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trigonometry
ratio
complex numbers
calculus
calculus computations

hope i have not asked this question before... ( i tend to make most of these
questions up myself, but i do always try a search on the forums first and see
if i have posted it before... ) sorry if i did, but the search came negative,
so i am posting the question... show :|

$ \sum_{n\equal{}1}^{\infty}\;\tan^{\minus{}1}\left(\frac{1}{n^{2}}\right)\;\equal{}\;\boxed{\tan^{\minus{}1}\left(\frac{1\minus{}\left(\frac{\tanh\left(\frac{\pi}{\sqrt 2}\right)}{\tan\left(\frac{\pi}{\sqrt 2}\right)}\right)}{1\plus{}\left(\frac{\tanh\left(\frac{\pi}{\sqrt 2}\right)}{\tan\left(\frac{\pi}{\sqrt 2}\right)}\right)}\right)}$

Another way of solving this is by realizing that $ \arctan\left(\frac{1}{n^{2}}\right)$ is the polar angle of the complex number $ 1\plus{}\frac{i}{n^{2}}$ and so

$ S \equal{}\sum_{n\equal{}1}^{\plus{}\infty}\arctan\left(\frac{1}{n^{2}}\right) \equal{}\sum_{n\equal{}1}^{\plus{}\infty}arg\left(1\plus{}\frac{i}{n^{2}}\right)$.

But the argument of a product of complex numbers is the sum of the arguments of the complex numbers, and so

$ S \equal{}\sum_{n\equal{}1}^{\plus{}\infty}arg\left(1\plus{}\frac{i}{n^{2}}\right) \equal{} arg\,\prod_{n\equal{}1}^{\plus{}\infty}\left(1\plus{}\frac{i}{n^{2}}\right)$.

Recall now that $ \frac{\sin x}{x}\equal{}\prod_{n\equal{}1}^{\plus{}\infty}\left(1\minus{}\frac{x^{2}}{n^{2}\pi^{2}}\right)$, and so let's solve the equation $ \frac{x^{2}}{\pi^{2}}\equal{}\minus{}i$:

$ x \equal{}\pi\sqrt{\minus{}i}\equal{}\pi\left[cis\left(\minus{}\frac{\pi}{2}\right)\right]^{1/2}\equal{}\pi cis\left(\minus{}\frac{\pi}{4}\right) \equal{}\frac{\pi (1\minus{}i)}{\sqrt{2}}$. Hence:

$ S \equal{} arg\,\prod_{n\equal{}1}^{\plus{}\infty}\left(1\plus{}\frac{i}{n^{2}}\right) \equal{} arg\,\prod_{n\equal{}1}^{\plus{}\infty}\left(1\minus{}\frac{[\pi (1\minus{}i)/\sqrt{2}]^{2}}{n^{2}\pi^{2}}\right) \equal{} arg\,\frac{\sin\left(\frac{\pi}{\sqrt{2}}\minus{}\frac{\pi}{\sqrt{2}}i\right)}{\frac{\pi}{\sqrt{2}}\minus{}\frac{\pi}{\sqrt{2}}i}$.

Now, as $ \sin(x\minus{}yi) \equal{}\sin x\cosh y\minus{}i\cos x\sinh y$, then

$ S \equal{} arg\,\frac{\sin\left(\frac{\pi}{\sqrt{2}}\right)\cosh\left(\frac{\pi}{\sqrt{2}}\right)\minus{}i\cos\left(\frac{\pi}{\sqrt{2}}\right)\sinh\left(\frac{\pi}{\sqrt{2}}\right)}{\frac{\pi}{\sqrt{2}}\minus{}\frac{\pi}{\sqrt{2}}i}$.

The angle of the ratio of two complex numbers is the difference of the angles of the two complex numbers, and so we have finally

$ S \equal{}\frac{\pi}{4}\minus{}\arctan\left[\frac{\tanh\left(\frac{\pi}{\sqrt{2}}\right)}{\tan\left(\frac{\pi}{\sqrt{2}}\right)}\right]$.


The values of $ x$ and $ y$ are always positive, and $ x^2$ and $ y$ vary inversely. If $ y$ is 10 when $ x$ is 2, then find $ x$ when $ y$ is 4000.

when two things vary inversely that means as one value goes up the other go down and the expression would be expressed as xy. however in this case it is expressed as (x^2)y. so if y is 10 and x is 2 the total value would be 40. then u set up the equation 40=x^2*4000.

x^2= 1/100
x=1/10


geometry
3D geometry
sphere
ratio
similar triangles

I have two spheres of radius 2, centered at M and N, and one sphere of radius 1, centered at O. All three spheres are externally tangent and lie on a flat table. M, N, and O lie on plane m. Triangle MNO has centroid G. A line is perpendicular to plane m and passes through G, intersecting the table at point P. Find the length of GP.

EDIT: i have my first answer in there..but then i realized i misread the quesion but it has all the information just not used correctly because i misreaqd the quesion but I get a final answer of
[hide="answer"]$\frac{5\sqrt{5}}{6}$

[hide]I get $\frac{5}{3}$ using similar triangles but Im bad at imagining these things..

I found that GO is $\frac{2\sqrt{5}}{3}$.  I also found the line GO going through O is $\sqrt{5}$ from the table so I used the simlar ratio of

$\frac{\frac{5\sqrt{5}}{3}}{x}=\sqrt{5}$ and $x=\frac{5}{3}$

EDIT: wait I misread the quesion, its a line perpendicular to the triangle going through G, not a line perpendicualr to the table going through G..let me edit..

Now that I read the quesioin correctly, I get $x=\frac{5\sqrt{5}}{6}$ using the same proportion I used before but instead having it equal to $2$ instead of $\sqrt{5}$[/hide][/hide]

[quote="Whatthedeuce"]sorry if my solution is difficult to understand. its not easy to explain this one without a picture
[hide]
MN=4
MO=NO=3
the median from O to MN has length sqrt(5)
and it moves 1 unit vertically and 2 units horizontally from O
OG=2/3 the median
so OG=2sqrt(5)/3
and G moves 2/3 units vertically and 4/3 units horizontally from O
so we add the 2/3 units to the 1 unit O is already suspended off of the plane
the answer is [b]5/3[/b][/hide][/quote]

uh I get 5/3 too initially but I misread the question..so I got a new answer of $\frac{5\sqrt{5}}{6}$


arithmetic sequence
geometry
perpendicular bisector
number theory
relatively prime

(1) The [i]digital sum[/i] of a number is the sum of its digits.  For how many of the positive integers $ 24\ \text{to}\ 125$ inclusive is the digital sum a multiply of $ 7$?

[hide="My Answer"]$ \boxed{12}$[/hide]

(2) The arithmetic sequence with first term $ 1$ has a common difference of $ 6$.  A second sequence begins with $ 4$ and has a common difference $ 7$.  In the range of $ 1$ to $ 100$, what is the largest number common to both sequences?

[hide="My Answer"]$ \boxed{67}$[/hide]

(3) Find the equation of the line which is the perpendicular bisector of the segment between $ P(2,3)$ and $ Q(\minus{}4,5)$.  Express your answer in the form $ Ax \plus{} By \plus{} C \equal{} 0$ where $ A$ is positive and $ A$, $ B$, and $ C$ are relatively prime integers.

[hide="My Answer"]$ \boxed{3x \minus{} y \plus{} 7 \equal{} 0}$[/hide]

Since you can do number 2 and 3 correctly, ill tell you how to solve #1.
Since you can only have two digit and three digit numbers, try to find ways to get seven by adding numbers.
Two Digit:
25
34
43
52
61
70
Three digit
106
115
124
So the answer is 9.

It asks for a multiple of 7, so 14 is also included. (21 won't work for 3 digit numbers starting with 1)
14:
59
68
77
86
95

So in total, it's 14.


If $a$ is $50 \%$ larger than $c$, and $b$ is $25 \%$ larger than $c$, then $a$ is what percent larger than $b$?

A. $20 \%$
B. $25 \%$
C. $50 \%$
D. $100 \%$
E. $200 \%$

[hide="solution"]
Write out the equations:
$a=\frac{3}{2}c\\b=\frac{5}{4}c\\c=\frac{4}{5}b$

Substituting, $a=\frac{6}{5}b$, or a is 20% larger than b. $\boxed{A}$[/hide]

[hide]
$a=\frac{3}{2}c$
$b=\frac{5}{4}c$
$c=\frac{4}{5}b$
Substituting, you see that
$a=\frac{6}{5}b$
So, $A$[/hide]


For $ a,b,c$ sides of an acute triangle we have the inequality:

$ a^4b^4c^4 \geq (a^2 \minus{} b^2)^2(b^2 \minus{} c^2)^2(c^2 \minus{} a^2)^2$

This is not true, for counter-example:

$ a\equal{}3,\, b\equal{}6,\, c\equal{}8.$

can_hang2007 communicated to me this nice solution

use \[ a^2 \ge \vert\ b^2\minus{}c^2 \vert\]

The given counter example is not true, since
$ 8^2 \ge 3^2\plus{}6^2 \rightarrow$ the triangle is obtuse

Since the triangle is acute, we have $ a^2\plus{}c^2\ge b^2$ and $ a^2\plus{}b^2\ge c^2$, i.e., $ a^2\ge|b^2\minus{}c^2|$. Similarly we have $ b^2\ge|c^2\minus{}a^2|$ and $ c^2\ge|a^2\minus{}b^2|$. Hence $ a^2b^2c^2\ge|(a^2\minus{}b^2)(b^2\minus{}c^2)(c^2\minus{}a^2)|$. Squaring, we have
\[ a^4b^4c^4 \geq (a^2 \minus{} b^2)^2(b^2 \minus{} c^2)^2(c^2 \minus{} a^2)^2,
\]
as desired.


P(x)=x^4-x^3-12x^2-2x+8. Find Q(x) with the least interger coefficients such that the roots of Q(x)=0 are the reciprocal of the roots of P(x)=0.

[hide]Just reverse the coefficients as taught in AoPS 2.

Q(x)=8^4-2x^3-12x^2-x+1[/hide]


function
analytic geometry
Pythagorean Theorem
geometry

Two problems, one I solved but not sure if it is right and another I don't know at all.

"An open-top box is a constructed from a 15 by 21 inch rectangular of cardboard by cutting out equal squares at each corner and then folding up the flaps. Express the Volume of V of the box as a function of X, the lenght of one side of each square."

I know volume for a rectangular solid is V=LWH. The height in this case is X. We now need to find L and the W. The L that I found is 21-2x, and the W I found is 15-2x. So the volume of this solid would be V=(21-2x)(15-2x)(x) right?

The Problem I can't seem to figure out for the life of me goes like this.

"Car A leaves an intersection traveling east at 60MPH. At the same time car B leaves the same intersection traveling north at 65MPH. Express the distance d(in miles) between the two cars as a function of t, where t denotes time in minutes after the cars leave the intersection."

So far, I figured this problem is like a problem on the Cartesian Plane, and setting it up like a distance problem of (x,y) and (h,k) but after that I am all baffled at the t, and how it is involved in the problem.

[quote="i_like_pie"]#1 is correct.

[hide="2"]Using the Pythagorean Theorem, here is the formula:

$D=\sqrt{\frac{60t}{60}+\frac{65t}{60}}$

$D=\sqrt{\frac{23t}{12}}$

$D=\frac{\sqrt{69t}}{6}$[/hide][/quote]

How about using squares under the root when trying to work with Pythagoras's? :lol:

[quote="Farcus"]how did you guys set up the problem and get the relationship set up?[/quote]
[hide] OK, basically the cars' routes form a right triangle, with Car A traveling on the shorter (60 mph) leg and Car B on the longer (65 mph) leg. Since we are working in minutes, we will convert 60 mph to 1 mile per minute. Also 65 mph is $\frac{65}{60}=\frac{13}{12}$ miles per minute. Thus the right triangle has legs $t$ and $\frac{13}{12}t$. Using the Pythagoean Theorem, we get $\sqrt{(t)^{2}+\left(\frac{13}{12}t\right)^{2}}=\sqrt{\frac{169+144}{144}t^{2}}=t\frac{\sqrt{313}}{12}$ [/hide]

[quote="daermon"][hide="2"]$t\frac{\sqrt{\frac{3600}{t^{2}}+\frac{4225}{t^{2}}}}{60}=\boxed{\frac{t\sqrt{313}}{12}}$[/hide][/quote]

Please do not just post an answer, especially when the problem asks for help, not the answer.  Even though the math does help, there is not a solution to go with it to show how you got your expression.

Thanks.


linear algebra
matrix
inequalities unsolved
inequalities

This problem was on Baltic Way in the early 2000:
If a(1).... a(n) is a sequence of positove reals for which
$ \begin{matrix}
\sum {{a}_{i}}^{3}\equal{}3\\
\sum {{a}_{i}}^{5}\equal{}5
\end{matrix}$
Prove that$ \sum {{a}_{i}}^{}\geq \frac{3}{2}$
When I read the offficial solution, it used several steps with Hölder and so on. I wonder if this isnt just a direct consequence of cauchy, since $ \begin{matrix}   (\sum {a}_{i})(\sum {{a}_{i}}^{5})\geq {(\sum {{a}_{i}}^{5/2}{{a}_{i}}^{1/2})}^{2}\\
\Rightarrow \sum {a}_{i}\geq \frac{9}{5}>\frac{3}{2}
\end{matrix}$ Am I wrong?

[quote="k-boy"]This problem was on Baltic Way in the early 2000:
If a(1).... a(n) is a sequence of positove reals for which
$ \begin{matrix} \sum {{a}_{i}}^{3} \equal{} 3 \\
\sum {{a}_{i}}^{5} \equal{} 5 \end{matrix}$
Prove that$ \sum {{a}_{i}}^{}\geq \frac {3}{2}$
When I read the offficial solution, it used several steps with Hölder and so on. I wonder if this isnt just a direct consequence of cauchy, since $ \begin{matrix} (\sum {a}_{i})(\sum {{a}_{i}}^{5})\geq {(\sum {{a}_{i}}^{5/2}{{a}_{i}}^{1/2})}^{2} \\
\Rightarrow \sum {a}_{i}\geq \frac {9}{5} > \frac {3}{2} \end{matrix}$ Am I wrong?[/quote]

Yeah i think it is. But then equaltiy will not hold.


inequalities calculus algebra polynomial derivative

We all know that $x+\frac{1}{x}\geq 2$ by AM-GM. But can you minimize this expression:
\[x+\frac{1}{x}+(x+1)+\frac{1}{x+1}\]
for $x \in \mathbb{R^{+}}$?

Can you generalize for this sum? :
\[f(x) = \sum_{k=0}^{n}\left( (x+k)+\frac{1}{x+k}\right) \]
(also for $x \in \mathbb{R^{+}}$)

Aaaah, I see what I missed.  I must be more sleep-deprived than I thought...

Anyway, thanks for that.

[hide="an idea"]$x+\frac{1}{x}+(x+1)+\frac{1}{x+1}=2x+1+\frac{2x+1}{x(x+1)}$, so we can just use AM-GM again.  I guess you could combine it into two numbers and do that for all of them, but it would get pretty nasty.  Even the solution for that one is nasty.  I got $\frac{x^{2}+x+1}{\sqrt{x^{2}+x}}\geq 2$, which I'm really having trouble dealing with without a calculator...

Ok, the calculator is telling me that gives $x\approx .61803$, which is not what Paladin8 got...[/hide]

[quote="tjhance"][quote="DanK"][quote="cincodemayo5590"]We all know that $x+\frac{1}{x}\geq 2$ by AM-GM.[/quote]

How do you prove this?  I can see that it's true, but I'm not seeing how AM-GM proves it.[/quote]

$\frac{\frac{1}{x}+x}{2}\ge \sqrt{(x)(\frac{1}{x})}$
$\frac{\frac{1}{x}+x}{2}\ge 1$
$\frac{1}{x}+x\ge 2$[/quote]
It can also be proven with trivial inequality.
$(x-1) \geq 0$
$x^{2}+1 \geq 2x$
$x+\frac{1}{x}\geq 2$

[quote="ch1n353ch3s54a1l"]Sorry, I don't get it...why can't you just AM-GM again making the minimum 4...then for the general case $2(n+1)$[/quote]

Well, you can show that $\text{expression}\ge 4$. But you don't know that it actually [i]reaches[/i] $4$. For example, I know

$x^{2}\ge-10$,

but that doesn't mean $-10$ is the minimum. Finding the minimum requires two things:

1. Find a lower bound.

2. Find the equality case.

In this case, you have found a lower bound of $4$ but you'll notice that there is no equality case.

[quote="breez"]It seems the trivial proof above only holds for $x$ 1 or greater.[/quote]

The first line should be $(x-1)^{2}\ge 0$.  The fact that he divided by $x$ means the proof holds for $x > 0$, which is the same scope as AM-GM anyway.  The result is different, but equally obvious, for $x < 0$.

[quote="paladin8"]In this case, you have found a lower bound of $4$ but you'll notice that there is no equality case.[/quote]

In fact, the equality case is $x = x+1 = 1$ which is clearly absurd.

[quote="t0rajir0u"][quote="DanK"]I got $\frac{x^{2}+x+1}{\sqrt{x^{2}+x}}\geq 2$[/quote]

You can't use AM-GM to get a new expression and then minimize that expression, since the new expression and the original one don't have to have the same equality case.[/quote]

I don't think that's quite what's going on in my work.  AM-GM gave me the inequality $\frac{2x+1+\frac{2x+1}{x(x+1)}}{2}\geq\sqrt{(2x+1)\left(\frac{2x+1}{x(x+1)}\right)}$, which means that the x-value satisfying the equality case minimizes the expression $2x+1+\frac{2x+1}{x(x+1)}$.  $\frac{x^{2}+x+1}{\sqrt{x^{2}+x}}\geq 2$ is simply a rearrangement of that original inequality that lets you solve for
$x$, so the x-value minimizing that expression should be the same, even if the minimum of the expression itself isn't.  Since all I did was solve for $x$, that should be the correct x-value to minimize the original expression.  If my value is correct (although it may not be, according to Paladin8 and JavaMan), that would make the minimum of the expression $\approx 4.4721$.

[quote="JavaMan"]You must've made a mistake. My calculator's showing the minimum ~0.77122988 which is what paladin8 got.[/quote]

I don't know what's going on then, because I just tried it again, and it gave me the same thing...

[quote="t0rajir0u"][quote="breez"]
It seems the trivial proof above only holds for $x$ 1 or greater.[/quote]

The first line should be $(x-1)^{2}\ge 0$. The fact that he divided by $x$ means the proof holds for $x > 0$, which is the same scope as AM-GM anyway. The result is different, but equally obvious, for $x < 0$.[/quote]

Isn't it a maximum for $x<0$?  Wouldn't it be $x+\frac{1}{x}\leq-2$?

@DanK: How is $\frac{x^{2}+x+1}{\sqrt{x^{2}+x}}\ge 2$ is rearrangement of the original inequality? What is the original inequality?

And btw, at $x \approx 0.7712$ we obtain a minimum of about $4.40\ldots$, so it's smaller than yours.

[quote="DanK"]I don't think that's quite what's going on in my work.  AM-GM gave me the inequality $\frac{2x+1+\frac{2x+1}{x(x+1)}}{2}\geq\sqrt{(2x+1)\left(\frac{2x+1}{x(x+1)}\right)}$, which means that the x-value satisfying the equality case minimizes the expression $2x+1+\frac{2x+1}{x(x+1)}$.  [/quote]

No.  That's exactly what I said.  This logic [b]isn't valid[/b].  For example, suppose I wanted to minimize $x+1$ for non-negative reals $x$.  Well, by AM-GM

$x+1 \ge 2 \sqrt{x}$

So by your logic the equality case of this AM-GM expression (which is when $x = 1$) minimizes the expression.  But it doesn't.  The actual minimum is at $x = 0$.

We can take the same expression and write it as $x+\frac{1}{2}+\frac{1}{2}$.  By AM-GM again,

$x+1 \ge 3 \sqrt[3]{ \frac{x}{4}}$

This inequality has the equality case $x = \frac{1}{2}$, which is both different from the first one and fails to correspond to the actual minimum.

[quote="DanK"]Isn't it a maximum for $x<0$?  Wouldn't it be $x+\frac{1}{x}\leq-2$?[/quote]

Yeah.

[quote="paladin8"]It appears that in the two-term case the minimum is attained by

$\frac{1}{2}\sqrt{3+2\sqrt{3}}-\frac{1}{2}\approx 0.77120$.

I doubt there is a nice way to do this. This is just the positive root of $f'(x) = 0$, where $f(x)$ is the sum. And it gets much, much uglier for $n > 2$.[/quote]

Do you mean $0.77123$ :wink:

Aaaah, I see.  So, is the only reason it works for the first case because the right side of the inequality becomes a constant?

[quote="paladin8"]@DanK: How is $\frac{x^{2}+x+1}{\sqrt{x^{2}+x}}\ge 2$ is rearrangement of the original inequality? What is the original inequality?[/quote]

The AM-GM inequality from the same post.  Of course, now that t0rajir0u pointed out the error in my solution, that doesn't really matter anyway...

So, does anyone know of a good way to do this without calculus?

[quote="DanK"]Aaaah, I see.  So, is the only reason it works for the first case because the right side of the inequality becomes a constant?[/quote]

More or less.

[quote="DanK"]So, does anyone know of a good way to do this without calculus?[/quote]

Doubtful.

[hide="The work (necessary to read)"]To truly solve the problem:
Set up $2x+1+\frac{2x+1}{x^{2}+x}\geq k$ as a polynomial ($k$ is the minimum):
\[\iff 2x^{3}+(3-k)x^{2}+(3-k)x+1\geq 0 \]
By a well known theorem, this must factor as the product of a square and a linear.
Furthermore, the square equals zero at the x which yields a minimum, determined to be $\frac{1}{2}\sqrt{3+2\sqrt{3}}-\frac{1}{2}\approx 0.77120=A$ by paladin.
So the square term is of the form constant$\cdot (x-A)^{2}$.
We set up as follows (letting $B=3-k$):
\[2x^{3}+Bx^{2}+Bx+1=(x-A)^{2}(ax+b)\frac{1}{bA^{2}}\]
where the constant was solved for by noting the unit terms.
Now a system must be set up and solved for comparing the coefficients in:
\[2x^{3}+Bx^{2}+Bx+1=(x^{2}-2xA+A^{2})(ax+b)\frac{1}{bA^{2}}= \]

\[=(ax^{3}+(b-2Aa)x^{2}+(aA^{2}-2Ab)x+bA^{2})\frac{1}{A^{2}b}\]
Right off the bat, $a=2A^{2}b$ so plug that in:
\[=(2A^{2}bx^{3}+(b-4A^{3}b)x^{2}+(2A^{4}b-2Ab)x+bA^{2})\frac{1}{A^{2}b}\]
thus $B=(1-4A^{3})b=(2A^{4}-2A)b$ (which is interesting, as $A$ can be solved as a root of $1-4A^{3}=2A^{4}-2A$ now if I didn't mess up)
Thus $b=\frac{B}{(1-4A^{3})}$. (which we don't actually need)
Our factorization is thus:
\[2x^{3}+Bx^{2}+Bx+1=(x-A)^{2}(2A^{2}x+1)\frac{1}{A^{2}}\]
[/hide]
So a proof would read as follows:
Consider the factorization
\[(x-A)^{2}(2A^{2}x+1)\frac{1}{A^{2}}\]
which is clearly positive. Upon expanding and factoring, this becomes:
\[2x^{3}+(3-k)x^{2}+(3-k)x+1\geq 0 \]
which implies that
\[x+\frac{1}{x}+x+1+\frac{1}{x+1}\geq k \]
with equality attained at $A$.
(The variables were defined in the work section).

A slicker method:
Note that the derivative of this polynom is $6x^{2}+2(3-k)x+3-k$. Then, $f''(A)$ is found which should immediately yield the linear term etc etc

EDIT: The reason this problem is messy and there is no "elementary" solution (technically all this is actually elementary except for the second derivative method) is that the term $x+1$ has no special algebraic significance in the case of polynomial inequalities. If this was over the integers, of course it would be an easy problem as would any polynomial, but that is the time when an $x+1$ term is easy to handle. Otherwise, no. You just have a general-case problem on your hands.

Well, that's what you get if you plug the $x$-value everyone else has been getting back into the original expression...

[quote="kunny"]Can anyone make a confirmation for my solution?

kunny[/quote]

It looks good to me. But we still can't avoid the calculus. me@home's solution works, but we would have to guess at the root (nearly impossible without knowing it already).


search
superior algebra
superior algebra unsolved

Hello, I have a problem with this:
We consider the norm $N: \mathbb{F}_{q^{n}}\rightarrow\mathbb{F}_{q}$. We fix $w$ a generator of $\mathbb{F}_{q}$ and I have to prove that it exists $z$ a generator of $\mathbb{F}_{q^{n}}$ which verify $N(z)=w$.

I have a solution but it uses the Dirichlet's theorem and I search something easier.

Thanks !

My solution using Dirichlet:
Let $a$ be a generator of $\mathbb{F}_{q^{n}}$. Then $N(a)=a^{(q^{n}-1)/(q-1)}$ is a generator of $\mathbb{F}_{q}$ then it exists $m$ verifying: $m$ is prime with $q-1$ and $a=w^{m}$.
Let $m'$ be the inverse of $m$ in $\mathbb{F}_{q}^{*}$.
Now we can choose $k$ such that $m''=m'+k(q-1)$ is a prime number greater than $q^{n}-1$ (consequently this number is prime with $q^{n}-1$). Then $N(z^{m''})=w$ and $z^{m''}$ is a generator of $\mathbb{F}_{q^{n}}$.
Is it correct ?

Please, I didn't have the solution yet. Have you ideas ?
Thanks


calculus
integration
logarithms
calculus computations

for (i)
$ \begin{array}{l} \smallint \frac {{{x^2} \plus{} 1}}{{{x^4} \plus{} 1}}\;dx\mathop \equal{} \limits_{\left( {1 \plus{} \frac {1}{{{x^2}}}} \right)dx \equal{} du}^{x \minus{} \frac {1}{x} \equal{} u} \equal{} \smallint \frac {1}{{{u^2} \plus{} 2}}\;du \equal{} \\
 \\
\mathop \equal{} \limits_{du \equal{} dt\sqrt 2 }^{u \equal{} t\sqrt 2 } \frac {{\sqrt 2 }}{2}\smallint \frac {1}{{{t^2} \plus{} 1}}\;du \equal{} \frac {{\sqrt 2 }}{2}\arctan t \plus{} c \equal{} \\
\equal{} \frac {{\sqrt 2 }}{2}\arctan \left( {\frac {{x \minus{} \frac {1}{x}}}{{\sqrt 2 }}} \right) \plus{} c \\
\end{array}$

for (ii)
$ \begin{array}{l} \smallint \frac {{{x^2} \minus{} 1}}{{{x^4} \plus{} 1}}\;dx\mathop \equal{} \limits_{\left( {1 \minus{} \frac {1}{{{x^2}}}} \right)dx \equal{} du}^{x \plus{} \frac {1}{x} \equal{} u} \equal{} \smallint \frac {1}{{{u^2} \minus{} 2}}\;du \equal{} \frac {1}{{2\sqrt 2 }}\smallint \left( {\frac {1}{{u \minus{} \sqrt 2 }} \minus{} \frac {1}{{u \plus{} \sqrt 2 }}} \right)\;du \\
 \\
\equal{} \frac {1}{{2\sqrt 2 }}\ln \left| {\frac {{u \minus{} \sqrt 2 }}{{u \plus{} \sqrt 2 }}} \right| \plus{} c \equal{} \frac {1}{{2\sqrt 2 }}\ln \left| {\frac {{x \plus{} \frac {1}{x} \minus{} \sqrt 2 }}{{x \plus{} \frac {1}{x} \plus{} \sqrt 2 }}} \right| \plus{} c \\
\end{array}$

for (iii) if we call the (i) integral as I and the (ii) integral as J, then (iii) can be calculated as $ \frac{1}{2}\left( {I \minus{} J} \right)$

Thanks. So for (ii), we can use the substitution [hide=" Click to reveal hidden content"] $ u\equal{}x\plus{}\frac{1}{x}$.[/hide]

I see you were replying while i was typing (was editing message to copmlete answer)  :D

Sorry mathxl, I was a bit hasty. I should have waited a while.


x(x2+11) + 6(x2+1)

is it still possible to factor this out? if yes, how?

It is $(x+1)(x+2)(x+3)$.

And, please, don't put such problems into the Open Questions.

I will move it to the Intermediate section.

i see, ok, thanks for correcting me and the quick reply -im a bit of a noob here ^^

ah nice, you've even clarified my previous post.  thank you!

Let's factorize $abcx^3-(a^2b+b^2c+c^2a)x^2+(ab^2+bc^2+ca^2)x-abc\ (abc\neq 0)$

[quote="probability1.01"]The roots are a/c, c/b, b/a, by Viete's.[/quote]
What's that? :?

[quote="probability1.01"]The roots are a/c, c/b, b/a, by Viete's.[/quote]

Why? :D

you can find the roots by depressed cubic.
for the equation $x^3+ax^2+bx+c=0$.
so devide the whole thing by $abc$,
$x^3 + ax^2 + bx + c = 0 $
now i define t, such that: $x = t - a/3. $
The substitution $ x = t - a/3 $ eliminates the quadratic term; in fact, we get the equation:
$t^3+Pt+Q=0$, where $P=b-(a ^2 /3)$, and $Q=c+[(2a^3-9b)/27]$
This is called the depressed cubic.

Suppose that we can find numbers u and v such that

$u^3-v^3=Q$, and $u.v=p/3 $
A solution to our equation is then given by

$t = v$ − $u$, (as can be checked by directly substituting this value for $t$ in the original equation above).
now, if you solve for "v" you get:
$v=P/(3u)$. know plug this into the $u^3-v^3=Q $ equation, you get:
$u=[(Q/2)-(  [(Q^ 2 /4)+(P^3/27)])]$^$(1/3)$
Since $ t=v$−$u$ and $t =x+a/3$, we find
$X=(P/3u)-u-(a/3) $
there are different valuess for $U$. so you will have different values for $X$
now in your equation
$a=-(a^2b+b^2c+c^2a)/abc$
$b=(ab^2+bc^2+ca^2)/abc$
$c=-1$
and if you put values for $a,b,c$ you get the roots are:
$a/c,c/b,c/a$

Guys, of course, I know your solutions. I just posted my problem to answer shobber's question. ;)


Exam if  a coffee mug and a doughnut are homeomorphic topological  spaces.

A picture (or [url=http://upload.wikimedia.org/wikipedia/commons/2/26/Mug_and_Torus_morph.gif]video[/url]) is worth a thousand words.

In your first post, at least one term is not so well-defined.

[quote="HilbertThm90"]We should be a little careful. That video seems to show they are homotopy equivalent.[/quote]
Looks pretty homeomorphism-y to me. I also have no idea how to write down the definition of a coffee cup, and just thought it was a motivation for handles.

[quote="Christiano89"]which of them?[/quote]
Think about it  :wink:

I might question what donut refers to, whether it's a standard torus, or a torus with two holes, or something entirely different.

Torus with two holes? I can't imagine sth like that. :maybe:

If there are two closed connected  surfaces with  only one hole, then exist a continuous transformation that makes them same. True or False? :maybe:

A [url=http://mathworld.wolfram.com/DoubleTorus.html]double torus[/url].

What are closed cohesive surfaces with one apenture?  Perhaps I don't know the terminology?

[quote="JRav"]A [url=http://mathworld.wolfram.com/DoubleTorus.html]double torus[/url]. [/quote]
Oh! I hadn't seen sth like that before.

[quote]What are closed cohesive surfaces with one apenture[/quote]
I am sorry very much. I edited my previous post. It was a fail. Now I believe that it is ok.

I guess that the right terminology  is:
[quote]If there are two closed connected surfaces with only one hole, then exist a continuous transformation that makes them same. True or False[/quote]


[quote="socrates"]is $3\sqrt{2}$ an exponent??

If so the power mean ineq may be helpful....[/quote]
Can You explain more clearly for me ???

We can factor $tg^{3 \sqrt{2}}$ out of $tg^{3 \sqrt{2}}{\dfrac{A}{2}}+ tg^{3 \sqrt{2}}{\dfrac{B}{2}} +tg^{3 \sqrt{2}}{\dfrac{C}{2}}$.

Please define your restrictions on your variables.  Without that, the problem cannot be solved.

This kind of optimization problem is a step too hard for the Intermediate forum.

[quote]We can factor $tg^{3\sqrt{2}}$[/quote]
Surely you can not! I suppose tg means tangens, and that is of course not associative.

I don't know if this is correct, but I think this is the solution:

By the AM-GM inequality, we know that:
$tg^{3 \sqrt{2}}{\dfrac{A}{2}}+tg^{3 \sqrt{2}}{\dfrac{B}{2}} +tg^{3 \sqrt{2}}{\dfrac{C}{2}}\geq 3\sqrt[3]{tg^{3 \sqrt{2}}{\dfrac{A}{2}}tg^{3 \sqrt{2}}{\dfrac{B}{2}}tg^{3 \sqrt{2}}{\dfrac{C}{2}}}$

With equality holds when $tg^{3 \sqrt{2}}{\dfrac{A}{2}} = tg^{3 \sqrt{2}}{\dfrac{B}{2}} = tg^{3 \sqrt{2}}{\dfrac{C}{2}}$

So the minimum of the expression is $3 tg^{3 \sqrt{2}}{30} = (\dfrac{3}{\sqrt{3}})^{3\sqrt{2}} = (3\sqrt{2})^{\sqrt{2}}$

Not sure how you came to that conclusion, not sure if its right, i think jensens is the way to go, if we assume its tangents and angles in a triangle...

my experience with Jensen is 0,0 , so could you please post your solution using Jensen?


Find the maximum value of real number $ k$ such that:

\[ \frac{(b\minus{}c)^2(b\plus{}c)}{a}\plus{}\frac{(c\minus{}a)^2(c\plus{}a)}{b}\plus{}\frac{(a\minus{}b)^2(a\plus{}b)}{c}\geq k(a^2\plus{}b^2\plus{}c^2\minus{}ab\minus{}bc\minus{}ca)\]
for all positive real numbers $ a,b,$ and $ c.$

[quote="usa"]Find the maximum value of real number $ k$ such that:
\[ \frac {(b \minus{} c)^2(b \plus{} c)}{a} \plus{} \frac {(c \minus{} a)^2(c \plus{} a)}{b} \plus{} \frac {(a \minus{} b)^2(a \plus{} b)}{c}\geq k(a^2 \plus{} b^2 \plus{} c^2 \minus{} ab \minus{} bc \minus{} ca)
\]
for all positive real numbers $ a,b,$ and $ c.$[/quote]
i think k max=2
after that,use S.O.S :D

Yes, $ b\equal{}c\equal{}1$ gives $ k\leq2.$


We know that $18$ cannot be expressed as $2$ consecutive postive numbers' sum, but it can be expressed in $2$ different ways as sum of n consecutive positive integers, $18=3+4+5+6$, and $\ 18=5+6+7$
Find the smallest positive integer such that cannot be expressed as $11$ consecutive integers' sum, but has $11$ different ways of being expressed as a sum of a bunch of positive consecutive integers.

Frank Morgan

This is a variation on problem AUS 3 proposed for the 1990 IMO.

http://www.kalva.demon.co.uk/short/soln/sh901.html


I'm just having trouble factoring this into the standard form.

Find the center, vertices, foci, and eccintricity of the following ellipse.
$4x^{2}-8x+y^{2}+4y-8=0$

uhh 2 things

1)use hide tags
2)ok you factored it, but you didn't answer the question...

Um was the second $4x$ supposed to be $4y$?  That seems to make the most sense, and that's how neelnal treated it...

Now, assuming that's true,

For an ellipse of the from $\frac{(x-h)^{2}}{a^{2}}+\frac{(y-k)^{2}}{b^{2}}=1$, The center is $(h,k)$, the major axis has length $2a$ with midpoint $a$, and the minor axis has length $2b$ with midpoint $b$.  I think that it was $\sqrt{a^{2}-b^{2}}$ for the distance from the foci to the center, with them placed on the major axis, and the eccentricity was $\frac{b}{a}$, but I'm not positive on those.

By the way, you can also have them of the form $\frac{(y-k)^{2}}{a^{2}}+\frac{(x-h)^{2}}{b^{2}}=1$, and then it has a vertical major axis rather than horizontal, but aside from that it's pretty much the same.  Hope that helps.

[hide]no, DanK, Carl meant that it's 4y, it's not a typo.  [/hide]


function
inequalities
calculus
derivative
logarithms
inequalities open

Can anyone explain the functions on the basis of whther they are concave and convex? Because in most of the inequalties that are solved by Jensen's inequality, I don't know how they judge that they are convex and concave functions ......Please can someone give a detalied explanation on how to undrdtand functions and their types PDF file or something, it would be really helpful.

The standard test is the second derivative test:  a twice-differentiable function is convex if $ f''(x) \ge 0$ for all $ x$ and concave if $ f''(x) \le 0$ for all $ x$.  So most "fast-growing functions" - power functions and exponentials - are convex, and their inverses - roots and logarithms - are concave.


Do you guys know any good jokes?  :rotfl:

sorry, i can't start,  :noo:  the only ones i know are retarded

CAN'T SOMEONE POST UP A GOOD JOKE?????????????????????????? :bomb:

I don't want to read this........ Ahhhhh, Not Looking, Not Looking!!!!!! Blah.

in the process of scrolling down, you probably caught a couple words. :lol:

i should have had a better subject for this

anyway, no one knows any jokes?

how'd u post twice in less than 10 minutes?

i double posted. sometimes that happens. i don't know why.
ok, as you can see, this topic didn't turn out the way i hoped.

Mine are gross.

So what is the difference between a truckload of bowling balls, and a truckload of babies?
[hide]
You can't unload the truckload of bowling balls with a pitch-fork.
[/hide]

yay! someone posted a joke(even though i didn't even get it).

i'll post one, but it's not good and it's against President Bush of USA.

[hide="click"]There were 5 president's on a plane; Lincoln, Washington, Jefferson, Clinton, and Bush. Lincoln threw a 5 dollar bill out the window and said, "yay! i made 1 person happy today!" Washington threw 2 5 dollar bills out the window and said "yay! i made 2 people happy today!" Jefferson threw 3 5 dollar bills and said "yay! i made 3 people happy today!" Clinton threw Bush out the window and said, " yay! i made the whole world happy today!"[/hide]

Dude, thats pretty funny actually. And hunter, that was gross..

Yeah, agreed.

[hide="Holy Water"]
There were three monks and a master. The master says to the three monks, "To drink the holy water tomorrow, you must first do something bad over the night." The next day, the first monk says that he robbed a bank, and the second monk said that he planted a bomb in his neighbors house. Suddenly the third monk starts laughing hysterically. They all ask, "Why are you laughing?" He says, "I urinated in the holy water..."
[/hide]
[hide="horse"]
This guy was traveling, and his horse got lost, so he bought another horse from a priest. The priest notes that this is a special horse: its starts moving when you say "thank god", and stops when you say "amen". He was traveling, when he saw a cliff that ends like this...
_
/

He sees it and screams "amen! amen!", and he stops at the very tip of the cliff. "Thank God!", he exclaims...    [/hide]

Holy Water Reaction:

$ H_{2}+O_{2}+Li(s) \rightarrow H_{2}O_{(holy)}$

[hide="Gross"]

Once there was a little boy in church. He had to go to the bathroom so he told his mother, ''Mommy, I have to wee-wee.''
The mother said, ''Son don't say wee-wee in church. Next time you have to wee-wee, say, 'whisper' because it is more polite.

The next Sunday, the little boy was sitting by his father this time, and once again, he had to go to the bathroom.

He told his father, ''Daddy I have to whisper.''

The father said, ''OK. Here, whisper in my ear.''

[u]$ \pi ss$ is replaced with wee-wee[/u][/hide]

Here's one I copied from somewhere else:

[hide="Pesky Squirrels"]There were three country churches in a small Texas town: Presbyterian, Methodist and Catholic.

Each church was overrun with pesky squirrels. One day, the Presbyterian church called a meeting to decide what to do about the squirrels.

After much prayer and consideration they determined that the squirrels were predestined to be there and they shouldn't interfere with God's divine will.

The Methodist group got together and decided that they were not in a position to harm any of God's creations.

So, they humanely trapped the squirrels and set them free a few miles outside of town. Three days later, the squirrels were back.

It was only the Catholics who were able to come up with the best and most effective solution:

They baptized the squirrels and registered them as members of the church. Now they see them only on Christmas and Easter.[/hide]

:rotfl:  :rotfl:

Man that was a harsh religious criticism but freaking hilarious!

Okay, here's another one i copied.

[hide=" :ninja: "]
That one about the Sox fan and Yanks fan

Late one foggy night a Yankees fan and a Red Sox fan collide head-on while driving across a bridge. Fortunately, both are unhurt, though their cars are pretty banged up.

"This is a sign," says the Yankees fan, "that we should put away our differences and live as friends instead of rivals."

"You're right," says the Red Sox fan. He pops open the trunk and takes out a bottle of bourbon. "Let's toast our newfound friendship."

The Yankees fan takes a big swig and hands back the bottle. "Your turn!"

"Nah," says the Sox fan, tossing the bottle into the river. "I think I'll just wait for the police to show up!"

Hehehe.[/hide]

OMG that one's priceless  :rotfl: I'll try and find some later.

[hide]

4 men board a plane.

When the plane is a maximum altitude, the men start up a lively chat.

man 1: What's everyone's occupation?
man 2: I'm a lawyer on vacation.
man 3: I'm a video game developer who's working for private people around the country.
man 4: I'm a plumber with an inheritance.
man 1: I'm George W. Bush.

men 2, 3, and 4 nod in consent and throw Bush out the window.[/hide]

Well,

Your momma is like a refrigerator, everyone sticks their meat inside.

Your momma is like Home Depot, 50 cents a screw.

Your momma is like a scooter, everyone is embarrassed to say they rid her.

Another one:

What is the difference between Ryan Seacrest and a vacuum?
[hide]
One sucks a lot and lives in the closet, and other is just a vacuum.
[/hide]

What is the difference between condoms and coffins
[hide]
They both hold something stiff but one is coming and the other is going.
[/hide]

Yo mama so bald that she took a shower and got brain-washed.

Yo mama so fat people jog around her for exercise

Yo mama so fat when she has wants someone to shake her hand, she has to give directions!

Yo mama so fat when she wears a yellow raincoat, people said "Taxi!"

Yo mama so stupid when your dad said it was chilly outside, she ran outside with a spoon

Yo mama so stupid that she tried to put M&M's in alphabetical order!

Yo mama so stupid when she read on her job application to not write below the dotted line she put "O.K."

Yo mama so clumsy she got tangled up in a cordless phone.

yo momma so stupid, she got stabbed in a shootout

i put them one by one

yo momma so fat, she fell down the grand canyon and got stuck.

yo momma so stupid, she thought DWI meant "Do without it."

yo momma so skinny, she fell through a straw when drinking her lemonade.

Hunter...priceless. :rotfl:  :rotfl:  :rotfl:

[hide="Democratic National Convention Schedule"]6:00 PM - Opening Flag Burning Ceremony
6:05 PM - Pledge of Allegiance to the U.N.
6:15 PM - Secular Prayers by Rev. Jesse Jackson and Rev. Al Sharpton
6:30 PM - Antiwar Concert by Barbara Streisand
6:40 PM - Ted Kennedy Proposes a Toast
7:00 PM - Tribute to France
7:10 PM - Collect Offerings for al-Zawahri Defense Fund
7:25 PM - Tribute to Germany
7:45 PM - Antiwar Rally (Moderated by Michael Moore)
8:25 PM - Ted Kennedy Proposes a Toast
8:30 PM - Terrorist Appeasement Workshop
9:00 PM - Roundtable Discussion of Taxes: "Calling for Higher Taxes on Others While You Pay None"
9:15 PM - Bill & Hillary Clinton Host a Seminar on "The Successful Selling of White House & Air Force One Mementos on eBay"
9:20 PM - Gay Marriage Ceremony (Both Male and Female Couples)
9:30 PM - * Intermission * Special Guest Soloist Jane Fonda
10:00 PM - Posting the Iraqi Colors by Sean Penn and Tim Robbins
10:10 PM - Reenactment of Kerry's Fake Medal Toss
10:20 PM - Howard Dean Screamfest 'Yeeearrrrrrrg!'
10:30 PM - Seminar: "The Boy Scouts and Other Paramilitary threats to National Security"
10:40 PM - Ted Kennedy Proposes a Toast
10:45 PM - Abortion Demonstration (NARAL)
11:00 PM - Multiple Gay Marriage Ceremony (Threesomes, Mixed and Same-Sex)
11:15 PM - 'Maximizing Welfare' Workshop
11:30 PM - 'Free Saddam' Pep Rally
11:50 PM - Ted Kennedy Proposes a Toast
12:00 AM - Kerry-Edwards 2004 Sealed With A Kiss
12:01 AM - Ted Kennedy Proposes a Toast
12:02 AM - Ted Kennedy Proposes a Toast[/hide]

[hide="Republican National Convention Schedule"]6:00 PM Opening Prayer, led by the Rev. Jerry Falwell
6:30 PM Pledge of Allegiance
6:35 PM Burning of Bill of Rights (excluding 2nd amendment)
6:45 PM Salute to the Coalition of the Willing
6:46 PM Seminar #1: Getting your kid a military deferment
7:30 PM First Presidential Beer Bong
7:35 PM Serve Freedom Fries
7:40 PM EPA Address #1: Mercury, it's what's for dinner
8:00 PM Vote on which country to invade next
8:10 PM Call EMTs to revive Rush Limbaugh
8:15 PM John Ashcroft Lecture: The Homos are after your children
8:30 PM Roundtable discussion on reproductive rights (MEN only)
8:50 PM Seminar #2: Corporations: the government of the future
9:00 PM Condi Rice sings "I Can't Help Lovin' Dat Man"
9:05 PM Second Presidential Beer Bong
9:10 PM EPA Address #2 Trees: the real cause of forest fires
9:30 PM Break for secret meetings
10:00 PM Second prayer, led by Cal Thomas
10:15 PM Lecture by Karl Rove: Doublespeak made easy
10:30 PM Rumsfeld demonstration: How to squint and talk macho
10:35 PM Bush demonstration of trademark deer-in-headlights stare
10:40 PM John Ashcroft demonstrates new mandatory Kevlar chastity belt
10:45 PM Clarence Thomas reads list of black Republicans
10:46 PM Third Presidential Beer Bong
10:50 PM Seminar #3: Education: a drain on our nation's economy
11:10 PM Hilary Clinton Pinata
11:20 PM Second John Ashcroft Lecture: Evolutionists: the dangerous new cult
11:30 PM Call EMTs to revive Rush Limbaugh again
11:35 PM Blame Clinton
11:40 PM Laura serves milk and cookies
11:50 PM Closing Prayer, led by Jesus Himself
12:00 AM Nomination of George W. Bush as Holy Supreme Planetary Overlord[/hide]

[hide][img]http://aurelienusa.free.fr/wordpress2/wp-content/uploads/cheque.jpg[/img][/hide]

Parallel lines never meet, unless you bend one or both of them.

HOW TO PROVE IT

proof by example:
The author gives only the case n = 2 and suggests that it contains most of the ideas of the general proof.

proof by intimidation:
"Trivial."

proof by vigorous handwaving:
Works well in a classroom or seminar setting.

proof by cumbersome notation:
Best done with access to at least four alphabets and special symbols.

proof by exhaustion:
An issue or two of a journal devoted to your proof is useful.

proof by omission:
"The reader may easily supply the details"
"The other 253 cases are analogous"
"..."

proof by obfuscation:
A long plotless sequence of true and/or meaningless syntactically related statements.

proof by wishful citation:
The author cites the negation, converse, or generalization of a theorem from the literature to support his claims.

proof by funding:
How could three different government agencies be wrong?

proof by eminent authority:
"I saw Karp in the elevator and he said it was probably NP-complete."

proof by personal communication:
"Eight-dimensional colored cycle stripping is NP-complete [Karp, personal communication]."

proof by reduction to the wrong problem:
"To see that infinite-dimensional colored cycle stripping is decidable, we reduce it to the halting problem."

proof by reference to inaccessible literature:
The author cites a simple corollary of a theorem to be found in a privately circulated memoir of the Slovenian Philological Society, 1883.

proof by importance:
A large body of useful consequences all follow from the proposition in question.

proof by accumulated evidence:
Long and diligent search has not revealed a counterexample.

proof by cosmology:
The negation of the proposition is unimaginable or meaningless. Popular for proofs of the existence of God.

proof by mutual reference:
In reference A, Theorem 5 is said to follow from Theorem 3 in reference B, which is shown to follow from Corollary 6.2 in reference C, which is an easy consequence of Theorem 5 in reference A.

proof by metaproof:
A method is given to construct the desired proof. The correctness of the method is proved by any of these techniques.

proof by picture:
A more convincing form of proof by example. Combines well with proof by omission.

proof by vehement assertion:
It is useful to have some kind of authority relation to the audience.

proof by ghost reference:
Nothing even remotely resembling the cited theorem appears in the reference given.

proof by forward reference:
Reference is usually to a forthcoming paper of the author, which is often not as forthcoming as at first.

proof by semantic shift:
Some of the standard but inconvenient definitions are changed for the statement of the result.

proof by appeal to intuition:
Cloud-shaped drawings frequently help here.[/hide]

[hide="Proud Lion"]

Once there was a lion. He went out hunting and saw a huge bull. After a long battle, he finally killed the bull and ate it. He was very proud and fulfilled, so he started roaring loudly, so everyone could hear him. A hunter heard him, came to where the lion was and shot him

Moral of the story: When you're full of bull, keep your mouth shut.

:-)
[/hide]
[hide="Oh God!"]

One day, a little boy asked his mom: "Mommy, does God live in our bathroom?" His mom said, "No, what gave you that idea?" The boy said, "Because every morning, when I'm in the bathroom, Daddy slams on the door and says: 'Oh God, are you still in there?!'
[/hide]
[hide="Headache"]

One day, a little girl had a stomach ache. She told her mother about it. Her mom said that it was only because she had nothing in it, and if she had something in it, it would feel better. So she ate, and it did feel better. Then she went to school. At school, here teacher was complaining of a headache. The little girl piped up, "That's only because you have nothing in it. If you had something in it, it will feel better"
[/hide]

[hide="At the Barber Shop"] One day, a man walks into a barber shop, wearing these huge headphones. He tells the barber to cut his hair, but not to take off the headphones, otherwise he'll die. The barber starts, but soon finds the headphones are getting in the way, so he takes them off. Within 5 minutes, the man slumps in his chair, dead. The barber puts on the headphones and he hears "breath in, breath out, breath in..."
[/hide]
[hide="Cannibal Tests"]
Once, there were 3 guys walking through a rainforest. While they were walking, they were ambushed and captured by a cannibal tribe. They were taken to the chief. The chief said, "Go into the forest and gather 10 fruits and bring them to me" So each guy went into the forest. The first guy came back with 10 apples. The chief then said, "Stuff them up your butt, without any expression, and you shall go free." The man got to the 4th apple before he burst out crying, so the cannibals killed and ate him. The second guy came back with 10 blueberries. The cannibal chief told him to do the same thing. The guy had 9 blueberries when he burst out laughing hysterically, so the cannibals killed and ate him. Later, the first two guys met up in heaven. The first guy said, "You were SO close! How come you started laughing?" The second guy replied, "I couldn't help it when I saw the other guy coming with pineapples!"
[/hide]

Your momma is so stupid, when she was filling out a form and she got to sex she put M, F and sometimes Wednesdays too.

You momma so fat, her blood type is Ragu.

Yo momma is so fat, she went on a seafood diet… Whenever she saw food she ate it.

Yo momma so stupid, she got locked inside a grocery store, and starved to death.

Yo momma is so fat, she stepped on a talking scale and it told her "One person at a time please"

yo momma is so stupid, she looked into a bowl of cherios and said "hey, i think i'm getting a message from the great above"

Yo momma is so hairy, when big foot saw her, he ran away screaming "BIGFOOT!"

Yo momma is so ugly, when she entered an ugly contest, they said "sorry, no professionals.

Yo momma is so stupid, she thinks jar jar comes after pickles pickles (thank you robot chicken)

Yo momma is so fat, when she jumped into the ocean, the whales start singing "WE ARE FAMILY... EVEN THOUGH YOUR BIGGER THAN ME!

Yo momma is so ugly, the boggie man looks for her under his bead every night.

[hide="might be offencive"]your momma's like a vacume cleaner, she sweeps, she sucks and gets laid in a closet.[/hide]

your mommas soo fat, when i was about to hit her while she was walking across the street, i swerved to the left and ran out of gas.

you momma is so fat, when she was in kindergarten, she sat by EVERY BODY

you momma is so old, she sat my Moses in first grade.

your momma is so stupid, she is why blonds are sterotypically stupid: all the smart ones cant mak eup for her stupidity.


function limit logarithms trigonometry ratio Support invariant

suppose that $ f: \mathbb{R} \to \mathbb{R}$ is a continuous function such that
$ f(x)\plus{}f(x\plus{}a)\plus{}f(x\plus{}b)\equal{}0$
for all $ x \in \mathbb{R}$ where $ 0<a<b$ are some fixed positive real numbers. Suppose moreover that $ \lim_{x \to \minus{}\infty} f(x)\equal{}0$. Does it follow that $ f$ is the null function ($ f(x)\equal{}0$ for all $ x \in \mathbb{R}$)

Not necessarily: pick $ a, b$ to be integers so that the equation $ t^a \plus{} t^b \plus{} 1 \equal{} 0$ has a root $ r \in {\bf R}_{ < \minus{} 1}$.  Define $ f(x) \equal{} \text{Re}(r^x)$ on the interval $ [0, b)$ and extend it via the given relation to the rest of the reals.  You should get a continuous solution that tends to 0 as $ x \to \minus{} \infty$.

$ r<0$, so you mean $ f(x) \equal{} Re(\exp(\ln(r) x ))$ where $ \ln$ is a branch of the logarithm ?  :maybe:

Yes.  Which I guess we can also write as $ f(x) \equal{} ( \minus{} r)^x\cos (\pi x)$.  And that function works as-is (i.e., none of this "define it on an interval and extend it" stuff is necessary).

Yes: let $ f(x) = ( - r)^x\cos \pi x$.  Then
\begin{align*}f(x) + f(x + a) + f(x + b) & = ( - r)^x\cos \pi x + ( - r)^{x + a} \cos \pi(x + a) + ( - r)^{x + b}\cos \pi(x + b) \\
& = ( - r)^x \left(\cos \pi x + ( - r)^a ( - 1)^a \cos \pi x + ( - r)^b ( - 1)^b \cos \pi x\right) \\
& = ( - r)^x \cos \pi x\left(1 + r^a + r^b\right) \\
& = 0. \end{align*}
To get from the first line to the second we use that $ \cos (x + n \pi) = ( - 1)^n \cos x$ for $ n \in \bf Z$.  Note that this relies quite heavily on the fact that $ a, b$ are integers, although we can extend it without trouble to the case where $ a, b$ are in the ratio $ p: q$, $ p, q \in {\bf Z}_{ > 0}$ and $ 1 + x^p + x^q$ has a root of the desired type.  In the case that $ 1 + x^p + x^q$ has a root in $ ( - 1, 0)$ we get a solution, but it goes to 0 as $ x \to \infty$ instead of $ x \to - \infty$.  If $ 1 + x^p + x^q$ doesn't have real roots (is this equivalent to $ q$ being even?) then we may still be able to do something using the complex roots (although in the case $ 1 + x + x^2$, where both roots are of absolute value 1, I was only able to come up with periodic solutions, not solutions tending to $ 0$ at either end; it wouldn't surprise me if this were a general result).  If $ a/b \not \in \bf Q$, then I've got nothing.

my idea was to show that $ f$ defines a tempered distribution (I think that $ f$ does not grow too fast) that satisfies $ [f] * (\delta_0 \plus{} \delta_{\minus{}a} \plus{} \delta_{\minus{}b}) \equal{} 0$. A fourier transform then shows that the support of $ F([T])$ is contained in those $ x \in \mathbb{R}$ that satisfy $ 1 \plus{} \exp(iax) \plus{} \exp(ibx)\equal{}0$ .. So the spectrum of $ f$ is very small and should not be hard to conclude. But I have some difficulties to do all this things properly  :(

There is quite a powerful theorem saying that if you have a shift-invariant closed (under  uniform convergence on compact sets) linear space of continuous functions, then it is just the closure of the linear span of exponential monomials $ x^ke^{\lambda x}$ ($ k\in\mathbb Z_ \plus{}$, $ \lambda\in\mathbb C$) contained in that space. Moreover, if the space is generated by a single function $ f$ and it is different from the entire $ C(\mathbb R)$, then all the exponential monomials contained in the space can be obtained by convolving the function $ f$ with certain compactly supported measures. Unfortunately, the proof is a bit too long to post it here though I'll try to see if I can extract the part needed just for solving this problem and post it :)

Applying this theorem to the shift-invariant space generated by $ f$, we conclude that if there exists any solution to the problem, then there exists an exponential monomial solving the problem as well. This is the Fourier analysis needed here (no, alekk, the functions solving the problem are not tempered distributions: the typical growth is exponential here). The rest is obvious: either $ 1 \plus{} e^{az} \plus{} e^{bz}$ has a complex root $ z$ with positive real part, in which case $ 2\Re e^{zx} \equal{} e^{zx} \plus{} e^{ \bar zx}$ is a counterexample or there is no such root, in which case the statement is true. It leaves open the question of how to "explicitly" describe all the pairs of reals for which the first possibility takes place but it is a different story.

thank  you :) I was just wondering what is the name of this theorem (or where I can find a proof on the net ?)

The theorem is due to L. Schwartz. If you want to try to search the relevant material online, the key words are "mean-periodic functions" and "spectral analysis/synthesis". If you manage to find some internet site containing a complete proof, post the url here (I spent about 20 minutes searching the web yesterday in the evening and found nothing :().

Now, as promised, the part of the proof that allows to solve the particular problem you posted. The reasoning below is an adapted version of the first step in Carleman's proof of the Schwartz theorem.

Suppose that $ f: \mathbb R\to\mathbb C$ satisfies the convolution equation $ f*\mu \equal{} 0$ where $ \mu$ is a finite linear combination of delta-measures. Then it is easy to see that $ f$ can grow at most exponentially in both directions. Split $ f \equal{} g \plus{} h$ where $ g$ is the part of $ f$ supported by $ [0, \plus{} \infty)$ and $ h$ is the part of $ f$ supported by $ ( \minus{} \infty,0]$. Define $ F(z)$ as $ \int g(x)e^{ \minus{} zx}\,dx$ when $ \Re z$ is large positive and as $ \minus{} \int h(x)e^{ \minus{} zx}\,dx$ when $ \Re z$ is large negative. For a compactly supported measure $ \nu$, let $ L\nu(z) \equal{} \int e^{ \minus{} zx}\,d\nu(x)$.

Now observe that, for every compactly supported measure $ \mu$  such that $ f*\mu \equal{} 0$, we have $ g*\mu \equal{} \minus{} h*\mu$ and this expression gives a compactly supported function. Considering this function as a density of some measure $ \nu$ and using the usual properties of convolution (basically we are talking just about the Laplace transforms here), we get
\[ F(z) \equal{} \frac {L\nu(z)}{L\mu(z)}\qquad(1)
\]
everywhere where both sides make sense, i.e., when $ |\Re z|$ is large. Thus, $ F(z)$ can be extended to a meromorphic function on the entire that can be represented as a ratio of 2 entire functions of exponential type. Also $ F\to 0$ as $ |\Re z|\to\infty$. The only possible poles of $ F$ are the zeroes of $ L\mu$.

Now comes the main trick. Assume that $ L\mu(\lambda) \equal{} 0$. Then we can consider the compactly supported measure $ \mu_\lambda$ whose density is $ e^{\lambda x}\int_{ \minus{} \infty}^x e^{ \minus{} \lambda t}\,d\mu(t)$. It is not hard to check that $ \mu_\lambda' \minus{} \lambda\mu_\lambda \equal{} \mu$  (in any reasonable sense) and $ L\mu_\lambda(z) \equal{} \frac {L\mu(z)}{z \minus{} \lambda}$ (or minus that :)). Now consider $ f*\mu_\lambda$. According to the first observation it should be a multiple of $ e^{\lambda x}$ (recall that $ f*\mu \equal{} 0$). If the coefficient is non-zero, we've got a purely exponential function satisfying the same convolution equation as $ f$ and decaying to $ 0$ at $ \minus{} \infty$. Assume that the coefficient is $ 0$. Then we can use $ \mu_\lambda$ instead of $ \mu$ in (1), which allows us to conclude that actually there is no pole at $ \lambda$ (if the zero was multiple, we'll have to use this trick several times, of course).

Thus, if we do not have a purely exponential solutions, we can conclude that $ F$ has no poles at all. Now comes a small bit of complex analysis. Since $ F$ is a ratio of two entire functions of exponential type, it cannot grow fast: it is of order $ 1$ at most. It is also bounded outside a strip of finite width around the imaginary axis. Hence, by Phragmen-Lindelof, it is bounded on the entire complex plane. Since it tends to $ 0$ as $ \Re z\to\infty$, it must be identically $ 0$. Thus, the Laplace transforms of $ g$ and $ h$ are identicaly $ 0$, whence $ f \equal{} 0$.


algebra
polynomial
Galois Theory
superior algebra
superior algebra theorems

I would like to understand the Galois groups. I know the definition: The Galois group of a polynomial $ f$ with respect to the coefficient field $ F$ is defined as the group of automorphisms of the splitting field of $ f$ that leaves $ F$ fixed. Now Maple can find Galois groups for irreducible polynomials over $ \mathbb{Q}$. But how can I find those if I have no computer available? And if $ f(x)\equal{}g_1(x)^{n_1}\cdots g_m(x)^{n_m}$ is the prime composition of $ f$ and $ \operatorname{Gal}(g_i)\equal{}G_i$, what can I say about $ \operatorname{Gal}(f)$ over $ \mathbb{Q}$?

do you know german? :-)
http://matheplanet.com/matheplanet/nuke/html/article.php?sid=1107

concerning the galois group of a polynomial which is not irreducible: we may assume that the polynomial is squarefree (ignore all exponents $ \geq 2$, because this doesn't change the galois group). then the splitting field is the compositum of the splitting fields of the irreducible factors. then you can use (if neccessary, inductively) the canonical isomorphism

$ Gal(E*E' / K) \cong Gal(E/K) \times_{Gal(E \cap E' / K)} Gal(E' / K)$.

[quote="-oo-"]do you know german? :-)[/quote]
Ich kan nur ein bischen Deutsch lesen aber vielen Danke für die WWW-Seite! :)


find all function on $ R$ such that :
$ f(xf(y))\equal{}f(xy)\plus{}x$ for all $ x$ and $ y$ in $ R$

Put in $ y \equal{} 0$. This gives $ f(xf(0)) \equal{} f(0) \plus{} x$ (*). Then $ x \equal{} \minus{}f(0)$ in (*) gives $ f(\minus{}f(0)^2) \equal{} 0$. Put $ x \equal{} 1, y \equal{} \minus{}f(0)^2$ into the original equation to get $ f(0) \equal{} 1$. Then put that in (*) to get $ f(x) \equal{} x \plus{} 1$.


solve: $ 2\sin 2x\minus{}3\sqrt{2}\sin x\plus{}\sqrt{2}\cos x\minus{}5\equal{}0$

There doesn't appear to be any easy way to solve this.


Express as a common fraction: $ \frac{0.1\overline6}{0.\overline4}$

[quote="GameBot"]Express as a common fraction: $ \frac {0.1\overline6}{0.\overline4}$[/quote]

$ 0.1\overline6 \equal{} \frac{1}{6}$ and $ 0.\overline4 \equal{} \frac{4}{9}$, so the answer is $ \frac{1}{6} \div \frac{4}{9} \equal{} \frac{9}{24} \equal{} \frac{3}{8}$.


For what value of $ n$ does $ 4(4^2 \plus{} 4^2\plus{} 4^2 \plus{} 4^2) \equal{} 2^n$?


MATHCOUNTS
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AMC 8

Since we can't reveal actual scores...talk about your PLACINGS at chapter.

I got 3rd individual, 2nd countdown, and 6th team.. :(
Darn, we epicly phailed
I was the only one doing the team questions
while the rest of my team just sang, slept, or broke pencils
I'm so MAD AT MY TEAM
cuz they didn't try their best
Fail. :(

[quote="Stargirl240"]Ours isn't until the 21st >.< My team most likely will be somewhat like yours...[/quote]
*gasp* STARGIRL IS BACK!!!!!!!!

hmm is your chapter getting stronger?
for mine, we are
it's like the 1st three finishers are tied
then tiebreakers brought me down :(

Megan!
I'm disappointed in you.  3rd in your chapter??? What are you thinking! >:O
And lol at your team...you should move to kennedy and guarantee us a 1st place in state this year. =P

[quote="suma_milli"]Guess what?  We had our school round [i]yesterday[/i].  My team's not that good but at least they're motivated.  I'm pretty sure I'll have to do most of the team round problems though.  They're good students and they always want to know how I got the answer so I have to explain really quickly before moving on.  This kind of makes us go slower but sometimes it's good because they catch any computational errors I make.

Yay guassy! ;D[/quote]
how did u do?

apparently, my chapter had three ties for first place, i was in there but lost to tiebreakers :(

Well, if you want, here's my full story: (copied from my blog)

-Last night, I went to bed pretty late and couldn't sleep cuz my mom's cell phone kept ringing. Stupid her. She shoulda turned it off and let me sleep.

-Sorry to tell you (pythag) but I crammed last night on AoPS and USAMTS.

-Two of my fingers were bleeding last night, and they still are right now. I got two paper cuts...Yuck.

-I haven't touched a single Mathcounts test in like 6 months already. But I still managed to get a high score at chapter and tied numerically with 1st and 2nd placers, but tiebreakers threw me off.

-My teammates epicly failed the worst. They didn't even try to do their best. They made random guesses on most of the problems.

-And they refused to work with me on the team round. So I had to do them alone while they just chatted, kept on singing, snored pretty loudly, and even snapped pencils.

-During the Countdown Round, some kids kept sneering right behind me whenever I got a correct answer, and they're like wtf how did u get teh answer so fast?????

-But at least I received the smiles from ppl at the end of the day...And you know what's funny? On my way home, while I was getting into my car with my mom, kids were banging on my car door screaming, "YAY GAUSSY! YAY GAUSSY!" in which Gaussy wwas just replaced with my real name, which I will not reveal unless I make Nationals.

-Last, but not least, our team phailed. And I have to go to state alone. The coach isn't going with me. And neither are the kids. Fail.

Aww... that doesn't sound very pleasant. I haven't been gone, you know =P *Sigh* just did last year's state >.< it's not that difficult anymore. I realized how stupid I was lol

I got first place in our school round, one of my friends made the team as an individual.

Yesterday I was practicing past state sprint questions untimed and I got everything except one.  So it's really the time that brings me down..hmmm..

Well, then speed is probably the most important thing you should work on.
Go on FTW or something.
Hopefully you should be able to make states and so we could all meet up (Gaussy, suma, stargirl, RussianRocket, avec_une_h's sister, and jgjgjmath)

[quote="suma_milli"]Yeah that's been my focus for a while now.[/quote]
 :D  :D  :D
Btw do you know any strong kids in your chapter who you think may be on the tough "Mishmosh For Nats?" list?

Here's the list so far (collaborated from yesterday, AoPS users, and the past)

THE MISHMOSH FOR NATS?:

James Tao
Patrick Liscio
Megan Chen
Mr. Emo Child whose name I don't remember (he's from my chapter)
Ravi Jagadeesan
Xiaoyu Li
avec_une_h's sister
Nikita Bondarenko
 :D

Oooo and guess who mr. green is? [hide]suma_milli![/hide]

lol yes I'll be the representative for :D at states :P

I don't think anybody from my chapter is that good.

how about those pplz who did well on amc 8? (got a perfect score)
weren't there some ppl who got perfects in the champaign area?

[quote="suma_milli"]Oh my god.  I looked for the amc 8 perfect scores and I saw a picture of Henry Carlton!  I knew him and I can't believe it.  Blarg I better get into Uni this year (which I probably will since last year was just a bleh year since I moved in the middle and could participate in barely anything, plus I didn't know about competitive math then.)

I barely knew Henry in 7th grade.  I didn't have any classes with him but I had scholastic bowl with him.  Everyone in our grade used to make fun of him (well not really he was just known as as "uncool").  He sat at the "geekiest" table, which he shouldn't have.  That was really just the table of "geeky" DUMB people.  All the smart people at my school are averagely popular and some popular.  I guess he was intimidated.  I thought he was really cool but just felt intimidated.  Still I barely talked to him in scholastic bowl too.  I never really got a chance to get to know him.  One time I tried to give him an air five but he said, "What's an air five?".  We told him but then he just walked away I think he was offended :( We laughed about how he knows so much math but didn't know what an air five was. Yeah.. I'll see him at mathcounts chapter I suppose,  well I better show him how much I've improved since 7th grade. At the end of the year he stopped coming to our school,  I don't know what happened because he should have kept coming til the year was over but my friend who did have classes said that he stopped coming in 4th quarter or something.  I never found out what happened..

I feel so bad after typing this post.  I mean I did have no classes with him so of course I never got to know him but gah I wish I had now.

A random cross country article he's in :
http://www.uni.uiuc.edu/og/sports/2008/10/subfreshman-wins-iesa-cross-country

It looks like he's doing good at Uni.[/quote]
Awww. :(  :(  :(
I wonder...is he as smart as you are?

btw
if you are to use ftw, please do not make it one of your hobbies.
and if you must, please do not make it your only hobby.
or i will have to sign on randomly and point out that you have no life.
yes, i've done it before. i'll confess.
BUT IT WASN'T ON MY ACCOUNT  :ninja:

Yes I'm sure he's as smart as I am, maybe smarter?

Aryth that was really random.. :P  :ninja:

[quote="xpmath"]Was I one of the people who was told that?

Because that is quite ironic.[/quote]

Maybe it was, or maybe it was diophantient.
And what is the irony in that?

Oh darn that is quite tricky. I wonder which one of you it was.

One reason is perhaps because I don't go on FTW at all compared to some others, just a lot recently because I just had chapter. However, perhaps that is not it... :wink:

Since I'm too lazy to go post in that other topic, good luck on the AMC A.

1st individual & CD(ish, we have weird CD...), 2nd team

how is your cd round weird?
but gj anyway and good luck at state
see ya there :D

Thanks, and same. Um... it's a little "different". Everyone is given two notecards to begin with, then they show the problem on the projector. You write your answer on the notecard, and you raise it as fast as you can so the spotter peoples can see. The first person to get it right gets a spot in the "Sweet Sixteen". Once all the spots are filled in the sixteen, the sixteen are shown a problem and they buzz in after the reader person is done reading the question and say their answer. Whoever gets it right first goes into the "Elite Eight". And so on for the "Final Four". Those four are the winners...

That is really strange. And not a Countdown round.  :huh:

I guess not... wish we actually had CD @ Chapter

Wait, so EVERYBODY participates in that "weird CD"? :maybe:

Yup... well, until the Sweet Sixteen is filled

:o  :D  :rotfl:

wow...really strange.

At my chapter, countdown is played with top 10 written finalists all against each other. We get 1 minute to answer each question, and whoever answers the most out of 20 problems wins.

got 3rd team and 3rd individual (beat 3 avery coonley ppl)
I am of the unknown  :ninja:

so you're from the avery coonley chapter? what chapter is that?

[quote="gauss1181"]also, stargirl: I forgot to ask you this but how small is your chapter?
and smileymonster ( :D ): how small is your chapter?[/quote]
Idk... it's not that small, I don't think. A normal sized chapter (whatever that is), I'd say =P

Err... uh... um... however many are in a normal chapter! I really don't know.

My chapter had around like 32-33 schools.

32-33? WOW.  I think ours is just a regular chapter or maybe it's small after yours.  Around 15-20 teams?

Mine's pretty large. In fact, it's probably one of the largest IL chapters.
I guess that's why the competition/state cutoff is so high.


f(x) and g(x) satisfies (1) and (2) for all real number x.

(1) $ f(x) \plus{} g(x) \equal{}(x^2\minus{}x\plus{}2)\sqrt{x^2\plus{}9x\plus{}134}$

(2) $ f(\minus{}x)\equal{}f(x), g(\minus{}x)\equal{}\minus{}g(x)$

find g(10)

Thanks.

replace $ x$ by $ \minus{}x$ to get $ f(x)\minus{}g(x)$ and after $ 2g(10)\equal{}(f(10)\plus{}g(10))\minus{}(f(10)\minus{}g(10))$.

HELLO , we can ansuer ur  probleme with this way
 we have  ;f(x)=f(-x) and g(-x)=-g(x) so g(x)=-g(-x)

and whene we calculate the f(x)+g(x) we find this result 18*92 so
f(10)+g(10)=18*92, this equvently for this  f(-10)-g(-10)=18*92 , i add in the part of this equation g(10), i find

f(-10)-g(-10)+g(10)=18*92+g(10) it equvelently for

f(-10)=18*92+g(10), because i said g(x)=-g(-x); and also i add again f(10) in too part, i find

f(-10)+f(10)=18*92+g(10)+f(10), it equivelently for this equation
2f(10)=18*92+18*92  so f(10)=18*92, so now i can find g(10)

we are  f(10)+g(10)=18*92 so  g(10)=0  and thanks, my name is zeraoulia rafik, algeria.


that should be equal to (1+1)^2003+(1+i)^2003+(1-1)^2003+(1-i)^2003. writing the second and last terms in polar form and using de moivre, I get  :rt2: e^( :pi: i/4)^2003+:rt2: e^( -:pi: i/4)^2003. This becomes 2^1001(:rt2:e^(3:pi:/4)+:rt2:e^(3:pi:/4)), or -2^1002. Adding on that 2^2003 from before, I get 2^2003-2^1002.

I believe it should be a quarter of that.

er ya sorry. so that would make it 2^2001-2^1000

[quote]that should be equal to (1+1)^2003+(1+i)^2003+(1-1)^2003+(1-i)^2003[/quote]

Could you explain how you got this please?

expand it and you should see how the terms cancel...

Hmm, I did, but the sum seemed to be 4*((2003 choose 0) + (2003 choose 2) + (2003 choose 4) + (2003 choose 6) + ... + (2003 choose 2000)), not 4*(2003 choose 0) + (2003 choose 4) + (2003 choose 8) + ... + (2003 choose 2000). For example, how do the (2003 choose 2) elements from the four expansions cancel?

Right. When you see something that sort of looks like the binomial theorem but with some terms missing, see if you can find a clever way of cancelling them out. Here you use 1+i and 1-i.

erm. You could always use the roots of unity, no? For example, if we had
C(n,0)+C(n,5)+C(n,10)+.....
could we write that as (1+1)^n+(1+fifth root of unity)^n+(1+another fifth root of unity)^n+...

Well that was essentially what you did for this problem, no? Except with the fourth roots of unity.

Note that we didn't quite use the fourth roots of unity. We used 1:pm:i. Those are :rt2: * eighth roots of unity.

how did mystic get (1+1)^2003+(1+i)^2003+(1-1)^2003+(1-i)^2003?  I understand that when you expand it it works, but how do you even think of that in hte first place?

I had seen this technique before the contest. I was dismayed that on the contest, they gave the hint to use 1:pm:i, since that gave most of it away.

no, Simon, I think we used the fourth roots of unity. We used:
(1+a fourth root of unity)^n+(1+another fourth root of unity)^n+two more terms like that.


How does one become a moderator on the forums??? im just wondering

um another, i just deleted my only blog, how can i make a new one, it doesnt have the option of create a new blog!!!

This is where you scroll down near the bottom of the page and look for the Questions, Suggestions, and Announcements or the Support forum.  Or use the search function...

Ah thanks!!! Sorry for putting this in the Mathcounts forum!


function
floor function
ceiling function

When a natural number is multiplied by each other, the result is a square number (as you probably know). How many positve square whole numbers are there between 1,000 and 2,000

[hide=" Question hint:"]What is the smallest square greater than 1,000? Largest square less than 2,000? Express both squares in the exponential form N^2[/hide]

[hide]The smallest square over 1000 is $32^2=1024$. The largest number under 2000 is $44^2=1936$. 44-32+1=13.[/hide]

[quote="kokosai"]How many square numbers are there between two positive reals $x$ and $y$?  Can you come up with a general formula?[hide="hint"]Look up the floor and ceiling functions if you don't know them already.[/hide][/quote]

[hide]$\lfloor \sqrt{y}\rfloor - \lceil \sqrt{x} \rceil +1$?[/hide]


algebra
polynomial
topology
linear algebra
matrix
linear algebra unsolved

Consider the set of matrices $A$ which are complex and verify $A^k=I$ for a fixed $k$. Find the connected components of this set. Now, some additional interesting questions: 1) what if we replace complex with real?
  2) The same problem but with $P$ a polynomial instead of $X^n-1$
  3) What are the connected components of the set of complex (real) matrices $A$ for which there exists $k$ such that $A^k=I$?

I can't help you I am afraid,  but I  am interested.  When you say connected, what kind of topology do you use?

I would assume the topology to be that given by $M_n(\mathbb b{C})\subset\mathbb{C}^{n^2}.$

:oops: And which is that?  I mean , probably the topology is induced by a matrix, but as I recall there are different approaches, and not all norms are induced by inner products?

Fredbel, the topology is the one inherited from $C^{n^2}$, the topology given by any norm.

I can only solve parts of it, like, for example, the complex cases of questions (1) and (3):

In both situations, the answer is the same: the connected components are the sets of matrices with the respective property which have the same characteristic polynomial.

To prove the necessity of the fact that the matrices in the same connected components have the same characteristic polynomial, consider the continuous map $f$ defined on the set $\mathcal S$ of matrices we're working with by $f(A)=(\mbox{tr}(A),\ldots,\mbox{tr}(A^n))$. The range of this map is countable (in problem (1) it's even finite), so it maps every connected component of $\mathcal S$ onto a single point in this range, i.e. all elements of $\mathcal S$ in a connected component have the same characteristic polynomial, since the characteristic polynomial of a matrix $M$ is uniquely determined by $f(M)$.

Conversely, take two matrices $A,B\in\mathcal S$ with the same characteristic polynomial $p$. They are diagonalizable, and can be written in the form $A=U^{-1}DU,\ B=V^{-1}DV$ for some invertible matrices $U,V$, where $D$ is a diagonal matrix whose diagonal entries are the roots of $p$. $\mbox{Gl}_n(\mathbb C)$ is path connected, so we can find a path from $U$ to $V$ which gives rise to a path from $U^{-1}$ to $V^{-1}$. Together, these paths define a path in an obvious manner from $A$ to $B$.


I was just going through the article on Probability and was working the practise probs....

1) Two mathematicians take a morning coffee break each day. They arrive at the cafeteria independently, at random times between 9 AM and 10 AM, and stay for exactly m minutes. Find
the value of m if the probability that either one arrives while the other is in the cafeteria is 40%.

I calculated m = 24 mins.... am i right?

2) An ’unfair’ coin has a 2/3 probability of turning up heads. If this coin is tossed 50 times,what is the probability that the total number of heads is even?
How do i do this one???

does anyone know where i can find solutions for the practise probs???

I can't think of a way to do either. :huh:

On number one, consider graphing it by considering geometric probability.  This is the reverse of the method normally used when asked for probability, so I'm not sure if it would be easy.  Technically it should work though...

I can't think of a way to do number 2 without a summation that I can't simplify.

2. let h be the probability of head, t be the probability of tail.
investigate the binomial expression:
$(s+t)^{50}$

the probability of even heads is just the sum of even terms in the above polynomial.

now notice:
$(s-t)^{50}$
alternates, let the probability of odd head be e, observer that:
$(s-t)^{50}+2e=(s+t)^{50}=1$
$e=\frac{1-(s-t)^{50}}{2}$

probability = 1-e

edit: forgot to subtract it from one.

[quote="mysmartmouth"]On number one, consider graphing it by considering geometric probability.[/quote]
Why geometric probability?

If the arrival times A and B are independent uniformly-distributed  between 9AM and 10AM and $p=P(|A-B|<m)$ (m in minutes), we can get (I can post it if you want):
$1-p=(\frac{60-m}{60})^{2}$,
$m=60(1-\sqrt{1-p})$.
If $p=40\%=0.4$, then $m=13.5242$.
If $p=0\%=0$, then $m=0$.
If $p=100\%=1$, then $m=60$.

[quote="Tim_Lou"]2. let h be the probability of head, t be the probability of tail.
investigate the binomial expression:
$(s+t)^{50}$

the probability of even heads is just the sum of even terms in the above polynomial.

now notice:
$(s-t)^{50}$
alternates, let the probability of even head be e, observer that:
$(s-t)^{50}+2e=(s+t)^{50}=1$
$e=\frac{1-(s-t)^{50}}{2}$[/quote]

In the expansion of $(s-t)^{50}$ ($s$ refers to probability of heads), the terms that are negative are where the number of heads is odd, aren't they? In which case you can instead add twice the probability of an odd number of heads, solve for that, and subtract that from $1$ to get the probability of even heads.

oh yeah... sry about that, it was 1 am what i posted that...

yeah, just substract the answer from one.

How to solve #1 geometrically and any simpler methods without binomial expansion for #2?

The first one was Problem 8 from the 1998 AIME exam

There's a solution (with graph) here:
[url]http://www.kalva.demon.co.uk/aime/aime98.html[/url]

And probably one in the Contests section of this site.

For the second question, I don't think there's an easier solution that doesn't use the Binomial Theorem.

[hide="2"]The generating function for the number of heads after one flip of the coin is $2x+1$ (The exponent of $x$ is the number of heads, and we get heads twice as often as tails).  Then the generating function for the number of heads after fifty flips of the coin is $P(x)=(2x+1)^{50}$.

We want to add up the even coefficients of $P(x)$.  To do this, we use the second roots of unity, $1$ and $-1$.  Consider $P(1)+P(-1)$.  For each term in the expansion of $(2x+1)^{50}$ of even degree $c_{2n}x^{2n}$, $P(1)+P(-1)$ gives $c_{2n}1^{2m}+c_{2n}(-1)^{2n}=2c_{n}$.  For each term in the expansion of $(2x+1)^{50}$ of odd degree $c_{2n+1}x^{2n+1}$, $P(1)+P(-1)$ gives $c_{2n+1}1^{2n+1}+c_{2n+1}(-1)^{2n+1}=0$.  Therefore, $P(1)+P(-1)$ gives double the sum of the even coefficients of $P(x)$, so $\frac{P(1)+P(-1)}{2}$ gives the sum of the even coefficients of $P(x)$.

$P(1)$ gives the sum of all coefficients of $P(x)$.  Therefore, the probability we want is $\frac{\frac{P(1)+P(-1)}{2}}{P(1)}=\frac{P(1)+P(-1)}{2P(1)}=\frac{3^{50}+(-1)^{50}}{2\cdot 3^{50}}=\frac{3^{50}+1}{2\cdot 3^{50}}$.[/hide]

Extension:  What is the probability that the number of heads is a multiple of three?


AMC USA(J)MO USAMO calculus email induction AIME

And the first round of winner calls have come in, it seems!

It looks like the cutoff was 23 :!:

Also--I heard that Probability1.01 didn't make it  :o --they discovered crippling flaws in his solutions to 2,4, and a minor flaw in 6.  But if he still has a shot at TST, the BUBALA wishes him good luck!

I'm still eagerly awaiting potential MOP calls...

WOOOOOO i got a call for usamo winner  :D  :D
Not what I expected, quite...

Eh "he still has a shot"? He got a $28$ lol. Much more than just a shot.

Anyways, Haitao and I are both winners, it would seem. $23$ for me, one more than I predicted, as well as one more than the middle of my predicted range before the test, and exactly the cutoff. Last year, I got a $19$, which was one more than I predicted, and was also exactly the Blue MOP cutoff. Yay for two point trends!

So are they gonna care if we are particularly late to the first day of the ceremonies, due to the TCHS tourney being that weekend as well?

And congrats to you, K8! I think it was even less expected for me, since you solved 4 problems iirc.

[quote="JSteinhardt"]
And congrats to you, K8! I think it was even less expected for me, since you solved 4 problems iirc.[/quote]

My solution to #2 was a sketch of a solution, and probably received at most 3 points. (at least I acknowledged in the proof that a step was missing, so maybe they took pity).  Assuming 1 point partial on the rest, that's a maximum of 25 points.  The cutoff for winner was like 28 last year...

[quote="JSteinhardt"]Eh "he still has a shot"? He got a $28$ lol. Much more than just a shot.
[/quote]

That's great news!!

But where did you hear this??

It's different from what I'd heard...

whoa this is unexpected. I overestimated the cutoff by 4 pts and my predicted score range was 17-26. I don't know what my score is yet though.

zhai zhai got a 28? That's not exactly what I heard  :maybe:

Why dont we just wait until he posts what he got...

Congratulations to all of you :)

when will be individual scoring be posted? does anyone have an idea?

This is pretty interesting because there are 8 returning winners this year. We already know of 3 winners who were not winner, not HM last year. Quite impressive :!:

@jason - I think the scores were posted one day after the first winner calls last year. So possibly tomorrow.

Why is no one expecting a 2005 IMO Silver Medalist and 2006 IMO Gold Medalist to even have any shot at winning?  :maybe: (I'm talking about Phumpong Watanaprakornkul)

[quote="JSteinhardt"]It also sounds like Adam Hesterburg, Eric Larson, and Delong Meng are winners, though I'm not positive on the first two.

How do we get eight btw? I'm counting:

Confirmed returning:
Brian Lawrence
Sherry Gong
Alex Zhai (uh I guess there's dispute about this?)
Arnav Tripathy
Adam Hesterburg

Confirmed new:
Haitao Mao
Jacob Steinhardt
Delong Meng
Eric Larson
Krishanu Sankar

So that's $10$ right now. And I'm guessing we are expecting Matt McCutchen, Bohua Zhan, and Zarathustra Brady to be winners since they were last year. Hm that's already $13$...I doubt anyone is lying, so probably there will be more than $12$ winners.[/quote]

According to http://www.unl.edu/amc/e-exams/e8-usamo/e8-1-usamoarchive/2006-ua/06usamo-winners.shtml we have 8 2006 winners this year:

Brian Lawrence
Alex Zhai
Sherry Gong
Richard McCutchen
Arnav Tripathy
Yi Han
Yakov Kogan
Tedrick Leung

We also have Zeb Brady, an IMO gold medalist. And there are also 6 new ones (I am moving Hesterburg to this category).

@Ignite - I don't think anyone's ruled Yod out. We are noting new confirmed and past winners, not predicting winners.

[quote="JSteinhardt"]It also sounds like Adam Hesterburg, Eric Larson, and Delong Meng are winners, though I'm not positive on the first two.

How do we get eight btw? I'm counting:

Confirmed returning:
Brian Lawrence
Sherry Gong
Alex Zhai (uh I guess there's dispute about this?)
Arnav Tripathy
Adam Hesterburg

Confirmed new:
Haitao Mao
Jacob Steinhardt
Delong Meng
Eric Larson
Krishanu Sankar

So that's $10$ right now. And I'm guessing we are expecting Matt McCutchen, Bohua Zhan, and Zarathustra Brady to be winners since they were last year. Hm that's already $13$...I doubt anyone is lying, so probably there will be more than $12$ winners.[/quote]

Was Bohua Zhan really a winner last year it says that he was a honorable mention? Unless he was selected after the Canada/RSI people were eliminated but he had 6 people ahead of him. Also, are we ruling out the fact that it is possible for one of Matt, Bohua, and Zarathustra to have a bad day?

didn't Zeb say he got like... less than a 14?

and i'm confused is this discussion regarding winners or those attending black MOP

Possibly he decided to care more about CompSci than Math, based on his performance on USACO.

[quote]Does anyone know what happened to Zeb?[/quote]
All of these tests have noise in them; on all of them, people go up and down, have better days and worse days. You can't look at a list of what happened last year and regard it as a prediction.

In this particular case, Zeb did not have the highest USAMO in 2007 in his immediate neigborhood - and by immediate neighborhood, I'm talking about less than a 10-mile radius. That's nothing new - he also didn't have the highest USAMO in 2006 in that same neighborhood. And that certain other individual's average AIME score over the last three years is nearly 3 questions per year better than Zeb's. And yet I seldom see you guys mention his name.

joseph chu?

completely random guess...

i think people mention zeb a lot because he was on the IMO team last year and got a gold medal

Jeff Manning?
Never mind. Joseph Chu is probably right. Though I believe zeb lives closer to jeff.

Based on looking from the winner's lists from this year and last year, there is only one person who was a Winner or HM both years from CA:

Tedrick Leung.

However, I am not sure if this person lives anywhere near Zeb; I have no knowledge of cities in California.

Also, Tedrick Leung is listed under two separate high schools each year...

Um, Tedrick Leung?  Considering that he has been a USAMO Winner 2 years in a row now . . .

Besides, doesn't Joseph Chu live a couple hundred miles from Zeb?

EDIT: dasherm, I think Tedrick does live quite close to Zeb.

Joseph Chu = Bay Area
Tedrick Leung = Southern California

Decide for yourself, I don't know where Zeb lives.

Zeb lives in Winnetka, and Tedrick lives in Reseda, which, according to Wikipedia, border each other.

Dude, Tedrick Leung is indeed pretty awesome. He solved that funky Gulf of Mexico lighthouse problem. Except a lot of times he is not ready... (Teddyyyyyyyyyyyyyyyyyyyyyyyyyyyy)

[quote="Kent Merryfield"]Yes, you figured it out. Winnetka, Reseda, and (for that matter), North Hollowood are all the names of neighborhoods within the San Fernando Valley in Los Angeles. (Jeff Manning actually lives in Pasadena, which is a little further east, although as a homeschooled student, he's listed on the AMC website with a community college that's also in the San Fernando Valley.)

A [url=http://www.artofproblemsolving.com/Wiki/index.php/Math_Day_at_the_Beach]couple of[/url] [url=http://www.lacitycollege.edu/academic/departments/math/mathdept/contest/pastwinners.htm]links[/url] to local contest results over that last five years or show should help make my point.

Oh, and Joseph Chu lives in San Jose, which is over 300 miles away, and wasn't a USAMO winner or HM in 2006.

And we've had [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=142333]part of this conversation[/url] before. See my comment near the bottom of the first page, about the letter "T", and the various reactions that followed.[/quote]

Joseph Chu was the top HM in 2006.

Oops, I missed that. Sorry, Joseph! But it's not my part of the state.

lol joseph's little brother failed to make it to red mop

a shame


function
real analysis
real analysis unsolved

let $ D_f$ denote the set of discontinuities of f. does there exist a function f such that the measure $ |D_f|\equal{}0$ and $ D_f\cap I$ is uncountable for every interval I.

Yes. Consider a Cantor set. In each open interval, place a smaller copy of the Cantor set. Then repeat this procedure on the resulting set. Do this infinitely many times. The characteristic function of that set should have the desired property.

your function does not have a lot of continuity point ..


function
logarithms
integration
geometry
rectangle
calculus
calculus computations

Let $N=1+\frac12+\frac13+...+\frac{1}{n}$ where $n\in\Bbb{Z}$.
By considering the function $y=\frac{1}{x}$, show:

$N<\ln{n}+1$

$1+\frac{1}{2}+\cdots+\frac{1}{n}\le 1+\int_{1}^{n}\frac{dx}{x}= \ln{n}+1$.

If you don't understand this, draw the graph and the rectangles corresponding to the original summation.


hai, :rotfl:  :D  :)
this problems is very hard for me
plz anyone can solve for me

I think the problem must be in the number theory forum :wink:


Please don't double post. And as was said in the other thread, people on AoPS won't do your homework for you.

[quote="i_like_pie"]people on AoPS won't do your homework for you.[/quote]

But we can help. So give hints.

[hide="HINT"]
Factor the denominator. Then find a common denominator (simliar to finding common denominators in integer fractions).
[/hide]

Hints are:
You need to simplify and keep on breaking down the numbers.
Use the cube method, the type that sort of looks like a battleship phase.
Try to get to your common denominator.

Where did you get this problem from? Was it from MATHCOUNTS or homework or what?  :starwars:

Oh yeah, i_like_pie is correct, you don't help on HW on AOPS, but you can do problems and classes for the heck of it (lol).  :lol:


\[ \sqrt{(144-x^{2})^{2}}=144-x^{2} \]
\[[(144-x^{2})^{2}]^{\frac{1}{2}}=144-x^{2}\]
\[(144-x^{2})^{\cancel2\cdot\frac{1}{\cancel2}}144-x^{2}\]
\[144-x^{2}=144-x^{2}\]

Since both sides of the equation are exactly the same, we cannot do anything else to the equation.

Therefore, $ \sqrt{(144-x^{2})^{2}}=144-x^{2} $ has $\boxed{\text{infinite solutions}}$; in other words, $x$ can equal $\boxed{\text{anything}}$.


algebra
polynomial
limit
real analysis
real analysis unsolved

Let $ f(x)$ be a polynomial of degree $ n$ such that $ f(x) \geq 0$ for all $ x \geq 0$, Show that for all $ x \geq 0$

$ f(x) \plus{} f'(x) \plus{} \cdots \plus{} f^{(n)}(x) \geq 0$

Let $ g(x)\equal{}e^{\minus{}x}\left(f(x)\plus{}f'(x)\plus{}\cdot\plus{}f^{(n)}(x)\right).$

Then $ g'(x)\equal{}e^{\minus{}x}\left(\minus{}f(x)\plus{}f^{(n\plus{}1)}(x)\right)\equal{}\minus{}e^{\minus{}x}f(x)$ since $ f^{(n\plus{}1)}(x)\equal{}0.$

Therefore, $ g'(x)\le0$ for $ x\ge 0.$

Note also that $ \lim_{x\to\infty}g(x)\equal{}0,$ since we have a decreasing exponential multiplied by a polynomial.

For $ 0\le x<y,$ we have $ g(x)\minus{}g(y)\ge0$ because $ g$ is decreasing. Now take the limit of that expression as $ y\to\infty$ to get that $ g(x)\ge0.$

Hence $ e^xg(x)\equal{}f(x)\plus{}f'(x)\plus{}\cdots f^{(n)}(x)\ge 0.$

An other approach . Let $ g(x) = f(x) + f'(x) + ... + f^{n}(x)$ then $ f(x) = g(x) - g'(x)\geq 0 ,\forall x\geq 0$ . Let $ x_o$ is the minimum point of $ g(x)$ on $ [0, + \infty )$ then by Fermat's theorem $ g'(x_0) = 0$ ,by the condition $ g(x_0)\geq g'(x_0)\geq 0$ . Therefore $ g(x)\geq g(x_0)\0 ,\forall x\geq 0$

Nice solution, just keep in mind the case where there is no absolute minimum:
$ \lim_{x \rightarrow \infty} g(x) \equal{} \minus{} \infty$
This is easily taken care of by noting that the coefficient of the leading degree in $ f(x)$ must be positive, and so $ f(x)$ and $ g(x)$ must both tend toward positive infinity.


Draw the perpendicular line $ d$ from $ A$ to $ E_bE_c$, which is just the $ A$-altitude. Let $ K$ be the midpoint of $ AE_b$. The circle of diameter $ KE_c$ meets the line $ d$ at $ E_a$. Construct $ H$ be the orthocenter of triangle $ E_aE_bE_c$. Lastly, construct A,B,C.


AMC 12 2001.
16. A spider has one sock and one shoe for each of its eight legs. In how many different orders can the spider put on its socks and shoes, assuming that, on each leg, the sock must be put on before the shoe?

A. 8! B.(2^8)8! C. (8!)^2 D. 16! /(2^8) E. 16!

[hide="solution"]
Suppose that there are n legs, then there are n socks and n shoes.

Imagine that the shoes and socks can be put on in any order: then the number of orders will be (2n)!.

There is an inherent symmetry to the orders in that for any given order where sock_i is put on before shoe_i, there is an identical ordering where the two have swapped places (shoe_i is put on instead of sock_i and sock_i is put on instead of shoe_i).  Only one of those orderings is valid since socks must be put on before shoes, so 1/2 of the orderings will have sock_1 put on before shoe_1.  Since there are n sock-shoe pairs, we need to divide by 2^n.

Thus, the answer would be (2n)! / 2^n.

Your guess for (8!)^2 is incorrect because it presumes that ALL socks are put on before ANY shoes, when that is not necessarily the case; nothing in the problem specifies this additional restriction.
[/hide]

label each foot:$ a,b,c,d,e,f,g,h$
then problem is equivalent to arranging 2 of each of the letters. (e.g. $ aabbccddeeffgghh$)

You just consider the first time a letter appears to be the event of putting on the sock and the second time putting on shoe. (i think this is same general argument used above, but maybe a little shorter?)

[hide]There are $ 16!$ ways to put on shoes and socks in any manner. BUt for each foot, we must multiply by $ \frac{1}{2}$ since there is a 1/2 chance that socks are put on first.
D[/hide]


LaTeX
logarithms
function
analytic geometry
conics
parabola
absolute value

[quote="amcavoy"]You may want to learn $\LaTeX$ as it will make your posts much easier to read.  I'll do a couple for you; they are all fairly similar.

3.  $\log_{3}x=-1\implies 3^{-1}=x=\frac{1}{3}$

7.  $x=\pm 7i\implies x^{2}=\left(\pm 7i\right)^{2}=-49\implies x^{2}+49=0$

Note that in some of your other problems, the vertex $\left(h,k\right)$ can be found from $y-k=a\left(x-h\right)^{2}$.[/quote]

I did those math symbol pictures before, but I totally forgot how to do them now. haha  :wacko:

Thanks for helping! I did #3 right!  :lol:

5) clearly, $-|x|\leq 0 \forall x$, so the vertex will be in $x=0$, because it's the maximum of $f(x)$

(1) Are you asking for $-\sqrt{-6} \cdot 5i$? Remember, by definition, $\sqrt{-1}=i$. So: $-\sqrt{-6} \cdot 5i=-i \sqrt{6} \cdot 5i=-(i \cdot i)(5\sqrt{6})=5\sqrt{6}$
(2) The 9/2 in the equation stretches the parent function vertically. This is because each point on the curve will have a y-coordinate 9/2 times farther from the x-axis.
(4) Here are two formulas: $\log{(ab)}=\log{a}+\log{b}$ and $\log{(a^b)}=b\log{a}$ So, $\log{(2x^2y^3z)}=\log{2}+\log{x^2}+\log{y^3}+\log{z}=\log{2}+2\log{x}+3\log{y}+\log{z}$
(5) The vertex is $(0,0)$, and for any absolute value function $a|x-h|+k$, the vertex is $(h,k)$.
(8) The vertex is $(1,2)$, and for any parabola $a(x-h)^2+k$, the vertex is $(h,k)$.


vector
analytic geometry
integration
calculus
function
trigonometry
geometry

problem:  Verify that Stokes' Theorem is true for the given vector field F and surface S.

F (x, y, z) = yi + zj + xk,

S is the hemisphere x^2+ y^2 + z^2 = 1, y>=0, oriented in the direction of the positive y-axis.

See that key word "Verify"? Parametrize the curve and surface, and calculate the integrals.
I'd recommend using some sort of polar coordinates to parametrize.

[quote="mli7"]How exactly do you do this?  can someone show it to me.  thanks.[/quote]

You need to show (my signature is a more general case of Stokes theorem).

$\int_{C}\bold{F}\cdot d\bold{R}= \iint_{S}\mbox{curl}\bold{F}\cdot \bold{N}ds$

So, $\bold{F}= y\bold{i}+z\bold{j}+x\bold{k}$.

And $C$ is around the boundary, $x^{2}+y^{2}=1$

And $S$ is the upper surface of $x^{2}+y^{2}+z^{2}=1$

Find $\mbox{curl}\bold{F}= \left| \begin{array}{ccc}\bold{i}& \bold{j}& \bold{k}\\ \frac{\partial}{\partial x}& \frac{\partial}{\partial y}& \frac{\partial}{\partial z}\\ x & y & z \end{array}\right|$

To find $\bold{N}$ that is the outward normal vector.
$\bold{N}= \frac{\bold{T}'}{\parallel \bold{T}\parallel }$ where $\bold{T}= \frac{\bold{R}'}{\parallel \bold{R}\parallel }$.

And now just compute the line and surface integrals after doing all those long steps to verify Stokes theorem.

[quote]Find $\mbox{curl}\bold{F}= | \begin{array}{ccc}\bold{i}& \bold{j}& \bold{k}\\ \frac{\partial}{\partial x}& \frac{\partial}{\partial y}& \frac{\partial}{\partial z}\\ x & y & z \end{array}|$[/quote]
That should be

Find $\mbox{curl}\bold{F}= | \begin{array}{ccc}\bold{i}& \bold{j}& \bold{k}\\ \frac{\partial}{\partial x}& \frac{\partial}{\partial y}& \frac{\partial}{\partial z}\\ y & z & x \end{array}|$

By explicit calculation, $\nabla\times\bold{F}=-(\bold{i}+\bold{j}+\bold{k})$

Now, parameterize the surface with $x$ and $z$. We have that

$d\vec{S}=\vec{n}dS= (\frac{x\bold{i}}{\sqrt{1-x^{2}-z^{2}}}+\bold{j}+\frac{z\bold{k}}{\sqrt{1-x^{2}-z^{2}}}) \,dx\,dz.$

Take that dot product of that with $\nabla\times\bold{F}=-(\bold{i}+\bold{j}+\bold{k})$ to get that the surface integral over the hemisphere is

$\iint_{\{x^{2}+z^{2}<1\}}-\frac{x+z}{\sqrt{1-x^{2}-z^{2}}}-1 \,dx\,dz$

But the integrals of odd functions in $x$ and in $z$ are zero over this symmetric set. The integral evaluates to $-\pi.$

Now: can you parameterize the boundary circle $(x^{2}+z^{2}=1,y=0)$ so as to get the same result for the line integral? (Be careful of the orientation - it would be easy to get that wrong in this case.)

Following up on my "polar coordinates" recommendation:
Parametrize using spherical coordinates $x=\cos\theta\cos\phi, z=\sin\theta\sin\phi, y=\cos\phi$ for $0\le \theta\le 2\pi, 0\le \phi\le\frac{\pi}2$. The standard outward normal to the sphere is $x\bold{i}+y\bold{j}+z\bold{k}$, and area on the sphere is $dA=\sin\phi\,d\theta\,d\phi$.
Our integral becomes $\int_{S}\text{curl}(F)\cdot dS=\int_{0}^{\pi/2}\int_{0}^{2\pi}-(\bold{i}+\bold{j}+\bold{k})\cdot (\cos\theta\cos\phi\bold{i}+\cos\phi\bold{j}+\sin\theta\sin\phi\bold{k}) \sin\phi\,d\theta\,d\phi$
$=\int_{0}^{\pi/2}\int_{0}^{2\pi}-(\cos\theta\cos\phi\sin\phi+\cos\phi\sin\phi+\sin\theta\sin^{2}\phi) \,d\theta\,d\phi$
Integrate $d\theta$; the $\cos\theta$ and $\sin\theta$ terms integrate to zero, leaving
$2\pi\int_{0}^{\pi/2}-\cos\phi\sin\phi\,d\phi=-\pi$

If you do this, it's natural to take $(\cos\theta,0,\sin\theta)$ as your parametrization of the circle. Unfortunately, it's in the wrong direction; can you see why? This can be fixed in several natural ways.

Note that jmerry's spherical coordinates are slightly non-standard, making the $y$ axis the distinguished axis rather than the more typical $z$ axis. That was an intentional choice on his part, a matter of matching the coordinate system to the needs of the problem.

Although it's natural to use spherical coordinates here, in my calculus class I recommend using Cartesian coordinates the way Kent did it above. The reason is that I want the students to use the general formula $\vec F(\vec r(u,v))\cdot (\vec r_{u}\times \vec r_{v})$ rather than memorize its special case for the sphere. Unfortunately, the formula is not easy to implement in $\theta$ and $\phi$.