calculus
integration
real analysis
real analysis unsolved

Hello Math professionals,

Actually I am a student in the 12 grade in germany, with the age of only 17
( => Your level of math is far beyond my imagination  :D )

While doing my homework ( don't laugh at me please  :blush: )
I found the following issue:

$ \int \!{x}^{4}{{\rm e}^{x}}{dx} \equal{} \left( {x}^{4} \minus{} 4\,{x}^{3} \plus{} 12\,{x}^{2} \minus{} 24\,x \plus{} 24 \right) {{\rm e}^{x}}$

Can you see the system?

Basically I tried to make a formula which now looks like this:
$ \int \!{x}^{n}t \left( x \right) {dx} \equal{} {x}^{n}\int \!t \left( x \right) {dx} \minus{} {{\it nx}}^{n \minus{} 1}\int \!\!\!\int \!t \left( x \right) {dx }\,{dx} \plus{} \left( n \minus{} 1 \right) {x}^{n \minus{} 2}\int \!\!\!\int \!\!\!\int \!t \left( x \right) {dx}\,{dx}\,{dx} \minus{} ...$

This seems to always be true, but of course I cannot proove it.
Now I ask you because if I tell my math teacher I solved it that way he will ask me for a proof  :)
and of course because i thought this can be interesting for you.

Thank you for your answers
Listing

hello
I m sorry but  :( R u not just using integration by parts .What to prove in this?
thank u

Yes, sorry my mistake. I didn't see it  :blush:


function
parameterization
parametric equation

The solution I saw was really elegant, but required very clever substitutions.  Surely, however, there are other solutions.

Find a general solution for
x^y=y^x

I have thought about this problem for a very long time. What do you mean by general solution. Do you mean of the form y=f(x)? The farthest I've gone to solving it is finding x^(1/x)=y^(1/y) and that 2,4 is a solution.

I believe Magnara is asking for a parametric solution.  That is, x = f(t) and y = g(t).  As I recall, there is one.

It seems my solution only works for x, y > 0.

[hide]Let y = xa, where a is a real number.

Then xxa = (xa)x

RS = xax

Since LS and RS have equal bases, their exponents must be equal as well; thus

xa = ax

xa - 1 = a

Thus x = a1 / (a - 1) and y = aa / (a - 1). I've tested it for a = 2 and 3 and it seems to work.[/hide]

Osiris: Your parametric equation does not include the case where x=y (and thus a=1), because then you would be dividing by 0 when you divide by a-1. It's a really neat solution though. I think that's the only missing case.

According to your solution,

[hide]

x = a1 / (a - 1) and y = aa / (a - 1)[/hide]

when x = y, a = 1 then it becomes undefined since you are dividing by 0?  Does it show there are infinite solutions when x=y?  (I know it's obvious that x=y works)

Edit: Oops, I must be slow in typing my reply then.

Ah yes, I forgot to take into account the trivial case.

Can you use Lambert's W function to express y in terms of x?
Does that count as a valid solution?

I meant a parametric solution.

I don't know what Lambert's W function is, but I saw a solution that gives an elementary form for x and y.

Here is a description of Lambert's W function:
[url]http://en.wikipedia.org/wiki/Lambert_W_function[/url]

I just learned about it from one of my friends.  I showed him this problem and he wrote the equation as
y = x...  (with a Lambert's W function)
He even gave me a graph of that function.
But I guess it's not really useful if you write it as a Lambert's W function, pretty ugly and hard to evaluate by hand.

The solution I saw involved substituting k*x=y.  Can anyone get it with that?

Ya, I tried the substitution kx=y a while ago, and you get EXACTLY the same thing as x^k=y (do you see why?), so I didnt bother to post it.

I thought about this problem for about eight hours yesterday when I was coming home from college. My best attempt was to try to find a function satisfying

[tex]\displaystyle f(x)^2=x+f(2x),[/tex]

but I couldn't think of any finite ones. Clearly it must have a [tex]\sqrt{x}[/tex] term, but then I couldn't find a way to make the smaller terms work out, so I was only able to find

[tex]\displaystyle f(x)=\sqrt{x+\sqrt{2x+\sqrt{4x+\sqrt{8x+\cdots}}}},[/tex]

which isn't very useful.

[quote="ComplexZeta"]I thought about this problem for about eight hours yesterday ...[/quote]

Wrong thread?

Hm you're right. Sleep deprivation is a bad thing...


A polyomino is a connected collection of unit squares where the unit squares create a continuous path from the first square to the last square.  Consecutive squares in the path must share a side.  The smallest polyomino that has one hole uses $ 7$ unit squares, as shown.  How many unit squares are in the smallest polyomino that has two holes? (A hole is a unit square that is not part of the polyomino but is surrounded on all sides by squares that are part of the polyomino.)

[asy]defaultpen(linewidth(0.8));
size(3cm,3cm);

fill(((0,0)--(3,0)--(3,1)--(0,1)--cycle),lightgray);

fill(((0,1)--(0,3)--(2,3)--(2,2)--(1,2)--(1,1)--cycle),lightgray);

fill(((2,1)--(2,2)--(3,2)--(3,2)--(3,1)--cycle),lightgray);

draw((0,0)--(3,0)--(3,1)--(0,1)--(0,0));

draw((1,0)--(1,1));

draw((2,0)--(2,1));

draw((0,1)--(0,3)--(2,3)--(2,2)--(1,2)--(1,1));

draw((0,2)--(1,2));

draw((1,3)--(1,2));

draw((2,1)--(2,2)--(3,2)--(3,2)--(3,1));[/asy]

You can start from the corner without a square, go 3 squares right, 2 squares down, and 1 squares left. So you add 6 to the 7 already there and get $ \boxed{13}$

OR
you can start from the top middle square
go one square up, two squares across, and one square down
now, we only add 4 to the original 7, to get an answer of $ \boxed{11}$

Your right. My answer isn't the smallest.

how do you show that 11 is the smallest answer, though?


inequalities
logarithms
number theory unsolved
number theory

$ p_k$ - k-th prime
consider the inequality $ p_{n \plus{} 1}\ln n > p_n\ln(n \plus{} 1)$
is it true for every $ n > 1$ ?
if not, is it true for sufficiently large $ n$ ?

Your inequality equavalent to
$ p_{n\plus{}1}\minus{}p_n>\frac{p_nln(1\plus{}\frac 1n)}{ln(n)}<\frac{p_n}{nln(n)}<2$. Therefore it is true for $ n>4$.


$\sum_{k=0}^{n}3^{k}* k * \binom{n}{k}$

wouldn't that mean the sum of the expression where k starts at 0 and goes up to n

such$3^{0}* 0* \binom{n}{0}+3^{1}* 1 * \binom{n}{1}....$

all the way until you get to "n" which is the last value of "k"
say n=5
$3^{5}* 5 * \binom{5}{5}$

Is that what you are asking?

so if n=5
$3^{0}* 0* \binom{5}{0}+3^{1}* 1 * \binom{5}{1}+3^{2}* 2 * \binom{5}{2}+3^{3}* 3 * \binom{5}{3}+3^{4}* 4* \binom{5}{4}+3^{5}* 5 * \binom{5}{5}$

so basically you're telling me that

$\sum_{k=0}^{n}3^{k}*k*\binom{n}{k}$ = $3^{0}*0 *\binom{n}{0}+...+3^{n}*n*\binom{n}{n}$

i edited to put my final result :lol:
oh and oops that is what i am saying have in mind i am human so i could have made a mistake

how did you just simplify it like that? :?:
i want to know...

[quote="Dr Sonnhard Graubner"]hello, it's
$\sum_{k=0}^{n}3^{k}\cdot k\cdot\binom{n}{k}=3\cdot n\cdot4^{n-1}$.
Sonnhard.[/quote]

Hi sonnhard

can you tell me how you got to that result pls ?

why did you write k=1 ? if k=0 in the first condition ?

[quote="funcia"]that one doesn't count because it is multiplied by $0$
but i am not sure how to simplify further...[/quote]

Hmm.. didn't OHO explain clearly how to simplify?

[quote="Lena"]Can you please explain to me what this means

$\sum_{k=0}^{n}3^{k}* k * \binom{n}{k}$[/quote]

Let us recall $(x+1)^{n}=\sum_{k=0}^{n}{n \choose k}x^{k}$.Lets modify this to get your result.
We have $\frac{d}{dx}(x+1)^{n}=n(x+1)^{n-1}$,but also
$\frac{d}{dx}(\sum_{k=0}^{n}{n \choose k}x^{k})=\sum_{k=0}^{n}{n \choose k}kx^{k-1}$ You can see now that your sum goes from $k=1$ ( as someone mentioned).Multiply it by x ,then we have ;
$xn(x+1)^{n-1}=\sum_{k=1}^{n}{n \choose k}kx^{k}$

Regards

I think I can explain, how did Dr Sonnhard got his result.

Let's have n lepperchauns. We want to give them hat's in 4 colours (white, red, green and blue) and also a bag of gold, to one of the lepperchauns in either red green or blue hat (we don't want to give the bag of gold to a lepperchauns in white hats). How can we do it? Well let's say we want at first to choose lepperchauns wchich we will give not-white hats, then give them hats, and then give the bag of gold to one of them. The not choosen lepperchauns will have white hats. So we do it. We choose k lepperchauns in $\binom{n}{k}$ ways, choose the one which we will give the bag of gold to in k ways, and then give the choosen lepperchauns hats in $3^{k}$ ways. As we should assume every possible value of k, we get by this way of counting:
$\sum 3 \cdot k\binom{n}{k}$.

The other way to count ,i s to firstly choose the lepperchaun which will get the bag of gold, then give him a hat in three ways (he cannot heve a white one!) and then give the other lepperchauns hats in $4^{n-1}$ ways.

So we have $\sum 3 \cdot k\binom{n}{k}= 3 \cdot n \cdot 4^{n-1}$.

I hope it is clear now. And sorry for my awful english  :P


function
geometric series
calculus
calculus computations

I have to find power series for the following functions and I'm a little confused about it, so any explanations would be greatly appreciated!

1) g(x) = 1/(1+x)
2) g(x) = x/(1-x)
3) g(x) = the square root of e^x

The first two do not even require any calculus.  Recall that the sum of a geometric series is given by

$a+ar+ar^{2}+... = \frac{a}{1-r}$

This gives us the power series

$f(x) = \frac{1}{1-x}= 1+x+x^{2}+...$

Which we can transform to obtain the following:

[b]1.[/b]  $f(-x) = \frac{1}{1+x}= 1-x+x^{2}-x^{3}+x^{4}+...$

[b]2.[/b]  $x f(x) = x+x^{2}+x^{3}+x^{4}+...$

As for the third,

$\sqrt{ e^{x}}= \left( e^{x}\right)^{ \frac{1}{2}}= e^{ \frac{x}{2}}$

Recall that $e^{x}= 1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+...$.  Then

[b]3.[/b]  $e^{ \frac{x}{2}}= 1+\frac{x}{2}+\frac{x^{2}}{2^{2}2!}+\frac{x^{3}}{2^{3}3!}+...$


i'm curious as to how one can find optimal solutions to the generalized fifteen problem. It kind of came up in myhead, and I'm really curious. I mean, how many moves will it take, how can we find them etc...?

[quote="Benjamin Hu"]I don't have any ways to compute the least number of moves it will take, but my strategy is to:
1. Solve most of it layer by layer until I get down to the last two rows.
2. Solve each successive "column" of two squares (from left to right) until I get down to the last three squares.
3. Shift the last three around and around until I get it.
I don't really know any other strategies other than solving it layer by layer until you get down to the last layer, then PLL. (permuting the last layer).[/quote]

Again, don't bring back old topics. This is the general method to solving a n by n cube. Solve the centers, pair the dedges and then solve like a 3x3. Parity will be same, I believe.

[quote="moogra"]This is the general method to solving a n by n cube.[/quote]  Are you aware that this thread has nothing to do with Rubik's cube?

And as I said in another thread, there is nothing bad with bringing back old topics!


algebra function domain LaTeX inequalities

y = x(x^2 - 1)(x^2 - 4)(x^2 - 9)/sqrt(x^2 - 16)

Please give the (real) domain in interval notation, please.

The only way for the denominator to be zero is $ x \equal{} \pm4$, so the domain is $ \boxed{x\in( \minus{} \infty, \minus{} 4)\cup( \minus{} 4,4)\cup(4, \plus{} \infty)}$.

Edit: OK...then it's just all real solutions to $ x^2>16\implies\boxed{x\in( \minus{} \infty, \minus{} 4)\cup(4, \plus{} \infty)}$.

I believe by "real" AyT meant "giving a real number as output", hence your interval is incorrect.

It's not defined when $ x^2 - 16\leq 0\implies x^2\leq 16\implies - 4\leq x\leq 4$.

So the domain is $ ( - \infty, - 4)\cup(4,\infty)$.

Edit: Beaten by math154  :mad: Shouldn't have spent so much time working out the $ \text{\LaTeX}$.

That's not the domain...those are the zeroes, roots, solutions, etc. of $ \frac{x(x^2\minus{}1)(x^2\minus{}4)(x^2\minus{}9)}{\sqrt{x^2\minus{}16}}\equal{}0$.

[quote="math154"]That's not the domain...those are the zeroes, roots, solutions, etc. of $ \frac {x(x^2 \minus{} 1)(x^2 \minus{} 4)(x^2 \minus{} 9)}{\sqrt {x^2 \minus{} 16}} \equal{} 0$.[/quote]
Wrong.  Zeroes are not listed as I gave them.

The domain is the union of all of the smallest x's to the largest x's,
which I showed in that hidden answer two (or more) posts back.

[quote="Arrange your tan"]The domain is the union of all of the smallest x's to the largest x's,
which I showed in that hidden answer two (or more) posts back.[/quote]

Where did you get this definition?

According to [url=http://www.artofproblemsolving.com/Wiki/index.php/Domain_(function)]this[/url], [url=http://en.wikipedia.org/wiki/Domain_(mathematics)]this[/url], and [url=http://mathworld.wolfram.com/Domain.html]this[/url], [b]the domain is the set of input values on which that function is defined[/b]. :wink:

I suppose that you could include $ 0,\pm1,\pm2,\pm3$ in the domain, but it's sort of iffy (i.e. depending upon what you want).

The question (as I have interpreted it) calls for a domain where the function is defined in $ \mathbb R$, so you wouldn't include 0, 1, etc in this case.

hello, for $ y(x)\equal{}\frac{x(x^2\minus{}1)(x^2\minus{}4)(x^2\minus{}9)}{\sqrt{x^2\minus{}16}}$ we have the condition $ x^2\minus{}16>0$, so the domain of definition is $ \minus{}\infty<x<\minus{}4 \cup 4<x<\plus{}\infty$.
Sonnhard.

This would be better if this was an inequality, I think it wouldn't cause much problem then.

Because all real values of x < -4 as well as all real values of x > 4
work fine to produce real values for y, then the domain must be
corrected to:

(-oo, -4) U {-3} U {-2} U {-1} U {0} U {1} U {2} U {3} U (4, oo)


limit
ratio
calculus
calculus computations

Determine which of the series diverges:

the sum of 1/square root of (n^2(n+2)) from n 1 to infinity
the sum of (((-1)^(n-1))*2^n)/n^2 from n 1 to infinity
the sum of e^-n from n 0 to infinity
the sum of (0,1)^n/n! from n 0 to infinity

what is the limit as n goes to infinity of n/((e*n)-e)

1. was probably meant to be $\sum_{n=1}^{\infty}\frac{1}{\sqrt{n^{2}(n+2)}}$. It converges by comparison with $\sum_{n=1}^{\infty}\frac{1}{n^{3/2}}$.

3. $\sum_{n=0}^{\infty}e^{-n}=\frac{1}{1-1/e}=\frac{e}{e-1}$.

4. I don't think that $(0,1)$ stands for $i$. The comma is probably a [url=http://en.wikipedia.org/wiki/Decimal_separator]decimal separator[/url]. The series converges by the Ratio Test.


x+2=5

(x+2)^2=25, 9x+18=45                                                  Square, multiply by 9
x^2+4x+4=25, 9x-2=25                                                  Expand, subtract 20
x^2+4x+4=9x-2                                                                   Set equal
x^2-5x+6=0                                                                     Subtract 9x-2
(x-2)(x-3)=0                                                                         Factor
x-2=0                                                                              Divide by x-3
x=2                                                                                Add 2
2+2=x+2=5                                                                     Subsitiute 2 for x

By using the transitive property in the last equation, 2+2=5.
[size=200]2+2=5??!!

[hide="The Big No-No"] Dividing by $ x\minus{}3$: From your first statement, $ x\equal{}3$ so you'd be dividing by 0, which is a no-no.[/hide]


OK- apart from the stuff we HAVE to bring to ARML, who wants to play Bughouse (like 2007 NYSML) or that-game-I-cannot-mention-sometimes-involving-POOs? Set and just plain playing cards, or mmaybe Risk is also good. Also, meet-up. Who wants to go to the Chinese restaurant or the pizzeria?

Off topic, but I love your thing about Groups, Rings, and Fields... Haha!

If all goes as planned, you'll see at ARML Penn State...

Oh, ha ha ha Matt.
BTW, did you know Matte means dull?  :lol:

I know it doesn't, the implication being that we're technically living at ARML for 2 or 3 days.

[quote="tan90=3/0"]chez toi does not mean what you said it means in french. i believe that it means "at your house."[/quote]
Hi,

I actually mean "with you".

-nbute


Is there any formula $ g*sin\alpha \equal{} a$?[/hide]

When a material point is moving on an inclined plane which is inclined in the angle $ \alpha$ to the horizon, the $ mg$ force can be separated in two vectors: one's intensity is $ mg \sin \alpha$ and the other's $ mg \cos \alpha$. The first one is the component of gravity pulling the body down, and the other is the "orthogonal" component (the one you use when calculating friction). If you take first one and divide it with mass, you get your formula for acceleration of a material point on inclined plane.

(my "physical" English is even worse than my "mathematical" English, so forgive me possible misinterpretation of some terms used here).

Thanks a lot  !

Your english is "udersandable"

Foreigners understand each other.   :)


If $y = 2x$ and $z = 2y$, then $x + y + z$ equals
\[ \text{(A)}\ x \qquad \text{(B)}\ 3x \qquad \text{(C)}\ 5x \qquad \text{(D)}\ 7x \qquad \text{(E)}\ 9x \]

[hide]
$x+2x+2(2x)=\framebox{7x}$ or $\framebox{D}$[/hide]

[hide="to show work to the people who don't get it"]y=2x and z=2y so substituting x in terms of y you get z = 4x so x + 2x + 4x = 7x  D[/hide]


Three friends have a total of 6 identical pencils, and each one has at
least one pencil. In how many ways can this happen?

If guy A has 4 pencils, then guys B & C can have the following: (1,1)
If guy A has 3 pencils, then guys B & C can have the following: (1,2) (2,1)
If guy A has 2 pencils, then guys B & C can have the following: (1,3) (2,2) (3,1)
If guy A has 1 pencil, then guys B & C can have the following: (1,4) (2,3) (3,2) (4,1)

add up the solutions, and the answer would come out to be [b]10[/b]

Notcie we give 3possibilities for the last 3 pencils (to A, to B, or to C)

so with repeated combinations forumla (i think thats what its called  :blush: )

We have $ {3\plus{}3\minus{}1}\choose{3}$=$ 5\choose{3}$=$ 10$

[quote="math154"]Wow mewto, I sent you a WHOLE EMAIL regarding this after states.

It's called Balls and Urns...

btw, how's my avatar.xD[/quote]

lol... is that a mewto?

i wasnt really good with memming my pokemon  :P

LOL if that is a mewtwo im a monkeys uncle.

Isn't that a character from this odd PBS show for little kids? I forget their name, it began with a "B" i think...

No, the new one looks like the Michelin Tires guy.

@mewto55555- You [b]ARE[/b] a monkey's uncle.  :rotfl:  :rotfl:  :rotfl:

ah-ha!  its a boobah!

ERNIE HAS SOLVED THE MYSTERY YET AGAIN!


modular arithmetic
induction
algebra
binomial theorem
number theory unsolved
number theory

1) Prove that if $ g$ is a primitive root mod $ p$ ($ p$ an odd prime) and $ g^{p\minus{}1}\not\equiv 1\pmod{p^{2}}$, then $ g$ is a primitive root mod $ p^{m}$.

I already know that if $ g$ is a primitive root mod $ p$ and $ g^{p\minus{}1}\not\equiv 1\pmod{p^{2}}$, then $ g^{(p\minus{}1)p^{m\minus{}2}}\not\equiv 1\pmod{p^{m}}$, and if $ g$ is a primitive root mod $ p$, then $ g$ belongs to the exponent $ h$ mod $ p^{m}$, where $ h\equal{}(p\minus{}1)p^{r}$ for some $ r$.  I'm not quite sure how to combine the two, though...

2) Prove that some odd numbers are primitive roots mod $ p^{m}$ for each odd prime $ p$ and each positive integer $ m$.

[hide="1"]
Induction of $ m$. Assume $ g$ is a generator for $ p^{m\minus{}1}$.

Since $ g^{(p\minus{}1)p^{m\minus{}2}}\not\equiv 1\mod p^{m}$ (as you stated) but $ g^{(p\minus{}1)p^{m\minus{}2}}\equiv 1\mod p^{m\minus{}1}$ and $ g$ is a generator mod $ p^{m\minus{}1}$. Then $ g^{(p\minus{}1)p^{m\minus{}2}}\equiv p^{m\minus{}1}r \plus{} 1$ for some $ r\ne 0$.

If $ g^k\equiv 1\mod p^m$, then $ k$ is divisible by $ (p\minus{}1)p^{m\minus{}2}$, so $ k\equal{}n(p\minus{}1)p^{m\minus{}2}$.

$ g^{n(p\minus{}1)p^{m\minus{}2}}\equal{}(g^{(p\minus{}1)p^{m\minus{}2}})^n\equiv (p^{m\minus{}1}r \plus{} 1)^n\equiv np^{m\minus{}1}r \plus{} 1$

(The last step is expanding with the binomial theorem and cancelling all terms divisible by $ p^{m}$). For $ np^{m\minus{}1}r \plus{} 1$ to be $ 1\mod p^m$, $ n$ must be divisible by $ p$, so the smallest exponent $ k$ is $ (p\minus{}1)p^{m\minus{}2}p\equal{}(p\minus{}1)p^{m\minus{}1}$.
[/hide]

[hide="2"]
If we prove that there is some generator $ g'$ such that $ g'^{p\minus{}1}\not\equiv 1\mod p^2$, which is the condition for part 1, then $ g'$ is a generator mod $ p^m$.

Let $ g$ be a generator mod p, and suppose $ g^{p\minus{}1}\equiv 1\mod p^2$ (Otherwise $ g'\equal{}g$ and we're done). Then expanding

$ (g\plus{}p)^{p\minus{}1}$ by the binomial theorem and cancelling terms divisible by $ p^2$ gives

$ (p\minus{}1)g^{p\minus{}2}p \plus{} g^{p\minus{}1}\equiv (p\minus{}1)g^{p\minus{}2}p \plus{} 1\not\equiv 1\mod p^2$.

Then we can say $ g'\equal{}g\plus{}p$, and we're done.
[/hide]

Thanks a lot.  However, what is a generator?  I have seen the term used but no one actually defined it.  However, I've noticed that when people use that term, they tend to not use "primitive root."  So are they the same thing?

In this context, yes.  "Generator" is more general and applies to any cyclic group.


Now it is better. But what you didn't understand ?

[quote="Arne"]I guess there is a typo in the denominator.[/quote]
yes it was a typo at the denominator, sorry. But now I corrected and the problem is correct

[quote="silouan"]Are you sure . Could you give me a counerexample ?[/quote]
Let $x_1\rightarrow 0$ and $x_2=...=x_n \rightarrow \infty$. The inequality becomes $36 \geq n$.

But we have $x_1x_2x_3...x_n=1$ . do you forget it ?

Vasc is correct. He is just taking the limit when $x_2, x_3, \cdots , x_n$ tend to infinity and $x_1=\frac{1}{x_2x_3\cdots x_n}$.

Because the first is not true, could you prove the following
[quote="silouan"]Let $n \geq 2$ and $x_1, \dots, x_n>0$ such that $x_1x_2....x_n=1$. Prove that \[ \sum_{i=1}^n\left(\frac{6}{x_i^2+4x_i+1}\right)^2\geq 2.  \][/quote]

Noone ?I think it is an interesting problem with a new idea.For a hint I will give the above. You can work as a harazi's problem. :lol:

Unsolved till now ?Why ?I think now it is correct .

now I think we should just prove it when:$n=2$(not hard)
for $n>2$ we can WLOG $x_1 x_2 \leq 1$
then it is true.

yes zhaobin nice job. Please post your solution,thanks


modular arithmetic
number theory unsolved
number theory

$2^{k}$ is the highest power of two that divides $ab + a - b - 1$.
We have that $a = \frac{2^{n} - 1}{b}$ $\Longrightarrow$ $\frac{2^{n}b - 2b + 2^{n} - 1 - b^2}{b} = \frac{(2^{n} - 1 - b)(b + 1)}{b} \equiv 0 \bmod 2^{k}$. Let $2^{h}$ be the highest power of two that divides $b + 1$. It is also the highest power of two that divides $2^{n} - 1 - b$. So $h < n$. And the highest power of two that divides $ab + a - b - 1$ is $2^{2h}$ cause $2^{h}$ comes from both factors. Then $2^{k} = 2^{2h}$.

If $ a \equal{} 1$ then $ ab \plus{} a \minus{} b \minus{} 1 \equal{} 0$ and there does not exist such a $ k$.
Let $ a \minus{} 1 \equal{} x$, $ b \plus{} 1 \equal{} y$, so $ x \ge 1$, $ y \ge 2$. Then $ k$ is the highest power of $ 2$ dividing $ xy$, and we have
$ 2^n \equal{} xy \minus{} x \plus{} y \equal{} x(y \minus{} 1) \plus{} y \ge x \plus{} y$.
Let $ 2^m$ be the highest power of $ 2$ dividing $ x$. Then since $ 2^n \ge x \plus{} y$, $ m < n$. Let $ x \equal{} 2^mr$, where $ r$ is odd. Substituting,
$ 2^n \equal{} 2^mry \minus{} 2^mr \plus{} y$. Thus, $ 2^m$ divides $ y$, so $ y \equal{} 2^mj$. Substituting,
$ 2^{n \minus{} m} \equal{} 2^mrj \minus{} r \plus{} j$
$ 2^{n \minus{} m} \plus{} r \equal{} j(2^m \plus{} 1)$
So the left side is odd since $ r$ is odd, but it also divisible by $ j$, so $ j$ is odd. So $ x$ and $ y$ both have $ m$ factors of $ 2$, so $ k \equal{} 2m$ is even.

easy to prove that $a,b$ are odd;and $ab+a-b-1=2^n+a-b-2=2^n+\frac{2^n-1}{b}-b-2=2^n+\frac{2^n}{b}-(b+1)^2$
if $2^n|a-b-2$,thern it's easy to prove $a=b+2$hence $2^n=(b+1)^2,2|n$
otherwice,$2^k||(b+1)^2$
hence $2|k$

[hide=Solution][quote=Balkan MO 2001 P1]Let $a,b,n$ be positive integers such that $2^n - 1 =ab$. Let $k \in \mathbb N$ such that $ab+a-b-1 \equiv 0 \pmod {2^k}$ and $ab+a-b-1 \neq 0 \pmod {2^{k+1}}$. Prove that $k$ is even.[/quote]
[b][color=#000]Solution:[/color][/b] We know, $v_2(ab+a-b-1)=k$. Clearly, $a,b$ are odd. Note, $ab+a-b-1=(a-1)(b+1)$, Let
$$\begin{cases}v_2(a-1)=x \\ v_2(b+1)=y \\ x+y=k \end{cases} \implies \begin{cases} a-1=k_12^x \implies a=k_12^x+1 \\ b+1=k_22^y \implies b=k_22^y-1 \end{cases}$$
So,
$$2^n-~1=ab \implies 2^n=1~+~(k_12^x+~1)(k_22^y-~1) = k_1k_22^{x+y}+~k_22^y-~k_12^x $$ $$\implies 2^n+k_12^x=k_22^y(k_12^x+1)$$
Now, maximum power of $2$ dividing $\text{ RHS }$ is $y$. If $x \geq n$ $\implies$ $y=n$, but then, we would have: $1+k_12^{x-n}=k_2(1+k_12^x)$, which is a clear contradiction! Hence, $n \geq x$ $\implies$ $x=y$ $\implies$ $2 \mid k$ $\qquad \blacksquare$
[/hide]

You can easily fing that $ v_2(a-1)$$=$$v_2(b+1)$


[hide]3992004[/hide]

This should be in the MathCounts section not here

[quote="hwenterprise"]All right, but how do you move it?[/quote]

You cant, but the mods might. Im not saying this problem is too hard for the Basics section, it just looks more like a mathcounts problem

okay, I get the answer, but not the formula. What's the formula?

[hide]I found the middle number of 1 and 1997 (1998/2)  then multiplied that by 1997.  That finds the average of all numbers in the parenthsies, then makes it all of the numbers.  Multiply that by 2 and add 1998 to get 3992004[/hide]

[quote="GoBraves"][quote="shinwoo"]DarkKnight, thanks for showing your works

the concept he used is adding backwards....

1 + 2 + 3
3 + 2 + 1
-----------
4 + 4 + 4

(4+4+4)/ 2

like this[/quote]

I think he used the sum of consecutive positive numbers formula-
n(n+1)/2

Unless thats what you mean by adding backwards[/quote]

Yes, I did use the n(n+1)/2 formula but it's like the same thing.

i always wondered how u would do this.  thank for showing me.

Thank you to all of you who showed your work...I appreciate it. It was a great help.

[hide]3992004. Simply apply the formula   1/2n(n+1)[/hide]


function
real analysis
real analysis unsolved

$\sum_{i=1}^{n}\ e^{^}(x^{i})$

Anyone know a closed form?

If you are looking for an elementary function $F(x,n)$ of two variables such that this sum is equal to $F(x,n)$ for all real $x$ and integer $n$, I believe there is none. If you insist, we can even try to prove it but that may take a lot of time and considerable effort unless someone has a bright idea of how it can be done right away.


let (xn) be a sequence of real numbers such that x0 and x1>0 and xn+2 = (xn+1+2)/(xn+2) for all n \geq 0. Proove that xn is convergent
[i]source i think ONM ROMANIA 1997[/i]

please edit your post and complete the problem's statement.

this is all what i have Valentin. The problem had smth else to?

well, if you look here, you can see that it can't be from ONM 2997

http://www.mathlinks.ro/viewtopic.php?t=2605

Can anyone tell me what is so important if the problem was  in ONM 1997 or not? Try to solve it!

I think I've found a solution, but I'm not sure:
    First of all, we have 2x_n+2<=x_n+1+2. Thus, 2^(k+2)x_k+2-2^k+1*x_k+1<=2^(k+2). Sum upp these and you will find that there exists k such that for all n>k we have x_n<3. Thus, for all n>k we have x_(n+2)>(2+x_(n+1))/5>2/5. So, for n>k+2 we have x_n>2/5. Then, for n>k+2 we have |x_(n+2)-1|=|x_(n+1)-x_n|/(2+x_(n+2)|<=(|x_(n+1)-1|+|x_n-1|)/(2+2/5). Take a_n=x_n-1. We have a_n+2<(a_n+1+a_n)/(2+2/5) for all n>k+2. Thus, since a_n is bounded and a_(n+2)<(a_n+a_(n+1))/2, a_n must be convergent. Let x be its limit. Since a_(n+2)<(a_n+a_(n+1))/(2+2/5), we must have x<=0. Since x>=0 ( clear), we must have x=0 and thus x_n tends to 1.


function
floor function
number theory
prime numbers
number theory theorems

$ \textbf{Conjecture: }$

If $ p, q, r$ are consecutive prime numbers such that $ p > q > r$ and if $ r$ is different from $ 2,3,5,7,11,13,19,23,31,47,83,89,113,199,1327$ then $ \left\lfloor\frac{pq}{r}\right\rfloor \equal{} p\plus{}q\minus{}r$.

Let $ p\equal{}r\plus{}a,q\equal{}r\plus{}b,a>b>0$, then $ [\frac{pq}{r}]\equal{}p\plus{}q\minus{}r$ equavalent to $ ab<r$.
If exist $ 0<c<1$, suth that $ p_{n\plus{}1}<p_{n}\plus{}c\sqrt{p_{n}}$, then it followed for $ p>N(c)$.


If ${ A=\{z}\in \mathbb{C}: \Re(z)>0\}$, and ${ f: A\rightarrow\mathbb{C}}$, with $ f(z)=\int_0^{+\infty} \!{e^{-zt^2}}\, dt$, prove that $ f$ is holomorphic.

Well, $ f'(z)\equal{}\int_0^\infty \minus{}t^2e^{\minus{}zt^2}\,dt$. That's an absolutely convergent integral, and everything works.

A variation on a standard theorem for use here:
Suppose $ f(z)\equal{}\int_I g(z,t)\,dt$, where $ g$ is differentiable in $ z$. If in addition $ |\frac{\partial g}{\partial z}(z,t)|\le h(t)$ for some integrable $ h$ on $ I$, $ f$ is differentiable and $ f'(z)\equal{}\int_I \frac{\partial g}{\partial z}(z,t)\,dt.$

We can't cover the whole half-plane at once with this, but differentiability is local. This method gives differentiability on each half-plane $ \text{Re}(z)>c$ for each $ c>0$, and taking the union gives the desired result.


calculus
derivative
trigonometry
function
calculus computations

At what points on the curve y - sin x + cos x, 0 </= x </= 2pi, is the tangent line horizontal.

I'm just wondering, for this question, would you take the derivative first and then get the critical numbers. And if so, how do you get critical numbers in these type of trig functions?[/code]

I'll assume you meant the curve $y = \sin x+\cos x,\, \, 0 \leq x \leq 2\pi$ and yes, do what you imagined to do, namely take the derivative w.r.t. x and set equal to zero, i.e. write $y^\prime = \cos x-\sin x =0$ so solve $\cos x =\sin x\Rightarrow \tan x =1$ or $x=\tan^{-1}(1)= \frac{\pi}{4}+k\pi,\, \, k=0,\pm 1, \pm 2,\ldots$, but we wanted $0 \leq x \leq 2\pi$ so we need only the solutions corresponding to $k=0,1$ i.e. $x=\frac{\pi}{4},\frac{5\pi}{4}$.


Determine el minimo valor de $(a+1)(b+1)(c+1)(d+1),$ dados $a,b,c,d \geq 0$ y

 $\frac{1}{a+1} + \frac{1}{b+1} + \frac{1}{c+1} + \frac{1}{d+1} = 1.$

(Sugerencia)

[hide="Hint"] Un inteligente cambio de variables y usar MA-MG[/hide]

Carlos Bravo :)
Lima -PERU

[hide]Sup. que $(a+1)(b+1)(c+1)(d+1) = x$
Expandiendo la suma de las fracciones tenemos que:
$\sum_{sym} {(a+1)(b+1)(c+1)} = x$
por $MA - MG$ se tiene que
$x\geq  4 \sqrt [4] {x^3}$ entonces se tiene que $x\geq 4^4$ elevando todo a la $4$ y dividendo entre $x^3$
La igualdad se da cuando $a=b=c=d=3$[/hide]


in ham yek hadde rahat vali ziba:
$\lim_{x \to \frac{\pi}{3}} \frac{\sqrt{\sin{(x)}}+\sqrt{\sin{(2x)}}}{\sin{(x)}+\sqrt{3}}$

[color=red]Latex edited by Mod.[/color]

[quote="ashwinrk_jain"]here is the answer[/quote]
tanx.(vali age mineveshti kheili behtar bood.) :D


Diophantine equation
number theory unsolved
number theory

prove that:
$a^2+b^2+c^2+d^2=abcd$ has infty solution in $\mathbb{N}$

I claim that there are already infinity many solutions with $c=d=2$:

to prove:
$a^2+b^2+8=4ab$ has infinity many solutions in integers.
But this can be written as $(a-2b)^2 - 3b^2=-4$ or after substitution $e^2-3b^2=-4$ which is a generalised Pell equation with trivial solution $e=4, b=2$, so it has infinity many, which was to be proven.


Εαν $ a,b,c \in \mathbb R^\plus{}$ με $ a\plus{}b\plus{}c\equal{}1$.Nδο:

$ \frac{1}{1\plus{}a}\plus{}\frac{1}{1\plus{}b}\plus{}\frac{1}{1\plus{}c}\plus{}6(ab\plus{}bc\plus{}ca) \le \frac{17}{4}$.

[hide="για τους γνωρίζοντες"]
Ρε παιδιά ο Pham Kim Hung στα mathlinks είναι ο hungkhtn????[/hide]

Σφιχτή (παρατηρείστε ότι πιάνουμε το μέγιστο στο κέντρο του τριγώνου), αλλά μάλλον εύκολη (βγαίνει με brute force βασικά):

[hide]Για να μη βάλω Lagrange, αρκεί να βρούμε το μέγιστο της συνάρτησης:
$ f(x,y) \equal{} \frac {1}{1 \plus{} x} \plus{} \frac {1}{1 \plus{} y} \plus{} \frac {1}{2 \minus{} x \minus{} y} \plus{} 6\left(x \plus{} y \plus{} xy \minus{} (x \plus{} y)^2\right)$
με $ 0\leq x,y \leq 1, x \plus{} y\leq 1$.

Με μία παραγώγιση και μία αφαίρεση, προκύπτει ότι τα στάσιμα σημεία που είναι στο εσωτερικό του π.ορισμού θα ικανοποιούν τη σχέση: $ \frac {1}{(1 \plus{} y)^2} \minus{} \frac {1}{(1 \plus{} x)^2} \equal{} 6(x \minus{} y)$ η οποία μπορεί να ισχύει μόνο όταν $ x \equal{} y$ (γιατί?). Έτσι, αρκεί να κοιτάξουμε τη συνάρτηση $ g(x) \equal{} f(x,x)$ η οποία και έχει μοναδικό στάσιμο σημείο στο $ x \equal{} 1/3$.

Από κριτήριο β' παραγώγου για $ x,y \equal{} \frac {1}{3}$, προκύπτει ότι το εν λόγω σημείο είναι μέγιστο (και δίνει τη ζητούμενη τιμή $ \frac {17}{4}$). Το μόνο που απομένει να ελέγξουμε είναι ότι στο σύνορο του π.ορισμού δεν εμφανίζεται μεγαλύτερη τιμή, πράγμα το οποίο είναι εύκολο.[/hide]

Cheerio,

Durandal 1707


Mr. Go has a weighted hexahedraon (cube) die. The number 1 is the least likely number to come up. The number 2 comes up twice as often as the number 1, the number 3 comes up three times as often as the number 1, and so on, up to the number 6 coming up 6 times as often as the number 1.

How often does the number 3 come up? Express your answer as a common fraction.

[hide]haha...

let the prob that 1 comes up be x.

x+2x+...+6x=1

x=1/21

3x=3/21=1/7[/hide]

you get $\frac{3}{1+2+3+4+5+6}$=probability of 3 coming up.  $\frac{3}{21}=\frac{1}{7}$ :idea:  our answer

Boy would Monopoly be fun! :D

Anyway, the probability of getting a 1 is $\frac{1}{x}$.  The probability of getting a 2 is $\frac{2}{x}$.  That means that the sum of the probabilities of all possible outcomes is $\frac{21}{x}=1$.  That means that x=21 and the answer is $\frac{3}{21}=\frac{1}{7}$.

[hide]roll a one, then roll two 2s, three 3s... how many dice rolls will u get?

1 + 2 + 3 + 4 + 5 + 6 = 21.

3 comes up thrice, so 3/21 = 1/7[/hide]


Find n to satisfy $3^{n}+63 =k^{2}$ (k $\in N$ )

[hide="solution"]For n=1mod2, that expression might just be a square.

For n=0mod2, it isn't a square since it's 2 mod 4.

The only n that worked(that I found) is 1.[/hide]

[hide]Works for $n = 4 \rightarrow 3^{4}+63 = 12^{2}$[/hide]

here is my solution
if n is odd , we easily see that $LHS \equiv 2(mod 4)$
so n is even , take $n=2m$
we get $63=(k-3^{m})(k+3^{m})$
notice $k-3^{m}\equiv k+3^{m}\equiv 0 (mod 3)$
so we obtain $k-3^{m}=3$ and $k+3^{m}=21$ , hence $m=2 ;n=4; k=12$

HTA:

Your second to last line does not hold if $m=0$
You lost $n=0$ as a second solution

oh , that's mistake , i didnt notice :D , thanks you

[hide]If n is odd, then $(-1)^{odd}+3\equiv 2\equiv k^{2}$ is untrue, no solutions.
If n is even, then $k=3k'$, and $3^{n-2}+7=k'^{2}\implies 7=k'^{2}-3^{n-2}\implies 7=(k'+3^{(n-2)/2})(k'-3^{(n-2)/2})$. Only solutions is $(n-2)/2=1, k'=4 \implies n=4, k=12$.

EDIT:  :oops: [quote="chickendude"]Your second to last line does not hold if $m=0$
You lost $n=0$ as a second solution[/quote]

$3^{n-2}+7=k'^{2}$ is not true iff n=0, so testing that seperately gives another solution, n=0, k=8.
[/hide]


Can someone please give me instructions on how to find limits of basic functions like (x^2-4)/(x+5) quickly(10 secs or so)?

(assuming you meant limit as x goes to infinity, otherwise theres not always the simplest rule:)
Divide top and bottom by the highest power of x you can find in that expression. Then when x goes to infinity, everything you don't care about cancels out to 0.

For this example, divide both the bottom and the top of (x 2 -4)/(x+5) by x 2 . Thus, you have (1-(4/x 2 ))/((1/x)+(5/x 2 ). The bottom approaches 0 as x goes to infinity, so anything over 0 is infinity, so the limit of this function as x goes to infinity is infinity.

"Highest powers" is too narrow a rule. To deal with limits quickly, what you really need to know is what grows (or shrinks) much faster than what else.

In Chinaboy's original problem, $\lim_{x\to\infty}\frac{x^2-4}{x+5}$, we see that $x^2$ gets much larger than 4 and $x$ gets much larger than 5, so we only need to look at $x^2$ and $x$. But since $x^2$ grows much faster than $x$, the limit is $\infty.$

But powers of $x$ and going to infinity aren't the only story. What about the following limits? You can do all of them quickly, if you have a clear understanding of what's bigger than what else.

1. $\lim_{x\to\infty}x^4e^{-x}.$

2. $\lim_{n\to\infty}\frac{n^4-4n^2-2^n}{n^3+n+2^n}.$

3. $\lim_{x\to0^+}\frac1{x^2}-\frac1x.$

4. $\lim_{n\to\infty}\frac{\log(n^2)}{\sqrt{n}}.$


a,b,c are the lenghts of triangle dimensions,
prove that:
                    3(ab+ac+bc)=<(a+b+c)²<4(ab+ac+bc)

[quote] $ a,b,c$ are sides of a triangle. Prove that $ 3(ab \plus{} bc \plus{} ca) \leq (a \plus{} b \plus{} c)^{2} < 4(ab \plus{} bc \plus{} ca)$[/quote]

[hide=" Solution"]
$ 3(ab \plus{} bc \plus{} ca) \leq (a \plus{} b \plus{} c)^{2} \iff a^{2} \plus{} b^{2} \plus{} c^{2} \geq ab \plus{} bc \plus{} ca$ which is true by Rearrangement Inequality.

$ (a \plus{} b \plus{} c)^{2} < 4(ab \plus{} bc \plus{} ca) \iff (a \plus{} b \minus{} c)(b \plus{} c \minus{} a) \plus{} (a \plus{} b \minus{} c)(c \plus{} a \minus{} b) \plus{} (b \plus{} c \minus{} a)(c \plus{} a \minus{} b) > 0$ which is true since $ a,b,c$ are sides of a triangle.  [/hide]

Other sollution :
$ (a\plus{}b\plus{}c)^2\,\geq\,3(ab\plus{}ac\plus{}bc)$

$ \iff \, (a\plus{}b\plus{}c)^2\minus{}3(ab\plus{}ac\plus{}bc)\,\geq\,0$
$ \iff \, \frac{1}{2}[(a\minus{}b)^2\plus{}(b\minus{}c)^2\plus{}(c\minus{}a)^2]\,\geq\,0$
wich is true.

and we have :
$ |a\minus{}b|\,\prec\,c$ and  $ |b\minus{}c|\,\prec\,a$   and   $ |c\minus{}a|\,\prec\,b$

so:

$ 4(ab\plus{}ac\plus{}bc)\minus{}(a\plus{}b\plus{}c)^2\equal{}c^2\minus{}(a\minus{}b)^2\plus{}a^2\minus{}(b\minus{}c)^2\plus{}b^2\minus{}(c\minus{}a)^2$

wish is also true.


[hide]Just subtract the parts away.
$ [BDF] \equal{} [ACE] \minus{} [BCD] \minus{} [DEF] \minus{} [ABF]$
$ [BDF] \equal{} \dfrac {(16)(12)}{2} \minus{}\dfrac{(9)(4)}{2} \minus{}  \dfrac {(5)(12)sin\theta}{2} \minus{} \dfrac {(3)(15)sin(90\minus{}\theta)}{2}$
$ [BDF] \equal{} 96 \minus{} 18 \minus{} 30sin\theta \minus{} \dfrac{45sin(90\minus{}\theta)}{2}$
$ [BDF] \equal{} 96 \minus{} 18 \minus{} \dfrac{(30)(3)}{5} \minus{} \dfrac{(45)(4)}{(5)(2)}$
$ [BDF] \equal{} 96 \minus{} 18 \minus{}18 \minus{} 18$
$ [BDF] \equal{} 42$
$ \dfrac {42}{96} \equal{} \dfrac {7}{16}$
[/hide]

Ooooooo that's good.

I was using the distance formula to find the length of each side of BDF and then using Heron's, but it took a long time. In the end I still came up with 7/16 though.

[hide="A Somewhat Longer Solution w/o Trig"]
We know that $ [DBF] \equal{} [ACE] \minus{} [DFE] \minus{} [BCD] \minus{} [ABF]$. If we draw the altitudes of the three triangles we need to subtract, we find that we have similar triangles.

$ [DBF] \equal{} \frac{12\cdot 16}{2} \minus{} \frac{3\cdot (16\cdot \frac{3}{4})}{2} \minus{} \frac{12\cdot (12\cdot \frac{1}{4})}{2} \minus{} \frac{9\cdot 4}{2}\\
\equal{} \frac{12(10) \minus{} 36}{2}\\
\equal{} \frac{84}{2}$

$ \frac{[DBF]}{[ACE]} \equal{} \frac{\frac{84}{2}}{\frac{192}{2}}\\
\equal{}\fbox{7/16 (E)}$
[/hide]

hey sinhera, could you possibly elaborate on the similar triangles part please

Seriously, just altitudes doesn't prove similar triangles...

Sure.

[hide="Details of the Similarity"]
Draw the altitude of $ \triangle AFB$ and $ \triangle DFE$, like so.

[asy]defaultpen(linewidth(.8pt)+fontsize(8pt));
dotfactor=3;

pair C = (0,0);
pair E = (16,0);
pair A = (0,12);
pair F = waypoint(E--A,0.25);
pair B = waypoint(A--C,0.25);
pair D = waypoint(C--E,0.25);

dot(A);dot(B);dot(C);dot(D);dot(E);dot(F);

label("$A$",A,NW);label("$B$",B,W);label("$C$",C,SW);label("$D$",D,S);label("$E$",E,SE);label("$F$",F,NE);

label("$3$",midpoint(A--B),W);
label("$9$",midpoint(B--C),W);
label("$4$",midpoint(C--D),S);
label("$12$",midpoint(D--E),S);
label("$5$",midpoint(E--F),NE);
label("$15$",midpoint(F--A),NE);

draw(A--C--E--cycle);
draw(B--F--D--cycle);

draw((foot(F,A,C))--F,linetype("4 4"));
draw((foot(F,C,E))--F,linetype("4 4"));[/asy]

It is clear that the altitudes are parallel to the legs of the right triangles (they are perpedicular to one leg and thus parallel to the other). By parallel lines we have equal angles so we get AA Similarity. You should be able to work on it from there.
[/hide]

From NICE's Areas document.

[hide=Solution]We use coordinates. Denote $C=(0,0)$. We immediately have $B=(0,9)$ and $D=(4,0)$. To find $F$, we drop an altitude from $F$ to $DE$, calling the contact point $G$. Note that by AA Similarity we get $\triangle EFG \sim \triangle EAC$, so $FG=3$ and $EG=4$. This means that $GD=8$ and hence $CG=12$. Therefore, $G=(12,3)$. By the Shoelace Formula, the area of $\triangle BDF$ is $42$. The area of $\triangle EAC$ is simply $\frac{(4+12)(9+3)}{2}=96$, so the answer is $\frac{42}{96}=\frac{7}{16} \implies \boxed{\text{(E)}}$.[/hide]

Hold up, isn’t this Wooga Looga Theorem?
[hide=sol]
We proceed using Wooga Looga Theorem. The ratio between the segments of the sides is consistent and it is $\frac{1}{3}$. The formula is $\frac{r^2-r+1}{(r+1)^2}$. Plugging in $r=\frac{1}{3}$ we get the answer is 7/16

Wait what is the Wooga Looga theorem again

The definition of the Wooga Looga theorem:
"If there is $\triangle ABC$ and points $D,E,F$ on the sides $BC,CA,AB$ respectively such that $\frac{DB}{DC}=\frac{EC}{EA}=\frac{FA}{FB}=r$, then the ratio $\frac{[DEF]}{[ABC]}=\frac{r^2-r+1}{(r+1)^2}$."

Oh, okay. Didn't realize it was actually legit.

[quote=TachyonPulse]In right triangle $ \triangle ACE$, we have $ AC \equal{} 12$, $ CE \equal{} 16$, and $ EA \equal{} 20$. Points $ B$, $ D$, and $ F$ are located on $ \overline{AC}$, $ \overline{CE}$, and $ \overline{EA}$, respectively, so that $ AB \equal{} 3$, $ CD \equal{} 4$, and $ EF \equal{} 5$. What is the ratio of the area of $ \triangle DBF$ to that of $ \triangle ACE$?
[asy]
size(200);defaultpen(linewidth(.8pt)+fontsize(8pt));
dotfactor=3;

pair C = (0,0);
pair E = (16,0);
pair A = (0,12);
pair F = waypoint(E--A,0.25);
pair B = waypoint(A--C,0.25);
pair D = waypoint(C--E,0.25);

dot(A);dot(B);dot(C);dot(D);dot(E);dot(F);

label("$A$",A,NW);label("$B$",B,W);label("$C$",C,SW);label("$D$",D,S);label("$E$",E,SE);label("$F$",F,NE);

label("$3$",midpoint(A--B),W);
label("$9$",midpoint(B--C),W);
label("$4$",midpoint(C--D),S);
label("$12$",midpoint(D--E),S);
label("$5$",midpoint(E--F),NE);
label("$15$",midpoint(F--A),NE);

draw(A--C--E--cycle);
draw(B--F--D--cycle);[/asy]$ \textbf{(A)}\ \frac {1}{4}\qquad \textbf{(B)}\ \frac {9}{25}\qquad \textbf{(C)}\ \frac {3}{8}\qquad \textbf{(D)}\ \frac {11}{25}\qquad \textbf{(E)}\ \frac {7}{16}$[/quote]

The inside triangle isn't right....


can any one tell me how can I add some mathematical statements to a .TEX file?

Thanx for help.

i didn't really understand your question... what exactly do you mean by adding mathematical statements to a $\text{\TeX}$ file ? ( Normally, all mathematical statements in $\text{\TeX}$ files show up once the files been compiled to dvi or pdf... )

Shouldn't this be moved to this forum?
[url]http://www.mathlinks.ro/Forum/index.php?f=123[/url] :maybe:

Amir.S, AOPS/Mathlinks has a site explaining lots of stuff about (La)TeX, I suggest you check it out :
[url]http://www.artofproblemsolving.com/LaTeX/AoPS_L_About.php[/url]

[quote="Amir.S"]now how can I creat a $\text{\TeX}$ file ?[/quote]

You could use some software like texnic center....  If you know tex well, though you could even write out your $\text{\TeX}$ code on notepad and save it as a $\text{\TeX}$ file.

[quote="karthik_k"]
....  If you know tex well, though you could even write out your $\text{\TeX}$ code on notepad and save it as a $\text{\TeX}$ file.[/quote]

 :) How can I find these codes?

This thing on AoPS itself should do for the basics.... :D

http://www.artofproblemsolving.com/LaTeX/AoPS_L_BasicFirst.php


[b]Theorem (Ceva’s Theorem for Chords).[/b] Let $A, B, C, D, E,$  and $F$  be six consecutive points around the circumference of a circle.  Then chords $AD, BE, CF$ concur  if and only if
\[ \begin{array}{|c|} \hline \\ ||AB|| \cdot|| CD|| \cdot ||EF|| = || BC|| \cdot || DE || \cdot || FA|| \\ \\ \hline \end{array}\; .  \]
 [b]References:[/b]
[1] H. S. M. Coxeter, [i]Problem 527[/i], The Mathematics Student, 28(1980)3.
[2] Stanley Rabinowitz, [i]The Seven Circles Theorem[/i],  Pi Mu Epsilon Journal, 8(1987) 441--449.

One implication is basically proven [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=60418]here[/url], and the converse follows immediately from it.


Let $A,\ B$ and $C$ be given points on a circumference $K$ such that the triangle $\triangle{ABC}$ is acute. Let $P$ be a point in the interior of $K$. $X,\ Y$ and $Z$ be the other intersection of $AP, BP$ and $CP$  with the circumference. Determine the position of $P$ such that $\triangle{XYZ}$ is equilateral.

One intersection P of the 3 Apollonius circles of the triangle $\triangle ABC$ (the 1st isodynamic point) always lies in the triangle interior, even if the triangle is obtuse. Invert the vertices A, B, C in a circle (P) centered at P and with arbitrary radius r. Then $A'B' = AB \cdot \frac{r^2}{PA \cdot PB},$  $B'C' = BC \cdot \frac{r^2}{PB \cdot PC},$ $C'A' = CA \cdot \frac{r^2}{PC \cdot PA}.$  Since P is on the B-Appolonius circle, $\frac{AB}{BC} = \frac{PA}{PC}$ and $\frac{A'B'}{B'C'} = \frac{AB}{BC} \cdot \frac{PB \cdot PC}{PA \cdot PB} = 1.$ Similarly, since P is on the C-Apollonius circle, $B'C' = C'A'.$ Thus the inverted triangle $\triangle A'B'C'$ is equilateral. But for any point P (not on the triangle circumcircle), the circumcevian triangle $\triangle XYZ$ of P, the pedal triangle $\triangle DEF$ with respect to P, and the triangle $\triangle A'B'C'$ inverted in an arbitrary circle centered at P are similar (for example, see  [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=70444]www.mathlinks.ro/Forum/viewtopic.php?t=70444[/url]).

i think there's a little easier solution. since $\angle ABP+\angle ACP=\angle YXZ=60$, it follows that the isogonals of $AP$ and $BP$ form an angle of $120$. same for $AP$ and $CP$, and $CP$ and $BP$, then the isogonal conjugate of $P$ must be $F$ (the Fermat-Torricelli point). we conclude that $P$ is the isogonal conjugate of $F$. the converse is also easy to prove.

As $XYZ$ is equilateral it follows that $m\angle APC=60+A$. Let $P'$ be the isogonal conjugate of $P$ wrt $\triangle ABC$ we easily get that $m\angle APC+m\angle AP'C=180+A$ (look my lemma at [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=109112[/url] post 4)so $m\angle AP'C=120$. Similary for the other vertices we get that $P'$ is the fermat point of $\triangle ABC$ so that $P$ is the first isodynamic point of $\triangle ABC$.


estw a,b,c > 0 kai abc=1 na apodeixtei oti :
a^3 + b^3 + c^3 + (ab)^3 + (bc)^3 + (ca)^2 >= 2(a^2*b + b^2*g + g^2*a)

to thema to brhka se ena biblio toy mpamph stergioy kai moy arese idiaitera.

[quote="Athinaios"]estw $ a,b,c > 0$ kai $ abc\equal{}1$ na apodeixtei oti :
$ a^3 \plus{} b^3 \plus{} c^3 \plus{} (ab)^3 \plus{} (bc)^3 \plus{} (ca)^3 \geq 2(a^2b \plus{} b^2c \plus{} c^2a)$

to thema to brhka se ena biblio toy mpamph stergioy kai moy arese idiaitera.[/quote]

Εχουμε από AM-GM: $ a^3\plus{}a^3\plus{}b^3\geq 3a^2b$ και προσθέτωντας κυκλικά προκύπτει

$ a^3\plus{}b^3\plus{}c^3\geq a^2b\plus{}b^2c\plus{}c^2a$  (1)

Επίσης από AM-GM παίρνουμε:

$ (ab)^3\plus{}(ab)^3\plus{}(ca)^3\geq 3a^3b^2c\equal{}3a^2b$ και προσθέτωντας κυκλικά προκύπτει:

$ (ab)^3 \plus{} (bc)^3 \plus{} (ca)^3\geq a^2b\plus{}b^2c\plus{}c^2a$ (2)

Από (1)+(2) το ζητούμενο :wink:

Πως προσθετουμε κυκλικα?  :huh:

ennoei oti kanei to antisoixo gia b^2 + b^2 + c^2>=3b^2c(me AM-GM) kai opws katalavaineis k meta prosthetei kata meli.wraia lysi pantws, an k sto mialo m eixa mia alli.  :)

Ο Δημήτρης εννοεί :

$ a^3 \plus{} a^3 \plus{} b^3 \ge 3a^2b$  (1)
$ b^3 \plus{} b^3 \plus{} c^3 \ge 3b^2c$  (2)
$ c^3 \plus{} c^3 \plus{} a^3 \ge 3c^2a$  (3)

Πρόσθεση κατά μέλη και ...  tsup!

$ 3a^2 \plus{} 3b^3 \plus{} 3c^3 \ge 3a^2b \plus{} 3b^2c \plus{} 3c^2a$

$ a^3 \plus{} b^3 \plus{} c^3 \ge a^2b \plus{} b^2c \plus{} c^2a$ ,     ...όμοια και για τις άλλες ανισότητες.

nai eis tin triti ennoousa alla panw stin viasini m egrapsa ^2.. :blush: sorry...

Το ότι $ a^3\plus{}b^3\plus{}c^3>\equal{}a^2b\plus{}b^2c\plus{}c^2a$ προκύπτει και από την Holder ως εξής:

$ a^3\plus{}b^3\plus{}c^3\equal{}(a^3\plus{}b^3\plus{}c^3)^{2/3}(b^3\plus{}c^3\plus{}a^3)^{1/3}>\equal{}a^2b\plus{}b^2c\plus{}c^2a$
[για εναλλακτική απόδειξη υπάρχει και η ανισότητα αναδιάταξης!]

και επίσης
$ (ab)^3\plus{}(bc)^3\plus{}(ca)^3\equal{}((ab)^3\plus{}(bc)^3\plus{}(ca)^3)^{2/3}((ac)^3\plus{}(ba)^3\plus{}(cb)^3)^{1/3}\equal{}((ab)^3\plus{}(bc)^3\plus{}(ca)^3)^{2/3}(1/b^3\plus{}1/c^3\plus{}1/a^3)^{1/3}>\equal{}a^2b\plus{}b^2c\plus{}c^2a$

με πρόσθεση αυτές οι 2 δίνουν το ζητούμενο!

Όχι μόνο απο τη Holder αλλά και απο την ανοσότηττα της αναδιάταξης. Έστω $ a\leq b\leq c\Leftrightarrow a^{2}\leq b^{2}\leq c^{2}$. Για τη μετάθεση $ (b,c,a)$ των $ a,b,c$ σύμφωνα με την ανισότητα της αναδιάταξης προκύπτει ότι $ a^{3} \plus{} b^{3} \plus{} c^{3}\geq a^{2}b \plus{} b^{2}c \plus{} c^{2}a$

[quote][για εναλλακτική απόδειξη υπάρχει και η ανισότητα αναδιάταξης!]
[/quote]

Αυτό εννοούσα εδώ! Και επίσης με την αναδιάταξης βγαίνει και το δεύτερο μέρος της ανισότητας!

kai egw aytin tn lysi eixa ipopsin m me tn anadiataksi. :wink: ..nice


This seems like a stupid question, but:

Am I allowed to bring and consume food and drink while taking the AIME?  I can get a good boost during practices from an energy drink, but I'm not sure if that counts as an unfair advantage or if its allowed. It doesn't seem to be mentioned in the FAQ on the MAA website.

I think my friend had coffee with him when he was taking the AMC 12.
Depends on where you take it, I guess.

I think you are allowed to; my friend brought a potato chip with him on AIME :D

Thanks for the information.  I guess that means I'm bringing my energy drink which I doubt contains modafinil, but worthawholebean makes an interesting observation, but of course any drugs come with serious risk.  I doubt any people good enough to qualify for USAMO would want to put their minds at jeopardy by taking drugs.  Still, it would be nice to see people care so much about math to take drugs for an exam.

[quote="worthawholebean"]This question has been asked a few times before; the answer that AMCDirector gives is that there's no AMC restriction on it.

On an interesting and slightly related note, does anyone foresee potential problems with "performance-enhancing" (mainly focus) drugs like Modafinil?[/quote]
Well I read in Discover that in a survey of college kids, 80% said they used drugs like Ritalin and Modafinil on tests before to help increase concentration and stuff...

I drink caffeine for AMC contests but I don't know how helpful it is, seeing that I feel like puking sometimes after too much of it :O I've aced some tests after drinking energy drinks, and I think most of the time it is that way, but I've also bombed a few tests while on caffeine too. My main motivation is that I am pretty much always sleepy due to chronic lack of sleep :(

However that same article said when they pulled random people and gave them stimulants and others placebos and had them take tests, ones who had taken the stimulant scored significantly better. Since drugs like those are so widely available, maybe the AMC would consider them a integrity threat? You can't exactly test for them though, so I suppose there isn't any way to stop it.

- You're full of adrenaline anyway. What's the point of adding outside stimulants on top of that?
- Stimulants can backfire. I recall one orchestra practice I had after having a lot of caffeine at dinner; I suddenly dropped off and lost all ability to focus about an hour in.

Now, I heartily recommend sugar and food in general, along with getting enough sleep before the test. Eating during the test is a distraction, so limit that.

I know I had a lot of caffeine one time before a contest for which I was really tired - I completely lost my ability to concentrate.

Well I think walking in calm and composed is best. Walking in anxious about results completely killed my AIME this year (although I still made USAMO most likely). I find that given I am calm, caffeine greatly improves my ability to think. If I am nervous, I start shaking, sweating, and feeling a burning sensation from taking caffeine.

Of course the natural threshold for your performance is set by how well you sleep and how well you eat, so those are priority (although I never sleep well...hmmm I think I see a correlation between sleep and contest scores).

Eating during say, the USAMO, I think could possibly be helpful as a breather when you're stuck on a problem, since it relaxes you somewhat. It certainly helps me get ideas when I go snacking.

I wouldn't advocate caffiene or stimulants before a test, especially a relatively long one like the AIME.  It's hard to predict when the stimulant will wear off and leave you with little energy left.  I use an energy drink as a boost once I get to around #11-12 for some more energy on the harder problems.

Like jmerry said, two things are what you need the most: good night sleep and mild breakfast (not heavy please).

Another thing to note is for people like me who don't drink coffee or caffeine at all (excluding soda), trying on those drinks will be extremely harmful on first couple times. So... I'm sure we are all smart to avoid it but if you are weak on caffeine, don't drink too much on your first AMC.  :rotfl:


floor function
combinatorics unsolved
combinatorics

How many ways are there of coloring the vertices of a regular $2n$-gon with n colors, such that each vertex is given one color, and every color is used for two non-adjacent vertices? Colorings are regarded as the same if one is obtained from the other by rotation.

An equivalent question  has been posted (but not answered) in the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=85715]Pre-Olympiad forum[/url].  I could have sworn I've solved it (or at least looked it up on OEIS) recently, but I can't find it, so either I'm wrong or it wasn't on the forum.  I'll get back to you (or someone else will solve it first ;) ).

This problem (under the form posted in that Pre-Olympiad topic) is quite well-known in combinatorics (due to Lucas?).
It can be solved with the principle of inclusion-exclusion, after using [b]Kaplansky's lemma[/b] :
The numbers $1, 2, ..., m \ \ (m \geq 3)$ are placed around a circle. For $1 \leq k \leq \lfloor m/2 \rfloor,$ let $a(k)$ be the number of subsets with k elements of $\{1, 2, ..., m\}$ which don't contain two adjacent elements on the circle. We then have :
\[a(k) = \frac{m}{k}\binom{m-k-1}{k-1}. \]

this problem has been solved in:
http://www.mathlinks.ro/Forum/viewtopic.php?t=4110


[hide]The only cubes that have no red are the cubes that you can't see.  That section is a 2x2x2 cube, so the answer is 8.[/hide]

[quote="bryantacademy"]The complete outside including the bottom of a wooden 4inch cube is painted red. The painted cube is then cut into one inch cubes. How many cubes have absolutely no red paint on any face?[/quote]
[hide]
The number of cubes you have are 64. Only the cubes in the middle have no paint. Remove the border of paint. Thus you have a 2x2x2 square. I believe that it is 8?[/hide]

anything on the inside, so (4-2)^3, or 8

its a 4 cube by 4 cube square, that is impossible!!! :furious: If you remove all the outer cubes, you must surely get a 2 cube by 2 cube square

how is it impossible? It needs to be a 2 by 2 by two or less!

take out outer 4...

inside cube = 2x2x2 = 8

[hide]Well, only the 2*2*2 part is going to be unpainted, so 8.[/hide]


What is the sum of the prime factors of the number represented by:
2^12-2^11+2^10-2^9+...+2:^2:-2

this was a target round, so i guess you could use a calc

I think the answer is... [hide]101[/hide]

MY bad

the correct answer is : [hide]30[/hide]

[hide]

Hmm, I did it a slow and tedious way, but got 30

[/hide]

sorry, i meant to say no to the answer of 101. yes [hide]30[/hide] is correcto

How did you do it? This was my method..

[hide]

I noted that

x^n -1/x-1 = x^n-1 + ..... 1

And found out what the number was, then factored it.

How did you guys do it? Is there a faster way?

[/hide]

SouthPaw: although your method is of course the better and requires more skill, most MATHCOUNTS contestants would have simply put it into their calcs, gotten 2730 out of it and then prime factored that....

Syntax Error wrote:What is the sum of the prime factors of the number represented by:
2^12-2^11+2^10-2^9+...+2:^2:-2

[hide]okay, so that equals 2^11+2^9+2^7+...+2=2(2^10+2^8+2^6+...+1)=2(4^5+4^4+4^3+4^2+4^1+1)

then i factored from there. it didnt take more than 20 seconds to get there (i think). but i dunno how far you could have factored before that. [/hide]

it would really help if you had a ti-89 to factor that i guess.

sigh, i guess its faster to factor after youve done the first step (2^11+2^9+2^7+...+2). but still, good to know how to get there, to make it easier, if you dont have that kind of...whatever

just wondering...syntax, did you get all this stuff from me? because I thought I was the only one weird enough to use 4's and all....hmmm...

im perdy sure that i told you that i had done it that way the first time i took the test in april/may

anyways, mystic terminator did it the same way. yah for him. and for the rest of us!

but i really didnt copy mystic. well i did, but i had done it the same way. YAH FOR EVERYONE!

I don't know. For a target round questions, I think it's a lot easier and guarantees no mistakes if u use a calc.

After all, 2^12 is pretty small, i've got it memorized to 2^16, but if u don't have it memorized, at least have 2^10 down good.


Find the sum of the coefficients in the polynomial that results when $(4x^2-3x-2)^{2001}$ is completely multiplied out.

Eh, am I missing something really obvious?

I ought to gag myself...   :blush:

I knew... that. :?

Thanks Joe.

Hmm why do we just need to plug in $x=1$ to find the solution

Take a polynomial $a_1x^{n}+a_2x^{n-1}+a_3x^{n-2}+\dots$

If you plug in one, we get $a_1+a_2+a_3+\dots$

Which is the sum of the coefficients.


Let $ a,b,c$ be none-gative numbers ,no two of which are zero.Prove that:
$ \frac{36(a^2\plus{}b^2\plus{}c^2)}{(a\plus{}b\plus{}c)^2}\plus{}\frac{5abc}{(a\plus{}b)(b\plus{}c)(c\plus{}a)} \ge \frac{49}{4}\plus{}(\frac{a}{a\plus{}b})^3\plus{}(\frac{b}{b\plus{}c})^3\plus{}(\frac{c}{c\plus{}a})^3$
I hope you will like it :)

I am again doubtful..... Again the inequyality seems to be unchanged by replacing a, b, c respectively with ka, kb, kc so $ abc\equal{}1$ is true.... AM I RIGHT??

PS: Dear quy, your inequalities are always nice , but I cannot solve MOST of them.... :rotfl:

[quote="Potla"]I am again doubtful..... Again the inequyality seems to be unchanged by replacing a, b, c respectively with ka, kb, kc so $ abc \equal{} 1$ is true.... AM I RIGHT??

PS: Dear quy, your inequalities are always nice , but I cannot solve MOST of them.... :rotfl:[/quote]
My friend,you can assume $ abc\equal{}1$,but I think it isn't nececsarry  with this inequality :)

[quote="Potla"]I am again doubtful..... Again the inequyality seems to be unchanged by replacing a, b, c respectively with ka, kb, kc so $ abc \equal{} 1$ is true.... AM I RIGHT??

PS: Dear quy, your inequalities are always nice , but I cannot solve MOST of them.... :rotfl:[/quote]
With $ abc\equal{}1$,you can solve it? :wink: Please post solution :oops:

I don't think it's a strong ineq.
We can prove easily the stronger:
$ \frac{81(a^2\plus{}b^2\plus{}c^2)}{4(a\plus{}b\plus{}c)^2} \plus{} \frac{5abc}{(a\plus{}b)(b\plus{}c)(c\plus{}a)} \geq\ 7 \plus{} \sum\ \frac{a^3}{(a\plus{}b)^3}$
quykhtn-qa1, can you solve this stronger ?  :)

[quote="nguoivn"]I don't think it's a strong ineq.
We can prove easily the stronger:
$ \frac {81(a^2 \plus{} b^2 \plus{} c^2)}{4(a \plus{} b \plus{} c)^2} \plus{} \frac {5abc}{(a \plus{} b)(b \plus{} c)(c \plus{} a)} \geq\ 7 \plus{} \sum\ \frac {a^3}{(a \plus{} b)^3}$
quykhtn-qa1, can you solve this stronger ?  :)[/quote]
I have a stronger inequality:
$ \frac{9(a^2\plus{}b^2\plus{}c^2)}{(a\plus{}b\plus{}c)^2}\plus{}\frac{5abc}{(a\plus{}b)(b\plus{}c)(c\plus{}a)} \ge \frac{13}{4}\plus{} \sum\ \frac {a^3}{(a\plus{}b)^3}$ :)
And,with my first inequality,I have a nice solution :)

[quote="quykhtn-qa1"][quote="nguoivn"]I don't think it's a strong ineq.
We can prove easily the stronger:
$ \frac {81(a^2 \plus{} b^2 \plus{} c^2)}{4(a \plus{} b \plus{} c)^2} \plus{} \frac {5abc}{(a \plus{} b)(b \plus{} c)(c \plus{} a)} \geq\ 7 \plus{} \sum\ \frac {a^3}{(a \plus{} b)^3}$
quykhtn-qa1, can you solve this stronger ?  :)[/quote]
I have a stronger inequality:
$ \frac {9(a^2 \plus{} b^2 \plus{} c^2)}{(a \plus{} b \plus{} c)^2} \plus{} \frac {5abc}{(a \plus{} b)(b \plus{} c)(c \plus{} a)} \ge \frac {13}{4} \plus{} \sum\ \frac {a^3}{(a \plus{} b)^3}$ :)
And,with my first inequality,I have a nice solution :)[/quote]
Yes, indeed, I had a mistake in my calculus. The best result is same to you:
$ \frac{9(a^2\plus{}b^2\plus{}c^2)}{(a\plus{}b\plus{}c)^2} \geq\ \frac{9}{4} \plus{} \sum\ \frac{a^3\plus{}b^3}{(a\plus{}b)^3}$
Of course, SOS solved it easily  :)

[quote="nguoivn"][quote="quykhtn-qa1"][quote="nguoivn"]I don't think it's a strong ineq.
We can prove easily the stronger:
$ \frac {81(a^2 \plus{} b^2 \plus{} c^2)}{4(a \plus{} b \plus{} c)^2} \plus{} \frac {5abc}{(a \plus{} b)(b \plus{} c)(c \plus{} a)} \geq\ 7 \plus{} \sum\ \frac {a^3}{(a \plus{} b)^3}$
quykhtn-qa1, can you solve this stronger ?  :)[/quote]
I have a stronger inequality:
$ \frac {9(a^2 \plus{} b^2 \plus{} c^2)}{(a \plus{} b \plus{} c)^2} \plus{} \frac {5abc}{(a \plus{} b)(b \plus{} c)(c \plus{} a)} \ge \frac {13}{4} \plus{} \sum\ \frac {a^3}{(a \plus{} b)^3}$ :)
And,with my first inequality,I have a nice solution :)[/quote]
Yes, indeed, I had a mistake in my calculus. The best result is same to you:
$ \frac {9(a^2 \plus{} b^2 \plus{} c^2)}{(a \plus{} b \plus{} c)^2} \geq\ \frac {9}{4} \plus{} \sum\ \frac {a^3 \plus{} b^3}{(a \plus{} b)^3}$
Of course, SOS solved it easily  :)[/quote]
YES,THANKS :)
@nguoivn:[hide] anh gioi qua',hinh nhu anh sap viet sach?,bao gio xong a.?' [/hide]


Prove that
$ f: \mathbb {R}\rightarrow \mathbb {R}$
$ f(x)\equal{}x^5\minus{}2x^2\plus{}x$ is a surjective function, onto mapping.

[hide]$ f$ is continuous on $ \mathbb R$, and $ \lim_{x\to \plus{} \infty}f(x) \equal{} \plus{} \infty$, $ \lim_{x\to \minus{} \infty}f(x) \equal{} \minus{} \infty$. Therefore $ f$ is surjective.[/hide]


The sum ∑(from n = 1 to ∞) arctan [n/(n^4 - 2n^2 + 2)] is equal to:
A) (arctan 2 + arctan 3)/4
B) 4. (arctan 1)
C) -π/16
D) 3π/16

More than one options may be correct.

hello, do you mean $ \sum_{n\equal{}1}^\infty\arctan\left(\frac{n}{n^4\minus{}2n^2\plus{}2}\right)$?
Sonnhard,

It does mean $ \sum_{n\equal{}1}^\infty\arctan\left(\frac{n}{n^4\minus{}2n^2\plus{}2}\right)$

I am eager to know the solution. :)  :)

It does mean that. But I dont know the solution. Anyone on AoPS???

[hide="Not a solution but may help"]Let's just assume for a moment the numerator is $ 4n$ instead of $ n$.

$ \begin{array}{rcl} \arctan \left( \frac{4n}{n^4 \minus{} 2n^2 \plus{} 2} \right) &\equal{}& \arctan \left( \frac{ (n\plus{}1)^2 \minus{} (n\minus{}1)^2 }{ 1 \plus{} (n^2\minus{}1)^2 } \right) \\ &\equal{}& \arctan ((n\plus{}1)^2) \minus{} \arctan ((n\minus{}1)^2) \\ \end{array}$

This telescopes getting $ \minus{}\arctan 0 \minus{} \arctan 1 \equal{} \minus{}\pi/4$, which is $ 4$ times one of the answers. Unfortunately you can't just yank $ 4$'s out of $ \arctan$ like that. I'm not sure how you'd introduce it either.[/hide]

hello, a numerical experiment with Maple ( i have calculated the sum to 100,
i.e. $ \sum_{n\equal{}1}^{100}\arctan\left(\frac{n}{n^4\minus{}2n^2\plus{}2}\right)$ )shows that the result is nearly $ 1.07220308423251547552748148278$. But neither the answers A, B, C nor D show approximately this result.
Sonnhard.

[b]MNrule[/b], are you sure you have stated the problem correctly?

[b]MellowMelon[/b], aren't you forgetting the limit?


calculus
integration
calculus computations

Let $ a,\ b,\ c$ be positive real numbers. Prove the following inequality.
\[ \int_1^e \frac {x^{a \plus{} b \plus{} c \minus{} 1}[2(a \plus{} b \plus{} c) \plus{} (c \plus{} 2a)x^{a \minus{} b} \plus{} (a \plus{} 2b)x^{b \minus{} c} \plus{} (b \plus{} 2c)x^{c \minus{} a} \plus{}(2a \plus{} b)x^{a \minus{} c} \plus{} (2b \plus{} c)x^{b \minus{} a} \plus{} (2c \plus{} a)x^{c \minus{} b}]}{(x^a \plus{} x^b)(x^b \plus{} x^c)(x^c \plus{} x^a)}\geq a \plus{} b \plus{} c.\]

I have just posted 500 th post.

[color=blue]Thank you for your cooperations, mathLinkers and AOPS users.[/color]

I will keep posting afterwards.

Japanese Communities Modeartor

kunny

Chikaya-san, i think there's a bit of a typo-error..anyway, it doesn't matter.


parameterization
number theory unsolved
number theory

prove that we can write any rational $r \in (0,1)$ with odd denominator in this form$\frac{xyz}{x^2+y^2+z^2}$

I think I saw it in Komal.

Anyway, it seems tough. I only have some trivial remarks about it. First of all, it's enough to prove it for numbers of the form $\frac 1{2n+1},\ n\in \mathbb N^*$, because then we can take $px,py,pz$ and prove it holds for $\frac p{2n+1}$ as well, so the problem is to show that for $n\in \mathbb N^*$, the equation $x^2+y^2+z^2=(2n+1)xyz$ has (non-trivial) solutions. This is equivalent to saying that for each $n\ge 1$ we can find $y,z\ne 0$ s.t. $(2n+1)^2y^2z^2-4y^2-4z^2$ is a perfect square. I have no idea how to do this..

Note that for $n=1$ this is the famous Markov equation $x^2+y^2+z^2=3xyz$. I've searched the Web it an attempt to find refrences to the problem we're interested in (we could call it a Markov equation with generalized parameter), but to no avail. For $n=1$ we can find an infinite set of solutions: $(x,y,z)=(F_{2n-1},F_{2n+1},1)$, where $F_n$ are the Fibonacci numbers (starting with $F_0=0,F_1=1$).

I dont think so.
It is not difficult to prove that the equation $x^2+y^2+z^2 = kxyz$ has a solution if and only if $k=1$ or $k=3$.
So, I think there must be another way.

Pierre.

I also found this fact when reading grobber's post, but unfortunately a friendly message " please come back again later, since the RAM memmory dosn't suffice" didn't let me to do this. But doesn't this imply that th problem is wrong?

You seem to be right.
Or maybe $x,y,z$ are supposed to be rationals and not integers?

So, Sam? What is your opinion about it?

Pierre.

No, the correct problem is that any such rational is the fractional part of a number of the form $\frac{xyz}{x^2+y^2+z^2} $ with x,y,z integers.


real analysis
function
integration
advanced fields
advanced fields unsolved

Let A be some subset of the line with positive lebesgue measure,is there any lipschitz function f such that f(A) is an interval?

I should have add ...non constant function.

Yes, for example $f(x)=\int_{0}^{x}\chi_{A}(t)\,dt$.

One can ask the same question in higher dimensions: given a set $A\subset\mathbb R^{n}$ of positive measure, is there a Lipschitz mapping $f\colon\mathbb R^{n}\to\mathbb R^{n}$ such that $f(A)$ is a (non-trivial) ball? This has been proved in dimension $n=2$ (Alberti-Csornyei-Preiss, not yet published). Remains open for $n>2$.

Your function does not work for arbitrary sets,take the irrationals for example and you ll have f(Ir)=Ir,its of course correct for compact sets.
I found out there is a proof in Laczkovich's paper 'Paradoxical decompositions using lipschitz functions',but its not of exercise level.


On another forum I saw that once you get past a certain amount of posts and have joined for a certain period of time, you get to make up your own ranking (under your post count ranking) through the profiles page. Have no idea how that's done.

Also, I've heard that moderators used to have their own rankings like administrators. What happened to that.

This has been mentioned, but we should also have a forum where posts don't count towards your post count (although they still can searched). On that other forum, it was the home to all the games. This can also solve the "post count deficits" which is when you post count says x posts, but when you click "find all posts by yyyyy" you usually get "search found x - k matches" where k is non-negative.

Just some ideas  :)

[quote="236factorial"]On another forum I saw that once you get past a certain amount of posts and have joined for a certain period of time, you get to make up your own ranking (under your post count ranking) through the profiles page. Have no idea how that's done.
[/quote]

Bah. Elitist :P .

On the other hand, I like your other suggestions.

These ideas have been discussed several times.

We don't care about post counts.  Why should they concern us at all?

The ranks are just for fun.  In our case, we add a mathematical theme -- the Millenium problems.  We're not about to spend time horsing around with them.  We have about a ginormozillion hours worth of educational projects ahead of us.  Users should remember that as entertaining as some aspects of this site might be, our goal is to build better and better educational resources for students.

That's why I put small ideas, they're not of high importance (because we can live without them). It's just when some of the administrators have too much free time (which I hope they do - life can't be all work and no play?  :D ), they can consider these.

I just noticed something.... all of these were about post counts  :?


A man drives along a main highway on which a regular service of buses is in operation. He notices that every three minutes he  meets a bus and that every six minutes a bus overtakes him. How often does a bus leave the terminal station at one end of the route ?

[quote="pleurestique"]

this problem's been posted before; I believe it's an old Russian problem.[/quote]

 :maybe:
It is an old Britain problem   :rotfl:  from [i]"Great book of math puzzles" by Philip Heafford[/i]
Originally published in Great Britain under the title "Mathematics for fun" 1959
The solution of these problem in the  book is arithmetic solution.
How high level  has be an old math text books !

[quote="treesamc-8"]sloution because every three min he sees a bus then every six min he is overtaken, then
um every two min a bus leaves the terminal because 6/3 is two[/quote]

silly, the answer is 4.

[quote="pleurestique"]let $ b$ be the speed of a bus and $ m$ the man's speed. since the man meets buses twice as fast as they overtake him, we get $ b \plus{} m \equal{} 2(b \minus{} m)$, which gives $ b \equal{} 3m$. when they travel in the same direction, the speed of the buses relative to the man is therefore $ 3m \minus{} m \equal{} 2m$, while the actual speed of the bus is $ 3m$, so if the man were stationary, buses would pass him every $ 6\cdot \frac {2}{3} \equal{} 4$ minutes.

this problem's been posted before; I believe it's an old Russian problem.[/quote]

very nice explanation!


geometry
logarithms
integration
calculus
rotation
calculus computations

Let R be the region bounded by the graph of $ y\equal{}\frac{lnx}{x}$, the $ x\minus{}axis$, and the line $ x\equal{}e$

a. find the area of the region R.
b. find the volume of the solid formed by revolving the region about the $ y\minus{}axis$

By parts: $ u\equal{}\ln x$, $ du\equal{}\frac{dx}{x}$, $ v\equal{}\ln x$, $ dv\equal{}\frac{dx}{x}$.  So $ \ln^{2} x\minus{}\int \frac{\ln x}{x}dx\equal{}\int \frac{\ln x}{x}dx$.  Move the integral on the left over and divide so $ \int \frac{\ln x}{x}dx\equal{}\frac{\ln^{2} x}{2}$.  The limits of integration are $ x\equal{}1$ and $ x\equal{}e$ so plugging these in, we get $ A\equal{}\frac{1}{2}$.

[quote="JRav"]By parts: $ u = \ln x$, $ du = \frac {dx}{x}$, $ v = \ln x$, $ dv = \frac {dx}{x}$.  So $ \ln^{2} x - \int \frac {\ln x}{x}dx = \int \frac {\ln x}{x}dx$.  Move the integral on the left over and divide so $ \int \frac {\ln x}{x}dx = \frac {\ln^{2} x}{2}$.  The limits of integration are $ x = 1$ and $ x = e$ so plugging these in, we get $ A = \frac {1}{2}$.[/quote]

This is my answer for part b, could someone check if this is correct?

it rotates around y-axis
use shell dx
$ V=\int_1^e {2{\pi}rh}dx = 2{\pi}\int_1^e {rh}dx$
where $ r=(x-1)$and $ h=f(x)=\frac {\ln {x}}{x}$
so: $ 2{\pi}\int_1^e {(x-1)(\frac {\ln {x}}{x})}dx$
=$ 2{\pi}\int_1^e {\ln {x}-\frac {\ln {x}}{x}}dx$
=$ 2{\pi}\int_1^e {\ln {x}}$$ -{2{\pi}\int_1^e {\frac {\ln{x}}{x}}}$
=$ 2{\pi}[xlnx-x]_1^e - 2{\pi}[\frac{(\ln {x})^2}{2}]_1^e$
=$ 2{\pi}[e\ln {e} -e-\ln {1} +1] - {2{\pi}[\frac {\ln {e^2}}{2}-\frac {\ln {1^2}}{2}]}$
=$ 2{\pi}[e-e+1] - 2{\pi}[1/2]$=$ 2{\pi}-{\pi}$=$ {\pi}$

^does this process seem correct?

=]

The process is correct, but you made an error early that you carried through the computation (and which made you do some extra work):
[quote="sneezywalrus"]$ r \equal{} (x \minus{} 1)$[/quote]
Other than that, everything was fine.  (Well, you dropped a couple of $ dx$s, but your actual computations were correct.)


geometry
circumcircle
trigonometry
trig identities
Law of Sines

[hide="Solution"]

Let's trig it!

$ s\equal{}\frac{a\plus{}b\plus{}c}{2}\equal{}R(\sin A\plus{}\sin B\plus{}\sin C)$
$ \sqrt3(R\plus{}r)\equal{}\sqrt3R(\cos A\plus{}\cos B\plus{}\cos C)$

So, it's left to prove $ \sin A\plus{}\sin B\plus{}\sin C\equal{}\sqrt3(\cos A\plus{}\cos B\plus{}\cos C)$. As $ A\equal{}60^\circ$, it's left to prove $ \sin B\plus{}\sin C\equal{}\sqrt3(\cos B\plus{}\cos C)$,with $ B\plus{}C\equal{}120^\circ$

which is easy if I didn't make a mistake. [/hide]

[hide="Solution 2"]
The ecuation $ f(x)\equal{} s\cdot x^3\minus{}(4R\plus{}r)\cdot x^2\plus{}s\cdot x\minus{}r\equal{}0$ has the solutions $ \left\{\ \tan\frac A2\ ,\ \tan\frac B2\ ,\ \tan\frac C2\ \right\}$.

Hence $ A\equal{}60^{\circ}\ \Longleftrightarrow\ f\left(\frac{\sqrt 3}{3}\right)\ \equal{}\ 0\ \Longleftrightarrow\ \boxed{s\equal{}\sqrt{3}(R\plus{}r)}$
[/hide]

Does anyone have a non-brute-forcing way of proving $ 1\plus{}r/R\equal{}\cos A\plus{}\cos B\plus{}\cos C$ (or maybe derive it)? I can only prove it by expressing everything in terms of the side lengths.
Also Mateescu Constantin how did you come up with that nice identity (involving tangents)? Thanks.

[quote="Mateescu Constantin"][hide="Solution 2"]
The ecuation $ f(x) \equal{} s\cdot x^3 \minus{} (4R \plus{} r)\cdot x^2 \plus{} s\cdot x \minus{} r \equal{} 0$ has the solutions $ \left\{\ \tan\frac A2\ ,\ \tan\frac B2\ ,\ \tan\frac C2\ \right\}$.

Hence $ A \equal{} 60^{\circ}\ \Longleftrightarrow\ f\left(\frac {\sqrt 3}{3}\right)\ \equal{} \ 0\ \Longleftrightarrow\ \boxed{s \equal{} \sqrt {3}(R \plus{} r)}$
[/hide][/quote]

Can someone explain how he did that ?

[quote="Mateescu Constantin"]Let $ ABC$ be a triangle. Prove that $ \angle A\equal{}60^{\circ} \Longleftrightarrow s\equal{}\sqrt{3}(R\plus{}r)$ ($ s$-semiperimeter, $ R$-radius of the circumcircle, $ r$-radius of the incircle).[/quote][b][u]Remarks[/u].[/b]

$\blacktriangleright$ The equivalence $\boxed{A=60^{\circ}\iff s=(R+r)\sqrt 3}$ is false because the implication

$\boxed{A=60^{\circ}\implies s=(R+r)\sqrt 3}$ is true and $\boxed {s=(R+r)\sqrt 3\implies A=60^{\circ}}$ is false.

$\blacktriangleright$ This equivalence $\boxed{60^{\circ}\in\{A,B,C\}\iff s=(R+r)\sqrt 3}$ is true. Indeed, prove easily that

$\left\{\begin{array}{c}
\tan\frac A2+\tan\frac B2+\tan\frac C2=\frac {4R+r}{s}\\\\
\tan\frac A2\tan\frac B2+\tan\frac A2\tan\frac B2+\tan\frac A2\tan\frac B2=1\\\\
\tan\frac A2\tan\frac B2\tan\frac C2=\frac rs\end{array}\right\|\implies$ $s\cdot \prod\left(1-\sqrt 3\cdot\tan\frac A2\right)=4\left[s-(R+r)\sqrt 3\right]$ .

[b][u]Other method[/u].[/b] Prove easily that $\boxed{x+y+z=0\implies \sum\sin x=-4\prod\sin\frac x2}\ (*)$ . Therefore, $s=(R+r)\sqrt 3\iff$

$\frac sR=\sqrt 3\cdot\left(1+\frac rR\right)\iff$ $\sum\sin A=\sqrt 3\cdot\sum\cos A\iff$ $\sum\sin\left(A-60^{\circ}\right)=0$ . For  $\left\{\begin{array}{c}
x:=A-60^{\circ}\\\
y:=B-60^{\circ}\\\
z:=C-60^{\circ}\end{array}\right\|$ , where

$x+y+z=0$ apply the identity $(*)$ . In conclusion, $s=(R+r)\sqrt 3\iff$ $\prod\sin\left(\frac A2-30^{\circ}\right)=0\iff$  $60^{\circ}\in\{A,B,C\}$ .

[hide="Another way"]
Put:
$y+z=BC, x+z=AC, x+y=AB$
where $x,y,z$ are some positive real numbers. Then we must have:
$y+z=2R\sin 60 = R\sqrt 3$ by the law of sines and $\tan 30 = \frac{1}{\sqrt 3}= \frac{r}{x}$.
Solving for $r$ and $R$ and noting that $x+y+z=s$ we get exactly the desired relationship.
[/hide]


How many positive integer solutions exist for $ 3(x\minus{}5)\le 7$

If 3(x-5) is less than or equal to 7, then divide by 3 and get x-5 is less than or equal to 7/3.  Add 5 to both sides and get x is less than or equal to 7/3+5=7/3+15/3=22/3.  The possible values of x are 1,2,3,4,5,6 and 7, so the answer is [b]7.[/b]


Hi,

I have been trying to create a letter using LaTeX and have come across a few problems. The code I have used is:

[code]\documentclass{letter}
\address{Something Road \\ England \\ M12 2PL}
\signature{Beans}

\begin{document}
\begin{letter}{md}
\opening{Dear Mr. Wolf}

Blah

\closing{Yours sincerely}
\end{letter}
\end{document}[/code]

The problem I face is that the "Yours sincerely" part is centred and I want it towards the left--can I do this?

Also, I don't want to write the senders address, but if I don't put anything down then my own address won't get printed! Help!

(I tried googling templates for letters, but alas I didn't come up with anything better).[/code]

[quote="stevem"]Hi, Beans

1. Don't bother with \closing just type Yours sincerely. Similarly you don't have to use \signature if you don't want to.
2. Instead of \begin{letter}{md} use \begin{letter}{}

Incidentally, the author of the TeX FAQ [url=http://www.tex.ac.uk/cgi-bin/texfaq2html?label=letterclass]Letters and the like[/url] also has had problems with the letter class so you are in good company. However it's worth sticking with for a normal bog-standard letter.[/quote] Thanks a lot--it worked a treat.

I was wondering if there was any latex package/template that you would recommend for creating a CV? (The joys of looking for a summer job...)

There are a number of CV packages on [url=http://tug.ctan.org/search.html#byDescription]CTAN[/url]. Unfortunately that's down at the moment but have a look when it's back. Do be aware that different countries tend to use different styles for their CVs.

CTAN is back again and you'll find a number of CV packages at http://tug.ctan.org/cgi-bin/search.py?metadataSearch=vitae or http://texcatalogue.sarovar.org/bytopic.html#applications

Thanks Steve! I hadn't thought about that and instead had been using Google (unsuccessfully might I add) and found some not very nice packages.


function
abstract algebra
group theory
Ring Theory
superior algebra
superior algebra unsolved

Recently, I have come up with an algebraic structure that I would like to call "Partial rings" with some interesting properties. They are similar to rings, as will be shown later. The rules for this structure are:

1) Group under addition
2) Another function, multiplication
3) mutiplication is distributive
4) multiplication is associative

5 (optional)) Noncommutative group under addition.

Some properties already found:
1) The identity under addition annihilates under multilplication (clearly)
2) The additive order of the product a*b divides the additive order of each a and b
3) The set of products is commutative
4) The set of products a*r, r is in R is a commutative group
5) If the order of the "partial ring" is squarefree, then the set of products is cyclic

Example:
[code]+    | e    a    2a   b    b+a  b+2a
_______________________________________
e    | e    a    2a   b    b+a  b+2a
a    | a    2a   e    b+2a b    b+a
2a   | 2a   e    a    b+a  b+2a b
b    | b    b+a  b+2a e    a    2a
b+a  | b+a  b+2a b    2a   e    a
b+2a | b+2a b    b+a  a    2a   e

*    | e    a    2a   b    b+a  b+2a
______________________________________
e    | e    e    e    e    e    e
a    | e    e    e    e    e    e
2a   | e    e    e    e    e    e
b    | e    e    e    b    b    b
b+a  | e    e    e    b    b    b
b+2a | e    e    e    b    b    b[/code]

Proof of properties:

1) 0*a = 0*a + 0*a + -(0*a) = (0+0)*a + -(0*a) = 0*a + -(0*a) = 0
2) ab+ab+ab+...+ab (o(a) times) = (a+a+a+a+...+a)b = eb = e, so o(ab)|o(a). Similar argument for b.
3) -ac + (a+b)(c+d) + -bd = -ac + (a+b)c + (a+b)d + -bd= -ac + ac + bc + ad + bd + -bd = bc + ad = -ac + a(c+d) + b(c+d) + -bd = -ac + ac + ad + bc + bd + -bd = ad + bc.
4) Commutative has already been shown. All we need show is closure and inverse to show it is a group. a*r1 + a*r2 = a*(r1+r2), so it is closed. a*r1 + a*-r1 = a*(r1 + -r1) = a*e = e, so we have inverse and therefore identity, so a*r is a group.
5) This one is long. It shall be proven later.

Other properties:
The 0th and 1st Isomorphism theorem works in these partial rings, that is:
0th: The kernel of any "partial ring" homomorphism is a partial normal ideal, that is, it is a normal subgroup under addition and is an ideal under multiplication. Any partial normal ideal N of R also has a homomorphism to R/N with kernel N.
The definition of a "partial ring" homomorphism is that it takes sums to sums and products to products.
1st: If you have an onto ring homomorphism from R to R' with kernel N, then R/N is isomorphic to R'.
I believe that this also implies the second and third isomorphism theorems, namely, H+N/N is isomorphic to H/H intersect N for any subring H and normal partial ideal N of R, and also that G/N1 = (G/N2)/(N1/N2) where N2 and N1 are normal partial ideals of G, and N2 is in N1.
Things found so far:
An 8-element partial ring with noncommutative addition and multiplication
A 32-element ring without identity such that the set of products is not a group (not closed under addition)
Can you guys think of anything else about it?
Note: I am uncool on scienceforums.net, so if you see it there, it's mine.
bmsiv[/code]

[quote="qleak"][quote="ZetaX"][quote="bmsiv"]The rules for this structure are:

1) Group under addition
2) Another function, multiplication
3) mutiplication is distributive
4) multiplication is associative[/quote]
Isn't this just the definition of a (noncommutative) ring (without unity).[/quote]

Not quite addition is not necessarily commutative.[/quote]
Correct, as shown by the group that I displayed (which is S3 under addition, which is noncommutative).
I believe that each ring can be simplified as follows:
Let N = {a: a*r = r*a = e for all r in R}. Then R/N is a ring (that is, R/N is commutative). This occurs for any partial ring that is a whole ring to begin with, and for any partial ring with order pq, p and q prime, p =/= q.
=Uncool-

There are currently two major theorems left that I have learned that deal with rings that do not include unity. The first one is that all of them can be embedded in a ring with unity. This clearly is not true for this structure, as if it were embedded, it would be commutative, as shown above. Therefore, let us look at the other theorem that I know of - the Chinese Remainder Theorem.

The Chinese Remainder Theorem, over rings, states the following:
(from wikipedia on chinese remainder theorem)
If R is a ring (with unity) and I1, ..., Ik are two-sided ideals of R which are pairwise coprime (meaning that Ii + Ij = R whenever i ≠ j), then the product I of these ideals is equal to their intersection, and the quotient ring R/I is isomorphic to the product ring R/I1 x R/I2 x ... x R/Ik

The proof uses the fact that R/A  x R/B is isomorphic to R/AnB, and that AnB is equal to AB, where the product is defined as the set of all finite sums of products of elements of the two rings.

The proof that R/A x R/B is isomorphic to R/AnB is the same. However, we can see that even in rings without unity, there are examples where AnB =/= AB. An example of this can be seen in the ring 6Z. Let A = B = 6Z. Then AnB = 6Z, but AB = 36Z. Therefore, all we have for both rings without unity and partial rings is that R/AnB = R/A x R/B. This is still a substantial result.

The only other proposition that may be important is how to get all of these from rings without unity. I have already explained how I believe you can.
=Bmsiv-

Finally found how to do it.
Let N = {x + y - x - y}. Then for any n in N, an = ax + ay - ax - ay = ax - ax + ay - ay = 0. One of the problems in the book by Herstein, this is a normal group. Clearly, it is a partial normal ideal, as all products are 0, which is in the group.

R/N is commutative, as for any a + N and b + N: a+N + b + N = (a+b)+N = (a + b - a - b + b + a) + N = (a+b-a-b)+N + (b+a)+N = N + (b+a) + N = N.

Therefore, we have that R can be created using a ring (without unity) and a normal group such that all the products of the normal group are 0.

A slight amendment to above, replace N with the commutator subgroup. After that, the partial ring is made trivial.


I went to page 26 of the blogs section, and the attached picture shows what was displayed.

Are there blogs that some users cannot see? Or is this a bug?

Ask one of the administrators. (For me, it says the same thing).

[quote="james4l"]My guess - the "754 blogs" it says enumerates all blogs, including blogs with the "Viewable only to friends/contribs/self" set. (Why someone would create a blog just to view themself is beyond me, but whatever.)[/quote]
An online journal.

Well, I don't see anything at all using the MathLinks board style.


Find positive integers a and b such that:
(cuberoota + cuberootb - 1) 2  = 49 + 20cuberoot6


real analysis
limit
real analysis solved

$a_n \; b_n$ reals sequences s.t. $\lim a_n = +\infty$

Prove that in any $]a;b[$ nonempty open  interval there exixt $x_0$

s.t. $\limsup_{n\rightarrow +\infty} \cos(a_nx_0+b_n)=1$

Nice problem, I hope the solution is good. Suppose the contrary and take the closed sets $ A_{k,p}={x\in (a,b), cos(a_nx+b_n)\leq 1-\frac{1}{p}}$ if ${n\geq k}$. Here k,p are natural numbers. The union of these sets contains an interval and this union is numerable thus by Baire's theorem one of them contains an interval. Thus there exists an interval non trivial I and two numbers k,p such that $ cos(a_nx+b_n)\leq 1-\frac{1}{p}$ for all $n\geq k$. This means that the length of the interval $ a_n I+b_n$ is always smaller than $2\pi$, which is impossible for large n.


[quote="ccu"]Two integers are randdomly selected from the set of integers greater than or equal to -5 and less than or equal to 5. The two numbers need not be different. What is the probability that the sum of the two integers is less than the product?[/quote][hide]

They must either be 1.both negative 2. both positive 3. negative and zero

so -5 -4 -3 -2 -1 0 1 2 3 4 5

The denominator is 55.

The top part is 1.10 2.10 3.5

So it is 15/55 which simplifies to 3/11. If I didn't make a mistake.[/hide]

I just did this problem, it was number 30 of some sprint round, and I got [hide]$\frac{50}{121}$[/hide] which was correct. :D [hide]The way you solve this is:
$11^2=121$, this is the denominator.  You get 25 possible ways to solve if they are all negative, 20 negative/positive/zero, and 5 only positive, or something like that.   $20+25+5=50, \frac{50}{121}=Answer$[/hide]

That's strange as I wouldn't put the denominator as 121 as that would count pairs like 4,5 twice (once as 5,4, once as 4,5) as well as any other pair like that. I would put a denominator of 66 as there are 11 pairs which only count once so subtract 11 from 121 and divide by 2 to get 55. Add the 11 back on and you have 66. Another looooooong painstaking way to get 66 is there are 11 pairings with -5, 10 with -4 (As -4,-5 is already counted), 9 with -3 (-3,-4 is already counted, so is -3,-5) etc. and if you add that up, you get 66.
[hide]
I got 29/66 as my answer.
6 combinations of the 11 pairings work out with -5, 5 combinations of the 10 pairings with -4, 4 of the 9 with -3, 3 of the 8 with -2, and 2 of the 7 with -1. That adds up to 20. Now none work with 0 (product = 0, sum = n), none work with 1. (product is always n, sum is always n+1) Only 3, 4, and 5 work for 2 giving you 23 total so far. All of the options of 3, 4, and 5, work giving you 6 more for a total of 29 making 29/66.
[/hide]

Well, the mathcounts answer key said that I was right...and they had my solution also... :D  EDIT:  You actually have to double count some pairs, as they represent different outcomes.  Example...picking 1 first and then -2 is different that picking -2 first and then 1.   There are therefore $11^2$  Possible outcomes

[hide]
There are 11 numbers, and thus $66$ combos (11+10+...+1)

You just have to list them out

-5 works with
-5,-4,-3,-2,-1,0

-4 works with
-4,-3,-2,-1,0

-3 works with
-3,-2,-1,0

-2 works with
-2,-1,0

-1 works with
-1,0

0 works with none

1 works with none

2 works with
3,4,5

3 works with
3,4,5

4 works with
4,5

5 works with
5

Total to get 29
$\frac{29}{66}$[/hide]

ccy, could you post the correct answer?  We have 2 different answers so far.

jhollenbeck, I am fairly sure that your answer is incorrect, because you said that there are 11*11=121 different combinations.

That would pair each number with each other number, but you would count each twice.
For example, for -5 you pair with -5,-4,-3,-2,-1,0,1,2,3,4,5
For -4, you pair with -5,-4,-3,-2,-1,0,1,2,3,4,5

As you can see, you pair -4 with -5 twice above.

So, you have to do it such that you pair -5 with -4,-3,-2,-1,0,1,2,3,4,5
You pair -4 with -4,-3,-2,-1,0,1,2,3,4,5
You pair -3 with -3,-2,-1,0,1,2,3,4,5
...
To remove doubles

actually, according to the answersheet, jhollenbeck's right. It was like 15 of some state sprint.


Consider a set $ S$ of $ 2006$ distinct numbers.A subset $ T$ of $ S$ is called stubborn if for every $ u,v$ (not necessarily distinct) in $ T$ ,the number $ u\plus{}v$ does not belong to $ T$. Prove that If $ S$ consists of $ 2006$ arbitrary positive integers then there exists a stubborn subset $ T$ of $ S$ having $ 669$ elements

Any solutions please   :blush:  :blush: ??


cSplash program held at NYU in March, where for a day they offer math and mathematical science courses to high school students free of charge at NYU.
[quote="[url=http://www.cims.nyu.edu/~csplash/index.php]cSplash[/url]"]
Courant Splash! is a one-day festival of classes in the mathematical and computer sciences, designed and taught by enthusiastic graduate and undergraduate students, faculty, and others associated with the Courant Institute. cSplash! is an opportunity to get your feet wet in subjects completely new to you. During the program, you can attend as many talks as you want on topics in pure math, applied math, computer science, and related fields. The program is free and is open to all students in grades 9-12 (or with equivalent mathematical background) with an interest in mathematical and computer sciences. Spaces are limited and are allocated on a first-come, first-serve basis so please register early to guarantee a spot.
[/quote]

Found out about it last year and went to it, it was quite fun.  :)

It was briefly mentioned in the NYC forum [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=54583303&t=75941]last year[/url].

Since no one is really posting here, I though I would mention that registration for classes starts in February.


I have a couple of questions about circle squaring sorry if this is in the wrong section.

I understand that it has been proven impossible with a pencil ruler and compass, but I wanted to find the simplest way to get as close as possible.

I think I can draw a square to match the area of a given circle as accuratly as I can draw another circle with the same area.

I found that if you draw a square within a circle the circle takes up 57% more area than the square, so would it not be possible to simple draw  a square with 57% more area? matching the area of the circle.

So you would square the circles radius, divide by two and square root the result, double this and square it to give you the area of the square drawn within the circle, just pythagoras really.

Then if this is a then (a/100*57 ) + a is the total area of the square matching the circle,

Depending on how many decimal places you want to work to you can draw a square with the same area as a given circle as accuratly as you can redraw another circle with the same area.

I have put together a very crude spreadsheet to demonstrate this.

http://www.mediafire.com/?zq3xtqnkm5m

The only problem is that it shows small differences between the area of the square produced and the circle is this due to the computer only working to a limited number of decimal places?

Any comments are welcome,

:)

I think this may be in the wrong section please could a moderator move it to the right place?


Find the greast value of $\frac{a}{c}+\frac{b}{d}+\frac{c}{a}+\frac{d}{b}-\frac{abcd}{(ab+cd)^{2}}$ , where $a,b,c,d$ are distinct reals satisfying $\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}=4$ and $ac=bd$.

[quote="VANCHANH123"][quote="toanhocmuonmau"]My solution.[/quote]
Chưa hết hạn gửi bài .. nên đừng thảo luận

[b]thay 4 bới k...............[/b][/quote]
No, It is exceed the limit  :wink: . I not so, I will not post.


geometry
geometric transformation
dilation
ratio

I need help on the following question and advice on how to approach similar types of the questions shown below or recommended sources to use. I could not seem to identify some of the substances.

Dilute hydrochloric acid was added to a metallic looking compound A (molecular weight 90.756). A colorless gas B with a characteristic, unpleasant odour was formed together with a pale green solution of the cation C.

   The gas B was burned in air to give another colourless gas D that turned yellow dichromate paper green. Mixing B and D gave a yellow solid element E. Depending on the mole ratios, E reacted with chlorine gas to give two chlorides, F and G in addition to hydrogen chloride (HCl).

    Addition of zinc metal to a sample of the green cation solution C gave a metal H (with electron configuration of [Ar]3d84s2.

Identify each of the substances and write balanced chemical equation for each reaction.

Thanks in advance.

[quote="Victornumber"]Addition of zinc metal to a sample of the green cation solution C gave a metal H (with electron configuration of [Ar]3d84s2.[/quote]

From the electron configuration we may conclude that H is nickel, and so $C = Ni^{2+}$. The observed reaction of C with zinc is in accordance with the reduction potentials of these two metals: $E^{0}Ni^{2+}/Ni =-0.23V$, $E^{0}Zn^{2+}/Zn =-0.76V$. This is also in accordance with the fact that Ni2+ ions in aqueous solution form the complex ions $[Ni(H_{2}O)_{6}]^{2+}$, which are responsable for the green color of their solution.

[quote="Victornumber"]Dilute hydrochloric acid was added to a metallic looking compound A (molecular weight 90.756). A colorless gas B with a characteristic, unpleasant odour was formed together with a pale green solution of the cation C.[/quote]

The unpleasant odour of the gas suggests that it might contains sulphur, and so compound A is NiS, nickel sulphide, which is in accordance with the given molecular weight. When this compound reacts with acid, there is the formation of sulphidric acid gas, $H_{2}S$, which is B.

[quote="Victornumber"]The gas B was burned in air to give another colourless gas D that turned yellow dichromate paper green. Mixing B and D gave a yellow solid element E.[/quote]

That sequence of reactions is known as [i]Claus Process[/i], which is used in oil industry to remove H2S from natural gas or from some destilation fractions of crude, turning it into elemental sulphur. The first reaction, where H2S (B) is burned in air, is

$2H_{2}S(g)+2O_{2}(g) \longrightarrow SO_{2}(g)+2H_{2}O(g)+S(l)$,

and so D is SO2, sulphur dioxide. The second reaction, where H2S reacts with SO2 (usually in the presence of a catalyst) is,

$2H_{2}S(g)+SO_{2}(g) \longrightarrow 3S(l)+2H_{2}O(g)$.

So E = sulphur. The "gas D that turned yellow dichromate paper green." part means that SO2 was able to reduce dichromate to the green ions Cr3+. Write this reaction as an exercise.

[quote="Victornumber"]Depending on the mole ratios, E reacted with chlorine gas to give two chlorides, F and G in addition to hydrogen chloride (HCl).[/quote]

Sulphur reacts with chlorine to form disulphur dichloride, $S_{2}Cl_{2}$, a yellow liquid used in the vulcanization of rubber. When S2Cl2 reacts with more chlorine, sulphur dichloride $SCl_{2}$ is formed as a red liquid. This last one was used in the production of [i]mustard gas[/i] by reaction with ethene.

[quote="Victornumber"]I need help on the following question and advice on how to approach similar types of the questions shown below or recommended sources to use.[/quote]

For example:

1. Knowledge of electron configurations and how to relate them with the Periodic Table.
2. Knowledge of redox reactions and how to balance them.
3. Common knowledge that, in general, sulphur compounds have bad odours.


trigonometry
algebra
functional equation
real analysis
real analysis unsolved

solve the differential equation:
f:R->R,
f(x)=f'(f(x))

 f(x)=x is valid, but i can't prove that :?

[quote="mrazvano"] f(x)=x is valid[/quote]
No, it's not. If $f(x)=x$, then $f'(x)=1,$ so $f'(f(x))=1,$ which is not $f(x).$

For a solution that is obviously valid, try $f(x)=\frac{x^2}2.$ But that's not the only solution. This is really more of a functional equation than a typical differential equation.

If $f$ is differentiable at $Im(f) = I$ an interval then:

 a. If $I$ is one point then $f = 0$
 b. If not for $u \in I$ we have $u= f'(u)$ then $f = \frac{u^2}{2} +c$ on $I$

 So, define $f$ as in b on an interval $I$.  The solution then depends on the way we expand $f$ contiuously on $\mathbb{R}$ and differentiably at the end points of $I$ so that $Im(f) = I$( but not very arbitrarily as it sounds )

- For example :
  $f(x)$ = $x^2$ for $x \geq 0$
  $f(x)$ = $x^4$  ( or $\sin^2(x)$ )for $x \leq 0$
   $I$ = $[0, \infty)$

 But it is quite hard to consider all cases!!!

I think, it is clear that $I\neq (a,b)$, there $a,b\in \mathbb{R}$, isn't it? Otherwise $Im(f)\neq I$.