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F(x) = {x}

What if X = 2.999999...

Since 2.999999.... = 3

Is the answer 3? Or 2?

Also, I've been wondering if the series 1.2.3,4,5,6... has more numbers or the series 2,4,6,8,10... has more numbers or do they have the same amount of numbesr?

Second part:
Lsat year at ARML, a professor told us some paradoxes involving infinity proving some things that should not be true, like what you stated.  The problem is that infinity is a concept, not a number, so you can't use it in this case to compare inequalities like we do with numbers. I.E., $2\infty>^{?}{\infty}$

[quote="xdl615x"]
Also, I've been wondering if the series 1.2.3,4,5,6... has more numbers or the series 2,4,6,8,10... has more numbers or do they have the same amount of numbesr?[/quote]
Consider set A to hold all the numbers in the first series (1, 2, 3...).  And set B contains all the numbers in the second series (2, 4, 6, and so on).  How many numbers does each set contain?  Infinitely many.  But the question is, are the two infinities the same?  Let's look at a simpler example first.

Let's say set A has just the numbers 1, 2, and 3.  Set B has 2, 4, and 6.  Do the two sets have the same amount of numbers?  Obviously yes, but how could you tell without counting them?  You would pair them up, one number from set A with one from set B.

So applying that to the previous case, can you match all the numbers from the first series to numbers from the second series?  You can tell by drawing a relationship between the two.  The $n^{th}$ term in set A would be designated as $n$.  So the relationship would be $2n$.  So yes, the amount of numbers in set A is equal to that in set B.

Another similiar example:

Suppose you have two line segments of different lengths.  Are the number of points in one line segment equal to the number of points in the other?

the way i see it, 2.999....is infinitely close to 3, but not 3 (if it were 3, then it would be written as "3"). i feel that $[2.999...]=2$

$2.999 \ldots < 3$, therefore $\lfloor 2.999 \ldots \rfloor = 2.$

However close $2.999 \ldots$ may get to $3$, it's still less than $3$.

[quote="4everwise"]2.999...= 3

You could consider .999... as being an infinate geometric sequence and solve for it with the formula $S=\frac{t_1}{1-r}$. This would show that .999... is equal to 1.
$2+1=3$[/quote]

Yes, this formula would....I never really knew the formula though  :blush:

You have $2.999999.....$
Take the $2$ out and let $x=.99999......$
Therefore $10x=9.9999.....$
So then subtract these and you'll get $9x=9$ so $x=1$
Therefore $2+1=3$

[quote="chess64"]Yes, but that's like saying $\frac{9}{9}=.999 \ldots$.

$\frac{1}{9}=.111 \ldots$
$\frac{2}{9}=.222 \ldots$
$\frac{3}{9}=.333 \ldots$
$\frac{4}{9}=.444 \ldots$
$\vdots$
$\frac{8}{9}=.888 \ldots$
$\frac{9}{9}=.999 \ldots$[/quote]

However, the number gets so close to the next integer that it pretty much is. For instance, in Chemistry, the electron is not counted in the atomic mass of the atom because the weight is so small that it is insignificant. Just like with this, the number gets so close that it doesn't really matter so it pretty much is the integer.

Okay, you're right; $\lfloor 2.999 \ldots \rfloor = 3.$

this again deals with infinity.  If there is a finite number of 9's then, $\lfloor 2.999... \rfloor=2$, but if there are infinite number of 9s, then we are essentially saying that $\infty-1=\infty$ because

[quote="Iversonfan2005"]
You have 2.999999.....
Take the 2 out and let x=.99999......
Therefore 10x=9.9999.....
So then subtract these and you'll get 9x=9 so x=1
Therefore 2+1=3[/quote]
If $x$ has y 9s after the decimal, 10x would have y-1 after it.
So infinity is a concept we really cannot grasp, not a number.  So if $2.999....$ has infinite 9s, then $\lfloor 2.999... \rfloor$ really is $3$

This has been talked about COUNTLESS of times on the forums. Search for 0.999... = 1 or something to that.

As for [2.999...] = 3

2.999.... =  3  so yeah definetly 3 thanks.

[quote="Iversonfan2005"]However, the number gets so close to the next integer that it pretty much is. For instance, in Chemistry, the electron is not counted in the atomic mass of the atom because the weight is so small that it is insignificant. Just like with this, the number gets so close that it doesn't really matter so it pretty much is the integer.[/quote]
hmm im not sure about your claim here on the Chemistry part, even though i understand what you are saying. although the mass of an electron is like 1/1800 of the mass of a proton, im not sure if you can neglect it...
ill have to ask amirhtlusa about this...

Yeah, it's true that you don't count it in the atomic mass.

[quote="amirhtlusa"][quote="Iversonfan2005"]
 For instance, in Chemistry, the electron is not counted in the atomic mass of the atom because the weight is so small that it is insignificant. Just like with this, the number gets so close that it doesn't really matter so it pretty much is the integer.[/quote]
not true :)[/quote]

What do you mean? They don't count that in the atomic mass... :?

this should go in chem forum but oh well:
that statement is not exactly true;  people believe that Atomic mass are defined only by protons, that is because protons are about $1837$times heavier than electorns. but $AMU$ (atomic mass unit)is defined as following:
one unified atomic mass unit is $1.66053873 * 10^{-27} $ kilogram, with a one-standard deviation uncertainty of ±$0.000 00013*10^{-27}$ kilogram.
now notice mass of proton is:
$m_p=1.6726 *10^{-27} $
do you see the differnce between the $AMU$ and $m_p$? they are not equal :)

sorry if this is off-topic.

[quote="furious"]for all practical purposes, 2.9999....=3, but i still feel that the greatest interger fuction of 2.999....=2 as you would take away the decimal part of the number like with any other number.[/quote]

Ugh, here we go again.

Do you know what a limit is?  (Obviously you should, I sat in your pre cal class the last few days watching you guys do limits).  Do you understand the notation $.\bar{9}.$?  Obviously not.

$.\bar{9}$ is defined as the limit of sum, $S_n=9^{-1}+9^{-2}+\cdots+9^{-n}$, where $n$ is an integer, as $n$ goes to infinity.  I.E. \[.\bar{9}=\lim_{n\to \infty} S_n=\lim_{n\to \infty} 9^{-1}+9^{-2}+\cdots+9^{-n}.\]

Thus it is a geometric series that equals 1.

First of all, to prove that $0.99999999...=1$, simply let $x=0.999999999...$.  Then $10x=9.9999999999...$ so $9x=9$ so $x=1$  Q.E.D.

Second, um, no this is a common misconception--but the rational numbers are the same order of infinity as the integers (in fact, the same order of infinity as the natural numbers)  :shocked: It's scary.

[quote]The number of rationals is a higher order of infinity than the number of integers, [/quote]
Not true.
BTW if you mean    [b] Real  numbers[/b]    then yes, it is an infinity of a higher order. But real numbers (which also contains non-rational numbers)  are not the same as rational numbers.  The order of set of all the rational numbers is same as  that of integers.

Hope this helps.

[quote="mathnerd314"][quote="K81o7"]Yeah. $0.9999...=1$.  Why?  $0.9999...-1=0.00000...=0$.  And if $n-x=0$, then $n=x$.
As for infinities...
You can even prove that the number of rationals is equal to the number of integers.  :D[/quote]

About 0.9999... = 1 because 0.9999... - 1 = 0.0000..., that's begging the question. You must assume that 0.9999... = 1 in order to conclude that 0.9999... - 1 = 0.

About infinities, there's all this transfinite math that one has to deal with. The number of rationals is a higher order of infinity than the number of integers, I guess. But for the average people (like me), infinity is infinity.

The rationals are a dense set, but the integers (or multiples of two) are not. Therefore the number of odd integers equals the number of total integers, but the number of integers doesn't equal the number of rationals.

Correct me if I'm wrong, please. I don't understand this either, and it's starting to make my head ache.[/quote]

Infinity is infinity, but if you can set up a 1-1 correspondence between the rationals and integers, then they are the same infinity.  If you can set up a infinity-1 correspondence, such as one between the irrationals and rationals, or the integers and reals, then they are different levels of infinity.  For instance, do this.  Suppose that you wrote all of the positive rationals on a sheet of paper.  The first in the top left corner is $\frac{1}{1}$, then going to the right, the numerator increases by one each time.  Going down, the denominator increases by 1 each time.  You CAN set up a 1-1 correspondence in the following way.  Assign the integer 1 to $\frac{1}{1}$.  Then assign 2 to 2/1.  Then assign 3 to 1/2.  Go along the diagonals like this.  Although the integers you're using increase at a rate much faster than the rate of increase of the rationals, it is still a 1-1 correspondence, because no matter what you do, you'll never run out of integers.
However, you can prove thar there are fewer rationals than irrationals by taking any rational, then showing that an infinite number of irrationals can be produced by simple changing digits and adding digits to the end.  It's complicated, and I won't attempt to explain it because I will probably fail.

For the proving that $0.9999...=1$...
You know that there are an infinite number of 9s....if we can define that...
So anyways, you could actually perform the following
$1-0.999...=0.0000......$
We can say that after an infinite number of 0s, equal to the number of 9s, there will be a 1.  However, there are an infinite number of 0s.  Therefore, the 1 will not come: it does not exist.  Therefore, $1=0.9999...$.

Your argument still wouldn't convince a skeptic.  They would just say that $.\bar{0}1$ is really small but still not 0.

Then the skeptic hasn't fully understood the meaning of $0.9999...$  :)

i donno how manytimes we have to discuss this:
$0.\bar{9}$ IS $1$.
this is just the basic concept of limit...
 ;)

I think it's also worth pointing out that the greatest integer function is [i]not[/i] the "drop the decimal" function -- just apply it to negative non-integers and you'll see.

Density is not a property that tells you very much about the number of something, just about how they are situated.

It is also interesting to note that not only do the set of rational numbers and the set of integers have the same cardinality (size), but the algebraic numbers (all numbers that are the solution to a polynomial equation with integer coefficents) and the integers also have the same cardinality.  This means that the transcendental numbers (not algebraic, like $e$ and $\pi$) account for the difference, even though we think of them as rare.  (All this, though, is much harder to prove than the fact that the rationals and the integers have the same cardinality, or that the reals and the integers do NOT have the cardinality.)

Proving that algrebraics have the same cardinality as the integers is not too difficult, even maybe for this forum.  You just have to prove some intermediate results about the cardinality of $Z^2$ (that is, an infinite square lattice).

[quote="mathnerd314"]Sorry! :blush: I guess I don't understand infinity. Or else I do understand it perfectly but I would pretend that I don't because it's a little difficult for this forum? :D Just kidding.

The 0.9999999... thing, though, is traditionally proven the way I saw bubala do it: x=0.99999... 10x=9.99999... 9x=9 x=1 qed.  :winner_first:[/quote]

Good lord...How many times are people going to do that exact same calculation?!?!?!? Anyway, this thing is really really getting old. $.999999........=1$ so give it up :mad:

[quote="Iversonfan2005"] $.999999........=1$ so give it up :mad:[/quote]
i just did yesterday. after years of arguing in favor of the contrary, joml88 convinced me that .99999...=1.

If you want to prove the set of irrational numbers is larger than the set of rational numbers, merely assume the opposite, ie that they have the same carnility. Now list the set of all irrational numbers, one under the other. Now we create a new number, by taking the first digit from the first number, the second digit from the second number, the third digit from the third, and so one. Now if we increase each of the digits by one, we have created a new number thtat is not on our list, as at least one of the digits is different. Contradiction.

I was just thinking, for the algebraic numbers, could we instead not just count the number of polynomials instead? Much easier done, i should think!

And finally, to keep the topic alive (im in love with infinity):

Take any 11 infinate decimals. Prove that among them, there are two that have the same corresponding digit in infinately many places.

My argument (which isn't a formal proof) is that there are 10 digits and you want 11 numbers. The best you could do to prevent having the same digit in infinitely many places is:

.000000000...
.111111111...
.222222222...
.333333333...
.444444444...
.555555555...
.666666666...
.777777777...
.888888888...
.999999999...

And the eleventh number must have a digit in common with one of these. Q.E.D.

Assume the contrary, that no two of them have the same digit in infinitely many places.  Thus there must exist some integer $n$ such that for any integer $k>n$ no two of the decimals share a digit.  But if we look at the digits in the $k$th decimal place for each of the numbers, there must be at least two that are the same by the pigeonhole principle.  Contradiction.  $\textbf{QED}$

If none of them share a single digit in a place, then
Take the tenths place digit of each number.  There are eleven, and each is a different digit.  Contradiction.  Thus, the original conditions are impossible.  This must happen for every single digit place.  Therefore, there are an infinite number of shared corresponding digits.  QED.


integration
calculus
calculus computations

This is fun.

1) $\int_{2}^{3}(3-x)f''(x)dx=7$ so by parts $u=3-x\Rightarrow du=-dx$ and $v=f'(x)\Rightarrow dv=f''(x)dx$ so $(3-x)f'(x)\big|_{2}^{3}+\int_{2}^{3}f'(x)dx=7$ and thus $-f'(2)+f(3)-f(2)=7\Rightarrow f(3)=10$. Possible sign error? Not good with keeping + and - straight.


What formulas should you memorize [size=200]AND[/size] understand for success in mathcounts?

I know some are

heron's
areas
perimeters (circ.)
pyth. theorom
gauss
-jorian

Try to look at this:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=73496


parameterization calculus derivative symmetry

15 champagne glasses are piled upon each other forming a tower, as illustrated by the figure.
Champagne was poured into the top glass until it wass full, and then the champagne started to flow into the next glass, which eventually became full, and then it ran over in the next glass and son on.

One continued to pour champagne, and by the time the sixth glass was being filled completeley, the tower started to swing seriously, but eventually after some more champagne it started to stabilize. Three whole bottles of champagne was consumed.

Determine the mass of an empty glass! You can assume that the glasses are relatively flat.

What do you mean by "started to swing seriously"?

You could skip "seriously". I meant that the tower got unstable and started to swing (much more than earlier and later).

Do we know anything about the dimensions of the glasses :?:

No ideas so far?  :(
This thread was not initiated to solve the problem, well of course someone should try,
but what is more interesting is the kind of methods used to do so. Especially
these kind of problems which does not imply directly how it is to be done.

So you're welcome to post any serious ideas on solving the problem.

I tried in that way:
The higher the center of mass of the tower and the champagne is, the more the tower swings. So, we could calculate the height of the center of mass and demand that it is biggest when six glasses are full. However at this point I got stuck, because I can't solve it without any more parameters of the glass given...

Very good, keep on.

[b]golduck[/b]
Its possible without any more information. Make some assumptions. No explicit though, I mean you don't have to approximate anything, but perhaps the total volume. But even
the total volume of champagne is given somehow, its three bottles.
Now make a model and apply some calculus and you're done with the maximization.
-------------------
What I am wondering is how you attack these kind of problems
and if anyone have another kind of solution.

I can't solve it, could you post your solution :?:

Hmm, here's my attempt at this...

Assume the glasses are cylindrical things stacked directly on top of each other, each having height $H$ and mass $m$. Let the top of the stack be $y=0$ and the y-axis is positive going down. Then the center of mass would be

$CM=\frac{\frac{15}{2}H\cdot 15m+\frac{n}{2}H\cdot nM}{15m+nM}$
where $n$ is the number of glasses filled with wine and M the mass of each glass of wine (mass of glass not included).

When $n=6$, the center of mass of the stack is highest, so taking the derivative of the above expression with respect to $n$ and setting to zero to find the maximum CM, we get

$\frac{2n}{225m+n^{2}M}-\frac{1}{15m+nM}=0$, solving for $n$ and taking the positive solution, we get

$n=\frac{-15+15\sqrt{m(m+M)}}{M}=6$
This equation leads to
$m=\frac{4}{5}M$

I assumed that the stack stabilized when all of the glasses were filled, and that each wine bottle contains 750mL of liquid, so the three bottles were around 2.25kg.
So $15M=2250g$, or $M=150g$
Thus $m=120g$

Is that even remotely close?

[quote="tzhu"]Assume the glasses are cylindrical things stacked directly on top of each other, each having height $H$ and mass $m$.[/quote]
The problem is: is this approximation good enough... :?:

Thanks a lot for the posts.

@[b]tzhu[/b]: Yes it is good.
@[b]golduck[/b]: Well, one should consider the aim of the precision of the problem.

Sorry for not posting a reply, but I will post my solution and a discussion
later. Unfortunately I haven't got the time now. Please be patient,  :P .

@[b]golduck[/b] Practically all champaign glasses have cylindrical symmetry. That dosent say they are cylinders, but that is enough to assume the $CM$ lies on the symmetry line for example on the
top of the foot.
But you were right on the groove assumption (well actually its enough but if one wants to be more precise... ) of [b]tzhu[/b] that they are place right on top if each other. In the problem it was stated that they are placed in each other, though it was a bit vague.

The solution below as some parameters say $\delta$ that gives the possibility to elaborate with
the $CM$ position so that the solution gets more precise. Also it assumes the glasses are placed IN each other.

[hide=" a similar one"]

Let the foot of one glass be of unit length and assume that the center of mass
lies one top of the foot. Then for the empty tower the $CM$ lies on heigth
$8$.
Let the mass of one glass be $m$. Let the $CM$ of a full glass be at the height
$\delta$ above the $CM$ for an empty glass. In the tower the $CM$ of [i]one[/i] glass then lies on height $15+\delta$. If two glasses are filled on $14.4+\delta$,
three glasses filled on $14+\delta$ and so on.

Call the mass of champagne in a glass $M$. $CM$ for $N$ filled glasses is at height
$h=\frac{15m\cdot 8+NM(15.5+\delta-0.5N)}{15m+NM}=\frac{120q+N(15.5+\delta-0.5N)}{15q+N}$
where $q=\frac{m}{M}$

Now differentiating yields and applying condition
$\frac{dh}{dN}=\frac{(15.5+\delta-N)(15q+N)-(120q+N()15.5+\delta-0.5N)}{15q+N}=0$
If this occurs at $N=6$ we get that
$q=\frac{18}{22.5+15\delta}$
Also we know by text that $M=\frac{3\cdot 750}{15}g=150g$
So assuming $\delta\in[0,0.5]$
one has $m\approx 100g$[/hide]


quadratics
calculus
inequalities
IMO Shortlist
inequalities unsolved

What is the maximal value of the expression $x_1+...+x_n-ax_1x_2...x_n$ where $x_1,...,x_n,a$ are real numbers and $ x_1^2+...+x_n^2=1$? For particular values of a and n=3 we find two nice problems, one from Imo Shortlist  and one, recent, from Vietnam. What about the general case?

Don't kill me for saying this, but wouldn't Lagrange Multipliers work fairly well here?

Let $f(x_1, x_2, \cdots, x_n) = x_1+...+x_n-ax_1x_2...x_n$ and $g(x_1, \cdots, x_n) = x_1^2+...+x_n^2=1$. If $(x_1, \cdots, x_n)$ maximizes $f$ under $g=1$, then $\nabla f = \lambda \nabla g$ for some $\lambda$. Thus
\[
1 - ax_1\cdots x_{i-1} x_{i+1} \cdots x_n =2  \lambda x_i
\]
for each $i$. Multiplying both sides by $x_i$ tells us that $- a x_1 x_2 \cdots x_n = 2 \lambda x_i^2 - x_i$ for each $i$. Thus $2 \lambda x_i^2 - x_i = c$ for some constant $c$. Since $2 \lambda x_i^2 - x_i$ is a quadratic, each $x_i$ can only take one of two values. Perhaps that can help to solve the problem.

I tried the Lagrange Method but I got stuck in following:

$\forall _{i \not = j}$

$(x_i-x_j)(b(x_i+x_j)-x_ix_j)=0$

where

$b=a \prod x_i$

Any ideas ?

[edit] I didn't see [b]billzhao[/b]'s post...

[quote="Pascual2005"]can you explaiin your method billzhao? Thanks![/quote]

Isn't it just the standard [url=http://mathworld.wolfram.com/LagrangeMultiplier.html]Lagrange Multipliers[/url] method, found in all multivariable calculus textbooks?

For the maximal value, and also for the minimal value, it is suffices to consider that $n-1$ of $x_i$ are equal.

Vasc, could you tell me why? I am new in this section!

I can only say that this hint is also valid to prove some difficult ineq. such as that one proposed by Harazi on this forum:
$\sum \frac 1{x_i} = n$ implies  $x_1\cdots x_n-1\geq \left(\frac{n-1}n \right)^{n-1} (x_1+...+x_n-n)$.

Another very difficult ineq. which can be proved by this hint is the following:

$(x_1+...+x_n-n)(\frac{1}{x_1}+...+\frac{1}{x_n}-n)+x_1...x_n+\frac{1}{x_1...x_n} -2 \ge 0$.

I like very much this inequality with $x_i$ positive numbers. :)

For the solution clearly,I suppose that $\sum{x_i}=n$

For the maximum,I will prove:

$P=x_1+x_2+...+x_n-ax_1x_2...x_n$

I will prove that:

$P \le max(n-a,\sqrt{n(n-1)})$ By Induction.

And this is the reason why we have result  that P max if $n-1$ numbers $x_i$ equal.


superior algebra
superior algebra theorems

For the set $ R: \equal{}\{\frac{a}{b} | a,b\in\mathbb{Z}$ are coprime, $ p \nmid b\}$, why is it that each nonzero ideal of $ R$ is principal? and, why is it equal to $ (p^e)$ for some $ e\in\mathbb{N}_0$?

Also, why does $ R$ have a quotient field that is a subfield of $ \mathbb{Q}$? What is it then?   :o

Thanks in advance!

Los

... Just look up [url=http://en.wikipedia.org/wiki/Localization_of_a_ring]ring localization[/url]. This is about the simplest example possible of that process.

As for the quotient field of $ R$- since $ R$ contains a copy of $ \mathbb{Z}$, its quotient contains the quotient field $ \mathbb{Q}$ of $ \mathbb{Z}$. There's an obvious containment the other way, so the quotient field is just the rationals. This is not really surprising.


function
trigonometry
search
inequalities theorems
inequalities

When $a+b+c=abc$, $ab+bc+ca=abc$ or $a^2+b^2+c^2+2abc=1$, we can subsitute $a,b,c$ by trigonometric functions. Which functions are these and, the most important, what are the restrictions for using them (for example, $a,b,c>0$) and the ranges of the arguments of the trig. function we get?
Thanks.

Hummmm, remember that famous equality about angles and tangents from the internal angles of a triangle?

$\tan{a_1}+\tan{a_2}+\tan{a_3} = \tan{a_1}\times \tan{a_2}\times \tan{a_3}, \mbox{when}\, a_1+a_2+a_3=\pi$.

well, well, we know we can compress all the real numbers into a semi-circle of radius 1, using the inverse of the tangent funcion. So, searching for real numbers that satisfy the restriction $a+b+c=abc$ is the same thing of searching for three angles which sum $\pi$. ;-) you have a new search, good luck. The rest of them is looking for some famous identities in trigonometric functions =].


Salut,

J'ai une demande à tous: pouvez vous changer vos signatures pour les rendre compatibles avec l'unicode? :)

Mersi.

[quote="Nuln"]C'est-à-dire?  :blush:  (unicode?)[/quote]... Tu voyez que ton signature semble étrange ? ... les accents sont montrés mal - éditez-les :D :)

[quote="Nuln"]Salut,

si je comprends bien, il n'est pas possible d'utiliser les accents?[/quote]Il est possible: éêâàè :D mais nous avons changé le encoder des lettres.

Juste récrire tois signature, et il paraîtra agréable avec les accents. (espérez que le ce se comprend  :blush: )


how would you prove $ \sum_{i\equal{}1}^{n}{i^2}\equal{}\frac{n(n\plus{}1)(2n\plus{}1)} {6}$

Well, if you already know it and just want to prove it, you can use induction.

This has been proven many times in AoPS/ML. Use search for the old topics discussed or ...

Use induction to prove it, or

Use the 'binomial' method:
\[ \begin{aligned}(n \plus{} 1)^2 & \equal{} n^2 \plus{} 2n \plus{} 1 \\
n^2 & \equal{} (n \minus{} 1)^2 \plus{} 2(n \minus{} 1) \plus{} 1 \\
(n \minus{} 1)^2 & \equal{} (n \minus{} 2)^2 \plus{} 2(n \minus{} 2) \plus{} 1 \\
\vdots & \equal{} \vdots \\
2^2 & \equal{} 1^2 \plus{} 2(1) \plus{} 1 \end{aligned}
\]
and then apply the formula $ \sum_{i \equal{} 1}^ni \equal{} \frac {n(n \plus{} 1)}2$

EDIT: Eh, oops, that's for deriving $ \sum_{i \equal{} 1}^ni \equal{} \frac {n(n \plus{} 1)}2$. Still, you can use the same binomial method to get the formula for $ \sum_{i \equal{} 1}^ni^2$ by expanding $ (n\plus{}1)^3\equal{}n^3\plus{}3n^2\plus{}3n\plus{}1$.

I discuss a general non-inductive technique [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=178381]here[/url].

Another way is to look at $ S \equal{} \sum_{r\equal{}1}^n r(r\plus{}1)$

We have $ S \equal{}  \sum_{r\equal{}1}^n r \plus{} \sum_{r\equal{}1}^n r^2$

Also, $ S \equal{} \sum_{r\equal{}1}^n T_r$ where $ T_r \equal{} \frac{1}{3} [r(r\plus{}1)(r\plus{}2) \minus{} (r\minus{}1)r(r\plus{}1)]$ i.e. a telescopic sum which easily computes to $ \frac{n(n\plus{}1)(n\plus{}2)}{3}$

Hence $ \sum_{r\equal{}1}^n r \plus{} \sum_{r\equal{}1}^n r^2 \equal{} \frac{n(n\plus{}1)(n\plus{}2)}{3}$

$ \Rightarrow \sum_{r\equal{}1}^n r^2 \equal{} \frac{n(n\plus{}1)(n\plus{}2)}{3} \minus{} \sum_{r\equal{}1}^n r \equal{} \frac{n(n\plus{}1)(n\plus{}2)}{3} \minus{} \frac{n(n\plus{}1)}{2}$

$ \equal{} \frac{n(n\plus{}1)(2n\plus{}1)}{6}$

That's not "another way"; that's the technique I outline.  You're implicitly using the Hockey Stick theorem.

\[ \sum^n_{j \equal{} k}\binom{j}{k} \equal{} \binom{n \plus{} 1}{k \plus{} 1}
\]
First I'll apply this method to derive the sum of $ j$

$ \sum^n_{j \equal{} 1} \binom{j}{1} \equal{} \sum^n_{j \equal{} 1} j \equal{} \binom{n \plus{} 1}{2} \equal{} \frac {(n \plus{} 1)!}{2!(n \minus{} 1)!} \equal{} \frac {n(n \plus{} 1)}{2}$

Now we consider the case $ k \equal{} 2$, and sum such binomials.

$ \binom{n \plus{} 1}{3} \equal{} \sum^n_{j \equal{} 2}\binom{j}{2} \equal{} \sum^n_{j \equal{} 2} \frac {j(j \minus{} 1)}{2}$

Expressed in closed form so we get somewhere,

$ \frac {1}{2}\sum^n_{j \equal{} 2} j^2 \minus{} j \equal{} \frac {n(n \plus{} 1)(n \minus{} 1)}{6}$

Observe on the LHS that for j=0,1, the expression is zero-- hence the sum from 2 to n is equal to the sum from 1 to n;

$ \sum^n_{j \equal{} 1}j^2 \equal{} \frac {n(n \plus{} 1)(n \minus{} 1)}{3} \plus{} \sum^n_{j \equal{} 1}j$

$ \sum^n_{j \equal{} 1}j^2 \equal{} \frac {n(n \plus{} 1)(n \minus{} 1)}{3} \plus{} \frac {n(n \plus{} 1)}{2}$

$ \sum^n_{j \equal{} 1}j^2 \equal{} \frac {n(n \plus{} 1)(2n \plus{} 1)}{6}$

Done!

This approach is equally powerful for deriving the sum of $ j^n$ for integer $ n$.  :wink:

So? It's not exactly the easiest thing to infer that approach from your writing, but I saw it pretty quick because when I read it I had already derived the formulas with that technique.  :)

Hey thanks, torajirou. I had a closer look at your post and learnt something new (Newton representation of Polynomials).

Use the binomial expansion to obtain:

$ (x + 1)^3 - x^3 = 3x^2 + 3x + 1$

We know that
 $ \displaystyle\sum_{x = 1}^{n} ((x + 1)^3 - x^3)$

is telescoping therefore, it is equal to $ (n + 1)^3 - 1 = n^3 + 3n^2 + 3n$.

but $ \displaystyle\sum_{x = 1}^{n} ((x + 1)^3 - x^3) = 3\displaystyle\sum_{x = 1}^{n} x^2 + 3\displaystyle\sum_{x = 1}^{n} (x) + 1$

so $ \displaystyle\sum_{x = 1}^{n} ((x + 1)^3 - x^3) = 3\displaystyle\sum_{x = 1}^{n} x^2 + 3\displaystyle\sum_{x = 1}^{n} (x) + 1 = n^3 + 3n^2 + 3n$

assume that $ \displaystyle\sum_{x = 1}^{n} x^2 = t$

so $ 3t + \frac{3n(n+1)}{2} + 1 = n^3 + 3n(n + 1)$

$ 6t = 2n^3 + 3n^2 + 3n$

$ 6t = n(n + 1)(2n + 1)$

$ t = \frac{n(n + 1)(2n + 1)}{6}$

so

$ \displaystyle\sum_{x = 1}^{n} x^2 = \frac{n(n + 1)(2n + 1)}{6}$

cognos599, there is a slight error in line 6... change +1 to +n


hmm i discovered that there's many style of letters in LaTeX.

for example, my signature is 'mathfrak' style.

Is there any more styles that we could use?

There is
$ \textbf{textbf\{\}}$
$ \textit{textit\{\}}$
Those are the only other ones I know though.   $ \ddot\smile$

\mathbb{}

$ \mathbb{HELLO WORLD}$

\mathcal{}

$ \mathcal{HELLO WORLD}$

\mathfrak{}
$ \mathfrak{mathfrak \{ \} }$.

$ \textsf{Sans\minus{}serif}$
[code]\textst{}[/code]
$ \texttt{Typewriter}$
[code]\texttt{}[/code]


I'm very weak at spanish and my midterm is coming soon. So, I want to see if there are any people who are excellent in spanish or can see if my writing of the spanish (I'm in spanish level I, the basic) is right or not. I'm hoping to get the perfect score on my spanish exam so any good suggestion is appreciated.

I tried to use google to find the site, say translator, and found it but it's not that good. Also, is there any spanish or just homework site like this one? (I know AoPS isn't homework site but it helps you out on the math)

If you don't like posting here but can help me, pm me.

Thank you! :)

I think that the [url=http://www.mathlinks.ro/Forum/index.php?f=255]Spanish Forums[/url] are a much better place to post this question :)

[quote="Silverfalcon"]Muy gracias. :D[/quote]

It's "muchas gracias"  ;)

What's the difference?

I know I feel dumb to say this but "muchas gracias" $\neq$ "muy gracias"?

We haven't learn the difference between muchas and muy yet..

"Muchas" (or muchos) means much/many.  "Muy" means "very".  "Muy gracias" translates as "very thanks" ;).

Oh.

Thanks.

Muchas gracias, how's that? :lol:

Silverfalcon, yo te puedo ayudar en español, es mi primer idioma.

(i can help you with spanish, it is my mother language)

I do have some sentence that I wrote for tomorrow's exam. Can anyone read this and see if I'm in right track? Just see if WHAT I HAVE is right. My class is still pretty much on basics so. :lol:

[story is based on the Pedro's family]

Esto es mi padre Pablo. El es muy inteligente y simpatico.
Esta es mi madre Ana. Ella es bonita y rubia.
Esto es mi abuelo Jose. El es comico y tiene el pelo de color cafe.
Esta es mi prima Marta. Ella es baja y morena.
Esta es mi hermana Susana. Ella tiene once años. Ella es muy traviesa y bonita.
Esta es mi abuela Maria. Ella es muy estricta pero ella es simpatica.
Esto es mi perro Pepe. (haha) Pepe es muy travieso y comico.
Esto es mi tio Manuel. El es juevo y tiene canas.
Esto es mi hermanastro Jake. El es comico y tiene los ojos azules.

[The guy named Keith has messy room]

We're talking about what he should do.

Debe organizar su cuarto.
Debe hacer la cama y poner la ropa en el armario.
Necesito limpiar la ventana.
Necesito pasar la aspiradora y encontrar su zapatillas de tenis.
Quiere pasar el rato con mis amigos.
Quiero descansar en mi cama y escuchar musica.

Faltan los acentos...
Pero ¿ qué edad tienes ? (los ejercicios están faciles, ¿ no ?)

Tengo la clase de espanol I.

Tengo quince anos pero este es mi primero ano por el espanol.

I know nothing in Spanish other than "no habla español", but...

Alt + 0 + 2 + 3 + 3 = é

yo hablo español, es mi primer idioma y no se acentos..jaja, pero no se dice "esto es mi abuelo..." y eso, sino "este es", esto es para cosas usualmente


function
limit
calculus
calculus computations

I am not sure what to think about this one.  As much background information as possible would be helpful.

[b]29)[/b] For all positive functions $ f$ and $ g$ of the real variable $ x$, let $ \sim $ be a relation defined by $ f \sim  g$ if and only if $ \lim_{x \to \infty}{\frac{f(x)}{g(x)}=1}$.

Which of the following is NOT a consequence of $ f \sim  g$?

a) $ f^2 \sim  g^2$
b) $ \sqrt{f} \sim  \sqrt{g}$
c) $ e^f \sim  e^g$
d) $ f\plus{}g \sim  2g$
e) $ g \sim  f$.

Sorry if you didn't post the rest of the question on purpose, but I think the choices are necessary on this to understand what you are looking for.

I think D and E are easy to eliminate, and then guessing something simple quickly led to letting $ f(x)\equal{}x$ and $ g(x)\equal{}x\plus{}1$, so $ \lim_{x\to\infty}\frac{e^f}{e^g}\equal{}\lim_{x\to\infty}\frac{e^x}{e^xe^1}\equal{}\frac{1}{e}$. So C!

[quote="HilbertThm90"]Which of the following is NOT a consequence of $ f \sim g$?

a) $ f^2 \sim g^2$
b) $ \sqrt {f} \sim \sqrt {g}$
c) $ e^f \sim e^g$
d) $ f \plus{} g \sim 2g$
e) $ g \sim f$.

Sorry if you didn't post the rest of the question on purpose, but I think the choices are necessary on this to understand what you are looking for.

I think D and E are easy to eliminate, and then guessing something simple quickly led to letting $ f(x) \equal{} x$ and $ g(x) \equal{} x \plus{} 1$, so $ \lim_{x\to\infty}\frac {e^f}{e^g} \equal{} \lim_{x\to\infty}\frac {e^x}{e^xe^1} \equal{} \frac {1}{e}$. So C![/quote]

a since $ lim\frac {f}{g} \equal{} 1$ then we have : $ lim\frac {f^2}{g^2} \equal{} lim\frac {f.f}{g.g}$=lim(f/g)lim(f/g)=1.1=1 so f^2~g^2
b. Because lim(f/g)=1 so  f(x).g(x) is positive when x goes to infinity
 so without lossing generality we can assume that f(x)> 0 and g(x)>0 when x-> infinity ( if f(x)<0 and g(x)<0 then f/g >0 when x-> infinity)
Since $ x\rightarrow \sqrt {x}$ is continuous so we have :
$ lim(\frac {\sqrt {f}}{\sqrt {g}} \equal{} \sqrt {limf/g}\equal{} \sqrt {1} \equal{} 1$ so $ \sqrt {f}\sim \sqrt {g}$

Sorry, I got interrupted while I was typing the question, and when I came back, I thought I typed it all and hit submit.  Yes, the answer choices are very necessary for the problem.

It is obvious letting f(x)=x and g(x)=x+1.

I don't know what it is.  This test is psyching me out.  These problems aren't hard, but when I first read them, I feel really intimidated.  They just aren't questions I am used to encountering in a format that I am not used to.  I also have the bad habit of wanting to use the most mechanical, cantankerous, formal methods possible when if I just use a little creativity, they're simple.

The fact that I have less than 3 minutes per problem also doesn't help since I feel like "just trying stuff" can waste time :|.

I know that this is easier said than done, but I took this test twice. The first time I didn't do all that well. I was like you. I psyched myself out and was really worried. I studied a lot for it. The second time I only did the one practice test they had a few days before. I went in really relaxed knowing that I could always rely on my first score if my second was worse. I did better by a long shot the second time

The best thing in my opinion is to just be calm. I went slower the second time. It is better to get less questions done and all of them right than to get all of them done, and miss some. Remember you are hurt by missing a question, but it remains neutral for no answer. Second, the test is standardized. If you go through the practice test and understand them, you are almost gauranteed to get variations on those questions.

So if you feel like you are spending longer than two or three minutes on a problem, then skip it and come back later if you have time. Remember it isn't really a hard test (there may be hard problems, though). It is meant to test competence at an undergrad level. If you can do these practice problems, then go in confident that you will do well. That is probably the best advice I could give. Don't doubt, panic, and rush.


function
real analysis
real analysis solved

Let x(n) sequence x(n)^2-x(n)+1=n/x(n) for n>0
Show that lim n/(x(n)^3)=1
--------------------------------------------------------
P(t) = t^3-t^2 +t
P'(t)=3t^2 - 2t+1 >0, P is strictly increasing function
P(0)=0, P(t) \geq 0 on [0;+oo[
lim_{t->+oo} P(t)=+oo

x(n) is define as the unique solution of the equation P(x(n))=n
P is bijective on [0;+oo(,  x(n) = P^(-1)(n)--->+oo
but 1- 1/x(n) +1/(x(n)^2) = n/(x(n)^3)
this prove that  lim n/(x(n)^3)=1

we have that x_n^3-x_n^2+x_n=n>=1 so we get that (x_n^2+1)(x_n-1)>=0 which gives that x_n>=1 for any n>=1.
from the fact that: x_(n+1)^3-x_(n+1)^2+x_(n+1)=n+1 we get that:
x_(n+1)^3-x_(n+1)^2+x_(n+1)=x_n^3-x_n^2+x_n+1 which can be rewritten like so:
(x_(n+1)-x_n)[x_(n+1)(x_(n+1)-1)+x_n(x_n-1)+x_n*x_(n+1)+1]=1 so x_n is incresing. we consider that it is bounded then we get that lim(n-->oo)x_n=00 so we rewrite the limit like so:
lim(x-->oo) n/x_n^3=lim(n-->oo) (n/x_n)/x_n^2=lim(n-->oo) (x_n^2-x_n+1)/x_n^2=1

cheers! :D :D


[quote="anirudh"]If $\sqrt{x+2}=2$, then $(x+2)^{2}$ equals

$\text{(A)}\ \sqrt{2}\qquad \text{(B)}\ 2 \qquad \text{(C)}\ 4 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 16$[/quote]
Even children in playschool know that $(x+2)^{2}= \left(\sqrt{x+2}\right)^{4}= 2^{4}= 16$, so it's $\mathrm{E}$.
Why do you post such trivial questions? They are good for topics like [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=81682]Pointless math[/url] and forums like [url=http://www.mathlinks.ro/Forum/index.php?f=300]Classroom math[/url].


6 children are playing dodgeball. each child has a ball. at the sound of a whistle, each randomly chooses another child to chase. what is the probability that at least one pair of children are chasing each other?

the problem is from  this website's are you ready for olympiad problem solving? pretest. the given answer is 1653/3125. but how is that obtained???

You could just brute force it out.

Consider the three cases:

Case 1: Exactly 1 pair.
$\binom{6}{2} * 5 * 4^3 = 4800$

Case 2: Exactly 2 pairs:
$\binom{6}{2} * \binom{4}{2} * 5 * 4 = 1800$

Case 3: Exactly 3 pairs:
$\binom{6}{2} * \binom{4}{2} * \binom{2}{2} = 90$

We have: $\frac{5^6 - 6690}{5^6}$ = $\frac{1787}{3125}$.

What did I do wrong?

i dont see how 5*4^3 covers the choices for the other 4 children in case 1

White horse king I think you over counted maybe you should PIE it.

The recurring answer I get with PIE is 3/5.
If right, this could be simply explained by saying 6C2=15 ways to choose a pair of children, *5^4 for the other childrens possibilities.
(15*5^4)/5^6  =  15/25   =  3/5.
But I found the problem in another form with the same 1653/3125 posted as the answer.

There are ${6\choose2}=15$ choices for the two children who are chasing each other. Each of the other $4$ children can chase any of the other $5$ children (this causes repetition, but we will account for this later), so there are $15\cdot5^4=9375$ ways this can occur.

Now, if there are two pairs of children chasing each other, this will be counted twice, so by PIE we must subtract the number of such cases from the previous total. There are again $15$ choices for the first group of two children, and now there are ${4\choose2}=6$ choices for the second group. We must divide by $2!=2$ because we select the groups in order, but they are actually a combination. The final $2$ children again have $5$ choices for children to chase. This gives us $\frac{1}{2}\cdot15\cdot6\cdot5^2=1125$ repetitions, so now we have $9375-1125=8250$ possibilities.

Finally, we must account for the case where there are three groups of children chasing each other. By PIE, we must add this to the number we have now. Again, we can choose $15$ pairs at first, and then we have a choice of $6$ pairs. Since there are $3$ pairs which can be arranged in any order, we must divide by $3!=6$. Thus, there are $\frac{1}{6}\cdot15\cdot6=15$ extra cases, for a total of $8250+15=8265$ possibilities.

Hence, the probability is $\frac{8265}{5^6}=\frac{1653}{3125}$.


linear algebra
matrix
algebra
polynomial
probability
linear algebra unsolved

Let $K$ be a finite field. How many nilpotent matrices of order $n$ are in $M_n(K)$?

$K=\mathbb{F}_{q}$.
There are $\mathcal{N}il(n)=q^{n^{2}-n}$ nilpotent matrices in $\mathcal{M}_{n}(K)$.

For a matrix $M\in \mathcal{M}_{n}(K)$, you can find $k \in \{0,1,\dots, n \}$ s.t. the minimal polynomial of $M$ could be written $X^{k}P(X)$ with $P(0)\not=0$.
Denote by $\mathcal{A}_{k}$ the set of such matrices.
We have $q^{n^{2}}=\mbox{Card}(\mathcal{M}_{n}(K)) = \sum_{k=0}^{n}\mbox{Card}(\mathcal{A}_{k})$.
The choice of a matrix in $\mathcal{A}_{k}$ is given by
1. A subspace of $K^{n}$ of dimension $k$
2. A matrix in $\mbox{GL}_{n-k}(K)$
3. A matrix in $\mathcal{M}_{k,n-k}(K)$
4. A nilpotent matrix in $\mathcal{M}_{k}(K)$

Denote by $\alpha_{n}= \prod_{i=1}^{n}(q^{i}-1)$.
There are
1. $\frac{\alpha_{n}}{\alpha_{k}\alpha_{n-k}}$ subspaces of $K^{n}$ of dimension $k$
2. $q^{\frac{(n-k)(n-k-1)}{2}}\alpha_{n-k}$ matrices in $\mbox{GL}_{n-k}(K)$
3. $q^{k(n-k)}$ matrices in $\mathcal{M}_{k,n-k}(K)$
4. $\mathcal{N}il(k)$ nilpotent matrices in $\mathcal{M}_{k}(K)$

So $q^{n^{2}}=\sum_{k=0}^{n}q^{\frac{(n-k)(n-k-1)}{2}}\frac{\alpha_{n}}{\alpha_{k}}\mathcal{N}il(k)$, which gives $\mathcal{N}il(n)=q^{n^{2}-n}$

PS: I think this is a theorem from Fine and Herstein. Fine, N. J.; Herstein, I. N. The probability that a matrix be nilpotent. Illinois J. Math. 2 1958 499--504


Hello everyone, got two more teasers for ya. Please be sure to show your work. Thanks and enjoy!

1. Adam scored 10 points in a basketball game. He could have scored with 1-point free throws, 2-point field goals or 3-point field goals. In how many different ways could he have scored his 10 points?

2. Mary is visiting New Orleans and wants to bring back T-shirts for all her relatives. She's found T-shirts she likes for 5 dollars, 10 dollars and 15 dollars. She has budgeted 40 dollars for the gifts. List all the ways Mary can spend 40 dollars or less on T-shirts.


[quote="mdk"]Let f(x) be a polynomial that takes on integer values for all x in Z. Can it be the case that f(n) is prime for all sufficiently large n?[/quote]

[hide="Only Constant Polynomials"]
Say $ f(x_{0})=p$ is prime. Show $ f(x_{0}+tp)=p(1+P(t))$ where $ P(t)\in \mathbb{Z}[t]$ for any integer $ t$. Since $ p$ is prime it follows that $ f(x_{0}+tp)=p$ for any $ t$. So $ f(x)$ must be constant polynomial.
[/hide]

ok, so how do I show that show $ f(x_{0}\plus{}tp) \equal{} p(1\plus{}P(t))$ where $ P(t)\in\mathbb{Z}[t]$ for any integer $ t$.

Assume that $ f(x) \equal{} a_{n}x^{n}\plus{}a_{n\minus{}1}x^{n\minus{}1}\plus{}\dots\plus{}a_{1}x\plus{}a_{0}$ is prime for all integers $ x\ge k$, where $ k$ is a sufficiently large integer. Let $ f(k) \equal{} p$ where $ p$ is prime. Then it is well known that $ dp\mid f(k\plus{}dp)\minus{}f(k)\Rightarrow f(k\plus{}dp)\minus{}p\equiv 0\pmod p\Rightarrow f(k\plus{}dp)\equiv 0\pmod p$ for all $ d\in\mathbb N$. This implies $ f(k\plus{}dp) \equal{} 0$ or $ p$ for all $ d$, which can be true only if $ f$ is a constant polynomial.


Define $ x\equal{}u^6,$ the conclusion follows.

Factor the following expressions.

$ A^2\minus{}B^2,\ A^3\minus{}B^3.$

Yes, both your suggestions work. My substitution was $ x \equal{} (t \plus{} 1)^{3}$ which gives the same solution but not as neat and easy as $ x \equal{} u^{6}$

$ \lim_{x \to 0} \frac {\sqrt [3]{x} \minus{} 1}{\sqrt {x} \minus{} 1} \equal{} \frac {0 \minus{} 1}{0 \minus{} 1} \equal{} \frac { \minus{} 1}{ \minus{} 1} \equal{} 1$.

However, you were obviously referring to $ L \equal{} \lim_{x \to 1} \frac {\sqrt [3]{x} \minus{} 1}{\sqrt {x} \minus{} 1}$. Make $ x \equal{} t^2$ to get

$ L \equal{} \lim_{t \to 1} \frac {t^{2/3} \minus{} 1}{t \minus{} 1}$ and this is, by definition, f'(1) with $ f(t) \equal{} t^{2/3}$. We have $ f'(t) \equal{} \frac{2}{3}t^{\minus{}1/3}$, and so $ L \equal{} f'(1) \equal{} \frac{2}{3}$.


search
inequalities
inequalities proposed

No one has a solution for this problem??? :(

It is a trival case of Cauchy-Swarch ineq
@VietNam: May be I know who you are.! :wink:

Oh,may be you know me because I'm in Vietnam.
But,first,you must solve this problem.

[quote="VIETNAM"]Let 's a,b,c>0 and $a^{2}+b^{2}+c^{2}=3$.
Find the minimum of P= $\sum_{cyc}\frac{a^{3}}{b^{2}+kbc+c^{2}}(k \in N,k \geq\ 2)$[/quote]
We will prove that for all $k \ge 0$:
\[\sum\frac{a^{3}}{b^{2}+kbc+c^{2}}\ge \frac{3}{k+2}\]
Applying Cauchy Inequalityand AM - GM inequality, we have
\[\sum\frac{a^{3}}{b^{2}+kbc+c^{2}}\ge\frac{(a^{2}+b^{2}+c^{2})^{2}}{\sum a(b^{2}+c^{2})+3kabc}= \frac{9}{\sum \sqrt{a^{2}(b^{2}+c^{2})}. \sqrt{b^{2}+c^{2}}+3kabc}\ge \frac{9}{\sqrt{\left(\sum a^{2}(b^{2}+c^{2})\right)(\left( \sum(b^{2}+c^{2}) \right)}+3k}\ge \frac{3}{k+2}\]
:)

The following is more interesting: $a,b,c \ge 0,a^{2}+b^{2}+c^{2}=1$. Prove that
\[\sum\frac{a^{3}}{b^{2}-bc+c^{2}}\ge \sqrt{2}\]
Actually, we can easily solve this problem by using Schur's Inequality. :)


[hide]The answer is 40/3. Basically we draw the diagram. If we call the point where the larger circle touches the radius 8 circle x, and another tangent point (it doesn't matter which) y, then we compute the distance from x to the center and the distance from y to the center, and they're both radii.[/hide]

Singular wrote:4 mutually tangent circles, 3 of which have radii 5, 5, 8, and one larger one.

Find the radius of the larger one.

u mean mutually externally tangent, right? If then, the 4th circle is actually smaller... I've seen this one before, here is how i did it before:

[hide]Let the radius of 4th circle be r, Let centers of the 3 circles with radii 5, 5, 8, r be O_1, O_2, O_3, O_4 respectively. Circle O_1 and O_2 tangent each other at point A. Triangle O_1O_2=10, O_2O_3=O_1 O_3=13, So O_3A=sqrt(13^2-5^2)=12. Now O_4O_1=O_4O_2=5+r,  O_4A=sqrt((5+r)^2-5^2) O_4O_3=8+r, so sqrt((5+r)^2-5^2)+8+r=12

solve for r, we get r=8/9[/hide]

ive seen the smaller one too beta, but i mean the larger one.

can someone be more specific about the larger one?  im trying to copy the solution above but i still dont get it.

Also, the problem said the 4th was larger.

ohh ok, it just reminded me of this AIME problem(1997, #4):

"4.  Circles radii 5, 5, 8, k are mutually externally tangent. Find k."

so I thought that's what you meant.... ok


i suggest posting this in the puzzles and brainteasers forum.

[quote="ritchjp"]the idea, I believe is to put in the appropriate symbols such +, -, etc. to make the equality true.[/quote]

if this is the case i am getting:

14 - 19 +7 +8/2 = 6

You have to use order of operations in order for this to work.


[url=http://olympiads.hbcse.tifr.res.in/uploads/rmo-2007]See the official solutions here...[/url]

Proof for part b;
We know 3>sqrt(8), since 9>8........(1).
We know 3>cbrt(8),since 27>8........(2)
We know 2>8^(1/4),since 16>8.........(3)
Adding (1),(2) and (3), we obtain the desired result.
For part c,
For natural numbers greater than 1,we know n^(1/4)<n^(1/3)<sqrt(n)
Thus n^(1/4)+n^(1/3)+sqrt(n)<3.sqrt(n)
Claim:3sqrt(n)<=n for all n greater than equal to 9.
Proof: Square both sides .
Hence we get,n^(1/4)+n^(1/3)+sqrt(n)<3.sqrt(n) <=n or
n^(1/4)+n^(1/3)+sqrt(n)<n QED


modular arithmetic
number theory solved
number theory

Let S={0,1,2,...,1994}. Given a,b element of S, s.t. (a,b)=1. Prove that we can find a sequence s_1,s_2,...,s_1995, such that s_(i+1)-s_i =  \pm a or  \pm b (mod 1995). for i=1,2,...,1994.

em... am i misunderstanding something?
what if s_i=1995i+a?

Maybe the problem is to write the elements of S in some order s_1, s_2,...,s_1995 such that the s_i's have the desired property.

Pierre.

Yes, you have to arrange elements of the set such that they have given property

[quote="harazi"]Nice and easy problem...[/quote]
What is your solution?  :lol:

Trust, but check, hmm? Let $ x$ be the first one such that $ xa$ is a multiple of $ 1995$ and let $ y=1995/x$. It is immediate to check that the map $ (m,n)\to  am+bn\pmod 1995$ is injective if $ m$ and $ n$ range in $ 0,x-1$ and $ 0,y-1$ respectively. Take now a sequence of $ y$ blocs of $ x$ numbers, the first bloc being $ a,a,...,a,b$, the next one $ -a,-a,...,-a,b$, the next one $ a,a,...,a,b$ and so on. Take $ s_{i}$ to be the sum of the first $ i$ terms of this chain, then $ s_{i}$ works.


Find all polynomials [i][b]P(x,y)[/b][/i] such that for every reals [i][b]a,b,c,d [/b][/i]we have:
 [i][b] P(a,b)P(c,d)=P(ac+bd,ad+bc)[/b][/i]

[url=http://www.kalva.demon.co.uk/balkan/bsoln/bsol882.html]there[/url] you go ! Balkan 1988 prob#2


The first term of a given sequence is 1, and each successive term is the sum of all the previous terms of the sequence. What is the value of the first term which exceeds 5000?

If you start listing,

$ 1, 1, 2, 4, 8, 16, 32,...$

It becomes clear that the sequence is just all powers of 2, starting on the second term. So,

$ 2^n > 5000$

We find that it's just $ 2^{13} \equal{} \boxed{8192}$.

[quote="GameBot"]The first term of a given sequence is 1, and each successive term is the sum of all the previous terms of the sequence. What is the [u][b][i]value[/i][/b][/u] of the first term which exceeds 5000?[/quote]

Since it says value, wouldn't the answer be $ \boxed{8192}$?


function
combinatorics unsolved
combinatorics

Sum_{i=0}^{k}{(p^i)*C(n+i,n)},where 0<p<1, k and n are all constants. C(n+i,n) means choosing n from (n+i).
When k=Infinity, there is solution. My question is when k is finite, what is the closed form solution of this sum?
Thanks very much!

Maple expressed this sum in terms of hypergeometric functions: \[\frac{1}{(1-p)^{n+1}}-p^{k+1}{n+k+1\choose n}{}_{2}F_{1}(1,n+k+2;k+2;p)\]

How do you get the formula? Just using Maple?


Each of the 15 student council members of Alamo Mid-school is responsible for telling exactly two other students about the upcoming school dance. Each of those two, in turn, is responsible for telling two other previously uninformed students. It takes 15 minutes for each round of students to be told. How many minutes will it take for 1,000 students to receive the information about the school dance?

I think you have to find the smallest value of $ n$ in the following equality:

$ 15 \times 2^n > 1000$

Then, the answer will be $ 15n$.

$ 2^n > 66.\overline{6}$

$ n$ is clearly $ 7$.

Thus, the answer is $ 15(7)$ or $ \boxed{105}$.

I just asked this question and was led here. The answer given in the review was 90, not 105.

The correct inequality is actually

$ 15(1 \plus{} 2 \plus{} 2^2 \plus{} 2^3 \plus{} \cdots \plus{} 2^n) \equal{} \frac {15(2^{n \plus{} 1}\minus{}1)}{2 \minus{} 1}$

$ \implies 2^{n \plus{} 1} > 67.\overline6\implies n\geq6\implies15\cdot6 \equal{} \boxed{90}$,

because the amount of students told in the $ n^{th}$ fifteen minute period is $ 15\cdot2^n$ (and we have to add the original $ 15$ student council members, as well).


function
topology
real analysis
real analysis unsolved

Prove that from each sequence of elements of the set of continous functions $ f: R \rightarrow R$
(with the $ \alpha \equal{} \frac {3}{4}$ holder norm
\[ \parallel{}f\parallel{}_\alpha \equal{} |f(0)| \plus{} \sup_{x,y\in[0,1],\ x\neq y} \frac {|f(x) \minus{} f(y)|}{|x \minus{} y|^\alpha}
\]
)
 we can choose a subsequence that forms cauchy sequence.

(Arzeli Ascoli)

I don't know how we call in englisch a metric space $ (X,d)$, that has following feature:
From every sequence of elements from that space we can choose a subsequence that forms cauchy sequence
that is why the text of this task is much longer than in my mother tounge

Please, be careful when stating problems. What is written hardly makes any sense. I, probably, could guess what was meant but it will be much better if you edit the formulation yourself.


AMC USA(J)MO USAMO AIME inequalities probability symmetry

First, I would like to make Alabama student more aware of the Mandelbrot competition.  It is a great competition that requires a greater depth of problem solving than other math leagues and local competitions.  Some Alabama students already compete on it.

The most interesting aspect of the Mandelbrot Competition is the team round.  Teams of students must write proofs.  If the whole proof where given as one problem, they would typically be as hard as an easier USAMO problem (sometimes even a hard USAMO problem).  However, broken up into parts, they are a great way to train for writing proofs.  The parts naturally break the proofs up into the different points of exploration that lead to deeper critical thinking.

I am not currently coaching a math team for the Mandelbrot.  If you students tell me what each next team test topic is, I will post problems on this forum relating directly to the topic so that you can practice writing proofs on each topic.  Students who take the time to do this are likely to increase their AIME scores and be far more likely to do well on the USAMO should they make it that far.

Post the next topic each month under this thread and good problems you will have.

Four of the sophomores, including me, Seva, and Evan (asdf2345), make one of our teams for the Mandelbrot team tests. I don't think our solutions were very good for the first round. Graph theory confuses me. :( Tipton never told us the topic, but I got it on the forums the day before. I'll have to get her to tell us next time.

I also did bad on the first round individual. :/

I'm not sure when we are taking the team test, but here's our topic.

[b]Round Two:[/b] Two positive integers are said to be [i]relatively prime[/i] if they have no common divisor other than 1. For instance, 35 and 66 are relatively prime, but 63 and 56 are not, because they are both divisible by 7. A nice result states that if $m$ and $n$ are relatively prime positive integers, then one can find multiples of $m$ and $n$ which differ by only 1. More precisely, there exist unique integers $a$ and $b$ such that $am - bn = 1$ with $0 \le a < n$ and $0 \le b < m$.

Topics: relatively prime integers, rational numbers, rounding, linear inequalities

Thanks much for the help.

The very first thing I would suggest is for somebody in the group to have taken the AoPS Introduction to Number Theory class which deals with linear Diophantine equations such as the one you mentioned.  Since you can't go back in time, I'll dig up some ideas that would be worth exploring in relation to this topic and post them later.

Incidentally, problem 2 from the 2004 USAMO dealt with exactly the subject that it sounds like you'll be tackling on this Mandelbrot team round (in a bit deeper way).  From what I've seen of the Mandelbrot (took it for 4+ years and mind problems from past competitions for my classes at AoPS), the team round is one of the absolute best ways to begin preparing for the USAMO (and really even the AIME).

[b]Round Three:[/b] Suppose that a certain procedure yields a variety of equally likely outcomes. For example, rolling two fair dice leads to $6^2 = 36$ outcomes. If there is a quantity that we can measure in each possible outcome, we may speak of the average, or [i]expected value[/i] of that quantity. Thus we may ask for the average total on the faces of the two dice. To compute expected value, simply average togethers all possible values of the quantity. In this way we find that expected total on two dice is $(2 + 3 + 3 + ... + 11 + 11 + 12)/36 = 7$.

Topics: expected value, counting techniques, parity

This is yet another topic that the Intermediate Counting and Probability class should help greatly with.

Since there is parity involved, Sam might have written a problem that involves symmetries (like even/odd symmetry around the expected value of one roll of one die).  This means looking for cases that are essentially the same calculation in order to do something interesting.

Here's a problem for anyone interested:

If one of the first $n$ Fibonacci numbers is selected at random, what is the probability that the number selected is greater than the expected value of the number selected?

Here's an exercise that might help with the third round of the Mandelbrot.

There are three basic types of probability problems: counting, geometry, and algebra.

Counting are the most common at the middle and high school competition levels.

Given an example of a probability problem solved through geometry and give an example of one solved purely through algebra.

MCrawford wrote:Given an example of a probability problem solved through geometry

Person A (Let's call him John) and Person B (Let's call him James) want to meet. James plans to arrive at the meeting place between 12:00 and 12:20, and will only stay 5 minutes before leaving. John will arrive between 12:00 and 12:25, and will only stay 10 minutes before leaving. What is the probably that they meet?

You can solve this by drawing a rectangle, with say sides of 20 and 25 (each unit representing a minute), and then an area of probability will occur; find the area of that over the area of the rectangle to get the probability they will meet:

[hide][tex]\frac{1}{2}[/tex] [/hide]

Great example!

Now, here's one you could solve via algebra.  Give a solution if you can.

Kay and Vincy play a game in which Kay begins by rolling a regular fair 6 sided die.  If Kay rolls a 6, she wins.  If not, it's Vincy's turn.  Likewise, when Vincy rolls a 6, she wins.  Otherwise, the game continues back and forth until somebody rolls a 6.

What is the probability Kay rolls the first 6?

MCrawford wrote:Kay and Vincy play a game in which Kay begins by rolling a regular fair 6 sided die.  If Kay rolls a 6, she wins.  If not, it's Vincy's turn.  Likewise, when Vincy rolls a 6, she wins.  Otherwise, the game continues back and forth until somebody rolls a 6.

What is the probability Kay rolls the first 6?

[hide]

Kay has many 'places' to win; either on the very first turn, the third turn, the fifth turn, etc. For her to win on the third turn, the first and second turns must 'fail' the roll, with a probability of 5/6 of failing. Thus, her probability can be expressed like so:

[tex]\displaystyle\left(\frac{1}{6}\right) + \displaystyle\left(\frac{5}{6}\right)^2\displaystyle\left(\frac{1}{6}\right) + \displaystyle\left(\frac{5}{6}\right)^4\displaystyle\left(\frac{1}{6}\right) + ...[/tex]

[tex]\displaystyle\left(\frac{1}{6}\right)\cdot\displaystyle\left[1 + \displaystyle\left(\frac{5}{6}\right)^2 + \displaystyle\left(\frac{5}{6}\right)^4 + ...\right][/tex]

[tex]\displaystyle\left(\frac{1}{6}\right)\cdot\displaystyle\left(\frac{36}{11}\right)[/tex]

[tex]\displaystyle\left(\frac{6}{11}\right)[/tex]

Kay has a probability of 6/11 of winning the game. (And innately has an unfair advantage over the second player due to her getting the first turn . Poor Vincy.).

[/hide]

Nice work, however, you could simplify this solution greatly.  The series sum is not necessary to calculate at all.  Do you see why?

Hey there everybody. Final Mandelbrot round next week.

[b]Round Four:[/b] Given a point $L$ outside a circle, there are two lines through $L$ tangent to the circle, and the distances from $L$ to the points of tangency are the same. Furthermore, inside any triangle there is a circle tangent to all three sides, called the [i]inscribed circle[/i]. The angle bisectors of each angle of the triangle pass through the center of this circle.

The measure of an angle inscribed in a circle is half the measure of its intercepted arc; this holds even when one side of the angle is tangent to the circle. If points $X$ and $Y$ lie on the same side of the line through $Z$ and $Z'$, the the four points are cyclic (lie on a single circle) if and only if $m\angle ZXZ' = m\angle ZYZ'$. If $X$ and $Y$ lie on opposite sides of line $ZZ'$, then the points are cyclic if and only if $m\angle ZXZ' + m\angle ZYZ' = 180$.

(A convex quadrilateral $ABCD$ has an inscribed circle if and only if $AB + CD = BC + AD$. If this circle is tangent to sides $\overline{AD}$ and $\overline{BC}$ at $U$ and $V$, then $\overline{UV}$,  $\overline{AC}$, and $\overline{BD}$ are concurrent. Finally, suppose that points $E_1$, $E_2$, and $E_3$ are on one line, while points $F_1$, $F_2$, and $F_3$ lie on another line. If lines $E_1 F_2$ and $E_2 F_1$ meet at $G_1$, $G_2$ and $G_3$ are collinear.)

Topics: inscribed angles, cyclic quadrilaterals, inscribed circle, (Pascal's theorem), (collinearity)

I would suggest going straight to the chapters in the AoPS books that deal with these subjects.

It's likely that you'll be doing a bit of angle chasing -- so make nice, big diagrams.

It's also likely that you'll be dropping tangents, possibly as radii, and the right angles involved could figure in to that angle chasing.

Good luck!


I apologize for asking this question in a maths forum, but I really didn't found a better one after searching for a long time.

I'm from Germany and I have do do an presentation about discrimination against blacks in the USA today. Because I live in a very rural area I hadn't very much contact with blacks by myself. I spent a long time on searching the internet for some information but yet, I didn't find many useful things.
So I had the idea to ask in a forum...

So my question is:
Is there still discrimination against blacks in the USA today, and if yes, in what forms does it appear?

I'd be very thankful if you could help me.

look in textbooks, like history or maybe gov books.
i know my has a chapter on it

I am going to give you over 200 years of history in 2 minutes.  Ready, Go!

In 1864, the Civil War over slavery (aka blacks) was raging.  The anti-slavery North beat the pro-slavery South, and all slaves were freed.  After the Civil War, blacks were treated like dirt, with the law not caring for them at all.  By the 1960's, a law was passed that said blacks and whites could be segregated, having seperate facilitites, as long as they were "seperate but equal."  These were Jim Crowe laws.  But it was not so.  When blacks really started going for equality, the KKK, or Ku Klux Klan, came up.  They were pro-white, and they burned and lynched houses of blacks and black people as well.  Some leading figures, such as police chiefs, were KKK Klansmen, and here was no fairness for blacks.  After Martin Luther King and his campaign for equal rights, a case known as Brown vs. the Board of Education came into the courts.  The Supreme Court ruled that seperate facilities were unfair, and the rest is history.  So now, most people do not openly hate or discriminate blacks.  But some people still harbor feelings of resent toward Black Americans.  A few are still even members of the KKK, though it is much more secret now.  As a general rule, the lower south you go, the more resentment and hatred there is toward blacks.  Now that is a stereotype, and while it is true, most Southerners are great people who would never think of mistreating blacks in any way.  But to answer your question, yes, there are still many people out there who discriminate against blacks.  Most will not admit it, however, and even fewer openly show it.

In retrospect, that was more history than you needed.  But I hope it gave you an idea of how it is.  That may have been obvious to you, but I wasn't sure how familiar you are with US History.  Please let me know if I can help you further.

Also, to answer another question you asked, there are many forms of discrimination against blacks.  Recently, a lawsuit was initiated against a Southern restaurant, with the plaintiffs claiming that blacks were served slower than whites, and, in some cases, blacks were denied service.  I don't know what happened to the case, but this is one of the most discreet and common forms of discrimination against blacks.  But it still does not happen very often, as far as we can tell.

However, there are lesser forms of discrimination.  Look at the survey [url=http://www.jointcenter.org/DB/table/NOP/NOP_97/Race%20relations/DISCRIM3.htm]here.[/url]  The results are very surprising.

[url=http://www.raceandhistory.com/cgi-bin/forum/webbbs_config.pl/noframes/read/1177]Here[/url] is another article, but it is not as helpful.  Let me know if I can help you any  further.

- MySmartMouth

[quote="yif man12"]Define discrimination.

The amount of people who think all blacks are stupid and worthless and should still be slaves are very small now, but still existent in the US. The KKK still exists in small parts here and there. They operate by the internet a lot now, and still sometimes come out. The town where my school is in has a few members of the KKK. One teacher recalled once when about 4 of them came out during a parade for MLK day, I think. They didn't do anything, though, since there were about 800 or so students and other people.

That kind of radical discrimination exists in small factions around the country. Discrimination exists more commonly as a semi-subconscious thing, such as a driver locking his or her doors when passing a group of african americans in the city, or someone just looking at an african american and making generalizations inside. Most of this exists internally though.

So discrimination against african americans definitely still exists. The larger discrimination has very few people, but many many people are subconsciously discriminating in their lives.[/quote]

KKK? What city? or state?

[quote="beta"]There will always be some people who will judge others by their color of skin, and there's nothing you can really do about it unfortunately.[/quote]

yeah someone on this forum i know is Indianist..... :(

First of all I want to thank you for helping me. I think I can use some information you gave me.
On searching the internet, I found an article about police brutality against blacks: http://www.marxist.com/usa/police_brutality600.html
Do you think this is a serious source?

I would suggest not relying too heavily on the history presented by mysmartmouth, as it (1) is a huge simplification of a number of rather complex events that happened over a very long period of time, and (2) contains a number of errors.  I suggest searching for information about the civil rights movement if you're looking for a historical view.  If you're looking for a modern-day view, reading about police brutality is a good place to look.  (I have not looked at the website you posted a link to, but I would suggest you look for newspaper or other sources -- Marxists are not always the most unbiased of observers.)  There is real racial tension that persists across this country -- there were major race riots in LA and NY in the 1990's, and racial tensions can easily be dug up in Chicago, Boston, and dozens of other cities across the North, not just in the South.  (I believe there were relatively recent, fairly major issues happening in Cincinati, but I don't know anything more substantial than that I saw some headlines.)

You might also consider looking at other, more subtle forms of economic discrimination that persist, such as housing discrimination.  One important thing to realize is that with the Civil Rights movement, overt discrimination has become almost entirely unacceptable in American society, so deciding what constitutes discrimination and whether or not it is practiced has become a much more difficult question.

[quote="LynnelleYe"]According to my English teacher, if it only happens inside a person, it's prejudice. Only when it translates into action does it become 'discrimination'.[/quote]

Well, I don't want to be rude, but you could suggest your English teacher to check his English dictionary, because

[quote="The American Heritage® Dictionary of the English Language"][b]Discrimination[/b]: Treatment or consideration based on class or category rather than individual merit; partiality or prejudice.
[b]Prejudice[/b]: An adverse judgment or opinion formed beforehand or without knowledge or examination of the facts. Detriment or injury caused to a person by the preconceived, unfavorable conviction of another or others.
[/quote]

Therefore, the discrimination do not need to be an action, a simple consideration would fit in the definition.

[quote="mysmartmouth"]KKK? What city? or state?[/quote]

Atlanta, Georgia or Robbinsville, North Carolina, just as two examples that I personally have seen. And I would bet that in the small towns in the South of the US (the so called [i]RedNex Towns[/i]) there are a lot more.

The KKK doesn't just exist in the south either. It exists throughout the country. I'm just building on what DJimenez said. Also, if you google for the KKK and recruitment, you may find sites recruiting for the KKK.

It's something that most people are pretty oblivious to, the existence of the KKK today. I was really surprised when one of my teachers mentioned it and said that there were even KKK in Wallingford, CT, though only a couple members.

to original poster:

Look at Southern Povery Law Center:  they are a watchdog group for hate groups. I might be too late, but it would be helpful to talk about 'modern' racism ------> racial profiling and the Rodney King 'incident.'[url]http://www.splcenter.org/[/url]

It's not the KKK exactly, but a similar organization, the Aryan Nations, operates out of... not the South, but OREGON. (It's NeoNazi, they don't discriminate in their discrimination! They just don't like anyone!).

Also, if you are doing a report on discrimination TODAY, you most definitely need this link:
http://www.birminghampledge.org/
The Birmingham Pledge was started in Birmingham, AL. However, it has been signed by people all over the country and the world. I recommend that all AoPS forum members sign it, if you can. By signing, you are promising to do all you can to fight prejudice and discrimination. It's a really great program that, among other things, has helped turned the South's image as "RedNex" (it's actually rednecks, which comes from men who worked in the fields all day and got sunburns on the back of their necks). While many people outside of the South still view it as backwards and black-hating, I have had more than one African-American, including my aunt, tell me that they feel safer from racial discrimination in Georgia, Alabama, or the Carolinas than they ever did in the North. Think about that. Racism is NOT just a Southern problem, and it never has been just a Southern problem (there were many race riots in the North because of blacks taking lower-paying jobs, although this is not generally covered in high school/middle school courses). The most important thing is to be aware of your own prejudices and to work to combat them.

I apologize for going so off course, but I feel someone has to speak up for those who live below the Mason-Dixon line. I am not saying that Southerners still don't have their problems, but they are working actively to fix them, which is more than a lot of people could say. Seeing as how I've heard both Richard and Mathew are from Alabama, I'd LOVE to hear their opinions on this matter.

To mysmartmouth, Do you think what you said is self-consistent?

"In 1864, the Civil War over slavery (aka blacks) was raging. The anti-slavery North beat the pro-slavery South, and all slaves were freed. After the Civil War, blacks were treated like dirt, with the law not caring for them at all"

Q: Why did south and north fight?
A: The north didn't want slavery, but south did.
Q: Who won?
A: North.
Q: Were the slaves set free after the war?
A: well.....

The only reason for the war is the south want to split from US.
At first, the north was beated black and blue, but the well developed industry provided inexhaustible ammunition, and promisesing abolition attracted many blacks into the army, thus the south got no chance to win, though they have general Lee, who was an abolitionist, general Grant, in the north side, owned quit some slaves.
The civila war was won partly BY PROMISESING abolition, not FOR abolition.

[quote="JBL"]I would suggest not relying too heavily on the history presented by mysmartmouth, as it (1) is a huge simplification of a number of rather complex events that happened over a very long period of time, and (2) contains a number of errors.[/quote]

He's definitely right about number 1, and most likely right about number 2!  Sorry!

[quote="Buffalo"]Q: Why did south and north fight?
A: The north didn't want slavery, but south did.
Q: Who won?
A: North.
Q: Were the slaves set free after the war?
A: well..... [/quote]

I know, that was a watered down, simplified, half of an idea that I gave.  Should I delete that post, or just leave it be?

I see no reason to change your post because someone or ones have criticized it.

Buffalo -- what's your alternative theory?  That the Southern states were righeous defenders of state's rights?  "The only reason for the war was that the South wanted to split from the North."  Well, sort-of: the South threatened to split of Lincoln was elected.  Why?  Because they were afraid he would abolish slavery, for the most part.  The preservation of the institution of slavery was absolutely central to that decision.  Would the North have gone for immediate abolition?  Probably not: painting the North as only anti-slavery is inaccurate, but painting the South as not pro-slavery would be far more inaccurate.


Okay, so this is the situation: I bought a new computer and I want to install WinEdt. On my previous computer, it worked well, but now I'm having some problems. I downloaded MiKTeX and WinEdt, but he still gives me an error:

Error starting PDFTeXify!
See WinEdt's TeX Configuration Wizard for detailed diagnosis of your TeX system.

So, I went to that Wizard, and there, I am told that

There are some problems with your TeX installation. Depending on your intended use of WinEdt and TeX you may have to install additional software, or missing TeX Components...

Anyone has an idea what I should do? I'm getting kinda desperate.

I don't have WinEdt but looking at Troubleshooting at http://www.winedt.com/ the Diagnosis tab of the Wizard should tell you what is missing and where it is looking so you could change the setting.

I also notice that it suggests reinstalling MikTeX then restarting WinEdt which sounds a good idea.

Well, I did the reinstalling like 5 times tonight. It didn't really help.

Is there a good substitute for WinEdt? (Which works like WinEdt.)

Try [url=http://www.toolscenter.org/]TexnicCenter[/url] which is used by many AoPS users

Oh sorry. I thought he needed something for pics

yeah texniccenter is good but acrobat 8.0+ doesn't work completely with it so i kept acrobat 7.0

Hmm, I downloaded TeXnicCenter, but I fail to see how to create a PDF  :D

In the box in the toolbar where it says LaTeX=>DVI change it so it says LaTeX=>PDF then Build your output and view it (I find it easier to use the icons on the toolbar to the right of the LaTeX=>PDF box, but you can use the Build menu or the key presses it says in that menu).

Hmm, the toolbar doesn't say anything over here  :|

I read something about "Output Profiles", but I didn't manage to create one...

What happens if you go to Build, Select Output Profiles? Do you get a list of profiles? TexnicCenter should have made at least 3 of them automatically.

Apparently, he didn't...I don't get it. I downloaded MiKTeX, so he should normally find everything he needs, no?

You do need to tell it where to find MikTeX. Go to Build, Define Output Profiles and click on Wizard which is at the bottom left. It will ask you where latex etc is. You need to press the button with ... to show it where that is. Assuming you've installed MikTeX to C:\Program Files then the directory it wants is C:\Program Files\miktex\bin. If you've installed it anywhere else then it will be anywhere_else\miktex\bin. Then TexnicCenter will make the Output Profiles for you.

This was asked when you first opened TexnicCenter but maybe you cancelled it?

It's not my lucky day.  :(  I can't see any C:\Program Files\miktex\bin

I installed MiKTex 2.5...should I try 2.4?

Edit: I think I fixed it.

Edit: OMG, finally  :D  Thank you, stevem.


geometry
3D geometry
sphere
calculus
calculus computations

I dont know how to start with this problem,  Any help is highly appreciated.

Find the flux of F =<x,y,z>/[(x^2+y^2+z^2)^3/2] across any closed oriented surface S with outward orientation that surrounds the origin.

Since it didn't tell you what the surface was, that's a strong hint that what the surface is doesn't matter.

Step 1: Find the flux through a sphere centered at the origin.

Step 2: Use the Divergence theorem to show that the flux through your arbitrary surface is equal to the flux through that sphere.


algebra
difference of squares
special factorizations

I know this may seem simple to some of you, but can someonw tell me the trick to solve the series questions that always go from n=0 to the current year. Such as

(-1)^2*n^2  from n=0..2008

any suggestions?

Are you asking $ \displaystyle\sum_{n \equal{} 0}^{2008}( \minus{} 1)^{2}(n^{2})$ or $ \displaystyle\sum_{n \equal{} 0}^{2008}( \minus{} 1)^{n}(n^{2})$.

For the first case, in general, the sum of the squares of the integers from $ 1$ to $ n$ is
\[ \displaystyle\sum_{k \equal{} 0}^{n}k^{2} \equal{} \frac {n(n \plus{} 1)(2n \plus{} 1)}{6}
\]
You can go about proving this either

$ \bullet$ using the Finite Difference Method (http://en.wikipedia.org/wiki/Finite_difference)
$ \bullet$ combinatorially by noting that $ k^{2} \equal{} 2\binom{i}{2} \plus{} \binom{i}{1}$, and then applying the hockeystick identity to evaluate the sum (http://www.artofproblemsolving.com/Wiki/index.php/Combinatorial_identity)

There's also numerous other ways but those are just a few to get you started.

If the latter is what you are asking, then consider the sum
\[ S_{n} \equal{} 1^{2} \plus{} 2^{2} \plus{} 3^{2} \plus{} 4^{2}\cdots \plus{} (n \minus{} 1)^{2} \plus{} n^{2} \qquad...(1)
\]
For even $ n$, note that we can also write
\[ 2^{2}(S_{n/2}) \equal{} 2^{2} \plus{} 4^{2} \plus{} \cdots \plus{} n^{2} \qquad...(2)
\]
Now with $ (1) \minus{} 2\cdot(2)$, you can get your desired sum.


calculus
integration
inequalities
number theory
least common multiple

How many integral values exist for $ x$ such that $ \frac14 < \frac{x}{5} < \frac23$?

hello, your inequality is equivalent with $ \frac{5}{4}<x<\frac{10}{3}$, so the only integral solutions are $ x\equal{}2$ or $ x\equal{}3$.
Sonnhard.

Um... I don't really get that... I have a different way to do it; or it's the same way explained... I'm sorry if it's the same thing!
$ 1/4<x/5<2/3$
An easy way to do it would to be to find their LCM of the denominators. The LCM of the numbers is 60 (or just a multiple I'm not sure...  :huh: ) so now the expression becomes $ 15/60<12x/60<40/60$ The only numbers in between 15 and 40 that are divisible by 12 are 24 and 36, so x can only be $ 2$ or $ 3$

$ x\equal{}2, x\equal{}3$

What Dr. Sonnhard Graubner did was he multiplied everything by 5.

Dr. Graubner, please include explanations in your posts from now on.

hello, yes it is correct, i'm multiplied the given inequality $ \frac{1}{4}<\frac{x}{5}<\frac{2}{3}$ by $ 5$, i thought that was clear,sorry. In the future i will give more explanations in my posts.
Sonnhard.


conics
hyperbola
algebra
polynomial
Vieta
Diophantine equation
number theory proposed

[i]In Math olympiad treasures 3.5.2[/i]
[quote]
prove that  \[ x^2\plus{}(x\plus{}1)^2\equal{}y^2\]

has infinite pos int solutions

[b]solution:[/b]
by explicitly constructing:
$ (x_0,y_0)\equal{}(3,5)$
\[ x_{n\plus{}1}\equal{}3x_n\plus{}2y_n\plus{}1\\ y_{n\plus{}1}\equal{}4x_n\plus{}3y_n\plus{}2\][/quote]

prove that this family of solutions is the only family of solutions.

This is a consequence of the [url=http://en.wikipedia.org/wiki/Dirichlet%27s_unit_theorem]Dirichlet unit theorem[/url], although this special case has an elementary solution.  First reduce to a Pell equation, then argue by contradiction.  (I believe this argument is in Engel somewhere.)

Also, x is a solution to the above iff there are integers m,n such that

$ x \equal{} \frac{1\minus{}i}{2}(m\plus{}ni\plus{}1)(m\plus{}ni\minus{}1)$

Can you please elucidate how you would apply the Dirichlet Unit Theorem?
Anyway, the idea behind this problem is that the pythagorean triplets are easily parametrized, so the problem reduces to finding integer points on the hyperbolas $ m^2\minus{}n^2\minus{}2nm\equal{}\pm1$, whereupon one uses the typical method of viete root jumping to derive a sequence of solutions that hits every solution.
Please show how this is a special case of a well known method :)

[quote="me@home"]Can you please elucidate how you would apply the Dirichlet Unit Theorem?[/quote]
Reduce to $ (2x \plus{} 1)^2 \minus{} 2y^2 \equal{} \minus{}1$.  The unit theorem implies that the group of units of $ \mathbb{Z}[ \sqrt{2} ]$ has rank $ 1$, so it remains to find a generator of norm $ \minus{}1$ and take its odd powers.

[quote="me@home"]Anyway, the idea behind this problem is that the pythagorean triplets are easily parametrized, so the problem reduces to finding integer points on the hyperbolas $ m^2 \minus{} n^2 \minus{} 2nm \equal{} \pm1$, whereupon one uses the typical method of viete root jumping to derive a sequence of solutions that hits every solution.[/quote]
Vieta jumping doesn't usually come with a guarantee that you'll get every solution.  Could you give more details?


calculus
integration
geometry
trigonometry
real analysis
real analysis unsolved

Consider the solid forming all of the points $P(x,y,z)$ which satisfy the condition in $xyz$ space,
$x^2+y^2 \leqq z^2,z^2 \leqq x,0 \leqq z \leqq 1$.Find the volume of this solid [b]without using double integral[/b].

Perhaps because recalling the formula for the area of a circular segment and finding the antiderivatives of $ z^2\arccos z$ and $ z^3\sqrt{1\minus{}z^2}$ are neither particularly interesting, nor entirely trivial. ;)

Seriously, if you have a nice solution and nobody has found it, post it yourself; if not, then don't bump a topic that is almost 3 years old. :)

Cut the solid by the plane $ z\equal{}\cos \theta$.


Does anyone know where I can find a list of the USAMO qualifiers from 1998.  I can't find it on the AMC website for some reason.

Back in 1998, we were not using the Web extensively and to my knowledge did not post those lists in any medium, students got contacted individually and that was the extent of it.

Although the records exist farther back than 1999, they are buried deep in our electronic archives and it  takes time and effort to bring them back to light.  The 1998 USAMO winners and Honorable Mentions are listed on page 190 of the 1998 Summary of Results and Awards.  If it is important to you, contact the AMC office separately at amcinfo@unl.edu and provide a postal mailing address and we can send you a copy of that page.

Steve Dunbar
AMC Director

I have that book at home. WarpedKlown, your profile says you're from Florida, so I'll guess that's what you're interested in. Here are the Florida USAMO qualifiers in 1998:

D. Moraseski, Dr. Phillips HS, Orlando (USAMO honorable mention, attended MOP)
W. Cheng, Leon HS, Tallahassee
A. Curtis, Hillsborough HS, Tampa
D. Wilson, King HS, Tampa
D. Brown, Vero Beach SR HS, Vero Beach
B. Richardson, Vero Beach SR HS, Vero Beach
A. Ockman, Winter Park HS, Winter Park

Oh wow haha I'm good friends with Dan Moraseski's brother (high school buddies), and Adam Ockman qualified?  0_o...wow I knew of them, both of them were before my time though.

*P.S. it's because I live in Winter Park, and both of those high schools happen to be in my county.*


prove that every positive integer is uniqely expressible in the form
$ 2^a\plus{}2^b\plus{}2^c\plus{}..........\plus{}2^m$
where $ 0\le a<b<......<m$


trigonometry
inequalities
geometry
function
trig identities
Law of Cosines
complex numbers

Label the angles, then using the fact: $l=d\sin{\theta}$, to transformt the original equation into a trig identity.

[quote="random_mathematician"]prove ptolemy's theorem

let ABCD be  cylic quad, a,b,c,d the sides
m,n, the diagonals

prove: ac+bd=mn[/quote]

Think of a certain trig identity.

And btw, I don't know what OMFG stands for. I wonder if you could tell me. It couldn't stand for anything bad, could it :P ?

[quote="Treething"]And btw, I don't know what OMFG stands for. I wonder if you could tell me. It couldn't stand for anything bad, could it :P ?[/quote]

http://www.acronymfinder.com is your friend.

I've never actually tried to prove Ptolemy's theorem (shame on me), but I would think that the law of cosines or sines would be pretty useful, especially since if A and C are oposite angles, C=180-A (degrees)

OMFG, why does everybody bash down Ptolemy with trigonometry? There is a quick synthetic proof at http://cut-the-knot.com/proofs/ptolemy.shtml (or http://groups.google.de/groups?hl=de&lr=&newwindow=1&threadm=e3850c89.0210062058.292d1793%40posting.google.com&rnum=1 for a do-it-yourself-solution).

  Darij

[quote="darij grinberg"]OMFG, why does everybody bash down Ptolemy with trigonometry?[/quote]

Because it looks like the most obvious approach :D

i have a proof with areas. But that concerns loads of lemma....i'll try to post it if i have time.

Ptolemy's theorem really is a trig identity- its original use was in as the angle-addition identity for building astronomical tables.

My favorite proof of Ptolemy's theorem uses complex numbers, and also proves the inequality $ab+cd\ge mn$ for any quadrilateral.

Consider a quadrilateral with vertices $A,B,C,D$. Let these vertices be represented by complex numbers $a,b,c,d$. It is an algebraic identity that
$(a-b)(c-d)+(b-c)(a-d)=(a-c)(b-d)$. By the triangle inequality, this tells us that $AB\cdot CD+BC\cdot AD\ge AC\cdot BD$, with equality if and only if $\frac{(a-b)(c-d)}{(b-c)(a-d)}$ is a positive real number. Establishing this fact when $A,B,C,D$ lie on a circle and are in that order is only a matter of angle-chasing; the argument of $\frac{a-b}{b-c}$ is the directed angle $\pi+\angle ABC$ and the argument of $\frac{c-d}{a-d}$ is the directed angle $\angle CDA$, so the argument of the product is $\pi+\angle ABC+\angle CDA=2\pi\equiv0$.

[quote="xxreddevilzxx"][quote="Treething"]And btw, I don't know what OMFG stands for. I wonder if you could tell me. It couldn't stand for anything bad, could it :P ?[/quote]

actually, it is somewhat bad. It has a swear word in it.  ;)[/quote]I think Treething knows very well what "OMFG" means, and he was just messing with you ...  :)

what does it mean? really don't know what you are talking about....
If it contains sth bad, can you please pm me the meaning of the four letters??

Trigonometry is not only the study of triangles. In fact, I even consider it as a branch of Algebra. It is only a tool in geometry, like complex number. For example, can you solve the this problem by geometry?
For $n\in\mathbb{N}$, $\sum_{i=1}^{3n}|\sin i|>\frac{8n}{5}$.

It can be proved using inversion:

http://mathcircle.berkeley.edu/BMC6/ps0405/inversionSJ04.pdf

http://matholymp.com/TUTORIALS/Ptolemy.pdf

[quote="joml88"]Trigonometry is just the study of triangles...I don't see why trigonometry and geometry should be considered different subjects.[/quote]

No, trigonometry is more generally a study of the exponential function (and its real and imaginary parts).  Trigonometry and geometry are different because reducing a geometric problem to a trigonometric problem almost inevitably hinges on using some sort of analytic consequence of the trigonometric functions and not purely geometric methods.

[quote="probability1.01"][quote="mecrazywong"][quote="probability1.01"]I know of a purely geometric proof of Ptolemy's, but it's hard to explain.  It involves shuffling the side lengths of the quadrilateral.  It also uses law of cosines.  The first step is dividing the diagonals into separate parts (where they intersect), and using law of cosines on the four triangles that are then obtained.[/quote]
You cannot say a proof is purely geometric if it uses cosine law.
Anyway, it's easy to prove by spiral similarity-even for the general case(inequality).[/quote]

Well in this case, the cosine is really just acts as a place holder.  There are no trig identities used.  One might as well use just similar triangles and the pythagorean theorem, but that would be a real pain.[/quote]
Of course it is if you consider the proof in this way, but personally I hate such "brute force" solution, especially when there are quite a number of elegant solutions.
Anyway, it will really be painful to prove the general Ptolemy's theorem using the same approach, though obviously it works.


Title is self explanatory.

The link is [url]http://www.octmtournament.org/octmattach%5Ctestresults%5C000000000073.pdf[/url]

You get more money for being tied for 4th in the OCTM than being 4th overall in the OHMIO.  And the difference between 4th and 3rd overall in the OHMIO is only 50 dollars of cash and a TI 89 Titanium. *mutters*

O well, I should be grateful I did as well as I did. :yup:

wow i did bad. only ranked 44.

i still think that last question was worded ambiguously.


search
function
combinatorics unsolved
combinatorics

let $S=\{x_1,x_2,\dots,x_n\}$be $n$ points in plan such that the distance betwen any two of them be at least $1$.
show that maximum there is $3n$ points that their distance be exactly $1$.

When will you start using the search function ¿
http://www.mathlinks.ro/Forum/viewtopic.php?t=44088
http://www.mathlinks.ro/Forum/viewtopic.php?t=67005


function
induction
algebra proposed
algebra

Let f:N->N an increasing function such that N-Imf is finite.Prove that there exists a,b form Z such that f(x)=x+a for every x>b.

I suppose f is strictly increasing (otherwise take f(n)=[n/2]).  Let us take k such that for all n>=k we have n is in Imf. Then we can find a sequence a_n such that f(a_n)=n for all n>=k. THus, n+1=f(a_n+1)>n=f(a_n), from where a_n+1>a_n. Thus, by induction we have a_n \geq a_k-k+n for all n \geq k. Thus, n \geq f(n+a_k-k) for all n>=k and thus f(n)-n \leq k-a_k for all n>=k. Thus the function g(n)=f(n)-n is increasing (not strictly!) and bounded above. From an p and a certain r we have g(n)=r for all n>=p.


vector
linear algebra
linear algebra unsolved

Hi!

Let's take a look at the symmetrical eigenvalue problem:
\[ Ax \equal{} \lambda x,A \in {\mathbb{R}^{n,n}}
\]
and A is symmetrical
Let
\[ {v^1},...,{v^n} \in {\mathbb{R}^n}
\]
be the eigenvectors and
\[ {\lambda _1},...,{\lambda _n} \in \mathbb{R}
\]
be the eigenvalues

If there are vectors
\[ u,w \in {\mathbb{R}^n},u \ne 0 \ and \ \mu \in \mathbb{R}
\]
so that:
\[ (A \minus{} \mu {I_n})u \equal{} w
\]
How do I show that there is a
\[ {\lambda _n} \in \sigma (A)
\]
so that:
\[ \left| {{\lambda _n} \minus{} \mu } \right| \leqslant \frac {{{{\left| w \right|}_2}}} {{{{\left| u \right|}_2}}}
\]
Regards,
\[ \Pi \Delta \Phi
\]

We assume that $ (v_i)$ is an orthonormal basis. $ u\equal{}\sum{u_iv_i}\not\equal{}{0}$ then $ w\equal{}\sum(\lambda_i\minus{}\mu)u_iv_i$. $ \frac{\parallel{}w\parallel{}^2}{\parallel{}u\parallel{}^2}\equal{}\frac{\sum(\lambda_i\minus{}\mu)^2u_i^2}{\sum{u_i}^2}\geq(\lambda_k\minus{}\mu)^2$ where $ |\lambda_k\minus{}\mu|\equal{}inf_i{|\lambda_i\minus{}\mu|}$.

Thank you very much for your help loup blanc!
I appreciate it alot.


[hide]
Call the distance d.  By the $\frac{distance}{rate}=time$ formula,
$\frac{d}{15}+\frac{d}{3}=3$[/hide]

[quote="sharkman"]Please, do not solve but simply set up the equation needed for me to answer this question.[/quote]

That's the whole point of the word problems by the way. :P

[hide="A faster way"] Solve $d=rt=15(3-x)=3x$[hide="Answer"] Answer is $x=2.5$[/hide][/hide]

[quote="sharkman"]Peter rode his bike from Avenue X to the beach at the rate of 15 miles per hour and then walked from the beach back to Avenue X along the same route at the rate of 3 miles per hour.  If the entire trip took 3 hours, how long did it take Peter to walk back to Avenue X?

Please, do not solve but simply set up the equation needed for me to answer this question.[/quote]

[hide]Since the distance is d and the speed is three, the first time is $\frac{d}{3}$.  The second time is $\frac{d}{15}$.  The sum is three, so the equation is $\frac{d}{3}+\frac{d}{15}=3$.

Hope that helps.[/hide]


Inspired by 236factorial's post, let's play some more:

Solve the following inequality:

$|x+3|+|x+5|<|1-2x|$  :ninja:

[hide]
I used trial and error to get $\boxed {x < -1.75}$
It continues forever because for every 1 you decrease x by (After -5), the left side goes up by 2 and the right side goes up by 2.[/hide]

[hide]I squared the equation twice and got the equation $3x^4-96x^3+184x^2+1140x+954>0$...ugly
Before I had $|x+3||x+5|<2x^2-20x-33$...[/hide]

[quote="10000th User"]That's sound reasoning but the answer is not complete. Work directly with the absolute values. :)[/quote]

Ok I edited my post, I found the new answer as you want to find the part where they balance and everything below that will work because it decreases at first, and then stays the same.

:ninja:  Uh... wow, ch1n353ch3s54a1l. Hmm, I guess I can give a hint. When working absolute values, a key step is to look for points where the expression inside an absolute value changes sign... So, in our problem, draw the real number line and you should be able to locate 3 critical points. Now work carefully for some $x$ in each of the intervals and see what you get :)

Remark for what I am referring to: a nice useful fact to know about absolute values is that

\[ |x|=\begin{cases}x\textrm{, if }x\ge0\\-x\textrm{, if }x<0\end{cases}  \]

Say you have something like $|x+1|$, then this means that $x+1$ changes signs at $x=-1$. Using above fact, you have

\[ |x+1|=\begin{cases}x+1\textrm{, if }x\ge-1\\-(x+1)\textrm{, if }x<-1\end{cases}  \]

Now do you see how this can help you solve the inequality? :)

[quote="10000th User"]That's sound reasoning but the answer is not complete. Work directly with the absolute values. :)[/quote]

The answer looks right to me... how is it incomplete?

[quote="hydro"]we can study 4 different cases, [1] $x\leq-5$, [2] $-5<x\leq-3$, [3] $-3<x\leq\frac{1}{2}$, [4] $x>\frac{1}{2}$

the result is $x\leq-\frac{7}{4}$[/quote]

Shouldn't the answer be $x<-\frac{7}{4}$, because when $x=-\frac{7}{4}$, the equation does not work out.

Well, if you saw that Ignite168 edited his previous post, then it is all clear why it was incomplete :)
And yes you are correct 236factorial, no equality.


The sum of two numbers is two more than their positive difference. What is the product of the least two positive integers that will satisfy these conditions?

[quote="5849206328x"]Suppose $ a$ and $ b$ are the two numbers and WLOG, let $ a\geq b$.  Then, \[ a\plus{}b\equal{}2\plus{}a\minus{}b \Rightarrow b\equal{}1\] There are no restrictions on $ a$, so to achieve the minimum, $ a\equal{}2$.  The product is obviously not $ \boxed{2}$.[/quote]

What's a WLOG?[quote="davidkim2106"][quote="5849206328x"]Suppose $ a$ and $ b$ are the two numbers and WLOG, let $ a\geq b$.  Then, \[ a\plus{}b\equal{}2\plus{}a\minus{}b \Rightarrow b\equal{}1\] There are no restrictions on $ a$, so to achieve the minimum, $ a\equal{}2$.  The product is obviously not $ \boxed{2}$.[/quote]

Obviously not...?[/quote]

Probably a typo or weird thought going into the user's brain :P

[quote="sjwon3789"]

What's a WLOG?
[/quote]

Without Loss Of Generality.

So the person who wrote the original solution just trolled all of us??? o.O [.>___<.]

This question is messed up, I got $a = 1$ and $b = 1$ making $ab = 1$

[quote="kz190"]The product is 3. The two numbers are 3 and 1.[/quote]
Wait why cant 1 and 2 work?
Why cant the answer be 2?

Well, I think 1 and 2 work, maybe I made a mistake

The answer is $ {2} $.(MC chapter countdown 1999)

[quote=bever209]but wut abt 1,1[/quote]

1-1=0 which is not positive.