function
algebra
functional equation
algebra proposed

$x=0 \Longrightarrow f(f(y)) = f^2(0) + y \Longrightarrow  f\  bijective$ .So there is $x_0$ such that $f(x_0)=0$ , so $f(f(y))=y$
$x=t \Longrightarrow f(tf(t)+f(y)) = f^2(t) + y $ ; $   x=f(t) \Longrightarrow f(f(t)t+f(y)) = t^2 + y$ , so  $f(t)=t $ or $f(t)=-t$

Suppose that there are $a\neq 0\neq b$ such that $f(a)=a , f(b)=-b)$ , so $ f(a^2-b)=a^2+b$
Two cases are possible:$  a^2+b=f(a^2-b)=a^2-b \Longrightarrow b=0 $ (contradiction) , or , $a^2+b=f(a^2-b)=-a^2+b \Longrightarrow a=0 $ (contradiction)

Finally the functional equation has only two solutions :$ f_1(t)=t\  ,\  f_2(t)=-t $
:cool:

It was also given in our pre TST this year  :lol:
Have a look at the third day problem 1 [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=32569]here[/url]  ;)

trivially,f is bijective.
so there exists $x$,$f(x)=1$
then $f(x+f(0))=f(x)$
hence $f(0)=0$
let $x=0,y=x$
then$f(f(x))=x$
let $y=0$,then $f(xf(x))=f^2(x)$
by letting $x=f(x)$,we het $f(xf(x))=f(f(x))=x^2$
hence $f^2(x)=x^2$,so $f(x)=x or -x$
if there exist non-zero x,y,$f(x)=x,f(y)=-y$
then $f(x^2-y)=x^2+y$
yielding $x=0$ or $y=0$
contradiction!
hence f can't be piecewice,yielding $f(x)=x$ or $f(x)=-x$.

Plugging in $x = 0$ we get $f(f(y)) = f(0)^2 + y$. Therefore, there exists a real number $r$ such that $f(r) = 0$ (specifically, $f(-f(0)^2)$). Then letting $x = r$ gives us $f(f(y)) = y$ for all $y \in \mathbb{R}$. This tells us that $f(0) = 0$ when we combine it with $f(f(y)) = f(0)^2 + y$. Now letting $y = 0$ we get
$$f(xf(x)) = f(x)^2$$
Replacing $x$ with $f(x)$ gives us
$$f(f(x)x) = x^2$$
So therefore $f(x) = \pm x$. Suppose we have $f(a) = a$ where $a \neq 0$. Then I will show that $f(x) = x$ for all $x \in \mathbb{R}$. Letting $x = a$ in the original FE gives us $f(a^2 + f(y)) = a^2 + y$. Letting $y = f(t)$ gives us $f(a^2 + t) = a^2 + f(t)$ and so we can let $t$ be anything we want. Suppose we had $f(a^2 + t) = -a^2 - t$ for some $t$. Then we have $-a^2 - t = a^2 + f(t)$ so then $f(t) = -2a^2 - t$. We therefore must have $f(t) = t$ which tells us that $t = -a^2$. That means the only time that $f(x) = -x$ is when $x = 0$.

So now we see that $f(x) = x$ and $f(x) = -x$ are the only solutions.

let $x=0 \implies f(f(y))=f^2(0)+y$ implies $f$ is bijective. $\exists!a$ such that $f(a)=0$, let $x=a \implies f(f(y))=y(2) \rightarrow f^2(0)=0 \rightarrow f(0)=0$. Let $y=0 \implies f(xf(x))=f^2(x)(1)$. Replaced $(1)$ and $(2)$ in the original equation we get $f(xf(x)+f(y))=f(xf(x))+f(f(y))$, let $x=f(x), y=f(y)$ we have $f(x^2+y)=f(x^2)+f(y) \leftrightarrow f(x+y)=f(x)+f(y)(*)$,for $x\geq 0, y\in \mathbb{R}$.From$(*)$ let $y=-x \rightarrow -f(x)=f(-x)$.For $x\leq 0$ we have $f(y)=f(y+x-x)=f(y+x)+f(-x)=f(x+y)-f(x) \leftrightarrow f(x+y)=f(x)+f(y)$,for $x\leq 0,y\in\mathbb{R}$ implies $f(x+y)=f(x)+f(y)$,for $x,y\in\mathbb{R}$.  Also from the original equation we have $f(xf(x)+f(y))\geq y\leftrightarrow f(x+y)\geq f(y)$, for $x\geq 0, y\in\mathbb{R} \implies f$ is increasing function. Because f is additive and increasing function implies $f(x)=ax$ for some $a\in\mathbb{R}$.Replaced in the original equation we have $a=1$ or $a=-1$. Thus $f(x)=x$ or $f(x)=-x$ is two functions satisfying the conditions of problem.

putting x = y =0,

f(f(0)) = f(0)^2

there exists a t such that , f(t) = 0;

setting x = t , y = 0;

f(f(0)) = 0;

so, f(0) = 0;

f(f(x)) = x;

setting x = f(x),

f(xf(x) + f(y)) = x^2 + y which means that f(x) = x or f(x) = -x

[u]case 1:[/u]

f(1) = 1;

putting x = 1 , y = x,

f(f(x) + 1) = x + 1 easy checking shows that f(x) = x for all real x;

[u]case 2:[/u]

f(1) = -1;

putting x = -1 , y = x;

f(f(x) - 1) = x + 1 easy checking shows that f(x) = -x;

from 2 cases,

f(x) = x for all real x

or , f(x) = -x for all real x

@fighter

Nice solution, brother. :coolspeak:
I have another idea.

Let $P(x,y)$ be $f(xf(x)+f(y))=f(x)^2+y$.Obviously $f$ is surjective.So let $f(c)=0$,then $P(c,y)\rightarrow
f(f(y))=y$.$P(f(x),y)\rightarrow f(xf(x)+f(y))=x^2+y$.Thus $f(x)^2=x^2\Leftrightarrow f(x)=\pm x$.

We suppose that ∃$a\neq 0$, ∃$b\neq 0$ s.t. $f(a)=a, f(b)=-b$.$P(a,b)\rightarrow f(a^2-b)=a^2+b$.Then $a^2-b=a^2+b\Leftrightarrow b=0$ or $-(a^2-b)=a^2+b\Leftrightarrow a=0$ which are both absurd.

Therefore $\forall x\in \mathbb R:f(x)=x$ or $\forall x\in \mathbb R:f(x)=-x$ which both satisfy the condition. :icecream:

can we find any solution using injectivity

Ans: $f(x)=x, f(x)=-x$. \\ It is easy to see that these satisfy the equation. \\ Let $P(x,y)$ be the given assertion, $P(x,-f(x)^2)\implies f(xf(x)+f(-f(x)^2))=0$, so there exists $k\in\mathbb{R}$ s.t $f(k)=0$. \\ $P(k,k)\implies f(0)=k$ and $P(0,0)\implies f(f(0))=f(0)^2$, so \[f(f(0))=f(k)=0=f(0)^2\implies f(0)=0.\] \\ Then $P(0,x)\implies f(f(x))=x$, so \[P(x,0)\implies f(xf(x))=f(x)^2\] and replacing $x$ by $f(x)$ gives \[f(xf(x))=f(f(x))^2=x^2,\] thus $f(x)^2=x^2\implies f(x)=\pm x$. Assume there exists $x,y\ne 0$ s.t $f(x)=x, f(y)=-y$, then \[P(x,y)\implies f(x^2-y)=x^2+y\implies x^2-y=x^2+y \ \ \textrm{or} \ \ y-x^2=x^2+y \implies x=0 \ \ \textrm{or} \ \ y=0\] which is a contradiction. Thus, $f(x)=x, -x$ are the only solutions. $\blacksquare$

Let $P(x,y)$ be the assertion of
$f(xf(x)+f(y))=f(x)^2+y$
$P(0,x)$:
$f(f(x))=f(0)^2+x$
there is a real number x and y for which
$f(x)=f(y)$
Let $x=f(x) $and $y=f(y)$
$f(f(x))=f(f(y))$
$f(0)^2+x=f(0)^2+y$
x=y thus it is injective.
 Again let $f(0)^2=c$
and set x=0 and y=y-c
$f(f(y-c))=y$ f is surjective too.thus  there exist a real number t such that $f(t)=0$
x=t,y=0
$f(f(0))=0$
$P(0,0)$ :
$f(0)=0$
$P(0,x) \rightarrow$ $f(f(x))=x$
$P(x,0) \rightarrow$ $f(xf(x))=f(x)^2$
set $x=f(x)$
we get $f(x)=x or -x$
which are solution indeed.
Is it right?

[hide]Let $P(x,y)$ be the assertion of $f(xf(x)+f(y))=f(x)^2+y$
$P(0,x) \Rightarrow f(f(x))=f(0)^2+x\quad\forall x\in\mathbb{R}$ implies that $f$ is injective and surjective too.
Let $u\in\mathbb{R}$ such that $f(u)=0$ and $a=f(0)$.Then
$P(u,u)\Rightarrow a=u$
$P(0,u)\Rightarrow a=a^2+u$
So we have $a^2=0$ so $f(0)=0$
$P(0,x)\Rightarrow f(f(x))=x\quad\forall x\in\mathbb{R}$
From $P(x,y)$ and $P(f(x),y$ we have that
$f(xf(x)+f(y))=f(x)^2+y=x^2+y\quad\forall x\in\mathbb{R}.$
$\Rightarrow f(x)^2=x^2\quad\forall x\in\mathbb{R}(1)$.
We assume that there $\exists a,b\in\mathbb{R},ab\neq 0$ such that $f(a)=a,f(b)=-b$.
$P(a,b)\Rightarrow f(a^2-b)=a^2+b$ so $a^2-b=a^2+b$ or $b-a^2=a^2+b$ implies that $ab=0$ which is absurd.
By $(1)$ we have that $f(x)=x$, $f(x)=-x$ that satisfy the condition.
We conclude that only two functions satisfy the problem
$$\boxed{f(x)=x\quad\forall x\in\mathbb{R}}\quad\boxed{f(x)=-x\quad\forall x\in\mathbb{R}}$$[/hide]

[quote=RaMlaF]Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that $\forall x,y \in \mathbb{R}$ we have \[f(xf(x)+f(y)) = f(x)^2 + y\][/quote]

Let $P(x,y)$ the assertion of the given FE.
$P(0,x)$
$$f(f(x))=f(0)^2+x \implies f \; \text{involution and bijective}$$
Hence exists a $c$ such that $f(c)=0$.
$P(c,x)$
$$f(f(x))=x \implies f(0)^2=0 \implies f(0)=0$$
$P(x,0)$
$$f(xf(x))=f(x)^2$$
$P(f(x),y)$
$$f(xf(x)+f(y))=x^2+y$$
Let compare this new equation with the older one.
$P(x,y)-P(f(x),y)$
$$f(x)^2=x^2 \implies f(x)=x \; \text{or} \; f(x)=-x$$
Assume that exists $a,b \ne 0$ such that $f(a)=a$ and $f(b)=-b$.
Note that $f$ is odd since $f(f(-x))=-x \implies f(x)=-f(-x)$ becuase if $f$ is even then $x=-x$ and that is not possible.
$P(a,b)$
$$\pm(a^2-b)=a^2+b \implies a=0 \; \text{or} \; b=0$$
Contradiction!!
Then the solutuons are:

$\boxed{f(x)=x \; \forall x \in \mathbb R}$

$\boxed{f(x)=-x \; \forall x \in \mathbb R}$

Thus we are done  :blush:

Let $P(x,y)$ denote the given assertion.

$P(0,y)$ gives that $f(f(y)) = f(0)^2 + y$ and so $f$ is injective and surjective.

Let $z$ be the unique number s.t $f(z) = 0$. Then, $P(z,y)$ gives that $f(f(y)) = y$ and so $f(0) = 0$

Also, $P(x,0)$ gives that $f(xf(x)) = f(x)^2$

Now, $P(f(x),f(y))$ gives that $f(xf(x) + y) = x^2 + f(y)$. Putting $y = 0$, we get $f(xf(x)) = x^2$ and so $f(x)^2 = x^2 \implies f(x) = \pm x$ for all $x$.

Suppose $f(a) = a$ and $f(b) = -b$ and $a, b \neq 0$. Then, $P(a,b)$ gives that $f(a^2 - b) = a^2 + b$. But the RHS is not $\pm$ of the LHS for any $a,b$ and so this is impossible.

So, the only solutions are $f(x) = x$ and $f(x) = -x$

@2below - Yeah that works too

My solution is way too easy, kindly check
My idea is same till we prove $f(f(y))=y$, after that replace $x$ by $f(x)$, LHS remains the same, compare RHS to get $f(x)^2=x^2$ and then last step same as @above just posted.

[quote=L567]Now, $P(f(x),f(y))$ gives that $f(xf(x) + y) = x^2 + f(y)$. Putting $y = 0$, we get $f(xf(x)) = x^2$ and so $f(x)^2 = x^2 \implies f(x) = \pm x$ for all $x$. [/quote]

Instead of $P(f(x),f(y))$ you could have done $P(f(x),y)$ the LHS would be exactly the same (as I have told in the post above) then and then there won't be any need to substitute $y=0$ ;)

Letting $f(0)=k$ we get that
$$f(f(x))=k^2+x \implies f  \text{is bijective}$$
Take $a$ such that $f(a)=0$ and we get upon putting $x=a$ that
$$f(f(x))= x \implies k=0$$
Now $f(xf(x)+f(y))$ can also be written as $f(f(x)f(f(x))+f(y)$ .Hence we get that
$$f(x)^2+y=x^2+y \implies \forall x , f(x) \in \{x,-x\}$$
The point wise trap can be easily sorted out by taking $f(x)=x$ and $f(y)=-y$ (both nonzero) but we will get that atleast one of them must be zero.Contradiction.
So $$f(x) \equiv  x$$
or $$f(x) \equiv -x$$

Nobody mentioned that this is [url=https://artofproblemsolving.com/community/c6h56415p24728138]Balkan 1997 and 2000[/url].


I'm pretty sure that any real number works except numbers in the range $(-999,999)$  Most integers would work because any difference between |x| and |x+1| is automatically wiped when one subtracts |x+2| and |x+3| from the two sides.  The absolute values for odd number-sums will always be either one greater, or one less, than corresponding even number-sums, depending on whether the number is negative or not.  However, if the number is in the range (-999, 999), then there will be a shifting point where the relation between odd number-sums and even number-sums will change, namely when they are around 0.  Therefore, the summation process is thrown out of whack, and it doesn't work.

What am I saying, the solution set should be $(-infinity, -999) U (0, infinity)$
Think about it.  Let's say that you plug in a random positive number.  Then you can replace the absolute value signs with parantheses.  Then, the x's disappear from both sides.  So then you are left with
0-2+4-6+8-...-998=1-3+5-7+9-...-999
This is true.  Simplify and you will get -500=-500.  So we know that any positive integer works.
Now try any negative number less than -999 (so that every term inside the absolute value signs is negative.  Then you will just get the opposite, 500=500.  So that works.
For any number in between -999 and 0, you'll get x+999 as positive, x as negative, and everything in between either one or the other, with one shifting point.  So you'll get something which looks like this on each side
414-412+410-...+2-0+2-4+...-584=413-411+409...+1-1+3-5+...-585
If the number is even, like in this case, then you'll get a bunch of 2's and -2's on both sides which cancel out.  However, you'll have something like is shown, a 2-0 on the left and a 1-1 on the right, which messes it up.  But then again, if you move it over to a multiple of 4, then it works, because the grouping works out and doesn't pair a negative term witha  positive one (negative and positive meaning the value inside the absolute value). Or, if it's a multiple of 4 plus 1, then the roles of the even and odd sides and pretty much switched except the starting point has been moved over by one.  Anything in between works as well.  So then I'll revise the answer to

$(-infinity, -999) U (0, infinity) U K(x: -999<x<0 [x=4x OR x=4x+1])$
Or you could just write it out...

I still don't see why it won't work with any positive number.  If it's a positive number, then you can change all of the absolute value signs into parantheses and you then can get rid of all of the x's.  Then you will have
$0-2+4-6+...-998=1-3+5-7+...-999$
which becomes
$-2-2-2-2-2-...-2=-2-2-2-2-2-...-2$
with 500 -2's on each side
Or, if x<-999, or in other words, the parts inside the absolute value signs are all negative, then you can change all of the absolute value signs to parantheses and change all of the signs.  Then you can cancel out all of the x's and get
$0+2-4+6-...+998=-1+3-5+7-...+999$
which becomes
$2+2+2+2+2+...+2=2+2+2+2+2+...+2$
with 500 2's on each side, which is true.
I'm not ENTIRELY sure which ones work in the interval (-999, 0) (that's what i meant in the other post) but at least these numbers satisfy the equation...


hello please can some one explain this in details thank you

How many strings of four decimal digits
(a) do not contain the same digit twice?
(b) end with an even digit?
(c) have exactly three digits that are 9’s?
(d) have at least one digit that is 9?
(e) have one digit that is 9 and do not contain the same digit twice?

I'll assume that these do not have to be 4-digit positive integers, but rather, 4-digit combinations

[hide="a"]
Our first digit can be anything from 0 to 9, so we have ten choices. Our second digit can be anything but the previous digit, so we have nine choices. The same principle applies for the next two digits, so they have eight and seven choices, respectively. $10\times9\times8\times7=5040$
[/hide]

[hide="b"]
for a number to be even, it must in 0, 2, 4, 6, or 8. There are no restrictions on the first three digits, so there are 10 choices for each of those, and five choices for the last digit. $10\times10\times10\times5=5000$
[/hide]

[hide="c"]
three digits must be nines, and the other one cannot be. There are four ways to place the digit that is not a nine, and nine choices (0,1,2,3,4,5,6,7,8) for that digit, but only one choice for each other digit. $9\times1\times1\times1\times4=36$
[/hide]

hopefully these will help you be able to solve the other two. just think about choices for digits and ways to arrange the digits

[quote="hyderman"]any one can solve d in a different way[/quote]
[quote="bpms"][quote="renatzu"][hide="d"]
First, assume that there is one 9 in the sequence: $\binom{4}{1}*9^{3}= 2916$
The $\binom{4}{1}$ comes from the fact that there is one $9$ which can be placed in 4 spots.  The $9^{3}$ comes from the fact that there are nine choices for each of the other 3 spots (anything except 9)

two 9s: $\binom{4}{2}*9^{2}= 486$
There are 2 9s to place in any of 4 spots $\Rightarrow \binom{4}{2}$, there are nine choices for the other two spots $\Rightarrow 9^{2}$

three 9s: $\binom{4}{3}*9^{1}= 36$

four 9s: $\binom{4}{4}9^{0}= 1$

$2916+486+36+1 = 3439$
[/hide]

[hide="e"]
There is one $9$ which can be placed in any of four spots.  Then there are $9$ choices for the next spot, $8$ choices for the third spot, and $7$ choices for the fourth spot (because they all have to be unique).

$\binom{4}{1}9*8*7=2016$[/hide][/quote]
Wouldn't complementary counting be easier for d?
[hide]No 9s gives $9^{4}=6561$ numbers. So our answer is 10000-6561=3439[/hide][/quote]


calculus
integration
algebra
polynomial
partial fractions
Galois Theory
complex numbers

I posted this integral with some others, but I think I underestimated it and it is not solved.
Find: $ \int \frac{3x^{4}\plus{}12x^{3}\plus{}6}{x^{5}\plus{}5x^{4}\plus{}16x\plus{}12} \,dx$

I haven't tried to evaluate it, but is it possible that it can be factored out?  Maybe method of partial fractions or integration by parts?  Just suggestions.

It is a very "ugly" integral. Only good real root of the denominator is -1. The other are two reals and two complex. Other suggestions?

$ x^5\plus{}5x^4\plus{}16x\plus{}12\equal{}(x\plus{}1)(x^4\plus{}4x^3\minus{}4x^2\plus{}4x\plus{}12)$ and the $ x^4\plus{}4x^3\minus{}4x^2\plus{}4x\plus{}12$-Galois group is $ S_4$ ie. there don't exist any algebraic relation between the 4 roots of this last polynomial (2 conjugate complex numbers and 2 real numbers).
The seeked antiderivative is in the following real form (without the absolute values in "log"): $ \minus{}3ln(x\plus{}1)\plus{}Aln(x^2\plus{}\alpha{x}\plus{}\beta)\plus{}Batn((2x\plus{}\alpha)/\sqrt{4\beta\minus{}\alpha^2})\plus{}Cln(x\plus{}\lambda)\plus{}Dln(x\plus{}\mu)$. There don't exist any simplification: $ A,B,C,D$ are distinct non zero real numbers and moreover $ \alpha,\beta$ are algebraic real numbers of degree 6 and aren't algebraically conjugate.
Conclusion: this antiderivative should look like an ectoplasm taken out from a Romero's film.

Thank you for your valuable opinion. I think you are right. It is very interesting integral. I found it not solved in sosmath.com. What is denoted with greek letters in your answer? How did you get this conclusion? Is there some special instrumentarium?

$ x^4\plus{}4x^3\minus{}4x^2\plus{}4x\minus{}12\equal{}(x^2\plus{}\alpha{x}\plus{}\beta)(x\plus{}\lambda)(x\plus{}\mu)$; we can write $ A,B,C,D,\alpha,\beta,\lambda,\mu$ with radicals (over $ \mathbb{Q}$) but it's a long way, the formulas are complicate and it isn't very interesting.

It is really interesting how stupid people can be when proposing it on a site and how stupid may be people like me proposing it twice :-).

I proposed this integral because I found it on sosmath.com. I have three similar integrals that use some "tricks" (not defined coefficients - 1 and algebraic manipulations and exotic substitution - 2) and integration software don't have good behaviour with them. Unfortunately I cannot find the "trick" with this integral.

Anyway it is a good new to understand that not all integrals may have good answers. Thank you again for the time spent with it.


ok...thanks to all i've managed to grab something out of inequalities now...now i face a couple of problems...

1)   (5x-6)/(x+6) < 1

a. -6 < x < 6
b. -6 < x < 0
c. -6 < x < 3
d. None of the Above

I solve it to get

x<3, x> -6 (or) x > 3, x< -6

This is answer choice c... but the given solution is d. none of the above....Why? How?

2) What is meant by the largest integral x that satisfies an inequality??? is it the largest value of x possilbe or the maximum value possible for the inequality?

3) How do you guys post sqr roots and other symbols in your post??? is there a tool for that???

1) I think that the original choice $C$ was:
$-6\leq{x}\leq3$,
because the choice $C$ you posted here works.
2)"Largest integral $x$" means "largest whole number"

3) To post square roots, go like this: \sqrt{x}, and surround that thing with dollar signs.

To make your math pretty, click on the "LaTeX" link at left.  The forums are equipped with a LaTeX renderer, so once you know how to format it, you just slap it in, between dollar signs.

The "largest integral something" is the largest something which is also an integer.

As for why you're wrong:  a shallow reason is that $\frac{5(-7)-6}{-7+6}= \frac{-41}{-1}= 41 > 1$.

That should be a hint about what your error was.

Here's another hint:  Let $x =-1$.  Then $\frac{1}{x}=-1 < 1$.  Does it follow that $1 < x$?  Why not?

[quote="fizsle"]Joel,

Still not clear.... -41/-1 > 1 yes...but the inequality says...<1 so this would be incorrect. again x cannot be equal to -6 cos that would make the denominator 0...???? Let me rephrase my question....what is the solution? and how?[/quote]

You're absolutely right, that part of my post was totally off the mark -- I'll take another look at it tomorrow.  Sorry!


algebra
system of equations
algebra unsolved

solve system of equations for $ x,y,z,t \in R$:

$  \{ \begin {array} {c} (x+y+z)^2=12t \\ (y+z+t)^2=12x \\ (z+t+x)^2=12y \\ (t+x+y)^2=12z \end {array}$

[quote="thanhnam2902"]solve system of equations for $ x,y,z,t \in R$:

$  \{\begin {array} {c} (x + y + z)^2 = 12t (1)\\
(y + z + t)^2 = 12x (2)\\
(z + t + x)^2 = 12y (3)\\
(t + x + y)^2 = 12z (4)\end {array}$[/quote]

I think this problem is quite simple
From these equations we have $ x,y,z,t \ge 0$. Assume that $ x \ge y$. From (2) and (3) we get
$ (y + z + t)^2 = 12x > 12y = (z + t + x)^2$, which implies $ y \ge x$, so $ x = y$. Similarly, we have $ x = y = z = t$. These equations become $ (3x)^2 = 12x$, which has two roots: $ x = 0$ and $ x = \frac{4}{3}$

So the system of equations has two solutions: $ x = y = z = t = 0$ and $ x = y = z = t = \frac{4}{3}$./.


I'm familiar with embedding images in a LaTeX document, but how does one go about drawing a lattice?  Am I to just use Photoshop and do it by hand?  Or is there software/LaTeX code that will produce an image like this for me

[img]http://upload.wikimedia.org/wikipedia/commons/thumb/b/b2/Lattice_of_partitions_of_an_order_4_set.svg/289px-Lattice_of_partitions_of_an_order_4_set.svg.png[/img]
(Taken from Wikipedia)

You'll need a drawing package. There is a discussion of these at [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=53217]Ultimate guide to inserting Pictures in LaTeX[/url]. If anyone has any experience of drawing the type of diagram above then do please post here.


linear algebra
matrix
floor function
advanced fields
advanced fields unsolved

Two players own n dollars each, and between them, an object of value $ c > 0$.  Each player makes a sealed bid, offering $ i$ dollars (where $ i$ is one of the integers from $ 0$ to $ n$) for the sole possession of the object.  The higher bidder gets the object and pays the other player whatever amount he had offered.  If both bid the same amount, the object is assigned, without a compensating side payment, to one of the players by tossing a coin, so that each in this case has an expected share of $ (1/2) c$ in the object.  Write off the payoff matrix and determine whether it has a saddle point.

The question basically asks if there exists a best strategy for either player.

It's easy, I know.  But it's fun :)

If the first player bids $ j$ and the second player bids $ k,$ then the expected payoff for the first player is $ c\minus{}j$ for $ k<j,$ $ c/2$ for $ k\equal{}j,$ and $ k\minus{}c$ for $ k>j.$ So that's the payoff matrix.

Fix $ j$ and look across that row looking for the minimum. That minimum would be $ \min\{c\minus{}j,c/2,j\plus{}1\minus{}c\}.$

At this point, let me assume that $ 1\le c\le n.$ (Other assumptions lead to messy computations.)

From drawing the graphs, we see that we can maximize $ f(j)\equal{}\min\{c\minus{}j,c/2,j\plus{}1\minus{}c\}$ by letting $ j\equal{}\lfloor c\rfloor,$ and that that maximum value would be $ \{c\},$ where we define $ \{x\}$ to be the distance to the nearest integer: $ \{x\}\equal{}\min(x\minus{}\lfloor x\rfloor,\lfloor x\rfloor\plus{}1\minus{}x).$

So each player bids $ \lfloor c\rfloor$ as an optimum strategy. But when they do so, the actual payoff each receives is the much more advantageous $ c/2.$

The cases $ c>n,0<c<1,$ and $ c\le0$ will require separate analysis.


trigonometry
calculus
limit
function
derivative
real analysis
real analysis unsolved

Let $ f(x)\equal{}x^2 \sin(\frac{1}{x})$ for $ x \not \equal{} 0$ and $ f(0)\equal{}0.$

a) Prove that $ f(x)$ is differentiable on $ \mathbb{R}$.
b) Prove that $ f'(x)$ is not continuous on $ \mathbb{R}$.

I need help on BOTH parts. I've tried this problem, but I think I am not being very explicit in my proofs. I am also very sure my proof in b) is not nice. Is there another way to do b) [possibly without using the fundamental theorem of calculus]?

Attempt:
a) Clearly $ f(x)$ is differentiable at $ x \not \equal{} 0$. [How do I show this?] In fact, $ f'(x)\equal{}2x \sin(x) \minus{} \cos(\frac{1}{x})$ [valid for $ x \not \equal{}0$]. At $ x\equal{}0$, $ f'(0)\equal{}\lim_{x \rightarrow 0} \frac{f(h)\minus{}f(0)}{h}\equal{}\lim_{h \rightarrow 0} h \sin(\frac{1}{h})\equal{}0$. Hence, $ f(x)$ is differentiable on $ \mathbb{R}$.

b) $ f'(x)\equal{}2x \sin(x) \minus{} \cos(\frac{1}{x})$. Let $ p(x)\equal{}2x \sin(x)$. Then $ p(0)\equal{}0$ because $ p(0)\equal{}2 \cdot \lim_{x \rightarrow 0} x \sin(\frac{1}{x})\equal{}0$. So $ p(0)\equal{}0$. So by FTC $ \exists P(x)$ s.t. $ P'(x)\equal{}p(x)$. So $ f'(x)\equal{}P'(x)\minus{}cos(\frac{1}{x})$. But then $ \cos(\frac{1}{x})$ is not a continuous function. But by FTC, $ P(x)$ is differentiable. Then $ P(x)$, $ f(x)$ are differentiable functions whose derivatives are not continuous.

Thanks.

I can follow those, and they look right. For (a) the standard differentiation rules that let you write down that formula also include a proof that the derivative exists.
For (b), we have that $ f'$ is the sum of the continuous function $ 2x\sin x$ and the discontinuous function $ \begin{cases}\minus{}\cos(\frac1x)&x\neq 0\\0&x\equal{}0\end{cases}$.


Euler
integration
limit
logarithms
real analysis
real analysis solved


ceiling function
induction
number theory unsolved
number theory

In integer sequence{$a_n$},$a_1=2$,$a_2=7$,and $-\frac{1}{2}<a_{n+1}-\frac{{a_n}^2}{a_{n-1}}\leq \frac{1}{2}$.
Prove:If $n\geq2$,$a_n$ is an odd number.

we have :

$-1/2 + \frac{a_n^2}{a_{n-1}} < a_{n+1} =< 1/2 + \frac{a_n^2}{a_{n-1}}$

But there is exactly one integer in $[-1/2 + \frac{a_n^2}{a_{n-1}} , 1/2 + \frac{a_n^2}{a_{n-1}}]$ , then :

$a_{n+1} = \displaystyle \left\lceil\ 1/2 + \frac{a_n^2}{a_{n-1}}\right\rceil = \displaystyle \left\lceil\ 1 + \frac{2a_n^2}{a_{n-1}}\right\rceil - \displaystyle \left\lceil\ \frac{a_n^2}{a_{n-1}}\right\rceil $ because $\displaystyle \left\lceil\ 2x \right\rceil = \displaystyle \left\lceil\ x \right\rceil + \displaystyle \left\lceil\ x+1/2 \right\rceil $

Finally : $a_{n+1} = 1 + \displaystyle \left\lceil\ \frac{2a_n^2}{a_{n-1}}\right\rceil - \displaystyle \left\lceil\ \frac{a_n^2}{a_{n-1}}\right\rceil $

Now, you must prove that $\displaystyle \left\lceil\ \frac{2a_n^2}{a_{n-1}}\right\rceil - \displaystyle \left\lceil\ \frac{a_n^2}{a_{n-1}}\right\rceil$ is even.

But I don't see how to do....

It seems to me that $a_{n+1}a_{n-1}-a_n^2=(-2)^{n-2}$.

If you can prove this, it will certainly establish the result.

And can you prove it with a complete solution?

Here's the outline of a proof.

Define $b_n$ by $b_1 = 2$, $b_2 = 7$, and $b_n = 3 b_{n-1} + 2 b_{n-2}$ for $n \ge 2$.  Clearly the $b_n$ are odd for all $n \ge 2$.  We will show that $a_n = b_n$ for all $n$.  The idea is to first verify that $b_n$ satisfies the identity that Diarmuid noticed, $b_{n+1} b_{n-1} - b_n^2 = (-2)^{n-2}$.  Then we can show that $a_n = b_n$ by induction on $n$.


symmetry calculus integration function Euler vector

[img]http://imgs.xkcd.com/comics/substitute.png[/img]

Credits: [url]http://xkcd.com/c135.html[/url]

Disregarding the hilarity, the 1st and 2nd math problems are actually pretty good. Problem one is fairly simple to solve, but what about problem number 2? Can anyone come up with a solution?

The following assumptions can be made:
The human will run radially outward in a straight line.
The human has infinite acceleration to 6 m/s.
The velociraptors can turn any angle at any velocity.
The velociraptors begin at rest and accelerate to their maximum speeds at 4 m/s/s.

Also, what would the solution be if the first assumption was ignored (meaning that you can take any path)?

I had some thoughts on this, but I'm not sure that they are correct and I'm wondering if there is a rigid solution to this. Thanks.

It doesn't seem very easy to me, since the raptors will all be running along different tractrixes.

cant this problem be solved using symmetry,instead of calculus?

first shift the top velociraptor upwards to account for the speed difference and give all three the same speed.now adjust the direction of the human in the north-east/north-west direction so as to to ensure that the top velociraptor and the one at the bottom get it simaltenously.
this is just a guess. if anyone knows the relevant equations pls
post them so that we can do the computation part and get rigorous method

I don't think that this can be solved without calculus. The fact that you are maximizing functionals (and integral of a function), rather than a function, makes it a variation calculus problem.

[quote="rishabh"]first shift the top velociraptor upwards to account for the speed difference[/quote]

Well, you don't know the time.  This also wouldn't work if your path was anything except straight towards or straight away from the top velociraptor, neither of which could be optimal.

How do the raptors move? Do they adjust their bearing instantaneously to run towards you at all times?

I'm gonna feel special and find a calculus solution to #1.

This equation is only valid if it has a solution at $t < 6.25$:

$40+\int6 \text{ dt}= \int4t \text{ dt}$

$40+6t = 2t^{2}$

$t^{2}-3t-20 = 0$

$t = \frac{3+\sqrt{89}}{2}\text{ seconds}$, which is under 6.25 (think about it, $\sqrt{89}< \sqrt{90.25}$).

In this time, you can go $(6)(\frac{3+\sqrt{89}}{2})$ meters, or simplified, $\boxed{9+3\sqrt{89}\text{ meters}}$.

The second problem is interesting.

The each raptor lie in the vertex of the triangle $ ABC$.

The initial angle the person need to run, is that such as is in direction of one median of the triangle.

Let's proof this:

Suppose you are in the poind $ X$ between the vertex $ A$ and $ B$ and that you don't run. Each raptor has the same velocity $ v$.

$ a\leq x\leq b$ , the time that each raptor will reach you is $ t_1\equal{}\frac{x\minus{}a}{v}$ , $ t_2\equal{}\frac{b\minus{}x}{v}$

the first assumintion is. I want the time of a raptor will reach me is maximum.

If we plot the time, we will see that $ t_1 \equal{} t_2$ (if one time is much great, the other is much small (you can thing that as "their product need to be maximum"))

this led to $ x\equal{}\frac{a\plus{}b}{2}$

so. the path of the person need to be, from the center and to the median.

and after? Suppose the stupid person go to the left slight. this is the pure stupid thing of ever. it will only make the distance of the person to on raptor be greater and smaller of another raptor.

so the person need to run is a straight line that passes trought the median.

This is very simply. and now, the true dificult question, how much time this person will survive?

The equation of movement of the person, is

$ x(t)\equal{}\begin{cases}
 \frac{a t^2}{2} & 0\leq t\leq t_a \\
 \frac{1}{2} a \left(2 t\minus{}t_a\right) t_a & t>t_a
\end{cases}$

$ t_a$ is the time that the person spend accelerationg, ( in general, very small (and since $ t$ is not to big, after that, the velocity of the person is approximately constant))

And the Raptor, mister Euler, you ask me, people, I will answer.

The Raptor are truely interesting reptiles. If you don't move, they go straight to you, without fear! Is people, they are fearless. But sir, what if I move? Wow. whata a interesting question.

If you move from $ A$ to $ B$ thei change their direction, and the vector velocity will point to you.

So, Recaling the Fermat Principle "The distance from a Raptor to You, is such as this distance is minimum".

by symmetry, I will solve the problem for the Raptor $ A$ and you crossing $ AB$. The Raptor $ C$ are dunce and dont enter in the problem.

The problem is solve thus, using an old tool of naviagation. If you know your position and you velocity now, you know your position after $ \Delta t$ seconds. if this interval is small, we will be vary happy.

Reader, Note the different of the navigation principle and the heisenberg principe (If you know your velocity, you are lost)

Create the vectors $ me_t\equal{}\{x(t),0\}$ , $ Raptor_t\equal{}\{X(t),Y(t)\}$ (Note that how fearfull those animals are, they have Big letters naming their functions)

$ R_{t\plus{}\Delta t}\equal{} R_t \plus{} \frac{Me_t \minus{} R_t}{\left| Me_t\minus{}R_t \right| } V t$

$ M_t\equal{}\{t v,0\}$

That perfectly explain the nature of the Raptors. The vector pointing to you is on the equations above, can you see that?.

But mister Euler, How can I solve this equation?! I will need a Calc Book for this!?

no my friend, is impossible to solve. but we can eat a banana


Hello,

There is on the top right the possibility to "View your posts", which is great (and by the way, this forum is very nice).

Once you click on that, it would great is we could select a number of posts (or all) and then have them all on a page (maybe in PDF).

Did anyone maintaining the forum think about that?
Thank you.

ZetaX,
One of the use is to get all the posts you started/contributed to (or just the one you select) in one click, and then archive these for your own use. Ideally in $ \text{\LaTeX}$ source code and PDF.
Then you can reuse for other purposes.

[quote="Valentin Vornicu"][quote="JC_math"]Then you can reuse for other purposes.[/quote]Such as? ...[/quote]
Such as teaching for instance. Not everyone carries a computer with an internet connection, or just printing and archiving your thouhgts/problems,... for those maybe "old fashion" who still like paper...

[quote="ZetaX"]Normally, most posts are completely useless for themselfes. They either depend on other posts or are just small informations. Thus extracting them just causes a lot of chaos.[/quote]
This is correct and not what I meant.
I didn't detail or explain enough. So allow me to detail more.

When you click on "View your posts" on the top right corner, you can see all the titles where you contributed. Then if you click on on of those titles you see the entire thread (not just what you wrote).
So the idea/suggestion is to keep the section "View your posts", when you are there, you could select some titles (or all), and then press a button that would say "Export", and choose the format ($ \text{\LaTeX}$, PDF,...).

HTH


LaTeX
number theory unsolved
number theory

[quote="ramakrishna"]find all primes p for which the quotient {(2^p-1)-1}/p  is a square[/quote]
$ LaTex$,OK
The result is $ p\equal{}2,p \equal{} 3: p \equal{} 7$

[quote="ramakrishna"]find all primes p for which the quotient {(2^p-1)-1}/p  is a square[/quote]
We want to find solutions to $ \frac {2^{p \minus{} 1} \minus{} 1}{p} \equal{} k^2$, with $ k,p \in \mathbb{N}_0$ and $ p$ a prime.

When we solve it, we want to get some ideas:
$ 2^{p \minus{} 1} \equiv 1 \bmod p$ for all primes $ p$
Maybe we could use Zsigmondy in some way?
We notice that $ p \minus{} 1$ is not a prime for $ p > 3$.
This factorising could be useful: $ (2^{\frac {p \minus{} 1}{2}} \minus{} 1)(2^{\frac {p \minus{} 1}{2}} \plus{} 1) \equal{} p \cdot k^2$
We could maybe use that $ 1 \plus{} 2 \plus{} 2^2 \plus{} ... \plus{} 2^{p \minus{} 2} \equal{} p \cdot k^2$.

The factorising seems as the best solution, so I'll go with that.
Since $ (2^{\frac {p \minus{} 1}{2}} \minus{} 1, 2^{\frac {p \minus{} 1}{2}} \plus{} 1) \equal{} 1$ we can split it into two cases:

case $ 2^{\frac {p \minus{} 1}{2}} \minus{} 1 \equal{} pa^2$, $ 2^{\frac {p \minus{} 1}{2}} \plus{} 1 \equal{} b^2$. Then we have:
$ 2^{\frac {p \minus{} 1}{2}} \equal{} (b \minus{} 1)(b \plus{} 1)$ (from the second equation) which gives $ b \equal{} 3$, and $ p \equal{} 7$. We see that $ p \equal{} 7$ is a solution.

case $ 2^{\frac {p \minus{} 1}{2}} \minus{} 1 \equal{} a^2$, $ 2^{\frac {p \minus{} 1}{2}} \plus{} 1 \equal{} pb^2$
Notice that $ p \equal{} 2$ and $ p \equal{} 3$ is a solution, and from here on assume $ p > 3$.
We notice that $ 2^{\frac {p \minus{} 1}{2}} \equal{} a^2 \plus{} 1$ But we know that $ a^2 \plus{} 1 \equiv 2 \bmod 4$, and so $ p \le 3$.

Thus all solutions is $ p \in \{2,3,7\}$


a)Prove the each identity:

1- sin(A-B)/cosAcosB + sin(B-C)/cosBcosC + sin(C-A)/cosCcosA=0

2- tanA + cotA = 2csc2A

b)Solve for all possible values of x:

1-sin2x-1= cos2x

[quote="erma12879"]Prove the each identity:

1- sin(A-B)/cosAcosB + sin(B-C)/cosBcosC + sin(C-A)/cosCcosA=0

2- tanA + cotA = 2csc2A

Solve for all possible values of x:

sin2x-1= cos2x[/quote]

Time for $\LaTeX$ converter!

1. $\frac{\sin (A-B)}{\cos A \cos B}+\frac{\sin (B-C)}{\cos B \cos C}+\frac{\sin(C-A)}{\cos C \cos A}=0$

whew that was a long one.

2. $\tan A+\cot A = 2 \csc 2A$

hmm hint for number one? convert the denominator to the same ($\cos A \cos B \cos C$) i haven't done much of trig  so im not sure if that will work

[hide]1) If we expand the top:

$0=\frac{\sin A\cos B-\sin B \cos A}{\cos A\cos B}+\frac{\sin B\cos C-\sin C\cos B}{\cos B\cos C}+\frac{\sin C\cos A-\sin A\cos C}{\cos C\cos A}$

By distributive this reduces to:

$0= \frac{\sin A}{\cos A}-\frac{\sin B}{\cos B}+\frac{\sin B}{\cos B}-\frac{\sin C}{\cos C}+\frac{\sin C}{\cos C}-\frac{\sin A}{\cos A}$

In this step it is clear how everything on the right reduces to $0$

2)
$\tan A+\cot A=\frac{2}{\sin (2A)}$

By working with the left side:

$\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A}= \frac{\sin^{2}A+\cos^{2}A}{\cos A \sin A}= \frac{1}{\cos A \sin A}=$

If we multiply the top and bottom by 2:

$\frac{2}{2\cos A \sin A}= \frac{2}{\sin (2A)}=$[/hide]

[hide="2"]
Starting with the left side, let $A = \theta$
$\tan{\theta}+\cot{\theta}$
$= \frac{\tan^{2}{\theta}+1}{\tan{\theta}}$
$= \frac{\frac{\sin^{2}{\theta}}{\cos^{2}{\theta}}+1}{\frac{\sin{\theta}}{\cos{\theta}}}$
$= \frac{ \frac{sin^{2}{\theta}+cos^{2}{\theta}}{\cos^{2}{\theta}}}{\frac{\sin{\theta}}{\cos{\theta}}}$
by the pythagorean identity this reduces to
$\frac{1}{\sin{\theta}\cos{\theta}}$

Now our right side
$2\csc{2\theta}$
$= 2\left(\frac{1}{\sin{2\theta}}\right)$
By the double angle identity
$= 2\left(\frac{1}{2\sin{\theta}\cos{\theta}}\right)$
$= \frac{1}{\sin{\theta}\cos{\theta}}$

L.S. = R.S. And our identity is proven.
[/hide]

Man, number two confused me for a while. I thought it read "two MINUS...," and as a result, I couldn't find a way to get ride of that damn 2 - ...

[hide="My Solution, after understanding 2"]

$\tan{A}+\cot{A}= \frac{\sin{A}}{\cos{A}}+\frac{\cos{A}}{\sin{A}}$

$\frac{\sin{A}}{\cos{A}}+\frac{\cos{A}}{\sin{A}}= \frac{\sin{A}^{2}+\cos{A}^{2}}{\cos{A}\cdot \sin{A}}$

$\frac{\sin{A}^{2}+\cos{A}^{2}}{\cos{A}\sin{A}}= \frac{1}{\cos{A}\sin{A}}$

$\frac{1}{\cos{A}\sin{A}}= \frac{1}{\frac{1}{2}\cdot \sin{2A}}$

$\frac{1}{\frac{1}{2}\cdot \sin{2A}}= \frac{2}{\sin{2A}}$

$\frac{2}{\sin{2A}}= 2\csc{2A}$

Yay!

[/hide]

Thanks guys, I really appreciate the help. I was also wondering though if anyone is able to solve the third question?
(sin2x-1=cos2x) I'm having un peu de trouble with it..
again, thanks :)
-Emma

[hide="The last"] $\sin 2x-1 = \cos 2x \implies$
$\sin 2x-\cos 2x = 1 \implies$
$\sin^{2}2x-2 \sin 2x \cos 2x+\cos^{2}2x = 1 \implies$
$1-\sin 4x = 1 \implies$
$\sin 4x = 0$ [/hide]

[size=150][color=darkred][b]Torajirou[/b],  you made a grave error : $a=b$ is no equivalently with $a^{2}=b^{2}$ !!
For example, $x=-\frac{\pi}{4}$ is no a solution for the our equation ![/color][/size]

[b][u]Proof.[/u][/b] $\sin 2x=1+\cos 2x$ $\Longleftrightarrow$ $2\sin x\cos x=2\cos^{2}x$ $\Longleftrightarrow$ $\cos x=0$ or $\tan x=1$ $\Longleftrightarrow$ $\boxed{\ x\in \left(\pi Z+\frac{\pi}{2}\right)\cup\left(\pi Z+\frac{\pi}{4}\right)\ }$.

@Virgil: I think that he squared the equaton a=b to get a^2=b^2. If a=b, then a^2=b^2. If a^2=b^2, then a=+-b

Well, I didn't claim reversibility, which is why all of the arrows are one-way (that and I was lazy).  A complete solution would have gone on to check that the solutions to the last line were solutions of the original problem.

[hide="Alternately"] $\cos 2x-\sin 2x =-1 \Leftrightarrow$
$\sqrt{2}\cos \left( 2x+\frac{\pi}{4}\right) =-1 \Leftrightarrow$
$\cos \left( 2x+\frac{\pi}{4}\right) =-\frac{\sqrt{2}}{ 2}$ [/hide]

I didnt even notice the third lol

[hide]
$\sin{2x}-1 = \cos{2x}$
$\Rightarrow 2\sin{x}\cos{x}-1 = 2\cos^{2}{x}-1$
$\Rightarrow 2\sin{x}\cos{x}= 2\cos^{2}{x}$
$\Rightarrow \sin{x}\cos{x}= \cos^{2}{x}$
$\Rightarrow \sin{x}= \cos{x}$

Thus

$x = \boxed{45^{\circ}+180n^{\circ}}$ or $\boxed{\pi n+\frac{\pi}{4}}$
[/hide]


[quote="zaya_yc"]
$ \frac {36}{25}\leq a^{2} \plus{} b^{2}\leq\frac {49}{25}$-----> $ \phi$[/quote]

I don't understand something. Why can't we have for example $ a^2 \plus{} b^2 \equal{} \frac{43}{25}$ ?

can we take $ a\equal{}cosx$ and $ b\equal{}cosa$ ?


Took USAMO for the first time.....[b]AS AN INTERNATIONAL[/b]

I really hate time zones.....12:30~5:00 EST was 1:30 AM ~ 6:00 AM in South Korea!!! Urgh!!!  :mad:

It really took my brains out of me....

4/29 Morning: Took school mid-terms

4/29 Night: Took USAMO (Actually, it was 4/30 1:30AM ~ 6:00AM)

Slept 1 hour

4/30 Morning: Took mid-terms

4/30 Night: Took USAMO (5/1 1:30AM ~ 6:00 AM)

Slept 1 hour

5/1 Morning: Took mid-terms.....

After qualification, I really worried about taking both the USAMO and mid-terms at once.

Well, it's over... :P  And I think I got satisfying results in the USAMO.

[b]Flunked mid-terms, though  [/b]:mad:

It was a REALLY special experience, taking a 4-30 math test when everyone else was sleeping  :P

My math teacher almost killed me....hehe!!

Are there any other people who took USAMO as an international? I see more South Koreans and some Taiwanese.....

I'm Canadian on the west coast and I had a nice sleep till 8:50am before the exam :D

I qualified for the USAMO, but I didn't participate.

But... the AIME answer sheet state that if you're not an US citizen, then you can't participate in the USAMO, even fi you're eligible. Then, why now, USAMO is an International Olympiad?


the error says
[quote]Underfull \hbox (badness 10000) in paragraph at lines 122--123)[/quote]

it says that when I use \ begin {paragraph}, then 1 line of text, then a double slash " \ \", then a return and then more text. so, what is the proper way to get a carriage return in texnic center? thx.

Just ignore it.  There is nothing inherently bad about a bad box... when I generate a PDF with TeXnic center, I generally get at least 10 or so bad boxes (although it depends on what I'm typesetting).  A bad box is just TeXnic center's way of complaining when a line isn't completely filled with text (or when there's too much text to fit in the line, usually as a result of a really long equation, but that's another matter).  Underfull boxes are nothing to worry about.

If it bothers you, though, you can try the \linebreak command instead of \\ and see if that works.


1. There are 12 men on a basketball team, and in a game, 5 of them play at any one time. If the game is 1 hour long, and if each man plays exactly the same amount of time, how many minutes does each man play?
Actually I don't really understand the problem, if you can sort of explain it to me, that would help me to solve it.

2. If c carpenters can build a garage in d days, how many days will it take e carpenters, working at the same rate, to build 2 garages?
On my book, it says 1 carpenter will take cd days. Though it makes sense because it takes a person longer than more people, mathematically speaking, c-->d, then 1=c/c=d/c, which is division. Why then it's multiplication?

Thanks.

Since 5 people are in the game at all time, the total 1 hour on the court is shared by 5 people at all times, so the sum of all hours played by the 12 players is 5 hours. Divide 5 hours by 12 to get 25 minutes per person.


[quote="indybar"]Let point $K$ be on side $BC$ of $\triangle ABC$. The incircle of $\triangle ABC$ is tangent to side $BC$ at point $D$. Point $Q$ is the center of the incircle of $\triangle ABK$ and $R$ is the center of the incircle of $\triangle ACK$. Show that $\angle QDR$ measures 90 degrees.[/quote]

This is equivalent to proving that the point D lies on the circle with diameter QR. And this was shown in http://www.cut-the-knot.org/htdocs/dcforum/DCForumID6/415.shtml , first post, Theorem 1 or in [url=http://groups.yahoo.com/group/Hyacinthos/message/8985?expand=1]Hyacinthos message #8985[/url].

  Darij

Darij , congr/tions!

You have done great work!You saved me much time.Thanks.

 Babis


induction
number theory unsolved
number theory

Prove that : $\forall n \in N$ , there exists $n$ natural numbers satisfying : sum of unintended $n-1$ numbers are divisible by the rest number .

Well I think, in this problem should be $n>2$ cos' for $n=2$ we would have numbers $x_1, x_2$ such that $x_1|x_2$ and $x_2|x_1$ which would imply that $x_1=x_2$ and we aren't interested in this situation :)
Now for $n=3$ we can take for example numbers $2,4,6$ We will prove our statement by induction... assume we have $n$ numbers $x_1,x_2,...,x_n$ forwhich our statement holds. Let $s_n=x_1+x_2+...+x_3$ then we have for each $i$ \[ x_i|s_n-x_i \] so $x_i|s_n$ Now let's take $x_{n+1}=s_n$ then $s_{n+1}=2s_n$ and so for each $1\leq i \leq n$, $x_i|s_{n+1}$ and $x_{n+1}=s_n|s_{n+1}$ . So by induction our statement is true for every $n\geq 3$


modular arithmetic
combinatorics open
combinatorics

Hello,

I'm trying to prove a primality test using a cycle of length $q-1$ of the Digraph under $x^{2}-2$ modulo a Mersenne prime, rather than using the tree of the same Digraph as does the old LLT test.

One phase deals about proving that $\alpha^{n}+\beta^{n}\equiv 2 \pmod{M_{q}}$, where $M_{q}=2^{q}-1$ (and thus with q prime) and $q \equiv 1 \pmod{4}$, $n= (M_{q}-1)/2$, $\alpha+\beta = 2^{q-1}$, and $\alpha \times \beta = 1$.
I've checked it is true with many and huge values of $q$.

After several steps to simplify the formula, I got:
$\alpha^{n}+\beta^{n}\equiv F(q)$ where $F(q) = 1+\sum_{i=1}^{2^{q-2}-1}(-1)^{i+1}\frac{2^{2i-1}}{i}{ 2^{q-1}-2-i \choose i-1}\pmod{M_{q}}$ .

But now I fail to find a proof ...

I checked with PARI/gp that this nice formula still computes the same thing than the first one. It's OK.
If $q=2^{a}$ then $F(q) = 2^{a-2}$ .
If q prime and $M_{q}$ prime, then $F(q) \equiv 2 \pmod{M_{q}}$ when $q \equiv 1 \pmod{4}$ or $F(q) \equiv 2^{q-1}-1 \pmod{M_{q}}$ when $q \equiv 3 \pmod{4}$ .

I need a proof for: $F(q) \equiv 2 \pmod{M_{q}}$ when $q \equiv 1 \pmod{4}$ .

Any idea or suggestion ?

Regards,

Tony

[quote="T.Rex"]One phase deals about proving that $ \alpha^{n} \plus{} \beta^{n}\equiv 2 \pmod{M_{q}}$, where $ M_{q} \equal{} 2^{q} \minus{} 1$ (and thus with q prime) and $ q \equiv 1 \pmod{4}$, $ n \equal{} (M_{q} \minus{} 1)/2$, $ \alpha \plus{} \beta \equal{} 2^{q \minus{} 1}$, and $ \alpha \times \beta \equal{} 1$.[/quote]
This is wrong in general.
First, we can get explicit expressions:
$ \alpha \equal{} 2^{q \minus{} 2} \plus{} \sqrt {2^{q \minus{} 2} \minus{} 1}$ and $ \beta \equal{} 2^{q \minus{} 2} \minus{} \sqrt {2^{q \minus{} 2} \minus{} 1}$ (or vice versa).

Second, it is easy to obtain that Jacobi symbol $ \left(\frac {2^{q \minus{} 2} \minus{} 1}{M_q}\right) \equal{} 1$, implying that both $ \alpha$ and $ \beta$ modulo $ M_q$ can be viewed as elements of $ \mathbb{Z}_{M_q}$. Therefore, $ \alpha^{2n}\equiv \beta^{2n} \equiv 1 \pmod{M_q}$ and, thus, either $ \alpha^n\equiv \beta^n \equiv 1 \pmod{M_q}$ (if $ \alpha$ is a square modulo $ M_q$) or $ \alpha^n\equiv \beta^n \equiv \minus{} 1 \pmod{M_q}$ (if $ \alpha$ is a non-square modulo $ M_q$).

In the former case we have $ \alpha^n \plus{} \beta^n \equiv 2 \pmod{M_q}$ while in the latter case we have $ \alpha^n \plus{} \beta^n \equiv \minus{} 2 \pmod{M_q}$. So, the statement $ \alpha^{n} \plus{} \beta^{n}\equiv 2 \pmod{M_{q}}$ is true only if $ \alpha$ is a square modulo $ M_q$.

It turns out that under specified restrictions, $ \alpha$ can be both square and non-square modulo $ M_q$. In particular,
for $ q \equal{} 5$, $ \alpha$ is a square
for $ q \equal{} 13$, $ \alpha$ is a non-square
for $ q \equal{} 17$, $ \alpha$ is a non-square
for $ q \equal{} 61$, $ \alpha$ is a non-square
for $ q \equal{} 89$, $ \alpha$ is a non-square
for $ q \equal{} 521$, $ \alpha$ is a square
for $ q \equal{} 2281$, $ \alpha$ is a non-square
for $ q \equal{} 3217$, $ \alpha$ is a square
for $ q \equal{} 4253$, $ \alpha$ is a non-square
for $ q \equal{} 9689$, $ \alpha$ is a square
for $ q \equal{} 9941$, $ \alpha$ is a non-square
etc.


geometry
circumcircle
projective geometry
algebra
polynomial
power of a point
radical axis

Let $ABC$ be a triangle and $m$ a line which intersects the sides $AB$ and $AC$ at interior points $D$ and $F$, respectively, and intersects the line $BC$ at a point $E$ such that $C$ lies between $B$ and $E$. The parallel lines from the points $A$, $B$, $C$ to the line $m$ intersect the circumcircle of triangle $ABC$ at the points $A_1$, $B_1$  and $C_1$, respectively (apart from $A$, $B$, $C$). Prove that the lines $A_1E$ , $B_1F$  and  $C_1D$  pass through the same point.

[i]Greece[/i]

I'll use the folowing:

Theorem (Desargues): Given a triangle $ABC$ and a line $d$ . Denote the ponints of intersection of the line with the sides of the triangle with $A_1,B_1,C_1$, and $A_2,B_2,C_2$ other three points on $d$. We have that iff there is an involution between $A_1,B_1,C_1$ and  $A_2,B_2,C_2$ then $AA_2,BB_2,CC_2$ are concurent.

Now take $O'$ the projection of $O$ on $m$.Denote by $D',E',F'$ points of intersection between $m$ and sides of $A',B',C'$. Aplying the lemma for the triangle $ABC$ and the line $m$ , and the simetry w.r.t $O'$ as the involution.  The conclusion follows imediatly.

How many points would I receive for such a solution? (With details completed). I mean if the theorem is admitted as known or i need to prove it.

[quote="xirti"]If the line passes trough $O$, the intersecion point is on the circumcircle. This is the second Papelier Theorem.[/quote]

The intersection point always lies on the circumcircle!

Using this observation, I leave it to you to find a one-line solution of the problem without using involutions ;) .

By the way, the involution argument you brought up in post #2 is more or less the proof of the Blaikie theorem at http://www.mathlinks.ro/Forum/viewtopic.php?t=38178 , since, as you can easily see, the problem follows from the Blaikie theorem. If I were a grader, I would definitely give you 7 points for your solution, but I don't know how well projective geometry is established among the ones that are actually reading your work...

  Darij

In the contest, probably I would proved that th using another desargue th  :D : that used by grobber in the cited post. Thank you for the sinthetic proof of the Balkie th!

My picture also showed that the intesection point lies on the circumcircle, but I was not sure that my picture is accurate enough  :D

The only thing I can see is that the intersection point is one endpoint ,denote it by $T$ and the other endpoint by $S$, of the diameter perpendicular to $m$ since whole picture is simetric about this diameter.
In oreder for the intersection to lie on the circle, we must have $ET\perp SA_1$ but keeping $m,B,C$ fixed and varying $A$ we observe that $E,T,S$ remains fixed and $A_1$ varies on the circle, hence the angle $SA_1E$ varies,hence it is not $90^\circ$ everytime. Please correct me if I'm wrong.

[Mistake corrected]

Can someone give a sinthetic proof?

I am really confused... The intersection point is (generally) not an endpoint of the diameter of the circumcircle perpendicular to m. In fact, the figure is not symmetric with respect to this diameter; only the triangles ABC and $A_{1}B_{1}C_{1}$ are.

Here is the promised simple solution of the problem: We work in the projective plane, so parallel lines intersect at the line at infinity. Let the line $A_{1}E$ meet the circumcircle of triangle ABC at a point P (apart from $A_{1}$). Applying the Pascal theorem to the cyclic hexagon $C_{1}PA_{1}ABC$, we see that the points $C_{1}P\cap AB$, $PA_{1}\cap BC$ and $A_{1}A\cap CC_{1}$ are collinear. But both points $PA_{1}\cap BC$ and $A_{1}A\cap CC_{1}$ lie on the line m (in fact, the point $PA_{1}\cap BC$ is the point E and thus lies on m, and the point $A_{1}A\cap CC_{1}$ is the infinite point of the line m, since it is the point of intersection of the two lines $A_{1}A$ and $CC_{1}$ which are both parallel to m). Hence, the point $C_{1}P\cap AB$ lies on the line m, too. Thus, this point coincides with the point $AB\cap m=D$. Hence, we see that the point D lies on the line $C_{1}P$; in other words, the point P lies on the line $C_{1}D$. Similarly, the point P lies on the line $B_{1}F$. Altogether, we see that the lines $A_{1}E$, $B_{1}F$ and $C_{1}D$ indeed pass through the same point: the point P (and this point P lies on the circumcircle of triangle ABC).

  Darij

So the problem was the Papelier theorem  ?Can I find it in an encyclopedeia  or at a link  ??

BTW, Darij your solution is nice and quite elementary..... :D

[quote="darij grinberg"]Applying the Pascal theorem to the cyclic hexagon $C_1PA_1ABC$[/quote]
I have a figure where the hexagon is $C_1APA_1CB$, and Pascal doesn't really work(at least the form of Pascal I know). What is the exact form of the Pascal's theorem? Can you apply in my case, where $C_1P$ and $AB$ are diagonals in the hexagon, not opposite sides?

In my picture the notatations is the same as you Frozen so Darij made some typo or he has a different picture .

BTW , my solution-hint .

Let the line $B_1P$   cuts the circle at  $P$ .The the quadrilaterals  $ADFP$  and

$ABEP$  are cyclic . After this  we obtain that  the points  $P,D, C_1$ are in the same line and also  the points  $E,A_1, P$ are in the same line .

Quite elementary e?? :D

Nice solution, Silouan!

[quote="Frozen"][quote="darij grinberg"]Applying the Pascal theorem to the cyclic hexagon $C_1PA_1ABC$[/quote]
I have a figure where the hexagon is $C_1APA_1CB_1B$, and Pascal doesn't really work(at least the form of Pascal I know). What is the exact form of the Pascal's theorem? Can you apply in my case, where $C_1P$ and $AB$ are diagonals in the hexagon, not opposite sides?[/quote]

Don't understand you... I use this here: http://www.mathlinks.ro/Forum/viewtopic.php?t=14775 . I think it is old news nowadays that the hexagon needs not be convex for the Pascal theorem, so what exactly is unclear?

  darij

[quote="darij grinberg"]Nice solution, Silouan!
[/quote]

Thank you very much Darij . I think this is the magic of Euclidian geometry .

BTW ,I think   Frozen  was confused  by  the  convexity of the  Exagonon because if we take the notatations clockwise then the exagonon is $C_1APA_1CB$ .
But if you said that you take the non-convex exagonon , then it is all good I think .

If it can be concave and with self intersections, it is clear

The proof of Pascal's theorem uses nowhere that the hexagon has to be convex. The proof merely uses algebraic products which are from Menelaus' Theorem. As long as you are consistent with any of the possible 60 cyclic orders, the theorem still holds.

This is why Pascal is such a powerful theorem.

Darij: I was about to post my wonderful two line theorem from Pascal's only to find you beat me. :roll: Oh well, its still really nice, although silouan's pure Euclidean solution is also beautiful.

However this problem has projective geometry written all over it and I wouldn't be surprised if there was another solution using Desargues or Menelaus.

Ahhh what a mess?

1-) Mr.Darij : how you can say that $AA_1$ and $CC_1$ are parallel.

2-) Mr.Silouan : please can you write your solution more accurate. Everytime you post good, elegant solutions but you write them in the short form. Please when you write a solution write them like in the contest.

Thank you all very much !!!

Davron

[quote="Davron"]1-) Mr.Darij : how you can say that $AA_1$ and $CC_1$ are parallel.[/quote]

By reading the problem? ;) (I have just edited it to make it a bit more clear, but I think it was already quite clear before...)

By the way, whoever came up with the idea to denote the points of intersection of a line with BC, CA, AB by E, F, D, respectively, should know that there are much more original and efficient ways to confuse people...

  Darij

Darij: That was the confusing part?! I kept losing track of the $_1$ in my working and at one point was looking desperately at where $D_1$ was!  :rotfl:

Also if you see the three quadrilaterals formed by the three segments, you have just to prove that they are cyclic and then the radical axis of the the their circumcircles will intersect.

I want to know if the following solution has any chance of being valid:

WLOG assume that the circumcircle is the unit circle and that $\ell \| Ox$, its equation being $y = \lambda$ ($D,E,F \in \ell$). Now just compute everything in terms of coordinates. $A_{1}D, B_{1}E, C_{1}F$ are concurrent iff a determinant (the one formed by the equations of the three lines) is zero. That's easily seen to be a third degree polynomial, so if we prove the conclusion for four different values of $\lambda$ we are done. First, we take $\ell$ to pass through $A,B,C$ (all these points have different ordinates, so the values of $\lambda$ are distinct). Here comes the part I'm not sure about: Can we also take $\ell$ to be line at infinity (in this case proving the concurrency is very easy, the sum of the elements of every column is $0$)?

Pascal's theorem kills it!
Assume $A_1E$ intersects the circle at point $S$. We apply the Pascal's theorem with points $B_1SA_1ACB$. Let $F'$ the intersection of $AC$ and $B_1S$. Then, $EF'$ is parallel to lines $AA_1$ and $B_1B$. But, $EF$ is parallel to these lines, so $F'=F$. Hence, $B_1F$ passes through $S$. With similar way we prove that $C_1D$ passes through $S$.


parameterization function probability trigonometry calculus derivative

We know that if I drop a pebble into a pond waves are created. But this is not very obvious to me. I think it is very strange.

I mean can't we do some calculations and derive that a point on the water surface is on SHM, that is if we knew the parameters can't we get the equation of displacement of a point against time.

I am asking that if a boy was going to throw a pebble into the pond for the first time in the world(!!!) how would you predict what's he going to see on the water surface(That is, propagation of waves of course)?

[quote="Carcul"]Start by answering this question: What is a wave?[/quote]
[url=http://en.wikipedia.org/wiki/Wave]Wikipedia's definition[/url].
According to my ninth grade textbook(Translation by me): An wave, created by the cumulative vibration of an elastic medium's particles, is a disturbance which carries energy through an elastic medium without causing any net displacement to its particles.

None helps me directly to answer my question. I didn't bring wave in my question. I just want an explanation of the phenomenon observed on the pond surface when the boy throws the pebble into it. If you call it wave, you call it wave, if you define it in some way, then ok, I want to know why that happens.

[quote="username718"]as the rock is dropped into the water, it would have to displace some water, which would then displace the water immediately next to it, and which would displace the water immediately next to it, and so on, in an outward fashion.[/quote]
What you mean in detail:The pebble creates a force on the neighboring water, but lack of room compels it to gain height. Gravity tries to pull it down creating a force on the neighboring water.....

But why doesn't the waters try to attain equal height which seems more likely to me, instead of dropping low, pushing the neighboring water higher up?

If this question is answered, then why doesn't a single ripple get created, spreading outward, instead of the succession of ripples?

the wave is progressive--each point is in SHM-moving up & down and not the whole system

ok basically Carcul directed u  in the right direction but most books here give a more intutive definition of waves which spoils the broth
[b][color=red]BEWARE[/color][/b]
This definition is absolutely dangerous and won't be of any use to u when u start reading EM waves where u have no medium!!!
To give u an intutive fell as to why waves are created in water .when u drop a pebble or wave ur hand or for that matter dip a staick and give it a little swing what u do is basically disturb a particle on water which due to it's inherent want to stay with it's normal surroundings(or mathematically the elasticity) pulls the neighbouring particlaes with it and in doing so it is only partly sucessfull and raises a height greater thna it's neighbour which inturn pulls it's neighbour and so one which creates an disturbance in the medium
Although each particle moves in an SHM vertically(or here in specific  circular waves are created    so that  particles vibrate to and fro) but it seems to us that they move outward although it is the disturbance or specifically the wavefunction that does the moving

[b]THE ACTUAL MEANING[/b]
here is how u define a wave
Any Property  P(\vec{r},t) of a medium which changes with time and space and if to it we can relatee a function $ \Psi(\vec{r},t)$ which determines the 'value' of this property in both space and time then the property $ P$ is said to be a wave or rather create a wave with the [b]Wavefunction[/b] $ \Psi(\vec{r},t)$

The property can be absolutely anything for example in an atom it could be the probability of finding a n eletron which has a correpoding probability wave funtion
Or it could be the vertical diplacement of a particle in a string or in a water
it also could be the number of lions in my house :D

But the point is not all of them satisfy the so called wave equation

Thanks Pardesi.
Now I have a few question.
1. In your definition there is nothing about periodicity and the first example is not periodic. Does that mean that waves need not be periodic?
2. Your first example(I'm ignoring the last one :D ) does not change with time. But in your definition the property is related with time. Or is time replaced by the distance from the nucleolus here?
3. The last line you wrote..."But the point is not all of them satisfy the so called wave equation"... Do you mean that the equation with sines and things that represent the periodic sine wave by the "so called wave equation"? Do you answer the first question by this line?

[quote="Akashnil"]
1. In your definition there is nothing about periodicity and the first example is not periodic. Does that mean that waves need not be periodic?[/quote]
all waves need not be periodic for example if u have a simple pendulum moving thru air then the frictional resistance slows is down and stops it .We can call the angular displacement of the pendulum as a wave have wave function say $ \Theta(t)$ the angular displacement.
Practiacally u alwyas have damping so even though they aren't periodic they can be called waves with suitable wavefunctions
[quote="Akashnil"]2. Your first example(I'm ignoring the last one :D ) does not change with time. But in your definition the property is related with time. Or is time replaced by the distance from the nucleolus here?[/quote]
no it's actually time dependent for example consider an $ \ce{He}$ atom with two electrons in a single shell(or more specifically  if u know what orbitals are then let both of them be in the s orbitals.Claerly the probability of an electron being at a spot sufficiently close to the another at that time is very less but at some other time it might be very high so in genral it is function of space and time.
[quote="Akashnil"]
3. The last line you wrote..."But the point is not all of them satisfy the so called wave equation"... Do you mean that the equation with sines and things that represent the periodic sine wave by the "so called wave equation"? Do you answer the first two questions by this line?[/quote]
well the wave equation only holds for those waves which are repetative or rather non-dissapative.
That is if u were to plot their wave functions with time then after some time u wud be drawing what u had drawn some time ago that is they keep repeating [b]non-disappatively[/b]
For example a wave on an [b]infinite string[/b] with [b]finite[/b] elasticity

they always need not be simple sine and cos function but they shud satify
$ \nabla ^{2} \Psi \equal{} \frac {\delta^{2}\Psi}{v^{2}\delta t^{2}}$
for example cylindrical waves don't ahve any simple representaion only way to represent them is to use [url=http://en.wikipedia.org/wiki/Bessel_Functions]Bessel Functions[/url]

[u]Two more:[/u]
1.What is dissapativity?
2.$ \nabla ^{2} \Psi \equal{} \frac {\delta^{2}\Psi}{v^{2}\delta t^{2}}$
I realize it will be troublesome to explain this to me(I understand simple derivatives maybe, but the second ones are a bit confusing, and I don't know the meaning of $ \nabla$ at all), still I am informing you that I don't understand this, in case there is some way to tell me.

sorry i shud have realized that but anyways
1)dissapative means those process where energy is being wasted say a pendulum oscillating in water becomes slow due to the friction by water here the enrgy is being 'wasted'
2)well $ \nabla^{2}\Psi\equal{}\frac{\delta^{2}\Psi}{\delta x^{2}}\plus{}\frac{\delta^{2}\Psi}{\delta y^{2}}\plus{}\frac{\delta^{2}\Psi}{\delta z^{2}}$ where we refer to double partial derivatives

Partial derivatives mean that you are differntiating with respect to one parameter alone keeping all the other parameters fixed.

[hide]Par, you can't expect everybody to be like you, finishing calculus in 8th itself[/hide]

it's just an operator like derivative it's called laplacian
whose meaning i have written :)


advanced fields
advanced fields unsolved

$ G$ is a topological group and $ H$ is a closed subgroup.Prove the following:
i)if $ H$and $ \frac{G}{H}$ be compact spaces then $ G$ is compact.
ii)if $ H$ and $ \frac{G}{H}$ be connected spaces then $ G$ is connected.

lemma: if $ F$ is a closed and $ K$ a compact subset of a topological group, then $ F*K$ is closed.
proof: let $ f_i k_i$ be a net in $ F*K$, which converges to $ g \in G$. since $ K$ is compact, the $ k_i$ have a convergent subnet. so we may assume $ k_i \to k$ for some $ k \in K$. then $ f_i \to g k^{ \minus{} 1}$ shows $ g k^{ \minus{} 1} \in F$ since $ F$ is closed, i.e. $ g \in F K$.

i) let $ F_i \subseteq G$ closed with f.i.p. (finite intersection property), we want to show that $ \cap_i F_i$ is nonempty. the projection $ p : G \to G/H$ is closed for if $ F \subseteq G$ is closed, we have that $ p^{ \minus{} 1}(p(F)) \equal{} F H$ is closed by the lemma above. thus the $ p(F_i) \subseteq G/H$ are closed with f.i.p.. since $ G/H$ is compact, there is some $ p(g) \in \cap_i p(F_i)$. then the $ gH \cap F_i$ have f.i.p. (I hope ... can someone check that?). the result follows since $ gH$ is compact.

(ii) let $ G$ be the union of two disjoint open subsets $ U$ and $ V$. now $ G$ is partionned by the connected subsets $ gH$, $ g \in G$, which lie in $ U$ or $ V$. thus $ UH$ and $ VH$ are disjoint and their images in $ G/H$ are also disjoint, open and cover $ G/H$. since $ G/H$ is connected, $ p(UH)$ or $ p(VH)$, i.e. $ U$ or $ V$ is nonempty. so $ G$ is connected.


inequalities
logarithms
inequalities proposed

[quote="cezar lupu"]Silouan,I think this inequality has been posted before. ;)[/quote]

Really? Could you give me a link?

Probably take log and then say hi to jensen. :P
$\dfrac{\sum m \ln m}{n} > \dfrac{n+1}{2} \ln \dfrac{n+1}{2}$

$(x ln x)'' = (lnx + 1)' = \dfrac{1}{x} > 0$

[quote="silouan"]If $n$ is a positive integer odd number prove the following

$1\cdot 2^2\cdot 3^3......\cdot n^n >\sqrt{(\dfrac{n+1}{2})^{n(n+1)}}$[/quote]

It was posted by jianuovidiu, if i am not wrong!


i scored a 123 :( if you guys know wayys to get better please give me tips. also, can you guys give me hints or clues on how to attempt numbers 12, 14,15, 20 21 24 25? i was really frustrated looking at your guys' scores too :(

14 you can just use the area law: 1/2*ab*sinC, and the sin(180-a)=sin(a) identity.
21-25 were discussed here: http://www.artofproblemsolving.com/Community/AoPS_Y_MJ_Transcripts.php?mj_id=49
Number 20 try to think about how many times each digit appears in each decimal place.

You can buy complete solutions from the AMC website.


trigonometry calculus logarithms function algebra polynomial geometry

Is it possible to solve the equations below without using any sort of guessing technique (Like Newton's Method)?

sin x = x
cos x = x
tan x = x

Oh, ok, so there's no nice method for doing it. Thanks.

[quote="JBL"]Well, you can graph y = sin(x) and y = x and use a little bit of calculus to prove the answer there.  The others don't have nice solutions as far as I know.  In general, it's tough to mix trancendental (exponents, logarithms, trigonometric functions) with polynomial functions and get anything nice.[/quote]
Hmmm... what calculus would you use to do that? I can't think of anything from what I know.

First graph it and you'll see why I say that, I think -- you would have to show that the one (obvious) point is unique, and the most straight forward way to do that is to use a derivative.

Newton's Method is a technique that you can use to approximate nasty function values near nice function values, provided you have a funtion with a reasonable derivative.  Basically, you just pretend that over small areas, it is linear.  Then you do that repeatedly.

I see that the intersection is at (0,0) from the graph, and I can intuitively understand why, but I don't understand how you would prove that it is the only intersection point using derivatives...or how you would find an intersection point using derivatives.

Derivative of y=x  --> y' = 1
Derivative of y=sin(x)  ---> y' = cos(x)

And then.... ?

Well you know that cos x <= 1.. and so, to use informal language, since the gradient of sin is always less than 1, its sort of bending down away from the line y=x.. if you know what I mean..

Well, yes, I see that. It makes sense to me that there is one (and only one) intersection. I was just wondering if there is some sort of "trick" or technique using derivatives that has some bearing on an intersection between two functions.

(You know, like finding mins/maxs with the first derivative and inflection points with the second derivative.)

You know that it intersects at 0.  However, you want to show that it doesn't intersect anywhere else [i]near[/i] 0.  The most straightforward way to do that is to show that sin(x) < x for small positive x, which you do by showing the derivative of sin is less than the derivative of x.

As for tan(x) = x, there are infinitely many solutions and I don't believe that there is any sort of general form/expression for them, although I might be wrong about that.  At least, I've never heard of any such thing.

Here's one way to make the argument formal.  Let f(x) = x - sin x.  Then f(0) = 0.  Also f'(x) = 1 - cos x :ge: 0, with equality for only a countable set of x.  Thus for x > 0,
f(x) = f(0) + Integral0x f'(t) dt > f(0) = 0.
Similarly, for x < 0, we have f(x) < 0.  Thus f has only one zero (at x = 0), which means that x = sin x has only one solution (at x = 0).

[quote]By Newton's method do you mean Newton's sums?[/quote]

No, see Joel's explanation.


This is from the 1999 AHSME.  Problem # 17.
  Let P(x) be a polynomial such that when P(x) is divided by x-19, the remainder is 99, and when P(x) is divided by x-99, the remainder is 19.  What is the remainder when P(x) is divided by (x-19)(x-99)?

I tried doing  synthetic polynomial division, but the numbers got really complicated.    :?

[hide="Setup"]
Let $P(x)=(x-19)(x-99)Q(x)+R(x)$, where $R(x)$ is the remainder when $P(x)$ is divided by $(x-19)(x-99)$.
Since $(x-19)(x-99)$ is a quadratic, the remainder $R(x)$ must be linear (or constant).
Therefore, we can let $R(x)=ax+b$.
So $P(x)=(x-19)(x-99)Q(x)+ax+b$.
[/hide]
[hide="Solution"]
Let $P(x)=(x-19)(x-99)Q(x)+ax+b$.
By remainder theorem, we have $P(19)=99$ (because when $P(x)$ is divided by $x-19$, the remainder is $99$).
So $P(19)=(19-19)(19-99)Q(19)+19a+b \Longleftrightarrow 19a+b=99$ (Note that $(x-19)(x-99)Q(x)$ becomes $0$ when $x=19$.)
Similarly, we have $P(99)=99a+b=19$.

So we have a 2 by 2 system: $19a+b=99$ and $99a+b=19$
Solving the system, we get $(a,b)=(-1,118)$

Therefore, the remainder is $\boxed{-x+118}$

[/hide]

[hide]

Let $P(x)=(x-19)(x-99)Q(x)+ (ax+b)$ (the remainder is $ax+b$ because it is of first degree !)
Using $P(19)=99$ and $P(99)=19$ we find $19a+b=99$ and $99a+b=19$
We solve that system and we find $a=-1$ and $b=118$
The remainder is then $-x+118$:) [/hide]


group theory
abstract algebra
linear algebra
matrix
superior algebra
superior algebra unsolved

Maybe the title is not correct, but I did not know what to write.  Well, the problem that I have is this:

Find a subgroup $ S$ of some group $ G$ such that there exists $ g \in G$ such that

$ g^{ \minus{} 1}Sg \subset S$, but $ g^{ \minus{} 1}Sg \neq S$.

Thanks in advance.

Some observations that I made were:

(1) $ G$ cannot be finite; otherwise $ g^{ \minus{} 1}Sg$ would be finite and because $ g^{ \minus{} 1}Sg \simeq S$ and $ g^{ \minus{} 1}Sg \subset S$, then $ g^{ \minus{} 1}Sg \equal{} S$, which contradicts what we want.
(2) $ S$ cannot be normal.   Obvious
(3) $ G$ cannot be abelian.   Obvious

Take $ G\equal{}\mathbb{Q}\times \mathbb{Z}$ with operation $ (r,m)(s,n)\equal{}(r\plus{}2^{\minus{}m}s, m\plus{}n)$ for all $ r,s\in\mathbb{Q},n,m\in\mathbb{Z}$.
Then $ G$ is a group (a [url=http://en.wikipedia.org/wiki/Semi-direct_product]semi-direct product[/url]) with subgroup $ S\equal{}\mathbb{Z}\times 0$ and $ g\equal{}(0,1)$ satisfies $ g^{\minus{}1}Sg\equal{}(2\mathbb{Z})\times 0\subsetneq S$.

[quote="olorin"]Take $ G \equal{} \mathbb{Q}\times \mathbb{Z}$ with operation $ (r,m)(s,n) \equal{} (r \plus{} 2^{ \minus{} m}s, m \plus{} n)$ for all $ r,s\in\mathbb{Q},n,m\in\mathbb{Z}$.
Then $ G$ is a group (a [url=http://en.wikipedia.org/wiki/Semi-direct_product]semi-direct product[/url]) with subgroup $ S \equal{} \mathbb{Z}\times 0$ and $ g \equal{} (0,1)$ satisfies $ g^{ \minus{} 1}Sg \equal{} (2\mathbb{Z})\times 0\subsetneq S$.[/quote]

Thank you very much.    However, how could you know such a great example?  :)

I didn't know, but simply constructed it using the semi-direct product.
You can also take $ G: \equal{}GL_2(\mathbb{Q})$ and $ S: \equal{}\begin{pmatrix}1&\mathbb{Z}\\&1\end{pmatrix}\le G$ and $ g: \equal{}\begin{pmatrix}1&\\&2\end{pmatrix}\in G$.
Then $ g^{\minus{}1}Sg\equal{}\begin{pmatrix}1&2\mathbb{Z}\\&1\end{pmatrix}\subsetneq S$.
This is pretty much the same example.


I installed latexrender and set up the paths ..etc

BUT it doesn't render any .gif files.

When I hit the render button I get a dos window that appears and disappears very quickly, so that I can't read it.

<Q?> where does latexrender place its output files??

And what is the problem/fix?

TIA

my .bat file for latexrender

@echo off

:: -------------------------------------------------
:: Please set the correct paths on your computer to
:: latex.exe, convert.exe, dvips.exe
:: below.
:: Latex programs may not be happy with paths of filenames
:: that incude spaces. If you need to use spaces then
:: surround paths with quotes; for example
:: set latex="C:\Program Files\miktex\bin\latex.exe"

::set latex=C:\texmf\miktex\bin\latex.exe
::set latex=C:\Program Files\MiKTeX 2.5\miktex\bin\latex.exe
::set convert=C:\Program Files\ImageMagick-6.3.3-Q16\convert.exe
::set convert=C:\ImageMagick\convert.exe
::set dvips=C:\texmf\miktex\bin\dvips.exe

:: -------------------------------------------------

:: Don't change anything below this line unless you know what you are doing!

%latex% --interaction=batchmode temp.tex
%dvips% -E temp.dvi -o temp.ps
%convert% -density 120  -trim -transparent "#FFFFFF" temp.ps %1

set latex=C:\Program Files\MiKTeX 2.5\miktex\bin\latex.exe
set convert=C:\Program Files\ImageMagick-6.3.3-Q16\convert.exe
set dvips=C:\Program Files\MiKTeX 2.5\miktex\bin\dvips.exe

You have made a bit of a mess with the batch file.
1. Where it says [b]Don't change anything below this line[/b] you should not be changing anything
2. You must not add comments [b]::[/b] to lines that will be ignored
3. You must use quotes round paths with spaces in them.
Based on what you've written the complete file should look like this:
[code]@echo off

:: -------------------------------------------------
:: Please set the correct paths on your computer to
:: latex.exe, convert.exe, dvips.exe
:: below.
:: Latex programs may not be happy with paths of filenames
:: that incude spaces. If you need to use spaces then
:: surround paths with quotes; for example
:: set latex="C:\Program Files\miktex\bin\latex.exe"

set latex="C:\Program Files\MiKTeX 2.5\miktex\bin\latex.exe"
set convert="C:\Program Files\ImageMagick-6.3.3-Q16\convert.exe"
set dvips="C:\Program Files\MiKTeX 2.5\miktex\bin\dvips.exe"

:: -------------------------------------------------

:: Don't change anything below this line unless you know what you are doing!

%latex% --interaction=batchmode temp.tex
%dvips% -E temp.dvi -o temp.ps
%convert% -density 120  -trim -transparent "#FFFFFF" temp.ps %1

set latex=
set convert=
set dvips=[/code]
The final lines are supposed to be blank to clean up the settings once the batch file has been run.

figured it out   forgot "  in path spec.

THX

It wasn't just that. The set lines came [i]after[/i] the call to LaTeX etc so couldn't work.

THX

I cleaned up the .bat file

One more thing.....

how do you improve on the sharpness of the output.....

I'm pasting the images into a .word document and the anti-aliasing features are overly pronounced.

Any ideas.

It's amazing how blurry it looks in Word yet when I tried it in other word processors like OpenOffice or even Wordpad it looks fine!

Anyway, there is a workaround. You change the density of the image by changing in latexrender.bat the density from 120 to 150 so you have [code]%convert% -density 150 -trim -transparent "#FFFFFF" temp.ps %1[/code]This makes the image bigger so then you shrink it inside Word to 50% or less which makes it much sharper. You can experiment with these values to get what you find satisfactory or else change to another word processor!

THXs - I will experiment,  actually I'm making most of the images larger not smaller.  Using them for high-school handouts and overheads etc.

so what does the trim option do?

I find converting them to .jpg works on sharpening the images.

From ImageMagick's documentation:
[quote]-trim
trim an image.

This option removes any edges that are exactly the same color as the corner pixels.[/quote]


[quote="linj"]Today, I go check website to buy the Calculator. I see one insterest infomation under the Add google, it said I culd get a calculator for free. Is there a trick?
Could any one give me an idea about it?[/quote]
Sorry, "culd" is "could"

um, i doubt that this is legit, getting a 100-150 item sounds to good to be true, so it probably is.


Probability with Combinations Part 2

We solve many probability problems by counting the number of successful outcomes and by counting the number of possible outcomes.  In some of these problems, we can either take order into account when counting outcomes, or we can ignore order entirely.  In this lesson, we see why we have to be consistent--we must either account for order in both counts, or ignore order in both counts.

[hide="My solution"]
We see that there is $\binom{10}{3}$ for there to be the three people chosen, so $\binom{10}{3}\implies \frac{10*9*8}{3*2}\implies \frac{\cancel{10*9}*8}{\cancel{3*2}}\implies 5*3*8\implies 40*3\implies 120$ possible ways.
$\frac{\text{3 democrats}}{120}$

There is $\binom{5}{3}$ ways to chose the three democrats.
We are going to have:
$\frac{\binom{5}{3}}{\binom{10}{3}}$ probability that three of them are democrats.
$\frac{\binom{5}{3}}{\binom{10}{3}}\implies \frac{\frac{5*4*3}{3*2*1}}{\frac{10*9*8}{3*2*1}}\implies \frac{\frac{5*\cancel{4*3}}{\cancel{3*2}*1}}{\frac{\cancel{10*9}*8}{\cancel{3*2}*1}}\implies \frac{10}{120}=\boxed{\frac{1}{12}}$ [/hide]


real analysis
calculus
integration
function
limit
trigonometry
search

Can anyone give me a hint to do the following?

If $ f\in L^1(\mathbb{R})$  then $ \int_{ \minus{} \infty}^{\infty}f(x) cos kx dx$ exists and converges to $ 0$ as $ k\to \infty$

This is a typical problem where approximation technique is applicable.

Let $ \mathcal{R}$ be the set of integrable functions $ f$ satisfying $ \lim_{n\to\infty} \int_{\mathbb{R}} f(x) \cos n x \, dx \equal{} 0$.

First, show that $ \mathcal{R}$ is a closed subspace of $ L^1 (\mathbb{R})$. That is, show that

(1) if $ f_1, \cdots, f_k \in \mathcal{R}$, then $ c_1 f_1 \plus{} \cdots c_k f_k \in \mathcal{R}$.
(2) if $ f_k \in \mathcal{R}$ and $ f_k \to f$ in $ L^1$ norm, then $ f \in \mathcal{R}$.

Then show that every characteristic function of a bounded interval is contained in $ \mathcal{R}$.

Since the set of all such functions forms a dense subset of $ L^1 (\mathbb{R})$, it follows that $ L^1(\mathbb{R}) \equal{} \mathcal{R}$.

Key words: Riemann Lebesgue Lemma.

One technical note: it's not the characteristic functions of bounded intervals that form a dense subset of $ L^1$ but rather the set of all finite linear combinations of such sets, which is to say the integrable step functions. That is implicit in sos440's proof, but it might be clearer to mention it explicitly.

There are many variants of this proof that use different "nice" dense subspaces of $ L^1.$ The fact that $ L^1$ has "nice" dense subspaces is very closely linked to the fact that $ L^1$ is separable.

That keyword doesn't work very well for searches on this site; "Riemann-Lebesgue" searches for Riemann [b]or[/b] Lebesgue, resulting in enormous numbers of false positives. Leaving out the dash makes it an "and" search, which cuts the false positives down to 150 or so.


1) do dayereye  [u]t1 [/u] VA [u]t2[/u] DAR [u]m[/u] VA [u]n[/u] MOTEGHTEAND . FARZE KONID [u]ab[/u] khate momas bar in dayereha be tartib dar  [u]a[/u] VA [u]b[/u] BASHAD . BE TORIKE [u]m[/u] BE [u]ab[/u] NAZDIKTAR AZ [u]n[/u] BE IN KHATE BASHAD . VA HAMCENIN FARZE KOND [u]cd[/u] KHATE MOVAZI [u]ab[/u] va gozarandeh az [u]m[/u] BASHAD KE [u]c[/u] ROYE DAYEREYE [u]t1[/u] ast . KHOTOTE [u]ac[/u] VA [u]bd[/u] dar[u]e[/u] , KHOTOTE [u]an[/u] VA [u]cd[/u] dar [u]p[/u] VA KHOTOTE [u]bn[/u] VA [u]cd[/u] DAR [u]q[/u] moteghateand  . [b]neshan dahid[/b] :

                                               \[ \boxed{EP \equal{} EQ}
\]

2) ertefahaye vared az rashaye  [u]A , B[/u] VA [u]C[/u] az mosalase hadeolazaviyeye  [u]abc[/u] , AZLAE MOGHABEL RA DAR[u] d , e[/u] , VA [u]f[/u] ghat mikonand  . khati ke az  [u]d[/u] BE MOVAZATE [u]ef[/u] rasm mishavad  ,[u]ac[/u] VA [u]ab [/u] ra dar  [u]q [/u]VA [u]r[/u] ghate mikonad va khate  [u]EF[/u] , [u]BC[/u] RA DAR [u]P[/u] GHAT MIKONAD . SABET KONID DAYEREYE MOHITIYE [u]PQR[/u] AZ VASATE BC MIGOZARAD .

[quote="SoNYDvDPro"]2) ertefahaye vared az rashaye  [u]A , B[/u] VA [u]C[/u] az mosalase hadeolazaviyeye  [u]abc[/u] , AZLAE MOGHABEL RA DAR[u] d , e[/u] , VA [u]f[/u] ghat mikonand  . khati ke az  [u]d[/u] BE MOVAZATE [u]ef[/u] rasm mishavad  ,[u]ac[/u] VA [u]ab [/u] ra dar  [u]q [/u]VA [u]r[/u] ghate mikonad va khate  [u]EF[/u] , [u]BC[/u] RA DAR [u]P[/u] GHAT MIKONAD . SABET KONID DAYEREYE MOHITIYE [u]PQR[/u] AZ VASATE BC MIGOZARAD .[/quote]
noghte vasate $ BC$ ra $ M$ bnamid,bayad sabet konim 4zelie $ PRMQ$ mohatie ya be ebarati kafie sabet konim $ DM.DP\equal{}DR.DQ$...

khob midoonim 4zelie $ BQCR$ mohatie pas $ \boxed{DB.DC\equal{}DR.DQ} (I)$...

az tarafi midoonim $ (PD,BC)\minus{}1$ pas tebghe rabete nioton:

$ MD.MP\equal{}MB^2$

$ \Rightarrow MD(MD\plus{}DP)\equal{}MB^2$

$ \Rightarrow MB^2\minus{}MD^2\equal{}MD.DP$

$ \Rightarrow (MB\minus{}MD)(MB\plus{}MD)\equal{}MD.DP$

$ \Rightarrow MD.DP\equal{}BD(MC\plus{}MD)\equal{}BD.DC$

$ \Rightarrow \boxed{MD.DP\equal{}BD.DC} (II)$

khob hala az $ (I),(II)$ natije mishe:

$ DM.DP\equal{}DR.DQ$

yeni hamoon chizi ke mikhastim...

[quote="SoNYDvDPro"]1) do dayereye  [u]t1 [/u] VA [u]t2[/u] DAR [u]m[/u] VA [u]n[/u] MOTEGHTEAND . FARZE KONID [u]ab[/u] khate momas bar in dayereha be tartib dar  [u]a[/u] VA [u]b[/u] BASHAD . BE TORIKE [u]m[/u] BE [u]ab[/u] NAZDIKTAR AZ [u]n[/u] BE IN KHATE BASHAD . VA HAMCENIN FARZE KOND [u]cd[/u] KHATE MOVAZI [u]ab[/u] va gozarandeh az [u]m[/u] BASHAD KE [u]c[/u] ROYE DAYEREYE [u]t1[/u] ast . KHOTOTE [u]ac[/u] VA [u]bd[/u] dar[u]e[/u] , KHOTOTE [u]an[/u] VA [u]cd[/u] dar [u]p[/u] VA KHOTOTE [u]bn[/u] VA [u]cd[/u] DAR [u]q[/u] moteghateand  . [b]neshan dahid[/b] :
\[ \boxed{EP \equal{} EQ}
\]
[/quote]
khob farz konid $ MN$ khatte $ AB$ ro dar $ T$ ghat kone...

midoonim mehvar aslie do dayere momas moshtarako nesf mikone,pas $ T$ vasate $ AB$ hastesh...

az tarafi chon $ AB\parallel PQ$ pas do mosalase $ \triangle NPQ,\triangle NAB$ motajanesan be markaze $ N$,hala chon $ T$ vasate $ AB$ hastesh natije mishe $ M$ niz noghte vasate $ PQ$ hastesh...

khob az tarafi farz konid $ O_1,O_2$ markaze $ T_1,T_2$ bashan,khob midoonim $ O_1A\perp AB$ hamchenin $ O_2B\perp AB$ hamchenin ghablan dashtim $ CD\parallel AB$ pas $ O_1A\perp CD\Rightarrow O_1A\perp MC$ dar natije khatte $ O_1A$ amood monassefe parekhate $ MC$ mishe pas $ AM\equal{}AC$

be tarighe moshabeh sabet mishe $ BM\equal{}BD$

pas do mosalase $ \triangle MAC,\triangle MBD$ motasaviosaghein hastand pas:

$ \angle ACM\equal{}\angle AMC$ va $ \angle BDM\equal{}\angle BMD$

hamchenin chon $ AB\parallel CD$ natije mishe:

$ \angle BAM\equal{}\angle AMC\equal{}\angle ACM\equal{}\angle EAB$ pas:

$ \boxed{\angle EAB\equal{}\angle BAM} (I)$

be tarighe moshabeh sabet mishe:

$ \boxed{\angle EBA\equal{}\angle ABM} (II)$

khob ba tavvajoh be $ (I),(II)$ natije mishe ke ghotre $ AB$ az 4zelie $ EAMB$ nimsaze do rase $ A,B$ hastesh pas in 4zeli kayte pas $ EM\perp AB$...

khob hala chon $ AB\parallel PQ$ natije mishe $ EM\perp PQ$ amma avvale esbat didim ke $ M$ vasate $ PQ$ hastesh pas $ EM$ amood monassefe $ PQ$ ast yeni $ EP\equal{}EQ$...


search
abstract algebra
group theory
superior algebra
superior algebra theorems

I feel that books on Algebra are way too easy in excercises. And I feel that if an Algebra enthusiat wants to become excellent in this theory he should have a book with Problems and Solutions in Algebra. That starts off easy and gets more complicated. I tried searching all over the internet for Problem books in Group and Field theory but I cant find anything. Does anybody know of a problem book in Algebra?

try [url=http://www.mathlinks.ro/Forum/search.php?search_author=grobber]this[/url]

but... as far as I know, no such books exist. I'm planning to put my favorite algebra problems in .tex someday. I'll let you know when that is :)

Maybe I can be of help if you are interested in commutative algebra (rings, modules, tensor products): Have you tried out [url=http://modular.math.washington.edu/scans/252/Atiyah-MacDonald/index.html]M. F. Atiyah, I. G. Macdonald, [i]Introduction to Commutative Algebra[/i][/url]? It is a rather short book and the actual text is pretty basic, but the exercises are rather challenging. See http://www.math.uchicago.edu/~allanaa/atiyah.pdf for unofficial solutions.

Can anyone point me to something similar for group theory? I have difficulties finding good exercises on the non-abelian case which don't require heavy artillery like Sylow...

  Darij

@darij: sylow is heavy artillery?! *g*
@sylow: my favorite is
http://www.mathlinks.ro/Forum/index.php?f=8 ;-) and http://www.math.ksu.edu/~dbski/book.pdf

[quote="-oo-"]@darij: sylow is heavy artillery?! *g*[/quote]
Are you talking about me  :rotfl: . I guess I should change my username to avoid confusing.

I agree with -00- what is so hard about the Sylow Theorems? I can explain them to you if you like, if you have a good understanding of homomorphisms and factor groups the Sylow theorems are very very basic.

[quote]
@sylow: my favorite is
http://www.mathlinks.ro/Forum/index.php?f=8 ;-) and http://www.math.ksu.edu/~dbski/book.pdf[/quote]
Thanks man!

By the way do you specialize in algebra? Because I want to tell you have outsanding skills with field theory.

Another question)Do you guys feel that ordinary textbooks on algebra are easy? And are merely a basic understanding of definitions?

[quote="sylow_theory"]I agree with -00- what is so hard about the Sylow Theorems? I can explain them to you if you like, if you have a good understanding of homomorphisms and factor groups the Sylow theorems are very very basic.[/quote]

I know them... I just call them heavy because it would have taken me 5 years to prove them on my own. But it seems I never was in good relation with combinatorical arguments...

(Yes, I was and am talking about the Sylow theorems, not about MathLinks nicknames. ;) )

  Darij

[quote="darij grinberg"]
I know them... I just call them heavy because it would have taken me 5 years to prove them on my own. But it seems I never was in good relation with combinatorical arguments...
[/quote]
When I first learned them, my author used a little different approach, which made it much simpler. You maybe learned them the classical way through the Class equation which overcomplicates things. So maybe that why you hate them.  :wink:

Do you think it is a good idea for me to post the Sylow Theorems? Meaning do you think it would be beneficial or a waste of time? Because I would be happy to make a post that explains them.

Huh, a proof easier than the one using actions on subsets and the class equation? It would be definitely no waste of time to have it posted, thanks!!

  darij

Please recomend to me some books with [b]Excercises in representations of group theory[/b] which has solutions!


[quote="math92"]Suppose a number $x$ satisfies $x^2-4x-1=0$.  How much greater is $x$ than its reciprocal?[/quote]

[hide]
since $x^2-4x-1=0$ , we have $1-\frac{4}{x}-\frac{1}{x^2}=0$ . Adding up both we get $(x+\frac{1}{x})(x-\frac{1}{x}-4)=0$ .So $x-\frac{1}{x}=4$ :D [/hide]

[quote="anirudh"][quote="math92"]Suppose a number $x$ satisfies $x^2-4x-1=0$.  How much greater is $x$ than its reciprocal?[/quote]
how do you do these kinds of questions[/quote]
look at shyong's solution  :roll:


Express the following radical in the form $ \frac {a \plus{} b\sqrt {c}}{d}$, where $ a, b, c, d \in \mathbb Z$:
\[ \sqrt [8]{2207 \minus{} \frac {1}{2207 \minus{} \frac {1}{2207 \minus{} \frac {1}{2207 \minus{} \ldots}}}}
\]

[hide="Hint"]Let the radical be $ x$. How can you write the terms in the radical in terms of $ x$?[/hide]


A certain three-digit number $n$ is reduced in value to a number $m$ when the hundreds and units digits are interchanged.  Given that $450<n-m<550$, what is $n-m$?

[hide](100x+10y+z)-(100z+10y+x)=99x-99z. From this equation you can automatically realize that the middle digit is irrelevant. Since n-m has to be between 450 and 550, the difference between x and z has to be 5 or else it would be too small or big. x-z=5. 99(x-z)=n-m. 99(5)=n-m. 495=n-m  [/hide]

Easier method:

Once you get $99x-99z$, you know that $m-n$ is divisible by 99.  Since the only number x such that $450<x<550$ and $99\mid x$ is 495, $n-m=495$.

Samath, you have an easy solution.  It is a good one.


I recently came across this beautiful problem, from the January 2004 issue of Crux:

Three disjoint circles $A_1$, $A_2$, and $A_3$ are given in the plane, none being interior to any other.  The common internal tangents to $A_j$ and $A_k$ are $\alpha_{jk}$ and $\beta_{jk}$.  If the $\alpha_{jk}$ are concurrent, prove that the $\beta_{jk}$ are concurrent.

[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=703[/img]

(I know, the diagram isn't quite labelled correctly, but it's pretty.)

My question is, can someone provide a sketch of the solution in Crux?  I'm curious to see if it's different from mine.

Let t = the line tangent to the first circle and going through J, the intersection of c2, a2.  We have to show it's tangent to the third circle.

By considering equal tangents, we get some equations (*).  We assume t is not tangent to the 3rd circle, and let t' be the tangent to the third circle passing through J and is not a2, M' be the intersection of t' with b1.  From equal tangents, we get similar equations to (*), where some contradictions follow through the triangle inequality.  So t is tangent to the third circle.

Thanks, Alex.  I don't know the details with the tangents, but it looks sufficiently different from my solution.  Where does someone send a Crux solution these days?

You can email it to crux-editors@cms.math.ca


Express each of the integers $ 25, 69, 81$ and $ 125$ as a sum of distinct odd primes.

Thanks Thunder365.

I had:

$ 25\equal{}3\plus{}5\plus{}17\equal{}5\plus{}7\plus{}13$.

$ 69 \equal{} 3\plus{}5\plus{}61\equal{}3\plus{}7\plus{}59\equal{}3\plus{}13\plus{}53\equal{}3\plus{}19\plus{}47\equal{}3\plus{}23\plus{}43\equal{}3\plus{}29\plus{}37$

$ 81\equal{}3\plus{}5\plus{}73\equal{}3\plus{}7\plus{}71\equal{}3\plus{}11\plus{}67$.

$ 125\equal{}5\plus{}13\plus{}107$

etc, etc.

It seems that every integer $ n>9$ can be written like this.


function
limit
calculus
calculus computations

Determine the values of $a,b$ for which $f(x)=\lim_{n\to\infty} \frac{x^{2n-1}+ax^2+bx}{x^{2n}+1}$ is continuous at all real numbers $x$.

For $|x|>1$, $f(x) = \frac{1}{x}$
$f(1) = \frac{1+a+b}{2}$
and
$f(-1) = \frac{a-b-1}{2}$.

From the continuity, we see that $f(1) = 1$ and $f(-1) = -1$. We get $a=0, b=1$.

For $|x|<1$, $f(x) = ax^2+bx = x$. Clearly that $f$ is continuous.


Was anyone else annoyed that the teams winning both games won by the second trivia question?  I mean, as much as that benefited me individually, I'm not sure we deserved to win a thousand dollars because Joy guessed that Perelman declined the Fields Medal.  :|

In any case, congratulations to Sam Keller (samath (I think)) for the most anticlimatic ending to the Square-Off round possible. :)

And Macky guessed that Norway offers the Abel prize - which was the key question in the other heat. (I actually knew both Perelman and Norway, but I don't think that many of the high school kids follow the news that much.)

I was annoyed by the number of trivia questions (and the number of non-mathematical questions, such as number 1). It was also annoying that my team didn't get Perelman because that was the only difference between Oregon and Washington's scores (I was the individual for that question and I got it right, but the team guessed randomly).

I knew that Abel was Norwegian, but then, I like trivia.

I guessed Perelman because it was the only name that sounded familiar (other than Terrance Tao) and I remembered that someone mentioned a guy that declined the Fields medal at a lecture a few weeks back.

Anyways, I was bugged more by the two #1's that were completely non-math related than by the math trivia, although I think that the math trivia should have come in the first four questions and therefore have been worth fewer points.

(and Sam Keller is samath).

The square-off round was samath, t0rajir0u, JAM, and litoxe27182818. And Sam had the quickest answer.

I'm guessing Las Vegas had a "Who wants to be a millionaire?" competition.

But, yeah,  I was pretty confused at first as well.  It would've been nice to preface the discussion since we didn't have that competition at other sites ;)

Agreed; that could be confusing. Here's a [url=http://ams.org/wwtbam/]link[/url] to a description of the competition. It's a game show format, with its own gadgets and its own quirky rules. It was staged at ARML Las Vegas on Friday evening after the team and power rounds were over - a time slot that doesn't really exist at the other three sites.

ah, i agree with t0rajir0u. very anticlimactic, even though sam's denying it. too bad WA didn't get the bonus question. I guess -someone- had other stuff on his mind. >_>

I think badly under pressure whether or not I've got prom the next night  :oops:  Recall that I've never gotten more than a 6 on individual.

um your prom was the next night, not that night ..

I wrote down the running scores for the second heat. Here's how it progressed:

$ \begin{tabular}{ccccc}Ques.&SoCal&SD&DSW&Col\\ 2.&450&350&450&450\\ 3.& 450&500&600&750\\ 4.&1050&1100&1200&1350\\ 5.&1800&1350&1950&2100\\ 6.&2400&1950&1950&2100\\ 7.&3450&2650&2300&3150\\ 8.&4650&2750&3850&4350\end{tabular}$

I have little cryptic notes beside a few of the questions. For #6, the note says "Norway" - so that was the question about the Abel Prize (with Macky answering it for us). For #7, the note says "41" - so the question was about how many Pythagorean triples have 41 as one member. And finally, for #7, the note says "7th root". The question was how many 7th root of unity are of the form $ a\plus{}bi$ with $ |a|<|b|.$ Jeff answered that one - but so did the Colorado player. It was the Norway question that swung the heat.

The thing I didn't like about the trivia questions were that most of the people that got them guessed.  I don't particularly like trivia any time, but it's definitely not a good thing to have on multiple choice questions.

ARGH!!!!!!!!!!
Can't believe we lost to Washington on a guess...
It should have been Oregon v. Washington in the Square-off. And I bet I could have beat Sam  :lol:

Hehehehe...
After SFBA got SMOTE (smited, maybe?) in the first round, me and a whole bunch of other SFBA'ers went to dinner at In'N'Out.

Seems like we made a good call to skip the other half of that trivial, pointless competition.

It seems like a lot of people are bashing this competition.

I'll admit I'm a bit biased here, but I actually enjoyed it (except for the part where Macky tried to gouge my eye out  :P ).

I think people are trying to hold this thing up to the same standards at the "top-tier" of math competitions AMC/AIME/ARML etc.). The thing is, WWTBAM isn't really trying to be something like this. Their main goal is to be entertaining, not to be a rigorous assessment of your math ability. This means that their competitions are going to have a very different feel than the competitions we're used to.

But that doesn't mean it's a bad competition. It does what it's intended to do. It's fun. Both to participate in, and to watch. It also fills a kind of boring empty space in the schedule at Las Vegas (besides the team and power rounds, there's usually not that much to do on Friday besides just sitting around playing Mafia/Super Smash Brothers). So, maybe the people that win this aren't going to be the people that do best at ARML (which, I notice was the case here...  :|  what [i]happened[/i] to you guys, Washington?). So, just don't count on necessarily being able to win, just do you're best, and have fun. It's a good way to chill out after the team/power round. And everyone gets some prizes, no matter how well you do, which certainly can't be said for most other competitions.

Humph.
Georgia didn't have Song or WWTBAM...
I don't even think they had T-Shirt contest... (maybe I missed it)

Does anyone know if they're going to do WWTBAM next year?

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=207909]What Paul Dreyer says about that.[/url]


group theory
abstract algebra
superior algebra
superior algebra unsolved

Determine all finite groups which have a faithful real two-dimensional representation.
How to do it?
I don't have a clue :blush:
thanks for any help

You're looking for finite subgroups of $ GL_2(\mathbb{R})$.  A recent post here contains the answer, but you should try doing it yourself first.

Steve


Friendly numbers are two numbers such that the divisors of one add to the other, eg 220 and 284. Is there and sort of formula for these?

I suppose you could use the formula for a number that is its own "friend", an even perfect number.  This formula is (2p-1)(2p-1), where 2p-1 is a mersenne prime.  I'm not sure about a general rule for two distinct numbers though.  Hopefully someone else can help you.