Calculate $ \sin{\frac {3\pi}{7}}\sin{\frac {6\pi}{7}}\sin{\frac {9\pi}{7}}$.

Using the identities $ \sin x \equal{} \sin (\pi\minus{}x)$ and $ \sin x \equal{} \minus{}\sin(\pi\plus{}x)$, the expression is equivalent to
\[ \minus{}\sin\frac{\pi}7\sin\frac{2\pi}7\sin\frac{3\pi}7,\]
which, by product-to-sum formulas, can be easily evaluated to $ \minus{}\frac{\sqrt 7}8$.

[quote="Yongyi781"]Using the identities $ \sin x = \sin (\pi - x)$ and $ \sin x = - \sin(\pi + x)$, the expression is equivalent to
\[ - \sin\frac {\pi}7\sin\frac {2\pi}7\sin\frac {3\pi}7,
\]
which, by product-to-sum formulas, can be easily evaluated to $ - \frac {\sqrt 7}8$.[/quote]

Thank you. You are a genius.  :)   But would you please show me the details? I was stuck somewhere.

$ \sin\frac {3\pi}7\sin\frac {6\pi}7\sin\frac {7\pi}7$
$ =- \sin\frac {\pi}7\sin\frac {2\pi}7\sin\frac {3\pi}7$
$ =\frac12(\cos{\frac{3\pi}7}-\cos{\frac{\pi}7)\sin{\frac{3\pi}7}}$
$ =\frac12\cos{\frac{3\pi}7}\sin{\frac{3\pi}7}-\frac12\cos{\frac{\pi}7}\sin{\frac{3\pi}7}$
$ =...$
I am lost... :(

Here's another way to do it.

First, by De Moivre's Theorem, $ \tan \frac{\pi}7\tan\frac{2\pi}7\cdots \tan\frac{6\pi}7\equal{}\minus{}7$, and since $ \tan x\equal{}\minus{}\tan(\pi/2\minus{}x)$, we have that $ \tan \frac{\pi}7\tan\frac{2\pi}7\tan\frac{3\pi}7\equal{}\sqrt 7$.

Next, we can prove that $ \cos\frac{\pi}7\cos\frac{2\pi}7\cos\frac{3\pi}7\equal{}\frac 18$. Take $ 8\sin\frac{\pi}7\sin\frac{2\pi}7\sin\frac{3\pi}7$ and multiply it by $ \cos\frac{\pi}7\cos\frac{2\pi}7\cos\frac{3\pi}7$. Using double-angle formula we get $ \sin\frac{2\pi}7\sin\frac{4\pi}7\sin\frac{6\pi}7\equal{}\sin\frac{\pi}7\sin\frac{2\pi}7\sin\frac{3\pi}7$, from which the result follows.

Therefore $ \sin\frac{\pi}7\sin\frac{2\pi}7\sin\frac{3\pi}7\equal{}\frac{\sqrt 7}8$.

[quote="mathwizarddude"]Eh, how did you get the product of tangents from De Moivre's theorem?[/quote]

Yeah...Please show us the details, again. :lol:

Can you think of what polynomials have the roots $ \sin{\left(\cdots\right)}$, $ \cos{\left(\cdots\right)}$?

I would say more, but I don't have time right now! Hopefully, you can understand this (or find another way) and prove a more generalized result (consider primes).  :)

to [b]Vworldv[/b], i am still waiting for that solution of yours

isnt de moivres theorem applicable only  to complex nos. ?how can we use that here? :!:  :?:

We can also use this $ \prod_{i \equal{} 1}^{m\minus{}1}\sin\frac{i\pi}{m}\equal{}\frac{m}{2^{m\minus{}1}}$   $ m\ge2$ to get the answer


linear algebra
matrix
algebra
polynomial
limit
IMC
college contests

Let $A$ be a real $4\times 2$ matrix and $B$ be a real $2\times 4$ matrix such that
\[ AB = \left(%
\begin{array}{cccc}
  1 & 0 & -1 & 0 \\
  0 & 1 & 0 & -1 \\
  -1 & 0 & 1 & 0 \\
  0 & -1 & 0 & 1 \\
\end{array}%
\right). \]
Find $BA$.

\[
\left(
\begin{array}{cc}
2 & 0\\
0 & 2\\
\end{array}
\right)
\]

Indeed, but I gues liyi didn't fell like posting the details :).

It's enough to observe that $A=\left (\begin{array}{c}\ A_1\\-A_1\end{array}\right ),B=\left (\begin{array}{c}\ B_1\\-B_1\end{array}\right )$, where $A_1,B_1\in \M_{2,2}(\mathbb R)$. We have $AB=\left ( \begin{array}{cc}\ A_1B_1&-A_1B_1\\-A_1B_1&\ A_1B_1\end{array}\right )$, which means that $A_1B_1=I_2\Rightarrow B_1A_1=I_2$, and this solves the problem because $BA=2B_1A_1$.

The problem is equivalent to find  $C$ be a real $4\times 2$ matrix and $D$ be a real $2\times 4$ matrix such that
\[ CD = \left(%
\begin{array}{cccc}
  1/2 & 0 & -1/2 & 0 \\
  0 & 1/2 & 0 & -1/2 \\
  -1/2 & 0 & 1/2 & 0 \\
  0 & -1/2 & 0 & 1/2 \\
\end{array}%
\right). \]

$(CD)^2=CD$ so $CD$ is a projector matrix

$2=rank(CD)=rank(C(DC)D)\leq rank(DC) $
but $DC \in M_2(R)$ it's rank is less or equal 2
so $rank(DC)=2$  and $DC$ is invertible matrix
$(DC)^3=D(CD)^2C=D(CD)C=(DC)^2$
so $DC =I_2$

We observe that $(AB)^2=2AB$
Multiplying at left with $B$ and at right with $A$ we get $(BA)^3=2(BA)^2\hfill{(1)}\\ $
and we know that if a matrix $C$ is a solution for a polynomial its eigenvalues are too, soo the eigenvalues of $BA$ veryfies the polynomial $x^3=2x^2$, results that its eigenvalues $(x_1,x_2)$ are in the set $\{0,2\}$.
We have that $2=\text{rank}(AB)^2=\text{rank}(A(BA)B)\le \text{rank}BA$ results that $BA $ is invertible  <b>(2)</b> so it can't have 0 as eigenvalue. Results that $\det BA=4, \text{tr}BA=4$.
From (1) and (2) we have
$(BA)^2=2BA$
and from Cayley-Hamilton equation:
$(BA)^2=4BA-4I_2$
Substracting these last two relations we get that $BA=2I_2$.

A similar problem (but easier) was proposed at the National Math Olympiad in Romania in 1994 at 11th grade:
Let $A\in M_{3,2}(C),B\in M_{2,3}(C)$ so that
$AB=\begin{pmatrix}
1 & -1 & 2\\
0 & 0 & 2\\
0 & 0 & 2\\
\end{pmatrix}$.
Find $\det BA$.

Seems that it is a problem with the matrix :-S ??

For the National Olympiad in Romania

$\chi_{AB}(x)=det(AB-xI_3)=-x.det(BA-xI_2)=-x.\chi_{BA}(x)$

$det(BA) = \lim _{x\rightarrow 0} \frac{-\chi_{AB}(x)}{x}=-\chi '_{AB}(0)$
the derivate of $\chi_{AB}$ evaluate at $x=0$

nice moubi  :D ... but how do you get $\det (AB-xI_3)=x\det(BA-xI_3)$ ??

and another thing.. the result must be $\det BA=2$ but $P_{AB}'(0)=-2$... :?


Let function $f: (0,+\infty)\to\mathbb{R}$ satisfy the conditions:
$a)$ $f$ is increasing function in $(0,\infty)$.
$b)$ $f(x)+\frac{1}{x}>0, \forall x>0$.
$c)$ $f(x)\cdot f(f(x)+\frac{1}{x})=1, \forall x>0$.
Calculate $f(1)$

Since f is strictly increasing, f is injective
Set $g(x)$ = $f(x)$ +$\frac{1}{x}$

$\large f(g(x)).f(f(g(x)) + \frac{1}{g(x)}) =1$ but $\large f(x).f(g(x)) =1$

Since f is injective $\large x = f(g(x)) + \frac{1}{g(x)} = \frac{1}{f(x)} + \frac{1}{\frac{1}{x} + f(x)}$

->$\large f(x)=\frac{1-\sqrt{5}}{2x}$

[quote="spix"]How do you know that $f$ is strictly increasing?[/quote]
Isn't it here?
[quote="Lovasz"]$a)$ $f$ is increasing function in $(0,\infty)$.[/quote]

The problem comes from 1997 Hellenic Math Olympiad and it says

$a)$  $f$ is strictly increasing;

 :)

[quote="spix"]Does increasing mean strictly increasing?  :roll:[/quote]

hmm... what's the difference between "increasing function" and "strictly increasing function"? :oops:

Increasing: $x<y\implies f(x)\leqslant f(y)$

Strictly increasing: $x<y\implies f(x)<f(y)$

If it were just increasing you couldn`t have made the statement that it is injective.


So, this year, I took a Flash class in school second semester. Actually, I didn't choose it, but my schedule was messed up. In April, I discovered the wonders of ActionScript (the Flash programming language) and made this for coding practice:
http://www.newgrounds.com/portal/view/313288
It's a simple pong game.
However, starting in May, I went further and spent hours coding, up to the point that up to yesterday, I hadn't played an FPS in weeks. However, yesterday, I finished my current latest Flash game, which I present to you:
http://www.newgrounds.com/portal/view/319977
Some computers have lag issues with it, if you do, try saving your game, quitting, clicking "Play this Game" again, and loading. I hope you enjoy it, it was fun making it.    :D

PS: The art was not done by me, but by a classmate in the Flash class who happens to be an artistic prodigy. Look at the graphics in the Pong game to see why I didn't want to ruin Nuke Patrol with my "art".

[quote="Ignite168"][quote="sponge008"]Wow, you're the first person I know of who actually beat it. By boring, do you mean the later levels were too easy, or was it just bad?[/quote]

Well I had a huge surplus of money near the last couple of levels, I just kept using gun #3 and gun #5 when I got them and they easily killed everything. My health wasn't even depleted halfway any of the time although they were about 60% toward finishing the bomb.[/quote]

I'm just back from a tennis tournament. I'll make the last 3 levels devilishly hard in a few hours.

I fixed the easiness problem. Good luck beating it now. :D

I liked the first game - simple and nice.

The second game had too many instructions which i didn't feel like reading.


geometry
3D geometry
tetrahedron
parallelogram
IMO Shortlist

In a given tedrahedron $ ABCD$ let $ K$ and $ L$ be the centres of edges $ AB$ and $ CD$ respectively. Prove that every plane that contains the line $ KL$ divides the tedrahedron into two parts of equal volume.

Let $p$ be the plane through $AB$ parallel to $CD$. Any plane $p'$ parallel to $p$ which intersects the segment $KL$ will intersect the tetrahedron in a parallelogram with center on the line $KL$. Any plane $q$ through $KL$ will pass through the center of the parallelogram and hence divide it into two equal parts. So suppose $q$ divides the tetrahedron into two parts $T$ and $T'$. We have established that the area of the intersection of $p'$ and $T$ is equal to the area of the intersection of $p'$ and $T'$. It follows from Cavalieri's principle that $T$ and $T'$ have equal volume.

(To see that $PQRS$ is a parallelogram, note that $PQ$ is parallel to $CD$, and $RS$ is parallel to $CD$, so $PQ$ and $RS$ are parallel. Similarly $QR$ and $PS$ are parallel to $AB$ and hence to each other.)

Arbitrary plane through $K,L$ cuts $AD,BC$ at $X,Y.$ It's clear that the locus of the points that bisect the segments with endpoints on $BC,AD,$ is the midplane $\lambda$ parallel to those passing through $BC,AD$ and parallel to each other, thus $KL \in \lambda$ $\Longrightarrow$ $KL$ bisects $\overline{XY}$ $\Longrightarrow$ $X,Y$ are equidistant from the plane $CDK.$ Since $[DLK]=[CLK],$ then tetrahedra $[DLKX]=[CLKY]$ are equivalent. If $V_1$ and $V_2$ denote the volumen of the pentahedra $LXKYBD$ and $LXKYCA,$ we have then

$V_1=[DKBYL]+[CLKY]=[DKBC]=\frac{_1}{^2}[ABCD]$

Similarly, $V_2=\frac{_1}{^2}[ABCD]$ $\Longrightarrow$ $V_1=V_2=\frac{_1}{^2}[ABCD]$


search
AoPSwiki
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I have two questions:

1) How do you prove the Chicken McNugget Theorem?
2) How did the Chicken McNugget Theorem get its name? :maybe:

Thanks in advance!

See for example [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=227374]here[/url], [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=413040436&t=205817]here[/url], the [url=http://www.artofproblemsolving.com/Wiki/index.php/Chicken_McNugget_Theorem]AoPSWiki[/url], [url=http://en.wikipedia.org/wiki/Chicken_McNugget_Theorem#McNugget_numbers]Wikipedia[/url] (includes source of name).

Very interesting.

Any practical example for two numbers other than the coin problem?

How do we actually extend it to three numbers or more? :|

Ninety-four bricks, each measuring $ 4"\times10"\times19"$  are to stacked one on top of another to form a tower 94 bricks tall. Each brick can be oriented so it contribues  4 or  10 or 19 to the total height of the tower. How many different tower heights can be achieved using all ninety-four of the bricks?

The only part I don't understand in Dr. math's proof is
6. Prove : Any number above a circled number and in the same column is
   impossible.

I guess because then you would have 2 circled numbers in the same column, but I still don't get it.
We must prove that if p,q relatively prime then it is impossible to have:
cp - q = zp + qr
where cp is the original number in the array and we are going back q to get to the number above.

Thanks!

Something to consider:

If your two numbers are $ a$ and $ b$, what's the relationship between the number $ c$ and $ ab \minus{} a \minus{} b \minus{} c$ if $ 0 < c\le ab \minus{} a \minus{} b$?

Nvm, I get it now.
For the one with the bricks though, the article on the Frobenius number only goes into 2 numbers. How would you use it for 3?

Well if c is less than ab-a-b then c = ka + zb where z<a and k<b
so this number is also attainable- how does that help? that just shows that if c = b it could work.

But do we necessarily have $ z,k\in\mathbb{N}$.

Will you please just post a proof Tenoreoz?

Also can someone explain how to do the chicks mick nugget in 3 numbers?


$ \dfrac{(\sqrt1)x^0}{\frac{1!}{0!}}\equal{}\fbox{1}$, since every part of the expression is 1.

To elaborate, we know that $\sqrt{1}=1,$ so $(1)x^0=x^0=1,$ since any number to the 0th power is 1.

Then we know that $1!=1,$ but some of you guys might not know that $0!=1$ also. So we simplify to get $\dfrac{1}{\dfrac{1}{1}}=\boxed{1}$.


I know Brown must have a great math department, but I don't know much about it.

So two questions:

1. How rigorous is the Brown undergrad math experience (classes, research, opportunities, etc)?
2. Are math majors in danger of running out of math classes if they fill their schedule with math classes, or are there lots of graduate classes that they can take (i.e. they'd never run out)?

I can't answer the first question, but as to the second, you're not going to run out of classes at any large college.  At Brown in particular, not only are there classes that overlap with graduate work, but you have the opportunity to [i]create your own classes[/i] (if that really became necessary).  :)

Ah, thanks a lot =)

I still welcome more feedback, to both Q1 and Q2.

I'm a math major at Brown. I do get the sense that our classes aren't the most rigorous at the undergraduate level. In general, the atmosphere at Brown is a relaxed one, and the math department doesn't tend to be an exception. Prerequisites for classes are not generally enforced, and so some classes will move more slowly to keep down with the students. Problem sets are probably not as challenging as they are at a place like Harvard, MIT, Princeton, Berkeley, or Chicago.

The department is relatively small and overall very friendly. There are a number of faculty members who are particularly interested in working with undergraduates, so most people don't have trouble finding a research supervisor. Brown also has a program for sponsoring student research programs in the summer.

There are plenty of graduate courses, so running out of classes shouldn't be a concern. Even if you were to only take math classes (which would be allowed, but it might not be the best idea), I don't think you'd run out. You can find the full list of courses for the upcoming semester [url='http://math.brown.edu/courses.html']here[/url]. The list of courses for the fall semester is linked from that page. Most courses beyond the introductory level are offered only once a year, so the two lists differ substantially.

If you have more questions, feel free to ask here or private message me.

There are also some Game Theory classes in the economics department and Logic/Set Theory-ish classes in the Philosophy department. The computer science and applied math departments are separate from the math department, so you can also look to them for classes and opportunities.


For which value of $ n\geq 3$ is it possible that there are $ n$ cities, each two connected by one-way roads, and from any city, it is possible to arrive to any other city going through at most one other city?

I did a construction with a odd regular gon. Cities are marked as vertices of 7 regular gon for example and we write for 1->2 if there is a road from 1 to 2. Start with 1: 1->2,1->4,1->6 (alternating). Then with 2: 2->3,2->5,2->7 . For n=3,5,7 I found that system works.

So does it work for all $ n$ odd?

What about $ n$ even?


here is the solution:
Let $z_{i}=x_{i} + i - 1$ with $1 \leq i \leq 8$. It is straighforward to check that
$x_{5} \leq 6 ,x_{6} \leq 5 , x_{7} \leq 4$ and $x_{8} \leq 4$
, so that $z_{1} \leq 8 , z_{2} \leq 9 ,z_{3} \leq 10 , z_{4} \leq 11 ,z_{5} \leq 10,z_{6} \leq 10 , z_{7} \leq 11$.
The value $11$ is possible for $z_{i}$ only when $i=4$ and we must have $(x_{1},...,x_{8})$= $( 8,8,8,8,0,0,0,0)$ or $i=8$ and we must have $(x_{1},...,x_{8})$=$(4,4,4,4,4,4,4,4)$.
In both case , the squares sum is 492 In a similar way , we find  that $x_{1} \geq 4, x_{2} \geq 4,x_{3} \geq 3,x_{4} \geq 2$ and so $z_{1} \geq 4,z_{2} \geq 5,z_{3} \geq 5,z_{4} \geq 5,z_{5} \geq 4,z_{6} \geq 5,z_{7} \geq 6,z_{8} \geq 7$.
Value 4 is possible only when $i=1$ or $i=5$
and we have $(x_{1},...,x_{8})$ = $( 8,8,8,8,0,0,0,0)$  or $i=8$ nad we must have $(x_{1},...,x_{8})$=$(4,4,4,4,4,4,4,4)$.

Otherwise, we must $5\leq z_{i}\leq 10$ for each i , and checking out the possibilities leads to square of sums less than 492


Find the maximum height reached by a projectile as a function of its initial speed and its firing angle?

$ x \equal{} v_0t\cos \theta\ , y \equal{} v_0t\sin \theta \minus{} \frac {1}{2}gt^2$

$ v_y \equal{} v_0\sin \theta \minus{}gt$, $ v_y\equal{}0\Longleftrightarrow t\equal{}\frac{v_0}{g}\sin \theta\ : \equal{}T$

$ \therefore y_{max}\equal{}v_0T\sin \theta \minus{}\frac{1}{2}gT^2\equal{}\boxed{\frac{v_0^2\sin ^ 2 \theta}{2g}}$


calculus integration trigonometry algebra function domain logarithms

#1)  Make the substitution $x=\sin u \Rightarrow dx = \cos u \ du$, then we have :

$\int\frac{\cos u \ du}{\sin^2 u \sqrt{1-\sin^2 u}} = \int \frac{\cos u \ du}{\sin^2 u\cos u} = \int \frac{du}{\sin^2 u}$

Now (unless you already recognize the above integral), make the substitution $u=2\tan^{-1} v \ du = \frac{2dv}{1+v^2}$, so now we have:

$\int \left(\frac{2dv}{1+v^2} \cdot \frac{1}{\sin^2 \left(2\tan^{-1} v\right)}\right) = \int \frac{v^2+1}{2v^2} \ dv = \frac{v^2-1}{2v}$

And now, $\frac{v^2-1}{2v} = \frac{\tan^2\frac{u}{2}-1}{2\tan\frac{u}{2}} = -\frac{1}{\tan u} = -\frac{1}{\tan \left(\sin^{-1} x\right)} = -\frac{\sqrt{1-x^2}}{x}$

So, we have $\int\frac{dx}{x^2\sqrt{1-x^2}} = -\frac{\sqrt{1-x^2}}{x} + C$

Are you sure #2 is written correctly?

[quote="Jimmy"]

Are you sure #2 is written correctly?[/quote]

Maybe it is $\int \frac{x}{\sin ^ 2 x}\ dx.$

[quote="kunny"]Is $\sqrt{1-\sin ^ 2 u}=\cos u?$[/quote]

Yes, because $u=\sin^{-1} x$, so on the domain of $u$, we have $\cos u = |\cos u|$, but good point I should have addressed that.

well, for 2) I may have oversimplified... the problem is given as:

∫ x * sin^-1(x) dx

and 3) remains as:

 ∫ (xdx) / [root(x^2+4x+13)]

Thanks so much!

Well, that's different! $\sin^{-1}x=\arcsin x$ is a completely different function than $(\sin x)^{-1}=\frac1{\sin x}.$

For $\int x\arcsin x\,dx,$ try letting $x=\sin \theta,$ so that $\theta=\arcsin x$ and $dx=\cos \theta\,d\theta.$ The integral becomes

$\int\theta\sin\theta\cos\theta\,d\theta=\frac12\int\theta\sin 2\theta\,d\theta$

Can you see how to proceed from there? It will require an integration by parts. We could have just integated by parts in the first step, but doing this substitution first may make things easier to see.

so Kent, I apply integration by parts to the last part? If this is the case, would u = theta and v = sin(2theta) = sin(2u)?

Also, can anyone help with the third problem that I posted?

Use the subtitution $x+2 = 3 \tan t$
\begin{eqnarray*}\int \frac{x}{\sqrt{x^2+4x+13}} \, dx &=& \int \frac{x}{\sqrt{(x+2)^2+9}} \, dx \\ &=& \int \frac{(3 \tan t - 2)3 \sec^2 t}{3 \sec t} \, dt \\ &=& \int (3 \tan t \sec t - 2 \sec t) \, dt \\ \end{eqnarray*}

\[ \int 3\tan t \sec t \, dt = \int \frac{ 3 \sin t}{\cos^2 t} \, dt = \int \frac{-3}{\cos^2 t} \, d(\cos t) = 6 \frac{1}{\cos t} + C  \]
and
 \[ \int 2 \sec t , dt = 2 \ln | \sec t + \tan t| +C  \]

Cahnge it to x.

[quote="SimpleComplexity"]Thanks so much Beginner... just a couple questions though... for my final answer, do I subtract the two results from each other? and how do I change it back to x when you made a substitution like that?[/quote]

$\tan t = \frac{x+2}{3}$  so we can take $\sin t = \frac{x+2}{\sqrt{(x+2)^2+9}}$ and $\cos t = \frac{3}{\sqrt{(x+2)^2+9}}$.
And I'm sure u know how to find $\sec t$  :)

Actually Kunny's subtitution is almost the same.

Alright, I worked through all of those problems, but I have another that I can't figure out:

∫ dx / (root((x+1)(x-3)))

Perhaps that factored denominator is confusing you?
$(x+1)(x-3)=(x-1)^2-2^2$. That sets up the trig substitution (or hyperbolic substitution) you need.

[hide="Without the method of the substitution !."] ${1.\blacktriangleright I=(0,1)\ : \ \int \frac{1}{x^2\sqrt{1-x^2}}\ dx=\int \frac{1}{x^3\sqrt{\frac{1}{x^2}-1}}}\ dx=$
$-\int \frac{\left (\frac{1}{x^2}-1\right)'}{2\sqrt{\frac{1}{x^2}-1}}\ dx=-\sqrt{\frac{1}{x^2}-1}+C=-\frac{\sqrt{1-x^2}}{x}+C\ .$
$2.\blacktriangleright I=\left(0,\frac{\pi}{2}\right)\ : \ \int \frac{x}{\sin^2x}\ dx=\int x(-\cot x)'\ dx=$
$-x\cot x+\int \cot x\ dx=-x\cot x+\ln(\sin x)+C\ .$
$3.\blacktriangleright \int x\arcsin x\ dx=\frac 12 x^2\arcsin x-\frac 12A\ ,$
$A=\int \frac{x^2}{\sqrt{1-x^2}}\ dx=\int x\left(-\sqrt{1-x^2}\right)'\ dx=$$-x\sqrt{1-x^2}+\int \frac{1-x^2}{\sqrt{1-x^2}}\ dx=$
$-x\sqrt{1-x^2}+\arcsin x-A\Longrightarrow A=-\frac 12x\sqrt{1-x^2}+\frac 12 \arcsin x+C\Longrightarrow$
$\int x\arcsin x\ dx=\frac 14 (2x^2-1)\arcsin x+\frac 14 x\sqrt{1-x^2}+C\ .$
$4.\blacktriangleright I=R\ : \ \int \frac{x}{\sqrt{x^2+4x+13}}\ dx=\frac 12\int \frac{\left(x^2+4x+13\right)'-4}{\sqrt{x^2+4x+13}}\ dx=$

$\int \frac{\left(x^2+4x+13\right)'}{2\sqrt{x^2+4x+13}}\ dx-2\int \frac{(x+2)'}{\sqrt{(x+2)^2+3^2}}\ dx=$

$\sqrt{x^2+4x+13}-2\ln \left(x+2+\sqrt{x^2+4x+13}\right)+C\ .$[/hide]


algebra
polynomial
calculus
algebra unsolved

Equation with Sum of two quotients each with differing square roots in the denominator

I am having tremendous problems with solving this equation for t:

[code]t/sqrt(t^2 + h^2) + (t - 2*a)/(sqrt((2*a - t)^2 + h^2)) = 0
[/code]

I've tried squaring both sides (after moving one term to the right of the =), removing the quotients via multiplication, and expanding the polynomials, and nothing seems to get me closer.

My book's solution guide and derive tell me that t=a, which does work, but I have no idea how I would get to this on my own.  Any algebraic manipulations I try just seem to make the darn thing more complication.

Can some one help me solve this, step by step?

Aside - Can anyone recommend a software that can show step by step instructions on how to solve equations like this?  I've tried a few but none that show steps (and some that don't, such as Maxima) can't seem to handle this one.

Thanks,

   DoctorChaotic

[A note to moderators on the placement of this topic: This actually is part of a calculus problem, but the algebra is the part that has given me trouble (I've verified that I've done the calc part correctly).  If I was supposed to place this in the calculus forum.  Also, maybe I should have put this in the "solved/proposed" problems section since I've provided the final answer, but since I'm looking for something step by step, I posted here]

$ \frac{t}{\sqrt{t^{2}\plus{}h^{2}}}\plus{}\frac{t\minus{}2a}{\sqrt{(2a\minus{}t)^{2}\plus{}h^{2}}}\equal{} 0$ second term to the other side
$ \frac{t}{\sqrt{t^{2}\plus{}h^{2}}}\equal{}\frac{2a\minus{}t}{\sqrt{(2a\minus{}t)^{2}\plus{}h^{2}}}$ squaring both sides
$ \frac{t^{2}}{t^{2}\plus{}h^{2}}\equal{}\frac{(2a\minus{}t)^{2}}{(2a\minus{}t)^{2}\plus{}h^{2}}$ passing denominators to the other side
$ (t^{2})((2a\minus{}t)^{2}\plus{}h^{2})\equal{}((2a\minus{}t)^{2})(t^{2}\plus{}h^{2})$
$ t^{2}. h^{2}\equal{} (2a\minus{}t)^{2}. h^{2}$

Thanks,  That helped greatly.  I didn't think of  teating (2a-t^2) as its own entity.


Ring Theory
superior algebra
superior algebra unsolved

Consider fields $ K$, $ L$, $ M$, $ E$ satisfying the inclusions $ K \subset L \subset E$ and $ K \subset M \subset E$. Suppose that $ L$ is algebraic over $ K$. Consider
\[ LM : \equal{} \left\{\sum_i x_i y_i : x_i \in L, y_i \in M\right\}.
\]
Prove that $ LM$ is the smallest subfield of $ E$ containing $ L$ and $ M$.

clearly $ LM$ is the smallest subring of $ E$ containing $ L$ and $ M$, so we only have to show that $ LM$ is a field. observe that $ LM$ is the union of the rings $ M[x_1,...,x_n]$ with $ x_i \in L$, which are fields since the $ x_i$ are algebraic over $ M$.


Let $ ABC$ be a triangle

Calculate $ S$ the area as $ f(a,b,c)$

Its sufficient to calculate $ r$ or $ R$ in function of $ a,b,c$

thanks


calculus
derivative
analytic geometry
graphing lines
slope
integration
function

#5: A graph of the derivative of $ g(x)$ is given, resembling the graph of $ y \equal{} \minus{} |x \minus{} a| \plus{} g(a)$ (sorry; I don't know how to draw graphs on asymptote).  But, the two rays of the graph (with common endpoints at $ (a, f(a))$) don't need to have slope $ 1$; the point is, it is some graph with two lines (the first with positive slope, the second with negative slope) meeting at some point $ (a, f(a))$ at a sharp corner.  From $ ( \minus{} \infty, a]$, the graph is a line with positive slope.  From $ [a, \plus{} \infty)$, the graph is a line with negative slope.  The question is if $ g$ is increasing at $ x \equal{} a$.

I'm not sure how a graph can be "increasing" at a certain point; I only know how it can be increasing on a certain interval.  Hopefully, you guys will make better sense of this than I can.

#6: If $ f(x)$ is continuous on $ [ \minus{} 1,3]$ and $ \minus{} 3\le f\prime(x)\le 2$ for all $ x\in( \minus{} 1,3)$, then what is the greatest possible value of $ f(3) \minus{} f( \minus{} 1)$?

The [i]mean[/i] [url=http://en.wikipedia.org/wiki/Mean_value_theorem]Value Theorem[/url] is gonna help you make your way through #6. See you!

:)

[quote="coquitao"]The [i]mean[/i] Value Theorem is gonna help you make your way through #6. See you!

:)[/quote]
Sorry, but, um, it's not going to help (at least not in any obvious way).

Read the problem carefully and note that the DERIVATIVE of $ f$ is between $ \minus{}3$ and $ 2$, not $ f$ itself.  Thus, the Mean Value Theorem isn't going to help in this case in any obvious way.
At first, I thought of using the Mean Value Theorem, but then I saw that $ \prime$ mark, changing the entire problem.

[quote="MathAndKnowledge"][quote="coquitao"]The [i]mean[/i] Value Theorem is gonna help you make your way through #6. See you!

:)[/quote]
Sorry, but, um, it's not going to help (at least not in any obvious way).

Read the problem carefully and note that the DERIVATIVE of $ f$ is between $ \minus{} 3$ and $ 2$, not $ f$ itself... [/quote]

So?

$ f(3) \minus{} f( \minus{} 1) \equal{} (3 \minus{} ( \minus{} 1))f^{\prime}(\xi)$ for some $ \xi \in ( \minus{} 1,3)$... Since $ f^{\prime}(\xi)$ is bounded above by $ 2$ it follows that...

Oh, wait!  I'm sorry... I didn't know the Mean Value Theorem until just now (I looked it up on the AoPS Wiki).  I just thought that "Mean Value Theorem" is another name for "Intermediate Value Theorem," but I was obviously wrong.

Anyways, I'm not allowed to use the Mean Value Theorem to solve #6, because I haven't learned that yet in my Calculus class.  Is there no other way to solve it?

[quote="MathAndKnowledge"]Oh, wait!  I'm sorry... I didn't know the Mean Value Theorem until just now (I looked it up on the AoPS Wiki).  I just thought that "Mean Value Theorem" is another name for "Intermediate Value Theorem," but I was obviously wrong.[/quote]

Just set up a link, too.

:)

[quote="MathAndKnowledge"]Is there no other way to solve it?[/quote]  Yes, there is.  What is $ f(3) \minus{} f(\minus{}1)$ in terms of $ f'$ (using some operation you've learned about in calculus class)?

Umm... I don't know which operation you're referring to, JBL.

The thing is, the answers to these problems are due in one week for me in my class.  In that week, we'll cover a lot more about the derivative (all we've done so far is two definitions of it, the power rule, the product rule, the quotient rule, and higher order derivatives).  But I know that we won't get to the Mean Value Theorem, which is why I hesitate to use it.
I've never had to write any thing involving $ f$ (such as $ f(3)\minus{}f(\minus{}1)$) in terms of $ f'$.  The expression $ f(3)\minus{}f(\minus{}1)$ only reminds me of four times the slope of a secant line passing through $ f$ at $ x\equal{}\minus{}1$ and $ x\equal{}3$.

I'm sorry if I sound like an ignoramus; the title of the topic, is, after all, "Some very basic questions that I'd like help on... ".  Can you please tell me the operation, JBL?

I believe JBL is referring to integration, but it seems you haven't covered that.

I'm almost certain this problem is intended to be done with MVT, but we can do it without it, albeit a bit less elegantly.  I'm going to assume that, in addition to what you mentioned, you have also encountered the following fact in class:
\[ \boxed{\text{If } g'(x)\geq 0 \text{ for } a < x < b \text{ then } g(x)\geq g(y) \text{ whenever } a < y\leq x < b}
\]
Now consider the following function
\[ g(x) \equal{} 2x \minus{} f(x)
\]
Since $ g'(x) \equal{} 2 \minus{} f'(x)\geq 0$, it then follows from the above fact that $ g(3) \minus{} g( \minus{} 1)\geq 0$, and hence $ f(3) \minus{} f( \minus{} 1)\leq 8$.  Notice the inequality becomes equality if we take $ f(x)\equal{}2x$.
[hide="remark"]Visual intuition is sometimes more useful than algebraic manipulation.  If you look at the graph of $ f$ on $ [ \minus{} 1,3]$ and draw two straight lines with slopes $ 2$ and $ \minus{} 3$ passing through the point $ ( \minus{} 1,f( \minus{} 1))$, what can you say about the graph of $ f$ relative to these lines?[/hide]

I've now solved both problems, and I'll show you guys my solution for #6 (since that was the only one that I received help on).  First, I'll note that whatever you guys suggested wasn't wrong, but was certainly inelegant and had too much work to it.
Integration, the Mean Value Theorem, and JoeBlow's method are all not needed.

The solution to #6 is dependent on one of the definitions of the derivative.  Of course, there's the two definitions where limits are used, but one of the other definitions is that the derivative is the rate of change of a function.
Since the rate of change is at most $ 2$ for our function $ f$ on the interval $ ( \minus{} 1,3)$, it will be what we use to maximize the difference $ f(3) \minus{} f( \minus{} 1)$.  From $ f(k)$ to $ f(k \plus{} 1)$, with $ k\in\{\minus{}1,0,1,2\}$, the difference or rate of change of $ f(k \plus{} 1) \minus{} f(k)$ is at most $ 2$, as stated earlier.  We can use this like a recursion:
$ f(3) \minus{} f( \minus{} 1) \equal{} f(2) \plus{} 2 \minus{} f( \minus{} 1) \equal{} f(1) \plus{} 2 \plus{} 2 \minus{} f( \minus{} 1) \equal{} f(0) \plus{} 2 \plus{} 4 \minus{} f( \minus{} 1) \equal{} f( \minus{} 1) \plus{} 2 \plus{} 6 \minus{} f( \minus{} 1) \equal{} \boxed{8}$.
The solution only takes about $ 10$ seconds of thought; it just looks long in writing.

[quote="MathAndKnowledge"]Of course, there's the two definitions where limits are used, but one of the other definitions is that the derivative is the rate of change of a function.[/quote]
 :o I think you need to review the definition of 'definition'.

Admittedly what passes for a 'proof' in the typical calculus class isn't exactly rigorous mathematics, but the argument you posted shows serious gaps in your understanding of even basic concepts.  You should ask your teacher what an acceptable solution to this problem would be.

To JoeBlow:
The derivative can be described in many ways.  When I said "the two definitions where limits are used," I meant $ \lim_{\triangle x\rightarrow0}\frac {f(x \plus{} \triangle x) \minus{} f(x)}{\triangle x}$ and $ \lim_{x\rightarrow a}\frac {f(x) \minus{} f(a)}{x \minus{} a}$.
Perhaps you need to review your calculus before telling me to review my definitions.  My main point is that you didn't know what I was talking about (since I didn't specify), yet you assumed that I was wrong.
Let me also say that I [b]never[/b] said that my solutions were rigorous or proofs.  Don't post a reply full of criticism, when you don't even specify what's wrong with my post.  It almost seems as if you're making things up when you say that I lack basic understanding of concepts (no proof of that was offered by you).
Also, my teacher is the one who came up with that solution, not me.  So don't say "You should ask your teacher what an acceptable solution to this problem would be." when you don't understand the whole situation... again, this shows how you assume things too often.
Continue this by PM with me (it seems like you're trying to pick a fight, but let's not do that on the forum).

To others:
Hopefully, you guys won't be as immediately hostile to my solution as JoeBlow was.  The point of this topic was for me to ask a question... now that it is answered, I'd appreciate it if you can PM me any of the flaws that you see in my solution.

[quote="MathAndKnowledge"]
Perhaps you need to review your calculus before telling me to review my definitions.

...

To others:
Hopefully, you guys won't be as immediately hostile to my solution as JoeBlow was.[/quote]

JoeBlow is much more of an 'authority' on calculus than you or your teacher, and he wasn't being hostile.

In fact, in the third post in this thread you are rude and condescending to coquitao even though he is correct and you are wrong.  The reason that I rarely post in threads like this is that I have no desire to answer a question correctly and completely and promptly be informed by someone who has been learning calculus for two weeks that what I did was 'wrong' and that they know better.


[quote="ehsan2004"]Find all function$f:R\rightarrow R$ such that for every $x\in R$ and $n\in N$, $f(x^n+x)\leq x \leq (f(x))^n+f(x)$. :o    :?:[/quote]

[hide="This question seems wrong"]
Let us fix n. If n is even, we first work on R^+.
$ x^n + x $ defines a bijection from $ R $ to $ R $ (or $R^+$ to $R^+$ if n is even).
Define $ g(x) = x^n + x $.
We get $ f(g(x)) \leq x \leq g(f(x)) $, so $ f(x) \leq g^{-1}(x) $ and $ g^{-1}(x) \leq f(x) $.
Thus $ f(x)=g^{-1}(x) $.
For n even and x negative I think the same sort of argument can be used.
[/hide]

[quote]enescu wrote:Is $n$ fixed?[/quote]
yes? ;)


probability
number theory
least common multiple

A bowl contains only red marbles and green marbles. The probability of selecting a red marble from the bowl is $\frac{1}{13}$. The probability of selecting a red marble from the bowl is $\frac{2}{5}$. There are fewer than $100$ marbles in the bowl. What is the probability of selecting, at random and without replacement, a green marble and then a red marble from the bowl on the first two selections? Express your answer as a common fraction.

[hide="solution"] 65 is the only number that is less than 100 and that is divisible by 5 and 13, so there are 65 marbles total.
Therefore, there are 5 reds and 26 greens.  The probability of first getting green is $\frac{2}{5}$.  Now there are 64 marbles.  The probability of now getting a red marble is $\frac{5}{64}$.  Multiply the two probabilities, and you get $\frac{1}{32}$.[/hide]

Edit:  Never mind, the problem makes $0$ sense.

Wait, if there 5 red and 26 green, thats 31 marbles and the whole bowl is either green or red and 21 is not 65, so thers a problem with this problem

wait.
how is there a 2/5 AND 1/13 chance of picking red.

#H34N1, we have two major problems.
1. Red has two probabilities
2. 2/5 + 1/13 = 1

[img]http://img511.imageshack.us/img511/5273/tempxr7.png[/img]

[hide]There are $65$ marbles because that is the LCM of $5$ and $13$.

$\frac{24}{65}\cdot\frac{15}{64}=\boxed{\frac{9}{104}}$[/hide]

That is what I had in the real contest, and that is correct.


How many edge disjoint cycles can a planar graph have?

In terms of what?

Consider a unique cycle, then it has only one...
Consider $n$ triangles all sharing a common vertex, then there are $n$...

Pierre.

can't we have a bound on the number of cycles a planar graph can have in terms of its number of vertices and its number of subgraphs?


In the figure (not to scale) the line segments $ AN$, $ AM$, and $ BC$ are tangent to a circle at points $ N$, $ M$, and $ P$, respectively. The points $ B$ and $ C$ are on segments $ AN$ and $ AM$, respectively. Angle $ BAC$ is 30 degrees and angle $ ABC$ is a right angle. If the length of $ AN$ is 12, what is the area of triangle $ ABC$?

Let $ L$ be a six digit base ten integer whose last digit is not 0, and let $ M$ be the number formed by moving the last digit to the from. For example, if $ L \equal{} 354268$, then $ M \equal{} 835426$. Find all possible $ L$ such that $ L \equal{} 3M$.

well your second question was recently answered in the HSB forum
(this is why the search function is useful)
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=238298]see here[/url]

[hide]$ AN \equal{} AM \equal{} 12$. $ BN \equal{} BP$ and $ PC \equal{} CM$, so the perimeter of triangle $ ABC$ is 24.

Triangle $ ABC$ is 30-60-90. If $ BC \equal{} s$, then $ AB \plus{} AC \plus{} BC \equal{} s(3 \plus{} \sqrt {3}) \equal{} 24$. Therefore, $ s \equal{} 12 \minus{} 4\sqrt {3}$.

$ AB \equal{} 12\sqrt {3} \minus{} 12$ and $ BC$ = $ 12 \minus{} 4\sqrt {3}$, so the area is simply their product divided by 2, which is $ \boxed{96\sqrt {3} \minus{} 144}$.[\hide][/hide]


calculus
integration
calculus computations

Find the arc length of the curve $ 30xy^2 \minus{} y^8 \equal{} 15$ from $ y\equal{}1$ to $ y\equal{}2$

It seems the problem's wrong. The integral can only be approximated.


whats the easiest way to graph functions, such as sin?  or even ones like $y = 3x^{3}+5$.


please can anybody solve these two sums i've forgotten.

1)convert 75 km/hr into m/sec.
2)convert 25m/sec into km/hr.

1)convert 75 km/hr into m/sec.

$ \dfrac{75km}{1hr}\equal{}\dfrac{75(1000)m}{1(60)sec}\equal{}\dfrac{75000m}{60sec}\equal{}\boxed{\dfrac{1250m}{1sec}}$

2)convert 25m/sec into km/hr.

$ \dfrac{25m}{1sec}\equal{}\dfrac{\frac{25}{1000}km}{\frac{1}{60}hr}\equal{}\dfrac{\frac{1}{40}km}{\frac{1}{60}hr}\equal{}\dfrac{\frac{30}{20}km}{1hr}\equal{}\dfrac{\frac{3}{2}km}{hr}\equal{}\boxed{\dfrac{1.5km}{hr}}$

If you are stuck, you should break it up into numerator and denominator. For example, a problem might make you convert 2 miles to feet. You know that there are 5280 of one in the other, but which one is it? Are tehre 5280 miles in a foot? No, or I could run a mile in less than 2 seconds. So that means there will be more feet than the equivalent length in miles, so you multiply 5280 on the feet side.

I think its wrong . 1 hr =3600 seconds and 60 minutes not seconds.


Hello everyone,

I'm new to the forum and may be this kind of a problem has been posted before, but I couldn't find anything in the archives, so here we go:

Find a real life situation that fits the formula (x-2)(x+3)=0

Anyone?

I think the question does not count as an unsolved algebra problem, in the same way that find a possible value for x does not.  Browse the other entries and you will quickly find that the emphasis is on proof and vagueness of any kind is discouraged!

In other words, if you really want to, post this in the Middle School section. This is the Olympiad section.

Hello,

I liked your advice and did post this in Middle School section.

If they can't answer it, I will post it in the High School section.

If they can't answer it, I will assume there's no one in this forum capable of solving this "simple" question.

Cheers!


Is there any high school or college campus in Massachusetts that is offering the 10B for students wishing to take it if their school only offers the A?

Fitchburg State.
http://www.unl.edu/amc/b-registration/b1-archive/2008-2009/CU2009/2009-CU-list.shtml

Last year Lexington allowed some visitors to take the B sitting with them, but I'm not sure about this year.  (Last year I allowed some visitors to my school, but I'm not doing it this year.)


Does anyone have any comments on the proof of Absolute Truth in my signature?

I personally agree with it logically, but disagree upon the principle.

Any suggestions? You can go to this site: http://members.chello.nl/p.cooijmans/essay/truth.htm

to see Paul's responses to common arguments against his proof by others who have encountered it.

[quote="paganinio"]logical mistake? not absolutely true is different from not true.
IMO, the sum of all relative truths=absolute truth[/quote]

I understand your point, but I think what Paul Cooijmans meant was that if there is anything we can be sure about, it would be this statement because there is no uncertainty within this statement. Unlike in real life, where we have many different factors that play into everything, and we can [i]never[/i] be 100% sure of anything, we know this statement to be true.

Therefore if even this statement is not absolutely true, then there is no absolute truth. But since by definition (apparently) this must be categorized under absolute truth...and here is where I disagree, much like you.

[quote="JBL"]Without stopping to think too much about absolute truth, it's worth pointing out that the validity of any logical argument relies on the assumption that our system of logic (or at least, our faculties to judge the validity of logical arguments within our system, or something) is an effective way to judge truth (or, in this case, absolute truth).  But there is no particular way that we can prove the validity of our abilities to prove.[/quote]

I agree completely. But there seems to be something wrong with his logic too...


Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ ab\plus{}ac\plus{}bc\equal{}3.$ Prove that
\[ a^2\plus{}b^2\plus{}c^2\plus{}15\geq6(a\plus{}b\plus{}c)\]

$ a^2\plus{}b^2\plus{}c^2\plus{}15 \ge 6(a\plus{}b\plus{}c)$
$ (a\plus{}b\plus{}c)^2\minus{}2(ab\plus{}bc\plus{}ca)\plus{}15 \ge 6(a\plus{}b\plus{}c)$
$ (a\plus{}b\plus{}c)^2\minus{}2\cdot 3\plus{}15 \minus{} 6(a\plus{}b\plus{}c) \ge 0$
$ (a\plus{}b\plus{}c)^2 \minus{} 6(a\plus{}b\plus{}c) \plus{} 9 \ge 0$
$ (a\plus{}b\plus{}c\minus{}3)^2 \ge 0$

Kouichi Nakagawa, your proof is nice!
The following inequality is true too.
Let $ a,$ $ b$ and $ c$ are non-negative numbers such that $ ab \plus{} ac \plus{} bc \equal{} 3.$ Prove that
\[ a^3 \plus{} b^3 \plus{} c^3 \plus{} 39\geq14(a \plus{} b \plus{} c)
\]


function
logarithms
integration
real analysis
real analysis solved

if $0<a<1$,can we know $\sum_{i=0}^{\to %Error. "infinit" is a bad command.
}a^{2^i}$

If there were a simple analytic expression, it would work for complex $a$ as well and the function $f(z)=\sum_i z^{2^i}$ ($z\in\mathbb D$) could be analytically continued through at least one point on the boundary of the unit disk (I do not want to formally define a "simple analytic expression" here but write any decent formula and you'll see what I mean). On the other hand, it is easy to show that there is no such continuation. So, in a sense, the answer is "No". In a sense, because you always can define this expression to be $F(a)$ and call it a new special function. There is even some chance that somebody has done it already for some expression that is related to yours and then you may happily say something like $f(a)=\theta_{1,2}'''(\ln a/{a^2+1})$  :) It is also not hard to show that your sum is comparable to $1+\ln\frac 1{1-a}$ (see the reference Myth gave for more precise statement).

[quote="Myth"]He-he...
You even don't know how many time some mathlinkers tried to solve the related problem.
See http://www.mathlinks.ro/Forum/viewtopic.php?t=4767[/quote]
 thank you.Myth you mean we haven't solven it??
and can you answer me how to Calculate $\int (1 \to infinity) x^u/u du $？ :)


Let be $ x,y,z\in \mathbb{R}_\plus{}$ such that $ xyz\equal{}1$ . Prove that :

$ \frac{1\plus{}xy}{1\plus{}z}\plus{}\frac{1\plus{}yz}{1\plus{}x}\plus{}\frac{1\plus{}zx}{1\plus{}y}\ge 3$

$ \sum \frac {1 \plus{} yz}{1 \plus{} x}\geq \sum \frac {1 \plus{} \frac {1}{x}}{1 \plus{} x} \equal{} \sum\frac {1}{x}\geq \frac {3}{\sqrt [3]{xyz}} \equal{} 3$.

Alternative Solution:

$ \boxed{1 \minus{} \frac {1}{z}\leq \ln z\leq z \minus{} 1\ (z > 0)}\Longrightarrow \frac {1}{z} \plus{} \ln z\geq 1$, the rest is left for you.

very easy,
by AM-GM,
$ S\equal{}\frac{1\plus{}xy}{1\plus{}z}\plus{}\frac{1\plus{}yz}{1\plus{}x}\plus{}\frac{1\plus{}zx}{1\plus{}y}\ge 3\sqrt[3]{\frac{(1\plus{}xy)(1\plus{}yz)(1\plus{}zx)}{(1\plus{}x)(1\plus{}y)(1\plus{}z)}}$
and we have $ (1\plus{}xy)(1\plus{}yz)(1\plus{}zx)\equal{}\left(1\plus{}\frac{1}{x}\right)\left(1\plus{}\frac{1}{y}\right)\left(1\plus{}\frac{1}{z}\right)\equal{}(1\plus{}x)(1\plus{}y)(1\plus{}z)$ then $ S \geq 3$


Given a circle center $O$, a secant to that circle ($l$), not passing through $O$, and a point $T$ outside the circle, on the secant, construct a circle tangent to $l$ at $T$ and tangent to the circle.

If you are just making a graph, then you can simply choose a point on the circle and then find $T$ on the secant line.

of course i can approximate it, but how would you construct it "exactly," the reason that i would want to know this is because constructions often give insight into solutions

[quote="Altheman"]constructions often give insight into solutions[/quote]
I agree. But I can only do it if we are given the secant line and tangent point, which is much easies than your original problem.

I have a not so elegant construction, which I expect could be improved.

Let $M$ be the foot of the perpendicular from $O$ to $l$ and let that perpendicular intersect the circle in points $P$ and $Q$ such that the order of the points is $P-M-O-Q$. In $T$ draw the perpendicular $p$ to $l$. Draw a tangent $TS$ from $T$ such that $S$, its interception point with the given circle, is on the same side of $l$ as $P$ (I take for granted that construction of a tangent is known.) On $p$ find $A$ such that $TS=TA$ and $A$ and $P$ are on the same side of $l$. Through $A$ draw a line parallel to $l$ until it intercepts $MP$ in $B$. Draw the line $PA$ and let it intersect $l$ in $F$. Draw the line $FB$ and let it intersect $p$ in $D$. Then $DT$ is the diameter of the desired circle.

Another such circle can be found if the construction is repeated on the other side of $l$.

[u]Proof of the construction[/u] (for one side only, since for the other the proof is similar):

If the desired circle is $(C,R)$, the radius of the given circle is $r$, $OT=d$ and $\angle OTM=\alpha$, then by the Cosine Law in $\triangle COT$

$(R+r)^2=d^2+R^2-2Rd\cos(\alpha+{\pi\over 2})\implies 2R=\frac{d^2-r^2}{r-d\sin\alpha}$

Since $MP=r-d\sin\alpha$ and $TS^2=d^2-r^2$, then $TA=MB=TS=\sqrt{d^2-r^2}$. By the construction, $\frac{MP}{TA}=\frac{MB}{TD}\implies TD=\frac{TA\cdot MB}{MP}=\frac{d^2-r^2}{r-d\sin\alpha}=2R$


[size=134]Does the sequence {$ \frac{n^n}{e^{n!}}$} converge or diverge?[/size]

The term in the denominator grows much faster than the numerator, so it converges to 0.

$ \lim_{n\to \infty} \sqrt[n]{t_n}\equal{}\lim_{n\to \infty} \bigg(\frac{n^n}{e^{n!}}\bigg)^{\frac{1}{n}}\equal{}\lim_{n\to \infty} \frac{n}{e^{(n\minus{}1)!}}\equal{}\lim_{n\to \infty} \frac{e}{(n\minus{}1)!e^{(n\minus{}1}!}\equal{}0$

$ \Leftrightarrow \lim_{n\to \infty} t_n\equal{}\lim_{n\to \infty} 0^n\equal{}0\cdot 0\cdot 0\cdots 0\equal{}0$


function
limit
logarithms
calculus
calculus computations

Let $f : \mathbb R \to \mathbb R$ be a function such that $\lim_{x \to \infty}f(x)$ exists and is equal to $a \in \left( 0, \infty \right)$ and $\lim_{x \to \infty}x [f(x)-a]$ exists and is equal to $b \in \mathbb R$.

Find $\lim_{x \to \infty}x \left[ f^{a}(x)-a^{f(x)}\right]$.

It seems to me that the *Calarasi limit* is:

$b a^{a}\left(1-\ln a\right)$.


complements: if the measure of two angles add up to 90 degrees the angles are called complementary angles

supplements: if the measure of two angles add up to 180 degrees, they are supplementary angles

heres a question that was on state mathcounts 03:

two complementary angles, A and B, have measures in the ratio of 7 to 23, respectively. what is the ratio of the complement of angle A to the complement of angle B? express your answer as a common fraction

sure there is an easier way to do this with some logic, but i want to see how you got there using some algebra and geometry!

supplements and complements show themselves in mc a good amount

I don't get it...isn't the "complement of the complement of angle A" equal to just angle A and isn't it given that the  "complement of angle B" is angle A?

you mean angle B for the complement of angle A

yes, its true. thats the easy way. so its just the inverse thingie so 23/7. but i wanted them to learn the other way. cause it may not be like that all the timo

[quote="Syntax Error"]you mean angle B for the complement of angle A[/quote]
Not really, it says "complement of the complement of angle A", so the answer is just 1:1 currently.

i edited it

i guess i made...a syntax error

loooooooool

my 1000th post!

Interesting, we both fixed it at the same time but only your change counted...
Anyways, congratulations!


A 10. Let $ n$ be a positive integer with $ n \geq 3$. Show that ${ n^{n^{n^{n}}}-n^{n^{n}}}$ is divisible by $ 1989$.


Find the sum of the following fractions:
1/18 + 1/54 + 1/108 + 1/180 + ... + 1/31860 . Express your answer as a common fraction in lowest terms.
a. 	29/270
b. 	59/540
c. 	1/9
d. 	19/180
e. 	61/540

By the way, my post count is 8^3!

[hide="Answer"]$\frac{1}{18}+\frac{1}{54}+\frac{1}{108}+...+\frac{1}{31860}=\frac{1}{18}\cdot \displaystyle\left(1+\frac{1}{3}+\frac{1}{6}+...+\frac{1}{1770}\right)$
Upon adding together the fractions in parentheses, we find a pattern. The denominators are all triangular numbers, or numbers in the form $\frac{n(n+1)}{2}$, where $n$ is an integer. When we add the reciprocal of a triangular number made with an odd $n$, we see that the resulting fraction is the two factors of the triangular number with the smallest difference. For example, the sum of the triangular reciprocals $1+\frac{1}{3}+\frac{1}{6}+...+\frac{1}{78}=\frac{24}{13}$. Then, $\frac{24}{13}+\frac{1}{91}=\frac{13}{7}$. $1770=\frac{n(n+1)}{2} \Rightarrow n=59$. Therefore, we take the two factors of $1770$ that differ by the smallest amount, those being $30\cdot 59$, since $59$ is prime. So, the fraction we are left with is $\frac{59}{30}$, and the answer to the question is $\frac{1}{18}\cdot \frac{59}{30}=\frac{59}{540}=\boxed{B}$.[/hide]


In the diagram, the two triangles shown have parallel bases. What is the ratio of the area of the smaller triangle to the area of the larger triangle? Express your answer as a common fraction.
[asy]path p = (0,0)--dir(-60)--dir(-120)--(0,0);
draw(p); draw(scale(3)*p);
label("4 cm",dir(-60)--dir(-120),S); label("10 cm",3*dir(-60)--3dir(-120),S);[/asy]

We use the theorem that states the ratio of two similar figures' areas is the square of the ratio of a pair of sides.  Thus, the ratio of the figures' areas is $ [\frac {4}{10}]^2 \equal{} [\frac {2}{5}]^2 \equal{} \boxed{\frac {4}{25}}$.


Tedadi karte ham andaze rooye yek mize mostatil shekl gharar darand be torike azlae anha ba azlae miz movazi ast. har kart ba yeku az $k$ rang,rangamizi shode.baraye har $k$ kart ba ranghaye mokhtalef,mitavan do ta az anha ra be yek sanjagh soorakh kard.sabet konid tamame karthaye ra ke daraye range moaiani hastan mitavan ba $2k-2$ sanjagh be ham vas kard.

farz kon vase k rang bargharar bashe. majmoeye k+1 ranga dar nazar begir.agar x karti bashe ke be labeye miz nazdiktare kartaee ke ba x eshterak darand ya paeen chape x a shamel mishan ya paeen rasta.
pas mishe ba do sanjagh be x vasl kard.hala tamame kartaee ra ke rangeshoon ba x motefavete va ba x eshterak nadaran ra dar nazar begir berahati mishe did ke tamame sharayete farze esteghra ra dare.pas bare har range delkha az in majmooe mishe kartasha ba 2k-2 sanjagh be ham vasl kard .do sanjagham ke ghablan dashtim .pas hale dige :wink:


[quote="Art of Owna"]I just saw this question somewhere.  I jsut want to clear up something.  ON a question if they ask like how many number less than 1000 are divisible by 2 or 3, that also is including the numbers that are divisible by 2 and 3 right?[/quote]

Yes. See [url]http://en.wikipedia.org/wiki/Logical_disjunction[/url]

yes.

in some problems, they will say find the number of integers divisible for 2 or 3, BUT NOT BOTH. then multiples of 6 wouldn't count.

ya thanks
that is what I thought (I actually got this question frm udayans post)
but someone said it didnt count so I wanted to be sure

But thanks guys.


[quote="Allnames"]Find $ x$ if $ lim_{n\to \infty} \frac {n^{1999}}{n^x \minus{} (n \minus{} 1)^x} \equal{} \frac {1}{2000}$[/quote]
 We have : $ \lim\limits_{n\to \infty} \frac {n^{1999}}{n^x \minus{} (n \minus{} 1)^x} \equal{}\lim\limits_{n\to \infty} \frac { n^{1999}}{ C^1_x n^{x \minus{} 1}}$Thus:$ x \equal{} 2000$.


function
integration
calculus
trigonometry
real analysis
real analysis unsolved

Can someone help me show that this function is even?
\[ \Phi(t) = \sum_{n = 1}^{\infty}\left\{2n^4\pi^2 e^{9t/2} - 3n^2 \pi e^{5t/2}\right\}\exp( - n^2 \pi e^{2t})
\]
Let's see, what have I tried . . . it's not obvious that $ \Phi(t) = \Phi( - t)$ and some of the terms are negative for negative $ t$, for example $ t = - 1$.  Also, if I let $ a = 2/3\pi$, then I can write it as:
\[ \sum_{n = 1}^{\infty}3n^2\pi e^{5/2t}\bigg(an^2e^{2t} - 1\bigg)e^{ - n^2\pi e^{2t}
}\]
I know . . .that's a pathetic attempt so far.  It's also positive and rapidly decreasing.  Any ideas?

Thanks,
Shaw

[edit] References to this function say see Titchmarsh which I've checked but he doesn't explain how to show it's even.

Wait a minute . . . just thought of something:  Maybe they do this:
Since:
\[ \Xi(t)\equal{}2\int_{0}^{\infty} \Phi(x)\cos(xt)dx\]

and it is known that $ \Xi(t)\equal{}\Xi(\minus{}t)$ then $ \Phi(x)$ must be an even function.  Is that all there is to it? Is there another way?

I'm sorry, probably I'm saying something very stupid because I have no paper with me right now, but I think that a good start would be to use Jacobi's identity for the theta function (the one used for instance to extend the zeta function) and to differentiate one and two times the respective identity.

Ok thanks.  I'll look at it.  I might be stupid for saying this, but I don't see how it necessarily follows that $ \Phi$ in the integral expression above is an even function if it's known that $ \Xi$ is even since we're only integrating from $ 0$ to $ \infty$.  I need to work with it I guess.

I got it.  Here it is for the record and if others are interested:
\begin{align*}
\int_0^{\infty}\Xi(t)\cos(xt)dt&=2\pi^2\sum_{n = 1}^{\infty}\left\{2n^4\pi^2 e^{-9x/2} - 3n^2 \pi e^{-5x/2}\right\}\exp( - n^2 \pi e^{-2x})\\
&=\pi^2\Phi(-x)\\
&=\int_0^{\infty}\Xi(t)\cos(-xt)dt\\
&=\pi^2\Phi(x)
\end{align*}
Thus $ \Phi$ is an even function.  It's in Titchmarsh's book on the zeta function, just not all in one place.


If:
    1      2
X --> Y ---> Z

and X is cyclopent-2-enone

1: ozonolysis
2: DMS

What are Y and Z

Ozonolysis followed by reduction is an oxidative cleavage of a double (or triple) bond to two keto groups at the double bond. The intermediate is a molozonide (?, don't remember completely?), which I think is cool.

The mechanism is also really fun too.

oh $ \$$#@!

haha whoops i drew the first unstable intermediate.

Y should be the ozonide, again with 3 oxygens.

Ok, now that's right.

You know, ozonolysis reactions can be very interesting (by the way, their mechanism is also very nice, since it goes through two consecutive 1,3-dipolar cycloadditions - Criegee mechanism). To give you a taste of what can happen, here are two exercises:

What about the second step with adding DMS to the ozonolysis product?

What about him? That step in only used when we want to hydrolize the 1,2,4-trioxolane into two carbonyl products. Sometimes we don't want to do that, and other times that's not needed (as in my exercises above).


integration
calculus
calculus computations

Let $ x_1,\ x_2$ be the solutions such that $ \{ \begin{array}{ll} \frac {dx_1}{dt} = - 3x_1 + x_2 & \quad \\
\frac {dx_2}{dt} = x_1 - 3x_2 + e^{ - t} & \quad \end{array} $ and $ x_1(0) = x_2(0) = 0$.

Evaluate $ \int_0^{\infty} (x_1 + x_2)\ dt$.

hello, solving your system we get
$ x_1(t)\equal{}\frac{1}{3}e^{\minus{}t}\minus{}\frac{1}{2}e^{\minus{}4t}C_1\plus{}e^{\minus{}2t}C_2$
$ x_2(t)\equal{}\frac{2}{3}e^{\minus{}t}\plus{}\frac{1}{2}e^{\minus{}4t}C_1\plus{}e^{\minus{}2t}C_2$
Inserting $ x_1(0)\equal{}0$ and $ x_2(0)\equal{}0$ we have $ C_1\equal{}\minus{}\frac{1}{3}$ and $ C_2\equal{}\minus{}\frac{1}{2}$.
So we have $ \int_{0}^{\infty}x_1(t)\plus{}x_2(t)\,dt\equal{}\int_{0}^{\infty}e^{\minus{}t}\,dt\equal{}1$.
Sonnhard.

hello, ohh, i think i have made a typo, it must be
$ \int_{0}^{\infty}x_1(t)\plus{}x_2(t)\,dt\equal{}\int_{0}^{\infty}e^{\minus{}2t}(\minus{}1\plus{}e^t)\,dt\equal{}\frac{1}{2}$.
Sonnhard.

That's correct. :)

Here is my solution:

Adding bothe sides gives $ (x_1 \plus{} x_2)' \equal{} \minus{} 2(x_1 \plus{} x_2) \plus{} e^{ \minus{} t}$.

$ \therefore x_1 \plus{} x_2 \equal{} e^{ \minus{} 2t}\left(\int e^{ \minus{} t}e^{2t}\ dt \plus{} C\right) \equal{} e^{ \minus{} t} \plus{} Ce^{ \minus{} 2t}$.

By initial condition: $ x_1(0) \plus{} x_2(0) \equal{} 0$, yielding $ C \equal{} \minus{} 1$.

$ \therefore \int_0^{\infty} (x_1 \plus{} x_2)\ dt \equal{} \int_0^{\infty} (e^{ \minus{} t} \minus{} e^{ \minus{} 2t})\ dt \equal{} \frac {1}{2}$.


Consider the six digit number: $ 444abc$

If the above number with $ a$, $ b$ and $ c$ being three non zero digits, what can be their value such that the six digit number is the minimum that is divisible by both $ 13$ and $ 17$?

[hide="With a calculator"]
Division shows that 443989 = 2009(!!!) * 13 * 17 and 444210 = 2010(!!!) * 13 * 17. Unfortunately there is a zero digit, so we add 13 * 17 again, yielding 444431, the minimum. a = 4, b = 3, c = 1.
[/hide]

If it were something like $ 4a4b4c$ you couldn't do this and would have to resort to moduli, but as the problem is, just divide, guess, and check is enough.

[hide] Clearly, since the number is divisible by both $ 13$ and $ 17$, it must be divisible by $ 13(17)\equal{}221$. since $ 221(2)\equal{}442$, we know that $ 442000$ is a multiple of $ 221$. In the same way, $ 2210$ is a multiple of $ 221$. Since they are both multiples of $ 221$, when we add them together, the result will also be a multiple of $ 221$. adding up, $ 442000\plus{}2210\equal{}444210$. However, we want our last $ 3$ digits to be non-zero, so we add $ 221$ to the above number and get $ 444210\plus{}221\equal{}444431$. Thus, $ a\equal{}4$, $ b\equal{}3$, and $ c\equal{}1$ are the minimum values.[/hide]


It is given string $S = s_1 s_2 \dots s_n$ such that (1<n<100000) and $s_i \in ['a', 'z']$. Given such a string matrix is formed from rotated vrsions of the string \[ s_1 s_2 \dots s_ {n-1} s_n \] \[ s_2 s_3 \dots s_n s_1 \] \[ \dots \] \[ s_n s_1 \dots s_{n-2} s_{n-1} \] Then the rows of the matrix are sorted in alphabetic order. Task is to write a program which reads the last column of the sorted matrix and writes the k-th row of the sorted matrix.

This is the Burrows-Wheeler transform.

http://en.wikipedia.org/wiki/Burrows-Wheeler_transform

I wrote an implementation once before in php that assumes extra information is known about which character is $s_n$.  The problem seems harder if this cannot be assumed.


If you see a function, how do you tell just by the look of it whether it is even or odd or neither?

Also, are there specific ways you can say, "Oh, let's write an odd one." and write an odd one on the first try?

I was wondering about this, because it seems I have to make a random equation and graph them to see. I know like $ f(x) = x+7$ would be neither, etc. but for others I don't know.

$ f: A\to R$ where $ A$ is a symetric set is:

- even if $ f(-x)=f(x),\forall x\in A$
- odd if $ f(-x)=-f(x),\forall x\in A$

cool. That was like in the previous example, but I didn't realize that was how you differentiated.

and just by looking at it...

Even functions are symmetric about the y-axis.

Odd functions are symmetric about the origin (in other words, if you take a point on the function and make a line from the point through the origin, you will hit another point on the opposite side of the origin that cooresponds to the point you initially started with... okay maybe that was even more confusing).

[quote="now a ranger"]Also, are there specific ways you can say, "Oh, let's write an odd one." and write an odd one on the first try?[/quote]

Given any function $ f(x)$:

- The function $ e(x) = f(x)+f(-x)$ is always even because $ e(-x) = f(-x)+f(x) = e(x)$.

- The function $ o(x) = f(x)-f(-x)$ is always odd because $ o(-x) = f(-x)-f(x) =-o(x)$.

More elementary ways:

- Any polynomial with only terms of even degree, such as $ x^{2}$, is even.

- Any polynomial with only terms of odd degree, such as $ x^{3}$, is odd.

[quote="t0rajir0u"]Any polynomial with only terms of even degree (and no constant term), such as $ x^{2}$, is even.[/quote]
Good point, but you can remove the restriction on "no constant term".


[quote="MCrawford"]Find the value of $x$ that makes the following statement true:
\[ 3x - 17 = 7. \][/quote]

[hide]3x=24

x=8[/hide]

[quote="MCrawford"]Find the value of $x$ that makes the following statement true:
\[ 3x - 17 = 7. \][/quote]

[hide]$3x-17=7$

$3x=24$

$x=8$[/hide]

[quote="MCrawford"]Find the value of $x$ that makes the following statement true:
\[ 3x - 17 = 7. \][/quote]

[hide="Answer"]$3x-17=7$=
$8$[/hide]

[quote="shinwoo"]math92, and jli

please avoid posting answers and solutions at Classroom Math forum because these are for like new middle schoolers, while you guys have been here long time and are more skilled in problem solvings than new peeps ;)[/quote]

If you don't post answers and solutions, what do you post?  ;)

[hide]Add both sides by 17 to obtain 3x=24
Divide by three
x=8[/hide]

[quote]
PostPosted: Sun Jul 24, 2005 9:41 am    Post subject:
shinwoo wrote:
math92, and jli

please avoid posting answers and solutions at Classroom Math forum because these are for like new middle schoolers, while you guys have been here long time and are more skilled in problem solvings than new peeps Wink

If you don't post answers and solutions, what do you post? Wink
[/quote]

I'm saying that you guys are far more advanced than classroom math peeps..

so give them some break

hwenterprise, you realized you probably made the strangest mistake of everyone?
24/3 = 6???

[quote="shinwoo"]math92, and jli

please avoid posting answers and solutions at Classroom Math forum because these are for like new middle schoolers, while you guys have been here long time and are more skilled in problem solvings than new peeps ;)[/quote]

  Just because they have been here for a long time, and they have a high rank, does not nessecarily mean they are above the classroom level.  Nevertheless, I think everyone is enitled to post solutions in any forum appropriate their age.

i agree with G-UNIT
i am still in middle school and classroom math is my level.  And you(shinwoo) said that this is for new peeps, but there are only about 10 new peeps in the classroom forum.  Do you just want me to post problems? :)

[quote=math92 wrote] Do you just want me to post problems? Smile[/quote]

actually that's a good idea

I've seen you answering problems as difficult as Getting Started and above..

so,,, what does that tell you?

same goes with

236factorial, jili, and G-UNIT


An automobile insurance company has compiled data from a survey of 1000 16-year-old
drivers during the year 2003. According to the results below, what percent of them
have had at least two accidents? Express your answer to the nearest tenth.
\begin{tabular}{ c  c}
   Total $ \#$ of Accidents & $ \#$ of 16-Year-Old Drivers \\ \hline
   0&124\\
   1&234\\
   2&346\\
   3&176\\
   4&120\\
\end{tabular}

$ 346 \plus{} 176 \plus{} 120 \equal{} 642$ had at least 2 accidents, and there were 1000 kids in total, so the answer is $ 64.2\%$.

$ \begin{tabular}{ c c} Total\#of Accidents & \#of 16 \minus{} Year \minus{} Old Drivers \\
\hline 0 & 124 \\
1 & 234 \\
2 & 346 \\
3 & 176 \\
4 & 120\end{tabular}$

EDIT: Uh ernie? ANd you accuse [i]me[/i] of spamming?


arithmetic sequence
algebra open
algebra

Show that if a, b, c are positive numbers such that a^2, b^2, c^2 are in arithmetic progression, then

1/(b+c), 1/(c+a), 1/(a+b)

are in arithmetic progression.

simple!
just do it reverse....start taking the second statement as true....solving that we will get first statement right..which proves our assumption of taking second statement right...

so 2.... 2/a+c = 1/b+c + 1/a+b...

           2/a+c =  (2b+a+c)  /  {ac+b(a+c)+b^2}

           2ac + 2ba +2bc +2b^2  =  2ab +2bc + 2ac + a^2 + c^2......

           2b^2    =  a^2  +  c^2 ........

hence proved...


[quote="Treething"]In Happy Joy Funland, the roller coaster Happiness has an average speed of 600 mph, and the ride is two minutes long. How long is the track?

The Shark-go-Round revolves at 15 rpm and has a radius of 40 feet. How much faster does a person on a shark on the outside of the Shark-go-Round move compared to a person halfway between the inside and the outside of the Shark-go-Round? (in mph)[/quote]

[hide="1"]
$\frac{600}{60}=10\\
10\cdot2=20$
miles long[/hide]

[hide]600 mph divided by an hour (or 60 minutes) would give you the distance (10 miles) for the roller coaster if it was 1 minute long. In this case, it is 2 minutes long, so  2(10 miles) = 20 miles long[/hide]

[hide="1"]20 miles. You go 10 miles in 1 minute. 600/60. You are on the ride for 2 minutes. 2*10=20 miles[/hide]
[hide="2"]Is it 204 6/11 mph faster?I found that you would go (ignoring pi) 1,440,000 feet in 1 hour on the outer loop, and 360.000 feet in one hour on the smaller loop. The difference is 1,080,000 feet. Changing that into miles, that's how I got the answer mph faster. [/hide]


Hi, I was working on some stoichiometry problems from an old chemistry book.  It got kind of tricky once I got to the later problems.  I was wondering if anyone here could please give me a hand.

Question: Antimony is alloyed with lead to increase the rigidity of components used in the construction of lead storage batteries.  $3.936g$ of a particular metallic alloy, compounded of only $Pb$ and $Sb$, can be quantitatively converted into a $4.577g$ mixture of the oxides $PbO_{2}$ and $Sb_{2}O_{4}$.  What was the percentage (by mass) of antimony in the alloy?

How can I approach this, by a balanced chemical equation?? I would greatly appreciate help with this.  Thanks again!

Of course, as in ANY stoichiometry, you need a balanced chemical equation.  This should not be that hard to make.  Once you have that, let the number of mols of Sb be $x$ and the number of mols of Pb be $y.$  Then you should be able to write a linear equation in $x,y$ that equals the total mass of Pb and Sb, 3.936g (hint: [hide]use the molar masses of these compounds[/hide]).

Similarly, based on $x$ and $y$ you should be able to find out the number of mols of each product; and from this you can write another linear equation.  After you're done with this, you have two equations, two variables, and can solve.

Hi gregx007, thanks for replying to my question.  I think that's what I'm stuck on getting the balanced chemical equation.  I was hoping if you can get me started by telling me what that equation would be.
After that, I think I can take it from there.  Thanks again.

There is no ONE balanced equation here, there are two, separate equations. One will tell you how much oxide was made from [b]y[/b] grams of lead, second - how much oxide was made from the [b]x[/b] grams of antimony. This, plus what gregx007 wrote.

Check out this [url=http://www.chembuddy.com/?left=balancing-stoichiometry-questions&right=stoichiometry-q7]stoichiometry calculations example[/url] - while it is slightly different question, it uses exactly the same idea to find out the answer.

Hi Borek and gregx007!   Thanks alot for the help you guys, I managed to solve it with your help.
Basically:
Let $x$ be the moles of lead, and $y$ be the moles of antimony.
From two chemical equations we have:
$Pb+O_{2}\rightarrow PbO_{2}$
$2Sb+2O_{2}\rightarrow Sb_{2}O_{4}$, thus $\frac{y}{2}=$ moles of $Sb_{2}O_{4}$
From the given information:
$207.2x+121.75y=3.936$ (1)
and
$239.2x+\frac{307.5}{2}y=4.577$ (2)
Solve for x and y.  The final answer is $\frac{121.75y}{3.936}\times 100\%$


I don't think you should be so harsh against Bush here. He has kept a good economy, despite the increase in unemployment. However, comparatively, I'd say the unemployment is only average. Besides, I also think a few things in the Middle East are out of his control.

Don't get me wrong though. The war in Iraq was clearly a mistake on his part.

Well said, but you have to realize that times change. With Iraq as a major event in which Bush spent much of the budget on, and with other factors, you may realize that maintaining a good economy would be difficult for anybody (with in the past year or so).

I think this country is just way too over-patriotic. We should all analyze the factors of a  certain decision before making it.

You say the deficit is at 400 billion.

Questions:
Where did all the money go?
Were any of the expenditures avoidable?
Did he spend money to actually help Americans?
What led to this depletion of money?
Why did Bush spend money the way he did?
etc.

I think just throwing data out won't convince if you haven't shown analysis.

Bush is the first president in the history of our country, and perhaps the first leader in the modern history of the world, to endorse tax cuts during a war.  That's where all the money has gone.  Meanwhile, he has slashed veteran's benefits, cut funding for daycare for poor families, and not funded No Child Left Behind, and in general forced states into a terrible economic situation above and beyond the problems caused by a depression.

Do I sense positive criticism... or an angry mob refuting an argument? Hmm. The latter.

I just want to know - is there anything that was done correctly during his term(that doesn't have a drastic downside)? Do people really despise Bush that much - if so - why do people support him? He is really liable to win the next election.

[quote="Rep123max"]The economy is terrible. Heres some comparison (I got these stats from bls.org). When Clinton inherited the presidency, unemployment was about 7.1%. He brought it down to a record low 3.8%. When Bush inherited the presidency (which he called a recession), it was 4.1%. It rose steadily to I believe 6.2% and is now 5.7%. He will be the first president since Harry Truman to have a net lose of jobs in his presidency. Then, when he made 308000 jobs last month which I will admit is good, and the Republicans boasted that it was the highest in 4 years which was part of Clinton's administration. The reason it was the highest was Clinton's unemployment was so low, he couldnt make any more jobs. Clinton made a total of about 2.5 million jobs compared to Bushs negatives.[/quote]

Granted, Clinton was president during the big dot-com boom, when the economy was great.  Bush became president during a time when the economy tanked, stocks were crashing, and so on.  So it's not really fair to compare America's economy under Clinton vs. under Bush, because the economic situation wasn't due to them.

Is it fair to point out that every single year Bush has been President, his economic council of advisors or whatnot have published analyses of the economy stating we would begin adding jobs immediately ... and been wrong by hundreds of thousands of jobs every single time?

First, I think the Bush is bad in foreign policy, the economy and just about everything else.  Still it's quite likely that 9-11 would have happen whoever was in office, it's quite possible that if someone else had been in office we wouldn't have alienated most of our closest allies, the economy isn't Bushes fault after all he does have the legeslature to contend wih(ohh wait they're on his side). Anyhow, WTF do I know, nothing.

[quote="Rep123max"]The economy is terrible. Heres some comparison (I got these stats from bls.org). When Clinton inherited the presidency, unemployment was about 7.1%. He brought it down to a record low 3.8%. When Bush inherited the presidency (which he called a recession), it was 4.1%. It rose steadily to I believe 6.2% and is now 5.7%. He will be the first president since Harry Truman to have a net lose of jobs in his presidency. Then, when he made 308000 jobs last month which I will admit is good, and the Republicans boasted that it was the highest in 4 years which was part of Clinton's administration. The reason it was the highest was Clinton's unemployment was so low, he couldnt make any more jobs. Clinton made a total of about 2.5 million jobs compared to Bushs negatives.[/quote]
It's generally accepted that wages and prices are sticky, and that economies are not perfectly competitive so they will not react quickly to return to a level of full employment and the long run. Many of the economic problems you attribute to Bush were the legacy of Clinton years, and many of the economic benefits you attribute to Clinton were a legacy of Bush Sr. You also fail to look at what Clinton did to the military - those cruise missiles needed to be replaced, and he didn't do it during his presidency, he just fired them at asprin factories. Bush was faced with a horrendous national disaster shortly after becoming president which wreaked havoc on the economy, destroying consumer faith, which takes time to build. You might say Bush should have prevented that disaster - BS. They had information saying "there's going to be something big." How does one prepare for that? There's nothing that can be done with that sort of intel. Clinton had several opportunities to have Bin Laden arrested during his presidency - if we knew what threat Bin Laden posed, wouldn't Clinton have had him arrested rather than stepping around the topic carefully making sure not to hurt anybody's feelings?

Get off your high Bush-bashing horse and do some research. [that applies to those regurgitating their parent's views, their neighbors views, or the drivel espoused by today's media, without having really looked into things]

[quote="Rep123max"]What I think is sad is that many Americans feel that an Iraqi life is not as important as an American life. When somebody says 400 Iraqi's died in one week, Americans dont care and thats the kind of attitude that makes the terrorists hate us.[/quote]

Does that give terrorists the right to crash planes into American buildings, or the right to mutilate American bodies and drag them through the streets acting as if the killing is a festival?

There's a reason why Americans care more about their countrymen than they do about Iraqis (aside from the fact that they [i]are[/i] our countrymen): because the Iraqis are the ones mutilating our dead and dancing for joy, whereas we're the ones who at least give terrorists a fair trial.

I'd like to appolgize to anyone I offended in my last post [hide]NOT![/hide] there's a reason sarcasam is called the  dark form of humor.  Second Quote:whereas we're the ones who at least give terrorists a fair trial. sorry this is the admistration that enacted the PATROIT (Uniting and Strengthening America by Providing Appropriate Tools Required to Intercept and Obstruct Terrorism) act.  this gives the FBI the privlige (read right) to look at a plethora of records Quote:`SEC. 501. ACCESS TO CERTAIN BUSINESS RECORDS FOR FOREIGN INTELLIGENCE AND INTERNATIONAL TERRORISM INVESTIGATIONS.
`(a)(1) The Director of the Federal Bureau of Investigation or a designee of the Director (whose rank shall be no lower than Assistant Special Agent in Charge) may make an application for an order requiring the production of any tangible things (including books, records, papers, documents, and other items) for an investigation to protect against international terrorism or clandestine intelligence activities, provided that such investigation of a United States person is not conducted solely upon the basis of activities protected by the first amendment to the Constitution.

`(2) An investigation conducted under this section shall--

`(A) be conducted under guidelines approved by the Attorney General under Executive Order 12333 (or a successor order); and
`(B) not be conducted of a United States person solely upon the basis of activities protected by the first amendment to the Constitution of the United States.
`(b) Each application under this section--

`(1) shall be made to--
`(A) a judge of the court established by section 103(a); or
`(B) a United States Magistrate Judge under chapter 43 of title 28, United States Code, who is publicly designated by the Chief Justice of the United States to have the power to hear applications and grant orders for the production of tangible things under this section on behalf of a judge of that court; and
`(2) shall specify that the records concerned are sought for an authorized investigation conducted in accordance with subsection (a)(2) to protect against international terrorism or clandestine intelligence activities.
`(c)(1) Upon an application made pursuant to this section, the judge shall enter an ex parte order as requested, or as modified, approving the release of records if the judge finds that the application meets the requirements of this section.

`(2) An order under this subsection shall not disclose that it is issued for purposes of an investigation described in subsection (a).

`(d) No person shall disclose to any other person (other than those persons necessary to produce the tangible things under this section) that the Federal Bureau of Investigation has sought or obtained tangible things under this section.

`(e) A person who, in good faith, produces tangible things under an order pursuant to this section shall not be liable to any other person for such production. Such production shall not be deemed to constitute a waiver of any privilege in any other proceeding or context.

First the court listed in Part 1B was established by FISA, take a guess at how many times they've rejected a petion by the FBI, that's right, not once.

Look at section 2, D and E this basically says that first the FBI CANOT say that they're invesegating under this act and if you get served a warrant under this act you are required by LAW to say it never happened.

There are quite a few disturbing things in Sec 2 of the USA PATRIOT ACT. this is just the most famous (and distressing).

Can you imagine how outraged the Iraqis would be if we burned, mutilated, and dragged Iraqi bodies through the streets of New York or Washington?  Their outrage would be justified, too.  So why aren't we justified in being angry that they've done that to us?

[quote="mathfanatic"]Can you imagine how outraged the Iraqis would be if we up and killed thousands of their civilians by crashing planes into their buildings, or by burning, mutilating, and dragging their bodies through the streets?  Their outrage would be justified, too.  So why aren't we justified in being angry that they've done that to us?[/quote]

When did Iraqis kill thousands of US civilians?  It is the United States that has killed more than 10,000 Iraqi civilians.  How many deaths does it take for Iraqi outrage to become justified?

[quote="Ravi B"]
When did Iraqis kill thousands of US civilians?  It is the United States that has killed more than 10,000 Iraqi civilians.  How many deaths does it take for Iraqi outrage to become justified?[/quote]

Well, considering that Saddam Hussein murdered roughly 1 million Iraqi civilians (60,000 alone from Baghdad), I would say that they shouldn't be outraged.

[quote="mathfanatic"]Can you imagine how outraged the Iraqis would be if we up and killed thousands of their civilians by crashing planes into their buildings, or by burning, mutilating, and dragging their bodies through the streets?  Their outrage would be justified, too.  So why aren't we justified in being angry that they've done that to us?[/quote]

Um, this is totally irrelevant.  The only American's impacted by Iraqi's are American soldiers fighting in another country.  That makes your comparison invalid on two grounds: (1) civilian deaths do not equate to military deaths, and (2) it's [i]our[/i] army in [i]their[/i] country.  If they were killing our soldiers [i]here[/i], then you would have a point.  As is, you have nothing.

[quote="zscool"]Well, considering that Saddam Hussein murdered roughly 1 million Iraqi civilians (60,000 alone from Baghdad), I would say that they shouldn't be outraged.[/quote]

Notably, we haven't had the time to do that much damage, so it's not a very good comparison.  I'd be interested to see the deaths-per-time for both the US and Saddam.  That little statistical point aside, what you are saying still doesn't make any sense.  It's the Hitler-Stalin argument: is Hitler less bad because Stalin also killed millions of people?  No -- Hitler was still bad.  Is America less bad for killing Iraqi civilians because Saddam did?  No -- it's still bad to kill civilians.
(Please note, for the record, that I have no desire to equate the US government with either Stalin or Hitler.  Any objections to my comments based on the grounds "Bush is nowhere near as bad as Stalin was, so you're wrong" will be proptly deleted in a massive abuse of moderator power.)

[quote="confuted"]They had information saying "there's going to be something big." How does one prepare for that? There's nothing that can be done with that sort of intel. Clinton had several opportunities to have Bin Laden arrested during his presidency - if we knew what threat Bin Laden posed, wouldn't Clinton have had him arrested rather than stepping around the topic carefully making sure not to hurt anybody's feelings?[/quote]

Well, here's the thing -- that's not what the information said.  It said, "Bin Laden Determined to Strike within the U.S."  Not only that, but it specifically pointed out a desire on his part to hijack planes.  Now, I don't know about you, but if I were President, and I heard that a proven terrorist group presenting a real threat was interested in hijacking planes, here's what I would not do:  I would not do nothing.  I would not wait for my intelligence services to send me the exact names of the hijackers, their current locations, and what exactly their mission plan was.  Why?  Because that's dumb -- because there is never intelligence like that.  Bush had intelligence, which may not have been perfectly explicit, but was certainly "actionable."  And he didn't action.

As for the Clinton administration, they worked very hard at putting bin Laden to justice.  I do not believe you are correct in saying there were several times they could have caught him -- I know there was one instance in which he was spotted by a Predator drone, but there was not enough time to actually send in anyone to get him before he was gone.  (By the way, I'd be interested to know what the Republican congress would have been shouting about if he [i]had[/i] done that.)  Anyhow, Clinton was far more serious about finding out about bin Laden and shutting him down then Bush was -- but the Bush administration ignored all of the intelligence on the Cole and on bin Laden that Clinton handed over.

[quote="confuted"]Get off your high Bush-bashing horse and do some research. [that applies to those regurgitating [...] the drivel espoused by today's media, without having really looked into things][/quote]

That last bit is a little surprising for you to say, given that the mainstream media consistently parrotted the Bush administration in calling for war.  There was fundamentally no anti-war sentiment in the major media even though lots and lots of people were pointing out that our inteligence was bad and that we shouldn't go to war.  (Remember hundreds of thousands of anti-war marchers being labelled a "focus group?")

Furthermore, most people here are in highschool or below, as are you, and they have the dignity not to insult your arguments as "regurgitating," so I would appreciate it if you would express the same politeness to them.

By the by, for those "Clinton did nothing" believers out there, there is a graphic towards the bottom of [url=http://www.msnbc.msn.com/id/4730946/]this[/url] page which should make clear that that was not at all the case.  You need Flash Player to view it, however.

Over the past year or so, I've really looked into these debates. I won't criticize the debates, because everyone has a justified view, but I have to point out one thing. Every discussion on economy, politics, wars, etc... all lead to one inevitable argument about the Bush administration and the war in Iraq.

It is just like on another website, where all discussions that get fired up leads to a comparison between Jews and Nazis.

Like confuted said - do more research than basic facts and figures - there is so much more to account for when dealing with the rather contraversial subjects of war and economy. Factors leading to the events - other influencing events during the same time, etc. Lay off Bush until you can make a very accurate compare and contrast with him and Clinton.

How would you feel if there were terroists in the U.S. and another country says I will spend 200 billion and send in trooops and annihilate these terroists in the U.S.? I will GUARANTEE that you will be scared to DEATH, and would be praying that the foreingers leave. REalizing how more dangerous LIFE has bcome and how basic necesities have dropped. The thing is the U.S. doesn't give a shiz about other countries populations. All they care is if any U.S. person dies. Look at Vietnam War. They sprayed all sorts of  chemicals, stuff in  Vietnam. It wasn't legal. They never released that they did it during the war untill many years later. Plus there was this massacre scandal where a U.S. trooop murdered a WHOLE village. And nothing was reported untill one U.S. soldier felt decades later that this should be reported. The U.S. army is reckless. They don't care about other people's lives. Imagine you are an Iraqi. Imagine how conditions have deterioted.

The real WMD and the terroists is the U.S. THey have been historically fillibustering and messing up countries. Murdering leaders that they don't like (like the democratically elected socialist Salvador Allende). Encouraging separation of communist countries and supporting separatist groups. INCLUDING giving BIN LADEN MONEY AND AID to fight the russians in Afghanistan. MY goodness how evil can the u.s. get.

[quote="churchilljrhigh"]How would you feel if there were terroists in the U.S. and another country says I will spend 200 billion and send in trooops and annihilate these terroists in the U.S.? I will GUARANTEE that you will be scared to DEATH, and would be praying that the foreingers leave. REalizing how more dangerous LIFE has bcome and how basic necesities have dropped. The thing is the U.S. doesn't give a shiz about other countries populations. All they care is if any U.S. person dies. Look at Vietnam War. They sprayed all sorts of  chemicals, stuff in  Vietnam. It wasn't legal. They never released that they did it during the war untill many years later. Plus there was this massacre scandal where a U.S. trooop murdered a WHOLE village. And nothing was reported untill one U.S. soldier felt decades later that this should be reported. The U.S. army is reckless. They don't care about other people's lives. Imagine you are an Iraqi. Imagine how conditions have deterioted.[/quote]

Hold on...why are you mixing Vietnam and Iraq together?  The part about chemicals is about Vietnam.  The village part is referring to Vietnam.  This discussion is about Iraq.

It's not fair to talk about how I'd be feeling if American terrorists were detonating bombs in the rest of the world, and another country spent $200 billion to kill them off, because it's (currently) merely a hypothetical situation.  It's just like me asking an animal-rights activist "what if dinosaurs come back and eat your children; will you still fight for their rights?" because it hasn't happened.

But following your hypothetical situation, I will argue that yes, I will be scared, and it is possible that I will get a gun to defend myself.  But if I kill an Iraqi invader, I will not burn it, mutilate it, and show it off to my friends.  That's the difference between - and I hate to use these pronouns - us and them.

Confuted!
[quote]

But following your hypothetical situation, I will argue that yes, I will be scared, and it is possible that I will get a gun to defend myself. But if I kill an Iraqi invader, I will not burn it, mutilate it, and show it off to my friends. That's the difference between - and I hate to use these pronouns - us and them.[/quote]

No. You are expressing the difference between you and that specific Iraqi(or group). There are many people in this country who would kill and burn an invader in rage or fear. I know you would not, but there is violence in our country too - and violent people act out in violent ways. There is somebody in this country who would do what that Iraqi did - I promise. You cannot blame an entire country for what a specific individual or group of individuals has done. Us and them is "What we think the perfect American is vs. what we think the worst Iraqi is" to the majority.

Let me throw something in there: During many genosides some Muslims and Iraqis and such actually helped the survivors of the genosides. Also the three Japanese people kidnapped last week were relesased [b]thanks to a Muslim group[/b]. Just wanted to let all you folks know it's not all black and white...but y'all probably already knew that eh? :lol:

But anyway the islamic bombings and shebangs are all nothing. When you look at history. You will see the THREE GREATEST MURDERING OF HUMANS are the two evil POWERS during world WAR TWO. First the jewish. Second, the russians (1/10 of russians died during world war 2), Third, the Chinese (japanese randomlly murdered people in CHina like bezerks (over 25 million during the WW2 period plus another 15 million pre world war 2.

Here's somthing a friend sent to me:
[quote]
Bush Haiku

This is a short poem made up entirely of actual quotations from George W. Bush. These have been arranged, only for aesthetic purposes, by Washington Post writer, Richard Thompson.

MAKE THE PIE HIGHER
I think we all agree, the past is over.
This is still a dangerous world.
It's a world of madmen and uncertainty
And potential mental losses.

Rarely is the question asked
Is our children learning?
Will the highways of the Internet
Become more few?

How many hands have I shaked?
They misunderestimate me.
I am a pitbull on the pantleg of opportunity.
I know that the human being
And the fish can coexist.

Families is where our nation finds hope,
Where our wings take dream.
Put food on your family!
Knock down the tollbooth!

Vulcanize society!
Make the pie higher!
Make the pie higher!

(Pass this on. Help cure mad Cowboy disease in the next election!)
[/quote]

[quote="Ragingg"]Hah... Don't think that China is so innocent churchill. Remember that China invaded Tibet murdering thousands...[/quote]

This is a quote from Microsoft Encarta.

More demonstrations by Tibetans took place, in addition to several [b]acts of terrorism[/b] against the Chinese. In 1995 a new conflict emerged in Tibet over the selection of the next Panchen Lama.

Lets just temporarily say that I belive everything that george bush says.  We must FIGHT terroism. We must annihilate terroism no Matter what the costs are. There are Cults of Mass Destruction (RMD) in Tibet. We must put a stop to them. (You can say that I'm wrong. But it doesn't matter I want to invade Tibet. I want to liberate Tibet from terroists.

[b]China was actually first very nice to Tibet. Giving Dalai Lama powers. A constitution for the special autonomous region was devloped. Plus, China made improvemnets in communications, transportation. BUt DALAI LAMA wanted Power. and he soon consorted with the U.S. CIA (who are famous for being a nuisance to many nations).  There were several rebellions supported and plotted by Dalai Lama. When of course you do those stuff. The chinese governement obviously is going to get pissed off you. [/b]

[quote="churchilljrhigh"]China was actually first very nice to Tibet.[/quote]

I think we can agree that invading another country, exiling its leader, engaging yourself in its religious affairs, and importing thousands of your own population into their territory in order to help destroy their culture are not "nice" things to do.

Terrorism is in the eye of the beholder.  And stop using your encyclopedia to prove political points.  It's just silly.

In Washington, the White House reported a carefully worded but searing critique Bush made to Spanish Prime Minister Jose Luis Rodriguez Zapatero.

In a phone call initiated by the Spanish premier, Bush "expressed his regret to president Zapatero about the decision to abruptly announce the pullout of Spanish troops from Iraq," said spokesman Scott McClellan.

In a veiled rebuke, Bush "stressed the importance of carefully considering future actions to avoid giving false comfort to terrorists or enemies of freedom in Iraq."

"The president urged that the Spanish withdrawal take place in a coordinated manner that does not put at risk other coalition forces in Iraq," the spokesman added.

Spain's Defence Minister Jose Bono said at the same time in Madrid that the pullout was already under way however.

[b]You see Bush has one of lowest standards of etiquette. Not only does he not thank Spain for spending considerable amount of money for 1500 troops. But instead rebukes Spain. Bush is one those spoiled brats who only takes compliments and thinks he is king **** on turd island being the president of the u.s. (well its probably true that bush is the most powerful man in the world but don't paint it on your face).[/b]


inequalities
induction
inequalities proposed

n are positive integers .Prove that:
$1+\frac{2}{n}\ geq \sqrt[n]{n}$

[quote="cdsppy"]n are positive integers .Prove that:
$1+\frac{2}{n}\geq \sqrt[n]{n}$
[/quote]

[quote="cdsppy"][quote="cdsppy"]n >1 positive integers .Prove that:
$1+\frac{2}{n}\geq \sqrt[n]{n}$
[/quote][/quote]

[quote="nayel"]$2\le (1+\frac{1}{n})^{n}<e$
and
$(1+\frac{1}{n})^{n}\to e$ as $n\to \infty$
where $e=2.718281828...$

can these help? :P[/quote]
And $\left(1+\frac1n\right)^{n}$ is increasing :wink:
You can simply use the Bernouilli inequality that can be proved easily by induction:[quote] $(1+\alpha)^{n}\geqslant 1+n\alpha$[/quote]


Okay, so I happened upon the really nice identity: The constant term of \[\sum_{i=0}^{n}\left(\sum_{j=i}^{j=2n-i}x^{j}\right)^{k}\] is $(n+1)^{k}$.

For instance, this claims that the constant term of \[(x^{-2}+x^{-1}+1+x+x^{2})^{2}+(x^{-1}+1+x)^{2}+1^{2}\] is $9$.

So how to prove it in a nice way? There's plenty unnice ways.

We could, erm, induct. But that's probably one of the worst ways to prove it.  :P

Er... the summation notation you have doesn't really seem to match the example you gave... but it seems to me as if something nice might happen if you use [hide="magic"] roots of unity. [/hide]

Theres already a stupid counting and induction type solution that's algebra extensive which I based on a previous post here... but I already tried roots of unity screens and some other creative stuff, even $N$ vector algebras etc...


function
floor function
logarithms
algebra proposed
algebra
functional equation

Find all functions $f: \mathbb{R}\rightarrow \mathbb{R}$ which satisfy
\[f(0)=2f^{2}(x)-f(2x) \quad \forall x \in \mathbb R.\]

[quote="Zaratustra"]Find all functions $f: \mathbb{R}\rightarrow \mathbb{R}$ who satisfy:

$f(0)=2f^{2}(x)-f(2x)$[/quote]

This is a rather frequent problem whose general form is $f(2x)=h(f(x))$ (here $h(x)=2x^{2}-a$)

In this case, you write :
1) $f(0)=h(f(0))$, so $f(0)$ is any fixed point of h. In this special case, $f(0)=0$ or $f(0)=1$
2) for $x>0$ $f(x)=g(\frac{\ln(x)}{\ln(2)})$ with $g(x)$ such that $g(x+1)=h(g(x))$, so $g(x+p)=h^{\circ p}(g(x))$ (where $f^{\circ p}(x)$ means $f\circ f\circ f\circ \ldots \circ f(x)$ (p times)).

So you just have to choose any function $a(x)$ defined on $(0,1]$ and then $g(x)=h^{\circ\lfloor x\rfloor}(a(x-\lfloor x\rfloor)$

But CAUTION : in order $h^{\circ p}(x)$ exists for any $p<0$, the best case is $h(x)$ bijective function. When this is not the case, special attention must be made on these "negative" compositions.

3) For $x<0$ $f(x)=g(\frac{\ln(-x)}{\ln(2)})$ with $g(x)$ such that $g(x+1)=h(g(x))$ ...

So the general answer is :
Let $a(x)$ and $b(x)$ be any functions defined on $(0,1]$. Then the solutions to the equation $f(2x)=h(f(x))$ are :

1) $f(0)$ is any fixed point of $h(x)$
2) For $x>0$, $f(x)=h^{\circ\lfloor\frac{\ln(x)}{\ln(2)}\rfloor}(a(\frac{\ln(x)}{\ln(2)}-\lfloor\frac{\ln(x)}{\ln(2)}\rfloor)$
3) For $x<0$, $f(x)=h^{\circ\lfloor\frac{\ln(-x)}{\ln(2)}\rfloor}(b(\frac{\ln(-x)}{\ln(2)}-\lfloor\frac{\ln(-x)}{\ln(2)}\rfloor)$

In our special case, there are two subtle modification :
1) Let $f(0)=a$ with either $a=0$, either $a=1$ and let $h(x)= 2x^{2}-a$

We need to define $h^{\circ-1}(x)$ and I suggest $h^{\circ-1}(x)=\sqrt{\frac{x+a}{2}}$ and so we need $a(x)\geq 0$ and $b(x)\geq 0$ (for $a=1$, this choice may lead to forget some solutions, I think).

2) For $x>0$, $f(x)=h^{\circ\lfloor\frac{\ln(x)}{\ln(2)}\rfloor}(a(\frac{\ln(x)}{\ln(2)}-\lfloor\frac{\ln(x)}{\ln(2)}\rfloor)$
3) For $x<0$, $f(x)=h^{\circ\lfloor\frac{\ln(-x)}{\ln(2)}\rfloor}(b(\frac{\ln(-x)}{\ln(2)}-\lfloor\frac{\ln(-x)}{\ln(2)}\rfloor)$

If the continuity is demanded, some conditions more on $a(x)$ and $b(x)$ exist.


modular arithmetic
number theory
Diophantine equation
relatively prime

Two quick questions:

1. How can you solve for variables in a case such as 2x is congruent to 4 (mod 5). (I can solve the diophantine equation 2x=5k+4 resulting in x=5m+2, where m is an integer but is there a faster way?)

2. Is there anyway to change modulus i.e. if you know x is congruent to 3 (mod 4) what is x congruent to (mod 7).

I think the second one is note possible, except if the second base is a multiple of the first, then I don't know.

For the first one, add 5 to the right until it is a multiple of 2, like $2x \equiv 4 + 10 \equiv 14 (\bmod\ 5)$, then you divide, since 2 and 5 are relatively prime, you get $x \equiv 7 \equiv 2 (\bmod\ 5)$.

For the first one, there is another way (because with larger numbers, pkerichang's solution can get very slow). We know that $5x \equiv 5 \pmod5$ and $2x \equiv 4 \pmod5$. Subtracting the second from the first gives $3x \equiv 1 \equiv 6 \pmod5$. Assuming that this wasn't nice enough for us to divide the 3 out, continue this process, subtract the 2x from the 3x to get $x \equiv 6-4 \equiv 2 \pmod5$.

For the second, x could be congruent to anything mod 7 because all we need to do is solve the system of congruences $x \equiv 3 \pmod4$ and $x \equiv k \pmod7$ for $0 \le k \le 6$ which all have a unique solution mod 28 by Chinese Remainder Theorem.

Other cases for the second say $x \equiv 1 \pmod2$ then what is x mod 4? Well, just consider what x is. It must be odd, so it could be 1 or 3 mod 4.

However, if the two moduluses are relatively prime it could be anything.