function
limit
integration
calculus
calculus computations

f is a continuous function on [0,1]. prove that:
$ lim_{n\rightarrow \infty}\frac1n\sum_{k\equal{}1}^n (\minus{}1)^kf(\frac k n)\equal{}0$

Define $ S_n\equal{}\sum_{k\equal{}1}^n(\minus{}1)^kf(\frac{k}{n})$. Since $ S_{2n\plus{}1}\minus{}S_{2n}$ is bounded it suffices to consider the limit $ L\equal{}\lim_{n\rightarrow\infty}\frac{S_{2n}}{2n}$. Direct compuation of $ L$ using that $ f$ is continuous yields:

$ L\equal{}\lim_{n\rightarrow\infty}\frac{1}{2}\left[\underbrace{\sum_{k\equal{}1}^nf(\frac{2k}{2n})\frac{1}{n}}_{\rightarrow \int_0^1f(x)dx}\minus{}\underbrace{\sum_{k\equal{}1}^nf(\frac{2k\minus{}1}{2n})\frac{1}{n}}_{\rightarrow \int_0^1f(x)dx}\right]\equal{}0$

Since $ f$ is continuous on a closed interval, it is bounded and uniformly continuous.

Therefore for each $ \epsilon > 0$ there exists $ \delta$, such that:
\[ \left|f(x) - f(y)\right| < \epsilon \quad \mathrm{for} \quad \left|x - y\right| < \delta .
\]
Hence for $ \frac {1}{n} < \delta$ we have $ \left|f(x) - f(x + \frac 1n)\right| < \epsilon$.

Now we choose $ \epsilon$, find $ \delta$ and for all $ n > \frac {1}{\delta}$ we know that (we may assume, that $ n$ is even $ (\ast)$):

\begin{eqnarray*} \left|\sum_{k = 1}^n ( - 1)^kf(\frac k n)\right| & = & \left| - f(\frac 1n) + f(\frac 2n) - \ldots - f(\frac {n - 1}{n}) + f(\frac n n) \right| \leq \\
& \leq & \left|f(\frac 2n) - f(\frac 1n)\right| + \left|f(\frac 4n) - f(\frac 3n)\right| + \ldots + \left|f(\frac nn) - f(\frac {n - 1}{n})\right| < \\
& < & \epsilon + \epsilon + \ldots + \epsilon = \\
& = & n \epsilon \end{eqnarray*}

Therefore for each $ \epsilon > 0$, there exists $ n_{0}$, such that for all $ n > n_{0}$ we have:
\[ \frac1n\left|\sum_{k = 1}^n ( - 1)^kf(\frac k n)\right| < \epsilon
\]
Hence $ \lim_{n\rightarrow \infty}\frac1n\sum_{k = 1}^n ( - 1)^kf(\frac k n) = 0$.

$ (\ast):$ Since $ |f(\frac nn)|\leq \sup{|f(x)|} = M$, $ \lim_{n \to \infty}\frac {f(\frac nn)}{n} = 0$. Hence we can omit $ f(1)$ if $ n$ is odd.

We know that if f is a continuous function on [0,1] then  $ lim_{n\rightarrow \infty}\frac1n\sum_{k \equal{} 1}^n f(\frac k n) \equal{} \int_{0}^{1}f(x) dx$

But here we get $ lim_{n\rightarrow \infty}\frac1n\sum_{k \equal{} 1}^n ( \minus{} 1)^kf(\frac k n) \equal{}0$. What is the interpretation of this?

for the answer to your question pls check my earlier response


Is acceleration= force/mass
 or
change in speed/time interval
or both??

Both, in a way. It is defined as the change in speed over a given time interval - or more precisely, a limit of this ratio whan the time interval approaches zero. But according to Newton's first law, $F=m\times a$ so it is also equal to the force acting on some object divided by that object's mass. (This only works in inertial reference frames.)

It is most definitely the 2nd one, as that is its definition.

The first one is true only under conditions in which Newton's Laws of motion apply.  The other guy did mention inertial reference frame as a requirement, but even that can be circumvented by use of "inertial forces" or "centrifugal forces".  In view of special relativity, it is actually never true.  However, the errors associated with low velocities is genearlly negligible to 10+ decimal places.


A sports conference has 14 teams in two divisions of 7.  How many games are in a complete season for the conference if each team must play every other team in its own division twice and every team in the other division once?

We have division A and division B. Each division has 7 teams.

Let's say we have Team 1 in division A. Team 1 has to play each of the other 6 teams twice, so that's 12 games for team 1. There are 7 teams in total, so the number of games is 7 times 12 = 84. However, we have to divide by 2, because we overcounted. But there are two divisions, so the number of games played inside the divisions is 84.

Team 1 also has to play the other 7 teams in division B once, so that's 7 extra games for each of the 14 teams. Again, we divide by two to correct for overcounting, to arrive at a result of 49.

84 + 49 = $ \boxed{133}$

I solved it a bit differently:

Pick any team.  It plays $ 2$ games against each of $ 6$ teams, for a total of $ 12$ inter-division games.  Additionally, it plays $ 1$ game against each of $ 7$ teams, for a total of $ 7$ games.  Adding, we get $ 19$ games played by this team.

There are $ 14$ teams, so there are $ 19 \times 14\equal{}266$ games.  However, each game has two teams, so we divide by $ 2$ to get $ \boxed{133}$.

[hide=My solution]
$2\cdot 2\cdot\binom{7}{2}+7\cdot 7=133$[/hide]

I thought we should hide our solution...


topology
real analysis
real analysis unsolved

Let $ L$ be some countable collection of lines in $ \mathbb{R}^3$. Prove that $ \mathbb{R}^3 \minus{} L$ is a connected set.

I have an idea but I'm not really sure about it, so if somebody could verify this I'd be greatful :)

I'd risk to say that this set $ \mathbb{R}^3\backslash L$ is even arc-wise connected. Suppose that we want to connect some point $ A$ with $ B$ using an arc. Lets look at all lines passing through $ A$. There are uncountably many of them so there must be one on which we can move out of $ A$ without touching any line from $ L$ (does it?). Now lets connect every point from this segment with $ B$ using lines. Again there are uncountably many of them so there must be one on which we can get to $ B$ without falling into $ L$ (again "?"). Sum of both of these segments (first <or part of it> when we "got out of" $ A$, second when we reached $ B$) is naturally the arc we've been looking for.


[quote="Glen Mann"]this was on the latest CNML and I just cant seem to get it >_<

I dont know too much about logarithms but this is bothering me that I cant atleast figure it out with my knowledge.

$\frac{\log_{2}{x}}{\log_{4}{2x}}= \frac{\log_{8}{4x}}{\log_{16}{8x}}$

I figured Id change it all into base two using the basechange and split up all the 2x etc

$\frac{(\log_{2}{x})(\log_{2}{4})}{\log_{2}{2}+\log_{2}{x}}= \frac{(\log_{2}{16})(\log_{2}{4}+\log_{2}{x})}{(\log_{2}{8})(\log_{2}{8}+\log_{2}{x})}$

but then I have no idea wat to do.. I tried to replace $\log_{2}$ with a variable and then solve but I realized this doesnt work  :huh:

I know what the [hide="answers"]2, $\frac{1}{16}$[/hide] are, but just cant get to them.. thanks  :([/quote]
You have the right step...

$\frac{(\log_{2}{x})(\log_{2}{4})}{\log_{2}{2}+\log_{2}{x}}= \frac{(\log_{2}{16})(\log_{2}{4}+\log_{2}{x})}{(\log_{2}{8})(\log_{2}{8}+\log_{2}{x})}$

Perhaps this helps? $2^{3}=8\Leftrightarrow\log_{2}8=3$... and of course

$2^{1}=2$, $2^{2}=4$, $2^{3}=8$, $2^{4}=16$...

And substituting $y=\log_{2}x$... Piece together my hints, I'm sure you will find this easy :)

huzzah!

I jsut took a shower and realized how $\log_{2}{8}$ etc.. all simplify really easily and all I have left is $\log_{2}{x}$

now that I got it this seems really easy.. doh!

if you guys know of anymore like this, I'd like to practice some more  :blush:

What did you get on the CNML in the first place? I got a 5. Stupid error on the last one...

3, I suck.. 5 is my best :o

next year I'll bring on the 6's  :roll:


If this is on one of the camp applications, my apologies and someone censor it:

Prove that the sum of the reciprocals of all positive integers, in any base :ge: 2, not containing the digit 1, converges.

For that matter, try this rather famous problem:

Prove that the sum of the reciprocals of the positive integers not containing the digit 9 (in base 10) converges to a number less than 90.

[quote="probability1.01"]Well first of all, it might help to know that the integral of 1/x from 1 to x (correct me if I'm wrong; I've never taken calc) is ln x.  Messing with this can set an upper bound for the sum of all positive integers.  Now just to subtract those with the digit 1...[/quote]

The sum diverges if you don't remove those with digit 1.

[quote="probability1.01"]Well first of all, it might help to know that the integral of 1/x from 1 to x (correct me if I'm wrong; I've never taken calc) is ln x.[/quote]
This is true, but ln(x)-> infinity as x -> infinity.  As Nukular said, the harmonic series (1/1 + 1/2 + ...) diverges.

I can show

 :Sigma: 1/n - 1/(10n-1)

definitely converges,

but I'm having troubles showing that it converges to a number less than 90

If there were some x's thrown in then I could just use upper bound and lower bound as given by the ratio test, but how do I create bounds on ones where there is no variable, just a series?

Wow JBL, good job breaking the 3000 mark. Now never post again so taht you don't spoil it ;)

It doesn't seem to me that

[tex]\displaystyle \sum \Bigg(\frac{1}{n} - \frac{1}{10n-1}\Bigg)[/tex]

converges. Did you mean something else maybe, or should I not be doing math at 4:00 in the morning?

Also note that that isn't all numbers not containing the digit 9 -- in fact, 1/(10n - 1) is exactly the set of integers with units digit equal to 9 -- you're missing the options for other digits being 9.  (For example, our sum excludes 93, but yours does not.)

first of all, there are 8*(9)^(n-1) n-digit numbers not containing the digit nine. each of these n-digit numbers is greater than or equal to 10^(n-1) (only one is the equality case...), thus the reciprocals of each of these n-digit numbers is less than or equal to 1/(10^(n-1)).
so, the sum of the reciprocals of all of the n-digit numbers that do not have a nine is less than:
8 * (9/10)^(n-1).
summing over all natural n, we see that the sum of the reciprocals of all of the numbers that don't have a 9 is less than 80.

JBL, I see your point. How do you know that the number of numbers with 9's in them is 8*(9)n^1?

[quote="scm007"]How do you know that the number of numbers with 9's in them is 8*(9)n^1?[/quote]

There are 9*(10)^(n-1) numbers with n digits, since we have 9 choices for the first (1-9), 10 for the others (0-9).

By the same reasoning, we have 8*(9)^(n-1) numbers with n digits that do not have any nines.  There are 8 choices for the first and 9 choices for the other n-1 digits.


trigonometry
algebra
polynomial
Vieta
complex numbers
algebra unsolved

Prove that

$\prod^n_{i=1}\cos\frac{i\pi}{2n+1}=2^{-n}$

Consider the equation $Re\left(\left(x+i\sqrt{1-x^2}\right)^{2n+1}\right)=1$.  Since the square roots cancel in the real part, this is a polynomial of degree $2n+1$ with integer coefficients.  The observation $\left(x+i\sqrt{1-x^2}\right)^2=2x^2-1$ leads to a simple inductive proof that the coefficient of $x^{2n+1}$ is $2^{2n}$. But now, obviously $x=1$ is a root, and by DeMoivre's Theorem, $\cos\left(\frac{i\pi}{2n+1}\right)$ is a double root of this equation for each $1\le i\le n$, so we have the complete factorisation.

Now we can apply Vieta's Theorem (sp.?) to find $\left(\prod_{i=1}^{n}\cos\left(\frac{i\pi}{2n+1}\right)\right)^2=2^{-2n}$, and the result follows.

[quote="Diarmuid"] But now, obviously $x=1$ is a root, and by DeMoivre's Theorem, $\cos\left(\frac{i\pi}{2n+1}\right)$ is a double root of this equation for each $1\le i\le n$, so we have the complete factorisation.

Now we can apply Vieta's Theorem (sp.?) to find $\left(\prod_{i=1}^{n}\cos\left(\frac{i\pi}{2n+1}\right)\right)^2=2^{-2n}$, and the result follows.[/quote]

I cant get this part  :(

Well, by DeMoivre's Theorem, if $x=\cos(\theta)$, then $Re\left((x+i\sqrt{1-x^2})^{2n+1}\right)$ $=Re\left((\cos(\theta)+i\sin(\theta))^{2n+1}\right)$ $=Re\left(\cos((2n+1)\theta)+i\sin((2n+1)\theta)\right)$ $=\cos((2n+1)\theta).$ Now if $\theta=\frac{i\pi}{2n+1}$, $\cos((2n+1)\theta)=cos(i\pi)=1$, so each of these are roots of the equation.

Does that make sense?

Ok I still cant get why $\cos \theta$ is double root and how you apply Vieta .... maybe that is too " imaginary " to me  :P

Anyway , I did it in this way

[hide]

Letting $\theta=\frac{\pi}{2n+1}$

Consider $C=\prod^{n}_{i=1}\cos i\pi$ , $S=\prod^{n}_{i=1}\sin i\pi$ then we have

$CS=\prod\frac{1}{2}\sin 2i\pi=2^{-n}\prod\sin 2i\pi$

It is sufficient to prove $S=\prod\sin 2i\pi=\prod \sin i\pi$

We can see that for even $i$ in $\prod\sin i\pi$ , all $\sin i\pi$ will be cancel with first $\frac{i}{2}$ term in $\sin 2i\pi$ .

For odd $i$ in $\prod\sin i\pi$ , all $\sin i\pi$ will be cancel with last $\frac{i}{2}$ term in $\sin 2i\pi$ . Hence $\prod\sin 2i\pi=\prod \sin i\pi$ and we are done .

 :)
[/hide]


Ok, I totally screwed up my alcumus account so I want to start again. Is there a way to reset my account? Like from the initial 60 questions and without being able to see the videos?

Alcumus is currently undergoing maintenance; maybe that's your problem here?

Additionally, you can't have your account reset.

i know its undergoing maintenence. but when its done with the maintenance, i REALLY need to start over. in the worst case ill just open a new account just for alcumus i guess...

No multiple accounts.

Why in the world do you absolutely need to start over?

i didnt take it seriously so i just answered the first 60 questions randomly. so my score is messed up. :D:D

so?
alcumus is a learning tool
if you start doing well later on and keep doing well, you'll be fine in no time

ok, i guess... but my score is like 13%. XDXDXD

Scores don't matter, its the learning thats the most important! Also, just try your best on the next problems you do. Maybe that'll help your score (my score is 30%).


geometry
inradius
inequalities
geometry unsolved

In triangle $ ABC$,let $ a,b,c$ be the lenght  of the sides,$ r$ the inradius an $ R$ the circumradius.
Prove that: $ \sum\frac{a^2bc}{(a\plus{}b)(a\plus{}c)}\leq \frac{9}{4}Rr$

[quote="math10"]In triangle $ ABC$,let $ a,b,c$ be the lenght  of the sides,$ r$ the inradius an $ R$ the circumradius.
Prove that: $ \sum\frac {a^2bc}{(a \plus{} b)(a \plus{} c)}\leq \frac {9}{4}Rr$[/quote]
The problem must be: $ \sum\frac {a^2bc}{(a \plus{} b)(a \plus{} c)}\leq \frac {9}{2}Rr$
$ \Leftrightarrow abc(\sum \frac {a(b \plus{} c)}{(a \plus{} b)(b \plus{} c)(a \plus{} c)})\leq \frac {9}{2}.\frac {abc}{4S}.\frac {S}{p} \equal{} \frac {9}{2}.\frac {abc}{2(a \plus{} b \plus{} c)}$
$ \Leftrightarrow \frac {2(ab \plus{} bc \plus{} ca)}{(a \plus{} b)(b \plus{} c)(c \plus{} a)}\leq \frac {9}{4(a \plus{} b \plus{} c)}$
$ \Leftrightarrow 8(ab \plus{} bc \plus{} ca)(a \plus{} b \plus{} c)\leq 9(a \plus{} b)(b \plus{} c)(c \plus{} a)$
$ \Leftrightarrow 8(3abc \plus{} \sum ab(a \plus{} b))\leq 9(\sum ab(a \plus{} b) \plus{} 2abc)$
$ \Leftrightarrow \sum ab(a \plus{} b)\geq 6abc$ (*)
Use AM-GM inequality then (*) is right!


function
integration
calculus
derivative
limit
real analysis
real analysis unsolved

Prove that the sequence $ (f_n)_{n \in N^{*}}$, where $ f_n: [0,1] \rightarrow [0, \infty)$, $ f_0$ is a continuous and arbitrary function, and
$ f_{n\plus{}1}(x)\equal{}\int_0^x \frac{1}{1\plus{}f_n(t)}dt$, for all $ x \in [0,1]$ and for all $ n \in N^{*}$, converges and also calculate its limit.

[b]The limit is the function[/b] $ \sqrt{1\plus{}2x}\minus{}1$.

[i]Proof[/i]: Let $ {\cal C}\equal{}\{g: [0,1]\rightarrow [0,1]\,|\,g\,\mbox{continuous}\}$ and define a map $ S$ on $ \cal{C}$ via $ S[g](x)\equal{}\int_0^x\frac{dt}{1\plus{}g(t)}$. If $ g$ is a fixed point of this map, its derivative $ g'$ exists and computes as $ g'(x)\equal{}\frac{1}{1\plus{}g(x)}\Longrightarrow [(1\plus{}g(x))^2]'\equal{}2\Longrightarrow g(x)\equal{}\sqrt{2x\plus{}c}\minus{}1$ for some $ c\in\mathbb{R}$. Now the identity $ g(0)\equal{}S[g](0)\equal{}0$ forces $ c\equal{}0$ and $ g(x)\equal{}\sqrt{1\plus{}2x}\minus{}1$. Moreover, for $ h\in{\cal C}$ we can compute:

$ \begin{array}{rl} \parallel{}S[h]\minus{}g\parallel{}\equal{} & \parallel{}S[h]\minus{}S[g]\parallel{}\equal{}\max_{x\in [0,1]}\{|S[h](x)\minus{}S[g](x)|\}\equal{} \max_{x\in [0,1]}\left\{\left|\int_0^x\frac{g(t)\minus{}h(t)}{[1\plus{}h(t)][1\plus{}g(t)]}\,dt\right|\right\}\\
\leq  &\max_{x\in [0,1]}\left\{\int_0^x\frac{|g(t)\minus{}h(t)|}{[1\plus{}h(t)][1\plus{}g(t)]}\,dt\right\}\leq\parallel{}h\minus{}g\parallel{}\cdot\max_{x\in [0,1]}\left\{\int_0^x\frac{1}{[1\plus{}h(t)][1\plus{}g(t)]}\,dt\right\}\\
 \leq & \parallel{}h\minus{}g\parallel{}\cdot\max_{x\in [0,1]}\left\{\int_0^x\frac{1}{1\plus{}g(t)}\,dt\right\}\equal{}\parallel{}h\minus{}g\parallel{}\cdot\max_{x\in [0,1]}\left\{S[g](x)\right\}\equal{}\parallel{}h\minus{}g\parallel{}\cdot g(1)\equal{}\parallel{}h\minus{}g\parallel{}\cdot (\sqrt{3}\minus{}1)\end{array}$

If we define $ S^n$ via $ S^0\equal{}id_{[0,1]}$ and $ S^{n\plus{}1}\equal{}S\circ S^n$ we conclude $ \forall h\in{\cal C}: \,\parallel{}S^n[h]\minus{}g\parallel{}\leq \parallel{}h\minus{}g\parallel{}\cdot(\sqrt{3}\minus{}1)^n$, i.e. $ S^n[h]\rightarrow g$ for $ n\rightarrow\infty$.

Since $ f_1\in{\cal C}$ and $ f_{n\plus{}1}\equal{}S^n[f_1]$ we get $ \lim_{n\rightarrow\infty}f_n\equal{}g$ (and obviously this is a uniform limit)


What fraction in the interval $ \frac{7}{10} \le x \le \frac{8}{11}$ has the smallest denominator?

Since $ \frac {7}{10} \equal{} 0.7$, and $ \frac {8}{11}\approx0.73$, the fraction with the smallest denominator between these two fractions is $ \boxed{\frac {5}{7}}\approx0.71$.

yeah, especially since the denominators of the original two fractions are relatively small.

[quote="mathemagician1729"]So here it would be helpful to memorize unit fractions?[/quote]

Well, here you had 10, which is simple enough.  And then you had $ 11$, which you could have rewritten in with denominator $ 99$ by multiplying numerator and denominator by 9.  Becaue it was $ 99$, the decimal is in the form $ .ababab...$ where $ a$ is the tens digit and $ b$ is the ones.  If there is no $ a$, it's simply $ .bbb...$.  Thus, the fraction must have been $ .727272...\equal{}\sim .73$

OF course, the $ 7$ would have been helpful to memorize.

Just curious as to what round this was.

In general, I'd guess+check numbers up from 2.

Well if you really wanted to know, then this problem would be very pointless because it was a target question. I could see this coming up easily on a sprint though.


modular arithmetic
number theory unsolved
number theory

find all pairs $p,q$ prime number , if:
$2^p+2^q = 0 \pmod {pq} $

ude this theorem
let $r$ be the smallest solution of $n^x \equiv -1 \pmod p$, then the solutions of $n^x \equiv -1 \pmod p$ are $(2k+1)r$, and all the solutions of $n^x \equiv 1 \pmod p$ are $2kr$

Have a look here :
http://www.mathlinks.ro/Forum/viewtopic.php?t=20918

Pierre.


We are given 3 lines defined by the equations y=x, y=3, and y=1. We are given the point P(2,0). What are the slopes of the lines that pass through P and intersect one of the given lines such that the point of interection is the midpoint of the segment connecting the intersection points of the other 2 given lines?

May want to hold off on answering this one until the weekend because I believe that it is part of an ongoing competition until that point.

mandelbrot problem, weren't the answers to that released 2 weeks ago?

Nah, the deadline to take the mandelbrot was last week.


geometry
geometric transformation
reflection
3D geometry
sphere
induction
superior algebra

This one seems really wierd (and I admit I don't know where to post it): there exists a partition of $R^{3}$ with circles such that any positive real number appears exactly once as a radius of these circles. Also, there exists such a partition where all circles are congruent.

The second one has been posted before [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=35242]here[/url].

I think I have a proof for the second one, and I wanted to post it in that other thread some time ago, but while typing it, I decided it was taking me way too long and I gave up :). I'll try again in this thread if I manage to solve the first one (but don't hold your breath :)).

Do you have a proof for either of them, harazi?

P.S.

As for the suitable forum where it should be posted, I think it definitely goes better in the "Advanced Fields" section than here.

I have proofs for them, but they are long and difficult. And sorry for posting it here, please some moderator move it...

uhm.. excuse me: circle ($D^{2}$) or circumference ($S^{1}$)?

[quote="harazi"]There exists a partition of $R^{3}$ with circles such that any positive real number appears exactly once as a radius of these circles. Also, there exists such a partition where all circles are congruent.[/quote]

Hopefully, this is going to be a correct proof of the following statement, which generalizes both problems:

Let $A$ be a set of cardinality $c$ (continuum cardinality), and let $f: A\to(0,\infty)$ be a function. Then the Euclidean space $\mathbb R^{3}$ is the union of a collection $\{\mathcal C_{a}\}_{a\in A}$ of disjoint circles such that the radius of $\mathcal C_{a}$ is $f(a)$ for all $a\in A$.

The first problem is the case $f$ bijective, while the second problem treats the case when $f$ is constant.

The construction will be by transfinite induction. Give $A$ the order type of $c$, regarded as an ordinal number (i.e. well-order it in such a way that for every element, there are fewer than $c$ smaller elements), and do the same for $\mathbb R^{3}$. We can now identify both $A$ and $\mathbb R^{3}$ with the ordinal number $c$. In particular, $0$ will denote the smallest element in both sets.

In step $0$, choose an arbitrary non-horizontal circle $\mathcal C_{0}$ of radius $f(0)$ passing through the point $0$. This takes care of the first step of the induction. Now assume we’ve reached step $\alpha$ (an ordinal number smaller than $c$). Let $P$ be the smallest point which has not yet been covered by a circle, and let $\Pi$ be a horizontal plane whose distance from $P$ is smaller than $f(\alpha)$. Now consider the circle $\mathcal C$ formed by the points in $\Pi$ situated at the distance $f(\alpha)$ from $P$. Choose a point $O_\alpha\in\mathcal C$ such that: (i) the reflection $Q$ of $P$ in $O_\alpha$ is not covered by any of the circles placed so far, and (ii) if $O_\beta,\ \beta<\alpha$ are the centers of the circles $\mathcal C_\beta$ placed so far, then $O_\alpha O_\beta$ is neither degenerate nor orthogonal to the plane in which $\mathcal C_\beta$ lies. Such a choice is possible, because up until now we’ve placed fewer than $c$ circles, all non-horizontal, so every horizontal plane cuts their union in fewer than $c$ points, and also, there are fewer than $c$ planes in which $\mathcal C_\beta$ lie.

As the notation suggests, $O_\alpha$ is going to be the center of the circle $\mathcal C_\alpha$, which has not yet been constructed. Consider the sphere centered at $O_\alpha$, of radius $f(\alpha)$ (i.e. the sphere having $PQ$ as a diameter). Every circle $\mathcal C_\beta,\ \beta<\alpha$ cuts this sphere in at most two points because of (ii). Moreover, none of these circles contain $Q$, because of (i). Since there are fewer than $c$ circles $\mathcal C_\beta,\ \beta<\alpha$, this means that there is a circle having $PQ$ as diameter and which doesn’t cut any of the aforementioned circles. This will be $\mathcal C_\alpha$, and the induction step is complete.


vector
linear algebra
matrix
abstract algebra

I have a question whihch asks me

Consider an arbitrary vector $ \vec{v}$ in $ \mathbb{R}^{3}$. Is the transformation
$ T(\vec{x}) \equal{}\vec{v}\times\vec{x}$ from $ \mathbb{R}^{3}$ to $ \mathbb{R}^{3}$ linear?

If so, find its matrix in terms of the compenents of the vector $ \vec{v}$.

I don't think I'm understanding the question. If I take a vector v and cross it with a vector x, isn't the matrix that transforms v to T(v) the vector x?
Any help to guide me to understanding is greatly appreciated.

Thanks
Ankur

[quote="ankur87"]If I take a vector v and cross it with a vector x, isn't the matrix that transforms v to T(v) the vector x?[/quote]
Of course not. The vector $ x$ is not a $ 3\times 3$ matrix.

In that case, I am somewhat confused as to how to proceed.

Here is what I dd to try to gain an understanding of the problem

I choose two matrices

$ \vec{v}\equal{}\begin{pmatrix}1\\ 3\\ 4\end{pmatrix}$ and $ \vec{x}\equal{}\begin{pmatrix}2\\ 5\\ 6\end{pmatrix}$

When I cross the two, I get the result:

$ \vec{v}\times\vec{x}\equal{}\begin{pmatrix}\minus{}2\\ 2\\ 1\end{pmatrix}$

However, now I don't know how to find a general formula for the transformation. Am I approaching the problem correctly?

When you compute the cross product of your two vectors, don't merely write the three final numbers.  Instead, write each of those three final numbers in terms of your original numbers.  That is, explain how you got those three numbers.  Then you will be in good position to determine the matrix.

If $ u\equal{}\begin{pmatrix}u_{1}\\u_{2}\\u_{3}\end{pmatrix}$ and $ x\equal{}\begin{pmatrix}x_{1}\\x_{2}\\x_{3}\end{pmatrix}$ then $ u\times x\equal{}\begin{pmatrix}u_{2}x_{3}\minus{}u_{3}x_{2}\\\minus{}u_{1}x_{3}\plus{}u_{3}x_{1}\\ u_{1}x_{2}\minus{}u_{2}x_{1}\end{pmatrix}$

That makes the matrix of the linear transformation (for fixed $ u$) this:

$ A\equal{}\begin{pmatrix}0&\minus{}u_{3}&u_{2}\\u_{3}&0&\minus{}u_{1}\\ \minus{}u_{2}&u_{1}&0\end{pmatrix}.$

Note that the matrix $ A$ is skew-symmetric: $ A^{T}\equal{}\minus{}A.$ If we reversed the order of the factors in the cross product we would pass to the transpose (negative) of this matrix. Note also that $ \det A\equal{}0.$ This must be so, since for any $ u$ one can find a nontrivial $ x$ such that $ u\times x\equal{}0.$

Thanks for all your help

I do not understand how Professor Merryfield came up with the matrix $ A$...
I do know the formula for the cross product, but I don't see how the matrix $ A$ was obtained.

I am just a beginner in matrix algebra and haven't yet gone over how to find the inverse of a matrix (which is something I will look up now).

Thanks for all the replies and helping me out even though I seem to be a slow learner...

If you see up to what the general form of $ u\times x$ is, then you want a matrix such that $ Ax\equal{}u\times x$. So you need a matrix such that

$ \left(\begin{matrix}a_{11}& a_{12}& a_{13}\\ a_{21}& a_{22}& a_{23}\\ a_{31}& a_{32}& a_{33}\\ \end{matrix}\right)\left(\begin{matrix}x_{1}\\ x_{2}\\ x_{3}\\ \end{matrix}\right)\equal{} u\times x$

(Didn't feel like typing out the vector component form of $ u\times x$ since it is already up).

Then multiply the left hand side out, equate components of both vectors, and find that $ a_{ij}\equal{}0$ for $ i\equal{}j$, $ a_{12}\equal{}\minus{}u_{3}$, $ a_{13}\equal{}u_{2}$ etc.

[quote="ankur87"]I am just a beginner in matrix algebra and haven't yet gone over how to find the inverse of a matrix (which is something I will look up now).[/quote]
But not because it has anything to do with this problem! None of the matrices in this problem are invertible, so knowing how to find inverses would be useless.

Invertible matrices give rise to invertible functions. That is, if $ A$ is invertible and we know that $ Ax\equal{}y$ for some specific $ y,$ then we know exactly what $ x$ is and we can write that unique vector as $ x\equal{}A^{\minus{}1}y.$

But if we know $ u$ and $ w$ and we know that $ u\times x\equal{}w,$ does that mean that we know $ x$? Most certainly not.

To argue this point: Suppose that $ u\times x\equal{}w$ and $ u\times y\equal{}w.$ Subtract these and use the fact that the cross product distributes over addition and subtraction:

$ u\times x\minus{}u\times y\equal{}w\minus{}w$
$ u\times(x\minus{}y)\equal{}0.$

Now, what does that tell us about $ x\minus{}y?$ It tells us that $ x\minus{}y$ is parallel to $ u.$ That's some information, but it still leaves infinitely many possibilities. In particular, we can't claim that $ x\equal{}y,$ so we don't really know $ x.$

This is an utterly standard argument in linear algebra: a linear transformation is one-to-one if and only if its kernel (a.k.a. null space) contains only the zero vector.

The matrix representing a linear transformation $ T$ has $ Te_{1}$, $ Te_{2}$, and $ Te_{3}$ as its columns. Here $ e_{1},e_{2},e_{3}$ are the standard basis vectors. It's not hard to calculate $ e_{1}\times x$, etc.

Thanks Professor Merryfield and HilbertThm90

I really appreciate the help and feel that I have a much better grasp of what it means to perform a linear transformation. I am going to do some extra problems from my book to test my understanding, and I'll bring up anything that confuses me.


geometry
geometric transformation
reflection
calculus
trigonometry
geometry solved

I've posted a similar problem ("4 dogs", in the "Proposed Problems" section).

A squint man is distance d away from his car. He sees everything with a deflection of 30 degrees. How long will be the path he follows to his car?

He walks on a logarithmic spiral, so I guess it's not that hard to get a solution using calculus. However, I saw this problem on a website (I can't remember the name), and the guy who wrote it there said that we could give a solution without calculus. Any ideas? I was thinking about decomposing speeds in 2 directions and stuff like that, but I don't have anything yet.

Thanhliem you must have answered a different question. What was it? :)

It's definitely something else :D:D How can this happen?

[quote="Thanhliem"][b]i think[/b]   a+b+c=(3 \sqrt 3)/2
            a^2+b^2+c^2+x^2+y^2+z^2=3
            a^2+b^2+c^2=2xyz+2
 a,b,c,d,x,y,z>0    a^2+b^2+c^2  \geq 9/4 ,x^2+y^2+z^2 \leq 3/4
    2xyz \geq 9/4-2  and (xyz)^2  \leq  (x^2+z^2+y^2)^3/ 27
      then  xyz=1/8
               a^2+b^2+c^2=(3 \sqrt 3)/2[/quote]

Let the man think, would you!
He saw your problem grobber and started thinking to what is written above. What is wrong with it? It may be a little weird, but who can tell how does the human mind work?

[quote="grobber"]
A squint man is distance d away from his car. He sees everything with a deflection of 30 degrees. How long will be the path he follows to his car?
[/quote]
All right, here goes my [i]calculus-free[/i] solution.   ;)

Since the man sees his car with a deflection of $30$ degrees, in any time he will move in a direction forming $30$ degrees with the line joining him with his car. Hence, the progress he actually makes to reach his car is always $\frac{\sqrt 3}{2}$ times the overall progress. Then, in the end he will reach his car, with an overall path of $\frac{2d}{\sqrt 3}$.
In other words, we may pretend that the man sees his car exactly where it is, but walks $\frac{\sqrt 3}{2}$ times slower. So, under this assumption his path will be $d$ long, while the real path will be $\frac{2}{\sqrt 3}$ times longer.

I guess it makes sense. If I remember correctly, using the properties of the logarithmic spiral one gets $\frac d{\cos {\alpha}}$, where $\alpha$ is the angle, which in our case is $\frac {\pi}6$. This is the same as your intuitive result. Thank you very much !


geometry
cyclic quadrilateral
geometry proposed

Let $ABCD$ be a quadrilateral inscribed in circle $(O)$. Let $S$ be the intersection of $AC$ and $BD$. Suppose $M, N, P, Q$ are the intersections of the bisectors of internal angles $A$ and $B$, $B$ and $C$, $C$ and $D$, $D$ and $A$. Prove that $MNPQ$ is a cyclic quadrilateral and the center lies on the line $OS$.


Points $P$ and $Q$ lie on the sides of $AB$ and $AC$ respectively of triangle $ABC$ and are distinct from $A$. The lengths $AP$ and $AQ$ are denoted by $x$ and $y$ respectively with the convention that $x>0$ if $P$ is on the same side of $A$ as $B$, and $x<0$ on the opposite side; and similarly for $y$. Show that $PQ$ passes through the centroid of the triangle if and only if

$3xy = bx+cy$

where $b = AC$ and $c = AB$.

since I'm really terrible with vectors, could someone post a solution using vectors and explain it? thanks!

First, let $M$ be the centroid. Then by elementary vector algebra, we have $\overrightarrow{AM}=\frac{\overrightarrow{AB}+\overrightarrow{AC}}{3}$. Next, since $P$ is on $AB$, there exists a real number $\beta$ such that $\overrightarrow{AP}=\beta\overrightarrow{AB}$; similarly, there exists a real number $\gamma$ such that $\overrightarrow{AQ}=\gamma\overrightarrow{AC}$

For a point $X$ to be lying on a line $PQ$, there must exist a real number $\alpha$ such that $\overrightarrow{AX}=\alpha\overrightarrow{AP}+(1-\alpha)\overrightarrow{AQ}$. If we put $M$ for $X$, we get the following equation:

$\overrightarrow{AM}=\alpha\beta\overrightarrow{AB}+(1-\alpha)\gamma\overrightarrow{AC}$, which gives

$\frac{\overrightarrow{AB}+\overrightarrow{AC}}{3}=\alpha\beta\overrightarrow{AB}+(1-\alpha)\gamma\overrightarrow{AC}$

Since $\overrightarrow{AB}$ and $\overrightarrow{AC}$ are linearly independent, the corresponding coefficients must be equal, hence the following must be satisfied:

${1\over 3}=\alpha\beta$ and ${1\over 3}=(1-\alpha)\gamma$

and from that we get $\alpha=\frac{1}{3\beta}$ and $1-\alpha=\frac{1}{3\gamma}$

which in turn gives

$\frac{1}{3\beta}+\frac{1}{3\gamma}=1$

Since $\beta={x\over c}$ and $\gamma={y\over b}$, substituting and rearranging gives the desired result.

Here is the solution using vectors.

Let $A$ be the origin, and let $\vec{c}= \vec{AB}, \vec{b}= \vec{AC}.$ Also, $AB=c, AC=b.$

So, $\vec{AP}= \frac{x}{c}\vec{c}$ and $\vec{AQ}= \frac{y}{b}\vec{b}$. Now, note if $G$ is the centroid of  $\triangle{ABC},$ we have $\vec{AG}= \frac13 (\vec{b}+\vec{c}+\vec{0}).$ (Remember, the position vector of $A$ is $\vec{0}$.)

Now the general vector equation of a line passing through two points $P$ and $Q$ is given by $\vec{\lambda}= (1-t)\vec{AP}+t\vec{AQ}$ for all $t \in \mathbb{R}$ ...(I)

So,
$PQ$ passes through the centroid,$G,$ of the triangle
$\Longleftrightarrow G$ lies on $PQ$
$\Longleftrightarrow \vec{AG}$ satisfies the equation of the line given in (I)
$\Longleftrightarrow \vec{AG}= (1-t)\vec{AP}+t\vec{AQ}$ for some $t \in \mathbb{R}$
$\Longleftrightarrow \frac13 (\vec{b}+\vec{c}) = (1-t)\frac{x}{c}\vec{c}+t\frac{y}{b}\vec{b}$
$\Longleftrightarrow \frac13 = \frac{(1-t)x}{c}$ and $\frac13 = \frac{ty}{b}$ ...(Note that $\vec{b}$ and $\vec{c}$ are independent vectors.)
$\Longleftrightarrow bx+cy = 3xy$ ...(eliminate $t$ above)

And, we are done. Hope the solution has a decent amount of clarity.


induction
inequalities
algebra solved
algebra

Can anyone prove the following by induction:

[color=green]Prove by induction that

(1/1^2)+(1/2^2)+...+(1/n^2) <= 2-(1/n)

for all integers n >= 1 [/color]

Thanks very much!!!

You don't need to post twice!

http://www.mathlinks.ro/Forum/viewtopic.php?t=72786

A little lemma: $\frac{n^2+n-1}{n(n-1)}\geq1$ for $n>1$

Proof: $2n\geq1\Rightarrow n-1\geq -n \Rightarrow n^2 + n - 1 \geq n^2 - n \Rightarrow \frac{n^2+n+1}{n(n-1)}\geq1$.

It's easy to see that the inequality holds for $n=1$. Assume it holds for $n-1$. Then,

$\sum_{i=1}^n\frac{1}{i^2}=\sum_{i=1}^{n-1}\frac{1}{i^2}+\frac{1}{n^2}\leq2-\frac{1}{n-1}+\frac{1}{n^2}=2-\frac{n^2+n-1}{n^2(n-1)}\leq2-\frac{1}{n}$.

The result follows.


[quote="GameBot"]How many paths are there from $ \textrm{A}$ to $ \textrm{B}$ passing through $ \textrm{C}$, if every step must be up or to the right?[asy]size(4cm,4cm);int w=6;int h=5;int i;pen p=fontsize(9);for (i=0; i<h; ++i){draw((0,i) -- (w-1,i));}for (i=0; i<w; ++i){draw((i, 0)--(i,h-1));}label("A", (0,0), SW, p);label("C", (1,2), NE, p);label("B", (w-1,h-1), NE, p);[/asy][/quote]

Let us use how to arrange letters in a word:

[asy]size(4cm,4cm);int w=6;int h=5;int i;pen p=fontsize(9);for (i=0; i<h; ++i){draw((0,i) -- (w-1,i));}for (i=0; i<w; ++i){draw((i, 0)--(i,h-1));}label("A", (0,0), SW, p);label("C", (1,2), NE, p);label("B", (w-1,h-1), NE, p);

label("$a$",(0,1.5),W);
label("$a$",(0,0.5),W);
label("$b$",(0.5,0),S);[/asy]

Basically we just broke the question down into (For point A to C): How many ways are there to arrange the letters in the word aab:

$ \dfrac{3!}{2!}=3$

Now we can do the same thing for C to B:

[asy]size(4cm,4cm);int w=6;int h=5;int i;pen p=fontsize(9);for (i=0; i<h; ++i){draw((0,i) -- (w-1,i));}for (i=0; i<w; ++i){draw((i, 0)--(i,h-1));}label("A", (0,0), SW, p);label("C", (1,2), NE, p);label("B", (w-1,h-1), NE, p);

label("$a$",(0,1.5),W);
label("$a$",(0,0.5),W);
label("$b$",(0.5,0),S);
label("$c$",(1.5,2),S);
label("$c$",(2.5,2),S);
label("$c$",(3.5,2),S);
label("$c$",(4.5,2),S);
label("$d$",(1,2.5),W);
label("$d$",(1,3.5),W);[/asy]

How many ways are there to arrange the letters in the word ccccdd:

$ \dfrac{6!}{4!2!}=15$

Now for each of the 3 ways there are to get from A to C, there are 15 ways to get to B.

$ 3\cdot 15=\boxed{45}$

This is sorta like what you did...but here goes.

There are $ \binom{3}{1}$ ways to go from A to C, because we make 3 moves and we need to choose which one is horizontal.  From C to B, there are $ \binom{6}{4}$ ways by similar reasoning.  Multiplying, we get $ 3*15\equal{}45$.

[font=times new roman][hide=My Solution]We can break this grid up into to grids - from A to C and from C to B. This means that for every path on Grid A-C, there is a number of paths on Grid B-C. So we simply have to multiply the number of paths for each grid. The first grid can be crossed in [hide=3]$\binom{2+1}{1}=3$[/hide] ways, while the second grid can be crossed in [hide=15]$\binom{2+4}{2}=15$[/hide] ways. Therefore, our answer is $3\cdot15=\boxed{45}.$ $$\definecolor{color}{rgb}{0.4, 0.0, 0.0} \scriptsize{\colorbox{color}{\color{white}{A pifinity solution}}}$$


function
inequalities solved
inequalities

Prove that $ \forall x,y,z \in [0,1]$ we have $ \sum \frac{x}{y+z+1} + \prod (1-x)\leq 1.$

The lhs is a convex function in each of the variables, so it reaches its maximum at the endpoints of [0,1], that is when x,y,z are in {0;1}.
It is easy to see that in that case the lhs is 1.

Pierre.


We are in the process of forming a new homeschool MOEMS team based in Cambridge, MA.  We will be meeting on Wednesdays at 2pm starting September 21.  Any homeschooled students in the Boston area are invited to join us.  If you are interested please contact me at the e-mail address below.

Andrew Sutherland
AndrewVSutherland@alum.mit.edu

You could post this here [url]http://www.artofproblemsolving.com/Forum/index.php?f=232[/url]
but actually people barely  ever look in the 'National and Local Communities'


integration
function
complex analysis
complex analysis unsolved

Suppose $ \mid \int^{}_{S^1} f(z)\,dz \mid \equal{} 2\pi M$ and $ \mid f(e^{i\theta})\mid \le M$ $ (0 \le \theta \le 2\pi)$.Prove that $ f(z)\equal{}M\bar{z}$.

$ 2\pi M\equal{}\mid \int^{}_{S^1} f(z)\,dz \mid \le \int^{}_{S^1}\mid f(z)\mid \,dz$
$ \equal{} \int^{2\pi}_{0} \mid f(e^{i\theta})  e^{i\theta} \mid \,d\theta \le \int^{2\pi}_{0} M  \,d\theta \equal{}2\pi M.$

squeeezing to equalities we see that $ \mid f(e^{i\theta}) e^{i\theta} \mid \equal{}M$, and thats $ \forall \theta$

cooler still, we get $ \mid f(z)z \mid\equal{}M$ now $ \forall z, \mid z \mid \equal{} 1$

let $ g(z)\equal{}f(z)z$. then $ \mid g(z) \mid \equal{} M$ on $ \mid z \mid \equal{}1$

we know that a function of constant modulus on the boundary of the unit disc must be constant itself on the unit disc by maximum modulus or something like that.

so $ g(z)\equal{}M e^{i\theta}$ for some $ \theta$

now we get $ f(z)\equal{}\tilde M/z , \mid \tilde M \mid \equal{} M$

this, however isnt holomorphic (neither is yours though).  if you wanted to derive your answer you can simply note that when $ \mid z \mid \equal{}1$ we have $ z \equal{} 1/\bar{z}$, and then just let $ g(z)\equal{}f(z)/\bar{z}$ with the same proof.

i hope this helps?


How many different points in the plane are at a distance 5 from the origin and have coordinates (m,n) where m and n are integers?

A.0
B.4
C.6
D.8
E.12

[hide="Answer"]
We have the 8 order/sign permutations of $ (\pm 3, \pm 4)$, and the 4 order/sign permutations of $ (\pm 5, 0)$, for a total of $ \boxed{12}$.
[/hide]

[hide="Systematic"]$ x^2\plus{}y^2 \equal{} 5^2 \implies x^2 \leq 25 \implies \minus{}5 \leq x \leq 5$
So now we have eleven cases to try out. However, the negative numbers are symmetric to the positive numbers.
We quickly see that $ 0$ produces $ \pm 5$, $ \pm 5$ produces $ 0$, and $ \pm 3$ and $ \pm 4$ produce each other as Pythagorean triples. In other words, it's only the $ x$ values $ \pm 1$ and $ \pm 2$ that are not used.
[/hide]

[hide]A Logical approach seems better for me:

We know that there are 4 points because any point 5 units away from the origin on either of the axises will work.

Next, we have to try to find a right triangle with integer lengths and hypotenuse 5 units (if you don't get how I reach this, look at the proof of the distance formula). If you know your Pythagorean Triplets, then you will easily find that the right triangle is a 3-4-5 triangle. That adds another four points because this triangle can be in all of the sectors of the coordinate plane. Therefore, there are:

$ 4\plus{}4\equal{}\boxed{8}$ different points, or D. [/hide]


integration
function
calculus
calculus computations

Let B be the region in the first quadrant of [tex] R^2 [/tex] bounded by [tex] xy=1,  xy=3,  x^2-y^2=1,  x^2-y^2=4.  [/tex]  Find [tex]  \int_B(x^2+y^2) [/tex] using the substitution [tex]u=x^2-y^2, v=xy [/tex].  Solution should use the Inverse Function theorem rather than solving for x and y explicitly.

Of course, that would be $u=x^2-y^2.$

Is this a problem you already know how to do or are you asking for help?

I bet it should be $u=x^2-y^2$. When making the change of variable, the first thing to do is to find the Jacobian. We have $u_x=2x$, $u_y=-2y$, $v_x=y$, $v_y=x$, so $J=|u_xv_y-u_yv_x|=2(x^2+y^2)$. So we are lucky because the Jacobian (up to the factor $2$) is already present in the integrand and the integral can be rewritten as $\frac12\iint_{1<u<4, 1<v<3}\, du\,dv=3$.

yeah that was a typo - it should be x instead of k.  thanks.


ai co ket qua khong tot thi vo dau de chung ta cung chia se nhe !
             tui la mot trong nhung so do  .toi uoc gi .... nhung bay gio tui dang phai chien dau voi 1 ki thi quan trong hon do la " nguong cua dai hoc "
               Vi vay  day la forum cua nhung ai giai " khuc khich " tro xuong den rot . Heheheh , nhung tui nghi rang chac chi co minh tui rot khi cac ban doc nhung trang nay .

chuc mung linkgreencold da lam moderator cua forum ASIA Vietnam
                      tu day toi hi vong rang forum nay se phat huy tac dung cua minh
 tui nghi day se la diem dung chan li tuong  sau nhung giay phut cang thang voi nhung bai toan

sao khong ai quan tam ah` , the  thi chung to chi co minh tui lam ai te hai thoi phai khong ?
                     cac bac lam tot the thi em hong dam noi chuyen nua dau

có lẽ những người xác định rớt như bạn họ đã lao vào ôn thi đại học mất rồi,mà cũng chẳng có gì buồn khi mình rớt cả,hãy đem nó làm động lực để tiến tới những tầm xa hơn


linear algebra
matrix
linear algebra unsolved

Let A be an nxn matrix with real entries, and let $d_{1}, d_{2}...d_{n}$ be the Euclidean norms of its column vectors.  Prove that $|det A| \le \prod_{i = 1}^{n}d_{i}.$


trigonometry
function
calculus
derivative
integration
calculus computations

A half-rectified sine wave is given by the $ 2\pi$-periodic function

$ f(t) =\left\{{\begin{array}{c}0 & (-\pi < t\leq 0),\ \sin{t}& (0 < t\leq\pi).\\end{array}}\right.$

Find the Fourier series of $ f(t)$.

Comments: this function is neither even nor odd. Hence if you do the Fourier series in real form, you will have both sine and cosine terms, or if you do in in complex form, $ \widehat{f}(k)$ will be neither even nor odd and neither pure real nor pure imaginary.

$ f(t)$ is continuous but not continuously differentiable. $ f'(t)$ has jump discontinuities at $ 0$ and $ \pi$ but is otherwise nice. From these facts, we predict that the Fourier coefficients will behave like $ O(k^{\minus{}2})$ as $ k\to\infty.$

I'll let someone else calculate, but the answers should be compared to these predictions.

----

Added:

Let me be more explicit about the even-odd thing. Any function can always be decomposed into and even part and an odd part: $ f(t) \equal{} f_{e}(t)\plus{}f_{o}(t).$ We can compute those as $ f_{e}(t) \equal{}\frac{f(t)\plus{}f(\minus{}t)}{2}$ and $ f_{o}(t) \equal{}\frac{f(t)\minus{}f(\minus{}t)}{2}$

In this case, $ f_{e}(t)\equal{}\frac{1}2\,|\sin t|$ and $ f_{o}(t)\equal{}\frac{1}2\,\sin t$

The Fourier series for the odd part is particularly easy: there's only one term. $ b_{1}\equal{}\frac{1}2$ and $ b_{k}\equal{}0$ for all other $ k.$

But there will be infinitely many nonzero cosine coefficients. Computing the $ a_{k}$ will be the bulk of the computation.

I'll go ahead and calculate it.

$ f(x) \equal{}\frac{1}{\pi}\plus{}\frac{1}{2}\sin t\plus{}\frac{2}{\pi}\sum_{k \equal{} 1}^{\infty}\frac{1}{1\minus{}4k^{2}}\,\cos(2kt)$

An important part of my calculation was noting that (using weak or distributional derivatives) $ f''(t)\plus{}f(t) \equal{}\delta(t)\minus{}\delta(t\minus{}\pi)$

[color=green][Edited to correct the coefficient of the sum.][/color]

The factor of 2 in the wrong place mistake was mine. The coefficient of the sum is $ \frac{2}{\pi},$ not $ \frac{1}{2\pi}$

This is a nice problem:

Find the Fourier series of the $ 2\pi$-periodic function:

$ f(t) =\left\{{\begin{array}{c}0 & (-\pi < t\leq 0),\ 1 & (0 < t\leq\pi).\ \end{array}}\right.$

Hence find the sum of the series

$ 1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\cdots$

The average value of that function is $ \frac{1}{2},$ so that will be the constant term.

$ f'(t) \equal{}\delta(t)\minus{}\delta(t\minus{}\pi)$ (The same object that appeared above.)

$ \widehat{f'}(k) \equal{} 1\minus{}(\minus{}1)^{k}\equal{}\begin{cases}2, & k\text{ odd}\\ 0, & k\text{ even}\end{cases}.$

Hence for $ k\ne 0,$ $ \widehat{f}(k) \equal{}\frac{2}{2\pi ik}$ if $ k$ is odd and zero otherwise.

That gives:

$ f(t) \equal{}\frac{1}{2}\plus{}\sum_{k\text{ odd}}\frac{2}{2\pi ik}\,e^{ikt}$

$ f(t) \equal{}\frac{1}{2}\plus{}\sum_{m \equal{} 0}^{\infty}\frac{2}{(2m\plus{}1)\pi}\,\sin((2m\plus{}1)t)$

[The "square wave," of which this is one form, is one of the most familiar of all Fourier series; I could have written this out from memory.]

The particular numerical sum you ask for calls for the evaluation of this series for $ t \equal{}\frac{\pi}{2}.$ We get

$ 1 \equal{}\frac{1}{2}\plus{}\frac{2}{\pi}\left(1\minus{}\frac{1}{3}\plus{}\frac{1}{5}\minus{}\frac{1}{7}\plus{}\cdots\right)$

Hence $ 1\minus{}\frac{1}{3}\plus{}\frac{1}{5}\minus{}\frac{1}{7}\plus{}\cdots \equal{}\frac{\pi}{4}$

For another calculation, apply Parseval's identity to this.

$ \|f\|_{2}^{2}\equal{}\frac{1}{2\pi}\int_{\minus{}\pi}^{\pi}|f(t)|^{2}\,dt \equal{}\frac{1}{2}$

$ \equal{}\sum_{k}|\widehat{f}(k)|^{2}\equal{}\frac{1}{4}\plus{}\sum_{k\text{ odd}}\frac{1}{|\pi ik|^{2}}$

$ \equal{}\frac{1}{4}\plus{}\frac{2}{\pi^{2}}\sum_{m \equal{} 0}^{\infty}\frac{1}{(2m\plus{}1)^{2}}$

This can be rearranged to $ 1\plus{}\frac{1}{3^{2}}\plus{}\frac{1}{5^{2}}\plus{}\frac{1}{7^{2}}\plus{}\cdots\equal{}\frac{\pi^{2}}{8}.$

One may in turn derive $ \zeta(2)$ from this.

Almost any function can be used for identities in this style; there are similar things that can be done with the function at the top of this topic.

One further comment about this square wave: although you defined this function to have the value $ 0$ at $ 0$ and the value $ 1$ at $ \pi,$ the Fourier series doesn't care. At both $ 0$ and $ \pi,$ the Fourier series sums to $ \frac{1}2,$ which is the midpoint of the jump discontinuity.

[quote="Kent Merryfield"]$ 1\plus{}\frac{1}{3^{2}}\plus{}\frac{1}{5^{2}}\plus{}\frac{1}{7^{2}}\equal{}\frac{\pi^{2}}{8}.$
[/quote]
I think you meant to include some $ \dots$; the left hand side is rational as written.

[color=green][Eh - read what I meant to say, not what I said, OK? - K.M.][/color]

Kent, I'm afraid much of what you write is lost on me. I only just started teaching myself fourier series yesterday, so I don't know about converting to complex forms. Are weak or distributional derivatives things like the impulse function?

I wrote up a fair amount of exposition on the subject in [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=166647]this recent topic.[/url]

"Impulse function" = "Dirac delta." And it's not a function, not literally.


Since many of the forums posting problems for other to solve, it will be nice to have the following options:

When I hit reply button, I will see a split screen with the area on top showing the original post (THE PROBLEM) and a my post area for my answer to the problem.

Right now, I have to have two browser windows open in order to see questions and replays.  Using the copy function does not work either as I cannot see the LeTex in its original format.

First please read the [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=20430]announcements regarding how to use the board[/url] and the [url=http://www.mathlinks.ro/Forum/faq.php]FAQ[/url].

You will thus find out that in order to get the latex formula of a post, it suffice to click (or hold the mouse above a) on the $\LaTeX$ (not L[b]e[/b]TeX ;) ) text. Further on the posts in that topic are presented in a topic review window below the post frame.

PS Not all people have 1600x1200 resolutions to be able too see two posting fields in one place ;)

[quote="zmli"]I agree with JLI and think it is a good suggestion.  The ability to copy Latex is nice, but requires extra step.

Screen resolution is really not the issue.  Your preview posting option actually takes more screen space than the split screen suggested by JLI.

If you are interested in making the site more useful, then, JLI's suggestion is a good one as it reduces the number of extra clicks.[/quote]Posting preview is not presented by default on the reply page. If you were referring to "posting [b]review[/b]" then that also contains the first post (and the last which is in most cases the one to which you are replying to).

I will analyze your ideas, but for now, I don't think it's applicable. However you put it, there will be an extra step.

[quote="paganinio"]if you really want the screen space to be utilized as much as possible, there would be 50% of the content cut off.
but I don't care.[/quote]We are speaking about posting reply format. What exactly could be taken off?
You must realize that we have some templates here, and if you don't need a certain info at one point, that doesn't mean nobody does.

In plus I really cannot grasp what  jli and zmli are asking. Maybe if they make a photop picture and attach it here, because otherwize it seems to be that they already have the tools to do what they like, but "there's an extra click involved".

For the record, there will always be an extra click involved, even if not for you, but for someone else ;)


calculus
integration
function
trigonometry
rotation
real analysis
real analysis unsolved

Let $ \Psi(x)$ be a positive even function decaying rapidly enough so that the following integral converges, then what is the value of:
\[ \int_{\minus{}\infty}^{\infty}\int_{\minus{}\infty}^{\infty}\Psi(s)\Psi(t)\sin\big[x(s\plus{}t)\big](s\minus{}t)^{2n}dsdt;\quad x\in\mathbb{R},\quad n\equal{}0,1,2,\cdots\]
I'm pretty sure it's zero.  My approach is since the integrand appears symmetrical along the axis $ y\equal{}\minus{}x$,  rotate the axes by $ \pi/4$ and integrate it over the new variables:
$ \displaystyle u\equal{}\frac{s}{\sqrt{2}}\plus{}\frac{t}{\sqrt{2}}$ and
$ \displaystyle v\equal{}\minus{}\frac{s}{\sqrt{2}}\plus{}\frac{t}{\sqrt{2}}$
then:
$ \displaystyle s\equal{}\frac{u}{\sqrt{2}}\minus{}\frac{v}{\sqrt{2}}$ and

$ \displaystyle t\equal{}\frac{u}{\sqrt{2}}\plus{}\frac{v}{\sqrt{2}}$
The integral in the new variable is then:\[ \int_{\minus{}\infty}^{\infty}\int_{\minus{}\infty}^{\infty}
(1)\Psi(s)\Psi(t)\sin\big[xu\sqrt{2}\big](\minus{}\sqrt{2}v)^{2n}dudv\] where I left the $ s$ and $ t$ in to make is easier to read.  Then I'm left with two even functions, a term always positive $ (\minus{}\sqrt{2}v)^{2n}$, the Jacobian which is the $ 1$ term, and the $ \sin(xu\sqrt{2})$ which is an odd function of u which would then make the integral zero (I think).   You guys agree?  Is there an easier way to evaluate this integral?

Thanks,
Shaw

I am completely confused since \[ \Psi(s)\Psi(t)\sin\big[x(s\plus{}t)\big](s\minus{}t)^{2n}\equal{}\minus{}\Psi(\minus{}s)\Psi(\minus{}t)\sin\big[x(\minus{}s\minus{}t)\big](\minus{}s\plus{}t)^{2n}.\]
Hence it is not surprise that the integral is zero.

Ok.  That makes sense then.   How about the rest of his paper:

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=208424]Zhang's proposed proof of the Riemann Hypothesis[/url]

Guess I'd like to know where the problem is if there is a problem which there must be a problem since no one's talking about it.  Anyway, it's an interesting subject for me.


trigonometry
algebra
polynomial
algebra unsolved

I can't even find $x_5$ in a nice form...

I don't know :(
A teacher give me that one,and he didn't have the solution.too
I have tried use the trig subbmition, but also failed :(

By using straightforward calculation, the desired term is 1.8793852415718...
(I have more decimals, but I suspect they're not accurate.)

I noticed that the higher terms approach a number, and that this number is approached even if the first and second terms are different (I'm not saying that unconditionally).

By using my conjecture that the terms quickly approach a number, I can substitute x for x(n-1), x(n), and x(n+1).
I eventually reach the polynomial equation (x^4)-3x-1=0.
One of its roots is (u/2)+(2/u) where u=(4+(4)(sqroot3)i)^(1/3).
(This is real and is 1.8793852415718...)
So, this the term x(infinity) and is another approximation for the 2005th term.

I have just come up with some observations and approximations.  I will be [u]very[/u] suprised if anyone arrives at a symbolic result.


The value of $ [x]$ is the greatest integer less than or equal to $ x$.
What is the arithmetic mean of the 10 members of the set \[ \left \{[\minus{}\pi],\left [\minus{}\frac{1}{2}\right ],[0],\left [\frac{1}{2}\right ],[0.689],\left [\frac{\pi}{4}\right ],\left [\frac{\pi}{3}\right ],[2],\left [\sqrt{5}\right ],[\pi]\right \}?\] Express your answer as a common fraction.

This reduces to finding the mean of
$ \{\minus{}4,\minus{}1,0,0,0,0,1,2,2,3\}$.

$ \{\minus{}4, \minus{}1, 0, 0, 0, 0, 1, 2, 2, 3\}$

To find the mean of these numbers, we add them up, then divide by the total number of numbers. Like this:

Step 1: $ \minus{}4 \plus{} \minus{}1 \plus{} 1 \plus{} 2 \plus{} 2 \plus{} 3 \equal{} 3$

We can ignore all those zero's when we add, because they won't make a difference on the sum.

Step 2: We count up the number of numbers(in this step, make sure you count the zero's!) which is 10. So we divide the sum by 10 $ \implies \boxed {\frac{3}{10}}$


quadratics
number theory
prime factorization
algebra

We use * as a multiplication sign if needed.
We use / or ÷ as a division sign.
x1/2 or sqrt(x) will be used to denote the square root of x.

To improve your score you can practice problems on the
Math Drills Home Page

Thank you for taking this test.
You can also select a subset from the displayed problems to work with.
1. The quantity 160 is equal to:

a. 	0
b. 	16
c. 	4
d. 	1
e. 	Undefined

2. The prime factorization of 198 is equal to one of following:

Which one?

a. 	2  *  32  *  11
b. 	22  *  32  *  11
c. 	2  *  33  *  11
d. 	22  *  32  *  112
e. 	2  *  32  *  5  *  11

3. Simplify completely the following expression and write it without negative exponents.

3 x-11 y-5

5 x-6 y-11

The answer is one of the following

		3 x5
	  a)
		5 y6

		3 y6
	  b)
		5 x5

		5 y6
	  c)
		3 x5

		3 y5
	  d)
		5 x5

		3 y6
	  e)
		5 x4

Which One?

a.
b.
c.
d.
e.

4. The quantity
- 2 w2 - w + 3   at   w = -2
is equal to:
a. 	-4
b. 	-5
c. 	-3
d. 	0
e. 	-6

5. After performing the indicated operations and simplifying, this expression

y2 -6 y + 8 		4
	  -
y2 - 16 		y + 4

is equal to one of the following:

		y - 6
	  a)
		y + 4

		y - 6
	  b)
		y - 4

		y + 6
	  c)
		y + 4

		y + 6
	  d)
		y - 4

		y - 4
	  e)
		y + 4

 6. After performing the indicated operations and simplifying, this expression

5 		w + 4 		10 w
	  +   		  +
w - 2 		w - 6 		w2 - 8 w + 12

is equal to one of the following:

		w - 19
	  a)
		w - 6

		w + 20
	  b)
		w - 6

		w + 19
	  c)
		w - 6

		19 w + 1
	  d)
		w + 6

		w + 19
	  e)
		w - 2

Which One?
a.
b.
c.
d.
e.

7. The equation of a line in general form is given by:

2 x + 3 y + 2 =0 .

Find the x -intercept of that line.

a. 	-2
b. 	0
c. 	-1
d. 	-3
e. 	1

8. What is the equation of the given line below ?

a. 	y=-2 x + 2
b. 	y=-3 x + 2
c. 	y=-4 x + 2
d. 	y=-2 x + 3
e. 	y=-2 x + 1

9. If for a quadratic equation of the form

a x2 + b x + c = 0 , one has b2 - 4 a c > 0

where a,b,c are real numbers, then the equation has:

a. 	Exactly two distinct real solutions
b. 	Exactly two distinct complex non-real solutions
c. 	Only one real solution
d. 	Only one complex non-real solution
e. 	One real solution and one complex non-real solution

10. The quantity (2 + 4 sqrt(2))2 equal to one of the following:

a. 	36
b. 	36 + 8 sqrt(2)
c. 	36 + 20 sqrt(2)
d. 	36 + 16 sqrt(2)
e. 	6 + 20 sqrt(2)

[quote="sheepwarrior"]???[/quote]

I second that :D

[quote="sheepwarrior"]???[/quote]

I triple that

None of these actually have the correct answers, do they?


Let $ ABCDE$ be a convex pentagon such that $ \angle BAC = \angle CAD = \angle DAE=\alpha$ and $ \angle ABC = \angle ACD = \angle ADE=\beta$. The diagonals $ BD$
and $ CE$ meet at $ P$.
Prove that the line $ AP$ bisects the side $ CD$.

It is ISL 2006 G3, see here: [url]http://www.mathlinks.ro/Forum/viewtopic.php?t=130813[/url]


I'm confused as to the solution, since I approached the problem differently and came up with a different solution:

[hide]Call the point at which line DE intersects line BC P. Call the point at which line DE intersects line AC Q, and call the point at which line AD intersects BC R. Triangle PQC is a 30-60-90 triangle. Since AR is the altitude of a equilateral triangle, of side length 4, AR has length 2 radical 3.

We can find the area of ARPQ, the common region, by subtracting the area of triangle PQC from triangle ARC. QC has length 4-2 radical 3, and PQ has length radical 3*(4 - 2 radical 3) = 4 radical 3 - 6. Area of triangle ARC is

(2 radical 3 * 2)/2 = 2 radical 3. Area of triangle PQC is :

( 4 - 2 radical 3) * ( 4 radical 3 - 6) /2=8 radical 3 - 12.

Area of ARPQ = 2 radical 3 - ( 8 radical 3 - 12) = 12 - 6 radical 3.

Problem is, this answer is half of the answer given in the solution book, which is 24 - 12 radical 3, which is obtained by finding the area of the triangles ARP and AQP. I'm pretty sure that my method is valid, but if anyone can find why, please reply. [/hide]

Your approach is fine, but you have an arithmetic error.  See if you can find it!


[u][b]Prove[/b][/u]
$f(y) = \frac{y}{3y-2}$
Prove (or disprove) that $f(N) = N/(10^{D})$ where $D$ is the number of digits of the integer $N$ only when $N = 333....334$.

[u][b]List[/b][/u]
List all the primes, $p$, where $2^{p}+p^{2}$ is also a PRIME.

[u][b]Find[/b][/u]
The two sides of an right triangle are $x$ and $y$; the hypotenuse being equal to $1$ unit. Find the maximum possible value of $1+x+y+xy$. Please, do NOT use calculus (maxima/minima). Then there is no fun in attempting this.

[quote="t0rajir0u"]The third's already been posted.

[hide="Find"] We have $x^{2}+y^{2}= 1$.  We also have

$2 \sqrt{ \frac{x^{2}+y^{2}}{2}}+\frac{x^{2}+y^{2}}{2}+1 = \boxed{ \sqrt{2}+\frac{3}{2}}\ge x+y+xy+1$

By QM-AM and AM-GM.  The maximum occurs when $x = y = \frac{ \sqrt{2}}{2}$.  [/hide]
Edit:  [hide="List"] $2^{3}+3^{2}= 17$.  For $p > 3$ we have $2^{p}+p^{2}\equiv 0 \bmod 3$. [/hide][/quote]

How to prove this...
[hide]For $p > 3$ we have $2^{p}+p^{2}\equiv 0 \bmod 3$[/hide]

Here is how you prove it:
[hide] $p^{2}$ can only be congruent to $1\bmod 3$ or $0\bmod 3$.  Only when $p$ is divisible by 3 is when $p^{2}$ is congruent to $0\bmod 3$.  However, since $p$ is prime, there is no such $p$ that is divisible by 3 and greater than 3.  Thus, $p^{2}\equiv1\bmod 3$.  Now, notice that $2^{p}\equiv1\bmod 3$ when p is even and $2^{p}\equiv2\bmod 3$ when p is odd.  Now, notice that above 3, since $p$ is prime, $p$ must always be odd.  Thus, $2^{p}\equiv2\bmod 3$ and $p^{2}\equiv1\bmod 3$ when $p$ is prime and above 3.  Thus, $2^{p}+p^{2}\equiv3\bmod 3\equiv 0\bmod 3$ [/hide]

[quote="xxxyyyy"][u][b]List[/b][/u]
List all the primes, $p$, where $2^{p}+p^{2}$ is also a PRIME.[/quote]

I remember a problem from a lecture I once did:

(Leningrad 1984) Find all primes a and b such that $a^{b}+b^{a}$ is also a prime.

[hide]Case 1: Assume a and b are both even. This means they're both 2, as 2 is the only even prime. But $2^{2}+2^{2}$ is not prime. Contradiction.

Case 2: Assume a and b are both odd. But an odd to an odd power is odd, and thus summing two of them will produce an even number. The only way the result can be prime is if it's 2, but that means a and b are both 1, which are not prime. Contradiction.

Case 3: Assume, without loss of generality, a is even and b is odd. Then, $a = 2$. Now we're at our original problem. $2^{b}+b^{2}$. Consider this mod 3. This is equivalent to $(-1)^{b}+b^{2}$, and since b is odd, the $(-1)^{b}$ is always -1.

So we are left with $-1+b^{2}\mod 3$. All squares are 0 or 1 mod 3.

Sub-Case 1: $b^{2}= 1 \mod 3$. Then $-1+b^{2}= 0 \mod 3$, which means $a^{b}+b^{a}$ is not a prime (as it's divisible by 3).

Sub-Case 2: $b^{2}= 0 \mod 3$. Then b is divisible by 3, and the only possible prime b is $\boxed{3}$.[/hide]

[quote="xxxyyyy"][u][b]Prove[/b][/u]
$f(y) = \frac{y}{3y-2}$
Prove (or disprove) that $f(N) = N/(10^{D})$ where $D$ is the number of digits of the integer $N$ only when $N = 333....334$. [/quote]
[hide="Easy"]
\[\frac{N}{3N-2}= \frac{N}{10^{D}}\implies N= \frac{10^{D}+2}{3}, \ \ D \geq 1\]
So we have $N$ as a function of $D$, with a range only of integers $33...34$. QED. [/hide]


Prove that for every integer n>2 there exists positive integers p,q such that n^2+q^2=p^2.

i dont no, i'm also getting same answer as pierre.

@ rust- yours is wrong i think. for example it doesnt work for n=8

the problem is actually very easy. i dunno why you guys think hard. the problem statement actually says that for every n>2 n^2 can be expressed as a difference of two positive squares. its obvious for odd n, and for even n, n^2=4m=(m+1)^2+(m-1)^2. :P


geometry
geometric transformation
reflection
ratio
limit

Maybe it's the same effect as:

You look at your reflection in a spoon (the concave side). First it is the right side up and small, but it gets larger and larger until it finally flips over and gets smaller and smaller again. Do I have it backwards?

I have glasses and I tried that, but I didn't notice anything.

Has anybody got an answer to this 'paradox'. Should I post mine!!!

Your answer? You already know the answer?!

[quote="Rushil"]A short-dighted person takes off his glasses and observes a fixed object through them, while moving the glasses away from his eyes. He is surprised to see that at first, the object looks smaller and smaller, but then becomes larger and larger. What is the reason for this???[/quote]Okay, here's my bit.

Magnification is the ratio of the height of an image to the height of its object, $\dfrac{h_i}{h_o} = M$

Because of the wonders of Euclidean geometry (vertical angles are congruent),

$\dfrac{h_i}{h_o} =\dfrac{s_i}{s_o}$

From Gauss's lens equation we know $\dfrac{1}{s_o}+\dfrac{1}{s_i}=\dfrac{1}{f}$

From which we derive, $s_i =\dfrac{s_of}{s_o-f}$

It follows then that $\dfrac{s_i}{s_o}=\dfrac{f}{s_o-f}$

The thing about a concave lens, used to correct nearsightedness, is that its focal length is negative. This implies the magnification though inverted with respect to a convex lens will never be undefined, because $s_o-f$ is always positive.

To begin with our myopic friend views an object such that $\lim_{s_o\to \infty}M = 0$.

As he moves his glasses closer to the object, $\lim_{s_o\to 0}M = -1$. (Again, the minus sign occurs because the image is inverted with respect to that formed by a convex lens.)

The point though is that as the lens gets closer to the object the image gets bigger.


[hide="evident step"]$x-\sqrt{x^{2}-3x-12}=3k, \ \ k\in Z$[/hide]
[hide="uderstandable step"]Consider $k$ as a parameter[/hide]
[hide="solution"]From the above we have
$x\ge 3k$  and $(x-3k)^{2}=x^{2}-3x-12$
The root $x=\frac{3k^{2}+4}{2k-1}$ of this equation we have to chek whether or not it satisfies the restriction $x\ge 3k$.
This gives $\frac{k^{2}-k-\frac{4}{3}}{k-\frac{1}{2}}\le 0$
If $k\ge 2$ or $k\le-1$ the numerator of the fraction is evidently positive. This gives all possible $k=1$  and all negative numbers[/hide]


ratio
geometry
3D geometry
geometric sequence
arithmetic sequence

actually, all these problems pratik (ragnarok23) has been posting recently are from our mock.

pratik and i (eli, the kid from summer camp, with the long hair, and ultamite ninja skilllllzzzz) wrote a mock (with sprint and targets) that we pwned our math class with! go us. and i wrote this problem btw.  i solved it less mathematically. =[. preet wins.
but actually, preet, i think youre wrong. actually im like 80% positive your wrong. and the problem stated r does not equal zero. so no arguing.

Problem:
[quote]where a does not equal zero and r does not equal zero  [/quote]

Preet:
[quote]If r=0 then the answer is 45. This problem probably needs to say "all the terms in the arithmetic sequence are positive" or something[/quote]

Solution:

[hide]but i did it like this.
so you have the first sequence.

a, ar, ar^2...
and the second:
0, x, 2x...

so a+0=1. so a=1.

so the first sequence is...
1, r, r^2...

so now, x+r=1 and 2x+r^2=2.  So...
2x+2r=2 and 2x+r^2=2.  So r^2=2r.  Since r does not equal 0, r=2.  therefore, x=-1.

so the first sequence is 1, 2, 4, 8...
and the 2nd is 0, -1, -2, -3...

so you then have 2^0+2^1...2^9=1023
and -1-2...-9= -((9)(10))/2=-45.
1023-45=[b]978[/b][/hide][/b][/quote]

Yeah but most of the problems were "borrowed" from other competitions.  :roll:

The formula for the sum of a finite geometric sequence is so easy to derive I do it everytime I see it.
So I was off by 1...

[quote="Walk Around The River"]pratik. mine werent borrowed. only yours. because youre not cool like me.  :ninja:[/quote]
You made like 3 problems.

[quote="bpms"][quote="ragnarok23"][quote="bpms"]Let me but in for a second. Out of these problems, I think that 5-7 of them were great. These plus other mediocre ones would've made a great mock. I'm trying to make mine as unique and distinct as possible but it is hard.
For the target I have one semi unique, one that I thought was unique, but I was beat to, and a couple others...[/quote]
Thanks. Which problems did you think were good? The circle one and the laser beam one, but what else?
I have a nice front cover I made for my mock. If you would like the code, pm me.
[quote="Walk Around The River"]i made like 10.

settle this over a ninja fight  :ninja:  :ninja:  on monday?[/quote]
You better believe it.[/quote]
The circle, The laser beam, this one and the cube one I enjoyed in particular. You still haven't confirmed the laser beam one...[/quote]
I did. I believe your answer is incorrect, but I may be stupid. It's probably the second.

[quote="nutz_for2.718281828"][quote="bpms"][quote="ragnarok23"]Given a geometric sequence with the first term $a$ and common ratio $r$, where $a\ne 0$ and $r\ne 0$ and an arithmetic sequence with the first term 0, a third sequence 1,1,2... is formed by adding the corresponding terms of the 2 given sequences. Find the sum of the first 10 terms of the third sequence.[/quote]
Why don't you make a mock :D
[hide]
Arithmetic:
0, x, 2x, ...
Geo:
a, ar, a$r^{2}$

So a=1.
Plugging we get geo to be
1, r, $r^{2}$
$r+x=1$
$r^{2}+2x=2$
$x=1-r$
$r^{2}-2r+2=2$
$r=2,0$
If r=0 then x=1.
If r=2 then x=-1
Anyways, we are looking for $(r+r^{2}+...+r^{9})+(x+2x+...+9x)$
This is equal to with the $\frac{r^{10}-r}{r-1}+45x$.
So if r=2 then the answer is 977.
If r=0 then the answer is 45. This problem probably needs to say "all the terms in the arithmetic sequence are positive" or something.[/hide][/quote]
It clearly says that r is not equal to 0.[/quote]
Did you not just see the above discussion? This has already been established.


hmmm....during my 8th grade year, at regional countdown, the guy handed out tootsie rolls before the countdown in an "unofficial" countdown. I solved nearly all the probs, got a lot of tootsie rolls, and got all woozy. I then did really well on the countdown.

At states, I brought my own tootsie rolls. I proceeded to do well on countdown.

At nats, I had no tootsie rolls. Some of you know what happened; some of you don't, and that's fine by me.

I will delete this thread in a couple of days but wanted to say: sugar is good. On an actual serious note, I would suggest bringing some form of high-sugar snack right before countdown. I don't know if it helped, but it seemed to work for me while I had it.

Ummm...Okay...
"Low blood sugar is baaaaaaad" :) , but too high and you might get seizures, especially when you're under a lot of stress such as competitions and such. mmkay? :lol:

In this vein, MCrawford's alma mater would give all their students grape juice right before a test for essentially this reason (back in my day, they may have stopped by the time MCrawford came around).  Does anyone know if the usefulness of a shot of sugar to quick thinking is myth or fact?

I'm pretty sure it's a proven fact. It's like using incredulously small amount of speed, I think. The sugar literally makes you a bit "high," speeding up your nerve cells.

Who (or What) is "alma mater"?

I think it does have a bit of truth to it.

``Alma mater'' is ``former educational institution,'' roughly.

Mmm...this is getting Twilight Zoney :)

I'll have to try that with my next test.


calculus
integration
function
calculus computations

evaluate: integral from 4 to 6 dx/{(x-3)[(x^2 - 6x + 8)^1/2]}

After easy transformation
\[ I \equal{} \int_{4}^{6}{\frac {\mathrm{d}x}{\left(x \minus{} 3\right) \sqrt {x^{2} \minus{} 6x \plus{} 8}}} \equal{} \int_{4}^{6}{\frac {\mathrm{d}x}{\left(x \minus{} 3\right) \sqrt {\left(x \minus{} 3\right)^{2} \minus{} 1}}} \textnormal{,}
\]
using appropriate substitution
\[ t \equal{} \left(x \minus{} 3\right)^{2} \minus{} 1 \textnormal{, } \mathrm{d}t \equal{} 2\left(x \minus{} 3\right) \mathrm{d}x
\]
we get
\[ I \equal{} \int_{0}^{8}{\frac {\mathrm{d}t}{2 \left(t \plus{} 1\right) \sqrt {t}}} \textnormal{.}
\]
Another substitution
\[ m \equal{} \sqrt {t} \textnormal{, } \mathrm{d}m \equal{} \frac {1}{2 \sqrt {t}} \mathrm{d}t \equal{} \frac {1}{2m} \mathrm{d}t
\]
gives
\[ I \equal{} \int_{0}^{2\sqrt {2}}{\frac {\mathrm{d}m}{m^{2} \plus{} 1}} \equal{} \arctan{2\sqrt {2}} \minus{} \arctan{0} \equal{} \arctan{2\sqrt {2}} \textnormal{,}
\]
and we're done. $ \checkmark$

We are not done, since this is an improper integral (singularity at x = 4) and you have to show first that it converges. You have only shown that, if it converges, then its value is $ \arctan(2 \sqrt{2})$.

Hey,
[quote="Carcul"]
We are not done, since this is an improper integral (singularity at x = 4) and you have to show first that it converges. You have only shown that, if it converges, then its value is $ \arctan(2 \sqrt {2})$.
[/quote]
Does that mean an improper integral is defined as a definite integral where one of the limits is either $ \pm \infty$ or a discontinuity (ie. the function is undefined at that point) of the function we are integrating?

In addition, does that mean for each improper integral we need to verify it converges?
Lastly, is the purpose of showing the improper integral converges to show that when the improper integral is evaluated it indeed is the evaluated value?

Thanks,

-[url=http://www.artofproblemsolving.com/Forum/profile.php?mode=viewprofile&u=33715]APSStudent[/url]

[quote="APSStudent"]Does that mean an improper integral is defined as a definite integral where one of the limits is either  or a discontinuity (ie. the function is undefined at that point) of the function we are integrating?[/quote]

We have an improper integral when we have one (or both) of the following two situations: (a) the interval of integration is not limited, i.e., the interval of integration is of the form $ [a, \plus{} \infty)$, $ (\minus{} \infty, a]$, or $ (\minus{} \infty, \plus{} \infty)$, $ a \in \mathbb{R}$; (b) the function to be integrated is not defined in some point belonging to the interval of integration.

[quote="APSStudent"]In addition, does that mean for each improper integral we need to verify it converges?[/quote]

Yes, otherwise we might be wasting time with an integral for which it does not make sense to atribute a real value, according to an appropriate definition of convergence of improper integrals.

[quote="APSStudent"]Lastly, is the purpose of showing the improper integral converges to show that when the improper integral is evaluated it indeed is the evaluated value?[/quote]

Yes, that's precisely the point.

a bit easier:

$ \int_{4}^{6}{\frac {\mathrm{d}x}{\left(x \minus{} 3\right) \sqrt {\left(x \minus{} 3\right)^{2} \minus{} 1}}} \equal{} arcsec 3 \minus{} arcsec 1 \equal{} arcsec 3$

which is equivalent to $ \arctan 2\sqrt {2}$


Let $a_{i}$ $i=1,2,...,k$ be coprime integers. let $A_{i}=\frac{\prod_{j=1}^{n}a_{j}}{a_{i}}$
find
a)$g(A_{1},A_{2},...,A_{k})$       (i.e. the largest natural number not representable as a nonnegative linear combination of the A_i 's)
b)$n(A_{1},A_{2},...,A_{k})$       (The number of natural numbers not representable)

giving a hint in a)
the answer is
$(n-1)a_{1}a_{2}\cdots a_{k}-\sum_{i=1}^{k}A_{i}$


.Let ADEF, a Diamond(lozenge,rhomb, I am not sure how to write that in english)such that D is in the side AB and F in the side BC.Calculate the perimeter of the diamond

ADEF is a rhomb.....

That's ok, we'll help you. So there's abcd, and then there's adef, which is a rhombus inside abcd? Is that right?


just go like this
$\frac{1}{3}(x^{3}+1+1) \geq x$
$\frac{1}{3}(y^{3}+1+1) \geq y$
$\frac{2}{3}(x^{3}+y^{3}+1) \geq 2xy$
by adding up those inequalities above , we get the desired result

Please do not revive old topics.  This one is 4 years old.


Solutions $f(x)=1$, and $f(x)=\frac{c}{c+x}$ where $c$ is a positive constant.

Suppose that for some $x$, $f(x)>1$.  Then setting $y=\frac{x}{f(x)-1}$ we see that $yf(x)=x+y=\frac{xf(x)}{f(x)-1}$, so, substituting in the original equation, $f(x)f\left(\frac{xf(x)}{f(x)-1}\right)=f\left(\frac{xf(x)}{f(x)-1}\right)$, a contradiction since $f(x)>1$.  So $f(x)\leq 1$ always.  Then $f$ is nondecreasing since $f(x+y)=f(x)f(yf(x))\geq f(x)$ for all positive $y$.

Suppose there exist reals $a<b$ such that $f(a)=f(b)$.  Then $f(a+y)=f(b+y)$ for all $y$, so $f(x)$ is periodic with period $b-a$ for all $x\geq a$.  As $f$ is nondecreasing, it must in fact be constant.  So $f(x)=l$, say, for all sufficiently large $x$; then substituting sufficiently large $x$, $y$ in the original equation gives $l^2=l$, so $l=1$.  Then, from the original equation, $f(y)=f(x+y)$ for all sufficiently large $x$, so $f(y)=1$ for all $y$.

Otherwise $f$ is strictly decreasing, and injective.  Substituting $z=yf(x)$ in the original equation, $f(x+\frac{z}{f(x)})=f(x)f(z)=f(z+\frac{x}{f(z)})$ for all positive reals $x$, $z$, and so $x+\frac{z}{f(x)}=z+\frac{x}{f(z)}$, and hence $\frac{z}{\frac{1}{f(z)}-1}=\frac{x}{\frac{1}{f(x)}-1}=c$, where $c$ is some constant.  So the function satisfies $x=c(\frac{1}{f(x)}-1)$, or $f(x)=\frac{c}{c+x}$, for all $x$.  Substituting back shows that this works for all positive $c$.


The world's tallest mountain, measured from its base, is Mauna Loa.  Only 13,680 feet of this 33,356-foot mountain are visible above the water; the rest of the mountain is under the Pacific Ocean.  Of the seven submersible vehicles in the graph, how many could get to the bottom of Mauna Loa on the ocean floor?
[metapost=3d2b7f1c008c5864c8e965fd3f595da5] save u,w;

 % Set the value of u to scale the entire image
 u:=0.7/6cm;
 w:=0.4cm;

 for i=0 upto 6:
 draw (0,6u*i)--(14w,6u*i);
 endfor;

 draw origin--(0,36u);
 fill (w,0)--(2w,0)--(2w,3u)--(w,3u)--cycle withcolor (0.6*white);
 draw (w,0)--(2w,0)--(2w,3u)--(w,3u)--cycle;
 fill (3w,0)--(4w,0)--(4w,3.3u)--(3w,3.3u)--cycle withcolor (0.6*white);
 draw (3w,0)--(4w,0)--(4w,3.3u)--(3w,3.3u)--cycle;
 fill (5w,0)--(6w,0)--(6w,9.775u)--(5w,9.775u)--cycle withcolor (0.6*white);
 draw (5w,0)--(6w,0)--(6w,9.775u)--(5w,9.775u)--cycle;
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 draw (7w,0)--(8w,0)--(8w,13u)--(7w,13u)--cycle;
 fill (9w,0)--(10w,0)--(10w,19.7u)--(9w,19.7u)--cycle withcolor (0.6*white);
 draw (9w,0)--(10w,0)--(10w,19.7u)--(9w,19.7u)--cycle;
 fill (11w,0)--(12w,0)--(12w,19.7u)--(11w,19.7u)--cycle withcolor (0.6*white);
 draw (11w,0)--(12w,0)--(12w,19.7u)--(11w,19.7u)--cycle;
 fill (13w,0)--(14w,0)--(14w,35.79u)--(13w,35.79u)--cycle withcolor (0.6*white);
 draw (13w,0)--(14w,0)--(14w,35.79u)--(13w,35.79u)--cycle;

 label.lft(btex $ 6000$ etex,(0,6u));
 label.lft(btex $ 12000$ etex,12u*up);
 label.lft(btex $ 18000$ etex,18u*up);
 label.lft(btex $ 24000$ etex,24u*up);
 label.lft(btex $ 30000$ etex,30u*up);
 label.lft(btex $ 36000$ etex,36u*up);

 fill bbox thelabel(btex \\tiny $ 3000$ etex,(1.5w,5u)) withcolor white;
 label(btex \\tiny $ 3000$ etex,(1.5w,5u));
 fill bbox thelabel(btex \\tiny $ 3300$ etex,(3.5w,5u)) withcolor white;
 label(btex \\tiny $ 3300$ etex,(3.5w,5u));
 fill bbox thelabel(btex \\tiny $ 9775$ etex,(5.5w,12u)) withcolor white;
 label(btex \\tiny $ 9775$ etex,(5.5w,12u));
 fill bbox thelabel(btex \\tiny $ 13000$ etex,(7.5w,15u)) withcolor white;
 label(btex \\tiny $ 13000$ etex,(7.5w,15u));
 fill bbox thelabel(btex \\tiny $ 19700$ etex,(9.5w,22u)) withcolor white;
 label(btex \\tiny $ 19700$ etex,(9.5w,22u));
 fill bbox thelabel(btex \\tiny $ 19700$ etex,(11.5w,22u)) withcolor white;
 label(btex \\tiny $ 19700$ etex,(11.5w,22u));
 fill bbox thelabel(btex \\tiny $ 35790$ etex,(13.5w,39u)) withcolor white;
 label(btex \\tiny $ 35790$ etex,(13.5w,38u));

 label(btex \\textbf{Greatest Depth Attained by Submersibles} etex,(7w,42u));
 label(btex Number of Feet etex,(0,0)) rotated 90 shifted (-4w,18u);

 label(btex Bathysphere etex,(0,0)) rotated 30 shifted (-w,-6u);
 label(btex DSRV-1 etex,(0,0)) rotated 30 shifted (2.2w,-4u);
 label(btex Cyana etex,(0,0)) rotated 30 shifted (4.5w,-4u);
 label(btex Alvin etex,(0,0)) rotated 30 shifted (6.4w,-4u);
 label(btex Deep Tow etex,(0,0)) rotated 30 shifted (8w,-5.7u);
 label(btex Argo - Jason etex,(0,0)) rotated 30 shifted (9.5w,-6u);
 label(btex Trieste etex,(0,0)) rotated 30 shifted (12.9w,-4u);
[/metapost]

Fix the Tex please.  ANyone understand?

I think the answer was 3 or 4 though...

It is 3. It's sort of like hwo you always guess C because C is the 3rd choice!

EDIT-WOOT! 200th most posts in all of AOPS!


geometry
incenter
rectangle
circumcircle
perpendicular bisector
geometry proposed

The incircle of triangle $ABC$ touches the sides $BC,CA,AB$ at the points $D,E,F$, respectively. Let $K$ be a point on the side $BC$ and let $M$ be the point on the line segment $AK$ such that $AM=AE=AF$. Denote by $L,N$ the incenters of triangles $ABK,ACK$, respectively.

Prove that $K$ is the foot of the altitude from $A$ if and only if $DLMN$ is a square.

[hide="Remark"]This problem is slightly connected to [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=344774#p344774]GMB-IMAR 2005, Juniors, Problem 2[/url]
[/hide]
[i]Bogdan Enescu[/i]

Suppose we manage to prove that $K,D,L,M,N$ always lie on a circle $\omega$, and that $LN$ is the perpendicular bisector of $MD$ (i.e. $LM=LD,\ NM=ND$). The result would then follow: $DLMN$ is a rectangle iff $\angle DLM=\frac\pi 2\iff\angle DKM=\frac\pi 2\iff\ K$ is the foot of the $A$-altitude of $ABC$. On the other hand, since $LM=LD,NM=ND,\ DLMN$ is a rectangle iff it's a square.

Now let's prove the first assertions: $K,D,L,M,N$ lie on a circle $\omega$, and $LM=LD$ ($NM=ND$ follows by symmetry).

$L$ lies both on the perpendicular bisector of $FD$ and on that of $FM$, so it must be the circumcenter of $FMD$. This proves $LM=LD$. Now, $\angle DLM$ is twice the angle $\angle DFM$, and $\angle DNM$ is twice the angle $\angle DEM$. A simple angle chase shows that $\angle DFM+\angle DEM=\frac\pi 2$, so $\angle LMN=\angle LDN=\angle LKN=\frac\pi 2$, as desired.

[quote="Arne"]That's a very nice problem, Enescu. Why didn't you propose it for a contest? :huh:[/quote]
Thank you! Actually, the problem, slightly different formulated, was considered as a possible Romanian IMO proposal for 2006. Unfortunately, we discovered a problem in AMM (1998) which asked to prove that $\angle{LDN}=90^{\circ}$. Therefore I could not propose it to a contest like IMO or RMO.
By the way, I was inspired by [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=51230]this problem[/url].


[quote="rnwang2"]a) Find the sum of the factors of 240.
b) Find the product of the factors of 240.

Do not just list the factors and add/multiply.[/quote]
[hide="1"]You prime factorize 240 to get $2^{4}\cdot3\cdot5$. Then you take each prime factor and do $(2^{0}+2^{1}+2^{2}+2^{3}+2^{4})(3^{0}+3^{1})(5^{0}+5^{1})=744$.[/hide]

[hide="2"]Since 240=2^4*5*3, it has 20 factors. Every 2 factors has a product of 240, so the product of the factors is $240^{10}$[/hide]

[hide="a little more exact"]
$6.3403381*10^{2}3$[/hide]

jorian

[quote="rnwang2"]a) Find the sum of the factors of 240.
b) Find the product of the factors of 240.

Do not just list the factors and add/multiply.[/quote]

[hide="Solutions"]
[hide="Problem 1"]
$240=2^{4}\times3\times5$.

$(2^{0}+2^{1}+2^{2}+2^{3}+2^{4})(3^{0}+3^{1})(5^{0}+5^{1})=\fbox{744}$ as stated in MC Toolbox.
[/hide]
[hide="Problem 2"]
Again, use MC toolbox. By MC Toolbox, you multiply the factors out  (cant explain well).

Anyway, you find that each 'pair' of factors multiplies to $240$. So, the answer is $240^{10}=\boxed{6.3403381\times 10^{23}}$.[/hide][/hide]

People appreciate it when you don't post the exact same solution right after 2+ other people have posted it.


is x mody the same thing as x + ny where n is an integer?

if so then if z is congruent to x mody then is x congruent to z mody?
-because z = x + ny so x = z - ny, which can be expressed as x is congruent to z mody

So if A is congruent to B modM and P is congruent to Q modM then it is implied that A + P is congruent to B + Q modM and B + Q is congruent to A + P modM .....but.....

A = B + nM and P = Q + nM so B = A - nM and Q = P - nM, which means B is congruent to A modM and Q is congruent to P modM...so does this imply A + P is congruent to A + Q modM and B + P modM, which means B + Q is congruent to B + P modM and A + Q modM, which gives back A + P is congruent to B + Q modM and B + Q is congruent to A + P modM...if that's even right....

eh can someone check if my reasoning is correct? I'm confused...

oh yea...and can someone explain this: if A is congruent to B modM then A^c is congruent to B^c modM when c is a positive integer....and also....why can't c be a negative integer
btw: this comes from AoPs v1, pg 44

An integer in the form $ x\plus{}ny$ is equivalent to $ x$ modulo $ y$.

Your second statement is correct by symmetry.

If you read your statements, you'll see you repeated yourself at the end.  :wink:

But yes, if $ a \equiv b$ and $ b \equiv c$, by transitivity, $ a \equiv c$. So $ p\equiv q$ coupled with $ a\plus{}p \equiv b\plus{}q$ implies $ a\plus{}q \equiv b\plus{}p$.

For your last statement, note that taking an integer to a negative power results in a fraction. Fractions aren't well-defined in modular arithmetic. Example: $ 1 \equiv 4$ modulo $ 3$, but $ 1^{\minus{}1} \equal{} 1$ and $ 4^{\minus{}1} \equal{} \frac{1}{4}$, but those two are clearly not equivalent modulo $ 3$.

ok thanks...there's one thing I don't completely understand:

so when you write something like A is congruent to B modN...just the B is modded right? but just the A being modded would yield the same thing, effectively?...or is the [b]whole[/b] congruency in modM - but does this yield both sides being modded or does it yield either A being congruent to B modN or B being congruent to A modN...and what's the difference, if there is one, when both sides are modded rather than just one side

No number is "modded", technically. Whenever you say two numbers are equivalent/congruent modulo $ n$, they just have the same remainder when divided by $ n$.


trigonometry
geometry
trapezoid
perpendicular bisector
angle bisector
geometry proposed

Given $ \triangle{ABC}$ with $ AB\equal{}AC$ and $ \angle{BAC}\equal{}100^o$. $ P$ is a point inside the triangle. $ \angle{PAC}\equal{}\angle{PCA}\equal{}20^o$.

Find $ \angle{PBA}$.

[geogebra]28604b1d8f27023e25bf7dda4fa6ef9457c55542[/geogebra]

Let angle PBA be x. By Ceva's theorem, $ \frac{\sin x}{\sin(40\minus{}x)}\cdot\frac{\sin 20}{\sin 80}\equal{}1$. It is easy to check that $ x\equal{}30$ satisfy the given conditions.

Take Q symmetrical of P w.r.t. perpendicular bisector of BC, see that AQP is an equilateral triangle, hence PQBC is an isosceles trapezoid with BQ = PQ = CP, consequently BP is angle bisector of angle QBC and angle ABP = 30 degs.

Best regards,
sunken rock


linear algebra
matrix
geometry
geometric transformation
rotation
linear algebra unsolved

Maybe, its a trivial question, but

Is the matrix AB positive semi-definite if both square matrices A and B are positive semi-definite?

If not, maybe this is still true if both A and B are symmetric real positive semi-definite?

Thanks in advance!

Counterexample: $A=\begin{bmatrix}9&0\\0&1\end{bmatrix}$, $B=\begin{bmatrix}5&4\\4&5\end{bmatrix}$ ($B$ is similar to $A$ by 45-degree rotation)

Starting with $x=\begin{bmatrix}-1\\2\end{bmatrix}$, we have $Bx=\begin{bmatrix}3\\7\end{bmatrix}$ and $ABx=\begin{bmatrix}27\\7\end{bmatrix}$, so $\langle x,ABx\rangle=-13$.


If the equation $x^{3}+(a-1)\sqrt{3}x^{2}-6ax+b=0$ has three real roots, prove that $|b|\leq |a+1|^{3}$.

Let $f(x) = x^{3}+(a-1)\sqrt{3}x^{2}-6ax+b$

Let $A,B,C$ be the three roots (not necessarily distinct)

$f(x) = (x-A)(x-B)(x-C) = x^{3}-(A+B+C)x^{2}+(AB+BC+CA)x-ABC$

So we have the system of the equations:

$\begin{array}{l}A+B+C = \sqrt{3}(1-a) \\ AB+BC+CA =-6a \\ ABC = b \end{array}\}$

$(A+B+C)^{2}= A^{2}+B^{2}+C^{2}+2(AB+BC+CA) \Rightarrow $

$(\sqrt{3})^{2}(1-a)^{2}= A^{2}+B^{2}+C^{2}+2(-6a) \Rightarrow $

$A^{2}+B^{2}+C^{2}= 3(a^{2}-2a+1)+12a = 3(a^{2}+2a+1)\Rightarrow $

$A^{2}+B^{2}+C^{2}= 3(a+1)^{2}$

Set $x=A^{2},\ y=B^{2},\ z=C^{2}$. Then $x,y,z\geq 0$ and $\sqrt{xyz}=|ABC|$

The problem is simplified to the following:

[color=brown]Given that $x+y+z = 3(a+1)^{2}$ find the maximum value of $\sqrt{xyz}$,or, equivalently, the maximum value of $xyz$[/color]

It is well known that the three variables have to be equal $x_{o}=y_{o}=z_{o}=(a+1)^{2}\Rightarrow $

$x_{o}y_{o}z_{o}=(a+1)^{6}\Rightarrow \sqrt{x_{o}y_{o}z_{o}}=|a+1|^{3}\Rightarrow \sqrt{xyz}\leq |a+1|^{3}\Rightarrow $

$|b|=|ABC|\leq |a+1|^{3}$


Let $ f: \mathbb R\to\mathbb R$ be a function that satisfies the following properties: $ \forall\,x,y\in\mathbb R\implies f(x\plus{}y)\equal{}f(x)\plus{}f(y)$ and $ f$ is continuous at zero. Prove that $ f$ is continuous.

first $ f(0)\equal{}f(0\plus{}0)\equal{}f(0)\plus{}f(0)\equal{}2f(0) \to f(0)\equal{}0$

let $ \epsilon >0$

then we know since f cont at 0

$ \exists \delta >0$ s.t. $ |x| < \delta \to |f(x) | < \epsilon$

then for such an epsilon, we claim the same delta works for all real y

$ |x\minus{}y| < \delta \to |(x\minus{}y) \minus{} 0| < \delta \to |f(x\minus{}y)| \equal{} |f(x)\minus{}f(y)|< \epsilon$

so f is continuous at y.

[quote="Μαύρο"]Let $ f: \mathbb R\to\mathbb R$ be a function that satisfies the following properties: $ \forall\,x,y\in\mathbb R\implies f(x \plus{} y) \equal{} f(x) \plus{} f(y)$ and $ f$ is continuous at zero. Prove that $ f$ is continuous.[/quote]
and if $ f$ is continuous at nay point we will have $ f$ is continuous
and then we will find $ f(x)\equal{}kx$


function
algebra
domain
limit
calculus
calculus computations

Suppose that $ f$ is continuous on $ [0,1]$ and that $ f(q) \equal{} 0$ for all $ q$ rational in $ [0,1]$. Prove that $ f(x) \equal{} 0$ for all $ x \in [0,1]$.

Do you know what it means if I say the rationals are dense in [0, 1]? If so, then nothing more needs to be said here; just apply that and the definition of continuity. If not, you need to learn what that means!

Do you mean infinitely dense? Since you can pick a numerator as large as you like, and any number ranging from $ 0$ to $ \infty$ as your denominator.

Can you provide a rigorous solution using the rationals are dense concept? So far my solution involves the $ \epsilon\minus{}\delta$ method which is what I think my professor is expecting.

The language of "arbitrarily close" is $ \epsilon$-$ \delta$ if we quantify it. Saying that the rationals are dense is saying that we can find rational numbers arbitrarily close to any given number- that for any $ x$ and any $ \epsilon>0$, there is a rational $ r$ with $ |x\minus{}r|<\epsilon$. If $ \frac1N<\epsilon$, we can do this with something that has denominator $ N$.

One idea would be to use the sequential definition of continuity: a function $ f$ is continuous at $ a$ iff for every sequence $ x_n$ in the domain of $ f$ such that $ \lim x_n\equal{}a,$ the $ \lim f(x_n)\equal{}f(a).$

Now the version of density we want is this: for every real number $ a\in[0,1],$ there is a sequence $ q_n$ of rational numbers in $ [0,1]$ such that $ \lim q_n\equal{}a.$

We know that for every $ x\in\mathbb{R},\quad\exists {x_n}\in\mathbb{Q}$ such $ x_n\longrightarrow x$.
We have $ \forall x_n, f(x_n)=0$ so, \begin{align*}f(x)&=f\left(\lim_{n\to\infty}x_n\right)\\&=\lim_{n\to\infty}f(x_n)=0\end{align*} because $ f$ is continuous.

oo that's a very nice solution... much nicer and simpler than my attempts at doing this using $ \epsilon\minus{}\delta$ method


linear algebra
matrix
algebra
polynomial
linear algebra unsolved

Let $ A$ be a square matrix of order $ 2$, satisfying the condition $ A^2\equal{}A$. Prove that $ AX\minus{}XA\equal{}0$,($ X$ being a square matrix of $ 2$), it is necessary and sufficient that there exists a square matrix $ X_0$ of order $ 2$, such that $ X\equal{}AX_0\plus{}X_0A\minus{}X_0$.

I assume that $ A$ is defined over a field. The minimal polynomial of $ A$ devides $ X(X \minus{} 1)$, hence $ A$ is diagonalizable. Thus assume WLOG that $ A$ is diagonal and compute the expressions to get your result.


limit
integration
logarithms
real analysis
calculus
calculus computations

[hide]
Express the sum $ \sum_{k\equal{}1}^n \frac{n\minus{}k}{n^2\plus{}nk\plus{}2007}$ like a Riemann sum:
\[ \frac{1}{n}\sum_{k\equal{}1}^n \frac{1\minus{}\frac{k}{n}}{1\plus{}\frac{k}{n}\plus{}\frac{2007}{n^2}}\]
The problem is, we have this annoying $ \frac{2007}{n^2}$ term in the denominator. Let's figure out whether it's significant:
\[ \frac{1}{1\plus{}\frac{k}{n}\plus{}\frac{2007}{n^2}}\equal{}\frac{1}{1\plus{}\frac{k}{n}}\minus{}\frac{2007}{n^2(1\plus{}\frac{k}{n})(1\plus{}\frac{k}{n}\plus{}\frac{2007}{n^2})}\]
So the error term is
\[ \frac{2007}{n^3} \sum_{k\equal{}1}^n \frac{1\minus{}\frac{k}{n}}{(1\plus{}\frac{k}{n})(1\plus{}\frac{k}{n}\plus{}\frac{2007}{n^2})}\]
and each term in the summation is nonnegative and bounded above by 1. Therefore, the error term is between $ 0$ and $ \frac{2007}{n^2}$, so it tends to 0.

So we're only concerned with
\[ \lim_{n\rightarrow \infty} \frac{1}{n}\sum_{k\equal{}1}^n \frac{1\minus{}\frac{k}{n}}{1\plus{}\frac{k}{n}}\equal{}\int_0^1 \frac{1\minus{}t}{1\plus{}t}\ dt\]
\[ \equal{}\int_0^1 \left( \frac{2}{1\plus{}t}\minus{}1\right) \ dt\equal{}\boxed{2\ln 2 \minus{}1}\]
[/hide]


Can anyone help me with this problem or point me in the direction of somewhere that can, i wuld be extreamly grateful.

Q) The range of data to be accepted by the program has been extended to allow numbers to be grouped within parenthesis eg. 01,{01,0,{10,1,{0,1},{1}}},1

Design a Turing machine which will check for correctly formatted data. In this version the restriction of 0 being only represented by a single symbol need not be adhered to. The TM should check for correctly matched {and} but need not detect misplaced commas (as in, eg. {0,}).

You should assume:

i) the data is initially positioned on an otherwise blank tape (i.e. all spaces) with the TM's reading head above the first character.

ii) All commas are legally placed but {and} may not be

iii) the TM will end in one of the following states:

  VALID DATA

  TOO MANY '}'

  TOO MANY '{'

with the obvious interpretation.

Let's consider a simpler problem: your input is guaranteed to be a string composed only of ( and ).  You want to check to make sure that it's valid, i.e. for each ) it has a corresponding ( which precedes it and for each ( there is a corresponding ) which follows it.  You should think about how [i]you[/i] would check that such a string is valid.  Then abstract it to a Turing Machine.  Once you have an answer to that question, answering your own question is only trivially more complicated.

Incidentally, the names "TOO MANY '{'" and "TOO MANY '}'" are really bad: what do you call the string "}{"?  It doesn't have too many of either bracket, it just has them in the wrong order.  (In fact, this will probably be classified under "TOO MANY '}'" -- once you figure out how to do what you're doing, I suspect that will be its natural home.  Unless your TMs read right-to-left, of course :).)

hi thanks for your reply it's greatly apprieciated, can u please ellaborate slightly, it's just i'm no good whatsoever at computer science. i really don't understand what i'm supposed to do

I'm not doing your homework for you  :wink:   Rather, read what I wrote.  Start with a high-level description of what it means for such a string to be "proper."  (This is what I meant by, "You should think about how [i]you[/i] would check that such a string is valid.")  Note that this requires thought, but no knowledge of computer science.  After you've written a decent description of what your TM needs to do, then you can think about translating it into TM language.

So is it that the TM will only accept 1's and not 0's?

I have no idea what you're talking about.


geometry
3D geometry
algebra unsolved
algebra

$a,b$ are $N^$ such that $a=b$ and $(a-b)^4=a^3-b^3$. Prove that $9a-1$ is cube.
($9a-1=k^3$ where k is $N^$)

:blush:  &a& and &b& are different. Sorry. So.. any ideas (and solutions)  :idea:

Let $d=\gcd(a,b).$ We have then
\[ d^{4}\left( x-y\right) ^{4}=d^{3}\left( x^{3}-y^{3}\right) ,  \]
or
\[ d\left( x-y\right) ^{4}=x^{3}-y^{3},  \]
where $x=a/d$ and $y=b/d$.

It follows that $\left( x-y\right) ^{4}$ divides $x^{3}-y^{3}.$ Let $t=x-y.$
Then $t^{4}$ divides $t^{3}+3t^{2}y+3ty^{2},$ that is, $t^{3}$ divides $t^{2}+3ty+3y^{2},$ so $t=1$ or $t=3.$ (recall that $(x,y)=1$, so $(x-y,y)=(t,y)=1$).

If $t=3,$ then $27$ divides $9+9y+3y^{2},$ hence $3$ divides $y.$ Writing $y=3z$ we get that $3$ divides $1+3z+3z^{2},$ a contradiction.

It follows that $t=1,$ hence $x=y+1$ and $d=\left( x+1\right) ^{3}-y^{3}.$
The solutions of the given equation are $a=\left( y+1\right) \left( \left( y+1\right) ^{3}-y^{3}\right)$ and $b=y\left( \left( y+1\right) ^{3}-y^{3}\right) ,$ with $y\in N.$

It is trivial to check that $9a-1=\left( 3y+2\right) ^{3},$ therefore a perfect cube.

thank you enescu. The solution you gave is beautiful

Find all integer numbers $(a,b,c)$ such that
$a^3+6ab+20, b^3+6bc+20, c^3+6ca+20$ are all perfect cubes
($a^3+6bc+20=k^3$, $k$ is $Z^$)

There is no one interested in this problem, but i think it`s beautiful  :?

Firstly - don't post three consecutive posts - there's an Edit button

Secondly - we don't post two problems in one topic - so there are greater chance that you'll receive answer if you just repost your question into new topic


counting derangement AoPSwiki articles ratio

[quote="pda"]In how many ways can the letters ABCD be arranged so that no letter occurs in its proper position? (In ABCD, all letters are in its proper position, but in BACD, the A and B are not in its proper position.)[/quote]
$ 9$-derangements

[quote="archimedes1"][hide]There are $ 3$ possible positions for A.  Then, A will take the original place of B,C, or D.  Let's assume WLOG that A takes the original place of B.  Then, B has 3 places to go:  original A, C, or D.  After this we'll have two letters left over , at least one of which is in its original place.  Thus, we'll have to switch the final two, and there's only $ 1$ way to do this.  So we have a total of $ 3\cdot3\cdot1=9$.[/hide][/quote]

Nice exposition, archimedes1.  For another way to solve this problem, you can also use:

[hide="hint"]the inclusion-exclusion principle[/hide]

As yalldontknowme stated, arrangements in which no element maintains its original position are known as "derangements."

Derangements have many interesting properties.  The number of derangements of n objects is denoted $ !n$ and called the "subfactorial of n."

It's interesting to explore the values of $ !n$ for small values of $ n$.   After exploring a bit you might want to read the [url=http://www.artofproblemsolving.com/Wiki/index.php/Derangement]aopswiki article on derangements[/url].

You might also want to investigate the ratio $ !n/n!$ for small values of $ n$ and make a conjecture about its trend as $ n$ gets larger.  Then you might want to read the [url=http://mathworld.wolfram.com/Derangement.html]Wolfram article on derangements[/url], which will answer that question as well as provide some other interesting information about derangements.


induction
limit
function
calculus
calculus computations

Let $\left\{ x_{n}\right\}_{n \geq 0}$ be a sequence defined as follows:
\[ x_{0}= 1, \, x_{1}= 2 \, \, \textrm{and}\, \, 12^{x_{n}}= 2 \cdot 9^{x_{n-1}}+10 \cdot 3^{x_{n-2}}.\]
Prove that it's convergent and find its limit.

By induction on $n$ we have $x_{n+1}>x_{n}$ for all $n$ and $x_{n}<3$ for all $n$. Therefore exist $L=\lim_{n\to\infty}x_{n}$, we have also $1=2.(\frac{3}{4})^{L}+10.(\frac{1}{4})^{L}\Rightarrow L=3$.

You NOT read my post.
[quote="didilica"]By induction you prove that $x_{n}\leq 3$ and that the sequence increases, hence it converges.

Let $l=\lim x_{n}$.

It follows that $12^{l}=2(9)^{l}+10(3)^{l}$ or $1=2\left(3/4\right)^{l}+10 \left(1/4\right)^{l}$. This equation has a uniques solution, namely $l=3$ since the right hand side of your equation is a decreasing function with range $(0,12)$  and the left hand side is a constant function.[/quote]
That is all what I wrote  :o

Check the time stamps on those posts - didilica was clearly working on the problem independently of N.T.TUAN, and was already typing it before the first post appeared. Then maybe when he saw the notice that another post had appeared, he saw that maybe he had a more thorough explanation of the argument, which made it worth posting. No one did anything wrong, so please don't get upset.