probability
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factorial
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probability and stats

Question 1: Suppose the number of events that occur in a given time is a Poisson r.v. with paramter lamda. If each event is classified as type i with probability P(sub i) i=1,2,...n where the sume of all Pi's is 1, independently of other events, show that the numbers of type i events occur, i=1,..,n, are independent Poisson r.v. with respective parameters P(i)*Lamda, i=1,...,n. For those of you with Sheldon First Course Seventh Ed, this is ch 6, theoretical exercise 2.

My main issue here is expressing the condition. I know we want the prob. of X1=say q1, X2=q2,... Xn=qn given the sum of Qn from 1 to n is the number of events, but I do not know how to express this. I got the right answer assuming you can multiply the given Poisson distribution by the P(i) terms times the factorial of the total number of events divided by the individual q factorial terms. I just need to figure this out.

Question 2: Same section ,theoretical ex. 6: Show by induction that X1+..._ Xn has a negative binomial distribution when the Xi, i=1,...,n are independent and identiclly distributed geometric r.v. Also, give a heuristic argument with no computations (easy).

I already did the base case and got a negative binomial r.v. However, when I go to do induction, I cannot make my result negative binomial r.b. I think the limits on my su may be wrong. Any ides?

Q1: Call $ N_i$ the numbers of type $ i$ events that occur, and $ N$ the total number of events (regardless of its type). Then
\[ P[N_i\equal{}m] \equal{} \sum_{k\equal{}m}^{\infty} P[N_i\equal{}m, N\equal{}k] \equal{} \sum_{k\equal{}m}^{\infty} P[N_i\equal{}m| N\equal{}k] P[N\equal{}k]\]
and also know that
\[ P[N_i\equal{}m| N\equal{}k] \equal{} \binom{k}{m} P_i^m (1\minus{}P_i)^{k\minus{}m} \]
and $ P[N\equal{}k] \equal{} \exp(\minus{} \lambda ) \frac{\lambda^k}{k!}$. You should be able to sum up the series to show that of course:
\[ P[N_i\equal{}m] \equal{}  \exp(\minus{} P_i \lambda ) \frac{(P_i \lambda)^m}{m!}\]

The independance also heavily uses the fact that there are Poison distributions:
\[ P[N_1\equal{}m_1, \ldots, N_n\equal{}m_m] \equal{} P[N_1\equal{}m_1, \ldots, N_n\equal{}m_m|N\equal{}m_1\plus{}\ldots\plus{}m_n]P[N\equal{}m_1\plus{}\ldots\plus{}m_n]\]
where
\[ P[N_1\equal{}m_1, \ldots, N_n\equal{}m_m|N\equal{}m_1\plus{}\ldots\plus{}m_n] \equal{} \frac{(m_1\plus{}\ldots\plus{}m_n)!}{m_1! \ldots m_n!} P_1^{m_1}\ldots P_n^{m_n}\]
\[ P[N\equal{}m_1\plus{}\ldots\plus{}m_n] \equal{} \exp(\lambda)\frac{\lambda^{(m_1\plus{}\ldots\plus{}m_n)}}{(m_1\plus{}\ldots\plus{}m_n)!}\]
so you should be able to prove that:
\[ P[N_1\equal{}m_1, \ldots, N_n\equal{}m_m] \equal{} P[N_1\equal{}m_1] P[N_2\equal{}m_2] \ldots P[N_n\equal{}m_n] \]

Is this OK ?

Thank you, and yes I understand everything. Any takers o nthe second problem?

Show us what you've got. It's hard to tell where you might be going wrong without seeing the sum.
If you do it right, the "hockey stick" identities will be involved.


By the way I think something is missing from the original problem.

A circular disk of negligible thickness and radius $ R$ has a charge $ Q$ distributed evenly across it. A mass $ m$ possessing a charge $ \minus{}q$ is directly above the centre of the circular disk at a distance of $ x$ ($ x<<R$). Find the period of oscillation. Let $ k$ be Coulomb's Constant.

I think it is supposed to be oscillating in Simple Harmonic Motion.

is there the presence of gravity(in the anti sense)? if it is'nt then i see no reason why the negative charge doesn't stick to the disc.
well what possibly could be the  question is time taken by them to collide

Does the circular disk have a hole at its center so that the tiny charge can oscillate up and down relative to the plane of the disk?

Ok, so we just have to compute the magnitude of the electric force applied by the disk at a small distance from its center.

well that logically should be that due to an infinite sheet carrying a positive charge with same  charge density that is $ \frac{\sigma}{2\epsilon_{o}}$

But then the motion isn't SHM. It is oscillatory motion all right with period
$ T \equal{} 8\sqrt\frac{\epsilon_{0}mx}{\sigma q}$

EDIT: Also, for a short interval of time, we cannot assume the hole to be small, though I have not included it in my calculation.


If $ {p}$ is prime and $ {p}\equal{}1 (mod4)$ such that $ {p}$ can be written as the sum of 4 numbers greater than zero $ {a,b,c,d}$,prove that  $ {a.d}$ cannot equal  $ {bc}$.

Let $ p>2$ be a prime, $ p\equal{}a\plus{}b\plus{}c\plus{}d$, with $ a,b,c,d$ positive integers. Assume $ ad \equal{} bc$.
We have $ a\plus{}d \equiv \minus{}(b\plus{}c) \pmod{p}$, so $ a^2 \plus{} 2ad \plus{} d^2 \equiv b^2 \plus{} 2bc \plus{} c^2 \pmod{p}$, hence $ a^2 \plus{} d^2 \equiv b^2 \plus{} c^2 \pmod{p}$. Then $ (a\plus{}b)(a\minus{}b) \equiv (c\plus{}d)(c\minus{}d) \pmod{p}$. But $ a\plus{}b \equiv \minus{}(c\plus{}d) \pmod{p}$ and $ a\plus{}c \equiv \minus{}(b\plus{}d) \pmod{p}$. Thus $ (a\plus{}b)(a\minus{}b\plus{}c\minus{}d) \equiv 0 \pmod{p}$, or $ 2(a\plus{}b)(a\plus{}c) \equiv 0 \pmod{p}$, absurd. We do not need  $ p \equiv 1 \pmod{4}$.


algebra
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superior algebra unsolved

Think of the elements of $\mathbb R^{m\times n}$ as matrices. Observe that $K=\{A\in S\colon \mathrm{rank}(A)=1\}$. This gives you a bunch of polynomials vanishing on $K$, such as the determinant of some $k\times k$-minor, for $2\le k\le\min(m,n)$.


greatest common divisor
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MATHCOUNTS
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Okay,
   So there is this type of problem I learned with a certain formula, but going along with the AOPS "Theory", I would rather understand why it occurs, rather than just knowing the formula.
Before I give the formula in question out, I'll try to see if I can stump anybody.  :rotfl:
Also, try not to solve the problem by making a list.

Patrick and Morris are playing a game in which they each chose a number out of a hat. Patrick's number is 9, while Morris's number is 11. The object of the game is to try and form every number greater than 0 by adding a combination of the two chosen numbers. (Obviously 1-8 are impossible, so disregard.) For example, the number 29 can be created by: 9+9+11.

What is the greatest number that  [b][i][u]can't[/u][/i][/b] be created by a combination of these numbers?

*I think the formula only works with relatively prime numbers*

[hide="proof of the formula... kinda"]
Consider the equation $ mx\plus{}ny\equal{}k$ where $ m>n$ and $ \text{GCF}(m,n)\equal{}1$.  In mod $ m$, this is \[ ny\equiv k\pmod{m}\] Letting $ y\equal{}1$ and adding $ m$ repeatedly gets us all numbers that are $ n$ in mod $ m$ that are greater than or equal to $ n$.

Similarly, letting $ y\equal{}2$ gets us all numbers $ 2n$ mod $ m$ that are greater than or equal to $ 2n$.  Note that the largest number that is either $ 0$, $ n$, or $ 2n$ in mod $ m$ that cannot be made is $ 2n\minus{}m$ (unless $ 2n<m$)

In general, for $ y\equal{}k$, the largest number that is $ kn$ in mod $ m$ that cannot be made is $ kn\minus{}m$ (do you see why?)

So if we let $ y\equal{}m\minus{}1$, we get $ (m\minus{}1)(n)\minus{}m\equal{}mn\minus{}m\minus{}n$.
[/hide]

In MATHCOUNTS, we call that the "Mcnugget Formula"  :rotfl:

[quote="yankeeprez"]*I think the formula only works with relatively prime numbers*[/quote]

If they aren't relatively prime, let their GCF be $ c$.  Then all numbers not divisible by $ c$ cannot be made.

When $ m$ and $ n$ are relatively prime, then the set $ \{ n,2n,3n, ..., (m\minus{}1)n \}$ when taken in mod $ m$ is the set $ \{ 1,2,3,...,m\minus{}1 \}$.  Without this the proof wouldn't quite work.

Maybe it's that I'm overtired, but what you guys said doesn't make sense to me. Stuff like "mod" is way past my abilities!
The formula I know is: Product-Sum (Of the Two Numbers.)

So basically, 9*11-(9+11)
So 99-20
79.

If you were to make a list (try with 5 and 7), you would get to the point where 23 would be the last number that doesn't work, and every number after that does.

So I am sorry if what I just said translates to what you two said, but I think my way is a little easier!
Thanks though!

[quote]mn-m-n[/quote]
yankeeprez, they simply proved this formula. :wink:

Wow, how did I miss that?! :blush:

I guess I just saw it as a whole and skipped over it! :rotfl:
Thanks!


function
integration
calculus
real analysis
real analysis solved

0<=a R is a monotonic function s.t. \int{a->b}xnf(x)dx=0 for all n>=k (n natural). Show that f(x)=0 for all x\in (a,b).

Dear Grobber, do you have a solution for this magnificient problem? I tried to solve it, but I failed. I like it very much and I'm also interested in a clever solution. Thank you very much.

Use the 'vanishing theorem'. Assume that the function is not zero and forcing the integral to vanish at the same time. You should get a contradiction.

Hi Hi harazi!!! :)
I'm sorry for my late reply!!! Yay!!! :blush:
Anyways, here is my solution but it depends on the 'vanishing' theorem and so I will post a topic in the
College Playground->Calculus-Analysis->Theorems&Formulas with proof!

Here we go:
Since f is monotonic throughout the interval [a,b], there is at most one zero in f. This fact can be shown by the Intermediate Value Theorem.
Let g(x)=xnf(x), for some fixed n. Clearly g(x) is also monotonic and has a zero at x=0.
Since a monotonic function can have at most one zero unless it is the zero function, we can assume that g(x) \geq 0 for all x\in[a,b].

By the 'vanishing' theorem, g(x)=0 in the interval (a,b). Since x\in(a,b) is positive, f(x)=0 in the same interval.

Now please refer to the 'vanishing' theorem in the Theorems & Formulas section for the complete solution!
 :D  :D  :D
3X.lich

In our problem f is not continuous, in the intermediate value theorem continuity is require.
http://mathworld.wolfram.com/IntermediateValueTheorem.html
In the vanishing theorem the function must be also continuous.

And why is x^n*f(x) monotonic? I don't think this is true!

There is a c s.t. f(c)=0. Then let c1<=c<=c2
 s.t. f(x)=/=0 for all x not in [c1,c2]. Just look at what happens to xnf(x) on the intervals [a,c1] and [c2,b] as n->00. I'll post the complete solution a bit later, because it wouldn't be nice to spoil it so soon :).

Here is a theorem for Moubinool why f is continuous 'almost everywhere'. Prove this theorem and you will know what I mean. My solution still holds true.

Theorem: Let f be monotonic on (a,b). Then the set of points of (a,b) at which f is discontinuous is at most countable.
 :D
If f is not continuous everywhere, the integrals wouldn't be integrable anyways.

The proof for that is pretty well-known (the countability of the discontinuity pts of an increasing function):

It's well-known that a monotonic function f:[a,b] has finite lateral limits in any point, and denote by d(x) the difference between the limit to the right and the limitto the left. d(x)>=0 because we assumed f to be increasing (it works, of course, for f decreasing as well). Then D (the set of discontinuity pts) =  \cup{n>=1}{x\in [a,b] | d(x)>=(f(b)-f(a))/n}. This is a countable reunion of sets, so in order to prove it's countable it suffices to show that every such set is at most countable. We thus want to show that the set An={x\in [a,b] | d(x)>=(f(b)-f(a))/n} is at most countable, which is pretty obvious (every such set is finite; try to figure out why).

f is monotone (for example increasing) => f is Riemann integrable
 => f is bounded

As grobber said there exist c in [a,b] s.t. f(c) = 0

let 0  \leq  a < c_1 < c_2 < b

such that f(x)=/=0 if x is not in [c_1,c_2]

0 =\int_[a,c_1]x^nf(x)dx+\int_[c_1,c_2]x^nf(x)dx+\int_[c_2,b]x^nf(x)dx

\int_[c_1,c_2]x^n|f(x)|dx  \geq ||\int_[a,c_1]x^nf(x)dx| - |\int_[c_2,b]x^nf(x)dx||

Let M = sup f > 0 > m = inf f ( f is increasing f is not 0 outside [c_1,c_2] )

\int_[c_1,c_2]x^n|f(x)|dx  \geq |m|\int_[a,c_1]x^ndx| - M|\int_[c_2,b]x^ndx||

M\int_[c_1,c_2]x^ndx \geq 1/(n+1). |m|c_1^(n+1)-a^(n+1)| - M|b^(n+1)-c_2^(n+1)||   (*)

(*) becomes

M.|c_2^(n+1) - c_1^(n+1)| \geq  |m|c_1^(n+1)-a^(n+1)| - M|b^(n+1)-c_2^(n+1)||   (**)

divides (**) by b ^(n+1)

M.|(c_2/b)^(n+1) - (c_1/b)^(n+1)| \geq  |m|(c_1/b)^(n+1)-(a/b)^(n+1)| - M| 1- (c_2/b)^(n+1)||

taking the limit when n-->+oo

0 \geq M >0 contradiction

This is exactly what I did, but now I like 3x.lich's solution better :D.

I think there's a nice theorem saying that on a compact interval a function f can be written as the limit of a sequence of polynomials.And with this one we get  \int {a->b}f 2 =0 => f=0
I dont' remember if the theorem states that f should be monotonic.If not you got a harder problem.

As far as i remember, f must be continuous.

I think that's yet another theorem of Weierstrass. From what I can remember, f must be continuous, and, anyway, the fact that we only have xn for n>=k might prevent us from a very good approximation (we migh need x, x2.. etc. in our approximations).

There was a Titeica contest where they proposed a problem which was really tough if you didn't know this particular theorem, but I can't remember the year (or the problem, for that matter :)).


abstract algebra
superior algebra
superior algebra solved

Hello , can anyone point me the mistake in that (small) proof
Let $G$ be a group of order $4$. Suppose $G$ not cyclic.
Hence $|Z ( G ) | = 2$ (we know that the center $Z ( G ) \neq e$ for any $p$-group) and all elements have period $2$ (otherwise if one element of $G$ has period $4$ then it generates the group $G$ contrary to assumption)
Thus there are exactly two elements that do not commute between them.
Let them be $a_1, a_2 .$ So $a_1 a_2 \neq a_2 a_1$. Multiply by $a_1$, then $a_2$ to obtain $a_2 a_1 a_2 a_1 \neq e$.
But $( a_2 a_1 )$ has itself period $2.$ Hence $( a_2 a_1 ) ( a_2 a_1 ) = e$. Contradiction.

The problem is that there is a group of order 4 not isomorphic to $G$ being the product of two cyclic groups.
So $G$ cannot be necessarily cyclic.

P.S: I'm tired so I apologize if I wrote a bunch of non-sense (which does not mean that I do not welcome any kind of replies :))

[quote="jmerry"]A group that isn't cyclic can be abelian (as this one is).[/quote]

Yes. Homever I proved that $G$ has order $4$ imply $G$ is cyclic.
I know this to be false. Hence I'm looking for the mistake.
If I proved something else, please point this out explicitely.
(I'm unable to figure this out at the moment)

You say that the center has just $2$ elements, which is wrong for $G= (\mathbb{Z} / 2 \mathbb{Z})^2$ (the center is not necessary cyclic when having $4$ elements, here you use the conclusion in the prove, thus a circular reasoning).

Okay I see that. $|Z(G)| = |G|$ (which is equivalent to $G$ abelian) does not imply that $G$ is cyclic.
Thus we can suppose $|G|$ cyclic and $|Z(G)|=4|$ in that particular case.
Thanks to all of you, ZetaX, jmerry ! :)


vector
function
advanced fields
advanced fields unsolved

Since continuity is automatic for "isolated points", we generally test for continuity only at accumulation or cluster points. So I will try to use the following fact:

Fact: A set M is closed iff it contains each of its accumulation (cluster) points.

Proof: If an accumulation point c were not in M, it would be a boundary point of M, but M must contain all its boundary points since it is closed, a contradiction. Going  the other way is not hard to prove.

It seems that $f$ is a linear functional on some topological vector space, possibly a Banach space. A clarification would help.

[quote="kingkong"]Let $f \in L(X, R)$. Prove that:
f is continiuos if and only if $M = \{ x \in X | f(x) \geq\ 1\}$ is closed in X[/quote]
Sorry.
Suppose f : X -> R is a linear function (not continiuos).


Find all $ m$ the inequality
$ m^2 \minus{} 2m(\cos x \minus{} \sin x) \plus{} 2\sin^2 x\geq 0$
was true for every $ x\in \mathbb{R}$

[hide="A start?"]
\[ m^2 \minus{} 2m(\cos{x} \minus{} \sin{x}) \plus{} 2\sin^2x \\
\equal{} (m \minus{} \cos{x} \plus{} \sin{x})^2 \plus{} 2\sin{x}\cos{x} \minus{} (\cos^2{x} \minus{} \sin^2{x}) \\
\equal{} (m \minus{} \cos{x} \plus{} \sin{x})^2 \plus{} \sin{2x} \minus{} \cos{2x}\ge0 \\
\Longleftrightarrow m\le\cos{x} \minus{} \sin{x} \minus{} \sqrt {\cos{2x} \minus{} \sin{2x}} \equal{} f(x) \\
\text{or} \\
m\ge\cos{x} \minus{} \sin{x} \plus{} \sqrt {\cos{2x} \minus{} \sin{2x}} \equal{} g(x)\]
for all $ x$ when $ \cos{2x} \minus{} \sin{2x}\ge0$ (otherwise the inequalities before taking the square root are trivially true). Since this is all periodic $ \bmod{2\pi}$, we only consider the intervals $ \left[ \minus{} \frac {3\pi}8,\frac\pi8\right]$ and $ \left[\frac {5\pi}8,\frac {9\pi}8\right]$ where $ \cos{2x} \minus{} \sin{2x}\ge0$.

We must prove four claims:
[list][*]On the interval $ \left[ \minus{} \frac {3\pi}8,\frac\pi8\right]$, $ \min{f(x)} \equal{} 0$.
[*]On the interval $ \left[\frac {5\pi}8,\frac {9\pi}8\right]$, $ \max{g(x)} \equal{} 0$.
[*]On the interval $ \left[\frac {5\pi}8,\frac {9\pi}8\right]$, $ \min{f(x)}\equal{}a\approx \minus{} 2.53516$.
[*]On the interval $ \left[ \minus{} \frac {3\pi}8,\frac\pi8\right]$, $ \max{g(x)}\equal{}b\approx2.53516$.[/list]

The first two can be proven without much difficulty, but I'm stuck on the other two.

Anyway, permissible values of $ m$ are $ ( \minus{} \infty,a]\cup\{0\}\cup[b,\infty)$.[/hide]


Hello,
Could anyone help me out here?  :)

Let $K_{1}, ..., K_{n}$ are subspaces of vector space $V$ and $V=U_{i=1}^{n}K_{i}$.
Show that $V=K_{r}$ for some $r$

That is not true.

Let $L$ be a field with cardinality two.

Let $V=L^{2}$ .

Take $K_{1}=\{(x,y),x=y\}\subseteq V$
$K_{2}=\{(x,y),x=0\}\subseteq V$
$K_{3}=\{(x,y),y=0\}\subseteq V$

[quote="koola"]I dont understand by your saying "it's not true".

Ohh, it maybe be my mistake since I used $K$ instead of $W$.
Since $K$ defines as field that I should have kept the $W$ as subspaces of a vector space $V$.

So would the proof be different now?[/quote]

Now you are confusing me.  Let's just do it properly and give me a full complete new version of the problem.  Define a field , spaces etc.. with nice letters  :)
I'm too confused now :(

Sorry my mistake,
The original question should be this:

Let $W_{1}, ..., W_{n}$ are subspaces of a vector space $V$ and $V=\cup_{i=1}^{n}W_{i}$
Show that $V=W_{r}$ for some $r$
Prove it when $K$ is infinite and give counterexample when $K$ is finite

Well then, I have already solved the second part of that problem (giving a counterexample for finite fields) in this topic.

But concerning the first part....  :o  it's so obvious, but while I do have some ideas for a proof, I can't really find something waterproof  :(

I'll assume first that $V$ is finite dimensional, of dimension $d\ge 1$, say. We can prove by induction on $d$ that an affine $d$-dimensional space $k^{d}$ over an infinite field $k$ cannot be covered by a finite number of affine hyperplanes $W_{1},\ldots,W_{n}$. The conclusion follows immediately from this.

For $d=1$ the assertion is just a restatement of the fact that $k$ is infinite, so it does indeed hold. Assume now that $d\ge 2$, and that we've shown it to be true for spaces of dimension $\le d-1$. Because $k$ is infinite, $k^{d}$ has infinitely many affine hyperplanes; we can thus find one, $W$, different from all the $W_{i}$. Those $W_{i}$ which intersect $W$ do so along hyperplanes of $W$, so we can use the induction hypothesis on $W$.

The infinite dimensional case follows from the finite dimensional one: every finite dimensional subspace $W$ of $V$ is contained in one of the $W_{i}$; if $w_{i}\ne 0$ is a vector in $V$ but not in $W_{i}$, then the space spanned by the $w_{i}$ is finite dimensional but not contained in any of the $W_{i}$, so we have a contradiction.


trigonometry
inequalities
logarithms
geometry
incenter
angle bisector

In a triangle $ ABC$, prove that

$ \tan \frac{A}{2} \tan \frac{B}{2} \tan \frac{C}{2} \leq \frac{\sqrt{3}}{9}$.

Consider the value of $ \tan \frac{A}{2}\tan \frac{B}{2}\plus{}\tan \frac{B}{2}\tan \frac{C}{2}\plus{}\tan \frac{C}{2}\tan \frac{A}{2}$.

Thanks kunny.

I make this $ 1$, by using $ \tan \frac{A}{2}\equal{} \sqrt{\frac{(s\minus{}b)(s\minus{}c)}{s(s\minus{}a)}}$ et cetera.

And then?

[By the way it works out nicely using Jensen's inequality for $ f(x) \equal{} \ln \tan x$.]

Thanks kunny. The answer was staring me in the face; don't know why I didn't see it. Thanks again.

You are welcome.  :lol:

[quote="AndrewTom"]$ \tan \frac {A}{2} \equal{} \sqrt {\frac {(s \minus{} b)(s \minus{} c)}{s(s \minus{} a)}}$ et cetera. [/quote]

Is this a common sense? In Japan, that fact usually doesn't come up in class.

I think that Japanese education of Geometry is out of date, so is Number theory.

Hi kunny.

I came across this in an old textbook: "Elementary Trigonometry" by C V Durrell and R M Wright, published in London by G Bell and Sons in 1927!

That's a pretty well-known and easily proven fact.

[hide="Proof"]
Take a triangle, draw the incenter, drop a perpendicular from the incenter to a side, and draw the angle bisector through the incenter. It's clear that the tangent of the half angle is $ \frac{r}{s\minus{}a}$ (or $ s\minus{}b$ or $ s\minus{}c$).

$ rs\equal{}\sqrt{s(s\minus{}a)(s\minus{}b)(s\minus{}c)}$ as two different expressions for triangle area, so $ r\equal{}\sqrt{\frac{(s\minus{}a)(s\minus{}b)(s\minus{}c)}{s}}$.

Substitute that in for $ r$ and you're done.
[/hide]

[quote="AndrewTom"]In a triangle $ ABC$, prove that

$ \tan \frac {A}{2} \tan \frac {B}{2} \tan \frac {C}{2} \leq \frac {\sqrt {3}}{9}$.[/quote]
[img]http://www.mathmontada.net/vb/uploaded/24_1263334418.gif[/img]

It is known $ \frac{r}{p}\equal{}\tan \frac{A}{2}\tan \frac {B}{2}\tan \frac{C}{2}$
$ p\ge 3\sqrt{3}r$
Hence $ \tan \frac{A}{2}\tan \frac{B}{2}\tan\frac{C}{2}\leq \frac{\sqrt{3}}{9}$


For a given arithmetic series, the sum of the first 50 terms is 4825, and the sum of the next 50 terms is 12325. find the first term of the arithmetic series.

[hide="SOLUTION (HELP!)"]
[b]1[/b]
With d as the common difference,
$\sum_{n-1}^{50}a_{n}=\sum_{n-1}^{50}(a_{1}+(n-1)d)$
which equals

[b]2[/b]
$50a_{1}+d\sum_{n-1}^{50}(n-1)=50a_{1}+d\frac{(50-1)50}{2}=4825$.
Similarly

[b]3[/b]
$\sum_{n-1}^{100}a_{n}$

[b]4[/b]
$=\sum_{n-1}^{100}(a_{1}+(n-1)d)$

[b]5[/b]
$=100a_{1}+d\sum_{n-1}^{100}(n-1)$

[b]6[/b]
$=100a_{1}+d\frac{(100-1)100}{2}$

[b]7[/b]
$=4825+12325=17150$.
Thus we have

[b]8[/b]
$50a_{1}+1225d=4825$ and $100a_{1}+4950d=17150$.

[b]9[/b]
Double the first of these and subtract from the second to get
$2500d=7500$, so $d=3$. Then $a_{1}=23$.

[/hide]

I think I need help with summation sign and how numbers keep coming in and out of it (identities)? Do you guys have a link for me to read and practice? or

for [b]2[/b] why are these two the same thing?
$d\sum_{n-1}^{50}(n-1)=d\frac{(50-1)50}{2}$

That bottom part should be $n=1$, not $n-1$.  It follows from the identity

$1+2+3+...+k = \frac{k(k+1)}{2}$

No.  Since $n$ is the "index" (the variable that changes in the summation), it has to stay inside the summation.

Anyway, this solution is "standard," but there's a bit of a shortcut you can employ.

[hide="Somewhat shorter solution"] The first $50$ terms are

$a, a+d, a+2d, ... a+49d$

The next $50$ terms are

$a+50d, a+51d, ... a+99d$

The sum of the difference between them is

$12325-4825 = 7500 = 50d+50d+...+50d = 2500d$

So $d = 3$.  Now,

$\sum_{n=0}^{49}(a+nd) = 50a+3 \cdot \frac{49(50)}{2}= 4825 \implies a = \boxed{23}$ [/hide]

My way:
[hide]Two equations, two variables
Let $a$ represent the first term of the series and $d$ represent the common difference, then we have:
$25(2a+49d)=193*25$
$25(2a+149d)=493*25$

$100d=300 \rightarrow d=3$
Plug $d$ back in to one of the equations to get
$a=23$[/hide]


Five sailors are marooned on an island.  So they can survive until the day that they are (hopefully) rescued, the sailors and a native monkey spend the day gathering coconuts, which are all put in a big pile at the end of the day, after which they all fall asleep.

During the night, one of the sailors gets up, and, not trusting the other sailors to distribute the coconuts evenly, decides to take his $\textstyle\frac{1}{5}$ of the pile of coconuts for his own then and there.  When he divides the coconuts into five equal piles, however, he has one left over, so he tosses it to the monkey.  He then hides his share, puts the rest of the coconuts in one pile again, and falls asleep.

Later in the night, one of the other sailors gets up, and, also not trusting the other sailors to distribute the coconuts evenly, decides to take his $\textstyle\frac{1}{5}$ of the pile to avoid getting shortchanged in the morning.  Like the first sailor, he finds that when he divides the pile into five equal piles, there is one coconut left over, so he tosses that to the monkey.  He hides his share, puts the rest of the coconuts in one pile again, and falls asleep.

As you have probably guessed, the other three sailors do the same thing.  Each time they split up the coconuts into five equal piles; each time they find one leftover coconut, which is tossed to the monkey; and each time they take their $\textstyle\frac{1}{5}$ of the remaining coconuts before putting the rest back into a big pile, hiding the coconuts that they took, and going back to sleep.

In the morning, when the sailors get up, they all pretend that they have taken none of the coconuts during the night and split up the remaining pile as they had planned; each sailor gets $\textstyle\frac{1}{5}$ of the pile.  This time, however, the division goes evenly, with no coconuts thrown to the monkey.  No coconuts were ever chopped up into pieces.

What is the least number of coconuts they could have gathered on their first day on the island?

Can you generalize?  If there are n sailors, and at each division the monkey gets k coconuts tossed to it (except at the final division,when it gets none), what is the least number of coconuts they could have gathered (in terms of n and k)?

Click [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=114985#114985]here[/url] for a link to an easier version of this problem.

[quote="Diogene"]The problem was proposed at the 9th IMO 1967 .
Actually, it's a much older problem (Nicolas Chuquet , 14th century) !

 :cool:[/quote]

I believe the original version you refer to probably has an extra coconut for the monkey at the final division (see the above link for that problem).  According to [i]The Colossal Book of Mathematics[/i], this version came about as part of a short story published in the October 9, 1926 issue of [i]The Saturday Evening Post[/i], with an added wrinkle to make the problem harder (no coconuts for the monkey at the final division).  Perhaps I'm wrong, though, and it is an old problem.

Do you have a solution?  [i]The Colossal Book of Mathematics[/i] describes this problem as "probably the most worked on and least often solved of all Diophantine brainteasers."


hello tell me u think mewto5555 is awesome

The closest option to "no" on the poll is "yes," so I'll have to go with that. And also, there is no mewto5555, only mewto55555, mewtoo55555, and mewto555555. And actually, the latter two were banned, so there is only mewto55555. And mewto55555 isn't awesome, either.

There is a mewto5555, who has been banned.

Please stop making pointless and spam threads like this. Thank you.

[quote="vallon22"]ok...
a few things i would like to point out

1. this is spam
there is no use to it
and worse yet, you created a new topic for this spam, filling up FTW other stuff, as well as the topic

2. you spelled mew's name wrong (there are five 5's)

3. it's pretty obvious you are just posting for a higher post count, otherwise, you wouldn't do this
do now post just for the heck of it
only post if you think it is necessary

the point of this post is to tell you pplz not to just post because you feel like it
remember, to chat on FTW, you not only need to have a high post count, you have to have a good post rating
posting spam will not get you a high rating[/quote]

look at the other thread made by falcon saying mew is awesome to find this.....
DO NOT SPAM AND DO NOT MAKE USELESS THREADS!!!!!

srry i made the pointless spam thread i got carried away i wouldnt mind if an administrator deleted this

also i was not making posts to increase the number of productive posts i was just having some fun and the topic on is mewto awesome i was trying to make a poll for the first time and i phailed so i made it in a new topic

can you please stop spamming on this topic. its done

i just spammed

also if you make pointless posts on forum it won't count as a productive post and doesn't help at all.

Yeah, post rating counts too.  Everybody please rate falcon's posts 1 cube.


trigonometry
invariant
algebra solved
algebra
indentity

Prove that $ \forall x\in \mathbb{R} $ , $ \cos ^7x+\cos ^7(x+\frac {2\pi}{3})+\cos ^7(x+\frac {4\pi}{3})=\frac {63}{64}\cos 3x $

We know that $cos^n(x)$ is a linear combination of $1,cos(x),cos(2x),\dots,cos(nx)$.
The expression on the left is invariant under $x \to x+ 2\pi/3$ so that only $1, cos(3x), cos(6x)$ appear.
The expression changes sign under $x \to x+\pi$ so that only $cos(3x)$ has a nonzero coefficient.
So the problem reduces to evaluation of the sum at some particular $x$, such as $x=0$, where the cosines are $1, -1/2, -1/2$ and the computation is trivial.

fleeting_guest, thank for your good solution ;)

:wow: I found  another good solution for very ugly identity:

[i]solution:[/i]
 Suppove that $A_n(x)=\cos^nx+\cos^n(x+\frac {2\pi}{3})+\cos^n(x+\frac {4\pi}{3})$ .
By useing the equation $4\cos^3x=\cos {3x}+\cos x$ so that $A_n(x)=\frac {\cos {3x}}{4}A_{n-3}(x)+\frac 34 A_{n-2}(x)$ then we will have
$A_0(x)=3 , A_1(x)=0 , A_2(x)=\frac 32 , A_3(x)=\frac 34 \cos {3x} , A_4(x)=\frac 98 ,
A_5(x)=\frac {15}{16}\cos {3x}, A_7(x)=\frac {63}{64}\cos {3x}$


What effect do you think the new AMC scoring system will have on the qualifying USAMO index?

Not much, maybe a few points. Most USAMO qualifiers (through Index) probably aren't worrying about how many AMC questions they left blank, but rather how many they got right. I wouldn't be surprised if it affected AIME qualification, though.

My guess is that it will significantly reduce the number of AIME qualifiers.  For the USAMO index, the effect probably shouldn't be that big - it'll probably decrease by around 5 points or so.

[quote]My guess is that it will significantly reduce the number of AIME qualifiers.[/quote]
My guess is that the AMC would like for there to be 10,000 to 15,000 AIME qualifiers. But they cannot pretest their exams and norm them for difficulty (like the ETS does with the SAT), so it can be a little hard to predict, so the number of AIME qualifiers can easily fall outside that range. As you may be aware, the number of AIME qualifiers in 2006 exceeded that range on the high side; a few years earlier (2001? 2002?) scores were so low that there would have been only about 3,000 AIME qualifiers except that the qualifying score was lowered to some number below 100.

The scoring system matters, but he difficulty of the actual test questions matters more.

Difficulty of questions is a factor, but I think that the old system made it too easy to qualify for the AIME, especially by taking the AMC12. All you had to do was answer 11 questions correctly and leave the rest blank, and the first 10-14 questions are usually freebies that appear on the AMC10 too.

I think it makes the value of the AIME appreciate.

It also weeds out the less gifted.  Though it makes me feel bad for the 6th and 7th graders who try and qualify for the AIME.  They'll probably get screwed.

Now we're only starting off with what, 37.5 points? As opposed to 62.5? Tyranny, I tell you!

I like the change, it lets me guess more.

[quote="Phelpedo"][quote="asdf4"]wut new scoring system? :blush:[/quote]

1.5 points for problems left blank (as opposed to 2.5).[/quote]

Could there be a reason for why this happened?

[quote]Could there be a reason why this happened?[/quote]
Go look at [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=24818]this thread[/url] from last year. Pay particular attention to Steve Dunbar's post #28 and to the discussion that followed it. Of course Steve is not making these decisions all by himself - there is a committee involved - but do note what kind of suggestions and opinions he was soliciting there.

I personally think it's a good thing because it makes the AIME feel more...err..what's the best term I should use, exclusive maybe..idk. I think there were way way too many AIME qualifiers this year and most of them got by by just attempting the first 12 or so problems making sure that they were all right. I agree with Pakman2012 that it weeds out the "less gifted," but then again, it does make USAMO index jump up too since those "less gifted" people won't be there to bring down the AIME average.

I always thought that we'd also get easier problems with the new scoring system.  Meaning the the ability of the average AIME taker should be about the same as before.

And although I do think it's too easy to qualify for the AIME (I think 20% of AMC 12 takers go in this year, which is way too much), letting more people in shouldn't affect the USAMO qualifying index.  The qualifying index is created by looking at the top x indeces, and not a percentage, so increasing the number of people who take the AMC 12 shouldn't affect the index at all, unless someone who didn't do so great on the AMC 12 managed to do very well on the AIME, which is rare.

I think Iversonfan meant that among AIME qualifiers, USAMO indices would on average be higher, not that the qualifying index or floor scores would be higher.

What makes you think that they are changing the scoring system.  I was just on the AMC website and there is no mention of a change.  The FAQs page explicitly states the scoring system as it always has been. Can someone clue me in as to what I'm missing?

Well I heard about it from the USAMO handbook; I'm sure it's somewhere on the AMC site.

I found it in Wikipedia.

Also, Professor Dunbar posted something saying that the scoring system will change but I don't know where.


Let $P$  an interior point of an equilateral triangle  $ABC$. Prove that there exists triangle with sides $PA , PB , PC$ .

 Babis

rotate the triangle about a vertex 60 degrees, you get the triangle...


trigonometry
vector
calculus
calculus computations

$ \parallel \vec{a} \times \vec{b} \parallel ^2 \equal{} \parallel \vec{a} \parallel^2 \parallel \vec{b} \parallel^2 \minus{} \vec{a} . \vec{b} ^2$

[quote="BlueVelvet"]Leaving from

$ |a\times b| \equal{} |a\parallel{}b|\sin\theta$ and
$ a\cdot b \equal{} |a\parallel{}b|\cos\theta$

we get

$ |a\times b|^2 \equal{} |a|^2|b|^2\sin^2\theta \equal{} |a|^2|b|^2(1 \minus{} \cos^2\theta)$
$ \equal{} |a|^2|b|^2 \minus{} |a|^2|b|^2\cos^2\theta \equal{} |a|^2|b|^2 \minus{} (a\cdot b)^2$[/quote]

is $ |a\times b|$ similar to $ a \times b$ or are you trying to say $ \parallel a\times b\parallel$

BlueVelvet is obviously using $ |a|$ as a synonym for $ \|a\|$ for a vector $ a.$ That's a pretty common notation - one shouldn't be surprised by it.

because as far as I know, the correct equation is :

$ a\times b \equal{} |a\parallel{}b|\sin\theta$

while BlueVelvet mentions

$ |a\times b| \equal{} |a\parallel{}b|\sin\theta$

it's totally different, one suggests the magnitude of the cross product and the other simply says the cross product

Everywhere $ a$ and $ b$ in my post were to be read as vectors, and $ |.|$ denotes the norm (which you denoted by $ \parallel{}.\parallel{}$)
[quote="EquinoX"]because as far as I know, the correct equation is :
$ a\times b \equal{} |a\parallel{}b|\sin\theta$
[/quote]This can't possibly be true: on the left you have a vector (the cross product of two vectors is again a vector) and on the right a number (scalar).
So indeed the equation $ |a\times b| \equal{} |a\parallel{}b|\sin\theta$ is correct.

What blue velvet did is not a "rigorous" proof, since he already used the property, namely the norm of the vector product represented by $ \sin ..$, which is derived from the property we want to prove, so of course it will be true.

The property we have to prove is actually a direct consequence of the Operator Norm we have defined on Euclidan space namely.
So if we have lets say two operators

\[ \textbf{A}\;,\; \textbf{B} \in \text{Lin}(N,N)\]

 and we have  the map

\[ N \times N \to \mathbb{R} \;\;\;\;\; (A,B) \mapsto \text{Tr}(A^{\dagger}B)\]

Now let $ A$ and $ B$ elements of a subspace of the maps on a three dimensional Euclidean space. Namely antisymmetric wrt the norm, on the space, that is:

\[ A,B \in N \wedge N\]

With this we have a form on euclidan space. We usually use the half of this, so :

\[ A\bullet B : \equal{} \frac12 \text{Tr}(A^{\dagger}B)\]

Now if, $ m_1,\, m_2,\, n_1 \, n_2 \in N$, then we have according to the above definition of the operator norm:

\[ (m_1\wedge n_1)\bullet (m_2\wedge n_2) \equal{} < m_1, m_2 > <n_1,n_2>\minus{} <m_1,n_2><m_2,n_1>\]

now if we have $ m_1\equal{}m_2\equal{}m$ and $ n_1\equal{}n_2\equal{}n$, then:

\[ |m\wedge n|^2\equal{} |m|^2 |n|^2 \minus{}<m,n>^2\]

So the property for the vectorial product follows...


AMC
USA(J)MO
USAMO
geometry
geometric transformation
reflection
rectangle

Let ABCD be a convex quadrilateral, the diagonals of which intersect orthogonally at E.
Prove that the reflections of E with respect to the sides of the quadrilateral are concyclic.

[hide="my solution"]By reflecting E across one of the sides of ABCD, say AB (without loss of generality), we are reflecting the right triangle ABE across AB, creating a rectangle EAE'B. We do this symmetrically for all four of the triangles formed by the diagonals, and we have a large rectangle formed by the reflected E's. Since all rectangles can be inscribed in a circle, the reflected E's are concyclic.[/hide]

[quote="Arvind_sn"]a rectangle EAE'B[/quote]

hmm?   :huh:

I do not know how to add a pic. Here is my solution:
If the refections are  $ I_1,J_1,K_1,L_1$ respectively then it is enough if we prove the concyclity of the orthogonal projections{$ I,J,K,L$} of the reflections onto the sides because of homothety with centre $ 0$ and  {ratio of homothety being $ 2$}.
 Now we can eaily prove that $ AIOL, IBJO, OJCK, LOKD$ as a result of which we use that the angles in the same segment of a circle are equal in each and then finally with the help of the right trinagles due to orthogonal intersections  prove that $ \angle ILK + \angle KJI = 180%Error. "degress" is a bad command.
$ resulting in concyclity of $ IJKL$ which proves the concyclity of $ I_1J_1K_1L_1$ as $ I_1J_1K_1L_1$ and $ IJKL$ are homothetic convex quadrilaterals.

Edit:Sorry for posting it after the link. But I was slow in latexing and did not see the message posted while I was typing.


One square foot is what fractional part of a square yard?

There are 3 feet in a yard. However, we have area, which is 2 dimensions, in the problem, so we must square this 1 dimensional statement. $ 3^2 \equal{} 9$ square feet in a yard. So, each foot is $ \boxed{1/9}$ of a yard.

It said I was [color=#f00]INCORRECT[/color] when I put that.

Sorry. I meant that I don't know how to pull up this question again to report. I did it yesterday, so I'm unsure of how to pull it up again to report and fix.


geometry
circumcircle
combinatorics proposed
combinatorics

Since this has a very strong geometric combinatorial flavor, I'm posting it here :):

Prove that given $n$ points in the plane, no three on a line and no four on a circle, we can find a unique triangulation having these points as vertices s.t. the circumcircle of any triangle in the triangulation has none of the $n$ points inside (by "inside" I mean strictly inside).

(a triangulation is a planar graph with the $n$ points as vertices, all of whose finite faces are triangles)

Edit:

I've edited the message to add the word "unique". A triangulation having this property is also uniquely determined :).

Well...the title you used gives the answer... :D

Pierre.

i've never seen this one... can somebody post a sol?


For all positive real numbers $ a$, $ b$ and $ c$,
\[ \sqrt {\frac {2a(b \plus{} c)}{(a \plus{} b)(a \plus{} c)}} \plus{} \sqrt {\frac {2b(c \plus{} a)}{(b \plus{} c)(b \plus{} a)}} \plus{} \sqrt {\frac {2c(a \plus{} b)}{(c \plus{} a)(c \plus{} b)}}\le\frac {a \plus{} b \plus{} c}{\sqrt [3]{abc}}
\]

[quote="tuandokim"]$ LHS \leq 3 \leq RHS$[/quote]
OK, you're right. I didn't realize it's that easy. Will be more careful next time.


Why does Isothermic process always do more work than adiabatic process - that is, the area under the p-V graph is always greater for Isothermic process?  If the two processes start at the same coordinates $ (V_0,p_0)$, then the pressure for adiabat decreases much faster than isotherms - how do we justify this?

Mathematical explanation:

Adiabatic process can be described in p-V as $ pV^{\gamma} \equal{} \text{const}$, and $ \gamma \equal{} \frac {\alpha \plus{} 1}{\alpha}$, so $ \gamma > 1$. In case of isothermal process, we have $ pV \equal{} \text{const}$. It's now obvious (just write it explicitly if you don't see) that first curve is steeper.

Proof for the adiabatic process formula is given
[url=http://en.wikipedia.org/wiki/Adiabatic_process#Derivation_of_continuous_formula]here[/url].

Physical explanation:

for physical explanation, consider both processes and see which energy is being transformed into work (adiabatic-> internal energy, isothermal->the added heat.)

Thanks for the source.
By the way, in compression, we know $ W$ is negative, but why is $ Q$ negative as well, and what sign is the change in internal energy ($ \Delta U$)?
Also, in expansion, why is $ Q$ positive, and what sign is $ \Delta U$?
The relation is $ Q \equal{} \Delta U \plus{} W$ by the way.

What process are we analyzing? Isothermal, or adiabatic?

If it is isothermal, then you don't have the change in internal energy at all, by definition. And then, the fact about the signs is pretty obvious (that's again mathematical approach, physical would include simple observation that, when compressing gasses, you have to deal with heating of it, so to keep the isothermal properties, need to take away $ Q$, or bring it to the system to make it expand in first place).

If the process is adiabatic, then there's no $ Q$, and the sign of $ \Delta U$ is same as sign of $ W$.

Hope I got it all right.

Hmm, I'm not sure, but I think it's for one cycle of an ideal gas heat engine, and $ W_{net}$ is determined by applying the first law of thermodynamics to the complete cycle.

So we have $ \Delta U_h \equal{} \plus{} |Q_h| \minus{} |W_{expansion}|$ eqn(1)
and $ \Delta U_c \equal{} \minus{} |Q_c| \plus{} |W_{compresson}|$ eqn(2)
and $ \Delta U_c \plus{} \Delta U_h \equal{} 0$.

So we get $ |W_{expansion}| \minus{} |W_{compression}| \equal{} W_{net} \equal{} |Q_h| \minus{} |Q_c|$

But WHY do we have $ \plus{} |Q_h|$ for eqn(1) and $ \minus{} |Q_c|$ for eqn(2)?  And what are the signs of $ \Delta U_h$ and $ \Delta U_C$?

Oh, it's the complete cycle, then. If that is so, lets try to give a good answer:

the plus in (1) is there because you're [b]heating [/b] the system, bringing the heat inside. The convention is to put '+' in front of incoming energy. Analogously, in (2), the [b]cooling[/b] of the system, taking heat away is the reason for '-'. This can be even more simplified, going in the exact physical analysis of compression/expansion.

In this analysis, we didn't take into account what processes are in the cycle. For answering your second questions, we'll assume it's a Carnot cycle: and you may try to generalize the approach for any cycle.

In Carnot cycle, both compression and expansion are consisted of an adiabatic and an isothermal process.

Expansion: work of the system is made $ W \equal{} Q$ and $ W \equal{} \Delta U$, so in this case, we're losing internal energy (system is cooled to the 'cold' temperature in the end of expansion)

Compression: the work is being made 'on the system', in the same manner so internal energy is gained.

The argument given above is pretty obvious from the TS diagram: change of internal energy is directly proportional to the change of temperature: so it's enough to observe the starting and the ending point of each part of cycle you're analyzing (the 'corner points').

[quote="wonderfuljack"]The above mathematical reasoning ( that PV^(gamma)=constant ) relies on the fact that we are dealing with an ideal gas.  However the principle that adiabatic processes do more work than isothermal ones is true in the more general case of non-ideal gases.  So our reasoning is a little incomplete.

How would one demonstrate this more general case mathematically?[/quote]

There is no need to over complicate this. In principle it is a straight forward question with a very simple answer.

The first law of thermodynamics expressed in differential form is this:

$ dU \equal{} dQ \minus{} dW$

For an adiabatic process $ dQ\stackrel{\triangle} \equal{} 0$ ; it follows that, $ dU \equal{} \minus{} dW_{ad}$.

Therefore, $ \minus{} dW_{ad} \equal{} dQ \minus{} dW \implies dW \equal{} dQ \plus{} dW_{ad}$.

This tells us that $ W \ge W_{ad}$ for any process operating between the same two states as an adiabatic process.

Edit: bad grammar (slap me!)

ignore this -- i'm just reposting an archived thread so i can see what the equations are (theyre all syntaxy eww)

Correct.

I'm not quite sure what you doing here, but you know that,

\left(\frac{\partial S}{\partial H}\right)_p =\frac{1}{T}

\left(\frac{\partial S}{\partial P}\right)_H =\frac{V}{T}

And you know that,

\left[\frac{\partial}{\partial P}\left(\frac{\partial S}{\partial H}\right)_p\right] = \left[\frac{\partial}{\partial H}\left(\frac{\partial S}{\partial P}\right)_H\right]

Can you see what you need to do next (it's a really simple step)?

subbing in becames..
(d/dp)(1/T)=(d/dH)(V/T)
i cant see how the |H and |p get switched around

[quote="wonderfuljack"]ignore this -- i'm just reposting an archived thread so i can see what the equations are (theyre all syntaxy eww)

Correct.

I'm not quite sure what you doing here, but you know that,

\[ \left(\frac{\partial S}{\partial H}\right)_p \equal{}\frac{1}{T}\]

\[ \left(\frac{\partial S}{\partial P}\right)_H \equal{}\frac{V}{T}\]

And you know that,

\[ \left[\frac{\partial}{\partial P}\left(\frac{\partial S}{\partial H}\right)_p\right] \equal{} \left[\frac{\partial}{\partial H}\left(\frac{\partial S}{\partial P}\right)_H\right]\]

Can you see what you need to do next (it's a really simple step)?

subbing in becames..
(d/dp)(1/T)=(d/dH)(V/T)
i cant see how the |H and |p get switched around[/quote]

Start with the definition:

\[ \mathrm{d}H\stackrel{\triangle}\equal{} T\mathrm{d}S \plus{} V\mathrm{d}p\]

Rearranging somewhat we get $ \mathrm{d}S \equal{} \frac{1}{T}\left(\mathrm{d}H \minus{} V\mathrm{d}p\right)$.

From differential calculus we know $ \mathrm{d}S \equal{}\frac{\partial S}{\partial H}\mathrm{d}H \plus{} \frac{\partial S}{\partial p}\mathrm{d}p$.

Now, $ \left.\right|_p$ means with $ p$ held constant ($ \mathrm{d}p \equal{} 0$), and likewise $ \left.\right|_H$ means with $ H$ held constant ($ \mathrm{d}H \equal{} 0$).

Comparing the right hand sides of the above expressions you can see immediately that

\[ \begin{array}{cc} \left(\frac{\partial S}{\partial H}\right)_p \equal{} \frac{1}{T} & \left(\frac{\partial S}{\partial p}\right)_H \equal{} \minus{}\frac{V}{T}\end{array}\]

Inspecting the above result, I believe there is a sign error in your second expression. (This could be due to how enthalpy, $ H$, is defined in your text. It would be a good idea to check it.)


inequalities
trigonometry
function
LaTeX
inequalities solved

If x, y, z are three positive reals, prove the inequality

$\left( \sqrt{x\left( y+z\right) }+\sqrt{y\left( z+x\right) }+\sqrt{z\left(x+y\right) }\right) \cdot \sqrt{x+y+z}$
$\geq 2\sqrt{\left( y+z\right) \left(z+x\right) \left( x+y\right) }$

When does equality occur?

(I have a nice solution.)

Darij

Congratulations for a beautiful problem. My solution is ugly, but it exists:
   Take x+y+z=1. Square the inequality $\sum\sqrt{x(1-x)}\geq 2\cdot\sqrt{(1-x)(1-y)(1-z)} $ and reduce it to $\sum xy-2xyz\leq\sum\sqrt{xy(y+z)(z+x)} $. But $\sum xy-2xyz\leq\sum xy$ and $\sum\sqrt{xy(x+z)(y+z)}\geq\sum xy+\sum z\cdot\sqrt{xy} $

Thanks! Not so ugly at all, BTW. As you can easily see, equality holds if and only if one of x, y, z is 0.

Here is my proof: Consider the triangle with sides $a = y + z, b = z + x, c = x + y$ and semiperimeter $s=\frac{a+b+c}{2}=x+y+z$. Then, our inequality becomes
\[\left( \sqrt{\left( s-a\right) a}+\sqrt{\left( s-b\right) b}+\sqrt{\left(s-c\right) c}\right) \cdot \sqrt{s}\geq 2\sqrt{abc},\]
or
\[\Large \sqrt{\frac{s\left( s-a\right) }{bc}}+\sqrt{\frac{s\left( s-b\right) }{ca}}+\sqrt{\frac{s\left( s-c\right) }{ab}}\geq 2.\]

If we call A, B, C the angles of our triangle, then this simplifies to

\[\cos \frac{A}{2}+\cos \frac{B}{2}+\cos \frac{C}{2}\geq 2\]

(due to the half-angle formulae $\cos \frac{A}{2} = \sqrt{\frac{s\left( s-a\right) }{bc}}$, $\cos \frac{B}{2} = \sqrt{\frac{s\left( s-b\right) }{ca}}$ and $\cos \frac{C}{2} = \sqrt{\frac{s\left( s-c\right) }{ab}}$), i. e., to

\[\sin \left( 90^\circ-\frac{A}{2}\right) +\sin \left( 90^\circ-\frac{B}{2}\right) +\sin \left( 90^\circ-\frac{C}{2}\right) \geq 2.\]

But $90^{\circ}-\frac{A}{2}$, $90^{\circ}-\frac{B}{2}$ and $90^{\circ}-\frac{C}{2}$ are the angles of an acute triangle (as one can easily see); hence, we must show that if A, B, C are the angles of an acute triangle, then
\[\sin A + \sin B + \sin C \geq 2.\]

(Actually, for any non-degenerate acute triangle, $\sin A + \sin B + \sin C > 2$, but I don't want to exclude degenerate cases.) Here is an elegant proof of this inequality by Arthur Engel: Since triangle ABC is acute, we have $A-B\leq C$, and thus $\cos \frac{A-B}{2}\geq \cos \frac{C}{2}$ (since the function $\cos$ is monotonically decreasing on $\left[0^{\circ},180^{\circ}\right]$), so that
\[\sin A+\sin B=2\underbrace{\sin \frac{A+B}{2}}_{=\cos\dfrac{C}{2}}\underbrace{\cos \frac{A-B}{2}}_{\geq \cos \dfrac{C}{2}}\]\[\geq 2\cos ^{2}\frac{C}{2}=1+\cos C\]

and $\sin A+\sin B+\sin C\geq 1+\cos C+\sin C\geq 2$. Here, we have used the very simple inequality $\cos C + \sin C \geq 1$ for any acute angle C. Proof complete.

(I admit that I did not find the proof while trying to solve the problem, but I rather constructed the problem while searching for a reasonable application of the $\sin A + \sin B + \sin C \geq 2$ inequality, but this doesn't matter afterwards...)

  Darij

[Thanks stevem: LaTeX corrected]

[quote="darij grinberg"](I use "^0" for "degrees".) [/quote] You can use \circ as in [tex]\sin \left( 90^\circ-\frac{A}{2}\right)[/tex]

[quote="darij grinberg"]If we call A, B, C the angles of our triangle, then this simplifies to

[tex]\cos \frac{A}{2}+\cos \frac{B}{2}+\cos \frac{C}{2}\geq 2[/tex],
[/quote]

Another way to prove this:

f(x)=cos(x) is concave on [0, pi/2], and (A/2, B/2, C/2) is majorized by (pi/2, 0, 0). The result follows from the Karamata inequality.

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=14975

Ok,here is another nice proof by darij and me:
notice:$\sqrt{\frac{x\left(y+z\right)\left(x+y+z\right)}{\left(y+z\right)\left(z+x\right)\left(x+y\right)}}\geq\frac{x\left(y+z\right)\left(x+y+z\right)}{\left(y+z\right)\left(z+x\right)\left(x+y\right)}$
then we get:
$\sum\sqrt{\frac{x\left(y+z\right)\left(x+y+z\right)}{\left(y+z\right)\left(z+x\right)\left(x+y\right)}}\geq\sum\frac{x\left(y+z\right)\left(x+y+z\right)}{\left(y+z\right)\left(z+x\right)\left(x+y\right)}=2\frac{\left(y+z\right)\left(z+x\right)\left(x+y\right)+xyz}{\left(y+z\right)\left(z+x\right)\left(x+y\right)} \ge 2$
QED.


Five people are sitting at a round table. Let f ≥ 0 be the number of people sitting next to at least one female, and let m ≥ 0 be the number of people sitting next to at least one male. Find the number of possible ordered pairs (f,m).

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=2092534566&t=57385[/url]

Please, use the search function before posting the problems!


for what $ a,b,c$ does this hold?

$ \sum \frac{a}{a\plus{}c} \ge \sum \frac{b}{a\plus{}c} \ge \sum \frac{c}{a\plus{}c}$

$ \sum\frac{a}{a\plus{}c}\ge\sum\frac{b}{a\plus{}c}\Longleftrightarrow$ $ \sum_{cyc}a^3\leq \sum_{cyc}ab^2$. But $ \sum_{cyc}a^{3}\geq\sum_{cyc}ab^{2}$ according to Muirhead(equality occurs when $ a\equal{}b\equal{}c$). So $ a\equal{}b\equal{}c$.

[quote="Flakky"]$ \sum\frac {a}{a \plus{} c}\ge\sum\frac {b}{a \plus{} c}\Longleftrightarrow$ $ \sum_{cyc}a^3\leq \sum_{cyc}ab^2$. But $ \sum_{cyc}a^{3}\geq\sum_{cyc}ab^{2}$ according to Muirhead(equality occurs when $ a \equal{} b \equal{} c$). So $ a \equal{} b \equal{} c$.[/quote]

Correct! :D

unfortunately for me, I realized the answer only a day after posting :(

[quote="Flakky"]$ \sum\frac {a}{a \plus{} c}\ge\sum\frac {b}{a \plus{} c}\Longleftrightarrow$ $ \sum_{cyc}a^3\leq \sum_{cyc}ab^2$. But $ \sum_{cyc}a^{3}\geq\sum_{cyc}ab^{2}$ according to Muirhead(equality occurs when $ a \equal{} b \equal{} c$). So $ a \equal{} b \equal{} c$.[/quote]
not Muirhead of course, it's only cyclic. But AM-GM $ 3\sum_{cyc}a^3 \equal{} \sum_{cyc}(a^3 \plus{} b^3\plus{} b^3)\ge\sum_{cyc}3ab^2$.  :!:


Το παρακατω προβλημα ελπιζω να το λυσεται και κυριως οι μικροι οι οποιοι πρεπει να ειναι καλα προετιμασμενοι.Το πρωτο ερωτημα ειναι σχετικα ευκολο το δευτερο καπως πιο απαιτιτικο.To τριτο ειναι γενικοτερου ενδιαφεροντος.....και πολυ δυσκολο. :wink:
Εστω η ακολουθεια Α_n:N->N.  Δινεται οτι Α_1=1,Α_2=2,Α_3=4,Α_4=8,Α_5=15,Α_6=26 κ.ο.κ..........
α)Βρειτε τον n-oστο ορο της ακολουθειας.
β)Βρειτε το αθροισμα ολων των ορων της ακολουθειας.
γ)αποδειξτε οτι η ακολουθεια εχει πεπερασμενο πληθος ορων που ειναι δυναμη του 2 και βρειτε τους.(οι λυσεις της εξισωσης του Ramanujan ειναι οροι της ακολουθειας.)

[quote="Αρχιμήδης 6"]Το παρακατω προβλημα ελπιζω να το λυσεται και κυριως οι μικροι οι οποιοι πρεπει να ειναι καλα προετιμασμενοι.Το πρωτο ερωτημα ειναι σχετικα ευκολο το δευτερο καπως πιο απαιτιτικο.
Εστω η ακολουθεια $a_{n}$ $: N\to N$.  Δινεται οτι
$a_{1}=1,a_{2}=2,a_{3}=4,a_{4}=8,a_{5}=15,a_{6}=26$ κ.ο.κ..........
α)Βρειτε τον n-oστο ορο της ακολουθειας.
β)Βρειτε το αθροισμα ολων των ορων της ακολουθειας.[/quote]

[hide]eykola pernume $a_{n}=2a_{n-1}-a_{n-2}+(n-2)$
diladi $a_{n}=3a_{n-1}-3a_{n-2}+a_{n-3}+1$
 $a_{n}=4a_{n-1}-6a_{n-2}+4a_{n-3}-a_{n-4}$
kai pernontas tin xaraktiristiki eksisosi tis akolu8ias..vgeni mia eksisosi tetartu va8mu,i opia ine telia tetarti dinami...klp..(as to telioisi kapios ...)

kati asxeto...akolu8ia grafete me i i me ei??...[/hide]

 :D

Καλα πας αλλα δεν ειναι και τοσες πολες οσες νομιζεις οι πραξεις.Αν δεν βαριεσαι λυσε την ασκηση. :wink:
Προς RAKAR:
[hide]Φιλε στην ορθογραφια ποτε δεν ημουν καλος[/hide]

sinexizontas apo eki pu ixa mini...i xaraktiristiki eksisosi dini
$r^{4}=4r^{3}-6r^{2}+4r-1$ <=> $(r-1)^{4}=0$ diladi $r_{1}=r_{2}=r_{3}=r_{4}=1$
 opote 8e ine $a_{n}=a+bn+cn^{2}+dn^{3}$
apo tus 4 protus orus pernume ena 4x 4 sistima ..
i to linume me to xeri
[hide]i me to  mathematica..
a={{1,1,1,1,1},{1,2,4,8,2},{1,3,9,27,4},{1,4,16,64,8}}
MatrixForm[a]
MatrixForm[RowReduce[a]][/hide]
pernume $a=0, b=\frac{4}{3},c=-\frac{1}{2},d=\frac{1}{6}$
diladi $a_{n}=\frac{4}{3}n-\frac{1}{2}n^{2}+\frac{1}{6}n^{3}$
gia to b)....olon ton oron..?tote 8a ine $+\infty$...
pantos ton n proton oron an 8es..to pernume eykola afu exume tus gnostus tipus gia to a8risma ton proton ari8mon ,to a8risma tetragonon ton n proton ari8mon +to aurisma kivon ton n proton ari8mon...
gia to c)..an kai den ksero pia ine i eksisosis ramanujan...(malon enoios ton ma8imatiko..+oxi ton ramanujan tu forum.... :P )mu fenete oti mono i proti 4 ori ine dinamis tu 2..(dn to exo apodiksi ayto omos akoma...ime sto yaksimo...)

a..kai kati alo...i akolu8ia etsi pos tin dinis..dn nomizo na ine kalos orismeni.... px. iparxun apira polionima 7 va8mu pu na dinun aytes tis eksi times pu dinis san orus tis akolu8ias.......sorry an kano la8os ..ala etsi mu fenete...

 :D

Καταρχην ζητω συγνωμη γιατι το σχολιο μου και η παρατηρηση για την εξισωση του Ramanujan δεν αφορουσαν αυτην την ακολουθεια αλλα την ακολουθεια (1,2,4,7,11,16,22,......n(n+1)/2).Πραγματι RAKAR σωστη η λυση σου και σορρυ που δεν εθεσσα κι αλλους περιορισμους στην ακολουθεια ωστε να μην παιρνουμε απειρες.

κατι μου λεει οτι οι τριγωνικοι αριθμοι που λεει ο Αρχιμηδης 6 συνδεονται με την ακολουθια στο πρωτο ποστ.Ραλη που διαβασες τη μεθοδο με την οποια λυνεις την ασκηση ?

kostas..den tin diavasa pu 8ena..ine aplos i 8eoria tis xaraktiristikis eksisosis mia akolu8ias pu perigrafete apo mia gramiki anadromiki sxesi...
olo to skeptiko mu itan na vro mia gramiki anadromiki sxesi...
sigekrimena ine i periptosi pu oles i rizes tis (4u va8mu)xaraktiristikis eksisosis..ine oles ises me $t$ tote
$a_{n}=(a+bn+cn^{2}+dn^{3})t^{n}$
apla edo ine $t=1$... :wink:

 :D

edit:gia more info gia tin xaraktiristiki eksisosi mias gramikis akolu8ias
des sel 400peripu..  (akolu8ies...)sto arxio pdf <paneyropaiki proetimasia gia ma8ites likiu (kapos etsi..)> strin istoselida tu cretanman..iparxun arketa xaraktiristika paradigmata...


abstract algebra
number theory proposed
number theory

1.   A number like $\mathfrak A$ is called [b]Square remainder[/b] $(\ module \ P)$ if
   There exist a $\mathfrak K \in \mathbb{Z} : \mathfrak K^2 \equiv \mathfrak A (\ module \ P)$

   [color=blue]Problem[/color]
    Let $\mathcal A$ be an integer such that it is square remainder by module every natural number.
    Prove that $\mathcal A$ is a perfect square.

2. A number like $\mathfrak A$ is called $n_{th}$[b]  remainder[/b] $(\ module \ P)$ if
   There exist a $\mathfrak K \in \mathbb{Z} : \mathfrak K^n \equiv \mathfrak A (\ module \ P)$

[color=blue]Problem[/color]
    Let $\mathcal A$ be an integer such that it is $n_{th}$ remainder by module every natural number.
    Prove that $\mathcal A$ is a perfect $n_{th}$ power.

1.: see http://www.mathlinks.ro/Forum/viewtopic.php?t=64322

But do you have purely elementary proves for both¿ Or at least something not using the $n$-th power reciprocity theorem¿

[hide="Solution 1:"]Assume $\mathcal A= P^{2k+1}N,gcd(P,N)=1$ where $P$ is one of the prime factors of $\mathcal A$(If $A$ is not a perfect square then such $P$ and $K$ exists), then there exist a number like $X$ such that

$X^2 \equiv P^{2k+1}N (\ module P^{2k+2}) \longrightarrow$

$X^2 =yP^{2k+2}+P^{2k+1}N \longrightarrow$

$X^2=P^{2k+1}(Py+N)$

But    $gcd(Py+N,P)=1$ contradiction .[/hide]

OK, I was thinking to complicated (reducing it to primes), and clearly the same as yours works for $n$-th powers.

But I would be really interested in any (elementary¿) solution to the problem in the link above and it's natural generalization.


Hi there, 1st post here so please forgive me if I am in the wrong place!

I have a question posed to me, which is as follows:

"a) Perform the following binary subtractions. Use two’s complement.
     001001001 – 001000111"

Now the problem is that I am getting "000000000" when I work it out. But I don't think this is right. I have been trying to understand two's complement from a text book and websites but I don't think I have the hang of it

Can anyone guide me to how this question would be answered, or more information which is in a clear concise manner based on twos complement

Thanks in advance :ninja:

I'll give the rule that I remember. Correct me if I'm wrong:

$A-B = A+C(B)$, where $C(B)$ is the twos complement of $B$. In this addition we should ignore the carry out in the highest digit.

The twos complement of $B$ can be found by the following rule: $C(B) = NOT(B)+1$.

This means:
- Invert each digit of B
- Add 1 to the result

So B becomes
$001000111 \to $
$110111000 \to $
$110111001$

The substraction

[code] 001001001
-001000111[/code]

turns to the addition

[code] 001001001
+110111001
------------------
 000000010[/code]

Notice that we ignore the highest carry out

[hide="general solution for a"]We can put them in base 10, and then subtract.

73-71=2, 10_2 in binary[/hide]

you could notice that the first five digits are exactly the same...

Thanks a lot for the help guys! I finished all the other questions with your help!!


Has anyone looked at the michigan problems before?
the contest is also sponsored by MAA so are the problems similar?

[url]http://www.svsu.edu/mmpc/graphics/exams/2004.I.problems.pdf[/url]

The Michigan Mathematics Prize Competition is sponsored by the Michigan Section of the MAA.
The AMC contests are sponsored by the national MAA organization.  The two competitions are distinct.

The official web page of the MMPC is
http://gauss.math.oakland.edu/mmpc/index.html
You can read the details there.

Steve Dunbar
MAA Director, American Mathematics Competitions


algebra
system of equations
algebra unsolved

This is a grade ten problem. I am no math whiz so I need some help.

A traffic helicopter pilot finds that with a tailwind her 120km trip away from the airport takes 30 minutes. On her return trip to the airport, into the wind it takes her 10 minutes longer. What is the speed of the helicopter? What is the speed of the wind?

I could solve the problem if I was given some direction as to what the system of equations was.

You should have posted in the physics forum :)
Anyway, welcome and have fun.

You must use the equation for constant velocity $d=v\times t$,
where $d$: displacement (km), $v$: velocity (km/h), $t$ time (h)
Translating the problem:
$120=(v_{h}+v_{w})\times\frac{30}{60}$,
$120=(v_{h}-v_{w})\times\frac{30+10}{60}$,
where $v_{h}$ is the velocity of the helicopter without wind, and $v_{w}$ is the velocity of the wind. The $60$ is there to convert minutes to hours.
Just solve the system of equations to reach the answer.

Now I'll ask you a question: why did you wrote "her" instead of "its"? I'm not very good at English. :maybe:
[Edit: OK, I've looked it up. According to Wikipedia (in the article "grammatical gender"), sometimes the pronouns "she/her" are used to refer to countries, ships or machines.]


trigonometry
real analysis
real analysis unsolved

How many solutions are there of this eq.
\[2=\sum_{n=1}^\infty (\sin{x})^{n}\]
and what are these?
$(0\leq x<2\pi)$

It's a geometric series. Just evaluate it.


abstract algebra
advanced fields
advanced fields theorems

Let C. be a complex of A-modules, x in A. Then we obtain an exact sequence of complexes 0->C.->C.(x)->C'. where C'. is the complex obtained by shifting the degrees up by 1.

So the first sentence in the proof is: We can identify

C_p(x) with C_p (+) C_{p-1}

and then everything becomes black  :rotfl:

matsumura,p127-128

I don't understand "the identification."...

probably I need to go through my homological algebra again, because it is most likely straightforward

ok, solved

I did not know the def of tensor product of complexes, so I made up the most obvious one myself (which was false:))


A man plays a game, whereby he stands $4$ feet behind a cliff and rolls a fair $6$-sided die. If it lands on any number from $1$ to $4$, he takes a step forward ($1$ foot); if it lands on $5$ or $6$, he steps one foot backwards. $4$ feet behind him is his family, who will grab him immediately and save him, allowing him to survive. If he continues the game ad infinitum, until he either falls off the cliff or is saved, what is the probability that he survives?

let p_{i} be the probability that he survives at position $i$, where i=-4 is safe, and i=4 is dead

$p_{-4}=1$, $p_{-3}=\frac{1}{3}*p_{-4}+\frac{2}{3}p_{-2}$, etc

solve the system for $p_{0}$


superior algebra
superior algebra unsolved

What is the number of partitions of the set {1,2,3,4} that contain 2 sets. The answer should be an integer from 1-9.


If a,b,c and h_a,h_b,h_c are the sides and heights of a triangle prove that

3(a^2+b^2+c^2) \geq 4((h_a)^2+(h_b)^2+(h_c)^2)

If I'm not wrong, all you have to do is to write h_a<=m_a and then square and sum up.

Actually Harazi is right here. This inequality is trivial. :D
Here is a better one, but also trivial. here it goes:

 $\dfrac{1}{4} (a+b+c)^{2}\geq h_{a}^{2}+h_{b}^{2}+h_{c}^{2}$ ;)

$l_a\geq h_a$ where $l_a$ is the bisector

isn't it ?? :D  :P


Twenty students bought tickets for a school party. All of the money recieved for these 20 tickets was used to purchase beverages. Then, and additional 10 students bought tickets. Rater than use this additional money to buy more refreshments, all 30 students recieved a $3.00$(three dollar) refund. How many dollars were used to buy beverages?

[hide]
Let's say each ticket costs x dollars. Then a total of 20x dollars was bought on refreshments. When 10 more students bought tickets, they spent 10x dollars. All 30 students also got 3 dollars back so the 10x dollars that they spent is equal to 90, so each ticket costs 9 dollars. We're looking for the cost of the beverages so multiply that by twenty. The answer is $180$.[/hide]

[hide]
we know that one person's cost is 9.00

we multiply that by 20 and get 180

the answer is $\$$180[/hide]
-jorian


Consider a finite sequence S=<a_1,a_2,...,a_n> of zeroes and ones.
We compute the following value f(S):

~  x:=2;
~  FOR k:=1 TO n DO
~~~      IF(a_k=0) THEN x:=x+1
~~~     IF(a_k=1) THEN x:=x*x
~  ENDFOR
~  f(S):=x;

Prove: The function f is an injection, that is,
S not equal T implies f(S) not equal f(T).

================================
Does anybody know the source of this problem?


What is the name of the theorem that says that if you draw any line through two shapes and the length of the intersection with each shape is equal (for all lines), the two shapes have equal area?


search
combinatorics unsolved
combinatorics

On a large, flat field $ n$ people are positioned so that for each person the distances to all the other people are different. Each person holds a water pistol and at a given signal fires and hits the person who is closest. When $ n$ is odd, show that there is at least one person left dry. Is this always true when $ n$ is even?

This result is well-known, I know I saw it in one of Darij's posts.

When n is odd we use induction. The claim is obviously true for only one person. Now for any other n, take the two closest people in the group. They will both squirt each other, so if the problem is true for the other n-2 people, then we are done. This completes the induction.

But if n is even, then we can just make pairs of people, where the people in each pair are very close but the pairs are set very far apart, and everyone in each pair will squirt the other member of the pair, so everyone gets wet.

You really should check to see if these have been posted before.  This one has:
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1472670372&t=159575
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1472670372&t=103974
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1472670372&t=130113

(I searched for the words "field" and "dry," so this only got posts that preserved the identical wording.)

How do we position them so that maximum of them survive ? And what is that number for a given $ n$ ?


[quote="dina013"]Fie ecuatia $2x^{2}-2mx+m^{2}-2m=0$, unde $m$ e real. Care este multimea valorior pe care le pot lua radacinile reale $x_{1}$, $x_{2}$ cand $m$ variaza?[/quote]

Could you please translate this in English to help you?

Alexandros

Sure. So, we have the equation:
     $2x^{2}-2mx+m^{2}-2m=0$ $m$ real.
When m varies, what values can $x_{1}$ and $x_{2}$ get?

$D=-4m^{2}+16m$. Then,

If $D>0$, that is $0<m<4$ then we have two solutions of the equation, namely $x_{1,2}=\frac{m\pm\sqrt{m(4-m)}{2}}$

If $D=0$, that is $m=0$ or $m=4$ then we have a root with multiplicity $2$ which is $x=0$ (if $m=0$) or $x=2$ (if $m=4$)

If $D<0$, that is $m>4$ or $m<0$ then no solutions of the equation exist.

Is that what you wanted?
Alexandros


Does anyone know when the Zonal Informatics Olympiad will be held in India. This is the First round of the Informatics Olympiad in india. What is the last date for registration? How do you register? What's the 'syllabus'?!

Hi Rushil,

ZIO (Zonal Informatics Olympiad) 2006 is on the 28th of this month. Registration closes on the 10th. Registration is through your school. The ZIO is a written test, and the next round, the INOI is a programming test. You can get more information including previous years' papers here: [url]http://www.iarcs.org.in/inoi[/url]

Harpreet

Thanks... I am really a lot out of touch from Programming...besides ZIO clashes with Chem National Olympiad...so I think I'll skip!!!


function
topology
complex analysis
real analysis
real analysis unsolved

Let $ X$ be measurable space, Y be topological space. Function $ f$ from X to Y is called measurable if $ f^{\minus{}1} (V)$ is measurable with all open set $ V\subset Y$.
Prove that :
1. Characterisitc function is measurable.
2. $ E\equal{}\{x : f(x) \equal{} 0\}$  is a measurable set.

I'm finding a book - solutions of Real & Complex Analysis of W. Rudin. Please help me if you know. Thanks alot.

[quote="kidmath"]Let $ X$ be measurable space, Y be topological space. Function $ f$ from X to Y is called measurable if $ f^{ \minus{} 1} (V)$ is measurable with all open set $ V\subset Y$.
Prove that :
1. Characterisitc function is measurable.
2. $ E \equal{} \{x : f(x) \equal{} 0\}$  is a measurable set.

I'm finding a book - solutions of Real & Complex Analysis of W. Rudin. Please help me if you know. Thanks alot.[/quote]

$ f$ from $ X$ to where ?  :maybe:

[quote="kidmath"]here $ f$ from $ X$ to $ \mathbb{R}$.[/quote]
There's a characterisitc function $ f$ which is not measurable, isn't it ?  :maybe:

[quote="Mr.Doe"]Yes, in simple situation take some unmeasurable set $ B\subseteq \mathbb{R}$ then $ 1_{B}$ isn't measurable.[/quote]

Yes that's why I ask him to check it again  :D


Given $ a,b,c,d>0$ and $ a^2\plus{}b^2\plus{}c^2\plus{}d^2\equal{}4$.Prove that:
$ \frac{a^3}{a^2\plus{}bc}\plus{}\frac{b^3}{b^2\plus{}cd}\plus{}\frac{c^3}{c^2\plus{}da}\plus{}\frac{d^3}{d^2\plus{}ab} \ge 2$ :)

$ LHS \ge \frac{16}{a^3\plus{}b^3\plus{}c^3\plus{}d^3\plus{}abc\plus{}bcd\plus{}cda\plus{}dab}$
And we have old and hard problem
$ a^3\plus{}b^3\plus{}c^3\plus{}d^3\plus{}abc\plus{}bcd\plus{}dca\plus{}dab \le 8$
And $ LHS \ge 2$

[quote="secrets"]$ LHS \ge \frac {16}{a^3 \plus{} b^3 \plus{} c^3 \plus{} d^3 \plus{} abc \plus{} bcd \plus{} cda \plus{} dab}$
And we have old and hard problem
$ a^3 \plus{} b^3 \plus{} c^3 \plus{} d^3 \plus{} abc \plus{} bcd \plus{} dca \plus{} dab \le 8$
And $ LHS \ge 2$[/quote]
Nice solution,secrets :)

Can you please tell me how come $ a^3\plus{}b^3\plus{}c^3\plus{}d^3\plus{}abc\plus{}bcd\plus{}dca\plus{}dab \le 8$?
 :blush:
Sorry that I could not solve this.  :noo:

[quote="Potla"]Can you please tell me how come $ a^3 \plus{} b^3 \plus{} c^3 \plus{} d^3 \plus{} abc \plus{} bcd \plus{} dca \plus{} dab \le 8$?
 :blush:
Sorry that I could not solve this.  :noo:[/quote]
 $ (a(a^2\plus{}bc)\plus{}b(b^2\plus{}cd)\plus{}c(c^2\plus{}da)\plus{}d(d^2\plus{}ab))^2 \le (a^2\plus{}b^2\plus{}c^2\plus{}d^2)((a^2\plus{}bc)^2\plus{}(b^2\plus{}cd)^2\plus{}(c^2\plus{}da)^2\plus{}(d^2\plus{}ab)^2)$
So,we will must prove that:
$ (a^2\plus{}bc)^2\plus{}(b^2\plus{}cd)^2\plus{}(c^2\plus{}da)^2\plus{}(d^2\plus{}ab)^2 \le (a^2\plus{}b^2\plus{}c^2\plus{}d^2)^2$
$ <\equal{}> a^2(b\minus{}c)^2\plus{}b^2(c\minus{}d)^2\plus{}c^2(d\minus{}a)^2\plus{}d^2(a\minus{}b)^2 \ge 0$
Which is obvious true :)
I think, now you understood

[quote="quykhtn-qa1"][quote="Potla"]Can you please tell me how come $ a^3 \plus{} b^3 \plus{} c^3 \plus{} d^3 \plus{} abc \plus{} bcd \plus{} dca \plus{} dab \le 8$?
 :blush:
Sorry that I could not solve this.  :noo:[/quote]
 $ (a(a^2 \plus{} bc) \plus{} b(b^2 \plus{} cd) \plus{} c(c^2 \plus{} da) \plus{} d(d^2 \plus{} ab))^2 \le (a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2)((a^2 \plus{} bc)^2 \plus{} (b^2 \plus{} cd)^2 \plus{} (c^2 \plus{} da)^2 \plus{} (d^2 \plus{} ab)^2)$
So,we will must prove that:
$ (a^2 \plus{} bc)^2 \plus{} (b^2 \plus{} cd)^2 \plus{} (c^2 \plus{} da)^2 \plus{} (d^2 \plus{} ab)^2 \le (a^2 \plus{} b^2 \plus{} c^2 \plus{} d^2)^2$
$ < \equal{} > a^2(b \minus{} c)^2 \plus{} b^2(c \minus{} d)^2 \plus{} c^2(d \minus{} a)^2 \plus{} d^2(a \minus{} b)^2 \ge 0$
Which is obvious true :)
I think, now you understood[/quote]
Thanks for your elaboration, quy, now I understand. :idea:


where is the message board I can not find my home work?

You should be able to see it now when you log in.  It is the forum right below the Classes Information forum that this thread is in.

I have the same problem as chuchen, where can I find my homework?

On the top of this page we are on you should see

"
[b]Forum index >> Online School >> Classes information[/b]
probability
"

Click on Online School
it should be in the section below your [b]Classes Information[/b] forum

Please let me know if you do not see the message board with the extra problems now.  Because you joined late, you weren't originally in the usergroup that has permission to see that message board.  You have been added to this group, so when you are logged in, you should see the message board.  There is also a link to this message board from the course homepage (click on My Classes above to reach the course homepage).


ratio
geometry
trapezoid
symmetry
projective geometry
geometry unsolved

Let $ {ABCD}$be inscribed in $ {\odot O}$ ,let $ {AB\cap DC \equal{} P}$ and $ {BC \cap AD \equal{} Q}$ $ {QE,QF }$ are tangent to $ {\odot O}$ prove that points$ {P,E,F}$ are collinear.

$ R \equiv BD \cap AC$ and $ X,Y$ are the intersections of line $ PR$ with the sidelines $ BC,AD.$ From the hamonic cross ratios $ (B,C,X,Q) \equal{} (A,D,Y,Q) \equal{} \minus{} 1,$ it follows that $ XY$ is the polar of $ Q$ with respect to $ (O).$ But the line joining the contact points of the tangents from $ Q$ to $ (O)$ is indeed the polar of $ Q$ $ \Longrightarrow$ $ P,E,F$ are collinear.

[hide="Alternate approach"]Project the line $PQ$ to infinity and the circle $(O)$ into another circle $(O').$ $ABCD$ goes to the isosceles trapezoid $A'B'C'D'$ with $B'C' \parallel A'D'$ and by obvious symmetry $E',F'$ become midpoints of the arcs $B'C'$ and $A'D'$ of $(O)'$ $\Longrightarrow$ $D',E',F'$ lie on the symmetry axis of $A'B'C'D'$ $\Longrightarrow$ $D,E,F$ are collinear.[/hide]

Let $ AC$ meet $ BD$ at point $ I \Longrightarrow QI$ is polar line of point $ P$ with circle $ (O) \Longrightarrow P$ lie on polar line of point $ Q$ with circle $ (O)$. But $ EF$ is polar line of point $ Q$. Hence point $ P$ lie on line $ EF$.
Q.E.D.


Someone please explain why $ \overline{OA}$ and $ \overline{ON}$ are equal in the following problem:

In $ \triangle ABC$, $ \angle C$ is a right angle. Point $ M$ is the midpoint of $ \overline{AB}$, point $ N$ is the midpoint of $ \overline{AC}$, and point $ O$ is the midpoint of $ \overline{AM}$. The perimeter of $ \triangle ABC$ is $ 112$cm, and $ ON\equal{}12.5$cm. What is the number of square centimeters in the area of quadrilateral $ MNCB$?

Draw $ \overline{NO}$. Consider $ \triangle ACM$. Since $ O$ and $ N$ are midpoints of $ \overline{AM}$ and $ \overline{AC}$ respectively, the length of $ \overline{ON}$ is half of the length $ \overline{MC}$. Since $ O$ is the midpoint, the length $ \overline{AO}$ is half of the length of $ \overline{MA}$. Since $ \triangle ABC$ is a right triangle and $ \overline{MC}$ is a median to the hypotenuse, $ MC \equal{} MA$. Thus, $ OA \equal{} ON$.

Is the answer to the question $ 1008$?

EDIT: I found a much easier way:

Consider $ \triangle ANM$. Since $ N$ and $ M$ are midpoints, $ \overline{NM}$ is parallel to $ \overline{CB}$, so $ \triangle ANM \sim  \triangle ACB$. Thus, $ \angle ANM$ is a right angle, so $ \triangle ANM$ is a right triangle. Since $ \overline{NO}$ is a median to the hypotenuse, it is equal in length to half the hypotenuse, which is also $ \overline{AO}$.


ratio
geometry
analytic geometry
greatest common divisor
trigonometry
geometry proposed

$\frac{FB}{CB}= ?$

A challenge to people how loves Geometry

in special for Shahzada and Kouichi Nakagawa

I did not want to solve this problem in [b]coordinate geometry[/b].
Can anyone solve this problem by [b][color=red]elementary[/color] geometry[/b]?

Nice problem ! Here is an equivalent enunciation :

[quote][color=darkred]Let $ ECFD$ be a convex quadrilateral for which $ m(\widehat {ECF}) \equal{} 135^{\circ}$ and $ m(\widehat {DFC}) \equal{} 120^{\circ}$ .

Denote $ A\in EF\cap DC$ and the circumcircle $ C(B,r)$ of $ \triangle ADE$ . Ascertain the ratio $ \frac {BF}{BC}$ .[/color][/quote]

[b][u]Proof[/u].[/b] If denote $ G\in EC\cap DF$ , then observe easily that the quadrilateral $ BEGD$ is cyclically ! I"ll come back soon.

[quote="Virgil Nicula"]If denote $ G\in EC\cap DF$ , then observe easily that the quadrilateral $ BEGD$ is cyclically ! I"ll come back soon.[/quote]

Yes, $ \angle EAD \equal{} 127.5^\circ$.
So $ \angle EBD \equal{} 2(180^\circ \minus{} \angle EAD) \equal{} 105^\circ$.
And, $ \angle CGF \equal{} 180^\circ \minus{} (\angle GCF \plus{} \angle GFC) \equal{} 75^\circ$.
Thus, $ \angle EBD \plus{} \angle CGF \equal{} 180^\circ \iff B, \ E, \ G, \ D \ \text{are concyclic.}$

But, I think that we cannot solve the problem only in this  :(

Suppose w.l.o.g. that $ EC=CF=FD=1$ . Denote $ \phi =\frac {\pi}{24}$ and $ \|\begin{array}{c}
x=m(\widehat {FED})\\\\
y=m(\widehat {CDE})\end{array}\|$ . Observe easily that $ 7\phi +5\phi =\frac {\pi}{2}$ ,

$ \|\begin{array}{c}
m(\widehat {CEF})=m(\widehat {CFE})=3\phi\\\\
m(\widehat {DCF})=m(\widehat {CDF})=4\phi\\\\
m(\widehat {BED})=m(\widehat {BDE})=5\phi\end{array}\|\ \ ,\ \  \|\begin{array}{c}
m(\widehat {DCE})=14\phi\\\\
m(\widehat {DFE})=13\phi\end{array}\|$ and $ \|\begin{array}{c}
m(\widehat {BEC})=x+8\phi\ \ ,\ \ m(\widehat {BDF})=y+9\phi\\\\
x+y=7\phi\ \ ,\ \ m(\widehat {BEC})+m(\widehat {BDF})=\pi\end{array}\|$ .

Apply [u][b]SINUS[/b]' theorem[/u] : $ \|\begin{array}{cccc}
\triangle CDE\ : & \frac {DE}{\sin\widehat {DCE}}=\frac {CE}{\sin \widehat {CDE}} & \implies & DE=\frac {\cos 2\phi}{\sin y}\\\\
\triangle DEF\ : & \frac {DE}{\sin\widehat {DFE}}=\frac {DF}{\sin \widehat {DEF}} & \implies & DE=\frac {\cos \phi}{\sin x}\end{array}\|$ $ \implies\ \boxed {\ \begin{array}{c}
DE=\frac {\cos\phi}{\sin x}\\\\
\cos 2\phi \sin x=\cos\phi \sin y\end{array}\ }\ \implies$

$ \cos 2\phi \sin x=\cos\phi \sin (7\phi -x)\Longleftrightarrow$ $ \cos 2\phi \tan x=\cos\phi (\sin 7\phi -\cos 7\phi \tan x)$ $ \Longleftrightarrow$ $ \boxed {\tan x=\frac {\cos\phi \cos 5\phi }{\cos 2\phi +\cos\phi \sin 5\phi }}\ \ (*)$ .

$ DE=2\cdot BE\cos\widehat {BED}\implies$ $ \frac {\cos\phi}{ \sin x}=2\cdot BE\cos 5\phi\implies$ $ \boxed {BE=BD=\frac {\cos \phi}{2\sin x\cos 5\phi}}$ . Apply [u]generalized [b]Pitagora[/b]'s theorem[/u] :

$ \|\begin{array}{cc}
\triangle BDF\ : & BF^2=BD^2+DF^2-2\cdot BD\cdot DF\cdot \cos\widehat {BDF}=\frac {\cos^2\phi}{4\sin^2x\cos^25\phi}+1-\frac {\cos\phi}{\sin x\cos5\phi}\cdot\sin (4\phi -x)\\\\
\triangle BEC\ : & BC^2=BE^2+EC^2-2\cdot BE\cdot EC\cdot \cos\widehat {BEC}=\frac {\cos^2\phi}{4\sin^2x\cos^25\phi}+1+\frac {\cos\phi}{\sin x\cos5\phi}\cdot\sin (4\phi -x)\end{array}\|$ $ \implies$

$ (\frac {BF}{BC})^2=\frac {\cos^2\phi(\sin^2x+\cos^2x)+4\sin^2x\cos^25\phi -4\cos \phi\cos 5\phi\sin x\sin (4\phi -x)}{\cos^2\phi(\sin^2x+\cos^2x)+4\sin^2x\cos^25\phi +4\cos \phi\cos 5\phi\sin x\sin (4\phi -x)}$ $ \implies$

$ (\frac {BF}{BC})^2=\frac {\cos^2\phi(\tan^2x+1)+4\tan^2x\cos^25\phi -4\cos \phi\cos 5\phi\tan x(\sin 4\phi -\cos 4\phi \tan x)}{\cos^2\phi(\tan^2x+1)+4\tan^2x\cos^25\phi +4\cos \phi\cos 5\phi\tan x(\sin 4\phi -\cos 4\phi \tan x)}$ $ \implies$

$ (\frac {BF}{BC})^2=\frac {\cos^2\phi-4\cos\phi\cos 5\phi \sin 4\phi\cdot\tan x+(\cos^2\phi +4\cos^2 5\phi +4\cos\phi\cos 5\phi cos 4\phi)\cdot \tan^2x}{\cos^2\phi+4\cos\phi\cos 5\phi \sin 4\phi\cdot\tan x+(\cos^2\phi +4\cos^2 5\phi -4\cos\phi\cos 5\phi cos 4\phi)\cdot \tan^2x}$ .

$ \|\begin{array}{c}
U=\cos^2\phi +4\cos^25\phi\\\\
V=4\cos\phi\cos 5\phi\cos 4\phi\\\\
M=4\cos \phi\cos 5\phi\sin 4\phi\\\\
N=\cos^2\phi\end{array}\|$ $ \implies$ $ (\frac {BF}{BC})^2=\frac {N-M\cdot \tan x+(U+V)\cdot \tan^2x}{N+M\cdot\tan x+(U-V)\cdot \tan^2x}$ .

Now we can ascertain $ M$, $ N$ , $ U$ , $ V$ and $ \tan x$ form $ (*)$ , where $ \phi=\frac {\pi}{24}$ . After ugly calculations obtain $ \frac {BF}{BC}= ...$ . :D :D :rotfl:

Thank you for your solution.
I solved it in a trig at the start, too.

1. Let $ G \in EC \cap DF$, Find $ GC$, $ GF$.
2. Find $ ED$.
3. Find $ BE\equal{}BD$.
4. Find $ BC$, $ BF$.
Last. Find $ \frac{BF}{BC}$.

(There may be how to solve others like Virgil Nicula, too.)

But, trig is [b]not[/b] elementary.
I think elementary geometry don't use "sine low", "cosine low" and coordinate etc.
I think that coordinate is a still elementary object than trig.
So I introduced coordinate solution.

[b]Will not the problem of most geometry be removed if we accept [color=red]trig[/color]?[/b]


A cylindrical oil tank, lying horizontally, has an interior length of $ 10$ feet and an interior diameter of $ 6$ feet.  If the rectangular surface of the oil has an area of $ 40$ square feet, the depth of the oil is:

$ \textbf{(A)}\ \sqrt{5} \qquad
\textbf{(B)}\ 2\sqrt{5} \qquad
\textbf{(C)}\ 3\minus{}\sqrt{5} \qquad
\textbf{(D)}\ 3\plus{}\sqrt{5} \\
\textbf{(E)}\ \text{either }3\minus{}\sqrt{5}\text{ or }3\plus{}\sqrt{5}$

[hide="Nonslick solution"]This solution is for those who wants to do it the long way (why?  :D )
The area of the oil surface is 40 with length of 10 so that means the width is 4. Cross-sectionally looking at the circle with radius of 3, we have a chord of length 4 so, this chord is $ \sqrt {3^2 \minus{} 2^2} \equal{} \sqrt5$ from the center and thus, the depth of the oil is either $ 3 \minus{} \sqrt5$ or $ 3 \plus{} \sqrt5$.[/hide]

[quote="ocha"][hide="solution - cheating"]
Well the rectangular surface will be the same in exactly two positions, so there must be two solutions
$ \longrightarrow \boxed{E}$  :rotfl: [/hide][/quote]

And if the rectangle just so happened to be a square?


factorial
number theory
prime factorization

What's a clear explanation, without math proof, of the fact that the number of trailing zeros in n factorial is the number of 5s, AND NOT 2s, in the prime factorization of n factorial. I know the technique but can't seem to grasp the idea why.

Same logic applies to problems like: If P = 64 factorial, then what is the largest postive integer n such that P is divisible by 12^n?
In this, 12 = 2^2 * 3. Why do we choose the lower 3s as the power n, instead of the abundant 2s.

The short answer is that the number of fives is more decisive than the number of twos, as factors, in deciding whether a number will end in even one zero. For example, two to the fifth power has NO zeros at the end. But note that what is REALLY decisive is how the twos and the fives interact. Five to the third power doesn't have any zeros at the end either. It's because two times five equals ten that each pairing of a factor of five with a factor of two will add another zero (power of ten) to the end of a number.

Does this help?

Hello, Done!

[quote]What's a clear explanation, without math proof,
of the fact that the number of trailing zeros in $n!$
is the number of 5s, AND NOT 2s, in the prime factorization of $n!$.
I know the technique but can't seem to grasp the idea why.[/quote]
Each factor of 5 in the prime factorization of $n!$  (when combined with a factor of 2)
will contribute a trailing zero.

We need to count the number of occurences of $5\times2$ in the prime facorization.

Factors-of-5 occur in every fifth number of $n!$.
Since factors-of-2 occur in every [u]second[/u] number of $n!$,
there will always be enough 2's to pair off with the 5's.

Hence, we need count only the factors-of-5.

yes..that is well-explained tokenadult and Soroban.

The thing I missed was the logic of how the two numbers 'pair'.

Thanks!

If the prime factorization includes x fives and y twos, whichever number (x or y) is smaller is the number of zeros at the end.

[quote="LuCy4EvA"]Whoa... I really don't get this trailing zero thing.. say the number 200, the prime factorization is 5^2*2^3, so how many trailing zeros would it have?[/quote]

Number of leading zeroes = number of times it can be divided by 10. 10 is 5*2. How many times can you divide by 5 and 2? You can divide by 5 twice, 2 three times. So you can divide by both of them twice - 2 leading zeroes.

The initial question was about factorials - then you have to count all the numbers from 1 up to that number, not just that number as it is.

Thanks triplem that makes much more since now.  I was wondering about that.

Hello, LuCy4EvA!

[quote]Whoa... I really don't get this trailing zero thing..
Say the number 200, the prime factorization is 5^2*2^3,
so how many trailing zeros would it have?[/quote]
I assume you are referring to 200! (200 factorial) . . .

Consider the product: 1 x 2 x 3 x ... x 199 x 200.

Every factor of 5 (together with a factor of 2) will create a trailing zero.

How many factors-of-5 are there in that product?
Here's my baby-talk explanation . . .

Every fifth number {5, 10, 15, ...} will have a factor of 5
There are: 200/5 = 40 of them.
. . .   So, there [b]40[/b] factors-of-5.

But wait . . . some of them {25, 50, 75, ...} contribute [b]two[/b] 5's.
How many are there?
There are: 200/25 = 8 of them.
. . . So, there are [b]8[/b] more 5's

And there are some {125, ...} which contribute [b]three[/b] 5/s.
How many are there?
There are:  200/125 = 1 of them (and some fraction).
. . . So, there is [b]1[/tex] more 5.

Hence, there are 40 + 8 + 1 = 49 factors-of-5 in 200! . . .
Therefore, 200! ends in 49 zeros.

I think Soroban forgot to mention that there are more factors of 2 in 200! than factors of 5. This is natural,  because we could just look at the even numbers less than 200, which are 100, and each gives us at least one factor of 2. Since Soroban already computed that there are exactly 49 factors of 5 in 200!, we can pair up each 5 with a 2, making a factor of 10, which is actually a trailing 0.

This actually happens each time we compute the number of 5's and 2's in n!. Proving it is a little bit harder, but we can  use the same tehnique Soroban used to find the number of 5's in 200!. Try it :)

i believe his was works because there are fewer fives but they along with the two's are needed to get a 10.  Therefore, the fives determine the number of ending zeros since it is the bigger of the 2 smallest primes needed to make 10.


[hide="."]Welcome to the game I like to call "Click to reveal hidden content..."  It is a game that a few of my friends asked me to do to pass the time to wait till next mafia round.  To play this game, you must do the following[hide=":"]Step 1: Type (hide=<your word here>), where the parentheses should be brackets and the <your word here> is your word[hide="."] Step 2: After your hide tag, make another one with a different word until you have a sentence. You can have more than 1 word to catch people off guard and maybe make them laughStep 3: For every hide tag you did, at the end of your post, put (/hide) for every hide tag, where the parentheses are bracketsplay![/hide][/hide][/hide]

ok, sounds fun!
[hide="my"]
[hide="tummybox"]
[hide="feels"]

[url=http://kevan.org/brain.cgi?zebarg]said[/url]

[/hide]
[/hide]
[/hide]

lol good one! that tricked me...seriously.

This was me b4 when you tricked me. :rotfl:

Then I was like that after zebarg "operated", except I wasn't moving...

[hide="the"]
[hide="fat"]
[hide="lady."]
WEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEE!!!!!111!!!one!!!!one\\!!!!!111\!!!!won!!!!!!1111!!!11!!!1111own!!1!1!!neon
[/hide]
[/hide]
[/hide]

go to kevan.org/brain ...  it makes it pretty simple (if you are asking what I think you are asking).

the way you link to it would be like this: (url=http://kevan.org/brain.cgi?yournamehere)Click me!(/url)

Wow, I haven't fallen for one of those since Treething introduced it to the site, but you got me.

[hide="This"][hide="seems"][hide="really"][img]http://www.jamesryman.com/archive/images/screaming%20zombiea.jpg[/img][/hide][/hide][/hide]

[quote="slamchess"][hide="i"]BOO [hide="am"]and [hide="da"]without furtherado...NO DA YODA[/hide][/hide][/hide][/quote]

[hide="Ha"][hide="Ha"][hide="thats"]that 1=3 not 2.[/hide][/hide][/hide]

[quote="la mafioso"]NOTE TO EVERYONE:

While it is indeed fun to trick people into eating a brain, it gets old if you do it too many times. While tricking is fun, please move on to cooler things soon.

I will do something really fun soon...kinda like a game i guess.[/quote]

Man, you're the one who's tricked people the last like 3 times...
 :?

[hide="i"]
[hide="keep"]
[hide="getting"]

[url=http://kevan.org/brain.cgi?math_92_]![/url]
[/hide][/hide][/hide]

[hide="This"][hide="is"][hide="getting"]And I didn't put a link to my zombie page! I must be sane![/hide][/hide][/hide]

Okay, here's something different.

[hide="Click here to read my story"]Okay, once upon a time a serial killer killed everyone, he comes to your house. You have a [hide="knife"]you stabbed your self on accident you die.[/hide], a [hide="gun"]The bullet knocks off a wall and hits you again[/hide], and [hide="a can of beans"]You eat the beans, and you fart, propelling yourself using Newton's third law of motion into the serial killer. He surrenders. You win![/hide]. What do you do?[/hide]

And if you need a tutorial on how to do this, ask me and i will be more than happy to tel.[/hide]

I think I'll try to eat the serial killer...

[quote="la mafioso"][hide="Wow"][hide="that"][hide="is"][url=http://kevan.org/brain.cgi?Sc0rpXLR]minutes[/url] [/hide][/hide][/hide][/quote]

 :wallbash_red:

My auto combo maker 1.0! Combo's must begin with 0, and have only 3 digits, but it's a work in progress to a better tomorrow!  :D

EDIT:It can now do 00...and 01... numbers. It is still 3 digits however, and once i add a digit it only then becomes 2.0.

[hide="click here to launch"]

[hide][hide="00"]Thank you for using la mafioso's auto combo maker 1.0. Your code choice is: 001Thank you for using la mafioso's auto combo maker 1.0. Your code choice is: 002Thank you for using la mafioso's auto combo maker 1.0. Your code choice is: 003Thank you for using la mafioso's auto combo maker 1.0. Your code choice is: 004Thank you for using la mafioso's auto combo maker 1.0. Your code choice is: 005Thank you for using la mafioso's auto combo maker 1.0. Your code choice is: 006Thank you for using la mafioso's auto combo maker 1.0. Your code choice is: 007Thank you for using la mafioso's auto combo maker 1.0. Your code choice is: 008Thank you for using la mafioso's auto combo maker 1.0. Your code choice is: 009[/hide][/hide][/hide][hide="01"][hide="011"]Thank you for using la mafioso's auto combo maker 1.0. Your code choice is: 011[/hide][hide="012"]Thank you for using la mafioso's auto combo maker 1.0. Your code choice is: 012[/hide][hide="013"]Thank you for using la mafioso's auto combo maker 1.0. Your code choice is: 013[/hide][hide="014"]Thank you for using la mafioso's auto combo maker 1.0. Your code choice is: 014[/hide][hide="015"]Thank you for using la mafioso's auto combo maker 1.0. Your code choice is: 015[/hide][hide="016"]Thank you for using la mafioso's auto combo maker 1.0. Your code choice is: 016[/hide][hide="017"]Thank you for using la mafioso's auto combo maker 1.0. Your code choice is: 017[/hide][hide="018"]Thank you for using la mafioso's auto combo maker 1.0. Your code choice is: 018[/hide][hide="019"]Thank you for using la mafioso's auto combo maker 1.0. Your code choice is: 019[/hide][/hide][/hide][/hide]Remember, it is a work in progress, and will be updated time to time. Thank you for trying it out!  :D


Solve the equation
$m\cos{x}-(m+1)\sin{x}=m\quad m \in \mathbb{R}$
finding $m$ such that the difference between two roots $x\prime$ and $x\prime\prime$ is equal to $\frac{\pi}{2}$.

Doing some ugly rearranging that I wasn't sure would work (and there are probably easier ways to solve), I got
$\cos x = \frac{m+1}{m} \sin x + 1$
I then did more ugliness, squaring both sides to get rid of the cos in exchange for a sin
$-\sin^2 x = y^2\sin^2 x + 2y \sin x$ where $y = \frac{m+1}{m}$
$\sin x = \frac{-2y}{y^2 + 1} = \frac{-2m(m+1)}{2m^2 + 2m + 1}$
Now we can use the fact the roots are $\frac \pi 2$ apart.  Since we only have sin left, and we know the differences between angles and the x-axis relate the absolute values of all the trig functions (I can't think of a better way to say that...), we can determine that one of the roots must be of the form $\frac \pi 4 x + \frac \pi 2 k$, and we know that the sin of all those angles are either $\pm \frac{\sqrt 2}{2}$.  So we can say
$\pm \frac{\sqrt 2}{2} = \frac{-2m(m+1)}{2m^2 + 2m + 1}$
Square to get rid of the plus/minus... even though it makes it very ugly.
$8m^4 + 16m^3 + 8m^2 = 4m^4 + 8m^3 + 8m^2 + 4m + 1$
$4m^4 + 8m^3 - 4m - 1 = 0$
At a loss of how to simplify this further... I graphed, and found zeros; the solutions are $\boxed{\pm \frac{\sqrt 2}{2} , -\frac{\sqrt 2}{2} \pm 1}$

I took this way:

Following the formulas $\sin{x} = \frac{2\tan{\frac{x}2}}{1+\tan^2{\frac{x}2}}$ and $\cos{x} = \frac{1 - \tan^2{\frac{x}2}}{1 + \tan^2{\frac{x}2}}$ we have:

$m\cos{x} - (m+1)\sin{x} = m$
$m\frac{1-\tan^2{\frac{x}2}}{1+\tan^2{\frac{x}2}} - (m+1)\frac{2\tan{\frac{x}2}}{1+\tan^2{\frac{x}2}} = m$
$m\left(1-\tan^2{\frac{x}2}\right) - 2(m+1)\tan{\frac{x}2} = m\left(1+\tan^2{\frac{x}2}\right)$
$m\tan^2{\frac{x}2} + (m + 1)\tan{\frac{x}2} = 0$
$\tan{\frac{x}2} = 0 \quad or \quad \tan{\frac{x}2} = -\frac{m+1}{m}$

The solutions for $\tan{\frac{x}2} = 0$ are $0,2\pi,4\pi,6\pi\dotsc$. Then I can infer that the solutions for $\tan{\frac{x}2} = -\frac{m+1}{m}$ [b]must[/b] be $\frac{\pi}2,\frac{5\pi}{2},\frac{9\pi}{2}\dotsc$, what implies in
$-\frac{m+1}{m} = 1 \quad \therefore \quad m = -\frac12$.

But I'm not sure about that...  :?

Those tans look very fun... well I did some graphing and comparing zeros... -0.5 works.  I compared the graph of the initial function to that of the one I derived, $\sin x = \frac{-2y}{y^2+1}$ where $y=\frac{m+1}{m}$, and the zeros are different.  That would explain how I got a wrong answer... although my algebra there looks right to me...


$ f,g$ are continuous function on $ (a,b)$ such that $ (f(x))^2\equal{}(g(x))^2\ne 0$ for every $ x\in (a,b)$.
Prove that either $ f(x)\equal{}g(x)$ for every $ x\in (a,b)$ or $ f(x)\equal{}\minus{}g(x)$ for every $ x\in (a,b)$.

obviously for every $ x\in (a,b)$, $ f(x) \equal{} g(x)$ or $ f(x) \equal{} \minus{} g(x)$. assume, for a contradiction, that there are $ x,y\in (a,b)$ such that $ f(x) \equal{} g(x)$ and $ f(y) \equal{} \minus{} g(y)$. we may assume WLOG that $ x < y$. now let $ r \equal{} \sup\{t\in [x,y]|\forall z \in [x,t], f(z) \equal{} g(z)\}$. by the definition of $ r$, for every natural $ n$, we have $ x_n\in (r \minus{} 1/n,r)$ and $ y_n\in (r,r \plus{} 1/n)$ such that $ f(x_n) \equal{} g(x_n)$ and $ f(y_n) \equal{} \minus{} g(y_n)$. but then $ \lim_{n\to \infty} |x_n \minus{} y_n| \equal{} 0$, yet $ \lim_{n\to \infty} |f(x_n) \minus{} f(y_n)| \equal{} \lim_{n\to \infty}|g(x_n) \plus{} g(y_n)| > 0$, contradicting the continuity of $ f$.

[quote="The Nguyen"]$ f,g$ are continuous function on $ (a,b)$ such that $ (f(x))^2 \equal{} (g(x))^2\ne 0$ for every $ x\in (a,b)$.
Prove that either $ f(x) \equal{} g(x)$ for every $ x\in (a,b)$ or $ f(x) \equal{} \minus{} g(x)$ for every $ x\in (a,b)$.[/quote]

We have that $ (f(x))^2 \equal{} (g(x))^2\ne 0$ for every $ x\in (a,b)$.

From above we have that $ |f(x)|\equal{}|g(x)|\ne 0$ for every $ x\in (a,b)$. So for any $ x \in (a,b)$ we have either $ f(x) \equal{} g(x)$ or $ f(x)\equal{}\minus{}g(x)$.

But $ f(x)$ is continuous over $ (a,b)$ and since it is not zero over that interval it is either $ f(x)>0$ for every $ x\in (a,b)$ or $ f(x)<0$ for every $ x\in (a,b)$. Infact, If there exist $ x_1$ and $ x_2$ in the interval $ (a,b)$ such that $ f(x_1)>0$ and $ f(x_2)<0$ for the "Intermediate value theorem" we should have that for some $ x_0$ in $ (a,b)$ it should be $ f(x_0)\equal{}0$ in contradiction with the given condition. The same applies to $ g(x)$ so we have that  either $ f(x) \equal{} g(x)$ for every $ x\in (a,b)$ or $ f(x) \equal{} \minus{} g(x)$ for every $ x\in (a,b)$.