linear algebra
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Give an example of a non-zero 2 x 2 matrix A with A^2 = 0 (0 matrix). Tried guessing but no such luck.

Where n<m and V and W are linear spaces with dim V = n and dim W = m, does there exist an invertible linear transformation T:V->W? If not, is there a proof?

Last, the condition for a linear transformation T:V->W to be invertible is that T be injective/surjective. But I don't understand what exactly the surjective part says. Couldn't you define W initially so that T(V) = W?

1. Here are two easy to remember examples:

$A=\begin{bmatrix}0&1\\0&0\end{bmatrix}.$ $B=\begin{bmatrix}1&-1\\1&-1\end{bmatrix}.$

Any nonzero $2\times 2$ matrix with both determinant zero and trace zero will work.

2. The proof has do do with the defintion of dimension. We have $V$ of dimension $n$ and $W$ of dimension $V,$ and $T: V\mapsto W$ linear.

Let $\{v_{1},\dots,v_{n}\}$ be a basis for $V.$ Since everything in $V$ can be written as a linear combination of these basis elements, it follows that everything in $T(V)$ (that is, the range of the mapping $T$) is in the span of $\{T(v_{1}),\dots,T(v_{n})\}.$ The span of $n$ vectors has dimension at most $n.$ (For that dimension to be $n,$ that set would have to be independent, which turns out to be the same as asking for $T$ to be one-to-one. If $T$ is not one-to-one, then the dimension of its range is less than $n.$) In any case, for $n<m,$ no set with $n$ elements can span an $m$-dimensional space. Hence $T(V)$ cannot be all of $W$ and the mapping is not onto (surjective).

3. You seem to be confusing [i]codomain[/i] with [i]range[/i]. You can (more or less, within reason) make whatever set you want to the codomain of a given function, but the range is what it is - you can't define it to be more or less. There's a difference between mapping [i]into[/i] and mapping [i]onto[/i].

Let's step away from linearity and leave the linear algebra context; we're talking pure "function-speak" here. Consider $f: \mathbb{R}\mapsto\mathbb{R}$ defined by $f(x)=x^{2}+1.$ This is properly defined for all real $x,$ and the values it takes on are real numbers. That's consistent with what we said when we said $f: \mathbb{R}\mapsto\mathbb{R}.$ This is a real-valued function whose domain is the whole real line; this maps the real numbers [i]into[/i] the real numbers. So the second $\mathbb{R},$ the one after the arrow - that's the codomain. Why that and not some smaller set? Pretty much because we chose to say so.

But this function does not map $\mathbb{R}$ [i]onto[/i] $\mathbb{R}.$ It's not surjective. In particular, there is no $x$ such that $f(x)=0.$ The actual range of the function is whatever the function says it is. (In this case, we can see that the range is $[1,\infty).$)

Consider $T: \mathbb{R}^{2}\mapsto\mathbb{R}^{3}$ defined by $T(x)=Ax$ where

$A=\begin{bmatrix}1&4\\2&-2\\4&7\end{bmatrix}.$

This is a one-to-one mapping. (How do I know that? Because I can see that $A$ has independent columns.) Does it map $\mathbb{R}^{2}$ onto $\mathbb{R}^{3}?$ No it doesn't. It's not surjective, so it's not invertible.

You want to redefine $W$ to be not all of $\mathbb{R}^{3}$ but rather the range of $T,$ which is some two-dimensional subspace (some plane) of $\mathbb{R}^{3}.$ Does $T$ map $\mathbb{R}^{3}$ onto $W?$ Yes it does. Does that make $T: \mathbb{R}^{2}\mapsto W$ invertible? Yes it does. But there's a price to be paid. What does it's inverse look like? Before we can even start to define $T^{-1}(w),$ we'll first have to determine whether $w$ belongs to the domain of $T^{-1}$ (which is the range of $T$.) If I gave you some more or less random 3-vector, how quickly could you determine whether that vector belonged to $W$?

[quote="Kent Merryfield"]$A=\begin{bmatrix}0&1\\0&1\end{bmatrix}.$ [/quote]

That's supposed to be $A=\begin{bmatrix}0&1\\0&0\end{bmatrix},$ of course.

[I edited the typo that JBL caught.]

I just finished grading the final exam for my "Introduction to Linear Algebra" exam. The results of one question on the exam are related to some of §outh§tar's concerns - and quite discouraging to me.

Here's the question:

[quote]Let $B=\begin{bmatrix}1&2&1\\1&0&-1\\-1&-2&-1\end{bmatrix}.$

(a) Suppose $T$ is the mapping from $\mathbb{R}^{3}$ to $\mathbb{R}^{3}$ given by $T(v)=Bv.$ Is $T$ a one-to-one mapping? Why or why not? Is $T$ an onto mapping? Why or why not?[/quote]
The single most popular set of answers - I think it was actually a majority of the class - was that $T$ [b]is not[/b] one-to-one but that it [b]is[/b] onto.

I was taken aback by the popularity of that answer. I am forced to conclude that "onto" (or "surjective" if you'd rather have it in Latin) is a trickier and more nuanced concept to learn than I had realized, and that my class really does not know what that word means.

[In case anyone is wondering, part (b) of the question was to find all eigenvalues and eigenvectors of $B.$ That didn't go particularly well, either.]

Thanks Kent, that clears things up.

In your first post, the very last example, I think you want to say W is in R(2) though, in order for T to be invertible. I think the main problem was I thought if dim V > dim W, we could make T:V->W invertible just by defining it to be T:V->T(V) and making it surjective (which seemed absurd). But when I asked my question I hadn't yet figured out that dim V = dim W was a necessary condition for 'T is invertible' to be equivalent to 'T is surjective'. My book usually does a good job on pointing out the subtleties beforehand but the relevant theorem made it seem as if dim V = dim W was merely a special case for which certain conditions were equivalent to T being invertible.

Here are three last ones which stumped me:

1) Show that R (the reals) is not a finite dimensional space when Q is the set of scalars.
2) Let V be the set of complex numbers regarded as a vector space over R. Find a function from V into V which is a linear function on the above vector space, but which is not a linear transformation on $C^{1}$, i.e., which is not complex linear.

3) An example in my book begins: If F is a subfield of the complex numbers (or a field of characteristic zero) and V is the space of polynomial functions, DE = I. D is the differention operator, E is the indefinite integral operator, and I is the identity operator. What's so special here about F being a field of characteristic zero?

[quote]In your first post, the very last example, I think you want to say W is in R(2) though, in order for T to be invertible.[/quote]
No, I don't want to say that. The values that this function takes on are not in $\mathbb{R}^{2}.$ They're in $\mathbb{R}^{3}.$ They happen to be confined to a particular two-dimensional subspace of $\mathbb{R}^{3},$ but that is not the same thing as being in $\mathbb{R}^{2}.$

The notion of "surjective" depends critically on the notion of "codomain" - that you establish what the domain and codomain of a function are before you find out anything else about the properties of the function. Yes, it's arbitrary. If we never set up a codomain but always talked only about range, then every function is alway surjective onto its range, and the very notion of "surjective" would become meaningless. But "surjective" is a useful mathematical concept, and it follows that "codomain" must also be useless.

Suppose $V$ is a finite dimensional vector space and $T$ is a linear mapping from $V$ into itself. (An "endomorphism," if you like words.) Such a mapping can be represented by a square matrix. And the central elementary theorems about matrices (and dimension, and rank, ... ) tell us that $T$ is one-to-one (injective) if and only if it is onto (surjective). Hence my displeasure with my class: for square matrices, the two concepts go hand in hand.

This theorem fails utterly for infinite dimensional vector space. It is common in the study of infinite dimensional vector spaces to introduce metrics or topologies. If you want to stay away from those, handle things strictly algebraically, but still have a "nice" space - let's say, a vector space with a countably infinite Hamel basis. Up to isomorphism, there's really only one such example, and you named one of it's many manifestations. So let's work with this example.

Let $\mathbb{P}=\mathbb{R}[x]$ be the set of all real-coefficient polynomials in a variable $x.$ (I called the coefficients real just to make things familiar for you - any field would be OK. And $x$ doesn't belong to anything - it's just a symbol, along for the ride.)

Let me introduce you two two famous examples: the right shift operator and the left shift operator.

Right shift: Let $R(p)(x)=xp(x).$

Left shift: If we write $p(x)=a_{0}+a_{1}x+a_{2}x^{2}+\cdots+a_{n}x^{n},$ then $L(p)(x)=\frac1x(p(x)-a_{0}).$

In other words, if $p=1+2x+3x^{2},$ then $R(p)=x+2x^{2}+3x^{3}$ and $L(p)=2+3x.$

The right shift operator is one-to-one (injective) but not onto (surjective). It's not onto because its range is restricted to polynomials with zero constant term.

The left shift operator is onto (surjective) but not one-to-one (injective). It's not injective because it has a nontrivial kernel: it maps all constant polynomials to zero.

$L\circ R$ - that is composing these operators as first the right shift followed by the left shift - is equal to the identity operator. That is, for every $p\in\mathbb{P},\ L(R(p))=p.$

But $R\circ L$ - composing these two operators in the opposite order - does not give the identity operator. (Try it to see what you get.)

Oops. I see what you mean about W. Thanks.

And as for R[x] is there nothing nothing which requires the field to be one of characteristic zero? My book performed the same example under 'dumbed down' terminology but specifically required the field to be of characteristic zero. I don't know anything about Hamel bases though. Maybe after I have finite dimensional spaces under my belt.  :blush:

$R \circ L$ gets rid of the constant term, so I suppose it acts as the identity on its range. Is it true anyway that a linear transformation between two infinite dimensional spaces is invertible if it sends a basis into (onto?) a basis? I could determine whether or not $R \circ L$ was the identity by checking its action on basis elements so maybe that idea works there. I'd guess if V is finite dimensional and W is infinite dimensional then T:W->V can never be invertible since there are 'too many' vectors in W for T to be injective. Or more simply its inverse can never be onto. :)

The technical word for "acts as the identity on its range" is "idempotent." There are two trivial idempotent operators: the identity, and the zero operator. Nontrivial idempotents are best visualized as projections.

On the other hand, at least in a pointwise sense, the left shift operator is nilpotent. That is, for any particular $p,$ the sequence $p,L(p),L^{2}(p),\dots$ is eventually zero. But there is no $n$ for which $L^{n}$ is the zero operator. Just another thing that can't happen in a finite dimensional space. In a finite dimensional space, it's easier to come up with one unambiguous definition of what we mean by "nilpotent."

Ah, I see why you needed a field of characteristic zero. My example of the left shift and right shift operators works in any field whatsoever. But the example you were talking about had the derivative and antiderivative operators on the polynomials. These are [i]almost[/i] the same as the left and right shift operators, but not quite the same.

The left shift maps $x^{n}$ to $x^{n-1}$ (and $1$ to $0$.)
The derivative maps $x^{n}$ to $nx^{n-1}$ (and $1$ to $0$.) The difference is the multiplication by $n.$

The right shift maps $x^{n}$ to $x^{n+1}.$
The antiderivative maps $x^{n}$ to $\frac1{n+1}\,x^{n+1}.$ The difference is the division by $n+1.$

On a field of characteristic zero, we're fine with that multiplication and division. But on a field of finite characteristic $p,$ then multiplication by $n$ is multiplication by zero whenever $p\mid n$ and division by $n+1$ is undefined whenever $p\mid (n+1).$

Just to add to the initial remarks, I figured I'd rehash what I've learned:

If V and W are finite dimensional linear spaces, a linear transformation T:V->W is invertible if and only if 1) T is injective and 2) dim V = dim W.

If V and W are infinite dimensional linear spaces, a linear transformation T:V->W is invertible if and only if 1) T is injective and 2) T is onto. In this case onto means W = T(V).

If one of V or W is infinite dimensional then no linear transformation T:V->W is invertible.

This clears things up since I'll be dealing mostly with finite dimensional spaces.

[quote="Kent Merryfield"]

[quote]Let $B=\begin{bmatrix}1&2&1\\1&0&-1\\-1&-2&-1\end{bmatrix}.$

(a) Suppose $T$ is the mapping from $\mathbb{R}^{3}$ to $\mathbb{R}^{3}$ given by $T(v)=Bv.$ Is $T$ a one-to-one mapping? Why or why not? Is $T$ an onto mapping? Why or why not?[/quote]
The single most popular set of answers - I think it was actually a majority of the class - was that $T$ [b]is not[/b] one-to-one but that it [b]is[/b] onto.

[/quote]

I understand how you show it is not 1-1 but how do you show that it is not onto?

For a matrix $A,$ the linear operation defined by $x\mapsto Ax$ is onto iff the system $Ax=y$ is consistent for all possible $y$ (of the appropriate size).

Computationally: row reduce $A$ until you reach row echelon form. The mapping is onto iff the echelon form has a pivot element in every row, and it is one-to-one iff the echelon form has a pivot element in every column.

Note that for square matrices, an accounting of pivot elements shows that the mapping is one-to-one if and only if it is onto.

The matrix $B$ given in my problem is row equivalent to a matrix in echelon form with two pivots. Hence, its rank is 2. By the rank-nullity theorem, it has a 1-dimensional null space - and such a mapping is one-to-one iff the null space is zero. Thus it's not one-to-one. The column space of $B$ (the span of the columns) is precisely the range of the the mapping. The statement that the rank of the matrix is 2 tells us that the range is a 2-dimensional subspace of $\mathbb{R}^{3}.$ Any 2-dimensional subspace must be a proper subspace of a 3-dimensional space - hence, the range is not everything, and the mapping is not onto.

[url=http://en.wikipedia.org/wiki/Pivot_element]Err...[/url]

For example, [url=http://en.wikipedia.org/wiki/Gaussian_elimination#The_general_algorithm_to_compute_ranks_and_bases]here[/url] the pivot elements are the 1's.


Three real numbers satisfy the following statements:
(1) the square of their sum equals to the sum their squares.
(2) the product of the first two numbers is equal to the square of the third number.

Find these numbers.

Let $x,y,z$ the numbers.

From hypothesis we have

$(x+y+z)^2=x^2+y^2+z^2$ and
$xy=z^2$

The first equality is equivalent to $xy+yz+zx=0$ which combined with the second one gives $z^2+yz+zx=0$ or $z(x+y+z)=0$ so either $z=0$ or $x+y+z=0$.

If $z=0$ then $x=0$ or $y=0$
If $x+y+z=0$ from the first part we get $x^2+y^2+z^2=0$ that is $x=y=z=0$

finally $(x,y,z)=(0,0,0),\ (0,k,0),\ (l,0,0)$

 :)


So I need the limit as n approaches infinity of

n^2 * {sin(6/n)}^2

I tried using L'Hopital's, but I couldn't get it to work.

Use the fact that $x-\frac{x^{3}}{6}\leq sin(x) \leq x$ whenever $x \geq 0$ (this you can show by studying the variations of the appropriate functions).

use the fact that as x -> 0, sin x ~ x.\

so as n->infinity, sin(6/n) ~ 6/n

so we'd get 6^2 = 36 by this.

oops. thx

[quote="julien_santini"]Very unlikely he has heard of equivalents in High School ... (maybe in a moroccan high school he would have, but definitely not in the US)[/quote]

was that directed to me?

I'm in a US high school...

Well, what I meant is that if he is in HS and he knows about equivalents, then very unlikely he'd need help on that question ... anyhow, it doesn't matter  :D

[quote="eggonabull"]
I tried using L'Hopital's, but I couldn't get it to work.[/quote]

Never mind .We have (substutation method)that  :

$\lim_{x \to \infty}(x \cdot \sin \frac{a}{x})= a \lim_{y \to 0}\frac{\sin y}{y}= a$


Do any of you post your homework on the blog provided by AoPs (because it has the LaTeX capabilities)? It would be actually pretty convenient because you could look at your school work from any place in the world. Or do you think that this is too time consuming?

SETTLE DOWN IT'S ONLY BEEN 4 HOURS!

No I don't do that and I never would have thought of it.

Wouldn't it be much easier to just use email+the texer?

I dont get how the "texer" and email would be easier....

you still have to type the code. And you cant include diagrams.  Its weird that nobody uses the LaTex  capabilities to keep organized.

o_O i heard that its against the rules to post your homework but i guess you can "preview" your post and copy the image using ways i won't go into right now and put it all on a Word document and email the doc to your teacher....

Or you could just get a TeX-rendering program for your computer for free and produce PDF's (or DVI's, I suppose)...

Looking at it from anywhere just isn't a meaningful advantage for homework- you're probably not going anywhere while you do it. If you're willing to spend the time to do your homework in $\LaTeX$, there's no reason not to just download an implementation and editor, so you can use the full typesetting capabilities.
I use TeXShop on a Mac myself.

(I'm a graduate student; the costs and benefits differ for a student who isn't as far along)
There are three major advantages to typesetting it:
1. It looks better. I'd rather not subject professors to my handwriting.
2. If I want to look back at an argument I used on some old homework problem, the original files are still there.
3. If an argument goes wrong, I can edit it without wasting paper.

[quote="Treething"]why not just write up your homework?[/quote]

Also, I use college rule paper, and its extremely hard showing work using the quadratic equation without messing up the format...or fractions and stuff...


Good morning,
I have some problems to resolve this exercice,please can you help me?
let  $ (T_n;\ n \geq 1)$ be  a  Poisson  process with intensity $ \lambda$ and $ (N_t;\ t\geq 0)$  is counting process associeted.
Proove that for any  $ t \ge 0$  fixed, proof that the process $ (T_{N_{t} \plus{} n} \minus{} t;\ n \geq 1)$ is a Poisson process  witn function intensity
$ s\rightarrow \lambda (t \plus{} s)$. :blush:

Ideas?
Thanks.


Any way without using trig ratios?

And 2 is not the answer given in my book.

I used another method and resulted with a different asnwer...

[hide]
We have $a^{2}-a+1=0$. Solving using the quadrative equation we get $a=\frac{1 \pm \sqrt{3}i}{2}$.
$(\frac{1+\sqrt{3}i}{2})^{2}=\frac{\sqrt{3}i-1}{2}\Rightarrow (\frac{\sqrt{3}i-1}{2})(\frac{1+\sqrt{3}i}{2})=-2$.
This means that $a^{3}=-2$, and do the same for $a=\frac{1-\sqrt{3}i}{2}$ and we get $a^{3}=2$. $1995/3=665$.

We get $a^{1995}+\frac{1}{a^{1995}}=\boxed{\pm (2^{665}+\frac{1}{2^{665}})}$[/hide]

While I'm beginning to doubt my book, I won't declare it wrong yet. So no, that answer doesn't fit the answer given.

Oh no I saw your [Potentially dangerous LaTeX] tag.

Can you give a constant value please? And I don't THINK that you need to use complex numbers. It's just supposed to be manipulating the expressions and substituting stuff in.

You know what? It's SUPPOSED to be -2.

Now can anyone show me how to get to that?

Maybe I made some mistake :(
why not trig, its is one of nicest part of contest math

[hide="the easier way with no complex"]$a^{2}=a-1$
$a^{4}=a^{2}-2a+1=-a$
$a^{3}=-1$
so the thing is equal to -2[/hide]
not really my idea, saw a similar problem before.

[quote="chesspro"][quote="jimhu"]
$(\frac{1+\sqrt{3}i}{2})^{2}=\frac{\sqrt{3}i-1}{2}\Rightarrow (\frac{\sqrt{3}i-1}{2})(\frac{1+\sqrt{3}i}{2})=-2$.
[/quote]

There's a miscalculation there.

[hide]$\frac{i\sqrt{3}-1}{2}\cdot \frac{i\sqrt{3}+1}{2}=-1$
Therefore, $a^{3}=-1$, and $a^{1995}+\frac{1}{a^{1995}}=-2$.[/hide][/quote]

Yea thans for catching the mistake

umm after posting the other day, i thought people would jump on it with the easy solution (it took some posts to get to finally)
well here's what i [hide="would have posted:"] \[\frac{a^{3}+1}{a+1}=0\] This is the given
So $a\not = 1, a^{3}=-1$ \[(a^{3})^{2N+1}=a^{1995}\ \ N\in \mathbb{N+}\\ \implies a^{1995}=-1 = a^{-1995}\] So the whole thing equals -2. As wanted. My mistake was a negative I should have picked up somewhere... challenge: find my mistake  :lol:
[/hide]

[quote="junggi"][quote="me@home"]* $a\not = 0$ (trivial) \[a^{2}-a+1=0 \implies a+\frac{1}{a}=1 \implies a^{1}+a^{-1}=2\cos \frac{\pi}{3}\] By DeMoivre's, \[a^{1995}+\frac{1}{a^{1995}}=2\cos \frac{1995 * 2 \pi}{3}=2\cos 0=\boxed{2}\] (I am using the fact that $\cos \theta+i \sin \theta=e^{i\theta}$, in conjunction with simple algebra to derive that $2\cos \theta =e^{i\theta}+e^{-i\theta}$)[/quote]

$a^{1995}+\frac{1}{a^{1995}}=2\cos \frac{1995 * 2 \pi}{3}=2\cos 0=\boxed{2}$ should be
$a^{1995}+\frac{1}{a^{1995}}=2\cos \frac{1995 * \pi}{3}=2\cos 0=\boxed{-2}$ :D
yay[/quote]

And $2 \cos{0}$ should be $2 \cos{\pi}$ :wink:.

Remember: Learn from my mistakes by deducing what I was trying to say and then learning from [b]that[/b]... :rotfl:


I'm a newbie to Latex and I keep trying to write 2^(AUBUC) in latex formula... but the (AUBUC) thing doesn't become an exponent, it's just normal size. I have the ^ sign and $ signs and {} brackets around the expression but it still doesn't work. How am I supposed to write it?? :D thanks!

[quote="ComplexZeta"]$2^{A\cup B\cup C}$ or $2^{(A\cup B\cup C)}$ should work, depending on which one you want.[/quote]

thank you!! :D


Let $ T$ be subset of natural numbers have $ 5$ other digits from $ \{1;2;3;4;5\}$.Prove that this subset can be divided to $ 2$ subsets $ A;B$ such that $ \sum_{x \in A} x^2\equal{}\sum_{x \in B} x^2$

[quote="lhplhp"]Let $ T$ be subset of natural numbers have $ 5$ other digits from $ \{1;2;3;4;5\}$.[/quote]

I don't understand this, what is "5 other digits" mean?

http://www.mathlinks.ro/Forum/viewtopic.php?t=150892

And also here
http://projectpen.wordpress.com/2008/05/13/10s-partitions/

Daniel


when do we start the discussion for round 4??? (I doubt anyone will have the time, tho...) after the math jam??

I bet after the math jam, seeing as the only time that wasn't the case was when the math jam was more than 2 days after the due date (and in that case discussion began 2 days after the due date). Anyway, this math jam is 2 days after the due date so...

However I hope we can discuss now lol  :roll: , I'm just dying to find out if my answers are correct and what other methods people used.
Maybe this time though they'll get things sorted out earlier (is that why they extend the curfew on discussion?) or something and we can discuss them sooner.

btw, when should we expect the final results???

There's no announcement saying not to, like there was before.  But still, we should probably wait for the mathjam.

Please wait until after the Math Jam tonight.

Final results will be available in mid-April.


Find the largest constant $k$ such that

$\frac{kabc}{a+b+c}\leq(a+b)^2+(a+b+4c)^2$

for all $a, b, c > 0$

Holding c and a+b constant, we have clearly a = b.  We then reduce to

$\frac{kb^2c}{2b+c} \le 4b^2 + (2b+4c)^2$

$kb^2c \le (2b+c)(8b^2+16bc+16c^2)$

$\frac{k}{8} b^2c \le (2b+c)(b^2+2bc+2c^2)$

$\frac{k}{8} b^2c \le 2b^3+4b^2c+4bc^2+b^2c+2bc^2+2c^3 = 2b^3+5b^2c+6bc^2+2c^3 = 2(b+c)^3-b^2c$

$\frac{k}{8} \le \frac{2(b+c)^3}{b^2c} - 1$

Now hold b+c constant.  Since $\sqrt[3]{2b^2c} \le \frac{b+b+2c}{3} \implies b^2c \le \frac{4}{27} (b+c)^3$ we conclude that the RHS is minimized to be $\frac{2}{\frac{4}{27}}-1 = \frac{25}{2}$.  Then it follows that k = 100.  Hope this is correct...


Many colleges' Part I applications are due soon, but on these applications, the colleges ask for SAT I scores already. I have taken the SAT in March and October, but my October scores won't be available until after the deadline for Part I passes. My March scores are low (2080), so should I submit my applications with my March scores, or wait until my October scores arrive and then apply with Part II? Thanks in advance!

I'm not completely sure, but if you send in your March scores, the colleges will have access to your October scores when they become available and take them into consideration. If you wait to send October scores, the colleges still see your March scores. Most colleges take the highest combined score and ignore the rest, but that might have changed since I applied.
Check with collegeboard.com and your colleges.

First, check with the colleges to see whether or not they will still accept the Oct. SAT scores for Part I.  If not, you should find out if you can apply to part II even if you don't make Part I.

Right. Definitely check with the college first. Afterwards, if you choose to apply with Part I, I believe the College Board offers some sort of rush grading service you can get (not for free) which you might want to consider.

Ok, I just emailed the college. The website says that if I do not submit Part I, I can always submit Part II, since Part I is only used to create my application file earlier and give me a discount on the application fee. If it doesn't work out, I'll just send in Part II and avoid any discrepancies between my Part I and Part II applications. Thanks guys!


Find all functions $ f: R\rightarrow R$ such that :$ f(x)\cdot f(y) \minus{}xy\equal{}f(x) \plus{}f(y) \minus{}1$ , $ x$ element $ R$

[hide="Solution"]Setting $ (x,y) \equal{} (k,k)$, we have:

$ f(k)^2 \minus{} 2f(k) \plus{} (1 \minus{} k^2) \equal{} 0 \Rightarrow f(k) \equal{} \left(2\pm \sqrt {4 \minus{} 4(1 \minus{} k^2)}\right)/2 \Rightarrow f(k) \equal{} 1\pm k$.

Alternatively we can solve for $ f(0)$ and consider the two cases.[/hide]

EDIT: $ f(k) \equal{} 1 \pm k$, not $ 1 \pm x$. Fixed.

EDIT2: I meant solve for $ f(0)$ if someone couldn't figure it out...

[quote="enndb0x"][quote="Vworldv"][hide="Solution"]Setting $ (x,y) \equal{} (k,k)$, we have:

$ f(k)^2 \minus{} 2f(k) \plus{} (1 \minus{} k^2) \equal{} 0 \Rightarrow f(k) \equal{} \left(2\pm \sqrt {4 \minus{} 4(1 \minus{} k^2)}\right)/2 \Rightarrow f(k) \equal{} 1\pm x$.

Alternatively we can solve for $ f(1)$ and consider the two cases.[/hide][/quote]

This don't seem to be the solution.If it is ,where are the fuctions ?
 I don't know why you did take $ x \equal{} y \equal{} k$,I think you should look either for the cases when $ x\neq y$[/quote]

You did not state that $ x, y$ must be distinct.
From the solution of Vworldv , $ f(x)\equal{}1 \pm x$ for all $ x$

Now put $ f(x_1)\equal{}1\plus{}x_1$ for some $ x_1 \neq 0$ we have $ y\equal{}1\plus{}y$ for all $ y$ also, so $ f(x)\equal{}1\plus{}x$ for all $ x$ is a solution after checking.
Now put $ f(x_1)\equal{}1\minus{}x_1$ for some $ x_1 \neq 0$ we have $ y\equal{}1\minus{}y$ for all $ y$ also, so $ f(x)\equal{}1\minus{}x$ for all $ x$ is a solution after checking.

So the functions are $ f(x)\equal{}1\plus{}x$ for all $ x$ or $ f(x)\equal{}1\minus{}x$ for all $ x$

Sorry, I wrote $ f(k) \equal{} 1 \pm x$, instead of either $ f(k) \equal{} 1 \pm k$ or $ f(x)\equal{} 1 \pm x$ (typo). This may have caused some confusion.

enndb0x, taking $ x\equal{}y\equal{}k$ yields a condition that $ f(k)$ must satisfy for all $ k$. Verifying that both solutions with the condition satisfy the original functional equation is easy.

In fact, your solution does not show that those are the only two solutions: after arriving at $ f(k) \equal{} 1 \pm k$, you are left the possibility that for some $ x$, $ f(x) \equal{} 1 \plus{} x$ while for some other $ y$, $ f(y) \equal{} 1 \minus{} y$.  The lines that stephencheng wrote are essential to the solution.

Well, yeah. That's just checking.. as I said... (I just posted a solution sketch; in an actual proof you have to verify everything.)

EDIT: I see that I rushed through my original solution, making two typos. By the way, the method of solving for $ f(0)$ that I mentioned is fully rigorous (without having to verify anything).

No, it's not what you said.  What you said was that there are two solutions and you have to check that they work.  In fact, what you did shows that there are uncountably many solutions, each having the property that for any given $ x$, $ f(x) \equal{} 1 \pm x$, but not necessarily having the property that for two different values of $ x$, the signs should agree.  It happens that the remaining steps are not very difficult, but they are not "just checking," either.

I agree that I rushed through my initial solution and didn't write the steps where I checked the functions, but these steps are trivial: $ (x\plus{}1)(y\minus{}1)\minus{}xy\equal{}(x\plus{}1)\plus{}(y\minus{}1)\minus{}1 \Rightarrow x\equal{}0$. In any case, I apologize for any ambiguity, and don't want to keep arguing.


digits $a_{1}$, $a_{2}$, .... $a_{1000000}$ are between 1 and 9. Prove that there are at most 100 perfect squares among the numbers

$a_{1}...a_{k}$, for $1\leq k\leq 1000000$

I haven't checked the calculations, but I'm pretty sure you can consider the cases k even and k odd separately. Then, for example, if x is a square in the sequence, then the next square in the sequence with the same parity of digits as x must have at least twice as many digits (why?).

im sorry, i don't understand. Could you please explain? Thank you!


Find all functions $f: R^{+}\rightarrow R^{+}$ such that

$(x+y)f(f(x)y)=x^{2}f(f(x)+f(y))$

I solved this only $2$ days ago.
[url]http://www.mathlinks.ro/Forum/viewtopic.php?t=138608[/url]


geometry
3D geometry
geometric transformation
rotation

Okay so here's this cone with vertex half angle $ \alpha$ and base radius $ R$. I place it on the ground and start rotating it about the vertex. It is placed slant edge horizontally. Now if it rolls with angular velocity $ \omega$ about it's vertex. if I have a particle sitting on the cone stationery w.r.t the cone  - what will it's net velocity/accelartion be. find both. Use whatever you need - mass , slant height and do this :)

I can't say for sure; maybe rotational inertia and torque play some sort of role.

Anyone got any leads? I think this was from Problems in General Physics by IE Irodov.

Assuming "net" in relation to the cone axis and the particle feeling comfortable sitting on

the rim of the base we'll have:

[hide]

   $ v \equal{} \omega \;cotg \alpha\; cos \alpha\;\;$   and   $ \;\;\;\;\;\;a \equal{} R{(\omega\; cotg \alpha\; cos \alpha)}^2$ [/hide]

otherwise, if it preferes the bases center, and in that case "net" is in relation to ground:

  [hide]
  $ v \equal{} \omega\; cotg \alpha \;cos \alpha$ (only the acceleration differs) $ a \equal{} {\omega}^2 R\; cos \alpha\; cotg \alpha$ [/hide]


Lady drives to work averaging 45km/hr. She drives home on the same route averaging 30km/hr. Having spent a total of 1hr behind the wheel traveling that day, how many km's did she clock up driving to work.

So...

Combined speeds = 75kmh so 45kmh/75kmh = 0.6 x 45 = 27km traveled in morning.

Is this right?

Why did you combine the speeds?

Anyway, let the distance to work be $d$.  She drove to work in $\frac{d}{45}$ hours and drove back in $\frac{d}{30}$ hours, so

$\frac{d}{45}+\frac{d}{30}= 1$

What if you just do this to find the average speed for the whole trip which is $\frac{45*30}{45+30}km/h$
Then $d = \frac{45*30}{45+30}km/h *1h = 18km$.


Let $a,b,c$ be positive reals such that $a^4+b^4+c^4=1$, prove that
$\sqrt{\frac{2a}{b^2+c^2}+\frac{2b}{c^2+a^2}+\frac{2c}{a^2+b^2}} \geq a\sqrt{a}+b\sqrt{b}+c\sqrt{c}$

The stonger inequality
[tex]\displaystyle \sqrt{\frac{2a}{b^2+c^2}+\frac{2b}{c^2+a^2}+\frac{2c}{a^2+b^2}} \geq \sqrt{3}\sqrt{a^3+b^3+c^3}[/tex]
is valid, too.

Thanks for your solution. I didn't think about Chebyshev.

By the way, is there a solution using Holder?

cauchy schwarz is enough.
$4(\sum \frac{a}{b^2+c^2}) \ge 2(\sum \frac{a}{b^2+c^2})(\sum a^2(b^2+c^2)) \ge (\sum a \sqrt{a})^2$
Take square root on both side.


Spoiler: [hide]The number of digits of any integer n is log n +1.  So the number of digits in these numbers combined is 2004log 2 +1 + 2004log 5 +1.  If we factored out a 2004 from the logs, then this would add an extra digit.  So 2004(log2 + log5) + 2 - 1 = 2005, which is the answer.[/hide]

[quote="rrusczyk"]How many digits total are written if we write $2^{2004}$, then write $5^{2004}$?[/quote]

$2005$


vector
geometry
parallelogram
absolute value

a)Let O,A1,A2,A3 be four distinct points in the plane so that:

|OA1|=|OA2|=|OA3|=|OA1+OA2+OA3|<>0. Show that there exists i,j in  {1,2,3} so that OAi+OAj=0.

b)Let n be a natural number, n>=3 and k>0. If O is a fixed point and the points A1,A2,..,An have the properties:
i)|OA1|=|OA2|=..=|OAn|<>0;
ii)|OA1+kOA2|=|OA2+kOA3|=...=|OA_(n-1)+kOAn|=|OAn+kOA1|, show that A1A2..An is a regulated poligon.

all the distances (OA1 for example) are vectors.

Okay, (a) isn't that bad.  [hide]Consider OA1 + OA2 + OA3.  If it's absolute value is the same as the absolute value for each of the other 3, then we have a rhombus.  That is, assume OA2 is not the negative of OA2 or OA3.  Then take points B, C, D, and E in the plane such that BC = OA1, CD = OA2, and DE = OA3.  Then BE = OA1 + OA2 + OA3, and we have that quadrilateral BCDE has four congruent sides and is thus a rhombus.  More importantly, it is a parallelogram, so two of the OAi's are negatives of each other.[/hide]

For part (b), is that supposed to be true for one fixed value of k or for all real k?

2) I'll take "true for all positive real k".

Let Am = (xm, ym) for all valid m. By property 1, xm2 + ym2 is constant for all valid values of m. Call this sum p.

By property 2, (xm + kxm+1)2 + (ym + kym+1)2 is constant. (suppose that m wraps around.) Expand to yield (k2 + 1)p + 2k(xmxm+1 + ymym+1) is constant. Thus xmxm+1 + ymym+1 is constant, which means that Am . Am+1 is constant. so any two angles between two pairs of adjacent vectors must be equal, hence we obtain a regular polygon. QED.

What if the vectors aren't restricted to 2-D space?  Can you say anything then?

:( I sort of assumed that - it was given in question a). I guess I just glanced over the question too quickly...

For three space though, the portion up to where I state that all dot products are equal is still valid. After that... I don't know...

I think you were right in assuming 2-D space, from the first question.  I was just curious if your proof says anything about higher-dimensional space, or if you can modify it so it does.


P is any point on a given line.A is a fixed point .

In AP or AP produced take a variable point such that AQ/AP= constant.

Prove that Q lies on a line

let P and P'be two points on L,and Qand Q'are on APand AP',inwich we have:

AP:AQ=AP':AQ'

now we claim that for each Q",we have Q,Q',Q" are colinear.

we have :triangles $AQ'Q$ and $AP'P$ are simillar .then for each point like Q',we know that QQ'and PP' are

parallel.

then i think it doesnt need to continue.because its clear...

another problem may be enjoyable here:find the locus of Q,in wich:AP.AQ=stable


calculus
probability
number theory
\/closed

How are the classes run? Are the topics introduced daily, or weekly?

Are you talking about normal math clases like Calculus, or clases more like the ones given here?

Here I believe since this is the Q&A Forum.

Depends on the course, I believe the more easier courses for competitions just goes over tons of problems that has some sort of connection. That's all I can say.

We hold our classes once a week for an hour and a half at a time.

We offer two types of classes.

One type offers a concentrated view of a subject such as number theory or counting (counting involves a lot of combinatorics, partition problems, probability, etc.).

The other type of class we offer is based on various problem solving competitions.  In those classes we look at a variety of subjects, but focus on methods of problem solving.

Really both types of courses are focused mainly on problem solving, though we hit more curriculum in the intro courses just to make sure the tools for problem solving are well understood.

If you are interested in how the classes are run, check out some of our Math Jams (or even the transcripts).  They are only one hour long, but they are similar to the way we run the rest of our courses.

I'm not sure what you mean by 'Are the topics introduced daily or weekly?', but I'll try to answer anyway.  The pace of our classes is significantly faster than the normal classroom.  We assume the students will work out the many examples we provide either in problem sets or the message board, so don't spend time doing the same type of problem over and over and over.  There is a little reinforcing repetition, but not nearly to the extent you'll find in a classroom with less avid students than typically are in our classes.


What would this limit evaluate to?  How would one do it?

gm(a, b) = limit((product(a + (b-a)* k / n, k = 0..n))^(1/n), n = infinity)

This is kind of a "geometric mean" of an interval, so my first guess would be sqrt(ab), but upon testing special cases I discovered this is not the case, as gm(1,2) = 4/e, or at least Maple says so.

Further trials gives:
gm(1,n) = n^(n/(n-1))/e
gm(a,b) seems to be: (b^b/a^a)^(1/(b-a))/e.
This is VERY VERY cool.

How do I prove it?

There is a proof using integral calculus.  Have you learned that subject yet?

As Ravi B said this is easily solved with calculus :

[hide] :Sigma: log (a+(b-a)k/n), k=1..n -> int (log(x),x=a..b) = b*log(b)-b-a*log(a)+a

This is a Riemann sum.

So limit is  : b^b/a^a * e^a/e^b[/hide]

Ah, that makes sense, I should have tried just taking a log and using the def. of the riemann integral :-P.

But something must have gone wrong there, because clearly the limit is less than b and greater than a, since (a+(b-a)*k/n) > a, and replacing the product with product(a,k=1..n)^(1/n) = a, gives us a as the limit.  We can also do the same with an upper bound, giving b>(a+(b-a)*k/n), so product(b,k=1..n)^(1/n)=b gives us an upper bound of b.  But b^b/a^a * e^(a-b) is not always less than b, for example with b = 7, a = 2, we have about 1387.

Ah, you used the definition wrong:

limit((product(a + (b-a)* k / n, k = 0..n))^(1/n), n = infinity)=
e^limit((1/(b-a))*((b-a)/n)*sum(log(a+(b-a)*k/n), k=0..n),n=infinity)=
e^((1/(b-a))integral(log(x),x=a..b))=e^((blogb-b-aloga+a)/(b-a)=
(b^b/a^a)^(1/(b-a))/e.


calculus
integration
trigonometry
calculus computations

Evaluate $ \int_0^{\frac {\pi}{2}} \sin |2x \minus{} a|\ dx$  for a real number $ a\in{[0,\ \pi]}$.

We note that
\[ |2x\minus{}a|\equal{}\left\{ \begin{array}{ll} a\minus{}2x & \quad \mathrm{for}\  0\leq x < \dfrac{a}{2}\\ 2x\minus{}a & \quad \mathrm{for}\ \dfrac{a}{2}\leq x \leq \dfrac{\pi}{2} \end{array}\right.
\]
Hence,
\[ \int_0^{\pi/2} \sin|2x\minus{}a|\ \mathrm{d}x \equal{}  \int_0^{a/2} \sin(a\minus{}2x)\ \mathrm{d}x \plus{} \int_{a/2}^{\pi/2} \sin(2x\minus{}a)\ \mathrm{d}x
\]
\[ \Rightarrow\ \int_0^{\pi/2} \sin|2x\minus{}a|\ \mathrm{d}x \equal{} \left|\minus{}\dfrac{\cos(a\minus{}2x)}{\minus{}2}\right|_0^{a/2} \plus{} \left|\minus{}\dfrac{\cos(2x\minus{}a)}{2}\right|_{a/2}^{\pi/2}
\]
Evaluating,
\[ \Rightarrow\ \int_0^{\pi/2} \sin|2x\minus{}a|\ \mathrm{d}x \equal{} \dfrac{1}{2}\minus{}\dfrac{\cos a}{2}\plus{}\dfrac{\cos a}{2} \plus{} \dfrac{1}{2}\equal{}1
\]


If a bird flies for 4 seconds at $|t-2|$ meters/sec, what is the distance it travels?

[hide="answer"]
$t=4$,
So $|4-2|=2$
So speed*time=distance
so,
Answer=$8$
I dont understand how this can be wrong :?  :(
[/hide]

what if he meant by the bird travels $|t-2|$ each seond

so after one second he is going at 1 m/s

after 2 s. he is going at 0 :D

and so on

im not sure

that is what is confusing in the back of the book the answer is $4$ when it is supposed to be $8$,
YOUR DOUBT he means what you mean, well that is how I understood it.

Essentially, you're looking for the area under the curve.  That speed is a function of $t$, so you can't just use $t = 4$ and multiply it by 4.  If you graph it, you'll see that you're basically looking for the area of two triangles.  They're identical:

$A = 2 \cdot \frac{2 \cdot 2}{2} = 4$

Total distance traveled is rate (speed) times time.  He traveled 4 meters.

Thats what i told riddler to do,plot a vt graph...told ya riddler :P

But dakryu,while drawin the graph,why do we bring the velocity down to 0m/s in the first triangle of the graph? :?

I was thinking what riddler was thinking, but I think dakyru was right, splitting it into two triangles (it took me a while to know what the t-2 was about).

[quote="dakyru"]That's just how the velocity is defined.  Let's say it's the speed of a runner.  His speed at $t = 0$ is 2 m/s and decreases linearly and at the instant it reaches 0, it begins to increase in a linear manner.  $|t - 2|$ describes the runner's instantaneous velocity.  Relationships between acceleration, velocity, and distance are explored in more detail when you take calculus in highschool/college.[/quote]#
uuuurrrrghhhh :wallbash:  :wallbash_red:  :wallbash:  :wallbash_red:
Why didnt I think of that!

[quote="IntrepidMath"][hide]
Hmmmmmm......
|4-2|=2

2*4=$8$
[/hide][/quote]
If you look back at the previous posts you will understand what we are talking about.

[quote="IntrepidMath"][hide]
Hmmmmmm......
|4-2|=2

2*4=$8$
[/hide][/quote]

Common mistake most of us would do :lol: (even i did the same)


geometry
combinatorics proposed
combinatorics

Prove that finite number of squares with sum of areas equal to $4$ can cover a square of with area $1$.


[quote="ll931110"]I don't know this below problem is actually in algebra, combinatoric or inequalities box, but I think it is  an algebra problem.

Problem:
Let $ u_0$ is a real number and the sequence $ u_n$, $ n \equal{} 1,2,...$ is defined by this following formula:
$ u_{n \plus{} 1} \equal{} u_n \minus{} 1$ or $ u_{n \plus{} 1} \equal{} \minus{} u_n$ for all $ n \equal{} 0,1,...$

Show that: If $ u_0 \equal{} u_{2009}$ then $ \minus{} 1004 \le u_0 \le 1004$./.[/quote]

Suppose $ u_k\neq 0$ $ \forall k\in[0,2009]$.
As a consequence, only 3 different steps can exist in the sequence :
Step S1: $ u_{n\plus{}1}\equal{}\minus{}u_n$ and so $ |u_{n\plus{}1}|\equal{}|u_n|$
Step S2: $ u_n>1$ and $ u_{n\plus{}1}\equal{}u_n\minus{}1$ and so $ |u_{n\plus{}1}|\equal{}|u_n|\minus{}1$
Step S3: $ u_n<0$ and $ u_{n\plus{}1}\equal{}u_n\minus{}1$ and so $ |u_{n\plus{}1}|\equal{}|u_n|\plus{}1$

If $ u_0\equal{}u_{2009}$, $ |u_0|\equal{}|u_{2009}|$ and so the number of steps S2 and S3 need to be equal
If $ u_0\equal{}u_{2009}$, the signs of $ u_0$ and $ u_{2009}$ are the same and so the number of steps S1 need to be even.

And so number(S1)+number(S2)+number(S3) is even, which is impossible since number(S1)+number(S2)+number(S3)=$ 2009$

So $ \exists k$ such that $ u_k\equal{}0$ and we need at least $ |u_0|$ steps to decrease $ |u_n|$ from $ |u_0|$ to $ 0$ and $ |u_0|$ steps to increase $ |u_n|$ from $ 0$ to $ |u_{2009}|\equal{}|u_0|$, so at least $ 2|u_0|$ steps in the $ 2009$ and so $ 2009\geq 2|u_0|$

Hence the result.

[quote="pco"]
Suppose $ u_k\neq 0$ $ \forall k\in[0,2009]$.
As a consequence, only 3 different steps can exist in the sequence :
Step S1: $ u_{n \plus{} 1} \equal{} \minus{} u_n$ and so $ |u_{n \plus{} 1}| \equal{} |u_n|$
Step S2: $ u_n > 1$ and $ u_{n \plus{} 1} \equal{} u_n \minus{} 1$ and so $ |u_{n \plus{} 1}| \equal{} |u_n| \minus{} 1$
Step S3: $ u_n < 0$ and $ u_{n \plus{} 1} \equal{} u_n \minus{} 1$ and so $ |u_{n \plus{} 1}| \equal{} |u_n| \plus{} 1$

If $ u_0 \equal{} u_{2009}$, $ |u_0| \equal{} |u_{2009}|$ and so the number of steps S2 and S3 need to be equal
If $ u_0 \equal{} u_{2009}$, the signs of $ u_0$ and $ u_{2009}$ are the same and so the number of steps S1 need to be even.

And so number(S1)+number(S2)+number(S3) is even, which is impossible since number(S1)+number(S2)+number(S3)=$ 2009$

So $ \exists k$ such that $ u_k \equal{} 0$ and we need at least $ |u_0|$ steps to decrease $ |u_n|$ from $ |u_0|$ to $ 0$ and $ |u_0|$ steps to increase $ |u_n|$ from $ 0$ to $ |u_{2009}| \equal{} |u_0|$, so at least $ 2|u_0|$ steps in the $ 2009$ and so $ 2009\geq 2|u_0|$

Hence the result.[/quote]

Great solution! Thanks for your help.  :)
(P/S: Obviously in the second case, $ u_0$ is an integer. I forgot it, so I misunderstood your solution 10 minutes ago)

[quote="ll931110"] (P/S: Obviously in the second case, $ u_0$ is an integer. I forgot it, so I misunderstood your solution 10 minutes ago)[/quote]

 :blush:  In fact, I misread the problem and considered $ u_0\in\mathbb Z$

With $ u_0\in\mathbb R$ the solution must be modified such that the first case is $ u_k$ never in $ [0,1)$ and the second case $ \exists k$ such that $ u_k\in[0,1)$


[hide="Outline"]Expand $ (x\plus{}ai)^7,(x\minus{}ai)^7$ by the binomial formula. Then establish $ (x\plus{}ai)^7\equal{}P\plus{}iQ,(x\minus{}ai)^7\equal{}P\minus{}iQ$ for some expressions $ P,Q$.[/hide]


inequalities
Cauchy Inequality
inequalities proposed

[quote="evarist"][color=purple] Let a,b,c are positive numbers.Prove that[/color]
  $\sqrt{\frac{5(a+b)(b+c)(c+a)}{8abc}-\frac{1}{2}}\geq\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}\geq2\sqrt{\frac{abc}{(a+b)(b+c)(c+a)}[\frac{5}{4}-\frac{2abc}{(a+b)(b+c)(c+a)}]+1}$[/quote]
For the left part:
\[\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}= \sqrt{\frac{a(c+a)}{(b+c)(c+a)}}+\sqrt{\frac{b(a+b)}{(c+a)(a+b)}}+\sqrt{\frac{c(b+c)}{(a+b)(b+c)}}\]
\[\le \sqrt{a(c+a)+b(a+b)+c(b+c)}\sqrt{ \frac{2(a+b+c)}{(a+b)(b+c)(c+a)}}\]
Thus it is sufficient to prove:
\[\frac{2(a^{2}+b^{2}+c^{2}+ab+bc+ca)(a+b+c)}{(a+b)(b+c)(c+a)}\le \frac{5(a+b)(b+c)(c+a)}{8abc}-\frac{1}{2}\]
\[16(a^{2}+b^{2}+c^{2}+ab+bc+ca)(a+b+c)abc \le 5(a+b)^{2}(b+c)^{2}(c+a)^{2}-4abc(a+b)(b+c)(c+a)\]
Can the last inequality proved by expansion?

[quote="evarist"][color=purple] Let a,b,c are positive numbers.Prove that[/color]
  $\sqrt{\frac{5(a+b)(b+c)(c+a)}{8abc}-\frac{1}{2}}\geq\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}\geq2\sqrt{\frac{abc}{(a+b)(b+c)(c+a)}[\frac{5}{4}-\frac{2abc}{(a+b)(b+c)(c+a)}]+1}$[/quote]
part a) is easy, so i'll post the solution of b) (if you want, i'll post my solution for a))
The inequality is equivalent to
\[\sum\sqrt{a(a+b)(a+c)}\ge 2\sqrt{abc\left(\frac{5}{4}-\frac{2abc}{(a+b)(b+c)(c+a)}\right)+(a+b)(b+c)(c+a)}\]
After squaring this inequality, we have to prove
\[\sum{a^{3}}+2\sum{(a+b)\sqrt{ab(a+c)(b+c)}+\frac{8a^{2}b^{2}c^{2}}{(a+b)(b+c)(c+a)}\ge 3\sum{ab(a+b)}+10abc }\]
Using Cauchy Inequality, we have $\sum{(a+b)\sqrt{ab(a+c)(b+c)}\ge \sum{(a+b)\sqrt{ab}(\sqrt{ab}+c)}}$

Hence, it suffices to prove that
\[\sum{a^{3}}+2\sqrt{abc}\sum{\sqrt{c}(a+b)}+\frac{8a^{2}b^{2}c^{2}}{(a+b)(b+c)(c+a)}\ge \sum{ab(a+b)}+10abc \]
From Schur Inequality, we have
\[\sum a^{3}+3abc \ge \sum{ab(a+b)}\]
So, we only need to prove
\[2\sqrt{abc}\sum{\sqrt{c}(a+b)}+\frac{8a^{2}b^{2}c^{2}}{(a+b)(b+c)(c+a)}\ge 13abc \]
But this follows from AM - GM. :)

[color=puprle]Could you post your stronger ineq  Can? My solution for this problem is very simple.I can " lam` manh no' " but it'sn't simple.Could you post a simple solution? Thanks.[/color]

[color=purple]Can, your solution for the above Ineq is true but it doesn't show the essence of problem.Two following problems are more hard:
  + With condition: $a+b+c\geq max{\frac{ab}{c},\frac{bc}{a},\frac{ab}{c}}$  $\sum\sqrt{\frac{a}{b+c}}\geq\sqrt{\frac{\prod(a+b)}{4abc}+\frac{18abc}{\prod(a+b)}+1}$
  +No condition
     $\sum\sqrt{\frac{a}{b+c}}\geq \sqrt{\frac{\prod(a+b)}{8abc}+\frac{28abc}{\prod(a+b)}}$
  Have you got any thing stronger and more beautiful [/color]


[quote="frt"][hide]
If we can prove that $ 6n^{5}+15n^{4}+10n^{3}-n$ is a multiple of 30 for all non-negative numbers $ n$, we are done.

...and there are a couple of different ways to prove that, I believe.[/hide][/quote]
[hide]
For example in this way:
$ 6n^{5}+15n^{4}+10n^{3}-n=n(n+1)(2n+1)(3n^{2}+3n-1)$
ofc $ n(n+1)(2n+1)$ is divisible by $ 6$, becouse for example $ \sum_{k=1}^{n}k^{2}=\frac{n(n+1)(2n+1)}{6}$, so now we must prove that $ n(n+1)(2n+1)(3n^{2}+3n-1)$ is divisible by $ 5$. We can do it for example in this way, let $ n=5k+r$, where $ r=0,1,2,3,4$. We put it into (*) and we show that for all $ n$ (*) is divisible by $ 5$.[/hide]

[hide="Or..."] Verify by induction that the above expression is equal to $ \sum_{k=1}^{n}k^{4}$. [/hide]


The dimensions of a peace of ice are 2x4x25 cm . This ice is swiming in the water . If we press the ice with a force 600 N ,  does the ice dive ?

[hide="Discussion"] [b]Archimede's Principle[/b] states, [i]An object immersed in a fluid will be buoyed up by a force equal to the weight of the fluid it displaces.[/i]

With that in mind, does the application of 600 N cause the ice cube to displace any more water than it displaces already?

If yes, then how much additional water is displaced? More than $ 6 \times 10^4\text{  }cm^3$?

If no, then is the $ \text{maximum buoyant force} > 600\text{ N}$? [/hide]


topology
function
advanced fields
advanced fields unsolved

If N={0,1,2,...} is in the cofinite topology (U open iff N \ U is finite), what is its fundamental group? Actually, I'm not even sure N is path connected in this topology...

It's not path connected; each path component is a single element.

The proof uses Baire category. Suppose we have a continuous function $f$ from $[0,1]$ to this space. Then the preimage of every closed set is closed. Consider the sets $S_{n}=f^{-1}(n)$. These sets are closed, since single points are closed in the cofinite topology. They are also disjoint and exhaust $[0,1]$; this means that one must be all of $[0,1]$ by the BCT.


From what little I know about modeling, I'm very interested in it. I like how applicable it is to the real world. One day I'll hopefully look into this field seriously as a career option. As for now to serve my curiosity, I have a few questions to ask to the AoPS community:

Is this a major in college or is it something you specialize in grad school?

What maths/preperation do you need to know for a succesful career in mathematical modeling?

What are the jobs one can work after a degree in modeling? How difficult is it to get these jobs in industry and academia? What other fields does it interact with and how?

What does this field truly entail in regards to what it is, how one goes about doing it, and the kind of lifestyle centered around it?

In addition, is it really as cool as I think it is? Can you solve a multitude of diverse, real-world problems with modeling?

Thanx

Wow, I think it would take 3 chapters of a book to answer these thoroughly.

Simply put, many if not most uses of mathematics in industry involve some aspect of mathematical modeling.  Typically what you'll do is take your modeling experience, tie it to some other interest like finance or biology or engineering, and there you have a career.  As for what to study in school, learn to program, study applied math, and take classes in whatever other areas you might wish to apply math to.

In biology, there is a lot of modeling going on, ie. cell interactions, protein folding and much much more. Try looking in to that.

after wandering around a bit here, i had to register to comment in this thread... modeling is a blast (and my current/future career, so I may be biased...  :oops: )

To me, modeling is simply experimentation with limiting physics.  Think of it this way- physical experiments use limited geometry, or simpler materials, or smaller/bigger scales to learn about a physical situation.  Generally, physical experiments can't change the physics though, since they occur in the same world.  Numerical experiments using models, however, can be designed to cut out certain physics, making things simpler in a different way, and allowing us to learn how adding certain new bits of physics changes the result, until we have a full simulation that accurately matches the real-world.

So to me, modeling is just a mindset that says we can learn the most when we start simple and build our way up to realistic.

As to wave hands at a few of your questions (three chapters might even be too little to do this well...):

- Is this a major in college or is it something you specialize in grad school?

It's generally not a major, though some Applied Mathematics and/or Mechanical Engineering departments may specialize in modeling.  I got into modeling after a very pure degree in Mathematics, however, so it's not necessary to have a "modeling" undergraduate degree.  Generally modeling takes two components- knowledge of the mathematics and solving the model, and knowledge of the physical situation you're trying to model.  Therefore, it takes a pretty broad knowledge base, and good modelers can easily pick up new problems and new physics.  Breadth is a good thing!

There are really two types of modelers- those coming at it from the science, learning modeling as a technique to answer their questions, and those coming at it from the mathematics, using tools to answer many different problems.  (I'm the latter... I like the ability to jump from field to field and constantly learn new things.)   Most of the latter type pick up the modeling techniques in graduate school- those without a graduate degree will often work with models, but most of the people designing models in academia and industry have advanced degrees.

- What maths/preperation do you need to know for a succesful career in mathematical modeling?

I guess I somewhat got into this already, but a wide breadth of knowledge is helpful.  On the math side, a strong background in Linear Algebra and computational methods for linear/nonlinear systems of equations are necessary (generally advanced undergrad LA and graduate Computational classes).  The rest depends on what exactly you're modeling- biological models often are discrete, meaning more linear algebra and some computer science (networks and the like).  Many industry applications such as fluid dynamics and solid mechanics depend upon continuous things, so ordinary and partial differential equations are extremely important.  And some models are stochastic or probabalistic, meaning statistics and probability.

On top of that, a working knowledge of general physics and more in-depth knowledge of your specific field of physics are necessary- you need to be able to at least talk intelligently with experts on the science side, picking their brains for intuition from observation and physical experiments.

- What are the jobs one can work after a degree in modeling? How difficult is it to get these jobs in industry and academia? What other fields does it interact with and how?

I can't answer this too well, but there are many jobs available for people who can look at a physical problem, break it down into components, understand those components individually, then combine them back to get intuition about the full system.  It's an extremely valuable skill.

- What does this field truly entail in regards to what it is, how one goes about doing it, and the kind of lifestyle centered around it?

I hope I've at least hinted at answers to this question in my other responses.  Modeling is as much an approach to problem solving as a field to work in.  Pretty much anyone- mathematicians and non-mathematicians- can benefit from using the approach.  Like any research, it's extremely exciting at times (when things are working and your models are giving answers that make sense) and extremely frustrating at times (when you're spending days debugging your most recent model or the model is spitting out nonsense data that you don't at all understand).

- In addition, is it really as cool as I think it is?  Can you solve a multitude of diverse, real-world problems with modeling?

The easy question- Most certainly! :D

As a pure mathematician (who did at least once teach a mathematical modeling course), I'm an outsider to this, but just to share a thought or two:

In school mathematics, students become accustomed to the notion that a mathematical problem has a right answer. In a true mathematical modeling problem, there's no such thing as a right answer; there are only partial answers, and improvements on the previous answers, and improvements on the previously improved answers. None of these answers are ever perfect.

The models in a beginning physics class or an elementary calculus or differential equations class are there to illustrate a principle. They're usually "first step" models in which you know you've ignored factors (friction, for instance) which would potentially have a fairly large influence on the actual behavior. In the first round of improvements, you would have to chase down the big things that you know you've been ignoring.

There can be, and often needs to be, and interplay between modeling and experimentation. How well does the model match what you see experimentally? Can you explain the differences?

When do you stop improving on an answer to a modeling problem? The end of the line is often signalled by some external factor: your report must be submitted by a certain deadline; the time has come to make the decision that this model was meant to inform; you personally have to move to a different project; there's only so much that you can do with the computing resources available.

[quote]There can be, and often needs to be, and interplay between modeling and experimentation.[/quote]

I would say that there [i]must[/i] be experimentation...if you want to know whether your model actually describes reality.

I'm paranoid of pseudoscience, and I sometimes fear that untested mathematical models could be one way for pseudoscience to creep into science.


Given $ a, b, c \ge 0$ satisfy $ a\plus{}2b\plus{}3c\equal{}4$. Prove that: $ (2a^2\plus{}bc)(2b^2\plus{}ca)(2c^2\plus{}ab) \le 32$
When does equality occurs? :)

With this ineq, equality occurs when $ (a,b,c) \equal{} (2, 1, 0)$.
Can_hang and me had 2 proofs by only Am-Gm for it. We'll post it soon.
Now, we'll wait for another proofs  :)

[quote="nguoivn"]Given $ a, b, c \ge 0$ satisfy $ a \plus{} 2b \plus{} 3c \equal{} 4$. Prove that: $ (2a^2 \plus{} bc)(2b^2 \plus{} ca)(2c^2 \plus{} ab) \le 32$
When does equality occurs? :)[/quote]
Below is two proofs from both of us:

[i][b]First proof. [/b][/i]Using the AM-GM Inequality, we have
\[ \begin{aligned} \dfrac{3}{2}[(a \plus{} 2b)(b \plus{} 2c)(c \plus{} 2a) \minus{} 9abc] & \equal{} \sum (a \plus{} 2b)(2c^2 \plus{} ab) \\
& \ge 3\sqrt [3]{\prod [(a \plus{} 2b)(2c^2 \plus{} ab)]}. \end{aligned}\]
Since $ (a \plus{} 2b)(b \plus{} 2c)(c \plus{} 2a) \minus{} 9abc \le (a \plus{} 2b)(b \plus{} 2c)(c \plus{} 2a),$ it follows that
\[ \dfrac{3}{2}(a \plus{} 2b)(b \plus{} 2c)(c \plus{} 2a) \ge 3\sqrt [3]{\prod[(a \plus{} 2b)(2c^2 \plus{} ab)]},\]
or
\[ \frac {(a \plus{} 2b)^2(b \plus{} 2c)^2(c \plus{} 2a)^2}{8} \ge (2a^2 \plus{} bc)(2b^2 \plus{} ca)(2c^2 \plus{} ab).\]
So, it suffices to prove that
\[ (a \plus{} 2b)(b \plus{} 2c)(c \plus{} 2a) \le 16.\]
By the AM-GM Inequality, we have
\[ \begin{aligned} (a \plus{} 2b)(b \plus{} 2c)(c \plus{} 2a) & \equal{} \frac {1}{4}\cdot (a \plus{} 2b)\cdot (4b \plus{} 8c)\cdot (c \plus{} 2a) \\
& \le \frac {1}{4}\left[\frac {(a \plus{} 2b) \plus{} (4b \plus{} 8c) \plus{} (c \plus{} 2a)}{3}\right]^3 \equal{} 16.\end{aligned}\]
Equality holds if and only if $ a \equal{} 2,$ $ b \equal{} 1$ and $ c \equal{} 0.$

[i][b]Second proof. [/b][/i]Notice that it suffices to prove the initial inequality for $ a \ge b.$ In this case, we have
\[ 2a^2 \plus{} bc \le 2a^2 \plus{} ac \le 2(a \plus{} c)^2,\]
and
\[ \begin{aligned}
(2b^2\plus{}ca)(2c^2\plus{}ab) &\le \frac{(2b^2\plus{}2c^2\plus{}ab\plus{}ac)^2}{4} \le \frac{[2(b\plus{}c)^2\plus{}a(b\plus{}c)]^2}{4}\\
&\equal{}\frac{(b\plus{}c)^2(a\plus{}2b\plus{}2c)^2}{4} \le \frac{(b\plus{}c)^2(a\plus{}2b\plus{}3c)^2}{4}\\ &\equal{}4(b\plus{}c)^2.
\end{aligned}\]
Therefore,
\[ (2a^2 \plus{} bc)(2b^2 \plus{} ca)(2c^2 \plus{} ab) \le 8(a \plus{} c)^2(b \plus{} c)^2.\]
On the other hand, by the AM-GM Inequality, we have
\[ (a \plus{} c)(b \plus{} c) \le \frac {[(a \plus{} c) \plus{} 2(b \plus{} c)]^2}{8} \equal{} \frac {(a \plus{} 2b \plus{} 3c)^2}{8} \equal{} 2.\]
Combining this with the above inequality, we get
\[ (2a^2 \plus{} bc)(2b^2 \plus{} ca)(2c^2 \plus{} ab) \le 8\cdot 2^2 \equal{} 32.\]


I want to print the page number of (say) the section in the header and the page number of the (say) article in the footer.  I am usiing fancyhdr package (irrelevant).

I have created in my .sty file a new counter to keep track of the book page number at the beginning of each section, and reset the page counter to 1 at the beginning of each section.

My header works fine, my footer doesnt.

\rfoot{page {\value{pagetally} + \value{page}}

does not work, nor anything else I've tried, such as:

\rfoot{page \number{\value{pagetally} + \value{page}}

or

\rfoot{page \numexpr{\value{pagetally}+ \value{page}}

(using \alphalph package - error message says these commands not available in horizontal mode)

Does anybody know how to do this?

Peter

\value is used as for inputting the value of the counter to another command like \addto counter. Try using \thepagetally and \thepage to output the number.

That simply creates the output  "page 18 + 3"

or whatever the counters happen to be, whereas what I want is "page 21" to be in the footer.  In order to add the values 18 and 3 together, it is necessary to turn them into values, what I need is a way to turn the number generated back into strings.

Peter

The [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=calc]calc package[/url] will allow you to do exactly that.


Θα ξεκινήσω με ένα ερώτημα του πενταλέπτου...

Ένας γρήγορος ορισμός πρώτα: θα λέμε ότι μία συνάρτηση ανήκει στο χώρο $ L^p[0,1]$ (όπου $ p > 0$) όταν $ \int_0^1 |f(x)|^p dx < \infty$. Π.χ. αν πάρουμε τη συνάρτηση $ f(x) \equal{} \frac {1}{x}$, τότε  $ f\in L^p[0,1]$ για όλα τα $ p < 1$ και $ f\notin L^p[0,1]$ οποτεδήποτε $ p\geq 1$.

Συναρτήσεις με παρόμοια συμπεριφορά υπάρχουν αρκετές, αλλά τί γίνεται αν θέλαμε να "κλείσουμε" το πάνω όριο των $ p$? Με άλλα λόγια, μπορούμε να φτιάξουμε μία συνάρτηση που να ανήκει στον $ L^p$ μόνο όταν $ p\leq 1$, δηλαδή τέτοια ώστε $ \int_0^1 |f|^p < \infty$ μόνο για $ p\leq 1$?

Cheerio,

Durandal 1707

PS: Please, όσοι το έχετε ξαναδεί, μην το χαλάσετε για όσους ίσως δεν το έχουν ξαναδεί. Αν δεν υπάρξει ενδιαφέρον από κανέναν, θα βάλω τη λύση σε μερικές μέρες...

Heh, και νόμιζα ότι είχε πεθάνει το thread. Τί ακριβώς έχεις στο μυαλό σου όμως?

Π.χ. $ \int_0^1 |\log x|^p dx \equal{} \Gamma (p \plus{} 1)$ και αυτό συγκλίνει για όλα τα $ p > \minus{} 1$ καθώς και σε όλα τα [i]ανοικτά[/i] διαστήματα της μορφής $ ( \minus{} n \minus{} 1, \minus{} n)$ :huh:

Edit: Επίσης, αφού για μεγάλα $ n$ έχουμε τελικά $ n>\log^p n$ δεν αποκλίνει αυτή η σειρά για κάθε $ p$?

Cheerio,

Durandal 1707

Γράψε λάθος ...Έδωσα παράδειγμα που να ανήκει στον $ c_0$ (δλδ των ακολουθιών που να συγκλίνουν στο 0) αλλα να μην ανήκει σε κανέναν $ l_p$ για $ p \leq 1$(για σειρές πάντα)

Meeeeh, δε βλέπω να υπάρχει πια ενδιαφέρον, οπότε here goes nothing:

Θεωρείστε τη συνάρτηση $ f_n(x) \equal{} \frac {2^n}{n^2}\mathbf{1}_{[1/2^n,2/2^n]}$, δηλαδή τη συνάρτηση που είναι 0 παντού εκτός από το διάστημα $ [\frac {1}{2^n},\frac {2}{2^n}]$ όπου είναι ίση με $ \frac {2^n}{n^2}$.

Τώρα πάρτε τη συνάρτηση $ f(x) \equal{} \sum_{n \equal{} 1}^\infty f_n(x)$ που είναι απλώς μια σκάλα με σκαλοπάτια ύψους $ 2^n/n^2$ και πλάτους $ 1/2^n$.

Το ολοκλήρωμα της $ f$ είναι απλώς $ \int_0^1 f(x) dx \equal{} \sum_{n \equal{} 1}^\infty \frac {2^n}{n^2}2^{ \minus{} n} \equal{} \sum \frac {1}{n^2} \equal{} \frac {\pi^2}{6} < \infty$.

Αντίθετα, για κάθε $ p > 1$ θα έχουμε: $ \int_0^1 f^p(x) dx \equal{} \sum_{n \equal{} 1}^\infty \frac {2^{n(p \minus{} 1)}}{n^2} \equal{} \infty$, δηλαδή $ f\notin\L^p$.

Πρέπει να γίνεται και με συναρτήσεις της μορφής $ x^a \log^b x$ αλλά δεν έχω κάτσει να δω τί σταθερές χρειάζονται αυτές...

Cheerio,

Durandal 1707


We know that $ f(ab)=af(b)+bf(a)$ and $ |f(x)|\leq 1$. Prove that $ f(x)=0$

Maybe you can use this: if $ \lim_{x\to \infty}g(x)=0$ and $ |f(x)|\leq M(constant)$, then we have $ \lim_{x\to \infty}(f(x)g(x))=0$

[quote="Twins"][quote="aviateurpilot"]$ \ |g(x)|=\frac{|f(x^{n})|}{n|x^{n}|}\le \frac{1}{n|x^{n}|}$
then  $ \forall x\in R^{*},\ g(x)=0$[/quote]
How do you get $ \forall x\in R^{*},\ g(x)=0$?[/quote]
sorry
$ \forall n\in N^{*},\ |g(x)|\le \frac{1}{n|x^{n}|}$
so $ |g(x)|=\lim_{n\to+\infty}|g(x)|\le \lim_{n\to+\infty}\frac{1}{n|x^{n}|}$
if $ |x|\ge1$ $ \lim_{n\to+\infty}\frac{1}{n|x^{n}|}=0$
then $ \forall x\in R-]-1,1[,\ g(x)=0$
and $ \forall x\in ]-1,1[-\{0\},\ g(x)=g(x\times \frac{1}{x})-g(\frac{1}{x})=0$

finally $ \forall x\in R^{*},\ g(x)=0$

[quote="aviateurpilot"]so $ |g(x)|=\lim_{n\to+\infty}|g(x)|$[/quote]
I'm sorry, still one question left.  Can you explain why $ |g(x)|=\lim_{n\to+\infty}|g(x)|$? I know little about functions, especially it has limit :blush:  Thank you!

If $ u_{n}=C(\text{constant})\forall n\geq 1$ then $ \lim_{n\to\infty}u_{n}=C$. You need a text book on Real analysis!  :wink:


[color=darkred]Consider the equation $ \sqrt [n]{x+1}-\sqrt [n] x=\frac{1}{n}$, where $ x\in [0,1]$ and $ n\in N\ ,\ n\ge 2$.
Prove that this equation has only one zero $ x_{n}\in (0,1)$ and $ \lim_{n\to\infty }x_{n}=\frac{1}{e-1}$, where $ e$ is the [b]Neper's number.[/b][/color]


Emily selects a number at random from the set $ \{1, 2, 3, 4, 5\}$ ten times (independently). The probability that she will choose an odd number at most three times can be written in the form $ \frac{2^n\cdot c}{5^m}$, where $ m$ and $ n$ are positive integers and $ c$ is a positive odd integer not divisible by $ 5$. Find the value of $ n\plus{}m\plus{}c$.

[hide="Solution"]
[list]
[*]There are $ 2^{10}$ strings with no odd numbers.
[*]With one odd number, there are $ {10 \choose 1}$ places to put the odd number, $ 3$ values for the number, and $ 2^9$ possibilities for the rest of the string, giving $ 3\times {10 \choose 1} \times 2^9 \equal{} 2^{10} \cdot 3 \cdot 5$.
[*]With two odd numbers, there are $ {10 \choose 2}$ places to put the odd number, $ 3^2$ values for the number, and $ 2^8$ possibilities for the rest of the string, giving $ 3\times {10 \choose 2} \times 2^8 \equal{} 2^{8} \cdot 3^4 \cdot 5$.
[*]With three odd numbers, there are $ {10 \choose 3}$ places to put the odd number, $ 3^3$ values for the number, and $ 2^7$ possibilities for the rest of the string, giving $ 3\times {10 \choose 3} \times 2^7 \equal{} 2^{10} \cdot 3^4 \cdot 5$. [/list]

Thus, the total probability is $ \frac {2^{10} \plus{} 2^{10} \cdot 3 \cdot 5 \plus{} 2^{8} \cdot 3^4 \cdot 5 \plus{} 2^{10} \cdot 3^4 \cdot 5}{5^{10}} \equal{} \frac {2^{8} \cdot 2089}{5^{10}}$, and $ n \plus{} m \plus{} c \equal{} 2107$.
[/hide]

hmm thats not exactly what i got, but i could be wrong!


linear algebra
matrix
vector
linear algebra unsolved

Let $ A$ be a $ m\times n$ matrix and $ B$ a $ n\times k$ matrix.
 Prove that if the columns of $ A$ are linearly independent then $ AB \equal{} 0$ implies $ B \equal{} 0$. Would this work for $ B$ having linearly independent columns($ AB \equal{} 0$ implies $ A \equal{} 0$). Would this work for linearly independent rows?

Reversing the order i like  transpose, so the counterpart involves $ B$ having independent rows. Imagine the case with $ B$ a single column ($ k\equal{}1$) to see why it's not columns.

jmerry ,are you saying that my implication works only if the columns of $ A$ are linearly independent or the rows of $ B$ are linearly independent and not the other ways around? I still don't know how to prove this though.

I think I got it. You just multiply with  $ ^{t}A$ to the right in $ AB\equal{}0$ and since $ ^{t}AA$ is invertible (not very hard to prove) you get $ B\equal{}0$.

If the columns of A are linearly independent then the homogeneous system $ Ax \equal{} 0$ admits only the trivial solution. But each of the columns of $ AB$ is a linear combination* of the columns of A, and since (1) the columns of A are linearly independent, and (2) each column of $ AB$ is equal to a zero vector, it follows that each coefficient in * must be zero. But this means that each column of B is completely made up of zeros. In other words, $ B \equal{} 0$.


Has anyone heard of CML (Continental Math League)?  I didn't see anything about it, so I was just wondering.  For those who don't know, it is a 6 question test, that you get 30 minutes on.  You take it 5 times a year and accumulate a score out of 30.   Your school also has a "team" score, consisting of the top 6 scores of each "meet".  One reason I mention this is because I got 30/30 last year.

I've heard about this in two ways: a) my sister in third grade curretnly does it and b) when I was in 5th grade, some kids in other schools in my town did it. It sounds a lot like Math Olympiads or Math League format wise, although many say it's much easier than Math Olympiads.

Umm I don't know much else about it since I've never done it but I thought it's only an elementary school contest. Maybe I'm wrong...

Yeah I do it. Its like the main competition for my math club.

[quote="djshowdown2"][quote="chess64"]Err... it seems exactly like the math league contest I participate in every once in a while...

Is this it? http://mathleague.com/[/quote]

I don't think its the same thing because my school does both the Illinois Math League and the Continental Math League.  I think I might have done it in junior high too, but I can't remember.

In high school, we do the CML Pythagorean division as freshman which fits the format you described, and seniors take the CML Calculus division which is 8 questions, 10 points (the last two are worth 2 points each.[/quote]
I agree.  The contest specified is 30 questions.  Yes, I do Pythagorean division (being an eighth grader), and unfortunately I only got 24/30 cumulative this year.


Euler
advanced fields
advanced fields theorems

Where Can I find the full proof to Fermat's Last Theorem?

Do you mean this (link was already given at http://www.mathlinks.ro/Forum/viewtopic.php?t=43482):
http://math.stanford.edu/~lekheng/flt/wiles.pdf

Hi keta
I was just so interested in the proof of the last theorem of the fermat but when i found the article of Andrew wiles and the last move for solving the fermat thoerem ,i found that hunderd of other thoerem had beed used in mr.wiles article and as he by himself has said it too that , he has gathered 300 years of mathematicions results so i have just understand a bit more, how much

                                        [b]Math is[/b] $beautiful$.

So three cheers or hazzards for Math :thumbup:  :thumbup:  :thumbup:

The book "Invitation to the Mathematics of Fermat-Wiles" by Yves Hellegouarch
is a textbook for a university classroom.  It is written for students who know
some abstract algebra.  In perhaps a year of study it is meant to bring the
student to a point of having some understanding of the foundations of the math
used by Wiles in his proof.  It might be possible for a dedicated amateur to go
through this book and make good progress on this.  This is the best resource
that I know to be able to do this.

[quote="Guía Rojo"][quote="ZetaX"]http://math.stanford.edu/~lekheng/flt/wiles.pdf[/quote]downloading... 3%  :blush:
thank you so much, ZetaX...  :)[/quote]

How many of you can understand till page 10??? I can't even number 1

well I think I found a proof for some cases. What do you suggest I shall do?

How many cases¿ And how difficult¿
If you have a nice and as most elementary proof for some bigger cases, that would be interesting. There is e.g. that one for regular primes, but that is also not elementary (but still a lot easier than Wiles proof).

Sorry for not answering so long but I didnt get a email...
what do you mean "how difficult"?
It is not the traditional cases (n=3,4,...)
Actually I could say that I proved the 2/3 of the theorem.
I mean that I found three cases and proved the 2.
Perhaps I am wrong but I do not think so.

Sorry I cant prove that I proved what I said that I proved without writing the proof which is something that you wouldnt advise me to do, right?
Anyway it is not quite right to say that you proved half or the 2/3 of a theorem, right?
I said that I am perhaps wrong.

The point is that 1/3 of the proof, which thatin didn't make, is maybe the hardest and most important step of the proof, and so it has much greater worth than the other 2/3. So, of course "it's not quite right to say you proved the 2/3 of a theorem".

Basically the real problem is that I do not know how I shall present it: as a prove of same case of FLT or a theorem that looks like FLT (TFT thatin first theorem perhaps? :rotfl: ) and through it some cases of FLT are proved. And obviously I do not know where I shall present it. Any help?

[quote="thatin"]I would rather like a more official way, a magazin etc[/quote]

Annals of Mathematics, Nature Magazine

Neither is really suitable, given that the annals of mathematics is reserved for the very, very best/most groundbreaking papers,and nature is not a mathematics journal at all.

Frankly, it is overwhelmingly likely that either thatin is completely wrong about what he claims, or it is well known already.

[quote="blahblahblah"]or it is well known already.[/quote]
I am afraid of that too, imagine someone going and make an  announcement like "hey I invented the wheel"! Does anyone know if there is a proof for cases different from the classic n=3, n=4 etc?

There is some history of it here:
http://www-groups.dcs.st-and.ac.uk/~history/HistTopics/Fermat's_last_theorem.html

The later proofs are really complicated though so if you have a "simpler" proof of a large fraction of the cases, then it might still be worth publishing (assuming it's not known).

Is Wiles' the only known proof about Fermat?

I think so José... you mean a proof of the whole flt, right?
Another thing that troubles me is what will happen if I send my proof and see it in the magazine with somebody's else name...

Just make sure the journal you send it to has a decent reputation. No major journal would do that.

[quote="José"][quote="thatin"]Actually I could say that I proved the 2/3 of the theorem.[/quote]

Sorry, but I don't believe you[/quote]
Ha,ha,ha! I'm more ready to belive thatin's proof for 2/3 cases than
this "proof" for all cases: http://arxiv.org/PS_cache/math/pdf/0510/0510081.pdf  :D

Actually ,"I" proved it too,but just for the half  of the cases  (with n=2,4,6,..   :lol: )

the proof for n=4 was found in Fermat's notes (I think by Euler)
just wait till December (I didn't say the year :rotfl: )

[quote="jmerry"][quote="Eros88"]Actually ,"I" proved it too,but just for the half of the cases (with n=2...)[/quote]
Now that's impressive.

I've seen quite a few discussions that completely ignore the even case. They shouldn't; the proof for $n=4$ is separate from the proof at primes. It's also pretty easy, and I'm sure Fermat knew it.[/quote]
It is easy to show that for $n>2 ,n=2k$  FLT holds.
Only you have to proove it for n=4 and the conclusion for all even cases is there .
n=prime -> much more difficult part of the proof.
The Indian (in the link Jose provided) certainly made  a mistake along the lines.
It is going to be nice exercise to find flaw in that paper .

[quote="Eros88"]It is easy to show that for n>2 ,n=2k FLT holds. [/quote]
I believe you. On the other hand, you initially included $n=2$ in that; that's what I was pointing out.

you proved flt? sorry but I do not believe you. I registered just  to post this: you will never be able to unterstand the proof, not to prove it. but if you did then why you don't post it here?

Nobody in this thread claims to have a proof of the entire theorem; the only claims here are proofs that apply to some exponents, and those results are probably easy.


geometry
geometric transformation
reflection

Evolution vs. Intellegent Design/Creation Debate
Just ask questions, and answer questions.
I'll start with a question.

Where are all the links between different species(like a bird-reptile)?
This is necessary for evolution.
There are many many fossils, so why are the no links. (Really, no links at all, not even one).

There are ore than enough links. Just because some creationists are to stubborn to even accept their existence doesn't mean they don't exist. For anything more: there was already a discussion about this. To make it short: ID is no science and thus has nothing to do in science (e.g.biology) classes; the whole idea is just a stupid (but because of stupidity of peoply partially successfull) attempt to bring religous ideas into science.

[quote="ZetaX"]There are [m]ore than enough links.[/quote]
Give me one.

[quote="ZetaX"]To make it short: ID is no science and thus has nothing to do in science (e.g.biology) classes; the whole idea is just a stupid (but because of stupidity of peoply partially successfull) attempt to bring religous ideas into science.[/quote]
You could consider evolution a religion. In fact it takes much, much, much... more faith to believe evolution than it does to believe creation. (ie. First there was nothing. Then more nothing got added to it. And that gave something. This somthing exploded, and out we popped.)
Most stuff taught in schools is already discarded by evolutionists.

EDIT: This debate is to prove that evolution cannot be true.

Intelligent design is basically a "compromise" among religious fanatics who want to advance religion in public schools.  It pretty much says "Evolution COULD be possible, but whatever happened, God did it."  Again, it's still trying to infuse religion in the public school systems.

Such religious nuts are the first to claim that separation of church and state is violated if something goes against them, but they're even quicker to violate it themselves.

[quote="jtw"]You could consider evolution a religion.[/quote]
Umm, no.  Religion requires faith in something with no evidence.  Hence the term "faith."  Evolution has scientific evidence supporting it.  If you choose to disregard or disbelieve the evidence, well you're the fool.

http://talkorigins.org/faqs/faq-transitional/part1a.html
http://www.holysmoke.org/cretins/transfos.htm
http://www.fossilmuseum.net/Evolution/transitionalfossils.htm
http://www.talkorigins.org/indexcc/CC/CC200.html (quoted below)
http://www.digisys.net/users/hoppnrmt/transitionfossils.htm
http://en.wikipedia.org/wiki/List_of_transitional_fossils

[quote="[url=http://www.talkorigins.org/indexcc/CC/CC200.html]www.talkorigins.org[/url]"]Claim CC200:
There are no transitional fossils. Evolution predicts a continuum between each fossil organism and its ancestors. Instead, we see systematic gaps in the fossil record.

Response:

   1. There are many transitional fossils. The only way that the claim of their absence may be remotely justified, aside from ignoring the evidence completely, is to redefine "transitional" as referring to a fossil that is a direct ancestor of one organism and a direct descendant of another. However, direct lineages are not required; they could not be verified even if found. What a transitional fossil is, in keeping with what the theory of evolution predicts, is a fossil that shows a mosaic of features from an older and more recent organism.

   2. Transitional fossils may coexist with gaps. We do not expect to find finely detailed sequences of fossils lasting for millions of years. Nevertheless, we do find several fine gradations of fossils between species and genera, and we find many other sequences between higher taxa that are still very well filled out.

      The following are fossil transitions between species and genera:

         1. [url=http://www.talkorigins.org/indexcc/CC/CC050.html]Human ancestry[/url]. There are many fossils of human ancestors, and the differences between species are so gradual that it is not always clear where to draw the lines between them.

         2. The horns of titanotheres (extinct Cenozoic mammals) appear in progressively larger sizes, from nothing to prominence. Other head and neck features also evolved. These features are adaptations for head-on ramming analogous to sheep behavior (Stanley 1974).

         3. A gradual transitional fossil sequence connects the foraminifera Globigerinoides trilobus and Orbulina universa (Pearson et al. 1997). O. universa, the later fossil, features a spherical test surrounding a "Globigerinoides-like" shell, showing that a feature was added, not lost. The evidence is seen in all major tropical ocean basins. Several intermediate morphospecies connect the two species, as may be seen in the figure included in Lindsay (1997).

         4. The fossil record shows transitions between species of Phacops (a trilobite; Phacops rana is the Pennsylvania state fossil; Eldredge 1972; 1974; Strapple 1978).

         5. Planktonic forminifera (Malmgren et al. 1984). This is an example of punctuated gradualism. A ten-million-year foraminifera fossil record shows long periods of stasis and other periods of relatively rapid but still gradual morphologic change.

         6. Fossils of the diatom Rhizosolenia are very common (they are mined as diatomaceous earth), and they show a continuous record of almost two million years which includes a record of a speciation event (Miller 1999, 44-45).

         7. Lake Turkana mollusc species (Lewin 1981).

         8. Cenozoic marine ostracodes (Cronin 1985).

         9. The Eocene primate genus Cantius (Gingerich 1976, 1980, 1983).

        10. Scallops of the genus Chesapecten show gradual change in one "ear" of their hinge over about 13 million years. The ribs also change (Pojeta and Springer 2001; Ward and Blackwelder 1975).

        11. Gryphaea (coiled oysters) become larger and broader but thinner and flatter during the Early Jurassic (Hallam 1968).

...[/quote]

[quote="JRav"][quote="jtw"]You could consider evolution a religion.[/quote]
Umm, no.  Religion requires faith in something with no evidence.  Hence the term "faith."  Evolution has scientific evidence supporting it.  If you choose to disregard or disbelieve the evidence, well you're the fool.[/quote]

Evolution has absolutely no scientific evidence supporting it.(If you find any show it to me. That is what this topic is for.)
[quote="JRav"]Religion requires faith in something with no evidence.[/quote]
Thus, [b]evolution is a religion[/b].

I guess I shouldn't say Intelligent Design, but I just wanted to [b]disprove[/b] evolution.

JBL:
I said give me one not one hundred  :lol: I can't get to that many.
But none these are valid.
They are just imagination or exaggeration or other.
Imagination:
Archaeologists find a tooth and from it construct a man-ape.
The tooth is later found to be from a pig.
Exaggeration:
Small "legs" are found in whales, later found to help in giving birth.

Archaeopteryx is a horrible example. Ask me for proof.
But it is one of the most popular.

Maybe you should do some research on your links and come back with the "best".

It took you less than five minutes to dismiss all of archeology.  This shows about the depth of thoughtfulness one might have expected.  Question for reflection now that this thread is locked: is there any more self-defeating argument than the explanation of the religious that we should dismiss evolution because it is a religion?


A sequence $a_n\in \mathbb{Q}$  having such property

$a_na_{n+1}+a_{n+2}a_{n+3}=1$  for every non-negative integer $n$ and

$a_0=1$ , $a_1=2$ , $a_2=3$ .  Find $a_{2005}$

 :)

[quote="frt"]If we put $n=0$, we get $a_0a_1+a_1a_2=1\cdot 2+2\cdot 3=8\ne 1$... :?[/quote]

Opps I didnt notice I make a typo there  :blush:  it should be
$a_na_{n+1}+a_{n+2}a_{n+3}=1$

I sort of messed up with this. I started with listing the first 5 or 6 terms of the sequence to look for a pattern. Then, I got something weird.

Generally, $a_{n+3}=\frac{1-a_na_{n+1}}{a_{n+2}}$

$a_3=\frac{1-1\cdot 2}{3}=-\frac{1}{3}$

$a_4=\frac{1-2\cdot 3}{-\frac{1}{3}}=15$

$a_5=\frac{1-3\cdot -\frac{1}{3}}{15}=0$

$a_6=\frac{1-(-\frac{1}{3})\cdot 15}{0}=\text{Undefined}$

From here, I can't go further... Did I do something wrong?


AMC AIME Support geometry search MIT college

I think that China will be the strongest country in the world in about 50-70 years........what do you think???????

[quote="h_s_potter2002"]No, Japan will overtake the US and all of the other countries in the next couple of years.[/quote]

Not sure about Japan. But what I'm sure about is that the USA will just collapse in less than 50 years.........

What do you mean by strongest?

Best army?
Biggest army?
Most populous nation?
Richest country?
Biggest economic influence

I wonder whether or not it will remain communist (remember, no communist nations of 'the first wave' after World War I are now still communist)

I'd say there's about a 97% chance china will overtake the US within 50 years.  We Americans can't keep living in luxury for long...
5% of world population and 25% of consumption...

[quote="drunner2007"]i think that right now the u.s. is in a critical situation. unless the education system is substantially improved, and more emphasis is placed on educating engineers, scientists, etc. the u.s. has a good chance of losing technical prominence. we're already behind japan is many respects.

china will be really strong regardless of how the u.s. does. i'd say there's about a 50% chance china will be ahead of the u.s. (in terms of economy, technology) in 50 years.[/quote]

Hmm, I recently I've heard many people say that.  What exactly do you mean though.  USA has a worse educational system for the average man and woman than in Chin/Japan, or it's not that worse in USA, but there are also MORE Chinese?

My topic about US knowledge of geography in Round Table is somewhat related.  Are Americans really that poorly educated.  Or are they still better educated than in Western European countries?

this post is kind of long  :roll: i meant for it to be kind of short but it kept getting longer.

i can't definitively tell you that the education system (as a whole) is worse in the U.S. since i have not had experience in other schools around the world in recent years.

however, i did attend an english elementary school for a few years, and from my experience there i noticed that it was far more disciplined/structured, and challenging than any public schools i have been to in the u.s. (with respect to grade level). i am hispanic, and several of my cousins have moved to america from latin american countries. a few have told me that schools in the u.s. are way too easy, especially in math!

i'm not saying all u.s. schools are bad, and in fact there are some great schools, most of which tend to be in suburban areas. but many, many american schools fail in that students are not interested in learning, discipline is lacking, and teaching is sub-par. in fact, in many schools there is a notion that being interested in classes and enjoying learning is nerdy (once again, not all schools, but many).

i think that the problems stem from the curriculum that students are taught, and the methods that teachers are encouraged to use when teaching. a lot of it doesn't work. many high school curriculum's are loaded with classes that have little relevance to the real/modern world, thus losing students' attention.

so overall i would contend that u.s. schools are not the worst, but are definitely in need of improvement, especially if the u.s. is to maintain international prominence. however, reforming schools is a really difficult matter.

[quote="fredbel6"]Are Americans really that poorly educated.  Or are they still better educated than in Western European countries?[/quote]Educated americans are [i]extremely[/i] well educated. Uneducated americans are [i]extremely[/i] uneducated.

[quote="drunner2007"]this post is kind of long  :roll: i meant for it to be kind of short but it kept getting longer.

i think that the problems stem from the curriculum that students are taught, and the methods that teachers are encouraged to use when teaching. a lot of it doesn't work. many high school curriculum's are loaded with classes that have little relevance to the real/modern world, thus losing students' attention.
[/quote]

Hmm, although I haven't seen much of real analysis or ring theory in high school (well I did see matrices and integrals...) I must say I am quite happy of my high school education.
I too had to study TEDIOUS demotivating stuff, like history : sum up five reasons for French Revolution....
It is only now that I am rediscovering such topics, and see how interesting they can be ( but don't get me started about history education in itself, which is my opinion biased, incomplete,boring,....)

Still my point is : sometimes being strict in classes ( I had 33 times 50 minutes of lessons a week) is good.  Like for studying French : memorising rules and vocabulary is THE way.  They are changing that now, result : the youngsters of only five years younger do much poorer in French, and can't spell even in Dutch (but they DO write essays all the time, do plays,...)

the best part of american education is the university system. that is where the u.s. makes up for a slack public schools.

I really have a hard time estimating China, so big, so different...

Does everyone have access to Mathlinks/Wikipedia,... ever.
How many people can afford it?

Is it true that the Western world knows more about Tiananmen square than they themselves? (I heard that the post Tiananmen generation is much more positive about the government)

And what about the languages?  Is the percentage of at least bilingual people higher than in USA?

Well, as I have said in other topics, China scares the European Union, as it is much cheaper to produce there (one factory after the other closes here, and announces almost simultaneously the coming of a new one in China).

One final question (I have also considered even making a topic about this) : what does its flag stand for?
One star and four smaller ones?  Five fingers?  Five parties?  Han Chinese and the four smaller groups of Tibetans, Manchurians, Mongolian and Uyghur people?

[quote="fredbel6"]
One final question (I have also considered even making a topic about this) : what does its flag stand for?
One star and four smaller ones?  Five fingers?  Five parties?  Han Chinese and the four smaller groups of Tibetans, Manchurians, Mongolian and Uyghur people?[/quote]
"The red colour of the flag symbolizes revolution; the stars take on the yellow colour in order to bring out their brightness on the red ground. The larger star represents the CPC, while the four smaller ones, the Chinese people. The relationship between the stars means the great unity of the Chinese people under the leadership of the CPC. "
http://fotw.vexillum.com/flags/cn.html

everyone's scared about how "uh-oh, china has 3 times as many engineers as we do!"
but they fail to realise that they have 5 times as many people as we do.  it's only natural that bigger developed countries should have more engineers.

I don't know the exact figures, but cars are more expensive in China, the government adds tax, or something. It's because they just don't want too many cars in the cities. It's true, I never saw anywhere you could park a lot of cars in Shanghai.

[quote="JBL"]"Public transportation system" is correct.  Instead of "maintainment," though, "maintenance."[/quote]
Thanks.
This is partly why i think English is harder to learn. It exhausts people's memory unnecessarily

I'm a native Chinese person, I've been going to Chinese school since I was four, and FIND THE LANGUAGE !%!@#%!@#%. I like spanish so much more than Chinese...

Anyway, right now I'm in China on vacation. I've been going between Beijing and Tianjin. The place I'm living is one of the middle-areas... you've got stink and dirt all over but there are people living happily.

Anyhow, I've got mixed feelings about China. The one crapfest about China is it's disregard for individuals and its medical system. Great things about China =

Bikes, and how easy it is to get around with them.
The clean places are great
Prices

Strange Fanatic, while I don't really have a gift for Chinese, I find the grammar very easy. Do you speak Chinese at home? That helps a lot with composing things.

I'm going to China from July 10th (we will depart on the US' 8th) to August 5th, while the rest of my direct family (sister and parents) were all born in China, and my sister has been there teaching English for about a year now, I have never been there. While I study the culture, speak the language at home, go to Chinese School, read about it here, I've never actually set foot there so it'll be an interesting experience. We are planning to stay with relatives.

By the way, looking at the recent posts, this topic is waay off topic, we are starting to discuss more US affairs and not how they compare to China.

Anyway, I'm guessing most of you have seen the famous pictures of cars lined up on a 6 lane street or so at a complete standstill in Beijing or somewhere. Chinese car ownership is a lot smaller than you think it is and biking is still probably the main method of transportation. Recently, McDonalds announced plans to build drive-throughs in China. Yes, they are just considering building drive-throughs now. That shows how little individual car ownership is (very small % of the population) compared to how families have two or three cars here.

I was at a chinese supermarket yesterday, and HOLY CRAP there were like a thousand bikes there and like 20 cars... the great thing was the parking space needed was minimal.

I can't stand living in China, I just don't like traveling in general =/. I speak Chinese at home, too. I guess I'm too Americanized now :D.

Perhaps a dumb question, but if bikes are so important, what are attitudes towards bike theft.  In Belgium the law is simple : it's a jungle, get robed or don't get robbed, steal or don't steal yourself.

Maybe out of topic, but what do you think about Chineese Supermarkets here in Argentina?

fredbel6: The amount of people in China is so huge that bike theft and theft in general can not really be cracked down. The reason I find it hard to live in China is that it's a very dog eat dog world. You have to look after yourself because no one else (ecept family) will. So if something is stolen it's really tough luck and be more careful next time.

The report by our channel, and the demonstration in Brussel, was last week, THAT remains a fact... :|

I also heard about the news. But it has been proved to be a nonsense.

[quote="Buffalo"][quote="fredbel6"]I would like to add somethig new to this discussion.

Last week Flemish public broadcasting channel Canvas interviewed Belgian Falun Gong members in Ghent and Brussels who claimed that in China, Falun Gong members were executed and had their organs sold.  The buyers included foreigners according to them.  They held demonstrations where they reenacted that practice (a little show thus).

A British politican( John Miller, but I am not sure though whether it was him, I don't have a source on his name anymore) said it needed to be investigated ( he went to China himself for that) and he claimed that if only one reporter managed to prove one such case of executing Falun Gong members to sell their organs, China would not host the 2008 olympics.

[url]http://www.taipeitimes.com/News/taiwan/archives/2006/03/21/2003298432[/url] is a related article.

What is your opinion on these accusations?[/quote]

You are hopelessly outdated.

On March 27 2006, Italy Prime Minister Silvio Berlusconi proved that China cook infants and make them plant food.
Isn't this much more shocking than yours? Are you feeling excited?

It's lucky for me that China has some power to protect herself, unlike Iraq.[/quote]

Jokes

Berlusconi has made a number of controversial jokes, often (perhaps inadvertently) causing offence to different groups and individuals, and being seen as embarrassing to his office and to Italian people which he represented. The jokes could also be seen, however, as a way of showing that he has not lost touch with the common Italian, as well as providing a stark contrast to the perceived dullness of many contemporary politicians.

On the Taipei article, it is more possible that this is just an article to spread the hatred of China and make more people support Taiwanese independence.

That whole executing Falun Gong members and selling their organs has been going on for quite a while now. I first heard of it over a year ago when I was walking into Chinese School (at Yale) and Falun Gong members were had easels with posters mounted on them showing graphic images and handing out reports both in English and Chinese to all passerbys about these organ camps. Whether it is true remains to be seen. The thing is, there hasn't really been a third party investigator. Falun Gong practioneers say it's real and happens daily, the government says it doesn't. Heck, maybe it does and the government doesn't know about it because it's very isolated.

Chinese Supermarkets not in China are just probably like the locale. You will never see only bikes at a Chinese Supermarket here in America.


Suppose that three intersecting rectangles are drawn on a flat surface. What is the maximum number of completely bounded areas, not further subdivided, that can thereby be formed, considering only the sides of the rectangles as boundaries? (The figure below illustrates two intersecting rectangles.)

[img]http://www.eskimo.com/~miyaguch/hoeflin/n3.gif[/img]

[hide="The most I have found so far is..."]
\[\boxed{19}\]
[/hide]

Im getting 21. Can post method. A google search leads me here and wolfram alpha does not cover that problem.

The answer should be $27$ right?

It is easy to see that two rectangles have a maximum of $9$ bounded regions so $3$ rectangles should have a maximum of $27$ bounded regions since there are $3$ distinct pairs of rectangles.

EDIT: Wow what a revive.

but remember, the ninth region is the one in the middle
so you're basically triple-counting it
and you forgot the outside region
so it's 26

Oh I see.
The problem states that the regions are completely bounded you would not count the one on the outside.