There is a house which has $N$ floors and there are $n$ eggs. The all of them are identical on their durability. It is known that the egg will be crashed if we shall dump it from the $d$-th floor or more higher one and will preserve their durability when damped from some lower floor. How many attempts is required to define this number $d$ in any case?
The egg would be crashed if we shall dump it from $N$-floor surely. For instance, if $n=1$ then evidently we need $N-1$ ettemps.

I don't see why $n$ matters...the question only has to do with the number of floors, $N$. I suppose an upper bound is the number of digits in the base-2 representation of N. Pick the middle floor, then depending on what happens, pick the middle of the bottom half or the top half, etc. each time you find out one of the base-2 digits.

$n$ does indeed matter. If you break all of your eggs, you have failed the experiment. The example given is if $n=1$, you have to start at the bottom floor and work your way up to be sure. If $N$ was 5, and $d$ was 1, if you used your method and dropped it from the 3rd floor, the egg would break and you would be left clueless to whether $d$ was 1 or 2.

Ok, I did interpret the problem incorrectly (I thought there was a particular floor that would cause the egg to break). However, I still don't see why $n$ matters. Suppose in the example that we are given $n=N-1$, then we only need one "try," but we still drop the egg $N-1$ times. That is, the answer or method does not have anything to do with n.

$n$ matters because if you have less eggs than you need for a specific optimum strategy (that is, $n = \infty$), then you need to use the eggs you already have "more carefully."

For example, if $n = 1$ the only permissible strategy is to drop from the first floor and up until the egg breaks, since you only have one egg to work with.

Edit:

[quote="Phelpedo"]Ok, I did interpret the problem incorrectly (I thought there was a particular floor that would cause the egg to break). [/quote]

There is.

[b]2 Moderators[/b].

Possibly [b]Combinatorics Proposed & Own Problems[/b]
 is more nice place for this problem?

[quote="t0rajir0u"]$n$ matters because if you have less eggs than you need for a specific optimum strategy (that is, $n = \infty$), then you need to use the eggs you already have "more carefully."

For example, if $n = 1$ the only permissible strategy is to drop from the first floor and up until the egg breaks, since you only have one egg to work with.
[/quote]
From your examples I still can't tell the difference. If you still know the results from failed experiments, then it's the same as if they hadn't failed and you had that many eggs to begin with. If you don't know results from failed experiments...then the problem becomes pretty much impossible.

[quote="besttate"]
If N was 5, and d was 1, if you used your method and dropped it from the 3rd floor, the egg would break and you would be left clueless to whether d was 1 or 2.

[/quote]

If d is 2 and I drop it from the 3rd floor, I am still clueless whether it is 1 or 2. I can determine the answer by dropping from the 1st floor and then the 2nd.

If d is 1, then I do the exact same thing with the next egg...

[quote="t0raji0u"]
[quote="Phelpedo"]Ok, I did interpret the problem incorrectly (I thought there was a particular floor that would cause the egg to break). [/quote]

There is.[/quote]
I meant I thought there was only one floor (as opposed to one and all higher ones).

hmmm... perhaps the best strategy is to drop the first one at floor N/2, then weather it breaks or not we use the next drop at floor N/4 or 3N/4 respectively.  Each time we split the amount left to check in half, then we use the last egg to start from the least open value and work our way up.  So given n and N we know that we can split N in half n-1 times worst case senario (if they all break each time), and then the floor d can be right underneath the upper bound of what we have to check with our last egg so thats at most N/(2^(n-1)) - 1 we have to check with the last one.  So thats n + N/(2^(n-1) - 2 total drops?

[b]nate:[/b]  Why does your number of tries increase with larger $n$?  It should decrease as $n$ approaches infinity, and the best-case scenario should be $\lceil \log_{2}N \rceil$.

[quote="Phelpedo"][quote="besttate"]
If N was 5, and d was 1, if you used your method and dropped it from the 3rd floor, the egg would break and you would be left clueless to whether d was 1 or 2.

[/quote]

If d is 2 and I drop it from the 3rd floor, I am still clueless whether it is 1 or 2. I can determine the answer by dropping from the 1st floor and then the 2nd.

If d is 1, then I do the exact same thing with the next egg...[/quote]

Let's go off of this case and make it very simple.  Say $N = 3$.

Now, let's look at what our strategy is like if $n = 2$.  When $n = 2$ we can afford to drop off of the second floor.  If egg 1 breaks, we can drop off of the first floor to determine whether the floor in question is $2$ or $1$.  If egg 1 does not break, we already know it must be the third floor.  Thus we require $\boxed{2}$ attempts at most.

This strategy fails when $n = 1$ because if we break on the second floor we have no way of discovering whether the floor in question is $2$ or $1$, so we need to drop from the first floor up, which requires $\boxed{3}$ attempts at most.

Hence the worst-case number of required attempts varies depending on $n$.

We drop from every $n$-th floor. When it breaks we use our $n-1$ remaining eggs to inspect the floors between the floor we just were at and the last known good floor. This gives $N/n+n-1$ as the number of tries.

As I said, the optimal case when $n > log_{2}N$ should give an answer of $\lceil log_{2}N \rceil$ by binary search.

[quote="t0rajir0u"]

Now, let's look at what our strategy is like if $n = 2$.  When $n = 2$ we can afford to drop off of the second floor.  If egg 1 breaks, we can drop off of the first floor to determine whether the floor in question is $2$ or $1$.  If egg 1 does not break, we already know it must be the third floor.  Thus we require $\boxed{2}$ attempts at most.

This strategy fails when $n = 1$ because if we break on the second floor we have no way of discovering whether the floor in question is $2$ or $1$, so we need to drop from the first floor up, which requires $\boxed{3}$ attempts at most.

Hence the worst-case number of required attempts varies depending on $n$.[/quote]

If we are allowed to drop from the first floor then the second, why can't we just drop from the second and then the first (like in the n=2 case).

[quote="Phelpedo"][quote="t0rajir0u"]Hence the [b]worst-case[/b] number of required attempts varies depending on $n$.[/quote]

If we are allowed to drop from the first floor then the second, why can't we just drop from the second and then the first (like in the n=2 case).[/quote]

In the worst case, $d = 1$ and dropping from the second floor would break the egg, leaving us unsure whether $d = 1$ or $d = 2$.

So we drop it from one...that's only two attempts (same as d=2) according to your previous post.

Ah.  You're correct there.  But let's re-analyze for $N = 5$.

With $n = 1$ the worst-case scenario comes if $d = 5$, in which case it will require $\boxed{4}$ tries (on floors $1, 2, 3, 4$) to ascertain the correct floor.  We will need to drop starting from the first floor.

With $n = 2$, however, we can drop at the third floor.  If it breaks, we test the first and second floors for a total of three tries.  If it does not break, we test the fourth floor for a total of two tries.  Hence we require $\boxed{3}$ tries in this case.

(I'm pretty sure this is correct this time.)


can anyone suggest good article on coordinate system.......A book or a good note will do.one may email me any article they possess.........

hello, here you will find an overview
http://www.colorado.edu/geography/gcraft/notes/coordsys/coordsys_f.html
Sonnhard.


The Gnollish language consists of 3 words, ``splargh,'' ``glumph,'' and ``amr.''  In a sentence, ``splargh'' cannot come directly before ``glumph''; all other sentences are grammatically correct (including sentences with repeated words).  How many valid 3-word sentences are there in Gnollish?

[hide="There is a much faster way then this, but..."]splargh amr glumph
splargh amr amr
splargh amr splargh
splargh splargh amr
splargh splargh splargh
amr amr amr
amr amr splargh
amr amr glumph
amr glumph splargh
amr glumph amr
amr glumph glumph
amr splargh amr
amr splargh splargh
glumph amr glumph
glumph amr splargh
glumph amr amr
glumph glumph amr
glumph glumph glumph
glumph glumph splargh
glumph splargh amr
glumph splargh splargh[/hide]

I got 20.

there are total $ 3^3$ ways not regarding grammar mistakes.

and there are 2*3=6 grammatically incorrect sentences.

so 27-6=21

i got 21.

Yeah, me too.  I thought 8 + 8 + 5 = 20. :P


linear algebra
matrix
advanced fields
advanced fields unsolved

Max logdet(S)
st.(1)LMI>=0 , LMI=[S,Phi*S,Phi1*S,Phi2*S;(Phi*S)',S-r*S,0,0;(Phi1*S),0,r*S,0;

(Phi2*S)',0,0,r*S];
(2)S>0;
(3)0<r<1;
(4)K*S*K'<=um
S is positive definite matrix variable;Phi,Phi1,Phi2,K,um is fixed.
r is scalar variable.

Because of the product of S and r, I can only maxmize the object by giving r

ie. 20 value between 0 and 1 and run the program in matlab 20 times and find

the largest objective in the 20 cycle. I find there is always a single r such that

the objective is the largest. But I want to proof it in maths , I can not make it

I try to use S prosedure but I can not.
I also try to proof its convexity, but it seems useless.
Do you have any better idea? Thanks very much

I removed your identical post. Please, do not post the same problem in two places simultaneously. As to the problem itself, do I understand right that $K$,$um$, and all $\Phi$'s are square matrices of the same size as $S$, that all matrices are with real entries, and that the comparision between $KSK'$ and $um$ should be understood in the sense that $um-KSK'$ is non-negative definite?

I am sorry.I won't next time.

Yes , you understand right. Do you have any suggestion?

Thanks


function
symmetry
algebra
domain
algebra unsolved

find all functional f:R*->R* such that
f(x*f(y))=f(x*y)+x;for all x,y in R* :(

 :(

 :)

This was our TST problem.
Put x=f(z),f(f(z)f(y))=f(f(z)y)+f(z)=f(zy)+f(z)+y=f(zy)+f(y)+z
hence f(x)=x+c.subsituting in original equation we'll get f(x)=x+1

I don't understand your last step: why $f(zy)+f(z)+y=f(zy)+f(y)+z$?

change y and z 's position in the equality f(f(z)f(y))=f(f(z)y)+f(z).
I think it is a useful way to let the equality have symmetry.

you put $x=f(z)$ how do you guarantee that $z$ exist?

[quote="hardsoul"]For all values of z there exist x,and the last equation f(zy)+f(z)+y=f(zy)+f(y)+z holds for any y,z element of R*[/quote]
We must prove that for every $x$ where exists $z$ such that $f(z)=x$ and not  x such that...

I agree with Hardsoul.

If f:R*-->R*with $x$, $y$, $z$ in R*,  then
$f(xf(y))=f(xy)+x$
implies
$f(f(z)f(y))= f(zf(y)) + f(y) = f(yf(z)) + f(z)$
which implies
$f(zy) + f(z)  + y = f(zy) + f(y) + z$
which implies
$f(z) - z = f(y) - y$
which can only be true if
$ f(x) = x + c$.
Thus, whenever the given identity is valid, the function must be in this form.  We can quickly verify that the only $c$ which works is $1$.

Sorry if it just seems as if I am repeating everything.  I just wanted to sort out the logic.

I don't agree:
Let A be the image f(R).
Why is x in A ?

(It's the same thing in IMO 1999 6/ )

[quote="hardsoul"]Why?[/quote]
Because you used in your solution $x=f(z)$ for [b]every[/b] x, but how do you know that for every x there exist z such that $f(z)=x$??? You must prove that f is sjurective.

P.S. I wrote the same as Igor just in other words.

[quote="Renshaw"]Put $y = \frac{1}{x}$ so that $f(xf(\frac{1}{x})) = f(1) + x$.  $f(1)$ is constant, so as x varies over R, so does $f$ of something.[/quote]
You have missed something: $f(xf(\frac{1}{x})) = f(1) + x$ , $x$ doesn't varies over $\mathbb{R}$ , $x$ varies over $\mathbb{R^{+}}$,  so  you can't get values which are less than $f(1)$.

I had thought that "x in R*" just meant x was real and not zero.  Does the R* actually mean $\mathbb{R}^+$?

Actually, nhat wote $\mathbb R^*$, and that's usually the notation for $\mathbb R\setminus\{0\}$, so we're not only working with positive numbers, but with all non-zero numbers. In this case, the surjectivity of the function is pretty clear, since $f(xy)+x$ can take any value we want it to take (which is just what Renshaw did, put into words).

I still don't think it is necessary that we prove $f$ to be surjective.

[quote]
Put x=f(z),f(f(z)f(y))=f(f(z)y)+f(z)=f(zy)+f(z)+y=f(zy)+f(y)+z
hence f(x)=x+c.subsituting in original equation we'll get f(x)=x+1
[/quote]

The point is that we don't care whether or not $f(z)$ gives each possible value of $x$.  The identity $f(xf(y))=f(xy)+x$ will hold if we put $f(z)$ in for $x$, because the range of $f$ is a subset of the range of $x$.   This is more of a "plugging in" than a substitution, for we plan not to let $x$ vary, but $z$.

Then we use the whole symmetry thing, and arrive at $f(z) - z = f(y) - y$.  Since both $z$ and $y$ are both free to vary thoughout the enitre domain (R* or whatever), this identity can only hold if $f(x) - x  = c $.  The original identity then gives $c = 1$.

I think we were all just confused about what it means to "put" f(z) = x.

Igor wrote:
"I don't agree:
Let A be the image f(R).
Why is x in A ? "
Dear Igor x can take any value of R*(positive reals) so why can't it take all the values that f(A) can take?
The set of the range of the f(A) is not greater than all R*.
Armo wrote:
"Because you used in your solution  for every x, but how do you know that for every x there exist z such that ??? You must prove that f is sjurective. "
Dear Armo I'm not trying to prove that f is surjective in the subsitution x=f(z) I'm just putting those values that f(z) can take.And in the  equation f(xf(y))=f(f(z)f(y)) for any z we can take x such that x=f(z).
I wrote the same as Renshaw wrote.

Sorry guys, I thought  R* means $\mathbb{R^{+}} $. And this is another reason to write your texts using LaTeX. :mad:

Never mind.I'm not very good at explaining,may be that's why I always got low points at contests . :)


geometry
circumcircle
trigonometry
congruent triangles

In a triangle $ABC$, $D$ is a point on $BC$ such that $AD$ is the internal bisector of $\angle A$. Suppose $\angle B = 2 \angle C$ and $CD =AB$. prove that $\angle A = 72^{\circ}$.

[hide="My solution"]Since $AD$ is bisector of  $\triangle ABC$ we have

$\frac{AB}{AC} = \frac{BD}{DC} \Rightarrow$

$AB \cdot DC = BD \cdot AC$

$AB^2 = BD \cdot AC$

Hence the segment $BA$ is tangent from $B$ to the circumcircle of  $\triangle ADC$

So we have $\angle DCA = \angle DAB \Rightarrow  \widehat A = 2\widehat C$

$\widehat A + \widehat B + \widehat C = 180 \Rightarrow$

$2\widehat C + 2\widehat C + \widehat C = 180 \Rightarrow \widehat C=36 \Rightarrow \widehat A=72$[/hide]

i don't have the time to type out a full solution, but another way goes like this

[hide]draw biscector of $B$, this intersect $AD$ at $I$, and $AC$ at $E$, let $BAD=y$, and $C=x$

$C=EBC$, so $CE=EB$, using SAS congurency, $\triangle ABE=\triangle DCE$ (1)
then $AE=ED$ by (1) so $EAD=EDA=y$,

$ECD=EAB=2y$ by (1)

looking at $\triangle ADC$, $180=4y+x$,
looking at $\triangle ABC$, $180=2y+3x$,
subtracting and simplifying gives $y=x$, so $180=5y \implies y=36 \implies \angle A =72^{\circ}$[/hide]

Altheman's solution is clear but I have a doubt about pontios' solution. My doubt is about the logical deduction that follows in the following lines:

[quote="pontios"]$ AB^2 \equal{} BD \cdot AC$

Hence the segment $ BA$ is tangent from $ B$ to the circumcircle of  $ \triangle ADC$[/quote]
What I understand is that the power of a point at $ B$ is $ AB^2\equal{}BD\cdot BC$. How can he deduce that's the circumcircle of $ \triangle ADC$ if it is not given that $ AC\equal{}BC$?

[quote="thundercoder"]Surprisingly, this question is a part of our high school syllabus.[/quote]
lol do the they still have this in the CBSE syllabus book? It used to be there when I was in india..

[quote="Altheman"][quote="10000th User"][quote="pontios"]$ AB^2 \equal{} BD \cdot AC$

Hence the segment $ BA$ is tangent from $ B$ to the circumcircle of  $ \triangle ADC$[/quote]
What I understand is that the power of a point at $ B$ is $ AB^2 \equal{} BD\cdot BC$. How can he deduce that's the circumcircle of $ \triangle ADC$ if it is not given that $ AC \equal{} BC$?[/quote]

This proposition is true. Just work backwards through the standard proof for power of a point for tangent-secant.[/quote]I just tried following his steps and unfortunately, I am unable to see why $ AB^2 \equal{} BD \cdot AC$ implies the power of $ B$ with respect to the circumcircle of $ \triangle ADC$.

I can see it if it is $ AB^2 \equal{} BD\cdot BC$, but not from $ AB^2 \equal{} BD \cdot AC$...

Okay nevermind, I read it as the power of a point theorem...i guess you are right then...

[quote="10000th User"]Altheman's solution is clear but I have a doubt about pontios' solution. My doubt is about the logical deduction that follows in the following lines:

[quote="pontios"]$ AB^2 \equal{} BD \cdot AC$

Hence the segment $ BA$ is tangent from $ B$ to the circumcircle of  $ \triangle ADC$[/quote]
What I understand is that the power of a point at $ B$ is $ AB^2 \equal{} BD\cdot BC$. How can he deduce that's the circumcircle of $ \triangle ADC$ if it is not given that $ AC \equal{} BC$?[/quote]

oops! you're right! It should be
$ AB^2 \equal{} BD \cdot BC \equal{}>$ [the segment $ BA$ is tangent from $ B$ to the circumcircle of  $ \triangle ADC$]

I don't remember what was my thought (it's more than 2 years..)
I guess that I misread and confused about $ AC$ and $ BC$, I'm sorry  :oops:

[hide]CAN BE DONE USEING SINE LAW[/hide]

Working on [b]Altheman[/b]'s little bit:
$\triangle CDE\cong\triangle BAE$ means that a rotation about $E$ will map the two congruent triangles, i.e. $\angle BEC=\angle AED$; with $DE=AE, BE=CE$ we see that, actually $\triangle BEC\sim\triangle AED\implies y=x$, hence $5y=180^\circ\iff\hat A=72^\circ$.

Best regards,
sunken rock

By angle bisector theorem ...... it directly implies $\Delta ABC \sim \Delta BDA  \implies \angle A=72^{\circ} $

Note that $\angle B =2C$,$\angle A=180-3C$,$\angle ADB = 90-\frac{C}{2}$.

Using Sin rules in $\triangle ADB$ ,

$\frac{AB}{\cos \frac{C}{2}} =\frac{AD}{\sin 2C}$.

Using sin rules in $\triangle ADC$ we have ,

$\frac{AD}{\sin C} = \frac{CD}{\cos \frac{3C}{2}}$.

We get the following results from two above equation,

$\cos \frac{C}{2} .\sin {C} = \sin {2C}.\cos \frac{3C}{2}$.

Get,$\sin \frac{3C}{2} =\sin \frac{7C}{2}$.

Get,$C=36$.

So,$A=72$.

[quote=problemsolve_AoPS]Sorry for reviving this old thread but I have a different solution.

Let $\angle BAD = y$ and $\angle ACB = x$, the bisector of $\angle B$ meet $AC$ at $E$.

Note that $\Delta ABE \cong \Delta DCE$. Therefore, $\angle EDC = 2y \implies \angle EDB = 180-2y \implies ABDE$ is cyclic. Considering segment $DE$, we get $x=y$.

Thus, $\angle A + \angle B + \angle C = 180^o \implies 2y+2x+x=180^o \implies \angle A = 2y = 72^o$[/quote]

That's really mindblowing bro...


"A horizontal and a vertical line intersect in exactly one point. Prove by induction that given n >= 1 horizontal
lines and n vertical lines, that the total number of intersections between horizontal and vertical lines is never
more than n^2."

Induction seems to be my weak point :( I can never figure it out,

Given $ n$ horizontal and $ n$ vertical lines, add an extra horizontal line.  How many extra intersections do you get?  Now add a vertical line.  Same thing.


algebra
polynomial
function
college
complex numbers

[i]Let f(x)=c[size=75]k[/size]x^k+...+c[size=75]0[/size] be a polynomial with c[size=75]k[/size] and c[size=75]0[/size] unequal to 0. Factor f(x) over the complex numbers as f(x)= c[size=75]k[/size](x-r1)^(m1) .(x-r2)^(m2)...(x-rn)^(mn), where the r sub n's are distinct nonzero complex numbers and m sub n's are positive integers. Then a sequence {a[size=75]n[/size]} satisfies the linear recurrence with characteristic polynomial f(x) if and only if there exists polynomials g1(n), g2(n), g3(n)...,gn(n) (numbers following g are subbed) with deg gi :le: mi-1 for i=1,2,...n such that
a[size=75]n[/size]=g1(n)(r1)^n+.....g[size=75]n[/size](n)(r sub n)^n
for all n. phew![/i]

just in case you were wondering whether this was a pointless challenge or a question, i was just wondering if this theorem had a name (because its quite useful and i was doing some proof-type questions with it).

It's just called the characteristic equation.

Once again, Simon takes the cake.  Or the cheese.  Eh, let's rephrase that:

[b]Once again, Simon takes the cheesecake![/b]

Just to make this a meaningful post, what is a characteristic polynomial?

If you have a linear recursion a_n+c_0*a_(n-1)+c_1*a_(n-2)+...=0, then the characteristic polynomial is x^n+c_0*x^(n-1)+c_1*x^(n-2)+...

i have another question about recurrences - what happens when the recurrence isnt linear? for example, what would happen if in some sequence {a[size=59]n[/size]}, a[size=59]n+k[/size] is multiplied or exponentiated or divided or logarithmed by n , instead of a constant? how would i find a general form for it then? would its characteristic polynom. be defined differently?

You can't (as far as I know) use characteristic polynomials in that case. You can use generating functions sometimes, but you'll probably have to solve a differential equation at the end. If you know how to do that, great. Otherwise, you're in a bit of trouble.

so, how would you find a formula for a[size=75]2n[/size]=a[size=75]n[/size]+n? the uc-berkely mathcircle packet talks about these inhomogenuous ones...but is there are another method?

There are infinitely many functions f: Z to R (or C or Z or whatever you want) satisfying that because you don't have any rules about the odd terms.

suppose we drag a constant terms behind a homogenous relation, something like:
a_n=a_(n-1)+a_(n-2)+k
how can i find an explicit formula?

Look at the first few terms of that particular example.  I think it is fairly clear what happens, no?  Just look for the pattern.

if we just look at the coefficient of k, we get:
1,2,4,7,12,20,33,54,88,143,...
i dont see the pattern...

Check out the first differences.  (That wasn't really as obvious as I made it sound, sorry.  But looking at differences is often a useful way of unlocking the pattern in a sequence.)

so the nth term would look something like:
u_n=a*(r1)^n+b*(r2)^n+(phi)^(n+1)+(phi-1)^(n+1)
??

Well, if we ignore the k part for a bit, we get a Fibonacci-type sequence, which is given by (a*:phi:^n + b*(-:phi:)^(-n)) for some real numbers a and b.  Then, the k will be entirely independent.  The coefficient of the k terms is partial sums of sequences of Fibonacci numbers, which you can evaluate using Binet's formula, or knowing the identity F(0) + F(1) + ... + F(n) = F(n + 2) - 1.  So in the case like that where the k can be summed independantly, it isn't bad at all.  It's when you start mixing things in together that it becomes difficult.


I need to learn more about generating functions. I have read the AopsWiki article on it, but it doesn't say much more than what it is. What are some good sites to learn more about them? Some example questions, explanations, solutions to example questions, etc. would also be helpful. Thanks in advance!

i am not sure if it is useful for you

   enter here  http://math.haiyanh.com/articles/generating-functions.html

Got this off of a google: http://en.wikibooks.org/wiki/HSE_Counting_and_Generating_functions

I only skimmed over it, so I'm not absolutely sure of its accuracy/reliability. But it did cover a lot of stuff.

One good problem to do when you think you've finally gotten some mastery of generating functions is to prove that the number of ways to partition a number into distinct parts is the same as the number of ways to partition it into odd parts. http://mathworld.wolfram.com/PartitionFunctionP.html has more about the partition function, and this problem is briefly discussed if you scroll to a bit above the middle of the page.

Go search the web or the forum for "generatingfunctionology," a book by the mathematician Herbert Wilf that he makes available for free online.

this time i am sure it will be useful.

[size=150][b]the book for Generating Functions[/b][/size]

download PDF here    http://math.haiyanh.com/articles/generating-functions.html


I will be applying to the OSU Ross Program this year.
However, I don't really know I'll be accepted or not...
I am solving the problems and can't solve two of them, but I'm still trying...

So, does anybody know how much people Ross accepts?
Because I will make other plans, if it's hard to get in.
Also, is there a date that the program announces people accepted?

Thanks.

I am an alum of Ross, and mathematically speaking, it was one of the most amazing experiences that I have had.

On the application it says that you don't have to solve the problems. Perhaps as important as actually presenting solutions is describing your thought process and demonstrating your ability to explore unknown and difficult math problems.

I believe there is no set number of people that Ross accepts, though the program may try to prevent enormous amounts of people from going. The rolling admission means that they tell you whether or not you were accepted once they receive your application, though it is probably beneficial to apply early in case the program fills up.

[quote="Myself"]I am an alum of Ross, and mathematically speaking, it was one of the most amazing experiences that I have had.[/quote]

This is true - I was a first year last year and now I know quite a bit of number theory. I'm also better at proving things and exploring mathematical topics due to Ross.

I believe that you'll get in - I remember not solving about one and a half problems last year.

Do apply early as you can though - I applied late April last year and got in but who knows this year.

I hear there about 30-40 people going this summer.

Keep working steadily on the problems, and show your steps and your thought processes. Ross is very worthwhile; it's a great program for learners who are serious about learning more math.


induction
superior algebra
superior algebra unsolved

Show that if $ R$ is a commutative ring and $ N \equal{} (a_1, \cdots, a_m)$ where the $ a_i$ are nilpotent then $ N$ is a nilpotent ideal.

I haven't thought this one through fully.  Doesn't this pretty much follow from the fact that the sum of nilpotent elements is nilpotent [use
that $ a^m\equal{}0$, $ b^n\equal{}0$ gives $ ra\equal{}0$ for all $ r$ and $ (a\plus{}b)^{m\plus{}n}\equal{}0$].  So,
it seems to me if $ a_i^{n_i} \equal{} 0$, we should think $ I^N\equal{}0$ where $ N \equal{} \sum n_i$?

Recall that the ideal $ I^n$ consists of finite sums of terms of the form $ x_1 \cdots x_n$ where the $ x_i \in I$. Of course, an ideal is nilpotent if there exists a positive integer $ n$ so that $ I^n \equal{} 0$. This is trickier than it appears because if $ I \equal{} (a_1, \cdots, a_m)$ then each of the above $ x_i$ is an $ R$-linear combination of the nilpotent $ a_i$.

Try induction on $ n$ and use the binomial formula on $ (x\plus{}y)^k\equal{}\dots$.

I have tried that, but the method you're suggesting isn't quite enough. Try applying it formally to the situation I just described and you'll see that the expressions get fantastically complicated. It makes me wonder if there's a less brute-force approach.

The approaches suggested thus far would be appropriate when showing that the nilradical is an ideal (i.e. $ R$-linear combinations of nilpotent elements are nilpotent) but this is a little more difficult. I should stress also that the ideal is finitely generated, which I think allows you to put an upper bound on the nil exponents of elements of the ideal.

OK, the trick I have in mind helps you to perform the induction and to show that the nilradical is an ideal: Let $ x^n\equal{}0$ and $ y^m\equal{}0$. Then $ (x\plus{}y)^{m\plus{}n}\equal{}0$ because the expansion of $ (x\plus{}y)^{m\plus{}n}$ is a sum of terms of the form $ x^ky^{m\plus{}n\minus{}k}$ in which the former vanishes for $ k\ge n$ and the latter for $ k<n$.

Does this help you to make the home run? ;)

I actually have the trick you're talking about in one of my attempts at this problem, but the question is somewhat more involved than showing the nilradical is an ideal. Essentially we're trying to produce a natural number $ n$ dependent on the nil exponents of the generators of $ I \equal{} (a_1, \cdots, a_m)$, sufficiently large so that for $ x_1, \cdots, x_n$ chosen from $ I$ the product $ x_1 \cdots x_n$ must vanish (cf. definition of $ I^n$ and nilpotent ideal above). I hope that clarifies what I'm asking. A direct attack will get nasty quickly even using induction since the $ x_i$ are $ R$-linear combinations of the $ a_i$.

That number is basically just the sum. Everything commutes, so you just bring those coefficients out front.

[quote="JustSee12"]Essentially we're trying to produce a natural number $ n$ dependent on the nil exponents of the generators of $ I \equal{} (a_1, \cdots, a_m)$, sufficiently large so that for $ x_1, \cdots, x_n$ chosen from $ I$ the product $ x_1 \cdots x_n$ must vanish (cf. definition of $ I^n$ and nilpotent ideal above). I hope that clarifies what I'm asking. A direct attack will get nasty quickly even using induction since the $ x_i$ are $ R$-linear combinations of the $ a_i$.[/quote]Honestly I cannot see where it get nasty if you perform a proof by induction: Let $ a_i^{n_i}\equal{}0$. Then for $ I_m\equal{}(a_1,\ldots,a_m)$ we get $ I_m^{N_m}\equal{}0$ (with $ N_m\equal{}n_1\plus{}\cdots \plus{} n_m$).

Proof by induction: This is easily verified for $ m\equal{}1$. Let $ m>1$, then $ I_m\equal{}I_{m\minus{}1}\plus{}Ra_m$. Now show that, for $ a\in I_{m\minus{}1}$, $ b\in R$, $ (a\plus{}ba_m)^{N_{m}}\equal{}0$ with the trick above. ... done. This is not too tricky. ;)

Note that basically the following is sufficient for the situation: Let $ J_1, J_2$ be nilpotent ideals in a commutative ring, i.e. $ J_i^{n_i}\equal{}0$, then $ (J_1\plus{}J_2)^{n_1\plus{}n_2}\equal{}0$.

Wonderful, I see it now. I do think that's a pretty subtle induction argument, but you're right, the above trick applies with $ m \equal{} N_{m\minus{}1}$ and $ n \equal{} n_m$.


i wanted to know what u guys think since there seems to be a controversy
i believe dumbledore asked snape to kill him

i think everything is the way it seems for once. I think snape wanted to kill dumbledore and succeded

Not dead. And I still think Snape is "good".

Uh...Dumbledore's "death" seemed too stupid on his part for it to be realistic. I don't know...that's just my opinion.

or maybe...dumbledore told snape that his position as volde's right=hand man would be of much more importance than dumble's own life...and that if given a choice b/w one...choose spyness

but either way...snape didn't like dumble too much lol...cuz didn't it say in book 5 that you needed to actually want to hurt somebody to use cruciatus? so you'd think it'd be the same for avada...specially since it prolly requires a lot more power

so in the end...snape did what he wanted...but i think he's still a good guy at heart...(saving harry from certain death at the end of book 6...saving harry from quirrell in book 1...bringing harry back from the forest stead of killing him in book 3...errr ignore book 5 lol)

i said this in the hbp thread too, but snape is always loyal to dumbledore, he is only loyal to dumbledore; dumbledore does not fear death like other people/wizards; dumbledore mentioned repeatedly that harry's life is worth much more than his own in the chapter before his death.

you need to be angry/filled with hatred to perform cruciatus or avada kedavra. Snape WAS very angry - at having to follow dumbledore's wishes out of loyalty but as a result be branded as a traitor and a [i]coward[/i]; also probably at harry for his ignorance and inability to master occlumency and silent spell-casting.

Dumbledore wanted to use Snape as a spy - most certain. Otherwise, the book has no room to um... expand. You know.

if anyone hasn't read the book yet, this title sort of gives away who "dies" ...  ;)

[quote="Derek"]My thoughts:

1. Dumblebore isn't dead.
2. Snape is a good person.
3. Snape doesn't want to kill Dumbledore.

No I haven't read the book but that's what I have in mind.[/quote]

oh ok so is this what the other party consists of? people who havent even READ THE BLOODY BOOK?!
if you actually read the two chapters leading up to his death, you'll be more convinced that dumbledore really died and that there's really no reason for rowling to want him to be secretly alive anyway.

go to [url=http://dumbledoreisnotdead.com/introduction.html]dumbledoreisnotdead.com[/url] thats all i have to say...

but tricking voldemort into actually not killing dumby is pretty hard even with skilled occlumency

I think we just spoiled the ending of the book for a lot of peoples.

He is not actually [i]bad[/i], only a bit misled. If Harry had a troubled childhood, Voldy had double that. We must understand his feelings.

Actually I think he is evvil, cause he's villain..... or he has a seriously messed up mind, which I think could be the case just as well

Well if you did read that site duimbledoreisnotdead.com,ou'll change your mind and say dumby's alive,read each and every thing.

I fet some of the things had been stolen from LORD OF THE RINGS.

I have better work thank you!
I know the kind of rubbish that is posted in these sites.
But the picture of Dumbledore is nice! :P

I agree that Dumbledore must be alive or Rowling would be booed on her 7th book...somehow he's alive. If he's not, I admit I am wrong

But isn't he technically dead since his picture got hung up in the office?

That gives me an idea... maybe harry will find a way to get dumbledore out of the portrait and dumbeldore will come back. But I still go with th idea of a horcrux.

He is technically dead in more ways than one. But vicseb could have a whole team of ex-Hogwarts headmasters like that!

i have read in a mugglenet article that the water of the lake might have been draught of living death and dd might b sleeping in his tomb
it didnt make much sense

Yes Snape's Avada Kedavra must put him to sleep somehow.

Yeah, but he fell out of the tower (which must've/could've? killed Dumbledore). However, you have to WANT to kill someone to do the Avada Kedavra curse. Maybe Snape just wanted to put Dumbledore into a swoon... I remember something in the fourth book about Moody saying that you have to feel rage. Maybe Snape "tried" to feel rage so that he could just make Dumbledore faint...  :ninja:

But wizards like Flitwick, McGonagall, Slughorn, Lupin, Moody and the others are more powerful than Snape and would have recognized a sign of life in A.D.

Or maybe they did? Maybe the funeral was just to deceive the Deah Eaters?

and only JKR knows who's RAB so let the matter rest

genius turned to madness due to the potion (or poison) and asked snape to kill him


$ \sum_{a\equal{}1}^{b}1\equal{}b$
$ \sum_{b\equal{}1}^{c}b\equal{}\frac{1}{2}c (c\plus{}1)$
$ \sum_{c\equal{}1}^{d}\frac{1}{2}c (c\plus{}1)\equal{}\frac{1}{6}d (d\plus{}1) (d\plus{}2)$
$ \sum_{d\equal{}1}^{e}\frac{1}{6}d (d\plus{}1) (d\plus{}2)\equal{}\frac{1}{24}e (e\plus{}1) (e\plus{}2) (e\plus{}3)$

$ \ldots$

By induction, we can afirm:

$ S \equal{}\sum_{x_{1}\equal{}1}^{s}\sum_{x_{2}\equal{}1}^{x_{1}}\ldots\sum_{x_{j}\equal{}1}^{x_{j\minus{}1}}1\equal{}\frac{1}{j!}\prod_{i\equal{}0}^{j\minus{}1}(s\minus{}i)$

for your problem $ j\equal{}9,s\equal{}6$

And by simple inspection of the formula we can afirm that $ S\equal{}0$ for $ s<j$

so the problem is just rather trick than a problem itself

[quote="lordykelvin"]
$ S \equal{}\sum_{x_{1}\equal{} 1}^{s}\sum_{x_{2}\equal{} 1}^{x_{1}}\ldots\sum_{x_{j}\equal{} 1}^{x_{j\minus{}1}}1 \equal{}\frac{1}{j!}\prod_{i \equal{} 0}^{j\minus{}1}(s\minus{}i)$
[/quote]

Don't you want $ \frac{1}{j!}\prod_{i \equal{} 0}^{j\minus{}1}(s\plus{}i)$?

The sum of a lot of ones will never become 0 ...

i had assumed that $ \plus{}\equal{}\minus{}$

[hide="Solution"]
For every ordered 9-tuple $ (x_{1},x_{2},x_{3},...x_{9})$ of [i]non-increasing[/i] integers between 1 and 6 inclusive, 1 will be added to the sum.  It is now a simple matter of using balls and urns to find how many of these there are.

$ \binom{9\plus{}6\minus{}1}{6\minus{}1}\equal{} 2002$

(Note: Balls and urns works here because any permutation of integers can be put into a non-increasing sequence in exactly one way)
[/hide]


Solve the following ecuation if n is natural: $\displaystyle \sin \frac{ \pi}{ (n+1)! } = \frac{1}{n!}.$

It is a classical result that if $\frac x {\pi}>  0$ is rational and that $sin(x)$ is rational then $\frac x {\pi} = 0, \frac 1 2 , \frac 1 6$.
It follows that the solutions of the given equation are $n=0,1,2$.

Pierre.


You've most likely seen the \mathbb-styled 1 they often denote for the identity map.

How do you write it? $ \mathbb{1}$ seems to produce something different, so I wondered.

See [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=comprehensive]The Comprehensive LATEX Symbol List[/url] page 65. Hence
[code]\usepackage{dsfont} or if you prefer \usepackage[sans]{dsfont}
...
$\mathds{1}$[/code]


Say that a=6-b. Then you have (6-b)b=14, 6b-b^2=14. b^2-6b+14=0, then just apply quadratic formula.

I'm guessing you want the values of $a$ and $b$.

$a + b = 6$

$a       = 6 - b$

$(6-b)b = 14$

$6b-b^2=14$

$b^2-6b+14=0$

The roots of this equation involve Complex Numbers so it shouldnt be in this forum, but the roots are $3 + i\sqrt5$ and $3 - i\sqrt5$

Ok, the first one works, although I prefer completing the square.

[quote="4everwise"]I know that this has been posted before, but how do you solve this?

$a + b = 6$
$ab = 14$

or something of the sort.[/quote]

I've devised my own little method for this kind of thing. Set a=b, so a=3, b=3, then do ab=9. Take 14-9=5, so you make $a=3+i\sqrt{5}$ and b=6-a
The part you add to a is the square root of -(ab-a-b)... May sound complicated, but it's simple to me and it works... I like making up my own methods for doing problems because:
a) They're faster (to me)
b) They're more understandable (to me)
c) Everyone likes doing things the way they want to do it... (anyone denying this is werid :))


As a member of Art of Problem Solving Forums, you must be very good at math, so your math grade at school is obvious to be really good.

But how do you do in other subjects? Such as:

-Science (Chemistry, Physics, Biology, Botany...)

-English (Language Arts, Writing, Critical Reading...)

-History (Social Studies...)

-Foreign Language (French, Latin, Spanish...)

-Fine Arts (Music, Art..)

[hide="My Personal Profile"]
[b]Grade Averages[/b]
Science- A+
English- A-/B+
History- B+/A-
Foreign Language- A+
Fine Arts- A+

As you can see, my weaknesses are English and History...  :blush:
[/hide]

mid quarter grades were (i'm in eigth grade)

[hide]tech ed, A-
strings, A
social studies, A
science, A
communication arts, A

geometry, 97.6

Tech ed's my only bad subject, but only because I miss alot of class because of geometry[/hide]

i'm good at most subjects. through 3 years of high school, i've only made one B - that was in English 3 H. i should make all A's this year as well. english is my weakest subject. i like sciences and history a lot. took 3 years of french and stopped as i quit enjoying the class. never took any fine arts classes.

I consider myself to be pretty good at.. most subjects:
[hide]Last Year Final Grades:
English: A (I think i have a great vocab)
Science: A+ (easy class)
Social Studies: B+ (messed up one semester)
Composition: B (didnt hand in 1 book report.. would explain)
Computers: A (but computers in my school isnt learning bout them, its just doing some projects for other teachers mostly. Otherwise, my friends come to me for tech help.)
Math: A - but i guess thats not another subject
Foreign Language (Hebrew): A
I cant think of any other subjects right now.

This year so far i have a :

100 in foreign language (hebrew)
104 (Religious test)
87 (religious)
95 (math - but it was a tough test, highest class, and was highest grade in the grade (half the grade failed it)
thats it so far i think.[/hide]

Well i only had to take one PE class in high school, but i did get an A. The sciences are very good subjects for me (chem is almost better than math), although i do not enjoy the memorization required for biology. History/Geography are also very good classes for me and I enjoy learning more about such topics. English is not class i am particularly talented in, but i work harder in it than in other classes (not that i work in any class, but anyway). I performed well in foreign languages (French), but the teacher had some mental problems (very stressed out, unbalanced) and i decided to not pursue it further. Visual arts are my second passion. I even spend time outside of class drawing for my own enjoyment, but other arts are not my favourite subjects.

[quote="G-UNIT"]LOL, why is it that everyone who has posted so far has failed to include there Physical Education marks?  I can proudly say the majority of my P.E. marks have been A's. :lol:[/quote]

I didn't put it because my school doesn't give a grade for Physical Education.

I don't know about others. :(

I didnt think that PE was really important to mention... my school's PE is easy.. i got an A - but im also on the basketball team and wrestling team.. soo...

i think that people haven't mentioned PE becasue the class can't be graded on physical/athetic abilities so it must be based on participation. since (almost) everyone participates, everyone gets A's in PE. since everyone get's A's, it's useless to mention.

I just love how everybody loves to inflate their own grades. Makes you feel very warm inside. And I also love excuses for "low" grades. Do we really care if everyone else flunked the class? I dunno. But still, it looks better when you try to impress other people.

me last year-
bc- A+
eng- A
spanish- A
AP World- A
Physics- A
Econ (macro + micro aps)- A, A+
AP Bio- A- (which was with the hardest teacher in the entire school and half the class got Cs or lower, and I got a 98 on the final when the mean score was a 72)
AP Suckup- A

[hide=" Shameless boosting of self"]Physics AP- A
Chemistry AP-  A
Business Law- A
Spanish 3- A
Composition- A

To tell you the truth, my hardest class is Chemistry AP, even though its a science. Its just too confusing for me.[/hide]

I personally hate chemistry.

And maybe if it made sense I'd like it.

Science = no problem
English = yes lots of problems.
History - I thought I sucked but I did not.
Japanese - We don't do anything.
Art - I am terrible.
PE - I'm ok.

I got along fine with all of my subjects, at least up till now
Except for the Comprensive Reading standardized test... I always in the B range
My Gym grade is unsteady too.

I did bad in english one year, history one year, and this class which was basically dissecting films like they were books, i did bad in that class.

[quote="bleumoose"][quote="yif man12"]I just love how everybody loves to inflate their own grades. Makes you feel very warm inside. And I also love excuses for "low" grades. Do we really care if everyone else flunked the class? I dunno. But still, it looks better when you try to impress other people.
AP Bio- A- ([b]which was with the hardest teacher in the entire school and half the class got Cs or lower, and I got a 98 on the final when the mean score was a 72[/b])
[/quote][/quote]

uhh... what's the point of that? Because I'm pretty sure I made it so that others would not take my grades seriously. Those are all made up. Well, not all made up. But partially made up. I keep this kind of info to myself. So guess away. Maybe one of them's right, maybe 6. Who knows? I know I do.

GPA: 4.0 ... says it all, doesn't it? or perhaps you need AP Test Average: 5 :D of course only one of those was not math or science...

Of course I don't consider myself a history or english expert by ANY means... I'm just good at snatching those A's.

Last year i earned a 4.0 (weighted and unweighted, no honors or AP classes cuz im a noob D:) during 2nd semester with Band, Teachers Aide (I should have taken German 3 honors but German is THAT boring that i couldn't do another year), Language Arts 3, US History, Algebra 2, and Physics.

Back in 7th grade i was in the honors track for both Language Arts (English) and Math. I excelled in honors language arts but struggled in honors 7th grade math (honors pre-algebra). Eventually i fell so behind that i stopped trying or caring. I passed with a C- but i needed a B- or higher to go onto Algebra 1 or an A- or higher to go onto Algebra 1 honors. So in 8th grade i continued to excel in honors language arts and i did very well in a second year of honors pre-algebra with a different teacher (thank god). So if you judge my math skill by what level of math i am taking as a senior (Trigonometry/Math Analysis....normal schools call it Precalculus), you'd probably say i am average (or possibly a little bit remedial since they are pushing back Algebra 1 to 8th grade so rapidly). However, in Algebra 1 in high school, i started to realize that i liked math and that i hated language arts so i took regular language arts instead because honors was just a pain in the ass and i couldn't stand to read as much as they expected you to. My math grades show how i gained an interest in math:

Algebra 1: B -
Geometry: B+
Algebra 2: A
Trig/math analysis and AP Stats: Hopefully A's :D

So even though my math level is average and most of my peers who were in my 7th grade math class are in Calculus BC right now, i'd say math is my best subject. I get A's in everything else but it's because college prep classes at my school are a joke. For example, i had a really easy chemistry teacher in 10th grade and i got A's both semesters but since she was easy, i didn't learn anything. All i remember from chemistry is how to balance basic equations and HCLBR FINO (Huckleberry fino....some gay pneumonic device for remembering the diatomic elements, H, Cl, Bf, F, I, N, O). Too bad i forgot what a diatomic element is!!!!!

Man i wish i tried harder in 7th grade honors pre-algebra D:

GPA: 4.0 Unweighted.

I guess that's pretty good.

I love english. It kind of contradictory to like english almost as much as math... but I do. I'm on our Newspaper Club, I'm a Literary Magazine Editor, and I do Speech and Debate.

Pre-Calculus A+
Physics A
Chemistry A+
English A+
Spanish A+
AP European History A

[hide="grades as of right now"]
Computers: A+
Social Studies: A like 98%
Science: A+
Math:A+
Industrial Tech: A
Gym: A (all based on effort and wearing your gym clothes)
English Language Arts: A+
I'm in 7th, and my school is really easy. :)
[/hide]

My grades are all right.  I like reading, but not writing, at least not to the school prompts.

grades:

AP Human Geography: 96
AP Biology: 98
Honors Algebra II/Trig: 104
Honors English 9: 98
Spanish I: 98
P.E. : A (don't know the percent)

i really like bio...i hate spanish, human geography is really boring, english is easy,p.e. is basically wear p.e. clothes and participate[/hide]

Afrikaans 90%
English 96%
Maths 100%
Additional maths 100%
Science 93%
Accountancy 99%
Music 95%
Music second instrument 90%
History 100%

In our country we have only 6 subjects, but I take 9, lol :rotfl: .
Funny that I do better in English than Afrikaans, since I am actually Afrikaans speaking.

The government wants to change the educational system in South Africa. In the new system, everyone who has 70%+ will get a "4", 40-69% will get a "3", 35-39% = "2" and less than 35% = "1". This will be the only mark that you receive. There will be no way for you to know whether you get 70% or 100% - you will get a 4. This is really stupid, isn't it? It sucks! :(

Im pretty good at other subjects, even though Im just a freshman and school started like a month ago. At my school we have a block schedule, so Im only taking 4 classes now, and next semester I'll take the other 4 classes but my grades right now are-

Honors Algebra II- A (greater than 95%)
Honors Biology- A(close to B...around 94%)
Honors Symphonic Band- A(Around 95%)
Honors Pre-Calc- A(Around 95%)

Next semester I will taking Honors English, Honors World History, Honors Symphonic Band II, and P.E. (not honors)

I have 98+ in mathematics and an average of 95+ in all of my subjects except my currect science class because they are talking about biology and stuff (bad memory causes a drop in marks).  Fortunately, I'll have physics, and all maths next semester, so I'll be good for next semester.  Besides, doign math in boring classes in fun.  :lol:

Masoud Zargar

i rule at science!
not good at LA or SS, tho
 :blush:

I used to be good last year in subjects like social,hindi computer but now that feeling of mugging stuff(Even in bio) is just fading away...I"m still good at english and Science.

I am good at math but i dont know why i fluctuate so much in xams coz of calculations n stuff,i get 98.5,then 81,then 94....li just finished my xams today...lets see what happens

[hide="Current Grades"]
Honors French IV: A
AP Calc BC:          A
Honors Orchestra: A
Honors Chemistry:A+
Ancient Philosophy: A
Honors British Lit: A
Honors World History: B+
 [/hide]

my grades for my first quarter... midterm for other classes

AP US History - B
AP Language (English) - B
AP Calculus AB - A
AP Calculus BC - A
AP Spanish - A
Theory of Knowledge - A
IB Philosophy - A

maths: 10
english(my 2nd language): 9
chemistry: 8
german(my3d language): 8
latin:7
history:7
italian:7
in italy the marks are
under 6: bad
6 good
7 better than 6
8 very good
9outstanding
10 more tha outstanding

[hide="my current grades"]
Honors Geometry: A
English: A-
Social Studies: A+
Science: A+
German: A+
Strings (violin): A+[/hide]

I'm really good at Math, French, and Science (my 3 honors). I'm pretty good at History/Social Studies, only not in honors cuz I don't wanna have way too much hw, and ok at writing (English),  though I have been improving my writing more recently. And in English, I'm really good at grammar.

Also, I'm a freshman in HS, and I've found it's easier to get A's at HS. In JH I had various B's here and there, mainly in English. Now at HS it's all A's.
Well, in French and Math I seemed to be the best person in my grade, but not as much in Science...now in HS science I (and various others) are taking biology, and I'm beating most ppl on our tests  :lol:

I especially hate projects though....95.5 test average in bio......got a 75 on a term project :(. I also forget to do homework a lot....got a couple B+ in french b/c of it.

In Math (all honors):
Algebra 1 was a joke, easy A+
But in geometry it was hard.....only a B
Now Algebra 2 is really easy (said to be hardest honors class at my school)

[hide="estimated grades"]
english: 102
science: 97
social studies: 96
spanish: 100

That's basically it.. there aren't a lot of grades for me [/hide]

[quote="pkothari13"]Everything is good to an extent.... but ENGLISH!!! ROAR!  :mad: Ever since high school... that subject... will be my doom....[/quote]We are in the same boat, that subject... will be my doom too....
[hide="click here to find out why"]I got a C on my recent English test :([/hide]

[hide]geometry: A-
reading language arts: A
social studies: A+
science: A
PE: A
French: A
math elective: A

I am surprised at my math grade......but I bombed a test, so it shows :( [/hide]

I'm good in everything except French, which is very hard, and the smartest class I'm in short of my math.

Well pretty good, I got clear  A+'s in Math, Science and social science and English and  barely scraped  an A+ in language(Sanskrit), (I didn't deserve it!)
  Ashwath

GPA: 4.0
currently taking math and science classes at a community college, cause i'm done with HS. :)

my science (ESPECIALLY Chemistry) is lot stronger than my math, my math is strong also

took the SATII this Oct for first time, Math: 800, Chem:800, Physics:770


linear algebra
matrix
vector
algebra
polynomial

I have a problem here:

Let matrix A = uv^T is a column times a row (a rank-1 matrix).
Note: v^T means transpose of v

1)By multiplying A times u, show that u is an eigenvector. What is its eigenvalue.

2) What are the other eigenvalues of A (and why?)

A = uv^T is a rank-1 matrix, so how do we show that u is an eigenvector if multiplying A times u?

We have  Au = Lu, (A - LI)u = 0,   where L represents lamda.

so how to show that det(A - LI) = 0 ?

You don't want to do that (look at the determinant). Instead, focus directly on the definition of eigenvector. Use matrix notation and the associativity of matrix multiplication.

$Au=(uv^{T})u=u(v^{T}u).$

Now, $v^{T}u$ is a $1\times 1$ matrix - that is, a scalar. To be exact, it's the inner product of $u$ and $v$. Since it is scalar, it can be written on either side of the vector it's multiplying. Hence:

$Au=(v^{T}u)u$ or $Au=\lambda u,$ where $\lambda = v^{T}u=u\cdot v.$

Beyond that, the fact that $A$ is a rank-1 matrix means that zero is an eigenvalue of it of multiplicity $n-1.$ So the eigenvalues of $A$ are $n-1$ copies of zero, and $u\cdot v.$ The characteristic polynomial of $A$ is $x^{n}-(u\cdot v)x^{n-1},$ and the minimal polynomial is $x^{2}-(u\cdot v)x.$

An interesting special case - what would happen if $u$ and $v$ were orthogonal (without either one being zero)? Try it with $u=\begin{bmatrix}1\\1\end{bmatrix}$ and $v=\begin{bmatrix}1\\-1\end{bmatrix}$ just to get the idea.

[hide="The results."]You get a rank-1 nilpotent matrix - in particular, you get a rank-1 matrix $A$ such that $A^{2}=0.$ Of course, all of its eigenvalues would be zero.[/hide]


I need a balancing equations program for my TI-89 Titanium.

Does anyone know of a program I could download (from like ticalc.org or something)?

Simple Equations like:

C3H8 + O2 ---> CO2 + H2O
Answer would be
C3H8 + 5O2 ---> 3CO2 + 4H2O

3 Carbons
8 Hydrogens
10 Oxygens

Does anyone know of a program I could download (from like ticalc.org or something?

The only I have seen that did that, did not give you a direct answer, it just told you how to set it up.

uhhhh just use matrices and the auto-solve thingy

also gonna point out that you make a program pretty easily if you use vectors in n dimensions, where n is the number of elements you're using (hey, an application of something i learned in multivariable calculus)


Have you seen the web site [url]http://www.kalva.demon.co.uk[/url] by John Scholes?  It's got a really nice collection of problems from the AIME, US Olympiads, International Olympiads, Putnam, and many more sources.  Anyway, I noticed that many of the pages there say that "to avoid possible copyright problems, I have changed the wording, but not the substance, of the problems".  Is that legal?  I would have thought that just changing the wording is not enough to avoid copyright infringement.  Just wondering...

My guess would be that he's a little bit iffy on the copyright question, but the math contest folk don't seem particularly likely to go suing people . . .

I bet he is violating the copyright laws, actually.  Because if you write a book similar enough to another work, that can be a violation.  So odds are good he's breaking the law.

It is a great site.  I should go practice for the Putnam . . .

Joel, I don't think that's quite right. It is not possible to patent mathematical expressions and formulas (circa copyright laws in the US, don't know about elsewhere)

Kalva has specifically modified the wording of each question as to change the actual statement of the problem but not the spirit of the problem.  It is well known in the competition circles that everyone "borrows" everyone else's problems from time to time, and it would be almost impossible to find a problem that one cannot find prior art for.

If you can copyright words, you can copyright math.  And if you went through Harry Potter, carefully changing the wording of every sentance while keeping the 'spirit' intact, that would probably be copyright infringement, too.  But, as I said, no one is going to bother about it anyways.

[quote="JBL"]If you can copyright words, you can copyright math.[/quote]

What one can actually copyright is termed "expression" by lawyers. Once I looked it up, after someone asked on the Internet whether cooking recipes are subject to copyright. The answer, which amazed me at the time, is no. The legal idea is that a mere recitation of facts doesn't have enough "expression" to be deserving of copyright protection. So a cook can write rhapsodically about how delicious food is, and all of those paragraphs might be protected by copyright, but the recipes in the cook's cookbook might not be. Moreoover, there are exceptions to general copyright protection related to the intended use of the copied material, and someone who adapts a FEW problems from a published source, for NON-PROFIT purposes, not in verbatim form, and credits the original source (which keeps readers interested in buying the original source, thus not destroying its market) will probably be found to have fit within the "fair use" doctrine, an explicit statutory exception in the United States copyright law.

All that having been said, I BUY sets of old problems from the AMC people at list price rather than photocopying them, because I think they ought to get fair value for their efforts in putting together contest programs. And I buy a lot of problem books--and notice that many problems recur in several of those books. I hope everyone here agrees that math problem books are, to quote Marcus Aurelius in a closely related context, something "on which money should be eagerly spent."


What non-zero, real value of $ x$ satisfies $ (5x)^4\equal{} (10x)^3$? Express
your answer as a common fraction.


What colleges did you all end up deciding on?  A combination of financial aid concerns and an awesome CPW experience convinced me to choose MIT.

Congratulations!  I'm sure you'll be very happy here.  :)

I'll be at MIT as well. See you guys there!

I'm going to Duke, but MIT is a good choice too.  :wink:

I could be wrong but, jonathanchou711, I thought you were in Middle School right now...

$ \text{Financial Aid} \bigcap \text{Good Weather} \Rightarrow \mathbb{CALTECH}$

[quote="LightEntropy"]I will be heading off to Cornell University for engineering. i look forward to it more than anything right now.[/quote]
So is my friend Brett

Hah, I'm jealous of all of you. (just found this thread linked through the other one)

I'm taking a gap year (and hopefully going to college after that), to a good degree due to financial concerns.

Anyways, I think this will be a good move for me in general, but I have to wonder if I'm literally the only USAMOan this year not heading to college (anyone know?). (Actually, I'm curious in general about how math competition results correlate with other things in life, like sports and language skills. A topic for the round table, I suppose)


Factoring yields $ 3^6(3\minus{}1)$, which takes the form of $ N\cdot 3^6$. Therefore, N = 3-1 =2.


Are there any programs for the Ti84+SE so that I can directly write in assembly code on the calculator?
I've searched on ticalc.org, and found ASM editor and on calc asm editor, etc., but none of them works well on my calculator.
Also I'd like to avoid opcoding if possible, or perhaps try it only after I get better with asm.
Could anyone give me some advice on where I can find such programs?
thx.

You can always look for emulators for ASM calculator.s


I saw this problem somewhere but it didnt give an answer so I want to make sure my solution is correct.
How many ways are there to get from the point (0,0) to (4,4) in 10 or less steps if for every step you can go either one unit north, south, west or east.  If you reach (4,4), you are done.
[hide="I got"]I got 2310, because it has to be 8 or 10 moves.  Ofr 8 moves, its just 8C4, for 10 moves it can be 5N,1S,4E or 5E,1W,4N.  FOr both of those combinations, there are 10!/(4!5!), but some of those you would already reach (4,4) before the 10 steps and there are 8C4*2 of them so (10!/(4!5!)-8C4*2)*2+8C4=2310.  I dont think I did anything wrong but i just want to be sure[/hide]

[hide="solution"]It will either take 8 or 10 steps to get there.

If he takes 8 steps, its just ${8\choose 4}=\frac{8\cdot 7\cdot 6\cdot 5}{4\cdot 3\cdot 2}=70$.

If he takes 10 steps he can go 5 west, 4 north and 1 east, or 5 north, 4 west and 1 south.

So $2\cdot{10\choose 5}\cdot 5=10\cdot \frac{10\cdot 9\cdot 8\cdot 7\cdot 6}{5\cdot 4\cdot 3\cdot 2}=2520$

$2590$

I think I calculated one of the combanatorics wrong. :maybe: [/hide]

I don't see anything wrong with your solution

Wait, jli, do you have to subtract something frmo your second case (10 moves) because in some of those cases for 10 moves, you would already have raeched (4,4), so you dont have to go anymore (for example if you go NNNNEEEEWE, you will already have reached (4,4) in 8 moves, but when you counted the ones for the 10 moves, you also incorperated that into the solution so you overcounted)

I was just wondering because everyone else got like really small answers (like 144, 210...)

ArtofOwna, you are correct (and you correctly point out jli's error).


So it turns out the MOP is June 12th through July 2nd. New York Regents exams are going to be held from June 16th to June 23rd. Some teachers will undoubtedly be giving final exams in June also. Surely this issue has come up for NYC kids in the past. How have people dealt with this?

The year I went, the timing was such that my last regents exam was on the same day I left and I got into Nebraska around midnight, so I didn't have that problem.  You should start talking to the school (Danny Jaye would be a good first bet) and to the MOP organizers right now about the opportunity to take the test, proctored, at MOP.

I wish I had the same issue.

bleh, I'm graduating.

How will your school let you miss so much school w/o failing your grade?

Next year if I make MOP (8th grade this year), I'd have the same problem.

Regents exams take place after classes end, so you only have to come in if you have exams to take.

Hooyoung ... my year, Danny Jaye had faxed them my history regents to Nebraska, where I took it ... 1 hr later than ny time

KD

Thanks for the responses. I will talk to Mr. Jaye tomorrow and try to work something out.

i second geehoon :(
so is hooyoung the only NYC perosn going? if so, congrats to hooyoung:)
otherwise, congrats to all who made MOP

Danny Zhu is going also, but i'm not surprised of course.

yea but didnt zhu not make it last year?

wow this thread is like a year old...

stop bumping old threads!


induction
combinatorics unsolved
combinatorics

Humanitarian organization has to deliver 18 tons of humanitarian aid which is distributed in n containers, where n≥35. They have 7 trucks each that can carry infinite number of containers whose total mass does not exceed 3 tons. It is known that these trucks can carry any 35 containers, from given n containers, in one tour. Prove that these trucks can carry all n containers in one tour.

It realy looks so easy but the complexity of its definition makes it nonattractive and tough.

It seems to me that this is a simple induction on $ n$. It obviously holds for $ n \equal{} 35$.
Suppose it holds for some $ n\geq35$ and let's prove that it holds for $ n \plus{} 1$. Choose
the least container with mass $ t\leq18/(n \plus{} 1)$. All the other containers can be put on
7 trucks by hypothesis. At least one truck has free space for $ \geq3 \minus{} \frac {18 \minus{} t}{7} \equal{} (3 \plus{} t)/7$
tones and this is not less than $ t$, because $ t\leq(3 \plus{} t)/7$ is equivalent with $ t\leq1/2$.
So we can put the last container in that truck.


Να βρεθούν όλοι οι όροι της $ (a_{n})$ με $ a_{n}\equal{} 1!\plus{}2!\plus{}3!\plus{}...\plus{}n!$ και     $ n\geq1$   που είναι τετράγωνα ακεραίων.

[quote="Protonios"]Δες εδώ: :wink:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=199152[/quote]

Δεν το πρόσεξα :blush:  ευχαριστώ για την υπόδειξη protonios :)


[quote="xxxxx"]Find the coefficient of the $a^{2}bc$ term in the expansion of $(a+2b+3c)^{4}$.[/quote]

$(a+2b+3c)(a+2b+3c)(a+2b+3c)(a+2b+3c)$

[hide]There is ${4\choose 2}=6$ ways to choose the $a^{2}$.  There are 2 ways left to choose the $b$ term but the coefficient in fron of all the $b$'s is 2 so $2\times 2=4$.  There is only one way to choose the $c$ term but the coefficient in front of all the $c$'s is 3 so $3\times 1=3$.  Therefore, $6\times 4\times 3=72$ i think.[/hide]

[hide="answer"]72[/hide]

I'm quite sure the answer is right.  Can anyone write the binomial expasion equation for this please?  (using combinatorics I think?)


To proof for any natural number n that:
$ a)323/a_{n}\equal{}20^{2n}\plus{}16^{2n}\minus{}3^{2n}\minus{}1$

$ b)24/a_{n} \equal{} n^6 \minus{} 3n^5 \plus{} 6n^4 \minus{} 7n^{3} \plus{} 5n^{2} \minus{} 2n$

Hi barcelona. Could you retype the first bit? Thanks.

Hi barcelona.

I've got to go but I guess both will succumb to atttack by mathematical induction, the second one since its factors seem to be

$ (n\minus{}1)n(n^{2}\minus{}n\plus{}1)(n^{2}\minus{}n \plus{} 2)$.

(By the way, I think \mid gives the divides sign: $ \mid$).

a) Of course it will succumb to induction, but I much rather prefer the following.

$ 323\equal{}17\cdot 19$. Now, $ 20^{2n} \plus{} 16^{2n} \minus{} 3^{2n} \minus{} 1 \equiv 3^{2n} \plus{} (\minus{}1)^{2n} \minus{} 3^{2n} \minus{} 1 \equiv 0 \pmod{17}$, and $ 20^{2n} \plus{} 16^{2n} \minus{} 3^{2n} \minus{} 1 \equiv 1^{2n} \plus{} (\minus{}3)^{2n} \minus{} 3^{2n} \minus{} 1 \equiv 0 \pmod{19}$.

For question (a) I also like to look at two different ways of "factoring" the problem.  I am not quite factoring but breaking it down to two differences of squares that can be factored, to show that 17 divides the number.  Then rearranging the terms and factoring two different differences of squares, to show that 19 divides it.  Like so:

$ (20^{n}\minus{}3^{n})(20^{n}\plus{}3^{n}) \plus{} (16^{n}\plus{}1^{n})(16^{n}\minus{}1^{n})$

Which shows that 17 is a factor.

or

$ (20^{n}\minus{}1^{n})(20^{n}\plus{}1^{n}) \plus{} (16^{n}\plus{}3^{n})(16^{n}\minus{}3^{n})$

Which shows that 19 is a factor.  In both cases you need to be careful that you make it clear what happens when n is odd vs when n is even.  Though that's not too tough.

Hence, their product 323 is a factor.

Hope that makes sense.


calculus
integration
logarithms
function
limit
geometry
geometric transformation

Here is an interesting integral formula:

Let $a$ and $b$ be real positive numbers such that $a<b$. Prove that

$\int_{0}^{1}\frac{f(ax)-f(bx)}{x}dx=f(0)\ln\frac{b}{a}-\int_{a}^{b}\frac{f(x)}{x}dx$.
Of course $f$ is a function for which the two integrals make sense.

The first integral have not sense for $x=0$ or I have lost something ?

Think about what happens  if $f(x)=\arctan(x)$, then both integrals make sense.
The integral does not depand on a set of measure 0, i.e. it is not important whether or not the integrand is or is not defined at a point, however for this problem $f$ is such a function for which the left hand side integral makes sense.
Didi

If $f$ is differentiable at zero, then $\lim_{x\to0}\frac{f(ax)-f(bx)}x=(a-b)f'(0).$

That would make this a proper integral at zero, if you merely extend the function being integrated by continuity.

didilica didn't explicitly say that $f$ was differentiable, but I would read his last sentence as permission to apply any reasonable regularity condition.

$f$ does not have to be a differentiable function, it is sufficient that $f$ is a continuous function on the interval $[0,1]$.
Didi

I think so : if $f(0)=0\ ,$ then prove easily; if $f(0)\ne 0\ ,$ then through the translation $f: =f-f(0)$ come back to the previous case and appears $f(0)\cdot \ln\frac{b}{a}\ .$

[url=http://www.mathlinks.ro/Forum/viewtopic.php?p=288711&highlight=specific#288711]Here[/url]'s a related problem. It took me quite a while to find it...

That is it.
This is my method too.
Yes, the formula resembles the famous Froullany integrals , the difference is that the interval is of finite form.


I know that there is a limit of 3 similies per post. But why does the website count a quote as a smilie too?:huh:

I don't know, but maybe because, after all, it continues being a smile

[quote="Inspired By Nature"]I know that there is a limit of 3 similies per post. [/quote]
To the best of my knoledge, this is just wrong. So far you may have as many of them as you wish :) :lol: :D  :P . It may change if the people continue to abuse them though :roll: Maybe Valentin can comment on what happened in your particular case.

:P: The maximal number of smilies allowed in a post depends of the size of the post.  ;)  There shouldn't be any problem unless you are really making posts consisting to 50% of smilies... :D

  darij ;) [hide=":lol:"][hide=":rotfl:"]This text is here to make it possible to have 6 smilies in this post. If it weren't here, the maximal number of smilies allowed would be 4.[/hide][/hide]

Actually, I remember in the SC forum someone asked why they had a smiley limit.  I didn't have that trouble, but that was because I'm a moderator.  So I think there currently is a limit but perhaps mods aren't subject to it :?:  I'm really not sure of the details though.

No, the limit is there for moderators, too.  Try making a post with 4 emoticons and nothing else.

If you want to use emoticons, put some content into your post.

Ok, then.  Clarification (after experimentation): mods aren't subject to the smiley limit in the forums that they moderate.

I think there is also a maximum limit of 10 smileys, no matter how much content is in your post.

What happened to the posts before the smiley limit with several thousand smileys?  I remember at some point in the never ending story, a bunch of people (including myself) posted enourmous armies of smileys.

you can use [size=0] tag to add unlimited hidden text and thus unlimited smilies. Or you can do like darij, but then you need a separate line.

[quote="Peter VDD"]you can use [size=0] tag to add unlimited hidden text and thus unlimited smilies.[/quote]

How funny! With size 0 you can see nothing

[quote="Valentin Vornicu"]Thanks for the input. I'll deal with all these hacks hopefully  :evilgrin:[/quote]In fact, you'd better first make that the thing count correctly... ;)

Anyway I'd suggest not using a smilie limit (sometimes the most appropriorate post [i]is[/i] just a smilie), but for people that post too many smilies, allow admins to just disable their smilies completely.

That would be just too much of a hassle. There will be a permission settings for this.

I suggest just putting a much looser limit. Then I can put a bunch of smilies into a message but cannot but huge waves of them.
:w00tb: :) :D :ninja:

[quote="José"]Maybe because it was at the Games and Fun Factory Forum, where the posts don't count[/quote]

Probably not because this was back in the fall of last year, before games and fun factory posts did not count.

[size=0]

 :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:

:ninja:  :ninja:  :ninja:  :ninja:  :ninja: indeed.

:rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl: in fact, no.

I don't know how you guys do it... :huh:
the website usd to tell me the simile limit was 3 no matter how long the message. Now it tells me 2. :(

[quote="Peter VDD"]:rotfl:  :rotfl:  :rotfl:  in fact, no.[/quote]By default moderators are not included in the spam settings, although I can always make an exception for you Peter if you like ;)

[quote="Peter VDD"]Hey sorry, I just thought it didn't work yet. :)
[usually mod status only affects forum you're moderating][/quote]Actually it's not the user_level that is checked, but rather being the [url=http://www.mathlinks.ro/Forum/memberlist.php?g=418]moderator usergroup[/url]

Can someone explain why there is a limit of 2 smilies for me? It's not as if I overuse them like Peter and adkamath over here. :(

[quote="Peter VDD"]It depends on the length of the post I think. The longer the post, the more smilies it can have.[/quote]Exactly. I came up with a formula for the normal number of smilies a post should contain.

I see, thanks. But how is Peter able to put so many smilies when his message is 3 words long. :huh:

Apparently the rule does not apply to moderators, which caused my test to fail. :)

dont worry i'll  find a crack soon........... :diablo:

i think you can post more now. i did 3, so it isn't 2  :D  :D  :D

you can post even more

Open up the hide to see

[hide]
 :)  :D  :lol:  :blush:  :maybe:  :)  :(  :o  :mad:
[/hide]

[quote="roadnottaken"]I'm curious to know what the formula is.  If you don't want to tell me, you don't have to, but I would like to know.[/quote]

Maybe you ought to quote me...

[quote="roadnottaken"]I'm curious to know what the formula is.  If you don't want to tell me, you don't have to, but I would like to know.[/quote]

Apparently, posts can't be edited in this forum.  This is a request [b][i]specifically to Valentin[/i][/b].  If you don't want to post the formula that is fine.  I would just appreciate it if you would let me know that that is your decision so that I don't end up waiting indefinately for a response that isn't coming.

I've made a habit of not posting any portions of my coding and I'll stick with ti :)


inequalities
linear algebra
matrix
probability
inequalities unsolved

Let $ x,y\in (0,1)$, s.t. $ x \plus{} y \equal{} 1$ and let $ m,n\ge 2$, where $ m,n\in \mathbb N$. Prove that:

$ (1 \minus{} x^m)^n \plus{} (1 \minus{} y^n)^m > 1$

I've seen this problem in slightly different form somewhere with a following solution

Let's randomly generate a $ m$x$ n$ matrix filled with ones and zeros. For any free cell in the matrix let $ x$ be the probability it is filled with $ 1$ and $ y$ be the probability $ 0$ appears at that place.
Of course $ x\plus{}y\equal{}1$.

Consider the following events:
 $ A$: in every row there is at least one $ 1$
$ B$: in every column there is at least one $ 0$

We have $ P(A)\equal{}(1\minus{}x^{m})^{n}$ and $ P(B)\equal{}(1\minus{}y^{n})^{m}$, but it is also obvious that $ P(A)\plus{}P(B)>1$


[url=http://www.google.com/virgle/index.html]Virgle[/url]

Would you go? Do you believe that if the two funding sources mentioned actually got together, humankind could go? How would we decide who to go in a fair manner? (And how would we define fair?)

I might go, [i]if[/i] I were married and my wife wanted to go too. I'm a fan of monogamy, myself, and actually sticking to marriage for a long time, and running away doesn't further that. I do believe the technology could be created, but I believe we don't really want to go - we haven't spent the money, we're too concerned about safety and the enviroment. And I believe it should be done privately, not by government, and the owners get to decide who goes. (But, then, there also needs to be clauses so that government there is independent of Earth, and eventually there is at least one soveriegn nation.) What'd ya'll think?

We need to learn to take care of Earth before we even think about doing anything like that.  Humans should not go around the solar system until they can act more civil on Earth.  This also goes for Moon colonies.  As of now, we have adulterated enough living space.

Virgle sounds interesting, and that would be quite an adventurous trip to take. It would be a wonderful experience and I would be honored to be among the first to further humanity's journey throughout the stars.  :)

@AstroPhys: I agree that while taking care of Mother Earth is an important part, however, in the unlikely event that it fails, we need a back-up. In this case, colonizing other planets would work.

[quote]# Phobos - Google now stores a full copy of the Internet on Mars as a physical backup, while Virgin is now the planet's major producer of cargo and crew ships and operates a large shipyard on the mineral-rich Martian moon Phobos.[/quote]

only good for 30 million years:

"Future destruction

Phobos' low orbit means that it will eventually be destroyed: tidal forces are lowering its orbit, currently at the rate of about 1.8 metres per century, and in 30-80 million years it will either impact the surface of Mars or (more likely) break up into a planetary ring. Given Phobos' irregular shape and assuming that it is a pile of rubble (specifically a Mohr-Coulomb body), it has been calculated that Phobos is currently stable with respect to tidal forces. But it is estimated that Phobos will pass the Roche Limit for a rubble pile when its orbital radius drops a little over 2,000 km to about 7,100 km. At this distance Phobos will likely begin to break up forming a short lived ring system around Mars. The rings themselves will then continue to spiral slowly into Mars.[24]"

[url=http://en.wikipedia.org/wiki/Phobos_%28moon%29#Future_destruction]http://en.wikipedia.org/wiki/Phobos_%28moon%29#Future_destruction[/url]

I'll do this properly this time:

[quote="solafidefarms"]
Would you go?
[/quote]

At a shot (assuming reasonable earthly communications (competing in contests etc)

[quote]
Do you believe that if the two funding sources mentioned actually got together, humankind could go?
[/quote]

No, Virgin and Google?

[quote]How would we decide who to go in a fair manner? (And how would we define fair?)[/quote]

Needs its own thread really, but I'll give it a shot.

1) Equal by country
2) Equal by gender
3) Equal by sexuality
4) Equal by intelligence
5) Equal by *insert property here*

[quote="SimonM"]Of course I would go at a shot.

Although have any of you checked the date (and can remember last years project :P)[/quote]

Based on the first three posts, I really do wonder whether they actually understand   :wink:

Yes, I understand it's an April Fools' joke. But I wish it were real. Gmail was announced April 1 too... ;)

There was another Google joke, too.

We were listening to NPR this morning, and Google said GMail was offering a new service that would allow e-mails to appear like they had been sent in time previously. You could even make them look like they had already been read....we were skepticle until we heard that it was an April Fools' Day joke. We didn't laugh. :P

Interestingly enough, the project is not completely impossible technically and might cost not much more than the Iraq war. Would I go? Probably I would prefer to do the tasks for the mission control. As to fairness, I really liked the "equal by intelligence" part :P. I wonder if we end up proclaiming that it is unfair that we use only highly trained and reasonably smart engineers to build bridges and doctors to perform heart surgeries. Should give everybody a chance to try, shouldn't we? :lol: Anyway, as I said, I would strongly prefer the Mars project to be a reality and the Iraq war an April joke but with our current politicians there is little chance for anything like that happening in my lifetime. So, I'll stay with pure mathematics. :)

Suppose this wasnt an April first joke...this would probably create some sort of controversy that more than likely will lead to nuclear warfare between people who support it and people who believe people were suppose to stay on earth and flourish here(so really extremeist..)why risk it?

[color=green]Moderator says: there is [i]never[/i] a good reason to block-quote the post immediately preceding yours.[/color]

"Probably"? I don't think so. A nuclear war would arise as a conflict between nations, not a conflict between different ideological groups within the same nation. And I don't think people who don't want to colonize space would feel so strongly about it as to violently oppose it. You might have a few crazies causing trouble, but you're not going to have war. Watch -- I bet there'll be a moon base by 2050 and that the people who oppose the project most will be politicians who want the money for themselves/their constituents/the companies that have them in their pockets.


topology
function
Support
advanced fields
advanced fields unsolved

$ M$ is $ C^r$ topological manifold,$ 1\le r$,$ A$ and $ B$ are two closed subset of $ M$
which they dont have intersection.Show that there is a $ C^r$ function $ f: M\longrightarrow[0,1]$
such that $ f(x)\equal{}1$ when $ x\in A$ and $ f(x)\equal{}0$ when $ x\in B$.

Assume A and B are nonempty. Put a metric d on M such that $ d(x,\cdot)^2$ is a $ C^r$ function for all $ x\in M$.
Let $ f(x)\equal{}\frac{d(x,B)^2}{d(x,A)^2\plus{}d(x,B)^2}$

I don't understand this. There are examples of smooth manifolds that don't have any metric. I don't see how to patch up the global structure to find the required function using metrics.

Let $ A'$ and $ B'$ be complements of $ A$ and $ B$. Then they form an open cover of $ M$. Thus there exists a $ C^r$ partition of unity $ \{\phi_a, \phi_b\}$ subordinate to this open cover (this is a standard fact). Since for all $ x\in M$ we have $ \phi_a(x) \plus{} \phi_b(x) \equal{} 1$ and the support of $ \phi_a$ is contained in $ A'$ and analogously the support of $ \phi_b$ is contained in $ B'$ we may take $ f \equal{} \phi_a$ which then obviously possesses the required properties.


LaTeX
number theory unsolved
number theory

prove that there exist 2002 positive integer $1< a_{1}<...<a_{2002}$ and $a_{1}.a_{2}...a_{2002} -1$ divise by $(a_{1}-1)...(a_{2002}-1)$

my Latex was successfully

have you idea? :( it is posted for a long time


[quote="spider_boy"]F:Z-->Z (Z is the set of integers.)

                F(x+F(y))=F(x)-F(y) for any x,y in Z.
 Find all solutions of the following function.[/quote]

[hide]

Let $x=y=0$ we have $F(F(0))=0$ . Now letting $F(0)=k$ for some integer $k$ .

Let $x=-F(y)$ , we obtain $F(0)=F(x)+x$  :arrow:  $F(x)=k-x$

Is it correct ?  :?
[/hide]

I think no. the second line is obscure to me...
 you must show that F is surjective.and i cant do it since april..!

Yes, it's an interesting problem.I'd like to know if there's a solution.


of course, that's assuming the time not set in millitary format...

[quote="math92"]A chime clock strikes 1 chime at one o'clock, 2  chimes at two o'clock, 3 chimes at three o'clock, and so forth.  What is the total number of chimes the clock will strke in a twelve-hour period?[/quote]

1+2+3+4+5+6+7+8+9+10+11+12= 13 + 13 + 13 + 13 + 13 + 13 = 78

[quote="Wickedestjr"][quote="Xantos C. Guin"]1+2+3+4+5+6+7+8+9+10+11+12+1= [b]79 ANS.[/b][/quote]

Isn't it 78. Where did you get that extra 1 at the end?[/quote]

Because before any time begins, it strikes $ 1$.

Oh okay. Thanks. I would have never gotten this one right.

Guys, please don't revive topics 3 years old.

EDIT: You don't have to keep replying though.

Sorry, I didn't notice it was 3 years old...

Wait a minute. It can be 78 or 79, because say the time starts at 12:30, then it adds up to 78, but if it begins at 1:00, then it adds up to 79.

guys
plz stop this discussion
if you are really inspired to find out whether the answer is 78 or 79, make a new thread


Post your highest rating ever.

First 14 to sign up participates in the tourney.

Sure I'll Join
Highest Rating: 1666

Is this on ftw or ftw2 i would prefer ftw

This cd tourney will be on FTW, not the test version.
And actually, I've increased the number of spots to 16.

[hide="Participants (2)"]
qazwsxedc
crazyzhang
[/hide]

YUSH count meh in. and what settings? cd, or 10/15-25 time

one on ones?
highest - 1552 (IKRS IT FAILS I always get up to mid 1500s then fail)
yays finally a tourney:D

@rizikin

Be sure to read the posts before posting. My post already answered your question.
Thanks!

-agent

I'll Join...

Highest Rating:1560 (yes i know i fail)

UPDATE!

[hide="Participants (5)"]
qazwsxedc - 1666
crazyzhang - 1304
rizikin - 1552
Zukex - 1265
steve123456 - 1560
[/hide]

I hope to get 16 participants by the end of January. I will be PM'ing all of the participants when there are no vacant spots left detailing the tournament.

I guess I'll join as well.

My highest rating is pathetic so no comment(1515).

OK. Starting form here, please copy and paste this list and add yourself when you sign up.

[hide="Participants (6)"]
qazwsxedc - 1666
crazyzhang - 1304
rizikin - 1552
Zukex - 1265
steve123456 - 1560
TwinDracula - 1515
[/hide]

UPDATE!
We will have
rizikin vs. steve123456
Zukex vs. crazyzhang
I might change it up a little if more participants sign up.

[hide="Participants (7)"]
qazwsxedc - 1666
crazyzhang - 1304
rizikin - 1552
Zukex - 1265
steve123456 - 1560
TwinDracula - 1515
gopack25 - 1399
[/hide]

UPDATE!
Ok. I have come to a verdict. Special is disqualified for memming too much, so qazwsxedc has won the first round! (we did the fifth game today because both Special and qazwsxedc were on)
We are now down to 9 contestants!
So we have
#1  CRICKET229 vs. TwinDracula (tomorrow, 8:30 PM CT)
#2  policecap vs. dragon96 (tomorrow, 8:15 PM CT)
#3  gopack25 vs. crazyzhang (tomorrow, 8:00 PM CT if crazyzhang and gopack25 aren't on FTW at the same time today)
#4  Zukex vs. musicnmath (tomorrow, 7:45 PM CT)
#5  qazwsxedc vs. Winner of game 1 (tomorrow, 8:45 PM CT)

[quote="agentcx"]Special is disqualified for memming too much[/quote]

WIN.

Wow.

qazwsxedc really got lucky because he got to automatically win his first 2 games.

[b]Learned life lesson:[/b] Never be afraid of people who have good memory because they might actually be disqualified.The situation might even turn out to be for the better like it did in this situation.

hehe i know i think i'm gonna lose though i just got worse at cd

LOL special got DQed

agrees with Twin :ninja:  :ninja:  :ninja:  RAWR

Ok, it's time for your UPDATE!
policecap won dragon96, so we are down to 8 participants. I'm rescheduling CRICKET229's game and musicnmath's game will be 15 minutes late.
#1 CRICKET229 vs. TwinDracula (tomorrow, 8:30 PM Central Time)
#2 policecap vs. winner of game 5 (tomorrow, 9 PM CT)
#3 gopack25 vs. crazyzhang (today, 8 PM CT)
#4 Zukex vs. musicnmath (today, 8 PM CT (don't worry, I can watch two games at once))
#5 qazwsxedc vs. winner of game 1 (tomorrow, 8:45 PM CT)

UPDATE!
crazyzhang and Zukex weren't here at 8 PM, so we had gopack25 play musicnmath. He won.
We are down to 5 participants.
#1 CRICKET229 vs. TwinDracula
#2 policecap vs. winner of game 3
#3 gopack25 vs. winner of game 4
#4 qazwsxedc vs. winner of game 1

remeber everybody goodluck and havefun :)

twin isn't on and neither is agent but cricket and i r on wat happenend to the scheduele

@qazwsxedc
We wait till rizikin's tourney is over. Then we continue with my tourney.

ok
 The message is too small. Please make the message longer before submitting.

agentcx...
i tried to pm you saying i dont know what is central time.
but i dont think it worked...
and also i was studying for a history test.

am i disqualified?

Nope you aren't. We wait until rizikin's team tourney is done.

I really do hope this is not considered as revival.

However, I heard from rizikin that his tournament is nearly dead.  So...is this tourney dead as well, or are we still continuing with this tourney???

Considering it's been about a month and nothing has happened, I'm going to assume it's dead.
@agent, if you would like to continue, plz pm me and I'll unlock the thread, or you can just start a new thread.


geometry
circumcircle
incenter
angle bisector
perpendicular bisector

Prove that for any triangle ABC, the intersection E of angle bisector of <C and the perpendicular bisector AB lies on the circumcircle of triangle ABC.

Let $I$ be the incenter and $O$ be the circumcenter. Let the intersection of $AI$ and the circumcircle be $D$, thus $BD=CD$, therefore $OD$ is the perpendicular bisector of $BC$.


function
real analysis
real analysis unsolved

$f: \mathbb{R} \to  \mathbb{R}$ a mesurable function that satisfies $f(x+y)=f(x)+f(y)$. Is it true that $f$ is linear.

Yes, it's true. It was discussed before, and I believe fedja posted a proof. Alternatively, you could try to prove it using [url=http://www.mathlinks.ro/Forum/viewtopic.php?highlight=additive&t=65985]this problem[/url] as a lemma.

It's true and the proof I know shows that $f$ is continuous at $0$ as so continuous everywhere

An interesting variation : if $f$ is Baire (limit of limit of .... of continuous functions) then $f$ is also linear.

the proof I know use the fact that if $A$ has strictly positive measure, the set $A-A=\{x-y|x,y \in A\}$ contains a neighboorhood of $0$.


trigonometry
analytic geometry
graphing lines
slope

a kangaroo can leap a distance of 10m on horizaontal ground. what's the minumum takeoff speed for such a leap?

It seems to me that you need to know the angle of takeoff.  Horizontal range of a projectile is dependent on angle and initial speed.

i think "kangaroo can leap" means maximum dist possible
that wud  give $ u \equal{} 10 m/s$ (taking $ g\equal{}10m/s$)

$ d\equal{}\frac{v^{2}\sin{2\theta}}{g}$ is the horizontal range of a projectile.  If you assume 10 is the maximum $ d$, then $ \theta\equal{}\frac{\pi}{4}$, so
$ 10g\equal{}v^{2}\Rightarrow v\equal{}\sqrt{10g}$

But I derived it :D

If you want to be completely rigorous...
[b]Exercise!![/b] Derive the horizontal and vertical range equations for a projectile.

Actually, it's a pretty interesting exercise.  Deriving a physics formula shows that you understand the basics.

[quote="rituraj007"]:oops:
tat's boring :mad:[/quote]

Kya yaar? That formula does not require more than 2 or 3 equations to derive....

It mostly relies on $ d\equal{}v_{0}t\plus{}\frac{1}{2}at^{2}$

[quote="hell_ever"]Kya yaar? That formula does not require more than 2 or 3 equations to derive....[/quote]
wat gud will writing equations of motion do?? :huh:

[hide] :P
this is sth better

General Eqn of projectile:$ y \equal{} xtan\theta \minus{} \frac {gx^2}{2u^{2}cos^{2}\theta}$

putting $ y \equal{} 0$ for horizontal range, $ x \equal{} \frac {u^2 sin2\theta}{g}$

differentiating and putting $ \frac {dy}{dx} \equal{} 0$ ,we get vertical range $ y \equal{} \frac {u^2 sin^{2}\theta}{2g}$

pls dont mind some typo's ,if any :P
 [/hide]

And where did you get that first equation from?  You need to derive that one, too.

well actually it will help if u r just given the Eqn of projectile :D
lyk $ y \equal{} ax \plus{} bx^2$
better 2 memorize it!! :P

You have to get the equation from somewhere.  It didn't just appear out of nowhere.  Or did it?

Hey Jrav,
deriving the equation of trajectory is damn simple.
U have [b][i]t=x/(ucostheta)
then u substitute this value for the vertical motion...u get it[/i].[/b]

I obviously know how to derive these equations. The point I'm making is that if you're going to just blindly use equations, you should be able to show where they come from.

well i cud hav done it easily (ofc as long as i m not outta my wits :(  :P )

but ppl wont expect me to do that :)

Seems like the projectile is long in motion....

Here's a more interesting problem for u guys... :)

Say, Sachin is batting on 94, and he decided to make his century with a sixer.....the ball was bowled at 148 kmph, and Sachin just [b]flicked[/b] it towards the sqaure-leg boundary....considering the square-leg boundary at 46 yards and a 5'11'' fielder fielding there....
My question is, if the flicking angle be nearly 50 degree, what ought to be the limit of height of Sachin to achieve his century???

Just find the height above which Sachin reaches the hundred, and below which he is caught at the deep....!!!! :wink:

Hope you like this one....

I have no idea how cricket is played so that question makes little sense to me.

i dont think u need 2 kno how cricket is played--its just lik baseball :P

PS--if we put in sachinn's actual height in place,i think he has very less chances of getting 3 figures :P  :rotfl:

There are basically 5 fundamental kinematic equations of classical mechanics, all of which can be derived from one another.  Once you know how to use these 5 equations, you can pretty much solve any projectile motion problem.
$ \bar{v}\equal{}\frac{d}{t}$
$ a\equal{}\frac{\Delta v}{t}$
$ v_{f}\equal{}v_{i}\plus{}at$
$ d\equal{}v_{i}t\plus{}\frac{1}{2}at^{2}$
$ v^{2}_{f}\equal{}v_{i}^{2}\plus{}2ad$

not necessarily.

u shud kno the basics of parabolic motion ,slopes etc.. to solve the good ones

correct coz without knowing the concept of parabolic path it becomes extremely difficult to know about projectile motion


We feel warmer, if we put our hand above the flame rather than to its side. Why? The distance between the flame and hand is same in both the situations.

[hide="short answer"]hot air travels  straight up[/hide]


Let $a,b,c$ be real positve numbers such that $a+b+c=1$, find the minimum of
\[ f=\frac1{a+1}+\frac1{b+1}+\frac1{c+1}. \]

Is this olympiad?

$f \ge \frac{9}{3+a+b+c} = \frac{9}{4}$?

Ermm... well you can always jensen it  :blush:

[quote="Nameless"][quote="Anh Cuong"]Oke, so try this
$f=\frac{1}{a^2+1}+\frac{1}{b^2+1}+\frac{1}{c^2+1}$  :)[/quote]
Find min or max? :huh:
If find max:
$\frac{1}{1+a^2}\le\frac{27}{50}(2-a)$[/quote]
So let find min  :)
To probability1.01 :  Um... how can we jensen it .
Another funny problem:
Give $a,b,c \geq 0$ where $ab+bc+ca=1$
Find minimum value of $f=\frac{1}{a^2+1}+\frac{1}{b^2+1}+\frac{1}{c^2+1}$  :)

$\displaystyle f=\sum\frac{1}{(a+b)(a+c)}=\frac{2\sum a}{\sum a-abc}\geq 2$ since $abc\geq 0$ :) .


trigonometry
inequalities
vector
abstract algebra
complex numbers
combinatorics unsolved
combinatorics

hey guys, prove this without using complex numbers...

Consider a polygonal line $P_0...P_n$ such that angle $P_0P_1P_2=P_1P_2P_3=...=P_{n-2}P_{n-1}P_n$ all measured clockwise. If $P_0P_1>...>P_{n-1}P_n$ show that $P_0$ and $P_n$ cannot coincide.

Let $a_i$ be the given lengths with $a_0>a_1>...>a_{n-1}$.
By looking at the ordinate's value when asuming that the first line lies on the $x$-axis, we get $\sum_{k=0}^{n-1} a_k \sin(\frac{2k\pi}{n})=0$, which contradicts what we get by summing up the inequalities $a_k \sin(\frac{2k\pi}{n}) + a_{n-k} \sin(\frac{2(n-k)\pi}{n}) = \sin(\frac{2k\pi}{n}) (a_k-a_{n-k}) > 0$ for $2k < n$.
(indeed this is more or less a solution using complex numbers or, more exactly, vectors ;) )

Cute problem you know.... Consider our segment line on a complex plane :) Such that  $P_0$ would be the center of this plane and let $Re$ axis would be so that $Im(P_1)=0$ and$Re(P_1)>0$ Let's take complex number $x$ such that $|x|=1$ and $arg(x)=\pi-\angle P_0 P_1 P_2$ Now let $l_i$ be the length of segment $P_i P_{i+1}$ So what is the form of our consecutive points??? Let's see... we have $P_1=l_0$ and we may create reccurency: $P_{i+1}=P_{i}+l_i x^i$( or something like that...) Which means that our point $P_n$ has the form: \[ P_n=l_0+l_1 x+l_2 x^2+...+l_{n-1}x^{n-1} \] Now all we have to is to show that this expression can't be equal $0$. There is some kind of summation formula ( it's called Abel's or something) from which we obtain \[ P_n=(l_0-l_1)(1)+(l_1-l_2)(1+x)+(l_2-l_3)(1+x+x^2)+...+l_{n-1}(1+x+x^2+...+x^{n-1}) \] Assuming that $P_n=0$ we obtain \[ (l_0-l_1)(\frac{1-x}{1-x})+(l_1-l_2)(\frac{1-x^2}{1-x})+(l_2-l_3)(\frac{1-x^3}{1-x})+...+l_{n-1}(\frac{1-x^n}{1-x})=0 \] But as $x\neq 1$ we have then \[ (l_0-l_1)+(l_1-l_2)+...+l_{n-1}=(l_0-l_1)x+(l_1-l_2)x^2+...+l_{n-1}x^n \] As this equality holds , it has to hold also for modules but here we have a contradiction cos' \[ |(l_0-l_1)x+(l_1-l_2)x^2+...+l_{n-1}x^n|<|(l_0-l_1)x|+|(l_1-l_2)x^2|+...+|l_{n-1}x^n|=(l_0-l_1)+(l_1-l_2)+...+l_{n-1} \] and we are done


Let's say that you were in the top four at states, but then your team bombed and got something like #45. Would you advance to nationals, and would your team advance with you?

Teams do not advance to Nationals.  Individuals advance to Nationals.

In the case you cite, you (as a top 4 individual) would advance to Nationals.

The remaining individuals on your team would not advance to Nationals, unless they were also among the top 4 at your State.


linear algebra
matrix
modular arithmetic
algebra
polynomial
linear algebra unsolved

Find an example of a $ 4\times 4$ real matrix $ A$ with these properties:

$ \bullet$ All of the entries of $ A$ are $ 1$ or $ \minus{}1.$

$ \bullet$ $ A$ is a circulant matrix: that is $ a_{jk}$ depends only on $ j\minus{}k\pmod{4}.$

$ \bullet$ The columns of $ A$ are orthogonal. (Which means that $ \frac12A$ is an orthogonal matrix.)

Then, having found such an example, determine both its characteristic polynomial and its minimal polynomial.

One such example is,
$ \begin{bmatrix}1&\minus{}1&1&1\\1&1&\minus{}1&1\\1&1&1&\minus{}1\\\minus{}1&1&1&1\end{bmatrix}$

its characteristic equation is,
$ x^4 \minus{} 4x^3 \minus{} 4x^2 \minus{} 4x \plus{} 16$

There are many other similarly constructed matrices with the required properties.

That characteristic polynomial factors as $ (x\minus{}2)^2(x^2\plus{}4).$ Note that $ A^4\equal{}16I,$ which is consistent with its minimal polynomial being $ (x\minus{}2)(x^2\plus{}4).$

You say there are "many other similarly constructed matrices." What is $ A^2$ for each of those other examples? Do they all have the same $ A^2?$

The other 'similarly constructed matrices' are, $ \minus{} A$ ,$ A^T$ and $ \minus{} A^T$
$ ( \minus{} A)^2$ = $ A^2$.

Also, $ A^4$=$ 16I$, And since $ \frac12A$ is orthogonal, $ AA^T \equal{} 4I$. So, $ 4AA^T \equal{} A^4$ Hence, multiplying bothsides by $ A^T$ we get, $ A^2 \equal{} {(A^T)}^2$.

Thus all of them have the same square, and that is,

   $ \begin{bmatrix}0 & 0 & 4 & 0 \\
0 & 0 & 0 & 4 \\
4 & 0 & 0 & 0 \\
0 & 4 & 0 & 0\end{bmatrix}$


Wow I never realized there was an AoPSWiki search engine for Firefox.  :)

It's automatically part of every MediaWiki-powered wiki.

See, here's a random wiki installation I made, and it has it too: http://problemsolving.4realhost.com/wiki/index.php/Main_Page