2$. Therefore there's only one positive integer $ n$ that generates a prime number.[/hide]

[hide=s]Factoring gives $(n - 1)(n-2)$. Since a prime only has one factor, one of the factors ($n-1$ or $n-2$) must be $1$. This is only true when $n = 2, 3$. If $n = 2$, then the expression gives $0$, but if $n = 3$, the expression gives two. So there is only one solution. $\textbf{B}$.[/hide]

[hide]parity arguments show it's always even for integer values so at most 1 just by that and $n=0$ produces the output 2 ...[/hide]

[hide=Sol]
We factor this as $(n-1)(n-2)$

From this, we see that the only prime number is $2$, as $\text{odd number} \times \text{even number}=\text{even number}$

Thus, our answer is $\textbf{(B)}$.

[hide=Cheap sol]
We know that quadratics have, no, one, or two solutions. We can guess one, or $\mathbf{B}$ because there are most likely two roots, one of which is prime.

[quote=michaelchang1][quote=Honestly]@2above - No, $2$ cannot be a solution because it makes the quadratic $0$, which is not prime.
[hide=real solution]
The solution is to note that the quadratic factors as $(n-2)(n-1)$, so the quadratic is only prime if one of the factors is $1$(in any other case $2$ is a factor). We note that this is true when $n-1=1 \implies n=2$ or $n-2=1 \implies n=3$. Clearly, if $n=2$, then the quadratic is $0$, so the only possible $n$ is $3$, leaving $\boxed{1}$ value for $n$.[/hide][/quote]

I meant that the only possible prime that the expression can equal is $2$, not as a solution.[/quote]

But I don't think your solution is completely rigorous, because it doesn't prove that there is an $n$ that gives $2$. Though you can clearly see it's $3$.

[hide]n^2-3n+2=n(n-3)+2
 if n=even,then n(n-3) is even
 if n=odd,then n-3 is even. 2 is also even.the only even prime number is 2 so the answer is:n=3.[/hide]

(n^2)-3n+2=(n-1)(n-2).
    For it to be prime, any of two factors should be 1 and the only case satisfying the condition is n=3.
Hence, there is only one possible value of n in positive integers generating a prime number...


Ρε παιδιά ξέρετε πού μπορώ να κατεβάσω κάποιο βίντεο με στιγμιοτυπα (στα Αγγλικά κατα προτίμηση ) της νίκης της ΑΕΚ επί της Μίλαν???
θα χρωστάω πο΄λλά
ευχαριτσώ εκ των προτέρων

εχω στα ελληνικα το [url]http://www.sportime.gr[/url]

Φίλοι μου χρόνια πολλά στους Νίκους μας κι αύριο στους συνονόματους  μου θα ήθελα να ξέρω πού μπορώ να κατεβάσω το Μια αγάπη έχω στην καρδία της ΑΕΚ ???? ευχαριστώ πολύ

σημ φίλε κώστα το σαιτ που μου έδωσες στο προηγούμενο ποστ δεν με στέλνει σε σελίδα που δουλεύει αν θες πρότεινε κάτι άλλο

Ambroσιε στα στελνω σε π.μ, σορυ  :blush:


geometry
circumcircle
3D geometry
prism
geometry solved

Let ABCDE be a, um, bicentric? pentagon (lyk, there exists a circle s.t. each side of the pentagon is tangent to that circle, and there also exists a circle passing through all 5 vertices of the pentagon). Let A' be the point where smaller circle touches DC. Similary define B', C', D' and E'.
Now prove AA', BB', CC', DD' and EE' are concurrent at a point X.
[ I'm sure that is true. The following two results might also be true, they r just a sorta general version of the bicentric quadrilateral problem, but dey mite wel be false:
(i) Define A'' to be the intersection of ED and BC, A''' to be the intersectin of BD and EC. Prove A, A', A'', A''' are collinear.
(ii) X, I and O are collinear. [ I is centre of smaller circle, O centre of larger circle. ]
(iii) By Poncelet's, you can, wel 'move around' the pentagon as the two circles are sorta, fixed. Prove X is fixed, i.e. it depends only on the two circles and not the pentagon. ]
So help in proving any of these 4 results would be appreciated (or lol help in disproving any of them.)
hmm .. maybe you can generalize some of these to a n-gon but dat really is really crazy to think abt .. mayb if there is lyk
a well-known infinite chain of thms abt this or sumfin lyk dat, i wud b interested.
so yah, thanx in advance 4 everyfin ~!!

hey all;

hm can ppl plz try 2 solve this problem? i dont want to sound pissd or anyfin, but this hasnt been getting much views for some reason, so I fink not many ppl have read this problem. So yah, plz do attempt to solve it.

What have happened this your language?

I ask you to write messages using clear english speech. I am sure it is not difficult for you.

You won't believe, but the answer to most parts of the problem is negative. The lines AA', BB', CC', DD', EE' don't concur in general, and thus the parts [b](ii)[/b] and [b](iii)[/b] don't make sense. Among the assertions of part [b](i)[/b], only the collinearity of the points A, A' and A''' is generally true, and it actually is a trivial consequence of Brianchon's theorem, applied to the (slightly degenerate) circumscribed hexagon ABCA'DE.

How did I disprove the other assertions? Well, if two circles o and i have radii R and r, respectively, and the distance between their centers is d, then there exists a pentagon with circumcircle o and incircle i if and only if $a^3+b^3+c^3=\left(b+c\right)\left(c+a\right)\left(a+b\right)$, where $a=\frac{1}{R+d}$, $b=\frac{1}{R-d}$, $c=-\frac{1}{r}$ (and if this condition is satisfied, then, by Poncelet's porism, there exist infinitely many such pentagons). This can be found on [url=http://mathworld.wolfram.com/PonceletsPorism.html]the MathWorld page "Poncelet's Porism"[/url] (although, as usual, the page has a mistake: $c=\frac{1}{r}$ should be $c=-\frac{1}{r}$, at least for the case of a pentagon). So we have a cubic equation in a, b, c. We can parametrize it by square roots expressions, arriving at

$a\left(s;\;t\right)=\frac12\left(s-\frac{1}{s}\sqrt{ts^3-t^3s+t^2s^2}\right)$;
$b\left(s;\;t\right)=\frac12\left(s+\frac{1}{s}\sqrt{ts^3-t^3s+t^2s^2}\right)$;
$c\left(s;\;t\right)=\frac12\left(s-t-\sqrt{2ts-3t^2+5s^2}\right)$

(I hope I didn't make mistakes here). I used the values s = 32 and t = 50 for making the sketch which confuted your statements.

Altogether, it was an interesting application of computer algebra!

  Darij


[hide="Solution"]We only have to worry about the highest one so, $2*4+3*3=8+9=17[/hide]

i dont think so shady

4everwise is quite correct 2x4+3x3=17

i was thinking first it should be $2^4+3^3$ = 43  :blush: , but no, i realised it wuz wrong..

[quote="shadysaysurspammed"]MY ans is 25 :huh:[/quote]

huh ? How did you get 25 ?  :huh:

[hide]
The highest power of x from the first term is 2*4=8.
x^8.
The highest power of x from the second term is 3*3=9.
x^9.

Thus, the highest power is x^(8+9)=x^17.

[/hide]

i'M STILL GETTIN 25 :rotfl: sORRY IF MY SOLUTION is stupid :blush:

See,in the firsst bracket,there's to the power 4.SO,first time when we square it,we get x^4 then squarin it more u get x^16?
SO, if we add them,it would be x^(16+9)=25

Correct me if i'm wrong

oops...i get it now...when u get x^4,u dont square it,u multiply it by 2,cause of the bracket thing.... :D

You supposed to multiply the exponents  :P , then add.

I'm surprised you did two seperate things for the two parenthesis. One you did 2^4, one you did 3*3  :roll:

[quote="PenguinIntegral"]What degree in x is
$(x^2+1)^4(x^3+3)^3?$[/quote]

[hide="Solution"]
2*4 + 3*3
8 + 9
[b]17[/b]
[/hide]


On a trip from the United States to Canada, Isabella took $ d$ U.S. dollars. At the border she exchanged them all, receiving $ 10$ Canadian dollars for every $ 7$ U.S. dollars. After spending $ 60$ Canadian dollars, she had $ d$ Canadian dollars left. What is the sum of the digits of $ d$?

$ \textbf{(A)}\ 5\qquad
\textbf{(B)}\ 6\qquad
\textbf{(C)}\ 7\qquad
\textbf{(D)}\ 8\qquad
\textbf{(E)}\ 9$

She received 10d/7 Canadian, and she spent 60 to finish with d, so 3d/7=60, or d=140, with sum of digits 5

A

She starts with $ d$ dollars, and after exchanging them she now has $ \frac{10d}{7}$.  After spending 60 Canadian dollars, she then has
$ 60\minus{}\frac{10d}{7}$.  This equals $ d$, since she now has $ d$ Canadian
dollars.  Hence, our equation is:
$ d\equal{}\frac{10d}{7}\minus{}60$.  Solving this, we see that $ d\equal{}140$, and the sum
of the digits is $ 5$.  $ \boxed {D}$


inequalities
trigonometry
geometry
cyclic quadrilateral
power of a point
radical axis
geometry unsolved

Let $ ABC$ be a triangle with $ AB\equal{}AC$.Let $ D$ be the foot of perpendicular from $ C$ to AB,and let $ M$ be the midpoint of $ CD$.Let $ AE \perp MB(E \in BM)$,$ AF \perp CE(F \in CE)$.Prove that $ AB \geq 3AF$

Let $ L$ be the midpoint of $ BC$.
We have $ BD.BA \equal{} BM.BE \equal{} BL.BC$ then $ LMEC$ is cyclic quadrilateral
$ \Rightarrow \angle MEC \equal{} \angle MLB \equal{} 180^o \minus{} \angle ABC$
So $ AC$ is a tangency of $ (BEC)$ we get $ \angle ACE \equal{} \angle CBE$
$ AB \equal{} AC\geq 3AF \Leftrightarrow sin \angle MBC \equal{} sin \angle ECA\leq \frac {1}{3}$
But $ \frac {sin^2 \angle MBC}{MC^2} \equal{} \frac {sin^2 \angle MCB}{MB^2}$
$ \Leftrightarrow sin^2 \angle MBC \equal{} sin^2 \angle MCB. \frac {MC^2}{MB^2}$
$ \equal{} \frac {BD^2}{BD^2 \plus{} 4DM^2}.\frac {DM^2}{BD^2 \plus{} DM^2} \equal{} \frac {x^2y^2}{(x^2 \plus{} y^2)(x^2 \plus{} 4y^2)}$
We need to prove that $ sin^2 \angle MBC\leq \frac {1}{9}$
$ \Leftrightarrow \frac {x^4 \plus{} 5x^2y^2 \plus{} 4y^4}{x^2y^2}\geq 9$
$ \Leftrightarrow \frac {x^2}{y^2} \plus{} 5 \plus{} \frac {4y^2}{x^2}\geq 9$
$ \Leftrightarrow \frac {x^2}{y^2} \plus{} \frac {4y^2}{x^2}\geq 4$ (It's right by AM-GM inequality)
We are done!

Undefine points $ M, E.$ Let $ P, L$ be midpoints of $ AB, BC$ and $ K$ midpoint of $ LC.$ $ (P), (L), (K)$ are circles with diameters $ AB, BC, LC.$ Let $ (P), (K),$ intersecting at $ L,$ meet again at $ E.$ Let $ BE$ cut $ (K)$ again at $ M.$ Then $ (BM \equiv BE) \perp AE$ and

$ \angle MCL \equal{} \angle MEL \equal{} \angle BEL \equal{} \angle BAL \equal{} \angle DAL \equal{} \angle DCL$ $ \Longrightarrow$ $ M \in CD.$

Since $ (K)$ is internally tangent to $ (L)$ at $ C,$ has half the diameter, and $ D \in (L)$ $ \Longrightarrow$ $ M$ is midpoint of $ CD.$ Let $ CE$ cut $ AB$ at $ G.$ $ LE$ is radical axis of $ (P), (K)$ and $ CE \perp LE$ $ \Longrightarrow$ $ (CG \equiv CE) \parallel KP$ and $ \frac {BG}{BP} \equal{} \frac {BC}{BK} \equal{} \frac {4}{3}$ $ \Longrightarrow$ $ BG \equal{} \frac {2}{3} BA$ $ \Longrightarrow$ $ AB \equal{} 3 AG \ge 3 AF.$

Denote $ x\equal{}m(\widehat{ACF})$ , $ AB\equal{}AC\equal{}a$ , $ L\in (BC)$ for which $ AL\perp BC$ and $ AL\equal{}h$ . Then easily that

$ \tan x\equal{}\frac {ah}{a^2\plus{}2h^2}\le \frac {1}{2\sqrt 2}$ . From this relation obtain $ \frac {AF}{AB}\equal{}\sin x\le\frac 13$ with equality iff $ \tan B\equal{}\frac {2h}{a}\equal{}\sqrt 2$ .


induction
graph theory
combinatorics proposed
combinatorics

$X$ is a finite set and $F \subseteq P(X)$ is such that $A \in F$ and $B \subseteq A \Rightarrow B\in F$.
Prove there exists an injective $ \sigma : F \to F$ so that $ \forall A \in F$ we have $\sigma (A) \cap A = \emptyset$.

I think nobody is interested in this beautiful problem :(  :(  :(  :(

This problem seems interesting..
Do you have a solution?
Did I get it right, is F(P) the set of all subsets?
This definition reminds me of matroids.

Well, I give you a hint,You can use induction on $|X|$

I have an idea to use Marriage Theorem.
Call a set maximal if there is no other set containing it.
Let A_1..A_n be maximal sets.
We must prove if we take k sets, there are at least k other sets disjoint to at least one of them.
One easily sees we must have a set S_i in A_i and all other sets in A_i to contain it (or else we could add them without inreasing the number of disjoint sets).
Now use induction on the number of elements has.
I mean take an element x, let m of S_i contain it, k-m not. One easily can now prove the induction step.

Well there is a simple solution using nothing except induction,If you want me I'll send it.

i guess that the empty set is also part of $F$.

or else you won't even have an even number of sets for such a pairing.

Of course you see for every $A$ we have $\emptyset \subset A$ :D  :D  :D  :D

of course,of course! :)
but what i meant to sayis:if you want to produce a matching,then you are actually getting an involution.which is not necessary from what you ask first.i was wondering if it would be possible to get such a map for which the empty set is actually mapped into itself....(then you see,the presence of the empty set in F makes no difference)
that might not be possible,though i am not sure why/why not.

Well of course there is no must that emptyset maps to itself

an inductive solution(induction on number of elements of X=n,say):in fact we prove that we can get an involution.

   for  n=1, quite trivial.

  for n>1, consider an element x in X such that at least one of the sets in F contains x. consider the subfamily F(x) of F of all those sets in F that contain x.write F = F(x) union F' (as a disjoint union).now there is a subfamily of F'(call it F'') and a bijective map g:F(x) -> F'',
                                                               A -> A\{x}.
(since the family is closed under subsets,this map is well defined.)Now the set F' also satisfies the hypothesis(since F does and F' is just that subcollection which doesn't contain x) and has fewer elements than F,so by induction there is an injective map(which is an involution also) h':F'->F' with the reqd properties.now define h:F \ F'' -> F \ F'' by:
           h(A):= h'(A) if A is not in F(x) and h'(A) is not in F''
           h(A):= h'(g(A)) if A is in F(x).and h'(A) is not in F''
           h(A):= g^(-1)(h'(A)) if h'(A) is in F''.
note that the 3 conditions are mutually exclusive so the map is well defined.

the claim is that this map satisfies that h(A) is disjoint from A for any A in
 F \ F''.the claim is routine to check(i hope i haven't goofed anywhere!actually verifying teh claim is routine becos of the assumption that h' is an involution) and in fact gives an involution too.then to get our final mapping,all we do is extend the map by taking another k:F''->F'' which satisfies the hypothesis(again induction.F'' also satisfies the same conditions as F and has fewer elements in the union of the sets in it),so finally the map we need is (h union k).

i think it is all right....

I have just accidentally found this ancient thread when searching for perfect matchings (the graph-theoretical kind of).

This is exactly the problem I posted at http://www.artofproblemsolving.com/Forum/viewtopic.php?f=42&t=421987 . So now I have three solutions each claiming to construct an involution, and I don't understand any of them. :(

iura's solution, even if correct (which I cannot judge, as I don't understand it beginning with "One easily sees we must have a set S_i in A_i and all other sets in A_i to contain it"), does not construct an involution. In fact, Hall's marriage theorem only works for bipartite graphs. We can "forcibly make $F$ bipartite" by doubling it (i. e., consider the bipartite graph with a disjoint union of $F$ and $F$ as vertex set), but then a matching will be a permutation, rather than an involution.

seshadri's solution loses me here:

[quote="seshadri"]h:F \ F'' -> F \ F'' by:
h(A):= h'(A) if A is not in F(x) and h'(A) is not in F''
h(A):= h'(g(A)) if A is in F(x).and h'(A) is not in F''
h(A):= g^(-1)(h'(A)) if h'(A) is in F''.[/quote]

What does h'(A) mean if A is in F(x)? Maybe the second line should be [i]h(A):= h'(g(A)) if A is in F(x) and h'(g(A)) is not in F''[/i]? But in this case, does the definition really cover all cases?

I have managed to find another idea for an inductive proof. It can be realized using the marriage lemma (so no involution) or with involution.

First proof (with the marriage lemma)

Choose $x\in X$ and let $X'=X\setminus \{x\}$. For any $A\subset X$ call $A'=A-\{x\}$ the "suppression" of $A$. Let $F'$ be the suppression of all sets in $F$, also let $F_1$ consist of all suppressions of all sets of $F$ that CONTAIN $x$.

It is immediate that both $F$ and $F_1$ satisfy the condition of the problem for the set $X'$, hence we can apply the induction step to them. It is also immediate that $F_1\subset F'\subset F$

Now choose $C$ a collection of $k$ sets in $F$. By the marriage lemma, it suffices to show there are $k$ sets in $F$ disjoint from at least one of them. Assume that $C$ contains $q$ (disjoint) pairs of the form $(A, A\cup\{x\})$. Then the suppression of $C$ will contain $k-q$ distinct sets, so there are at least $k-q$ sets in $F'$ disjoint from some of them - by the induction assumption for $F'$. However since the suppression of $C$ contains at least $q$ sets from $F_1$, by the induction assumption for $F_1$, at least $q$ of those at least $k-q$ sets belong to $F_1$.

We now obtain at least $k-q$ sets not containing $x$ that are disjoint from something in $C$, and at least $q$ of those can be extended by adding $x$ to them. Therefore the conclusion follows.

NB: those $q$ sets in $F_1$ can be extended by adding $x$ to them because each of them is disjoint from some set $A$ not containing $x$ -belonging to one of the $q$ pairs of the form $(A, A\cup \{x\})$ - and adding $x$ keeps it disjoint from $A$.

Second proof (with pairing)

We now seek to partition the sets in $F$ into pairs of disjoint sets. Keep the notations from the first proof - what we need is that $F'$ and $F_1$ satisfy the condition and thus the conclusion of the problem for $X'$ (by the inductive assumption).

For simplicity, color the sets in $F_1$ red; the other sets in $F'$ we color blue. There is then a full matching $M_1$ of the red sets, and a full matching $M'$ of the red and blue sets (we connect two sets if they are disjoint).

We deal with edges connecting sets of a different color as follows:

Say $B_1$ is blue and is connected to $R_1$ which is red via $M'$. Then $R_1$ is connected to some other red vertex $R_2$ via $M_1$. $R_2$ is connected to some other red vertex $R_3$ via $M'$ and we keep going. Eventually we will either meet a red set the second time or meet a blue set again. The first case is impossible since it yields a vertex of degree 3 (in graph theory language). In the second case we meet a blue set $B_2$ distinct from $B_1$ (as the only way to reach $B_1$ is via $R_1$).

We have thus decomposed the sets in $F'$ into following groups:

a) pairs $R_1,R_2$ of red sets connected via $M'$. In this case, the inverse images of these sets under suppression are divided into disjoint pairs $(R_1, R_2\cup \{x\})$ and $(R_2, R_1\cup \{x\})$.

b) pairs $B_1, B_2$ of blue sets connected via $M'$. The inverse images of these are just $B_1$ and $B_2$ which are disjoint.

c) an alternating path in $M'$ and $M_1$ of the form $B_1-R_1-R_2-\ldots-R_k-B_2$ (note that $k$ is even but we do not need this). The inverse images of these under suppression can be divided into pairs as follows: $(B_1, R_1\cup \{x\}), (R_1, R_2\cup \{x\}), \ldots, (R_{k-1}, R_k\cup \{x\}), (R_k, B_2)$.

This finishes the proof.


[quote="kunny"][hide]We can set $kP(k)-1=(k-1)(k-2)(k-3)\cdots (k-999).$ Plugg $k=1001$ into this equality, we have $1001P(1001)-1=1000!\Longleftrightarrow P(1001)=\frac{1000!+1}{1001}.$[/hide][/quote]

Is it necessary that the leading coefficient of polynomial to be equal to 1(as you have taken).  If we take a different leading coefficient, what would have been the answer?  I am getting a different answer from yours, irrespective of the leading coefficient in P(x).

[quote="kunny"][hide]We can set $kP(k)-1=(k-1)(k-2)(k-3)\cdots (k-999).$ Plugg $k=1001$ into this equality, we have $1001P(1001)-1=1000!\Longleftrightarrow P(1001)=\frac{1000!+1}{1001}.$[/hide][/quote][hide]Under these circumstances, $P(k)$ is in fact not a polynomial.[/hide]

Oh, thanks!

Then a = 1/999!

P(1001) = 1

as you stated.

[quote="vidyamanohar"]If P(x) is a polynomial of degree 998 such that P(k)=1/k is true for k=1, 2, 3,...999  then find the value of P(1001)[/quote]
Try following generalization: suppose that   $x_0,x_1,...,x_n$ are mutual distinct real numbers, i.e.   $x_{\alpha } \ne x_{\beta }$ for $0\le \alpha \ne \beta \le n .$
Let  $\{y_0,y_1,....,y_n\}\subset {\mathbb R}$ and $P(x)$ be (the) polynomial satisfying   $P(x_j)= y_j$  for $j\in \{0,1,...,n\} .$
[color=green] [b] If $w\not\in \{x_0,x_1,...x_n\},$ find $P(w)\;$ [/b][/color]
[b]Solution.[/b] $P(x)$ is  well-defined [ i.e.  exists and it's unique] , more precisely  P(x)  is  the so called  Lagrange  (interpolation) polynomial :
\[ (*) \; \; \; \; P(x)=\sum\limits_{k=0}^{n}\frac{\omega(x)}{(x-x_k)\omega^{\prime}(x_k)}y_k \; .  \]
Further [b] select  $x: = w$ in (*) [/b] .  I have used notation :  $\left\{\begin{array}{l} \omega(x): =\prod\limits_{k=0}^{n}(x-x_k)\\ \omega^{\prime}(x_k): =\prod\limits_{j=0,j\ne k }^n (x_k-x_j)\\ \end{array}\right. \; .$


I need information about of Wavelet's Tranformed, I need aplications in imagines procces and other aplicattions; and somethings properties of that, please I need that in Spanish... Tell me where Can I look for it?

Er.... Spanish 3 skills don't fail me now!

Yo lo busque en google...

http://www.google.com/search?as_q=wavelet+transform&num=10&hl=en&btnG=Google+Search&as_epq=&as_oq=&as_eq=&lr=lang_es&as_ft=i&as_filetype=&as_qdr=all&as_occt=any&as_dt=i&as_sitesearch=&safe=images

espero que ayuda.

 :blush:

I hope I didn't manage to mangle that too badly.

lol... you basically said "I will wait for you to help me" and you didn't even do that correctly.

Well esperar means both to help and to wait. She tried to say "I hope that helps", but yeah.  Can't often translate directly from English to Spanish with the meaning staying the same.


function
superior algebra
superior algebra unsolved

I was read an algebra book and I found the definition of function field $F/K$ of one variable. One of the conditions says that $K$ is algebraically closed in $F$. I have learnt that the algebraic closure of given field is unique so does it mean that if $G$ is an algebraic closure of $F/K$, then $G\subset F$ or what?

There is no algebraic extension of $K$ contained in $F$, e.g. $K(X)$, where $K$ is not necessarily algebraically closed.

"$K$ is algebraically closed in $F$" means just that every element of $F\setminus K$ is transcendental over $K$. (The set of all elements of $F$ which are algebraic over $K$ is called the algebraic closure of $K$ in $F$, and $K$ being algebraically closed in $F$ means that it equals its algebraic closure in $F$.) owk


inequalities
graph theory
combinatorics unsolved
combinatorics

Let $n$ be an integer greater than 1. For a positive integer $m$, let $S_{m}= \{ 1,2,\ldots, mn\}$. Suppose that there exists a $2n$-element set $T$ such that
(a) each element of $T$ is an $m$-element subset of $S_{m}$;
(b) each pair of elements of $T$ shares at most one common element;
and
(c) each element of $S_{m}$ is contained in exactly two elements of $T$.

Determine the maximum possible value of $m$ in terms of $n$.

Imagine a bipartite graph where the first set of vertices are the elements of $[mn]$ and the second set of vertices are the subsets of $T$.

Two vertices are adjacent if the corresponding element is in the corresponding subset.

Let $d_{i}$ denote the degree of vertex $i$ in the first set of vertices.  Note that
\[\sum_{i=1}^{mn}d_{i}=2mn\]
and
\[\sum_{i=1}^{mn}\binom{d_{i}}{2}\leq \binom{n}{2}\]
By convexity (or Cauchy's Inequality), $\sum_{i=1}^{mn}\binom{d_{i}}{2}$ is minimized when the $d_{i}$'s are all equal.  So $\sum_{i=1}^{mn}\binom{d_{i}}{2}\geq \sum_{i=1}^{mn}\binom{2}{2}=mn$.  Combining this with a previous inequality gives $m\leq\frac{1}{n}\binom{n}{2}$.


[i]Αν είχαμε το σύστημα που εφαρμόζουν στη Ρουμανία (πχ) , όπου οι ασκήσεις των διαγωνισμών προτείνονται από διάφορους πολίτες, μαθηματικούς , μαθητές  , μηχανικούς κλπ , τότε θα πρότεινα για τον Αρχιμήδη την παρακάτω άσκηση. Αυτό βέβαια με την ελπίδα ότι δε θα είχε και κάποιος άλλος την ίδια ιδέα !
 Εσείς θα μου πείτε αν λέει κάτι ή αν πρόκειται για ξενέρωτο πρόβλημα. [/i]

[b] ΑΣΚΗΣΗ[/b]

   Δύο κύκλοι τέμνονται στα σημεία $ K,L$.Έστω $ AB$  η κοινή εφατπομένη των δύο κύκλων που  είνα πλησιέστερη στο $ K$. Ένας τρίτος κύκλος $ (\Omega)$ που έχει το κέντρο του στην ευθεία $ AB$ και περνάει από τα $ K,L$  τέμνει το τμήμα $ AB$ σε σημείο $ C$. .Αν $ P$ είναι τυχαίο σημείο του κύκλου $ (\Omega)$ , να αποδείξετε ότι $ \angle CPA \equal{}\angle CPB$.

Εξαιρετική άσκηση . Κομψή και πολύ όμορφη . Αν είναι δική σας κ.Μπάμπη η κατασκευή συγχαρητήρια .
Έστω ότι η ΑΒ τέμνει τον κύκλο (Ω) στο Ν .Αφού $ CP\bot PN$ αρκεί να δείξουμε ότι τα σημεία (A,C,B,N) σχηματίζουν αρμονική τετράδα.
Έστω τώρα ότι η KL τέμνει την ΑΒ στο Μ . Τότε ΜΑ=ΜΒ από το θεώρημα της δύναμης σημείου ως προς κύκλο και επιπλέον
$ MA^{2}\equal{}MK\cdot ML\equal{}MC\cdot MN$ άρα από το αντίστροφο του θεωρ Newton τα (A,C,B,N) σχηματίζουν αρμονική τετράδα QED .

Γιαυτούς που δεν το γνωρίζουν
θεωρ Newton
Έστω τέσσερα σημεία Α ,Β, C,D pano se mia eutheia me auti ti diataksi kai M meso tou AC .Tote i tetrada einai armoniki an kai mono an $ MA^{2}\equal{}MB\cdot MD$

[quote="silouan"]Εξαιρετική άσκηση . Κομψή και πολύ όμορφη . Αν είναι δική σας κ.Μπάμπη η κατασκευή συγχαρητήρια .
Έστω ότι η ΑΒ τέμνει τον κύκλο (Ω) στο Ν .Αφού $ CP\bot PN$ αρκεί να δείξουμε ότι τα σημεία (A,C,B,N) σχηματίζουν αρμονική τετράδα.
Έστω τώρα ότι η KL τέμνει την ΑΒ στο Μ . Τότε ΜΑ=ΜΒ από το θεώρημα της δύναμης σημείου ως προς κύκλο και επιπλέον
$ MA^{2}\equal{} MK\cdot ML \equal{} MC\cdot MN$ άρα από το αντίστροφο του θεωρ Newton τα (A,C,B,N) σχηματίζουν αρμονική τετράδα QED .

Γιαυτούς που δεν το γνωρίζουν :

[color=red][b]Θεώρημα Newton[/b][/color]

Έστω τέσσερα σημεία Α ,Β, C,D pano se mia eutheia me auti ti diataksi kai M meso tou AC .Tote i tetrada einai armoniki an kai mono an $ MA^{2}\equal{} MB\cdot MD$[/quote]

 [b]Εύγε ! [/b]

 Το πρωί την έφτιαξα , προσπαθώντας να βρω κάτι άλλο. Αφού την έλυσες με τον ίδιο τρόπο  , για τον οποίο την έφτιαξα , αξίζεις  μπράβο.
 Λοιπόν , θα το πω δημόσια. Μόλις διάβασα τη λύση σου , δάκρυσα από λύπη γιατί από καθαρή ατυχία ( είναι να μην σε ψάχνει η κακή στιγμή :( ) έχασες τόσες μονάδες από το Θ2 στην ΙΜΟ και μαζί με αυτές και το μετάλλιο. Ας με συγχωρέσουν οι άλλοι φίλοι. Αλλά το ίδιο θα έκανα και για αυτούς , αν τους συνέβαινε μια τόσο μεγάλη ατυχία - εύχομαι όμως να μην τους συμβεί αυτό , αλλά να πάρουν μετάλλια και να συγκινηθώ από  τη χαρά που θα νοιώσω. ).

 Μπάμπης

[quote="stergiu"]   Το πρωί την έφτιαξα , προσπαθώντας να βρω κάτι άλλο. [/quote]

Πολύ ωραία η άσκηση και ευχαριστώ για τα καλά σας λόγια .. :)

Ωραία άσκηση....χεχεχε έξυπνη λύση!!!!

Συγχαρητήρια και για την imo Silouane! Κρίμα που έχασες το μετάλλειο για λίγο.... :(

Oρίστε ακόμα μια λύση σε αυτή την πολύ όμορφη και κομψή άσκηση.

Θα δείξουμε ότι ο κύκλος $ \Omega$ (με κέντρο $ O$) είναι ο απολλώνιος κύκλος του τμήματος $ AB$ με λόγο $ \frac{CA}{CB}\equal{} k$, οπότε για κάθε σημείο του $ P$ θα ισχύει $ \frac{PA}{PB}\equal{}\frac{CA}{CB}$, δηλαδή $ \angle CPA \equal{}\angle CPB$. Για τούτο αρκεί να δείξουμε ότι υπάρχουν τρία σημεία $ X$ του κύκλου ώστε $ \frac{XA}{XB}\equal{} k$.Tα σημεία αυτά είναι τα $ C$ (προφανώς), $ K$ και $ L$.
Έστω ότι η $ OK$ τέμνει τους δύο κύκλους στα $ E, F$. Τότε $ OB^{2}\equal{} OE*OK, OA^{2}\equal{} OK*OF$,άρα $ OA^{2}*OB^{2}\equal{} OK^{2}*OE*OF$. Όμως λόγω ομοιοθεσίας (με κέντρο το $ O$ που μεταφέρει τον έναν κύκλο στον άλλο) θα είναι $ \frac{OE}{OK}\equal{}\frac{OK}{OF}$, οπότε $ OK^{2}\equal{} OE*OF$ και τελικά $ OK^{2}\equal{} OA*OB$, δηλαδή $ \angle BKO \equal{}\angle BAK$. Επομένως $ \angle CKA \equal{}\angle OKC\minus{}\angle BKO \equal{}\angle OCK\minus{}\angle BAK \equal{}\angle CKB$ και $ \frac{KA}{KB}\equal{}\frac{CA}{CB}$.
Tώρα, αν $ M$ είναι το μέσο του $ AB$, τότε θα ισχύουν $ \frac{MA}{AK}\equal{}\frac{ML}{AL},\frac{MB}{BK}\equal{}\frac{ML}{BL}$, απ' όπου έπεται ότι $ \frac{LA}{LB}\equal{}\frac{KA}{KB}\equal{}\frac{CA}{CB}\equal{} k$, που είναι το ζητούμενο.


Hi guys,
I found this very interesting problem...

-------------------------
THE PLUG GAUGES
This interesting shop problem, which is more difficult
than appears at first glance, was originated by Edward C.
Varnum, Mathematician, Barber-Colman Co., Rockford, Ill.

A foreman noticed an inspector checking a 3" hole with a
2" plug and a 1" plug and suggested that two more gauges be
inserted to be sure that the fit was snug. If the new gauges
are to be just alike, what should their diameters be?

-----------------------------

Gabriele  :)

http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1330100458&t=250684

I searched for the words "plug gauge*"

Thank you...

I didn't see it...  :blush:


modular arithmetic
number theory
algebra proposed
algebra

Prove that for every natural $a,$ there exist infinitely many naturals $n$ such that $a \cdot 10^{n}-1$ is composite.

I broke it up into various cases, where $n \equiv 1, 2, 3, 4, 5 \bmod{6}.$ I'm not sure how to prove for when $a \equiv 3 \bmod{6},$ though  :( Any other solutions would also be greatly helpful to me  :)

if $a-1$ has an odd prime factor $p$ then choose $n=k(p-1)$ and you have $a*10^{n}-1\equiv a-1\equiv 0$ mod $p$. if it doesn't have a prime factor then it is etiher $0,1$ or $2^{m}$. in any of those cases it is easy to get that $7\not|a$. since the numbers $a,10a,10^{2}a,10^{3}a,10^{4}a,10^{5}a$ all give different residues modulo $7$ and none are divisible by $7$ then one of them(lets say $10^{l}a$) is congruent to $1$ modulo $7$. then any $n$ of the form $6k+l$ satisfies...
sorry,just found a mistake.
if $a-1$ is of the form $2^{b}5^{c}$ i cannot use little fermats theorem to get the result...so i still need to check that...

[hide]
Let $p$ be a prime divisor of $10a-1$ and consider $n=1+k(p-1)$ for all positive integers $k$. Then we have
\[a\cdot 10^{n}-1=10a-1\equiv 0\pmod{p}\]
(having $10^{n}\equiv 10\pmod{p}$ works from Fermat's Little Theorem unless $p=2$ or $p=5$, but this can't happen), and $a\cdot 10^{n}-1>10a-1\geq p$. Therefore $a\cdot 10^{n}-1$ is composite and we get infinitely many composites.
[/hide]

can't believe i didn't finish this myself. thanks scorpius  :wink:


group theory
abstract algebra
superior algebra
superior algebra unsolved

how about this

third isomorphism theorem gives you that

(G/N) /  (M/N) is isomorphic with G/M

so we have group/twoelementnormalsubgroup is cyclic

now consider a group A with twoelementnormalsubgroup B such that A/B is cyclic
a group with two elements (e,u) is normal iff u commutes with all of A

so every a in A can now, for fixed g, be written as g^k*u=u*g^k
and for b, the same : g^l*v=v*g^l

so a*b=g^k*u*g^l*v=g^(k+l)*u*v=g^l*g^k*v*u=g^l*v*g^k*u=b*a

i hope the proof works

works perfect. pretty analogue to the other one thus :)

Suppose that $G$ is a cyclic group and $N \lhd G$. Can we prove that $G/N$ is cyclic?

(Or is this false?)

It is of course false... it is even hardly ever true? :)

$\{1,-1\} \unlhd Q_8$ but $\{1,i,j,k\}$ is far from what you would call cyclic... ;)

Peter VDD counterexample is wronge since $Q_8$ is not cyclic.
Yes, Arne's conjecture is true, since when $G$ is a cyclic group and $f: G \mapsto H$ is a homomorphism of groups, then $f(G)$ is a cyclic subgroup of $H$. Then just take $H=G/N$ and you are done.
(Prove: Let be $\zeta$ anelement that generates $G$, then $f(\zeta )$ generates $f(G)$)

[quote="fredbel6"]third isomorphism theorem gives you that

(G/N) /  (M/N) is isomorphic with G/M[/quote]
Seems like you wrote $M$ instead of $N$ and vice versa.

And is it possible to apply the fact that if $G$ is cyclic, then $G/N$ is cyclic? Or doesn't that give a quick way to do this problem? I guess not, since then we would somehow get that $G/M$ is cyclic, while we only need to prove that it is Abelian?

I'm still struggling a little bit on this problem.

No I don't really think I made a mistake there?
http://planetmath.org/encyclopedia/ThirdIsomorphismTheorem.html

Basically it says you can 'simplify' denominator and nominator, as if you were working with fractions

I can't think of any other solutions?
What is it you feel uncomfortable with?

[quote="Arne"]Let $G$ be a group, let $M \lhd G$, $N \lhd G$ and $M \subset N$ such that $G/N$ is cyclic and $[N: M] = 2$. Show that $G/M$ is Abelian.[/quote]

We know that $M \subset N$ and $[N : M] = 2$ which suggests considering the group $N/M$, you are working with $M/N$. Basically, it's just a matter of notation: you just wrote $M$ where you should use $N$, and vice versa. You proved that $G/N$ is Abelian, while you should prove that $G/M$ is Abelian, and you used the fact that $G/M$ is cyclic, while it is only given that $G/N$ is cyclic. But if you just swap $M$ and $N$ everywhere in your proof, there is no problem anymore. It's not a big deal, just a detail :)

ooooo, yes i see now, careless of me !!
In my post I am consistent but not with your statement of problem :o

[quote="ZetaX"]Peter VDD counterexample is wronge since $Q_8$ is not cyclic.[/quote]Ah, didn't read that the $G$ had to be cyclic as well... now it makes sense :)


:blush: An acute triangle ABC is inscribed in a circle. Two altitudes AM and CN are extended to meet the circle at P and Q respectively. Given that PQ:AC=7:2, then sin angle ABC is equal to?

this is not a homework-solving problem forum, but in this case the problem is posted in the right place, and has a nice statement, so we'll make an exception :D ;)

anyway are you absolutely sure about the statement of the problem?

Hmm... is there a definite value for sin (ABC)? All I can prove is that AB = BC

The statement of the problem is wrong. PQ:AC can't be greater than 2. But if it were less than 2 then sin <ABC would have a definite value. And AB does not necessarily equal BC.

A hint: Look for a way to use the law of sines.


Show that if s_0=a and s_n=2s_(n+1)+b for n in P, then s_n=2^n(a)+(2^n-1)b for every n in N,

You made a typo:

The question should read:

Show that if $ s_{0}= a$ and $ s_{n}= 2s_{n-1}$ then $ s_{n}= 2^{n}a+(2^{n}-1)b$

i.e. $ n-1$ instead of $ n+1$

Answer:

Lets see if conjecture holds for $ n= 1$:

$ s_{1}= 2s_{0}+b = 2a+b$ from recurrence relation.

$ s_{1}= 2^{1}a+(2^{1}-1)b = 2a+(2-1)b = 2a+b$

$ \Longrightarrow$ conjecture holds for $ n = 1$

Lets assume the conjecture holds for $ n = k$

$ \Longrightarrow s_{k}= 2^{k}a+(2^{k}-1)b$
but $ s_{k+1}= 2s_{k}+b$
$ \Longrightarrow s_{k+1}= 2(2^{k}a+(2^{k}-1)b)+b$
$ = 2^{k+1}a+2^{k+1}b-2b+b$
$ = 2^{k+1}a+2^{k+1}b-b$
$ = 2^{k+1}a+(2^{k+1}-1)b$

$ \Longrightarrow$ conjecture holds for $ n = k+1$ completing the induction step.


Solve in $ \mathbb{R}$ the equation: $ [\sin x] \cdot \{ \sin x \} \equal{}\sin x.$

[quote="moldovan"]Solve in $ \mathbb{R}$ the equation: $ [\sin x] \cdot \{\sin x \} \equal{} \sin x.$[/quote]
If 0=<sinx=<1 then [sinx]=0.Thus, sinx=0
If -1=<sinx<0 then [sinx]=-1 and {sinx}=1-sinx
the equation equivalent to -(1+sinx)=sinx then sinx=-1/2


Support geometry AMC USA(J)MO USAMO calculus

Hello,

My 17 year old junior in HS is starting to visit colleges.  He has always had a gift for math and his SAT's support that (780 math but only 620 verbal).  While math comes easy to him, he says he has no interest in math or engineering as a field of study.  His primary interest is history and he has been looking at liberal arts colleges.

As his dad, I truly believe he should pursue a career he will enjoy so that it will be more than just "a job."  However, I do get concerned that he will someday discover that history majors have to find a job, and that job won't always be in that field.

I realize he is just 17 and will likely change his mind about many things over time. Should I push him towards schools that also offer math/engineering so that if he does change his mind he can change majors without having to change schools?

Anybody have any experience in this area and what did your child end up pursuing?  To me, it seems a shame to waste a gift, but if he truly has no interest, then what liberal arts fields would help him use his math skills?

Thanks,

I suggest you read a little about Paul Zeitz.  Here's a link

[url]http://www.msri.org/activities/events/9798/bamo/Old/zeitz.html[/url]

First on the USAMO, won the Westinghouse.

Majored in history at Harvard.

Taught high school for a few years, then went back for his PhD in mathematics, wrote a very popular text book for problem solvers, now is a very well-respected professor at USF.

My advice: let your son do what he wants.  Most decent liberal arts schools have solid math programs.  Maybe not engineering programs, but people rarely 'discover' engineering after a few years (though they do go the other way, as this Chemical Engineering major can tell you).

[quote="rrusczyk"]My advice: let your son do what he wants.[/quote]
As someone whose business is being a college academic advisor, I must agree strongly with this. I run into students all the time who are pursuing a particular major not because they want to but because their parents expect it of them. If the student's heart isn't in his or her studies, it all has a way of not working out, either through lack of academic success or just from being generally unhappy.

Now since I do represent a math department and I do recruit majors, I would perhaps (hypothetically) take an interest in your son. Let's see: you're an undeclared major and really don't know what you want to do, but you do have a passing score on the AP Calculus exam. How about you just take Calculus II now, while your memories are fresh? No pressure, no expectation beyond that. And if that goes OK, maybe you take one more course. I'm not going to draw up any 4-year plans for you that specify everything - that would be ridiculous. You've got the whole rest of your schedule to explore what your interests really are.

What would I say to you, the father? I'd tell you to back off. He's got to do this himself. I don't even want you to come pay me a visit. Legally, I can't even tell you how your son is doing - that's the way the rules are written.

Hi - Here is another perspectiive offered by my wife. I am cutting and pasting it.  Hope it is helpful.

 [quote]As a parent of two sons  who both scored very well on standardized tests and who are now very happy and successful in highly ranked universities, I offer my perspective on your situation in the hope that it will be helpful to you.

You will be responsible for the cost of paying for the education of your children, and you obviously care very much for your kids and want them to be both successful and able to support themselves.  You seem like a very dedicated parent and you will be the best advisor for your own children, in my opinion.

Many kids choose to double major these days, with  one major in a math/science field and another in something very different from that.  In addition, the highly ranked schools (both universities and liberal arts colleges) have populations of students with very diverse interests, and even a student who has chosen to be a math major is not necessarily focused exclusively on math, as some people might  imagine that they would be.  Our sons and their peers are focused on the sciences, but also have a breadth of talents and interests that span the humanities, social sciences, and the arts.

If your son is a Junior, I would recommend that he consider the “Operation Catapult? program that The Rose Hulman Institute of Technology offers in June or July.  This is a well-planned program that offers a venue for kids with math and science talent to meet and exchange ideas.  This is valuable because many kids who have math talent have not had the opportunity to meet others like themselves and to discover the diversity of interests and talents within their group.

I would also recommend that your son schedule stay-over visits to some schools that he is interested in to see if he likes the environment.  Most of all, I strongly recommend that you continue to stay involved with your kids because you care more for them than any outside advisor can, and you will be there for the kids for the long haul.

[/quote]

Hello

[quote="rrusczyk"]My advice: let your son do what he wants. [/quote]

I am partially agree and partially disagree, and this is the opinion of a 24 years Ph.D. candidate in mathematics. When I was a freshman, my father didn't allow me to be a Math major and that's it. As many parents he tought that being a math major I would never get a better job that as Instructor (what is not a bad job, but my father considered it as a second cathegory job, even when I think that in a rational world, the teachers and professors would be the most important citizens) then, I had to take a double major, and fortunatelly I got a good deal, and it took me just five semesters to end my doble major (Math and Software Engineering). The truth is that Software engineering allowed me to survive while I got the money to apply to grad School here in the US, but it was not really what I wanted to do.

On the other hand, I learned that I shouldn't take everything in a hurry. Yes, It took me just two years and a half to finish my two degrees, but now I think that I would like to take acting classes, or economy, or linguistics while I was an undergrad, but I never did. I suggest you to encourage your child to take a variety of courses. Not only history, but he could also take courses in art, literature, sciences, law, or whatever he wants. Encourage him to think that he needs to have a goal, but he must enjoy the journey as much as the destination (or even more). Those are, I think , beautiful years, he should expand his horizons as much as he wants.

On the other hand, let him be whatever he decides to be at the end. Yes, you pay his education, what definetely shows you care about your son (besides, you are concerned about his election), but he is the one who would have to live with his election, and not you.

I remember that when I was in my senior year in High School, I considered a friend of mine a great math prospect. He was brilliant. Nevertheless, he decided to study Biology, and now he is happy studing insects and plants in the forest in central america. I didn't understand his decition at first, but I have been in the forest with him, and if I enjoy my field, he loves his.

Best,


Hello,

I'm a student who will be attending MathPath 2009. I am really excited for the summer. I wanted to know anything anyone who has attended before had as suggestions. In particular, I wanted to know if people had any information one could benefit from knowing before MathPath. Is there anything I really should bring? How should I balance time between math and fun time?, etc. Thanks everyone!

(Mathpath '08)

You should bring

* A fan. No air conditioning.

* A bathing suit if you swim.

* like 3 notebooks.

* About 200-250 dollars spending money

Basically, after dinner is free time until around 10 PM. You should hang out with friends at that point, and have fun. You'll be doing math from 7:30-5:00 (not counting meals), so don't worry that you aren't doing enough math.

Also, how old are you? If you're 11 (like I was), and you are trying to get into a game of ping-pong or pool or w/e, be prepared to be cut by some 14 year olds. And if you don't like music, do NOT go down to the rec room.

You should bring some snacks, like chips or pretzels or something [WARNING: Don't show anyonhe, or they won't last very long].

Very important: DO NOT BE LAST IN LINE DURING MEALS. MAKE SURE YOUR GROUP MOVES. If you don't follow this guideline, don't say I didn't warn you when you get 5 minutes to eat.

Hope this helped.

BOGTRO,  you assume that the setup at MathPath is always the same.  Not so.  For instance, you say to bring a fan.  But last time we were at Colorado College, the college provided fans for everyone.

Pinkmuskrat, you will just have to wait until you get the Precamp Briefing to get answers to all your questions.  I appreciate your eagerness, but take a deep breadth!

Ah. Sorry. That info was based on Mathpath '08, which was in Vermont. So it might be different.

[b]BOGTRO wrote:[/b][quote]And if you don't like music, do NOT go down to the rec room. [/quote]

BOGTRO, i take it you don't like music?? :P  :roll:


Can someone please give me some basic instructions on how to draw graphs (graph theory) using asymptote?

You can find on Asymptote --> links :
Beginnings of a graph theory module and example
http://www.cs.uwaterloo.ca/~hy3chan/graphtheory2.asy
http://www.cs.uwaterloo.ca/~hy3chan/graphtest.asy

But I don't know if it is OK with asymptote 1.78.

O.G.


AMC
AIME
trigonometry
AoPSwiki
geometry
trig identities
Law of Sines

After a great deal of problem writing/inventing, deliberation, and endless editing and answer checking, we've finally prepared a Mock AIME! This was co-authored by myself, [b]Zhero[/b], and [b]andersonw[/b], with assistance from [b]dysfunctionalequations[/b].

No signups are required. Just download the file, take the test within a continuous 3-hour period, and submit your answers in [url=http://spreadsheets.google.com/viewform?hl=en&formkey=dDBSUUp4ZXVYV3BlMzcxM296dUl4d1E6MA]this form[/url]. Because of all the other mock competitions going on around now, we'll give you two weeks to turn it in (until Feb. 6). If this isn't enough time for too many people, we can always push the date back. Whenever the testing period is over, we'll release the results and solutions.

Some notes: we intentionally made our exam start out somewhat harder than a regular AIME so that we could cram in a few more interesting problems. We had a pool of around 25 problems to cram into 15 spots! (Some weren't very good at all though...) With our surplus problems, we might release a second mock AIME around a month from now. (Probably not though.)

We've had a lot of fun making this and hope that you'll enjoy taking it!

Edit:
Obviously, we can't stop you from cheating, but we hope nobody will. Not there is much of a point to cheating on a mock competition, anyways.
Also, please no [public] discussion about this until the testing period is over.
[size=75]
[color=green]Note: the updated test is on post 12.[/color][/size]

Hmm I looked at the document again, and it was kinda weird-looking, so I made the margins larger. This version is otherwise exactly the same.

[quote="andersonw"]I would like to say that these are very good quality problems (they were written, edited, checked, and rechecked by a blue MOPper, multiple red MOPpers, and one almost-red MOPper :P). So try them out!

And of course, just to make sure there is no confusion, the normal AIME rules apply. Graph paper, ruler, compass, and protractor are allowed, but no calculators or other outside sources of help (other people, books, internet, etc.).[/quote]

wOW this is gonna be harrdd.

I don't want to give too much away, but I'll say that we tried to make the test's difficulty as close to a real AIME as possible. It may not have turned out that way, as a few problems are trickier/more annoying than on an actual AIME, and we may have messed up on the difficulty order in a few places. However, if difficulty is defined as the amount of creativity/insight required, I would say that the mock AIME did a good job of being similar in difficulty to a real AIME.

Also, for the amount of times the files have been downloaded, there has been a much less impressive number of people who have submitted answers. :(

Edit: Also see Brut3Forc3's post below. :P

[quote="SnowEverywhere"]Hello. I'm just wondering, how much harder would you say this test is than a normal AIME? Also, these problems are all really good. They are professional (i.e. as though they could appear on a real AIME).[/quote]
I think it may me significantly harder than an actual AIME, at least for the low-numbered questions. This test isn't really meant to be an indicator of performance, but rather just a good source of practice.

Hey everyone. We realize that the deadline is far away, and some of you wish to know your score before it is over. We have thus decided to PM initial results to users that submit their answers. Final results will be posted after the deadline is over. (The deadline may be extended if we don't get many submissions.)

Your initial score report will consist only of your score. We will try to send initial score reports as soon as we realize that answers have been submitted. Please use your AoPS username when filling out the answer form so that we will be able to send a PM to you.

[quote="andersonw"]I don't want to give too much away, but I'll say that we tried to make the test's difficulty as close to a real AIME as possible. It may not have turned out that way, as a few problems are trickier/more annoying than on an actual AIME, and we may have messed up on the difficulty order in a few places. However, if difficulty is defined as the amount of creativity/insight required, I would say that the mock AIME did a good job of being similar in difficulty to a real AIME.

Also, for the amount of times the files have been downloaded, there has been a much less impressive number of people who have submitted answers. :([/quote]
I think it could be because some people download it for use later, like when they have free time. I for one am going to download it soon, but I probably won't do it until next weekend.

Uh oh, a couple of errors:
#6 has an issue with the ambiguous meaning of "another". Replace it with "a different".
Another problem uses $ A \in S_n$ when it should be $ A \subseteq S_n$. (You'll know which problem I'm talking about when you see it.)
Attached is a new version fixing these errors. Sorry for any inconvenience.

[quote="Ihatepie"]What does that new sign mean btw?[/quote]
It means that $ A$ is a subset of $ S_n$. (This is pretty clear in the context of the problem.)

[quote="Brut3Forc3"]Another problem uses $ A \in S_n$ when it should be $ A \subseteq S_n$.[/quote]

I didn't know they were different. :P

Just a reminder: the deadline is today at midnight. If someone wants an extension, we might be able to do that. Not that you can't take the test and check your answers after the deadline has passed, anyways...

Ditto, I completely forgot about this one after worrying so much about getting the Mock AMC answers in on time.

I can also do this one tomorrow.

We really don't have a problem with extending it a full week. What matters to us more is that people enjoy our test.

In any case, as of now we've only written up half the solutions. :P

I just found this today, so seeing as it looks like the deadline has been extended I submitted my answers. Great test, thanks to all the writers!

You should wait until we have finished writing the solutions.

Hopefully 1)that will be before aime and 2)I will still remember by then.

Yeah, the solutions have been written up by now. Someone wants to take it officially though, so we're waiting a day for him.

Okay, we're finally done! Attached is the solutions pdf.

[b]Statistics:[/b]
We had 20 participants who scored an average of 4.90 on this test. Of these 20, 15 disclosed their previous score. We had an average of 7.33 previous year scores from these 15, and these 15 averaged 4.87 on this test. So our test is roughly 1.5 times the difficulty of a regular AIME, if we assume proportionality. (Yes, that is a bad assumption, but the point is that this is significantly harder than a regular AIME.)

[i]Problem distribution:[/i]
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|} \hline Problem & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 \\
\hline \# Correct & 13 & 7 & 12 & 17 & 5 & 1 & 3 & 11 & 13 & 6 & 4 & 0 & 2 & 3 & 1 \\
\hline \end{tabular}

[i]Score distribution:[/i]
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|} \hline Score & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 \\
\hline \# People & 0 & 1 & 1 & 6 & 3 & 2 & 2 & 1 & 2 & 1 & 0 & 1 & 0 & 0 & 0 & 0 \\
\hline \end{tabular}

[b]High scorers:[/b]
1st: dnkywin with 11 points
2nd: alkjash with 9 points
3rd: azjps, Bacteria with 8 points

You may freely discuss the problems now. :)

Gah I failed to read (#11 - find $ a$? etc), but the test was nice. The last two looked deceivingly difficult.

For #11, another equivalent approach is to use Extended Law of Sines to find $ 2R \equal{} \frac{a}{\sin BAC} \equal{} \frac{BC}{\sin BDC} \equal{} \frac{CD}{\sin DBC} \equal{} \frac{CD}{\cos ABC} \Longrightarrow CD \equal{} 2R \cos ABC$; then $ AF \plus{} \cdots \plus{} EA \equal{} 4R \sum_{cyc} \cos ABC$. Also, $ \sum_{cyc} \cos ABC \equal{} 1 \plus{} \frac rR$ (identity, equivalent to Carnot).

[quote="azjps"]Gah I failed to read (#11 - find $ a$? etc), but the test was nice. The last two looked deceivingly difficult.

For #11, another equivalent approach is to use Extended Law of Sines to find $ 2R \equal{} \frac {a}{\sin BAC} \equal{} \frac {BC}{\sin BDC} \equal{} \frac {CD}{\sin DBC} \equal{} \frac {CD}{\cos ABC} \Longrightarrow CD \equal{} 2R \cos ABC$; then $ AF \plus{} \cdots \plus{} EA \equal{} 4R \sum_{cyc} \cos ABC$. Also, $ \sum_{cyc} \cos ABC \equal{} 1 \plus{} \frac rR$ (identity, equivalent to Carnot).[/quote]

How do you do this with Carnot?

[hide="what Ive seen"]
let $ d_a$ denote $ OA'$, where $ A'$ is the midp of BC, and cyclic for $ d_b,d_c$. Then by Ptolemy's on cyclic quad $ AB'OD'$ and cyclic,

\[ (a \plus{} b \plus{} c)(\sum d_a) \equal{} 2K \plus{} 2sR \Longleftrightarrow \sum d_a \equal{} r \plus{} R\].[/hide]

Nice solution azjps.

Carnot: $ R\cosh = d_a$, I believe.

What did you guys all think of the problems in general? How about the difficulty?

I thought the problems were great (as are the solutions).

They were certainly more difficult than normal, but man I'd LOVE to have like 10 of these.  Practice with quality problems that are harder than normal is perhaps the best practice possible.

The annoying thing to me though, was that the order of difficulty was inconsistant.  Number 1 was easy, but number 2 i couldn't get.  Then 3 and 4 were not too bad.  Then you had to jump to 8 and 9 before it was "easy" again.  I'd imagine it's difficult to determine just how hard a given problem is, especially if you wrote it, but I found myself jumping around not having a good idea of which problems I had a legitamate chance of solving (and after #2 I thought it was hopeless  :)  )

But that aside, thanks for this and I wish there were lots more like it.

I agree with the above, even though I only got 3,8,9 right, but I missed the others before 8 due to careless errors. I suppose I ought to check my methods/or computation more.

Thanks for all the feedback. I'm glad that you enjoyed our tests.  :D

After seeing some of the results and hearing what some of you guys are saying, I realize that the ordering of the problems may have been a bit... off. I apologize for this. As pt90 has mentioned, this is in part attributable to the fact that we wrote the problems ourselves. I welcome opinions on how the problems should have been correctly ordered.

How did you all feel about the amount of computation involved? I've heard from various people that there were some problems with too much of it. Any ones in particular that you'd like to mention?

[quote]The last two looked deceivingly difficult. [/quote]
#14 was placed there as a joke. You can probably guess why. :P

[quote="Zhero"]How did you all feel about the amount of computation involved? I've heard from various people that there were some problems with too much of it. [/quote]

I had more answers with mistakes than answers without mistakes. :P

[quote="dysfunctionalequations"]This may be biased, but I don't think there was excessive computation on any of the problems.[/quote]
I agree, with same caution of bias. Some of my problems are the big offenders, but they figuring out how to quickly calculate it became as big of a portion of the problem as figuring out what to calculate. (That is, they punished bashing but were very managable with smarter methods. This is especially true of my $ \frac{n\plus{}x}{x}$ problem...

[quote="Bacteria"]Yeah, I think the computation was at a good level and the problems were balanced and a good challenge. On a side note I see that I'm the only one who got #15  :lol: , though it was a half-guess:
[hide="15 how I did it"]
First, similarily to the official solution, I figured out that n is a multiple of 2 and not 4. Then I noticed if n is a member of X and p is a prime greater than n then pn is a member of X. Also, I realized that the best number was probably a product of distinct primes because that would normally give more divisors (ie if x and y are prime then xy has 4 divisors while x^2 has 3). I started with 2*3=6, then since 7>6 I got 42. But adding two more primes using my method would give a number larger than 2010, so I pick the largest prime p such that 42p<=2010. This happens to be 47 so my number was 42*47=1974 which gave an answer of 584.
[/hide][/quote]
Nice. :P That's actually around the same way I guessed the answer. (I simply asked myself this question without any solution in mind, so I had to solve it too!) Of course, I had to prove it after... It turned out to have a decent solution rather than the bashbashbash one I had been fearing, so we added it to the test!

I still don't see how I was the only one who got the n+x / x thing? lol

I think it was a pretty good test, but it was harder than an actual AIME because there weren't any problems that IMO could be solved by any random person who got 99 on the AMC12. By that I mean like last year's number 1, or something. Basically, not as many free points.

I didn't get a chance to look at the later problems, because I was stuck making computation mistakes that gave me answers over 1000 :P

With some easier problems thrown in the beginning, I'd say it'd be almost just like a real AIME.

I was one off on the n+x/x thing

Missing #1 is not good

I dunno if I should say this but...I was the one with the lowest score yay!

Yeah...missing a lot of questions by misreads (#1, among other beginning ones) and arithmetic errors was pretty fail, but I think it helps a lot to have a good, little-harder-than-normal and arithmetically-inclined AIME to take for practice. I really think these type of AIME's will help a lot for the real thing.

Okay, Zhero showed me that an AoPSWiki page exists for this now... hooray for fame! Anyway uhh I'll just upload the .tex's to make it easier to make those pages if anyone ever does it now >.>

Edit: yes I could do this myself but I'm lazy!

If anyone thought 2020 AIME I #9 was easy:

[hide]
#2 is essentially a beefed up version of 2020 AIME I #9. Goes to show how even though this mock is 10 years old (and I mocked this a long time ago back in the summer of 2019), it can still help for some problems nowadays
[/hide]

all problems have been posted in separate threads in HSM,  I just created the related post collection

[url=https://artofproblemsolving.com/community/c3691213_2010_mock_aime_i]here[/url]

enjoy  / start solving

more USA mocks [url=https://artofproblemsolving.com/community/c2439870_usa_mocks]here [/url] (also mentioned [url=https://artofproblemsolving.com/community/c5h3200761p29225908]here[/url])
more AIME mocks [url=https://artofproblemsolving.com/community/c2439872_aime_mocks]here[/url],
more User-created contests [url=https://artofproblemsolving.com/community/c2435719_user_created_contests]here[/url] (contains all above)


topology
real analysis
real analysis unsolved

Let $ \gamma: [0,1]\rightarrow [0,1]\times [0,1]$ be a mapping such that for each $ s,t\in [0,1]$
\[ |\gamma(s) \minus{} \gamma(t)|\leq M|s \minus{} t|^\alpha
\]
in which $ \alpha,M$ are fixed numbers. Prove that if $ \gamma$ is surjective, then $ \alpha\leq\frac12$

Divide the square into $ N^2$ small squares in the usual way, take the centers of these squares to be $ Q_i$ and let $ P_i$ be preimages of $ Q_i$. Then we get $ P_iP_j\geq C/N^{x}$ with $ x\alpha\equal{}1$. Now, it's clear that because all $ P_i$ are on $ [0,1]$ one of the segments is smaller than $ 2/N^2$ and taking $ N$ large we get $ x\geq 2$.

this all is true with condition $ \alpha > 0$
Generalization:
if $ \gamma : [0,1] \to [0,1]^{r}$
with above condition then $ 0 < \alpha \leq \frac {1}{r}$
similari by the same way.
Divide the square into $ N^{r}$ ...e.t.c.
And if there exist map $ \gamma: [0,1] \to [0,1]^{\infty}$ then I think statetment is not true. (but I don't  understand what does it means: statement is not true  :lol: ):maybe: .
maybe there is no  surjective $ \gamma$..

Well if there's no surjective $ \gamma$ then it's true, albeit vacuously  :P.  Can there be a surjective $ \gamma: [0,1]\to[0,1]^\infty$?  Where's the topologist in the house?

[quote="Omid Hatami"]There is a surjective map $ \gamma: [0,1]\longrightarrow [0,1]^{\mathbb N}$, but there can not be a surjective map $ \gamma: [0,1]\longrightarrow [0,1]^{\mathbb R}$, because $ |\mathbb R^{\mathbb R}| > |\mathbb R|$.[/quote]
yes, but is it obviusly? $ |\mathbb R^{\mathbb R}| > |\mathbb R|$
At first I don't understand what is uniformal  metric in $ [0,1]^{\infty}$  :maybe: , I can Understand metric for some paralelepiped infinity dimensional  , but for square  :( ...

I didn't talk about having a metric on $ [0,1]^{\mathbb N}$, but maybe we can consider the metric induced by the box topology ($ \sum\frac e{d_i}{2^i}$) or consider the supremum norm.


The fraction halfway between $ \frac{1}{5}$ and $ \frac{1}{3}$ (on the number line) is

\[ \textbf{(A)}\ \frac{1}{4} \qquad
\textbf{(B)}\ \frac{2}{15} \qquad
\textbf{(C)}\ \frac{4}{15} \qquad
\textbf{(D)}\ \frac{53}{200} \qquad
\textbf{(E)}\ \frac{8}{15}
\]


The sum of three consecutive even integers is $ \minus{}6$. What is the least of the three numbers?

we know that the numbers are consecutive even integers
let's call the middle number $ x$, so the higher and lower numbers are $ (x\plus{}2)$ and $ (x\minus{}2)$ respectively

we know that $ (x\minus{}2)\plus{}x\plus{}(x\plus{}2)\equal{}\minus{}6$

so, $ 3x\equal{}\minus{}6$, and $ x\equal{}\minus{}2$

we want to find the least of the three numbers, so we want $ (x\minus{}2)$

$ (x\minus{}2)\equal{}\minus{}2\minus{}2\equal{}\minus{}4$

our answer is $ \boxed{\minus{}4}$


if $f$ is differentiable in the interval $[a,b]$
and for $\any x \in (a,b)$
we have $f'(x)>\frac{f(x)-f(a)}{x-a}$
prove:
$f'(b)>f'(a)$

sorry,can anybody help me.
And sorry,I can do nothing follows with liyi's hint :(

$f$ is not necessarily convex, but your result still holds.

Set $g(x)=\frac{f(x)-f(a)}{x-a}$. $g$ is derivable on $]a,b[$ and $g'(x)=\frac{f'(x)(x-a)-(f(x)-f(a))}{(x-a)^{2}} > 0$. As a result $g$ is strictly increasing on $]a,b[$ and $f'(x)>\frac{f(x)-f(a)}{x-a} > f'(a)$ there, by taking $x \rightarrow a$. Now applying Darboux's theorem gives the result (if $f'(b) \leq f'(a)$ then $f'$ does not verify the intermediate value property on $[\frac{a+b}{2},b]$ since $g$ is strictly increasing, which would be absurd).


geometry
trapezoid
calculus
integration
3D geometry
probability
AMC

[color=cyan]So, at the request of darkquantum, I have dug up (stolen outright) a bunch of problems that were given as tests to choose the active members of the Stuyvesant High School junior and senior math team two years ago.  Many of these are probably stolen from elsewhere, so you may have seen them before.  They are in no particular order, so it is recommended you look through for the problems most appropriate for your level:

1. In trapezoid ABCD, AB is parallel to CD and the area of triangle ABC divided by the area of triangle ACD is 1/3. If E and F are midpoints of BC and AD respectively, find the numerical value of the area of region ABEF divided by the area of region EFDC.

2. Given three positive numbers a, b, and c, such that  a/b = b/c = c/a, find the numerical value of  (a + b + c)/(a - b + c).

3. Will is given a list of 10 positive integers, which together have exactly 9 distinct values.  Will finds that by adding 9 of the numbers, he can make any integral sum from 86 to 94, inclusive. Find the integer that occurs twice.

4. In triangle ABC, E is the midpoint of BC and D is a point on side AC. If AC = 1, angle BAC = 60 degrees, angle ACB = 20 degrees, and angle DEC = 80 degrees, then evaluate numerically the area of triangle ABC plus twice the area of triangle CDE. Do not include  trigonometric expressions.

5. Compute the number of positive integers less than 1,000,001 that are either a perfect square or a perfect cube, but not both.

6. Solve for all real x:  (2x:^2:  x +4):^2:  6(2x:^2:  x) = 19.

7. Solve for all x:  (x - 3):^3: + (x - 7):^3: = (2x - 10):^3:

8. How many pairs, (x, y), of positive integers satisfy
x:^3: = y:^3: - 64 ?

9.  Let P be a point inside regular pentagon ABCDE, such that triangle APE is equilateral.  Find the degree measure of obtuse angle BPD.

10. During a bike marathon, Hubert was already 3/8 of the way across a narrow bridge.  A train traveling at 60 miles per hour was approaching the bridge from behind him.  Hubert, using his crazy math skills, calculated that if he were to continue biking forwards at x miles per hour, he and the train would meet at the farther end of the bridge.  Hubert also figured out that if he were to bike backwards at x miles per hour, he and the train would meet at the closer end of the bridge.  In both cases, Hubert gets hit by the train just as he thinks hes escaped.  Compute x.

11. AB is a chord of the larger of two concentric circles and is tangent to the smaller circle at T.  Given AB = 8, find the area of the region bounded by the two circles.

12. Circle O_1 of radius 105 is tangent to both sides of a 60 degree angle.  For all n greater than 1, Circle O_(n+1) is tangent to circle O_n and both sides of the angle, and is smaller than circle O_n.  Find the sum of the radii of all of the circles O_i.

13. The sum of two positive numbers multiplied by the first of the two is 84.  The difference of the two multiplied by the second is 16.  Find the largest and smallest possible sums for the two numbers.

14. A fair coin is tossed 10 times.  Compute the probability that heads does not occur on consecutive tosses.

15. tan(x) = (1 + sin(40) - cos(40)) / (1 + sin(40) + cos(40)).  Find all values of x if 0 :le: x < 360 and all angles are measured in degrees.[/color]


modular arithmetic
function
number theory
relatively prime

How would you do that?  (Other than guess and check with multiples of 11)

Start with trying to find a multiple of 5 in which the sum of its digits is 2 less than 9 (since two greater than this number is a multiple of 9). Then, i think it's guess and check.

[hide]You start with 33 as the highest number since 11|33, and 5|33-3.  Then keep adding 5*11=55, until you get 9|33+55x-1 or 9|32+55x, which makes x=4, since 32=5 mod 9. and 55=1 mod 9. 33+55*4=253.  Then you keep adding 55*9=495, until you get 7|253+495x-2, 7|251+495y, which makes y=3.  Since 251= 6 mod 7, and 495= 5 mod 7.   253+495*3=1738, which is the highest number.  So the 4 numbers are 1735,1736,1737,1738.[/hide]

[hide]

I used guess and check but there could be a 'law' you could follow to make solving the problem easier.

There has to be a multiple of 9 2 more than a multiple of 5 and a multiple of 7 1 more than a multiple of 5.

I, then, used guess and check and got 1735,1736,1737,1738. [/hide]

Find an $x$ such that

$x \equiv 0 \pmod {11}$
$x \equiv 1 \pmod {9}$
$x \equiv 2 \pmod 7$
$x \equiv 3 \pmod 5$

And since the moduli are conveniently pairwise relatively prime, this can easily be done by using the Chinese Remainder Theorem.

Using a function I wrote to solve such systems I came up with the answer -43307 which by adding a bunch of $11 \cdot 9 \cdot 7 \cdot 5$ turns into 1738

[quote="anirudh"][quote="jsc"]Find four consecutive whole numbers such that the first one is a multiple of 5, the second is a multiple of 7, the third is a multiple of 9, and the largest is a multiple of 11 ...  Is this an old one ?  It's nice anyway...

 :whistling:[/quote]

This isn't very hard with modular algebra

$x\equiv 0\ (\text{mod 5})$

$x\equiv 1\ (\text{mod 7})$

$x\equiv 2\ (\text{mod 9})$

$x\equiv 3\ (\text{mod 11})$

$5\cdot7\cdot9\cdot11=3465$[/quote]
WEll, you did absolutely nothing, again. All you did was copy tarquins. You have no answer or solution, its just spam

[quote="236factorial"]Did he give us an answer? Is 3465 the answer?[/quote]
we are looking for 4 consecutive numbers, not just one


Prove that no perfect squares exist for $ 2^n \minus{} 7$, where n is any integer greater than 181.

The standard solution method for the [url=http://en.wikipedia.org/wiki/Ramanujan-Nagell_equation]Ramanujan-Nagell equation[/url] uses algebraic number theory; see page 50 [url=http://www.math.sc.edu/~filaseta/gradcourses/TheMath784Notes.pdf]here[/url].

Very enlightening. These notes are very useful. Where is this from?


function
algorithm
limit
real analysis
real analysis unsolved

Give another proof of the Intermediate Value Theorem by completing the following argument:

If  is a continuous real-valued function on the closed interval $[a,b]$ in $R$ and $f(a) < \gamma < f(b)$, then $f( sup\{x\in[a,b]: f(x) \le \gamma\}) = \gamma$.

I know what the intermediate value theorem is: If $a,b \in R, a < b$, and $f$ is a continuous real-valued function on the closed interval $[a,b]$, then for any real number $\gamma$ b/w $f(a)$ and $f(b)$ there exists at least one point $c\in(a,b)$ such that $f(c) = \gamma$. The proof using connectedness is really easy, but what can I get from using this statement?

By the least upper bound axiom, the $sup$ exists since the given set has $b$ as an upper bound, and is nonempty since $a$ belongs to it.
Now by continuity of $f$, $f(sup(\{x \in [a,b]: f(x) \leq \gamma\}))$ cannot be $< \gamma$, otherwise you can find $\epsilon>0$ such that $f(sup(\{x \in [a,b]: f(x) \leq \gamma\})+\epsilon) \leq \gamma$, which contradicts the definition of the $sup$. It cannot be $>\gamma$ either (why ?). This solves the question.

You may think about the fact that the usual proof using connectedness is not easier at all, because it requires you to characterize the connected sets of $\mathbb{R}$, which takes much longer explanation than the above ...

There's also the proof by algorithm: successive bisection.
WLOG, $f(a)<\gamma<f(b)$. Let $a_{0}=a,b_{0}=b$. For each $n\ge 0$, look at $f\left(\frac{a_{n}+b_{n}}2\right)$. If $f\left(\frac{a_{n}+b_{n}}2\right)\ge \gamma$, let $a_{n+1}=a_{n},b_{n+1}=\frac{a_{n}+b_{n}}2$. If $f\left(\frac{a_{n}+b_{n}}2\right)< \gamma$, let $a_{n+1}=\frac{a_{n}+b_{n}}2, b_{n+1}=b_{n}$.
For each $n$, $f(a_{n})<\gamma\le f(b_{n})$. The sequence $a_{n}$ is increasing and bounded above, so it has a limit $L$. Since $b_{n}-a_{n}=2^{-n}(b-a)$, $\lim b_{n}=L$ as well. Now $\gamma\ge \lim f(a_{n})=f(L)=\lim f(b_{n})\ge \gamma$, and $f(L)=\gamma.$


A large house contains a TV set in each room that has an odd number of doors. There is
only one entrance to the house. Show that it is always possible to enter this house and get
to a room with a TV.

so this is the solution:

"Consider the house as a graph where each vertex represents a room, and each edge
represents a door between two rooms. An additional vertex will represent the area outside
the house (connected to one of the room though the entrance to the house). Without loss
of generality, we can consider only the rooms connected to the outside, i.e., the connected
component of this graph with the edge representing the entrance.
We know that the sum of the degrees of all vertices is even in any graph (being twice the
number of edges). Since there is just one entrance to the house, the degree of the vertex
representing the outside area is 1. Therefore, the sum of the degrees of all the vertices
representing rooms connected to the entrance is odd. Therefore there is at least one room
with an odd number of doors connected to the entrance. By the assumption of the problem,
such a room has a TV. So it is always possible to a reach a room with a TV."
this explanation makes little sense to me.

--vertex will represent the area outside the house? (what? how? can someone draw a picture?)

--, the degree of the vertex representing the outside area is 1?  "Therefore, the sum of the degrees of all the vertices representing rooms connected to the entrance is odd." (but 2*1=2 is even)

Can someone make this clearer or provide a different solution?

http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1955848122&t=23648

I searched for "TV doors."  If I'd known this was from Zeitz, it would have been even easier to find.


logarithms
calculus
calculus computations

A wall absorbs radiation from a substance with $\frac{dS}{dx}=-0,08S$
S=intensity and x=thickness

how thick should the wall be so that 70% of the radiation is absorbed?

would appreciate some assistance :blush:
what I need to understand is howe does one get to$0,30C=Ce^{-0,08x}$ I just dont get that why 0,30C

thanks

[Notation: since I'm American, I'll say $0.30$ instead of $0,30.$]

$C$ is the intensity on the near side of the wall; in differential equations terms, it's an initial condition.

The formula $S=S(x)=Ce^{-0.08x}$ is the solution to the differential equation. $S(x)$ is the intensity after passing through a thickness $x$ of wall.

So the question is this: what thickness of wall (what value of $x$) would be sufficient for the intensity to be 30% of the original intensity. But 30% of the original intensity is $0.30C.$ Hence we want to solve the following equation for $x:$

$Ce^{-0.08x}=0.30C.$

Note that the particular value of $C$ does not matter. Our first step in solving will be to divide by $C.$ Our second step will be to take the natural logarithm of both sides of the equation.

This has exactly the same structure as a half-life problem, except that we are trying to got down to 30% rather than 50%.

so $C$ is really the begining value of the wall, i.e the thickness. So I guess I understand how that works, but frankly, I still think I would guess that it would be $0.7C$ but now I see that because we want to know when the original value of $C$ gone down with $70\%$ the result would be$0.30C$.
Hence $0.30C=Ce^{-0.08x}$

thanks


advanced fields
advanced fields unsolved

Let K be a compact subset of a metric space M. Show that $ \bar{A}$ is compact

You've posted a batch of about 20 problems here, some (perhaps all) of which require just a straightforward application of the definitions, and many of which with typos (like this one) or unclear terms. They tend to look very much like homework problems, or routine exercises from a textbook. I think your chances of getting helpful responses here would increase a lot if you A. explain what these problems are and what you're looking for, B. Clarify the definitions you're working with (otherwise at least one problem is just a repetition of the standard definition) C. make an effort to write them down without error, and [b]most importantly[/b] D. explain what you tried, what you've thought of, where you got stuck and what you need help with.

Indeed, a lot of them have stupid mistakes that would have already been avoided by either reading again or more understanding. Homework problems are dicouraged as long as the original poster doesn't show some own work.

[quote="ZetaX"]Homework problems are dicouraged as long as the original poster doesn't show some own work.[/quote]Do you mean [i]does[/i] show some own work?

Well, "discouraged" can be a bit misleading, so a better version of that sentence:
Homework problems are only welcome if the original poster shows some work.

Okay, we're on the same page.  I forgot the discouraged part.

Either way, we both agree that homework problems should show some attempt.


Hi!

Let $a,b,c$ be three postive real numbers. Can you prove that :
\[\left({a\over b}+{{c}\over{(abc)^{{1\over 3}}}}\right)^{2}+\left({b\over c}+{{a}\over{(abc)^{{1\over 3}}}}\right)^{2}+\left({c\over a}+{{b}\over{(abc)^{{1\over 3}}}}\right)^{2}\geq 12.\]

This is just AMGM
\[\sum\left(\frac{a}{b}+\frac{c}{\sqrt[3]{abc}}\right)^{2}\geq \sum\left(2\sqrt{\frac{ac}{b\sqrt[3]{abc}}\right)^{2}= 4\sum\frac{ac}{b\sqrt[3]{abc}}\geq 12\sqrt[3]{\prod \frac{ac}{b\sqrt[3]{abc}}}=12 }\]


invariant
superior algebra
superior algebra unsolved

Determine the number of solutions of the equation $x^{2}=1$ in $Z_{2004}$. If $x_{1},x_{2},...,x_{k}$ are these solutions, compute $x_{1}\cdot x_{2}\cdot x_{3}\cdot ... \cdot x_{k}$.

$\mathbb{Z}_{2004}^{\times}$ is isomorphic to $\mathbb{Z}_{4}^{\times}\times\mathbb{Z}_{3}^{\times}\times\mathbb{Z}_{167}^{\times}\cong\mathbb{Z}_{2}\times\mathbb{Z}_{2}\times\mathbb{Z}_{166}$

All you have to do is find out, how many elements of order 2 exist in this group.

This of course is a visible thing but I don't think it solves part two of the problem.

The isomorphism $\mathbb{Z}_{2004}^{\times}\cong\mathbb{Z}_{4}^{\times}\times\mathbb{Z}_{3}^{\times}\times\mathbb{Z}_{167}^{\times}$ is the one given by the chinese remainder theorem, so you can go it backwards and compute the concrete elements you're looking for.

More directly:
The group of elements of order 2 is isomorphic to $(\mathbb{Z}_{2})^{3}$. The product of the elements of this group is an element which is invariant under isomorphism, so it must be the identity

If you replace 2004 with any $n$, the only possible answers are $\pm 1$.