Hello, I was wondering if there is any chance of USAMO graders messing up?

Because I solved 1, 2, 4, 5 and wrote up very clean solutions. And I firmly believe they are correct solutions. And they are clean, trust me I know how to write clean solutions...

Somehow I got 17 pts. :mad:

I guess this is a pointless post and a pointless topic, but I am extremely disappointed.

All that can be said is that you probably had an incorrect solution in your set of 4. Maybe you could post your solutions and we can help you identify the loss of points?

[quote="Ignite168"]Didn't you make these exact same posts either last year or two years ago?[/quote]
I guess so. But 2 yrs ago I had very little olympiad experience. I got IMO silver last yr so I know how to write up proper solutions. :mad: No way I could get sth as low as 17 :mad:
Well I did not scan them or anything. Like for 2 I used the symmedian argument and continued did the thing posted by Delong on the forum, solutions for 4 and 5 were identical to the solutions on the website, 1 was just inductive construction which easily worked...


$ x,y,z \in R$ and
$ x\plus{}y\plus{}z\equal{}5$
$ x^{2}\plus{}y^{2}\plus{}z^{2}\equal{}9$  .
$ x^{2}y\plus{}y^{2}z\plus{}z^{2}x$     find  min and max.


function
floor function
inequalities
quadratics
modular arithmetic

1)Prove that for all $n\in \mathbb{N}$,

$\lfloor \sqrt{n}+\sqrt{n+1}\rfloor=\lfloor \sqrt{4n+2} \rfloor$

2)For any +ve integer $n$,prove that

$\lfloor \frac{n+1}{2} \rfloor+\lfloor \frac{n+2}{4} \rfloor+\lfloor \frac{n+4}{8} \rfloor+\cdots+\lfloor \frac{n+2^{n-1}}{2^n} \rfloor=n$

alright, I've got 2 its not great but here goes

[hide]
ok base case (1+1)/2  = 1

ok now assume it works for some n.

so for n+1 in order for any of the terms to increase from n to n+1 this means that n+1 must be 2^b(mod2^(b+1)).

now to prove that every n+1 must fall into one and only one of those as to not increase by 2.

lets assume the contrary and n+1 = 2^(a+1)k + 2^a = 2^(c+1)p +2^c.  where k and p are integers.

rearanging we get 2^(a-c) = (2p +1)/(2k+1). and its impossible to divide two odd numbers and get an even number so we get a = c which is a contradiction.

so now to prove that every n+1 must fall into one equivalence class 2^k(mod 2^(k+1))...Im not sure how to do this.  dang i was close too.[/hide]

uhh #1 asks to show the relation is true for all $n$ such that $n$ is a natural number right? cuz in that case i think i have it...

[hide]

let $p^2 \le n < (p+1)^2$ for some $n$ and some $p$ such that both $n$ and $p$ are natural numbers

then, our equation is turned into:

$\lfloor p + \sqrt{p^2+1}\rfloor = p + \lfloor \sqrt{p^2+1}\rfloor = \lfloor \sqrt{4p^2 + 2}\rfloor$

now we compare $\lfloor \sqrt{4p^2 + 2} \rfloor$ with $\lfloor \sqrt{4p^2} \rfloor = 2p$ and see that those two are the same for all natural numbers $p$

then we compare $\lfloor \sqrt{p^2+1} \rfloor = \lfloor \sqrt{p^2}\rfloor = p$ and see that this is again true for all natural numbers $p$

hence our equation just becomes $p+p = 2p$ which is obviously true

hmmm does that work?

[/hide]

[hide="1"]Suppose $k=\lfloor\sqrt{n}+\sqrt{n+1}\rfloor$.  Then

$k\le\sqrt{n}+\sqrt{n+1}<k+1$.

Since $\sqrt{n}+\sqrt{n+1}>0$, all parts of the inequality are nonnegative.  Squaring, we obtain

$k^{2}\le 2n+1+2\sqrt{n(n+1)}<(k+1)^{2}$

Also, we have $2\sqrt{n(n+1)}=\sqrt{4n^{2}+4n}=\sqrt{(2n+1)^{2}-1}$.  Therefore, the quantity inside the radical is not a perfect square, and the radical is not an integer.  We can replace the radical with the greatest integer less than it without affecting the value of $k$.  Therefore, we have

$k^{2}\le 2n+1+\sqrt{(2n+1)^{2}}-1<(k+1)^{2}$
$k^{2}\le 4n+1<(k+1)^{2}$

Now, suppose $j=\lfloor\sqrt{4n+2}\rfloor$.  As above, we have

$j\le\sqrt{4n+2}<j+1$

Again, we can square to obtain

$j^{2}\le 4n+2<(j+1)^{2}$

Suppose $k\ne j$.  Then $4n+2$ is a perfect square, which is impossible since $2$ is not a quadratic residue $\pmod{4}$.  Therefore, $k=j$, and $\lfloor\sqrt{n}+\sqrt{n+1}\rfloor=\lfloor\sqrt{4n+2}\rfloor$ for all $n$.[/hide]

[quote="gotztahbeazn"]then, our equation is turned into:

$\lfloor p + \sqrt{p^2+1}\rfloor = p + \lfloor \sqrt{p^2+1}\rfloor = \lfloor \sqrt{4p^2 + 2}\rfloor$
[/quote]

I think in that step you basically said that $\lfloor\sqrt{n}+\sqrt{n+1}\rfloor=\lfloor\sqrt{n}\rfloor+\lfloor\sqrt{n+1}\rfloor$, which isn't necessarily true.

[quote="gotztahbeazn"]then, our equation is turned into:

$\lfloor p + \sqrt{p^2+1}\rfloor = p + \lfloor \sqrt{p^2+1}\rfloor = \lfloor \sqrt{4p^2 + 2}\rfloor$
....
hmmm does that work?[/quote]

No.  That isn't what the equation is turned into.  You had the right first step, though.

matt, your criticism would be right if $p = \sqrt n$ (not necessarily an integer) -- because $p$ is an integer, he can in fact make that transformation.  The problem was plugging in $p^2$ for $n$.

[quote="gotztahbeazn"]let $p^2 \le n < (p+1)^2$ for some $n$ and some $p$ such that both $n$ and $p$ are natural numbers[/quote]

In his solution, $p\le \sqrt{n}<p+1$, and $p=\lfloor\sqrt{n}\rfloor$.  So didn't he basically do $\lfloor\sqrt{n}+\sqrt{n+1}\rfloor=\lfloor\sqrt{n}\rfloor+\lfloor\sqrt{n+1}\rfloor=p+\lfloor\sqrt{n+1}\rfloor$?  I'm confused... :huh:

yea i think i see wat i did wrong... hmm

[quote="gotztahbeazn"][quote="matt276eagles"]

$k^{2}\le 2n+1+\sqrt{(2n+1)^{2}}-1<(k+1)^{2}$
$k^{2}\le 4n+1<(k+1)^{2}$
[/quote]

nice solution btw, but do you mean $k^2 \le 2n+1 + \sqrt{(2n+1)^2 - 1} < (k+1)^2$?

and from there since ur taking the floor of everything you can, without affecting the value of $k$, turn that into:

$k^2 \le 2n+1 + \sqrt{(2n+1)^2} = 4n+2 < (k+1)^2$

and ur done arent u?[/quote]

I had $k^{2}\le 2n+1+\sqrt{(2n+1)^{2}-1}<(k+1)^{2}$, and the floor of the radical is $\sqrt{(2n+1)^{2}}-1$, not $\sqrt{(2n+1)^{2}}$.  Switching it with $\sqrt{(2n+1)^{2}}$ would be like taking the ceiling, which might affect the value of $k$.

1)

look at the LHS, it is equivalent to:

$2\sqrt{n}+\frac{1}{ \sqrt{n+1}+\sqrt{n} }$ (1)

the RHS is:

$2\sqrt{n}+\frac{1}{\sqrt{n+\frac{1}{2}}+\sqrt{n}}$ (2)

now it seems intuitively obvious, but that is no proof...perhaps that extra half could push the bottom expression to the next integer...i might finish it tomorrow, i have to go to bed now...:P

EDIT:...

now, i will address that the case where after the floor is applied, it will not push it into the next integer...

clearly, for large positive integers, (1) < (2)

suppose $\lfloor (1) \rfloor = q$, (note $q\in\mathbb{Z}^+$ then if they were not to be equal, $\lfloor (2) \rfloor = q+1$, then we get the set of inequalities

$q\le \sqrt{n}+\sqrt{n+1} < q+1$ (3)
$q+1\le \sqrt{4n+2} < q+2$ (4)

$q^2\le 2n+1+2\sqrt{n(n+1)}<(q+1)^2$ (this is (3)^2 )

$(q+1)^2\le 4n+2<(q+2)^2$ (this is (4)^2 )

subtract those to get

$2q+1 \le 2n+1-2\sqrt{n(n+1)}<2q+3$

$q\le n-\sqrt{n(n+1)}<q+1$

$q\le \frac{-n}{n+\sqrt{n(n+1)} }<q+1$

which is obviously not true...contradiction, so the LHS and RHS have to be the same...


Say I solved for values of $ x$ in my solution.  To check my answer and to eliminate extraneous solutions, I resubstitute my $ x$'s into my original equation.  Do I have to mathematically show that an equation is true by manipulating the variables? Or can I simply substitute my answer for $ x$ into the original equation and say that $ x \approx n$ by using a calculator? Obviously the former would be much more rigerous but can the latter method be used to get a 5?

either way would probably get you a 5, but its definitely better to show it algebraically - or you could prove some bounds for the solution which eliminate the other possibilities.

Showing that two quantities are "approximately equal" using a calculator is not rigorous.  I cannot say how much deduction (if any) there would be... it would likely vary with the problem.

By USAMTS Rules, if you use a computer, or calculator, etc., you must prove that whatever you are using is performing the steps correctly. That is, you must prove that your calculator is verifying the equation correctly, and that's basically the same as verifying it yourself.

[quote="USAMTS Rule Number 9"]Except where otherwise noted in problems, participants may use calculators and computers to solve problems. In this case, the participant must explain the program that was used to solve the problem, and must prove that the method used in the program does solve the problem. Symbolic manipulation calculators and mathematical programs such as Mathematica, Maple, and Matlab are considered programs for this purpose. Therefore, if a key step of a participant's solution uses these tools, the participant must demonstrate that the tool completed the step correctly.[/quote]

[quote="Ketvoman"]On the contrary of using approximate values.  Can I say in my solution that I pluged my solution into the original equation and typed that whole thing into a calculator.  And then I use a test function on a calculator to define whether the equation is true or not.  Is that rigerous?[/quote]

The problem with calculators is that they cannot do precise calculations with long decimals. Although two expressions may turn out equal on a calculator, you can't be sure that they aren't say, 10^-20 apart.

For example, last year I bashed round 4 problem 5. i computed all of the side lengths in terms of trig functions of multiples of pi/7, and had to proved that some messy trig expression was equal to some other messy trig expression. before i did so, i plugged them both into mathematica, which is very precise, but still rounds. they appeared to be equal, but i still had to come up with an algebraic proof (which i did, it wasn't exactly nice).

Ok so just show everything algebraically if you can


Looking at this Mini Map will help you understand my question:

http://img209.imageshack.us/my.php?image=examplexw9.png

- The Green Dot is where I started (Coordinates right now arn't important).
- The Blue Dot was my first move which was (723, 345).
- I'm the little white square that I outlined with red, which is (748, 298).
- From where I am now, the Blue Dot is at or around (807, 303).

From the Green Dot to the Blue Dot, is how I got (723, 345).

So, how do I obtain the (X, Y) of the Blue Dot from where I am now?

It's extremely rude to post the same thing in multiple fora.

Well I'm not getting much help in the other forum so I thought I would increase my chances in two. How is that rude?

Discussion of the same problem should be concentrated in the same thread.  It wastes everyone's time to see the same question posted twice, and the second post is essentially spam.  You should think harder about what forum you post the question in originally instead of posting in multiple forums.

You also shouldn't expect answers to come quickly or promptly.  Beggars can't be choosers.

You're clearly in Falador (splling?) I used to play that game, but that's all you need to know to play :)


linear algebra
matrix
algebra
polynomial
superior algebra
superior algebra solved

Does anyone have some interesting problems regarding the nth power of a matrix? If yes, please post them here or send me a private msg with them!!!

thank you very much! :D :D tschuesss!!!

eh.. well I see that nobody posted a problem :(
I will post one then:
Compute the nth power of this matrix:
[1 1 0 0]
[0 1 1 0]
[0 0 1 1]
[1 0 0 1]

cheers! :D

[quote="Lagrangia"]
Compute the nth power of this matrix:
[1 1 0 0]
[0 1 1 0]
[0 0 1 1]
[1 0 0 1]

[/quote]
Let M the given matrix M=I + A where I is the identity matrix in M4(R)
A=
[0 1 0 0]
[0 0 1 0]
[0 0 0 1]
[1 0 0 0]

A=
[0 0 1 0]
[0 0 0 1]
[1 0 0 0]
[0 1 0 0]

A^3=
[0 0 0 1]
[1 0 0 0]
[0 1 0 0]
[0 0 1 0]

A^4=I

M^n = (I+A)^n =\sum_{k = 0 to n}C(n,k)A^k
=\sum C(n,4k)A^4k+\sum C(n,4k+1)A^(4k+1) +\sum C(n,4k+2)A^(4k+2) + \sum C(n,4k+3)A^(4k+3) now use the expression of A,A^3,A^4 to get the coefficient of M^n

Moubinool

A problem related to this one, was in GM. (a friend gave me this problem)
:)

Here is it:
consider the matrices:
A=
[1 1 0 0]
[0 1 1 0]
[0 0 1 1]
[1 0 0 1]

B=
[1 1 1 1]
[1 1 1 1]
[1 1 1 1]
[1 1 1 1]

C=
[1 0 -1 0]
[0 1 0 -1]
[-1 0 1 0]
[0 -1 0 1]

they are all matrices in M4(R)
a) Show that the sequence (Xn)n>=1 in M4(R) given by the relation:
Xn=2^(1-n/2)A^n-2^(n/2-1)B for any n natural is periodical.
b)For which values of n we have Xn=C?

cheers! :D :D

[quote="Lagrangia"]A problem related to this one, was in GM
consider the matrices:
A=
[1 1 0 0]
[0 1 1 0]
[0 0 1 1]
[1 0 0 1]

B=
[1 1 1 1]
[1 1 1 1]
[1 1 1 1]
[1 1 1 1]

C=
[1 0 -1 0]
[0 1 0 -1]
[-1 0 1 0]
[0 -1 0 1]

they are all matrices in M4(R)
a) Show that the sequence (Xn)n>=1 in M4(R) given by the relation:
Xn=2^(1-n/2)A^n-2^(n/2-1)B for any n natural is periodical.
b)For which values of n we have Xn=C?
[/quote]

a) Incredible the period is 8, X(n+8)=X(n) for n \in N\{0}
b) X(8k)=C where k \in N\{0}

Moubinool Omarjee

1)The characteristic polynomial of A is det(A-xI)=x^4-4x^3+6x-4x
eigenvalues are 0; 2; 1+i= 2^(1/2)e^(i \pi /4); 1-i=2^(1/2)e^(-i \pi /4)

Canonical Jordan form of A is
[1/4  1/4  1/4  1/4]  [0 0  0  0]     [1 -1 1 -1]
[-1/4 1/4  i/4  -i/4]  [0 2   0  0]    [1  1 1  1]
[1/4 1/4  -1/4 -1/4] [0 0 1+i 0]    [1 -i -1  i]
[-1/4 1/4 -i/4  i/4]   [0 0 0   1-i]   [1 i -1  -i]

Taking the n power we get

A^n = (2^n)/4.B + 2^(n/2-1)F_n where F_n is the following matrix
F_n=
[cos(npi/4)  sin(npi/4)  -cos(npi/4)  -sin(npi/4)]
[-sin(npi/4)  cos(npi/4)  sin(npi/4)   -cos(npi/4)]
[-cos(npi/4)  -sin(npi/4)  cos(npi/4)  sin(npi/4)]
[sin(npi/4)   -cos(npi/4)   -sin(npi/4)  cos(npi/4)]

From now we get X_n=2^(1-n/2)A^n -2^(n/2 -1)B in term of F_n

X_n =2^(1- n/2 +n-2)B +F_n -2^(n/2 -1)B=F_n

now X_n is 8 period like cos(npi/4), sin(npi/4)

oubinol hack omarje


function
algebra
domain
topology
real analysis
real analysis theorems

why is $ f^* = \sup\{g: g\in \mathcal{L}\}$ Borel measurable? Here ${ \mathcal{L} = \{g\in C[a,b]: g <= f\}}$, and $ f$ is a bounded real function on $ [a,b]$.

I don't know where to start-is it Borel measurable simply because $ f^*$ is an open set, and all open sets are Borel-measurable (since they are the complements of closed sets, being B-measurable)? Something seems sloppy here.

Thanks,

Los

Start with recalling that $ C([a,b])$ is separable ;). Also $ f^*$ is a [b]function[/b], not a [b]set[/b], so what you wrote about its openness makes little sense if any :(.

Let $ \mu$ be a set function whose domain is a class $ \mathcal{A}$ of subsets of a set $ X$, and whose values are nonnegative extended reals. We can consider our function $ f^*$ to be a set function (whose domain is a class $ C[a,b]$ of continuous functions on a set $ X \equal{} \mathbb{R}$). Right. Now our $ \mu \equal{} f^*$ is countably additive because $ \mu (\bigcup_{k \equal{} 1}^\infty A_k) \equal{} \sum_{k \equal{} 1}^\infty \mu (A_k)$--or in our notation:

$ f^*([a_1,a_2]\cup [a_2,a_3]\cup \cdots [a_{n_1},a_n]\cup) \equal{} f^*([a_1,a_2]) \plus{} f^*([a_2,a_3]) \plus{} \cdots f^*([a_{n \minus{} 1},a_n])$, which holds as $ n \rightarrow \infty$.

Now a measure on $ X$ is a countably additive nonnegative set function--whose domain is a $ \sigma$-algebra of subsets on $ X$.

But the $ \sigma$-algebra generated by the closed subsets of $ X$ is precisely the Borel-algebra, and so $ f^*$ is then Borel-measurable!

Is this correct?

Los

I'm afraid you are thoroughly confused about the conditions of the problem. At least, what you wrote in your second post is incompatible with what you wrote in the first. The first post reads (as far as I can understand it):

Let $ f: [a,b]\to\mathbb R$ be any bounded function. Consider the class $ \mathcal L$ of continuous functions $ g: [a,b]\to\mathbb R$ such that $ g\le f$ pointwise on $ [a,b]$. Now, for every $ x\in[a,b]$, define $ f^*(x)\equal{}\sup_{g\in\mathcal L}g(x)$. You'll get a real-valued function $ f^*$ defined on $ [a,b]$. Prove that this function is Borel measurable.

Your second post is much less comprehensible, so I'll not try to figure out what exactly you meant there, but it is clearly something completely different.

hallo fedja, yes the second post does seem to be incomprehensible. It was supposed to be an answer to your hint, that: Start with recalling that $ C([a,b])$ is separable, but now i see that it is way off. I would appreciate any further direction.

Thanks,

Los

Actually the next step is the full solution: every continuous function is Borel measurable, separability implies that the supremum can be taken over a countable set and, finally, a supremum of any countable family of Borel measurable functions is Borel measurable. That's it. :)

1. Why is every continuous function Borel measurable? Does this have something to do with the fact that the domain of a continuous function is a $ \sigma$-algebra?

2. why does seperability imply the supremum can be taken over a countable set?

I see :roll:... OK, I'll answer your questions with questions.

1) What is the definition of a measurable function?

2) What does separability mean?

:lol:
function is measurable if  $ f^{ \minus{} 1}$ of set from algerba lays in Algebra.
if $ f$ is continuous  then $ f^{ \minus{} 1}$ of open is open , hence lays in borel algebra.


If $a,b,c>0$  with $abc=1$  then prove that

$\sqrt{\frac{a+b}{b+1}}+\sqrt{\frac{b+c}{c+1}}+\sqrt{\frac{c+a}{a+1}}\geq 3$

Please post full solutions.Thanks

Just use $x+y+z\geq 3\sqrt[3]{xyz}$ so we need t o prove that $\prod(a+b)\geq \prod(a+1)$,so $(\sum a)(\sum ab)\geq 3+\sum a+\sum ab$,so $(a+b+c-1)(ab+bc+ca-1 )\geq 4$ ,obvious. :)

Well, I thought that this problem has a nice idea, but I was wrong. :( Anyway, my solution is the same with Vlad's one  :D . In fact, I might add something. I don't know who created this problem, as Vlad showed the inequality reduces to this one: \[ (a+b+c)(ab+bc+ca)\geq 3+a+b+c+ab+bc+ca,  \]
which is equivalent with \[ (a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\geq 3+a+b+c+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}.  \]

Some time ago, I posted here on this forum an inequality which appeared in G.M. 1959 or 1969 which states that:

 (*)  If $a,b,c >0$ such that $abc=1.$  Prove that the following inequality holds:

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

 Well, strange, the same inequality (*), appears in 2004 RMNO shortlist due to Harazi. Anyway, (*) follows immediately from AM-GM and guess what: (*) can be rewritten as:

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

 Compare this one with the one obtained by Vlad  and you will see that (*) is stronger. ;)
 So, I guess Harazi created the mathlinks contest problem. :)


[quote="Ceva Theorem"]This problem was adopted from the CSU Math Day competition in Colorado.

What is the area of the largest octagon that can be inscribed is a square of side $1$?[/quote]
Regular octagon?

Wow I am curious. How do you do this one? does it say it is a regular octagon? because i don't know

[quote="arsemesomething"]hope i did all my calculations right:

[hide]
i calculated the total angles in the octagon, then the angle each side in the octagon subtends with respect to the center of the square, then use the cosine law to find the side of the octagon, knowing that each side makes an isosceles with the center, the two equal sides being 1/2. then sine to find the height, then area, then total area.

$\text{Area}=\frac{5}{2}\sqrt{\frac{1}{2}(1-\cos{36})}\sin{72}\doteq 0.734731565$
[/hide]

edit: if it weren't a regular octagon, the octagon can be made to be as large as the area of the square.    :o[/quote]
What does $\doteq$ mean and how do you make it fill up the whole square if it's irregular? Then it would [i]be[/i] a square.

You don't have to resort to trig. I got a different answer, sidelength s,

$\frac{2s}{\sqrt{2}}+s = 1$
Once you get s, you can subtract the four corner triangles from 1.

I got $2(\sqrt{2}-1)$

thats greater than 1 the area of the square so i think  thats not possible

ok first assume that one side of the octagon is the length x

because the exterior angles of an octagon are 45 degrees this means you have a 45 45 90 triangle

so then that mean x+(2x)/sqrt(2)=1
solve for x and we get x=sqrt(2)/(2+sqrt(2))

then we subtract from 1 4 triangles with base and height x

so 1-2(sqrt(2)/(2+sqrt(2)))^2

witch is about 0.65685424949238019520675489683879  XD MS calculator goes out a lot of places.

(sorry i don't know latex :[ )

[hide]We form an octagon by cutting off the four corners of the square. Call the side length of the octagon $x$. Thus, using 45-45-90 triangle properties, the side lengths of the triangles are $\frac{x\sqrt{2}}{2}$. The area of the octagon would be the area of the square minus the area of the for triangles. This is $1-4\left(\frac{x\sqrt{2}}{2}\div 2\right)=1-x^{2}$ To find $x$, we use the fact that the three segments make up the side of the square. $x\sqrt{2}+x=1\rightarrow x=\sqrt{2}-1$. Thus, the area is $1-(\sqrt{2}-1)^{2}=2(\sqrt{2}-1)$.[/hide]

[quote="arsemesomething"]god darn it  :mad: i thought octagon meant a 10-sided figure  :blush:[/quote]
:rotfl:

That's a decagon.

[quote="JavaMan"]Are you sure their base and height are x? Perhaps they're $\frac{x}{\sqrt{2}}$.[/quote]

i said 1 side of the octagon is x

and because i said 1=x+2x/sqrt(2)

and that is 1 side of the octagon + 2 bases of the triangle i had that in the answer but i didn't specifically point that out sorry for any confusion

sorry it took me so long to respond and yes, it is a regular octagon.
[hide]

Here's the solution:

$x+2(\frac{x}{\sqrt{2}}) = 1$
$Area = 1-x^{2}= 1-\frac{1}{(\sqrt{2}+1)^{2}}$
$= \frac{2}{1+\sqrt{2}}= 2(\sqrt{2}-1) = 2\sqrt{2}-2$
[/hide]

[quote="i_like_pie"]What does $\doteq$ mean and how do you make it fill up the whole square if it's irregular? Then it would [i]be[/i] a square.[/quote]

One question at a time:

$\doteq$ is another way of saying "approximately equal to".

And yes, you couldn't fill up the whole square, but you could get arbitrarily close.


[url=http://img91.imageshack.us/my.php?image=vectorkc8.jpg][img]http://img91.imageshack.us/img91/9968/vectorkc8.th.jpg[/img][/url]

if the magnitude of v is 8 then if I want to resolve this into components should it be:

8*cos(320)i - 8*sin(320)j

or should it be

8*cos(40)i + 8*sin(40)j

For the i coefficients both are OK and for the j coefficients the same would be truth if you interchange the sign...

A similar question:

Give the components of the velocity vector for wind blowing at 12 km/hr toward the southeast.

12*cos(45) - 12*sin(45) is my answer, why is this wrong?

Maybe it's the way you're putting it; coul be $ (6 \sqrt2, \minus{}6 \sqrt2)$, or can you post the answer?


number theory
greatest common divisor
algebra
polynomial
Vieta
modular arithmetic
quadratics

find all natural number a,b,c such that
$ a\plus{}b\plus{}c\equal{}\sqrt{abc}$

[quote="tkrass"]find all natural number a,b,c such that
$ a \plus{} b \plus{} c \equal{} \sqrt {abc}$[/quote]
Solutions so far...

[hide]
Case I)

Without loss of generality,
let $ a \equal{} b \equal{} c$.

$ a \plus{} a \plus{} a \equal{} \sqrt {aaa}$

$ 3a \equal{} \sqrt {a^3}$

$ (3a)^2 \equal{} (\sqrt {a^3})^2$

$ 9a^2 \equal{} a^3$

$ a^3 \minus{} 9a^2 \equal{} 0$

$ a^2(a \minus{} 9) \equal{} 0$

$ a^2 \equal{} 0$
$ a \equal{} 0$ . . . Omit this because it is not a positive integer.

$ a \minus{} 9 \equal{} 0$
$ a \equal{} 9$

$ (9,9,9)$

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

Case II)

Without loss of generality,
let $ b \equal{} a \ne c$.

$ a \plus{} a \plus{} c \equal{} \sqrt {aac}$

$ 2a \plus{} c \equal{} \sqrt {a^2c}$

$ (2a \plus{} c)^2 \equal{} (\sqrt {a^2c})^2$

$ 4a^2 \plus{} 4ac \plus{} c^2 \equal{} a^2c$

$ c^2 \plus{} 4ac \minus{} a^2c \plus{} 4a^2 \equal{} 0$

$ c^2 \plus{} (4a \minus{} a^2)c \plus{} 4a^2 \equal{} 0$

The discriminant $ B^2 \minus{} 4AC \equal{}$

$ (4a \minus{} a^2)^2 \minus{} 4(1)(4a^2) \equal{}$

$ 16a^2 \minus{} 8a^3 \plus{} a^4 \minus{} 16a^2 \equal{}$

$ a^4 \minus{} 8a^3 \equal{}$

$ a^2(a^2 \minus{} 8a)$.

The discriminant must be a perfect square in hopes
that $ c$ can be an integer.

$ a^2 \minus{} 8a \equal{} 0$

$ a(a \minus{} 8) \equal{} 0$

$ a \equal{} 0$  . . . This cannot be used.

$ a \equal{} 8$
Then $ b \equal{} 8$

Then $ c \equal{} 16$

The order can be changed.

$ (8,8,16)$
$ (8,16,8)$
$ (16,8,8)$

Also, for other possibilities,

$ a^2 \minus{} 8a$
$ a^2 \minus{} 8a \plus{} 16 \minus{} 16$
$ (a \minus{} 4)^2 \minus{} 16 \equal{} k^2$, for $ k$ a positive integer
$ (a \minus{} 4)^2 \minus{} k^2 \equal{} 16$

$ (a \minus{} 4 \minus{} k)(a \minus{} 4 \plus{} k) \equal{} 16$

Possibilities:

$ a \minus{} 4 \plus{} k \equal{} 8$
$ a \minus{} 4 \minus{} k \equal{} 2$
--------------------
$ 2k \equal{} 6$
$ k \equal{} 3$

with $ a \equal{} 9$

Then $ b \equal{} 9$

$ c \equal{} \frac {45 \pm 27}{2}$

$ c \equal{} 9$ . . . Omit this from this case because $ c$ is different from
$ a$ and $ b$, but it is a solution in Case I).

$ c \equal{} 36$

The order can be changed.

$ (9,9,36)$
$ (9,36,9)$
$ (36,9,9)$

$ a \minus{} 4 \plus{} k \equal{} \minus{} 2$
$ a \minus{} 4 \minus{} k \equal{} \minus{} 8$
----------------------
$ 2k \equal{} 6$

with $ a \equal{} \minus{} 1$ . . .  Omit this because $ a$ is a positive integer.

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

Solutions so far for all variables equal to one another and for two of the
variables equal to each other with the third variable different:

$ (9,9,9)$
$ (8,8,16)$
$ (8,16,8)$
$ (16,8,8)$
$ (9,9,36)$
$ (9,36,9)$
$ (36,9,9)$

[/hide]

Doesn't look like this question is easy.

[hide]
Squaring gives $ a^2 \plus{} b^2 \plus{} c^2 \plus{} 2ab \plus{} 2bc \plus{} 2ca \equal{} abc \quad (*)$.

Let
\[ t \equal{} \text{max}\,\big\{\text{gcd}\,(a,b),\text{gcd}\,(b,c),\text{gcd}\,(c,a)\big\},\]and suppose wlog $ t \equal{} \text{gcd}\,(a,b)$. Then $ (*) \Longrightarrow a | b^2 \plus{} 2b \plus{} c^2 \Longrightarrow t | c$, so it follows that the pairwise $ \text{gcd}$ of $ a,b,c$ is $ t$. Define $ a' \equal{} a/t,\, b' \equal{} b/t,\, c' \equal{} c/t$. At most one of $ a',b',c'$ may share a common factor with $ t$, wlog $ c'$. Then
\[ t(a' \plus{} b' \plus{} c') \equal{} \sqrt {t^3a'b'c'}\ \Longrightarrow\ a' \plus{} b' \plus{} c' \equal{} \sqrt {ta'b'c'}.\]Thus $ a' \equal{} p^2, b' \equal{} q^2, c' \equal{} r^2/t$ and $ tp^2 \plus{} tq^2 \plus{} r^2 \equal{} tpqr\quad (\star)$.

By Vieta jumping (have not verified whether or not it gives complete solutions), if $ (p,q,r)$ is a solution with respect to a given $ t$, then so are $ (qr \minus{} p,q,r),(p,pr \minus{} q,r)$ and $ (p,q,tpq \minus{} r)$, if an initial solution can be found with that $ t$. [hide="Concrete examples"]As a concrete example, take $ t \equal{} 9$. If $ 9 | r,$ then $ 3 | p^2 \plus{} q^2$, but $ \text{gcd}\,(p,t) \equal{} \text{gcd}\,(q,t) \equal{} 1$, so $ p^2 \plus{} q^2 \equiv 2 \pmod{3}$, contradiction. Thus, taking $ r \mapsto r/3$, if suffices to solve the equation
\[ 3pqr \equal{} p^2 \plus{} q^2 \plus{} r^2.\]By Vieta jumping, if $ (p,q,r)$ is a solution, then $ (p,q,3pq \minus{} r)$ is a solution, and all such solutions can be generated in this way from the initial solution $ p \equal{} q \equal{} r \equal{} 1$; examples being $ (p,q,r) \equal{} (1,1,2),(1,2,5),(1,5,13),(1,13,34),(2,5,29)$, etc. This gives an infinite class of solutions, where $ \{a,b,c\} \equal{} \{9p^2,9q^2,9r^2\}$.

Another example is provided above by [i]Arrange Your Tan[/i], with $ t \equal{} 8$, $ (p,q,r) \equal{} (1,1,4)$, and casework can probably show that all such solutions for $ t \equal{} 8$ can be generated by that triple. [/hide]

Note: If $ \text{gcd}\,(c,t) \equal{} 1$, then $ t \equal{} (t')^2$ and $ r : \equal{} \sqrt {c}$, giving $ p^2 \plus{} q^2 \plus{} r^2 \equal{} t'pqr$. Suppose $ 3 \nmid p,q,r$; then $ 3 | p^2 \plus{} q^2 \plus{} r^2 \Longrightarrow 3 | t' \equal{} \sqrt {\text{gcd}\,(p,q,r)}$, contradiction. It follows that $ 3$ divides into at least one of $ p,q,r$, wlog $ p$; if $ 3 \nmid q,r$, then $ q^2 \plus{} r^2 \equiv 1,2 \pmod{3}$, but $ 3 | p(p \minus{} t'qr) \equal{} q^2 \plus{} r^2$, contradiction. Thus $ 3 | p,q,r \Longrightarrow 3 | t$, contradiction.

Some criterion on $ t$ can be obtained by quadratic reciprocity from $ (\star)$, eg $ \text{gcd}\,(r^2/t,t)$ is not divisible by $ 4$ or any number of the form $ 4k \plus{} 3$.
[/hide]

[hide="So the solutions are .."]
$ t \equal{} 5$: $ \{a,b,c\} \equal{} \{5,20,25\}, \{5,45,25\}, \{405,20,25\}, \ldots$

$ t \equal{} 8$: $ \{a,b,c\} \equal{} \{8,8,16\}, \{8,72,16\}, \{8,72,400\}, \ldots$

$ t \equal{} 9$: $ \{a,b,c\} \equal{} \{9,9,9\}, \{9,9,36\}, \{9,36,225\},\{9,225,1521\}, \{36,225,7569\}, \ldots$.

I believe that these are the only values of $ t$ that work.
[/hide]

Edit: @dgreenb801, should be solve-able by Pell's equations.

I have it down to seeing when the following can be perfect squares, for natural $ b$:
$ 3b^2 \minus{} 36b$
$ 21b^2 \minus{} 196b$
$ 5b^2 \minus{} 36b$
If someone can solve this then I will post the whole solution.

(I'm curious to see your solution)

For the first, solutions exist; $ x \plus{} y\sqrt {3} \equal{} (3 \plus{} 2\sqrt {3})(2 \plus{} \sqrt {3})^n \Longrightarrow b \equal{} 6y \plus{} 6$, for $ x,y,n \in \mathbb{N}$, that is $ b \equal{} 18, 48, 162 \ldots$

The second probably has no solutions.

For the third, solutions exist: $ x \plus{} y\sqrt {5} \equal{} (27 \plus{} 9\sqrt {5})(9 \plus{} 4\sqrt {5})^n \Longrightarrow b \equal{} \frac {18 \plus{} x}{5}$, that is $ b \equal{} 9, 1521, \ldots$; other families exist, starting with $ b\equal{}36$, $ b\equal{}225$, etc.

[quote="azjps"]
[/hide]

[hide="So the solutions are .."]
$ t \equal{} 5$: $ \{a,b,c\} \equal{} \{5,20,25\}, \{5,45,25\}, \{405,20,25\}, \ldots$

$ t \equal{} 8$: $ \{a,b,c\} \equal{} \{8,8,16\}, \{8,72,16\}, \{8,72,400\}, \ldots$

$ t \equal{} 9$: $ \{a,b,c\} \equal{} \{9,9,9\}, \{9,9,36\}, \{9,36,225\},\{9,225,1521\}, \{36,225,7569\}, \ldots$.

I believe that these are the only values of $ t$ that work.
[/hide]

[/quote]

Your ordering of the values of the variables as shown and your ellipses don't
account for all the permuting of the values.

It's rather hard to systemically write out the solutions, see the attached (poorly drawn) tree, which just shows solutions divisible by $ 5$. [Note; the solutions are written by sets which accounts for all permutations.]

Sorry, my proof was wrong, I assumed that if $ a \ge b \ge c$, then $ (a \plus{} b \plus{} c)^2 \ge 9ab$ which is not always true. This led to $ c \le 9$, and we must have $ (a \plus{} b \plus{} c)^2 > 4ab$ so I plugged in $ c \equal{} 5$,$ 6$,$ 7$,$ 8$, and $ 9$, got quadratics in $ a$ and found their discriminants.


I have recently updated installed the newest acrobat reader 8.0 and subsequently uninstalled the older 5.0 version.

Now my Latex with TeXniCenter doesn't work anymore, it still refers to the old 5.0 directory. How do I fix that?

Could you please give the directories?

You could fix that at Tools>Options, I believe.

If the problem persists, re-install TeXniCenter

TexnicCenter needs to know where the new Adobe Reader is. Tell it by going to Tools - Define Output Profiles - LaTeX=>PDF (and also LaTeX=PS=>PDF if you have it) - Viewer Tab - Path of Executable - Press button with ... on it and find where AcroRd32.exe is - OK


function
integration
calculus
derivative
logarithms
calculus computations

Let $f: \mathbb{R+}\longrightarrow\mathbb{R}$ be a differentiable function with f(1)=3 and satisfying:

$\int^{xy}_{1}f(t)\,dt = y\int^{x}_{1}f(t)\,dt+x \int^{y}_{1}f(t)\,dt$  $\forall x,y \in \mathbb{R+}.$

Find f(x).

Taking partial derivatives first with respect to $x$ and then with respect to $y$, we get
$yf(xy)=yf(x)+f(y)$
$f(xy)+xyf'(xy)=f(x)+f(y)$

Since $f(x)+f(y)$ depends only on $xy$, let's set $g(t)=f(e^{t})$ for $t\in\mathbb{R}$. We have $g'(t)=e^{t}f'(e^{t})$, so our equation becomes $g(a+b)+g'(a+b)=g(a)+g(b)$ (using $a=\ln x$ and $b=\ln y$). Setting $b=0$, we get $g(a)+g'(a)=g(a)+3$, so $g'(a)=3$ for all $a$, and $g(s)=3s+3$ for all $s$.
This makes $f(t)=3\ln t+3$ for all $t$

Verifying that this actually works:
$\int_{1}^{xy}3\ln t+3\,dt=\left[3t\ln t\right]_{t=1}^{t=xy}=3xy\ln(xy)$
$y\int_{1}^{x}3\ln t+3\,dt+x\int_{1}^{y}3\ln t\,dt=y\left[3t\ln t\right]_{t=1}^{t=x}+x\left[3t\ln t\right]_{t=1}^{t=y}$
$= 3xy\ln x+3xy\ln y$
Since $\ln(xy)=\ln x+\ln y$, the two sides are equal. We have that $f(1)=3$ as well, and we are done.


let A,B be comlex n*n matrices s.t.(AB-BA)*A=A*(AB-BA).show AB-BA is nilpotent. :)

Assume the following lemma known (it's probably a dozen times on that forum):
$M\in \mathcal{M}_{n}(\mathbb{C})$ is nilpotent iff $\text{tr}(M^{k})=0$ for all $k>0$.
Now, denote $[A,B]=AB-BA$.
Check that $[A,B]^{k}=[A,B[A,B]^{k-1}]$, for all $k>0$.
Hence $\text{tr}([A,B]^{k})=0$ for all $k>0$, and $AB-BA$ is nilpotent.


geometry
3D geometry
tetrahedron
rotation
area of a triangle
Heron\u0027s formula
algebra

The sides of $\triangle ABC$ measure 11,20, and 21 units. We fold it along $PQ,QR,RP$ where $P,Q,R$ are the midpoints of its sides until $A,B,C$ coincide. What is the volume of the resulting tetrahedron?

Yes, sir!

If you can find that it's 45 cubic unit then you're cool.

It involves solving a series of equations with a series of unknowns.

It came out to be exactly 45 cubic units.

Dien

Picture:$AB=11,BC=20,CA=21$ and $PQ=10,QR=5.5,RP=10.5$
$AD \bot BC, BE \bot CA \quad AD \cap BE = T $
$\triangle APQ$ rotates to $PQT$, $\triangle CQR$ rotates to $QRT$ and $\triangle BRP$ rotates to $RPT$
$\Longrightarrow $ tetrahedon $(T,PQR)$
Heron's formula: area $\triangle PQR = 7.5 \sqrt{13}$
Height $BF =\frac{10\sqrt{13}}{7}$ and $PF = \frac{27}{14} \quad \quad (1)$
Heron's formula: area $\triangle ABC = 30 \sqrt{13}$
Height $AD =3.\sqrt{13}$ and $BE = \frac{20\sqrt{13}}{7}$
$AE=\frac{27}{7}$ and $BD = 2$
System:
	\[ \left\{\begin{array}{ll}AT^{2}=(\frac{27}{7})^{2}+ET^{2}\\ (\frac{20\sqrt{13}}{7}-AT)^{2}=(2)^{2}+(3\sqrt{13}-ET)^{2}\end{array}\right. \]
$\Longrightarrow ET=\frac{162.\sqrt{13}}{91}$;$TF=ET-BF=\frac{32\sqrt{13}}{91}\quad \quad (2)$
(1) and (2): $PT = \frac{1}{14}.\sqrt{\frac{13573}{13}}$
Height tetrahedron $=\sqrt{5.5^{2}-\frac{13573}{14^{2}.13}}=\frac{18}{\sqrt{13}}$
Volume tetrahedron $=\frac{1}{3}.7.5 \sqrt{13}.\frac{18}{\sqrt{13}}=45$

[hide]Another possible system of equations to set up uses the fact that the resulting tetrahedron is isosceles (its non-adjacent edges are equal in length). Such a tetrahedron can be placed in a parallelepiped (with rectangular faces) so that the edges form the diagonals of the figure.

Furthermore, the tetrahedron makes up one-third of the area of the parallelepiped, so we can find the volume of the tetrahedron if we know the volume of the parallelepiped. If we call the side lengths of the parallelepiped $a$, $b$, and $c$, then diagonals (which we know measure $\frac{11}{2}$, $10$, and $\frac{21}{2}$ can be expressed as follows:

$a^2 + b^2 = \left(\frac{21}{2}\right)^2$
$b^2 + c^2 = 10^2$
$c^2 + a^2 = \left(\frac{11}{2}\right)^2$

To find $a^2$, $b^2$, and $c^2$, we can add the three equations and subtract individual ones, yielding $a^2 = \frac{81}{4}$, $b^2 = \frac{360}{4}$, and $c^2 = \frac{40}{4}$. This means that $abc = 135$ and $\frac{1}{3}abc = 45$.[/hide]


linear algebra
matrix
superior algebra
superior algebra solved

It's simple. You just subtract one column from another until you reach In. Then all you have to do is to apply the same operations for the lines of Into reach the inverse.

The proof for this method is related to writing an operation on the columns as the product with a matrix. I can post it if you want, but I'll need some time to remember. But you can see it works! :)

I see you solved with operation on lines and columns, good job blondu.
Here another proof with more tools:  nilpotent matrix, formal power serie

N=
(0  1  0  ..                  ...)
(0  0  1  ..                  ...)
...................................
(0  0  0 ......................1)
(0  0  0 ......................0)

N is nilpoten N^n = 0, N^(n-1)=/=0
A can be express in terms of N
A= \sum _(k=1 to n)k(k+1)N^(k-1) =  \sum _(k \geq 1)k(k+1)N^(k-1)

this expression invite us to consider the formal power serie

f(X)= \sum _(k \geq 1)k(k+1)X^(k-1)
=d/dX ( \sum _(k \geq 1)(k+1)X^k )
= d/dX (  d/dX (\sum _(k \geq 0 )X^(k+1) )
=d^2/dX^2 ( 1/(1-X) )
=2/(1-X)^3

so f(X)(1-X)^3 = 2 (1)

Formally from  (1) you check that
1/2.A(I-N)^3 = I_n
A^(-1) = 1/2.(I_n- N)^3

Your proof is more elementary blondu :)

Cool stuff with the power series, but can this be done on matrixes?
I've never seen it before.:)


Did Any of you here of the Larouche movement. I think it is crazy!!!!!! Today, on campus there were some representatives of the movement, they are volunteers. Can you belive me what the guy told me (he apprently was in my high school before). He said that Gauss fundamental theorem is related to economics and philosophy and that axioms cant be used, and so is Riemann's phd disseration, said how Gauss said some mathematicians were wrong, etc etc. He said the american economy will soon be bankrupt!! Larouche by the way, was a democratic presidential candidate. I also got some brochure booklets where its talks about the bad things Bush and Schwarznegger are doing, motivated by economist George Schultz. This all looks insane too me!!! What do you guys think? Have you heard of this?

WELL!!!!!!!!!!! Will somebody type something, this is a very interesting topic. Let me give you some links, one of them talks so badly about Cauchy, how dare they talk like that about great mathematicians. Ok, the links are http://www.larouchepub.com/lar/2005/3213cauchy_fraud.html

The following is a general link where all the Larouch stuff is.
http://www.larouchepub.com/

Criticizing mathematicians--bad.

Criticizing Bush--good.

LOL, I agree with you, but I dont know if his movement is good, even though I think Bush isnt a good president like Clinton was.

It's probably just somehting that no one has even heard of, and those who have probably don't listen.

Go beat'em up Hello! Take them down! As it says in Halo 2, "quell this heresy, Arbiter."

This is one of those randomly generated academic 'papers', right?

No, not even close. It is a movement led by Democrat Lyndon Larouche, I have 2 times met them encouraging people to engage in the same views. Go to the website I made in the link in the 1st or 2nd post and read it.

I was being sarcastic.  But of course it's all nonsense, a bunch of vaguely intelligent sounding academic doublespeak without any basis in real mathematics.  It doesn't really assert anything, so what can it mean?

A bunch of crazy lunatics ready to slash tires and sabotage voting booths at a moment's notice.  Hmm.. such a group [i]could[/i] be useful...


function
advanced fields
advanced fields unsolved

Give an example of unbounded function on complete and not precompact space $ X$.(From $ X$ to $ R$).

Or show that exist such function.
Any idea?


Very, very nice problem!

_

A convex polygon is called balanced if for any interior point P, the sum of the distances from this point to the lines containing the sides of polygon is a constant. Two polygons $A_{1}A_{2}..A_{n}$ and $B_{1}B_{2}..B_{N}$ have mutually parallel sides. Prove that  $A_{1}A_{2}..A_{n}$ is balanced if and only if $B_{1}B_{2}..B_{N}$ is balanced!

Errr... isn't that kind of trivial? Although I do not recall ever seeing this problem at MOP.

i see this problem in the section 'Selected problems for MOSP 2003 and 2004' of the archive 'MOSP 2006 Homework', precisely the problem 4.41!

my solution is not trivial.  :huh:

but, if you have one easy way to solve, please post it!

It will become trivial if you can prove:
http://www.mathlinks.ro/Forum/viewtopic.php?t=119086

However, I can't quite understand Little Gauss's solution there...


If Log base y of x + Log base x of y = -2. What is xy?

[hide="Logarithmic Frenzy"]Through [url=http://en.wikipedia.org/wiki/Logarithmic_identities#Changing_the_base]COB[/url], we find that

$ \log_{x} y \equal{} \frac{\log_{y}y}{\log_{y} x} \equal{} \frac{1}{\log_{y} x}$.

Making the substitution $ u \equal{} \log_{y} x$, we get

$ u \plus{} \frac{1}{u} \equal{} \minus{}2 \Longleftrightarrow \\
u^{2} \plus{} 1 \equal{} \minus{}2u \Longleftrightarrow \\
u^{2} \plus{} 2u \plus{} 1 \equal{} (u \plus{} 1)^{2} \equal{} 0 \Longleftrightarrow \\
u \equal{} \minus{}1$

Thus, $ y^{\minus{}1} \equal{} x$, and  $ \boxed{xy \equal{} 1}$[/hide]


Let $ P$ be any point on the plane and a $ ABC$ be a triangle.Draw perpendicular lines $ PD,PE,PF$ to the sides $ BC,CA,AB$ respepectively.(i.e. $ DEF$ is the pedal triangle) Again draw perpendicular lines $ AX,BY,CZ$ on the sides $ EF,FD,DE$ respectively.
Prove that,
[b](a)[/b] $ AX,BY,CZ$ are concurrent on a point $ P'$.
[b](b)[/b] If circumcircle of $ \triangle ABC$ is $ \omega$ and $ AP \cap \omega\equal{}Q$ and $ AP' \cap \omega\equal{}Q'$.Then, $ BC \parallel QQ'$.

call D' the projection of A on EF and equaly define E' and F'. The quad AFPE is cyclic so $ \angle FAP \equal{} \angle FEP \equal{} 90 \minus{} \angle FEA \equal{} \angle D'AE$. So AD', BE' and CF' concur on the isogonal conjugate of P. Then it's also true the (b) clearly.

p.s. for the part (a) it's sufficent to say that DEF and ABC are orthologic

gabriel wat do u mean by orthologic????the term is new to me.can u descrbe kindly???

solution
part a is a direct conclusion of a lemma which is stated in plane eucledean geometry by ukmt.u know moon :lol:
part b
<AFK=<APE
so AFX and APE triangles are similar
so BQ'=CQ
so <BCQ'=<CQ'Q
the result follows :D
ishfaq

1st part was just a hint to the problem and I certainly know that there is a problem in plain euclidean geometry book(UKMT) equivalent to the part 1.
And the problem is more or less straight forward as you know I can't make hard problems :P

[color=darkred][u][b]Orthologic:[/b][/u]
Two triangles $ ABC$ and $ A'B'C'$ are said to be [b]orthologic [/b] if perpendiculars from $ A, B, C$ to $ B'C', A'C', A'B'$ are concurrent. The point of concurrency is known as the [b]orthologic center[/b] of $ \triangle ABC$ with respect to $ \triangle A'B'C'$. [/color]

I have 2 solutions:

1st one like [b]¬[ƒ(Gabriel)³²¹º]¼ [/b] using orthologic part and isogonal conjugate.

2nd one like Ishfaq:
Using the perpendicular fact in a triangle and then, simple angle chasing.


quadratics
LaTeX
algebra
quadratic formula

[quote="Temperal"]What are the roots to $ 3x^{4}\minus{}22x^{3}\plus{}57x^{2}\minus{}63x\plus{}24$?[/quote]

[hide="Hint"]
Use the $ p/q$ method.
[/hide]

[hide="Solution"]
Factors of 3: 1, 3
Factors of 24: 1, 2, 3, 4, 6, 8, 12, 24

Possible solutions: $ \pm 1$, $ \pm 2$, $ \pm 3$, $ \pm 4$, $ \pm 6$, $ \pm 8$, $ \pm 12$, $ \pm 24$, $ \pm\frac{1}{3}$, $ \pm\frac{2}{3}$, $ \pm\frac{4}{3}$, $ \pm\frac{8}{3}$

Simply check to see each possible solution to weed out any extraneous ones.
[/hide]

I think that's right.  :blush:

[quote="7h3.D3m0n.117"][hide="Solution"]
Factors of 3: 1, 3
Factors of 24: 1, 2, 3, 4, 6, 8, 12, 24

Possible solutions: $ \pm 1$, $ \pm 2$, $ \pm 3$, $ \pm 4$, $ \pm 6$, $ \pm 8$, $ \pm 12$, $ \pm 24$, $ \pm\frac{1}{3}$, $ \pm\frac{2}{3}$, $ \pm\frac{4}{3}$, $ \pm\frac{8}{3}$

Simply check to see each possible solution to weed out any extraneous ones.
[/hide]

I think that's right.  :blush:[/quote]

Your solution is not a solution.  It doesn't even contain the answer.  Also, you only have listed the possible rational roots, and there may not be any rational roots of the quartic (a quartic doesn't necessarily have any real roots, either).

The roots generated by Mathematica are NOT pretty...

[hide="a root"]$ x=\frac{11}{6}-\frac{1}6\sqrt{7-15(\frac{2}{-37+\sqrt{1869}})^{\frac{1}3}+3(\frac{-37+\sqrt{1869}}{2})^{\frac{1}3}}-$

$ \frac{1}2\sqrt{\frac{14}{9}+\frac{5}3(\frac{2}{-37+\sqrt{1869}})^{\frac{1}3}-\frac{1}3(\frac{1}2(-37+\sqrt{1869}))^{\frac{1}3}-\frac{-34}{9\sqrt{7-15(\frac{2}{-37+\sqrt{1869}}})^{\frac{1}3}+3(\frac{-37+\sqrt{1869}}{2})^{\frac{1}3})}}$[/hide]

[quote="7h3.D3m0n.117"]
[hide="Solution"]
Factors of 3: 1, 3
Factors of 24: 1, 2, 3, 4, 6, 8, 12, 24

Possible solutions: $ \pm 1$, $ \pm 2$, $ \pm 3$, $ \pm 4$, $ \pm 6$, $ \pm 8$, $ \pm 12$, $ \pm 24$, $ \pm\frac{1}{3}$, $ \pm\frac{2}{3}$, $ \pm\frac{4}{3}$, $ \pm\frac{8}{3}$
[/hide][/quote]

The problem with $ p/q$ is that your listed solutions are only the [i]possible, real, rational[/i] ones. [i]If[/i] there are any irrational or imaginary answers they would not be included.

According to Descarte's Rule of Signs you can find the number of positive and negative real roots in the following manner:

$ 3x^{4}\minus{}22x^{3}\plus{}57x^{2}\minus{}63x\plus{}24$ has several sign changes $ 3x^{4}$ is positive and $ \minus{}22x^{3}$ is negative, so count $ 1$ sign change. $ \minus{}22x^{3}$ is negative and $ \plus{}57x^{2}$ is positive, so count $ 2$ sign changes. $ \plus{}57x^{2}$ is positive and $ \minus{}63x$ is negative, so count $ 3$ sign changes. $ \minus{}63x$ is negative and $ \plus{}24$ is positive, so count $ 4$ sign changes.

As $ 4$ is highest power and the number of positive real roots there is no need to check for negative real roots.

This means that there are $ 4$, $ 2$, or no real, positive roots. All missing positive roots are replaced by a pair of complex roots in the usual paired manner.

[quote="Temperal"]There's a much easier way to solve this.[/quote]
Could you give me a hint?

[quote="b-flat"]Your solution is not a solution.  It doesn't even contain the answer.  Also, you only have listed the possible rational roots, and there may not be any rational roots of the quartic (a quartic doesn't necessarily have any real roots, either).[/quote]

Yeah I realize....my mom pushed me off the comp before I could finish.

Also, I found (by graphing) that neither of the possible rational roots are actual roots. (Remember according to the Rational Test that if that there was one rational root, then finding the other, rational, irrational, or imaginary would be easy.)

Is there an  [i]Ir[/i]rational Test?  :lol:

[quote="Temperal"][quote="mihail911"]at first glance i would also try using the p/q method. i understand that this method overlooks rational and complex roots, however, once you have found the real roots, can't u just solve for those complex/irrational roots the old-fashioned way-factoring the complementary factor and/or using the quadratic formula (that is if the complementary factor is a quadratic)?[/quote]

The is quite close.  :wink:[/quote]

Only problem is that finding the real roots is very difficult because there are no rational roots (all of the roots are very similar to the one in my previous post), according to Mathematica and 7h3.D3m0n.117.  I have no idea how someone would be expected to find roots like that.

@7h3.D3m0n.117:  oh, ok.

@Temperal:  could you be a little more specific with your hint, please?

Oh, oops.  :blush:  I mistyped the problem. I'm sure you'll find it MUCH easier after I fix this.

[quote="Temperal"]What are the roots to $ 3x^{4}\minus{}22x^{3}\plus{}57x^{2}\minus{}62x\plus{}24$?[/quote]

[hide="solution"]Simply trying a few simple values $ x\in\{\minus{}2,\minus{}1,0,1,2\}$, we see that $ x\equal{}1$ and $ x\equal{}2$ are roots.  Thus, we have:

$ 3x^{4}\minus{}22x^{3}\plus{}57x^{2}\minus{}62x\plus{}24\equal{}$
$ (x\minus{}1)(x\minus{}2)(3x^{2}\minus{}13x\plus{}12)\equal{}$
$ (x\minus{}1)(x\minus{}2)(x\minus{}3)(3x\minus{}4)$.

Thus, the roots are $ \{1,\frac{4}3,2,3\}$.[/hide]

Much easier  :)

does that mean now u can change the title of the post from not so easy quartic to completely do-able quartic? :wink:

[quote="archimedes1"]Didn't Mathematica spurt out some incredibly un-simplifiable answer? :?:[/quote]

What's Mathematica?

[quote="13375P34K43V312"]archimedes: he changed the problem.

temperal: mathematica=god of brute force[/quote]

i.e. its basically $ (\text{TI\minus{}}89)^{1337^{1337}}$

[quote="mihail911"]ok..i still don't get it :blush: . is it a book?[/quote]

computer program

Oh, I see. http://www.wolfram.com/  So it's a souped-up calculator program.

Hint: [hide]Use Synthetic Division[/hide]

Solution:
[hide]Well looking at this quickly and trying to plug in values right away we see that 1 returns a value of 0 so 1 is a solution. So
1; 3, -22, 57, -62, 24
__0,   3 ,-19,  38,-24
__3, -19, 38, -24, 0
So now we have (x-1)(3x^3-19x^2+38x-24)

Plugging in numbers again we see that 2 is also a solution so
2; 3,-19,38,-24
__0, 6 ,-26, 24
__3, -13, 12, 0
Now we have (x-1)(x-2)(3x^2-13x+12) All we need to do is factor and we get (x-1)(x-2)(3x-4)(x-3)

So the solutions are: 1,2,3,4/3[/hide]

(Sorry but I don't know how to write in Latex yet so it may be a little hard to read)


If two numbers are chosen from the set $ \{ 0, \minus{}5, \minus{}8, 4, 7 \}$, and one is subtracted from the other, what is the smallest possible result?

the smallest result would occur 7 is subtracted  from 8, this would look like
$ \minus{}8\minus{}7$
thus the answer is [b]15[/b]

I believe you mean $ \minus{}15$.

$ \minus{}8\minus{}7 \implies \minus{}8\plus{}(\minus{}7) \implies \boxed{\minus{}15}$


First part: For a fixed n,

[tex]\displaystyle b_n = \sum_{m=1}^n a_m[/tex]

[tex]\displaystyle = \sum_{m=1}^n \sum_{k=1}^m \frac{1}{k}[/tex]

[tex]\displaystyle = \sum_{k \le m \le n} \frac{1}{k}[/tex]

[tex]\displaystyle = \sum_{k=1}^n \sum_{m=k}^n \frac{1}{k}[/tex]

[tex]\displaystyle = \sum_{k=1}^n \frac{n+1-k}{k}[/tex]

[tex]\displaystyle = (n+1) \sum_{k=1}^n \frac{1}{k} - \sum_{k=1}^n 1[/tex]

[tex]\displaystyle = (n+1)a_n-n[/tex]

[tex]\displaystyle = (n+1)a_{n+1}-(n+1)[/tex]

So, [tex]\displaystyle b_{1988} = 1989a_{1989}-1989[/tex]

Second part:

[tex]\displaystyle c_n = \sum_{k=1}^n \frac{b_k}{k+1}[/tex]

[tex]\displaystyle  = \sum_{k=1}^n \frac{(k+1)a_{k+1}-(k+1)}{k+1}[/tex]

[tex]\displaystyle  = \sum_{k=1}^n a_{k+1} - \sum_{k=1}^n 1[/tex]

[tex]\displaystyle = (b_{n+1}-a_1)-n[/tex]

[tex]\displaystyle = b_n+a_{n+1}-(n+1)[/tex]

[tex]\displaystyle = (n+2)a_{n+1}-(2n+2)[/tex]

So [tex]\displaystyle c_{1988} = 1990a_{1989}-3978[/tex]


limit
trigonometry
LaTeX
calculus
calculus computations

$ \lim_{x \to \plus{} \infty} \left(\frac{x\plus{}2}{x\plus{}3} \right )^{\frac{1}{sin \frac{1}{x}}}$

hello, we have $ \lim_{x\to\plus{}\infty}\left(\frac{x\plus{}2}{x\plus{}3}\right)^{\frac{1}{\sin(\frac{1}{x})}}\equal{}e^{\minus{}1}$.
Sonnhard.

$ \lim_{x \to \plus{} \infty} \left(\frac{x\plus{}2}{x\plus{}3}\right)^{\frac{1}{\sin\frac1x}} \equal{} \lim_{x \to \plus{} \infty} \left[\left(1 \plus{} \frac{\minus{}1}{x\plus{}3}\right)^{x\plus{}3}\right]^{\frac{1}{(x\plus{}3) \sin \frac1x}}$

$ \equal{} exp \left[ \minus{} \lim_{x \to \plus{} \infty} \left(\frac{\sin\frac1x}{\frac1x} \plus{} 3 \sin \frac1x\right)^{\minus{}1} \right] \equal{} exp [\minus{}(1 \plus{} 0)] \equal{} \frac1e$

This can also be done by the method I have given in limit 4.
lim(x->infinity){ 1 -1/x+3}^1/sin(1/x)
= e^lim{ -1/(x-3)(sin(1/x)}
=1/e

hello kabi, why don't you use $ \text{\LaTeX}$ in your post?
Sonnhard.

Hey Dr. I m don't have Latex.I have to buy the C.D to install it as the speed is not good here.
I know its really hard to understand my post.
I m really sorry. :(

hello, but you can use the $ \text{\LaTeX}$-editor from this forum.
Sonnhard.


你好!! 恭喜发财!!!

Never said I could speak the language.
I'm a purely stupid American through and through.

How do you get those awesome chinese characters that say which country you're from?

[quote="kstan013"]我是从美国

btw 1=2 you said you [b]were[/b] an American living in the US, according to goooooogle[/quote]

That's not good grammar (the chinese).

我也是中国人在美国!

(just fyi: i can speak chinese but not write it, so I used a translator for that. It's probably really stupid.)

Haha! I am full 100% Chinese.  But...I am just another American inside so I have no idea what those symbols mean.  The only Chinese I know is spoken, not written.

我也是台灣人! :D 但是只會講國語,不會台語. :(

大韓民國

hahahahahaha
i know how to type THAT much...

[quote="you-know-who"]我也是台灣人! :D 但是只會講國語,不會台語. :([/quote]

yeah me too. i understand Taiwanese though!!! :D  :D

dis is da azn invazn... man
i like da azn sensazn
im full chines"!!!!  -___-

你是大陆人么？
好像我 是 在这 的 魏 一 一 大陆人。

how do you get chinese characters?
wo shi shang hai ren, dan shi wo sheng zai Maryland. wo bu hui shuo shang hai hua...

You have to download it i think. try going to control panel and then the language settings on ur computer.

我是北京人， 但是我没趣去看奥运会  :(

我是在美国生的一个中国人。

我会说中文，写中文，读中文。

好多台湾人啊！
我是生在美国的中国人
父母是陕西人
你们又回去过吗？

我是中国人，可是我生活在美国。我每个星期五去中文学校。

我的妈妈是生在上海, 我的爸爸是生在厦门.  :lol:  Yay!

[quote="jonathanchou711"]Wo shi zhong guo ren. Wo bu zhi dao ru guo wo shi shang hai hai shi blah blah blah...  :lol:

wo sheng zhai oklahoma. mei guo ren chi tai duo le.  :wink:

wo shi huan shu shue.[/quote]

[i]Wrong[/i]. On so many counts. Here's what it should read (corrections in bold):

[quote="jonathanchou711"]Wo shi zhong guo ren. Wo bu zhi dao ru guo wo shi shang hai [b]ren[/b] hai shi blah blah blah...  :lol:

wo (sheng zhai) [what is this?] oklahoma. mei guo ren chi tai duo le.  :wink:

wo [b]x[/b]i huan shu [b]x[/b]ue.[/quote]

我是大陆来的。我从吉安出生的，可是你们也能说我是东北人，应为我的家庭也在东北也在东南。

我的中文很差。

[quote="jonathanchou711"]wo sheng zhai oklahoma. mei guo ren chi tai duo le.  :wink:
[/quote]

哈哈哈. *giggles*

[quote="r15s11z55y89w21"]wo mei jue de na me hao xiao
我没觉得那么好笑。
(I didn't think it was that funny.)[/quote]
Ying wei ni shi mei guo ren. Wo shi jia na da ren, er qie wo jue de hen hao xiao.
应为你是美国人。我是加拿大人，而且我觉得很好笑。
(Because you're American. I'm Canadian, and I think it's hilarious.)

Wo gao su nin men yi ge xiao hua.
我告诉您们一个笑话。
(I'll tell you a joke.)

Nü ren: Lao gong, ni jue de wo hao kan hai shi nan kan?
女人：老公，你觉得我好看还是难看？
(Woman: Dear, do you think I'm pretty or ugly?)

Lao gong: Ni liang ge dou you. Ni shi hao nan kan.
老公：你两个都有。你是好难看。
(Husband: You're a bit of both. You're pretty ugly.)

Hao xiao de shi zhe ge xiao hua neng zai guo yu he ying yu dou ke yi jiang!
好笑的是这个笑话能在国语和英语都可以讲！
(The funny part is that the joke can be told in both English and Chinese!)

[quote="echen"][quote="guineapiggies"] i understand Taiwanese though!!![/quote]

Dont they speak Mandarin in Taiwan. I know they have some native languages but I have never met anyone who can speak it. I can speak a little Cantonese but not much...[/quote]

Yea, as you can tell, I am from Taiwan. They do speak mandarin there. They also speak Taiwanese. It's kind of sad that I can't understand it though.  :blush:

only Chinese people are ninjas. :ninja:  :ninja:  :ninja:

[quote="PowerOfPi"]I'm asian, but not Chinese or Indian. (9 18 1 14 - decode that to find the country I am from)[/quote]

iran

UTRXBG key asia

you can get the Chinese characters from Google translator

Chinese-American from California. Not to be racist... but a banana, I guess you could say. :P

您應該使用傳統的中國，因為它看起來票友，人們會更加深刻的印象。


modular arithmetic
induction
combinatorics proposed
combinatorics

Consider a $n$x$n$ table such that the unit squares are colored arbitrary in black and white, such that exactly three of the squares placed in the corners of the table are white, and the other one is black. Prove that there exists a $2$x$2$ square which contains an odd number of unit squares white colored.

[hide="My solution from the contest"]Assume the contrary. Denote $a_{i,j}$ the colour on the $i$-th row and $j$-th column $\forall i,j \in \overline{1,n}$ - $0$ if black and $1$ if white. Furthermore, make the notation $A_{m,p}=a_{m,p}+a_{m+1,p}+a_{m,p+1}+a_{m+1,p+1}$, which is therefore even, $\forall m,p \in \overline{1,n-1}$. We get the contradiction:
$0 \equiv$ $\sum_{1\le i,j\le n-1}A_{i,j}\equiv$ $4\sum_{1\le i,j\le n}a_{i,j}-2\sum_{1 \le i \le n}(a_{1,i}+a_{i,1}+a_{n,i}+a_{i,n})+$ $a_{1,1}+a_{1,n}+a_{n,1}+a_{n,n}\equiv$ $a_{1,1}+a_{1,n}+a_{n,1}+a_{n,n}\equiv$ (from the hypothesis) $3 \equiv 1 \pmod{2}$.[/hide]

Let us try to color the table in such a way that all the 2 x 2 squares have an even number of white unit squares. We start coloring the first two rows. If the first column has the two squares of the same color, then so has the second, and easy induction shows that it holds for the entire "2-row". If the first column has the two squares of different colors, then so have the second and the others.

Take the first square and the last square of the bottom row. Assume that they are of the same color. The previous discussion shows that the first square and the last square of the second row must be both black or both white. The same then applies to the third column an so on. We see that we must have an even number of white corners. With a similar reasoning, if the first square and last square of the bottom row were of different colors then so would be the first square and last square of the second row and so on. Again the number of white corners is even, what finishes the problem.

Remark: this proof works for an m x n table.


Let $k\geq 1$ be a fixed integer. We define the sequence $(a_n)$ by:
\[
a_1=k+1,\qquad a_{n+1}=a_n^2-ka_n+k\,\,\hbox{for}\,\, n\geq 1.
\]
Prove that if $m\neq n$ then $a_m$ and $a_n$ are relatively prime.

Michael

it was on polish 53 OM 2002
Prove and use  this $a_{n+1}=a_1a_2....a_n +k$

if $k=2$ then $a_1=3 ,a_{n+1}=(a_n-1)^2$ Therefore $a_2=4, a_3=9$ so $a_1$ and $a_3$ are not relatively prime ?????

of course
$a_{n+1}=a_n^2-ka_n+k$ isn't it?


integration
logarithms
real analysis
real analysis unsolved

Here is an idea:

Let $ S$ be the double sum. Then

$ S=\sum_{n,m\geq 1}^{\infty}(-1)^{n+m-2}nm\int_{0}^{1}x^{n+m-1}dx\int_{0}^{1}y^{m+n-1}dy$

$ =\int_{0}^{1}\int_{0}^{1}\sum_{n=1}^{\infty}(-1)^{n-1}n(xy)^{n}\sum_{m=1}^{\infty}(-1)^{m-1}m(xy)^{m-1}dxdy$

which is equal, after some calculations, to:

$ =\int_{0}^{1}\int_{0}^{1}\frac{xy}{(1+xy)^{4}}dxdy$,

and after straightforward calculations

$ =\int_{0}^{1}\left(\frac{-x^{2}-3x-3}{3(1+x)^{3}}+\frac{2+x}{2(1+x)^{2}}\right)dx$.

It is simple to show that the preceding sum equals $ \frac{1}{6}\left(\ln 2-\frac{1}{4}\right)$.

Now here is a generalization.

Let $ k\geq 2$ be a natural number. Find the sum:

$ \sum_{m_{1},\cdots, m_{k}\geq 1}^{\infty}\frac{(-1)^{m_{1}+\cdots+m_{k}}m_{1}m_{2}\cdots m_{k}}{\left(m_{1}+m_{2}+\cdots+m_{k}\right)^{k}}$


So how many of you here have cleared the first round???? :D

[quote="VARUNARASU"]i havent cleared the first round. im so upset. :([/quote]

bcos u didnt write eh? :D :D :rotfl:

[quote="anitsahu"]phoenix has
all of u who think him to be me :rotfl:  :rotfl:[/quote]
isnt it obvious that its u who has risen from the ashes like a phoenix??? :rotfl:

hello everybody
i have qualified pmt
according to the great albert einstein
i didnt appear
and i still managed to qualify :rotfl:  :rotfl:

[quote="rituraj007"]the cut off fell 2 189 from 285--and in such an easy paper--i m in 11th and i cud've got over that NOW--and most of u wud have got atleast 397/400 in P&C so......[/quote]

you r not in 11th  :)  :D
now nobody in india is in 11th
nearest to 11th are me and my friends :wink:

nobody in india is eleventh?????? :rotfl:

[quote="skand"][quote="rituraj007"]the cut off fell 2 189 from 285--and in such an easy paper--i m in 11th and i cud've got over that NOW--and most of u wud have got atleast 397/400 in P&C so......[/quote]

you r not in 11th
now nobody in india is in 11th
nearest to 11th are me and my friends :wink:[/quote]

i m just a month adrift 11th!!!!!!!! :P  :P

i m 2 months younger 2 get admitted in 11 (if i manage 2 pass) :P  :P
and 1 year and sum months if i fail
maybe :o

and yes if u cudn't clear pmt screening dis time
then it would be a joke won't it[/hide]

i got through, 386/600,in phy,chem 232/300.
bad, isnt it? :maybe:

if u're getting 154/300 in bio,u must certainly b an IIT aspirant?????? :lol:

yeah, I am pretty much a medical aspirant as an IIT one, I think got around 50% in IIT this year. As that isnt much, ill have to prepare for med too!! :juggle:


Let $k\in \{3,4,6\}$. Show that for every prime $p\equiv 1\mod k$ there is an integer $x$ with $x^{2}\leq 4p$ and $x\equiv{{2\frac{p-1}k}\choose{\frac{p-1}k}}\mod p$.

Comment: There do not seem to be bounds for other values of $k$. Indeed, for $k=5$ and $p=10091$, the residue is $5794\equiv-4297$ which still lies quite in the middle.

I have completely elementary proofs for these statements; you don't need to use higher methods.

Hint(for $k=3$, but the other cases are similar): Write $4p=a^{2}+27b^{2}$ with $a\equiv 2\mod 3$. Prove that $a\equiv{{\frac{2(p-1)}3}\choose{\frac{p-1}3}}$.


I need to learn geometry.  On the AMC12 problems 18, 19, 22, and 24 were geometry.  And I didn't get any of them.  You could even argue that  problem 20 was geometry too (the what is the probability that A+B=C problem).  I really need a way to learn geometry before the AIME.  What do people recommend?

--Dan

AoPS 1, AoPS 2, Geometry Revisited (in that order, bitte)

someone said that Challenging Problems in Geometry is good, too.

These are already good suggestions.  I would also suggest any of the problem books that include past AHSME (old AMC) exams and past AIME problems.  There are an enormous number of good geometry problems with solutions within those old exams and exam books.


I posted this in the G&FF forum and my blog, but people will listen here.

[youtube]cXz6UJL5oDA[/youtube]

It's very inspirational and I hope you take action (I did).

-Jorian Hoover

Three posts on AoPS, and one IM, I think I've had enoguh of this video.

i find it interesting that the general populace finds these "help [insert child suffering from cancer]" videos so much more touching than the messages about poverty, starvation, and genocide in africa

must be the skin color


Find the integer solutions of
\[5x^2 - 14y^2 = 11z^2\]

Pierre.

It is a really easy one, but they don't want it in the intermediate forum  :maybe:

Pierre.

consider the equation (mod 7)
$2x^2 \equiv 3z^2$ So $x^2 \equiv 5y^2$ but $a^2 \equiv 5$ has no solution
So $ 7|x,y,z$ and Now with infinite decent problem is obvious.
Solution $x=y=z=0$

That's funny, I have the same proof but modulo 4.
Does anyone have other modulii ? :)

Pierre.

what can you cancel out with 4 from
x2=2y2+3z2 (mod 4)

If $(x,y,z) \neq (0,0,0)$ is a solution :
First you may assume that $x,z$ are coprime because if a prime $p$ divides both of them then $p^2$ divides $14y^2$ and then $p$ divides $y$ (since 14 is squarefree), and then you may consider $(\frac x p , \frac y p , \frac z p)$ which is also a solution.

It follows that $x,z$ cannot be both even. But from the initial equation, they have the same parity, thus they both are odd.
It follows that $14y^2 = 5-11 = 2$ mod[8], so that $7y^2  = 1$ mod[4], that is $y^2 = 3$ mod[4] which is impossible.

Pierre.


bacheha yek ketabe  khoob baraye number theory moareffi konin albate be joz mirzakhani va farghi nemikone english bashe ya persian

ketabe srepinski(albate ketabe darsesh na masale)
albate ye meghdar sangine va english ham hast.

masele bargozideye nazariyeye adad
mahdi safa,baghestani

martabeye 3 be peimaneye che adadi 12 mibashad? :?:

adad mmored nazar bayad yeky az avamel aval [img]http://www.mathlinks.ro/Forum/latexrender/pictures/10070cc613af76e2203bc5c949c1e900.gif[/img]
bashad
[img]http://www.mathlinks.ro/Forum/latexrender/pictures/e07da438d3a5b9a0e37e5626f47b44bb.gif[/img]
ama martabe 3 be peymane avamel [img]http://www.mathlinks.ro/Forum/latexrender/pictures/b1fe089121f784dc05b5d71dfa1d5f61.gif[/img] be tartib hadaksar 6,4 ast
be rahati mitavan did ke martabe 3 be peymane73 ,12ast

I think i have this book.thank u davron
in iran there is a perfect book about number theory. ;)

[quote="Davron"]Number Theory : Arthur Engels "Art of problem solving", number theory section very good book, i have this one if you want i can send it for you :-)
for your first steps it is very good for learning basic strategies ...

davron[/quote]
thanks for your guidance I have this book but i think we have some better books in number theory than ART too ;)

[quote="Omid Hatami"][size=125]
يك جزوه ي مفيد از مرحوم "رضا صادقي" هست،ِ البته شايد نتوني گير بياري
[/size][/quote]

hamooni ke hamash mas'ale ast?

in jozve ro man daram.az aghaye karimi gereftam.agar forsat kardam scanesh mikonam va baratoon miferestam. :)


The formula $ x_{n\plus{}1} \equal{} \frac {x_{n}^{3} \minus{} 9x_{n}}{16}$ represents some sequence. Show that we can find $ a$, such that the sequence converges whenever $ x_{0} \in \left[ \minus{} a, a\right]$ and diverges otherwise. Find $ a$ and corresponding limit.

Show also that for adequately chosen $ x_{0}$, the numbers $ i\sqrt {7}, \minus{} i\sqrt {7}, \pm\frac {(\sqrt {41}\pm i\sqrt {23})}{2}$ can be adherent values of the sequence.

[quote="tmrfea"] represents some sequence. [/quote]

No, it doesn't, there is no sequence in the given equality :wink: . Please correct your problem.

$ \lim_{n\to\infty} x_n \equal{} \left\{\begin{array}{ll} 0\ (0 < a < 5) & \quad \\
5\ (a \equal{} 5)\ & \quad \\
\plus{} \infty\ \ \ (a>5) & \quad \end{array} \right.$

In fact

$ \lim_{n\to\infty} x_n \equal{} \left\{\begin{array}{ll} 0 & (\minus{}5 < a < 5)  \\
5 & (a \equal{} 5)  \\
\minus{}5 & (a\equal{}\minus{}5) \\
\plus{} \infty & (a>5)   \\
\minus{} \infty & (a<\minus{}5) \end{array} \right.$

However, here exists a standard way with proving monotonicity and boundness of the sequence:

  [list]1. $ \left|x_{n\plus{}1}\right| \leq \left|x_{n}\right| <5$ for $ |x_{0}| \in [0,5)$[/list]
   [list]2. $ 5< x_{n} < x_{n\plus{}1}$ for $ x_{0} \in (5,\infty)$[/list]
    [list]3. $ x_{n\plus{}1} < x_{n} <\minus{}5$ for $ x_{0} \in (\minus{}\infty,\minus{}5)$[/list][/list]

BUt is it possible to find out a common formula for $ x_{n}$ given $ x_{0}$.


I got some more information regarding the GPA. According to this information, I have to describe the nature of the high school and the class rank. In that case, I will have to say that my school does not give ranks. If I do so, will the colleges use the usual formula of SAT+ GPA *800? Or will they look primarily at the SAT score (which is perhaps the only qualification I have)?
And I want to know which colleges care about the USAMO (I managed to make it to the IMO selection camp once, which is more or less equivalent to the USAMO).

"SAT + GPA*800" is a CSU-only formula; the UC equivalent is different and places somewhat more emphasis on test scores and somewhat less emphasis on grades. For particular CSU issues, it is probably best that you communicate with me directly rather than post here.

Actually, I posted on this forum instead of a private message because I wanted information about the UCs too. But do the UCs
have sufficient scholarships for out-state students? Basically being rejected and being refused a scholarship will amount to the same thing in my case. Of course, I am going to fill the FAFSA but I know little about federal loans

I feel ya abel, I made mistakes for 2 years straight. GPA is spiking now, but.... lets just say I'm going to run out of gas unless i get consecutive 4.8's in my next to semesters.


aya rabeteye zir dorost ast?
if $lim$ $\frac{f(x)}{g(x)}=1$ (when $x \to a$) then $lim$ $\frac{g(x)}{f(x)}=1$ (when $x \to a$)

[quote="boxedexe"]$\\\lim_{x\rightarrow a}\frac{f(x)}{g(x)}=\frac{\lim_{x\rightarrow a}f(x)}{\lim_{x\rightarrow a}g(x)}=1\\\implies \lim_{x\rightarrow a}f(x)=\lim_{x\rightarrow a}g(x)\\\implies \lim_{x\rightarrow a}\frac{g(x)}{f(x)}=1$

Masoud Zargar[/quote]
eival. :D


Joe must determine the greatest whole number of feet from which a ceramic ball can be dropped without breaking. He has two identical ceramic balls which he can drop from any whole numbered height he wants. If he must determine this height with no more than 12 drops, what is the greatest height for which he can determine this with certainty?

A$ 20\minus{}25$ft B$ 26\minus{}40$ft C$ 41\minus{}50$ft D$ 51\minus{}75$ft E more than $ 75$ ft

I know that the answer is E but I don't understand why.
Thank you for your help.

[hide]
Try it for a few small values like 3 instead of 12, and you find that the maximum value you can determine with certainty is $ 2^{12\minus{}1}\minus{}1\equal{}$...or somewhere close to that
[/hide]


posted many times ... use $9(x+y)(x+z)(y+z)\geq 8(x+y+z)(xy+xz+xy)$ and $(xy+xz+yz)^2\geq 3xyz(x+y+z)$ ...


In how many ways can a 2*n board be tiled with n 1*2 dominoes?

[quote="madness"]i think this problem can be solved by recursion  :D
Fibonacci series and that's how my friend did the inoi problem  :D[/quote]

A mathematician always thinks like a mathematician ...

what do you mean Vihang? I am a mathematician or is my friend one?  :D

Obviously fibinocci series. Similar to previous year ZIO problem. :D

f(n) = f(n-1) + f(n-2)


Vieta algebra polynomial conics parabola

Just a problem I made up for my school's math club:

$ P(x) \equal{} x^3 \plus{} ax^2 \plus{} bx \plus{} c$ is a polynomial with real coefficients, and two roots, one of which is x=3. Given that the graph of the polynomial is not tangent to the x-axis at any positive values of x, what is the maximum value of $ a \plus{} c$ ?

To avoid confusion, $ a\plus{}c$ is not a typo, that is the question.

[hide=" :) "]FACT: A cubic with 2 distinct real roots must be of the form $ (x\minus{}r)(x\minus{}r)(x\minus{}q)$

FACT: If a cubic has a double-root at a real value, it must be tangent a the $ x$ axis at that value.

Then, the double root of this cubic is not $ 3$, so $ P(x) \equal{} (x\minus{}3)(x\minus{}r)^2$ for some $ r \leq 0$

$ a \equal{} \minus{}3\minus{}2r$
$ b \equal{} 6r\plus{}r^2$
$ c \equal{} \minus{}3r^2$

$ a\plus{}b\plus{}c \equal{} \minus{}2r^2\plus{}4r\minus{}3$, which attains its maximum at $ r\equal{}1$.

Since $ r \leq 0$, and this parabola has a negative leading coefficient, it is maximized at the $ r$ value nearest $ 1$, or $ r\equal{}0$

Hence the equation is $ P(x) \equal{} x^3 \minus{} 3x$, and $ a\plus{}b\plus{}c \equal{} \minus{}3$[/hide]

Don't click the smiley unless you've thought about the problem

Beautifully explained. I would not take points off for answering an alternate question, but we are looking for the maximum of $ a\plus{}c$, not $ a\plus{}b\plus{}c$.


Putnam
algebra
polynomial
linear algebra
matrix
college contests

Problem B1.
=============================
Do there exist polynomials a(x), b(x), c(y), d(y) such that
1 + x y + x^2 y^2 = a(x) c(y) + b(x) d(y)
holds identically?
=============================

Suppose such polynomials a,b,c,d exist.
Take x=0. Then 1=a(0)b(y)+c(0)d(y).
At least one of a(0) or c(0) is not 0. Suppose c(0)<>0. Then
d(y)=1/c(0)-(a(0)/c(0))b(y).
Replace in 1+xy+(xy) 2 =a(x)b(y)+c(x)d(y) (1) and get
1+xy+(xy) 2 =(a(x)-c(x)a(0)/c(0))b(y)+c(x)/c(0).
Denote (a(x)-c(x)a(0)/c(0))=A(x) and c(x)/c(0)=C(X). Notice that A(0)=0 and C(0)=1.
We have 1+xy+(xy) 2 =A(x)b(y)+C(x). Taking y=0, we get 1=A(x)b(0)+C(x), or C(X)=1-A(x)b(0).
So 1+xy+(xy) 2 =1+A(x)(b(y)-b(0)).
Denote B(y)=b(y)-b(0).
We have xy+(xy) 2 =A(x)B(y). So x|A and y|B. Denote a1(x)=A(x)/x and b1(y)=B(y)/y. We get 1+xy=a1(x)b1(y).
I suppose we are working in C. Take t to be a root of a1. This root exists since a1 is not constant because 1+xy is not constant in x.
We have 1+t*y=a(t)b(y)=0. So 1+t*y=0 for all y in C. This is a contradiction.

Another suggestion: even though no matrix was given in the problem, find a matrix anyway. Let $a(x)=a_0+a_1x+a_2x^2$ and make similar notations for the other polynomials. Let the coefficients of a polynomial in $x$ and $y$ (with no power of either $x$ or $y$ exceeding 2) be given as a 3 by 3 matrix. Then the condition given in the problem can be written as:

$\left[\begin{array}{cc}a_0&b_0\\\ a_1&b_1\\\ a_2&b_2\end{array}\right] \left[\begin{array}{ccc}c_0&c_1&c_2\\\ d_0&d_1&d_2\end{array}\right] = \left[\begin{array}{ccc}1&0&0\\\ 0&1&0\\\ 0&0&1\end{array}\right]$

You can't do that, because the rank of the product of two matrices is no greater than the larger of the ranks, and the 3 by 3 identity has rank > 2.

I just wanted to bump this thread because Simon (ComplexZeta) featured it during yesterday's jam.

I am rather fond of the matrix solution.

Note that this proof would work over any field, even a field of characteristic 2.

yesterday's jam? how come I didn't know that?


quadratics
induction
modular arithmetic
number theory unsolved
number theory

The sequences $(a_n),(b_n)$ are defined by $a_0=1,b_0=4$ and for $n\ge 0$
 \[a_{n+1}=a_n^{2001}+b_n,\ \ b_{n+1}=b_n^{2001}+a_n\]
Show that $2003$ is not divisor of any of the terms in these two sequences.

Suppose that $2003|a_k$ or $b_k$ for some $k$ and that $k$ is the smallest index among these two sequences such that $2003$ divides one term. Wlog we can assume that $2003|a_k$. As 2003 is prime we can rewrite our recurrences for $a_n,b_n$ not divisible by 2003 as $a_{n+1} \equiv \frac{1}{a_n}+b_n$ and $b_{n+1} \equiv \frac{1}{b_n} +a_n$ which leads to $(a_{n+1}-b_n)a_n \equiv 1$ and $(b_{n+1}-a_n)b_n \equiv 1$ (everything modulo 2003 of course). Now $a_k \equiv 0$ implies $b_{k-1}a_{k-1} \equiv -1$ (*). We get also that for $l < k$, $a_l a_{l-1} \equiv b_l b_{l-1}$. So $\frac{a_l}{b_l} \equiv 4$ or $\frac{1}{4}$. These leads with (*) to $a_{k-1} ^2 \equiv -4$ or $-\frac{1}{4}$ - contradiction in both cases. Hence 2003 does not divide any term of these two sequences

Use Induction.  We can easily check that $a_{0}$ and $a_{1}$ aren't divisible by 2003.  Let us assume $2003$ doesn't divide $a_{k-1}$ and $a_{k-2}$.

For sake of contradiction, assume that $2003|a_{k}$. Then \begin{eqnarray}a_{k}&\equiv& a_{k-1}^{2001}+b_{k-1}\equiv 0\left(\mod 2003\right) \nonumber\\ a_{k-1}a_{k}&\equiv&a_{k-1}^{2002}+a_{k-1}b_{k-1}\equiv 0\left(\mod 2003\right)\nonumber\\ &\equiv& 1+a_{k-2}^{2002}+b_{k-2}^{2002}+\left(a_{k-2}b_{k-2}\right)^{2001}+\left(a_{k-1}b_{k-1}\right)\equiv 0\left(\mod 2003\right)\nonumber\\ xa_{k-1}a_{k}&\equiv& x^{2}+3x+1\equiv 0\left(\mod 2003\right)\nonumber \end{eqnarray} where $x=a_{k-2}b_{k-2}$.  Thus, there exists an integer $k$ such that \[x^{2}+3x+1=2003k\] but $x$ is an integer so $9-4(1-2003k)$ must be a perfect square, but that implies 5 is a square modulo 2003. By Quadratic Reciprocity, \[\left(\frac{5}{2003}\right)\left(\frac{2003}{5}\right)=\left(-1\right)^{\frac{5-1}{2}\cdot\frac{2003-1}{2}}=1\] and $\left(\frac{2003}{5}\right)=\left(\frac{3}{5}\right)=-1$, so $\left(\frac{5}{2003}\right)=-1$, contradiction. Thus, 2003 doesn't divide $a_{k}$.

Same arguments can be used to prove 2003 doesn't divide $b_{k}$.

Assume for contradiction that $2003$ is the divisor of some term in one of the sequences. Let $k$ be the minimal positive integer such that $2003$ divides one of $\{ a_k, b_k \}$ where $k \ge 1$. We will now prove by induction that there exists a positive integer $x_i$ such that $x_{i}^{2}  \equiv a_i \cdot b_i \pmod{2003}$ for all $0 \le i \le k-1$. Note that the claim holds for $i=0$ and now assume that the claim holds for a certain $i$ where $0 \le i \le k-1$. Now note that since $2003$ is prime,
\[x_{i}^2 \cdot a_{i+1} \cdot b_{i+1} \equiv (a_{i}^{2002} + a_i \cdot b_i) \cdot (b_{i}^{2002} + a_i \cdot b_i) \equiv (a_i \cdot b_i + 1)^2 \pmod{2003}.\]
Since $k$ is the minimal positive integer such that $2003$ divides one of $\{ a_k, b_k \}$, it follows that $2003$ does not divide either $a_i$ or $b_i$ since $0 \le i \le k-1$. Therefore $2003$ does not divide $x_i$ and hence there exists a positive integer $m$ such that $m\cdot x_i \equiv 1 \pmod{2003}$ since $2003$ is prime. Hence
\[a_{i+1} \cdot b_{i+1} \equiv m^2 \cdot x_{i}^2 \cdot a_{i+1} \cdot b_{i+1} \equiv \left( m(a_i \cdot b_i + 1) \right)^2 \pmod{2003}.\]
This completes the induction. Now if $2003$ divides $a_k$, it follows that
\[0 \equiv a_{k-1}a_k \equiv a_{k-1}^{2002} + a_{k-1} b_{k-1} \equiv 1 + x_{k-1}^2 \pmod{2003}.\]
However, since $2003 \equiv 3 \pmod{4}$, this is a contradiction since $-1$ is not a quadratic residue modulo $2003$. A similar contradiction arises if $2003$ divides $b_k$. Hence no term in either sequence is divisible by $2003$.

(All congruences are modulo $2003$.)

Suppose to the contrary, let $n_0$ be the minimal such $n$, then $\forall n \in \{ 1, \cdots , n_0-1 \}$ $gcd ( a_n , 2003) = gcd (b_n, 2003) = 1$ (as $2003$ is prime) whence by Fermat's Little Theorem, $\forall m \leq n_0,$

$$a_{m} \equiv a_{m-1}^{-1} + b_{m-1} \pmod{2003}, b_{m} \equiv a_{m-1} + b_{m-1}^{-1} \pmod{2003} \hspace{1 mm} \cdots (1)$$
$$\implies 1 \equiv b_{m-1} b_{m-1}^{-1} \equiv (a_m-a_{m-1}^{-1})(b_m-a_{m-1}) \iff a_mb_m \equiv a_ma_{m-1}+b_ma_{m-1}^{-1} \hspace{1mm} (\forall m \leq n_0) \cdots (2)$$

Now we claim that $a_{n_0} \equiv b_{n_0} \equiv 0 \pmod{2003}$. To show this we insert $m:=n_0$ into $(2)$ to obtain $a_{n_0}a_{n_0-1} \equiv -b_{n_0}a_{n_0-1}^{-1}$. Now since $2003$ is prime $2003 | a_{n_0}b_{n_0}$ $\implies 2003|a_{n_0}$ or $2003| b_{n_0}$. This means both sides of the previous congruence are $0$ and by the minimality of $n_0$ (if $2003 | a_{n_0}$ then $b_{n_0}a_{n_0-1}^{-1} \equiv 0 \implies b_{n_0} \equiv 0$ as $a_{n_0-1} \in U_{2003}$ and if $2003| b_{n_0}$, then $2003| a_{n_0}a_{n_0-1} \implies 2003 | a_{n_0}$ otherwise $2003| a_{n_0-1}b_{n_0-1}$ contradicts the minimality of $n_0$), the claim follows.

Next we observe that inserting $m:=n_0$ into $(1)$ we get $a_{n_0-1} b_{n_0-1} \equiv -1 ...(3)$ whence employing $(2)$ with $m := n_0-1$ we obtain
$$-1 \equiv a_{n_0-1}b_{n_0-1} \equiv a_{n_0-1}a_{n_0-2}+b_{n_0-1} a_{n_0 - 2}^{-1} \equiv a_{n_0-1}a_{n_0-2} - (a_{n_0-1}a_{n_0-2})^{-1} \pmod{2003}$$
$$ \iff (a_{n_0-1}a_{n_0-2})^2 + a_{n_0-1}a_{n_0-2} -1 \equiv 0 \iff (2a_{n_0-1}a_{n_0-2} + 1)^2 \equiv 5 \pmod{2003}$$

This means $5 \equiv 1 \pmod{4}$ is a quadratic residue modulo $2003$, so that by the Law of Quadratic Reciprocity, $3 \equiv 2003 \pmod{5}$ is also a quadratic residue modulo $5$, an immediate contradiction!

This completes the proof.

We rearrange the definitions to become $a_{n+1} = b_n^{2001}+a_n = \frac{a_nb_n+1}{b_n}$ and $b_{n+1} = a_n^{2001}+b_n = \frac{a_nb_n+1}{a_n}$. We claim that for all $n$ we have $b_n \equiv 4a_n$. The base case is given. Then, otherwise we have
\[\frac{b_{n+1}}{a_{n+1}} = \frac{a_n^{2001}+b_n}{b_n^{2001}+a_n}\equiv \frac{b_n+a_nb_n^2}{a_n+a_n^2b_n}=\frac{b_n}{a_n}=4 \pmod{2003}\]
AFTSOC that at some point we have $a_nb_n+1$ is divisible by 2003. Consider the first time this happens. Then,
\[-1\equiv a_nb_n \equiv 4a_n^2 \pmod{2003}\]
This implies that -1 is a quadratic residue mod 2003, but since 2003 is 3 mod 4, this is a contradiction, so we will never have $2003\mid a_nb_n+1$, so $a_{n+1},b_{n+1}$ will never be divisible by 2003.

We work modulo the prime $2003$. If $a_n \not \equiv 0$ then we have $a_{n+1}=a_n^{-1}+b_n$ and likewise if $b_n \not \equiv 0$ we have $b_{n+1}=b_n^{-1}+a_n$, hence $a_{n+1}b_{n+1}=a_nb_n+(a_nb_n)^{-1}+2$. Thus if some minimal $k$ has $a_kb_k=0$ we must have $a_{k-1}b_{k-1}$ be a root of $x^2+2x+1$ in $\mathbb{F}_{2003}$, hence $a_{k-1}b_{k-1} \equiv 1$. But then $a_{k-2}b_{k-2}$ is a root of $x^2+3x+1$ in $\mathbb{F}_{2003}$, implying that $3^2-4(1)=5$ is a quadratic residue modulo $2003$ by the quadratic formula. On the other hand, by quadratic reciprocity we have
$$1=\left(\frac{5}{2003}\right)\left(\frac{2003}{5}\right)\left(\frac{5}{2003}\right)\left(\frac{3}{5}\right)=-\left(\frac{5}{2003}\right),$$
so $5$ is not a quadratic residue. $\blacksquare$


Hi all,

Just wondering, is it possible to put math into sections and so forth without latex giving you all of the warnings? I know the warnings don't really matter but it's more for my peace of mind.

Sample code:

[code]
\documentclass{article}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{fancyhdr}
\usepackage[pdftex]{graphicx}
\headheight 35pt
\pdfpagewidth 8.5in
\pdfpageheight 11in
\setlength{\topmargin}{-35pt}
\setlength{\parindent}{0pt}
\setlength\oddsidemargin{-0.5in}
\setlength\evensidemargin{-0.5in}
\setlength\textwidth{7in}
\setlength\textheight{8in}
\pagestyle{fancy}
\usepackage[pdftex]{hyperref}
\begin{document}

\tableofcontents
\newpage

\subsection{\ensuremath{\dfrac{dy}{dx} = f(x, y)}}

\end{document}

[/code]

I've never had warnings triggered by subsections.  I would check to see where the warnings are actually coming from.  Perhaps there is another error in there.

The problem is not LaTeX but the PDF. It's not possible to put maths, colours etc in the PDF bookmarks and the warnings are telling you this:
[quote]Package hyperref Warning: Token not allowed in a PDFDocEncoded string,
(hyperref)                removing `math shift' on input line 21.[/quote]
Hyperref has a solution which allows you to use alternative words in your bookmarks using \texorpdfstring so for example you can use
(the first option goes in the document, the second is for the bookmark)
[code]\subsection{\texorpdfstring{$\dfrac{dy}{dx} = f(x, y)$}{dy//dx=f(x,y)}} [/code]or
[code]\subsection{\texorpdfstring{$\dfrac{dy}{dx} = f(x, y)$}{given gradient}} [/code]or whatever you fancy.

Ah, that makes sense. I forgot about the bookmark sidebar.

Thanks!


limit
real analysis
real analysis unsolved

Find $\displaystyle \lim_{n\to\infty} \sum_{k=1}^{n} (\frac{k}{n})^n$

For $n>t$, the sequence $\left(\frac{n-t}n\right)^n$ increases (I hope :)) to $e^{-t}$., which would mean that the general term of our sequence is less than $1+e^{-1}+\ldots+e^{-(n-1)}$.

On the other hand, $\frac{n^n}{n^n}$ goes to $1$, $\frac{(n-1)^n}{n^n}$ goes to $e^{-1}$, and so on, which means that for any $k$, we can find $n_k$ s.t. the terms of our sequence are larger than $1+e^{-1}+\ldots+e^{-k}$ for all $n\ge n_k$.

All that should show that the limit is $1+e^{-1}+e^{-2}+\ldots=\frac 1{1-e^{-1}}=\frac e{e-1}$.

How do you prove the sequence is inscreasing? I proved it with Bernoulli's inequality. Maybe you have another     proof. Please,Groober post it!

I did not prove it, I just observed it :). That's why that "I hope" is there. Besides, we don't need the monotonicity of the sequence at all. We only need $\left(\frac{n-t}n\right)^n<e^{-t}$, which follows from $(1-x)^{-\frac 1x}<e$ for all $0<x<1$, or $e^{-x}>1-x$, which really is well known and easy to prove ;).

The sequence $S_n=\sum_{k=0}^{n-1}\bigg(1-\frac kn\bigg)^n$ is increasing. Infact we have:
\[S_n=\sum_{k=0}^{n-1}\bigg(1-\frac kn\bigg)^n<\sum_{k=0}^{n-1}\bigg(1-\frac k{n+1}\bigg)^{n+1}<\\
\\
\bigg(\sum_{k=0}^{n-1}\bigg(1-\frac k{n+1}\bigg)^{n+1}\bigg)+\bigg(1-\frac n{n+1}\bigg)^{n+1}=S_{n+1}\]
Then: $S_n<S_{n+1}$ and $S_n<\frac e{e-1}$ as Grobber showed.
But you can write:
\[\lim_{n\to +\infty}\sum_{k=0}^{n-1}\bigg(1-\frac kn\bigg)^n=\sum_{r=0}^{+\infty}\lim_{n\to+\infty}\bigg(1-\frac rn\bigg)^n=\sum_{r=0}^{+\infty}e^{-r}=\frac e{e-1}\]

Nene how do you justify that last use of the limit?  :maybe:

I think grobber's solution is very well ."we don't need the monotonicity of the sequence at all."very beautiful!!


trigonometry
geometry
inradius
circumcircle
geometry theorems

Some one provide me a link please for trigonometric formulas in the TRIANGLE like:

$ \sum cosA \equal{} 1 \plus{} 4 \prod sin(\frac {A}{2})$
$ \sum cos^2A \equal{} 1 \minus{} 2 \prod cosA$

Thanks  :wink:

[quote="N.T.TUAN"]Please see in 103 Trigonometry Problems of Titu and Feng.[/quote]

I can't buy it  :(  because I dont use Master card

I was asking just about a link on a website or on mathlinks  :D

You may find on the net book for free. It is a good book. But for this particular question I've a better offer. See the attachment.

[quote="FOURRIER"]

I can't buy it  :(  because I dont use Master card

I was asking just about a link on a website or on mathlinks  :D[/quote]
You can go to my page , and see download section //http://trungtuan.wordpress.com/

you can also prove them by using the following facts:

$ \cos A\plus{}\cos B\plus{}cos C\equal{}\frac{R\plus{}r}{R}$

$ \cos^2 A\plus{}\cos^2 B\plus{}\cos^2 C\equal{}\frac{6R^2\plus{}4Rr\plus{}r^2\minus{}p^2}{2R^2}$

$ \cos A.\cos B.\cos C\equal{}\frac{p^2\minus{}(2R\plus{}r)^2}{4R^2}$

$ \sin \frac{A}{2}.\sin \frac{B}{2}.\sin \frac{C}{2}\equal{}\frac{r}{4R}$

where $ p,r,R$ denote the semiperimeter,inradius and the circumradius of the trianghle respectively...

[quote="N.T.TUAN"]
You can go to my page , and see download section //http://trungtuan.wordpress.com/[/quote]
I'm sorry but I think sharing these files is illegal.

A good problem
$ \text{cos}x \plus{} \text{cos} y \plus{} \text{cos} x \plus{} \text{cos} (x \plus{} y \plus{} z) \equal{} 4cos\frac {x \plus{} y}{2}cos \frac {y \plus{} z}{2} \plus{} cos\frac {z \plus{} x}{2}$
Samilar for sin.

It's illegal but valuable. There are many places on the net you can find them. One more ... is not a big problem.

[quote="N.T.TUAN"]
You can go to my page , and see download section //http://trungtuan.wordpress.com/[/quote]

Very Nice Page  :lol:

Thanks :)

Maybe we should now move this topic into the infamous "Filesharing Computations and Tutorials" forum for frequent posters? ;) Because it doesn't look like someone is like discussing trigonometric formulas here...

  darij


geometry
geometric transformation
rotation
integration
calculus
function
analytic geometry

Irodov problem no. 1.260

http://irodovsolutions.blogspot.com/2008_12_15_archive.html

I have a confusion about how to compute the work done by friction.

THe solver used Kinetic work energy theorem, which yielded [m(wr)^2]/6.  I solved for the time, and used the time to compute the linear displacement, and then used W= F.dx and obtained  [m(wr)^2]/18.

Why am I wrong?  :(   I used the same technique Mystic terminator used to solve some similar rotational problem in the USA phO thread.

In which case should you use W=F.dx instead of W= delta (KE rotational + KE translational)

Thanks! :)

Well, imagine a rotating cylinder on an axis (so it cannot move anywhere) that you touch with your hand until it stops eventually. How would you compute the work of the friction force in this case? The "linear displacement" here is $ 0$, isn't it? ;)

Can you take a look at this then?
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=70112&start=60

In post 69, MysticTerminator solves a rotational motion and computes the Work using F.dx.  I used delta KE= work, and obtained a result 3 times greater.

I mean, doesn't friction also contribute to the KE rotational as well as translational in the problem in the link above?  If so, the work should be delta KE rotation+translational.  What Mystic computed looks like only the translational KE change (which explains why it's 3 times smaller than what I got)

Or am I understanding it wrong? :huh:

[quote="integration"] [b][color=blue]@1[/color][/b] I have a confusion about how to compute the work done by friction.
...
In which case should you use W=F.dx instead of W= delta (KE rotational + KE translational)

Thanks! :)[/quote]

[quote="integration"][b][color=blue]@ 3[/color][/b] I mean, doesn't friction also contribute to the KE rotational as well as translational in the problem in the link above? If so, the work should be delta KE rotation+translational. What Mystic computed looks like only the translational KE change (which explains why it's 3 times smaller than what I got)
[/quote]

[i]The [b]work-energy theorem[/b] states that the net work done on a particle by external forces equals the change in kinetic energy of the particle.[/i] [172, Serway 3ed]

With the above in mind, the total net work on a given particle (body) can be partitioned as:

$ W_{net} \equal{} W_{nc} \plus{} W_{c}$,  where $ W_{nc}$ is the work done by non-conservative forces, and $ W_{c}$ is the work done by conservative forces.

Conservative forces are forces that can be expressed as $ \vec{F} \equal{} \minus{}\nabla U$.

Friction is a [b]non-conservative[/b] force.

$ W_{net} \equal{} W_{nc} \plus{} W_{c} \implies \Delta K \equal{} W_{nc} \minus{}\Delta U \\ \\
\therefore W_{nc} \equal{} \Delta K \plus{} \Delta U.$

The work done by friction is equal to the change in total kinetic energy plus the change in total potential energy.

[quote="integration"]Can you take a look at this then?
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=70112&start=60
[/quote]
That was also wrong (but gives you quite a clear idea of what is really going on: imagine that the rope is pulled by a donkey and ask yourself what is the displacement of [i]the donkey[/i] ;))

:?:   So is Terminatormystic's solution wrong then?

In the case of the USA PhO problem in the post above (no.69 in the link above), Wnc= delta (KE)+ delta (U).
Since delta (U)= 0, then delta KE= Wfriction (not F.dx)

I have another question (I made up a scenario):

Suppose you have identical 2 disks, one on top of each other, rotating in opposite directions at equal magnitude of angular velocity.

Friction occurs between both disks, and because of conservation of angular momentum the disks come to a rest.  If they ask for the work done by friction, is it delta KE rotation for both disks, which would be :

(I[w^2])/ 2 *2=(I[w^2])

If they ask what is the wok done by friction by the first disk on the second disk, would it be the delta KE computed above, or half of that delta KE?

Thanks.

[quote="integration"]:?:   So is Terminatormystic's solution wrong then?[/quote]

I don't know. I didn't look at the PhO problem. I was more answering your question about how to compute work done by friction.

[quote="integration"]In the case of the USA PhO problem in the post above (no.69 in the link above), Wnc= delta (KE)+ delta (U).
Since delta (U)= 0, then delta KE= Wfriction (not F.dx)[/quote]

If $ \Delta U \equal{} 0$ then $ W_{nc} \equal{} \Delta K$, yes.

However, I am not sure that $ \Delta U \equal{} 0$ in this case. Typically $ \Delta U \equal{} \int_{\theta_1}^{\theta_2} \minus{}I\,\vec{\alpha}(\theta)\cdot\textrm{d}\vec{\theta}$ for a problem of this sort.

[quote="integration"]I have another question (I made up a scenario):

Suppose you have identical 2 disks, one on top of each other, rotating in opposite directions at equal magnitude of angular velocity.

Friction occurs between both disks, and because of conservation of angular momentum the disks come to a rest.  If they ask for the work done by friction, is it delta KE rotation for both disks, which would be :

(I[w^2])/ 2 *2=(I[w^2])

If they ask what is the wok done by friction by the first disk on the second disk, would it be the delta KE computed above, or half of that delta KE?[/quote]

The answer is embedded in your question.

$ \textrm{Work}$ $ \equiv$ $ \left(\textrm{Net external force }\right)$ $ \cdot$ $ \left(\textrm{ distance over which the force is applied.}\right)$

If you consider both disks as part of your system, the force produced by one upon the other is an [b]internal force.[/b]

Internal forces do no work.

Considering only one of the disks, the force (torque) produced by the other is external and therefore does work upon the one under consideration.

On the other hand, if you want the work done by friction on the system of both disks then use the total kinetic energy of both disks.

To Dr. No:

Isn't the integral you computed for delta (U) the rotational Kinetic energy though?  I thought delta (U) would only be potential energy (like -GMm/r and 1/2 kx^2)

If it's rotational kinetic energy, then I included in KE already.

[quote="integration"]To Dr. No:

Isn't the integral you computed for delta (U) the rotational Kinetic energy though?[/quote]

For a conservative force field only, i.e. without friction, yes - or rather it is minus the change in rotational kinetic energy.

But strictly speaking, no, it is not just the (change in) rotational kinetic energy.

For a conservative force $ ,\,\vec{F} \equal{} \minus{} \nabla U \rightarrow \minus{} \frac {1}{r}\frac {\partial U}{\partial \theta} \equal{} F_{\theta}.$

$ \Rightarrow \minus{} \frac {\partial U}{\partial\theta} \equal{} r F_{\theta} \equal{} \tau \equal{} I\,\alpha.$

$ \therefore \Delta U \equal{} \int_{\theta_1}^{\theta_2} \minus{} I\,\alpha(\theta)\,\mathrm{d}\theta.$

It so happens that $ W_{c} \equal{} \int_{\theta_1}^{\theta_2} F_{\theta}\, r\,\mathrm{d}\theta \equal{} \minus{} \Delta U.$

By the work-energy theorem $ \Delta K \equal{} W_{nc} \plus{} W_{c} \equal{} W_{nc} \minus{} \Delta U.$

If there are no non-conservative forces (friction) then $ W_{nc} \equal{} 0 \rightarrow \Delta K \equal{} \minus{} \Delta U.$

$ \Delta K \equal{} \minus{} \Delta U \Rightarrow \Delta K \plus{} \Delta U \equal{} 0$    (This is Conservation of Energy.)

What I think might be confusing you is that where there is friction, energy is [u]not[/u] conserved.

This is what one means when one says, "Friction is a non-conservative force."

[quote="integration"]I thought delta (U) would only be potential energy (like -GMm/r and 1/2 kx^2)[/quote]

Angle is a position variable just like $ x$ or $ r$ are position variables. Potential energy is a function of position.

[quote="integration"]If it's rotational kinetic energy, then I included in KE already.[/quote]

If there is no friction involved then $ W_{nc} \equal{} \Delta K \plus{} \Delta U \equal{} 0 \implies \Delta K \equal{} \minus{} \Delta U \equal{} \int I\,\alpha\,\mathrm{d}\theta.$

Oh thanks a lot! I just got it, you just write in angle form, which made it slightly confusing, but that's the same way we derive W= F.dx=- delta (U) for conservative force, from which we can get all the fancy -GMm/r gravitational potential energy and so forth.

Do you have a general test for conservative forces?  I know that taking the closed loop integral of F.dx=0 for such force fields, but the only ones I know is the harmonic motion force field, and inverse square force fields as conservative...

1. What common other fields will be conservative as well?

2. You say that when there is friction, energy is not conserved, but do you mean kinetic friction?  Because static friction can't do any work, thus energy is conserved b/c integral Ffriction. dx=0  :huh: which is how i guess our physics teacher can use KE rotational + KE translational= mgh for rolling bodies?

[quote="hell_ever"]Here's an interesting question:

Is work-energy theorem applicable in a non-inertial frame of reference? If so, what "extra factor" must we take into account? If not, why not?[/quote]

That is an interesting question.  I believe you can just take into account the work done by the pseudoforces, and all should be well.

[quote="integration"]Oh thanks a lot! I just got it, you just write in angle form, which made it slightly confusing, but that's the same way we derive W= F.dx=- delta (U) for conservative force, from which we can get all the fancy -GMm/r gravitational potential energy and so forth.[/quote]

I'm sorry. I did not intend to make this more difficult than it is already. However, polar coordinates, i.e. "Angle form," are the most suitable coordinates to use for problems concerning rotational motions.

[quote="integration"]Do you have a general test for conservative forces?[/quote]

I do, or I should say there is one.

Conservative forces are forces which can be expressed as the gradient of some scalar potential function (or in keeping with proper sign convention, minus the gradient of some scalar potential function). This condition is what leads to the closed path integral equaling zero.

That is $ \vec{F}$ is conservative if $ \vec{F} \equal{} \minus{}\textrm{grad}(U).$

Now, $ \textrm{curl}(\textrm{grad}(Q)) \equal{} 0$ everywhere, where $ Q$ is an arbitrary scalar field.

Therefore, if $ \textrm{curl}(\vec{F}) \equal{} 0$ everywhere then $ \vec{F}\equal{} \minus{}\textrm{grad}(U)$, and if that is the case then $ \vec{F}$ is conservative.

[quote="integration"]1. What common other fields will be conservative as well?[/quote]

I believe you covered the most common ones:  Gravity, the Coulomb force, the SHO (Simple Harmonic Oscillator)... In  quantum mechanics there is the square well, the 3-D square well (i.e. particle in a box), and many others.

[quote="integration"]2. You say that when there is friction, energy is not conserved, but do you mean kinetic friction?[/quote]

Yes.

[quote="integration"]Because static friction can't do any work, thus energy is conserved b/c integral Ffriction. dx=0  :huh:[/quote]

Exactly.

[quote="integration"]...which is how i guess our physics teacher can use KE rotational + KE translational= mgh for rolling bodies?[/quote]

Your physics teacher uses "KE rotational + KE translational= mgh" because $ U(h) \equal{} mgh$ is an excellent approximation for the gravitational potential of a small body, close to the surface of the earth, and because $ W_{\text{friction}} \equal{}\int_{\theta_1}^{\theta_2} f(r,\theta)\, r\,\mathrm{d}\theta$, where $ f(r,\theta)$ is the force of (kinetic) friction, is often impossible to calculate in closed form. Typically one assumes that friction is negligible. To extent that this is true, one obtains accurate solutions.


Two missiles are initially 5000 km apart. They travel along a straight line directly towards one another, one traveling at 2000 km/ hour and the other traveling at 1000 km/ hour. How many kilometers are they apart one minute before impact?

[hide]If you combine their speeds, 2000+1000 means they go 3000mph together. Since this question is in minutes, we find that they go 50mpm. Since they go 50 miles each minute, they were 50 miles apart before they crashed.[/hide]

First, write two equations.

$ d_1 = 2000t$

$ d_2 = 1000t$

They meet when the distances traveled add up to $ 5000$. Therefore,

$ d_1 + d_2 = 2000t + 1000t = 3000t = 5000$

$ t = \frac{5}{3} = 1 \ hour \ 40 \ minutes$.

A minute before would be $ 1 \ hour \ 39 %Error. "minutes" is a bad command.
\ = \frac{33}{20}$.

$ d_1 = 2000 \cdot \frac{33}{20} = 3300$

$ d_2 = 1000 \cdot \frac{33}{20} = 1650$

So the distance between them is $ 5000 - 3300 - 1650 = \boxed{50 \ km}$

[quote="modularmarc101"]First, write two equations.

$ d_1 = 2000t$

$ d_2 = 1000t$

They meet when the distances traveled add up to $ 5000$. Therefore,

$ d_1 + d_2 = 2000t + 1000t = 3000t = 5000$

$ t = \frac {5}{3} = 1 \ hour \ 40 \ minutes$.

A minute before would be $ 1 \ hour \ 39 %Error. "minutes" is a bad command.
\ = \frac {33}{20}$.

$ d_1 = 2000 \cdot \frac {33}{20} = 3300$

$ d_2 = 1000 \cdot \frac {33}{20} = 1650$

So the distance between them is $ 5000 - 3300 - 1650 = \boxed{50 \ km}$[/quote]

See thats the solution i thought of, its just i though that the two times would be different. But it occurs to me that i was wrong. Thankyou for the insight  :lol:

The times are the same because if one of the missles moves for $ x$ amount of time, the other missile is also moving for that time.


One of the fastest supercomputers in the world is NEC's Earth Simulator, which operates at a top-end of 40 teraflops (forty trillion operations per second).   How long would it take this computer to perform 250 million calculations?

If a calculation = an operation......

10x6.25^-8 seconds?