Propose a synthesis of 2-methyl-2-pentanol using acetone as the only organic starting material.

Does it mean that the only Organic material we can use is acetone?

How long is it meant to take because my way (as usual) is abnormally long.

1. Treat acetone with $ NaBH_4 / ethanol$ (followed by acid) to cause a reduction to $ \text{propan \minus{} 2 \minus{} ol}$
2. Treat $ \text{propan \minus{} 2 \minus{} ol}$ with concentrated sulphuric acid to dehydrate it to $ \text{propene}$
3. Hydroboration ($ BH_3,THF$ followed by $ H_2O_2/OH^{ \minus{} }$) to obtain $ \text{propan \minus{} 1 \minus{} ol}$
4. Treat with $ PBr_3/ether$ to yield $ \text{1 \minus{} bromopropane}$
5. Treat $ \text{1 \minus{} bromopropane}$ using $ Mg/ether$ (Grignard Reagent) and react with acetone followed by $ H_3O^{ \plus{} }$.

If there is a better or shorter way then would you mind giving me a few hints so I can try again?

P.S. When you say the 'only organic starting material'. If I use an organic material as a solvent, is that allowed?

Your synthesis is correct, good job. Just a remark: in your last step, in the part "followed by H3O+", we should be carefull in the acid used to protonate the alkoxide, or dehydration occur instead. For tertiary alkoxides like in this case, ammonium chloride is often used.

[quote="BanishedTraitor"]If I use an organic material as a solvent, is that allowed?[/quote]

Yes.

Can you give some examples to what acids to use for protonation and what for dehydration?

For protonation I already gave an example:

[quote="Carcul"]For tertiary alkoxides like in this case, ammonium chloride is often used.[/quote]

For dehydration, sulphuric acid is often used, but others can also do the job.

[quote="VARUNARASU"]cant we use aldol condensation foll by wolff kishner reduction.??[/quote]

Possibly (I wrote the process before learning Aldo Reactions  :roll: ).

Just a note, in practice, the Aldol product is easily dehydrated to give an enone product. This is because the aldol reaction requires a base and the temperature needed for aldol dehydration is just slightly higher than aldol condensation (hence the name as $ H_2O$ is given off during the process in real life). So after Wolff-Kishner reduction we should really carry out $ \alpha\minus{}\beta$ conjugate addition with $ H_2O$.

BUT ALDOL PRODUCT OF ACETONE CAN BE OBTAINED IN VERY GOOD YIELD IF WE USE Ba(OH)2. AND CONCERNING REDUCTION OF KETO GROUP, CANT WE USE A DITHIOACETAL FOLL BY RED WITH H2,/Ni(MOZINGO METHOD).

[quote="Carcul"][quote="VARUNARASU"]cant we use aldol condensation foll by wolff kishner reduction.??[/quote]

I wouldn't go that way, since aldol condensations are not favourable for ketones (unless they are intramolecular). Also, Wolff-Kishner reduction uses basic conditions, and we have an -OH group on the molecule.[/quote]

Again, sorry about reviving an old topic. What if we used Clemmensen Reduction instead?

Chlorination of alkanes is not a selective process...

[quote="hell_ever"]I'm not aware of much adavanced organic synthesis processes. But what about this:

as already mentioned, after synthesizing propene,
---> Hydrogenate to propane.
---> React with Cl2 in sunlight to get monochloropropane
---> React with lithium di-(isopropyl) copper to get 2-methyl pentane.
---> Brominate to get 2-methyl pentan-2-ol
---> React with water to get reqd. product

Is it correct? :maybe:[/quote]
wow that's a perfect 11th calss student answer.....just learn in bits from Carcul...u surely will become very good :)

But isn't Br2 selective for 3 degree hydrogen? If not, why? If so why again (as I'm not aware  :wink: )??

Molecular bromine is more selective than chlorine, due to its lowest reactivity. That's why can use radical bromination to differentiate between tertiary, secundary, and primary carbons. Chlorine is very reactive and so is not selective.

So, there is no way to obtain 1-chloropropane as mjaor product??

starting from propane
1.use bromination
2.elimnation using alc.KOH to get propene
3.Hydroboartion oxidation to get propan-1-ol
4.then use $ \ce{SOCl2}$

@pardesi

Was that comment supposed to be sarcastic or appreciative or neither? :)

In which section can i look this up in morrison if it is there?? (the use of SOCl2)

ofcourse that was appreciative...u have to attempt in order to learn things won't enter to ur mind automatically...
read the methane+alkane chapter it has an excellent account of the free radical reaction infact the methane chapter tells u what mechanism is all about in organic...it's the best there read it  :)

i think its a bit early for you to start working on problems on synthesis(no offense meant) :D . perhaps you could try the easier questions over here, like acidity n stuff.

[quote="Carcul"]Clemmensen reduction uses strongly acidic conditions, so dehydration would probably occur. Reduction using thioacetals would be more appropriate, since it used neutral conditions.[/quote]

I can't find much information on Thioacetal Reduction. Can someone post some? Also, if possible, the mechanism of the reduction.

Thanks.

-What is it used for
-Reagents involved
-Mechanism
-Reaction condition
-Any limitations (e.g. will it be interfered by the presence of another functional group)

Thanks.

Thioketals have two very important applications:

1) Umpolung of aldehyde carbonyls (look [url=http://en.wikipedia.org/wiki/Umpolung]here[/url] for an introduction)

2) Reduction of carbonyls to alkanes

The reduction of carbonyls with thioketals can be simply made with ${ \ce{H2}/\ce{Ni}}$, which are neutral conditions. Therefore, this method is important for the reduction of a carbonyl group in a molecule having acid- and base-sensitive groups.

I thought we were talking about Thioacetal Reduction  :maybe: .

???

That's what I described above. Unless you didn't noticed that thioacetals and thioketals are two names for the same type of compounds.


abstract algebra
algebra
polynomial
function
domain
Ring Theory
superior algebra

The standard argument for classifying maximal ideals of $ R\equal{}\mathbb{Z}[x]$ is sort of long and tedious. You deal with content and factoring and all sorts of things. Now that I know about Krull dimension, I was wondering if there was a nicer way of proving the first step, "all maximal ideals of $ R$ intersect $ \mathbb{Z}$ non-trivially".

My proposed outline is something like the following. Since $ \mathbb{Z}$ commutative and Noetherian of dimension 1, $ R$ is commutative and Noetherian of dimension 2.

Let $ M$ be a maximal ideal of $ R$. If $ M\cap\mathbb{Z}\equal{}\{0\}$, then $ M\equal{}(f_1, \ldots , f_k)$ where each $ f_i$ is a polynomial of degree at least 1.

Can I somehow conclude that $ dim(R/M)$ only went down by 1, and hence is not a field, and hence $ M$ was not maximal?

This would be much slicker in my mind if it works, and its conclusion gets you all the way to $ R/M$ is a finite field.

In general, since $ \mathbb{Z}$ has infinitely many primes, any maximal ideal in $ \mathbb{Z}[x]$ intersects $ \mathbb{Z}$ non-trivially.

I don't know about a Krull dimension argument, which avoids content, etc.  A nice argument is if $ M$ is a maximal ideal of $ \mathbb{Z}[x]$ that misses $ \mathbb{Z}$, then $ \mathbb{Z}$ embeds in the field $ \mathbb{Z}[x]/M$, so $ \mathbb{Q}$ does too, implying $ \mathbb{Q}$ is a finitely generated $ \mathbb{Z}$-algebra.

Steve

Oh. That's neat. I'm not sure how that embedding implies $ \mathbb{Q}$ is a finitely generated $ \mathbb{Z}$-algebra, though.

Well, $ \mathbb{Z}$ is Noetherian, $ \mathbb{Z}[x]/M$ is a f.g. $ \mathbb{Z}$-algebra, and is also a finite extension of $ \mathbb{Q}$ (by, say, the weak Nullstellansatz). Thus $ \mathbb{Q}$ is a f.g. $ \mathbb{Z}$-algebra (see, for example, Prop. 7.8 in Atiyah-Macdonald).

Steve

THEOREM.   TFAE for D a Krull domain (e.g. a UFD or Dedekind domain)

(i)   D[x] has a max ideal M with M/\D = 0

(ii)  D is a PID with finitely many primes

See my sci.math post [1] for much more, including a proof.
These results are special cases of results at the basis of
an elegant abstract approach to the Hilbert Nullstellensatz
discovered circa 1950 by Goldman, Krull and Zariski.

--Bill Dubuque

[1] sci.math, 22 Apr 2009, -- Maximal ideals in Z[x] and Z[x, y]
http://google.com/group/sci.math/msg/b00f5e1f7397842f
http://google.com/groups?selm=y8zfxg12ckl.fsf%40nestle.csail.mit.edu

Ah. Thanks. It is really blurry to me as to what officially counts as part of the Nullstellensatz. I usually refer to that finiteness result as the Artin-Tate Lemma.


Just out of curiousity.  The poll options may seem to be overly specific :lol:.
I have a couple of observatories nearby in local universities and colleges. Though I have not been to all of them, I do frequent one of them.

Never been to one, unless SMU has one, but I have visited SLAC.

What is SLAC?

Also, who is the other person who voted? :wink:

GMRT and Giravli IUCAA telescopes near Pune, India.

I live in the middle of nowhere. It's great to be able to see the sky at night, but I have to drive three hours to get to an observatory. Also, our town just got its first Wal-Mart, and the lights spill out over the spot I used to watch meteor showers. :mad:

[quote=Junior_Hat]A school that I aspire to be in has one at its roof. It's a very functional one![/quote]

Is it Phillips Andover? ;)

[quote=CoolCarsOnTheRun]andover op
exeter op

anyways
ive been to two, one at greenwich (london, called the royal observatory), and one at LA (forgot what, but I actually looked through the telescope to see a faraway star or smt like that[/quote]

I think the one in LA that your talking about is Griffith Observatory.

[quote=Junior_Hat]A school that I aspire to be in has one at its roof. It's a very functional one![/quote]

I'm accepted in the school I talked about earlier!

P.S. I don't live in America and the school is a math and science school! (Literally, 'math and science' is in the name of the school!)

P.P.S. I'm not telling the name of the school because it'd give away my location.  :P

[quote=nkouevda]I've been to an observatory somewhere near San Jose, but I'm not sure what's it's called. I actually got to stand inside the big round white thing. :lol:[/quote]

If anyone is curious, it's called the [url=https://www.ucolick.org/main/visit/info.html]Lick University[/url] hosted by University of California (in San Jose).


we know that there are infinitely many primes. an easy proof for this is euclidean proof that:

suppose thare finite many primes and the last prime is p then

2.3.5.7.11.......p= A. So we say that A+1 is prime. My question : is A-1 prime?

If A+1 is prime, then we can write

2.3.5.7.11......p.A+1=K   ,    so K+1 is again prime. So if K-1 is again prime, we can say that A+1 and A-1 or K+1 and K-1 are twin primes, so there are infinitely many twin primes. Is this theory correct? If it's true, we can prove that twin primes are infinite. I think my mistake is K+1 and K-1 are not primes. What do you think?

Firstly, your assumption that $ A\plus{}1$ is prime is false. The proof that there are infinitely many primes is that the number $ A\plus{}1$ is either prime or it is the product of primes that have not already been seen.

Therefore, $ A\minus{}1$ is not necessarily prime.

If it were that easy, people would have figured it out a long time ago, I think.  :P

This is actually an extremely common false proof.  Incidentally, the first counter-example is $ 2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13 \plus{} 1 \equal{} 59 \cdot 509$.


$ \sum_{n \equal{} 1}^{\infty} (q^{n}cosnx)$   ;  |q|<1.   Find the sum without using complex numbers.

Why?  Any solution you find will be equivalent to the complex numbers solution, but less clear and/or less motivated.


function
combinatorics proposed
combinatorics

Given a set $A$ and a function $f: A\rightarrow A$, denote by $f_{1}(A)=f(A)$, $f_{2}(A)=f(f_{1}(A))$, $f_{3}(A)=f(f_{2}(A))$, and so on, ($f_{n}(A)=f(f_{n-1}(A))$, where the notation $f(B)$ means the set $\{ f(x) \ : \ x\in B\}$ of images of points from $B$).
Denote also by $f_{\infty}(A)=f_{1}(A)\cap f_{2}(A)\cap \ldots = \bigcap_{n\geq 1}f_{n}(A)$.

a) Show that if $A$ is finite, then $f(f_{\infty}(A))=f_{\infty}(A)$.

b) Determine if the above is true for $A=\mathbb{N}\times \mathbb{N}$ and the function
\[f\big((m,n)\big)=\begin{cases}(m+1,n) & \mbox{if }n\geq m\geq 1 \\ (0,0) & \mbox{if }m>n \\ (0,n+1) & \mbox{if }n=0. \end{cases}\]

a) We have $...\subseteq f_{i+1}(A)\subseteq f_{i}(A)\subseteq ...\subseteq A$ therefore there exists $k_{0}$ such that $f_{k_{0}}(A)=f_{k_{0}+1}(A)=...$(Note : $|A|<\infty$), and we're done!.

It is very likely there is misprint in part b). Probably, the supposed function is

$ f\big((m,n)\big)\equal{}\begin{cases}(m\plus{}1,n) & \mbox{if }n\geq m\geq 1 \\
    (0,0) & \mbox{if }m>n \\
    (0,n\plus{}1) & \mbox{if }m\equal{}0 \end{cases}$

In this case $ f_{\infty}(A)\equal{}\{(0,n) | n \in \mathbb{N} \}$ and the answer is false.

Yes, I checked and Fcheck is right, that's the correct statement


geometry
incenter
circumcircle
Euler
geometry proposed

Let $I$ be the incenter of a non-equilateral triangle $ABC$, and $D,E,F$ be the tangency points of the incircle with the sides $BC,CA,AB$, respectively. Prove that the orthocenter of $\triangle DEF$ lies on the line $OI$, where $O$ is the circumcenter of triangle $ABC$.

we prove that the euler line of $\triangle XYZ$ passes through $O$...
we can prove this using inversion with center $I$(take the incircle the circle of inversion)
then the inverse of $A$ is the intersection of $AI,EF$ which is the midpoint of $EF$...
hence the circumcircle will become the [b]9-points[/b] circle of $\triangle DEF$ so the center of these circles must be colinear with the center of inversion,so $I,O,N'$ are colinear where $N'$ is the center of the [b]9-point[/b] circle of $\triangle DEF$ so the euler line of $\triangle DEF$ passes through $O$ and we are done...
note that $A'$ is the inverse of $A$ because $AY,AZ$ are tangent to the circle and $EF$ is the polar of point $A$ so $A'$ is the inverse of $A$...


Find an expression for sin5x as a fifth-degree polynomial in the variable sinx.

You can use the general eqn for $ ({\sin x}) ^n $ as given by Demiovre's theoerm

[hide="I got"]
$16\sin^5 x-20\sin^3 x+5\sin x$

Is it correct?[/hide]

I can use Demoivre's Theoerm for this question?
Can you set it up for me?


Hi

Let $ z,a,b \in C (a\neq b)$ correspond to the points P, A, B of the Argand Plane. Let $ C_\lambda$ be the locus of P defined by $ \frac{PA}{PB}\equal{}\lambda$, where $ \lambda$ is a fixed positive real constant. Show that $ C_\lambda$ is a circle when $ \lambda\neq 1$.

Thanks

Answer of Reynaldo Gil Pons:
Let $ P'$ the point belong the segment $ AB$ such that $ \frac {AP'}{BP'}\equal{}\lambda$.
Let $ P''$ the point outside the segment $ AB$, but in the line $ AB$,  such that $ \frac {AP''}{BP''}\equal{}\lambda$.
It is easy to prove that the angle $ P'PP''\equal{}90^o$ always. Then the locus of the problem is the circumference $ C_\lambda$ with diameter $ P'P''$.

First we note that the general equation of a circle in the Argand plane is
\[ z\overline{z} \plus{} \overline{a}z \plus{} a\overline{z} \plus{} b \equal{} 0 \qquad (a\in \mathbb{C},\, b\in\mathbb{R})
\]
This circle has the center $ \minus{} a$ and the radius $ r \equal{} \sqrt {a\overline{a} \minus{} b}$
For the problem, we have been given $ \dfrac{|z \minus{} a|}{|z \minus{} b|} \equal{} \lambda$. Tranposing and squaring we obtain
\[ |z \minus{} a|^2 \equal{} \lambda^2|z \minus{} b|^2
\]
$ \Rightarrow \ |z|^2 \plus{} |a|^2 \minus{} z\overline{a} \minus{} \overline{z}a \equal{} \lambda^2 |z|^2 \plus{} \lambda^2|b|^2 \minus{} \lambda^2 z\overline{b} \minus{} \lambda^2 \,\overline{z}\,b$
$ \Rightarrow z\overline{z} \plus{} \left(\dfrac{\lambda^2}{1 \minus{} \lambda^2}\,\overline{b} \minus{} \dfrac{1}{1 \minus{} \lambda^2}\,\overline{a}\right)z \plus{} \left(\dfrac{\lambda^2}{1 \minus{} \lambda^2}\,b \minus{} \dfrac{1}{1 \minus{} \lambda^2}\,a\right)\overline{z} \plus{} \dfrac{|a|^2 \minus{} \lambda^2|b|^2}{1 \minus{} \lambda^2} \equal{} 0$
which is a circle with center at the complex point
\[ \dfrac{1}{1 \minus{} \lambda^2}\,a \minus{} \dfrac{\lambda^2}{1 \minus{} \lambda^2}\,b
\]
and the radius
\[ \left|\dfrac{\lambda}{1 \minus{} \lambda^2}\right| |a \minus{} b|
\]

This circle is called the Apollonius circle of the segment AB. You can use the angle bisector theorem to prove gilcu3's statement (try it!)


limit
real analysis
real analysis solved

Thank you for your reply, amfulger. :)

I managed to solve this somehow or other as follows.

$\sum_{n=1}^{\infty} \frac{n^2-2}{n!}=\lim_{n\to\infty} \sum_{k=1}^n \frac{k^2-2}{k!}$

$=\lim_{n\to\infty} \sum_{k=1}^n \left\{\frac{k+1}{(k-1)!}-\frac{k+2}{k!}\right\}=\frac{2}{0!}-\frac{n+2}{n!}=2$

kunny


Build a bijective function $f: [0;1]\to \mathbb R$.

It's really well-known.

Note that $g(x) = \frac1{x-\frac12}-2$ is a bijection from $\left( \frac12, 1 \right]$ to $\left[ 0, \infty \right)$.

Also, $h(x) = \frac1{x-\frac12}+2$ is a bijection from $\left[ 0, \frac12 \right)$ to $\left(-\infty, 0 \right]$. Define $a_{n}= h^{-1}\left(-\frac1{n}\right)$, for all integers $n \geq 1$.

We define $f$ as follows:

- $f(x) = g(x)$ if $x > \frac12$;
- $f(x) = h(x)$ if $x \in \left[ 0, \frac12 \right) \setminus \left\{ 0, a_{1}, a_{2}, a_{3}, \ldots \right\}$;
- $f(x) =-1$ if $x = \frac12$;
- $f(x) =-\frac12$ if $x = 0$;
- $f \left( a_{n}\right) =-\frac1{n+2}$ for all $n \geq 1$.

[quote="Zamfirmihai"]Build a bijective function $f, f: [0;1]\to \mathbb R$[/quote]

I have a rather similar solution.
Since it is obvious to have bijection between $(0,1)$ $\leftrightarrow$ $\mathbb{R}$, the problem is to find a bijective function between $[0,1]$ and $(0,1)$. I suggest the following one :

Let $A=\{2^{-n}\forall n \in \mathbb{N}\cup\{0\}\}$
$f(0)=\frac{1}{2}$
$f(x)=\frac{x}{4}$  ${\forall x\in A}$
$f(x)=x$ everywhere else (in $(0,1]-A$)

--
Patrick


calculus
integration
function
trigonometry
Galois Theory
real analysis
real analysis unsolved

I am trying to find the integral of (1/((1 + (x^4))^.5)) between 0 and 1 but as of now have had no luck. Does anyone have a suggestion?

The function $ \frac{1}{\sqrt{1\plus{}x^4}}$ doesn't have an elementary antiderivative. For the particular definite integral, there might be some tricky substitution that ends up giving something related to the original integral. (For example, the substitution $ u\equal{}\frac{\pi}{2}\minus{}x$ on $ \int_0^\frac{\pi}{2} \frac{1}{1\plus{}\tan ^3 x}\ dx$.) For this one, though, I can't find any choice that appears to be nice.

[quote="scorpius119"]The function $ \frac {1}{\sqrt {1 \plus{} x^4}}$ doesn't have an elementary antiderivative. [/quote]

I've yet to study integration with any rigor, what's necessary and sufficient for a function to have an elementary antiderivative?

There is no simple condition (other than the obvious and circular one).
See [url=http://en.wikipedia.org/wiki/Differential_Galois_theory]here[/url] for some basic information about the associated theory.


[size=150]Let [/size]\[ \frac{1}{a} \plus{} \frac{1}{b} \plus{} \frac{1}{c} \equal{} \frac{1}{{a \plus{} b \plus{} c}}\]
Prove that
\[ \frac{1}
{{a^n }} \plus{} \frac{1}
{{b^n }} \plus{} \frac{1}
{{c^n }} \equal{} \frac{1}
{{a^n  \plus{} b^n  \plus{} c^n }}\]
Where $ n$ is a given odd and positive integer.
Is it correct for even positive integers?

[quote="madadpoor"][size=150]Let [/size]
\[ \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c} \equal{} \frac {1}{{a \plus{} b \plus{} c}}\]
Prove that
\[ \frac {1} {{a^n }} \plus{} \frac {1} {{b^n }} \plus{} \frac {1} {{c^n }} \equal{} \frac {1} {{a^n \plus{} b^n \plus{} c^n }}\]
Where $ n$ is a given odd and positive integer.
Is it correct for even positive integers?[/quote]

$ \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c} \equal{} \frac {1}{{a \plus{} b \plus{} c}}$
$ \Leftrightarrow \frac{ab\plus{}bc\plus{}ca}{abc}\equal{}\frac{1}{a\plus{}b\plus{}c}$
$ \Leftrightarrow 2abc\plus{}a^2b\plus{}ab^2\plus{}b^2c\plus{}bc^2\plus{}c^2a\plus{}ca^2\equal{}0$
$ \Leftrightarrow (a\plus{}b)(b\plus{}c)(c\plus{}a)\equal{}0$

Put $ a\equal{}a^n, b\equal{}b^n, c\equal{}c^n$ we get

$ \frac {1} {{a^n }} \plus{} \frac {1} {{b^n }} \plus{} \frac {1} {{c^n }} \equal{} \frac {1} {{a^n \plus{} b^n \plus{} c^n }}$
$ \Leftrightarrow (a^n\plus{}b^n)(b^n\plus{}c^n)(c^n\plus{}a^n)\equal{}0$

For $ n$ odd, $ (a^n\plus{}b^n)(b^n\plus{}c^n)(c^n\plus{}a^n)\equal{}(a\plus{}b)(b\plus{}c)(c\plus{}a)Q(a,b,c)\equal{}0$ if $ (a\plus{}b)(b\plus{}c)(c\plus{}a)\equal{}0$, so the first part is done.

For $ n$ even , put $ a\equal{}b\equal{}1, c\equal{}\minus{}1$ we get it's wrong, done also.

[quote="madadpoor"][size=150]Let [/size]
\[ \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c} \equal{} \frac {1}{{a \plus{} b \plus{} c}}\]
Prove that
\[ \frac {1} {{a^n }} \plus{} \frac {1} {{b^n }} \plus{} \frac {1} {{c^n }} \equal{} \frac {1} {{a^n \plus{} b^n \plus{} c^n }}\]
Where $ n$ is a given odd and positive integer.
Is it correct for even positive integers?[/quote]

[b]Near problem[/b]

$ \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c} \equal{} \frac {1}{{a \plus{} b \plus{} c}}$
$ \Longrightarrow \forall n \in \text{odd positive integer  s.t.  }\frac {1}{a^n} \plus{} \frac {1}{b^n} \plus{} \frac {1}{c^n} \equal{} \left(\frac {1}{{a \plus{} b \plus{} c}}\right)^n$

[quote="Kouichi Nakagawa"][b]Near problem[/b]

$ \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c} \equal{} \frac {1}{{a \plus{} b \plus{} c}}$
$ \Longrightarrow \forall n \in \text{odd positive integer s.t. }\frac {1}{a^n} \plus{} \frac {1}{b^n} \plus{} \frac {1}{c^n} \equal{} \left(\frac {1}{{a \plus{} b \plus{} c}}\right)^n$[/quote]

[b]Solution[/b]

$ \frac {1}{a} \plus{} \frac {1}{b} \plus{} \frac {1}{c} \equal{} \frac {1}{{a \plus{} b \plus{} c}} \iff (b\plus{}c)(c\plus{}a)(a\plus{}b)\equal{}0$.
Therefore $ b\plus{}c\equal{}0$ or $ c\plus{}a\equal{}0$ or $ a\plus{}b\equal{}0$.

If $ a\plus{}b\equal{}0 \to b\equal{}\minus{}a$. So,
LHS$ \equal{}\frac{1}{c^n}$,
RHS$ \equal{}\frac{1}{c^n}$.

The others are similar, too.


problem 2:
there are [i]n[/i] points in a given plane. for any three points [i]A,B,C,[/i] from this collection, the angle ABC is formed. then among all these angles the least one is chosen. what is the greatest possible value of such an angle (as a function of [i]n[/i])? [i]Note: if three points belong to a line then some angles they form are equal to 0.[/i]

Hm. I am not quite sure I get this problem either - considering I'm doing it with a 103 degree fever, but I believe the angle ABC is 180/n where n is the number of points. So, f(n) = 180/n is my answer and this only works if n :ge: 3. I figured that the largest smallest angle would occur in the case of regular polygons after connecting the n points.  So, such an angle in each case would be the interior angle( 180(n-2) / n ) and that angle is equally intersected by connecting that vertex with the other points - so you divide by (n-2).

[quote="random%0"]maybe the "angle ABC" was meant as "triangle ABC"? That would seem to make sense, and the question essentially asks what is the largest angle measure of the smallest angle in a triangle, which would be 60, in the case of an equilateral[/quote]

No, "angle" was correct. Basically to reword the problem, you have all these points in the plane, and if you take some points A, B, and C (in that order), you get an angle ABC. Do that for all possible groups of 3 points, you now have heaps of angles. When you take the smallest one, since theres so many its gonna be pretty small right? So whats the biggest that value could be?


Can anyone describe the (n-1)EV method to solve inequalities?

Take a look at [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=33456]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=33456[/url] ;)


logarithms
limit
calculus
calculus computations

2) $\sum\left(\ln 2-\left(\frac{1}{n+1}+\cdots+\frac{1}{2n}\right)\right]=\infty$.
To see this, simply observe, assuming  I made no computational mistakes, that

$\lim n \left[\ln 2-\left(\frac{1}{n+1}+\cdots+\frac{1}{2n}\right)\right]=\frac{1}{4}$, hence the sum diverges.

1) It converges.

Observe that

$\frac{\arctan x_{n}}{x_{n}}=nx_{n}^{1/3}$ it follows that

$n^{3}x_{n}\rightarrow 1$ hence $x_{n}\approx \frac{1}{n^{3}}$, thus
the sums $\sum x_{n}$ and $\sum \frac{1}{n^{3}}$ have the same nature.

3) I think it converges.

Here is a method of attack.

$\sum_{n=1}^{\infty}\frac{(-1)^{[\sqrt{n}]}}{n}=\sum_{k=1}^{\infty}\left(\sum_{k^{2}\leq n<(k+1)^{2}}\frac{(-1)^{[\sqrt{n}]}}{n}\right)$

$=\sum_{k=1}^{\infty}\left(\sum_{k^{2}\leq n<(k+1)^{2}}\frac{(-1)^{k}}{n}\right)=\sum_{k=1}^{\infty}(-1)^{k}b_{k}<\infty$,

where

${b_{k}=\sum_{k^{2}\leq n<(k+1)^{2}}\frac{1}{n}}$ which is of order $\frac{1}{k}$,

hence your sum is a Leibniz sum.

Ravi B,

More precisely, it should be a Leibniz series, i.e., a series of the form

$\sum_{n\in N}(-1)^{n}a_{n}$, where $a_{n}$ is a nonincreasing sequence of positive numbers which converges  to 0.

Thanks, didilica.  But have you shown that $b_{k}$ is nonincreasing?

It's enough to notice that $b_{n}=\frac2n+O(\frac1{n^{2}})$

[quote="didilica"]well, we have that

$b_{k}\leq \frac{(k+1)^{2}-k^{2}-1}{k^{2}}=\frac{2k+2}{k^{2}}$, hence it decreases, since it converges to 0.[/quote]
This doesn't prove that $b_{k}$ decreases.

Okay, RIP's estimate finishes the problem, because $b_{k}$ splits into a decreasing part and an absolutely convergent part.

didilica, just showing that $b_{k}$ is $O(\frac{1}{k})$ is not enough.  For example, if $b_{k}$ was $\frac{2}{k}$ for $k$ even, and $\frac{1}{k}$ for $k$ odd, then the series wouldn't converge.

and $b_{k}\geq \frac{2k+2}{(k+1)^{2}}$, more precisely

$\frac{2}{k+1}\leq b_{k}\leq \frac{2}{k}+\frac{2}{k^{2}}$.
It follows from the definition of ${b_{k}}$, this is in fact RIP`s estimate.

[quote="didilica"]3) I think it converges.

Here is a method of attack.

$ \sum_{n \equal{} 1}^{\infty}\frac {( \minus{} 1)^{[\sqrt {n}]}}{n} \equal{} \sum_{k \equal{} 1}^{\infty}\left(\sum_{k^{2}\leq n < (k \plus{} 1)^{2}}\frac {( \minus{} 1)^{[\sqrt {n}]}}{n}\right)$
[/quote]

May I know how to justify the equality?


[b]Problem[/b]:  Suppose ABC is a triangle such that $ 2BC^2 \equal{} AB^2 \plus{} AC^2$
Let G is centroid and O is center of circumcircle of triangle ABC.Prove that  : (BCG) tangent to ( AGO)    :D

Suppose $ BG,CG$ again meets$ (O)$ at $ P,Q$,respectively.Prove  that $ PQ$ tangent to Euler's circle of triangle $ ABC$

p/s: This triangle is very beautiful   :thumbup:  :blush:   :D


rotation
analytic geometry
complex numbers
geometry
angle bisector
geometry solved

Two right-angel triangles, ABC and A'B'C' are inscribed in circle with center O (AB and A'B' -- diameters).
Let AC :cap: A'C' = P, BC :cap: B'C' = Q, BB' :cap: CC' = R.
Prove, that OP :perp: QR.

It's easily done with complex numbers, but I'd like to see more geometric solution.

Can you please post solution with complex numbers? :)

Rotate A'B' until it matches AB.   P -> A , Q -> B ,  B'B becomes tangent    RB _|_ OP

 The angle bisectors of P and R are also orthogonal.

   Col. Pestich

Armo, of course.

Let the circle have center 0 and radius 1.
Then for each z lying on it we have $z\cdot \bar z=1$.
If complex coordinate of B, C, B', C' are a, b, c, d respectively, then A and A' have coordinates -a, -c respectively.

If points x, y, z, t lie on the unit circle, the intersection of XY and ZT is $\frac{xy(z+t)-zt(x+y)}{xy-zt}$. Using this formula to find p, q, r, we have:
$p=\frac{ab(d-c)-cd(b-a)}{ab-cd}$;
$q=\frac{ab(d+c)-cd(b+a)}{ab-cd}$;
$r=\frac{ac(d+b)-bd(c+a)}{ac-bd}$.
So, if we want to prove OP :perp: QR, we have to show that $\frac{r-q}{p}$ is purely imaginary, wich is equivalent to
$\frac{r-q}{p}=\overline{\left(\frac{r-q}{p}\right)}=\frac{\bar r-\bar q}{\bar p}$. The last equality is easy to prove by direct calculation :)

[quote="pestich"]The angle bisectors of P and R are also orthogonal.[/quote]
What do you mean by this?

>
  Let AC x A'C' = P, BC  x  B'C' = Q, BB'  x  CC' = R.
>

   Angle bisector of < APA'   and angle bisector of < CRB  are perpendicular to
   each other.

  Maj. Pestich

[color=blue][b]Problem.[/b] Let ABC and A'B'C' be two right-angled triangles inscribed in one given circle k with center O; hereby, we assume that the right angles of the triangles ABC and A'B'C' are at the vertices C and C', respectively, so that the segments AB and A'B' are diameters of the circle k.

Let $AC\cap A^{\prime}C^{\prime}=P$, $BC\cap B^{\prime}C^{\prime}=Q$ and $BB^{\prime}\cap CC^{\prime}=R$.

Prove that $OP\perp QR$.[/color]

[i]Solution.[/i] Remember a well-known property of polars of points with respect to circles:

[b]Lemma 2.[/b] If X, Y, Z, W are four points on a circle u, then the point $XY\cap ZW$ lies on the polar of the point $XZ\cap YW$ with respect to the circle u.

Now, in order to solve the problem, let $U=BC^{\prime}\cap CB^{\prime}$.

Applying Lemma 2 to the points B, C, C' and B' lying on the circle k, we see that the point $BC\cap C^{\prime}B^{\prime}$ lies on the polar of the point $BC^{\prime}\cap CB^{\prime}$ with respect to the circle k. In other words, the point Q lies on the polar of the point U with respect to the circle k. On the other hand, applying Lemma 2 to the points B, B', C' and C lying on the circle k, we see that the point $BB^{\prime}\cap C^{\prime}C$ lies on the polar of the point $BC^{\prime}\cap B^{\prime}C$ with respect to the circle k. In other words, the point R lies on the polar of the point U with respect to the circle k.

Altogether, we see that the points Q and R both lie on the polar of the point U with respect to the circle k. Hence, the line QR is the polar of the point U with respect to the circle k. Thus, this line QR is perpendicular to the line OU (since the point O is the center of the circle k).

Since the segments AB and A'B' are diameters of the circle k, both of them pass through the center O of the circle k. In other words, $A^{\prime}B^{\prime}\cap AB=O$.

Applying the Pascal theorem to the cyclic hexagon $BC^{\prime}A^{\prime}B^{\prime}CA$ (inscribed into the circle k), we see that the points $BC^{\prime}\cap B^{\prime}C$, $C^{\prime}A^{\prime}\cap CA$ and $A^{\prime}B^{\prime}\cap AB$ are collinear. But $BC^{\prime}\cap B^{\prime}C=U$, $C^{\prime}A^{\prime}\cap CA=P$ and $A^{\prime}B^{\prime}\cap AB=O$; thus, we see that the points U, P and O are collinear. In other words, the line OU coincides with the line OP. And hence, our result that the line QR is perpendicular to the line OU can be rewritten as follows: The line QR is perpendicular to the line OP.

So the problem is solved.

  Darij


An automobile radiator contains 16 liters of antifreeze and water. This mixture is 30% antifreeze. How much of this mixture should be drained and replaced with pure antifreeze so there will be 50% antifreeze?

There is 3/10*16=24/5 antifreeze so there are 56/5 water

If you want 8 liters of antifreeze you also want 8 liters of water.  For every liter you drain, you will be draning 3/10 antifreeze and 7/10 water (it still is porprotional).  So 56/5-7/10x=8, solving x=32/7


Please proof this inequality:

For any $a_{1},a_{2},...,a_{n}$ $\in$ $\Re^{+}$, let $A_{n}$=(1/n)*$\sum a_{i}$ and $G_{n}=\sqrt[n]{\prod a_{i}},$
then $(n+1)(A_{n+1}-G_{n+1})$$\geq$ $n(A_{n}-G_{n})$
and the equality holds if and only if $a_{n+1}=G_{n}$

[quote="dennishk"]Sorry that I really don't understand about your answer.
Could you explain it more clearly?[/quote]
in brief, clear the denominator then weighted amgm.


It would be really cool if you guys could find to way so that we can watch blog comments.


limit
calculus
integration
logarithms
trigonometry
real analysis
calculus computations

Can you please elaborate and explain in detail?

These are each Riemann sums for some integral (OK, you need to take a logarithm in the second).
Here's #1:
Consider the integral $ \int_0^1 \sin^{2k} \left(\frac{\pi}{2}x\right)$, and the $ n$-part partition of intervals $ [x_{r\minus{}1},x_r]$ with $ x_r\equal{}\frac{r}{n}$. The sum $ \frac1n\sum_{r\equal{}1}^n \sin^{2k}\frac{r\pi}{2n}$ is the right-endpoint Riemann sum for this integral and partition. As $ n\to\infty$, the partitions become arbitrarily fine and the sums converge to the integral.


function
geometry
calculus
integration
probability
expected value

How do you solve the following game theory problem: A company has a value V which is uniformly distributed between 0 and 1. you are planning to place a bid B for the company. If B is smaller than V, then your bid loses and you get nothing; if B is larger than V, you get to purchase the company at price B, and the company will end up being worth 1.5 * V. What price B should you bid to maximize your profit?

Please only post actual AMC questions in the AMC forum.

Just to make sure I properly understand: if $ B < V$ then you get 0.  If $ V \leq B < \frac{3}{2}V$, you'll end up turning a profit, and if $ B > \frac{3}{2} V$ you'll end up with a loss, right?  More specifically, if $ V \geq B$ then the net value of the auction to you is 0, while if $ V \leq B$ then the net value is $ \frac{3V}{2} \minus{} B$.  For any fixed value $ B$, this describes a function $ P(V)$ that you can graph.  The signed area between the graph and the $ V$-axis ("signed area" being what you get when you take a conventional definite integral) is exactly the expected value you want.  Write this down (you don't need calculus to compute it) -- it should (of course) be a function of $ B$.  Now try to maximize this function.

You might post your workings and then others can check your work for you.

It seems to me that the maximum average profit as V ranges from 0 to 1 is 0, when $ B \equal{} 0$.
If $ V>B$, the profit is $ \minus{}B$; if $ V\leq B$, the profit is $ 3V/2\minus{}B$.

Case 1: $ B\in[0,1]$
$ \int_0^B (3V/2\minus{}B)dV\minus{}\int_B^1B dV\equal{} (3B^2/4\minus{}B^2)\minus{}(B\minus{}B^2)\equal{}3B^2/4\minus{}B$ - maximum is 0 when B=0

Case 2: B>1
$ \int_0^1(3V/2\minus{}B)dV\equal{}3/4\minus{}B<0$

is this correct?

No, if $ V > B$ then the net value should be 0, as I wrote -- you get your bid back if you don't win anything.

OK I thought that you won't get you money back...
so in that case we only have the first integral in case 1, which yields the answer as 0 still...

Yes, that's right -- the multiplier $ \frac{3}{2}$ would have to be at least 2 in order for this game to be worth playing.

The question is asking us to maximise 1.5v-B


Since I was having a little trouble and AoPS is my go-to site, I thought I would ask.

How do you use extended header files such as genlib.h in your program?
The course I take told me to:

"Move the header files:
		  	   genlib.h
                           simpio.h
                           strlib.h
                           random.h

    provided by us into the directory:

    C:\Dev-C++\Include"

But I don't really get how to move the files to that place...

Hope to get a response as soon as possible

Ok nevermind problem solved.
Reading through the compilation errors, it said that "simpio.h" was not found in a certain file, so I just copy/pasted into that file and it all worked.  After I fixed syntax, program works fine.


function integration trigonometry calculus derivative algebra polynomial

Hi I need help
Problem:
Find the Fourier transform of
\begin{eqnarray*} f(x) &=& 1-x^2 \,\,\, |x|<1 \\ &=& 0 \,\,\,\,\,\,\,\,\,\,\, \,\,\,\,\,|x|>1 \end{eqnarray*}
it is even function so
\[\begin{eqnarray*} F_c(\alpha) &=& \int_{0}^1 \sqrt{\frac{2}{\pi}} (1-x^2) \cos \alpha x \, dx \\ &=& \sqrt{\frac{2}{\pi}} \left(\int_{0}^1 \cos \alpha x - \int_{0}^1 x^2 \cos \alpha x \, dx \right) \\ &=& \sqrt{\frac{2}{\pi}} \left(\int_{0}^1 2x \frac{\sin \alpha x}{\alpha} \, dx \right) \\ &=& \sqrt{\frac{2}{\pi}} \left( \frac{-2 \cos \alpha}{\alpha^2} + \frac{2 \sin \alpha }{\alpha^3} \right) \\ &=& 2 \sqrt{\frac{2}{\pi}} \left(\frac{\sin \alpha - \alpha \cos \alpha}{\alpha ^3} \right) \end{eqnarray}\]
The book answer give
$2 \sqrt{\frac{2}{\pi}} \left(\frac{\alpha \cos \alpha - \sin \alpha }{\alpha ^3}\right)$. Reverse the sign.  :?
Then the second part of problem ask us to evaluate
\[ \int_{0}^{\infty} \left( \frac{x \cos x - \sin x}{x^3} \right) \cos \frac{x}{2} \, dx  \] of course using my method i got $- \frac{3 \pi}{16}$, and again the book give $\frac{3 \pi}{16}$  :?
any help?
thanks.

Short answer: you're right, even though I'm having trouble following the steps of your integral.

Let's make an independent check. The Fourier transform, evaluated at zero, is the integral of the original function. This function clearly has a positive integral, so its Fourier transform evaluated at zero should be positive. [color=green][See technical note below.][/color]

Let's look at what you have: a positive constant times $\frac{\sin\alpha-\alpha\cos\alpha}{\alpha^3}.$

$\frac{\sin\alpha-\alpha\cos\alpha}{\alpha^3}= \frac{\left(\alpha-\frac{\alpha^3}6+O(\alpha^5)\right)-\alpha\left( 1-\frac{\alpha^2}2+O(\alpha^4)\right)}{\alpha^3}$

$=\frac{\left(-\frac16+\frac12\right)\alpha^3+O(\alpha^5)}{\alpha^3}= \frac13+O(\alpha^2)$

Since this is positive at $\alpha=0,$ you have the correct sign.

Note: the actual Fourier transform of this compactly-supported function is continous at zero; in fact, it is an entire function (given by a power series with an infinite radius of convergence). Using the formula which isn't defined at zero is a notational convenenience, but shouldn't lead you to thinking there's an actual singularity there.

[color=green]Technical note:[/color]

You've got the $2\pi$ in the definition of the Fourier transform in a different place than I'm used to having it, so I'll have to adjust. I looks like you're using this defintion:

$\hat{f}(\alpha)=\frac1{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(x)e^{-i\alpha x}\,dx$

in which case $\hat{f}(0)=\frac1{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(x)\,dx.$

Let's do a quick sensibility check:
On the first, look at the case $\alpha=\frac{\pi}2$. We are integrating a positive function, so the answer should be positive. $\frac{\frac{\pi}2\cdot0-1}{\left(\frac{\pi}2\right)^3}$ is negative, so the book is wrong, and you're right.

The same sign considerations show that $\frac{x\cos x-\sin x}{x^3}$ is negative near $x=0$ (actually from $-\pi$ to $\pi$); since $\frac1{x^3}$ goes to zero pretty fast, the integral should be negative. The $\cos\frac x2$ factor makes things unpredictable, but I think we're okay. Also, this probably directly used the preceding, so sign mistakes would carry over.

The short answer: You're right, the book is wrong.

Here's how I would actually calculate the Fourier transforms. I'm going to use three things:

1. The differentiation formula $\widehat{f'}(\alpha)=i\alpha\hat{f}(\alpha).$ We usually use that in the form $\hat{f}(\alpha)=\frac{\widehat{f'}(\alpha)}{i\alpha}.$

2. A willingness to extend that differentiation formula to distributions (generalized functions). For piecewise-polynomial functions, the more derivatives we get to take, the better.

3. Familiarity with one well-known already-computed example: the "box function." Let $b(x)=1$ for $x\in[-1,1]$ and zero otherwise. Then

$\hat{b}(\alpha)=\frac1{\sqrt{2\pi}}\int_{-1}^1e^{-i\alpha x}\,dx=\frac2{\sqrt{2\pi}} \,\frac{\sin\alpha}{\alpha}.$

Our example is

$f(x)=\begin{cases}1-x^2&-1<x<1\\0&\text{otherwise}\end{cases}$

$f'(x)=\begin{cases}-2x&-1<x<1\\0&\text{otherwise}\end{cases}$

$f''(x)=2\delta(x+1)+2\delta(x-1)-2b(x)$

$\widehat{f''}(\alpha)=\frac2{\sqrt{2\pi}}\int_{-\infty}^{\infty}(\delta(x+1)+\delta(x-1)) e^{-i\alpha x}\,dx-2\hat{b}(\alpha)$

$=\frac2{\sqrt{2\pi}}\left(e^{-i\alpha}+e^{i\alpha}\right)-\frac4{\sqrt{2\pi}} \,\frac{\sin\alpha}{\alpha}$

$=\frac4{\sqrt{2\pi}}\left(\cos\alpha-\frac{\sin\alpha}{\alpha}\right) =\frac4{\sqrt{2\pi}} \,\frac{\alpha\cos\alpha-\sin\alpha}{\alpha}$

Then $\hat{f}(\alpha)=\frac{\widehat{f''}(\alpha)}{(i\alpha)^2} = \frac4{\sqrt{2\pi}} \,\frac{\sin\alpha-\alpha\cos\alpha}{\alpha^3}.$

This is in full agreement with beginner's original result.

Many thanks . Wow that's really good method Kent.


induction
real analysis
real analysis solved

A sequence will be made in order as follows.

$(1)$ The first sequence is $1,1$.
$(2)$ The $(n+1)$-th sequence will be made as follows.
First inserting $0$ to adjacent terms of the $n$-th sequence,then adding the first term of the $n$-th sequence to all terms of the sequence which we have made. For example,the second sequence will be obtained $2,1,2$ by adding $1$ to each term of the sequence $1,0,1$,the third sequence $4,2,3,2,4$.
Find the number of the term and the sum of the $n$-th sequence.

Probably almost everybody can... If I haven't made an idiotic mistake somewhere, the number of terms is $2^{n-1}+1$, the first term is $2^{n-1}$, the sum of terms is $2^{n-1}+2\frac {4^{n-1}-1}3+1$. The proof is by induction (or by summing up a couple of geometric progressions). The problem is a little bit too simple and straightforward to be really interesting. I guess that's why nobody bothered to answer. But I may be wrong  :) Thanks for posting it anyway!

Thank you for your reply, fedja. :)

You are right!

kunny


Let $ABC$ be a right angled triangle at $A$. Denote $D$ the foot of the altitude through $A$, and $O_{1}$, $O_{2}$ the incenters of triangles $ADB$ and $ADC$. The circle with centre $A$ and radius $AD$ cuts $AB$ in $K$ and $AC$ in $L$.
Show that $O_{1}$, $O_{2}$, $K$ and $L$ are on a line.

What a nice but old problem!

http://www.mathlinks.ro/Forum/viewtopic.php?t=57295

[hide="Ok, well here is my solution anyway"]
$\angle AKL = \angle ALK = 45$ degrees
Triangles $A{O_{1}}K$ and $A{O_{2}}D$ are congruent, because of $AO_{1}$ common side, $KA=DA$ radii in circle and $\angle KAO_{1}= \angle DAO_{1}$ from angle bisector $AO_{1}$.
Similarly triangles $A{O_{2}}D$ and $A{O_{2}}L$ are congruent.
Thus $\angle AKL = \angle AKO_{1}=45$ degrees, and $\angle ALK =\angle ALO_{2}=45$ degrees, and hence $K$, $O_{1}$, $O_{2}$ and $L$ are all on the same line.[/hide]


symmetry
geometry
angle bisector
exterior angle
perpendicular bisector

Triangle $\triangle ABC$ has $AC=BC$ and $\angle ACB=106^\circ$. $M$ is a point inside the triangle such that $\angle MAC =7^\circ$ and $\angle MCA=23^\circ$. Find $\angle CMB$.

[hide]
Construct point $M'$ by symmetry such that $\angle MCM' = 60 ^\circ$. Since everything is symmetrical $\triangle MCM'$ is equilateral. This means $\angle MCA = \angle M'CB = 23 ^\circ$ and $\angle M'BC = 7^\circ \Rightarrow \angle CM'B = \angle MM'B = 150^\circ$. Now it's quite obvious that $\triangle CM'B \cong \triangle MM'B.$
$\angle CMM' = 60^\circ$ and $\angle M'MB = 23^\circ\Rightarrow \angle CMB = 83^\circ$

I don't think I explained very well.
[/hide]

Construct it by symmetry. It just like $M$ but on the other side of the triangle.

[hide="Solution"] Find point $X$ on line $AM$ such that $\angle XBC = 7^\circ$. Because $CA = CB$, $X$ is on the angle bisector of $\angle C$.

Thus,  $\angle BCX = 53^\circ$ and $\angle XCM = 53^\circ-\angle ACM = 30^\circ$. Now notice that $\angle XMC = 7^\circ+23^\circ = 30^\circ$, by the exterior angle theorem thing.

We're almost done now. $XB$ is perpendicular to $MC$. However, we also know that $XM = XC$, as $\triangle MXC$ is isosceles, so it is also the perpendicular bisector of $MC$. Therefore, $\angle CMB = \angle MCB = 83^\circ$.[/hide]

[quote="stupidkid"]
Now it's quite obvious that $\triangle CM'B \cong \triangle MM'B.$
[/quote]

Please give the explicit proof of this.

$CM' = MM'$ since both are sides of an equilateral triangle, $\overline{BM'}$ is shared and $\angle CM'B= \angle MM'B$ therefore congruency by $SAS$.


trigonometry
LaTeX
function
logarithms
complex analysis

Derive the following formulas for 0 < θ < 2pi:

1 + cos θ + cos 2θ + ... + cos (nθ) = cos(nθ/2) sin[(n+1)θ/2] / sin(θ/2),

sin θ + sin 2θ + sin 3θ + ... + sin (nθ) = sin(nθ/2) sin[(n+1)θ/2] / sin(θ/2).

It's obvious, note the first sum as $ S_1$ and the second as $ S_2$. Then for every $ \theta\ne 2k\pi$:
\[ S_1\plus{}i S_2\equal{}\sum_{k\equal{}0}^{n}(\cos{k\theta}\plus{}i\sin{k\theta})\equal{}\sum_{k\equal{}0}^{n}e^{ik\theta}\equal{}\\
\equal{}\frac{e^{i(n\plus{}1)\theta}\minus{}1}{e^{i\theta}\minus{}1}\]
and because $ \sin{x}\equal{}\frac{e^{ix}\minus{}e^{\minus{}x}}{2i}$, then:
\[ S_1\plus{}i S_2\equal{}\frac{e^{i(n\plus{}1)\theta}\minus{}1}{e^{i\theta}\minus{}1}\equal{}\frac{e^{i(n\plus{}\frac{1}{2})\theta}\minus{}e^{\minus{}\frac{i\theta}{2}}}{e^{\frac{i\theta}{2}}\minus{}e^{\minus{}\frac{i\theta}{2}}}\equal{}
e^{\frac{in\theta}{2}}\frac{e^{\frac{i(n\plus{}1)\theta}{2}}\minus{}e^{\minus{}\frac{i(n\plus{}1)\theta}{2}}}{e^{\frac{i\theta}{2}}\minus{}e^{\minus{}\frac{i\theta}{2}}}\equal{}e^{\frac{in\theta}{2}}\frac{\sin{\frac{(n\plus{}1)\theta}{2}}}{\sin{\frac{\theta}{2}}}\equal{}\cos{\frac{n\theta}{2}}\frac{\sin{\frac{(n\plus{}1)\theta}{2}}}{\sin{\frac{\theta}{2}}}\plus{}i\sin{\frac{n\theta}{2}}\frac{\sin{\frac{(n\plus{}1)\theta}{2}}}{\sin{\frac{\theta}{2}}}\]
Because $ S_1$ and$ S_2$ are real, we have:
\[ S_1\equal{}\cos{\frac{n\theta}{2}}\frac{\sin{\frac{(n\plus{}1)\theta}{2}}}{\sin{\frac{\theta}{2}}}\\
S_2\equal{}\sin{\frac{n\theta}{2}}\frac{\sin{\frac{(n\plus{}1)\theta}{2}}}{\sin{\frac{\theta}{2}}}\]

We can easily show that these equations by induction.

[quote="mathwizarddude"]Derive the following formulas for 0 < θ < 2pi:

1 + cos θ + cos 2θ + ... + cos (nθ) = cos(nθ/2) sin[(n+1)θ/2] / sin(θ/2),
[/quote]

Base case n=0:
$ 1 \equal{} cos(0) \frac {sin(\frac {\theta}{2})}{sin(\frac {\theta}{2})}$

Assume n, prove n+1:
$ 1 \plus{} cos(\theta) \plus{} cos(2\theta) \plus{} ... \plus{} cos(n\theta) \plus{} cos((n \plus{} 1)*\theta) \equal{} cos (\frac {n\theta}{2})\frac {sin(\frac {(n \plus{} 1)\theta}{2})}{sin(\frac {\theta}{2})} \plus{} cos((n \plus{} 1)*\theta)$

Now show that $ sin (\frac {n\theta}{2})\frac {sin(\frac {(n \plus{} 1)\theta}{2})}{sin(\frac {\theta}{2})} \plus{} cos((n \plus{} 1)\theta) \equal{} cos (\frac {(n \plus{} 1)\theta}{2})*\frac {sin(\frac {(n \plus{} 2)\theta}{2})}{sin(\frac {\theta}{2})}$

[quote="mathwizarddude"]Derive the following formulas for 0 < θ < 2pi:
sin θ + sin 2θ + sin 3θ + ... + sin (nθ) = sin(nθ/2) sin[(n+1)θ/2] / sin(θ/2).[/quote]

Base case n=1:
$ sin(\theta) \equal{} sin(\frac {\theta}{2})\frac {sin(\frac {2\theta}{2})}{sin(\frac {\theta}{2})}$

Assume n, prove n+1:
$ sin(\theta) \plus{} sin(2\theta) \plus{} .... \plus{} sin(n\theta) \plus{} sin((n \plus{} 1)\theta) \equal{} sin (\frac {n*\theta}{2})*\frac {sin(\frac {(n \plus{} 1)*\theta}{2})}{sin(\frac {\theta}{2})} \plus{} sin((n \plus{} 1)*\theta)$

Now show that $ sin (\frac {n\theta}{2})\frac {sin(\frac {(n \plus{} 1)\theta}{2})}{sin(\frac {\theta}{2})} \plus{} sin((n \plus{} 1)\theta) \equal{} sin (\frac {(n \plus{} 1)\theta}{2})*\frac {sin(\frac {(n \plus{} 2)\theta}{2})}{sin(\frac {\theta}{2})}$

Man, LATEX is a PITA

Notes for making that prettier: \sin and similar expressions look better for standard functions: $ \sin$. For large parentheses around fractions, use \left( and \right) - you can replace ( and ) with other delimiters like ] and | as well. Left and right signs should balance; use \left. or \right. for an invisible delimiter.
Examples:
\sin \left( \frac1x \right) : $ \sin\left(\frac1x\right)$
\left.\frac{\ln x}{x^2}\right|_1^2=\frac{\ln 2}{4} : $ \left.\frac{\ln x}{x^2}\right|_1^2\equal{}\frac{\ln 2}{4}$


limit
trigonometry
calculus
calculus computations

If $ \lim_{x\rightarrow\infty}(\sqrt {x^2 \minus{} x \plus{} 1} \minus{} ax \minus{} b) \equal{} 0,$ then

for $ k > \equal{} 2,$ Find

$ \lim_{n\rightarrow\infty} \sec^{2n}(k! \pi b)$

Please explain in detail.

$ 0 = \lim_{x\rightarrow\infty}{(\sqrt {x^2 - x + 1} - ax - b)} = \lim_{x\rightarrow\infty}{\frac {x^2 - x + 1 - (ax + b)^2}{\sqrt {x^2 - x + 1} + ax + b}} = \\
= \lim_{x\rightarrow\infty}{\frac {(1 - a^2)x^2 - (2ab + 1)x + (1 - b^2)}{\sqrt {x^2 - x + 1} + ax + b}}\Rightarrow\{\begin{array}{c}a = 1 \\
b = - \frac {1}{2}\end{array}.$

$ \mbox{If } b = - \frac {1}{2}\mbox{, then: }\sec^2(k!\pi b) = \frac {1}{\cos^2(k!\pi b)} = \frac {1}{\cos^2{\frac {k!\pi}{2}}} = 1\Rightarrow \sec^{2n}(k!\pi b)=1.$


[quote="t0rajir0u"]A typo.  Of course, it should be $ \frac{a^{n+1}-1}{a-1}$.[/quote]
this is right

[quote="i_like_pie"][quote="abacadaea"]the formula is $ \frac{a^{n+1}-1}{n-1}$. a being 2 and n being $ \log_{2}{65536}$ whatever that is. then plug the numbers in.[/quote]
That's incorrect. Plugging in $ a=2$ and $ n=16$, we get a fraction: $ \frac{2^{16+1}-1}{16-1}=\frac{131071}{15}$. Obviously the sum of some amount of integers can't be non-integral.[/quote] woops i meant $ a-1$ not $ n-1$:oops:


trigonometry
function
calculus
integration
derivative
geometric series
calculus computations

How can we find the power series for $ \sin x$ by using integration? Is it possible? Or being more specific, can you provide proofs of the aforesaid representation for such function?

Compute the derivatives of sin(x) and use the maclaurin expansion and the power series will follow? I don't see how to do anything like the trick of integrating 1/(1+x^2) (geometric series) to derive the arctanx power series.


find all $f: \mathbb{N}\rightarrow\mathbb{N}$ such that for every $m,n\in \mathbb{N}$, $f(m)+f(n)|m+n$.

[hide]Letting $m=n=1$, $2f(1)|2\Rightarrow f(1)|1\Rightarrow f(1)=1$.

For $n>1$, use strong induction to show $f(n)=n$. A well-known theorem states that there exists a prime between $n$ and $2n$, so there exists $m<n$ such that $m+n$ is a prime, $p$.

If $f(m)+f(n)$ divides $p$, then $f(m)+f(n)$ is either 1 or $p$. But $f(m)+f(n)\geq 1+1=2$, so it can't be 1. Therefore $f(m)+f(n)=p$. Since $f(m)=m$, $f(n)=p-m=n$.

Therefore, the only such function is $\boxed{f(n)=n}$.[/hide]


geometry
3D geometry
geometric transformation
rotation

Each face of a cube is to be painted one of six different colors, and each color is used once. How many ways are there to paint the cube? Colorings that are equivalent by rotation are not considered distinct.

I've seen this posted at least three times before, though I don't remember by whom.

Let $A,B,C,D,E,F$ be distinct 6 colors, if $A$ is colored on a cirtain side, the ways  of coloring on its opposite side is $_{5}C_{1}=5.$ For example, assumed that we select $B,$ making the axis of rotation passing through these 2 sides and coloring on the 4 sides rest, since colorings  that are equivalent by rotation are not considered distinct, this problem is circular permitation. Turning over up side down doesn't over lapping other coloring. Thus ways of coloring on the rest 4 sides is $\frac{4!}{4}=6.$

I was suspicious about the solution given in Posamentier's book, which says the answer is 30. So you think the correct answer is 90?


conics
parabola
quadratics
algebra
absolute value

Al G. Brah dives off the high diving board.  His distance from the surface of the water varies quadratically with his distance from the board.  THe particular e quation the best fits Al G. Brah's dive is $y=-5x^2+9x+20$, where y is the height of the dive in feet aboce the water and x is the number of feet Al G. Brah has jumped horizontally from the diving board.

How high is the diving board?
What is the highest Al G. Brah gets above the diving board?
How far out does Al G. Brah hit the water?

I forgot how to find the highest value of y... :blush:

[hide="Solution"]The diving board must be 20 feet high, since when $x=0$, $y=20$.  To find the highest (or lowest, whichever exists) value on a parabola, we find the vertex, whose abscissa will be half-way between the abscissas of the roots of the equation.  This must be at $\begin{displaystyle}\frac{-b}{2a}\end{displaystyle}$, because that would be the average of $\begin{displaystyle}\frac{-b\pm\sqrt{b^2-4ac}}{2a}\end{displaystyle}$.  In this case, that is $\begin{displaystyle}\frac{9}{10}\end{displaystyle}$.  However, what we want is the maximum height, which would be the corresponding ordinate, namely, $\begin{displaystyle}\frac{481}{20}=24\frac{1}{20}\end{displaystyle}$.

Now, all we need is where he hit the water, which would be the only positive root to the equation, namely, $\begin{displaystyle}\frac{9+\sqrt{481}}{10}\end{displaystyle}$.
Q.E.D.[/hide]

[hide]

The graph has to be a parabolla, because this is a quadratic equation. y has to be 20 to be the highest it can get (at the very top of the parabolla) then x = 0. So the diving board is 20 units high (math92 didn't give any measurements). Al G. Brah jumps 0 units.

To get the third part of the question, you replace y with 0 and solve the equation.
The formula to solve a quadratic equation is $  x = \frac{-b \pm \sqrt{-4ac}}{2a}  $
So the answer is $   \frac{-9 \pm \sqrt{400}}{2(-5)} = \frac{-9 \pm \sqrt{400}}{-10}  $
So the answer is -40.9 or -39.1 or the absolute value of either.[/hide]

[quote="pascal_1623"]
The formula to solve a quadratic equation is $  x = \frac{-b \pm \sqrt{-4ac}}{2a}  $[/quote]
no its: $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$


Jonah signed a loan contract that requires a down payment of $1500 and payments of$200 a month for 10 years. The interest rate is 9% per annum, compounded monthly.

a) A = $1500 + 120months ($200x1.09) = $27660 The cash value of the contract is$27660.

I was wondering if i did this correctly because i'm not sure if compounded means it is the interest rate for 200 or towards the increased amount multiplied with the interest rate.

Hmm... I'm a little confused over your question but...

you already know that the total amount of money you have to pay is
$1500+ 200\cdot 12 \cdot 10=25500$

To find how much money you're actually taking out on a loan would be:

$x(1+\frac{.09}{12})^{12 \cdot 10}=25500-1500$

$x+1500 \approx 11290.495$ (I believe)

Is that what you're asking?


If $ a\plus{}b\plus{}c\equal{}0$ prove that $ a^{2}(b\plus{}c)\plus{}b^{2}(a\plus{}c)\plus{}c^{2}(a\plus{}b)\equal{}\minus{}5abc$

[quote="Elio (n)"]If $ a \plus{} b \plus{} c \equal{} 0$ prove that $ a^{2}(b \plus{} c) \plus{} b^{2}(a \plus{} c) \plus{} c^{2}(a \plus{} b) \equal{} \minus{} 5abc$[/quote]

$ a^{3}\plus{}b^{3}\plus{}c^{3}\minus{}3abc\equal{}(a\plus{}b\plus{}c)(a^{2}\plus{}b^{2}\plus{}c^{2}\minus{}ab\minus{}ac\minus{}bc)$ ------> $ a\plus{}b\plus{}c\equal{}0$ ---> $ a^{3}\plus{}b^{3}\plus{}c^{3}\equal{}3abc$

$ a^{2}(b \plus{} c) \plus{} b^{2}(a \plus{} c) \plus{} c^{2}(a \plus{} b)\equal{}a^{2}(\minus{}a)\plus{}b^{2}(\minus{}b)\plus{}c^{2}(\minus{}c)\equal{}\minus{}(a^{3}\plus{}b^{3}\plus{}c^{3})\equal{}\minus{}3abc\not\equal{}\minus{}5abc$

Well in this case since $ a\plus{}b\plus{}c\equal{}0$ than $ a^{3}\plus{}b^{3}\plus{}c^{3}\minus{}3abc\equal{}(a\plus{}b\plus{}c)(a^{2}\plus{}b^{2}\minus{}ab)$


The last thread I believe was solely about Saddam Hussein so here's a new thread about the entire war.

I still disagree with the war because no matter what pro came out of this whole war for the Iraqi people on the extreme long run Iraq will just get all it's wealth sucked up by the U.S. (specifically, The Bush/Cheney Texas Oil Co. :() Here's a visual aid:

Short run: (Bad) Iraqi people are more violent and disorderly, lots of murder and such because of differences, American/foreign soldiers dying, etc.
Long Run: (Good) The people will settle down, (hopefully) there'll be a democratic government, and Iraqi will have more than enough money to run things smoothly
Extreme long run: (Bad) U.S. will start sucking up Iraqi wealth like a leech just like Hussein did (especially Bush/Cheney), traditional Iraqi life will be gone (modernization and internationalization will make Iraq full of McDonalds and Starbucks), people will immigrate to Iraq too much, they will never fit into the Muslim society of the world again, and overall Iraq will become an American colony.

Now some of the opinions here are a bit radical but I think it captures the general idea...

My opinion on the war is that it was primarily waged out of fear (some parts rational and some parts irratoinal). I believe that Bush had genuine motives of protecting the country when he made his decision - though perhaps clouded by his personal feelings too. Whether that decision was well measured will be the source of endless debate.

There was a series of events before the invasion of Iraq that I think scared the administration greatly. The (still unresolved) anthrax attacks, missing viles of bubonic plague from labs in Virgina, the visit of a Small Pox scientist from the former Soviet Union to Iraq... lots of probably unrelated incidents that I think put biological terrorist attacks high on the minds of Bush and his advisors.

I think his decision was also influenced by the outcome in Afghanistan, and how (relatively) smoothly that war and occupation had gone. I think he believed that Iraq would go similarly. So it was seemingly a no-lose situation in the administration's mind. On the one hand, if they were right that Iraq was a threat - they would have eliminated it. And on the other hand, if they were wrong and Iraq wasn't a threat - who cares! No one could prove it because no one would believe Saddam anyway, and we would at least reap the perceived benefits of establishing a new form of government and a potential long term middle eastern ally to replace Saudi Arabia.

Whether the invasion was a good idea? Ultimately, I don't think so because it sets a really bad precedent for foreign policy. Not that our foreign policy was all that coherent to begin with, but I think this sets a particularly ill conceived one. The doctrine of pre-emption leads to more instability than it creates. It allows states to basically attack each other based on perceived threats.

If that were adopted by other nations you would have worldwide chaos. India and Pakistan would have obliterated each other a long time ago under such a doctrine. And basically it is a doctrine based on the principle of fear. Whatever happened to the saying, "We have nothing to fear but fear itself?" Decisions based on fear are not always the most well thought out or the most rational.

Some good will probably come out of the war - particularly for many of the Iraqi people (but also some bad consequences). I don't think it makes America safer in any way, though. Partly because I do not believe that Iraq was a threat to us to begin with; and secondly, because if they did have WMD they are now unaccounted for and possibly in the hands of people that they otherwise wouldn't have been. Not to mention that this has desensitized our allies to our fight against terrorism, and it has incited many more people against us (even within our own borders).

[quote="gauss202"]Whether the invasion was a good idea? Ultimately, I don't think so because it sets a bad precedent for foreign policy. Not that our foreign policy was all that coherent to begin with, but I think this sets a particularly ill conceived one. Some good will come out of it for some people - in particular for some Iraqi people. But I don't think it makes America safer in any way. Partly because I do not believe that Iraq was a threat to us to begin with. And secondly, because if he did have WMD they are now unaccounted for and possibly in the hands of people that they wouldn't have otherwise been.[/quote]
You have the right to think the war in Iraq was a mistake. However, I find your reasoning to be a bit off the wall. It sets a bad precedent in our foreign policy? bah humbug. What about the other wars that have been fought by our troops in countries where we didn't necessarily belong? For a short list, off the top of my head: Somalia, Boznia-Herznogovena (sp?), Yugoslavia, Afghanistan, the Gulf War. In order for the War in Iraq to have set a precedent, it would have had to have been the first of its kind.

I think this war was very distinct from those. Those, for the most part, were humanitarian missions to stop ongoing genicides. This war was not for that reason. It was explicitly sold as a war of pre-emption - which to the best of my knowledge, is unique in modern times.

[quote="Tare"][url=http://www.freerangegraphics.com/flash/fl_cycle.html]Here[/url]'s a little something from freerangegraphics

Food for thought: It seems as though the newly established Department of Homeland Security is doing everything it can to make America on guard and secure. However they do so by creating fear with the help of the conservative medias. If Saddam Hussein united and created a stable government by striking fear into his own people so that they are less rebellious how different are Bush any different from Hussein?[/quote]
[i][u]Conservative[/u][/i] media? [i][u]Conservative[/u][/i]? Since when is the media in this country [i][u]conservative[/u][/i]? FOX News is the closest thing I've seen to conservative US media, and even they don't serve as much of a bulwark against the legions of liberal reporters of the other stations.

As for your conjecture that Saddam Hussein and President Bush are morally equivalent - I suggest you read the news and a few history books before you continue on that train of thought. However, I can do a quick table for you:

[b]Saddam:[/b]
Gassed thousands of Kurds with chemical weapons
Responsible for thousands of grave human rights violations. [url=http://www.un.org/Overview/rights.html]Reference[/url] (yes, Tare, I've read that on many occasions)
Ruled by force
Fixed elections
Dictator
Invaded Kuwait
...and Iran
...and shot missiles at Israeli apartments
(this list is not exhaustive)

[b]Bush:[/b]
Elected by the people (don't try to contest this)
Invaded Afghanistan... to remove terrorists. Was welcomed by the people of the country.
Invaded Iraq... to remove terrorists and overthrow a dictator. Was welcomed by the people of the country.
Is acting in your interest whether you admit it or not.
(this list is not exhaustive)

Now, with a fever of 100.4F, I'm going to sleep for a while.

[quote="JBL"]Somalia and Boznia-Herznogovena -- a war or major conflict was already taking place, and teh U.S. stepped in in an attempt to restore order.
Yugoslavia -- similar to above, plus all the troops were NATO troops, meaning there was agreement among several nations for the military action.
Afghanistan -- reasonable ties existed between Osama bin Laden and the Taliban.  Given the chance to give up bin Laden to avoid war, the Taliban refused.
The Gulf War -- war already begun by Iraqi invasion of Kuwait.  International cooperation in military action.

Note also that in all examples other than Afghanistan, we didn't set about occupying the nation where our troops ended up.  At most there were NATO or UN troops there, not American troops.

Iraq 2 -- none of the above conditions apply.  (I mean, okay, there are a couple of other nations with troops in Iraq.  Ignoring Britian, however, I have to say that the list of remaining nations does not strike me as similar to the list of most powerful countries on earth.)  Most notably, there was no military or pseudo-military event (incl. civil war or major act of terrorism) that was linked in any way to the invasion.

That's the precedent.[/quote]
Yugoslavia had NATO... Afghanistan had a coallition... Iraq 2? "a couple of other nations"? Try 63.

[quote="confuted"]Is acting in your interest whether you admit it or not.[/quote]

If I don't agree that he's acting in my interest, then he isn't.

As for the point about conservative media -- gimme a break, both of you.  By and large, the media in this country is remarkably non-partisan -- they report what they think will sell papers.  It's easy to bring claims of media bias on either side.  Personally, I feel the media tends to be a bit conservative (Fox news, by the way, is practically propaganda, as is nearly everything else Rupert Murdoch owns.  The New York Post is everything but an official RNC newsletter.  Anyhow ...)  but I have an admitted leftist bias myself.  It's worth noting that this country as a whole is far more centrist now than in the past -- there aren't any more people running on segregationist lines, and there aren't any more people running on communist lines.  I happen to feel we've shifted rightward from 40 years ago, and would be happy to fashion examples supporting my claim.  I also know people on the right feel exactly the opposite, and can also provide examples of their own.  This is a discussion not worth having.

As for the 63 nations:  how many of them have contributed anything more than their signature?  In an effort to raise money from nations for Iraq, we raised -- get this -- all of 2 billion dollars.  We're spending more than that every month.  Compare that to our share of the NATO missions.

More significant, however, is the lack of any [i]ongoing[/i] conflict.  And don't give me that nonsense about the Iraqi genocide as a conflict that we got involved in, either -- Donald Rumsfeld was in Baghdad shaking Saddam Hussein's hand when that was going on.

I object to people putting up links to effective internet petitions on this site.  The "food for thought" that Tare linked to is [i]not[/i] just some little idea meant to get you thinking; instead, it is a full-fledged advertisement encouraging people to support a certain view on American policy and providing people with a means of showing their support.  I don't know what "click here" does because I haven't tried doing so, but I don't like it...

I don't think we should go about posting links that advertise political, religious, etc. beliefs.  Just think what could happen when the 2004 elections roll around!  (I really don't want to see links on http://www.artofproblemsolving.com to "Bush 2004 Campaign Videos" or "Support Lieberman for President, 2004" or "Dean 2004" or whoever it'll be...)

:)

Wow maybe because of this heated argument you got a fever :P
But really if you're sick then get some rest, this argument won't go anywhere for a while.

Oh [b]please[/b]! I'd say some of the Bush side's a bad thing:
"Invaded Afghanistan... to remove terrorists. Was welcomed by the people of the country."
Yeah except for the fact that many casualties were caused and the anti-U.S. feelings grew stronger. Again if you consider that a warm welcome then you do have a point there.

"Invaded Iraq... to remove terrorists and overthrow a dictator. Was welcomed by the people of the country."
That is what the media has shown us, yes, but now that we have other sources other than the TV and the radio I'm pretty sure you can dig up some article about what it was really like when the U.S. arrived. Most locked their doors and didn't go outside while others were happy for a while then really sad at the state their country was in: Homes blew up, everything [b]but[/b] the oil wells blown up (now I wonder who's order that was :evil:) not to mention the fact they practically ignored most of the request of help of the looting of several important buildings yet they did nothing and it'll take years to clean that up. Again if you completely trust the media to be liberal (which is what they and the politicians [b]say[/b]; have you have any thoughts about whetehr or not they could be working together?)

As for the list of inhumane murder you can add that Bush killed thousands of civillians (can't quite say they were innocent bystanders since they could've been Bin Laden/Hussein supporters but at least most of them didn't take up arms in Iraq; besides the U.S. killed 'em anyway just like the British didn't care which colonist they killed whether they're a loyalist or a patriot)
Also speaking of human rights violation have you read anything about what the U.S. do to people who are even accused of being a terrorist and they become convinced that they in fact are? I don't blame you if you don't know this because it's hard to find with it being hidden so well but maybe you should at least give it a go and see where that'll get you.

Heh 63 African countries are nothing to a couple of developed nations.

Don't be silly.  The 63 countries weren't all African.  And besides, who are you to call African countries "nothing"?  I'm sure the leaders and inhabitants alike of such countries would be pretty insulted by that one.

Just 'yoking...still I believe he didn't get a lot of European support...(as in power not number)

There's a lot of confusion between when you're joking and when you're not.  For instance:

Name It!  Ways to Torture Saddam Hussein
- Tare says it was all in fun
- confuted and mathfanatic are highly disturbed.  Someone removes it.

The 63 countries supporting the Iraqi reformation are largely African, and are "nothing" compared to developed countries.
- Tare was 'yoking.  (This sounds a lot like IW.OAV: good or bad?)
- an insult?

Methinks you should either
1) make a big distinction between "joking" and "serious", the way MysticTerminator does
2) be serious all the time, the way confuted does

I think the second option is preferable, since if you continue to joke you will make off-color statements and inevitably won't be believed when you do try to be serious.

and if youre joking anyways, dont put it. once again, this is for serious discussions

[quote="mathfanatic"]
Methinks you should
1) make a big distinction between "joking" and "serious", the way MysticTerminator does
2) be serious all the time, the way confuted does
3) joke all the time, the way MysticTerminator does
I think the second option is preferable, since if you continue to joke you will make off-color statements and inevitably won't be believed when you do try to be serious.[/quote]

i like the third option the best...

ok now i'm serious.
come on don't joke on this forum. go be silly in etc.
tare, there is a huge difference between Bush and Hussein. stop being silly. read confuted's post. he's right on every point about saddam. for bush, bush was not elected by the people. none of our presidents are elected by the people. they are elected by the electoral college, and even then we had the whole thingie. but come on, if you still care about that, you are a loser. on confuted's second point about bush, there really were not that many casualties. now you can of course say (OMG ONE CASUALTY IS TOO MUCH) but only if your IQ is less than that of a stick of celery. bin Laden clearly did plan the 9-11 attacks, and if we did not invade Afghanistan, there would certainly be more terrorist attacks by him. personally, i do not understand why we are in iraq before capturing bin laden. clearly that should be our main objective. but it's been shunted to the side...

confuted's third point and tare's subsequent response: i agree with tare that our media, although some might call it liberal (which i do not wholly agree with), is not anti-American. They will most likely show pictures depicting american glory. power outages/food outages/general hardship was widespread from my little political knowledge. and then we have the whole looting issue. hm. perhaps in the long run it'll be better we ousted saddam, but why now? we already have our hands full with bin Laden.

confuted's fourth point: complexzeta pretty much got that cleaned up...

right. people are going off topic too much. if the thread moves to another political topic, i see no problem as the conversation should by all means flow, but when we're talking about all this irrelevant crap (not a swear word, so live with it) it gets annoying

My understanding is that more than 3000 civilians have been killed in the war on Afghanistan.  More than 8000 civilians have been killed in the war on Iraq.  To put those numbers in perspective, their total is more than four times the number of people killed in the September 11 attacks.  Not many casualties?  Only if you view September 11 as having few victims.

[url=http://www.cnn.com/2003/WORLD/asiapcf/12/26/japan.troops.iraq/index.html]Japan's troops are on their way to Iraq.[/url]


calculus
integration
logarithms
calculus computations

anybody want to help me with this one ? i can't seem to do it...  :(  a student asked me and when i couldn't do it, he laughed at me.  :huh:  apparently he knows how to do it. he said he transformed it into an integral first.

show

$ \left(\frac{2}{1}\right)^{\frac{1}{8}}\cdot \left(\frac{2^2}{1\cdot 3}\right)^{\frac{3}{16}}\cdot \left(\frac{2^3\cdot 4}{1\cdot 3^3}\right)^{\frac{6}{32}}\cdot \left(\frac{2^4\cdot 4^4}{1\cdot 3^6\cdot 5}\right)^{\frac{10}{64}}
\cdot \left(\frac{2^5\cdot 4^{10}\cdot 6}{1\cdot 3^{10}\cdot 5^5}\right)^{\frac{15}{128}}$
$ \cdot \left(\frac{2^6\cdot 4^{20}\cdot 6^6}{1\cdot 3^{15}\cdot 5^{15}\cdot 7}\right)^{\frac{21}{256}}
\cdot \left(\frac{2^7\cdot 4^{35}\cdot 6^{21}\cdot 8}{1\cdot 3^{21}\cdot 5^{35}\cdot 7^7}\right)^{\frac{28}{512}}
\cdot \left(\frac{2^8\cdot 4^{56}\cdot 6^{56}\cdot 8^8}{1\cdot 3^{28}\cdot 5^{70}\cdot 7^{28}\cdot 9}\right)^{\frac{36}{1024}}$
$ \cdot\quad\ldots\quad\ldots\quad\;\equal{}\;\boxed{e^{\dfrac{7\zeta(3)}{24\zeta(2)}}}$

i think i see it, just take the ln of the expression and simplify it to get 7zeta(3)/24zeta(2)

not a lot of step between the two

[quote="kenn4000"][b]i think i see it[/b], just take the ln of the expression and simplify it to get 7zeta(3)/24zeta(2)
[/quote]

no you don't. there are only a handful of people on AoPS that can truly solve this, and you   ain't one of them. sorry, kiddo.

I believe that the following helps:

Let $ a_n$ be the n-th term.  Then $ a_n=\left(\frac{2^{\binom{n}{0}}4^{\binom{n}{3}}6^{\binom{n}{5}}\hdots}{3^{\binom{n}{2}}5^{\binom{n}{4}}\hdots}\right)^{\frac{2n-1}{2^{n+2}}}$.  Then it follows that
the product $ P$ satisfies $ \ln P=\sum_{n=1}^{\infty}\frac{2n-1}{2^{n+2}}\left[\sum_{k=1}^{\infty}\binom{n}{2k-1}\ln 2k-\binom{n}{2k}\ln(2k+1)\right]$

Let
\[ \kappa \equal{} \prod_{n \equal{} 1}^{\infty}\left(\prod_{k \equal{} 0}^{n}(k \plus{} 1)^{( \minus{} 1)^{k \plus{} 1}\binom{n}{k}}\right)^{\frac {{\scriptscriptstyle \frac {1}{2}}n(n \plus{} 1) } {2^{n \plus{} 2}}}
\]
Then
\[ \log\kappa \equal{} {\scriptstyle \frac {1}{8}}\sum_{n \equal{} 1}^{\infty}\frac {n(n \plus{} 1)}{2^n}\sum_{k \equal{} 0}^{n}( \minus{} 1)^{k \plus{} 1}\binom{n}{k}\log (k \plus{} 1)
\]

\[ \equal{} {\scriptstyle \frac {1}{8}}\sum_{n \equal{} 1}^{\infty}\frac {n(n \plus{} 1)}{2^n}\sum_{k \equal{} 0}^{n}( \minus{} 1)^{k \plus{} 1}\binom{n}{k}\int_{0}^{\infty}\left(e^{ \minus{} t} \minus{} e^{ \minus{} t(k \plus{} 1)}\right)\frac {dt}{t}
\]

\[ \equal{} {\scriptstyle \frac {1}{8}}\sum_{n \equal{} 1}^{\infty}n(n \plus{} 1)\int_{0}^{\infty}e^{ \minus{} t}\left(\frac {1 \minus{} e^{ \minus{} t}}{2}\right)^n\frac {dt}{t}
\]
One may check that for $ 0\le t\le \infty$, we have $ 0\le \frac {1 \minus{} e^{ \minus{} t}}{2}\le \frac {1}{2}$ and that $ \frac {2x}{(1 \minus{} x)^3} \equal{} \sum_{n \equal{} 1}^{\infty}n(n \plus{} 1)x^n$ for $ |x| < 1$ giving
\[ \log\kappa \equal{} \int_{0}^{\infty}e^{ \minus{} t}\frac {1 \minus{} e^{ \minus{} t}}{\left( 1 \plus{} e^{ \minus{} t}\right)^3}\cdot\frac {dt}{t} \equal{} \minus{} 7\zeta^{\prime}( \minus{} 2)
\]
(according to my handy portable table of integrals, namely [url=http://www.hostsrv.com/webmab/app1/MSP/quickmath/02/pageGenerate?site=quickmath&s1=calculus&s2=integrate&s3=basic]quickmath.com[/url])

To be continued...

It's interesting to see:
why :
\[ \zeta'(\minus{}2n)\equal{}\frac{(\minus{}1)^n\zeta(2n\plus{}1)(2n)!}{2^{2n\plus{}1}\pi^{2n}}\]
and also one simple question:
why:
\[ \int_{0}^{\infty}e^{ \minus{} t}\frac {1 \minus{} e^{ \minus{} t}}{\left( 1 \plus{} e^{ \minus{} t}\right)^3}\cdot\frac {dt}{t} \equal{} \minus{} 7\zeta^{\prime}( \minus{} 2)\]


(xy+yz+xz)[1/(x+y)^2 + 1/(y+z)^2 + 1/(y+z)^2] e mai mare sau egal cu 9/4. x,y,z pozitive strict

e chiar "celebra" Iran 96, postata de multe ori pe forum, uite aici un link: [url]http://www.mathlinks.ro/Forum/viewtopic.php?p=10586#10586[/url].

nu stiam ca a mai fost postata :?:  mersi de link

ma revansez cu alta inegalitate:

x+y+z=1. x,y,z>0

sa se dem ca: 3/2 (x^2/y^2 + y^2/z^2 + z^2/x^2 +1) e mai mare sau egal cu x/y+2 sqrt(y/z) + 3 * radical de ordinul 3 din z/x si mai mare sau egal cu 6

Daca am inteles bine avem de aratat ca:\[\frac{3}{2}(\sum \frac{x^{2}}{y^{2}}+1)\geq \frac{x}{y}+2\sqrt{\frac{y}{z}}+3\sqrt[3]{\frac{z}{x}}\geq 6\]
A doua inegalitate reiese f usor aplican medii.
Deoarece $\frac{x}{y}+2\sqrt{\frac{y}{z}}+3\sqrt[3]{\frac{z}{x}}\leq \sum \frac{x}{y}+3$ (din inegalitatea mediilor) , este suficient sa aratam ca $\frac{3}{2}(\sum \frac{x^{2}}{y^{2}}+1) \geq \sum \frac{x}{y}+3$ $\leftrightarrow$ $\sum \frac{x^{2}}{y^{2}}+1\geq \frac{2}{3}\sum \frac{x}{y}+2$ $\leftrightarrow$ $\sum \left (\frac{x}{y}-\frac{1}{3}\right )^{2}\geq \frac{4}{3}$
Dar din CBS avem $3\sum \left (\frac{x}{y}-\frac{1}{3}\right )^{2}\geq \left ( \sum \frac{x}{y}-1\right )^{2}$ $\rightarrow$ este suficient sa aratam ca $\left ( \sum \frac{x}{y}-1\right )^{2}\geq 4$ $\leftrightarrow$ $\sum \frac{x}{y}\geq 3$ . QED
Se dovedeste ca relatia $x+y+z=1$ este inutila.

cum reiese a doua inegalitate?
prima am inteles cum ai rezolvat-o

AM-GM $\rightarrow$ $\frac{x}{y}+2\sqrt{\frac{y}{z}}+3\sqrt[3]{\frac{z}{x}}\geq 6\sqrt[6]{\frac{x}{y}\cdot \left ( \sqrt{\frac{y}{z}\right )^{2}\cdot \left( \sqrt[3]{\frac{z}{x}}\right )^{3}}}=6$ , sper ca e clar acum :lol:

ma mai revansez cu una: x,y,z>0. atunci:

x/(y+z) + y/(x+z) + z/(x+y) + 9/2 e mai mic sau egal cu (x+y)/z + (y+z)/x + (x+z)/y

problema precedenta am luat-o de la un exercitiu care avea subunctele i, ii, iii si cea care am postat-o eu era ii. ipoteza x+y+z =1 trebuia doar pt celelalte puncte, insa ea era postata ca o ipoteza pt toate cele 3 subuncte. nu-i asa ca nu am fost destul de clar? :D

De fapt are loc inegalitatea $\sum{\frac{x+y}{z}}\geq  4\sum{\frac{x}{y+z}}\geq  \frac{9}{2}+\sum{\frac{x}{y+z}}$.

Pentru prima se foloseste Cauchy: $z (\frac{1}{x}+\frac{1}{x}) \geq  \frac{4z}{x+y}$, si pentru a 2-a Nesbitt.

De fapt are loc chiar si inegalitatea mai tare decat prima, gasita de mine si de Cezar:
$\sum{\frac{x+y}{z}}+\frac{16xyz}{(x+y)(y+z)(z+x)}\geq  4\sum{\frac{x}{y+z}}+2$.

Astept solutiile voastre :).


Let $H$ be the orthocenter of triangle $ABC$ with traces $D,E,F$ on the sidelines $BC,CA,AB$ respectively.Let $M,N,P$ be the mid-points of $EF,FD,DE$.Contruct $B',C'$ such that $N,P$ are mid-points of $BB',CC'$ respectively.Prove that:$HM\perp B'C'$

Not really difficult by using complex number/vector :D

You can show the following which is easier to prove:

If we call $H'$ the projection of $H$ to $EF$, then $B', H', C'$ are collinear. Moreover, the line $B'H'C'$ is the polar of the point $M$ with respect to the incircle of $DEF$.


This is my first time participating in the USAMTS, and I have a question. If I have a partial solution, i.e. I can solve about half the problem, should I submit it, or would I be better off not answering the question at all?

Thanks.

I think its best to leave your half-solution in because then you might at least get 1 point rather than no points at all.

If you have a partial solution, then by all means, please submit it.  You'll likely get some partial credit, and the graders may be able to show you (in their comments) how to extend your partial solution to a complete solution.

Right, thanks, I'll do that if I can't find a way to complete my solution by the 12th.


floor function
combinatorics proposed
combinatorics

Let $ n,k$ be positive integers. Determine the largest $ m$ such that for any set $ S$ of $ m$ permutations on $ \{1,...,n\}$, there exists a permutation which agrees with each element of $ S$ in at most $ k$ places

Well, here a few small values of $ m(n, k)$:

$ m(3,0) \equal{} 1$, $ m(3, 1) \equal{} m(3, 2) \equal{} 5$.
$ m(4, 0) \equal{} 2$, $ m(4, 1) \equal{} 3$, $ m(4, 2) \equal{} m(4, 3) \equal{} 23$.

In general, $ m(n, n \minus{} 2) \equal{} m(n, n \minus{} 1) \equal{} n! \minus{} 1$ since any permutation not in $ S$ disagrees with every member of $ S$ in at least two places.
Also $ m(n, 0) \leq \left\lfloor\frac{n}{2}\right\rfloor$: split the permutation into two unequal halves and apply a long cycle to the larger half separately to get a set of $ \left\lfloor\frac{n}{2}\right\rfloor \plus{} 1$ permutations that can't be avoided.  I feel this should be optimal.
Haven't thought further yet.

[quote="JBL"]Also $ m(n, 0) \leq \left\lfloor\frac {n}{2}\right\rfloor$: split the permutation into two unequal halves and apply a long cycle to the larger half separately to get a set of $ \left\lfloor\frac {n}{2}\right\rfloor \plus{} 1$ permutations that can't be avoided.[/quote]Can you elaborate this a little more? I thought $ m(n, 0)\equal{}n$

Like if you have
12543
35421
24315
(this is chosen randomly) you can complete it with
43152
51234

No?

I was suggesting
12345
23145
31245
as a triple with no fourth that avoids all of them.  (You didn't require that the set $ S$ consist of permutations with no values in common; was that intended?)


[hide]We put 21 objects in a line, and then split them into three groups using two dividers.  There are 20 possible locations for the dividers, so the answer is $\binom{20}{2}=190$.[/hide]

[quote="superwilson"]if x, y, z are positive integers, compute the number of solutions (x, y, z) that solve the equation x+y+z=21.[/quote]
Let's look at the following picture:
-----|--------------|--
The number of -'s before the first | represents $x$, the number of -'s between the |'s represents $y$, and the number of -'s after the  second | represents $z$. This is a one-to-one correspondence between every solution to $x+y+z=21$ and every configuration of the picture.
There are 21 -'s, so there are 20 spaces in between them that we can place the 2 |'s in, the total number of ways is ${20 \choose 2}=\frac{20 \times 19}{2 \times 1}=\boxed{190}$, the answer.

Nice solution....now...extending the problem....what would be the solution & how - if (i) none of them is zero (ii) if they need to take unique values (iii) if both (i) & (ii) need to be satisfied...??? :?:

with zeroes you get 2 more places for the divider (both ends), and you can also use one divider (y=0)...
it should be
${22\choose{2}}+22$

If we want nonnegative solutions (zero is allowed), then we can again take twenty objects and two dividers.  This time we can arrange the dividers however we want - we can put two dividers between the same pair of adjacent objects, and we can put one or more dividers at the end.  We can take care of this problem by thinking about it as a sequence of 22 "things,"  two of which are dividers.  Therefore, the answer is $\binom{22}{2}=231$.

[quote="fizsle"]Oops my question was wrong.... how many ways with '0' and with duplicate values?[/quote]

[hide]
by using dividers and sticks
I think it is $\binom{22}{2}$ which is ... 231
[/hide]


In order to complete a large job, $ 1000$ workers were hired, just enough to complete the job on schedule.  All the workers stayed on the job while the first quarter of the work was done, so the first quarter of the work was completed on schedule.  Then $ 100$ workers were laid off, so the second quarter of the work was completed behind schedule.  Then an additional $ 100$ workers were laid off, so the third quarter of the work was completed still further behind schedule.  Given that all the workers work at the same rate, what is the minimum number of additional workers, beyond the $ 800$ workers still on the job at the end of the quarter, that must be hired after three-quarters of the work has been completed so that the entire project can be completed on schedule or before?

$ AIME$ $ \#5$

Suppose the whole project involved building 4 similar buildings, each of which required a time of $ T$ to complete. The expected time for the total project is then $ 4T$.

During the second quarter there are only 900 men to do the work, so a time of $ \frac {10}{9}T$ is required (the time taken is inversely proportional the the number of workers.

During the third quarter there are only 800 men to do the work, so a time of $ \frac {5}{4}T$ is required.

To finish the work on schedule requires the fourth building to be constructed within time $ 4T - \frac {10}{9}T - \frac {5}{4}T = \frac {23}{36}T$.

Since $ \frac {10}{8}T \times \frac {23}{45} = \frac {23}{36}T$, we need $ 800 \times \frac {45}{23} = 1565.2174$ workers. The next highest integer is $ 1566$, and so our answer is $ 1566 - 800 = \boxed{766}$.


I've seen this come up, and tested it myself. It seems that if you post a link at the end of your post, and someone quotes it, a </td> tag comes up in the quoted text.

Original:
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1100797#1100797

Test Forum:
http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1100822#1100822


Hi
I need to insert *.dwg in my document (autocad drawing). Naturally, Latex doesn't know the extension .dwg.  I tried to use advanced batch converter to covert my drawing but it doesn't work with *.dwg.

I also tried to insert a *.jpg and it didn't work neither. When converted in *.eps, nothing.  The only format working is *.pdf but then I have trouble sizing.

thank you

ledurt

Please search for this image including problem in this forum, it has been discussed before. For example in this topic: http://www.mathlinks.ro/Forum/viewtopic.php?highlight=image+latex+png&t=28002 ;)


The European equivalent of 8$\frac{1}{2}$' by 11' paper is called A4 paper, and its dimensions are 210 mm wide by 297 mm long. Its dimensions were chosen to make the ratio of length to width as close as possible to (sq.rt. n), where n is an integer. WHat is the value of n?

[hide]The ratio of length to width is $\frac{99}{70}$. The decimal is about $1.414$, which is close to $\sqrt{2}$. Therefore, $n=\boxed{\boxed{\boxed{\boxed{2}}}}$[/hide]

YUP, your right!
guess you should take back your words "why am i always wrong"! :D  :lol:

read the problem, it asked for the value of n.
so your answer should be 2, not sq rt 2. :D

YESYESYES :first:  :first:
I am right for once


modular arithmetic
algebra unsolved
algebra

3. Show that $ \sqrt{x}$ of the number $ x\equal{}\overline{11\dots 122\dots 25}$ is an integer, where $ 1$ appears $ n$ times and $ 2$ appears $ (n\plus{}1)$ times. Besides, find the exact value of $ \sqrt{x}$.

$ x \equal{} \overline{11\dots 122\dots 25}$
$ x \equal{} 11\dots 1 \times 10^{n\plus{}2} \plus{} 22\dots 2 \times 10 \plus{} 5$

Use the property $ \overline{11\dots 1} \equal{} \frac{10^n \minus{} 1}9$, where $ n$ denotes the number of ones:

$ x \equal{} \frac{10^n \minus{} 1}9 \times 10^{n\plus{}2} \plus{} \frac{2(10^{n\plus{}1} \minus{} 1)}9 \times 10 \plus{} 5$

$ x \equal{} \frac{10^{2n \plus{} 2} \minus{} 10^{n\plus{}2} \plus{} 2 \times 10^{n\plus{}2} \minus{} 20 \plus{} 45}9$
$ x \equal{} \frac{10^{2n \plus{} 2} \plus{} 10^{n\plus{}2} \plus{} 25}9$
$ x \equal{} \frac{(10^{n\plus{}1} \plus{} 5)^2}9$

$ \sqrt{x} \equal{} \frac{10^{n\plus{}1} \plus{} 5}3$

As $ 10^{n\plus{}1} \plus{} 5 \equiv 0 \pmod{3}$, we have that $ \sqrt{x}$ is always an integer.

In fact, $ \sqrt{x} \equal{} \frac{10^{n\plus{}1} \plus{} 5}3 \equal{} \frac{10^{n\plus{}1} \minus{} 1 \plus{} 6}3 \equal{} \frac{10^{n\plus{}1} \minus{} 1}3 \plus{} \frac63 \equal{} 3\frac{10^{n\plus{}1} \minus{} 1}9 \plus{} 2$

Use the property again to get $ \sqrt{x} \equal{} 33\dots 3 \plus{} 2 \equal{} \overline{33\dots 35}$, where there are exactly $ n$ threes.


The rickroll is a different color than the hide tags.

:rotfl:  :rotfl:  :rotfl:

that is so hilarious.

dang I almost got rickrolled again.

Oh so gauss's suspicions about stevenmeow being bobby shen were right.

[quote="gauss1181"]Well I didn't know your voice was so high pitched so I thought you were pythag :rotfl:  :rotfl:  :rotfl:[/quote]

DUDE HOW COULD YOU MISTAKE SOMEONE SO UN-AWESOME FOR ME.

I did call Max's room about 1000 times to ask about the mafia game results but everyone just ignored me.

Dude...

[quote="Temperal"]dude david you only called twice?

also was it you who called my room, calvin? or was it david? or bleh?[/quote]

I only called max.

[quote="pianoforte"]Dude, don't play with a girl's heart. That's terrible[size=59]-y funny[/size]. I hope gauss learns her lesson, that men are all arseholes.

Except mew. He's too beastly.

PS. what are you kids doing up at 2 AM? STOP MAKING BABIES. Mathcounts is competitive enough already. If there's a huge boom in mathcounts skill in 11-14 years, I shall be *very* suspicious.

PPS. who doesn't have a crush on bobby shen? he's adorable.[/quote]

Is bobby shen ever active on AoPS?
Also, i never knew that girls liked boys smaller than herself.

Bobby Shen used to be active before suffering a fatal "accident."

He will be active again once he returns from New York.

Man you kids sound like you had fun at Nats...
What happened to the CA team? They are like the only team that is...legi- nah I won't say it that's too mean to everyone else.

David was rude, Lewis failed, Michael failed, Kevin failed.

[quote="dnkywin"](since she apparently has a crush on stevenmeow)[/quote]

Dude, who doesn't. He's really cute.

If I was a girl, I would love Bobby Shen.

His voice is really cool too.

If I was a girl, I would love Max.

Unfortunately I'm not a girl :(

Also unfortunately Max already has a gf :(

How was disney world... :(

Oh God. Bobby's so dreamy...

His eyes, his face, his nose, his smile...

His voice most of all...

Oh Bobby, if I only I could confess my love for you.

I am thy Juliet, and thou art my Romeo. Such it is, and forever will be.

[quote="alex960531"]Ummm...... is this like a love topic now?[/quote]

A word of advice: Do not read Romeo and Juliet at 12:30 am. It has serious and disastrous side effects including but not limited to the evocation of certain romantic rues similar to those used in Shakesperian plays, predominantly like those in Romeo and Juliet.

Btw, I was totally kidding. I still think he's cute though.

[quote="alex960531"]veexbo, are you a male or a female?[/quote]

Another thing, I am not reading Romeo and Juliet for enjoyment. Blargh, darn LA.
Actually, Romeo and Juliet is really quite entertaining and I really like the story.

I am male.

First male that likes shakespeare!!!!!!!!!!!!!

can somebody plz change the title to 'Who thinks bobby shen is hot'

[quote="ra5249"][quote="alex960531"][quote="ra5249"]that would be silly, because then everyone would fit in that category?
how about just change it to "everybody <3's bobby shen"[/quote]

nononononononono, we should make it Bobby Shen is cute club since  that is true     :roll:[/quote]

see, therein lies your problem. the idea of a "club" is the idea of exclusion; that is, there are people who aren't members of it. clearly this is a contradiction.[/quote]

I'm not good with specific definitions of each words!!!!!!!!!!  :wink:

Then how should I word it then?

[quote="1=2"][quote="ra5249"]i confess to the world my crush on bobby shen[/quote]

hi kevin![/quote]

ra5249's real identity revelaed?    :ninja:

TPAM:
Hi Chen....

[color=#F8F8F8]seriously, there are like a million people in the US with last name Chen... [/color]

[quote="resurrection"][quote="dnkywin"]Since this is the forum that everyone visits:

Most hilarious thing:

At like 2AM on Sunday (at nats):

I decide to call gauss1181's room (since she apparently has a crush on stevenmeow) and I use my pythag011 voice to tell her that "me" and stevenmeow are in room 1105 (a random empty room).

A few minutes later, we call again, and say, "why haven't you come yet?". And she goes, "I went but no one answered the door," in a whiny voice.

And then, that morning, I go up to her and go like, "I heard you got prank-called last night."

She then rants about how David Yang is bad and how his coach needs to control him....

 :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:  :rotfl:

BTW:

[hide="pythag011"]David Yang[/hide]

[hide="stevenmeow"]Bobby Shen[/hide]

[url=http://smouch.net/lol/][color=blue]gauss1181's real name[/color][/url][/quote]

omg. i seriously just got rickrolled. why do i fail at life so much lmao. i even read the second comment -_-

oh. and dude. bobby shen loves me back. yeah. I WIN.[/quote]

Well, aren't you lucky.

[quote="dnkywin"][color=#F0F0F0]I love invisible text...
Anyways, I would also like to add that Max's winning strategy at Masters Round was playing poker the night before.[/color][/quote]

Is that the win where I lost even though I had AA?


It¡s well known the following formula,
\[
B_{n,m}=B_{n-1,m-1}+B_{n,m-1}, \qquad B_{n,m}=\binom n m,
\]
the question, find the general expression of $f_{m,n}$ such that satisfies an analogous recurrence relation, i.e.
\[
f_{n,m}=f_{n-1,m-1}+f_{n,m-1},\quad {\rm with} \quad f_{1,1}=0, \ f_{1,2}=A\in \mathbb{Z}, \quad f_{2,2}=1.
\]

Best Regards.

RobertuX

If $f_{1,2}$ really right?  Anyway just draw up a Pascal-type triangle and count the number of ways each entry contributes to the ones lower down.


I have a physics problem which went:

When a shower is turned on in a closed bathroom, the splashing of the water on the bare tub can fill the room's air with negatively charged ions and produce an electric field in the air as great as 1000 N/C. Consider a bathroom with dimensions 2.8 m x 2.4 m x 3.0 m. Along the ceiling, floor, and four walls, approximate the electric field in the air as being directed perpendicular to the surface and as having a uniform magnitude of 963 N/C. Also, treat those surfaces as forming a closed Gaussian surface around the room's air. What are (a) the volume charge density r and (b) the number of excess elementary charges e per cubic meter in the room's air?

i can't figure out how to get (a).  obviously it will bbe N/(C^3) but i'm not sure where to go.  If you are wondering (b) is 1.18*10^11

Hope you guys can help me

Because the electric field is perpendicular to all surfaces, the electric flux can be calculated as E * A where A is the surface area of the bathroom. Set this equal to Q/Eo as per the definition of Gauss's law where Eo is 8.85 * 10^-12.

-963 * (2.8 * 2.4 * 2 + 2.4 * 3 * 2 + 2.8 * 3 * 2) = Q/Eo

Q = -3.8 * 10^-7 C

This is the total charge of the room. To find charge density we divide this by the volume (2.8 * 2.4 * 3.0) and get     -1.89 * 10^-8 C/m^3

For part B, we simply divide the amount in part (a) by the charge on an electron (-1.6 * 10^-19) to get 1.18 * 10^11 e/m^3 excess.

In manipulating Gauss's law, its helpful to know that almost all the problems you get will involve the electric flux equalling 0 or E*A.

Along the same lines I have two more Gauss' Law Question

1-Two large metal plates of area 0.96 m2 face each other. They are 5.4 cm apart and have equal but opposite charges on their inner surfaces. If the magnitude E of the electric field between the plates is 45 N/C, what is the magnitude of the charge in coulombs on each plate?

2-Charge of uniform volume density  = 4.5 µC/m3 fills a nonconducting solid sphere of radius 2.8 cm. What is the magnitude of the electric field (a) 1.4 cm and (b) 3.7 cm from the sphere's center?

I would appreciate anyone's help. Thanks.

For the problem: Charge of uniform volume density  = 4.5 µC/m3 fills a nonconducting solid sphere of radius 2.8 cm. What is the magnitude of the electric field (a) 1.4 cm and (b) 3.7 cm from the sphere's center?

What does 'R' represent and what does 'r' represent? I solved part (b) thinking that R = .037 m (3.7 cm) and my answer was incorrect.

Thanks again.

[color=blue]Here's my solution[/color]
[quote="jeter123"]Along the same lines I have two more Gauss' Law Question

1-Two large metal plates of area 0.96 m2 face each other. They are 5.4 cm apart and have equal but opposite charges on their inner surfaces. If the magnitude E of the electric field between the plates is 45 N/C, what is the magnitude of the charge in coulombs on each plate?[/quote]
[color=blue]I think we don't need the distance between two plates
$\large E=\frac{Q}{\epsilon_0 S}$
$\large \Longrightarrow Q=E\epsilon_0 S$[/color]

[quote="jeter123"]2-Charge of uniform volume density  = 4.5 µC/m3 fills a nonconducting solid sphere of radius 2.8 cm. What is the magnitude of the electric field (a) 1.4 cm and (b) 3.7 cm from the sphere's center?[/quote]
[color=blue]
Using Gauss's law we have
$\large \frac{q}{\epsilon_0}=\frac{E}{S}$
$\large \Longrightarrow E=\frac{\rho }{3\epsilon_0}r$ for the magnitude of the electric field inner the sphere

where $0\leq r\leq R$ and

$E=\frac{\rho}{3\epsilon_0}\frac{R^3}{r^2}$ for the magnitude of the electric field outer the sphere

where $R\leq r$[/color]


i just want to post a website  :lol:

http://www.yourname.youaremyfriend.com

http://www.yourname.youaremighty.com

by your name i mean put your real name

hmmm I've also seen some bad ones
they kind of died though
yourname.youarelame.com
this one is still alive
yourname.justgotowned.com


limit
inequalities
algebra unsolved
algebra

I'd say $1$.

[hide="Proof"]
I think I remember this:
$\lim_{n\rightarrow +\infty}\sqrt[n]{a_n}=\lim_{n\rightarrow +\infty}\frac{a_n}{a_{n-1}}$ if every $a_i>0$ and the last limit exists and is finite.
Please correct me if I'm wrong. Has it got a name? (I think it's a corollary of the limit of a geometric mean).

In this case,
$a_n=\sqrt[3]{\binom{3n}{3k}}$.
$\frac{a_n}{a_{n-1}}=\sqrt[3]{\frac{\binom{3n}{3k}}{\binom{3n-3}{3k}}}\rightarrow 1$,
because
$\frac{\binom{3n}{3k}}{\binom{3n-3}{3k}}=\frac{(3n)(3n-1)(3n-2)}{(3n-3k)(3n-3k-1)(3n-3k-2)}\rightarrow 1$.
[/hide]

I think your proof it's fine.

Another way is to use the inequality

$\\ \left( \frac{n}{k} \right) ^k \leq \binom{n}{k} \leq \left( \frac{e \cdot n}{k} \right) ^k$ .

 :)

[quote="lordWings"]But then, you'd still have to prove that
$\lim_{n\rightarrow +\infty}\sqrt[n]{n}=1$,
and then you'd have to use the same property I used. Or is there another way?[/quote]

This is theory  ;)

Yes there another way to show it.

$\\ x_{n}= \sqrt[n]{n}-1 \Rightarrow (1+x_{n})^{n}=n$

and from binomial expansion $n \geq \frac{n(n-1)}{2}x^{2}_{n}$

therefore $0 \leq x_{n} \leq \sqrt{ \frac{2}{n-1} }$.

 :)

[quote="Ramanujan"]and from binomial expansion $n \geq \frac{n(n-1)}{2}x^{2}_{n}$[/quote]
Arg! I can't get this step. :?  Please someone help me step by step and then let me feel stupid.

[quote="lordWings"]
Arg! I can't get this step. :?  Please someone help me step by step and then let me feel stupid.[/quote]

 :lol:

 It is $x_{n}>0$ and

$\\ n=(1+x_{n})^{n}= 1+x_{n}+ \frac{n(n-1)}{2}x^{2}_{n} +... > \frac{n(n-1)}{2}x^{2}_{n}$   .

Hope it is ok now . :)

Thanks, now I can happily feel stupid  :P
In this step,
$n=(1+x_{n})^{n}= 1+x_{n}+ \frac{n(n-1)}{2}x^{2}_{n} +... > \frac{n(n-1)}{2}x^{2}_{n}$,
I was not sure of the inequality because I was stubborn not to assume $x_n>0$, even when it was obvious because $\sqrt[n]{n}>1$ with $n>1$.
To overthink or not to overthink. That is the question.