Αν $ x\plus{}y\plus{}z\equal{}1$ να αποδείξετε ότι:

$ 0\leq xy\plus{}yz\plus{}zx\minus{}2xyz\leq \frac{7}{27}$

[quote="Athinaios"]den exeis pou anikoun oi $ x,y,z$ ?[/quote]

Δεν καταλαβαίνω που είναι το λάθος. Έχουμε ότι

$ xy \plus{} yz \plus{} zx\leq\frac {7}{27} \plus{} 2xyz\Leftrightarrow 2(xy \plus{} yz \plus{} zx)\leq \frac {14}{27} \plus{} 4xyz\Leftrightarrow x^{2} \plus{} y^{2} \plus{} z^{2} \plus{} 2(xy \plus{} yz \plus{} zx)\leq \frac {14}{27} \plus{} 4xyz \plus{} x^{2} \plus{} y^{2} \plus{} z^{2}\Leftrightarrow (x \plus{} y \plus{} z)^{2}\leq \frac {14}{27} \plus{} 4xyz \plus{} x^{2} \plus{} y^{2} \plus{} z^{2}\Leftrightarrow 1\leq \frac {14}{27} \plus{} 4xyz \plus{} x^{2} \plus{} y^{2} \plus{} z^{2}\Leftrightarrow x^{2} \plus{} y^{2} \plus{} z^{2} \plus{} 4xyz\geq \frac {13}{27}$  (1).

Αρκεί λοιπόν να αποδείξουμε τη σχέση (1).

Όμως $ 3(x^{2} \plus{} y^{2} \plus{} z^{2})\geq (x \plus{} y \plus{} z)^{2}\Leftrightarrow x^{2} \plus{} y^{2} \plus{} z^{2}\geq \frac {1}{3} \equal{} \frac {9}{27}$      (2).

Οπότε τώρα έχουμε ότι ότι $ x^{2} \plus{} y^{2} \plus{} z^{2} \plus{} 4xyz\geq \frac {9}{27} \plus{} 4xyz$ και πρέπει να αποδείξουμε  από τη σχέση (1) ότι $ 4xyz\geq \frac {4}{27}\Leftrightarrow xyz\geq \frac {1}{27}$   (3).

Αλλά $ x \plus{} y \plus{} z\geq 3\sqrt [3]{xyz}\Leftrightarrow xyz\leq \frac {1}{27}$   (4)

Στη σχέση (4) η φορά της ανίσωσης έπρεπε να βγαίνει αντίθετη απο αυτή που έχουμε τώρα για να ισχύει η (3). Γιατί όμως βγαίνει λάθος? :(

Φιλε μου Γιωργο και εγω νομιζω πως δεν εχεις κανει καποιο λαθος σε πραξεις. Απλα η προσεγγιση σου δεν δουλευει.
Νομιζω συγκεκριμενα οτι χαλαει στη σχεση 2 που γραφεις. Χρησιμοποιεις αυτη την πολυ γενικη ανισοτητα μαζι με το δεδομενο σου
οποτε δεν φτιαχνεις την καλυτερη δυνατη σχεση που θα μπορουσε. Αυτο ειναι ενα κομματι που ομορφαινει τις ανισοτητες.
Μπορει ομως να κανω και λαθος.

estw oti einai mi arnitikoi pragmatikoi
exw gia to prwto
 $ xy \minus{} xyz \plus{} yz \plus{} zx \minus{} xyz \equal{} xy(1 \minus{} z) \plus{} yz \plus{} zx(1 \minus{} y) \equal{} \equal{} xy(x \plus{} y) \plus{} yz \plus{} zx(x \plus{} z ) > \equal{} 2(xy)*riza (xy) \plus{} yz \plus{} 2xz*riza(xy) > \equal{} 0, pou isxyei$

Φιλε Athinaios ξανακοιταξε την προσεγγιση σου
Κοιτα τα εξης
1ον) $ x^2\plus{}y^2>\equal{}2xy$ και οχι $ x\plus{}y>\equal{}2xy$
2ον) Εκει που εχεις $ xyz<\equal{}1/27$ πολλαπλασιαζοντας με - αλλαζει η φορα

Παρολαυτα ωραιες σκεψεις :)

giwrgo dn prepei na exeis kanei lathos stis prakseis apla einai i omorfia ayti stis anisotites ot dn vgenun panta me ola. par'ola ayta emena m arese i prosegisi p kaneis (pou ftiaxneis to $ (x\plus{}y\plus{}z)^2$) :wink:

meta tis episimanseis tou alexandrou nomizw oti tha simfwnisw me ton giwrgo gia tn fora  :P

Γνωστή είναι ΙΜΟ 1984 pr 1 . Πρέπει να πεις νομίζω ότι είναι μη αρνητικοί οι αριθμοί . Για τη λύση
 (ομοιγενοποίηση και μετά Schur)  :)

Κατά πρώτον συγγνώμη που ξέχασα να γράψω ότι οι αριθμοί είναι μη αρνητικοί.

Κατά δεύτερον Γίωργο κάνεις το λάθος ότι [u]μικραίνεις πάρα πολύ την ανισότητα[/u]. Αν το κοιτάξεις είναι σαν να λες π.χ. ότι για να αποδείξεις την $ x^2\plus{}y^2\plus{}2xy\geq 0$ επειδή $ x^2\plus{}y^2\geq0$ πρέπει $ 2xy\geq 0$ δηλαδή οι $ x,y$ ομόσημοι, που στην πραγαμτικότητα όμως δεν είναι αναγκαστικό (εύχομαι να είναι κατανοητό το παράδειγμα μου...).

Επίσης sry αλλά δεν γνώριζα ότι ήταν πρόβλημα IMO (κι άρα γνωστή)  :blush: .

Τέλος εγώ στη λύση μου περιληπτικά αυτό που έκανα είναι (για το 2ο μέλος της ανισότητας...το 1ο είναι αρκετά εύκολο και η ισότητα ισχύει γαι την τριάδα $ (1,0,0)$ κι όχι για $ (1/2,1/2,0)$ :wink: ):

Θεωρώ $ z$ τον μικροτερο κι άρα μικρότερο από $ \frac{1}{3}$. Epeita αποδεικνύω ότι:
$ f(x,y,z)\leq f(\frac{x\plus{}y}{2},\frac{x\plus{}y}{2},z)$ όπου $ f(x,z,y)\equal{}xy\plus{}yz\plus{}zx\minus{}2xyz$.

Telos apodeiknyo oti:
$ f(\frac{x\plus{}y}{2},\frac{x\plus{}y}{2},z)\leq \frac{7}{27}$, που αποδεικνύεται σχετικά εύκολα (με ανάλυση) αφού είναι μίας μεταβλητής (αφού $ x\plus{}y\equal{}1\minus{}z$, $ 0\leq z\leq \frac{1}{3}$).

Αν θέλετε περισσότερες λεπτομέρειες πείτε μου ευχαρίστως :)

[quote="bilot4"]Κατά πρώτον συγγνώμη που ξέχασα να γράψω ότι οι αριθμοί είναι μη αρνητικοί.

Κατά δεύτερον Γίωργο κάνεις το λάθος ότι [u]μικραίνεις πάρα πολύ την ανισότητα[/u]. Αν το κοιτάξεις είναι σαν να λες π.χ. ότι για να αποδείξεις την $ x^2 \plus{} y^2 \plus{} 2xy\geq 0$ επειδή $ x^2 \plus{} y^2\geq0$ πρέπει $ 2xy\geq 0$ δηλαδή οι $ x,y$ ομόσημοι, που στην πραγαμτικότητα όμως δεν είναι αναγκαστικό (εύχομαι να είναι κατανοητό το παράδειγμα μου...).

Επίσης sry αλλά δεν γνώριζα ότι ήταν πρόβλημα IMO (κι άρα γνωστή)  :blush: .

Τέλος εγώ στη λύση μου περιληπτικά αυτό που έκανα είναι (για το 2ο μέλος της ανισότητας...το 1ο είναι αρκετά εύκολο και η ισότητα ισχύει γαι την τριάδα $ (1,0,0)$ κι όχι για $ (1/2,1/2,0)$ :wink: ):

Θεωρώ $ z$ τον μικροτερο κι άρα μικρότερο από $ \frac {1}{3}$. Epeita αποδεικνύω ότι:
$ f(x,y,z)\leq f(\frac {x \plus{} y}{2},\frac {x \plus{} y}{2},z)$ όπου $ f(x,z,y) \equal{} xy \plus{} yz \plus{} zx \minus{} 2xyz$.

Telos apodeiknyo oti:
$ f(\frac {x \plus{} y}{2},\frac {x \plus{} y}{2},z)\leq \frac {7}{27}$, που αποδεικνύεται σχετικά εύκολα (με ανάλυση) αφού είναι μίας μεταβλητής (αφού $ x \plus{} y \equal{} 1 \minus{} z$, $ 0\leq z\leq \frac {1}{3}$).

Αν θέλετε περισσότερες λεπτομέρειες πείτε μου ευχαρίστως :)[/quote]

Εχεις δίκιο Δημήτρη...τα έγραφα στο σχολείο..συγχώρεσε με! Αλλά πιστεύω ότι η ABC είναι αρκετά κομψη για αυτή την ανισότητα...Σου βγάζει και μέγιστη και ελάχιστη αμέσως.

[quote="Nick Rapanos"]Λόγω της ABC, για το μέγιστο και το ελάχιστο θα πρέπει δύο από τις τρεις μεταβλητές να είναι ίσες (Αποδεικνύεται ότι η $ f(xyz)$ είναι γνήσια φθίνουσα). Οπότε θέτοντας $ x \equal{} y \equal{} a$ και $ z \equal{} 1 \minus{} 2a$ το πρόβλημα ανάγεται σε μελέτη συνάρτησης μίας μεταβλητής. Το ελάχιστο επιτυγχάνεται όταν $ \{x,y,z\} \equal{} \left\{\frac {1}{2},\frac {1}{2},0\right\}$ και το μέγιστο όταν $ \{x,y,z\} \equal{} \left\{\frac {1}{3},\frac {1}{3},\frac {1}{3}\right\}$, φυσικά λαμβάνοντας υπόψιν και όλες τις μεταθέσεις αυτών λόγω της συμμετρικότητας.[/quote]
Pε παιδιά ποιά είναι η ανισότητα ABC :?:

Δες εδώ το συννημένο αρχείο!

[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=189754[/url].

Δημήτρη, δεν είναι ακριβώς αυτή η abc principle...αυτό το article είναι μόνο η αρχή της κατα κάποιον τρόπο!


solve: [tex] a^2+p^n=b^4 [/tex]
where [tex]a,b,n \in \mathbb{N}[/tex], [tex]p[/tex] is a prime number and [tex]\frac {p^2-1}8[/tex] is an odd number

$(a,b,n)=(0,p^k,4k)$ are the only solutions.
$Proof:$
$\frac{p^2-1}{8}$ is an odd number, so $p>2$.
Let $(a,b,n)$ a solution; we may assume $a>0$.
We have $b^2-a=p^k$ and $b^2+a=p^m$ with $0\leq k<m$ and $k+m=n$.
$2b^2=p^k(1+p^{m-k})\Rightarrow k=2q$ and $1+p^{m-k}=2\beta^2\Rightarrow2$ is a square $\mod p$ $\Rightarrow\left(\frac{2}{p}\right)=(-1)^{\frac{p^2-1}{8}}=1$. Contradiction.


induction
number theory
prime numbers
algebra unsolved
algebra

Find all integers n>2 for which:
 (2n)!=(n-2)!*n!*(n+2)!

Here is my solution:

Between (2n)! and (n+1)! can't be any prime numbers because (2n)!=(n+2)!*(n+3)*.....*(2n).
If any of (n+3), (n+4),...,(2n) is prime P you can't get P=a*b where a€n! and b€(n-2)! (€-is element of)

If you want to see proof that 3 and 5 are the only number that has no primes between (n+2) and 2n go to http://www.mathlinks.ro/Forum/viewtopic.php?t=187791

You can check that 3 and 5 works for upper equation:
3: 1!*3!*5!=5!*2*3=5!*6=6!
5: 3!*5!*7!=7!*2*3*4*5*2*3=7!*2*4*3*3*2*5=7!*8*9*10=10!

This is what I came up with. I'm sure there are some easier solutions for this problem.

Use mathematical induction to prove that there are no solution for $ n>10$.


[quote="benndamann33"]somehow my setspace package got deleted. I download the package setspace.sty but don't know what directory to place it in? anyone know where packages go?[/quote]

You should really ask this questoin in a latex forum. However, I assume you pose this question cause you are working under linux. Do from a console

locate *.sty

for example in my red hat disrtibution a good place to place latex packages for generic uses is

/usr/share/texmf/tex/latex/misc/

One thing you need to know is that the path where the system searches for latex packages is stored in an environment variable which if I recall correctly is called TEXINPUTS (do info latex). You can modify this variable in order to add some directory of your convenience, for instance , since I don't want to be root all the time I have a directory in my user home called

~/usr/share/..

and I put there my newly acquired packages

Ursula


this game is great plus its free!
[url=http://www.gangstawar.com/index.php?act=registration&referal=c8r7497342]www.gangstawar.com/index.php?act=registration&referal=c8r7497342[/url]

or this might be the right link... :blush:  [url=http://www.gangstawar.com/index.php?act=speclink&speclink=x13s497342]www.gangstawar.com/index.php?act=speclink&speclink=x13s497342[/url]

ummm, lol, why don't you just post it, and we can copy paste?

i like the game it is pretty fun to play.


This is problem 40 in AoPS volume 1:

$ K$ takes 30 minutes less time than $ M$ to travel a distance of 30 miles. $ K$ travels $ \frac13$ mile per hour faster than $ M$. If $ x$ is $ K$'s rate of speed in miles per hour, then find K's time for the distance in terms of $ x$.

After trying for a while and ending up with nothing I looked at the solution and it said that all the information about $ M$ was useless and that, because rate X time = distance, time X $ x$ = 30, or time = $ \frac {30}{x}$.

[hide="What I tried"]
Let $ M$'s rate be $ y$. Since it said the relationship between $ x$ and $ y$, we know:

$ x \equal{} y \plus{} \frac {1}{3} \implies y \equal{} x \minus{} \frac13$

From the first sentence we know:

$ \frac {30}{x} \plus{} .5 \equal{} \frac {30}{y}$

$ 30y \plus{} .5 xy \equal{} 30x$

Now, we can substitute back in what we had from the first equation:

$ 30(x \minus{} \frac13) \plus{} \frac12(x)(x \minus{} \frac13) \equal{} 30x$

$ 30x \minus{} 10 \plus{} \frac {x^2}{2} \minus{} \frac {x}{6} \equal{} 30x$

$ \minus{} 10 \plus{} \frac {x^2}{2} \minus{} \frac {x}{6} \equal{} 0$

$ x^2 \minus{} \frac {x}{3} \minus{} 10 \equal{} 0$

$ x \equal{} \frac {\frac 13 \pm \sqrt {\frac 19 \minus{} 4(1)( \minus{} 10)}}{2(1)}$

$ x \equal{} \frac {\frac 13 \pm \sqrt {\frac {361}{9}}}{2}$

Since the rate must be positive:

$ x \equal{} \frac {10}{3}$

Now, we have:

$ time \equal{} \frac {30}{\frac {10}{3}}$

$ \boxed{time \equal{} 9}$
[/hide]

Is there anything wrong with my solution? Or am I just doing extra work?

[quote="PowerOfPi"]
$ \minus{} 10 \plus{} \frac {x^2}{2} \minus{} \frac {x}{6} \equal{} 0$

$ x^2 \minus{} \frac {x}{3} \minus{} 10 \equal{} 0$

[/quote]

You forgot to multiply $ \minus{}10$ by $ 2$.

Oops. But a numerical answer can still be found (it might look ugly).

*sigh*... I have that problem with AoPS books too... It's their fatal flaw... Some of the solutions expect you to just all of a sudden "get" it...

But this one is quite obvious... You can find both K&M's times like this: K=x, so:

(x)(t)=30, t=30/x which is super easy...

For M:

(x-1/3)t=30, t=(30/x)-90

this problem is WAY to easy to be in AoPS volume 1....

[quote="batteredbutnotdefeated"]
this problem is WAY to easy to be in AoPS volume 1....[/quote]
Would you say that if you were the one posting the question? ;)

This problem isn't difficult, it's just that you have to know what's useful information and what isn't.

What I mean is, why can't I just solve for x?

PowerOfPi, based on the question you asked in the first post [quote]Is there anything wrong with my solution? Or am I just doing extra work?[/quote]
The answer is that you are doing extra work. The problem state to find the time for K's distance [b]in terms of x. [/b]

So, they weren't expecting numerical values.

But you could also say the answer is (where $ n$ is the numerical answer): $ n\plus{}0x$

I maybe misunderstanding your question, but what is the point of representing the answer in the form: $ n \plus{} 0x,$ a coefficient of $ 0$ makes that the same as $ n,$ which is not what the problem wants.

@PowerOfPi, are you asking if there is a way to find a numerical answer to this question even though the problem doesn't ask for it?

I know there is a numerical answer, but I got the answer to my question.


Let x and y be odd integers so |x| is not equal to |y|. Suppose that every integer is coloured either red, green, blue, or yellow. Show that there are two integers of the same colour whose difference has one of the following values: x, y, x+y, x-y.

Warning - a solution can be found at

[hide]http://www.kalva.demon.co.uk/short/soln/sh99c6.html[/hide]

although I recommend you to work on the problem first.


abstract algebra
trigonometry
function
limit
real analysis
geometry
3D geometry

$P_{r}( \theta)=\frac{1}{2 \pi}\frac{1-r^{2}}{1-2r \cos( \theta)+r^{2}}$ is the Poisson kernel ($0<r<1$). Prove that if $f \in L^{1}([-\pi;+\pi])$ is an integrable function, then for almost every $x \in [-\pi;+\pi]$:
$\lim_{r \to 1^{-}}(f * P_{r})(x)=f(x)$

Rough outline of the approach I was taught:

This can all be relaxed a little from "radial" to "nontangential."

First: this is true everywhere for $f$ continuous. So it what follows we will write an integrable function as a sum of a continuous function and a function of small $L^{1}$ norm.

Now consider the "bad" function of small $L^{1}$ norm. Look at it's "oscillation," meaning the difference between the liminf and limsup of the very limit we're trying to compute. Show that that oscillation is majorized by a constant times a maximal function based on this operation. (This maximal function - the nontangential maximal function - is simply the sup over an appropriate cone of $|f*P_{r}(\theta)|.$)

Now show that the nontangential maximal function is bounded by a constant multiple of the Hardy-Littlewood maximal function. Since this is weak-type 1-1, the oscillation, which is what we are interested in, is also weak-type 1-1.

That will lead to the argument for pointwise a.e. convergence.

---

The famous result of Eli Stein is that for a family of convolutions operators to have a pointwise a.e. limit, the appropriate maximal function must satisfy a weak-type inequality.

Thank you. I think I did almost the same, introducing the maximal function $P^{*}(x)=\sup_{0<r<1}|(f*P_{r})(x)|$ and majorize it with the Hardy-Littlewood maximal function. Nevertheless, I don't see how to generalize my approach to a cone ?

With the maximal function  $f^{*}(x)=\sup_{0<r<1}\sup_{|y-x|<1-r}|(f*P_{r})(y)|$.

yep, I had understood that, but with this new maximal function, I can't prove a maximal inequality  :maybe:

Let $\tilde P_{r}(\theta): =\sup_{|t-\theta|\le 1-r}P_{r}(t)$. The "new" (nontangential) maximal function is dominated by $\sup_{0<r<1}(|f|*\tilde P_{r})(x)$, which in turn should be dominated by the Hardy-Littlewood maximal function.


floor function
number theory
prime factorization

In how many zeros does the number represented by $ 10!$ end?

Either have it memorized, or note that it's just the number of $ 5$'s in the prime factorization (we need $ 10$'s, but there are already many more $ 2$'s than $ 5$'s), which is $ \left\lfloor\frac {10}5\right\rfloor \plus{} \left\lfloor\frac {10}{5^2}\right\rfloor \plus{} \cdots \equal{} \boxed2 \plus{} 0 \plus{} 0 \plus{} \cdots$.

[quote="math154"][b]Either have it memorized,[/b] or note that it's just the number of $ 5$'s in the prime factorization (we need $ 10$'s, but there are already many more $ 2$'s than $ 5$'s), which is $ \left\lfloor\frac {10}5\right\rfloor \plus{} \left\lfloor\frac {10}{5^2}\right\rfloor \plus{} \cdots \equal{} \boxed2 \plus{} 0 \plus{} 0 \plus{} \cdots$.[/quote]

[quote="nt811581147"]Memorization gives 10!=3,628,800 which has 2 zeroes.[/quote]

Lol? :P


How many of you get less than 8 hours of sleep every night?

I get about 7.75 :)

No. Actually, I get $7.\overline{9}$ hours of sleep

Let's see...

I usually get $9: 00$ hours of sleep, but today, I waked up at $5: 45$, so it is $8: 45$ just for today.

School nights or other occasion to get up early: less than 8 hours
Otherwise: more than 8 hours

uhh <7-ish hours weekdays, 12-ish weekends and summer

I'm in exams.  I cannot sleep more than that.  And the bad thing is that I feel tired when at work and awake when I must sleep. :(  :mad:  :mad:

lol that sucks. i have that too. i get around 10 hours. every once in a while my friend invites me for a sleepover so i get zero sleep (my record was 30 minutes of sleep becuase his mom came in lol).

around seven hours a day on average?

i think i need some more

7-9 on weekdays, 12-14 on weekends
got 10 hours the past 2 days bc arml


A car is averaging 50 miles per hour. If the car maintains this speed, how many minutes less would a 450-mile trip take than a 475-mile trip?

If the car is going at $ 50$ miles per hour, the time it takes to travel $ 1$ mile would be $ \frac{1}{50 \frac{mi}{h} \times 1 mi$ *. Thus, it takes $ 0.02$ hours, or $ 0.02(60)=1.2$ minutes** to drive $ 1$ mile. The difference between $ 450$ and $ 475$ miles is $ 475-450=25$ miles. Thus, the difference in time in the two trips is $ 25(1.2)=\fbox{30}$ minutes.

*We take the reciprocal so that miles would cancel out, leaving a pure time unit.
**We use this conversion since there are 60 minutes in an hour.


hello, do you mean
$ (x\plus{}y\plus{}z)^3\equal{}3u$
$ (y\plus{}z\plus{}u)^3\equal{}3v$
$ (z\plus{}u\plus{}v)^3\equal{}3x$
$ (u\plus{}v\plus{}x)^3\equal{}3y$
$ (v\plus{}x\plus{}y)^3\equal{}3z$
Sonnhard.


I saw the Purser's theorem from Mathworld ( http://mathworld.wolfram.com/PursersTheorem.html ). I was wondering whether someone knows a criterion of choosing those +/- signs in the theorem. And a proof of the theorem would be nice too.

Let t, u, and v be the lengths of the tangents to a circle K from the vertices of a triangle with sides of lengths a, b, and c. Then the condition that K is tangent to the circumcircle of the triangle is that

$\pm at \pm bu \pm cv = 0$

If the circle K is reduced to the tangency point (circle of zero radius) with the triangle cirrcumcircle, Purser's theorem is reduced to Ptolemy's theorem. Hence, it the circle K is tangent to the triangle circumcircle between the vertices B, C, then at = bu + cv, etc. More general theorem:

4 circles $C_1,\ C_2,\ C_3,\ C_4$ are tangent to a single circle C or to a single line (in this order) iff $t_{12}t_{34} + t_{23}t_{41} = t_{13}t_{24}$, where $t_{ij}$ are the segment lengths of the common external tangents of the circles $C_i,\ C_j$, if these 2 circles have the same tangencies with the circle C (both external or both internal) and $t_{ij}$ are the segment lengths of the common internal tangents of the circles $C_i,\ C_j$, if these 2 circles have different tangencies with the circle C (one external, the other internal).

Purser's theorem from Mathworld is a special case of the more general theorm, when 3 of the 4 circles  $C_1,\ C_2,\ C_3,\ C_4$ are reduced to points.

[b]Purser's theorem (necessity):[/b] Let (O) be the circumcircle of the triangle $\triangle ABC$ with the sides a = BC, b = CA, c = AB and let t = AT, u = BU, v = CV be the tangent lengths form the vertices A, B, C to another circle (K). If the circle (K) is tangent to the circle (O) between the vertices B, C, then at = bu + cv.

Assume that the circle (K) is tangent to the circle (O) at a point T on the arc BC not containing the vertex A. Let $\rho,\ R$ be the radii of the circles (K), (O). Let A', B', C' be the intersection of the lines AT, BT, CT with the circle (K) other than the tangency point T. The circles (O), (K) are centrally similar with the similarity center at the tangency point T. Consequently,

$\frac{TA'}{AT} = \frac{TB'}{BT} = \frac{TC'}{CT} = \frac \rho R$

The power of the point A to the circle (K) is

$t^2 = AT \cdot AA' = AT (AT + TA') = AT^2 \left(1 + \frac{TA'}{AT}\right) = AT^2 \left(1 + \frac \rho R\right)$

which implies

$t = AT \sqrt{1 + \frac \rho R}$

and similarly,

$u = BT \sqrt{1 + \frac \rho R},\ \ v = CT \sqrt{1 + \frac \rho R}$

Since the points A, B, T, C are concyclic in this order, by Ptolemy's theorem,

$AT \cdot BC = AB \cdot CT + BT \cdot CA$

Multiplying the last equation by $\sqrt{1 + \frac \rho R}$ and substituting a = BC, b = CA, c = AB, $t = AT \sqrt{1 + \frac \rho R},$ etc.,

$at = cv + bu$

Sufficiency can be proved similarly as sufficiency of Ptolemy's theorem.


Show that there are infinite solutions $(a,b,c)$, with $a, b, c \in N$, satisfying:

$2a^2 + 3b^2 -5c^2=1997$

This isn't really a formal proof, but...

Note that 2c^2 + 3c^2 - 5 c^2 = 0.
Thus, this problem can be rewritten as 2(x+c)^2+3(y+c)^2-5(c^2)=1997, where x and y are both greater than c.  This can be further simplified into 2d^2+3e^2=1997.  I'm fairly certain there are going to be two numbers that satisfy this equation...correct me if i'm wrong though.

Hope this helps.

[quote="falconwing64"]2d^2+2e^2=1997.  I'm fairly certain there are going to be two numbers that satisfy this equation...correct me if i'm wrong though.

Hope this helps.[/quote]

one side is even, the other is odd, so there are no solutions

odd+even=odd

P.S.
5 and 31 work for d and e.

sry, didnt' notice my mistake until just now  :blush:  ...see above for edit.


$\frac{1}{\sqrt{11-2\sqrt{30}}}-\frac{3}{\sqrt{7-2\sqrt{10}}}-\frac{4}{\sqrt{8+4\sqrt{3}}}$
if u can get the texer to take i would appreciate it but the main problem is i don't know how to solve it the answers say the answer is zero  :huh:
but i don't know how to solve the problem

HELP!!! :(

[size=75][color=orange][LaTeXed by moderator solafidefarms][/color][/size]

[quote="runnerspd8"]1/sqrt of the quantity 11-2root30
minus
3/sqrt of the quantity 7-2root10
minus
4/sqrt of the quantity 8+4root3

sorry its not in LATEX
the TEXER wouldn't take it
if u can get the texer to take i would appreciate it but the main problem is i don't know how to solve it the answers say the answer is zero  :huh:
but i don't know how to solve the problem

HELP!!! :([/quote]
for the first(I can infer the others) is it 2*sqrt10 or sqrt10*2?

Do you want this?

$\sqrt{11-2\sqrt{30}} - 3\sqrt{7-2\sqrt{10}} - 4\sqrt{8+4\sqrt3}$

Correct it please  :)

Use conjugate radicals.

I think...
Look in chapter one of art of problem solving, p11.

Like this?

$\displaystyle{\sqrt[4]{3}}$

$\frac{1}{\sqrt{11-2\sqrt{30}}}=\frac{\sqrt{11+2\sqrt{30}}}{121-120}=\sqrt{11+2\sqrt{30}}$

$-\frac{3}{\sqrt{7-2\sqrt{10}}}=-\frac{3\sqrt{7+2\sqrt{10}}}{\sqrt{49-40}}=-\sqrt{7+2\sqrt{10}}$

$-\frac{4}{\sqrt{8+4\sqrt{3}}}=-\frac{4\sqrt{8-4\sqrt{3}}}{\sqrt{64-48}}=-\sqrt{8-4\sqrt{3}}$

If you have no idea what just happened, tell me. :)

Now when we add them up, we get

$\sqrt{11+2\sqrt{30}}-\sqrt{7+2\sqrt{10}}-\sqrt{8-4\sqrt{3}}$

Time to denest the radicals.

$\sqrt{8-4\sqrt{3}}=x+y\sqrt{3}$
$8-4\sqrt{3}=x^2+3y^2+2xy\sqrt{3}$
$8=x^2+3y^2$
$-4\sqrt{3}=2xy\sqrt{3}$

$-2=xy$
$8=x^2+3y^2$
$x=\sqrt{6}, y=-2/sqrt{6}=-\frac{\sqrt{6}}{3}$

:wacko:
Where did you get this problem from again? I definitely don't like it... :angry:

[quote="236factorial"]WAIT!!! chess64, are you sure you rationalized the denominators correctly?![/quote]

What are you talking about? :whistling:

BTW, don't double post ;)

Nah, you fixed it. What were you thinking?
Sorry about the double post  :blush:

Where in the AoPS book do you learn to denest radicals? I know I have seen it before, but I have never actually applied it and forgot how to do it.

[quote="236factorial"]i think this should be bumped.
By the way, chess64 rationalized the denominators in the first step.[/quote]

Whoa, stop spamming.  Yes, I consider bumping a topic after [b]16 minutes[/b] spam.

I think by denesting he means rationalizing the denominator.

[quote="236factorial"]What's denesting?[/quote]

Read chess64's post.

[quote="Gyan"][quote="chess64"]$\frac{1}{\sqrt{11-2\sqrt{30}}}=\frac{\sqrt{11+2\sqrt{30}}}{121-120}=\sqrt{11+2\sqrt{30}}$

$-\frac{3}{\sqrt{7-2\sqrt{10}}}=-\frac{3\sqrt{7+2\sqrt{10}}}{\sqrt{49-40}}=-\sqrt{7+2\sqrt{10}}$

$-\frac{4}{\sqrt{8+4\sqrt{3}}}=-\frac{4\sqrt{8-4\sqrt{3}}}{\sqrt{64-48}}=-\sqrt{8-4\sqrt{3}}$
... And lot of good work not repeated here .,,.,

Where did you get this problem from again? I definitely don't like it... :angry:[/quote]

Of course when   one also notices that:

$\sqrt{11+2\sqrt{30}} = \sqrt 6 + \sqrt 5$ and
$\sqrt{7+2\sqrt{10}}= \sqrt 5 + \sqrt 2$ and
$\sqrt{8-4\sqrt{3}} = \sqrt 6 - \sqrt 2 $
 One need  not be that  :mad: (  :angry)   one can even :) :)
 :) :) when things simplify to $0$[/quote]

Where would one seek information on how to aptly notice these special cases?

[quote]Where would one seek information on how to aptly notice these special cases?[/quote]
Just Notice that $(\sqrt x + \sqrt y)^2 =  (x+y) + 2 \sqrt {xy} $
for example in the first case ; solve for $(x,y)$ when $x+y=11$ and $xy=30$ IOW find two factors of $30$ whose sum is $11$ etc..

Another standard trick which  one should know (Chess64 mentioned it above but it can be simlified) is - how to remove radicals from the denominator:

eg: $\frac 1 {\sqrt 6 - \sqrt 5 } = \frac  1 {\sqrt 6 - \sqrt 5 }\cdot \frac {\sqrt 6 + \sqrt 5 } {\sqrt 6 + \sqrt 5 }
=  \frac  {\sqrt 6 + \sqrt 5 }{6-5} =   \sqrt 6 + \sqrt 5 $ etc ...

Hope this is helpful


[hide] a+3x/2a + a-x/6a - 2x+a/3a
(3a+9x+a-x-4x+2a)/6a
6a+4x/6a
a)1+2x/3a

b-c/bc = c-a/ca + a-b/ab
ab-ac/abc=bc-ab/abc+ac-bc/abc
ab-ac=bc-ab+ac-bc
ab-ac=-ab+ac
2ab=2ac
ab=ac
b=c and/or a=0

6/x-2y + 4/2y-x
-(6/x-2y)+4/2y-x
-6/2y-x+4/2y-x
-2/2y-x[/hide]

so can anione check if the answer posted is correct?


calculus integration trigonometry logarithms function algebra domain

Calculate the following indefinite integrals.

[1] $\int \frac{\sin x\cos x}{1+\sin ^ 2 x}dx$

[2] $\int x\log_{10} x dx$

[3] $\int \frac{x}{\sqrt{2x-1}}dx$

[4] $\int (x^2+1)\ln x dx$

[5] $\int e^x\cos x dx$

[quote="amirhtlusa"]4(using by parts)
$\frac{3ln{|x|}(x^3+3x)-x^3-9x}{9}$[/quote]

In this case, | | can be omitted.

[quote="amirhtlusa"][quote="kunny"][quote="amirhtlusa"]4(using by parts)
$\frac{3ln{|x|}(x^3+3x)-x^3-9x}{9}$[/quote]

In this case, |\ | can be omitted.[/quote]
no, i don't think so... ;)[/quote]

Why? :?

amirhtlusa - The original function only has domain [tex](0, \infty)[/tex]

[quote="kunny"][quote="amirhtlusa"]just looking through them, 5 seems the easiest:
$\frac{e^x(\sin{x}+\cos{x})}{2}$[/quote]

Yes, in this case integral by parts doesn't need.[/quote]

I have another method

Its a little long-winded - I would not use it on this problem, but I post it because it is a useful technique in some problems

Write

$\int e^x \cos(x)\,dx=\mathbb{R}\left[\int e^x e^{ix}\,dx\right ]$

[quote="kunny"][quote="amirhtlusa"]2
$\frac{(\log_{10}|x|)x^2}{2}-\frac{x^2}{4\ln10}$[/quote]

 :([/quote]
hmm i get the same thing, again
here is how you i did it.
let $u=\log_{10}x, dv=xdx$ then do by parts:
you get:
$\frac{x^2}{2}(\log_{10}x-\frac{1}{2\ln{10}})$ ;)


Determine the smallest value of the expression
$ T\equal{}\frac{ab\plus{}bc}{a^2\minus{}b^2\plus{}c^2}\plus{}\frac{bc\plus{}ca}{b^2\minus{}c^2\plus{}a^2}\plus{}\frac{ca\plus{}ab}{c^2\minus{}a^2\plus{}b^2}$ where $ a,b,c$ are three sides of a triangle $ ABC$ and $ abc\equal{}1$

[quote="peine"]I found that $ T_{min} \equal{} 3$[/quote]

Why ?????????
With a=b=c=1 =>T=6


IMO Shortlist
geometry
3D geometry
sphere
Volume
IMO Longlist

Determine the volume of the body obtained by cutting the ball of radius $R$ by the trihedron with vertex in the center of that ball, it its dihedral angles are $\alpha, \beta, \gamma.$

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

I remember solving this before the crash.

The ratio between this volume and the volume of the sphere is equal to the ratio between the area determined by the trihedron on the sphere and the area of the sphere. The region is a spherical triangle with angles $\alpha,\beta,\gamma$, and its area is thus $(\alpha+\beta+\gamma-\pi)\cdot R^2$. The volume we are looking for must then be $\frac{\alpha+\beta+\gamma-\pi}{4\pi}\cdot \frac{4\pi\cdot R^3}3=\frac{(\alpha+\beta+\gamma-\pi)\cdot R^3}3$.


Find all pairs $(m;n)$ of rational numbers such that $n^4-6n^2m^2+8m^4=0$

Hi, dear!

My solution:

        $\large{n^4-6n^2m^2+8m^4\,=\,0\,\Longleftrightarrow\,(n^2-4m^2)(n^2-2m^2)=0\Longrightarrow{n=+-2m\,\,\textrm{or}\,\,n=+-m\sqrt{2}\,\,\,\forall{m,n\,\,\in{\textrm{R}}}}}$

You forgot the negative cases:
$n=\pm 2m$ or $n=\pm m\sqrt{2}$ ;)

Edit: The problem actually specifies that $m$ and $n$ are rational numbers, so the answer should be $n=\pm 2m$.


Prove that $ \forall x,y,z>0$ we have:

$ \sum \frac{x}{(x+y)(x+z)} \leq \frac{9}{4(x+y+z)}$.

made by Nedelcu C.T.

cheers! :D :D

By multiplication with 4(y+z)(z+x)(x+y)(x+y+z), your inequality is equivalent to

  4(x(y+z) + y(z+x) + z(x+y)) * (x+y+z) <= 9(y+z)(z+x)(x+y).

But

  4(x(y+z) + y(z+x) + z(x+y)) * (x+y+z)
= 4(x(y+z)(x+y+z) + ...)
= 4((x(y+z)x + x(y+z)(y+z)) + ...)
= 4((x^2 (y+z) + x (y+z)^2) + ...)
= 4((yz^2 + ...) + (yz^2 + ...) + 6 xyz)
= 4(2(yz^2 + ...) + 6 xyz)
= 8(yz^2 + ...) + 24 xyz,

and

  9(y+z)(z+x)(x+y)
= 9((yz^2 + ...) + 2 xyz)
= 9(yz^2 + ...) + 18 xyz.

So your inequality transforms to

  8(yz^2 + ...) + 24 xyz <= 9(yz^2 + ...) + 18 xyz,

or

  6 xyz <= yz^2 + ... .

Now, this is equivalent to

  0 <= (yz^2 + ...) - 6xyz
<==>  0 <= (yz^2 - xyz) + (zx^2 - xyz) + (xy^2 - xyz) + (zy^2 - xyz) + (xz^2 - xyz) + (yx^2 - xyz)
<==>  0 <= yz(z-x) + zx(x-y) + xy(y-z) + zy(y-x) + xz(z-y) + yx(x-z)
<==>  0 <= y(z-x)(z-x) + z(x-y)(x-y) + x(y-z)(y-z)
<==>  0 <= y(z-x)^2 + z(x-y)^2 + x(y-z)^2,

what is trivial. Hence, your inequality is proven. Equality occurs if and only if x = y = z.

There SHOULD be a simpler proof !

  Darij

Let's go for a simple proof :
Let 2a = x+y, 2b = y+z and 2c = z+x.
Then, a,b,c are the sides of a triangle, and the inequality reduces to :
(a+c-b)(a+c+b)/(4ac) + (a+b-c)(a+b+c)/(4ab) + (b+c-a)(b+c+a)/(4ab)  \leq  9/4.
The LHS is (cos(A) + cos(B) + cos(C))/2 + 3/2.
Thus the inequality is equivalent to cos(A) + cos(B) + cos(C)  \leq  3/2, which is well-known to be true, with equality iff ABC is equilateral.

Pierre.


Let $ ABC$ be a triangle and $ XYZ$ be a points. Then, the perpendicular lines frome $ X,Y,Z$ on the lines $ AB,BC,AC$ are concurrent if and only if \[ AX^2\minus{}XB^2\plus{}BY^2\minus{}YC^2\plus{}CZ^2\minus{}ZA^2\equal{}0\]

[b]I need some online sources for this lemma, and if you can, I need also the proof.[/b]

First assume the lines intersect at point $D$. Then,
\[AX^2-BX^2=AD^2-BD^2\]
Cyclic sum them and they cancel each other to get 0.

Now assume the other direction. Let $X'$ be such that the perpendiculars of $X',Y,Z$ concurrent at $D$. Then from the above result we get $AX^2-BX^2=AX'^2-BX'^2$ and thus $X\equiv X'$.


From what I know Mandelbrot round 2 is tmr...  for people in bc canada anyway lol

what kind of problems seem to appear in the mandelbrot? I've done most of the questions on the official website.. may some of you post some questions that tend to relate to the topics on the mandelbrot? thanks i hope this is the right forum to post it

thanks,

agent ';..;'

I recently posted some mandelbrot problems:

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=63267
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=62844

Here are some team test questions

1. Show that for any $g$ the function $f(m) = m$ is a solution to $f(g(m)) = g(f(m))$.
2. If $g(m) = m^2$, describe all posible functions $f: \mathbb{N} \rightarrow \mathbb{N}$ that are solutions of \[ f(g(m)) = g(f(m)) \]

4. if $g: \mathbb{N} \rightarrow \mathbb{N}$ is one-to-one and onto, prove there is a one-to-one, onto function $f: \mathbb{N} \rightarrow \mathbb{N}$, besides the trivial solution $f(m) = m$, which satisfies $g(f(m)) = f(g(m))$


quadratics
calculus
derivative
inequalities
inequalities proposed

If $a, b, c, d, e \ge 0$, prove that

$3\sum_{cyc}a^{2}+\sum_{cyc}ab \ge 4\sum_{cyc}ac$

 :huh:

All right, it is true. I solve it by an extremely ugly method: it is a quadratic in $a$, so I minimize it by taking the derivative and solving the resulting linear equation. The resulting expression is a quadratic in $b$, and so on. After we minimize in $a, b, c$, we are left with a constant times $(d-e)^{2}$.

So: does anyone have a reasonably nice solution?

Also, what are opinions about for $n$ nonnegative variables,
$(n-2)\sum_{cyc}a^{2}+\sum_{cyc}ab \ge (n-1)\sum_{cyc}ac$
 :?:  I am sure this can't hold for arbitrary $n$.

And are $(3, 1, 4)$ in the original ineq in fact the best we can do?

[quote="Xevarion"]  I am sure this can't hold for arbitrary $n$.

[/quote]For even $n$ it doesn't hold. Put $a_{2i}=1,a_{2i+1}= 0$. But it might well hold for odd $n$. Did you try your ugly method for $n=7$ :blush: ? If the outcome is $C(x_{6}-x_{7})^{2}$ you may be lucky of course :P

A nice generalisation I just crated is:
let $n=2k+1$ then $3\sum_{cyc}x_{i}^{2}+\sum_{cyc}x_{i}x_{i+1}\geq 4\sum_{cyc}x_{i}x_{i+k}$ which for k=2 turns out to be identical to your original inequality  :wink:
The proof is very very simple !! :)


limit Euler calculus function Princeton college logarithms

[quote="ffrogg"]Show that   $ \lim\limits_{n\to \infty}\sum\limits_{k=1}^{n}\left(\frac{k}{n}\right)^n =\frac{e}{e-1} \;  . $ [/quote]
In 1982 I.J.Schoenberg-[11]  stated the problem to prove that    $\; (1)\; \; \; \;      \lim_{n\to \infty }\mathcal{S}_{n}=\displaystyle \frac{1}{e-1}, $ where
 \[ \begin{array}{|c|} \hline
 \mathcal{S}_{n}=\displaystyle \frac{1}{(n+1)^n}\sum\limits_{k=1}^{n}k^{n}\\
\hline \end{array}  \; , ( n=1,2,...). \]
We mention that problem (1)  was previously proposed in K. Knopp  [5]p.102 . But it's seems that the question was proposed for the first time by by Wolstenholme in [15] .
 [b]The complete asymptotic expansion[/b]
\[
\mathcal{S}_{n}\sim \sum_{k=0}^{\infty }\frac{\beta_{k}}{\left( n+1\right)
^{k}} \; ,  \left( n\to \infty \right).\]
is given in [1] .
The latter equation means that, for all $p=0,1,2,\ldots,$ there holds
 \[
\mathcal{S}_{n}=\sum_{k=0}^{p}\frac{\beta_{k}}{\left( n+1\right)^{k}}+
o\left(n^{-p}\right) \qquad \left( n\to \infty \right).
\]
Schoenberg's result (1)  is the special case $p=0$ with $ \beta_{0}=(e-1)^{-1}.$ There were many contributions to Schoenberg's problem [12] . J.H.Van Lint gave a short proof of (1)  using the dominated convergence (i.e., Tannery's Theorem, see  Bromwich [2] ). I.J. Schoenberg gave a solution using
well-known results for Bernoulli numbers and the Euler-Maclaurin formula. Among the solutions there was also a proof of  $\beta_{1}=-(1/2)\,e\,(3-e)\,(e-1)^{-3}$ due to D. Dijkstra. It's poossible to determine all coefficients $\beta_{k}$ in an explicit form. It turns out that combinatorial numbers such as Stirling numbers will play an important role.
In [1] was also investigated the asymptotic form of the sequence  having the terms
\[
\begin{array}{||c||}
\hline\hline
{\mathcal L }_n:=\displaystyle \sum\limits_{k=1}^{n}\left(\frac{k}{n}\right)^n \\
\hline\hline
\end{array}\; ,\; n\in {\mathbb N}^* .;
\]
[b]References, perhaps of interest ;[/b]
[1] Ulrich Abel , Mircea Ivan and Alexandru Lupas , [i] The asymptotic behaviour of a sequence considered by    I. J. Schoenberg[/i] ,The Journal of Analysis  vol.  8 (2000) 179-191.
[2] T. J. J. A. Bromwich,  [b]An Introduction to the Theory of  Infinite Series,[/b] Macmillan, 1965
[3] M.Ivan, [b]A sequence considered by I.~J. Schoenberg  ? [/b] Proc. of the third Romanian-German Seminar on Approximation Theory (RoGer-98), Sibiu, June 1-3,~1998, pp. 22--23.
[4] C. Jordan,  [i]Calculus of finite differences,[/i] Chelsea, New York, 1965
[5] K. Knopp, [b]Theorie und Anwendung der unendlichen Reihen, [/b] Grundlehren der Mathematischen Wissenschaften 2, Springer-Verlag Berlin,1922.
[6] A. Lupas , [i] A sequence convergent to Napier's constant[/i], available on Internet.
 [7] R. Mullin and G.-C. Rota,  [i]Theory of binomial enumeration,[/i] Graph Theory and its Applications. Academic Press,1970, pp. 167--213.
[8] E. D. Rainville,  [i]Special Functions,[/i] Chelsea Publ.Co., New York,1971.
[9] J. Riordan, [i]Combinatorial identities,[/i] Wiley, New  York,1968.
 [10] J. Riordan, 1980, [i]An Introduction to Combinatorial Analysis,[/i] Princeton Univ. Press, Princeton, New  Jersey,1980.
 [11] I.J. Schoenberg,  [b]Problem 640,[/b] Nieuw Arch. Wiskd., III. Ser.~30,(1982), p. 116.
[12] H. van Lint, D. Dijkstra e.a.,[b]  Solutions to Problem 640,[/b] Nieuw Arch. Wiskd., III. Ser.~30, (1982) pp. 334--337.
 [13] Wolstenholme , [b]Problem 19[/b] in Bromwhich's monograph [2]  , page 528.

[quote="flip2004"]
[b]The asymptotic behaviour of the sequence : [/b]  $  \; (*)\; \; \; \; \; \begin{array}{||c||} \hline\hline
{\mathcal L }_n:=\displaystyle \sum\limits_{k=1}^{n}\left(\frac{k}{n}\right)^n \\  \hline\hline \end{array}\; ,\; n\in {\mathbb N}^* .;  $  [b] ( see [1] ). [/b] [/quote]
[b]1. The Stirling numbers:[/b]  Let $n$ be a positive integer. If, for all $x\in {\mathbb R},$
\[ x(x-1)\ldots(x-n+1)=\sum_{k=0}^{n} s(n,k)x^k,\;
s(0,0)=1,\;s(0,k)=0\;(k\in\mathbb{N})\; \; \; \;
\]
and  $ \displaystyle x^n=\displaystyle \sum\limits_{k=1}^{n} S(n,k) x(x-1)\ldots(x-k+1),\;\; S(0,0)=1,\;S(0,k)=0\;(k\in\mathbb{N})\; \; \; \; ,\;  $ then the coefficients $s(n,k)$ and $S(n,k),$ $n,k \ge 0,$ are called  the [b]Stirling numbers of the first and second kind [/b], respectively (see [9]--[10] ).

[b]2. The Jordan coefficients $J(k,i)$ :[/b]
[b]Definition 1. [/b] [i]  The [b]Jordan coefficients[/b] $\mathcal{J}({k},{i})$ are defined by means of the generating function
\[
\left( \log {(1-x)+x}\right) ^{i}=x^{i}\sum\limits_{k=0}^{\infty }\mathcal{J}(%
{k},{i})\frac{x^{k}}{k!} \; \; \; \; \;  (i=0,1,\ldots;\; \; \; \;
-1\le x<1).
\]
[/i]
[b]Lemma  1. [/b] [i] For $k\in \mathbb{N}$ and $i=0,\ldots, k,$ we have  $ \displaystyle \mathcal{J}({k},{i})=\displaystyle \sum\limits_{\nu =0}^{i}\frac{(-i)_{\nu }}{(k+1)_{\nu }} |s(k+\nu, \nu )|.  $
[/i]
Note that $ \begin{array}{ll} \mathcal{J}({k},{i})=0 & \; \; \; \;  (i=1,2,\ldots;\; \; \; \;  k=0,1,\ldots,i-1),\\
\mathcal{J}({1},{0})=1,\; \; \; \;  \mathcal{J}({k},{0})=0,& \; \;
\; \; (k=1,2,\ldots). \end{array} $
As a consequence , the Jordan coefficients satisfy the binomial identity $ \; \mathcal{J}({k},{i+j})=\sum\limits_{\nu =0}^{k}{{{{{\binom{k}{\nu }}}}}}
\mathcal{J}({\nu },{i})\mathcal{J}({k-\nu },{j}). $
 The coefficients $\mathcal{J}({k},{i})$ were for the first time investigated by Jordan-[4]( see pp. 150--152) .

[b]3. The numbers $b_n$ :[/b] In order to put in evidence the asymptotic behaviour of the sequence $\mathcal{L}_{n}$ we need a class of some special numbers.
[b]Definition 2. [/b] [i] The polynomials $t_{n},$ given by $ \displaystyle t_{n}(x)=\displaystyle  \sum\limits_{k=0}^{n}S(n,k)\,x^{k},\; \; \; \;
n=0,1,\ldots, $ are called  [b]Touchard polynomials[/b] or [b]exponential polynomials.[/b]  [/i]
 Their generating function is  $\displaystyle  e^{x(e^{t}-1)}=\displaystyle \sum\limits_{k=0}^{\infty }t_{k}(x)\,\frac{t^{k}}{k!}. $
In the detailed proof , we make use of the following remarkable representation known as  [b]Dobinski formula[/b] (see, e.g.,  Mullin-[7] )
\[
t_n(x)=e^{-x}\sum_{k=0}^{\infty} \frac{k^n}{k!}\,x^k.
\]
[b]Definition 3 .[/b] [i] Define the sequence $(b_{n})_{n\geq 0}$ by  the identity  $   \displaystyle  \frac{e}{e-e^{t}}=\displaystyle \sum\limits_{j=0}^{\infty  }b_{j}\frac{t^{j}}{j!}, \; \; \; \; \;  (t<1). $
[/i]
 The first terms of the sequence are:  $\displaystyle b_{0}=\displaystyle \frac{e}{e-1},\; \; \; \;   b_{1}=\displaystyle \frac{e}{(e-1)^{2}},\; \; \; \; b_{3}=\displaystyle \frac{e(e+1)}{(e-1)^{3}}.  $
[b]Lemma 2. [/b] [i] Some properties of the sequence $(b_{n})_{n\geq  0}, $ are : $ \displaystyle b_{j}=\displaystyle \sum\limits_{k=0}^{\infty }e^{-k}k^{j}\; \; \; \; \;  (j=0,1,\ldots ); $
\[
b_{j+1}=\frac{e+1}{e-1}b_{j}+\sum\limits_{\nu =1}^{j-1}{{{j\choose
\nu }}} b_{\nu }b_{j-\nu } \; \; \; \; \; (j=2,3,\ldots );
\]
\[
 \sum\limits_{k=0}^{n-1}e^{-k}k^{j}=b_{j}+\mathcal{O}(e^{-n}n^j)\; \; \;
\; \; (j=0,1,\ldots ;\; \; \; \;  n\to \infty);
\]
\[
b_{j}=\frac{e}{e-1}\int\limits_{0}^{\infty }e^{-y}t_{j}\left(
\frac{y}{e-1} \right)dy = \frac{e}{e-1}\sum\limits_{k=0}^{j}\frac{k!}{(e-1)^{k}}S(j,k)\;
\; , \; \;  (j\in {\mathbb N}). \]
[/i]
[b] 4. Asymptotic expansions : [/b]  Using above notions it's proved  (see [1]) :
[b]THEOREM. [/b] [i]  The sequence (*) has the complete asymptotic expansion $\displaystyle \begin{array}{|c|} \hline \\
\displaystyle \mathcal{L}_{n}\sim \displaystyle  \sum\limits_{j=0}^{\infty }\frac{\alpha _{j}}{n^{j}} \\
\\ \hline  \end{array}  $  with coefficients
\[
\begin{array}{|c|}
\hline
\\
 \displaystyle \alpha _{j}=\displaystyle \frac{1}{j!}\sum_{i=0}^{j}
\frac{\mathcal{J}(j,i)b_{i+j}}{i!}\\
\\
\hline
\end{array}\; \; \; \; \;  (j\in {\mathbb N} ).
\]
Moreover  $ \displaystyle \begin{array}{|c|} \hline
 \\
 \displaystyle \mathcal{L}_{n} =\displaystyle
\sum\limits_{j=0}^{p}\frac{\alpha_{j}}{n^{j}}+
\mathcal{O}\left(\left( \frac{\log n}{n} \right)^{p+1}\right)\\
\\
\hline \end{array}.  $
[/i] Note that $\mathcal{O}\left(\left( \frac{\log n}{n} \right)^{p+1}\right)$ carries on more information than  $o\left(n^{-p}\right).$

For the convenience  we list the following values :
\[
\begin{array}[t]{||c|c|c||}
\hline\hline
j & b_j &   \alpha_j\\
\hline\hline
&&\\
0&\displaystyle \frac{e}{e-1}&\displaystyle \frac{e}{e-1} \\
&&\\
 \hline
&&\\
  1&\displaystyle \frac{e}{{{(e-1)}^2}}&
                 \displaystyle -\frac{e (1+e)}{2 {{(e-1)}^3}} \\
                 &&\\
\hline &&\\
 2&\displaystyle \frac{e (1+e)}{{{(e-1)}^3}}&\displaystyle
             -\frac{e (1+5 e) \big(-11-2 e+{e^2}\big)}{24 {{(e-1)}^5}} \\
             &&\\
\hline
&&\\

 3&\displaystyle \frac{e \big(1+4
 e+{e^2}\big)}{{{(e-1)}^4}}&\displaystyle
                  \frac{-e \big(21+365 e+502 {e^2}-138 {e^3}-35 {e^4}+5 {e^5}\big)}{48 {{(e-1)}^7}}\\
      &&\\
\hline\hline
\end{array}
\]


Compute the sum $ S_n\equal{}\displaystyle\sum_{k\equal{}1}^{n}\frac{(\minus{}1)^{k\minus{}1}}{(n\minus{}k)!(n\plus{}k)!}$.


hello, the plural version of axis is axes.
Sonnhard.


Quite a few questions here. What I am trying to do is make a template (I think that's what they're called - headers/footers/margins/etc.) for math solutions that I am writing. I am using the fancyhdr package to deal with all of this. To make my life easier (as I plan to type up 100s of solutions eventually) I am going to make a template. How I would like it to work (and how it is basically working is as follows:
[code]
%... in my "main" file
%begin{solution}{problem#}{problem prompt}
\begin{solution}{3}{Solve for x...}
%... all the stuff for my solution
\end{solution}
[/code]
And that would produce a page where at the top (centered, and some other formatting) it would say "Problem 3: Solve for x..." and then it would have a black line across the page, a little more formatting, and then it would print the rest of my solution.

Here is the snippet form my style file: (note that "qnumber" is the question number (argument 1 in the solution environment), totalpages is *supposed* to be a counter for the total number of pages, and "page" is a counter inside fancyhdr... i think)
[code]
\newenvironment{solution}[2]{
\setcounter{qnumber}{#1}
\setcounter{page}{1}
\begin{center}
Problem \theqnumber : #2
% other formatting
\end{center}
testing}
{end stuff for solution \theqnumber \addtocounter{totalpages}{page}}[/code]
Notice that I have a random "testing" after the first part of my template and "end stuff for solution 3" inside the ... end replace... thingy.

Anyway, I run into a couple of problems. How the above snippet ends up... compiling?... is that I will have one page with "Problem 3:... ... testing", a page break, then my solution, another page break, and finally "end stuff for problem 3". This is not supposed to happen. Why do I get these page breaks? How do I get rid of them?

Also, the totalpages counter never increments. At all. Starts at 0, ends at 0. Why is it not incrementing? (I know that \thepage is incrementing because in each page's footer I have \thepage in \rhead, and it does go 1, 2, 3,...)

Also, I am trying to count both the total number of pages and the number of pages within each solution (for multi-solution documents). How would I implement counters for the number of pages in [i]each[/i] solution? (I have no idea how to implement this).

And as far as my understanding goes, to render the number of pages in each solution in my final document, I simply *\renew* the environment, right?

I hope this makes some sort of sense. If something is clear as mud, just reply and I will try to clear it up.

---- APPENDED ----

Oh, and one more question... how would I achieve formatting like this in the header:

[code]
_________
|       | Name
| [some | Date
| image]| School
|_______| Other random fact

[/code]
Where the image is displayed using \includegraphics*

This is probably a simple problem, but I am still having trouble...  :blush:
[/code]

It's late here :wink: so just a quick answer:
Finding the total number of pages is tricky but fortunately the [url=http://tug.ctan.org/cgi-bin/ctanPackageInformation.py?id=lastpage]lastpage package[/url] does this for you. Indeed that is how fancyhdr produces page N of M type footers.

lastpage works by adding a label whose number is the number of pages so needs more than one pass through LaTeX (like all labels this goes into the aux document). Maybe you could copy its method and use it for solutions. I must admit to not having thought this through so leave it as an exercise for the reader!

PS Would you mind putting some spaces in the title of this thread - without them PHP can't wrap the title and so causes a very wide table in http://www.artofproblemsolving.com/Forum/index.php?f=123 that needs horizontal scrolling. Thanks.

I'm not sure I follow everything here, even though I am now more awake!
1. I don't get the unwanted page breaks you get so something else is doing it. Perhaps you could attach a complete tex document?
2. The problem with totalpages is because you aren't telling Latex how much to increase the counter by. So you need
\addtocounter{totalpages}{1}
or
\stepcounter{totalpages}
3.[quote]And as far as my understanding goes, to render the number of pages in each solution in my final document, I simply *\renew* the environment, right?[/quote]Eh? I really don't follow this. You may have to make more than one pass to get everything correct, but it's not unusual for Latex to need 3 passes to get all its refences, citations etc correct; \renewenvirenoment is just for redefining the environment.
4. To find the number of pages in each solution you could use the calc package. This allows you to do arithmetic on counters, so you could subtract the last page number of a solution from its first page number (+1 of course).
5. With the image in the header with name, date etc next to it, you could use a table with 2 columns or perhaps minipages or parboxes.

okay, just got back from school and the Texas Math League contest...

Sounds like I missed a few concepts completely. First of all, how else do you force multiple passes (obviously not using renewenvironment  :blush: )?

That is probably my problem with totalpages (the counter), because I thought I was using multiple passes and I obviously wasn't (I had an \addtocounter{totalpages}{\thepage} in my code).

Attached is a simplified version of the style and program structure that I am using. I still don't know what's up with the page breaks.

I will start working on using the \lastpage package. But I would really like to learn how to do all of that myself, and I think I could do it with knowledge of multiple passes... (and once again, those page breaks are bugging me)

Anyway stevem, if you ever get the time, I really appreciate the help you have given me (and everybody else here on this forum).

And sorry for not including spaces in the thread title (thank you mod for fixing that). I guess I get spoiled from 1600x1200. :(

1. To do multiple passes you just compile again. If you are using the command line just reissue the same command or use a batch file. In TexnicCenter just press the Build button again. It's possible that some tex editors let you tell it to do a number of passes - which it will do just by repeating the compilation command. The log usually tells you if another pass is needed.

2. In style_test.tex you have \newenvironment{solution} followed by \renewenvironment{solution}. What happens here is that the second definition overwrites the first (which is what \renew does) so the first one is never used.

3. If you want to use the first definition then remove the \renewenvironment{solution} definition and you'll need to change \addtocounter{totalpages}{page} to \addtocounter{totalpages}{\thepage} otherwise you'll get an error.

4. Your method of finding the totalpages won't work. Latex reads style_test.tex, sees the line \setcounter{totalpages}{-1} followed by \rfoot{Page \thepage\ of \thetotalpages} and so writes Page n of -1 in all the footers. Later on, totalpages is increased but then the footer has been defined. When the compilation is over all values are lost so the second pass doesn't help. That's why lastpage.sty uses the .aux file to store the value between passes.
As you see it's tricky because Latex has to write Page n of m on the first page before it knows what the value of m is.
You'll find more information on this problem at [url=http://www.tex.ac.uk/cgi-bin/texfaq2html?label=nofm]Page numbering “<n> of <m>”[/url] and [url=http://www.tex.ac.uk/cgi-bin/texfaq2html?label=howmanypp]How many pages are there in my document?[/url]

5. And now for the mystery page breaks. It is \include that does it in the line
\newcommand{\prob}[2]{\begin{solution}{#1}{#2}\include{Problem#1}\end{solution}}
In fact the \include command issues a \clearpage command both before and after it so getting lots of page breaks.
If you use \input instead then you'll find solutions_test.pdf is just one page long instead of three.

Hope that helps.

Wow... thanks so much stevem! (are you part of the TeX committee or something, or do you just typeset a lot of stuff?)

If we were in the java.sun.com forums, you would definitely get your duke dollars  :lol: (lost? don't worry about it)

Speaking of Java, I wish $\LaTeX$ made as much sense as java... that would be nice

once again, thanks for your help!

Glad that helped.

LaTeX is an example of [url=http://www.vivtek.com/litprog.html]literate programming[/url] (where the program is written to be readable to a human and where the documentation is built into the source code) and is much older than languages like Java. Thus it looks and works quite differently from Java so takes some getting used to. But it's definitely worth it, particularly for mathematics where it has no equal.

LaTeX is simple enough to get going with but it has great depth to it. If you want to see real experts at work then have a look at the [url=http://groups.google.co.uk/group/comp.text.tex]comp.text.tex[/url] newsgroup.

Well, I don't want to clutter up the forum with all kinds of threads by me, so on to my next problem.

Looking at the lastpage.sty file, It seems pretty clear how to deal with the page N of M that I am trying to implement. But I have the following line in my environment definition:

[code]
\newenvironment{solution}[2]{
%other stuff not applicable
}
{
\write\@auxout{\string\newlabel{LastPage\theqnumber}{{}{\thepage}}}
\clearpage }
[/code]
What I am trying to do is place a "LastPage2" label on the last page of solution 2, etc.

Everything seems to work correctly when I try "\pageref{LastPage2}" for instance (LaTeX correctly looks for the LastPage2 label and realizes it's not there :D). But when I do actually try to write the LastPage\theqnumber label (as my code above outlines) I get some kind of "Improper \spacefactor" error ... along with some "runaway text," but last time I checked my text wasn't going anywhere  :maybe:  ... any ideas?

Is there something that I am again totally missing about \@auxout?

Also, where did you learn to TeX well? I've been googling for quite some time and I can't find any sites (besides actual source files) that mention @auxout.

Unfortunately you can't easily copy code from a style file into a tex file because of the way \@ is used. See [url=http://www.tex.ac.uk/cgi-bin/texfaq2html?label=atvert]\spacefactor complaints[/url] and [url=http://www.tex.ac.uk/cgi-bin/texfaq2html?label=atsigns]\@ and @ in macro names[/url].
Perhaps you could try adapting lastpage.sty and creating your own style file (which must either reside in the same directory as your tex file or you need to refresh the latex file database). See [url=http://www.tex.ac.uk/cgi-bin/texfaq2html?label=writecls]Learning to write LaTeX classes and packages[/url] which links to documentation for writing classes and packages.

As for learning about LaTeX, the references above to the [url=http://www.tex.ac.uk/cgi-bin/texfaq2html?introduction=yes]UK TeX FAQ[/url] prove that it is a valuable resource for solving advanced problems. But despite the wealth of resources on the web, far and away the best way to learn about LaTeX is to use books. Here are a few recommendations with links to Amazon:
[list][*][url=http://www.amazon.com/Guide-LaTeX-4th-Helmut-Kopka/dp/0321173856/sr=8-1/qid=1161947605/ref=pd_bbs_sr_1/102-6787070-3364907?ie=UTF8&s=books]Guide to LaTeX[/url] by Kopka & Daly. This is a superb book suitable for beginners to intermediate. I would suggest that anyone using LaTeX should have this book handy.
[*][url=http://www.amazon.com/Companion-Addison-Wesley-Tools-Techniques-Computer/dp/0201362996/sr=1-1/qid=1161947729/ref=pd_bbs_sr_1/102-6787070-3364907?ie=UTF8&s=books]LaTeX Companion[/url] by Frank Mittelbach & others. This is [b][i]the[/i][/b] reference book for intermediate to advanced users. Whilst it does sometimes overlap with Kopka & Daly it goes into LaTeX much more deeply in its massive 1100+ pages.
[*][url=http://www.amazon.com/LaTeX-Document-Preparation-System-2nd/dp/0201529831/sr=1-1/qid=1161947940/ref=pd_bbs_sr_1/102-6787070-3364907?ie=UTF8&s=books]LaTeX: A Document Preparation System[/url] by Leslie Lamport (hence the [i]La[/i] of [i]La[/i]TeX). I haven't read this book but many people think of it as a very important and authoritative book.
[*][url=http://www.amazon.com/TeXbook-Donald-E-Knuth/dp/0201134489/sr=1-1/qid=1161948224/ref=pd_bbs_1/102-6787070-3364907?ie=UTF8&s=books]The TeXbook[/url] by Donald E. Knuth is the classic textbook about the way TeX (rather than LaTeX) works by the author of TeX. If you want to get a deep understanding of TeX then this is the book to read - but beware it is only suitable for advanced users who want to get to grips with the anatomy of TeX - its eyes, mouth and stomach - yes these are real words Donald Knuth uses to help you understand how TeX operates.[/list]
I would hope that any good University library would have these books.

aahhh... thanks stevem

that should wrap up the questions for now... :D


I have a cool idea. Let's all participate in a chess game.
Everybody who wants to join the game just write one move for the team whose turn is to go next. Try to bold the moves and write the moves for white in red and for black in blue.
WriteHave fun! Smile Mr. Green

1. d4 d5
[color=red][b]2. c4[/b][/color]


counterexample: $n=3, x=1, y=2, z=4 \Rightarrow LHS = 64.5 < 73 = RHS$

no can you read it more carefully
x>=y>=z>0
thanks

i think it isn't too difficult
please solve it now , thanks

[quote="nttjuvwamsncc"]let x>=y>=z>0 and n _ integer > 0
who can prove that
(x^n)y/z+(y^n)z/x+(z^n)x/y >= (x^n)+(y^n)+(z^n)[/quote]

you mean \[\sum_{cyc}\frac{x^{n}y}{z}\geq \sum_{cyc}x^{n}\] ?

well at first glance i see it should work with monotonic progressions

no it was wrong
x/y>1, yes , y/z>1  also
but z/x , no no

if its true doesn't it follow by generalized schur? seems like that kind of problem

well thank you very much arqady because of your proof
i want to see if there are better solution of this problem than mine
here is my solution
dividing two sides by $y^{n}$
$\Leftrightarrow \frac{x^{n}}{y^{n-1}z}+\frac{z}{x}+\frac{z^{n}x}{y^{n+1}}\geq \frac{x^{n}}{y^{n}}+\frac{z^{n}}{y^{n}}+1$
multiply the two sides with  $\frac{xz}{y^{2}}$
and consider $X=\frac{x}{y}$ and $Z=\frac{z}{y}$
we have
$X^{n+1}+Z^{2}+Z^{n+1}X^{2}\geq XZ(X^{n}+Z^{n}+1)$
$\leftrightarrow X^{n+1}+Z^{2}+Z^{n+1}X^{2}-XZ(X^{n}+Z^{n}+1) \geq 0$
simplify it then consider f(X) = ... (with $X \geq 1$ and $Z \leq 1$)
calculate f'(X) and f''(X)
$f''(X) \geq 0$ then $f'(X) \geq f(1) \geq 0$
thus $f(X) \geq f(1) \geq 0$
that's what we must prove

please give me more wonderful solution for this problems thank you


function
inequalities
analytic geometry
graphing lines
slope
algebra proposed
algebra

we have a function $ {f(x)\equal{}2x\minus{}\frac{2}{x^2}\plus{}\frac{a}{x}}$ and $ {a \in (0,4]}$. find all $ {k \in R}$ for any $ {x1,x2 \in R\plus{}}$, the inequality: abs($ {f(x1)\minus{}f(x2)}$) >=k*abs($ {x1\minus{}x2}$) always holds.

the condition given implies that the modulus of the slope of the function should be greater than or equal to k  [LMVT by choosing x1 and x2 very close)].

Now it cannot be that the slope at one point is >=k and another point <=-k as the function is differentiable so it must have passed from a region of modulus(slope)<k
So,

writing the final equation we have either

2+(4/x^3)-(a/x^2)>=k or

2+(4/x^3)-(a/x^2)<=-k

now solving the first one

as x>0 we multiply x^3 to both the sides

(2-k)x^3 -ax+4>=0

clearly if 2-k=<0 the thing is <0 for large x  as a is also finite)

for its minimum value when 2-k>0

3(2-k)x^2=a

now replacing this value of x we get the condition and as we know the limits on a we get the desired limits on k ...

similarly for the -k one ..


geometry
3D geometry
algebra
factorization
sum of cubes

Find the smallest square which is the sum of more than three consecutive cubes (excluding the cube $ 1$).

[hide="start"]So it's well known that the sum of cubes from $ 1^3$ to $ n^3$ is $ \left(\frac{n(n\plus{}1)}2\right)^2$.  We can think of the sum of cubes from $ (k\plus{}1)^3$ to $ n^3$ as $ 1^3\plus{}2^3\ldots n^3\minus{}(1^3\plus{}2^3\ldots k^3)\equal{}\left(\frac{n(n\plus{}1)}2\right)^2\minus{}\left(\frac{k(k\plus{}1)}2\right)^2$.  Now we need this to be a square, say $ m^2$.  Then $ \frac{k(k\plus{}1)}2$, $ m$ and $ \frac{n(n\plus{}1)}2$ are a Pythagorean triple.[/hide]


The cover on the AMC states that the MAA reserves the privilege to "reexamine" students and schools that they believe to have cheated or obtained an unfair advantage as to distort the results. For example, if all 80 people at one school took the AMC 12 and got 150, then it would seem reasonable for MAA to investigate how such a coincidence could have occured. My question is that if the school took the B, does the AMC committee give them another AMC test, say AMC C, that they have to take to prove their merit and honesty?

No idea, but I'd say that's the assumption.

Haha, I was just about to say we all got 150's, all 80 of us....

And we don't want a retake...........

We don't have math club. The people here just 'know' the stuff.

JK.  :lol:

Heh, you should ask someone from AAST ...

Hahahaha...yes indeed. I'm not sure the topic would be welcome to any AASTers though. You could try nonetheless.

The AAST story is discussed [url=http://artofproblemsolving.com/Forum/viewtopic.php?highlight=AAST+cheat&t=11870]here[/url] a little bit.


Find the exact value without the use of a calculator and show all work: $\frac{sin10+sin30+sin50}{cos10+cos30+cos50}$.

your method of doing it gets u the right answer but how did u manage to evaluate the $cos20$ or did u factor out the $sin30$?

[hide]
Ok, Expanding on my last step, we have:

$\frac{\frac 12 + \cos20}{\frac{\sqrt3}{2}+\sqrt3(\cos20)}$  To visualize my next steps more clearly, let p=numerator and q=denominator.  (we have $\frac pq$)

First lets make p and q simpler:

$p=\frac 12 + \cos20= \frac{\cos20+1}{2}$
${q=\frac{\sqrt3}{2}+\sqrt3(\cos20)}=\frac{2(\cos20)\sqrt3+\sqrt3}{2}$

Noticing that dividing p and q cancels out their denominator of 2, leaving behind $\frac{2\cos20+1}{2\cos20\sqrt3+\sqrt3}$
Rationalizing the denominator separately gives $\frac{\sqrt3}{3(2\cos20+1)}$ and substituting back gives:

$\frac{(2\cos20+1)\sqrt3}{3(2\cos20+1)}=\frac{\sqrt3}{3}$

In essence, I never calculated the value of $\cos 20$, becuase they would cancel out.

[/hide]

Yeah, I noticed, the $\frac{\sqrt3}{3}$ looked familiar.

i think he factored out the 2cos(20)+1

edit: whoops must've left this post on from a while ago


Hiya,

I'm totally new to this board and dont know where i should post this. anyhow my question is so simple.  :?

if i have ordinal data say (gender and work type) and would like to test if there are differences between them. Can i use two independent sample t-test or chi-squred test?  or what to do ( means in such case have no meaning  :blush: )

the hypotheses were as follows

\[H_0: \mu_F \geq \mu_M\]
\[H_1: \mu_F < \mu_M\]

all the best to you

I think you get the same p-values, so it's your preference. But based on the given hypotheses, you should use 2-sample t-test.
If it were
$H_0:\mu_F\text{ and }\mu_M\text{ are independent.}\\H_a:\mu_F\text{ and }\mu_M\text{ are not independent.}$
then I would use the $\chi^2$ test for independence.

And welcome, this is probably the best place to put such a question. I've already asked for a statistics section, but I assume they've been very busy.


group theory
abstract algebra
superior algebra
superior algebra solved

let $C=\left\langle a \right\rangle$ be cyclic group of order 60, let $D=\left\langle a^{35} \right\rangle$, find all the left cosets of $D$ in $C.$

  so would it be something like this... $D=\left\{e,a^{35}\right\} $, $aD=\left\{a,a^{36}\right\},$ $a^{2}D=\left\{a^{2},a^{32}\right\},$ ....  untill i reach $a^{35}$  ?

No.

How many elements are in the subgroup generated by $a^{35}$?

Of course I know the answer to that question, and I know it wasn't provided in the problem. Figure it out yourself!

Yes, the order of $D$ is 12. Now how many cosets does it have?

Now try this for some more groups/subgroups and make a conjecture relating:

order of the group, order of the subgroup, and number of cosets of that subgroup...you should see something nice


inequalities
number theory
inequalities proposed

given the condition \[ a_{n\plus{}2}\equal{}\frac{a_{n\plus{}1}\plus{}a_n}{\gcd(a_n,a_{n\plus{}1})}\], find all $ a_1,a_2\in\mathbb{N}$ s.t. the sequence is periodic.

We will prove that if an infinite sequence $ \{a_i\}_{i \equal{} 1}^{\infty}$ is bounded and for all $ n\in\mathbb{N}$:
                              $ a_{n \plus{} 2} \equal{} \frac {a_{n \plus{} 1} \plus{} a_n}{\gcd(a_n,a_{n \plus{} 1})}$
then $ a_i \equal{} 2$,for all $ i\in\mathbb{N}$.
Proof:
If we have some coprime terms $ a_i,a_{i \plus{} 1}$,then obviously $ \{a_i\}_{i \equal{} 1}^{\infty}$ is unbounded.Thus $ \forall i\in\mathbb{N}$,$ \gcd(a_i,a_{i \plus{} 1})\geq 2$,hence $ a_{n \plus{} 2}\leq\frac {a_n \plus{} a_{n \plus{} 1}}{2}\leq\max\{a_{n \plus{} 1},a_{n}\}$.Hence
$ \max\{a_{n \plus{} 1},a_n\}\leq\max\{a_{n \plus{} 2},a_{n \plus{} 1}\}\leq\max\{a_{n \plus{} 3},a_{n \plus{} 2}\}\leq\dots$,hence starting from some $ k$,we have
$ \max\{a_{k \plus{} 1},a_k\} \equal{} \max\{a_{k \plus{} 2},a_{k \plus{} 1}\} \equal{} \max\{a_{k \plus{} 3},a_{k \plus{} 2}\} \equal{} m$.
If $ a_{k \plus{} 1}\neq a_k$,then $ a_{k \plus{} 2},a_{k \plus{} 3} < m$,hence $ a_k \equal{} a_{k \plus{} 1}$,but then $ a_{k \plus{} 2} \equal{} 2$,thus $ a_{k \plus{} 3} \equal{} 2$,so
$ \forall n\geq k$,we have $ a_n \equal{} 2$.
Also we know that $ 2 \equal{} \frac {a_{k \minus{} 1} \plus{} 2}{2}$,hence $ a_{k \minus{} 1} \equal{} 2$,so $ a_n \equal{} 2$,for all $ n\in\mathbb{N}$.
Now it is enough to prove that if $ \{a_i\}_{i \equal{} 1}^{\infty}$ is periodic,then its bounded,but the last fact is obvious.
So the answer is $ a_1 \equal{} a_2 \equal{} 2$.
P.S:I guess this is N.T problem,not an inequality :wink:

nice solution, thank you Erken!

sorry for posting in wrong place, I mistaked browser windows...
please move it to number theory or olympiad section


I know this may not be in the difficulty of Intermediate (because I don't know the difficulty of this problem), so I charge the moderators to move it as they see neccessary.

We know the values of things such as i^4 and i^6 and so on, but what is i^i?

$i^i$ is a real number :D

Use this $e^{\pi i}=-1$

[hide]by euler's theorem, we have that $i=e^{i\pi(\frac{1}{2}+2k)}$, so

$i^i = e^{-\pi(\frac{1}{2}+2k)}$

where $k \in \mathbb{Z}$

so basically, there are infinite answers[/hide]

[quote="Hamster1800"][quote="10000th User"]What [b]are[/b] $\log i$? :D[/quote]
[hide]
Well that depends on the base. Base e is probably easiest.

We know $e^{i\pi}=-1$
and
$-1^\frac12=i$
so
$(e^{i\pi})^\frac12 = i$
and thus
$\ln i = \frac{i\pi}{2}$
HOWEVER, since $e^{ix} = \cos x + i\sin x$
and the period for both $\cos$ and $\sin$ is $2\pi$, we have to include the other solutions
$\ln i = \frac{i\pi}{2} + 2ik\pi$ where $k \in \mathbb{Z}$ <-- I said that right, right? (I haven't used this notation much)

When I don't specify a base, assume that it can be any base (because it can be)

Now, for base $a$, we just substitute in $a^{\log_a{e}} = a^{\frac{\log e}{\log a}}$
This comes out nicest in base $e$: $a^{\frac{1}{\ln a}}$
so:
$\log_a{i} = (\frac{1}{\ln a})(\frac{i\pi}{2} + 2ik\pi)$
so for $a=10$
$\log i = \frac{\frac{i\pi}{2}+ 2ik\pi}{\ln {10}}$

What do you think? :D[/hide][/quote]
 :thumbup: I use logarithm with base $e$ as this: $\log=\log_e=\ln$ :)

ok, thanks guys, that answered my question.


calculus
derivative
complex analysis
complex analysis unsolved

Let f: D(0,1)-->C (where C is the complex plane) be analytic.

(a) Suppose f(z) is in D(0,1) for all z in D(0,1). Show that |f'(z)|[u]<[/u]1
(b) Suppose |f(z^(2))|[u]>[/u]|f(z)| for all z in D(0,1). Show that f is constant on D(0,1).

(a) is false: see $ z^n$, and take a derivative near the edge. We do have that $ |f'(0)|\le 1$, with equality only if $ f(z)\equal{}cz$ with $ |c|\equal{}1$.
(b): If you repeatedly square a number, it goes to zero...

oops! you're right about part (a). i meant:

Let f: D(0,1)-->C (where C is the complex plane) be analytic.

(a) Suppose f(z) is in D(0,1) for all z in D(0,1). Show that |f'(0)|[u]<[/u]1
(b) Suppose |f(z^(2))|[u]>[/u]|f(z)| for all z in D(0,1). Show that f is constant on D(0,1).


Suppose a,b and c are positive real numbers. Prove that

(a + b +c)/3 ≤ √[(a^2 + b^2 + c^2)/3] ≤ [(ab/c + bc/a + ca/b)/3]

For each of the inequalities, find conditions on a, b and c such that equality holds.

[hide="First Inequality"]
\[(a+b+c)^{2}\stackrel{\text{Cauchy}}{\leq}(a^{2}+b^{2}+c^{2})(1^{2}+1^{2}+1^{2}) \implies \frac{a+b+c}{3}\leq \sqrt{\frac{ a^{2}+b^{2}+c^{2}}{3}}\]
Equality is at $a=b=c$.[/hide]

[hide="Hint for Second Inequality"]
Cauchy, then AM-GM.[/hide]

[hide="Second Inequality"]
\[\left( \frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}\right)^{2}\stackrel{\text{Cauchy}}{\geq}(a^{2}+b^{2}+c^{2}) \left( \frac{b^{2}}{c^{2}}+\frac{c^{2}}{a^{2}}+\frac{a^{2}}{b^{2}}\right) \]
\[(a^{2}+b^{2}+c^{2}) \left( \frac{b^{2}}{c^{2}}+\frac{c^{2}}{a^{2}}+\frac{a^{2}}{b^{2}}\right) \stackrel{ \text{AM-GM}}{\geq}3(a^{2}+b^{2}+c^{2}) \]
\[\implies \left(\frac{1}{3}\right) \left(\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}\right)\geq \sqrt{\frac{a^{2}+b^{2}+c^{2}}{3}}\]
Equality from Cauchy and AM-GM is $a=b=c$.[/hide]


Let me preface this by the fact that my book is not very rigorous about Riemann surfaces.

So I want to put together a Riemann surface of the function

$ f(x)\equal{}[(z\minus{}1)(z\plus{}1)]^\frac{1}{2} z^\frac{1}{3}$

So around $ 1$ and $ \minus{}1$ the power factor is $ e^{2 \pi i \frac{1}{2}m}\equal{}\minus{}1$ and around $ 0$ the power factor is $ e^{2 \pi i \frac{1}{3}m}$. So I cut the complex plane from $ \minus{}1$ to $ 1$ and from $ 0$ to $ \infty$. Then for gluing it I took six sheets and I glued the twelve sides of the $ \minus{}1$ to $ 1$ slit and glued each of the six sheets to each other with those into a sphere. Then I glued the sides of the $ 0$ to $ \infty$ slit together to make three handles around the side of the sphere.

Did I do that right? If not then how is it supposed to be done?

Out of pure curiosity, what book is this?

I'm using Gamelin's book. In the section on power functions we did a problem that had $ \frac {1}{2}$ and $ \frac {1}{3}$ powers and I decided to make up a function that had both in it. So I came up with the present function. So it's not actually in the book. At the very end of the book there's a whole chapter on Riemann surfaces that looks a lot better than what we have right now.


[quote="can_hang2007"]Let $ a,b,c$ be nonnegative real numbers such that $ a \plus{} b \plus{} c \equal{} 2$. Prove that
\[ \sqrt {9 \minus{} 8bc} \plus{} \sqrt {9 \minus{} 8ca} \plus{} \sqrt {9 \minus{} 8ab} \ge 7
\]
With equality holds if and only if $ (a,b,c) \equal{} (2/3,2/3,2/3)$ or $ (a,b,c) \equal{} (1,1,0).$
:)[/quote]
See also here:
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=156002

[quote="shfdfzhjj"][quote="can_hang2007"]Let $ a,b,c$ be nonnegative real numbers such that $ a \plus{} b \plus{} c \equal{} 2$. Prove that
\[ \sqrt {9 \minus{} 8bc} \plus{} \sqrt {9 \minus{} 8ca} \plus{} \sqrt {9 \minus{} 8ab} \ge 7
\]
With equality holds if and only if $ (a,b,c) \equal{} (2/3,2/3,2/3)$ or $ (a,b,c) \equal{} (1,1,0).$
:)[/quote]
Wei Dongyi have a nice proof: http://www.mathlinks.ro/Forum/viewtopic.php?p=1091404#1091404 :)[/quote]
Thanks, Hanjingjun. :) My solution used the old my technique: [b][i]Extension on the tangent line technique [/i][/b](That is the technique to establish the inequality in the form: $ f(x) \le ax^2\plus{}bx\plus{}c$ where $ x$ is an element of the compact $ D$). From now, our inequality will have no root and it will be easier to prove. :)


Find all pairs of positive integers $(m,n)$ for which it is possible to paint each unit square of an $m$x$n$ chessboard either black or white in such a way that, for any unit square of the board, the number of unit squares which are painted the same color as that square and which have at least one common vertex with it (including the square itself) is even.

I think I solved it.

First we'll show that all chessboard's that are (2n+1) x (2k+1) are bad.

Such a chessboard has an odd number of unit squares. Let's suppose we colored this chessboard good. Then Let's define the number S(a,b), which for the unit square (a,b) is the number of his neighbours (I mean all unit squares having at least one common vertex with square (a,b) , NOT including the square (a,b) itself) that has the same colour as (a,b). Then, for avery a and b S(a,b) is odd. So if we sum all of the S(x,y), we'll get an odd number, because there is an odd number of unit squares on the chessboard. This is a contraddiction, because all neighbourhoods were counted twice (from the perspective of the first and the second neighbour).

Coloring a chessboard 2k x n is easy, I'm too lazy to post it.

Ok, ok, I'll post it. The coloring looks something like this:

101010101010101
101010101010101
010101010101010
010101010101010
101010101010101
101010101010101

I think the pattern is clear  :lol:


Prove that there are infinitely many primes $ p$ such that the numerator of

$ \frac{1}{1}\plus{}\frac{1}{2}\plus{}\cdots\plus{}\frac{1}{\frac{p\minus{}1}{2}}$ is divisible by p.

[quote="Nak"]Prove that there are infinitely many primes $ p$ such that the numerator of

$ \frac {1}{1} \plus{} \frac {1}{2} \plus{} \cdots \plus{} \frac {1}{\frac {p \minus{} 1}{2}}$ is divisible by p.[/quote]
It is open problem, because
\[ \sum_{k \equal{} 1}^{(p \minus{} 1)/2}\frac 1k \equiv \frac {2 \minus{} 2^p}{p}\mod p.
\]
Known only 2 primes $ 2^{p \minus{} 1}\equiv 1\mod p^2,p < 10^{15}$, they are $ p \equal{} 1093,3511$.

[quote="Rust"]
It is open problem, because
\[ \sum_{k \equal{} 1}^{(p \minus{} 1)/2}\frac 1k \equiv \frac {2 \minus{} 2^p}{p}\mod p.
\]
Known only 2 primes $ 2^{p \minus{} 1}\equiv 1\mod p^2,p < 10^{15}$, they are $ p \equal{} 1093,3511$.[/quote]
Where I can find the proof that $ \sum_{k \equal{} 1}^{(p \minus{} 1)/2}\frac 1k \equiv \frac {2 \minus{} 2^p}{p}\mod p?$


invariant
combinatorics proposed
combinatorics

The number from $ 1$ through $ 2008$ are written on a blackboard.Every second,Dr.Math erases four numbers of the form $ a,b,c,a\plus{}b\plus{}c$,and replaces them with the numbers $ a\plus{}b,b\plus{}c,c\plus{}a$.Prove that this can continue for at most $ 10$ minutes.

[hide="Hint1"]The sum of the numbers and the sum of their squares is invariant of the operations.[/hide]
[hide="Hint2"]Cauchy Schwartz  :wink: [/hide]

[i]as [b]Albanian Eagle [/b]has hinted,[/i] notice that $ a \plus{} b \plus{} c \plus{} (a \plus{} b \plus{} c) \equal{} (a \plus{} b) \plus{} (b \plus{} c) \plus{} (c \plus{} a)$, and $ \sum a^2 \plus{} (a \plus{} b \plus{} c)^2 \equal{} \sum(a \plus{} b)^2.$

denote $ \sum x \equal{} 1004 \times 2009$ as the sum of the numbers on the blackboard, $ \sum x^2 \equal{} 1004 \times 2009 \times 1339 \equal{} 1339 \sum x$ as the sum of the squares of the blackboard, and $ n$ as the current number of terms in the blackboard.

by Cauchy-Schwarz, $ (\sum x)^2 \le n\sum x^2 \Rightarrow n\ge \frac {1004\times 2009}{1339}\ge 1506.$

therefore the process will last at most $ 2008 \minus{} 1506 \equal{} 502$ seconds (which is less than 10 minutes).

please check my computation, in case of errors :ddr:


algebra
polynomial
function
algebra proposed

prove that there does not exist a polynomial function f(a,b,c,d) so that
f(a,b,c,d)>0 iff a,b,c and b,c,d and c,d,a and d,a,b are each sides of triangles. :)

Consider the locus $ f(a, b, c, d) \equal{} 0$.  By the Intermediate Value Theorem, this is precisely the set of planes $ a \plus{} b \equal{} c, d; b \plus{} c \equal{} d, a; c \plus{} d \equal{} a, b; d \plus{} a \equal{} b, c$.  Moreover, each of these factors must have odd multiplicity; we can assume without loss of generality that that multiplicity is $ 1$.  We therefore have

$ f(a, b, c, d) \equal{} k(a \plus{} b \minus{} c)(a \plus{} b \minus{} d)(b \plus{} c \minus{} d)(b \plus{} c \minus{} a)(c \plus{} d \minus{} a)(c \plus{} d \minus{} b)(d \plus{} a \minus{} b)(d \plus{} a \minus{} c)$

where $ k$ is a polynomial that is positive almost everywhere, so can be a constant without loss of generality.  But clearly $ f(a, b, c, d) \equal{} f( \minus{} a, \minus{} b, \minus{} c, \minus{} d)$; contradiction.


I think this can probably be done by induction on intervals or length 1, except for the lower cases where the bound is probably tighter. I did it once when trying to solve a China West MO problem (2001) tho the bound was kinda loose so it worked


calculus
integration
logarithms
function
inequalities
real analysis
calculus computations

(1) For real numbers $x,\ a$ such that $0<x<a,$ prove the following inequality.
\[\frac{2x}{a}<\int_{a-x}^{a+x}\frac{1}{t}\ dt<x\left(\frac{1}{a+x}+\frac{1}{a-x}\right). \]
(2) Use the result of $(1)$ to prove that $0.68<\ln 2<0.71.$

[hide="partial solution"]
since $\frac{1}{x}$ is a decreasing function we have

$RHS<\int<LHS$ where RHS is the right hand riemann sum and LHS is the left hand reimann sum.

$\int_{a-x}^{a+x}\frac{dt}{t}<\frac{1}{a-x}+\frac{1}{a-x+1}+\dots+\frac{1}{a+x-1}$   $(1)$

$(1)<2x\left(\frac{1}{a+x-1}\right)<x\left(\frac{1}{a-x}+\frac{1}{a+x}\right)$

for the other inequality we have

$\frac{1}{a-x+1}+\frac{1}{a-x+2}+\dots+\frac{1}{a+x}<\int_{a-x}^{a+x}\frac{dt}{t}$    $(2)$

$(2)>2x\left(\frac{1}{a}\right)=\frac{2x}{a}$

so we have

$\frac{2x}{a}<\int_{a-x}^{a+x}\frac{dt}{t}<x\left(\frac{1}{a-x}+\frac{1}{a+x}\right)$
[/hide]

any hint on how to solve the second part of the equation?


group theory
abstract algebra
geometry
3D geometry
tetrahedron
icosahedron
geometric transformation

Looking for some insight/general hints.  Any takers?

Suppose $H$ is a finite subgroup of $SO(\mathbb{R}^{3})$ and $K$ is a subgroup of $H$ such that	 $|H|=2|K|$.

1.  Let $H^{*}=H\cup\lbrace-T|T\in H\rbrace$.  Prove that $H^{*}$ is a subgroup of $O(\mathbb{R}^{3})$ whose intersection with $SO(\mathbb{R}^{3})$ is $H$.

2.  Let $H]K=K\cup\lbrace-T|T\in H-K\rbrace$.  Prove that $H]K$ is a subgroup of $O(\mathbb{R}^{3})$ whose intersection with $SO(\mathbb{R}^{3})$ is $K$

3.  Let $J$ be the group of rotational symmetries of the tetrahedron whose vertices are at $(1,1,1),(1,-1,-1),(-1,-1,1)$ and $(-1,1,-1)$.  Show that this tetrahedron is regular.  Is $J^{*}$ the group of symmetries of this tetrahedron?  Prove your answer.  Is there a subgroup $W$ of $SO(\mathbb{R}^{3})$ such that $W]J$ is the group of symmetries of this tetrahedron?  Prove your answer.

4.  What can you say along these lines about the relationship between the group $I$ of rotational symmetries of an icosahedron and the group of all its symmetries?

I've been able to do 1.  However, for 2, I need to show that the set is closed under multiplication.  The only case I'm having trouble with is when two elements $T_{1},T_{2}\in H-K$ exist such that $(-T_{1})(-T_{2})\in H]K$.  Do I need to do something with the coset $K$ and the representatives $T_{1}$ and $T_{2}$?

You guessed right. The quotient $H/K$ is a cyclic group of order $2$. Thus, for any $T_{1},T_{2}\in H\setminus K$ you have $T_{1}T_{2}\in K$.

Is there a subgroup $W$ of $SO(\mathbb{R}^{3})$ such that $W]J$ is the group of symmetries of this tetrahedron?  Prove your answer.

I am having an extremely difficult time understanding this.  Won't $W]J$, as it is defined in the problem, contain point inversions?  And since tetrahedrons cannot be mapped to themselves (solely by) point inversions, then $W]J$ will not be the group of symmetries for the tetrahedron, right?

The professor told us not to worry about this problem since we haven't dealt with symmetries/geometrical representation at all this semester, but this problem has been bothering me for the last few days and I'm desperately trying to understand the underpinnings of the ideas involved.

I don't see [url=http://mathworld.wolfram.com/Inversion.html]inversions[/url] here: only rotations and reflections in hyperplanes (and their compositions). Both can map  a tetrahedron into itself.

Hmm.  The way I'm picturing all of this is by imagining that if the vertex, $a$, located at $(1,1,1)$ is inverted (so $x$ maps to $-x$), then the resulting tetrahedron (with $a$ located at $(-1,1,1)$) will not be a map to the original tetrahedron.

I apologize for my naivete ahead of time.  There's a good chance I'm just not understanding everything given my limited background in the area.  If my understanding/interpretation is incorrect, does that mean it [i]is[/i] possible for $W]J$ to be the group of all symmetries?

I misunderstood what you meant by inversion. Your approach seems to show that $J^{*}$ is not the group of symmetries of the tetrahedron, since it contains the map $x\mapsto-x$.
However, the group $W]J$ does not have to contain such elements. Indeed, it contains $-T$ for each $T\in W\setminus J$; this means that $T$ must [b]not[/b] map the tetrahedron to itself, so that $-T$ can do it.
I don't know if such $W$ exists.


Question from 07-08 School Handbook:

The point A(5/2, 0) is reflected over the line y=1/2 x to the point A'. What are the coordinates of A'? Express any non-integer coordinate as a common fraction?

How would you do this problem? Any help is greatly appreciated.

thie is the only thing i can think of.
draw a line though A perpendicular to y=1/2x.  then A' must also be on that line an equal distance away from the line


If x, y and z are unequal real numbers whose sum is 0 and whose product is 2, determine the value of x^3 + y^3 + z^3.

[hide="Solution"]It is easy to prove that if $a+b+c=0$, then $a^3+b^3+c^3=3abc$. Since $abc=2$, $a^3+b^3+c^3=6$.[/hide]

this is the identity

     $\ x+y+z = 0$ then $\ x^3 + y^3 + z^3 = 3xyz$
hence the answer is 6.

those solutions are all fine...but i think that it requires you to know an obscure identity...substituting $x=-y-z$ and expanding works just as well


I need some hints to find all the functions $ f:\mathbb R\to\mathbb R$, such that $ f(xf(y))\plus{}f(yf(x))\equal{}2xy$. I can only prove that $ f(x)\equal{}x$ and $ f(x)\equal{}\minus{}x$ are solutions. Thanks :)


[quote="nguyenphihung"]Mọi người thử sức xem :
[size=34][FONT=Times New Roman]Cho $\large a,b,c >0$ thoả $\large \sqrt{a}=\frac{1-c}{1+b}$[/FONT][/size]
[size=34][FONT=Times New Roman]CMR :[/FONT][/size]
[size=34][FONT=Times New Roman]$\large (ab+c)(c+\frac{1}{b})(1+\frac{b}{c}) \geq 3(1-\sqrt{\frac{abc}{3}})^{2}$[/FONT][/size][/quote]
Làm sao bé tí như vậy?  :lol:

Chắc là đề như thế này
Mọi người thử sức xem :
Cho$\large a,b,c >0thoả \large \sqrt{a}=\frac{1-c}{1+b}$
CMR
$\large (ab+c)(c+\frac{1}{b})(1+\frac{b}{c}) \geq 3(1-\sqrt{\frac{abc}{3}})^{2}$

Trông bài toán này chẳng có chút ý nghĩa gì cả. Kết quả lại chẳng đẹp.

[size=34][font=Times New Roman]Vậy thêm bài này nha , bài đó quá xấu :
Cho a,b,c là độ dài của một tam giác không tù . Chứng minh rằng .
$\large \frac{ \sqrt{a^{2}+b^{2}-c^{2}}}{b+c-a}+\frac{ \sqrt{b^{2}+c^{2}-a^{2}}}{c+a-b}+\frac{ \sqrt{c^{2}+a^{2}-b^{2}}}{a+b-c}\geq 3$[/font][/size]

Cho a,b,c là độ dài của một tam giác không tù . Chứng minh rằng :[size=34][font=Times New Roman]
$\large \frac{ \sqrt{a^{2}+b^{2}-c^{2}}}{b+c-a}+\frac{ \sqrt{b^{2}+c^{2}-a^{2}}}{c+a-b}+\frac{ \sqrt{c^{2}+a^{2}-b^{2}}}{a+b-c}\geq 3$[/font][/size]

cái này kô có gì
Ta dùng BĐT
$\prod(a+b-c)\le \prod(a^{2}+b^{2}-c^{2})$

Gõ thế này [code]Cho a,b,c là độ dài của một tam giác không tù . Chứng minh rằng :[size=4][font=Times New Roman]
$\large \frac{ \sqrt{a^{2}+b^{2}-c^{2}}}{b+c-a}+\frac{ \sqrt{b^{2}+c^{2}-a^{2}}}{c+a-b}+\frac{ \sqrt{c^{2}+a^{2}-b^{2}}}{a+b-c}\geq 3$[/font][/size][/quote][/code]
nhẽ nào lại không ra bé tí như vậy. Phải thế này này:

Cho a,b,c là độ dài của một tam giác không tù . Chứng minh rằng :
$\large \frac{ \sqrt{a^{2}+b^{2}-c^{2}}}{b+c-a}+\frac{ \sqrt{b^{2}+c^{2}-a^{2}}}{c+a-b}+\frac{ \sqrt{c^{2}+a^{2}-b^{2}}}{a+b-c}\geq 3$

[code]Cho a,b,c là độ dài của một tam giác không tù . Chứng minh rằng :
$\large \frac{ \sqrt{a^{2}+b^{2}-c^{2}}}{b+c-a}+\frac{ \sqrt{b^{2}+c^{2}-a^{2}}}{c+a-b}+\frac{ \sqrt{c^{2}+a^{2}-b^{2}}}{a+b-c}\geq 3$[/code]

Cách giải của bạn sai rồi , dùng cái BĐT của bạn thì làm sao chứng minh được , còn cả dâú căn thức mà ?

viết nhầm chút , vế trái của bổ đề có căn thức. BĐT này có trong cuốn tin á.

Bạn có thể nhớ nhầm chứ hình như dấu ngược lại mà ?


algebra
polynomial
quadratics
quadratic formula

How do i factor and find the imaginary roots for this polynomial?
$ x^{3}\plus{}6x^{2}\plus{}13x\plus{}10\equal{}0$

Hint:  Factor it into the product of a linear and quadratic, and then find the roots of the quadratic.

[hide="Solution"]Set the expression equal to $ (x\plus{}a)(x^2\plus{}bx\plus{}c)$ and mutiply out and compare coefficients.

$ x^3\plus{}6x^2\plus{}13x\plus{}10\equal{}(x\plus{}a)(x^2\plus{}bx\plus{}c)$

$ x^3\plus{}(a\plus{}b)x^2\plus{}(c\plus{}ab)x\plus{}ac\equal{}x^3\plus{}6x^2\plus{}13x\plus{}10$

Now we have $ a\plus{}b\equal{}6$, $ c\plus{}ab\equal{}13$, $ ac\equal{}10$.

Solving these yields: $ a\equal{}2$, $ b\equal{}4$, $ c\equal{}5$.

$ x^3\plus{}6x^2\plus{}13x\plus{}10\equal{}(x\plus{}2)(x^2\plus{}4x\plus{}5)\equal{}0$

Thus the complex roots of the equation are the complex roots of $ x^2\plus{}4x\plus{}5\equal{}0$.

Quadratic formula shows that these are $ x\equal{}\minus{}2\minus{}i$, $ x\equal{}\minus{}2\plus{}i$.[/hide]

[hide="Alternatively"]
The terms are separated with + signs, so we know a root is negative. Trying $ x \equal{} \minus{}1$ gives us 2, so it has to be smaller. $ x \equal{} \minus{}2$ gives us 0, so we know this polynomial is divisible by $ x \plus{} 2$. Divide and you get $ (x\plus{}2)(x^{2}\plus{}4x\plus{}5)\equal{}0$, and from there it's the same.
[/hide]

[quote="Problemsolver2011"]How do i factor and find the imaginary roots for this polynomial?
$ x^{3} \plus{} 6x^{2} \plus{} 13x \plus{} 10 \equal{} 0$[/quote]

Let P(x)=x^3+6x^2+13x+10

Since P(-2)=0 therefore x+2 is a factor of P(x)

By long division, We get P(x)=A(x)Q(x)+R(x) [where R(x)=ax+b]

P(x)=(x+2)(x^2+4x+5)

P(x)=(x+2)[(x+2)^2-i^2]

P(x)=(x+2)(x+2-i)(x+2+i)

Let P(x)=0 therfore x=-2,-2+i,-2-i

EDIT: Didn't see the above solutions lol ah well, another one can't hurt

[quote="Problemsolver2011"]How do i factor and find the imaginary roots for this polynomial?
$ x^{3} \plus{} 6x^{2} \plus{} 13x \plus{} 10 \equal{} 0$[/quote]

Trial and error for rational root. $ (x \plus{} 2)(x^2 \plus{} 4x \plus{} 5) \equal{} 0 \Rightarrow x \equal{} \{ \minus{} 2, \minus{} 2 \plus{} i, \minus{} 2 \minus{} i\}$

[quote="isabella2296"][hide="Alternatively"]
The terms are separated with + signs, so we know a root is negative. Trying $ x \equal{} \minus{} 1$ gives us 2, so it has to be smaller. $ x \equal{} \minus{} 2$ gives us 0, so we know this polynomial is divisible by $ x \plus{} 2$. Divide and you get $ (x \plus{} 2)(x^{2} \plus{} 4x \plus{} 5) \equal{} 0$, and from there it's the same.
[/hide][/quote]

To divide it, you can use synthetic division, which speeds up isabella's process even more.

I believe that's what she did. If not, then you're right.

it was a very simple factor to obtain just using long division, if it was anymore difficult i would have used synthetic method aswell


This is out of AOPS Vol I:

Given that $ x$ is directly proportional to $ y$ and to $ z$ and is inversely proportional to $ w$, and that $ x \equal{} 4$ when ($ w$ , $ y$ , $ z$) $ \equal{}$ ($ 6$ , $ 8$ , $ 5$) , what is $ x$ when $ ( w, y, z)$ $ \equal{}$ $ (4, 10, 9)$ ?

I don't know if I'm just completely misunderstanding the question, but I can't get a sensible answer...

I start out with $ \frac{x}{y}$ $ \equal{}$ $ k$ where $ k$ is a constant

Then plugging in the first set of numbers $ x \equal{} 4$ and $ y \equal{} 8$ we get $ k \equal{} \frac{4}{8}$ $ \Rightarrow$ $ k \equal{} \frac{1}{2}$

However when I plug in the second value for $ y$, I get $ \frac{x}{10} \equal{} k$ plugging $ k$ in I get $ x \equal{} 5$

This contradicts the answer I get when I do the same thing for $ w$ and $ z$....

[hide="Hint"]The fraction should include every variable in it to be able to find the constant, so think about which variables should go on the top or bottom, according to whether they are inversely or directly proportional[/hide]

u  see it   has been given that x  is  directly  proportional to y and z.

    but it does not mean that x/y  and x/z  have the  same  value.

here the `k`  u hv\ae given is called as constant of proportionality.   this is  not the same in both the cases.

so u cannot plugin the value of `k`   in other cases.

Yeah I know that $ k$ is different for the different variables... I just thought it was suppose to be 3 separate equations, instead of 1.  How can you tell to use all the info to form a single equation?

actually [hide="i thought"]$ \frac{xw}{yz}$[/hide]

[quote="yaofan"]actually [hide="i thought"]$ \frac {xw}{yz}$[/hide][/quote]
That's the same thing  :wink: (assuming your equation equals k)

Wait a minute. Z and w don't have anything to do with the values of x and y right?

^ That's what I thought when I looked at the problem, however in their answer they make it a single equation - I'm still confused as to how you decide if the problem is talking about a single or multiple equations.

You should just be able to eliminate z and w. I don't get why the solutions include them.

The question has to involve every variable in order to get the constant $ k$. Now depending on whether the variables are inversely or directly proportional  they will go in either the denominator or numerator. The equation to find the constant $ k$ would be $ \frac{xw}{yz}\equal{}k$. Now just plug in every thing to find the constant and use everything else to find the final variable.

Didn't get that. So w and z are related to x in some way?

yes  mathchampion  , they all are interrelated  by a single equation  as explained above.


calculus derivative integration trigonometry geometry rectangle function

When's the latest someone could turn it in?

[quote="JS1527"]I just finished and sent in (I swear) my calculus placement test for college, and there were a few problems I was stuck on.  Would it be OK for me to ask AoPS for help, just because I am curious about how to do them?  I sent in the test today and I can't change my answers.  I'll wait before I post anything...[/quote]
Bah humbug, you can ask me on AIM if you want.

There's a Calculus forum in the undergraduate forum.

hspotter - Yea but it's not really a competition style question!  Anyway, no telling moderators what to post (or not to post) in their own forum!  ;)

Magnara - It was due on something like July 6th but I got an extension b/c I am learning calculus over the summer (well I just finished).

Okay, I am going to go ahead and post them since I am really curious.  I can always delete it later if people think it's a problem...

This one I guessed on:

*Determine the anti-derivative: [tex]\int{\frac{1}{x^2*\sqrt{4-x^2}}[/tex]

I don't know what to do here!  I tried all sorts of stuff...trig substitution, integration by parts, u-substitution...

This one I must have never covered in my class even though we did do differential equations...

* The general solution to the differential equation [tex]y''+6y'+9y=0[/tex] has  which of the following forms where [tex]c[/tex] and [tex]k[/tex] are arbitrary constants?
(a) [tex]ce^{-3x}+ke^{3x}[/tex]
(b) [tex]ce^{-3x}+kxe^{-3x}[/tex]
(c) [tex]c*\cos{(3x)}-k*\sin{(3x)}[/tex]
(d) [tex]ce^{-x}*\sin{(3x)}+ke^{-x}*\cos{(3x)}[/tex]
(e) None of the above.

On topics we covered, but I was still confused about it:

*The area of a rectangle is held fixed at 100 square meters although the lengths of the sides vary with time.  At some instant, the longer side is 20 meters long and is increasing at a rate of 2 m/s.  How fast is the length of the shorter side changing?
(a) -5 m/s
(b) -2 m/s
(c) -0.5 m/s
(d) 0.5 m/s
(e) 2 m/s

Thanks for your insight...

For the last one,

The shorter side must be 5 at the moment in time we are concerned with. Now, the formula for the area of a rectangle is A = lw. Here l = 5 and w = 20. Differentiating with respect to time, we have:

(dA/dt) = w*(dl/dt) + l*(dw/dt) (by the product rule)

0 = 20*(dl/dt) + 5*2

dl/dt = -0.5 (c).

[quote="liyi"][quote="JS1527"]
*Determine the anti-derivative: [tex]\int{\frac{1}{x^2*\sqrt{4-x^2}}[/tex]
[/quote]

Hmm. You forgot a $dx$ there, didn't you?
This one is not difficult, indeed.[/quote]
Where did you learn that trickery? It's unlike anything I've ever seen.

[quote="JS1527"]* The general solution to the differential equation [tex]y''+6y'+9y=0[/tex] has  which of the following forms where [tex]c[/tex] and [tex]k[/tex] are arbitrary constants?
(a) [tex]ce^{-3x}+ke^{3x}[/tex]
(b) [tex]ce^{-3x}+kxe^{-3x}[/tex]
(c) [tex]c*\cos{(3x)}-k*\sin{(3x)}[/tex]
(d) [tex]ce^{-x}*\sin{(3x)}+ke^{-x}*\cos{(3x)}[/tex]
(e) None of the above.
[/quote]

It has already been solved, but alternatively, couldn't one take advantage of a multiple choice test and reverse engineer the answer; i.e. differentiate each possibility and substitute the derivatives into the differential equation?

Okay, to be honest I don't get how liyi solved either of them.  Yea I did forget the dx though.  I have never seen this method either, confuted.  Can you explain what you did (esp for the diff equation one)?

I got the problem Rosen solved, thanks.

Mildorf - I should have done that, but I didn't think to do it since we only spent 3 days on differential equations, and it was mostly applications of them like logistic differential equations and putting an equation in a certain form (not solving, you just memorized what the solution looked like).

On the differential equation: if you really didn't know, because you'd never learned this material, the fact that it's multiple choice would give you another option: take the derivatives of all of the proposed solutions and see if they do satisfy the differential equation. That would be inefficient and time-consuming, but it would work.

On the integral: liyi's solution is indeed inventive and clever. The person who wrote the problem wasn't looking for inventive and clever. You say you tried trig substitutions, but let's try again:
Let $x=2\sin\theta$
Then $dx = 2\cos\theta d\theta$
and $\sqrt{4-x^2}=2\cos\theta$
Then $\displaystyle \int\frac{dx}{x^2\sqrt{4-x^2}} = \int\frac{2\cos\theta d\theta}{(4\sin^2\theta)(2\cos\theta)} = \frac14 \int\csc^2\theta d\theta$
You know what the integral of $\sec^2\theta$ is, so it won't be hard to guess that $\displaystyle \frac14 \int\csc^2\theta d\theta = -\frac14\cot\theta+C = -\frac{\sqrt{4-x^2}}{4x} + C$
Personally, I never do that last step, converting the trig function back to a function of $x$, without drawing an appropriately labeled right triangle.

I've been teaching calculus for many years. Sure there are many ways to solve this. What you want is a method you can remember and for which the motivations of the intermediate steps are clear.

[quote="Kent Merryfield"]
On the integral: liyi's solution is indeed inventive and clever, (and has a typographical error on the last line). [/quote]
thanks. I've fixed it.

[quote="JS1527"]hspotter - Yea but it's not really a competition style question!  Anyway, no telling moderators what to post (or not to post) in their own forum!  ;)[/quote]

I wasn't exactly telling you what not to post.  I was just simply stating a fact. :blush:

[quote="liyi"][quote="JS1527"]* The general solution to the differential equation [tex]y''+6y'+9y=0[/tex] has  which of the following forms where [tex]c[/tex] and [tex]k[/tex] are arbitrary constants?
(a) [tex]ce^{-3x}+ke^{3x}[/tex]
(b) [tex]ce^{-3x}+kxe^{-3x}[/tex]
(c) [tex]c*\cos{(3x)}-k*\sin{(3x)}[/tex]
(d) [tex]ce^{-x}*\sin{(3x)}+ke^{-x}*\cos{(3x)}[/tex]
(e) None of the above.
[/quote]
okay. $x^2+6x+9=0$ has root $x=-3$ with mulplicity 2.
thus the general solution is $y = c_1e^{-3x} + c_2x e^{-3x}$.[/quote]

Why do you use that equation?  I thought factoring might have something to do w/it, but derivatives are not like x^2=x*x...  How did what you get from the equation follow to give you the general solution?

Thanks Kent Merryfield, I must have made an error when I did it...I should have been able to figure that out...

[quote="Kent Merryfield"] You know what the integral of $\sec^2\theta$ is, so it won't be hard to guess that $\displaystyle \frac14 \int\csc^2\theta d\theta = -\frac14\cot\theta+C = -\frac{\sqrt{4-x^2}}{4x} + C$
[quote]

Right, I am not sure about how to do that w/o a table formula.  I know the integral of sec(x)^2 because d/dx(tanx)=sec(x)^2, so integrating gives you tangent, but I don't know why your guess worked...

Do you know $(\cot x)' = -\csc^2 x$?  :D

[quote="liyi"]I suggest the people who ask about differential equations read some fundamental books.
Hereby I write some notes on the diff eqn.

A second order linear homogeneous ordinary differential equation with constant coefficients is \[a y'' + b y' + c y = 0\]
If we have two linear independent solutions $y_1, y_2$, then all its solutions can be written as $c_1 y_1 + c_2 y_2$.

The characteristic equation is $a \lambda^2 + b\lambda + c = 0$
The solutions of the characteristic equation is $\lambda_1$ and $\lambda_2$.

If $\lambda_1\neq\lambda_2$, we take $y_i=e^{\lambda_ix}$, $i=1,2$.
If $\lambda_1=\lambda_2=\lambda$, we take $y_1=e^{\lambda x}$ and $y_2 = xe^{\lambda x}$[/quote]

hrm. I certainly didn't know this right after taking Calculus BC (and not after taking calculus III either). I might have known (vaguely) the solution when the two roots are not the same, but certainly not when the two roots were the same. I know this only because I took differential equations...I am quite surprised it appeared on a calculus placement test. Maybe the question writer WANTED you to differentiate each answer and see if it worked ?

[quote]I know the integral of sec(x)^2 ..., but I don't know why your guess worked...[/quote]

A way to organize your memory of the calculus of trig functions: the formulas come in three bunches. There's the "sin-cos" bunch of formulas, which are few and simple. There's the "tan-sec" bunch of formulas, one of which you just quoted. Note that it you try hard enough, you can express all of these without using any names other than "tan" and "sec". You do need to have these memorized, particularly the integral of the secant which would be quite hard to find in any systematic way.

Finally, there's the "cot-csc" bunch of formulas. My advice to my students is not to try to separately memorize these. Just follow this general principle: each of the "cot-csc" formulas is the "same" as the corresponding "tan-sec" formulas, (replacing tan by cot and sec by csc), except that there's always a sign that changes. If there's any question about which sign changes, you can always check the antiderivative formulas by differentiation.

Material on second order constant coefficient homogenous ordinary differential equations isn't universally part of a second semester college calculus course, but it's reasonably common. (It is in ours.) liyi didn't even get into the question of how to interpret the answer should the roots of that characteristic polynomial be non-real complex numbers. My suggestion to the person who started this thread is to consider starting in college with that second semester course - there's a lot to learn there, and it sets up a number of courses to follow.

[quote="Kent Merryfield"]liyi didn't even get into the question of how to interpret the answer should the roots of that characteristic polynomial be non-real complex numbers.[/quote]

hmm ??
$e^\lambda x$ works for complex numbers, say $\lambda_1 = a+bi$ and $\lambda_2 = a-bi$, according to the formula above,
the solution is $c_1 e^{(a+bi)x} + c_2 e^{(a-bi)x} = c_1e^{ax}(\cos bx+i\sin bx) + c_2e^{ax}(\cos bx-i\sin bx) = k_1 e^{ax} \cos bx + k_2 e^{ax}\sin bx$.

I knew you knew the answer to that, liyi. Following up on what you just posted, the four multiple choice answers in the original problem are the answers to the following differential equations:
(a)  $y''-9y=0$
(b)  $y''+6y'+9y=0$ (which is what was asked)
(c)  $y''+9y=0$
(d)  $y''+2y'+10y=0$