theres some algebra thing where $ x^{3}\plus{}y^{3}\plus{}z^{3}\equal{}3xyz$ if $ x\equal{}y\equal{}z$ or something. i dont know if that will help in any way but who cares

[quote="Elio (n)"]$ {x}^{3}\plus{}{y}^{3}\plus{}{z}^{3}\minus{}3xyz$[/quote]
This shouldn't be in the Moems forum
Here's the solution, nonetheless

[hide="answer"]$ (x\plus{}y\plus{}z)(x^{2}\plus{}y^{2}\plus{}z^{2}\minus{}xy\minus{}xz\minus{}yz)\equal{}\frac{1}{2}(x\plus{}y\plus{}z)((x\minus{}y)^{2}\plus{}(y\minus{}z)^{2}\plus{}(x\minus{}z)^{2})$[/hide]
[hide="motivation"]Consider when $ x\plus{}y\plus{}z\equal{}0$
$ x\equal{}\minus{}y\minus{}z$
$ \minus{}(y\plus{}z)^{2}\plus{}y^{3}\plus{}z^{3}\plus{}3yz(y\plus{}z)$
$ 0$
So we have that $ x\plus{}y\plus{}z$ must be a factor of this polynomial. Dividing it out, we get $ (x\plus{}y\plus{}z)(x^{2}\plus{}y^{2}\plus{}z^{2}\minus{}xy\minus{}xz\minus{}yz)$. We have if x=y=z then this is zero. Notice that it resembles a symmetric sum of $ (x\minus{}y)^{2}$. Indeed, the symmetric sum is double what we need, so we can further factor this as
$ \frac{1}{2}(x\plus{}y\plus{}z)((x\minus{}y)^{2}\plus{}(y\minus{}z)^{2}\plus{}(x\minus{}z)^{2})$ (this form is slightly easier to work with).[/hide]

1. You can use $ x^{3}\plus{}y^{3}\equal{} (x\plus{}y)^{3}\minus{}3xy(x\plus{}y)$.
2. Consider the cubic function with respect to $ x$, $ f(x) \equal{} x^{3}\minus{}3(yz)x\plus{}(y^{3}\plus{}z^{3})$.
3. Consider the cubic equation with respect to $ t$ with roors $ x,\ y,\ z$, say $ (t\minus{}x)(t\minus{}y)(t\minus{}z) \equal{} 0$.
4. Split $ 3xyz$ into $ xyz\plus{}xyz\plus{}xyz$.
5. You can use $ (x\plus{}y\plus{}z)^{3}\equal{}x^{3}\plus{}y^{3}\plus{}z^{3}\plus{}3(x\plus{}y)(y\plus{}z)(z\plus{}x)$.

This is Adam Hesterberg's factorization.

I'm sure it existed before him though :P, so we can't really call it his... like SFFT is fine, because it's just his favorite, but giving possession of a factorization to someone seems rather odd, if not illegal :!:  :wink:.


function
complex numbers
calculus
calculus computations

In general, if $ f(x)$ is a function then $ f^{ \minus{} 1}(x)$ does not equal $ \frac {1}{f(x)}$. Are there any functions for which equality does in fact hold?

In that case, $ f^{\minus{}1}(x)\equal{}f(x)$

The function $ f(x) \equal{} \frac{1\plus{}x}{1\minus{}x}$ is very close to what you want.

And, if we are interested in functions on the extended complex numbers, we can alter that slightly to get $ f(x) \equal{} \frac {ix \minus{} 1}{i \minus{} x}$.  But in fact there are lots of these functions.  Focusing only of $ {\bf R}_{ > 0}$, divide those numbers larger than 1 into pairs.  Then if $ a$ and $ b$ are paired together, let $ f(a) \equal{} b$, $ f(b) \equal{} 1/a$, $ f(1/a) \equal{} 1/b$ and $ f(1/b) \equal{} a$, with $ f(1) \equal{} 1$.  Whether there is a "nice" such function on $ {\bf R}_{ > 0}$, I don't know.  Here's a nice function on $ {\bf R}_{ > 0} \smallsetminus\{1/2, 2\}$: let
\[ f(x) \equal{} \begin{cases}\frac {x \minus{} 1}{x}, & 2 < x \\
\frac {x}{x \minus{} 1}, & 1 < x < 2 \\
1 \minus{} x, & 1/2 < x < 1 \\
\frac {1}{1 \minus{} x}, & 0 < x < 1/2.\end{cases}
\]
Fedja posed the general existence question [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1352951#1352951]here[/url].

Not surprisingly, something like this has appeared before. See also [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=113408]here[/url].

Any such function on all of $ (0,\infty)$ must have infinitely many discontinuities.


kk, Will and I were talking, and the other thread died. Sooo....

we wanna have a seminar reunion!  Seminars from any year!!  Hehehe.  Now....when....I'm in waterloo on FRIDAY!!!!

[quote="Elyot"]ZOMFG. YES [/quote]

[b]ROFLCOPTER[/b]

Yea, but on a serious note, lets start planning something.  I think some people aren't moving in until the 3rd or the 4th.  Of course, that doesn't mean we can't all meet up on Friday, but if we want to have an official seminar reunion, it might not be a bad idea to wait a few days extra.

I'm in!

whenever you guys are all here!!

[quote="Sunny"]Hmm, What about that thursday where we all have free time, or the weekend of the 9th?[/quote]

Elyot has co-op.  Weekend is better.

[quote="jpark"]darn.. jenn has to go play for the church on the sunday, so if it's going to be the weekend, it has to be later than probably 2 or 3pm.

however, ralph is planning to show up sometime on thursday as well. elyot, do you start work during the orientation week?[/quote]

Yes, I start on the Tuesday.  However, my evenings are free.
Since you posted this like 2 minutes ago, why don't you get on MSN and we can discuss!

[quote="Inspired By Nature"]I have a few questions, what does kk, ZOMG, and ROFLCOPTER mean? ZOMFG is prbably an inappropriate variation of ZOMG. :rotfl:[/quote]
Could i know what they mean? :P


How do you solve this -

5x^6 + 4x^5 + 15x^4 + 8x^3 + 2x^2 + 34x + 19 = 0 ?

Or anything similar to this?

[quote="Bugi"][hide="Very useful"]
Divide by x, and get $ x\mid 19$ [/hide][/quote]

Okay, if I divide by x, I get

5x^5 + 4x^4 + 15x^3 + 8x^2 + 2x + 34 + 19/x = 0 ?

?... That didn't really help.

[hide]
All the factors of the constant divided by all the factors of the leading coefficient of the greatest power of x are all the possible solutions excluding imaginary ones.

So the possible solutions are:
$ \pm 5, \pm 19, \pm \frac{19}{5}$

Test those until you get one that works, then synthetically divide.

You may have imaginary solutions though.

[/hide]

[quote="math92"][hide]
All the factors of the constant divided by all the factors of the leading coefficient of the greatest power of x are all the possible solutions excluding imaginary ones.

So the possible solutions are:
$ \pm 5, \pm 19, \pm \frac {19}{5}$

Test those until you get one that works, then synthetically divide.

You may have imaginary solutions though.

[/hide][/quote]

I still don't really understand it.

Hello, divide by x and everything is integer, so 19/x is integer too!

Nineteen is prime, so does that mean that x has to be 1, or 19?

Yeah true, but do any of those work? Maybe x isn't a real number. Or is it? :wacko:

Hmm, well we know that it has to be negative, because all of the co-effecients are positive.

Or, rather than trying to do all of this work, Wickedestjr, you could just plug in the values, like the last 3 people have said.

And to explain what math92 said, if you have a function $ f(x)$, and the leading coefficient (the number in front of the highest power of $ x$) is $ p$, and the constant (the number with no variable at the end) is $ q$, then all possible solutions will be $ \frac{p_f}{q_f}$, where $ _f$ denotes the number being a factor of $ p$ or $ q$, respectively.

Note: This doesn't mean that all $ \frac{p_f}{q_f}$ will be solutions, but it's a useful trick to sort out some potential solutions.

As for what bugi said, it's already been explained.

[quote="lgd0612"]19/x has to be an integer

x value are not just 1,19,-1,-19

x can be any 19/a (a is an integer but not 0)

19/x = 19/(19/a) = 19*a/19 = a[/quote]

If a is something like 2, or 3, then when you divide 19 by it, you won't get an integer. Is there something I'm missing?

But we know that a cannot be 2 or 3 because of this:

[quote]And to explain what math92 said, if you have a function $ f(x)$, and the leading coefficient (the number in front of the highest power of $ x$) is $ p$, and the constant (the number with no variable at the end) is $ q$, then all possible solutions will be $ \frac{p_f}{q_f}$, where $ _f$ denotes the number being a factor of $ p$ or $ q$, respectively.

Note: This doesn't mean that all $ \frac{p_f}{q_f}$ will be solutions, but it's a useful trick to sort out some potential solutions.[/quote]

[quote="fishythefish"]But we know that a cannot be 2 or 3 because of this:

[quote]And to explain what math92 said, if you have a function $ f(x)$, and the leading coefficient (the number in front of the highest power of $ x$) is $ p$, and the constant (the number with no variable at the end) is $ q$, then all possible solutions will be $ \frac {p_f}{q_f}$, where $ _f$ denotes the number being a factor of $ p$ or $ q$, respectively.

Note: This doesn't mean that all $ \frac {p_f}{q_f}$ will be solutions, but it's a useful trick to sort out some potential solutions.[/quote][/quote]

Oh oops, didn't see that.

If this is a Diophantine (solutions are integers) then try values 1,19,-1,-19 and see do they work.
If not, then it's much more difficult.

Well, like I said before it is not 1, or 19 because all teh coeffecients are positive, so it is -1 or -19.

If we plug -1 into the equation, we get 5. If we do -19, we get
-12380495 + 521284 + -102885 + 2888 + (-38) - 1  which equals;
-11959247, so neither of those two work.

 :huh: What does this mean?

Well we know it is not a natural number, because it has to be negative. We also know that it isn't a whole number, because it is not a natural number, and it is obviously not 0. We know that it is not an integer. We know that it is not rational, because the only rational numbers that could work are integers or fractions, but fractions obviously don't work, and we already found out that it is not an integer. So here are the possibilities;

The Roots are Irrational
The Roots are Imaginary

Any suggestions?

The roots are definitely negative. Note that all the terms are positive if $ x$ is positive.

Next, the argument that $ x \mid 19$ does not hold if the roots are irrational, which is what it is in this case. That argument only holds if we are trying to find integer solutions.

The Rational Root Test does not work here because the roots are definitely not rational. (The expression cannot be factored into anything useful.)

To solve general equations like this, the only method you can rely on is a solver. (Yeah, I'm sorry. Unless it's a special form that can be factored nicely [this isn't one of them], you just have to use a solver.)

There are exacty 6 solutions. All of which are irrational, and two of them are negative, and the other four are complex.

So, what are the solutions, and how did you find them?

There are many ways to solve this polynomial equation, or at least get all possible information by using some theorems

$ P(x) \equal{} 5x^6 \plus{} 4x^5 \plus{} 15x^4 \plus{} 8x^3 \plus{} 2x^2 \plus{} 34x \plus{} 19 \equal{} 0$

Given are the possible rational roots $ \pm 5, \pm 19, \pm \frac{19}{5}$, we can use the Theorem on the Bounds of the Real Zeros of Polynomials:

$ (i)$ if we use $ r > 0$ by synthetic division and the numbers at the bottom row are all positive, then $ r$ is an upper bound of the greatest zero of $ P$.

$ (ii)$ if we use $ r < 0$ by synthetic division and the numbers at the bottom row are alternately plus and minus, then $ r$ is a lower bound of the greatest zero of $ P$.

Of course if this is not the case, the test fails.

It means that if we have $ r$ as an upper bound (or a lower bound), we may not try to use the roots greater than (lower than) $ r$

Use this theorem so that you will not bother plugging EACH rational root. ^^

Oh, so this had turned out to be a difficult problem then huh?

Well, IMHO it turned out to be a rather not very interesting problem. The problem in which the solutions could have been found using RRT or some clever substitution - if the equation had solutions representable in "closed form" would be of interest. Problems like this: not really.

It isn't difficult-it's kind of inappropriate.  :lol:


The algebraic expression $ (3x\minus{}2y)^2 \minus{} x(x\plus{}4y)$ can be written in the form $ Ax^2 \plus{} Bxy \plus{} Cy^2$. What is the value of the sum $ A\plus{}B\plus{}C$?

If the two algebraic expressions are equivalent, then they must be equal for all values of $ x$ and $ y$. Set $ x\equal{}y\equal{}1$. Then $ (3\minus{}2)^2\minus{}1(5)\equal{}\boxed{\minus{}4}\equal{}A\plus{}B\plus{}C$.


[quote="frost13"]In Mr. Peterson's gym class they are playing dodge ball with several different sized balls.  Mike picks one ball with radius 7 inches and one ball with radius 12 inches.  Mike measures the longest circumference of each ball by wrapping  a string around each such that the circle created by the string would have the same center as the sphere.  He then takes the two pieces of string and cuts each into eighths.  He then leaves all the pieces on the floor.  Bob comes and picks up 3 of the shorter pieces and 5 of the longer pieces.  What is the total length on the string Bob picks up?  Express your answer as a common fraction in terms of $\pi$[/quote]

[hide]
$3(\frac{14\pi}{8})+5(\frac{24\pi}{8})=5.25\pi+15\pi=20.25\pi=\frac{81}{4}\pi$

so is
$\frac{81\pi}{4}$ the answer (in inches)?[/hide]

-jorian

[quote="frost13"]In Mr. Peterson's gym class they are playing dodge ball with several different sized balls.  Mike picks one ball with radius 7 inches and one ball with radius 12 inches.  Mike measures the longest circumference of each ball by wrapping  a string around each such that the circle created by the string would have the same center as the sphere.  He then takes the two pieces of string and cuts each into eighths.  He then leaves all the pieces on the floor.  Bob comes and picks up 3 of the shorter pieces and 5 of the longer pieces.  What is the total length on the string Bob picks up?  Express your answer as a common fraction in terms of $\pi$[/quote]
[hide="Not sure of answer, but then never sure of answer"]So the circumference of the 7 inch one is $14\pi$ and the 12 is $24\pi$. So dividing each into eighths, we now have $\frac{7}{4}\pi$ and $3\pi$. So three times the smaller one and 5 times the longer ones look likes this:$\frac{7}{4}\pi\cdot3+3\pi\cdot5=\frac{81}{4}\pi$.[/hide]
Edit:jhredsox beat me by a few seconds. :(

remember to put in common fraction like i did

jorian

[hide]okay so:

the circumfrence of the ball w/ radius 12 inches is 24 pi.
the circumfrence of the ball w/ radius 7 is 14 pi.

Cutting them gives 3 pi * 8 and 7/4 pi * 8

Picking up 5 longer pieces = 15 pi
Picking up 3 shorter pieces = 21/4 pi

Total: [b]81pi/4[/b][/hide]


modular arithmetic
combinatorics proposed
combinatorics

In a competition, there were $2n+1$ teams. Every team plays exatly once against every other team. Every match finishes with the victory of one of the teams. We call cyclical a 3-subset of team ${ A,B,C  }$  if $A$ won against $B$, $B$ won against $C$ , $C$ won against $A$.

(a) Find the minimum of cyclical 3-subset (depending on $n$);
(b) Find the maximum of cyclical 3-subset (depending on $n$).

isn't the minimum 0?
let the teams be 1,2,...,2n+1
suppose i beats j if i>j then there is no cycle..

Ok Albanian Eagle... You've done the easy part; let's complete the solution!!

I think the maximum is equal to $\frac{6n^{3}+n^{2}-n}{6}$
[hide]we count the number of triangles in the directed graphs so we get that:
$\binom{2n+1}{3}=T+S$ where $T$ is the desired triangles and $S$ are the other triangles...
now every triangle in $S$ is like this:$A$ has won from $B,C$ so we can count $S$ acording to vertex $A$ of it...
so $S=\sum_{i=1}^{2n+1}\binom{d_{i}}{2}$ where $d_{i}$ is the number of games that vertex $i$ has lost...
now we know that $\sum_{i=1}^{2n+1}d_{i}=\binom{2n+1}{2}=n(2n+1)$so we get that
$S=\frac{\Sigma d_{i}(d_{i}-1)}{2}=\frac{\Sigma d_{i}^{2}-\Sigma d_{i}}{2}=\frac{\Sigma d_{i}^{2}-n(2n+1)}{2}$
so $S$ is minimum iff $\Sigma d_{i}^{2}$ is minimum,but we know that $\Sigma d_{i}^{2}=\frac{(d_{1}+...+d_{2n+1})^{2}}{2n+1}=n^{2}(2n+1)$ so we get that:
$T=\binom{2n+1}{3}-S\leq \binom{2n+1}{3}-\frac{n^{2}(2n+1)+n(2n+1)}{2}=\frac{6n^{3}+n^{2}-n}{6}$[/hide]

I've done it the same way, you've just made some mistakes(for $ n\equiv 2\pmod{3}$ the number isn't even an integer). I think the result is something like:$ \frac{2n^{3}+3n^{2}+n}{6}$, and it is reached when there are exactly $ n$ edges pointing to each vertex.

lol, what a shame...
http://www.artofproblemsolving.com/community/c6h130647p740398

This problem looked awfully familiar to me... turns out it's similar to https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_12B_Problems/Problem_20   :-D


The vertices of a rhombus are the midpoints of the sides of an isosceles trapezoid.  The sum of the lengths of the parallel sides of the trapezoid is 5 times the distance between these sides.  The area of the trapezoid is 80 square units larger than the area of the rhombus.  Find the area and perimeter of the rhombus.

$A_{rhombus} = \displaystyle\frac{h b_{avg}}{2} = \displaystyle\frac{A_{trapezoid}}{2}$, where $b_{avg}$ denotes the average of the bases of the trapezoid, which is the diagonal of the rhombus parallel to the bases.  Because $A_{trapezoid} - A_{rhombus} = 80$, it follows that $A_{rhombus} = 80, A_{trapezoid} = 160$.

The sum of the lengths of the parallel sides is twice the average of the bases, so $2b_{avg} = 5h$, which gives us $A_{trapezoid} = 160 = hb_{avg} = \frac{5}{2} h^2$, giving us $h = 8, b_{avg} = 20$.  The perimeter can be found by observing that one side length of the rhombus, $s$, is the hypotenuse of a right triangle with $\frac{h}{2} = 4, \frac{b}{2} = 10$ as the other two sides, giving us $s = \sqrt{16 + 100} = 2 \sqrt{29}$ (Edit: Fixed; sorry) and $p = 4s = 8 \sqrt{29}$.

Very close to what I got for perimeter.

Check the square root of 116.

(good job)

---

Here's another similar question:

One diagonal of a rhombus is 4/3 times as long as the other diagonal of the rhombus.  The vertices of the rhombus are the midpoints of the sides of an isosceles trapezoid ABCD, with parallel bases, AB and DC, parallel to the longer diagonal of the rhombus, and with AB shorter than DC.  A line segment, AT, which is perpendicular to DC at T, passes through the midpoints of two adjacent sides of he rhombus.  If the rhombus has a perimeter of 100 feet, find the perimeter of isosceles trapezoid ABCD.


What is the speed and direction of propagation of the following wave (A,B and C are constants):

$ y(x,t)$=$ A*$exp($ Bx^2\plus{}BC^2 t^2 \minus{}2BCxt$)

[quote="Nihat"]What is the speed and direction of propagation of the following wave (A,B and C are constants):

$ y(x,t)$=$ A*$exp($ Bx^2 + BC^2 t^2 - 2BCxt$)[/quote]

[hide="Solution"]

First, notice that $ y (x,t) = A \exp (B[x -Ct]^2 )$.

If $ y(x,t) = f(x - Ct)$ then $ y(x,t)$ is a forward traveling wave.

Next,

\[ \frac{dy}{dt} = \frac{\partial y}{\partial t} + \frac{\partial y}{\partial x}\frac{dx}{dt}\]

\[ \frac{dy}{dt} = 2B[x - Ct](\frac{dx}{dt} -C)A\exp(B[x-Ct]^2)\]

[/hide]

[i]dy/dt[/i] is not speed of propogation...

dy/dx is only the wave velocity....
dy/dt is the particle velocity.


For real numbers $n$, let $f$ be a function with the property that $f(n+1)=f(n)f(n-1)$. If $f(0)=a^{2}b$ and $f(1)=\frac{1}{ab}$, where $a$ and $b$ are nonzero real numbers, find $f(15)$.

[hide]I brute forced it to get $f(15)=\frac{a^{154}}{b^{243}}$. the coefficients turn out to be fibonacci numbers as you go on...[/hide]


How can that be?

You stated:

a + ar + ar^2...

If a=1, and r=3, you would get

1 + 1*3 + 1*(3^2) + ... + 3^(n-1) = 4 + 3^2 + 3^3 .... + 3^(n-1)

     NOT                                               3 + 3^2 + 3^3 .... + 3^(n-1)

It seems you forgot to add the "1" that precedes the 1*3?

[quote="parentrob"]Also, there seems to be another error if a=1 as you stated:

a - a(r^n) = 1 - (1*3^n),
    NOT         3 - (1*3^n)

It seems you are using a=3 and a=1 at the same time?!?[/quote]

Okay, he forgot to subtract a at the end.
[tex]\frac{a-ar^n}{1-r}=a+ar+ar^2+...ar^{n-1}[/tex]
subtract [tex]a[/tex] on both side to get
[tex]\frac{ar^n-ar}{r-1}=ar+ar^2+...ar^{n-1}[/tex]
sub a=1, r=3 into it to get the equation.

[tex]\frac{a-ar^n}{1-r}=a+ar+ar^2+...ar^{n-1}[/tex]

subtract [tex]a[/tex] on both sides to get

[tex]\frac{ar^n-ar}{r-1}=ar+ar^2+...ar^{n-1}[/tex]

How does subtracting "a" from the left side change the numerator and denominator as described above? I don't understand this.

I just figured out how subracting "a" yields the result on the left.

You need to subtract [a(1-r)]/(1-r) from the left side of the equation. This yields a - an^r -[a(1-r)] in the numerator, all over 1-r.

[a - an^r - a -(-ar)]/(1-r) = (ar - an^r)/(1-r)

Factor out a (-1)
=-1(an^r - ar) / [-1(r-1)]

-1/-1=1, so it drops out, leaving:

(an^r - ar) / (r-1)

Then it works out.

Thanks!


[b]1.[/b] Find the sum of the first $ 4$ prime numbers. Then take that number and divide it by the 23rd prime number. Express your answer as a common fraction.

[b]2.[/b] Find the difference between the number of degrees in an interior angle of a pentagon and an exterior angle of a hexagon.

[b]3.[/b] What is the sum of interior angles in an $ n$-gon? ($ n$= number of sides of a polygon) How about the [i]sum[/i] of [i]exterior angles[/i] of an $ n$-gon?

[b]4.[/b] For what values of A is the five digit number 3124A a multiple of $ 6$?

[b]5.[/b] My best buddy Joe and I race to McDoodle's from my house. The distance between my house and McDoodles's is $ 8$ miles. If Joe runs at a constant rate of $ 15$ mph and I can get there in $ 24$ minutes, how many more [i]hours[/i] does Joe take to get to McDoodle's? Express your answer as a common fraction.

Good Luck! :D

[quote="garfielddisco123"]Q1. 17/83

Q2. 48

Q3. (n-2)180, 360

Q4. 2

Q5. (Why is it always Joe?) 2/15[/quote]

Sorry, but you apparently didn't listen to the instructions today.... :wink:
Can you please hide your answers? There's a "Hide" button near the top by where it says Message body.

And also, it'd be nice if everyone could post the solutions so that others can get help from your solutions.


probability
ARML
combinatorics unsolved
combinatorics

Sally and Ilene are playing a game using a standard deck of 52 cards in a pile face down in front of them. They take the top card from the pile and put it on the pile of turned cards face up, repeating until someone wins. Sally wins if 2 sapdes are turned over before a heart; Ilene wins if a heart is turned over before 2 spades. You walk in after the second card, the Queen of clubs, is turned completely covering the first card so that you cannot see it. What is the probability that Sally wins the game?

what? Chicago ARML?
No, I'm not.

plz someone at least give a hint for this problem, my friend is stuck with this problem for like 2 days

WLY ... we had this exact same problem for homeowrk for the Chicago ARML team, so I can help.
[hide]
Look at the probability in 2 cases:
CASE 1: the first card was a spade.
CASE 2: the first card was not a spade.

Note that it is impossible for the first card to have been a heart.  If that were the case, the game would have ended then.  Nor can it be the queen of clubs.  This leaves 38 possibilities for the first card.  13 are spades and 25 are not spades.

CASE 1:
If the first card was a spade, then Sally needs one more spade to come up before a heart in order to win.  There are 12 spades and 13 hearts left in the deck, so there is a 12/25 chance of this happening.  Multiply this by the probability the first card was a spade and you get.

13    12      84
--- x ---- = -----
38    25     475

CASE 2:
First off, Sally needs one spade to come up before the first heart.  There are equal spades and hearts, so the probability of this is 1/2.

Then, she needs a second spade to come up before a heart.  At this point, there are 12 spades and 13 hearts, so the probability of that happening is 12/25.

Multiplying these by the probability that the first card was not a spade leaves:

25    1     12     3
--- x --- x --- = ---
38    2     25    19

Adding the two probabilities from case 1/and case 2 yeilds

 3      78     153
--- + ---- = ------
19    475    475

JB[/hide]


calculus
integration
function
calculus computations

I'm learning integral of multi-variable function and I can't get through this one:

Prove that the function:
$ g : ]0, + \infty[ \rightarrow \mathbb{R}$
$ g(x) = \int^{ + \infty}_1{\frac {e^{ - t} - e^{ - tx}}{t}dt}$

is continuous in $ ]0, + \infty[$
Thank you!

PS: This is just an example for the continuity theorem in my book, but I can't solve it because I'm totally new with multi-variable function.

PS2: I finally found a way to solve it using the theorem in the book. Thanks for your attention.


i personally think that ocean's 12 is pretty good...:)

Um... 11 was awesome, 12 kinda sucked, and 13 was pretty good.


Is there a finite set $A$ of natural numbers such that if $n\in A$ then $A$ contains at least one of the numbers $n+1,n-1$ and also the sum of the inverses of the elements in $A$ is integer?


[quote="rachid"]Let $ \triangle ABC$ and $ O$ a point from $ [BC]$ different from $ B$ and $ C$. $ (D)$ pass trough $ B$ and $ (D')$ trough $ C$

such that $ (D) \parallel (D').$The line that passes trough $ O$ parallel to $ AB$ intersect $ (D')$ at $ I$.The line that passes trough

$ O$ parallel to $ AC$ intersect $ (D)$ at $ J.$Prove that $ A,I,J$ are collinear.[/quote]
[hide="Solution"]
[b]Lemma[/b]: Consider trapezoid $ ABCD$ with bases $ AB$ and $ CD$.  Let $ E$ and $ F$ be on $ AB$ and $ CD$ respectively such that $ \frac {AE}{EB} \equal{} \frac {DF}{CF}$.  Let $ AD$ and $ BC$ intersect at $ X$.  Then, $ X$, $ E$, and $ F$ are collinear.

[i]Proof[/i]: Let $ XE$ meet $ CD$ at $ F'$.  Since $ AB\parallel CD$, we have that $ \triangle AEX\sim \triangle DF'X$ and $ \triangle BEX\sim \triangle CF'X$.  It follows that $ \frac {AE}{DF'} \equal{} \frac {XE}{XF'} \equal{} \frac {BE}{CF'}$, so $ \frac {AE}{EB} \equal{} \frac {DF'}{CF'}$ and so $ F \equal{} F'$, so $ X$, $ E$, and $ F$ are collinear.

Now for the problem.  Let $ AB$ and $ (D')$ meet at $ X$ and $ (D)$ and $ AC$ meet at $ Y$.  Since $ JO\parallel AC$ and $ IO\parallel AB$, we see that $ \triangle BJO\sim \triangle BYC$ and $ \triangle COI\sim \triangle CBX$.  Thus, $ \frac {CI}{IX} \equal{} \frac {CO}{BO} \equal{} \frac {JY}{BJ}$.  From the Lemma, we see that $ A$, $ I$, and $ J$ are collinear. [/hide]


geometry
3D geometry
sphere
tetrahedron
probability
algebra
polynomial

[quote="ggbg90"]where can i find solutions of these?[/quote]

You can find a few solutions to these problems searching the forum for 'IMO Shortlist 1981'.

[b][size=150]IMO 1981 ShortListed Problems: (Posted in Forums)[/size][/b]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=366641&#p366641]1.(BEL)[/url][b] IMO4[/b] [b]a.)[/b] For which $n>2$ is there a set of $n$ consecutive positive integers such that the largest number in the set is a divisor of the least common multiple of the remaining $n-1$ numbers?

[b]b.)[/b] For which $n>2$ is there exactly one set having this property?

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2019619#p2019619]2. (BUL)[/url] A sphere $S$ is tangent to the edges $AB,BC,CD,DA$ of a tetrahedron $ABCD$ at the points $E,F,G,H$ respectively. The points $E,F,G,H$ are the vertices of a square. Prove that if the sphere is tangent to the edge $AC$, then it is also tangent to the edge $BD.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2019621#p2019621]3. (CAN)[/url] Find the minimum value of
\[\max(a + b + c, b + c + d, c + d + e, d + e + f, e + f + g)\]
subject to the constraints

(i) $a, b, c, d, e, f, g \geq 0,$

(ii)$ a + b + c + d + e + f + g = 1.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2019625#p2019625]4. (CAN)[/url] Let $\{fn\}$ be the Fibonacci sequence $\{1, 1, 2, 3, 5, \dots\}. $

(a) Find all pairs $(a, b)$ of real numbers such that for each $n$, $af_n +bf_{n+1}$ is a member of the sequence.

(b) Find all pairs $(u, v)$ of positive real numbers such that for each $n$, $uf_n^2 +vf_{n+1}^2$ is a member of the sequence.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2019628#p2019628]5. (COL)[/url] A cube is assembled with $27$ white cubes. The larger cube is then painted black on the outside and disassembled. A blind man reassembles it. What is the probability that the cube is now completely black on the outside? Give an approximation of the size of your answer.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2019630#p2019630]6. (CUB)[/url] Let $P(z)$ and $Q(z)$ be complex-variable polynomials, with degree not less than $1$. Let
\[P_k = \{z \in \mathbb C | P(z) = k \}, Q_k = \{ z \in \mathbb C | Q(z) = k \}.\]
Let also $P_0 = Q_0$ and $P_1 = Q_1$. Prove that $P(z) \equiv  Q(z).$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=366648&sid=163c428c143c1e1067c5ec38241e4b22#p366648]7. (FIN)[/url][b] IMO6[/b] The function $f(x,y)$ satisfies: $f(0,y)=y+1, f(x+1,0) = f(x,1), f(x+1,y+1)=f(x,f(x+1,y))$ for all non-negative integers $x,y$. Find $f(4,1981)$.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=366639&#p366639]8. (FRG)[/url][b] IMO2[/b] Take $r$ such that $1\le r\le n$, and consider all subsets of $r$ elements of the set $\{1,2,\ldots,n\}$. Each subset has a smallest element. Let $F(n,r)$ be the arithmetic mean of these smallest elements. Prove that: \[ F(n,r)={n+1\over r+1}. \]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2019633#p2019633]9. (FRG)[/url] A sequence $(a_n)$ is defined by means of the recursion
\[a_1 = 1, a_{n+1} = \frac{1 + 4a_n +\sqrt{1+ 24a_n}}{16}.\]
Find an explicit formula for $a_n.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2019634#p2019634]10. (FRA)[/url] Determine the smallest natural number $n$ having the following property: For every integer $p, p \geq n$, it is possible to subdivide (partition) a given square into $p$ squares (not necessarily equal).

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2019635#p2019635]11. (NET)[/url] On a semicircle with unit radius four consecutive chords $AB,BC, CD,DE$ with lengths $a, b, c, d$, respectively, are given. Prove that
\[a^2 + b^2 + c^2 + d^2 + abc + bcd < 4.\]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=366642&#p366642]12. (NET)[/url] [b]IMO3[/b] Determine the maximum value of $m^2+n^2$, where $m$ and $n$ are integers in the range $1,2,\ldots,1981$ satisfying $(n^2-mn-m^2)^2=1$.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2019638#p2019638]13. (ROM)[/url] Let $P$ be a polynomial of degree $n$ satisfying
\[P(k) = \binom{n+1}{k}^{-1} \qquad \text{ for } k = 0, 1, . . ., n.\]
Determine $P(n + 1).$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2019640#p2019640]14. (ROM)[/url] Prove that a convex pentagon (a five-sided polygon) $ABCDE$ with equal sides and for which the interior angles satisfy the condition $\angle A \geq \angle B \geq \angle C \geq \angle D \geq \angle E$ is a regular pentagon.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=366638&#p366638]15. (GBR)[/url] [b]IMO1[/b] Consider a variable point $P$ inside a given triangle $ABC$. Let $D$, $E$, $F$ be the feet of the perpendiculars from the point $P$ to the lines $BC$, $CA$, $AB$, respectively. Find all points $P$ which minimize the sum \[ {BC\over PD}+{CA\over PE}+{AB\over PF}.  \]

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2019643#p2019643]16. (GBR)[/url] A sequence of real numbers $u_1, u_2, u_3, \dots$ is determined by $u_1$ and the following recurrence relation for $n \geq 1$:
\[4u_{n+1} = \sqrt[3]{ 64u_n + 15.}\]
Describe, with proof, the behavior of $u_n$ as $n \to \infty.$

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=366643&#p366643]17. (USS)[/url] [b]IMO5[/b] Three circles of equal radius have a common point $O$ and lie inside a given triangle. Each circle touches a pair of sides of the triangle. Prove that the incenter and the circumcenter of the triangle are collinear with the point $O$.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2019647#p2019647]18. (USS)[/url] Several equal spherical planets are given in outer space. On the surface of each planet there is a set of points that is invisible from any of the remaining planets. Prove that the sum of the areas of all these sets is equal to the area of the surface of one planet.

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2019648#p2019648]19. (YUG)[/url] A finite set of unit circles is given in a plane such that the area of their union $U$ is $S$. Prove that there exists a subset of mutually disjoint circles such that the area of their union is greater that $\frac{2S}{9}.$


Hey, I was wondering if anyone knew exactly how the new countdown system worked for MathCounts? The year that I went to Nationals was the first year they did it, but since I didn't make countdown I didn't really pay much attention to how things worked. We're trying to format a practice competition here to help students prepare for MathCounts, so I need to figure out how the new Countdown system works.

Thanks,
Chen

Anyways...

Last year(06-07), they had 10 people seated in a row. The 9th and 10th ranked person would each sit at a table. The announcer would put the question on a projector and project it onto a huge screen. Next, the announcer would read the problem out loud on a microphone. The 2 people that were competing would have to buzz in (buzz the buzzer) when they got the answer, they had 3 seconds to say the answer. If they said it right, they got 1 point. If they said the wrong answer, the other person would have 45 seconds to buzz in the right answer (note: if you get an answer wrong, you may not answer again for that question. You can answer again on the next question). After 3 questions, whoever has the most points moves on to face the next person (in this case, the 8th person). If for some reason, you are tied after 3 questions, the announcer will give you more questions, the first person to get the first question right to break the tie wins. (The announcer will say "I declare a sudden victory situation!"
If you didn't understand my explanation, you can also visit [url=http://www.mnspe.org/mathcounts/2006statetests/510-06C&SGenInstruct_NEW.pdf]here[/url]

Hope that helps!:lol:

-Nexmus

[quote="xxazurewrathxx"]Ah sweet, that pdf really helped.

I went to Nationals in 2004 with Florida.[/quote]

Thats not how Nats work, thats only for state and chapter with that pdf...

[quote="Nexmus"]But he's talking about practicing for state and chap.[/quote]

many states dont follow that, but I believe florida is official countdown so they do

The PDF also talks about unofficial ones too... :)


limit
real analysis
real analysis solved

Prove that there exist sets A and B in R s.t. they are dense everywhere in R and A+A=B.

I suppose A and B are disjoint. Or else this problem does not stick to its title.

Ooops! For got this: A \cap B= \emptyset  and A \cup B = R.

Ok I came up with this really wild idea in 10 minutes but I'm not sure it's correct :

A_n is the union of intervals of the form ( (3k+1)/2^n , (3k+2)/2^n ] for k in Z (closed at one end and open at the other!)

B_n is the union of intervals of the form (  (3k-1)/2^n , (3k+1)/2^n ] for k in Z

It's clear that A_n  \cup  B_n = R , A_n  \cap B_n= \emptyset , A_n + A_n = B_n

Take A =  \lim (n->oo) A_n , B =  \lim (n->oo) B_n and we're done. I'm not sure about the lim part!! Can anyone tell me why it's wrong if it is?

This looks really cool, and I have a feeling it's Ok. Someone showed me a different type of solution, but it involves some vectorial spaces (nothing too tough, but it's still much uglier than yours). Maybe I'll post it when I have more time (posting from school).

Ok, here it goes (I didn't find this, it was shown to me):

Let A1 be  \cup {n in Z} [3n+1,3n+2)  \cap Q and let B1 be the complementary of A1 in Q. it's easy to see that A1+A1=B1.

Now let (a_i){i in I} be a base of R over Q, so any x in R can be written as  \sum x_i*a_i uniquely. Let i0 be an indice in I. A={ \sum x_i*a_i | x_i0 is in A1} and B is the complementary of A in R. I think it's not that hard to prove these sets work.


pigeonhole principle
combinatorics proposed
combinatorics

For $n$ people, if it is known that

(a) there exist two people knowing each other among any three people, and

(b) there exist two people not knowing each other among any four people.

Find the maximum of $n$.

Here, we assume that if $A$ knows $B$, then $B$ knows $A$.

That problem is unsolved for several days!!

You can post your solution if you want! :)

I found $n \leq 6$.Now I cant, but i will post the solution here.It is easy by using túran's theorem.

It was posted here: http://www.mathlinks.ro/Forum/viewtopic.php?p=376517#p376517

But there is no solution either.

Now i cant again because today is christimas, but as soon as possible I will post my solution.Happy C/ for you, bye bye^^

I recall this being a problem Zuming gave in some packet or other. The answer is 8: if you arrange 8 points in a circle and connect any two that have at most 1 point between them, it will give a graph that works. If we have 9 vertices, we first prove a vertex A can not have 6 or more edges. Indeed, if it did, then it's a well-known result that among 6 vertices connected to A, there are some 3, X, Y, and Z, such that either

1) XYZ forms a triangle, and no two of AXYZ are unacquainted, or
2) No pair in XYZ is connected, which violates condition 1.

Furthermore, there can be at most 3 vertices not connected to A, for if there were 4, some two of them, say X and Y, would not know each other. But then no two of A, X, and Y know each other.

Then each person must know exactly 5 people. Let the people be A, B, C, D, E, F, G, H, and I. WLOG A knows B, C, D, E, and F. If B knows three of {C, D, E, F}, WLOG C, D, and E, then WLOG C knows D. But then A, B, C, and D all know each other. If B knows each of G, H, and I, since G, H, and I all know each other, we have a similar problem. Since B knows exactly 5 people, he knows exactly two of {C, D, E, F} and two of {G, H, I}. Analogous facts apply to C, D, E, and F.

Among B, C, D, E, and F, by pigeonhole, some two of them, WLOG B and C, know the same set of people from {G, H, I}, WLOG G and H. But if B knows C, then B, C, G, and H all know each other, which is contradiction. Yet if B does not know C, none of B, C, and I know each other, another contradiction.

So basically, contradiction is EVERYWHERE, and n must be at most 8.

[quote]Then each person must know exactly 5 people.[/quote]

We can end immediately after this line: if each of 9 people know exactly 5 others, that's 45 relations.  But relations come in multiples of 2.  (In other words, the number of odd-degree vertices in any graph must be even.)

:blush:

I knew there must be a shortcut. The solution was much shorter when I first solved it a couple months ago.


Is it just my browsers, or are all the file links for the contests not working?

And if it's my browser, how can I fix it?

It didn't work for me either, I have Firefox 1.5.0.6. :(


Sorry if I don't restate the problem correctly, but I remember it went something like
Suppose I give you two coordinates $A(u,v)$ and $B(s,t)$ ($s,t,u,v>0$ with $u\geq{s}$ and $v\leq{t}$) and a pt. $C$ on either the x-axis or y-axis. Now draw $AC$ and $BC$ and compute the sum of their lengths. Where should $C$ be to minimize this sum?

You could probably differenciate to solve.


I'm definitely with Macintosh. What are you with?

windows is a computer type?

and dells are computer types? arent they just a company that makes pcs?

well Im definitely with pcs, but macs are cool too, but only the newest ones. the old macs suck

Macs are the best because of there intense awesomeness. Userfriendly, easy to use, fast, awesome, and the new ones can have pc stuff installed on them :10: .

Hmeh, if I have to choose between a Windoze computer and a Smac, I'd choose a smac. But where's the Linux and the BSD and the plain UNIX choices? For Linux is much better and nicer and easier than Windows, and it's almost as nice as OSX.

I don't really understand what's with all the "MAC IS SO MUCH BETTER THAN WINDOWS" people, it's good, it's probably a bit better, but it's not that much better, and then there's the thing with far more programs being windows-compatible.

[quote="solafidefarms"]Linux is much [...] easier than Windows.[/quote]
If Linux was easier than Windows, then more people would use it. Probably the main reason that so many people use Windows is that it is pretty easy to use. Whereas I think that Linux is overcomplicated on purpose so that only a small number of people use it. It's like a cult.

I use Windows mostly, but I do use Linux for some things, and I use Cygwin a lot as well. But I don't use Linux too often, because I can't connect to my wireless router through it. When I tell it to connect, it acts like it was successful, but it still doesn't work. With Windows, I just hit connect, type in my password, and everything's good. And with Windows, if it doesn't work, I can at least find help online, as opposed to Linux, where everyone assumes that if you are using Linux, you must be a Linux guru, which makes no sense. But aside from that, Linux isn't a bad OS. I like to use it on the university computers, where it's already all set up and I don't have to worry about problems like configuring Internet access.

What I want to know is, why do so many people hate Windows? Besides the fact that it's not free, as in freedom or free beer, but most programs aren't. Everyone says it crashes a lot, which may have been true for 95 and 98, but my computer running XP has only crashed a couple times in the year and a half I've been using it.

But I'd say my favorite operating system is Dell. :P

Firstly, I dont know what you mean by a "computer type", but what the heck.

I don't understand mac people. IMO mac os is like windows, an OS with alot of flashy and *cute* effects (I saw a mac that would change window by spanking the monitor on the side :|) which are eating unnecessary CPU.
If I want prestanda, I would use a linux dist. or freebsd (which I prefer). If I want something cute to look at in min freebsd I'll set a pic of my gf as background image. End of story.

If we didn't have windows, the black sheep would be mac os. It behaves like windows and would be treated as windows (not to forget that microsoft saved mac once in the past).

There are some easy linux dists. Ubuntu is a popular one and I also find it "cute" :) .

Many of my friends stubbornly say that Macs are better for stability and use-friendliness...
Personally, I find Windows much more user-friendly and easier to use, and I haven't seen much difference in stability.  The main difference is the computer/monitor setup, and I prefer having them separate...

I guess "computer type" is OS type then...
Windows is NOT stable.  Every few times I open some uncommon prog, Windows displays a nice message saying that needs to close...    Macs are rather confusing sometimes, and right click is horrible.  Linux is cool.  :)

two things are wrong with this poll

1. Macs and Windows are operating systems, dell isnt

2. Dell and macs are hardware, windows is just the operating system.

The only annoying thing about macs is the mouse.

Linux is fun

But I only have windows :P

WINDOWS is better because I actually like being able to run programs that came out in the last FIVE years

I'm looking at YOU MACINTOSH

for an os SOLARIS all the way
its like a combination of all the os's
u get the programing ablility of linux the cool bar thingie of OSX and the ease of use of window
google it and try!

PC or SuSE Linux, depending on what you meant... :huh:

I've been using linux for a while now and haven't had any of the stupid bugs in windows. I also didn't have any trouble configuring my wireless connection. And also it's very easy to install software that you want because (a) it's all free and (b) you just search for what you want in a HUGE repository of software.

Oh and macss are pretty weird... mac os x is better than windows but barely.

yeah one more thing

[quote="jmadsen"]And with Windows, if it doesn't work, I can at least find help online, as opposed to Linux, where everyone assumes that if you are using Linux, you must be a Linux guru, which makes no sense.[/quote]

there are support forums for every (major at least) linux distribution

whoops. I can't edit my post since 24 hours has passed. :mad:

nutz: All you posted was that you couldn't edit the post. Anybody who's realized that change knows it's obvious. Can you clarify the poll?

Seriously, Dell isn't the best at anything. It's like AOL. The goal is just to lure people who may not know much about the market and load tons of crap onto them so they're buried in it and trapping them that way.

Explains a lot of why they have a partnership.

My laptop is a dell and I love it.  When I was looking for laptops, it was the best laptop in terms of specs per dollar.  Here at MIT, I'd have to guess that about half the laptops people own are Dells for the same reason (cheap, but relatively powerful).  I don't use macs very often, so I can't really say anything about them.

My computer setup is a 13" macbook hooked up to a 24" Dell 2407WFP (with an Apple Wireless Keyboard and a Logitech MX1000).  It's amazing, esp. with Boot Camp/Parallels if you need that sort of thing (I don't).

I have a PC at home, but OS X is so much better than XP, and since I don't play many games I would consider myself firmly in the Mac camp for now.

[quote="jmadsen"][quote="solafidefarms"]Linux is much [...] easier than Windows.[/quote]
 Everyone says it crashes a lot, which may have been true for 95 and 98, but my computer running XP has only crashed a couple times in the year and a half I've been using it.

But I'd say my favorite operating system is Dell. :P[/quote]

ive had a mac for 5 years and i hasn't crashed yet. Maybe i've been lucky but still, I play a bunch of games and i have hundreds maybe even thousands of things o the comp

emachine offers the best all-around capabitilies in this modern world of technology.

Buy one today, get one free!

[quote="Totally Zealous"]Windows is getting more and more like mac.
The new windows explorer 7 is almost exactly like firefox.
The new windows vista is almost exactly like vista.[/quote]

Iff I remember correctly Windows Explorer is a file manager. Perhaps you are talking about Internet Explorer? Otake likes Internet Explorer. That is why I dumped him some years ago. Still, I believe that Firefox is not a Mac exclusive nor originally a Mac product? I thought that Safari was the Mac browser. Perhaps Internet Explorer is growing like unto Safari and that is the source of your confusion?

Furthermore, it is my earnest desire that things be similar to themselves. That was Otake's other problem. He was not similar to himself, causing a space-time warp. I could never tell whether I was talking to him or to his time-travelled self. Fortunately Windows Vista is indeed similar to Vista. Else we would have space-time-bent computers. Imagine a Beowulf cluster of these!

-Your lovely, talented, and humble Mango Parfait.

(Mango Parfait is a writer for TUX magazine. She does question-and-answers in an extraordinarily humorous manner; this is my best (though still poor) emulation of her style... :rotfl:)

Um i just noticed something about ur poll:
Dell-Computer manufactorer
Windows-computer operating system
Mac-computer manufactorer, operating system, and McDonald's meal.
The choices are independent.

Windows are my favorite

Macs i think suck. our school's are extremely slow, and thats without internet access..

i used ubuntu for a month over the summer, and absolutely loved it
however, i couldn't live without iTunes+iPod, so i switched back to windows;
we have macs at school, the interface is a blast, and apple aesthetics is just unbeatable, i might switch to a mac since thats where iTunes truly is :D

windows though i have a linux environment running on a computer too

Meh, kubuntu's very good, windows is ok, especially since there's a lot of progs for just windows, though it crashes often.  Mac X is interesting, but all other macs are horrible... Of course, all this depends on your needs. :wink:  As for manufacturers, i think dell makes the best desktops while ibm makes the bes laptops(fingerprinting! :)  ).

Windows is better because IMO it's easy to use and highly customizable the only down is that needs more security than macs. But windows just seems more fun... than mac. And clearly theres more programs  and games for windows than for mac.

pc's are good, because it supports almost every program, have ok graphics-macs actually have better on a g5, and are good all around.

Macs are better than Windows, but they cost more. I hate how my friend has a Windows and says that it is better than a Mac because he just got his comp a year ago and I got mine 5 years ago...

mac is way better, although the price difference may draw people towards windows.

i use a macbook, and it is flawless.  no viruses, it doesnt freeze.

my pc runs windows and it is always freezing and doing stupid stuff.

the only advantage to windows is if you are doing work related stuff and using microsoft excel, which you can get for mac anyways...

also, the built in camera/video chat on my macbook is awesome, as well as iphoto and imovie.

windows ftw

maybe i just hold a grudge against macs cause the only one's i've ever used are the ones at my school... they made me sick

but i'm still more used to windows and i can do much more with them

Windows often has more software and hardware support than Linux.

I'm not saying Linux is bad but I've tried kubuntu and my sound never worked (also occured with people using the same laptop as mine). I couldn't get my wireless internet working, and when i did, it was dodgy and i couldn't get wireless security working.

Not always true........my thinkpad installed Kubuntu Linux flawlessly....
And Windows has a lot of weird errors, for example, on my friend's dv9000z, he had to reinstall the OS 3 times, and Media Center took 6 hours each time... Then when system restore showed up, it was: 1. You can delete the restore files and proceed to windows normally. Period.  Plus when he reinstalled xp pro, it had not single driver somehow.  The ram rendering was 1 Hz :)

Well Dell has the best computers I believe (next to Alienware lol) but Windows has the most and best software. and Macs just...

#462310 +(6465)- [X]

< robT> Name ONE thing that your windows comp can do that my MAC cant
< bawss> Right click.

source: from bash.org

[quote="JavaMan"]Yeah, I was using a Mac at the state library, found it pretty awkward having only a single mouse button. What's with that?[/quote]

Hawhaw... what about a middle-button? :lol:

That's because the original person decided that it was simpler.  Imagine you had a blank keyboard.  Every time you type the key for "a", you get a "u", etc.  Basically, there's only one button to push.  I think right-click is option+click, but I'm not sure.

I like the Aqua GUI.  That's the only thing about Macs.

With Linux, there are various ways to have a Mac-ish theme.

[quote="jmadsen"]
If Linux was easier than Windows, then more people would use it. Probably the main reason that so many people use Windows is that it is pretty easy to use. Whereas I think that Linux is overcomplicated on purpose so that only a small number of people use it. It's like a cult. [/quote]
Try Gnome or KDE.  The real reason that so many people use Windows is 1. Marketing and 2. Everyone else uses it (just ask me!).  The overcomplication is only in the past (compare it to using DOS instead of Windows).  I use Fedora Core 6 (and Windows) and am very inexperienced at Linux, but I can still do everything I can do on Windows.
[quote="jmadsen"]
And with Windows, if it doesn't work, I can at least find help online, as opposed to Linux, where everyone assumes that if you are using Linux, you must be a Linux guru, which makes no sense. [/quote]
Wrong.  Even then, you can just read forums.
[quote="jmadsen"]
What I want to know is, why do so many people hate Windows? [/quote]
1. It's not open source.  2. Steve Ballmer, the CEO of Microsoft, hates Google and iPods.  3. Windows doesn't give as much control over the OS as Linux.

Go Linux!

Anything but macs.  I program and any real programmer hates macs.  And none of those are "computer types".  :furious:

[quote="tjhance"]lol yeah... Dell is a computer type but Windows and Macs are both operating systems... the fact that the choices are not mutually exclusive make this a strange poll[/quote]

i think what he ment by Macs are Apples and ive never seen Microsoft make computers that have Windows on them...

:?:
Choices don't really make sense, but Gentoo (Linux) is the best.


function
trigonometry
calculus
derivative
algebra
difference of squares
special factorizations

Let $ f,g: [a,b] \to \mathbb{R}$ be continuous functions.If $ f(a) \neq 0, f(b)\equal{}0,g(a)\equal{}0 ,g(b) \neq 0$ and $ f(x)\plus{}g(x) \neq 0 ,\forall x\in [a,b]$ prove that $ \exists c \in (a,b)$ such that $ f(c)\equal{}g(c)$

I think you break it up into cases (depending on endpoints of f and g) and use mean value theorem.

Maybe you already knew that and are looking for a more elegant way :?:

suppose that there is no $ c \in [a,b]$ such that $ f(c) \equal{} g(c)$ and we define two continious functions $ u$ and $ v$ such that $ u(x) \equal{} f(x) \minus{} g(x) ,$ and $ v(x) \equal{} f(x) \plus{} g(x)$.since for every real number $ x$ on $ [a,b]$ we have :$ u(x) \neq 0$ and $ v(x) \neq 0$ then $ u$ and $ v$ has a fixed sign .

if $ u < 0$ and $ v < 0$ then $ f \equal{} \frac {u \plus{} v}{2} < 0$ , but we see a contradiction with $ f(b) \equal{} 0.$

if $ u > 0$ and $ v > 0$ then $ f \equal{} \frac {u \plus{} v}{2} > 0$,but $ f(b) \equal{} 0$.

if  $ u > 0$ and $ v < 0$ then $ g \equal{} \frac {v \minus{} u}{2} < 0$ but $ g(a) \equal{} 0.$
and if $ u < 0$ and $ v > 0$ then $ g \equal{} \frac {v \minus{} u}{2} > 0$ but $ (a) \equal{} 0.$
we obtain in all cases a contradiction .so there is $ c \in [a,b]$ such that $ f(c) \equal{} g(c)$ and sine $ (a) \neq g(a)$ and $ f(b) \neq g(b) , c$ must be in $ (a,b)$

[quote="§outh§tar"]I think you break it up into cases (depending on endpoints of f and g) and use mean value theorem.[/quote]

We only know that $ f$ and $ g$ are continuous; one of the hypotheses of the Mean Value Theorem is that the function is differentiable everywhere on the interval's interior. We could even have $ f$ and $ g$ be functions which don't have derivatives anywhere inside $ [a, b]$; it's pathological but such functions do exist.

[quote="Ravi B"]Let $ h(x) \equal{} f(x)^2 \minus{} g(x)^2$.  Then $ h(a)$ is positive and $ h(b)$ is negative.  So $ h$ has a zero.  Now use the difference-of-squares factorization.[/quote]
nice solution  :)

[quote="ZetaX"]@hadjimurad: With mean value theorem, he actually means the intermediate value theorem^^[/quote]

Thanks for the clarification. I get the names mixed up all the time. :blush:

Shame on you for presuming I'm male though.  :P

[quote="ZetaX"]@hadjimurad: With mean value theorem, he actually means the intermediate value theorem^^[/quote]

Ah, I should have figured!

[quote="bookworm_vn"]GM 2003 means...?  :?:[/quote]

"[b]G[/b]azeta [b]M[/b]atematica" as it is in Romanian , and it is a math magazine .


calculus
integration
trigonometry
logarithms
calculus computations

This came up in a problem and my computer is giving me something different than what I came up with.  Could someone tell me where / if I went wrong?  Thanks.

$\int\frac{x\,dx}{x^{2}+x+1}=\int\frac{x\,dx}{\left(x+\frac{1}{2}\right)^{2}+\frac{3}{4}}$

I said that $x=\frac{\sqrt{3}}{2}\tan\theta-\frac{1}{2}\implies dx=\frac{\sqrt{3}}{2}\sec^{2}\theta\,d\theta$.  This transforms the integral into

$\int\frac{\left(\frac{\sqrt{3}}{2}\tan\theta-\frac{1}{2}\right)\left(\frac{\sqrt{3}}{2}\sec^{2}\theta\,d\theta\right)}{\frac{3}{4}\sec^{2}\theta}=\frac{2}{3}\int\left(3\tan\theta-\sqrt{3}\right)\,d\theta$

$=-2\ln\left(\cos\theta\right)-\frac{2\sqrt{3}}{3}\theta$

$=-2\ln\left(\frac{1}{\sqrt{\frac{4}{3}\left(x^{2}+x+1\right)}}\right)-\frac{2\sqrt{3}}{3}\arctan{\left(\frac{\sqrt{3}}{3}\left(2x+1\right)\right)}$

$=\boxed{\ln\left(\frac{4}{3}\left(x^{2}+x+1\right)\right)-\frac{2\sqrt{3}}{3}\arctan{\left(\frac{\sqrt{3}}{3}\left(2x+1\right)\right)}}$

But Mathematica is saying that the answer is

${\frac{1}{2}\ln\left(x^{2}+x+1\right)-\frac{\sqrt{3}}{3}\arctan{\left(\frac{\sqrt{3}}{3}\left(2x+1\right)\right)}}$

Does anyone see my mistake?  Thanks.

This is the way I would have done this problem:

\[ \begin{aligned} \int\frac{x\ dx}{x^2+x+1}&=\frac12\int\frac{2x+1\ dx}{x^2+x+1}-\frac12\int\frac{dx}{x^2+x+1}\\ &=\frac12\ln(x^2+x+1)-\frac12\cdot\frac2{\sqrt3}\int\frac{\frac2{\sqrt3}\ dx}{1+\left(\frac2{\sqrt3}(x+\frac12)\right)^2}\\ &=\frac12\ln(x^2+x+1)-\frac1{\sqrt3}\arctan\left(\frac2{\sqrt3}\left(x+\frac12\right)\right) \end{aligned} \]

[quote="amcavoy"]$\int\frac{\left(\frac{\sqrt{3}}{2}\tan\theta-\frac{1}{2}\right)\left(\frac{\sqrt{3}}{2}\sec^{2}\theta\,d\theta\right)}{\frac{3}{4}\sec^{2}\theta}=\frac{2}{3}\int\left(3\tan\theta-\sqrt{3}\right)\,d\theta$[/quote]

This part has a mistake. The integral should be $\int\left(\tan\theta-\frac1{\sqrt3}\right)\ d\theta$

This should give you

$\int\left(\tan\theta-\frac1{\sqrt3}\right)\ d\theta$

$=-\ln|\cos\theta|-\frac{\sqrt3}3\theta$

$=-\ln\left|\frac1{\sqrt{1+\left(\frac2{\sqrt3}\left(x+\frac12\right)\right)^2}}\right|-\frac{\sqrt3}3\arctan\left(\frac2{\sqrt3}\left(x+\frac12\right)\right)$

Same answer :)

Ahh I see, that should have been a $\frac{1}{3}$ instead of a $\frac{2}{3}$.  Thanks!


calculus
integration
limit
trigonometry
calculus computations

(1) Evaluate $ \int_1^{3\sqrt{3}} \left(\frac{1}{\sqrt[3]{x^2}}\minus{}\frac{1}{1\plus{}\sqrt[3]{x^2}}\right)\ dx.$

(2) Find the positive real numbers $ a,\ b$ such that for $ t>1,$ $ \lim_{t\rightarrow \infty} \left(\int_1^t \frac{1}{1\plus{}\sqrt[3]{x^2}}\ dx\minus{}at^b\right)$ converges.

Setting $ \root{3}\of{x^{2}}\equal{}\tan ^{2}u$

$ \int_{1}^{3\sqrt{3}}\left( \frac{1}{\root{3}\of{x^{2}}}\minus{}\frac{1}{1\plus{}\root{3}\of{x^{2}}}\right) dx$

$ \equal{}\int_{1}^{3\sqrt{3}}\frac{1}{\left( \root{3}\of{x}\right) ^{2}\left( 1\plus{}\left( \root{3}\of{x}\right) ^{2}\right) }dx$

$ \equal{}\allowbreak \int_{\frac{\pi }{4}}^{\frac{\pi }{3}}3\,du\equal{}\allowbreak \frac{\pi }{4}$

Setting $ \root{3}\of{x^{2}}\equal{}\tan ^{2}u$

$ \int_{1}^{t}\frac{1}{1\plus{}\root{3}\of{x^{2}}}dx\equal{}\allowbreak \int_{\frac{\pi }{4}}^{\tan ^{\minus{}1}\root{3}\of{t}}3\tan ^{2}u\,du\equal{}\allowbreak 3\root{3}\of{t}\minus{}3\tan ^{\minus{}1}\root{3}\of{t}\plus{}\frac{3}{4}\pi \minus{}3$

Hence for limit to exist, $ a\equal{}3,b\equal{}\frac{1}{3}$

It's been a while, but what inspired the tangent substitution?

$ (x^2 \plus{} a^2)^{\frac {n}{m}}$ implies the substitution of $ x \equal{} a\tan \theta$ for $ a > 0,\ \minus{} \frac {\pi}{2} < \theta < \frac {\pi}{2}$.


I have a pouch with pens and pencils in it. At the end of math class, the ratio of pens to pencils is $ 9: 7$ While walking from math class to photography class, I found 3 pens and pencil. After I put them in my pouch, the ratio of pens to pencils was $ 7: 5$. How many pencils did I have at that point?

We have the equation \[ \frac{9x \plus{} 3}{7x \plus{} 1} \equal{} \frac{7x}{5x}\] Cross multiplying gives us $ 45x^2 \plus{} 15x \equal{} 49x^2 \plus{} 7x \implies \minus{}4x^2 \plus{} 8x \equal{} 0 \implies \minus{}4x(x \minus{} 2) \equal{} 0$ or $ x \equal{} 2$. Thus he had $ 5 \cdot 2 \equal{} \boxed{10}$ pencils at that point.

Although if you check that, we get $ 11/9\equal{}9/7$.

Let $ x$ be the number of pens after finding the stuff and $ y$ be the number of pencils at that time.  We have
\[ \frac{x\minus{}3}{y\minus{}1}\equal{}\frac{9}{7}\] \[ \frac{x}{y}\equal{}\frac{7}{5}\]
Solving this gets you $ y\equal{}\boxed{15}$.


Is there any way to have LaTeX in spoiler mode?

I know a lot of people solve problems using LaTeX, but that makes it hard for people who are also trying to solve the problem to not see what they have written.

So far, no. If you can think of any good way that makes it possible, please say..


In unit square $ABCD$, unit circles with centers $A,B,C,D$ are drawn. Find the area of the union of these circles.

[hide]Only consider the stuff inside the square. Let the desired retion have area $a$, then the other two types of congruent regions be $b$ and $c$. Use the areas of the quarter circles and the area of the square to write two equations in $a,b,c$, then find another computable area (less obvious) so that you can write a third equation.[/hide]

[hide]ALl you need to do is find the area of the thing outside of the square, as you know the square area is 1.

The thing outside of the square has 4 of the little intersections between the two circles.  The area of one of them is just the area of 2 60 degree sectors minus the equilateral triange so it is $\frac{\pi}{3}-\frac{\sqrt3}{4}$.
So the area of the thing outside is just 4 3/4 (270 degree) circles minum 4*area of one of those intersections, so it is $3\pi-4(\frac{\pi}{3}-\frac{\sqrt3}{4})$, which is $\frac{5\pi}{3}-\sqrt{3}$, add one to get the area of the square so it is $\frac{5\pi}{3}-\sqrt{3}+1$. [/hide]

Extension: find the area of the intersection using the same lengths and the same 4 circles (circle squared now you can do it)

o.0 for some reason I keep making these typos...it should be intersection. I think Art of Owna did the intersection anyway, so yeah.

hah, wow, my solution was really messy...

I did find the intersection?  I thought I found the union....

Uh..my math vocab is bad, intersection is the one thats usaally smaller than union right?

I didnt read your hint (that gave it away-since its inside the square it must be intersection)...oh well, since your question said union, I found the union....ill find the intersection later

[quote="CircleSquared"]:wallbash:
I deleted my post about intersection after someone rated it spam.

I'm too lazy to retype it, so here's a
[hide="hint"]
Use PIE.  Yay!
[/hide][/quote]
well its not your fault the question was typed up wrong........intersection..umm yea use PIE..use it like 4 times on things that intersect only once-twice+3 times-4 times..too lazy to do..this is spam..yeah.


geometry
vector
incenter
geometry proposed

[color=darkblue]Given are the triangle $ABC$ and the its incircle $C(I)$ which touches the its sides in the points $D\in BC$, $E\in CA$, $F\in AB$. Denote the midpoints $M$, $N$, $P$ of the sides $[BC]$, $[CA]$, $[AB]$ and the intersections $X\in MP\cap DE$, $Y\in MN\cap DF$. Prove that $\{X,Y\}\subset AI$ and the line $AI$ is perpendicular on the lines $BX$, $CY$.[/color]

All is base on the following lemma
line $IA\cap MN=\{X\}$ then $XE\perp IC\Rightarrow X,E,D$ are colinear.
I only have a solution by vector
$\vec{XE}=\vec{XP}+\vec{PE}=\frac{c}{b}\vec{CA}+\vec{PA}+\vec{AE}=\frac{c}{b}\vec{CA}+\frac{1}{2}\vec{BA}+\frac{s-a}{b}\vec{AC}$
$\vec{DE}=\vec{DC}+\vec{CE}=\frac{s-c}{a}\vec{CB}+\frac{s-c}{b}\vec{CA}$
Now to prove $\vec{XE}\cdot \vec{DE}=0$ is only calculation.
But it look ugly, sorry I don't have time to think about a synthetic solution!

This topic:
http://www.mathlinks.ro/Forum/viewtopic.php?t=23677
has most properties of the points mentioned (see my post there).

  Darij

First, we will use the following [b]lemma[/b]: Let $ABC$ a triangle and $D,E,F$ the points where the incircle cuts $BC,CA,AB$. Let there $L \in BI \cap EF$, then $LB \perp LC$.

proof of [b]lemma[/b]: Let $T \in EF \cap BC$, therefore $(TBDC)$ is harmonically, so $L(TBDC)$ is a harmonic pencil, therefore $LB \perp LC \Longleftrightarrow \angle{TLB}=\angle{DLB}\Longleftrightarrow \angle{FLB}=\angle{DLB}$.

but $\triangle{FBL}\equiv\triangle{DBL}$, because $FB=BD,BL=BL$, and $\angle{FBL}=\angle{DBL}$, so $\angle{FLB}=\angle{DLB}\Rightarrow LB\perp LC$, and the lemma is proved.

now returning to our problem, we can see that for proving $X$ is on $AI$, we can also asume that $X$ is the intersection of $AI$ and $DE$, and we will prove that $X \in MP$, and the same thing for $Y$.

but now $AI \cap DE ={X}$, therefore (from the lemma) $AX\perp BX$, but $P$ is midpoint of $AB$, so in the right-triangle $AXB (XP)$ is the median, therefore $XP=AP \Longleftrightarrow \angle{AXP}=\angle{XAP}\Longleftrightarrow \angle{AXP}=\angle{XAC}\Longleftrightarrow XP\| AC$, but $MP \| AC$, therefore $X \in MP$.

now for $Y$ is the same thing, $YN$ is median in right-triangle $AYC$, so $AN=YN$, therefore $YN \| AB$ , but $MN \| AB$ , so $Y \in MN$.

[b]remark[/b]: the quadrupoint $(AXA'Y)$, where $A'$ is the foot of the bisector from $A$, is harmonically, and even more: if we draw the circle with the diameter $XY$, and denote $K,L$ the points were the circle cuts $BC$, then $X$ , $Y$ are the incenter, and respectively the excenter of the triangle $AKL$.

Very nice this problem and the its last proof !

Another way to prove pohoatza's really nice lemma:
Once you know that $\angle DLI = \angle DLB = \angle BLF = \angle AFE-\angle ABI = \angle ACI$, observe that $\angle ILD = \angle ICD$, therefore $ILCD$ is cyclical but $\angle IDC = 90$, then $\angle ILC = \angle BLC = 90$.  :)

This is a very useful lemma .
This lemma could be also used to prove this very vry nice problem proposed by Russia
http://www.mathlinks.ro/Forum/viewtopic.php?p=257883#p257883

Indeed. Another important application of this lemma should be the BMO 2005 exercise [url]http://www.mathlinks.ro/Forum/viewtopic.php?p=224997#p224997[/url], which follows immediatelly from this.

Assume $AB > AC$. Let $S \equiv CY \cap AB$.  Since $MN \parallel AB$ we have that $\triangle DMY \sim \triangle DBF$ and since $BF=BD$ we get $MY=MD$.

Since $c > b$ we see that $BD = \frac{a+c-b}{2} > \frac{a}{2} = BM$ and so $MD= \frac{c-b}{2}$.

Clearly, $MN$ bisects $CS$, i.e $CY=YS$.  From this we have $BS= 2MY=2MD=c-b$. So $BS+ AS= c => AS= b$. So $\triangle ASC$ is isosceles with $AC=AS$. Since $Y$ is midpoint of $CS$ we have that $AY$ bisects $\angle SAC$. Hence $Y$ lies on $AI$ and $AI$ is perpendicular to $CY$.

Because of our assumption $Y$ lies inside the triangle and $X$ lies outside. To prove point $X$ lies on $AI$ define $G \equiv BX \cap AC$.

Because of the similarity of triangles  $\triangle MDX$ and $\triangle CDE$ we get $MD=MX$. In a similar way $CG = c-b$ and so $AG= c$. So $\triangle ABG$ is isosceles and the conclusion follows.

Dear Mathlinkers,
this nice situation with more developpements has been studied on my site
http://perso.orange.fr/jl.ayme  vol. 4 An unlikely concurrence, revisited and generalized
and also, a generalization in English in vol. 10, Another unlikely concurrence.
Sincerely
Jean-Louis


limit
trigonometry
calculus
calculus computations

$ \lim_{x\to1} \frac{\tan\frac{x\pi}{2}}{x\minus{}1}$

please.

It should be $ \minus{}\infty$. If you approach from the left, then the numerator goes to infinity while the denominator stays negative (i.e. the whole thing is approaching negative infinity).

If you approach from the right, then the numerator approaches negative infinity, but the denominator stays positive.

Try graphing the $ \tan(\frac{\pi x}{2})$. The numerator goes to zero on both, so it is not an indeterminate form or anything. Both numerator and denominator are making the limit tend $ \minus{}\infty$.


Let $\mathcal U = \left\{ \left( x,y \right) | x,y \in \mathbb Z, \ 0 \leq x,y < 4 \right\}$.

(a) Prove that we can choose $6$ points from $\mathcal U$ such that there are no isosceles triangles with the vertices among the chosen points.

(b) Prove that no matter how we choose $7$ points from $\mathcal U$, there are always three which form an isosceles triangle.

[i]Radu Gologan, Dinu Serbanescu[/i]

Can one compute the number of the vertices we must pick for an arbitrary $n$ (not just $n=4$) s.t. the problem is still true?

the solution to (a) is
${(0,0), (0,1), (2,0), (3,1), (2,3), (3,3)}$

To prove (b), we first need to note that the total number of different distances from one of the four center points to any leagal point is only five: $1, \sqrt{2}, \sqrt{5}, 2, 2\sqrt{2}$
By the pigeon hole theorem, no point can be placed in the four center places. Notice that a maximum of two points can be placed in one of the four corners. This leaves five points to be placed on edges, and these will always form an isosceles triangle with the size $\sqrt{5}$ appearing twice. To see this, look at the two squares $(0,1), (1,3), (2,0), (3,2)$ and $(0,2), (1,0), (2,3), (3,1)$.
By the pigeon hole theorem, at least one of them will contain three or more points, creating an isosceles triangle.


Support probability articles rotation calculus topology

Death is natural. I'm not really scared of it. I could be a good thing for another person in your life (like some guy competing with you in MC) or a bad thing to another person.(like your mom)

[quote="13375P34K43V312"]Alot of people, including me, are afraid of death. Is it really a bad thing? Is death actually good? What's your opinion?[/quote]

You can look at it through religion/belief or science

Death isn't bad, or good. It's just natural. If you are afraid of death, then "Why?"

OOO OOO I'm actually more afraid of immortality than death. In my opinion... If everyone else does it, I want to do it too. (Yes even jump off a bridge)

Well, once you die, you will never come back to life (supposedly), so that's why some people are scared of it.

I don't know what I think--I'm a little afraid of it, but everyone does die, so it's not that bad.  I just hope I don't die from a car crash or something like that.

I'm not afraid of death, to me, it's God's will. But I am afraid of the pain that might come along with death.

[quote="Treething"]I'm planning on being immortal, so far so good.[/quote]

Ahhh enjoy it

[quote="Treething"]I'm planning on being immortal, so far so good.[/quote]

that's amirhtlusa's signature

[quote="Treething"]I'm planning on being immortal, so far so good.[/quote]

That's an old quote/joke, various people have said it.

As CS Lewis (Oxford Don, wrote Narnia series -- an English guy that was a national icon during and after WW II for the English -- we Americans liked him a lot too) said, "You have never met a mortal."  I think it was in his book Mere Christianity.

He went on to explain, as anyone who believes the Bible should be able to tell you, we are all immortal.  With very rare exceptions, all will die, but that is merely the death of your human body.  Who you are will continue, residing in Sheol until the final day, when based on the judgment of your life, you will either go to Hell where you will be in constant, conscious pain and punishment, removed from the presence of God.  It is also described as being burned, but not consumed for all of eternity.  Those who meet God's standard will be go to heaven, being in perfect bliss, in God presence for all of eternity.

There is one problem -- God doesn't grade on a curve.  His standard is perfect.  One sinful act or thought and you fail.  (We all fail because of inherited sin, but we also fail spectacularly on our own.)  The Bible say all fail, no one is good, not even one.  God then send his son, also a part of God to pay for our sins and who lived a sin free life which God will apply to our sinful existence and treat us as if we were his son.  (Let's not get into a discussion of the triune God, but suffice it to say that God is greater than any of us and unlike a human who has multiple personalities, all of his three personalities are real and present at the same time.)

There is a second problem -- we are all so sinful that none of us have the ability to seek God and ask for salvation, except for those he reaches out and heals our hearts to allow us to accept the offered narrow path that leads to salvation.  Only those God elects will go to heaven.  Most people will go to hell.  Those that God elected are commanded to spread his gospel message, even though none can accept the message absent his changing their hearts.  (This election occurred before the creation started.  We don't want to go there, but suffice it to say that none have done anything to deserve the election -- it is an amazing act of Grace by God alone.)

Short, but the bottom line is that you can fear death as the end of this life, but don't buy the line that you will not persist as you path death.  Just like none re-enter their mother's womb after birth, with very few exceptions, none come back after passing death.

It is fearful, but Jesus demonstrated that he has the keys of death and we will all pass death with our existence complete, just loose these bodies until we receive the new ones at the end of time.

It may be too much for most of you, but I am commanded by my God to present the Gospel message where opportunities are presented.

Because of these reasons, I do not fear death.  Like Paul, I look forward to being in the presence of Jesus.

Death is synonymous with birth. As soon as we are born, we are faced with certain death. There is no reason to fear death. Life is just one stage. As long as death is not planned, it should be seen as a stage of life where one moves on to greater places

[quote="Ignite168"]This is the main reason I'm a theist.

When there is no life anymore, what will happen? All our creations and progress in physics, chemsitry, those things that took so long to discover, our different cultures, etc. will be lost forever, and we can't be sure at all of life on other planets.

I think there must've been something to start this all, to make this intricate structure of life, and that something must happen once it's all over.[/quote]
What is theist?

[quote="mustafa"][quote="kstan013"]SCIENCE TEACHER: Look all around you. Do you see God? No. So there must not be one.
LITTLE GIRL: Look at this man. Do you see his brain? No. So he must not have one.[/quote]

This is definetly going in my sig :rotfl:  :rotfl:

I don't think that people need proof to believe in God as much as we [i]need[/i] God. It makes us feel better to think that we aren't just some strange accident, and that someone cares and watches over us.[/quote]
Another (somewhat incorrect, but you get the point!) quote:
The world is so intricate and perfect that it could not have come about by chance. A higher being must have been involved in this great gift we call life- Robert Gallup

Having been inches from death on a two or three occassions due to health issues, I am not scared of it.  The unknown is far more scary, the unknown in life that is.

~lyra

[quote="kstan"]There are plenty of miracles in the Bible.

SCIENCE TEACHER: Look all around you. Do you see God? No. So there must not be one.
LITTLE GIRL: Look at this man. Do you see his brain? No. So he must not have one.[/quote]

I'd love you to realize how silly this is an argument to me. I mean, don't get me wrong, I mean no offense, what I mean is, don't you think I've thought of the possibility that there actually is a God but I don't see it, just like your example? Once checked I don't see it, this probably was the first thing that came to my mind ages ago. That's why in the first place I don't say I'm positive about  God's nonexistence, I just say I don't think so. In the other hand, the teacher statement is silly in itself and it is silly to put those words in the mouth of a science teacher since in science you are perfectly used to accept the existence of things you don't see, even that you don't detect by any means. So what I mean is I think my reasons for not believing in God are stronger than just because I don't see it. Nevertheless, that I don't see proof or need for the existence of God is also a supporting fact.

[quote="wizard"]Those waiting for proof of God need to realize that at least the Bible says that salvation is by faith alone and if there ever were proof there would be no need for faith. Thus, a proof of God isn't there and can't be found.[/quote]

This is the eternal discussion that doesn't get anywhere. I wonder, who put this lack of faith in my mind? Did I do it? Because I certainly had no intention of going to hell. But every atempt of mine to believe in God, or in hell ends in failure. I must conclude I was born doomed.

[quote]The world is so intricate and perfect that it could not have come about by chance. A higher being must have been involved in this great gift we call life- Robert Gallup[/quote]

Well, so say you or Gallup. This is already the act of faith you talk about. You believe in this sentence beacuse you believe in God and not the other way around. Because if you don't there's no reason not to question the veracity of it. Besides, who says it came about by chance? I'm worried about it too but I work on alternative explanations. Nevertheless from some point on, like nanoseconds after God's act of creation some billions of years ago, do we agree that we don't need to call him back again in order to explain everything we see today?

Ursula

[quote="ursula"]This is the eternal discussion that doesn't get anywhere. I wonder, who put this lack of faith in my mind? Did I do it? Because I certainly had no intention of going to -. But every atempt of mine to believe in God, or in - ends in failure. I must conclude I was born doomed.[/quote]
This is exactly my feeling as well. Everyone I have seen who was not indoctrinated to be religious when very young but converted and became devout when older (to any religion) has undergone some sort of "religious experience" which was neither planned nor voluntary. (Well they don't complain about it, but they weren't asked in advance.) I don't think one can make oneself have faith. I don't think it is possible to force yourself to have a "religious experience." Maybe you just need to be 'ready' -- but how do we become ready? There is a limit to how actively we will be willing to pursue this when we don't actually have faith yet anyway; it's quite irrational.

[hide="Bit out of topic but interesting nonetheless"] I just thought of supporting a bit my previous claim that scientists believe in things they don't detect. I was thinking of a good beautiful physics example. Almost everyone is convinced that gravitational waves exist. This is because they emerge as a natural prediction of general relativity and almost everyone is convinced, that except for very extreme regimes like when our universe was born(not clear what means to be born since t is nopt clear what time is), that general relativity is fine. However the closest thing we have to detection of gravitational waves is this: We see a binary system, two stars, one of which emits pulsed radio waves that we detect. The frequency of the pulse appears to increase, that is, we receive those pulsed emissions more often with time. The stars are rotating around their center of gravity and we interpret this frequency growth as an indication that they are getting closer to each other. Just like when you tie a stone to a thread and make it rotate around your hand, if you shorten the thread, the stone will rotate faster.
We in the other hand predict that this system will emit gravitational waves and we calculate how much energy will be loss in this emission. This loss of energy we claim, is responsible for the loss of total energy of the binary system and hence responsible for the stars coming close together. Our prediction for the frequency growth matches the curve we measure and hence we have deduced the existence of gravitational waves.
This is really the best there is about detecting gw. This discovery is worth a Nobel Prize. People believed in gw also before this was discovered. Hope I said something interesting
[/hide]

Ursula

I feel kinda awkward going back to more-or-less the original topic, but...

I fear death because, with the afterlife being irrelevent to this world and to me, everything I've thought, felt, planned, learned, and struggled for will be deleted, and I may not have time to share or to pass anything of my self on to others. I've put way too much into life to be happy at the prospect of its loss. This is only fear of an early death, of course--I think that, if I have enough time to establish my presence in the world and to be of benefit and note, that'll be enough.

I once thought about writing a story, from the perspective of someone who is forced, briefly, to work with someone who was widely disliked. This unpopular person then willingly risks her life and loses it, leaving with her partner only the message, "Remember that I had a soul." She had hoped to touch others in her life, but she wasn't able to do so in time. Many people don't know the real me, don't see the person behind me, and few enough fully realize that others are human, just like them. Consider the great historical oppressions and tragedies, often the result of one group of people being "dehumanized" in the eyes of another. To die without anyone truly feeling a connection to you is like never having existed at all, as far as the world will feel, and to die faceless, without anyone knowing who you really, specifically were is almost the same.

That said, across generations and geography, there are always people LIKE you. You, specifically, can be deleted, but who you are isn't lost because your characteristics aren't confined to one body. They're just combined in different ways, and it's often considered that variance and trying out new combinations leads to progress. Fear of death, which is a sort of resistance to change, can lead to a failure to see potential and opportunities--you only see what is falling apart, behind, not that which is growing and branching out.

Eventually, humanity will be completely gone, but I'm not convinced that that means anything. Certainly, to us, it can have little bearing on how we live.

Celebrate and make the most of life, filling yourself with the world and your world with you as much as possible, because then you lose the least by death, or so I think to myself.

I wouldn't think of death with the finality that religion imposes upon us. It's more like an oppurtunity for continuity, like a tangent graph, without its discontinuities it'll be dull to do calculus with it.
And even if we accept the point of view of certain religions and believe death to be the end of existence rather than life, why do we have to fear it, it's not like we'll feel it. In that sense immortality would be a greater curse than death or ceased existence.
In certain aspects it also depends on how you've done in life, how much you are happy with and how much you regret.
......
WAIT, isn't this a math forum/?!?!?

The secret to immortality is to not die.

Here is the summary:

We are born
We live our lives
We die

If we(our bodies) weren't meant to die, then we(our bodies) could talk to George Washington, which we(our bodies) can't. Therefore, we(our bodies) are meant to die.

death is just another way for God to take us into his heart.  Even if your not catholic, God will love you all the same.

as a little child (around 4-11, I think), I was terrified of death and dying; it used to keep me up at night, largely because the two posibilities I was aware of were:

1. Hell- I was not raised in a religion but was exposed to religious ideas, including 'if not believe in God, then go to Hell'

2. Absolute destruction.  Cesation of being.

begining in 7th grade, my fear of death began to disapate.  this is partially the result of the death of one of my friends during the summer before 7th grade.  Nowdays, it very very very rarely bothers me.

to me death would be a good thing. It means going to a better existence than this one will ever be. With the world problems we have today, it has got to be better. That doesn't mean that I want it to come quickly. There are too many things I want to experience yet.

I wanna go back and read all yer widdou arguments later, but I have a feeling I won't, unfortunately.... maybe someone could remind me :roll:

anyhow, as far as fear goes:  it ain't rational.  MY fear of death never was.  and now it never will be.  Why? Pyrrho.  Honestly, since the underlying metaphysical reality of the world, which may or may not exist in a pataphysical sense is, in either case, unknoweable, there is no reason to suspect anything negative awaits after death.

on a side note, I think death is the ultimate irony, since I think (opperative word here: think): no matter what you expect, it is impossible to have any conception of the lack of conception that awaits you.

I had a dream I died.  It began with me being unconcious, then moved to a stage where the only think I was concious of was my total unconciousness.  Then I heard the last note of Jesus Christ Pose by Soundgarden.  Then I think I jumped off into some crazy dream.


In  a jar of base area $A$, there are two immiscible, non-viscous , zero- surface tension... et cetera, et cetera liquids of density $p$ and $2p$ , filled to a height of
$h$ and $2h$ respectively. At the interface of these liquids rests a small sphere of volume $V$ with half its volume in either liquid. Estimate the total force on the base of the jar. don't give me just the method, its the final answer that matters.

what happened to everybody?
 is this too easy or something?
 i was thinking i'd get a reply as soon as i posted it!


integration
logarithms
calculus
calculus computations

solve for this homogeneous differential equation
(x^2+y^2)dx +xydy=0

Just subs $u= \frac{y}{x}$ or $\frac{x}{y}$ if you want. And you'll get separable equation.

still dont know how to do it.. please show me all the way through plz

Ok though i still have 15 minutes before class begin.
We have \[ \frac{dy}{dx} =- \frac{x^2+y^2}{xy} = - \frac{x}{y}-\frac{y}{x}  \]

With subtitution $u=\frac{y}{x}$. So $y=ux$ and $\frac{dy}{dx}=x \frac{du}{dx}+u$ And our ODE is now \[ x \frac{du}{dx}+u = -u - \frac{1}{u}  \]
Or \[ x \frac{du}{dx}= -2u - \frac{1}{u}  \]
this is separable. equivalent with.
\[ \frac{1}{-2u - \displaystyle \frac{1}{u}} \, du=\frac{dx}{x}  \]
Integrated both side.

\begin{eqnarray*}\int \frac{1}{-2u - \displaystyle\frac{1}{u}} \, du &=& \int \frac{u}{-2u^2 - 1} \, du \\ &=& -\frac{1}{4} \int \frac{1}{2u^2 + 1} \, d(2u^2) \\ &=& -\frac{1}{4} \ln|2u^2+1| +C_1 \end{eqnarray*}
And \begin{eqnarray*}\int \frac{dx}{x}= \ln |x| +C_2\end{eqnarray*}
So \[ \ln|2u^2+1|= -4 \ln |x| +C  \]
where $C=-4(C_1+C_2)$.Now back subtitution $u=\frac{y}{x}$. Yield \[ \ln \left|2 \frac{y^2}{x^2}+1 \right|= -4 \ln |x| +C  \]
If you want you can simplify the expression. I hope i didn't make any typo.


Is an intermediate geometry course offered? The reason I'm asking is because
http://www.artofproblemsolving.com/Classes/AoPS_C_ClassesS.php
Doesn't have info on one.

[quote="rrusczyk"]No, and we probably won't have one for quite a while.  Too many other things on the todo list right now...[/quote]
Dang. Do you know when the book is coming out, sir?

Lol sir, you can just call him Mr. Rusczyk I assume.

He told me, I believe, towards the end of 2007, not positive. He said Intro to Algebra would be out in like January, and Interm C/P would be out in like March, Interm Algebra in like June. Note the likes.


I proposed this inequality and gave a solution by putting $ab=t$, $0<t\leq \tfrac12$. It was published in a Vietnamese journal.


oops. :blush: How do you edit the poll? I can't seem to be able to do that.

oh well [b]HOPEFULLY THEY'LL SEE THIS AND DELETE MICHIGAN OUT OF THE POLL[/b] *cough*

I don't know if it was new, but Puerto Rico had the best pins, and they had 3.  Oh, and congrats on making nats 3 striaght years, kstan

thanks rd. and yes puerto rico had that pin back in '06 and in '05 they had a pin with a PICTURE of a frog on it...so no that's not a new one.

I remember a very nice Puerto Rican frog pin from 1996. They've been around for a while.

I think I have one of the frog pins from 2001.

The frogs are pretty good though, way better than some other pins that I will not name.

[quote="13375P34K43V312"]um michigan had the exact same pin
but i would have to say massachusetts because we had an approximately $\frac{1}{\epsilon}$ percent improvement from last year.[/quote]

Of course they were better-they said MA.

[quote="Ignite168"]Why is Connecticut not on the poll?  :maybe:  They're full of surprises.

I don't really think the free plastic thingies MA had last year could really count as pins so anything is an improvement. RI had good ones, did they even have pins last year?[/quote]They had them 2 years ago...it was a Rhode Island Red...but last year I didn't get one so they must not have had them (either that or I suck at pin trading). Is Connecticut really full of surprises if they come back with the exact same pins year after year? :rotfl:


Let ${\mathbb Q}^{+}$ be the set of positive rational numbers. Construct a function $f:{\mathbb Q}^{+}\rightarrow{\mathbb Q}^{+}$ such that \[f(xf(y)) = \frac{f(x)}{y}\] for all $x, y \in{\mathbb Q}^{+}$.

We show first that $ f(1) \equal{} 1$. Taking $ x \equal{} y \equal{} 1$, we have f(f(1)) = f(1). Hence $ f(1) \equal{} f(f(1)) \equal{} f(1 f(f(1)) ) \equal{} \frac{f(1)}{f(1)} \equal{} 1.$

Next we show that $ f(xy) \equal{} f(x)f(y)$. For any y we have $ 1 \equal{} f(1) \equal{} f(\frac{1}{f(y)} f(y)) \equal{} f(\frac{1}{f(y)})/y$, so if $ z \equal{} \frac{1}{f(y)}$ then $ f(z) \equal{} y$. Hence $ f(xy) \equal{} f(xf(z)) \equal{} \frac{f(x)}{z} \equal{} f(x) f(y)$.

Finally, $ f(f(x)) \equal{} f(1 f(x)) \equal{} \frac{f(1)}{x} \equal{} \frac{1}{x}$.

We are not required to find all functions, just one. So divide the primes into two infinite sets S = {p_1, p_2, ... } and T= {q_1, q_2, ... }. Define f(p_n) = q_n, and f(q_n) = 1/p_n. We extend this definition to all rationals using $ f(xy) \equal{} f(x) f(y); f(p_{i_1}p_{i_2}...q_{j_1}q_{j_2}.../(p_{k_1}...q_{m_1}...)) \equal{} p_{m_1}...q_{i_1}.../(p_{j_1}...q_{k_1}...)$. It is now trivial to verify that $ f(x f(y)) \equal{} \frac{f(x)}{y}$.


find all positive integers x,y which
x^3=y^2+2
or
x^2=y^3+2
-> (x,y)=(3,5) is the only answer for the first equation

please tell me the proof of this problem[/quote][/code]

Any reasonable proof would use algebraic number theory. So would you be pleased with such a proof¿

$ x^3\equal{}y^2\plus{}2$
yields
$ 8x^3\plus{}y^4\equal{}(y^2\plus{}4)^2$.

Is there some solution formula of
$ a^3\plus{}b^2\equal{}c^2$?

It's more convenient to work with the problem as given.  $ y^2 \plus{} 2 \equal{} (y \plus{} \sqrt{\minus{}2})(y \minus{} \sqrt{\minus{}2})$, so working in $ \mathbb{Z}[ \sqrt{\minus{}2} ]$ we can show that $ x$ must be the norm of $ a \plus{} b \sqrt{\minus{}2}$ for some $ a, b$, hence

$ (a \plus{} b \sqrt{\minus{}2})^3 \equal{} y \pm \sqrt{\minus{}2}$.

The rest is computation.


geometry
3D geometry
tetrahedron
inradius
sphere
geometry proposed

[b](a)[/b] [Problem for class 11]

Let r be the inradius and $r_a$, $r_b$, $r_c$ the exradii of a triangle ABC. Prove that $\frac{1}{r}=\frac{1}{r_a}+\frac{1}{r_b}+\frac{1}{r_c}$.

[b](b)[/b] [Problem for classes 12/13]

Let r be the radius of the insphere and let $r_a$, $r_b$, $r_c$, $r_d$ the radii of the four exspheres of a tetrahedron ABCD. (An [i]exsphere[/i] of a tetrahedron is a sphere touching one sideface and the extensions of the three other sidefaces.)

Prove that $\frac{2}{r}=\frac{1}{r_a}+\frac{1}{r_b}+\frac{1}{r_c}+\frac{1}{r_d}$.

I am really sorry for posting these, but else, Orl will probably post them. This time, we really did not have any challenging problem on the DeMO. But at least, the problems were simple enough that I solved all of them. ;)

  Darij

I think that the first equality is well known, I am going to look for a proof in a second  :lol:

As for (a), using area.
Let the corresponding sides of the triangle $ABC$ be $a, b, c$ respectively, $S$ be the area of triangle $ABC$,
then $S= \frac{1}{2}br_a+\frac{1}{2}cr_a-\frac{1}{2}ar_a$,

so $r_a=\frac{2S}{b+c-a}$, that is $\frac{1}{r_a}=\frac{b+c-a}{2S}$.

Likely, $\frac{1}{r_b}=\frac{c+a-b}{2S}$, $\frac{1}{r_c}=\frac{a+b-c}{2S}$.

$\frac{1}{r_a}+\frac{1}{r_b}+\frac{1}{r_c}=\frac{a+b+c}{2S}$.

While $S=\frac{1}{2}r(a+b+c)$, so $\frac{1}{r}=\frac{a+b+c}{2S}$.

So, $\frac{1}{r}=\frac{1}{r_a}+\frac{1}{r_b}+\frac{1}{r_c}$.

As for (b), using volume. the proof is similar.
Let the corresponding face areas of the tetrahedron $ABCD$ be $a, b, c, d$ respectively, $V$ be the volume of tetrahedron $ABCD$,
then $V= \frac{1}{3}br_a+\frac{1}{3}cr_a+\frac{1}{3}dr_a-\frac{1}{3}ar_a$,

so $r_a=\frac{3V}{b+c+d-a}$, $\frac{1}{r_a}=\frac{b+c+d-a}{3V}$.

Likely, $\frac{1}{r_b}=\frac{c+d+a-b}{3V}$, $\frac{1}{r_c}=\frac{d+a+b-c}{3V}$, $\frac{1}{r_d}=\frac{a+b+c-d}{3V}$.

$\frac{1}{r_a}+\frac{1}{r_b}+\frac{1}{r_c}+\frac{1}{r_d}=\frac{2(a+b+c+d)}{3V}$.

While $V=\frac{1}{3}r(a+b+c+d)$, so $\frac{1}{r}=\frac{a+b+c+d}{3V}$.

So, $\frac{2}{r}=\frac{1}{r_a}+\frac{1}{r_b}+\frac{1}{r_c}+\frac{1}{r_d}$.

I had posted a problem which maybe have the same idea as this one:

http://www.mathlinks.ro/Forum/viewtopic.php?t=6363

[quote="darij grinberg"]

I am really sorry for posting these, but else, Orl will probably post them.
  Darij[/quote]

There are always grade 8-10 problems.  ;)


I made up this problem and think I know the answer, but I don't know how to solve it:

Let $ a$ and $ b$ be positive integers greater than $ 1$. Find the smallest positive value of $ 1000 \minus{} a^b$.

$ a^b$ must be as great as possible and less than $ 1000.$ Now just look at perfect powers less than $ 1000.$ The greatest value is when $ (a,b)\equal{}(31,2),$ so the answer is \[ 1000\minus{}31^2\equal{}1000\minus{}961\equal{}\boxed{39}.\]

Not much of a solution though.

[quote="tennis123"]so it would just be guess and check?[/quote]
Yes.

[quote="21 7 15"]a=33 b=2
the number is 969[/quote]
$ 33^2 \equal{} 1089$.

Well... Heres a relatively quick guess-and-check.

First, find the greatest number $ a$ such that $ a^2 < 1000$. The greatest $ a$ is 31.

Now, we find the smallest number such that $ a^3 > 1000$. This smallest $ a$ is 10.

Thus, it can't be any numbers between $ 10$ and $ 30$ inclusive.

Then just check to find the greatest powers of each of the numbers $ 2$ to $ 9$, without checking any perfect squares (since we can write them as a power of their square root).

The answer turns out to be $ a \equal{} 31$, $ b \equal{} 2$ and $ 1000 \minus{} a^b \equal{} 39$.

You guys forgot that b could equal 1 but I guess that is implied that it can't equal 1

Good point. I fixed the problem. Also, I have a similar question, but I will put that in another thread.

For any problem related to this, i.e. finding any power $ a^b$ closest to some given positive integer $ N$, at least brute force would yield the following bound for $ a$: $ 2\le a\le\lfloor\sqrt{N}\rfloor$. From this, you'll have to check all of those powers that are closest to $ N$. Another robust bound for $ b$ immediately yields $ 2\le b\le\lfloor\log_{2}N\rfloor\plus{}1$
For example, find the closest power to 126. Certainly, $ 5^3\equal{}125$ is the closest one but if we changed the number to 127, it would be $ 2^7\equal{}128$.

This is insanely similar to a problem on the 2009 MATHCOUNTS National Sprint (it was like, find the largest perfect power less than 400).

This  problem is pretty because 1000 is a pretty small number but when we are talking about million can we formulate a proof so that the square is always the highest.


superior algebra
superior algebra unsolved

Here is simple problem about simple extension fields. I am posting it because this sometimes is a useful thing to know about extension fields and you might use it.

Let $n,m\geq 0$ and $n,m\in \mathbb{Q}$ such as $\sqrt{n}+\sqrt{m}\not = 0$. Show that:
$\mathbb{Q}(\sqrt{n}+\sqrt{m}) = \mathbb{Q}(\sqrt{n},\sqrt{m})$.

Because $\sqrt{m},\sqrt{n}\in\mathbb{Q}(\sqrt{m},\sqrt{n})$ we have $\sqrt{m}+\sqrt{n}\in\mathbb{Q}(\sqrt{m},\sqrt{n})$, so $\mathbb{Q}(\sqrt{m}+\sqrt{n})\subseteq \mathbb{Q}(\sqrt{m},\sqrt{n})$.(1)

We have $\sqrt{m}+\sqrt{n}\in \mathbb{Q}(\sqrt{m}+\sqrt{n})$ and $\sqrt{m}-\sqrt{n}=\frac{m-n}{\sqrt{m}+\sqrt{n}}\in \mathbb{Q}(\sqrt{m}+\sqrt{n})$ therefore $\sqrt{m},\sqrt{n}\in \mathbb{Q}(\sqrt{m}+\sqrt{n})$, so $\mathbb{Q}(\sqrt{m},\sqrt{n})\subseteq \mathbb{Q}(\sqrt{m}+\sqrt{n})$. (2)

From (1) and (2) we are done!

[quote="N.T.TUAN"]Because $ \sqrt{m},\sqrt{n}\in\mathbb{Q}(\sqrt{m},\sqrt{n})$ we have $ \sqrt{m}+\sqrt{n}\in\mathbb{Q}(\sqrt{m},\sqrt{n})$, so $ \mathbb{Q}(\sqrt{m}+\sqrt{n})\subseteq \mathbb{Q}(\sqrt{m},\sqrt{n})$.(1)

We have $ \sqrt{m}+\sqrt{n}\in \mathbb{Q}(\sqrt{m}+\sqrt{n})$ and $ \sqrt{m}-\sqrt{n}=\frac{m-n}{\sqrt{m}+\sqrt{n}}\in \mathbb{Q}(\sqrt{m}+\sqrt{n})$ therefore $ \sqrt{m},\sqrt{n}\in \mathbb{Q}(\sqrt{m}+\sqrt{n})$, so $ \mathbb{Q}(\sqrt{m},\sqrt{n})\subseteq \mathbb{Q}(\sqrt{m}+\sqrt{n})$. (2)

From (1) and (2) we are done![/quote]

This is best viewed as an example of the Primitive Element Theorem, e.g. see http://google.com/groups?selm=y8z1xciwk3t.fsf%40nestle.csail.mit.edu

[quote="billwgd"]
This is best viewed as an example of the Primitive Element Theorem, e.g. see http://google.com/groups?selm=y8z1xciwk3t.fsf%40nestle.csail.mit.edu[/quote]

But why! It is a relatively simple problem.

[quote="sylow_theory"][quote="billwgd"]
This is best viewed as an example of the Primitive Element Theorem, e.g. see http://google.com/groups?selm=y8z1xciwk3t.fsf%40nestle.csail.mit.edu
[/quote]
But why! It is a relatively simple problem.[/quote]

My point is that knowing the basic strategy of the $ {\bf Primitive\; Element\; Theorem}$ immediately points the way toward the solution. Namely, here it suffices to find $ r \neq 1\in Q$ such that $ \sqrt{m}+r\sqrt{n}\in Q(\sqrt{m}+\sqrt{n}),$ with obvious solution $ r=-1.$ That completes the proof. One needn't prove any further inclusions nor manipulate any further equations. If it's still not crystal clear, you may find it helpful to review the simple proof of the PET referenced in my above sci.math post. By now it's overkill, but the point was worth emphasis.


The integer 216 is expressed as the sum of six consecutive odd integers. What is the largest of these integers?

Divide $ 216$ by $ 6$ (the number of terms added to make $ 216$) to find the median (also the mean) of the six numbers. $ \frac{216}{6}\equal{}36$. Now, to find the largest one, find the $ \frac{6}{2}\equal{}3$rd smallest odd number above $ 36$. Counting up, the answer is... $ 37$, $ 39$, $ \fbox{41}$.


calculus
integration
calculus computations

Is there any calculater or website computes indefinite integral?

Try this : http://wolframalpha.com/

Or this : http://integrals.wolfram.com/