$P, Q$ are on side $BC$ of an equalateral $\triangle{ABC}$ such that $BP=PQ=QC=\frac13 BC$. A circle is constructed with diameter $BC$. The extensions of $AP$, $AQ$ meet the circle at $X$, $Y$.
Prove: arc BX = arc XY = arc YC.

[hide]$\angle C=\angle B=\angle A= \frac{\pi}{3}$
We call the centre of the circle O.
$CQ=2OQ$, $AC=2OY$, and $\angle AQC=\angle OQY \Rightarrow \Delta AQC$ ~ $\Delta YQO$
\[\Rightarrow \angle OCY= \frac{\pi}{3}\]

We can prove on the same way that $\angle OBX=\frac{\pi}{3}$

\[\Rightarrow \angle XOY=\pi-\angle OBX-\angle OCY=\frac{\pi}{3}=\angle OCY=\angle OBX\]

So it follows that arc CY= arc XY= arc BX

QED!  :) [/hide]

Dear Jan,

  is this a correct kritirium  for the similarity of two triangles ? The equal angles must be between the proportional sides. Or not?

   Babis

You are absolutely right stergiu. I was a little too optimistic there. Still, it's no problem, we can still prove that they are similar:

In triangle $AQC$: \[\frac{AC}{\sin{Q}}=\frac{\frac{AC}{3}}{\sin{CAQ}}\]
So $\sin{CAQ}=\frac{\sin{Q}}{3}$

In triangle $OQY$: \[\frac{OY}{\sin{Q}}=\frac{\frac{OY}{3}}{\sin{QYO}}\]
So $\sin{QYO}=\frac{\sin{Q}}{3}$

\[\Rightarrow \sin{QYO}=\sin{CAQ} \Rightarrow \angle QYO=\angle CAQ \vee \angle QYO = \pi - \angle CAQ\]

Because $\angle CAQ < \frac{\pi}{3}$ and because $\angle QYO < \frac{\pi}{2}$, it follows that \[\angle QYO=\angle CAQ\]

So still QED after all  :)
Thanks for correcting me!

Jan , very well!

   I have made the same trigonometric solution in a slightly different  way. Yours is a little better , becouse I computed  $\sin CAQ $ ,and this is not necessary! I have posted the same problem in the Olympiad (solved)section to find and no trigo solution. I think I have  seen in the past a solution by producing AB , AC and consindering line XY or  a tangenr parallel to BC. We' ll see !!!

Babis

Jan , Some thoughts(and a solution):

  If AC meets the circle at D and line DO meets the circle at E , then OE // AB and since AB = 2 OE , we take BZ = 2ZO , where Z is the comon point of AE and OB. But BP = 2  PO , so E and X coinside. We have angle BOX = angle COD = 60 , so BX = R , the radius and we are done.

 babis

Jan,

Why it is $AC=2OY$ ? (In your first post here)

Thank you

[quote="levi"]$\triangle ABC,\ \triangle A'B'C',\ A=max\{ A,B,C\} ,\ A'=max\{ A',B',C'\}.$ [b]Then[/b]

$1^{\circ}.\ \ A=A',\ AB=A'B',\ BC=B'C'\Longrightarrow \triangle ABC\equiv \triangle A'B'C'.$

$2^{\circ}.\ \ A=A',\ \frac{AB}{BC}=\frac {A'B'}{B'C'}\Longrightarrow \triangle ABC\sim\triangle A'B'C'.$[/quote]

Levi, how did you obtain this?

[quote="stergiu"]Well , instead Jan , I' ll answer to you.

   We have $AC = BC = 2 OY $, since triangle ABC is equilateral  ,  O is the midpoint of BC and OY is radius. I think to have helped you.

Babis[/quote]

Oh! stupid of me  :oops:

Thank you babis!

[quote="stergiu"]Levi , thanks.I had never seen these nice observations , although they are not standard theory in geometry text books.This means also that the first  solution of Jan is correct.
  Jan , sorry!

Babis[/quote]

how did you prove Levi's observations?

Peerfect _radio ,

 I have not proven them untill now!I did not find time. The first was a little known , with a different formulation and has not difficult proof( I have it . take two cases) ect). The second observation is new to me ! Probably it is not difficult. I ' ll try it , just when I ' ll find a little time and i ' ll send a pdf file.

Babis


Let $ABC$ be a triangle with a right-angle at $B$. Let $M$ and $N$ be the midpoints of $AB$ and $AC$ respectively. If the lines $BN$ and $CM$ are perpendicular and $BC = 1$, what is the value of $CM^2$?

 :ewpu: help!

[hide]Since it says perpendicular, put the triangle on the coordinate system.

Let $B(0,0), \ C(-1,0),$ and $A(0,x)$.  Thus, $M(0,\frac{x}2), \ N(-\frac12,\frac{x}2)$.

The slope of MC is $\frac{x}{2}$
The slope of NB is $\frac{\frac{x}{2}}{-\frac12}=-x$

Since they are perpendicular, $\frac{2}{x}=x \implies x=\sqrt2$

So then $CM^2=1+\frac12=\boxed{\frac{3}{2}}$[/hide]

Isn't the slope of $NB$ $\frac{-2}{x}$  :?

The actual answer is supposedly $\boxed{\frac{3}{2}}$. I can see how he got that gradient now  :blush:

[quote="fireemblem13"]i was with u till the last sentence. i dont see where the 1 and the 1/2 come from. after u know x = $\sqrt{2}$
and it is the height of CBM, u can get CM
$CM^2$ = $CB^2 + BM^2$
$CM^2$ = $1 + \sqrt{2}^2$
$CM^2$ = $1 + 2$
$CM^2$ = $3$[/quote]

Little error there, $BM = \frac{\sqrt{2}}{2}$. And his reasoning is correct :)

[quote]Isn't the slope of $NB \frac{-2}{x}$[/quote]
i think so
slope of $MC = \frac{x}{2}$
since it's perpendicular to BN,
$\frac{-2}{x} = BN$

[quote]Little error there [/quote]
i see what u mean

I kinda skipped a few steps.  Sorry :oops:


let $ a,b,c\geq 0$. prove that
$ \frac{4a}{a\plus{}b}\plus{}\frac{4b}{b\plus{}c}\plus{}\frac{4c}{c\plus{}a} \plus{}\frac{ab^2\plus{}bc^2\plus{}ca^2\plus{}abc}{a^2b\plus{}b^2c\plus{}c^2a\plus{}abc}\geq 7$

[quote="VIMF"]let $ a,b,c\geq 0$. prove that
$ \frac {4a}{a \plus{} b} \plus{} \frac {4b}{b \plus{} c} \plus{} \frac {4c}{c \plus{} a} \plus{} \frac {ab^2 \plus{} bc^2 \plus{} ca^2 \plus{} abc}{a^2b \plus{} b^2c \plus{} c^2a \plus{} abc}\geq 7$[/quote]

ha ...

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


Έχω μπερδευτεί πολύ με τα ιδιοδιανύσματα σε πίνακα 3x3. Επειδή την τετάρτη δίνω γραμμική άλγεβρα χρειάζομαι κάποιος να βρει τα ιδιοδιανύσματα στον ακόλουθο πίνακα πολύ πολύ αναλυτικά εξηγώντας το κάθε στάδιο:

Α=1 -3 3
   3 -5 3
   6 -6 4

Οι ιδιοτιμές του είναι -2 και 4.

Επίσης θέλω κάποιος να μου πει έναν εύκολο τρόπο για να αποφασίζω πότε ένας πίνακας είναι διαγωνοποιήσιμος και πότε όχι. Εδώ για παράδειγμα γιατί είναι? Παρακαλώ βοηθείστε έχω αγχωθεί πολύ! Ευχαριστώ πολύ.

[quote="laracroftheoriginal"]Έχω μπερδευτεί πολύ με τα ιδιοδιανύσματα σε πίνακα 3x3. Επειδή την τετάρτη δίνω γραμμική άλγεβρα χρειάζομαι κάποιος να βρει τα ιδιοδιανύσματα στον ακόλουθο πίνακα πολύ πολύ αναλυτικά εξηγώντας το κάθε στάδιο:

Α=1 -3 3
   3 -5 3
   6 -6 4

Οι ιδιοτιμές του είναι -2 και 4.
[/quote]

Sorry, μόνο ένα παράδειγμα αντέχω να γράψω ;)

Πάρε την ιδιοτιμή 4, οπότε αν το $ \mathbf{v} \equal{} (x,y,z)^\text{T}$ είναι ιδιοάνυσμα, θα έχουμε:
\[ A\mathbf{v} \equal{} 4\mathbf{v}\implies x \minus{} 3y \plus{} 3z \equal{} 4x, 3x \minus{} 5y \plus{} 3z \equal{} 4y, 6x \minus{} 6y \plus{} 4z \equal{} 4z
\]
Επειδή η ιδιοτιμή $ 4$ έχει πολλαπλότητα $ 1$, θα έχεις μία ελεύθερη μεταβλητή, οπότε θέτεις π.χ. $ x \equal{} 1$ και λύνεις το σύστημα... και σου προκύπτει το ιδιοάνυσμα $ (1, 1, 2)^\text{T}$.

Ομοίως για την άλλη ιδιοτιμή, βγάζεις τα $ ( \minus{} 1,0,1)$ και $ (1,1,0)$.

[quote]Επίσης θέλω κάποιος να μου πει έναν εύκολο τρόπο για να αποφασίζω πότε ένας πίνακας είναι διαγωνοποιήσιμος και πότε όχι. Εδώ για παράδειγμα γιατί είναι? Παρακαλώ βοηθείστε έχω αγχωθεί πολύ! Ευχαριστώ πολύ.[/quote]

Ok, quick'n'dirty (αν και όχι 100% ακριβές): δεν θέλεις να υπάρχουν μιγαδικές ιδιοτιμές.

Edit: δεν είναι σωστό έτσι όπως το έγραψα (thanks re Dem)! Το σωστό είναι ότι πρέπει τα ιδιοδιανύσματα του πίνακα να παράγουν το χώρο σου. Sorry for any confusion :(

Hope this helps,

Durandal 1707

[quote="laracroftheoriginal"]Ευχαριστώ πολύ Durandal για την υπομονή σου. Έχω ακόμα μερικές απορίες. Στους πραγματικούς για να διαγωνοποιείται γράφουν οι σημειώσεις μου οτι πρέπει να υπάρχει μια βάση των ιδιοδιανυσμάτων. Αυτό μπορώ να το δώ αν η διακρίνουσα είναι διάφορη του 0; Επίσης τα ιδιοδιανύσματα που θα πάρω πρέπει να αποτελούν βάση;Και στην ιδιοτιμή 2 γιατί βγάζουμε 2 ιδιοδιανύσματα; συγγνώμη για τα πολλά μου ερωτήματα αλλά δε θέλω να χάσω το δεκάρι. :)[/quote]

Lara, sorry δεν το είχα γράψει σωστά στο προηγούμενο post :oops:

0. Οι σημειώσεις σου είναι σωστές, ξέχνα αυτό που είχα γράψει για τη διακρίνουσα (άλλο πράγμα είχα στο μυαλό μου, γι'αυτό το έσβησα). Sorry αν σε μπέρδεψα, I was way too sloppy :oops:

1. Αν μπορέσεις να βρεις $ n$ γραμμικά ανεξάρτητα ιδιοδιανύσματα, έχεις βάση. Αν τα γράψεις στήλη με στήλη παίρνεις έναν πίνακα $ P$ και ο $ P^{ \minus{} 1}AP$ είναι διαγώνιος; τα στοιχεία του είναι οι ιδιοτιμές του $ A$.

2. Αν ο πίνακας σου έχει $ n$ διαφορετικές ιδιοτιμές, τότε όλα καλά, είναι διαγωνοποιήσιμος.

3. Αλλιώς, σκέψου ότι ιδιοτιμές είναι ρίζες του χαρακτηριστικού πολυωνύμου και έχουν η καθεμία την αλγεβρική πολλαπλότητά τους (π.χ. οι μονές έχουν πολ/τα 1, οι διπλές 2, κοκ). Τώρα, κάθε ιδιοτιμή μπορεί να έχει το πολύ τόσα γραμμικά ανεξάρτητα ιδιοδιανύσματα (σκέψου πόσες σχέσεις προκύπτουν και για πόσους αγνώστους στο σύστημα που πας να λύσεις).

4. Η γεωμετρική πολλαπλότητα μιας ιδιοτιμής είναι η διάσταση του χώρου που παράγεται από τα ιδιοδιανύσματα που αντιστοιχούν σε αυτήν την ιδιοτιμή. Όταν η γεωμετρική πολ/τα είναι ίση με την αλγεβρική όλα είναι καλά. Αν δε συμβαίνει αυτό, ο πίνακας σου έχει ένα μηδενοδύναμο μέρος (μηδενοδύναμοι είναι οι πίνακες με $ A^k \equal{} 0$) που σου απαγορεύει να διαγωνοποιήσεις τον πίνακα και που εμφανίζεται ως block στη μορφή Jordan του πίνακα.

Hope this is better (και Dem, thanks for the catch!) :)

Cheerio,

Durandal 1707

Durandal ευχαριστώ και πάλι. Η βοήθεια σου είναι πολύτιμη. Έχω απορία μόνο σε ένα παράδειγμα. Ο πίνακας

Α=[2 1 0
    0 1 -1
    0 2 4]

με ιδιοδιανύσματα [x, 0, 0] και [x, x, -2x] γιατί δεν είναι διαγωνοποιήσιμος;

Επίσης στις σημειώσεις ο πίνακας

Β=[1 -3 3
    3 -5 3
    6 -6 4]

με χαρακτηριστικό πολυώνυμο -(x+2)^2 (x-4) γράφει ότι είναι διαγωνοποιήσιμος. Αυτό πως γίνεται αφού η ιδιοτιμή -2 έχει πολλαπλότητα 2;

Ευχαριστώ και πάλι.

Ο πίνακας Α έχι χαρακτηριστικό πολυώνυμο $ \minus{}(x\minus{}2)^2(x\minus{}3)$. Θα είναι διαγωνοποιήσιμος αν και μόνο αν υπάρχουν δύο (γραμμικά ανεξάρτητα) ιδιοδιανύσματα για την ιδιοτιμή 2 και ένα για την ιδιοτιμή 3. Αλλά για την ιδιοτιμή 2, έχουμε Α - 2Ι =

0  1   0
0 -1 -1
0  2   2

και μετά από μερικές πράξεις βρίσκουμε ότι τα μόνο ιδιοδιανύσματα του Α για την ιδιοτιμή 2 είναι της μορφής (χ,0,0). Η γεωμετρική πολλαπλότητα της ιδιοτιμής 2 είναι 1 ενώ η αλγεβρική της πολλαπλότητα είναι 2. Άρα ο Α δεν είναι διαγωνοποιήσιμος.

Για τον Β, έχουμε Β+2Ι=

3 -3 3
3 -3 3
6 -6 6

άρα ο Β έχει 2 γραμμικά ανεξάρτητα ιδιοδιανύσματα για την τιμή -2. Π.χ. τα (1,1,0) και (0,1,1). Σε αυτήν την περίπτωση και η αλγεβρική και η γεωμετρική πολλαπλότητα του -2 είναι 2.

[quote="laracroftheoriginal"]Τώρα έχει ξεκαθαρίσει κάπως το τοπίο. Τη γεωμετρική πολλαπλότητα όμως πως τη βρίσκω; Πρώτη φορά την ακούω. ευχαριστώ. :)[/quote]

Αν λ είναι ιδιοτιμή του Α, και Ε(λ) είναι το σύνολο όλων των ιδιοδιανυσμάτων του Α με ιδιοτιμή λ, (όλες οι λύσεις δηλαδή του (Α-λΙ)χ = 0) τότε το Ε(λ) είναι διανυσματικός χώρος και η διάστασή του είναι η γεωμετρική πολλαπλότητα του λ.

Ισχυέι πάντα ότι γεωμετρική πολλαπλότητα <= αλγεβρική πολλαπλότητα, και ο Α είναι διαγωνοποιήσιμος αν και μόνο αν ισχύει η ισότητα για όλες τις ιδιοτιμές του Α.

[quote="laracroftheoriginal"]Τώρα μπορώ να πώ οτι τα κατανόησα. Έχω όμως μια τελευταία απορία. Στον πίνακα

Α=[1 -3 3
     3 -5 3
     6 -6 4]

οι σημειώσεις γράφουν οτι για την ιδιοτιμή -2 ο ιδιόχωρος είναι ο x(1,1,0)+y(1,0,-1)
ενώ εγώ βρίσκω τον y(1,-1,0)+z(-1,0,1) δηλ η γραμμική θήκη <(1,-1,0),(-1,0,1)>. Τελικά ποιά είναι η σωστή λύση; Και αν είναι αυτή των σημειώσεων πώς τη βρίσκει; Ευχαριστώ θερμά και πάλι.[/quote]

Το ότι έχεις διαφορετική βάση για ένα χώρο (τον ιδιόχωρο) δε σημαίνει ότι αλλάζει ο ιδιόχωρος αυτός καθαυτός.

Cheerio,

Durandal 1707

ok. ευχαριστώ πολύ  :)


hi people ...
we all know the magic software (Matlab), the powerful tool for simple and complex computations ...

my question: is there any authorized certification that is givien upon passing a   cenrtian exam ...?

plz help and thx


An object is moving away from the light source at a rate of 4 units/s.  Find the rate of change of the intensity of illumination when it is 2 units from the light source.

I'm getting different answers.

rate of change = k(1/unit² second)???


search
combinatorics unsolved
combinatorics

We got fixed numbers $ n$ and $ k$.. How many permutations of the numbers 1,2,...n that contain exactly k increasing sequnces of max length there are?
For example permutation [color=brown]5[/color][color=blue]23[/color][color=green]146[/color] has exactly 3 increasing sequnces.

If I understand you correctly, you are asking how many permutations of $ [n]$ have $ k \minus{} 1$ descents?  The answer is the Eulerian number $ A(n, k \minus{} 1)$ -- here are some posts on the forum that mention them:
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1166327478&t=214478
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1166327478&t=213847
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1166327478&t=185691
http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=1166327478&t=170097


geometry
perimeter
calculus
inequalities
advanced fields
advanced fields unsolved

1)Prove that for a given volume, the minimum surface area is that of a sphere.

2) Prove that for a given perimeter, the maximum area is that of a circle.

P.S: Not sure if this belongs to calculus forum.

check this link http://www.cut-the-knot.org/do_you_know/isoperimetric.shtml

This belongs into Advanced Fields actually. Moved.

  darij

For a circle, there is an elementary proof of the isoperimetric inequality:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=167173

Darij, Thanks!! And Thanks alchemist & JBL.

Any idea for the other one?

I actually encountered these in physics and wanted a good mathematical proof for this. :D


Euler
analytic geometry
geometry
circumcircle
similar triangles
perpendicular bisector

How could one go about proving that the Euler line exists without a coordinate-bash?

One nice approach is with vectors.  If the origin is chosen to be the circumcenter and the three vertices are $ A, B, C$, then the orthocenter is $ A \plus{} B \plus{} C$ and the centroid is $ \frac{A \plus{} B \plus{} C}{3}$; those three points are clearly collinear.  It remains to prove the claim about the orthocenter, which is a pretty straightforward geometric argument.

[b]T0rajir0u[/b]; $ 10101$ -- what a perfect number of posts!

[hide="another solution"]
 [geogebra]dda756ff57eb4de3f95d10a5c03b90f8a75bc9d6[/geogebra]

Take a circle with circumcenter $ O$ and centroid $ C$ and extend $ OC$ to $ H$, such that $ OC: OH \equal{} 1: 2$

Now if we draw the median $ AM_a$, It passes through $ C$ and $ AC: CM_a \equal{} 2: 1$

Since $ \angle ACH \equal{} \angle ACM_a$ we have two similar triangles $ \bigtriangleup AHC \sim \bigtriangleup OCM_a$

Now since the perpendicular bisector of $ BC$ passes through $ O$ and $ M_aO \parallel{} HA \rightarrow HA$ is perpendicular to $ BC$. Or $ HA$ is the altitude of $ BC$.

If we do this for all sides of the triangle then it is clear that $ H$ is the othocenter.
Therefore $ OCH$ is collinear.

[/hide]

Next Questions:
1) Prove the nine-point Circle (that is: the three feet of the altitudes and the three medians of $ \bigtriangleup ABC$,  as well as the three midpoints between the othocenter and each vertex are all cyclic)

2) Prove that the nine-point center also lies on the Euler Line...

Almost all proofs except the most elementary are too difficult for High School Basics, and the Euler Line is a bit above elementary high school topics, so it belongs in a higher forum.

Moved to Intermediate.


algebra
polynomial
superior algebra
superior algebra unsolved

Prove that $\left(\sum_{k=1}^{n}(Y^{k}-1)X^{k-1}\right)-X^{n}-X^{n+1}$ is an irreducible polynomial in $\mathbb Z[X,Y]$ for odd $n\geq 3$.

If there were not the $-X^{n}$ or the $-X^{n+1}$ term, we could apply Eisenstein with $Y-1$. So perhaps one must substitute $X$ and $Y$?

I think $n\geq 3$ is enough without the odd condition.


easy...

If $ a, b, c$ are non-negative real numbers, then

$ \frac{a^2}{b^2} \plus{} \frac{b^2}{c^2} \plus{} \frac{c^2}{a^2} \ge 2(\frac{a^2}{c^2} \plus{} \frac{c^2}{b^2} \plus{} \frac{b^2}{a^2}) \minus{}3$.

[quote="filomen"]easy...

If $ a, b, c$ are non-negative real numbers, then

$ \frac {a^2}{b^2} \plus{} \frac {b^2}{c^2} \plus{} \frac {c^2}{a^2} \ge 2(\frac {a^2}{c^2} \plus{} \frac {c^2}{b^2} \plus{} \frac {b^2}{a^2}) \minus{} 3$.[/quote]

I think inequality is wrong.
Try $ a\equal{}3$ and $ b\equal{}c\equal{}1$.. The LHS $ \equal{}\frac{91}{9}$
and RHS $ \equal{}\frac{182}{9}\minus{}3 > \frac{91}{9}$


trigonometry
Duke
college
algorithm
vector
Pythagorean Theorem
geometry

Solve for $t$, where $\frac{-\pi}{4}\le t \le \frac{\pi}{4}$:

$\sin t+\cos t = \frac{\sqrt{6}}{2}$

The answer is above. There is an identity table that will help a lot in basic trigonometry. Altohugh they do get complex later, knowing the basics like the one above used will help a lot in the future.

OK, I should have seen that.  What confused me is how Duke did it on their solutions manual: ([url=http://www.math.duke.edu/dumu/contests/problems/index.html]1999 individual[/url])

It's $\sin t+\cos t = \sqrt{2}\sin \left( t+\frac{\pi}{4}\right)$ .

I have a partial solution but don't know how to finish.  I drew a right triangle with hypotneuse 1 and legs x and y and t as the angle between 1 and y.  using the given information and the pythagorean theorem, we have $x+y=\frac{\sqrt{6}}{2}$ and $x^{2}+y^{2}=1$.  solving gives $x=\frac{\sqrt{6}+\sqrt{2}}{4}$ and $y=\frac{\sqrt{6}-\sqrt{2}}{4}$ (or vice versa).  but how would one go about finding $\arcsin{\frac{\sqrt{6}+\sqrt{2}}{4}}$?

$\arcsin{\frac{\sqrt{6}+\sqrt{2}}{4}= 15^\circ}$  but that's just something I knew off the top of my head.  Not sure how to actually get there, (I can derive sin 15 though, so it's probably not hard to do it backwards).

I know how to derive $\sin{15^\circ}$, but how would you do that in reverse?  :maybe:

There's really no algorithm for that, you just kinda have to know it.

$\sin t=\frac{\sqrt{6}+\sqrt{2}}{4}$
$\sin^{2}t=\frac{8+4\sqrt{3}}{16}$
$1-2\sin^{2}t=-\frac{\sqrt{3}}{2}$
$=\cos 2t$

thats how you can solve it, then get rid of the extraneous roots

[quote="t0rajir0u"]We can generalize:

$A \cos t+B \sin t = \sqrt{A^{2}+B^{2}}\cos (t-\arctan \frac{B}{A})$[/quote]

Um...is this a proof for it?

[hide]
$A \cos t+B \sin t$

If we multiply by $\frac{\sqrt{A^{2}+B^{2}}}{\sqrt{A^{2}+B^{2}}}$, we get:

$(\frac{\sqrt{A^{2}+B^{2}}}{\sqrt{A^{2}+B^{2}}})(A \cos t+B \sin t)$

Now, if we set $\cos\theta = \frac{A}{\sqrt{A^{2}+B^{2}}}$, we see that $\sin\theta = \frac{B}{ \sqrt{A^{2}+B^{2}}}$.

Substituting back, we get : $(\sqrt{A^{2}+B^{2}})(\cos\theta \cos t+\sin\theta \sin t) \rightarrow (\sqrt{A^{2}+B^{2}})\cos(t-\theta)$

Recall, $\cos\theta = \frac{A}{\sqrt{A^{2}+B^{2}}}$, and $\sin\theta = \frac{B}{ \sqrt{A^{2}+B^{2}}}$;therefore, $\tan\theta = \frac{B}{A}\rightarrow \theta = \arctan \frac{B}{A}.$

Therefore, $A \cos t+B \sin t = \sqrt{A^{2}+B^{2}}\cos (t-\arctan \frac{B}{A})$

[/hide]

Why do people post repeats of solutions that are posted right before them?

*Cough*nutz_for2.718281828,erik-the-red*Cough*

'Cause ppl want to post them before checking if they were right. To prove to themselves that they know it.

[quote="13375P34K43V312"]Is there a complex number argument for that?[/quote]

Close.  "That" is actually a statement of two equivalent forms of the dot product (of the vectors $\left< A, B \right>, \left< \cos \theta, \sin \theta \right>$).


function logarithms algorithm probability parameterization calculus

i've been trying to find the definition for lamda for a while now. what does lambda mean in math and how is it used?

It's usually just used as a variable.  In particular, it is often used to denote eigenvalues.

Frequently also used to denote a scalar when dealing with vectors.

Lambda makes me think of Lagrange multipliers  :D

Well , lamda is a greek letter.This $\lambda$ denotes eigenvalues. Different people use it for differnetn things.
[u]Babis[/u]

And capital Lambda denotes Mangold's function in number theory.

thanks for the responses, but where can i read an introduction to lambda and its different uses? and how might it be used in a logarithm algorithm to generate exponentially distributed random numbers in a computer science program?

You must be joking.  It is JUST a variable.  In a given context it can have some meaning, but your question is essentially the same as asking 'where can I find an article on $x$ and its various uses'.  You need to specify more details for your question to have any meaning.

Maybe it means the Poisson distribution?  This is a discrete probability distribution which describes the number $X$ of randomly-occurring "rare events" which happen in a continuous interval.  It's based on the exponential series and $\lambda$ is usually used as the parameter for the mean number of events in an interval.
\[ P(X=n)=\frac{e^{-\lambda}\lambda^n}{n!} \]

Perhaps you mean the $\lambda$ calculus?

In computer science, the lambda calculus is a formal system designed to investigate function definition, function application, and recursion. It was introduced by Alonzo Church and Stephen Cole Kleene in the 1930s; Church used the lambda calculus in 1936 to give a negative answer to the Entscheidungsproblem. The calculus can be used to cleanly define what a computable function is. The question of whether two lambda calculus expressions are equivalent cannot be solved by a general algorithm, and this was the first question, even before the halting problem, for which undecidability could be proved. Lambda calculus has greatly influenced functional programming languages, especially Lisp.

SCNR ;)

Wavelength is also denoted by $\lambda$.


calculus
integration
geometry
calculus computations

Hi everyone,

Could someone please help me with this interval?

Find the area under the curve:

4x+y squared=12
y=x
(-6,-6) (2,2)

I solved for x and got -1/4y squared+3

I then solved the integral -6 to 2 -1/4y squared+3-ydy

Once I plugged in my values I got a negative value.

5/3-18+6+1

but it can't be a negative area

Could someone please help me with this?

Thank you very much

Is this what you were trying to solve?
$ \int^{2}_{\minus{}6}\sqrt{12\minus{}4x}dx$

Thank you

No, I think I'm trying to solve the integral from -6 to 2 y+1/4y squared+3 dy.

Does that look right? I wasn't given the integral. Based on the rest of the information we have to come up with the integral.

Thank you


What is the mean value of the five terms of the arithmetic sequence whose first term is 7 and whose last term is 17?

The average of the terms in an arithmetic sequence is the average of the first and last terms, or in this case, $ \frac{7\plus{}17}2\equal{}\boxed{12}$.


$ What$ $ is$ $ the$ $ remainder$ $ when$ $ 1234567890$ $ is$ $ divided$ $ by$ $ 99?$
[hide="Answer By Long division :( "]$ 72$[/hide]

This can be found easily without long division if you are familiar with the mechanics of modular arithmetic. I think it can also be solved even faster with Chinese Remainder Theorem, but I am not very good at applying it.
[hide="My Solution"]
(All the variables that I will use and implicitly define are integers.)

Let $ x \equal{} 1234567890$. We have $ x \equiv 0\pmod{9}$ and $ x\equiv 6\pmod{11}$ using basic modular arithmetic techniques. Therefore, we have $ x \equal{} 9m$ and $ x \equal{} 11n \plus{} 6$ which implies that $ 9m \equal{} 11n \plus{} 6\implies 2n \equiv 3\equiv 12\pmod{9}\implies n\equiv 6\pmod{9}$. So we have $ n \equal{} 9k \plus{} 6$. Substituting this back, we get $ x \equal{} 11n \plus{} 6 \equal{} 11(9k \plus{} 6) \plus{} 6\equal{} 99k \plus{} 66 \plus{} 6 \equal{} 99k \plus{} 72.$ Therefore, $ x \equal{} 1234567890$ leaves a remainder of $ \boxed{72}$ when divided by $ 99$.
[/hide]

Here is another, potentially faster solution that should work a bit more nicely:

[hide]
Notice that $ 10^2\equiv 1\pmod{99}$.  Thus, $ a*100^k\equiv a\pmod{99}$ and we can write $ 1,234,567,890 \equal{} 90 \plus{} 78\cdot100 \plus{} 56\cdot100^2 \plus{} 34\cdot100^3 \plus{} 12\cdot100^4$
$ \equiv90 \plus{} 78 \plus{} 56 \plus{} 34 \plus{} 12\equiv \minus{} 9 \minus{} 21 \plus{} 90 \plus{} 12\equiv 72\pmod{99}$.[/hide]

@sunehra's solution:  What you are doing is using the Chinese Remainder Theorem; CRT basically says that the method you went through will in fact produce a solution.

Oh wow, I feel dumb now. I read about it on wikipedia again and it finally sinked in that CRT only tells you about the existence and a characteristic of the solution(s) to a system of linear congruences. By the way, nice solution.


factorial
function
real analysis
real analysis unsolved

The other day I calculated that $\sum_{i=1}^{\infty}\frac{1}{(k*i)!}=\frac{1}{k}*( e^{(}r_{1})+e^{(}r_{2}) ...+e^{(}r_{k}) )$ where $r_{j}=e^{^}(\frac{2*\pi*i*j}{k})$

But the method I used does not seem to produce the correct answer for $\sum_{i=1}^{\infty}\frac{1}{(k*i+b)!}$

How would you approach this sum?

[quote="DavidShor"]The other day I calculated that $\sum_{i=1}^{\infty}\frac{1}{(k*i)!}=\frac{1}{k}*( e^{(}r_{1})+e^{(}r_{2}) ...+e^{(}r_{k}) )$ where $r_{j}=e^{^{}}(\frac{2*\pi*i*j}{k})$

But the method I used does not seem to produce the correct answer for $\sum_{i=1}^{\infty}\frac{1}{(k*i+b)!}$

How would you approach this sum?[/quote]

The proof for the first sum is easy.  Let $\zeta_{n}= \exp(2\pi i/n)$ be a primitive n(th) root of unity, and observe that for integer m, construct the sum

$\sigma(m;n) = \sum_{k=1}^{n}\zeta_{n}^{km}$.

Then $\sigma(n;n) = n$, and $\sigma(m;n) = 0$ if $m \ne n$.  Now note that

$\exp(z) = \sum_{k=0}^\infty \frac{z^{k}}{k!}$,

so

$\sum_{k=1}^{n}\exp(\zeta_{n}^{k}) = \sum_{m=0}^\infty \frac{1}{m!}\sum_{k=1}^{n}\zeta_{n}^{km}= \sum_{m=0}^\infty \frac{n}{(mn)!}$.

Dividing by $n$ throughout gives the desired result.   If we wished to extend this to summands of the form 1/(mn+b)!, then we need to construct a similar function to $\sigma$ except its sums vanish for all integers not of the form mn+b.

[quote="atomicwedgie"] $\sum_{m=0}^\infty \frac{1}{m!}\sum_{k=1}^{n}\zeta_{n}^{km}= \sum_{m=0}^\infty \frac{n}{(mn)!}$.
[/quote]
Maybe I don't understand something..
we have:
 $\sum_{m=0}^\infty \frac{1}{m!}\sum_{k=1}^{n}\zeta_{n}^{km}=\frac{1}{0!}\sigma(0,n)+\frac{1}{1!}\sigma(1,n)+...+\frac{1}{n!}\sigma(n,n)+...=\frac{1}{n!}\sigma(n,n)=\frac{1}{(n-1)!}$ because $\sigma(p,n)=0$ when $p$ is not equal to $n$.
How did you get: $\sum_{m=0}^\infty \frac{n}{(mn)!}$
 :maybe:

[color=red]Unnesessary quote removed by moderator: it never makes sense to quote the preceeding post in full.[/color]
I should have instead written $\sigma(m;n) = n$ when $m$ is a multiple of $n$, rather than when $m = n$.


NYSML (the New York State Math League) will take place on Saturday, April 8, 2006.  It will be held at Nyack High School in Rockland County.

(Thanks to Ravi B for this information.)

Who here is going, and with what team?

There is a person down my street that is going (and I live in the Vestal School district, wherever that is).

According to the new-and-improved [url=http://www.nysml.org]NYSML web site[/url], the member math leagues around the state are:

Albany Area Math Circle
Buffalo Public High School Math League
Dutchess-Ulster-Sullivan-Orange Math League
Genesee Valley Math League
Ithaca High School Math Team
Monroe County Math League ([b]my team!  :winner_second: way back in 1988[/b])
[i]Nassau County Interscholastic Math League[/i]
[i]New York City Interscholastic Math League[/i]
Niskayuna High School Math Team
Onondaga County Math League
Rockland County Math League
Southern Tier Interscholastic Math League
[i]Suffolk County Interscholastic Math League[/i]
Westchester-Putnam Math League

[i]I've italicized out the downstate teams.[/i]

ok, so just a comment, the url is misspelled (so the link doesn't work) it's http://www.nysml.org/
i will hopefully be there with the NYC team as assistant coach. good luck to everyone :)

[quote="frankthetank"]Whoa, I'm from Monroe County![/quote]I'm actually not.  :blush:  I'm from Genesee County (Batavia to be specific), but we were way too small to have our own team.

I'll be there (Monroe County team, in case you wanted to know.)

i'm like, really excited about it!!! i'm from monroe county

[quote="jazzy200084"]I'll be going on the DUSO team! Hopefully I'll improve upon my terrible NYSML performance last year and win the Buzz Competition...[/quote]

Whoa, someone else from DUSO? Hi.

I am not going to either NYSML or ARML this time around due to silly conflicts :-(

[quote="Adunakhor"]How many people are on each team for the different leagues?[/quote]

I believe the standard is for 15-man teams to participate in each of these competitions.  I wouldn't be surprised if each of the organizations send multiple teams--New York City has done that pretty much all of the time.

As for NYSML, I will certainly be there as an assistant coach for one of the New York City teams for my second year running.  Last year was a fantastic experience, and I'm looking forward to repeating it.

--KC

Thanks for the reply. I just got the registration form from my teacher yesterday; I'll be going with one of the teams from Westchester-Putnam  :)

I think my school is from the Southern Tier Interscholastic Math League.

The results of 2006 NYSML have been posted on the [url=http://www.nysml.org/]NYSML web site[/url].

Congrats to NYC A for winning Division A, Westchester-Putnam A for being the highest scoring upstate team, Albany Math Circle for winning Division B, and DUSO B for being the most improved team.

Darn! I didn't see this topic until after NYSML! Well, I was the youngest person on the Albany Area Math Circle team.

same here, and I'm the second youngest.
(but I didn't want to go to ARML anyway)


function
limit
complex analysis
complex analysis unsolved

Let $ A(z)$ be the branch of the arc-tangent function that is analytic everywhere except for $ {z: Re z=0,|Im z|\geq 1}$ and that has $ A(0)=\pi$. find $ lim Re [A(t+i)]$ as $ t\rightarrow 0^{+}$ ( t approaches 0 through positive reals)

\begin{align*} ArcTan(z) & = \frac {i}{2}log\left(\frac {i + z}{i - z}\right) \\
& = \frac {i}{2}\left[ln\left|\frac {i + z}{i - z}\right| + i\left\{Arg(\frac {i + z}{i - z}) + 2n\pi\right\}\right] \end{align*}

$ ArcTan(0) = \pi\Rightarrow n = - 1$

Then wish:

$ \begin{align*} \mathop\lim\limits_{t\to 0^{ + }}Re\left[ArcTan_{ - 1}(t + i)\right] & = \mathop\lim\limits_{t\to 0^{ + }} - 1/2Arg\left\{\frac {2i + t}{ - t} - 2\pi\right\} \\
& = - 1/2\left[Arg( - 1 - i\infty) - 2\pi\right] \\
& = - 1/2( - \pi/2 - 2\pi) = 5\pi/4$

I believe the limit depends on the path (news to me if I'm right):

$ \mathop\lim\limits_{z\to i^{\gamma}} Re[ArcTan_{ - 1}(z)] = a;\quad a\in(\pi/2,3\pi/2)$

For example, what must the straight-line contour $ \gamma(t)$ be for:

$ \mathop\lim\limits_{z\to i^{\gamma}} Re[ArcTan_{ - 1}(z)] = 9\pi/8$;

Or even better:  Find $ f(\theta)$ such that:

$ \mathop\lim\limits_{z\to i^{\gamma}} Re[ArcTan_{ - 1}(z)] = f(\theta)$ where $ \theta$ is the angle $ \gamma(t)$ makes as it passes through $ i$.

Finally, just scrap the -1 and do it for any sheet:

$ \mathop\lim\limits_{z\to i^{\gamma}} Re[ArcTan_{n}(z)] = f(\theta,n)$

I believe the following then is true:

$ \{f(\theta,n), \forall n\in\mathbb{Z}\} = \mathbb{R}$

[edit] corrected last statement to include all n


Umm...If you're at that point, I think you're fine...

A complete and correct solution will get a 5. If you make a generalization but your original solution was incorrect, I doubt you're going to get a 5.

What do you actually mean by generalizations in solutions?

[quote="pythag011"]Will generalizations make you more likely to get a 5?[/quote]No.


$ x_1,x_2, \cdots ,x_n \geq 0,$
$ \Longrightarrow (n\minus{}1)(x_1^n \plus{} x_2^n \plus{} \cdots \plus{}x_n^n)\plus{}nx_1x_2\cdots x_n  \geq (x_1,x_2, \cdots ,x_n)(x_1^{n\minus{}1} \plus{} x_2^{n\minus{}1} \plus{} \cdots \plus{}x_n^{n\minus{}1}).$

[quote="jcc0107"]$ x_1,x_2, \cdots ,x_n \geq 0,$
$ \Longrightarrow (n \minus{} 1)(x_1^n \plus{} x_2^n \plus{} \cdots \plus{} x_n^n) \plus{} nx_1x_2\cdots x_n \geq (x_1,x_2, \cdots ,x_n)(x_1^{n \minus{} 1} \plus{} x_2^{n \minus{} 1} \plus{} \cdots \plus{} x_n^{n \minus{} 1}).$[/quote]
What do you mean?I think it must be:
$ \Longrightarrow (n\minus{}1)(x_1^n \plus{}x_2^n\plus{} \cdots \plus{} x_n^n)\plus{}nx_1x_2\cdots x_n \ge (x_1\plus{}x_2\plus{}...\plus{}x_n)(x_1^{n\minus{}1}\plus{}x_2^{n\minus{}1}\plus{}...\plus{}x_n^{n\minus{}1})$
It is Suranji 'inequality :wink:


What is the sum of the integer solutions of $ |x \plus{} 3| < 4$?

hello, for $ x\geq\minus{}3$ we get $ x<1$ and for $ x<\minus{}3$ we have $ x>\minus{}7$ and all integer solutions are
$ L\equal{}\{\minus{}6;\minus{}5;\minus{}4;\minus{}3;\minus{}2;\minus{}1;0\}$
Sonnhard.

Thus the sum of these is $ \boxed{\minus{}21}$.

hello, sorry, i have forgotten to sum up all solutions, thank you for doing this.
Sonnhard.


AMC AIME USA(J)MO USAMO geometry \/closed

Although the challenge problems in a chapter are much harder, are they still solvable using techniques from that chapter?

They are solvable using techniques from the chapter, and they are much harder than most of the problems in the chapter.

I am working through the Intermediate series.
(Alg 3, Counting)

I saw lots of AIME problems, some USAMO, a few counting problems IMO.

I asked something similar before:
was wondering if the Intermediate series helps with USAMO and other olympiad.
People said no because supposedly the USAMO/Olympiad Problems in there were "old" problems that wouldn't help with today's level

Of course they will help you build a foundation, but only EASY olympiad problems appear in the Intermediate series, and they are the HARDEST problems in the book.

What about the IMO problems?
I swear I saaw one in Alg 3 and a few in Interm C/P

It's what AIME said. Only the easiest olympiad problems appear in the series. Maybe there was a particularly easy IMO problem that year.

Does anyone know how to prep. for the really really hard AMC 12 and AIME problems?

That's correct -- our Intermediate series, along with Volume 2, will help with these.

Wouldn't Vol. 2 be best as review, since it is cover a lot of topics, it doesn't go into as much detail about each one.

Also, is there going to be Interm NT any time soon?

[quote="AlphaBetaTheta"]Probably not.
Does anyone know how to prep for the aime geometry problems?[/quote]

Intro to Geometry Ch. 17 is pretty hard, I heard if you master Intro you should get most of the nontrig AIME problems

yeah i agree with AIME15USAMO
the aime problems are 2 to 3 times harder than the back of the book

[quote="AIME15USAMO"]Sorry to revive...

but are you sure Introduction to Geometry covers all AIME geometry? It doesn't seem like it's a very difficult book. But then again, I'm only about half way through the book.[/quote]

I said ALMOST all... not all.


Elmo makes N sandwiches for a fundraiser. For each sandwich he uses B globs of peanut butter at 4 cents per glob and J blobs of jam at 5 cents per glob. The cost of the peanut butter and jam to make all the sandwiches is 2.53. Assume that B, J, and N are all positive integers with N>1. What is the cost of the jam Elmo uses to make the sandwiches?

[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=75346]This problem has already been discussed in the AMC-Competition section of AoPS forum[/url]

 :)


John has to order Pizza boxes for a restaurant. Only two sizes of boxes are available. Boxes are square in shape and the diamater must be 2cm longer then the diameter of the pizza. What size of boxes should he order if the diamater of the longer pizza is 40 cm and its area is 3 times the smaller one.

If there are only two sizes of boxes, John should order those two sizes, because he has no other choice. Either his pizzas are just the right size for the available boxes, or he will have to start making pizzas of a size to fit the boxes.

The problem should have stated that John makes pizzas in only two sizes. (Those are the sizes of the pans he has). The large pizza is 40 cm in diameter and its area is 3 times the area of the small one. Boxes can be made to order in any size he wants.

Obviously John will order 42-cm boxes for his large pizzas.

You are expected to figure out the diameter of the small pizza. Remember that for similar figures, the ratio of the areas is the square of the ratio of the lengths (diameter, or whatever linear measurement you have).

If I were John, I would round up to the next integer, because the box manufacturer may be confused by an order for boxes of side length $ 40 \frac {\sqrt{3}}{3} \plus{} 2$.

[quote="KMST"]If there are only two sizes of boxes, John should order those two sizes, because he has no other choice. Either his pizzas are just the right size for the available boxes, or he will have to start making pizzas of a size to fit the boxes.

The problem should have stated that John makes pizzas in only two sizes. (Those are the sizes of the pans he has). The large pizza is 40 cm in diameter and its area is 3 times the area of the small one. Boxes can be made to order in any size he wants.

Obviously John will order 42-cm boxes for his large pizzas.

You are expected to figure out the diameter of the small pizza. Remember that for similar figures, the ratio of the areas is the square of the ratio of the lengths (diameter, or whatever linear measurement you have).

If I were John, I would round up to the next integer, because the box manufacturer may be confused by an order for boxes of side length $ 40 \frac {\sqrt {3}}{3} \plus{} 2$.[/quote]

Thanks for the help, I think I understand a bit now.

I guess you did:

40^2   3
____=  _
x^2     1

40^2 = 3x^2
40^2/3= x^2

square root of (40^2/3) gives 40 and 1 on square root of 3, which is square root of 3 on 3. And since the diameter has to be two more, we get 40 square root 3/ 3 + 2.

But if boxes are SQUARE in shape, how could they have diameters? That's the part I dont understand..

I think by diameter they meant diagonal. Definitely a typo.


find all integers $ x$ such that :

$ x^3 \plus{} (x\plus{}1)^3 \plus{}  ... \plus{} (x\plus{}7)^3 \equal{} y^3$

[hide="Hint"]If $ m^3 < n^3 < (m \plus{} k)^3$ and $ m,n\in\mathbb{Z},k\in\mathbb{N}$ then $ n\in\{m \plus{} 1,\cdots, m \plus{} k \minus{} 1\}$.[/hide]

$ x^3 \plus{} (x \plus{} 1)^3 \plus{} ... \plus{} (x \plus{} 7)^3 \equal{}(2x\plus{}7)^3\plus{}126x\plus{}441\equal{} y^3$, therefore $ \minus{}14\le x\le 7$.
If $ (x,y)$ solution. then $ (\minus{}x\minus{}7.\minus{}y)$ is solution too. Obviosly $ (\minus{}5,\minus{}6),(\minus{}4,\minus{}4),(\minus{}3,4),(\minus{}2,6)$ are solutions.


a) Prove that
\[ \dfrac{1}{2} \plus{} \dfrac{1}{3} \plus{} ... \plus{} \dfrac{1}{2^{2n}} > n,
\]
for all positive integers $ n$.

b) Prove that for every positive integer $ n$ we have $ \min\left\{ k \in \mathbb{Z}, k\geq 2 \mid \dfrac{1}{2} \plus{} \dfrac{1}{3} \plus{} \cdots \plus{} \dfrac{1}{k}>n \right\} > 2^n$.

[quote="Ahiles"]
b) Prove that for every positive integer $ n$ we have $ \min\left\{ k \in \mathbb{Z}, k\geq 2 \mid \dfrac{1}{2} \plus{} \dfrac{1}{3} \plus{} \cdots \plus{} \dfrac{1}{k} \right\} > 2^n$.[/quote]

Is there some typo? I can't understand.

I think the last part of the second condition is supposed to read $ \frac{1}{2} \plus{} \frac{1}{3} \plus{} ... \plus{} \frac{1}{k} > n$.  Or something along those lines.

[quote="t0rajir0u"]I think the last part of the second condition is supposed to read $ \frac {1}{2} \plus{} \frac {1}{3} \plus{} ... \plus{} \frac {1}{k} > n$.  Or something along those lines.[/quote]

If so, I would formulate this statement as "prove the inequality $ \frac12\plus{}\frac13\plus{}\dots\plus{}\frac1{2^n}\le n$", it's bit less sophisticated.  :)

Sorry guys I was hasted. The statement has been modified, so now it makes sense :D[/i]

Two problem have same method .
We only prove
$ \sum_{k\equal{}2}^{2^n}\frac{1}{k}\leq n$ (1)
It is quite easy from
$ \frac{1}{2}<\sum_{k\equal{}2^{m}}^{2^{m\plus{}1}\minus{}1}\frac{1}{k}<1$
Add these inequalities we have (1)
So problem claim .


Hi all,

I would like to ask for your participation in an evaluation I'm doing. It is about the perception of problem difficulty in analysis (sequence problems).
You have to download the attached file that contains 18 sequence problems. You should rate the difficulty of each on a scale from 1 to 5.
It would be useful for me if you would solve the problems too. Since I know that not everybody has time for that, I would appreciate even if you would do only the rating part. Your answers should be mailed directly to my email address (it is made available) in order to not influence the responses of others. The test i thought for college students.
I am sending the attached file in two versions, as doc Word file and as a zip of the doc file.
Any contribution is welcomed, thanks in advance.

IP

[quote="IPelczer"]You have to download the attached file that contains 18 sequence problems.[/quote]
 :( What will happen to me if I don't?

Hi,

well, maybe your sleep will be perturbed  :)

Of course, you are not forced to participate, you are kindly invited to do so.

Regards,
IP


trigonometry
real analysis
real analysis solved

Let us consider the sequence $ x_n=|sin n|^n$. Does the series $\sum_{n>1} {\frac{x_n}{n^2\cdot x_{n+1}}}$ converge?

I am sure we can't state that terms tend to zero.

Hmm... It is really obvious. I don't know whether I should repeat common argumentation, that we can find infinitely many $n$ s.t. $|\sin n|\leq \frac{c}{n}$.

Myth, I am sorry, but I still do not understand. Why from here it follows that the general term does not tend to 0? I am sorry, I am completely blind. :(

If $|\sin(n+1)|<\frac{c}{n+1}$ then $|\sin n|\geq \frac{1}{3}$ for sufficietly large $n$, so
\[\frac{x_n}{n^2x_{n+1}}\geq \frac{(n+1)^{n+1}}{n^2\cdot 3^n \cdot c^{n+1}}.\]
What is wrong?

[quote="harazi"]Nothing, it is just me who is idiot. Sorry for the stupid posts. Anyway, what am I doing on this site anymore? Probably I must convince absolutely everyone that I am completely idiot.[/quote]
Probably I can say it about myself too.  :D

THus I was not the only man who complicated a lot this problem. :(  This gives me 2 more days to live... ;)


arithmetic sequence
combinatorics proposed
combinatorics

Consider the set $ A \equal{} \{ 0,1,2,...,n \}$ and $ S_n$ the set: $ \{ (a_1,a_2,a_3)|a_1,a_2,a_3 \in A$ with $ |a_1 \minus{} a_2| \equal{} |a_2 \minus{} a_3| \}$

Find the number of the elements of $ S_n$.

Basically, $ S_n$ is the set of all triples that are either arithmetic progressions or palindromes.  The only intersection between these two categories is the set of constant triples, so we'll count those with the arithmetic progressions and not with the palindromes.

The set of all (non-constant) palindromes is defined by $ (a_1, a_2, a_1)$, so the number of palindromes is $ (n\plus{}1)n \equal{} n^2\plus{}n$.

Any arithmetic progression is defined by $ (a_1, \frac{a_1\plus{}a_3}{2}, a_3)$, and in order for the middle term to be an integer, we need $ a_1, a_3$ to have the same parity (they could possibly be equal).

If $ n$ is even, there are $ \frac{n\plus{}2}{2}$ even integers and $ \frac{n}{2}$ odd integers, so we have $ \left(\frac{n\plus{}2}{2}\right)^2 \plus{} \left(\frac{n}{2}\right)^2 \equal{} \frac{n^2\plus{}2n\plus{}2}{2}$ arithmetic progressions.

If $ n$ is odd, there are $ \left(\frac{n\plus{}1}{2}\right)^2\plus{}\left(\frac{n\plus{}1}{2}\right)^2 \equal{} \frac{n^2\plus{}2n\plus{}1}{2}$ arithmetic progressions.

Adding in the palindromes, we have
$ n$ even: $ \frac{3n^2\plus{}4n\plus{}2}{2}$
$ n$ odd: $ \frac{3n^2\plus{}4n\plus{}1}{2}$

or perhaps more concisely,
$ \frac{1}{2}\left(3n^2\plus{}4n\plus{}\frac{3}{2}\plus{}\frac{1}{2}(\minus{}1)^n\right)$


I think it is really a hard problem. I have some ideas but I'm still working on it!!!! :huh:  :huh:


[quote="weatherbottle"]Source: Massachusetts Mathematics League March 2003

Given $P(x) = x^4 + bx^3 + cx^2 + dx + e$ for $b, c, d, e$ real values. If $P(2) = 0, P(1) = -39,$ and $P(a) = P(-a)$ for any real number $a,$ find the other three zeroes of $P(x)$.[/quote]

[hide="Solution"]
The function is even, so $b=0$ and $d=0$. So now we have
$P(x) = x^4 + cx^2 + e$

Substitute our two known points,
$0 = 16 + 4c + e$
$-39 = 1 + c + e$

Subtract the second equation from the first,
$39 = 15 + 3c$
$c = 8$
It follows that $e = -48$

So our polynomial is
$P(x) = x^4 + 8x^2 - 48$
$P(x) = (x^2 - 4)(x^2 + 12)$

So the four roots are $\pm 2, \pm 2i \sqrt{3}$

[/hide]


function calculus derivative limit trigonometry search Putnam

Let $ f$ be an infinitely differentiable real-valued function defined on the real numbers. If $ f(1/n)\equal{}\frac{n^{2}}{n^{2}\plus{}1}, n\equal{}1,2,3,...,$ Compute the values of the derivatives of $ f^{k}(0), k\equal{}0,1,2,3,...$

[quote="mdk"]Let $ f$ be an infinitely differentiable real-valued function defined on the real numbers. If $ f(1/n) \equal{}\frac{n^{2}}{n^{2}\plus{}1}, n \equal{} 1,2,3,...,$ Compute the values of the derivatives of $ f^{k}(0), k \equal{} 0,1,2,3,...$[/quote]

Here it is useful to know the "sequence definition" of continuity.

[b]Definition:[/b] Let $ f: D\mapsto\mathbb{R}$ be a function. We say $ f$ is [b]continous[/b] at $ x_{0}\in D$ if and only if for every sequence $ \{x_{n}\}\in D$ converging to $ x_{0}$ we have $ \{ f(x_{n})\}$ converges to $ f(x_{0})$.

Now we can find what $ f(0)$ has to be. Notice that $ x_{n}\equal{}\frac{1}{n}$ is a sequence which converges to $ 0$ so $ f(0) \equal{}\lim f(x_{n}) \equal{}\lim\frac{n^{2}}{n^{2}\plus{}1}\equal{} 1$ by continuity.

The derivative is given by,
$ \lim_{x\to 0}\frac{f(x)\minus{}f(0)}{x\minus{}0}\equal{}\lim_{x\to 0}\frac{f(x)\minus{}1}{x}$

It is useful to know the "sequence definition" of limit.

[b]Definition:[/b] Let $ f: D\mapsto\mathbb{R}$ be a function defined on an open interval $ I$ containg $ c$ we say $ \lim_{x\to c}f(x) \equal{} L$ if and only if for each $ \{ x_{n}\}\in I\minus{}\{ c\}$ we have $ \lim f(x_{n}) \equal{} L$.

Now we need to compute that limit,
$ \lim_{x\to 0}\frac{f(x)\minus{}1}{x}\equal{} L$
We know that it exists, by the condition that $ f$ is infinitely differentiable.
So by the definition above if $ x_{n}\equal{}\frac{1}{n}$ then $ \lim f(x_{n}) \equal{} L$.
Thus,
$ \lim\frac{\frac{1}{n}\minus{}1}{\frac{1}{n}}\equal{}\minus{}1$.

Can you continue....

[hide="Hint"]
$ f''(0) \equal{}\lim_{x\to 0}\frac{f'(x)\minus{}f'(0)}{x\minus{}0}$  :wink:
[/hide]

EDIT: I messed up minorly on my 2nd defintion. But Im not gonna change it, so it wouldnt get messy.

Take $ f(x)\equal{}\frac{1}{1\plus{}x^{2}}$ for example.

If $ |x|<1$ then it has the series representation $ 1\minus{}x^{2}\plus{}x^{4}\minus{}x^{6}\plus{}...$

Therefore $ \frac{f^{(k)}(0)}{k!}\equal{}\cos\frac{k\pi}{2}$

This exact problem has been posted before; I looked, but I didn't know what to search for to find it efficiently.

Yeah, I was thinking it was a Putnam or something, but I also don't know what to search for (or where) to find it.

This is from Putnam 1992, see http://www.unl.edu/amc/a-activities/a7-problems/putnam/-pdf/1992.pdf ... By the way, Dr. Doe, your solution is incomplete, since you need to prove that from the fact that the given function $ f$ and your power series agree at $ x \equal{} 1/n$, for all positive integers $ n$, it follows that all derivatives at $ 0$ agree, too.

My given function $ f$ and my power series do not only agree at $ x\equal{}\frac{1}{n}$ for $ n\in\Bbb{Z}^{>1}$ .
They agree for all real numbers between $ \minus{}1$ and $ \plus{}1$ , hence all derivates agree, too.

But don't you need to prove then that they agree at all real numbers between $ \minus{}1$ and $ 1$? ;)

[I'm not even sure whether that is necessarily true... I don't see any obvious reason for this to be true :)]

Hmm... Im not exactly sure what Dr. Doe meant but [url=http://www.kalva.demon.co.uk/putnam/psoln/psol924.html]this[/url] solution is similar to his somehow.

Well, yes, and the biggest part of the proof is proving "unicity" (starting at "Suppose $ g$ is another such function...").

Maybe we are talking at cross purposes.

My chosen function $ f(x): \equal{}\frac{1}{1\plus{}x^{2}}$ obviously agrees with $ 1\minus{}x^{2}\plus{}x^{4}\minus{}x^{6}\plus{}...$ for all $ x\in ]\minus{}1,1[$ .

I didn't claim that every arbitrary infinitely differentiable real-valued function $ f$ with $ f(n)\equal{}\frac{1}{1\plus{}\left(\frac{1}{n}\right)^{2}}$
agrees with $ 1\minus{}x^{2}\plus{}x^{4}\minus{}x^{6}\plus{}...$ for all $ x\in]\minus{}1,1[$ , because that's not true.

For example $ f(x)\equal{}\frac{1}{1\plus{}x^{2}}\plus{}\left\{\begin{matrix}\exp\left(\frac{1}{x\minus{}\frac{1}3}\minus{}\frac{1}{x\minus{}\frac{1}4}\right) &\frac{1}4<x<\frac{1}3\\ \\ 0 &\text{otherwise}\end{matrix}\right.$

is infinitely differentiable and complies with $ f(n)\equal{}\frac{1}{1\plus{}\left(\frac{1}{n}\right)^{2}}$

but isn't $ 1\minus{}x^{2}\plus{}x^{4}\minus{}x^{6}\plus{}...$ for all $ x\in]\minus{}1,1[$ .

The derivatives $ f^{(k)}(0)\equal{}\lim_{n\to\infty}\frac{\sum_{j\equal{}0}^{k}(\minus{}1)^{k\minus{}j}{k\choose j}f\left(\frac{j}{n}\right)}{\left(\frac{1}{n}\right)^{k}}$ are well defined and unique.

Therefore it is up to me how I choose $ f$ because the choise doesn't concern the values $ f^{(k)}(0)$ .

A function that doesn't agree anywhere else (but still has the right derivatives at zero): $ \frac{1}{1\plus{}x^{2}}\plus{}\sin\frac{\pi}{x}e^{\minus{}1/x^{2}}$, filled in to be 1 at zero.


integration
calculus
function
trigonometry
algebra
rational function
calculus computations

[quote="brad sue"][quote="ISU20CpreE"]The problem is that there is a u on top of my fraction so i can't apply the rule to integrate that i learned.

This is the part i got stuck:

$\int\frac{x-3} {\sqrt{6x-x^2}}$[/quote]

I believ that oyu wnat to compute the integral.

you can let u=6x-x^2. so du=6-2x= (1/2)*(3-x)= -(1/2)*(x-3)
so yournew  integral is:

$\frac{-1} {2}$ $\int\frac{du} {\sqrt{u}}$

It is easy from here.[/quote]

you are right. i got it
but can you help me out in this one:

$\int\frac{\sqrt{x-2}} {x+1}$

but i have to use $u=\sqrt{x-2}$

$\int\frac{\sqrt{x-2}} {x+1} \,dx$

(Rule #1: [b]always[/b] include the $dx$. It becumes an important part of the bookkepping.)

You want to make the substitution $u=\sqrt{x-2}$ but this doesn't go through completely as a "natural" substitution. What now? Try forcing it as a "backwards" substitution. (jmerry said something to you about "backwards" substitutions in another topic.)

Let $u=\sqrt{x-2}$ but solve for $x$ in terms of $u.$

$x=u^2+2$ so $x+1=u^2+3$
$dx=2u\,du.$

$\int\frac{\sqrt{x-2}} {x+1} \,dx=\int\frac{u}{u^2+3}\,2u\,du$

$=\int\frac{2u^2}{u^2+3}\,du$

Our substitution has now transformed this into a rational function. As a rational function, it's an improper fraction - that is, the degree of the numerator is not smaller than the degree of the denominator. That calls for long division:

$\frac{2u^2}{u^2+3}=2-\frac6{u^2+3}.$

We now use the formula $\int\frac{du}{u^2+a^2}=\frac1a\tan^{-1}\left( \frac ua\right)+C$ to get

$=\int\frac{2u^2}{u^2+3}\,du=2u-2\sqrt{3}\tan^{-1}\left(\frac u{\sqrt{3}}\right) +C$

$=2\sqrt{x-2}-2\sqrt{3}\tan^{-1}\left(\sqrt{\frac{x-2}3}\right)+C.$


In a raffle 20 tickets are sold. Two prizes will be given.  A student buys 2 tickets.  What is the probability that this student wins at least one prize?

Why can't I use the binomial probability formula here?

$2C0 \left(\frac{2}{20}\right)^{0}\left(\frac{18}{20}\right)^{2}$

I don't understand.  Cards aren't mentioned in the rest of the problem...

I think he means that the red ace is a winning raffle ticket, which makes the answer 37/190

wait the binomial probability formula is only for cards?  I thought you use it for finding the probability of independent events.

Ok number one-is my answer correct
Number two- Can you rephrase what you just said about binomial probablility

Theres already a problem with your first solution- you have 20 in the denominator twice, tahts implyin you can replace tickets, but clrealy you cannot- if you alreayd have a1, theres 19 remaining.

opps  :blush: .. I copied the wrong problem. I'm so sorry..  ok I edited the red ace part.

[quote="myc"]In a raffle 20 tickets are sold. Two prizes will be given.  A student buys 2 tickets.  What is the probability that at least one of the probability that this student wins at least one prize?

Why can't I use the binomial probability formula here?

$2C0 \left(\frac{2}{20}\right)^{0}\left(\frac{18}{20}\right)^{2}$[/quote]
2 prizes will be given. they don't have to be to different tickets.

the chance that the student loses both prizes is $\frac{18}{20}\frac{18}{20}$. it is not $\frac{18}{20}\frac{17}{19}$
simplified, it is $\frac{81}{100}$
so the answer is $\frac{19}{100}$
what i dont get is what it means by "the probability that at least one of the probability that"

myc seems to have edited his post incompletely.  It should just read "What is the probability that the student wins at least one prize?"

[quote="ProtestanT"]2 prizes will be given. they don't have to be to different tickets.[/quote]

That's not the way I remember raffles working...

[quote="ProtestanT"]the chance that the student loses both prizes is $\frac{18}{20}\frac{18}{20}$. it is not $\frac{18}{20}\frac{17}{19}$[/quote]

Er, no, actually, I think it is.

if you're picking both of the raffles simultaneously than isn't the probability of picking the non-winning raffle, $\frac{18}{20}$  for both raffles.  You're not picking first raffle and then the second raffle. There are 20 choices when your are picking both of the raffles. but $1-\frac{18}{20}*\frac{17}{19}$ is the correct answer.

[quote="myc"]if you're picking both of the raffles simultaneously than isn't the probability of picking the non-winning raffle, $\frac{18}{20}$  for both raffles. [/quote]

Are you sure you know how a raffle works?

There are $20$ tickets, each of which has some particular identification number.  Two of these numbers are chosen beforehand to have prizes.

When you buy two tickets, you are buying them [b]without replacement.[/b]  (That's kind of how buying tickets works.)


Solve for $ cosx$:
$ \frac{cos4x}{cosx}\equal{}2cos3x\minus{}2$.

$ cos 4x\equal{}2cos 3x \cdot cosx \minus{} 2 cosx \equal{} cos4x \plus{} cos2x \minus{}2cos x$

$ 0\equal{}cos2x\minus{}2cosx\equal{}2cos^2x\minus{}2cosx \minus{}1$

How do you know that
\[ 2 \cos 3x \cdot \cos x \equal{} \cos 4x \plus{} \cos 2x?
\]

$ 2cos x \cdot cosy\equal{}cos(x\minus{}y)\plus{}cos(y\plus{}x)$

I multiplied by $ \cos{x}$, and then added $ \sin{3x}\sin{x}$ to both sides.
The quadratic gives values of $ \boxed{\frac {1\pm \sqrt {3}}{2}}$

Yes, you're right. So the answer is

$ \boxed{cosx\equal{}\frac{1\minus{}\sqrt{3}}{2}}$

Wow, I made this problem and posted it on an earlier thread of mine. -__-

Here's how I did it: we get that $ \cos 4x \equal{} 2 \cos x \cos 3x \minus{} 2 \cos x$; since $ \cos(4x) \equal{} \cos(3x \plus{} x) \equal{} \cos x \cos 3x \minus{} \sin 3x \sin x$. Rearranging, we get that $ 2 \cos x \equal{} \cos 3x \cos x \plus{} \sin 3x \sin x \equal{} \cos(3x \minus{} x) \equal{} \cos(2x)$. Letting $ u \equal{} \cos x$, we see that $ 2u^2 \minus{} 2u \minus{} 1 \equal{} 0$. We then solve the quadratic.


Ring Theory
superior algebra
superior algebra unsolved

Find a surjective ring homomorphism $ f: A \to B$ such that $ f(rad(A)) \neq rad(B)$. Here $ rad(X)$ is the intersection of all maximal ideals in $ X$. It is in general true that $ f(rad(A)) \subset rad(B)$.

The only thing I found was the canonical projection

$ f: \mathbb{Z} \to \mathbb{Z}/4\mathbb{Z}$

which is not very impressive. I could not find other examples...

usually, $ rad(a)$ denotes the radical of an ideal $ a$ in $ A$. in particular, $ rad(A)\equal{}A$. you probably mean $ rad(0)\equal{}nil(A)$ with $ rad(A)$. ;)

what is an impressive counterexample? it's quite clear that $ f(nil(A))\equal{}nil(B)$ is false in general, e.g. when $ A$ is reduced but $ B$ is not. this can be simply achieved: take some prime ideal $ p$, which is not idempotent, and  consider $ A \to A/p^2$.


[quote="Mashimaru"]Find all function $ f(x): (0; + \infty)\to(0; + \infty)\#$, strictly increase on $ (0; + \infty)\#$ and satisfy: $ f(\frac {x^2}{f(x)}) = x,\forall x\in(0; + \infty)\#$[/quote]

Edited: Find all function $ f:$$ R_ +   \to R_ +$ strictly increase on $ R_ +$ and satisfy: $ f\left( {\frac {{x^2 }}{{f\left( x \right)}}} \right) = x,\forall x \in R_ +$


The three distinct real solutions of x^3-2x^2-3x+5=0 are r, s, and t. Compute the value of 1/(rs)+1/(rt)+1/(st).

Please use spoiler.

Solution:

[hide]1/(rs)+1/(rt)+1/(st)

(r+s+t)/(rst)

Applying Viete's Formula to x^3-2x^2-3x+5=0

r+s+t=2

rst = -5

(r+s+t)/(rst) = -2/5[/hide]

Hm...in my precalc class, we just finished polynomials.  I didn't know that "x^3-2x^2-3x+5=0

r+s+t=2
rst = -5"
was actually a formula!  I thought that it was a simple manipulation of (x-r_1)(x-r_2)...(x-r_n) for the polynomial f(x) of degree n.  What does Viete's formula actually say?

-interesting_move

Vite's Formula is what you get when you multiply out (x-r_1)(x-r_2)(x-r_3)...(x-r_n) and equate terms.

Yea, these are pretty straightforward, with Viete's formula, or whatever u call it.


anyone know when the CMO and APMO will take place this year? The AIME 1 will be on Mar. 14, and AIME 2 will be on Mar 28.

APMO will be probably be Mar 13~14, and CMO Mar 28~30, so just wondering if the EXACT dates will coincide this year....(I hope not)

Anyone know the dates of CMO and APMO yet? Thanks!

I dunno either. Btw aren't they supposed to send a letter as an invitation to CMO, and they tell you the date of the contest too.
I remember i received this letter in December last school year.

[quote="APMO Committee"][b]The 2007 Asian Pacific Mathematics Olympiad (APMO) is to be held in the afternoon of Monday March 12 for participating countries in the Americas, and in the morning of Tuesday March 13 for participating countries on the Western Pacific.[/b][/quote]
See http://www.kms.or.kr/competitions/apmo/

[quote="andy_tok"]CMO Mar 28 9:00~12:00 local time[/quote]
Thank you very much.


Adam ran across the field at a speed of 5 yards per second and walked back along the same path at a speed of 2 yards per second. For how many seconds did he run if the total round-trip time was three and a half minutes?

3.5 minutes=210 seconds. How is this 60?

He ran for $ \frac{2}{2\plus{}5}\equal{}\frac{2}{7}$ of the time, which is $ 210 \times \frac{2}{7}\equal{}\boxed{60}$ seconds.  Remember, $ \text{run} \neq \text{walk}$.  :)

ooooooow.  I made the same mistake.  Thanks AIME :blush:  :) 15!


[quote="lightrhee"]Now the team is released so I can discuss this. :)
Everyone in the team is grade 12(Farzan is also grade 12, right?). So, it will be difficult to pass on Canmoo :moose:  to the next IMO team.
What do we do with Canmoo? Who should keep him?[/quote]

Yup, this year we sure have a graduating team. Maybe you should keep Canmoo for now, and ceremoniously award him to the next year's team at the summer training camp. Or you could transfer him now to someone who is likely to be on the next year's team (e.g. Lin Fei or rem). But besides those two suspects, I really really wonder who the remaining team members will be next year ... Probably lots of grade 10s.

[quote="YawningLion"][quote="lightrhee"]
What do we do with Canmoo? Who should keep him?[/quote]

Along with Canmoo, you could pass down Elyot's Canadian flags to the next team! I can imagine you solemnly handing over a huge pile of stuff to an overwhelmed grade 10 genius: Canmoo, flags, last year's problem set, random books of problems, a deck of cards, instructions on playing mafia, a blank sheet of instructions for playing mao, assorted IMO t-shirts, a rubik's cube, etc, etc etc.[/quote]
 :rotfl: The Mao instruction was the best part. :D
Is Richard going to IOI again? If he does, he could take it to IOI and keep him until June, 2007(if he doesn't mind) and hand Canmoo over to the next team at the seminar(since he is going to U of W).

You can always keep canmoo and hand him over to the next group of IMO people.  Given that you'll be at Waterloo, I'm sure things can be worked out.

[quote="lightrhee"][quote="Conan777"]Sorry....i'm new here....may i ask what is Canmoo? a 450g what?[/quote]
Canmoo is the official(?) mascot of the Canadian IMO team. We bought him last year. He is a very cute moose(my avatar is a picture of him).
__________________________________________________
EDITED: forgot to mention that he is a 'stuffed' moose :D[/quote]

hahaha...... :rotfl:  :rotfl:  :rotfl:
u know what? i thought it's some living thing ...XD
u guys were talking about 'him' all the time:P

:D   Take Canmoo to Waterloo for the year!!!!!  THen I can steal him again!!!!   :D

[quote="YawningLion"][quote="Conan777"][quote="lightrhee"]
Canmoo is the official(?) mascot of the Canadian IMO team. We bought him last year. He is a very cute moose(my avatar is a picture of him).
__________________________________________________
EDITED: forgot to mention that he is a 'stuffed' moose :D[/quote]

hahaha...... :rotfl:  :rotfl:  :rotfl:
u know what? i thought it's some living thing ...XD
u guys were talking about 'him' all the time:P[/quote]

He comes alive, but only on contests, to give ingenious hints to his lucky keeper. And no one will ever suspect him, because he's stuffed. That's because they don't know that he's stuffed with knowledge  :roll: .[/quote]

So who's keeping him right now? :P

[quote="YawningLion"]That's because they don't know that he's stuffed with knowledge  :roll: .[/quote]
That's a good one. :rotfl:

[quote="YawningLion"]So David, did you decide how to transfer Canmoo to the next generation of IMOers?

Seminar is in a week! Woohoo![/quote]
No. But I do have a question to ask. Can you keep Canmoo after the seminar and bring it to the training camp? Because I found flying with Canmoo little hard. :D
By the way, I met Adrian in Calgary last weekend. He was also wondering about who gets Canmoo. He also said to me, "You actually weighed him? You must have been really bored when you did that."(something along that line) :rotfl:

[quote="lightrhee"][quote="YawningLion"][quote="lightrhee"]
No. But I do have a question to ask. Can you keep Canmoo after the seminar and bring it to the training camp? Because I found flying with Canmoo little hard. :D
By the way, I met Adrian in Calgary last weekend. He was also wondering about who gets Canmoo. He also said to me, "You actually weighed him? You must have been really bored when you did that."(something along that line) :rotfl:[/quote]

Hehe  :D ... what does Adrian do these days?

Why is it hard to fly with Canmoo? After all, he does weigh only 480 grams. Is it because you have to carry him around in a dignified way, not by the leg or something?

I'll take Canmoo from you at Seminar. It'll be fun to be the Canmoo Keeper, at least temporarily. Also, I'll try doing problems with the moose sitting on my lap, and see whether my ingenuity skyrockets. If it does, then I'll buy a fake Canmoo to give back to you, and keep the real one for the IMO  :ninja: .[/quote]
He is not something I can put in my baggage, and I have to carry him on board.
:pilot: + :moose: =  :juggle:[/quote]

hahah why actually bought it though?

[quote="billzhao"]Just out of curiosity, where can you buy a stuffed moose here in Toronto?[/quote]

Aha, so you also want a lucky mascot-advisor for the IMO ... soon the whole team will get equipped with personal meese, and Canada will rock the contest. I could imagine every team sitting on the contest with their national mascots on their tables. Americans have a cow, right?

Where to buy one ... a toy store, I guess (Toys R Us?). Or, even better, you could go to a forest and catch a live one (preferably a cub). We could teach it math from an early age!

Hahaha.  I might have to make a trip into Banff and buy a bunch of Math Meese so the team can all have one.  I know I'll be getting one just for me, hehehe because he's cute!!!

[quote="lightrhee"][quote="billzhao"]Just out of curiousity, where can you buy a stuffed moose here in Toronto?[/quote]
I don't know about Toronto, but there is [url=http://maps.google.ca/maps?hl=en&q=moose%20factory%2CON&sa=N&tab=wl]Moose Factory[/url] in Ontario.  :lol:[/quote]
how great!
human now produce animals industrially :rotfl:

So did David actually bring the moose to the seminar? I heard from Yufei that no moose was sighted at the banquet and I got quite worried as a result :( . Please tell me that Canmoo is still with you and living quite well :moose:.

I will be going to the IOI agian this year. So could I PLEASE keep the moose after the IMO? I'll be going to Waterloo next year, so I doubt I'll have any problems giving Canmoo to the next IMO team at the Seminar in 07'.

To all those who aspire to steal a moose: DAVID HAS A BLACK BELT, so don't get between him and the moose. (unless you want result analogous to having a real moose sitting on you)

David: How dare you claim that a moose that runs cars over weights only 480 grams? I demand the additoin of a k in front of the g. :moose:

[quote="rem"]Here is the great story about the canmoo moosenapped at seminar.
At the Tuesday banquet, canmoo was left on a table all alone. It was then seated at one of the tables in Ian Vanderburg's seat( i hope I spelled the name right). The canmoo was then taken away by Lino and left in his room. :ninja:
The next day canmoo disappearance was noticed by most students at seminar. Everyone deperately looked for it but no luck. David and Jenn even made signs "Find Canmoo!" with the drwaing of the moose on it, the contact info (contact David) and the reward(a night with Canmoo :P ). A ransom note was given to David saying he had to leave the Sunlife Cup in the hallway at 8 pm if he wants the moose back. :moose:
The moose was given back at the Thurday banquet. In the very end, a draw was made, and one of the staff won(i think it was Mr. Brown), and he got a big box wrapped in David's COMC papers :D . The canmoo was in the box, and David received it after it was taken out of the box in front of everyone.

THE END[/quote]

David, you're suppose to be the moose guard......poor Canmoo :moose:

This is what? Attempt no.12 in stealing the moose? Glad to hear that Canmoo surviived all those attempts. He is definitely one lucky moose.

I do not think it's David's fault. In my opinion the only way the moose would not get stolen is if David always held it in his hands(even when slleping). I think moose would get stolen anyway.
Even though the moose got stolen, the moosenapping event made seminar much more exciting and fun. :)

Canadian mascot the CANMOO beats all the other mascots in my opinion. :o

[quote="billzhao"]Just out of curiosity, where can you buy a stuffed moose here in Toronto?[/quote]

So, Yufei, did you succeed in buying a moose? Because I have never seen anything of the sort in Toronto stores (they only have monkeys and silly pink horses). Our Canmoo is so unique ...

Richard, why is your display picture an upside-down beaver? I would think that, of all things, it would be a penguin!

[quote="Richard Peng"][quote="Inspired By Nature"]YawningLion-Americans have a cow, right?

 :rotfl: lol. You mean eagle.[/quote]
Americans have a cow as the mascot of their IOI team. (see link)
http://oldweb.uwp.edu/academic/mathematics/usaco/2006/camp06/day7/target3.html
http://oldweb.uwp.edu/academic/mathematics/usaco/2006/camp06/day7/target4.html
http://oldweb.uwp.edu/academic/mathematics/usaco/2006/camp06/day7/target7.html

However, what they have for their IMO team is still unofficial, a grasshopper maybe?

The strangest mascot I've seen was some random plant by the French team last year.
http://www.ioi2005.pl/gallery/view_photo.php?set_albumName=Mascots&id=aai
Checklist for mascot-knapping:
http://www.ioi2005.pl/gallery/view_album.php?set_albumName=Mascots[/quote]
 :rotfl: Cool! Cow, really? Thats wierd... :P

The US mascot is Alex Zhai...

WE STOLE THE MOOSE


algebra
polynomial
function
domain
induction
Ring Theory
quadratics

Does the Fundamental Theorem of Algebra extend to multivariable polynomials (or some known subclass of them)?

Yes.  Every polynomial ring over a unique factorization domain is a unique factorization domain, so the result follows by induction on the number of variables.  (The proof is by the one-variable division algorithm.)

Since the FTA has several different consequences for univariate polynomials, I guess I should specify what aspects of the FTA I'm asking extend to multivariate polynomials.

1. I already know that polynomial rings over UFDs are UFDs (but I don't think that really has anything to do with FTA, does it?)
2. I also know that multivariate polynomials of degree n can have more than n solutions (in $ \mathbb{C}$) - so in this way, multivariate polynomials are already dissimilar to univariate polynomials.

But what I don't know is:
3. Does every multivariate polynomial in k variables (over $ \mathbb{R}$ or $ \mathbb{C}$) have at least one solution in $ \mathbb{C}^k$?
4. Are the absolute irreducible factors over $ \mathbb{C}$ of multivariate polynomials always linear? Are the irreducible factors over $ \mathbb{R}$ always quadratic?

I may have confused the fundamental theorem of algebra with the fundamental theorem of arithmetic  :oops:

[quote="gauss202"]Does every multivariate polynomial in k variables (over $ \mathbb{R}$ or $ \mathbb{C}$) have at least one solution in $ \mathbb{C}^k$?[/quote]
Yes.  By specializing all the variables except one to arbitrary numbers (or even better, algebraically independent of the coefficients of the polynomial as well as each other) one guarantees with probability $ 1$ that the remaining polynomial in one variable is non-constant.  If you want to make this argument more precise, again induct on the number of variables by showing that the coefficient of, say, the leading term in $ x$ is non-vanishing.  This is also a corollary of the [url=http://en.wikipedia.org/wiki/Hilbert's_Nullstellensatz]weak Nullstellensatz[/url].  Alternately, you may prefer appealing to the [url=http://en.wikipedia.org/wiki/Inverse_function_theorem]inverse function theorem[/url].

[quote="gauss202"]Are the absolute irreducible factors over $ \mathbb{C}$ of multivariate polynomials always linear? Are the irreducible factors over $ \mathbb{R}$ always quadratic?[/quote]
No and no.  For example, the polynomial $ x^n \minus{} y \in \mathbb{C}[x, y]$ is irreducible for all $ n$ by Eisenstein's criterion on $ \mathbb{C}[y][x]$.  (Note that Eisenstein's criterion makes sense over any PID.)


WOW something practical! It's good too.
My friend called me a few minutes ago on his break. [hide="This is what he wanted to know"]"Hey man I've got a really good math problem for you."
"OK, go for it."
"I was thinking about license plates and was wondering [b]how many 3 character combinations made up of any 3 random letters of the alphabet are there? It doesn't matter if the letters are the same.[/b]"

Yay! My two weakest fields.  :(

Of course, I'll want to know how many possible plates there actually [i]are[/i], not just half (of the plate). Here is my immediate reaction, please check it and let me know if I did it right, these really are what I'm worst at.
[hide="All letters are different"]Well obviously $ 3!$, also known as $ 6$
And, with a possibility degeneration of 1 less digit after every turn we have:
$ (n)(n-1)(n-2)(3!) = 3!(26\times 25\times 24) = 93600$[/hide][hide="Two digits are identical"]Well, there are 26 letters that may be Identical, after that, 1 turn of digit restriction, now with only 3 permutations : $ 3(n)(n-1)=3(26\times 25) = 1950$
Total permutations : $ 95550$[/hide][hide="Three digits are identical"]Only $ n$, or $ 26$ possibilites.[/hide]
To answer his question, I find $ 95576$ 3 letter permutations.
[hide="Finishing it off"]
After that, plates are simple, the same restrictions in permutation but this time $ n$ is equal to $ 10$, not $ 26$. Soooooo,
$ 3!(n)(n-1)(n-2)+3(n)(n-1)+n = 4320+270+10 = 4600$

To find the number of possible American license plates is simple because the numbers and letters are separate, making it a 2 character string. [i]simply[/i], $ 4600\times 95576 = 439,649,600$[/hide]
Hmmmm I guess that means we'll have to make a new system soon.
Well, maybe not, they could just reverse whether the numbers or letters go first, making twice as many possible plates.
[/hide]

Uhhh I forgot to say, but what I'm really asking here is:

Are those numbers right? I always seem to think I know the answer, then get them horribly wrong.

[hide]
For when there are 3 identical numbers should be:
$ 26P1=\boxed{26}$

For when there are 2 identical
$ 26P2=26\cdot25=\boxed{650}$

For when there are no identical numbers it is:
$ 26P3=26\cdot25\cdot24=\boxed{15600}$

So the answer is: $ 26+650+15600=\boxed{\boxed{16276}}$
[/hide]

This is assuming that the order matters.

Yes, I could tell, but we wanted not to know how many possible sets of letters exist, but in how many ways those sets could be arranged in total.

[hide="What I learned from the problem"]the letters are actually factorial systems. note that $ k!n(n-1)(n-2)....(n-k)= \frac{k!n!}{(n-k)!}$, where $ k$ is the number of letters that cannot be identical. That's a formula by the way. It can be used to easily count anything with the same characteristics. Be careful though, to apply that to every different number of identicals, also, the $ k!$ is flawed within that statement, but the rest holds true.[/hide]

I learned a lot from that one, but this is still my weakest subject.

igul, you are overthinking this question.

This is the most basic counting question.  DO NOT make three cases, there is only ONE case.

IF order matters, then its just 26^3

because first choice, i have 26 to choose from, second i have 26.. (i can repeat), third i have 26.


Attachments don't work any more

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=39305&start=500, also a post i made in intermediate trig online class dosn't show attachments! this only happens to images.!

Sorry, but I don't see a post of you there, and the one of I_like_pie shows his attachments to me  :maybe:

i won't be abel to see this, but i hope u can

First of all, please don't shout.

I think this is a local problem on your end.

I can see the attachment just fine, in both Firefox (1.5) and IE (6.0).  If you close your browser and restart it, can you still not see the attachments?  And can you not see [i]any[/i] attachments, or just certain ones?

i can't see jpgs for sure, i can see pdfs and docs

If you're using IE6, sometimes images don't display properly.  Try resizing your browser window -- this sometimes tricks IE6 into doing what it should do.

If you're using a good broswer (like Firefox, Opera, or IE7 -- though the jury is still out on IE7 I suppose), then I can't explain it.  Does the image still show up as a clickable link?  If you click the link, can you then see the image?

Totally unrelated note -- you need to change your signature to not have a deceptive off-site link in it.

no. for pdfs and docs yes. for jpgs no. and i use firefox

EDIT: I fixed it. I disabled Zone Alarm Ad blocker, and it works ow :blush:


conics
probability
ellipse
real analysis
ratio
geometry
3D geometry

This is a really cool old problem from the problem department in the Pi Mu Epsilon journal.

Given six real constants $ a, b, c, d, e,$ and $ f$, not all zero, a conic section $ C :
ax^2 \plus{} bxy \plus{} cy^2 \plus{} dx \plus{} ey \plus{} f \equal{} 0$ is determined. What is
the probability that $ C$ is an ellipse?

[quote="t0rajir0u"]What's the probability distribution assumed on the reals?[/quote]

uniform

[quote="t0rajir0u"]There is no uniform probability distribution on the reals.  (As far as I know.  The best we can do is to assign a probability that a point chosen from some Lebesgue-measurable region $ R$ in $ \mathbb{R}^n$ is within a smaller Lebesgue-measurable region $ r \in R$ by finding the ratio of the measures.)

Idea: we can scale $ a, b, c, d, e, f$ so that they are all within the unit 6-sphere and then we can sensibly define the probability we're looking for as the 6-volume of some region in 6-space determined by an inequality involving the conic section discriminant, divided by the volume of the unit 6-sphere.  I'm not sure if the "uniform probability distribution on the reals" (whatever that means) is preserved in this case.  Perhaps we should scale so they are [b]on[/b] the unit 6-sphere?[/quote]

I believe u mean unit 5-sphere which is a subset of $ R^6$. but yes that idea is correct. $ (a,b,c,d,e,f)$ is selected based on a uniform distribution on $ S^5$. You can then use the discriminant to find the volume on the surface of the sphere corresponding to ellipses. (You are correct that taking the volume occupied in the sphere will not preserve the prob. dist.)

There are two problems here though.
1) You have to take into account imaginary ellipses.
2) The Integral is horrendous. I dont think numerical approximations are allowed.

Another idea could be to reduce all equations to their canonical form (e.g: $ \pm \frac {x^2}{a^2}\pm\frac {y^2}{b^2} \equal{} 1$) by action of translation and rotation. That may make measuring the probability easier perhaps...

We have an ellipse for $ b^2 \minus{} ac < 0$ and hyperbola for $ b^2 \minus{} 4ac > 0.$ Coefficients $ d, e, f$ are irrelevant. To avoid the problem that probability density of a uniform distribution of reals is zero everywhere, we contain $ a, b, c$ in an arbitrary finite box symmetrical around the origin, for simplicity $ |a|, |b|, |c| < 1.$ Thus $ a, b, c$ are values of random variables $ A, B, C$ with uniform distribution in $ ( \minus{} 1, \plus{} 1).$ The sign of $ b$ is irrelevant, hence we can consider $ B$ as random variable in $ (0, 1).$ The signs of $ a, c$ are opposite with the probability $ P_{a} \equal{} \frac {_1}{^2}$ and in such case, $ b^2 \minus{} 4ac > \minus{} 4ac > 0$ $ \Longrightarrow$ we get a hyperbola for sure. Thus we can consider the random variables $ A, C$ with the same sign, i.e., uniform in $ (0, 1),$ for splitting the remaining probability between hyperbolas and ellipses. Let $ u(a)$ be the probability density of a random variable uniform in $ (0, 1)$: $ u(a) \equal{} 1$ for $ a \in (0, 1)$ and zero elsewhere. Let $ U(a)$ be its distribution function: $ U(a) \equal{} 0$ for $ a \le 0,$  $ U(a) \equal{} a$ for $ a \in (0, 1)$ and $ U(a) \equal{} 1$ for $ a \ge 1.$

Let $ S \equal{} B^2$ be a random variable such that $ s \equal{} b^2$ and $ F_S(s)$ its distribution function. This is the probability that $ B^2 < s$ or $ B < \sqrt {s},$ hence $ F_S(s) \equal{} U(\sqrt {s}) \equal{} \sqrt {s}$ for $ s \in (0, 1).$ The probability density is $ f_S(s) \equal{} \frac {\text dF_S}{\text ds} \equal{} \frac {1}{2 \sqrt {s}}$ for $ s \in (0, 1)$ and zero elsewhere.

Let $ P \equal{} AC$ be a random variable such that $ p \equal{} ac$ and $ F_P(p)$ its distribution function. This is the probability that $ AC < p.$ For a given $ a,$ the probabilty of $ C < \frac {p}{a}$ is $ U(\tfrac {p}{a}).$ Integrating over all possible values of $ a,$

$ F_P(p) \equal{} \int_0^1 U(\tfrac {p}{a}) \cdot u(a)\ \text da \equal{} \int_0^p \text da \plus{} \int_p^1 \frac {p\ \text da}{a} \equal{} p \minus{} p \ln p$

for $ 0 < p < 1.$ Again, the probability density is $ f_P(p) \equal{} \frac {\text dF_P}{\text dp} \equal{} \minus{} \ln p$ for $ 0 < p < 1$ and zero elsewhere.

Let $ D \equal{} B^2 \minus{} 4AC \equal{} S \minus{} 4P$ be a random variable such that $ d \equal{} b^2 \minus{} 4ac \equal{} s \minus{} 4p$ and $ F_D(d)$ its distribution function. This is the probability that $ S \minus{} 4P < d.$ For a given $ p,$ the probabilty of $ S < d \plus{} 4p$ is $ F_S(d \plus{} 4p).$ Integrating over all possible values of $ p,$

$ F_D(d) \equal{} \int_0^1 F_S(d \plus{} 4p) \cdot f_P(p)\ \text dp$ (convolution)

But what we actually need is the probability of $ D \equal{} B^2 \minus{} 4AC < 0$ (getting an ellipse), which is the value of $ F_D(0)$:

$ F_D(0) \equal{} \int_0^1 F_S(4p) \cdot f_P(p)\ \text dp \equal{} \int_0^{1/4} \sqrt {4p} ( \minus{} \ln p)\ \text dp \plus{} \int_{1/4}^1 ( \minus{} \ln p)\ \text dp \equal{}$

$ \equal{} \left[\tfrac{1}{3} 4p^{3/2} \left(\tfrac {2}{3} \minus{} \ln p \right) \right]_0^{1/4} \plus{} \left[p (1 \minus{} \ln p)\right]_{1/4}^1 \equal{}$

$ \equal{} \frac {1 \plus{} 3 \ln 2}{9} \plus{} \frac {3 \minus{} 2\ln 2}{4} \equal{} \frac {31 \minus{} 6 \ln 2}{36}$

The total probability of getting an ellipse is half of this, $ \frac {31 \minus{} 6 \ln 2}{72} \doteq 0.3728.$ Needless to say that I tested the result with Excel, getting about 37.300 ellipses out of 100.000 random conics.

I believe you have found the probability of getting a real ellipse or an imaginary one.

The question asks the probability for getting only a real ellipse.

[quote="Jordan86"]... Given six real constants $ a, b, c, d, e,$ and $ f$, not all zero, a conic section $ C : ax^2 \plus{} bxy \plus{} cy^2 \plus{} dx \plus{} ey \plus{} f \equal{} 0$ is determined. What is
the probability that $ C$ is an ellipse?[/quote]

I do not see anything about the ellipse being real in the posted problem.

Im sorry about that guys. I have the habit of calling only real ellipses an ellipse.

The original question only asked for the probability of a real ellipse. This does make the problem a little more difficult.

[quote="yetti"]We have an ellipse for $ b^2 \minus{} ac < 0$ and hyperbola for $ b^2 \minus{} 4ac > 0.$ Coefficients $ d, e, f$ are irrelevant. To avoid the problem that probability density of a uniform distribution of reals is zero everywhere, we contain $ a, b, c$ in an arbitrary finite box symmetrical around the origin, for simplicity $ |a|, |b|, |c| < 1.$ [/quote]
I considered this, but I don't think that this is the same probability distribution as the even distribution on $ S^5$.  Since $ (a, b, c, d, e, f)$ essentially represents a point in $ \mathbb{P}^5$, I'd prefer my interpretation and call this one suspect.

There is nothing wrong with the uniform distribution $ |a|, |b|, |c| < 1.$

Well, not in and of itself; I just wonder if it represents the problem accurately.  The problem itself is somewhat unclear.

It does not matter if uniform distribution in $ ( \minus{} 1, \plus{} 1)$ of several independent random variables is generated by the coordinates of points in the volume of a cube, or by the direction cosines of points on the surface of a sphere.

Exactly 50% of the ellipses are real and 50% imaginary. The conic $ ax^2 \plus{} bxy \plus{} cy^2 \plus{} dx \plus{} ey \plus{} f \equal{} 0$ is ellipse if

$ \det \mathbf K \equal{} \left| \begin{array}{cc} a & b/2 \\
b/2 & c \\
\end{array}\right| > 0$

and this ellipse is real if

$ \det \mathbf A \equal{} \left| \begin{array}{ccc} a & b/2 & d/2 \\
b/2 & c & e/2 \\
d/2 & e/2 & f \\
\end{array}\right| < 0$

Expanding by the 3rd column,

$ \det \mathbf A \equal{} f \det \mathbf K \plus{} \frac {_1}{^4} (bde \plus{} cd^2 \minus{} ae^2).$

$ d, e, f$ are values of random variables $ D, E, F$ with uniform distribution in $ ( \minus{} 1, \plus{} 1).$ After excluding hyperbolas by $ \det \mathbf K > 0,$ $ a, c$ are values of random variables $ A, C$ with equal probability densities symmetrical WRT zero and $ b$ is value of a random variable $ B$ with probabilty density symmetrical WRT zero. It follows that the probability density of $ BDE \plus{} CD^2 \minus{} AE^2$ is also symmetrical WRT zero, which means that the value $ f$ of $ F$ will change the the sign of $ \det \mathbf A$ from that of $ \det \mathbf K > 0$ with probability equal to $ \frac {_1}{^2}.$ Probability of getting a real ellipse is therefore $ \frac {31 \minus{} 6 \ln 2}{144} \doteq 0.1864.$ Again, this was verified by generating 20.000.000 random conics (just a minute, but not with Excel, this would overstretch it). The real/all ratio for ellipses was $ 0.499994 \pm 0.000187.$

Wow thats pretty cool. I did not think half of all ellipses are real.

By the way how did you run that program in excel?

Insert formula $ \equal{} 2*RAND() \minus{} 1$ into cell $ A1$

Select cell $ A1$ and drag-copy up to cell $ F1$ (6 random coefficients $ a, b, c, d, e, f$ uniform in (-1, +1))

Insert formula $ \equal{} A1*C1 \minus{} B1*B1/4$ into cell $ H1,$ skipping $ G1$ as a separator ($ \det \mathbf K$).
Insert formula $ \equal{} F1*H1 \plus{} (B1*D1*E1 \plus{} C1*D1*D1 \minus{} A1*E1*E1)/4$ into cell $ I1$ ($ \det \mathbf A$).
Insert formula $ \equal{} (H1 > 0)*1$ into cell $ J1$ (test for ellipse, multiplication by 1 formats the cell as number 1/0, instead of boolean TRUE/FALSE).
Insert formula $ \equal{} (I1 < 0)*J1$ into cell $ K1$ (test for real ellipse).

Select cells $ A1: J1$ and drag-copy up to cells $ A2: J2$.

Insert formula $ \equal{} J1$ into cell $ L1$ (initialize sum of ellipses)
Insert formula $ \equal{} K1$ into cell $ M1$ (initialize sum of real ellipses).
Insert formula $ \equal{} L1 \plus{} J2$ into cell $ L2$ (sum of ellipses).
Insert formula $ \equal{} M1 \plus{} K2$ into cell $ M2$ (sum of real ellipses).

Select cells $ A2: M2$ (2nd row) and drag-copy up to cells $ A1000: M1000,$ as a test.

Number of ellipses (out of 1000 conics) shows in the cell $ L1000,$ should be roughly 373.
Number of real ellipses (out of 1000 conics) shows in the cell $ M1000,$ should be roughly 186.

If everything is OK, select cells $ A1000: M1000$ (last row) and drag-copy up to cells $ A10000: M10000.$

Number of ellipses (out of 10000 conics) shows in the cell $ L10000,$ should be roughly 3730.
Number of real ellipses (out of 10000 conics) shows in the cell $ M10000,$ should be roughly 1860.

To run the program again, enter something (like 1) into an arbitrary empty cell. This causes all $ RAND()$ in the sheet to be generated again.
I also used similar code for 20.000.000 conics, but in C++, not Excel.

Taking the volume of a box is an invalid model of the space.

You may take the surface area of the box for the correct answer.

In fact any shape homeomorphic to a sphere can be used, if we take its surface area.

The only shape wherein you may take the volume is the sphere. The reason is that each scaled coordinate gets the same weight. In contrast to the box where the edges get the most weight, and the midsection of one face the least.

The program you wrote agrees with your answer yetti, because the points are taken out of a box.

I do believe that real and imaginary ellipses occupy the same area, but I am skeptical of your proof.

[quote="Ish"]Taking the volume of a box is an invalid model of the space..[/quote]
The problem was not to model space.

[quote="Ish"]You may take the surface area of the box for the correct answer.
In fact any shape homeomorphic to a sphere can be used, if we take its surface area.[/quote]
Both incorrect.

[quote="Ish"]The only shape wherein you may take the volume is the sphere. The reason is that each scaled coordinate gets the same weight. In contrast to the box where the edges get the most weight, and the midsection of one face the least.[/quote]
Incorrect, does not give independent random variables with uniform distribution.

[quote="Ish"]The program you wrote agrees with your answer yetti, because the points are taken out of a box...[/quote]
The program agrees with the answer becuse it is the correct answer. (One little omission in the last post, which is of no consequence.)

[quote="Ish"]I do believe that real and imaginary ellipses occupy the same area, but I am skeptical of your proof.[/quote]
You are welcome to find your own.

[quote="yetti"]

and this ellipse is real if

$ \det \mathbf A \equal{} \left| \begin{array}{ccc} a & b/2 & d/2 \\
b/2 & c & e/2 \\
d/2 & e/2 & f \\
\end{array}\right| < 0$
[/quote]

Im sorry for responding to this post so late.
The above statement is not true.
$ x^2 \plus{} y^2 \minus{} 1 \equal{} 0$ and
$ \minus{} x^2 \minus{} y^2 \plus{} 1 \equal{} 0$ are the same conic sections(real ellipse)  but have different determinants.

Your proof is incorrect.
I think idea that the probability of getting and imaginary or real ellipse is the same is correct.

That is the small omission of no consequence I mentioned above. The ellipse is real not when $ \det \mathbf A > 0,$ but when $ \frac {\det \mathbf A}{a \plus{} c} > 0.$ $ a, c$ have to be of the same sign to get ellipse in the 1st place (otherwise $ \det \mathbf K < 0$ for sure), with equal probability both positive or both negative. Note that $ \det \mathbf A$ depends only on $ a^2, c^2 > 0$ and on $ ac > 0$ for an ellipse. After eliminating hyperbolas by $ \det \mathbf K > 0,$ $ f$ flips the sign of $ \det \mathbf A$ from that of $ \det \mathbf K > 0$ with probability 50% and $ a \plus{} c$ again flips the sign of $ \frac {\det \mathbf A}{a \plus{} c}$ from that of $ \det \mathbf A$ with probability 50%, so it is of no consequence.

Furthermore, this part of the calculation is rather trivial. The tricky part is to calculate the probability $ \frac {31 \minus{} 6 \ln 2}{72}$ of getting any ellipse (real or imaginary).


trigonometry
inequalities unsolved
inequalities

Let real sequence {xn} n=1,... satisfy that:
   x1=2; x(n+1) *(1-2xn) = 2+xn ( n>=1)

Is there exist n different m such that xn=xm???

Is that
$x_1=2$
$x_{n+1}(1-2x_n) = 2+x_n, n \ge 1$?
[hide]
Notice
$x_{n+1}=\frac{x_n+2}{1-2x_n}$, the tan addition formula.  Thus, we have \[ \tan^{-1}{(x_{n+1})}=\tan^{-1}{(x_n)}+\tan^{-1}{2} \]

So let's say $\theta$ is the angle in $(0, \pi)$(though it doesn't have to be) s.t $\tan{\theta}=2$.  Then $x_n=\tan{n\theta}$.  And I believe $\frac{\pi}{\theta}$ is irrational, so there do [b]not[/b] exist $m, n, m \neq n, x_m=x_n$
[/hide]


AMC
AIME
modular arithmetic
number theory
relatively prime
least common multiple

I'm stumped, and can't figure them out. Help!

1. Let $T=(9^k: k$ is an integer, $0 \le k \le 4000)$. Given that $9^{4000}$ has 3817 digits and that its first digit is 9, how many elements of $T$ have 9 as their leftmost digit?

2. The sets $A= (z: z^{18} =1)$ and $B= (w: w^{48}=1)$ are both sets of complex roots of unity. The set $C= (zw)$ is also a set of complex roots of unity. How many distinct elements are in $C$?

[quote="WarpedKlown1335"][hide]1
9
81
729
6561
...

Whenever there are two n-digit powers of 9, you'll have a number starting with a 9.  When there's only one n-digit power of 9, there won't be a number starting with 9.  This is because 9ab...mn = 1cd...yz * 9.

You know there are 3817 digits in 9^4000.  If x is the number of pairs of powers of 9 with the same number of digits and y is the number of powers of 9 with distinct numbers of digits, then 2x + y = 4001 and x + y = 3817.  Solving, we get x = 184.

Since there are 184 pairs of powers of 9 with the same amount of digits, there are 184 elements of T with 9 as the leftmost digit.[/hide][/quote]

So basically, any power of 9 that begins with 9 also has a number of one less power beginning with 1 only with the same # of digits? Cool solution. :)

[quote="scorpius119"][hide="2"]The least common multiple of 18 and 48 is 144, so if $z^{18}=1$ and $w^{48}=1$, then $(zw)^{144}=1$. In addition $zw$ can be written in the form $\zeta^k$, where $\zeta$ is a complex number, $\zeta^{144}=1$ but $\zeta^d \neq 1$ for any positive integer $d<144$, and $k$ is a nonnegative integer less than or equal to 143.

Let $z=\zeta^{8a}$, $w=\zeta^{3b}$ (since $z$ and $w$ must have these forms. Then $8a+3b$ is divisible by 144. Since this can be written $3(a+b)+5b\equiv 0\pmod{144}$ and 5 and 144 are relatively prime, this will work for any $a+b$ so $zw$ can be $\zeta^k$ for any $0\leq k\leq 143$. Therefore there are $\boxed{144}$ different elements.[/hide][/quote]

I don't get the least common multiple part. [quote]Since this can be written $3(a+b)+5b\equiv 0\pmod{144}$ and 5 and 144 are relatively prime, this will work for any $a+b$ so $zw$ can be $\zeta^k$ for any $0\leq k\leq 143$.[/quote] Could you also explain this part in more detail?

"Whenever there are two n-digit powers of 9, you'll have a number starting with a 9. When there's only one n-digit power of 9, there won't be a number starting with 9."

sorry but could you explain this more clearly to me..

If an n-digit number starts with a 9, it could also be expressed as 9*10^n + c.
Dividing that by 9, we get 10^n + c/9.

In our scenario, the number is also a power of 9, meaning it's a multiple of 9, so c/9 will be a whole number as well.  10^n + c/9 will clearly be a string of n digits as well.

[quote="1234567890"][quote="scorpius119"][hide="2"]The least common multiple of 18 and 48 is 144, so if $z^{18}=1$ and $w^{48}=1$, then $(zw)^{144}=1$. In addition $zw$ can be written in the form $\zeta^k$, where $\zeta$ is a complex number, $\zeta^{144}=1$ but $\zeta^d \neq 1$ for any positive integer $d<144$, and $k$ is a nonnegative integer less than or equal to 143.

Let $z=\zeta^{8a}$, $w=\zeta^{3b}$ (since $z$ and $w$ must have these forms). Then $8a+3b$ is divisible by 144. Since this can be written $3(a+b)+5b\equiv 0\pmod{144}$ and 5 and 144 are relatively prime, this will work for any $a+b$ so $zw$ can be $\zeta^k$ for any $0\leq k\leq 143$. Therefore there are $\boxed{144}$ different elements.[/hide][/quote]

I don't get the least common multiple part. [quote]Since this can be written $3(a+b)+5b\equiv 0\pmod{144}$ and 5 and 144 are relatively prime, this will work for any $a+b$ so $zw$ can be $\zeta^k$ for any $0\leq k\leq 143$.[/quote] Could you also explain this part in more detail?[/quote]

[hide="the first part"]From the least common multiple of 18 and 48 being 144, then $z^{144}=1$ and $w^{144}=1$. Multiplying, $(zw)^{144}=1$.[/hide]

Yes, the second part wasn't very clear at all. This might make more sense:

[hide]To show that it will work for all $k$, show that $\zeta$ has the form $zw$ for some $z$ and $w$, because if $\zeta$ is in the set, so will $\zeta^2$, $\zeta^3$, and so on up to $\zeta^{143}$.

$z=\zeta^{8a}$ and $b=\zeta^{3b}$. Then $zw=\zeta^{8a+3b}$ so that $\zeta$ is in the set if there is a solution to $8a+3b\equiv 1\pmod{144}$. And this does have a solution $a=17$ $b=3$.

Therefore, $\zeta$ is in the set, so $\zeta^k$ is for $0\leq k\leq 143$.[/hide]


limit
logarithms
calculus
integration
function
real analysis
real analysis unsolved

is true that $\sum_{n=0}^{+\infty} |a_n|$ (and $a_n$ is a succession not defined for cases) is convergent if and only if $\lim_{n\to\infty}nlog(n)a_n=0$?

in the case of affermative answer, I'd like to see the demonstration. Thanks :D

could you show me in base to that criterion $\sum_{n}\frac{1}{nln(n)ln(ln(n))}$ diverges? and if i add other logaritm to denominator for example as $\frac{1}{nln(n)ln(ln(n))ln(ln(ln(n)))}$? diverges or converges?

Integral test: they all meet the criteria (positive decreasing, continuous)

$\sum_{n} |a_n|$ is convergent if and only if $\lim_{n\to\infty}na_n=0$

under the conditions:

i) $a_n$ is not a succession defined for cases;

ii) let $f_{k+1}(n)=ln(f_k(n))\ ,\ f_0(n): =ln(n)$
$a_n\ne\frac{1}{n\prod_{k\in \ P}f_k(n)}$ where $P\subseteq\mathbb{Z}^+$?

I think you're trying for something you can't have. There does not exist any function $f(n)$ such that $\sum_n^{\infty}|a_n|$ converges iff $\lim_{n\to\infty}f(n)a_n=0.$

You try to get around that by imposing all sorts of conditions, including "$a_n$ is not a succession defined for cases." I don't exactly know what that means, but I think it's an attempt to confine the discussion to elementary functions (the usual formulas of calculus) and to forbid piecewise-defined functions (an unnatural distinction).

Even if you could pile enough conditions in there to make something true, the result would be so narrow as to be useless in practice.

[quote="Kent Merryfield"]I think it's an attempt to confine the discussion to elementary functions (the usual formulas of calculus) and to forbid piecewise-defined functions (an unnatural distinction)[/quote]

ok ;) ,sorry ,I have literally translated the text in Italian :( , however it is one curiosity of mine.
I don't think that it would be so useless if it were true, but is it true or not? thanks