geometry
trigonometry
circumcircle
Ross Mathematics Program
ratio
analytic geometry
perpendicular bisector

If median passing thru vertex A, symedian passing thru vertex B,
   and altitude thru vertex C  are concurrent, then this altitude is
   parallel to line connecting Brocard points of this triangle.

   Col. Pestich.

Problem is not stated correctly, it should be:
If the median passing through the vertex $A$, symedian through the vertex $B$, and the altitude passing through the vertex $C$ of a triangle $\triangle ABC$ are concurrent, then the altitude passing through the vertex $B$ (not $C$!) is parallel to the line connecting the two Brocard points of the triangle.

[b]Lemma:[/b] A median through one vertex of a triangle, symmedian through another vertex and one of the Brocard rays from the remaining vertex of a triangle are concurrent.

Let $L, M, N$ be the midpoints of the sides $a = BC, b = CA, c = AB$. Let $U \in BC, V \in CA, W \in AB$ be feet of symmedians through the vertices $A, B, C$. Let $Z$ be the 2nd Brocard point of $\triangle ABC$, so that $\angle ZCB = \angle ZAC = \angle ZBA = \theta$ is the Brocard angle. Let $CZ$ cut $AB$ at $F$. Since $\angle ZBF = \angle ZBA = \angle ZCB = \theta$ $\Longrightarrow$ $AFB$ is tangent at $B$ of a circle passing through $Z, C$ $\Longrightarrow$ $FB^2 = FZ \cdot FC$ $\Longrightarrow$ triangles $\triangle BCF \sim \triangle ZBF$ are similar $\Longrightarrow$ $\frac{ZF}{BZ} = \frac{BF}{BC}$ and $\angle BZF = \angle B$ of $\triangle ABC$. Since $\angle ZAC = \angle ZCB = \theta$ $\Longrightarrow$ $BC$ is tangent at $C$ of a circle passing through $Z, A$ $\Longrightarrow$ $\angle FZA = \angle C$  of $\triangle ABC$. Using the sine theorem for the triangles $\triangle ABZ, \triangle AFZ$,

$\frac{BZ}{AB} \cdot \frac{AF}{ZF} = \frac{\sin \widehat {FZA}}{\sin \widehat {BZA}} = \frac{\sin C}{\sin A} = \frac{c}{a}$

Substituting for $ZF = BF \cdot \frac{BZ}{BC}$, we get

$\frac c a = \frac{BZ}{AB} \cdot \frac{AF}{ZF} = \frac{AF}{BF} \cdot \frac{BC}{AB} = \frac{AF}{BF} \cdot \frac a c$ $\Longrightarrow$ $\frac{AF}{BF} = \frac{c^2}{a^2}$

By Ceva's theorem,

$\frac{\overline{BL}}{\overline{CL}} \cdot \frac{\overline{CV}}{\overline{AV}} \cdot \frac{\overline{AF}}{\overline{BF}} = -\frac{a^2}{c^2} \cdot \frac{c^2}{a^2} = -1$

Consequently, median $AL$, symmedian $BV$ and Brocard ray $CZF$ are concurrent at a point $X$.

By the condition of the problem, the altitude through the vertex $C$ is identical with the Brocard ray $CZ \equiv CF$. Let $D, E$ be the feet of altitudes $AD, BE$ through the vertices $A, B$. From similarity of the triangles $\triangle CBF \sim \triangle ABD$, it immediately follows that the 1st Brocard point $Y$ lies on the altitude $AD$, i.e., this altitude is also a Brocard ray. Since $\angle BCF = \angle BAD = \theta$ (Brocard angle), the angle $\angle B = 90^\circ - \theta$.

Let $O$ be circumcenter of the triangle $\triangle ABC$. Since the angle $\angle BOM = \angle B$, the angle $\angle MBO = 90^\circ - \angle B = \theta$. Hence, the circumcenter $O$ lies on the Brocard ray $CY$ of the 1st Brocard point $Y$. The circumcenter $O$ also lies on the Brocard circle (together with the 2 Brocard points $Y, Z$ and the symmedian point). The triangle $\triangle YOZ$ is isosceles and the angle $\angle YOZ = 2 \theta$ (double the Brocard angle). Hence, the angles $\angle OZY = \angle OYZ = \frac{180^\circ - 2 \theta}{2} = 90^\circ - \theta = \angle B$. Since the angle $\angle YOM = \angle B$ as well, it follows that the perpendicular bisector $OM$ of the side $CA$ is tangent to the Brocard circle at the point $O$. Since the angles $\angle ZYO = \angle YOM$ are both equal to the angle $\angle B$, lines $ZY, OM$ cut by the transversal $YO$ are parallel. And of course, altitude $BE$ through the vertex $B$ to the side $CA$ is parallel to perpendicular bisector $OM$ of the side $CA$, i.e., also to the line $ZY$ connecting the 2 Brocard points $Z, Y$.

[quote="yetti"]Nice problem, unfortunately, it is not stated correctly. It should be:

If the median passing through the vertex $A$, symedian through the vertex $B$, and the altitude passing through the vertex $C$ of a triangle $\triangle ABC$ are concurrent, then the altitude passing through the vertex $B$ (not $C$!) is parallel to the line connecting the two Brocard points of the triangle.[/quote]

There is more to say: The line connecting the two Brocard points of the triangle ABC is the perpendicular to the side CA through the foot of the B-symmedian.

I have actually proved a more general result:

[color=blue][b]Theorem.[/b] Let the B-symmedian of a triangle ABC meet the side CA at a point E, and let W and W' be the two Brocard points of triangle ABC such that < WBC = < WCA = < WAB = w and < W'CB = < W'AC = < W'BA = w, where w is the Brocard angle of triangle ABC. Then, the following assertions are all equivalent:

[b](1)[/b] The A-median, the B-symmedian and the C-altitude of triangle ABC are concurrent.
[b](2)[/b] The point W' lies on the C-altitude of triangle ABC.
[b](3)[/b] The point W' lies on the perpendicular to the line CA at the point E.
[b](4)[/b] The C-median, the B-symmedian and the A-altitude of triangle ABC are concurrent.
[b](5)[/b] The point W lies on the A-altitude of triangle ABC.
[b](6)[/b] The point W lies on the perpendicular to the line CA at the point E.
[b](7)[/b] We have $\tan C \tan A = \tan^2 B$.
[b](8)[/b] We have w = 90° - B.
[b](9)[/b] We have $c^4+a^4=\left(c^2+a^2\right)b^2$.
[b](10)[/b] The line WW' is perpendicular to the line CA.[/color]

[i]Proof.[/i] What we need for solving the problem is that assertion [b](1)[/b] implies assertions [b](3)[/b] and [b](6)[/b], but we will prove the equivalence of all ten assertions:

Let D be the midpoint of the side BC of triangle ABC, so that the line AD is the A-median of triangle ABC; let F be the point where the line CW' meets the side AB. Then, < FCB = < W'CB = w.

By the sum of angles in triangle BFC, we have < BFC = 180° - < FBC - < FCB = 180° - B - w. Now, the point W' lies on the C-altitude of triangle ABC if and only if $CW^{\prime}\perp AB$, what is equivalent to < BFC = 90°, what is equivalent to 180° - B - w = 90° (according to < BFC = 180° - B - w), what simplifies to w = 90° - B. Thus, we have proven the equivalence [b](2)[/b] $\Longleftrightarrow$ [b](8)[/b].

In post #2, Yetti proved that the symmedian from one vertex of a triangle, the median from another, and the line joining the third vertex with the appropriate Brocard point are concurrent. This fact can also be found with proof in [url=http://www.amazon.com/exec/obidos/tg/detail/-/0883856395/qid=1092737678/sr=8-1/ref=sr_8_xs_ap_i1_xgl14/104-6043065-0271943?v=glance&s=books&n=507846]Ross Honsberger, [i]Episodes in Nineteenth and Twentieth Century Euclidean Geometry (New Mathematical Library)[/i], Washington 1995[/url], Chapter 10, §6. Anyway, applying this fact to triangle ABC, we see that the symmedian BE, the median AD, and the line CW' are concurrent. In other words, the lines AD, BE and CF are concurrent. Thus, the Ceva theorem yields $\frac{BD}{DC}\cdot\frac{CE}{EA}\cdot\frac{AF}{FB} = 1$; but since the point D is the midpoint of the segment BC, we have $\frac{BD}{DC}=1$, and thus we get $\frac{CE}{EA}\cdot\frac{AF}{FB}=1$, so that $\frac{CE}{EA}=\frac{FB}{AF}$, or, equivalently, $\frac{CE}{EA}=\frac{BF}{FA}$, what immediately yields EF || BC after Thales.

From EF || BC, we conclude < W'FE = < W'CB. But since < W'CB = < W'AC, we have < W'FE = < W'AC; in other words, < W'FE = < W'AE. Thus, the points W', A, F, E lie on one circle, so that < AFW' + < AEW' = 180°. Hence, < AFW' = 90° holds if and only if < AEW' = 90°. But < AFW' = 90° is equivalent to $CW^{\prime}\perp AB$, what is equivalent to the assertion that the point W' lies on the C-altitude of triangle ABC. On the other hand, < AEW' = 90° is equivalent to the assertion that the point W' lies on the perpendicular to the line CA at the point E. Thus, the point W' lies on the C-altitude of triangle ABC if and only if the point W' lies on the perpendicular to the line CA at the point E. So we have proven that [b](2)[/b] $\Longleftrightarrow$ [b](3)[/b].

We know that in the triangle ABC, the B-symmedian BE, the A-median AD and the line CW' are concurrent. Hence, the A-median, the B-symmedian and the C-altitude of triangle ABC are concurrent if and only if the C-altitude coincides with the line CW'; but this obviously holds if and only if the point W' lies on the C-altitude of triangle ABC. Thus, we have proven the equivalence [b](1)[/b] $\Longleftrightarrow$ [b](2)[/b].

Let T be the foot of the C-altitude of triangle ABC. Then, in the right-angled triangle CTA, we have $AT=CT\cdot\cot\measuredangle CAT=CT\cdot\cot A$. Similarly, the right-angled triangle BTC yields $TB=CT\cdot\cot\measuredangle TBC=CT\cdot\cot B$. Thus, $\frac{AT}{TB}=\frac{\cot A}{\cot B}$ and $c=AB=AT+TB=CT\cdot\cot A+CT\cdot\cot B=CT\cdot\left(\cot A+\cot B\right)$. If $\Delta$ is the area of triangle ABC, then, since $\text{area of a triangle }=\frac12\cdot\text{ side }\cdot\text{ corresponding altitude}$, we have $\Delta=\frac12\cdot c\cdot CT$, and thus $CT=\frac{2\Delta}{c}$. Thus, $c=CT\cdot\left(\cot A+\cot B\right)=\frac{2\Delta}{c}\cdot\left(\cot A+\cot B\right)$. Consequently, $c^2=2\Delta\left(\cot A+\cot B\right)$. Similarly, $a^2=2\Delta\left(\cot B+\cot C\right)$. Thus, $\frac{a^2}{c^2}=\frac{\cot B+\cot C}{\cot A+\cot B}$.

By the Ceva theorem, the A-median AD, the B-symmedian BE and the C-altitude CT of triangle ABC are concurrent if and only if $\frac{BD}{DC}\cdot\frac{CE}{EA}\cdot\frac{AT}{TB} = 1$. But as we know, $\frac{BD}{DC}=1$ and $\frac{AT}{TB}=\frac{\cot A}{\cot B}$, and since a symmedian of a triangle divides the opposite side in the ratio of the squares of the two adjacent sides, we have $\frac{CE}{EA}=\frac{a^2}{c^2}$, what, using $\frac{a^2}{c^2}=\frac{\cot B+\cot C}{\cot A+\cot B}$, becomes $\frac{CE}{EA}=\frac{\cot B+\cot C}{\cot A+\cot B}$. Hence, the equation $\frac{BD}{DC}\cdot\frac{CE}{EA}\cdot\frac{AT}{TB} = 1$ rewrites as $1\cdot\frac{\cot B+\cot C}{\cot A+\cot B}\cdot\frac{\cot A}{\cot B} = 1$. This is equivalent to $\frac{\cot A+\cot B}{\cot B+\cot C}=\frac{\cot A}{\cot B}$. By rearranging the terms, we rewrite this as $\frac{\cot A+\cot B}{\cot A}=\frac{\cot B+\cot C}{\cot B}$, or, equivalently, as $1+\frac{\cot B}{\cot A}=1+\frac{\cot C}{\cot B}$. But this is clearly equivalent to $\frac{\cot B}{\cot A}=\frac{\cot C}{\cot B}$. Since $\cot x=1/\tan x$ for any x, this becomes $\frac{1/\tan B}{1/\tan A}=\frac{1/\tan C}{1/\tan B}$, what simplifies to $\frac{\tan A}{\tan B}=\frac{\tan B}{\tan C}$, or, equivalently, to $\tan C\tan A=\tan^2 B$. Altogether, we see that the A-median AD, the B-symmedian BE and the C-altitude CT of triangle ABC are concurrent if and only if $\tan C\tan A=\tan^2 B$. Thus, the equivalence [b](1)[/b] $\Longleftrightarrow$ [b](7)[/b] is shown.

So we have proved the equivalences [b](2)[/b] $\Longleftrightarrow$ [b](8)[/b], [b](2)[/b] $\Longleftrightarrow$ [b](3)[/b], [b](1)[/b] $\Longleftrightarrow$ [b](2)[/b] and [b](1)[/b] $\Longleftrightarrow$ [b](7)[/b]. By a symmetric permutation of the vertices A, B, C of triangle ABC, these equivalences become [b](5)[/b] $\Longleftrightarrow$ [b](8)[/b], [b](5)[/b] $\Longleftrightarrow$ [b](6)[/b], [b](4)[/b] $\Longleftrightarrow$ [b](5)[/b] and [b](4)[/b] $\Longleftrightarrow$ [b](7)[/b]. Hence, the eight assertions [b](1)[/b], [b](2)[/b], [b](3)[/b], [b](4)[/b], [b](5)[/b], [b](6)[/b], [b](7)[/b], [b](8)[/b] are all equivalent. In order to complete the proof of the theorem, it will be now enough to show that these assertions are equivalent to [b](9)[/b] and to [b](10)[/b]. We will do this by proving the equivalences [b](7)[/b] $\Longleftrightarrow$ [b](9)[/b] and [b](7)[/b] $\Longleftrightarrow$ [b](10)[/b].

First let's show that [b](7)[/b] $\Longleftrightarrow$ [b](9)[/b]: As we showed above, $c^2=2\Delta\left(\cot A+\cot B\right)$ and $a^2=2\Delta\left(\cot B+\cot C\right)$. Similarly, $b^2=2\Delta\left(\cot C+\cot A\right)$. Hence,

$c^2+a^2-b^2=2\Delta\left(\cot A+\cot B\right)+2\Delta\left(\cot B+\cot C\right)-2\Delta\left(\cot C+\cot A\right)$
$=4\Delta\cot B$.

Thus, $\cot B=\frac{c^2+a^2-b^2}{4\Delta}$, so that $\tan B=\frac{1}{\cot B}=\frac{4\Delta}{c^2+a^2-b^2}$. Similarly, $\tan C=\frac{4\Delta}{a^2+b^2-c^2}$ and $\tan A=\frac{4\Delta}{b^2+c^2-a^2}$. Hence, the equation $\tan C\tan A=\tan^2 B$ is equivalent to $\frac{4\Delta}{a^2+b^2-c^2}\cdot\frac{4\Delta}{b^2+c^2-a^2}=\left(\frac{4\Delta}{c^2+a^2-b^2}\right)^2$. This immediately simplifies to $\left(a^2+b^2-c^2\right)\left(b^2+c^2-a^2\right)=\left(c^2+a^2-b^2\right)^2$. But since $\left(a^2+b^2-c^2\right)\left(b^2+c^2-a^2\right)-\left(c^2+a^2-b^2\right)^2=-2\left(\left(c^4+a^4\right)-\left(c^2+a^2\right)b^2\right)$, this is equivalent to $c^4+a^4=\left(c^2+a^2\right)b^2$. Thus, $\tan C\tan A=\tan^2 B$ is equivalent to $c^4+a^4=\left(c^2+a^2\right)b^2$. This proves that [b](7)[/b] $\Longleftrightarrow$ [b](9)[/b].

Remains to establish [b](7)[/b] $\Longleftrightarrow$ [b](10)[/b]. For that aim, we consider the points X, Y, Z where the tangents to the circumcircle of triangle ABC at the points A, B, C meet the lines BC, CA, AB, respectively. It is well-known that these points X, Y, Z lie on one line, which is called the Lemoine axis of triangle ABC and is perpendicular to the Brocard axis of triangle ABC. On the other hand, by the properties of the Brocard points, the Brocard axis of triangle ABC is perpendicular to the line WW' joining the two Brocard points W and W'. Hence, the line XYZ is parallel to the line WW'. Thus, the line WW' is perpendicular to the line CA if and only if the line XYZ is perpendicular to the line CA.

Let S be the orthogonal projection of the point Z on the line CA. Then, the line XYZ is perpendicular to the line CA if and only if S = Y. But S = Y is trivially equivalent to $\frac{AY}{CY}=\frac{AS}{CS}$. Now, by http://www.mathlinks.ro/Forum/viewtopic.php?t=6557 post #4 Lemma 2, we have $\frac{AY}{CY}=\frac{c^2}{a^2}$; in other words, $\frac{AY}{CY}=1:\frac{a^2}{c^2}=1:\frac{\cot B+\cot C}{\cot A+\cot B}=\frac{\cot A+\cot B}{\cot B+\cot C}$. On the other hand, in the right-angled triangle ASZ, we have $AS=ZS\cdot\cot\measuredangle ZAS=ZS\cdot\cot A$, and in the right-angled triangle CSZ, we have $CS=ZS\cdot\cot\measuredangle ZCS$. Since the line CZ is tangent to the circumcircle of triangle ABC at the point C, we have < ZCA = < ABC; in other words, < ZCS = B, and thus, $CS=ZS\cdot\cot B$. Hence, $\frac{AS}{CS}=\frac{ZS\cdot\cot A}{ZS\cdot\cot B}=\frac{\cot A}{\cot B}$. Altogether, this yields that the equation $\frac{AY}{CY}=\frac{AS}{CS}$ becomes $\frac{\cot A+\cot B}{\cot B+\cot C}=\frac{\cot A}{\cot B}$. As we already saw above, this is equivalent to $\tan C\tan A=\tan^2 B$.

Combining all these steps, we get a proof of the fact that the line WW' is perpendicular to the line CA if and only if $\tan C \tan A = \tan^2 B$. In other words, we have shown that [b](7)[/b] $\Longleftrightarrow$ [b](10)[/b]. This finally completes the proof of the theorem.

  Darij

I noticed the line connecting the Brocard points passing through the foot V of the symmedian BV through the vertex B to the line CA, wondered about it, but since I did not need it, I passed on.

yetti

[b]Problem:[/b] If median  $ AM_a,$ symmedian $ BK_b$ and altitude $ CH_c$ of $ \triangle ABC$  are concurrent, then the altitude  $ BH_b$  is parallel to the line connecting the two Brocard points $ \Omega_1, \Omega_2$ of the $\triangle ABC.$

Let us use barycentric coordinates with respect to $ \triangle ABC.$

$ AM_a \equiv y \minus{} z \equal{} 0 \ , \ BK_b \equiv c^2x \minus{} a^2z \equal{} 0 \ , \ CH_c \equiv S_Ax \minus{} S_By \equal{} 0$

These three lines are concurrent if and only if

$\left[\begin{array}{ccc}0 & 1 & \minus{} 1 \\
 \\
c^2 & 0 & \minus{} a^2 \\
 \\
S_A & \minus{} S_B & 0\end{array}\right] \equal{} 0 \Longrightarrow a^2 \plus{} c^4 \equal{} a^2b^2 \plus{} b^2c^2$

Coordinates of the infinite point $B_{\infty}$ of $\perp AC$ are given by

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

Using the upper condition we get: $ B_{\infty } (a^2(b^2 \minus{} c^2): b^2(c^2 \minus{} a^2): c^2(a^2 \minus{} b^2))$

But note that the infinite point of the line connecting the two Brocard points $ \Omega_1,\Omega_2$  is precisely $ T_{\infty} (a^2(b^2 \minus{} c^2): b^2(c^2 \minus{} a^2): c^2(a^2 \minus{} b^2))$ $\Longrightarrow$ $ \Omega_1\Omega_2 \parallel BH_b.$

[quote="yetti"][b]Lemma:[/b] A median through one vertex of a triangle, symmedian through another vertex and one of the Brocard rays from the remaining vertex of a triangle are concurrent.[/quote]
$ AM_a \equiv y \minus{} z \equal{} 0 \ , \ BK_b\equiv c^2x \minus{} a^2z \equal{} 0 \ ,\ C\Omega_2 \equiv c^2x \minus{} a^2y \equal{} 0$

$ \left[\begin{array}{ccc}0 & 1 & \minus{} 1 \\
 \\
c^2 & 0 & \minus{} a^2 \\
 \\
c^2 & \minus{} a^2 & 0\end{array}\right] \equal{} a^2c^2 \minus{} a^2c^2 \equal{} 0 \Longrightarrow AM_a \cap BK_b \cap C\Omega_2 \neq \emptyset$


Prove that,
$ \sum (a\minus{}b)(a\minus{}c)(ka\minus{}1) \geq 0$
for all $ a,b,c,k\geq 0$

[quote="peine"]Prove that,
$ \sum (a \minus{} b)(a \minus{} c)(ka \minus{} 1) \geq 0$
for all $ a,b,c,k\geq 0$[/quote]
Try $ k\equal{}0.$

[quote="arqady"][quote="peine"]Prove that,
$ \sum (a \minus{} b)(a \minus{} c)(ka \minus{} 1) \geq 0$
for all $ a,b,c,k\geq 0$[/quote]
Try $ k \equal{} 0.$[/quote]
what about $ k\geq 1$ ?

for $ k \ge  \frac{1}{min(a,b,c)}$ it's true by Vornicu-Shur ineq

[quote="peine"]Prove that,
$ \sum (a \minus{} b)(a \minus{} c)(ka \minus{} 1) \geq 0$
for all $ a,b,c,k\geq 0$[/quote]
It's wrong even If $ k \geq 1$

BTW,if $ a,b,c \geq \frac{1}{k}$ where $ k > 0$ the inequality is true.See here:

http://www.mathlinks.ro/viewtopic.php?t=254184


how do you make a box around some text like $\boxed{\text{this}}$ with html?

How do you align a group of text to the right side with html?

I suggest you buy a book on html
or go to a html teaching website.
[url]http://www.w3schools.com/[/url] is a good html/javascript/blabla teaching website thingy.

to your question, I don't know b/c I suck at html.

htmlgoodies.com is a good site for html, its different than bb code though

For having a "box" around something, use the code:

[code]
 Stuff that's supposed to be in the box
[/code]

It is really CSS (Cascading Style Sheets) that does it, not HTML, but CSS is usually better to use. The span tag itself does not do anything, but the style attribute. That's where the CSS is. I would recomender learning it, one good place mentioned is http://www.w3schools.com

As to how to align text to the right, similar thing using CSS

[code]

Stuff to align

[/code]


Prove that for all positive integers n, $29^n+4$ is never prime

If $n$ is even then $5\mid 29^n+4$, and if $n$ is odd then $3\mid 29^n+4$.

Myth, it is not helpful to give answers to questions which are quite easy for you. Especially when the answers are not hidden. For students who are trying to learn, the answers to questions are not the most important thing -- it is the process of finding the answer that is the most important.

I need little coaching thanks, Irish IMO team 2004 in Athens, I just like posting up cool problems to get second opinions...

I didn't mean you specifically seamous. I just mean for people in general who can learn from working on the problem.


I want to write equation for a sawtooth...

a) Starts at $(a, b)$ in $XY$ plane
b) Maximum height of sawtooth is $c$

What would be the equation of this sawtooth curve...

I think it has to be derived using Fourier transofrms... but could not get it..

PS: The figure may not be accurate, the sawtooth are evenly placed etc etc....

[quote="Altheman"]$y-b=c*\sin^{-1}[\sin (x-a)]$[/quote]

how do you get it?

btw, isnt it same as: $y-b=c*(x-a)$

and how about the period of the sawtooth wave? (say = $d$)

Summary:
a) Starts at $(a, b)$
b) Height = $c$
c) Period = $d$

[quote="xxxyyyy"]isnt it same as: $y-b=c*(x-a)$
[/quote]

nope, the arcsin and the sin don't cancel out due to the horizontal shift the a term provides

i actually just knew that since i was messing around with graphs on my calc a while ago and discovered this

Yes, you can derive a Fourier Series for a sawtooth, but the details are rather annoying.  I wouldn't bother.  What do you need it for?


calculus
integration
number theory unsolved
number theory

Find all integral solutions of $8^{m}-2^{m}=n(2n-1)(2n+2)$

$8^{m}-2^{m}=2^{m}(2^{m}-1)(2^{m}+1)=n(2n-1)2(n+1)$.
m=0 give $n=0$ or $n=-1$.
$ord_{2}(2n(n+1))\ge 2$, therefore $m\ge 2$.
In these case $n=2^{m-1}c$ or $n=2^{m-1}c-1$, were c is odd.
It is easy to chek, that these equation had not other solutions.

[quote="lucio"]I am sorry
 What does" $ord_{2}(2n(n+1)) \geq 2$" mean ?[/quote]
it describes the exponent of $2$ in the number $2n(n+1)$ .  :wink:


Show that in a convex hexagon with area $S,$ we can always find a diagonal which divides the hexagon in a pentagon in a triangle, the triangle having area
$\leq \frac{1}{6}S$.

[color=red][Moderator edit: Also discussed at http://www.mathlinks.ro/Forum/viewtopic.php?t=27331 .][/color]

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]


function
algebra
domain
logarithms
functional equation

[b]Problem.[/b] [i]Find all continuous functions $ f,g,h: \mathbb{R}\rightarrow\mathbb{R}$ such that
\[ f(x+y)=g(x)h(y).\][/i]

[quote="boxedexe"][b]Problem.[/b] [i]Find all continuous functions $ f,g,h: \mathbb{R}\rightarrow\mathbb{R}$ such that
\[ f(x+y)=g(x)h(y). \]
[/i][/quote]

[hide="Needs little work"]
Let $ y=0$ so we get,
$ f(x)=g(x)h(0)$
Let $ x=0$ so we get,
$ f(y)=g(0)h(y)$

Now we get to casework.

[b]Case 1:[/b] If $ h(0)=0$ then $ f(x) = 0$ on $ \mathbb{R}$. From $ f(x+x)=g(x)h(y)\implies 0 = g(x)h(x)$ means $ g(x) = 0$ or $ h(x)=0$. So all functions $ 0,g,h$ with the property that for any $ x$ we have $ g(x)=0$ or $ h(x)=0$ (but not necessarily both). This list seems indescribable.  :maybe:

[b]Case 2:[/b] If $ g(0)=0$ then $ f(y)=0$ on $ \mathbb{R}$. And using a similar approach it means functions $ 0,g,h$ with the property is Case 1.

[b]Case 3:[/b] If $ h(0),g(0)\not =0$ it means $ f(x+y)=cf(x)f(y)$ where $ c=\frac{1}{h(0)g(0)}$ and so $ f(x) = e^{x+c}$ since its continous.

[/hide]

Your solution set is incomplete.  Here's what I did.

[b]Solution.[/b] If $ g,h$ vanishes at any point in their domain, then $ f\equiv 0$.  Therefore, the solutions set $ (f,g,h)\equiv (0,g,h)$ is valid iff either $ g$ or $ h$ or both has/have at least one point at which it/they vanish.

Suppose that $ g(x),h(x)\neq 0\ \forall x\in\mathbb{R}$.  Let $ g(0)=a,h(0)=b$.  Set $ y=0$ to obtain $ f(x)=bg(x)\implies g(x)=b^{-1}f(x)$.  Set $ x=0$ to obtain $ h(x)=a^{-1}f(x)$.  Then $ (ab)^{-1}f(x+y)=((ab)^{-1}f(x))((ab)^{-1}f(y))$.  Now, let $ q(z)=(ab)^{-1}f(z)$.  It follows that our functional equation is equivalent to $ q(x+y)=q(x)q(y)$.  Since $ f>0$, $ q>0$ for all $ x\in\mathbb{R}$, and so $ q(x)=\left[q\left(\frac{x}{2}\right)\right]^{2}>0$.  Let $ p(x)=\log_{c}q(x),\ c>0$.  Then $ p(x+y)=p(x)+p(y)$.  It is easy to prove that $ p(x)=tx\ \forall x\in\mathbb{Q},\ t\in\mathbb{R}$.  Since $ \mathbb{Q}$ is dense in $ \mathbb{R}$ (very easy to prove), $ p(x)=tx\ \forall x\in\mathbb{R}$ (using continuity).

It follows that $ f(x)=abq(x)=abc^{p(x)}=abd^{x},\ g(x)=ad^{x},\ h(x)=bd^{x}$ for some $ a,b,d\in\mathbb{R},\ d>0$. $ \Box$


limit
trigonometry
calculus
calculus computations

How about not thinking about when to use L'Hôpital and thinking about the problem instead. Power series are a more flexible tool anyway:

$\prod_{k=1}^{n}\sin(kx)=\prod_{k=1}^{n}\left(kx-\frac{k^{3}x^{3}}6+O(x^{5})\right)$

$=n!x^{n}\prod_{k=1}^{n}\left(1-\frac{k^{2}x^{2}}6+O(x^{4})\right)$

$=n!x\left(1-\left(\sum_{k=1}^{n}\frac{k^{2}}6\right)x^{2}+O(x^{4})\right)= n!x^{n}\left(1-\frac{n(n+1)(2n+1)x^{2}}{36}+O(x^{4})\right)$

If we put that expression into the limit we see that the limit is

$\frac{n!n(n+1)(2n+1)}{36}.$

Or you can use this method:

Let $L_{n}$ be your limit. Then, observe that

$L_{n}=(n-1)!\lim_{x\rightarrow 0}\frac{nx-\sin nx}{x^{3}}+\lim_{x\rightarrow 0}\frac{(n-1)!x^{n-1}-\sin x\cdots \sin(n-1)x}{x^{n+1}}\lim_{x\rightarrow 0}\frac{\sin nx}{x}$,

In other words you get the following recurrence relation:

$\frac{L_{n}}{n!}=\frac{n^{2}}{6}+\frac{L_{n-1}}{(n-1)!}$, and the result follows.

didilica's solution was what i expected. thanks ! kent's solution is far to complicated for me because i haven't yet learned about series

[quote="Kent Merryfield"]
$\prod_{k=1}^{n}\sin(kx)=\prod_{k=1}^{n}\left(kx-\frac{k^{3}x^{3}}6+O(x^{5})\right)$

$=n!x^{n}\prod_{k=1}^{n}\left(1-\frac{k^{2}x^{2}}6+O(x^{4})\right)$

$=n!x^{n}\left(1-\left(\frac{1}{6}\sum_{k=1}^{n}k^{2}\right)x^{2}+O(x^{4})\right)$

$=n!x^{n}\left(1-\frac{1}{6}\;\left(\frac{n(n+1)(2n+1)}{6}\right)\;x^{2}+O(x^{4})\right)$

If we put that expression into the limit we see that the limit is

$\frac{n!n(n+1)(2n+1)}{36}.$[/quote]

that's a neat solution!   :read:


Anyone else going to the April 19 GPML?
I haven't really practiced GPML-style problems, so I'm going to get owned (as always). I never do well on contests where speed is a significant factor.

YAY!!!! ANOTHER IOWAN!!!!   :D
Yeah, I'll be there.  I'm aiming for 2nd freshman.  (Lucy Chen's probably gonna beat me, but I think, after weeks of mad crazy AIME studying, I can beat Shi-Ke.  I hope)

EDIT:  The comp's at my school, hehehe
EDIT (again, I like editing  :lol: ):  Yeah, I know the questions are below AIME level, but I just got a 116 on last year's Jefferson sprint.  Yay.  Of course with my luck, I'll probably miss questions 1-5 and get less than 100...

Thou art lying
Shi-Ke is friggin awesome at math
and he's going to pwn lucy chen at state this time
and he's talking in third person.

I'm trying to get through 30-50 problems for practice for nationals (mathcounts)
so I really have not done much studying for gpml

oh well

Shi-Ke, you FINALLY got an account.  Anyway, we moved most of the discussion to "Random discussion of Randomness"

STATE IS TOMORROW!!!
Yay.
Where's everyone going for lunch?  I personally have no idea (I'm freeloading a ride from one of our juniors)

EDIT (State's finished):
Nice job Lawrence!  Man, I screwed up Target...  Missed the first 2 questions, and on the last one read $ 3 : 1$ as $ 2 : 1$  Oh well.  See you at regionals!


topology
real analysis
real analysis solved

$F$ is a countable closed set. Show that $F$ has an isolated point.

We choose an element of $F$ at step $0$. Then, at step $1$, we choose an element which is within $\varepsilon$ from the first element and order the two in an increasing order. Then, at step $n+1$, if the minimal length between two points in step $n$ is $\ell$, then choose $2^n$ elements of $F$ each of which is within $\frac {\ell}{2^n}$ from one of the $2^n$ points from step $n$ and then order the $2^{n+1}$ points in an increasing order.

Then, for each binary sequence, we may choose a point of $F$ as follows: if the first digit is $0$ then we choose the first element from step $1$, then look at the first half of elements from the next step and repeat this procedure, and if the first digit is $1$, we choose the second point from step $1$ and then move on to the next step, looking only at the second half of the points on this step. Each time we move on to the next step we choose another point, so the point we will finally choose will be the limit of the points we choose at steps $1,2,\ldots$. This is possible because the set $F$ is closed, so each limit point belongs to $F$.

This means that we can define an injective map from the binary sequences (a set which has cardinality $c$) to $F$, so $F$ is uncountable if $F$ is closed and has no isolated points, which is what we wanted.

Is this correct? I'm really unsure about this..

hmm.
The hint in the book is:

suppose $F$ has no isolated point. thus for $x_n\in F$, $(\mathbb{R}\setminus\{x_n\})\cap F$ is dense in $F$. then $(\bigcup_{i=1}^n \mathbb{R}\setminus\{x_i\})\cap F$ is dense in $F$. a contradiction...

i just wonder how does it get $(\bigcup_{i=1}^n \mathbb{R}\setminus\{x_i\})\cap F$ is dense in $F$?

More often this result is stated as:

Every perfect set is uncountable.

There are proofs in many books on real analysis and maybe topology.

I think the answer to lyi's question is: Baires theorem.

I only know the following Baire Theorem:

$E\subset \mathbb{R}^n$ is a $F_\sigma$ set, i.e, \[ E=\bigcup_{k=1}^\infty F_k, \] $F_k$ are closed sets. If each $F_k$ has no interior points, then $E$ has no interior points.

How to apply it?

[quote="liyi"]
How to apply it?[/quote]
Here you need a more general version for an arbitrary complete metric space. Note that a closed subset  $F$ of $\mathbb R$ (or of any other complete metric space) may be considered as such space. If $F$ is countable, it is a countable union of a one point sets. These sets are, clearly, closed. Therefore, at least one of them contains an inner point (in topology of $F$), but it can happen only if the only point in this set is an isolated point of $F$.


Show that for any integer n > 10000, there are integers a, b such that n < a 2  + b 2  < n + 3*n^1/4.

I'm pretty sure this was posted before. Try searching for it.


Arqady, I think you have considered that $x,y,z$ are non-negative numbers. But they are [b]reals[/b]. ;)

Vasc, this inequality looks like another inequality of yours. and the solutions are done in the same way.

The question is: do you have an identity proving it all?

Do we have a Stefan-type solution for this one?

My main solution is arqady's solution. But I have also a Stefan's type solution, namely:

$\sum(xy-2yz+zx-3x^2+3y^2)^2 \geq 0$.

May I ask what's a "Stefan's type solution"?  :?

Darij mentioned Stefan some time ago. This Stefan gave a nice solution to a Vasc inequality, and the name is now used to characterize an one-line solution to an inequality through an impossible to derive identity. (or something like this)

Oh ok, cool :D
Thanks for the piece of knowledge ;)

[quote="Vasc"]Cool, Gladkich! :ninja:[/quote]

Is that the way you derived the inequality? :o

[quote="Vasc"][quote="Soarer"] Is that the way you derived the inequality? :o[/quote] It is possible. I don't remember very well.
But another Stefan's solution type is
$\sum(2x^2-y^2-z^2-xy+yz)^2 \geq 0$
 :lol:[/quote]
cool! :ninja:


...is everyone going to wear their AoPS/Mathlinks t-shirts the first day of MOSP? :)

I don't have either  :|

Will I become an outcast the first day?  :lol:

We can be outcasts together  :( ... unless AoPS decides it worth the advertising to send us all free T-shirts ;)

Huh, that's not going to win you style points.

You should wear last year's MOP T-shirts on the first day of MOP, duh.

[quote="MysticTerminator"][quote="jrshoch"]I have national mathcounts shirts but idk if those still fit me...

And Aops does have a tshirt i saw someone else wearing it.  It was one of those "you know you're a _____ if: [list of things that math nerds do, numbered in base 12]"[/quote]

NO MATHCOUNTS = TERRIBLE

wow, wearing the MOP shirt ... that's an amazingly good idea. strange that that didn't even occur to me . . .[/quote]

Or you could, like, wear a normal shirt like I'm going to.

Would it be OK to wear a shirt from another summer math program, you think?


In a triangle ABC having the  (length) sides $ a,b,c $  and the  medians $ m_a,m_b,m_c\; ,\;  $ prove that
\[
\begin{array}{|c|}
\hline
\\
 \displaystyle \frac{3}{2}\max{(a,b,c)} \le
 m_a+m_b+m_c\le \frac{3\sqrt{3}}{2}\max(a,b,c)
\\
\\
\displaystyle \max{(m_a,m_b,m_c)} \le  s\le
\sqrt{3}\max(m_a,m_b,m_c)
\\
\\
\hline
\end{array} \; \; ,\; \; \;  (2s:=a+b+c).
\]

[quote="Diarmuid"]I can do some of this...

For each of the medians, we have $m_a=\frac{\sqrt{2b^2+2c^2-a^2}}{2}$.

Applying the power mean inequality, we get $m_a+m_b+m_c\le\frac{3}{2}\sqrt{a^2+b^2+c^2}$, and the first RHS follows.

From the same formula, we can conclude that $m_a<\frac{b+c}{2}<s$, which proves the second LHS.

Also, if $a\ge b\ge c$, then the largest median is $m_c=\frac{\sqrt{2a^2+2b^2-c^2}}{2}\ge \frac{\sqrt{3}}{2}\sqrt{\frac{a^2+b^2+c^2}{3}}\ge\frac{a+b+c}{2\sqrt{3}}$, which proves the RHS.

Finally, the first LHS - I'm not sure about this one, but certainly it would be sufficient to show that $m_a+m_b+m_c\ge \frac{3}{4}(a+b+c)$.  This was an Irish MO problem this year, but unfortunately I never got around to solving it...

There may be an easier way to prove this slightly weaker result, but I'm not sure.[/quote]
  Indeed , the first LHS is very easy . Geometric solution :
 Let $ AA' , BB' , CC' $ be the medians .
  It's easy to show : $ AA' + \frac{a}{2} > b ; BB'+ \frac{b}{2}> c ; CC' +\frac{c}{2} > a $  ;)
   Could you please tell me what Alex collection is ? And who is Alex ?  :)


trigonometry
algebra
polynomial
rational numbers

Find all $x$ and $y$ which are rational multiples of $\pi$ with $0<x<y<\frac{\pi}{2}$ and $\tan x+\tan y =2$.

Let me complete the solution:
Firstly,if $\tan x+\tan y=2$,then
\[\sin(x+y)=2\cos x\cos y=cos(x+y)+\cos(x-y)\Longrightarrow \sqrt{2}(\sin(x+y-\frac{\pi}{4}))=\cos(x-y) \Longrightarrow 2(\sin(x+y-\frac{\pi}{4}))^{2}=1-\sin^{2}(x-y) \Longrightarrow \cos(2(x+y)-\frac{\pi}{2})=\cos^{2}(x-y)\]
Let $z=x-y,t=2(x+y)-\frac{\pi}{2}$
then $\cos^{2}z=\cos t$
Then we only need to find all pairs of (z,t) such that $z,t$ both are
rational multiples of $\pi$ and $\cos^{2}z=\cos t$

[b]Lemma1[/b],Define ${p_{n}(x)=\prod_{1\le k \le \frac{n}{2},gcd(k,n)=1}(x-\cos \frac{2k\pi}{n})},(n\ge 3)$ then we
firstly proved $p_{n}(x)$ is an polynomial with rational
coeffients,and it is irreducible(Do I spell correctly?I have not a
dictionary here...).

[b]Proof of Lemma 1[/b]:Let $f_{n}(x)$ be the polynomial with the property $f_{n}(\cos x)=\cos nx$.
then we can prove that $f_{2n+1}(x)-1=(x-1)h_{2n+1}(x)^{2},f_{2n}=(x-1)(x+1)h_{2n}(x)^{2}$
where $h_{n}(x)=\prod_{1\le k< \frac{n}{2}}{(x-\cos{\frac{2k\pi}{n})}}$
and we can easily find that $\prod_{d|n}{p_{d}(x)}=h_{n}(x)$
then $p_{n}(x)=\prod_{d|n}{(h_{d}(x))^{\mu(\frac{n}{d})}}$ is a polynomial.
We consider ${p_{n}(\frac{t+\frac{1}{t}}{2})=\prod_{1\le k \le \frac{n}{2},gcd(k,n)=1}(\frac{t+\frac{1}{t}}{2}-\cos \frac{2k\pi}{n})}=$,$\frac{\Phi_{n}(t)}{t^{\frac{\phi(n)}{2}}}$
becuase $\Phi_{n}(t)$ is irreducible,$p_{n}(x)$ is also irreducible.

 [b]Lemma 2[/b]IF a polynomial $f(x)\in Q[x]$ is
irreducible, and  $f(x^{2})$ is reducible,then there exist $g(x)\in Q[x]$ such that $f(x^{2})=\pm g(x)g(-x)$.

 Suppose that $z=\frac{u}{v}\pi ,t=\frac{m}{n}\pi,(n,v\ge 3)$
and Let $q(x)=p_{n}(x^{2})$,then $p_{v}(\cos z)=0,p_{n}(\cos t)=0,q(\cos z)=0$ by Lemma 1 ,$p_{v}(x)|q(x)$ by Lemma 2,$p_{v}(x)=q(x) ,or, p_{v}(x)p_{v}(-x)=\pm q(x)$
But both of these two case leads to contraduction,becuase $q(x)$ has at least one non-real root,but $p_{v}(x)$ does not.

I hope my solution is correct.Could somebody tell me if it is correct?I am not so confident.......

@Peter,could you help me delet the first two post?

[quote="Hawk Tiger"]

@Peter,could you help me delet the first two post?[/quote]

I have deleted the posts :smile: ,though I am not Peter :grin: .

[quote="Hawk Tiger"]Let me complete the solution:
Firstly,if $ \tan x + \tan y = 2$,then
\[ \sin(x + y) = 2\cos x\cos y = cos(x + y) + \cos(x - y)\Longrightarrow \sqrt {2}(\sin(x + y - \frac {\pi}{4})) = \cos(x - y) \Longrightarrow 2(\sin(x + y - \frac {\pi}{4}))^{2} = 1 - \sin^{2}(x - y) \Longrightarrow \cos(2(x + y) - \frac {\pi}{2}) = \cos^{2}(x - y)
\]
Let $ z = x - y,t = 2(x + y) - \frac {\pi}{2}$
then $ \cos^{2}z = \cos t$
Then we only need to find all pairs of (z,t) such that $ z,t$ both are
rational multiples of $ \pi$ and $ \cos^{2}z = \cos t$

[b]Lemma1[/b],Define ${ p_{n}(x) = \prod_{1\le k \le \frac {n}{2},gcd(k,n) = 1}(x - \cos \frac {2k\pi}{n})},(n\ge 3)$ then we
firstly proved $ p_{n}(x)$ is an polynomial with rational
coeffients,and it is irreducible(Do I spell correctly?I have not a
dictionary here...).

[b]Proof of Lemma 1[/b]:Let $ f_{n}(x)$ be the polynomial with the property $ f_{n}(\cos x) = \cos nx$.
then we can prove that $ f_{2n + 1}(x) - 1 = (x - 1)h_{2n + 1}(x)^{2},f_{2n} = (x - 1)(x + 1)h_{2n}(x)^{2}$
where $ h_{n}(x) = \prod_{1\le k < \frac {n}{2}}{(x - \cos{\frac {2k\pi}{n})}}$
and we can easily find that $ \prod_{d|n}{p_{d}(x)} = h_{n}(x)$
then $ p_{n}(x) = \prod_{d|n}{(h_{d}(x))^{\mu(\frac {n}{d})}}$ is a polynomial.
We consider ${ p_{n}(\frac {t + \frac {1}{t}}{2}) = \prod_{1\le k \le \frac {n}{2},gcd(k,n) = 1}(\frac {t + \frac {1}{t}}{2} - \cos \frac {2k\pi}{n})} =$,$ \frac {\Phi_{n}(t)}{t^{\frac {\phi(n)}{2}}}$
becuase $ \Phi_{n}(t)$ is irreducible,$ p_{n}(x)$ is also irreducible.

 [b]Lemma 2[/b]IF a polynomial $ f(x)\in Q[x]$ is
irreducible, and  $ f(x^{2})$ is reducible,then there exist $ g(x)\in Q[x]$ such that $ f(x^{2}) = \pm g(x)g( - x)$.

 Suppose that $ z = \frac {u}{v}\pi ,t = \frac {m}{n}\pi,(n,v\ge 3)$
and Let $ q(x) = p_{n}(x^{2})$,then $ p_{v}(\cos z) = 0,p_{n}(\cos t) = 0,q(\cos z) = 0$ by Lemma 1 ,$ p_{v}(x)|q(x)$ by Lemma 2,$ p_{v}(x) = q(x) ,or, p_{v}(x)p_{v}( - x) = \pm q(x)$
But both of these two case leads to contraduction,becuase $ q(x)$ has at least one non-real root,but $ p_{v}(x)$ does not.

I hope my solution is correct.Could somebody tell me if it is correct?I am not so confident.......

@Peter,could you help me delet the first two post?[/quote]

Thank you for brilliant lemma 1.
BUT I don't think $ Q(X) = \pm P_v(X)P_v( - X)$
Lemma ONLY tell you $ P_v(X)P_v( - X)$ divides $ Q(X)$
Maybe $ Q(X) = P_v(X)P_v( - X)R(X)R( - X)$
Right?

[quote="hxy09"]Lemma ONLY tell you $ P_v(X)P_v( \minus{} X)$ divides $ Q(X)$
Maybe $ Q(X) \equal{} P_v(X)P_v( \minus{} X)R(X)R( \minus{} X)$[/quote]Thenyou write $ Q(X) \equal{} \big(P_v(X)R(X)\big)\big(P_v( \minus{} X)R( \minus{} X)\big)$, so his claim for $ g(X)\equal{}P_v(X)R(X)$? (Your claim misses the $ \pm$ though)

[quote="Peter"][quote="hxy09"]Lemma ONLY tell you $ P_v(X)P_v( \minus{} X)$ divides $ Q(X)$
Maybe $ Q(X) \equal{} P_v(X)P_v( \minus{} X)R(X)R( \minus{} X)$[/quote]Thenyou write $ Q(X) \equal{} \big(P_v(X)R(X)\big)\big(P_v( \minus{} X)R( \minus{} X)\big)$, so his claim for $ g(X) \equal{} P_v(X)R(X)$? (Your claim misses the $ \pm$ though)[/quote]

Sorry I think you didn't really catch my question.Lemma 2 is obviously true.What I mean is $ q(x)$ has at least one non real root but $ p_v(x)$ doesn't.[b]HOWEVER[/b] it doesn't lead to contradiction.You see, $ R(x)$ and$ R( \minus{} x)$ might have non real root.

I'm not sure I understand you.

His argument (at least how I interpret it) is that $ p_v$ is irreducible, $ p_v|q$, and $ \deg(q)\equal{}2\deg p_v$ by definition. Hence if $ g(x)\equal{}p_v(x)R(x)$, then $ \deg(R)\equal{}0$...

[quote="Peter"]I'm not sure I understand you.

His argument (at least how I interpret it) is that $ p_v$ is irreducible, $ p_v|q$, and $ \deg(q) \equal{} 2\deg p_v$ by definition. Hence if $ g(x) \equal{} p_v(x)R(x)$, then $ \deg(R) \equal{} 0$...[/quote]

Sorry,but I think $ \deg(q) \equal{} 2\deg p_n$ by definition not $ \deg(q) \equal{} 2\deg p_v$ right?

        Solution:Take derivtive to find $tanx+tany$ is monotonic increasing as $x$ or $y$ increases.
        So the only solution is $x=y=\frac {\pi}{4}$


Given a sequence $ (u_n)$ defined $ u_1 \equal{} 1,u_2 \equal{} 4,u_3 \equal{} 9,u_n \equal{} 15u_{n \minus{} 2} \minus{} 4u_{n \minus{} 3}(n\ge 4)$. Prove that $ u_n$ is [b]not[/b] a prime for every natural numbers $ n$.

Sorry, but is there a mistake? $ u_2$ and $ u_3$ certainly are not prime...

Yes, I want to say that $ u_n$ is [b]not[/b] a prime.  :blush:

hello, $ u_5\equal{}119$ and this is a prime-number.
Sonnhard.

[quote="Dr Sonnhard Graubner"]hello, $ u_5 \equal{} 119$ and this is a prime-number.
Sonnhard.[/quote]
It is divisible by 7  :)

hello, yes it is divisible by seven, i have calculated with 113, sorry for my misstake.
Sonnhard.


[quote="eagerkill"]Let $a,b,c$ is possitve number. Prove that
  $\frac{a(b+c)}{b^{2}+bc+c^{2}}+\frac{b(c+a)}{c^{2}+ca+a^{2}}+\frac{c(a+b)}{a^{2}+ab+b^{2}}$ $\geq 2$[/quote]
Look here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=25553


Given a,b,c positive number . prove that

$a^{3}b+b^{3}c+c^{3}a \geq a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}$

1) "look" is a nonsense topic title, it says pretty much nothing about the topic. a possible title would be "a^3b + b^3c + c^3a >= a^2b^2 + b^2c^2 + c^2a^2" or "SUM a^3b >= SUM a^2b^2".

2) your inequality doesn't hold for arbitrary positive a, b, c but it holds for a, b, c being sidelengths of a triangle:

http://www.mathlinks.ro/Forum/viewtopic.php?t=55327

  darij


Simon used the distributive property. For example, if you have the expression $ a-(a-b)$, you can rewrite it as $ a+-(a-b)$, because $ a-b=a+-b$. Then, $ a+-(a-b)=a+-1 \cdot (a-b)$. Now, applying the distributive property, we have $ a+(-1 \cdot a)+(-1 \cdot -b)=a+(-a)+(b)=a-a+b=\boxed{b}$.

[quote="SimonM"][quote="leroyjenkens"]I'm really confused on some things. I've checked out this page and some of the answers don't seem to make sense to me.
http://www.math.com/school/subject2/practice/S2U2L5/S2U2L5Pract.html

Like 5mn-n(6m-2m^2). They simplified it into -mn+2m^2n. How did they get that? Why did the minus sign turn into a plus sign? And how did the 5 turn into a 2?

I thought you couldn't use the associative or commutative properties if a minus sign is in the equation.
Thanks.[/quote]

\begin{align*} 5mn - n(6m - 2m^2) & = 5mn - [n(6m - 2m^2)] \\
& = 5mn - (6mn - 2m^2n) \\
& = 5mn - 6mn + 2m^2n \\
& = - mn + 2m^2n \end{align*}

Can you tell me which line you have a problem with?[/quote]

Ok on the 3rd line, why did the minus sign between 6mn - 2m^2n turn into a plus sign?

And on the last line, why does it not show -1mn instead of showing -mn? Where did the 1 go?

123456789, sorry but you kind of confused me. I just started algebra. Where did the -1 come from?

I also have another quick question.

If 6b-5b=1b, then why does 6b-6b=0? Where does the "b" go in the second equation? Shouldn't the answer be 0b?

[quote="t0rajir0u"]$ 6b \minus{} 6b$ is [b]always[/b] zero.  A number subtracted from itself is nothing.

Here's the thing about minus signs and parentheses:

- Minus signs "distribute" over parentheses because putting a minus sign in front of an expression is the [b]same thing as multiplying by $ \minus{} 1$.[/b]  Thus, $ \minus{} (3 \plus{} 4) \equal{} \minus{} 3 \minus{} 4 \equal{} \minus{} 7$.
- Subtracting a negative number is the [b]same thing as adding a positive number.[/b]  Thus, $ \minus{} (3 \minus{} 4) \equal{} \minus{} 3 \boxed{ \plus{} 4} \equal{} 1$.

I think both of those points confused you.  Is it clearer now?  If not, try replacing $ m, n$ by small numbers and using PEMDAS to convince yourself that the two expressions are the same.[/quote]
Sorry about the 6b-6b question, that was a dumb question.

So the minus sign before the parentheses changes the numbers inside to negative and changes the plus sign to a minus? Or did you just change the plus to a minus because -3-4 is the same thing as -3+(-4)?
[quote]I think both of those points confused you. Is it clearer now?[/quote]
Well when Simon was solving the equation step by step, what confused me was that the minus sign wasn't next to the parentheses until he pulled the "n" into the parentheses himself and that left the minus sign next to the parentheses.
So since it wasn't originally there, I think that's what was confusing me. Maybe I'm putting too much thought into it.
[quote]If not, try replacing M,N by small numbers and using PEMDAS to convince yourself that the two expressions are the same.[/quote]
Oh, ok so would the M and N be different numbers? And would they multiply with each other and then multiply with the coefficient?
Like for example 6mn can be 6,5,4 with m=5 and n=4. How would I multiply that?

Thanks for the being patient with me guys, I think I'm slowly getting it.
[quote]Also, for certain coefficients (the number in front of the letter, or variable), you do not need to show the variable. For any coefficient of 0, you do not need to show the coefficient because 0 times anything is nothing. Also, for any coefficient of 1, you do not need to write it out, because anything times 1 is itself. Beware: With -1, however, be sure to write the correct sign.[/quote]
Thanks. And with -1, it just changes the variable to a negative? Like -1b would just be -b?

If you have $ 6mn$, using PEMDAS, you know to multiply from left to right, so you have $ \left(6 \cdot m\right) \cdot n$. For example, if $ m\equal{}5$ and $ n\equal{}4$, $ 6mn\equal{}\left(6 \cdot m \right) \cdot n\equal{}\left(6 \cdot 5\right) \cdot 4\equal{}\left(30\right) \cdot 4 \equal{} 120$.

And yes, a $ \minus{}1$ changes a variable to negative, because $ \minus{}1b$ means $ \minus{}1 \cdot b$, which means $ \minus{}b$. Another example; you can write $ 1x$, or just $ x$, because one times itself is just $ x$. Think about it. If you have $ x\equal{}3$, and $ 1x$, $ 1\cdot x\equal{}x$, right?

Well, actually, multiplication is commutative (I'm not going to argue about vectors), which just means that you can multiply 6mn like this: (6*m)*n

or like this: 6*(m*n)

and can also be written as 6nm, but the variables are normally listed in alphabetical order.

Ok thanks, that makes sense.
Regarding Simon's post above; he pulled the "n" into the parentheses himself, which left the minus sign next to the parentheses. Now that the minus sign is next to the parentheses, that means the sign of everything in the parentheses is changed. But my question is the minus sign wasn't originally next to the parentheses; Simon pulled the "n" into the parentheses himself, which changed the equation and left the minus sign next to the parentheses. How is he allowed to do that?
Thanks.

Edit: Oh and another thing that's been bugging me is when I see an expression with subtraction in it and people simplify it by moving around the terms. Doesn't that mess up the final answer?
For example, on this page: http://www.wtamu.edu/academic/anns/mps/math/mathlab/beg_algebra/beg_alg_tut11_simp.htm

Halfway down the page, he has an expression on example 3.
5x² + 7x - 2x² - 10x + 5.
Ok, that expression has minus signs in it, and since the associative property and commutative property doesn't work for subtraction, how is he able to switch around the terms and simplify it?
Here's a picture of how he simplified it: http://www.wtamu.edu/academic/anns/mps/math/mathlab/beg_algebra/beg_alg_tut11ex3b.gif
He moved the subtraction around. How was he able to get away with that?

Thanks again.

Remember that that n was a negative, though. So it was equal to -1*n. Then it was just the associative property. Actually, some people will just multiply the whole thing out, negative sign and all.

e.g.
$ 3\minus{}5n(6n\plus{}2)$
$ \equal{}3\minus{}30n^2\minus{}10n$
$ \equal{}\minus{}30n^2\minus{}10n\plus{}3$

It's easy to think about if you think of it as multiplying by each term. You just have to remember that each sign is part of the following coefficient. Same with your second question.

[quote]Remember that that n was a negative, though. So it was equal to -1*n. [/quote]
So in the equation 5mn - n, the n is negative? Why? It's just being subtracted.
[quote]You just have to remember that each sign is part of the following coefficient. Same with your second question.
[/quote]
I'm sorry, I still don't understand. In this equation for example: 5x² + 7x - 2x²
He turned it around and switched it to 5x² - 2x² + 7x. It turns out to be the same answer, but I thought you couldn't move subtraction around because it could potentially change the answer.

[quote="The Zuton Force"]5x² + 7x - 2x² = 5x² + 7x + (-2x²)

Since these are just being added, we can add them in any order (Commutative Property of Addition)

= 5x² + (-2x²) + 7x =  5x² -2x² + 7x[/quote]
Ok that makes sense. So any equation with subtraction in it, you can change them all into addition by changing the sign of the following term?
Thanks for being so patient with me guys, I don't know why this is so hard for me to understand. You'd think the simplifying part would be the simple part, but apparently not. At least for me.
I still have a few more questions.
With that first example 5mn - n(6m - 2m²):
Simon made the n go inside the parentheses, which left the minus sign use the distributive property for everything in the parentheses. But I don't understand some things about that.
I thought you were supposed to drop the parentheses first, which would mean the minus sign wouldn't be able to use the distributive property for everything in the parentheses, because the parentheses are gone.

Edit: And I have another question with an example.
For example in this equation: 3x² + x(-5x + 4)
It gets simplified like this 3x² + (-5x) + 4x?
And then 3x² + (-1x)?
Or did I do that all wrong?

EDIT: actually, that was me being dumb.
3x² + x(-5x + 4) becomes 3x^2 + [b](-5x^2)[/b] + 4x, because the x distributes. Sorry, I wasn't looking at it correctly.

As for the negative sign distributing, think of it as $ \minus{} 1n(6m \minus{} 2m^2)$. Now, if the $ n$ distributes, you have $ \minus{} 1(6mn \minus{} 2m^2n)$, but then you have to distribute the $ \minus{} 1$ as well.

[quote="davidyko"]For your second example/question, yes, that is correct.[/quote]
Ok the reason I asked is because that's how I would simplify it with my knowledge, but the website I go to has it simplified as
-2x² + 4x. Are both of these simplifications acceptable?
[quote]As for the negative sign distributing, think of it as $ \minus{} 1n(6m \minus{} 2m^2)$. Now, if the $ n$ distributes, you have $ \minus{} 1(6mn \minus{} 2m^2n)$, but then you have to distribute the $ \minus{} 1$ as well.[/quote]
Ok, but when the N distributes, won't the parentheses disappear? Because that's the first order of operations in simplifying; removing the parentheses.

Thanks.

I take what I said back; see my edited previous post.
As for removing the parentheses, the reason why they don't disappear is because, well, you haven't completely distributed everything. You've distributed the $ n$ part of $ \minus{}1n$, but you still have to do it for the $ \minus{}1$ part. Alternatively, you could just distribute $ \minus{}1n$, i.e. multiplying every term inside by $ \minus{}1n$.

[quote="davidyko"]EDIT: actually, that was me being dumb.
3x² + x(-5x + 4) becomes 3x^2 + [b](-5x^2)[/b] + 4x, because the x distributes. Sorry, I wasn't looking at it correctly.[/quote]
NP. So when the x distributes among the -5x and the 4, the -5x becomes -5x². Ok that makes sense. Now that it's -5x², it can combine with the 3x² since they're now "like terms".
Ok thanks for clearing that up for me.
[quote]As for removing the parentheses, the reason why they don't disappear is because, well, you haven't completely distributed everything. [/quote]
Ok, so when the numbers or variables get distributed among the terms in the parentheses, they leave their current position and go inside the parentheses?
But the minus sign just stays where it's at and affects the terms in the parentheses?
[quote="davidyko"]You've distributed the $ n$ part of $ \minus{} 1n$, but you still have to do it for the $ \minus{} 1$ part. Alternatively, you could just distribute $ \minus{} 1n$, i.e. multiplying every term inside by $ \minus{} 1n$.[/quote]
So the N is actually -1N? Why does the subtraction sign turn the N into a negative N? I think I'm just getting mixed up with seeing subtraction signs being treated as negative signs and I don't know why.

[quote="leroyjenkens"]So the N is actually -1N? Why does the subtraction sign turn the N into a negative N? I think I'm just getting mixed up with seeing subtraction signs being treated as negative signs and I don't know why.[/quote]
Remember in the original equation N was being subtracted. So taking a look at that first part, $ 5mn\minus{}n$. n was being subtracted, so that equation would be the same as $ 5mn\plus{}(\minus{}n)$.

[quote="ckck"][quote="leroyjenkens"]So the N is actually -1N? Why does the subtraction sign turn the N into a negative N? I think I'm just getting mixed up with seeing subtraction signs being treated as negative signs and I don't know why.[/quote]
Remember in the original equation N was being subtracted. So taking a look at that first part, $ 5mn \minus{} n$. n was being subtracted, so that equation would be the same as $ 5mn \plus{} ( \minus{} n)$.[/quote]
I thought about that, but when Simon did it, he kept the subtraction sign instead of turning it into an addition sign.

Edit:
I've been checking out other websites and I encountered an example that I don't understand.
Here's how he ordered them:
3x – (2 – x)
= 3x + (–1)[2 + (–x)]
= 3x + (–1)(2) + (–1)(–x)
= 3x – 2 + x
= 4x – 2

Why didn't the (-1)2 turn the 2 into a negative 2?
And why did the 3x get turned into a 4x by adding a negative x?

Edit 2: I also ran into another question. 3x(4x)= 7x² correct?

[quote="leroyjenkens"][quote="ckck"][quote="leroyjenkens"]So the N is actually -1N? Why does the subtraction sign turn the N into a negative N? I think I'm just getting mixed up with seeing subtraction signs being treated as negative signs and I don't know why.[/quote]
Remember in the original equation N was being subtracted. So taking a look at that first part, $ 5mn \minus{} n$. n was being subtracted, so that equation would be the same as $ 5mn \plus{} ( \minus{} n)$.[/quote]
I thought about that, but when Simon did it, he kept the subtraction sign instead of turning it into an addition sign.[/quote]
They're the same thing.

OK. Here's a breakdown of multiplying by -n.

$ \minus{}n(3n\plus{}5)$

$ \equal{}(\minus{}1)(n)(3n\plus{}5)$  Because -n = -1*n

$ \equal{}(\minus{}1)[(n)(3n\plus{}5)]$ Associative property

$ \equal{}(\minus{}1)(3n^2\plus{}5n)$

$ \equal{}\minus{}3n^2\minus{}5n$

In algebra, prepare to interchange negative signs and subtraction signs, so to speak. Because subtracting is the same as adding the additive inverse (the negative).

[quote="jbano"][quote]I've been checking out other websites and I encountered an example that I don't understand.
Here's how he ordered them:
3x – (2 – x)
= 3x + (–1)[2 + (–x)]
= 3x + (–1)(2) + (–1)(–x)
= 3x – 2 + x
= 4x – 2

Why didn't the (-1)2 turn the 2 into a negative 2?
And why did the 3x get turned into a 4x by adding a negative x?
[/quote]
try it this way
3x-(2-x)
=3x+(-1)[2+(-x)]
=3x+(-1)(2)+(-1)(-x)
=3x+(-2)+(x)                                                                                (negative times a negitive is positve i.e. -1(-2)=2)
=4x+(-2)
=4x-2

[quote]Edit 2: I also ran into another question. 3x(4x)= 7x² correct?[/quote]

yes your right

hope it helps[/quote]
For the $ (3x)(4x)\equal{}7x^{2}$, you are multiplying them, not adding them. So $ (3x)(4x)\equal{}(3\cdot4)(x\cdot x)\equal{}12x^{2}$

[quote]jbano wrote:
Quote:
I've been checking out other websites and I encountered an example that I don't understand.
Here's how he ordered them:
3x – (2 – x)
= 3x + (–1)[2 + (–x)]
= 3x + (–1)(2) + (–1)(–x)
= 3x – 2 + x
= 4x – 2

Why didn't the (-1)2 turn the 2 into a negative 2?
And why did the 3x get turned into a 4x by adding a negative x?

try it this way
3x-(2-x)
=3x+(-1)[2+(-x)]
=3x+(-1)(2)+(-1)(-x)
=3x+(-2)+(x) (negative times a negitive is positve i.e. -1(-2)=2)
=4x+(-2)
=4x-2

Quote:
Edit 2: I also ran into another question. 3x(4x)= 7x² correct?

yes your right

hope it helps

For the , you are multiplying them, not adding them. So [/quote]

sorry silly mistake

I think your biggest problem right now is this: you are having trouble understanding that adding a negative is the same thing as subtracting the positive.

To clarify, here's an example:

3x+(-1)

=3x-1

Do you see why?

This applies to all of your calculations.

Try this: the next time you're multiplying by a negative, just think of it as changing the sign in front of each term.


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First, I apologize if this be not the right forum,
and also the syntax of my expressions, I'm Hispanic,
I need help with the following:

[b]My turn to simulate a collective Risks Model, what I have is this:[/b]

A process of ruin is defined as = U (t) = u + ct - S (t), t> = 0, where U (t) is the insurance capital at time t. u = U (0) initial capital, c, (constant) = a premium income per unit time.
S (t) = X1 + X2 + ... + XN (t) with N (t) the number of claims up to time t, and Xi, the value of the ith claim, assumed nonnegative.

In this case, I think it all boils down to ensure that U (t) is a vector graphics, that will indicate, on a graph against time capital,e to be the core capital, and the X axis over time, whether it has reached a state of ruin or not.For here I come. What I have is this, and this is where my confusion begins:

Take S = x1N1 + x2N2 + ... + XmNm for a number of complaints xi, assuming that all xi are different, and denoting Neither frequency of the claimed amount xi,ie the number of terms in S with value xi.
Then I proceeded to calculate the distribution of the sum S = x1N1 + x2N2 + ... + XmNm, convolution using Panjer recursion,
as follows: I assumed, arbitrarily, that S is distributed as a compound Poisson,where for the sum S = XiNi, the number Ni ~ Poisson (λi).

And that's what I have, by way of example, it would be something like this: For an arrival process claim,
which follows a Poisson distribution with average λ claims, suppose λ = 4 y Pr[X = 1, 2, 3] = 1/4, 1/2, 1/4. Then the sum S = x1N1 + x2N2 + ... + XmNm has been calculated by convolution,
using Panjer recursion, ie, we calculated: Pr [S = x].

for example, the calculation of P [S = x] to X = 3 has given us, in turn: f(0) = f(1) = e^-4, f(2) = 5/2*e^-4, f(3) = 19/6*e^-4.

So I'm reading this way: The probability that there are three claims is equal to e^-4,
you have 1 = e ^ -4, which has 2 = 5 / 2 * e ^ -4, and has 3 = 19 / 6 * e ^ -4.

I wonder if I did well this interpretation of the model. On the other hand, then the sum S = x1N1 + x2N2 + ... + XmNm, what would it be? the sum of each of these chances?, ie it is the distribution of the sum?,
I think if so, the insurer could estimate the probability of ruin.

Well, that's all, I accept opinion.

A greeting and thank you very much

[quote="extranger"]
So I'm reading this way: The probability that there are three claims is equal to e^-4,
you have 1 = e ^ -4, which has 2 = 5 / 2 * e ^ -4, and has 3 = 19 / 6 * e ^ -4.

I wonder if I did well this interpretation of the model.
[/quote]

No, $ S$ is the total claim amount, not the number of claims $ ( \equal{} N_1 \plus{} N_2 \plus{} \ldots \plus{} N_m)$.

So the probability that the total claim amount is 0 is $ e^{ \minus{} 4}$, probability that the total claim amount is 1 is $ e^{ \minus{} 4}$, probability that the total claim amount is 2 is $ \frac {5}{2}e^{ \minus{} 4}$ and the probability that the total claim amount is 3 is $ \frac {19}{6}e^{ \minus{} 4}$.

[quote="extranger"]

On the other hand, then the sum S = x1N1 + x2N2 + ... + XmNm, what would it be? the sum of each of these chances?, ie it is the distribution of the sum?,
I think if so, the insurer could estimate the probability of ruin.

[/quote]

I'm not sure if I understand what you're asking. S is the distribution of the sum, which is the total claim amount (as above). You're adding up the total cost of claims of size $ x_1 (\equal{} x_1n_1)$, the total cost of claims of size $ x_2 (\equal{} x_2n_2)$ ... up to the total cost of claims of size $ x_m (\equal{} x_mn_m)$, with $ n_1$, $ n_2$, ... $ n_m$ being the respective numbers of claims.

You may wish to look at the paper
Dickson, D.C.M. (1995), "A review of Panjer's recursion formula and its applications", British Actuarial Journal , 1(1), pp. 107-124

Here they show numerical methods for estimating the probability of ruin $ \psi(u)$.

Hello, first of all thank you very much for your response,
helped me solve some doubts.
now I'd appreciate one more thing:
in this part.

A process of ruin is defined as = U (t) = u + ct - S (t), t> = 0, where U (t) is the insurance capital at time t. u = U (0) initial capital, c, (constant) = a premium income per unit time.
S (t) = X1 + X2 + ... + XN (t) with N (t) the number of claims up to time t, and Xi, the value of the ith claim, assumed nonnegative.

I think that lack little to simulate the risk model, but also confuses me that the "premium", Panjer calculate it suffice to replace in the formula? or still need more?.
what would be the calculations for the example given : ?

I deeply appreciate ...

greetings

[quote]
I think that lack little to simulate the risk model, but also confuses me that the "premium", Panjer calculate it suffice to replace in the formula? or still need more?.
what would be the calculations for the example given : ? [/quote]

I'm sorry, I don't understand what you're asking.

To simulate this process, we need a few parameters:
$ u$ - initial surplus
$ c$ - premium rate
$ \lambda$ - rate for the Poisson process $ N(t)$
$ P(x)$ - the distribution function for the claims $ X_i$

$ u$ and $ ct$ are deterministic, so we're only interested in when claims are made; that is, we want to know about the distribution of $ S$.
We can simulate the times of claims (since the time between Poisson events is Exponential), and then also the value of the claim. At each claim we check if the surplus is below zero, and if so the process is ruined.

Repeat to get multiple sample paths.

We wouldn't use Panjer here because the distribution of $ N(t)$ changes as $ t$ increases.

Hello, you're right in everything you say. 	Since the last inconvenience would help me understand one more thing.  I am using this book:

You can download it. I enclose a picture that would solve what I want to do.

 I think that would be a good simulation. The question I have is:

Here: rgamma (n, shape = alpha) / alpha, I can use other distribution, any other?.

I think it would only replace in the algorithm.

The other question is: is because it is an "gamma" they do?:

$ E[U(t)] \equal{} u \plus{} \theta . \lambda.tE[X]$ and $ Var[U(t)] \equal{} \lambda.t.E[X^2]$.

Note also that the algorithm is Wi <- rexp(n)/lab; Ti <- cumsum(Wi). Ans The expression means that calculates the exponential, you think they do because "The times between arrivals in a Poisson process is exponential?

Also, could you explain what you mean by "The claims are gamma $ (\alpha, \alpha)$ distributed with $ \alpha$ = 2". In short, I need to do me the great favor to explain the mathematics of this problem, I undertake to modify the algorithm.

Thank you very much so help!!

I am not completely convinced by the R code, but to answer your questions:

1) Yes, the gamma distribution is used for the claims distribution. You can replace this with whatever distribution you want (whether discrete or continuous or mixed).

2)   These are true for any distribution, provided that $ N(t)$ is a Poisson process.
Note that $ c \equal{} (1 \plus{} \theta)\lambda E[X]$ and $ E[S] \equal{} E[N(t)]E[X]$ so that
$ \\
E[U(t)] \equal{} E[u \plus{} ct \minus{} \displaystyle\sum_{i \equal{} 1}^{N(t)}X_i] \\
\equal{} u \plus{} (1 \plus{} \theta)\lambda t E[X] \minus{} \lambda t E[X] \\
\equal{} u \plus{} \theta \lambda t E[X]$

and $ Var[U(t)] \equal{} Var[u \plus{} ct \minus{} \displaystyle\sum_{i \equal{} 1}^{N(t)}X_i] \equal{} Var[S]$

The last line is because $ u$ and $ ct$ are deterministic functions; the only random component comes from the compound sum.

Also, because $ N(t)$ is Poisson distributed with rate $ \lambda t$, we have:
$ \\
E[S] \equal{} \lambda t E[X] \\
Var[S] \equal{} \lambda t E[X^2] \\
E[(S \minus{} E[S])^3] \equal{} \lambda t E[X^3]$

These last three lines can be proved by considering the cumulant generating function of $ S$.

3) Yes, that's why they simulate the exponential distribution.

4) Gamma($ \alpha, \alpha$) is just their chosen claims distribution. As in point (1) above, this can be replaced with the claims distribution of your choice - this is the distribution of $ X_i$.

Hi! Many thanks for all your support, I appreciate it greatly.

I'm better now to write my work. You program in R? Why do you say that he is not convinced of the code?

I'd like to know

The value of $ c \equal{} 1 \plus{} \theta.\lambda.E(X)$ is always the same?, that is standard for these problems of the collective risk?

I've noticed that if I increase the value of $ \theta$ ,	in the program R, there is never ruin; on the contrary decreased if there is debris.

And there's something I do not understand this code: N <- rep(Inf, nSim), and later say: N <- N[N<Inf]; resulting vector with a length of 745.

And they say that is: The 745 finite ruin claim numbers N[N<Inf] have a maximum of 255. But, yes there numbers greater than infinity? where they come from?

Thank you very much again

code:

[quote="extranger"]
I'm better now to write my work. You program in R? Why do you say that he is not convinced of the code?
[/quote]

I understand a bit of R but not in great detail. I haven't had a chance to copy the code and look at it in R; perhaps I will have time a little later today.
It seems to me that they are simulating the value of the process after the first 400 claims, and doing this 10000 times, but are adding up the claims of these 10000 simulations, which doesn't make sense. Perhaps I am just misinterpreting the code and it would become apparent that I am wrong if I tried running it. I am not completely familiar with all of the functions being used, so my error may also be occurring there.

[quote]
The value of $ c \equal{} 1 \plus{} \theta.\lambda.E(X)$ is always the same?, that is standard for these problems of the collective risk?

I've noticed that if I increase the value of $ \theta$ ,	in the program R, there is never ruin; on the contrary decreased if there is debris.
[/quote]

$ c \equal{} (1 \plus{} \theta)\lambda E[X]$ is the rate of premiums received per time period. This is $ (1 \plus{} \theta)E[S(1)] \equal{} (1 \plus{} \theta)E[N(1)]E[X]$. $ \theta$ is the relative security loading for the insurer. Of course if it increases, then ruin will decreases - you're getting more premiums which offset the small claims! And if it decreases, then you have less premiums to cover your claims, so you will get ruined more often.

[quote]

And there's something I do not understand this code: N <- rep(Inf, nSim), and later say: N <- N[N<Inf]; resulting vector with a length of 745.

And they say that is: The 745 finite ruin claim numbers N[N<Inf] have a maximum of 255. But, yes there numbers greater than infinity? where they come from?

[/quote]

I haven't looked at the results so I can't say.

Hi, how I feel now with this new course, but I want to know how they calculate this value:

When they say:

"Therefore $ \Psi(u) \approx{} \Psi_{400}(u)$, Hence a good estimate of the ultimate ruin probability is"

$ \Psi(u) \approx{} 7.5$%

According to the book, the ruin probability is calculated as:

$ \Psi(u) \equal{} \Psi(0)*e^{ \minus{} Ru}$, where $ R \equal{} \displaystyle\frac {\theta}{(1 \plus{} \theta).\mu_1}$ is the coefficient of adjustment, and $ \Psi(0) \equal{} \displaystyle\frac {1}{1 \plus{} \theta}$

I replaced with the values that are there for

$ \equal{} 7.5, \theta \equal{} 0.3$, and if I am wrong, the sample say $ \mu_1 \equal{} E(X) \equal{} 1$.

As I say, replacing these values in $ \Psi(u) \equal{} \Psi(0)*e^{ \minus{} Ru}$

But do not get that value of 7.5%

Get is 13%

what shall I do wrong?

Thank you very much for your help again

if you help me clarify this.

greetings

$ \Psi(0) \equal{} \dfrac{1}{1\plus{}\theta}$ and $ R \equal{} \dfrac{\theta}{(1 \plus{} \theta)\mu_1}$ only if claims are exponential.

Ok. thank you very much, but how do the calculation there?

greetings

Hi, I found this in the book:

We can approximate the ruin probability by a gamma distribution:

$ \Psi(u) \equal{} \displaystyle\frac {1}{1 \plus{} \theta}(1 \minus{} G(u, \alpha, \beta)), u \geq{0}$

I do not know how to calculate it, I really appreciate your help,

many greetings

What have you tried to do? Show me your working.

Use the expressions for $ E[L_j]$ (equation 4.73), $ E[L^2]$ (equation 4.76) to solve for the parameters  in equation (4.80). $ mu_k$ are the non-central moments of the claims distribution - which is gamma$ (\alpha, \alpha)$ in this case, so $ \mu_k \equal{} \dfrac{\Gamma(\alpha \plus{} k)}{\Gamma(\alpha)\alpha^k}$

Hi, sorry,
I can not find these equations in the book, http://www.ebookslab.info/modern-actuarial-risk-theory-using-r.html , page (108) and (109)

I only know that says

We can approximate the ruin probability by a gamma distribution:

$ \Psi(u) \equal{} \displaystyle\frac {1}{1 \plus{} \theta}(1 \minus{} G(u, \alpha, \beta)), u \geq{0}$

on the other hand say:

To fit the first two moments, the parameters α and β of the gamma cdf G(u;α,β )
must meet the following conditions:

$ E[L] \equal{} \displaystyle\frac {1}{1 \plus{} \theta} \displaystyle\frac {\alpha}{\beta}$
$ E[L^2] \equal{} \displaystyle\frac {1}{1 \plus{} \theta} ( \displaystyle\frac {\alpha^2}{\beta^2} \plus{} \displaystyle\frac {\alpha}{\beta^2})$

the equations that appointments can not find them (I mean that I listed are some examples)

Equation (4.80) that appointments
is this?:

$ E[L] \equal{} \displaystyle\int_{0}^{\infty} \Psi(u).du$

$ E[L^2] \equal{} \displaystyle\int_{0}^{\infty } 2u\Psi(u).du$?

It also says that

$ E[L_j^k] \equal{} \displaystyle\frac {\mu_{k \plus{} 1}}{\mu_1(k \plus{} 1)}$

the truth I'm a mess with it

greetings

I am using the version of Modern Actuarial Risk Theory without R, perhaps the equation numbers are different there.

We have:
$ E[L] \equal{} \dfrac{\mu_2}{2\theta\mu_1}$ and $ Var[L] \equal{} \dfrac{\mu_3}{3\theta\mu_1} \plus{} \dfrac{\mu_2^2}{4\theta^2\mu_1^2}$ so $ \dfrac{1}{2} E[L^2] \equal{}  \dfrac{\mu_3}{6\theta\mu_1} \plus{} \dfrac{\mu_2^2}{4\theta^2\mu_1^2}$

Then equate these with the conditions you have stated to solve for $ \alpha$ and $ \beta$: these arfe the parameters of your approximating gamma distribution.


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calculus computations

Find:

(1) $ \lim_{x \to \minus{} \infty} \left(\frac{\sin^3x}{x} \plus{} \cos^2x \plus{} 3^{1\minus{}x}\right)$

(2) $ \lim_{x \to 0} \tanh x \cdot \arccos \left(\frac1x\right)$

(3) $ \lim_{x \to 1^\minus{}} \arccos x \cdot \cos \left(\frac{1}{\ln x}\right)$

(1)
sin^3(x) is a bounded function then lim( sin^3(x)/x)(when x----> -infinity )=0
Lim3^(1-x) )(when x----> -infinity ) = infinity .Since cos^2(x) is a bounded function :
Lim( sin^3(x)/x + cos^2(x)+3^(1-x) ) = + infinity
(2)
 lim( tanh(x))(when x----> 0 )=0
arccos(1/x) is a bounded function then lim ( tanh(x)*arcos(1/x)) (when x----> 0 )=0
(3)
cos(1/ln(x)) is a bounded function and lim ( arcos(x)) (when x----> 0 )=0  so
lim ( arcos(x)* cos(1/ln(x)) ) (when x----> 0 )=0

$ \lim_{x\to{1}}\arccos(x)\cos(\frac {1}{\ln(x)})$

Let $ \frac {1}{\ln(x)} = u\Rightarrow{x = e^{\frac {1}{u}}}$

So we have

$ \lim_{u\to{ - \infty}}%Error. "acos" is a bad command.
(e^{\frac {1}{u}})\cos(u)$

Now we can see that

$ \forall{u}\in\mathbb{R}\quad{ - 1\leq\cos(u)\leq{1}}$

$ \Rightarrow\lim_{u\to{ - \infty}}\arccos(e^{\frac {1}{u}})\leq\arccos(e^{\frac {1}{u}})\cos(u)\leq\arccos(e^{\frac {1}{u}})$

$ \therefore\lim_{u\to{ - \infty}}\arccos(e^{\frac {1}{u}) = 0\leq\lim_{u\to{ - \infty}}\arccos(e^{\frac {1}{u}})\cos(u)\leq\lim_{u\to{ - \infty}}\arccos(e^{\frac {1}{u}}) = 0}$

$ \therefore\lim_{u\to{ - \infty}}\arccos(e^{\frac {1}{u}})\cos(u) = 0$

$ \boxed{\therefore\lim_{x\to{1^ - }}\arccos(x)\cos(\frac {1}{\ln(x)}) = 0}$

Ill come back later and do the rest if someone else hasnt. I made the change not because it was neccesarry but because it was more aesthetically pleasing and intuitive to me.

I don't know about you guys but I think arccos is only defined on [-1,1] so 2 is meaningless.

I guess (2) is to be interpreted as $ \lim_{x\to 0}i\tanh x \cosh ^{ \minus{} 1}\left(\frac {1}{x}\right)$, though I don't believe that limit exists (though both the left and right limits do).

Edit: Oops, wrong again, the limits do agree from both sides do agree and are indeed $ 0$ as christ said.

If x is bigger than one then the arccos of that value will just be imaginary...because if the x value is imaginary it turns into cosh which is unbounded


floor function
calculus
calculus computations

How to solve $ \sum\limits_{i\equal{}1}^{n\minus{}1}{\left\lfloor \frac{im}{n} \right\rfloor }$ ?

It makes sense to me to call it $ \sum_{i\equal{}0}^{n\minus{}1}\left\lfloor \frac{im}n\right\rfloor.$ Note that the $ i\equal{}0$ term is always zero, so this is no change.

Just to get an idea, let's look at a set of examples: say, $ n\equal{}6$ and vary $ m.$

$ m\equal{}1: \qquad 0\plus{}0\plus{}0\plus{}0\plus{}0\plus{}0\equal{}0$

$ m\equal{}2: \qquad 0\plus{}0\plus{}0\plus{}1\plus{}1\plus{}1\equal{}3$

$ m\equal{}3: \qquad 0\plus{}0\plus{}1\plus{}1\plus{}2\plus{}2 \equal{} 6$

$ m\equal{}4: \qquad 0\plus{}0\plus{}1\plus{}2\plus{}2\plus{}3\equal{}8$

$ m\equal{}5: \qquad 0\plus{}0\plus{}1\plus{}2\plus{}3\plus{}4\equal{}10$

$ m\equal{}6: \qquad 0\plus{}1\plus{}2\plus{}3\plus{}4\plus{}5\equal{}15$

Can you see any patterns there?

As for what would happen for larger $ m:$ denote the sum by $ S(m,n).$

Suppose $ m\equal{}kn\plus{}j,$ with $ 0\le j<n.$ Then

$ S(kn\plus{}j,n)\equal{}\sum_{i\equal{}0}^{n\minus{}1}\left\lfloor \frac{ikn\plus{}ij}{n}\right\rfloor\equal{}\sum_{i\equal{}0}^{n\minus{}1}ik\plus{}\sum_{i\equal{}0}^{n\minus{}1}\left\lfloor\frac{ij}{n}\right\rfloor$

$ \equal{}\frac{kn(n\minus{}1)}2\plus{}S(j,n).$

So $ S(28,6)\equal{}4\cdot 15\plus{}S(4,6)\equal{}60\plus{}8\equal{}68.$

There are hints all over this. Can you put it together?


If $ \sin x\plus{} \cos x\plus{} \tan x\plus{} \csc x\plus{} \sec x \plus{}\cot x\equal{}7$
and $ \sin 2x \equal{} a\minus{}b\sqrt{7}$ for $ a,b \in N$, then find the ordered pair $ (a,b)$

[hide="Answer"]$ (a,b)\equal{}(22,8)$[/hide]

[hide="Proof"]
Set $ u \equal{} \sin x$ and $ v \equal{} \cos x$ so $ (u \plus{} v)^{2} \equal{} 1 \plus{} 2uv$
so the given quetsion is
$ u \plus{} v \plus{} \frac {1}{u} \plus{} \frac {1}{v} \plus{} \frac {u}{v} \plus{} \frac {v}{u} \equal{} 7$
thsi gives $ (1 \plus{} u)(1 \plus{} \frac {1}{v}) \plus{} (1 \plus{} v)(1 \plus{} \frac {1}{u}) \equal{} 9$
hence we have $ (1 \plus{} u)(1 \plus{} v)(u \plus{} v) \equal{} 9uv$
thus we  have on expansion $ \frac {7uv \minus{} 1}{uv \plus{} 1} \equal{} u \plus{} v$
thus $ (7uv \minus{} 1)^{2} \equal{} (uv \plus{} 1)^{2}(1 \plus{} 2uv) \Rightarrow 49x^{2} \minus{} 14x \plus{} 1 \equal{} x^{2} \plus{} 2x \plus{} 2x^{3} \plus{} 2x \plus{} 4x^{2} \plus{} 1$
where $ x \equal{} uv$ note that the $ 1$ on both sides cancel out and assuming $ x \neq 0$ we have
$ x^{2} \minus{} 22x \plus{} 9 \equal{} 0$ thus $ \boxed{\sin 2x \equal{} 2uv \equal{} 2x \equal{} 22 \minus{} 8\sqrt {7}}$[/hide]

very nice solution pardesi  :)   :clap2:


linear algebra
combinatorics proposed
combinatorics

I get this one from linear algebra course, but I think it is quite interesting.
 Consider the permutations of $n$ elements.
 We are allowed to swap positions between two elememts i.e
 if we have $1,2,3,4$ and swap $2,4$ we get $1,4,3,2$

 a) Prove that there exists two permutations which we can't change one into another in less than $n-1$ swaps.
 b) Solve the general case: determine the minimum number of swaps need to change a given permutation in to another given one!!

Note: the special case a) also has a special solution, other than the one derived from b)

Do the identical permutation and the cycle $(1,2,\ldots,n)$ provide a proper example for the first one?

By this I mean the arrangements $1,2,\ldots,n-1,n$ and $2,3,\ldots,n,1$.

Here's something else which I think is correct, but haven't actually proved it, although it's very similar to what I said above. If that's correct, this should also be correct, I think.

Given permutations $a,b$, the minimal number of transpositions "to get from $a$ to $b$" is, in fact, the minimal number of transpositions used to write $a^{-1}b$, I think, so we can write $a^{-1}b$ as a product of disjoint cycles in a unique way up to a change in the order of the cycles. Assume these cycles have lengths $i_1,\ldots,i_k$ (also take into consideration cycles of length $1$, i.e. fixed elements). In this case, I believe the answer to be $i_1-1+i_2-1+\ldots+i_k-1=i_1+\ldots+i_k-k$.

Is this true?

I.e. your answer is $n-k$. It is my opinion too. I think, each transposition increase number of cycles by at most 1.

Indeed,
1) if we take $a$ and $b$ from distinct cycles, then we will join them;
2) if we take $a$ and $b$ from the same cycle, then we separate it into two cycles.

Right?

Yes!! But if we consider the special case a) alone:
Takes Grobber's example: $1,2,..n$ and $2,..,n,1$, and try ro prove that it must take at least $n-1$ transposes!! Then probably we may not think about the cycles, but instead in another direction. That's why there is a special solution for this case. This is the interesting point of the problem.
 Infact, some of my friends find the a) case more difficult  :D


Slove this equation: $ x\sqrt{1\plus{}\frac{1}{x}}\plus{}\frac{1}{x}\sqrt{x\plus{}2}\equal{}1\plus{}\sqrt{x\plus{}\frac{2}{x}\plus{}3}$

Note that $ (1 \plus{} \frac {1}{x})(x \plus{} 2) \equal{} x \plus{} \frac {2}{x} \plus{} 3$.

Rearranging a little yields:
$ \sqrt {1 \plus{} \frac {1}{x}}(x \minus{} \sqrt {x \plus{} 2}) \equal{} \frac {1}{x}(x \minus{} \sqrt {x \plus{} 2})$.

Putting all terms on one side,
$ (\sqrt {1 \plus{} \frac {1}{x}} \minus{} \frac {1}{x})(x \minus{} \sqrt {x \plus{} 2}) \equal{} 0$.

Thus, $ \sqrt {1 \plus{} \frac {1}{x}} \equal{} \frac {1}{x}$ or $ x \equal{} \sqrt {x \plus{} 2}$.

Solving the first equation gives:
$ \sqrt {1 \plus{} \frac {1}{x}} \equal{} \frac {1}{x}\rightarrow1 \plus{} \frac {1}{x} \equal{} \frac {1}{x^2}\rightarrow x^2 \plus{} x \minus{} 1 \equal{} 0\rightarrow x \equal{} \frac { \minus{} 1\pm\sqrt {5}}{2}$.

Solving the second equation gives:
$ x \equal{} \sqrt {x \plus{} 2}\rightarrow x^2 \minus{} x \minus{} 2 \equal{} 0\rightarrow(x \minus{} 2)(x \plus{} 1) \equal{} 0\rightarrow x \equal{} 2 \ \text{and}\ x \equal{} \minus{} 1$.

Thus there are four solutions: $ \boxed{x \equal{} {\frac {\sqrt {5} \minus{} 1}{2}}, {\frac { \minus{} \sqrt {5} \minus{} 1}{2}}, \\
2, \text{and}\minus{}1}$.

Thank. It's nice solution. Equation $ \Longleftrightarrow (\sqrt {1 \plus{} \frac {1}{x}} \minus{} \frac {1}{x})(x \minus{} \sqrt {x \plus{} 2}) \equal{} 0$ is nice.


[quote="andrewmath"]I was just doing the 2002 State Sprint Round and there's one I have no idea how to do:

What is the remainder when 2^87 + 3 is divided by 7?[/quote]

[hide]You should see that $2^{3}=8$ has a remainder of $1$ when divided by 7.

Calculating the remainders only,
$2^{87}+3={(2^{3})}^{28}\cdot 2^{3}+3 = 1^{27}\cdot 8+3=11=4$.

(Modular arithmetic is used but I don't know how to make the ≡ sign in LaTeX)

So the remainder is 4.[/hide]

[quote="andrewmath"]The actual exponent was 87, but what you did does work for that too. The answer is supposed to be 4.[/quote]

Sorry, I don't know why I changed that number;

I will edit.

[hide]i got 4, you do 2^87/7 remainder 1 because 2^3 = remainder 1 and 87 is divisible by 3. so you add 1 to 3 and get 4. correct me if i am wrong.[/hide]

[quote="vamathletes"][quote="undefined117"][hide]
or, for people like me who don't know those special little tricks (are they?), there is a pattern for the modulus of those terms:

$2^{1}+3 \equiv 3\mod{7}$

$2^{2}+3 \equiv 0\mod{7}$

$2^{3}+3 \equiv 4\mod{7}$

$2^{4}+3 \equiv 3\mod{7}$

$2^{5}+3 \equiv 0\mod{7}$

$\cdots$

so it repeats every three terms, so

$\frac{87}{3}\equiv 0\mod{3}$

so the third number in that pattern is $4$.
[/hide][/quote]

 wouldn't $2^{1}+3 \equiv 3\mod{7}$ be $2^{1}+3 \equiv 5\mod{7}$?

I'm not too good at modular arithmetic though, so I'm probably wrong.[/quote]

yup you are right...the pattern starts out as

$2^{1}+3 \equiv 5 \pmod 7$ so the pattern is $5,0,4,5...$

[quote="vamathletes"][quote="undefined117"][hide]
or, for people like me who don't know those special little tricks (are they?), there is a pattern for the modulus of those terms:

$2^{1}+3 \equiv 3\mod{7}$

$2^{2}+3 \equiv 0\mod{7}$

$2^{3}+3 \equiv 4\mod{7}$

$2^{4}+3 \equiv 3\mod{7}$

$2^{5}+3 \equiv 0\mod{7}$

$\cdots$

so it repeats every three terms, so

$\frac{87}{3}\equiv 0\mod{3}$

so the third number in that pattern is $4$.
[/hide][/quote]

 wouldn't $2^{1}+3 \equiv 3\mod{7}$ be $2^{1}+3 \equiv 5\mod{7}$?

I'm not too good at modular arithmetic though, so I'm probably wrong.[/quote]

haha oops, my bad :oops:


Use the following standard enthalpies of formation to estimate the N-H bond energy in ammonia: N(g), 472.7 kJ/mol; H(g), 216.0 kJ/mol; NH3(g) , -46.1 kJ/mol.

The first thing that's confusing me is that my teacher said that the enthalpy of formation of any element in it's standard state is 0. The values that are given to me in the problem are not diatomic so they're not bonded and the elements are also in thei standard state... Is tehre something simple that I'm just overlooking?

Your teacher is correct. An "element in its standard state" does not mean the element is not bonded with anything else (it does not mean that its just a single atom); standard state means how it is naturally found in nature (I think, I am not that good with chemistry either, but I believe that is how it works). For example, $ N$(g) is not the standard state of nitrogen; $ N_2$(g) is the standard state of nitrogen - nitrogen is found as the molecule $ N_2$ in nature, and $ N_2$ has an enthalpy of formation of 0 kJ/mol.


1. Let n be an integer with n> =2 . Show that n does not divide 2^n -1

2. Suppose that k >=2 and n[1], n[2], n[3],...,n[k]>=1 be natural number having the property
n[2]  | 2^n[1]  - 1
n[3]  | 2^n[2]  - 1
...
n[k]  |  2^n[k-1]  - 1
n[1]  |  2^n[k]   -1
show that n[1]=n[2]=...=n[k]=1

1. Consider a set of all such $n$, i.e., $N = \{ n : n\mid (2^n-1)\}$, and suppose it is not empty. Let $P$ be a set of prime divisors of elements of $N$. Let $p$ be the smallest elements of $P$. Then $p\mid n$ for some $n\in N$. Since $n\mid (2^n-1)$, we have also $p\mid (2^n-1)$ that implies $ord_p(2) \mid n$, where  $ord_p(2)$ is the [url=http://mathworld.wolfram.com/MultiplicativeOrder.html]multiplicative order[/url] of $2$ modulo $p$. Obviously, $ord_p(2)<p$ and since $p$ is the smallest prime divisor of $n$, we have $ord_p(2)=1$. In other words, $p\mid (2^1-1)=1$ that is impossible. A contradiction proves that $N$ is empty.

2. Assume that $n_1 n_2 \dots n_k>1$. Let $p$ be the smallest prime divisor of $n_1 n_2 \dots n_k$. Without loss of generality we assume that $p\mid n_1$. Then $p\mid 2^{n_k}-1$ implying that $ord_p(2) \mid n_k$. Since $ord_p(2)<p$ and $p$ is the smallest prime divisor, we have $ord_p(2)=1$ that is impossible. Therefore, $n_1 n_2 \dots n_k=1$ meaning that $n_1 = n_2 = \dots n_k = 1$.


[hide]
WLOG, let $|a| \le |b| \le |c|$. We have $|b+c| \le |a| \le |b|$, so $b,c$ have opposite sign (either can be zero also). Similarly, $|a+c| \le |b| \le |c|$ so $a,c$ have opposite sign (or zero). WLOG, assume $a,b \le 0$ and $c \ge 0$ (multiplying each term by $-1$ does not change the problem).

CASE 1: $a+b+c > 0$

Then $a+c > -b$ and both sides are positive so $|a+c| > |b|$, contradiction.

CASE 2: $a+b+c < 0$

Then $a+b < -c$ and both sides are negative so $|a+b| > |c|$, contradiction.

Hence we must have $a+b+c = 0$.[/hide]

Squaring $|a| \ge |b + c|$, we get $a^2 \ge b^2 + 2bc + c^2$.  Similarly, $b^2 \ge a^2 + 2ac + c^2$ and $c^2 \ge a^2 + 2ab + b^2$.  Adding, we get $0 \ge a^2 + b^2 + c^2 + 2ab + 2ac + 2bc = (a + b + c)^2$.  Therefore, $a + b + c = 0$.


function
integration
probability
advanced fields
advanced fields unsolved

Let $X_1,...,X_n$ be i.i.d. r.v.s. with continous distribution functions F.
Let $X= \max\{X_1,...,X_n\}$ and $Y=\min\{X_1,...X_n\}$. Prove the following statements.

i)
\[P\{Y>y|X=x\}=\frac{(F(x)-F(y))^{n-1}}{F(x)^{n-1}}, y<x\]
ii)
\[P\{X_k\leqq x|X=t\}=\frac{n}{n-1}\frac{F(t)}{F(x)}, x<t\]
\[P\{X_k\leqq x|X=t\}=1, x\geqq t\]
iii)
\[E(X_k|X=t) = \frac{n-1}{n \cdot F(x)} \int_{-\infty}^{t} ydF(y) + \frac{t}{n}\]

[quote]i)

\[P\{Y>y|X=x\}=\frac{(F(x)-F(y))^{n-1}}{F(x)^{n-1}}, y<x\][/quote]
For $x$ to be the maximum, one of the $X_j$ must equal $x.$ Suppose WLOG that $X_n=x.$

Now, what is the probablilty that $x$ is the maximum? For that to occur, every one of $X_1\,\dots,X_{n-1}$ must be less than or equal to $x.$ The probability of this happening is $F(x)^{n-1}.$

Now, given that, what it the probability that the minimum is greater than $y$? For that to happen, all $n-1$ of the random variables must lie in the interval $(y,x].$ The probability that one specific $X_j$ lies in that interval is $F(x)-F(y).$ For all $n-1$ of them to be in there, the probability is $(F(x)-F(y))^{n-1}.$

The result now follows by the definition of conditional probability.

[quote]ii)

\[P\{X_k\leqq x|X=t\}=\frac{n}{n-1}\frac{F(t)}{F(x)}, x<t\]

\[P\{X_k\leqq x|X=t\}=1, x\geqq t\][/quote]
Suppose $x<t.$ Choose the $k$ for which you will consider $X_k.$ With probability $\frac1n,$ we have chosen the maximum value and $X_k=t,$ in which case $X_k\le x$ would be false.

With probability $\frac{n-1}n,$ $X_k$ is not the maximum. Then we know that $X_k\le t$ and what we need to know is $P(X_k\le x\,|\,X_k\le t)=\frac{P(X_k\le x)}{P(X_k\le t)}=\frac{F(x)}{F(t)}.$

The original statement is not correct; the correct version is the reciprocal:

\[P\{X_k\le x|X=t\}=\frac{n-1}n\frac{F(x)}{F(t)}\]

The second part is just the meaning of "maximum."

The correct formula for iii) is this:

\[E(X_k|X=t)=\frac{n-1}{nF(t)}\int_{-\infty}^tx\,dF(x)+\frac tn\]

This is simply what you get integrating the result of ii), with the reminder that $P(X_k=t)=\frac1n.$


[quote=Obel1x]
$ 3^3 \equiv 1$ mod $ 7$
[/quote]
$3^6 \equiv 1$ mod $ 7$

Just use Fermat's Little Theorem and some modular bashing and you should get:

$ 2^{20} + 3^{30} + 4^{40} + 5^{50} + 6^{60} \equiv 4 + 1 + 4 + 4 + 1 = 14 \equiv {0} \pmod{7}$


induction
number theory
prime numbers
calculus
calculus computations

If $ (x_n)$ is a cauchy sequence prove that there exists a subsequence $ (x_{n_k})$ sucht that for all $ k > 0$ we have that

$ |x_{n_{k \plus{} 1}} \minus{} x_{n_k}| < \frac {1}{2^k}$.

[hide]I first show that for each k we can find a subsequence such that for all $ l<k$ we have that $ |x_{n_{l \plus{} 1}} \minus{} x_{n_l}| < \frac {1}{2^l}$.
which can be easily proven. So because the principle of induction we have then proven that that is true for all $ k > 0$.

Is this a sketch of a perfect proof ?[/hide]

Generally, to have a finite sequence of arbitrary length with some property is not the same as having an infinite sequence with that property. (For example, there are arbitrary long arithmetic progressions in the set of prime numbers, but there are no such infinite progressions).

[hide="I would define"]$ n_1$ to be the smallest integer such that $ |x_i\minus{}x_j|<\frac{1}{2^1}$ for all $ i,j\ge n_1$ and then $ n_{k}$ to be the smallest integer such that $ n_{k}>n_{k\minus{}1}$ and $ |x_i\minus{}x_j|<\frac{1}{2^{k}}$ for all $ i,j\ge n_{k}$[/hide]

Are you sure it will work, for example if we have found $ n_2$, but from $ n_2\plus{}1$ the sequence changes so strong that we cannot find an $ n_3$ such that $ x_{n_4}\minus{}x_{n_3}<\frac{1}{2^3}$, because $ n_3$ is fixed now.


A compressed air car for a school project...  It's been crippled but nothing wood glue can't fix.

is that a pinewood derby car i see on the top right?

jorian

Yep.  It got 2nd... :(

There is also an old Texas License Plate from when I lived there, and the Art of Problem Solvings, and Intro to Geometry...

They changed the trophies! I feel betrayed :( . The old ones were much cooler.

Some of my rookie football cards.  From Top left clockwise.
Philip Rivers, QB NC State, SD Chargers.
Shaun Alexander, HB Alabama, Seattle Seahawks.
LaDanian Tomlinson, HB TCU, SD Chargers.
Drew Brees, QB Purdue, New Orleans Saints via Trade.
And DeAngelo Hall, CB Virginia Tech, Atlanta Falcons.


calculus
integration
logarithms
calculus computations

[quote="kunny"]$p,q$ satisfies $px+q\geq \ln x$ at $a\leq x\leq b\ (0<a<b)$.

Find the value of $p,q$ for which the following definite integral is minimized and then the minimum value.

\[\int_a^b (px+q-\ln x)dx\][/quote]

What are the 'variables' on which this definite integral is assumed to be dependent ?

[quote="kunny"]You are to regard $a,b$ as constant numbers. :)[/quote]

As $\int\limits_a^b {\log x} \,\,dx$ does not depend on p and q, the part we may have to minimize is

$\int\limits_a^b {(px + q)} \,\,dx$ which is equal to

$p\frac{{(b^2  - a^2 )}}{2} + q(b - a)$ and is of the form $pA+qB$, with A and B constants.  For this to be minimized,  I think we need to have a relation between p and q to proceed.

[quote="kunny"]Sorry for my delay reply,vidyamanohar.

You are to argue the case that $y=px+q$ is the tangent of $y=\ln x$.

kunny[/quote]

The minimum is attained at $p = \frac{2}{{b + a}},\,\,q =  - 1 + \log (b + a)\,$.

and the minimum value is $(b - a)\log \left( {\frac{{b + a}}{2}} \right) + \int\limits_a^b {\log xdx}$

The answer of $p$ is right, but the rest is wrong. :(

To vidyamanohar:

Here is my solution.

Since $y=\ln x$ is concave up, we can auusme, rather than without loss of generality, that $y=px+q$ is the tangent line of $y=\ln x$ at $a\leq x\leq b$.

Now the tangent line to $y=\ln x$ at the point $(t,\ln t)\ (a\leq t\leq b)$  is $y=\frac{1}{t}(x-t)+\ln t$.

Thus $I=\int_a^b \left\{\frac{1}{t}x+\ln t-1-\ln x\right\}dx$

$=(b-a)\left[\frac{a+b}{2t}+(\ln t-1)\right]+(b-a)-(b\ln b-a\ln a)$.

Let $f(t)=\frac{a+b}{2t}+\ln t-1$, $f'(t)=\frac{1}{t^2}\left(t-\frac{a+b}{2}\right)$,

so for $t<\frac{a+b}{2}$,we have $f'(t)<0$, for $t>\frac{a+b}{2}$, we have $f'(t) >0$.

Therefore $f'(t)$ is minimal and minimized when $t=\frac{a+b}{2}$.

Consequently the desired answer is

$p=\frac{1}{t}=\boxed{\frac{2}{a+b}},q=\ln t-1=\boxed{\ln \frac{a+b}{2}-1},I=\boxed{(b-a)\ln \frac{a+b}{2}+b-a-b\ln b-a\ln a}$.

kunny


Hello

[quote][size=75]There are really only 40 verses that are in doubt in the New Testament, it only comes up to around 40 verses; that means that 99.9% of the New Testament is been verified, has been clarified, using manuscript evidence.[/size]

First of all, at the most basic level, the missionaries own mathematics do not add up. There are 7,947 verses in the New Testament (Table III). If 40 verses are in doubt then the percentage of the New Testament that is "verified" comes to: [100-40/7947*100] = 99.5%. In fact for the missionary's mathematics to be correct, we would have to inflate the number of verses of the New Testament from 7,947 to nearly 40,000[/quote]

okay if 99.5 % = 7907 verses then how do we find out how many verses = 99.9 %?

99.9 percent - 99.5 % = 0.4 what do i  do next?

knowing that 99.5% equals 7907, we divide by 199 and multiply by 200 to find how many verses there are in all, or 7947. then, divide by 1000 to find 0.1%, or 8 verses. so, 99.9% is equal to 7947-8, or 7939 verses.

how exactly is 99.9 percent = to 7947 verses? if 99.6% = 10, 000 verses then how is 99.9 % = to 7947?
the missionary said above that only 40 verses are doubtful in the  new testament. 10,000 - 40 = 9960 / 10000 =0.996 * 100 = 99.6 % 99.6 % = 10000 verses and here you're saying 99.9% = 7947?

that's not what i said. if you look at my post, i said that 99.9% is equal to 7939 verses.

also, how is 99.6% equal to 10000?


Let $ a,b,c \in R^{ \plus{} }$,Proof that $ \sum_{cyc}\frac {a}{a^2 \plus{}2bc} \leq \frac {a \plus{} b \plus{} c}{ab \plus{} bc \plus{} ca}$

[quote="apollo"]Let $ a,b,c \in R^{ \plus{} }$,Proof that $ \sum_{cyc}\frac {a}{a^2 \plus{} bc} \leq \frac {a \plus{} b \plus{} c}{ab \plus{} bc \plus{} ca}$[/quote]
 $ a\equal{}b\equal{}c\equal{}1$ not true  :)
I think it's better is
Let $ a,b,c \in R^{ \plus{} }$,Proof that $ \sum_{cyc}\frac {2a}{a^2 \plus{} bc} \leq \frac {3(a \plus{} b \plus{} c)}{ab \plus{} bc \plus{} ca}$

Sorry,It is $ \sum_{cyc}\frac{a}{a^2\plus{}2bc} \leq \frac{a\plus{}b\plus{}c}{ab\plus{}bc\plus{}ca}$

see my posts at http://www.mathlinks.ro/viewtopic.php?t=173359.... for a proof using the Vornicu Schur inequality


[quote="Michael Niland"]Denote the probability of an event E by P(E).

Show that for any events A and B

IP(A$ \cap$B) - P(A).P(B)I $ \leq$ 0.25[/quote]
It's obvious if $ A\cup B\equal{}\Omega,$ where $ \Omega$ is a sample space. :wink:


In this game, one person posts a series of words that can be made by taking the letters from another word.

For instance, "cable" "stable" "stretch"
are the clues. The answer is "stretchable"

And after solving the word, you post one of your own and the chain goes on.

So, I'll start:
Clues:
"dose" "goose" "hog" "dug" "due"

Yup, just find a possible solution and give more clues for the next person. Here, I'll make another:

"cult" "cat" "cut" "call" "or"


integration
function
calculus
limit
absolute value
calculus computations

we consider the sequence $ u_{n}\equal{}\int_{0}^{\frac{1}{2}}\frac{sin^{2}(\pi nx)}{tan(\pi x)}dx.$

study the serie defined by $ \sum_{n\geq 0}\frac{u_{n}^{\alpha}}{n^{\beta}}$ where $ (\alpha,\beta)\in\mathbb{R}^{2}.$

hi!
[color=darkblue][i]I presente my demonstration and I hope I have not made some catastrophic mistakes.

it is clear that the function $ x\in\ (0,\frac{1}{2})\to\frac{sin(\pi nx)^{2}}{tan(\pi x)}dx$  extends by continuity on $ 0$ and on $ \frac{0}{2}.$

This means that $ u_{n}$ is well defined for every $ n\in\mathbb{N}$.

we put ${ v_{n}=\int_{0}^{\frac{1}^{2}}}\frac{sin(\pi nx)^{2}}{\pi x}dx$ for every $ n\in\mathbb{N}.$

operate the change of variables $ z=\pi nx$  in the integral:

$ v_{n}=\frac{1}{\pi}\int_{0}^{\frac{n\pi}{2}}\frac{sin(z)^{2}}{z}dz=\frac{1}{2\pi}\int_{0}^{\frac{n\pi}{2}}\frac{1-cos(2 z)}{z}dz=\frac{1}{2\pi}\int_{0}^{n\pi}\frac{1-cos(u)}{u}du$

let $ C=\frac{1}{2\pi}\int_{0}^{n\pi}\frac{1-cos(u)}{u}du.$

it comes that:
$ v_{n}=C+\int_{\pi}^{n\pi }\frac{du}{u}du-\frac{1}{2\pi}\int_{\pi}^{n\pi}\frac{cosu}{u}du$

an intergation by parts gives:

$ \int_{\pi}^{n\pi}\frac{cosu}{u}du=\left[\frac{sinu}{u}\right]_{\pi}^{n\pi}+\int_{\pi}^{n\pi}\frac{sinu}{n^{2}}du=\int_{\pi}^{n\pi}\frac{sinu}{n^{2}}du$

or the function $ u: \to\frac{sinu}{u},$  increased  in absolute value by $ u\to\frac{1}{u^{2}},$  is intregable on $ [\pi,\infty)$,so it converges.

it remains near $ +\infty$, $ v_{n}=\frac{ln n}{2\pi}+\epsilon(n)$ with $ \lim_{n\to\infty}\epsilon(n)=0$
 therefore $ v_{n}\sim _{n\to\infty}\frac{ln n}{2\pi}$

or, $ u_{n}-v_{n}=\int_{0}^{\frac{1}{2}}sin(\pi nx) f(x)dx$ with $ f(x)=\frac{1}{tan(\pi x)}-\frac{1}{\pi x}$

il is classical to demonstrate that the $ f$ is continuous on $ 0$ and $ \frac{1}{2}$,so $ f$ is continuous on the compact $ [0,\frac{1}{2}]$,it is as a result borned,as a consequence: $ |u_{n}-v_{n}|\le\frac{\parallel{}f\parallel{}_{\infty}}{2}.$.

Sence, $ u_{n}\sim _{n\to\infty}\frac{ln n}{2\pi}.$

For $ (\alpha,\beta)\in\mathbb{R}^{2}$,we have:${ \frac{u_{n}^{\alpha}}{n^{\beta}}\sim _{n\to\infty}\frac{(ln n)^{\alpha}}{(2\pi)^{\alpha}}n^{\beta}}.$

by the theorem of BERTRAND,the serie of terme general $ \frac{u_{n}^{\alpha}}{n^{\beta}}$ converge if and only if $ \beta>1$ or $ \beta=1$ and $ \alpha\ge\-1.$[/i][/color]
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Radouane.


Does there exist non-zero integers $ a,b$ ,such that $ a2^n \plus{} b5^n$ is a perfect square for every integer $ n$?

Posted only [url=http://www.mathlinks.ro/viewtopic.php?t=231375]yesterday[/url], same forum....


Ring Theory
superior algebra
superior algebra solved

I am trying to find a subring S of a ring R such that S has a multiplicative identity and R not.
is taht possible?

Define on $\mathbb{Z} \times \mathbb{Z}$ addition as ever and multiplication as $(a,b)(c,d): =(ac,2bd)$.
Then there is no identity in the whole ring, but the subring $\mathbb{Z}$ identified with the elements of type $(a,0)$ has one.