inequalities
geometry
trigonometry
geometry unsolved

in $ \Delta ABC$ $ \angle A \ge \angle B \ge \angle C$, $ AD$ is the height and $ D$ is on $ BC$, $ BE$ is the median and $ E$ is on $ AC$, $ CF$ is the bisector of $ \angle C$ and $ F$ is on $ AB$.  let $ P$, $ Q$, $ R$, be the intersection of $ AD$  $ BE$  $ CF$, let $ S_1$ be the area of $ \Delta PQR$ and $ S_2$ be the area of $ \Delta ABC$. prove that $ \frac {S_1}{S_2} < \frac {1}{6}$

could somebody please give me some ideas on how to solve this problem?

thanks!!

does somebody have an idea on this problem?

i tried menelaus's theorem, but it was really long and complicated and i don't think that quite solved this problem

please share your ideas!!

after some calculations i found that the original problem turn into the following inequality
$ \frac {a}{2a \plus{} b} \plus{} \frac {\tan C}{\tan B \plus{} 2\tan C} \plus{} \frac {ab \tan B}{(a \plus{} b)(a\tan B \plus{} b\tan C)} > \frac {5}{6}$
but how do i proceed with this? does anybody have an idea?

$ \boxed {\ \begin{array}{c} \triangle\ ABC \\
\ \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \equal{} \\
\ M\in (BC)\ ,\ \frac {MB}{MC} \equal{} m \\
 \\
N\in (CA)\ ,\ \frac {NC}{NA} \equal{} n \\
 \\
P\in (AB)\ ,\ \frac {CA}{CB} \equal{} p\end{array}\ }$ $ \implies\ \sigma [XYZ] \equal{} \frac {(1 \minus{} mnp)^2}{(1 \plus{} m \plus{} mn)(1 \plus{} n \plus{} np)(1 \plus{} p \plus{} pm)}\cdot \sigma [ABC]$ .

In the your particular case obtain $ \frac {\sigma [PQR]}{\sigma [ABC]}\equal{}\frac {b(a\cos C\minus{}c\cos B)^2}{a(a\plus{}b)(1\plus{}\cos C)(a\plus{}c\cos B)}$ .

thanks a lot. what you gave me is really helpful!


group theory
abstract algebra
superior algebra
superior algebra unsolved

Is anyone of know another solution to show that the only subgroup of index $ 2$ in $ S_n$ is $ A_n$ without depending on $ A_n$ is simple group whenever $ n\geq 5$.

 :)

A subgroup of $ G$ of index 2 is the kernel of a nontrivial homomorphism $ \phi$ from $ G$ to $ \{\minus{}1,1\}$. The symmetric group is generated by transpositions, so the images of the transpositions determine ny such homomorphism.
Since $ \{\minus{}1,1\}$ is abelian, $ \phi(h^{\minus{}1}gh)\equal{}\phi(h^{\minus{}1})\phi(g)\phi(h)\equal{}\phi(g)$ and any two conjugate elements in $ S_n$ have the same image. Transpositions are all conjugate to each other, so their images are all the same. That leaves two possibilities:
- For any transposition $ g$, $ \phi(g)\equal{}\minus{}1$. In this case, the kernel of $ \phi$ is $ A_n$.
- For any transposition $ g$, $ \phi(g)\equal{}1$. In this case, $ \phi$ is trivial and its kernel is $ S_n$.


Can someone please help me out with these questions from
[url=http://math.rice.edu/~lanius/fractals/Sier_Ques/]this site?[/url]

Also, I am having trouble with drawing the triangle on that grid paper that is provided. Anyone's help will be appreciated. Thanks!!  :)

[url=http://math.rice.edu/~lanius/fractals/]math.rice.edu/~lanius/fractals/[/url]

IM assuming by "steps" they mean drawing a smaller triangle in.
So naturally, 1-(fraction of shaded stuff)=fraction of not shaded stuff.
In the first step, the shaded stuff is 1/4
in the next, its 1/4+3/16
In the next, its 1/4+3/16+9/64.
so the total formula for the shaded parts would be
1/4+3/16+9/64+...+3^(n-1)/4^n


Two people, A and B, are playing a game. They each have a white card and a black card. At the same time, they put either their white or black card on the table. If there are 2 black cards, A gets 5 dollars. If there are 2 white cards, A gets 1 dollar. If there is a white card and a black card, B gets 2 dollars. What is A's best strategy for winning as much money as possible?

Ok, first I need to make a few assumptions. I'm assuming that information is public, i.e. each player knows the other's payouts. I'm also assuming that the money is coming from some 3rd party, not from the players, and that the payouts listed are the only ones that occur (i.e. A gets 0 if its black and white and B gets 0 if its both white or both black), then let this be a diagram of the payouts:

A\B
-----B------W
B  (5,0)  (0,2)
W(0,2)  (1,0)     <--- in each (a,b), a is the payout to A and b is the payout to B

Using maximin, playing black is obviously a dominant strategy for player A. However, since player B sees this, he/she will play white because either strategy yields the same two payouts anyway, and player A's maximin strategy is to play black.  Now realizing this, player A will be forced to play white as well. It stands to reason, therefore, that WW --> (1,0) is an equilibrium point for this game and will be the end result if both players play rationally. I'm pretty sure this is correct, however I could use some verification.

The money does come from the players, not a 3rd party. The players put the cards on the table at the same exact time, so they don't know what the other puts down ahead of time. I am looking for: what fraction of the time should A put down black? White?  ex: 1/8 black and 7/8 white.

[hide="Hint"]Let the probability that player A puts down black be p.[/hide]


calculus
function
derivative
real analysis
real analysis unsolved

Hi there. I'm working through "Forgotten Calculus" (Barbara Lee Bleau) and am stuck on an optimisation problem. I'll give the question, show my attempt, and then reveal her (stated, not worked) answer. Would appreciate guidance.
-----------------
a) the question

The price equation for a new product is
p= 1,600 - 4x
where p = price and x = # of units produced.
The cost function is 40x + 2,000.
Find the maximum profit and the price which needs to be charged to achieve it.
-------------------
b) I'm using an approach which has worked well with a number of similar-looking problems.

What's to be maximized? Profit

Profit = revenues - costs
= (units * price) - (units * costs)
= (x (1600-4x)) - (x (40x + 2000))
= (1600x - 4x^2) - (40x^2 + 2000x)
= 1600x - 4x^2 - 40x^2 - 2000x
= -400x - 44x^2

Even though I'm already struggling to see how this product will make anything but losses, I seek out maxima via the first derivative

Profit ' = -400 -88x
0 = -400 - 88x
400 = -88x
-400 = 88x
-4.545454 = x

Plugged back into the original function, it means the more units produced, the more money lost. Making a profit seems impossible as it would mean selling a negative number of units.

------------------------

Her answer: "A price of $824 leads to a maximum profit of$148,544."

 I cannot reach this. Where am I going wrong?

That cost function is clearly intended to be total, not per unit.


geometry
circumcircle
perpendicular bisector
geometry unsolved

Let $ (K)$ be the circle with diameter $ CD$. Then clearly $ A \in (K)$
Let $ F$ be the second point of intersection between the line $ AB$ and the circle $ (K)$
The point $ F$ is known (easily constructible)

$ KF\equal{}KA$ so $ K$ is on the perpendicular bisector of $ FA$

$ K$ is the midpoint of $ CD$, so $ KC\perp KO \Rightarrow KB\perp KO$. So $ K$ belongs on the circle with diameter $ BO$

[size=125]Construction[/size]

Let $ BC'D'$ be an arbitrary secant of the circle $ (O)$. We draw the circumcircle of $ C'D'A$. The line $ BA$ intersects this circle at the points $ A$ and $ F$

Let $ m$ be the perpendicular bisector of $ AF$

Let $ w$ be the circle with diameter $ BO$

Then $ K \in m \cap w$ (we can have $ 2$, $ 1$ or $ 0$ solutions)


function
geometry
3D geometry
analytic geometry
graphing lines
slope
real analysis

Suppose f is absolutely continuous on [a,b], g is absolutely continuous on [c,d], c<f<d, then how to prove g(f(x)) is also absolutely continuous?

[quote="§outh§tar"]I once heard that absolutely continuous functions take measure zero sets into measure zero sets.

So[/quote]

So what?

I can prove what you say. Then...???

Can you prove the converse? That matters too; it says that "$ f(S)$ has measure zero whenever $ S$ has measure zero" is a valid definition of absolute continuity. With that definition, this composition property is immediately obvious.

If only that converse were true without the extra assumption that $ f$ is of bounded variation... ;) (for a counterexample to jmerry's claim, consider a sawtooth function whose teeth accumulate to $ 0$ and have the divergent sum of heights).

All right, let's turn that counterexample into a counterexample to the original claim. Let $ g$ be the piecewise linear function on $ [0,1]$ connecting the origin and the points $ \left(\frac1n,\frac{(\minus{}1)^n}{n^2}\right)$ for natural numbers $ n$ and $ f$ be the cube root function $ f(x)\equal{}x^{\frac13}$ for $ \minus{}1\le x\le 1$. This $ g$ is absolutely continuous; it's even Lipschitz, as the absolute slopes of the segments start at $ \frac52$ and converge to $ 2$. This $ f$ is also absolutely continuous; we can squeeze together intervals at the center to increase variation, and it's uniformly continuous.

Now, $ f\circ g$ connects the points $ \left(\frac1n,\frac{(\minus{}1)^n}{n^{\frac23}}\right)$. It's not of bounded variation, so it's not continuous.


Let $ P$ be a given point inside circle $ (O)$. $ AB$ and $ CD$ are its chords such that $ AB \perp CD$ at $ P$.
Find the minimum and maximum value of $ AB.CD$

[hide]
Let the radius of $ O$ be $ r$, and let $ p \equal{} OP^2$. Let $ E,F$ be the feet of perpendiculars from $ O$ to $ AB$, $ CD$ respectively and let $ a \equal{} OE^2, b \equal{} OF^2$; then $ OEPF$ is a rectangle, and we have (Pythagorean Theorem) $ p \equal{} a\plus{}b$. Then $ AB \cdot CD \equal{} 4\left(\sqrt{r^2 \minus{} a}\sqrt{r^2 \minus{} b}\right) \equal{} 4\sqrt{r^4 \minus{} pr^2 \plus{} ab}$, and by AM-GM, $ 0 \le ab \le \frac{p^2}{4}$. Thus $ 4r\sqrt{r^2 \minus{} p} \le AB \cdot CD \le 4r^2 \minus{} 2p$.
[/hide]


inequalities
function
integration
complex analysis
complex analysis unsolved

Let $f(z)$ be analytical function and $M = \sup \{ z: |z|=1, \Re(z) \geq 0 \}$,$m = \sup \{ z: |z|=1, \Re(z) \leq 0 \}$. Prove that $|f(0)| \leq \sqrt{Mm}$

$|f(0)|=\frac{1}{2 \pi} | \int_0^{2 \pi} f(e^{it}) dt | \leq \frac{m+M}{2}$.
To have a better bound we want to have $m$ almost equal to $M$. It suffices to consider the function $h(z)=f(z)f(-z)$ and to apply what we have just shown to $h$ and the result will follow.

The same conclusion is true if one cuts the unit circle into $n$ equal sectors i.e. $|f(0)| \leq (M_1M_2..M_n)^{1/n}$.(same proof with $h(z)=f(z)f(wz)f(w^2z)..f(w^{n-1}z)$ ). But what can one say if the sectors do not have equal lengths ?

Very nice, Alekk.
Btw, we can also use a bit more complicated function $f(z) conj( f( -\bar z))$, here $conf(f)$ means the conjugate of  function $f$.


For each integer $n\geq0$, let $S(n)=n-m^2$, where $m$ is the greatest integer with $m^2\leq n$. Define a sequence by $a_0=A$ and $a_{k+1}=a_k+S(a_k)$ for $k\geq0$. For what positive integers $A$ is this sequence eventually constant?

[hide]
clearly for each $A$ that is a perfect square the sequence is constant, because $a_0=A$ $a_1=A+(A-m^2)$, but for the conditions we put on $m$ it must be $m=A$, so $a_1=a_k=A$ $\forall k\geq 0$. Now we have to prove that this is the only case. Since the sequence is constant
$a_0=a_1 \implies A=2A-m^2 \implies A=m^2$ [/hide]

We claim that only the squares work.
Suppose that there is another $A$ that works, then the sequence must eventually contain a square, i.e. there is a positive integer $m^2 <n <(m+1)^2$ in the sequence such that
$n+S(n) =(m+k)^2$ for some positive integer $k$. If $k=1$, we get $2n=m^2 +(m+1)^2$, which is impossible because of parity reasons.
 If $k\geq 2$, we have
$$n =\frac{m^2   +(m+k)^2}{2}= m^2 +km +\frac{ k^2 }{2} \geq (m+1)^2 >n,$$
also impossible.

An easy problem
We define $f(n) = n+n - m^2 = n +S(n) $ .So , clearly $f(a_k) = a_{k+1}$ .Now let's say for some $i$ the sequence ${a_n}_{n\geq i} $ is a constant sequence . so $a_{i+1}=f(a_i) = 2a_i - m^2 \implies a_i = m^2$ Hence for the sequence to be constant , a perfect square has to appear .

Claim: If $n$ is a non perfect square then $f(n)$ is also a no perfect square .
Proof : Let $n = m^2 +k$ where $0<k\leq 2m $.So , $f(n) = 2n -m^2 = m^2 +2k $ .Now note that $m^2<f(n)\leq m^2+4m <(m+2)^2$.So , if $f(n)$ is a perfect square then $f(n) = (m+1) ^2 \implies 2k = (m+1)^2 - m^2 $which is  the contradiction .
By the above claim we have shown that if $a_n$ is eventually constant then   $a_0$ cannot be a non perfect square  and so to be a perfect square is a necessary condition and it is trivially sufficient too .


Lookout station A is 15 km west of station B.  The bearing from A to a fire directly south of B is S 37°50' E.  How far is the fire from B?

Thank you

[quote="cincodemayo5590"][hide="Easy"]
Forms a right triangle, $h=\frac{15}{\cos{52^\circ 10'}}$[/hide][/quote]

Hi Cinco!  Thanks.

Can you extend that out for me?   :)


Suppose R is the radius of convergence of the power series: $\sum_{k=0}^{\infty}a_{k}(x-x_{0})^{k}$.
Show that:
a) The power series converges absolutely for all x such that $|x-x_{0}|< R$

b) The power series diverges for all x such that $|x-x_{0}| > R$

c) If $0 < r < R$ then $f_{n}(x) = \sum_{k=0}^{\infty}a_{k}(x-x_{0})^{k}$ converges uniformly to $f(x) = \sum_{k=0}^{\infty}a_{k}(x-x_{0})^{k}$ on $[x_{0}-r, x_{0}+r]$.

d) $f(x)$ defined above is continuous on $(x_{0}-R, x_{0}+r)$

This is a theorem in my textbook. Does any one have some hints for b and d? I used Weierstrass M-test for part c.

[quote="akech"]Suppose R is the radius of convergence of the power series: $\sum_{k=0}^{\infty}a_{k}(x-x_{0})^{k}$.
Show that:
a) The power series converges absolutely for all x such that $|x-x_{0}|< R$
[/quote]
$R=1/L$ where $L=\limsup |a_{n}|^{1/n}$. (And when $L=0$ or $L=+\infty$ those are special cases, $R=+\infty$ and $R=0$ respectively.)

By root test, $|x-x_{0}|<R$ shall converge absolutely.

[quote]
b) The power series diverges for all x such that $|x-x_{0}| > R$
[/quote]
By root test, $|x-x_{0}|>R$ shall diverge.

What I dont understand is how you can ask (a) and (b) isnt that just the definition of what it means "radius of convergence"?
[quote]
c) If $0 < r < R$ then $f_{n}(x) = \sum_{k=0}^{\infty}a_{k}(x-x_{0})^{k}$ converges uniformly to $f(x) = \sum_{k=0}^{\infty}a_{k}(x-x_{0})^{k}$ on $[x_{0}-r, x_{0}+r]$.

d) $f(x)$ defined above is continuous on $(x_{0}-R, x_{0}+r)$

This is a theorem in my textbook. Does any one have some hints for b and d? I used Weierstrass M-test for part c.[/quote]

Dont have time to do (c) and (d) by for (c) I think you mean the "open interval".

Choose $0<R_{1}<R$ if $R>0$ and then show it is convergent uniformyl on $[-R_{1},R_{1}]$ by Weierstrass Test.

Same thing with (d).

When I come back after sweeping floors I might have time to respond.

[quote="akech"]Well, some theorems are taken as definitions after they are proven (so perhaps some books may take (a) and (b) as definitions while others consider them as theorems to be proved).
For part (c) it should be closed because $0 < r < R$.[/quote]
You're right, I just was posting really fast.

(Use Weierstrass Test).


trigonometry
geometry
circumcircle
ratio
greatest common divisor
angle bisector

Let ABC be a triangle for which B=A+60. I note $N\in (AC),\ M\in (AB)$, such that
$4m(\widehat {ACM})=C\ and\ m(\widehat{ABN})=30.\ Ascertain\ m(\widehat {CMN}).$

[color=blue][b]Lemma[/b][/color] Let $\triangle ABC$ be a triangle with the sides a = BC, b = CA, c = AB and internal angles $\alpha = \angle A, \beta = \angle B, \gamma = \angle C$. If $\beta - \alpha = \angle B - \angle A = 60^\circ$, then the length $l_C$ of the internal bisector of the angle $\gamma = \angle C$ is equal to $\l_C = \frac{abc}{b^2 - a^2}$.

[color=blue][b]Proof:[/b] [/color] In a general triangle $\triangle ABC$, the length $l_C$ of the internal bisector of the angle $\gamma = \angle C$ is given by

$l_C = \sqrt{ab\left(1 - \frac{c^2}{(a + b)^2}\right)}$

Hence, we have to show that if $\beta - \alpha = 60^\circ$, then

$(?)\ \ \frac{abc}{b^2 - a^2} = \sqrt{ab\left(1 - \frac{c^2}{(a + b)^2}\right)}$

$(?)\ \ \frac{a^2b^2c^2}{(b^2 - a^2)^2} = \frac{ab [(a + b)^2 - c^2]}{(a + b)^2}$

$(?)\ \ \frac{abc^2}{(b - a)^2} = (a + b)^2 - c^2$

$(?)\ \ abc^2 = (b^2 - a^2)^2 - c^2(b - a)^2$

$(?)\ \ 0 = (b^2 - a^2)^2 - c^2(b^2 - ba + a^2)$

$(?)\ \ b^2 - a^2 = c \sqrt{b^2 - ba + a^2}$

Using the extended sine theorem $2R = \frac{a}{\sin \alpha} = \frac{b}{\sin \beta} = \frac{c}{\sin \gamma}$, where R is the triangle circumradius,

$b^2 - ba + a^2 = 4R^2 (\sin^2 \beta - \sin \beta \sin \alpha + \sin^2 \alpha)$

Substituting

$\sin^2 \beta = \sin^2 (\alpha + 60^\circ) = \sin^2 \alpha \cos^2 60^\circ + \cos^2 \alpha \sin^2 60^\circ + 2 \sin \alpha \cos \alpha \sin 60^\circ \cos 60^\circ =$

$= \frac 1 4 \sin^2 \alpha + \frac 3 4 \cos^2 \alpha + \sin \alpha \cos \alpha \sin 120^\circ =$

$= \frac 1 4 \sin^2 \alpha + \frac 3 4 \cos^2 \alpha + \sin \alpha \cos \alpha \sin 60^\circ$

$\sin \beta \sin \alpha = \sin (\alpha + 60^\circ) \sin \alpha = (\sin \alpha \cos 60^\circ + \cos \alpha \sin 60^\circ) \sin \alpha =$

$= \frac 1 2 \sin^2 \alpha + \sin \alpha \cos \alpha \sin 60^\circ$

$b^2 - ba + a^2 = 4R^2 \left(\frac 1 4 \sin^2 \alpha + \frac 3 4 \cos^2 \alpha - \frac 1 2 \sin^2 \alpha  + \sin^2 \alpha\right) =$

$= 3R^2 (\sin^2 \alpha + \cos^2 \alpha) = 3 R^2$

Using $\sin \gamma = \sin (180^\circ - \gamma)$ and $180^\circ - \gamma = \alpha + \beta = 2 \alpha + 60^\circ$, the equation in question becomes

$(?)\ \ 4R^2 (\sin^2 \beta - \sin^2 \alpha) = 2 \sqrt 3\ R^2 \sin \gamma$

$(?)\ \ \sin^2 \beta - \sin^2 \alpha = \frac{\sqrt 3}{2} \sin (\beta + \alpha)$

$(?)\ \ \frac 1 4 \sin^2 \alpha + \frac 3 4 \cos^2 \alpha + \sin \alpha \cos \alpha \sin 60^\circ - \sin^2 \alpha = \frac{\sqrt 3}{2} \sin (2\alpha + 60^\circ)$

$(?)\ \ \frac 3 4 (\cos^2 \alpha - \sin^2 \alpha) + \frac{\sqrt 3}{4} \cdot 2\sin \alpha \cos \alpha = \frac{\sqrt 3}{2} (\sin 2\alpha \cos 60^\circ + \cos 2\alpha \sin 60^\circ)$

$(?)\ \ \frac 3 4 \cos 2\alpha + \frac{\sqrt 3}{4} \sin 2\alpha = \frac{\sqrt 3}{4} \sin 2\alpha + \frac 3 4 \cos 2\alpha$

Thus the equation in question has been transformed into an obvious identity, hence, it is true.
_____________________________________________________________________________

Now we apply the above lemma to the original problem. Let the internal bisector of the angle $\gamma = \angle BCA$ intersect the opposite side c = AB at a point F. The triangle $\triangle NBC$ is isosceles with NC = BC, because the angles $\angle CBN = \angle CNB$ are equal:

$\angle CBN = \angle CBA - \angle NBA = \beta - 30^\circ = (\alpha + 60^\circ) - 30^\circ = \alpha + 30^\circ$

$\angle CNB = \angle BAN + \angle NBA = \alpha + 30^\circ$

Therefore, the internal bisector of the angle $\gamma = \angle BCA \equiv \angle BCN$ is perpendicular to the opposite side BN of the isosceles triangle $\triangle NBC$, $l_C \equiv CF \perp BN$ and CF cuts BN in half. Consequently, the quadrilateral BFNC is a kite and the triangle $\triangle NBF$ is also isosceles with NF = BF. The internal angle bisector CF cuts the opposite side AB in the ratio of the adjacent sides. Therefore,

$\frac{AF}{BF} = \frac{CA}{BC} = \frac{b}{a}$

$NF = BF = AB \frac{BF}{BF + AF} = AB \frac{BC}{BC + CA} = \frac{ac}{a + b}$

Likewise, the internal bisector CM of the angle $\angle FCA$ cuts the opposite side AF of the triangle $\triangle AFC$ in the ratio of the adjacent sides. Therefore,

$\frac{AM}{FM} = \frac{CA}{CF} = \frac{b}{l_C} = \frac{b(b^2 - a^2)}{abc} = \frac{b^2 - a^2}{ac}$

Let the internal bisector of the angle $\angle ANF$ intersect the opposite side of the triangle $\triangle AFN$ at a point M'. Again, this bisector cuts the side AF in the ratio of the opposite sides.

$\frac{AM'}{FM'} = \frac{NA}{NF} = \frac{CA - CN}{NF} = \frac{CA - BC}{NF} = \frac{(a + b)(b - a)}{ac} = \frac{b^2 - a^2}{ac}$

Thus the bisectors CM, NM' of the angles $\angle ACF, \angle ANF$ cut the segment AF in the same ratio, which means that the points $M' \equiv M$ are identical. Once we established that the line NM bisects the angle $\angle ANF$, it is a simple angle chase to find the angle $\angle CMN$, which can be done in various ways. For example,

$\angle BMC = \angle CAM + \angle ACM = \alpha + \frac \gamma 4 = \alpha + \frac{180^\circ - (\alpha + \beta)}{4} =$

$= \alpha + \frac{180^\circ - (2\alpha + 60^\circ)}{4} = \frac \alpha 2 + 30^\circ$

$\angle BMN = \angle NAM + \angle ANM = \alpha +  \frac{\angle ANF}{2} = \alpha + \frac{180^\circ - \angle FNC}{2} =$

$= \alpha + \frac{180^\circ - \angle FBC}{2} = \alpha + \frac{180^\circ - \beta}{2} = \alpha + \frac{180^\circ - (\alpha + 60^\circ)}{2} = \frac \alpha 2 + 60^\circ$

As a result,

$\angle CMN = \angle BMN - \angle BMC = \left(\frac \alpha 2 + 60^\circ\right) - \left(\frac \alpha 2 + 30^\circ\right) = 30^\circ$

Great ! It is formidably. I am come dumb. You have a impressive mastery of the calculus. I tired following you.
For instance, here is a syntetic (no metric) solution of the following wellknown (another) problem.
**************************************************************************************** :spider:
[b]Problem.[/b] Let ABC be a isosceles triangle for which
$(AB=AC),\ m(\widehat{BAC})=20,\ E\in (AC),$ $\ m(\widehat{ABE})=30,\ F\in (AB),\ m(\widehat{ACF})=20.$
Ascertain $m(\widehat{AFE}).$
---------------------------------------------------------------------------------------------------------------------------
[b]Solution.[/b] Let $G\in (AB),\ m(\widehat{BCG})=20.$ Therefore, BC=CG==GF because $m(\widehat{GFC})=m(\widehat{GCF})=40.$
From $m(\widehat{CBE})=m(\widehat{BEC})=50$ results BC=CE and from CE=GC, $m(\widehat{GCE})=60$ results GE=GC=GF.
From $m(\widehat{CGE})=60$ results $m(\widehat{GEF})=m(\widehat{GFE})=70.$ The conclusion is $m(\widehat{AFE})=110$ and $m(\widehat{AEF})=50.$
****************************************************************************************
See if you can find a syntetic solution (this is the beauty !). Good luck !

[quote="woulgar"]Construct equilateral triangle on MN and use Morley 's Theorem.

    W.[/quote]
Welcome back, military man :trampoline:. Would you like to have your tail trisected again?

Yetti

I don't understand what you say. You with me talk ?
Woulgar is magnificient ! What beautiful solution !

No, I am talking to Woulgar = Pestich = Golland = ... = Leon Trotsky = ... = Nona P. McEculitz :rotfl: [url=http://www.cut-the-knot.org/htdocs/dcforum/DCForumID6/400.shtml#5](click)[/url] = Sveta Nedavala :rotfl: [url=http://www.cut-the-knot.org/htdocs/dcforum/DCForumID6/400.shtml#7](click)[/url] = ... (the list is virtually endless, this is just one thread).

Yetti

The sucession of the messages baffled me. Sorry !. Later I will offer my syntetic solution of the initial problem.

[quote="levi"]...Woulgar is magnificient ! What beautiful solution ![/quote]
Do you really think so ? :? :? :? Let's put the "solution" to a test. You certainly can drop a normal from the point N to the bisector CM of the angle $\angle ACF$ intersecting the bisector CF of the angle $\angle ACB$ at a point L and the triangle $\triangle LMN$ is obviously isosceles with MN = ML. However, there is no obvious reason, why this isosceles triangle should be equilateral. Conversely, if you construct an equilateral triangle $\triangle LMN$ on the segment MN, the vertex L falls on the bisector CF, but again, there is no obvious reason for this, apart from the proof that $\angle CMN = 30^\circ$, which I provided.

According to the "solution" by Woulgar = Pestich = Golland = ..., this equilateral triangle should be Morley's triangle of some other triangle, somehow related to the original triangle $\triangle ABC$. The vertices N, L lie on the lines CA, CF, intersecting at the point C, so the point C should be one vertex of this related triangle and the lines $CN \equiv CA, CL \equiv CF$ the trisectors of its angle at the vertex C. Assume that the angle $\beta > 90^\circ$ (the case $\beta < 90^\circ$ is handled similarly) and construct a line through the vertex C intersecting the line AB at a point $A_1$, such that

$\angle A_1CN = \angle NCL = \angle B_1CL$

It is easy to show that the point N lies on the bisector of the angle $\angle BA_1C$. From the triangle $\triangle A_1BC$, we have

$\angle BCA_1 = \frac 3 2 \gamma = \frac 3 2 [180^\circ - (\alpha + \beta)] = \frac 3 2 [180^\circ - (2\alpha + 60^\circ)] = 180^\circ - 3\alpha$[color=white].[/color] $= 360^\circ - 3\beta$

$\angle CA_1B = 180^\circ - \left(\beta + \frac 3 2 \gamma\right) = 180^\circ - (\alpha + 60^\circ + 180^\circ - 3\alpha) = 2\alpha - 60^\circ$[color=white].[/color] $= 2 \beta - 180^\circ$

Using the sine theorem for this triangle,

$\frac{CA_1}{a} = \frac{CA_1}{BC} = \frac{\sin \widehat{A_1BC}}{\sin \widehat{CA_1B}} = \frac{\sin \beta}{\sin(2\beta - 180^\circ)} = -\frac{\sin \beta}{\sin 2\beta} = -\frac{\sin \beta}{2 \sin \beta \cos \beta}$[color=white].[/color] $= -\frac{1}{2 \cos \beta}$

$\frac{BA_1}{a} = \frac{BA_1}{BC} = \frac{\sin \widehat{BCA_1}}{\sin \widehat{CA_1B}} = \frac{\sin (360^\circ - 3\beta)}{\sin(2\beta - 180^\circ)} = \frac{\sin 3\beta}{\sin 2\beta} = \frac{3\sin \beta - 4\sin^3 \beta}{2 \sin \beta \cos \beta} =$

$= \frac{3 - 4 \sin^2 \beta}{2 \cos \beta} = \frac{4 \cos^2 \beta - 1}{2 \cos \beta} = 2 \cos \beta - \frac{1}{2 \cos \beta}$

$\frac{AA_1}{CA_1} = \frac{BA_1 - BA}{CA_1} = -\frac{2 \cos \beta}{a} \left(2a \cos \beta - \frac{a}{2 \cos \beta} - c\right)$[color=white].[/color] $= 1 - 4 \cos^2 \beta + 2 \cos \beta \cdot \frac c a$

Earlier, we have shown that the triangle $\triangle BCN$ is isosceles with $CB = CN = a$. Now we have to show that

$(?)\ \ \frac{AA_1}{CA_1} = \frac{AN}{CN} = \frac{CA - CN}{CN} = \frac{CA - CB}{CB} = \frac{b - a}{a} = \frac b a - 1$

because then the bisector of the angle $\angle CA_1A$ cuts the opposite side CA of the triangle $\triangle A_1AC$ at the point N.

$(?)\ \ 1 - 4 \cos^2 \beta +  2 \cos \beta \cdot \frac c a= \frac b a - 1$

$(?)\ \ 2(1 - 2 \cos^2 \beta) + 2 \cos \beta\ \frac{\sin \gamma}{\sin \alpha} = \frac{\sin \beta}{\sin \alpha}$

$(?)\ \ 2 \sin \alpha (1 - 2 \cos^2 \beta) + 2 \cos \beta \sin (\alpha + \beta) = \sin \beta$

$(?)\ \ 2 \sin \alpha (1 - 2 \cos^2 \beta) + 2 \cos \beta (\sin \alpha \cos \beta + \cos \alpha \sin \beta) = \sin \beta$

$(?)\ \ 2 \sin \alpha (1 - \cos^2 \beta) + 2 \cos \beta \cos \alpha \sin \beta = \sin \beta$

$(?)\ \ 2 \sin \alpha \sin^2 \beta + 2 \cos \beta \cos \alpha \sin \beta = \sin \beta$

$(?)\ \ 2 \sin \beta (\sin \alpha \sin \beta + \cos \alpha \cos \beta) = \sin \beta$

$(?)\ \ 2 \sin \beta \cos (\beta - \alpha) = \sin \beta$

$(?)\ \ 2 \sin \beta \cos 60^\circ = \sin \beta$

which is an obvious identity. Hence, the vertices N, M lie on the lines $A_1N, A_1B$ intersecting at the point $A_1$, so the point $A_1$ should be another vertex of the related triangle and the lines $A_1N, A_1M \equiv A_1B$ the trisectors of its angle at the vertex $A_1$. Now we construct a line through the vertex $A_1$ intersecting the line CB at a point $B_1$, such that

$\angle CA_1N = \angle NA_1M = \angle B_1A_1M$

So far so good. The point $B_1$ has to be the remaining vertex of the related triangle $\triangle A_1B_1C$. But now the lines $B_1M, B_1L$ should trisect its angle at the vertex $B_1$. First, we notice that the angle $\angle A_1B_1C = 90^\circ$ is right. Indeed,

$\angle B_1CA_1 = 180^\circ - 3\alpha$

$\angle CA_1B_1 = \frac 3 2 \angle CA_1B = \frac 3 2 (2 \alpha - 60^\circ) = 3\alpha - 90^\circ$

$\angle A_1B_1C = 180^\circ - (\angle B_1CA_1 + \angle CA_1B_1) = 180^\circ - [(180^\circ - 3\alpha) + (3\alpha - 90^\circ)]$[color=white].[/color] $= 90^\circ$

Therefore, the trisectors of the angle $\angle A_1B_1C$ form the angles of $30^\circ$ with each other and with the triangle sides $B_1A_1, B_1C$. But this is not the case for the lines $B_1M, B_1L$. It is sufficient to show that the angle $\angle CB_1M \equiv \angle BB_1M \neq 60^\circ$. We are going to calculate the sides of the triangle $\triangle BB_1M$. Since CM obviously bisects the angle $\angle BCA_1$, it divides the opposite side $BA_1$ of the triangle $\triangle A_1BC$ in the ratio of the adjacent sides. Hence,

$BM = \frac{BC \cdot BA_1}{BC + CA_1} = a^2 \left(2 \cos \beta - \frac{1}{2 \cos \beta}\right) \cdot \frac{2 \cos \beta}{2a \cos \beta - a} =$

$= a\ \frac{4 \cos^2 \beta - 1}{2 \cos \beta - 1} = a (2 \cos \beta + 1)$

Since the angle $\angle A_1B_1C \equiv \angle A_1B_1B = 90^\circ$ is right

$BB_1 = BA_1 \cos (180^\circ - \beta) = -a \left(2 \cos \beta - \frac{1}{2 \cos \beta}\right) \cos \beta = -\frac a 2 (4 \cos^2 \beta - 1)$

$\frac{BB_1}{BM} = -\frac 1 2 \cdot \frac{4 \cos^2 \beta - 1}{2 \cos \beta + 1} = -\frac 1 2 (2 \cos \beta - 1) = \frac 1 2 - \cos \beta$

But if the angle $\angle BB_1M$ was equal to $60^\circ$, as required for Morley's equilateral triangle, then using the sine theorem for the triangle $\triangle BB_1M$ would yield

$\frac{BB_1}{BM} = \frac{\sin \widehat{BMB_1}}{\sin \widehat{BB_1M}} = \frac{\sin (\beta - 60^\circ)}{\sin 60^\circ} =$

$= \frac{\sin \beta \cos 60^\circ - \cos \beta \sin 60^\circ}{\sin 60^\circ} = \frac{1}{\sqrt 3} \sin \beta - \cos \beta$

Thus we see that the lines $B_1M, B_1L$ can be the trisectors of the angle $\angle A_1B_1C$ only if $\frac{1}{\sqrt 3} \sin \beta = \frac 1 2$ or $\sin \beta = \frac{\sqrt 3}{2}$ or $\beta = 120^\circ$ (because we assumed $\beta > 90^\circ$). But this implies $\alpha = \beta - 60^\circ = 60^\circ$, [color=white].[/color]$\gamma = 180^\circ - (\alpha + \beta) = 0$ and the original triangle $\triangle ABC$ degenerates into a line segment. For $\beta < 90^\circ$, we could similarly deduce $\beta = 60^\circ,\ \alpha = \beta - 60^\circ = 0,\ \gamma = 120^\circ$ and again, the original triangle $\triangle ABC$ would degenerate into a line segment. In conclusion, the angles

$\angle A_1B_1M \neq \angle MB_1L \neq \angle CB_1L$

cannot be equal, the equilateral triangle $\triangle LMN$ is [b]not[/b] Morley's triangle of the related triangle $\triangle A_1B_1C$ and the whole setup crashes down. Rest assured that all trigonometric formulas in this entire thread check with the Sketchpad. At the first glance, the lines $B_1M, B_1L$ look like the trisectors of the angle $\angle A_1B_1C$. But if something quacks as a duck, it does not necessarily have to be a duck, for example, it can be a drake (Woulgar = Pestich = Golland = ... surely remembers this bit). It might be useful to read the article [url=http://www.cut-the-knot.org/triangle/Morley/Morley.shtml]Doodling and Miracles[/url] by A. Bogomolny. I am tired of shooting down these also-proofs by Woulgar = Pestich = Golland = ... the magnificent and you would be too, if you were me. If you did not get fooled by him, I would not be doing it.

Yetti

I note $G\in (AB),m(\widehat{ACG})=60,u=m(\widehat{BAC}).$ From B=60+u,C=120-2u results $m(\widehat{AGC})=120-u,m(\widehat{BGC})=m(\widehat{GBC})=60+u.$Thus,BC=GC. But $m(\widehat{NBC})=m(\widehat{BNC})=30+u.$ Therefore BC=NC. Results NC=GC=GN. From $m(\widehat{GCM})=30+\frac 12 u=m(\widehat{GMC})$results GC=GM=GN, $m(\widehat{MGN})=60-u, m(\widehat{GMN})=60+\frac 12 u.$ In conclusion, $m(\widehat{CMN})=m(\widehat{GMN})-m(\widehat{GMC})=60+\frac 12 u-(30+\frac 12 u)=30.$


prove that $ [a,b)$ is not homeomorphic to $ (a,b)$.

I want to show by contradiction, so I construct a homeomorphism between the two intervals: $ f: [a,b) \rightarrow(a,b)$. For simplicity, $ [a,b) \equal{} M$ and $ (a,b) \equal{} N$. Since $ f$ is a homeomorphism and $ N$ connected interval, then $ M\equal{}f^{\minus{}1}(N)$ is a connected interval containing endpoint $ a$. Then M and N are homeomorphic. But then $ M\minus{}\{a\}$ is connected and $ N\minus{}\{f(a)\}$ is disconnected which contradicts the fact that if $ M$ is compact and homeomorphic to $ N$, then $ N$ is connected.

I think this is right, but I wanted a second opinion before I had to turn it in.


I was wondering if $ S_1$ and $ S_2$ are connected sets in $ \mathbb{R}^n$. does it necessarily follow that $ S_1\cap S_2$  is connected?

Clearly false if $ n\ge2.$

For instance, let $ S_1$ be the $ x$ axis in $ \mathbb{R}^2$ and let $ S_2$ be some circle centered at the origin.


Show that there are infinitely many primes.

Quite a well-known and old problem.  :smile:

Suppose for contradiction that there exist a finite number of primes $p_1, p_2, \dots, p_n$.

Let $Q=p_1p_2 \dots p_n+1$

All integers greater than 1 have a (unique) decomposition into primes, but $Q>1$ is not divisible by any prime- [b][color=red]contradiction[/color][/b].

[quote="Ilthigore"]Lol, I can just imagine somebody saying "Dirichlet's Theorem states that there exist infinitely many primes of the form ak+b for constant a,b. Taking this set, we have a proof that there exist infinitely many primes."[/quote]Basicly that's the same discussion as we had [url=http://www.problem-solving.be/pen/viewtopic.php?t=110]here[/url]. It depends on wether you are trying to establish results as a research result, or by elementary means.

I always liked to look for elementary proofs of special cases of Dirichlet's theorem. Small moduli like $4$ (consider divisors of $n^2+1$ and $4n-1$) and $3$ (consider $x^2+x+1$ and $3x-1$) are well known, and so is $1 \mod n$ for arbitrary $n$ (consider cyclotomic polynomials; for $n$ prime some nice proofs are given by some Baltic Way problems).
But if one wants to get more, it get's complicated (there is a polynomial that works out for primes $\equiv a  \mod n$ iff $a^2 \equiv n$; especially $a=-1$ being the next easiest case after the already mentioned $a=1$ works).

Here I give the verbatim proof of this resultgiven by Furstenberg in 1955. Its a topological proof.
His proof was:" In this note we would like to offer an elementary 'topological' proof of the infinitude of the prime numbers. We introduce a topology into the space of integers S, by using the AP as a basis. It is not difficult to verify that this actually yields a topological space. In fact, under this topology, S may be shown to be normal and hence metrizable. Each AP is closed as well as open, since its complement is the union of other APs (having same CD). As a result , the union of any finite number of APs is closed. Consider now the set A=union of A1, where A1 consists of all multiples of p, and p runs through the set of primes. The only number not belonging to A are -1 and 1, and since the set {-1,1} is clearly not an open set, A cannot be closed. Hence A is not a finite union of closed sets which proves that there are infinity of primes".

Golomb developed this further and wrote an interesting short paper in 1959.

- Actually, every proof of E-24 can be regarded as a proof of the infinitude of primes. Well, as long as that E-24 proof doesn't take for granted what we wished to settle in the first place. ;)
- There is also this beautiful proof given by Pólya whose main ingredient is the pairwise coprimality of the Fermat numbers.
- A proof  which you are not likely to encounter in most textbooks is the one that resorts to the Chinese Remainder Theorem. In fact, this proof is nothing more than a disguised presentation of the well-known euclidean argument yet, you can't blame a man for dreaming...  :lol:
- I came up with the following wacky proof several months ago: fix $ n \in \mathbb{N}$. For every $ k \in \mathbb{Z}^{ \plus{} }$, Bertrand's Postulate allows us to ascertain the existence of a prime number $ p_{k}$ in the interval $ I_{k} \equal{} (2^{k} \cdot n,2^{k \plus{} 1} \cdot n]$. Since the intervals $ I_{k}$ are pairwise disjoint and there are infinitely many of them, the veracity of E-12 follows at once. How do you like this proof-proposal of mine?

Lemme add another proof. Suppose that $p_i \quad (1\leq i\leq s)$ are the only primes. But then $$\infty=\sum_{j\geq 1}{\dfrac{1}{j}}=\prod_{i=1}^{s}{\sum_{k=0}^{\infty}{\dfrac{1}{p_i ^k}}}=\prod_{i=1}^{s}{\dfrac{p_i}{p_i-1}}$$ The last product however is finite $\therefore$ Contradiction .


rotation
geometry
geometric transformation

I think this problem has an easy solution, but I can't find it.

Given are [i]n[/i] points in space: A0, A1, ... , An with An = A0.  Pick a point B0 and then pick points B1, B2, ... , Bn such that for each k = 1, 2, ... , n the line segments Ak-1Ak and  Bk-1Bk have common midpoints.  Show that B0 = Bn provided that the number of points [i]n[/i] is even

Let the midpoint of AkAk+1 be Mk. For notational purposes, denote H[P] to be the half turn of the plane around point P. I will prove that the composit tranformation

T := H[M1] H[M2]...H[Mn]

is the identity transformation, which will prove that the image of B0 (which is Bn) is itself.

Ok. On each half turn, the plane gets rotated 180 degrees, and since there are an even number of half turns since n is even, the whole transformation boils down to a translation. But, notice that H[M0](A0)=A1, H[M1](A1)=A2, and so on... So, T(A0)=A0, so T is a translation with a fixed point, so it is the identity as desired. [tex]\blacksquare[/tex]

really clever solution, I really need to review isometries.

Just one thing, the problem asks about n points in space, and you keep talking about the plane, but I see that the proofs would essentially be the same.

oops, i misread it as n points in the plane... anyway, the proof would still work in space


[quote="sks119"]Two cyclists are riding along the same straight road. Alice, riding a light racing bike, is maintaining a constant speed of 41 km/h, while Bob is on a mountain bike riding at a constant speed of 20 km/h. They are both traveling in the same direction. At exactly 10:00 AM, Alice passes a road sign marked: ``Abbotsford: 28 km''. At 10:10 AM, Bob passes a sign marked ``Abbotsford: 15 km''. Assuming the road signs are accurate, when does Alice pass Bob? How far are they from Abbotsford at that instant?[/quote]
[hide]
Alice passes Bob at 10:27 am.

In the ten minutes between when Alice passes the 28 km sign, and when Bob passes the 15 km sign Alice has traveled $41\text{ km/h }\times \frac{10\text{ minutes}}{60\text{ minutes/h}}= 6.83\text{ km}$.

Alice's position at this time is: $28\text{ km}-6.83\text{ km}= 21.17 \text{ km}$.

The distance between Alice and Bob at this time is $21.17\text{ km}-15\text{ km}= 6.17\text{ km}$.

Beginning at 10:10 am, Alice's position by the time she passes Bob is $v_{A}t$. Bob's position is $6.17\text{ km}+v_{B}t$. At the moment they pass one another their positions are equal.

$v_{A}t = 6.17\text{ km}+v_{B}t$

$\Rightarrow (v_{A}-v_{B})t = 6.17\text{ km}$

$\Rightarrow t =\frac{6.17\text{ km}}{(v_{A}-v_{B})}$

$\Rightarrow t =\frac{6.17\text{ km}}{(41\text{ km/h}-20\text{ km/h})}= 0.2938\text{ hrs}$

$0.2938\text{ hrs}\times 60\text{ min/h}= 17.6\text{ minutes}$

$10: 10\text{ am}+17\text{ minutes}= 10: 27\text{ am}$.

The distance to Abbotsford at that instant is: $28\text{ km}-41\text{ km/h}\cdot \frac{27.6\text{ minutes}}{60\text{ min/h}}= 9.12\text{ km}$
[/hide]


ceiling function
geometry
circumcircle
greatest common divisor
national olympiad

Here are the questions from this years SAMO.  There is a senior round which is written by grade 10 to 12 (in our school system thats about 16 - 18 year olds) and then the junior paper is written by grades 8 and 9.  The top 100 seniors and juniors in the initial multiple choice rounds get to write these papers.
The english is on the second page of the pdf.  Actually I think there is a typo in questions 3, if I remember right this is correct :

1. Let a = 1111...111 and b = 111...1111 where a has forty ones and b has twelve ones.  Determine the greatest common divisor of a and b.

2. Fifty points are chosen inside a convex polygon having eighty sides such that no three of the fifty points lie on the same straight line.  The polygon is cut into triangles such that the vertices of the triangels are just the fifty points and the eighty vertices of the polygon.  How many triangles are there?

3. Find all real numbers x such that $x \lceil x \lceil x \lceil x \rceil \rceil \rceil = 88$

4  Let $A_1$ and $B_1$ be two points on the base AB of isosceles trianlge ABC (with $\hat{C} > 60^\circ$) such that $A_1 \hat{C} B_1 = B \hat{A}C$.  A circle externally tangent to the circumcircle of triangel $A_1 B_1 C$ is tangent, also to rays $CA$ and $CB$ at points $A_2$ and $B_2$ respectively.  Prove that $A_2 B_2 = 2AB$.

5.  For $n \geq 2$, find the number of integers $x  (0 \leq x < n )$ such that $x^2$ leaves a remainder of 1 when divided by n.

6. $a_1, a_2$ and $2_3 $ are distinct positive integers, such that
 $a_1$ is a divisor of $a_2 + a_3 + a_2 a_3$
$a_2$ is a divisor of $a_3 + a_1 + a_3 a_1$
$a_3$ is a divisor of $a_1 + a_2 + a_1 a_2$

Prove that $a_1, a_2$ and $a_3$ cannot all be prime.

I think it would be better to have these in separate topics in the apropriate sections.

1. $\gcd(a,b)=\gcd(b,a-kb)$ for all positive integers $a$, $b$ and $k$. First $a=b(10^{28}+10^{16}+10^4)+1111$ implies $\gcd(a,b)=\gcd(b,1111)$ and second $b=1111(10^8+10^4+1)$ implies $\gcd(b,1111)=\gcd(1111,0)=1111$. So, $\gcd(a,b)=1111$.
I hope that I finally posted a right answer :( .

A small remainder: you can post up to 3 attachements in one post ;)


trigonometry
limit
calculus
calculus computations

$lim_{x \to 0} \frac{1-cosx \sqrt{cos2x} \sqrt[3]{cos3x}}{x^2}=?$

really nice solution,i really enjoyed it. :D


In the figure shown, a perpendicular segment is drawn from B in rectangle ABCD to meet diagonal AC at point X.  Side AB is 6 cm and diagonal AC is 10 cm.  How many centimeters away is point X from the midpoint M of the diagonal AC?  Express your answer as a decimal to the nearest tenth.

[asy]draw((0,0)--(25,0)--(25,15)--(0,15)--cycle,linewidth(1));
draw((0,0)--(25,15),linewidth(1));
draw((7.2,4.3)--(0,15),linewidth(1));
dot((12.5,7.5));
label("$M$",(12.5,7.5),N);
label("$A$",(0,0),S);
label("$B$",(0,15),N);
label("$C$",(25,15),N);
label("$D$",(25,0),S);
label("$X$",(7.2,4.3),SE);[/asy]

All of the main points of the solve are drawn and included in the attached figure, all you have to do is use the additional information to solve the problem, if you still cant get it then look at my hint.

[hide]If you still cant get it then try the Pythagorean Theorem on the two  right triangles with a y value and solve for y.  [/hide]


Hello,

I'm wondering what software is most suitable for making presentations when mathematicians give talks with the aid of a screen.

One example would be powerpoint, but the math equations would not be as pretty.

I have heard that mathematica has certain feature to create Presentations, but I don't know exactly how.

Also when there is no screen, let's say you have to make a big poster to give your presentations, how do people make these?

There are several $ \text{\LaTeX}$ presentation packages (which have the distinct advantage of making the math look really nice); look up prosper, beamer, and beamerposter (for your last question), but there are others.

Some people use [url=http://latex-beamer.sourceforge.net/]beamer[/url], a documentclass for LaTeX. The output is a PDF file. Good compatibility with other software.

Others (e.g. me) use something like foiltex (also a LaTeX documentclass) and print the result on transparencies. Even better compatibility with other software. :) I also leave spaces here and there, to be filled in during the presentation (with a wet-erase marker). The result is (hopefully) more engaging than a fully pre-packaged presentation.

Or you can simply write on transparencies, if your handwriting is sufficiently legible.

thanks!!

But if you make a really big poster, usually how do you print these out?

I guess you have to take it to the Kinkos or something.

as mentioned, beamer is really good (does anyone actually use prosper?)

[quote]I guess you have to take it to the Kinkos or something.[/quote]
I would use Kinko's; you can order prints online by sending them a PDF and then either have them sent to you or pick up at the store.

On our campus, an office affiliated with the College of Natural Sciences and Mathematics has a large-format printer suitable for making posters. If you're on a college campus, check to see what the experimental scientists do; they're often the ones with the greatest interest in this style of presentation.

Or if you're at a lab, I know Sandia has at least one gigantic (4 foot wide at least) printer, and I'd imagine other commercial major labs would too.


[quote="tkhtn"]Let $a,b,c\ge 0$ .
Prove that: $\sum \frac{a(b+c)}{a^2+bc}\ge 2$.[/quote]
This is hungkhtn's inequality.
$\sum \frac{a(b+c)}{a^2+bc}\ge 2\Leftrightarrow(a-b)^2(a-c)^2(b-c)^2+2a^2b^2c^2+\sum_{sym}a^3b^2c\geq0.$

i think it is quite difficult to end up with such simple expression after long calculation. is there any step by step way? i mean any reasonable way?


Simone has a collection of coins consisting of 1 quarter, 1 nickel and 1 penny. Using at least two coins from the collection, how many different sums of money are possible?

Each different way of picking coins results in a different value.  There are $ \binom{3}{2}$ ways to choose $ 2$ coins and $ \binom{3}{3}$ ways to choose $ 3$ coins.  This is $ 3\plus{}1\equal{}\boxed{4}$.

[hide="Or"]
You could list them:

QP
PN
QN
QPN

($\boxed{4}$ ways)
[/hide]

But what if your values were 2, 4, 5, 1? You have two ways to make six. I was wondering if the solution to the problem was foolproof for any type of problem like this. It may work now because we only have three values and we have to use at least two, but what about 4 values? Or five?


For every $ n \ge 2$, $ a_n \equal{} \sqrt [3]{n^3 \plus{} n^2 \minus{} n \minus{} 1}/n$. What is the least value of positive integer $ k$ satisfying $ a_2a_3\cdots a_k > 3$ ?

$\textbf{(A)}\ 100  \qquad\textbf{(B)}\ 102  \qquad\textbf{(C)}\ 104  \qquad\textbf{(D)}\ 106  \qquad\textbf{(E)}\ \text{None}$

[hide]Since $ a_n\equal{}\sqrt[3]{{(n\minus{}1)(n\plus{}1)^2\over n}}$, the cube of the stated product is

$ P^3(k)\equal{}\frac{(k\minus{}1)!\left({(k\plus{}1)!\over 2}\right)^2}{k!^3}\stackrel{\mathrm{simple\ rearrangement}}{\equal{}\ \dots\ \equal{}}{(k\plus{}1)^2\over 4k}$

We need $ {(k\plus{}1)^2\over 4k}>27\iff {k^2\plus{}2k\plus{}1\over k}>108\iff k\plus{}{1\over k}>106$

so obviously $ k_{\min}\equal{}106$ and the answer is $ D$.[/hide]


Let f:[a,b]->R be a differentiable function s.t. [f(x)]2 + [f'(x)]2 >0 for all x \in [a,b]. Show that the function has only finitely many distinct roots.

Sort of easy, but I like it.

Suppose that f have an infinite distincts roots x(n) \in [a,b]
f(x(n))=0, [a,b] is compact by Bolzano-Weierstrass there is a
subsequence x(g(n)) wich converge to d \in [a,b] when n-->+oo

f(d)=0

(f(x(g(n)) - f(d))/(x(g(n))-d) = 0 taking the limit when n-->+oo

f'(d)=0

So we have found d such that (f(d))^2+(f'(d))^2=0 contradict the hypothesis

I guess I did the same thing. I got that the set of the roots is discrete, and that's not possible if there are infinitely many roots. Thx for the solution.

If your solution is different I am interested, there is always something
to learn in different proof.

It's not really different, I just expressed everything you said in different words: we assume the set of the roots isn't discrete, so there is a root which has infinitely many roots in any vicinity, and i proved in the same way you did that f' in that specific point is also 0, so I got a contradiction, meaning that the set of the roots is indeed discrete.


geometry
3D geometry
sphere
analytic geometry
trigonometry
calculus
integration

Assume the following constellation:
The origin of a three-dimensional cartesian coordinate system (that means: x-, y- and z-axis are orthogonal on each other) is set in an arbitraty point in the interior of a sphere. Now, each axis has two points of intersection with the sphere >>totally: 6 points on the periphery on the sphere, so that each octant of the coordinate system receives a spheric triangle on the sphere. If we paint those 8 triangles alternately like a chessboard, how can I proove, that the half surface of the sphere is coulored white?

im not sure if this would work but we can reduce it to the case

$ (x\minus{}a)^2\plus{}(y\minus{}b)^2\plus{}(z\minus{}c)^2\equal{}1$ for some $ a^2\plus{}b^2\plus{}c^2<1$

then do a substitution to $ x\equal{}\cos \phi \sin \theta \plus{}a, y\equal{} \sin \phi \sin \theta \plus{} b, z\equal{} \cos \theta \plus{}c$
with $ \phi \in [\minus{}\frac {\pi}{2},\frac{\pi}{2}]$ and $ \theta \in [0,2\pi]$

then maybe look for some limits and set up an ugly integration?

Let the z-axis cut the sphere at $ E, F.$ Planes through $ E, F$ perpendicular to the z-axis slice of 2 identical sperical caps with exactly complementary colors. Slice the remaining part of the sphere by planes perpendicular to the z-axis into disks of height $ \text dz$ for integration. The coordinate origin $ O$ projects along the z-axis to the interior of each disk. It is now sufficient to show that if 2 perpendicular lines $ x, y$ intersecting at a point $ O$ inside a circle with center $ P,$ divide it into 4 arcs $ AB, BC, CD, DA,$ then $ AB \plus{} CD \equal{} BC \plus{} DA$ for the arc lengths. This is equivalent to $ \angle APB \plus{} \angle CPD \equal{} \angle BPC \plus{} \angle DPA,$ which is true because
$ \angle APB \plus{} \angle CPD \equal{} 2 (\angle ACB \plus{} \angle CBD) \equal{} 2 (\pi \minus{} \angle BOC) \equal{} \pi.$

I don't know, if you understood me right. I can try again, to say, what I want:
Through a point in the interior of a sphere we put three planes, standing perpendicular on each other. These planes cut up the spherical surface into 8 curvilinear(spherical) triangles. The triangles are colored alternately black and white (Chessboard). One proves that exactly then half of the spherical surface is white.

Perhaps, I haven't understand you right. Maybe, then you can answer a little bit more extended? [I don't know, if there is the possibility to create a sketch of the subject matter.]
Thanks a lot.

EDIT: I think, I could get your message. But how to transform your idea into 3 dimensions? The case you described works good for two dimensions.

Again, the z-axis cuts the sphere at $ U, L$ (for upper and lower) Plane $ \mathcal P_K$ perpendicular to the z-axis through the sphere center $ K$ cuts the segment $ UL$ at its midpoint. Therefore, spherical caps cut off by planes $ \mathcal P_U, \mathcal P_L$ perpendicular to the z-axis through $ U, L$ are congruent. The yx- zx-planes cut  the spherical surface of the 2 caps in identical arc pattern. Since the black-white colors are exchanged on the xy-plane cutting the segment $ UL$ in its interior, the sperical caps have complemetary colors.

$ S_B(U) \equal{} S_W(L)$ (upper cap black surface = lower cap white surface)

$ S_W(U) \equal{} S_B(L)$ (upper cap white surface = lower cap black surface)

Summing the black color surface of the 2 caps: $ S_B(U) \plus{} S_B(L) \equal{} S_W(L) \plus{} S_W(U)$

Sums of black and white color surfaces of the 2 caps are equal.

For the remaining spherical frustrum, integrate. $ R$ is the sphere radius. Using spherical coordinates $ (r, \varphi, \theta)$ with origin at the sphere center. (Cylindrical coordinates $ (\varrho, \varphi, z)$ with the same origin would serve as well). $ \theta \in ( \minus{} \frac {_{\pi}}{^2}, \plus{} \frac {_{\pi}}{^2})$ is angle from the plane $ \mathcal P_K$ perpendicular to the z-axis through the sphere center. Integration limits of $ \theta$ are $ \pm \theta_m$ to the upper and lower caps. $ \varphi$ is rotation angle around the line parallel to the z-axis through the sphere center. Intergration limits of $ \varphi$ for the black color are $ (A(\theta), B(\theta))$ and $ (C(\theta), D(\theta))$ dependent on $ \theta.$ These are intersections of the yz, zx-planes with a circle on the sphere at the angle $ \theta.$

Integrating the black color surface of the spherical frustrum:

$ S_{B} \equal{} R^2 \int_{ \minus{} \theta_m}^{ \plus{} \theta_m} \cos \theta\ \text d\theta \left( \int_{A(\theta)}^{B(\theta)} \text d\varphi \plus{} \int_{C(\theta)}^{D(\theta)} \text d\varphi \right) \equal{}$

$ \equal{} R^2 \int_{ \minus{} \theta_m}^{ \plus{} \theta_m} \cos \theta\ \text d\theta \left( \pi \right) \equal{} 2 \pi R^2 \sin \theta_m$

Integrating the white color surface of the spherical frustrum:

$ S_{W} \equal{} R^2 \int_{ \minus{} \theta_m}^{ \plus{} \theta_m} \cos \theta\ \text d\theta \left( \int_{B(\theta)}^{C(\theta)} \text d\varphi \plus{} \int_{D(\theta)}^{A(\theta)} \text d\varphi \right) \equal{}$

$ \equal{} R^2 \int_{ \minus{} \theta_m}^{ \plus{} \theta_m} \cos \theta\ \text d\theta \left( \pi \right) \equal{} 2 \pi R^2 \sin \theta_m$

Integrals of the black and white color surfaces of the spherical frustrum are also equal.

Oh, that sounds great. I hope, my Englisch can cope with your explanations. Thanks a lot.

$ x = \cos \theta \sin \phi + a, y = \sin \theta \sin \phi + b, z = \cos \phi + c$
with $ \phi \in [0,\pi]$ and $ \theta \in [0,2\pi ]$

then notice that the octants will only part the interval of $ \phi$ at one point $ \phi_o$ and the interval $ [0, 2\pi]$ at  four points call them $ \theta_1, \theta_2, \theta_3, \theta_4$ from smallest to biggest.

then the surface area of the colors in the octants are

$ O_1 = \int^{\pi}_{\phi_o}\left(\int^{\theta_1}_{0}d\theta + \int^{2\pi}_{\theta_4}d\theta\right) \sin(\phi)d\phi = \left(1 + \cos(\phi_o)\right)(\theta_1 - \theta_4 + 2\pi)$

$ O_2 = \int^{\phi_o}_{0}\left(\int^{\theta_1}_{0}d\theta + \int^{2\pi}_{\theta_4}d\theta\right) \sin(\phi)d\phi = \left(1 - \cos(\phi_o)\right)(\theta_1 - \theta_4 + 2\pi)$

$ O_3 = \int^{\pi}_{\phi_o} \int^{\theta_2}_{\theta_1}\sin (\phi)d\theta d\phi = \left(1 + \cos(\phi_o)\right)(\theta_2 - \theta_1)$

$ O_4 = \int^{\phi_o}_{ 0} \int^{\theta_2}_{\theta_1}\sin (\phi)d\theta d\phi = \left(1 - \cos(\phi_o)\right)(\theta_2 - \theta_1)$

$ O_5 = \int^{\pi}_{\phi_o}\int^{\theta_3}_{\theta_2}\sin (\phi)d\theta d\phi = \left(1 + \cos(\phi_o)\right)(\theta_3 - \theta_2)$

$ O_6 = \int^{\phi_o}_{ 0}\int^{\theta_3}_{\theta_2}\sin (\phi)d\theta d\phi = \left(1 - \cos(\phi_o)\right)(\theta_3 - \theta_2)$

$ O_7 = \int^{\pi}_{\phi_o}\int^{\theta_4}_{\theta_3}\sin (\phi)d\theta d\phi = \left(1 + \cos(\phi_o)\right)(\theta_4 - \theta_3)$

$ O_8 = \int^{\phi_o}_{ 0}\int^{\theta_4}_{\theta_3}\sin (\phi)d\theta d\phi = \left(1 - \cos(\phi_o)\right)(\theta_4 - \theta_3)$

inspecting
$ \Delta = O_1 - O_2 - O_3 + O_4 + O_5 - O_6 - O_7 + O_8 =$

$ (O_1 - O_2) - (O_3 - O_4) + (O_5 - O_6) - (O_7 - O_8) =$

${ 2\cos(\phi_o) \{(\theta_1 - \theta_4 + 2\pi) - (\theta_1 - \theta_2) + (\theta_3 - \theta_2) - (\theta_3 - \theta_4})\} =$

$ 4\cos(\phi_o) \left(\theta_1 + \theta_3 - \theta_2 - \theta_4 + \pi \right)$

all thats left to show is that
$ \theta_1 + \theta_3 + \pi = \theta_2 + \theta_4$
for the corresponding $ \theta_i$ when two perpendicular lines intersect inside of a circle

but this is always true, since really all 4 $ \theta_i$ are parametrized by $ z = p + iq$ in the unit disc $ |z| < 1$

$ e^{i\theta_1} = p + i\sqrt {1 - p^2}$, $ e^{i\theta_3} = p - i\sqrt {1 - p^2}$, $ e^{i\theta_2} = \sqrt {1 - q^2} + iq$, $ e^{i\theta_4} = - \sqrt {1 - q^2} + iq$

then $ e^{i(\theta_1 + \theta_3 + \pi - \theta_2 - \theta_4)} = - \frac {p^2 + (1 - p^2)}{ - q^2 - (1 - q^2)} = 1$

so we know $ \theta_1 + \theta_3 + \pi - \theta_2 - \theta_4 = 2k\pi$, for some integer k.
luckily the only possible k is 0, and we are then done!

$ k = 0 \to \Delta = 0 \to$ the difference of areas is zero $ \to \mu(white) = \mu(black)$

Okay, I see, that both possibilities are quite similar, thanks a lot for you commitment. I hope, that I can consider this problem as solved!


induction
function
algebra unsolved
algebra

Let $\, |U|, \, \sigma(U) \,$ and $\, \pi(U) \,$ denote the number of elements, the sum, and the product, respectively, of a finite set $\, U \,$ of positive integers.  (If $\, U \,$ is the empty set, $\, |U| = 0, \, \sigma(U) = 0, \, \pi(U) = 1$.) Let $\, S \,$ be a finite set of positive integers. As usual, let $\, \binom{n}{k} \,$ denote $\, n! \over k! \, (n-k)!$. Prove that \[ \sum_{U \subseteq S} (-1)^{|U|} \binom{m - \sigma(U)}{|S|} = \pi(S)  \] for all integers $\, m \geq \sigma(S)$.

[hide="hint"]What does the left side remind you of?[/hide]

[hide="Solution"] Let $ \{x_1,x_2,...,x_n\}$ denote the elements of $ S$ with $ n \equal{} |S|$.

Consider a set $ R$ of $ m$ elements, with $ n$ mutually disjoint subsets $ y_1,y_2,...,y_n$ where $ |y_i| \equal{} x_i$.

We wish to count the number of $ n$-element subsets of $ R$ that contain at least one element from all of the $ y_i$'s.

Since the $ y_i$ are mutually disjoint, and there are $ n$ of them, each such subset of $ R$ consists of exactly one element from each $ y_i$, which means the number of such subsets is $ x_1 x_2...x_n \equal{} \pi{(S)}$.

As well, we will count the number of such subsets of $ R$ using the Principle of Inclusion-Exclusion. If $ \{x_{i_1},...,x_{i_k}\} \equal{} U \subseteq S$, then the number of $ n$-element subsets of $ R$ that do not contain any element of $ y_{i_1},...,y_{i_k}$ is equal to $ \binom{m \minus{} \sigma{(U)}}{n}$. By PIE, the number of $ n$-element subsets of $ R$ that contain at least one element from every $ y_i$ is
\[ \sum_{U \subseteq S} ( \minus{} 1)^{|U|} \binom{m \minus{} \sigma(U)}{n}
\]
and we are done.[/hide]

[hide="Solution"]

{Lemma 1:} We first prove, by induction:

  $ \displaystyle\sum_{U \subseteq S} (\minus{}1)^{|U|} \dbinom{m \minus{} \sigma(U) \minus{} 1}{|S| \minus{} 1} \equal{} 0 \forall m$ where variables and functions are described in the problem statement.

For our base case, we set $ |S| \equal{} 1$. Then, suppose $ S \equal{} \{ a \}$. Then the left hand side is equal to:

  $ (\minus{}1)^0 \dbinom{m \minus{} 1}{1 \minus{} 1} \plus{} (\minus{}1)^1 \dbinom{m \minus{} a \minus{} 1}{1 \minus{} 1}$

  $ \equal{} 1 \minus{} 1 \equal{} 0$

This completes the base case.

Now, suppose that for some $ |S| \equal{} k$, the statement is true; we wish to show it is true for $ |S| \equal{} k\plus{}1$ as well.

Notice that the subsets of a $ k\plus{}1$-element set can be generated from one of its $ k$-element subsets by taking all subsets of the $ k$-element subset, in addition to the union of each subset of the $ k$-element subset with the other element of the $ k\plus{}1$ element subset.

Suppose a set $ S^\prime$ has $ k\plus{}1$ elements, including the $ k$ elements of the set $ S$. The other element of $ S^\prime$ is $ y$.

Then we can say:

  $ \displaystyle\sum_{U \subseteq S^\prime} (\minus{}1)^{|U|} \dbinom{m \minus{} \sigma(U) \minus{} 1}{|S^\prime| \minus{} 1} \equal{} \left( \displaystyle\sum_{U \subseteq S} (\minus{}1)^{|U|} \dbinom{m \minus{} \sigma(U) \minus{} 1}{|S^\prime| \minus{} 1} \right) \plus{} \left( \displaystyle\sum_{U \subseteq S} (\minus{}1)^{|U| \plus{} 1} \dbinom{m \minus{} \sigma(U) \minus{} 1 \minus{} y}{|S^\prime| \minus{} 1} \right)$

  $ \equal{} \displaystyle\sum_{U \subseteq S} (\minus{}1)^{|U|} \left( \dbinom{m \minus{} \sigma(U) \minus{} 1}{|S^\prime| \minus{} 1} \minus{} \dbinom{m \minus{} \sigma(U) \minus{} 1 \minus{} y}{|S^\prime| \minus{} 1} \right)$

Now we double-count as follows: Suppose there are $ m \minus{} \sigma(U) \minus{} 1$ people divided into two groups, one containing $ y$ people and the other containing the rest. Then the quantity

  $ \left( \dbinom{m \minus{} \sigma(U) \minus{} 1}{|S^\prime| \minus{} 1} \minus{} \dbinom{m \minus{} \sigma(U) \minus{} 1 \minus{} y}{|S^\prime| \minus{} 1} \right)$

is the number of ways we can choose $ |S^\prime| \minus{} 1$ people such that at least one of the people is from the group containing $ y$ people.

We can alternatively do this by choosing one person from the group containing $ y$ people and then choosing the $ |S^\prime| \minus{} 2$ other people from the remaining $ m \minus{} \sigma(U) \minus{} 2$ people. Thus,

  $ \displaystyle\sum_{U \subseteq S} (\minus{}1)^{|U|} \left( \dbinom{m \minus{} \sigma(U) \minus{} 1}{|S^\prime| \minus{} 1} \minus{} \dbinom{m \minus{} \sigma(U) \minus{} 1 \minus{} y}{|S^\prime| \minus{} 1} \right)$

  $ \equal{} \displaystyle\sum_{U \subseteq S} (\minus{}1)^{|U|} y \dbinom{(m\minus{}1) \minus{} \sigma(U) \minus{} 1}{|S^\prime|\minus{}2}$

But we can simply plug $ m\minus{}1$ into $ m$ in our inductive hypothesis (since the statement is true for all applicable $ m$) and find that

  $ y \displaystyle\sum_{U \subseteq S} (\minus{}1)^{|U|} \dbinom{(m\minus{}1) \minus{} \sigma(U) \minus{} 1}{|S^\prime|\minus{}2} \equal{} 0$

so we have completed the inductive step.

  $ \Box$

 {Lemma 2:} Now we use the previous result:

  $ \displaystyle\sum_{U \subseteq S} (\minus{}1)^{|U|} \dbinom{m \minus{} \sigma(U) \minus{} 1}{|S| \minus{} 1} \equal{} 0$

By Pascal's Identity, this can be written as:

  $ \displaystyle\sum_{U \subseteq S} (\minus{}1)^{|U|} \left( \dbinom{m \minus{} \sigma(U)}{|S|} \minus{} \dbinom{m \minus{} \sigma(U) \minus{} 1}{|S|} \right) \equal{} 0$

  $ \displaystyle\sum_{U \subseteq S} x (\minus{}1)^{|U|} \left( \dbinom{m \minus{} \sigma(U)}{|S|} \minus{} \dbinom{m \minus{} \sigma(U) \minus{} 1}{|S|} \right) \equal{} 0$

  $ \displaystyle\sum_{U \subseteq S} (\minus{}1)^{|U|} x \dbinom{m \minus{} \sigma(U) \minus{} 1}{|S|} \equal{} \displaystyle\sum_{U \subseteq S} x (\minus{}1)^{|U|} \dbinom{m \minus{} \sigma(U)}{|S|}$

We use the same double-counting scenario as above to reach:

  $ \displaystyle\sum_{U \subseteq S} (\minus{}1)^{|U|} \left( \dbinom{m \minus{} \sigma(U)}{|S| \plus{} 1} \minus{} \dbinom{m \minus{} \sigma(U) \minus{} x}{|S| \plus{} 1} \right) \equal{} \displaystyle\sum_{U \subseteq S} x (\minus{}1)^{|U|} \dbinom{m \minus{} \sigma(U) \minus{}1}{|S|}$

  $ \Box$

Now we will prove the main statement of the problem, by induction.

For our base case, we set $ |S| \equal{} 1$. Then, suppose $ S \equal{} \{ a \}$. The left hand side equals:

  $ (\minus{}1)^0 \dbinom{m}{1} \plus{} (\minus{}1)^1 \dbinom{m\minus{}a}{1} \equal{} m \minus{} (m\minus{}a) \equal{} a$

Then the base case holds.

Suppose that for some $ |S| \equal{} j$, the statement holds true; we wish to prove it holds true for $ |S| \equal{} j\plus{}1$. This implies:

  $ \displaystyle\sum_{U \subseteq S} (\minus{}1)^{|U|} \dbinom{m \minus{} \sigma(U)}{|S|} \equal{} \pi(S) \forall m$

We can generate subsets of a set $ S^\prime$ with $ j\plus{}1$ elements exactly as we did before. Suppose that $ S^\prime$ has the same elements as $ S$ does, except $ S^\prime$ also includes the element $ x$. Then,

  $ \displaystyle\sum_{U \subseteq S^\prime} (\minus{}1)^{|U|} \dbinom{m \minus{} \sigma(U)}{|S^\prime|} \equal{} \displaystyle\sum_{U \subseteq S} (\minus{}1)^{|U|} \dbinom{m \minus{} \sigma(U)}{|S^\prime|} \plus{} \displaystyle\sum_{U \subseteq S} (\minus{}1)^{|U| \plus{} 1} \dbinom{m \minus{} \sigma(U) \minus{} x}{|S^\prime|}$

  $ \equal{} \displaystyle\sum_{U \subseteq S} (\minus{}1)^{|U|} \left( \dbinom{m \minus{} \sigma(U)}{|S| \plus{} 1} \minus{} \dbinom{m \minus{} \sigma(U) \minus{} x}{|S| \plus{} 1} \right)$

But by our second lemma, this is equal to:

  $ \equal{} \displaystyle\sum_{U \subseteq S} (\minus{}1)^{|U|} x \dbinom{m \minus{} \sigma(U) \minus{} 1}{|S|}$

  $ \equal{} x \displaystyle\sum_{U \subseteq S} (\minus{}1)^{|U|} \dbinom{(m\minus{}1) \minus{} \sigma(U)}{|S|}$

By our inductive hypothesis, this is equal to:

  $ \equal{} x \pi(S)$

Since $ x$ is the element of $ S^\prime$ that is not an element of $ S$, the product of $ x$ with the product of the elements of $ S$ is equal to the product of the elements of $ S^\prime$. Thus,

  $ \displaystyle\sum_{U \subseteq S^\prime} (\minus{}1)^{|U|} \dbinom{m \minus{} \sigma(U)}{|S^\prime|} \equal{} \pi(S^\prime)$

given that

  $ \displaystyle\sum_{U \subseteq S} (\minus{}1)^{|U|} \dbinom{m \minus{} \sigma(U)}{|S|} \equal{} \pi(S)$

This concludes our inductive step.

By induction, we have proved:

  $ \displaystyle\sum_{U \subseteq S} (\minus{}1)^{|U|} \dbinom{m \minus{} \sigma(U)}{|S|} \equal{} \pi(S)$

as desired.

[/hide]

[quote=WolfusA]Can somebody explain what's going on above? Especially what is $[x^0]$?[/quote]
For a polynomial $P(x)$ and a natural number $n$, the number $[x^n]P(x)$ is the coefficient of $x^n$ in $P(x)$.


Is it possible to resurrect inadvertently deleted polls?  I edited the original message associated with the "[b]Fields of Study Motivated by Math Contests[/b]" poll on the AMC forum and the poll disappeared. [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=100328[/url]

I wasn't aware of this poll deleting feature.  It might be wise to warn the OP of a poll that the poll will be destroyed if the original message is edited (as the edit button is clicked.)

It might even make sense to prohibit the OPs from deleting the polls.  Most of the posts in a poll thread implicitly or explicitly reference the poll.  Therefore, the content of such threads will be difficult to interpret without the poll.

Thanks,

Greg

The poll should not be destroyed. It's a bug, that I will take care of once I am back from the IMO.


MATHCOUNTS
calculus
vector
geometry
inequalities
pigeonhole principle
probability

Yup, choose 1 that you'd want the most, just as a reference.

I only raised the general topics, so if you could say which subject you want to learn in particular, that'll be great. (For example, you could say "modular arithmetic" instead of just "arithmetic")

There was a similar poll about this a while ago...

Yes, I know.
1.) This is limited only to MATHCOUNTS level and maybe a bit beyond, but it won't include calculus and fractals and such.
2.) The last poll was just for fun while this is for reference as to what the people want to learn about, a la Rep123Max and Idiot_without_a_village

Anyways, do you think vector's appropriate for middle-schoolers? I'm learning it this year, so I thought it might be appropriate, but I don't know about others. Any thoughts?

When I saw the title, I thought this was a poll to settle the dispute between Fierytycoon and me over who was hotter...but as I realize, it's one of dem mathy things.  So I voted.

By the way, isn't modular arithmetic classified as number theory?

As far as I know, vectors is high school stuff...i had it in middle school too (but that was after "skipping" out of 3 levels of math)...but generally its high school pre-calcish

Proofs dont exactly appear in MC...

Well ... it's difficult to choose.

Note that arithmetic  number theory.

I like geometry, inequalities (and I'm not bad in solving both types of problems - for a Belgian at least). I also like number theory very much but I'm not so good at number theory (unfortunately, i'll have to practice a bit more). I'm really bad at combinatorics (pigeonhole is an exception) (I got 0 at this year's IMO question 1) But some combinatorics problems are just magnificent.

hmm...my worst area is probly probability (although geometry is catching up fast)

[quote="MithsApprentice"]Proofs dont exactly appear in MC...[/quote]
[quote="Arne"]Note that arithmetic  number theory.[/quote]
I know I know, I'm talking MC leveled problems, not just the stuff on MC.

[quote="Arne"]I like geometry, inequalities (and I'm not bad in solving both types of problems - for a Belgian at least). I also like number theory very much but I'm not so good at number theory (unfortunately, i'll have to practice a bit more). I'm really bad at combinatorics (pigeonhole is an exception) (I got 0 at this year's IMO question 1) But some combinatorics problems are just magnificent.[/quote]
Mmm, I agree.

[quote="Syntax Error"]cover it all foo[/quote]
Translation? :)

It'd be sad if no one was interested after I post lots of stuff, like what happened with my recent failure :cry:

Hmm...interesting, proofs are the number 1 topic...OK, "The people hath spoken!" :)

Ummm..proofs about what? Proofs can be about any of the other topics.

Mmm, I tried putting most of the gemetric ones at the other poll, but I haven't thought about others...hmm, I assumed people were talking about geometric proofs...

What type of proof were you (plural) thinking of when you voted for proofs?

Proofs...generally mean proofs from all branches of math...

Yeah...that's a pretty big topic though Mith...

I need to get a sense of what category proof we're talking about since I'm doubting I know everything...


vector
combinatorics unsolved
combinatorics

a) 100 different real numbers are written on a blackboard. Show that you can find 8 numbers whose mean does not equal the mean of any 9 numbers on the blackboard.
b) Show that if 100 numbers are such that for any 8 you can find 9 with the same mean, then all the numbers must be equal.

I proved a).
I want the solution b).

P.S) In T.O.Towns, they were integers, but I want about real numbers.

I think it is the nice problem
what is the solution for (a)
thnk you  :blush:

But, if you know how to prove it for integers, then it also holds for rational with an appropriate scaling.
Thus, it holds for real numbers, using that $\mathbb{R}$ is a vector space over $\mathbb{Q}$, and using the result above for each coordinate.

Pierre.

For part B, maybe you can work your way up by first comparring the smallest set of 8 numbers to the smallest set of 9 numbers, and concluding that the smallest 9 numbers must all be equal or something.  Maybe.

[quote="cocoowner"]solution (a)
sum smallest 8 numbers. That's all.[/quote]
It is a nice solution


function
real analysis
real analysis unsolved

Let $ f_n = n \chi_{[\frac{1}{n},\frac{2}{n}]}$ on $ L_p(\mathbb{R},B.m)$,$ B$ is the borel subset of $ \mathbb{R}$.

Show that the sequence of function is not uniform convergence outside a set of zero measure(that means,there does not exist a set of measure zero,on the complement of which $ (f_n)$ is uniformly convergent.)

I have no idea how to start the proof.Can anyone give me any hints?

You should just write "Let $ f_n = n \chi_{[\frac {1}{n},\frac {2}{n}]}$"; the $ L^p$ stuff has nothing to do with it. Hints: 1. First show $ f_n$ converges pointwise to $ 0$ everywhere on $ R$. 2.  $ f_n\rightarrow 0$ uniformly on $ R - E$ would imply $ f_n < 1$ on $ R - E$ for $ n$ large enough, contradiction.


how do you carryout the foll conv.
3 heptyne to 1-pentanamine?

Why do you post each message two times???
 :spam:

Please try not to double post.  It really accomplishes nothing.  Also, please read Rule 1 in the "Welcome" message [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=179608]here[/url].

[quote="VARUNARASU"]3 heptyne to 1-pentanamine?[/quote]

[hide="Answer"]1) 3-heptyne + KNHCH2CH2CH2NH2 -----> 1-heptyne

2) 1-heptyne + KMnO4/base/heat -----> $ CH_3CH_2CH_2CH_2CH_2CO_2H$

3) $ CH_3CH_2CH_2CH_2CH_2CO_2H \plus{} HN_3/H_2O/H^\plus{}$ -----> $ CH_3CH_2CH_2CH_2CH_2NH_2$[/hide]


I'm writing a document that will include a very long array.  I've been using the tabular environment, and I just reached the point at which the array is longer than a single page.  LaTeX does not automatically insert a page-break.  Is there any way to do this within a single array, or must I manually break in into several pieces?

The [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=longtable]longtable package[/url] could be what you need.


[quote="Allnames"]Let $ a,b,c > 0$$ ab \plus{} bc \plus{} ca \equal{} 1$
Prove that $ \sum \sqrt {a^3 \plus{} a}\ge 2\sqrt {a \plus{} b \plus{} c}$[/quote]
$ \sum_{cyc} \sqrt {a^3 \plus{} a}\ge 2\sqrt {a \plus{} b \plus{} c}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc} \sqrt {a^3 \plus{} a^2b\plus{}a^2c\plus{}abc}\ge 2\sqrt {(a \plus{} b \plus{} c)(ab\plus{}ac\plus{}bc)}\Leftrightarrow$
$ \Leftrightarrow\sum_{cyc}(a^3 \plus{} a^2b\plus{}a^2c\plus{}abc)\plus{}2\sum_{cyc}\sqrt{(a^2(a\plus{}b\plus{}c)\plus{}abc)(b^2(a\plus{}b\plus{}c)\plus{}abc)}\ge$
$ \geq 4(a \plus{} b \plus{} c)(ab\plus{}ac\plus{}bc).$
But $ \sqrt{(a^2(a\plus{}b\plus{}c)\plus{}abc)(b^2(a\plus{}b\plus{}c)\plus{}abc)}\ge ab(a\plus{}b\plus{}c)\plus{}abc.$
Id est, it remains to prove that
$ \sum_{cyc}(a^3 \plus{} a^2b\plus{}a^2c\plus{}abc\plus{}2(ab(a\plus{}b\plus{}c)\plus{}abc))\geq4(a \plus{} b \plus{} c)(ab\plus{}ac\plus{}bc),$ which is Schur.

[quote="Allnames"]Let $ a,b,c > 0$$ ab \plus{} bc \plus{} ca \equal{} 1$
Prove that $ \sum \sqrt {a^3 \plus{} a}\ge 2\sqrt {a \plus{} b \plus{} c}$[/quote]

Follow Chebyshev we have :

$ LHS\equal{}\sum  \sqrt {a}\sqrt {a^2 \plus{} 1} \ge\ \frac{1}{3}(\sum\sqrt {a})(\sum \sqrt {a^2 \plus{} 1})$

And we have $ (\sum\sqrt {a})^2(\sum \frac{1}{a}) \ge\ 27$

$ \rightarrow \sum \sqrt {a} \ge\ 3\sqrt{3abc}$

Follow Minkopski we have :

$ \sum \sqrt {a^2 \plus{} 1} \ge\ \sqrt{(a\plus{}b\plus{}c)^2\plus{}9}$

Need prove $ 3abc[(a\plus{}b\plus{}c)^2\plus{}9] \ge\ 4(a\plus{}b\plus{}c)$

I think it 's easy  :P

[quote="onlylove_math"]

Need prove $ 3abc[(a \plus{} b \plus{} c)^2 \plus{} 9] \ge\ 4(a \plus{} b \plus{} c)$

I think it 's easy  :P[/quote]
Try $ a\rightarrow0^\plus{}.$ :wink:

[quote="Allnames"]Let $ a,b,c > 0$ and $ bc \plus{} ca \plus{} ab \equal{} 1.$ Prove that

$ \sqrt {a^3 \plus{} a} \plus{} \sqrt {b^3 \plus{} b} \plus{} \sqrt {c^3 \plus{} c}\ge 2\sqrt {a \plus{} b \plus{} c}.$[/quote]$ 4(c \plus{} a)(a \plus{} b)(a \plus{} b \plus{} c)^3(bc \plus{} ca \plus{} ab)$

$ \minus{} a\left[ 7bc(b \plus{} c)\plus{}4ca(c\plus{}a)\plus{} 4ab(a\plus{}b) \plus{}6abc \right]^2$

$ \equal{} \frac {b(a \plus{} b)}{4}(a \minus{} b)^2\left[b(16a^2 \plus{} 29ac \plus{} 6c^2) \plus{} 16a^3 \plus{} 6a^2c \plus{} 17ac^2 \plus{} 18c^3\right]$

$ \plus{} \frac {c(a \plus{} c)}{4}(a \minus{} c)^2\left[c(16a^2 \plus{} 29ab \plus{} 6b^2) \plus{} 16a^3 \plus{} 6a^2b \plus{} 17ab^2 \plus{} 18b^3\right]$

$ \plus{} \frac {bc}{4}(b \plus{} c \minus{} 2a)^2\left[a^3 \plus{} 3a^2(b \plus{} c) \plus{} 3a (b^2 \plus{} bc \plus{} c^2) \plus{} 10bc(b \plus{} c)\right]\geq0\Longrightarrow$

$ \sum{\sqrt {{\frac {a^3 \plus{} a}{a \plus{} b \plus{} c}}}} \equal{} \sum{\sqrt {{\frac {a(c \plus{} a)(a \plus{} b)}{(a \plus{} b \plus{} c)(bc \plus{} ca \plus{} ab)}}}}$

$ \geq \sum{\frac {a[7bc(b \plus{} c)\plus{}4ca(c\plus{}a)\plus{} 4ab(a\plus{}b) \plus{}6abc ]}{2(a \plus{} b \plus{} c)^2(bc \plus{} ca \plus{} ab)}} \equal{} 2.$

See also : http://www.mathlinks.ro/viewtopic.php?t=206627

[quote=sqing][quote=Allnames]Let $ a,b,c > 0$ , $ ab \plus{} bc \plus{} ca \equal{} 1$
Prove that$\sqrt {a^3 + a} + \sqrt {b^3 + b} + \sqrt {c^3 + c}\geq2\sqrt {a + b + c}$[/quote]
[url=http://www.artofproblemsolving.com/community/c6h206627p1136956]Iran TST 2008[/url]
The following inequality is also true.
Let $a,b,c \geq 0 $ and $ ab \plus{} bc \plus{} ca \equal{} 1$ . prove that$$\sqrt{a^3\plus{}a}\plus{}\sqrt{b^3\plus{}b}\plus{}\sqrt{c^3\plus{}c} \geq \sqrt{(a\plus{}b\plus{}c)^3}.$$[/quote]

The following inequality is also true.
Let $a,b,c \geq 0 $ and $ ab \plus{} bc \plus{} ca \equal{} 1$ . prove that$$\sqrt{a^3\plus{}a}\plus{}\sqrt{b^3\plus{}b}\plus{}\sqrt{c^3\plus{}c} \leq \sqrt{(a\plus{}b\plus{}c)^3}\frac{2\sqrt{3}}{3}.$$

[quote=sqing]
The following inequality is also true.
Let $a,b,c \geq 0 $ and $ ab \plus{} bc \plus{} ca \equal{} 1$ . prove that$$\sqrt{a^3\plus{}a}\plus{}\sqrt{b^3\plus{}b}\plus{}\sqrt{c^3\plus{}c} \geq \sqrt{(a\plus{}b\plus{}c)^3}.$$[/quote]
It was here:
http://www.artofproblemsolving.com/community/c6h145158

[quote=xzlbq]
The following inequality is also true.
Let $a,b,c \geq 0 $ and $ ab \plus{} bc \plus{} ca \equal{} 1$ . prove that$$\sqrt{a^3\plus{}a}\plus{}\sqrt{b^3\plus{}b}\plus{}\sqrt{c^3\plus{}c} \leq \sqrt{(a\plus{}b\plus{}c)^3}\frac{2\sqrt{3}}{3}.$$[/quote]

By C-S,we have
\[[\sum{\sqrt{(a^3+a)}}]^2=[\sum{\sqrt{(a(a+b)(a+c))}}]^2\le{\sum{a}\sum{(a+b)(a+c)}}\]
Hence,we need to prove that
\[\sum{a}\sum{(a+b)(a+c)}\le{\frac{4}{3}(a+b+c)^3}\]
\[<=>\frac{4}{3}(a+b+c)^3-\sum{a}\sum{(a+b)(a+c)}=\frac{1}{3}(a^3+b^3+c^3-3abc)\ge{0}\]

[quote=sqing]Let $a,b,c \geq 0 $ and $ ab + bc + ca = 1$ . prove that$$\sqrt{(a+b+c)^3}\leq \sqrt{a^3+a}+\sqrt{b^3+b}+\sqrt{c^3+c} \leq \frac{2\sqrt{3}}{3}\sqrt{(a+b+c)^3}.$$
[b]Very nice.[/b][/quote]

Let $a,b,c \geq 0 $ and $ ab + bc + ca = 1$ . prove that
$$\sqrt{(a+b+c)^2}\leq \sqrt{a^2+a}+\sqrt{b^2+b}+\sqrt{c^2+c} \leq \sqrt{1+\sqrt{3}}\sqrt{(a+b+c)^2}.$$

More stronger is:

$$\sqrt{(a+b+c)^2}\leq \sqrt{a^2+a}+\sqrt{b^2+b}+\sqrt{c^2+c} \leq \sqrt{(a+b+c)(a+b+c+3)}.$$

[quote=xzlbq]

Let $a,b,c \geq 0 $ and $ ab + bc + ca = 1$ . prove that
$$\sqrt{(a+b+c)^2}\leq \sqrt{a^2+a}+\sqrt{b^2+b}+\sqrt{c^2+c} \leq \sqrt{1+\sqrt{3}}\sqrt{(a+b+c)^2}.$$

More stronger is:

$$\sqrt{(a+b+c)^2}\leq \sqrt{a^2+a}+\sqrt{b^2+b}+\sqrt{c^2+c} \leq \sqrt{(a+b+c)(a+b+c+3)}.$$[/quote]

I think these 2 are lot more easier as compared to the one posted by sqing.

[quote=sqing]Let $a,b,c \geq 0 $ and $ ab + bc + ca = 1$ . prove that$$a+b+c\leq \sqrt{a^2+a}+\sqrt{b^2+b}+\sqrt{c^2+c}
 \leq \sqrt{1+\sqrt{3}}\left(a+b+c\right).$$[/quote]
The following inequality is also true
[img]http://s16.sinaimg.cn/middle/0018zOAxgy72B87ngGH8f&690[/img]

Let $a,b,c \geq 0 $ and $ ab \plus{} bc \plus{} ca \equal{} 1$ . prove that$$\sqrt{a^3\plus{}a}\plus{}\sqrt{b^3\plus{}b}\plus{}\sqrt{c^3\plus{}c} \leq \sqrt{(a\plus{}b\plus{}c)^3}\frac{2\sqrt{3}}{3}.$$
The following inequality is also true
[img]http://s6.sinaimg.cn/middle/0018zOAxgy72B8wtq8Re5&690[/img]


A high school is fairly sports-minded. 1/2 of the students play soccer. 1/5 of the students play basketball. There are 99 students playing soccer and/or basketball, including 24 that play both. What is the total enrollment of this high school?
 :huh:  :huh:  :huh:

Please include how to solve, not just answer.

[hide]1/2*1/5 is the number of students that play both. 1/10 of the students play both, 24 play both, the total enrollment is $240$[/hide]

kstan ur wrong that would mean that 120 kids play soccer but it says 99 kids play soccer or basketball

[quote="pontios"][quote="polarpan523"]A high school is fairly sports-minded. 1/2 of the students play soccer. 1/5 of the students play basketball. There are 99 students playing soccer and/or basketball, including 24 that play both. What is the total enrollment of this high school?[/quote]

[hide]Are you sure the problem is correct? I found:

$63 \frac{6}{7}$ students play [u]only[/u] soccer

$11 \frac{1}{7}$ students play [u]only[/u] basketball

$76 \frac{5}{7}$ students play nothing[/hide][/quote]

HAHA. $\frac{5}{7}$ of a person. Thats nasty.

[quote="pontios"][quote="polarpan523"]A high school is fairly sports-minded. 1/2 of the students play soccer. 1/5 of the students play basketball. There are 99 students playing soccer and/or basketball, including 24 that play both. What is the total enrollment of this high school?[/quote]

[hide]Are you sure the problem is correct? I found:

$63 \frac{6}{7}$ students play [u]only[/u] soccer

$11 \frac{1}{7}$ students play [u]only[/u] basketball

$76 \frac{5}{7}$ students play nothing[/hide][/quote]

I foound something like thaT BUT I thought I made a mistake maybe the teacher gave a trick question or something maybe I was right...


geometry
area of a triangle
geometric inequality
IMO Shortlist
point set

Consider in a plane $ P$ the points $ O,A_1,A_2,A_3,A_4$ such that \[ \sigma(OA_iA_j) \geq 1 \quad \forall i, j \equal{} 1, 2, 3, 4, i \neq j.\]  where $ \sigma(OA_iA_j)$ is the area of triangle $ OA_iA_j.$  Prove that there exists at least  one pair $ i_0, j_0 \in \{1, 2, 3, 4\}$ such that \[ \sigma(OA_iA_j) \geq \sqrt{2}.\]

Funny result!

Let $a, b, c, d$ denote the lengths of $OA_1, OA_2, OA_3, OA_4$. WLOG, let $A_1, A_2, A_3, A_4$ be labelled in counter-clockwise order as viewed from $O$. Let $\alpha = \angle A_1OA_2, \beta = \angle A_2OA_3, \gamma = \angle A_3OA_4$. In this way, we have that $\angle A_4OA_1 = 360 - \alpha - \beta - \gamma.$

Rather than using that ugly $\sigma$ to denote area, we will use the simple $[\triangle ABC]$ to denote the [i] unsigned [/i] area of $\triangle ABC.$ The following relations are derived from a well-known triangle area formula:

$$[\triangle OA_1A_2] = \left| \frac12 a b \sin \alpha \right| ,$$

$$[\triangle OA_2A_3] = \left | \frac12 bc \sin \beta \right|,$$

$$[\triangle OA_3A_4] = \left|\frac12 cd \sin \gamma\right|,$$

$$[\triangle OA_4A_1] = \left|\frac12 da \sin (\alpha + \beta + \gamma)\right|,$$

$$[\triangle OA_1A_3] = \left|\frac12 ac \sin(\alpha + \beta)\right|,$$ and

$$[\triangle OA_2A_4] = \left|\frac12 bd \sin(\beta + \gamma)\right|.$$

Hence, we obtain that:

$$[\triangle OA_1A_2] \cdot [\triangle OA_3A_4] = \frac14 abcd \left|\sin\alpha \cdot \sin \gamma\right|,$$

$$[\triangle OA_2A_3] \cdot [\triangle OA_4A_1] = \frac14 abcd \left|\sin \beta \cdot \sin (\alpha + \beta + \gamma)\right|,$$ and

$$[\triangle OA_1A_3] \cdot [\triangle OA_2A_4] = \frac14 abcd \left| \sin (\alpha + \beta) \cdot \sin (\beta + \gamma) \right|.$$

[b] Lemma. [/b] $\sin \alpha \cdot \sin \gamma + \sin \beta \cdot \sin (\alpha + \beta + \gamma) = \sin (\alpha + \beta) \cdot \sin (\beta + \gamma).$

[b] Proof. [/b] This is a straightforward application of the sum-to-product formulas, and the proof is ommitted.

$\blacksquare$

As a result of the lemma, we have that:

$$\pm [\triangle OA_1A_2] \cdot [\triangle OA_3A_4] \pm [\triangle OA_2A_3] \cdot [\triangle OA_4A_1] \pm [\triangle OA_1A_3] \cdot [\triangle OA_2A_4] = 0,$$

for some choice of the signs. WLOG, there is one $+$ sign and two $-$ signs.

Since all areas are at least one, each term in the above is of absolute value at least $1$. This implies that the term with the $+$ sign before it must have absolute value at least $2$, and therefore one of the multiplicands has absolute value $\ge \sqrt2$, as desired.

$\square$


function
algebra
domain
LaTeX
polynomial
quadratics
quadratic formula

Check if function [img]http://www.mathlinks.ro/Forum/latexrender/pictures/33ba42eb5a3e566748db9c42af04e6ef.gif[/img] (with domain?) [img]http://www.mathlinks.ro/Forum/latexrender/pictures/d15b68f9b155384ee04b4f40f3cf5345.gif[/img]. Is a different value function (I don't know how it is named in english, but... by symbols : [img]http://www.mathlinks.ro/Forum/latexrender/pictures/12396aa0c4bfe325367e5135e406401e.gif[/img].
I don't know how I can transform f(x).

-------------------------------------
If I made too much errors tell me I will try to write it better.
There is a example of checking if $f(x)=x^{2}$ is a different value function.
$f(x)=x^{2}$
$x_{1}=1\text{ and }x_{2}=-1$
we have
$x_{1}\not=x_{2}\wedge g(x_{1})=g(-1)=1=g(1)=g(x_{2})$
I think it should be done in te same way, but I don't know how to transform $\frac{x^{2}}{x-2}$ to simpler function.

[size=75][color=red]b-flat:  I tried to fix your $\LaTeX$, but I am not sure of what you mean on some parts.[/color][/size]

I think what you mean is this:

Check if $f(x) = \frac{x^{2}}{x-2}, x \in (4, \infty)$ is a one-to-one function, that is:

$f(x_{1})=f(x_{2}) \implies x_{1}= x_{2}$ on this domain.

[hide="Hint"] Polynomial division. :) [/hide]

[quote="t0rajir0u"]I think what you mean is this:

Check if $f(x) = \frac{x^{2}}{x-2}, x \in (4, \infty)$ is a one-to-one function, that is:

$f(x_{1})+f(x_{2}) \implies x_{1}= x_{2}$ on this domain.

[hide="Hint"] Polynomial division. :) [/hide][/quote]

i think you mean $f(x_{1})=f(x_{2})$? anyway, you want to isolate x, use the quadratic formula

Yes there should be f(x1)=f(x2), I will fix it.
Hmm... I must check if this function is a fuction that for every y value have different x value. I don't know how it is named in english.
1.
f(x)=x is right because for every y value there is different x value.
2.
f(x)=|x| is wrong because for x=1 and x=-1 there the same y value, y=1.
Now I must check function [img]http://www.mathlinks.ro/Forum/latexrender/pictures/33ba42eb5a3e566748db9c42af04e6ef.gif[/img] with domain [img]http://www.mathlinks.ro/Forum/latexrender/pictures/d15b68f9b155384ee04b4f40f3cf5345.gif[/img] but I don't know how.

[quote="Marcin121"] but I don't know how.[/quote]

See if you can find $x_{1}, x_{2}$ such that $x_{1}\neq x_{2}$ but $f(x_{1}) = f(x_{2})$, or see if you can show that they don't exist.


Let $ f(n)$ be the integer closest to $ n^{\frac{1}{4}}$. Find \[ \sum_{k\equal{}1}^{1995} \frac{1}{f(k)}.\]


The quotient below is a bit small and somewhat tedious to read.  However when I use \displaystyle in TeXnicCenter it's larger and more readable.  \displaystyle here though does not make it larger.  Is there something I can use here to make it more easily readable?

\[ b_n=\{\begin{array}{ccc}\displaystyle \frac{(-1)^{(3+n)/2}2 (2^{n-1}-1)B_{1+n}}{(1+n)!}-\frac{2}{\pi^{n+1}} & \mbox{for} & \text{n odd} \\
0 & \mbox{for} & \text{n even} \end{array}\]

Thanks,
Shaw

\[ b_n \equal{} \begin{cases}\cfrac{( \minus{} 1)^{\frac {3 \plus{} n}{2}}2 (2^{n \minus{} 1} \minus{} 1)B_{1 \plus{} n}}{(1 \plus{} n)!} \minus{} \cfrac{2}{\pi^{n \plus{} 1}} & \text{for } n \text{ odd} \\
0 & \text{for } n \text{ even} \end{cases}
\]
Some notes:
1. In your code there was a dot missing after the final \right;
2. you should be using the cases environment (notice how 0 is left aligned and there's no gap after 'for');
3. you can use \text instead of \mbox. The n should be in maths mode;
4. this forum does fractions rather differently - the workaround to get a larger fraction is to use \cfrac to replace \frac;
5. in a document you can use \text{for $ \$$n$ \$$ odd} instead of \text{for } n \text{ odd} but not in this forum.


calculus
integration
trigonometry
real analysis
real analysis unsolved

Here is an interesting identity of Ramanudjan.

The following identity holds:

$\sqrt{\alpha}\int_{0}^{\infty}\frac{e^{-x^{2}}}{ch (\alpha x)}dx=\sqrt{\beta}\int_{0}^{\infty}\frac{e^{-x^{2}}}{ch(\beta x)}dx$, where $\alpha \beta =\pi$.

yes, ch stands for the hyperbolic cosine.
$ch(x)=\frac{e^{x}+e^{-x}}{2}=cosh(x)$


do there exist $n$ such that $2n+1, 3n+1$ and $4n+1$ all are primes? i came this far that $n$ must be a multiple of $6$.

why isn't anyone replying? is it really that hard? :huh:  maybe, but i dunno.

[quote="Rust"]First $n=330$, second $n=690$,...[/quote]

how did you find them? just by putting in values?

Yes it is certain (up to computer checks) but as I know open to be proven.