Suppose that 2 defective Radio have been included in a shipment of 7 Radio, the buyers begin to test the 7 Radio one at a time.

(i)What is the probability that the last defective Radio is found on the fourth test?

(ii)What is the probability that no more than 4 Radio need to be tested to locate both of the defective Radio?

(iii)When given the exactly one of the two defective Radio has been located in the first two test, what is the probability that the remaining defective Radio is found in the third or fourth test?

(iv)Let the radom variable Y represent the number of defective Radio found after 3 Radio have been tested. Compute the probabilities for each value of Y.

(v)Let the random variable Z represent the number of the test on which the last defective Radio is located. Compute the probabilities for each value of Z.

(vi)If the cost of testing a Radio is Rm2 and the cost of repairing a defective is Rm4, find the expected total cost for testing and repairing the lot.

if the last defective radio is found on the 4th test we have $ \frac{3!}{2!} \equal{} 3$ ways to have 1 defective radio in the first three , out of a total of $ \frac{7!}{5!\cdot2!} \equal{} 21 \equal{} \frac{3}{21} \equal{} \frac{1}{7}$?

I do it same way as yours... but my friend said it is 1/5. How do other ppl think? I hope there is someone can teach me how to solve it.

ii. There are $ \displaystyle\binom{4}{2}\equal{}6$ ways to permute the two radios in the first four radios, as opposed to $ \displaystyle\binom{7}{2}\equal{}21$ overall ways, for a probability of $ \frac{6}{21}\equal{}\frac{2}{7}$.

iii. The last radio is in one of the last five test, so the probability is $ \frac{2}{5}$.

iv. $ Y\equal{}0$ means all the defective radios are in the last 4, and since there are $ \displaystyle\binom{4}{2}\equal{}6$ ways for the defective radios to be in the last four, and 21 total, the probability is $ \frac{6}{21}\equal{}\frac{2}{7}$.  Y=2 means all the defective radios are in the first 3, and since there are $ \displaystyle\binom{3}{2}\equal{}3$ ways for the defective radios to be in the first 3, and 21 total, the probability is $ \frac{3}{21}\equal{}\frac{1}{7}$.  Since Y=0,1,or 2, the probability of Y=1 is $ 1\minus{}\frac{2}{7}\minus{}\frac{1}{7}\equal{}\frac{4}{7}$.

how to get the answer for part (iii)?i can't get the answer 2/5...


Find the greatest value of positive integer $ x$ , such that the number

  $ A\equal{} 2^{182} \plus{} 4^x \plus{} 8^{700}$

 is a perfect square .

[hide]$ 2^{182}\plus{}4^{x}\plus{}8^{700}\equal{}2^{182}\plus{}2^{2x}\plus{}2^{2100}$.

If $ x\ge 91$, we can factor the expression as $ 2^{182}(1\plus{}2^{2x\minus{}182}\plus{}2^{1918})$.  Since $ 2^{182}$ is a perfect square, it is sufficient to consider $ 1\plus{}2^{1918}\plus{}2^{2x\minus{}182}$.

If $ x\equal{}2008$ the expression is $ 1\plus{}2^{1918}\plus{}2^{3834}\equal{}(1\plus{}2^{1917})^{2}$.

If $ x>2008$ then $ (2^{x\minus{}91})^{2}<1\plus{}2^{1918}\plus{}2^{2x\minus{}182}<(1\plus{}2^{x\minus{}91})^{2}$, so our expression cannot be a perfect square.

Therefore, (not surprisingly  :) ), the answer is 2008.[/hide]


calculus
integration
function
conics
ellipse
calculus computations

Evaluate the integral C of the function:

[(x^2 *y + y^3)*e^(xy) + x + y]/(x^2 + y^2) dx + [(x^3 + x*y^2)*e^(xy) - x + y]/(x^2 + y^2) dy

in the following cases:

(a) The curve C is given by the parametric equations:

x = sin(pi*t/2)*e^(t*sin^2(pi*t))
y = cos(pi*t/2)*e^(t*cos^2(pi*t))

with t E [0, 1]

(b) C is the ellipse 3x^2 + 2y^2 = 6 oriented in the counter-clockwise direction.

[quote="mathbeauty"]Evaluate the integral C of the function:

$ \frac{(x^2 y \plus{} y^3)e^{xy}\plus{} x \plus{} y}{x^2 \plus{} y^2} dx \plus{} \frac{(x^3 \plus{} xy^2)e^{xy}\minus{} x \plus{} y}{x^2 \plus{} y^2} dy$

in the following cases:

(a) The curve C is given by the parametric equations:

$ x \equal{} \ sin(\frac{\pi t}{2})e^{tsin^2(\pi t)}$
$ y \equal{} \ cos(\frac{\pi t}{2})e^{tcos^2(\pi t)}$

with $ t \in[0, 1]$

(b) $ C$ is the ellipse $ 3x^2 \plus{} 2y^2 \equal{} 6$ oriented in the counter-clockwise direction.[/quote]

On rearranging one can see the expression simplifying too

$ e^{xy}(ydx\plus{}xdy)\plus{}\frac{xdx\plus{}ydy}{x^2\plus{}y^2}\plus{}\frac{ydx\minus{}xdy}{x^2\plus{}y^2}$

$ \equal{}e^{xy}d(xy)\plus{}\frac{1}{2}\frac{d(x^2\plus{}y^2)}{x^2\plus{}y^2}\minus{}\frac{d(\frac{y}{x})}{1\plus{}\frac{y^2}{x^2}}$

On integrating one gets

$ e^{xy}\plus{}\frac{1}{2}log_{e}(x^2\plus{}y^2)\minus{}tan^{\minus{}1}(\frac{y}{x})\plus{}c$

Beyond this I am not able to simplify


Compute the largest integer $ n$ with $ 0<n\leq 72$ such taht the triangle with sides $ n$, $ 2n-3$, $ 3n-9$ has integer area.

If you think this belongs in Intermediate, feel free to move it..it was an old ARML relay problem..but its not that hard if you know what the area formula for a triangle is..and you dont need fancy formulas like Herons.

[quote="JavaMan"][hide="Herons"]
$ a = n, b = 2n-3, c = 3n-9, s = \frac{a+b+c}{2}$
$ A = \sqrt{s(s-a)(s-b)(s-c)}= 3|n-3|\sqrt{2(n-2)}$

For some integer $ d$, we must have the following to get an integer area:
$ \therefore 2(n-2) = d^{2}$
$ d$ must be divisible by 2, since the LHS is divisible by 2.
Let $ d = 2e$
$ n-2 = 2e^{2}\implies n = 2e^{2}+2$
Choosing the largest $ n$, which occurs when $ e = 52$ gives us the largest area of $ 1470$
[/hide][/quote]

yea thats right...i get the same thing (just checking my answer because I dont have answer key).  I didntuse Herons (used altitude and you get two right traingles, solved the height in terms of n).  But I also got 52. basically i drew the attitude to the side $ 3n-9$ and you get $ \sqrt{2(n-2)}$ is the height and if thats an itegner, and lessthan 72, n=52.

Just tried that method, works out too :)


analytic geometry
graphing lines
slope
function

[b]Let f(x) = x^2 - 3x - 4. [/b]

[b]Find the slope, m(h), of the secant line between the points A(1, f(1)) and B(1 + h, f(1 + h)). [/b]

This is what I've done so far.

f(1) = 1^2 - 3(1) - 4
f(1) = -6
So, A is the point (1, -6)

f(1 + h) = (1 + h)^2 - 3(1 + h) - 4
f(1 + h) = 1 + 2h + h^2 - 3 - 3h - 4
f(1 + h) = h^2 - h - 6
So, B is the point (1 + h, h^2 - h - 6)

Neways, how do I go about finding the slope? Was this work even necesary?

you're fine so far

now use the [hide="slope formula"]

$\frac{y_{1}-y_{2}}{x_{1}-x_{2}}$ is your desired slope.

substituting, we get

$\frac{(h^{2}-h-6)-(-6)}{(1+h)-1}=\frac{h^{2}-h}{h}=h-1$[/hide]

[hide]

We have the points $A(1,-6)$ and $B(1+h,h^{2}-h-6)$

The slope of the line between these points is
$\frac{(h^{2}-h-6)+6}{(1+h)-1}=\frac{h^{2}-h}{h}=h-1$

so the function $m(h)$ is $h-1$

which is actually first degree

[/hide]

yeah I really wasn't thinking then and forgot to reduce my fractions. edited.

so h-1 is the slope of the secant line between the points A and B?

cause that is what is seems to me, but for some reason I have the feeling I am wrong!

haha


topology
advanced fields
advanced fields unsolved

I'm going to assume that the reader is already familiar with the construction of Hausdorff measures.

Normally, the Hausdorff measure is defined in terms of coverings by arbitrary sets. So, we can define a similar measure by only allowing our covering sets to be balls. Otherwise the definition is the same as for Hausdorff measure. This measure is known as the spherical measure (for obvious reasons).

We can easily see that the $s$-dimensional Hausdorff measure and the $s$-dimensional spherical measure are equivalent measures: if $E \in \mathbb{R}^{n}$, then

$\mathcal{H}^{s}(E) \le \mathcal{S}^{s}(E)$
because balls are one possible covering, and the Hausdorff measure takes the inf, but also

$\mathcal{S}^{s}(E) \le 2^{s}\mathcal{H}^{s}(E)$
because for an arbitrary covering, we can cover each covering set with a ball of twice the diameter.

Can we obtain a better constant than $2^{s}$ for general sets $E$? What restrictions could we put on $E$ to make the constant better (like maybe to make it 1)?

one can improve a little the constant thanks to http://www.mathlinks.ro/Forum/viewtopic.php?t=124475  :)  What's the better constant ??

Well clearly for general $E$ your/grobber's result is the best possible, right? For example you could have a Sierpinski type thing. I think Besicovitch discusses such an example somewhere in his paper in Annals of Math 98, 1928 (though I haven't read that paper... :P).

I don't know an answer to my second question. I just thought of it as I was typing, and I haven't worked on it yet. But now that I give it a moment's thought, I expect it's actually a stupid question -- the Sierpinski example shows that even if your set is pretty nice and easy to study, it could still screw up the spherical measure thing.

what do you mean by "Sierpinski type thing" ?

Oh sorry, well the basic Sierpinski triangle starts with an equilateral triangle and deletes smaller equilateral triangles. So the best covering of it would be with equilateral triangles, which suggests that we couldn't do better than $2/\sqrt{3}$ times the Hausdorff measure if we only used balls. I think I read somewhere that Besicovitch discussed this in his paper in the Annals.

Presumably, you could extend the Sierpinski construction to higher dimensional Euclidean space, too.

http://en.wikipedia.org/wiki/Sierpinski_triangle


Each team has 2 days to decide what it's next move will be. All team members should send their moves to their team captain, and if at least 2/3 (4 people) agree on the move, then it should be done. The move should be posted in algebraic notation (e2-e4, Nb1-Nc3, etc.). I will post a picture of the current position after each move. Please try to decide as fast as possible to keep the game moving, but don't rush if you need the time. Make sure at least 2/3 of your team agrees with the move you have decided on to avoid arguments and such. No more players as of now for this game unless somebody stops playing. If anybody leaves for a while, their team should continue without them, but play may be halted/new players may be accepted if a team has less than 4 players at any point.

To sign up for Game 2, or have any questions, go [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=156818]here[/url].

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

[b]Current Board Position as of 9/18/07[/b]

[img]http://img2.freeimagehosting.net/uploads/a988b36cd8.jpg[/img]

[b]It is white's move.[/b]

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

White:

[list]nayel (captain)
now a ranger
Nerd_of_the_Ages
BoesFX
shaggy75
mz94[/list]
Black:

[list]cincodemayo5590 (captain)
Temperal
7h3.D3m0n.117
i_like_pie
goldendomer
pieterminate[/list]
Good luck to everybody and let the game begin! :)

White, you have 2 days to decide what your first move will be.

As a guide to everyone, please use a chessboard image generator to display the board [b]after each ply[/b], such as [url=http://www.apronus.com/chess/wbeditor.php]this one[/url]. Click to move the pieces on the board as usual, then click "Make Diagram" on the top. Right click the image that's produced, copy the [b]whole[/b] image url, and place the image after your move statement.  :)

The board right now:
[url=http://www.apronus.com/chess/diagramik.php?a1=whiterookondark.gif&a2=whitepawnonlight.gif&a3=darksquare.gif&a4=lightsquare.gif&a5=darksquare.gif&a6=lightsquare.gif&a7=blackpawnondark.gif&a8=blackrookonlight.gif&b1=whiteknightonlight.gif&b2=whitepawnondark.gif&b3=lightsquare.gif&b4=darksquare.gif&b5=lightsquare.gif&b6=darksquare.gif&b7=blackpawnonlight.gif&b8=blackknightondark.gif&c1=whitebishopondark.gif&c2=whitepawnonlight.gif&c3=darksquare.gif&c4=lightsquare.gif&c5=darksquare.gif&c6=lightsquare.gif&c7=blackpawnondark.gif&c8=blackbishoponlight.gif&d1=whitequeenonlight.gif&d2=whitepawnondark.gif&d3=lightsquare.gif&d4=darksquare.gif&d5=lightsquare.gif&d6=darksquare.gif&d7=blackpawnonlight.gif&d8=blackqueenondark.gif&e1=whitekingondark.gif&e2=lightsquare.gif&e3=darksquare.gif&e4=whitepawnonlight.gif&e5=darksquare.gif&e6=lightsquare.gif&e7=blackpawnondark.gif&e8=blackkingonlight.gif&f1=whitebishoponlight.gif&f2=whitepawnondark.gif&f3=lightsquare.gif&f4=darksquare.gif&f5=lightsquare.gif&f6=darksquare.gif&f7=blackpawnonlight.gif&f8=blackbishopondark.gif&g1=whiteknightondark.gif&g2=whitepawnonlight.gif&g3=darksquare.gif&g4=lightsquare.gif&g5=darksquare.gif&g6=lightsquare.gif&g7=blackpawnondark.gif&g8=blackknightonlight.gif&h1=whiterookonlight.gif&h2=whitepawnondark.gif&h3=lightsquare.gif&h4=darksquare.gif&h5=lightsquare.gif&h6=darksquare.gif&h7=blackpawnonlight.gif&h8=blackrookondark.gif]Here.[/url]

I can't get the in-post images to work. What's the code for it again?  :)

[b]Edit:[/b]

Black's first move:
[b]c7-c5[/b]

[url=http://www.apronus.com/chess/diagramik.php?a1=whiterookondark.gif&a2=whitepawnonlight.gif&a3=darksquare.gif&a4=lightsquare.gif&a5=darksquare.gif&a6=lightsquare.gif&a7=blackpawnondark.gif&a8=blackrookonlight.gif&b1=whiteknightonlight.gif&b2=whitepawnondark.gif&b3=lightsquare.gif&b4=darksquare.gif&b5=lightsquare.gif&b6=darksquare.gif&b7=blackpawnonlight.gif&b8=blackknightondark.gif&c1=whitebishopondark.gif&c2=whitepawnonlight.gif&c3=darksquare.gif&c4=lightsquare.gif&c5=blackpawnondark.gif&c6=lightsquare.gif&c7=darksquare.gif&c8=blackbishoponlight.gif&d1=whitequeenonlight.gif&d2=whitepawnondark.gif&d3=lightsquare.gif&d4=darksquare.gif&d5=lightsquare.gif&d6=darksquare.gif&d7=blackpawnonlight.gif&d8=blackqueenondark.gif&e1=whitekingondark.gif&e2=lightsquare.gif&e3=darksquare.gif&e4=whitepawnonlight.gif&e5=darksquare.gif&e6=lightsquare.gif&e7=blackpawnondark.gif&e8=blackkingonlight.gif&f1=whitebishoponlight.gif&f2=whitepawnondark.gif&f3=lightsquare.gif&f4=darksquare.gif&f5=lightsquare.gif&f6=darksquare.gif&f7=blackpawnonlight.gif&f8=blackbishopondark.gif&g1=whiteknightondark.gif&g2=whitepawnonlight.gif&g3=darksquare.gif&g4=lightsquare.gif&g5=darksquare.gif&g6=lightsquare.gif&g7=blackpawnondark.gif&g8=blackknightonlight.gif&h1=whiterookonlight.gif&h2=whitepawnondark.gif&h3=lightsquare.gif&h4=darksquare.gif&h5=lightsquare.gif&h6=darksquare.gif&h7=blackpawnonlight.gif&h8=blackrookondark.gif]Image here.[/url]

2nd move(white): [b]Ng1-f3[/b].

[img]http://img2.freeimagehosting.net/uploads/e6975b95d6.jpg[/img]

Could you guys make a new post for this, cause I'd like to play...

You can post all your thoughts and suggestions [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=156818]here[/url]. This thread is only for the moves of game 1. Besides if you want to play you can join in the next game, because this one's already started. Also, you can join if someone else refuses to continue.

ps. Black, you have one more day to make your next move.

Sorry this is late!

Black's move:
[b]e7-e6[/b]

[img]http://i61.photobucket.com/albums/h70/cincodemayo5590/chess.jpg[/img]

Move 3(White): [b]Nb1-c3[/b]

[img]http://img2.freeimagehosting.net/uploads/4854498037.jpg[/img]

sorry for being late but most of our team members weren't responding.

im in on game 2

i havent been on this forum before

If you want to play in future games, please post [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=894960#894960]here[/url]. This topic is reserved for moves only.

Black's Move, Turn 3:

[b]Nb8-c6[/b]

[img]http://i61.photobucket.com/albums/h70/cincodemayo5590/chess-1.jpg[/img]

I personally think Nf6 might have been better now, but oh well.

[b]Temperal[/b], PLEASE, keep comments off of this thread. It is for [b]game moves[/b] only. And especially keep comments on strategies off of this thread. Keep those to PM with your TEAM MEMBERS.

Oy.

Current Board:

[img]http://i61.photobucket.com/albums/h70/cincodemayo5590/chess-1.jpg[/img]

4th move(White): [b]d2-d4[/b]

[img]http://img2.freeimagehosting.net/uploads/325806d8d8.jpg[/img]

The games will be held on the Wiki, with an article dedicated to each game.

Please post all further questions/comments about the upcoming tournament in the designated thread (or in the discussion section of the Wiki project page), a link to which I included in my previous post. Make sure, however, that you read over all existing questions/comments to avoid repeating something already posted.

[quote="i_like_pie"]
Please post all further questions/comments about the upcoming tournament in the designated thread (or in the discussion section of the Wiki project page), a link to which I included in my previous post. Make sure, however, that you read over all existing questions/comments to avoid repeating something already posted.[/quote]

Read.

Nice way to finish the game. :)  It was real fun wasn't it? Thanks to all members of both teams who played till the end. Personally I am very sorry for being unable to tune in for the last few days. :(  So how are we gonna celebrate our victory mates? :roll:

You can celebrate when (if) you win the tournament. :D

Oh we will win the tournament! So all teams, beware of team gamma! :P

*is bewaring*

Team Beta doesn't exactly seem like the strongest team; we'll probably lose the first round....  :(

@ Temperal - It's Swiss format, so you play the first four games, and then the teams that advance to the next rounds are decided.

Can I play? My rating in real life is aprox 1100 and my school team is 4th in NY State(after PS 318), and I've won a few tournaments.

It seems like a nice game, dependin on the point of view. anyway, beware of team alpha, we r goin to rule

Lol nice, Charles. Did you get owned at Riderwood?  :P

I think black's demise started with the move early on that lost us tempo. We could have been in a nice solid Sicilian position were it not for a pawn move I believe.

Move 3 Nc3 is only for Sicilian O'kelly! Play d4!

I didn't agree with black's second move: e6. There is no development and the pawn serves no purpose.

Um, I believe this was the move, as it cost a pawn:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=159073#897992

No, the knight was a good move but it was one move late. The reason? e6.


superior algebra
superior algebra unsolved

I'm looking for a nice example of a non-split extension that would help motivate the definition of a split extension. But I'm not an expert in group extensions.

I'm using the definition of extension by short exact sequences and our eventual goal is the theory of p-groups. So an example that is a p-group would be nice. Any suggestions welcome.

Cheers,

-Q

Classical example: The quaternion group $Q_{8}$. It is the smallest group that is not a semidirect product.


If a regular $ n$-gon is inscribed in a circle of radius 1 with sides $ S_1,S_2,S_3,...$ what is the product of $ \prod_{i\equal{}1}^{n} |S_i S_1|$? It seems that this product does NOT depend on $ n$...

I can do the case $ n\equal{}4$ which of course is easy. But the case $ n\equal{}3$ seems impossible. Where do I begin?

Thanks!

hm. can't i just use some orthic triangle thing? How would you use complex numbers here?

for the case of $ n \equal{} 3$, go to this page,

http://mathworld.wolfram.com/Altitude.html

they give the relations between the sides of a triangle, their altitudes, and the radius of the circle circumscribed about that triangle.  Also, note that $ a \equal{} b \equal{} c$ and $ sinA \equal{} sinB \equal{} sinC \equal{} \frac{\sqrt{3}}{2}$

This should help you compute that answer.

this plainly has something to do with $ n^th$ roots of unity of the $ n$-gon.

After a little more work, I found this.

The interior angle of an n-gon is equal to $ \frac {(n \minus{} 2)\pi}{n}$ in radians.  Then, for n > 3, you can divide the polygon into right triangles.  The hypotenuse is the radius = 1, and the leg we want is half the length of one side.  hence, we can find the length of any side with the formula

$ \frac {S_n}{2} \equal{} cos(\frac {(n \minus{} 2)\pi}{2n})$  The reason we have 2n in the denominator this time is because the angle it split in half when we draw our triangles.  Then we have

$ S_n \equal{} 2cos(\frac {\pi}{2} \minus{} \frac {\pi}{n}) \equal{} 2(cos(\frac {\pi}{2})cos(\frac {\pi}{n}) \plus{} sin(\frac {\pi}{2})sin(\frac {\pi}{n}))$

$ S_n \equal{} 2sin(\frac {\pi}{n})$

Since the regular n-gon has equal sides, then the product over them is simply $ S_n^n$, where $ S_n$ is the length of one side of an n-gon.

Hence we have $ 2^{n}sin^n(\frac {\pi}{n})$

Let me know if I made a mistake


Find all f:R to R such that :
  (x-y)f(x+y) - (x+y)f(x-y) = 2y(x^2-y^2) for all x,y in R

Let $z=x+y,t=x-y$ then $tf(z)-zf(t)=(z-t)zt$, t=1 give $f(z)=az+z(z-1)$, were a=f(1).


http://www.artofproblemsolving.com/Forum/viewtopic.php?t=18642

There are positive integers $k$, $n$, and $m$ such that

\[ \frac{19}{20} < \frac{1}{k} + \frac{1}{n} + \frac{1}{m} < 1. \]
What is the smallest possible value of $k+n+m$?

Can you do this one without guess-and-check?

Billy

I would try to set k, n, m equal, and as small as possible (i.e. the fractions are as large as possible, close to 1). But the last time I tried to do that, someone else proposed a good solution with Am-Gm (arithmetic-geometric mean inequality), so try that too. :P


A triangle has lattice points as vertices and contains no other lattice points. Prove that its area is $\frac{1}{2}$.

Pick's Theorem trivializes the problem ( doesn't it?)

Method 1.
 is by simple application of Pick theorem.
$ A(\Delta)\equal{}  i(\Delta) \plus{}  \frac{1}{2} b(\Delta) \minus{} 1$
so
$ A(\Delta)\equal{}  \frac{3}{2} \minus{}1\equal{} \frac{1}{2}$

Method 2
 is as follows:
There is a triangle $ \Delta OPQ$ with $ O(0,0)$ , $ P(x_1,y_1)$ and $ Q(x_2,y_2)$.
We know that points  $ P(x_1,y_1)$  and $ Q (x_2,y_2)$ are visible from the origin $ O(0,0)$.
So, $ (x_1, y_1)\equal{}1$ , $ (x_2,y_2)\equal{}1$. That means that  $ \frac{x_1}{y_1}$, and $ \frac{x_2}{y_2}$ are irreductible.
  So,  $ \frac{x_1}{y_1}$ and $ \frac{x_2}{y_2}$ are the terms of Farey sequence.
But if $ \frac{x_1}{y_1}$, and $ \frac{x_2}{y_2}$ are two successive terms of Farey sequence, then
 $ x_1 \cdot y_2 \minus{} x_2 \cdot y_1\equal{}1$.
From the other part  a surface of an arbitrary  trangle is given by the formula
$ A(\Delta)\equal{} \frac{1}{2} \mid x_1 \cdot y_2 \minus{} x_2 \cdot y_1 \mid$
so
$ A(\Delta OPQ ) \equal{}  \frac{1}{2}$   and it is done.


linear algebra
matrix
Ring Theory
superior algebra
superior algebra solved

Let I be an ideal of a ring R. Show that M_n(R), the set of matrices with coefficients in I is an ideal of M_n(R).

I dont know how to show this - any help would be appreciated.

Routine, use defined of ideal of a ring and multiply two matrix :wink:

[quote="amfulger"]Is it true that any ideal of $M_{n}(R)$ is $M_{n}(I)$ for some ideal $I$ of $R$?
I think it is.[/quote]Unfortunately it is not true in general. The ring has to be unitary for this proposition to be true.

I am used to rings having unity. Sorry about that.

Can you post prove of statement that, amfulger and The Captain?

Let $I\subset R$ be an ideal of $R$. I'm assuming you can do the $M_{n}(I)\subset M_{n}(R)$ is a subring part. It is a ring and a subset. Now let $A=(\alpha)_{ij}\in M_{n}(I)$ and $B=(\beta)_{ij}\in M_{n}(R)$, then
$(AB)_{ij}=\sum_{r=1}^{n}\alpha_{ir}\beta_{rj}$. By this you can say that each entry of $AB$ is in $I$, since you multiply $\alpha_{ir}\in I$ and $\beta_{rj}\in R$ to get something in $I$ (because $I$ an ideal). Then sum $r=1$ to $r=n$, which is just adding things in $I$, to get your final entry in $I$. So each entry of $AB$ is in $I$, so $AB\in M_{n}(I)$.

Reverse the multiplication to see that it is a two-sided ideal (which I'm assuming by the generic word ideal) and possibly work that out the above a little more rigorously for a final proof, but that is the idea.

@N.T.TUAN Let $J$ be an ideal of $M_{n}(R)$. What you are supposed to show is that $J=M_{n}(I)$ for some $I\triangleleft R$. The natural choice is to let $I$ be the set consisting of all entries of elements of $J$. Clearly $J\subset M_{n}(I)$. Now you should use the fact that $R$ is unitary. I won't do this part for you! Thats quite easy if you know [i]how[/i] elementary matrices work.

A ring without i (the multiplicative identity ) is sometimes called a rng. Why the blazes?


linear algebra
matrix
vector
algebra unsolved
algebra

Hi friend,

I want to proof  the set of orthognal matrices are compact?Who can help me.

They form a closed and bounded set in $\mathbb R^{n^2}$.

Identify the matrix $(a_{ij})$ with the vector $(a_{11},a_{12},\ldots,a_{1n},a_{21},\ldots,a_{nn})$ in $\mathbb R^{n^2}$. We have to show that the set consisting of these vectors is closed and bounded.

It's clearly bounded, since the norm of each such vector is $\sqrt n$. It's also closed, because if a $(A_k)_k$ of orthogonal matrices tends to another matrix $A$, then, since we have $A_kA_k^T=I_n,\ \forall k$, we must also have $AA^T=I_n$, meaning that $A$ is also orthogonal.

Now, since in Euclidean spaces compactness of a set is equivalent to the set being closed and bounded, we're done.

Sir,

Thank you for your explanation.I undrestood why they are bounded  :D  but the closeness of them are not clear to me   :(  (sory if I am not intelligent enough.Actually I am an engineer not a mathematician)  :blush:

First of all, you need not call me sir :).

Think about it like this:

Consider a sequence $A_n=(a_{ij}^{(n)})_{i,j}$ of orthogonal matrices which tends to the matrix $A$. Let $i,j\in\overline{1,n}$, and let $v_i^{(n)},v_j^{(n)}$ be the $i$'th and $j$'th rows of $A_n$, and let $v_i,v_j$ be the $i$'th and $j$'th rows of $A$. Since the usual scalar product is continuous, it means that from $v_i^{(n)}\to v_i,v_j^{(n)}\to v_j$ we get $(v_i^{(n)},v_j^{(n)})\to(v_i,v_j)\ (*)$, where $(x,y)$ is the scalar product of $x$ and $y$. Since $A_n$ is orthogonal for all $n$, we have $(v_i^{(n)},v_j^{(n)})=\delta_{ij}\ (**)$, where $\delta_{ij}=1$ if $i=j$ and $0$ otherwise. From $(*),(**)$ we get $(v_i,v_j)=\delta_{ij},\ \forall i,j\in\overline{1,n}$, which means that $A$ is orthogonal.


A bar of mass $ M$ and length $ L$ is leaned against a vertical wall and a block, as shown in the figure. The distance between the wall and the block is $ d$. If the bar is in equilibrium, find its angle of inclination $ \theta$.

[img]http://img522.imageshack.us/img522/1489/prob1ri2.jpg[/img]

I guess I should use torques and the like, but I got confused with the normal forces... :maybe: Thanks in advance for your help. :)

We have there the equilibrium of a rigid body. There are two conditions that must be met in order for a rigid body to be in equilibrium:

1) The sum of all forces applied to the body must equal zero.
2) The sum of the torques of all forces applied to the body, relative to any point, must be zero.

So begin by asking yourself "What are the forces applied to the body?".

Side question: is there friction between the wall and the bar?

There is no friction. I forgot to point out that! :oops:

It's all clear now. My answer is $ \theta \equal{}\arccos\sqrt [3]{\frac{2d}{L}}$. Thanks for your enlightening comment, Carcul. :wink:

[quote="pardesi"]ar u sure that there is no friction between the rod and the block and block and the ground.if so then this system can't be in equilibrium.because the normal force of the rod on the block which has a horizontal component makes it move forward unopposed by any friction[/quote]

Yes, but the normal force exerted by the wall on the rod also has a horizontal component... :)


Let A(r) denote the number of lattice points inside a circle with the radius r with it's middle in origo. Evaluate k = \lim_{r \to \infty} A(r)/r^2 then let B(r) = A(r) - kr^2. For what values of a does B(r)/r^a converge when r goes to infinity? (The smaller a you can get, the better)

[hide]I haven't solved it yet, but k = pi, right? (since the number of lattice points is approximately equal to the area of the circle and becomes more accurate for large r). If this is right then B(r) is just the amount of error in the approximation of A(r) with its area.[/hide]

<b>Conjecture</b>: let P be a region* of k-space (in our problem, P is the unit disc in R^2). Define A(n) to be the number of lattice points in nP, that is, P stretched by a factor of n in all directions. Then,

[tab]A(n) ~ c * n^k + 1/2 * d * n^{k-1}, where c is the k-dimensional volume of P, and d is the (k-1)-dimensional (codimension 1) volume of the boundary of P.

Assume the validity of this conjecture. We have in particular, for our case,

[hide]A(n) ~ Pi * n^2 + 1/2 * 2Pi * n.

Thus, B(r)/r^a diverges if a < 1, converges to Pi if a = 1, and diverges if a > 1.[/hide]

Someone slap me if I'm wrong.

* The region should probably have some nice properties like measurability and the likes.

One question about it: why the 1/2 in front of the d * n^(k-1)?

Boris's conjecture was too optimistic.  It is known that |B(r)| :le: Cr46/73 for some constant C.  It is also known that the exponent can't be replaced by 1/2.  The best exponent is not yet known.  See [url]http://mathworld.wolfram.com/GausssCircleProblem.html[/url].

Thanks Ravi. I think the conjecture [b]is[/b] true for polygonal regions and unions thereof, however. But I've been wrong in the past.

So I'm "new" to the AOPS page, in some ways. (You wouldn't have thought it by my join date.) Are unsolved problems common posts here?

froufrou4189, if you think about taking two regions and joining them at their boundary, each region should get about half of the points "near" the boundary to make everything work out. Here're some fun examples for cases where this formula [i]does[/i] work:
A. Unit line segment [0, 1] stretched by a factor of n has n+1 lattice points within it. This is 1 * n^1 because of the length 1 of the segment plus 1/2 * 2 * n^0 because of the segment's dimension zero volume of 2.
B. Unit cube [0,1]^3 stretched by a factor of n has n^3 + 3n^2 + 3n + 1 lattice points in it. Note the coefficients of 1 and 1/2 * 6.
C. This isn't the same, but it is related, I believe. The sum of the first n k-th powers is
S(n) = \sum_{i=1}^{n}{i^k} ~ 1/(k+1) * n^{k+1} + 1/2 * n^k. The 1/2 coefficient is really cool; few people know about it. The 1/(k+1) follows from calculus.


Hi all, the IBEROAMERICAN OLYMPIAD 2005 Will be held in Cartagena, Colombia next week, I will post the problems here and discuss results.

Today the seminar ended.

We had Talks by Patricia Fauring, Titu Andreescu, Andy Liu, Maria Losada and Eduardo wagner

It was a great academical activity and i Will post something about it later

pascual it would be great if you could post the topics and the most challenging, interesting and beautiful problems disucssed (of couse when you have time), and thanks!

Hi,

I posted Iberoamerican Olympiad problems (first day) [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=53808]here[/url].


Divide 18 balls including 5 red balls, 6 blue balls and 7 yellow balls for 9 students, each student receives two balls which are not in the same color. A and B are two of these students. What is the probability that A and B have the same pair of balls?

pick 7 students to get a yellow ball. then the only possible way for no one to get two of the same color is to give four of them blue balls and three of them red balls.. leaving two pairs of red/blue.  thus there is only one way do distribute these pairs..

4 YB 3 YR 2 RB.

so.. the probability of getting two students to match..

$ P\equal{}\frac{4}{9}\cdot\frac{3}{8}\plus{}\frac{3}{9}\cdot \frac{2}{8}\plus{}\frac{2}{9}\cdot \frac{1}{8}\equal{}\frac{20}{72}\equal{}\frac{5}{18}$


A cannon and a supply of cannon balls are inside a sealed railroad car with length $L$ . The cannon is placed inside the left most of the railroad car and fires to the right resulting the car recoiling to the left . The cannon balls remain in the car after hitting the far wall . Show that no matter how the cannon balls are fired ,the railroad car cannot travel more than $L$ ,assuming it starts from rest .

Since there's no outside force acting on the railroad car, it's center of gravity won't move relative to the outside world. It can, however, move relative to the car and the distance the center of gravity has traveled relative to the car will be equal to the distance the car has traveled relative to the outside world (because the center of gravity remains stationary realtive to the outside world, remember?). Also, the center of gravity of the car can't be outside of the car so the center of gravity can travel at most $L$ relative to the car.

bus ,

I still cant figure out any connection between your explaination and the cannon ball thing in the question  :(  . How is it that the center of gravity can travel is at most $L$ relative the the car . $L$ is not the length travelled by the car but instead it is the length of the car .

Shyong,
the center of mass of the car as whole can NOT move at all. It is just that we can rearrange things inside the car so that the center of gravity is on one side or more on the other. But the center of gravity of the car must be INSIDE the car. So, you can imagine it as having a fixed point C, which is the center of gravity of the car and which does NOT move relative to the ground. You can move the car, but C must remain inside. Therefore the range of possible movement of the car is between "having C on the left side" and "having C on the right side" ... moving inside this range, the car cannot move more than L.

Kubus ,

which means if the cannon balls is fired to the right , its center of gravity will somehow swift to the right . And the more cannon ball you fire , the more it is swifted to right until the maximum possible of the center of gravity can travel relative to the car which is the right most inside the car? Is this what you are trying to say?

Well , im thinking that say if a person is standing on a skateboard and try to fire his rifle to the right , will his skateboard keeps moving to the left assuming the road is smooth . If it is true , why couldnt the same situation being applied to the above question ?

This can't be applied because when the cannon ball hits the opposite side of the car it stops the car. And what kubus said is what I tried to tell you but much better explained, I'm sorry for being so confusing :).

Does this one work?

Consider the cannonball first.  since displacement=velocity*time, then let
$L=vt$ or $t=\frac{L}v$

Now, similarly, consider the carriage, for some displacement "x", and a velocity "u", the time travelled will be same (when the cannonball collides with the wall again, the train will stop, and this must be true due to the law of conservation of momentum).  So, $x=ut$ or $t=\frac{x}u$

equating, $\frac{L}v=\frac{x}u$ or $x=\frac{Lu}v$
But, $u \leq v$, since the momentum of the carriage will include the mass of the cannonball too (and if we consider the cannonball's mass negligible, then the distance travelled is "L")

Note: "u" is in fact negative, and so the displacement of the carriage is to the left.

Hm, something like that, yes... but you need to take into account that we can fire more than 1 ball. There's no difference really, it's just that this may not be a completely correct way to express this.... :roll:


A jar contains 60 marbles, each colored either green or yellow, all the same size. There are twice as many green marbles as yellow in the jar. How many yellow marbles must be added to the jar so the probability of choosing a green marble is $ \frac12$?

There are 40 green and 20 yellow marbles.  Then, you must add 20 yellow marbles for the probability of choosing a green marble to be 1/2.


Let a,b,c,d be non-negative real numbers such that
$a+b+c+d=1$
Show that
$abc+bcd+cda+dab\le \frac{1}{27}+\frac{176}{27}abcd$

There are a plenty of similar inequalities of the forum.
Check http://www.mathlinks.ro/Forum/viewtopic.php?t=28455


calculus
integration
function
calculus computations

Define the function $f(x)=\begin{cases}\ \ \ \ \ \ 1\ \ \ \ (-1\leq x\leq 2)\\-x+3\ \ (\ 2\leq x\leq 3)\end{cases}$

Evaluate $\int_{-1}^{3}|1-x|f(x)\ dx.$

Typing
\[\int_{-1}^{3}\text{Abs}[1-x] \star \text{Piecewise}\left[ \Big\{ \left\{ 1,-1 \leq x \leq 2 \right\}, \left\{-x+3, 2 < x \leq 3 \right\}\Big\}\right] \, dx \]
into Mathematica (5.2) I get the answer $\frac{19}6$.


inequalities
function
inequalities proposed

Prove that $ \frac{a^4\plus{}b^4\plus{}c^4}{ab\plus{}bc\plus{}ca}\plus{}\frac{3abc}{a\plus{}b\plus{}c}\ge \frac{2}{3}(a^2\plus{}b^2\plus{}c^2)$ for $ a,b,c>0$

[quote="N.N.Trung"]Prove that $ \frac {a^4 \plus{} b^4 \plus{} c^4}{ab \plus{} bc \plus{} ca} \plus{} \frac {3abc}{a \plus{} b \plus{} c}\ge \frac {2}{3}(a^2 \plus{} b^2 \plus{} c^2)$ for $ a,b,c > 0$[/quote]
Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ ab\plus{}ac\plus{}bc\neq0.$
This inequality is fifth degree. Thus, it equivalent to $ f(w^3)\geq0,$ where $ f$ is linear function
and $ a\plus{}b\plus{}c\equal{}3u,$ $ ab\plus{}ac\plus{}bc\equal{}3v^2,$ $ abc\equal{}w^3.$
Hence, for the proof enough to check two cases:
1)$ c\equal{}0,$ $ b\equal{}1$;
2)$ b\equal{}c\equal{}1.$

Multiplying by $ 3(ab \plus{} bc \plus{} ca)$ yields that the inequality is equivalent to

$ 3\sum_{cyc}a^4 \plus{} \frac {9abc(ab \plus{} bc \plus{} ca)}{a \plus{} b \plus{} c} \ge 2(a^2 \plus{} b^2 \plus{} c^2)(ab \plus{} bc \plus{} ca)$
Using schur inequality we obtain

$ 2\sum a^4 \plus{} 4abc(a \plus{} b \plus{} c) \ge 2(a^2 \plus{} b^2 \plus{} c^2)(ab \plus{} bc \plus{} ca)$

And thus it enough to check that :

$ \sum a^4 \plus{} \frac {9abc(ab \plus{} bc \plus{} ca)}{a \plus{} b \plus{} c} \ge 4abc(a \plus{} b \plus{} c)$

$ \sum(a^4\plus{}b^4\minus{} 2a^2b^2) \plus{}\sum(a^2b^2 \minus{}2ab^2c \plus{}b^2c^2)\plus{}\frac{6abc\left[3(ab\plus{}bc \plus{}ca) \minus{} (a \plus{}b \plus{}c)^2\right]}{a \plus{}b \plus{}c}\ge 0$

or equivalently
$ \sum_{cyc}\frac {(a \minus{} b)^2\left[(a \plus{} b)^2 \plus{} c^2)(a \plus{} b \plus{} c) \minus{} 6abc\right]}{a \plus{} b \plus{} c}$$ \ge \sum_{cyc}\frac {(a \minus{} b)^2\left[(a^2 \plus{} b^2 \plus{} c^2)(a \plus{} b \plus{} c) \minus{} 6abc\right]}{a \plus{} b \plus{} c} \ge 0$

Acording to $ (a \plus{} b \plus{} c)(a^2 \plus{} b^2 \plus{} c^2) \ge 9abc > 6abc$


calculus
integration
probability and stats

Please, check the following thoughts:
I want to find  first and second moments of $ \xi\equal{}\int\limits_{0}^{T} W_t dt$
$ \mathbb{E}[\xi]\equal{}\int\limits_{0}^{T}E[W_t]dt\equal{}0$, (we can change the order of integration because of $ W_t$ is in $ L_2$)
$ \mathbb{E}[\xi^2]\equal{}\mathbb{E}[\int\limits_{0}^{T} W_t dt \cdot \int\limits_{0}^{T} W_s ds]\equal{}\int\limits^{T}_{0}\int\limits^{T}_{0}\mathbb{E}[W_t W_s]dtds\equal{}\int\limits^{T}_{0}\int\limits^{T}_{0} min\{t,s\}dtds\equal{}...$
$ \mathbb{E}[\xi^2]\equal{}\mathbb{E}[T^2W_T^2\minus{}2TW_T\int\limits^{T}_{0}t dW_t\plus{}\left(\int\limits^{T}_{0}t dW_t\right)^2]\equal{}...$
the firts and the third  summand can be easily calculated.How can  the second be calculated?

\[ E[W_T\int\limits^{T}_{0}t dW_t] \equal{} E[\int_0^T 1 dW_t \cdot \int_0^T t dW_t] \equal{} \langle \int_0^T 1 dW_t \cdot \int_0^T t dW_t \rangle \equal{} \int_0^T t \, dt\]

can you please explain the last equality? :blush:
ah, $ dW_t dW_t\equal{}dt$ ,am I right?

1: if you have two continuous martingales $ M$ and $ N$, then $ MN\minus{}\langle M,N \rangle$ is also a (local) martingale so that
\[ E[M_t N_t]\equal{}E[\langle M,N \rangle_t \minus{} M_0 N_0]\]

2: if $ M,N$ are two continuous martingales and $ A_t, B_t$ are two previsible processes (+appropriate integrability conditions):
\[ \langle \int_0^t A_s \, dM_s, \int_0^t B_s \, dN_s \rangle \equal{} \int_0^t A_s B_s \, d\langle M_s, N_s \rangle \]

so that in your case I just used the fact that $ \langle W, W \rangle_t \equal{}  t$. Does this help ?

yes, I' ll read smthg about it and everythig will be clear.thanks a lot.


quadratics
arithmetic sequence
algebra
difference of squares
special factorizations

I know that the Chapter Target is really easy but not for me so please help.

could you post the problems, maybe how far uve gotten?

3.The product of two consecutive negative integers is 2550. What is the sum of the two integers?

4.What is the positive difference between the sum of the first 20 positive multiples of 5 and the sum of the first 20 positive, even integers?

Hint for 3 (I forget the formula for number 4)
3.[hide]so you'd have two integers, who lets pretend r positive. so x(x+1)=2550 as they are consecutive. solve as a quad formula[/hide]

edit: heres the formula for 4

[hide]N/2 times (first term plus last term) where n is the amount of terms) ts only for arithmetic sequences[/hide]

[quote="vallon22"]BTW, on target, you can use calcs, so for number 3, you can guess and check to get the answer
for number 4, you could use a calc and add up all those, then subtract

[hide="number three"]-101,
the numbers would be -50 and -51 because -51 times -50 is 2550 and the sum is -101
and yes, it did ask for negative[/hide]

[hide="number four"]i'm thinking 830, but i'm not sure
if you do an arithmetic sequence, then the first 20 multiples of 5 would be 5, 10....100
so, the mean of the numbers would be (100+5)/2 = 52.5
we take the mean and multiply it by the number of number to get the sum 52.5 times 20 = 1050

the first 20 even integers would be 2, 4.....20 so, the mean would be (2+20)/2 = 11
and 11 times 20 is 220

1050 - 220 = 830
correct me if i'm wrong[/hide][/quote]

But if you guess and check it might take longer than the alloted 6 minutes to do guess and check and add manually.

Formulas may take longer to memorize, but they'd be much faster. Though Guess and check for number 3 might be easier if you can estimate a place to start

ya, well
i provided an explanation for number 4

also
[hide="what i did for chapter target"]for number 4, i actually did it out
i added everything up and subtracted
i actually made it too  :D [/hide]

i was extremely lazy this afternoon and didn't even try those problems  :D

But when i did (after posting this) they seemed pretty simple  :lol:

[hide="Solutions"]

#3. We have x(x+1)=2550. From here, you can just pull out your calculator and take the squareroot of 2550 then guess and check from there. (except with negative numbers). Factoring the quadratic is pretty much the same. You get -50 and -51. Adding those gives -101.

#4. Remember that the sum of the first n numbers is n(n+1)/2. So, the first 20 positive multiples of 5 is 5*20(21)/2 and the first 20 even integers is 2*20(21)/2. Subtracting, 3*20*21/2=630, I believe. [/hide]

Vallon, I think your #4 is wrong, since the first  20 even integers isn't 2, 4, 6,...,20, but 2, 4, 6,...,40. Else, that would work

can you maybe use difference of squares or prime factorization for number 3?

You can't do a difference of squares for #3 since the numbers only differ by 1. However, since 2550 has only 12 factors, you can brute force it if you've got enough time.
[hide="fast way for #4"]
$ \sum_{a\equal{}1}^{20} 5a \minus{} \sum_{b\equal{}1}^{20} 2b \equal{} 5\sum_{a\equal{}1}^{20} a \minus{} 2\sum_{b\equal{}1}^{20} b \equal{} 3\sum_{c\equal{}1}^{20} c \equal{} \frac{3\times20\times21}{2} \equal{} 630$.[/hide]

the sum of the first twenty multiples of 5 equals 5 multiplied by the sum of the first 20 natural numbers, which is given by $ \dfrac{n(n\plus{}1)}{2}$  So, the sum of the first 20 multiples of 5 is equal to $ \dfrac{5*20*21}{2}\equal{}1050$  The sum of the first $ n$ even numbers is $ n^2\plus{}n$ so the sum of the first 20 even numbers is $ 20^2\plus{}20\equal{}400\plus{}20\equal{}420$

Our answer is 1050-420=630!

[quote="Poincare"]I don't really see how you got that dude  :maybe:[/quote]
$ \sum_{a \equal{} 1}^{20} 5a \equal{} 5 \plus{} 10 \plus{} 15 \plus{} ... \plus{} 100$, so it's the sum of the first $ 20$ multiples of $ 5$. $ \sum_{b \equal{} 1}^{20} 2a \equal{} 2 \plus{} 4 \plus{} 6 \plus{} ... \plus{} 40$, so it's the sum of the first $ 20$ even integers.
We are looking for the positive difference of these two, or $ \sum_{a \equal{} 1}^{20} 5a \minus{} \sum_{b \equal{} 1}^{20} 2b$.
You can factor out a $ 5$ from the first sum and a $ 2$ from the second sum to get $ 5\sum_{a \equal{} 1}^{20} a \minus{} 2\sum_{b \equal{} 1}^{20} b$. Then, since $ \sum_{a \equal{} 1}^{20} a \equal{} \sum_{b \equal{} 1}^{20} b$, the answer is equal to $ 3\sum_{c \equal{} 1}^{20} c \equal{} 630$. $ c$ is just another dummy variable.
Hope that helps.

On number 3 it doesnt relaly matter if their negative, just multiply ur sum by -1 at end.

Thx but what if your calculator can't do summations cause i have a scientific

EDIT:
Never mind i can just use the formula  :oops:


Given the set $ S_n \equal{} \{1,2,...,n\}$. Find out in how many ways can one divide $ S_n$ into $ 3$ subsets $ A_n,$ $ B_n,$ $ C_n$ such that:

$ (i)$  $ |A_n\cap B_n\cap C_n|$ $ \ge$ $ 1$.

$ (ii)$  $ A_n\cup B_n\cup C_n \equal{} S_n$.

$ (iii)$  None of three subsets $ A_n$, $ B_n$, $ C_n$ contains any two consecutive integer numbers.

According to i), there exists an integer $ m$ that belongs $ A_n\cap B_n\cap C_n$. But then neither $ m \minus{} 1$ nor $ m \plus{} 1$ can be an element of $ A_n\cup B_n\cup C_n \equal{} S_n$ (otherwise they would violate property iii).
Therefore, the answer is $ 1$ if $ n \equal{} 1$ and $ 0$ if $ n > 1$.


Let ABC be a triangle and X,Y,Z 3 points inside the triangle
so that: BAZ=CAY; ABZ=CBX; BCX=ACY  (i'm talking about angles)

Prove that (AX),(BY),(CZ) are concurrent.

Just write the concurrence conditions for AX, BX, CX; AY, BY, CY; AZ, BZ, CZ, according to Ceva's theorem, but using sines of angles instead of segments. Then by multiplying all 3, you'll get exactly what you want, Ceva's relation in angles for AX, BY, CZ.

the reciprocal relationship follows easily. If AD, BE and CF are 3 cevian lines concurrent in M and <DAB = <ECB and <FBA = <ECA then also <DaC = < FBC. Also uses Ceva's theorem for sines.


calculus
integration
limit
trigonometry
real analysis
absolute value
calculus computations

Prove that $\lim_{n\to\infty} \int_{\frac{1}{n^2}}^{\pi} \sqrt{x}\cos (nx) dx=0.$

$0\le \lim_{n\to\infty} \int_{\frac{1}{n^2}}^{\pi} \sqrt{x}\cos (nx) dx = \lim_{n\to\infty} \int_0^{\pi} \sqrt{x}\cos (nx) dx-\lim_{n\to\infty} \int_0^{\frac{1}{n^2}} \sqrt{x}\cos (nx) dx$

First integral converge to 0 from Riemann's Lemma, and the second converges since, from mean value theorem it's modulus is equal with
$|\frac 1 {n^2} \sqrt{\theta_n}\cos n\theta_n|\le |\frac 1 {n^2} \sqrt{\pi}|\to 0$

That solution works, but surely the Riemann-Lebesgue lemma isn't part of the desired solution.

you can do a kind of proof of riemann lemma by integrating by parts :P

Let's just do the integration by parts:

$\int_{\frac1{n^2}}^{\pi}\sqrt{x}\cos(nx)\,dx$

$=\left.-\frac1n\sqrt{x}\sin(nx)\right|_{\frac1{n^2}}^{\pi}+\frac1{2n} \int_{\frac1{n^2}}^{\pi} \frac1{\sqrt{x}}\sin(nx)\,dx$

The boundary terms evaluate to $\frac1{n^2}\sin\left(\frac1n\right)$ which easily goes to zero.

The absolute value of the integral is bounded by

$\frac1{2n}\int_{\frac1{n^2}}\frac1{\sqrt{x}}\,dx=\left.\frac1n\sqrt{x}\right|_{\frac1{n^2 }}^{\pi}$

$=\frac1n\left(\sqrt{\pi}-\frac1n\right)$ and that tends to zero.

This isn't a delicate question; we have more than we need at every step.

That's right, Dr.Kent Merryfield.This is easy problem. :lol:


Hi there. I am sitting for the SAT exam in June and currently i am studying physics. Although the questions are generally easy, one should know lots of material. So could you please name what branches of physics does this exam covers?
I know:
-mechanics
-electricity
-waves
-kinematics
....
-what else?

I also know that optics are examine but what exactly?

Well technically kinematics is part of mechanics :P .  You should know:
-Mechanics
-Electricity
-Magnetism
-Thermo
-Fluids
-Waves
-Optics
-Nuclear

These are broad areas there shouldn't be anything outside of the physics B/C curriculum.

Yo. I'm taking the exam in may. http://www.sparknotes.com/testprep/books/sat2/physics/ has good material on the exam. Optics is mostly how images are formed from mirrors and thin lenses, and focal length/image distance, and snell's law. It's treated very superficially.

Straight from the college board, page 33: [url=http://www.collegeboard.com/prod_downloads/prof/counselors/tests/sat/2006-07-SAT-subject-tests-preparation-booklet.pdf]here[/url]

The SAT II focuses more on qualitative and intuitive facts in a superficial way than on quantitative things. Any math that's on there is usually fairly light, and requires no calculus.

Still, you have 80 questions (or is it 75?) in an hour, so you should work fairly quickly through the beginning of the test. This means it helps to understand the physics with much greater depth than what you will be tested on. Prep books useless for learning the material, since they are just as superficial as the tests. The best approach is to learn the material properly, if you have time, and just use the test prep books for practice tests.

It's a worthwhile investment; it never hurts to know more physics, and when you go to college, you can fly through intro physics, or if you want, try for a more analytical or honors version of intro physics.

Thanks guys :)

Another question, how long after the test date do the results come. I mean, if my results at June are not good enough and i retake the tests in October(4th?) will time allow me to send the new results at universities on early decision aplication?
??

You'll have to check with the particular school.  Most schools will receive the score in time but nothing is guaranteed.  Check with the school.

You'd usually get the scores in about 3 weeks, and then up to 1 week to send, so figure a month total? You may want to contact the College Board and ask them about their timeline. Also, some universities have a separate deadline for application receipt and test score receipt, since the test scores can roll in while they are still busy sorting out the applications.


Heres round 2 of the AoPS Member of the Month competition.
Remember, you can still participate in round 1. Its not over till the end of the month. (September 30)
Also, you can fix your answer and get more points. Why? Because there are some problems in which you can earn more points for effort or ingenuity.

Question 1
Correct answer: 1/4 points each (no partial credit on this one)
If x:^4:=1, what does x equal?
If x:^4:=-1, what does x equal?

Question 2
Correct Answer: 1 point for part a, 3 for part b.
Bob needed money. So he called on his friend, Paul, who works at a lottery company. Paul advises Bob to buy Lucky Ducky, an ordinary 3 by 3 scratch card in which you can scratch three places at a time, because on this weeks Lucky Ducky, the You Win! square is on the same place in every card. So Bob goes out to buy the scratch cards. (FYI, a scratch card is a method of gambling, usually sold in Wal-Mart and such.)

Bob is bad at memorizing things, but he doesnt want to write down which squares he scratched before. He knows that he might have one or two duplicates, but certainly not all three (of the scratches)

a.) How many cards are required to guarantee that hell win?
b.) Generalize this problem with x by x scratch card, y numbers of scratches allowed on one card, and z number of duplicates Bob might have.

Question 3
1 point for a, 3 points for b
2 points for each correct way of getting the answer for a, 2 for b.
[img]http://www.artofproblemsolving.com/wforum/images/avatars/2083862793f63854177311.jpg[/img][img]http://www.artofproblemsolving.com/wforum/images/avatars/2083862793f63854177311.jpg[/img][img]http://www.artofproblemsolving.com/wforum/images/avatars/2083862793f63854177311.jpg[/img]
[img]http://www.artofproblemsolving.com/wforum/images/avatars/2083862793f63854177311.jpg[/img][img]http://www.artofproblemsolving.com/wforum/images/avatars/2083862793f63854177311.jpg[/img][img]http://www.artofproblemsolving.com/wforum/images/avatars/2083862793f63854177311.jpg[/img]
[img]http://www.artofproblemsolving.com/wforum/images/avatars/2083862793f63854177311.jpg[/img][img]http://www.artofproblemsolving.com/wforum/images/avatars/2083862793f63854177311.jpg[/img][img]http://www.artofproblemsolving.com/wforum/images/avatars/2083862793f63854177311.jpg[/img]
(it might look better on one style than the other; I recommend looking at it from both styles)
a.) How many triangles are in the above diagram?
b.) If this is a 3 by 3 cross-squares (I defined [img]http://www.artofproblemsolving.com/wforum/images/avatars/2083862793f63854177311.jpg[/img]as a "cross-square"), how many triangles are in an x by x cross-square?

...and yes, Syntax Error/Mr. Rusczyk/Mr. Crawford, I know I'm crowding up the forum. I'll make all of these a Normal (except for one) on October 1st.

[hide]1.  for x^4 = 1, x = 1, -1, i, -i

for x^4 = -1, x =  :sqrt: i, - :sqrt: i

[/hide]

For question 2, [hide]he can scratch up to two that he has before[/hide], right?

Correct. (for both)

and I know I sound stupid, but I'd like everything to be in Spoiler, because this is a competition after all. "Thanks much!" :) (no offence Mr. R)


I'm not sure where to put this, and this seemed like the best place.  I've been looking for something like this but haven't found it, and I haven't been able to doodle it out myself :P

Suppose I have a 60-day history of prices for 5 different types of candy (A, B, C, D, E). I want to plot a bar graph for each candy's price today, with candy on the x-axis and price on the y-axis (I know how to do this  :P ); then I want to overlay on top of this graph a histogram for each candy, in terms of the price frequency over the
past 60 days. Each of the 5 histograms would be oriented sideways, so that histogram bars go sideways and they are lined up with the y-axis (price) and are on the same scale. The point is so that you can see how the price today compares with the price over the past 60 days.

Thanks for any help!  :)

Hey,
Use the Matlab forums.  I usually post their for matlab questions and they give great responses too (the people who respond are like professional programmers):

http://newsreader.mathworks.com/WebX?14@@/comp.soft-sys.matlab

good luck.

Did you try the HOLD ON instruction? It should work...


Show that 24 is the largest integer divisible by all the integers less than its square root.


calculus derivative algebra function domain limit Asymptote

1. For the Mean Value Theorem, If y=f(x) is continuous at every point of the closed interval [a,b] and differentiable at every point of its interior (a,b), then there is at least one point c in (a,b) at which f'(c)=f(b)-f(a)/b-a.  I have a question here: i know that [a,b] means a closed interval, whereas, (a,b) means an open interval, but why here it says continuous in closed interval but didfferentiable in open interval? What's the difference here? Why not both closed intervals?
2. Find the critical points of f(x)=x^3-12x-5. Find the function's local and absolute extreme values. I know how to find its local values, but how to find absolute extrema, in other words, how do I know if the local value that I found is the absolute extreme?  And in a question, if it asks you to find extrema, does it mean local or absolute though absolute is local.
3. According to the First Derivative Test for local extrema, if f' doesn't change sign at c, then f has no local extreme value at c.  But for a question on my book, f(x)=(4-x^2)^(-1/2),  the critical point is 0, but i think it doesn't have local extreme because the derivative doesn't change sign ,but my book says it has minimum value at x=0.
Thanks for help.

To answer your second question, you know that local extrema are those within a specified domain.  If you're not given boundaries, then any extrema you find must be absolute extrema.  Since you didn't give any specified domain, any extrema you find must be local (as well as absolute).

On #1: We state it that way because we don't need differentiability at the endpoints- it's true even for functions that aren't differentiable there. Also, differentiability at the endpoint of an interval of definition requires us to write that special case into our definition of differentiation.

On #2: The domain is $ (\minus{}\infty,\infty)$, and we always need to check the endpoints when looking for absolute extrema. What are $ \lim_{x\to\infty}f(x)$ and $ \lim_{x\to\minus{}\infty}f(x)$?

[quote="MIA6"]
Why a function is required to be continuous in a closed interval but didfferentiable in open interval? What's the difference here? Why not both closed intervals?
[/quote]

Define: $ f(x) \equal{} |x|$
Then $ f(x)$ is continuous in $ [0,3]$ and differentiable in $ (0, 3)$ but not in $ [0,3]$

[quote="jmerry"]

On #2: The domain is $ ( \minus{} \infty,\infty)$, and we always need to check the endpoints when looking for absolute extrema. What are $ \lim_{x\to\infty}f(x)$ and $ \lim_{x\to \minus{} \infty}f(x)$?[/quote]

So we are actually looking for the asymptote?


the equation
$ \frac{1!}{2003!} \plus{} \frac{2!}{2004!} \plus{} \frac{3!}{2005!}  \plus{}\frac{4!}{2006!} \plus{}...\plus{}\frac{2003!}{4005!} \plus{}\frac{2004!}{4006!} \equal{}A.(\frac{1!}{2002!} \minus{}\frac{2005!}{4006!})$
holds for some rational number $ A$. find the value of $ A$ and express your answer in the form $ \frac{a}{b}$ where $ a$ and $ b$ are positive integers <4000


Where can i find the solution and proof for poincare conjecture???

Follow links on [url=http://math.berkeley.edu/~lott/ricciflow/perelman.html]this page[/url]. Now, whether you'll be able to understand any of it is another big question :P.

I am finding a teacher to help me to understand it :)  hehe xP

I thought I posted a response here.  I guess I forgot to hit submit.  Anyway...

fedja's link is a great list of resources, but it is probably more than what you are looking for.

The proof of the Poincare Conjecture is written up in gory detail, together with a lot of background material, in Morgan and Tian's book, which you can find here.

http://arxiv.org/abs/math/0607607

Another good resource which is far less comprehensive but more approachable (and abbreviated) is Terry Tao's lecture notes.

http://en.wordpress.com/tag/285g-poincare-conjecture/

Although it would take a lot of background and effort to understand the proof, a decent background in differential geometry and PDE is enough to learn quite a bit *about* the proof.


[i]Find all integral solutions $(x,y,z)$ to \[\frac{1}{x+y}=\frac{z^{z}+1}{xy}\] such that $z^{z}+1<1111111111$ is in the set of primes $\mathbb{P}$.[/i]

Masoud Zargar

EDIT: The original statement was ${1\over x}+{1\over y}=\frac{1}{z^{z}+1}$, which is of course equivalent to the current one.

SOLUTION: First we note that $z\leqslant 9$ since $10^{10}>\frac{10^{10}-1}{9}$

Now let's analyze the general case of the equation ${1\over x}+{1\over y}={1\over p}$ where $p$ is a given prime. We have $y=\frac{px}{x-p}=\frac{px-p^{2}+p^{2}}{x-p}=p+\frac{p^{2}}{x-p}$. From this it's obvious that possible values for $x-p$ are $1, p, p^{2}$, giving the following solutions: $(p+1,p+p^{2}),(2p,2p), (p^{2}+p,p+1)$.

Therefore, it remains to find all values of $z$ such that $z^{z}+1$ is a prime. First, all odd values greater than $1$ are discarded since then $z^{z}+1$ is a composite even. Out of the remaining ones, we manually check that $z=1, z=2$ and $z=4$ give prime values - respectively $2, 5, 257$.

As for $z=6$, by Fermat's Little we have $6^{12}-1\equiv 0\pmod{13}$, and

\begin{eqnarray*}6^{12}-1 &=& (6^{6}-1)(6^{6}+1)\\ &=& (6^{3}-1)(6^{3}+1)(6^{6}+1)\\ &=& (6-1)(6^{2}+6+1)(6+1)(6^{2}-6+1)(6^{6}+1)\\ &=& 5\cdot 43\cdot 7\cdot 31\cdot (6^{6}+1)\end{eqnarray*}

but because neither of the first four factors is divisible by $13$, we get $13|6^{6}+1$, hence it's composite.

For $z=8$ we have $8^{8}+1=2^{24}+1=(2^{8}+1)(2^{16}-2^{8}+1)$, hence it's also composite.

Therefore, the initial equation has solutions for $z\in\{1,2,4\}$ and those are

$(3,6,1),(4,4,1),(6,3,1);(6,30,2),(10,10,2),(30,6,2);(258, 66306, 4),(514,514,4),(66306,258,4)$

I had a similar solution. And for the $6^{6}+1$-part: just use $6^{6}+1 = 36^{3}+1 \equiv (-3)^{3}+1 \equiv-26 \equiv 0 \mod{13}$. But yeah, doesn't really matter.  :)  Nice problem.

I have to slap my forehead and ammend the proof:

Fermat's Little and mods aren't needed at all, since $6^{6}+1=36^{3}+1=(36+1)(36^{2}-36+1)$.


limit
integration
real analysis
real analysis unsolved

Dear everyone.

I have a not-bad problem related to calculus.
please read the following.

in [0,1], f(x)$ \in C^{1}$. prove that $ \lim_{n\to\infty}{\left[ \sum_{k \equal{} 1}^{n}{f\left(\frac {k}{n}\right)} \minus{} n\int_{0}^{1}{f(x)dx} \right] } \equal{} \frac {1}{2}(f(1) \minus{} f(0))$.

May I ask your nice answers ?  :blush:
thanks for reading my post.

Define $ F(t) \equal{} \int_0^t f(s) ds$ an use the Taylor formula with 2 terms on $ [k/n,(k\plus{}1)/n]$.


calculus
derivative
function
complex analysis
complex analysis unsolved

I saw this residue answear in 1 test but without showing the way and i can't seem to understand how they got to that answear

Res [ exp (2/3z+i)  ,  -i/3 ] = 2/3

my best guess it they just ignored to exp and derivative the bottem by z and got 2/3 , is it just like that ? then where is the refrence for -i/3 ?

tnx in advance for any reply

I assume you mean $e^{\frac{2}{3z+i}}$? Note that the residue is $a_{-1}$ in the Laurent series.  To determine the Laurent expansion around $z=-i/3$ use the Taylor series for $e^{z}=1+z+z^{2}/2!+z^{3}/3!+z^{4}/4!...$, plugging in $z=\frac{2}{3z+i}$.  The $a_{-1}$ term is simply $\frac{2}{3z+i}= \frac{2}{3}\frac{1}{z+i/3}\implies a_{-1}= \frac{2}{3}$

Something I would like to verify, though, is that you cannot use residue calculus to integrate about a contour around this point, since it's an essential singularity.  Can somebody confirm or deny that?  If so, what's the use of characterizing the residue for mathematical purposes?

The residue theorem can handle essential singularities; we can write any analytic function $f$ on a punctured disk around $a$ as $f(z)=\frac{r}{z-a}+g'(z)$, where $g$ is analytic on that punctured disk and $r$ is the residue. Integrating $g'$ around any closed curve gives zero by the fundamental theorem, regardless of singularities inside.

This can be extended to a function with a finite set of isolated singularities- after we subtract off the residue terms at each point, the remainder has an antiderivative.


IMO 1998 Shortlist. Define $a_{n + 1} = 1 - \sum_{i = 1}^na_i$.
We need to prove that $\prod_{i = 0}^{n + 1}(1 - a_i) \geq n^{n + 1} \prod{i = 0}^{n + 1}a_i$.
However, $1 - a_1 = a_2 + \dots + a_{n + 1} \geq n\sqrt[n]{a_2\dots a_{n + 1}}$.
Multiplying this inequality and the $n$ similar inequalities, we get the result immediately.

Very nice solution, but you probably meant
$\prod_{i = 0}^{n + 1}(1 - a_i) \geq n^{n + 1} \prod_{i = 0}^{n + 1}a_i$.


In my friends dream she has moved to Hawaii.  And since it is her dream she lives in a mansion close to the beach. She demands only the most luxurious of accommodations. So for her bath, she had glacial ice from the Swiss Alps shipped to her mansion. There, the ice was carved into a perfect sphere, exactly 2 meters in diameter. The ice sphere melted into her hexagonal bath measuring 1.5 meters on a side, and 0.8 meters deep.

If the island heat uniformly melts the surface of the ice at a rate of 1.5mm of depth per minute, how many minutes will it take for her bath to be completely full? Please round up to the nearest minute, and assume the volume does not change between solid and liquid states.

I probably made a mistake, but I don't think the bath will ever fill.

The volume of the bath is $\frac{3\sqrt{3}}{2}s^2h=2.7\sqrt{3}=4.677$.
The volume of the ice is $\frac{4}{3}\pi r^3=\frac{4}{3}\pi=4.189$.
Thus, the volume of the ice is not enough to fill the bath.

that is right it would never fill :)

She was very upset about that so she went back to sleep and
this time, there were three ice spheres: one was 0.9 meters in diameter; one was 1.5 meters in diameter; and the last was 1.8 meters in diameter. They all started melting at the same time. Also, the water in the bath evaporated at a rate of 0.1mm of depth per minute.

This time how many minutes will elapse before the bath is completely full? Please round up to the nearest minute.

great!  thanks for giving me the heads up!


[quote="nebula42"]Let a rectangular prism have side lengths $l$, $w$, and $h$. Given that $lw=6$, $wh=8$, and $hl=12$, find the sum of the lengths of all of the edges of the rectangular prism.[/quote]

[hide]Volume$=\sqrt{lw*wh*hl}=\sqrt{6*8*12}=24$

Divide that by each of the given equations to get:
$h=4$
$l=3$
$w=2$

Answer: $4(4+3+2)=\boxed{36}$
[/hide]

[size=75][color=darkred]Try to avoid putting text into LaTeX as much as possible. Thanks.
-nebula42[/color][/size]

[quote="SplashD"][quote="nebula42"]Let a rectangular prism have side lengths $l$, $w$, and $h$. Given that $lw=6$, $wh=8$, and $hl=12$, find the sum of the lengths of all of the edges of the rectangular prism.[/quote]

[hide]Volume$=\sqrt{lw*wh*hl}=\sqrt{6*8*12}=24$

Divide that by each of the given equations to get:
$h=4$
$l=3$
$w=2$[/hide]

[size=75][color=darkred]Try to avoid putting text into LaTeX as much as possible. Thanks.
-nebula42[/color][/size][/quote]

You did all the work, but you forgot to give the answer to the question.  :wink:  :P

It was just one word that I put in $\LaTeX$ and I fixed my answer now.

[quote="lotrgreengrapes7926"][quote="nebula42"]Let a rectangular prism have side lengths $l$, $w$, and $h$. Given that $lw=6$, $wh=8$, and $hl=12$, find the sum of the lengths of all of the edges of the rectangular prism.[/quote]
[hide] Assuming they all have integer side lengths, $lw=2\cdot3$, Since $wh=8$, $w=2$, $l=3$, $h=4$. The sum of all the edge lengths is $4(2+3+4)=4\cdot9=\boxed{36}$. [/hide][/quote]
You don't have to assume that they have integer side lengths.  :wink: There is a method to solve it without assuming that. Try to see if you can find it.
[hide="Hint"]Try finding the volume of the rectangular prism.[/hide]

Well, yes, what SplashD did. But the numbers looked so nice that it seemed like integers.

[hide]We can find the volume first and get that $lwh=24$, so we get w=2, l=3, and h=4.

So the answer is $4(2+3+4)=36$

$36$[/hide]

 :ninja:


Buna ziua Dle Bogdan Enescu ,tocmai pe aczel ma uitam
din notitele lui hojoo lee si nu inteleg demonstratia lui aczel.
apreciez daca puteti da niste explicatii.
chiar ma uit de vreo 2 zile incoace pe ea si tot n-am reusit
sa-mi dau seama.

Este vorba de proprietăţile funcţiei de gradul 2. Astfel, să considerăm
\[f(x)=(a_{1}x-b_{1})^{2}-(a_{2}x-b_{2})^{2}-\ldots-(a_{n}x-b_{n})^{2}.\]
Avem $f\left(\frac{b_{1}}{a_{1}}\right)\le 0$, iar, pe altă parte, coeficientul lui $x^{2}$ este $a_{1}^{2}-a_{2}^{2}-\ldots-a_{n}^{2}>0.$ Rezultă că ecuaţia $f(x)=0$ are soluţii reale, aşadar $\Delta \ge 0.$ Scriind explicit aceasta, obţinem inegalitatea cerută.

multumesc dle Enescu
ati fost de mare ajutor :)

Ce este de detaliat,e doar: $ac+bd \le \sqrt{(a^2+b^2)(c^2+d^2)}$.

multumesc ,eu am mai folosit si (m^2-n^2)(p^2-q^2)<=(mp-nq)^2


videos
number theory unsolved
number theory

find all integers satisfying
21x + 20y + 12xy = 18

We have: $ x\equal{}2m$ for some $ m\in Z$. Then:
\[ 21m\plus{}10y\plus{}12my\equal{}9\]
\[ 21m\plus{}2y(6m\plus{}5)\equal{}9\]
\[ 2y(6m\plus{}5)\equal{}9\minus{}21m\]
\[ 2y\equal{}\frac{9\minus{}21m}{6m\plus{}5}\equal{}\frac{\minus{}3(6m\plus{}5)\minus{}3(m\minus{}8)}{6m\plus{}5}\equal{}\minus{}3(1\plus{}\frac{m\minus{}8}{6m\plus{}5})\]
If $ m\equal{}8$, $ y$ is not integer. So here must be:
\[ |m\minus{}8|\geq |6m\plus{}5|\]
\[ (m\minus{}8)^2\geq (6m\plus{}5)^2\]
\[ m^2\minus{}16m\plus{}64\geq 36m^2\plus{}60m\plus{}25\]
\[ 35m^2\plus{}76m\minus{}39\leq 0\]
\[ \minus{}5\leq m\leq 0\]
Checking this interval we get: $ m\equal{}\minus{}1$.
In this case we get the answer $ (x,y)\equal{}(\minus{}2,\minus{}15)$.

Another way is to remark that the given equation is equivalent to $ (3x\plus{}5)(4y\plus{}7)\equal{}53$.
Since $ 53$ is prime, it reduces to solve the four systems
$ 3x\plus{}5\equal{}\minus{}53,4y\plus{}7\equal{}\minus{}1$
$ 3x\plus{}5\equal{}\minus{}1,4y\plus{}7\equal{}\minus{}53$
$ 3x\plus{}5\equal{}1,4y\plus{}7\equal{}53$
$ 3x\plus{}5\equal{}53,4y\plus{}7\equal{}1$

Pierre.

[quote="pbornsztein"]Another way is to remark that the given equation is equivalent to $ (3x \plus{} 5)(4y \plus{} 7) \equal{} 53$.
Since $ 53$ is prime, it reduces to solve the four systems
$ 3x \plus{} 5 \equal{} \minus{} 53,4y \plus{} 7 \equal{} \minus{} 1$
$ 3x \plus{} 5 \equal{} \minus{} 1,4y \plus{} 7 \equal{} \minus{} 53$
$ 3x \plus{} 5 \equal{} 1,4y \plus{} 7 \equal{} 53$
$ 3x \plus{} 5 \equal{} 53,4y \plus{} 7 \equal{} 1$

Pierre.[/quote]

Isn't this an instance of SFFT(Simons Favorite factoring trick)? :lol:

[quote="pbornsztein"]Who Is Simon? :?:

Pierre.[/quote]
It is just the plain ol' trick of adding some number on either sides, to factorize the expression. i am mot a good explainer... You can watch Richard Rusczyk's video there, he does it quite well.

[quote]Simon's Favorite Factoring Trick (abbreviated SFFT) is a special factorization first popularized by AoPS user Simon Rubinstein-Salzedo. [/quote]

More of it on: http://www.artofproblemsolving.com/Wiki/index.php/Simon%27s_Favorite_Factoring_Trick

[quote="pbornsztein"]Who Is Simons? :?:

Pierre.[/quote]

Simon is the user 'ComplexZeta'.


topology
function
advanced fields
advanced fields unsolved

not able to figure this one out (from "Topology of Surfaces"-Kinsey Excercise 3.10)
Suppose X is a topological space with closed subsets A and a B s.t. $ X \equal{} A \cup B$. Let $ f: X \rightarrow Y$ and $ g: X \rightarrow Y$ be  continuous functions s.t. $ \forall x\in A\cap B, f(x) \equal{} g(x)$. Define
$ f\cup g: X \rightarrow Y$ as:

$ f(x), x\in A$
$ g(x), x\in B.$

Prove $ f\cup g$ is continuous and give an example to show that the condition that each of A and B is closed is necessary

The condition is necessary because you could take $ X\equal{}[0,1]$ and $ Y\equal{}\mathbb{R}$ and the function $ f(x)\equal{}0$ and $ g(x)\equal{}1$ with $ A\equal{}[0,\frac{1}{2})$ and $ B\equal{}[\frac{1}{2},1]$. Then $ f\cup g$ is not continuous at $ x\equal{}\frac{1}{2}$.

concerning the continuity: show that preimages of closed sets are closed.


How do you raise your rank?  I know that it is based on the number of posts.  What number of posts raises your rank?  No I'm not going to spam a lot to get my rank up.

also, the stars below your name are related to the number of your posts.

[quote="xxreddevilzxx"]by number of posts. There have been a ton of posts about that here. just search it. please do not overpost. the admin might stop your count from going up. those useless posts are called junks.[/quote]You're right. I can easily disable any person that tries to spam around just to increase his post count ;)

Also before posing a problem, try to seach for it. [b]Most likely[/b] it has been posted before :)


invariant
combinatorics unsolved
combinatorics

There are $n+1$ containers arranged in a circle. One container has $n$ stones, the others are empty. A move is to choose two containers $A$ and $B$, take a stone from $A$ and put it in one of the containers adjacent to $B$, and to take a stone from $B$ and put it in one of the containers adjacent to $A$. We can take $A = B$. For which $n$ is it possible by series of moves to end up with one stone in each container except that which originally held $n$ stones.

[hide="Answer (too lazy to describe construction)"]When n+1 is odd or divisible by 4, you can construct a way to do it. If n+1 is 2 mod 4, then number the containers 1 through n+1. The parity of the total number of stones in even-numbered containers is invariant, and since the original and desired final states have different parity, this case fails.[/hide]

If $ n$ is even the answer is obvious.

If $ n$ is odd:
Let $ C_1$, $ C_2$, .., $ C_{n\plus{}1}$ be the containers:
$ C_1$ is the originally held with n stones.
Let $ S_{odd}$ be the sum of the number of stones that are in all the containers $ C_i$ with
$ i$ odd.
Let $ S_{even}$ be the sum of the number of stones that are in all the containers $ C_i$ with
$ i$ even.
See that after every move the parity of $ S_{odd}$ and $ S_{even}$ don`t change.
Then $ S_{odd}$ is always odd and $ S_{even}$ is always even.
Then if $ n$ is odd it is impossible by series of moves to end up with one stone in each container except that which originally held $ n$ stones.


Two Canadian IMO reports, one by team leader Dr. Recio and one by contestant Elyot Grant, have been published in the [url=http://journals.cms.math.ca/Notes/]CMS Notes[/url] December 2005.

I have attached the pages containing the articles.

Thanks, Yufei. It is always hilarious to see your surreptitious advertising for Mr. Potter.  :D

It is funny to see that at first they don't try to mention any web site: "...Several internet math problems clubs, some of them recently created, have greatly contributed to the currently fast growing interest of groups of students toward the IMO..." But finally it was said by Elyot Grant in his IMO report.

o_O I can't read the pdf for some reason.

Can anyone post it here?

What does it mean, "by Elyot Grant, [b]Samuel Beatty Contestant[/b] to the 2005 IMO" ? (emph added)


if the matrix $ I \minus{} AB$ is nonsingular, prove that the matrix $ I \minus{} BA$ is nonsingular, where $ I$ is the identity matrix.

$ (I\minus{}AB)$ is non singular implies that $ ABX\equal{}X$ iff $ X\equal{}O$. Consider $ BAX\equal{}X$ Multiplying by $ A$, we get $ AB(AX)\equal{}AX$ This means $ AX\equal{}O$. Hence, $ BAX\equal{}O$, Which means,$ X\equal{}O$. Hence, $ (I\minus{}BA)$ is also non singular.

It is not just that $ (I\minus{}AB)$ is nonsingular iff $ (I\minus{}BA)$ is nonsingular. It is true that these two have the same determinant. Further, it is possible to prove that they have the same eigen equation.

If $ A$ is nonsingular, then, $ A(I\minus{}BA)A^\minus{}1 \equal{} (I\minus{}AB)$ hence, they have the same determinant. If both $ A$ and $ B$ are singular, then, consider a secquence of nonsingular matrices approaching $ A$. (One way to do this is, take $ A_k\equal{}(A\minus{}a_kI)$, where $ a_k$ is a nonzero secquence approaching 0, none of them being eigen values of $ A$.) The equality $ det(I\minus{}BA_k) \equal{} det(I\minus{}A_kB)$ holds for all k. Hence, taking the limits, we get $ det(I\minus{}BA)\equal{}det(I\minus{}AB)$.

A similar procedure will show that the have the same eigen equation too.

$ I\minus{}AB$ is nonsingular means $ AB$ has no eigenvalues equal to $ 1$. $ AB$ has the same eigenvalues as $ BA$ if they have the same sizes.

Even if they don't have the same size, the extra eigenvalues are zero.

There's a nice argument posted [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=116836]here[/url], which doesn't rely on anything being square.


Find all integers $ x, y$ and $ z$ such that

$ x^{2}\plus{}y^{2}\plus{}z^{2} \equal{} 2(yz\plus{}1)$ and $ x\plus{}y\plus{}z\equal{}4018$.

[hide]Hint: prove that $ x^2\equal{}1$[/hide]

Yes, this appears here: http://www.bmoc.maths.org/solutions/bmo1-2010/

I'm wondering if there are other, possibly neater, ways of doing this question.

$ x^2 \plus{} y^2 \plus{} z^2 \equal{} 2(yz \plus{} 1)\leq y^2 \plus{} z^2 \plus{} 2$, yielding $ |x|\leq \sqrt{2}\Longleftrightarrow x\in\mathbb{Z} \equal{} 0,\ \pm 1$.

Another Aprroach:

$ x^2 \plus{} y^2 \plus{} z^2 \equal{} 2(yz \plus{} 1)\Longleftrightarrow (y \minus{} z)^2 \equal{} 2 \minus{} x^2$

$ \therefore |y \minus{} z| \equal{} \sqrt {2 \minus{} x^2}\ (|x|\leq 1)$.

$ \therefore y \equal{} \frac {y \plus{} z}{2} \plus{} \frac {y \minus{} z}{2}$

$ z \equal{} \frac {y \plus{} z}{2} \minus{} \frac {y \minus{} z}{2}$

I can see how

$ (y\minus{}z)^{2} \ge 0$

$ \implies y^{2}\plus{}z^{2} \ge 2yz$

$ \implies y^{2} \plus{} z^{2} \plus{}1 \ge 2yz \plus{}1$,

but not why $ y^{2}\plus{}z^{2} \ge 2yz\plus{}2$.

Sorry, I made a typo.
I have just edited it.

[quote="kunny"]How about this problem?

Find all postive integers $ x,\ y,\ z$ such that $ x^2 \plus{} 2y^2 \plus{} 2z^2 \minus{} 2xy \minus{} 2xz \plus{} 2yz \minus{} 5 \equal{} 0$.

2001 Kyoto University entrance exam/Science, 2nd Exam[/quote]
$ \Longleftrightarrow (x\minus{}y\minus{}z)^2\plus{}y^2\plus{}z^2\equal{}5$. There is only one decomposition of $ 5$ into 3 squares, thus

$ \{(x\minus{}y\minus{}z)^2,y^2,z^2\}\equal{}\{0,1,4\}$, thus we can choose both $ y\neq z\in \{0,1,2\}$ and for each of these 6 pairs $ (y,z)$ we get  1 or 2 positive values for $ x$.

Altogether this yields 8 solutions: 301 310 321 312 302 102 320 120.

We are to find positive integers $ x,\ y,\ z$. :(

[quote="AndrewTom"]Find all integers $ x, y$ and $ z$ such that

$ x^{2} \plus{} y^{2} \plus{} z^{2} \equal{} 2(yz \plus{} 1)$ and $ x \plus{} y \plus{} z \equal{} 4018$.[/quote]No one posted full solution, so:

[hide]$ x^2\plus{}(y\minus{}z)^2\equal{}2$

$ x^2\equal{}1,(y\minus{}z)^2\equal{}1$

(i) $ x\equal{}1,y\minus{}z\equal{}1$
$ y\plus{}z\equal{}4017$
$ y\equal{}2009$
$ z\equal{}2008$

(ii) $ x\equal{}\minus{}1,y\minus{}z\equal{}1$
$ y\plus{}z\equal{}4019$
$ y\equal{}2010$
$ z\equal{}2009$

(iii) $ x\equal{}1,y\minus{}z\equal{}\minus{}1$
$ y\plus{}z\equal{}4017$
$ y\equal{}2008$
$ z\equal{}2009$

(iv) $ x\equal{}\minus{}1,y\minus{}z\equal{}\minus{}1$
$ y\plus{}z\equal{}4019$
$ y\equal{}2009$
$ z\equal{}2010$

Therefore $ (x,y,z)\in\{(1,2009,2008),(\minus{}1,2010,2009),(1,2008,2009),(\minus{}1,2009,2010)\}$, easy to check that these solutions are valid.[/hide]

kunny, how does $ y\equal{}\frac{y\plus{}z}{2} \plus{}\frac{y\minus{}z}{2}$ and $ z\equal{}\frac{y\plus{}z}{2} \minus{} \frac{y\minus{}z}{2}$ help solve the original problem?

I just showed $ y\plus{}z, y\minus{}z$ gives the answer.


step one: x2 - 2x -1
step two: (x - 1)^2 - 2

I got this from a site, it didn't explain anything, I just got the answer, anyone show me how step one got to step two?

Thank you all.

$x^2-2x-1=x^2-2x+1-2=(x-1)^2-2$.
What was the original problem?


[quote="Arvind_sn"]Hello, my name is Arvind. I am starting to do olympiad problems. I  am posting one of my solutions to a problem, can you guys edit it and suggest some things i should add?

Problem:
Prove that the values of $ x$ for which $ x \equal{} (x^2 \plus{} 1)/198$ lie between $ 1/198$ and $ 197.99494949\cdots$.

$ f(\frac {1}{198}) > 0$. $ f(1) < 0$. So, we know there is a root of $ f(x)$ between $ \frac {1}{198}$ and $ 1$.

We find that $ 197.994949\cdots$ is $ 198 \minus{} \frac {1}{198}$. $ f(198 \minus{} \frac {1}{198}) < 0$.We can manipulate the equation to yield
$ x^2 \minus{} 198x \plus{} 1 \equal{} 0$

Since it's a quadratic, it must have two roots. Let's let
$ f(x) \equal{} x^2 \minus{} 198x \plus{} 1$

We know that for two numbers $ a$ and $ b$, $ a < b$, if $ f(a) < 0$, and $ f(b) > 0$, or if $ f(a) > 0$, and $ f(b) < 0$ then between $ a$ and $ b$ there must be a root.

$ f(198) > 0$. There must be a zero between these two also.

Any help would be appreciated.
Thanks![/quote]
how old are you ???????????

im 14...is that bad?

and can u explain wat u mean by defining f(x)?

I said let f(x)=..........

You use $ f(x)$ in the first line of your solution without previously defining $ f(x)$, right?

Oh, sorry.

I did a lot of copying and pasting in this, I probably switched some lines... :blush:

Thanks very much for pointing that out!

Also, can I post other problems that I have solved in this thread for review, or should I make new threads?

You proved that $ f(x)$ has a root between $ 198 \minus{} 1/198$ and 198, but the problem asks you to prove that all of the roots of $ f(x)$ are between $ 1/198$ and $ 198 \minus{} 1/198$.  This means that something is wrong with your solution.  :wink:

About your proof-writing, there are a few key steps in your proof that you don't explain, such as proving that $ 197.99494949\ldots$ is equal to $ 198 \minus{} 1/198$, and proving that $ f(198 \minus{} 1/198) < 0$.  These are non-trivial steps that graders would expect to see if this problem appeared on an olympiad.  In fact, the second of these is not even true.  The only other things I would point out are that you never defined $ f(x)$ and that you stated the fact about roots and sign changes after you had already used it once.

You could probably just keep posting in this thread if you have more proofs that you want reviewed.

Wow, thanks a lot.

That really helps me with my writing style, because I actually do explain trivial steps. Thanks for pointing out that I don't really have to write them, it saves time! :)

And thank you for correcting my solution...I must have made a stupid mistake...

I was actually trying to say that there are steps in that solution that you should have explained more clearly than you did.  :oops:

Deciding which parts of a problem are important enough to explain in your proof is one of the difficult parts of writing a proof.  You should definitely read [url=http://www.artofproblemsolving.com/Resources/AoPS_R_A_HowWrite.php]How to Write a Solution[/url] if you haven't already (I think part 7 talks about this issue).

Regarding the exercise you posted, clearly if $ x$ satisfies the equation, then $ x>\frac{1}{198}$.  Also, the sum of the 2 roots is $ 198$ (by Vieta).  And the conclusion follows.


Find all funtion $ f$ from positive real numbers to real numbers such that
\[ xf\left (x \plus{} \frac {1}{y}\right ) \plus{} yf(y) \plus{} \frac {y}{x} \equal{} yf\left (y \plus{} \frac {1}{x}\right ) \plus{} xf(x) \plus{} \frac {x}{y}\]
holds for all positive reals $x$ and $y$.

[quote="mathVNpro"]Find all funtion $ f$ from positive real numbers to real numbers such that:
\[ xf\left (x \plus{} \frac {1}{y}\right ) \plus{} yf(y) \plus{} \frac {y}{x} \equal{} yf\left (y \plus{} \frac {1}{x}\right ) \plus{} xf(x) \plus{} \frac {x}{y}\]
[/quote]

Let $ f(x) \equal{} g(x) \plus{} x$ and the let $ P(x,y)$ be the new equation :

$ P(x,y)$ : $ xg(x \plus{} \frac 1y) \plus{} yg(y) \equal{} yg(y \plus{} \frac 1x) \plus{} xg(x)$
Let $ g(1) \equal{} a$

$ P(\frac 1x,\frac 1y)$ $ \implies$ $ \frac 1xg(\frac 1x \plus{} y) \plus{} \frac 1yg(\frac 1y) \equal{} \frac 1yg(\frac 1y \plus{} x) \plus{} \frac 1xg(\frac 1x)$

Multiplying by $ xy$, we get $ yg(\frac 1x \plus{} y) \plus{} xg(\frac 1y) \equal{} xg(\frac 1y \plus{} x) \plus{} yg(\frac 1x)$

Adding to $ P(x,y)$, we get new assertion $ Q(x,y)$ : $ yg(y) \plus{} xg(\frac 1y) \equal{} xg(x) \plus{} yg(\frac 1x)$

$ Q(x,1)$ $ \implies$ $ g(\frac 1x) \equal{} a(x \plus{} 1) \minus{} xg(x)$

Replacing $ g(\frac 1x)$ by $ a(x \plus{} 1) \minus{} xg(x)$ and $ g(\frac 1y)$ by $ a(y \plus{} 1) \minus{} yg(y)$ in $ Q(x,y)$, we get :

$ yg(y) \plus{} x(a(y \plus{} 1) \minus{} yg(y)) \equal{} xg(x) \plus{} y(a(x \plus{} 1) \minus{} xg(x))$ and so :

$ (1 \minus{} x)(yg(y) \minus{} a) \equal{} (1 \minus{} y)(xg(x) \minus{} a)$

So, $ \forall x,y\ne 1$ : $ \frac {yg(y) \minus{} a}{1 \minus{} y} \equal{} \frac {xg(x) \minus{} a}{1 \minus{} x}$ and so $ \frac {xg(x) \minus{} a}{1 \minus{} x} \equal{} c$ and :

$ \forall x\ne 1$ : $ g(x) \equal{} \frac {a \plus{} c(1 \minus{} x)}x$ and since this equality is still  true when $ x \equal{} 1$, we get :

$ g(x) \equal{} \frac {\alpha x \plus{} \beta}x$ $ \forall x$ and it is easy to check back that this indeed matches $ P(x,y)$

Hence the solution $ f(x) \equal{} x \plus{} g(x) \equal{} \boxed{\frac {x^2 \plus{} \alpha x \plus{} \beta}x}$


algebra
polynomial
function
linear algebra
matrix
linear algebra unsolved

Let $f(X)=f(...,x_{ij},...)$ -nonzero polynomial of $n^2$ variables,$x_{ij}\in K$ with coefficients from $\mathbb{Z}$ or another field.$f$ is supposed to be a function of the matrix $X=(x_{ij})$, such that $f(XY)=f(X)f(Y)$ for all $X,Y \in M_n(K)$, Prove that $f(X)=(detX)^m$ fro some non-negative integer $m$

Here's a proof for complex polynomials (I'll use a limit argument, so this proof doesn't work for any field).

Since $f$ is clearly constant on each similarity class, it suffices to determine it for Jordan cells (by Jordan cell I mean something a little bit different from the usual; here a Jordan cell of order $k$ means a matrix of order $n$ having a Jordan cell of order $k$ in the usual sense occupying the first $k$ positions the main diagonal and having the main diagonal filled with $1$'s everywhere else).

First of all, since $P_i(x)$, defined as the image through $f$ of the matrix $A_i(x)$ having $x$ on $(i,i)$, $1$ everywhere else on the main diagonal, and $0$ everywhere else, is a multiplicative polynomial, it must be of the form $x^{m_i}$ for some $m_i$ (this is a well-known result), and since $A_i(x),A_j(x)$ are similar, all the $m_i$'s are equal to some $m$. This shows that $f=\det^m$ for diagonal matrices.

Now take a Jordan cell $A$ (as defined above) of order $k$, which has some values on the diagonal and $1$ right below the first $k-1$ positions in the main diagonal. It is similar to the matrix having the exact same entries, except that it has $\alpha\ne 0$ instead of $1$ below the main diagonal (for any $\alpha$). By taking $\alpha\to 0$, we see that $f(A)$ is equal to the value of $f$ in the matrix obtained by keeping only the main diagonal of $A$, and this value is $\det(A)^m$, according to the paragraph above, and I guess this finishes it.

I think I have another approach. We know that if $detA=1$ then there exist $X,Y$ such that $ A=XYX^{-1}Y^{-1}$. Thus, if the field is commutative, we have $ f(A)=1$. Next, classical results on the structure of the linear group show that any invertible matrix $A$ is the product of some transvections, after a diagonal matrix which has 1 on the diagonal, except the last position where we have $ detA$ and again some transvections.  This shows that $ f(A)=f(T(A))$ where $ T(A)=diag(1,1,...,1,detA)$. Since $ f(A)$ is a polynomial in the elements of $A$, we see that $ F(T(A))$ is a polynomial in $detA$ and thus $ f(A)$ is a polynomial $ H(detA)$ for invertible $A$. But $ H(x^2)=H^2(x)$ (since $ f(A^2)=f^2(A)$) and this easily implies that $ H(x)=X^m$ for a certain $m$. It remains to see what happens when A is not invertible. It is not difficult to see that if $f(T)=0$ then $ F(X)=0$ for any matrix X equivalent to T. Since any non-invertible matrix is equivalent to a nilpotent matrix (whose image under f is clearly 0), we have finished the case of non-invertible matrices also.

Thank you both,Grobber and Harazi
I'll show the following proof, maybe it is a bit simplier, maybe not ...
Firstly it is not difficult to figure out with $f$ when $n=1$, all such polynomials in this case are of the form $f(x)=x^m$
Then define $g(x)=f(xI_n)$, it is readily that $g(xy)=g(x)g(y)$, from where it follows that $g(x)=x^m$.
Thus since $XadjX=det(X)I$, $f(X)f(adjX)=f(det(X)I)=g(detX)=(detX)^m$ and it is left to use the fact that the determinant of the matrix as the function of it's $n^2$ elements is irredicuble over the field we are working.
From this fact it follows that $f(X)=c(det(X))^m$ and the rest is readily.


function
calculus
derivative
LaTeX
calculus computations

hi could anyone help me with these problems

1. find any points of inflection, maxima and minima for the following function:

f(x) = (x^3/3) - 2x^2 + 4x

and state whether the points of inflection are stationary or non-stationary

2. consider the following function

f(x) = 80x + (5 + 4x)^2 - (2x^3/3)

show that f has a non-stationary point of inflection at the value of x which maximizes the first-order derivative of f

any help would be appreciated.

thanks

Use the definitions:

A critical point $\left(x,f\left(x\right)\right)$ occurs when $f'\left(x\right)=0$.  This is a local maximum if $f''\left(x\right)<0$ and a local minimum if $f''\left(x\right)>0$.

A [i]possible[/i] point of inflection occurs when $f''\left(x\right)=0$, and when $f''\left(x+\Delta x\right)$ and $f''\left(x-\Delta x\right)$ have opposite parity.

Excuse me,
how to input the equations or formulas in a pretty way?

Using LaTeX -- if you already know how, put your code between two dollar signs.  If you don't know how, the link that says LaTeX on it (in AoPS mode, on the left sidebar) contains everything you need to know.  (You don't need to download anything -- that's just if you want to create LaTeX documents on your computer.)


What do you prefer: Summer olympic games or winter olympic games??? why???

[hide="My opinion"] I prefer summer olympic games, I think the sports are better, and another thing should be that here in Argentina we don't practise a lot of winter sports.[/hide]

[quote="karthik_k"][hide]I prefer the summer olympics....out here in India, there are really few people who're into the winter olympics......[/hide][/quote]

The same as in Argentina!!!!

[hide="what I think"]I like the summer games the best and I don't really know why.[/hide]

summer of course... the real sports are played then

i mean come on, figure skating and curling are two of the biggest sports in winter olympics...

figure skating is so subjective and curling... well at least the ice gets cleaned :rotfl:

i like both.  i like gymnastics, figure skating, diving, and ice dancing the best. maybe summer, becuase i can watch more because there isnt school

First of all, curling is very cool.  Second of all, how can you honestly respect the Summer Olympics after the addition of beach volleyball?  (Okay, I admit, snowboarding is no better, but still, beach volleyball?)

Both olympics have their questionable events. But summer olympics is way cooler because it has so many more events, or at least that's how it has always seemed to me. I didn't bother to check, but considering the number of medals awarded I think I'm right.


inequalities
function
algebra
rational function

Recall that $m^{2}_{a}=\frac{1}{4}(2b^{2}+2c^{2}-a^{2})$
So the inequality (1) reduces to:
$\sum_{cyc}\frac{a^{2}}{2b^{2}+2c^{2}-a^{2}}\geq 1$
Because this inequality is homogeneous, we can assume that $a^{2}+b^{2}+c^{2}=1$. Then we need to prove:
$\sum_{cyc}\frac{a^{2}}{2-3a^{2}}\geq 1$
Now, $\frac{a^{2}}{2-3a^{2}}\geq \frac{6a^{2}-1}{3}\Leftrightarrow 18a^{4}-12a^{2}+2\geq 0 \Leftrightarrow 2(3a^{2}-1)^{2}\geq 0$ which is true.
Therefore, $\sum_{cyc}\frac{a^{2}}{2-3a^{2}}\geq \sum_{cyc}\frac{6a^{2}-1}{3}=1$, because $a^{2}+b^{2}+c^{2}=1$.
The result follows.

nice proof; i see your substitution transforms a rational function in a,b,c into a rational function in one variable; this makes sense to me

once you go past that, i lost your motivation, how do you come up with the idea to set that greater than (6a^2-1)/3

what about some other function

i see the idea that that sums to the correct thing based on 1, but i feel like there would be other functions that you could use...like (a^n)/(a^n+b^n+c^n) n=2 or something like that

Standard inequalities trick; actually, the assumption $\sum a^{2}=1$ does nothing but allow for nicer convention. The method + motivation is isolated fudging standard technique, look at some of hte old mildorf posts in inequlaities section.
It is really straightforward, actually :wink: just choosing a suitable linear function. (okay its not the main technique, i.e. there's a constant thrown in to the typical $\frac{a^{r}}{a^{r}+b^{r}+c^{r}}$)

That $\frac{4}{3}$ in the second one needs to be replaced by $\frac{2}{\sqrt{3}}$ (consider the equilateral triangles).
[hide="2"]
Like what rem stated above, $2m_{a}=\sqrt{2b^{2}+2c^{2}-a^{2}}$ so that the inequality is equivalent to
\[\sum_\text{cyc}\frac{\sqrt{3}a^{3}}{\sqrt{2b^{2}+2c^{2}-a^{2}}}\geq a^{2}+b^{2}+c^{2}\]
Note that AM-GM gives
\[\sqrt{(3a^{2})(2b^{2}+2c^{2}-a^{2})}\leq \frac{3a^{2}+2b^{2}+2c^{2}-a^{2}}{2}=a^{2}+b^{2}+c^{2}\]
Applying for similar terms,
\[\sum_\text{cyc}\frac{\sqrt{3}a^{3}}{\sqrt{2b^{2}+2c^{2}-a^{2}}}\geq \sum_\text{cyc}\frac{3a^{4}}{a^{2}+b^{2}+c^{2}}\]
Now it just remains to prove $3(a^{4}+b^{4}+c^{4})\geq (a^{2}+b^{2}+c^{2})^{2}$, which is easy.
[/hide]

[quote="Altheman"]once you go past that, i lost your motivation, how do you come up with the idea to set that greater than (6a^2-1)/3
[/quote]
It's very often in this type of problems that one has to come up with a bound for a term. So just playing around with this bound and making sure the "cyclic sum of this bound" adds up to 1 I got this expression.


I am trying to find a viable route of the production of 3,3' Dinitro diphenyl sulfone using a Friedel craft reaction between m-nitrobenzene sulfonic acid (to sulfonyl chloride using thionyl chloride) and nitrobenzene. Since nitrobenzene is inert is this synthesis even possible?

Also is the synthesis of 3,3' Diamino diphenyl sulfone possible from a friedel-craft reaction between Metanillic acid and aniline?

Fridel-Crafts reactions with deactivated rings are not viable. How about nitrating diphenylsulfone?

Thanks for your response. That we have done and it works as well. Isolation of isomers 3,4' and 3,5' is an issue and cost is another. DPS is not that cheap. Im trying to explore if aniline can be introduced in the equation and it can be friedel-crafted for two reasons:
1. Reduction will not be required
2, I get it as a waste/by product stream

Im also trying to explore the possibility of friedel-crafting
Aniline and Chlorobenzene sulfonyl chloride without the formation of sulfonamides?

Anilines cannot also undergo Fridel-Crafts reactions. Here's a route that might work:

1) Prepare m-dinitrobenzene by nitrating benzene and then by nitrating nitrobenzene.

2) Reduce only one of the nitro groups of m-dinitrobenzene to get m-nitroaniline.

3) Diazotize m-nitroaniline and treat with SO2/CuCl2 to get m-nitrobenzenesulfonyl chloride (A).

4) Treat nitrobenzene (from step 1 or from a bottle available in the lab) with Br2/Fe to get m-bromonitrobenzene.

5) Treat m-bromonitrobenzene with magnesium metal in THF and inert atmosphere to get the corresponding Grignard reagent (B).

6) Finnaly, react A with B and a good yield of 3,3'-dinitrodiphenylsulphone should be obtained.

Thanks again for the route I will try it in the lab
The other thing I am going to try

Benzene sulfonic acid to m-nitrobenzene sulfonic acid
m-nbsa to m-nitrobenzene sulfonyl chloride (thionyl chloride/ chlorosulfonic acid + cucl2)
m-nbsc friedel craft with nitrobenzene to give 3,3' Dinitro diphenyl sulfone
Reduce H2 to 3,3' Diamino diphenyl sulfone

I think it should work?

[quote="Carcul"]

2) Reduce only one of the nitro groups of m-dinitrobenzene to get m-nitroaniline.

[/quote]

you could use (nh4)Sx or even HCOOH,Et3N,Pd tho the first one wud be better for selective reduction.(just in case you dont know)
[quote="Carcul"]
1) Prepare m-dinitrobenzene by nitrating benzene and then by nitrating nitrobenzene.

2) Reduce only one of the nitro groups of m-dinitrobenzene to get m-nitroaniline.

3) Diazotize m-nitroaniline and treat with SO2/CuCl2 to get m-nitrobenzenesulfonyl chloride (A).
[/quote]
cant we directly nitrate benzene sulfonic acid???

and how abt this?
nitrobenzene+ bromination(mono) to give meta bromo nitrobenzene
foll by addition of mg with THF as solvent to give corr organometallic
+ reaction with half eq. SO2Cl2 foll by red.

[quote="carcul"]
One more time, Fridel-Crafts reactions are not feasible with deactivated rings (such as nitrobenzene).
[/quote]
and can you carry out a F&C reaction bw benzene and m-nbsc and then nitrate the product? nitration will take place on the ring with no substituent as no2 is deactivating??

[quote="VARUNARASU"]cant we directly nitrate benzene sulfonic acid???[/quote]

Why couldn't we?

I think that even the -SO3H group gets replaced by -NO2 :maybe:  :maybe:

i think $ \ce{SO3H}$ will only go away bcos of ipso attack and hence may or may not get attacked at that posn  :maybe:

we know that -SO3H is good leaving group(infact sulphonation is a reversible reaction) so that's why I mentioned it.

Yes SO3H- is a good leaving group but not SO3H+. and during nitration, so3h cannot be replaced by no2+, as it wud have to leave as so3h+ or SO3. also if so3h was to be  replaced so easily by no2, we wudnt have used so3h as a protecting group for para position during nitration. of course if you OH or NH2, there is a possibility of so3h being replaced. i know sulfonation is a reversible reaction, but you have to use dil. acid or steam and excess of water for desulfonation, whereas during nitration u use conc. h2so4.


(I have no idea how difficult this is, move it please if necessary)

Find $\sum_{k=2}^\infty (\zeta{(k)}-1)$
(or , in other words:)
$\sum_{k=2}^\infty \sum_{j=2}^\infty \frac{1}{j^{k}}$.

[hide]
If we just change the order of summation we have

$\sum_{j=2}^{\infty}\sum_{k=2}^{\infty}\frac{1}{j^{k}}$,

which has a geometric series. So it becomes

$\sum_{j=2}^{\infty}\frac{\frac{1}{j^{2}}}{1-\frac{1}{j}}= \sum_{j=2}^{\infty}\frac{1}{j(j-1)}$,

which telescopes to

$\sum_{j=2}^{\infty}\left(\frac{1}{j-1}-\frac{1}{j}\right) = 1$.[/hide]

This is actually quite simple.

[hide="Hint"]Reverse the order of sums[/hide]

EDIT: Beaten by a split second ;)


function
analytic geometry
graphing lines
slope

There is a function $ f(x)$ which goes through $ (1,2)$. For two real numbers $ a$ and $ b$, points $ (a, f(a) \plus{} a), (b, f(b) \plus{} b)$, and $ (0,ab)$ lies on the same line. Find the function $ y \equal{} f(x)$.

The slopes between any two points are the same so

$ \frac {f(a) \minus{} f(b) \plus{} a \minus{} b}{a \minus{} b} \equal{} \frac {f(b) \plus{} b \minus{} ab}{b}$

$ \frac {f(a) \minus{} f(b)}{a \minus{} b} \equal{} \frac {f(b)}{b} \minus{} a$

We are given $ f(1) \equal{} 2$ so let $ b \equal{} 1$. Then

$ \frac {f(a) \minus{} f(1)}{a \minus{} 1} \equal{} f(1) \minus{} a$

$ \frac {f(a) \minus{} 2}{a \minus{} 1} \equal{} 2 \minus{} a \Rightarrow f(a) \equal{} (2 \minus{} a)(a \minus{} 1) \plus{} 2$

So $ f(x) \equal{} (2 \minus{} x)(x \minus{} 1) \plus{} 2$.