1.  Αν μια συνάρτηση είναι συνεχής και άρτια στο διάστημα [-π,π] και ισχύει'
     $ \int_0^α f(x)cos(nx) dx = 0$ όπου α=π για κάθε n=0,1,2,3,...
     Tότε να δειχθεί ότι f(x)=0 στο διάστημα [-π,π]

2.  Αν μια συνάρτηση είναι συνεχής στο [α,β] και ισχύει'
     $ \int_α^β f(x)x^n dx = 0$   για κάθε n=0,1,2,3,...
     Tότε να δειχθεί ότι f(x)=0 στο διάστημα [α,β]

Συγνώμη παιδιά αυτό ήθελα να γράψω, αλλά μπερδεύτηκα!

1.  Αν μια συνάρτηση είναι συνεχής και άρτια στο διάστημα [-π,π] και ισχύει'
     $ \int_{0}^{a} f(x)cos(nx) dx \equal{} 0$ όπου α=π για κάθε n=0,1,2,3,...
     Tότε να δειχθεί ότι f(x)=0 στο διάστημα [-π,π]

2.  Αν μια συνάρτηση είναι συνεχής στο [α,b] και ισχύει'
     $ \int_{a}^{b} f(x)x^n dx \equal{} 0$   για κάθε n=0,1,2,3,...
     Tότε να δειχθεί ότι f(x)=0 στο διάστημα [α,b]

Ok spinos, πρέπει να κάνω μία ερώτηση για διευκρίνιση: τι είδους λύσεις ψάχνεις? Ρωτάω επειδή και οι δύο λύνονται εύκολα αν γνωρίζεις θεωρία ανάλυσης Fourier, αλλά, αν όχι, υπάρχουν αρκετά λεπτά σημεία (σύγκλισης κλπ) που ναι μεν δεν είναι δύσκολο να λυθούν/κατανοηθούν (ιδίως αφού μιλάμε για ένα συμπαγές διάστημα), αλλά χρειάζεται δουλίτσα. Επίσης, αν δεν έχεις δει ποτέ στη ζωή σου την ιδέα και άρα αναζητούμε μία λύση που να μην έχει να κάνει με ανάλυση Fourier (γιατί δεν είναι και πολύ τίμιο να λέμε ότι ξαναεφήυραμε τον τροχό ;) ), πες μου για να αρχίσω να ψάχνω για πιο στοιχειώδεις ("λυκειακές") λύσεις.

Τώρα, χρησιμοποιώντας ιδέες από ανάλυση Fourier, η βασική ιδέα (πχ για την (1) που θέλει και λιγότερο γράψιμο), είναι να παρατηρήσεις ότι $ \int_{ \minus{} \pi}^\pi \cos(mx)\cos(nx) dx \equal{} \pi\delta_{mn}$ οπότε η βάση $ \left\{\frac {1}{\sqrt {\pi}}cos(mx)\right\}_{m \equal{} 0}^\infty$ είναι ορθοκανονική. Έτσι, στην τοπολογία ομοιόμορφης σύγκλισης θα έχουμε: $ f(x) \equal{} \sum_{n \equal{} 0}^\infty b_n \cos (nx)$ (*) οπότε, πολλαπλασιάζοντας τα δύο μέλη με $ cos(mx)$ και ολοκληρώνοντας βρίσκουμε ότι:
\[ \int_{ \minus{} \pi}^\pi f(x) cos(mx) \equal{} \sum_{n \equal{} 0}^\infty \pi b_n \delta_{mn} \equal{} \pi b_m
\]
απ'όπου και προκύπτει $ b_m \equal{} 0$ για κάθε $ m$. Συνεπώς, ξανά από την (*), έπεται ότι $ f(x) \equal{} 0$.

Η 2 βγαίνει με ακριβώς ανάλογο τρόπο, απλώς βολεύει να κάνεις Gram-Schmidt για να κάνεις τη βάση $ \{x^k\}$ ορθοκανονική.

Cheers,

Durandal 1707

Αγαπητέ Durandal, καταρχήν σ΄ευχαριστώ πολύ για το ενδιαφέρον σου να με βοηθήσεις'
Για να σου δώσω να καταλάβεις, εδώ μου ζητάει να χρησιμοποιήσω το θεώρημα του Parseval για σειρές fourier

Θεώρημα'
Για όλες τις συνεχείς συναρτήσεις f(x) περιόδου 2π, ισχύει
$ \frac{1}{2}a_0^2\plus{} \sum_{n\equal{}1}^{\infty} (a_n^2\plus{}b_n^2) \equal{}\frac{1}{p} \int_{\minus{}p}^{p} f^2(x) dx$ (όπου το p είναι το γνωστό π, δεν μπορούσα να το γράψω στα ελληνικά)

(   $ a_n\equal{}\frac{2}{p}\int_{0}^{p} f(x)cos(nx) dx$          (p=π)    και     $ b_n\equal{}0$     όταν η f(x) είναι άρτια )

Βάση αυτού του θεωρήματος μου ζητάει να του αποδείξω"

Aν η f(x) είναι συνεχής και άρτια στο [-π,π] και ισχύει
$ \int_{0}^{p} f(x)cos(nx) dx\equal{}0$ ( το p=π πάλι) για n=0,1,2,3,...
τότε να δειχθεί ότι f(x)=0 στο [-π,π]

Αν και είμαι ολίγον τι άσχετος όσον αφορά την αλγεβρική τοπολογία, με τη χρήση του θεωρήματος που δίνεις είναι:

μελετώντας την $ g(x) \equal{} f(x)$ για κάθε $ x \in [ \minus{} \pi, \pi]$ με $ g$ να ορίζεται μόνο σε αυτό το διάστημα,
η g είναι άρτια στο διάστημα αυτό, αφού είναι και η f
η g είναι περιοδική με περίοδο $ 2\pi$, αφού $ g( \minus{} \pi \plus{} 2\pi) \equal{} g(\pi) \equal{} g( \minus{} \pi)$ και $ g( \pi \minus{} 2\pi) \equal{} g(\minus{}\pi) \equal{} g(\pi)$

$ \frac {a_0^2}{2} \plus{} \sum_{n \equal{} 1}^{\infty}(a_n^2 \plus{} b_n^2) \equal{} 0 \Leftrightarrow \frac {1}{\pi}\int_{ \minus{} \pi}^{\pi} g^2(x) dx \equal{} 0 \Leftrightarrow$
$ \int_{ \minus{} \pi}^{\pi} g^2(x) dx \equal{} 0$
έστω $ g(x) \not \equal{} 0$ τότε
$ g^2(x) > 0 \Rightarrow \int_{ \minus{} \pi}^{\pi} g^2(x) > 0 \Leftrightarrow$ άτοπο
άρα $ g(x) \equal{} 0$
άρα $ f(x) \equal{} g(x) \equal{} 0$ για κάθε $ x \in [ \minus{} \pi, \pi]$

P. S. To π γράφεται ως \pi.

Ummm, maniopas, τα παραπάνω είναι (συναρτησιακή) ανάλυση, όχι αλγεβρική τοπολογία  :wink:

Τώρα, spinos, αφού ξέρεις ανάλυση Fourier, όλα αλλάζουν (είχα ιδρώσει μια στιγμή γιατι δεν έβλεπα στοιχειώδη λύση στο πρόβλημά χωρίς κάτι που να μοιάζει με Fourier). Ουφ!!!  :)  :)

Φυσικά, αν η Parseval θεωρείται γνωστή, η λύση είναι αυτή του maniopas. Διαισθητικά, αυτό που συμβαίνει είναι το εξής: οι συντελεστές Fourier είναι όλοι 0. H Parseval, σαν πυθαγόρειο θεώρημα που είναι, σου λέει ότι το $ L^2$ μέτρο της $ f$ είναι ίσο με το $ \ell^2$ μέτρο $ \sum_{n \equal{} 0}^\infty a_n^2$ των συντελεστών. Έτσι, αφού η $ f$ προκύπτει να έχει νόρμα 0 και είναι και συνεχής, θα πρέπει να είναι αναγκαστικά η μηδενική συνάρτηση.

Για την 2, αρκεί να σκεφτείς ως εξής: η συναρτησιακή βάση $ \{x^k\}_{k \equal{} 0}^\infty$ παράγει τον χώρο $ L^2[a,b]$ (αν δεν το ξέρεις αυτό, προκύπτει από το θεώρημα Stone-Weierstrass: κάθε συνεχής συνάρτηση προσεγγίζεται ομοιόμορφα από ένα πολυώνυμο και οι συνεχείς συναρτήσεις είναι πυκνές στον $ L^2$).
Με ορθογωνιοποιήση Gram-Schmidt, μπορούμε να πάρουμε μία καινούρια, ορθοκανονική βάση $ \{u_j\}$ από την $ \{x^k\}$ και, αφού οι συντελεστές της $ f$ είναι 0 ως προς την πρώτη, θα είναι όλοι 0 και ως προς την δεύτερη (ορθοκανονική) βάση.
Τώρα, αρκεί να εφαρμόσεις τη γενική μορφή της Parseval ($ \|f\|_{L^2}^2 \equal{} \sum_{j \equal{} 0}^\infty\left < u_j|f\right >^2$, όπου $ u_j$ ορθοκανονική βάση) για την $ \{u_j\}$: $ \|f\|_{L^2} \equal{} \sum_{j \equal{} 0}^\infty 0 \equal{} 0$

Νόρμα 0 $ \implies$ τελείωσες εφόσον η $ f$ είναι συνεχής :)

Cheers,

Durandal 1707


inequalities
number theory unsolved
number theory

Find all pairs of positive integers $ (x,y)$ such that $ x^y\plus{}y\equal{}y^x\plus{}x$

WLOG $ y > \equal{} x$ then $ y \equal{} x \plus{} n$ or $ x^{x \plus{} n} \plus{} n \equal{} (x \plus{} n)^x$.
If $ n \equal{} 0$ then the solution is $ (x, x)$ where $ x$ is any positive integer.
If $ n \equal{} 1$ then $ (1, 2)$, $ (2,3)$ satisfy the equation. For $ x > \equal{} 3$ the following inequality holds: $ x^{x \plus{} 1} \plus{} 1 > (x \plus{} 1)^x$ (we can probably prove this by induction).
If $ n \equal{} 2$ then $ (1,3)$ satisfy the equation. For $ x > \equal{} 2$ the inequality holds.
If $ n > \equal{} 3$ then the inequality holds for all $ x$.

Finally we have: $ {(x,x), (1,2), (2,1), (2,3), (3,2), (1,3), (3,1)}$


using a small subtitutions,we change the problem to:
[tex]\sum (\frac{a_i}{s-a_i})^\frac{n-1}{n}\geq\ \frac{n}{\sqrt[n]{(n-1)^(n-1)}}[/tex],where [tex]s=a_1+a_2+...a_n[/tex]
But [tex](\frac{a_i}{s-a_i})^\frac{n-1}{n}=\frac{a_i}{(s-a_i)^\frac{n-1}{n}.(a_i)^\frac{1}{n}}\geq\frac{na_i}{s.(n-1)^\frac{(n-1)}{n}}(by AM-GM)[/tex].We are done


abstract algebra
linear algebra
linear algebra unsolved

This is what that looks like on my computer:

[img]http://adudzik.com/problem.png[/img]


Solve the system \begin{eqnarray*} \\ (x+y)(x+y+z) &=& 18 \\ (y+z)(x+y+z) &=& 30 \\ (x+z)(x+y+z) &=& 2A \end{eqnarray*} in terms of the parameter $A$.

[quote="10000th User"]shyong: Case (ii) doesn't make sense. Try substituting in the values into the original system, it's false[/quote]

opps , you are right  :blush:  case (ii) doesnt work


How many positive three-digit prime numbers less than 200 have two digits the same?

[quote="AstroPhys"][hide]
Finding a list of primes
$101 \ , \ 103 \ , \ 107 \ , \ 109 \ , \ 113 \ , \ 127 \ , \ 131 \ , \ 137 \ , \ 139 \ , \ 149 \ , \ 151 \ , \ 157 \ , \ 163 \ , \ 167 \ , \ 173 \ , \ 179 \ , \ 181 \ , \ 191 \ , \ 193 \ , \ 197 \ , \ 199$
So $7$ prime numbers.  :maybe:
Is there a different approach?
[/hide][/quote]

Did you miss "have two digits the same"?

[hide="Solution"]We only need to consider

$113, 115, 117, 119$

The second dies, it is a multiple of $5$.
The third dies, it is a multiple of $13$.
The fourth dies, it is a multiple of $7$.

There is only one left it is $113$.

$111, 133, 155, 177, 199$
(the rest are even)

The first dies, it is a multiple of $3$.
The second dies, it is a multiple of $7$.
The middle dies, it is a multiple of $5$.
The fourth dies, it is a multiple of $3$.

There is only one left, it is $199$.

So the numbers are $113, 199$.

So $\boxed{\textrm{2 primes}}$.[/hide]

Edited.

[quote="b-flat"][quote="AstroPhys"]Did does it say they have to be next to each other?[/quote]

No, of course it does not.

[hide="full solution"]There are two 3 cases.  The number must be of the form $1d1$, $11d$, or $1dd$.

CASE 1:  $1d1$
Possible numbers (all multiples of 3 are immediately skipped):
101, 121, 131, 151, 161, 181, 191
$121=11^{2}$
$161=7\times23$
Of these, 101, 131, 151, 181, 191 are prime, for 5 total.

CASE 2:  $11d$
Possible numbers (all multiples of 3 are immediately skipped, as are multiples of 5):
113, 119
$119=7\times17$, so only one prime is added in case 2.

CASE 3:  $1dd$
Possible numbers (ditto):
133, 199
$133=7\times19$, so only one prime is added in case 3.

Therefore, there are $5+1+1=7$ total such primes.[/hide][/quote]
To quicken up the search, note that all primes must be 1 or -1 mod 6 (except 2 and 3).

I did that.  I eliminated all even numbers $\{0,2,4\}\pmod{6}$, and all multiples of 3 $\{0,3\}\pmod{6}$, leaving only $\{1,5\}=\{1,-1\}\pmod6$.


Since many people seem to have forgotten, in the good old days, people usually didn't really say explicitly whether they were making mock AMC 12s or mock AMC 10s.  They were just mock AMCs.  So you don't have to lose sleep over whether your competition is a mock AMC 10 or a mock AMC 12, because it doesn't have to be that specific (of course, it can if you want it to).  (Besides, as the question-writer, you are in a uniquely bad position to determine how hard the test is—it's very difficult to guage a problem's difficulty accurately without solving it on your own.)  And you certainly don't need to lose any sleep over whether [i]other[/i] people's mock competitions are mock AMC 10s or AMC 12s.  In fact, you don't even need to snipe at them about it in pointless conversations in which everybody is miscommunicating.

(I would have posted this in an already existing thread, but it seemed to belong in a few of them.  I also included this content in a stickied thread, but I figured more people would read it here.)

aww I didn't know that.... Well I wasn't here in the good old days

Good point  :lol:


[quote="mathsrox"]Find the number of integral solutions to the following equation:
1/x+1/y=1/12[/quote]
Note neither $ x$ or $ y$ can be equal to 12.  Rewrite the equation as $ y=\frac{12x}{x-12}=12\bigg( 1+\frac{12}{x-12}\bigg)$.  Then, we have that
\[ |y| \le 12+\bigg| \frac{12}{x-12}\bigg| \le 24\]
since the latter fraction is bounded by 12 for integral $ x$.  This leaves $ <48$ possible solutions.  Try them and you will have your solution.   :)

[hide="Solution"]$ \frac{1}{x}+\frac{1}{y}= \frac{1}{12}\iff y = \frac{12x}{x-12}$

We want positive integral solutions; that is, if we force $ y$ to be a positive integer, then $ x$ must also be a positive integer.

Thus, $ x-12 | 12x \implies (x-12) | 12x-12(x-12) \implies x-12 | 144$

$ 144 = 2^{4}\times 3^{2}\implies \sigma_{0}(144) = (4+1) \times (2+1) = 15$

Hence, there are $ 15$ positive values of $ x-12$, and thus $ 15$ values of $ x$. Since $ x > 12 \implies y > 0$, there is no contradiction, and there are $ \boxed{15}$ ordered pairs $ (x,y)$ in the solution set.[/hide]
[hide="Answer"]
There are $ \boxed{15}$ ordered pairs $ (x,y), \; x,y \in \mathbb{N}$. In particular, they are:

$ (13,156), \;(14,84), \;(15,60), \;(16,48), \;(18,36)$
$ (20,30), \;(21,28), \;(24,24), \;(28,21), \;(30,20)$
$ (36,18), \;(48,16), \;(60,15), \;(84,14), \;(156,13)$
[/hide]

[hide]
$ xy=12x+12y$
$ (x-12)(y-12)=144$
x-12 can be any factor of 144 and generate a unique y.
144 has 15 factors.
$ \boxed{15}$[/hide]

Where did it say that they had to be positive integers?

\[ \boxed{30}\]

Ubemaya I think is closest here (why did we assume only positive integer solutions in previous posts?). But there's one extraneous solution in his set. It should be 29.
[hide]If you go from Diophantient's equation, what's wrong with $ (-12)(-12) = 144$? Don't forget we got to that equation by multiplying by $ xy$.[/hide]

Oops... I just multiplied by 2 and didn't bother to check for erroneous solutions :D .

[quote=javierlc2000]This problem is from de PROMYS application problem set... Please don't post more solutions[/quote]

Mouahhhhhh ! It has been posted ten years ago !
What is this "PROMYS" which requires a silencious period of more than ten years  :rotfl:  ?

[quote=algebra_star1234]this is a 10 year old thread....[/quote]

(I think the point is that before, people would have to look through the forums to find this post, and thus have the solution to the PROMYS problem. Now, they can stumble upon it by chance, which could seriously discredit their solution if it's similar to any solutions on this post. And PROMYS knows about it, which they might find out from someone.)

Then again, if you've seen a relatively classic problem before, I don't see a problem with putting a classic solution, although the PROMYS people may want something more original.


Verify that the equations X^2 - 34*y^2 = ±33 has integer solutions, use a naive method

Deduce, from this,  a rational solution to X^2 - 34*y^2 = -1.

..................................

Clearly naive:
$1^2-34 \cdot 1^2 =-33$
$13^2-34 \cdot 2^2 =33$

Now multiply these to get $81^2- 34 \cdot 15^2 = -33^2$, and dividing through $33^2$ will give a rational solution.


$ u_n\equal{}Re(1\plus{}i)(2\plus{}i)^n$, therefore $ lim \equal{}0$.

[quote="negative"]Consider the sequence $ (u_n)(n \equal{} 1,2,...)$ given by the following recursive formula
$ u_1 \equal{} u_2 \equal{} 1;u_{n \plus{} 1} \equal{} 4u_n \minus{} 5u_{n \minus{} 1}$ for all $ n \geq 2$
Prove that $ lim_{n \to \plus{} infty}(\frac {u_n}{a^n})$ for all real number $ a > \sqrt {5}$[/quote]
Hey,Rust dont show it more detail,here is a problem of Mathematics and Youth Magazine (377)

hello, prove that $ u(n)\equal{}\frac{3}{5}e^{\frac{n\ln(5)}{2}}\cos(n\arctan(\frac{1}{5}))\minus{}\frac{1}{5}e^{\frac{n\ln(5)}{2}}\sin(n\arctan(\frac{1}{2}))$.
Sonnhard.


Let $ a, b$ and $ c$ be the sides, and $ r$ in-radius, of an triangle $ ABC$. Let $ a , b, c \ge 2$. Prove or disprove that

$ \frac{(a^3 \plus{} c)(b^3 \plus{} a)(c^3 \plus{} b)}{a \plus{} b\plus{} c}\ge 500r^2$ .

________
 :oops:

[quote="Ligouras"]Let $ a, b$ and $ c$ be the sides, and $ r$ in-radius, of an triangle $ ABC$. Let $ a , b, c \ge 2$. Prove or disprove that

$ \frac {(a^3 \plus{} c)(b^3 \plus{} a)(c^3 \plus{} b)}{a \plus{} b \plus{} c}\ge 500r^2$ .

________
 :oops:[/quote]

This inequality equivolent to
$ (a^3 \plus{} c)(b^3 \plus{} a)(c^3 \plus{} b) \geq 1000rS$
$ <\equal{}>$
$ \frac{abc}{4S}(a^3 \plus{} c)(b^3 \plus{} a)(c^3 \plus{} b) \geq 2r \cdot 125abc$
$ <\equal{}>$
$ R(a^3 \plus{} c)(b^3 \plus{} a)(c^3 \plus{} b) \geq 2r \cdot 125abc$
$ <\equal{}>$
let $ a\equal{}2x$, $ b\equal{}2y$, $ c\equal{}2z$ then
$ R(8x^3 \plus{} 2z)(8y^3 \plus{} 2x)(8z^3 \plus{} 2y) \geq 2r \cdot 1000xyz$ Let's prove this one and our solution will finish:
By AM-GM we have:
$ (8x^3 \plus{} 2z)(8y^3 \plus{} 2x)(8z^3 \plus{} 2y) \geq 1000(xyz)^{\frac{26}{10}} \geq 1000xyz$
and we have $ R \geq 2r$
So by multiplying this to inequality we will have desired result.


Prove that if $a,b,c\geq0$, then
\[|(a-b)(b-c)|+|(b-c)(c-a)|+|(c-a)(a-b)|\geq3(ab+bc+ca)-(a+b+c)(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}).\]

I understand the last step but I do not know how you started with?

[quote="kunny"]$(a+b+c)^{2}-6(ab+bc+ca)+(a+b+c)(\sqrt{ab}+\sqrt{bc}+\sqrt{ca})$
$\geq 6(a\sqrt{bc}+b\sqrt{ca}+c\sqrt{ab})\geq 0$
$\Longleftrightarrow 3(ab+bc+ca)-(a+b+c)(\sqrt{ab}+\sqrt{bc}+\sqrt{ca})\leq |a^{2}+b^{2}+c^{2}-(ab+bc+ca)|\leq |(a-b)(b-c)|+|(b-c)(c-a)|+|(c-a)(a-b)|.$[/quote]
I think there may something wrong in Kunny's proof.
But it is also a good approach:
First,we have:
 \[|a^{2}+b^{2}+c^{2}-(ab+bc+ca)|\leq |(a-b)(b-c)|+|(b-c)(c-a)|+|(c-a)(a-b)|\]
Then it is suffcient to prove:
\[a^{2}+b^{2}+c^{2}+(a+b+c)(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}) \ge 4(ab+bc+ca)\]
Substitute $a=x^{2},b=y^{2},c=z^{2}$.
Then it is equivalent to:
\[x^{4}+y^{4}+z^{4}+(x^{2}+y^{2}+z^{2})(xy+yz+zx) \ge 4(x^{2}y^{2}+y^{2}z^{2}+z^{2}x^{2})\]

but use Schur's ineuality we have:
\[x^{4}+y^{4}+z^{4}+x^{2}yz+y^{2}zx+z^{2}xy \ge \sum_{sym}x^{3}y \ge 2(x^{2}y^{2}+y^{2}z^{2}+z^{2}x^{2})\]
 :)


calculus
integration
function
trigonometry
calculus computations

Are you sure???
I don't thinks such an elementary function exist.
The definite integral $\int_{0}^{\pi/4} x \tan x \, dx$ in
Still unnsolved.

I thinks the definite $\int_{0}^{\frac{\pi}{2}} x \, \tan x \, dx$ is posible to find

I don't know, that is why i post this here.
Agree with you :)

Quick note:  $\int_{0}^{\frac{\pi}{2}} x \, \tan x \, dx=\infty$

The integrand is comparable to $\frac{\frac{\pi}2}{\frac{\pi}2-x}$ as $x\to\frac{\pi}2$  and we can show the divergence of the integral by comparison.


trigonometry
inequalities proposed
inequalities

Let $ \{\alpha,\beta,\gamma\}\subset\left(\minus{}\frac{\pi}{2},\frac{\pi}{2}\right)$ such that $ \tan\alpha\plus{}\tan\beta\plus{}\tan\gamma\equal{}0.$ Prove that:
$ \frac{1}{3\minus{}\tan{\frac{\alpha}{2}}}\plus{}\frac{1}{3\minus{}\tan{\frac{\beta}{2}}}\plus{}\frac{1}{3\minus{}\tan{\frac{\gamma}{2}}}\geq1.$

Let $ x\equal{}\tan \frac{\alpha }{2},\ y\equal{}\tan \frac{\beta }{2},\ z\equal{}\tan \frac{\gamma }{2}$ with $ |x|<1,\ |y|<1,\ |z|<1,$

We are to prove that $ 3(x\plus{}y\plus{}z)\plus{}xyz\geq 2(xy\plus{}yz\plus{}zx)$ under the condition of $ x\plus{}y\plus{}z\plus{}xyz(xy\plus{}yz\plus{}zx)\equal{}x(y^2\plus{}z^2)\plus{}y(z^2\plus{}x^2)\plus{}z(x^2\plus{}y^2).$


Let $ U_1, U_2, \ldots, U_n$ be $ n$ iid random variables uniformly distributed on $ [0,1]$. Rearrange them in increasing order:
$ U_{i_1} \leq U_{i_2} \leq \ldots \leq U_{i_n}$
and define $ M_n \equal{} \max_{1 \leq j \leq n\minus{}1} U_{i_{j\plus{}1}} \minus{} U_{i_j}$
Can one compute the law of $ M_n$ ?

I would try to prove that the random variables $ T_k \equal{} \frac {U_{(k \plus{} 1)} \minus{} U_{(k)}}{1 \minus{} U_{(k)}},\ k \equal{} 0\dots n \minus{} 1$ are independent, then the solution follows easily. But the first part is a bit technical, although intuitively clear.

i don't know if it helps but in Exercise 22 and 23 in Feller's book "an introdution to probability theory and its applications Vol II" (page 42) we have the following formula :

$ P(L_1>x_1,...,L_{n\plus{}1}>x_{n\plus{}1})\equal{}max(t^{\minus{}n}(t\minus{}x_1\minus{}...\minus{}x_{n\plus{}1})^n,0)$
Where $ L_k\equal{}U_{i_{k\plus{}1}}\minus{}U_{i_k}$ and $ U_{i_{n\plus{}1}}\equal{}t$ and $ U_j$ is uniformly distributed in $ [0,t]$

taking $ t\equal{}1$ we have
$ P(L_1>x_1,...,L_{n\plus{}1}>x_{n\plus{}1})\equal{}max((1\minus{}x_1\minus{}...\minus{}x_{n\plus{}1})^n,0)$

Apparently the result is known as the Withworth's Formula ( refernce Kendall and Moran (1963) "Geometric Probability" Griffin, London)
and it is the following :

$ P(M_n > x) \equal{} \sum_{k\geq 1;kx < 1}(\minus{}1 )^{k \plus{} 1}(1 \minus{} kx)^{n \minus{} 1}C^n_k$


This is an easy CML question:

Richie has 6 large marbles and 14 small marbles. One large marble weighs the same as 3 small marbles. If the weight of all 20 marbles is 80 ounces, then the weight of 2 large marbles plus 4 small marbles is _____________ ounces.

This is equivalent to:

$ l \equal{} 3s$ (1)
$ 6l \plus{} 14s \equal{} 80\implies3l \plus{} 7s \equal{} 40$ (2)

Find $ 2l \plus{} 4s$.

From here, there are two ways to proceed (that come to mind):

[hide="1"]
This is the "standard way": solve for the two variables, and then plug-and-chug.

So, substituting (1) into (2), we get $ 3(3s) \plus{} 7s \equal{} 16s \equal{} 40$, so $ s \equal{} \frac52$.

Plugging this into (1), we get $ l \equal{} 3\cdot\frac52 \equal{} \frac {15}2$, and

$ 2l \plus{} 4s \equal{} 2\cdot\frac {15}2 \plus{} 4\cdot\frac52 \equal{} 15 \plus{} 10 \equal{} \boxed{25}$.
[/hide]

[hide="2"]
Multiply equation (2) through by $ 5$ to get $ 15l \plus{} 35s \equal{} 200$.

Now, add the equation $ l \minus{} 3s \equal{} 0$ to the left side of this to get

$ 16l \plus{} 32s \equal{} 200$, and divide both sides by $ 8$ to get

$ 2l \plus{} 4s \equal{} \boxed{25}$.
[/hide]

Edit: Sort of beaten, but you can still refer to number two for a different way.

[quote="mewto55555"]Alternatively, this is the same as 6x3+14=32 small marbles, so 1 small weighs 5/2 so one large weighs 15/2 so 15+5+5=25[/quote]

That's the same solution I posted.


Why are there spaces around your nick in topic author field (viewforum)? This gives weird effects, as in the attachement.

Oh boy, I have reached my maximum upload quota of 0 bytes, you'll have to fix another bug first. ;)

Nope,

Sorry, you have reached your maximum Upload Quota Limit of 0 MB

still.

The attachment rules says 0Bytes too, so here is the link

[url]http://www.mathlinks.ro/Forum/attach_rules.php?f=149[/url]

probably the attachment feature is not back yet...

i just edited a post by someone else and the attachment disappeared. if you edit it, however, you see it's there, it just doesn't show up when viewing the thread. probably it's related to the same...

  darij

It seems normal to me  :? Is the problem still valid now?

Two oddities I just noticed. First the emotions are super small, and the topic review window seems to be showing the whole topic window, not just the posts :huh:. There's no real link, just click "post reply".

yep, it says: Sorry, you have reached your maximum Upload Quota Limit of 0 MB.

Topic review is fixed. I really don't understand that attachments part though :(

I've just created a test user and it doesn't show this behaviour  :|

[quote="Valentin Vornicu"]I really don't understand that attachments part though :([/quote]

apparently it's fixed now (the post i mentioned above shows up with the graphic now), thanks.

  darij

[quote="Peter VDD"][quote] Sorry, you have reached your maximum Upload Quota Limit of 0 MB[/quote]

Maybe refresh the cache? :? This is kind of weird.[/quote]So it's still happening for you?

[quote]Sorry, you have reached your maximum Upload Quota Limit of 0 MB[/quote]Most certainly.

Some people seem to be able to attach files now. mine still says... you-know-what

[quote="Inspired By Nature"]The topics on the main forum (getting started, etc) keep collapsing randomly everytime a come back to that Page. Is this normal?[/quote]They do not collapse randomly (obviously). They are guided by cookies (and if not cookies, by your profile settings).

I strongly suggest you enable cookies for browsing this site.

Peter: there's no caching done for the Attach Quotas. I really have absolutely no idea why only a couple of you have this problem ....  :?


Hi,

I've searched Google for this but nowadays, Google is almost dangerous as viruses itself with all those fake information out there. Anyway, my computer is now being attacked by Spyware called ErrorSafe. The problem is though, even when I run full scan by Norton program, I can't seem to delete this virus. It says that it's in use! I have no idea why it says that I really want to delete this virus because it's plain annoying.

Help?

P.S. If I knew how to track IP Address, I want to know who is the stupid idiot sending me this crap.  :mad:

First of all, don't use Norton.  Norton is a piece of s---.  Use BitDefender instead.  Also download spybot: search and destroy.  That is a free anti-spyware program that is very effective and usually gets all the spyware off (and it can find many times the number of spyware that Norton finds).  If that doen't work, I don't know what you should do.  A couple months ago I had a bunch of viruses on my computer and I couldn't find a way to get them off, so I had to do a system recovery which made me have to reinstall a ton of programs including s.p. 2.

This should keep it dormant for a while if the above don't work:

1. go to run and type 'regedit' without the apostrophes
2. then click on search (its somewhere) and type ErrorSafe.
3. delete everything with errorsafe in it as long as it isn't system files

I ran Search and Destroy (Spybot) and it didn't come up.

Here's a site that has helped me a lot with viruses/spyware:

http://www.siteadvisor.com/

It's not an anti-virus program. Instead, before going to website X, you can enter the address of website X into siteadvisor.com, and they'll give you a report about X, giving you information about any spyware, malware, annoying spam, etc.

I think that the technology was developed by some MIT students, but they sold it to McAfee.

I know that it doesn't help at the moment, but it might help you out in the future  :wink:

Thank you so much for the spybot tip... my internet is running like 10x faster w/ all the bad stuffs removed!!!
 :coolspeak:

Even Lavasoft's Ad-aware SE does a good job. ( Don't use it alone though, use it along with Spybot....)

And as far as an anti-virus goes, I would recommend AVG Free... that thing does some really good work.

Try using system restore.

It's Start>All Programs>Accessories>System Tools>System Restore.

"Restore your computer to an earlier date"...

cool


Do you know the formula for this:
$1^8+2^8+...+n^8$

you consider a polynome $p$ with $degp = 9$ and $p(n+1)-p(n) = n^8$
and deduce $1^8+2^8+...+n^8$  ;)

According to [url=http://mathworld.wolfram.com/PowerSum.html]Mathworld[/url], the formula is

$\sum_{k=1}^{n} k^8=\frac{1}{90}(10n^9+45n^8+60n^7-42n^5+20n^3-3n)$


linear algebra
matrix
linear algebra unsolved

need a simple proof that for a positive definite matrix:
All diagonal entries are real and positive.

Consider $ \langle Ae_i,e_i\rangle$ (or $ e_i^TAe_i$ in different notation), where $ e_i$ is a standard basis vector.

in the same spirit, show that if $ S$ is positive definite then
$ \det(S) \leq \Pi_{i\equal{}1}^n S_{i,i}$

Replying to alekk:

Let $ B\equal{}\text{diag}(\sqrt{S_{1,1}},\dots,\sqrt{S_{n,n}}).$ Let $ U\equal{}B^{\minus{}1}SB^{\minus{}1}.$

Then $ U$ is also positive definite, $ \det(S)\equal{}\det(U)\cdot\prod_{i\equal{}1}^nS_{i.i},$ and $ U_{i,i}\equal{}1$ for each $ i.$

Let $ \lambda_1,\dots,\lambda_n$ be the eigenvalues of $ U.$ Note that $ \sum_{j\equal{}1}^n\lambda_j\equal{}\text{trace}(U)\equal{}n$ and that each $ \lambda_j>0.$

By AM-GM, $ \det(U)\equal{}\prod_{j\equal{}1}^n\lambda_j\le
\left(\frac1n\sum_{j\equal{}1}^n\lambda_j\right)^n\equal{}1^n\equal{}1.$

Since $ \det(U)\le1,$ it follows that $ \det(S)\le\prod_{i\equal{}1}^nS_{i,i}.$


1  $ \alpha ,\beta ,\gamma \in \ (0,\frac {\pi}{2}),$
proof that: $ \frac {sin^{2}\alpha}{1 \plus{} sin(\beta)sin(\gamma)} \plus{} \frac {sin^{3}\beta}{1 \plus{} sin(\gamma)sin(\alpha)} \plus{} \frac {sin^{5}\gamma}{1 \plus{} sin(\alpha)sin(\beta)} < 2$

2    define $ f: N^{*} \rightarrow N^{*}$     $ \forall m,n \in N^{*}$
$ f(f(m) \plus{} f(n)) \equal{} m \plus{} n ,$     find the all possible value of  $ f(2007)$

THX!

if f(1)=0 u can easily see that it's impossible.(note that if f(m)=f(n) then n=m)
if f(1)=c>1 we have f(f(1)+f(c-1))=c=f(1) => 1=f(1)+f(c-1) contradiction;
if f(1)=1 then f(n)=n.


Greetings,

I am in my 50's and I received a bachelor's degree with a major in math in 1971.

I am thinking that when I retire, I would like to go back to school and get another bachelor's degree, this time in statistics.  I am not just interested in the statistics major - there are lots of other subjects I never studied the first time around.

I certainly have lots of math background, but it is not current and much of it has been forgotten.

I have a number of questions:

1.  I would be neither a transfer student nor someone right out of high school.  Do you think a state university would want me as a student?

2.  Would I need references and SATs?

3.  What do you think I would have  to do with math courses?  I can not currently do a lot of calculus problems, for example, but if I had to retake calculus I think I would have an unfair advantage over people taking it for the first time.

4.  Any other advice?

I think that at least no. 3 is not really a problem here. It's not like you are competing with other students in your Calculus class like in an Olympiad, so you knowing from ahead some stuff cannot hurt at all, in fact it can be very good for the class.

A state university would definitely want you as a student, I think. I considered getting a second undergraduate degree (not enough) when I got my professional school degree, and I considered it again a while ago. There are definitely colleges that are open to applicants pursuing second undergraduate degrees.

But one thought I have is that you might be able to get into a master's program in statistics with the background you already have. Let's see what other people responding to this thread have to say.

To clarify, I am looking for non-statistics liberal arts courses too.  That's why I am considering an entire bachelor's degree, not just a few statistics courses.

As an example, as an undergraduate decades ago with my head in mathematics, I took the minimum required English courses.  Since then I have written a few fictional short stories, and not surprisingly I realize I know nothing about creative writing. So, I am looking forward to taking lots of non-math and statistics electives.

I am curious about the letters of recommendation and SAT stuff.  Do I need to take the SAT?  And would letters from employers be helpful?

I think it probably best if you emailed some admission officers down at the universities you're thinking of applying to. They should be able to provide you with much more relevant answers than any suppositions we might make here.

Of course it would be great if you'ld tell us afterwards what they said, for future reference :)

I agree with Valentin -- You should definitely talk to admissions officers, who will also be able to tell you what kind of information they'll want about you.  I imagine you're something of a special case.

you might be able to take the classes you are interested without necessarily having to satisfy the requirements of a degree, which would provide more flexibility and may or may not still provide you with what you're looking for.

Hello,

[quote="kentliu"]why would you go back to school? i cant wait to get out.. its stressful.. also, i highly doubt u'll need to take SAT.. it would be very weird to go into a testing room with all these juniors and a 50 year old.. lol[/quote]

Sometimes people just love to learn. You think that the school is stressful, well, I would consider that you prepare yourself psicologically for the real world, for the real jobs, because almost everywhere job is HIGHLY more stressful than school, I tell you that by own experience (well, may be being a TA or RA while you are in Grad School would be an exception)

Eitherway, I agree with tokenadult, you should may be consider a master program instead of a bachelor. In the master program you probably will have the oportunity to focus in what you care the most with more emphasis than in a Bachelor program.

You may want to consider a Doctorate Program, but it is a little more painful. Some schools have age limitations, but I believe most don't. I met a guy in his early 60s who made it to the Ph.D program in Math in Standford not long ago (I believe he was admited by the 2000 or 2001 fall, not sure) and the guy was from engineering.

Look around, think carefully, and decide what you want to do. Ahh! for Master Programs probably you wouldn't need SAT, but you'll need GRE.

Best regards,

you dont have to go to school to learn.. i love to learn too, i just dont like the system.. and i agree with that jobs are more stressful.. but he's retiring, why not just relax and enjoy life..

Mabye to joeorez, relaxing and enjoying life includes going to college and learning.

When I was in college I was afraid of taking a course that I didn't think I could get a good grade in, especially outside of math.

Once I am retired, that won't be important.  Neither will I stress over getting an assignment in on time.  What are they going to do, put these things on my permanent record, affecting my future employment?

I realize college is stressful, but work is much more stressful.  I think college the second time around would be great fun.  There are certainly a lot of subjects that I never studied the first time that seem more interesting now.

Plus, computers, food, movies, and the gym will be much cheaper than I could pay elsewhere.  Can I get free beer at fraternity parties?

Many state universities have open admissions policies, or have minimal entrance requirements (23 on the ACT for example).  However, if you already have a bachelor's degree, it's less likely that you'll be allowed to enroll for a second one in a very related field.

You'll certainly be able to enroll as a nondegree student and take whatever classes you choose.  If you don't care about the certification which comes with a degree, that's a good option... you won't have to take any courses you don't want to, and can take as many courses at a time as you wish, etc.

Your presence in the classroom will be appreciated by the traditionally-aged undergrads (though perhaps not immediately) as well as your instructors.  On the other hand, I doubt you'll be welcome at frat parties---the traditionally-aged undergrads will not want to socialize with you---and I've not heard of free beer at a state school, just at private institutions.


When I tried to use \boxed on the Texer, it gives me error and not work.

\fbox works but I like \boxed better since \boxed let me do:

$\boxed {\frac{a+b}{2}}$ while \fbox gives $\boxed{\frac{a+b}{2}}$ with nothing of the box.

Only reason I want to use \boxed or \fbox is to point out the final answer on my contest but unfortunately, \boxed doesn't work.

Can anyone tell me alternate code or tell me why it's not working?

Thanks.

in texnic center, the command \boxed work well.  What are error messages?

the other way to do it is, according to me, not pretty but anyway...
\begin{tabular}{|c|}\hline
$\frac{2}{a+b}$
\\ \hline
\end{tabular}

ledurt

Hi,
The \boxed command automatically puts you in Math Mode. So you should use \boxed for Math Mode and \fbox for Text Mode. I have never used nor seen the \fboxed command you have used in your post. I get an error when using it in MikTeX/WinEdt.
SSN.


modular arithmetic
algebra
number theory
Divisibility

Let $p > 3$ be a prime. Prove that if \[ \sum_{i=1 }^{p-1}{1\over i^p} = {n\over m},  \] with $\gdc(n,m) = 1$, then $p^3$ divides $n$.

Sorry, but posted at least two times before.

I would like to see the links, because i suspected it was well known

Hint: group $1/a^p+1/(p-a)^p$, use the binomial theorem and rule out the $p^2$ factor, we are left with a sum mod $p$, and we must show it is $0$ mod $p$, this sum is related to the sum of squares mod $p$ which is $0$.

Well, sometimes I think it's easier to put up a solution than searching...

I suspected that it was well known too. But I solved it in a slightly different way.

I used the extended binomial expansion, that is,
\[ (a + b)^\alpha = \sum_{k=0}^\infty{\alpha \choose k}a^{\alpha-k}b^k \]
for $\alpha$ real.

We define ${\alpha \choose k}$ for $\alpha$ real and $k$ nonnegative integer as ${\alpha \choose 0} = 1$ and
\[ {\alpha\choose k} = \dfrac{\alpha(\alpha-1)(\alpha-2)\ldots (\alpha - k + 1)}{k!} \]
for $k > 0$.

So
\[ a^{-p} + (p-a)^{-p} = a^{-p} + \sum_{k=0}^\infty {-p\choose k}(-a)^{-p-k}p^k \equiv a^{-p} - a^{-p} + {-p\choose 1}a^{-p-1}p - {-p\choose 2}a^{-p-2}p^2 \pmod{p^3} \]

It's not hard to see that $p$ divides ${-p\choose 2}$, so
\[ a^{-p} + (p-a)^{-p} \equiv -a^{-p-1}p^2\pmod{p^3} \]

OK, it looks like "cheating" but it's not. You can check by hand that, indeed,
\[ (p-a)^p \cdot (-a^{-p}-a^{-p-1}p^2) \equiv 1 \pmod{p^3} \]

In fact, using the "regular" binomial expansion,
\[ (p-a)^p \equiv -a^p + {p\choose 1}a^{p-1}p \pmod p \]
and
\[ (p-a)^p \cdot (-a^{-p}-a^{-p-1}p^2) \equiv (-a^p + a^{p-1}p^2)\cdot (-a^{-p}-a^{-p-1}p^2) \equiv 1 + a^{-1}p^2 -a^{-1}p^2\equiv 1 \pmod{p^3} \]

Thus we only need to prove that
\[ \sum_{k=1}^{p-1}a^{-p-1}\equiv 0\pmod p \]
or, recalling that $a^{-p-1} \equiv (a^{-1})^{p+1} \equiv (a^{-1})^2\pmod p$,
\[ \sum_{k=1}^{p-1}(a^{-1})^2\equiv 0\pmod p \]

This is a standard problem in number theory. Let $S = \sum_{k=1}^{p-1}(a^{-1})^2\bmod p$. Thus, since $\mathbb{Z}/p\mathbb{Z}^* = \{1,2,\ldots, p-1\}$ is a reduced residue system, $2^{-2}\mathbb{Z}/p\mathbb{Z}^* = \mathbb{Z}/p\mathbb{Z}^*$ and
\[ 2^{-2}S \equiv S\pmod p\iff S \equiv 0 \pmod p, \]
since $p \neq 3$ and we are done.

[quote="cyshine"]Let $p > 3$ be a prime. Prove that if
\[ \sum_{i=1}^{p-1}{1\over i^p} = {n\over m},  \]
with $\gdc(n,m) = 1$, then $p^3$ divides $n$.[/quote]

Raise Wolstenholme's congruence (mod $p^2$) to the $p$'th power.

Other solution:

Let $P(x)=(x-1^p)(x-2^p)...(x-(p-1)^p)=x^{p-1}+x^{p-2}A_1...+xA_{p-2}+\prod_{i=1}^{p-1}i^p$

By a standard result and since $a^p=a$ modp, we have that $A_1,A_2...A_{p-2}$ are multiples of $p$.

Now put take $P(p^2)-\prod_{i=1}^{p-1}i^p$, we will show that $p^5$ divides it.

Iff $p^5$ divides it, then $p^5$ divides the sum $p^{2(p-1)}+p^{2(p-2)}A_1...+p^2A_{p-2}$ which implies that $p^3$ divides $A_{p-2}$ (this implies of course what we wanted)

Now we have that $(p^2-a^p)(p^2-(p-a)^p)=p^4-p^2(a^p+(p-a)^p)+a^p(p-a)^p$

But $p^5$ divides $p^4-p^2(a^p+(p-a)^p)$ by a simple binomial expansion of $(p-a)^p$.

So we get that $(p^2-a^p)(p^2-(p-a)^p)=p^4-p^2(a^p+(p-a)^p)+a^p(p-a)^p=a^p(p-a)^p$(mod$p^5$)

So we get that $P(p^2)-\prod_{i=1}^{p-1}i^p$ is a multiple of $p^5$ like we wanted.

The previus post inmediately generalizes to further results.

[quote="Pascual2005"]why can you raise it? i mean, i know exponentation rule...but this?[/quote]

I think you are right, that it is not so simple.   I had in mind that raising to the power $p$ tends to improve congruences mod $p^k$ to congruences modulo $p^{k+1}$, but did not check carefully the details.   If it does work I will repost with the details.

This is equivalent to wishing to prove that \[2\sum_{i=1}^{p-1}\frac{1}{i^p}\equiv 0\pmod{p^3}\] treating each of the numbers as inverses and using $\gcd(p,2)=1$.

We now notice $\frac{1}{j^p}+\frac{1}{(p-j)^p}=\frac{(p-j)^p+j^p}{j^p(p-j)^p}$.  Therefore \[2\sum_{i=1}^{p-1}\frac{1}{i^p}=\sum_{i=1}^{p-1}\frac{(p-i)^p+i^p}{i^p(p-i)^p}\]

By lifting the exponent $v_p\left( (p-i)^p+i^p\right)=v_p\left( p-i+i\right)+v_p(p)=2$.  Therefore we can factor a $p^2$ out of the numerator to give us \begin{align*} \sum_{i=1}^{p-1}\frac{(p-i)^p+i^p}{i^p(p-i)^p}=p^2\sum_{i=1}^{p-1}\frac{\frac{(p-i)^p+i^p}{p^2}}{i^p(p-i)^p}  \equiv 0\pmod{p^3}\\\iff  \sum_{i=1}^{p-1}\frac{\frac{(p-i)^p+i^p}{p^2}}{i^p(p-i)^p}\equiv 0\pmod{p}\end{align*}

Notice that $i\equiv -\left(p-i\right)\pmod{p}$ so hence we now need \[\sum_{i=1}^{p-1}\frac{\frac{(p-i)^p+i^p}{p^2}}{i^{2p}}\equiv 0\pmod{p}\] since we can take the negative out of the denominator.

We consider \[\sum_{i=1}^{p-1}\frac{\frac{(\frac{p+1}{2})^p+(\frac{p-1}{2})^p}{p^2}}{i^{2p}}+\sum_{i=1}^{p-1}\frac{\frac{i^p+(p-i)^p-(\frac{(p+1)}{2})^p-(\frac{(p-1)}{2})^p}{p^2}}{i^{2p}}\]

We desire to show each part is divisible by $p$.  For the first sum notice that we want $\sum_{i=1}^{p-1}\frac{1}{i^{2p}}\equiv \sum_{i=1}^{p-1}i^{2p}\pmod{p}$ since each element from $1$ to $p-1$ has a distinct inverse including $1^{-1}\equiv 1\pmod{p}$ and $(p-1)^{-1}\equiv (p-1)\pmod{p}$.

By Fermat's Little Theorem we have \[\sum_{i=1}^{p-1}i^{2p}\equiv \sum_{i=1}^{p-1}i^2\pmod{p}\equiv \frac{(p-1)(p)(2p-1)}{6}\equiv 0\pmod{p}\] since $p\neq 2,3$.

Now we desire to prove that $\frac{i^p+(p-i)^p-\left(\frac{(p+1)}{2}\right)^p-\left(\frac{p-1}{2}\right)^p}{p^2}\equiv 0\pmod{p}$.  Since $\gcd(p,2)=1$ we want \begin{align*} 2^pi^p+2^p(p-i)^p-(p+1)^p-(p-1)^p\equiv 0\pmod{p^3} \\ 2^p\left[ i^p+\binom{p}{1}p(-i)^{p-1}+(-i)^{p}\right]-\left[p\binom{p}{1}+1+p\binom{p}{1}(-1)^{p-1}+(-1)^{p}\right]\equiv 0\pmod{p^3} \\ 2^p\left[p^2\right]-2p^2\equiv0\pmod{p^3} \end{align*}

The second step follows from the fact that $p^2\binom{p}{2}\equiv 0\pmod{p^3}$ and every $i\ge 3$ we have $p^i\binom{p}{i}\equiv 0\pmod{p^3}$, and the third step follows from $p$ being odd.

Now, $p^2\left(2^p-2\right)\equiv 0\pmod{p^3}$ is true by Fermat's Little Theorem therefore we are done $\blacksquare$.

Hopefully this isn't a fakesolve :P
Someone please check. :)

[b]Solution. [/b] It's equivalent to show, similarily as above, that $$2\sum_{i=1}^{p-1} \frac{1}{i^p} \equiv 0 \pmod{p^3}$$ which is, by pairing
\begin{align*}
2\sum_{i=1}^{p-1} \frac{1}{i^p} =\sum_{i=1}^{p-1} \frac{i^p+(p-i)^p}{i^p(p-i)^p} &= \sum_{i=1}^{p-1} \frac{\left[\dbinom{p}{0}p^p - \dbinom{p}{1}p^{p-1}i + \ldots - \dbinom{p}{p-2}p^2i^{p-2}+ \dbinom{p}{p-1}pi^{p-1}\right]}{i^p(p-i)^p} \\
&\equiv \sum_{i=1}^{p-1}\frac{\dbinom{p}{p-1}pi^{p-1}}{i^p(p-i)^p} \pmod{p^3} \\
&= \sum_{i=1}^{p-1} \frac{p^2i^{p-1}}{i^p(p-i)^p}=p^2 \sum_{i=1}^{p-1}\frac{1}{i(p-i)^p}  \\
\end{align*} Therefore, it suffices to show
\begin{align*}
\sum_{i=1}^{p-1} \frac{1}{i(p-i)^p} \equiv 0 \pmod{p} &\iff \sum_{i=1}^{p-1} \frac{1}{-i^{p+1}} \equiv 0 \pmod{p} \\
&\iff \sum_{i=1}^{p-1} -\frac{1}{i^2} \equiv 0 \pmod{p} \iff \sum_{i=1}^{p-1} -i^2 \equiv 0 \pmod{p} \\
\end{align*}
but last statement clearly holds since $$-\frac{(p-1)(p)(2p-1)}{6} \equiv 0 \pmod{p}$$ which finishes the problem. $\blacksquare$

"Wolstenholme
take me home to the place I belong
West Virginia
mountain mamma
 take me home
Wolstenholme."

Shout out to John Denver  :love:

A huge motivation for solving this problem is Wolstenholme’s Theorem, which is stated as follows
------
[b]Theorem (Wolstenholme)[/b]: Let $p>3$ be prime.  Then if
$$\sum_{i=1}^{p-1}\frac{1}{i}=\frac{m}{n}$$
for $\gcd(m,n)=1$ we have $p^2~|~m$.
------
An alternative statement is simply that
$$\sum_{i=1}^{p-1}\frac{1}{i}\equiv 0\pmod{p^2}.$$
A proof of this theorem will provide some motivation on how we should approach this problem.  We have
\begin{align*}
2\sum_{i=1}^{p-1}\frac{1}{i}&=\sum_{i=1}^{p-1}\left(\frac{1}{i}+\frac{1}{p-i}\right)\\
&=\sum_{i=1}^{p-1}\frac{p}{i(p-i)}\\
&\equiv p\sum_{i=1}^{p-1}-\frac{1}{i^2}\\
&\equiv-p\sum_{i=1}^{p-1}i^2\\
&=-\frac{p^2(p-1)(2p-1)}{6}\\
&\equiv 0\pmod{p^2}
\end{align*}
which completes the proof.  So how does this motivate our solution? We notice that the problem statement is equivalent to proving that
$$2\sum_{i=1 }^{p-1}{1\over i^p}\equiv 0\pmod{p^3}$$
so we are basically proving Wolstenholme, but with a couple of more special conditions to think about.  Using an identical argument we see that
\begin{align*}
2\sum_{i=1 }^{p-1}{1\over i^p}=\sum_{i=1}^{p-1}\frac{(p-i)^p+i^p}{i^p(p-i)^p}\\
\end{align*}
and using the fact that
$$(p-i)^p=\sum_{k=0}^{p}\binom{p}{k}p^{p-i}(-i)^k$$
our sum simplifies to
\begin{align*}
\sum_{i=1}^{p-1} \frac{i^p+(p-i)^p}{i^p(p-i)^p} &\equiv \sum_{i=1}^{p-1}\frac{\dbinom{p}{p-1}pi^{p-1}}{i^p(p-i)^p}\\
&\equiv\sum_{i=1}^{p-1} \frac{p^2i^{p-1}}{i^p(p-i)^p}\\
&\equiv p^2 \sum_{i=1}^{p-1}\frac{1}{i(p-i)^p}\pmod{p^3}.
\end{align*}
Similar to how we did it in proving Wolstenholme, we have to prove that
$$\sum_{i=1}^{p-1}\frac{1}{i(p-i)^p}\equiv 0\pmod{p}$$
since that is easier to prove and generalizes to $p^3$.  Using the fact that $p-i\equiv -i\pmod{p}$ again and some smooth slight of hand we get
\begin{align*}
\sum_{i=1}^{p-1}\frac{1}{i(p-i)^p}&\equiv \sum_{i=1}^{p-1}-\frac{1}{i^{p+1}}\pmod{p}\\
&\equiv \sum_{i=1}^{p-1}-\frac{1}{i^{2}}\pmod{p}\\
&\equiv -\sum_{i=1}^{p-1}i^2\\
&\equiv 0\pmod{p}\\
\end{align*}
which finishes just like Wolstenholme.

Let $p > 3$ be a prime. Prove that if $$\sum_{i=1 }^{p-1}{1\over i^p} = {n\over m}, $$ with $ gcd\,\, (n,m) = 1$, then $p^3$ divides $n$.

[quote=parmenides51]Let $p > 3$ be a prime. Prove that if $$\sum_{i=1 }^{p-1}{1\over i^p} = {n\over m}, $$ with $ gdc\,\, (n,m) = 1$, then $p^3$ divides $n$.

[hide]posted for the latex, in order to fix a post collection, this hide shall be removed soon[/hide][/quote]

Hi!
I will like to add a minor correction. It's actually $\gcd (n,m)$ and not $gdc(n,m)$, the original post has a typing mistake there.

Let $S=\frac{n}{m}.$ We want to show that $2S = 0$ modulo $p^3$. To do this notice that
$$2S=\sum_{i=1}^{p-1}\frac{1}{i^p}+\frac{1}{(p-i)^p}=\sum_{i=1}^{p-1}\frac{i^p+(p-i)^p}{i^p(p-i)^p}.$$ See that, by Newton's Binomial, that $$i^p+(p-i)^p\equiv -\binom{p}{2}p^2i^{p-2}+\binom{p}{1}pi^{p-1} \mod p^3.$$ But as obviously $p\mid \binom{p}{2}$, we obtain that the expresion is just $p^2i^{p-1}$ modulo $p^3$. So now, showing that our sum is $0$ modulo $p^3$ is equivalent to showing that the following is $0$ modulo $p$ $$\sum_{i=1}^{p-1} \frac{i^{p-1}}{i^p(p-i)^p}\equiv \sum_{i=1}^{p-1} \frac{1}{-i^2}\equiv-\sum_{j=1}^{p-1}j^2\equiv -p \frac{(p+1)(2p+1)}{6},$$ which is $0$ modulo $p$ for $p>3$, which ends the problem.


AMC
AIME
USA(J)MO
USAMO
geometry
trigonometry
AMC 10

Do not discuss the Alternate AIME until after 8AM CST Wednesday.  We will delete any discussion of the test before that time, including comments about difficulty or speculation about scores.

[quote="Silverfalcon"]what's "Alternate AIME "?[/quote]

The alternate AIME is the second date for administering the [url=http://www.unl.edu/amc/e-exams/e7-aime/aime.html]American Invitational Mathematics Examination[/url], which happens tomorrow (Tuesday 6 April 2004). The AIME test is, as the name suggests, by invitation only. High-scorers on the AMC 10 or AMC 12 are invited to take each year's AIME, on either of two dates. People with a high combined "index" score on the AMC 10/12 and the AIME will be invited to take the [url=http://www.unl.edu/amc/e-exams/e8-usamo/usamo.html]United States of America Mathematical Olympiad[/url] (USAMO), which is a rare honor. High scorers on the USAMO will be invited to attend, expenses-paid, the [url=http://www.unl.edu/amc/a-activities/a6-mosp/mosp.html]Mathematical Olympiad Summer Program[/url] (MOP), an intensive summer course in mathematics. And some of the MOPpers will go on to represent the United States in the International Mathematical Olympiad.

Good luck to everyone taking the AIME on 6 April 2004. Please don't discuss the test until after the deadline mentioned in this thread.

Can someone post all the problems on the alternate AIME after
8AM CST Wednesday?

If you can wait an extra half day or so, you will be able to see all the problems in the Math Jam. It's not very convenient for someone to type out all the problems. I suppose if someone wants to do that, then it's ok (unless it breaks some AMC rule).

Can I ask that discussion of the AIME ALternate be held until at least 12:00 noon EDT on Wednesday, April 7.  We have some far-flung locations taking AIME Alternate at late times, and I ask that no answers in particualr, and no discussion in general take place before then.

I hope this won't interfere with the Math Jam.

Steve Dunbar
AMC Director

The Math Jam isn't until Wednesday evening so time constraints will not be a problem for our schedule.

click on the "Math Jams" link under "Community" on lefthand side of your screen...

tomorrow night's math jam will discuss today's AIME

Ok umm... I would take the AIME I.  The AIME II is generally harder, or at least to me that is.  Then again I'm not sure if I can even take the USAMO, it's at one in the morning for me!

[quote="probability1.01"]What is the difference between taking the AIME A and AIME B for qualifying for the USAMO?

Are scores compared with other takers of the contest or participants of both?

Also, about how well does one need to score to qualify for the USAMO given an AMC 12 score of 130?  Will 10 (resulting in a 230 index score) be sufficient?

Does it help to be in a lower grade?

Sorry about all the questions... your replies are much appreciated.[/quote]

For your first question, theoretically taking either test should not affect your score. (However that is often the case as one test may contain more problems in your stronger areas and/or one test is slightly harder than the other test.) The AIME Alt (aka. AIME B) is for students who have a schedule conflict on the original AIME date (i.e. spring break or something). It should not make a huge difference on your USAMO qualification.

All scores are compared to participants of both test iirc.

A USAMO index of 230 is more than sufficient. If you're in 9th or 10th grade, sometimes a 200 or slightly below will even be okay.

Yes, 10th graders and below are given a slight advantage in that there are some "reserved" spots in the USAMO for them.

I certainly hope they decide to somehow alter results slightly to compensate for this ridiculously difficult AIME B...

[quote="xirtamax"]Yes AIME Alt. was much harder than the first one.
IT didn't have many trig problems.....
It wasn't anything like AIME problems before...[/quote]

It was an unusual test...yes...

No trig whatsoever...

I guess the best way to say it is "simple and elegant"...

I thought the AIME B was actually about the same difficulty.  I just got really confused by the last problem, but the other problems I think I know how to solve.

When can we share/compare answers?

[quote]All scores are compared to participants of both test [/quote]

Nope, this never happens (according to Simon).

Yeah...you're right...nevermind...lol

I think they only are compared to people who took the same test as you.

[quote="MithsApprentice"][quote="MCrawford"]You can compare your answers with ours at this evening's Math Jam.[/quote]

Out of the good of your hearts, would either you or Mr. Rusczyk post the official AoPS AIME B Answers so that those of us who are waiting for scores won't go crazy at the anxiety?[/quote]

Nah, we're mean like that.

Actually, we got a very late start with this AIME and had a lot of other work in the meantime.  We are still writing up our solutions.  I will however say that we disagree with several answers from the lists that I've seen posted here so far.

[quote="**********"]wow, definitely bad news...[/quote]

not necessarily so...at least, not for you...

he only disagrees with "several" of your answers...at least u know you got almost all of them right...


The distance between any two of six given points in  the plane is at least 1. prove that the distance between some two points is at least ((5+5^1/2)/2)^1/2

I think this problem needs to be reformulated. What precisely do you mean? Six points on a line with distance 1 between them is a counterexample to this?

It is asked for 'at least' not for 'at most'...

Pierre.


Does anyone know any methods for partial fraction decomposition other than using the method involving systems of linear equations.

hello, i hope this will help you
http://www.math.ubc.ca/~feldman/m101/partdecomp.pdf
Sonnhard.


I know there's an option under the profile so that it organizes the topics by "Topic Status" but that's either the oldest unread ones or the moved ones depending on whether you choose ascending or descending. I was wondering if there's an option that allows you to organize your posts so that unread topics are kept at the top of the the page above the unread ones and then they're organized by date of last entry. If not, could that be an added option?

yea. that would be a nice option. i support.


The square table $ 10\times 10$ is divided in squares $ 1\times1$. In each square $ 1\times1$ is written one of the numers $ \{1,2,3,...,9,10\}$. Numbers from any two adjacent or diagonally adjacent squares are reciprocal prime. Prove, that there exists a number, which is written in this table at least 17 times.

I think I've seen this problem before,maybe it's from russia or something.

It was a bad translation.  The intended phrase was "relatively prime" or "coprime."

[quote=bravado]Suppose that each number is written at most $ 16$ times. Consider a $ 2 \times 2$ square - it can contain at most one even number, thus there're at most $ 25$ even numbers.
Moreover, a $ 2 \times 2$ square can have at most one number from the set $ \{3,9\}$. This means that the total number of $ 3$'s and $ 9$'s does not exceed $ 25$.
We're finally left with the numbers $ 1,5,7$ which,by our assumption, are each used at most $ 16$ times. In conclusion, we can use at most $ 25 \plus{} 25 \plus{} 3\cdot{16} \equal{} 98 < 100$ numbers, which is a contradiction. Thus there has to be a number that is used at least 17 times.[/quote]

Nice sol! I did the same thing.


:lol: What is your mobile ringtone ?
Do you often to change it ?
It is easy , just download mp3 to ringtone gold at http://www.oursdownload.com , I have using it for a long time , it works very good .
Why not to try now ?

:spam:  I don't even have a mobile phone.  :P


calculus limit function derivative articles

[i]Find the real number $ p$ so that $ \lim_{x \to \infty} x^p (\sqrt [3]{x \plus{} 1} \plus{} \sqrt [3]{x \minus{} 1} \minus{} 2 \sqrt [3]{x})$ converges to some non-zero real number.[/i]

Is this solution even legit, and if not, what is a legit way of solving it?

[hide="What I Did"]$ f(x) : \equal{} \sqrt [3]{x}$, then we have that:
$ f(x \plus{} 1) \plus{} f(x \minus{} 1) \minus{} 2f(x) \equal{} \frac {\left( \frac {f(x \plus{} 1) \minus{} f(x)}{1} \right) \minus{} \left( \frac {f(x) \minus{} f(x \minus{} 1)}{1} \right)}{1}$ $ \approx \frac {f' \left( x \plus{} \frac {1}{2} \right) \minus{} f' \left(x \minus{} \frac {1}{2} \right) }{1} \approx f''(x)$

Since f'' is on the order 1/3 - 2 = -5/3, then [b]p = 5/3[/b].[/hide]

The implicit lemma here is that the forward difference of a differentiable monotonic function has the same growth rate as its derivative.  See [url=http://en.wikipedia.org/wiki/Finite_difference#Relation_with_derivatives]the Wiki article on finite differences[/url].

You can use the $ a\minus{}b\equal{}\frac{a^3\minus{}b^3}{a^2\plus{}ab\plus{}b^2}$ with $ a\equal{}\frac{1}{\sqrt{x}}$ and $ b\equal{}\frac{1}{\sqrt{x\minus{}1}}$, etc.

[quote]
[hide="What I Did"]$ f(x) : \equal{} \sqrt [3]{x}$, then we have that:
$ f(x \plus{} 1) \plus{} f(x \minus{} 1) \minus{} 2f(x) \equal{} \frac {\left( \frac {f(x \plus{} 1) \minus{} f(x)}{1} \right) \minus{} \left( \frac {f(x) \minus{} f(x \minus{} 1)}{1} \right)}{1}$ $ \approx \frac {f' \left( x \plus{} \frac {1}{2} \right) \minus{} f' \left(x \minus{} \frac {1}{2} \right) }{1} \approx f''(x)$

Since f'' is on the order 1/3 - 2 = -5/3, then [b]p = 5/3[/b].[/hide][/quote]

I'm pretty sure that this can be rigorized using the mean value theorem--instead of x+1/2 and x-1/2, we have a and b, for some a btwn x and x+1, and some b btwn x and x-1. a-b varies between 0 and 2, and so long as it doesn't go to 0 as x goes to infinity, we get the correct answer. I'm not quite sure how to prove that it doesn't go to 0, but I haven't thought about it for too long.

Too bad that I didn't realize that 1+1-2=0...

Doing binomial expansion on $(x+1)^{\frac{1}{3}}+(x-1)^{\frac{-1}{3}} +2\sqrt[3]{x}$  it is clear that the highest power of $x$ that reamisn after simplification is $x^{\frac{-5}{3}}$ Hence ans is $\boxed{\frac{5}{3}}$


geometry
parallelogram
geometry proposed

The opposite sidelines of quadrilateral $ ABCD$ intersect at points $ P$ and $ Q$. Two lines passing through these points meet the side of $ ABCD$ in four points which are the vertices of a parallelogram. Prove that the center of this parallelogram lies on the line passing through the midpoints of diagonals of $ ABCD$.

dear April  is it true ? or where I misunderstood?

Dear [b]planegeometry[/b], I attach here the original problem's statement and its solution, could you please check it for me? Thanks a lot!

sorry.dear [b]April[/b]
the right figure is like this: :blush:

[b][size=100][color=DarkBlue]LEMMA. – A quadrilateral $ ABCD$ is given and le $ E,\ F$ be, two arbitrary points on $ AC,\ BD$ respectively, such that $ EF\parallel CD$ $ ($ suppose the point $ E,$ between the points $ A,\ Y,$ where $ Y\equiv AC\cap BD$ $ ).$ Prove that $ PQ\parallel EF\parallel CD,$ where $ P\equiv AB\cap DE$ and $ Q\equiv CD\cap AF.$[/color][/size][/b]

[b][size=100]PROOF.[/size][/b] – We consider the triads of the points $ A,\ P,\ D$ and $ B,\ Q,\ C,$ on the line segments $ AD,\ BC$ respectively and applying the [b][size=100]Pappos theorem[/size][/b], we have that the points $ X\equiv CP\cap DQ$ and $ Y\equiv AC\cap BD$ and $ Z\equiv AQ\cap BP,$ are collinear.

From this collinearity, based on the [b][size=100]Desarques theorem[/size][/b], we have that the triangles $ \bigtriangleup CEP$ and $ \bigtriangleup DFQ$ are perspective and so, we conclude that the line segments $ CD,\ EF,\ PQ$ are concurrent at infinity, because of $ EF\parallel CD.$

Hence, we conlude that $ PQ\parallel EF\parallel CD$ and the proof of the [b][size=100]Lemma[/size][/b] is completed.

$ \bullet$ Return now in the configuration of the proposed problem, let $ EFKL$ be, the parallelogram as it has been stated in its formulation, with $ EF\parallel KL$ and $ EL\parallel FK.$

We consider the quadrilateral $ PQLK$ and the points $ E,\ F$ on $ QK,\ PL$ pespectively, of which $ EF\parallel KL.$ So, based on the above [b][size=100]Lemma[/size][/b], we have that $ EF\parallel BD\parallel KL$ and similarly, $ EL\parallel AC\parallel FK.$

Applying now, the problem in the topic [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=134116]Locus of center parallelogram[/url] of [b][size=100]N.T.TUAN[/size][/b], we conclude that the point $ O\equiv EK\cap FL,$ lies on the line segment $ MN,$ where $ M,\ N$ are the midpoints of the segments $ BD,\ AC$ respectively and the proof of the proposed problem, is completed.

Kostas Vittas.


Find the minimum $ k$ that it is possible to split a cube into $ n$ non-overlapping cubes for every $ n\ge k$

[quote=bballlegend]11 year bump, nice[/quote]

checking my old posts


Which weighs more, a pound of feathers or a pound of gold?

[hide="Hint"]The title gives a clue, sort of.[/hide]

[hide="the right answer"]
A pound of feathers, because a pound of gold is in troy weight, with a pound only 12 ounces. :P
[/hide]

[hide="Actually"]While a troy pound is lighter than a normal pound, a troy ounce is, I think, heavier than a normal ounce, so it's more than just 12 ounces. ;)[/hide]

But, you did get the right answer. :P

Oh. But the drams are the same, aren't they? Or is it the grains that are the same?

An avoirdupois ounce is about 16 drams, right? (Correct me if I'm wrong) How much is a troy ounce?

I was going to say a pound of feathers, but for the reason that feathers have more trapped air pockets and thus weighs more  :huh:

I didn't realize the troy weights part though...

[hide="the CORRECT answer"]they are the same weight, because they are both one pound[/hide]


inequalities
linear algebra
matrix
linear algebra unsolved

Prove following inequalities:

1. If $A, B$ are matrices dimensions $n \times m$ prove that
\[ \det (AB^T)^2 \leq \det AA^T \det B B^T. \]

2. If $A, B$ are symetric matrices dimensions $n \times n$, prove that
\[ \mbox {tr} {ABAB} \leq \mbox {tr} {A^2B^2}. \]

Greetings!

Hint for the first question :) :
assume that $m\leq n$ and consider the matrices $A'=\begin{pmatrix} A & 0_{n,n-m} \end{pmatrix}$and $B'=\begin{pmatrix} B & 0_{n,n-m} \end{pmatrix}$ and then $A\"=\begin{pmatrix} A' & 0_n\\ I_n & I_n \end{pmatrix}$and $B\"=\begin{pmatrix} B' & 0_n \\ I_n & I_n \end{pmatrix}$

Regarding the second exercice you use the Cauchy Schwarz inequality :
for $U$ and $V$ two real matrices :$tr(U^{t}V)^2\leq tr(U^{t}U)^2tr(V^{t}V)^2$ and you just have to put $AB$ instead of $U$ and $BA$ instead of $V$ .


function
probability
ARML
induction
analytic geometry
graphing lines
slope

These are from the 2003 Georgia ARML tryout-- as you've probably deduced, I'm going out for the team this year and practicing on old tests. Please help!

11. Find the following sum: 1/1 + 1/3 + 1/15 + 1/35 + 1/63 +....+1/323 +1/399.

[I see the pattern for denominators: 1x3, 3x5, 5x7,...19x21, but I'm not sure how to sum it all.]

A. 3/2   B. 31/21   C. 32/21   D. 33/23    E. 16/11

16. If  	3 f(X) - f(1/x) = 1/x,   Then f(x)=

A. 3x/8 - 1/8x
B. x/8-3/8x
C. x/8+3/8x
D. 8x/3 + 8/3x
E. 8x/3 - 3/8x

17. Suppose that f(1)= -2 and f`(x) is less than or equal to 3 for all values of 1<x<3. Determine the largest possible value of f(3).

A. 4      B. 6      C. 11      D. 0      E. -4

27. Find ABC, where (9x^2 - x - 18)/(x^3 - x^2 - 6x) = A/x +B/(x-3) + C/(x+2)

A. -9     B. 528/5    C. 9     D. 2688/5     E. 24

42. Suppose that in a best-of-seven games World Series (the first team to win four games wins the Series), the Atlanta Braces have a 3 in 5 chance of winning in each of the games. What is the probability that they will win the series in 6 games?

A. 3/5    B. 81/625    C. 216/625    D. 486/625     E. 648/3125

#11

[hide]

we write the sum as:

$1 + \sum_{n=0}^{9} \frac{1}{(2n+1)(2n+3)}$

partial fraction decomp gives:

$1 + \frac{1}{2} \cdot \sum_{n=0}^{9} \frac{1}{2n+1} - \frac{1}{2n+3}$

which telescopes and gives:

$1 + \frac{1}{2} \cdot \Big( 1 - \frac{1}{21}\Big) = \frac{31}{21} \implies \boxed{B}$

[/hide]

#16

[hide]

we're given:

$3f(x) - f(1/x) = 1/x$

replace $x$ with $1/x$ (we're just working with cyclic functions)

now you have: $3f(1/x) - f(x) = x$

divide by $3$ and add with the first equation to get:

$f(x) = \frac{3}{8} \Big( \frac{1}{x} + \frac{x}{3}\Big) \implies \boxed{C}$

[/hide]

uhh not sure on #17?

[hide]

so right off the bat im assuming the function is linear... cuz im not sure how to deal with this problem if it can be anything

so you have $f'(x) = a$, integrating gives $f(x) = ax + c$, we know $f(1) = -2$, so $c = -2 - a$

Now, $f(3) = 2a - 2$ for $a \le 3$, hence largest value of $f(3)$ occurs when $a=3$ and $f(3) = 4 \implies \boxed{A}$

[/hide]

42

[hide]
The braves must win 4 and lose 2.  The probability of this happening is $\left( \frac 3 5 \right)^4 \left( \frac 2 5 \right)^2 = \frac{3^4 2^2}{5^6}$
They also must win the last game, or else they would have won the series in 5 or less games.  So then we find the arrangements of WWWLL, which is $\frac{5!}{3!2!} = 10$  Then the final probability is $\frac{3^4 2^3}{5^5} = \frac{648}{3125} \implies \boxed E$
[/hide]

[quote="bthings_2000"]42. Suppose that in a best-of-seven games World Series (the first team to win four games wins the Series), the Atlanta Braces have a 3 in 5 chance of winning in each of the games. What is the probability that they will win the series in 6 games?

A. 3/5    B. 81/625    C. 216/625    D. 486/625     E. 648/3125[/quote]
[hide]The Braces (Braves???) must win the last game. Then they must also win three other games of the remaining five. This occurs with probability \[ \left(\frac{2}{5}\right)^2\left(\frac{3}{5}\right)^4=\frac{324}{15625} \] There are $\binom{5}{3}=10$ ways to pick the three games the team wins. Hence, the answer is just \[ 10\left(\frac{324}{15625}\right)=\frac{648}{3125}. \] So the answer is E?[/hide]

Answer to #11:
[hide]If we look at the first few sums, we see a pattern. $S_1=1$, $S_2=\frac{4}{3}$, $S_3=\frac{7}{5}$, $S_4=\frac{10}{7}$. We notice that the numerator of each reduced $S_n$ increases by $3$ each time, starting at $1$, and the denominator is the $n$-th odd number. Hence, we hypothesize that $S_n$ is given by $S_n=\frac{3(n-1)+1}{2n-1}$. Using the formula, the desired sum is $S_{11}=\frac{31}{21}\Rightarrow\framebox{B}$.[/hide]

NOTE: The formula for $S_n$ can be proved by induction, but in a competition where time is an issue and a proof is not required, the above solution suffices.

Do you want to see the induction?

Thanks so much to all you guys so far. This forum owns for learning stuff!!!!!!!!!!!

Everyone is correct so far according to the key, even gotztahbeazn on 17 (you weren't sure, though I'd assume the function could be anything...)

Arjunlandes, the induction would be cool, if you have time. Thanks again!

hmm does anyone have a nice solution to 17 tho? one that maybe doesnt make the assumption that f(x) is linear

and also bthings_2000 i think u might have made a typo on #27, cuz your denominator has like x^3 - x^2 x 6x  or something

[hide="27"]Since the product of the denominators on the right side is $x^3-x^2-6x$, we just need to set the numerators equal.  When we add the fractions on the right, we get \[ Ax^2-Ax-6A+Bx^2+2Bx+Cx^2-3Cx=9x^2-x-18  \]Now we just split this up by powers of $x$.  Since the only term without an $x$ is $-6A$, \[ -6A=-18\Rightarrow A=3  \]Next, for the coefficients of $x^2$, \[ -3+B+C=9\Rightarrow B+C=6  \]and for the coefficients of $x$, \[ -3+2B-3C=-1\Rightarrow 2B-3C=2  \]Using substitution, \[ B=-C+6\Rightarrow -5C=-10\Rightarrow C=2  \] \[ B=-2+6=4  \]Therefore, \[ ABC=3*4*2=24  \]So the answer is $\boxed{E}$.[/hide]

By the way, could you tell me what the Mean Value Theorem is?

If a function $f(x)$ is differentiable on the interval $[a,b]$, then there exists $c\in(a,b)$ such that \[ \frac{f(a)-f(b)}{a-b}=f'(c). \] In other words, the function has a tangent line with the same slope in that interval as the secant line of that interval.


Support
AMC
USA(J)MO
USAMO
number theory

Is it necessary to proof every lemma you use in a proof?

I suppose this may vary from place to place, but in a high level competition you are supposed to do it?

I was reading "Excursion in number theory" and in the 3 chapter it says:
[quote]We need two preliminary theorems, or 'lemmas', before we can proceed. Someone once defined lemma as 'the hard part of the proof', and he was not far of the mark. Often many pages of tough preliminary reasoning are necessary before a proof can be pushed to its conclusion.[/quote]

And I am curious... What do you guys think a proof is?

Personally, putting it in a few words. I think it is to go from a point to another or saying something in an unambiguos way. To do this using logic (though the essential logic seems a rather obscure topic) seems pretty common...

In most cases, it's best to prove lemmas if you want a full high score on an olympiad problem.

You see, lemmas technically are some little "theorems" that you come up with in the middle of your proof. Obviously, you don't just get them in simple steps; you have to use other famously known theorems (larger image ones) to arrive at a conclusion, which is basically what a lemma is all about. You have to actually SHOW how you got to your lemma (since it's not generally well known by the mathematical society, only in a specific problem in most cases--I'm not sure if this is actually true but this is what I believe).

A proof is something where you always have to support your arguments with well-known theorems and solutions in order to make conclusions in a problem. They must be clear, concise, accurate, and easy to follow, if you want to guarantee yourself a good score on olympiads.

In competitions like the USAMO, if a lemma has a commonly known name, you can probably quote it without proving it.
Otherwise, if your lemma is non-trivial, prove it.


If there are 100 people in a room and each person shakes hands once with every other person in the room, then how many handshakes would have taken place?

[quote="abacadaea"]$ \sum_{k \equal{} 1}^{100\minus{}1}{k}$[/quote]

Correct, but that can be simplified.

$ \sum_{k \equal{} 1}^{99}\equal{}\frac{99(99\plus{}1)}{2}\equal{}\frac{99\cdot 2\cdot 50}{2}\equal{} 100(50)\minus{}1(50) \equal{}\boxed{4950}$

nice job guys wow you do know da trick....
u use the sum trick: if there are n people, then 1+2+3+..+(n-1) handshakes would take place.

mods plz lock this...its already solved

[quote="mihail911"]isn't that the same as doing

$ \binom{100}{2}$?[/quote]
Yes, because you are choosing 2 people out of 100 to shake hands.

The computation for that would be

$ \frac{100!}{2!98!}\equal{}\frac{(100)(99)(98!)}{2(98!)}\equal{}\frac{100(99)}{2}$ which brings you back to my original post (which I fixed).

I am just curious.

I know how to solve it, but did you guy already know the rule for the problem(sum trick)?

yeah. its n(n+1)/2 for the sum of all the numbers from 1 to n.

and you know if there are 100 people, person 1 person 2....person 100, that person one shakes hands with all 99 other ppl. person 2 shakes hands with everyone except person one because he has already shaken hands with him. so thats 98. person 3 shakes hands with 97, 4 96 and so forth.


For all nonnegative real numbers $ a,b$ and $ c$, no two of which are zero,
\[ \frac{1}{(a\plus{}b)^2}\plus{}\frac{1}{(b\plus{}c)^2}\plus{}\frac{1}{(c\plus{}a)^2}
\ge\frac{(a\plus{}b\plus{}c)^4}{4(a^2\plus{}b^2\plus{}c^2)(ab\plus{}bc\plus{}ca)^2} \]

[quote="ductrung"]For all nonnegative real numbers $ a,b$ and $ c$, no two of which are zero,
\[ \frac {1}{(a \plus{} b)^2} \plus{} \frac {1}{(b \plus{} c)^2} \plus{} \frac {1}{(c \plus{} a)^2} \ge\frac {(a \plus{} b \plus{} c)^4}{4(a^2 \plus{} b^2 \plus{} c^2)(ab \plus{} bc \plus{} ca)^2}
\]
[/quote]

The following is strongest:

\[ \sum \frac {1}{2a^2\plus{}3ab\plus{}2b^2} \ge\frac {(a \plus{} b \plus{} c)^4}{7(a^2 \plus{} b^2 \plus{} c^2)(ab \plus{} bc \plus{} ca)^2}.
\]

[quote="ductrung"]For all nonnegative real numbers $ a,b$ and $ c$, no two of which are zero,
\[ \frac {1}{(a \plus{} b)^2} \plus{} \frac {1}{(b \plus{} c)^2} \plus{} \frac {1}{(c \plus{} a)^2} \ge\frac {(a \plus{} b \plus{} c)^4}{4(a^2 \plus{} b^2 \plus{} c^2)(ab \plus{} bc \plus{} ca)^2}
\]
[/quote]
This inequality follows from
$ (a^2 \plus{} b^2 \plus{} c^2) \left( \frac {1}{(b \plus{} c)^2} \plus{} \frac {1}{(c \plus{} a)^2} \plus{} \frac {1}{(a \plus{} b)^2}\right) \ge \left( \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b}\right)^2,$
and
$ \frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b} \ge \frac {(a \plus{} b \plus{} c)^2}{2(ab \plus{} bc \plus{} ca)}.$
:)
[quote="Vasc"][quote="ductrung"]For all nonnegative real numbers $ a,b$ and $ c$, no two of which are zero,
\[ \frac {1}{(a \plus{} b)^2} \plus{} \frac {1}{(b \plus{} c)^2} \plus{} \frac {1}{(c \plus{} a)^2} \ge\frac {(a \plus{} b \plus{} c)^4}{4(a^2 \plus{} b^2 \plus{} c^2)(ab \plus{} bc \plus{} ca)^2}
\]
[/quote]

The following is strongest:
\[ \sum \frac {1}{2a^2 \plus{} 3ab \plus{} 2b^2} \ge\frac {(a \plus{} b \plus{} c)^4}{7(a^2 \plus{} b^2 \plus{} c^2)(ab \plus{} bc \plus{} ca)^2}.
\]
[/quote]
Very nice, Vasc.  :lol:

Ok, it is just Holder inequality:
\[ \sum_{cyc} a^2\sum_{cyc} a(b\plus{}c)\sum_{cyc} a(b\plus{}c)\sum_{cyc}\frac{1}{(b\plus{}c)^2}\ge (a\plus{}b\plus{}c)^4\]
But what about the following:
\[ \frac{1}{(a\plus{}b)^2}\plus{}\frac{1}{(b\plus{}c)^2}\plus{}\frac{1}{(c\plus{}a)^2}
\ge\frac{3(a\plus{}b\plus{}c)^2}{8(ab\plus{}bc\plus{}ca)}\left(\frac{1}{ab\plus{}bc\plus{}ca}\plus{}\frac{1}{a^2\plus{}b^2\plus{}c^2}\right)\]

[quote="ductrung"]Ok, it is just Holder inequality:
\[ \sum_{cyc} a^2\sum_{cyc} a(b \plus{} c)\sum_{cyc} a(b \plus{} c)\sum_{cyc}\frac {1}{(b \plus{} c)^2}\ge (a \plus{} b \plus{} c)^4
\]
But what about the following:
\[ \frac {1}{(a \plus{} b)^2} \plus{} \frac {1}{(b \plus{} c)^2} \plus{} \frac {1}{(c \plus{} a)^2} \ge\frac {3(a \plus{} b \plus{} c)^2}{8(ab \plus{} bc \plus{} ca)}\left(\frac {1}{ab \plus{} bc \plus{} ca} \plus{} \frac {1}{a^2 \plus{} b^2 \plus{} c^2}\right)
\]
[/quote]
My proof (See at my attachment) :)


limit
calculus
derivative
calculus computations

The class that I TA for has a limit on their L'Hospital Rule homework that I cannot seem to find a good step-by-step explanation of.  That is, I need the steps that exists in showing
\[ \lim_{x\rightarrow\infty}xe^{\frac {2}{x}} \minus{} x \equal{} 2.
\]
Can anyone help me?  Thanks in advance.

Write $ y \equal{} \frac{1}{x}$ and rewrite the limit as

$ \lim_{y \to 0} \frac{e^{2y} \minus{} 1}{y}.$

This is even simpler than l'Hopital's - it's a derivative.

Thanks.  I feel silly for not seeing that.

[quote="t0rajir0u"]Write $ y \equal{} \frac {1}{x}$ and rewrite the limit as

$ \lim_{y \to 0} \frac {e^{2y} \minus{} 1}{y}.$

This is even simpler than l'Hopital's - it's a derivative.[/quote]
or then its type $ \frac{0}{0}$ so we can Lhopital it anyway and get 2(1) = 2

Yes, we can, but the definition of a derivative is so much more basic than the use of l'Hopital's.  I know of a frankly depressing number of calculus students who can't recognize limits that are actually definitions of derivatives.  Sure, this is just the simplest case of l'Hopital's, but if you can't recognize these types of limits for what they really are then you haven't really understood the definition of a derivative.


Omidvaram Bacheha khob dade bashan. Saole 3 va 6 sakhto technical bodan.
[quote="Omid Hatami"]The only thing I know from Iranian Team is that all of them solved problem 2.[/quote]
Malome, Ma to hendese ghaviyim. :D

[quote="dashmiz"]my result:
1)A complete and i think B my solution is wrong :( .
2)i solved it by simson line.
3)0
and i solved 4 & 5 and i just wrote for 6.
i think iran will get 2 gold ,3 or 4 silvers.[/quote]
Khase Nabashi Dashmiz.

manam omidvaram ke bachehamoon khoob dade bashan...
tebghe hamoon linkai ke gozashti,mese inke emsal soala kheili sakht boodan nesbat be salaye ghablo kheili az adamey ghavi emsal kharab kardan,omidvaram ke bachehamoon hamashoon khoob dade bashano movafagh beshan...  :lol:

[quote="BaBaK Ghalebi"][quote="Omid Hatami"]And here is the Iranian Final Results:
[code]
        1    2    3    4    5    6    SUM
IRN1    3    7    0    7    7    2      26
IRN2    6    1    0    7    6    0      20
IRN3    7    7    2    7    6    1      30
IRN4    3    7    0    7    5    2      24
IRN5    7    0    0    7    1    0      15
IRN6    7    7    0    7    7    0      28
                                        143
[/code][/quote]
be tartib:
Mohammad Reza Takapoui
Shayan Dashmiz
Arman Fazeli Chaghooshi
Seyedmahyar Sefidgaran
Seyedhesam Firouzi
Saeed Hadikhanloo[/quote]
bebin fekr mikonam eshtebah gofti tartib in bayad bashe:
1:shayan dashmiz
2:arman fazeli
3:seyyed hesam firouzi
4:saeed hadikhanloo
5:seyyedmahyar sefidgaran
6:mohamad reza takapooi
http://www.imo2007.edu.vn/index.php?module=ViewRank
chon age page IMO negah koni mibini shomareye bacheha intori hast!

kheili kheili ajibe ke iran har sal zaiiftar mishe makhsoosan ke... :huh:

be salamati $ 13^{th}$ ham ke shodimo...

mage timhaii ke emtize mosavi daran ham rotbe nistan ,agar in tor bashe iran nohome.

are hagh ba toe,ba in tafasir iran shode $ 9^{th}$ vali bazam be nazare man teame iran mitoone rotbehaye kheili behtari biare. :|

firoozi tala
takapooi noghre
dashmiz noghre
hadikhanloo noghre
fazeli boronze
sefidgaran boronze


Let $n$ and $q$ be integers with $n \ge 5$, $2 \le q \le n$. Prove that $q-1$ divides $\left\lfloor \frac{(n-1)!}{q}\right\rfloor $.

[quote="nicetry007"]For $ q \;<\; n$, $ \;q(q-1)\; |\; (n-1)!$. Hence, it is enough discuss the problem for $ q = n$.

If $ n$ is composite, then $ n = a \cdot b$ and $ gcd(n-1,a) = gcd(n-1,b) = gcd(n-1,n) = 1$. Both $ a$ and $ b$ are less than $ n-1$. Hence, $ ab(n-1)\; |\; (n-1)!$.

We are left with the case of $ n$ being a prime.

Let $ n = p$. We know that $ (p-1)! =-1 (mod\; p)$.

In other words, $ (p-1)! = (p-1)(p-2)! = pk-1 = p(k-1)+p-1 \Rightarrow (p-1)\left[(p-2)!-1\right] = p(k-1) \Rightarrow (p-1) \;| \; k-1$.

But $ \lfloor\frac{(p-1)!}{p}\rfloor = k-1$. Hence, $ (p-1) \;| \; \lfloor\frac{(p-1)!}{p}\rfloor$[/quote]

for $q<n$ is obvious and for $n=q$ only note that $\left\lfloor \frac{(q-1)!}{q}\right\rfloor =\frac{(q-1)! +1}{q} -1$


Who here took Part 2 of the University of Maryland Math Competition today? How do you think you did?

I think I got 1, 2, and 3 completely correct and a bit of partial credit on #5 . . . maybe 100/150?

Okay, I saw the questions, and here is what I think.

1-Easy
2-Easy/Medium
3-Medium
4-Medium/Hard
5-Medium/Hard

hmmm #3 is actually pretty easy if you write $f_{n}(m) = \underbrace{\sqrt{m+\sqrt{m+\sqrt{m+\cdots}}}}_{n\text{ square roots}}= \sqrt{m+f_{n-1}(m)}$ because if $f_{n-1}(m)$ doesn't work than neither does $f_{n}(m)$ so just prove $f_{2}(m)$ doesn't work.

edit: i mean it's "pretty easy", not "pretty" :P

Wait, how did you hear that? Aren't they putting the results up in January? :maybe:

[quote="E^(pi*i)=-1"]Finally someone posted here . . .

Yeah, #3 is pretty easy if you use chess64's method (which if you have done problems like this before should look familiar).

#4 is very hard (of course, I am terrible at geometry).  I know someone who spent over half the test bashing it with calculus.

#5 I'm mad about because I had the right idea but wasn't able to figure out exactly how to work the argument before time ran out.

(#1 and #2 are really easy.)

I hope ~100 will get me an honorable mention along with a 64 on Round 1.  I heard Brian Lawrence got a perfect on Round 2  :ninja: .[/quote]

I think he is assuming Brian Lawrence got a perfect not necessarily.

I heard from Tim Peng (who goes to Montgomery Blair) that Brian Lawrence solved all 5 problems correctly.  Of course, he may have had style\rigor points subtracted, but that's unlikely.

Well I mean it's not surprising at all. He got a perfect on the USAMO and this about 100 times easier than the USAMO :)

Well, the first two are about 100 times easier, but the others are easier but maybe 10 times easier.


For every n the sum of n terms of an arithmetic progression is $2n+3n^2$. What is the rth term of the sequence in terms of r?

Okay so first we plug in $n=1$.  This yeilds that $a_1=5$

Plugging in $n=2$, we find that the second term $a_2=11$.

Therefore the common difference is 6 and the starting term is five.

Therefore, in my notation, $a_n=5+6(n-1)$

You can use the relationship $S_n-S_{n-1}=a_n$ as well.
[hide="In this case..."]$a_n=S_n-S_{n-1}=2n+3n^2-2(n-1)-3(n-1)^2=2+6n-3=6n-1$

$a_1=5$ (which satisfies $a_n=6n-1$ also)
[/hide]

The good thing about this method is, it works for any type of sequence ;)


calculus
integration
LaTeX
algebra
polynomial
real analysis
real analysis unsolved

f_n(x)=(x^(n+1)-1)/(x-1), n natural. prove that if I(alpha)=integral(alpha to alpha+1)f_n(x)dx has min value, then it is not a whole number.

In $\LaTeX$:

$f_{n}(x)=\frac{(x^{(n+1)}-1)}{(x-1)}$, $n$ natural. prove that if $I(\alpha)=\int_\alpha^{\alpha+1}f_{n}(x)dx$ has min value, then it is not a whole number.

That doesn't address the ambiguity in the language of the statement; what is "it"?

Since $I(\alpha)$ is a polynomial of degree $n$, it has a minimum iff $n$ is even.  But in this case $I(\alpha)>0$ for all $\alpha$, and $I(-1)<1$, so the minimum can't be an integer.

10x but i don't get why I(alpha)>0 when n+1 is even

If $n$ is odd then a polynomial of degree $n$ does not have a minimum because the equation $f(x)=y$ has at least one real solution $\forall y \in R$ thus it can be arbitrarily large and small.

yeah
but i don't understand this part of Eigenray' solution
"But in this case I(alpha)>0 ..."

If $n$ is even show that $f_{n}(x)>0$ for all $x$.


$a+b+c+abc=4$ prove that :

$\frac {a}{\sqrt {b+c}}+\frac {b}{\sqrt {a+c}}+\frac {c}{\sqrt {a+b}}\geq \frac {\sqrt {2}}{2}(a+b+c)$

Come on, not again this one. Please check proposed problems section, I solved it there.

sorry But I couldn't find it :(  ;)  :mad:

I posted this one. ;) .Just use Cauchy two times and $a+b+c\geq ab+bc+ca$.Anyway take a look around "Inequalities solved problems" ;)

 By the way this is not hard at all. ;)

Yes i am agrre with harazi. this problem has solved long ago.
here is a link to the  [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=45996]solution[/url].  ;)

harazi your solution for this problem is very nice but i can not understand that how did you get inequalities in your solution?(i mean how did you know you should use that inequalities?) :?:

The answer is simple: the problem was created by Cezar, so it must be solved using Cauchy :D . I just applied Cauchy till I got an inequality that I knew and I knew then I'm on the good way.

Excuse me, Harazi, but do you think that all my problems can be solved using Cauchy. Man, i'm beggining to think that you have something with me. :( . I don't know maybe, you will tell me. :?

Yes, since all the problems you posted and that I've seen and which were not completely obvious were consequences of Cauchy. No, I have nothing with you, but this is the truth. I don't want to offend you if this is what you think and I apologize for giving you this impression. But the truth is the truth.


calculus
integration
logarithms
trigonometry
function
geometry
calculus computations

[quote="NaZAf"]Demonstrate that $\int_{\frac{-1}{2}}^{\frac{1}{2}}cos(x)\ln\left (\frac{1+x}{1-x} \right)dx \ = \ 0$ but [b]without evaluating the integral[/b].[/quote]

$\cos x$ is even function and $\ln \left(\frac{1+x}{1-x} \right)$  is odd function (by replacing $x$ with $-x$ you will get $- \ln \left(\frac{1+x}{1-x} \right) )$ . So the product $\cos x \, \ln \left(\frac{1+x}{1-x} \right)$ is the odd function. It follow that the graph is symetric about the origin, and the area between $\left[- \frac{1}{2}, - \frac{1}{2} \right]$ is $0$. Ortogonality.

[quote="blahblahblah"]Orthogonality?[/quote]

 :P  You didn't mean i refer to orthogonality relation of $\cos x$ and $\sin x$ (in somekind of inner product) .

Ok this is maybe more clear "Like orthogonality". One word forget.

\[ \boxed {\ \int_{-a}^a f(x)\ dx=\int_0^a [f(x)+f(-x)]\ dx\ }  \]A particular case : $\int_{-a}^a f(x)\ dx= \left \{ \begin{array}{cc} 2\cdot\int_0^a f(x)\ dx & \mathrm {if\ f\ is\ even}\\\\ 0 & \mathrm {if\ f\ is\ odd} \end{array}\right.$

1. $cos(x)$ is even $\Longrightarrow cos(x) = cos(-x)$

2. let  $f(x) = \ln\left(\frac{1+x}{1-x}\right) = \ln(1+x) - \ln(1-x), \ \ \forall x\epsilon \left[\frac{-1}{2},\frac{1}{2}\right]$
    $f(x)$ is odd $\Longleftrightarrow f(-x)=-f(x) \ \Longleftrightarrow \ f(x)=-f(-x)$

$-f(-x) = -\ln\left(\frac{1-x}{1+x}\right) = -ln(1-x) + ln(1+x) = \ln\left(\frac{1+x}{1-x}\right) = \mathbf{f(x)}$
$\therefore f(x)$ is odd,$\ \ \forall x\epsilon \left[\frac{-1}{2},\frac{1}{2}\right]$

3. $\int_\frac{-1}{2}^\frac{1}{2}cos(x)\ln\left(\frac{1+x}{1-x}\right)dx = \int_\frac{-1}{2}^0cos(x)\ln\left(\frac{1+x}{1-x}\right)dx + \int_0^\frac{1}{2}cos(x)\ln\left(\frac{1+x}{1-x}\right)dx$

$\int_\frac{-1}{2}^0cos(x)\ln\left(\frac{1+x}{1-x}\right)dx = -\int_0^\frac{-1}{2}cos(x)\ln\left(\frac{1+x}{1-x}\right)dx$

$\ \ \ \ \texttt{let u = -x} \Longrightarrow \text t{du = -dx}$

$= \ -\int_0^\frac{-1}{2}cos(u).-\ln\left(\frac{1+u}{1-u}\right).-du = -\int_0^\frac{1}{2}cos(u)\ln\left(\frac{1+u}{1-u}\right)du = -\int_0^\frac{1}{2}cos(x)\ln\left(\frac{1+x}{1-x}\right)dx$

$\boxed{\therefore \int_\frac{-1}{2}^\frac{1}{2}cos(x)\ln\left(\frac{1+x}{1-x}\right)dx =-\int_0^\frac{1}{2}cos(x)\ln\left(\frac{1+x}{1-x}\right)dx + \int_0^\frac{1}{2}cos(x)\ln\left(\frac{1+x}{1-x}\right)dx = 0}$


[quote="spider_boy"]Find all functions $f: R \to R$ satisfying
$f(x+y)=f(x)+f(y)$ and $f(\displaystyle\frac{1}{x})=\frac{f(x)}{x^2}$ for all $x,y$ in $R$.[/quote]

Um I am perhaps missing something. Can't we just quote that one theorem thingy about additive functions and get f(x) = kx then verify that the second condition is also satisfied?

Clearly w have lim y -> 0 f(x+y) = f(x) because f(x+y) = f(x) + f(y), and as y -> 0, we have f(y) = f(\frac{1}{1/y}) = \frac{f(1/y)}{(1/y)^2} = y^2f(1/y). Additivity should guarantee that f(1/y) is at most linear in 1/y. So we have continuity and are done.

Or is this not rigorous?

[quote="JSteinhardt"]
Um I am perhaps missing something. Can't we just quote that one theorem thingy about additive functions and get f(x) = kx then verify that the second condition is also satisfied?

Clearly w have lim y -> 0 f(x+y) = f(x) because f(x+y) = f(x) + f(y), and as y -> 0, we have f(y) = f(\frac{1}{1/y}) = \frac{f(1/y)}{(1/y)^2} = y^2f(1/y). Additivity should guarantee that f(1/y) is at most linear in 1/y. So we have continuity and are done.
[/quote]

I don't think you can talk about limits this way, they might not exist...

@JSteinhardt: You still assume continuity in $0$.

@Nameless: That's good. It follows $f(x)=kx + \frac{k-1}{1}$ where $k=f(1)$. But $f(x+y)=f(x)+f(y)$ implies $k=1$. So $f(x) \equiv x$ is the only solution.

That's not quite right- after all, $f(x)=kx$ is a solution for all $k$.

Nameless' method, corrected:
$f\left(\frac{x}{x+1}\right)=f(1)-f\left(\frac1{x+1}\right)=f(1)-\frac{f(x+1)}{(x+1)^2}$
$=f(1)\cdot\left(1-\frac1{(x+1)^2}\right)-\frac{f(x)}{(x+1)^2}= f(1)\cdot\frac{x^2+2x}{(x+1)^2}-\frac{f(x)}{(x+1)^2}$
$f\left(\frac{x}{x+1}\right)=f\left(\frac{x+1}{x}\right)\cdot\frac{x^2}{(x+1)^2}= \left(f(1)+f\left(\frac1x\right)\right)\cdot\frac{x^2}{(x+1)^2}$
$=f(1)\cdot\frac{x^2}{(x+1)^2}+\frac{f(x)}{x^2}\cdot\frac{x^2}{(x+1)^2}= f(1)\cdot\frac{x^2}{(x+1)^2}+\frac{f(x)}{(x+1)^2}$

Setting the two expressions equal, $\frac{2xf(1)}{(x+1)^2}=\frac{2f(x)}{(x+1)^2}$ for all $x$ not equal to 0 or -1, and $f(x)=xf(1)$ for these values. It is easy to verify that $f(0)=0$ and $f(-1)=-f(1)$, so $f(x)=xf(1)$ for all $x$ and the linear functions $f(x)=kx$ are the only solutions.


Hmm.
I predict:
First: Mattew Tan (made nats last year)
Second: Ani
Third: GSB
Fourth: LHM

I predict I barely make countdown. Yay.

THe scores are already out

xpmath won.

gj xp

Dude, will you stop saying that when we know the compeition is on Saturday. I know I won't win.

Yeah, I won't win. K?
...hmm, I better make nats though.

A possible make up date is Saturday March 29.

Yeah, my coach just told me that. Heh, more time.

i predict.

1) matthew tan
2) #H34N1
3) xpmath (beast)
4) George S. Bashour.

dude wtf

i can't win

edit: and i think 1=2 is matthew tan

1=2 is Matthew Babbit or something. Anyway, he lives in New York.

GSB tells me that someone claiming to be "Truant523" is Matthew Tan.

Official Predictions:

1) Henry Zi, Sycamore
2) Siddhant Dogra, Ottawa Hills
3) George Bashour, Birchwood
4) Matthew Tan, forgot-his-school

I hope the one person from my old school who made it to states makes it to nationals in order to carry on the tradition... But I don't really think that will be.

What school is that? Sycamore?

SYCAMORE SHOULD BURN LOL

Hmm... Well, it's in Fairfield county, so i'm guessing Central Ohio.

Haha, I'm from Reynoldsburg.  But I don't know much about our Mathcounts teams, or if we even have a team this year.

[quote="tenniskidperson3"]:lol: No, not sycamore... Pickerington.  I agree about sycamore...  But anyways, the first year our school allowed mathcounts, someone made it to nats.  This is the second year.[/quote]

I believe Matthew tan was from Pickerton :lol:

Nah, look at
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=137938&start=20[/url]

Oh lol yeah.

I don't see pickerton though :huh:

well, it's called "Ridgeview Jr. High" in Pickerington, but they got the wrong county. It's not in franklin.

When you get back from it today, post as soon as you get here!
 :P

Good luck to everyone competiting in ohio state today! =D

Results:
1st- Dylan Lee (wrong spelling I think)
2nd- Clayton Cardinal
3rd- Ryan Welch (not sure about the last name)
4th- Me.
Sycamore won team.

Heh, due to the fact that everyone in my team besides me (well, me somewhat) failed pretty badly, we came 16th. Oh well.

Congrats, xpmath!
How did H34N1 and GSB do?

GSB got 5th.

Yo failed.

I BOMBED (and I mean BOMBED) targets.

Around 7th or something...not in the top 6.

[quote="thealphafish8"]OH YEAH.
What did I say about Ryan Welch???? Hmm.. But I think he just got a bit confuzzled at the Countdown.
HE USUALLY BLOWS EVERYONE AWAY!
But Ryan's got class. Poor guy in the wheelchair. Ugh. He HAD to hurt himself THIS WEEK. Ugh.
SYCAMORE GOT CLASS!!!!
GO AVES :D

I came in 10th place!!! Last year I was in 126th place or something like that...

:D[/quote]

OK what was this Ryan Welch's AMC/AIME score

this is thealphafish8's new screenname.

idk about ryan's amc scores. I'll have to ask him. But he was school champ for amc8 and amc10. so yeah. Hey Sid? if you're still replying to this thread that is so old that its rotting, how did you do at nats? ryan got 192nd place... he's still a champ at our school.

Eh? Ryan came 97th with the same score I had, but he had two less sprint than me.


A fisherman can see seven aquatic creatures in a lake --- four crocodiles, a catfish, and two giant squid. If he does not wish to fish two crocodiles consecutively, in how many orders can he catch all seven creatures?

Call the crocodiles "C", the catfish "F", and the giant squids "S".
Clearly, the order must go C_C_C_C. There are 3 ways to make such an order.

Yes, it has to go C_C_C_C....but there are 6 ways to permute the catfish and squids, and 24 ways to permute the crocodiles; 6*24=144 shouldn't that be the answer? or am i makign a stupid mistake.... :blush:

bob is correct if you can differentiate between the crocodiles, and brut3 is correct if you cannot differentiate between two members of the same species. In a game I would go with Bob's answeer, but I don't know, I haven't memmed this one...


Prove the inequality:
$ (1 \minus{} a)(1 \minus{} b)(1 \minus{} c) \ge (1 \minus{} \sqrt[3]{{abc)}}$
Determine the cases of equality. :oops:

(3,3,3) gives $ \minus{}8 \geq \minus{}2$

[quote="nojack"]Prove the inequality:
$ (1 \minus{} a)(1 \minus{} b)(1 \minus{} c) \ge (1 \minus{} \sqrt [3]{{abc)}}$
Determine the cases of equality. :oops:[/quote]Are the variables in the interval $ [0,1]$?

[quote="Mewto55555"](3,3,3) gives $ \minus{} 8 \geq \minus{} 2$[/quote]
Sorry !

The correct problem :)

[quote]Let a,b,c =<1.Prove the inequality:
$ (1 \minus{} a)(1 \minus{} b)(1 \minus{} c) \ge (1 \minus{} \sqrt [3]{{abc)}}$
Determine the cases of equality. :oops:[/quote]

When $ a\equal{}b\equal{}c\equal{}0.5$ the inequality does not hold.

[quote="nojack"]Prove the inequality:
$ (1 \minus{} a)(1 \minus{} b)(1 \minus{} c) \ge (1 \minus{} \sqrt [3]{{abc)}}$
Determine the cases of equality. :oops:[/quote]What if the inequality is reversed, ie $ (1 \minus{} a)(1 \minus{} b)(1 \minus{} c) \le (1 \minus{} \sqrt [3]{{abc)}}$? Is it always true? Now that should be fun unless we see another counterexample :P

Assuming $ a,b,c \in [0,1]$ because otherwise $ a \equal{} b \equal{} c \equal{} \minus{} 1$ breaks it:

[hide]Expand. $ \sqrt [3]{abc} \plus{} ab \plus{} bc \plus{} ca \leq a \plus{} b \plus{} c \plus{} abc$. Use AM-GM to reduce it to $ 3ab \plus{} 3bc \plus{} 3ca \leq 2a \plus{} 2b \plus{} 2c \plus{} 3abc$, which is equivalent to $ \sum_{cyc} a(2 \minus{} b)(1 \minus{} c) \geq 0$, which is clearly true.

As for equality, we clearly need $ a \equal{} 0$ or $ c \equal{} 1$ and the cyclic equivalents, which eventually gives $ (0,0,0)$ and $ (1,1,1)$ as the only ones.[/hide]


Find $x$ such that $\begin{equation*} \frac{\sin \left( 3x\right) \cdot \cos \left( 60 -4x\right) + 1}{\sin \left( 60 -7x\right) -\cos \left( 30 +x\right) +m}=0,
\end{equation*}$ where $m$ is a given real number.

The given expression can be written as sin(3x)sin(30+4x)+1/[-2sin3xsin(30+4x)+m]. This expression reduces to -1 when m=-2 in which case there is no solution. So lets consider the case when m <> -2.

Then for the expression to be equal to 0 we must have:
sin(3x)sin(30+4x)= -1 now we have for any real x -1 <= sin 3x <= 1 and similarly -1<= sin(30+4x) <= 1 hence the required condition occurs when one of the two is -1 and other 1. So lets consider this case we have

Let sin 3x= 1 and sin(30+4x)= -1. we have x= (4n+5):pi:/6 where n \varepsilon [b]W[/b] for all m \varepsilon [b]R[/b].

If we assume sin3x= -1 and sin(30+4x)= 1 then this will give us no solution as we will have x= (4n+3):pi:/6=(6m+1):pi:/12 which is not possible as 2(4n+3)<> 6m+1 for any real n&m.


[quote="kunny"]Let $ a,\ b,\ c$ be distinct real numbers. Prove that

$ |\frac{b-c}{c-a}|(|a-b|+|b-c|)|+|\frac{c-a}{a-b}|(|b-c|+|c-a|)|+|\frac{a-b}{b-c}|(|c-a|+|a-b|)$

$ +|a-b\parallel b-c\parallel c-a|+1\geq 8\sqrt{|a-b\parallel b-c\parallel c-a|}$.[/quote]
Twice AM-GM! :wink:


Does anyone know what the last three digits of 3^(4798) are or how to find them?

the answer is $ 889$, notice that $ 3^{100}\equiv 1(1000)$ then $ 3^{4800}\equiv1(1000)$
We have to solve $ 9 x \equiv 1 (1000)$ obtaining $ x\equal{}889$


Hi.

Could someone explain the easiest way to compute the composition of permutations?

For example:

$ (1342)(3645)(1623)$ .

So I start with 1 and: (where $ (1342)2\equal{}1$ means 2 goes to 1 in that cycle.)

$ (1342)(3645)(1623)1\equal{}(1342)(3645)6\equal{}(1342)4\equal{} 2$ so 1 goes to 2.

$ (1342)(3645)(1623)2\equal{}(1342)(3645)3\equal{}(1342)6\equal{} 6$ so 2 goes to 6.

Is this a proper way to calculate the composition? If so, how do I continue from 6? or do I continue with 3 ?

Thanks.

[quote="RealTime"]Is this a proper way to calculate the composition? If so, how do I continue from 6? or do I continue with 3 ?[/quote]  Yes, assuming the class/text you're doing doesn't compose in the opposite direction  :wink:

$ (1342)(3645)(1623)6 \equal{} (1342)(3645)2 \equal{} (1342)2 \equal{} 1$.  Etc.


find all integer solutions $(x,y)$ to
$x^4+2x^3+2x^2+2x+1=y^2$

[hide]The left side factors into $(x+1)^2(x^2+1)$, so in order to equal $y^2$, $x^2+1$ must be a perfect square or $x+1=0$.
Let $x^2+1=a^2\Rightarrow (a+x)(a-x)=1$. For the product to equal 1, both factors must be 1 or -1, so $x=0$. Then the only solutions are $(0,1),\ (-1,0)$.[/hide]

i do not think that you factored the left hand side correctly.

Dont forget about $(x+1)^2=0$
Yielding $(-1,0)$

oops forgot that they were being multiplied  :rotfl:  sorry :blush:

One can also say taht the left side is bounded such that

$n^4\le LHS \le (n+1)^4$

And there is no perfect fourths in between two consecutive perfect fourths, so the only solutions are when the $LHS = x^4$ or $(x+1)^4$.

it's asking for squares, not fourths...

but that can work by setting bounds at $(x^2+x)^2=x^4+2x^3+x^2$ and $(x^2+x+1)^2=x^4+2x^3+3x^2+2x+1$. The first is less by $(x+1)^2$ and the second is greater by $x^2$. One of these will have to be 0.