find all polynominal P such that :
$P(P(.....(P(x))......))=x$

Obviosly $deg(P(x))=1$ and $P(0)=0$, therefore $P(x)=exp(\frac{2\pi ki}{n})x$.

Can you explain this a little bit further?

Just for completeness: if $a=1$, then $b=0$.

But if you don't understand something, don't ask for help but tell what you don't understand.

I think mathboy_cntHDVN not understand complex fomulae.
it has in many book in vietnamese :lol:


Solve the equation $x^3+2y^3+5z^3=0$ in integers.

try mod 7.  cubes are 0, 1 or -1 only.

EDIT:  sorry i went through it quickly and didnt notice an extra soln.

Why do you say '0,else,else no '¿ ($0^3+2\cdot 1^3 + 5 \cdot 1^3 \equiv 0 \mod 7$)

On the other hand, mod 9 actually works. When you see cubes, think mod 7 and mod 9 first.

If $t\in\mathbb Z$ then $t^3\equiv0,\pm1(mod9).$
Hence, $x\equiv y\equiv z\equiv0(mod3).$
Hence, if $(x,y,z)$ is solution of the equation, then $(\frac{x}{3},\frac{y}{3},\frac{z}{3})$ is solution of the equation.
Hence, if $x^2+y^2+z^2\neq0$ then  $\forall i\in\mathbb N$  $(\frac{x}{3^i})^2+(\frac{y}{3^i})^2+(\frac{z}{3^i})^2\in\mathbb N.$ Contradiction.
Hence, $x=y=z=0.$
Well $\{(0,0,0)\}.$ :)


modular arithmetic
algebra
polynomial
induction
number theory unsolved
number theory

Does someone know how to prove that:
$\binom{2p}{p} \equiv 2  \pmod{p^3}$ where $p$ is a prime number such that $p \ge 5$ ?

Here it goes: $\displaystyle \binom{2p}{p}=\sum_{k=0}^p\binom{p}{k}^2$. We need to show that $\displaystyle p^3|\sum_{k=1}^{p-1}\binom{p}{k}^2$. Notice that $\displaystyle \frac {\binom{p}{k}^2}{p^2}\equiv \frac 1{k^2}\pmod p$ (we work in $\mathbb Z_p$, so the meaning of $k^{-1}$ is clear). All we need to show now is that $\displaystyle \sum_{k=1}^{p-1}\frac 1{k^2}=0$, and this is easy to show for $p\ge 5$ by using the fact that $\displaystyle \frac 1k$ are the roots of the polynomial $x^{p-1}-1$ and the first and second Viete sums, which are both zero.

more generally,

[tex] \binom{a \cdot p}{b \cdot p} \equiv \binom{a}{b}  \pmod{p^3}[/tex] for [tex]p \ge 5[/tex]

which i'm not quite sure how to prove, but it may be an induction argument using the base case Grobber proved.

figured out the proof, and too lazy to write it out but it uses the identity:

[tex]\displaystyle \binom{ap}{bp} = \sum_{k_1+k_2+...+k_a=bp} \binom{p}{k_1}\binom{p}{k_2}...\binom{p}{k_a}[/tex]

and does reduce down to grobbers proof.

Sorry for my ignorance,but how can I  proof that
 C(p,k)^2/p^2=1/k^2 mod p

(C(p,k) the combinatorialnumber, 0<k<p integer)

The original problem is:
(MOP 2000) Prove that [tex] \binom{qp}{p} \equiv q \pmod{p^3}[/tex] where p is a prime number such that [tex]p \ge 5[/tex]

thank u...

I was trying to pove C(p,k)/p=\pm 1/k mod p using wilsons theorem, or something like that......well I just got confuse... jeje


trigonometry
quadratics
geometry
circumcircle
incenter
ratio
angle bisector

ABC is a right triangle with angle ABC = 90°.  D is a point on AB such that BCD = DCA.  E is a point on BC such that BAE = EAC.  If AE = 9 inches and CD = 8 r00t2 inches, find AC.

AE, CD are bisectors of $\angle A, \angle C.$ Denote their feet K, M instead of E, D and denote a = BC, b = CA, c = AB. We have $k = AK = 9,\ m = CM = 8 \sqrt{2}.$ Using the sine theorem for $\triangle AKC,\ \triangle AMC,$

$\frac{k}{b}= \frac{\sin C}{\sin \left(C+\frac{A}{2}\right)},\ \ \frac{m}{b}= \frac{\sin A}{\sin \left(A+\frac{C}{2}\right)}$

Eliminating b, we get $\frac{k}{m}= \frac{\sin C}{\sin A}\cdot \frac{\sin \left(A+\frac{C}{2}\right)}{\sin \left(C+\frac{A}{2}\right)}$

Eliminating $\angle C = 90^\circ-\angle A$ leads to

$\frac{k}{m}= \frac{\cos A}{\sin A}\cdot \frac{\sin \left(45^\circ+\frac{A}{2}\right)}{\cos \frac{A}{2}}= \frac{1-\tan^{2}\frac{A}{2}}{2 \tan \frac{A}{2}}\cdot \left(1+\tan \frac{A}{2}\right) \cdot \frac{\sqrt{2}}{2}$

This is a cubic equation for $x = \tan \frac{A}{2}.$ Putting $\frac{k}{m}= \frac{9}{8 \sqrt 2},$

$(1-x^{2})(1+x) = 2x \sqrt{2}\cdot \frac{k}{m}= \frac{9x}{4}$

$4x^{3}+4x^{2}+5x-4 = 0$

Before using other methods, we can look for a rational root. By the rational root theorem, the rational root (if any) has to be $\frac{p}{q},$ where p divides the absolute term 4 and q divides the coefficient 4 at $x^{3},$ or $\pm 4,\ \pm 2\ \pm 1\ \pm \frac{1}{2},\ \pm \frac{1}{4}.$ Starting with positive numbers in decending order, we quickly find that $\frac{1}{2}$ is a root. Factoring it out by the long division,

$4x^{3}+4x^{2}+5x-4 = (2x-1)(2x^{2}+3x+4) = 0$

The remaining quadratic factor has no real roots, hence $x = \tan \frac{A}{2}= \frac{1}{2},$ $\tan A = \frac{2 \tan \frac{A}{2}}{1-\tan^{2}\frac{A}{2}}= \frac{4}{3}= \frac{a}{c}.$ Therefore, the legs of the right triangle $\triangle ABC$ are $a = 4 \eta,\ c = 3 \eta$ and its hypotenuse $b = 5 \eta.$  Calculating one of its internal angle bisector lengths, say k, yields,

$k = \frac{\sqrt{bc[(b+c)^{2}-a^{2}]}}{b+c}= \frac{3 \sqrt{5}}{2}\cdot \eta = 9$

As a result, $\eta = \frac{6}{\sqrt{5}},\ \ b = 5 \eta = 6 \sqrt{5}.$

[color=darkblue][b]Lemma.[/b] Denote $AE=l_{a}$ , $CD=l_{c}\ .$ Then $\boxed{\ \frac{l_{a}}{l_{c}}=\frac{c}{a}\cdot\sqrt{\frac{b+a}{b+c}}\ }\ \ (*)\ .$

[b]Proof.[/b] $\{\begin{array}{c}2\cos^{2}\frac{A}{2}=1+\cos A=1+\frac{c}{b}\Longrightarrow\cos\frac{A}{2}=\sqrt{\frac{b+c}{2b}}\\\\ 2\cos^{2}\frac{C}{2}=1+\cos C=1+\frac{a}{b}\Longrightarrow\cos \frac{C}{2}=\sqrt{\frac{b+a}{2b}}\end{array}$ $\Longrightarrow\{\begin{array}{c}c=l_{a}\cdot \cos\frac{A}{2}\\\\ a=l_{c}\cdot\cos\frac{C}{2}\end{array}\Longrightarrow \rho\equiv\frac{l_{a}}{l_{c}}=\frac{c}{a}\sqrt{\frac{b+a}{b+c}}\ .$

[b]Remark.[/b] $\{\begin{array}{c}x\equiv\tan \frac{A}{2}=\frac{\sin A}{1+\cos A}=\frac{\frac{a}{b}}{1+\frac{c}{b}}\Longrightarrow x=\frac{a}{b+c}\\\\ y\equiv\tan\frac{C}{2}=\frac{\sin C}{1+\cos C}=\frac{\frac{c}{b}}{1+\frac{a}{b}}\Longrightarrow y=\frac{c}{b+a}\end{array}$ and $\frac{C}{2}=\frac{\pi}{4}-\frac{A}{2}\Longrightarrow$ $y=\frac{1-x}{1+x}$ and $\tan A=\frac{2x}{1-x^{2}}.$

In conclusion, $\rho=\frac{c}{a}\cdot\sqrt{\frac{c}{a}\cdot\frac{b+a}{c}\cdot\frac{a}{b+c}}=$ $\frac{1}{\tan A}\cdot\sqrt{\frac{1}{\tan A}\cdot \frac{\tan\frac{A}{2}}{\tan\frac{C}{2}}}=$ $\frac{1-x^{2}}{2x}\cdot\sqrt{x\cdot\frac{1+x}{1-x}\cdot\frac{1-x^{2}}{2x}}\Longrightarrow$ $\rho=\frac{(1-x^{2})(1+x)}{2x\sqrt 2}$ $\Longrightarrow$

$\boxed{\ x^{3}+x^{2}+(2\rho\sqrt 2-1)-1=0\ }\ .$ In the our case, $l_{a}=9$ and $l_{c}=8\sqrt 2\ .$ For $\rho=\frac{l_{a}}{l_{c}}=\frac{9}{8\sqrt 2}$The previous equation becomes

$4x^{3}+4x^{2}+5x-4=0\ ,$ i.e. $(2x-1)(2x^{2}+3x+4)=0$ $\Longrightarrow$ $x=\frac{1}{2}$ and $y=\frac{1}{3}$ $\Longrightarrow$ $\{\begin{array}{c}b+c=2a\\\ b+a=3c\end{array}$ $\Longrightarrow$ $\frac{a}{4}=\frac{c}{3}=\frac{b}{5}\ .$

The triangle $ABC$ is similarly with the triangle $A'B'C'\ (\ 4\ ,\ 5\ ,\ 3)$ in which the length $l'_{a}$ of the bisector from the angle $\widehat{B'A'C'}$ is equally to $\frac{3\sqrt 5}{2}\ .$

Therefore $,\ \frac{a}{4}=\frac{c}{3}=\frac{b}{5}=\frac{l_{a}}{l'_{a}}=\frac{9}{\frac{3\sqrt 5}{2}}=\frac{6\sqrt 5}{5}$ $\Longrightarrow$ $\boxed{\ b=6\sqrt 5\ }\ .$[/color]

Does anyone have a pure geometric solution without trigonometry? When I first looked at the problem I thought knowing the angle bisector thm and Menaleus thm should be able to kill the problem, but I was totally wrong...

[hide="Without trigonometry,"]

$ AE$ cuts the circumcircle $ (O)$ again at $ X.$ $ I, I_a$ are incenter and A-excenter. Power of $ E$ to $ (O)$ is

$ \overline{EA} \cdot \overline{EX} \equal{} \overline{EB} \cdot \overline{EC} \equal{} \minus{} \frac {a^2bc}{(b \plus{} c)^2}.$

Cross ratio $ G(I_a, I, E, A) \equal{} \minus{} 1$ is harmonic and $ X$ midpoint of $ II_a$ $ \Longrightarrow$

$ \frac {\overline{AE}}{\overline{EX}} \equal{} \frac {\overline{AE}}{\overline{IE}} \cdot \frac {\overline{IE}}{\overline{EX}} \equal{} \frac {\overline{AE}}{\overline{IE}} \cdot \left(\frac {\overline{EI_a}}{\overline{EX}} \minus{} 2\right) \equal{} \frac {\overline{AE}}{\overline{IE}} \cdot \left(\frac {\overline{EA}}{\overline{EI}} \minus{} 2\right) \equal{} \frac {(b \plus{} c)^2 \minus{} a^2}{a^2}\ \Longrightarrow$
________________________________________________________________________[/hide]

$ AE^2 \equal{} \frac {bc((b \plus{} c)^2 \minus{} a^2)}{(b \plus{} c)^2},\ CF^2 \equal{} \frac {ba(b \plus{} a)^2 \minus{} c^2)}{(b \plus{} a)^2}$

$ \frac {AE^2}{CF^2} \equal{} \frac {c}{a} \cdot \frac {b \plus{} c \minus{} a}{b \plus{} a \minus{} c} \cdot \frac {(b \plus{} a)^2}{(b \plus{} c)^2}.$

For right $ \angle B,$ $ b^2 \equal{} a^2 \plus{} c^2$ $ \Longrightarrow$ $ (b \plus{} c \minus{} a)(b \plus{} a) \equal{} c(a \plus{} b \plus{} c)$ $ \Longrightarrow$

$ \frac {AE^2}{CF^2} \equal{} \frac {c^2}{a^2} \cdot \frac {b \plus{} a}{b \plus{} c} \equal{} \frac {81}{128}.$

Assume $ (a, b, c)$ is proportional to a Pythagorean triple $ (2mn,\ m^2 \plus{} n^2,\ m^2 \minus{} n^2),$ with $ a$ proportional to an even integer. (This is no different from looking for a rational root of the above cubic.)  $ \Longrightarrow$

$ \frac {AE^2}{CF^2} \equal{} \frac {(m^2 \minus{} n^2)^2}{(2mn)^2}\cdot \frac {(m \plus{} n)^2}{2m^2} \equal{} \frac {81}{128}\ \Longleftrightarrow$

$ \frac {(m \plus{} n)^2 (m \minus{} n)}{m^2n} \equal{} \frac {9}{4} \Longleftrightarrow\ m \equal{} 2,\ n \equal{} 1\ \Longleftrightarrow\ (a, b, c) \propto (4, 5, 3).$ The triple exists, let $ \eta$ be the proportionality ceofficient.

$ AE^2 \equal{} \frac {45}{4} \eta^2 \equal{} 81\ \Longrightarrow\ \eta \equal{} \frac {6}{\sqrt {5}}\ \Longrightarrow\ b \equal{} 5 \eta \equal{} 6 \sqrt {5}.$


[quote="Redriver"]Give  $a,b,c>0.$
Prove that : $2(ab+ac+bc)( \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}) \leq (a+b+c)( \frac{ab}{c}+\frac{bc}{a}+\frac{ac}{b})$[/quote]
Expand all, note that $LHS=(a(b+c)+bc)\sum_{cyc}{a \over b+c}$, we get: \[2\sum_{cyc}{abc \over a+b}+2\sum_{cyc}a^{2}\leq \sum_{cyc}ab+\sum_{cyc}a^{2}\left({b \over c}+{c \over b}\right) \\ \iff \sum_{cyc}{a^{2}\over bc}(b-c)^{2}\ge \sum_{cyc}ab\left({2c \over a+b}-1\right) \\ \iff \sum_{cyc}\left({a^{2}\over bc}+{a^{2}\over (a+b)(a+c)}\right)(b-c)^{2}\ge 0.\]


There is a small square centered inside a large square... The large square is 3x-1 by x+8 and the smaller square is x-5 by 2x-1 find x if the area of the space between the large and small square 141units^2.

 i need some help i've been working on this problem for a while and can't seem to come to the correct answer... please and thank-you!!!

[hide]
This is just the differences of the area between the two squares. So we have $(3x-1)(x+8)-(x-5)(2x-1)=144$. We reject the negative value, since it would lead to a negative side length in the square, and get that $x_{1}= \sqrt{443}-17$. But this is less than $5$, so the side with length $x-5$ will be negative. Thus, there is $\boxed{ \text{No Solution}}$.[/hide]

[quote="i_like_pie"][hide]It's a square so $3x-1=x+8$, or $x=4.5$. However, this doesn't work in the other equation, so $\boxed{\texttt{no solution}}.$[/hide][/quote]

Oh duh, its a square. Right.  :P


[quote="luimichael"]Find the maximum value of k such that $ \binom{2008}{1000}$ is divisible by $ 21^k$, where k is a natural number.[/quote]

We have $ ord_p(\binom{2008}{1000})\equal{}ord_p(2008!)\minus{}ord_p(1000!)\minus{}ord_p(1008!)$ $ \equal{}\sum_{k\equal{}1}^{\plus{}\infty}\lfloor\frac{2008}{p^k}\rfloor$ $ \minus{}\sum_{k\equal{}1}^{\plus{}\infty}\lfloor\frac{1000}{p^k}\rfloor$ $ \minus{}\sum_{k\equal{}1}^{\plus{}\infty}\lfloor\frac{1008}{p^k}\rfloor$

So $ ord_3(\binom{2008}{1000})\equal{}\lfloor\frac{2008}{3}\rfloor \plus{}\lfloor\frac{2008}{9}\rfloor \plus{}\lfloor\frac{2008}{27}\rfloor$ $ \plus{}\lfloor\frac{2008}{81}\rfloor \plus{}\lfloor\frac{2008}{243}\rfloor$ $ \plus{}\lfloor\frac{2008}{729}\rfloor$ $ \minus{}\lfloor\frac{1008}{3}\rfloor \minus{}\lfloor\frac{1008}{9}\rfloor \minus{}\lfloor\frac{1008}{27}\rfloor$ $ \minus{}\lfloor\frac{1008}{81}\rfloor \minus{}\lfloor\frac{1008}{243}\rfloor$ $ \minus{}\lfloor\frac{1008}{729}\rfloor$ $ \minus{}\lfloor\frac{1000}{3}\rfloor \minus{}\lfloor\frac{1000}{9}\rfloor \minus{}\lfloor\frac{1000}{27}\rfloor$ $ \minus{}\lfloor\frac{1000}{81}\rfloor \minus{}\lfloor\frac{1000}{243}\rfloor$ $ \minus{}\lfloor\frac{1000}{729}\rfloor$

And $ ord_3(\binom{2008}{1000})\equal{}669\plus{}223\plus{}74$ $ \plus{}24\plus{}8 \plus{}2$ $ \minus{}336\minus{}112\minus{}37$ $ \minus{}12\minus{}4 \minus{}1$ $ \minus{}333\minus{}111\minus{}37$ $ \minus{}12\minus{}4\minus{}1$ $ \equal{}0$

And so $ k\equal{}0$

I think the answer should be k = 2.

Edit: My calculation is incorrect.
The answer is the last post is correct.


Euler
Gauss
Support
number theory theorems
number theory

You are hilarious!!! :D  :D  :rotfl:

So you meant to say you've solved all these conjectures which have previuosly confounded legendary minds like Euler, Lagrange, Gauss etc and think we believe you!!!!

Have you proved Riemann's hypothesis, or maybe your greatness can find/ disprove an odd perfect prime etc!!!

With all due respect, O king of bluffing.

 :D
Bomb

I UNDERSTAND YOUR .....
IF IT COULD NOT BEEN DONE YEARS AGO THEN IT CNNOT BE DONE NOW...

 I REST MY CASE.

I HOPE THE WORLD WILL CONSIDER MY WORK SERIOUSLY.
THANKS

I dont want to be scheptical, but it sounds kind of promising. Although it would be really nice that all that you have a proof for all of that, but its sort of "too good to be true". So lets wait and see

Best

Well, the second claim is true for $n\le10000$, as I just verified on my computer.

Can U tell us more?For example what did U use to prove GoldBach conjecture?Did U use Analytic or algebric Number Theory.Please give us more information.

You shouldn't believe people shouting around to have proven three difficult unsolved problems, as long as they don't give any evidence...

[quote="BONZION"]IN GHANA IT IS QUITE DIFFICULT GETTING THE RIGHT CHANNEL IN ORDER TO PUBLISH ONES FINDINGS.
IF NYONE COULD HELP ME PUBLISH THESE FINDINGS IN A WELL-ESTABLISH JOURNAL I WILL BE VERY GRATEFUL[/quote]If you really proved goldbach's conjecture CORRECTLY and if you really have a CORRECT elementary proof to Fermat's last theorem, then just publish them anywhere. As soon as they're found correct, all big world journals will publish them instantly.

It seems [i]very[/i] unlikely though. :roll:

:o  :|  :huh:  :? Can you explain what you/he wants to say by this¿ Or tell us the background¿

Ok, I'll give you the benefit of the doubt - I've clicked on the 'Watch this topic' link so that I'll be able to read your findings next year but please, don't waste my time...

[quote="BanishedTraitor"][quote="BONZION"]1. THERE EXISTS [b]AT LEAST TWO[/b] PRIMES NUMBERS BETWEEN N*N AND (N+1)*(N+1), WHERE N IS A NATURAL NUMBER.

2. THERE EXISTS [u]AT LEAST ONE[/u] TWIN-PRIME PAIR BETWEEN N*N AND (N+2)*(N+2), WHERE N IS AN ODD NATURAL NUMBER.
(THEREFORE THE TWIN PRIME CONJECTURE IS TRUE)

3. THE GOLDBACH CONJECTURE IS TRUE.

USING A VERY POWERFUL TOOL I HAVE DISCOVERED THE BLUEPRINT OF THE WORLD OF PRIMES. I HOPE TO PROVIDE THE PROOF OF THE ABOVE CONJECTURES IN MY THESIS.

THE TOOL : PRIMY SPACES

WATCH OUT FOR THE RESULTS NEXT YEAR JUNE (2006) AFTER I HAVE DEFENDED MY WORK.

I ALSO HAVE AN ELEMENTRY PROOF OF FERMAT'S LAST THEOREM.
IN GHANA IT IS QUITE DIFFICULT GETTING THE RIGHT CHANNEL IN ORDER TO PUBLISH ONES FINDINGS.
IF NYONE COULD HELP ME PUBLISH THESE FINDINGS IN A WELL-ESTABLISH JOURNAL I WILL BE VERY GRATEFUL[/quote]

And you are telling us this because? Personally I don't really care if you did find an elementary proof of Fermat's Last Theorem. Wile's proof of the Taniyama-Shimuara Conjecture/Theorem (and hence Fermat's Last Theorem) is far more significant.[/quote]

banished; I am not competing with anybody. Personally, I believe that to every problem there is an easier way around it. Please do not line me against the great Wile. We are here to share knowledge. If you do not agree with me, i'm cool. :(  :(  :(

[quote="mgb"]I have discovered five proofs of Reimann Hypothesis. I will publish them on this board in 2010, 2011, 2012, 2013 and 2014. Watch this space... :roll:[/quote]

Oh well done  :lol:  I will be waiting  :roll:

P.S. I found $ 7$ proofs of the Hodge Conjecture all involving elementary mathematics. $ 7$ is of no significance but it is a prime number!

The Riemann hypothesis is part of my homework assignment :) Most of my class solves it quite routinely.

[quote="BONZION"]I guess you have all taken sometime to ponder over my statements concerning the primes.
I have finished defending my work and (GOD-WILLING) make the results known to all and sundry very soon. Maybe my style is not the best. forgive me. I will provide an external link to the results on this forum.[/quote]

Hey BONZION i'm ready to support you okay, in my mind you're more intellectual than any others those laughed to you, don't mind them, you know, feel it better by trusting in yourself. Thanks. I'm Persian

I cant believe this topic has resurfaced. Can you give us a link to some of your work? Maybe a teaser or two?

Trolls feeding trolls, do we already have an ASCII art for that? darij


Two tour guides are leading six tourists.  The guides decide to
split up. Each tourist must choose one of the guides, but with the
stipulation that each guide must take at least one tourist.  How
many different groupings of guides and tourists are possible?

2^6(because each tourist can choose a guide)-2 for the formats where 1 guide has 6 and the other has 0. (The guides are distinguishable).Ans=62

The first guide can take any combination of tourists except all the tourists or none of the tourists. Therefore the number of possibilities is $\binom{6}{1}+\binom{6}{2}+\binom{6}{3}+\binom{6}{4}+\binom{6}{5}=6+15+20+15+6=62.$
OR
If each guide did not need to take at least one tourist, then each tourist could choose one of the two guides independently. In this case there would be $2^6=64$ possible arrangements. The two arrangements for which all tourists choose the same guide must be excluded, leaving a total of $64-2=\boxed{62}$ possible arrangements.

I don't know how this is classified as a complementary counting problem. I didn't start Combinations yet how come I got this problem in Alcumus?

I would count this as complementary counting, because I used $2^6-2=\boxed{62}$ for counting the cases I didn't want.

AMC[hide]10A 2007[hide]#12[hide]The first guide can take any combination of tourists except all the tourists or none of the tourists. Therefore the number of possibilities is \[
\binom{6}{1}+\binom{6}{2}+\binom{6}{3}+\binom{6}{4}+\binom{6}{5}=6+15+20+15+6=62.
\] OR

If each guide did not need to take at least one tourist, then each tourist could choose one of the two guides independently. In this case there would be $2^6=64$ possible arrangements. The two arrangements for which all tourists choose the same guide must be excluded, leaving a total of $64-2=\boxed{62}$ possible arrangements.[/hide][/hide][/hide]

I still don't understand - why 64? Also what does independently mean? (in this context)

There are 2 options for each tourist, so there are $2*2*2*2*2*2=64$ combos. In the context, 'independently' refers to all of them being able to choose the same thing regardless of what other people chose.

[hide]Every single combination of $\binom6n$ is possible but we must subtract $2$ from that for the cases where there is 6 tourists with one guide so answer is $64-2=62$

[hide=AMC Solution]Each tourist has to pick in between the $2$ guides, so for $6$ tourists, there are $2^6$ possible groupings. However, since each guide must take at least one tourist, we subtract the $2$ cases where a guide has no tourist. Thus the answer is $2^6 - 2 = 62$. [/hide]


geometry trapezoid ratio LaTeX rectangle

Ok so I have a math tournament tomorrow. From now until midnight, I should be posting questions I dont know how to do on this thread:

I have the answers but not solutions

#1. If you know the sides of a trapezoid can you find the height? Say if the sides are 1, 2, 3, and 4 units long for example (the actual problem is a 1, 2, 3, 4 trapezoid)

#2. In circle O, chords AC and BD are prependicular and intersect at E. If DE=3, BE=8, and CE=4, find the length of the radius of circle O.

#3. Quarter circle XOY of circle O contains circle A so that XO, YO, (both line segments), and arc XY are tangent to circle A. If the radius of circle O is 3, find the radius of circle A.

#4. The length of the hypotenuse of a right triangle is h, and the radius of the inscribed circles is r. The ratio of the area of the circle to the area of the triangle is :?

#1  I am not sure you can make a trapezoid with it, but you can make a quadrilateral for sure.
#2  Um, what's D?
#3  [hide="hint"]This is going to be hard to do with latex but oh well. So you have the circle inscribed inside the quarter circle. Draw the radius of circle [b]O[/b] so that it passes right through the diameter of circle [b]A[/b]. Call the radius of circle A, x. Then you call draw the radius of the circle A such that it is perpendicular to radius OX and OY. This forms a square with diagonal x root 2. So this means that the radius of circle A equals x + x root 2. So you solve 3 = x + xroot2. I will let you do that. [/hide]
#4 [hide="hint"]Almost the same as #3, first you draw the radius of the circle such that it is perpendicular with all three sides. From here you have a square with diagonal r root 2. Then the perpendicular line from the base is r + r root 2. From here you can solve for the ratio  [/hide]

Sorry if my English is bad, tell me and I will try to clarify.

#5. In Rectangle ABCD, a point M is selected inside the rectangle. If AM=4, BM=5, and CM=6, find the length of DM


USAMTS geometry perimeter function ratio trigonometry quadratics

How did you do??

Also, what did you guys get for the general formula for f(n,k) for the last question of Euclid?
I got a formula

$ f(n,k) \equal{} 2k^{n\minus{}2k}$

But apparently, this does not work for f(7,3)...

I thought about adding 1 to this, but then f(6,3) does not fit in... :(

ya i dont think the answer will be that simple. and also the relation they give u in part b is a degree 2 linear recurrsion so im guess there will be more to the formula.
anyways, can someone post the questions to hypatia?

The Hypatia's problems are very easy in this year, even much easier than they were last year. Though I got owned on part c) #4, it's a used problem on USAMTS last year(problem 5 round 2, which I also got owed when I did USAMTS :oops: ). The rest is easy, I think. Does anybody consider the case when the vertical line is outside of the triangle ABC? I intended to consider all 3 cases, but saw that it's too complicated for such problem like #3, so I just assumed the vertical line is inside the triangle ABC. And on #4 part b), did anyone get "ugly" answers? Like $ radical \plus{} radical \plus{} 48$? I checked my calculation a bit, so I hoped it was right.

I agree with ghjk the Hypatia was really easy I am expecting at least a 38.

[quote]And on #4 part b), did anyone get "ugly" answers? Like ? I checked my calculation a bit, so I hoped it was right.[/quote]

I got the same answer.
By the way, what was the answer for 4c)? I want to check if my answer is correct.

are you sure ICAE wrote the euclid/hypatia yet?

just in case, don't discuss euclid/hypatia until it's absolutely safe to do so cuz i remember they wrote hypatia on saturday last year.

to azn_rocker and ghjk: can you edit your posts?

o ya i think we can talk about it cause ICAE people did it last weekend. anyways can someone post the hypatia question it? i wanta do it!!!

[quote="Complex_Ninja"]Is it okay to post the Euclid contest yet?
And what were your answers for 9a and 9b
I got 64 for 9a and 8 for 9b[/quote]

Did you miss $ \frac{1}{4}$ for 9a?

I would scan the questions sheet, but my scanner is broken  :mad:

Yeah, the Hypatia was really easy but i messed up 4c. During the test I figured out how to do it but I thought no, that won't work so I didn't do it.  :mad:
For 9a I got 64 and I think the perimeter of RUT in 9b is always 8. I got a formula for #10 very quickly; I'll post it once I finish downloading the stuff I need for latex. I'm pretty sure that its right; I was able to do 10a and 10b with it.

In my opinion, this contest was almost the same in difficulty as last year's, and I fear that the # 10's are becoming a little bit formulaic, as they always seem to involve some sort of recursive function definition, a proof of the recursion, and finding the general formula to determine some value of the sequence.

Ahh well, I should have done better, but I wasted a lot of time on #8, and had little time left for #10...

Overall, I hope for like 89 - 95.... what about everyone else?
  :maybe:

Hmm for number 10 I didn't use recursion and found a formula really quickly. I'm hoping for a 87-95 on it.

[quote="Complex_Ninja"][quote="Darmani"][quote="Complex_Ninja"]Is it okay to post the Euclid contest yet?
And what were your answers for 9a and 9b
I got 64 for 9a and 8 for 9b[/quote]

Did you miss $ \frac {1}{4}$ for 9a?[/quote]
Ya but sub 1/4 for x. You will get a sequence that is like this -2 , 0 , 0........ So basically then 0/0 = 0/-2 ???????[/quote]
well if u read the definition of a geometric sequence, it saids each term after the 1st is mutiplied by a constant. in this case the constant is 0. therefore x=1/4 is a solution

Rotate the square 90 degrees counterclockwise about the top left hand corner. Now find the 2 congruent triangle right there with the angle at the top left hand corner being 45 degrees.

Conclude that the length of the hypotenuse is the same as the sum of the lengths on the right and bottom face which are not the 2 arms of the right triangle. Therefore the sum of the hypotenuse and the 2 sides is just the bottom side+right side=8.

Then again, I used tangent bash on the contest  :roll:

[quote="ghjk"]Anybody minds posting $ \#9b$? I would like to try doing that question :) .[/quote]
You didn't do 10 yet ;). But here 9b

Sorry for no diagram, i think it is pretty obvious as to how it will look.

PQRS is a square  sides of length 4. Points T and U are on sides QR and RS respectively such that (anlge)UPT = 45deg. Determine the maximum possible perimeter of triangle RUT.

So.. anybody know when the Hypatia/Euclid results will be released?

I just did 10c so I'm pretty sure that the answer is n=49, k=6.

[quote="zjzjimbo"]I'm not sure, but I think you can ask your teacher and see if the marks are on that waterloo website, cuz it says "Unofficial Euclid, Fryer, Galois and Hypatia results are being imported."
it just says "being imported"..dunno if they've finished it or not..[/quote]
ya just tell ur teacher that the unofficial scores are out and tell her to go get it.

[quote="ghjk"][quote="eddyliu"]someone want to explain why the hypatia gold cutoff just drop from last year. i swear the last question is alot easier than last year.[/quote]
Perhaps the students are worse than past students were :wink: . And what did u get? I "temporarily" got a bronze but still wait for my protest result(so mad about #3's score :( )[/quote]
i didnt write it cause im in gr12. but in 45min i solved everything so i guess i would have got 38-40 this year. and last year i should have got 40 but careless mistakes... so i got a 34 instead

[quote="CEM_College_Champ"]On a side note, were there problems where calculus could have been of any help?[/quote]
uhhh no not this year. but there is one question where vectors came in handy.

The person who marked my 8b is going to have a headache... my final answer was an expression that involved at least 12 2's, seven square root signs, and quite a few horizontal lines.

And our school didn't write FGH because of an ordering error. WTF?!?

Did anyone receive the prizes for Hypatia this year? I thought they would give out the medals too but they didn't :( . Did anyone get medals last year? Well, I really should have gotten gold instead of silvers, if not because of some problems in the grading system(I still think this is the only issue).

^^
I don't think they give out medals for Hypatia, but I still got gold. I failed Euclid though, 66. How is that possible I answered 8 questions fully. I made a mistake on 7b.


inequalities
function
Putnam
invariant
geometry
geometric transformation

Given that $\{ x_1,x_2,...,x_n\} =\{1,2,...,n\}$,

find the largest possible value of

 $x_1x_2+x_2x_3+....+x_{n-1}x_n+x_nx_1$ in terms of $n$. ( $n\geq 2$ ).

Rearrangement inequality?

So I think it's just [hide]$\displaystyle\sum^{n-1}_{i=1}{i(i+1)}= \frac{n(n-1)(n+1)}{3}$[/hide]

both wrong... vulgar, you missed the $x_nx_1$ term, Jan, a much sharper bound is possible, your maximum can never be attained.

Maybe try a greedy algorithm.

[hide="better hint"]how much is n(n-1)+n(n-2)+(n-1)(n-3)+(n-2)(n-4)+(n-3)(n-5)+...+6.4+5.3+4.2+3.1+1.2?[/hide]

It is an interesting question -- the maximum is very nicely behaved, but the minimum jumps around a little depending on the parity of n.

The answer (unproven) for those who are interested is [hide]$\frac{n^3 + 6n^2 + 5n + 12}6$ for the maximum[/hide]

[quote="luimichael"]JBL,

You answer is incorrect.

[hide="Official answer"]$\frac{2n^3+3n^2-11n+18}6$[/hide][/quote]

That's absolutely right, and I made at least two dumb mistakes.  The first is that I started my function shifted by 2, but I also either made some arithmetic error or just turned a $2{n\choose 3}$ into an $n\choose3$ for no particular reason.

On the other hand, my point about the difference between the maximum and minimum is still valid.  This question is from Putnam, don't recall which year.

[hide="Some ideas"]
The problem is equivalent to maximizing $T = (x_1 + x_2)^2 + (x_2 + x_3)^2 + ... + (x_1 + x_n)^2$.  Because this value is invariant under translation of the indices, WLOG $x_1 = 1$.  Then we can see that $x_{i} = i$ majorizes any other permutation, so the maximum should occur there.  [/hide]

No, it is as my hint indicates, and as the result luimichael posts... why would your $T$ be maximum when $x_i=i$??  :|

[b]Problem:[/b] Find the maximum of $x_1x_2+x_2x_3+...+x_{n-1}x_n+x_nx_1$ where $\{x_1,...,x_n\}=\{1,...,n\}$.

[b]Solution.[/b]

Using Chebyshev's inequality, we have that $x_1x_2+x_2x_3+...+x_nx_1\leq \frac{(x_1+...+x_n)^2}{n}=n\left(\frac{n+1}{2}\right)^2$

This just shows that the maximum cannot exceed $n\left(\frac{n+1}{2}\right)^2.$
Masoud Zargar

[quote="boxedexe"]$n\left(\frac{n+1}{2}\right)^2-1$[/quote]What is that supposed to mean? Your bound is also way too weak...

I just noticed where I went wrong, so nevermind my post. :blush:


geometry
3D geometry
Pythagorean Theorem

Find the number of squares whose sides have integer length and whose vertices are four points on the attached grid.  The distance between two adjacent vertical or horizontal dots is 1.

*The grid and my answer is attached.*

[hide="ANSWER"]140[/hide]

[size=75]ATTACHMENT SLIGHTLY EDITED AFTER ZHANGPEIJIN'S POST[/size]

Um, it's a 7 by 7 not 8 by 8.
So you need to subtract 64.

I think that what Cube did is right - $ \boxed{142}$.

Was that a typo?  The answer is $ \boxed{140}$.  It's basically finding $ \sum_{n\equal{}1} ^{7} n^2$

[quote="alanchou"]Was that a typo?  The answer is $ \boxed{140}$.  It's basically finding $ \sum_{n \equal{} 1} ^{7} n^2$[/quote]

But aren't there two additional squares?  $ \sum_{n \equal{} 1} ^{7} n^2$ only calculates the number of squares with their sides strictly being horizontal or vertical.  By the Pythagorean Theorem, there are two more squares with sides five.

So is the answer $ 140$ or $ 142$?

That summation just counts the "upright" squares. There are two more squares that are "sideways" with length 5 (3-4-5 right triangle) that just fit in the grid. So the answer is [b]142[/b]

oh yeah, right.

although technically, the answer is an infinite amount since there are an infinite number of points in each of the black dots :P


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                                                                               Thank you!

Well as you can see, this is primarily a math forum, and there are just a couple high schoolers posting on this subforum, so don't expect much. But we'll answer.


calculus integration parameterization conics ellipse function trigonometry

Find the average value of $ f(x,y,z) \equal{} z^2 \plus{} xe^y$ on the curve C obtained by intersecting the elliptic cylinder $ x^2/5 \plus{} y^2 \equal{} 1$ by the plane $ z \equal{} 2y$.

1)  The term "elliptic curve" is reserved for beasts of a rather different nature.

2)  What is the definition of "average value" in this context?  It can only be some kind of integral, but any "average" you're going to get depends on the parameterization you use.

I suppose it's the average with respect to arclength, as in $ \int_C f\,ds$ divided by the length of $ C$. Of course, the length of ellipse is not an elementary function, and parametrizing it by arclength is a pain. So we must be clever.

Yea mlok that's the integral we're looking for. I don't even know how to get the equation for the elliptic curve...

The average value of the function $ f\left( {x,y,z} \right) \equal{} z^2 \plus{} xe^y$ on the curve $ C$ finded by the formula:

$ \begin{gathered}
\overline {f\left( {x,y,z} \right)}  \equal{} {{\int\limits_C {f\left( {x,y,z} \right)ds} } \mathord{\left/
 {\vphantom {{\int\limits_C {f\left( {x,y,z} \right)ds} } {\int\limits_C {ds} }}} \right.
 \kern\minus{}\nulldelimiterspace} {\int\limits_C {ds} }} \equal{} {{\int\limits_C {\left( {z^2  \plus{} xe^y } \right)ds} } \mathord{\left/
 {\vphantom {{\int\limits_C {\left( {z^2  \plus{} xe^y } \right)ds} } {\int\limits_C {ds} }}} \right.
 \kern\minus{}\nulldelimiterspace} {\int\limits_C {ds} }},{\text{ where }}ds \equal{} \sqrt {\left( {x'} \right)^2  \plus{} \left( {y'} \right)^2  \plus{} \left( {z'} \right)^2 } dt.
\end{gathered}$

Now we will present the curve in a parametric form:

$ \begin{gathered}
\left\{ \begin{gathered}
  x \equal{} \sqrt 5 \cos \left( t \right), \hfill \\
  y \equal{} \sin \left( t \right), \hfill \\
  z \equal{} 2\sin \left( t \right). \hfill \\
\end{gathered}  \right.
\end{gathered}$

Where $ t$ is parameter and $ t \in \left[ {0;{\text{ }}2\pi } \right].$
So we have

$ \begin{gathered}
  ds \equal{} \sqrt {\left( {\frac{d}
{{dt}}x\left( t \right)} \right)^2  \plus{} \left( {\frac{d}
{{dt}}y\left( t \right)} \right)^2  \plus{} \left( {\frac{d}
{{dt}}z\left( t \right)} \right)^2 } dt \equal{} \sqrt {\left( {\frac{d}
{{dt}}\sqrt 5 \cos \left( t \right)} \right)^2  \plus{} \left( {\frac{d}
{{dt}}\sin \left( t \right)} \right)^2  \plus{} \left( {\frac{d}
{{dt}}2\sin \left( t \right)} \right)^2 } dt \equal{}  \hfill \\
   \equal{} \sqrt {5\sin ^2 \left( t \right) \plus{} \cos ^2 \left( t \right) \plus{} 4\cos ^2 \left( t \right)} dt \equal{} \sqrt {5\left( {\sin ^2 \left( t \right) \plus{} \cos ^2 \left( t \right)} \right)} dt \equal{} \sqrt 5 dt. \hfill \\
\end{gathered}$

$ \begin{gathered}
  \overline {f\left( {x,y,z} \right)}  \equal{} {{\int\limits_C {\left( {z^2  \plus{} xe^y } \right)ds} } \mathord{\left/
 {\vphantom {{\int\limits_C {\left( {z^2  \plus{} xe^y } \right)ds} } {\int\limits_C {ds} }}} \right.
 \kern\minus{}\nulldelimiterspace} {\int\limits_C {ds} }} \equal{} {{\int\limits_0^{2\pi } {\left( {4\sin ^2 \left( t \right) \plus{} \sqrt 5 \cos \left( t \right)e^{\sin \left( t \right)} } \right)} \sqrt 5 dt} \mathord{\left/
 {\vphantom {{\int\limits_0^{2\pi } {\left( {4\sin ^2 \left( t \right) \plus{} \sqrt 5 \cos \left( t \right)e^{\sin \left( t \right)} } \right)} \sqrt 5 dt} {\int\limits_0^{2\pi } {\sqrt 5 dt} }}} \right.
 \kern\minus{}\nulldelimiterspace} {\int\limits_0^{2\pi } {\sqrt 5 dt} }} \equal{}  \hfill \\
   \equal{} \frac{2}
{\pi }\int\limits_0^{2\pi } {\sin ^2 \left( t \right)dt}  \plus{} \frac{{\sqrt 5 }}
{{2\pi }}\int\limits_0^{2\pi } {\cos \left( t \right)e^{\sin \left( t \right)} dt}  \equal{} \frac{1}
{\pi }\int\limits_0^{2\pi } {\left( {1 \minus{} \cos \left( {2t} \right)} \right)dt}  \plus{} \frac{{\sqrt 5 }}
{{2\pi }}\int\limits_0^{2\pi } {d\left( {e^{\sin \left( t \right)} } \right)}  \equal{}  \hfill \\
   \equal{} \frac{1}
{\pi }\left. {\left( {t \minus{} \frac{{\sin \left( {2t} \right)}}
{2}} \right)} \right|_0^{2\pi }  \plus{} \left. {\frac{{\sqrt 5 }}
{{2\pi }}e^{\sin \left( t \right)} } \right|_0^{2\pi }  \equal{} \left( {\frac{{2\pi }}
{\pi } \minus{} 0} \right) \plus{} \left( {\frac{{\sqrt 5 }}
{{2\pi }} \minus{} \frac{{\sqrt 5 }}
{{2\pi }}} \right) \equal{} 2. \hfill \\
\end{gathered}$

So the problem was rigged to make $ C$ a circle... What if it was not?

In any case we can dispose of the term $ xe^y$ at once: its average is zero on any curve that is symmetric about the $ yz$-plane.

We can consider the curve $ C$ as the trajectory of a moving point ($ t$ is parameter, for example time). When this point is moving, its describes the projection of ellipse (coordinates $ x \equal{} \sqrt 5 \cos \left( t \right),{\text{ }}y \equal{} \sin \left( t \right)$ - are coordinates projection). In this case $ z \equal{} 2y \equal{} 2\sin \left( t \right)$ is applikata of moving this point. In other words, in any value of $ t$ we know the situation point. Next, we can calculate the function $ f\left( {x,y,z} \right) \equal{} z^2 \plus{} xe^y$ at that point.

P.S. I'm sorry for my English.


Do we have to prove that an empty set is the subset of any set? Or do you regard this as a definition?

The definition of $ A\subseteq B$ is that $ \forall x (x\in A\Rightarrow x\in B)$. One property of implication $ \Rightarrow$ is that if the statement on the left is false, the implication is always true. If $ A$ is the empty set, then $ x\in A$ is always false. We conclude that the implication is always true and that $ A\subseteq B$ regardless of $ B$.

To answer your question, you can't say it is by definition, but it's almost trivial to prove.

Or you can say that it follows directly from the DEFINITION of implication.

[quote="Math^_^"]Just wondering, what is the difference between axioms and definitions?[/quote]
The way I think as axioms are "rules of the game", when you create a mathematical theory you need definitions and axioms, i.e. the rules you can use. Think of it as a game, the definitions are the peices (or whatever) and the rules you follow are the axioms.

Now the mathematical theory used over here is called "set theory" and the axioms are called "Zermelo-Frankael Axioms". The existence of an empty set is one of them.

But this should not belong in this forum.

Sometimes the words are used interchangeably.  This is especially true when one speaks, for example, of the "field axioms" (or similar).  This really means the defining properties of a field.  They are called the field axioms because (in sylow's terminology) they define the rules of the game for working over a field.  When you say, "Let $ k$ be a field," you are saying, "Suppose that the field axioms apply when dealing with $ k$."  I think it's worth keeping in mind that this is all meta-mathematical: the content of the mathematics is not affected by whether you call them field axioms, the definition of a field, or something else entirely.  The real usefulness here is in communicating with others about mathematics in a mutually meaningful way.

[quote="sylow_theory"]The existence of an empty set is one of them.[/quote]
Not always. In fact, the most formal theory has the existence of the empty set as a theorem, following from the Axiom of Infinity (which says a set exists) and then performing the Axiom of Replacement on that set with a contradictory statement (which means no elements are in the resulting set).


Given a triangle ABC, prove the following two statements are equivalent:

(1) A' is the intersection of BC with the incircle of triangle ABC
(2) A' is on BC such that the incircles of triangles ABA' and ACA' are tangent.

[hide="Solution of an old nice problem"]The incircle of $ABC$ touches $AB, AC$ in $Q, R$ respectively. The incircle of $AA'C$ touches $AA', A'C, CA$ in $T, M, S$ respectively while the incircle of $AA'B$ touches $AA', A'B, BA$ in $U, N, P$ respectively. Let $s_{1}$ be the semiperimeter of $AA'C$ and $s_{2}$ the one of $AA'B$.

Then $s_{1}= CS+SA+A'T = AC+A'T$ and $s_{2}= AP+PB+A'U = AB+A'U$. So the two incircles are tangent iff $T \equiv U$ that is $A'U = A'T$.
But \[A'U = A'T \Leftrightarrow s_{1}-s_{2}= AC-AB \Leftrightarrow \frac{2s-AB-A'B+AA'}{2}-\frac{2s-AC-A'C+AA'}{2}= AC-AB \Leftrightarrow \frac{AC-AB+A'C-A'B}{2}= AC-AB \Leftrightarrow A'C-A'B = AC-AB \Leftrightarrow 2A'C = 2s-2AB \Leftrightarrow A'C = AQ+BQ+XC-AB \Leftrightarrow A'C = XC\]
where $X$ is the contact point of the incircle with $BC$. Hence the circles are tangent iff $A'$ is the intersection of $BC$ with the incircle.[/hide]


Find all polynomials $P(x)$ with real coefficients that satisfie the condition :
$P(x)P(2x^2-1)=P(x^2)P(2x-1)$ for all $x \in R$.

You can show that 0 and 1 are the only possible roots for P ( otherwise r^2 or 2r-1 is also a root, and you can show that there exist infinitely many roots for P then, meaning P = 0 ).Whence P has the form a*x^k*(x-1)^j, but it is easy to check we must have k = 0.

Answer: P = a*(x-1)^j, where a is real and j is a nonnegative integer.


ratio
trigonometry
geometry proposed
geometry

Points $D,E$ lie on side $AB$ of $\triangle ABC$ and satisfy $\frac{AD}{DB}\times\frac{AE}{EB} = (\frac{AC}{CB})^2$. Prove that $\angle ACD = \angle BCE$

This is Steiner's theorem about isogonal conjugate lines, isn't it? :)

$m(\angle DCA)=m(\angle ECB)=x,\ m(\angle DCB)=m(\angle ECA)=y\Longrightarrow$

$\frac{DA}{DB}=\frac{CA}{CB}\cdot \frac{\sin x}{\sin y},\ \frac{EA}{EB}=\frac{CA}{CB}\cdot \frac{\sin y}{\sin x}\Longrightarrow \frac{DA}{DB}\cdot \frac{EA}{EB}=\left(\frac{CA}{CB}\right)^2.$


calculus
function
limit
trigonometry
integration
search
calculus computations

From my current Calculus II course:

4. Let $ f(x)$ be the continuous function such that $ f(x)\equal{}\frac{1\minus{}e^{\minus{}x}}{x}$ for $ x\ne0.$

Find $ f(0)$ and $ f'''(0).$

5. Find $ \lim_{n\to\infty}\frac{(n^7\plus{}n^5)^{1/7}\minus{}n}{(n^5\plus{}2n^3)^{1/5}\minus{}n}.$

[quote="mlok"]#5 is hardcore. My Calc 2 midterm had $ \lim_{n\to\infty}\frac {3^n}{4^n \plus{} 1}$, and only $ 62$ students out of $ 145$ got it right.  :([/quote]

wouldn't that just be 0? Or is there some trick I'm missing?

You're not missing anything. It's a simple question of size - hence mlok's frowny face. I'm not even sure what the usual wrong answers would be in this case.

[quote="mlok"]#5 is hardcore. My Calc 2 midterm had $ \lim_{n\to\infty}\frac {3^n}{4^n \plus{} 1}$, and only $ 62$ students out of $ 145$ got it right.  :([/quote]Funny. In Romania you can't finish highschool if you don't know how to solve that limit :)

[quote="Kent Merryfield"]I'm not even sure what the usual wrong answers would be in this case.[/quote]
$ \frac34$, naturally. Don't you see that $ n$s cancel out?

Actually, the overall scores on that exam were decent (average a little over $ 70$), and there were much harder problems there. Perhaps they thought the problem could not be that simple.

Yes, both of the last two posts have correct answers, and in both cases, that was the reasoning I expected from my students.

[quote="Kent Merryfield"][quote="mlok"]#5 is hardcore.[/quote]
I haven't done the grading yet, but I just flipped through the stack of papers, and I got 12 out of 31 with the right answer, and another three or four that were off by a factor of $ 2.$ Mostly first-semester freshmen with AP Calculus AB credit. But mostly it's that I have been specifically pushing uses of the binomial series.

Actually this is the question some students were grousing to me about because I didn't put anything like it on the practice exam:

6. Identify the [b]exact[/b] values of the sums of these convergent series:

(a) $ 1 \plus{} 1 \plus{} \frac12 \plus{} \frac16 \plus{} \frac1{24} \plus{} \cdots \equal{} \sum_{n \equal{} 0}^{\infty}\frac1{n!}$

(b) $ \pi \minus{} \frac {\pi^3}6 \plus{} \frac {\pi^5}{120} \minus{} \frac {\pi^7}{7!} \plus{} \cdots \equal{} \sum_{k \equal{} 0}^{\infty}\frac {( \minus{} 1)^k\pi^{2k \plus{} 1}}{(2k \plus{} 1)!}$

(c) $ \frac1{\sqrt {3}}\left(1 \minus{} \frac1{3\cdot3} \plus{} \frac1{5\cdot3^2} \minus{} \frac1{7\cdot3^3} \plus{} \cdots\right) \equal{} \frac1{\sqrt {3}}\sum_{k \equal{} 0}^{\infty}\frac {( \minus{} 1)^k}{(2k \plus{} 1)3^k}$

(I wasn't trying to make those [i]too[/i] mysterious. These aren't the kinds of series misan posts.)[/quote]

Hey Kent,
(a) looks like 'e' and (b) looks like $ \sin(\pi)$
Am I right? and by the way (c) is mysterious to me :(

Edit: I worked out (c), is it $ \frac{\pi}{6}$??

[quote="Kent Merryfield"]Find $ \lim_{n\to\infty}\frac {(n^7 \plus{} n^5)^{1/7} \minus{} n}{(n^5 \plus{} 2n^3)^{1/5} \minus{} n}.$[/quote]
[b]In Romania[/b] : $ a_n \equal{} \frac {(n^7 \plus{} n^5)^{1/7} \minus{} n}{(n^5 \plus{} 2n^3)^{1/5} \minus{} n} \equal{}$ $ \frac 12\cdot\frac {\left(1 \plus{} \frac {1}{n^2}\right)^{\frac 17} \minus{} 1}{\left(1 \plus{} \frac {1}{n^2}\right) \minus{} 1}\cdot \frac {\left(1 \plus{} \frac {2}{n^2}\right) \minus{} 1}{\left(1 \plus{} \frac {2}{n^2}\right)^{\frac 15} \minus{} 1}\ .$

Using the remarkable limit of function $ \lim_{x\to 1}\frac {x^a \minus{} 1}{x \minus{} 1} \equal{} a\ ,\ a > 0$ obtain $ a_n\rightarrow \frac {5}{14}\ .$

( [color=darkred]Similarly with [b]Carcul[/b][/color] )

isomorphism: yes, that is $ \sin\pi,$ which of course means that it is zero.

c) is $ \arctan(\frac13),$ which is why it is $ \frac {\pi}6.$

[color=green][Edit: Typo. That should be $ \arctan\left(\frac1{\sqrt{3}}\right).$[/color]

[quote="Kent Merryfield"]isomorphism: yes, that is $ \sin\pi,$ which of course means that it is zero.

c) is $ \arctan(\frac13),$ which is why it is $ \frac {\pi}6.$[/quote]

Yes I did that :D

First two are directly definitions and for the third one
I noticed it is of the form $ \int ({1 \minus{} x^2 \plus{} x^4 \minus{} x^6 \plus{}.....}) \ dx \equal{} \int \frac1{1\plus{}x^2} \ dx \equal{} \tan^{\minus{}1} x$ where $ x \equal{} \frac{1}{\sqrt 3}$. If $ \tan^{\minus{}1} x$ series was popular, I would have directly got it. :(

Which year students are you teaching this? ( Is it some engineering math course?)

[quote="Kent Merryfield"]It's a second-semester calculus course, taken by engineering/science/math majors, and it's the equivalent of the AP Calculus BC exam. Most of my students were freshmen who had AP Calculus AB credit. And I gave them [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?search_id=877993624&t=145324]this[/url] as a class handout and told them to memorize it.[/quote]

Ya thought so :)
I studied these things in my first or second semester(I am an engineering student). We were almost always given really tricky questions and so it was fun to have these classes. Do you have some exam questions posted here?
I would like to try them. Haven't solved nice Calculus problems for a couple of years now.

Thanks a lot :D


Ive got this DE: 8(du/dx)+4(du/dy)=9u   anyone could help me?
I separate it with u=X(x)Y(y)
then I get: 8( X'(x)/X(x) =-4 ( X(x)Y'(y)) +9X(x)Y(y)
If it would be just:  8( X'(x)/X(x) =-4 ( X(x)Y'(y)) I could write = constant, but with that 9X(x)Y(y)     how to solve?

Then I got conditions: u(0,y)=33*e^-y +50*e^4y

Please help me solve this.

[quote="DanielM99"]I've got this DE: 8(du/dx)+4(du/dy)=9u   anyone could help me?
I separate it with u=X(x)Y(y)
then I get: 8( X'(x)/X(x) =-4 ( X(x)Y'(y)) +9X(x)Y(y)
[/quote]
The correct formula is  $8\frac{X'(x)}{X(x)}=-4\frac{Y'(y)}{Y(y)}+9$ (you divide by $X(x)Y(y)$, right?). Now it should be clear what to do next.


quadratics
geometry
geometric transformation
reflection
algebra unsolved
algebra

A monic quadratic trinomial with integer coefficients assumes prime values at three successive integer points. Prove that it assumes a prime value in at least one other integer point.

Without loss of generality, $ P(\minus{}1), P(0), P(1)$ are prime.  But since they are distinct $ \bmod 3$ one must be equal to $ 3$.  And either one or two of them are even, hence either one or two are equal to $ 2$.

[b]Case:[/b]  $ P(\minus{}1) \equal{} 2, P(0) \equal{} 3, P(1) \equal{} 2$.  Then $ P(x) \equal{} 3 \minus{} x^2$, which is not monic.

[b]Case:[/b]  $ P(\minus{}1) \equal{} 3, P(0) \equal{} 2$ or $ P(0) \equal{} 2, P(1) \equal{} 3$ (actually the same case).  Then $ P(x) \equal{} x^2 \plus{} 2$, which assumes many other prime values such as $ P(3) \equal{} 11$.

Something seems wrong here. Take $ P(x) \equal{} x^2 \plus{} x \plus{} 5$.
$ P(\minus{}1) \equal{} 5$
$ P(0) \equal{} 5$
$ P(1) \equal{} 7$
$ P(2) \equal{} 11$
$ P(3) \equal{} 17$

Doesn't violate the conditions of the problem, of course, but... where did those assertions about mods come from?

Whoops.  Not sure what happened there.   :oops:  Based off of that example, we could let $ P(x) \equal{} x^2 \plus{} x \plus{} p$ where $ (p, p \plus{} 2)$ is a twin prime pair, which is a few more solutions than I'd anticipated... does the OP intend "distinct primes"?

Edit:

[b]Case:[/b]  The linear coefficient is even.  $ P(x)$ therefore alternates parities, so it takes on the value $ 2$.  Again WLOG either $ P(\minus{}1) \equal{} P(1) \equal{} 2$ (which gives $ P(x) \equal{} x^2 \plus{} 1$, which violates the problem conditions) or $ P(0) \equal{} 2$ (which gives $ P(x) \equal{} x^2 \plus{} 2kx \plus{} 2, P(1) \equal{} 2k \plus{} 3, P(\minus{}1) \equal{} 3 \minus{} 2k$ and the lone solution $ k \equal{} 0$).

[b]Case:[/b]  The linear coefficient is odd.  Then there exists an odd integer $ a$ such that $ P(x) \equal{} P(a \minus{} x)$ for any $ x$.  Given three consecutive prime values, reflect them to get a fourth.  In your example, $ a \equal{} \minus{}1$ and $ P(\minus{}2) \equal{} 7$.


real analysis
vector
real analysis unsolved

Hi all

I am working on this problem from Rudin. It asks to classify all complex measure on the unit circle with periodic Fourier coefficients. I know that such a measure must be singular with respect to Lebesgue measure by the Rie-Lesb. lemma and Radon-Nikodym. I also know that any finite combination of point mass measures on rational points works. I suspect that these are all such measures, but I don't know how to go about proving it. I would greatly appreciate it if anyone could give me a hint on how to proceed, as I have been stuck on this for a while now, but please just give a hint or hide your solution as this is a homework problem I want to solve myself.

Thanks

[hide="Hint."]Fix $ N,$ think finite-dimensional.[/hide]

[hide="Solution, mostly."]Fix $ N$ and consider all possible sequences that are periodic of period $ N.$ (We're considering them as possible Fourier transforms of measures.) The set of all such sequences is an $ N$-dimensional vector space. The set of all linear combinations of point masses at points of the form $ \frac {k}{N}$ times the length of your fundamental interval is also an $ N$-dimensional vector space. The Fourier transform maps that set of measures onto that set of sequences - in fact, this is a well-known invertible linear operator, the discrete Fourier transform.

Given that, what you want follows from uniqueness. You need to know that if $ \widehat{\mu}(k) \equal{} \widehat{\nu}(k)$ for all $ k,$ then $ \mu \equal{} \nu$ as measures. Have you proved that?[/hide]


Prove that for any positif integer $ n$, there exists a prime number $ p$ such that the sum of its digits is greater than $ n$ .

I think we can prove that there are too few numbers with digit sum $ \leq n$ so that as we go to infinity, there would be a contradiction to the prime number theorem (or more elementary estimates if you prefer)

Dirchlets theorem guarantees that there are infinitely many primes ending in any string of digits that does not end in a 2 or a 5. The result immediately follows.

Cheers,

Rofler


what is the minimum of this:

$\frac{(X^2+Y+1)(Z+1)(9Y+Z)}{XYZ}$

i got ;$2{}^2{}^1$

i mean maximum is...


ceiling function
inequalities proposed
inequalities

Assume $x_{1},x_{2},\dots,x_{n}\in\mathbb R^{+}$, $\sum_{i=1}^{n}x_{i}^{2}=n$, $\sum_{i=1}^{n}x_{i}\geq s>0$ and $0\leq\lambda\leq1$. Prove that at least $\left\lceil\frac{s^{2}(1-\lambda)^{2}}n\right\rceil$ of these numbers are larger than $\frac{\lambda s}{n}$.

It's strange because I got the result:
$\left\lceil \frac{s^{2}(1-\lambda)^{2}}{n+n\alpha^{2}-2s\alpha}\right\rceil$
where $\alpha=\frac{s\lambda}{n}$
 :maybe:

This has been sitting without a solution for more than a decade. Here is a probabilistic proof. I'll use Cantelli's inequality:
[b]Thm (Cantelli):[/b] Let $X$ be a random variable, with standard deviation $\sigma$. Then, for any $t<0$:
$$
\mathbb{P}(X-\mathbb{E}[X]>t)\geqslant 1-\frac{\sigma^2}{\sigma^2+t^2}.
$$
Now, equipped with this, let $X$ be a random variable uniformly distributed on $S=\{x_1,x_2,\dots,x_n\}$, that is, $\mathbb{P}(X=x_i)=1/n$ for each $i $. Note that, $\mathbb{E}[X^2]=1$, and $\mathbb{E}[X]=s/n$; and therefore, $\sigma^2 = 1-\frac{s^2}{n^2}$, where $\sigma$ is the standard deviation of $X$. With this, observe the following chain:
$$
\mathbb{P}(X\geqslant \frac{\lambda s}{n}) = \mathbb{P}(X\geqslant \lambda \mathbb{E}[X]) = \mathbb{P}(X-\mathbb{E}[X]\geqslant  (\lambda-1)\mathbb{E}[X]) \geqslant \frac{(1-\lambda)^2\mathbb{E}[X]^2}{(1-\lambda)^2\mathbb{E}[X]^2+\sigma^2}.
$$
Now, inserting $\mathbb{E}[X]=s/n$, and $\sigma^2 = 1-s^2/n^2$, we have:
$$
\mathbb{P}(X\geqslant \frac{\lambda s}{n}) \geqslant \frac{(1-\lambda)^2 s^2}{(1-\lambda)^2s^2 + n^2-s^2}>\frac{(1-\lambda)^2s^2}{n^2}.
$$
Thus, the 'number' of $x_i$'s above this level is nothing but $n$ times the lower bound (due to the fact that the probability mass is $1/n$), that is, there is at least $>\frac{(1-\lambda)^2s^2}{n}$ values of $x_i$ with the desired condition, from which the ceil is obtained.


what are the hyperbolic trigonometric functions used for? Anyone can directly see the use of basic trigonometric functions in geometry problems but i can't seem to figure out what the hyperbolic trigonometric functions are used for.

See http://en.wikipedia.org/wiki/Sinh for more on hyperbolic functions.

Look at the picture in the article JRav linked to see how they can be interpreted geometrically.

I know they come up a lot in calculus because they are linked to the exponential function. Writing functions in terms of the hyperbolic trigonometric functions can make them easier to deal with, because it is then easy to apply identities analogous to the circular trigonometric identities to simplify them. See [url=http://en.wikipedia.org/wiki/Osborne%27s_rule#Similarities_to_circular_trigonometric_functions]here[/url].


trigonometry
function
limit
real analysis
real analysis solved

[b]Find , with at least four decimals. the roots of equation
\[ \tanh{x}=\tan{x}\; \; \; ,\; \; x\in \left(\pi,\frac{5\pi}{4}\right)\; .  \]
[/b]
 Proposed by  Heinrich D\"orrie  in ,, Praktische Algebra", Verlag von R. Oldenburg ,M\"unchen, 1953, page 193.

This seems like a rather trivial numerical exercise. Iterating the function $x_{n+1}=\tan^{-1}(tanh(x_n+\pi))$, (starting at 1) I found that $\lim x_n = .7850$, so the desired solution is $x=3.9266023$. Uniqueness in the range $\pi<x<\frac54\pi$ is obvious.

Of course, I used a spreadsheet to get this; the calculation would have been more interesting in 1953.


A triangle ABC has an area of 20 square centimeters. We extend the side AC to obtain a point M such that A is between M and C and AM=(1/2) AC . We extend the side AB to obtain a point N such that B is between N and A and BN=(1/2)AB . We extend the side BC to obtain a point P such that C is between P and B and CP=(1/2)BC . How many square centimeters are in the area of the triangle MNP ?

$\frac{S_{ACP}}{S_{ABC}}=\frac{CP}{BC}=\frac{1}{2}\Longrightarrow S_{ACP}=10$,
$\frac{S_{MAP}}{S_{ACP}}=\frac{MA}{AC}=\frac{1}{2}\Longrightarrow S_{MAP}=5$ ,
so $S_{MCP}=15$. by the same way we have $S_{AMN}=S_{NBP}=15$, thus: $S_{MNP}=3\times 15+20=65$. :)


For a finite set X of integers, let |X| denote the cardinality of X, and let X - X denote {x-x'| x, x' from X}. Show that if A, B are subsets of {1, 2, ..., n} with |A||B| >= 2n-1, then (A-A)  \cap (B-B) contains a positive element. Here n>1.

This is very nice prb. I dont have the solution.

[i]AMM S5 by R.L.Graham[/i]

Consider the pairs (x,y) with x in A and y in B. There are |A||B|>=2n-1 such pairs. But the sum x+y of each of these pairs takes on values from 2 to 2n.

If both 2 and 2n are taken, then there exist the pairs (1,1) and (n,n) so n-1 belongs to (A-A) \cap (B-B).

If not both 2 and 2n are taken, then the sum x+y of the pairs (x,y) takes on at most 2n-2 values. Because there are at least 2n-1 pairs, there exist two pairs (x1,y1) and (x2,y2) such that x1+y1=x2+y2, so a=|x1-x2|=|y1-y2| belongs to (A-A) \cap (B-B).

Merry Christmas!

Very nice solution. Thank you Andrei. Merry Christmas and s novym godom!

Namdung


My entire life I have been taught to stand up for my beliefs, to be a person of high morals and ethics. That's why I feel obligated to discuss the advantages of two-parent families, the essential role of individual and family responsibility, the need for uniform standards of civil behavior, and the primacy of the work ethic. Before I begin, let me point out that were he alive today, Hideki Tojo would be Mr. Bubka's most trustworthy ally. I can see Tojo joining forces with Bubka to help him sully a profession that's already held in low esteem. Above all, his proposed social programs are nothing short of flippant. He is differentiated from your average malodorous mountebank by virtue of the fact that he wants to spread hatred, animosity, and divisiveness. This moral issue will eventually be rendered academic by the fact that Bubka intends to create a new social class. Self-pitying fiends, disgraceful, unholy pissants, and blasphemous, hectoring varmints will be given aristocratic status. The rest of us will be forced into serving as their cohorts. He is not as randy or saturnine as you might think. He's more so.

I plan to work within the system to persuade my fellow citizens that Bubka is essentially describing a situation that does not exist, not because I lack the courage for more drastic steps, but because Bubka is always trying to change the way we work. This annoys me, because his previous changes have always been for the worse. I'm positive that Bubka's new changes will be even more illiberal, because he wants to harvest what others have sown. Faugh. Although the proper definition of "ultramicrochemistry" is hotly disputed, as our society continues to unravel, more and more people will be grasping for straws, grasping for something to hold onto, grasping for something that promises to give them the sense of security and certainty that they so desperately need. These are the types of people Bubka preys upon. A small child really couldn't understand that his attendants are just as bad as he is, if not worse. But any adult can easily grasp that his claims are a load of claptrap. Regular readers of my letters probably take that for granted, but if I am to punish him for his amateurish threats, I must explain to the population at large that when he says that my bitterness at him is merely the latent projection of libidinal energy stemming from self-induced anguish, that's just a load of spucatum tauri.

I mean, really. Bubka thinks that doing the fashionable thing is more important than life or liberty. Of course, thinking so doesn't make it so. I've always thought that I must defend my honor, and hearing the rubbish that he spews forth proves it beyond all doubt. This is well illustrated in what remains one of the most divisive issues of our day: elitism. Before I leave this issue, let me share an interesting finding from a recent poll: Four out of five people surveyed maintain that Bubka teaches workshops on onanism. Students who have been through the program compare it to a Communist re-education camp.

Bubka is like a magician who produces a dove in one hand, while the other hand is busy trying to condone universal oppression. If my own experience has taught me anything, it's that he holds onto power like the eunuch mandarins of the Forbidden City -- sterile obstacles to progress who leave behind a legacy of perpetual indebtedness in developing countries. I recently overheard a couple of primitive yobbos say that Bubka is a martyr for freedom and a victim of escapism. Here, again, we encounter the blurred thinking that is characteristic of this Bubka-induced era of slogans and propaganda. While some of his tractates are very attractive on the surface and are undoubtedly entertaining, they ultimately serve to give me reason to run around like a chicken with its head cut off. This is not Nazi Germany or Soviet Russia, where the state would be eager to scupper my initiative to take up the all-encompassing challenge of freedom, justice, equality, and the pursuit of life with full dignity. Not yet, at least. But he insists that coercion in the name of liberty is a valid use of state power. How can he be so blind? Very easily. Basically, if anything, Bubka truly believes that the few of us who complain regularly about his expedients are simply spoiling the party. It is just such peremptory megalomania, invidious egoism, and intellectual aberrancy that stirs Bubka to develop a credible pretext to forcibly silence his opponents. It's mudslinging hooligans like Bubka that lead to the destruction of the human race. How much more illumination does that fact need before Bubka can grasp it? Assuming the answer is "a substantial amount", let me point out that if Bubka honestly believes that some of my points are not valid, I would love to get some specific feedback from him.

Bubka wants us to believe that he commands an army of robots that live in the hollow center of the earth and produce earthquakes whenever they feel like shaking things up a bit on the surface. How stupid does he think we are? There aren't enough hours in the day to fully answer that question, but consider this: All Bubka really wants is to hang onto the perks he's getting from the system. That's all he really cares about. Despite some perceptions to the contrary, he is inherently gin-swilling, pusillanimous, and crotchety. Oh, and he also has a bleeding-heart mode of existence. Bubka's faculty for deception is so far above anyone else's, it really must be considered different in kind as well as in degree. While the concept of broad-based peace and social justice coalitions remains desirable, Bubka seeks scapegoats for his own shortcomings by blaming the easiest target he can find, that is, fickle spivs. His most recent deeds are irreverent, in bad taste, and inappropriate. No wonder that if we don't soon tell him to stop what he's doing, he will proceed with his inane imprecations, considerably emboldened by our lack of resistance. We will have tacitly given him our permission to do so.

If Bubka can give us all a succinct and infallible argument proving that he is the one who will lead us to our great shining future, I will personally deliver his Nobel Prize for Spiteful Rhetoric. In the meantime, I do not propose a supernatural solution to the problems we're having with Bubka. Instead, I propose a practical, realistic, down-to-earth approach that requires only that I stop this insanity. Confuse, disorient, and disunify if you like, Bubka, because I simply don't care. He's like the man behind the curtain in the Wizard of Oz. Pull back the curtain of chauvinism and you'll see a villainous Philistine hiding behind it, furiously pulling the levers of Trotskyism in a vicious attempt to create a Frankenstein's monster. That sort of discovery should make any sane person realize that for Bubka's belligerent plans to succeed, he needs to "dumb down" our society. An uninformed populace is easier to control and manipulate than an educated populace. Before long, schoolchildren will stop being required to learn the meanings of words like "physicophilosophical" and "superphlogistication". They will be incapable of comprehending that if we let Bubka cause one-sided personal attacks to be entered into historical fact, all we'll have to look forward to in the future is a public realm devoid of culture and a narrow and routinized professional life untouched by the highest creations of civilization. Though I don't doubt the depth of Bubka's sentiments, it's rather the form of his expressions that I find both sententious and capricious.

Bubka not only lies, but he brags about his lying to his lackeys. His secret agents' thinking is fenced in by many constraints. Their minds are not free because they dare not be.

Forgive me if I ramble; I'm really upset, as I think you can tell. A great many of us don't want Bubka to descend to character assassination and name calling. But we feel a prodigious societal pressure to smile, to be nice, and not to object to his unrealistic, snotty editorials. He says that honesty and responsibility have no cash value and are therefore worthless. This is at best wrong. At worst, it is a lie. I once managed to get Bubka to agree that he is not only woefully mean-spirited, but terribly mealymouthed. Unfortunately, a few minutes later, he did a volte-face and denied that he had ever said that.

Bubka is obviously trying to make widespread accusations and insinuations without having the facts to back them up, and unless we act now, he'll unquestionably succeed. His goal is to use scapegoating as a foil to draw anger away from more accurate targets. How lame-brained is that? How abusive? How passive-aggressive?

If Bubka thinks his lamentations represent progress, he should rethink his definition of progress. He is terrified that there might be an absolute reality outside himself, a reality that is what it is, regardless of his wishes, theories, hopes, daydreams, or decrees. Bubka's intimations are more than just truculent. They're a revolt against nature. Bubka knows how to lie. It's too bad he doesn't yet understand the ramifications of lying. Okay, I've written enough for one letter, so let me just finish by saying that the rambunctious aspect of Mr. Bubka's put-downs will create a stir between inaniloquent hostes generis humani and the pouty public at large.

I was reading to see if you had anything of interest or value, and this sentence pretty much guarantees that you didn't write that post by yourself.

[quote="ShoeFactory"] ...Regular readers of my letters probably take that for granted...[/quote]

Seriously, though, some people in this forum have something against bubka... I can't understand why.

He's one of the smarter members who posts in G&FF, and though you could argue that he spams, many members spam in this forum, and no offense meant to anyone, but there are plenty of members I can name for whom I'd rather have bubka posting instead of them... [size=9][color=white]grammar might be a bit incorrect in that last sentence[/color][/size]

You found the online complaint generator.

Congratulations.


Find the largest positive real $ k$, such that for any positive reals $ a,b,c,d$, there is always:
\[ (a\plus{}b\plus{}c) \left[ 3^4(a\plus{}b\plus{}c\plus{}d)^5 \plus{} 2^4(a\plus{}b\plus{}c\plus{}2d)^5 \right] \geq kabcd^3\]

[quote="Altheman"]Taking $ a\equal{}b\equal{}c\equal{}\frac{1}{3}$ and $ d\equal{}1$, we get $ k\le k_{max}$. Now lets prove this.

Wlog $ a\plus{}b\plus{}c\equal{}1$. By AM-GM,

$ 3^4(1\plus{}d)^5\plus{}2^4(1\plus{}2d)^5\ge 3^4(2\sqrt{d})^5\plus{}2^4(3\sqrt[3]{d^2})^5$

$ \equal{}6^4\cdot 5\cdot \left( \frac{2}{5}\cdot d^{\frac{5}{2}}\plus{}\frac{3}{5}\cdot d^{\frac{10}{3}}\right)\ge 6^4\cdot 5\cdot d^3\ge 6^4\cdot 5\cdot \frac{abc}{27} d^3$

So $ k\equal{}\frac{6^4\cdot 5}{27}$[/quote]

Is this right???

at last step by AM-GM
$ 6^4\cdot 5 \cdot\ 27 ?? $

[quote=Altheman]Taking $ a\equal{}b\equal{}c\equal{}\frac{1}{3}$ and $ d\equal{}1$, we get $ k\le k_{max}$. Now lets prove this.

Wlog $ a\plus{}b\plus{}c\equal{}1$. By AM-GM,

$ 3^4(1\plus{}d)^5\plus{}2^4(1\plus{}2d)^5\ge 3^4(2\sqrt{d})^5\plus{}2^4(3\sqrt[3]{d^2})^5$

$ \equal{}6^4\cdot 5\cdot \left( \frac{2}{5}\cdot d^{\frac{5}{2}}\plus{}\frac{3}{5}\cdot d^{\frac{10}{3}}\right)\ge 6^4\cdot 5\cdot d^3\ge 6^4\cdot 5\cdot \frac{abc}{27} d^3$

So $ k\equal{}\frac{6^4\cdot 5}{27}$[/quote]

I don't understand where this addressed that $\frac{6^4\cdot5}{27}$ is the maximum? Can you explain more Altheman?

The proof is composed into two parts - proving that $k_{max}$ is indeed the maximum, and proving that $k_{max}$ satisfies the inequality.

The first line shows that by plugging $a=b=c=\frac{1}{3}$ and $d=1$, we get $k \le \frac{6^4 \cdot 5}{27}$.
The second part shows that $k=\frac{6^4 \cdot 5}{27}$ satisfies the inequality.

Therefore, we can conclude that $k_{max}$ is equal to $\frac{6^4 \cdot 5}{27}$, as desired.

How do you even approach these types of questions where you have to sort of "guess" the inequality you want to prove? Like, how would you know to plug in $(a,b,c,d)=(\frac 1 3, \frac 1 3, \frac 1 3, 1)$?


Assuming that there is always a prime between n  and 2n-7  for n>9 , prove that, with the exception of 1, 4 and 6, every natural number can be written as the sum of distinct primes. (Hint : Use induction.)


How are you supposed to pronounce Schleiden and Schwann ??  { The people who proposed the cell theory.......} By the way , I've seen varied spellings in various places , so I was wondering as to which one is the right spelling.....  :)

well, if you spelled it correctly, you'd say Schleiden "shhhh-leye(as in "eye"ball with the long i sound)-den"

and Schwann "Shhh-juan" like telling Juan to be quiet.

Generally, if you visit dictionary.com you can hear pronunciations of the looked up word.
[quote]"When you perform a word search on Dictionary.com, you'll be returned results for your query from a number of dictionary sources including The American Heritage Dictionary and the Dictionary.com Unabridged. Both of these dictionaries contain many audio pronunciation files designated by this audio pronunciation icon:"[/quote]
Look for the image of a speaker(should be right next to the word you look up) and click on in it for pronunciation.

Here are the pages on Schwann and Schleiden:
[url]http://dictionary.reference.com/browse/Schleiden[/url]
[url]http://dictionary.reference.com/browse/Schwann[/url]

-arb95


Three friends, Ralph, Emerson and Waldo, each select a number from the set
{1,2,3,11,12,13,21,22,23} and remove it. Then they add their three numbers together.
If they put their numbers back and repeat this process for all possible combinations,
how many different sums can they get?

There are 84 total combinations, but I can't figure out the number of repeats of the sums.
Thank you!  :D

9C3= 84 right.
9*8*7/6=84
But the only problem is that there is some overcounting; so are u sure that the answer is 84.

He knows that, he's asking for how to count the overcount.

I meant that the total number combinations is 84. Some of the sums are repeated, so I was asking if anyone knew how to find the repeats.

[hide]The answer is 29, if anyone wanted to know.  :D [/hide]


limit
trigonometry
calculus
integration
calculus computations

i think integration would be helpful here but i can't really put it on paper...

No, integration would not be helpful.

The real question: exactly how close is that quantity inside the cosine to $ \pi n?$

Is this limit $ \frac {\pi}{3}$ ?

For Svejk's.

Yes, that is correct. Now, what does that tell you about the original limit?

Answer is 0.25 , you can use asynptotic formula!

The asymptotic formula ?

I just plugged $ x \equal{} \frac{\pi}{3}$ into $ \cos^2(x)$ and got $ \frac{1}{4}$. What asymptotic formula? :)

[quote="Svejk"][hide="hint"]Use the formula $ \cos(x \plus{} k\pi) \equal{} ( \minus{} 1)^k\cos x ,\forall x\in\mathbb{R} \text{ and } \forall k\in\mathbb{N}$.Therefore

$ \lim_{n\to\infty}\cos^2\pi(\sqrt [3]{n^3 \plus{} n^2 \plus{} 2n}) \equal{} \lim_{n\to\infty}\cos^2\pi(\sqrt [3]{n^3 \plus{} n^2 \plus{} 2n} \minus{} n)$.

Now, can you find $ \lim_{n\to\infty} (\sqrt [3]{n^3 \plus{} n^2 \plus{} 2n} \minus{} n)\pi$ ?  :wink: [/hide][/quote]
yeh,
$ \lim_{n\to\infty}cos^2 (\sqrt [3]{n^3 \plus{} n^2 \plus{} 2n} \minus{} n)\pi \equal{} \lim_{n\to\infty} cos^2 \pi \left( \frac{n^2\plus{}2n}{\sqrt [3]{(n^3 \plus{} n^2 \plus{} 2n)^2}\plus{}n\sqrt[3]{n^3\plus{}n^2\plus{}2n} \plus{}n^2} \right) \equal{}cos^2 \frac{\pi}{3} \equal{}\frac{1}{4}$


[quote="illmaticman"]Hi I am currently a grade 12 student, and am wondering if there are any good books that help with proofs, and have alot of problems involving them. I would like this to prepare for Geometry & Discrete math which i have next semester, thanks.[/quote]
Please don't post the same topic twice. Thanks.


geometry
3D geometry
function
logarithms
integration
calculus
calculus computations

Hello, i am having some difficulties solving the following problems using newton's method:

1)What does the sequence defined by
$ x_0\equal{}1,\;\;\;\;\;x_{k\plus{}1}\equal{}\frac{1}{2}x_k\plus{}\frac{1}{x_k}$
converge to?
and

2)Devise a subroutine using only simple operations that finds, using newton's method, the cube root of some input number z.

Do u have any ideas?
Thanks

i only need some help to start..i know how to use newton to find roots of  equations, but i haven't seen such problems before...

If I say that "$ a$ is a cube root of $ z$," what's the corresponding equation?

[quote="georgech"]Thanks, i have done the second question. But i am not sure how to proceed with the first one, which needs the convgence of the sequence. How can i do something like tha without using matlab? thanks[/quote]

[hide]Recall that Newton's method for finding roots of a function f(x) is given by

$ x_{k \plus{} 1} \equal{} x_k \minus{} \frac {f(x_k)}{f'(x_k)}$.

Hence we must solve a differential equation to find the function f.  We must have

$ x_k \minus{} \frac {f(x_k)}{f'(x_k)} \equal{} \frac {x_k}{2} \plus{} \frac {1}{x_k}$,

or

$ \frac {f(x_k)}{f'(x_k)} \equal{} \frac {x_k}{2} \minus{} \frac {1}{x_k} \equal{} \frac {x_k^2 \minus{} 2}{2x_k}$.

Now we use the fact that $ \frac {f'(x)}{f(x)} \equal{} \frac {d}{dx}\left[\log f(x) \right]$:

$ \frac {d}{dx_k}\left[ \log f(x_k) \right] \equal{} \frac {f'(x_k)}{f(x_k)} \equal{} \frac {2x_k}{x_k^2 \minus{} 2}.$

Integrating,

$ \log f(x_k) \equal{} \int \frac {2x_k}{x_k^2 \minus{} 2} \, dx_k \equal{} \log |x_k^2 \minus{} 2|$

$ f(x_k) \equal{} C(x_k^2 \minus{} 2)$

where C is a constant of integration.  Since the roots of $ f(x) \equal{} C(x^2 \minus{} 2)$ are the same for all nonzero values of C we know that the given recursion relationship, if it converges, must converge to either $ \sqrt {2}$ or $ \minus{} \sqrt {2}$.  But given $ x_0 \equal{} 1$, it is obvious that $ x_k > 0$ for all $ k$, and therefore the sequence converges* to $ \sqrt {2}$.

*Note.  The only thing that remains is to show that the sequence is in fact convergent, which I leave as an exercise to the reader.[/hide]

That was a pretty impressive reverse-engineering of the question, atomicwedgie.  The natural way to do that question has nothing to do with Newton's method: if $ x_k \to L$ then $ x_{k \plus{} 1} \to L$ and so $ L \equal{} \frac {1}{2}L \plus{} \frac {1}{L}$.

Think about graph $ y \equal{} \frac {1}{2}x \plus{} \frac {1}{x}$ and $ y \equal{} x$. The limit we want to find is the fixed point.

This is one well-known method, but I don't know this is the Newton's method. The book [Mathematical Methods for Physicists] by Arfken will help.


In the heptagon ABDFGEC, the difference between each successive angle is constant, and the difference between the largest angle and the smallest is $ 54^ \circ$. What are the measures of the 7 angles associated with heptagon ABCFGEC?

I didn't include in image because it seemed irrelevent. (It's obviously not a regular heptagon).

[hide]Let $ x^\circ$ be the measure of the smallest angle and $ d^\circ$ be the difference between each successive angle.  The measure of the 2nd smallest angle is $ (x \plus{} d)^\circ$.  The next after that is $ ((x \plus{} d) \plus{} d)^\circ$, or $ (x \plus{} 2d)^\circ$.  The measures of the remaining angles are $ (x \plus{} 3d)^\circ, (x \plus{} 4d)^\circ, (x \plus{} 5d)^\circ,$ and $ (x \plus{} 6d)^\circ$.

The difference between the largest and smallest angles is $ 54^\circ$.
So, $ (x \plus{} 6d) \minus{} x \equal{} 54 \Rightarrow 6d \equal{} 54 \Rightarrow d \equal{} 9$.

The sum of the interior angles of a polygon is $ (n \minus{} 2)180^\circ$, where $ n$ is the number of sides of the polygon.  So, $ (7 \minus{} 2)180^\circ \equal{} 900^\circ$.

Adding all the interior angles together, you get $ (7x \plus{} 21d)^\circ$.
$ 7x \plus{} 21d \equal{} 900 \Rightarrow 7x \plus{} 21(9) \equal{} 900 \Rightarrow x \equal{} \frac {711}{7}$.

Thus, the measures of the 7 angles are $ \frac {711}{7}^\circ, \frac {774}{7}^\circ, \frac {837}{7}^\circ, \frac {900}{7}^\circ, \frac {963}{7}^\circ, \frac {1026}{7}^\circ,$ and $ \frac {1089}{7}^\circ$.
[/hide]


I'm not sure if I get what you're saying so tell me if I'm right and then I will explain [hide="my answer"]$5, \frac{9}{5},1$[/hide]

I edited the post, for better understanding.

Now, how did you solve the problem?

[hide="solution"]
for
$m=1$
the original equation become

$x^{3}-4x^{2}+4x=0$, so $x=0$ is a root.

then $x^{2}-4x+4=0$ , $x=2$.

once that ${0,2}\in \mathbb{N}$, $m=1$ is true.

Now, the question is: Is there only one possible value for m, that satisfy the condition?

[/hide]

yeah but, the idea is not estimate any value.

I mean, how can I find the value of $m$ algebraicly

[quote="Line"]the idea is not estimate any value.

I mean, how can I find the value of $m$ algebraicly[/quote]

like I said, do not estimate.

how can I find an algebraicly way to solve it?

[quote="nutz_for2.718281828"]Here's a neat trick:

[hide="solution"]We can sum the coeficciants:

1+m-5+9-5m+6m-6=-1+2m

When that equals 0, 1 will be a root. So m=1/2 is a good solution.

When x=2:

8+4m-20+18-10m+6m-6=0

So, no matter what m is, 2 will be a root.[/hide]

Nice proof, eh?[/quote]

I don't think m=1/2 is a good solution because we have

$x^{3}-\frac{9}{2}x^{2}+\frac{13}{2}x-3=0$

so $x=1,2$ are root, but the last root is $\frac{3}{2}$ and $\frac{3}{2}\notin \mathbb{N}$

Here is my solution.
[hide] Call the roots a,b, and c.  Since $(x-a)(x-b)(x-c)=x^{3}-x^{2}(a+b+c)+x(ab+bc+ac)-abc$, we have that $a+b+c=5-m$, $ab+bc+ac=9-5m$, and $abc=-6(m-1)=6-6m.$  Now, since $a, b,$ and $c$ are natural, we have that $a,b,$ and $c$ are integers and are greater than or equal to one.  Since the sum of integers is an integer, we have that $a+b+c=5-m$ is an integer.  Therefore, $5-m$ is an integer, meaning that $m$ is an integer.  Now since abc must be greater than or equal to 1, we have that $6-6m$ is greater than or equal to 1.  Therefore, m is less than or equal to 5/6.  Therefore, we have that m must be 0 or below, so m is a nonpositive, integral number.  Furthermore, we can add nutz_for2.718281828's idea that $x=2$ is always a solution.  Dividing $x^{3}+x^{2}(m-5)-x(9-5m)+6(m-1)$ by x-2, we get $x^{2}+x(m-3)-3(m-1)$.  This implies that $x=((3-m)+-\sqrt{(m-3)^{2}+12m-12})/2=((3-m)+-\sqrt{m^{2}+6m-3})/2.$  Since the discriminant of this value must a perfect square (because $x$ is an integer), we can say that $m^{2}+6m-3=x^{2}$.  Therefore, $m^{2}+6m-3-x^{2}=0.$  Therefore, $m=(-6+-\sqrt{36+12+4x^{2}})/2=-3+-\sqrt{12+x^{2}}$.  We can say the discriminant of this is also a perfect square because m is an integer.  Therefore, $x^{2}+12$ is also perfect square.  Say $x^{2}+12=y^{2}$.  This implies $y^{2}-x^{2}=(x+y)(x=y)=12$.  Since x and y are integers, we can take all of the integral solutions to this.  Finally, we can conclude that the only integral solution is that x must be 2 and y must be 4.  This implies $m^{2}+6m-3=4$, so $m^{2}+6m-7=0$.  This means that m=-7 or m=1.  We know that m is negative, so $m=-7$[/hide]


calculus
integration
trigonometry
logarithms
calculus computations

Calculate the integral
\[ I \equal{} \int_{0}^{\frac {\pi}{2}} ln(1 \plus{} tanx)dx
\]

Put $ t\equal{}\tan x$, then $ dx\equal{}\frac{dt}{1\plus{}t^2}$, and so our integral will be:

\[ \int_0^{\infty}\frac{\ln(1\plus{}t)}{t^2\plus{}1}dt\]

Now define:

\[ I(\lambda)\equal{}\int_0^{\infty}\frac{\ln(1\plus{}\lambda t)}{1\plus{}t^2}\]

Differentiating under the integral sign (of course we can...):

\[ \frac{dI(\lambda)}{d\lambda}\equal{}\int_0^{\infty}\frac{t}{(1\plus{}\lambda t)(1\plus{}t^2)}dt\equal{}\frac{\pi}{2}\frac{\lambda}{1\plus{}\lambda^2}\minus{}\frac{\ln\lambda}{1\plus{}\lambda^2}\]

Now we can see that $ I(\lambda\equal{}0)\equal{}0$, hence our original integral is:

\[ I(\lambda)\equal{}\frac{\pi}{2}\int_0^1\frac{\lambda}{1\plus{}\lambda^2}d\lambda\minus{}\int_0^1\frac{\ln\lambda}{1\plus{}\lambda^2}d\lambda\equal{}\frac{\pi\ln2}{4}\plus{}G\]

Where $ G$ is Catalan's constant.

So our original integral is:

\[ \int_0^{\frac{\pi}{2}}\ln(1\plus{}\tan x)\,dx\equal{}\frac{\pi\ln2}{4}\plus{}G\]

[quote="yeahnigga"]Put $ t \equal{} \tan x$, then $ dx \equal{} \frac {dt}{1 \plus{} t^2}$, and so our integral will be:
\[ \int_0^{\infty}\frac {\ln(1 \plus{} t)}{t^2 \plus{} 1}dt
\]
Now define:
\[ I(\lambda) \equal{} \int_0^{\infty}\frac {\ln(1 \plus{} \lambda t)}{1 \plus{} t^2}
\]
Differentiating under the integral sign (of course we can...):
\[ \frac {dI(\lambda)}{d\lambda} \equal{} \int_0^{\infty}\frac {t}{(1 \plus{} \lambda t)(1 \plus{} t^2)}dt \equal{} \frac {\pi}{2}\frac {\lambda}{1 \plus{} \lambda^2} \minus{} \frac {\ln\lambda}{1 \plus{} \lambda^2}
\]
Now we can see that $ I(\lambda \equal{} 0) \equal{} 0$, hence our original integral is:
\[ I(\lambda) \equal{} \frac {\pi}{2}\int_0^1\frac {\lambda}{1 \plus{} \lambda^2}d\lambda \minus{} \int_0^1\frac {\ln\lambda}{1 \plus{} \lambda^2}d\lambda \equal{} \frac {\pi\ln2}{4} \plus{} G
\]
Where $ G$ is Catalan's constant.

So our original integral is:
\[ \int_0^{\frac {\pi}{2}}\ln(1 \plus{} \tan x)\,dx \equal{} \frac {\pi\ln2}{4} \plus{} G
\]
[/quote]
Thank for your help,dear. But your solution is too hard for me :( ( I am only a high school student).
Who has other understandable ones (for me)?


linear algebra
matrix
vector
linear algebra unsolved

"Solve the two systems

x+2y-2z=1
2x+5y+z=9     (2)
x+3y+4z=9

x+2y-2z=9
x+5y+z=9  (5)
x+3y+4z=-2

by doing elimination on a 3x5 augmented matrix and then performing two back substitutions." How must i approach it?

Im so sorry the coefficient of x of 5th equation is 2. So The equations (2) and (5) are same. (2x+5y+z=9)

By doing elimination on a 3x4 augmented matrix and then performing two back substitutions.  :wink:

What part of this instruction (other than the typo) do you find confusing?

must i write that 5 equality to the same table?

The $ 3 \times 4$ means that the matrix you use should have three rows and four columns.  What they want you to do is use matrix notation as you manipulate the equations.  So for the first system,
\begin{align*} x + 2y - 2z & = 1 \\
2x + 5y + z & = 9 \\
x + 3y + 4z & = 9, \end{align*}
they want you to write
$ \left[\begin{array}{ccc|c} 1 & 2 & - 2 & 1 \\
2 & 5 & 1 & 9 \\
1 & 3 & 4 & 9 \end{array}\right]$
and do row operations on that matrix and backsolve to solve the system.

[quote="ereneren"]"Solve the two systems

x+2y-2z=1
2x+5y+z=9
x+3y+4z=9

x+2y-2z=9
x+5y+z=9   (5)
x+3y+4z=-2

by doing elimination on a 3x5 augmented matrix and then performing two back substitutions." How must i approach it?[/quote]

Im so sorry the coefficient of x of 5th equation is 2. So The equations are same.

The coefficients of the variables are the same in both problems. The $ 3\times 5$ matrix would have the coefficients in a $ 3\times 3$ matrix, augmented with two vectors, so both problems could be solved in parallel. In other words, take what benzi did and augment the other vector.


Euler
number theory
number theory unsolved

http://www.geocities.com/jacobin2001/primitiveroots.jpg

comes from elementary number theory, by g jones.

that symbol over the a is the euler totient.

These statements is true only for case, when $m$ is prime or degree of prime. Exactly for $m=p^{k},2p^{k}$.

also true for m=4, but not for 2 to any higher power.
m has a primitive root implies this.

Try proving in cases,  m=p^k , m=2(p^k) or maybe there's another way...  [/img]

okay. So I tried to prove for when m=p^alpha going the reverse way for part a.

i need another hint as to why the condition implies that a is a reside mod p.
 :(


Find the next 3 terms in this sequence:

1, 6, 36, 216, 1296, $\cdots$

[quote="Drunken_Math"]Find the next 3 terms in this sequence:

1, 6, 36, 216, 2196, $\cdots$[/quote]

did you mean 1296 for the next term
[hide="if you did"]If you did the next 3 terms would be
7776, 46656, 279936
*6*6*6[/hide]


does the conjecture
a=b modulo m
a=c modulo n
=>a=b+c modulo mn

have a proof??or it is valid for only when c=0

Your conjecture is a bit too specific, but more generally, we have the [url=http://mathworld.wolfram.com/ChineseRemainderTheorem.html]Chinese Remainder Theorem[/url]. There actually is a formula for the solution, but it's quite a bit more complex than $b+c$.

Well, your conjecture is clearly false when $m = n, b = c$.  Testing easy cases is a good preliminary test of a conjecture.

I think it was clear already that his conjecture was false as the bases multiply while the residues add; this is absurd (also adding the residues doesn't even allows for a majority of the residues mod mn where one of m,n is large... :P )


The number $ 10^{2007}$ is written on the blackboard. $ A,B$ play a game in which the players in turn perform one of the following two operations:
(i) replaces a number $ x$ on the blackboard by two integers $ a,b$ greater than $ 1$ with $ ab\equal{}x$.
(ii) strikes off one or both of two equal numbers on the blackboard.

The player who cannot perform any more moves loses. Who wins if $ A$ starts and both players play their best?

[quote="cancer"]a nice game[/quote]
Yes, nice!
[hide="hint1"]The first "evident" thougt which was visited my mind was wrong. I've have checked - the same came to my friend.[/hide]
[hide="hint2"]To understand that the 1st evident step helped this: (ii) strikes off [b]one or both [/b]of two equal numbers on the blackboard.  [/hide]
[hide="1st step"]$ 10^{2007} \longrightarrow 2^{2007}, \ \ 5^{2007}$ [/hide]
[hide="answer"]A will win[/hide]
It remember me very old nice problem and it hard to say whether this was the matter which helped me to solve or contrary i've connected this with that problem after solution.


geometry
AMC
AIME
probability
number theory
AIME I

Hello, hello.
Greetings, greetings.

I am the mod of this tournement of 1000.
This round is open to [i]everybody[/i], so grab your ticket now!

The round will consist of 40 total questions, 10 each from the category of:
Algebra
Geometry
Number Theory
Counting and Probability.

Now I must give credit to our question writers:

1) BOGTRO - Algebra
2) Isabella2296 - Geometry
3) AIME15 - NT/C&P
4) Dragon96- NT/C&P

After this round, only the top x people will be invited to the next tourney.
After that only y more.
After that only z more.
After that only a more.
After that only b more.
After that.. etc. etc.

Until we have the tournament of the champion, which everyone is invited to watch.

Join now!!!

1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
25.
26.
27.
28.
29.
30.

(If more space is needed, so be it.)

There is no deadline... just when I feel like cutting people off.

I guess we'll have to add new rules, which are somewhat implied.

Graders cannot join...

(It's like my mic that is called the Snowball, and one of the warnings is: DO NOT PUT IN FRIDGE)

[quote="Dojo"]This round is open to [i]everybody[/i], so grab your ticket now![/quote]

Then:

[quote="Dojo"]Graders cannot join...[/quote]

I SENSE LYING!

/me still wants to join...:)

I'm not a grader, just a question writer.

Okay seriously, you are writing the questions, and you are grader.

Don't make myt change my avie to  :mad: .

JOIN.

[hide=".."] meh, not long enough. [/hide]

OMG, BOGTRO is writing problems.

I join.

Update: Pythag011 is writing problem solving questions as well.

1. Twin Prime Conjecture
2. Wickedestjr
3. ernie
4. nikeballa96
5. akhil0422
6. Discrete_Math
7. vamathletes
8. athunder
9. FantasyLover
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
25.
26.
27.
28.
29.
30.

MORE MORE MORE!!!

I don't think you are going to get 30 people. I honestly think you will be lucky to get 20, but we'll see.

Same, 1 submission from leewongjao. Hurry up everyone else.

[quote="AIME15"]I only have 1.  However, I don't have an answer key...[/quote]
You wrote the questions.

That doesn't mean I have answers to the problems.  I still have to work them out.

I have recieved one more submission.

My test ins't THAT hard!!

[quote="Dojo"]Okay.

Tentative due date: Fri 27th[/quote]

I assumed this was the due date. I was planning on doing these probs. some time next week.... probably after the AMC B, just because of time constraints....

Thanks to our mods, we'll likely only have 2 (or maybe 3) divisions this round.

Scores:

Algebra:

[color=green]vamatheletes score 10/10
dragon96 9/10
007math 9/10
wickedestjr 8/10[/color]
THE CUT:
[color=red]Twin Prime Conjecture
ernie
nikeballa96
akhil0422
Discrete_Math
athunder
FantasyLover
12markkram34
shentang
theprodigy
nt811581147
dafjbomb
SuperMathBoy
AIME15
leewongjao
suma_milli
akhilupneja
ghjk
qwertythecucumber
ReseesPieces
AIME_is_hard[/color]

[quote="Dojo"]Thanks to our mods, we'll likely only have 2 (or maybe 3) divisions this round.

Scores:

Algebra:

[color=green]vamatheletes score 10/10
dragon96 9/10
007math 9/10
ernie 9/10
wickedestjr 8/10[/color]
THE CUT:
[color=red]Twin Prime Conjecture
nikeballa96
akhil0422
Discrete_Math
athunder
FantasyLover
12markkram34
shentang
theprodigy
nt811581147
dafjbomb
SuperMathBoy
AIME15
leewongjao
suma_milli
akhilupneja
ghjk
qwertythecucumber
ReseesPieces
AIME_is_hard[/color][/quote]
Updated.

So what's the possible third division?
Geometry or Number Theory?

Probably geometry.

Due date is feb 27th.

Sorry guys.  I think I will have to quit.  I have a lot of work on my hands, and so I just don't think I can do the problems.  Sorry:(

Well, the deadline's over... is this going to continue?

I had my answers for both categories awhile ago but I lost them. Anyways I am too busy with schoolwork and preparing for states. So I am going to drop out. Sorry.

I guess this tourney can be officially declared dead. *sigh*


vector
linear algebra
matrix
abstract algebra
complex numbers
IMC
college contests

What is the maximal dimension of a linear subspace $ V$ of the vector space of real $ n \times n$ matrices such that for all $ A$ in $ B$ in $ V$, we have $ \text{trace}\left(AB\right) \equal{} 0$ ?

Well I just get this far: $\text{dim}V\leq \left[\frac{n^2}{2}\right]$ but I don't bother to construct $V$ with $\text{dim}(V)=\left[\frac{n^2}{2}\right]$.

Consider $V^*=\{A^*|A\in V\}$. Then $V^*$ is also a vector subspace. Consider the Frobenius inner product on the space $M_n(\mathbb{C})$ defined by $\langle A, B \rangle=\text{tr}(B^*A)$. The condition given in the problem is equivalent to saying that $V\perp V^*$ with respect to this inner product. Since $V\oplus V^*$ is a subspace of $M_n(\mathbb{C})$ and $\text{dim}(V)=\text{dim}(V^*)$, we get the desired inequality.

This is completely wrong. Just take n=2, you'll see dim(V) = 1 there and not 2 as you pretend.

The only correct answer is $\frac{n\cdot(n-1)}{2}$. An example is the upper/lower triangular matrices with zero diagonal.
Proof of the upper bound goes like transposing and adding, to get to 2*dim(V) <= n^2 - dim(diag), so dim(V) <= n(n-1)/2. Somehow. Didn't fully get that part :)

In fact the question doesn't explicit specify which field the vector space of $n\times n$ matrices is over. Maybe your answer is right for $F=\mathbb{R}$(I am not certain as I didn't think over it). But if $F=\mathbb{C}$, certainly there exists an example $V\subset M_2(\mathbb{C})$ where $\text{dim}(V)=2$. Simply consider $V$ spanned by
$\left\{\begin{pmatrix}
0&1\\
0&0
\end{pmatrix},
\begin{pmatrix}
1&1\\
0&i
\end{pmatrix}\right\}$

Yes, I think the problem stated real matrices, sorry.

Over the real matricies it is clear that sum(a[i,i]^2) must be zero, thus all diagonal elements must be zero as well. However over the complex numbers I think your answer is correct, since it is exactly the same problem, but on the diagonal each pair of (0,0) can also be (1,i).

Consider the subspace of symmetric matrices which intersect with any such $V$ by the zero matrix,so $\dim(V){\leq}n^{2}-\frac{n(n+1)}{2}=\frac{n(n-1)}{2}$.An example of $V$ with dimension $\frac{n(n-1)}{2}$ is the subspace of upper triangular matrices with $0$-s on the diagonal.This is the official solution.

I have another solution for this problem (I'm surprised noone posted it yet, it's just straight-forward Linear Algebra II).

It is easy to check that $ (A, B) : \equal{} \text{tr}(AB)$ is a symmetric bilinear form on $ Mat_n(\mathbb R)$. Thus it turns the space of all real matrices into an orthogonal space. A corollary of Witt's theorem then tells us that the following representation exists and is unique up to isometry: $ Mat_n(\mathbb R) \equal{} L_0 \oplus L_{an} \oplus L_h$ where $ L_0$ is the kernel of the form, $ L_{an}$ and $ L_h$ are the anisotropic and hyperbolic parts of the space and the direct sums are orthogonal. Now $ L_0 \equal{} 0$ since for any non-zero matrix $ A$ one can easily construct $ B$ such that $ (A, B) \neq 0$ (just let $ B$ have a single one in an appropriate position). Now $ L_{an}$ contains all diagonal matrices since no diagonal matrix can be isotropic (for obvious reasons) so $ \dim L_{an} \ge n$ and hence $ \dim L_h \le n(n\minus{}1)$. Then what we are asked in the problem is a dimension of a maximal isotropic subspace (which is well-defined - does not depend on $ V$ as long as it's maximal, another consequence of Witt's theorem). But any maximal isotropic subspace $ V$ must have a hyperbolic counterpart (a well-known lemma) in $ L_h$ (in $ L_h$, since there are no isotropic vectors in $ L_{an}$) and thus $ 2\dim V \le dim L_h \le n(n\minus{}1)$ and so $ \dim V \le \frac{n(n\minus{}1)}{2}$. The example of upper diagonal matrices shows that this bound can indeed be obtained.


Suppose the sum converged

It would certainly be larger than sum(k=2 to infinity) $\frac{2}{(k+1)(ln(k+1))}$

But if the above converged then so would the integral (from 2 to infinity) of

$\frac{2}{(k+1)(ln(k+1))}$

This integral is easily found to diverge

Who are you by the way?

It is the classic log of log. OK that was a pointless post... I just wanted to post here to seem North Carolinian. NORTH CAROLINA @@ (?)


An often asked problem : which is larger: $e^{\pi} \ or \ \pi^e$ ?

And, the answer : $e^{\pi} > \pi^{e}$ :)

[quote="boxedexe"]Can you please prove this.  I'm very eager to know the proof.

Masoud Zargar[/quote]
One of the  special property of $e$ is that  $x^{1/x}$ has a maximum when $x=e$, thus, if   $e^{anything} > (anything)^e$ (as long as [b]anything[/b] is other than $e$)

Straight forward is to use calculus to find maximum of the function $x^{1/x}$ and verify that it is at $x=e$.

(Above follows from $e^{1/e} > \pi^{1/\pi}$)

There is a very slick proof for this one using the Taylor series expansion for $e$:

Note that $e^x = \frac{x^0}{0!} + \frac{x^1}{1!} + \frac{x^2}{2!} + \cdots > 1 + x$ for $x > 0$.

Let $x = \frac{\pi}{e} - 1$, which is greater than 0, thus
\[ e^{\frac{\pi}{e} - 1} > 1 + \frac{\pi}{e} - 1 = \frac{\pi}{e}. \]
Multiply through by $e$ and we have
\[ e^{\frac{\pi}{e}} > \pi \ \ \Rightarrow \ \ e^{\pi} > \pi^e. \]

Very nice, Of course, one has $e^x \ge 1+x$ for  negative $x$ too,  (though not that obvious from the series) so this can be used for the case even when $a/e <1$